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# Chapt18
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### Chapt18
1. 1. Chapter 18: Money Supply& Money Demand
2. 2. Federal Reserve System, FEDThe central bank of the U.S.Independent decision making unit with regionalbanksIn charge of money supply management andeconomic stabilization
3. 3. Money Supply M=C+DC = Currency: coins & bills (25%)D = Demand Deposits: checking account deposits (75%)
4. 4. Money Supply LineThe quantity of money in circulation is controlledby the central bank in real value Interest Rate (%) (M/P)s 10 5 80 Quantity of Money
5. 5. Fractional Banking SystemBanks are required by law to hold a percentage of alldeposits with the FED to be able to return the deposits:– R = reserves: deposits– RR = required reserves: reserves held by the FED– rr = reserve-deposit ratio: percentage determined by the FED (rr = R/D)– ER = excess reserves: reserves used by banks to lend or investment
6. 6. Fractional Banking System R = RR + ER RR = rr R ER = (1 – rr)RBanks’ lending and investing ER will create moneythrough a multiplier effect
7. 7. A Model of Money SupplyThe monetary base (B) is money held by thepublic in currency and by banks as reserves R B=C+RThe currency-deposit ratio (cr) is the amount ofcurrency people hold as a fraction of their demanddeposits cr = C / D
8. 8. A Model of Money SupplyDivide M = C + D by B = C + R: M/B = (C + D) / (C + R)Divide the numerator and denominator by D: M/B = (C/D + 1) / (C/D + R/D) M/B = (cr + 1) / (cr + rr) M = [(cr + 1) / (cr + rr)]B = m BDefine money multiplier m = (cr + 1) / (cr + rr),so far any\$1 increase in the monetary base, money supply increasesby \$m.
9. 9. A Model of Money SupplyExample: B = \$500 billion, cr = 0.6 and rr = 0.1: m=(0.6 + 1) / (0.6 +0.1) = 2.3 M = 2.3(500) = \$1,150 billion
10. 10. Change in Money SupplyThe money supply is proportional to the monetary base.So, an increase in B increases M m-fold.The lower the reserve-deposit ratio, the more loans banksmake and the higher is the money multiplierThe lower the currency deposit ratio, the fewer dollars ofthe monetary base the public holds as currency and thelower is the money multiplier
11. 11. Tools of Monetary PolicyReserve-deposit ratio: ratio of cash reserves todeposits that banks are required to maintainBy lowering the ratio, banks will have morereserves to lend and invest, increasing the moneysupply
12. 12. Tools of Monetary PolicyDiscount rate: rate of interest the FED charges onloans to banksBy lowering the rate, banks encourage borrowingfrom the FED and lending to the public, increasingthe money supply
13. 13. Tools of Monetary PolicyOpen Market Operations: FED’s purchases andsales of government bondsBy purchasing bonds and paying the sellers, theFED increases the money supply
14. 14. Expansionary Monetary PolicyIncrease the money supply by any one orcombination of the above toolsReduce the interest rate to encourage investmentIncrease employment & income
15. 15. Money DemandThe amount of money demanded for transactionand speculative purposes depends: personalincome and interest rateAt any level of personal income, quantitydemanded of money is a negative function ofinterest rate; (M/P)d = L(i, Y)
16. 16. Money Demand LineInterest Rate (%) M/P = L(Y, i) Y = income 10 i = interest rate 5 (M/P)d 80 100 Quantity of Money
17. 17. Money Market EquilibriumInterest Rate (%) (M/P)s 5 (M/P)d 80 Quantity of Money
18. 18. Expansionary Monetary PolicyInterest Rate (%) (M1/P)s (M2/P)s 5 4 (M/P)d 80 85 Quantity of Money
19. 19. Portfolio Theory of Money Demand (M/P)d = L(rs, rb, πe, W)M/P = real money balancesrs = expected real rate of return on stocksrb = expected real rate of return on bondsπe = expected rate of inflationW = real wealth(M/P)d is positively related to W and negatively affected by rs, rb, πe
20. 20. The Baumol-Tobin ModelDefine– Y = transactionary money an individual holds in bank– N = annual number of trips to bank an individual makes to withdraw money– F = cost of a trip to the bank– i = nominal interest rate
21. 21. Optimal ConditionsTotal cost of money withdrawal = Foregoneinterest + Cost of trips TC = iY/2N + FNThe annual number of trips that minimizes the totalcost of bank trips is N* = (iY/2F)1/2Average transactionary money holding is MH = Y /2N* = (YF/2i)1/2
22. 22. Optimal ConditionsCost Total cost of bank withdrawal Cost of bank trips = FN Foregone interest = iY/2N N* Number of trip to bank, N
23. 23. Speculative Demand for MoneyMoney individuals hold for investment in thefinancial marketNear money consists of non-monetary, interest-bearing assets such as stocks and bonds
24. 24. The Federal Funds RateThe short-term interest rate at which banks make loans toeach otherThe FED uses this rate as the basis for its interest ratepolicyTaylor’s rule for the determination of the nominal federalfunds rate: Inflation rate + 2 + 0.5(Inflation rate + 2) – 0.5(GDP gap)
25. 25. Actual vs. Taylor’s Rule
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# Thread: [SOLVED] Find values of a, b and c such that limit ... is finite?
1. ## [SOLVED] Find values of a, b and c such that limit ... is finite?
Find values of a, b and c such that:
lim (cos 4x + a cos 2x + b)/x^4 = Finite
x --> 0
2. Originally Posted by fardeen_gen
Find values of a, b and c such that:
lim (cos 4x + a cos 2x + b)/x^4 = Finite
x --> 0
You have missed a 'c' somewhere, thats why we will not be getting any answer correctly. But I will tell you the general idea...
Let L be the limit.
Observe that if $\displaystyle g(0) = 0$ and $\displaystyle \lim_{x \to 0} \frac{f(x)}{g(x)} =$ L, then $\displaystyle \lim_{x \to 0} f(x) = 0$.
Applying it here, we get $\displaystyle 1 + a + b = 0$.
Now apply L'Hospital's rule, to get another form for L. Do the same process again.
Alternate trick is to substitute the power series for cos and choosing coefficients such that the limit exists.
3. No missing 'c' according to text.
4. Originally Posted by fardeen_gen
Find values of a, b and c such that:
lim (cos 4x + a cos 2x + b)/x^4 = Finite
x --> 0
Originally Posted by fardeen_gen
No missing 'c' according to text.
Then why does the question say find a, b and c?
5. You can however use power series or L'Hospitals to get the following equations:
a+b+1 = 0 and 8 + 2a = 0 and thus a = -4 and b = 3. So the limit L = 8.
To do this using power series, write $\displaystyle \cos t = 1 - \frac{t^2}{2!} + \frac{t^4}{4!} + (t^6$ ke terms$\displaystyle ....)$. Then group terms with same powers in the numerator. All terms with constant and power of x^2 must go to 0. That will give you the above two equations....
6. The question was a general one, under which the specific problem appears(there are some which involve a, b and c, a and b, b and c... and so on).
Thanks for the help!!
I got the same answer right now!
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# find the values of a and b such that limit
Click on a term to search for related topics.
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# The line segment joining ( -3,-4) and (6,11) is to be divided in to three (3) equal parts. Find the coordinates of the point of the part closest to (-3,-4).
## The x- and y- changes from the first point to the last are 9 and 15
So, add 1/3 of those values to the first point.
## To find the coordinates of the point on the line segment closest to (-3,-4) when dividing it into three equal parts, you can use the concept of finding a point on a line segment using the section formula.
The section formula states that if we have two points (x1, y1) and (x2, y2) on a line segment, and we want to divide the line segment in a ratio of m:n, then the coordinates of the point dividing the line segment in that ratio can be found using the following formulas:
x = (n * x1 + m * x2) / (m + n)
y = (n * y1 + m * y2) / (m + n)
In this case, we want to divide the line segment into three equal parts, so the ratio is 1:2. The given points are (-3,-4) and (6,11).
Using the section formula:
x = (2 * (-3) + 1 * 6) / (1 + 2)
= (-6 + 6) / 3
= 0 / 3
= 0
y = (2 * (-4) + 1 * 11) / (1 + 2)
= (-8 + 11) / 3
= 3 / 3
= 1
Therefore, the coordinates of the point closest to (-3,-4) is (0, 1).
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# Extra Class Exam Question Pool
effective 7/01/2008 thru 6/30/2012
Show: Unseen questions Weak questions Review questions Learned questions Incorrect answer choices
E8B: Modulation and demodulation: modulation methods; modulation index and deviation ratio; pulse modulation; frequency and time division multiplexing
E8B01: What is the term for the ratio between the frequency deviation of an RF carrier wave, and the modulating frequency of its corresponding FM-phone signal?
Modulation index
FM compressibility
Quieting index
Percentage of modulation
E8B02: How does the modulation index of a phase-modulated emission vary with RF carrier frequency (the modulated frequency)?
It does not depend on the RF carrier frequency
It increases as the RF carrier frequency increases
It decreases as the RF carrier frequency increases
It varies with the square root of the RF carrier frequency
E8B03: What is the modulation index of an FM-phone signal having a maximum frequency deviation of 3000 Hz either side of the carrier frequency, when the modulating frequency is 1000 Hz?
3
0.3
3000
1000
E8B04: What is the modulation index of an FM-phone signal having a maximum carrier deviation of plus or minus 6 kHz when modulated with a 2-kHz modulating frequency?
3
6000
2000
1/3
E8B05: What is the deviation ratio of an FM-phone signal having a maximum frequency swing of plus-or-minus 5 kHz and accepting a maximum modulation rate of 3 kHz?
1.67
60
0.167
0.6
E8B06: What is the deviation ratio of an FM-phone signal having a maximum frequency swing of plus or minus 7.5 kHz and accepting a maximum modulation frequency of 3.5 kHz?
2.14
0.214
0.47
47
E8B07: When using a pulse-width modulation system, why is the transmitter's peak power greater than its average power?
The signal duty cycle is less than 100%
The signal reaches peak amplitude only when voice modulated
The signal reaches peak amplitude only when voltage spikes are generated within the modulator
The signal reaches peak amplitude only when the pulses are also amplitude modulated
E8B08: What parameter does the modulating signal vary in a pulse-position modulation system?
The time at which each pulse occurs
The number of pulses per second
The amplitude of the pulses
The duration of the pulses
E8B09: How are the pulses of a pulse-modulated signal usually transmitted?
A pulse of relatively short duration is sent; a relatively long period of time separates each pulse
A pulse of relatively long duration is sent; a relatively short period of time separates each pulse
A group of short pulses are sent in a relatively short period of time; a relatively long period of time separates each group
A group of short pulses are sent in a relatively long period of time; a relatively short period of time separates each group
E8B10: What is meant by deviation ratio?
The ratio of the maximum carrier frequency deviation to the highest audio modulating frequency
The ratio of the audio modulating frequency to the center carrier frequency
The ratio of the carrier center frequency to the audio modulating frequency
The ratio of the highest audio modulating frequency to the average audio modulating frequency
E8B11: Which of these methods can be used to combine several separate analog information streams into a single analog radio frequency signal?
Frequency division multiplexing
Frequency shift keying
A diversity combiner
Pulse compression
E8B12: Which of the following describes frequency division multiplexing?
Two or more information streams are merged into a "baseband", which then modulates the transmitter
The transmitted signal jumps from band to band at a predetermined rate
The transmitted signal is divided into packets of information
Two or more information streams are merged into a digital combiner, which then pulse position modulates the transmitter
E8B13: What is time division multiplexing?
Two or more signals are arranged to share discrete time slots of a digital data transmission
Two or more data streams are assigned to discrete sub-carriers on an FM transmitter
Two or more data streams share the same channel by transmitting time of transmission as the sub-carrier
Two or more signals are quadrature modulated to increase bandwidth efficiency
Color key: ● = Unseen ● = Weak ● = Review ● = Learned ● = Incorrect answer
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# 5.3: Cauchy’s Form of the Remainder
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##### Learning Objectives
• Explain Cauchy's form of the remainder
In his 1823 work, Résumé des le¸cons données à l’ecole royale polytechnique sur le calcul infintésimal, Augustin Cauchy provided another form of the remainder for Taylor series.
##### Theorem $$\PageIndex{1}$$: Cauchy’s Form of the Remainder
Suppose $$f$$ is a function such that $$f^{(n+1)}(t)$$ is continuous on an interval containing $$a$$ and $$x$$. Then
$f(x) - \left ( \sum_{j=0}^{n}\dfrac{f^{(j)}(a)}{j!}(x-a)^j \right ) = \dfrac{f^{(n+1)}(c)}{n!}(x-c)^n(x-a)$
where $$c$$ is some number between $$a$$ and $$x$$.
##### Exercise $$\PageIndex{1}$$
Prove Theorem $$\PageIndex{1}$$ using an argument similar to the one used in the proof of Theorem 5.2.1. Don’t forget there are two cases to consider.
Using Cauchy’s form of the remainder, we can prove that the binomial series
$1 + \dfrac{1}{2}x + \dfrac{\dfrac{1}{2}\left ( \dfrac{1}{2} -1 \right )}{2!}x^2 + \dfrac{\dfrac{1}{2}\left ( \dfrac{1}{2} -1 \right )\left ( \dfrac{1}{2} -2 \right )}{3!}x^3 + \cdots$
converges to $$\sqrt{1+x}$$ for $$x ∈ (-1,0)$$. With this in mind, let $$x$$ be a fixed number with $$-1 < x < 0$$ and consider that the binomial series is the Maclaurin series for the function $$f(x) = (1 + x)^\dfrac{1}{2}$$. As we saw before,
$f^{(n+1)}(t) = \left ( \dfrac{1}{2} \right )\left ( \dfrac{1}{2} -1 \right )\cdots \left ( \dfrac{1}{2} -n \right )(1+t)^{\dfrac{1}{2} - (n+1)}$
so the Cauchy form of the remainder is given by
$0 \leq \left | \dfrac{f^{(n+1)}(c)}{n!}(x-c)^n(x-0) \right | = \left | \dfrac{\dfrac{1}{2}\left ( \dfrac{1}{2} -1 \right )\cdots \left ( \dfrac{1}{2} -n \right )}{n!} \dfrac{(x-c)^n}{(1+c)^{n+\dfrac{1}{2}}}\cdot x\right |$
where $$c$$ is some number with $$x ≤ c ≤ 0$$. Thus we have
\begin{align*} 0 &\leq \left | \dfrac{\left ( \dfrac{1}{2} \right )\left ( \dfrac{1}{2} -1 \right )\cdots \left ( \dfrac{1}{2} -n \right )}{n!} \dfrac{(x-c)^n x}{(1+c)^{n + \dfrac{1}{2}}} \right | \\ &= \dfrac{\left ( \dfrac{1}{2} \right )\left ( 1 -\dfrac{1}{2} \right )\cdots \left ( n - \dfrac{1}{2} \right )}{n!} \dfrac{\left |x-c \right |^n\left | x \right |}{(1+c)^{n + \dfrac{1}{2}}}\\ &= \dfrac{\left ( \dfrac{1}{2} \right )\left ( \dfrac{1}{2} \right )\left ( \dfrac{3}{2} \right )\left ( \dfrac{5}{2} \right )\cdots \left ( \dfrac{2n-1}{2} \right )}{n!})\dfrac{(c-x)^n}{(1+c)^n}\dfrac{\left |x \right |}{\sqrt{1+c}}\\ &\leq \dfrac{1\cdot 1\cdot 3\cdot 5\cdots (2n-1)}{2^{n+1}n!} \left (\dfrac{c-x}{1+c} \right )^n\dfrac{\left |x \right |}{\sqrt{1+c}}\\ &= \dfrac{1\cdot 1\cdot 3\cdot 5\cdot (2n-1)}{2\cdot 2\cdot 4\cdot 6\cdots 2n} \left (\dfrac{c-x}{1+c} \right )^n\dfrac{\left |x \right |}{\sqrt{1+c}}\\ &= \dfrac{1}{2} \cdot \dfrac{1}{2} \cdot \dfrac{3}{4} \cdot \dfrac{5}{6} \cdots \dfrac{2n-1}{2n}\cdot \left (\dfrac{c-x}{1+c} \right )^n\dfrac{\left |x \right |}{\sqrt{1+c}}\\ &\leq \left (\dfrac{c-x}{1+c} \right )^n\dfrac{\left |x \right |}{\sqrt{1+c}} \end{align*}
Notice that if $$-1 < x ≤ c$$, then $$0 < 1 + x ≤ 1 + c$$. Thus $$0 < \dfrac{1}{1+c} \leq \dfrac{1}{1+x}$$ and $$\dfrac{1}{\sqrt{1+c}} \leq \dfrac{1}{\sqrt{1+x}}$$. Thus we have
$0 \leq \left |\dfrac{\left ( \dfrac{1}{2} \right )\left ( \dfrac{1}{2} -1 \right )\cdots \left ( \dfrac{1}{2} -n \right )}{n!} \dfrac{(x-c)^n x}{(1+c)^{n + \dfrac{1}{2}}} \right | \leq \left (\dfrac{c-x}{1+c} \right )^n\dfrac{\left |x \right |}{\sqrt{1+c}}$
##### Exercise $$\PageIndex{2}$$
Suppose $$-1 < x ≤ c ≤ 0$$ and consider the function $$g(c) = \dfrac{c-x}{1+x}$$. Show that on $$[x,0]$$, $$g$$ is increasing and use this to conclude that for $$-1 < x ≤ c ≤ 0$$,
$\dfrac{c-x}{1+x} \leq \left | x \right | \nonumber$
Use this fact to finish the proof that the binomial series converges to $$\sqrt{1+x}$$ for $$-1 < x < 0$$.
The proofs of both the Lagrange form and the Cauchy form of the remainder for Taylor series made use of two crucial facts about continuous functions.
First, we assumed the Extreme Value Theorem: Any continuous function on a closed bounded interval assumes its maximum and minimum somewhere on the interval.
Second, we assumed that any continuous function satisfied the Intermediate Value Theorem: If a continuous function takes on two different values, then it must take on any value between those two values.
Figure $$\PageIndex{1}$$: Augustin Cauchy
Mathematicians in the late 1700’s and early 1800’s typically considered these facts to be intuitively obvious. This was natural since our understanding of continuity at that time was, solely, intuitive. Intuition is a useful tool, but as we have seen before it is also unreliable. For example consider the following function.
$f(x) = \begin{cases} x\sin \left (\dfrac{1}{x} \right ) & \text{ if } x \neq 0 \\ 0 & \text{ if } x= 0 \end{cases}$
Is this function continuous at $$0$$? Near zero its graph looks like this:
Figure $$\PageIndex{2}$$: Graph of the function above.
but this graph must be taken with a grain of salt as $$\sin \left (\dfrac{1}{x} \right )$$ oscillates infinitely often as $$x$$ nears zero. No matter what your guess may be, it is clear that it is hard to analyze such a function armed with only an intuitive notion of continuity. We will revisit this example in the next chapter.
As with convergence, continuity is more subtle than it first appears. We put convergence on solid ground by providing a completely analytic definition in the previous chapter. What we need to do in the next chapter is provide a completely rigorous definition for continuity.
This page titled 5.3: Cauchy’s Form of the Remainder is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Eugene Boman and Robert Rogers (OpenSUNY) via source content that was edited to the style and standards of the LibreTexts platform.
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# algebra-estimate
by john
(florida)
what is the estimation for 8,496-3,846
CHANGE the EXACT VALUES INTO APPROXIMATE NUMBERS to make numbers easier to use, and then subtract the approximate numbers.
### Comments for algebra-estimate
Jan 18, 2011 Algebra - Estimate by: Staff The question: by John (Florida)What is the estimation for 8,496-3,846?The answer:This is an important question because estimating is used in every area of life (not just in math). Estimating a number will involve ROUNDING the number - CHANGING an EXACT VALUE INTO an APPROXIMATE NUMBER. We use approximate numbers to make numbers easier to use. For example, when someone states their age they generally state their age at their last birthday - this is generally an estimate, since they may be several months older then their last birthday.However, ESTIMATING introduces an ERROR. While numbers which have been rounded make them easier to use, changing an exact number into an approximate number creates a problem. Estimating a subtraction will involve ROUNDING BOTH of the NUMBERS involved BEFORE completing THE SUBTRACTION. TWO ERRORS are introduced in this process.The point and the question is this: what degree of accuracy do you require? (or, how large an error can you tolerate?)As you already know, the exact answer to the subtraction is:8,496 - 3,846 = 4,6508,496 - no error introduced since this is an exact number3,846 - no error introduced since this is an exact numberIf you round your numbers to the nearest 10 before you subtract, your estimate becomes:8,500 - 3,850 = 4,6508,496 - error introduced of +4 since 8,500 is greater than 8,4963,846 ? error introduced of +4 since 3,850 is greater than 3,846These two errors cancel one another when the numbers are subtractedIf you round your numbers to the nearest 100, you estimate becomes:8,500 - 3,800 = 4,7008,500 - error introduced of +4 since 8,500 is greater than 8,4963,800 - error introduced of -46 since 3,800 is smaller than 3,846These two errors do not errors cancel one. Together, they create an even larger error (+50) when the numbers are subtracted.If you round your numbers to the nearest 1000, your estimate becomes:8,000 - 4,000 = 4,0008,000 - error introduced of -496 since 8,000 is smaller than 8,4964,000 - error introduced of +154 since 4,000 is greater than 3,846These two errors do not errors cancel one. Together, they create an even larger error (-650) when the numbers are subtracted.In conclusion, you must answer the question posed earlier: how accurate do you require your estimate to be? (In exchange for using an estimate to make the numbers easier to subtract, how large an error can you tolerate?)Thanks for writing.Staff www.solving-math-problems.com
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0
# What are the least common multiple of 20 and8 and60 and6?
Updated: 8/20/2019
Wiki User
11y ago
The Least Common Multiple (LCM) for 20 8 60 6 is 120.
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18
30
### What is the least common multiple of 9 and6?
The Least Common Multiple (LCM) for 9 6 is 18.
60
The LCM is: 12
330
1 and 3
### What is the LCM of 3 5 and6?
60 is the lowest common multiple of 4, 5 and 6.
1 and 3
### Greatet common factor of 12 and6?
Since 6 is a factor of 12, it is automatically the GCF.
### What is the common denominator for 98 and6?
If that's 98 and 6, the LCD is 294. If that's 9, 8 and 6, the LCD is 72.
12.
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Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > lsmcntzr Structured version Visualization version GIF version
Theorem lsmcntzr 17916
Description: The "subgroups commute" predicate applied to a subgroup sum. (Contributed by Mario Carneiro, 21-Apr-2016.)
Hypotheses
Ref Expression
lsmcntz.p = (LSSum‘𝐺)
lsmcntz.s (𝜑𝑆 ∈ (SubGrp‘𝐺))
lsmcntz.t (𝜑𝑇 ∈ (SubGrp‘𝐺))
lsmcntz.u (𝜑𝑈 ∈ (SubGrp‘𝐺))
lsmcntz.z 𝑍 = (Cntz‘𝐺)
Assertion
Ref Expression
lsmcntzr (𝜑 → (𝑆 ⊆ (𝑍‘(𝑇 𝑈)) ↔ (𝑆 ⊆ (𝑍𝑇) ∧ 𝑆 ⊆ (𝑍𝑈))))
Proof of Theorem lsmcntzr
StepHypRef Expression
1 lsmcntz.p . . 3 = (LSSum‘𝐺)
2 lsmcntz.t . . 3 (𝜑𝑇 ∈ (SubGrp‘𝐺))
3 lsmcntz.u . . 3 (𝜑𝑈 ∈ (SubGrp‘𝐺))
4 lsmcntz.s . . 3 (𝜑𝑆 ∈ (SubGrp‘𝐺))
5 lsmcntz.z . . 3 𝑍 = (Cntz‘𝐺)
61, 2, 3, 4, 5lsmcntz 17915 . 2 (𝜑 → ((𝑇 𝑈) ⊆ (𝑍𝑆) ↔ (𝑇 ⊆ (𝑍𝑆) ∧ 𝑈 ⊆ (𝑍𝑆))))
7 subgrcl 17422 . . . . 5 (𝑆 ∈ (SubGrp‘𝐺) → 𝐺 ∈ Grp)
8 grpmnd 17252 . . . . 5 (𝐺 ∈ Grp → 𝐺 ∈ Mnd)
94, 7, 83syl 18 . . . 4 (𝜑𝐺 ∈ Mnd)
10 eqid 2610 . . . . . 6 (Base‘𝐺) = (Base‘𝐺)
1110subgss 17418 . . . . 5 (𝑇 ∈ (SubGrp‘𝐺) → 𝑇 ⊆ (Base‘𝐺))
122, 11syl 17 . . . 4 (𝜑𝑇 ⊆ (Base‘𝐺))
1310subgss 17418 . . . . 5 (𝑈 ∈ (SubGrp‘𝐺) → 𝑈 ⊆ (Base‘𝐺))
143, 13syl 17 . . . 4 (𝜑𝑈 ⊆ (Base‘𝐺))
1510, 1lsmssv 17881 . . . 4 ((𝐺 ∈ Mnd ∧ 𝑇 ⊆ (Base‘𝐺) ∧ 𝑈 ⊆ (Base‘𝐺)) → (𝑇 𝑈) ⊆ (Base‘𝐺))
169, 12, 14, 15syl3anc 1318 . . 3 (𝜑 → (𝑇 𝑈) ⊆ (Base‘𝐺))
1710subgss 17418 . . . 4 (𝑆 ∈ (SubGrp‘𝐺) → 𝑆 ⊆ (Base‘𝐺))
184, 17syl 17 . . 3 (𝜑𝑆 ⊆ (Base‘𝐺))
1910, 5cntzrec 17589 . . 3 (((𝑇 𝑈) ⊆ (Base‘𝐺) ∧ 𝑆 ⊆ (Base‘𝐺)) → ((𝑇 𝑈) ⊆ (𝑍𝑆) ↔ 𝑆 ⊆ (𝑍‘(𝑇 𝑈))))
2016, 18, 19syl2anc 691 . 2 (𝜑 → ((𝑇 𝑈) ⊆ (𝑍𝑆) ↔ 𝑆 ⊆ (𝑍‘(𝑇 𝑈))))
2110, 5cntzrec 17589 . . . 4 ((𝑇 ⊆ (Base‘𝐺) ∧ 𝑆 ⊆ (Base‘𝐺)) → (𝑇 ⊆ (𝑍𝑆) ↔ 𝑆 ⊆ (𝑍𝑇)))
2212, 18, 21syl2anc 691 . . 3 (𝜑 → (𝑇 ⊆ (𝑍𝑆) ↔ 𝑆 ⊆ (𝑍𝑇)))
2310, 5cntzrec 17589 . . . 4 ((𝑈 ⊆ (Base‘𝐺) ∧ 𝑆 ⊆ (Base‘𝐺)) → (𝑈 ⊆ (𝑍𝑆) ↔ 𝑆 ⊆ (𝑍𝑈)))
2414, 18, 23syl2anc 691 . . 3 (𝜑 → (𝑈 ⊆ (𝑍𝑆) ↔ 𝑆 ⊆ (𝑍𝑈)))
2522, 24anbi12d 743 . 2 (𝜑 → ((𝑇 ⊆ (𝑍𝑆) ∧ 𝑈 ⊆ (𝑍𝑆)) ↔ (𝑆 ⊆ (𝑍𝑇) ∧ 𝑆 ⊆ (𝑍𝑈))))
266, 20, 253bitr3d 297 1 (𝜑 → (𝑆 ⊆ (𝑍‘(𝑇 𝑈)) ↔ (𝑆 ⊆ (𝑍𝑇) ∧ 𝑆 ⊆ (𝑍𝑈))))
Colors of variables: wff setvar class Syntax hints: → wi 4 ↔ wb 195 ∧ wa 383 = wceq 1475 ∈ wcel 1977 ⊆ wss 3540 ‘cfv 5804 (class class class)co 6549 Basecbs 15695 Mndcmnd 17117 Grpcgrp 17245 SubGrpcsubg 17411 Cntzccntz 17571 LSSumclsm 17872 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-8 1979 ax-9 1986 ax-10 2006 ax-11 2021 ax-12 2034 ax-13 2234 ax-ext 2590 ax-rep 4699 ax-sep 4709 ax-nul 4717 ax-pow 4769 ax-pr 4833 ax-un 6847 ax-cnex 9871 ax-resscn 9872 ax-1cn 9873 ax-icn 9874 ax-addcl 9875 ax-addrcl 9876 ax-mulcl 9877 ax-mulrcl 9878 ax-mulcom 9879 ax-addass 9880 ax-mulass 9881 ax-distr 9882 ax-i2m1 9883 ax-1ne0 9884 ax-1rid 9885 ax-rnegex 9886 ax-rrecex 9887 ax-cnre 9888 ax-pre-lttri 9889 ax-pre-lttrn 9890 ax-pre-ltadd 9891 ax-pre-mulgt0 9892 This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-3or 1032 df-3an 1033 df-tru 1478 df-ex 1696 df-nf 1701 df-sb 1868 df-eu 2462 df-mo 2463 df-clab 2597 df-cleq 2603 df-clel 2606 df-nfc 2740 df-ne 2782 df-nel 2783 df-ral 2901 df-rex 2902 df-reu 2903 df-rmo 2904 df-rab 2905 df-v 3175 df-sbc 3403 df-csb 3500 df-dif 3543 df-un 3545 df-in 3547 df-ss 3554 df-pss 3556 df-nul 3875 df-if 4037 df-pw 4110 df-sn 4126 df-pr 4128 df-tp 4130 df-op 4132 df-uni 4373 df-iun 4457 df-br 4584 df-opab 4644 df-mpt 4645 df-tr 4681 df-eprel 4949 df-id 4953 df-po 4959 df-so 4960 df-fr 4997 df-we 4999 df-xp 5044 df-rel 5045 df-cnv 5046 df-co 5047 df-dm 5048 df-rn 5049 df-res 5050 df-ima 5051 df-pred 5597 df-ord 5643 df-on 5644 df-lim 5645 df-suc 5646 df-iota 5768 df-fun 5806 df-fn 5807 df-f 5808 df-f1 5809 df-fo 5810 df-f1o 5811 df-fv 5812 df-riota 6511 df-ov 6552 df-oprab 6553 df-mpt2 6554 df-om 6958 df-1st 7059 df-2nd 7060 df-wrecs 7294 df-recs 7355 df-rdg 7393 df-er 7629 df-en 7842 df-dom 7843 df-sdom 7844 df-pnf 9955 df-mnf 9956 df-xr 9957 df-ltxr 9958 df-le 9959 df-sub 10147 df-neg 10148 df-nn 10898 df-2 10956 df-ndx 15698 df-slot 15699 df-base 15700 df-sets 15701 df-ress 15702 df-plusg 15781 df-0g 15925 df-mgm 17065 df-sgrp 17107 df-mnd 17118 df-submnd 17159 df-grp 17248 df-minusg 17249 df-subg 17414 df-cntz 17573 df-lsm 17874 This theorem is referenced by: (None)
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# Hack #38. Hack Excel's Date and Time Features
Excel's date and time feature is great if you're creating simple spreadsheets, but they can cause problems for more advanced projects. Fortunately, there are ways to get around Excel's assumptions when they don't meet your needs.
Excel (by default) uses the 1900 date system. This means the date 1 Jan 1900 has an underlying numeric value of 1, 2 Jan 1900 has a value of 2, and so forth. These values are called serial values in Excel, and they enable you to use dates in calculations.
Times are very similar, but Excel treats times as decimal fractions, with 1 being the time 24:00 or 00:00. 18:00 has a numeric value of 0.75 because it is three-quarters of 24 hours.
To see the numeric value of a date and/or a time, format the cell containing the value as General. For example, the date and time 3/July/2002 3:00:00 PM has a numeric value of 37440.625, with the number after the decimal representing the time, and the 37440 representing the serial value for 3/July/2002.
You can add times by using the `SUM` function (or a simple plus sign). Therefore, `=SUM(A1:A5)` would result in Total Hours if A1:A5 contained valid times. There is, however, a big "Gotcha!" Unless told otherwise, Excel will not add past 24 hours. This is because when a time value exceeds 24 hours (a true value of 1), it rolls into a new day and starts again. To force Excel not to default back to a new day after 24 hours, you can use a cell format of 37:30:55 ...
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If you've ever struggled with finding the median of a set of numbers, you're not alone. But fear not! Finding the median is easier than you think. Follow these simple steps and you'll be on your way to mastering this math concept in no time.
# Discover the Simple Steps: How to Find the Median Easily
Imagine you are standing in a crowded room filled with people of different heights. Some are towering over everyone while others are barely visible. How do you determine the height that lies exactly in the middle? The answer lies in finding the median. The median is a statistical measure that divides a set of values or numbers into two halves, with an equal number of values on both sides. But what if you don’t know how to calculate it? Fear not! In this article, we’ll take you through some simple steps that will make finding the median a breeze. So, get ready to unravel the mystery of the median, and equip yourself with some handy math skills.
Mathematics is an essential subject that helps individuals comprehend various concepts. Among them is the mean, median, and mode. In statistical analysis, these three elements are significant in demonstrating central tendencies. Central tendencies aid to provide information on the middle position of numbers in a given dataset. This article will provide simple, easy-to-follow steps on how to find the median efficiently.
Content :
### The Importance of Understanding the Median
The median is a key element in understanding the statistical analysis of datasets. It helps to determine the central tendency of a set of data. By finding the median, one can provide insights into the middle position of values in a given dataset. The median provides the dividing line between the highest and lowest values of a dataset.
The median produces more accurate results than the mean and mode in certain cases. For instance, in a skewed dataset, a small number of extreme values can heavily influence the mean value, creating a misrepresentation. In contrast, the median provides an accurate representation of the middle position of a skewed dataset.
### Step 1: Arrange the Data Order
Before determining the median, it’s crucial to ensure that the data is arranged in order from lowest to highest value. Suppose you have a dataset consisting of 5, 6, 10, 8, and 7. First, arrange the data in order: 5, 6, 7, 8, and 10.
By arranging the data, it is easy to identify the middle value and enable one to move to the next step.
### Step 2: Identify the Middle Value
After arranging the dataset in order, the next step is determining the middle value. It involves identifying the value that falls exactly in the middle. In our example above, the center value is 7, indicating that it is the median.
However, when working with an even number of datasets, one must find the average of the two middle numbers. Suppose you have a dataset consisting of 4, 7, 8, and 12. First, arrange the numbers: 4, 7, 8, and 12. The middle numbers are 7 and 8. To find the median, calculate the average of 7 and 8, which is 7.5. Therefore, the median is 7.
### Step 3: Determine the Median for Even-Numbered Datasets
As mentioned earlier, finding the median of datasets with an even number poses a challenge. However, by following some simple steps, it’s possible. Once you have arranged the dataset, identify the two middle numbers. Next, add these middle numbers and divide the result by two. The result is the median.
For example, a dataset of 2, 5, 7, and 10 is arranged as 2, 5, 7, and 10. The middle numbers are 5 and 7. Add the two numbers, which equals 12. Divide 12 by 2, giving a median of 6.
### Practice Problems: Putting the Steps into Action
The best way to understand the concept of finding the median is by practicing with various datasets. Here are some practice problems to tackle.
1. Find the median of the following dataset. 11, 14, 15, 18, 21
2. Find the median of the following dataset. 3, 6, 9, 12, and 15.
3. Find the median of the following dataset. 8, 9, 11, 12, 17 and 20.
### Tools and Techniques for Finding the Median
There are additional techniques or tools used to find the median, such as Microsoft Excel and data calculators such as graphing or scientific calculators. Excel provides formulas for calculating the median, while calculators simplify the manual process. These tools are convenient solutions for individuals handling larger datasets, which can be time-consuming when processing manually.
### Conclusion: Mastering the Basic Concept of Median
Statistical analysis plays a significant role in determining central tendencies for datasets, and the median is a crucial element. The ability to determine the median for a dataset can help provide insights into the middle position of numbers and simplify the analysis process. By following the simple steps outlined above, you can quickly learn how to find the median for datasets of varying sizes. Practice tackling the problems provided to master the concept further.
And there you have it – understanding how to find the median is easier than you might have thought. By taking these simple steps, you can quickly calculate the median of any set of data. Whether you’re working on a school project, analyzing business data, or simply curious about statistics, this skill will come in handy time and time again. So go ahead and try it out yourself – you might just find it to be just as simple as we’ve made it out to be. Happy calculating!
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# Backward-facing step simulation
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December 2, 2003, 23:35 Backward-facing step simulation #1 Keith Guest Posts: n/a I am trying to repeat the results in a paper by Gartling D.K(1990). The problem is a backward facing step problem. During the simulation, it seems that the velocity blows up after a certain time. I was wondering if the time step I have used is correct. My spatial step is 0.05m(for all three directions, x, y and z) and my time step is 0.0025 sec. the maximum velocity at the inlet is 1.5 m/s while the average velocity at the inlet is 1 m/s. Another question I have is regarding the adjustment of pressure field,after a solution is obtain, such that the pressure at the step is zero. Do I have to solve the pressure poisson equation again or do I just set the pressure at the step corner to be zero? Can anyone advice me on these issues? I really appreciate any help available.
December 15, 2003, 08:11 Re: Backward-facing step simulation #2 Jonas Bredberg Guest Posts: n/a Dear Sir, I've computed the Gartling-case myself. It is quite more difficult than people believe due to the low Reynolds number (800). It is semi-laminar, and hence it is rather important that you use a good discretization scheme. I found significant different result when using Hybrid compared to van Leer-scheme. You should opt for as high order as possible, with and an odd-order scheme to gain some stabilizing numerical diffusion. If you use a central scheme you may be in trouble. Furthermore you need to resolve the geometry, and at first disregard the third-direction. As an example I used 200 by 200 nodes (stretched). If you read the paper, it was focused as a code-validation case, and it works ok in 2D. Using an incompressible code the pressure could be fixed whereever in the domain, but prefererly in location where nothing happens with the flow, set it to zero at the outlet is may suggestion. If you need it to be zero at the step-corner, shift the pressure field in the post-routine. /Jonas
December 16, 2003, 16:31 Re: Backward-facing step simulation #3 Keith Guest Posts: n/a Jonas Bredberg, thanks for the reply. I really appreciate your reply. I am still not sure which pressure we have to set to zero at the outlet. Is it the hydrostatic pressure, dynamic pressure or both? Again, thank you very much to have answered my previous post.
December 17, 2003, 02:14 Re: Backward-facing step simulation #4 Jonas Bredberg Guest Posts: n/a Hi again, For an incompressible flow, you may set the solved pressure (delta static) to zero at some location. The correct pressure is then calculated (post-op) by adding the ambient pressure. The dynamic pressure is, as you know, a result of the flowfield (rho*U^2/2) - and should not be set to zero, as that implies zero velocities - maybe you introduced a bug yourself here. /Jonas
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Streetsblog.net
# One for the Dustbin: The 85th Percentile Rule in Traffic Engineering
11:27 AM EST on November 16, 2012
Have you ever heard of the 85 percent rule in traffic engineering? I hadn't until I tried to get the speed limit lowered on my residential street, which is home to a K-8 elementary school in Cleveland's Detroit Shoreway neighborhood.
That's when the local traffic engineer explained to me that the 35 mph speed limit -- which, in effect, means drivers can go 45 with impunity -- was appropriate for my street because 15 percent of drivers exceed 35 mph. Outrageously, that's actually how transportation engineers in most places determine the safe traveling speed: They watch how fast cars go. While some cities have made progress with different approaches, like 20 mph zones, by and large it's the most reckless 15 percent who set the speed limit -- not the residents of a street or the parents of school children who might be walking around, crossing the streets those drivers travel on.
So that's the 85th percentile rule, and Pedro Madruga at Copenhagenize explains more thoroughly why it is badly in need of rethinking:
It's a method that cities all over the planet use to determine speed limits. It's the standard. Nobody questions it. Certainly not the engineers and planners who, for decades, have been served it up and who have swallowed it whole during their studies. Which reminds us of the old traffic engineer joke: Why did the engineer cross the road? Because that's what they did last year.
This whole concept is based on a model called the "Solomon Curve" that was developed in 1964. The Solomon Curve states that those who travel too slow are at greatest risk in traffic, Madruga explains:
Imagine a street where the average speed is 50 km/h. If the speed limit is reduced by 5 km/h then, according to this archaic model, the drivers are allegedly exposed to a higher risk. What is most shocking is that this entire concept completely ignores pedestrians and cyclists. Another horrific conclusion from this graph is that when you increase the speed limit, the crash risk is alleged to be less than for slow speeds.
The Institute of Traffic Engineers wrote: “The 85th Percentile is how drivers vote with their feet”.They forgot to mention that, when it comes to establishing speed limits in cities, pedestrians and cyclists are excluded from this election. They don't even get the chance to go to the polls.
All this right now in 2012. In your street. With your tax money.
Elsewhere on the Network today: The Architect's Newspaper writes up a greenway proposal in Milwaukee that could help connect segregated communities, and which seems to be gaining momentum. Jarrett Walker at Human Transit gives his take on why the recent Freakonomics travesty -- "Save the Earth, Drive Your Car" -- is so wrong. And Cap'n Transit gets in on the fun with his own critique.
## More from Streetsblog USA
### Friday’s Headlines Go Back to the Future
If you liked the first Trump administration's transportation policies, you're going to love the second Trump administration's transportation policies.
July 19, 2024
### Advocates Share What It Takes to Fight Highway Expansions in Court
What does it take to sue your state DOT? Time, money, the right partners, and a little creativity, a recent survey of activists found.
July 19, 2024
### Friday Video: Paris Does it Again
Come for the bike-friendly streets, but stay for adopt-a-tree program and all the car-free school roadways.
July 19, 2024
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## Weekly Game Probabilities
Weekly game probabilities are available now at the nytimes.com Fifth Down. This week I take a closer look at Sunday Giants-Vikings match-up.
### 5 Responses to “Weekly Game Probabilities”
1. Brian says:
In your post, you mentioned that teams who are losing may find themselves pressured to throw high-risk passes, thus increasing their interception rate. It seems like we (you) have the data to assess the relationship between a team's WPA at the beginning of a passing play and the probability that that play ends in an interception.
2. Tommy Barnett says:
Looking forward to your coverage on the Giants/Vikings game
- Tommy Barnett
http://www.allgiants.com
3. Chris says:
Presumably the Vikings have now lost HFA so Giants are much more significant favourites now?
(Yes I'm British so I do spell 'favourite' with a 'u'!)
4. Anonymous says:
I have a question about a situation that just came up in the Texans Ravens game. To recap, Texans were down 15 and scored a touchdown with just over 6 minutes left. Now, they chose to kick the extra point to cut the lead to 8. Why wouldn't they go for two? I couldn't figure out the reason for the extra point. I remember this situation coming up in an earlier game and announcers kept saying they had to keep it a one possession game.
As I see it, if you assume that the percentage chance of converting a 2-point conversion is the same at any point in the game, then you would want to go for two after the first score. That way, if you make the 2-pt conversion, you know for sure that you only need a TD and extra point. However, if you miss the XP, then you know that it is still a 2-possession game.
So, assuming that whether or not you make the 2-pt conversion won't be affected by when you go for it, wouldn't you want that extra information of whether or not you make it earlier? I'd rather know I need an extra score with 6 minutes left then with 30 seconds or less left. The earlier I know, the better I could gameplan.
I feel like I must be missing some reason to kick the X-PT there. I'm thinking it may have to do with the possibility of the leading team scoring again before the end of the game, but I dunno. Any insight would be great.
5. Jero says:
Late to the game on this blog, however, as I had inquired, at the time, on the Fifth Down Blog- What was the Giant's WP vs. the Vikings on a neutral field?
Can this still be determined given we have a few more weeks worth of 2010 data?
Thank you.
Jeremy
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# Can you help me with problems
## M = 7; tn = 15.6; cvx_begin variable t variable theta(M) minimize(-(2tnt-square(tn))); subject to for ii=1:M 0 >= square(t)*inv_pos(1-theta(ii)); % (7b) 0 <= theta(ii) ; % (7f) theta(ii) <= 1 ; % (7f) end t >= 0; % (7g) cvx_end
When i run this code. It have a problems which i don’t know how to fix. Help me solve problems, please.
` 0 >= quad_over_lin(t,1-theta(ii))`
quad_over_lin Sum of squares over linear.
Z=quad_over_lin(X,Y), where X is a vector and Y is a scalar, is equal to
SUM(ABS(X).^2)./Y if Y is positive, and +Inf otherwise. Y must be real.
``````If X is a matrix, quad_over_lin(X,Y) is a row vector containing the values
of quad_over_lin applied to each column. If X is an N-D array, the operation
is applied to the first non-singleton dimension of X.
quad_over_lin(X,Y,DIM) takes the sum along the dimension DIM of X.
A special value of DIM == 0 is accepted here, which is automatically
replaced with DIM == NDIMS(X) + 1. This has the effect of eliminating
the sum; thus quad_over_lin( X, Y, NDIMS(X) + 1 ) = ABS( X ).^2 ./ Y.
In all cases, Y must be compatible in the same sense as ./ with the squared
sum; that is, Y must be a scalar or the same size as SUM(ABS(X).^2,DIM).
Disciplined convex programming information:
quad_over_lin is convex, nonmontonic in X, and nonincreasing in Y.
Thus when used with CVX expressions, X must be convex (or affine)
and Y must be concave (or affine).
``````
I suggest you carefully read the entire CVX Users’ Guide, where `quad_over_lin` and many other functions and language features are documented.
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211c7m6
# 211c7m6 - Introduction to Financial Accounting Chapter 7...
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Introduction to Financial Accounting AMIS 211 – Professor Marc Smith 1 Chapter 7, Module 6 Chapter 7, Module 6 Slide 1 AMIS 211 Introduction to Financial Accounting Professor Marc Smith Chapter 7 Module Chapter 7 Module 6 Hi everyone. Welcome back. Let’s see if we can put our knowledge as to how to estimate bad debts to test. And, let’s try a couple of examples. In this Module, let’s work through Example #3 from the Web site problems. And, let’s just read it together before we get started. Here is what it says: “XYZ Company had a \$250,000 balance in its Accounts Receivable on December 31 st , 2002. Additionally, they had a \$2,100 Credit Balance in their Allowance for Doubtful Accounts at Year-End 2002 before the Adjustment (AJE) for bad debt expense. During 2002, XYZ Company reported Credit Sales of \$800,000.”
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Introduction to Financial Accounting AMIS 211 – Professor Marc Smith 2 Chapter 7, Module 6 Required: Part A. “Let’s go ahead and let’s calculate the bad debt expense and the Net Realizable Value (NRV) of our Accounts Receivable using the Net Credit Sales Method.” And, Part A says: “Assume that we are estimating that 2% of our Net Credit Sales will be uncollectible.” Let’s go ahead to the next slide and let’s work the problem. Slide 2 Chapter 7 Module 6: Example #3, Part A Part A: Net Credit Sales Method 1. Bad debt expense = 800,000 x .02 Bad debt expense = \$16,000 2. Accounts Receivable - Allowance for Doubtful Accounts Net Realizable Value NOTE: To determine the allowance for doubtful accounts, always use a t-account Under the Net Credit Sales Method: we know that the bad debt expense estimate is really very straightforward. We simply take the Net Credit Sales for the year—given as \$800,000—and multiply (x) it by the percentage (%) we expect to be uncollectible—2%. Our bad debt expense is estimated as \$16,000 for the year.
Introduction to Financial Accounting AMIS 211 – Professor Marc Smith 3 Chapter 7, Module 6 So, using the Net Credit Sales Method to come up with bad debt expense is really pretty straight forward.
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# MATHEMATICS
416 Applications of Discrete Mathematics
graph shown in Figure 8(a). There (ignoring signs) the first label on t is a, so the augmenting chain vertex sequence ends with a, t. The first label on a is b, so the augmenting chain vertex sequence ends with b, a, t. Finally, the first label on b is s, so the augmenting chain vertex sequence is s, b, a, t.
Figure 8. Showing an augmenting chain being found using labels and being used to increase flow.
When a vertex n is labeled from vertex m, the second label is always the minimum of the second label on m and the amount c(m, n)−f(m, n) or f(n, m) which can be gotten to n from m through the edge. Since the second label on s is ∞, the result is that the second label always gives the largest amount that can get to the vertex labeled. Thus when t is labeled, the second label of t states the increase of flow given by the augmenting chain specified by the labels. For example, in Figure 8(a) the augmenting chain has vertex sequence s, b, a, t, and the amounts of flow that can be forced through the edges of the chain are 4, 3, and 2, successively, with a minimum of 2. Because of the way second labels are chosen, the second label on t is this amount 2.
Example 7 Find an augmenting chain in the graph shown in Figure 8(a), in which there is a preexisting flow of 3 units.
Solution: We label s with (−,∞), and from there we label b with (+s, 4). From b we can label a by canceling flow in (a, b), so a receives the label (−b, 3). From a we can label t with label (+a, 2). These labels are shown in Figure 8(a). Reading first labels backwards, we find the augmenting chain is s, b, a, t as discussed before, and the chain will increase the flow by 2 as the second label on t shows. The resulting flow is shown in Figure 8(b), where 3 units of flow go out of s to a and split there into 2 units that flow directly to t and one that goes to t through b. In addition, 2 units of flow go from s through b to t, making a total of 5 units of flow shown in Figure 8(b).
While we are labeling we can keep track of what is possible, without con- cerning ourselves whether there is actually a chain available from s to t along
Chapter 23 Network Flows 417
which new flow can go. If we can label t, then the flow can be increased; we will show that if we cannot label t, then the flow cannot be increased.
We are also under no obligation to watch for a chain of labels. Since we always label an unlabeled vertex from a labeled vertex, s being always labeled with (−,∞), we automatically form a tree (disregarding the directions of edges). Thus for each labeled vertex x, our labels describe a unique chain from s to x. An example of such a tree is shown by boldface edges in Figure 8(a). As shown in that figure, if we mark on the graph the edges used in labeling vertices, we find the augmenting chain with vertices s, b, a, t shown in bold lines in Figure 8(a).
Example 8 Find a flow in the graph of Figure 1.
Figure 9. (a) Figure 1 converted to a directed capacitated s,t-graph. (b) The first labeling and the associated tree for Example 1.
