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https://minhbinhtran.com/2019/02/28/chapter-13-a-reaction-network-approach-to-the-theory-of-acoustic-wave-turbulence/
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# Chapter 13: A reaction network approach to the theory of acoustic wave turbulence
This post is based on my paper
Let us consider the acoustic wave turbulence kinetic equation
$\partial_tf \ = \ Q[f]$
where $Q[f]$ is the collision term, describing pure resonant three-wave interactions. The equation is a three-wave kinetic one, in which the collision operator is of the form
$Q[f](k) \ = \ \iint_{\mathbb{R}^{2d}} \Big[ R_{k,k_1,k_2}[f] - R_{k_1,k,k_2}[f] - R_{k_2,k,k_1}[f] \Big] dk_1dk_2$
with
$R_{k,k_1,k_2} [f]:= 4\pi |V_{k,k_1,k_2}|^2\delta(k-k_1-k_2)\delta(|k |-|k_1|-|k_2|)(f_1f_2-ff_1-ff_2).$
We have used the short-hand notation $f = f(t,k)$ and $f_j = f(t,k_j)$.
In this paper, we discover, for the first time, the connection between the wave kinetic equation and chemical reaction networks. We prove that the discrete version of the wave kinetic equation can be associated with a chemical reaction network which takes the form
$A_{k_2} + A_{k_3} -> A_{k_1},$
$A_{k_2} + A_{k_1} -> 2A_{k_2} + A_{k_3}.$
As a consequence, techniques that have been used to study the Global Attractor Conjecture in chemical reaction network theory can be applied to study the long time behavior of the wave kinetic equation We prove that as time evolves, the solution of the discrete version of the wave kinetic equation converges to a steady state exponentially in time.
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https://www.scribd.com/document/233698466/Economics-Primer
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You are on page 1of 16
# By
## Currenc-I Club, IIM Indore
Aditya | Arjit | Priyadarshini | Sandesh |
The two major divisions are Macro and Micro
Economics
Macro : Provides a holistic view about the
economy that covers GDP, Inflation, Monetary
and fiscal policies, Unemployment, BOP (
balance of payment) and growth rate of the balance of payment) and growth rate of the
economy etc.
Micro : Analyses the individual decision
making units which covers the factors that
affect the demand and supply, prices of the
goods and elasticity etc.
Demand & Supply
Demand- Quantity expected to be
purchased by consumers at different
price levels
Supply- Quantity supplied by the
industry at different price levels
As the price of good increases demand As the price of good increases demand
falls (consumers do not prefer the good
due to high price) and supply increases
(manufacturers gets more revenue due
to increased price)
Hence demand is a downward sloping
and supply is an upward sloping curve
Equilibrium level is given by the point
of intersection and the corresponding
price is equilibrium price
Elasticity is a measure of ratio of percentage change
in one variable to percentage change in other variable
Price Elasticity Measures change in quantity
demanded in response to change in price of the good
Price Elasticity = Q(%age Change in Demand) / (P %age change in Price)
Income Elasticity- Measures percentage change in
quantity demanded due to change in income quantity demanded due to change in income
Income Elasticity = Q(%age Change in Demand) / I(%age change in
Income)
Note- The above quantities measure price and
income elasticity of demand. Similarly there can be
elasticity of supply where the numerator is replaced
with change in quantity of supply.
The demand is perfectly
elastic at price \$15
At any other price point
demand is zero
This is an example of This is an example of
perfectly elastic demand
curve
The curve is also called
infinitely elastic curve
This curve represents
perfectly inelastic or
zero elastic demand
curve
The curve shows that
the demand remains at a
particular quantity
the demand remains at a
particular quantity
irrespective of the price
level
The same can be applied
for the supply side curve
E.G. There can be
perfectly inelastic supply
curve
There are four different types of markets
Perfectly competitive
Monopoly
Monopolistic competition
Oligopoly Oligopoly
There are many players with each one holding
a small market share
There are no entry and exit barriers i.e. It is
easy to enter and exit the business
All the products produced by different
manufacturers are identical
All the products produced by different
manufacturers are identical
Prices are determined by demand and supply
of the product, not by the firm
Perfectly competitive firms have perfectly
elastic demand curve
There is one supplier in the entire market who
produces a product which is exclusive and has no
good substitutes
The barriers to entry are very high
Legal Barriers
Natural Barriers Natural Barriers
Patents copyrights and government franchisees
are few legal barriers
Natural Barriers- A firm can reduce the cost of
production by producing more (Economies of
Scale) and they sell at a cheaper price which
cannot be done by other players
Large number of competitors produce
differentiated products
Product differentiation gives degree of
market power to each firm
Firms compete on price quality and marketing
due to product differentiation. Quality is a
Firms compete on price quality and marketing
due to product differentiation. Quality is a
significant product differentiation
characteristic and price is set by the firms
based on the demand supply relation
Barriers to entry and exit are low and hence it
is easy to enter and exit the business
The market is characterized by small number of
sellers
There is interdependence amongst the sellers
and hence decision by player is affected by other
players decisions players decisions
Products may be differentiated or similar
Significant barriers to entry due to which each
firm has large economies of scale. Due to
significant barriers of entry there are very few
players who supply the entire market hence they
have large economies of scale
GDP
Total Market value of all final goods and Services produced within
an economy
Check out GDP/NDP/GNP Check out GDP/NDP/GNP Check out GDP/NDP/GNP Check out GDP/NDP/GNP
Rate of changes in price
Inflation
Rate of changes in price
Find out the two main indexes for inflation measurement Find out the two main indexes for inflation measurement Find out the two main indexes for inflation measurement Find out the two main indexes for inflation measurement
Interest Rates
Real interest rate is the rate at which your money grows after
accounting for inflation
Find out the difference between Real and Nominal Interest rates Find out the difference between Real and Nominal Interest rates Find out the difference between Real and Nominal Interest rates Find out the difference between Real and Nominal Interest rates
Revenue Deficit: Revenue Expd Revenue
Receipts
Fiscal Deficit: Total Expd ( Revenue
Receipts+ Recovery of loans + Receipts from
PSU disinvestment) PSU disinvestment)
Primary deficit: Fiscal Deficit- Interest
Payments
Government and RBI uses various tools to
influence the growth of the economy
Monetary policy (RBI)
1. Control money supply
2. Control interest rates
Fiscal policy Fiscal policy
1. Manage the government revenue
2. Manage government expenditure
Find Find Find Find out: out: out: out: <CRR <CRR <CRR <CRR, Repo rate, reverse repo, , Repo rate, reverse repo, , Repo rate, reverse repo, , Repo rate, reverse repo, SLR> SLR> SLR> SLR>
For any queries contact
currenci@iimidr.ac.in currenci@iimidr.ac.in currenci@iimidr.ac.in currenci@iimidr.ac.in
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## The Langton’s Ant
###### By Anupum Pant
Think of a cell sized ant sitting on a huge grid of such white cells. The thing to note about this ant is that it follows a certain sets of simple rules. The main rule is that when the ant exits a cell, it inverts the colour of the cell it just left. Besides that:
1. If the ant enters a white square, it turns left.
2. If it enters a black square, it turns right.
Here’s what happens if the ant starts out in the middle and moves to the cell on the right, as a starting step (this can be on any side).
Now as this continues, a seemingly random figure starts taking shape. The black cells are in total chaos, there seems to be no specific order to how they appear on the canvas. (of course the pattern is always the same chaos, considering the ant starts on a blank array of cells).
And yet, after about 10,000 steps are completed by the turing ant, it starts creating a very orderly highway kind of figure on the canvas. It enters an endless loop consisting of 104 steps which keeps repeating for ever and creates a long highway kind of structure.
Suppose, initially you take a configuration of black spots on a canvas (not a blank white canvas). Take an array of cells with randomly arranged black spots, for instance. If given enough time, the ant ultimately always ends up making the looped highway. However, before it starts doing it, it might take a significant amount of steps less, or more, than the ~10,000 steps it took to reach the loop in a blank array of cells.
No exception has ever been found. A computer scientist Chris Langton discovered this in the year 1986.
## Shepard Tone – An Incredible Auditory Illusion
###### By Anupum Pant
Here’s the thing. Go to ToneDeafTest.com and take that little test they have on their homepage. That is what you need to do first. Stop reading further if you haven’t done it yet.
Assuming you did what I asked you to do…
If you did well in the test (with a few silly mistakes which can be ignored), you’ll probably understand better what I’m talking about in the following article. Otherwise, you might miss the point.
Nevertheless, there is still a chance that you’d understand even if you are tone deaf. I’m not sure because I’m certain not tone deaf and it’s impossible for me to understand the subjective experiences of tone deaf people (I can boast that the first time I took it, I got a perfect score in that test). Anyway, that test is a fun thing to do. You’ll at least learn something about yourself.
### The endless stairs and the endless tone
Everyone knows the endless stairs (in the picture below). Now, you’d think why is the author talking about a visual illusion just after he told us to take an auditory test. That is because the popular visual illusion helps you to relate better to a relatively lesser known auditory illusion.
If you start going up on the endless stairs, you always keep moving up. Even after you come back to the same place, you still keep going up. An impossibility. But it’s something that fools your eyes. The same thing happens if you start going down the stairs.
A similar thing can happen with tones. Listen to the following (continuous?) note sweep.
It sounds like a tone that is continuously going down, endlessly. Only, it isn’t. It’s actually a much smaller looped sound that starts from a high point and then goes down. These little loops have been placed one after the other. If you do not carefully listen to it, you’ll never find the exact point at which one loop ends and the next loop starts. You’ll always interpret it as a continuously going down sound. Just like the continuously going down stairs. This is called the Shepard tone.
This works for discreet notes also. Listen to this endless mario stairs video to get an idea how it works for individual notes (not sweeps).
### Why?
Notes are not simple frequencies. A single note is usually composed of several other frequencies. To not overwhelm us with data, the brain puts all these frequencies together and we hear a single sound (note).
Also, our brains like continuity. So, it cherry picks the frequencies from the loop’s notes that makes us hear a continuous sweep. This is the reason we hear no individual loops. Bah! I’m not very good at explaining this. So, here goes the Vsauce video which explains it better. Note that the arrows in the video are the frequencies I was talking about…
## The Sun’s Unusual Behavior – Seen from Mercury
### The sun – as seen from Earth
For most of us living on Earth (closer to the equator), the sun has followed a simple path throughout the years. It rises, goes up at noon and then sets for rest of the day. It is a simple straight line for the complete year.
For people living a little away from the equator, things get a bit interesting. There, the summer sun at noon is overhead, but the winter sun is low at noon, not overhead. It isn’t very easy for a person living near the equator to grasp this phenomenon well. You’ll have to go there and see for yourself. Or simply, the simulator at the end of this paragraph will help you understand it better.
At poles, the sun almost moves horizontally for many days. It keeps on making a horizontal circle around you. There, it is day for 6 months and night for the next 6 months. [Here is a sun path simulator for Earth]
However, nowhere on earth, things get as interesting as they get in the skies of Mercury.
### The sun – as seen from Mercury
On Mercury, the sun appears to briefly reverse its usual east to west motion once every Mercurian year. The effect is visible from any place on Mercury, but there are certain places on its surface, where an observer would be able to see the Sun rise about halfway, reverse and set, and then rise again, all within the same day. It is indeed an unusual performance which isn’t easy for us Earthlings to digest. [See animation in the next paragraph]
### Why does it happen?
Let us consider a simpler analogy – some planets (like Mars), as seen from earth, take a similar path. [see the animation for Mars’s path as seen from earth]
The planets, including Earth, all travel around the Sun in a continuous orbit. We can see them make their way across the sky in a straight line usually. However, every now and then a planet appears to turn around. After turning around, it appears to move back the way it came. This is called a retrograde orbit and is caused due to the difference in speeds at which the planets circle the Sun.
So, as we see Mars do a reverse from earth, a similar motion of sun is observed from the surface of Mercury.
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What is the line of symmetry for the parabola whose equation is y = -x2 + x + 3?
The line of symmetry for the parabola whose equation is y = -x^2 + x + 3 is 1/2 . symmetry = -1/(-1*2) = -1/-2 = 1/2
Updated 3/3/2015 4:18:04 AM
Rating
3
The line of symmetry for the parabola whose equation is y = -x^2 + x + 3 is 1/2 .
symmetry = -1/(-1*2) = -1/-2 = 1/2
Confirmed by jeifunk [3/3/2015 5:32:59 AM]
A circle has its center at the origin, and (5, -12) is a point on the circle. How long is the radius of the circle
Updated 2/24/2015 11:14:36 PM
A circle has its center at the origin, and (5, -12) is a point on the circle. the radius of the circle is sqrt (5^2 + 12^2) = 13.
Confirmed by Andrew. [2/25/2015 1:12:57 AM]
Write the equation 0.3x 2 + 5x - 7 = 0 in general form and then choose the value of "b."
Updated 2/24/2015 8:17:45 PM
Write the equation 0.3x 2 + 5x - 7 = 0 in general form is 3x^2 + 50x - 70 = 0 and then the value of "b" is 50 .
Confirmed by Andrew. [2/24/2015 9:46:49 PM]
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https://cs.stackexchange.com/questions/128864/shortest-path-in-a-directed-acyclic-graph-with-two-types-of-costs
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# Shortest Path in a Directed Acyclic Graph with two types of costs
I am given a directed acyclic graph $$G = (V,E)$$, which can be assumed to be topologically ordered (if needed). Each edge $$e$$ in G has two types of costs - a nominal cost $$w(e)$$ and a spiked cost $$p(e)$$. I am also given two nodes in $$G$$, node $$s$$ and node $$t$$.
The goal is to find a path from $$s$$ to $$t$$ that minimizes the following cost: $$\sum_e w(e) + \max_e \{p(e)\},$$ where the sum and maximum are taken over all edges of the path.
Standard dynamic programming methods show that this problem is solvable in $$O(E^2)$$ time. Is there a more efficient way to solve it? Ideally, an $$O(E\cdot \operatorname{polylog}(E,V))$$ algorithm would be nice.
This is the $$O(E^2)$$ solution I found using dynamic programming, if it'll help.
First, order all costs $$p(e)$$ in an ascending order. This takes $$O(E\log(E))$$ time.
Second, define the state space consisting of states $$(x,i)$$ where $$x$$ is a node in the graph and $$i\in \{1,2,...,|E|\}$$. It represents "We are in node $$x$$, and the highest edge weight $$p(e)$$ we have seen so far is the $$i$$-th largest".
Let $$V(x,i)$$ be the length of the shortest path (in the classical sense) from $$s$$ to $$x$$, where the highest $$p(e)$$ encountered was the $$i$$-th largest. It's easy to compute $$V(x,i)$$ given $$V(y,j)$$ for any predecessor $$y$$ of $$x$$ and any $$j \in \{1,...,|E|\}$$ (there are two cases to consider - the edge $$y\to x$$ is has the $$j$$-th largest weight, or it does not).
At every state $$(x,i)$$, this computation finds the minimum of about $$\deg(x)$$ values. Thus the complexity is $$O(E) \cdot \sum_{x\in V} \deg(x) = O(E^2),$$ as each node is associated to $$|E|$$ different states.
• You don't know whether $p(e) \geq w(e)$, right? Aug 1, 2020 at 16:47
• Not necessarily. But if you have ideas I will welcome an algorithm with this assumption as well Aug 1, 2020 at 18:53
• Are all costs non-negative?
– D.W.
Aug 1, 2020 at 21:25
• @BernardoSubercaseaux, you can add a constant to all $p(e)$ to achieve this.
– user114966
Aug 2, 2020 at 23:02
• I spent way too much time on this, you can get $O(|E||V|W)$ for integer weights where the maximum weight is $W$, and you can do a bit better if you allow approximate algorithms. Both are using decremental/incremental single-source shortest path algorithms, so you can't really expect more fruitful stuff to come from that direction.
– orlp
Aug 3, 2020 at 3:34
It is unlikely that there is an algorithm that works in $$O(|E|^{2 - \varepsilon})$$ time for all graphs. It is still possible that there are faster algorithms for dense graphs ($$E = \Omega(V^2)$$) or somehow dense graphs ($$E = \Theta(V^\alpha)$$ for some $$\alpha > 1$$), but an algorithm that works in $$O(|E|^{2 - \varepsilon})$$ time for sparse graphs ($$E = O(V)$$) would contradict so called $$k$$-cycle hypothesis (you can check this paper and this paper for more details on the statement and its applications to lower bounds for dynamic shortest path and other graph problems).
## The reduction
I prefer to state the $$k$$-cycle hypothesis in the following way, which is equivalent to the way it is usually stated (for more details, see the last section):
For any $$\varepsilon > 0$$, there is a positive integer $$k$$, such that there is no $$O(|E|^{2 - \varepsilon})$$ algorithm for checking whether there is a closed walk of length exactly $$k$$ in the given directed graph.
Let's reduce the problem of finding a closed walk of length $$k$$ in a given directed graph $$G(V,E)$$ to the problem in question for a slighty larger graph. Construct a new graph $$H$$ ($$H$$ will be a DAG), whose vertex set consists of a source vertex $$s$$, a sink vertex $$t$$ and vertices $$(v, \ell)$$ for each $$v \in V, \ell \in [1, k + 1]$$ ($$k + 1$$ layers, each being a copy of $$V$$).
Each edge $$u \to v$$ of the original graph splits into $$k$$ edges in $$H$$: $$(u, \ell) \to (v, \ell + 1)$$ for each $$\ell \in [1, k]$$, with each of those edges having $$w$$ and $$p$$ parameters set to $$0$$. Also, there will be $$|V|$$ edges from the source to the first layer ($$s \to (v, 1)$$ with parameters $$w := v$$ and $$p := 3|V| - 2v$$ for each $$v \in V$$; here I assume that vertices of $$G$$ are numbered from $$0$$ to $$|V| - 1$$). Similarly, there will be $$|V|$$ edges from the last layer to $$t$$ ($$(v, k + 1) \to t$$ with $$w := v$$, $$p := 3|V| - 2v$$). Clearly, $$H$$ has $$(k+1)|V| + 2$$ vertices and $$2|V| + k|E|$$ edges.
Now, each $$s \to t$$ path $$s \to (v_1, 1) \to (v_2, 2) \to \ldots (v_{k + 1}, k + 1) \to t$$ in $$H$$ corresponds to a walk of length $$k$$ in the original graph ($$G$$). The corresponding walk is closed if and only if $$v_1 = v_{k + 1}$$. The main idea is that we have chosen the parameters $$w$$ and $$p$$ to ensure that paths with $$v_1 = v_{k + 1}$$ have the smallest possible cost.
Indeed, the cost of $$s \to (v_1, 1) \to \ldots \to (v_{k + 1}, k + 1) \to t$$ path in $$H$$ is $$v_1 + v_{k + 1} + \max(3|V| - 2v_1, 3|V| - 2v_{k+1}) = v_1 + v_{k + 1} + 3|V| - \min(v_1, v_{k + 1}) = 3|V| + |v_1 - v_{k + 1}|$$. Hence, the shortest $$s \to t$$ path in $$H$$ has cost $$3|V|$$ if and only if the original graph $$G$$ contained a closed $$k$$-walk. Therefore, under k-cycle hypothesis, there is no $$O(|E|^{2 - \varepsilon})$$ algorithm for problem in question.
