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http://www.jiskha.com/display.cgi?id=1386372704
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Math
posted by .
A 5 inch by 8 inch rectangle is enlarged to a similar rectangle whose smallest side measures 9 inches. What is the length in inches of the diagonal of the enlarged rectangle?
• Math -
D1/D2 = 5/9
D1 = √(5^2 + 8^2) = √89
so √89/D2 = 5/9
5D2 = 9√89
D2 = 9√89/5
notice that the diagonal increased by a factor of 9/5 or the same as the increase in the sides.
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Free Version
Easy
# 2x2 Matrix Inverse - 2, 3, 1, 2
LINALG-VXRYQE
Which of the following is the inverse of the matrix:
$$\begin{pmatrix} 2&3\\\ 1&2\end{pmatrix}$$
A
$\begin{pmatrix}2&3\\\ 1&2\end{pmatrix}$
B
$\begin{pmatrix}2&-3\\\ -1&2\end{pmatrix}$
C
$\begin{pmatrix}-2&1\\\ 3&-2\end{pmatrix}$
D
$\begin{pmatrix}-2&3\\\ 1&-2\end{pmatrix}$
E
$\begin{pmatrix}2&-1\\\ -3&2\end{pmatrix}$
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:: A Theory of Sequential Files
:: by Hirofumi Fukura and Yatsuka Nakamura
::
:: Copyright (c) 2005-2017 Association of Mizar Users
theorem Th1: :: FILEREC1:1
for p, q, r, s being FinSequence holds
( ((p ^ q) ^ r) ^ s = (p ^ (q ^ r)) ^ s & ((p ^ q) ^ r) ^ s = (p ^ q) ^ (r ^ s) & (p ^ (q ^ r)) ^ s = (p ^ q) ^ (r ^ s) )
proof end;
theorem Th2: :: FILEREC1:2
for f being FinSequence holds f | (len f) = f
proof end;
theorem Th3: :: FILEREC1:3
for p, q being FinSequence st len p = 0 holds
q = p ^ q
proof end;
theorem Th4: :: FILEREC1:4
for D being non empty set
for f being FinSequence of D
for n, m being Element of NAT st n <= m holds
len (f /^ m) <= len (f /^ n)
proof end;
theorem Th5: :: FILEREC1:5
for D being non empty set
for f, g being FinSequence of D st len g >= 1 holds
mid ((f ^ g),((len f) + 1),((len f) + (len g))) = g
proof end;
theorem Th6: :: FILEREC1:6
for D being non empty set
for f, g being FinSequence of D
for i, j being Element of NAT st 1 <= i & i <= j & j <= len f holds
mid ((f ^ g),i,j) = mid (f,i,j)
proof end;
theorem :: FILEREC1:7
for D being non empty set
for f being FinSequence of D
for i, j, n being Element of NAT st 1 <= i & i <= j & i <= len (f | n) & j <= len (f | n) holds
mid (f,i,j) = mid ((f | n),i,j)
proof end;
theorem :: FILEREC1:8
for a being set
for D being non empty set
for f being FinSequence of D st f = <*a*> holds
a in D
proof end;
theorem :: FILEREC1:9
for a, b being set
for D being non empty set
for f being FinSequence of D st f = <*a,b*> holds
( a in D & b in D )
proof end;
theorem Th10: :: FILEREC1:10
for a, b, c being set
for D being non empty set
for f being FinSequence of D st f = <*a,b,c*> holds
( a in D & b in D & c in D )
proof end;
theorem :: FILEREC1:11
for a being set
for f being FinSequence st f = <*a*> holds
f | 1 = <*a*>
proof end;
theorem :: FILEREC1:12
for a, b being set
for D being non empty set
for f being FinSequence of D st f = <*a,b*> holds
f /^ 1 = <*b*>
proof end;
theorem :: FILEREC1:13
for a, b, c being set
for D being non empty set
for f being FinSequence of D st f = <*a,b,c*> holds
f | 1 = <*a*>
proof end;
theorem Th14: :: FILEREC1:14
for a, b, c being set
for D being non empty set
for f being FinSequence of D st f = <*a,b,c*> holds
f | 2 = <*a,b*>
proof end;
theorem :: FILEREC1:15
for a, b, c being set
for D being non empty set
for f being FinSequence of D st f = <*a,b,c*> holds
f /^ 1 = <*b,c*>
proof end;
theorem :: FILEREC1:16
for a, b, c being set
for D being non empty set
for f being FinSequence of D st f = <*a,b,c*> holds
f /^ 2 = <*c*>
proof end;
theorem :: FILEREC1:17
for f being FinSequence st len f = 0 holds
Rev f = f
proof end;
theorem Th18: :: FILEREC1:18
for D being non empty set
for r being FinSequence of D
for i being Element of NAT st i <= len r holds
Rev (r /^ i) = (Rev r) | ((len r) -' i)
proof end;
theorem Th19: :: FILEREC1:19
for D being non empty set
for f, CR being FinSequence of D st not CR is_substring_of f,1 & CR separates_uniquely holds
instr (1,(f ^ CR),CR) = (len f) + 1
proof end;
theorem :: FILEREC1:20
for D being non empty set
for f, CR being FinSequence of D st not CR is_substring_of f,1 & CR separates_uniquely holds
f ^ CR is_terminated_by CR
proof end;
notation
let D be set ;
synonym File of D for FinSequence of D;
end;
definition
let D be non empty set ;
let r, f, CR be File of D;
pred r is_a_record_of f,CR means :: FILEREC1:def 1
( ( CR ^ r is_substring_of addcr (f,CR),1 or addcr (f,CR) is_preposition_of ) & r is_terminated_by CR );
end;
:: deftheorem defines is_a_record_of FILEREC1:def 1 :
for D being non empty set
for r, f, CR being File of D holds
( r is_a_record_of f,CR iff ( ( CR ^ r is_substring_of addcr (f,CR),1 or addcr (f,CR) is_preposition_of ) & r is_terminated_by CR ) );
theorem Th21: :: FILEREC1:21
for D being non empty set
for r being FinSequence of D holds
( ovlpart ((<*> D),r) = <*> D & ovlpart (r,(<*> D)) = <*> D )
proof end;
theorem Th22: :: FILEREC1:22
for D being non empty set
for CR being FinSequence of D holds CR is_a_record_of <*> D,CR
proof end;
theorem :: FILEREC1:23
for D being non empty set
for a, b being set
for f, r, CR being File of D st a <> b & CR = <*b*> & f = <*b,a,b*> & r = <*a,b*> holds
( CR is_a_record_of f,CR & r is_a_record_of f,CR )
proof end;
theorem Th24: :: FILEREC1:24
for D being non empty set
for f, CR being File of D holds f ^ CR is_preposition_of
proof end;
theorem :: FILEREC1:25
for D being non empty set
for f, CR being File of D holds addcr (f,CR) is_preposition_of
proof end;
theorem Th26: :: FILEREC1:26
for D being non empty set
for r, CR being File of D st CR is_postposition_of r holds
0 <= (len r) - (len CR)
proof end;
theorem Th27: :: FILEREC1:27
for D being non empty set
for CR, r being File of D st CR is_postposition_of r holds
proof end;
theorem Th28: :: FILEREC1:28
for D being non empty set
for CR, r being File of D st r is_terminated_by CR holds
proof end;
theorem :: FILEREC1:29
for D being non empty set
for f, g being File of D st f is_terminated_by g holds
len g <= len f by FINSEQ_8:def 12;
theorem Th30: :: FILEREC1:30
for D being non empty set
for f, CR being File of D holds
( len (addcr (f,CR)) >= len f & len (addcr (f,CR)) >= len CR )
proof end;
theorem Th31: :: FILEREC1:31
for D being non empty set
for f, g being FinSequence of D holds g = (ovlpart (f,g)) ^ (ovlrdiff (f,g))
proof end;
theorem :: FILEREC1:32
for D being non empty set
for f, g being FinSequence of D holds ovlcon (f,g) = (ovlldiff (f,g)) ^ g
proof end;
theorem :: FILEREC1:33
for D being non empty set
for CR, r being File of D holds addcr (r,CR) = (ovlldiff (r,CR)) ^ CR
proof end;
theorem Th34: :: FILEREC1:34
for D being non empty set
for r1, r2, f being File of D st f = r1 ^ r2 holds
( r1 is_substring_of f,1 & r2 is_substring_of f,1 )
proof end;
theorem Th35: :: FILEREC1:35
for D being non empty set
for r1, r2, r3, f being File of D st f = (r1 ^ r2) ^ r3 holds
( r1 is_substring_of f,1 & r2 is_substring_of f,1 & r3 is_substring_of f,1 )
proof end;
theorem Th36: :: FILEREC1:36
for D being non empty set
for CR, r1, r2 being File of D st r1 is_terminated_by CR & r2 is_terminated_by CR holds
CR ^ r2 is_substring_of addcr ((r1 ^ r2),CR),1
proof end;
theorem Th37: :: FILEREC1:37
for D being non empty set
for f, g being File of D
for n being Element of NAT st 0 < n & g = {} holds
instr (n,f,g) = n
proof end;
theorem :: FILEREC1:38
for D being non empty set
for f, g being File of D
for n being Element of NAT st 0 < n & n <= len f holds
instr (n,f,g) <= len f
proof end;
theorem Th39: :: FILEREC1:39
for D being non empty set
for f, CR being File of D holds CR is_substring_of ovlcon (f,CR),1
proof end;
theorem Th40: :: FILEREC1:40
for D being non empty set
for f, CR being File of D holds CR is_substring_of addcr (f,CR),1
proof end;
theorem Th41: :: FILEREC1:41
for D being non empty set
for f, g being FinSequence of D
for n being Element of NAT st g is_substring_of f | n,1 & len g > 0 holds
g is_substring_of f,1
proof end;
theorem :: FILEREC1:42
for D being non empty set
for f, CR being File of D ex r being File of D st r is_a_record_of f,CR
proof end;
theorem :: FILEREC1:43
for D being non empty set
for f, CR, r being File of D st r is_a_record_of f,CR holds
r is_a_record_of r,CR by Th28;
theorem :: FILEREC1:44
for D being non empty set
for CR, r1, r2, f being File of D st r1 is_terminated_by CR & r2 is_terminated_by CR & f = r1 ^ r2 holds
( r1 is_a_record_of f,CR & r2 is_a_record_of f,CR )
proof end;
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# Enumerations of the rationals with summable gaps $(q_i-q_{i-1})^2$
Here is a question from my undergraduate days which I never knew the answer to. I just want to know if anyone can offer me a hint.
Consider the rationals in $[0,1]$. Does there exist a (bijective) enumeration of these rationals $q_1,q_2,\ldots$ such that the sum $\sum\limits^\infty_{i=1} (q_i-q_{i-1})^2$ is finite?
• Presumably, you mean a $1-1$ enumeration? Oct 21, 2012 at 17:09
• Yes, of course. Oct 21, 2012 at 17:14
Summary:
• There exists a (bijective) enumeration $(q_i)_{i\geqslant0}$ of the rationals in $[0,1]$ such that, for every $a\gt1$, the series $\sum\limits_i|q_i-q_{i-1}|^a$ converges.
• This result is optimal in the sense that no (bijective) enumeration $(q_i)_{i\geqslant0}$ of the rationals in $[0,1]$ is such that the series $\sum\limits_i|q_i-q_{i-1}|$ converges.
Let us show the first point. For every $n\geqslant0$ and $0\leqslant k\lt 2^n$, define the interval $I_{2^n+k}$ as $$I_{2^n+k}=\left\{\begin{array}{ll}(k\cdot2^{-n},(k+1)\cdot2^{-n}]&\text{if n is even},\\ (1-(k+1)\cdot2^{-n},1-k\cdot2^{-n}]&\text{if n is odd}.\end{array}\right.$$ Thus, for every $n\geqslant0$, $(I_{2^n+k})_{0\leqslant k\lt 2^n}$ is a partition of $(0,1]$ into $2^n$ intervals of length $2^{-n}$. For every $i\geqslant 2^n$, if $x$ is in $I_i$ and $y$ in $I_{i+1}$, then $|x-y|\leqslant c\cdot2^{-n}$, probably for $c=2$ and certainly for $c=42$.
Define recursively $(q_i)_{i\geqslant0}$ as follows: let $q_0=0$ and, for every $n\geqslant0$ and $0\leqslant k\lt 2^n$, let $q_{2^n+k}$ denote the rational in $I_{2^n+k}$ not already in $\{q_i\mid i\lt 2^n+k\}$ whose reduced fraction has minimal denominator and, if several such rationals exist, minimal numerator.
Then $(q_i)_{i\geqslant0}$ enumerates the rationals in $[0,1]$. Furthermore, for every $n\geqslant0$, if $2^n\leqslant i\lt 2^{n+1}$, then $$|q_i-q_{i-1}|\leqslant c\cdot2^{-n}.$$ Hence the slice of the sum $\sum\limits_i|q_i-q_{i-1}|^a$ for $2^n\leqslant i\lt 2^{n+1}$ is at most $$2^n\cdot(c\cdot2^{-n})^a=c^a\cdot2^{-(a-1)n}.$$ Since $a\gt1$, summing these shows that the complete series converges. QED.
The fact that the series $\sum\limits_i|q_i-q_{i-1}|$ diverges for every enumeration is easy. Define recursively the sequence $(N(n))_{n\geqslant0}$ as follows: let $N(0)=0$ and, for every $n\geqslant0$, let $N(n+1)$ denote the smallest integer $i\geqslant N(n)$ such that $$|q_i-q_{N(n)}|\geqslant\tfrac12.$$ Then $N(n)$ is finite for every $n$ (why?) and the triangular inequality yields $$\sum\limits_i|q_i-q_{i-1}|\geqslant\sum\limits_n|q_{N(n)}-q_{N(n-1)}|\geqslant\sum\limits_n\tfrac12,$$ which is infinite.
• n=0, k=1, I think we get the interval (1,2]? Oct 21, 2012 at 17:21
• Call a term of the enumeration high if it is in (3/4,1) and low if it is in (0,1/4). There are infinitely many highs and lows. A subsequence of $(q_n)$ is such that every $q_{\varphi(2n)}$ is high and every $q_{\varphi(2n+1)}$ is low. And |high$-$low| is always at least 1/2. QED.
– Did
Oct 22, 2012 at 5:25
• I am extremely glad to seethis answer at last got a 'good answer' badge after some promotion here: math.stackexchange.com/questions/671076/… Feb 11, 2014 at 14:53
• @Lost1 Hmmm... I was wondering what caused the sudden renewed interest for this answer. Thanks for the appreciation.
– Did
Feb 11, 2014 at 15:26
• I guess the "and certainly for $c = 42$" bit was just to check that we were all paying attention. When I saw it I got suspicious and started to work on my own solution - taking 'harmonic $1/n$ steps'. +1 for a sense of humor! Sep 7, 2017 at 4:01
[EDIT: The argument below traverses all of the rational numbers instead of just the rational numbers in $$[0,1]$$.]
You can define the sequence $$u_n$$ in such a way that it moves from 0 to 1 in steps of $$\frac{1}{1!}$$, from to 1 to -1 in steps of $$\frac{1}{2!}$$, and from -1 to 2 in steps of $$\frac{1}{3!}$$ etc.
To get a good intuition for this solution, you can start by solving a different problem, which is enumerating all numbers with a finite decimal expansion. This can be done by moving from 0 to 1 in steps of size 0.1, from 1 to -1 in steps of 0.01, and then from -1 to 2 in steps of 0.001 etc. Since you want it to be a bijection, you simply skip over any point you've previously visited. Call this sequence $$v_n$$. The summation $$\sum_{r=1}^{\infty} (v_r - v_{r-1})^2$$ can be upper-bounded as $$4(10 \times 0.1^2 + 20 \times 0.01^2 + 30 \times 0.001^3 + \ldots) = 4 \sum_{r=1}^{\infty} 10r \times 10^{-2r}$$ which is clearly convergent. The multiplication by 4 came from the fact that every time you skip a point, the difference squared multiplies by 4, so to give an upper bound multiply the whole thing by 4.
Clearly, that's going to miss out on every rational number with an infinite decimal. So instead of using steps of 0.1, 0.01 etc. go in steps of $$\frac{1}{1!}$$, $$\frac{1}{2!}$$, $$\frac{1}{3!}$$ etc. Again, skip points you've already visited. Now construct an upper bound for $$\sum_{r=1}^{\infty} (u_r - u_{r-1})^2$$, which is $$4( 1 \times 1! \times \frac{1}{1!^2} + 2 \times 2! \times \frac{1}{2!^2} + 3 \times 3! \times \frac{1}{3!^2} + \ldots) = 4 \sum_{r=1}^{\infty} r \times r! \times \frac{1}{r!^2} = 4 \sum_{r=1}^{\infty} \frac{1}{(r-1)!} =4e$$ which is finite.
EDIT:
There is nothing special about the rational numbers in this problem.
Theorem: If a subset $$S$$ of the real numbers $$\mathbb R$$ is countable and dense in $$\mathbb R$$, then $$S$$ is traversable with summable gaps.
(Remark: In particular, this means that the rational numbers can be replaced by a more general set like the real algebraic numbers and the same result would be true.)
Proof: Let $$s_i$$ be an arbitrary enumeration of $$S$$. Let $$\text{boundary}_n$$ be a sequence that moves from $$0$$ to $$1$$ in steps of $$0.1$$; from $$1$$ to $$-1$$ in steps of $$0.01$$; from $$-1$$ to $$2$$ in steps of $$0.001$$; etc. Let $$\text{traversal}_n$$ be the first element of $$s_i$$ in between $$\text{boundary}_n$$ and $$\text{boundary}_{n+1}$$ that hasn't previously been visited. By first element, we mean the element $$s_i$$ with the smallest value of $$i$$ that meets our condition. The sequence $$\text{traversal}_n$$ then visits every element of $$S$$ exactly once, and the upper bound on the sum is at most $$4\sum_{n=0}^\infty r10^{-r} < 0.5$$; the multiplication by $$4$$ comes from assuming the largest possible distance between $$\text{traversal}_n$$ and $$\text{traversal}_{n+1}$$.
$$\blacksquare$$
This is the 'harmonic two-step' solution
Let $\mathbb{Q}_1 = [0,1] \cap \mathbb{Q}$.
Proposition 1: There exist an injective enumeration $q: \mathbb{N} \to \mathbb{Q}_1$ satisfying
$\tag 1 \text{For all } n \gt 0 \text{, } \; \; \; \frac{1}{2n} \le |q_{n} - q_{n-1}| \le \frac{1}{n}$ $\tag 2 \text{For all } n \gt 1 \text{, } \; \; \; 0 \lt q_{n} \lt 1$ Proof: Exercise.
Proposition 2: For any given $\hat q \in \mathbb{Q}_1$ there exist an injective enumeration $q: \mathbb{N} \to \mathbb{Q}_1$ satisfying (2) and
$\tag {1'} \text{For all } n \gt 0 \text{, } \; \; \; |q_{n} - q_{n-1}| \le \frac{1}{n}$ $\tag 3 \text{There exists a } m \in \mathbb{N} \text{ with } q_m = \hat q$ Proof: Exercise.
Proposition 3: There exist a bijective enumeration $q: \mathbb{N} \to \mathbb{Q}_1$ satisfying
$\tag {1'} \text{For all } n \gt 0 \text{, } \; \; \; |q_{n} - q_{n-1}| \le \frac{1}{n}$
Proof (critics might justly call it hand-waving)
Set $q_0 = 0$ and $q_1 = 1$ and with $n = 2$ just take a step of $-\frac{1}{2}$ to the left, and set $q_2 = .5$. For your next $\frac{1}{3}$ step, you can go left or right, so continue going left, setting $q_3 = q_2 - \frac{1}{3} = \frac{1}{6}$. You can continue taking these harmonic steps to your hearts content, but if you hit on a duplicate just start bisecting the step as specified by (1).
Looking at your favorite bijective correspondence $\mathbb{N} \equiv \mathbb{Q}_1$, you notice that you still haven't 'stepped on' the fifth number $\hat q$ on that list. But, with $n$ large enough, you can take harmonic (1)-adjusted steps towards that number ending on either an exact hit (uhh, very low odds) or straddling it. But instead of taking a straddle step, just adjust it for a direct hit. You can then continue taking harmonic (1)-adjusted steps to get to any number in $\mathbb{Q}_1$ still needing a '$q$ hit' check-mark.
So you can make further adjustments to 'get-em all', completing this do-si-do dance step proof.
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4 Replies Latest reply: Oct 17, 2011 10:56 AM by Onur Civelek
# Set Analysis
Helle Community,
I have read the tutorial to set analysis of Tom Mackay. Unfortunately I did not become really smart to solve my problem .
How can I aggregating the below table is always the sum of the 12 months in the chart?
Calculation of the values in Table 1-12 sums 20253.88 etc... see disired table.
Thank you !
• ###### Set Analysis
I think you could just use a sum with appropriate total qualifier:
I assume you have three dimensions in your chart like Dim1, Dim2, Dim3 (where Dim1 has values like F&C, Dim2 like Personalaufwand... and Dim3 are the Months).
Then you could use
sum(total<Dim1,Dim2> VALUEFIELD)
where VALUEFIELD is the field containing the numbers your blue and green values are based on.
Hope this helps,
Stefan
• ###### Re: Set Analysis
Hi Stefan,
Thank you for your support. I have five dimensions in my diagram. I tried your example like:
In the diagram there are five dimensions. I tried your example on the following:
sum(total< PD_OBERBEREICH, PD_KOSTENGRUPPE, PD_CAROTKONTONAME> PD_WERT )
But the values are not always correct. See Screenshot
Or see my qvw file
• ###### Re: Set Analysis
One question:
why do you divide your sum 1-12 months by 10?
I noticed that you have several records per combination of your dimensions, sometimes 10, sometimes 5, sometimes more than 10.
Maybe try this:
sum(total< PD_OBERBEREICH, PD_KOSTENGRUPPE, PD_CAROTKONTONAME> PD_WERT ) / count(PD_WERT)
to get an average value.
Stefan
• ###### Re: Set Analysis
Hi Stefan,
many thanks for the tip. Now it fits.
I thought at first that the number formatting deviates one decimal place.
But you have correctly identified the problem. The records are more than 5 or 10 times, therefore the correct solution is to divide by the quantity.
Thanks again
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# MML Student Access Kit for Ad Hoc Valuepacks & THOMAS CALC EARLY TRANS SINGLE&MML&SSM PKG (7th Edition) View more editions Solutions for Chapter 4.5 Problem 42EProblem 42E: Applying l’Hôpital’s Rule Use l’Hôpital’s Rule to find th...
• 6616 step-by-step solutions
• Solved by professors & experts
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Chapter: Problem:
Applying l’Hôpital’s Rule
Use l’Hôpital’s Rule to find the limits
Step-by-Step Solution:
Chapter: Problem:
• Step 1 of 4
Consider
Observe that using method takes to indeterminate form like
So, to avoid this form, first simplify this expression to as small as possible
Use the identities and in this,
• Chapter , Problem is solved.
Corresponding Textbook
MML Student Access Kit for Ad Hoc Valuepacks & THOMAS CALC EARLY TRANS SINGLE&MML&SSM PKG | 7th Edition
9780321721792ISBN-13: 0321721799ISBN: Authors:
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# Using 'solve' command in MATLAB R2008a
as a part of a code i have written to solve an equivalent system in vector mechanics, i have reached the point where i need to solve 3 equations to find x, y, z, (the placement of the Vector Ftot)
i know the total momentum, which i call Mtot, and i know the total force, which is Ftot, what i am looking for is R which i call (x, y, z) while i know that RxFtot=Mtot, so what i have done is
.....
Code:
syms x y z
R=[x y z];
RF=cross(R,Ftot);
now what i want to do somehow is get matlab to solve the 3 equations i get
Code:
solve(RF(1)=Mtot(1),RF(2)=Mtot(2),RF(3)=Mtot(3))
but matlab doesnt recognise the contents of RF as an equation or Mtot as an answer, it wants me to type an actual equation which is not what i need since this is a code which needs to work for every system.
how can i use the solve function for the contents of a matrix?
hopefully you guys understood what im trying to do, if not this is my entire code, at the end you will see i get 3 equations to find x,y,z where in most cases only 2 will be linearly independant, so what i would like to do is get to some kind of equation wher i can plug in any x of my choice and matlab will give me the other 2,
Code:
nf=0;
nm=0;
nf=input('Enter number of Forces :');
nm=input('enter number of Moments :');
disp('------------------------------')
dim=['x' 'y' 'z'];
F=zeros(nf,3); %code for building Force matrix
for i=1:nf
for j=1:3;
prompt=fprintf(1,'Enter magnitude (including direction) of F%d%s ',i,dim(j));
F(i,j)=input(':');
end
disp('------------------------------')
end
M=zeros(nm,3); %code for building Moment matrix
for i=1:nm;
for j=1:3;
prompt=fprintf(1,'Enter magnitude (including direction) of M%d%s ',i,dim(j));
M(i,j)=input(':');
end
disp('------------------------------')
end
Ftot=[0 0 0]; %code for calculating total Force
for i=1:nf;
Ftot=Ftot+F(i,:);
end
Mm=[0 0 0];%code for calculating total moment excludinf moment from forces
for i=1:nm;
Mm=Mm+M(i,:);
end
disp('------------------------------')
R=zeros(nf,3);%code for building matrix of moments due to moved forces
for i=1:nf;
for j=1:3;
prompt=fprintf(1,'Enter placement R%s of F%d',dim(j),i' );
R(i,j)=input(':');
end
disp('------------------------------')
end
Mf=[0 0 0]; %code for calculating moment due to moved forces
for i=1:nf;
Mf=Mf+cross(R(i,:),F(i,:));
end
Mtot=Mm+Mf; %Total System Moment
clc
disp('------------------------------')
disp('EQUIVALENT SYSTEM AT SPECIFIED POINT') %display of Equivalent system
fprintf(1,'\nF = %f*i + %f*j +%f*k\n',Ftot)
fprintf(1,'|F| = %f\n\n',norm(Ftot))
fprintf(1,'M = %f*i + %f*j +%f*k\n',Mtot)
fprintf(1,'|M| = %f\n\n',norm(Ftot))
Wrench=(Mtot*normr(Ftot)')*normr(Ftot); %calculating wrench
Mp=Mtot-Wrench;%calculating moment perpendicular to Ftotal
syms x y z; %building vector for x,y,z placement equations
R=[x y z];
Mperp=cross(R,Ftot);%Mperp=moment perpendicular to force
disp('------------------------------')
disp('SIMPLEST EQUIVALENT SYSTEM AT SPECIFIED POINT')%display of simplest system
fprintf(1,'\nF = %f*i + %f*j +%f*k\n',Ftot)
fprintf(1,'|F| = %f\n\n',norm(Ftot))
fprintf(1,'Wrench = %f*i + %f*j +%f*k\n',Wrench)
fprintf(1,'|Wrench| = %f\n\n',norm(Wrench))
disp('The following equations define the placement of the simplest equivalent system')
for n=1:3
fprintf(1,'(%d) %f=',n,Mp(n))
disp(Mperp(n))
end
disp('------------------------------')
nvn
Homework Helper
Dell: Perhaps try X=linsolve(RF,Mtot), or perhaps X=RF\Mtot. Let us know if this doesn't help.
in my code, RF is called Mperp,
so...
Code:
RR=Mperp;
X=linsolve(RF,Mtot),
[COLOR="Red"]??? Error using ==> linsolve
First and second arguments must be single or double.[/COLOR]
didnt work, neither did X=RF\Mtot, but there is something in it, it gives me a matrix with 9 values, some of which seem right but i cannot see how i could use this to get a conclusive answer.
is there no way to use the solve function for the contents of a matrix/vector, once i get my 3 equations and manually use the solve function it works perfectly, for example, if
A=[2*x, x+y, z+y+2*x] and B=[2, 3, 6]
i have gotten to the point where MATLAB displays
2*x=2
x+y=3
z+y+2*x=6
but cannot advance from that, from there i need to manually plug in solve('2*x=2','x+y=3','z+y+2*x=6') to get solutions, but this is not good for my general code, as every time i get 3 different solutions, i would like to be able to solve(A,B) or solve(A=B) or even solve(A(1)=B(1),A(2)=B(2),A(3)=B(3))
Päällikkö
Homework Helper
Well, this is really an ugly brute force approach, but it's what first came to my mind. Using eval, assuming A is symbolic, and B numeric (otherwise you'd need to flip the char and num2str-stuff).
str = [];
for i = 1:length(A)
str = [str ',' char(A(i)) '=', num2str(B(i))];
end
str = str(2:end);
sol = eval(['solve(''' str ''')']);
If you have the relevant toolboxes (and maybe Maple), you could use the maple-command, which makes symbolic stuff real easy.
nvn
Homework Helper
Dell: Try the following after you have computed Ftot and Mtot, and after you have input R(1), R(2), or R(3). In other words, you must give the program either the x, y, or z component of the unknown position vector R. Let ii equal the index number of the R component given by the user; i.e., ii = 1, 2, or 3. Please clean up my syntax, in case I did not use semicolons correctly. Let us know whether or not it works.
% check orthogonality.
if(abs(dot(Ftot,Mtot))>1.0e-6)
...print "Error: Mtot is not perpendicular to Ftot. Mtot must be ";
...print "perpendicular to Ftot to be a valid cross product.";
...exit;
end;
% solve for R.
if(Ftot(ii)~=0.0)
...U=cross(Ftot,Mtot)/dot(Ftot,Ftot);
...V=Ftot*(R(ii)-U(ii))/Ftot(ii);
...R=U+V;
...print R;
else print "No solution found.";
end;
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Question: In a study of the reaction 3Fe s 4H2O g
In a study of the reaction 3Fe(s) + 4H2O(g) ⇌ Fe3O4(s) + 4H2(g) at 1200 K, it was observed that when the equilibrium partial pressure of water vapor is 15.0 torr, the total pressure at equilibrium is 36.3 torr. Calculate Kp for this reaction at 1200 K.
View Solution:
Sales1
Views286
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# Testing for Nil
I think there would be a flaw with that code. I do not wish to overwrite any good values in the array with “0” since I save/update the array every day to the HD. It needs to maintain all existing values. I need to test it yet but changing the array to a variant may be answer for my code.
1 Like
Just set to -1 if not valid – Then you will be solid unless that is an issue then the Variant route is solid also.
Thanks Mike, it’s all good to help with the steep learning curve I’m on.
…and do not forget the leap year… day !
Trouble maker!
Convenient
Actually, I don’t think it’s an issue since the array will be 366 days and a new array is only triggered when the system clock changes the year. I am only interested in daily totals to override my watering system and a yearly total for grins. Not a 366th day, who cares it’s “0”
No, experienced oldster who forgot things everytime
Now, think at the Jewish Calendar (new year occured to them (Roch Hachana ended yesterday) and at the muslim calendar. I do not even talk about the different dates in use in India.
That would be trouble making
Agreed…
I need to take this a step further. I saved the RainSensorArray(day) array as a variant array and later loaded it back in on startup. Each value is 0.0. All looks fine until I use this For Each statement to get a total on the variant array:
Var values() As Variant = RainSensorArray
Var sum As Variant
For Each d As Variant In values
sum = sum + d
Next
I receive a nil exception for the sum=sum+d code. Would 0.0 = 0.0 + 0.0 cause a nil?
The array values look okay in the debugger for startup and all show as 0.0. Values = Variant(366) and d = 0.0
Changing sum to an integer seems make things happier. Not sure I understand yet.
When you dim sum as a variant, it starts out nil. That would cause the first iteration to throw a nilobject. If you defined it as
``````Var sum as Variant = 0.0
``````
That would probably make it happy as well. But it’s best to avoid variants where possible.
1 Like
You nailed it, Tim. Sounds like a bug but what do I know…
That’s just how variants work. And why you should avoid them. Too much head-scratching involved.
I’m just about bald from this already so what’s a few more hairs. I’ll keep the variant taboo in mind. The project is coming along nicely but I can already see a rewrite to the rewrite, well some.
I think a dictionary would be ideal to store your readings rather than an array. The dictionary using a key/value pair so you don’t need a reading for every day and you can use the HasKey & Lookup functions to check if a key (day) exists or set a default value using Lookup.
It is also easy to save using the generatejson & parsejson functions to serialise & deserialise the data.
3 Likes
Good point, Wayne! I do like using dictionaries but I’m not familiar with Json . I have another method to construct that will keep track of the lowest 24-hour temperatures. I am trying to reason this out now. This may be an opportunity to learn some Json commands with a dictionary. I need to be able to continue with these 24 hr low temps even when the program has been stopped for a which. Yes, the blank entries. The 24hr low needs to be a “running low temp”. Need a challenge?
when i first saw dim sum… i thought of this https://en.wikipedia.org/wiki/Dim_sum instead…
3 Likes
You’re making me hungry.
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https://www.gradesaver.com/textbooks/math/geometry/CLONE-68e52840-b25a-488c-a775-8f1d0bdf0669/chapter-3-section-3-2-corresponding-parts-of-congruent-triangles-exercises-page-137/12
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## Elementary Geometry for College Students (6th Edition)
Proof for the problem: 1. $\angle R$ and $\angle V$ are right $\angle$s (1. Given) 2. $\triangle RST$ and $\triangle VST$ are right triangles. (2. A triangle that has one right angle is a right triangle) 3. $\overline{ST}\cong\overline{ST}$ (3. Identity) 4. $\overline{RT}\cong\overline{VT}$ (4. Given) 5. $\triangle RST\cong\triangle VST$ (5. HL)
1) First, it is given that $\angle R$ and $\angle V$ are right $\angle$s. So, $\triangle RST$ and $\triangle VST$ are right triangles. 2) It is also given that $\overline{RT}\cong\overline{VT}$ 3) By identity, we find that $\overline{ST}\cong\overline{ST}$ Now we see that the leg and hypotenuse of right $\triangle RST$ are congruent with the leg and hypotenuse of right $\triangle VST$. So we would use HL to prove triangles congruent. Now we would construct a proof for the problem: 1. $\angle R$ and $\angle V$ are right $\angle$s (1. Given) 2. $\triangle RST$ and $\triangle VST$ are right triangles. (2. A triangle that has one right angle is a right triangle) 3. $\overline{ST}\cong\overline{ST}$ (3. Identity) 4. $\overline{RT}\cong\overline{VT}$ (4. Given) 5. $\triangle RST\cong\triangle VST$ (5. HL)
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https://questions.examside.com/past-years/jee/question/if-length-of-tangent-at-any-point-on-the-curve-yfx-intecepte-jee-advanced-2005-marks-4-grkndgus4zmrdwg0.htm
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1
IIT-JEE 2005
Subjective
+4
-0
If length of tangent at any point on the curve $$y=f(x)$$ intecepted between the point and the $$x$$-axis is length $$1.$$ Find the equation of the curve.
2
IIT-JEE 2004
Subjective
+4
-0
A curve $$'C''$$ passes through $$(2,0)$$ and the slope at $$(x,y|)$$ as $$\,{{{{\left( {x + 1} \right)}^2} + \left( {y - 3} \right)} \over {x + 3}}$$. Find the equation of the curve. Find the area bounded by curve and $$x$$-axis in fourth quadrant.
3
IIT-JEE 2003
Subjective
+4
-0
A right circular cone with radius $$R$$ and height $$H$$ contains a liquid which eveporates at a rate proportional to its surface area in contact with air (proportionality constant $$= k > 0$$. Find the time after which the come is empty.
4
IIT-JEE 2001
Subjective
+10
-0
A hemispherical tank of radius $$2$$ metres is initially full of water and has an outlet of $$12$$ cm2 cross-sectional area at the bottom. The outlet is opened at some instant. The flow through the outlet is according to the law $$v(t)=0.6$$ $$\sqrt {2gh\left( t \right),}$$ where $$v(t)$$ and $$h(t)$$ are respectively the velocity of the flow through the outlet and the height of water level above the outlet at time $$t,$$ and $$g$$ is the acceleration due to gravity. Find the time it takes to empty the tank. (Hint: From a differential equation by relasing the decreases of water level to the outflow).
EXAM MAP
Medical
NEET
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https://pupman.com/listarchives/1995/september/msg00077.html
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# Re: Capacitor calculations
• To: tesla-at-grendel.objinc-dot-com
• Subject: Re: Capacitor calculations
• Date: Tue, 26 Sep 1995 10:41:16 EDT
• >Received: from ns-1.csn-dot-net (root-at-ns-1.csn-dot-net [199.117.27.21]) by uucp-1.csn-dot-net (8.6.12/8.6.12) with SMTP id IAA00878 for <tesla-at-grendel.objinc-dot-com>; Tue, 26 Sep 1995 08:44:06 -0600
```> I have seen several postings recently regarding calculating the
> correct capacitor size for a given transformer. I assume these values
> are coming from a program such as TESLADP. I tried this for my pole
> pig and get a value of .064 mfd. What does this mean? I would
> guess that this is the maximum value this power supply can charge
> effectively in a tesla tank circuit?
I'm not familiar with the Tesla programs either , but I have seen
capacitor selection described as a problem in impedence matching (i.e.
- for the most energy transfered, the impedence of the load should
match the impedence of the source).
In English, for a transformer with a given voltage (=V) and current (=I)
rating, you first calculate the impedence that will load the transformer to
it's rated capacity (R=V/I). For the maximum output from a transformer,
the capacitor you choose should have a capacitive reactance (=Xc)
equal to the transformer's impedence (=R) at your line frequency (f =
60Hz here in the good old USA, 50Hz elsewhere). The capacitive
reactance can be found using Xc = 1/(2 * pi * f * C) where C =
capacitance in Farads. From the desired Xc that you computed, it's
simple to find C = 1/(2*pi*f*Xc) to fully load your transformer. Since this
formula is for the steady state condition and a capacitive discharge
Tesla coil is not running at steady-state, the value of the capacitor
found above can be increased by 50% to 100%.
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https://www.includehelp.com/golang/convert-64-bit-float-number-into-32-bit-float-number.aspx
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Golang program to convert 64-bit float number into 32-bit float number
Here, we are going to learn how to convert 64-bit float number into 32-bit float number in Golang (Go Language)?
Submitted by Nidhi, on March 14, 2021 [Last updated : March 03, 2023]
Converting a 64-bit float number into 32-bit float number in Golang
Problem Solution:
In this program, we will convert the 64-bit float number into a 32-bit float number using the float32() function and print the result on the console screen.
Program/Source Code:
The source code to convert 64-bit a float number into a 32-bit float number is given below. The given program is compiled and executed successfully.
Golang code to convert a 64-bit float number into 32-bit float number
```// Golang program to convert 64-bit float number
// into 32-bit float number
package main
import "fmt"
func main() {
var floatNum64 float64 = 76.28
var floatNum32 float32 = 0
floatNum32 = float32(floatNum64)
fmt.Println("32-Bit float number: ", floatNum32)
}
```
Output:
```32-Bit float number: 76.28
```
Explanation:
In the above program, we declare the package main. The main package is used to tell the Go language compiler that the package must be compiled and produced the executable file. Here, we imported the fmt package that includes the files of package fmt then we can use a function related to the fmt package.
In the main() function, we two variables floatNum64, floatNum32 that are initialized with 76.28, 0 respectively.
After that, we converted the floatNum64 variable into a 32-bit float number and assigned it to the floatNum32 variable using float32() function, and print the result on the console screen.
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Verbal Reasoning - Reasoning quiz 9
Verbal Reasoning - Reasoning Online Quiz 9
Verbal Reasoning - Reasoning quiz 9 is a free online quiz challenge under Verbal Reasoning - Reasoning category. There are 277 free online quiz challenges available in Reasoning category
A man starts walking in the morning facing the sun. After sometime, he turned to his left. Later he again turned to his left. The direction in which the man is moving now is:
Starting from a point X, Ravi walked 20 m towards South. He turned left and walked 30m. He then turned left and walked 20 m. He again turned left and walked 40 m and reached at a point Y. How far and in which direction is the point Y from the point X?
Kiwi , Eagle , Emu , Penguin , Ostrich
Jonny left for his office in his car.He drove 15 km towards north and then 10 km towards west. He then turned to the south and covered 5 km.Further,he turned to the east and moved 8 km.Finally ,he turned right and drove 10 km .How far and in which direction is he from his starting point?
Directions: In the following question, select the related letters from the given alternatives?
MAN : PDQ :: WAN : ______
Kumar walks 10 meters in front and 10 meters to the right. Then every time turning to his left walks 5, 15 and 15 meters respectively. How far is he now from his starting point?
2 + 3 = 10 8 + 4 = 96 7 + 2 = 63 6 + 5 = 66 9 + 5 = ?
ATA is a place which is located 2km away in the north-west direction from the capital P. R is another place that is located 2 km away in the north-west direction from the capital P. R is another place that is located 2km away in the south-west direction from K. M is another place and that is located 2km away the north- westdirection from R. T is yet another place that is located 2km away in the southwest direction from M. In which direction is T located in relation to P?
Looking at the portrait of a man, Naveen said, "His mother is the wife of my father's son. I have no brothers or sisters." At whose portrait was Harsh was looking ?
A walks 10 m in front and 10m to the right. Then every time turning to his left he walks 5, 15 and 15m respectively. How for is he now from his starting point?
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https://rdrr.io/cran/seqtest/man/seqtest.cor.html
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# seqtest.cor: Sequential triangular test for Pearson's correlation... In seqtest: Sequential Triangular Test
## Description
This function performs the sequential triangular test for Pearson's correlation coefficient
## Usage
```1 2``` ```seqtest.cor(x, k, rho, alternative = c("two.sided", "less", "greater"), delta, alpha = 0.05, beta = 0.1, output = TRUE, plot = FALSE) ```
## Arguments
`x` initial data, i.e., Pearson's correlation coefficient in a sub-sample of k observations. `k` number of observations in each sub-sample. `rho` a number indicating the correlation coefficient under the null hypothesis, ρ.0. `alternative` a character string specifying the alternative hypothesis, `delta` minimum difference to be detected, δ. `alpha` type-I-risk, α. `beta` type-II-risk, β. `output` logical: if `TRUE`, output is shown. `plot` logical: if `TRUE`, an initial plot is generated.
## Details
Null and alternative hypothesis is specified using arguments `rho` and `delta`. Note that the argument k (i.e., number of observations in each sub-sample) has to be specified. At least k = 4 is needed. The optimal value of k should be determined based on statistical simulation using `sim.seqtest.cor` function.
In order to specify a one-sided test, argument `alternative` has to be used (i.e., two-sided tests are conducted by default). That is, `alternative = "less"` specifies the null hypothesis, H0: ρ >= ρ.0 and the alternative hypothesis, H1: ρ < ρ.0; `alternative = "greater"` specifies the null hypothesis, H0: ρ <= ρ.0 and the alternative hypothesis, H1: ρ > ρ.0.