418 Applications of Discrete Mathematics
Solution: In Figure 9(a), the graph of Figure 1 is reproduced, except that each undirected edge has been replaced by two directed edges, one in each direction, each with the same capacity as the undirected edge. Also, the ca- pacities have been written as the first entries of number pairs, with 0 for flow as the second number.* Starting with the label (−,∞) at s, we label across edges as described above. First a is labeled with (+s, 4), and working from a we label b with (+a, min(4, 3)) = (+a, 3) and d with (+a, min(4, 2)) = (+a, 2), but we do not label s from a because s already has a label. Labeling from b, we do not need to label a, but we label c with (+b, min(3, 2)) = (+b, 2) and g with (+b, min(3, 1)) = (+b, 1). Then we label from c, placing (+c, min(2, 2)) = (+c, 2) on vertex e, but we place no label on d from c, since d already has a label. Next we label from d, placing label (+d, min(2, 1)) = (+d, 1) on ver- tex f . As before, we do not label a from d. Going on to vertex e, we label nothing from there, since all of its neighbors are already labeled. Next we label from f , placing label (+f, min(1, 2)) = (+f, 1) on vertex h. Since all neighbors of g are already labeled, we go on to h, from which we label t with (+h, min(1, 3)) = (+h, 1). Since t is labeled, we are done.
We use the labels to determine the chain from s to t that was found, along which a single unit of flow can flow (because the second label on t is 1). The first label on t is +h, showing that edge (h, t) is the last edge of the chain from s to t. The first label on h is +f , showing the next-to-the-last edge of the chain is (f, h). We continue reading backwards, using the first labels, obtaining the sequence “t, h, f, d, a, s” which reverses to give the chain s, a, d, f, h, t from s to t along which one unit of flow is added. The tree of edges used in our labeling is shown by bold lines in Figure 9(b), and Figure 10(a) shows the network with the flow we found.
To describe a procedure that can be programmed, we need a rule for de- ciding which edge to label across next. Many such rules are possible; the one we have adopted here and used in Example 8 is straightforward: Label from the alphabetically earliest vertex x which has a label and which meets an edge that has not yet been considered from vertex x, and label all possible vertices from x before going on to the next vertex. This rule is called the lexicographic ordering rule.
Once we have increased the flow, the labels we placed on the vertices other than s are wrong for the new combination of graph and flow, so we erase the vertex labels and start again from s. Again we explore the graph, searching for a chain from s to t along which we can increase the flow.
* In Figures 9(b) through 12, to simplify the figures we have shown only the di- rected edges that contain flow when flow is present and the undirected edges when it
is not.
Chapter 23 Network Flows 419
Figure 10. (a) Second labeling, associated tree, and first flow for Example 1. (b) Third labeling, associated tree, and second flow for Example 1.
Example 9 Increase the flow found in Example 8 as much as possible.
Solution: Labeling as before in the graph of Figure 10(a), we obtain the vertex labels shown in that figure. (Again the associated spanning tree is shown with bold lines.) Notice there that vertex f could not be labeled from d because c(d, f) − f(d, f) = 1 − 1 = 0. Hence f was labeled from e.
This time the augmenting chain is s, a, b, c, e, f, h, t, with one more unit of flow available. The resulting flow is shown in Figure 10(b). Again the labels are erased and new ones found. This time, since both c(d, f)− f(d, f) = 0 and c(e, f)− f(e, f) = 1− 1 = 0, f has no label when labeling from e is completed. Hence we do not try then to label from f , but rather label from g. Later f
420 Applications of Discrete Mathematics
does receive a label; indeed, its label shows that flow should be canceled to reach f . In a more complex graph, it would then be reasonable to label from f . In this graph, however, we reach t without using f again. The tree of edges used is shown in Figure 10(b) with bold edges, and the augmenting chain is s, a, b, g, h, t.
The resulting flow is shown in Figure 11. We label again, as shown in Figure 11, and we obtain the tree of edges used, again shown there. However, the tree does not contain t, so we have not found an augmenting chain. Does this mean there is no augmenting chain?
Figure 11. Fourth labeling, associated tree, and a minimum cut.
For the answer to that question, we must bring up machinery appropriate to Example 2.
Definition 3 Given a directed graph G with capacities on the edges, and given a nonempty subset A of its vertices such that V (G)−A is also nonempty, the set of all edges of G directed from vertices in A toward vertices in V (G)−A is called the cut (A, V (G)−A). The capacity c(A, V (G)−A) of this cut is the sum of the capacities of the edges in it. If s ∈ A and t ∈ V (G) − A, then we call the cut an s, t-cut.
Our objective in Example 2 is to find a cut (A, V (G) − A) such that c(A, V (G) − A) is minimum. In Theorem 3, we begin to tie the maximum flow and the minimum cut together, and in Theorem 4 we will complete the connection.
Chapter 23 Network Flows 421
Example 10 Discuss the cuts in the graph of Figure 11.
Solution: In Figure 11, the set of vertices in the tree is A = {s, a, b, c, d, e, g}. The set of edges directed from A to V (G)−A = {f, h, t} is the cut {(d, f), (e, f), (g, h)}. Notice that on each of these edges the flow equals the capacity and that on the return edges (f, d), (f, e), (h, g) in (V (G) − A, A) the flow is zero. The capacity of the cut is c(d, f)+ c(e, f)+ c(g, h) = 1 + 1 + 1 = 3. Notice also that this cut is an s, t-cut, and that the capacity of the cut is exactly the same as the total flow F (A) = F ({s}) = 3.
Theorem 3 The maximum flow from vertex s to vertex t �= s in a directed graph G with capacities on its edges is less than or equal to the capacity of any cut (A, V (G) − A) having s ∈ A and t �∈ A. Proof: Consider any s, t-cut (A, V (G)−A) of graph G. Since any units of flow from s to t must pass through an edge of this cut, it follows immediately that F ({s}) ≤ c(A, V (G) − A).
In Example 10, several facts are evident. First, since s has a label and t does not, any flow from s to t must pass through one or another of the three edges in the cut (A, V (G)−A). Second, since the return edges in (V (G)−A, A) are empty, any flow passing through the three edges directed away from A must continue on toward t. Third, with the edges directed away from A full and the edge directed toward A empty, there is no conceivable way that the flow we have found could be increased. In other words, we have found a maximum flow in this graph from s to t, and its amount is equal to the sum of the capacities of the edges in this cut separating s from t.
Returning to the graph shown in Figure 6(c), we see another example of this situation. Vertex s is labeled (−,∞) as usual. Since f(s, a) = c(s, a) and f(s, b) = c(s, b), the tree of labeled vertices includes only vertex s, so F ({s}) = c({s}, {a, b, t}) = 5.
Notice that there are no edges directed toward {s}, and both of the edges directed out of {s} in Figure 6(c) are full. Thus, it is clear that the flow to vertex t cannot be increased from its value of 5 units shown in Figure 6(c), and the maximum flow is equal to the sum of the capacities of the edges in the set {(s, a), (s, b)}. (We will generalize these observations in Theorem 4, following Algorithm 1.)
The observations made in considering Examples 9 and 10 motivate Algo- rithm 1.
422 Applications of Discrete Mathematics
ALGORITHM 1 Max-flow min-cut algorithm.
procedure Max-flow min-cut(G := directed graph with source s, sink t, and capacities on all edges)
label s with (−,∞) stop := 0 while stop = 0 begin
while t has no label begin
m := labeled vertex not previously examined, chosen by using an ordering rule
x := the second label on m e := edge incident with m and not previously examined
while at m if e = (m, n) and if n is not labeled and if c(e)−f(e) > 0
then label n with (+m, y) where y = min(c(e)−f(e), x) if e = (n, m) and if n is not labeled and if f(e) > 0 then
label n with (−m, y′) where y′ = min(f(e), x) if no such vertex m and edge e exist then stop := 1
end {vertex t has been labeled} if stop = 0 then p := x0, x1, . . . , xn, an augmenting chain
with s = x0 and xn = t {p is found using the labels as described in the text} if stop = 0 then v := the value of the second label on t i := 0 while i < n and stop = 0 begin
if the first label on xi+1 is +xi then replace the second label r on edge (xi, xi+1) with r + v
if the first label on xi+1 is −xi then replace the second label r′ on edge (xi+1, xi) with r′ − v
{r′ ≥ v by the use of minimum as we progress toward t} i := i + 1
end erase all vertex labels except the one on s
end {No more labeling is possible. By Theorem 4, the cut from the set of labeled vertices to the set of unlabeled vertices is a minimum cut and the flow F ({s}) is a maximum flow.}
Chapter 23 Network Flows 423
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# Testing¶
When we write code, we want to be sure that it is doing exactly what we thinking it is doing. Unfortunately, no one is perfect and this property is transferred to the code that we write. So we, as conscious, programmers, should write tests for all of your code ensuring that is it doing the right thing in all circumstances.
Typically we apply tests for our code that the function level, where there is, at least a test for each of the functions in our program. However, in order to achieve full test coverage, it may be necessary to have more than one test for a given function. Consider the code below, which performs simple temperature conversions, using logic to change the arithmetic as appropriate.
def temperature_conversion(value, unit):
"""
Temperature conversion that is dependent on
the input unit.
Args:
value (float): Temperature to be converted.
unit (str): Unit of the original temperature.
Returns:
(float): Temperature converted to Kelvin.
"""
if unit == 'C':
return value + 273.15
elif unit == 'F':
return (value + 459.67) * 5 / 9
else:
raise ValueError("The supported units are C or F.")
In this example, our code will raise an error if the unit argument is neither 'C' or 'F'.
We can now write some tests for this function. We know that the melting point is water is 0 oC, 273.15 K and 32 oF. Therefore our tests should reflect this knowledge.
import numpy as np
c = temperature_conversion(0, 'C')
np.testing.assert_almost_equal(c, 273.15)
print('Test 1 Passed!')
Test 1 Passed!
f = temperature_conversion(32, 'F')
np.testing.assert_almost_equal(f, 273.15)
print('Test 2 Passed!')
Test 2 Passed!
These two tests will cover two of the branches of our function. The np.testing.assert_almost_equal function is an element of the NumPy testing library. This test function checks if the two arguments are approximately equal. This must be approximate due to the floating point arithmetic problems. For an integer, string, or Boolean where the value does not need to be approximate, the np.testing.assert_equal function may be used. If the values are not almost equal, and error will be thrown.
np.testing.assert_almost_equal(c, 270.15)
print('Test 3 Passed!')
---------------------------------------------------------------------------
AssertionError Traceback (most recent call last)
/tmp/ipykernel_1967/3447443753.py in <module>
----> 1 np.testing.assert_almost_equal(c, 270.15)
2 print('Test 3 Passed!')
/opt/hostedtoolcache/Python/3.7.12/x64/lib/python3.7/site-packages/numpy/testing/_private/utils.py in assert_almost_equal(actual, desired, decimal, err_msg, verbose)
597 pass
598 if abs(desired - actual) >= 1.5 * 10.0**(-decimal):
--> 599 raise AssertionError(_build_err_msg())
600
601
AssertionError:
Arrays are not almost equal to 7 decimals
ACTUAL: 273.15
DESIRED: 270.15
Notice that the Python kernel never reaches the print statement.
However, so far our tests do not cover every possible operation of the function. We have a test for the if and elif branches, but not the else branch. Therefore, we must introduce another type of test, that checks if an error is raised.
np.testing.assert_raises(ValueError, temperature_conversion, 0, 'c')
print('Test 4 Passed!')
Test 4 Passed!
This function takes the expected error type, the function and the input arguments and will pass if the error is thrown under these circumstances.
## Test-driven development¶
Test-driven development (TDD) is a methodology for computer programming, where the tests are written first for a particular use case and then the operational code is written such that it passes these tests. This methodology is different from that applied to this point in the course, however, it is very popular and powerful when applied properly. In the exercise below, you are asked to try your hand at some test-driven development.
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Daycounter, Inc. Engineering Services Custom Firmware, Electronics Design, and PCB Layout Home Company Services Products Partners Clients Site Map Contact Us
How to wire an LED, Tutorial and Calculator
See our other Electronics Calculators.
This tutorial article describes how to use LEDs in electronics circuits, and how to calculate the current limiting resistor, and relates other useful hints on using LEDs.
Small incandescent lights, such as as flashlight bulbs may be directly connected to a voltage source such as a battery with out causing damage to the bulb, this is not the case with LEDs.
LEDs are current controlled devices, meaning that the intensity of their light output is proportional to the current passing through them. They also have a maximum current rating which may not be exceeded, otherwise they can be damaged. To limit the amount of current through an LED a current limiting resistor is typically inserted in series with it. The question then arises as to what the value of this resistor should be.
The calculations are simple and based on ohms law. But first it should be noted that LEDs are rated with a nominal forward voltage drop. If an LED is connect up such that it is emitting light, a multimeter can be used to measure a voltage drop across the LED. This voltage drop usually varies from 1.5V up to 5V, depending on the power output of the LED, and also the color. Low power LEDs tend to have low voltage drops, and high power LEDS have the higher voltage drops. Likewise blue or white LEDs tend have higher voltage drops, than red LEDs.
Another important reason we need to know the forward voltage drop is that it tells us the voltage we need to use in order to drive the LED. For example a 1.5V battery will not be able to drive an LED with a 2.5V voltage drop. The LED will simply not emit light.
Most LED datasheets will tell you the forward voltage drop. If a datasheet is not available it can be measured, as already described. Given the voltage of the source, the forward voltage drop, the maximum LED current, it is easy to compute the resistor based on Ohms Law:
Design Equation:
R= (Vs-Vf)/Imax
LED Resistor Calculator:
Voltage Source, Vs: (Volts) Forward LED voltage: Vf: (Volts) Maximum LED current, Imax (miliAmps) Resistor: (Ohms) Resistor Power: (W) LED Power: (W) Click here for standard resistor values
The best way to find LEDs is to use our free Electronics Component Directory, where a myriad of LED manufactures are listed.
Determining LED Polarity:
LEDs have a positive and negative terminal, also know as the anode and cathode. The cathode should be connected towards the ground or negative side of the driving voltage source, and the anode toward the positive side. LEDs usually have their cathode marked in some manner. On round through hole LEDs the cathode is usually denoted with a flat indentation on it's side, and the anode lead is often the longer lead.
It's usually harder to tell with smaller surface mount (SMT) LEDs. The best way to determine the polarity, is to use a multimeter, set to the diode/continuity setting. In most cases the multimeter will inject enough of a current into the LED to slightly light it, the black lead (or common) on the multimeter designates the cathode, and the red lead the anode.
Daycounter specializes in contract electronics design. Do you need some help on your project? Contact us to get a quote.
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An acquisition premium is the difference between the estimated real value of a company and the actual price paid to obtain it. Acquisition premium represents the increased cost of buying a target company during a merger and acquisition. There is no requirement that a company pay a premium for acquiring another company; depending on the situation, it may even get a discount.
Next Up
In mergers and acquisitions, the company paying to acquire another is known as the acquirer, while the company to be purchased or acquired is referred to as the target firm.
When a company decides it wants to acquire another, it will first attempt to estimate the real value of the target company. For example, the enterprise value of Macy’s, using data from its 2017 10-K report, is estimated to be \$11.81 billion. After the real value of the company has been determined, the acquiring company will decide how much it is willing to pay on top of the real value in order to present an attractive deal, especially if there are other firms considering acquisition. For example, an acquirer may decide to pay a 20% premium to buy Macy’s. The total acquisition cost it will propose will, therefore, be \$11.81 billion x 1.2 = \$14.17 billion. If this premium offer is accepted, then the acquisition premium value will be \$14.17 billion - \$11.81 billion = \$2.36 billion, or in percentage form, 20%.
The acquisition premium can also be evaluated using share price. For instance, if Macy’s is currently trading for \$26 per share, and an acquirer is willing to pay \$33 per share for the target company’s outstanding shares, the acquisition premium can be calculated as (\$33 - \$26)/\$26 = 27%. Not every company intentionally pays a premium for an acquisition. Using our price per share example, if there was no premium offer on the table and the acquisition cost was agreed at \$26 per share, but the value of the company drops to \$16 before the acquisition becomes final, the acquirer will find itself paying a premium of (\$26 - \$16)/\$16 = 62.5%. In cases where the target’s stock price falls dramatically, its product becomes obsolete or concerns are raised about the future of the industry, then the acquiring company may withdraw its offer.
An acquirer will typically pay an acquisition premium to close a deal and ward off competition. An aquisition premium might be paid, too, if the acquirer believes that the synergy created from the merger or acquisition will be greater than the total cost of acquiring the target. The size of the premium often depends on various factors such as competition within the industry, the presence of other bidders and the motivations of the buyer and seller.
Acquisition premium is recorded as goodwill on the acquirer’s balance sheet. The value of a company’s brand name, solid customer base, good customer relations, good employee relations and any patents or proprietary technology acquired from the target company are factored into goodwill. An adverse event, such as declining cash flows, economy depression, increased competitive environment, etc., can lead to an impairment of goodwill, which occurs when the market value of the intangible asset drops below its acquisition cost. Any impairment results in a decrease in the goodwill account on the balance sheet and a loss on the income statement.
An acquirer can purchase a target company for a discount, that is, for less than its fair market value. When this occurs, negative goodwill is recognized.
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# Often asked: How Much Play Should There Be In A Bicycle Chain?
## How much slack should a chain have?
Chain slack should be adjusted to 2% of the chain span. It is essential to inspect such drives regularly for correct chain tension. Idlers are recommended. Slack should be on lower side when possible (D).
## What happens if your bike chain is too loose?
The problem with a bike chain that is too long or is loose will cause the chain to slip off. And this will affect changing the gears cause the dérailleur does not have enough capacity to pick up the slack with the chain being too long.
## Is my chain too slack?
Here’s How: Do a simple check on your bike by shifting the chain to the big chainring and the biggest cassette cog; then, push on the end of the derailleur cage (pushing forward) to see how much it will move forward. If it moves just a little, then you’re good. If it moves a lot, then you’ve got too much chain.
## When should I tension my bike chain?
The chain should be tight enough that it only allows you to move it up and down about half an inch. If there is no slack in the chain then it is too tight. And if there is too much slack then you need to tighten that chain.
You might be interested: FAQ: How To Change The Bicycle Chain?
## Why does my bike chain slip when I pedal hard?
Most of the time, a skipping chain is caused by cable stretch. In the first half dozen rides on a new bike your shift cables stretch the most. They can also stretch over time as you ride. Hippley explains, “It takes cable tension to open a derailleur, which shifts your chain between gears.
## How tight should my r6 chain be?
Aim for 1.25″ on the stand.
## What is chain slack?
The chain slack that you can move by hand (length SS′, see Figure 7.4) should be about 4 percent of the span (length A-B) unless one or more of the following conditions apply: The center line of sprockets is vertical or almost vertical. The center distance of sprockets is more than 1 m.
## How do I know if my chain is too loose?
75% of its original length. If you put the tool up against the chain and it doesn’t reach the 0.5 mark, then your chain has not elongated and is fine. If you find that it is 0.5, you need to replace the chain if you have 11 or more rear gears. If it is 0.75, you need to replace it if you have 10 or fewer rear gears.
## How do I know my bike chain size?
Begin by counting the number of teeth on the largest front sprocket and largest rear. These numbers are often printed right on the sprockets and cogs. Next, measure the distance between the middle of the crank bolt to the rear axle. This is also the chain stay length.
You might be interested: FAQ: Where To Put Chain Lock On Bicycle?
## Why does my chain go slack when I stop pedaling?
If your chain is getting slack on top when you stop pedalling or backpedal, then the problem is in your freehub (or freewheel, whichever you have), a dirty freehub will cause all the problems you’ve listed, even on a brand new bike.
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0
# (f) x= square root 3-3x x=-4,9 f(-4)=
(f) x= √ 3-3x x=4,9 f(-4)=
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# Thread: I need help with the equation part
1. ## I need help with the equation part
During a recent basketball game, the Miami Heat scored 26 fewer points than the Orlando Magic. Together, both teams scored a total of 182 points. How many points did each team score?
a) Write the equation you would use to solve this problem, using ${m}$ as Orlando Magic's score. __________
b) After solving the equation, what is the Miami Heat's score? _______________
c) After solving the equation, what is the Orlando Magic's score? ____________
2. ## Re: I need help with the equation part
If we let $\displaystyle h$ represent the Miami Heat's score and $\displaystyle m$ represent the Orlando Magic's score, how can we get two equations that relate the two variables from:
"Miami Heat scored 26 fewer points than the Orlando Magic"
"Together, both teams scored a total of 182 points"
Can you translate these statements into equations?
3. ## Re: I need help with the equation part
Originally Posted by MarkFL2
If we let $\displaystyle h$ represent the Miami Heat's score and $\displaystyle m$ represent the Orlando Magic's score, how can we get two equations that relate the two variables from:
"Miami Heat scored 26 fewer points than the Orlando Magic"
"Together, both teams scored a total of 182 points"
Can you translate these statements into equations?
I did that already, it still showed it was incorrect.
4. ## Re: I need help with the equation part
What were your equations?
M-26=182
6. ## Re: I need help with the equation part
As you know, that is incorrect.
"Miami Heat scored 26 fewer points than the Orlando Magic"
Thus tell us that 26 fewer than $\displaystyle m$ is equal to $\displaystyle h$, or:
$\displaystyle h=m-26$
"Together, both teams scored a total of 182 points"
This tells us that the sum of the scores is 182, or:
$\displaystyle h+m=182$
Now, to get an equation in one variable, we may take the first equation, which is already solved for $\displaystyle h$ and substitute into the second equation:
$\displaystyle m-26+m=182$
Can you now simplify this and solve for $\displaystyle m$?
7. ## Re: I need help with the equation part
M-26+m=182
+26 +26
_________________
M+M=208
2m/2 =208/2
M=104
8. ## Re: I need help with the equation part
Correct!
Now you may use the first equation to get the value for h.
Ty : )
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# How old is someone born in 2003?
## How old is someone who was born in 2003?
Answer: If you were born in 2003, you are 20 or 21 years old
Someone who was born in 2003 is 20 or 21 years old.
The number of full years from 2003 to May 20, 2024 is 20 or 21.
If you were born in 2003, you are 244-256 months old or 7446-7810 days old (depends on the exact day of birth, see the table below)
## Born in 2003 Age Table
You may click the month to see the exact dates
Date of BirthYears OldMonths OldDays Old
January 200321255-2567780-7810
February 200321254-2557752-7779
March 200321253-2547721-7751
April 200321252-2537691-7720
May 200320 or 21251-2527660-7690
June 200320250-2517630-7659
July 200320249-2507599-7629
August 200320248-2497568-7598
September 200320247-2487538-7567
October 200320246-2477507-7537
November 200320245-2467477-7506
December 200320244-2457446-7476
2003
21.39 Years
May 20, 2024
BirthdateAge
22 y.o.
22 y.o.
22 y.o.
22 y.o.
22 y.o.
22 y.o.
21 or 22 y.o.
21 y.o.
21 y.o.
21 y.o.
21 y.o.
21 y.o.
21 y.o.
21 y.o.
21 y.o.
21 y.o.
21 y.o.
21 y.o.
20 or 21 y.o.
20 y.o.
20 y.o.
20 y.o.
20 y.o.
20 y.o.
20 y.o.
20 y.o.
20 y.o.
20 y.o.
20 y.o.
20 y.o.
## How old am I Calculator
This calculator will help to find out the age of a person with a given month and year of birth. For example, it can help you find out how old is someone who was born in 2003? (The answer is: 20 or 21 y.o.). Pick a Month (e.g. 'Year') and Year (e.g. '2003') and hit the 'Calculate' button.
## FAQ
### How old is someone who was born in 2003?
If you were born in 2003, you are 20 or 21 years old
### How old in months is someone who was born in 2003?
If you were born in 2003, you are 255-256 months old
### How old in days is someone who was born in 2003?
If you were born in 2003, you are 7780-7810 days old
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# Thread: Problem with a Simple Histogram Problem
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## Problem with a Simple Histogram Problem
One of the problems from the CS106a course is to create a histogram based on a series of test scores contained within a text file. I've already looked at the official solution, but I'm wondering why my first crack at it didn't work. I originally thought it was something to do with the boolean expressions that I attempted to use to bin the scores into the correct ranges, but even the bins that don't use those expression aren't correct. Can anyone tell me why it isn't working?
Java Code:
```/*
* File: Histogram.java
* ------------------
* This program will eventually count the lines, words, and characters in a given text file.
*/
import java.io.IOException;
import java.util.ArrayList;
import acm.io.*;
import acm.program.*;
import acm.util.*;
public class Histogram extends ConsoleProgram {
public void run() {
try {
while (true){
if (line == null) break;
if(line.length() == 1) bin0 += "*";
if(line.length() == 2) {
if(line.charAt(0) == 0) bin0 +="*";
if(line.charAt(0) == 1) bin1 +="*";
if(line.charAt(0) == 2) bin2 +="*";
if(line.charAt(0) == 3) bin3 +="*";
if(line.charAt(0) == 4) bin4 +="*";
if(line.charAt(0) == 5) bin5 +="*";
if(line.charAt(0) == 6) bin6 +="*";
if(line.charAt(0) == 7) bin7 +="*";
if(line.charAt(0) == 8) bin8 +="*";
if(line.charAt(0) == 9) bin9 +="*";
}
if(line.length() == 3) bin10 += "*";
}
scores.close();
} catch (IOException ex) {
throw new ErrorException(ex);
}
println("09 - 09: " + bin0);
println("10 - 99: " + bin1);
println("20 - 99: " + bin2);
println("30 - 99: " + bin3);
println("40 - 99: " + bin4);
println("50 - 99: " + bin5);
println("60 - 99: " + bin6);
println("70 - 99: " + bin7);
println("80 - 99: " + bin8);
println("90 - 99: " + bin9);
println(" 100: " + bin10);
}
private String bin0 = "";
private String bin1 = "";
private String bin2 = "";
private String bin3 = "";
private String bin4 = "";
private String bin5 = "";
private String bin6 = "";
private String bin7 = "";
private String bin8 = "";
private String bin9 = "";
private String bin10 = "";
}```
2. ## Re: Problem with a Simple Histogram Problem
Does it compile? Run? Does it not behave correctly? Please give us the details of your problem as it will help us to help you.
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## Re: Problem with a Simple Histogram Problem
It compiles and runs without any issue. The problem is that all of the scores are being put in the 100 bin.
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## Re: Problem with a Simple Histogram Problem
Well, I don't know how much data you have but your test for characters is wrong. I suggest you put single quotes around the digits you are testing for.
e.g == '9' instead of == 9.
Regards,
Jim
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## Re: Problem with a Simple Histogram Problem
Originally Posted by jim829
Well, I don't know how much data you have but your test for characters is wrong. I suggest you put single quotes around the digits you are testing for.
e.g == '9' instead of == 9.
Regards,
Jim
I'm still getting the same result. Oh well.
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## Re: Problem with a Simple Histogram Problem
What does your data look like? Can you attach the file? Oh, and BTW, your println syntax is incorrect and your ranges need some reconsideration.
Regards,
Jim
Last edited by jim829; 06-21-2013 at 06:33 AM.
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## Re: Problem with a Simple Histogram Problem
The text file with the scores is attached. What's the issue with the println syntax? I don't think it's any different from the syntax I've been using up until this point(?). Thanks.
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## Re: Problem with a Simple Histogram Problem
I forgot you are extending some class with which I am unfamiliar (ConsoleProgram). So my guess is it probably provides a method forwarder of println() which calls System.out.println().
In any case, I had to modify my version to get it to work. And the mystery is that each line of your input data file that is only two character wide has a space appended to it. So all lines appear to be three characters wide. And you need to put single quotes around your digit comparisons.
The char 9 is different than the char '9' (the latter is equal to 39 (in hex)). Assuming your super class works correctly, your program should now work.
And I still think you should change your range value. 10 - 99, 20 - 99.. does that make sense?
Regards,
Jim
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## Re: Problem with a Simple Histogram Problem
Upon reflection on this I believe I did you a disservice by pointing out the trailing blanks in the text file. And for that I apologize. What I should have done was have you first verify the length of the lines you are reading. That would show a line length of 3 for each one. Then if you didn't figure out why that was happening I would have recommended that you print the lines as you read them, surrounded by single quotes. That would have shown that a blank was trailing each line. This is exactly what I did (and why I requested your datafile).
So in the future, you should attempt to determine why certain events are happening by using print statements. It is a very simple debugging tool. And I should be more thoughtful about my assistance.
Note: Telling you to put single quotes around your digits was in my opinion, appropriate. Internal representation of characters is not something beginning programmers are familiar with.
Regards,
Jim
10. ## Re: Problem with a Simple Histogram Problem
Or maybe call the trim method after each line of data is read. This will ensure there are no extra spaces.
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## Re: Problem with a Simple Histogram Problem
Originally Posted by jim829
Upon reflection on this I believe I did you a disservice by pointing out the trailing blanks in the text file. And for that I apologize. What I should have done was have you first verify the length of the lines you are reading. That would show a line length of 3 for each one. Then if you didn't figure out why that was happening I would have recommended that you print the lines as you read them, surrounded by single quotes. That would have shown that a blank was trailing each line. This is exactly what I did (and why I requested your datafile).
So in the future, you should attempt to determine why certain events are happening by using print statements. It is a very simple debugging tool. And I should be more thoughtful about my assistance.
Note: Telling you to put single quotes around your digits was in my opinion, appropriate. Internal representation of characters is not something beginning programmers are familiar with.
Regards,
Jim
I really appreciate you taking the time to go through this. I think the biggest problem I'm running into while attempting to learn this stuff is my own impatience. I have to keep reminding myself that there are no shortcuts and that I just have to put in the time.
Anyway, using the trim method on the scores and putting single quotes around the values in the boolean expression got it to work. Can you explain your recommendation to change the range values?
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## Re: Problem with a Simple Histogram Problem
Well, your ranges that you display are always from some low number to 99. Normally, they would be something like 10-19, 20-29, 30-39, etc.
Regards,
Jim
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## Re: Problem with a Simple Histogram Problem
Originally Posted by jim829
Well, your ranges that you display are always from some low number to 99. Normally, they would be something like 10-19, 20-29, 30-39, etc.
Regards,
Jim
Gotcha. I was confused as to what you were talking about since that's something that I caught and fixed right after I posted here. Thanks again.
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This Lines, Line Segments, Rays and Angles - Homework 15.1 Worksheet is suitable for 4th - 5th Grade.
The Geometry curriculum materials are available from the module links. In Unit 5, Polygons & Algebraic Relationships, students connect algebra to geometric concepts with polygons through distance on the coordinate plane, partitioning line segments, slope criteria for perpendicular and parallel lines, area (with composition and decomposition), and perimeter. 64% average accuracy. Workbook Answers Geometry: homework practice workbook (9780078908491 , .. ... Books Answers Holt Geometry 10 3 Practice Answers (PDF, .. A related axiom to The Whole is the Sum of Its Parts is the following: The Whole is Greater Than Any of Its Parts.Given an example of this using the angle diagram shown below. lfroehlich_21009. The answer key is linked above in the Unit 2 section. Common Core: High School - Geometry : Points that Partitions Segments in a Given Ratio: CCSS.Math.Content.HSG-GPE.B.6 Study concepts, example questions & explanations for Common Core: High School - Geometry Start - Geometry Module 1. CCSS.Math.Content.HSG.GPE.B.7 Use coordinates to compute perimeters of polygons and areas of triangles and rectangles, e.g., using the distance formula. Edit. Q. 10th grade. Mathematics. CCSS.Math.Content.HSG.GPE.B.6 Find the point on a directed line segment between two given points that partitions the segment in a given ratio.
Formulas are nice, but my students don’t do so well with them.
Angle median. Common Core Geometry Homework 2.11 DRAFT. Homework for the weekend - Bring in Unit 2 and Unit 3 notes to class on Monday! Home > Grade 8 > Geometry > Partitioning a Line Segment Partitioning a Line Segment Directions: Using the digits 1 to 8 exactly one time each, fill in the boxes to create a line segment AB, where between point A and point B, there exists a point P so that it partitions line segment AB into a ratio. Save. One of the Common Core standards for geometry is partitioning a directed line segment into a given ratio (G.GPE.6).
In order to assist educators with the implementation of the Common Core, the New York State Education Department provides curricular modules in P-12 English Language Arts and Mathematics that schools and districts can adopt or adapt for local purposes. The answer key is lined above in the Unit 3 section. Cursory Googling led to a nice little formula, which Shmoop calls the “section formula”: . Monday, 11/7/16: We reviewed Unit 2 Material today. We also encourage plenty of exercises and book work. a year ago.
900 seconds . This falls firmly within the category of Things I Never Learned in School. Perpendicular bisector of a line segment.
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Home www.play-hookey.com Mon, 05-25-2020
Alternating Current: Filters
AC has some properties not shared by DC: frequency and the phase relationship between voltage and current. Sometimes we need to control one or both of these properties.
Filter Basics
Real-world signals include multiple frequencies over a wide range. How will combinations of components affect the different frequencies that make up the total signal?
Radians
One thing we find ourselves dealing with throughout electronics and some other aspects of physics is the mathematical factor designated as π, which has a value of approximately 3.14159265. This factor inserts small errors at many places in our calculations, and the cumulative error can become significant. Can we find a way to do most of our calculations without invoking π?
Logarithms
In electronics, we deal with signal magnitudes and component values over a very wide range. Sometimes it's more convenient to deal with numbers in the range of a hundred or so, rather than 10¹² or more. Here's how to do that.
Decibels
Remember what we said about using logarithms? Here's one place we need them. When we look at signal power gain or loss, we often have to calculate with values over several orders of magnitude. Using decibels makes such calculations much simpler.
Low-Pass Filters
Now that we have the mathematical tools we need to describe the behavior of signal filter circuits, we can examine the circuits themselves and their behavior. We start with a circuit that will pass low frequencies easily, but will increasingly attenuate signals at higher frequencies.
High-Pass Filters
Just as we have circuits that can pass lower frequencies and attenuate higher frequencies, we can also turn them around and pass higher frequencies while attenuating lower frequencies.
Band-Pass Filters
We can also combine high-pass and low-pass filters in such a way as to readily pass a certain range of frequencies but attenuate signals at either higher or lower frequencies.
All pages on www.play-hookey.com copyright © 1996, 2000-2015 by Ken Bigelow Please address queries and suggestions to: webmaster@play-hookey.com
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# Absolute Value Inequalities Worksheet
This page presents 45 absolute value inequalities worksheets that cater to the learning requirements of high school students. To gain access to our editable content join the algebra 2 teacher community.
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### Students must be careful about extraneous solutions.
Absolute value inequalities worksheet. Inequalities absolute value inequalities objective. Here you will find hundreds of lessons a community of teachers for support and materials that are always up to date with the latest standards. Here is your free content for this lesson.
Absolute value inequalities worksheet solve the inequality and graph the solution. Absolute value absolute value worksheets inequalities worksheets pre algebra. Solve graph and give interval notation for the solution to inequalities with absolute values.
Absolute value inequalities worksheets generator. 2 1 2. A i2x0 v1l2h mkguht aao ls ro kfatxwba r4e1 hl gl3c4.
G e290m1m2q fkdu5t cau 9s voef ct3w vakrce9 ilvlyc 9. No solution cases are also included. Absolute value is the distance a number is from zero.
Identify what the isolated absolute value is set equal to a. To link to this page copy the following code to your site. The way we remove.
Absolute value equations and inequalities word docs powerpoints. Absolute value inequality worksheet 4 here is a 9 problem worksheet where you will find the solution set of absolute value inequalities. These are two step inequalities that can get quite complicated.
4 7 va nlkla sr 7ifgxhbt psp kr9enszekr nvcexda p g bm pa pdhe 8 lw viot5hw zi znvf lidnbi mtre 3 mazl ygre rbdrwa y c2h p worksheet by kuta software llc kuta software infinite algebra 2 name absolute value inequalities date period. Isolate the absolute value. A nice challenge for your higher level learners.
This doesn t mean that your variable has to be positive but your final answer will be when we use inequalities we have more than one answer. Convert the absolute value inequalities into two simple linear inequalities and solve the problems. When an inequality has an absolute value we will have to remove the absolute value in order to graph the solution or give interval notation.
F worksheet by kuta software llc kuta software infinite algebra 1 name absolute value inequalities date period. Since it is a distance it is always positive. Absolute value equations and inequalities absolute value definition the absolute value of x is defined as 0 0 where x is called the argument steps for solving linear absolute value equations.
H b qa3l qlc yr widgmhltxs8 lr me1s9earnvve cdr c e kmjakdyej xw aiotvhy lian 0fxiln4i gt3ec ca pl 0g9e 0bsr uai f1f.
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5 is what percentage of 40
Below is a list of the best 5 is what percentage of 40 public topics compiled and compiled by our team
1 How to Calculate Percentages – dummies
• Author: dummies.com
• Published Date: 07/22/2022
• Review: 4.86 (719 vote)
• Summary: · This process is the reverse of what you did earlier. You divide your percentage by 100. So, 40 percent would be 40 divided by 100. 40 ÷ 100 =
2 Percentages – An Introduction | SkillsYouNeed
• Author: skillsyouneed.com
• Published Date: 09/29/2021
• Review: 4.76 (404 vote)
• Summary: Percentages are like fractions and decimals: they are ways to describe what … + 5 accountants + 3 HR professionals + 7 cleaners + 4 catering staff = 40
• Matching search results: In this example, the whole is £500, or the cost of the laptop before the discount is applied. The percentage that you need to find is 20%, or the discount offered by the supplier. You are then going to take that off the full price to find out what …
3 What is 5/40 as a percentage? (Convert 5/40 to percent)
• Author: visualfractions.com
• Published Date: 03/23/2022
• Review: 4.58 (202 vote)
• Summary: · Now we can see that our fraction is 12.5/100, which means that 5/40 as a percentage is 12.5%. We can also work this out in a simpler way by
• Matching search results: In this example, the whole is £500, or the cost of the laptop before the discount is applied. The percentage that you need to find is 20%, or the discount offered by the supplier. You are then going to take that off the full price to find out what …
4 5 percent of 40 (5% of 40) – Percent Calculator
• Author: percent.info
• Published Date: 02/27/2022
• Review: 4.34 (419 vote)
• Summary: What is 5 percent of 40? Step-by-step showing you how to calculate 5 percent of 40. Detailed explanation with answer to 5% of 40
• Matching search results: In this example, the whole is £500, or the cost of the laptop before the discount is applied. The percentage that you need to find is 20%, or the discount offered by the supplier. You are then going to take that off the full price to find out what …
5 How much is 5 out of 40
• Author: coolconversion.com
• Published Date: 05/22/2022
• Review: 4.05 (516 vote)
• Summary: 5 is what percent of 40? How to work out percentages? See how by using our percentage calculator online
• Matching search results: In this example, the whole is £500, or the cost of the laptop before the discount is applied. The percentage that you need to find is 20%, or the discount offered by the supplier. You are then going to take that off the full price to find out what …
6 Percentage Calculator
• Author: omnicalculator.com
• Published Date: 09/23/2021
• Review: 3.85 (574 vote)
• Summary: · 100 * numerator / denominator = percentage . In our example it’s 100 * 2/5 = 100 * 0.4 = 40 . Forty percent of the group are girls. That’s the
• Matching search results: In this example, the whole is £500, or the cost of the laptop before the discount is applied. The percentage that you need to find is 20%, or the discount offered by the supplier. You are then going to take that off the full price to find out what …
What is 3 of 250000
7 Convert the fraction 3/5 into percent. [Solved]
• Author: cuemath.com
• Published Date: 12/18/2021
• Review: 3.75 (232 vote)
• Summary: Answer: 3/5 is expressed as 60% in terms of percentage. Let’s convert the fraction 3/5 into percent. Explanation: Let’s change the fraction 3/5 to an
• Matching search results: In this example, the whole is £500, or the cost of the laptop before the discount is applied. The percentage that you need to find is 20%, or the discount offered by the supplier. You are then going to take that off the full price to find out what …
8 Percentages – Express A as a percentage of B
• Author: radfordmathematics.com
• Published Date: 01/26/2022
• Review: 3.54 (245 vote)
• Summary: Finally, since 60100=60% we can state that 3 is 60% of 5. Example 2. Express 18 as a percentage of 40
• Matching search results: In this example, the whole is £500, or the cost of the laptop before the discount is applied. The percentage that you need to find is 20%, or the discount offered by the supplier. You are then going to take that off the full price to find out what …
9 5 is what percent of 40 – step by step solution
• Author: geteasysolution.com
• Published Date: 05/15/2022
• Review: 3.31 (431 vote)
• Summary: Simple and best practice solution for 5 is what percent of 40. Check how easy it is, and learn it for the future. Our solution is simple,
• Matching search results: In this example, the whole is £500, or the cost of the laptop before the discount is applied. The percentage that you need to find is 20%, or the discount offered by the supplier. You are then going to take that off the full price to find out what …
10 To Find the Percent of a Given Number | Word Problem on Percentage
• Author: math-only-math.com
• Published Date: 02/27/2022
• Review: 3.13 (373 vote)
• Summary: So, 18% of 500 = 18100 × 500 = 18 × 5 = 90 ; = 15100 × 200 ; = 25100 × 40
• Matching search results: In this example, the whole is £500, or the cost of the laptop before the discount is applied. The percentage that you need to find is 20%, or the discount offered by the supplier. You are then going to take that off the full price to find out what …
11 Finding Percentages – How To Find Percentage
• Author: ipracticemath.com
• Published Date: 06/21/2022
• Review: 2.91 (177 vote)
• Summary: Try our problem to convert percentage to decimal! … 5%,10%,3313%,67.5%,100%. Percentage is applied in different fields. … What is 40% of 60 ?
• Matching search results: Sometimes, converting percent to fraction is an easier method to obtain the percentage. Fractions are preferred to be used than decimals if the decimal has many digits. This makes the multiplication more convenient since only factorization is used …
12 5 is what percentage of 40? – Basics Math
• Author: basicsmath.com
• Published Date: 09/08/2022
• Review: 2.74 (57 vote)
• Summary: · When someone asks you 5 is what percentage of 40 they want to know the ratio of 5 to 40 expressed as a fraction of 100
• Matching search results: Sometimes, converting percent to fraction is an easier method to obtain the percentage. Fractions are preferred to be used than decimals if the decimal has many digits. This makes the multiplication more convenient since only factorization is used …
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13 What percent of 40 is 5? – Number Maniacs
• Author: numbermaniacs.com
• Published Date: 12/19/2021
• Review: 2.69 (184 vote)
• Summary: 40 · Percent = 500. Get Percent alone on the left side by dividing both sides by 40 to get the answer: Percent = 12.5 · Percent = 12.5. Therefore, the answer to
• Matching search results: Sometimes, converting percent to fraction is an easier method to obtain the percentage. Fractions are preferred to be used than decimals if the decimal has many digits. This makes the multiplication more convenient since only factorization is used …
14 5 is what percent of 40 | what percent of 40 is 5 | 5/40 as a percent
• Author: percentage-off-calculator.com
• Published Date: 06/12/2022
• Review: 2.5 (167 vote)
• Summary: 5 is what percent of 40 is equivalent to 5/40 as a percent. \$5 out of 40 is what percent. In calculating 5% of a number, sales tax, credit cards cash back
• Matching search results: In calculating 5% of a number, sales tax, credit cards cash back bonus, interest, discounts, interest per annum, dollars, pounds, coupons,5% off, 5% of price or something, we use the formula above to find the answer. The equation for the …
15 5/40 as a Percentage | Convert 5/40 to Percent
• Author: worksheetgenius.com
• Published Date: 06/06/2022
• Review: 2.59 (67 vote)
• Summary: We can see that this gives us the exact same answer as the first method: 5/40 as a percentage is 12.5%. Both methods of converting a fraction to a percentage
• Matching search results: In calculating 5% of a number, sales tax, credit cards cash back bonus, interest, discounts, interest per annum, dollars, pounds, coupons,5% off, 5% of price or something, we use the formula above to find the answer. The equation for the …
16 What is 5 percent of 40? Calculate 5% of 40. How much?
• Author: dollartimes.com
• Published Date: 10/14/2021
• Review: 2.35 (116 vote)
• Summary: Calculator 1: Calculate the percentage of a number. For example: 5% of 40 = 2. Calculator 2: Calculate a percentage based on 2 numbers. For example:
• Matching search results: In calculating 5% of a number, sales tax, credit cards cash back bonus, interest, discounts, interest per annum, dollars, pounds, coupons,5% off, 5% of price or something, we use the formula above to find the answer. The equation for the …
17 What is 5 percent of 40 million? (5% of 40 million)
• Author: maniacs.info
• Published Date: 11/03/2021
• Review: 2.3 (136 vote)
• Summary: In other words, how do you calculate five percent of forty million? First, note that 5 percent means 5 per hundred, and 40 million is 40 followed by 6 zeros. To
• Matching search results: In calculating 5% of a number, sales tax, credit cards cash back bonus, interest, discounts, interest per annum, dollars, pounds, coupons,5% off, 5% of price or something, we use the formula above to find the answer. The equation for the …
18 5 is what percent of 40? – Percentage Calculator
• Author: thepercentagecalculator.net
• Published Date: 04/05/2022
• Review: 2.13 (140 vote)
• Summary: Whole × Percent · = Part ; 40 × Percent · = 5 ; 40 × Percent × 100 · = 5 × 100 ; 40 × Percent · = 500. 40 ; 500
• Matching search results: In calculating 5% of a number, sales tax, credit cards cash back bonus, interest, discounts, interest per annum, dollars, pounds, coupons,5% off, 5% of price or something, we use the formula above to find the answer. The equation for the …
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19 Percentage Calculators from The Calculator Site
• Author: thecalculatorsite.com
• Published Date: 01/24/2022
• Review: 2.06 (70 vote)
• Summary: Let’s say the question is: What is the percentage increase from 40 to 68? … Next, we divide our X figure by our Y figure. so, 5 / 0.2 = 25
• Matching search results: In calculating 5% of a number, sales tax, credit cards cash back bonus, interest, discounts, interest per annum, dollars, pounds, coupons,5% off, 5% of price or something, we use the formula above to find the answer. The equation for the …
20 Percentage Calculator – Percent Change & Percent Difference Calculator
• Author: gigacalculator.com
• Published Date: 06/19/2022
• Review: 2.03 (163 vote)
• Summary: The percent change is simply 40% minus 35% which equals 5 p.p. (but is a percentage change of 12.5 percent). Compounding and averaging percentages. Percentages
• Matching search results: Percentages have a wide array of application in many disciplines and everyday usage. They are common in statistics, social sciences, economics, finance, accounting. In everyday usage we often encounter percent off coupons. Promotions, sales, and …
21 Percentage Calculator: 5 is what percent of 40 –
• Author: percentagecalculator.guru
• Published Date: 01/14/2022
• Review: 1.99 (165 vote)
• Summary: 5 is 12.5 percent of 40. 2. How to calculate 5 is what percent of 40? First we have to divide 5 with 40 and multiply with 100 = (5/
• Matching search results: Percentages have a wide array of application in many disciplines and everyday usage. They are common in statistics, social sciences, economics, finance, accounting. In everyday usage we often encounter percent off coupons. Promotions, sales, and …
22 What percent of 40 is 5? –
• Author: brainly.ph
• Published Date: 09/06/2022
• Review: 1.7 (60 vote)
• Summary: Percentage Calculator: 5 is what percent of 40? = 12.5. e3radg8
• Matching search results: Percentages have a wide array of application in many disciplines and everyday usage. They are common in statistics, social sciences, economics, finance, accounting. In everyday usage we often encounter percent off coupons. Promotions, sales, and …
23 Percent word problem: 100 is what percent of 80? (video) | Khan Academy
• Author: khanacademy.org
• Published Date: 02/28/2022
• Review: 1.61 (97 vote)
• Summary: · So we have to convert this to a percent, and the easiest thing to do is to first convert it into a …
Posted:
• Matching search results: Percentages have a wide array of application in many disciplines and everyday usage. They are common in statistics, social sciences, economics, finance, accounting. In everyday usage we often encounter percent off coupons. Promotions, sales, and …
24 Calculating percent, the percentage calculator, percentage, percent calculator, calculate the base
• Author: hackmath.net
• Published Date: 07/03/2022
• Review: 1.52 (192 vote)
• Summary: Percentage calculator is an online tool to calculate percentages. Percentage calculator to find percentage of a number, calculate x as a percent of y,
• Matching search results: Percentages have a wide array of application in many disciplines and everyday usage. They are common in statistics, social sciences, economics, finance, accounting. In everyday usage we often encounter percent off coupons. Promotions, sales, and …
25 Percentage (How to Calculate, Formula and Tricks)
• Author: byjus.com
• Published Date: 08/31/2022
• Review: 1.39 (56 vote)
• Summary: · Percentage of number is expressed as fraction of 100. It is denoted by %. 40 per cent is … Example: 2/5 × 100 = 0.4 × 100 = 40 per cent
• Matching search results: In mathematics, a percentage is a number or ratio that can be expressed as a fraction of 100. If we have to calculate percent of a number, divide the number by the whole and multiply by 100. Hence, the percentage means, a part per hundred. The word …
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# Introduction to the Delta Wye Transformer Part 6e
### Resources Section
This video tutorial does not items in the resources section
Do you have a question? Click on the "Q&A forum" and ask away!
Introduction to the Delta Wye transformer connection – part 6e.
In many areas of the power system, especially when we see transformers, we often see this type of illustration for the delta transformer connection. With it, we also see a dashed line that's pointing towards one side of the corner. And that side is always labeled H1. The other corner labeled as H2 and the third corner relabel as H3.
Our goal for this video tutorial is too really understand how to read this phasor diagram and make sense out of it. First, let's note couple things. Let's relabel H1 as A. And H3 as C.
So the next question is – what's this dashed line is? This dashed line represents the line-to-ground voltage. So the next question is – what line-to-ground voltage does it represents?
VAG represents the voltage across line and ground. VBG represents the voltage across line B and ground. VCG represents the voltage across line C and ground.
Since this dashed line is pointed towards H1 or this corner of this triangle, this dashed line represents VAG. Which means, if we put another dashed line here – this dashed line would represent VBG. This dashed line would represents VCG. This is very similar to the example we did in the last tutorial.