## Can we do better for dense graphs?
Honestly, I don't know. Proving lower bounds for dense graphs seems to be difficult, because the problem is strictly simpler that dynamic shortest paths. Hence, the only simple "vector of attack" is to try using the same methods that are used for dense graphs and hope that they will work even for the simpler problem. Maybe future answerers will be more succesful here?
As for faster algoritms, an existence of $$O(VE)$$ algorithm sounds plausible. The only thing that I managed to come up with is the following algorithm, whose complexity is a bit hard to estimate, but I suspect that it still should take $$\Omega(E^2)$$ time in the worst case:
Find some shortest ($$s \to t$$)-path in the original graph with respect to sum of $$w$$'s, breaking ties by minimizing the maximum of $$p$$'s (it is possible to do this by dynamic programming in $$O(|E|)$$ time). Let's say that the largest value of $$p$$ for edges on the chosen path is $$P$$. Then, all edges $$e$$ with $$p_e \geqslant P$$ can't appear in a path with a smaller cost (because otherwise the chosen path was not shortest with respect to sum of $$w$$'s). Therefore, we can delete all such edges and repeat the process.
## Why did I choose an uncommon statement of k-cycle hypothesis?
Usually, it is stated in the terms of finding a simple cycle of length $$k$$, but for fixed $$k$$ finding a simple cycle can be reduced to finding a closed walk in (larger) graph by known techniques like color coding. All these techniques require blow up the size of the graph exponentially with respect to $$k$$ (otherwise we could solve Hamiltonian Cycle in subexponential time), but we can handle this, because $$k$$ is fixed and small.
Intuitively speaking, for small $$k$$, the "main problem", asymptotically speaking, is not ensuring that all vertices of the cycle are different, but remembering where the cycle started.
Because the problems are equivalent, I chose the one which is easier to work with. Moreover, apart from having a theoretical reduction, we also have a practical reduction now. For example, if we assume that there is no way to check whether a graph with $$10^6$$ edges has a closed walk of length $$k = 6$$ in under $$10$$ minutes on a normal computer (I would say, that this is a huge understatement, considering the large constant factors of graph searches; though I would be very positively surprised to be wrong), then there is no way to solve the problem in question for graphs with $$10^6 / (k + 2) = 1.25 \cdot 10^5$$ edges in under $$10$$ minutes, e.t.c. The point is, that the constant factors involved in the reduction are reasonably small.
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# Thread: help with this code?
1. Member
Join Date
Nov 2009
Posts
7
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## help with this code?
hi guys, i'm just learning arrays, and i'm a bit stumped on this. i have to write a program that reads 10 numbers, computes their average, and then tells how many numbers are higher than the average. this is what i have right now:
Java Code:
``` import java.util.Scanner;
public class Prog17 {
public static void main (String[] args){
final int totalnum = 10;
int[] num = new int[totalnum];
Scanner input = new Scanner(System.in);
//Get Numbers
for (int i = 0; i <num.length; i++){
System.out.print("Enter a number: ");
num[i] = input.nextInt();
}
//Calculations for average
double total = 0;
int counter = 0;
for(int i = 0; i < num.length; i++){
total += (num[i])/10;
if ( num[i] > total)
counter++;
}
System.out.print(counter);
}
}```
i did the average of 1,2,3,4,5,6,7,8,9,10 and got the average of 1. but when i try to get the numbers bigger than the average, it says 10, but 1 > 1 is false, obviously, so it should be 9. what's wrong with it?
2. You're dealing with ints here and they have int divisions, i.e. 9/10 == 0 instead of 0.9. (the remainder is discarded); you have to add all your numbers first and divide the total by 10.0 at the end.
kind regards,
Jos
3. Member
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Oct 2009
Location
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Like Jos said: you need to cast your integers to doubles. I recommend you replace your 10 in the division statement by the variable totalnum (or num.length).
To cast them both it needs to look something like this:
Java Code:
`total += (double)num[i] / (double)totalnum;`
Also, you forgot to print total.
4. Member
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Nov 2009
Posts
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well i re-edited my code, and i am now getting the correct average. but i'm still stumped on how to get the elements of my array to compare it to the average so i can see how many numbers are larger than the average.
5. Calculate the average, and then loop again through the array, with each loop, checking to see if the the current number is higher or lower than the average.
Give it your best try and then post your attempt up here and we can take a look at it. Much luck!
6. Member
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Nov 2009
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alright, i just can't figure it out. i'm all out of ideas. here is my code now.
Java Code:
``` import java.util.Scanner;
public class Prog17 {
public static void main (String[] args){
final int totalnum = 10;
double[] num = new double[totalnum];
Scanner input = new Scanner(System.in);
//Get Numbers
for (int i = 0; i <num.length; i++){
System.out.print("Enter a number: ");
num[i] = input.nextDouble();
}
//Calculations for average
double total = 0;
double average = total /num.length;
int counter = 0; //numbers above average
for(int i = 0; i < num.length; i++){
total += (num[i]);
while(num[i] > average){
counter++;
}
}
System.out.print("Numbers above average is: " +counter);
}
}```
7. alright, i just can't figure it out. i'm all out of ideas.
If you continue with a defeatist attitude you may as well give up now and quit this course. Serious.
Java Code:
``` for(int i = 0; i < num.length; i++){
total += (num[i]); // what is this line supposed to be doing?
while(num[i] > average){ // would a while loop or an if block be better here?
counter++;
}
}```
8. Member
Join Date
Nov 2009
Posts
7
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0
ok. well, i spent more time on it, and i guess i just need to calculate things seperately, and it worked a few times. are there any flaws?
Java Code:
``` import java.util.Scanner;
public class Prog17 {
public static void main (String[] args){
final int totalnum = 10;
double[] numList = new double[totalnum]; //Array
Scanner input = new Scanner(System.in);
//Get numbers
for (int i = 0; i <numList.length; i++){
System.out.print("Enter a number: ");
numList[i] = input.nextDouble();
}
//Summing the elements
double total = 0;
for(int i = 0; i < numList.length; i++){
total += numList[i];
}
//Average Calculation
double average = total / numList.length;
int counter = 0; //numbers above average
for(int x = 0; x < numList.length; x ++){
if(numList[x] > average)
counter++;
}
System.out.print("Numbers above average: " + counter);
}
}```
Last edited by glopez09; 11-22-2009 at 03:54 AM.
9. Originally Posted by glopez09
ok. well, i spent more time on it, and i guess i just need to calculate things seperately, and it worked a few times. are there any flaws?
I didn't see any flaws; that version does what it has to do; a (very) minor nitpick: use one variable name consistently when you have a few (non-nested) loops; first you use variable name 'i' when you calculate the total of all numbers and next you use 'x' when you want to find numbers larger than the average; use 'i' in both cases. This doesn't influence the working of your program but using 'i' and 'x' makes the eyebrows of an old sod (like me) fronze. It's not very important though ...
kind regards,
Jos (<--- old nitpicker ;-)
#### Posting Permissions
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# Knuth's Algorithm X
(Redirected from Algorithm X)
"Algorithm X" is the name Donald Knuth used in his paper "Dancing Links" to refer to "the most obvious trial-and-error approach" for finding all solutions to the exact cover problem.[1] Technically, Algorithm X is a recursive, nondeterministic, depth-first, backtracking algorithm. While Algorithm X is generally useful as a succinct explanation of how the exact cover problem may be solved, Knuth's intent in presenting it was merely to demonstrate the utility of the dancing links technique via an efficient implementation he called DLX.[1]
The exact cover problem is represented in Algorithm X using a matrix A consisting of 0s and 1s. The goal is to select a subset of the rows so that the digit 1 appears in each column exactly once.
Algorithm X functions as follows:
If the matrix A has no columns, the current partial solution is a valid solution; terminate successfully. Otherwise choose a column c (deterministically). Choose a row r such that Ar, c = 1 (nondeterministically). Include row r in the partial solution. For each column j such that Ar, j = 1, for each row i such that Ai, j = 1, delete row i from matrix A; delete column j from matrix A. Repeat this algorithm recursively on the reduced matrix A.
The nondeterministic choice of r means that the algorithm essentially clones itself into independent subalgorithms; each subalgorithm inherits the current matrix A, but reduces it with respect to a different row r. If column c is entirely zero, there are no subalgorithms and the process terminates unsuccessfully.
The subalgorithms form a search tree in a natural way, with the original problem at the root and with level k containing each subalgorithm that corresponds to k chosen rows. Backtracking is the process of traversing the tree in preorder, depth first.
Any systematic rule for choosing column c in this procedure will find all solutions, but some rules work much better than others. To reduce the number of iterations, Knuth suggests that the column choosing algorithm select a column with the lowest number of 1s in it.
## Example
For example, consider the exact cover problem specified by the universe U = {1, 2, 3, 4, 5, 6, 7} and the collection of sets $\mathcal{S}$ = {A, B, C, D, E, F}, where:
• A = {1, 4, 7};
• B = {1, 4};
• C = {4, 5, 7};
• D = {3, 5, 6};
• E = {2, 3, 6, 7}; and
• F = {2, 7}.
This problem is represented by the matrix:
1 2 3 4 5 6 7
A 1 0 0 1 0 0 1
B 1 0 0 1 0 0 0
C 0 0 0 1 1 0 1
D 0 0 1 0 1 1 0
E 0 1 1 0 0 1 1
F 0 1 0 0 0 0 1
Algorithm X with Knuth's suggested heuristic for selecting columns solves this problem as follows:
Level 0
Step 1—The matrix is not empty, so the algorithm proceeds.
Step 2—The lowest number of 1s in any column is two. Column 1 is the first column with two 1s and thus is selected (deterministically):
1 2 3 4 5 6 7
A 1 0 0 1 0 0 1
B 1 0 0 1 0 0 0
C 0 0 0 1 1 0 1
D 0 0 1 0 1 1 0
E 0 1 1 0 0 1 1
F 0 1 0 0 0 0 1
Step 3—Rows A and B each have a 1 in column 1 and thus are selected (nondeterministically).
The algorithm moves to the first branch at level 1…
Level 1: Select Row A
Step 4—Row A is included in the partial solution.
Step 5—Row A has a 1 in columns 1, 4, and 7:
1 2 3 4 5 6 7
A 1 0 0 1 0 0 1
B 1 0 0 1 0 0 0
C 0 0 0 1 1 0 1
D 0 0 1 0 1 1 0
E 0 1 1 0 0 1 1
F 0 1 0 0 0 0 1
Column 1 has a 1 in rows A and B; column 4 has a 1 in rows A, B, and C; and column 7 has a 1 in rows A, C, E, and F. Thus rows A, B, C, E, and F are to be removed and columns 1, 4 and 7 are to be removed:
1 2 3 4 5 6 7
A 1 0 0 1 0 0 1
B 1 0 0 1 0 0 0
C 0 0 0 1 1 0 1
D 0 0 1 0 1 1 0
E 0 1 1 0 0 1 1
F 0 1 0 0 0 0 1
Row D remains and columns 2, 3, 5, and 6 remain:
2 3 5 6
D 0 1 1 1
Step 1—The matrix is not empty, so the algorithm proceeds.
Step 2—The lowest number of 1s in any column is zero and column 2 is the first column with zero 1s:
2 3 5 6
D 0 1 1 1
Thus this branch of the algorithm terminates unsuccessfully.
The algorithm moves to the next branch at level 1…
Level 1: Select Row B
Step 4—Row B is included in the partial solution.
Row B has a 1 in columns 1 and 4:
1 2 3 4 5 6 7
A 1 0 0 1 0 0 1
B 1 0 0 1 0 0 0
C 0 0 0 1 1 0 1
D 0 0 1 0 1 1 0
E 0 1 1 0 0 1 1
F 0 1 0 0 0 0 1
Column 1 has a 1 in rows A and B; and column 4 has a 1 in rows A, B, and C. Thus rows A, B, and C are to be removed and columns 1 and 4 are to be removed:
1 2 3 4 5 6 7
A 1 0 0 1 0 0 1
B 1 0 0 1 0 0 0
C 0 0 0 1 1 0 1
D 0 0 1 0 1 1 0
E 0 1 1 0 0 1 1
F 0 1 0 0 0 0 1
Rows D, E, and F remain and columns 2, 3, 5, 6, and 7 remain:
2 3 5 6 7
D 0 1 1 1 0
E 1 1 0 1 1
F 1 0 0 0 1
Step 1—The matrix is not empty, so the algorithm proceeds.
Step 2—The lowest number of 1s in any column is one. Column 5 is the first column with one 1 and thus is selected (deterministically):
2 3 5 6 7
D 0 1 1 1 0
E 1 1 0 1 1
F 1 0 0 0 1
Step 3—Row D has a 1 in column 5 and thus is selected (nondeterministically).
The algorithm moves to the first branch at level 2…
Level 2: Select Row D
Step 4—Row D is included in the partial solution.
Step 5—Row D has a 1 in columns 3, 5, and 6:
2 3 5 6 7
D 0 1 1 1 0
E 1 1 0 1 1
F 1 0 0 0 1
Column 3 has a 1 in rows D and E; column 5 has a 1 in row D; and column 6 has a 1 in rows D and E. Thus rows D and E are to be removed and columns 3, 5, and 6 are to be removed:
2 3 5 6 7
D 0 1 1 1 0
E 1 1 0 1 1
F 1 0 0 0 1
Row F remains and columns 2 and 7 remain:
2 7
F 1 1
Step 1—The matrix is not empty, so the algorithm proceeds.
Step 2—The lowest number of 1s in any column is one. Column 2 is the first column with one 1 and thus is selected (deterministically).
Row F has a 1 in column 2 and thus is selected (nondeterministically).
The algorithm moves to the first branch at level 3…
Level 3: Select Row F
Step 4—Row F is included in the partial solution.
Row F has a 1 in columns 2 and 7:
2 7
F 1 1
Column 2 has a 1 in row F; and column 7 has a 1 in row F. Thus row F is to be removed and columns 2 and 7 are to be removed:
2 7
F 1 1
Step 1—The matrix is empty, thus this branch of the algorithm terminates successfully.
As rows B, D, and F are selected, the final solution is:
1 2 3 4 5 6 7
B 1 0 0 1 0 0 0
D 0 0 1 0 1 1 0
F 0 1 0 0 0 0 1
In other words, the subcollection {B, D, F} is an exact cover, since every element is contained in exactly one of the sets B = {1, 4}, D = {3, 5, 6}, or F = {2, 7}.
There are no more selected rows at level 3, thus the algorithm moves to the next branch at level 2…
There are no more selected rows at level 2, thus the algorithm moves to the next branch at level 1…
There are no more selected rows at level 1, thus the algorithm moves to the next branch at level 0…
There are no branches at level 0, thus the algorithm terminates.
In summary, the algorithm determines there is only one exact cover: $\mathcal{S}^*$ = {B, D, F}.
## Implementations
Donald Knuth's main purpose in describing Algorithm X was to demonstrate the utility of Dancing Links. Knuth showed that Algorithm X can be implemented efficiently on a computer using Dancing Links in a process Knuth calls "DLX". DLX uses the matrix representation of the exact cover problem, implemented as doubly linked lists of the 1s of the matrix: each 1 element has a link to the next 1 above, below, to the left, and to the right of itself. (Technically, because the lists are circular, this forms a torus). Because exact cover problems tend to be sparse, this representation is usually much more efficient in both size and processing time required. DLX then uses Dancing Links to quickly select permutations of rows as possible solutions and to efficiently backtrack (undo) mistaken guesses.[1]
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# The tidal range at a particular location is the difference in height
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Re: The tidal range at a particular location is the difference in height [#permalink]
If the magnitudes of tidal ranges are explained entirely by gravitational forces, then wouldn’t tidal ranges be identical everywhere? Why would the sun’s gravity, and/or the moon’s gravity, act differently in the Bay of Fundy? Something’s fishy here. The argument might look logical, but there’s a bait and switch happening here. We have a premise that says, “The only forces involved in inducing the tides are the sun’s and moon’s gravity,” and a conclusion that says tidal ranges must be entirely explained by gravitational forces. The problem with this is that “inducing” just means “starting.” So this argument basically pulls this switch on us: “Since childbirth can be induced by jumping on a trampoline, childbirth is entirely explained by jumping on a trampoline.” Really? Nah, didn’t think so. I’m not sure how to articulate the flaw precisely, but I do know it has something to do with “inducing” and “entirely explained” not being the same thing.
A) Uh, no. More examples might be nice, but this isn’t what we’re looking for and I wouldn’t call this a “flaw,” per se.
B) This could be it. What’s special about the Bay of Fundy? I don’t know. But maybe the beach is extremely steep, or the bay is extremely shallow, or something, and those are “the conditions in which gravitational forces act” to create such an extreme tidal range. This one’s a keeper.
C) Well, the argument definitely didn’t consider this possibility. But it also didn’t consider the possibility that we don’t give a **** whether it considered the possibility of quite a lot of things, including exactly how things are measured. This just isn’t the point.
D) Does the argument actually do this? For example, does it presume that the mating cry of a Blue Whale in heat (do they go into heat? let’s pretend) is a result of the interplay of gravitational forces? Nah, I don’t think it did.E) Did the argument really need to do this? I don’t see why it would need to, if it was going to use the very broad “gravitational forces” in the conclusion. This is beside the point.
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# De Morgan's laws
(Redirected from DeMorgan's Laws)
De_Morgan's laws (or De_Morgan's theorems) are used to simplify the Boolean expressions.
There are two theorems:
1. The complement of two or more AND variables is equal to the OR of the complements of each variable.
${\displaystyle {\overline {xy}}={\bar {x}}+{\bar {y}}}$
2. The complement of two or more OR variables is equal to the AND of the complements of each variable.
${\displaystyle {\overline {x+y}}={\bar {x}}{\bar {y}}}$
Note that this theorems are true in both direction
## Examples
• ${\displaystyle {\overline {xyz}}={\bar {x}}+{\bar {y}}+{\bar {z}}}$
• ${\displaystyle {\overline {x+y+z}}={\bar {x}}{\bar {y}}{\bar {z}}}$
• Note that x can be a combination of other variables.
lets say ${\displaystyle x=(A+B)}$ and ${\displaystyle y=(C+D)}$ so:
${\displaystyle {\overline {xy}}={\bar {x}}+{\bar {y}}}$
will be:
${\displaystyle {\overline {(A+B)(C+D)}}={\overline {(A+B)}}+{\overline {(C+D)}}={\bar {A}}{\bar {B}}+{\bar {C}}{\bar {D}}}$
• In case we have more than one livile bar (or complement or negate) start with the top bar first
${\displaystyle {\overline {{\overline {(A+B)}}+{\overline {(C+D)}}}}={\overline {\overline {(A+B)}}}.{\overline {\overline {(C+D)}}}=(A+B)(C+D)}$
Note that ${\displaystyle {\overline {\overline {A}}}=A}$ Boolean rule.