The main characteristic of the sequential triangular test is that there is no fixed sample size given in advance. That is, for the most recent sampling point, one has to decide whether sampling has to be continued or either the null- or the alternative hypothesis can be accepted given specified precision requirements (i.e. type-I-risk, type-II-risk and an effect size). The sequence of data pairs must we split into sub-samples of length k >= 4 each. The (cumulative) test statistic `Z.m` on a Cartesian coordinate system produces a "sequential path" on a continuation area as a triangle. As long as the statistic remains within that triangle, additional data have to be sampled. If the path touches or exceeds the borderlines of the triangle, sampling is completed. Depending on the particular borderline, the null-hypothesis is either accepted or rejected.
## Value
Returns an object of class `seqtest`, to be used for later update steps. The object has following entries:
`call` function call `type` type of the test (i.e., correlation coefficient) `spec` specification of function arguments `tri` specification of triangular `dat` data `res` list with results
## Author(s)
Takuya Yanagida [email protected],
## References
Schneider, B., Rasch, D., Kubinger, K. D., & Yanagida, T. (2015). A Sequential triangular test of a correlation coefficient's null-hypothesis: 0 < ρ ≤ ρ0. Statistical Papers, 56, 689-699.
`update.seqtest`, `sim.seqtest.cor`, `seqtest.mean`, `seqtest.prop`, `print.seqtest`, `plot.seqtest`, `descript`
``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18``` ```#-------------------------------------- # H0: rho = 0.3, H1: rho != 0.3 # alpha = 0.05, beta = 0.2, delta = 0.2 seq.obj <- seqtest.cor(0.46, k = 14, rho = 0.3, delta = 0.2, alpha = 0.05, beta = 0.2, plot = TRUE) seq.obj <- update(seq.obj, c(0.56, 0.76, 0.56, 0.52)) #-------------------------------------- # H0: rho <= 0.3, H1: rho > 0.3 # alpha = 0.05, beta = 0.2, delta = 0.2 seq.obj <- seqtest.cor(0.46, k = 14, rho = 0.3, alternative = "greater", delta = 0.2, alpha = 0.05, beta = 0.2, plot = TRUE) seq.obj <- update(seq.obj, c(0.56, 0.76, 0.66)) ```
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# Binary searching the turning point of a function
The problem
You are given a list of $n$ positive integers. Your task is to split the list into $k$ subarrays so that the largest subarray sum is minimized.
The input
On the first line you have two integers n and k; the size of the list and the number of subarrays. On the second line you have n numbers: x1, x2, ..., xN.
The limits
$1 \leq n \leq 10^5$
$1 \leq k \leq n$
$1 \leq x \leq 10^9$
An example
Input
5 3
2 4 7 3 5
Output:
8 (the subarrays are [2,4], [7], [3,5])
My solution
Let's define a function isPossible, that returns true if it is possible to divide the list into $k$ subarrays so that the maximum sum is the parameter sum. The function loops through the user-inputted array and pushes elements into a vector as long as the elements do not exceed the parameter sum. When the vector can't take the next element without exceeding sum, a new vector is created. All these created vectors are pushed into a vector and if the size of the vector holding all the vectors is less than or equal to $k$, return true. Else return false.
The method vectorsum is a shortcut to getting the sum of the elements inside a vector.
After that I handle the input in main() and binary search the turning point of isPossible.
What went wrong?
I can handle the small test cases easily, but with larger numbers I get a "time limit exceeded"-error. However the time limit is 1 second and I don't think the test even took so long, so I suspect there might be something off with my data types. Any other improvements are accepted, too!
#include "stdafx.h"
#include <iostream>
#include <vector>
#include <numeric>
#include <math.h>
using namespace std;
int arraysize;
int numberofsubarrays;
long long usernumber;
vector<long long> userArray; //Used in main()
vector<vector<long long>> subarrays; //Used in isPossible()
int vectorsum(vector<long long> userVector) {
int ans = accumulate(userVector.begin(), userVector.end(), 0);
return ans;
}
bool isPossible(long long n) {
subarrays = {};
subarrays.push_back({});
for (auto u : userArray) {
subarrays.back().push_back(u);
if (vectorsum(subarrays.back()) <= n) {
continue;
}
else {
subarrays.back().pop_back();
subarrays.push_back({ u });
}
}
if ((long long)subarrays.size() > numberofsubarrays) {
return false;
}
else {
return true;
}
}
int main()
{
cin >> arraysize >> numberofsubarrays; //Input handling
for (int u = 0; u < arraysize; u++) {
cin >> usernumber;
userArray.push_back(usernumber);
}
long long left = 1;
long long middle;
long long right = pow(10, 10);
while (left + 1 != right) { //Binary search
middle = ceil((left + right) / 2);
/*cout << left << " " << middle << " " << right << endl;*/
if (isPossible(middle)) {
right = middle;
}
else {
left = middle;
}
}
cout << left + 1 << endl;
return 0;
}
Don't use using namespace std; Writing std:: a few times isn't that bad, and it helps keep the std lib types identifiable in your code.
Don't use globals so much. It is much better to limit the scope of variables to where they are needed and pass them around (possibly by reference) when needed. Usernumber for example is only used when filling the array, So something like this is much better:
for (int u = 0; u < arraysize; u++) {
long long number;
std::cin >> number;
userArray.push_back(number);
}
vectorsum has a vector passed by value, this will copy the entire vector. instead you'll want to pass by const ref:
int vectorsum(const std::vector<long long>& userVector) {
int ans = std::accumulate(userVector.begin(), userVector.end(), 0);
return ans;
}
You don't really need this function at all though, you can keep a running total instead and save yourself the loop.
You keep rebuilding the vectors over and over again, you don't even need to do that. Instead you only need to keep track of the sum of the current subarray you are testing and how many subarrays you have built already.
You can also keep track of the largest subarray sum and pass it back as new upperbound. This will reduce the iteration time once the options have been narrowed down somewhat because then you will never try 2 numbers that will create the same partition.
long long isPossible(const std::vector<long long> &user_array, long long max_sum, int max_number_subarrays) {
long long running_sum = 0;
long long largest_sum = 0;
int subarray_count = 0;
for (auto current : userArray) {
if (running_sum + current <= max_sum) {
running_sum += current;
} else {
if(running_sum > largest_sum)
largest_sum = running_sum;
running_sum = current;
subarray_count++;
if(subarray_count > max_number_subarrays)
return -1; //early out
}
}
return largest_sum;
}
and in the while loop.
while (left + 1 != right) { //Binary search
middle = ceil((left + right) / 2);
long long result = isPossible(userArray, middle, numberofsubarrays)
if (result >= 0) {
right = result;
}
else {
left = middle;
}
}
std::cout << right << endl;
You can initialize left with the largest number in the input Because if it was less than at least one sub array sum will be larger than it.
right can be initialized with the sum of the array where every element is in a single subarray.
int left = 0;
int right = 0;
for (int u = 0; u < arraysize; u++) {
long long number;
std::cin >> number;
userArray.push_back(number);
if(left < number)
left = number;
right+= number;
}
• Thanks for all your helpful tips. I'm left with two questions: 1. sum = current;? What is sum? It is not declared anywhere. 2. Argument list for class template "std::vector" is missing? How do I fix that? Commented Jul 4, 2018 at 14:47
• 1. I forgot to change that one when I renamed the variable. and 2. I forgot to add the template param to the argument in isPossible. Commented Jul 4, 2018 at 15:10
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Tagged Questions
Prime numbers are natural numbers greater than 1 not divisible by any smaller number other than 1. This tag is intended for questions about, related to, or involving prime numbers.
86 views
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Small primes congruent to $a$ mod $p$.
Let $p$ be a prime and $a$ be an integer such that $0 \lt a \lt p$. Is there a prime number, $q$, congruent to $a$ mod $p$ such that $q\lt p^2$? I have checked that this is true for the first $3000$...
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Euclid Mullin Sequence
Consider the Sequence as follows. Let $a_1 = 2$, $a_n$ be the largest prime divisor of $P_n = 1 + {\prod_{i = 1}^{n - 1} a_{i}}$ Then we obtain a sequence of prime numbers How do you show that 5 ...
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Examples of Weil's explicit formula
In Bombieri, PROBLEMS OF THE MILLENNIUM: THE RIEMANN HYPOTHESIS, Clay Mathematics Institute (2000), from page 8, V. Further evidence: the explicit formula the author tell us that there is a ...
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Which primes $p$ divide $q^q-1$ for a prime divisor $q$ of $p-1$
I am looking for (a formula) for all the primes $p$ less than or equal to $X$ with the following criteria: There is at least one prime $q$ dividing $p-1$ such that $p$ divides $q^q-1$. $7$, for ...
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Can this function be a new test for primality?
The following function returns always 0 only if a number is not prime. $$H(x)=\prod_{i=2}^{x-1}\left\{\left[\sum_{k=1}^{x/i}(-i)\right]+x\right\}$$ what do you think? Bye!
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For each prime $p>3$ there are non twin primes $q,r$ with $p^3=2q+r$
Define $\mathbb P'=\{n\in\mathbb P|n-2,n+2\notin \mathbb P\}$. Conjecture: Given a prime $p>3$, then $\exists q,r\in\mathbb P':p^3=2q+r.$ Tested for the first 10000 primes. The solutions ...
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Maximum length of a string that has no substring divisible by a prime number $p$ is $p-1$?
What is the maximum length of a string of nonzero digits that has no substring that is divisible by a given prime number? I want to find a string of length n which has no substring divisible by the ...
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multiples (of primes) coverage formula
I apologize in advance if my explanation is not clear. Please let me know if clarification is required and I will do my best to fix it! I am attempting to find an explicit formula (in terms of ...
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Conjecture: Every prime number is the difference between a prime number and a power of $2$
Conjecture: $\forall p\in\mathbb P\exists q\in\mathbb P\exists n\in \mathbb N: q-p=2^n$ Verified for the 100 first primes.
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A statement about divisibility of relatively prime integers
I'm solving a problem, and the solution makes the following statement: "The common difference of the arithmetic sequence 106, 116, 126, ..., 996 is relatively prime to 3. Therefore, given any three ...
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Does any sum of twin primes, where the sum is greater than 12, also represents the sum of 2 other distinct primes?
I was in the midst of proving a conjecture when I came across an observation that led me to forming a potentially new conjecture. The conjecture goes as follows: Any given sum of twin primes (...
Prime conjecture containing primorial: the difference between the primorial $n\#$ and the smallest prime $p > n\# + 1$ is always a prime
Help me find the exact conjecture statement. What I roughly remember is that it stated that the difference between primorial $n\#$ (product of first $n$ primes) and "some" larger number than the ...
A conjecture about the prime function $p_n$: $p_m \cdot p_n >p_{m \cdot n}$
While testing my system Zet for computational mathematics I find possible relations now and then. The latest is: Conjecture: For all $(m,n)\in\mathbb Z_+^2$ except $(3,4),(4,3) \text{ and } (4,4)$...
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# Transistors
Discussion in 'General Electronics Discussion' started by TechNotes, Sep 5, 2011.
1. ### TechNotes
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I am trying to find a good resistance value for the base and collector of a npn transistor. I have no idea how to do this so please help. I have a MPS650GOS transistor. The page I ordered from, the datasheet, and the schematic is below.
The schematic I am working from: http://www.instructables.com/files/orig/FJ4/7RE1/GICYB6CI/FJ47RE1GICYB6CI.png
Datasheets:
http://parts.digikey.com/1/parts/1007871-trans-npn-ss-2a-40v-to92-mps650g.html
Thanks!
2. ### jackorocko
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What is your required current? Divide this by the HFE (transistor DC gain) and you will know your base current. If you know the base voltage, you should just be able to use Ohms law to figure the resistor needed to provide the current. Don't forget, you will need to make sure the transistor is on, so take into account the base-emitter junction voltage drop.
edit: Formula is (Vb - Vbe)/Ib
Last edited: Sep 5, 2011
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4. ### jackorocko
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Base-emitter junction is typically 0.7V for a silicon based transistor. Transistor DC gain is 'HFE' found in the datasheet(I looked, it is anywhere from 40 to 75 based on Ic, it is always best to use the low number which gives you some extra breathing room). The base voltage is what your circuit will be providing to the base of the transistor. Since all your transistors are being powered by the MEGA32, it would probably be listed in the MEGA32 datasheet.
Vb - Voltage applied to the Base of the transistor.
Vbe - base-emitter junction voltage drop, explained above
Ib - Base current, you will need to figure this out on your own, based upon the current you need divided by the HFE of the transistor.
Last edited: Sep 5, 2011
5. ### TechNotes
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Thank you so much! This really helped
6. ### TechNotes
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I have another question. How do you find the value of the resistor on the collector using the same transistor and schematic as stated above?
7. ### rogerk8
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This depends on what you want to do. Normally you want the transistor to be able to swing equally posively and negatively so you leave half of the rest of the voltage (the voltage over the emitter-resistance in a high gain CE-configuration is normally low and may be disregarded) over the transistor and the rest over the collector-resistance. Using Ohm's law it is then easy to calculate the resistance needed. Hope this will help. KR/Roger
8. ### TechNotes
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Sep 5, 2011
Sorry, I really am a newbie. What is high gain CE-configuration? And what do you mean by I want the transistor to be able to swing equally positively and negatively? I also don't see how using Ohm's Law will help. Please explain.
Thanks!
9. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator
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By CE, Roger means "Common Emitter". This is the fairly common connection you see for a transistor where (for an NPN transistor) the emitter is grounded (often via a relatively small resistance), the input is to the base of the transistor (often via a resistor) and the load or a load resistor is placed between the collector and the positive supply rail.
Here is a very simple example:
In practice, unless the transistor is simply used to switch something on and off, some form of biasing is required. Biasing turns the transistor part way on. This means that an audio dignal input to the base of the transistor can turn the transistor on more (with positive half cycles) and off more (negative half cycles).
Without the biasing, only half of the waveform would be amplified. For audio signals this would produce huge amounts of distortion.
A very simple form of biasing is to attach a resistor between the base and some positive bias voltage.
The disadvantage of this circuit is that the output voltage is highly dependant on the transistor characteristics and the temperature. In addition any DC voltage on the input will affect the bias, and the DC voltage on the output will affect the bias of any following stage.
It is clear that there is no positive and negative output. The output MUST always be positive with respect to ground. The output is more properly described as AC with a DC bias, or perhaps DC with ripple .
In practice, you might use a voltage divider to set the bias point as this makes the amplifier more stable. An emitter resistor provides negative feedback (which can be bypassed with a capacitor to provide higher gain at higher -- audio -- frequencies). Capacitors are used between stages to isolate the DC bias between each stage. It is the capacitor that allows you to have a truly AC (absent of DC bias) input or output.
So eventually you get a design that looks like this:
Not all single transistor amplifier stages look like this, there are many variations. In addition there are other ways to eliminate DC bias that do not require capacitors.
However, understanding a basic common emitter amplifier will help you in a huge number of cases where you need to use a transistor,
Oh, and you are right in questioning the statement about ohms law. In determining what value resistors to use there is more to it than just ohms law.
For more information, you might like to check this out. Page 2 is where it starts talking about the common emitter amplifier.
Last edited: Sep 8, 2011
10. ### rogerk8
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Better late than never...I like your explanation of my kind of stupid way of trying to help. It even learned me some. And you're right, there's lots more to properly bias a transistor than Ohm's Law. It is kind of delicate actually. But mostly the gain is close to the Rc/Re-ratio. It is however not exactly this because it depends on input driving impedance (Rs), hie, Re and hfe. I have studied this closely like the amateur I am and have come to this conclusion. Yet my Germanium pre-amp using a single transistor works just fine (Rc/Re=10 but actual gain ended at 7). Take care! BR/Roger
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Alejandrox
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# Trapezoid is 6 feet by 5 feet by 1.5 feet what is area
The area would be $\frac{1}{2} h (b _{1}+ b_{2})$. So 5+1.5 is 6.5. 6times6.5 is 39. 39 divided by 2 is 19.5. 19.5 feet squared is the area.
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# Who came up with the laws of conservation of momentum?
Who came up with the laws of conservation of momentum? I'm more specifically interested in the conservation of angular momentum.
• Hi, you might want to have a look at this page to improve your question: hsm.stackexchange.com/help/how-to-ask – VicAche Jun 27 '15 at 17:02
• Angular momentum isn't momentum. – Ben Crowell Jun 28 '15 at 12:12
• Considering that angular momentum isn't momentum, this is really two questions in one. Also, you really should add some research to show your efforts, as explained in the link provided by VicAche. – Cicero Jun 29 '15 at 14:17
As to the first part of your question, conservation of momentum comes directly from Newton's Laws, so technically Newton himself came up with conservation of momentum, and others followed Newton's logic to notice that in certain scenarios that other conservation conditions exist (energy, angular momentum, mass, etc).
Descartes wrote in *Le traité du monde et de la lumière *:
…when one of these bodies pushes another, it cannot give the 2nd any motion, except by losing as much of its own motion at the same time… (AT, XI, 41)
Motion being "le transport d'une partie de la matière, ou d'un corps" this is probably the first enonciation of the conservation law for linear momentum. The book was published in 1664, some 30 years after it was written, and meanwhile the Principles of philosophy published in 1644 stated:
Lorsqu'une partie de la matière se meut deux fois plus vite qu'une autre, et que cette autre est deux fois plus grande que la première, nous devons penser qu'il y a tout autant de mouvement dans la plus petite que dans la plus grande ; et que toutes fois et quantes que le mouvement d'une partie diminue, celui de quelque autre partie augmente à proportion.
Introducing a more refined "quantity of movement" the book however stated some erroneous propositions.
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# Need help with a triangle's length side/ angle relation.
• Dyatlov
In summary, to find the length of a side in a triangle, you can use the Pythagorean theorem or trigonometric ratios. To find the measure of an angle, you can use the law of cosines or law of sines, or trigonometric ratios for a right triangle. The lengths of the sides in a triangle are related by the Pythagorean theorem or the law of cosines/sines. The angle-side-angle congruence criterion can be used to solve for a missing side or angle in a triangle. It is possible to find the length of a side or measure of an angle in a triangle with limited information, but additional information may be needed.
Dyatlov
Hello.
I am trying to wrap my head around where from did he got the x = sin of theta equation at the 32:44 mark of the video: .
Isn't sine of theta x over the hypotenuse in the diagram ?
Dyatlov said:
Isn't sine of theta x over the hypotenuse in the diagram ?
Yes it is, and the length of the hypotenuse (=the pendulum arm) seems to be 1.
## 1. How do I find the length of a side in a triangle?
To find the length of a side in a triangle, you can use the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. You can also use trigonometric ratios to find the length of a side in any triangle.
## 2. How do I find the measure of an angle in a triangle?
To find the measure of an angle in a triangle, you can use the law of cosines or the law of sines, depending on the information you have about the triangle. You can also use trigonometric ratios to find the measure of an angle in a right triangle.
## 3. What is the relationship between the lengths of the sides in a triangle?
In a right triangle, the lengths of the sides are related by the Pythagorean theorem. In any triangle, the lengths of the sides are related by the law of cosines or the law of sines. Additionally, the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.
## 4. How do I use the angle-side-angle (ASA) congruence criterion to solve for a missing side or angle in a triangle?
The angle-side-angle (ASA) congruence criterion states that if two triangles have two congruent angles and the included side between the angles is also congruent, then the triangles are congruent. This can be used to solve for a missing side or angle in a triangle by setting up an equation and solving for the unknown variable.
## 5. Is it possible to find the length of a side or measure of an angle in a triangle if I only know the lengths of two other sides or measures of two other angles?
Yes, it is possible to find the length of a side or measure of an angle in a triangle if you know the lengths of two other sides or measures of two other angles. You can use the law of cosines or the law of sines to solve for the missing side or angle. However, you may need to use additional information, such as the sum of the angles in a triangle being 180 degrees, to solve for the unknown variable.
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# "lambda" term usage in programming
could any one please let me know what is the relation between "lambda" and anonymous functions in programming?
in other words why we say lambda function to an anonymous function?
I am here trying to understand the reason for use of the term
• not sure why forum has made two minuses for this question. A considerable amount of knowledge sharing been happened, so IMO this post not deserve -2 Jul 31, 2013 at 12:44
• Good answers doesn't mean good question. You could have found the answer easily by googling for lambda notation history, e.g. math.stackexchange.com/q/64468. Aug 2, 2013 at 6:16
• @Kaveh If you carefully read the question , you would notice the question is in the programming context not in mathematics. And people won't get same question in their head even they might lead to a same answer. On the other hand it is intellectually not fair to categories questions as bad or good , and as most would agree a question would NEVER be bad. Aug 2, 2013 at 8:05
• I have read the question. You seem to be new to Stack Exchange. I haven't down voted your question, I just explained why your question has got 4 down votes and 1 close vote. On cstheory a question can be considered bad for several reasons, e.g. when the answer can be found easily by searching as is in this case. I would suggest you check help center. Aug 2, 2013 at 8:22
There is a popular formal system called Lambda calculus. It is Turing complete and has great importance in the theory of programming languages. Lambda calculus is based on anonymous functions with very simple syntax. You start a new function with a $\lambda$, list the arguments, place a dot, and write down the return value. For example, $\lambda x.x$ is the identity function, $\lambda x.xx$ applies the functional argument $x$ to itself.
Since programming language designers know Lambda calculus very well, it seemed like a good idea to call anonymous functions in their language lambdas.
• Great explanation, many thanks for that."λx.xx applies the functional argument x to itself." does this mean if x ==2 , then return is 4? or you meant something else? Jul 24, 2013 at 11:13
• could you please suggest a resource such as a book to gather knowledge for Lambda Calculus (in perspective of programming) ? I have seen many advance book that could make someone really tired ! Jul 24, 2013 at 11:21
• λx.xx, or rather λx.(x x), means "apply x to x." Or think of it as x(x). For example, if arg = λx.(x x), then (λx.(x x) arg) = (λx.(x x) λx.(x x)), which has reached a fixed point. It can't be simplified further. Here the argument is a function and the return result is an application. Note that application is as important as function in lambda calculus. Jul 24, 2013 at 13:16
• Oh I see lets say I want to convert following function into lambda, public int sum(int x , int Y){ return x+y; } what would be the representation using lambda? Jul 24, 2013 at 13:22
• λxy.(+ x y). Here + is a function and it's applied to two arguments. Jul 25, 2013 at 15:40
By the way, why did Church choose the notation “λ”? In [Church, 1964, §2] he stated clearly that it came from the notation “xˆ” used for class-abstraction by Whitehead and Russell, by first modifying “xˆ” to “∧x” to distinguish function- abstraction from class-abstraction, and then changing “∧” to “λ” for ease of printing. This origin was also reported in [Rosser, 1984, p.338]. On the other hand, in his later years Church told two enquirers that the choice was more accidental: a symbol was needed and “λ” just happened to be chosen.
-- History of Lambda-Calculus and Combinatory logic, F. Cardone and J. R. Hindley, 2009
The reason lambda is used is because of the lambda-calculus, where λx.E (for any expression E) is used to denote the function which takes a value a and returns E[a/x], meaning "the result of substituting a for x in E". Thus, (λx.x+1)(5) = 5+1 = 6.
This however doesn't answer the question why specifically λ was chosen for this role. The answer requires a little bit of history, but if you're interested, here it is.
In the Principia Mathematica, the notation "ẑP(z)" was introduced for "the set of all z such that P(z)" (which we would now write as {z:P(z)}). The notation was quite popular for a while after, and hence was the one Alonzo Church (the creator/discoverer of the λ-calculus) was familiar with.
Alonzo Church, was interested in a view of mathematics where sets were interpreted as functions from individuals to truth-values. Thus rather than writing 6 ∈ ẑ(z > 5), we could just write "[ẑ(z > 5)](6)", which would have the same value as "(6 > 5)", and hence TRUE. Viewed this way, there should be no difference between an expression like "ẑ(z > 5)" and "ẑ(z+5)", as they both represent functions which can accept values. "[ẑ(z + 5)](6)" would then have the same value as "(6+5)", which is 11.
The long and short of it is, when he first started writing papers on the formal properties of these systems, there were some difficulties with typesetting "ẑ", so "ẑ" became "^z", which then became "λz" for stylistic reasons.
Later, when students would ask him how he decided on the letter "λ", he used to claim "eenie meanie minee moe".
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## Wednesday, December 26, 2012
### Teaching Conversions to the Right Brained Child
Right brained children work better when they can see things in pictures. When converting from one unit of measure to another, this image will help ALL children remember, but especially right-brained children who might not remember it otherwise.
This image "says" that 1 gallon = 4 qts. 1 qt = 2 pts. 1 pt. = 2 cups, etc. Quick...how many pints in a gallon? How many cups in a gallon? How many cups in 4 gallons? All can be solved quickly by using this picture. Kids might have 2-3 days of unit conversions before moving on to something else. If they draw this 2-3 times, they will remember it on their own, and eventually, will be able to see it in their heads.
On a test, such as a state assessment, they cannot bring this in with them, but they can jot it down real quick when a conversion problem pops up. Now if you remember, there might be 2-3 of these on the whole test. Why bother to teach them this for just a couple of answers? Well, I've needed this info in real life...so it IS one of those things that is helpful to just know.
#### 1 comment:
Anonymous said...
This really helped me =)
Thank you so much!
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# Someone Else’s Homework: Was It Hard? An Umbrella Search
I wanted to follow up, at last, on this homework problem a friend had.
The question: suppose you have a function f. Its domain is the integers Z. Its rage range is also the integers Z. You know two things about the function. First, for any two integers ‘a’ and ‘b’, you know that f(a + b) equals f(a) + f(b). Second, you know there is some odd number ‘c’ for which f(c) is even. The challenge: prove that f is even for all the integers.
My friend asked, as we were working out the question, “Is this hard?” And I wasn’t sure what to say. I didn’t think it was hard, but I understand why someone would. If you’re used to mathematics problems like showing that all the roots of this polynomial are positive, then this stuff about f being even is weird. It’s a different way of thinking about problems. I’ve got experience in that thinking that my friend hasn’t.
All right, but then, what thinking? What did I see that my friend didn’t? And I’m not sure I can answer that perfectly. Part of gaining mastery of a subject is pattern recognition. Spotting how some things fit a form, while other stuff doesn’t, and some other bits yet are irrelevant. But also part of gaining that mastery is that it becomes hard to notice that’s what you’re doing.
But I can try to look with fresh eyes. There is a custom in writing this sort of problem, and that drove much of my thinking. The custom is that a mathematics problem, at this level, works by the rules of a Minute Mystery Puzzle. You are given in the setup everything that you need to solve the problem, yes. But you’re also not given stuff that you don’t need. If the detective mentions to the butler how dreary the rain is on arriving, you’re getting the tip to suspect the houseguest whose umbrella is unaccounted for.
(This format is almost unavoidable for teaching mathematics. At least it seems unavoidable given the number of problems that don’t avoid it. This can be treacherous. One of the hardest parts in stepping out to research on one’s own is that there’s nobody to tell you what the essential pieces are. Telling apart the necessary, the convenient, and the irrelevant requires expertise and I’m not sure that I know how to teach it.)
The first unaccounted-for umbrella in this problem is the function’s domain and range. They’re integers. Why wouldn’t the range, particularly, be all the real numbers? What things are true about the integers that aren’t true about the real numbers? There’s a bunch of things. The highest-level things are rooted in topology. There’s gaps between one integer and its nearest neighbor. Oh, and an integer has a nearest neighbor. A real number doesn’t. That matters for approximations and for sequences and series. Not likely to matter here. Look to more basic, obvious stuff: there’s even and odd numbers. And the problem talks about knowing something for an odd number in the domain. This is a signal to look at odds and evens for the answer.
The second unaccounted-for umbrella is the most specific thing we learn about the function. There is some odd number ‘c’, and the function matches that integer ‘c’ in the domain to some even number f(c) in the range. This makes me think: what do I know about ‘c’? Most basic thing about any odd number is it’s some even number plus one. And that made me think: can I conclude anything about f(1)? Can I conclude anything about f at the sum of two numbers?
Third unaccounted-for umbrella. The less-specific thing we learn about the function. That is that for any integers ‘a’ and ‘b’, f(a + b) is f(a) + f(b). So see how this interacts with the second umbrella. f(c) is f(some-even-number) + f(1). Do I know anything about f(some-even-number)?
Sure. If I know anything about even numbers, it’s that any even number equals two times some integer. Let me call that some-integer ‘k’. Since some-even-number equals 2*k, then, f(some-even-number) is f(2*k), which is f(k + k). And by the third umbrella, that’s f(k) + f(k). By the first umbrella, f(k) has to be an integer. So f(k) + f(k) has to be even.
So, f(c) is an even number. And it has to equal f(2*k) + f(1). f(2*k) is even; so, f(1) has to be even. These are the things that leapt out to me about the problem. This is why the problem looked, to me, easy.
Because I knew that f(1) was even, I knew that f(1 + 1), or f(2), was even. And so would be f(2 + 1), that is, f(3). And so on, for at least all the positive integers.
Now, after that, in my first version of this proof, I got hung up on what seems like a very fussy technical point. And that was, what about f(0)? What about the negative integers? f(0) is easy enough to show. It follows from one of those tricks mathematics majors are told about early. Somewhere in grad school they start to believe it. And that is: adding zero doesn’t change a number’s value, but it can give you a more useful way to express that number. Here’s how adding zero helps: we know c = c + 0. So f(c) = f(c) + f(0) and whether f(c) is even or odd, f(0) has to be even. Evens and odds don’t work any other way.
After that my proof got hung up on what may seem like a pretty fussy technical point. That amounted to whether f(-1) was even or odd. I discussed this with a couple people who could not see what my issue with this was. I admit I wasn’t sure myself. I think I’ve narrowed it down to this: my questioning whether it’s known that the number “negative one” is the same thing as what we get from the operation “zero minus one”. I mean, in general, this isn’t much questioned. Not for the last couple centuries.
You might be having trouble even figuring out why I might worry there could be a difference. In “0 – 1” the – sign there is a binary operation, meaning, “subtract the number on the right from the number on the left”. In “-1” the – sign there is a unary operation, meaning, “take the additive inverse of the number on the right”. These are two different – signs that look alike. One of them interacts with two numbers. One of them interacts with a single number. How can they mean the same thing?
With some ordinary assumptions about what we mean by “addition” and “subtraction” and “equals” and “zero” and “numbers” and stuff, the difference doesn’t matter much. We can swap between “-1” and “0 – 1” effortlessly. If we couldn’t, we probably wouldn’t use the same symbol for the two ideas. But in the context of this particular question, could we count on that?
My friend wasn’t confident in understanding what the heck I was getting on about. Fair enough. But some part of me felt like that needed to be shown. If it hadn’t been recently shown, or used, in class, then it had to go into this proof. And that’s why I went, in the first essay, into the bit about additive inverses.
This was me over-thinking the problem. I got to looking at umbrellas that likely were accounted for.
My second proof, the one thought up in the shower, uses the same unaccounted-for umbrellas. In the first proof, the second unaccounted-for umbrella seemed particularly important. Knowing that f(c) was odd, what else could I learn? In the second proof, it’s the third unaccounted-for umbrella that seemed key. Knowing that f(a + b) is f(a) + f(b), what could I learn? That right away tells me that for any even number ‘d’, f(d) must be even.
Call this the fourth unaccounted-for umbrella. Every integer is either even or odd. So right away I could prove this for what I really want to say is half of the integers. Don’t call it that. There’s not a coherent way to say the even integers are any fraction of all the integers. There’s exactly as many even integers as there are integers. But you know what I mean. (What I mean is, in any finite interval of consecutive integers, half are going to be even. Well, there’ll be at most two more odd integers than there are even integers. That’ll be close enough to half if the interval is long enough. And if we pretend we can make bigger and bigger intervals until all the integers are covered … yeah. Don’t poke at that and do not use it at your thesis defense because it doesn’t work. That’s what it feels like ought to work.)
But that I could cover the even integers in the domain with one quick sentence was a hint. The hint was, look for some thing similar that would cover the odd integers in the domain. And hey, that second unaccounted-for umbrella said something about one odd integer in the domain. Add to that one of those boring little things that a mathematician knows about odd numbers: the difference between any two odd numbers is an even number. ‘c’ is an odd number. So any odd number in the domain, let’s call it ‘d’, is equal to ‘c’ plus some even number. And f(some-even-number) has to be even and there we go.
So all this is what I see when I look at the question. And why I see those things, and why I say this is not a hard problem. It’s all in spotting these umbrellas.
## Author: Joseph Nebus
I was born 198 years to the day after Johnny Appleseed. The differences between us do not end there. He/him.
## 3 thoughts on “Someone Else’s Homework: Was It Hard? An Umbrella Search”
1. Given that a – b is a combination of three symbols and the minus sign is an operation ,we can build the plus operation a + b from the minus operation, as a – (0 – b)
The plus operation is commutative and associative, but the minus operation is neither.
What a shame, as we don’t need the plus sign at all.
I’ve got more on this, later.
Like
1. Ooh, that’s a nice, slick observation. It puts me in mind of how the NAND logical operator can be used to produce all the common basic operations.
Liked by 1 person
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Plotting functions on top of datapoints in R
Is there a way of overlaying a mathematical function on top of data using ggplot?
``````## add ggplot2
library(ggplot2)
# function
eq = function(x){x*x}
# Data
x = (1:50)
y = eq(x)
# Make plot object
p = qplot(
x, y,
xlab = "X-axis",
ylab = "Y-axis",
)
# Plot Equation
c = curve(eq)
# Combine data and function
p + c #?
``````
In this case my data is generated using the function, but I want to understand how to use `curve()` with ggplot.
-
2 Answers
You probably want `stat_function`:
``````library("ggplot2")
eq <- function(x) {x*x}
tmp <- data.frame(x=1:50, y=eq(1:50))
# Make plot object
p <- qplot(x, y, data=tmp, xlab="X-axis", ylab="Y-axis")
c <- stat_function(fun=eq)
print(p + c)
``````
and if you really want to use `curve()`, i.e., the computed x and y coordinates:
``````qplot(x, y, data=as.data.frame(curve(eq)), geom="line")
``````
-
Given that your question title is "plotting functions in R", here's how to use `curve` to add a function to a base R plot.
Create data as before
``````eq = function(x){x*x}; x = (1:50); y = eq(x)
``````
Then use `plot` from base graphics to plot the points followed by `curve` with the `add=TRUE` argument, to add the curve.
``````plot(x, y, xlab = "X-axis", ylab = "Y-axis")
curve(eq, add=TRUE)
``````
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Binary logic. Truth tables
11.3.3.3 build truth tables AND, OR, NOT, NAND, NOR, XOR
Binary logic. Truth tables
Binary logic
At the most elementary level, an elecrtonic device can only recognise the presence or absence of current or voltage. Either electricity is present or isn't. This is a switch - on or off, True or False, 1 or 0. With a computer's semiconductor, the voltage at the input and output terminals is measured and is either high or low; 1 or 0. Computers comprise billions of these switches and manipulating these sequences of Ons and Offs can change individual bits.
Electronic logic gates can take one or more inputs and produce a single output. This output can become the input to another gate and a complicated cascaded sequence of logic gates can be implemented to from a circuit in, for example, the CPU.
Logic gates and truth tables
There are a number of different logic gates that are each designed to perform a different operation in terms of output. These are: NOT, AND, OR, XOR, NAND and NOR gates.
source of gif: vivaxsolutions.com
What is a truth table?
A truth table is a table representing the output boolean values of a logical expression based on their entries. The table thus presents all the possible combinations of the input logical variables (generally 0 / FALSE and 1 / TRUE) and the result of the equation as output.
NOT gate
The NOT gate is represented by the symbol below and inverts the input. The small circle denotes an inverted input.
Using 1s and 0s as input to a gate, its operation can summarised in the form of a truth table.
NOR gate Truth table
Q = NOT A
Input A Output Q
1 0
0 1
AND gate
AND gate Truth table
Q = A AND B
Input A Input B Output Q
1 1 1
1 0 0
0 1 0
0 0 0
The boolean expression for AND is written: Q = A * B where * represents AND.
The truth table reflects the fundamental property of the AND gate: the output of A AND B is 1 only if input A and input B are both 1.
OR gate
OR gate Truth table
Q = A OR B
Input A Input B Output Q
1 1 1
1 0 1
0 1 1
0 0 0
The boolean expression for OR is written: Q = A + B where + represents OR.
The truth table reflects the fundamental property of the OR gate: the output of A OR B is 1 if input A or input B is 1.
Creating logic gate circuits
Multiple logic gates can be connected to produce an output based on multiple inputs
This citcuit can be represented by Q = (NOT A) OR (B AND C) or Q = -A + (B * C) and shown using the truth table below:
Input A Input B Input C D = NOT A E = B AND C Q = D OR E
1 1 1 0 1 1
1 1 0 0 0 0
1 0 1 0 0 0
1 0 0 0 0 0
0 1 1 1 1 1
0 1 0 1 0 1
0 0 1 1 0 1
0 0 0 1 0 1
XOR gate
The XOR (ex-or) gate stands for exclusive OR, meaning that the output will be true if one or other input is true, but not both. Compare this to the OR gate? Which outputs true if either or both inputs are true.
XOR gate Truth table
Q = A XOR B
Input A Input B Output Q
1 1 0
1 0 1
0 1 1
0 0 0
The Boolean algebraic expression is written: Q = A ⊕ B where ⊕ the represents XOR, and is the equivalent of Q = (A * (-B)) + ((-A) * B). This gate similar to the OR gate but excludes the condition where A and B are both true.
NAND gate
The NAND gate is an combination of the AND and NOT gates, which inverts the output of the AND gate. Having a single type of NAND gate that can that can perform two separate functions can help to reduce development costs if a NAND gate is cheaper than separate AND and NOT gates.
NAND gate Truth table
Q = NOT (A AND B)
Input A Input B Output Q
1 1 0
1 0 1
0 1 1
0 0 1
The Boolean algebraic expression id written: .
NOR gate
NOR gate Truth table
Q = NOT (A OR B)
Input A Input B Output Q
1 1 0
1 0 0
0 1 0
0 0 1
The Boolean algebraic expression is written: . This gate only produces an output of true when both inputs are false.
Questions:
1) Explain what is boolean data type.
2) Build from memory the truth tables AND, OR, NOT, NAND, NOR, XOR.
3) What difference between AND and OR?
4) What difference between AND and NAND?
5) What difference between OR and XOR?
Exercises:
Ex. 1
Ex. 2
Test "Boolean logic"
Ex. 3
Ex. 4
Ex. 5
Ex. 6
Logic gates - Inteactive QUIZ
Exam questions:
Draw a truth table for the following circuit (Marks: ):
• Input A Input B Input C D = B OR C E = A AND D Q = NOT E
1 1 1 1 1 0
1 1 0 1 1 0
1 0 1 1 1 0
1 0 0 0 0 1
0 1 1 1 0 1
0 1 0 1 0 1
0 0 1 1 0 1
0 0 0 0 0 1
Категория: Boolean logic | Добавил: bzfar77 (11.09.2020)
Просмотров: 727 | Теги: and, Or, Nor, Logic, logic gate, nand, XOR, not, Boolean, truth tables | Рейтинг: 5.0/1
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# How to deal with inverse distributed load ramp starts and ends somewhere along the beam?
I know how to solve it by moment line methode but i want to use singularity function here and i'm very confused. My strategy consist of writing the correspond singularity function for each element but i have absolutely no clue how can i write the right singularity function for the inverse ramp distributed load at the right side?
Are you familiar with the Dirac delta function, Heaviside step function and the Ramp function?
You could use Laplace Transforms together with those three handy functions:
Dirac Delta function
$$\delta(x)= \left\{\begin{array}{l}+\infty\qquad&x=0\\0&x\neq0\end{array}\right.$$ The Dirac Delta function is equal to infinity, at $x=0$, and $0$ everywhere else. Respectively $\delta(x-a)=+\infty$ if $x=a$.
With the Euler-Bernoulli beam theory, where $$EI\frac{d^4w}{dx^4}=q(x)$$ a static interpretation of the Dirac function is, that a point load $Q$ at point a represents a distributed load with infinite density, thus can be written as $q(x)=Q\cdot \delta(a-x)$
Heaviside step function
The step function is the integral of the Dirac Delta function $$H(x)=\int_{-\infty}^x\delta(s)ds$$ or $$H(x)= \left\{\begin{array}{l}1\qquad&x\geq0\\0&x<0\end{array}\right.$$ Thus, a distributen load, with magniuted $q$, between $x=a$ and $x=b$ can be written as $q(x)=q\left[ H(x-a)-H(x-b) \right]$
Ramp function
The ramp function again is the integral of the Heaviside function. $$R(x)= \left\{\begin{array}{l}x\qquad&x\geq0\\0&x<0\end{array}\right.$$ Thus, a distributied load , starting at $x=a$, with a slope $m$ can be written as $q(x)=m\cdot R(x-a)$
The following distributed load, for example:
This load can be written as: $$q(x)=\frac{q}{b-a}\left[R(x-a)-R(x-b)\right]-\frac{q}{d-c}\left[R(x-c)-R(x-d)\right]$$ please see attached file for the derivation
Further explanation of the Dirac Delta function
Take, for example the point force $Q$. That force can be approximated by a distributed load $Q=q\cdot dx$, therefore $q=\frac{Q}{dx}.$ Now, as a point load acts on an interval $dx\to0$, we can say $\lim_{dx\to0} q=+\infty$
This is also represented, if you look at a shear force diagram. We know, that $V'(X)=q(x)$, and we know that a point load $Q$ (external force or support) at point $a$ causes a "jump" in the shear force diagram, according to its magnitude, this can be represented by the Heaviside function. $V=Q\cdot H(x-a)$, now $$V'(x)=q(x)=(Q\cdot H(x-a))'=Q\cdot H'(x-a)$$ As explained above, the derivative of the Heaviside function is the Dirac delta function, therefore $$q(x)=Q\cdot \delta(x-a)$$
• Thanks for your elegant answer. Could you please explain more about the static interpretation of Dirac function ? the rest is clear.
– user14407
Jan 23, 2018 at 21:16
• I added some further interpretation/explanation to the Dirac Delta function at the end of my answer, I hope my explanation is comprehensible. Jan 24, 2018 at 11:28
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## How to specify Maximum Likelihood
For technical questions regarding estimation of single equations, systems, VARs, Factor analysis and State Space Models in EViews. General econometric questions and advice should go in the Econometric Discussions forum.
Moderators: EViews Gareth, EViews Moderator
cuongnh
Posts: 70
Joined: Thu Dec 11, 2008 10:09 am
Contact:
### How to specify Maximum Likelihood
Dear,
Could someone help me in how to specify for using maximum likelihood in Eviews? I dont understand how to define beta, alpha, scale in the specification. in the case of my analysis, I have series of prices of different commodities which are used as dependent variable and the remaining as independent ones.
EViews Gareth
Fe ddaethom, fe welon, fe amcangyfrifon
Posts: 12602
Joined: Tue Sep 16, 2008 5:38 pm
### Re: How to specify Maximum Likelihood
You're going to have to be way more specific than that.