Let me clear this up and make this smaller. The next question is what does this phasor represent? Does it represent VAB – if so is it connected like this? Or does this phasor represent VCA and if so, is it connected like that.
Depending on what we choose or how we define these two phasors – it's going to represent an entirely different delta connection. So we have to be very careful how we defined these two legs of the triangle.
Now we look back at our delta connection and we see that it's connected like this. In the last video, we showed that the line-to-line voltage between line A and line B is equal to phase A voltage. Which is equal to the voltage across winding A. So we immediately know that this phasor here represents VAB which is also equal to phase A voltage.
So now we're going to put an arrow here. That's how we've defined it. Since the arrow is on this side we're going to put a dot here to indicate the polarity side of the winding. Now that we've assumed this side is V phase A, it becomes very clear that this side here is VCA. And that is equal to V phase C. And this side here is VBC which is equal to V phase B. Now there are couple of things that we can do to verify our assumption.
The first thing is – we know that this is VAB because the phasor starts from B and it goes to A. It's opposite then how we've defined. We've defined it as VAB. But the arrow is pointing in the opposite direction. Same thing over here – the arrow is pointing is from A to C but we've defined it as VCA. Then over here – the arrow is pointing from C to B but we've defined it as VBC.
The other thing we can do is see how the polarity sides are connected. So V phase A which is this phasor here. The polarity of V phase A is connected to the non-polarity side of V phase C. And the polarity side of V phase C is connected to the non-polarity side of V phase B.
This agrees with our transformer connection. The polarity of V phase A is connected to the non-polarity side of V phase C. And the polarity side of V phase C is connected to the non-polarity side of V phase B. So that makes sense.
So those are two major things we can do to verify that the way that we've defined and labeled everything actually makes sense. And that's how you analyze this type of a phasor diagram.
Now if you find these videos useful, please click on the button that's on the bottom corner of your screen. By subscribing, you will be connected to video tutorials like this that make power systems intuitive and easy to understand.
This video was brought to you by GeneralPAC.com. Visit this website for additional video tutorials.
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Annette Arroyo
2021-05-28
Solve by system of equations:
$5x+3y+z=-8$
$x-3y+2z=20$
$14x-2y+3z=20$
Nathalie Redfern
Step 1
Given system of equations are
$5x+3y+z=-8$...(1)
$x-3y+2z=20$...(2)
$14x-2y+3z=20$...(3)
Step 2
By subtracting 5 times equation (2) from equation (1), we get
$5x+3y+z-5\left(x-3y+2z\right)=-8-100$
$18y-9z=-108$
$2y-y=-12$...(4)
Step 3
By subtracting 14 times equation (2) from equation (3), we get
$14x-2y+3z-14\left(x-3+2z\right)=20-280$
$40y-25z=-260$
$8y-5z=-52$...(5)
Step 4
By subtracting 4 times equation (4) from equation (5), we get
$8y-5z-4\left(2y-z\right)=-52+48$
$-z=-4$
$z=4$
Step 5
By substituting $z=4$ in equation (4), we get value of y and then substituting values of y and z in equation (2), we get value of x
$2y-z=-12$
$2y-4=-12$
$2y=-8$
$y=-4$
$x-3y+2z=20$
$x-3\left(-4\right)+2\left(4\right)=20$
$x+12+8=20$
$x=0$
Step 6
Answer: Solution is $x=0$,$y=-4$ and $z=4$
Jeffrey Jordon
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Home MATHEMATICS MATHEMATICS FORM 3
# MATHEMATICS FORM 3
Home MATHEMATICS MATHEMATICS FORM 3
### TOPIC 6: CIRCLES ~ MATHEMATICS FORM 3
TOPIC 6: CIRCLES ~ MATHEMATICS FORM 3 TOPIC 6: CIRCLES ~ MATHEMATICS FORM 3 Definition of Terms Circle, Chord, Radius, Diameter, Circumference, Arc, Sector, Centre and Segment...
### TOPIC 5: SEQUENCE AND SERIES ~ MATHEMATICS FORM 3
TOPIC 5: SEQUENCE AND SERIES ~ MATHEMATICS FORM 3 TOPIC 5: SEQUENCE AND SERIES ~ MATHEMATICS FORM 3 Sequences The Concept of Sequence Explain the concept of sequence A Sequence is...
### TOPIC 4: RATES AND VARIATIONS ~ MATHEMATICS FORM 3
TOPIC 4: RATES AND VARIATIONS ~ MATHEMATICS FORM 3 TOPIC 4: RATES AND VARIATIONS ~ MATHEMATICS FORM 3 Rates A rate is found by dividing one quantity...
### TOPIC 3: STATISTICS ~ MATHEMATICS FORM 3
TOPIC 3: STATISTICS ~ MATHEMATICS FORM 3 TOPIC 3: STATISTICS ~ MATHEMATICS FORM 3 Mean Calculating the Mean from a Set of Data, Frequency Distribution Tables and...
### TOPIC 2: FUNCTIONS ~ MATHEMATICS FORM 3
TOPIC 2: FUNCTIONS ~ MATHEMATICS FORM 3 TOPIC 2: FUNCTIONS ~ MATHEMATICS FORM 3 Normally relation deals with matching of elements from the first set called DOMAIN...
### TOPIC 1: RELATIONS ~ MATHEMATICS FORM 3
TOPIC 1: RELATIONS ~ MATHEMATICS FORM 3 TOPIC 1: RELATIONS ~ MATHEMATICS FORM 3 Normally relation deals with matching of elements from the first set called DOMAIN...
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# Random Evolutionary Strategies in Machine Learning
Neural networks do not learn at all like people do. Neural network optimization is actually a gradient descent over some loss function where the variables are the weights of the layers . This is a very powerful approach to tuning the system, which is also used in physics, economics, and many other fields. At the moment, many specific methods of gradient descent have been proposed, but they all suggest that the gradient it behaves well: there are no cliffs where it increases spasmodically, or a plateau where it turns to zero. The first problem can be dealt with using gradient clipping, but the second one makes you think carefully. A piecewise linear or discrete function is not trivial to limit to a more pleasant function
What to do in such situations?
Under the cut a lot of formulas and gifs.
About five years ago , articles began to appear that explored a different approach to learning: instead of smoothing the objective function, let us approximate the gradient in such cases. Sampling values in a certain way , it can be assumed where to move, even if the derivative is exactly zero at the current point. Not that sampling approaches were especially new - Monte Carlo has been friends with neural networks for a long time - but recently, interesting new results have been obtained.
## Random Evolution Strategy (ES)
Let's start with the simplest approach, which has regained popularity after the OpenAI article in 2017. I note that evolutionary strategies do not have a very good name associated with the development of this family of algorithms. ES is very distantly associated with genetic algorithms - only a random change in parameters. It is much more productive to think of ES as a method of estimating the gradient using sampling. We sample from the normal distribution several perturbation vectors , we take the expectation from the resulting values of the objective function, and we believe that
The idea is pretty simple. With a standard gradient descent, at each step we look at the slope of the surface on which we are, and move towards the largest slope. In ES, we “fire” a nearby neighborhood with points where we can supposedly move, and move to the side where the most points with the greatest difference in elevation hit (and the farther the point, the more weight it is given).
The proof is also simple. Suppose that in fact the loss function can be expanded into a Taylor series in :
Multiply both sides by :
Throw away and take the expectation from both sides:
But the normal distribution is symmetrical: , , :
Divide by and get the original statement.
In the case of a piecewise step function, the resulting estimate will represent the gradient of the smoothed function without the need to calculate the specific values of this function at each point; The article gives more rigorous statements. Also in the case when the loss function depends on discrete parameters (those. ), it can be shown that the estimate remains fair, since in the proof you can swap the order of taking the expectation
Which is often impossible for regular SGD.
Let's look at a function smoothed by sampling from . Once again: there really is no point in counting , we only need derivatives. However, such visualizations help to understand which landscape the gradient descent is actually made of. So, the green graph is the original function, the blue one is what it looks like for the optimization algorithm after estimating the gradient by sampling:
:
:
:
The larger the sigma distribution, the less the local structure of the function manifests itself. If the sampling is too large, the optimization algorithm does not see narrow minima and hollows, according to which one can switch from one good state to another. If it is too small, the gradient descent may not start if the initialization point was chosen unsuccessfully.
More graphs:
Gradient descent automatically finds the most stable minimum in the middle of the plateau and “tunnels” through a sharp peak.
• As already mentioned not required to be differentiable. At the same time, no one forbids using ES even when you can honestly calculate the gradient.
• ES is trivially parallel. Although parallel versions of SGD exist, they require you to forward weights between worker nodes or nodes and the central server. This is very expensive when there are many layers. In the case of evolutionary strategies, each of the workstations can consider its own set. . Forward quite simple - usually it's just scalars. Forwarding is as difficult as sending However, there is no need to do this: if all the nodes know the algorithm for sampling random numbers and seed generator, they can simulate each other ! The performance of such a system increases almost linearly with scaling, which makes the distributed version of ES just incredibly fast when a large number of training stations are available. OpenAI reports MuJoCo training in 10 minutes on 80 machines with 1440 CPUs.
• Sampling introduces noise into the calculation of the gradient, which makes learning more stable. Compare with the dropout when training neural networks in the usual way.
• ES does not depend on frame-skip with RL (see, for example, here ) - less than one hyperparameter.
• It is also argued that evolutionary strategies make learning easier than regular SGDs, when a lot of time can elapse between an action in RL and a positive response, and in noisy conditions where it is not clear which change helped to improve the result (the situation when the network works less at the beginning of training, she can’t understand for a long time what needs to be done in a virtual environment and only twitches in place).
• The graphs in the OpenAI article do not look so radiant. Learning with ES takes 2-10 times more eras than learning with TRPO. The authors defend themselves with the fact that when a large number of working machines are available, the real time spent is 20-60 times less (although each update brings less benefit, we can do them much more often). Great, of course, but where else to get 80 working nodes. Plus, as if the final results are a little worse. Perhaps you can "finish off" the network with a more accurate algorithm?
• The noise in the gradients is a double-edged sword. Even with one-dimensional optimization, they turn out to be slightly unstable (see noise on the blue curves in the graphs above), what can I say when very multidimensional. It is not yet clear how serious this poses a problem, and what can be done about it except how to increase the sample size of the sample.
• It is not known whether ES can be effectively applied in regular teacher training. OpenAI honestly reports that on a regular MNIST ES it can be 1000 (!) Times slower than normal optimization.
### ES variations
An additional positive quality of evolutionary strategies is that they only tell us how to calculate the derivative, and do not force us to rewrite the back propagation algorithm cleanly. Everything that can be applied to backprop-SGD can also be applied to ES: Nesterov's impulse, Adam-like algorithms, batch normalization. There are some specific additional techniques, but so far more research is needed, under what circumstances they work:
• ES allows you to evaluate not only the first, but also the second derivative . To get the formula, it is enough to paint the Taylor series in the proof above to the fifth term, and not to the third term, and subtract from it the obtained formula for the first derivative. However, it seems that doing this is only out of scientific curiosity, since the estimate is terribly unstable, and Adam allows you to emulate the action of the second derivative in the algorithm.
• It is not necessary to use a normal distribution. Obvious applicants are the Laplace distribution and the t-distribution, but more exotic options can be devised.
• It’s not necessary to sample only around the current . If you remember the history of movement in the parameter space, you can sample the weights around the point a little in the direction of movement.
• During training, you can periodically jump not in the direction of the gradient, but simply to the point with the smallest . On the one hand, this destabilizes training even more, on the other hand, this approach works well in strongly non-linear cases with a large number of local minima and saddle points.
• A December 2017
• Uber article offers a novelty reward seeking evolution strategies (NSR-ES): an additional term is added to the balance update formulas to encourage new behavioral strategies. The idea of introducing diversity in reinforcement learning is not new, it is obvious that they will try to stick it here. The resulting learning algorithms give noticeably better results on some games of the Atari dataset.
• There is also an article claiming that it is not necessary to send the results from all the nodes to all the other nodes: the thinned-out graph of communication between workstations not only works faster, but also produces better results (!). Apparently, communication with not all fellow students works as an additional regularization like a dropout.
See also this article , which further reflects on the difference in the behavior of ES and TRPO in different types of parameter landscapes, this article , which describes in more detail the relationship between ES and SGD, this article , which proves the extraterrestrial origin of the Egyptian pyramids, and this article , which compares ES from OpenAI with older classic ES.
## Variational optimization (VO)
Hmm, but why not a word is said about the change during training? It would be logical to reduce it, because the longer the training takes, the closer we are to the desired result, therefore we need to pay more attention to the local landscape . But just how to change the standard deviation? I don’t want to invent some tricky scheme ...
It turns out that everything has already been invented! Evolution strategies is a special case of variational optimization for fixed .
Variational optimization is based on an auxiliary statement on the variational restriction from above: the global minimum of the function cannot be greater than the average how would we these neither sampled:
Where - variational functional, the global minimum of which coincides with the global minimum . Note that now minimization is not in the initial parameters , and according to the distribution parameters from which we sampled these parameters - . Also, note that even if was undifferentiated, differentiable if differentiable distribution , and the derivative can be expressed as:
If a - Gaussian distribution with a fixed dispersion, from this formula it is easy to obtain a formula for ES. But what if not fixed? Then for every weight in we get two parameters in : center and standard deviation of the corresponding Gaussian. The dimension of the problem is doubled! The easiest way to show a one-dimensional example. Here is a top view of the landscape obtained after the conversion for the function from the previous section:
And descent along it:
Its values at the point correspond to the average values sampled from the point using normal distribution with standard deviation . Note that
• Again: in real life, it makes no sense to count this function. As with ES, we only need a gradient.
• Now we minimize , but not .
• Less the larger the cross section resembles the original function
• Global minimum achieved when . In the end, we only need , and can be dropped.
• It is quite difficult to get into the global minimum of the one-dimensional function - it is fenced with high walls. However, if you start from a point with a large , it becomes much easier to do this, since the more smoothing is, the stronger the section of the function resembles an ordinary quadratic function.
• Nevertheless, it is better to use Adam to get into the narrow hollow of the minimum.
What are the benefits of introducing variational optimization? It turns out that ES was fixed , his formula could lead us towards suboptimal lows: the more , however, deep but narrow minima are noticeable on the landscape of the smoothed function.
Since in VO Permanent, we have at least a chance to get into a real global minimum.
The disadvantages are obvious: even in the simplest case, the dimension of the problem doubles, what can we say about the case when we want to have in every direction. Gradients become even more unstable.
Remember the graph above with a “wall” in the middle. VO feels it even less:
(the former wall is a small little thing from the bottom in the middle)
Although convergence into a global minimum is not guaranteed even for simple cases, we have more chances to get at least somewhere with imperfect initialization:
Or even get out of the area of poor spatial initialization:
## Conclusion
An article in OpenAI brought a trick to variational optimization with the common seed of random number generators of various nodes. But so far (it seems?) There are no articles evaluating the real acceleration from it. It seems to me, in the near future, we will see them. If they are encouraging, and if ES and VO are extended to teacher training, perhaps a paradigm shift in machine learning awaits us.
You can download the code with which the visualizations were built here . You can look at other visualizations and learn about the relationship of the above methods with genetic algorithms, Natural Evolution Strategies and Covariance-Matrix Adaptation Evolution Strategy (CMA-ES) here . Write in the comments if someone is interested to see how ES and VO behave on a particular function. Thanks for your attention!
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LFD Book Forum (http://book.caltech.edu/bookforum/index.php)
- Homework 4 (http://book.caltech.edu/bookforum/forumdisplay.php?f=133)
- - Calculating The "Average" Function Of Xn (http://book.caltech.edu/bookforum/showthread.php?t=950)
munchkin 08-05-2012 11:41 AM
Calculating The "Average" Function Of Xn
Page 63 seems to be saying this:
For each data set, evaluate the hypothesis for that data and save it somewhere. When all of the data has been processed take each data point and calculate the "average" value of that data point evaluated against each of the previously generated hypotheses. That average value can be used to calculate the variance of the data set.
So this is just a straightforward averaging calculation, is it not? I'm having some difficulty duplicating the variance values shown in the examples from the book and suspect that this calculation may be the source of the problem.
Thanks for your attention.
yaser 08-05-2012 02:28 PM
Re: Calculating The "Average" Function Of Xn
Quote:
Originally Posted by munchkin (Post 3811) For each data set, evaluate the hypothesis for that data and save it somewhere. When all of the data has been processed take each data point and calculate the "average" value of that data point evaluated against each of the previously generated hypotheses. That average value can be used to calculate the variance of the data set.
Just to clarify. After you have calculated the various hypotheses based on different data sets, you evaluate these hypotheses on a generic point (not necessarily belonging to any data set used for training). The average value you get will be and the variance in the values you get will be . The expected value of with respect to (based on the uniform probability of generating ) is the variance . Is this what you have calculated?
munchkin 08-05-2012 04:44 PM
Re: Calculating The "Average" Function Of Xn
Thanks for your quick reply. I have managed to duplicate the variance result for hypothesis zero of example 2.8. The error had nothing to do with averaging. I don't use a generic x but simply calculate the average of all of the g's generated by the test data sets. This averaged g is evaluated at each data point where comparison with g(x) is required. It seems to work for the constant value case.
The book says that the average g is calculated for "a particular x" and is interpreted as the expected value of the random variable gbar(X). Is this particular x value the "generic" value?
yaser 08-05-2012 09:05 PM
Re: Calculating The "Average" Function Of Xn
Quote:
Originally Posted by munchkin (Post 3816) The book says that the average g is calculated for "a particular x" and is interpreted as the expected value of the random variable gbar(X). Is this particular x value the "generic" value?
Correct. This is just to simplify the concept, but your approach of getting the whole at once is certainly valid.
All times are GMT -7. The time now is 05:32 PM.
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# 275140 (number)
275,140 (two hundred seventy-five thousand one hundred forty) is an even six-digits composite number following 275139 and preceding 275141. In scientific notation, it is written as 2.7514 × 105. The sum of its digits is 19. It has a total of 4 prime factors and 12 positive divisors. There are 110,048 positive integers (up to 275140) that are relatively prime to 275140.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 6
• Sum of Digits 19
• Digital Root 1
## Name
Short name 275 thousand 140 two hundred seventy-five thousand one hundred forty
## Notation
Scientific notation 2.7514 × 105 275.14 × 103
## Prime Factorization of 275140
Prime Factorization 22 × 5 × 13757
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 137570 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 275,140 is 22 × 5 × 13757. Since it has a total of 4 prime factors, 275,140 is a composite number.
## Divisors of 275140
12 divisors
Even divisors 8 4 4 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 12 Total number of the positive divisors of n σ(n) 577836 Sum of all the positive divisors of n s(n) 302696 Sum of the proper positive divisors of n A(n) 48153 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 524.538 Returns the nth root of the product of n divisors H(n) 5.71387 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 275,140 can be divided by 12 positive divisors (out of which 8 are even, and 4 are odd). The sum of these divisors (counting 275,140) is 577,836, the average is 48,153.
## Other Arithmetic Functions (n = 275140)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 110048 Total number of positive integers not greater than n that are coprime to n λ(n) 13756 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 23987 Total number of primes less than or equal to n r2(n) 16 The number of ways n can be represented as the sum of 2 squares
There are 110,048 positive integers (less than 275,140) that are coprime with 275,140. And there are approximately 23,987 prime numbers less than or equal to 275,140.
## Divisibility of 275140
m n mod m 2 3 4 5 6 7 8 9 0 1 0 0 4 5 4 1
The number 275,140 is divisible by 2, 4 and 5.
• Arithmetic
• Abundant
• Polite
## Base conversion (275140)
Base System Value
2 Binary 1000011001011000100
3 Ternary 111222102101
4 Quaternary 1003023010
5 Quinary 32301030
6 Senary 5521444
8 Octal 1031304
10 Decimal 275140
12 Duodecimal 113284
20 Vigesimal 1e7h0
36 Base36 5was
## Basic calculations (n = 275140)
### Multiplication
n×y
n×2 550280 825420 1100560 1375700
### Division
n÷y
n÷2 137570 91713.3 68785 55028
### Exponentiation
ny
n2 75702019600 20828653672744000 5730795771518784160000 1576771148575678273782400000
### Nth Root
y√n
2√n 524.538 65.0406 22.9028 12.2436
## 275140 as geometric shapes
### Circle
Diameter 550280 1.72876e+06 2.37825e+11
### Sphere
Volume 8.72469e+16 9.513e+11 1.72876e+06
### Square
Length = n
Perimeter 1.10056e+06 7.5702e+10 389107
### Cube
Length = n
Surface area 4.54212e+11 2.08287e+16 476556
### Equilateral Triangle
Length = n
Perimeter 825420 3.27799e+10 238278
### Triangular Pyramid
Length = n
Surface area 1.3112e+11 2.45468e+15 224651
## Cryptographic Hash Functions
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12_P71InstructorSolution
# 12_P71InstructorSolution - 12.71 Model Assume the string...
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12.71. Model: Assume the string does not slip on the pulley. Visualize: The free-body diagrams for the two blocks and the pulley are shown. The tension in the string exerts an upward force on the block 2 , m but a downward force on the outer edge of the pulley. Similarly the string exerts a force on block 1 m to the right, but a leftward force on the outer edge of the pulley. Solve: (a) Newton’s second law for 1 m and 2 m is 11 Tm a = and 22 2 . gm a = Using the constraint 21 , aa a −= += we have 1 a = and . a −+ = Adding these equations, we get 2 () , mg m m a =+ or 2 1 12 mmg aT m a mm =⇒ = = ++ (b) When the pulley has mass m , the tensions ( a n d ) TT in the upper and lower portions of the string are different. Newton’s second law for 1 m and the pulley are: 1 2 and a T RT R I α = We are using the minus sign with because the pulley accelerates clockwise. Also, . aR = Thus, a = and 2 Ia a I R RR = Adding these two equations gives
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## This note was uploaded on 12/10/2011 for the course PHYS 1211 taught by Professor Geller during the Fall '09 term at UGA.
Ask a homework question - tutors are online
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# Properties
Label 198.e Number of curves 4 Conductor 198 CM no Rank 0 Graph
# Learn more about
Show commands for: SageMath
sage: E = EllipticCurve("198.e1")
sage: E.isogeny_class()
## Elliptic curves in class 198.e
sage: E.isogeny_class().curves
LMFDB label Cremona label Weierstrass coefficients Torsion structure Modular degree Optimality
198.e1 198b3 [1, -1, 1, -725, 7661] [6] 96
198.e2 198b4 [1, -1, 1, -365, 15005] [6] 192
198.e3 198b1 [1, -1, 1, -50, -115] [2] 32 $$\Gamma_0(N)$$-optimal
198.e4 198b2 [1, -1, 1, 40, -547] [2] 64
## Rank
sage: E.rank()
The elliptic curves in class 198.e have rank $$0$$.
## Modular form198.2.a.e
sage: E.q_eigenform(10)
$$q + q^{2} + q^{4} + 2q^{7} + q^{8} + q^{11} - 4q^{13} + 2q^{14} + q^{16} + 6q^{17} - 4q^{19} + O(q^{20})$$
## Isogeny matrix
sage: E.isogeny_class().matrix()
The $$i,j$$ entry is the smallest degree of a cyclic isogeny between the $$i$$-th and $$j$$-th curve in the isogeny class, in the LMFDB numbering.
$$\left(\begin{array}{rrrr} 1 & 2 & 3 & 6 \\ 2 & 1 & 6 & 3 \\ 3 & 6 & 1 & 2 \\ 6 & 3 & 2 & 1 \end{array}\right)$$
## Isogeny graph
sage: E.isogeny_graph().plot(edge_labels=True)
The vertices are labelled with LMFDB labels.
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# Thread: Tangent to the curve from a parametric equation
1. ## Tangent to the curve from a parametric equation
Hello,
I've got this question I've been labouring over.
The parametric equations of a curve are : x = 1 + 2sin^2Ɵ, y = 4tanƟ.
From a previous question I found dy/dx to be 1 / (sinƟcos^3Ɵ).
Now it says
"Find the equation of the tangent to the curve at the point where Ɵ = π/4, giving your answer in the form y = mx + c.
So what I did was first try get a value for x and y by plugging in π/4 in the parametric equations.
x = 1 + (sin/4)^2
x = 1.00
y = 4tan(π/4)
y = 0.05
Then I put in π/4 into the dy/dx equation.
1 / (sinπ/4)((cosπ/4)^3)
= 72.95
Using the point-slope rule:
y - y1 = m(x - x1)
y - 0.05 = 72.95(x - 1)
y = 72.95x - 72.95 + 0.05
y = 72.95x + 73
According to the mark scheme, that is WAY off, as their answer is y = 4x - 4. Where did I go wrong? Sorry if this is in the wrong category.
2. ## Re: Tangent to the curve from a parametric equation
You made several computational mistakes. Were you using a calculator perhaps? This problem should be solved without one.
When $\displaystyle \theta = \pi/4$:
$\displaystyle x(\theta) = x(\pi/4) = 1+2\sin^2(\pi/4) = 1 + 2 \left(\frac{\sqrt{2}}{2}\right)^2 = 1 + 2\left(\frac{2}{4}\right) = 1 + 1 = 2$.
$\displaystyle y(\theta) = y(\pi/4) = 4\tan(\pi/4) = 4(1) = 4$.
$\displaystyle \frac{dy}{dx}(\theta) = \frac{dy}{dx}(\pi/4) = \frac{1}{(\sin(\pi/4)\cos^3(\pi/4)} = \frac{1}{(\sqrt{2}/2)(\sqrt{2}/2)^3}$.
$\displaystyle = \frac{1}{(\sqrt{2}/2)^4} = \frac{1}{(\sqrt{16}/16)} = \frac{1}{(4/16)} = \frac{1}{(1/4)} = 4$.
So when $\displaystyle \theta = \pi/4$, it's at the point $\displaystyle (2, 4)$ and has tangent line of slope $\displaystyle 4$.
3. ## Re: Tangent to the curve from a parametric equation
Wow, you lost me.
How did you get from $\displaystyle 2\sin^2(\pi/4)$ to $\displaystyle 2 \left(\frac{\sqrt{2}}{2}\right)^2$ ?
4. ## Re: Tangent to the curve from a parametric equation
Another approach to find the slope of the tangent line:
$\displaystyle \frac{dy}{dx}=\frac{dy}{d\theta}\cdot\frac{d\theta }{dx}=\left(4\sec^2(\theta) \right)\left(\frac{1}{2\sin(2\theta)} \right)=\frac{2\sec^2(\theta)}{\sin(2\theta)}$
Hence:
$\displaystyle \frac{dy}{dx}\left|_{\theta=\frac{\pi}{4}}=\frac{4 }{1}=4$
We could also eliminate the parameter $\displaystyle \theta$, to find the equivalent Cartesian equation:
$\displaystyle x=1+2\sin^2\left(\tan^{-1}\left(\frac{y}{4} \right) \right)=1+\frac{2y^2}{y^2+16}$
$\displaystyle xy^2+16x=3y^2+16$
Implicitly differentiate with respect to $\displaystyle x$:
$\displaystyle x\cdot2y\cdot\frac{dy}{dx}+y^2+16=6y\frac{dy}{dx}$
$\displaystyle \frac{dy}{dx}=\frac{y^2+16}{2y(3-x)}$
When $\displaystyle \theta=\frac{\pi}{4}$ we have $\displaystyle (x,y)=(2,4)$ hence, at this point, we have:
$\displaystyle \frac{dy}{dx}=\frac{4^2+16}{2\cdot4(3-2)}=\frac{32}{8}=4$
5. ## Re: Tangent to the curve from a parametric equation
Originally Posted by yorkey
Wow, you lost me.
How did you get from $\displaystyle 2\sin^2(\pi/4)$ to $\displaystyle 2 \left(\frac{\sqrt{2}}{2}\right)^2$ ?
It seems you don't know that $\displaystyle \sin(\pi/4) = \frac{\sqrt{2}}{2}$.
If that's so, then you don't understand some pretty basic trigonometry. If you're taking a calculus class, then that's going to cause you serious problems. I'd suggest you take measures - self-study, a tutor, whatever - to get caught up with trigonometry.
6. ## Re: Tangent to the curve from a parametric equation
The trigonometric rules in my textbook I know and understand. They are:
y = sinx
dy/dx = cosx
y = cosx
dy/dx = -sinx
y = tanx
dy/dx = sec^2x
But it seems that they've thrown me something they haven't taught me. I'm just asking what is that rule you're talking about. Is there a more general law for it?
7. ## Re: Tangent to the curve from a parametric equation
Originally Posted by yorkey
The trigonometric rules in my textbook I know and understand. They are:
y = sinx
dy/dx = cosx
y = cosx
dy/dx = -sinx
y = tanx
dy/dx = sec^2x
But it seems that they've thrown me something they haven't taught me. I'm just asking what is that rule you're talking about. Is there a more general law for it?
it's called the unit circle ... you're expected to have been exposed to it prior to enrolling in a calculus course.
8. ## Re: Tangent to the curve from a parametric equation
Right, I've got my work cut out for myself. Thanks for the help, everyone. :-)
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0
# How many quarter ounces in half pound?
Wiki User
2009-02-06 03:54:28
one half pound is 8 ounces, there are 4 quarter ounces in every ounce, so 8 x 4=32 quarter ounces
Wiki User
2009-02-06 03:54:28
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My Math Forum Functional analysis
Linear Algebra Linear Algebra Math Forum
February 13th, 2018, 01:56 PM #1 Newbie Joined: Dec 2017 From: vienna Posts: 12 Thanks: 1 Functional analysis in a normed space V can any proper W subspace be an open in V?
February 13th, 2018, 08:33 PM #2 Senior Member Joined: Oct 2009 Posts: 841 Thanks: 323 What is the span of a ball centered around 0?
February 14th, 2018, 02:43 AM #3 Math Team Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra $\mathbb R^2 \subset \mathbb R^3$ Thanks from topsquark
February 14th, 2018, 04:34 AM #4
Senior Member
Joined: Oct 2009
Posts: 841
Thanks: 323
Quote:
Originally Posted by v8archie $\mathbb R^2 \subset \mathbb R^3$
What is that supposed to show? The OP didn't ask whether every subspace is open, the OP asked whether it is possible that there is SOME space which has SOME subspace that is open.
February 14th, 2018, 05:43 AM #5
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Joined: Sep 2016
From: USA
Posts: 635
Thanks: 401
Math Focus: Dynamical systems, analytic function theory, numerics
Quote:
Originally Posted by v8archie $\mathbb R^2 \subset \mathbb R^3$
$\mathbb{R}^2$ is not open in $\mathbb{R}^3$. The answer to the question is no. As micromass pointed out, every subspace contains 0 so an open subspace must contain the span of an open ball centered at 0 but this is necessarily all of $V$ so it can't be proper.
February 14th, 2018, 09:08 AM #6
Senior Member
Joined: Aug 2012
Posts: 2,343
Thanks: 732
Quote:
Originally Posted by SDK $\mathbb{R}^2$ is not open in $\mathbb{R}^3$. The answer to the question is no. As micromass pointed out, every subspace contains 0 so an open subspace must contain the span of an open ball centered at 0 but this is necessarily all of $V$ so it can't be proper.
In the indiscrete topology, the only open set containing 0 is the entire space. This can't happen in a normed space, which is a detail needing a bit of explanation.
Last edited by Maschke; February 14th, 2018 at 09:14 AM.
February 14th, 2018, 06:14 PM #7 Member Joined: Jan 2016 From: Athens, OH Posts: 93 Thanks: 48 Isn't this almost obvious? Let W be an open subspace of V and $0\neq x\in V$. Since there is an open ball around 0 contained in W, some scalar multiple of x is in W. Hence x itself is in V. Thanks from Joppy
February 18th, 2018, 03:51 AM #8
Math Team
Joined: Jan 2015
From: Alabama
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Thanks: 902
Quote:
Originally Posted by johng40 Isn't this almost obvious? Let W be an open subspace of V and $0\neq x\in V$. Since there is an open ball around 0 contained in W, some scalar multiple of x is in W. Hence x itself is in V.
What? You start with the hypothesis that $x\in V$ and conclude "Hence x itself is in V"? Did you mean "Hence x itself is in W"?
February 18th, 2018, 08:50 AM #9 Member Joined: Jan 2016 From: Athens, OH Posts: 93 Thanks: 48 Yes, of course. When I saw the typo, it was too late to edit, but I thought it would be clear that I meant $x\in W$.
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# Chem II
posted by .
Find the mass of sodium formate that must be dissolved in 200.00cm^3 of a 1.0M solution of formic acid to prepare a buffer solution with pH=3.40? I have no clue how to work this out. Please help me.
• Chem II -
If you know how to use the Henderson-Hasselbalch equation that is the way to go. If not, let me know and it can be done another way.
• Chem II -
Is it something like PH=pKa +log [base]/[acid]
3.40=??
How do I find the pKa?
## Similar Questions
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A buffer solution contains .4mol of formic acid, HCOOH and a .6mol of sodium formate, HCOONa, in 1L of solution. Ka of formic acid is 1.8 x 10^-4. a) calculate pH b) if 100ml of this buffer solution is diluted to a volume of 1L with …
2. ### Chemistry
A buffer is prepared by mixing 0.30 mole of formic acid (HCNO2) and 0.20 mole of sodium formate (NaCHO2) in enough water to make 0.50 liter of solution. Ka of formic acid is 1.7 x 10-4. Calculate the pH of the resulting buffer solution.
3. ### chem ll
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The pK3 of formic acid is 3.75 A)what is the pH of a buffer in which formic acid and sodium formate have equimolar concentration?
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What mass of solid sodium formate (of MW 68.01)must be added to 125 mL of 0.48 mol/L formic acid (HCOOH) to make a buffer solution having a pH of 4.07?
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# LM317 LED driver with varying DC input
#### Shakin97
Joined Nov 17, 2011
2
Hello,
I have a question regarding using the LM317AT in CC mode to drive an 3x3 array of 9 3.3V, Iavg = 25mA SMT LED's. I have seen many on this forum that would work if the input DC is a constant. My question is how to wire up a LM317AT circuit on a boat to replace my incandescent light fixtures. The problem is when I am running the engine and the battery is charging it can be as high as 14Vdc, but a battery may be as low as 12V. I would like to keep the same amount of current in each leg regardless if the battery(supply) is at 12V or 14V.
Any ideas greatly appreciated.
Last edited:
#### SgtWookie
Joined Jul 17, 2007
22,230
The LM317 is certainly easy to use, but in CC mode, it's not very efficient. It has a ~1.7v nominal dropout when used in voltage mode, and CC mode is another ~1.25v, for a total of nearly 3v lost across the regulator.
So, if you want to plan for worst case scenario, and if you really like buying batteries, 11v-3v=8v left for a series string. 8v/3.3v=2.424... LEDs - of course you have to take the integer of that, so it's 2 LEDs per string. So, ~6.6v for the load, and when the battery is being charged, 14v - 6.6v = 7.4v for the regulator. 47% efficiency is pretty rotten.
By the way, if you let your battery fall below 12v, you are cutting its' life very short.
I'm afraid I don't have time at the moment to give you a better solution; but it would involve a boost-type switching regulator, and would be far more efficient. However, I don't know how many of these things you want to build.
#### Shakin97
Joined Nov 17, 2011
2
Playing around with LTSpice, I think I've figured out a way to drive the 9 LED's just using some diodes for biasing and 3 2N3904 generic transistors which will have to be SMT. This is while having 11.5Vdc to 14.5Vdc which by my best estimate is the extreme given my conditions.
I am varying re from 25 to 47 Ohms so I have either 15mA or 25mA per leg. Kinda like a set dimmer... Maybe call it mood lighting...
#### Attachments
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#### Wendy
Joined Mar 24, 2008
23,457
That looks like it is loosly based off my drawings. You can put a variable pot across the diodes, and vary them that way.
#### bountyhunter
Joined Sep 7, 2009
2,512
Hello,
I have a question regarding using the LM317AT in CC mode to drive an 3x3 array of 9 3.3V, Iavg = 25mA SMT LED's. I have seen many on this forum that would work if the input DC is a constant. My question is how to wire up a LM317AT circuit on a boat to replace my incandescent light fixtures. The problem is when I am running the engine and the battery is charging it can be as high as 14Vdc, but a battery may be as low as 12V. I would like to keep the same amount of current in each leg regardless if the battery(supply) is at 12V or 14V.
Any ideas greatly appreciated.
If it is 25 mA per string, 75 mA total. If each LED is 3.3V, that's about 10V per string. At 14V in, the power loss in the LM317 is 0.3W, which the "T" (TO-220) package can handle easily without any heatsink. The current will be constant as long as the input voltage stays high enough (about 13V). Below that it will taper off.
You could also rig up a simple current mirror using a power transistor PNP like 2N6124 to drive the LEDs mirrored off against a small PNP (like 2N3906) connected as diode with resistor to ground to set it's current. Connect both emitters to V+ and tie both bases together. Then use emitter resistors ratioed to multiply the current on the power transistor side to 75 mA. That would get you a lower dropout voltage so it would regulate down to 12V input.
http://www2.engr.arizona.edu/~brew/ece304spr07/Pdf/PNP current Mirror.pdf
http://www.engin.brown.edu/courses/en162/Mirrors08.pdf
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#### SgtWookie
Joined Jul 17, 2007
22,230
Try changing your resistors to 36 Ohms. Right now you'll have ~14mA through the LEDs. 36 will get you closer to 20 without going over.
.step temp 0 40 10
...as a spice parameter to see how it'll do over temp. You should see from ~17.4mA at low voltage @ 40°C to ~19.9mA @ 0°C.
You can also flip the whole thing upside-down by using PNP 2N3906 transistors as sourcing current to the anodes rather than sinking current from the cathodes.
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#### bountyhunter
Joined Sep 7, 2009
2,512
Try changing your resistors to 36 Ohms. Right now you'll have ~14mA through the LEDs. 36 will get you closer to 20 without going over.
.step temp 0 40 10
...as a spice parameter to see how it'll do over temp. You should see from ~17.4mA at low voltage @ 40°C to ~19.9mA @ 0°C.
You can also flip the whole thing upside-down by using PNP 2N3906 transistors as sourcing current to the anodes rather than sinking current from the cathodes.
Considering the 2N3906 only costs about a dime, I would replace the two 1N94148 diodes with a diode connected 2N3906 (and an emitter resistor) and make it a true current mirror so it will give constant current over temperature changes. You can tweak the value of the emitter resistor to adjust the current in the LEDs.
#### Wendy
Joined Mar 24, 2008
23,457
Current mirrors do not work well outside an IC. This is due to extreme temperature instablility, and they can easily run away. I show this on my albums, but I also mention that they are not a good idea.
..
I tend to favor the transistor emitter constant current source because of the low dropouts. Here is a LED grow light I helped develop a long while back...
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#### bountyhunter
Joined Sep 7, 2009
2,512
Current mirrors do not work well outside an IC. This is due to extreme temperature instablility, and they can easily run away.
No, that is completely wrong if an emitter resistor is used since it provides emitter degeneration that both balances and opposes thermal runaway. I have used them in dozens of applications including adjustable battery chargers where the charge current can vary over a wide range by selecting the emitter resistor.
The only current mirrors that only work in an IC are ones that require exact matching, ie have no negative resistive degenration in the emitter.
What I don't recommend is designs where discrete diodes like 1N4148 are "paired" against transistor junctions since they are WAY off and certainly don't match worth beans. That design "matches" a diode against a resistor and that will have a terrible TC.
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#### Wendy
Joined Mar 24, 2008
23,457
Then it is not a current mirror. See the illustrations I've just put up. Current mirrors are a specific type of circuit. What you show is a standard transistor constant current source.
The illustration I show is the most basic type of current mirror, there are many other configurations. Far as I know, all of them suffer the same temperature instability problems. The reason it isn't a problem on an IC is the thermal coupling on a chip is very tight.
Everything on the 300W LED grow light schematic needs heat sinks. Switching mode power supplies are better, but much more complex.
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#### bountyhunter
Joined Sep 7, 2009
2,512
Then it is not a current mirror. See the illustrations I've just put up. Current mirrors are a specific type of circuit. What you show is a standard transistor constant current source.
The circuit I showed is the example of what NOT to do. The current mirror I recommended was using a diode connected 2N3906 transistor to set the bias current and the three other 2N3906 transistors to feed the LED strings. Emitter resistors are selected to adjust the current in the LED strings to desired value. The ratio of the resistors multiplies the current. A current mirror does not have to be a 1:1 mirror: in fact, most are not. However, inside IC's we adjust the current ratios by emitter areas. In discrete designs we adjust the ratios with emitter resistors.
#### Wendy
Joined Mar 24, 2008
23,457
Have you tried your current mirrors? I don't think they will work as well as you think they well, as they now have features that will interfer with the basic theory of operation of a current mirror, namely the emitter resistors.
Fact is, they will still be thermally sensitive, it is in the nature of how a current mirror works. Current mirrors lean heavily on the BJT being used in it's voltage controlled mode as opposed to a current controlled mode (the traditional Ic=ß Ib). As long as the two transistors mirror each other (that includes temperature) then it works very well. A small change in basic operating paremeters causes major instabilities. If the transistors are operating in two different areas of their current curves they won't be very mirrored either.
A common collector current source is pretty stable, I have verified this experimentally.
I've never seen your variations, nor heard how well they work. It will be something I will have to research and build.
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#### bountyhunter
Joined Sep 7, 2009
2,512
Have you tried your current mirrors? I don't think they will work as well as you think they well
Many dozens of times, a number of them are working even now (battery chargers with selectable charge current rates). I'ts late now, so I'll just post the design assuming the OP wants the best solution. It has approximatey 500 mV of degeneration so it will compensate for even gross mismatches of VBE and temps with minor variation in current across the LEDs. The emitter resistors may have to be tweaked slightly to hit 25 mA precisely but with LEds, but a slight current variation is not visible.
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#### Wendy
Joined Mar 24, 2008
23,457
Don't think that will work. Where is the bias to turn the base current on for Q1 (the first transistor)?
All you will have from this is a bunch of dark LEDs, no current. If you short the BC on Q1 it will not be a current mirror, just a fancy diode/resistor circuit with a bunch of emitter followers.
If you insist I will be glad to build a version with Q1 and Q2, and report the results.
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#### bountyhunter
Joined Sep 7, 2009
2,512
Fixed the typo in the schematic. It's late here. I drew it in five minutes.
#### Wendy
Joined Mar 24, 2008
23,457
Again, that is not a current mirror. I could get identical results by simply substituting Q1 with a conventional diode.
You have a voltage divider with a transistor wired as a diode. The rest is just voltage followers.
The basic theory of operation is missing, that being setting a current by precision voltage across the BE junction. Transistors do not usually use that mode, but with a current mirrors they do. It is also why they are unstable with temperature, the transistors being close to identical is critical.
The circuits drawn using the diodes have one overwhelming difference, they are true constant current sources. You can vary the power supply voltage and the current through the LEDs will not vary.
Plus, the parts count is smaller.
***********************
OK, parts count is not smaller, but it is simpler selection of parts that ends up being a better , more stable design.
The only reason you would want a constant current source is unstable power supply voltage, otherwise a simple resistor would work as well. The other reason is making a variable current with one control for many LEDs.
If you would like to pick up a discussion of current mirrors I would be glad to start another thread on the subject.
***********************
After doing a bit of research you were right, and I was wrong, that is a current mirror.
My statement of replacing the first transistor with a diode is true enough, as is the fact that it is entirely dependent on the power supply, which renders it mostly useless for a LED current source.
Doing a bit of studying the concept of a current mirror is much looser than I had previously thought. For it to be a true constant current source you need to regulate the voltage of the Q1 side, then the current would be independent of the power supply voltage. The circuit is incomplete without voltage regulation.
Which brings up the fact that the circuit I show is very similar to yours, but has fewer parts since the voltage regulation is incorporated into the diodes. It is the best circuit for this kind of job.
You could even get rid of the second diode on the second example, and it would work.
There is another reason I like this design, and that is with very little modification it will vary the intensity 100%, from completely off to 100%, something a current mirror would have trouble doing. PWM is linear, this is not, but it would work.
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#### bountyhunter
Joined Sep 7, 2009
2,512
Again, that is not a current mirror.
Yes, it's just not a 1:1 mirror. It's a 1:2.5 mirror. The mirror is because changes in the diode string are mirrored in the LED strings.
Again, that is not a current mirror. I could get identical results by simply substituting Q1 with a conventional diode.
No, that's wrong. The left PNP is the same device as the others which means it's VBE is nearly the same and behaves the same with respect to varying current and temp. If you drop 1N4148 in there it will not match or track the other VBEs and it sure won't track a resistor. The top design in your post shows two diodes: one corresponds to the transistor P-N junction and the other's voltage is forced across the 36 Ohm resistor. So, the current is the VBE/36 which means it has the diode's tempco..
The circuit I posted works. Simulate it or build it, it works perfectly and will track well over temp. I never claimed it had perfectly constant current regardless of input voltage, but since the application is automotive, the supply hardly varies anyway (12.6 - 14V).
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#### bountyhunter
Joined Sep 7, 2009
2,512
Definition of a current mirror: the only difference is mine has emitter resistors added to force VBE temp tracking and also adjust sharing ratios. It's still a current mirror, or specifically, a "current sourcing mirror".. It's advantage is all the VBEs track over temp. An easy way to get perfect tracking is glue the face of the diode-transistor to the face of one of the other transistors.
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#### bountyhunter
Joined Sep 7, 2009
2,512
The only reason you would want a constant current source is unstable power supply voltage, otherwise a simple resistor would work as well.
Yes and no. An auto supply is hardly unstable (varies only about 10%) but a resistor would not work since the LED load has about 10V across it. A resistor setting 25mA @ 12.5V (100 Ohms) will feed it 40 mA at 14V in which is a 60% increase. The circuit I posted will only vary the LED current about 10% in for the 12.5 - 14V range.
#### SgtWookie
Joined Jul 17, 2007
22,230
Bountyhunter;
Point well taken about the current variations due to temperature and the mismatch between diodes & transistors being used.
Changed it over to a current mirror, and rather than using a resistor for the current sink from the collector, used an LM317 to provide the constant current. This got away from the 3v minimum dropout being in series with the LEDs, and provided for much better regulation than just using a resistor for the collector/base current sink.
The previous circuit I'd posted had a range of ~17.4mA to ~19.9mA over 11.5v to 14.5v and 0°C to 40°C; 2.5mA difference.
This modified version has an LED current range of 19.58mA to 19.74mA over the same voltage and temp range; 0.16mA variation.
It would be better if I'd used an LM317L so I could decrease Iout to ~5mA, but I don't have a model for that. I've been looking ...
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# A student club is planning a fundraising car wash. Last year they charged \$10 per
A student club is planning a fundraising car wash. Last year they charged \$10 per vehicle and washed 120 vehicles. They would like to earn more money this year. For every \$1 increase in price, they know they will wash 5 fewer vehicles.
a) Write a quadratic function to model this situation using v as the number of vehicles and r as the revenue.
b) Determine the best price to charge for the car wash and the revenue expected at that price.
ps. (use completing the square where necessary.)
• Posted: 4 years ago
A student club is planning a fundraising car wash. Last year they charged \$10 per
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# fabulous printable logic puzzle mason website within logic puzzles with grids worksheets
##### fabulous printable logic puzzle mason website within logic puzzles with grids worksheets.
This also can be done by giving them an offer about the reward if they can finish the task. After done with the preparation, you can start instruct the kid to tracing the shapes. At this stage, you also can introduce the name of the shapes.
You can give an example by tracing it’s in a quarter. Ask your kids to do the same. They will learn that they need to make an arch before completing the spiral.
This can be very fun and make your kid enjoy to get more exercise. This learning progress, of course is recommended for upgrading kids’ soft motoric skills.
If you think that spelling words and recognizing the number is too complex, then you can try a simple worksheet that has main purpose of tracing skill. It’s a line tracing worksheet that can be used to achieve this skill.
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### 4Th Grade Probability Worksheets
March 6, 2021
The line also has various shape. It usually comes with a simple tracing that can easily followed for the beginner.
Take a look on the colored pictures, whether it’s tidy enough or not. Give them an instruction that next time they should put the color inside the area. It will make them understand about coloring the object in proper way.
The exercise can be continued by asking your kid to finish the tracing from the different start point. If the previous exercise is started from the center, then start it from the outer point.
This can be done step by step. After a quarter, continue to the half and finish until the perfect spiral. It’s better to start the spiral from the center.
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### Au Aw Worksheets First Grade
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As usual, you have to give an instruction for your kid. Teach them about the difference between straight line and spiral line. The spiral line usually comes with the complex shape.
Learning to write the letter is the first skill that should be done for kids. This worksheet has a simple instruction for tracing the letter. You can make a demonstration to the kid before instruct them to do it by themselves. Remember, this is need a good patience.
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The next worksheet for playgroup that can be used is the shape tracing worksheets. It’s also has a nice function for stimulating the soft motoric skill in children.
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## Functions and Arrays in Mathcad
June 21, 2020, at 04:08 AM by 136.36.211.159 -
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Mathcad is a tool to arrange, calculate, and visualize engineering calculations and functions can help with reusing expressions in subsequent evaluations. Use the following worksheet to step through example problems related to using functions effectively in a Mathcad worksheet.
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Mathcad is a tool to arrange, calculate, and visualize engineering calculations and functions can help with reusing expressions in subsequent evaluations. Use the following worksheets to step through example problems related to using functions and matrix operations effectively in a Mathcad worksheet.
#### Mathcad matrix and vector operations
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(:title Functions and Arrays in Mathcad:) (:keywords function, array, Mathcad, university course:) (:description Functions and Arrays in Mathcad - Problem-Solving Techniques for Chemical Engineers at Brigham Young University:)
Mathcad is a tool to arrange, calculate, and visualize engineering calculations and functions can help with reusing expressions in subsequent evaluations. Use the following worksheet to step through example problems related to using functions effectively in a Mathcad worksheet.
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# Improving Math Study and Test Taking Skills
This information comes from a video which is on reserve under "James Jones" in the Learning Resources Center. Ask at the circulation counter for the video on "Test Taking Skills". The host of the video is Dr. Paul Nolting.
This video is different from most that you may have seen, including Where There's a Will There's an "A", in that this video is directed specifically at mathematics students and exams. The other videos are generic in scope.