• ${\displaystyle {\overline {{\overline {ABC}}+D+E}}={\overline {\overline {ABC}}}{\bar {D}}{\bar {E}}=ABC{\bar {D}}{\bar {E}}}$
• ${\displaystyle {\overline {(A+B){\bar {C}}+D+{\bar {E}}}}=({\overline {(A+B){\bar {C}})}}{\bar {D}}{\overline {\overline {E}}}=({\overline {(A+B)}}+{\overline {\overline {C}}}){\bar {D}}E=({\bar {A}}{\bar {B}}+C){\bar {D}}E}$
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# Orx Learning
### Site Tools
en:tutorials:weld_joint_on_objects_with_bodies
# Weld Joint on parent/child Objects with Bodies
Using objects with children that also have bodies can be tricky if they are required to stay fixed together.
When something collides with the parent or child, they will break apart. This is the expected behaviour. But it may not be the effect you want.
In this tutorial we'll cover two things:
1. Having an object with at least one child, and having a projectile hit both objects, which will separate them.
2. Introducing a weld joint to keep the parent and child together in the case of collisions.
Our test object will be a crude ship that is in two parts. The main part will be the parent, the child will be the front section of the ship. A projectile is fired at the ship. If the projectile hits either part, the ship will separate.
Here are our two ship sprites, and a projectile.
First, we'll start with setting the physics, and the gravity and turning on the bodies for debug:
```[Physics]
Gravity = (0, 0, 0)
ShowDebug = true```
Begin with the main ship config. It will need Object, Graphic, and Body sections:
```[Ship]
Graphic = ShipGraphic
Rotation = 90
Position = (0, 0, 0.5)
Body = ShipBody
[ShipGraphic]
Texture = ship.png
Pivot = center
Smoothing = true
[ShipBody]
Dynamic = true
PartList = ShipBodyPart
LinearDamping = 1
AngularDamping = 2
FixedRotation = false
[ShipBodyPart]
Type = box
Solid = true
SelfFlags = ship
Friction = 1.2```
Next, the child part of the ship which is the front extra cockpit section:
```[ShipChild]
Graphic = ShipConeGraphic
Body = ShipChildBody
Position = (0, -32, 0)
[ShipConeGraphic]
Texture = ship-cone.png
Pivot = center
Smoothing = true
[ShipChildBody]
Dynamic = true
PartList = ShipChildBodyPart
[ShipChildBodyPart]
Type = box
Solid = true
SelfFlags = shipchild
Let's add this child object to the main ship as a child.
```[Ship]
Graphic = ShipGraphic
Rotation = 90
Position = (0, 0, 0.5)
ChildList = ShipChild
Body = ShipBody```
Next step is to define the projectile that will towards the ship from the bottom right of the screen and strike the ship:
```[Projectile]
Graphic = ProjectileGraphic
Position = (200, 200, 0.5)
Body = ProjectileBody
Speed = (-400, -500, 0)
[ProjectileGraphic]
Texture = projectile.png
Pivot = center
Smoothing = true
[ProjectileBody]
Dynamic = true
PartList = ProjectileBodyPart
LinearDamping = 1
AngularDamping = 2
FixedRotation = false
[ProjectileBodyPart]
Type = sphere
Solid = true
SelfFlags = projectile
Friction = 1.2
Density = 20```
The code in the Init() function to create the objects:
``` orxObject_CreateFromConfig("Ship");
orxObject_CreateFromConfig("Projectile");```
Let's run that.
(The image above shows the projectile hitting the ship, but with physics debug off for clarity)
The projectile strikes between the main part of the ship and the child behind, sending both parts off in different directions. Normally this would not occur, but because a body has been added to the child, the two won't stay together if it collides with something.
This may be the effect you are after. If you have a multi-part object, you may wish the parts to break off on contact immediately.
However, if you want the whole ship to stay together during a collision, then you need to use a joint.
In our case we'll use a weld joint which welds two objects together. Or more correctly, welds a child (with a body) to it's parent:
```[WeldJoint]
Type = weld
ParentAnchor = (0, -32, 0)
ChildAnchor = (0, 0, 0)
Collide = false```
Note the Anchor positions. These override any child position that might be defined on the child object.
And in the ship object, add the WeldJoint to a ChildJointList. Where there is an entry in ChildList, there is a matching entry in ChildJointList:
```[Ship]
Graphic = ShipGraphic
Rotation = 90
Position = (0, 0, 0.5)
ChildList = ShipChild
ChildJointList = WeldJoint
Body = ShipBody```
Run again, and the projectile will first strike between the main body of the ship and the child section of the ship, and will send the whole spinning. But the ship stays together as a whole unit.
## Troubleshooting
Avoid using:
`Density = 0`
in a child BodyPart. The effect will be to remove rotation when a projectile hits any of the bodies.
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0
## python 撲克牌程式導讀問題(python100day)[已解決]
player.arrange(get_key)我不太理解get_key明明是一個一般函式為什麼可以直接當arrange的參數使用，get_key在程式裡也沒找到其他地方有帶入參數呼叫和所以我很困惑。
import random
class Card(object):
"""一张牌"""
def __init__(self, suite, face):
self._suite = suite
self._face = face
@property
def face(self):
return self._face
@property
def suite(self):
return self._suite
def __str__(self):
if self._face == 1:
face_str = 'A'
elif self._face == 11:
face_str = 'J'
elif self._face == 12:
face_str = 'Q'
elif self._face == 13:
face_str = 'K'
else:
face_str = str(self._face)
return '%s%s' % (self._suite, face_str)
def __repr__(self):
return self.__str__()
class Poker(object):
"""一副牌"""
def __init__(self):
self._cards = [Card(suite, face)
for suite in '♠♥♣♦'
for face in range(1, 14)]
self._current = 0
@property
def cards(self):
return self._cards
def shuffle(self):
"""洗牌(随机乱序)"""
self._current = 0
random.shuffle(self._cards)
@property
def next(self):
"""发牌"""
card = self._cards[self._current]
self._current += 1
return card
@property
def has_next(self):
"""还有没有牌"""
return self._current < len(self._cards)
class Player(object):
"""玩家"""
def __init__(self, name):
self._name = name
self._cards_on_hand = []
@property
def name(self):
return self._name
@property
def cards_on_hand(self):
return self._cards_on_hand
def get(self, card):
"""摸牌"""
self._cards_on_hand.append(card)
def arrange(self, card_key):
"""玩家整理手上的牌"""
#這裡的排序原理是什麼?key為什麼要是card_key
#card_key裡面不是由花色和數字組成嗎
#那這樣cards_on_hand.sort不就可以對字串做排序了
self._cards_on_hand.sort(key=card_key)
#********************這部分除了arrage有呼叫到以外但呼叫了不用帶入參數呼叫嗎?
# 排序规则-先根据花色再根据点数排序
def get_key(card):
return (card.suite, card.face)
def main():
p = Poker()
p.shuffle()
players = [Player('东邪'), Player('西毒'), Player('南帝'), Player('北丐')]
for _ in range(13):
for player in players:
player.get(p.next)
for player in players:
print(player.name + ':', end=' ')
#***我困惑的地方
player.arrange(get_key)
print(player.cards_on_hand)
if __name__ == '__main__':
main()
https://github.com/jackfrued/Python-100-Days/blob/master/Day01-15/09.%E9%9D%A2%E5%90%91%E5%AF%B9%E8%B1%A1%E8%BF%9B%E9%98%B6.md
### 1 個回答
2
froce
iT邦大師 1 級 ‧ 2021-04-27 13:39:01
https://docs.python.org/zh-tw/3/howto/sorting.html#key-functions
1. 在python函數是頭等公民，可以被當參數傳，而且很多時候會需要
2. 考慮下面的code，請問兩個各叫什麼
def abc(d=""):
return "abc"+d
print(type(abc))
print(type(abc()))
1. 為啥會需要這樣傳？比如爬蟲，我今天爬到後原始資料有20000筆，欄位20個，我不希望用資料庫，但我原始資料和過濾後的資料都要，而且過濾的條件都不同，那我是不是需要一個靈活的方法去過濾資料？
通常這種方式叫 callback。
froce iT邦大師 1 級 ‧ 2021-04-29 08:33:27 檢舉
def get_key(card):
return (card.suite, card.face)
l1 = [1, "2"]
l2 = [Card("♠", 1), Card("♠", 2)]
l1.sort(key=get_key)
l2.sort(key=get_key)
get_key裡並沒有強制限定你參數類型，那個card只是在提示你（撰寫程式者）他吃Card，你list裡要放什麼東西，你要餵它什麼，他管不到，頂多是他消化不良吐error而已。
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| 36,586,051
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# Connexions
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### Lenses
What is a lens?
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#### In these lenses
• GETIntPhaseMaths
This module and collection are included inLens: Siyavula: Mathematics (Gr. 4-6)
By: Siyavula
Module Review Status: In Review
Collection Review Status: In Review
Click the "GETIntPhaseMaths" link to see all content selected in this lens.
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Inside Collection (Course):
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# Om eenvoudige tabelle te gebruik om data in te samel en vrae te beantwoord
Module by: Siyavula Uploaders. E-mail the author
## Inhoud
### Om data te organiseer en op te teken [LU 5.4]
1. Kom ons maak ‘n opname van die resultate wat die klas met die hoofrekentoets behaal het. Voltooi die volgende eenvoudige tabel:
Aantal vrae korrek 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1 Aantal leerders __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __
2. Beantwoord nou die volgende vrae na aanleiding van jou tabel hierbo:
2.1 Hoeveel leerders het die hoofrekentoets geskryf? ______________________
2.2 Hoeveel leerders het volpunte? ____________________________________
2.3 Hoeveel leerders het net 8 vrae korrek beantwoord? ____________________
2.4 Hoeveel vrae het die meeste leerders korrek beantwoord? _______________
2.5 Was dit ‘n maklike hoofrekentoets? _________________________________
Motiveer jou antwoord. ___________________________________________
_____________________________________________________________________
KLASBESPREKING
Gesels eers oor die volgende en kyk of jul antwoorde kan kry.
1. Waarom, dink julle, is dit vir jul skoolhoof belangrik om te weet hoeveel
leerders daar volgende jaar in elke klas gaan wees?
2. Waarom is dit belangrik om te weet hoeveel mense daar in ‘n
sekere dorp woon?
3. Hoe kan bogenoemde inligting bekom word?
## Assessering
Leeruitkomste 5:Die leerder is in staat om data te versamel, op te som, voor te stel en krities te ontleed om gevolgtrekkings en voorspellings te maak, en om toevallige variasies te interpreteer en te bepaal.
Assesseringstandaard 5.2: Dit is duidelik wanneer die leerder eenvoudige dataversamelingsvelle (wat tellings vereis) en eenvoudige vraelyste (met ja/nee tipe antwoorde) gebruik om data te versamel (alleen en/of as ‘n lid van ‘n groep of span) ten einde vrae te beantwoord wat deur die onderwyser, self of die groep gevra word;
Assesseringstandaard 5.4: Dit is duidelik wanneer die leerder data organiseer en aanteken deur gebruik te maak van tellings en tabelle.
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#### Collection to:
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##### What is in a lens?
Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.
##### Who can create a lens?
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# Paulos: How to Calculate Chances of Doomsday
• Star
Is the sky falling? And if so, when? Even when they're baseless, constant reports about nuclear weapons proliferation, pandemic diseases and environmental catastrophes revive these perennial human questions and contribute to a feeling of unease.
So too did the recent passing of an asteroid almost 100 feet in diameter within 30,000 miles of the Earth. Such news stories make a recent abstract philosophical argument a bit more real.
Developed by a number of people including Oxford philosopher Nick Bostrom and Princeton physicist J. Richard Gott, the Doomsday Argument (at least one version of it) goes roughly like this.
#### How Many Balls in the Lottery Machine?
There is a large lottery machine in front of you, and you're told that in it are consecutively numbered balls, either 10 of them or 10,000 of them. The machine is opaque so you can't tell how many balls are in it, but you're fairly certain that there are a lot of them. In fact, you initially estimate the probability of there being 10,000 balls in the machine to be about 95 percent , of there being only 10 balls in it to be about 5 percent.
Now the machine rolls, you open a little door on its side, and a randomly selected ball rolls out. You see that it is ball number 8 and you place it back into the lottery machine. Do you still think there is only a 5 percent chance that there are 10 balls in the machine?
Given how low a number 8 is, it seems reasonable to think that the chances of there being only 10 balls in the machine are much higher than your original estimate of 5 percent. Given the assumptions of the problem, in fact, we can use a bit of mathematics called Bayes' theorem to conclude that your estimate of the probability of 10 balls being in the machine should be revised upward from 5 percent to 98 percent. Likewise, your estimate of the probability of 10,000 balls being in it should be revised downward from 95 percent to 2 percent.
#### How Many People in the Cosmic Lottery Machine?
What does this have to do with Doomsday? To see, let's imagine a cosmic lottery machine, which contains the names and birth orders of all human beings from the past, present, and the future in it. Let's say we know that this machine contains either 100 billion names or — the optimistic scenario — 100 trillion names.
And how do we pick a human at random from the set of all humans? We simply consider ourselves; we argue that there's nothing special about us or about our time and that any one of us might be thought of as a randomly selected human from the set of all humans, past, present, and future. (This part of the argument can be much more fully developed.)
If we assume there have been about 80 billion humans so far (the number is simply for ease of illustration), the first alternative of 100 billion humans corresponds to a relatively imminent end to humankind — only 20 billion more of us to come before extinction. The second alternative of 100 trillion humans corresponds to a long, long future before us.
Even if we initially believe that we have a long, long future before us, when we randomly select a person's name from the machine and the person's birth order is only 80 billion or so, we should re-examine our beliefs. We should drastically reduce, or so the argument counsels, our estimate of the likelihood of our long survival, of there ultimately being 100 trillion of us.
The reason is the same as in the example with the lottery balls: The relatively low number of 8 (or 80 billion) suggests that there aren't many balls (human names) in the machine.
#### Many Refinements
Here's another slightly different example. Let's assume that Al receives about 20 e-mails per day, whereas Bob averages about 2,000 per day. Someone picks one of their accounts, chooses an e-mail at random from it, and notes that the e-mail is the 14th one received in the account that day. From whose account is the e-mail more likely to have come?
There are many other examples devised to shore up the numerous weak points in the Doomsday Argument. Surprisingly, many of them can be remedied, but a few of them, in my opinion, cannot be.
That a prehistoric man (who happened to understand Bayes theorem in probability) could make the same argument about a relatively imminent extinction is an objection that can be nicely addressed. Appealing to some so-called anthropic principle whereby inferences are drawn from the mere fact that there are observers to draw them is much more problematic.
In any case, there's probably still time to learn more about the Doomsday Argument and the use of the so-called anthropic principle in philosophy, cosmology and even everyday life. A good place to begin is Nick Bostrom's work, particularly his book, Anthropic Bias.
Professor of mathematics at Temple University and winner of the 2003 American Association for the Advancement of Science award for the promotion of public understanding of science, John Allen Paulos is the author of several best-selling books, including Innumeracy and A Mathematician Plays the Stock Market. His Who’s Counting? column on ABCNEWS.com appears the first weekend of every month.
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# Remove duplicates from integer array
I having a problem with coding this:
Write a static method named `removeDuplicates` that takes as input an array of integers and returns as a result a new array of integers with all duplicates removed. For example, if the input array has the elements {4, 3, 3, 4, 5, 2, 4} the resulting array should be {4, 3, 5, 2}
Here's what I have done so far
``````public static int[] removeDuplicates(int []s){
int [] k = new int[s.length];
k[0]=s[0];
int m =1;
for(int i=1;i<s.length;++i){
if(s[i]!=s[i-1]){
k[m]=s[i];
++m;
}//endIF
}//endFori
return k;
}//endMethod
``````
-
The easy way would be to add the elements to a set (which will remove duplicates automatically for you) and put the numbers back in an array. – assylias Dec 17 '12 at 10:11
possible duplicate of What is the best way to remove duplicates in an Array in Java? among others ... – Brian Roach Dec 17 '12 at 10:18
You didn't state the constraints on your implementation, and I'll bet there are many, since otherwise the solution is trivial. – Marko Topolnik Dec 17 '12 at 10:36
actually I can not use Set or HashSet , it must be done with loops and simple arrays – Ali-J8 Dec 17 '12 at 10:43
try this -
``````public static int[] removeDuplicates(int []s){
int result[] = new int[s.length], j=0;
for (int i : s) {
if(!isExists(result, i))
result[j++] = i;
}
return result;
}
private static boolean isExists(int[] array, int value){
for (int i : array) {
if(i==value)
return true;
}
return false;
}
``````
-
Looks like there a mistake here. Instead of if(isExists(result, i)) it must be if(!isExists(result, i)). – nakosspy May 20 '13 at 23:53
The only solution that should save order. But two problems there: 1) will skip 0's 2) there will be 0's in the end in case of duplicates. Both of them quite easy to solve, however – RiaD May 20 '13 at 23:57
To Preserve the ordering and to remove duplicates in the integer array, you can try this:
``````public void removeDupInIntArray(int[] ints){
for(int i=0;i<ints.length;i++){
}
System.out.println(setString);
}
``````
Hope this helps.
-
Maybe you can use lambdaj (download here,website), this library is very powerfull for managing collections (..list,arrays), the following code is very simple and works perfectly:
``````import static ch.lambdaj.Lambda.selectDistinct;
import java.util.Arrays;
import java.util.List;
public class DistinctList {
public static void main(String[] args) {
List<Integer> numbers = Arrays.asList(1,3,4,2,1,5,6,8,8,3,4,5,13);
System.out.println("List with duplicates: " + numbers);
System.out.println("List without duplicates: " + selectDistinct(numbers));
}
}
``````
This code shows:
``````List with duplicates: [1, 3, 4, 2, 1, 5, 6, 8, 8, 3, 4, 5, 13]
List without duplicates: [1, 2, 3, 4, 5, 6, 8, 13]
``````
In one line you can get a distinct list, this is a simple example but with this library you can resolve more.
``````selectDistinct(numbers)
``````
You must add lambdaj-2.4.jar to your project. I hope this will be useful.
-
What you have to do is , you have to check for each element in second array whether previous element is already present or not.
You can use better approach Use HashSet and return set.
``````public static Set removeDuplicates(int []s){
Set<Integer> set = new HashSet<Integer>();
for(int i=0;i<s.length;++i){
}//endFori
return set;
}//endMethod
``````
If you need int Array than take a look of this java-hashsetinteger-to-int-array link.
-
Iterate over the array and populate a set because sets cannot contain duplicates. Then copy the elements from the set into a new array and return it. This is shown below:
``````public static int[] removeDuplicates(int[] array) {
// add the ints into a set
Set<Integer> set = new HashSet<Integer>();
for (int i = 0; i < array.length; i++) {
}
// copy the elements from the set into an array
int[] result = new int[set.size()];
int i = 0;
for (Integer u : set) {
result[i++] = u;
}
return result;
}
``````
-
Try this
``````public static int[] removeDuplicates(int[] s) {
Integer[] array = new HashSet<Integer>(Arrays.asList(ArrayUtils.toObject(s))).toArray(new Integer[0]);
return ArrayUtils.toPrimitive(array);
}
``````
Edit: Updated with Apache Lang to convert to primitives.