What is the likelihood function you're trying to maximise? What is the data you're trying to use? What are the parameters you're trying to estimate?
What is beta, alpha and scale?
cuongnh
Posts: 70
Joined: Thu Dec 11, 2008 10:09 am
Contact:
### Re: How to specify Maximum Likelihood
I would like to analyze the response of prices of rice on the changes in prices of crude oil (some dummies are included on the right hand side). Price data are daily basis, which is from Jul 2, 2007 to end of Mar, 2009. I am totally new to ML method, I am reading the User's Guide of Eviews to understand how to estimate using this method. I hope this method may help me to reduce the serial correlation in the results by using OLS.
As I read in the book, like below:
Parameter Names
In the example above, we used the coefficients C(1) to C(5) as names for our unknown parameters. More generally, any element of a named coefficient vector which appears in the specification will be treated as a parameter to be estimated.
In the conditional heteroskedasticity example, you might choose to use coefficients from three different coefficient vectors: one vector for the mean equation, one for the variance equation, and one for the variance parameters. You would first create three named coefficient vectors by the commands:
coef(3) beta
coef(1) scale
coef(1) alpha
You could then write the likelihood specification as:
@logl logl1
res = y - beta(1) - beta(2)*x - beta(3)*z
var = scale(1)*z^alpha(1)
logl1 = log(@dnorm(res/@sqrt(var))) - log(var)/2
I dont understand how to create coefficient vetors, and what "z" means in the specification.
Please explain me. Thank you very much!
EViews Gareth
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Posts: 12602
Joined: Tue Sep 16, 2008 5:38 pm
### Re: How to specify Maximum Likelihood
beta, scale and alpha are coefficient vectors in the workfile. They're just vectors that store parameters.
You can create them, just as the manual says, by typing "coef(3) beta" in the EViews command window.
"Z", as mentioned in the manual, is a series in the workfile.
Perhaps before attempting to use the LogL object, it might be worth learning a little on maximum likelihood estimation (MLE) from an econometrics textbook. Learning MLE from reading the EViews manual is going to be tough.
cuongnh
Posts: 70
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Contact:
### Re: How to specify Maximum Likelihood
Thank you. In the textbook, z is define as a series in the workfile (where x, y, and z are the observed series (data)). in the example: y_t=b1+b2x_t+b3z_t+e_t, Let's say y is dependent variable, x, z are independent ones. But in the
series res = y - c(1) - c(2)*x - c(3)*z
series var = c(4) * z^c(5)
series logl1 = log(@dnorm(res/@sqrt(var))) - log(var)/2
"z" is treated in the var function. What is the difference between x and z. Since I have price of oil (independent variable) and dummy varibles, how can I define which is z?
EViews Gareth
Fe ddaethom, fe welon, fe amcangyfrifon
Posts: 12602
Joined: Tue Sep 16, 2008 5:38 pm
### Re: How to specify Maximum Likelihood
The particular likelihood function that the example is estimating has the error term's variance as a function of Z (if you look at equation 32.1). Hence the LogL object has the variance as a function of Z.
cuongnh
Posts: 70
Joined: Thu Dec 11, 2008 10:09 am
Contact:
### Re: How to specify Maximum Likelihood
Dear Gareth, I am sorry for asking you too many times, I dont really understand.
In the equation (22.1), it mentions that, but I dont know why it is the function of Z, not X, while both X and Z are independent variables. I need to define exactly in the case of mine which is Z.
EViews Gareth
Fe ddaethom, fe welon, fe amcangyfrifon
Posts: 12602
Joined: Tue Sep 16, 2008 5:38 pm
### Re: How to specify Maximum Likelihood
As I said above, I think you need to have a deeper understanding of MLE before you can understand how to estimate MLE in EViews.
I can't possibly tell you which variable should be "z". Do you even know if the particular likelihood function this specific example is estimating is the particular likelihood function you wish to estimate?
If the answer to that is "yes", then you (and only you) should already know which variable is "z".
cuongnh
Posts: 70
Joined: Thu Dec 11, 2008 10:09 am
Contact:
### Re: How to specify Maximum Likelihood
Thanks. I read to understand more MLE.
cuongnh
Posts: 70
Joined: Thu Dec 11, 2008 10:09 am
Contact:
### Re: How to specify Maximum Likelihood
Dear Gareth, I hope you are still with me. I tried to estimate using logl, but it popped up with this warning: Missing values in @logl series at current oeeficients at observation 1. The specification used to estimate is as follows:
@logl loglrice
res=price-beta(1)-beta(2)*poil-beta(3)*dricq-beta(4)*sdmep-beta(5)*sdvxb-beta(6)*dths
var = scale(1)*poil^alpha(1)
loglrice = log(@dnorm(res/@sqrt(var))) - log(var)/2
Could you please explain me this and how to solve this problem?
cuongnh
Posts: 70
Joined: Thu Dec 11, 2008 10:09 am
Contact:
### Re: How to specify Maximum Likelihood
Dear Friends, could any one help me this?
I tried to estimate using logl, but it popped up with this warning: Missing values in @logl series at current oeeficients at observation 1. The specification used to estimate is as follows:
@logl loglrice
res=price-beta(1)-beta(2)*poil-beta(3)*dricq-beta(4)*sdmep-beta(5)*sdvxb-beta(6)*dths
var = scale(1)*poil^alpha(1)
loglrice = log(@dnorm(res/@sqrt(var))) - log(var)/2
Could you please explain me this and how to solve this problem?
Did you use forum search?
Posts: 1518
Joined: Thu Nov 20, 2008 12:04 pm
### Re: How to specify Maximum Likelihood
Looks like you are having an initial value problem. You should try different starting values for your coefficients, especially the ones that are used in the variance specification (i.e. scale and alpha). There have been numerous discussions on this issue, so you might want to do a simple search in the forum.
cuongnh
Posts: 70
Joined: Thu Dec 11, 2008 10:09 am
Contact:
### Re: How to specify Maximum Likelihood
trubador wrote:different starting values for your coefficients, especially the ones that are used in the variance specification (i.e. scale and alpha).
What do you mean by "starting value for your coefficients"?
Did you use forum search?
Posts: 1518
Joined: Thu Nov 20, 2008 12:04 pm
### Re: How to specify Maximum Likelihood
If you have not assigned any values to scale(1) and alpha(1), then EViews assumes 0 as their starting values, which can create problems in your variance equation. So it may be a good idea to supply different values, before running the estimation.
cuongnh
Posts: 70
Joined: Thu Dec 11, 2008 10:09 am
Contact:
### Re: How to specify Maximum Likelihood
I just changed the value "0" to "1" for both alpha and scale. It worked. But in the result sheet, the standard error, z-stats and probability are all not available (NA). WHat is the problem?
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# Design of tensile members
We have seen design examples to determine the tensile design strength for structural steel sections. Let us explore on how to design a tensile member and learn the simple design steps.
## What are the basic information needed to start the design?
To design a tensile member, the following things are required:
1. Length of the member
2. End plate connections
3. Load it has to support $($also called design tension load, $T)$
Now a tension member should be designed such that it’s design tensile strength, $T_d$ should be greater than design factored load, $T$ i.e., $T_d>T$.
## Design: an AB process → Assumption and Backing
It is obvious that the designer has to select a cross-section such that $T_d>T$ (for the member to resist load)
For tensile members, the design tensile strength $T_d$ is minimum of $T_{dg}, T_{dn}, T_{db}$.
The table above shows the main assumption and design overview.
Thus, a designer has to adopt following two steps:
1. Step A: Assumption → Assume a cross-section. Design the end connection for it.
2. Step B: Backing → Use design equations of IS 800:2007 to compute $T_d$ such that $T_d>T$
Let us look at the design steps with a design example.
## Step-by-step procedure of design of tension member
Design Example: Design a 2.5 m long single section angle acting as a tension member in a bridge truss. The member is subjected to a factored load of 280 kN. Assume the gusset plate thickness as 16 mm and the diameter of bolts to be used as 20 mm of grade 8.8. Assume Fe 410 grade of steel.
Step 1: Calculate the required area of cross-section taking $T_{dg} = T$
Here,
$T=280\; kN$ and $T_{dg}=A_gf_y/\gamma_{m0}$
Hence, Area required $A_g=T_{dg}\gamma_{m0}/f_y$
$= 280\times 10^3\times 1.1/250=1232\; mm^2$
Step 2: Adopt a trial section from SP 6:Part 1 for the required gross-area $A_g$
Let us try ISA $90 \times 60 \times 10$ mm. Hence we have,
Gross area, $A_g=1401\; mm^2$
## Design of end connection (Number of bolts required, Length of weld)
The design equations to determine the strength of bolt, $V_{ ds}$ is given in article.
$V_{ ds} =Min (V_{dsb}, V_{dpb})$
$V_{ dsb}$ for M20 8.8 grade bolt
$$V_{dsb}=\frac{f_{ub}}{\sqrt{3}\gamma_{mb}}(n_nA_n+n_sA_s)$$
Here, for grade 8.8 bolt, $f_{ub} =800 \; MPa$
$n_n=1;\; A_n=0.78\times \frac{\pi}{4}\times (20)^2=245\; mm^2$
Hence, $V_{dsb}=(800\times 245)/(\sqrt{3}\times 1.25)=90.53\;kN$
$V_{ dpb}$ for design problem
$$V_{dpb}=2.5 k_bdt\frac{f_{u}}{\gamma_{mb}}$$
where,
$$k_b=Min (\frac{e}{3d_0},\frac{p}{3d_0}-0.25, \frac{f_{ub}}{f_u}, 1.0)$$
## IS code recommendation
In a design problem of tension members, $p,e$ will not be provided. In such a case, use IS 800: 2007 provisions to meet minimum requirements and select $p,e$.
## Application of recommendation
Therefore,
$$k_b=Min (\frac{e}{3d_0},\frac{p}{3d_0}-0.25, \frac{f_{ub}}{f_u}, 1.0)$$
$$k_b=Min (\frac{40}{3\times 22},\frac{60}{3\times22}-0.25, \frac{800}{410}, 1.0)$$
Hence,
$$k_b= 0.606$$
Thus,
$$V_{dpb}=2.5 k_bdt\frac{f_{u}}{\gamma_{mb}}$$
$$V_{dpb}=2.5 \times 0.606\times 10\times 20\times\frac{410}{1.25}=99.38\; kN$$
Hence, $V_{ds} =Min (V_{ dsb }, V_{ dpb}) =90.53 \; kN$
Thus, number of bolts required, $n =280 /90.53 =3.1 \simeq 4$
Thus, provide $4$ M20 bolts of grade $8.8$ at a pitch of $60 \; mm$ and edge distance $40\; mm$.
Step 4: Determine $T_{dn}, T_{db}$ to check if $T_d>T$
For $T_{ dn}$
As per the end plate design, we arrived at a requirement of $4$ bolts of grade $8.8$ at a pitch of $60 \; mm$ and edge distance $40\; mm$.
Therefore, $A_{nc}=(90-20-5)\times10=650\; mm^2$ [Refer to article]
Now, $A_{go}=(60-5)\times10=550\; mm^2$
As per clause 6.3.3 IS 800 :2007,
$$T_{dn}=0.9A_{nc}f_u/\gamma_{m1}+\beta A_{go}f_y/\gamma_{m0}$$
where,
$\beta=$ $1.4-0.076(w/t)(f_y/f_u)(b_s/L_c)\leq(f_u\gamma_{m0}/f_y\gamma_{m1})\geq 0.7$
For the current scenario,
$w=$ outstand leg width $=60\;mm$;
$b_s=$ shear lag width $=60+45-10=95\;mm$;
$L_c=$ Length of end connection $=3\times60=180\;mm$;
$A_{nc} =$ Net area of the connected leg $=650\;mm^2$;
$A_{go} =$ Gross area of the outstanding leg $=550\; mm^2$;
$t=$ Thickness of the leg $=10\; mm$
Hence, $\beta=1.253$
Therefore, $T_{dn} =349.13 \;kN$ $>300\; kN$ . Hence OK.
For $T_{ db}$
Refer to figure below
$A_{vg} =220 \times 10= 2220\;mm^2$
$A_{vn}= [220- (3.5\times 22)]\times 10 = 1430\; mm^2$
$A_{tg} = 45\times 10 = 450 \; mm^2$
$A_{tn} =(45 – 0.5\times 22) \times 10 = 340\; mm^2$
Hence,
$$T_{db_1}=\frac{A_{vg}f_y}{\sqrt{3}\gamma_{m0}}+\frac{0.9A_{tn}f_u}{\gamma_{m1}}$$
$$T_{db_1}=\frac{A_{vg}f_y}{\sqrt{3}\gamma_{m0}}+\frac{0.9A_{tn}f_u}{\gamma_{m1}}$$
$$T_{db_1}=\frac{2220\times250}{\sqrt{3}\times1.10}+\frac{0.9\times340\times410}{1.25}= 391.67\; kN$$
$$T_{db_2}=\frac{0.9A_{vn}f_u}{\sqrt{3}\gamma_{m1}}+\frac{A_{tg}f_y}{\gamma_{m0}}$$
$$T_{db_2}=\frac{0.9\times1430\times410}{\sqrt{3}\times1.25}+\frac{450\times250}{1.10}=346\; kN$$
Hence, $T_{db}=346\; kN > 280 \; kN$
## Conclusions
In this section we have learned the following key points:
• Failure modes: We have learned the different failure modes for the members subjected to tension. Like failure by net area rupture, gross section yielding, and block shear failure.
• Net area: In this article, we have learned how to calculate the net area of the cross-section.
## Android Apps
At eigenplus, our goal is to teach civil engineering students about structural analysis and design starting from the fundamental principles. We do this with the help of interactive android applications and accompanying web articles and videos.
Our apps have helped more than 400 thousand students across the world to understand and learn the concepts of structural engineering. Check out our apps on the google play store.
This article was crafted by a group of experts at eigenplus to ensure it adheres to our strict quality standards. The individuals who contributed to this article are:
# Author
## Kamal Patel
PhD
He is a computational scientist with a doctorate in advanced composites design.
PhD
M.Tech
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Explore BrainMass
# Single Phase Circuit
Not what you're looking for? Search our solutions OR ask your own Custom question.
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
A resistor and a capacitor are connected in series across a 150 V a.c supply. When the frequency is 40 Hz the current is 5 A, and when the frequency is 50 Hz the current is 6 A. Find the resistance and capacitance of the resistor and capacitor respectively.
If they are now connected in parallel across the 150 V supply, find the total current and its power factor when the frequency is 50 Hz.
© BrainMass Inc. brainmass.com March 5, 2021, 12:33 am ad1c9bdddf
https://brainmass.com/engineering/power-engineering/single-phase-circuit-516568
#### Solution Preview
A resistor and a capacitor are connected in series across a 150 V a.c supply. When the frequency is 40 Hz the current is 5 A, and when the frequency is 50 Hz the current is 6 A. Find the resistance and capacitance of the resistor and capacitor respectively.
If they are now connected in parallel across the 150 V supply, find the total current and its power factor when the frequency is 50 Hz.
Solution: R C
I
V = Vosinωt = Vosin2Πft
Part I
Case 1: Frequency f = 40 Hz, Angular frequency = ω = 2Πf = 2Πx 40 = 251.33 rad/sec
Reactance of the capacitor = XC = 1/jωC = 1/j(251.33C)
Impedance of the R-C series combination = Z = R + XC = R + 1/j(251.33C) = R - j ...
#### Solution Summary
This solution is provided within an attached Word document and clearly demonstrates how to solve this physics based problem. All calculations are included and diagrams also accompany this solution.
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Prediction Market Based Electoral Map Forecast
Slashdot points to ElectoralMap.net, a site which aggregates the Intrade prediction market results for individual states in the fall U.S. Presidential election, to predict aggregate electoral college totals and the election winner. I’m bookmarking it as an alternative to media coverage of the race, which will put their own spin on likely outcomes. Note the arrows at the bottom of the map will allow you to track the changes in sentiment over the weeks and months leading up to the election.
A somewhat more elaborate alternative is Electoral Vote Predictor, which uses polls rather than prediction markets. However, they are almost always the same, to tell the truth. I followed this site closely in the 2004 election. Site operator Andrew Tanenbaum includes a frank discussion of the accuracy of the 2004 predictions.
Robin Hanson argued earlier that closely following the news is not that great an idea, and I have to agree. It is something of a bad habit, but I find myself doing it anyway. We have this instinct that choosing our Leader is as important to our lives as it was when we were a tribe of two dozen, and that we have similar influence over the result. Following elections and participating in politics activates these vestigial tribal instincts in much the same was as sports, with similarly futile results. It can be seen as another form of “information porn“, exciting and titillating but ultimately unproductive.
I wish they used a different algorithm in the ElectoralMap site. It appears that the methodology is to take the individual state Intrade predictions, and assign all the electoral votes for a state to the party that has a 55% or greater predicted probability of carrying the state. States with less than a 55% probability for either party are shown as undecided. What I would like to see instead is to multiply the number of electoral votes for a state times the probability that the party will carry that state, and sum those results per party. This way, states that are seen as solidly preferring one party will be given greater weight towards that party than those which are more uncertain.
Then to make it even more useful, run a Monte Carlo simulation where each state is randomly assigned to one party or the other, with the odds taken from the prediction market, to create a simulated election result. Repeat this many times to calculate the mean and standard deviation for possible outcomes. In this way, confidence intervals can be given for the predictions. It’s possible that we might see the predicted electoral votes stay roughly the same for some time, but the confidence interval might narrow or widen. I don’t know if there is some statistical trick to compute the C.I. without actually running simulated elections, just from the distribution of probability estimates, but if so, that would be even better. My suspicion is that realistic confidence intervals are going to show that the election will be too close to call and could easily go either way – at least, that would have been the case in the last few elections.
Why not do this programming myself? Well, I might have the ability to do that. Intrade has an API, but it’s not clear if it allows retrieving the necessary data. I wouldn’t know how to display it on a colorful map, but the math should be simple enough, if you can retrieve the market prices, so producing a textual result would be straightforward. Maybe I will try to work on it in my spare time.
We don’t often post these kinds of pointer articles here, although they dominate many blogs. In that earlier posting about the news, Robin wrote, “I avoid posts that should not be nearly as interesting a year before or after.” This one does not really qualify, although there may be some historical value to be able to go back and look at how market sentiment evolved over time.
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# Welcome to the LEAP Q&A Forum
## SAT - Math - Speed Time Distance
### At which air temperature will the speed of a sound wave be closest to 1,000 feet per second?
a = 1,052 1.08t
The speed of a sound wave in air depends on the air temperature. The formula above shows the relationship between a, the speed of a sound wave, in feet per second, and t, the air temperature, in degrees Fahrenheit.
At which air temperature will the speed of a sound wave be closest to 1,000 feet per second?
we need to find t when a = 1000
1000 = 1052 + 1.08t
-52 = 1.08 t
t = -52/1.08 = 48.14 Farehnheit
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# R/equifreq.R In lfl: Linguistic Fuzzy Logic
#### Documented in equifreq
```#' Return equifrequent breaks
#'
#' If both `left` and `right` equal to `"none"`, the function returns a vector of `n` values from `x`
#' that divide the range of values in `x` into `n - 1` equidistant intervals. If the `left` (resp. `right`)
#' argument equals to `"infinity"`, `-Inf` (resp. `Inf`) is prepended (resp. appended) to the result. If
#' it equals to `"same"`, the first (resp. last) value is doubled. See [fcut()] for what such vectors
#' mean.
#'
#' If the `left` (resp. `right`) argument equals to `"infinity"`, `-Inf` (resp. `Inf`) is prepended
#' (resp. appended) to the result. If it equals to `"same"`, the first (resp. last) value is doubled.
#' Such functionality is beneficial if using the result of this function with e.g. the [fcut()] function:
#' `Inf` values at the beginning (resp. at the end) of the vector of breaks means that the fuzzy set
#' partition starts with a fuzzy set with kernel going to negative (resp. positive) infinity; the doubled
#' value at the beginning (resp. end) results in half-cut (trimmed) fuzzy set.
#'
#' @param x A numeric vector of input values
#' @param n The number of breaks of `x` to find (`n` must be at least 2)
#' @param left The left border of the returned vector of breaks: `"infinity"`, `"same"` or `"none"`
#' (see the description below)
#' @param right The right border of the returned vector of breaks: `"infinity"`, `"same"` or `"none"`
#' (see the description below)
#' @return A vector of equifrequent breaks
#' @seealso [equidist()], [fcut()]
#' @author Michal Burda
#' @keywords models robust
#' @export
equifreq <- function(x,
n,
left = c('infinity', 'same', 'none'),
right = c('infinity', 'same', 'none')) {
left <- match.arg(left)
right <- match.arg(right)
.mustBeNumericVector(x)
.mustBeNumericScalar(n)
.mustBe(n >= 2, "'n' must be greater or equal to 2")
.mustBeCharacterScalar(left)
.mustBeCharacterScalar(right)
i <- seq(from = 1,
to = length(x),
length.out = n)
i <- round(i)
res <- sort(x)[i]
if (left == 'infinity') {
res <- c(-Inf, res)
} else if (left == 'same') {
res <- c(res[1], res)
}
if (right == 'infinity') {
res <- c(res, Inf)
} else if (right == 'same') {
res <- c(res, res[length(res)])
}
res
}
```
## Try the lfl package in your browser
Any scripts or data that you put into this service are public.
lfl documentation built on Sept. 8, 2022, 5:08 p.m.
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Factor Analysis And Confirmatory Factor Analysis Essay
Factor analysis is an arithmetical technique used to describe variability regarding observed variables with regard to lower number of unobserved variables. Factor analysis looks for such joint variations in response to unobserved hidden variables. The observed variables are molded as linear combinations of potential factors including the error terms.
Information attained regarding interdependence between observed factors can later be utilized to reduce the set of variables within a dataset. Factor analysis originated in psychometrics and is applied in behavioral sciences operations research and applied sciences which deal with large quantities of data. In psychology, factor analysis is in most cases associated with intelligence search. Factor analysis has been used to search for factors within a broad range of spheres such as character, beliefs and attitudes.
Factor analysis isolates the underlying variables that make clear the data. There are two types of factor analysis; principal factor analysis and common factor analysis. The factors generated by principal factor analysis are theoretical as being as liner combinations of variables whereas those generated by common factor analysis are theoretical latent variables. Computationally, the main difference is that the diagonal relationship matrix is substituted with common variables in common factor analysis.
Factor analysis is performed through examining the pattern of connection between the observed variables. Variables which are highly related have a likelihood of being influenced by factors such as those which are moderately unrelated and have a more likelihood of being influenced by different factors.
Principal component analysis is the most widespread factor analysis. Principal factor analysis seeks for a linear combination of measures in such a way that the maximum difference is extracted form the measures. It then removes the difference and search for a second liner a combination that explains the maximum proportion of the remaining variance.
Conducting a Confirmatory Factor Analysis
The main purpose of a Confirmatory Factor Analysis is to establish the ability of a prearranged variable model to fit within an observed set of data. Among the normal uses of Confirmatory Factor Analysis include; establishing the weight of a single factor representation compares the ability of two differing models to account for the same set of data, test the significance of particular factor loading, test the connection between two or more factor loadings and also to evaluate the convergent and discriminate strength of a set of measures.
Conducting a Confirmatory Factor Analysis
The six stages involved include;
Describing the factor model which is the first thing required to be done accurately to define the model one wants to test. This involves choosing the number of factors and defining the nature of loadings between measures and factors. The loading can be fixed at zero or any other constant number or allowed to vary within specified constraints.
Collect the measurements through measurement of variables on same experimental units.
Obtain a correlation matrix by getting the correlation between each of the variables.
Fit the model into data by selecting a method to obtain the estimates of factor loadings which were free to vary. The normal model-fitting method is the Maximum likelihood estimation that needs to be used unless the measures serious lack multivariate normality. In such a case one can use Asymptotically distribution free estimation.
Evaluation of model adequacy s done when the factor model is fit the data, the factor loading are selected to minimize the difference between the correlation matrix implied by the model and the actual observed matrix. The amount of difference after the best parameters have been selected can be used as a measure as to how reliable the reproduction is with the data.
The commonly used assessment of model adequacy is the X2 goodness of fit test. Null hypothesis for this test holds that the model sufficiency for the data, while the other is that there a significant level f differences. Regrettably, this test is highly sensitive to sample size since, tests used in testing large samples generally lead to a rejection of null hypothesis, even when factor model is suitable. Other statistics like the Tucker-Lewis index, compare the fitness of planned model to a null representation. These statistics show less sensitivity to sample size.
By comparing these two models with other model one can is able observe the difference between their X 2 statistics which is almost equal to X2 distribution. About al individual factor loading tests can be compared to reduced and full factor models. In situations where there is no comparison of full and reduced models, use of Root mean square error of approximation is recommended which is n estimation of discrepancy per degree of freedom within the model.
References
DeCoster, J. (1998). Overview of Factor Analysis. Retrieved on August, 16, 2010 from http://www.stat-help.co/notes.html
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# Analysis of DHS data
I have a DHS survey data set and need some simple analysis done ASAP. Hello,
The following is the data analysis I was looking to obtain for my study.
1. Categorize each variable to see how many women are reporting physical, emotional or sexual violence in each, and as well as a combination of all three among the ever married sample.
2. information regarding wealth index and the number of participants within each range among the ever married sample.
3. obtain some crude data (from within data file) looking at prevalence of spousal violence among married/ever married women in each category against the different wealth index.
4. The prevalence of spousal violence against married/ever married women will be calculated with corresponding 95% CI among five sub groups of household wealth from lowest to highest; then corresponding odds ratios to be calculated.
5. The same analysis taking into account certain confounders. The following confounders to be controlled for are age, marital status, number of living children, religion, education, urban/rural environment and employment.
6. A logistic regression will then be used to examine association between the exposure and outcome after controlling for confounders.
Thanks,
Miguel
Compétences : Statistiques SPSS
Concernant l'employeur :
N° du projet : #8524343
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The simple binomial theorem of degree 2 can be written as:
${(x+a)}^2=x^2+2xa+a^2 \ \ldots (1)$
Replacing $a$ by $(n+a)$ where $x, n, a \in \mathbb{R}$ , we can have
${(x+(n+a))}^2= x^2+2x(n+a)+{(n+a)}^2$
or, ${(x+n+a)}^2 =x^2+2xn+2ax+{(n+a)}^2$
Arranging terms in a way that
${(x+n+a)}^2 =ax+{(n+a)}^2+x^2+2xn+ax=ax+{(n+a)}^2+x(x+2n+a)$
Taking Square-root of both sides
or,
$x+n+a=\sqrt{ax+{(n+a)}^2+x(x+2n+a)} \ \ldots (2)$
Take a break. And now think about $(x+2n+a)$ in the same way, as:
$x+2n+a =(x+n)+n+a$ .
Therefore, in equation (2), if we replace $x$ by $x+n$ , we get
$x+2n+a=(x+n)+n+a=\sqrt{a(x+n)+{(n+a)}^2+(x+n)((x+n)+2n+a)}$
or, $x+2n+a=\sqrt{a(x+n)+{(n+a)}^2+(x+n)(x+3n+a)} \ \ldots (3)$
Similarly, $x+3n+a=\sqrt{a(x+2n)+{(n+a)}^2+(x+2n)(x+4n+a)} \ \ldots (4)$
and also, $x+4n+a=\sqrt{a(x+3n)+{(n+a)}^2+(x+3n)(x+5n+a)} \ \ldots (5)$
Similarly,
$x+kn+a=\sqrt{a(x+(k-1)n)+{(n+a)}^2+(x+(k-1)n)(x+(k+1)n+a)} \ \ldots (6)$ where, $k \in \mathbb{N}$ .
Putting the value of $x+2n+a$ from equation (3) in equation (2), we get:
$x+n+a=\sqrt{ax+{(n+a)}^2+x\sqrt{a(x+n)+{(n+a)}^2+(x+n)(x+3n+a)}} \ \ldots (7)$
Again, putting the value of $x+3n+a$ from equation (4) in equation (7), we get
$x+n+a =\sqrt{ax+{(n+a)}^2+x\sqrt{a(x+n)+{(n+a)}^2+(x+n)\sqrt{a(x+2n)+{(n+a)}^2+(x+2n)(x+4n+a)}}} \ \ldots (8)$
Generalizing$the result for$ k$-nested radicals:$ x+n+a =\\ \sqrt{ax+{(n+a)}^2+x\sqrt{a(x+n)+{(n+a)}^2+(x+n)\sqrt{a(x+2n)+{(n+a)}^2+(x+2n)\sqrt{\ldots+(x+(k-2)n)\sqrt{a(x+(k-1)n)+{(n+a)}^2+x(x+(k+1)n+a)}}}}} \ \ldots (9)$This is the general formula of Ramanujan Nested Radicals up-to$ k$roots. Some interesting points As$ x,n$and$ a$all are real numbers, thus they can be interchanged with each other. i.e., etc. Putting$ n=0$in equation (9) we have$ x+a =\sqrt{ax+{a}^2+x\sqrt{ax+{a}^2+x\sqrt{ax+{a}^2+x\sqrt{\ldots+x\sqrt{ax+{a}^2+x(x+a)}}}}} \ \ldots (11)$or just,$ x+a =\sqrt{ax+{a}^2+x\sqrt{ax+{a}^2+x\sqrt{ax+{a}^2+x\sqrt{\ldots}}}} \ \ldots (12)$Again putting$ x=1 \ a=0$in (9)$ 1+n =\sqrt{{n}^2+\sqrt{n^2+(1+n)\sqrt{{n}^2+(1+2n)\sqrt{\ldots+(1+(k-2)n)\sqrt{{n}^2+1+(k+1)n}}}}} \ldots (13)$Putting$ x=1 \ a=0$in equation (8)$ 1+n =\sqrt{{n}^2+\sqrt{{n}^2+(1+n)\sqrt{{n}^2+(1+2n)(1+4n)}}} \ \ldots (14)$Again putting$ x=a=n$=n(say) then$ 3n=\sqrt{n^2+4{n}^2+n\sqrt{2n^2+4{n}^2+2n\sqrt{3n^2+4{n}^2+3n\sqrt{\ldots+(k-1)n\sqrt{kn^2+4{n}^2+(k+3)n^2}}}}}$or,$ 3n=\sqrt{5{n}^2+n\sqrt{6{n}^2+2n\sqrt{7{n}^2+3n\sqrt{\ldots+(k-1)n\sqrt{(k+4)n^2+(k+3)n^2}}}}} \ \ldots (15)$Putting$ n=1$in (15)$ 3=\sqrt{5+\sqrt{6+2\sqrt{7+3\sqrt{\ldots+(k-1)\sqrt{(2k+7)}}}}} \ \ldots (16)$Putting$ x=n \in \mathbb{N}$and$ a=0$in (9) we get even numbers$ 2n =\sqrt{{n}^2+n\sqrt{{n}^2+2n\sqrt{{n}^2+3n)\sqrt{\ldots+(k-1)n\sqrt{(k-1)n)+{n}^2+(k+2)n^2}}}}} \ \ldots (17)$Similary putting$ x=n \in \mathbb{N}$and$ a=1$in (9) we get a formula for odd numbers: $\tiny \dpi{150} 2n+1 =\sqrt{n+{(n+1)}^2+n\sqrt{2n+{(n+1)}^2+2n\sqrt{3n+{(n+1)}^2+3n\sqrt{\ldots+(k-1)n\sqrt{kn+{(n+1)}^2+(k+2)n^2+n}}}}} \ \ldots (18)$ or, $\tiny \dpi{150} 2n+1 =\sqrt{n+{(n+1)}^2+n\sqrt{2n+{(n+1)}^2+2n\sqrt{3n+{(n+1)}^2+3n\sqrt{\ldots+(k-1)n\sqrt{(k+3)n^2+(k+3)n+1}}}}} \ \ldots (19)$ Comments? ## Published by Gaurav Tiwari A designer by profession, a mathematician by education but a Blogger by hobby. Loves reading and writing. Just that. ### 7 Comments 1. wow! i’m impressed!…and you’re reading…my blog? uh…i’m flattered…hope 2012 brings you more math puzzles to solve… 🙂 Reply 2. this is very helpful to me and you are doing a great job . thank you Reply 3. hi..i am utkarsh.i have been working on a formula and i am stuck in nested radicals. basically, i want to find out value of sqrt(2+sqrt(2+sqrt(2…………….sqrt(2) for x of times,for example, for x=3, i want value of sqrt(2+sqrt(2+sqrt(2+sqrt(2)))) would you please help me? Reply 4. by the way,are you left-handed,your hand writing is similiar to mine! Reply 5. Dear Utkarsh! Thanks for reading the post. Before I comment, I would like to mention that Ramanujan Nested Radical formulas are proposed for infinite number of radicals in a number. When, there are finite number of nested radicals, the exact numerical value is calculated by an advanced calculator. Let me be clear.$ \sqrt {2}$always means$ \sqrt {2}$or approximately 1.4142… Similarly$ \sqrt {2+\sqrt{2}}$has its own numerical value. And so on. As we increases the number of squareroots, the value tends to 2 (not exactly 2). But when infinite terms are considered, the numerical values cam be easily calculated using algebraic equations. Let$ N= \sqrt {2+\sqrt{2+\sqrt{2+ \ldots +\sqrt{2}}}}$upto infinte terms$ N= \sqrt {2+N}$or,$ N^2-N-2=0\$.
The non-negative solution of above quadratic equation is the numerical value of the nested radical (i.e., N=2).
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``` A POINT MASS M MOVING WITH VELOCITY U TOWARDS THE CENTRE OF THE DISC COLLIDES WITH A DISCOF MASS M AND RADIUS R RESTING ON A ROUGH HORIZONTAL SURFACE. ITS COLLISION IS PERFECTLY INELASTIC . FIND ANGULAR VELOCITY OF SYSTEM AFTER PURE ROLLING STARTS.
A. 2U/7R B. 7U/2R C. 5U/2R D. 2U/5R
```
7 years ago
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``` The answer is A. 2U/7R.
Considering the conservation of momentum, we get the velocity of the combined mass to be U/2.
Consider the angular velocity of the system just after the collision about the point of contact of the disc.
2M * U/2 * R = M U R
Angular momentum will be conserved about this point because the frictional force will pass through the point of contact, and will thus not provide any torque. The moment of Inertia of the combined mass will be (1/2MR^2 + MR^2). Thus, applying conservation of angular momentum about the point of contact,
M U R = (1/2 MR^2 + MR^2)* w + 2M V R
When pure rolling starts, w= V R
So,
M U R = 3/2 MR^2 w +2Mw.
After simplification, we get the required result.
```
7 years ago
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https://brainmass.com/computer-science/cpp/program-that-reads-a-student-s-name-together-with-his-or-her-test-scores-33216
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Purchase Solution
# program that reads a student's name together with his or her test scores.
Not what you're looking for?
Write a program that reads a student's name together with his or her test scores. The program should then compute the average test scores for each student and assign the appropriate grade. The grade scale is as follows: 90-100, A; 80-89, B; 70-79, C; 60-69, D; 0-59, F.
Your program must use the following functions:
1. A void function, calculateAverage, to determine the average of the five test scores for each student. Use a loop to read and sum the five test scores. (This function does not output the average test score. That task must be done in the function main.)
2. A value-returning function, calculateGrade, to determine and return each student's grade. (This function does not output the grade. That task must be done in the function main.)
Test your program on the following data. Read the data from a file and send the output to a file. Do not use any global variables. Use the appropriate parameters to pass value in and out of functions.
Please see attached for full question.
##### Solution Summary
Write a program that reads a student's name together with his or her test scores
##### Solution Preview
calculateAverage, to determine the average of the five test scores for each student has to be int or double value returning function. Because it will return the average value to the main function.
#include <iostream>
#include <iomanip>
#include <fstream>
#include <string>
using namespace std;
double calculateAverage(double test1, double test2, double test3, double test4, double test5);
void main()
{
string studentName;
double test1, test2, test3, test4, test5;
double average=0.0;
int studentCount=0;
double sumAverage=0.0;
double cAverage=0.0;
int ...
##### Inserting and deleting in a linked list
This quiz tests your understanding of how to insert and delete elements in a linked list. Understanding of the use of linked lists, and the related performance aspects, is an important fundamental skill of computer science data structures.
##### Basic UNIX commands
Use this quiz to check your knowledge of a few common UNIX commands. The quiz covers some of the most essential UNIX commands and their basic usage. If you can pass this quiz then you are clearly on your way to becoming an effective UNIX command line user.
##### Word 2010: Tables
Have you never worked with Tables in Word 2010? Maybe it has been a while since you have used a Table in Word and you need to brush up on your skills. Several keywords and popular options are discussed as you go through this quiz.
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https://www.coursehero.com/tutors-problems/Applied-Mathematics/12416481-Let-A-1-A-n-be-mm-matrices-Suppose-also-that-B-and-C-are-mm/
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View the step-by-step solution to:
# Let A 1 ,.,A n be mm matrices. Suppose also that B and C are mm matrices. Given that B Span(A 1 ,.,A n ) and B+C Span(A 1 , . , A n ), show that C...
Let A1,...,An be m×m matrices. Suppose also that B and C are m×m matrices. Given that B ∈ Span(A1,...,An) and B+C ∈ Span(A1, . . . , An), show that C ∈ Span(A1, . . . , An).
I don't understand where to start the proof and how I'm supposed show C is in the span given the other information.
It is shown that , C ∈ S p a n ( A... View the full answer
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https://stats.stackexchange.com/questions/269053/how-to-select-hyperparameters-for-svm-regression-after-grid-search
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# How to select hyperparameters for SVM regression after grid search?
I have a small data set of $150$ points each with four features. I plan to fit a SVM regression for the reason that the $\varepsilon$ value gives me the possibility of define a tolerance value, something that isn't possible in other regression techniques.
I have run cross-validated grid search on the $\gamma$ and $C$ values, at different values of $\varepsilon$. For different combinations for $\varepsilon$, $\gamma$, and $C$, I receive similar scores (as indicated in grid and results) .
Question: How do I define a criterion on improving the selection of hyper-parameters and making a rational model for my data set?
Results:
**Epsilon = 0.06**
The best parameters are {'C': 48.939009184774889, 'gamma': 0.03562247890262444} with a score of 0.64
**Epsilon = 0.09**
The best parameters are {'C': 48.939009184774889, 'gamma': 0.03562247890262444} with a score of 0.64
**Epsilon = 0.11**
The best parameters are {'C': 48.939009184774889, 'gamma': 0.03562247890262444} with a score of 0.66
**Epsilon = 0.14**
The best parameters are {'C': 48.939009184774889, 'gamma': 0.03562247890262444} with a score of 0.67
**Epsilon = 0.17**
The best parameters are {'C': 48.939009184774889, 'gamma': 0.03562247890262444} with a score of 0.66
**Epsilon = 0.19**
The best parameters are {'C': 48.939009184774889, 'gamma': 0.03562247890262444} with a score of 0.65
**Epsilon = 0.22**
The best parameters are {'C': 48.939009184774889, 'gamma': 0.03562247890262444} with a score of 0.64
**Epsilon = 0.25**
The best parameters are {'C': 14873.521072935118, 'gamma': 0.00072789538439831537} with a score of 0.65
**Epsilon = 0.27**
The best parameters are {'C': 621.0169418915616, 'gamma': 0.0038566204211634724} with a score of 0.65
**Epsilon = 0.30**
The best parameters are {'C': 4175.3189365604003, 'gamma': 0.0012689610031679235} with a score of 0.66
• This seems to be a very relevant question pertaining to the mission statement of this group. I would really like to see some positive interaction. Commented Mar 23, 2017 at 3:26
Though I haven't fully understood the problem, I am answering as per my understanding of the question.
Have you tried including Epsilon in param_grid Dictionary of Grid_searchCV.
I see you have only used the C and gamma as the parameters in param_grid dict.
Then i think the system would itself pick the best Epsilon for you.
Example:
from sklearn.svm import SVR
import numpy as np
n_samples, n_features = 10, 5
np.random.seed(0)
y = np.random.randn(n_samples)
X = np.random.randn(n_samples, n_features)
parameters = {'kernel': ('linear', 'rbf','poly'), 'C':[1.5, 10],'gamma': [1e-7, 1e-4],'epsilon':[0.1,0.2,0.5,0.3]}
svr = svm.SVR()
clf = grid_search.GridSearchCV(svr, parameters)
clf.fit(X,y)
clf.best_params_
output: {'C': 1.5, 'epsilon': 0.1, 'gamma': 1e-07, 'kernel': 'poly'}
• GridSearchCV can be done only using two parameters. So I had to use Gamma and C for the grid search but I changed the value of epsilon for each run of GridSearchCV Commented Mar 27, 2017 at 12:55
• No you can add any number of parameters.I have tried. once check the edit in the answer for the code. Commented Mar 27, 2017 at 13:18
• Have you tried this. Did my answer meet your requirements? Commented Mar 28, 2017 at 5:10
• I tried it, the GridSearchCV does work. I'm running the program on a large range of values to get finalize my parameters. Thank you for the input. Commented Mar 28, 2017 at 5:12
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https://physics.stackexchange.com/questions/271338/could-all-the-energy-in-all-the-photons-in-the-universe-account-for-dark-matter
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# Could all the energy in all the photons in the universe account for dark matter? [duplicate]
I was hit with fridge logic, and I am curious:
Is it possible that the gravitational influence of photons inside of galaxies (And all throughout the universe) could account for dark matter?
Photons would be most concentrated close to the core and inner galaxy... And have a lessening concentration as they expanded away from galaxies due to the inverse square law..
I am struggling to understand the math involved, because I've not yet reached that level in my education.... But I'm really curious how the total photons stack up vs the measured dark matter in the universe?
• Dark matter moves relatively slowly; otherwise it wouldn't cluster around galaxies. Photons don't move slowly; they move at the speed of light. – Peter Shor Aug 3 '16 at 2:15
• Isn't the observation of dark matter kind of vicarious and indirect? How would we get an even remotely accurate measurement of the speed at which dark matter of traveling through the universe..? I am stating that it maybe instead that concentrations of photons may instead act as dark matter... not really the photons themselves... but the collective whole. – Daniel Drake Aug 3 '16 at 2:34
• The very word 'dark' in dark matter means that we can't see it. Photons are exactly what we can see. If there were enough to contribute significantly to mass, we would notice. – knzhou Aug 3 '16 at 3:20
• We can measure the distribution of dark matter due to its gravitation. No massless field can be confined in ways like that. If dark matter is matter, then it has to be massive and cold. If it isn't, then we simply don't have a model that works. – CuriousOne Aug 3 '16 at 3:37
• Possible duplicates: physics.stackexchange.com/q/45387/2451 , physics.stackexchange.com/q/34516/2451 and links therein. – Qmechanic Aug 3 '16 at 22:12
The problem here is the contrasting effects the two forms of matter have.
Photons are massless; they create negligible gravitational fields.
Dark matter, on the other hand, is about 85-90% of all mass in the universe. It's responsible for holding galaxies at constant rotation rates past certain radii (see Galactic Rotational Velocity Curves).