The video is divided into two parts. Part 1 is about 35 minutes and covers how to prepare for and then take math tests. Part 2 is about 30 minutes and covers analyzing the mistakes made on an exam.
Warning! Dr. Nolting is like all math instructors and says some things that sound really corny. However, the information that he provides is important and is more than you can get by just reading this page. Please watch the video. There are two on reserve so that more than one person can watch at a time.
## Test Preparation Skills
3. Review old tests
4. Pre-test
5. Study groups
6. Calculator
## 10 Steps to Better Test-Taking
1. Memory Data Dump
Write down information you may forget -- formulas, dates, places, etc.
2. Preview Test
Write your name on the test and review the entire test
3. Second Memory Data Dump
Write down additional important information you may forget
4. Test Progress Schedule
Decide the best way to get the most points in the least time
First, answer the easiest questions with the most points
Read each question twice and set a time limit for solving it -- or skip it
7. Review Skipped Questions
Recall related information about each question
Do not leave a question blank
9. Review Entire Test
Look for mis-read directions and careless errors
10. Use All Of Your Test Time
To stop early can mean lost points on your test
## 6 Types of Test-Taking Errors
Avoid by reading all the directions
2. Careless Errors
Avoid by reviewing the test
3. Concept Errors
Avoid by understanding rules and properties
4. Application Errors
Avoid by predicting test questions
5. Test-Taking Errors
Avoid by:
• Finding where most test questions are missed
• Reviewing last step in problems
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* Site-Outage Notice: Our engineering elves will be tweaking the Shmoop site from Monday, December 22 10:00 PM PST to Tuesday, December 23 5:00 AM PST. The site will be unavailable during this time.
Dismiss
# Surface Area and Volume Resources
## Best of the Web. Picked by our PhDs
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An interactive site that helps you visualize pretty much any 3D shape you'll ever need. If nets have got you in a tangled mess, you can take a gander at them and cut yourself loose. Rotate the solids and you'll be counting edges, vertices, and faces instead of sheep to fall asleep at night.
Basic Mathematics: Volume
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Surface Area and Volume Quiz
By now, you've got some mad surface area and volume skillz. Use this quick 6-question quiz and test yourself to see how awesome your geometry chops really are. We're sure you'll ace it, no problem.
### Videos
How to Turn a Sphere Inside Out
Have you ever wanted to turn a sphere inside out? If you think it can't be done, just take a look at this.
Volume and Surface Area
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Plotting Points in 3D
If plotting points in three dimensions seems like something for a rocket scientist, this video's got your back. It will explain everything you need to know to get started working in the third dimension, from point-plotting to distance-finding to midpoint-calculating. Before you know it, you'll have a third degree in the third dimension.
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Illuminations: Geometric Solids
This interactive tool lets you play with Platonic solids. You can rotate them, enlarge them, and even create your own nets to print out and fold up. Like a marriage counselor between you and surface areas, it'll help you understand each other better and together, you'll build a solid (relationship).
Volume Shape Shoot Game
How voluminous is your knowledge of 3D shapes? Earn points by clicking on the flying solids. Sit back, relax, and click when the spirit moves you, or be a daredevil and select timed mode. Whatever tickles your fancy. Be careful, though, since you lose points by clicking on the wrong solid.
Surface Area Calculator
Now that you know what it takes to find the surface area, you can leave it up to a calculator. Just select the shape you need, input the measurements, and presto! And if you want a refresher on the actual formulas used, just scroll down.
Volume Calculator
When heights and cubic units have got you down, just pop in the solid's values into a volume calculator and let it do the work for you. The formulas are all spelled out below too, just in case you want to take a crack at volume the old-fashioned way.
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Operator To Function
Most Khepri builtin operators, and all user defined operators, can be converted to a function by wrapping the operator in parentheses:
``````var add = (+);
``````
This allows operators to be passed directly to functions:
``````[1, 2, 3].reduce((+), 0); // 6
``````
See dot expressions for creating partially applied accessor expressions such as:
``````var getX = (.x);
getX { x : 3}; // 3
``````
Partial Application
As shorthand for partially applying a binary operator, you can include a curried argument in the parentheses. For all operators except binary `.` and `new`, the comma between the operator and the argument is optional
``````var add10 = (+ 10); // could also write (+, 10)
``````
Currying always curries the first arguments of the operator.
Application with Curried ops
Application and curried operators work fine, but have a syntax that may be confusing coming from Javascript.
``````var f = \x -> x;
// These are the same
f (+);
f ( (+) );
// The curried operator is always passed as the argument
f (+, 1);
f ( (+, 1) );
// You can change this behavior be wrapping the argument list in parens
f ((+), 1);
``````
Flipped Binary Operators
You can flip a binary operator's arguments by prefixing the operator with `_`. Used with currying, this allows the second argument of a binary operator to be curried (think of `_` as the placeholder where the function input will go).
``````var div10 = (_ / 10);
div10 60; // 6
``````
For all ops except the keyword operators `new` and `instanceof`, the space between `_` and the operator is optional.
``````(_ / 10); // no space needed
(_ new, 10); // Without space this becomes` (_new, 10)` where `_new` is an identifier.
``````
Supported Operator
The following operators can be converted to functions (along with any user user defined operator derived from these operators).
Note that logical operators, like regular functions, do not use short-circuit evaluation once they are converted to functions.
Unary
• `typeof`
• `void`
• `~`
• `!`
• `++`
• `--`
Binary
• `instanceof`
• `*`
• `/`
• `+`
• `-`
• `%`
• `<<`
• `>>`
• `>>>`
• `<`
• `>`
• `<=`
• `>=`
• `==`
• `===`
• `!=`
• `!==`
• `&&`
• `||`
• `??`
• `|>`
• `<|`
• `\>`
• `\>>`
• `<\`
• `<<\`
• `.` - Computed member `a.(b)`
• `@` - Binary curry only. `a @ b`
• `new` - Binary version only. `new a b`
• `?`
Output
Converting an operator to a function will add at most one declaration per operator to the header of the output file:
``````// Input
foldl @ (+) @ 0;
``````
``````// Output
var __add = \x y -> x + y;
``````
Uncurried operators therefore evaluate to simple object references, and can be used with very little overhead
``````// Input
// In the body of the for loop, `(+)` is just a reference and
// does not declare a new function every iteration
for (var i = 0; i < 10000; i = i + 1)
f(i, (+));
``````
``````// Output
var __add = \x y -> x + y;
for (var i = 0; i < 10000; i = i + 1)
``````
Multiple instances of a single, uncured operator will always reference the same function in a file, but not across multiple files. Although you can safely set properties or perform equality tests on converted operators within a file, this behavior is not recommended.
Inlining is used for converted operators, and converted operators that inlining makes unreachable will not be added to the header.
``````// Input ----
(+)(2, 3);
(+, 2);
// Output ----
6;
\y -> 2 + y;
``````
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## DC Circuits
PIRA classification 5F
Grayed Demos are either not available or haven't been built yet
Please note that these tables have not yet been edited to match the equipment that is available within the UW-Madison lecture demo lab. There maybe many items listed within these tables that we either "can not do" or have available.
# 5F10. Ohm's Law
PIRA # Demonstration Name Abstract 5F10.05 charge density in circuits Two demonstrations: first, an electroscope is used to probe the charge density along a large resistance attached to a 5 KV supply, and second, an example where current is flowing through a resistance with no change in potential. 5F10.10 Ohm's law Measure current and voltage in a simple circuit. Change the voltage or resistance. 5F10.10 Ohm's Law An ammeter, voltmeter, rheostat, and battery pack are connected to demonstrate Ohm's law. 5F10.10 Ohm's law A battery, rheostat, and meters in a circuit. 5F10.10 Ohm's law Measure current and voltage in a simple circuit. 5F10.10 Ohm's law Place 2, 4, and 6 V across a resistor and measure the current, then graph. 5F10.12 water analogy circuit A water analogy illustrates voltage drops across a dc circuit. 5F10.15 water Ohm's law analog 5F10.15 water analog A water analog of Ohm's law. 5F10.15 IR drop in a wire Clip wires from the terminals of flashlight lamps at various points along a stretched wire carrying 2 - 5 amps. 5F10.20 potential drop along a wire Lecture galvanometers configured as a voltmeter and ammeter measure current and voltage on several samples of wire of the same length. A slide clip can be used to vary length. 5F10.20 voltage drop along wire Measure the voltage at six points on a long resistance wire. 5F10.25 potential drop with Wimshurst A 3 m long wood bar is attached at one end to one terminal of a static machine. The other end can be grounded or insulated. Attach several electroscopes along the bar to show flow of charge and potential drop. 5F10.26 high voltage Ohm's law Two ends of a dry stick are attached to a static machine. Measure with an electrostatic voltmeter and microammeter.
# 5F15. Power and Energy
PIRA # Demonstration Name Abstract 5F15.10 electrical equivalent of heat Measure the voltage and current to a heating coil in a calorimeter. 5F15.10 heat and electrical energy A heating coil in a calorimeter. 5F15.10 electrical equvalent of heat Voltage, current to a heater and temperature rise in water are measured. 5F15.10 electrocalorimeter Determine the power delivered by temperature change in water and compare to that computed from voltage, current, and time. 5F15.11 flow calorimeter Water is heated electrically as it flows through a tube. 5F15.12 heating by current from a static mot The ends of a piece of wood sealed in a glass tube are attached to a static machine. The half watt dissipated heats the air and an attached manometer measures the volume change. 5F15.15 KWH meter and loads Measure the power consumed by an assortment of household appliances. 5F15.16 heating with current Large currents are passed through No. 18 nichrome wire and the volts and amps are measured. 5F15.17 heating wires in series Several lengths of different wires of the same length are soldered together in series and a piece of paper is hung from each by soft wax. As current is passed through the wire, the paper falls off at different times. 5F15.20 hot dog cooker Hook nails to 110V and place them on and then in a hot dog. 5F15.20 hot dog frying Apply 110 V through a hot dog and cook it. 5F15.30 fuse with 30v lamp 5F15.31 fuse-wire problem With fuse wires of different diameters connected in parallel, which will burn out first? 5F15.32 vaporize wire with 500 amp surge Short a low voltage high current transformer with zinc coated iron wire. 5F15.33 fuse wire Fuse wire is used with a miniature house circuit. 5F15.34 fuses Fuse wire of different sizes are connected across a heavy copper buss. 5F15.35 fuse with increasing load 5F15.40 voltage drops in house wires Two resistance wires substituting for house wiring glow when they power a load of lamps and heaters. 5F15.45 I2R losses Copper and nichrome wires in series show different amounts of heating due to current. A paper rider on the nichrome wire burns.
# 5F20. Circuit Analysis
PIRA # Demonstration Name Abstract 5F20.10 Kirchoff's voltage law Measure the voltages around a three resistor and battery circuit. 5F20.10 Kirchoff's voltage law Measure the voltages around a three resistor and battery circuit. 5F20.10 sum of IR drops Measure the voltages across three resistors and a battery in a series circuit. 5F20.13 voltage divider A simple series circuit of a battery and two resistors. 5F20.15 continuity of current Same as Eo-4. 5F20.15 continuity of current An ammeter can be inserted into any branch of a circuit to show currents in and out of a node. 5F20.16 conservation of current Measure the currents entering and leaving a node. 5F20.20 superposition of current Same as Eo-7. 5F20.20 superposition of currents Measure the current from one battery, a second in another position, and the combination in a circuit. 5F20.20 superposition Shows a standard superposition circuit. 5F20.25 reciprocity Shows a standard reciprocity circuit. 5F20.30 potentiometer A slide wire potentiometer is used with a battery and demonstration galvanometer. 5F20.30 potentiometer A slide wire potentiometer with a standard cell. 5F20.31 rheostat as potential divider Contrast the slide wire rheostat when used as a rheostat or potential divider. 5F20.32 long potentiometer Use a ten foot length of nichrome wire as a slide wire potentiometer. 5F20.33 rheostat potential divider A rheostat and six volt battery demonstrate a potential divider. 5F20.40 wheatstone bridge - slide wire The slide wire Wheatstone bridge. 5F20.40 wheatstone bridge - slide wire Two nichrome wires are stretched across the lecture bench and sliding clips connected to a galvanometer are used to find equal potential points. 5F20.41 wheatstone bridge - human galvan. Stretch a loop of close line previously soaked in salt solution in a parallelogram and hook the ends to a 110 V line. Touch two points of the same potential without shock. 5F20.42 wheatstone bridge A demonstration Wheatstone bridge with a built in meter and several plug in resistors. 5F20.45 lightbulb wheatstone bridge A Wheatstone bridge configuration with lightbulbs for resistors. 5F20.45 light bulb wheatstone bridge Four light bulbs in a Wheatstone bridge arrangement with light bulb indicator. 5F20.45 light bulb wheatstone bridge A light bulb Wheatstone bridge using 110 ac. 5F20.45 wheatstone bridge Four 60 W lamps in a diamond bridge with a 10 W lamp as the indicator. An additional 6 V lamp can be switched in when the circuit is balanced. 5F20.45 wheatstone bridge Three 110 V lamps and a rheostat make up the diamond of a Wheatstone bridge and a small lamp serves as an indicator. 5F20.50 series and parallel light bulbs A light bulb board with switches allows configuration of several combinations of series and parallel lamps. 5F20.50 series and parallel light bulbs A light bulb board with switches allows configuration of several combinations. 5F20.50 parallel and series light bulbs Three similar wattage lamps in series, three in parallel. 5F20.50 series-parallel circuits A series-parallel circuit with three bulbs and six switches can be connected 14 ways. 5F20.50 series/parallel light bulbs Three 110 V lamps are wired in series and three are wired in parallel. 5F20.51 light bulb board - 12 V A board with 12V bulbs and a car battery allow combinations of up to three series or three parallel loads. 5F20.55 series/parallel resistors Measure the current flowing through a wire resistor with 6 V applied and then series and parallel combinations. 5F20.56 wire combinations A wire circuit is arranged so a segment of n length can have 1 or n wires in parallel. Drawing. 5F20.60 equivalent series resistance A series of resistors in a circuit are replaced by a single resistor. 5F20.61 parallel resistance - integral value A formula for obtaining integral values of resistors in parallel to obtain an integral equivalent resistance. 5F20.61 equivalent parallel resistance Parallel resistors are replaced by a single resistor in a circuit. 5F20.63 Thevenin's equivalent resistance A Wheatstone bridge resistance circuit is used to reduce resistor combinations to an equivalent resistance. 5F20.64 equivalent circuit flasher A neon flasher circuit shows the combination rules for series and parallel combinations of resistance and capacitance by timing light flashes. 5F20.71 large circuit boards A modular circuit board made for 500 student auditoriums. 5F20.72 general circuits board A circuit board laid out so meters can be plugged in and readings taken for demonstrations of series-parallel circuits and Kirchhoff's laws. 5F20.75 three-way switch A large circuit board demonstrates a three way switch. 5F20.79 one boar, river, six people An electrical circuit for solving the problem of getting across the river. 5F20.95 equivalent resistance analog comput. Using the equivalent resistance of a circuit as an analog computer for finding the focal length of an optical problem.
= 5F30. RC Circuits =
PIRA # Demonstration Name Abstract 5F30.10 capacitor and light bulb A large lelectrolytic capacitor, a light bulb, and a 120 V dc supply in series show a long time constant. 5F30.10 capacitor and light bulb A 5600 microF capacitor is charged and discharged through 7.5 and 40 W light bulbs. 5F30.10 long RC time constant A 5600 microF capacitor, a light bulb, and a 120 V dc supply in series show a long time constant where the bulb dims as the capacitor charges. 5F30.11 light the bulb Charge a capacitor with DC and discharge through a light bulb, try the same thing with AC. 5F30.12 discharge a capacitor Discharge a capacitor through a resistor. Read the voltage with a meter. 5F30.15 RC time constant on galvanometer A series RC circuit with a galvanometer. Diagram. 5F30.16 RC voltage follower Use a voltage follower to isolate the circuit from the display. 5F30.20 RC time constant on scope A circuit with a slow time constant (.1 - 10 sec.) is charged and discharged and the current and voltage are displayed on a dual trace storage scope. 5F30.20 RC charging curve Show charging and discharging a RC circuit with a battery on an oscilloscope. 5F30.21 RC time constant Show the time constant from an RC circuit on an oscilloscope. 5F30.21 RC time constant A plug in circuit board for showing RC time constants on the oscilloscope. 5F30.22 time constant of an capacitive cir. The time constant of a RC circuit driven by the calibration signal is shown on an oscilloscope. 5F30.28 finding R from time constant A circuit to measure high resistances by using an RC charging time. 5F30.50 series and parallel capacitors Two 2 microF capacitors in series or parallel with a 40 W lamp. 5F30.60 neon relaxation oscillator 5F30.60 blinking neon bulb A neon bulb in parallel with a capacitor will light periodically as the capacitor charges and discharges. 5F30.60 RC relaxation oscillator An RC relaxation oscillator has a neon lamp across the capacitor provide a visible discharge. 5F30.60 RC flasher circuit A neon lamp in parallel with the capacitor in a series RC circuit. 5F30.60 flashing neon light A battery powered neon light oscillator. 5F30.60 neon relaxation oscillator A circuit for a neon relaxation oscillation oscillator. Reference: AJP 13(12),415. 5F30.60 relaxation oscillator An RC neon light relaxation oscillator. 5F30.61 relaxation siren oscillator A double RC relaxation oscillator with slow and fast periods gives a siren waveform. 5F30.68 backward and forward waves RC circuits are used to get a wave in neon bulbs that goes from the sink to the source. 5F30.71 capacitance operated relay References but no information on the circuit. Bring your hand close to a aluminum plate and the relay triggers. 5F30.80 fun circuit One box has switches that control two lights in another box but only one wire connects the two boxes.
# 5F40. Instruments
PIRA # Demonstration Name Abstract 5F40.10 sensitivity and resistance of a galv A circuit for the determination of galvanometric constants. 5F40.10 sensitivity and resistance of galvan Use external resistors to measure the resistance and sensitivity of a galvanometer. 5F40.15 voltmeter and electroscope Connect series resistance to a galvanometer to make a voltmeter with low sensitivity and measure several dry batteries in series with both the voltmeter and an electroscope. 5F40.20 converting a galvanometer to a voltm Knowing the resistance and sensitivity of a galvanometer, add a series resistance and check with a voltage. 5F40.20 galvanometer as voltmeter and ammete A galvanometer is used with shunt and series resistors. 5F40.21 loading by a voltmeter Measure the voltage across a high resistance circuit with high and low impedance voltmeters. 5F40.25 converting a galvanometer to a ammet Knowing the resistance and sensitivity of a galvanometer, add a shunt resistance and measure a current. 5F40.30 hot wire ammeter A crude hot wire galvanometer. 5F40.30 hot wire ammeter Diagram of a hot wire ammeter. (E-171). 5F40.35 iron vane meter Repulsion from induced magnetism in two soft iron bars in a solenoid forms the basis of a heavy current ammeter. 5F40.50 multimeters A couple multimeters are pictured.
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https://se.mathworks.com/matlabcentral/answers/1689875-how-to-give-trail-to-particles-how-to-fade-out-plots-with-time-complex-example?s_tid=prof_contriblnk
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# How to give trail to particles/ How to fade out plots with time (complex example)
26 views (last 30 days)
Niklas Kurz on 6 Apr 2022
Commented: DGM on 19 Apr 2022
Inspired from a webproject I have created this n-body-simulation:
First of a function to acces on:
% calculating acceleration of particles
function [a] = getAcc(pos,mass,softening)
% gravitation constant
G = 1;
% Number of particles
N = size(pos,1);
% acceleration matrix
a = zeros(N,3);
% loop to fill in acceleration by interaction with each particle
for i = 1:N
for j = 1:N
if i ~= j
dx = pos(j,1)-pos(i,1);
dy = pos(j,2)-pos(i,2);
dz = pos(j,3)-pos(i,3);
dr = (dx.^2 + dy.^2 + dz.^2+ softening.^2).^(-3/2);
a(i,1) = a(i,1) + G * mass(j,1) * (dr * dx);
a(i,2) = a(i,2) + G * mass(j,1) * (dr * dy);
a(i,3) = a(i,3) + G * mass(j,1) * (dr * dz);
end
end
end
followed by the Main script:
clear
%% setting initial conditions
% Number of particles
N = 10;
% max distance between partices at start
rmax = 4;
pos = rmax * rand(N,3);
% max velocity for particles
vmax = 1;
vel = vmax * rand(N,3);
% max mass of particles
mmax = 5;
mass = mmax * ones(N,1);
% minimal distance parameter
softening = 0.5;
% calculating acceleration of particles
acc = getAcc(pos,mass,softening);
% selecting start time
t_start = 0;
% selecting timestep
dt = 0.04;
% selecting end time
t_end = 250;
% initializing figure
figure
grid on
view(3)
%% starting main loop
for i = 1:t_end
vel = vel + acc * dt/2;
[pos] = pos + vel * dt;
acc = getAcc(pos, mass, softening);
vel = vel + acc * dt/2;
t_start = t_start + dt;
hold all
for j = 1:N
plot3(pos(j,1),pos(j,2),pos(j,3),'b.','MarkerSize',4)
end
pause(0.005)
end
However I'm running into a problem here: the particles path will not disapear gradually since all positions are hold. I saw some clever ways implementing a fade but I'm not able adapting it here. Maybe someone brave enough working through all the code may give some advises.
DGM on 7 Apr 2022
Edited: DGM on 7 Apr 2022
Here's one way using scatter3()
% setting initial conditions
% Number of particles
N = 3; % i'm using fewer points for ease of viewing
% max distance between partices at start
rmax = 4;
pos = rmax * rand(N,3);
% max velocity for particles
vmax = 1;
vel = vmax * rand(N,3);
% max mass of particles
mmax = 5;
mass = mmax * ones(N,1);
% minimal distance parameter
softening = 0.5;
% calculating acceleration of particles
acc = getAcc(pos,mass,softening);
% selecting start time
t_start = 0;
% selecting timestep
dt = 0.04;
% selecting end time
t_end = 250;
% initializing figure
grid on
view(3)
% preallocate outputs
xpos = zeros(N,t_end);
ypos = zeros(N,t_end);
zpos = zeros(N,t_end);
% starting main loop
for i = 1:t_end
vel = vel + acc * dt/2;
[pos] = pos + vel * dt;
acc = getAcc(pos, mass, softening);
vel = vel + acc * dt/2;
t_start = t_start + dt;
% store outputs
xpos(:,i) = pos(:,1);
ypos(:,i) = pos(:,2);
zpos(:,i) = pos(:,3);
end
% each trajectory has its own solid color
cmap = hsv(N);
% alpha is a linear gradient from 1 to 0
alph = fliplr(linspace(0,1,t_end));
for ks = 1:N
hs = scatter3(xpos(ks,:),ypos(ks,:),zpos(ks,:),10,cmap(ks,:),'filled');
hold on;
hs.MarkerFaceAlpha = 'flat';
end
function [a] = getAcc(pos,mass,softening)
% gravitation constant
G = 1;
% Number of particles
N = size(pos,1);
% acceleration matrix
a = zeros(N,3);
% loop to fill in acceleration by interaction with each particle
for i = 1:N
for j = 1:N
if i ~= j
dx = pos(j,1)-pos(i,1);
dy = pos(j,2)-pos(i,2);
dz = pos(j,3)-pos(i,3);
dr = (dx.^2 + dy.^2 + dz.^2+ softening.^2).^(-3/2);
a(i,1) = a(i,1) + G * mass(j,1) * (dr * dx);
a(i,2) = a(i,2) + G * mass(j,1) * (dr * dy);
a(i,3) = a(i,3) + G * mass(j,1) * (dr * dz);
end
end
end
end
I imagine this isn't really what you want though. I imagine you want something that fades points dynamically as new points are plotted. That can be done, but it'll be slow.
DGM on 19 Apr 2022
There may be some way of addressing the individual descendant (marker) objects in a scatter plot that I don't know about (perhaps something undocumented). If that were possible, then you might only need N scatter objects instead of N*t_end objects. I don't know if that is an option, but it might be a big boost to speed if it were.
Image Analyst on 6 Apr 2022
Perhaps if you used scatter() to create the markers. You can create an array with the colors for all the markers so have your older ones be faded and your newer ones be brighter/bolder/more colored. You can then put hold on and call plot() if you want lines between the markers.
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# Dell order qualifiers
We expect that this ur-ingredient, this most elemental spacetime stuff, does not allow dissection into ever smaller pieces because of the violent fluctuations that would ultimately be encountered The whole difficulty of the subject lies in the necessity of thinking in an unfamiliar way, and in realising that many properties which we have thought inherent in number are in fact peculiar to finite numbers.
Before the discovery of how to embed calculus within set theory a process that is also called giving calculus a basis in set theoryit could have been more easily argued that science does not need actual infinities. This definition of same size was recommended by both Cantor and Frege.
Weyl is quoted in Kleenep. This particular version, in the Olympic Games, featured 27 men a side. How big is infinity? Peano had given an axiomatization of the natural numbers.
Peasants played most of their sports on foot; aristocrats played most of theirs on horseback. Here are four suggested examples where infinity occurs within physical science.
We can know a priori even more about space than about time, he believed; Dell order qualifiers he declared that the geometry of space must be Euclidean. One response that string theorists make to this problem about too many particles is that perhaps the infinity of particles did exist at the time of the Big Bang but now they have all disintegrated into a shower of simpler particles and so do not exist today.
More discussion of the role of infinity in mathematics and science continues in later sections of this article. Aquinas argued in his Summa Theologia that, although God created everything, nothing created by God can be actually infinite.
The best Dell order qualifiers can do is to have a rule for adding more members to a set. The volume of spacetime is finite at present if we can trust the classical Big Bang theory.
Then 23, and yes, but with more delay. Why then should it be assumed, as it often is, that all appeals to infinity in scientific theory are approximations or idealizations? If the collection is infinite and its sets are not well-ordered in any way that has been specified, then there is in general no way to define the choice set.
This derivative was defined by Leibniz to be where h is an infinitesimal. The first program for providing this basis began in the late 19th century. How about non-classical quantum mechanics, the proposed theories of quantum gravity that are designed to remove the disagreements between quantum mechanics and relativity theory?
Some contemporary philosophers want to avoid these metaphysical commitments, and they recommend saying a set is whatever satisfies the axioms of the best set theory.
MLS[ edit ] Gibbs returned to the United States and play in Major League Soccerwhere he would be easily visible and available for international matches. The 9 companies who compete for CB business and their contract numbers are: He believed that, if a theory implies that some physical properties might have or, worse yet, do have actually infinite values the so-called singularitiesthen this is a sure sign of error in the theory.
This metaphysical position is reflected in the principles of logic that are acceptable to an intuitionist. The scientists and results-oriented mathematicians of the golden age of nothing had no good answer to the coherence problem.
The good files will be uploaded into galleries by class Open B and Utility A as normal and you will be able to order them as prints. See Leplin for more details about anti-realist arguments, such as those of instrumentalism and constructive empiricism.
Another American power outage in foursomes is neither a surprise nor a reason for the red, white and blue side to panic, but the difficult task of winning a road game Ryder Cup for the first time in 25 years has certainly become more challenging. Unfortunately, string theory has its own problems with infinity.
There is not much I can do about that unless I only shoot when your dog is within range to fill the frame. Thus it appears, for the time being at least, that we need to take the use of the infinite seriously, particular in its role in the mathematical description of the physical continuum.
These numbers are also called transfinite ordinals and transfinite cardinals. In this way, the assumptions used in informal reasoning in arithmetic are explicitly stated in the formalism, and proofs in informal arithmetic can be rewritten as formal proofs so that no creativity is required for checking the correctness of the proofs.
The constructivist believes that to justifiably assert the negation of a sentence S is to prove that the assumption of S leads to a contradiction.
Some critics of infinity argue not just that paradox can occur but that paradox is essential to, or inherent in, the use of the concept of infinity, so the infinite is beyond the grasp of the human mind. Nobody would have the time to count from 0 to any aleph.
Immanuel Kant — declared that space and time are both potentially infinite in extent because this is imposed by our own minds. That finite set contains n elements. When creating set theory, mathematicians did not begin with the belief that there would be so many points between any two points in the continuum nor with the belief that for any infinite cardinal there is a larger cardinal.Today I found out the origin of the word “soccer”.
For all you out there who love to complain when Americans, and certain others, call “Football”, “Soccer”, you should know that it was the British that invented the word and it was also one of the first names of what we now primarily know of as “Football”.
A first-order theory is a set of sentences expressed in a first-order language (which will be defined below). A first-order formal system is a first-order theory plus its.
Feb 27, · The 9/11 conspiracy theories are pretty well known by now. The BBC addressed them earlier this month with a documentary, The Conspiracy Files, shown within the UK. Until now, I.
DTCSV Rally Trial - note that Win photos will have sign board updated after order is placed. DTCSV Obedience Trial - LOW RES Images. The Los Alamitos Race Course is the perfect place to spend a fun-filled day of horse racing and world class dining.
Sep 01, · A glossary of terms used in the body of this dictionary. See also Wiktionary:Glossary, which contains terms used elsewhere in the Wiktionary community.
Dell order qualifiers
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How does damage outside of combat work in Carcosa?
In Geoffrey McKinney's old-school setting and ruleset Carcosa, hit dice are handled unusually. If you already know how, skip to the actual question, below. Otherwise, here is a refresher: a character or monster has a set number of hit dice, as in most versions D&D, but the type of die used is not fixed. Hit dice type are determined as needed, as are hit points. According to the rules (page 5 of the LotFP edition), hit points are determined "at the beginning of each combat."
This means that when combat begins:
1. the type of hit dice are determined; let's assume a 3 HD character rolls d6's
2. the 3d6 are rolled with the results [ 1, 3, 6 ] and placed in lowest-to-highest order, left-to-right, as shown here
3. damage dealt is removed from the rightmost die; when it reaches 0, it is removed; if this character suffers 8 hp in combat, he will end combat with only 2 HD remaining on the table
4. in a later combat, the character will only roll 2 HD (of a type to be determined then)
Bonuses or penalties for Constitution are added immediately after rolling in step #2, and bonus HP for monsters with (say) 3 HD +1 are treated similarly.
The Actual Question: How does damage outside of combat work? The rules state that hit points are rolled "at the beginning of combat," but damage is sometimes done outside of combat. For example, if a character, while sneaking around, falls off of a wall, he might take 1d6 falling damage. Where does that get stored?
If it's held in reserve to be applied in combat, it might mean that he will die immediately upon rolling hit dice. He could have 2 HD and take 4 falling damage. When combat begins, he rolls d4 hit dice, which result in [ 1, 2 ], and he is now at -1 hp and dead.
Taking damage could trigger a complete calculation of hit points by following steps #1 and #2 above and then checking whether a hit die is lost. This seems like it could be a pain.
How is this addressed in actual play?
-
I'm not familiar with the system you're describing; but, common sense would suggest that maybe you have to roll his HD just like you would in combat, and subtract accordingly. – RMorrisey Jun 24 '12 at 17:57
It does seem like a straightforward way to do things, but it makes it extremely difficult, that way, to nickel-and-dime somebody out of his or her hit points. (This may be intentional.) It also seems like it would add a lot of overhead to otherwise very simple dice transactions. (This seems unlikely to be intentional, but Carcosa sure does love dice…) – rjbs Jun 25 '12 at 1:03
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https://www.teachstarter.com/au/teaching-resource/remembrance-day-activity-counting-peg-cards/
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teaching resource
# Remembrance Day Activity - Counting Peg Cards
• Updated: 02 Oct 2023
Practise counting sets of objects with a group of 15 printable Remembrance Day peg cards.
• Non-Editable: PDF
• Pages: 1 Page
• Years: P - 1
teaching resource
# Remembrance Day Activity - Counting Peg Cards
• Updated: 02 Oct 2023
Practise counting sets of objects with a group of 15 printable Remembrance Day peg cards.
• Non-Editable: PDF
• Pages: 1 Page
• Years: P - 1
Practise counting sets of objects with a group of 15 printable Remembrance Day peg cards.
## Remembrance Day Activities for Primary Students
Are you looking for things to do on Remembrance Day? We have an engaging set of clip cards to help your students practise counting with one-to-one correspondence. This hands-on activity will have students look at groups of objects (up to 15) and count them.
To use this Remembrance Day Activity, students will look at the pictures on each card and count them. They will then clip a clothes peg on one of the answer choices. Afterwards, students will flip the card over to see if the clothes peg matches the correct answer (indicated by a teacher-drawn star on the back of each card).
Through this activity, students will show they can count objects up to fifteen
## Using Your Remembrance Day Activity
A team of dedicated, experienced educators created this resource to support your maths lessons.
In addition to individual student work time, use this easy maths centre to enhance learning through small groups or whole class lessons.
If you have a mixture of above and below-level learners, check out these suggestions for keeping students on track with the concepts:
### 🆘 Support Struggling Students
Help students who need support by inviting them to reference posters or anchor charts from previous lessons. This activity can also be completed in a small group and a handful of counters.
➕ Challenge Fast Finishers
For students who need a bit of a challenge, encourage them to create more peg cards by drawing objects to count.
👋 Exit Ticket
Use these cards as a formative assessment after your lesson. Pick a random assortment of cards and project them on the board for the whole class to see. Students can record their answers on paper, sticky notes, or notebooks.
## Easily Prepare This Remembrance Day Resource for Your Students
Use the dropdown icon on the Download button to choose between this resource’s printable PDF or editable Google Slides version.
To keep the task cards out of pockets or under desks, punch a hole in each corner to place them on a binder ring.
Don’t stop there! We’ve got more activities to fill your Remembrance Day lesson plans!
### teaching resource
#### What is Remembrance Day? Teaching PowerPoint
Guide your students to discover the origin of Remembrance Day and how it is celebrated in Australia with a teaching slide deck.
Years: F - 6
### teaching resource
#### Remembrance Day Art Activity - Group Poster
Use this collaborative art activity to create and display in your classroom on Remembrance Day.
9 pagesYears: 1 - 6
### teaching resource
#### Remembrance Day & Anzac Day Craft Template - Soldier Writing
Create the perfect Remembrance Day or Anzac Day Art Project with our Soldier writing and craft activity.
Years: P - 2
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https://math.stackexchange.com/questions/1003918/computation-on-hyper-surface-z-x2y2
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# computation on hyper surface $z=x^2+y^2$
I have problem with following exercise
Consider the hypersurface $M$ parametrized by $z=x^2+y^2$.
Endow this with the Riemannian metric induced from the $\mathbb{R}^3$.
Compute the sectional curvature.
• Bonnet-Myers theorem says that if the (Ricci) curvature is bounded from below by a positive constant then the manifold is compact. So, the given surface can't have curvature bounded from below by a positive constant. – mfl Nov 3 '14 at 10:15
• Which parts do you have problem with? – user99914 Nov 3 '14 at 10:28
• @John, How one can compute the sectional curvature of above? and i guess this calculation gives positive curvature, but this hypersurface is not compact. – phy_math Nov 3 '14 at 10:32
• Do you know the metric of this surface? – user99914 Nov 3 '14 at 10:35
• No.... Are you taking a course? – user99914 Nov 3 '14 at 10:40
I'd like to know my calculation process is done correctly.
Note $z=x^2+y^2$, in $\mathbb{R}^3$, by introducing spherical coordinate such that \begin{align} x= r\cos{\theta}, \quad y= r\sin{\theta} , \quad z=r^2 \end{align} we have \begin{align} x^2 +y^2 = r^2= z \end{align} and doing calculation with chain rule \begin{align} \partial_{r} = \cos(\theta)\partial_x + \sin(\theta) \partial_{y} +2r \partial_z\end{align} \begin{align} \partial_{\theta} = -r\sin(\theta) \partial_x + r\cos(\theta) \partial_y \end{align} Thus \begin{align} ds^2 = (1+4r^2)dr^2 + r^2d\theta^2 \end{align} and sectional curvature How can i use sectional curvature formula for this case?
I know
Note i used the formula for sectional curvature for geodesic polar coordinate $i.e$ \begin{align} ds^2 =dr^2 +f(r,\theta)^2 d\theta^2 \qquad K=-\frac{1}{f}\frac{\partial^2 f}{\partial r^2} \end{align}
• I might be overlooking something, but it appears to me as though the surface that you have parametrized does not satisfy $z = x^2 + y^2$. Namely, I see that you have $x^2 + y^2 = \sin^{2}\theta_{1} \cos^{2}\theta_{2} + \sin^{2}\theta_{1} \sin^{2}\theta_{2} = \sin^{2}\theta_{1}$. By my calculations, this surface satisfies $x^2 + y^2 + z^1 = 1$ (i.e. is a sphere of radius 1). If you are going to introduce new coordinates to address the problem, you might consider spherical coordinates with $x = r\cos \theta$, $y = \r\sin \theta$ and $z = r^2$. – THW Nov 3 '14 at 18:12
• It might be more instructive to set your surface up as a Monge surface with $(u, v)\mapsto (u, v, f(u, v))$ and derive the Gaussian curvature for a general Monge patch. – THW Nov 3 '14 at 18:20
• @THW Thanks, I edited the process – phy_math Nov 4 '14 at 3:16
• @THW, The informative for a general Monge patch helps me a lot. THanks – phy_math Nov 4 '14 at 3:35
• I figure out what i did wrong in the above, and how to do this calculation correctly and obtain desired curvature. But still i am confused of considering Bonnet-Myers theorem. Maybe next week i will summarize my calculation and post it below. – phy_math Nov 4 '14 at 11:43
This method was using Monge surface parametrization, which was introduced by @THW. in my former question.
Endowment of $\mathbb{R}^3$, we can do Gauss approach as follows. \begin{align} \mathbf{x}(r,\theta)=(f(r),r\cos(\theta), r\sin(\theta)) \end{align} where $f(r)=r^2=z$. \begin{align} &\mathbf{x}(r,\theta)=\left(f(r),r\cos(\theta), r\sin(\theta) \right) \quad \mathbf{x}_r = \left(f'(r), \cos(\theta), \sin(\theta) \right) \\ & \mathbf{x}_{rr} = \left( f''(r),0,0 \right) \quad \mathbf{x}_\theta = \left( 0,-r\sin(\theta), r\cos(\theta) \right) \\ &\mathbf{x}_{\theta\theta} = \left( 0,-r\cos(\theta), -r\sin(\theta) \right) \quad \mathbf{x}_{r\theta} = \left(0, -\sin(\theta), \cos(\theta)\right) \end{align} Here $'$ denote the $r$-derivative. Then \begin{align} I=ds^2 =(\mathbf{x}_r du + \mathbf{x}_\theta dv) \cdot(\mathbf{x}_r dr + \mathbf{x}_\theta d\theta) = Edr^2 + 2F drd\theta+ G d\theta^2 \end{align} with $E = 1 +(f')^2$, $F=0$, $G=r^2$. Note \begin{align} &\mathbf{x}_r \times \mathbf{x}_\theta = \left(r, -f'(r) r \cos(\theta), -f'(r) r \sin(\theta) \right) \\ &||\mathbf{x}_r \times \mathbf{x}_\theta|| = \sqrt{EG-F^2} = r\sqrt{1 + (f')^2} \\ & U=\frac{\mathbf{x}_r \times \mathbf{x}_\theta}{||\mathbf{x}_r \times \mathbf{x}_\theta||}= \frac{\left(1, -f' \cos(\theta), -f' \sin(\theta) \right)}{\sqrt{1 + (f')^2}} \end{align} and \begin{align} &e= \mathbf{x}_{rr} \cdot U=\frac{f''}{\sqrt{1 + (f')^2}}, \qquad f=\mathbf{x}_{r\theta}\cdot U=0 , \qquad g=\mathbf{x}_{\theta\theta} \cdot U=\frac{f'r}{\sqrt{1 + (f')^2}} \\ &\kappa_\mu = \frac{e}{E} =\frac{f''}{\sqrt{\left(1 +(f')^2 \right)^3}}, \quad \kappa_{\pi} = \frac{g}{G}=\frac{f'}{r\sqrt{1+(f')^2}} \end{align} Thus we obtain Gauss Curvature \begin{align} K = \kappa_\mu \kappa_\pi =\frac{f'f''}{r\left( 1 +(f')^2 \right)^2}\stackrel{f =r^2 }=\frac{4}{\left( 1 +4r^2 \right)^2} >0 \end{align}
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https://physics.stackexchange.com/questions/261874/solution-for-schr%C3%B6dinger-equation-for-constant-box-potential/261892
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# Solution for Schrödinger equation for constant box potential?
It is known that in a box potential, when we set $V = 0$ inside and $V = \infty$ on the boundaries, the solution to the equation $$- \frac{\hbar}{2m} \bigg( \frac{\partial^2}{\partial x^2} + \frac{ \partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} \bigg) \psi(x,y,z) = E \psi(x,y,z)$$ is given by $$\psi_{n_x, n_y, n_z} (x,y,z) = C \sin( \frac{n_x \pi x}{L}) \sin (\frac{n_y \pi y}{L}) \sin(\frac{n_z \pi z}{L}).$$ Here $L$ is the dimension of the box. The energies are $E_n = \frac{\hbar^2 \pi^2 }{2mL^2} n^2$ with $n^2 = n_x^2 + n_y^2 + n_z^2$. Now, I was wondering, what if the potential inside the box is not zero, but we let it be $V = V_0$, a constant? What is the solution of the Schrödinger equation then? And what are the energy eigenvalues?
Does this amount to adding a phase factor to the solution (the product of sines)?
• Still the same. Only that $E_n = \frac{\hbar^2n^2\pi^2} {2mL^2} + V_0$ – philip_0008 Jun 10 '16 at 20:56
• I see. But how did you deduce that? – Kamil Jun 10 '16 at 21:07
The equation will become $$- \frac{\hbar}{2m} \bigg( \frac{\partial^2}{\partial x^2} + \frac{ \partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} \bigg) \psi(x,y,z) = (E-V_0) \psi(x,y,z)$$ And the solutions are the same: $$\psi_{n_x, n_y, n_z} (x,y,z) = C \sin( \frac{n_x \pi x}{L}) \sin (\frac{n_y \pi y}{L}) \sin(\frac{n_z \pi z}{L}).$$ And Energy: $$(E_n - V_0) = \frac{\hbar^2\pi^2} {2mL^2}n^2$$ $$E_n = \frac{\hbar^2\pi^2} {2mL^2}n^2 + V_0$$
Path to the solution:
By separation of variables: $$\frac{1}{X}\frac{d^2X}{dx^2} + \frac{1}{Y}\frac{d^2Y}{dy^2} + \frac{1}{Z}\frac{d^2Z}{dz^2} = -\frac{2m}{\hbar^2}(E-V_0)$$ $$\frac{d^2X}{dx^2} = -k_x^2X; \frac{d^2Y}{dy^2} = -k_y^2Y; \frac{d^2Z}{dz^2} = -k_z^2Z$$ with $$(E_n - V_0) = \frac{\hbar^2} {2m}(k_x^2 + k_y^2 + k_z^2)$$ Solution:
$X(x) = A_x\sin k_xx + B_x\cos k_xx$ and so on.
as usual, $B = 0$ because $X(0) = 0$ because of infinite potential at boundaries.
also, $X(L) = 0$ (infinite potential) means $\sin k_xL = 0$ or
$k_x = n_x\pi/L$ and so with the others.
So still the same: $$\psi_{n_x, n_y, n_z} (x,y,z) = A_xA_yA_z \sin( \frac{n_x \pi x}{L}) \sin (\frac{n_y \pi y}{L}) \sin(\frac{n_z \pi z}{L}).$$ $$\psi_{n_x, n_y, n_z} (x,y,z) = C \sin( \frac{n_x \pi x}{L}) \sin (\frac{n_y \pi y}{L}) \sin(\frac{n_z \pi z}{L}).$$ $$(E_n - V_0) = \frac{\hbar^2\pi^2} {2mL^2}(n_x^2+n_y^2 + n_z^2)$$
• This is because $V_0$ is a constant – philip_0008 Jun 10 '16 at 21:11
• Very nice little derivation. +10. – Gert Jun 10 '16 at 23:05
• I got this from solution manual (Griffiths) :) but without the $V_0$. – philip_0008 Jun 11 '16 at 8:35
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Find the two square roots of each number. Exploring Square Roots and Irrational Numbers COURSE 3 LESSON 4-8 a. 81 9 9 = 81 –9 (–9) = 81 The two square.
Presentation on theme: "Find the two square roots of each number. Exploring Square Roots and Irrational Numbers COURSE 3 LESSON 4-8 a. 81 9 9 = 81 –9 (–9) = 81 The two square."— Presentation transcript:
Find the two square roots of each number. Exploring Square Roots and Irrational Numbers COURSE 3 LESSON 4-8 a. 81 9 9 = 81 –9 (–9) = 81 The two square roots of 81 are 9 and –9. b. 1 36 1616 1616 –– = 1 36 The two square roots of are and –. 1616 1616 1 36 1616 1616 = 1 36 4-8
Estimate the value of – 70 to the nearest integer. Exploring Square Roots and Irrational Numbers COURSE 3 LESSON 4-8 Since 70 is closer to 64 than it is to 81, – 70 –8. 4-8
The math class drops a small ball from the top of a stairwell. They measure the distance to the basement as 48 feet. Use the formula d = 16t 2 to find how long it takes the ball to fall. Exploring Square Roots and Irrational Numbers COURSE 3 LESSON 4-8 48 = 16t 2 Substitute 48 for d. d = 16t 2 Use the formula. = t 2 Divide each side by 16. 48 16 3 = t 2 Simplify. 4-8
(continued) Exploring Square Roots and Irrational Numbers COURSE 3 LESSON 4-8 It takes about 1.7 seconds for the ball to fall 48 ft. 3 = tFind the positive square root. Use a calculator.31.7320508 Round to the nearest tenth.1.7 t 4-8
Identify each number as rational or irrational. Explain. Exploring Square Roots and Irrational Numbers COURSE 3 LESSON 4-8 d. 6.36366366636666... Irrational; the decimal does not terminate or repeat. Rational; the decimal repeats.a. –9.3333 b. 4 7979 Rational; the number can be written as the ratio. 43 9 Irrational; 90 is not a perfect square. c. 90 4-8
1.Find the two square roots of 400. 2.Estimate 34 to the nearest integer. 3.Using d = 16t 2, find how long it takes a skydiver to fall 676 ft from an airplane. 4.Is rational or irrational? Explain. Exploring Square Roots and Irrational Numbers 20 and –20 6 COURSE 3 LESSON 4-8 6.5 s Rational; it can be written as. 8585 64 5 4-8
Download ppt "Find the two square roots of each number. Exploring Square Roots and Irrational Numbers COURSE 3 LESSON 4-8 a. 81 9 9 = 81 –9 (–9) = 81 The two square."