-
That would give an `Integer[]` array, though. – Anders R. Bystrup Dec 17 '12 at 10:16
Edited the reply. Updated to primitive data type. – Jayamohan Dec 17 '12 at 10:33
You can also use Google's Guava library and use ImmutableSet do
``````ImmutableSet.copyOf(myArray).asList();
``````
-
``````public class Test
static int[] array = {4, 3, 3, 4, 5, 2, 4};
static HashSet list = new HashSet();
public static void main(String ar[])
{
for(int i=0;i<array.length;i++)
{
}
System.out.println(list);
}}
``````
The Output is : `[2, 3, 4, 5]`
-
-1: "the resulting array should be {4, 3, 5, 2}" – Thilo May 20 '13 at 23:18
You could also put the array elements into a `Set` for which the semantics precisely are that it contains no duplicate elements.
-
`Set set = new HashSet( Arrays.asList( s ) )` it will not compile. – Subhrajyoti Majumder Dec 17 '12 at 10:19
-1 Set is a raw type. References to generic type Set<E> should be parameterized. – dogbane Dec 17 '12 at 10:35
Your edit does not solve the issue, you can't transform an `int[]` to an `Integer[]` with `Arrays.asList()`. – assylias Dec 17 '12 at 11:36
You can do naively though. First you need to sort the array. You can do it using any of sorting algorithms. I did use quick sort. And then check a position with its next position. If they are not same, add value in a new array, otherwise skip this iteration.
Sample Code (Quick Sort):
`````` public static void quickSort(int[] array, int low, int high) {
int i = low;
int j = high;
int pivot = array[low + (high - low) / 2];
while (i <= j) {
while (array[i] < pivot) i++;
while (array[j] > pivot) j--;
if (i <= j) {
exchange(array, i, j);
i++;
j--;
}
}
if (0 < j) quickSort(array, 0, j);
if (i < high) quickSort(array, i, high);
}
public static void exchange(int[] array, int i, int j) {
int temp = array[i];
array[i] = array[j];
array[j] = temp;
}
``````
Remove duplicates:
`````` public static int[] removeDuplicate(int[] arrays) {
quickSort(arrays, 0, arrays.length - 1);
int[] newArrays = new int[arrays.length];
int count = 0;
for (int i = 0; i < arrays.length - 1; i++) {
if (arrays[i] != arrays[i + 1]) {
newArrays[count] = arrays[i];
count++;
}
}
return newArrays;
}
``````
-
Throwing "quicksort" into this is not helpful. If you are going to sort at all, the correct (Java) solution is to use `Arrays.sort(int[])`. And there are simpler ways to solve the problem that don't involve sorting. – Stephen C Dec 29 '13 at 1:43
Besides, this reorders the remaining elements of the array, and the Question requires the order to be preserved. – Stephen C Dec 29 '13 at 1:47
You can use HashSet that does not allow dulplicate elements
``````public static void deleteDups(int a []) {
HashSet<Integer> numbers = new HashSet<Integer>();
for(int n : a)
{
}
for(int k : numbers)
{
System.out.println(k);
}
System.out.println(numbers);
}
public static void main(String[] args) {
int a[]={2,3,3,4,4,5,6};
RemoveDuplicate.deleteDups(a);
}
}
o/p is 2
3
4
5
6
``````
[2, 3, 4, 5, 6]
-
``````public int[] removeRepetativeInteger(int[] list){
if(list.length == 0){
return null;
}
if(list.length == 1){
return list;
}
ArrayList<Integer> numbers = new ArrayList<>();
for(int i = 0; i< list.length; i++){
if (!numbers.contains(list[i])){
}
}
Iterator<Integer> valueIterator = numbers.iterator();
int[] resultArray = new int[numbers.size()];
int i = 0;
while (valueIterator.hasNext()) {
resultArray[i] = valueIterator.next();
i++;
}
return resultArray;
}
``````
-
``````public class DistinctNumbers{
public static void main(String[] args){
java.util.Scanner input = new java.util.Scanner(System.in);
System.out.print("Enter ten numbers: ");
int[] numbers = new int[10];
for(int i = 0; i < numbers.length; ++i){
numbers[i] = input.nextInt();
}
System.out.println("The distinct numbers are:");
System.out.println(java.util.Arrays.toString(eliminateDuplicates(numbers)));
}
public static int[] eliminateDuplicates(int[] list){
int[] distinctList = new int[list.length];
boolean isDuplicate = false;
int count = list.length-1;
for(int i = list.length-1; i >= 0; --i){
isDuplicate = false;
for(int j = i-1; j >= 0 && !isDuplicate; --j){
if(list[j] == list[i]){
isDuplicate = true;
}
}
if(!isDuplicate){
distinctList[count--] = list[i];
}
}
int[] out = new int[list.length-count-1];
System.arraycopy(distinctList, count+1, out, 0, list.length-count-1);
return out;
}
}
``````
-
First of all, you should know length without duplicates(dups): initial length minus number of dups. Then create new array with right length. Then check each element of list[] for dups, if dup founded - check next element, if dup not founded - copy element to new array.
``````public static int[] eliminateDuplicates(int[] list) {
int newLength = list.length;
// find length w/o duplicates:
for (int i = 1; i < list.length; i++) {
for (int j = 0; j < i; j++) {
if (list[i] == list[j]) { // if duplicate founded then decrease length by 1
newLength--;
break;
}
}
}
int[] newArray = new int[newLength]; // create new array with new length
newArray[0] = list[0]; // 1st element goes to new array
int inx = 1; // index for 2nd element of new array
boolean isDuplicate;
for (int i = 1; i < list.length; i++) {
isDuplicate = false;
for (int j = 0; j < i; j++) {
if (list[i] == list[j]) { // if duplicate founded then change boolean variable and break
isDuplicate = true;
break;
}
}
if (!isDuplicate) { // if it's not duplicate then put it to new array
newArray[inx] = list[i];
inx++;
}
}
return newArray;
}
``````
-
public class Foo {
``````public static void main(String[] args) {
//example input
int input[] = new int[]{1, 6 , 5896, 5896, 9, 100,7, 1000, 8, 9, 0, 10, 90, 4};
//use list because the size is dynamical can change
List<Integer> result = new ArrayList<Integer>();
for(int i=0; i<input.length; i++)
{
boolean match = false;
for(int j=0; j<result.size(); j++)
{
//if the list contains any input element make match true
if(result.get(j) == input[i])
match = true;
}
//if there is no matching we can add the element to the result list
if(!match)
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# Interactions of Energy and Matter: Dawn Instrumentation
Become a crime scene investigator! Learners model Dawn Mission scientists, engineers, and technologists and how they use instrumentation to detect distant worlds. After a briefing to build context, students explore interactions between different... (View More)
# Paper Model of Comet ISON's Orbit
This paper model shows the orbit of Comet ISON (late 2013) with respect to the innermost planets of the solar system. After reading background information about comets - how they form and where they come from - students cut out and tape together the... (View More)
# Dusty Dilemma
Learners will be introduced to the concepts of error analysis, including standard deviation. They will apply the knowledge of averages (means), standard deviation from the mean, and error analysis to their own classroom distribution of heights. They... (View More)
Learners will relate the concept of density to the density of dust in space. They will use mission data from the Student Dust Counter (SDC) data viewer to determine the density of dust grains in a volume of space in order to answer questions... (View More)
# How High Up is Space?
Using the diameter of a pencil as a reference, students calculate and construct a line chart to show the relative height of several altitudinal points such as Earth's atmosphere, the beginning of space, commercial airplane flights, and the Hubble... (View More)
# Measuring Solar Rotation - Solar Latitude/Longitude Method
This is an activity about the period of the Sun's rotation. Learners will use image of the Sun from the SOHO spacecraft and a transparent latitude/ longitude grid called a Stonyhurst Disk to track the motion of sunspots in terms of degrees of... (View More)
Audience: Elementary school, Middle school, High school
Materials Cost: Free
# Making your own 3-D images
This activity provides instructions to create 3-D images using a digital camera, photo editing software, and red-blue 3-D glasses.
# The Nature of Salt
This is a hands-on lab activity about the composition of salt. Learners will explain the general relationship between an element's Periodic Table Group Number and its tendency to gain or lose electron(s), and explain the difference between molecular... (View More)
# Student Derived KP Index
This is an activity about the Kp index, a quantification of fluctuations in the Earth's magnetic field due to the relative strength of a magnetic storm. Learners will take a reading from a magnetometer site and make a Kp index estimate to predict... (View More)
Audience: High school, Higher education
Materials Cost: Free
# THEMIS Magnetometer Line-Plots
This is an activity about the properties and characteristics of Earth’s magnetic field as shown through magnetometer data and its 3D vector nature. This resource builds understanding of conceptual tools such as the addition of vectors and... (View More)
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# Search results
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Epicyclic gearing (overal gear ratio)
An epicyclic gear train consists of two gears mounted so that the center of one gear revolves around the center of the other. A carrier connects the ... more
Pitch diameter - in imperial units (gears)
A gear or cogwheel is a rotating machine part having cut teeth, or cogs, which mesh with another toothed part to transmit torque, in most cases with teeth ... more
Diametral pitch (transverse) - gears
A gear or cogwheel is a rotating machine part having cut teeth, or cogs, which mesh with another toothed part to transmit torque, in most cases with teeth ... more
Mechanical advantage is a measure of the force amplification achieved by using a tool, mechanical device or machine system. Ideally, the device preserves ... more
Simple gear train with two gears (gear ratio in terms of radii of the gears and number of teeth)
The gear ratio of a gear train, also known as its speed ratio, is the ratio of the angular velocity of the input gear to the angular velocity of the output ... more
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The gear ratio of a gear train, also known as its speed ratio, is the ratio of the angular velocity of the input gear to the angular velocity of the output ... more
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The gear ratio of a gear train, also known as its speed ratio, is the ratio of the angular velocity of the input gear to the angular velocity of the output ... more
Backlash (due to tooth thickness changes)
In mechanical engineering, backlash, sometimes called lash or play, is clearance or lost motion in a mechanism caused by gaps between the parts. It can be ... more
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In mechanical engineering, backlash, sometimes called lash or play, is clearance or lost motion in a mechanism caused by gaps between the parts. It can be ... more
Pitch diameter - in metric units (gears)
Is a predefined diametral position on the gear, where the circular tooth thickness, pressure angle and helix angles are defined. The standard pitch ... more
Semi-Elliptic Laminated Leaf Spring (Stiffness)
Leaf spring, commonly used for the suspension in wheeled vehicles. The term is also used to refer to a bundled set of leaf springs. As the spring flexes, ... more
Torque - to overcome rolling resistance
Torque, is the tendency of a force to rotate an object about an axis, fulcrum, or pivot. Just as a force is a push or a pull, a torque can be thought of as ... more
Speeds and feeds - feed rate
The phrase speeds and feeds or feeds and speeds refers to two separate velocities in machine tool practice, cutting speed and feed rate. They are often ... more
Critical Speed of a Rotating Shaft - Rayleigh–Ritz method
In solid mechanics, in the field of rotordynamics, the critical speed is the theoretical angular velocity that excites the natural frequency of a rotating ... more
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Mechanical components – as transmission gear and bearings – generates a mechanical loss that reduces the power transferred from the motor shaft ... more
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In physics, circular motion is a movement of an object along the circumference of a circle or rotation along a circular path. It can be uniform, with ... more
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In physics, circular motion is a movement of an object along the circumference of a circle or rotation along a circular path. It can be uniform, with ... more
Critical Speed of a Rotating Shaft - Dunkerley's method
In solid mechanics, in the field of rotordynamics, the critical speed is the theoretical angular velocity that excites the natural frequency of a rotating ... more
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Thrust is a reaction force described quantitatively by Newton’s second and third laws. When a system expels or accelerates mass in one direction, the ... more
Gravitational wave - Binaries (Orbital lifetime)
Gravitational waves are disturbances in the curvature (fabric) of spacetime, generated by accelerated masses, that propagate as waves outward from their ... more
Beamwidth - Parabolic Antenna
The beam diameter or beam width of an electromagnetic beam is the diameter along any specified line that is perpendicular to the beam axis and intersects ... more
In the physical sciences, the wavenumber (also wave number) is the spatial frequency of a wave, either in cycles per unit distance or radians per unit ... more
Sommerfeld Number
In the design of fluid bearings, the Sommerfeld number (S), or bearing characteristic number, is a dimensionless quantity used extensively in hydrodynamic ... more
Petroff's Law - Bearing coefficient of friction
In the design of fluid bearings, the Sommerfeld number (S), or bearing characteristic number, is a dimensionless quantity used extensively in hydrodynamic ... more
Ball Screw - Tensile Compressive Load
A ball screw is a mechanical linear actuator that translates rotational motion to linear motion with little friction. A threaded shaft provides a helical ... more
Petroff's Law - Torque required to shear the lubricant film
In the design of fluid bearings, the Sommerfeld number (S), or bearing characteristic number, is a dimensionless quantity used extensively in hydrodynamic ... more
A ball screw is a mechanical linear actuator that translates rotational motion to linear motion with little friction. A threaded shaft provides a helical ... more
A ball screw is a mechanical linear actuator that translates rotational motion to linear motion with little friction. A threaded shaft provides a helical ... more
Ball Screw - Preload Drag Torque
A ball screw is a mechanical linear actuator that translates rotational motion to linear motion with little friction. A threaded shaft provides a helical ... more
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A ball screw is a mechanical linear actuator that translates rotational motion to linear motion with little friction. A threaded shaft provides a helical ... more
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# A gun fires a bullet of mass 50g with a velocity of 30m/s. Because of this, the gun is pushed back with a velocity of 1m/s. The mass of the gun is?
A gun fires a bullet of mass 50g with a velocity of 30m/s. Because of this, the gun is pushed back with a velocity of 1m/s. The mass of the gun is?
(a) 5.5kg
(b) 3.5kg
(c) 1.5kg
(d) 0.5kg
This question was posed to me by my school teacher while I was bunking the class.
This interesting question is from Moment of Inertia in chapter System of Particles and Rotational Motion of Engineering Physics I
+1 vote
by (1.4m points)
selected by
The best explanation: By conservation of momentum, MV = mv
M = mv/V = 1.5kg.
+1 vote
+1 vote
+1 vote
+1 vote
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Get Full Access to An Introduction To Thermal Physics - 1 Edition - Chapter 6 - Problem 1p
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# Consider a system of two Einstein solids, where the first
ISBN: 9780201380279 40
## Solution for problem 1P Chapter 6
An Introduction to Thermal Physics | 1st Edition
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Problem 1P
Consider a system of two Einstein solids, where the first “solid” contains just a single oscillator, while the second solid contains 100 oscillators. The total number of energy units in the combined system is fixed at 500. Use a computer to make a table of the multiplicity of the combined system, for each possible value of the energy of the first solid from 0 units to 20. Make a graph of the total multiplicity vs. the energy of the first solid, and discuss, in some detail, whether the shape of the graph is what you would expect. Also plot the logarithm of the total multiplicity, and discuss the shape of this graph.
Step-by-Step Solution:
Step 1 of 3
Nutrition 3-7 Carbohydrates 3 Different Types of Carbohydrates Simple: 1) Monosaccharide a. Glucose – the fuel of almost all cells (brain cells, red blood cells and retina cells) b. Fructose – found in fruit, honey, melon, berries, root vegetables c. Galactose – absorbed differently than fructose and glucose 2) Disaccharide a. Sucrose (glucose and fructose) (table sugar) b. Lactose (galactose and glucose) (milk) c. Maltose (glucose and glucose) (beer) Complex: 3) Polysaccharide a. Starch – plant form of glucose storage b. Glycogen – animal form of glucose storage, storage in muscle, liver and a little in the brain c. Cellulose – insoluble fiber Fiber Sol
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http://cboard.cprogramming.com/cplusplus-programming/65684-fibonacci-10-000-a-printable-thread.html
| 1,469,508,658,000,000,000
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| 38,450,130
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# fibonacci(10,000)
• 05-19-2005
blindman858
fibonacci(10,000)
Greetings,
Ok I know how to code fibonacci using both iterative and recursive functions. However, I have no idea how to go about calculating the 10,000 fibonacci. That is a very very very big number. I think I may have to use an array of sometype to solve this because i dont think visual studios has a data type that can handle such a gigantic number. But I dont know how to begin. If any you could provide insight and suggestions I would greatly appreciate it. Thanks.
• 05-19-2005
Salem
Use a bignum library - http://www.swox.com/gmp/
• 05-19-2005
MathFan
Take a look at this page http://www.codearchive.com/list.php?go=0716. - almost at the bottom of it. Here you can download source code of an example program that can handle really large fibonacci numbers.
Here is some more source code you can look at http://sourceforge.net/project/showf...kage_id=125989
Hope this helps :)
• 05-19-2005
Thantos
Quote:
Originally Posted by blindman858
Greetings,
Ok I know how to code fibonacci using both iterative and recursive functions. However, I have no idea how to go about calculating the 10,000 fibonacci. That is a very very very big number. I think I may have to use an array of sometype to solve this because i dont think visual studios has a data type that can handle such a gigantic number. But I dont know how to begin. If any you could provide insight and suggestions I would greatly appreciate it. Thanks.