Photons, on the other hand, provide a repulsive force that we call radiation pressure.
In terms of contribution to the energy density of the universe:
Let the critical density $\Omega_c=1$. This is the energy density of flat space, and provides an accurate description of the large-scale universe by current measurements.
$\Omega_m$ is the energy density contribution of barionic matter. It's approximately equal to $0.3$. Of that $0.3$, visible matter contributes about $0.05$.
$\Omega_\Lambda$ is the energy density contribution attributed to the cosmological constant $\Lambda$, and is about $0.7$, or about 70% of the energy in the universe.
You might have noticed that $\Omega_m + \Omega_\Lambda \approx 1 = \Omega_c$. Where's the light/radiation?
$\Omega_r$, the energy density contribution of radiation, is $\approx 0.00001$.
• Photons have energy. Any energy/mass contributes to gravitation. – Peter Shor Aug 3 '16 at 2:12
• You're right, but not even close to the same magnitude. I'll edit. – hebetudinous Aug 3 '16 at 2:19
• As I understood, if all the light from all the stars in a 300 light year radius shone for ten years, it'd be enough mass to create a black hole. If all the light was in one spot, that is. Those stars have been shining for alot longer than 10 years.. The galaxy alone is 100,000 light years accross? Thats alot of "black holes" worth of gravitational influence that cannot be seen, floating around in the galaxy? – Daniel Drake Aug 3 '16 at 2:54
• "$Ω_m$ is the energy density contribution of barionic matter" : no, of the matter. We don't know its nature and cannot qualify it this way. You must remove barionic (and one the 2 another hand, do you have 2 other hands ? :) ). Then, be more clear on the density of photons : you said that its effect was neglictible but this needs details or a reference. Planck mission give a lot of values, including an evaluation of this one. – user46925 Aug 3 '16 at 3:01
• @DanielDrake: The energy density of light created by one star can be easily calculated and it is tiny. It doesn't even increase the total mass-energy: whatever mass-energy is in the radiation was lost by the star, so the total gravitational mass of the system is exactly the same. – CuriousOne Aug 3 '16 at 3:57
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Cody
# Problem 166. Kaprekar numbers
Solution 1016666
Submitted on 14 Oct 2016 by Massimo Zanetti
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 16; tf_correct = false; assert(isequal(kap(x),tf_correct))
2 Pass
x = 704; tf_correct = false; assert(isequal(kap(x),tf_correct))
3 Pass
x = 9 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 9 tf_correct = logical 1
4 Pass
x = 45 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 45 tf_correct = logical 1
5 Pass
x = 55 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 55 tf_correct = logical 1
6 Pass
x = 99 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 99 tf_correct = logical 1
7 Pass
x = 297 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 297 tf_correct = logical 1
8 Pass
x = 703 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 703 tf_correct = logical 1
9 Pass
x = 999 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 999 tf_correct = logical 1
10 Pass
x = 2223 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 2223 tf_correct = logical 1
11 Pass
x = 2728 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 2728 tf_correct = logical 1
12 Pass
x = 4950 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 4950 tf_correct = logical 1
13 Pass
x = 5050 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 5050 tf_correct = logical 1
14 Pass
x = 7272 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 7272 tf_correct = logical 1
15 Pass
x = 7777 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 7777 tf_correct = logical 1
16 Pass
x = 9999 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 9999 tf_correct = logical 1
17 Pass
x = 17344 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 17344 tf_correct = logical 1
18 Pass
x = 22222 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 22222 tf_correct = logical 1
19 Pass
x = 77778 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 77778 tf_correct = logical 1
20 Pass
x = 82656 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 82656 tf_correct = logical 1
21 Pass
x = 95121 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 95121 tf_correct = logical 1
22 Pass
x = 99999 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 99999 tf_correct = logical 1
23 Pass
x = 142857 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 142857 tf_correct = logical 1
24 Pass
x = 148149 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 148149 tf_correct = logical 1
25 Pass
x = 181819 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 181819 tf_correct = logical 1
26 Pass
x = 187110 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 187110 tf_correct = logical 1
27 Pass
x = 208495 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 208495 tf_correct = logical 1
28 Pass
x = 318682 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 318682 tf_correct = logical 1
29 Pass
x = 329967 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 329967 tf_correct = logical 1
30 Pass
x = 351352 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 351352 tf_correct = logical 1
31 Pass
x = 356643 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 356643 tf_correct = logical 1
32 Pass
x = 390313 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 390313 tf_correct = logical 1
33 Pass
x = 461539 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 461539 tf_correct = logical 1
34 Pass
x = 466830 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 466830 tf_correct = logical 1
35 Pass
x = 499500 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 499500 tf_correct = logical 1
36 Pass
x = 500500 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 500500 tf_correct = logical 1
37 Pass
x = 533170 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 533170 tf_correct = logical 1
38 Pass
x = 538461 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 538461 tf_correct = logical 1
39 Pass
x = 609687 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 609687 tf_correct = logical 1
40 Pass
x = 643357 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 643357 tf_correct = logical 1
41 Pass
x = 648648 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 648648 tf_correct = logical 1
42 Pass
x = 670033 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 670033 tf_correct = logical 1
43 Pass
x = 681318 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 681318 tf_correct = logical 1
44 Pass
x = 681319 tf_correct = false assert(isequal(kap(x),tf_correct))
x = 681319 tf_correct = logical 0
45 Pass
x = 681320 tf_correct = false assert(isequal(kap(x),tf_correct))
x = 681320 tf_correct = logical 0
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Вы находитесь на странице: 1из 6
Quest for Space Experiment #1: Heat, Humidity, and Radiation High Tech High Chula Vista 1945 Discovery Falls Dr, Chula Vista, CA 91915 Ms. Meghan White and Ms. Kara Quinlan Project Manager: Angelica Luceo Mechanical Engineer: Jordi Molina Computer Programmer: Iker Reyes Astrobiologist/Electrical Engineer: Cesar Monarrez Documentarian: Hana Suiter
Space is a vast void with a lot of discoveries waiting to be found. Heat, Humidity, and Radiation are all affected and spread in different ways while in space due to many factors. Some of these factors include the lack of conduction, convection, and more assets to assist th
Our Research
1. What is heat transfer? What is the difference between conduction, convection, and radiation?
Heat transfers when an object of a higher temperature meets with an object with a lower temperature. The heat from the higher temperature spontaneously flows to the colder one. The flow continues until both objects reach the same temperature or thermal equilibrium. The difference between conduction, radiation, and convection is the way heat is the way heat is transferred from object to object. In conduction, the heat is transferred from direct contact while convection and radiation transfers heat without direct contact. Convection uses air molecules or liquids to transfer heat while radiation uses waves and particles.
1. How is heat transferred in space?
Since there is almost no convection and conduction in space, the heat moves throughout space mostly from radiation, making radiation the main cause of heat transferring in space. This is due to the molecules in the atmosphere not being able to spread and conduct through each other, making heat transfers throughout space almost
impossible through conduction or convection which leaves the heat waves to create warmth in that area from radiation.
1. How are spacesuits engineered to retain/expel heat and transfer energy?
Without spacesuits astronauts wouldn’t be able to live for more than fifteen seconds in space. This is why space suits are designed to keep the astronauts safe while completing extravehicular activities (activities astronauts perform outside of a spacecraft and out of Earth’s atmosphere). Seeing that space’s temperature varies in different locations, the suits are able to insulate heat with heating elements built into them. They also are supplied with water to insure that the astronauts stay hydrated while on a mission.
3. Does black or white best absorb heat? Why?
Black colored objects absorb more heat than white objects. This is caused by the light spectrum and electromagnetic radiation that light has. When light hits an object, the object is actually absorbing a select amount of colors and reflecting the rest and our eyes detect the colors that are being reflected. Black is created when all of the colors are being absorbed, storing more energy in the object and thus absorbing more heat. White is created when all of the colors are being reflected, causing most of the heat in the object to be reflected as well.
2.
What is the black body radiation theory? How does it apply to our experiment?
Black body radiation theory states that all objects and substances absorb all frequencies of light. Due to sciences thermodynamic laws, the amount of light that the object absorbed must be equal to the amount it emits/radiates. This applies to our experiment because our experiment
will calculate the amount of energy an object will absorb from light, so acknowledging this theory
is crucial to understanding the results of our experiment.
3. What are the heat transfer coefficients? How does it apply to our experiment?
Heat transfer coefficients are the factors when calculating the difference the solid object
will have on the convection of a fluid and the affect a liquid’s heat will have on the conduction of
a solid object. This information is important to our project because we have to make sure the conduction from the resistors or wires don’t fry anything in the circuit.
4. What is Planck’s law? How does it apply to our experiment?
Planck’s law states that radiation isn’t distributed continuously, but is
Part 2: Why our Research Matters
1. Why is our research of how heat moves on Earth and in Space needed? What is the relevance of our experiment (heat and humidity) to humans, our lives, the earth?
Our research of how heat moves on Earth and in Space is needed because it allows us to understand the way energy moves throughout space. This can teach us th
2. What will be the impact of our research? For our school? HTHCV Pueblo? Community? Scientific Community?
Our research impacts
Works Cited
Directions: For each source, cite in MLA format and explain what information you learned from the source
Your Sources MLA Works Cited What I used/learned from this resource 1 Sixth Edition Giancoli Physics Giancoli, Douglass, Sixth Edition Physics, Pearson Prentice Hall, 2005 Conduction, convection, and radiation 2. How Astronauts Forbes: We found out how astronauts space suits are able to insulate warmth within using heaters Stay Warm And Safe In The Deep Cold Of Space 3. NASA/Marshall NASA: How heat travels through space. Conduction, convection, radiation, and diffusion Space Processing Headline: Chuck - heat transfer
4. Temperature Northwestern University: System 5 6 7 8 9 10 Directions: Electrical Engineer/Computer Programmer
1. Use your check off sheet to take out all of your electrical materials and make sure you have
everything
2. Set-up your Time Lapse Filming of your process with the Computer Programmer
3. Use your circuit board lab sheet to build your circuit board
4. Draw/Model out your circuit board below and make sure your model has the following items:
A. Voltage, Resistance, and Current Identified
B. Resistors, Light Bulbs, male/female wires, switches
5. Explain each step of your process with electronics you used and how it works
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https://www.stat.math.ethz.ch/pipermail/r-help/2009-February/380442.html
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[R] Linear model: contrasts
Ian Fiske ianfiske at gmail.com
Fri Feb 6 18:38:16 CET 2009
```The problem comes from mixing up general linear model (LM) theory to compute
B with the classical anova estimators. The two methods use different
approaches to solving the normal equations. LM theory uses any generalized
inverse of X'X to solve the normal equations. Yours comes from ginv() which
uses SVD. On the other hand, classical anova theory uses imposed contraints
to give the system a unique solution. I'm not sure which generalized
inverse would correspond to the one that imposes the classical anova
constraints, but it must not be the one resulting from SVD in your example.
Without the imposed anova constraints (called cell means), you won't have
the interactions summing to zero as you want.
hope that helps,
Ian
Stefan Kraemer-2 wrote:
>
> Hey,
> I am modelling a linear regression Y=X*B+E. To compute the effect of
> “group” the B-values of the regressors/columns that code the interaction
> effects (col. 5-8 and col. 11-14, see below) have to be weighted with
> non-zero elements within the contrast "Group 1" minus "Group 2" (see
> below). My first understanding was that the interaction effects add up to
> zero in each group. Why do they get non-zero contrast weights? Do you
> recommend any literature discussing this topic?
>
> Many Thanks,
> Stefan
>
> The columns of matrix X dummy-codes
> 1. Data of Group1 (G1)
> 2. Data of Group2 (G2)
> 3. Data condition A1 (factor A)
> 4. Data condition A2 (factor A)
> 5. Data in cell A1G1
> 6. Data in cell A2G1
> 7. Data in cell A1G2
> 8. Data in cell A2G2
> 9. Data condition B1 (factor B)
> 10. Data condition B2 (factor B)
> 11. Data in cell B1G1
> 12. Data in cell B2G1
> 13. Data in cell B1G2
> 14. Data in cell B2G2
> 15. Mean
>
> Example:
> #create X
> dimnames = list( paste("sub",c(1:8)),
> c("Group1","Group2","A1","A2","A1G1","A2G1", "A1G2","A2G2","B1","B2",
> "B1G1","B2G1", "B1G2","B2G2","mean" ))
> X=matrix(nrow=8,ncol=15,dimnames=dimnames)
> X[,1]= c(rep(1,4), rep(0,4))
> X[,2]=c(rep(0,4),rep(1,4))
> X[,3]=c(1,1,0,0,1,1,0,0);
> X[,4]=c(0,0,1,1,0,0,1,1);
> X[,5]=c(1,1,0,0,0,0,0,0);
> X[,6]=c(0,0,1,1,0,0,0,0);
> X[,7]=c(0,0,0,0,1,1,0,0);
> X[,8]=c(0,0,0,0,0,0,1,1);
> X[,9]=c(1, 0, 1, 0, 1, 0,1, 0);
> X[,10]=c(0, 1, 0, 1, 0, 1, 0, 1);
> X[,11]=c(1, 0, 1, 0,0, 0, 0, 0);
> X[,12]=c( 0, 1, 0, 1,0, 0, 0, 0);
> X[,13]=c( 0, 0, 0, 0, 1, 0, 1, 0)
> X[,14]=c(0, 0, 0, 0, 0, 1, 0, 1)
> X[,15]=c(1,1,1,1,1,1,1,1)
> #y data
> Y=c(1.2,5.1,1.7,7,6.1,8,1.1,2.5)
>
> library(MASS)
> ginv()
> B=(ginv(t(X)%*%X) %*% t(X)) %*% Y
>
> #contrast
> c1=rep(0,15)
> c1[1]=1
> c1[c(5,6,11,12)]=1/2
> c2=rep(0,15)
> c2[2]=-1
> c2[c(7,8,13,14)]=-1/2
>
> #Group effect
> mean(Y[1:4])-mean(Y[5:8])
> c1%*%B+c2%*%B
>
> --
>
>
> --
> Jetzt 1 Monat kostenlos! GMX FreeDSL - Telefonanschluss + DSL
> für nur 17,95 Euro/mtl.!* http://dsl.gmx.de/?ac=OM.AD.PD003K11308T4569a
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
>
--
View this message in context: http://www.nabble.com/Linear-model%3A-contrasts-tp21871201p21876891.html
Sent from the R help mailing list archive at Nabble.com.
```
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https://www.geeksforgeeks.org/building-and-visualizing-sudoku-game-using-pygame/?ref=rp
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Open In App
Building and visualizing Sudoku Game Using Pygame
Sudoku is a logic-based, combinatorial number-placement puzzle. The objective is to fill a 9Ă—9 grid with digits so that each column, each row, and each of the nine 3Ă—3 subgrids that compose the grid contain all of the digits from 1 to 9.
We will be building the Sudoku Game in python using pygame library and automate the game using backtracking algorithm.
Features Implemented :
• Game Interface to Play
• Auto solving
• Visualization of auto solving i.e., Backtracking Algorithm visualization
• Options: Reset, Clear game
Implementation Steps :
1. Fill the pygame window with Sudoku Board i.e., Construct a 9×9 grid.
2. Fill the board with default numbers.
3. Assign a specific key for each operations and listen it.
4. Integrate the backtracking algorithm into it.
5. Use set of colors to visualize auto solving.
Instruction:
• Press ‘Enter’ To Auto Solve and Visualize.
• To play the game manually,
Place the cursor in any cell you want and enter the number.
• At any point, press enter to solve automatically.
Below is the Implementation :
Python3
`# import pygame library``import` `pygame` `# initialise the pygame font``pygame.font.init()` `# Total window``screen ``=` `pygame.display.set_mode((``500``, ``600``))` `# Title and Icon``pygame.display.set_caption(``"SUDOKU SOLVER USING BACKTRACKING"``)``img ``=` `pygame.image.load(``'icon.png'``)``pygame.display.set_icon(img)` `x ``=` `0``y ``=` `0``dif ``=` `500` `/` `9``val ``=` `0``# Default Sudoku Board.``grid ``=``[`` ``[``7``, ``8``, ``0``, ``4``, ``0``, ``0``, ``1``, ``2``, ``0``],`` ``[``6``, ``0``, ``0``, ``0``, ``7``, ``5``, ``0``, ``0``, ``9``],`` ``[``0``, ``0``, ``0``, ``6``, ``0``, ``1``, ``0``, ``7``, ``8``],`` ``[``0``, ``0``, ``7``, ``0``, ``4``, ``0``, ``2``, ``6``, ``0``],`` ``[``0``, ``0``, ``1``, ``0``, ``5``, ``0``, ``9``, ``3``, ``0``],`` ``[``9``, ``0``, ``4``, ``0``, ``6``, ``0``, ``0``, ``0``, ``5``],`` ``[``0``, ``7``, ``0``, ``3``, ``0``, ``0``, ``0``, ``1``, ``2``],`` ``[``1``, ``2``, ``0``, ``0``, ``0``, ``7``, ``4``, ``0``, ``0``],`` ``[``0``, ``4``, ``9``, ``2``, ``0``, ``6``, ``0``, ``0``, ``7``]`` ``]` `# Load test fonts for future use``font1 ``=` `pygame.font.SysFont(``"comicsans"``, ``40``)``font2 ``=` `pygame.font.SysFont(``"comicsans"``, ``20``)``def` `get_cord(pos):`` ``global` `x`` ``x ``=` `pos[``0``]``/``/``dif`` ``global` `y`` ``y ``=` `pos[``1``]``/``/``dif` `# Highlight the cell selected``def` `draw_box():`` ``for` `i ``in` `range``(``2``):`` ``pygame.draw.line(screen, (``255``, ``0``, ``0``), (x ``*` `dif``-``3``, (y ``+` `i)``*``dif), (x ``*` `dif ``+` `dif ``+` `3``, (y ``+` `i)``*``dif), ``7``)`` ``pygame.draw.line(screen, (``255``, ``0``, ``0``), ( (x ``+` `i)``*` `dif, y ``*` `dif ), ((x ``+` `i) ``*` `dif, y ``*` `dif ``+` `dif), ``7``) ` `# Function to draw required lines for making Sudoku grid ``def` `draw():`` ``# Draw the lines`` ` ` ``for` `i ``in` `range` `(``9``):`` ``for` `j ``in` `range` `(``9``):`` ``if` `grid[i][j]!``=` `0``:` ` ``# Fill blue color in already numbered grid`` ``pygame.draw.rect(screen, (``0``, ``153``, ``153``), (i ``*` `dif, j ``*` `dif, dif ``+` `1``, dif ``+` `1``))` ` ``# Fill grid with default numbers specified`` ``text1 ``=` `font1.render(``str``(grid[i][j]), ``1``, (``0``, ``0``, ``0``))`` ``screen.blit(text1, (i ``*` `dif ``+` `15``, j ``*` `dif ``+` `15``))`` ``# Draw lines horizontally and verticallyto form grid `` ``for` `i ``in` `range``(``10``):`` ``if` `i ``%` `3` `=``=` `0` `:`` ``thick ``=` `7`` ``else``:`` ``thick ``=` `1`` ``pygame.draw.line(screen, (``0``, ``0``, ``0``), (``0``, i ``*` `dif), (``500``, i ``*` `dif), thick)`` ``pygame.draw.line(screen, (``0``, ``0``, ``0``), (i ``*` `dif, ``0``), (i ``*` `dif, ``500``), thick) ` `# Fill value entered in cell ``def` `draw_val(val):`` ``text1 ``=` `font1.render(``str``(val), ``1``, (``0``, ``0``, ``0``))`` ``screen.blit(text1, (x ``*` `dif ``+` `15``, y ``*` `dif ``+` `15``)) ` `# Raise error when wrong value entered``def` `raise_error1():`` ``text1 ``=` `font1.render(``"WRONG !!!"``, ``1``, (``0``, ``0``, ``0``))`` ``screen.blit(text1, (``20``, ``570``)) ``def` `raise_error2():`` ``text1 ``=` `font1.render(``"Wrong !!! Not a valid Key"``, ``1``, (``0``, ``0``, ``0``))`` ``screen.blit(text1, (``20``, ``570``)) ` `# Check if the value entered in board is valid``def` `valid(m, i, j, val):`` ``for` `it ``in` `range``(``9``):`` ``if` `m[i][it]``=``=` `val:`` ``return` `False`` ``if` `m[it][j]``=``=` `val:`` ``return` `False`` ``it ``=` `i``/``/``3`` ``jt ``=` `j``/``/``3`` ``for` `i ``in` `range``(it ``*` `3``, it ``*` `3` `+` `3``):`` ``for` `j ``in` `range` `(jt ``*` `3``, jt ``*` `3` `+` `3``):`` ``if` `m[i][j]``=``=` `val:`` ``return` `False`` ``return` `True` `# Solves the sudoku board using Backtracking Algorithm``def` `solve(grid, i, j):`` ` ` ``while` `grid[i][j]!``=` `0``:`` ``if` `i<``8``:`` ``i``+``=` `1`` ``elif` `i ``=``=` `8` `and` `j<``8``:`` ``i ``=` `0`` ``j``+``=` `1`` ``elif` `i ``=``=` `8` `and` `j ``=``=` `8``:`` ``return` `True`` ``pygame.event.pump() `` ``for` `it ``in` `range``(``1``, ``10``):`` ``if` `valid(grid, i, j, it)``=``=` `True``:`` ``grid[i][j]``=` `it`` ``global` `x, y`` ``x ``=` `i`` ``y ``=` `j`` ``# white color background\`` ``screen.fill((``255``, ``255``, ``255``))`` ``draw()`` ``draw_box()`` ``pygame.display.update()`` ``pygame.time.delay(``20``)`` ``if` `solve(grid, i, j)``=``=` `1``:`` ``return` `True`` ``else``:`` ``grid[i][j]``=` `0`` ``# white color background\`` ``screen.fill((``255``, ``255``, ``255``))`` ` ` ``draw()`` ``draw_box()`` ``pygame.display.update()`` ``pygame.time.delay(``50``) `` ``return` `False` `# Display instruction for the game``def` `instruction():`` ``text1 ``=` `font2.render(``"PRESS D TO RESET TO DEFAULT / R TO EMPTY"``, ``1``, (``0``, ``0``, ``0``))`` ``text2 ``=` `font2.render(``"ENTER VALUES AND PRESS ENTER TO VISUALIZE"``, ``1``, (``0``, ``0``, ``0``))`` ``screen.blit(text1, (``20``, ``520``)) `` ``screen.blit(text2, (``20``, ``540``))` `# Display options when solved``def` `result():`` ``text1 ``=` `font1.render(``"FINISHED PRESS R or D"``, ``1``, (``0``, ``0``, ``0``))`` ``screen.blit(text1, (``20``, ``570``)) ``run ``=` `True``flag1 ``=` `0``flag2 ``=` `0``rs ``=` `0``error ``=` `0``# The loop thats keep the window running``while` `run:`` ` ` ``# White color background`` ``screen.fill((``255``, ``255``, ``255``))`` ``# Loop through the events stored in event.get()`` ``for` `event ``in` `pygame.event.get():`` ``# Quit the game window`` ``if` `event.``type` `=``=` `pygame.QUIT:`` ``run ``=` `False` ` ``# Get the mouse position to insert number `` ``if` `event.``type` `=``=` `pygame.MOUSEBUTTONDOWN:`` ``flag1 ``=` `1`` ``pos ``=` `pygame.mouse.get_pos()`` ``get_cord(pos)`` ``# Get the number to be inserted if key pressed `` ``if` `event.``type` `=``=` `pygame.KEYDOWN:`` ``if` `event.key ``=``=` `pygame.K_LEFT:`` ``x``-``=` `1`` ``flag1 ``=` `1`` ``if` `event.key ``=``=` `pygame.K_RIGHT:`` ``x``+``=` `1`` ``flag1 ``=` `1`` ``if` `event.key ``=``=` `pygame.K_UP:`` ``y``-``=` `1`` ``flag1 ``=` `1`` ``if` `event.key ``=``=` `pygame.K_DOWN:`` ``y``+``=` `1`` ``flag1 ``=` `1` ` ``if` `event.key ``=``=` `pygame.K_1:`` ``val ``=` `1`` ``if` `event.key ``=``=` `pygame.K_2:`` ``val ``=` `2` ` ``if` `event.key ``=``=` `pygame.K_3:`` ``val ``=` `3`` ``if` `event.key ``=``=` `pygame.K_4:`` ``val ``=` `4`` ``if` `event.key ``=``=` `pygame.K_5:`` ``val ``=` `5`` ``if` `event.key ``=``=` `pygame.K_6:`` ``val ``=` `6`` ``if` `event.key ``=``=` `pygame.K_7:`` ``val ``=` `7`` ``if` `event.key ``=``=` `pygame.K_8:`` ``val ``=` `8`` ``if` `event.key ``=``=` `pygame.K_9:`` ``val ``=` `9` ` ``if` `event.key ``=``=` `pygame.K_RETURN:`` ``flag2 ``=` `1` ` ``# If R pressed clear the sudoku board`` ``if` `event.key ``=``=` `pygame.K_r:`` ``rs ``=` `0`` ``error ``=` `0`` ``flag2 ``=` `0`` ``grid ``=``[`` ``[``0``, ``0``, ``0``, ``0``, ``0``, ``0``, ``0``, ``0``, ``0``],`` ``[``0``, ``0``, ``0``, ``0``, ``0``, ``0``, ``0``, ``0``, ``0``],`` ``[``0``, ``0``, ``0``, ``0``, ``0``, ``0``, ``0``, ``0``, ``0``],`` ``[``0``, ``0``, ``0``, ``0``, ``0``, ``0``, ``0``, ``0``, ``0``],`` ``[``0``, ``0``, ``0``, ``0``, ``0``, ``0``, ``0``, ``0``, ``0``],`` ``[``0``, ``0``, ``0``, ``0``, ``0``, ``0``, ``0``, ``0``, ``0``],`` ``[``0``, ``0``, ``0``, ``0``, ``0``, ``0``, ``0``, ``0``, ``0``],`` ``[``0``, ``0``, ``0``, ``0``, ``0``, ``0``, ``0``, ``0``, ``0``],`` ``[``0``, ``0``, ``0``, ``0``, ``0``, ``0``, ``0``, ``0``, ``0``]`` ``]`` ``# If D is pressed reset the board to default`` ``if` `event.key ``=``=` `pygame.K_d:`` ``rs ``=` `0`` ``error ``=` `0`` ``flag2 ``=` `0`` ``grid ``=``[`` ``[``7``, ``8``, ``0``, ``4``, ``0``, ``0``, ``1``, ``2``, ``0``],`` ``[``6``, ``0``, ``0``, ``0``, ``7``, ``5``, ``0``, ``0``, ``9``],`` ``[``0``, ``0``, ``0``, ``6``, ``0``, ``1``, ``0``, ``7``, ``8``],`` ``[``0``, ``0``, ``7``, ``0``, ``4``, ``0``, ``2``, ``6``, ``0``],`` ``[``0``, ``0``, ``1``, ``0``, ``5``, ``0``, ``9``, ``3``, ``0``],`` ``[``9``, ``0``, ``4``, ``0``, ``6``, ``0``, ``0``, ``0``, ``5``],`` ``[``0``, ``7``, ``0``, ``3``, ``0``, ``0``, ``0``, ``1``, ``2``],`` ``[``1``, ``2``, ``0``, ``0``, ``0``, ``7``, ``4``, ``0``, ``0``],`` ``[``0``, ``4``, ``9``, ``2``, ``0``, ``6``, ``0``, ``0``, ``7``]`` ``]`` ``if` `flag2 ``=``=` `1``:`` ``if` `solve(grid, ``0``, ``0``)``=``=` `False``:`` ``error ``=` `1`` ``else``:`` ``rs ``=` `1`` ``flag2 ``=` `0` ` ``if` `val !``=` `0``: `` ``draw_val(val)`` ``# print(x)`` ``# print(y)`` ``if` `valid(grid, ``int``(x), ``int``(y), val)``=``=` `True``:`` ``grid[``int``(x)][``int``(y)]``=` `val`` ``flag1 ``=` `0`` ``else``:`` ``grid[``int``(x)][``int``(y)]``=` `0`` ``raise_error2() `` ``val ``=` `0` ` ` ` ``if` `error ``=``=` `1``:`` ``raise_error1() `` ``if` `rs ``=``=` `1``:`` ``result() `` ``draw() `` ``if` `flag1 ``=``=` `1``:`` ``draw_box() `` ``instruction() ` ` ``# Update window`` ``pygame.display.update() ` `# Quit pygame window ``pygame.quit() `` `
Output:
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{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
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https://oeis.org/A114939
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crawl-data/CC-MAIN-2022-49/segments/1669446711151.22/warc/CC-MAIN-20221207085208-20221207115208-00659.warc.gz
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A114939 Number of essentially different seating arrangements for n couples around a circular table with 2n seats avoiding spouses being neighbors and avoiding clusters of 3 persons with equal gender. 5
0, 1, 7, 216, 10956, 803400, 83003040, 11579823360, 2080493573760, 469031859192960, 129727461014726400, 43176116371928601600, 17025803126147196057600, 7850538273249476117913600 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,3 COMMENTS Arrangements that differ only by rotation or reflection are excluded by the following conditions: Seat number 1 is assigned to person (a). Person (a)'s spouse (A) can only take seats with numbers <=(n+1). If (A) gets seat n+1 (i.e. sits exactly opposite to her/his spouse) then person (B) can only take seats with numbers <= n. LINKS M. A. Alekseyev, Weighted de Bruijn Graphs for the Menage Problem and Its Generalizations. Lecture Notes in Computer Science 9843 (2016), 151-162. doi:10.1007/978-3-319-44543-4_12; arXiv:1510.07926 [math.CO], 2015-2016. FORMULA See Alekseyev (2016) and the PARI code for the formula. a(n) = A258338(n) / (4*n). EXAMPLE a(2)=1 because the only valid arrangement is aBAb. a(3)=7 because the only valid arrangements under the given conditions are: abAcBC, aBAcbC, aBcAbC, aBcACb, acAbCB, acBAbC, aCAbcB. MATHEMATICA a[1] = 0; a[n_] := (n-1)!/4 Sum[(-1)^j(n-j)! SeriesCoefficient[ SeriesCoefficient[Tr[ MatrixPower[{{0, 1, 0, y^2, 0, 0}, {z y^2, 0, 1, 0, y^2, 0}, {z y^2, 0, 0, 0, y^2, 0}, {0, 1, 0, 0, 0, z}, {0, 1, 0, y^2, 0, z}, {0, 0, 1, 0, y^2, 0}}, 2n]], {y, 0, 2n}] , {z, 0, j}], {j, 0, n}]; Array[a, 14] (* Jean-François Alcover, Dec 03 2018, from PARI *) PROG (PARI) { a(n) = if(n<=1, 0, (-1)^n*(n-1)!*2^(n-1) + n! * polcoeff( polcoeff( [0, 2*y*z^3 + z^2, -3*y*z^5 - 4*z^4 + ((-2*y^2 - 1)/y)*z^3, 6*y*z^7 + (4*y^2 + 11)*z^6 + ((8*y^2 + 4)/y)*z^5 + 3*z^4] * sum(j=0, n-1, j! * [0, 0, 0, -z^6 + z^4; 1, 0, 0, ((y^2 + 1)/y)*z^5 - 2*z^4 + ((-y^2 - 1)/y)*z^3; 0, 1, 0, ((2*y^2 + 2)/y)*z^3 + z^2; 0, 0, 1, -2*z^2]^(n+j) ) * [1, 0, 0, 0]~, 2*n, z), 0, y) / 2 ); } CROSSREFS Cf. A114938, A137729, A137730, A137737, A137749, A258338. Sequence in context: A193836 A193877 A193186 * A193224 A319538 A290974 Adjacent sequences: A114936 A114937 A114938 * A114940 A114941 A114942 KEYWORD nonn,nice AUTHOR Hugo Pfoertner, Jan 08 2006 EXTENSIONS a(4)-a(7) corrected, formula and further term provided by Max Alekseyev, Feb 15 2008 STATUS approved
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Last modified December 7 05:41 EST 2022. Contains 358649 sequences. (Running on oeis4.)
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The Super Factorial Primes
# The Super Factorial Primes
First the Definition of Super Factorial: N s! = N (N+1) (N+2)... (N+N-1)
This is similar to the rising factorial: (a)_k is the kth rising factorial of a.
Also know as the "Pochhammer Symbol"
a * (a+1) * ... (a+k+1) or
e.g. (5)_3 = 5 * 6 * 7
So Super Factorial is the same as (a)_a, e.g. 3 s! = (3)_3
Therefore a super factorial prime is a super factorial plus or minus one which is also a prime number.
Also the formula N s! = (2N-1)! / (N-1)!
1 s! + 1 = 2
2 s! - 1 = 5
3 s! - 1 = 59
3 s! + 1 = 61
4 s! - 1 = 839
5 s! + 1 = 15,121
6 s! + 1 = 332,641
11 s! + 1 = 14,079,294,028,801
12 s! + 1
14 s! - 1
15 s! - 1
17 s! - 1
24 s! - 1
36 s! + 1
41 s! - 1
54 s! + 1
94 s! + 1
108 s! + 1
146 s! + 1
Back to Main
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An ammeter is always connected in ____ with the circuit, whereas a voltmeter is connected in ____ with the circuit. a. parallel, series b. parallel, parallel c. series, parallel d. series, series
FindFindarrow_forward
Automotive Technology
7th Edition
ERJAVEC + 1 other
Publisher: Cengage,
ISBN: 9781337794213
Solutions
Chapter
Section
FindFindarrow_forward
Automotive Technology
7th Edition
ERJAVEC + 1 other
Publisher: Cengage,
ISBN: 9781337794213
Chapter 16, Problem 1MC
Textbook Problem
8 views
An ammeter is always connected in ____ with the circuit, whereas a voltmeter is connected in ____ with the circuit.a. parallel, seriesb. parallel, parallelc. series, paralleld. series, series
To determine
The correct options for filling the gap in the given statement.
Explanation of Solution
Given Information: "An ammeter is always connectedwith the circuitand a voltmeter is connected with the circuit"
An ammeter is an electrical device used to measure the current through the circuit. The ammeter is always connected in series.Connectingan ammeter in parallelseries will cause short circuit as the ammeter is a low impedance electrical device. A short circuit will system failure in appliance. So, the ammeter is always connected in series. The voltmeter is an electrical device used to measure the potential difference in the circuit...
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Explore BrainMass
# confidence interval
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
An Econometrician has estimated the inverse demand relation as:
P = a + b Q + e and found that a = 400, b = -2.75, e (a) = 8, e (b) = 0.75.
Find the approximmate 95% confidence interval for the true values of a and b.
https://brainmass.com/statistics/confidence-interval/inverse-demand-relation-problem-55670
#### Solution Preview
Assuming that we have a large sample size, say, > 30 observations, then we can use z
statistic to construct the 95% confidence interval.
Interval ...
#### Solution Summary
The expert determines the confidence interval.
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Need help with Hagen Kh/Gh test
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10-18-2006, 04:07 AM #1
LoJack
Algae Grower
Join Date: Oct 2006
Location: Winnipeg, Manitoba, Canada
Posts: 127
# Need help with Hagen Kh/Gh test
I hate these stupid color tests.
I am doing the Kh test, and your supposed to add a single drop of reagent, and then add a drop at a time until your solution turns from blue to "yellow/lime" color.
On the back, where you are supposed to match the color, its got a blue square for start, and a yellow square for finish. At 8 drops my water turns from blue to a greenish color, and at 10 drops its the yellow on the back, so what is my Kh?
Anyone who uses this tests please give me some insight as to what is accurate.
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10-18-2006, 05:00 AM #2 Left C Planted Tank Guru PTrader: (19/100%) Join Date: Nov 2003 Location: Burlington, NC Posts: 5,009 Your answer is on the back of the last page in the directions. Well here goes: The yellow color is your end color for the KH test. You multiply the number of drops used to get the yellow color by 10. It was 10 drops in your case. This gives you 100 mg/L or 100 ppm. If you want to convert ppm to degrees, you multiply your ppm value by 0.056 to get degrees. So 100 x 0.056 = 5.6 degrees. The GH test has a little different formula. As an example, let's say that you used 10 drops to reach the end color of light blue. For the GH test, you multiply the number of drops used to reach the end color by 20 instead of 10 like the KH test. In this example, this gives you 200 mg/L or 200 ppm. To convert to degrees, you again multiply by 0.056. So, 200 x 0.056 = 11.2 degrees. You can get degrees from ppm another way. Divide your ppm by 17.86 to get degrees. Note that both KH and GH are measured in CaCO3 equivalents. I hope that this helps.
10-22-2006, 09:55 PM #3 LoJack Algae Grower PTrader: (0/0%) Join Date: Oct 2006 Location: Winnipeg, Manitoba, Canada Posts: 127 thanks a lot ... you helped more than you know __________________ Is Hokey Pokey Really What It's all about ?
10-22-2006, 10:48 PM #4 deadmonkey Algae Grower PTrader: (0/0%) Join Date: Oct 2006 Location: Moscow, ID Posts: 28 I don't have a KH tester, but mh GH tester says to keep adding drops until it changes from yellow to green (shaking each time). Once it changes to green, count the drops and look at the GH chart for the different fish you have to make sure it's in the right range. On my kit you multiply the drops by 17.9 and get your Parts Per Millon (ppm) or percent GH in the tank. I would guess by the lst post that KH has a similar method. For example, it took mine 13 drops times 17.9 is roughly 230 ppm. This is actually rather high. Frequent water changes should fix this though eventually. as far as accuracy, you are only going to be accurate to the nearest drop.
10-23-2006, 01:29 AM #5
Left C
Planted Tank Guru
Join Date: Nov 2003
Location: Burlington, NC
Posts: 5,009
Quote:
Originally Posted by deadmonkey I don't have a KH tester, but mh GH tester says to keep adding drops until it changes from yellow to green (shaking each time). Once it changes to green, count the drops and look at the GH chart for the different fish you have to make sure it's in the right range. On my kit you multiply the drops by 17.9 and get your Parts Per Millon (ppm) or percent GH in the tank. I would guess by the lst post that KH has a similar method. For example, it took mine 13 drops times 17.9 is roughly 230 ppm. This is actually rather high. Frequent water changes should fix this though eventually. as far as accuracy, you are only going to be accurate to the nearest drop.
It sounds like you are using AP's KH/GH test kit. It measures in degrees. The Hagen KH/GH test kit uses a different indicator solution. That's why the colors are different and one measures in ppm and the other measures in degrees.
If you round 17.86 degrees up to one decimal place, you'll have 17.9 degrees.
You can have two ways to calculate ppm and degrees. If you take 1 and divide it by 17.86 you get 0.056. Also, if you take 1 and divide it by 0.056 you get 17.86.
Then, 17.86 ppm per degree and 0.056 degrees per ppm are your constants.
So, let's says you have 5 degrees of hardness. You can multiply 5 degrees by 17.86 and get 89.3 ppm.
Here's the other way to do it. Let's use 5 degrees of hardness again. You can divide 5 degrees by 0.056 and also get 89.3 ppm.
Let's deal in ppm this time. Lets say that you have 100 ppm. You can divide 100 ppm by 17.86 and you get 5.6 degrees. The other way is to multiply 100 ppm by 0.056 and you get 5.6 degrees.
Do you understand what I did in these calculations?
Here's a tip. Both of these test kits use 5 ml of water for the test. You can be a little more accurate if you can find a test tube that will hold 10 ml of water. When you get the end color, you divide the number of drops by two. This may help if you have trouble deciding where the end point actually is. I believe that Tetra and Red Sea use 10 ml test tubes in some of their kits. (You can also use 15 ml, 20 ml, etc. also. Just be sure to divide by the number of drops correctly.)
LaMotte's alkalinity test kit is more precise and it's about \$20. It's a good one though. It's color change end point is rather abrupt. That helps it to be more accurate. LaMotte KH/Alkalinity Test Kit 4491- DR / Direct Reading Titrator Method - Marine Depot - Marine and Reef Aquarium Super Store
10-23-2006, 01:35 AM #6
Left C
Planted Tank Guru
Join Date: Nov 2003
Location: Burlington, NC
Posts: 5,009
Quote:
Originally Posted by LoJack thanks a lot ... you helped more than you know
You're welcome.
I know that when you got the yellow/lime color, it was confusing. That means that you were almost to the end point. You had to add two more drops to reach the end point.
01-19-2013, 10:33 PM #7
steak
Algae Grower
Join Date: Sep 2007
Location: South FL
Posts: 144
Quote:
Originally Posted by Left C Your answer is on the back of the last page in the directions. Well here goes: The yellow color is your end color for the KH test. You multiply the number of drops used to get the yellow color by 10. It was 10 drops in your case. This gives you 100 mg/L or 100 ppm. If you want to convert ppm to degrees, you multiply your ppm value by 0.056 to get degrees. So 100 x 0.056 = 5.6 degrees. The GH test has a little different formula. As an example, let's say that you used 10 drops to reach the end color of light blue. For the GH test, you multiply the number of drops used to reach the end color by 20 instead of 10 like the KH test. In this example, this gives you 200 mg/L or 200 ppm. To convert to degrees, you again multiply by 0.056. So, 200 x 0.056 = 11.2 degrees. You can get degrees from ppm another way. Divide your ppm by 17.86 to get degrees. Note that both KH and GH are measured in CaCO3 equivalents. I hope that this helps.
This allowed me to use my test. I lost my booklet and no longer had the color chart, or the directions. I got on here looking for this information specifically. Thank you
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PDA
heathmichael86
09-18-2013, 07:56 AM
I'm trying to understand the science of nutrition.
180 lbs, with 25% body fat
Goal:
195 lbs, with 15% body fat
This is an increase of 29 lbs of lean mass, and a decrease of 14 lbs of fat.
What should my caloric intake be?
I've seen caloric calculators, but it wasn't exactly clear to input current weight vs target weight, or if the output was to maintain or to grow.
>> Also, how do I account for strength training 5 days a week?
Do you guys have a recommended timeline? What can I expect in 12 weeks?
Please give me some numbers, thanks!!
chamelious
09-18-2013, 07:58 AM
I'm trying to understand the science of nutrition.
180 lbs, with 25% body fat
Goal:
195 lbs, with 15% body fat
This is an increase of 29 lbs of lean mass, and a decrease of 14 lbs of fat.
What should my caloric intake be?
I've seen caloric calculators, but it wasn't exactly clear to input current weight vs target weight, or if the output was to maintain or to grow.
>> Also, how do I account for strength training 5 days a week?
Do you guys have a recommended timeline? What can I expect in 12 weeks?
Please give me some numbers, thanks!!
You should read the stickys for the numbers you want :)
In 12 weeks, if you're cutting, maybe a loss of 12lbs ish?
By the way at 6 1 and 180lbs i can't understand how you can be 25% BF.