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# ec2302 question bank with answers
What is the reason that FIR filter is always stable? The representation of continuous signal amplitude by a fixed digit produce an error, which is known as input quantization error. 67. 23. 15 Aug, 1947 = (1946 years + Period from 1.1.1947 to 15.8.1947). 13. This effect compresses the magnitude and phase response. 62. What is memory system and memory less system? To prevent overflow, the signal level at certain points in the digital filter must be scaled so that no overflow occurs in the adder. For EC6502 DSP Previous Year Question Papers – Click here. The sum of two floating-point numbers is carried out by shifting the bits of the mantissa. A signal is said to be periodic if x(n+N)=x(n), A signal is said to be non- periodic if x(n+N)≠x(n), 18. R(ax1(t)+bx2(t))=ax1(t)+bx2(t). 13.For the second order filter Draw the direct form II realization and find the scaling factor S0 to avoid over flow, I= --------------------------------------- (1-r2)(1-2r2cos2ø =r4). 91.which realization is less sensitive to the process of quantization? CS1401 INTERNET COMPUTING Two Marks Questions With... CS55- PROGRAMMING PARADIGMS Questions Bank 2014. What is a continuous and discrete time signal? Such limit cycles have been referred to as over flow oscillations. 24. ( fs ≥ 2fh ). 78.what do you understand by a fixed-point number? 2. Derive the condition of FIR filter to be linear in phase. Ans: The expression is N=log (λ /€) 1/2/log (1/k) ½. 15. For floating point number the error made by rounding a number to b bits satisfy the inequality. 15. EC6 502 PDSP Syllabus notes download link is provided and students can download the EC 6502 Syllabus and Lecture Notes and can make use of it. Software-Programmable Wait-State Generators, 8. 66. Note : This is the Read More … 40.What is meant by impulse invariant method? Efficient realization of FIR filter exists as both recursive and non-recursive structures. What are the desirable characteristics of the windows? The difference signal e(n)=xq(n)-x(n) is called A/D conversion noise. What are its advantages? EC2302 Digital Signal Processing – May/June 2016 Regulation 2008. In some cases, you likewise reach not discover the broadcast digital signal processing question bank with answers that you are looking for. Search "" Common help topics. The transfer function of the Butterworth filter is given by. d) Filter order required in multirate application is low. For EC6502 DSP Question Bank/2marks 16marks with answers – Click here. When cascade from realization is preferred in FIR filters? CS2252 Microprocessor And Microcontrollers All Uni... EC2203 Digital Electronics Two Marks With Answers. Processing Question Bank With Answersremained in right site to start getting this info. What condition on the FIR sequence h(n) are to be imposed n order that this filter can be called a liner phase filter? microphone, photocell. 2.The low pass FIR filter designed will have first side lobe peak of –53 dB, The main lobe width ,the peak side lobe level can be varied by varying the. The theory of processing signals at different sampling rates is called multirate signal processing. .(e.g.) EC2202 – DATA STRUCTURES AND OBJECT ORIENTED PROGR... EC2201-ELECTRICAL ENGINEERING Question Bank. 83.what are the three-quantization errors to finite word length registers in digital filters? SUBJECT NAME: DIGITAL SIGNAL PROCESSING. A stationary random process is said to be white noise if its power density, Spectrum is constant. Where b is the number of bits and xt is the truncated value of x. State the steps to design digital IIR filter using bilinear method, Substitute s by 2/T (z-1/z+1), where T=2/Ώ (tan (w/2) in h(s) to get h (z), For smaller values of w there exist linear relationship between w and .but for larger values of w the relationship is nonlinear. (ii) Summarize the properties of DFT. Even signal: continuous time signal x(t) is said to be even if it satisfies the condition x(t)=x(-t) for all values of t. Odd signal: The signal x(t) is said to be odd if it satisfies the condition x(-t)=-x(t) for all t. In other words even signal is symmetric about the time origin or the vertical axis, but odd signals are anti-symmetric about the vertical axis. 9. 74.What is the principle of designing FIR filter using frequency sampling method? 82.what are the advantages of floating point arithmetic? 68.What are the disadvantage of Fourier series method ? sk=acos¢k+jbsin¢k,where ¢k=∏/2+(2k-1)/2n)∏, 34. A discrete time signal is often derived from the continuous time signal by sampling it at a uniform rate. Therefore this method is not much preferred. At first the signal x(t) is sampled at regular intervals to produce a sequence x(n) is of infinite precision. Show that the discrete time system described by the input-output relationship, For a sys to be linear R{a1x1[n]+b1x2[n]}=a1y1[n]+b1y2[n] L.H.S:R{ a1x1[n]+b1x2[n] }=R{x[n]} /x[n] → a1x1[n]+b1x2[n], = a1 nx1[n]+b1 nx2[n] ------------------- (1), R.H.S: a1y1[n]+b1y2[n]= a1 nx1[n]+b1 nx2[n]-------------------- (2), Equation(1)=Equation(2) Hence the system is linear. One possible way of finding an FIR filter that approximates H(ejω)would be to truncate the infinite Fourier series at n= (N-1/2).Abrupt truncation of the series will lead to oscillation both in pass band and is stop band .This phenomenon is known as Gibbs phenomenon. The numerator polynomial value depends on the value of n. If n is odd: put s=0 in the denominator polynomial. The origin digit produce an error, which is known as input quantization error Question ma2265... Butterworth filter is always stable because all its poles are at the origin its density! Cookies to ensure that we give you the best experience on our website to! The eight point DFT +1 and -1 having a dynamic range 2, spectrum constant... Software Questions Bank 2014 input as 2marks & 16marks for all three types number! C1+C2 )... ge6151 Computer Programming the CONDITIONAL EXPRES... ge6151 Computer Programming Unit STRUCTURES! Frequencies and attenuate particular range of values called the dead band of the sample data sequence (. Download ec2302 digital signal processing an... CS55- Programming PARADIGMS Questions Bank 2014 changes... Bank with answers and sufficient condition for linear phase FIR filter Collins … 1 z is. Regulation 2008 a sequence with its shifted version, and this indicates how fast the signal changes linear...... CS2301 SOFTWARE Engineering Questions Bank 2014 for scaling in the rectangle window ) =xq n... Be white noise if its power density, spectrum is constant the stop band or the pass.. Internet COMPUTING Two Marks Questions with ans... CS2303 THEORY of processing signals at different rates! - Question Bank digital materials for Engineering students without spammy contents and redirects correlation of a with! The exponent p=limt→∞1/t∫x2 ( t ) +bx2 ( t ) dt, integration limit is from-T/2 to +T/2 )! Group of signals with each signal in the denominator polynomial Hilbert transforms differentiators! Noise Variance and steady state I/P noise power assume a1=0.7 and a2= 0.8and find o.p off... 91.Which realization is less sensitive to the right place to get all semester Anna University Regulation ECE. Should be narrow and plot amplitude response curve n. 4 they may be viewed as group signals... Truncation error satisfies { 1,1,2,2,3,3 } and, 19 order n, major, minor axis an! As follows, Let f1 = M1 * 2c1 and f2 = M2 * 2c2 output a! Is from-T/2 to +T/2 aunewsblog.net is the relationship between truncation error satisfies function of w, which are uniformly.... ; fh -- - samplinf frequency ; fh -- - samplinf frequency ; fh -- - highest signal.! Described by the input output relation y [ n ] time invariant system responds idenditically matter. For different orders of filter the convolution of the frequency response are called IIR filter considering all the infinite of! Representing a decimal into binary level and n. 2 amplitude and phase delay satisfying... Then f3 = f1 * f2 = ( M1 * M2 ) 2 ( c1+c2 ) xt is correlation. This site we will assume that you are looking for Previous Year Question Papers – Click here in! Reason that FIR filter design PROGR... EC2201-ELECTRICAL Engineering Question Bank ) part – b Marks! ) dt, integration limit is from-T/2 to +T/2 of processing signals at different sampling rates is A/D. Of coefficients in digital Computer frequency is decimated and input is bi reversed output! Rec - Question Bank - 1st Edition ; GE2211 Environmental Science and Engineering certain flow. A building block for the linear phase FIR filter is causal if its response. And derive sampling rate conversion by a fixed digit produce an error, which amplify particular range of frequencies attenuate! ) will yield constant group may June 2014 Exam Results, B.E Civil - semester -! Find H ( n ) is called A/D conversion noise + Period from 1.1.1947 to 15.8.1947 ) noise be... ( EC6502 Principles of digital signal processing SOFTWARE Two Mark Questions with... CS55- Programming PARADIGMS Two Questions. Are happy with it in natural order output is referred to as single input single output system conver in realizable!, 1 's complement & sign magnitude, 1, 2 's complement & sign,... Are the Two numbers are equal and then adding the mantissas it by ( )... Sensitive to the same value for the designer to select the side lobe level and n... Non-Periodic signals input is natural order output is referred to as over flow oscillations carried out by shifting bits. 23. a ) discuss on efficient transversal structure for decimator and interpolator before its actual occurrence the addition of sequences! Republishing of the binary sequence of 0s and 1s design lo pass and bandpass filter level can be varied varying. Relationship between truncation error satisfies given by provide Study Material and Education News and Notification from Tamil Nadu Colleges a! Own right, in that they may be viewed as group of signals with each signal in rectangle..., B.E Civil - semester 5 - Question Bank Collection ec2302 question bank with answers Anna University ) signal... You continue to use this site we will assume that you are happy with it stages are there for point... The Two numbers are equal and then adding the mantissas this indicates how the... Order required in multirate application is low Wi... CS2302 -COMPUTER NETWORKS Questions Bank 2014 Deterministic signal is a selective... /€ ) 1/2/log ( 1/k ) ½ x > 0 is stored thus saving memory Aug, 1947 = 1946... Of converting analog to digital filter is always stable because all its poles are at the.! Require more time to spend to go to the ebook inauguration as capably as search for.! 1+E2 ) 1/2 sample x ( n ) =0 for n <.... Is carried out in floating point arithmetic defined as, 21 Computer HEADER! Matter when the input of the Two kinds of limit cycle are confined to a of! Cs55- Programming PARADIGMS Questions Bank 2014 Notes Syllabus all 5 units are provided below as single input single output continuous... ) +bx2 ( t ) the side lobe level of the sequence by using the following methods fixed... Is not preferred in FIR filters, 27 estimation is the necessary and sufficient condition for linear characteristics..., Susan Collins … 1 is N=log ( λ /€ ) 1/2/log ( )! With it bits excluding sign bit of operations needed to recover the input of the number! Central lobe of the output signal DSP 2marks-16marks and H2 when connected in and! System output process is said to be white noise if its impulse response is anti symmetrical satisfying... Truncation error satisfies from-T/2 to +T/2 transforms and differentiators sequences x ( n ) =0 for n <,. Given the sequence by using the DIF – frequency is decimated and input bi. Side ’ s lobes of the frequency response, n. Convert H ( n ) for... Function and its down sampled signal efficient realization of FIR filter, the symmetrical impulse response is summable. Methods of converting analog to digital filter to be white noise if its impulse response having even number bits... Cs2304-System SOFTWARE Questions Bank 2014 - Two Marks Questions with... CS55- Programming PARADIGMS Questions Bank 2014 all poles. That Deterministic signals may be viewed as group of signals with each signal in the rectangle window behavior DSP., 27 arithmetic & only one exponent per block is stored thus saving memory as recursive! Result of quantization on pole location a number to the right until the exponents of chebyshev!, mantissa is multiplied using fixed-point arithmetic and the exponents are added both recursive non-recursive... Response is absolutely summable in the rectangle window in decimal briefly about various number representation in digital filters lead problems! +1 and -1 having a dynamic range 2 filter design in phase out the addressing modes supported C5X... Lines and Wave guides with Qu... Anna University Regulation 2013 ECE EC6502 Previous... Scaling in the digital signal processing operations CS2301 SOFTWARE Engineering Questions Bank 2014 band frequency selective device, A/D that. An error, which are uniformly spaced is to provide digital materials for Engineering students without spammy contents redirects... By rounding a number to b bits satisfy the inequality Fifth semester bit reversed format 19.given (! Their impulse response having even number of bits and xt is the number of samples Two Mark Questions...... Guides - REC - Question Bank Searching for Anna University Question Paper for B.E Electronics and Communication Engineering semester. Is from-T/2 to +T/2 the C5X DSP on-chip peripherals available are as follows: 3 z ) which binary. Out as follows, Let f1 = M1 * 2c1 and f2 = ( 1946 years Period! The mapping from s plane to z plane is many to one will assume that you are with! Random signal bits given the sequence by using the following methods response are non-zero for different orders of.... Representing a decimal into binary window to the ebook inauguration as capably as search for them DFT! } ans: Deterministic signal: random signal is a signal about which there is no certainty with to... Is 0.045-0.075z-2 -.159 Z-3-0.22Z-4+0.75Z-5-.22Z-6 -0.159Z-7 - power and steady state the cascade from realization is less sensitive to process! Ant symmetrical impulse response should be narrow map to the ebook inauguration as capably as search them. Effect in multiplication FFT algorithm if x < 0, then for sign magnitude signal Answer... Thus there are an infinite number of bits and xt is the reason that FIR filter using the –. Question Bank/2marks 16marks with answers link that we provide here and check out the.. * 2c1 and f2 = M2 * 2c2 signal processing, the duration of impulse having! 84.How the multiplication & addition are carried out as follows: 3 i.e., 2, 1, 's. Error satisfies is bit reversed format output in natural order output is bit reversed format output in order... Sequence xq ( t ) is called multirate signal processing – May/June 2016 Regulation Anna. Down sampling of samples can be used Answer is provided below output during a limit cycle occur as result! Be handled is divided into blocks changes in their value converting analog to digital filter implementation variety. In block point arithmetic & only one exponent per block is stored thus ec2302 question bank with answers.! Second order filter write the difference equation random process pole locations will changed...
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### Home > PC > Chapter 11 > Lesson 11.1.2 > Problem11-25
11-25.
Using the results of the previous problems, write the equation $y = 3 − 2x$ in polar form.
Use $x = r \cos\left(θ\right)$ and $y = r \sin\left(θ\right)$.
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Alvenchrst
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Discussion Starter · ·
I'm just curious how many weeks you were when you had your third baby? I have heard that baby 1 is usually about 41ish weeks, baby 2 is usually about 40 weeks, and baby 3 can be a wildcard.<br><br>
I was induced at 40w0d with #1<br>
Am currently 35 weeks with #3<br><br>
Please share how many weeks you were with all 3 (or more!)<br><br>
Thanks <img alt="" class="inlineimg" src="http://www.mothering.com/discussions/images/smilies/smile.gif" style="border:0px solid;" title="smile">
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#1 = 41w2d<br>
#2 = 39w2d<br>
#3 = 38w<br><br>
Good luck!
Mama Poot
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Both of my boys were born at 40 weeks exactly, each born one day before their due dates, the first in the hospital, the second at home. I was never induced, went into labor naturally both times. I am pregnant with my 3rd, and I don't anticipate it being any different this time, but we'll have to see!
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38<br>
40<br>
42
DreamsInDigital
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41+4<br>
41+3<br>
42+1<br>
40+4<br><br>
We'll see what happens with #5!
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41+5<br>
33+0<br>
39+2
cappuccinosmom
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ds1--39 w 4(?) days<br>
ds2--42 w 2 days<br>
ds3--42 w 2 days
ameliabedelia
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DD1 38 + 4<br>
DD2 38 + 4<br>
DS1 38 + 4<br><br>
(notice a pattern here ) <img alt="" class="inlineimg" src="http://www.mothering.com/discussions/images/smilies/lol.gif" style="border:0px solid;" title="lol">
sweeetpea
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#1 - 34 w 5d<br>
#2 - 39w 6d<br>
#3 - 41w 2d (and boy, was I ready!)
Graceoc
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1st 40 wks + 1 day<br>
2nd 40 wks + 3 days<br>
3rd 40wks + 4 days<br>
4th 39wks + 1 day <img alt="" class="inlineimg" src="http://www.mothering.com/discussions/images/smilies/thumb.gif" style="border:0px solid;" title="thumbs up">
shanetedissac
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#1 42 wks<br>
#2 40 wks<br>
#3 41 wks<br><br>
All babies were born on their own time without interventions.<br><br>
Be well,
spamama
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#1 (DS): 36w, 1d<br>
#2 (DD): 39w, 1d<br>
#3 (DS): 37w, 3d
yeahwhat
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I'm very consistent.<br>
1 39+6<br>
2 40+1<br>
3 39+6<br>
4 40+1<br>
5 39+5 (but hey, she was born at 9:30pm, she almost completed the pattern <img alt="" class="inlineimg" src="http://www.mothering.com/discussions/images/smilies/orngtongue.gif" style="border:0px solid;" title="Stick Out Tongue"> )
mama_ani
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#1 - induced 11 days over<br>
#2 - naturally 26 days over<br>
#3 - naturally 28 days early (VERY complicated pregnancy though, bedrest from 19 weeks on)<br>
#4 - 1 day "late"<br>
#5 - currently 37 weeks exactly! So we shall see!
MsElle07
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#1: 39w1d<br>
#2: 39w1d<br>
#3: 41w<br><br>
My labor was also longer for #3. It really is a wild card! <img alt="" class="inlineimg" src="http://www.mothering.com/discussions/images/smilies/smile.gif" style="border:0px solid;" title="smile">
HarperRose
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39w5d<br>
40w 1hr 43m (due the 18th, born at 1:43 am on the 19th)<br>
43w 1d
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40w3d<br>
42w<br>
40w2d
DoomaYula
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34w2d (induction for twins with TTTS and IUGR in one twin)<br>
40w<br>
41w<br><br>
I seem to get longer with each pregnancy!
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#1: 39w5d<br>
#2: 39w4d<br>
#3: 40w0d
Mama Poot
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<div style="margin:20px;margin-top:5px;">
<div class="smallfont" style="margin-bottom:2px;">Quote:</div>
<table border="0" cellpadding="6" cellspacing="0" width="99%"><tr><td class="alt2" style="border:1px inset;">
<div>Originally Posted by <strong>MsElle07</strong> <a href="/community/forum/post/10774628"><img alt="View Post" class="inlineimg" src="/community/img/forum/go_quote.gif" style="border:0px solid;"></a></div>
<div style="font-style:italic;">#1: 39w1d<br>
#2: 39w1d<br>
#3: 41w<br><br>
My labor was also longer for #3. It really is a wild card! <img alt="" class="inlineimg" src="http://www.mothering.com/discussions/images/smilies/smile.gif" style="border:0px solid;" title="smile"></div>
</td>
</tr></table></div>
<br>
I've heard this from a lot of women. One woman in my mom's group just had her second recently and the labor was several hours longer, and the woman who started the group said her 3rd took longer because he was 1.5lbs bigger than her first two babes! I hope I keep with my pattern of the labor being shorter each time. My first labor was 20 hours, my second was 8 hours.
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How to convert 2 amps to watts. Also, explore tools to convert megawatt or volt ampere to other power units or learn more about power conversions. How many ampere in 1 watt/volt? Amps to watts calculator Ohms calculations. The unit, defined as one joule per second, measures the rate of energy conversion or transfer. The megawatt [MW] to volt ampere [V*A] conversion table and conversion steps are also listed. The user must fill one of the two fields and the conversion will become automatically. Other units of power include ergs per second (erg/s), horsepower (hp), and foot-pounds per minute . 1 watt is defined as the energy consumption rate of one joule per second. The resistance R in ohms (Ω) is equal to the voltage V in volts (V) divided by the current I in amps (A): Watts (W) / volts (V) / amps (A) / ohms (Ω) calculator. V = I×R, where V is the difference of potentials, I is the electrical current, and R is the resistance. The answer is 1. In the SI system, power is measured in watts (W). We assume you are converting between farad [international] and ampere second/volt. 1.5 Ampere to Biot 8.5 Volt/ohm to Milliampere 3 Volt/ohm to Megaampere 7900 Milliampere to Ampere 2.81 Ampere to Ampere 62 Watt/Volt to Volt/ohm 35 Abampere to Watt/Volt 0.23 Milliampere to Milliampere 9 Volt/ohm to Kiloampere 160 Volt/ohm to Volt/ohm 9360 Ampere to Watt/Volt Ampere: Amperage is a term often used by electricians, and means electrical current, measured in amperes, or amps. We assume you are converting between ampere and watt/volt. Watt is an SI unit of electric power and is denoted by 'W'. Instant free online tool for megawatt to volt ampere conversion or vice versa. Walaupun tidak mungkin "mengubah" watt menjadi ampere, Anda masih bisa menghitung ampere menggunakan hubungan antara ampere, Watt, dan volt. This tool converts watts to volt-ampere (w to va) and vice versa. 1 watt is equal to 1 joule per second (J/s). Example: 2.4KW watts are being sent at 120 volts. 1 watt = 1 volt-ampere. The formula for Watt’s law stays the same, just as long as you express the wattage in watts (your sum will go wrong if you use ‘5W’ to mean ‘5KW’; you need to use 5000W instead). The watt (symbol: W) is a unit of power. How to convert electric current of 2 amps (A) to electric power in watts (W). You can calculate (but not convert) the watts from amps and volts: 1 farad is equal to 1 ampere second/volt. 1 watt is equal to 1 joule per second (J/s). millivolt to watt/ampere (mV—W/A) measurement units conversion. The ampere is the SI unit for electrical current, or the amount of electrical charge that flows through a conductor in a given time. Other units of power include ergs per second (erg/s), horsepower (hp), and foot-pounds per minute . Note that rounding errors may occur, so always check the results. Cara Mengubah Watt Menjadi Ampere. Conversion Watt to Volt-ampere. You can view more details on each measurement unit: ampere or watt/volt The SI base unit for electric current is the ampere. In the SI system, power is measured in watts (W). You can view more details on each measurement unit: farad or ampere second/volt The SI derived unit for capacitance is the farad. Here is a useful online Amps to Watts Calculator which helps you in converting electrical unit Ampere (A) To Watt (W) (Amperage to Wattage). 1 ampere is equal to 1 ampere, or 1 watt/volt. , dan volt conversion steps are also listed watt/volt the SI system, power is measured in,... Include ergs per second ( J/s ) and means electrical current, and R is the farad to... In amperes, or 1 watt/volt `` mengubah '' watt menjadi ampere, or amps view more on... Ampere is equal to 1 joule per second in amperes, or 1 watt/volt = I×R, where V the. ( hp ), horsepower ( hp ), horsepower ( hp ) horsepower... Also, explore tools to convert megawatt or volt ampere to other power units or more! The unit, defined as the energy consumption rate of one joule per second ( erg/s,... Volt-Ampere ( W to va ) and vice versa is measured in (... To electric power in watts ( W ) / ohms ( Ω ).... ( J/s ) of 2 amps ( A ) to electric power and is denoted by ' W ' to. Megawatt or volt ampere to other power units or learn more about power conversions =,... The resistance fields and the conversion will become automatically example: 2.4KW watts are being sent at 120.! Instant free online tool for megawatt to volt ampere [ V * ]!: ampere or watt/volt the SI base unit for capacitance is the electrical current, and per... Mw ] to volt ampere conversion or transfer by ' W ' =,. [ MW ] to volt ampere to other power units or learn more power. '' watt menjadi ampere, or 1 watt/volt on each measurement unit: ampere or watt/volt the SI unit... The user must fill one of the two fields and the conversion become... Megawatt or volt ampere conversion or vice versa or amps for electric current is the.! I is the farad SI base unit for electric current is the ampere rounding errors may occur so... The difference of potentials, I is the difference of potentials, I is resistance! Measurement unit: ampere or watt/volt the SI system, power is measured in amperes, 1... ) to electric power in watts ( W ) is A term often used by electricians and. ), horsepower ( hp ), and watt naar ampère per minute user fill. ( W ) 2.4KW watts are being sent at 120 volts SI base for. Unit: farad or ampere second/volt the SI derived unit for electric current 2! / amps ( A ) / ohms ( Ω ) calculator and vice versa menghitung ampere menggunakan hubungan antara,... Errors may occur, so always check the results measured in watts ( W va... On each measurement unit: ampere or watt/volt the SI base unit for current... One of the two fields and the conversion will become automatically and the conversion will become automatically, V. Become automatically electricians, and foot-pounds per watt naar ampère mungkin `` mengubah '' watt menjadi,! ( J/s ) tools to convert megawatt or volt ampere [ V * A ] conversion and. Unit, defined as one joule per second system, power is measured in (... Ampere [ V * A ] conversion table and conversion steps are also listed 1 joule second... Conversion will become automatically errors may occur, so always check the.. Ampere: Amperage is A unit of power include ergs per second ( erg/s ), and foot-pounds per.., power is measured in watts ( W to va ) and vice versa (. / ohms ( Ω ) calculator or vice versa denoted by ' W ' in amperes, or 1.... Horsepower ( hp ), and R is the farad conversion steps are also listed watts! Capacitance is the farad bisa menghitung ampere menggunakan hubungan antara ampere, watt, dan volt: W ) between... R is the difference of potentials, I is the ampere millivolt to watt/ampere ( mV—W/A ) measurement conversion! Ergs per second ( erg/s ), and foot-pounds per minute between and... Tidak mungkin `` mengubah '' watt menjadi ampere, Anda masih bisa menghitung ampere menggunakan hubungan antara ampere,,. Rounding errors may occur, so always check the results units of power watt, dan.. Power and is denoted by ' W ' convert megawatt or volt ampere to other power units learn! Si system, power is measured in watts ( W ) / amps ( A ) / amps ( )! Means electrical current, measured in watts ( W to va ) and vice.. Conversion will become automatically the unit, defined as one joule per second ( J/s ) the watt symbol., explore tools to convert electric current of 2 amps ( A ) to power. Hp ), and R is the ampere millivolt to watt/ampere ( ). Si derived unit for capacitance is the farad masih bisa menghitung ampere menggunakan antara... Or ampere second/volt the SI derived unit for electric current of 2 amps A! By ' W ' fill one of the two fields and the conversion will become.! Will become automatically each measurement unit: farad or ampere second/volt the SI base unit for capacitance is the.! Watts are being sent at 120 volts include ergs per second, measures the rate of energy conversion transfer. Or ampere second/volt the SI system, power is measured in amperes or. I is the ampere V * A ] conversion table and conversion steps also..., explore tools to convert electric current of 2 amps ( A ) volts. Units conversion explore tools to convert electric current is the electrical current and! Is denoted by ' W ' ampere or watt/volt the SI base unit for capacitance is the current.: 2.4KW watts are being watt naar ampère at 120 volts amperes, or.. To convert megawatt or volt ampere conversion or transfer electric current is the resistance walaupun tidak ``... ) measurement units conversion of power include ergs per second ( erg/s ), and electrical. ' W ' V * A ] conversion table and conversion steps are also listed ampere, watt, volt. Measured in watts ( W ) / amps ( A ) / volts ( V /... ( A ) / amps ( A ) / amps ( A ) to electric power in watts ( to... To electric power in watts ( W to va ) and vice versa fields and the conversion will automatically! Ampere [ V * A ] conversion table and conversion steps are also listed,! Ampere conversion or vice versa units or learn more about power conversions denoted by ' '. The watt ( symbol: W ) is A unit of electric power in watts W. Watt/Volt the SI base unit for electric current is the electrical current, and foot-pounds per minute base for. Watt ( symbol: W ) / volts ( V ) / amps ( A /. Watt menjadi ampere, or amps rate of energy conversion or vice versa two fields and the conversion become. Watts are being sent at 120 volts current of 2 amps ( A ) to electric power watts! Antara ampere, Anda masih bisa menghitung ampere menggunakan hubungan antara ampere, or.. The megawatt [ MW ] to volt ampere conversion or vice versa joule! Is denoted by ' W ' to watt/ampere ( mV—W/A ) measurement units conversion energy... ] to volt ampere conversion or vice versa megawatt or volt ampere [ V A. To 1 joule per second, measures the rate of one joule per (... Or 1 watt/volt per second, measures the rate of energy conversion or vice versa measures the rate of joule! Foot-Pounds per minute 2 amps ( A ) to electric power in watts ( W ) / volts V! To watt/ampere ( mV—W/A ) measurement units conversion watt is defined as one joule per second the two fields the! Volt ampere conversion or vice versa tool for megawatt to volt ampere V! Measurement unit: ampere or watt/volt the SI derived unit for capacitance is the ampere as the energy consumption of... Other power units or learn more about power conversions 1 watt is defined as the energy consumption rate of conversion... Electric current of 2 amps ( A ) to electric power in watts ( W ) the two fields the... V * A ] conversion table and conversion steps are also listed convert megawatt or ampere... / volts ( V ) / amps ( A ) / volts ( V /. Anda masih bisa menghitung ampere menggunakan hubungan antara ampere, Anda masih bisa menghitung menggunakan... Electric power in watts ( W to va ) and vice versa A term often used by electricians, foot-pounds! Other power units or learn more about power conversions ), and R is the current... Rate of energy conversion or vice versa ( V ) / volts ( V ) amps... '' watt menjadi ampere, or amps online tool for megawatt to volt ampere conversion or.. Of potentials, I is the farad Anda masih bisa menghitung ampere menggunakan hubungan antara ampere, Anda bisa. Steps are also listed ) is A unit of power include ergs per second ( erg/s ) and. Or vice versa term often used by electricians, and R is the electrical current, means... The SI system, power is measured in amperes, or amps other power units or learn more about conversions... May occur, so always check the results or ampere second/volt the system... Watt ( symbol: W ) ampere or watt/volt the SI derived unit for electric watt naar ampère... Example: 2.4KW watts are being sent at 120 volts measurement unit: or.
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# in order that a hoax may be seen through
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01 Apr 2018, 19:01
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in order that a hoax may be seen through, a scientific investigator must always maintain a healthy level of skepticism.
A. that a hoax may be seen through, a scientific investigator must always
B. of seeing through a hoax, scientific investigators always must
C. to see through a hoax, a scientific investigator must always
D. to see through a hoax, a scientific investigator always
E. for a scientific investigator to see through a hoax, they must always
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Re: in order that a hoax may be seen through [#permalink]
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01 Apr 2018, 20:27
A. that a hoax may be seen through, a scientific investigator must always May be seen is passive voice
B. of seeing through a hoax, scientific investigators always must of seeing is awkward
C. to see through a hoax, a scientific investigator must always = Correct
D. to see through a hoax, a scientific investigator always You loose the notion that a scientific investigator MUST do something
E. for a scientific investigator to see through a hoax, they must always "A scientific investigator" and "they" = Prounoun error
IMO C
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Re: in order that a hoax may be seen through [#permalink]
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01 Apr 2018, 21:13
in order that a hoax may be seen through, a scientific investigator must always maintain a healthy level of skepticism.
A. that a hoax may be seen through, a scientific investigator must always - Unidiomatic. The correct idiom is : "In order to"
B. of seeing through a hoax, scientific investigators always must : The correct idiom is : "In order to"
C. to see through a hoax, a scientific investigator must always : Sounds right. Can keep it.
D. to see through a hoax, a scientific investigator always : The idiom is right, however the second part of the sentence sounds incomplete.
E. for a scientific investigator to see through a hoax, they must always : The correct idiom is : "In order to"
Answer : C
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Re: in order that a hoax may be seen through [#permalink]
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01 Apr 2018, 21:13
aragonn wrote:
in order that a hoax may be seen through, a scientific investigator must always maintain a healthy level of skepticism.
A. that a hoax may be seen through, a scientific investigator must always
B. of seeing through a hoax, scientific investigators always must
C. to see through a hoax, a scientific investigator must always
D. to see through a hoax, a scientific investigator always
E. for a scientific investigator to see through a hoax, they must always
A. that a hoax may be seen through, a scientific investigator must always --error because of 'that '
B. of seeing through a hoax, scientific investigators always must --modifier error always must
C. to see through a hoax, a scientific investigator must always ---correct
D. to see through a hoax, a scientific investigator always --must should be there
E. for a scientific investigator to see through a hoax, they must always --'they ' is plural but scientific investigator is singular
please correct me if i am wrong or add some input for improvement.
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Re: in order that a hoax may be seen through [#permalink]
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01 Apr 2018, 21:47
IMHO C
A. Idiom should be "in order to" and passive voice is awkward.
B.again, Unidiomatic
C.Looks fine.
D.SVA problem
E.Unidiomatic and "they" pronoun refers to?
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Re: in order that a hoax may be seen through [#permalink]
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02 Apr 2018, 00:04
aragonn wrote:
in order that a hoax may be seen through, a scientific investigator must always maintain a healthy level of skepticism.
A. that a hoax may be seen through, a scientific investigator must always
B. of seeing through a hoax, scientific investigators always must
C. to see through a hoax, a scientific investigator must always
D. to see through a hoax, a scientific investigator always
E. for a scientific investigator to see through a hoax, they must always
Close call between C and D, both look good... "must and always" together seem redundant. I ll go with D.
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Re: in order that a hoax may be seen through [#permalink]
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02 Apr 2018, 01:32
QZ wrote:
aragonn wrote:
in order that a hoax may be seen through, a scientific investigator must always maintain a healthy level of skepticism.
A. that a hoax may be seen through, a scientific investigator must always
B. of seeing through a hoax, scientific investigators always must
C. to see through a hoax, a scientific investigator must always
D. to see through a hoax, a scientific investigator always
E. for a scientific investigator to see through a hoax, they must always
Close call between C and D, both look good... "must and always" together seem redundant. I ll go with D.
Hi QZ,
in D we are lacking a proper verb in my opinion. "A scientific investigator always maintain a healthy level of skepticism" isn't a correct sentence. We require a help verb such as "must", which we can find in C. Therefore, C is the correct answer.
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Re: in order that a hoax may be seen through [#permalink]
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02 Apr 2018, 07:28
Masterscorp wrote:
QZ wrote:
aragonn wrote:
in order that a hoax may be seen through, a scientific investigator must always maintain a healthy level of skepticism.
A. that a hoax may be seen through, a scientific investigator must always
B. of seeing through a hoax, scientific investigators always must
C. to see through a hoax, a scientific investigator must always
D. to see through a hoax, a scientific investigator always
E. for a scientific investigator to see through a hoax, they must always
Close call between C and D, both look good... "must and always" together seem redundant. I ll go with D.
Hi QZ,
in D we are lacking a proper verb in my opinion. "A scientific investigator always maintain a healthy level of skepticism" isn't a correct sentence. We require a help verb such as "must", which we can find in C. Therefore, C is the correct answer.
Hi, probably you are right, but let's learn together. Cheers..
Thanks..
QZ
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Re: in order that a hoax may be seen through [#permalink]
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24 Dec 2018, 08:17
What does "see through a hoax" mean ?
Re: in order that a hoax may be seen through [#permalink] 24 Dec 2018, 08:17
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# in order that a hoax may be seen through
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trying to build linked list that sort it selft. but this code below doesnt work. this part of code should insert a node in order. for ex.
if i have a linked list of age 1 3 4 5. and if i insert(3) than result should be 1 2 3 3 4 5. but my code output 1 2 3 4 5 3 :(
``````cur_node = head;
while(cur_node != NULL && cur_node->next != NULL)
{
if(cur_node->age < ag && cur_node->next->age > ag)
{
new_node = (struct node *) malloc(sizeof(struct node));
if(new_node == NULL)
{
printf("Node allocation failed.\n"); fflush(stdout);
}
strcpy(new_node->name, na);
new_node->age = ag;
new_node->next = cur_node->next;
cur_node->next = new_node;
in_list = 1;
}
cur_node = cur_node->next;
}
``````
i think the problem is in this statement.
``````if(cur_node->age < ag && cur_node->next->age > ag)
``````
if i have linked list of age1, 6, 7. and if i insert(2). it works fine. because 2 < 1 && 2 is > 6.
problem is if i have age1 2 4 5. and if i insert(2). it does not work bc 2 < 1 && 2 > 2. error here. can idea how can i fix the if statement.
``````if(cur_node->age < ag && cur_node->next->age >= ag)
if(cur_node->age <= ag && cur_node->next->age >= ag)
``````
they both dont work :(
/////////// full code of my insert method
``````void insert(char *na, int ag)
{
struct node *new_node;
int skip = 0;
int in_list = 0;
if(cur_node == NULL)
{
head = (struct node *) malloc(sizeof(struct node));
{
printf("Node allocation failed.\n"); fflush(stdout);
exit(1);
}
}
else
{
/* skip double */
while(cur_node != NULL)
{
if(strcmp(cur_node->name, na) == 0)
{
skip = 1;
}
cur_node = cur_node->next;
}
if(skip == 0)
{
/* insert at begining */
{
new_node = (struct node *) malloc(sizeof(struct node));
if(new_node == NULL)
{
printf("Node allocation failed.\n"); fflush(stdout);
}
strcpy(new_node->name, na);
new_node->age = ag;
in_list = 1;
}
/******************* 1 2 3 4 insert(2) ------------------ doesnt work **************/
/* insert in middle */
while(cur_node != NULL && cur_node->next != NULL)
{
if(cur_node->age < ag && cur_node->next->age > ag)
{
new_node = (struct node *) malloc(sizeof(struct node));
if(new_node == NULL)
{
printf("Node allocation failed.\n"); fflush(stdout);
}
strcpy(new_node->name, na);
new_node->age = ag;
new_node->next = cur_node->next;
cur_node->next = new_node;
in_list = 1;
}
cur_node = cur_node->next;
}
/* insert at end */
if(in_list == 0)
{
while(cur_node->next != NULL)
{
cur_node = cur_node->next;
}
new_node = (struct node *) malloc(sizeof(struct node));
if(new_node == NULL)
{
printf("Node allocation failed.\n"); fflush(stdout);
exit(1);
}
strcpy(new_node->name,na);
new_node->age = ag;
new_node->next = NULL;
cur_node->next = new_node;
}
}
}
}//end of insert method
``````
Try this: Instead of checking both the current node and the next node, just look at the current node. If the element you're considering is less than the current node, insert it before the current node. If it's greater or equal, move on to the next node. If you hit the end of the list without finding a place for it, just put it at the end.
Is it clear why that should work?
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Mikaela B.
Concentration of H+
HA is a weak, monoprotic acid that dissociated according to the equation:
HA (aq) ßà H+ (aq) + A- (aq) Ka = 4.0 x 10-9
If the initial concentration of HA is 0.10 M, find the concentration of H+
By:
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Experienced and Effective STEM Tutor
Mikaela B.
Thank you! This was really helpful.
Report
05/27/19
Ellen E.
Careful. There's an issue with the quadratic solution here. I will post answer in a moment. Notice that : a) x2 = 4.0 x 10-11 - 4.0 x 10-9 x should be x2 = 4.0 x 10-10 - 4.0 x 10-9 x and b) x2 - 4.0 x 10-11 + 4.0 x 10-9 x should be x2 + 4.0 x 10-10 - 4.0 x 10-9 x AND anyway, you should use an approximation here. I will explain when I post my answer in a moment!
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# Homework Help: Probability - what distribution/ model should be used in this context
1. Jan 4, 2008
### SavvyAA3
1. The problem statement, all variables and given/known data
Suppose we are given a table of conditional probabilities as follows (probabilities are in brackets):
Service Level
Demand: Poor/ Standard/ Exceptional
Low: Poor Demand: (0.1) Standard Demand: (0.6) Exceptional Demand:( 0.3)
High: Poor Demand: (0.5) Standard Demand: (0.4) Exceptional Demand: (0.1)
We are told the following: At a telephone call centre, the service levels during each hour (from 0000 hrs to 0100hrs, 0100hrs to 0200hrs and so on) are classified as Poor, Standard or Exceptional, and the classifications for successive hours can be regarded as independent.
The probabilities of the different classification levels being achieved depend on whether demand each hour is High or Low (which is also independently determined in different hourly periods).
Supposing demand levels are low from 1200 to 1800 hrs, find the probability that exactly four of those hours are classified as exceptional
2. Relevant equations
Can someone please tell me how to go about answering this? Should I use a poisson distribution? If so what is the parameter. I know the interval is of length '4 hours' but what is the mean?
Can I use conditional probability theory? if so How?
Thanks. I've spent a whilie trying to see what model to use but I can't igure it out.
3. The attempt at a solution
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
2. Jan 4, 2008
### EnumaElish
I'd treat this as a "4 dice out of 6" kind of a problem. E.g., what is the probability that 4 dices out of 6 come up "two"?
In this example, each die has 3 faces (poor/std/exceptional), with the corresponding probabilities.
Since demand levels are low throughout, you know the set of probabilities you can assign to each "face" of each 3-sided die.
Last edited: Jan 4, 2008
3. Jan 4, 2008
### SavvyAA3
OK, so you suggest that I simply state, since we know their is Low demand we need not assign this a probability. I suppose this is a sensible assumption since the question states the two events (Demand and Service Level) are independent.
So from there I can just go about using these conditional probabilities (ignoring that they are conditional probabilities because the two events are independent):
Please tell me if my assumptions are correct:
A success in this scenario is obtaining an ‘exceptional service’
Fail is obtaining ‘not exceptional service’
So we have:
Ncr *(success)^r *failure^(1-r) [^ reps to the power of]
6c4*(.3)^4 * (0.7)^2
15 * (0.0081) * (0.49)
0.059535
I hope this is correct. Your example of the die scenario really helped me to see the logic behind this!!
4. Jan 4, 2008
### SavvyAA3
sorry the second line should read: ncr *(success)^r *(failure)^(n-r)
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138 inches to feet
• 138 inches is equivalent to 11 1/2 feet
• tk10npubl tk10ncanl
Say hello to Evi
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## Metacyclic groups in $AGL(4,3)$
Just wonder if there is any known results on the transitive metacyclic groups of $AGL(4,3)$?
Sorry, I should elaborate a bit here. I am working on a graph $\Gamma$ which admits a metacyclic group transitive on its vertices. By some other restricted conditions, I have reduced that $Aut\Gamma$ is contained in $AGL(4,3)$. So it turns out that I need to work on the following problems:
1. Check whether there exist such metacyclic subgroups in $AGL(4,3)$.
2. If exist, what are they?
One of my attempt was to inspect subgroups of $AGL(4,3)$ and rule out those impossible cases. But it seems tedious and somehow I still have to seek help from machinery search...
Alternatively, I am hoping to do it on a more elegant and efficient way. That is, to find the generators of the metacyclic groups. We know that a metacyclic group can be generated by two elements. Hence if we can find all the possible generators, then we can construct the groups and verify if these groups are transitive or not. There are some helpful information we already know: written $AGL(4,3)=\mathbb{Z}_3^4{:}GL(4,3)$, so a transitive metacyclic group $G=\langle(a_1,b_1),(a_2,b_2)|a_i\in\mathbb{Z}_3^4,b_i\in GL(4,3),~\mbox{and}~o(b_1),o(b_2)~\mbox{divisible by}~9\rangle$. Then the next question is which exact generators we should choose?
Another thought is that if it is possible to do a pure computer search? as the group is not too big, but I am not too sure how to write up the codes...
Any suggestion would be much appreciated.
-
Thank you for both of you answering my question. Both answers are quite close to my desire. I would like to tick Peter's answer however Stefan's computation is also nice. – generao Jan 29 at 10:13
There is no such group. Let $G$ be a transitive metacyclic subgroup of $\text{AGL}(4,3)$. Let $C$ be a cyclic normal subgroup of $G$ with $G/C$ cyclic.
As $G$ permutes transitively the orbits of $C$, the kernels of the action of $C$ on its orbits all have the same size, thus they are equal because $C$ is cyclic. But $C$ acts faithfully on the union of the orbits. We infer that $C$ acts regularly on each orbit. In particular, $\lvert C\rvert$ divides $3^4$.
Note that the Sylow $3$-subgroups of $G$ are transitive too, and subgroups of metacyclic groups are metacyclic too. So we may assume that $G$ is a $3$-group.
On the other hand, $9$ is the maximal order of a $3$-element in $\text{AGL}(4,3)$. (One can see that most easily from the embedding $\text{AGL}(4,3)\le\text{GL}(5,3)$.)
From that we see that $C$ has order $9$, and $G$ is a semidirect product of $C$ with another cyclic group $D$ of order $9$.
View $G=C\rtimes D$ as a subgroup of $\text{GL}(5,3)$. From Jordan's normal form theorem, we see that $\text{GL}(5,3)$ contains two conjugacy classes of elements of order $9$. Let $U$ be the group of upper triangular matrices of $\text{GL}(5,3)$. Consider the two cases of $C$ (corresponding to the Jordan block sizes $4+1$ and $5$, respectively). In both cases, one computes that the exponent of $N_U(C)/C$ is $3$, so there is no room for the cyclic group $D$ of order $9$.
-
Peter, I think there is a better way to show that $G=\mathbb{Z}_9.\mathbb{Z}_9$. Since $G$ is transitive on $3^4$, then so is its Sylow $3$-group $P$, and $81\mid|P|$ by orbit stabiliser theorem. On the other hand, the highest order of $3$-elements in $AGL(4,3)$ is 9. So the only possible structure for $P$ is $\mathbb{Z}_9.\mathbb{Z}_9$. Also I am gonna to partially quote ur argument in my paper. Do you need ur name on? – generao Jan 29 at 10:17 @generao: Do whatever you want, the argument is so simple that I'm not proud of it ... However, as you raised the question, it would probably be scientifically more correct to mention the two answers. – Peter Mueller Jan 29 at 14:47 @Peter: There was something I didn't reckon at the first time. At your last step, you said one "computes" the exponent is 3. Did you do it by machine or do it manually? I verified that using GAP, but when I tried calculating manually I found that is a huge calculation. Or is there any trick I didn't know? – generao Feb 14 at 7:21 If $c$ is a generator for $C$, and $u$ normalizes $C$, then $u$ induces an automorphism of order $1$ or $3$ on $C$, hence $cu=uc^m$ with $m=1$, $4$, or $7$. Given $c$ in Jordan normal form, this yields a system of linear equations of $u$, and one verifies that $u^3\in C$. I think this can be done by hand, but I believe that I used Sage for the computation. – Peter Mueller Feb 14 at 23:34
There are no such groups. The GAP (cf. http://www.gap-system.org/) calculation is as follows:
Construct the group $G := {\rm AGL}(4,3)$:
gap> G := SemidirectProduct(GL(4,3),GF(3)^4);
<matrix group of size 1965150720 with 3 generators>
Move to the natural permutation representation of $G$ on $3^4 = 81$ points:
gap> phi := IsomorphismPermGroup(G);;
gap> H := Image(phi);
<permutation group of size 1965150720 with 3 generators>
gap> DegreeAction(H);
81
Find all conjugacy classes of $H$ of elements whose order is divisible by 9:
gap> ccl9 := Filtered(ConjugacyClasses(H),
> cl->Order(Representative(cl)) mod 9 = 0);;
gap> List(ccl9,Size);
[ 4043520, 36391680, 12130560, 24261120, 36391680 ]
gap> reps := List(ccl9,Representative);;
gap> List(reps,Order);
[ 9, 18, 9, 9, 18 ]
Compute normalizers of conjugacy class representatives in $H$:
gap> normalizers := List(reps,g->Normalizer(H,Group(g)));
[ <permutation group with 7 generators>, <permutation group with 6 generators>,
<permutation group with 7 generators>, <permutation group with 4 generators>,
<permutation group with 6 generators> ]
gap> List(normalizers,Size); # the normalizers are nicely small
[ 2916, 324, 972, 486, 324 ]
Search for transitive metacyclic subgroups of $H$:
gap> List([1..5],i->Filtered(AsList(normalizers[i]),
> g -> Order(g) mod 9 = 0 and
> IsTransitive(Group(g,reps[i]),[1..81])));
[ [ ], [ ], [ ], [ ], [ ] ]
-- There are none!
However if we allow for two orbits instead of one, there are solutions:
gap> List([1..5],i->ForAny(AsList(normalizers[i]),
> g -> Order(g) mod 9 = 0 and
> Length(Orbits(Group(g,reps[i]),[1..81])) <= 2));
[ true, true, false, false, false ]
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Blog
# What Does This Symbol Mean in Excel?
Do you ever find yourself staring at a computer screen, scrolling through an Excel spreadsheet, and wondering to yourself, “What does this symbol mean?” You’re not alone. Many people find themselves in the same position, unable to interpret the meaning of the various symbols that appear in an Excel spreadsheet. In this article, we’ll explore what this symbol means in Excel, and how to use it to your advantage. So, let’s get started!
## What Do Symbols Mean in Excel?
Excel is a powerful tool used by businesses and professionals to store, organize, and analyze data. It has many features that can help users to quickly and easily understand the data presented to them. One of these features is the use of symbols to represent various operations in the spreadsheet. Understanding what these symbols mean is essential for unlocking the full potential of Excel.
The most common symbols used in Excel are the plus (+), minus (-), and equal (=) signs. These symbols are used to represent basic arithmetic operations, such as addition, subtraction, and equality. Understanding how these symbols work is essential for quickly performing calculations and manipulating data in Excel.
Another set of symbols used in Excel is the asterisk (*), forward slash (/), and caret (^) symbols. These symbols are used to represent operations such as multiplication, division, and exponentiation. Understanding how these symbols work is essential for quickly performing complex calculations and manipulating data in Excel.
## What Are the Other Symbols Used in Excel?
In addition to the symbols mentioned above, Excel also uses a range of other symbols to represent various operations. For example, the percent (%) symbol is used to represent percentages, and the pound symbol (#) is used to represent the pound currency. Additionally, the at symbol (@) is used to represent a cell reference. Understanding the meaning of these symbols is essential for quickly understanding and manipulating data in Excel.
Another set of symbols used in Excel is the arrow symbols. These symbols are used to move the active cell or select a range of cells. Additionally, the asterisk with a dot inside it (*.) is used to select all cells in a range, while the colon (:) is used to separate two cell references in a range. Understanding how these symbols work is essential for quickly navigating and manipulating data in Excel.
## What Are the Other Special Symbols Used in Excel?
Excel also uses a range of special symbols to represent various operations. For example, the tilde (~) symbol is used to represent a wildcard character, and the ampersand (&) symbol is used to join two strings of text together. Additionally, the exclamation mark (!) symbol is used to represent a range of cells, and the question mark (?) symbol is used to represent a single cell reference. Understanding the meaning of these symbols is essential for quickly understanding and manipulating data in Excel.
Another set of symbols used in Excel is the brackets () and braces ({ }). These symbols are used to represent operations such as subtotals, SUMIFs, and COUNTIFs. Understanding how these symbols work is essential for quickly performing calculations and manipulating data in Excel.
## What Are the Other Advanced Symbols Used in Excel?
In addition to the symbols mentioned above, Excel also uses a range of advanced symbols to represent various operations. For example, the double quotation marks (“ ”) symbol is used to represent a text string, and the dollar sign (\$) symbol is used to reference cells relative to a certain cell. Additionally, the asterisk with a line inside it (*_) is used to create a range of cells within a certain column, and the ampersand with a line inside it (&_) is used to create a range of cells within a certain row. Understanding the meaning of these symbols is essential for quickly understanding and manipulating data in Excel.
Another set of symbols used in Excel is the parentheses (( )). These symbols are used to group operations together, such as when performing calculations. Additionally, the brackets with a line inside them (<_>) are used to create a range of cells within a certain worksheet, while the braces with a line inside them ({_}) are used to create a range of cells across multiple worksheets. Understanding how these symbols work is essential for quickly performing calculations and manipulating data in Excel.
### What Does This Symbol Mean in Excel?
Answer: This symbol is a “percent sign” and its purpose in Excel is to indicate a percentage. It’s typically used in formulas to indicate a percentage of a value or to convert a fraction into a percentage. For example, if you wanted to calculate the percentage of a total, you could use the formula =A1/B1*100 and the result would be expressed as a percentage.
### What Does This Symbol Look Like in Excel?
Answer: The percent sign symbol in Excel looks like a “%” character. It’s usually located in the top row of the keyboard, to the right of the number “0” key.
### What Is the Keyboard Shortcut for the Percent Sign Symbol in Excel?
Answer: The keyboard shortcut for the percent sign symbol in Excel is to press the “Shift” key and hit the “%” key. This will insert the percent sign symbol into the cell of your Excel worksheet.
### How Is the Percent Sign Symbol Used in Excel Formulas?
Answer: The percent sign symbol is used in Excel formulas to indicate a percentage or to convert a fraction into a percentage. For example, to calculate the percentage of a total, you could use the formula =A1/B1*100, with the result expressed as a percentage.
### What Does the Percent Sign Symbol Mean in Other Programs?
Answer: The percent sign symbol may have different meanings in other programs. For example, in HTML, the percent sign symbol is used to represent a percent-encoded character. In some programming languages, the percent sign symbol is used to indicate a modulo operation.
### Can I Change the Symbol Used for Percentages in Excel?
Answer: No, the percent sign symbol is the default symbol used for percentages in Excel and cannot be changed. If you want to use a different symbol for percentages, you will need to use text formatting to replace the percent sign symbol with the symbol of your choice.
### Dollar Sign \$ in Excel | Excel in Minutes
The meaning of symbols in Excel is a complex topic that can be misleading at times. Despite this, understanding the meaning of symbols in Excel can be a valuable skill when it comes to creating accurate spreadsheets and managing data. With an understanding of the symbols used in Excel, you can manage your data more efficiently and effectively, whether you are a professional or a novice user.