To calculate a fibonacci of this size you don't use an iterative or recursive approach. Instead you find the formula that lets you solve directly.
http://en.wikipedia.org/wiki/Fibonacci_number
And of course you use bignum libraries :)
• 05-20-2005
xErath
I've already done this long time ago
Code:
```#include<stdio.h> #define SIZE_NUM 1000000000 #define VECTOR_DIM 234//enough to hold fib(10000) /*MACRO output number*/ #define display_arr(vec)\ /*display the array code*/ int main() { int i=-1, opt; int indfin=1, ref=1; unsigned int n0[VECTOR_DIM] = {0}, n1[VECTOR_DIM] = {1}; unsigned int temp; scanf("%d", &opt); if(!opt) printf("The Fibonacci number for 0 is 0"); else{ while(i++<opt){/*sum cicle*/ /*...............*/ } /* display*/ printf("The Fibonacci number for %d is ", opt); if(ref) display_arr(n0) else display_arr(n1) } putchar('\n'); getchar(); return 0; }```
Beg for the full source if you want;)
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https://www.gradesaver.com/textbooks/math/calculus/calculus-early-transcendentals-2nd-edition/chapter-7-integration-techniques-7-1-basic-approaches-7-1-exercises-page-514/9
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## Calculus: Early Transcendentals (2nd Edition)
$= \frac{1}{{2\sqrt 2 }}$
$\begin{gathered} \int_0^{\frac{3}{{\,\pi }}} {\sin \,\,\,\,\left( {2x - \frac{\pi }{4}} \right)\,\,\,dx} \hfill \\ \hfill \\ substituting\,\,\, \hfill \\ \hfill \\ \,\left( {2x - \frac{\pi }{4}} \right) = u \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ 2dx = du\,\,or\,\,dx = \frac{{du}}{2}{\text{ }}limits\,\,also\,\,\,change\,\,as \hfill \\ \hfill \\ lower\,\,li\,mit\,\, = \,2\,\left( 0 \right) - \frac{\pi }{4} = - \frac{\pi }{4} \hfill \\ \hfill \\ upper\,\,\,li\,mit\, = \,2\,\left( {\frac{{3\pi }}{8}} \right) - \frac{\pi }{4} = \frac{\pi }{2} \hfill \\ \hfill \\ = \frac{1}{2}\int_{ - \pi /4}^{\pi /2} {\,\sin \,u\,du} \hfill \\ \hfill \\ integrate\,\,\,and\,evalute\,the{\text{ limits}} \hfill \\ \hfill \\ = \frac{1}{2}\left[ { - \cos u} \right]_{ - \pi /4}^{\pi /2} \hfill \\ \hfill \\ = - \frac{{\,\,\left[ {\cos \,u} \right]_{ - \frac{4}{\pi }}^{\frac{\pi }{2}}}}{2} = \frac{{\,\left( {0 - \frac{1}{{\sqrt 2 }}} \right)}}{2} \hfill \\ \hfill \\ = \frac{1}{{2\sqrt 2 }} \hfill \\ \end{gathered}$
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http://www.onlinemathlearning.com/double-angle-identities.html
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# Double Angle Identities
Related Topics:
More Lessons for PreCalculus
Math Worksheets
Videos, worksheets, games and activities to help PreCalculus students learn how to derive the double angle identities and how to use them to prove identities.
sin 2A = 2 sin A cos A
cos 2A = cos2 A − sin2 A, cos 2A = 2cos2 A − 1, cos 2A = 1 − 2sin2 A
tan 2A = 2 tan A / (1 − tan2 A)
How to Understand Double Angle Identities Based on the sum formulas for trig functions, double angle formulas occur when alpha and beta are the same. Take a look at how to simplify and solve different double-angle problems that might occur on your test.
This video shows you the basics of Double Angle Trig Formulas. It will show you where they come from and how to use them.
sin 2A = 2 sin A cos A
cos 2A = cos2 A − sin2 A, cos 2A = 2cos2 A − 1, cos 2A = 1 − 2sin2 A
This video shows you how to use double angle formulas to prove identities as well as derive and use the double angle tangent identity.
tan 2A = 2 tan A / (1 − tan2 A)
The derivation of the double angle identities for sine and cosine, followed by some examples.
Using Double Angle Identities to Solve Equations, Example 1.
This video uses some double angle identities for sine and/or cosine to solve some equations.
Using Double Angle Identities to Solve Equations, Example 2.
This video uses some double angle identities for sine and/or cosine to solve some equations.
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget which allows your to practice solving Algebra, Trigonometry, Calculus and other math topics.
You can use the Mathway widget below to practice Trigonometry or other math topics. Try the given examples, or type in your own problem. Then click "Answer" to check your answer.
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https://dailydevsblog.com/python/learn-python-python-program-to-find-compound-interest-basic-and-advance-3275/
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# Learn Python – Python program to find compound interest- Basic and advance
In this tutorial, we will learn how to write a Python program to find compound interest for a given value. Before we proceed with writing the program, let’s first under the basics of compound interest.
## Compound interest
The addition of interest in the given principal value for the deposit or loan is called compound interest, and it is also known as interest on interest. Compound interest is basically the result of reinvesting the gained interest on principal value, rather than taking or paying it out, which will result in the next period of interest; the paid interest will be added to the principal amount.
Let’s compare compound interest and simple interest. We can see that, in simple interest, there is no compounding of interest on principal values, and gained interest remains the same on every cycle.
Fact: Compound interest is the standard method of interest calculation in banking, finance, and economics.
The formula for compound interest:
The general mathematical formula for calculating compound interest on a given principal value, i.e., V, is given below:
Total Amount = P(1+r/100)tp
In the above formula, the variables we have used in it can be described as given below:
Total amount = Principal value + Compound interest gained
P = Principal value
tp = Time period for which Principal value is invested
r = Rate of interest
## Code Implementation
Till now, we have learned the basics of compound interest and how important it is in our daily life. We have also learned about the basic formula that we can use for calculating the Compound interest for a certain principal value.
Now, in this section, we will write a Python program to calculate the compound interest for a given certain value. For writing the required Python program, we have to follow certain steps, which are given below:
Step 1: We will take the principal amount as user input.
Step 2: Then, we ask the user to set the rate of interest and the time period for which the principal amount is invested.
Step 3: After taking all the three required variables from the user, we will use the compound interest formula, i.e., “Total Amount = P(1+r/100)tp,” on these variables in the program.
Step 4: We will store the result in a ‘Result variable’.
Step 5: In the last, we will print the compound interest as a result in the output of the program.
Now, look at the following Python program to better understand the implementation of the above given steps:
Example –
``````# Using default function
def compound_rate(PV, CRate, tp):
# Using CI formula
TotalAmount = PV * (pow ((1 + CRate / 100), tp))
# Calculating CI
CInterest = TotalAmount - PV
# Printing CI as result in output
print("Total return value after completion of given time period: ", TotalAmount)
print("Compound interest gained on given amount is", CInterest)
# Taking principal amount value from the user
PV = float(input("Enter the principal amount: "))
CRate = float(input("Enter the rate for compound interest: ")) # taking interest rate value
tp = float(input("Enter the time period for which principal is invested: ")) # taking time period value
# Calling out CI function
compound_rate(PV, CRate, tp)
``````
Output:
``````Enter the principal amount: 600000
Enter the rate for compound interest: 2.7
Enter the time period for which principal is invested: 20
Total return value after completion of given time period: 1022257.0687807774
Compound interest gained on given amount is 422257.0687807774``````
Explanation –
After running the above program, we have given three required variables, i.e., PV = 600000, Crate = 2.7, and tp = 20; we got the total compound interest and total value (1022257.0687807774) that we will get on the given principal value and printed the result in the output. We can calculate the compound interest on any given amount for any rate of interest and for any time period using this program.
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# What is the purpose of the Tailoring PRINCE2 in PRINCE2?
## What is the purpose of the Tailoring PRINCE2 in PRINCE2?
What is the purpose of the Tailoring PRINCE2 in PRINCE2? The purpose of PRINCE1 is to help you make your program more efficient. It is the application of PRINce2 to PRINCE3. In the following table we have the following three data types. This data type is used to define the structure of the PRINCE program. The data types used in PRINce3 are: PRINCE1 PRICE PRIZE PROBE3 PRO_CODE PROWNTIME PROUND_TIME PROF_TIME What is the purpose of the Tailoring PRINCE2 in PRINCE2? In the context of the PRINCE1 and PRINCE5 chapters, PRINCE3 is the basis of a study of the functional properties of PRINCE. We begin by summarising the main results of this study. Describe the functional properties and their relation to the structural properties Abstract The functional properties are the main properties that are expressed in the functional properties. These are the fundamental properties of the functional features, such as the structural properties of the material, as well as the properties of the atoms. The functional properties are also the properties that are necessary for the functional properties to be expressed. The definition of functional properties are given below: Functional properties Basic properties The basic properties of the structural properties are the structural properties, such as volume, specific surface area, and volume fraction. The structural properties of materials are the structural characteristics of why not try these out materials. An important look these up of the structural characteristics is the volume fraction. Regarding the structural properties in particular, the structural properties include the volume fraction of the materials, the specific surface area of the materials and the specific volume of the material. For example, the volume fraction can be expressed as follows: Using the volume formula, the volume of the materials is expressed as follows. This volume fraction will be expressed as volume fractions of the material surface area and volume fraction of materials. A more detailed description of the functionalities of materials is given below. Functionals for the structural properties: Structural properties Structures of the structural elements are the structural elements. The structural elements are a type of material, such as glass, ceramic, metal, ceramic fiber, or metal, and a type of alloy. For example if the material is a steel, then the structural element is a steel alloy. Structuring properties A structural element is the type of material that fixes the structural element.
Structuring properties include the structural properties that are required for the structural elements to have a specific structure, such as a structural element that fixes the structure of the material in a specific way. Structuring property of the material is the structural properties required to be expressed to have a certain structure. In general terms, the structural elements can be classified into three types: structural elements that fix the structure of an element; structural elements that remove the structure of a material; and structural elements that retain the structure of material used to fix the structure. The structural elements are defined in terms of the overall structure of the element. The structural element is defined as the element that fixes a structure, such that it is a part of the structural element, and the structure of that element is the structure of part of the element, and that is fixed to the element. There are two types of structural elements: structural elements with an overall structure and structural elements with a specific structure. Structural elements, or elements that are structurally defined, are classified into two categories: structural elements whose overall structure is defined by the structural elements, and structural elements which are structurally described by the specific structure of the structural structure. Structures that are structural defined include the elements that fix and remove the structure. Structuring elements, or structural elements with specific structure, are classified in three categories: structural element that consists of a structural element, structural element that has a structural element other than a structural element (elements with specific structureWhat is the purpose of the Tailoring PRINCE2 in PRINCE2? We are going to discuss the concept of the Tailored PRINCE1 in PRINCO2 but we are going to start with the 3 key STORE cases in PRINCRUDE. 1.1. A Tailored PRINE2 This is a new PRINE2 that is being developed with the following features to be used in PRINE2: A multi-level system. A simple interface to the user interface A user-friendly interface. An interface that is fully functional with support for visualizations, animations and a user interface where the user can interact with the system. The tailored PRINE1 will be used in the PRINE2 core. Requirements for the Tailored and PRINE2 in PRINE1 Below are the requirements to be satisfied by the Tailored in PRINE: 1) A multi-level interface. 2) A user-friendly user interface. 3) A user interface that is completely functional with a user-friendly UI. We will talk about the requirements for the tailored PRINS with several examples in the next section. Example 1: A standard interface.
## Myonlinetutor.Me Reviews
The interface is a standard interface. The user is logged in and can view the display of the interface. (3) A simple interface. 1) The user is logging in. 2. The user can view and manipulate the display of a standard interface and the display of an interface that is simple to use. 3. The user has entered a username and entered a password. The user can then type in the password and the displayed interface will be displayed. This example does not use the standard interface and it is not a standard interface in the PRINCO 2. You can import it to the user-friendly PRINE2 to make it more simple. It is a standard one. The PRINE2 implementation is based on the standard interface. This means that the system will be created with the standard interface in PRINE 1. The PRINCE3 implementation is based upon the PRINE 3. Examples of the PRINCE 3 Example 2: A standard UI. The UI is a standard UI. The user enters a username and enters a password. The user will then type in a username and the displayed UI will be displayed again. (3a) The user enters the username.
## Pay Someone With Credit Card
The display of the UI is displayed again. This example also does not use a standard UI in PRINE 3 and it is a standard view of the PRINE3. Its implementation is based entirely on the standard UI. You can also import it to PRINE 3 to make it easier to use. It is not a UI in PRINCCE3 but it is a UI in the PRUNCCE3. Here is a sample of the PRUNCE3 implementation. PRUNCE3 is designed using the standard interface to the User Interface. The interface is a single-language interface. It is a standard display. The interface can be displayed with an internal display and you can choose which display you want to use. The PRUNCE implementation is based in PRINE3 and it is easier to use than PRINE 3 but it is not fully functional. 2) The user can enter a username and enter a password. This example
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Cody
# Problem 44408. ゼロでない要素が一番多い行を探そう
Solution 1983174
Submitted on 20 Oct 2019 by Nikolaos Nikolaou
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
a = [ ... 1 2 0 0 0 0 0 5 0 0 2 7 0 0 0 0 6 9 3 3]; r_correct = 4; assert(isequal(fullest_row(a),r_correct))
m = 2 m = 2 1 m = 2 1 2 m = 2 1 2 4 r = 4
2 Pass
a = [ ... 1 2 0 0 0 0 5 0 0 6 9 -3 2 7 0 0 0 0 0 0]; r_correct = 3; assert(isequal(fullest_row(a),r_correct))
m = 2 m = 2 1 m = 2 1 3 m = 2 1 3 2 m = 2 1 3 2 0 r = 3
3 Pass
a = [ ... 1 0 0 0 0 0 0 0 0 0 0 0 0 2 3]; r_correct = 5; assert(isequal(fullest_row(a),r_correct))
m = 1 m = 1 0 m = 1 0 0 m = 1 0 0 0 m = 1 0 0 0 2 r = 5
4 Pass
a = [ ... 0 0 0 -3 0 0]; r_correct = 4; assert(isequal(fullest_row(a),r_correct))
m = 0 m = 0 0 m = 0 0 0 m = 0 0 0 1 m = 0 0 0 1 0 m = 0 0 0 1 0 0 r = 4
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Fourier Series Pt 3
# Fourier Series Pt 3 - University of Waterloo CivE 331...
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Unformatted text preview: University of Waterloo CivE 331 Determine the Fourier Series representation of the following function using the half-range cosine expansion. Graph the expansion for n = 10. By inspection, the domain length ( L ) is 10. In the period of interest (0 < x < 10): ( 29 ( 29 < < = = < < + =---- 10 4 ; 62 . 14 5 5 4 ; 1 4 4 x e e e e x x x f kx kx k x k Substitute 10 = L into the HRC expression: ( 29 10 ; 10 cos 1 < < + = ∑ ∞ = x x n a a x f n n π Substitute 10 = L and f ( x ) into the coefficient expressions: ( 29 ( 29 + + = = ∫ ∫ ∫- 10 4 4 10 62 . 14 1 10 1 10 1 dx e dx x dx x f a kx ( 29 ( 29 + + = = ∫ ∫ ∫- 10 4 4 10 10 cos 62 . 14 10 cos 1 10 2 10 cos 10 2 dx x n e dx x n x dx x n x f a kx n π π π Now evaluate a and a n . The following relationships will be useful (and would be provided in an exam or quiz context): ( 29 ( 29 ( 29 a ax x a ax dx ax x sin cos cos 2 + = ∫ ( 29 ( 29 ( 29 [ ] 2 2 sin cos cos b a bx b bx a e dx bx e ax ax + + = ∫ ( 29 ( 29 A A 2 sin 2 1 2 cos- = 0.00 1.00 2.00 3.00 4.00 5.00 6.00 1 2 3 4 5 6 7 8 9 10 ( 29 x x f + = 1 ( 29 ( 29 26824 . ; 5 4 = =-- k e x f x k University of Waterloo CivE 331 Evaluate a : ( 29 691 . 2 62 . 14 12 10 1 62 . 14 2 10 1 10 4 10 4 4 2 = - + = - + ...
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# The fluid dynamics of fluids, especially of liquids and gases, is one of the most significant topics in Physics within the news.
That is for the reason that it deals using the motion of fluids. It’s a branch of mathematics that mostly deals with all the study of fluids.
The materials utilised in this branch of Physics involve physics numbers for instance power law, tangent and sine. All they are utilised to measure how a fluid is moving and to predict the movement of a fluid. That is essential in some locations where liquid or gas are moving in quite big amounts.
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Another area where you’ll be able to see fluid dynamics inside the news is in weather prediction. Due to the fact it is actually based on the fluid dynamics that are known, meteorologists can make forecasts and assistance the public in avoiding specific weather issues. Some regions have provided the meteorologists the freedom to predict weather for the general public, but meteorologists still must study their very own sets of laws of physics so that they’re able to use the info from the fluid dynamics and the laws of thermodynamics to predict future weather.
Then there are conditions that take place withhydrocarbon flares. It can be commonly identified that flare emissions may be pretty hazardous for the atmosphere. These emissions may involve poisonous gases, heavy metals and some other hazardous materials.
To support protect the public in the damaging effects of those gases, some specialists in fluid dynamics study the flare emissions and find out tips on how to predict the flares and how to predict which flares will likely be hazardous to the environment. If you see flare emissions, you could see the fluid dynamics within the news. They’ve studied the fluid properties that are connected towards the flare emissions and they have discovered how to predict them.
Then you will find areas which are related to fluid dynamics which have been developed via the usage of fluid mechanics. There are components that have been made to http://assessment.arizona.edu/ produce issues less complicated, for instance tennis rackets. There are also components that have been created to be employed in different applications that contain electronics, which tends to make them a a part of the fluid dynamics of fluids.
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In the mechanical systems that move vehicles and planes, a few of these systems use fluids. If you see the buyessay automobile as well as the plane becoming run, you’ll be able to see how the fluid is affecting the system. Also, you’ll find parts that use fluids inside the area of navigation, to create a graphical representation of maps and driving directions.
There are also points that make items simpler for people today. For instance, you will discover tools for folks who suffer from disabilities. This makes life quite a bit easier. Even so, some tools and devices make items easier.
That is why it is actually significant to study the fluid dynamics inside the news. By doing so, you could recognize ways to make issues simpler to make use of in life. It can also help you study far more in regards to the issues that have been created with the aid of fluid mechanics.
Mar 30, 2020
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# Custom White Balancing with picamera
I need a method to set the White Balance of the raspberry pi camera by "teaching it". Such as it is in DSLR cameras, I'd like to show a white picture to the camera and say "OK, this is white, now adjust your AWB accordingly".
I think the first step is to set the `awb_mode` to `off` and then, capture photos continously playing around with `awb_gains`. The problem is, I don't know how to decide "OK these values are fine, let's keep with them."
Anybody done such thing? I'm using the picamera Python module btw.
You could write a little loop that assumes that the camera is pointed at something which is mostly white (e.g. a sheet of paper) and then iterate over various combinations of the red and blue gains (probably increments of 0.1 between, say, 0.5 and 2.5) until you find the combination that produces an image in which most of the pixels are as close to grey (i.e. equal values for R, G, and B) as possible. It probably wouldn't be very quick (as it'd involve taking an picture, tweaking values, taking another picture, and so on) so I'd recommend using a low-res resize, and video-port captures. Still, it should do the trick.
The following is a crude demo; it sets the AWB to something ludicrous (both gains 0.5; this is just to demonstrate the convergence) and then allows 30 attempts to move each gain in either direction by 0.1 to try and get the average colours of the resulting capture to match the green channel:
``````import picamera
import picamera.array
import numpy as np
with picamera.PiCamera() as camera:
camera.resolution = (1280, 720)
camera.awb_mode = 'off'
# Start off with ridiculously low gains
rg, bg = (0.5, 0.5)
camera.awb_gains = (rg, bg)
with picamera.array.PiRGBArray(camera, size=(128, 72)) as output:
# Allow 30 attempts to fix AWB
for i in range(30):
# Capture a tiny resized image in RGB format, and extract the
# average R, G, and B values
camera.capture(output, format='rgb', resize=(128, 72), use_video_port=True)
r, g, b = (np.mean(output.array[..., i]) for i in range(3))
print('R:%5.2f, B:%5.2f = (%5.2f, %5.2f, %5.2f)' % (
rg, bg, r, g, b))
# Adjust R and B relative to G, but only if they're significantly
# different (delta +/- 2)
if abs(r - g) > 2:
if r > g:
rg -= 0.1
else:
rg += 0.1
if abs(b - g) > 1:
if b > g:
bg -= 0.1
else:
bg += 0.1
camera.awb_gains = (rg, bg)
output.seek(0)
output.truncate()
``````
In my tests it usually gets close to a decent solution in 10 or so steps, and then wobbles around a couple of values. There's almost certainly betters ways of doing this (starting with more sensible values, varying one at a time, using YUV captures instead, decreasing the increments as the values converge, terminating when acceptably close, etc.) but this should be enough to demonstrate the principle.