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## TPTP Problem File: SYO493^6.p
View Solutions - Solve Problem
```%------------------------------------------------------------------------------
% File : SYO493^6 : TPTP v7.1.0. Released v4.0.0.
% Domain : Syntactic
% Problem : Ted Sider's S5 quantified modal logic wff 19
% Version : Especial.
% English :
% Refs : [Sid09] Sider (2009), Logic for Philosophy
% Source : [Sid09]
% Names :
% Status : CounterSatisfiable
% Rating : 0.33 v6.2.0, 0.00 v5.4.0, 0.67 v5.0.0, 0.33 v4.1.0, 0.00 v4.0.0
% Syntax : Number of formulae : 74 ( 0 unit; 37 type; 33 defn)
% Number of atoms : 258 ( 38 equality; 139 variable)
% Maximal formula depth : 11 ( 6 average)
% Number of connectives : 150 ( 5 ~; 5 |; 8 &; 124 @)
% ( 0 <=>; 8 =>; 0 <=; 0 <~>)
% ( 0 ~|; 0 ~&)
% Number of type conns : 182 ( 182 >; 0 *; 0 +; 0 <<)
% Number of symbols : 41 ( 37 :; 0 =)
% Number of variables : 91 ( 3 sgn; 30 !; 6 ?; 55 ^)
% ( 91 :; 0 !>; 0 ?*)
% ( 0 @-; 0 @+)
% SPC : TH0_CSA_EQU_NAR
%------------------------------------------------------------------------------
%----Include axioms for modal logic S5
include('Axioms/LCL013^0.ax').
include('Axioms/LCL013^6.ax').
%------------------------------------------------------------------------------
thf(f_type,type,(
f: mu > \$i > \$o )).
thf(g_type,type,(
g: mu > \$i > \$o )).
thf(prove,conjecture,
( mvalid
@ ( mimplies
@ ( mexists_ind
@ ^ [X: mu] :
( mbox_s5 @ ( mor @ ( f @ X ) @ ( g @ X ) ) ) )
@ ( mbox_s5
@ ( mor
@ ( mforall_ind
@ ^ [X: mu] :
( f @ X ) )
@ ( mexists_ind
@ ^ [X: mu] :
( g @ X ) ) ) ) ) )).
%------------------------------------------------------------------------------
```
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Math
posted by Martha
You have 48 yellow blocks and 40 green blocks. What is the greatest number of identical towers that you can build using all 88 blocks?
1. MathMate
For a particular tower built, all 48 yellow blocks can be rearranged in 48! ways, and similarly, the 40 green blocks can be arranged in 40! different ways.
So the number of identical towers that can be built is 48!*40! This is approximately equal to 1.0*10^19.
However, if the number of different towers is required, the number is 88! if all blocks are of different colours. Since we have found 48!*40! identical towers for each tower that we build, so there are 88!/(40!48!) different towers using the above 88 blocks. This evaluates to "only" 18312575054317505569702710 different towers, or 1.8*10^25 different towers.
2. Shaniqua
You can just build as many identical towers as you're little heart can handle, and don't you let nobody tell you otherwise honey.
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# 7.6: Population Growth and the Logistic Equation
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##### Motivating Questions
• How can we use differential equations to realistically model the growth of a population?
• How can we assess the accuracy of our models?
The growth of the earth's population is one of the pressing issues of our time. Will the population continue to grow? Or will it perhaps level off at some point, and if so, when? In this section, we look at two ways in which we may use differential equations to help us address these questions.
Before we begin, let's consider again two important differential equations that we have seen in earlier work this chapter.
##### Preview Activity $$\PageIndex{1}$$
Recall that one model for population growth states that a population grows at a rate proportional to its size.
1. We begin with the differential equation
$\frac{dP}{dt} = \frac12 P\text{.} \nonumber$
Sketch a slope field below as well as a few typical solutions on the axes provided.
2. Find all equilibrium solutions of the equation $$\frac{dP}{dt} = \frac12 P$$ and classify them as stable or unstable.
3. If $$P(0)$$ is positive, describe the long-term behavior of the solution to $$\frac{dP}{dt} = \frac12 P\text{.}$$
4. Let's now consider a modified differential equation given by
$\frac{dP}{dt} = \frac 12 P(3-P)\text{.} \nonumber$
As before, sketch a slope field as well as a few typical solutions on the following axes provided.
5. Find any equilibrium solutions and classify them as stable or unstable.
6. If $$P(0)$$ is positive, describe the long-term behavior of the solution.
### $$\PageIndex{1}$$ The earth's population
We will now begin studying the earth's population. To get started, in Table $$\PageIndex{1}$$ are some data for the earth's population in recent years that we will use in our investigations.
Table $$\PageIndex{1}$$. Some recent population data for planet Earth.
Year 1998 1999 2000 2001 2002 2005 2006 2007 2008 2009 2010
Pop
(billions)
$$5.932$$ $$6.008$$ $$6.084$$ $$6.159$$ $$6.234$$ $$6.456$$ $$6.531$$ $$6.606$$ $$6.681$$ $$6.756$$ $$6.831$$
##### Activity $$\PageIndex{2}$$
Our first model will be based on the following assumption:
The rate of change of the population is proportional to the population.
On the face of it, this seems pretty reasonable. When there is a relatively small number of people, there will be fewer births and deaths so the rate of change will be small. When there is a larger number of people, there will be more births and deaths so we expect a larger rate of change.
If $$P(t)$$ is the population $$t$$ years after the year 2000, we may express this assumption as
$\frac{dP}{dt} = kP \nonumber$
where $$k$$ is a constant of proportionality.
1. Use the data in the table to estimate the derivative $$P'(0)$$ using a central difference. Assume that $$t=0$$ corresponds to the year 2000.
2. What is the population $$P(0)\text{?}$$
3. Use your results from (a) and (b) to estimate the constant of proportionality $$k$$ in the differential equation.
4. Now that we know the value of $$k\text{,}$$ we have the initial value problem
$\frac{dP}{dt} = kP, \ P(0) = 6.084\text{.} \nonumber$
Find the solution to this initial value problem.
5. What does your solution predict for the population in the year 2010? Is this close to the actual population given in the table?
6. When does your solution predict that the population will reach 12 billion?
7. What does your solution predict for the population in the year 2500?
8. Do you think this is a reasonable model for the earth's population? Why or why not? Explain your thinking using a couple of complete sentences.
Our work in Activity $$\PageIndex{2}$$ shows that that the exponential model is fairly accurate for years relatively close to 2000. However, if we go too far into the future, the model predicts increasingly large rates of change, which causes the population to grow arbitrarily large. This does not make much sense since it is unrealistic to expect that the earth would be able to support such a large population.
The constant $$k$$ in the differential equation has an important interpretation. Let's rewrite the differential equation $$\frac{dP}{dt} = kP$$ by solving for $$k\text{,}$$ so that we have
$k = \frac{dP/dt}{P}\text{.} \nonumber$
We see that $$k$$ is the ratio of the rate of change to the population; in other words, it is the contribution to the rate of change from a single person. We call this the per capita growth rate.
In the exponential model we introduced in Activity $$\PageIndex{2}$$, the per capita growth rate is constant. This means that when the population is large, the per capita growth rate is the same as when the population is small. It is natural to think that the per capita growth rate should decrease when the population becomes large, since there will not be enough resources to support so many people. We expect it would be a more realistic model to assume that the per capita growth rate depends on the population $$P\text{.}$$
In the previous activity, we computed the per capita growth rate in a single year by computing $$k\text{,}$$ the quotient of $$\frac{dP}{dt}$$ and $$P$$ (which we did for $$t = 0$$). If we return to the data in Table $$\PageIndex{1}$$ and compute the per capita growth rate over a range of years, we generate the data shown in Figure $$\PageIndex{3}$$, which shows how the per capita growth rate is a function of the population, $$P\text{.}$$
From the data, we see that the per capita growth rate appears to decrease as the population increases. In fact, the points seem to lie very close to a line, which is shown at two different scales in Figure $$\PageIndex{4}$$.
Looking at this line carefully, we can find its equation to be
$\frac{dP/dt}{P} = 0.025 - 0.002P\text{.} \nonumber$
If we multiply both sides by $$P\text{,}$$ we arrive at the differential equation
$\frac{dP}{dt} = P(0.025 - 0.002P)\text{.} \nonumber$
Graphing the dependence of $$dP/dt$$ on the population $$P\text{,}$$ we see that this differential equation demonstrates a quadratic relationship between $$\frac{dP}{dt}$$ and $$P\text{,}$$ as shown in Figure $$\PageIndex{5}$$.
The equation $$\frac{dP}{dt} = P(0.025 - 0.002P)$$ is an example of the logistic equation, and is the second model for population growth that we will consider. We expect that it will be more realistic, because the per capita growth rate is a decreasing function of the population.
Indeed, the graph in Figure $$\PageIndex{5}$$ shows that there are two equilibrium solutions, $$P=0\text{,}$$ which is unstable, and $$P=12.5\text{,}$$ which is a stable equilibrium. The graph shows that any solution with $$P(0) \gt 0$$ will eventually stabilize around 12.5. Thus, our model predicts the world's population will eventually stabilize around 12.5 billion.
A prediction for the long-term behavior of the population is a valuable conclusion to draw from our differential equation. We would, however, also like to answer some quantitative questions. For instance, how long will it take to reach a population of 10 billion? To answer this question, we need to find an explicit solution of the equation.
## Solving the logistic differential equation
Since we would like to apply the logistic model in more general situations, we state the logistic equation in its more general form,
$\frac{dP}{dt} = kP(N-P)\text{.}\label{yri}\tag{$$\PageIndex{1}$$}$
The equilibrium solutions here are $$P=0$$ and $$1-\frac PN = 0\text{,}$$ which shows that $$P=N\text{.}$$ The equilibrium at $$P=N$$ is called the carrying capacity of the population for it represents the stable population that can be sustained by the environment.
We now solve the logistic equation ($$\PageIndex{1}$$). The equation is separable, so we separate the variables
$\frac{1}{P(N-P)}\frac{dP}{dt} = k\text{,} \nonumber$
and integrate to find that
$\int \frac{1}{P(N-P)}~dP = \int k~dt\text{.} \nonumber$
To find the antiderivative on the left, we use the partial fraction decomposition
$\frac{1}{P(N-P)} = \frac 1N\left[\frac 1P + \frac 1{N-P}\right]\text{.} \nonumber$
Now we are ready to integrate, with
$\int \frac 1N\left[\frac 1P + \frac 1{N-P}\right] ~dP = \int k~dt\text{.} \nonumber$
On the left, observe that $$N$$ is constant, so we can remove a factor of $$\frac{1}{N}$$ and antidifferentiate to find that
$\frac 1N (\ln|P| - \ln|N-P|) = kt + C\text{.} \nonumber$
Multiplying both sides of this last equation by $$N$$ and using a rule of logarithms, we next find that
$\ln\left|\frac{P}{N-P}\right| = kNt + C\text{.} \nonumber$
From the definition of the logarithm, replacing $$e^C$$ with $$C\text{,}$$ and letting $$C$$ absorb the absolute value signs, we now know that
$\frac{P}{N-P} = Ce^{kNt}\text{.} \nonumber$
At this point, all that remains is to determine $$C$$ and solve algebraically for $$P\text{.}$$
If the initial population is $$P(0) = P_0\text{,}$$ then it follows that $$C = \frac{P_0}{N-P_0}\text{,}$$ so
$\frac{P}{N-P} = \frac{P_0}{N-P_0}e^{kNt}\text{.} \nonumber$
We will solve this equation for $$P$$ by multiplying both sides by $$(N-P)(N-P_0)$$ to obtain
\begin{align*} P(N-P_0) &= P_0(N-P)e^{kNt}\4pt] &= P_0Ne^{kNt} - P_0Pe^{kNt}\text{.} \end{align*} Solving for $$P_0Ne^{kNt}\text{,}$$ expanding, and factoring, it follows that \begin{align*} P_0Ne^{kNt} &= P(N-P_0) + P_0 Pe^{kNt}\\[4pt] &= P(N-P_0 + P_0e^{kNt})\text{.} \end{align*} Dividing to solve for $$P\text{,}$$ we see that \[ P = \frac{P_0Ne^{kNt}}{N-P_0 + P_0e^{kNt}}\text{.} \nonumber
Finally, we choose to multiply the numerator and denominator by $$\frac{1}{P_0}e^{-kNt}$$ to obtain
$P(t) = \frac{N}{\left(\frac{N-P_0}{P_0}\right) e^{-kNt} + 1}\text{.} \nonumber$
While that was a lot of algebra, notice the result: we have found an explicit solution to the logistic equation.
##### Solution to the Logistic Equation
The solution to the initial value problem
$\frac{dP}{dt} = kP(N-P), \ P(0) = P_0\text{,} \nonumber$
is
$P(t) = \frac{N}{\left(\frac{N-P_0}{P_0}\right) e^{-kNt} + 1}\text{.}\label{gnE}$
For the logistic equation describing the earth's population that we worked with earlier in this section, we have
$k=0.002, N= 12.5, \ \text{and} \ P_0 = 6.084\text{.} \nonumber$
This gives the solution
$P(t) = \frac{12.5}{1.0546e^{-0.025t} + 1}\text{,} \nonumber$
whose graph is shown in Figure $$\PageIndex{6}$$.
The graph shows the population leveling off at 12.5 billion, as we expected, and that the population will be around 10 billion in the year 2050. These results, which we have found using a relatively simple mathematical model, agree fairly well with predictions made using a much more sophisticated model developed by the United Nations.
The logistic equation is good for modeling any situation in which limited growth is possible. For instance, it could model the spread of a flu virus through a population contained on a cruise ship, the rate at which a rumor spreads within a small town, or the behavior of an animal population on an island. Through our work in this section, we have completely solved the logistic equation, regardless of the values of the constants $$N\text{,}$$ $$k\text{,}$$ and $$P_0\text{.}$$ Anytime we encounter a logistic equation, we can apply the formula we found in Equation \ref{gnE}.
##### Activity $$\PageIndex{3}$$
Consider the logistic equation
$\frac{dP}{dt} = kP(N-P) \nonumber$
with the graph of $$\frac{dP}{dt}$$ vs. $$P$$ shown in Figure $$\PageIndex{7}$$.
1. At what value of $$P$$ is the rate of change greatest?
2. Consider the model for the earth's population that we created. At what value of $$P$$ is the rate of change greatest? How does that compare to the population in recent years?
3. According to the model we developed, what will the population be in the year 2100?
4. According to the model we developed, when will the population reach 9 billion?
5. Now consider the general solution to the general logistic initial value problem that we found, given by
$P(t) = \frac{N}{\left(\frac{N-P_0}{P_0}\right)e^{-kNt} + 1}\text{.} \nonumber$
Verify algebraically that $$P(0) = P_0$$ and that $$\lim_{t\to\infty} P(t) = N\text{.}$$
## Summary
• If we assume that the rate of growth of a population is proportional to the population, we are led to a model in which the population grows without bound and at a rate that grows without bound.
• By assuming that the per capita growth rate decreases as the population grows, we are led to the logistic model of population growth, which predicts that the population will eventually stabilize at the carrying capacity.
This page titled 7.6: Population Growth and the Logistic Equation is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Matthew Boelkins, David Austin & Steven Schlicker (ScholarWorks @Grand Valley State University) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
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# Cost of Goods Sold (COGS): Calculation & Example
Instructor: Yuanxin (Amy) Yang Alcocer
Amy has a master's degree in secondary education and has taught math at a public charter high school.
After reading this lesson, you'll be able to calculate the cost of goods sold for any business. You'll learn the formula to use as well as how this number benefits you tax-wise.
## Cost of Goods Sold
Every business needs to know or be able to calculate its cost of goods sold come tax time. We define the cost of goods sold (COGS) as the cost to make all the products that sold. If a business is making the calculation at the end of the year for that particular tax year, then the cost of goods sold will be the cost to make all the products that sold that year.
It's important to know or be able to calculate this number because this number will actually give you a very useful tax benefit. The IRS lets you deduct your cost of goods sold from your gross receipts, thus decreasing the amount of taxable income your business has. This means that you'll have less tax to pay.
## Formula
Because the IRS is giving you this benefit, it also has stipulated a certain way to calculate your cost of goods sold.
The formula the IRS uses is this one:
(beginning inventory) + (cost to make more products) - (ending inventory) = COGS
## Using the Formula
To use this formula, you first add your beginning inventory and whatever costs you spent on making more products during the year. Then you subtract your ending inventory. The resulting number is your cost of goods sold. The IRS actually provides you a little table that helps you think about all the costs that go into the making of your products.
So, say for example your toy business has a beginning inventory of \$5,000. During the year, you make more toys at a cost of \$50,000. At the end of the year, you are left with \$3,000 worth of inventory. Plugging these numbers into the formula, you get this:
\$5,000 + \$50,000 - \$3,000
Evaluating this, you get a COGS of \$5,000 + \$50,000 - \$3,000 = \$52,000. This is the number you give to the IRS, and this is the number that the IRS lets you deduct from your gross receipts.
## Example
Let's look at another example.
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### Register to view this lesson
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Recognized HTML document
APPENDIX B. 221
of maintaining trustworthy life-histories that shall be of medical service in afterlife to the person who keeps them. The Record shows how the life histories of members of the same family may be collated and used to forecast the development in mind and body of the younger generation of that family. Both works are intended to promote the registration of a large amount of information that has hitherto been allowed to run to waste in oblivion, instead of accumulating and forming stores of recorded experience for future personal use, and from which future inquirers into heredity may hope to draw copious supplies.
B.
PROBLEMS BY J. D. HAMILTON DICx8ON, FELLOW AND TUTOR OF
ST. PETER'S COLLEGE, CAMBRIDGE.
(Reprinted front Pree. Royal Soe., No. 242, 1886, p. 63.)
Problem 1.-A point P is capable of moving along a straight line FOP, making an angle tan-12 with the axis of y, which is drawn through 0 the mean position oaf P'; the probable error of the projection of P on Oy is 122 inch : another point p, whose mean position at any time is P, is capable of moving from P parallel to the axis of x (rectangular co-ordinates) with a probable error of 1.50 inch. To discuss the "surface of frequency " of p.
1. Expressing the " surface of frequency " by an equation in x, y, z, the exponent, with its sign changed, of the exponential which appears in the value of z in the equation of the surface is, save as to a factor,
e + (3x - 2y)2 (2) (1-22), 9(1 50)2'
hence all sections of the "surface of frequency" by planes parallel to the plane of xy are ellipses, whose equations may be written in the form,
(1 22)2 + (9(l. 0)22 - C, a constant . (2)
2. Tangents to these ellipses parallel to the axis of y are found,
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Email us to get an instant 20% discount on highly effective K-12 Math & English kwizNET Programs!
#### Online Quiz (WorksheetABCD)
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### Grade 3 - Mathematics2.2 Division Vocabulary
Explanation: Addition, subtraction, multiplication and division are called operations. Subtraction is the inverse of addition. Multiplication is the inverse of division. Multiplication is repeated addition. Division is repeated subtraction. Zero divided by any number (except 0 equals 0) Number divided by any 0 is not defined Any number(except 0) divided by itself equals 1. Any number divided by 1 equals that number. The answer to the division problem is called the quotient. The number you are dividing is called the dividend. The number you are dividing by is called the divisor. Example: What times 5 equals 20? From 5 tables, we have 5 * 4 = 20 Answer: 4 Directions: Answer the following questions. Explain the division vocabulary in your own words. Also, write the times tables from one through ten.
Q 1: You cannot divide by 0 (ex: 5/0 is impossible).TrueFalse Q 2: Any number (except 0) divided by itself equals 1.FalseTrue Q 3: Division and multiplication are inverse operations.FalseTrue Q 4: Division is a repeated _________.additionmultiplicationarithmeticsubtraction Q 5: Any number divided by 1 equals that number.FalseTrue Q 6: Any number divided by the same number is equal to 1.TrueFalse Question 7: This question is available to subscribers only! Question 8: This question is available to subscribers only!
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## MA/CS 371 - Lab 5 More Numerical Integration - The Composite Trapezoid Rule (CTR) Versus The Composite Simpson's Rule (CSR)
### Section 1: Introduction
In this lab you will:
• Write a practical implementation of the CSR
• Compare the convergence behavior of the CSR and the CTR.
If you have not already done so, create a directory for this lab and copy the lab 5 files into it. If you start off in your home directory you can do the following:
``` mkdir ~/cs371/lab5
cd ~/cs371/lab5
cp ~cs371/lab5/* .
```
and (as always) don't forget to type that last period.
### Section 2: Topics for Lecture
Today in lecture I'm going to discuss the flops command which is used to count the number of "floating-point operations" that MATLAB performs. Comparing the "flops count" of two different algorithms is one way of determining which is more efficient.
### Section 3: Today's Function
Today's special guest function is one with some flat spots and some sharp curves:
``` x sin(x)
f(x) = sin(e ) - e
```
This week we are going to estimate the integral of f(x) from -10 to 3 with the Composite Simpson's Rule and compare the results to the Composite Trapezoid Rule.
First off, I recommend that you plot f(x) on the interval [-10,3] just to see what it's doing. Make sure to use enough points that you can see how the function wiggles around towards the upper end of the interval.
### Section 4: Implementing the Composite Trapezoid Rule
The trap.m file given to you for this lab is the one that I wrote for the previous lab assignment. For this lab, you should modify trap.m in the following ways:
• Approximate func(x) instead of 1-sin(x).
• Integrate from -10 to 3 instead of 0 to pi/2.
• Add another column to your table which shows how many flops are required for each number of partitions.
### Section 5: Implementing the Composite Simpson's Rule
Using your trap.m function as a template, make a new file simpson.m which will approximate the integral with the CSR. (Note: Problem 5.4.1 in the book contains the formula for the CSR.) Use the same number of intervals as in trap.m and format your output the same way.
Recall that the CSR requires an even number of intervals (or an odd number of evaluation points.) You will need to change the vector q in simpson.m to accommodate this.
### Section 6: Submitting Your Work
``` ~cs371/submit lab5
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All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
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Family Murder Muse (Posted on 2014-04-09)
Murder occurred one evening in the home of a married couple Mr. Hahn and Mrs. Hahn and their son Oliver and their daughter Claire. One member of the family murdered another member, the third member witnessed the crime, and the fourth member was an accessory after the fact.
1. The accessory and the witness were of the opposite sex.
2. The oldest member and the witness were of the opposite sex.
3. The youngest member and the victim were of the opposite sex.
4. The accessory was older than the victim.
5. Mr. Hahn was the oldest member.
6. The killer was not the youngest member.
From the above clues, determine which one of the four – Mr. Hahn, Mrs. Hahn, Oliver or Claire - was the killer?
No Solution Yet Submitted by K Sengupta No Rating
Comments: ( Back to comment list | You must be logged in to post comments.)
Ambiguity: 2 solutions possible | Comment 3 of 4 |
There is an ambiguity in who is the second and who is the third oldest person. From the conditions of the logic puzzle alone there are 2 conclusions possible that we can draw by pure logic (or a programm can draw).
We don't have the information that Oliver is Mrs. Hahn's own (blood) son. So there is no necessity to presuppose that she must be older than Oliver. You can confirm this for example by a Prolog program. Therefore, the second solution is:
Mr Hahn oldest male accessory
Oliver 2nd oldest male victim
Mrs Hahn 3rd oldest female killer
Claire youngest female witness
Posted by braun on 2014-12-14 15:39:27
Search: Search body:
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# Chem 1046
posted by on .
Consider the reaction:
I2(g) + Cl2(g) 2 ICl(g) Kp = 81.9 at 25°C
Calculate ΔGrxn for the reaction at 25°C under each of the following conditions:
(a) standard conditions
(b) at equilibrium
(c) PICl = 2.48 atm; PI2 = 0.321 atm; PCl2 = 0.219 atm
• Chem 1046 - ,
a) dG = -RTlnK
b) dG at equilibrium is zero
c) I suppose c means immediately after those conditions are used and before it re-establishes equilibrium. Substitute the values, solve for Qsp and recalculate dG.
• Chem 1046 - ,
Note: for c remember
dG = dGo + RTlnQ
• Chem 1046 - ,
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# 1895. Largest Magic Square
• Time: $O(mnk^2)$
• Space: $O(mn)$
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 class Solution { public: int largestMagicSquare(vector>& grid) { const int m = grid.size(); const int n = grid[0].size(); // prefixRow[i][j] := prefix sum of first j nums in i-th row vector> prefixRow(m, vector(n + 1)); // prefixCol[i][j] := prefix sum of first j nums in i-th col vector> prefixCol(n, vector(m + 1)); for (int i = 0; i < m; ++i) for (int j = 0; j < n; ++j) { prefixRow[i][j + 1] = prefixRow[i][j] + grid[i][j]; prefixCol[j][i + 1] = prefixCol[j][i] + grid[i][j]; } for (int k = min(m, n); k >= 2; --k) if (containsMagicSquare(grid, prefixRow, prefixCol, k)) return k; return 1; } private: // Returns true if grid contains any magic square of size k x k. bool containsMagicSquare(const vector>& grid, const vector>& prefixRow, const vector>& prefixCol, int k) { for (int i = 0; i + k - 1 < grid.size(); ++i) for (int j = 0; j + k - 1 < grid[0].size(); ++j) if (isMagicSquare(grid, prefixRow, prefixCol, i, j, k)) return true; return false; } // Returns true if grid[i..i + k)[j..j + k) is a magic square. bool isMagicSquare(const vector>& grid, const vector>& prefixRow, const vector>& prefixCol, int i, int j, int k) { int diag = 0; int antiDiag = 0; for (int d = 0; d < k; ++d) { diag += grid[i + d][j + d]; antiDiag += grid[i + d][j + k - 1 - d]; } if (diag != antiDiag) return false; for (int d = 0; d < k; ++d) { if (getSum(prefixRow, i + d, j, j + k - 1) != diag) return false; if (getSum(prefixCol, j + d, i, i + k - 1) != diag) return false; } return true; } // Returns sum(grid[i][l..r]) or sum(grid[l..r][i]). int getSum(const vector>& prefix, int i, int l, int r) { return prefix[i][r + 1] - prefix[i][l]; } };
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 class Solution { public int largestMagicSquare(int[][] grid) { final int m = grid.length; final int n = grid[0].length; // prefixRow[i][j] := prefix sum of first j nums in i-th row int[][] prefixRow = new int[m][n + 1]; // prefixCol[i][j] := prefix sum of first j nums in i-th col int[][] prefixCol = new int[n][m + 1]; for (int i = 0; i < m; ++i) for (int j = 0; j < n; ++j) { prefixRow[i][j + 1] = prefixRow[i][j] + grid[i][j]; prefixCol[j][i + 1] = prefixCol[j][i] + grid[i][j]; } for (int k = Math.min(m, n); k >= 2; --k) if (containsMagicSquare(grid, prefixRow, prefixCol, k)) return k; return 1; } // Returns true if grid contains a magic square of size k x k. private boolean containsMagicSquare(int[][] grid, int[][] prefixRow, int[][] prefixCol, int k) { for (int i = 0; i + k - 1 < grid.length; ++i) for (int j = 0; j + k - 1 < grid[0].length; ++j) if (isMagicSquare(grid, prefixRow, prefixCol, i, j, k)) return true; return false; } // Returns true if grid[i..i + k)[j..j + k) is a magic square. private boolean isMagicSquare(int[][] grid, int[][] prefixRow, int[][] prefixCol, int i, int j, int k) { int diag = 0; int antiDiag = 0; for (int d = 0; d < k; ++d) { diag += grid[i + d][j + d]; antiDiag += grid[i + d][j + k - 1 - d]; } if (diag != antiDiag) return false; for (int d = 0; d < k; ++d) { if (getSum(prefixRow, i + d, j, j + k - 1) != diag) return false; if (getSum(prefixCol, j + d, i, i + k - 1) != diag) return false; } return true; } // Returns sum(grid[i][l..r]) or sum(grid[l..r][i]). private int getSum(int[][] prefix, int i, int l, int r) { return prefix[i][r + 1] - prefix[i][l]; } }
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 class Solution: def largestMagicSquare(self, grid: List[List[int]]) -> int: m = len(grid) n = len(grid[0]) # prefixRow[i][j] := prefix sum of first j nums in i-th row prefixRow = [[0] * (n + 1) for _ in range(m)] # prefixCol[i][j] := prefix sum of first j nums in i-th col prefixCol = [[0] * (m + 1) for _ in range(n)] for i in range(m): for j in range(n): prefixRow[i][j + 1] = prefixRow[i][j] + grid[i][j] prefixCol[j][i + 1] = prefixCol[j][i] + grid[i][j] # Returns true if grid[i..i + k)[j..j + k) is a magic square. def isMagicSquare(i: int, j: int, k: int) -> bool: diag, antiDiag = 0, 0 for d in range(k): diag += grid[i + d][j + d] antiDiag += grid[i + d][j + k - 1 - d] if diag != antiDiag: return False for d in range(k): if self._getSum(prefixRow, i + d, j, j + k - 1) != diag: return False if self._getSum(prefixCol, j + d, i, i + k - 1) != diag: return False return True # Returns true if grid contains a magic square of size k x k. def containsMagicSquare(k: int) -> bool: for i in range(m - k + 1): for j in range(n - k + 1): if isMagicSquare(i, j, k): return True return False for k in range(min(m, n), 1, -1): if containsMagicSquare(k): return k return 1 # Returns sum(grid[i][l..r]) or sum(grid[l..r][i]). def _getSum(self, prefix: List[List[int]], i: int, l: int, r: int) -> int: return prefix[i][r + 1] - prefix[i][l]
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### Stefano Di Maio (@dimaios)
Automatic control specialist
## Re: 50Hz and harmonics filtering
Hi Fred, thanks for your reply.The approch is good but the problem f the sawtooth wave step response.As explained before I've realized the...
## Re: 50Hz and harmonics filtering
Hi Rick, that was one of the first attempts, unfortunately I've to filter both odd and even multiple of the fondamental. Best regards
## Re: 50Hz and harmonics filtering
As explained before the signal s(t) to be analyzed is a random process with spectrum support within [0..1 kHz]. Grid harmonics are summed to this signal up to the...
## Re: 50Hz and harmonics filtering
"I am a firm believer that the easiest signal processing is the signal processing that you don't have to do!"Let's put that one in a frame ! :)
## Re: 50Hz and harmonics filtering
I try to answer to your questions:Does the desired AC signal consist of single or multiple frequencies? The signal s(t) to be analyzed is a random process with...
## Re: 50Hz and harmonics filtering
The best algorithm for achieving what you want (of which I am aware) is called the "shielded cable algorithm." :)Cables in the plant are shielded. :)At first......
## Re: 50Hz and harmonics filtering
Is the 50 Hz a power main signal? Yes, but consider that frequency is not fixed but time variable. Alarm condition are signaled outside limits [49.9 - 50.1] Hz....
## Re: 50Hz and harmonics filtering
The sampling rate is about 3kHz and we are trying to attenuate both odd and even harmonics.The harmonics can extend a lot into the spectrum ( up to 700-800 Hz ).The...
## 50Hz and harmonics filtering
I'm facing a problem trying to filter 50Hz and harmonics from a signal s(t).I've implemented several techniques like Kalman filtering, optimized FIR and IIR filters...
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++ed by:
Kevin Ryde
and 1 contributors
# NAME
Math::PlanePath::RationalsTree -- rationals by tree
# SYNOPSIS
`````` use Math::PlanePath::RationalsTree;
my \$path = Math::PlanePath::RationalsTree->new (tree_type => 'SB');
my (\$x, \$y) = \$path->n_to_xy (123);``````
# DESCRIPTION
This path enumerates reduced rational fractions X/Y > 0, ie. X and Y having no common factor.
The rationals are traversed by rows of a binary tree which represents a coprime pair X,Y by steps of a subtraction-only greatest common divisor algorithm which proves them coprime. Or equivalently by bit runs with lengths which are the quotients in the division-based Euclidean GCD algorithm and which are also the terms in the continued fraction representation of X/Y.
The SB, CW, AYT, HCS, Bird and Drib trees all have the same set of X/Y rationals in a row, but in a different order due to different encodings of the N value. The L tree has a shift which visits 0/1 too.
The bit runs mean that N values are quite large for relatively modest sized rationals. For example in the SB tree 167/3 is N=288230376151711741, a 58-bit number. The tendency is for the tree to make excursions out to large rationals while only slowly filling in small ones. The worst is the integer X/1 for which N has X many bits, and similarly 1/Y is Y bits.
See examples/rationals-tree.pl in the Math-PlanePath sources for a printout of all the trees.
## Stern-Brocot Tree
The default `tree_type=>"SB"` is the tree of Moritz Stern and Achille Brocot.
`````` depth N
----- -------
0 1 1/1
------ ------
1 2 to 3 1/2 2/1
/ \ / \
2 4 to 7 1/3 2/3 3/2 3/1
| | | | | | | |
3 8 to 15 1/4 2/5 3/5 3/4 4/3 5/3 5/2 4/1 ``````
Within a row the fractions increase in value. Each row of the tree is a repeat of the previous row as first X/(X+Y) and then (X+Y)/Y. For example
`````` depth=1 1/2, 2/1
depth=2 1/3, 2/3 X/(X+Y) of previous row
3/2, 3/1 (X+Y)/Y of previous row``````
Plotting the N values by X,Y is as follows. The unused X,Y positions are where X and Y have a common factor. For example X=6,Y=2 has common factor 2 so is never reached.
`````` tree_type => "SB"
10 | 512 35 44 767
9 | 256 33 39 40 46 383 768
8 | 128 18 21 191 384
7 | 64 17 19 20 22 95 192 49 51
6 | 32 47 96
5 | 16 9 10 23 48 25 26 55
4 | 8 11 24 27 56
3 | 4 5 12 13 28 29 60
2 | 2 6 14 30 62
1 | 1 3 7 15 31 63 127 255 511 1023
Y=0 |
----------------------------------------------------
X=0 1 2 3 4 5 6 7 8 9 10``````
The X=1 vertical is the fractions 1/Y which is at the left of each tree row, at N value
`` Nstart = 2^depth``
The Y=1 horizontal is the X/1 integers at the end each row which is
`` Nend = 2^(depth+1)-1``
Numbering nodes of the tree by rows starting from 1 means N without the high 1 bit is the offset into the row. For example binary N="1011" is "011"=3 into the row. Those bits after the high 1 are also the directions to follow down the tree to a node, with 0=left and 1=right. So N="1011" binary goes from the root 0=left then twice 1=right to reach X/Y=3/4 at N=11 decimal.
## Stern-Brocot Turn Sequence
Since each row is fractions of increasing value the path goes from the Y axis across and down to the X. Each row is further from the origin than the previous row and doesn't intersect any other row. The X/(X+Y) first half is an upward "shear" of the X,Y points in the previous row. The second half (X+Y)/Y is a shear to the right. For example,
`````` N=8 to N=11
previous row
sheared up X,X+Y
depth=2 N=4to7
| | 9--10 . depth=3 N=8to15
| | / | .
| | 8 11 .
| | .
| 4---5 | . 12--13 N=12 to N=15
| \ | . | previous row
| 6 | . 14 sheared right
| | | . / as X+Y,Y
| 7 | 15
| |
+--------------- +----------------``````
The sequence of turns left or right is unchanged by the shears. So at N=5 the path turns towards the right and this is unchanged in the sheared copies at N=9 and N=13. The angle of the turn is different, but it's still to the right.
The first and last points of each row are always a turn to the right. For example the turn at N=4 (going N=3 to N=4 to N=5) is to the right, and likewise at N=7. This is because the second point in the row such as N=5 is above a 45-degree line down from N=4, and similarly the second last such as N=6 (since the row is symmetric).
The middle two points in each row for depth>=3 are always a turn to the left. For example N=11 and N=12 shown above are the first such middle pair and both turn to the left. The middle two are transposes across the leading diagonal and so make a 45-degree line. The second-from-middle points are above that line (N=10 and N=13).
The middle left turns are copied into successive rows and the result is a repeating pattern "LRRL" except for the first and last in the row which are always right instead of left. So,
`````` RRRL,LRRL,LRRL,LRRL,LRRL,LRRL,LRRL,LRRR # row N=32 to N=63
condition turn
--------- ----
if N=3 left
otherwise if N=2^k or N=2^k-1 right # first and last of row
otherwise if N=0 mod 4 left # LRRL pattern
N=1 mod 4 right
N=2 mod 4 right
N=3 mod 4 left``````
Pairs N=2m-1 and N=2m can be treated together by taking ceil(N/2),
`````` if N=3 left
otherwise if Nhalf=2^k right
otherwise if Nhalf=0 mod 2 left
Nhalf=1 mod 2 right
where Nhalf = ceil(N/2)``````
## Stern-Brocot Mediant
Writing the parents between the children as an "in-order" tree traversal to a given depth has all values in increasing order (the same as each row individually is in increasing order).
`````` 1/1
1/2 | 2/1
1/3 | 2/3 | 3/2 | 3/1
| | | | | | |
1/3 1/2 2/3 1/1 3/2 2/1 3/1
^
|
next level (1+3)/(1+2) = 4/3 mediant``````
New values at the next level of this flattening are a "mediant" (x1+x2)/(y1+y2) formed from the left and right parent. So the next level 4/3 shown is left parent 1/1 and right parent 3/2 giving mediant (1+3)/(1+2)=4/3. At the left end a preceding 0/1 is imagined. At the right end a following 1/0 is imagined, so as to have 1/(depth+1) and (depth+1)/1 at the ends for a total 2^depth many new values.
## Calkin-Wilf Tree
`tree_type=>"CW"` selects the tree of Calkin and Wilf,
Neil Calkin and Herbert Wilf, "Recounting the Rationals", American Mathematical Monthly, volume 107, number 4, April 2000, pages 360-363.
As noted above, the values within each row are the same as the Stern-Brocot, but in a different order.
`````` N=1 1/1
------ ------
N=2 to N=3 1/2 2/1
/ \ / \
N=4 to N=7 1/3 3/2 2/3 3/1
| | | | | | | |
N=8 to N=15 1/4 4/3 3/5 5/2 2/5 5/3 3/4 4/1``````
Going by rows the denominator of one value becomes the numerator of the next. So at 4/3 the denominator 3 becomes the numerator of 3/5 to the right. These values are Stern's diatomic sequence.
Each row is symmetric in reciprocals, ie. reading from right to left is the reciprocals of reading left to right. The numerators read left to right are the denominators read right to left.
A node descends as
`````` X/Y
/ \
X/(X+Y) (X+Y)/Y``````
Taking these formulas in reverse up the tree shows how it relates to a subtraction-only greatest common divisor. At a given node the smaller of P or Q is subtracted from the bigger,
`````` P/(Q-P) (P-Q)/P
/ or \
P/Q P/Q``````
Plotting the N values by X,Y is as follows. The X=1 vertical and Y=1 horizontal are the same as the SB above, but the values in between are re-ordered.
`````` tree_type => "CW"
10 | 512 56 38 1022
9 | 256 48 60 34 46 510 513
8 | 128 20 26 254 257
7 | 64 24 28 18 22 126 129 49 57
6 | 32 62 65
5 | 16 12 10 30 33 25 21 61
4 | 8 14 17 29 35
3 | 4 6 9 13 19 27 39
2 | 2 5 11 23 47
1 | 1 3 7 15 31 63 127 255 511 1023
Y=0 |
-------------------------------------------------------------
X=0 1 2 3 4 5 6 7 8 9 10``````
At each node the left leg is X/(X+Y) < 1 and the right leg is (X+Y)/Y > 1, which means N is even above the X=Y diagonal and odd below. In general each right leg increments the integer part of the fraction,
`````` X/Y right leg each time
(X+Y)/Y = 1 + X/Y
(X+2Y)/Y = 2 + X/Y
(X+3Y)/Y = 3 + X/Y
etc``````
This means the integer part is the trailing 1-bits of N,
`````` floor(X/Y) = count trailing 1-bits of N
eg. 7/2 is at N=23 binary "10111"
which has 3 trailing 1-bits for floor(7/2)=3``````
N values for the SB and CW trees are converted by reversing bits except the highest. So at a given X,Y position
`````` SB N = 1abcde SB <-> CW by reversing bits
CW N = 1edcba except the high 1-bit``````
For example at X=3,Y=4 the SB tree has N=11 = "1011" binary and the CW has N=14 binary "1110", a reversal of the bits below the high 1.
N to X/Y in the CW tree can be calculated keeping track of just an X,Y pair and descending to X/(X+Y) or (X+Y)/Y using the bits of N from high to low. The relationship between the SB and CW N's means the same can be used to calculate the SB tree by taking the bits of N from low to high instead.
See also Math::PlanePath::ChanTree for a generalization of CW to ternary or higher trees, ie. descending to 3 or more children at each node.
## Andreev and Yu-Ting Tree
`tree_type=>"AYT"` selects the tree described (independently is it?) by Andreev and Yu-Ting.
D. N. Andreev, "On a Wonderful Numbering of Positive Rational Numbers", Matematicheskoe Prosveshchenie, Ser. 3, 1, 1997, pages 126-134 http://mi.mathnet.ru/mp12
Shen Yu-Ting, "A Natural Enumeration of Non-Negative Rational Numbers -- An Informal Discussion", American Mathematical Monthly, 87, 1980, pages 25-29. http://www.jstor.org/stable/2320374
Their constructions are a one-to-one mapping between integer N and rational X/Y as a way of enumerating the rationals. This is not designed to be a tree as such, but the result is the same 2^level rows as the above trees. The X/Y values within each row are the same, but in a different order.
`````` N=1 1/1
------ ------
N=2 to N=3 2/1 1/2
/ \ / \
N=4 to N=7 3/1 1/3 3/2 2/3
| | | | | | | |
N=8 to N=15 4/1 1/4 4/3 3/4 5/2 2/5 5/3 3/5``````
Each fraction descends as follows. The left is an increment and the right is reciprocal of the increment.
`````` X/Y
/ \
X/Y + 1 1/(X/Y + 1)``````
which means
`````` X/Y
/ \
(X+Y)/Y Y/(X+Y)``````
The left leg (X+Y)/Y is the same the CW has on its right leg. But Y/(X+Y) is not the same as the CW (the other there being X/(X+Y)).
The left leg increments the integer part, so the integer part is given by (in a fashion similar to CW 1-bits above)
`````` floor(X/Y) = count trailing 0-bits of N
plus one extra if N=2^k``````
N=2^k is one extra because its trailing 0-bits started from N=1 where floor(1/1)=1 whereas any other odd N starts from some floor(X/Y)=0.