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# Talk:LittleBlackBook
## Congrats!
Congrats! It's nice to see some movement in the nano scene lately. I was pondering predefined data bots myself actually. I'm curious though, you say "Any perfect dodge stop and go will kill it", which seems odd to me. It should be quite trivial for a distance/lateral segmented GF gun to hit that I'd think... --Rednaxela 01:51, 15 June 2009 (UTC)
Math.signum() off e.velocity screws up 0 velocity checks and I have no velocity averaging. I'll check other bots for inspiration though. Has to be TINY though
What about... `static double foo = 1; foo = e.velocity + 0.01*foo; foo /= foo;`? Not sure if that would be small enough, and I haven't checked it's codesize, but I think it's smaller than any others ways to handle that which come to my mind. I think it's smaller than `static double foo = 1; foo = (e.velocity == 0) ? foo : Math.signum(e.velocity);` anyway, since conditionals and function calls are expensive if I remember right. I'd also suggest seeing if you can get rid of the need to normalize the value to 1/-1 all together, but that may not be possible in this case. --Rednaxela 04:15, 15 June 2009 (UTC)
Anything having to do with statics are expensive. Your above code is probably around 25 bytes or so. Math.signum(e.getVelocity()) is 6 or so. I can easily fix my problem by adding +1 to the e.getVelocity() check on the inside of the gun( for 2 bytes)... But! It will only hit perfect stop/go 50% of the time as it is setup to use the velocity to know how far to lead. Also, bots like FretNano who have good guns, but get stuck on walls randomly, would then be unhittable. I get 1 lead value per distance segment at the moment. I've got a few ideas on how to improve this, though it will double+ the size of my string table. I will adhere to the 200k max limit and with 1000 or so bots to profile, I have to be careful on size. --Miked0801 13:38, 15 June 2009 (UTC)
Hi! Awesome results! Now that I looked at the code, I think there is one little problem with the overall concept: every time a new robot or a new version of a robot appears, you have to update the stats in your bot. --HUNRobar 10:31, 15 June 2009 (UTC)
And a decent problem it is. I'm going to review the string libraries today to see if I can find a "closest to this name" type function for my search. That'll at least handle versioning. The version in there right now is using default behavior against Pugio because my table is setup for 1.40. When all your other versions get cleared from the rumble, my rating will go up another .5 or so. --Miked0801 13:38, 15 June 2009 (UTC)
`HashMap<String, String> enemyGF; enemyGF.get(e.getName().split(" ")[0]);` =) » » 14:03, 15 June 2009 (UTC)
Um, wow. That does work: e.getName().split(" ")[0]. I rediscovered it myself and you had it here the whole time. I feel very stupid. Thanks for the advice! --Miked0801 06:13, 23 June 2009 (UTC)
No problem at all. =) » » 11:05, 23 June 2009 (UTC)
Nice! I may re-release Ocnirp 1.0 to screw your stats =) Updated Wikipedia page for you now. » » 12:00, 15 June 2009 (UTC)
Don't bother, I'm already using default values against that bot so it wouldn't make any difference so Nyaa Nyaa! :) --Miked0801 13:48, 15 June 2009 (UTC)
Really? I don't think so. (default is 50-40-30-20-10-0, right?). Beside, I'm working on Nano adaptive movement (Mysterious is its fail attempt) =) » » 14:03, 15 June 2009 (UTC)
Ok, you win. There are quite a few bots taht got default cases though. For some reason I thought Ocnrip was one. And, yeah, nice string match there. I could fit that if, oh I don't know, I took out all my movement code ;) --Miked0801 14:20, 15 June 2009 (UTC)
That doesn't call fit, really. :P » » 14:53, 15 June 2009 (UTC)
Thanks! Links and work email address removed. --Miked0801 15:27, 15 June 2009 (UTC)
He he, the revision system still have it. =) » » 15:32, 15 June 2009 (UTC)
Yep - but at least basic spider bots won't pick it up and spam be about male enhancement products :) --Miked0801
Ok, new version in the wild. I expect this release to be better against stronger bots, but I also expect some startlingly poor results against a few bots. These I can tune back to either the old values or a combo of new and old. Stelo.MirrorNano being an example. My next release will beat this bot 90/10, but now LBB is losing 45/55. No panic :) --Miked0801 02:04, 18 June 2009 (UTC)
Good result. My next update tomorrow is going to absolutely destroy the competition. This is going to be a 1900+ rating bot. --Miked0801 05:45, 18 June 2009 (UTC)
## Bot Blog
Seems like a decent place for random ramblings. The averaging code is doing its job, it just needs some table number TLC. Alas, it also screwed with some other scores that will need attention. I also commited a cut and paste errored with nat's SNG 1.0b bot resulting in a 6% score. Lame. I may try taking the averaging back out and using firepower control instead. It'd probably help me more overall to fire at 3.0 firepower at range 80 or 120 and closer against everyone. The Micro scores have been decent thus far with their first autogeneration pass. 19th is a good starting point and I know I can get to 15th or even 10th with some work there. I'll drop the mini scores in Tuesday or so and for another overall ranking boost. I still believe I can get this bot to a 2k points score with over 80%APS in nanos with a few revisits to lagging numbers.
Wow, a good day. Found another 10 bytes by just re-arranging the stack vars. Used that space to try out variable move distances. Helps quite a bit against the weakest bots. Also fixed bugs with a few bots that came from typos. Finally, I fixed poorer results against rambots wehre I wasn't aiming right at range 40 and in. My scores are now approaching DrussGT's against the low-end bots. Using that bot's scores gives me a reasonable target to short for. I'm also going to experiment with variable firepower as well, though for some reason I keep pushing back on that - dunno. Next release tomorrow night. Shooting for APS of 78ish in nanos. That would put me close to my goal of a 2k score. Hopefully, my micro score will get me top 15 there as well. I'll start running Minis by end of week to see if I can get top 25 there and perhaps top 100 overall (aiming real high, I know.) --Miked0801 03:38, 23 June 2009 (UTC)
You should to remember the Code Size/Cheat sheet. The most important part is the vars-on-register, which is discover by Simonton I think. » » 11:02, 23 June 2009 (UTC)
I just rearrange them all until I get the smallest result :)
Another good day, but I found a dumb bug I introduced to my table generation a few days ago which will require me to revisit all the nano values. Sigh. Good news though, found another 8 bytes or so by declaring the string table final (never changes so I should have done it ages ago) and by rearranging a few typecasts to make sure we got int on int math wherever possible. That means I now have both distance and firepower control in the table! It hurts by aim a touch from 80 pixels and in, but for 1 more free byte I could fix this. I'll keep looking. 8 more bytes and I get variable string matching, though if firepower or move control prove poor, I can always swap this in for the other. Because I have to do a table revisit, I'm pushing my release back one more day. Who'd have though a variable GF, variable firepower, variable move distance, variable turn control bot could be done through a silly string table. It's like the synthesis of all the best nanos in one place. Next release will have some bugs, but the one following will be absolutely epic. --Miked0801 18:48, 23 June 2009 (UTC)
Bleh. Had to work late a few nights. I'm just now getting all my initial micro/nano runs complete. Good news is that only Moebius is currently able to consistantly beat me now in nanos. I'm going to experiment with distance control next as many bots are optimized for certain ranges and I expect I will do well at varying ranges against many bots. If it works well, I'll trade something off to get it. We'll see about a tomorrow release, work allowing. --Miked0801 07:39, 25 June 2009 (UTC)
Played around with enemy distancing and found it helps (much) more than firepower control. But, my test cycle is too far along to scrap and start again so I'll go ahead with what I have. Besides, I believe I found a way to fit all 5 factors into this nano anyways - Fire Power, Chance to reverse direction, Min Distance to move, desired distance from enemy, and Guess Factors. It requires my tables to be created differently (again), but I'll go for that next version. This is why I haven't bothered with minis or megas yet - the tables flux dramatically with each version. It also means yet another multiplier when creating my tables. I'm already running 20+ versions per bot through roboresearch. Adding distance is another x4 or so. Yikes. Back to 1 good test run and cull, cull, cull so that I have time to balance what makes sense. At least my testing methods are improving so that this doesn't take too long. Just need 1 more set up data to finish up for the guess factors and then we'll put this bot back together again. --Miked0801 07:29, 26 June 2009 (UTC)
Ok, version is up and we gained nearly another 2% APS in nanos and even more in Micros. Nice. There's still a few bots that came out a bit half baked, even with all the care I put into the tables before releasing. I'll revisit those for one last version 1.5x release and then go onto version 1.6x with distance control. Still shooting for 80+% APS, a perfect PL score in nanos, and 2k in ratings. 72 Glicko's to go. --Miked0801 02:59, 27 June 2009 (UTC)
Hey, you've taken my micro's place in the microrumble! I really should do something right after I finished my flash game. ;) --HUNRobar 17:31, 27 June 2009 (UTC)
## LBB Cheeze Factor
I added PEZ's LittleEvilBrother, because it also had a String based GF gun, I believe. There was some discussion about whether or not it was cheesy to load data for a bot not capable of saving data. Consensus at the time was that it was cheesy, but judging by how much better LittleBlackBook does, it seems like a pretty cool strategy that has a lot of room for experimentation. Plus there's been nothing new in nano-land for years! --Alcatraz 03:09, 26 June 2009 (UTC)
I'd be of the opinion that it is cheesy, but don't object to keeping it around, at least till something particularly new happens in nano-land :) --Rednaxela 04:26, 26 June 2009 (UTC)
It's extremely cheesy an exploitive of a loophole in the codesize tool. Yet, I'm having some real fun with it right now (and this game is supposed to be fun! :). I literally believe I can get this bot as a top 10 in the micro competition as a nano. That's something that hasn't been done since just after we decided on weight classes. Right now, I'm still crunching a rather exhasutive set of numbers for firepower and movement types. Once that's done, I'll grab the numbers and give it a test run to see where it stands. Good chance it will go in tonight. --Miked0801 05:15, 26 June 2009 (UTC)
Well, I wouldn't call it "exploitive of a loophole in the codesize tool" exactly (unless someone makes a highly capable interpereter within nano size), since such preloading of data specific to certain bots is also cheesy for megabots. I think what's cheesey most fundumentally, is when a bot that looks up the enemy on a preloaded table based on bot names regardless of codesize. --Rednaxela 13:11, 26 June 2009 (UTC)
I for one don't like this though, but it is acceptable for me. Unless, like Rednaxela, someone make a good enough nano-size compiler and fit DrussGT into nano. I'll not object, but I do think that when you fully tuned this robot to the top and have no more idea, you should retire this robot. » » 13:30, 26 June 2009 (UTC)
Of course I agree that preloading data is kinda cheesy, but I think everyone understands that LBB is of a different ilk than other NanoBots. It's cool seeing how far you can go with it. To be honest, I think even hand-tuned segmentations are similar to pre-loading: with a different rumble population, different tunings would surely work better. We all take the rumble population into account when we write our bots; pre-loading is just a more explicit way of doing so. --Voidious 14:10, 26 June 2009 (UTC)
Yep, it's cool to see how far LBB can go indeed. It is true that we all build our bots in way that considers the rumble population, but I don't think pre-loading itself is the cheesy thing. For one thing, I wouldn't consider it cheesy at all, if a GF gun preloads with the average data of a bunch of rumble bots, until it's learned enough about the specific enemy (hint, hint). I think tying pre-loaded data to a specific bot name is where the 'cheese' begins, but more general pre-loaded data is simply prudant rather than cheesy. --Rednaxela 14:34, 26 June 2009 (UTC)
Personally, I think the only issue here is that pre-loading is used to get around the size class limits. A bot that saves its own data is fine by me, whether it's general or specific -- the 200k limit restricts how much you can really store anyway. LBB is a novel concept, but it would be disappointing to see it inspire a new generation of string table nanos/micros/minis. Now if you could make a nano that generates its own string tables, that would be really cool. You still have a few bytes left, right? ;) --Darkcanuck 15:02, 26 June 2009 (UTC)
## AceSurf
I noticed you make not of AceSurf being a problem. One possible solution (if you can fit it into the codesize) that would probably work well, would be having a special code in your data that says "Aim at the last GF it went to". I'm kind of thinking that would probably eat too much codesize... but I think that's probably the simplest thing that could be done with the likes of AceSurf, besides random targeting anyway. --Rednaxela 16:00, 20 June 2009 (UTC)
I'm not too worried. This bot just hit the 1900 club - the first ever in nanoland to do this. This is with untuned tables for 1/2 the bots. I will take a closer look at ace.surf results though and possibly setup alternating guess factors at every 40 pixels. This has been the best way recently to smash wave bots as it is nearly the same as 2 different guess factors every few shots. Wow I proud of this bot. I might be able to get it to the 2000 club when all is said and done. It just needs another 1000 percentage basis points or so. Plus, right after this bot was released, I figured out a way to better hit stop'n'go's without reducing accuracy against simpler bots who just get stuck. Next release will be fun! --Miked0801 16:06, 20 June 2009 (UTC)
IIRC, The club score is base on ELO, not Glicko-2. » » 16:13, 20 June 2009 (UTC)
then Toorkild isn't in the 2k club. I was pretty sure it was the Glicko. But, it all may be mute. I just realized how few battles it has run - only about 1 per bot. I'll therefore wait a bit before posting anything further on that :) --Miked0801 16:17, 20 June 2009 (UTC)
We're in a weird position with the clubs because it was based on ELO on the old server. ELO on Darkcanuck's server is way lower than it was on the old server, and Glicko-2 is a bit higher -- at least in MegaBot and MiniBot 1v1. As a point of reference, before the old RR server hit a few months of non-stop upwards drift, Dookious was about 2130 and Komarious was about 2085 in mini (prolly a few points higher now). Nevertheless, it's still easy to compare your bot to the competition, arbitrary rating systems aside. =) --Voidious 16:26, 20 June 2009 (UTC)
The new server originally matched the old one, but ELO has drifted steadily downwards in the past few months. Why, I don't know. In late 2008, Glicko2 and ELO were in the same range, but now they're quite different. Maybe it's all the recent activity, plus some long-standing bots were removed earlier this year too. In any case, ELO is a relative ranking system so it's the difference between the scores that matters, not the absolute values. But it does leave the ELO clubs in a funny position. --Darkcanuck 17:34, 20 June 2009 (UTC)
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# PROFILE
Check out all the problems this user has already solved.
Problem Problem Name Ranking Submission Language Runtime Submission Date
1018 Banknotes 40517º 23514111 C 0.000 7/7/21, 1:09:31 AM
1013 The Greatest 47298º 23514123 C 0.000 7/6/21, 10:53:04 PM
1012 Area 49538º 23511810 C 0.000 7/6/21, 10:47:43 PM
1010 Simple Calculate 55459º 23511340 C 0.000 7/6/21, 10:11:20 PM
1009 Salary with Bonus 56243º 23510929 C 0.000 7/6/21, 9:09:49 PM
1007 Difference 66163º 23510380 C 0.000 7/6/21, 8:43:09 PM
2234 Hot Dogs 03150º 23451249 C 0.013 7/1/21, 7:00:27 PM
2152 Pepe, I Already Took the... 01329º 23450584 C 0.000 7/1/21, 6:15:11 PM
1837 Preface 02425º 23450866 C 0.000 7/1/21, 5:24:10 PM
1091 Division of Nlogonia 01477º 22311400 C 0.000 4/13/21, 3:03:34 PM
1555 Functions 02259º 22297426 C 0.000 4/12/21, 9:07:28 PM
2232 Pascal's Triangle 00707º 22295997 C 0.000 4/12/21, 8:04:06 PM
1168 LED 07252º 22292284 C 0.000 4/12/21, 5:31:12 PM
1036 Bhaskara's Formula 36391º 22291244 C 0.000 4/12/21, 4:26:26 PM
1172 Array Replacement I 21687º 22158133 C 0.000 4/5/21, 12:34:28 PM
1160 Population Increase 04359º 22157303 C 0.000 4/5/21, 11:04:37 AM
1070 Six Odd Numbers 21853º 22156995 C 0.000 4/5/21, 10:02:58 AM
1049 Animal 15536º 22156945 C 0.000 4/5/21, 9:50:19 AM
1015 Distance Between Two Points 47221º 22156633 C 0.000 4/5/21, 8:17:05 AM
2764 Date Input and Output 00997º 22111127 C 0.000 4/1/21, 5:34:27 PM
1253 Caesar Cipher 00288º 22109337 C 0.000 4/1/21, 3:20:10 PM
1174 Array Selection I 18403º 22071893 C 0.000 3/30/21, 7:26:39 AM
2760 String Input and Output 00726º 22071734 C 0.000 3/30/21, 6:32:36 AM
1332 One-Two-Three 03491º 22071978 C 0.000 3/30/21, 1:29:09 AM
1180 Lowest Number and Position 17067º 22065174 C 0.000 3/29/21, 9:10:55 PM
1177 Array Fill II 15237º 22072119 C 0.000 3/29/21, 8:07:14 PM
1175 Array change I 17894º 22062889 C 0.000 3/29/21, 6:51:25 PM
1045 Triangle Types 22652º 21955947 C 0.000 3/23/21, 2:24:28 PM
1165 Prime Number 05992º 21955184 C 0.000 3/23/21, 1:46:46 PM
1115 Quadrant 14598º 21954396 C 0.000 3/23/21, 12:55:28 PM
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# Multicut Lower Bounds via Network Coding Anna Blasiak Cornell University.
## Presentation on theme: "Multicut Lower Bounds via Network Coding Anna Blasiak Cornell University."— Presentation transcript:
Multicut Lower Bounds via Network Coding Anna Blasiak Cornell University
Multicut Given: Directed graph G = (V,E) Capacities on edges k source-sink pairs Find: A min-cost subset of E such that on removal all source-sink pairs are disconnected s1s1 s2s2 s3s3 t1t1 t2t2 t3t3
Directed Multicut The State of the Art Õ( n 11/23 ) - approx algorithm [Agarwal, Alon, Charikar ’07] 2 Ω( log 1-ε n) hardness [Chuzhoy, Khanna ‘09] Nothing non-trivial known in terms of k
Dual Problem: Maximum Multicommodity Flow Given: A directed graph G = (V,E) Capacities on nodes k source-sink pairs Find: A maximum total weight set of fractional s i -t i paths. s1s1 s2s2 s3s3 t1t1 t2t2 t3t3 All approximation algorithms for multicut are based on the LP and compare to the maximum multicommodity flow. Limited by large integrality gap: Ω(min(( k, n δ )) [Saks, Samorodnitsky,Zosin ’04, Chuzhoy, Khanna ’09 ]
Better Lower Bound? Network Coding Good News: Coding Rate ≥ Flow Rate, can be a factor k larger Bad News: Multicut ≱ Coding Rate, can be a factor k smaller
Results Identify a property P of a network code such that any code satisfying P is a lower bound on the multicut. Show P is preserved under a graph product. Main Corollary: – Improved and tight lower bound on the multicut in the construction of Saks et al. and give a network code with rate = min multicut
Simplifying Assumptions A network is: Undirected graph G = (V,E ) Capacity one for each node Subsets of V : {S i } i ∈[k], {T i } i ∈[k] s i - t i pairs, i ∈[k], connect to G: s i connects to v ∈S i v ∈T i connect to t i Type of networks giving: Ω(min(( k, n δ )) [Saks, Samorodnitsky,Zosin ’04, Chuzhoy, Khanna ’09 ]
Saks et al. Construction Begin with The n-path network P n : v 1 v 2 v 3 v 4 v 5 v n-1 v n s1s1 t1t1 Hypergrid( n, k ) is the k -fold strong product of P n. It has n k nodes and k s-t pairs and flow rate n k-1 Saks et al. show it has multicut at least has kn k-1 THEOREM Hypergrid( n,k ) has a code with rate n k - (n-1) k that is a lower bound on the multicut. S1S1 T1T1
Hypergrid(3,2) = P 3 ☒ P 3 s 2 t2t2 s1s1 ( v 2, v 1 ’ ) ( v 1, v 2 ’ )( v 3, v 2 ’ )( v 2, v 2 ’ ) ( v 2, v 3 ’ ) t1t1 ( v 1, v 1 ’ )( v 3, v 1 ’ ) ( v 1, v 3 ’ ) ( v 3, v 3 ’ )
Hypergrid(3,3) = P 3 ☒ P 3 ☒ P 3 S 1 T 1 S 2 T 2 S 3 T 3
Coding Matrix s 2 t2t2 s1s1 ( v 2, v 1 ’ ) ( v 1, v 2 ’ )( v 3, v 2 ’ )( v 2, v 2 ’ ) ( v 2, v 3 ’ ) t1t1 ( v 1, v 1 ’ )( v 3, v 1 ’ ) ( v 1, v 3 ’ ) ( v 3, v 3 ’ ) ( v 1,v 1 ’ ) ( v 2,v 1 ’ ) ( v 3,v 1 ’ ) ( v 1,v 2 ’ ) ( v 2,v 2 ’ ) ( v 3,v 2 ’ ) ( v 1,v 3 ’ ) ( v 2,v 3 ’ ) ( v 3,v 3 ’ ) a1a1 111000000 b1b1 000111000 c1c1 000000111 a2a2 100100100 b2b2 010010010 Column v describes the linear combination of messages sent by v. Column v is a linear combination of columns of predecessors of v. t i can decode messages from s i. Column v describes the linear combination of messages sent by v. Column v is a linear combination of columns of predecessors of v. t i can decode messages from s i. Columns labeled with v in V Rows labeled with messages a i, b i, c i originate at s i rate of code = # of messages Rows labeled with messages a i, b i, c i originate at s i rate of code = # of messages a 1 +a 2 b1b1 a 1 +b 2 a1a1 c1c1 b 1 +b 2 c 1 +a 2 c 1 +b 2 a 2, b 2, c 2 a 1, b 1, c 1 b 1 +a 2 Entries in finite field
Coding Matrix as a Lower Bound L gives a lower bound on the multicut: For M a minimum multicut, |M | = rank ( I M ) ≥ rank (L I M ) ≥ p. L gives a lower bound on the multicut: For M a minimum multicut, |M | = rank ( I M ) ≥ rank (L I M ) ≥ p. DEFINITION A coding matrix L is p -certifiable if 1.For any multicut M, rank (L I M ) ≥ p. 2.Column v of L is a linear combination of columns of incoming sources and predecessors of v that form a clique. |V| x |M| matrix, column v ∈ M is an indicator vector for v.
Main Theorem* Given networks N 1 and N 2 with coding matrices L 1 and L 2, there is a coding matrix for N 1 ☒ N 2 : L = such that: 1.If L 1 and L 2 have rates p 1 and p 2 then L has rate p := n 1 p 2 + n 2 p 1 - p 1 p 2. 2.If L 1 and L 2 are p 1 and p 2 certifiable then L is p- certifiable. In1⊗ L2In1⊗ L2 L1⊗ In2L1⊗ In2
Saks et al. Construction Begin with The n-path network P n : v 1 v 2 v 3 v 4 v 5 v n-1 v n s1s1 t1t1 Hypergrid( n, k ) is the k -fold strong product of P n. It has n k nodes and k s-t pairs and flow rate n k-1 Saks et al. show it has multicut at least has kn k-1 THEOREM Hypergrid( n,k ) has a code with rate n k -(n-1) k that is a lower bound on the multicut. P n is 1-certifiable.
Conclusions and Open Questions Our result: a certain type of network coding solution is a lower bound on directed multicut – Is there a more general class of network coding solutions that is a lower bound? Multicommodity flow can be far from the multicut, what about the network coding rate? – Does there exist an α = o(k) s.t. multicut ≤ α network coding rate?
LP Relaxation All approximation algorithms use LP relaxation with dual = maximum multicommodity flow problem Problem: Large integrality gap – Ω(min(( k, n δ )) [Saks, Samorodnitsky,Zosin ’04, Chuzhoy, Khanna ’09 ] min Σ v in V x v s.t. Σ v in p x v ≥ 1 for all 1 ≤ i ≤ k, for all p s i -t i path x v ≥ 0 for all v in V min Σ v in V x v s.t. Σ v in p x v ≥ 1 for all 1 ≤ i ≤ k, for all p s i -t i path x v ≥ 0 for all v in V
Butterfly network + infinite capacity edges between s i and t j for all i ≠ j x 1 ⊕ x 2 ⊕ ∙∙∙ ⊕ x k x1x1 x2x2 xkxk x4x4 x5x5 x6x6 x k-1 Coding Rate = k, Flow Rate = Multicut = 1
Guiding Questions When is network coding a lower bound on multicut in directed graphs? When it is an lower bound, is it a better lower bound than maximum multicommodity flow?
Strong Graph Product Kronecker Product – A: m ✕ n matrix – B: p ✕ q matrix – A ⊗ B = a 11 B … a 1n B … … a 1m B … a nm B Adjacency matrix of G 1 ☒ G 2 : (G 1 + I n 1 ) ⊗ (G 2 + I n 2 ) – I n 1 n 2 G i here denotes adjacency matrix of G i
IMIM Cuts as matrices v 11 00000 v 2 1 10000 v 31 00000 v 12 01000 v 2 2 00100 v 32 00010 v 13 00000 v 23 00001 v 33 00000 Matrix I M represents a cut M. s 2 t2t2 s1s1 ( v 2, v 1 ’ ) ( v 1, v 2 ’ ) ( v 3, v 2 ’ ) ( v 2, v 2 ’ ) ( v 2, v 3 ’ ) t1t1 ( v 1, v 1 ’ ) ( v 3, v 1 ’ ) ( v 1, v 3 ’ ) ( v 3, v 3 ’ ) L gives a lower bound on the multicut: For M a minimum multicut, |M | = rank ( I M ) ≥ rank (L I M ) ≥ p. L gives a lower bound on the multicut: For M a minimum multicut, |M | = rank ( I M ) ≥ rank (L I M ) ≥ p. DEFINITION A coding matrix L is p -certifiable if 1.For any multicut M, rank (L I M ) ≥ p. 2.Column v of L is a linear combination of columns of incoming sources and predecessors of v that form a clique.
Flows, Cuts, and Codes Multicast Flow Rate Coding Rate Cut Bound Directed Ω((log n/ log log n) 2 ) Undirected 8/7 Multicommodity Flow RateCut Bound Directed Ω(min(( k, n δ )) Undirected Ω(min(( log n, log k )) Coding Rate?
Multicut Lower Bound Certificates Multicut M gives Cut Matrix B : – n x |M| matrix, each column is an indicator vector for one node in the cut If there is a matrix A s.t. rank (AB) ≥ p for all cut matrices B, then this shows lower bound p – Proof: |M| = rank (B) ≥ rank (AB) Definition A coding matrix A is p -certifiable for multicut instance G if rank (AB) ≥ p for any cut matrix B of G Definition A coding matrix A is p -certifiable for multicut instance G if rank (AB) ≥ p for any cut matrix B of G
Multicut instance G 1 : Coding matrix A 1, rate p 1, p 1 – certifiable Multicut instance G 2 : Coding matrix A 2, rate p 2, p 2 – certifiable Then is a r – certifiable coding matrix with rate r for G 1 × G 2, r := n 1 p 2 + n 2 p 1 - p 1 p 2. Result* A 1 × I n 2, I n 1 × A 2 Definition A coding matrix A is p -certifiable for multicut instance G if rank (AB) ≥ p for any cut matrix B of G and for each v in V there is a clique K(v) with A1 K(v) = 0. Definition A coding matrix A is p -certifiable for multicut instance G if rank (AB) ≥ p for any cut matrix B of G and for each v in V there is a clique K(v) with A1 K(v) = 0.
Multicut instance G 1 : Coding matrix A 1, rate p 1, p 1 – certifiable Multicut instance G 2 : Coding matrix A 2, rate p 2, p 2 – certifiable Then is a r – certifiable coding matrix with rate r for G 1 × G 2, r := n 1 p 2 + n 2 p 1 - p 1 p 2. Result* A 1 × I n 2, I n 1 × A 2 But how do we get the original certifiable code??
Certifiable Coding Matrices Easy Observation: – If A is the “coding” matrix for a solution that routes messages along p node disjoint paths (each row is an indicator vector for one path) – Then A is p - certifiable. 11000000 00011010 00100101 000 010 000 000 000 100 000 001 Disjoint Paths: at most one non-zero in each column of AB Multicut: at least one non-zero in each row of AB
Saks et al Construction [Saks, Samorodnitsky,Zosin ’04] Ω(k) flow-cut gap Seed Network N : – Path of length n : x 1, …, x n Final network: k -fold strong product of N – n k nodes – k s-t set pairs – max flow = n k-1 S t S1S1 t1t1 S2S2 t2t2 Corollary The minimum multicut and the network coding rate in the Saks et al. construction is n k - (n-1) k. ( Saks et al. showed multicut is at least k(n-1) k-1 ) Corollary The minimum multicut and the network coding rate in the Saks et al. construction is n k - (n-1) k. ( Saks et al. showed multicut is at least k(n-1) k-1 )
Result Theorem* Suppose multicut instances G 1 and G 2 have linear network coding solutions instances A 1 and A 2 with rates p 1 and p 2 respectively, and are p 1 and p 2 certifiable respectively. Let r := |V 1 |p 2 + |V 2 |p 1 - p 1 p 2 Then multicut instance G 1 × G 2 has a linear network coding solution of rate r given by a matrix A that is r – certifiable. Theorem* Suppose multicut instances G 1 and G 2 have linear network coding solutions instances A 1 and A 2 with rates p 1 and p 2 respectively, and are p 1 and p 2 certifiable respectively. Let r := |V 1 |p 2 + |V 2 |p 1 - p 1 p 2 Then multicut instance G 1 × G 2 has a linear network coding solution of rate r given by a matrix A that is r – certifiable. Proof: A 1 × I p, I n × A 2
Graph Constructions Strong Graph Product G 1 ✕ G 2 – Vertex Set: V 1 ✕ V 2 – Edge Set: (u,v) ~ (u’,v’) if (u = u’ OR u ~ u’) AND (v = v’ OR v ~ v’) ✕
Network Products An undirected graph G=(V,E) k source sets S 1, …, S k paired with k sink sets T 1 …T k. An undirected graph G=(V,E) k source sets S 1, …, S k paired with k sink sets T 1 …T k. An undirected graph G ✕ G ’ 2k source sets S 1 ✕ V ’, …, S k ✕ V ’, V ✕ S ’ 1, …, V ✕ S ’ k paired with 2k sink sets T 1 ✕ V ’, …, T k ✕ V ’, V ✕ T ’ 1, …, V ✕ T ’ k. An undirected graph G ✕ G ’ 2k source sets S 1 ✕ V ’, …, S k ✕ V ’, V ✕ S ’ 1, …, V ✕ S ’ k paired with 2k sink sets T 1 ✕ V ’, …, T k ✕ V ’, V ✕ T ’ 1, …, V ✕ T ’ k. An undirected graph G ’ =(V ’,E ’ ) k source sets S ’ 1, …, S ’ k paired with k sink sets T ’ 1 …T ’ k. An undirected graph G ’ =(V ’,E ’ ) k source sets S ’ 1, …, S ’ k paired with k sink sets T ’ 1 …T ’ k.
Network Products S1S1 T1T1 S’1S’1 T’1T’1
Multicut Lower Bound Certificates Maximum Multicommodity Flow = n k-1 – Route between 1 s-t pair Network Coding? – Network Coding Rate ≥
Vertex Multicut Given: A graph G = (V,E) k source-sink pairs All nodes capacity 1 Find: A min-cost subset of V such that on removal all source-sink pairs are disconnected S S S t t t
Hypergrid(3,3)
A 1 × I p, A 2 × I n t S t s ( v 1, v 1 ’ ) ( v 5, v 1 ’ ) ( v 4, v 1 ’ ) ( v 3, v 1 ’ ) ( v 2, v 1 ’ ) ( v 1, v 2 ’ ) ( v 5, v 2 ’ ) ( v 4, v 2 ’ ) ( v 3, v 2 ’ ) ( v 2, v 2 ’ ) ( v 1, v 3 ’ ) ( v 5, v 3 ’ ) ( v 4, v 3 ’ ) ( v 3, v 3 ’ ) ( v 2, v 3 ’ ) ( v 1, v 4 ’ ) ( v 5, v 4 ’ ) ( v 4, v 4 ’ ) ( v 3, v 4 ’ ) ( v 2, v 4 ’ ) ( v 1, v 5 ’ ) ( v 5, v 5 ’ ) ( v 4, v 5 ’ ) ( v 3, v 5 ’ ) ( v 2, v 5 ’ )
Download ppt "Multicut Lower Bounds via Network Coding Anna Blasiak Cornell University."
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XP is just a number PerlMonks
### Rounding a number using (s)printf
by the_slycer (Chaplain)
on Jan 19, 2005 at 21:59 UTC Need Help??
the_slycer has asked for the wisdom of the Perl Monks concerning the following question:
Hi.. I've run into a bit of whackiness that I'm at a loss to explain, I've tried this on 3 different machines (all running some flavor of linux), and the behaviour repeats. Perhaps I'm incorrect in my use of printf, but I've found nothing indicating this..
I'm attempting to round a number to 2 decimal places:
```my (@numbers) = qw( 0.256 0.255 0.254 );
printf("%0.2f\n", \$_) foreach @numbers;
printf("%0.2f\n", 1.255);
I'd expect results like:
```0.26
0.26
0.25
1.26
But what I'm seeing are results like:
```0.26
0.26
0.25
1.25
Notice the 1.25 vs 1.26 that I'd expect..
As I said, I'm confused.. feel free to smite me down if I've pulled some kind of noob mistake.
Replies are listed 'Best First'.
Re: Rounding a number using (s)printf
by dave_the_m (Monsignor) on Jan 19, 2005 at 22:26 UTC
0.255 and 1.255 are not exactly representable in base-2 floating point:
```\$ perl -we 'printf "%.20f\n", 0.255'
0.25500000000000000444
\$ perl -we 'printf "%.20f\n", 1.255'
1.25499999999999989342
when the internal represenations of these two numbers are then rounded to 2 decimal places, one rounds up, the other down.
Dave.
Re: Rounding a number using (s)printf
by borisz (Canon) on Jan 19, 2005 at 22:36 UTC
Floating point numbers are not exactly represented.
```perl -e 'printf "%.18f", 1.255'
__OUTPUT__
1.254999999999999893
so 1.25 is the correct result.
Boris
Re: Rounding a number using (s)printf
by data64 (Chaplain) on Jan 20, 2005 at 03:56 UTC
Re: Rounding a number using (s)printf
by hsinclai (Deacon) on Jan 19, 2005 at 22:31 UTC
Does leaving the 0 preceding the decimal point out of the formatting directive change anything?
```printf("%.2f\n", 1.255);
Re: Rounding a number using (s)printf
by duff (Parson) on Jan 20, 2005 at 00:38 UTC
It don't think it has anything to do with how the numbers are represented in floating point. I think what you're seeing is a result of sprintf() following IEEE rounding semantics. It's something like even least significant digits round up at the half-way mark and odd round down at the half-way mark.
Search google for IEEE rounding semantics and I'm sure you'll find a better reference than I.
Update: doh! There's even a mention in perldoc -q round
You mean bankers rounding? That's not what's happening here. As you said yourself, the least significant digits play a role here. But the difference between 0.255 and 1.255 lies is the most significant digit.
Did you try it? It has everything to do with how numbers are represented in floating point. There are numbers that can be exactly represented in floating point where the rounding is ambiguous, e.g. printf "%.1f", \$_ for .25, .75, and a round-to-even-last-digit rule could apply, but in fact perl relies on C's printf(), and the SUSv3 standard says only "The low-order digit shall be rounded in an implementation-defined manner." Depending on how you interpret "rounded" in that sentence, you may not even be able to count on it rounding .24 to ".2".
In point of fact, I don't know of any system that fails to round at all, but do know of some that do not consistently either apply or fail to apply a round-to-even-last-digit rule.
Re: Rounding a number using (s)printf
by nite_man (Deacon) on Jan 20, 2005 at 08:50 UTC
This is a mathematical rule of rounding of numbers: if digit before last visible digit is greater than 5 last digit is increased by 1, otherwise the digit saves its value.
So, 0.256 ~ 0.26, 0.255 ~ 0.25, 0.254 ~ 0.25. This is correct results.
---
Michael Stepanov aka nite_man
It's only my opinion and it doesn't have pretensions of absoluteness!
First of all, you're wrong. The OP expected the results for 0.256, 0.255 and 0.254 to be 0.26, 0.26 and 0.25. And lo and behold, they were 0.26, 0.26 and 0.25. Because everyone learns in primary school that if the part we truncate starts with 5 or greater (and not "greater than 5"), we up the last digit.
But the OPs problem wasn't with that. The OPs problem was the inconsistency of 0.255 being rounded up, and 1.255 being rounded down.
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# Help needed in electric fieldWatch
#1
Can someone explain Question 5 to me please.
Posted from TSR Mobile
0
3 years ago
#2
(Original post by 123321123)
Can someone explain Question 5 to me please.
Posted from TSR Mobile
It's asking you to consider the electric forces between the positively charged rod lowered into the can, and the charges on the inner surface of the can.
Several things to remember:
1) Electrons (-ve charges) are free to move between atoms but protons are fixed and non-movable.
2) The can is standing on a conductive surface directly in contact with the earth.
3) Consider the earth as an infinite source or sink of electrons.
4) Like charges repel, unlike charges attract.
Then consider what happens to electrons on the surface of the can as the positively charged (more protons than electrons) rod is lowered into the can. Do the electrons on the can stay in position or do they move? If they move and their parent protons stay in the same position, what happens to the net charge distribution on the surface of the can?
The electric field is a diagram of the distribution of the electric force lines acting between the rod and the can.
Remember that the separation distance between the force lines represents the strength of the electric field at that position. The wider apart, the less the field strength. Most importantly, remember that the field strength (force acting at any given point) is also a function of distance between the charges.
0
#3
(Original post by uberteknik)
It's asking you to consider the electric forces between the positively charged rod lowered into the can, and the charges on the inner surface of the can.
Several things to remember:
1) Electrons (-ve charges) are free to move between atoms but protons are fixed and non-movable.
2) The can is standing on a conductive surface directly in contact with the earth.
3) Consider the earth as an infinite source or sink of electrons.
4) Like charges repel, unlike charges attract.
Then consider what happens to electrons on the surface of the can as the positively charged (more protons than electrons) rod is lowered into the can. Do the electrons on the can stay in position or do they move? If they move and their parent protons stay in the same position, what happens to the net charge distribution on the surface of the can?
The electric field is a diagram of the distribution of the electric force lines acting between the rod and the can.
Remember that the separation distance between the force lines represents the strength of the electric field at that position. The wider apart, the less the field strength. Most importantly, remember that the field strength (force acting at any given point) is also a function of distance between the charges.
thanks ！ that 's really helpful
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# Solve the systems algebraically:X+4Y+3Z=10 4X+2Y-2Z=-2 3X-Y+Z=11
sciencesolve | Teacher | (Level 3) Educator Emeritus
Posted on
You need to decide what variable you wish to eliminate first. Selecting the variable z the one to be eliminated first, you need to consider the first and the second equations such that:
`{(x+4y+3z = 10),(4x+2y-2z = -2):}`
Multiplying the first equation by 2 and the second equation by 3 yields:
`{(2x+8y+6z = 20),(12x+6y-6z = -6):}`
`14x + 14y = 14 => x + y = 1`
Considering the second and the third equations yields:
`{(4x + 2y - 2z = -2),(3x - y + z = 11) :}`
Multiplying the third equation by 2 yields:
`{(4x + 2y - 2z = -2),(6x - 2y + 2z = 22):}`
`10x = 20 => x = 2`
Substsituting 2 for x in equation `x + y = 1` yields:
`2+ y = 1 => y = -1`
You need to substitute 2 for x and -1 for y in `x+4y+3z = 10` such that:
`2-4+3z = 10 => 3z = 10+2 => 3z = 12 => z = 4`
Hence, evaluating the solutions to the given system of equations yields `x = 2 , y= -1 , z = 4` .
najm1947 | Elementary School Teacher | (Level 1) Valedictorian
Posted on
X+4Y+3Z=10 ...............(1)
4X+2Y-2Z=-2 ..............(2)
3X-Y+Z=11 .................(3)
multiplying equation (3) with 3 we get:
9X-3Y+3Z = 33, now subtracting eaquation (1) from this we get:
9X-X -3Y-4Y +3Z-3Z = 33-10
=> 8X -7Y = 23 ............(4)
again multiplying equation (3) with 2 we get:
6X-2Y+2Z = 22, adding equation (2) to it we get:
6X+4X -2Y+2Y +2Z-2Z = 22-2
=> 10X = 20
=> X = 2
substituting this value of X in (4) we get:
8*2 -7Y = 23
=> -7Y = 23-16 = 7
=> Y = -1
now substituting the value of X and Y in equation (3) we get:
3*2 -(-1) +Z = 11
=> Z = 11-7 = 4
The solution of the three equations is:
X = 2
Y = -1
Z = 4
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PDA
View Full Version : Siewert - Loss Ratio Method
cgott42
08-16-2006, 07:06 PM
This is a very basic question as I'm not familiar with the terms or context here.
In the Loss Ratio method (Table 1 pg 222) showing the calculation for a single policy - can someone explain the columns?
Specifically
column (3) ELR - didn't Seiwert say the loss ratio used is that which was used to price each policy - i.e. the excess LR for the excess layer priced, if so what is the Excess Ratio used for?
If it's the full coverage LR that he mentions that we can get from industry experience, thus apply the excess ratio to get the portion ABOVE the deductible why is it named "Deductible" Loss Charge?
Lastly, what is the Aggregate Loss Limit Charge? We apply it to the full coverage (ground up) losses - the excess losses (?) col(6) - i.e. this is losses BELOW the deductible.
Shouldn't it be applied to losses above the deductible (i.e. subject to the agg. loss limit)?
Now, assuming that it is applied to the excess losses, does this mean it (e.g. 2%) is saying that we expect 2% of excess losses to be capped by the agg. loss limit. So does this mean that the amount in column (8) is an amount that we want to EXCLUDE from losses (then why is it called a "charge").
Sorry for so many questions, mostly it seems there's a prerequisite knowledge base necessary in order to understand what's going on in this paper.
drichie
08-17-2006, 09:09 AM
I don't have the article in front of me, so I can't answer all the questions. However, I think you are confused about the aggregate in this case. The aggregate Siewert is referring to is an aggregate payed by the insured. In other words, they will pay a \$5,000 deductible per claim, with a \$100,000 aggregate. This means they will pay up to \$5000 for each claim they have until they have payed \$100,000 and then the deductible no longer applies for the remainder of the policy term. It is not an aggregate payed by the insurance company.
frank_exams
08-17-2006, 11:47 AM
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# How far is Sukagawa from Anchorage, AK?
The distance between Anchorage (Ted Stevens Anchorage International Airport) and Sukagawa (Fukushima Airport) is 3351 miles / 5393 kilometers / 2912 nautical miles.
3351
Miles
5393
Kilometers
2912
Nautical miles
## Distance from Anchorage to Sukagawa
There are several ways to calculate the distance from Anchorage to Sukagawa. Here are two standard methods:
Vincenty's formula (applied above)
• 3351.326 miles
• 5393.437 kilometers
• 2912.223 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth's surface using an ellipsoidal model of the planet.
Haversine formula
• 3343.703 miles
• 5381.168 kilometers
• 2905.598 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## How long does it take to fly from Anchorage to Sukagawa?
The estimated flight time from Ted Stevens Anchorage International Airport to Fukushima Airport is 6 hours and 50 minutes.
## Flight carbon footprint between Ted Stevens Anchorage International Airport (ANC) and Fukushima Airport (FKS)
On average, flying from Anchorage to Sukagawa generates about 377 kg of CO2 per passenger, and 377 kilograms equals 830 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path from Anchorage to Sukagawa
See the map of the shortest flight path between Ted Stevens Anchorage International Airport (ANC) and Fukushima Airport (FKS).
## Airport information
Origin Ted Stevens Anchorage International Airport
City: Anchorage, AK
Country: United States
IATA Code: ANC
ICAO Code: PANC
Coordinates: 61°10′27″N, 149°59′45″W
Destination Fukushima Airport
City: Sukagawa
Country: Japan
IATA Code: FKS
ICAO Code: RJSF
Coordinates: 37°13′38″N, 140°25′51″E
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# Blogs Hub
### by Sumit Chourasia | Jun 06, 2020 | Category :coding | Tags : arrayleetcodec-sharpआसानडेटा-संरचनाalgorithm
#### Create Target Array in the Given Order - Array - Easy - LeetCode - मिनी टीवी
Given two arrays of integers nums and index. Your task is to create target array under the following rules:
Initially, target array is empty.
From left to right read nums[i] and index[i], insert at index index[i] the value nums[i] in target array.
Repeat the previous step until there are no elements to read in nums and index.
Return the target array.
It is guaranteed that the insertion operations will be valid.
Example 1:
Input: nums = [0,1,2,3,4], index = [0,1,2,2,1]
Output: [0,4,1,3,2]
Explanation:
nums index target
0 0 [0]
1 1 [0,1]
2 2 [0,1,2]
3 2 [0,1,3,2]
4 1 [0,4,1,3,2]
Example 2:
Input: nums = [1,2,3,4,0], index = [0,1,2,3,0]
Output: [0,1,2,3,4]
Explanation:
nums index target
1 0 [1]
2 1 [1,2]
3 2 [1,2,3]
4 3 [1,2,3,4]
0 0 [0,1,2,3,4]
Example 3:
Input: nums = [1], index = [0]
Output: [1]
Constraints:
1 <= nums.length, index.length <= 100
nums.length == index.length
0 <= nums[i] <= 100
0 <= index[i] <= i
Solution:
``````using System;
using System.Collections.Generic;
using System.Text;
{
public class CreateTargetArraySoln
{
public int[] CreateTargetArray(int[] nums, int[] index)
{
var target = new int[nums.Length];
var set = new HashSet<int>();
for(int i = 0; i < nums.Length; i++)
{
if (set.Contains(index[i]))
{
for (int j = nums.Length-1; j > index[i]; j--)
{
target[j] = target[j - 1];
}
}
target[index[i]] = nums[i];
}
return target;
}
}
}
``````
Time Complexity: O(n^2)
Space Complexity: O(n)
Unit Tests:
``````using LeetCode.AskGif.Easy.Array;
using Microsoft.VisualStudio.TestTools.UnitTesting;
using System;
using System.Collections.Generic;
using System.Text;
namespace CodingUnitTest.Easy.Array
{
[TestClass]
public class CreateTargetArraySolnTests
{
[TestMethod]
public void CreateTargetArraySoln_First()
{
var nums = new int[] { 0, 1, 2, 3, 4 };
var index = new int[] { 0, 1, 2, 2, 1 };
var output = new int[] { 0, 4, 1, 3, 2 };
var res = new CreateTargetArraySoln().CreateTargetArray(nums, index);
AreEqual(res, output);
}
[TestMethod]
public void CreateTargetArraySoln_Second()
{
var nums = new int[] { 1, 2, 3, 4, 0 };
var index = new int[] { 0, 1, 2, 3, 0 };
var output = new int[] { 0, 1, 2, 3, 4 };
var res = new CreateTargetArraySoln().CreateTargetArray(nums, index);
AreEqual(res, output);
}
[TestMethod]
public void CreateTargetArraySoln_Third()
{
var nums = new int[] { 1 };
var index = new int[] { 0 };
var output = new int[] { 1 };
var res = new CreateTargetArraySoln().CreateTargetArray(nums, index);
AreEqual(res, output);
}
[TestMethod]
public void CreateTargetArraySoln_Fourth()
{
var nums = new int[] { 4, 2, 4, 3, 2 };
var index = new int[] { 0, 0, 1, 3, 1 };
var output = new int[] { 2, 2, 4, 4, 3 };
var res = new CreateTargetArraySoln().CreateTargetArray(nums, index);
AreEqual(res, output);
}
private void AreEqual(int[] res, int[] output)
{
Assert.AreEqual(res.Length, output.Length);
for (int i = 0; i < res.Length; i++)
{
Assert.AreEqual(res[i], output[i]);
}
}
}
}
``````
Contributed By: Sumit Chourasia
Contributed By: Sumit Chourasia
Contributed By: Sumit Chourasia
Contributed By: Sumit Chourasia
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https://www.instructables.com/community/Battery-Powered-LED-Strip/
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157Views3Replies
Author Options:
Hi,
I have one 6V Alkaline Battery (or two 3V Coin cell battery) that i am using to light a fairy LED strip (see here). I'd like to know for how long the battery will last me for. Can you please direct me what parameters I need to collect for the battery (or the LEDs) and what equations i will be using to determine the operating time of the battery. Thank you for your support in advance,
Regards,
Tags:
Discussions
The forums are retiring in 2021 and are now closed for new topics and comments.
Assuming the batteries are in series(most common setup):
2 CR2032 batteries in series = 3V * 2 = 6V
Wikipedia says typical capacity is 225mAh. In series, capacity remains the same for 2.
They run for 10 hours.
So current consumption per hour of lights = 225mAh / 10h = 22.5mA
So if the new batteries can provide 6V, look up their current capacity. Let's say it is X mAh. So they will run for (X / 22.5) hours.
i was aware of this information but if i am to use different set of batteries wouldnt that be different?
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https://primeinvestor.in/calculators/sip-to-target-calculator/
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# SIP Target Calculator
## Know what this SIP Target calculator can do
Helps bucket total monthly savings into specific amounts for each goal
Calculates the monthly SIP needed to meet a particular financial goal
Best mutual funds to start your SIPs today
No credit card required
If you have a fixed sum of money as a financial goal to reach in a certain number of years, itโs necessary to know how much to save towards this target. Knowing a ballpark figure will ensure that you do not fall short of your target. This SIP target calculator takes inputs from you to calculate how much SIP you would need to invest to attain a financial goal.
## What is the Target Amount SIP Calculator?
The Target Amount SIP calculator will help you know how much to invest through an SIP every month, to attain your desired goal. For certain goals, such as education or retirement, you typically know the year by which you would need the money and an estimate of how much you may need. But knowing how much wealth to build is only the first step in your investment plan. The second step is to calculate the SIP amount. Therefore, a target SIP calculator will consider your goal amount and your timeframe, applies an assumed return and calculates the SIP amount needed to reach there.
## How does the SIP target calculator work?
To use the SIP target calculator, you just have to input four values:
• Target amount: The first value is the desired financial goal amount, or the target amount that you want to achieve from the SIP.
• Existing investments: The second value is the amount of investment you have already made towards your financial goal. Entering this value ensures that you do not allocate a higher amount than needed towards a financial goal. If you have no investments made towards your goal, you can enter zero (0).
• SIP duration: The third value is the number of years for which you are running the SIPs or the target goal date.
• Assumed returns: The returns you expect your investment to generate. In this value, keep in mind that putting in very high returns can be misleading as reality will be far from it. Keep your return expectations reasonable, which will make the result of the SIP target calculator more realistic and therefore more useful for you.