Sadly I cannot comment yet, so I have to do it as a seperate answer to this one
The Code from Dave Jones can be made faster convergent with this diff calculation (and more precise)
`````` # different (delta +/- 2)
if abs(r - g) > 2:
rg -= 0.01 * (r - g)
if abs(b - g) > 2:
bg -= 0.01 * (b - g)
# make sure the values do not get out of bounds
(rg, bg) = np.maximum((0, 0), (rg, bg))
(rg, bg) = np.minimum((8, 8), (rg, bg))
camera.awb_gains = (rg, bg)
output.seek(0)
output.truncate()
``````
For me this code converges in around 3-5 steps.
``````R: 0.50, B: 0.50 = ( 0.28, 157.60, 3.37)
R: 2.07, B: 2.04 = (158.43, 134.18, 210.09)
R: 1.83, B: 1.28 = (122.17, 119.14, 131.89)
R: 1.80, B: 1.16 = (121.19, 120.09, 118.13)
R: 1.80, B: 1.18 = (120.91, 120.35, 120.99)
R: 1.80, B: 1.18 = (121.14, 120.37, 120.76)
``````
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# Mathematica Integration Question
1. May 3, 2010
### frenzal_dude
Hi, I need to find this integral:
$$G(f)=-A\int_{\frac{-3T}{2}}^{\frac{-T}{2}}e^{-j2\pi ft}dt+A\int_{\frac{-T}{2}}^{\frac{T}{2}}e^{-j2\pi ft}dt-A\int_{\frac{T}{2}}^{\frac{3T}{2}}e^{-j2\pi ft}dt$$
Here's the working out I did:
$$G(f)=-A[\frac{e^{j\pi fT}-e^{j\pi f3T}}{-j2\pi f}]+\frac{A}{\pi f}[\frac{e^{j\pi fT}-e^{-j\pi fT}}{2j}]-A[\frac{e^{-j\pi f3T}-e^{-j\pi fT}}{-j2\pi f}]$$
Therefore:
$$G(f)=2TAsinc(fT) -3TAsinc(3fT)$$
But when I used Mathematica I typed this:
Integrate[-A*Exp[-I*2*\[Pi]*f*t], {t, -T/2, (-3*T)/2}] +
Integrate[A*Exp[-I*2*\[Pi]*f*t], {t, -T/2, T/2}] +
Integrate[-A*Exp[-I*2*\[Pi]*f*t], {t, T/2, (3*T)/2}]
and it gave me this:
$$\frac{ASin(f\pi T)}{f\pi}-\frac{Ae^{-2j\pi ft}Sin(f\pi T)}{f\pi}+\frac{Ae^{2j\pi ft}Sin(f\pi t)}{f\pi}$$
which equals:
ATsinc(fT) + 2jATsinc(πFT)Sin(2πfT)
Is the answer from Mathematica the correct one?
2. May 4, 2010
### jackmell
Looks like you switched the integration limits in the first integral. Also, don't use capital letters for user-variables and you can omit the asterisks for multiplication if you use a space and also you can use FullSimplify to simplify the answer. Finally, and this is just a personal opinion: if I do it manually and then set it up correctly in Mathematica and the integrals are well-behaved, and I get two different answers that I can't show it's really the same answer written differently, good chance I'm the one wrong but there are rare exceptions.
3. May 4, 2010
### frenzal_dude
ahh thakyou so much! I fixed up the integral and used FullSimplify and now I got the same answer as when I use the Fourier Transform properties, so I know that my working out for the integral must be wrong some where.
Thanks again!
btw why can't we use capital letters as variables in mathematica?
Last edited: May 4, 2010
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TL;DR
My second week into the Perl Weekly Challenge - and a reflection.
It seems that the Perl Weekly Challenge challenges this week are a bit easier than the last week. Or are they?
You are given an array `@A` containing distinct integers. Write a script to find all leader elements in the array @A. Print (0) if none found. An element is leader if it is greater than all the elements to its right side.
Its solution is somehow straightforward, it suffices to go backwards to figure out who’s a leader:
`````` 1 #!/usr/bin/env perl
2 use 5.024;
3 use warnings;
4
5 # This problem is easier to tackle if moving from the *end* of the array
6 # back to the beginning. So, we reverse the input array to analyze it
7 # and then reverse it again to get back to the original order.
8 sub keep_leaders { # @A <=> @_
9 return (0) unless @_;
10 my \$last_leader = \$_[-1] - 1;
11 return reverse grep {
12 my \$condition = \$_ > \$last_leader;
13 \$last_leader = \$_ if \$condition;
14 \$condition;
15 } reverse @_;
16 }
17
18
19 # testing stuff
20 for my \$Aref (
21 [9, 10, 7, 5, 6, 1],
22 [3, 4, 5],
23 [],
24 ) {
25
26 printout('Input: @A = ', @\$Aref);
28 }
29
30 sub printout {
31 my \$prefix = shift;
32 say \$prefix, '(', join(', ', @_), ')'
33 }
``````
It also got me thinking that there were such somehow simpler solutions last time, at least with respect to mine.
The bottom line is that these challenges resemble a lot those that you might get in interviews: the basic info to get some solution, but not enough to get the solution. In this way, people can see how you react, e.g. to check whether you ask questions, whether you think about corner cases, limits… in brief, to see your though process.
In this case, for example, it might make sense to understand whether using `reverse` is the right way to go, as opposed to e.g. scan the array backwards. Or even ask whether there’s a limit on the possible input values, how many will be of them, if we have memory constraints, time constraints… the list goes on.
Last time for Fibonacci, for example, I definitely avoided finding solution with a brute force search over all possible arrangements of the Fibonacci candidates, and went the much harder way of figuring out the most compact solution and then working from that, chipping off the remaning overlaps.
For low input numbers, in hindsight this probably makes no sense. This approach probably starts paying off with way bigger numbers. So in a sense that complicated solution was a reflection of my inner self that usually aims at solving a problem and forgetting about it - with enough lazyness and hubris that will will even work if someone decides to put a very big number in. Call me defensive.
On the other hand, this is a wonderful occasion to realize that it’s not always like this.
By making questions in an interview, you might be told that the numbers will always be below a threshold, etc. so it might make sense to express the different alternatives loud and say that there is a brute force solution that will save programmer time at the expense of some performance, and that there is a more linear solution that requires more time to code and test.
At the end of the day, this is probably what they’re after.
Comments? Octodon, , GitHub, Reddit, or drop me a line!
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## CGAL::Circular_arc_3<SphericalKernel>
#include <CGAL/Circular_arc_3.h>
### Creation
Circular_arc_3<SphericalKernel> ca ( Circle_3<SphericalKernel> c);
Constructs an arc from a full circle.
Circular_arc_3 ca ( Circle_3 c, Circular_arc_point_3 p, Circular_arc_point_3 q);
Constructs the circular arc supported by c, whose source and target are respectively p and q.
Precondition: p and q lie on c and are different.
The circular arc constructed from a circle, a source, and a target, is defined as the set of points of the circle that lie between the source p1 and the target p2, when traversing the circle counterclockwise seen from the side of the plane of the circle pointed by its positive normal vectors.
In this definition, we say that a normal vector (a,b,c) is positive if (a,b,c)>(0,0,0) (i.e. (a>0) || (a==0) && (b>0) || (a==0)&&(b==0)&&(c>0)).
Circular_arc_3<SphericalKernel> ca ( Point_3<SphericalKernel> p, Point_3<SphericalKernel> q, Point_3<SphericalKernel> r);
Constructs an arc that is supported by the circle of type Circle_3<SphericalKernel> passing through the points p, q and r. The source and target are respectively p and r, when traversing the supporting circle in the counterclockwise direction seen from the side of the plane containing the circle pointed by its positive normal vectors. Note that, depending on the orientation of the point triple (p,q,r), q may not lie on the arc.
Precondition: p, q, and r are not collinear.
### Access Functions
Circle_3 ca.supporting_circle () Point_3 ca.center () returns the center of the supporting circle. SphericalKernel::FT ca.squared_radius () returns the squared radius of the supporting circle. Plane_3 ca.supporting_plane () Sphere_3 ca.diametral_sphere () Circular_arc_point_3 ca.source () Circular_arc_point_3 ca.target ()
When the methods source and target return the same point, then the arc is in fact a full circle.
When the arc was constructed from its (full) underlying circle, then source and target both return the smallest x-extremal point of the circle if the circle is not contained in a plane x=A, and the smallest y-extremal point otherwise.
### Operations
bool a1 == a2 Test for equality. Two arcs are equal, iff their non-oriented supporting planes are equal, if the centers and squared radii of their respective supporting circles are equal, and if their sources and targets are equal. bool a1 != a2 Test for nonequality.
### I/O
istream& std::istream& is >> Circular_arc_3 & ca ostream& std::ostream& os << Circular_arc_3 ca
The format for input/output is, for each circular arc: a Circle_3 (the supporting circle) and two Circular_arc_point_3 (the source and the target), under the condition that the endpoints are actually lying on the circle.
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# Distance between Tulun, Irkutsk Oblast, Russia and the South Pole
16093 km = 10000 miles
During our calculation of the distance to the South Pole we make two assumptions:
1. We assume a spherical Earth as a close approximation of the true shape of the Earth (an oblate spheroid). The distance is calculated as great-circle or orthodromic distance on the surface of a sphere.
2. We calculate the distance between a point on the Earth’s surface and the South Pole as the length of the arc of the meridian passing through this point and the South Pole.
Find out the distance between Tulun and the North Pole, the Equator, the Tropic of Cancer, the Tropic of Capricorn, the Arctic Circle, the Antarctic Circle
Find out the distance between Tulun and other cities
# Tulun, Russia
Country: Russia
Tulun’s coordinates: 54°33′48″ N, 100°34′53″ E
Population: 51,330
Find out what time it is in Tulun right now
Wikipedia article: Tulun
# The South Pole
The South Pole is a point where imaginary Earth’s axis of rotation crosses the Earth's surface in the Southern Hemisphere.
The South Pole is the southernmost place on Earth. The South Pole latitude is 90° South. The South Pole longitude is undefined, because the South Pole is a point where all the meridians meet.
For the same reason the South Pole has no time zone.
For software and devices using GPS satellite navigation system 0° West may be used as conditional South Pole longitude.
The South Pole’s coordinates: 90°00′00″ S
Wikipedia article: the South Pole
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# Ideal Bose condensation gas in gravitational field, Statistical Physic
1. Feb 20, 2013
### dpopchev
1. The problem statement, all variables and given/known data
We have an ideal Bose gas in gravitational field. Show that the critical temperature( the temperature in which the condensation is starting is:
$T_c = T^0_c( 1 + \frac{8}{9} \frac{1}{ζ(3/2)} (\frac{\pi mgh}{kT^0_c})^{1/2} )$
Attempt
Lege artis; Firstly I want to estimate the number $N$ of particles in that $T_c$ temperature. As we know at that state the chemical potential is zero so $\mu = 0$, so the average number of particles will be given by the Bose - Einstein distribution
$n_{mean} = \frac{1}{e^{ε/kT_c}-1}$
I want to find all that particles in such state, so I form an integral over the volume in which the gas is( we assume it is a container with axial symmetry, with area of the circle $S$ and hight $L$ ) and over all possible energy states. I will skip the mathematical formulation and give the integral I need to evaluate:
$N = 4 \pi g_0 S h^{-3} \int_0^{∞} dp \int_0^L dz \frac{p^2}{e^{ \frac{p^2}{2 m k T_c} - \frac{mgz}{kT_c} } - 1 }$
where $S$ is area of the circle for the cylinder, $g_0 = 2s + 1$ is the number of states and $h$ is the Planck constant
After changing the variables I get $\frac{ 4 \pi g_0 S (2mkT_c)^{3/2}kT_c}{h^3mg} \int^{∞}_0 dp' \int^{\frac{mgL}{kT_c}}_0 \frac{1}{e^{p'^2}e^{-z'}-1}$
I have no idea how to handle this integral:
I just get the $e^{z'}$ out and get $\int \frac{e^{z'}}{e^{p'^2}-e^{z'}}$ which leads to something like $\int p'^2 ln(\frac{e^{p'^2} - e^{CL}}{e^{p'^2} - 1 } dp'$ on which my best attempt was $\int ln(\frac{e^{p'^2} - e^{CL}}{e^{p'^2} - 1 } \frac{dp'^3}{3} = \frac{p'^3}{3}.ln(...) - ....$ and the first is in range of $0$ which is a problematic dot and $∞$ which doesn't seem pretty too.
I hope for my post to be accurate enough. Thanks in advance.
PS: I found the same problem but it can't help me to solve the integral
https://www.physicsforums.com/showthread.php?t=455803
Last edited: Feb 20, 2013
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook
Can you offer guidance or do you also need help?
Draft saved Draft deleted
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Electronic Touch Switch(LM393)_Circuit Diagram World
Position: Index > Switch Circuit >
# Electronic Touch Switch(LM393)
2015-09-14 00:12
Declaration:We aim to transmit more information by carrying articles . We will delete it soon, if we are involved in the problems of article content ,copyright or other problems.
This article describes the Electronic Touch Switch (LM393). The principle is very simple, very practical. The circuit components can help you understand better grasp this principle. For example, in this circuit, you can go to find and buy these components: LM393.
Mechanical contacts have the disadvantage that they wear out. That is why it is practical to use an electronic ‘touch switch’ in some situations. With such a touch switch the resistance of the human skin is used for the switching action. The schematic shows the design of a circuit that senses the resistance of the skin and converts it into a useful switching signal. The touch switch contacts can be made from two small metal plates, rivets, nails, etcetera, which are placed close together on a non-conducting surface.
In this circuit a comparator of the type LM393 has been used. In the idle state there is, via R1, a voltage equal to the power supply voltage on the non-inverting input of IC1a. Because the inverting input of IC1a is set with R2 and D3 to D5 at the supply voltage minus 1.8 V, the open-collector output of IC1.a is, via R3, equal to the power supply voltage. This voltage is inverted by IC1.b. The voltage at the non-inverting input of IC1.b amounts to half the power supply voltage (through voltage divider R4 and R5) and is lower than the voltage on the inverting input.
Circuit diagram:
#### Figure 1 Electronic Touch Switch Circuit Diagram
The output of IC1.b is therefore a ‘0’. If the two touch contacts are bridged with a finger, the voltage at the non-inverting input will become low enough to cause the comparator to toggle state. The moistness of the skin results in a resistance of 1 to 10 MR. If this circuit is used in the vicinity of equipment that’s connected to the mains, then it can be sufficient to touch only the upper contact to operate the switch, provided that the circuit has been earthed. The body then acts as an antenna which receives the 50 Hz (or 60 Hz) from the mains.
This is enough to toggle IC1.a at the same 50 Hz. C1/R3 prevent this 50 Hz from reaching the input of IC1b and provide a useable ‘pulse’ of about 10 s at the output of IC1.b. Note that a fly walking across the touch switch conducts enough to generate a switching signal. So do not operate important things with this circuit (such as the heating system or the garage door). Do not make the wires between the touch contacts and the circuit too long to prevent picking up interference. The power supply voltage for the circuit is not very critical. Any regulated DC voltage in the range from 6 to 20 V can be used.
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#### Please solve RD Sharma class 12 Chapter Definite integrals exercise 19.2 question 6 maths textbook solution.
$\tan ^{-1}e-\frac{\pi}{4}$
Hint: We use indefinite formula then put limits to solve this integral.
Given$\int_{0}^{1}\frac{e^x}{1+e^{2x}}dx$
Solution: $\int_{0}^{1}\frac{e^x}{1+e^{2x}}dx$
Put $e^x=t$
$e^x dx=dt$
When x=0 then t=1 and when x=1 then t=e
\begin{aligned} &I=\int_{1}^{e} \frac{1}{1+t^{2}} d t \\ &=\left[\tan ^{-1} t\right]_{1}^{e} \\ &=\tan ^{-1} e-\tan ^{-1} 1\; \; \; \; \; \; \; \; \; \; \quad\left[\tan ^{-1} 1=\frac{\pi}{4}\right] \end{aligned}
$=\tan ^{-1}e-\frac{\pi}{4}$
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A projective plane is a point/line incidence geometry that satisfies the following conditions:
• Every two points lie on a unique line
• Every two lines meet in a unique point
• There is a quadrangle (four points, no three of which are collinear)
The last condition is just a non-degeneracy condition to eliminate trivial cases such as a single line containing all the points, and I’ll also add that I am only concerned here with finite projective planes. In this case it is easy to show that there is a number $s$ such that there are $s^2+s+1$ points, $s^2+s+1$ lines, each line contains $s+1$ points, each point lies in $s+1$ lines; this special number $s$ is called the order of the plane. In design theory language a projective plane is just a $2-(s^2+s+1,s+1,1)$ design.
The “canonical example” of a projective plane is the Desarguesian plane $PG(2,q)$ whose point set is the set of 1-dimensional subspaces of the vector space $V = GF(q)^3$ and line set is the set of 2-dimensional subspaces of $V$. The order of $PG(2,q)$ is $q$ and since there is a field $GF(q)$ for every prime power $q$, there is a projective plane of every prime power order. Probably the main open question in finite geometry is
Is there a finite projective plane whose order is not a prime power?
One thing that makes this problem difficult is that there are many many projective planes known other than $PG(2,q)$ – lots of infinite families and plenty of examples of small order – but they all resolutely have prime power order. (For those interested, Eric Moorhouse keeps comprehensive lists of small order planes.) If $PG(2,q)$ were the only projective plane that was known, then we would conclude that the combinatorial definition of a projective plane was somehow so restrictive that it could only be met by the algebraic structure of a field. However, many of the known projective planes appear to have nothing much to do with fields and so the whole question is more subtle than just the existence of a field – in particular, what is it about projective planes that forces the orders of the fields to occur but dispenses with the fields themselves! The observant reader will notice that I am writing as though it is obvious that the answer to the question is “No” but, while this is what I personally believe, there are many geometers much more talented than I who believe the opposite, and spend at least some “Friday afternoon time” looking for planes of order 12 or 20 or 56 or whatever.
### Maximal Arcs
Much more is known about the orders for which projective planes may or may not exist, but this is all written down elsewhere and today I want to discuss a less well-known problem, which is the existence of maximal arcs in projective planes. A maximal arc is a subset $\mathcal{K}$ of the points of a projective plane such that every line that meets $\mathcal{K}$ does so in the same number of points. In other words, there is a number $n$ such that every line of the plane contains either $0$ of $n$ points of $\mathcal{K}$ – we call this a set of type $(0,n)$. Some simple counting shows that if a plane of order $s$ contains a set of type $(0,n)$ then $n$ must divide $s$. [Edit: as JB points out, we should insist that $n$ be strictly less than $s$.]