The Y/(X+Y) right leg forms the Fibonacci numbers F(k)/F(k+1) at the end of each row, ie. at Nend=2^(level+1)-1. And as noted by Andreev, successive right leg fractions N=4k+1 and N=4k+3 add up to 1,
`````` X/Y at N=4k+1 + X/Y at N=4k+3 = 1
Eg. 2/5 at N=13 and 3/5 at N=15 add up to 1``````
Plotting the N values by X,Y gives
`````` tree_type => "AYT"
10 | 513 41 43 515
9 | 257 49 37 39 51 259 514
8 | 129 29 31 131 258
7 | 65 25 21 23 27 67 130 50 42
6 | 33 35 66
5 | 17 13 15 19 34 26 30 38
4 | 9 11 18 22 36
3 | 5 7 10 14 20 28 40
2 | 3 6 12 24 48
1 | 1 2 4 8 16 32 64 128 256 512
Y=0 |
----------------------------------------------------
X=0 1 2 3 4 5 6 7 8 9 10``````
N=1,2,4,8,etc on the Y=1 horizontal is the X/1 integers at Nstart=2^level=2^X. N=1,3,5,9,etc in the X=1 vertical is the 1/Y fractions. Those fractions always immediately follow the corresponding integer, so N=Nstart+1=2^(Y-1)+1 in that column.
In each node the left leg (X+Y)/Y > 1 and the right leg Y/(X+Y) < 1, which means odd N is above the X=Y diagonal and even N is below.
The tree structure corresponds to Johannes Kepler's tree of fractions (see Math::PlanePath::FractionsTree). That tree starts from 1/2 and makes fractions A/B with A<B by descending to A/(A+B) and B/(A+B). Those descents are the same as the AYT tree and the two are related simply by
`````` A = Y AYT denominator is Kepler numerator
B = X+Y AYT sum num+den is the Kepler denominator
X = B-A inverse
Y = A``````
## HCS Continued Fraction
`tree_type=>"HCS"` selects continued fraction terms coded as bit runs 1000...00 from high to low, as per Paul D. Hanna and independently Czyz and Self.
This arises also in a radix=1 variation of Jeffrey Shallit's digit-based continued fraction encoding. See "Radix 1" in Math::PlanePath::CfracDigits.
If the continued fraction of X/Y is
`````` 1
X/Y = a + ------------ a >= 0
1
b + ----------- b,c,etc >= 1
1
c + -------
... + 1
--- z >= 2
z``````
then the N value is bit runs of lengths a,b,c etc.
`````` N = 1000 1000 1000 ... 1000
\--/ \--/ \--/ \--/
a+1 b c z-1``````
Each group is 1 or more bits. The +1 in "a+1" makes the first group 1 or more bits, since a=0 occurs for any X/Y<=1. The -1 in "z-1" makes the last group 1 or more since z>=2.
`````` N=1 1/1
------ ------
N=2 to N=3 2/1 1/2
/ \ / \
N=4 to N=7 3/1 3/2 1/3 2/3
| | | | | | | |
N=8 to N=15 4/1 5/2 4/3 5/3 1/4 2/5 3/4 3/5``````
The result is a bit reversal of the N values in the AYT tree.
`````` AYT N = binary "1abcde" AYT <-> HCS bit reversal
HCS N = binary "1edcba"``````
For example at X=4,Y=7 the AYT tree is N=11 binary "10111" whereas HCS there has N=30 binary "11110", a reversal of the bits below the high 1.
Plotting by X,Y gives
`````` tree_type => "HCS"
10 | 768 50 58 896
9 | 384 49 52 60 57 448 640
8 | 192 27 31 224 320
7 | 96 25 26 30 29 112 160 41 42
6 | 48 56 80
5 | 24 13 15 28 40 21 23 44
4 | 12 14 20 22 36
3 | 6 7 10 11 18 19 34
2 | 3 5 9 17 33
1 | 1 2 4 8 16 32 64 128 256 512
Y=0 |
+-----------------------------------------------------
X=0 1 2 3 4 5 6 7 8 9 10``````
N=1,2,4,etc in the row Y=1 are powers-of-2, being integers X/1 having just a single group of bits N=1000..000.
N=1,3,6,12,etc in the column X=1 are 3*2^(Y-1) corresponding to continued fraction 0 + 1/Y so terms 0,Y making runs 1,Y-1 and so bits N=11000...00.
## HCS Turn Sequence
The turn sequence left or right following successive X,Y points is the Thue-Morse sequence.
`````` count 1-bits in N+1 turn at N
------------------- ---------
odd right
even left``````
This works because each row is two copies of the preceding. The first copy is (X+Y)/Y so just a shear. This is N=10xxxxx introducing a 0-bit at the top of N and so count 1-bits unchanged. The second copy is Y/(X+Y) so a shear and then transpose. This is N=11xxxxx introducing a further 1-bit at the top of N and transpose changes turns left<->right.
For the last point of a row and the first of the next the points are
`````` N binary
--------
second last 11110 Lucas L[n]/L[n+1] eg. 4/7
last 11111 Fibonacci F[n]/F[n+1] eg. 8/13
first 100000 d+1 / 1 eg. 6/1
second 100001 2d-1 / 2 eg. 9/2``````
The second last of a row 11110 is a pair of Lucas numbers and the last of a row 11111 is a pair of Fibonacci numbers bigger than those Lucas numbers. Plotting the examples shows the layout,
`````` 13 | __* Fib
| __/ / [Right]
| __/ /
| / /
7 | * /
| Luc /
| /
2 | / ___* 2nd
1 | 1st *---
| [Left]
+--------------------------
4 6 8 9``````
The Lucas and Fibonacci pairs are both on a slope roughly Y=X*phi for phi=(1+sqrt(5))/2 the golden ratio. The first and second points of the next row are then off towards X=d+1 and hence a right turn at the last of the row and it corresponds to N+1 = binary "100000" having an odd number of 1-bits (a single 1-bit). Then at the first of the next row the turn is left corresponding to N=1 = binary "100001" having an even number of 1-bits (two 1-bits).
The cases for the middle of a row, where the two copies of the previous row meet, behave similarly,
`````` middle prev 1011110 Lucas L[n+1]/L[n]
middle end 1011111 Fibonacci F[n+1]/F[n]
middle first 1100000 1 / d+1
middle second 1100001 2 / 2d-1``````
These points are like a transpose of the first/last shown above, though the Lucas and Fibonacci pairs are one step further on. The "middle end" 1011111 turns to the right, corresponding to N+1=1100000 having even 1-bits, and then at the "middle first" 1100000 turn left corresponding to N+1=1100001 having odd 1-bits.
## Bird Tree
`tree_type=>"Bird"` selects the Bird tree,
It's expressed recursively, illustrating Haskell programming features. The left subtree is the tree plus one and take the reciprocal. The right subtree is conversely the reciprocal first then add one,
`````` 1 1
-------- and ---- + 1
tree + 1 tree``````
which means Y/(X+Y) and (X+Y)/X taking N bits low to high.
`````` N=1 1/1
------ ------
N=2 to N=3 1/2 2/1
/ \ / \
N=4 to N=7 2/3 1/3 3/1 3/2
| | | | | | | |
N=8 to N=15 3/5 3/4 1/4 2/5 5/2 4/1 4/3 5/3``````
Plotting by X,Y gives
`````` tree_type => "Bird"
10 | 682 41 38 597
9 | 341 43 45 34 36 298 938
8 | 170 23 16 149 469
7 | 85 20 22 17 19 74 234 59 57
6 | 42 37 117
5 | 21 11 8 18 58 28 31 61
4 | 10 9 29 30 50
3 | 5 4 14 15 25 24 54
2 | 2 7 12 27 52
1 | 1 3 6 13 26 53 106 213 426 853
Y=0 |
----------------------------------------------------
X=0 1 2 3 4 5 6 7 8 9 10``````
Notice that unlike the other trees N=1,2,5,10,etc in the X=1 vertical for fractions 1/Y is not the row start or end, but instead are on a zigzag through the middle of the tree binary N=1010...etc alternate 1 and 0 bits. The integers X/1 in the Y=1 vertical are similar, but N=11010...etc starting the alternation from a 1 in the second highest bit, since those integers are in the right hand half of the tree.
The Bird tree N values are related to the SB tree by inverting every second bit starting from the second after the high 1-bit,
`````` Bird N=1abcdefg.. binary
101010.. xor, so b,d,f etc flip 0<->1
SB N=1aBcDeFg.. to make B,D,F``````
For example 3/4 in the SB tree is at N=11 = binary 1011. Xor with 0010 for binary 1001 N=9 which is 3/4 in the Bird tree. The same xor goes back the other way Bird tree to SB tree.
This xoring is a mirroring in the tree, swapping left and right at each level. Only every second bit is inverted because mirroring twice puts it back to the ordinary way on even rows.
## Drib Tree
`tree_type=>"Drib"` selects the Drib tree by Ralf Hinze.
It reverses the bits of N in the Bird tree (in a similar way that the SB and CW are bit reversals of each other).
`````` N=1 1/1
------ ------
N=2 to N=3 1/2 2/1
/ \ / \
N=4 to N=7 2/3 3/1 1/3 3/2
| | | | | | | |
N=8 to N=15 3/5 5/2 1/4 4/3 3/4 4/1 2/5 5/3``````
The descendants of each node are
`````` X/Y
/ \
Y/(X+Y) (X+Y)/X``````
The endmost fractions of each row are Fibonacci numbers, F(k)/F(k+1) on the left and F(k+1)/F(k) on the right.
`````` tree_type => "Drib"
10 | 682 50 44 852
9 | 426 58 54 40 36 340 683
8 | 170 30 16 212 427
7 | 106 18 22 24 28 84 171 59 51
6 | 42 52 107
5 | 26 14 8 20 43 19 31 55
4 | 10 12 27 23 41
3 | 6 4 11 15 25 17 45
2 | 2 7 9 29 37
1 | 1 3 5 13 21 53 85 213 341 853
Y=0 |
-------------------------------------------------------
X=0 1 2 3 4 5 6 7 8 9 10``````
In each node descent the left Y/(X+Y) < 1 and the right (X+Y)/X > 1, which means even N is above the X=Y diagonal and odd N is below.
Because Drib/Bird are bit reversals like CW/SB are bit reversals, the xor procedure described above which relates Bird<->SB applies to Drib<->CW, but working from the second lowest bit upwards, ie. xor binary "0..01010". For example 4/1 is at N=15 binary 1111 in the CW tree. Xor with 0010 for 1101 N=13 which is 4/1 in the Drib tree.
## L Tree
`tree_type=>"L"` selects the L-tree by Peter Luschny.
It's a row-reversal of the CW tree with a shift to include zero as 0/1.
`````` N=0 0/1
------ ------
N=1 to N=2 1/2 1/1
/ \ / \
N=3 to N=8 2/3 3/2 1/3 2/1
| | | | | | | |
N=9 to N=16 3/4 5/3 2/5 5/2 3/5 4/3 1/4 3/1``````
Notice in the N=9 to N=16 row rationals 3/4 to 1/4 are the same as in the CW tree but read right-to-left.
`````` tree_type => "L"
10 | 1021 37 55 511
9 | 509 45 33 59 47 255 1020
8 | 253 25 19 127 508
7 | 125 21 17 27 23 63 252 44 36
6 | 61 31 124
5 | 29 9 11 15 60 20 24 32
4 | 13 7 28 16 58
3 | 5 3 12 8 26 18 54
2 | 1 4 10 22 46
1 | 0 2 6 14 30 62 126 254 510 1022 2046
Y=0 |
-------------------------------------------------------
X=0 1 2 3 4 5 6 7 8 9 10``````
N=0,2,6,14,30,etc along the row at Y=1 are powers 2^(X+1)-2. N=1,5,13,29,etc in the column at X=1 are similar powers 2^Y-3.
## Common Characteristics
The SB, CW, Bird, Drib, AYT and HCS trees have the same set of rationals in each row, just in different orders. The properties of Stern's diatomic sequence mean that within a row the totals are
`````` row N=2^depth to N=2^(depth+1)-1 inclusive
sum X/Y = (3 * 2^depth - 1) / 2
sum X = 3^depth
sum 1/(X*Y) = 1``````
For example the SB tree depth=2, N=4 to N=7,
`````` sum X/Y = 1/3 + 2/3 + 3/2 + 3/1 = 11/2 = (3*2^2-1)/2
sum X = 1+2+3+3 = 9 = 3^2
sum 1/(X*Y) = 1/(1*3) + 1/(2*3) + 1/(3*2) + 1/(3*1) = 1``````
Many permutations are conceivable within a row, but the ones here have some relationship to X/Y descendants, tree sub-forms or continued fractions. As an encoding of continued fraction terms by bit runs the combinations are
`````` bit encoding high to low low to high
---------------- ----------- -----------
0000 1111 runs SB CW
0101 1010 alternating Bird Drib
1000 1000 runs HCS AYT``````
A run of alternating 101010 ends where the next bit is the oppose of the expected alternating 0,1. This is a doubled bit 00 or 11. An electrical engineer would think of it as a phase shift.
## Minkowski Question Mark
The Minkowski question mark function is a sum of the terms in the continued fraction representation of a real number. If q0,q1,q2,etc are those terms then the question mark function "?(r)" is
`````` 1 1 1
?(r) = 2 * (1 - ---- * (1 - ---- * (1 - ---- * (1 - ...
2^q0 2^q1 2^q2
1 1 1
= 2 * (1 - ---- + --------- - ------------ + ... )
2^q0 2^(q0+q1) 2^(q0+q1+q2)``````
For rational r the continued fraction q0,q1,q2,etc is finite and so the ?(r) sum is finite and rational. The pattern of + and - in the terms gives runs of bits the same as the N values in the Stern-Brocot tree. The RationalsTree code can calculate the ?(r) function by
`````` rational r=X/Y
N = xy_to_n(X,Y) tree_type=>"SB"
depth = floor(log2(N)) # row containing N (depth=0 at top)
Ndepth = 2^depth # start of row containing N
2*(N-Ndepth) + 1
?(r) = ----------------
Ndepth``````
The effect of N-Ndepth is to remove the high 1-bit, leaving an offset into the row. 2*(..)+1 appends an extra 1-bit at the end. The division by Ndepth scales down from integer N to a fraction.
`````` N = 1abcdef integer, in binary
?(r) = a.bcdef1 binary fraction``````
For example ?(2/3) is X=2,Y=3 which is N=5 in the SB tree. It is at depth=2, Ndepth=2^2=4, and so ?(2/3)=(2*(5-4)+1)/4=3/4. Or written in binary N=101 gives Ndepth=100 and N-Ndepth=01 so 2*(N-Ndepth)+1=011 and divide by Ndepth=100 for ?=0.11.
In practice this is not a very efficient way to handle the question function, since the bit runs in the N values may become quite large for relatively modest fractions. (Math::ContinuedFraction may be better, and also allows repeating terms from quadratic irrationals to be represented exactly.)
## Pythagorean Triples
Pythagorean triples A^2+B^2=C^2 can be generated by A=P^2-Q^2, B=2*P*Q. If P>Q>1 with P,Q no common factor and one odd the other even then this gives all primitive triples, being primitive in the sense of A,B,C no common factor ("PQ Coordinates" in Math::PlanePath::PythagoreanTree).
In the Calkin-Wilf tree the parity of X,Y pairs are as follows. Pairs X,Y with one odd the other even are N=0 or 2 mod 3.
`````` CW tree X/Y
--------
N=0 mod 3 even/odd
N=1 mod 3 odd/odd
N=2 mod 3 odd/even``````
This occurs because the numerators are the Stern diatomic sequence and the denominators likewise but offset by 1. The Stern diatomic sequence is a repeating pattern even,odd,odd (eg. per "Odd and Even" in Math::NumSeq::SternDiatomic).
The X>Y pairs in the CW tree are the right leg of each node, which is N odd. so
`````` CW tree N=3 or 5 mod 6 gives X>Y one odd the other even
index t=1,2,3,etc to enumerate such pairs
N = 3*t if t odd
3*t-1 if t even``````
2 of each 6 points are used. In a given row it's width/3 but rounded up or down according to where the 3,5mod6 falls on the N=2^depth start, which is either floor or ceil according to depth odd or even,
`````` NumPQ(depth) = floor(2^depth / 3) for depth=even
ceil (2^depth / 3) for depth=odd
= 0, 1, 1, 3, 5, 11, 21, 43, 85, 171, 341, ...``````
These are the Jacobsthal numbers, which in binary are 101010...101 and 1010...1011.
For the other tree types the various bit transformations which map N positions between the trees can be applied to the above N=3or5 mod 6. The simplest is the L tree where the N offset and row reversal gives N=0or4 mod 6.
The SB tree is a bit reversal of the CW, as described above, and for the Pythagorean N this gives
`````` SB tree N=0 or 2 mod 2 and N="11...." in binary
gives X>Y one odd the other even``````
N="11..." binary is the bit reversal of the CW N=odd "1...1" condition. This bit pattern is those N in the second half of each row, which is where the X/Y > 1 rationals occur. The N=0or2 mod 3 condition is unchanged by the bit reversal. N=0or2 mod 3 precisely when bitreverse(N)=0or2 mod 3.
For SB whether it's odd/even or even/odd at N=0or2 mod 3 alternates between rows. The two are both wanted, they just happen to switch in each row.
`````` SB tree X/Y depth=even depth=odd
---------- ---------
N=0 mod 3 odd/even even/odd
N=1 mod 3 odd/odd odd/odd <- exclude for Pythagorean
N=2 mod 3 even/odd odd/even``````
# FUNCTIONS
See "FUNCTIONS" in Math::PlanePath for behaviour common to all path classes.
`\$path = Math::PlanePath::RationalsTree->new ()`
`\$path = Math::PlanePath::RationalsTree->new (tree_type => \$str)`
Create and return a new path object. `tree_type` (a string) can be
`````` "SB" Stern-Brocot
"CW" Calkin-Wilf
"AYT" Andreev, Yu-Ting
"HCS"
"Bird"
"Drib"
"L"``````
`\$n = \$path->n_start()`
Return the first N in the path. This is 1 for SB, CW, AYT, HCS, Bird and Drib, but 0 for L.
`(\$n_lo, \$n_hi) = \$path->rect_to_n_range (\$x1,\$y1, \$x2,\$y2)`
Return a range of N values which occur in a rectangle with corners at `\$x1`,`\$y1` and `\$x2`,`\$y2`. The range is inclusive.
For reference, `\$n_hi` can be quite large because within each row there's only one new X/1 integer and 1/Y fraction. So if X=1 or Y=1 is included then roughly `\$n_hi = 2**max(x,y)`. If min(x,y) is bigger than 1 then it reduces a little to roughly 2**(max/min + min).
## Tree Methods
Each point has 2 children, so the path is a complete binary tree.
`@n_children = \$path->tree_n_children(\$n)`
Return the two children of `\$n`, or an empty list if `\$n < 1` (ie. before the start of the path).
This is simply `2*\$n, 2*\$n+1`. Written in binary the children are `\$n` with an extra bit appended, a 0-bit or a 1-bit.
`\$num = \$path->tree_n_num_children(\$n)`
Return 2, since every node has two children. If `\$n<1`, ie. before the start of the path, then return `undef`.
`\$n_parent = \$path->tree_n_parent(\$n)`
Return the parent node of `\$n`. Or return `undef` if `\$n <= 1` (the top of the tree).
This is simply Nparent = floor(N/2), ie. strip the least significant bit from `\$n`. (Undo what `tree_n_children()` appends.)
`\$depth = \$path->tree_n_to_depth(\$n)`
Return the depth of node `\$n`, or `undef` if there's no point `\$n`. The top of the tree at N=1 is depth=0, then its children depth=1, etc.
This is simply floor(log2(N)) since the tree has 2 nodes per point. For example N=4 through N=7 are all depth=2.
The L tree starts at N=0 and the calculation becomes floor(log2(N+1)) there.
`\$n = \$path->tree_depth_to_n(\$depth)`
`\$n = \$path->tree_depth_to_n_end(\$depth)`
Return the first or last N at tree level `\$depth` in the path, or `undef` if nothing at that depth or not a tree. The top of the tree is depth=0.
The structure of the tree means the first N is at `2**\$depth`, or for the L tree `2**\$depth - 1`. The last N is `2**(\$depth+1)-1`, or for the L tree `2**(\$depth+1)`.
## Tree Descriptive Methods
`\$num = \$path->tree_num_children_minimum()`
`\$num = \$path->tree_num_children_maximum()`
Return 2 since every node has 2 children so that's both the minimum and maximum.
`\$bool = \$path->tree_any_leaf()`
Return false, since there are no leaf nodes in the tree.
# OEIS
The trees are in Sloane's Online Encyclopedia of Integer Sequences in various forms,
`````` tree_type=SB
A007305 X, extra initial 0,1
A047679 Y
A057431 X,Y pairs (initial extra 0/1,1/0)
A007306 X+Y sum, Farey 0 to 1 part (extra 1,1)
A153036 int(X/Y), integer part
A088696 length of continued fraction SB left half (X/Y<1)
tree_type=CW
A002487 X and Y, Stern diatomic sequence (extra 0)
A070990 Y-X diff, Stern diatomic first diffs (less 0)
A070871 X*Y product
A007814 int(X/Y), integer part, count trailing 1-bits
which is count trailing 0-bits of N+1
A086893 N position of Fibonacci F[n+1]/F[n], N = binary 1010..101
A061547 N position of Fibonacci F[n]/F[n+1], N = binary 11010..10
A047270 3or5 mod 6, being N positions of X>Y not both odd
which can generate primitive Pythagorean triples
tree_type=AYT
A020650 X
A020651 Y (Kepler numerator)
A086592 X+Y sum (Kepler denominator)
A135523 int(X/Y), integer part,
count trailing 0-bits plus 1 extra if N=2^k
tree_type=HCS
A229742 X, extra initial 0/1
A071766 Y
A071585 X+Y sum
tree_type=Bird
A162909 X
A162910 Y
A081254 N of row Y=1, N = binary 1101010...10
A000975 N of column X=1, N = binary 101010...10
tree_type=Drib
A162911 X
A162912 Y
A086893 N of row Y=1, N = binary 110101..0101 (ending 1)
A061547 N of column X=1, N = binary 110101..010 (ending 0)
tree_type=L
A174981 X
A002487 Y, same as CW X,Y, Stern diatomic
A047233 0or4 mod 6, being N positions of X>Y not both odd
which can generate primitive Pythagorean triples
A000523 tree_n_to_depth(), being floor(log2(N))
A059893 permutation SB<->CW, AYT<->HCS, Bird<->Drib
reverse bits below highest
A153153 permutation CW->AYT, reverse and un-Gray
A153154 permutation AYT->CW, reverse and Gray code
A154437 permutation AYT->Drib, Lamplighter low to high
A154438 permutation Drib->AYT, un-Lamplighter low to high
A003188 permutation SB->HCS, Gray code shift+xor
A006068 permutation HCS->SB, Gray code inverse
A154435 permutation HCS->Bird, Lamplighter bit flips
A154436 permutation Bird->HCS, Lamplighter variant
A054429 permutation SB,CW,Bird,Drib N at transpose Y/X,
(mirror binary tree, runs 0b11..11 down to 0b10..00)
A004442 permutation AYT N at transpose Y/X, from N=2 onwards
(xor 1, ie. flip least significant bit)
A063946 permutation HCS N at transpose Y/X, extra initial 0
(xor 2, ie. flip second least significant bit)
A054424 permutation DiagonalRationals -> SB
A054426 permutation SB -> DiagonalRationals
A054425 DiagonalRationals -> SB with 0s at non-coprimes
A054427 permutation coprimes -> SB right hand X/Y>1``````
The sequences marked "extra ..." have one or two extra initial values over what the RationalsTree here gives, but are the same after that. And the Stern first differences "less ..." means it has one less term than what the code here gives.
http://user42.tuxfamily.org/math-planepath/index.html
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##### Notes
This supplemental quiz accompanies the Grade 1 Daily Math Measurement & Data Review printable. Use on the fifth day of Daily Math Practice - Operations Review for progress monitoring.
##### Print Instructions
NOTE: Only your test content will print.
To preview this test, click on the File menu and select Print Preview.
See our guide on How To Change Browser Print Settings to customize headers and footers before printing.
Print Test (Only the test content will print)
## Measurement & Data Review Quiz - Grade 1
1.
A tree is taller than a flagpole. The flagpole is taller than a traffic sign. So, the tree is than the traffic sign.
1. taller
2. shorter
2.
Look at the pencils:
A.
B.
C.
Which orders the pencils longest to shortest?
1. A, C, B
2. B, C, A
3. C, A, B
3.
About how many cubes long is the shovel?
1. 5
2. 6
3. 7
4. 8
4.
About how many long is the pencil?
1. 2
2. 5
3. 10
4. 20
5.
Which clock shows nine o'clock?
6.
What time does the clock show?
1. 12:00
2. 12:30
3. 1:00
4. 1:30
7.
Do the clocks show the same time?
1. no
2. yes
8.
Students picked from three activities. The table shows the students and the activities they picked.
ActivityStudents
blocksKara Sam Nora
puzzlesEthan Hiro Maya Jonah
drawingLuis Cane
How many students picked activities?
1. 3
2. 4
3. 7
4. 9
9.
Maleki asks 20 students which chore they do at home. The table shows the number of students who do each chore.
ChoreNumber
of Students
wash dishes7
fold laundry5
vacuum8
How many more students vacuum than fold laundry?
1. 1
2. 3
3. 5
4. 8
10.
The number and type of snacks that the students in Ms. Hughes' class ate on Friday are shown below.
Apples
Carrots
Pears
Muffins
How many total snacks were eaten by the class on Friday?
1. 3
2. 5
3. 7
4. 9
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682 views
### Differentiation: How to use the chain rule
If y is a function of u, which itself is a function of x, then
dy/dx=(dy/du) x (du/dx)
Differentiate the outer function and multiply by the derivative of the inner function.
To illustrate this rule, look at the example below:
y=(2x+3)10
in which y=u10 and u=2x+3
Now,
dy/du=10u9=10(2x+3)9
du/dx=2
The chain rule then gives
dy/dx=(dy/du) x (du/dx) = 10(2x+3)9(2) = 20(2x+3)9
2 years ago
Answered by Nicolas, an A Level Maths tutor with MyTutor
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### You may also like...
#### Other A Level Maths questions
The equation of a curve is x(y^2)=x^2 +1 . Using the differential, find the coordinates of the stationary point of the curve.
What are the roots of 3x^2 + 13x + 4 ?
How do you find the gradient of a parametric equation at a certain point?
How to get A and A* in Maths?
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# Complex Numbers Basics!
• Nov 26th 2012, 08:08 PM
Tutu
Complex Numbers Basics!
Hi I seem to not understand what the question is asking.
If z=2cis(delta),
write -z* in polar form.
I understand how to do this, my -z* was -2cos(delta) + i2sin(delta)
This is the second quadrant, and since
the coordinates of z are (2cos(delta), 2sin(delta))
the coordinates of -z* are (-2cos(delta), 2sin(delta)),
Can I can safely say that from z to-z*, it is a rotation of pi/2 about the origin?
Thus, my answer will be 2cis( delta + pi/2 ).
However, the answer is 2cis(pi - delta), which I understand because it still refers to the second quadrant, but I really wonder what is wrong with my answer. Looking at the coordinates, isn't it a pi/2 rotation? Unless my coordinates are wrong..
Sorry also, could you help me with this, Express sin(delta) - icos(delta) in polar form? The only thing I can gather is that the modulus equals to 1 and arctan(pi/2-delta).
Thank you so much!
• Nov 26th 2012, 08:57 PM
Prove It
Re: Complex Numbers Basics!
Quote:
Originally Posted by Tutu
Hi I seem to not understand what the question is asking.
If z=2cis(delta),
write -z* in polar form.
I understand how to do this, my -z* was -2cos(delta) + i2sin(delta)
This is the second quadrant, and since
the coordinates of z are (2cos(delta), 2sin(delta))
the coordinates of -z* are (-2cos(delta), 2sin(delta)),
Can I can safely say that from z to-z*, it is a rotation of pi/2 about the origin?
Thus, my answer will be 2cis( delta + pi/2 ).
However, the answer is 2cis(pi - delta), which I understand because it still refers to the second quadrant, but I really wonder what is wrong with my answer. Looking at the coordinates, isn't it a pi/2 rotation? Unless my coordinates are wrong..
Sorry also, could you help me with this, Express sin(delta) - icos(delta) in polar form? The only thing I can gather is that the modulus equals to 1 and arctan(pi/2-delta).
Thank you so much!
No it's not a rotation of \displaystyle \begin{align*} \frac{\pi}{2} \end{align*} about the origin.
Notice that if \displaystyle \begin{align*} z = r\,\textrm{cis}\,(\theta) \end{align*}, then \displaystyle \begin{align*} \overline{z} = r\,\textrm{cis}\,(-\theta) \end{align*}. This represents a reflection about the Real axis.
Then \displaystyle \begin{align*} -\overline{z} = -r\,\textrm{cis}\,(-\theta) \end{align*}, which represents a complete reflection about both the Imaginary and Real axes. So by symmetry, in the second quadrant you end up with the same angle as in the first quadrant but swept out in a clockwise direction from the negative real axis. If we were to measure this angle from the positive real axis in the anticlockwise direction (as we normally do), then it's the same as subtracting that reference angle from \displaystyle \begin{align*} \pi \end{align*}.
Say our angle had been \displaystyle \begin{align*} \frac{\pi}{6} \end{align*}. When we perform the transformations to \displaystyle \begin{align*} z = r\,\textrm{cis}\,\frac{\pi}{6} \end{align*} to get \displaystyle \begin{align*} z = -r\,\textrm{cis}\,\frac{\pi}{6} \end{align*}, you should find you get \displaystyle \begin{align*} r\,\textrm{cis}\,\frac{5\pi}{6} \end{align*}, which is NOT a rotation of \displaystyle \begin{align*} \frac{\pi}{2} \end{align*} from \displaystyle \begin{align*} \frac{\pi}{6} \end{align*}.
To answer your second question, you should know that \displaystyle \begin{align*} \sin{(\theta)} = \cos{\left( \frac{\pi}{2} - \theta \right)} \end{align*} and \displaystyle \begin{align*} \cos{(\theta)} = \sin{\left( \frac{\pi}{2} -\theta \right)} \end{align*}. So
\displaystyle \begin{align*} z &= \sin{(\delta)} - i\cos{(\delta)} \\ &= \sin{(-\delta)} + i\cos{(-\delta)} \\ &= \cos{\left[ \frac{\pi}{2} - \left( -\delta \right) \right]} + i\sin{\left[ \frac{\pi}{2} - \left( -\delta \right) \right]} \\ &= \cos{\left( \frac{\pi}{2} + \delta \right)} + i\sin{\left( \frac{\pi}{2} + \delta \right)} \\ &= \textrm{cis}\left( \frac{\pi}{2} + \delta \right) \end{align*}
• Nov 26th 2012, 09:24 PM
Tutu
Re: Complex Numbers Basics!
Hi thank you so much! I understand the first part!
For the second question, does that mean that we can have sin(-delta) even though in the question sin(delta) - icos(delta), the sin is positive? I was stuck there since I thought we have to manipulate it in such a way that it fits the question.
Thank you so very much!
• Nov 26th 2012, 09:37 PM
Prove It
Re: Complex Numbers Basics!
Quote:
Originally Posted by Tutu
Hi thank you so much! I understand the first part!
For the second question, does that mean that we can have sin(-delta) even though in the question sin(delta) - icos(delta), the sin is positive? I was stuck there since I thought we have to manipulate it in such a way that it fits the question.
Thank you so very much!
Actually I think I may have made a mistake in part 2.
\displaystyle \begin{align*} \sin{(\delta)} - i\cos{(\delta)} &= -\left[ -\sin{(\delta)} + i\cos{(\delta)} \right] \\ &= - \left[ \sin{(-\delta)} + i\cos{(-\delta)} \right] \\ &= - \left\{ \cos{\left[ \frac{\pi}{2} - (-\delta) \right]} + i\sin{\left[ \frac{\pi}{2} - (-\delta) \right]} \right\} \\ &= - \left[ \cos{\left( \frac{\pi}{2} + \delta \right)} + i\sin{\left( \frac{\pi}{2} + \delta \right)} \right] \end{align*}
• Nov 26th 2012, 09:51 PM
Tutu
Re: Complex Numbers Basics!
Thank you!
The answer is cis( delta - pi/2 ), so from your last working, all I have to do is to multiply the negative in
-( cos(pi/2 + delta) + isin(pi/2 + delta) )
= ( -cos(pi/2 + delta) - isin(pi/2 + delta) )
= ( cos(-pi/2 + delta) + isin(-pi/2 + delta) ) ?
then wouldn't it be ( cos(-pi/2 - delta) + isin(-pi/2 - delta) ) when I multiply the negative in?
Thank you so much!
• Nov 27th 2012, 08:46 PM
Prove It
Re: Complex Numbers Basics!
Quote:
Originally Posted by Tutu
Thank you!
The answer is cis( delta - pi/2 ), so from your last working, all I have to do is to multiply the negative in
-( cos(pi/2 + delta) + isin(pi/2 + delta) )
= ( -cos(pi/2 + delta) - isin(pi/2 + delta) )
= ( cos(-pi/2 + delta) + isin(-pi/2 + delta) ) ?
then wouldn't it be ( cos(-pi/2 - delta) + isin(-pi/2 - delta) ) when I multiply the negative in?
Thank you so much!
First of all, be VERY careful with your brackets. You were trying to write \displaystyle \begin{align*} \cos{\left[ -\left( \frac{\pi}{2} + \delta \right) \right]} + i\sin{\left[ -\left( \frac{\pi}{2} + \delta \right) \right] } \end{align*}, not as you had it.
It's also incorrect. Remember that while \displaystyle \begin{align*} \sin{(-\theta)} = -\sin{(\theta)} \end{align*}, we have \displaystyle \begin{align*} \cos{(-\theta)} = \cos{(\theta)} \end{align*}, NOT \displaystyle \begin{align*} -\cos{(\theta)} \end{align*}.
What we really have to do is remember that by negating an entire complex number, we are really rotating it by \displaystyle \begin{align*} \pi \end{align*}, or if you like, reflecting it in both the real and imaginary axes. So if we subtract \displaystyle \begin{align*} \pi \end{align*} from both angles, you should get your answer in the desired form :)
• Nov 28th 2012, 08:14 PM
Tutu
Re: Complex Numbers Basics!
Thanks! That way, I'd get
-[cos(delta - (pi/2) + isin(delta - (pi/2)]
= - cis (delta - (pi/2)
But since modulus cannot be negative, it automatically becomes,
= cis (delta - (pi/2), is that right?
How do you know to negate the complex number? The answer negated the complex number, but the question did not indicate that such had to be done..is it intuitive?
Thank you so so m
• Nov 29th 2012, 12:49 AM
Prove It
Re: Complex Numbers Basics!
Quote:
Originally Posted by Tutu
Thanks! That way, I'd get
-[cos(delta - (pi/2) + isin(delta - (pi/2)]
= - cis (delta - (pi/2)
But since modulus cannot be negative, it automatically becomes,
= cis (delta - (pi/2), is that right?
How do you know to negate the complex number? The answer negated the complex number, but the question did not indicate that such had to be done..is it intuitive?
Thank you so so m
No, it does not become \displaystyle \begin{align*} \textrm{cis}\left( \delta - \frac{\pi}{2} \right) \end{align*}. Like I said, to negate the complex number you need to rotate it by \displaystyle \begin{align*} \pi \end{align*} units about the origin. Therefore, your angle should become \displaystyle \begin{align*} \delta - \frac{\pi}{2} + \pi = \delta + \frac{\pi}{2} \end{align*} as required.
• Nov 29th 2012, 03:21 AM
Tutu
Re: Complex Numbers Basics!
But the answer is cis( delta - pi/2 ). To negate a complex number, you can choose to -pi or +pi right?
How do I know that I have to negate a complex number?
Thank you!
• Dec 5th 2012, 10:15 AM
Tutu
Re: Complex Numbers Basics!
Can I.ask, for the.negative in front of cis(pi/2+delta), can I just ignore it since cis.is never negative?
• Dec 5th 2012, 11:26 AM
Plato
Re: Complex Numbers Basics!
Quote:
Originally Posted by Tutu
Can I.ask, for the.negative in front of cis(pi/2+delta), can I just ignore it since cis.is never negative?
Suppose that $z=r\text{cis}(\theta)$ then $-z=r\text{cis}(\theta+\pi)$. Notice that $|z|=|-z|=r$ and $-z$ is the reflection of $z$ through the origin, which is a half-turn about the origin.
• Dec 5th 2012, 12:11 PM
Deveno
Re: Complex Numbers Basics!
here is the way i look at it:
-(z*) = (-1)(z*) = (-1)*(z)* = (-z)*...it doesn't matter if we negate and then conjugate, or conjugate and then negate. suppose we conjugate first.
this turns cis(δ) into cis(-δ). now when we take the negative, we rotate by a half-turn, or π radians. this turns cis(-δ) into cis(π-δ).
or, we could negate first, turning cis(δ) into cis(π+δ). conjugating this gives us cis(-π-δ) = (1)(cis(-π-δ)) = cis(2π)cis(-π-δ) = cis(2π+(-π-δ)) = cis(π-δ), as before.
(using the fact that cis(x)cis(y) = cis(x+y), and that cis(2π) = cis(0) = 1).
• Dec 5th 2012, 12:38 PM
Tutu
Re: Complex Numbers Basics!
But the answer is cis( delta - pi/2 ), is it wrong?thank you'!!
• Dec 5th 2012, 01:10 PM
Plato
Re: Complex Numbers Basics!
Quote:
Originally Posted by Tutu
Hi I seem to not understand what the question is asking.
If $z=2\text{cis}(\delta)$, write $-\overline{z}$ in polar form.
That was in fact the original posting then the answer is:
$-\overline{z}=2\text{cis}(-\delta +\pi)$
Example:
$\begin{gathered} z = 2\text{cis}\left( {\frac{{2\pi }}{5}} \right) \hfill \\ \overline z = 2\text{cis}\left( { - \frac{{2\pi }}{5}} \right) \hfill \\ - \overline z = 2\text{cis}\left( {\frac{{3\pi }}{5}} \right) \hfill \\ \end{gathered}$
The is the reason it works.
\begin{align*}\overline{\text{cis}(\delta)}& =\overline{\cos(\delta)+i\sin(\delta)}\\ &=\cos(\delta)-i\sin(\delta) \\&=\cos(-\delta)+i\sin(-\delta) \\ &=\text{cis}(-\delta)\end{align*}.
• Dec 5th 2012, 01:14 PM
Deveno
Re: Complex Numbers Basics!
well, i *have* been know to make mistakes. let's pick a value for delta, and see what happens: we'll use δ = π/3.
so z = cis(π/3) = cos(π/3) + i sin(π/3) = 1/2 + i(√3/2).
then z* = 1/2 - i(√3/2), and -z* = -1/2 + i(√3/2). by inspection, we see this is cis(2π/3) (a primitive cube root of unity) (note it is in the second quadrant)
so which formula is right: π-δ, or δ-π/2?
π-δ = π - π/3 = 2π/3. this agrees with what we found.
δ-π/2 = π/3 - π/2 = -π/6, which is in the fourth quadrant. this is not right. re-read your ORIGINAL post. i stand by what i posted.
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# Properties
Label 175.l Modulus $175$ Conductor $175$ Order $10$ Real no Primitive yes Minimal yes Parity odd
# Related objects
Show commands for: Pari/GP / SageMath
sage: from sage.modular.dirichlet import DirichletCharacter
sage: H = DirichletGroup(175, base_ring=CyclotomicField(10))
sage: M = H._module
sage: chi = DirichletCharacter(H, M([4,5]))
sage: chi.galois_orbit()
pari: [g,chi] = znchar(Mod(6,175))
pari: order = charorder(g,chi)
pari: [ charpow(g,chi, k % order) | k <-[1..order-1], gcd(k,order)==1 ]
## Basic properties
Modulus: $$175$$ Conductor: $$175$$ sage: chi.conductor() pari: znconreyconductor(g,chi) Order: $$10$$ sage: chi.multiplicative_order() pari: charorder(g,chi) Real: no Primitive: yes sage: chi.is_primitive() pari: #znconreyconductor(g,chi)==1 Minimal: yes Parity: odd sage: chi.is_odd() pari: zncharisodd(g,chi)
## Related number fields
Field of values: $$\Q(\zeta_{5})$$ Fixed field: 10.0.2564544677734375.1
## Characters in Galois orbit
Character $$-1$$ $$1$$ $$2$$ $$3$$ $$4$$ $$6$$ $$8$$ $$9$$ $$11$$ $$12$$ $$13$$ $$16$$
$$\chi_{175}(6,\cdot)$$ $$-1$$ $$1$$ $$e\left(\frac{2}{5}\right)$$ $$e\left(\frac{3}{10}\right)$$ $$e\left(\frac{4}{5}\right)$$ $$e\left(\frac{7}{10}\right)$$ $$e\left(\frac{1}{5}\right)$$ $$e\left(\frac{3}{5}\right)$$ $$e\left(\frac{2}{5}\right)$$ $$e\left(\frac{1}{10}\right)$$ $$e\left(\frac{1}{10}\right)$$ $$e\left(\frac{3}{5}\right)$$
$$\chi_{175}(41,\cdot)$$ $$-1$$ $$1$$ $$e\left(\frac{1}{5}\right)$$ $$e\left(\frac{9}{10}\right)$$ $$e\left(\frac{2}{5}\right)$$ $$e\left(\frac{1}{10}\right)$$ $$e\left(\frac{3}{5}\right)$$ $$e\left(\frac{4}{5}\right)$$ $$e\left(\frac{1}{5}\right)$$ $$e\left(\frac{3}{10}\right)$$ $$e\left(\frac{3}{10}\right)$$ $$e\left(\frac{4}{5}\right)$$
$$\chi_{175}(111,\cdot)$$ $$-1$$ $$1$$ $$e\left(\frac{4}{5}\right)$$ $$e\left(\frac{1}{10}\right)$$ $$e\left(\frac{3}{5}\right)$$ $$e\left(\frac{9}{10}\right)$$ $$e\left(\frac{2}{5}\right)$$ $$e\left(\frac{1}{5}\right)$$ $$e\left(\frac{4}{5}\right)$$ $$e\left(\frac{7}{10}\right)$$ $$e\left(\frac{7}{10}\right)$$ $$e\left(\frac{1}{5}\right)$$
$$\chi_{175}(146,\cdot)$$ $$-1$$ $$1$$ $$e\left(\frac{3}{5}\right)$$ $$e\left(\frac{7}{10}\right)$$ $$e\left(\frac{1}{5}\right)$$ $$e\left(\frac{3}{10}\right)$$ $$e\left(\frac{4}{5}\right)$$ $$e\left(\frac{2}{5}\right)$$ $$e\left(\frac{3}{5}\right)$$ $$e\left(\frac{9}{10}\right)$$ $$e\left(\frac{9}{10}\right)$$ $$e\left(\frac{2}{5}\right)$$
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http://www.thefullwiki.org/Index_ellipsoid
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• Wikis
Index ellipsoid: Wikis
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Note: Many of our articles have direct quotes from sources you can cite, within the Wikipedia article! This article doesn't yet, but we're working on it! See more info or our list of citable articles.