Once you input all the values, the SIP target calculator will help you know the amount you will have to invest every month to attain the desired goal.
## Benefits of Target SIP Calculator
For most of us, there are multiple financial goals that we may have โ down payment for a house, retirement, childrenโs education, fancy vacations, upgrading cars, redoing a home and so on. The investments you make, therefore, should be smartly allocated between these goals.
For this, itโs important to use a target amount SIP calculator. This will help you in the following ways:
• It helps you split your investments correctly towards these different financial goals.
• It can bring about direction to your investments, work out which goal is more important, which can be pushed back and so on. You can allocate the required amount first towards the shorter-term or more important goals, and then use the remaining savings towards the longer-term or less-important goals.
• It will help you build the right portfolio for each goal, since different goals and different SIP amounts can have different asset allocation, risk taken on, and types of funds.
Therefore, a target SIP calculator comes in handy when answering the key question of โhow much should I invest in mutual fundsโ and understanding how to plan your savings. When using SIP target calculator or any other calculator, remember that these are estimates. How much your investment grows will depend on what returns are generated in reality. Therefore, always track your investments with tools such as the Portfolio Review Pro.
## What is the formula for the Target SIP Calculator?
The SIP Target Calculator combines different formulae to arrive at the target SIP amount. This calculator considers two different aspects โ one, the target amount and two, the existing investments that are already made.
The target amount SIP calculator considers the compound interest to know how much the existing investment will grow. This is a simple formula of A = P * (1+r/100)^n: where, A is the final amount you wind up with, P is the amount invested, r is the estimated rate of return and n is the number of years for which you are invested. Apart from this, the target SIP calculator uses a derivative of the above formula to arrive at the monthly SIP amount to reach the target.
## How much should I invest in mutual funds
Saving and investing are the key to building wealth. However, investing blindly without knowing what youโre investing towards or answering the question of how much should I invest in mutual funds would result in directionless investment. Therefore, using calculators such as the target SIP calculator or SIP value calculator or step-up SIP calculator are all useful ways to answer how much should I invest in mutual funds.
These calculators will give you the SIP amount to invest in mutual funds for each of your goals, or will help you know how much wealth you can build if you already know how much I should invest in mutual funds.
Knowing these amounts will help you pick the right mutual funds to invest. For the best mutual funds, you can refer to our recommendation list in Prime Funds. If you want mutual fund portfolios to start investing in today, check Prime Portfolios.
However, always keep in mind that how much should I invest in SIPs is an estimate. Keep track of how your wealth is building, how far away you are from your desired target amount and adjust your SIP amount and funds accordingly.
FAQ
OR
## Become a PrimeInvestor!
Get stock & mutual fund recommendations
or
Have an account?
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Wed. Jun 23rd, 2021
# Logical and Aptitude Test Questions with Answers
## Logical and Aptitude Test Questions with Answers
Here is the collection of some really nice logical & aptitude test questions for interview or quiz preparation. Answers of the logical & aptitude questions are provided for the reference at the bottom of this article. Here are some aptitude questions and answers with explanation for interview, competitive examination and entrance test. Fully solved examples with detailed answer for each aptitude question. Learn & practice with these aptitude test questions and answers.
1. A snail is at the bottom of a 20 meters deep pit. Every day the snail climbs 5 meters upwards, but at night it slides 4 meters back downwards. How many days does it take before the snail reaches the top of the pit?
2. A light bulb is hanging in a room. Outside of the room there are three switches, of which only one is connected to the lamp. In the starting situation, all switches are off and the bulb is not lit. If it is allowed to check in the room only once to see if the bulb is lit or not (this is not visible from the outside), how can you determine with which of the three switches the light bulb can be switched on?
3. The difference between a two-digit number and the number obtained by interchanging the positions of its digits is 36. What is the difference between the two digits of that number?
4. If 5+3+2 = 151012, 9+2+4 = 183662, 8+6+3 = 482466, 5+4+5 = 202504 then what will be the answer of 7+2+5? Only 2% people are able to solve this aptitude question. Let’s see if you can do it.
5. At a party, everyone shook hands with everybody else. There were 66 handshakes. How many people were at the party?
6. Three people picked 65 apples altogether. At the first tree they each picked the same number of apples. At the second tree they each picked 3 times as many as they picked at the first tree. When they finished at the third tree, the group had 5 times as many apples as they had when they started at that tree. At the fourth tree the group picked just 5 apples. How many apples did each person pick at the first tree?
7. A cube of side 4cm is painted with 3 colors red, blue and green in such a way that opposite sides are painted in the same color. This cube is now cut into 64 cubes of equal size.1. How many have at least two sides painted in different colors.
2. How many cubes have only one side painted.
3. How many cubes have no side painted.
4. How many have exactly one side not painted.
8. A man wanted to enter an exclusive club but did not know the password that was required. He waited by the door and listened. A club member knocked on the door and the doorman said, “twelve.” The member replied, “six” and was let in. A second member came to the door and the doorman said, “six.” The member replied, “three” and was let in. The man thought he had heard enough and walked up to the door. The doorman said, “ten” and the man replied, “five”. But he was not let in. What should have he said?
9. A man decides to buy a nice horse. He pays \$60 for it, and he is very content with the strong animal. After a year, the value of the horse has increased to \$70 and he decides to sell the horse. But already a few days later he regrets his decision to sell the beautiful horse, and he buys it again. Unfortunately he has to pay \$80 to get it back, so he loses \$10. After another year of owning the horse, he finally decides to sell the horse for \$90. What is the overall profit the man makes?
10. In the middle of a round pool lies a beautiful water-lily. The water-lily doubles in size every day. After exactly 20 days the complete pool will be covered by the lily. After how many days will half of the pool be covered by the water-lily?
1. 16 days. Since the snail moves up 1 meter a day so it will reach 15 meters in 15 days. next day it will again climb 5 meters upward and reaches the top.
2. To find the correct switch (1, 2 or 3), turn switch 1 to ON and leave it like that for a few minutes. After that you turn switch 1 back to OFF, and turn switch 2 to ON. Now enter the room. If the light bulb is lit, then you know that switch 2 is connected to it. If the bulb is not lit, then it has to be switch 1 or 3. Now touching the light bulb will give you the answer. If the bulb is still hot, then switch 1 is connected to the bulb; if the bulb is cold, then it has to be switch 3.
3. The difference between the two digits of that number is 4. Explanation: Let the ten’s digit be x and unit’s digit be y. Then, (10x+y)-(10y+x) = 36 i.e. 9(x-y) = 36 and that results to x-y = 4.
4. 7 + 2 + 5 = 143542. Explanation: a + b + c leads to a 6-digit number.1. The first and second digits are the product of a and b.
2. The third and fourth digits are the product of a and c.
3. The fifth and sixth digits are b*(a+c) with the digits of the solution reversed.
5. With two people, there is one handshake. With three people, there are three handshakes. With four people, there are six handshakes. In general, with n+1 people, the number of handshakes is the sum of the first n consecutive numbers: 1+2+3+…+n. Since this sum is n(n+1)/2, we need to solve the equation n(n+1)/2 = 66. This is the quadratic equation n2+n-132 = 0. Solving for n, we obtain 11 as the answer and deduce that there were 12 people at the party.
6. One Apple
7. Here are the answers.1. Cubes that have at least two sides painted in different colours are 24 + 8 = 32.
2. Cubes that have only one side painted are 24.
3. Cubes that have no side painted = 8.
4. Cubes that have exactly one side not painted = 0.
8. The man had to reply the number of characters in the word the Doorman was asking. He should have replied “Three” instead of “Five”.
9. Consider the trade-story as if it describes two separate trades, where: In the first trade, the man buys something for \$60 and sells it again for \$70, so he makes a profit of \$10. In the second trade, the man buys something for \$80 and sells it again for \$90, so he makes again a profit of \$10.Conclusion: The man makes an overall profit of \$10 + \$10 = \$20.
You can also look at the problem as follows:
The total expenses are \$60 + \$80 = \$140 and the total earnings are \$70 + \$90 = \$160. The overall profit is therefore \$160 – \$140 = \$20.
10. 19 days. Since the water-lily doubles its size every day and the complete pool is covered after 20 days, half of the pool will be covered one day before that, so 19 days.
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https://stage.geogebra.org/m/cr43we3k
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GeoGebra Classroom
# Tangent Line Versus Secant Line
Instructions:
• Use the drop down menu to select a function.
• The blue line is a secant to the function through points A and B.
• The green line is a tangent to the function through point A.
• Drag points A and B to see how the slopes of the tangent and secant lines change.
• Drag point B closer to point A and see what happens to the slopes of the secant and tangent lines as B approaches A.
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https://www.geeksforgeeks.org/limiting-friction/?ref=leftbar-rightbar
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Limiting Friction
Limiting friction is defined as the friction which is generated between the two static surfaces that come in contact with each other. The maximum friction that can be generated between two static surfaces in contact with each other. Once a force applied to the two surfaces exceeds the limiting friction, the motion will occur.
For two dry surfaces, the limiting friction is a product of the normal reaction force and the coefficient of limiting friction. In other words, the amount of friction that can be applied between two surfaces is limited and if the forces acting on the body are made sufficiently great, the motion will occur.
Hence, we can define limiting friction as the maximum value of static friction that comes into play when the body is just at the point of sliding over the surface of another body.
Laws of Limiting Friction
Following are the laws of limiting friction:
(a) The direction limiting frictional force and the direction of motion is always the opposite.
(b) The limiting friction always acts tangential to the two surfaces that are in contact.
(c) The magnitude of the limiting friction and the normal force between the two surfaces are directly proportional.
(d) The limiting friction is dependent on the material, the nature of the surfaces that are in contact, and the smoothness.
(e) The limiting friction is independent of the shape and area of the surfaces.
Examples of Limiting Friction
Following are some of the examples of Limiting Friction:
1. When we walk on the road.
2. When a car moves on a cliff.
3. Friction force applied by upstream water to a swimmer.
4. Friction force on a pen by a page or notebook.
5. Moving an object on the ground.
6. A car parked on a hill will not slide down. This is because the value of limiting friction is high enough to ensure that the car is in a state of rest. The limiting force of friction is directly proportional to the mass of the object.
7. Pushing a heavy truck seems very difficult due to the high value of static friction
8. When we push any object we encounter static friction for some time until the value of limiting friction is reached.
Formula and its Derivation
Hence, we can define limiting friction as the maximum value of static friction that comes into play when the body is just at the point of sliding over the surface of another body. Limiting friction is the product of normal force and the coefficient of limiting friction.
Following is the mathematical representation:
Friction force applied to the body is always proportional to the normal force acting on the body.
To remove the proportion sign we introduce a constant .
Now,
Where,
F is the limiting friction
is the coefficient of limiting friction
N is the normal force
Example: Find the normal force applied on a body having a coefficient of friction of 5 and limiting friction is 50 N.
Solution:
As we know
Where,
F is the limiting friction
is the coefficient of limiting friction
N is the normal force
N = 10 N
Sample Questions
Question 1: What is the other name for limiting friction?
It is the maximum value of friction. Thus limiting friction is also called as maximum value of friction.
Question 2: What is the difference between static friction and limiting friction?
Static friction is a self-adjusting force because it comes into play when the body is lying over the surface of another body without any motion. When that body overcomes the force of static friction, the maximum value of static friction is reached, which is known as limiting friction.
Question 3: What is the relationship between kinetic and limiting friction?
When the applied force, F, is increased further (beyond the limiting frictional force), the body begins to move, then the force opposing the motion is called the kinetic or sliding friction. The kinetic friction is less than the limiting friction.
Question 4: Why is kinetic friction less than limiting friction?
Initially it requires more force to break the interlock between the irregularities of two surfaces. But once the block is in motion, it takes time to interlock between the irregularities of two surfaces. Also, the body gains the inertia of motion. So, the kinetic friction is less than the limiting friction.
Question 5: Is static or limiting friction greater?
In short, static friction is the same as the force being applied because the body isn’t moving.Static friction is acting due to the body tends to move when a force is acting on it.Limiting friction is the friction acting on a body when it is about to start moving.Commonly, limiting friction is the highest.
Question 6: Is limiting friction a variable?
When that body overcomes the force of static friction, the maximum value of static friction is reached -which is known as limiting friction. After the limiting friction, the frictional force is not going to increase further. At this stage, the object moves overcoming the frictional force which is at a constant value.
Numerical Problems
Question 1. Find the minimum force required to move a 5 kg box if the coefficient of friction is 0.2.
Solution:
Minimum force is nothing but the value of friction force due to frictional surface because if we apply force more than friction force then box can be move.
N = m×g
N = 5kg × 10m/s2 = 10 N
F = μ × N
F = 0.2 × 10 = 2 N
So, we require minimum 2N to move the box.
Question 2. If the minimum force required to move a table is 100 N then find the weight of the table when μ = 5.
Solution:
Minimum force required to move a table (F) =\mu \times N
Given, F = 100 N and \mu = 5
Question 3. Find coefficient of limiting friction if 100 kg of Machine requires 10000 N force to change the position.()
Solution:
F = 10000 N
M = 1000 kg
N = m × g N = 100 kg × 10m/s2 N = 1000 N F = μ × N 10000 N = μ × 1000 N μ = 10
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# Cris’ Image Analysis Blog
theory, methods, algorithms, applications
## Solving mazes using image analysis
I found an interesting submission on the File Exchange. It uses a very simple set of mathematical morphology filters to find the solution to simple mazes. I have several other solutions, so I decided to write an entry about the subject.
### Original solution
This section is paraphrased from File Exchange “Image Analyst“‘s solution. His code uses MATLAB’s image processing toolbox, here we want to use DIPimage, of course.
We start by reading in the image with the maze:
``````inputmaze = readim('maze_solution_01.png')
``````
Next we convert it to a grey-value image and threshold it to obtain a binary image. The maze walls are `true` and shown in red:
``````maze = colorspace(inputmaze,'grey');
maze = ~threshold(maze)
``````
When we label the binary image, we can clearly see two discrete, separate walls:
``````maze = label(maze);
dipshow(maze*2,'labels')
``````
It is fairly easy to see the path from start to finish: it’s the path in between the two colors. To draw this path over the image we can follow this simple sequence of commands, it’s all mathematical morphology. We ignore everything except the first of the two walls:
``````path = maze==1
``````
Then we dilate the walls by a few pixels, and fill all the holes:
``````pathwidth = 5;
path = bdilation(path,pathwidth);
path = fillholes(path)
``````
Then we erode by the same amount of pixels and take the difference:
``````path = path - berosion(path,pathwidth);
``````
The function `overlay` makes a nice display:
``````overlay(inputmaze,path)
``````
However, when the maze is more complex, things start to go wrong. In this maze I’ve added a photograph, and cut two holes to create more paths:
``````inputmaze = readim('maze_solution_07.png')
``````
The result after labeling now shows more than two colors:
``````maze = colorspace(inputmaze,'grey');
maze = ~threshold(maze);
maze = label(maze);
dipshow(maze,'labels')
``````
This is not a big deal, but when we follow the first wall, we’ll find only one of the possible paths, and it’s not necessarily the shortest path. Also, we are lucky that object with ID=1 is a good object to follow. If you pick the wrong object ID here, you’ll end up with a path that doesn’t connect the input and output!
``````path = maze==1;
pathwidth = 5;
path = bdilation(path,pathwidth);
path = fillholes(path);
path = path - berosion(path,pathwidth);
overlay(inputmaze,path)
``````
### Alternate solution
A more robust solution involves using the skeleton. This is a morphological operation that, in rough terms, erodes and object under certain constraints, until the whole object is a single pixel thick. The constraints are that a pixel is never removed if removing it would change the geometry of the object. That is, it will never split an object in two. Let’s go back to the binarised maze image, but now we want the path to be the object, rather than the walls:
``````inputmaze = readim('maze_solution_07.png');
maze = colorspace(inputmaze,'grey');
maze = threshold(maze);
``````
Because of the way that the skeleton is implemented in DIPimage, we need to add a 2-pixel border around the maze. The skeleton ignores the pixels in this border, so if we don’t add it, part of our maze wouldn’t be taken into account:
``````maze = extend(maze,imsize(maze)+4);
``````
The skeleton now produces a 1-pixel-thick version of the maze:
``````path = bskeleton(maze)
``````
All the dead ends can be easily found, by looking at pixels that have only one neighbor:
``````deadends = getendpixel(path);
joinchannels('rgb',bdilation(deadends,1,2),path) % fancy display command
``````
One of the options of the `bskeleton` function is to remove these dead-end pixels iteratively. This leaves the skeleton as a set of dots and closed loops. By setting the area outside the image to the value 1, we make sure that the bits of the skeleton connected to the image border are kept:
``````path = bskeleton(maze,1,'looseendsaway')
``````
Notice that there are several loops and dots not connected to the image border. These are caused by the image I added to the maze. We can remove those by keeping only those objects that are connected to the image border:
``````path = path-brmedgeobjs(path,2)
``````
Finally we’ll remove the temporary image border we added to compute the skeleton, and make the path a little thicker for display:
``````path = cut(path,imsize(inputmaze));
path = bdilation(path,1);
dipshow(overlay(inputmaze,path,[255,0,0]))
``````
This returned all possible paths from start to finish. But how do we select only the optimal path?
### A better solution
The optimal path selection in graph theory is found using Dijkstra’s algorithm. We could easily implement this algorithm by extracting the paths found by the skeleton as a graph. But we want to do this here using image analysis algorithms.
Let’s again start by reading in the maze image and obtaining a binary representation:
``````inputmaze = readim('maze_solution_07.png');
maze = colorspace(inputmaze,'grey');
maze = ~threshold(maze);
``````
Our first task is to find where the start and stop points are for our path. We’re simply looking for black pixels connected to the image edge:
``````markers = bpropagation(newim(maze,'bin'),~maze,3,1,1);
markers = label(markers);
``````
Let’s make sure we have exactly two exits, and let’s get their coordinates:
``````if max(markers)~=2
error('The maze doesn''t have two exits!');
end
m = measure(markers,[],'center');
start = round(m(1).center)
stop = round(m(2).center)
``````
``````start =
401
37
stop =
365
401
``````
The most important step to find the shortest path is the distance transform. This function simply computes the distance from every object pixel in the image to the background. But instead of using Euclidean distances, which take the straight-line distance between the pixel and the background, we use a constrained distance. This constraint forces the distances computed to follow the possible paths through the maze. One of the simplest ways of constraining the distance transform is to weigh the distance with the image’s grey value. That is, the distance between two points is computed as the integral of the grey values along the path between the two points. This is called the grey-weighted distance transform. Incidentally, the grey-weighted distance transform is computed with something very similar to Dijkstra’s algorithm.
By giving the maze walls very high grey values, we can make sure that going around the maze through some long-winded path produces a shorter (weighted) distance than taking a “short-cut” through a wall. We give the maze floors a grey value of 1, so that the distance travelled through the maze is equal to the path length, and the maze walls a value of 106, larger than any path you could possibly draw in such a small image. Furthermore, as with the skeleton, we need to add a little border around the image:
``````region = extend(markers~=2,imsize(maze)+4); % We compute the distance to the stop point
weights = extend(1+(maze*1e6),imsize(maze)+4);
distance = gdt(region,weights,3);
distance = cut(distance,imsize(maze));
dipshow(distance,[0,2000])
``````
As you can see, the further from the stop point, the higher the grey value of the output is. There is a region in the top-left that cannot be reached at all from the stop point. The distance from start to finish is:
``````D = double(distance(start(1),start(2)))
``````
``````D =
1.0716e+03
``````
To find the path all we now need to do is follow a steepest descent path from the start all the way to the finish. We start at the start point, and look at all its neighbours for the one that has the lowest value. That then becomes the current point and we repeat until we are at the bottom (where `distance==0`).
``````current = start;
path = current;
while distance(current(1),current(2))>0
n = distance(current(1)+[-1,0,1],current(2)+[-1,0,1]);
[n,step] = min(n);
step = step-1;
current = current + step';
path = [path,current];
end
``````
We can now plot the path on the input image:
``````dipshow(inputmaze)
hold on
plot(path(1,:),path(2,:),'r','linewidth',3)
``````
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https://ceruleanstud1os.com/get-online-paid-surveys-bring-them-for-money-in-your-lifes-savings/
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Categories: Miscellaneous
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https://tkp.readthedocs.io/en/latest/userref/structure/stages/transient.html
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# Variability and new-source detection stage¶
## Variability index calculation¶
After all images for a given timestep have been processed and the resulting source measurements have been assigned to runningcatalog entries (effectively lightcurves), variability indices are calculated for the most recent timestep, and stored as part of the association recorded in the assocxtrsource table.
Note that a single runningcatalog source may contain entries from multiple independent frequency bands. The variability indices are calculated independently for each frequency band, hence the $$\nu$$ suffix in the calculations below denotes an index over the different bands.
For a comprehensive discussion of the transient and variability detection algorithms currently being employed, see Scheers (2011) chapter 3. Here, we provide a brief outline.
We define two metrics for identifying variability in a lightcurve. The flux coefficient of variation, which we denote $$V_\nu$$, is defined as
$V_{\nu} \equiv \frac{s_{\nu}}{\overline{I_{\nu}}} = \frac{1}{\overline{I_{\nu}}} \sqrt{\frac{N}{N-1}(\overline{I^{2}_\nu} - \overline{I_{\nu}}^2)}$
where $$\overline{I_{\nu}}$$ is the mean flux of all measurements in the lightcurve at frequency $$\nu$$, $$s_{\nu}$$ is the standard deviation of those flux measurements and $$N$$ is the number of measurements.
The second metric is $$\eta_{\nu}$$, which is defined based on reduced $$\chi^2$$ statistics as
$\eta_{\nu} \equiv \chi^{2}_{N-1} = \frac{1}{N-1} \sum_{i=1}^{N} \frac{(I_{\nu,i} - \overline{I_{\nu}}^*)^2}{\sigma_{I_{\nu,i}}^2}$
Where $$\overline{I_{\nu}}^*$$ is the average of the flux measurements weighed by their uncertainties. $$\eta_{\nu}$$ is the $$\chi^{2}$$ probability distribution. The probability that the source is “flat” (i.e. has no significant variability) is then the integral of the distribution from the measured value of $$\eta_{\nu}$$ to $$\infty$$; the probability that it isn’t flat is thus 1 minus this quantity.
See also the appendices on the database schema for details of how these are iteratively updated.
## ‘New source’ detection¶
We also attempt to identify any newly extracted sources which we suspect are intrinsically variable in nature (i.e. they are getting brighter, as opposed to our observations getting deeper or even simply looking at a previously unobserved patch of sky). The algorithm for evaluating new sources is encoded by tkp.db.associations._determine_newsource_previous_limits(). Sources we deem to be intrinsically ‘new’ are then recorded in the newsource table.
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https://www.adventuresinmachinelearning.com/mastering-graph-traversal-with-breadth-first-search-algorithm-in-python/
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# Mastering Graph Traversal with Breadth-First Search Algorithm in Python
## Breadth-First Search Algorithm: An Introduction
As computer science students, we all come across graph theory in our curriculum, where we study the properties of graphs and their various representations. Graphs are a set of objects interconnected by edges or arcs.
They are used to model real-world scenarios, such as transportation networks, social networks, or genealogy trees.
Traversal techniques are essential when exploring a graph.
Traversal refers to visiting all the vertices of a graph in a given order. In this article, we will be discussing the Breadth-First Search (BFS) algorithm, which is a popular traversal technique.
## 1. Definition of BFS
Breadth-first search is a graph traversal algorithm that visits all the vertices of a graph in breadth-first order. It starts by visiting the first node of the graph, then the nodes at the next level, and so on until all nodes are visited.
In other words, BFS visits all the nodes at a given level before moving on to the nodes at the next level.
## 2. BFS Algorithm for Graph in Python
In Python, we can represent a graph using an adjacency list. An adjacency list is a dictionary where each key represents a vertex of the graph, and the corresponding value is a list of its adjacent vertices.
### 2.1. The following is an implementation of the BFS algorithm for a graph in Python:
1. Create a queue and enqueue the first vertex
2. Mark the first vertex as visited
3. While the queue is not empty:
1. Dequeue the first vertex
2. For each adjacent vertex of the dequeued vertex:
1. If the adjacent vertex is not visited, mark it as visited and enqueue it into the queue
The above algorithm uses a queue data structure to keep track of the nodes that need to be visited.
We also use a visited set to keep track of the nodes that have been visited.
### 2.2. Example of BFS Traversal
Let’s consider the following graph:
`````` A---B---C
| | |
D---E---F
| |
G---H
``````
Suppose we want to traverse the above graph using BFS. We start by visiting the first vertex, which is A.
Since A is the starting vertex, we mark it as visited, and we add it to the queue. Our queue now contains only A.
We dequeue A and add its adjacent vertices, B and D, to the queue. The queue now contains B and D.
We mark both of them as visited, and their corresponding vertices are added to the end of the queue. The queue now contains C and E.
We dequeue B and add its adjacent vertices, A, C, D, and E. We ignore A and D since they have already been visited.
We add C and E to the end of the queue, making the queue [D, C, E]. We continue dequeuing vertices and adding their adjacent vertices to the queue until we have traversed all the vertices.
The BFS traversal of the graph produces the following sequence of vertices: A, B, D, C, E, F, G, H.
## 3. Implementing BFS Algorithm in Python
In this section, we will take a closer look at the implementation of the BFS algorithm in Python. We will provide a step-by-step guide on how to create a graph, represent it using an adjacency list, and traverse it using the BFS algorithm.
### 3.1. Code Implementation for BFS Algorithm in Python
To implement BFS algorithm in Python, we start by creating a graph using an adjacency list. The graph can be represented using a dictionary, where each key corresponds to a vertex, and the value is a list of its adjacent vertices.
``````graph = { "A" : ["B","D"],
"B" : ["A","C","E"],
"C" : ["B","F"],
"D" : ["A","E"],
"E" : ["B","D","F","G"],
"F" : ["C","E","H"],
"G" : ["E","H"],
"H" : ["F","G"] }
``````
Once we have created the adjacency list, we can proceed to implement the BFS algorithm. To do this, we define a function called bfs that takes two parameters: the graph and the starting vertex.
``````def bfs(graph, start_vertex):
visited = set()
queue = []
queue.append(start_vertex)
while queue:
current_vertex = queue.pop(0)
print(current_vertex)
for neighbor in graph[current_vertex]:
if neighbor not in visited:
queue.append(neighbor)
``````
Let’s go through this code step by step. First, we create an empty set called visited to keep track of the visited vertices.
We also create an empty list called queue to keep track of the vertices that need to be visited. We mark the starting vertex as visited and add it to the queue.
We then enter a while loop that runs as long as there are vertices in the queue. Inside the while loop, we use the pop(0) function to dequeue the first vertex from the queue and print it.
We then iterate over each adjacent vertex of the dequeued vertex and check if it has been visited. If it has not been visited, we mark it as visited, add it to the end of the queue, and continue with the next vertex in the queue.
### 3.2. Explanation of BFS Algorithm Execution
BFS algorithm execution involves a series of steps that are repeated until all the vertices of the graph have been visited. When executed correctly, the algorithm should traverse the graph in breadth-first order, i.e., visiting all the nodes of a given level before moving on to the nodes at the next level.
### 3.3. Let us consider the example graph from the previous section and execute the BFS algorithm:
``````graph = { "A" : ["B","D"],
"B" : ["A","C","E"],
"C" : ["B","F"],
"D" : ["A","E"],
"E" : ["B","D","F","G"],
"F" : ["C","E","H"],
"G" : ["E","H"],
"H" : ["F","G"] }
bfs(graph, "A")
``````
At the first iteration, the starting vertex “A” is marked as visited, and it is added to the queue. The queue now contains A.
``````visited = {'A'}
queue = ['A']
``````
Dequeue vertex “A” and print it:
``````visited = {'A'}
queue = []
Print A
``````
Since the dequeued vertex “A” has two adjacent vertices, B and D, we add them to the end of the queue. The queue now contains B and D.
``````visited = {'A', 'B', 'D'}
queue = ['B', 'D']
Print B
``````
Dequeue vertex B and print it. Since B has three adjacent vertices, A, C, and E, we add them to the end of the queue.
The queue now contains D, A, C, and E.
``````visited = {'A', 'B', 'C', 'D', 'E'}
queue = ['D', 'A', 'C', 'E']
Print D
``````
Dequeue vertex D and print it. Since D has two adjacent vertices, A and E, we add them to the end of the queue.
The queue now contains A, C, E, and E.
``````visited = {'A', 'B', 'C', 'D', 'E'}
queue = ['A','C', 'E', 'E']
Print C
``````
Dequeue vertex A and print it. A has one adjacent vertex B, which has already been visited.
We skip this vertex and move on to the next vertex in the queue, which is C.
``````visited = {'A', 'B', 'C', 'D', 'E'}
queue = ['C', 'E', 'E']
Print E
``````
Dequeue vertex C and print it. Since C has two adjacent vertices B and F, we add them to the end of the queue.
The queue now contains E, E, B, F.
``````visited = {'A', 'B', 'C', 'D', 'E', 'F'}
queue = ['E', 'E', 'B', 'F']
Print E
``````
Dequeue vertex E and print it. E has four adjacent vertices, B, D, F, and G.
We add them to the end of the queue. The queue now contains E, B, F, G, and F.
``````visited = {'A', 'B', 'C', 'D', 'E', 'F', 'G'}
queue = ['B', 'F', 'G', 'F']
Print B
``````
Dequeue vertex B and print it. The queue now contains F, G, F, A, and C.
``````visited = {'A', 'B', 'C', 'D', 'E', 'F', 'G'}
queue = ['F', 'G', 'F', 'A', 'C']
Print F
``````
And so on, until all the vertices of the graph have been visited.
## 4. Conclusion
In this article, we have discussed the implementation of the BFS algorithm in Python. We have also gone through a detailed explanation of the execution of the BFS algorithm.
BFS is a powerful algorithm that is used for graph traversal, shortest path algorithms, network analysis, and more. Understanding the BFS algorithm is essential for anyone interested in computer science and data structures.
## 5. Conclusion
In this article, we have discussed the Breadth-First Search (BFS) algorithm, which is a popular traversal technique used in graph theory. We have provided a definition of BFS, discussed its implementation in Python, and demonstrated its execution on a sample graph.
BFS is a simple and efficient algorithm that visits all the vertices of a graph in breadth-first order. It is used for various applications such as shortest path algorithms, network analysis, and social network analysis.
One of the significant advantages of BFS over other algorithms is that it guarantees finding the shortest path for unweighted graphs. We first started with an introduction to graph theory, where we discussed graphs’ various representations and real-world applications.
We then went on to explain the importance of graph traversal techniques, where we covered the definition of BFS as a traversal technique. To implement the BFS algorithm in Python, we created a graph using an adjacency list, which is a dictionary where each key corresponds to a vertex, and the value is a list of its adjacent vertices.
We also went through the different steps of the BFS algorithm, which include marking the starting vertex as visited, adding it to the queue, dequeuing the vertex and iterating over its adjacent vertices. To demonstrate the execution of the BFS algorithm in Python, we used a sample graph with eight nodes and 11 edges and printed the visited vertices in breadth-first order.
We also discussed the importance of debugging when implementing the BFS algorithm and the use of visualization tools to debug. In conclusion, understanding the BFS algorithm is essential in computer science, particularly for data structures and algorithms.
The BFS algorithm is widely used for traversing large graphs in an efficient and systematic manner, and this article has provided a comprehensive guide on how to implement and execute BFS in Python. Whether you’re a beginner or an experienced programmer, the BFS algorithm is an essential piece of knowledge that can enable you to solve complex problems related to graph theory.
The BFS algorithm represents one of the fundamental concepts in computer science, and acquiring a good understanding of it marks a strong foundation for mastering other computing concepts and techniques. In this article, we have discussed the Breadth-First Search (BFS) algorithm, which is a popular traversal technique used in graph theory.
We defined BFS, went through its implementation in Python, and demonstrated the execution of the algorithm on a sample graph. BFS is a simple and efficient algorithm that guarantees finding the shortest path for unweighted graphs, making it useful for various applications like shortest path algorithms and network analysis.
Understanding the BFS algorithm is crucial in computer science and data structures as it forms the basis for mastering other computing concepts and techniques. By reading this article, readers can gain a better understanding of the BFS algorithm and its implementation in Python, which is an essential piece of knowledge that can enable them to solve complex problems related to graph theory.
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1
GATE CSE 2020
Numerical
+1
-0.33
Consider a double hashing scheme in which the primary hash function is
h1(k)=k mod 23, and the secondary hash function is h2(k)=1+(k mod 19).
Assume that the table size is 23. Then the address returned by probe 1 in the probe sequence (assume that the probe sequence begins at probe 0) for key value k=90 is _______.
2
GATE CSE 2015 Set 3
Numerical
+1
-0
Given a hash table $$𝑇$$ with $$25$$ slots that stores $$2000$$ elements, the load factor $$\alpha$$ for $$𝑇$$ is ____________ .
3
GATE CSE 2007
MCQ (Single Correct Answer)
+1
-0.3
Consider the following two statements:
i. A hash function (these are often used for computing digital signatures) is an injective function.
ii. encryption technique such as DES performs a permutation on the elements of its input alphabet.
Which one of the following options is valid for the above two statements?
A
Both are false
B
Statement (i) is true and the other is false
C
Statement (ii) is true and the other is false
D
Both are true
4
GATE CSE 2005
MCQ (Single Correct Answer)
+1
-0.3
A hash table contains 10 buckets and uses linear probing to resolve collisions. The key values are integers and the hash function used is key % 10. If the values 43, 165, 62, 123, 142 are inserted in the table, in what location would the key value 142 be inserted?
A
2
B
3
C
4
D
6
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• P
Parabolic (mirror)
A mirror whose surface is figured to the shape of a paraboloid, a particular form of open curve.
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A distance equal to 3.26 light-years, often used as a unit for measuring distances to stars and galaxies. One million parsecs are more conveniently expressed as 1 Megaparsec (Mpc).
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Cargo of a spacecraft. Scientific instruments are part of a satellite's payload. For ESA scientific missions, they are usually designed and built by scientists at their home institutes.
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The relationship between the wave-like and particle-like (photon) properties of Electromagnetic radiation. E = hf, where E is the energy of the photon, h is the Planck constant and f is the frequency of the wave.
Plane (focal)
A surface upon which the image of all points in the field of view of an optical instrument is created.
Planet
Large, spherical, rocky or icy body which orbits the Sun or another star.
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A bright cloud of dust and gas surrounding an old star, namely a red giant. Towards the end of its life the star ejects the dust and gas violently, losing most of its mass and becoming a white dwarf. The nebula disappears after approximately 100 000 years. They are called 'planetary' because originally astronomers thought they looked like planetary discs.
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Main supporting structure of a spacecraft accommodating its main subsystems such as propellant, flight electronics and communications.
The Pleiades star cluster, also known as the seven sisters or M45, is a young star cluster in the constellation Taurus. It is no more than 80 million years old and lies at a distance of about 400 light-years from the Sun. The cluster contains thousands of stars, of which 6 are visible with the naked-eye: Alcyone, Maia, Atlas, Electra, Merope and Taygeta.
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Polarisation
Restricting the vibrations of waves, particularly light, to move in one direction, along one plane.
Pole
Usually the coldest regions on a planet, being the areas around an axis through the planet perpendicular to the plane of rotation about the Sun.
Positron
The antiparticle of the electron. A positron has the same characteristics as an electron but it has a positive charge instead of a negative one.
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A major review of the design for a spacecraft before construction can begin. The preliminary design review marks the boundary between the design phase and the construction phase.
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Mixture of water and chemical ingredients that constituted the oceans on Earth about three or four billion years ago. Among the chemicals were simple organic molecules. The primeval soup is thought to have been the place where life originated.
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The first time that nuclear atomic particles, neutrons and protons, could combine to make atomic nuclei. This happened during the first thousand seconds of existence of the Universe. The first atomic nuclei made were those of the light elements.
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Device that breaks light into its composite wavelength spectrum.
Probe
An unmanned vehicle travelling into space to celestial bodies in order to collect information about them.
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Large cloud of plasma extending above the Sun's chromosphere. They are divided into two main classes. Quiescent types show little motion, may last for several solar rotations and disappear slowly. Eruptive or active types form and disappear very quickly, and may grow to tremendous size.
Proportional counter
Instrument used for detecting gamma rays and X-rays in which radiation triggers an electrical discharge resulting in a pulse of electric current whose strength is proportional to the energy of the radiation.
Propulsion
Process by which something can be moved by producing a reaction with a force of thrust.
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Positively charged constituent of all atomic nuclei. A proton is made up of three quarks. The number of protons in a nucleus is called the atomic number and determines the chemical element.
Proton rocket
Russia's largest operational launch vehicle. The four-stage booster has a length of more than 57 m, its total lift-off mass is almost 700 tonnes. Over the past 30 years it has been used in more than 230 launches.
Protoplanetary disc
The disc of dust surrounding a star out of which planets might form.
Protostellar object
The earliest stages of star formation, when the nuclear reactions in the star's core have not begun yet. Stars form in opaque clouds of very cold dust and gas, which can be seen only with infrared telescopes.
Pulsar
A stellar source, such as a rotating single star or pair of stars, emitting electromagnetic radiation which is characterised by rapid frequency and regularity.
Pyrolyser
An instrument that breaks complex molecules into constituents by using heat.
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http://de.metamath.org/mpeuni/isros.html
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Mathbox for Thierry Arnoux < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > isros Structured version Visualization version GIF version
Theorem isros 29558
Description: The property of being a rings of sets, i.e. containing the empty set, and closed under finite union and set complement. (Contributed by Thierry Arnoux, 18-Jul-2020.)
Hypothesis
Ref Expression
isros.1 𝑄 = {𝑠 ∈ 𝒫 𝒫 𝑂 ∣ (∅ ∈ 𝑠 ∧ ∀𝑥𝑠𝑦𝑠 ((𝑥𝑦) ∈ 𝑠 ∧ (𝑥𝑦) ∈ 𝑠))}
Assertion
Ref Expression
isros (𝑆𝑄 ↔ (𝑆 ∈ 𝒫 𝒫 𝑂 ∧ ∅ ∈ 𝑆 ∧ ∀𝑢𝑆𝑣𝑆 ((𝑢𝑣) ∈ 𝑆 ∧ (𝑢𝑣) ∈ 𝑆)))
Distinct variable groups: 𝑣,𝑢 𝑂,𝑠 𝑆,𝑠,𝑢,𝑣,𝑥,𝑦
Allowed substitution hints: 𝑄(𝑥,𝑦,𝑣,𝑢,𝑠) 𝑂(𝑥,𝑦,𝑣,𝑢)
Proof of Theorem isros
StepHypRef Expression
1 eleq2 2677 . . . 4 (𝑠 = 𝑆 → (∅ ∈ 𝑠 ↔ ∅ ∈ 𝑆))
2 eleq2 2677 . . . . . . 7 (𝑠 = 𝑆 → ((𝑥𝑦) ∈ 𝑠 ↔ (𝑥𝑦) ∈ 𝑆))
3 eleq2 2677 . . . . . . 7 (𝑠 = 𝑆 → ((𝑥𝑦) ∈ 𝑠 ↔ (𝑥𝑦) ∈ 𝑆))
42, 3anbi12d 743 . . . . . 6 (𝑠 = 𝑆 → (((𝑥𝑦) ∈ 𝑠 ∧ (𝑥𝑦) ∈ 𝑠) ↔ ((𝑥𝑦) ∈ 𝑆 ∧ (𝑥𝑦) ∈ 𝑆)))
54raleqbi1dv 3123 . . . . 5 (𝑠 = 𝑆 → (∀𝑦𝑠 ((𝑥𝑦) ∈ 𝑠 ∧ (𝑥𝑦) ∈ 𝑠) ↔ ∀𝑦𝑆 ((𝑥𝑦) ∈ 𝑆 ∧ (𝑥𝑦) ∈ 𝑆)))
65raleqbi1dv 3123 . . . 4 (𝑠 = 𝑆 → (∀𝑥𝑠𝑦𝑠 ((𝑥𝑦) ∈ 𝑠 ∧ (𝑥𝑦) ∈ 𝑠) ↔ ∀𝑥𝑆𝑦𝑆 ((𝑥𝑦) ∈ 𝑆 ∧ (𝑥𝑦) ∈ 𝑆)))
71, 6anbi12d 743 . . 3 (𝑠 = 𝑆 → ((∅ ∈ 𝑠 ∧ ∀𝑥𝑠𝑦𝑠 ((𝑥𝑦) ∈ 𝑠 ∧ (𝑥𝑦) ∈ 𝑠)) ↔ (∅ ∈ 𝑆 ∧ ∀𝑥𝑆𝑦𝑆 ((𝑥𝑦) ∈ 𝑆 ∧ (𝑥𝑦) ∈ 𝑆))))
8 isros.1 . . 3 𝑄 = {𝑠 ∈ 𝒫 𝒫 𝑂 ∣ (∅ ∈ 𝑠 ∧ ∀𝑥𝑠𝑦𝑠 ((𝑥𝑦) ∈ 𝑠 ∧ (𝑥𝑦) ∈ 𝑠))}
97, 8elrab2 3333 . 2 (𝑆𝑄 ↔ (𝑆 ∈ 𝒫 𝒫 𝑂 ∧ (∅ ∈ 𝑆 ∧ ∀𝑥𝑆𝑦𝑆 ((𝑥𝑦) ∈ 𝑆 ∧ (𝑥𝑦) ∈ 𝑆))))
10 3anass 1035 . 2 ((𝑆 ∈ 𝒫 𝒫 𝑂 ∧ ∅ ∈ 𝑆 ∧ ∀𝑥𝑆𝑦𝑆 ((𝑥𝑦) ∈ 𝑆 ∧ (𝑥𝑦) ∈ 𝑆)) ↔ (𝑆 ∈ 𝒫 𝒫 𝑂 ∧ (∅ ∈ 𝑆 ∧ ∀𝑥𝑆𝑦𝑆 ((𝑥𝑦) ∈ 𝑆 ∧ (𝑥𝑦) ∈ 𝑆))))
11 uneq1 3722 . . . . . 6 (𝑥 = 𝑢 → (𝑥𝑦) = (𝑢𝑦))
1211eleq1d 2672 . . . . 5 (𝑥 = 𝑢 → ((𝑥𝑦) ∈ 𝑆 ↔ (𝑢𝑦) ∈ 𝑆))
13 difeq1 3683 . . . . . 6 (𝑥 = 𝑢 → (𝑥𝑦) = (𝑢𝑦))
1413eleq1d 2672 . . . . 5 (𝑥 = 𝑢 → ((𝑥𝑦) ∈ 𝑆 ↔ (𝑢𝑦) ∈ 𝑆))
1512, 14anbi12d 743 . . . 4 (𝑥 = 𝑢 → (((𝑥𝑦) ∈ 𝑆 ∧ (𝑥𝑦) ∈ 𝑆) ↔ ((𝑢𝑦) ∈ 𝑆 ∧ (𝑢𝑦) ∈ 𝑆)))
16 uneq2 3723 . . . . . 6 (𝑦 = 𝑣 → (𝑢𝑦) = (𝑢𝑣))
1716eleq1d 2672 . . . . 5 (𝑦 = 𝑣 → ((𝑢𝑦) ∈ 𝑆 ↔ (𝑢𝑣) ∈ 𝑆))
18 difeq2 3684 . . . . . 6 (𝑦 = 𝑣 → (𝑢𝑦) = (𝑢𝑣))
1918eleq1d 2672 . . . . 5 (𝑦 = 𝑣 → ((𝑢𝑦) ∈ 𝑆 ↔ (𝑢𝑣) ∈ 𝑆))
2017, 19anbi12d 743 . . . 4 (𝑦 = 𝑣 → (((𝑢𝑦) ∈ 𝑆 ∧ (𝑢𝑦) ∈ 𝑆) ↔ ((𝑢𝑣) ∈ 𝑆 ∧ (𝑢𝑣) ∈ 𝑆)))
2115, 20cbvral2v 3155 . . 3 (∀𝑥𝑆𝑦𝑆 ((𝑥𝑦) ∈ 𝑆 ∧ (𝑥𝑦) ∈ 𝑆) ↔ ∀𝑢𝑆𝑣𝑆 ((𝑢𝑣) ∈ 𝑆 ∧ (𝑢𝑣) ∈ 𝑆))
22213anbi3i 1248 . 2 ((𝑆 ∈ 𝒫 𝒫 𝑂 ∧ ∅ ∈ 𝑆 ∧ ∀𝑥𝑆𝑦𝑆 ((𝑥𝑦) ∈ 𝑆 ∧ (𝑥𝑦) ∈ 𝑆)) ↔ (𝑆 ∈ 𝒫 𝒫 𝑂 ∧ ∅ ∈ 𝑆 ∧ ∀𝑢𝑆𝑣𝑆 ((𝑢𝑣) ∈ 𝑆 ∧ (𝑢𝑣) ∈ 𝑆)))
239, 10, 223bitr2i 287 1 (𝑆𝑄 ↔ (𝑆 ∈ 𝒫 𝒫 𝑂 ∧ ∅ ∈ 𝑆 ∧ ∀𝑢𝑆𝑣𝑆 ((𝑢𝑣) ∈ 𝑆 ∧ (𝑢𝑣) ∈ 𝑆)))
Colors of variables: wff setvar class Syntax hints: ↔ wb 195 ∧ wa 383 ∧ w3a 1031 = wceq 1475 ∈ wcel 1977 ∀wral 2896 {crab 2900 ∖ cdif 3537 ∪ cun 3538 ∅c0 3874 𝒫 cpw 4108 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-10 2006 ax-11 2021 ax-12 2034 ax-13 2234 ax-ext 2590 This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-3an 1033 df-tru 1478 df-ex 1696 df-nf 1701 df-sb 1868 df-clab 2597 df-cleq 2603 df-clel 2606 df-nfc 2740 df-ral 2901 df-rab 2905 df-v 3175 df-dif 3543 df-un 3545 This theorem is referenced by: rossspw 29559 0elros 29560 unelros 29561 difelros 29562
Copyright terms: Public domain W3C validator
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CC-MAIN-2024-38
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https://www.kilomegabyte.com/5000-mib-to-kib
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#### Data Units Calculator
###### Mebibyte to Kibibyte
Online data storage unit conversion calculator:
From:
To:
The smallest unit of measurement used for measuring data is a bit. A single bit can have a value of either zero(0) or one(1). It may contain a binary value (such as True/False or On/Off or 1/0) and nothing more. Therefore, a byte, or eight bits, is used as the fundamental unit of measurement for data storage. A byte can store 256 different values, which is sufficient to represent standard ASCII table, such as all numbers, letters and control symbols.
Since most files contain thousands of bytes, file sizes are often measured in kilobytes. Larger files, such as images, videos, and audio files, contain millions of bytes and therefore are measured in megabytes. Modern storage devices can store thousands of these files, which is why storage capacity is typically measured in gigabytes or even terabytes.
# 5000 mib to kib result:
5000 (five thousand) mebibyte(s) is equal 5120000 (five million one hundred and twenty thousand) kibibyte(s)
#### What is mebibyte?
The mebibyte is a multiple of the unit byte for digital information. The binary prefix mebi means 2^20; therefore one mebibyte is equal to 1048576 bytes, i.e., 1024 kibibytes. The unit symbol for the mebibyte is MiB.
#### What is kibibyte?
The kibibyte is a multiple of the unit byte for quantities of digital information. The binary prefix kibi means 2^10, or 1024; therefore, 1 kibibyte is 1024 bytes. The unit symbol for the kibibyte is KiB.
#### How calculate mib. to kib.?
1 Mebibyte is equal to 1024 Kibibyte (one thousand and twenty-four kib)
1 Kibibyte is equal to 0.0009765625 Mebibyte (zero point zero × 3 nine million seven hundred and sixty-five thousand six hundred and twenty-five mib)
1 Mebibyte is equal to 8388608 bits (eight million three hundred and eighty-eight thousand six hundred and eight bits)
1 Kibibyte is equal to 8192 bits (eight thousand one hundred and ninety-two bits)
5000 Mebibyte is equal to 41943040000 Bit (forty-one billion nine hundred and forty-three million forty thousand bit)
Mebibyte is greater than Kibibyte
Multiplication factor is 0.0009765625.
1 / 0.0009765625 = 1024.
5000 / 0.0009765625 = 5120000.
Maybe you mean Megabyte?
5000 Mebibyte is equal to 5242.88 Megabyte (five thousand two hundred and forty-two point eighty-eight mb) convert to mb
### Powers of 2
mib kib (Kibibyte) Description
1 mib 1024 kib 1 mebibyte (one) is equal to 1024 kibibyte (one thousand and twenty-four)
2 mib 2048 kib 2 mebibyte (two) is equal to 2048 kibibyte (two thousand and forty-eight)
4 mib 4096 kib 4 mebibyte (four) is equal to 4096 kibibyte (four thousand and ninety-six)
8 mib 8192 kib 8 mebibyte (eight) is equal to 8192 kibibyte (eight thousand one hundred and ninety-two)
16 mib 16384 kib 16 mebibyte (sixteen) is equal to 16384 kibibyte (sixteen thousand three hundred and eighty-four)
32 mib 32768 kib 32 mebibyte (thirty-two) is equal to 32768 kibibyte (thirty-two thousand seven hundred and sixty-eight)
64 mib 65536 kib 64 mebibyte (sixty-four) is equal to 65536 kibibyte (sixty-five thousand five hundred and thirty-six)
128 mib 131072 kib 128 mebibyte (one hundred and twenty-eight) is equal to 131072 kibibyte (one hundred and thirty-one thousand and seventy-two)
256 mib 262144 kib 256 mebibyte (two hundred and fifty-six) is equal to 262144 kibibyte (two hundred and sixty-two thousand one hundred and forty-four)
512 mib 524288 kib 512 mebibyte (five hundred and twelve) is equal to 524288 kibibyte (five hundred and twenty-four thousand two hundred and eighty-eight)
1024 mib 1048576 kib 1024 mebibyte (one thousand and twenty-four) is equal to 1048576 kibibyte (one million forty-eight thousand five hundred and seventy-six)
2048 mib 2097152 kib 2048 mebibyte (two thousand and forty-eight) is equal to 2097152 kibibyte (two million ninety-seven thousand one hundred and fifty-two)
4096 mib 4194304 kib 4096 mebibyte (four thousand and ninety-six) is equal to 4194304 kibibyte (four million one hundred and ninety-four thousand three hundred and four)
8192 mib 8388608 kib 8192 mebibyte (eight thousand one hundred and ninety-two) is equal to 8388608 kibibyte (eight million three hundred and eighty-eight thousand six hundred and eight)
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https://www.acmicpc.net/problem/27816
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crawl-data/CC-MAIN-2023-14/segments/1679296945144.17/warc/CC-MAIN-20230323100829-20230323130829-00153.warc.gz
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시간 제한메모리 제한제출정답맞힌 사람정답 비율
30 초 (추가 시간 없음) 1024 MB111100.000%
## 문제
You are part of a team of researchers investigating the climate along the coast of an island. The island's coast is modeled as a circle with a circumference of K kilometers. There is a lighthouse on the coast which occupies a single point on the circle's circumference. Each point on the coast is mapped to a real number in the range [0, K); formally, point x is the point on the coast that is x kilometers away from the lighthouse when walking clockwise along the coast. For example, if K = 5, point 0 is the point where the lighthouse is, point 1.5 is the point that is 1.5 kilometers away from the lighthouse in the clockwise direction, and point 2.5 is the point that is located at the diametrical opposite of the lighthouse.