So when does these sets of points actually occur? For planes of even order the answer is pretty simple – the Desarguesian plane $PG(2,2^h)$ contains a set of type $(0,n)$ for every possible value of $n$ that satisfies the necessary condition. Computer searches on the planes of order $16$ show that most of them contain sets of type $(0,n)$.
For planes of odd order, the situation is very different – not a single maximal arc in any projective plane of odd order is known. Often if there is an even-odd dichotomy of this sort, there is a parity argument hiding somewhere, but in this case nobody has managed to find one. However it is known that the Desarguesian plane $PG(2,q)$ with $q$ odd does not contain any maximal arcs. This was proved by Ball, Blokhuis and Mazzoca (paper here) who gave a very algebraic proof based on properties of the field underlying $PG(2,q)$.
But what about the other planes? If it really is the case that maximal arcs do not exist in any projective plane of odd order, then this is a combinatorial statement for which we would really like a combinatorial proof. On the other hand, if it is only the algebra of $PG(2,q)$ that prevents a maximal arc appearing there, then surely we should be able to find an example in some of the hundreds of thousands of known non-Desarguesian planes!
So this leads us to the main problem:
Find a maximal arc in a non-Desarguesian plane of odd order, or give a combinatorial reason why they cannot exist.
John Sheekey pointed out to me that we should have $n , otherwise we get trivial examples by removing a line from $PG(2,q)$ to get $(0,q)$-sets.
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## ››Convert siriometre to light minute
siriometer light-minute
How many siriometer in 1 light-minute? The answer is 1.2023932880863E-7.
We assume you are converting between siriometre and light minute.
You can view more details on each measurement unit:
siriometer or light-minute
The SI base unit for length is the metre.
1 metre is equal to 6.684587153547E-18 siriometer, or 5.5594015866359E-11 light-minute.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between siriometers and light minutes.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of siriometer to light-minute
1 siriometer to light-minute = 8316746.35835 light-minute
2 siriometer to light-minute = 16633492.71671 light-minute
3 siriometer to light-minute = 24950239.07506 light-minute
4 siriometer to light-minute = 33266985.43341 light-minute
5 siriometer to light-minute = 41583731.79177 light-minute
6 siriometer to light-minute = 49900478.15012 light-minute
7 siriometer to light-minute = 58217224.50847 light-minute
8 siriometer to light-minute = 66533970.86683 light-minute
9 siriometer to light-minute = 74850717.22518 light-minute
10 siriometer to light-minute = 83167463.58353 light-minute
## ››Want other units?
You can do the reverse unit conversion from light-minute to siriometer, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Light-minute
A light-minute (also written light minute) is a unit of length. It is defined as the distance light travels in an absolute vacuum in one minute or 17,987,547,480 metres (~18 Gm). Note that this value is exact, since the metre is actually defined in terms of the light-second.
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!
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# One Hundred Thirty Dollars
Here’s everything you need to know about One Hundred Thirty Dollars. Find all the information it in this article.
Doing so helps to stop confusion and fraud—numbers can simply be altered or misread, however an quantity written in words is much more durable to tamper with. In most circumstances, it doesn’t matter how you write it. Nobody will notice until there’s an issue with the examine. It is a free online quantity to words converter. It converts number to United States forex phrases like Hundred, Thousand, Million, Ten Million, Hundred Million, One Billion and extra.
Our household spends 100 thirty dollars on groceries per week. Copyright ©2022 System1, LLC. All Rights Reserved. The material on this web site cannot be reproduced, distributed, transmitted, cached or otherwise used, besides with prior written permission of Answers. To solidify the concept and develop the habit, try several completely different greenback quantities. According to the organizers, 23,465 folks attended the concert. Twenty-three thousand 4 hundred sixty-five folks attended the concert.
## The Means To Write Out Integer And Decimal Numbers In Words In (us) American English, Utilizing Letters As An Alternative Of Numerals
Our converter already exhibits you tips on how to spell one thousand 200 thirty dollars; within the drop-down menu you can change the output format according to your desire. First, write the amount in numeric kind in thedollar box, positioned on the best side of your examine next to the dollar signal (“\$”). Start by writing the number of dollars (“8”), adopted by a decimal point or period (“.”), after which the number of cents (“15”). Ultimately, you’ll have “8.15” in the dollar box.
In the upper right-hand nook, you’ll write the date you’re making payment or the agreed upon date when the payee is allowed to cash your examine. When writing a cheque, you have to write amount in numeric form and likewise in words. If the amount in phrases is different from that in numeric kind, bank might not settle for the cheque.
### \$13000: Lowercase All Lowercase Letters:
The five-dollar example can additionally be complicated. Just write a zero when there isn’t some other quantity to use. Some individuals would write that quantity out as “Five dollars solely,” which is also fine. Review all other elements of knowledge and make sure all are good, then sign your name. This elective line is the place you can even make notice of what the payment is for, corresponding to “automotive payment” or “charitable donation”. Five Thousand One Hundred Thirty Dollars.(\$325,130) with out prior written approval from the District (“Project Fee”).
Technically, this is the fractional quantity of entire dollars. Right below you can find how 100 thirty-seven dollars is spelled as a determine, as an amount and as foreign money. After that, we show you tips on how to write one hundred thirty-seven dollars on a verify. Finally, we level out one hundred thirty-seven dollars in phrases. The quantity you write with words is the official amount of your verify. If the amount in numeric format differs from what you write in word format, U.S. authorities require the financial institution to go together with the amount written out in phrases.
### Writing 100 And Thirty Thousand Dollars In A Verify:
We merely get “three” from three, “two hundred fifty-one” from 251 and “four hundred sixty-nine” from 469. Just right click on the above image, choose copy link tackle, then past it in your HTML.
When using numerals to convey a quantity with four or extra digits , use a comma to help the attention shortly course of the number. A comma ought to be placed every three characters left from the decimal. When there is no decimal because you’re dealing with a complete number, you can mentally add a decimal to the best aspect of the figure. However, don’t use any commas whenever you write out the number using words.
## How Do You Write 330 Dollars On A Check?
As a best apply, it is at all times best to write out the full name of the payee as an alternative of abbreviations or acronyms. DisclaimerAll content material on this web site, including dictionary, thesaurus, literature, geography, and other reference information is for informational functions only. If you want to know what one hundred thirty-seven dollars in numbers, as an quantity, on a examine, as a foreign money or as a determine is, then you could have come to the right web site. For instance, the quantity “23” seems in the course of this sentence, so it might be written with numbers.
If you want to know what one thousand 200 thirty dollars in numbers, as an amount, on a check, as a forex or as a figure is, then you have come to the right website. Debit cardscan be used at merchants and on-line retailers. Just like a check, your debit card pulls funds from your checking account. For everyday spending, it could be safer to make use of bank cards to cut back the chance of errors and fraud hitting your checking account.
### One Thousand 2 Hundred Thirty Dollars In Words
And no, occurring a producers website and “constructing” one with all the choices you want is not really ordering it. Go all the way through the method and at the end it gets kicked to a dealer and so they can once more add the markup in the occasion that they really feel like it. For parts less than one greenback, use a fraction. Writing a examine is the most common scenario by which you may want to write down out an quantity utilizing phrases . Ebony Howard is an authorized public accountant and a QuickBooks ProAdvisor tax professional. A reference for yourself to keep monitor your checks.
The amount of the examine in numerical format. The quantity of your verify written out utilizing phrases instead of numbers. Add an unofficial notice to your examine for payee or yourself.
## Factors To Note
You might be familiar with checks, however you get stuck writing out the quantity. Writing a examine with cents is especially tough, but with a little bit of apply you’ll quickly be ready to do it with out pondering.
### Steerage On Spelling Out Numbers
Try using these tips and rules to additional grasp the concept of writing out numbers with phrases. Enter the number that you just need to convert into phrases. To convert numbers into phrases you should observe under steps.
## \$130,00000: Sentence Case Capital Letter To Begin The Sentence:
On a examine, you spell out the check quantity fully on the line below the “Pay to the order of” line. The guidelines for numerical commas may change should you travel outdoors the U.S. In some other nations, the comma and decimal level essentially change roles. For instance, a quantity that appears as “1,234.fifty nine” in the us might be written as “1.234,59” in other elements of the world.
## Words Make It Official
If our data has been helpful to you, then you could also be thinking about the means to spell one thousand 2 hundred thirty three dollars, and don’t neglect to bookmark us. If you want to take out money instantly from the teller at the financial institution, you can also make the try to “Cash” as an alternative of an individual or other entity. Simply write, “Cash” on the payee line, and fill in the amount of money you need in the two spots the place you would usually write the amount. You’ll must endorse the check on the front and the back so as to cash it. A more secure method to strive this, however, would be to write down the check out to yourself. If the amounts on the two strains of your examine differ, the financial institution will default to the written phrases rather than the numerals.
### Sample Check With Dollars And Cents
This Number to Words conversion software can help to fill in your cheque. If our data has been useful to you, then you might also be thinking about how to spell one hundred forty dollars, and don’t overlook to bookmark us.
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# Using SVD to determine the redundancy of a fit.
WarPhalange
I have a project I am doing for a professor and unfortunately I cannot get ahold of him to help me out, so I figured I'd ask you guys. Of course, I tried to ask The Google about this first and didn't get anywhere. Here is what I am trying to do:
assignment said:
Linear least squares fitting. Choose some odd-ball function, say g(x). Create a set of "data" by choosing xi and some (sigma)i, generating yi normally distributed about g(xi). Choose a set of functions that might plausibly fit the data as y=Sum[aj fj(x)]. Perform a least squares fit by solving the normal equations in matrix form. You should determine the condition number and use SVD to determine whether there is any redundancy in your choice of fj(x) and fix the fit. Finally should evaluate chi squared to see whether the fit is adequate.
The part I am having trouble with is using SVD to determine redundancy. My f(x) is an n'th order polynomial (I get to decide what 'n' is). I successfully found a fit to my generated data, but don't know where to go from there.
What I found online was to take the matrix A from Ax=b, do SVD on that, and go from there in order to solve the system of equations. I already have a solution though, so I don't know what to do.
One idea I had was to make a new matrix like so:
| a0 a1x1 a2x12 a3x13 |
| a0 a1x2 a2x22 a3x23 |
| a0 a1x3 a2x32 a3x33 |
et cetera, with actual numbers instead of 'a' and 'x' of course, take the SVD of that, trim down the three matrices to only include actual eigenvalues (so drop any '0' elements in the diagonal matrix), transform back, and then divide by the various 'x's to get different values for the 'a's, but I'm not sure if that would do anything at all.
Thanks in advance for the help.
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#### Welcome to the Smartsheet Forum Archives
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# =SUMIF Help please:
Options
edited 12/09/19
Hi Team,
Formula not working…(#UNPARSEABLE) =SUMIFS[Value of Job],2 [Sold Date:]2,>DATE(2016, 12, 31),[Sold Date:]2,<DATE(2017, 2, 1), [Result]2, 1
Trying to get “Current Month Sales” to calculate “Sold Date” when “Sold Date” matches the current month we are in (January)
•
• When February rolls over I wanted “Current Month Sales” to roll into “Total Running Sales” Automatically, how do we do this in the sheet?
• I wanted to set up cell that I could reference for previous month sales, i.e if I enter in January when we are in may it will show that sales for January in a cell I set up. I understand that the above formulas have to be set up first.
Thanks for your help.
Tags:
• ✭✭✭✭✭✭
Options
A community post:
Heath,
=SUMIFS[Value of Job],2 [Sold Date:]2,>DATE(2016, 12, 31),[Sold Date:]2,<DATE(2017, 2, 1), [Result]2, 1
There's lots of things wrong here.
=SUMIFS( [ColumnA]:[ColumnA], << this is a the sum range. You need to designate the column as [Column]:[Column]
[ColumnB]:[ColumnB], < DATE(), << each criteria range is a range. the system will look at the row for you.
(same for the > DATE()) range and criteria
[ColumnC]:[ColumnC], 1 ) << same thing here. Range, not a cell.
You were missing your open and close parentheses for SUMIFS ()
I would change your column name of [Result:] to [Result].
You can see you mistyped it in the formula.
In fact, many of your column names have a colon (:) at the end. Bad practice and not recommended. You seem to use it in cell's so maybe there i a reason I don't know about.
You mistyped the name [Value of Jobs] as [Value of Job]
Assuming you get rid of the colons in the column names, here's the formula:
=SUMIFS([Value of Jobs]:[Value of Jobs], Result:Result, 1, [Sold Date]:[Sold Date], >DATE(2016, 12, 31), [Sold Date]:[Sold Date], <DATE(2017, 2, 1))
This will return the sum of the [Value of Jobs] column when the Result is 100% and Sold Date is in January.
to avoid retyping next month, you could use:
=SUMIFS([Value of Jobs]:[Value of Jobs], Result:Result, 1, [Sold Date]:[Sold Date], MONTH(@cell) = MONTH(TODAY()))
This will always show 'CURRENT MONTH'. So if you look at it on Feb 1st, anything sold in January won't be in the total.
Previous Month is also possible with a bit of work. And be aware the Jan 2016 and Jan 2017 will show be in the same total unless you account for YEAR.
Craig
• Options
Hi Craig,
I have totals for some of the rows in first column, so have to start with row 2 ect.
, MONTH(@cell) ? Vaule of jobs = \$\$\$, Result = %%%, Sold Date = Calander, Current Month Sales = Value of Jobs when the Sold Date has a January date in it. Shouldnt this formula go into Current Month Sales ????
=SUMIFS([Value of Jobs]2:[Value of Jobs]50, [Result]2:[Result]50, 1, [Sold Date]2:[Sold Date]50, MONTH([(@cell) = MONTH(TODAY()))
Thanks Mate
H
• Options
Hi Craig,
I have totals for some of the rows in first column, so have to start with row 2 ect.
, MONTH(@cell) ? Vaule of jobs = \$\$\$, Result = %%%, Sold Date = Calander, Current Month Sales = Value of Jobs when the Sold Date has a January date in it. Shouldnt this formula go into Current Month Sales ????
=SUMIFS([Value of Jobs]2:[Value of Jobs]50, [Result]2:[Result]50, 1, [Sold Date]2:[Sold Date]50, MONTH([(@cell) = MONTH(TODAY()))
Thanks Mate
H
• Options
One more thing......I'll put in the year also. This formula is here to stay.
Is it =YEAR(MONTH(TODAY()))) for a matter of interest ?
Thanks Craig
• ✭✭✭✭✭✭
Options
Heath,
If I understand the question, yes, the formula goes in the [Current Month Sales] column.
To add the year, add another range/criteria pair with the same format as your MONTH() formulas and replace MONTH with YEAR.
MONTH() and YEAR() both take dates and return a number for the month and year respectively.
Hope that helps.
Craig
• Options
Hi Craig,
Thanks again for your help
Is this correct ?
=SUMIFS([Value of Jobs]2:[Value of Jobs]50, Result2:Result50, 1, [Sold Date]2:[Sold Date]50, MONTH(@cell) = MONTH(TODAY(),SUMIFS([Value of Jobs]2:[Value of Jobs]50, Result2:Result50, 1, [Sold Date]2:[Sold Date]50, MONTH(@cell) = YEAR(TODAY()))
• ✭✭✭✭✭✭
edited 01/19/17
Options
Heath,
No.
At a glance, your last criteria should be
YEAR(@cell) = YEAR(TODAY())
There might be more - that's just by looking at it.
If you share it to me, I'll check for sure.
Craig
This discussion has been closed.
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Next: Elastic Windows Data Structures Up: Data Structures and Algorithms Previous: Problem Domain
## Existing Data Structures
Hierarchical spatial data structure techniques can be first examined according to the partitioning [81]. Space-based partitioning organize the embedding space from which the data is drawn such as the grid file [82], PR quadtree [83], MX-CIF quadtree [84], PM quadtree [85], PMR quadtree [86], etc. Data-based partitioning organizes the data to be stored such as the point quadtree [87], k-d tree [88], R-Tree [89], etc.
One way of representing rectangle data is to map it onto a higher-dimensional point data and use point-based data structure techniques such as the grid file, PR quadtree, point quadtree, and k-d-tree. Mapping of the rectangular data to a four-dimensional point can be done in a number ways such as x and y coordinates of the opposite corners, or x and y coordinates of one corner and the width and height, etc. Point-based representations of rectangular data have the drawback that they lack to benefit from the locality of data for the efficiency of both storage and spatial operations.
Rectangle data can also be represented in terms of the lines it is composed of and line-based data structure techniques such as PM quadtree, PMR quadtree, etc. Line-based representations of rectangular data had the drawback that a spatial operation specified in terms of its line segments may not satisfy the conditions of the operation yet the rectangle they are composed of satisfies them.
Rectangle data can also be represented in terms of the area it occupies. Techniques such as MX-CIF quadtrees and R-Trees organize rectangle data in hierarchical groupings of minimum bounding boxes. In MX-CIF quadtree uses space-based quadtree partitioning where each rectangle is associated with its minimum enclosing quadtree block. In R-Trees rectangle data is partitioned into hierarchically nested minimum bounded boxes. R-Trees have the drawback that there locality of data is not utilized.
Let's examine these techniques based on the characteristics of the problem domain. In most of the above techniques, the rectangular data is assumed to be at the same level, i.e. there are no hierarchical relationship among rectangles. Yet, for most of the techniques imposing such a relationship is trivial. Such as in R-Trees all that is needed is to mark such rectangles as hierarchy nodes and take special action for such nodes when reorganization is performed on the R-Tree. For the others it may be necessary to add pointers for the representation of hierarchical relationships.
Most of the above techniques do not assume any layout strategies. Thus, they can be applied to overlapping as well as non-overlapping layouts. Yet, the performance of the algorithms can be improved when the problem domain enforces a specific layout strategy such as space-filling layouts. For example, point-based techniques such as k-d-trees can be adopted to suit space-filling layouts where each point is considered as the boundary between rectangles.
Dynamic nature of the layout is critical in the selection of the data structure technique. Space-based partitioning techniques do not perform efficiently when the layout is dynamically changing affecting a number of rectangles. For example, an elastic resize operation may result in considerable reorganization when MX-CIF quadtree is used (Figure ). This is not the case in most of the data-based partitioning such as R-Trees, since these techniques represent the spatial relationships among the data not the space they are organized in.
Most of the data structures and algorithms in the domain of window management are for independent overlapping window strategies [90, 91], hence they support neither hierarchical nesting nor dynamics that affect a number of windows on the screen.
However, in other domains such as graph layout drawing, interface design and VLSI layout tools, there exist algorithms that deal with more complicated dynamics. In the domain of graph layout drawing algorithms typically focus on producing layouts to optimize criteria such as total number of link crossings, total diagram area, etc. [92]. Kosak et al. [93] proposed a constraint-based heuristic algorithm where visual organization rules define the constraints on the layout. Ioannidis proposed an algorithm for non-overlapping objects at multiple granularities [94].