Encyclopedia
From Wikipedia, the free encyclopedia
In optics, an index ellipsoid is a diagram of an ellipsoid that depicts the orientation and relative magnitude of refractive indices in a crystal.
The equation for the ellipsoid is constructed using the electric displacement vector D and the dielectric constants. Defining the field energy W as
$8\pi W= \frac{D^2_1}{\varepsilon_1} + \frac{D^2_2}{\varepsilon_2} + \frac{D^2_3}{\varepsilon_3}.$
and the reduced displacement as
$R_i= \frac{D_i}{\sqrt{8\pi W}},$
then the index ellipsoid is defined by the equation
$\frac{R_1^2}{\varepsilon_1} + \frac{R_2^2}{\varepsilon_2} + \frac{R_3^2}{\varepsilon_3} = 1.$
The semiaxes of this ellipsoid are dielectric constants of the crystal.
This ellipsoid can be used to determine the polarization of an incoming wave with wave vector $\vec s$ by taking the intersection of the plane $\vec R \cdot \vec s =0$ with the index ellipsoid. The axes of the resulting ellipse are the resulting polarization directions.
Indicatrix
An important special case of the index ellipsoid occurs when the ellipsoid is an ellipsoid of revolution, i.e. constructed by rotating an ellipse around either the minor or major axis, when two axes are equal and a third is different. In this case, there is only one optical axis, the axis of rotation, and the material is said to be uniaxial. When all axes of the index ellipsoid are equal, the material is isotropic. In all other cases, in which the ellipsoid has three distinct axes, the material is called biaxial.
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poker-royal777.com
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VIP rakeback
# Rake in poker: what is it and how it is calculated
Every professional poker player knows about such a term as rake in poker. However, beginners often face difficulties when calculating the commission and have almost no idea what a raked hand is. In this article, we will try to understand these issues.
## What is rake in poker
Rake in poker is a certain amount (commission) that the room takes from the pot after the hand. The fee for providing a platform for the game is the main income of the company.
The process in which a resource takes at least a minimum commission is called a raked hand. For example, in Hold’em and Omaha, all hands are considered raked hands, except those finished on the preflop.
The amount of the commission in most cases depends on:
• the selected limit – the higher it is, the higher the percentage;
• the number of participants at the table – the more participants, the greater the CAP (CAP is the maximum percentage in one hand. Accordingly, the higher the indicator, the less profitable the game is in a particular resource);
• a type of poker.
The rake percentage also depends on the type of game: on average, the percentage in cash is 3-5% of the pot, and in tournaments – 5-10% of the buy-in.
## Rake collection methods
Rake in poker rooms is charged in different ways. We present the main methods of collecting commission – from the rarest to the most common.
• Dealt.
It is the rake collection method, in which each player who takes the cards pays an equal amount of commission. It does not matter whether the poker player has discarded the cards or not. This method belongs to the most outdated methods of calculating the rake.
For example: out of 5 players, three of them discarded their cards on the preflop, the rest went all-in. The rake will be divided equally between all participants at the table.
• Simple Contributed.
With this system, the rake in poker will be charged only from those poker players who have deposited funds to the pot. This method has become a replacement for Dealt, but it also cannot be called objective.
• Weighted Contributed
It is the most popular and the most objective method for paying rake today. Under the terms of this system, each player receives as much rake as the money he has deposited to the pot. Here we would like to suggest poker players that many poker rooms working on such a system organize rake races. How to participate in these exciting competitions, you can find out in our section about rake racing.
• Actual
The winner pays the rake at the end of the game.
• Source based rake
The main factor in this method is the factor of playing for real money or for winning funds. Even though all participants will pay the same, the poker player who plays for the winning funds will receive only half of the rake. By deposit, we mean the first deposit and the sum of all deposits and cashouts.
For example: one poker player will play for the deposited funds, the other – for the won ones. Both participants will deposit \$5, and the total rake will be \$10. Since half of the commission will be calculated using the WC system (Weighted Contributed), each player is guaranteed to receive \$2.5. Next, the depository factor comes into effect, thanks to which the player receives \$7.5 rake for money, and his opponent – only \$2.5.< / em>
• True value rake
Payouts are based on an analysis of the player’s skill level: the number of wins, hands played, etc. Amateurs are awarded a higher percentage than defeated.
For example: you pay \$5, which is distributed as follows: \$ 2 gets a regular, \$3 – an amateur player.
On our website you will find reviews of rooms with the best rake in the poker industry. In addition, when registering from Poker-Royal777, you will be guaranteed favorable Rakeback conditions in all networks and rooms, which you can read about here.
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# the managers of tomorrow will not be bigger man than his father was before him
1230 Words5 Pages
LOYOLA COLLEGE (AUTONOMOUS), CHENNAI – 600 034
B.Com. DEGREE EXAMINATION – COMMERCE
SIXTH SEMESTER – APRIL 2013
CO 6606 - ADVANCED CORPORATE ACCOUNTING
Date: 30/04/2013
Time: 1:00 - 4:00
Dept. No.
Max. : 100 Marks
SECTION A
(10*2=20marks)
1) What is Purchase Consideration as per AS 14?
2) What is Rebate on bills discounted? How is it dealt in a Bank’s Final statements?
3) List the methods for the calculation of Purchase Consideration
4) Raman Ltd agrees to purchase the business of Krishnan Ltd on the following terms:
• For each of the 10000 shares of Rs.10 each in Krishnan Ltd,2 shares in Raman
Ltd of Rs.10 each will be issued at an agreed value of Rs.12 per share.In addition,Rs.4 per share
On the date the profit and loss account of
S ltd had a credit balance of Rs.1000 and in reserve Rs. 3000.
Prepare a consolidated balance sheet from the following.
BALANCE SHEET AS ON 31-12-96
LIABILITIES
RS
RS
ASSET
RS
RS
Share
100000
50000
Sundry asset
60000
63000 capital(Rs. 10 each Reserve
10000
5000
Investment- 4000 65000 shares in s ltd
Profit and loss 10000
4000
account
Sundry
5000
4000
creditors
125000
63000
125000
63000
18) The XYZ Electricity Company decided to replace some parts of its plant by an improved plant. The plant to be replaced was built in 2003 for Rs 54, 00,000. It is estimated that it would now cost Rs 80,00,000 to build a new plant of the same size and capacity. The cost of the new plant as per the improved design was Rs1,70,00,000 and in addition, material belonging to the old plant valued at Rs 5,50,000 was used in the construction of the new plant. The balance of the old plant was sold for Rs 3,00,000. Compute the amount to be
Capitalised. Also pass the necessary journal entries and Replacement Account.
SECTION C
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https://git.dev.opencascade.org/gitweb/?p=occt.git;a=blame;f=dox/user_guides/voxels_wp/voxels_wp.md;h=e577e983acf48a8ab1805f02e0b64a61c90fdb70;hb=bf62b306abca0232dec6bc1dfac1c9fdb603b42b
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CommitLineData
bf62b306 1Voxel Package {#occt_voxels_wp}
2========================
3
4@tableofcontents
5
6@section occt_voxels_wp_1 Introduction
7
8 A voxel is a sub-volume box with constant scalar/vector value.
9 The object in voxel representation is split into many small sub-volumes (voxels)
10 and its properties are distributed through voxels.
11
12 Voxels are used for analysis and visualization of 3D-dimensional distribution of data.
13 Medicine (mainly, tomography), computational physics (hydrodynamics, aerodynamics, nuclear physics)
14 and many other industries use voxels for 3D data visualization and analysis of physical processes.
15
16 To produce a voxel representation the 3D space is split by equal intervals
17 along the main orthogonal coordinate axes to obtain nx x ny x nz voxels (small cubes):
18
19@image html voxels_wp_image003.jpg "A cube of 3 X 3 X 3 = 9 voxels."
20@image latex voxels_wp_image003.jpg "A cube of 3 X 3 X 3 = 9 voxels."
21
22 The data are attached to each voxel and remain the same within the voxel.
23 It means that we obtain the 3D space with discrete data distribution.
24
25 The number of voxels used in a calculation can vary.
26 An average model contains several tens of millions of voxels.
27 Such a great amount of data requires special algorithms of computation,
28 data containers keeping data in memory and visualization tools.
29
30 Open CASCADE Technology provides several basic data containers for voxels
31 with fast access to the data and optimal allocation of data in memory.
32
33 Also, a special visualization toolkit allows visualizing voxels
34 as colored or black/white points and cubes, displaying only the voxels
35 visible from the user's point of view.
36
37@image html voxels_wp_image004.jpg "A shape and its voxel representation"
38@image html voxels_wp_image005.jpg "A shape and its voxel representation"
39@image latex voxels_wp_image004.jpg "A shape and its voxel representation"
40@image latex voxels_wp_image005.jpg "A shape and its voxel representation"
41
42In these images a boundary representation is displayed to the left. In the center and to the right there are 3D discrete representations (or 3D discrete topology). Any solid shape can be translated into a voxel representation.
43
44@section occt_voxels_wp_2 Data structure
45
46 The data structure to store the voxels data is a special class which gives
47 fast access to the data of each voxel and allocates the data in an optimal way in the memory of a computer.
48
49 Fast access to the data is provided by means of bit-wise operators on the indices of internal arrays.
50
51 The optimal data allocation is reached through division
52 of the whole data set into data subsets and keeping only non-zero pieces of data in memory.
53
54 A voxel can contain different data types,
55 but presently Open CASCADE Technology implements only several basic ones:
56 * 1 bit or Boolean data type – a voxel contains a flag: 0 or 1 (false or true).
57 * 4 bits or Color data type – a voxel contains a value occupying 4 bits.
58 It is an integer in the range of 0 .. 15. The data can be divided into 16 subsets and displayed by Color-voxels.
59 * 4 bytes or Float data type – a voxel contains a floating-point data type.
60
61 In addition, the data structures provide methods for calculation of a center point
62 by voxel indices and a reverse task – fast search of a voxel by a point inside the cube of voxels.
63
64@section occt_voxels_wp_3 Algorithms
65
66 There are two service classes implemented for data structures of voxels:
67
68 * Boolean operations – provides simple boolean operations on cubes of voxels (fuse and cut).
69 * Voxelization – the conversion of a geometrical model into its voxel representation.
70
71### Boolean operations
72
73Fusion and cutting of two cubes of voxels are performed the class *Voxel_BooleanOperations*. The cubes should have the same size and be split into voxels in the same way.
74* <i>::Fuse()</i> summarizes the values of the corresponding voxels and limits the result by the upper limit (if succeeded).
75* <i>::Cut()</i> subtracts the values of the corresponding voxels and limits the result by zero.
76
77### Voxelization
78
79A class *Voxel_Convert* converts a *TopoDS_Shape* into one of the voxel data structures filling the solid shape by non-zero values.
80
81The algorithm of voxelization generates only 1-bit or 4-bit voxels. Other data types may be obtained by conversion of voxels from one type to another.
82
83Voxelization of a shape is performed by means of computation of intersection points between lines filling the volume and triangulation of the shape. The lines are parallel to main orthogonal axes and can intersect the shape from different sides: along +X, +Y and/or +Z axes.
84
85The algorithm can run in multi-threaded mode (the number of threads is unlimited). The user can see an integer value indicating the progress of computation.
86
87@section occt_voxels_wp_4 Visualization
88
89 Visualization of voxels is not a simple task due to a great amount of data used for 3D analysis.
90
91 Open CASCADE Technology allows visualization of a cube of voxels in two modes:
92 * Points – the centers of voxels as 3D points.
93 * Boxes – the voxels as 3D cubes of adjustable size.
94
95 A degenerated mode displays only the points (boxes) visible
96 from the point of view of the user for transformation operations (zoom, pan and rotate).
97
98 To focus on a particular part of the model non-relevant voxels can be erased.
99 The displayed region is defined by six co-ordinates along X, Y and Z axes .
100
101 It is possible to display the voxels from a particular range of values (iso-volume):
102
103@image html voxels_wp_image006.jpg "Iso-volume of a shape"
104@image latex voxels_wp_image006.jpg "Iso-volume of a shape"
105
106The voxels are displayed by means of "direct drawing in Open GL" technology or "user draw" technology. Therefore, some visualization files are compiled within Open CASCADE Technology, but the files of "direct drawing" are compiled by the end-user application.
107
108It is necessary to include the files *Voxel_VisData.h*, *VoxelClient_VisDrawer.h* and *VoxelClient_VisDrawer.cxx* into the visualization library of the application (containing all files of *OpenGl* package) and call the method *Voxel_VisDrawer::Init()* from the application before the visualization of voxels.
109
110@section occt_voxels_wp_5 Demo-application
111
112 A demonstration application has been created to show OCCT voxel models.
113 This is a test demo application because it includes a set of non-regression tests
114 and other commands for testing the functionality (accessible only through TEST pre-processor definition).
115
116 The *File* menu allows creation of canonical shapes (box, cylinder, sphere, torus) or loading of shapes in BREP format:
117
120
121The menu *Converter* voxelizes the shape. Two types of voxels can be obtained: 1-bit or 4-bit voxels.
122 * 1-bit voxels are displayed in white color on black background.
123 * 4-bit voxels use 16 colors filling the model in a special way for demonstrative purposes:
124
125@image html voxels_wp_image008.jpg "Demo-application. Voxelization"
126@image latex voxels_wp_image008.jpg "Demo-application. Voxelization"
127
128 The converter uses two threads (two processors, if available) to perform voxelization.
129
130 The menu *Visualization* offers two modes of visualization: Points and Boxes,
131 allows defining the size of points and boxes (quadrangles),
132 the minimum and the maximum displayed color, and the boundaries of the bounding box for displayed voxels:
133
134@image html voxels_wp_image009.jpg "Demo-application. Visualization"
135@image latex voxels_wp_image009.jpg "Demo-application. Visualization"
136
137 The last menu, *Demo* contains a demo-command for running waves of 4-bit voxels:
138
139@image html voxels_wp_image010.jpg "Demo-application. Running waves"
140@image latex voxels_wp_image010.jpg "Demo-application. Running waves"
141
142@section occt_voxels_wp_6 Future development
143
144In the future OPEN CASCADE plans to develop the platform of voxels in the following directions:
145 * Data structure:
146 * Extension of the list of basic data types.
147 * Development of a deeper hierarchy of voxels (for example, octree – division of a voxel into 8 sub-voxels).
148 * Development of a doxel (4D voxels where the fourth co-ordinate is the time, for example).
149
150 * Algorithms:
151 * Conversion of a voxel model into a geometrical model (a reversed operation to voxelization).
152
153 * Visualization:
154 * Optimization of visualization (mainly, the speed of visualization).
155 * New shapes of voxel presentation in the 3D Viewer and new approaches to visualization.
156 * Selection of voxels.
157
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# Math 3 assignment 9- trigonometry
## 20 questions
Show Answers
See Preview
• Multiple Choice
Please save your changes before editing any questions.
30 seconds
1 pt
What is the ratio for Sine?
Hyp/Opp
Opp/Hyp
Opp/Adj
Adj/Opp
• Multiple Choice
Please save your changes before editing any questions.
1 minute
1 pt
What is the correct trigonometric ratio?
9/41
40/41
9/40
40/9
• Multiple Choice
Please save your changes before editing any questions.
1 minute
1 pt
What is the correct equation to solve for x ?
tan( 46 °) = 17/x
sin ( 46º ) = x/17
tan (46º ) = x/17
cos (46º) = x/17
• Explore all questions with a free account
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## 255263900000000
255,263,900,000,000 (two hundred fifty-five trillion two hundred sixty-three billion nine hundred million) is an even fifteen-digits composite number following 255263899999999 and preceding 255263900000001. In scientific notation, it is written as 2.552639 × 1014. The sum of its digits is 32. It has a total of 18 prime factors and 324 positive divisors. There are 100,176,960,000,000 positive integers (up to 255263900000000) that are relatively prime to 255263900000000.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 15
• Sum of Digits 32
• Digital Root 5
## Name
Short name 255 trillion 263 billion 900 million two hundred fifty-five trillion two hundred sixty-three billion nine hundred million
## Notation
Scientific notation 2.552639 × 1014 255.2639 × 1012
## Prime Factorization of 255263900000000
Prime Factorization 28 × 58 × 53 × 48163
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 18 Total number of prime factors rad(n) 25526390 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 255,263,900,000,000 is 28 × 58 × 53 × 48163. Since it has a total of 18 prime factors, 255,263,900,000,000 is a composite number.
## Divisors of 255263900000000
324 divisors
Even divisors 288 36 18 18
Total Divisors Sum of Divisors Aliquot Sum τ(n) 324 Total number of the positive divisors of n σ(n) 6.48944e+14 Sum of all the positive divisors of n s(n) 3.9368e+14 Sum of the proper positive divisors of n A(n) 2.00291e+12 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1.5977e+07 Returns the nth root of the product of n divisors H(n) 127.446 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 255,263,900,000,000 can be divided by 324 positive divisors (out of which 288 are even, and 36 are odd). The sum of these divisors (counting 255,263,900,000,000) is 648,943,718,521,896, the average is 20,029,127,114,87.,333.
## Other Arithmetic Functions (n = 255263900000000)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 100176960000000 Total number of positive integers not greater than n that are coprime to n λ(n) 3130530000000 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 7940417740018 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 100,176,960,000,000 positive integers (less than 255,263,900,000,000) that are coprime with 255,263,900,000,000. And there are approximately 7,940,417,740,018 prime numbers less than or equal to 255,263,900,000,000.
## Divisibility of 255263900000000
m n mod m 2 3 4 5 6 7 8 9 0 2 0 0 2 3 0 5
The number 255,263,900,000,000 is divisible by 2, 4, 5 and 8.
• Abundant
• Polite
• Practical
• Frugal
## Base conversion (255263900000000)
Base System Value
2 Binary 111010000010100101000010000101010101111100000000
3 Ternary 1020110210222202002022110121112
4 Quaternary 322002211002011111330000
5 Quinary 231424220431400000000
6 Senary 2302522341120142452
8 Octal 7202450205257400
10 Decimal 255263900000000
12 Duodecimal 24767a2b767a28
20 Vigesimal 14ib4i8f0000
36 Base36 2ihem7cagw
## Basic calculations (n = 255263900000000)
### Multiplication
n×i
n×2 510527800000000 765791700000000 1021055600000000 1276319500000000
### Division
ni
n⁄2 1.27632e+14 8.5088e+13 6.3816e+13 5.10528e+13
### Exponentiation
ni
n2 65159658643210000000000000000 16632908587934493119000000000000000000000000 4245781114499651658079104100000000000000000000000000000000 1083794645833527630882738621071990000000000000000000000000000000000000000
### Nth Root
i√n
2√n 1.5977e+07 63435.1 3997.12 761.023
## 255263900000000 as geometric shapes
### Circle
Diameter 5.10528e+14 1.60387e+15 2.04705e+29
### Sphere
Volume 6.96718e+43 8.1882e+29 1.60387e+15
### Square
Length = n
Perimeter 1.02106e+15 6.51597e+28 3.60998e+14
### Cube
Length = n
Surface area 3.90958e+29 1.66329e+43 4.4213e+14
### Equilateral Triangle
Length = n
Perimeter 7.65792e+14 2.8215e+28 2.21065e+14
### Triangular Pyramid
Length = n
Surface area 1.1286e+29 1.96021e+42 2.08422e+14
## Cryptographic Hash Functions
md5 c94bf4353a0aff7f2eabe2452f277df3 00c7bb2a689895b32af1ed7124f5224a61b48acc 4dce2a02cd02e52f61e03426f617fa5f7e402edd40e0b304def26c463085f801 a9d251e7e7e533bd9812acc0e871c893fdabe7b3f03a502a640ea94787883787b2a2dc96426b7bc3c33f6d6ffb8f8ebae005e82d893919fb759e51100c98c491 2ab640a2132d75c287509de0ac4ece1cd826afd6
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# How Much Resin Is In My Water Softener
## How Much Resin Is In My Water Softener
### How many cubic feet of resin are there in a water softener?
What is normally sold as a 32,000-grain system contains 1 cubic foot of water softener. If you look at the resin manufacturer’s specifications, you’ll find that under ideal laboratory conditions, one cubic foot of resin can remove approximately 32,000 grains of hardness.
### How much resin should a water softener contain?
Depending on the size, the amount of resin should be between 0.64 and 2.00 cubic feet. Getting the right amount will eliminate a potential source of water softener problems.
### Also, what does the number of grains in the softener mean?
Grain per gallon is the industry standard method for discussing water hardness. If you lower mg / L down to 17.1 it will be converted to GPG. A durum wheat is 1/7000 of a pound. When buying a plasticizer, it is very important to know the hardness of the water.
### How do I also know the grain capacity of my water softener?
Use the following formula to calculate the correct size:
1. Multiply the number of members of your family by 70 (liters of water consumed per day, national average).
2. Multiply the answer by the water hardness of the grain per gallon (to convert mg / L or ppm to grain, divide by 17.1).
3. This is the number of grains per day.
### How difficult is it to remove a plasticizer?
Calcium and magnesium are often referred to as hardness minerals. Sometimes plasticizers are even used to remove iron. Plasticizers can remove up to five milligrams per liter (5 mg / l) of dissolved iron. The water softeners can be operated automatically, semi-automatically or manually.
### How often should I replace the resin in the plasticizer?
You can typically get 10 years or more on city water and 510 years on well water with iron, manganese, or high organic pollution. The things that increase the degradation of the resin are the coating or oxidation of the resin.
### Does my water softener need to be repaired?
Other water softeners require regular maintenance to maintain a standard quality level. Assuming you keep the water softener full of salt, you don’t have to fix it over its life.
15 years
15 years
### What happens if you connect an inverted softener?
Yes, a water softener with a back channel blows resin throughout the house. The muzzle is designed upstream or downstream so that it is difficult to return. Go back but the device is in bypass. If the water pressure is not too high, then everything is fine. but you need to fix it asap.
### Can I put vinegar in the water softener?
Use a cup of bleach, e.g. Vinegar in 45 liters of clean water. Gently mix and let the solution sit for 15 minutes. Pour it into the water softener and scrub again with a brush. For greater efficiency, do not fill the salt to the brim with descaler.
### How to rinse resin from a water softener?
How to wash resin beads from plumbing. Bypass the water softener. You will see two pipes leading to the descaler, an inlet and an outlet for the water. Empty and rinse the kettle. Turn off the power and cold water to the water heater. Rinse the taps with hot and cold water. Use equipment that uses water to rinse the beads.
### What size do I need the fabric softener for a family of 4?
Determine the right fabric softener based on your hardness and the number of people in the house. Number of people in the house Grain hardness Per gallon 1 to 2 people * 3 to 4 people * 510 GPG 32,000 grains 32,000 grains 1120 GPG 32,000 grains 32,000 grains 2130 GPG 32,000 grains 40,000 grains
### What setting should the water softener be water?
Softener hardness setting The hardness range is 1 to 99 grains per gallon.
### What if the water softener is too big?
If the water softener is too big for your home, it won’t do enough work to trigger regeneration in time. If a water softener doesn’t regenerate, it won’t be as effective and bacteria can grow in the tank. You can also calculate the amount of water you use in relation to the capacity of the plasticizer.
### What is the flow rate through the water softener?
This rate is determined by the amount of resin with which a plasticizer is formed. At hardness levels no higher than 855 mg / L 50 grains per gallon, a plasticizer can produce approximately 5 gpm of fresh water for every cubic foot of resin used to make the plasticizer.
### How much salt should I put in the softener?
It is recommended to purchase water softener salt with a salt content of approximately 99.5% or higher. All plasticizers can use potassium chloride in place of salt.
### What is the good hardness of the water?
The general guidelines for water classification are: 0 to 60 mg / L (milligrams per liter) if calcium carbonate is classified as soft 61 to 120 mg / L medium hard 121 to 180 mg / L if hard and over 180 mg / LL as very hard.
### How much resin is in my water softener?
Depending on the size, the amount of resin should be between 0.64 and 2.00 cubic feet. Getting the right amount will eliminate a potential source of water softener problems.
### How much does it cost to replace the resin in the softener?
Instructions for the procedure can be found in the user manual and contact a professional if the task seems too difficult or if the device is still subject to a service contract. Resins typically cost between 80 and 120, and the service can cost between 200 and 300.
### Also, how do I know the grain capacity of my water softener?
Use the following formula to calculate the correct size:
1. Multiply the number of members of your family by 70 (liters of water consumed per day, national average).
2. Multiply the answer by the water hardness of the grain per gallon (to convert mg / L or ppm to grain, divide by 17).
3. This is the number of grains per day.
### How often should I replace the resin in the plasticizer?
You can typically get 10 years or more on city water and 510 years on well water with iron, manganese, or high organic pollution. The elements that make the degradation of the resin more likely are the coating or oxidation of the resin.
### What is resin made of to soften water?
Resin is a synthetic polystyrine sulfonate-based plastic that is molded into pearls that look like brown sugar.
### Does my softener need to be overhauled?
Other water softeners require regular maintenance to maintain a standard quality level. Assuming you store the water softener in a salt-filled block, you don’t have to fix it over its lifetime.
### Does the resin in the rinse water decompose?
Yes, water softening resins are actually just small plastic balls. Since these pearls are made of tough, tough plastic, they should never be bad in theory.
### How to rinse resin from a water softener?
How to rinse the plumber’s resin beads
### Can I replace the resin in the water softener?
Changing the resin can take up to 4 hours of your time and save between 100 and 150 in professional fees. First, you need to determine the amount of resin required for the water softening system. You should also be careful to use a good quality resin.
15 years
### What happens if you put the plasticizers upside down?
Yes, a water softener in the back channel spreads the resin throughout the house. The muzzle is designed upstream or downstream so that it is difficult to return. Go back but the device is in bypass. If the water pressure is not too high, then everything is fine. but you need to fix this asap.
### How long do water softeners last?
Depending on the type and quality, water softeners can last from 10 to 20 years. On average, a single electric softener will last up to 12 years while a Kinetico system will last up to 20 years.
### What if there is no more salt in the descaler?
If the softener does not contain salt, it will not be able to remove hard minerals and iron from the softener resin. However, a lack of salt in the water softener can cause long-term damage to the taps and even overflow the salt reservoir.
### What chemicals are used in water softeners?
Chemicals used for water softening include calcium hydroxide (slaked lime) and sodium …
### Can chlorine damage a water softener?
Effects of Chlorine
### What size of fabric softener do I need for a family of 4?
Determine the right water softener based on hardness and the number of people in your home.
### In which position should the softener be placed?
Setting the hardness of the water softener
### What is the good hardness of the water?
The general guidelines for water classification are: 0 to 60 mg / L (milligrams per liter), as calcium carbonate is classified as soft 61 to 120 mg / L as medium hard, 121 to 180 mg / L as hard and over 180 mg / L as very hard.
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# Algebra 2
posted by .
I am having trouble with understanding log.
I don't know what to do when it says log x=3 or when it has log with a little 5 x=2
## Similar Questions
1. ### math
write as a single logarithm: -2logbase3(1/x)+(1/3)logbase3(square root of x) please show the steps to solving this. thanx. remember that 1 log (AxB) = log A + Log B (same base) 2 log (A/B) = log A - log B 3 log A^n = n log A use these …
2. ### Algebra 2
how do not understand how to do this Log X + Lob (X-3) = 1 I know I do this Log X(X-3)=1 then I do this Log X^(2) - 3X = 1 then I do this 2 Log (X-3X) = 1 then 2 Log (-2X) = 1 then (2 Log (-2X) = 1)(1/2) Log (-2X) = 1/2 then (Log (-2X) …
3. ### Mathematics
Prove that log a, log ar, log ar^2 is an a.p. Is the following below correct?
Which of the following expressions is equal to log (x sqrt-y)/z^5 A. log x + log (1/2) + log y– log 5 – log z B. log [x + (1/2)y – 5z] C. log x + (1/2)log y – 5 log z d. [(1/2) log x log y]/(5 log z)
1) use the properties of logarithms to simplify the logarithmic expression. log base 10 (9/300) log - log 300 log 9 = 2 log 3 log 300 = log 3 + log 100 = log 3+2 I just do not know how to put these together now!
6. ### math
Logarithm!!! Select all of the following that are true statements: (a) log(2x) = log(2) + log(x) (b) log(3x) = 3 log(x) (c) log(12y) = 2 log(2) + log(3y) (d) log(5y) = log(20y) – log(4) (e) log(x) = log(5x) – log(5) (f) ln(25) …
7. ### Trigonometry
This is a logs question If u=x/y^2, which expression is equivalent to log u?
8. ### college algebra
Write expression as one logarithm and simplify if appropriate. log 3√x + log x^4 - log x^3 4 log (x+3) - 5 log (x^2+4) + 1/3 log y I have these who problems but I don't know where to start. HELP Please.
9. ### Algebra 2
-Explain the difference between log base b (mn) and ( log base b of m)(log base b of n). Are they equivalent?
10. ### Precaculculus
Suppose you are told that log(2)=0.3562 and log(3)=0.5646. All of them with the base of 'a'. Find: i) log(6) ii) log(9) Solutions i) log(6)= log (3)(2) = log 3 + log 2 = 0.5646 + 0.3562 = 0.9208 ii)log(9)= log 3^2 = 2 log 3 = 2 (0.5646) …
More Similar Questions
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+0
# How many hours would I swim in a year if.
0
380
1
I swim:
-10 months out of a year = 43 weeks out of a year
-2 hours on Monday, Thursday, and Friday
-4 hours on Wednesday
-1.5 hours on Tuesday and Saturday
Jul 8, 2017
edited by Guest Jul 8, 2017
#1
+7347
+1
Monday: 2 hrs
Tuesday: 1.5 hrs
Wednesday: 4 hrs
Thursday: 2 hrs
Friday: 2 hrs
Saturday: 1.5 hrs
Sunday: 0 hrs
Total: 13 hours in a week
If you swim 13 hours in a week for 43 weeks, that is
13 hours, done 43 times
13 * 43 = 559 hours
Jul 9, 2017
#1
+7347
+1
Monday: 2 hrs
Tuesday: 1.5 hrs
Wednesday: 4 hrs
Thursday: 2 hrs
Friday: 2 hrs
Saturday: 1.5 hrs
Sunday: 0 hrs
Total: 13 hours in a week
If you swim 13 hours in a week for 43 weeks, that is
13 hours, done 43 times
13 * 43 = 559 hours
hectictar Jul 9, 2017
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0
# What equals 572 doing times?
Wiki User
2016-09-01 12:00:23
1 x 572, 2 x 286, 4 x 143, 11 x 52, 13 x 44, 22 x 26 = 572
Wiki User
2016-09-01 12:00:23
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# The Kirchoff's law for a.c. is applicable upon consideration of -
8 views
The Kirchoff's law for a.c. is applicable upon consideration of -
a) Phasor sum b) Algebraic sum c) Progression sum d) Linear sum
The Kirchoff's law for a.c. is applicable upon consideration of - Phasor sum
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The values of L and C for a low pass filter with cut off frequency of 2.5 KHz and operating with a terminated load resistance of 450 Ohm are given by (a) 57.3 mH; 0.283 μF (b) 28.66 μH; 0.14 μF (c) 114.64 mH; 0.566 mF (d) 50.23 mH; 0.632 mF
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# 1 understand the expected pattern of
Some of the categories within which these behaviors are seen include: Health and Social Care Course: Consequently, priests were expected to be able to figure out how to double the area of a relatively complex geometric figure.
As a practical example, consider counting from one to one hundred by ones. You can also create basic counting games by counting aloud how long it takes to set the table, put away toys, or put on pajamas. Generally there is a theme to the book, such as counting bugs or counting crayons. Outcome 4 - Understand the importance of early intervention to support the speech, language and communication needs of children and young people.
Math and Language The language demands of mathematics are extensive. Pi in the Sky Given that the much simpler problem of doubling the area of a square presupposes knowledge of the Pythagorean theorem written down by Pythagoras around B.
Correspondingly, the center of the mandala represents order and the realm of the gods, while the outside represents the chaos of the material world. Can you see how to double the area of the altar?
Just as a child can predict that a red bead will come next after seeing a string with a red bead, blue bead, green bead, red bead, blue bead, green bead pattern, a child will be able to make accurate predictions about other things or events that occur with regularity. Finally, memory for rules is also critical for success in math.
I talked about the different transitions and the affects these can have on children and young people. Sometimes a student understands the underlying concept clearly but does not recall a specific term correctly. Learn more about the specific math and number awareness skills your child will be expected to have at the beginning of preschool and at the beginning of kindergarten.
Importance of Hands-On Learning Math learning is most exciting for children when hands-on manipulatives fancy teacher-speak for small objects that can be easily handled or manipulated are incorporated.
Another example of math and religion interacting is the mandala -- a design used for meditation. Levine points out that "Math is full of sequences. Children are more likely to experience success in math when any neurodevelopmental differences that affect their performance in mathematics are dealt with promptly -- before children lose confidence or develop a fear of math.
In early years settings, when children are first settling in, children will of course have their parents but the effects of separation anxiety are such that they will also need a relationship within the setting.
In addition to understanding the meaning of specific words and sentences, children are expected to understand textbook explanations and teacher instructions. Additional information about milestones and K math curriculum is available on The National Council of Teachers of Mathematics Web site.
For specific information about the range of skills and concepts in school mathematics, please visit the Principles and Standards for School Mathematics on the National Council of Teachers of Mathematics Web site.CYP Core - Understand the expected pattern of development for children and young people from birth - 19 years.
Explain the sequence and the rate of each aspect of development from birth to 19 years The sequence of child development means the expected development of a child from birth to 19 years.
Child development refers to. 1 Understand the expected pattern of development for children and young people from birth to 19 yrs Explain the sequence and rate of each aspect of development from birth to 19 yrs Explain the difference between sequence of development and rate of development and why the.
Understand the potential effects of transitions on children and young people.
Learning Outcomes. All will know the main stages of child and young person development. Most will understand and be able to describe young person’s physical, communication, social and emotional development The expected pattern of development.
Expected Course Outcomes Acctg 1: BUS EXPECTED COURSE OUTCOMES (1) Understand, recognize and distinguish the various types of financial investments as mirrored. the ability to organize according to a pattern that is appropriate to the discipline. Tools: quarterly report, e.
Silly me, and Xcode, it's the Code Auto-Completion feature that confuses me. The little icons in code shown in the first screenshot aren't nice visual decorations that match with my variable names (although I'd love these smart "emojis" decorating my code).
Child development: months. At birth a baby does not know or understand anything. They quickly learn to recognise the smell and voice of the person who feeds them and holds them most often but they do not know this is their 'mother'.
1 understand the expected pattern of
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# Thread: Sketch a graph for all (x,y) that are related to the point (1,0)
1. ## Sketch a graph for all (x,y) that are related to the point (1,0)
Consider the real plane, ℝ². Define the relation S on ℝ² by:
(x,y)S(u,v) when x² + y² = u² + v²
(a) Sketch a graph for all (x,y) that are related to the point (1,0)
My attempt:
I think the answer is x=1 since the graph of x=1 intersects the point (1,0) and all circles with a radius of 1 or higher.
Or do I need to sketch a graph that intersects (1,0) and ALL circles of any given radius.
Or it something more simple that i'm not thinking about
Thanks!
2. ## Re: Sketch a graph for all (x,y) that are related to the point (1,0)
Originally Posted by yoman360
Consider the real plane, ℝ². Define the relation S on ℝ² by: (x,y)S(u,v) when x² + y² = u² + v²
(a) Sketch a graph for all (x,y) that are related to the point (1,0)
Question: Is it true that $(1,0)\mathcal{S}(0,1)$ true?
What about $(1,0)\mathcal{S}\left( {\tfrac{{\sqrt 2 }}{2}, - \tfrac{{\sqrt 2 }}{2}} \right)~?$.
And if $a^2+b^2=1$, is it true that $(1,0)\mathcal{S}(a,b)$ true?
3. ## Re: Sketch a graph for all (x,y) that are related to the point (1,0)
Originally Posted by Plato
Question: Is it true that $(1,0)\mathcal{S}(0,1)$ true?
yes, because 1² + 0² = 0² + 1² ==> 1 = 1
Originally Posted by Plato
What about $(1,0)\mathcal{S}\left( {\tfrac{{\sqrt 2 }}{2}, - \tfrac{{\sqrt 2 }}{2}} \right)~?$.
No because 1 = 1
Originally Posted by Plato
And if $a^2+b^2=1$, is it true that $(1,0)\mathcal{S}(a,b)$ true?
oh i see so the answer is the sketch of the graph x²+y² = 1
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# Introduction
We are now in RC1 phase of Julia 1.11. One small but important addition it is making IdSet a public type. Today I want to discuss when this type is useful.
The code was tested under Julia 1.11 RC1.
# Equality in Julia
There are three basic ways to test equality in Julia:
1. the == operator;
2. the isequal function;
3. the === operator.
I have ordered these comparison operators by their level of strictness.
The == operator is most loose. It can return true, false or missing. The missing value is returned if any of the compared values are missing (or recursively contain missing value). For floating point numbers it assumes that 0.0 is equal to -0.0 and that NaN is not equal to NaN. Let us see the last case in action as it can be surprising if you have never seen this before:
julia> NaN == NaN
false
Next is isequal that is more strict. It guarantees to return true or false. It treats all floating-point NaN values as equal to each other, treats -0.0 as unequal to 0.0, and missing as equal to missing. It compares objects by their value, not by their identity. So, for example, two different vectors having the same contents are considered equal:
julia> v1 = [1, 2, 3]
3-element Vector{Int64}:
1
2
3
julia> v2 = [1, 2, 3]
3-element Vector{Int64}:
1
2
3
julia> isequal(v1, v2)
true
Finally we have ===, which is most strict. It returns true or false. However, true is returned if and only if the compared values are indistinguishable. They must have the same type. If their types are identical, mutable objects are compared by address in memory and immutable objects (such as numbers) are compared by contents at the bit level. Therefore the v1 and v2 vectors we created above are not equal when compared with ===:
julia> v1 === v2
false
You might ask about NaN. We saw that we talked about before. Here the situation is complicated. They can be equal or be not equal. Since NaN values are immutable === compares them on bit level. So we have:
julia> Float16(NaN) == Float32(NaN)
false
julia> isequal(Float16(NaN), Float32(NaN))
true
julia> Float16(NaN) === Float32(NaN)
false
julia> Float16(NaN) == Float16(NaN)
false
julia> isequal(Float16(NaN), Float16(NaN))
true
julia> Float16(NaN) === Float16(NaN)
true
Thus, you have to be careful. Each of the three comparison methods I discussed have their uses and it is well worth learning them.
# Sets in Julia
Standard sets in Julia, created using the Set constructor use isequal to test for equality. Therefore we have:
julia> Set([v1, v2])
Set{Vector{Int64}} with 1 element:
[1, 2, 3]
We see that v1 and v2 got de-duplicated because they are equal with respect to isequal since they have the same contents. This is often what the user wants.
However, sometimes we want to track actual objects (irrespective of their contents). This is especially important when working with mutable structures. In this case IdSet is useful:
julia> IdSet{Vector{Int}}([v1, v2])
IdSet{Vector{Int64}} with 2 elements:
[1, 2, 3]
[1, 2, 3]
Note that we needed to specify the type of the values stored in IdSet. As an exception the IdSet() is allowed (not requiring you to pass the stored object type specification) and in this case an empty IdSet{Any} is created.
# Conclusions
Now you might ask when in practice IdSet is most useful. I needed it in my coding practice most often when I worked with nested mutable containers that potentially could contain circular references. In such case using IdSet allows you to easily keep track of the list of mutable objects already seen and avoid an infinite loop or stack overflow if you e.g. use recursion to work with such a deeply nested data structure.
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MATH PRE-TEST- Ch. 1
starstarstarstarstarstarstarstarstarstar
by Mary Francisco
| 4 Questions
Note from the author:
Order of Operations and exponents
1
1
12 + 12/ 6
4
14
20
6
2
1
What is the prime factorization of the number 30
6 x 5
5 x 3 x 2
15 x 2
(7.5 + 7.5) x 2
3
1
Solve: 15 + 15 / 3 x 4 -2
15
33
10
38
4
1
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# The Pythagorean Theorem Was Probably Already Used by Stonehenge Before Its Discovery
A new book claims that Stonehenge was built by expert astronomers using complex geometry that was not thought to exist in the Stone Age.
A new book named ‘Megalith: Studies in Stone’ re-examined the ancient geometry associated with Neolithic monuments and concluded that the people who constructed them centuries ago understood the concepts of solar and lunar eclipse cycles quite well. As a result of this ingenuity, they used complex geometry and recorded it all on massive stone calendars.
Robin Heath, the Megalith expert, went on to propose that the famous British landscape and the prehistoric site of Stonehenge are linked to the Pythagorean triangle. Not just this, but builders have been utilizing the discovery of the sum of the squares of the base and perpendicular adding up to the square of the hypotenuse in a right-angled triangle (a² + b² = c²) for millennials now to come up with these prehistoric marvels.
The book further demonstrates how there is a rectangle of 4 Sarsen stones in one of the earliest incarnations of Stonehenge that dates back to 2750 BC. These stones when split diagonally in half form a perfect Pythagorean triangle of the measurements 5:12:13.
In addition to this, the eight lines radiating from the triangles and the rectangle align perfectly with the most important dates present in the Neolithic calendar. They include the winter and summer solstices and the autumn and spring equinoxes as reported by Telegraph.
Editor and contributor of Telegraph, John Matineau further cleverly states, "People often think of our ancestors as rough cavemen, but they were also sophisticated astronomers. They were applying Pythagorean geometry over 2,000 years before Pythagoras was born.”
Woodhenge which stands almost two miles from Stonehenge in the north-east direction was apparently also built using the Pythagorean theorem with a 12:35:37 triangle. Furthermore, the book reinstates the fact that there were several stone circles on the site that weren’t entirely circular.
However, their geometry is derived from the Pythagorean theorem using whole numbers of the Megalithic yards that equal 2.72 feet. This, the book further comments, was most probably laid out with the use of pegs and ropes.
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# Math
posted by .
Through a given point on a line, there is one and only one line perpendicular to the given line.
The hints that it gave was "Do not limit your thinking to just two dimensions". If it just a regular line there will only be one perpendicular line right?
• Math -
hold a pencil up. A real pencil. I bet you can make perpendiculars in many directions, because the pencil is in a three dimensional space.
• Math -
I don't get what you mean by holding up a pencil. I hold up one and I don't see anything?
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Breaking News
News & Analysis
Energy harvesting conference shows you may be able to get something for (almost) nothing
11/17/2010 02:06 AM EST
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Page 1 / 2 > >>
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re: Energy harvesting conference shows you may be able to get something for (almost) nothing
12/10/2010 4:09:12 PM
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thank you Bill, interesting gadget...my problem with wind powered devices is what do you do if there is no wind? use your own lungs??? ;-) Kris
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re: Energy harvesting conference shows you may be able to get something for (almost) nothing
12/10/2010 4:07:15 PM
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To @Splinter60089: thank you for your comment...I understand efficiency is secondary in energy harvesting but you still have to make sure you harvest enough energy for the gadget in question to work...so efficiency determines what cane powered up and what not...as an example, I had a sun powered calculator several years ago that would only occasionally work continuously...Kris
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re: Energy harvesting conference shows you may be able to get something for (almost) nothing
12/10/2010 2:04:12 PM
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Kris, As mentioned on the very top of the article energy harvesting (or scavenging) is usage of small amounts of wasted power for supplying objects that require negligible amounts of energy for performing. Therefore, it is not reasonable to talk about efficiency or cost. In many cases efficiency of 5% would suffice. Usage of wind or tide-wave generators etc. is not energy harvesting. It is usage of free energy sources.