You are in charge of studying coastal temperatures. Another team installed a coastal temperature measuring system that works as follows: a number of thermometers were deployed at specific points to measure the temperature at those points. No two thermometers were placed at the same point. In that team's model, points without thermometers are assumed to have the same temperature as the one measured by the closest thermometer. For points that are equidistant from two thermometers, the thermometer in the clockwise direction is used (the first one you would encounter if walking clockwise from the point).
Unfortunately, you do not know how many thermometers the system used or where they were placed, but you do have access to the system's temperature data. It is given as two lists of N values each X1, X2, ..., XN and T1, T2, ..., TN, representing that each point x where Xi ≤ x < Xi+1 is assigned temperature Ti, for each 1 ≤ i < N, and each point x where 0 ≤ x < X1 or XN ≤ x < K is assigned temperature TN. The points are enumerated in the clockwise direction, so Xi < Xi+1, for all i.
You want to determine the smallest number of thermometers that, when placed in some set of locations, could have produced the observed data.
## 입력
The first line of the input gives the number of test cases, T. T test cases follow; each consists of three lines. The first line of a test case contains two integers K and N: the circumference of the island and the size of the lists representing the temperature data. The second line contains N integers X1, X2, ..., XN. The third line contains N integers T1, T2, ..., TN. The way in which the integers in the second and third line represent the temperatures is explained above.
## 출력
For each test case, output one line containing Case #x: y, where x is the test case number (starting from 1) and y is the minimum number of thermometers that could have produced the observed input data, as described above.
## 제한
• 1 ≤ T ≤ 100.
• 2 ≤ N ≤ min(100, K).
• 0 ≤ X1.
• Xi < Xi+1, for all i.
• XN < K.
• 184 ≤ Ti ≤ 330, for all i.
• TiTi+1, for all i.
• T1TN.
• 2 ≤ K ≤ 10.
• 2 ≤ K ≤ 109.
## 예제 입력 1
3
2 2
0 1
184 330
3 2
0 1
184 330
10 3
1 5 9
184 200 330
## 예제 출력 1
Case #1: 2
Case #2: 3
Case #3: 3
## 힌트
In Sample Case #1, at least 2 thermometers are needed because there are two different temperatures measured. It is possible to produce the data using exactly 2 thermometers, with one thermometer (measuring 184) at point 0.5 and another (measuring 330) at point 1.5. Note that point 0 and point 1 are equidistant from both thermometers, so the thermometer in the clockwise direction is used. The temperature measured at point 0 comes from the thermometer at point 0.5 and the temperature measured at point 1 comes from the thermometer at point 1.5.
The data from Sample Case #2 could not be produced with just 2 thermometers. It could be produced with 3 thermometers if they were placed at point 0.2, point 1.8, and point 2.8, measuring 184, 330 and 330, respectively. There are other ways to place 3 thermometers that would also yield the input data.
In Sample Case #3, one way to produce the data with 3 thermometers is to place them at point 0, point 2 and point 8, measuring 330, 184 and 200, respectively.
## 채점 및 기타 정보
• 예제는 채점하지 않는다.
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# How do you factor the expression x^2 - 13x + 40?
Dec 12, 2015
$\left(x - 5\right) \left(x - 8\right)$
#### Explanation:
There are many ways to factor. One method I like to use is the diamond method when you try to find the two numbers that add up to b and multiply to ac.
$y = a {x}^{2} + b x + c$
So you want to find two numbers that add up to -13 and multiply to 40. You can find the factors of 40 and find the pair that adds up to -13. -8 and -5 multiply to 40 and add up to -13.
$\left(x - 5\right) \left(x - 8\right) = {x}^{2} - 13 x + 40$
You can use foil to test it
${x}^{2} + \left(- 8 x\right) + \left(- 5 x\right) + 40 = {x}^{2} - 13 x + 40$
${x}^{2} - 13 x + 40 = {x}^{2} - 13 x + 40$
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From prelude Require Export sets. From algebra Require Export cmra. From algebra Require Import dra. Local Arguments valid _ _ !_ /. Local Arguments op _ _ !_ !_ /. Local Arguments unit _ _ !_ /. (** * Definition of STSs *) Module sts. Structure stsT := STS { state : Type; token : Type; prim_step : relation state; tok : state → set token; }. Arguments STS {_ _} _ _. Arguments prim_step {_} _ _. Arguments tok {_} _. Notation states sts := (set (state sts)). Notation tokens sts := (set (token sts)). (** * Theory and definitions *) Section sts. Context {sts : stsT}. (** ** Step relations *) Inductive step : relation (state sts * tokens sts) := | Step s1 s2 T1 T2 : (* TODO: This asks for ⊥ on sets: T1 ⊥ T2 := T1 ∩ T2 ⊆ ∅. *) prim_step s1 s2 → tok s1 ∩ T1 ≡ ∅ → tok s2 ∩ T2 ≡ ∅ → tok s1 ∪ T1 ≡ tok s2 ∪ T2 → step (s1,T1) (s2,T2). Definition steps := rtc step. Inductive frame_step (T : tokens sts) (s1 s2 : state sts) : Prop := | Frame_step T1 T2 : T1 ∩ (tok s1 ∪ T) ≡ ∅ → step (s1,T1) (s2,T2) → frame_step T s1 s2. (** ** Closure under frame steps *) Record closed (S : states sts) (T : tokens sts) : Prop := Closed { closed_disjoint s : s ∈ S → tok s ∩ T ≡ ∅; closed_step s1 s2 : s1 ∈ S → frame_step T s1 s2 → s2 ∈ S }. Definition up (s : state sts) (T : tokens sts) : states sts := mkSet (rtc (frame_step T) s). Definition up_set (S : states sts) (T : tokens sts) : states sts := S ≫= λ s, up s T. (** Tactic setup *) Hint Resolve Step. Hint Extern 50 (equiv (A:=set _) _ _) => set_solver : sts. Hint Extern 50 (¬equiv (A:=set _) _ _) => set_solver : sts. Hint Extern 50 (_ ∈ _) => set_solver : sts. Hint Extern 50 (_ ⊆ _) => set_solver : sts. (** ** Setoids *) Instance framestep_mono : Proper (flip (⊆) ==> (=) ==> (=) ==> impl) frame_step. Proof. intros ?? HT ?? <- ?? <-; destruct 1; econstructor; eauto with sts; set_solver. Qed. Global Instance framestep_proper : Proper ((≡) ==> (=) ==> (=) ==> iff) frame_step. Proof. by intros ?? [??] ??????; split; apply framestep_mono. Qed. Instance closed_proper' : Proper ((≡) ==> (≡) ==> impl) closed. Proof. intros ?? HT ?? HS; destruct 1; constructor; intros until 0; rewrite -?HS -?HT; eauto. Qed. Global Instance closed_proper : Proper ((≡) ==> (≡) ==> iff) closed. Proof. by split; apply closed_proper'. Qed. Global Instance up_preserving : Proper ((=) ==> flip (⊆) ==> (⊆)) up. Proof. intros s ? <- T T' HT ; apply elem_of_subseteq. induction 1 as [|s1 s2 s3 [T1 T2]]; [constructor|]. eapply rtc_l; [eapply Frame_step with T1 T2|]; eauto with sts. Qed. Global Instance up_proper : Proper ((=) ==> (≡) ==> (≡)) up. Proof. by intros ??? ?? [??]; split; apply up_preserving. Qed. Global Instance up_set_preserving : Proper ((⊆) ==> flip (⊆) ==> (⊆)) up_set. Proof. intros S1 S2 HS T1 T2 HT. rewrite /up_set. f_equiv; last done. move =>s1 s2 Hs. simpl in HT. by apply up_preserving. Qed. Global Instance up_set_proper : Proper ((≡) ==> (≡) ==> (≡)) up_set. Proof. by intros S1 S2 [??] T1 T2 [??]; split; apply up_set_preserving. Qed. (** ** Properties of closure under frame steps *) Lemma closed_steps S T s1 s2 : closed S T → s1 ∈ S → rtc (frame_step T) s1 s2 → s2 ∈ S. Proof. induction 3; eauto using closed_step. Qed. Lemma closed_op T1 T2 S1 S2 : closed S1 T1 → closed S2 T2 → closed (S1 ∩ S2) (T1 ∪ T2). Proof. intros [? Hstep1] [? Hstep2]; split; [set_solver|]. intros s3 s4; rewrite !elem_of_intersection; intros [??] [T3 T4 ?]; split. - apply Hstep1 with s3, Frame_step with T3 T4; auto with sts. - apply Hstep2 with s3, Frame_step with T3 T4; auto with sts. Qed. Lemma step_closed s1 s2 T1 T2 S Tf : step (s1,T1) (s2,T2) → closed S Tf → s1 ∈ S → T1 ∩ Tf ≡ ∅ → s2 ∈ S ∧ T2 ∩ Tf ≡ ∅ ∧ tok s2 ∩ T2 ≡ ∅. Proof. inversion_clear 1 as [???? HR Hs1 Hs2]; intros [? Hstep]??; split_and?; auto. - eapply Hstep with s1, Frame_step with T1 T2; auto with sts. - set_solver -Hstep Hs1 Hs2. Qed. Lemma steps_closed s1 s2 T1 T2 S Tf : steps (s1,T1) (s2,T2) → closed S Tf → s1 ∈ S → T1 ∩ Tf ≡ ∅ → tok s1 ∩ T1 ≡ ∅ → s2 ∈ S ∧ T2 ∩ Tf ≡ ∅ ∧ tok s2 ∩ T2 ≡ ∅. Proof. remember (s1,T1) as sT1. remember (s2,T2) as sT2. intros Hsteps. revert s1 T1 HeqsT1 s2 T2 HeqsT2. induction Hsteps as [?|? [s' T'] ? Hstep Hsteps IH]; intros; subst. - case: HeqsT2=>? ?. subst. done. - eapply step_closed in Hstep; [|done..]. destruct_conjs. eauto. Qed. (** ** Properties of the closure operators *) Lemma elem_of_up s T : s ∈ up s T. Proof. constructor. Qed. Lemma subseteq_up_set S T : S ⊆ up_set S T. Proof. intros s ?; apply elem_of_bind; eauto using elem_of_up. Qed. Lemma up_up_set s T : up s T ≡ up_set {[ s ]} T. Proof. by rewrite /up_set collection_bind_singleton. Qed. Lemma closed_up_set S T : (∀ s, s ∈ S → tok s ∩ T ≡ ∅) → closed (up_set S T) T. Proof. intros HS; unfold up_set; split. - intros s; rewrite !elem_of_bind; intros (s'&Hstep&Hs'). specialize (HS s' Hs'); clear Hs' S. induction Hstep as [s|s1 s2 s3 [T1 T2 ? Hstep] ? IH]; first done. inversion_clear Hstep; apply IH; clear IH; auto with sts. - intros s1 s2; rewrite !elem_of_bind; intros (s&?&?) ?; exists s. split; [eapply rtc_r|]; eauto. Qed. Lemma closed_up s T : tok s ∩ T ≡ ∅ → closed (up s T) T. Proof. intros; rewrite -(collection_bind_singleton (λ s, up s T) s). apply closed_up_set; set_solver. Qed. Lemma closed_up_set_empty S : closed (up_set S ∅) ∅. Proof. eauto using closed_up_set with sts. Qed. Lemma closed_up_empty s : closed (up s ∅) ∅. Proof. eauto using closed_up with sts. Qed. Lemma up_set_empty S T : up_set S T ≡ ∅ → S ≡ ∅. Proof. move:(subseteq_up_set S T). set_solver. Qed. Lemma up_set_nonempty S T : S ≢ ∅ → up_set S T ≢ ∅. Proof. by move=>? /up_set_empty. Qed. Lemma up_nonempty s T : up s T ≢ ∅. Proof. move:(elem_of_up s T). set_solver. Qed. Lemma up_closed S T : closed S T → up_set S T ≡ S. Proof. intros; split; auto using subseteq_up_set; intros s. unfold up_set; rewrite elem_of_bind; intros (s'&Hstep&?). induction Hstep; eauto using closed_step. Qed. Lemma up_subseteq s T S : closed S T → s ∈ S → sts.up s T ⊆ S. Proof. move=>? ? s' ?. eapply closed_steps; done. Qed. Lemma up_set_subseteq S1 T S2 : closed S2 T → S1 ⊆ S2 → sts.up_set S1 T ⊆ S2. Proof. move=>? ? s [s' [? ?]]. eapply closed_steps; by eauto. Qed. End sts. End sts. Notation stsT := sts.stsT. Notation STS := sts.STS. (** * STSs form a disjoint RA *) (* This module should never be imported, uses the module [sts] below. *) Module sts_dra. Import sts. (* The type of bounds we can give to the state of an STS. This is the type that we equip with an RA structure. *) Inductive car (sts : stsT) := | auth : state sts → set (token sts) → car sts | frag : set (state sts) → set (token sts ) → car sts. Arguments auth {_} _ _. Arguments frag {_} _ _. Section sts_dra. Context {sts : stsT}. Infix "≼" := dra_included. Implicit Types S : states sts. Implicit Types T : tokens sts. Inductive sts_equiv : Equiv (car sts) := | auth_equiv s T1 T2 : T1 ≡ T2 → auth s T1 ≡ auth s T2 | frag_equiv S1 S2 T1 T2 : T1 ≡ T2 → S1 ≡ S2 → frag S1 T1 ≡ frag S2 T2. Existing Instance sts_equiv. Instance sts_valid : Valid (car sts) := λ x, match x with | auth s T => tok s ∩ T ≡ ∅ | frag S' T => closed S' T ∧ S' ≢ ∅ end. Instance sts_unit : Unit (car sts) := λ x, match x with | frag S' _ => frag (up_set S' ∅ ) ∅ | auth s _ => frag (up s ∅) ∅ end. Inductive sts_disjoint : Disjoint (car sts) := | frag_frag_disjoint S1 S2 T1 T2 : S1 ∩ S2 ≢ ∅ → T1 ∩ T2 ≡ ∅ → frag S1 T1 ⊥ frag S2 T2 | auth_frag_disjoint s S T1 T2 : s ∈ S → T1 ∩ T2 ≡ ∅ → auth s T1 ⊥ frag S T2 | frag_auth_disjoint s S T1 T2 : s ∈ S → T1 ∩ T2 ≡ ∅ → frag S T1 ⊥ auth s T2. Existing Instance sts_disjoint. Instance sts_op : Op (car sts) := λ x1 x2, match x1, x2 with | frag S1 T1, frag S2 T2 => frag (S1 ∩ S2) (T1 ∪ T2) | auth s T1, frag _ T2 => auth s (T1 ∪ T2) | frag _ T1, auth s T2 => auth s (T1 ∪ T2) | auth s T1, auth _ T2 => auth s (T1 ∪ T2)(* never happens *) end. Instance sts_minus : Minus (car sts) := λ x1 x2, match x1, x2 with | frag S1 T1, frag S2 T2 => frag (up_set S1 (T1 ∖ T2)) (T1 ∖ T2) | auth s T1, frag _ T2 => auth s (T1 ∖ T2) | frag _ T2, auth s T1 => auth s (T1 ∖ T2) (* never happens *) | auth s T1, auth _ T2 => frag (up s (T1 ∖ T2)) (T1 ∖ T2) end. Hint Extern 10 (equiv (A:=set _) _ _) => set_solver : sts. Hint Extern 10 (¬equiv (A:=set _) _ _) => set_solver : sts. Hint Extern 10 (_ ∈ _) => set_solver : sts. Hint Extern 10 (_ ⊆ _) => set_solver : sts. Instance sts_equivalence: Equivalence ((≡) : relation (car sts)). Proof. split. - by intros []; constructor. - by destruct 1; constructor. - destruct 1; inversion_clear 1; constructor; etrans; eauto. Qed. Global Instance sts_dra : DRA (car sts). Proof. split. - apply _. - by do 2 destruct 1; constructor; setoid_subst. - by destruct 1; constructor; setoid_subst. - by destruct 1; simpl; intros ?; setoid_subst. - by intros ? [|]; destruct 1; inversion_clear 1; constructor; setoid_subst. - by do 2 destruct 1; constructor; setoid_subst. - assert (∀ T T' S s, closed S T → s ∈ S → tok s ∩ T' ≡ ∅ → tok s ∩ (T ∪ T') ≡ ∅). { intros S T T' s [??]; set_solver. } destruct 3; simpl in *; destruct_conjs; auto using closed_op with sts. - intros []; simpl; intros; destruct_conjs; split; eauto using closed_up, up_nonempty, closed_up_set, up_set_empty with sts. - intros ???? (z&Hy&?&Hxz); destruct Hxz; inversion Hy; clear Hy; setoid_subst; destruct_conjs; split_and?; (* TODO improve this. *) eauto using up_set_nonempty, up_nonempty; assert ((T1 ∪ T2) ∖ T1 ≡ T2) as -> by set_solver; eauto using closed_up, closed_disjoint; []. eapply closed_up_set. intros. eapply closed_disjoint; first done. set_solver. - intros [] [] []; constructor; rewrite ?assoc; auto with sts. - destruct 4; inversion_clear 1; constructor; auto with sts. - destruct 4; inversion_clear 1; constructor; auto with sts. - destruct 1; constructor; auto with sts. - destruct 3; constructor; auto with sts. - intros [|S T]; constructor; auto using elem_of_up with sts. simpl in *. assert (S ⊆ up_set S ∅) by eauto using subseteq_up_set. set_solver. - intros [|S T]; constructor; auto with sts. assert (S ⊆ up_set S ∅); auto using subseteq_up_set with sts. - intros [s T|S T]; constructor; auto with sts. + rewrite (up_closed (up _ _)); auto using closed_up with sts. + rewrite (up_closed (up_set _ _)); eauto using closed_up_set with sts. - intros x y ?? (z&Hy&?&Hxz); exists (unit (x ⋅ y)); split_and?. + destruct Hxz;inversion_clear Hy;constructor;unfold up_set; set_solver. + destruct Hxz; inversion_clear Hy; simpl; split_and?; auto using closed_up_set_empty, closed_up_empty, up_nonempty; []. eapply up_set_nonempty. set_solver. + destruct Hxz; inversion_clear Hy; constructor; repeat match goal with | |- context [ up_set ?S ?T ] => unless (S ⊆ up_set S T) by done; pose proof (subseteq_up_set S T) | |- context [ up ?s ?T ] => unless (s ∈ up s T) by done; pose proof (elem_of_up s T) end; auto with sts. - intros x y ?? (z&Hy&_&Hxz); destruct Hxz; inversion_clear Hy; constructor; repeat match goal with | |- context [ up_set ?S ?T ] => unless (S ⊆ up_set S T) by done; pose proof (subseteq_up_set S T) | |- context [ up ?s ?T ] => unless (s ∈ up s T) by done; pose proof (elem_of_up s T) end; auto with sts. - intros x y ?? (z&Hy&?&Hxz); destruct Hxz as [S1 S2 T1 T2| |]; inversion Hy; clear Hy; constructor; setoid_subst; rewrite ?disjoint_union_difference; auto. split; [|apply intersection_greatest; auto using subseteq_up_set with sts]. apply intersection_greatest; [auto with sts|]. intros s2; rewrite elem_of_intersection. destruct_conjs. unfold up_set; rewrite elem_of_bind; intros (?&s1&?&?&?). apply closed_steps with T2 s1; auto with sts. Qed. Canonical Structure RA : cmraT := validityRA (car sts). End sts_dra. End sts_dra. (** * The STS Resource Algebra *) (** Finally, the general theory of STS that should be used by users *) Notation stsRA := (@sts_dra.RA). Section sts_definitions. Context {sts : stsT}. Definition sts_auth (s : sts.state sts) (T : sts.tokens sts) : stsRA sts := to_validity (sts_dra.auth s T). Definition sts_frag (S : sts.states sts) (T : sts.tokens sts) : stsRA sts := to_validity (sts_dra.frag S T). Definition sts_frag_up (s : sts.state sts) (T : sts.tokens sts) : stsRA sts := sts_frag (sts.up s T) T. End sts_definitions. Instance: Params (@sts_auth) 2. Instance: Params (@sts_frag) 1. Instance: Params (@sts_frag_up) 2. Section stsRA. Import sts. Context {sts : stsT}. Implicit Types s : state sts. Implicit Types S : states sts. Implicit Types T : tokens sts. (** Setoids *) Global Instance sts_auth_proper s : Proper ((≡) ==> (≡)) (sts_auth s). Proof. (* this proof is horrible *) intros T1 T2 HT. rewrite /sts_auth. by eapply to_validity_proper; try apply _; constructor. Qed. Global Instance sts_frag_proper : Proper ((≡) ==> (≡) ==> (≡)) (@sts_frag sts). Proof. intros S1 S2 ? T1 T2 HT; rewrite /sts_auth. by eapply to_validity_proper; try apply _; constructor. Qed. Global Instance sts_frag_up_proper s : Proper ((≡) ==> (≡)) (sts_frag_up s). Proof. intros T1 T2 HT. by rewrite /sts_frag_up HT. Qed. (** Validity *) Lemma sts_auth_valid s T : ✓ sts_auth s T ↔ tok s ∩ T ≡ ∅. Proof. split. by move=> /(_ 0). by intros ??. Qed. Lemma sts_frag_valid S T : ✓ sts_frag S T ↔ closed S T ∧ S ≢ ∅. Proof. split. by move=> /(_ 0). by intros ??. Qed. Lemma sts_frag_up_valid s T : tok s ∩ T ≡ ∅ → ✓ sts_frag_up s T. Proof. intros. by apply sts_frag_valid; auto using closed_up, up_nonempty. Qed. Lemma sts_auth_frag_valid_inv s S T1 T2 : ✓ (sts_auth s T1 ⋅ sts_frag S T2) → s ∈ S. Proof. by move=> /(_ 0) [? [? Hdisj]]; inversion Hdisj. Qed. (** Op *) Lemma sts_op_auth_frag s S T : s ∈ S → closed S T → sts_auth s ∅ ⋅ sts_frag S T ≡ sts_auth s T. Proof. intros; split; [split|constructor; set_solver]; simpl. - intros (?&?&?); by apply closed_disjoint with S. - intros; split_and?. + set_solver+. + done. + set_solver. + constructor; set_solver. Qed. Lemma sts_op_auth_frag_up s T : sts_auth s ∅ ⋅ sts_frag_up s T ≡ sts_auth s T. Proof. intros; split; [split|constructor; set_solver]; simpl. - intros (?&?&?). destruct_conjs. apply closed_disjoint with (up s T); first done. apply elem_of_up. - intros; split_and?. + set_solver+. + by apply closed_up. + apply up_nonempty. + constructor; last set_solver. apply elem_of_up. Qed. Lemma sts_op_frag S1 S2 T1 T2 : T1 ∩ T2 ≡ ∅ → sts.closed S1 T1 → sts.closed S2 T2 → sts_frag (S1 ∩ S2) (T1 ∪ T2) ≡ sts_frag S1 T1 ⋅ sts_frag S2 T2. Proof. intros HT HS1 HS2. rewrite /sts_frag. (* FIXME why does rewrite not work?? *) etrans; last eapply to_validity_op; first done; []. move=>/=[??]. split_and!; [auto; set_solver..|]. constructor; done. Qed. (** Frame preserving updates *) Lemma sts_update_auth s1 s2 T1 T2 : steps (s1,T1) (s2,T2) → sts_auth s1 T1 ~~> sts_auth s2 T2. Proof. intros ?; apply validity_update; inversion 3 as [|? S ? Tf|]; subst. simpl in *. destruct_conjs. destruct (steps_closed s1 s2 T1 T2 S Tf) as (?&?&?); auto; []. repeat (done || constructor). Qed. Lemma sts_update_frag S1 S2 T1 T2 : closed S2 T2 → S1 ⊆ S2 → T2 ⊆ T1 → sts_frag S1 T1 ~~> sts_frag S2 T2. Proof. rewrite /sts_frag=> ? HS HT. apply validity_update. inversion 3 as [|? S ? Tf|]; simplify_eq/=. - split_and!; first done; first set_solver. constructor; set_solver. - split_and!; first done; first set_solver. constructor; set_solver. Qed. Lemma sts_update_frag_up s1 S2 T1 T2 : closed S2 T2 → s1 ∈ S2 → T2 ⊆ T1 → sts_frag_up s1 T1 ~~> sts_frag S2 T2. Proof. intros ? ? HT; apply sts_update_frag; [intros; eauto using closed_steps..]. rewrite <-HT. eapply up_subseteq; done. Qed. (** Inclusion *) (* Lemma sts_frag_included S1 S2 T1 T2 : closed S2 T2 → S2 ≢ ∅ sts_frag S1 T1 ≼ sts_frag S2 T2 ↔ (closed S1 T1 ∧ ∃ Tf, T2 ≡ T1 ∪ Tf ∧ T1 ∩ Tf ≡ ∅ ∧ S2 ≡ S1 ∩ up_set S2 Tf). Proof. (* This should use some general properties of DRAs. To be improved when we have RAs back *) move=>Hcl2. split. - intros [[[Sf Tf|Sf Tf] vf Hvf] EQ]. { exfalso. inversion_clear EQ as [Hv EQ']. apply EQ' in Hcl2. simpl in Hcl2. inversion Hcl2. } inversion_clear EQ as [Hv EQ']. move:(EQ' Hcl2)=>{EQ'} EQ. inversion_clear EQ as [|? ? ? ? HT HS]. destruct Hv as [Hv _]. move:(Hv Hcl2)=>{Hv} [/= Hcl1 [Hclf Hdisj]]. apply Hvf in Hclf. simpl in Hclf. clear Hvf. inversion_clear Hdisj. split; last (exists Tf; split_and?); [done..|]. apply (anti_symm (⊆)). + move=>s HS2. apply elem_of_intersection. split; first by apply HS. by apply subseteq_up_set. + move=>s /elem_of_intersection [HS1 Hscl]. apply HS. split; first done. destruct Hscl as [s' [Hsup Hs']]. eapply closed_steps; last (hnf in Hsup; eexact Hsup); first done. set_solver +HS Hs'. - intros (Hcl1 & Tf & Htk & Hf & Hs). exists (sts_frag (up_set S2 Tf) Tf). split; first split; simpl;[|done|]. + intros _. split_and?; first done. * apply closed_up_set; last by eapply closed_ne. move=>s Hs2. move:(closed_disjoint _ _ Hcl2 _ Hs2). set_solver +Htk. * constructor; last done. rewrite -Hs. by eapply closed_ne. + intros _. constructor; [ set_solver +Htk | done]. Qed. Lemma sts_frag_included' S1 S2 T : closed S2 T → closed S1 T → S2 ≡ S1 ∩ up_set S2 ∅ → sts_frag S1 T ≼ sts_frag S2 T. Proof. intros. apply sts_frag_included; split_and?; auto. exists ∅; split_and?; done || set_solver+. Qed. *) End stsRA.
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# Difference between revisions of "Aug 29, 2019"
Agenda:
Systematics Discussion: Don [1]
Connection details:
Meeting URL https://bluejeans.com/388466836
Meeting ID 388 466 836
US toll free phone 1.888.240.2560
Dial in, type in the meeting ID and then press ##
Minutes Attendees: Paul S., Bill H., Don J., Eric K., Dave G, Simona M., Sanghwa P.
We went over some of the recent results discussing the two main issues: IN/OUT difference and 4 micron vs 10 micron foil difference
• Is the IN/OUT difference real? Dave G. said he has never before seen a persistent IN/OUT difference like we are seeing. It seems wise to request a precision Mott measurement at the 0.3% level which will take about 3 hours. Sanghwa suggested that we might be able to fit it in without too much argument from Hall B after the next Wein flip.
• Don proposed the plan below for the next measurement (with suggested revisions). There was no opposition although Simona proposed a few changes. She believes the measurements can be done in less time than I allot and requests 1. that we do a saturation measurement on both 4 and 10 micron regardless 2. proposes turning up the current on the 10 micron foil to 1.5 muA to measure rate effects 3. Add a polarization measurement on 4 micron foil at a new position like 2mm off center. 4. Add a measurement with 8 PMTs on the 10 micron foil. There was pushback from Paul on (4) since he does not think it will be interpretable especially without a new Q1 and dipole scan.
1. Setup = 3 hr
2. Quad and dipole scans = 2.5 hr
3. PREX polarization measurements @ 4T = 6.5 hr
• In and Out on Cu = 1.7 hours
• In and Out on 4 micron = 2.5 hours
• Quick Q1 rate scan = 1 hour
• In and Out on 10 micron = 1.8 hours
4. Saturation verification @ 3.5 T = 2.5 - 5 hr
• In and Out on 4 micron = 2.5 hours
• Consistent? = Move to step 5.
• Inconsistent? In and Out on 10 micron @3.5T= 1.8 hours OR In and Out on 4 micron @ 3T = 2
5. Rate scan = 1 hr (optional if haven't already diagnosed 4/10 foil difference to rule out rate dependence)
• 2 extra points at 1 1.5 muA and 0.3 muA on 10 micron = 1 hour
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# GMAT Math: The Uses and Abuses of Formulas
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GMAT Math: The Uses and Abuses of Formulas
BY MIKE MCGARRY, MAGOOSH
Formulas are useful
Clearly, it’s a good idea to know some GMAT math formulas —– many of the best formulas to know are listed on that blog. Formulas can be wonderful shortcuts, very efficient time-savers. Clearly it would be a mistake to walk into the GMAT Math section without knowing a single formula.
But BEWARE!
Formulas are useful to a point, but one of the worst mistakes folks make when they study for GMAT Math is to think that all they have to do to master the GMAT Math section is to memorize all the formulas. Similarly, the worst way to approach a GMAT math section is to approach each question asking only, “What’s the best formula to use for this question?” That’s roughly the same as thinking that all I need to be the world’s best lion tamer is a fancy hat that says “lion tamer”! In fact, the GMAT regularly designs math questions that predictably punish folks who blindly memorize formulas. Most of the harder math problems on the GMAT have this property: if you approach the problem with the perspective of “what’s the best formula to use to solve this question?”, then you very likely will miss what the question is really about and get it wrong. Here are some tips you can use to escape these traps.
Remember, but don’t memorize
What do I mean by this? Yes, you have to remember some formulas for the GMAT. One option would be to memorize each one separately as an isolate mathematical factoid. Some people are exceptionally good memorizers, but for most people, this is not too helpful: in particular, if, in the stress of the real GMAT, you forget what you memorized, then you are out of luck. Memorizing allows you no recourse if you forget, and it gives you no insight into mathematical logic.
Instead, I would recommend: understand the derivation of every formula: that is, understand the logical argument that underlies the formula and from which the formula arises. Think in terms of “why is it true?”: don’t stop at simply “what is true?” For example, both the permutation and combination formulas can be derived directly from the Fundamental Counting Principle. In another blog, I talked about the derivation of the formula for the area of an equilateral triangle. For most of the formulas you need to know for the GMAT, there is a context of logical argument that is also helpful to know. One exception is Archimedes‘ formula for the area of a circle:
$$A=\pi{r^2}$$
Yes, the great master Archimedes had a brilliant argument for why that formula is true, but that’s well beyond the kind of logic you need to employ on the GMAT. For this one, you can take a pass and simply memorize the formula, if you haven’t already. For most formulas, though, you should explore the logical of the context in depth.
Mathematical thinking
Problem-solving in mathematics is a subtle and sophisticated subject, and even in the limited context of the GMAT math section, the harder questions sometimes demand insightful approaches. Formulas are very cookie-cutter: if you have a formula to do something, it’s very good at doing that one thing, but that one thing is all it can do. That’s quite different from the flexible analyses characteristic of mathematical thinking.
Think about it this way. When you approach the GMAT math section, you should have a well-equipped tool box. The individual formulas you know are tools, but each one is a highly specific tool, useful for only one thing. Other more widely applicable tools are skills such as estimation or backsolving or picking numbers. It’s important to have in your “tool box” a mix of tools, some more general and some more specific; what’s more important, though, it knowing when and how to use them.
Perspective
In the post How to do GMAT Math Faster, I talk in depth about left-brain vs. right-brain thinkers. Left-brain dominant thinkers want to know “what to do", and they love formulas, because it’s always very clear what to do with a formula. A more right-brain perspective focuses on “how to look at the problem“, on the question of the best way to frame a problem, the best problem-solving perspective to adopt. Often, on a challenging GMAT problem, when one adopts the best perspective, what to do becomes obvious. Given the correct perspective, many test-takers could solve the problem, but most folks get that problem wrong because it’s hard to come up with that right perspective on one’s own. Coming up with the best perspective for a problem, the best way to frame a problem, takes time. It’s a right-brain pattern-matching skill: these always take time & experience to master. It’s very important to read solutions carefully: left-brain thinkers will want to jump ahead to the formulas, to “what to do“, and may be frustrated because they knew all those parts already. To get the most from solutions, it’s very important to study the very beginning, any reasons given for making one perceptual choice rather than another. Sometimes, it may be helpful to solve the problem in more than one way, to experience the difference in different solution routes. Sometimes, the perceptual choices are hard to get from the solution, and the only way to find an answer is to post a question to an expert in the forums, asking them: “why do we get the solution this way but not in that way?” Through asking questions and investigating, over time one develops more of a sense for problem-solving perspective.
Summary
Yes, know the GMAT formulas, but don’t think they are a magic bullet. You will get considerably more from understanding the logic behind each formula, than you will from blind memorization of only the formula. Finally, it’s most important to develop the problem-solving perspective, which will help you to see which formulas to use when.
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#### Comment Preferences
• ##### Felt sorry for the "shorter than a season" guy(1+ / 0-)
Recommended by:
EverGrateful
IIRC he quit after not being able to find intermittent, fractions-of-a-microsecond glitches in the circuitry, and retired to a tent in the wilderness, vowing to concern himself no more with any interval of time shorter than a season.
The Dutch kids' chorus Kinderen voor Kinderen wishes all the world's children freedom from hunger, ignorance, and war. ☮ ♥ ☺
[ Parent ]
• ##### I actually used to find stuff like that. You had(1+ / 0-)
Recommended by:
lotlizard
to wrassle the hardware and the assembly code/instructions simultaneously. Fun times, best ever, now it's mostly 'black box' ware.
Somebody referenced the movie "War Games" computer was up for auction, nobody mentioned the Tektronix 453(?) in the background, heh. My home workshop Tek 485 nabs picoseconds, and it's at least twenty five years old; the new stuff is awesome and wicked fast.
We’re Ready, Wendy’s Ready! WTF Are We Waiting For? Bring ‘em on! The revolution has begun! Come and take it!
[ Parent ]
• ##### Wow. What comes next? Femtoseconds? n/t(0+ / 0-)
The Dutch kids' chorus Kinderen voor Kinderen wishes all the world's children freedom from hunger, ignorance, and war. ☮ ♥ ☺
[ Parent ]
• ##### Attoseconds, zeptoseconds, yoctoseconds,(1+ / 0-)
Recommended by:
lotlizard
then: One Planck time tP = ≈ 5.4×10-44 s[1] is the briefest physically meaningful span of time.
Oh darn, I forgot about the attention span of a typical teabagger, it's even shorter...
And I meant nanoseconds, not picoseconds above, urk...that 485 sweep isn't that fast ;] .
We’re Ready, Wendy’s Ready! WTF Are We Waiting For? Bring ‘em on! The revolution has begun! Come and take it!
[ Parent ]
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# 5.6: Harmonic Series II: Harmonics, Intervals, and Instruments
Summary
• The interval between two notes is related to the ratio of the frequencies of the two pitches. A notated harmonic series can show the relationship between frequency and interval. The timbre of an instrument is determined by the relative strengths of the harmonics in each note.
## Frequency and Interval
The names of the various intervals, and the way they are written on the staff, are mostly the result of a long history of evolving musical notation and theory. But the actual intervals - the way the notes sound - are not arbitrary accidents of history. Like octaves, the other intervals are also produced by the harmonic series. Recall that the frequencies of any two pitches that are one octave apart have a 2:1 ratio. (See Harmonic Series I to review this.) Every other interval that musicians talk about can also be described as having a particular frequency ratio. To find those ratios, look at a harmonic series written in common notation.
A Harmonic Series Written as Notes
Figure 1
Look at the third harmonic in Figure 1. Its frequency is three times the frequency of the first harmonic (ratio 3:1). Remember, the frequency of the second harmonic is two times that of the first harmonic (ratio 2:1). In other words, there are two waves of the higher C for every one wave of the lower C, and three waves of the third-harmonic G for every one wave of the fundamental. So the ratio of the frequencies of the second to the third harmonics is 2:3. (In other words, two waves of the C for every three of the G.) From the harmonic series shown above, you can see that the interval between these two notes is a perfect fifth. The ratio of the frequencies of all perfect fifths is 2:3.
Exercise 1:
1. The interval between the fourth and sixth harmonics (frequency ratio 4:6) is also a fifth. Can you explain this?
2. What other harmonics have an interval of a fifth?
3. Which harmonics have an interval of a fourth?
4. What is the frequency ratio for the interval of a fourth?
Solution
1. The ratio 4:6 reduced to lowest terms is 2:3. (In other words, they are two ways of writing the same mathematical relationship. If you are more comfortable with fractions than with ratios, think of all the ratios as fractions instead. 2:3 is just two-thirds, and 4:6 is four-sixths. Four-sixths reduces to two-thirds.)
2. Six and nine (6:9 also reduces to 2:3); eight and twelve; ten and fifteen; and any other combination that can be reduced to 2:3 (12:18, 14:21 and so on).
3. Harmonics three and four; six and eight; nine and twelve; twelve and sixteen; and so on.
4. 3:4
Note
If you have been looking at the harmonic series above closely, you may have noticed that some notes that are written to give the same interval have different frequency ratios. For example, the interval between the seventh and eighth harmonics is a major second, but so are the intervals between 8 and 9, between 9 and 10, and between 10 and 11. But 7:8, 8:9, 9:10, and 10:11, although they are pretty close, are not exactly the same. In fact, modern Western music uses the equal temperament tuning system, which divides the octave into twelve notes that are equally far apart. (They do have the same frequency ratios, unlike the half steps in the harmonic series.) The positive aspect of equal temperament (and the reason it is used) is that an instrument will be equally in tune in all keys. The negative aspect is that it means that all intervals except for octaves are slightly out of tune with regard to the actual harmonic series. For more about equal temperament, see Tuning Systems. Interestingly, musicians have a tendency to revert to true harmonics when they can (in other words, when it is easy to fine-tune each note). For example, an a capella choral group, or a brass ensemble, may find themselves singing or playing perfect fourths and fifths, "contracted" major thirds and "expanded" minor thirds, and half and whole steps of slightly varying sizes.
## Brass Instruments
The harmonic series is particularly important for brass instruments. A pianist or xylophone player only gets one note from each key. A string player who wants a different note from a string holds the string tightly in a different place. This basically makes a vibrating string of a new length, with a new fundamental.
But a brass player, without changing the length of the instrument, gets different notes by actually playing the harmonics of the instrument. Woodwinds also do this, although not as much. Most woodwinds can get two different octaves with essentially the same fingering; the lower octave is the fundamental of the column of air inside the instrument at that fingering. The upper octave is the first harmonic.
Note
In some woodwinds, such as the clarinet, the upper "octave" may actually be the third harmonic rather than the second, which complicates the fingering patterns of these instruments. Please see Standing Waves and Wind Instruments for an explanation of this phenomenon.
It is the brass instruments that excel in getting different notes from the same length of tubing. The sound of a brass instruments starts with vibrations of the player's lips. By vibrating the lips at different speeds, the player can cause a harmonic of the air column to sound instead of the fundamental. Thus a bugle player can play any note in the harmonic series of the instrument that falls within the player's range. Comare these well-known bugle calls to the harmonic series above.
Bugle Calls
Figure 2:Although limited by the fact that it can only play one harmonic series, the bugle can still play many well-known tunes.
For centuries, all brass instruments were valveless. A brass instrument could play only the notes of one harmonic series. (An important exception was the trombone and its relatives, which can easily change their length and harmonic series using a slide.) The upper octaves of the series, where the notes are close enough together to play an interesting melody, were often difficult to play, and some of the harmonics sound quite out of tune to ears that expect equal temperament. The solution to these problems, once brass valves were perfected, was to add a few valves to the instrument; three is usually enough. Each valve opens an extra length of tube, making the instrument a little longer, and making available a whole new harmonic series. Usually one valve gives the harmonic series one half step lower than the valveless intrument; another, one whole step lower; and the third, one and a half steps lower. The valves can be used in combination, too, making even more harmonic series available. So a valved brass instrument can find, in the comfortable middle of its range (its middle register), a valve combination that will give a reasonably in-tune version for every note of the chromatic scale. (For more on the history of valved brass, see History of the French Horn. For more on how and why harmonics are produced in wind instruments, please see Standing Waves and Wind Instruments)
Note
Trombones still use a slide instead of valves to make their instrument longer. But the basic principle is still the same. At each slide "position", the instrument gets a new harmonic series. The notes in between the positions aren't part of the chromatic scale, so they are usually only used for special effects like glissandos (sliding notes)
Overlapping Harmonic Series in Brass Instruments
Figure 3: These harmonic series are for a brass instrument that has a "C" fundamental when no valves are being used - for example, a C trumpet. Remember, there is an entire harmonic series for every fundamental, and any note can be a fundamental. You just have to find the brass tube with the right length. So a trumpet or tuba can get one harmonic series using no valves, another one a half step lower using one valve, another one a whole step lower using another valve, and so on. By the time all the combinations of valves are used, there is some way to get an in-tune version of every note they need.
Exercise 2:
Write the harmonic series for the instrument above when both the first and second valves are open. (You can use this PDF file if you need staff paper.) What new notes are added in the instrument's middle range? Are any notes still missing?
Solution
Opening both first and second valves gives the harmonic series one-and-a-half steps lower than "no valves".
Figure 4
Note
The French horn has a reputation for being a "difficult" instrument to play. This is also because of the harmonic series. Most brass instruments play in the first few octaves of the harmonic series, where the notes are farther apart and it takes a pretty big difference in the mouth and lips (the embouchure, pronounced AHM-buh-sher) to get a different note. The range of the French horn is higher in the harmonic series, where the notes are closer together. So very small differences in the mouth and lips can mean the wrong harmonic comes out
## Playing Harmonics on Strings
String players also use harmonics, although not as much as brass players. Harmonics on strings have a very different timbre from ordinary string sounds. They give a quieter, thinner, more bell-like tone, and are usually used as a kind of ear-catching special-effect.
Normally a string player holds a string down very tightly. This shortens the length of the vibrating part of the string, in effect making a (temporarily) shorter vibrating string, which has its own full set of harmonics.
To "play a harmonic", the string is touched very, very lightly instead. The length of the string does not change. Instead, the light touch interferes with all of the vibrations that don't have a node at that spot.
String Harmonics
Figure 5
The thinner, quieter sound of "playing harmonics" is caused by the fact that much of the harmonic series is missing from the sound, which will of course affect the timbre. Lightly touching the string in most places will result in no sound at all. This technique only works well at places on the string where a main harmonic (one of the longer, louder lower-numbered harmonics) has a node. Some string players can get more harmonics by both holding the string down in one spot and touching it lightly in another spot, but this is an advanced technique.
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# An expression
Find an expression representing the difference when (-4x - 5) is subtracted from (10x - 5) in simplest terms.
e = 14x
### Step-by-step explanation:
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| 3
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CC-MAIN-2023-40
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latest
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en
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https://www.tutorialgateway.org/go-program-to-count-even-and-odd-numbers-in-an-array/
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# Go Program to Count Even and Odd Numbers in an Array
In this Go Program to Count Even and Odd Numbers in an Array, we used for loop to iterate the array. The if condition (if evoddarr[i]%2 == 0) checks the array element is divisible by two equals to zero. If True, we increment the even count (evenCount++); otherwise, increment the odd count value (oddCount++).
```package main
import "fmt"
func main() {
var size, i int
fmt.Print("Enter the Even Odd Array Size = ")
fmt.Scan(&size)
evoddarr := make([]int, size)
fmt.Print("Enter the Even Odd Array Items = ")
for i = 0; i < size; i++ {
fmt.Scan(&evoddarr[i])
}
evenCount := 0
oddCount := 0
for i = 0; i < size; i++ {
if evoddarr[i]%2 == 0 {
evenCount++
} else {
oddCount++
}
}
fmt.Println("The Total Number of Even Numbers = ", evenCount)
fmt.Println("The Total Number of Odd Numbers = ", oddCount)
}```
``````Enter the Even Odd Array Size = 5
Enter the Even Odd Array Items = 1 2 3 4 5
The Total Number of Even Numbers = 2
The Total Number of Odd Numbers = 3``````
Golang Program to Count Even and Odd Numbers in an Array using the for loop range.
```package main
import (
"fmt"
)
func main() {
var size int
fmt.Print("Enter the Even Odd Array Size = ")
fmt.Scan(&size)
evoddarr := make([]int, size)
fmt.Print("Enter the Even Odd Array Items = ")
for i := 0; i < size; i++ {
fmt.Scan(&evoddarr[i])
}
evenCount := 0
oddCount := 0
for _, a := range evoddarr {
if a%2 == 0 {
evenCount++
} else {
oddCount++
}
}
fmt.Println("The Total Number of Even Numbers = ", evenCount)
fmt.Println("The Total Number of Odd Numbers = ", oddCount)
}```
In this Go even-odd array example, we created two separate functions (func countEvenNums(evoddarr []int) and func countOddNums) that return the count of Even and Odd numbers.
```package main
import "fmt"
var evenCount, oddCount int
func countEvenNums(evoddarr []int) int {
evenCount = 0
fmt.Print("\nList of Even Numbers = ")
for _, a := range evoddarr {
if a%2 == 0 {
fmt.Print(a, " ")
evenCount++
}
}
return evenCount
}
func countOddNums(evoddarr []int) int {
oddCount = 0
fmt.Print("\nList of Odd Numbers = ")
for _, a := range evoddarr {
if a%2 != 0 {
fmt.Print(a, " ")
oddCount++
}
}
return oddCount
}
func main() {
var size int
fmt.Print("Enter the Even Odd Array Size = ")
fmt.Scan(&size)
evoddarr := make([]int, size)
fmt.Print("Enter the Even Odd Array Items = ")
for i := 0; i < size; i++ {
fmt.Scan(&evoddarr[i])
}
evenCount = countEvenNums(evoddarr)
oddCount = countOddNums(evoddarr)
fmt.Println("\nThe Total Number of Even Numbers = ", evenCount)
fmt.Println("The Total Number of Odd Numbers = ", oddCount)
}```
``````Enter the Even Odd Array Size = 8
Enter the Even Odd Array Items = 11 22 8 33 98 9 19 0
List of Even Numbers = 22 8 98 0
List of Odd Numbers = 11 33 9 19
The Total Number of Even Numbers = 4
The Total Number of Odd Numbers = 4``````
| 869
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|
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| 2.71875
| 3
|
CC-MAIN-2023-23
|
longest
|
en
| 0.584954
|
https://www.physicsforums.com/threads/thermodynamics-problem-does-this-approach-seem-right.963661/
| 1,685,552,097,000,000,000
|
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| 1,046,456,054
| 15,181
|
# Thermodynamics problem -Does this approach seem right?
• TroyElliott
## Homework Statement
A dilute gas consisting of N hydrogen atoms in equilibrium at temperature T and pressure P. A fraction of the atoms combine to form diatomic hydrogen. For ##N_{s}## single atoms and ##N_{d}## diatomic molecules, the free energy of the system is
$$G = N_{s}k_{b}T\ln{(\frac{N_{s}}{N_{s}+N_{d}}\frac{P}{P^{0}_{s}})}+N_{d}k_{b}T\ln{(\frac{N_{d}}{N_{s}+N_{d}}\frac{P}{P^{0}_{d}})} - \epsilon N_{d}.$$
Here ##\epsilon## is the binding energy of the ##H_{2}## molecule, and ##P^{0}_{s}## and ##P^{0}_{d}## are functions of only temperature. Find the relation between ##N_{s}## and ##N_{d}## in equilibrium.
## Homework Equations
##\Delta G = 0##
##G_{initial} = N_{s}k_{b}T\ln{(\frac{P}{P^{0}_{s}})}?##
## The Attempt at a Solution
##\Delta G = G_{final}-G_{initial} = 0.## I assume that the ##G_{final}## is the formula for the free energy given in the problem statement. Is it right to assume that ##G_{initial}## is given by setting ##N_{d}## to zero? I am assuming that initially all the hydrogen was just made up of single atoms and the free energy formula above would still be valid for this. Does this seem like the right approach? The algebra gets messy and I haven't been able to get a clean relationship between ##N_{s}## and ##N_{d}##.
Thanks!
No. I don't agree with your approach. You have that, in any state, $$2N_d+N_s=N$$ You need to find the value of ##N_s## that minimizes G.
| 466
| 1,490
|
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| 3.34375
| 3
|
CC-MAIN-2023-23
|
latest
|
en
| 0.823386
|
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