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re: Energy harvesting conference shows you may be able to get something for (almost) nothing
11/25/2010 9:17:15 AM
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Thanks for the information
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re: Energy harvesting conference shows you may be able to get something for (almost) nothing
11/19/2010 8:41:38 PM
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@iniewski: You bring up a good point: at microscale, wind loses out to other means like PV, temperature delta and vibration. One can intuitively guess that the PV, motion/vibration and thermal will be the ones to tap into for self-powering nodes. TI has a good white paper: http://www.ti.com/corp/docs/landing/cc430/graphics/slyy018_20081031.pdf It would seem that light and thermal handily beat vibration/motion in energy harvesting. One can get to as much as 10mW/cm^2 these two, per the table in the above doc. @new2coding: the 2X to 3X cost delta between wind and solar includes storage for wind. How ever, the real cost gets murky when you look at the transmission costs. It is very dependent on the location of the energy generating assets and the ease of access to the consumer. One can have PV modules on their roof but to have equivalent energy source from wind may not be feasible in an urban environment. Nice discussions, all! Dr. MP Divakar
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re: Energy harvesting conference shows you may be able to get something for (almost) nothing
11/19/2010 6:15:36 PM
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To @DrQuine: why is light man made in your comment? we are mostly using sunlight for harvesting energy....man made light would produce very little energy if captured (100x less), it would never be econamically vialable...back water wheels: there were many in the 19-th centure! but few if any survived, it would be amazing is they come from the ushes! Kris
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re: Energy harvesting conference shows you may be able to get something for (almost) nothing
11/19/2010 3:05:42 PM
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There is a long history of harvesting energy from the local environment using water wheels and windmills. The advent of devices that require very little power now makes it possible to harvest energy from small local phenomena that didn't used to be considered obvious sources of power. At the same time, if these available energy sources are manmade (electrical fields, light), we should be considering how to reduce the wasted energy they are dispersing.
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re: Energy harvesting conference shows you may be able to get something for (almost) nothing
11/18/2010 6:39:12 PM
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Hi @new2coding, I think payback time doesn't matter here much...we are off-grid, trying to harvest energy "free of charge"...the key metric in my mind is how much energy you can extract in a given amount of time...Kris
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re: Energy harvesting conference shows you may be able to get something for (almost) nothing
11/18/2010 6:14:46 PM
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thank you @docdivakar...no, I was not aware that wind is cheaper than photovoltaics...is that still true at the micro level? I am having hard time imaging small size wind turbine while small size photodiode works just fine...Kris
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re: Energy harvesting conference shows you may be able to get something for (almost) nothing
11/18/2010 6:02:04 PM
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@Bill Schweber: thanks for a nice summary, I hope there is a followup more in-depth article on the promising (lower cost) technology as well. For the explosive (billions) sensor growth that HP, Ericson and others are predicting by 2020, this is a must have piece of the solution. I took a quick look at the Etesian website (www.etesian-tech.com), their products seem to be priced for B2B market. It would be nice to find products for the consumer market that exploit this but I realize that is currently the hot innovation topic! @iniewski: I am sure you already know this, the cost/\$ for wind energy is lower than solar by 2x to 3x. A lot needs to happen in increasing efficiency of solar (particularly the thin film) and cost; only then we will see an explosion of product innovations using energy harvesting for the consumer market. PrintedElectronics conference is coming up soon in the Silicon Valley, I plan to attend and perhaps post a followup comment in EE Times. Dr. MP Divakar
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# Interior and Exterior Angles of Polygons
Objective: I know how to calculate the interior and exterior angles of polygons.
The formula to find the sum of the interior angles of any polygon is
sum of angles = (n - 2)180° , where n is the number of sides of the polygon.
The sum of exterior angles of any polygon is 360º.
Fill in all the gaps, then press "Check" to check your answers. Use the "Hint" button to get a free letter if an answer is giving you trouble. You can also click on the "[?]" button to get a clue. Note that you will lose points if you ask for hints or clues!
Calculate the sum of the interior angles of the following polygons.
a 10-sided polygon
a 12-sided polygon
a 15-sided polygon
an 18-sided polygon
Find the number of sides of a regular polygon with an exterior angle of
45°
36°
30°
12°
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
We hope that the free math worksheets have been helpful. We encourage parents and teachers to select the topics according to the needs of the child. For more difficult questions, the child may be encouraged to work out the problem on a piece of paper before entering the solution. We hope that the kids will also love the fun stuff and puzzles.
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Lesson 5: Adding and subtracting fractions with unlike denominators word problems
# Subtracting fractions word problem: tomatoes
Learn how to solve word problems involving the addition and subtraction of fractions with unlike denominators. Watch examples being worked out step-by-step, and practice applying the same techniques to solve similar problems. Created by Sal Khan.
## Want to join the conversation?
• This could have been solved much easier. What does it take for 2 7/8 to get to 3? 1/8. What does it take for 3 to get to 3 1/4? 1/4. Therefore the answer is 1/4 + 1/8 which is 3/8.
• For this case, that is an excellent way to do it. I commend you for seeing it that way.
I think many can't see the shortcuts yet, and the long way helps demonstrate the individual steps of the path.
Over time, with much practice, people will also find the fast way like you have already mastered.
• I'm confused. How did 3 1/4 turn into 12/4+1/4?
• 12/4 is equal to 3, so either way of writing it is valid
• What is the point of fractions?(I'm serious).
• Think of fractions like this: Suppose you solve a 2x2 Rubik's Cubes in half a second, somebody else solves it in 1/3 of a second, and another person solves it in 11/12 of a second. You want to find the average all the 3 solves. You would have to use knowledge about fractions, or else you won't be able to find the average of the solves, or even express the solve time!
• i feel like sal is making this too confusing, does any body agree with me?
• If you are confused, I recommend going back into the curriculum into something you're a bit more comfortable with, then, when you feel ready, you can get back to what was challenging before. Chances are, it will be a lot easier the second time.
• 0/0 is undefined, meaning that it can be a lot of different things, and therefore doesn't make sense as a fraction. As Kim says above, dividing by 0 is a nonsensical concept. Still, there can be times when 0/0 pops up in legitimate math questions, and it's useful to understand what it's close to, if not what it actually is (which is no particular number).
For example, imagine that I have some number of tomatoes (call this number n) and I'm always going to divide this number of tomatoes amongst that number of friends. In other words, each friend will always get n/n tomatoes. It's somewhat obvious that each friend will always get 1 tomato (for example, if I have 5 tomatoes and divide them among 5 friends, each one gets one). But what if I have 0 tomatoes and divide them amongst 0 friends? The answer doesn't make sense. However, it's useful to still know that if there were any friends, they would each get 1. So in this case, even though 0/0 is an undefined fraction, it acts like 1. In calculus, they would say that "the limit of n/n as n approaches 0 is 1."
Now let's imagine the same scenario, except where I have twice as many tomatoes, (2n), and I'm dividing them amongst n friends. Each friend gets 2n/n tomatoes. In this case, each friend will always get 2 tomatoes. But imagine I have 0 friends and 0 tomatoes. Again, the fraction 0/0 pops up, and it's still undefined, but this time it mostly means "2," whereas previously it mostly meant "1." Again, those who study calculus would say "the limit of 2n/n as n approaches 0 is 2."
We can think of similar examples for all possible numbers, which is why 0/0 is undefined. However, we can imagine that it's more of a context-dependent fraction than a fraction itself, and sometimes the number 0/0 can mean something, even though, in general, it doesn't mean anything.
• Does anyone else think he over explained a bit? I know this concept already but he could've shortened the video down
• I have a question? i worked it out on paper and got 3/8 what if the whole number is lesser than the fraction then what do i do.
• What do you mean that the whole number is less than the fraction. Whole numbers are like: 0, 1, 2, 3, 4, 5, etc. The only whole number that can be smaller than a fraction is 0. Or do you have an improper fraction where the numerator is larger than the denominator? If you still need help, I suggest you post the actual numbers and the problem you are trying to do so that someone can help you.
• why chery tomatoes am I right
• Many times in these questions, they will add a little more info, to see how if you can get find your answer or get confused with the numbers. It's really all about how your brain processes the information
• Sal, your artistic skill on computer is amazing.....
I bet this comment cant get 10 upvotes
• can anybody please solve this for me?
Q.
50-21x2
------------
18x6-4
• The numerator:
50 minus 21 times 2 is equal to:
-> 50 - 42 (because 21*2=42)
-> 8 (because 50-40 = 10 -2 = 8
Denominator:
18 times 6 minus 4
(because there is no "parentheses," Order of Operations dictates that we multiply before subtracting, thus):
18 * 6 = (10*6) + (8*6) - 4 =
= (60 + 48) - 4
= (60 + 40) + (8) - 4
= 108 - 4
= 104
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# viscosity of binder at elevated temperature using brookfield viscometer
Post on 22-Jan-2018
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## Engineering
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1. 1. Viscosity of Binder at Elevated Temperature Using Brookeld Viscometer Priyansh Singh 16th February 2016 Priyansh Singh | 16th February 2016 0/33 Viscosity of Binder at Elevated Temperature Using Brookeld Viscometer Priyansh Singh | 16th February 2016 1/33
2. 2. 1 What is viscosity? Introduction Use of Viscosity 2 Different Methods Empirical Methods Absolute Methods 3 Brookeld Viscometer Introduction Working Testing What is viscosity? - Introduction Table of contents Priyansh Singh | 16th February 2016 2/33
3. 3. Viscosity Measure of the resistance to deformation of a uid under shear stress. Viscosity is measure of internal friction of a uid. This friction becomes apparent when a layer of uid is made to move with respect to other layer. Greater the friction greater the amount of force required to cause movement. Which is known as shear. What is viscosity? - Introduction What is viscosity? Priyansh Singh | 16th February 2016 3/33
4. 4. fluid. This friction becomes apparent when a layer of fluid is made to move in relation to another layer. The greater the friction, the greater the amount of force re- quired to cause this movement, which is called shear. Shearing occurs whenever the fluid is physically moved or distributed, as in pouring, spreading, spraying, mix- ing, etc. Highly viscous fluids, therefore, require more force to move than less viscous materials. A A V2 V1 dv dx F Figure 4-1 Isaac Newton defined viscosity by considering the model represented in Figure 4-1. Two parallel flat areas of fluid of the same size A are separated by a distance dx and are moving in the same direction at different velocities V1 and V2. Newton assumed that the force required to maintain this difference in speed was proportional to the difference in speed through the liquid, or the velocity gradient. To express = viscosity = = s s . The fundamental unit of viscos poise. A material requiring a s dyne per square centimeter to p of one reciprocal second has a vis or 100 centipoise. You will encou surements expressed in Pascal- milli-Pascal-seconds (mPas); t International System and are som erence to the CGS designations. is equal to ten poise; one milli-Pas to one centipoise. Newton assumed that all mater temperature, a viscosity that is shear rate. In other words, twice th the fluid twice as fast. As we shall see, Newton was o 4.3 Newtonian Fluids This type of flow behavior whic for all fluids is called, not surprisin is, however, only one of several ty you may encounter. A Newtonian graphically in Figure 4-2. Grap relationship between shear stres Newtons Law Newton assumed that the force required to maintain this difference in speed was proportional to the difference in speed through the liquid, or the velocity gradient What is viscosity? - Introduction Viscosity Priyansh Singh | 16th February 2016 4/33
5. 5. = V iscosity dv dx = Shear rate F A = Shear stress F A = dv dx = Shear Stress Shear Rate What is viscosity? - Introduction Viscosity Priyansh Singh | 16th February 2016 5/33
6. 6. 1 What is viscosity? Introduction Use of Viscosity 2 Different Methods Empirical Methods Absolute Methods 3 Brookeld Viscometer Introduction Working Testing What is viscosity? - Use of Viscosity Table of contents Priyansh Singh | 16th February 2016 6/33
7. 7. Usefull Behaviour Binder Grading What is viscosity? - Use of Viscosity NECESSITY of Viscosity Measurement Priyansh Singh | 16th February 2016 7/33
8. 8. IS 73 : 2013 Table 1 Requirements for Paving Bitumen (Clause 6.2) Paving GradesSl No. Characteristics VG10 VG20 VG30 VG40 Me (1) (2) (3) (4) (5) (6) i) Penetration at 25C, 100 g, 5 s, 0.1 mm, Min 80 60 45 35 IS 1 ii) Absolute viscosity at 60C, Poises 800-1 200 1 600-2400 2 400-3 600 3 200-4 800 IS 1 iii) Kinematic viscosity at 135C, cSt, Min 250 300 350 400 IS 1 iv) Flash point (Cleveland open cup), C, Min 220 220 220 220 IS 1 v) Solubility in trichloroethylene, percent, Min 99.0 99.0 99.0 99.0 IS 1 vi) Softening point (R&B), C, Min 40 45 47 50 IS 1 vii) Tests on residue from rolling thin film oven test: a) Viscosity ratio at 60C, Max b) Ductility at 25C, cm, Min 4.0 75 4.0 50 4.0 40 4.0 25 IS 1 IS 1 precautions mentioned therein.All these samples from individual containers shall be stored separately. 7.4 Number of Tests 8 PACKING AND MARKING 8.1 Packing What is viscosity? - Use of Viscosity Binder Grading Priyansh Singh | 16th February 2016 8/33
9. 9. Effect of Processing By viscosity measurement the effective change in binder can be assessed. Formulation changes Potential change in asphalt behavior can be accessed by viscosity measurement. Aging Phenomena Binder aging can be assessed by change in viscosity measurement. Production Temperature The mixing and compaction temperatures of the asphalt concrete can be determined by viscosity measurement. What is viscosity? - Use of Viscosity NECESSITY of Viscosity Measurement Priyansh Singh | 16th February 2016 9/33
10. 10. What is viscosity? - Use of Viscosity Determination of mixing and compaction Temperatures Priyansh Singh | 16th February 2016 10/33
11. 11. 1 What is viscosity? Introduction Use of Viscosity 2 Different Methods Empirical Methods Absolute Methods 3 Brookeld Viscometer Introduction Working Testing Different Methods - Empirical Methods Table of contents Priyansh Singh | 16th February 2016 11/33
12. 12. These methods determines the viscosity without addressing the constitutive relationship. Capillary Viscometer Relation with other test. Different Methods - Empirical Methods Empirical Methods Priyansh Singh | 16th February 2016 12/33
13. 13. 1 What is viscosity? Introduction Use of Viscosity 2 Different Methods Empirical Methods Absolute Methods 3 Brookeld Viscometer Introduction Working Testing Different Methods - Absolute Methods Table of contents Priyansh Singh | 16th February 2016 13/33
14. 14. These methods uses the basic stress strain relationship to determine viscosity of material. Rotational Viscometer. Shear Rheometer. Different Methods - Absolute Methods Absolute Methods Priyansh Singh | 16th February 2016 14/33
15. 15. 1 What is viscosity? Introduction Use of Viscosity 2 Different Methods Empirical Methods Absolute Methods 3 Brookeld Viscometer Introduction Working Testing Brookeld Viscometer - Introduction Table of contents Priyansh Singh | 16th February 2016 15/33
16. 16. Brookeld Viscometer - Introduction Brookeld Viscometer Priyansh Singh | 16th February 2016 16/33
17. 17. STEPPER MOTOR CLUTCH DIAL PIVOT SHAFT PIVOT CUP GUARDLEG SPINDLE SAMPLE CONTAINER HOUSING GEAR TRAIN POINTER CALIBRATED SPIRAL SPRING JEWELLED BEARING Figure 3-1 over 80 years of c sound foundation f a starting point from can be explored. 3.4.1 Record Ke We recomm tion always be measurement; cessory), rotatio sions, sample te sample prepara or not the spind Forms supplied Viscometer are 3.4.2 The Spind Examine ea corroded or dam dimensions, a f Since all spindle Brookeld Viscometer - Introduction Components of Brookeld Viscometer Priyansh Singh | 16th February 2016 17/33
18. 18. 1 What is viscosity? Introduction Use of Viscosity 2 Different Methods Empirical Methods Absolute Methods 3 Brookeld Viscometer Introduction Working Testing Brookeld Viscometer - Working Table of contents Priyansh Singh | 16th February 2016 18/33
19. 19. Brookeld Viscometer - Working Principle of Rotational Viscometer Priyansh Singh | 16th February 2016 19/33
20. 20. It measures the torque required to rotate an immersed element (Spindle) in a uid. Spindle is driven by a motor through a calibrated spring. Deection (tension) in spring is indicated by digital display. For a given viscosity, viscosity drag or resistance to ow is proportional to spindle speed of rotation. Brookeld Viscometer - Working Principle of Rotational Viscometer Priyansh Singh | 16th February 2016 20/33
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# A Tutorial on the OpenOffice Calc AVERAGE Function
OpenOffice Calc's built-in AVERAGE function makes calculating the average of a column easy.
Calc is the spreadsheet component of Apache's free Open Office suite of programs.
## How an Average Is Calculated
An average is calculated by adding a group of numbers and dividing the total by the count of those numbers.
The example here calculates the average for the values in column C, which comes out to 13.5. If you were calculate this manually, you'd add all the numbers, and divide the sum by 6 (11 + 12 + 13+ 14 + 15 + 16 = 81; 81 ÷ 6 = 13.5). Instead of finding this average manually, however, you can tell OpenOffice Calc to do it for you using the AVERAGE function:
=AVERAGE(C1:C6)
Should any of the values in the column change, the average update appropriately.
## The AVERAGE Function's Syntax and Arguments
In OpenOffice and other spreadsheet programs such as Excel and Google Sheets, a function's syntax refers to the layout of the function and includes the function's name, brackets, and arguments. The syntax for the AVERAGE function is:
=AVERAGE (number 1; number 2; ...number30)
Up to 30 numbers can be averaged by the function.
A function's arguments are the numbers affected by the function:
• Argument number 1 (required)—the data to be averaged by the function
• Argument number 2; ... number30 (optional)—additional data that can be added to the average calculations.
The arguments can contain:
If the data you want to average is spread out in individual cells in the worksheet rather than in a single column or row, enter each cell reference into the dialog box on a separate argument line (for example, C5, E4, G2).
## Example: Find the Average Value of a Column of Numbers
1. Enter the following data into cells C1 to C6: 11, 12, 13, 14, 15, 16.
2. Select cell C7, where the results will be displayed.
3. Select Insert > Function from the menu along the top of the screen.
4. Choose Statistical from the Category list.
5. Select Average, and click or tap Next.
6. Highlight cells C1 to C6 in the spreadsheet to enter this range into the dialog box in the number 1 argument line; Click OK to complete the function and close the dialog box.
If you can't see the cells you need to click, drag the dialog box out of the way.
The number 13.5 should appear in cell C7, which is the average for the numbers you entered in cells C1 to C6. When you select cell C7, the complete function =AVERAGE (C1:C6) appears in the input line above the worksheet
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Just Keep the Change (Posted on 2003-06-15)
In a certain country, it takes a minimum of exactly 5 coins to make 13 cents, and a minimum of exactly 4 coins to make 14 cents.
What are the denominations of the coins (of value less than 15 cents) in this country?
Bonus: What country is it?
See The Solution Submitted by DJ Rating: 4.1818 (11 votes)
Comments: ( Back to comment list | You must be logged in to post comments.)
Solution | Comment 12 of 13 |
There cannot be a 7-cent coin because you could make 14 cents with 2 coins (7, 7). Also, there cannot be a 13-cent coin or a 14-cent coin.
Suppose there is a 1-cent coin. Then, there cannot be a 4 because you could make 13 cents with 4 coins (1, 4, 4, 4). Also, there cannot be a 6 because you could make 13 cents with 3 coins (1, 6, 6). There cannot be a 10 because of (1, 1, 1, 10). There cannot be an 11 because of (1, 1, 11). There cannot be a 12 because of (1, 12).
Suppose there is a 1 and a 5. Then, there cannot be a 2 because you could make 13 cents with 4 coins (1, 2, 5, 5). There cannot be a 3 because you could make 13 cents with 3 coins (3, 5, 5). There cannot be an 8 because you could make 13 cents with 2 coins (5, 8). There cannot be a 9 because you could make 14 cents with 2 coins (5, 9). Therefore, if there is a 1-cent coin and a 5-cent coin, then those are the only coins less than 15 cents. However, it is impossible to make 14 cents with 4 of these coins. Therefore, there cannot be a 1 and a 5.
Suppose there is a 1 and an 8. Then, there cannot be a 2 because of (1, 2, 2, 8). There cannot be a 3 because of (3, 3, 8). The only other possibility for another coin is 9, but you cannot make 14 cents with 4 coins that are either 1, 8, or 9 cents. Therefore, there cannot be a 1 and an 8.
Suppose there is a 1 and a 9. Then, there cannot be a 2 because of (2, 2, 9). There cannot be a 3 because of (1, 3, 9). Therefore, 1 and 9 are the only coins, but you cannot make 14 cents with 4 of these coins. Therefore, there cannot be a 1 and a 9. If there is a 1, then there can only be a 2 or a 3. However, the greatest sum of 4 coins is 12<14. Therefore, there cannot be a 1-cent coin.
Since, you can make 13 cents with 5 coins, there must be a coin less than 3 cents. It must be 2 cents. Now, there cannot be a 9 because of (2, 2, 9). There cannot be a 10 because of (2, 2, 10). There cannot be an 11 because of (2, 11). There cannot be a 12 because of (2, 12).
Suppose there is a 3-cent coin. Then, there cannot be a 4 because of (3, 3, 3, 4). There cannot be a 5 because of (3, 5, 5). There cannot be a 6 because of (2, 2, 3, 6). There cannot be an 8 because of (3, 3, 8). Therefore, 2 and 3 are the only coins. However, you cannot make 14 cents with 4 of these coins. Therefore, there cannot be a 3-cent coin.
Since you can make 13 cents, there must be an odd coin. It must be 5 cents. Now, there cannot be a 4 because of (2, 2, 4, 5). There cannot be a 6 because of (2, 5, 6). There cannot be an 8 because of (2, 2, 8). Therefore, 2 and 5 are the only coins. You make 13 cents with (2, 2, 2, 2, 5). You make 14 cents with (2, 2, 5, 5).
Posted by Math Man on 2014-07-12 11:32:24
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### Home > GC > Chapter 4 > Lesson 4.2.2 > Problem4-66
4-66.
Examine pentagon $SMILE$ at right. Do any of its sides have equal length? How do you know? Be sure to provide convincing evidence. You might want to copy the figure onto graph paper.
Draw a slope triangle for each side of the pentagon. Now can you determine the lengths of the sides?
By the Pythagorean Theorem, if the corresponding legs of two right triangles are congruent, then their hypotenuses will be congruent.
Use the eTool below to check your work.
Click the link at right for the full version of the eTool:
4-66 HW eTool (Desmos)
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• anonymous
A stationery shop has a sale on address books and calendars. Two of the address books and one calendar cost a total of $10. One of the address books and three calendars cost a total of$15. How much does one calendar cost? a. $6 b.$3 c. $1 d.$4 Please select the best answer from the choices provided
Mathematics
• Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
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RVUC- Statistics Short Note (2).doc
# Solution steps 1 calculate µ and x µ birr 15000 x
• Notes
• 29
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Solution: Steps: 1. Calculate µ and x µ = Birr 15,000 x = δ/√n= 2000/√30 = Birr 365.15 2. Calculate Z for X X X X X X X Z 05 . 2 365 000 , 15 750 , 15 750 , 15 Z 3. Interpret the results There is a 2.02% chance that the average earning being more than Birr 15, 750 annually in a group of 30 tellers. CHAPTER TWO 6
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7 | P a g e HYPOTHESIS TESTING Basic concepts The word hypothesis is made up of two Greek roots The word hypothesis consists of two words: Hypo + thesis = Hypothesis ‘Hypo’ means tentative or subject to the verification and ‘Thesis’ means statement about solution of a problem Hypotheses are conjectural statements that are amenable to empirical investigation Example: The Statement that “ Sixty Percent or more of the residents of Addis Ababa city attend worship services at least once a week” It is a statement of purported fact and can therefore be tested A statement, which is a value judgment, will not be considered as good hypothesis. Example: Psychology is a more important subject than Sociology. - It is the assumption we wish to test. - It is denoted by (H 0 ) - It always includes the equality sign - It refers to the view that the observations result purely from chance. - It is initially assumed to be true. H 0 : There is no significant interaction effect of schedule of reinforcement and extroversion on learning outcomes. H 0 : There is no significant relationship between intelligence When the null hypothesis is rejected, the outcome is side to be “statistically significant’’ When the null hypothesis is not rejected, then the outcome is said to be “not statistically significant” and achievement of students. Alternative Hypothesis It describes what you will conclude if you reject the null hypothesis. It is accepted if the sample data provide us with enough statistical evidence that the null hypothesis is false. is a statement that would be accepted if our sample data provide us with ample evidence that the null hypothesis is false or has to be rejected. 7
8 | P a g e - It is the opposite of what is stated in the null hypothesis . It is denoted by Ha or H 1 .It always includes the inequality sign - The assumption that the observations are the result of influences by some non- random cause. Types of Errors (Decision Errors) There are four possible outcomes of any hypothesis test, two of which are correct decisions and two of which are incorrect. The incorrect ones are called type I and type II errors. Types of Errors in hypothesis testing The following table presents the possible conclusions and errors in performing a test. State of Nature (Null hypothesis) Decision H 0 :True H 0 : False Accept H 0 correct Decision Type II error Reject H 0 Type I error Correct Decision Type I Error - Occurs when we reject a null hypothesis that is true.
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# integration (squareroot)
• May 30th 2010, 11:10 PM
gomes
integration (squareroot)
What would be the best/quickest way to integrate
squareroot(1-x^2) ?
• May 30th 2010, 11:49 PM
earboth
Quote:
Originally Posted by gomes
What would be the best/quickest way to integrate
squareroot(1-x^2) ?
1. Use integration by parts:
$\int(\sqrt{1-x^2}) dx = \int(1 \cdot \sqrt{1-x^2}) dx$
2. In the second step keep in mind that
$\int \left(\frac1{\sqrt{1-x^2}}\right) dx = \arcsin(x)$
• May 30th 2010, 11:51 PM
tonio
Quote:
Originally Posted by gomes
What would be the best/quickest way to integrate
squareroot(1-x^2) ?
Substitute $\sin t=x$ ...(Wink)
Tonio
• May 30th 2010, 11:54 PM
tonio
Quote:
Originally Posted by earboth
1. Use integration by parts:
$\int(\sqrt{1-x^2}) dx = \int(1 \cdot \sqrt{1-x^2}) dx$
2. In the second step keep in mind that
$\int \left(\frac1{\sqrt{1-x^2}}\right) dx = \arcsin(x)$
I don't think the above helps: $u=\sqrt{1-x^2}\Longrightarrow u'=-\frac{x}{\sqrt{1-x^2}}\,,\,\,v'=1\Longrightarrow v=x$ , so we get the integral $\int\frac{x^2}{\sqrt{1-x^2}}\,dx$ ...which not only is not an arcsine but it doesn't look easy at all...
Unless, of course, I missed something.
Tonio
• May 31st 2010, 12:08 AM
Prove It
Quote:
Originally Posted by gomes
What would be the best/quickest way to integrate
squareroot(1-x^2) ?
Trigonometric or Hyperbolic substitution works well...
Let $x = \sin{\theta}$ so that $dx = \cos{\theta}\,d\theta$.
The integral becomes
$\int{\sqrt{1 - x^2}\,dx} = \int{\sqrt{1 - \sin^2{\theta}}\,\cos{\theta}\,d\theta}$
$= \int{\cos{\theta}\cos{\theta}\,d\theta}$
$= \int{\cos^2{\theta}\,d\theta}$
$= \int{\frac{1}{2}\cos{2\theta} + \frac{1}{2}\,d\theta}$
$= \frac{1}{4}\sin{2\theta} + \frac{1}{2}\theta + C$
$= \frac{2\sin{\theta}\cos{\theta}}{4} + \frac{\theta}{2} + C$
$= \frac{\sin{\theta}\cos{\theta} + \theta}{2} + C$.
Remembering that $x = \sin{\theta}$, that means $\cos{\theta} = \sqrt{1 - x^2}$ and $x = \arcsin{\theta}$ and substituting, we find
$\int{\sqrt{1 - x^2}\,dx} = \frac{x\sqrt{1 - x^2} + \arcsin{x}}{2} + C$.
• May 31st 2010, 09:40 AM
tom@ballooncalculus
Quote:
Originally Posted by tonio
Unless, of course, I missed something.
Tonio
I think earboth is going to re-write
$\frac{-x^2}{\sqrt{1 - x^2}}$
as
$\frac{1 - x^2 - 1}{\sqrt{1 - x^2}}\ =\ \sqrt{1 - x^2}\ -\ \frac{1}{\sqrt{1 - x^2}}$
and then solve the top row of...
http://www.ballooncalculus.org/asy/p...tOneMinus1.png
... for I. Which is no longer, really.
Key:
Spoiler:
http://www.ballooncalculus.org/asy/maps/parts.png
... is lazy integration by parts, doing without u and v.
http://www.ballooncalculus.org/asy/prod.png
... is the product rule, straight continuous lines differentiating downwards (integrating up) with respect to x.
Full size:
http://www.ballooncalculus.org/asy/p...otOneMinus.png
__________________________________________________ ___
Don't integrate - balloontegrate!
Balloon Calculus; standard integrals, derivatives and methods
Balloon Calculus Drawing with LaTeX and Asymptote!
• May 31st 2010, 01:43 PM
gomes
Thanks everyone, really appreciate it. :)
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Javascript sorting algorithm for counting sort instances
• 2020-03-30 02:34:17
• OfStack
Counting sort is a stable sorting algorithm. Count sort USES an additional array Count_arr, where the ith element is the number of elements whose median value in the array Arr to be sorted is equal to I. The elements in Arr are then sorted into the correct positions based on the array Count_arr.
There are four steps:
1. Find the largest and smallest elements in the array to be sorted
2. Count the number of occurrences of each element of value I in the array, and store the ith term of the array Count_arr
4. Reverse traversal of the original array: place each element I in the Count_arr(I) entry of the new array, subtracting Count_arr(I) from 1 for each element
Example:
``````
function countSort(arr, min, max) {
var i, z = 0, count = [];
for (i = min; i <= max; i++) {
count[i] = 0;
}
for (i=0; i < arr.length; i++) {
count[arr[i]]++;
}
for (i = min; i <= max; i++) {
while (count[i]-- > 0) {
arr[z++] = i;
}
}
return arr;
}
// test
var i, arr = [];
for (i = 0; i < 100; i++) {
arr.push(Math.floor(Math.random() * (141)));
}
countSort(arr, 0, 140);
``````
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HW1_Solutions2011
# HW1_Solutions2011 - curves be at these same strains in a...
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Homework 1 Solutions 1. Solution: Solution: Solution: Solution:
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ME 382 Fall 2010 HW1 Solutions Page 6 Solution:
ME 382 Fall 2010 HW1 Solutions Page 7 Below is a true stress versus true strain curve from an ME395 lab, in which a tensile test is conducted on an annealed sample of brass. The modulus of brass is 105 GPa. Calculate the slope of the elastic unloading curve in an engineering (nominal) stress-strain curve for the annealed brass below at a nominal strain of (a) 2% and at a nominal strain of (b) 25%. What would the slopes of the unloading
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Unformatted text preview: curves be at these same strains in a true stress-strain curve? (You may neglect the small effects of elastic deformation in your answer.) 4. Solution: ± ² ³ ´µ¶ · , ¸¹ º ± ² ³ » Elastic recovered strain, ¸¹ º , is the same value for true and engineering. ME 382 Fall 2010 HW1 Solutions Page 8 5. Solution: Solution: ME 382 Fall 2010 HW1 Solutions Page 9 ME 382 Fall 2010 HW1 Solutions Page 10 6. Solution: ME 382 Fall 2010 HW1 Solutions Page 11 Solution: ME 382 Fall 2010 HW1 Solutions Page 12 7. Solution: ME 382 Fall 2010 HW1 Solutions Page 13 316 MPa, -196 MPa...
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# How do I know the input current of a LED driver?
I want to driver 3 power LEDs and depending on the solution I choose I find something strange.
If I have this situation :
I want 330 mA flowing in my LEDs. I choose a 10 V DC-DC converter. The forward voltage of the LEDs is 3 V. I set the current through the resistor with the remaining 1 V by choosing R = 3 ohms. 330 mA under 10 V equals to 275 mA under 12 V, so the input current of the DC-DC converter is 275 mA. So far so good.
Now let's consider this situation, where I use this LED driver : http://www.ti.com/lit/ds/symlink/tl4242-q1.pdf and choose its resistor so that it outputs a constant 330 mA :
I want to know the input current of the DC-DC converter. To do so I need to know the output current of the DC-DC, which is the input current of the LED driver. In the datasheet they say the LED driver consumes around 12 mA when its output voltage is 6.6 V. First : How does this value change when the output voltage changes ? Why don't they give that information ? Finally if the output current of the DC-DC converter is 12 mA, then the input current of the DC-DC converter is 10 mA.
In the first situation the system draws 275 mA, in the second situation is draws 10 mA.
Is there something I am doing wrong ?
Thanks.
• why are you using a DC-DC converter with the LED driver? Feb 13, 2020 at 16:26
• The datasheet is telling you how much current the driver itself consumes internally in addition to whatever it's pushing thru the LEDs. Feb 13, 2020 at 16:34
• I will need to light 1 branch of 3 LEDs, 1 branch of 2 LEDs, and 1 branch of 1 LED. Given that my power supply is 12 V, I am afraid the voltage dropout for the branch of 1 LED might cause too much power dissipation (12 V to 4 V) Feb 13, 2020 at 16:35
• That DCDC converter serves no purpose, so you can remove it. Feb 13, 2020 at 17:16
• Wouldn't it be useful if I want to light only one LED from a 12V power supply to first reduce the voltage to 4 V with a DC-DC converter to handle power dissipation ? Feb 13, 2020 at 22:12
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How to determine the human resources I need given financial and operational constraints?
I have a project and I want to minimize the costs. I am are responsible for the inspection of 1000 kilometers of sewer grid in Canada. My goal is to provide time high quality inspection reports. I tried to define the problem using optimization but I haven't used for a year. I want to know what would be the optimised number of people I should hire.
The people I have to pay are robots drivers and inspectors.
• I have a \$\\$90000\$ budget, two months on site and two weeks off site before.
• robots drivers take 6 pictures of every pole under different point of views and upload them on my cloud. The drivers upload their pics once a week.
• A well trained driver can inspect 15miles of sewer grid per day. Training lasts for one day, you can train two pilots in parallel.
• A photo inspector performs a quality check of the pictures on the Sterblue cloud. Sterblue asks the drone pilots to reinspect any poles not passing the quality checks. You can assume 10% of the poles have to be re-inspected.
• My AI on the cloud detects the defects. The customer expects 95% accuracy.
• The inspector performs a quality check of the detections found by the AI. They cost \$300 a day and can handle 30miles/day for image quality review, 30miles/day to review the work of the AI.
• I have a free operations team that generates the inspection reports. 0,5 day is needed to prepare a report for 100 miles of grid.
Where x_1 is the number of robots drivers, x_2 the number of inspectors, y_1 the time spent by robot drivers, y_2 the time spent by inspectors.
I know that inspection can't start before reports, and that I haven't found a way to write the 95% accuracy constraints.
Can you help me improve the problem so I take into account every constraints to determine the number of people I should hire ?
I know I should do a Gantt diagram as well but I don't know yet where the major steps and dependencies are.
• Can you create separate resource groups for each type of worker (men, machines) and convert their productivity into FTE? I think that would allow you to track their work load so you can distribute it more evenly or find spots where certain resource will be overloaded. Epicflow does that sort of analysis, but I'm not sure if it fits your situation. – Igor Gorbenko Jun 25 '19 at 10:36
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# 6.2 The bohr model (Page 3/9)
Page 3 / 9
Since Bohr’s model involved only a single electron, it could also be applied to the single electron ions He + , Li 2+ , Be 3+ , and so forth, which differ from hydrogen only in their nuclear charges, and so one-electron atoms and ions are collectively referred to as hydrogen-like atoms. The energy expression for hydrogen-like atoms is a generalization of the hydrogen atom energy, in which Z is the nuclear charge (+1 for hydrogen, +2 for He, +3 for Li, and so on) and k has a value of 2.179 $\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}$ 10 –18 J.
${E}_{n}=\text{−}\frac{k{Z}^{2}}{{n}^{2}}$
The sizes of the circular orbits for hydrogen-like atoms are given in terms of their radii by the following expression, in which ${\text{α}}_{0}$ is a constant called the Bohr radius, with a value of 5.292 $×$ 10 −11 m:
$r=\phantom{\rule{0.2em}{0ex}}\frac{{n}^{2}}{Z}\phantom{\rule{0.2em}{0ex}}{a}_{0}$
The equation also shows us that as the electron’s energy increases (as n increases), the electron is found at greater distances from the nucleus. This is implied by the inverse dependence on r in the Coulomb potential, since, as the electron moves away from the nucleus, the electrostatic attraction between it and the nucleus decreases, and it is held less tightly in the atom. Note that as n gets larger and the orbits get larger, their energies get closer to zero, and so the limits $n\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\infty$ $n\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\infty ,$ and $r\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\infty$ $r\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\infty$ imply that E = 0 corresponds to the ionization limit where the electron is completely removed from the nucleus. Thus, for hydrogen in the ground state n = 1, the ionization energy would be:
$\text{Δ}E={E}_{n\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\infty }-{E}_{1}=0\phantom{\rule{0.2em}{0ex}}+\phantom{\rule{0.2em}{0ex}}k=k$
With three extremely puzzling paradoxes now solved (blackbody radiation, the photoelectric effect, and the hydrogen atom), and all involving Planck’s constant in a fundamental manner, it became clear to most physicists at that time that the classical theories that worked so well in the macroscopic world were fundamentally flawed and could not be extended down into the microscopic domain of atoms and molecules. Unfortunately, despite Bohr’s remarkable achievement in deriving a theoretical expression for the Rydberg constant, he was unable to extend his theory to the next simplest atom, He, which only has two electrons. Bohr’s model was severely flawed, since it was still based on the classical mechanics notion of precise orbits, a concept that was later found to be untenable in the microscopic domain, when a proper model of quantum mechanics was developed to supersede classical mechanics.
## Calculating the energy of an electron in a bohr orbit
Early researchers were very excited when they were able to predict the energy of an electron at a particular distance from the nucleus in a hydrogen atom. If a spark promotes the electron in a hydrogen atom into an orbit with n = 3, what is the calculated energy, in joules, of the electron?
## Solution
The energy of the electron is given by this equation:
$E=\phantom{\rule{0.2em}{0ex}}\frac{-k{Z}^{2}}{{n}^{2}}$
The atomic number, Z , of hydrogen is 1; k = 2.179 $\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}$ 10 –18 J; and the electron is characterized by an n value of 3. Thus,
$E=\phantom{\rule{0.2em}{0ex}}\frac{-\left(2.179\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-18}\phantom{\rule{0.2em}{0ex}}\text{J}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{\left(1\right)}^{2}}{{\left(3\right)}^{2}}\phantom{\rule{0.2em}{0ex}}=-2.421\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-19}\phantom{\rule{0.2em}{0ex}}\text{J}$
The electron in [link] is promoted even further to an orbit with n = 6. What is its new energy?
−6.053 $×$ 10 –20 J
what are oxidation numbers
pls what is electrolysis
Electrolysis is the process by which ionic substances are decomposed (broken down) into simpler substances when an electric current is passed through them. ... Electricity is the flow of electrons or ions. For electrolysis to work, the compound must contain ions.
AZEEZ
thanks
Idowu
what is the basicity of an atom
basicity is the number of replaceable Hydrogen atoms in a Molecule. in H2SO4, the basicity is 2. in Hcl, the basicity is 1
Inemesit
how to solve oxidation number
mention some examples of ester
do you mean ether?
Megan
what do converging lines on a mass Spectra represent
would I do to help me know this topic ?
Bulus
oi
Amargo
what the physic?
who is albert heistein?
Bassidi
similarities between elements in the same group and period
what is the ratio of hydrogen to oxulygen in carbohydrates
bunubyyvyhinuvgtvbjnjnygtcrc
yvcrzezalakhhehuzhbshsunakakoaak
what is poh and ph
please what is the chemical configuration of sodium
Sharon
2.8.1
david
1s²2s²2p⁶3s¹
Haile
2, 6, 2, 1
Salman
1s2, 2s2, 2px2, 2py2, 2pz2, 3s1
Justice
1s2,2s2,2py2,2
Maryify
1s2,2s2,2p6,
Francis
1s2,2s2,2px2,2py2,2pz2,3s1
Nnyila
what is criteria purity
cathode is a negative ion why is it that u said is negative
cathode is a negative electrode while cation is a positive ion. cation move towards cathode plate.
king
CH3COOH +NaOH ,complete the equation
compare and contrast the electrical conductivity of HCl and CH3cooH
The must be in dissolved in water (aqueous). Electrical conductivity is measured in Siemens (s). HCl (aq) has higher conductivity, as it fully ionises (small portion of CH3COOH (aq) ionises) when dissolved in water. Thus, more free ions to carry charge.
Abdelkarim
HCl being an strong acid will fully ionize in water thus producing more mobile ions for electrical conduction than the carboxylic acid
Valentine
differiante between a weak and a strong acid
david
how can I tell when an acid is weak or Strong
Amarachi
an aqueous solution of copper sulphate was electrolysed between graphite electrodes. state what was observed at the cathode
write the equation for the reaction that took place at the anode
Bakanya
what is enthalpy of combustion
Bakanya
Enthalpy change of combustion: It is the enthalpy change when 1 mole of substance is combusted with excess oxygen under standard conditions. Elements are in their standard states. Conditions: pressure = 1 atm Temperature =25°C
Abdelkarim
Observation at Cathode: Cu metal deposit (pink/red solid).
Abdelkarim
Equation at Anode: (SO4)^2- + 4H^+ + 2e^- __> SO2 + 2H2O
Abdelkarim
Equation : CuSO4 -> Cu^2+ + SO4^2- equation at katode: 2Cu^2+ + 4e -> 2Cu equation at anode: 2H2O -> 4H+ + O2 +4e at the anode which reacts is water because SO4 ^ 2- cannot be electrolyzed in the anode
Niken
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CC-MAIN-2020-50
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latest
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en
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