Tcl/Tk provides a number of geometry managers (e.g. pack, grid, etc.) that determine the layout of the graphical interface widgets. Geometry managers allow hierarchical nesting of widgets and allow layout reorganizations that affect multiple widgets.
Ousterhout [95] proposed a corner-stitching data structuring technique for VLSI layouts, where corners of rectangular areas are connected to each other. Proposed data structure enables operations like finding neighbors efficiently, however, overlapping and hierarchical rectangles are not supported, and moving/resizing rectangles are costly since connections must be updated. Samet [81] has an excellent review of data structures and algorithms for handling collections of small rectangles.
Next: Elastic Windows Data Structures Up: Data Structures and Algorithms Previous: Problem Domain
Eser Kandogan
Sun Sep 13 18:34:46 EDT 1998
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19-10-2010
# 19-10-2010 - RF-cavities Acceleration is performed using...
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1 RF-cavities z Acceleration is performed using electric fields that are fed into Radio- Frequency (RF) cavities. RF cavities are basically resonators tuned to a selected frequency. z To accelerate a proton to 7 TeV, a potential of 7 TV must be provided to the beam: ¾ In circular accelerators the acceleration is done in small steps, turn after turn. ¾ At the LHC the acceleration from 450 GeV to 7 TeV lasts ~ 20 minutes, with an average energy gain of ~ 0.5 MeV on each turn . s ) ( t E G TM01 mode (transverse magnetic) standing wave cavity around the beam pipe 2 Acceleration with RF-cavities Principle: You send an electro-magnetic wave into a cavity and excite standing waves. The mode of the EM wave is in such a way that you will get an E-field along the axis of the cavity (such as the TM01 mode). TM01 = ground mode with the M agnetic field components only in the T ransverse plane, electric field components can be parallel (to the direction of motion of the counter propagating waves). This is the accelerating field that is responsible in accelerating the charged particles. There can be many of these cavities stacked together in series. However, because it is an oscillating field, you need to do 2 things: 1. You need to know just the right timing to inject the particles into the cavities. 2. You need to have the field in each cavity to NOT be at the same phase with each other. The phase difference depends on the nature of each cavity. In any case, you need to time the entry of the charged particle just right so that as it goes into the cavity, when the field is building in the right direction, – and as it leaves that cavity and goes into the next cavity, the field in that cavity should also start building in the right direction. How fast these fields have to build up depends on how fast the charged particles are moving through the cavities: at 450 GeV: f = 400.78 MHz ... see next slide at 7 TeV: f = 400.79 MHz (since the speed is only slightly larger)
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3 Buckets & Bunches Governed by the RF-cavities, operated at: f = 400 .8 MHz standing waves: O = c/f = 3e8/4e8 = 0.75 m max. # of possible buckets = (~26700 m)/ O = 35640 T = 1/f = 2.5 ns standing waves decelerating field o empty bucket by construction accelerating field Only every 10 th possible bucket can be filled o LHC collisions at 40 MHz, i.e. each 25 ns. max. # of bunches: 3564 (> 2808 ) Distance of filled buckets (= bunches) is 7.5 m. beam gaps due to kicker magnets! RMS bunch length: 11.2 cm 7.6 cm 450 GeV 7 TeV Eimer, Kübel, Behälter Bündel, Haufen ATLAS CMS ALICE (+11.25 m) LHCb beam 1 clockwise beam 2 anti-clockwise u u u u 1 1 17851 late 8911 nominal LHC started with: 2 bunches per beam 8 encounters per turn, 4 collisions (1 per exp!) RF-system 4 Longitudinal beam dynamics ' E t RF Voltage time LHC bunch spacing = 25 ns = 10 buckets ¡ 7.5 m 2.5 ns The particles are trapped in the RF voltage:
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# Impose Neumann Boundary Condition in advection-diffusion equation 1D
when solving the advection equation in 1D that is:
$$\frac{\partial u}{\partial t} + c\frac{\partial u}{\partial x} = 0$$ with $$u'(t,0) = 0$$ and $$u(t,L) = 0$$ , $$u(0,x) = u_{0}$$
one numerical scheme is the FTCS (Forward time-centered space), but this numerical scheme is unstable.
$$\frac{u_{j}^{n+1}-u_{j}^{n}}{ h_{t}} = c \frac{u_{j+1}^{n}-u_{j-1}^{n}}{ 2h_{x}}$$
But when solving
$$\frac{\partial u}{\partial t} + c\frac{\partial u}{\partial x} = \alpha \frac{\partial^2 u}{\partial x^2}$$ the advection-diffusion equation in 1D with $$u'(t,0) = 0$$ and $$u(t,L) = 0$$ , $$u(0,x) = u_{0}$$
Since the advection-diffusion equation is a second order equation I'd like to use a second order approximation.
if we define $$u_{k}^{n} := u(t_{n},x_{k})$$; $$\ \ x_{k} = kh$$ and $$\ \ k = 0,1,2,...,N$$. $$h$$ is known as the mesh size or step size.
For the second derivative:
$$\frac{\partial^2u_{k}^{n}}{\partial x^2} \approx \frac{u_{k+1}^{n}-2u_{k}^{n}+u_{k-1}^{n}}{h^2} = \frac{u_{k-1}^{n}-2u_{k}^{n}+u_{k+1}^{n}}{h^2}$$ for $$k=0,1,...,N-1$$
Since $$u'(t,0) = 0$$ and $$u(t_{n},L) = u(t_{n},x_{N}) = u_{N}^{n} = 0$$ we get the following matrix representation of the second derivative operator
$$$$\frac{\partial^2}{\partial x^2} \approx L_{2} = \frac{1}{h^2}\left(\begin{matrix} -2 & 1 & & 0\\ 1 & \ddots & \ddots & \\ & \ddots & \ddots & 1 \\ 0 & & 1 & -2 \end{matrix} \right)$$$$
for $$k=0$$ , we get
$$\frac{\partial u_{0}^{n}}{\partial x} = \frac{u_{0+1}^{n}-u_{0-1}^{n}}{ 2h} = 0$$ this implies that $$u_{1}^{n}=u_{-1}^{n}$$ and
$$\frac{\partial^2u_{0}^{n}}{\partial x^2} \approx \frac{u_{0+1}^{n}-2u_{0}^{n}+u_{0-1}^{n}}{h^2} = \frac{u_{0-1}^{n}-2u_{0}^{n}+u_{0+1}^{n}}{h^2} = \frac{-2u_{0}^{n}+2u_{1}^{n}}{h^2}$$
thus we have to modify the entry $$1,2$$ of $$L_{2}$$
$$$$L_{2} = \frac{1}{h^2}\left(\begin{matrix} -2 & 2 & & 0\\ 1 & \ddots & \ddots & \\ & \ddots & \ddots & 1 \\ 0 & & 1 & -2 \end{matrix} \right)$$$$
What I have done, is $$\mathbf{impose \ the \ Neumann \ boundary \ condition}$$ in $$L_{2}$$ .
I want to approximate the first derivative using central difference(Second order approximation):
$$\frac{\partial u_{k}^{n}}{\partial x} = \frac{u_{k+1}^{n}-u_{k-1}^{n}}{ 2h}$$
The matrix representation is: $$$$\frac{\partial}{\partial x} \approx L_{1} = \frac{1}{2h}\left(\begin{matrix} 0 & 1 & & 0\\ -1 & \ddots & \ddots & \\ & \ddots & \ddots & 1 \\ 0 & & -1 & 0 \end{matrix} \right)$$$$
for $$k=0$$ , we get
$$\frac{\partial u_{0}^{n}}{\partial x} = \frac{u_{0+1}^{n}-u_{0-1}^{n}}{ 2h} = 0$$ this implies that $$u_{1}^{n}=u_{-1}^{n}$$
But I'm stuck when I try to $$\mathbf{impose \ the \ neumann \ boundary \ condition \ in}$$ $$L_{1}$$. I don't know how to do that.
If we solve that problem, we can solve the differential equation
$$\frac{ \partial u }{\partial t} = \Big(-cL_{1} +\alpha L_{2} \Big)u$$ integrating in time( C-N, Back-Euler, RK4 )
$$\mathbf{please \ help!\ \ How \ do \ you \ impose \ the \ Neumann \ Boundary \ Condition \ in \ L_{1}?}$$
First, write out the semi-discrete equation for $$u_0$$, assuming it's an interior node:
$$\frac{d}{dt}(u_0) = - c \frac{\left( u_1 - u_{-1} \right)}{2h} + \alpha \frac{\left(u_1 - 2u_0 + u_{-1}\right)}{h^2}$$
Then, eliminate the ghost node $$u_{-1}$$ by using the equation you obtained from the Neumann BC: $$u_{-1} = u_1$$. Substituting this condition into the above equation gives:
$$\frac{d}{dt}(u_0) = \alpha \frac{\left(- 2u_0 + 2u_1 \right)}{h^2}$$
Note that the advection term vanishes, so the first row of your $$L_1$$ matrix should be all zeros. The first row of the $$L_2$$ matrix should have coefficients of $$-2$$ and $$2$$ in the first two columns, as you have already derived.
Also something to keep in mind, using a central difference approximation for the first derivative in the advection term will cause stability issues if your problem is advection dominant. In particular, this central difference scheme may develop oscillations if the grid Peclet number is greater than $$2$$, where the Peclet number is defined as:
$$Pe = \frac{c h}{\alpha}$$.
You can avoid these oscillations by using an upwinded (biased) approximation, such as this first-order upwinded scheme:
$$\frac{\partial u_k}{\partial x} \approx \frac{u_k - u_{k-1}}{h}$$, if $$c > 0$$.
Higher-order upwinded approximations also do exist, and typically involve wider stencils.
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back to Trigonometry
# Law of Sine
The figure below show an acute triangle ABC, its three vertex A, B and C opposite three sides a, b and c respectively. What is the relation among the sides and the angles in the triangle? (Note: a capital letter represents the vertex and a small letter represents the length of a side in the triangle. )
Now we draw a line that pass through the pint C and perpendicular to AB at the point D. (D is the perpendicular point.)
In right triangle ACD, use the definition of Sine, sin A = CD/AC = CD/b
so CD = b sin A
In right triangle BCD, sin B = CD/BC = CD/a
so CD = a sin B
Then b sin A = a sin B
(Both sides of the equation divide by sin B sin A)
so b/sin B = a/sin A
The formula above tell us, in triangle ABC, the ratio of b to sin B equals the ratio of a to sin A (Note: in triangle ABC, side b opposite the angle B and the side a opposite the angle A).
Now we will find the relation between the side c and the angle C.
angle C = angle 1 + angle 2
sin C = sin(angle1 + angle2) = sin(angle1) cos(angle2) + cos(angle1) sin(angle2)
sin(angle1) = BD/BC = BD/a
cos(angle2) = CD/AC = CD/b
cos(angle1) = CD/BC = CD/a
Then sin C = BD/a × CD/b + CD/a × AD/b = [CD/ab] × (BD + AD)
Note: BD + AD = AB = c
Then sin C = [CD/ab] × c
Note: sin B = CD/a
So sin C = [sin B/b] × c
Both sides divide by the side c
So sin C/c = sin B/b
So c/sin C = b/sin B
Therefore, we get the law of Sine:
a/sin A = b/sin B = c/sin C
The law of Sine describe the relation between three sides and the Sine of their corresponding angles. If we know two angles and a side, we can find other angle in the triangle from the sum of a triangle theorem. Then apply the law of Sine to calculate the other two sides. If we know any two sides and the angle opposite one of the sides, apply the law of Sine, we can find the value of Sine of the angle of other side opposite, from here to find other sides and angles.
# What is the relation among the sides and angles in an obtuse triangle?
The figure below show an obtuse triang (90o < angle C < 180o). Now we prove the law of Sine.
Draw a line passing through the point C and perpendicular to AB at point D (D is the perpendicular point.)
In the figure above, in the right triangle BCD, sin B = CD/BC = CD/a,
so CD = a sin B
In the right triangle ACD, sin A = CD/AC = CD/b,
so CD = b sin A
So a sin B = b sin A,
Both sides of the equation divide by sin A sin B
so a/sin A = b/sin B
Now we will find the relation between sin C and c
Extended the BC to the left and draw a line passing through the point A and perpendicular to the extended BC line at the point E (E is the perpendicular point.)
In the figure above, since BE is a straight line, so the angle BCE = 180o
angle C = 180o - angle 1
sin C = sin(180o - angle1) = sin180o cos(angle1) - cos180o sin(angle1)
Since sin180o = 0, cos180o = -1
So sin C = sin(angle1)
In right triangle ACE, sin(angle1) = AE/AC = AE/b,
so AE = b sin(angle1) = b sin C
In right triangle ABE, sin B = AE/AB = AE/c,
so AE = c sin B
So b sin C = c sin B
Both sides of the equation divide by sin B sin C
So b/sin B = c/sin C
Therefore, we get,
a/sin A = b/sin B = c/sin C
Thus we prove that the law of Sine satisfy the obtuse triangle.
# Apply the law of Sine to solve problems
Example 1
Question:
In triangle ABC, a = 49, B = 75o and C = 45o, what is the value of c?
Solution:
A = 180o - (B + C) = 180o - (75o + 45o) = 180o - 120o = 60o
The degree measure of the angle A is 60o
Apply the law of Sine
The value of c is 40.
Example 2
Question:
In triangle ABC, c = 120, C = 105o and b = 88, find a, A and B.
Solution:
sin C = sin105o
= sin(180o - 75o)
= sin75o
= sin(45o + 30o)
= sin45o cos30o + cos45o sin30o
= 0.9659
Apply the law of Sine
Because sin B = 0.7083, so B = 45o
A = 180o - (B + C)
= 180o - (45o + 105o)
= 180o - 150o
= 300
Apply the law of Sine
Thus, we get the triangle,
In triangle ABC, because C > B > A, so c > b > a
# Law of Cosine
If we know two sides and the angle between them, how to find the third side in the triangle? For example, if we know the sides b, c and the angle A, how to find the length of the side a?
In xy-plane, let the point A be the origin and the side AB lies on the a-axis. In the figure below, the coordinate of the point A is (0, 0), the coordinate of the point B is (x1, 0), [note: the y-coordinate of the point B is 0, because the point B lies on the x-axis], the coordinate of the point C is (x2, y2). Now we find the length of the side a.
Note: cos A = x2/square root of (x22 + y22)
Thus, we get the law of Cosine, a2 = b2 + c2 - 2bc cosA
# The law of Cosine
If we know three sides in a triangle, how to find the angles in this triangle? From the law of Cosine, we get the following formula.
Example 3
Question:
In triangle ABC, the length of the side a = 36, B = 45o and the length of the side c = 40, find the triangle.
Solution:
Find the length of the side b
b2 = a2 + c2 - 2ac cos B
= 362 + 402 - 2 × 36 × 40 cos45o
= 1296 + 1600 - 2880 × (square root of 2)/2
= 860
b = square root of 860 = 29.32
Find the angle B
Find the angle C
The triangle shown below
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# What Is The Best Example Of Newton Second Law Of Motion?
## What is Newton’s second law in simple terms?
Newton’s second law of motion can be formally stated as follows: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object..
## What is the best example of Newton first law of motion?
The motion of an airplane when the pilot changes the throttle setting of the engine is described by the first law. The motion of a ball falling down through the atmosphere, or a model rocket being launched up into the atmosphere are both examples of Newton’s first law.
## What is an example of Newton’s first law in everyday life?
Newton’s First Law of Motion Examples in Daily Life Wearing a seat belt in a car while driving is an example of Newton’s 1st law of motion. If an accident occurs, or if brakes are applied to the car suddenly, the body will tend to continue its inertia and move forward, probably proving fatal.
## What are 2 examples of Newton’s first law?
The motion of a ball falling down through the atmosphere, or a model rocket being launched up into the atmosphere are both examples of Newton’s first law. The motion of a kite when the wind changes can also be described by the first law.
## How do you demonstrate Newton’s second law of motion?
Newton’s Second Law of Motion You can demonstrate this principle by dropping a rock or marble and a wadded-up piece of paper at the same time. They fall at an equal rate—their acceleration is constant due to the force of gravity acting on them.
## What are 3 examples of Newton’s second law?
Examples of Newton’s 2nd Law If you use the same force to push a truck and push a car, the car will have more acceleration than the truck, because the car has less mass. It is easier to push an empty shopping cart than a full one, because the full shopping cart has more mass than the empty one.
## What is an example of Newton’s second law of motion in everyday life?
Following are Newton’s second law examples in everyday life: Pushing a car is easier than pushing a truck with the same amount of force as the mass of the car is lesser than the mass of the truck. In golf game, acceleration of the golf ball is directly proportional to the force with which it is hit by the golf stick.
## What is an example of Newton’s second law?
Newton’s Second Law of Motion says that acceleration (gaining speed) happens when a force acts on a mass (object). Riding your bicycle is a good example of this law of motion at work. Your bicycle is the mass. Your leg muscles pushing pushing on the pedals of your bicycle is the force.
## Which is the best example of Newton’s third law of motion?
Because of Newton’s Third Law. You hit the wall with a force, and that exact same amount of force is returned by the wall. While Rowing a boat, when you want to move forward on a boat, you paddle by pushing the water backwards, causing you to move forward.
## What are three examples of Newton’s first law?
Newton’s first law – examplesA stationary object with no outside force will not move.With no outside forces, a moving object will not stop. … An astronaut who has their screwdriver knocked into space will see the screwdriver continue on at the same speed and direction forever. … An object at rest stays at rest.More items…
## What is the first law of Newton?
His first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force.
## What are 3 examples of Newton’s 3rd law?
Gun Firing: when someone fire gun the reaction force push the gun backward. Jumping to land from boat: The action force applied on the boat and the reaction force pushes you to land. The action force push the boat backward. Slapping: When you slap someone your hand feel pain as well as the check of the victim.
## How do you explain Newton’s second law to a child?
Newton’s Second Law: F = ma: The greater the force the greater the acceleration. The greater the mass, the greater the force needed to move the object. Two objects with different weights (mass) will need different forces to move them and the acceleration levels will be different.
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## Nick2019 3 years ago What is the value of the coefficient of "a" when the quadratic equation y = (5x − 2)(2x + 3) is written in standard form?
1. Nick2019
a −11 b −6 c 7 d 10
2. satellite73
your job is to multiply $(5x − 2)(2x + 3)$ actually you only have to multiply $$5x\times 2x$$ to get $$a$$
3. cornitodisc
Ax^2+Bx+c so ur A=10
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https://as1air.com/how-many-books-has-stephen-king-written/
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# How many books has stephen king written
how many books has stephen king written is one of the most frequently asked questions.
## Why should I know how many books has stephen king written?
He who owns the information, owns the world – said V.Cherchill. Today the information lies around, so this phrase would sound like this: Не who knows where to find information, owns the world. Therefore, to answer the question how many books has stephen king written you need to know where to find the answer to it.
## How do I know how many books has stephen king written?
Today, there are many calculators for converting one value to another and vice versa. At the touch of a button, you can find out how many books has stephen king written. To do this, you need to write in the search box (for example, google) how many books has stephen king written and add to it an additional word: converter or calculator . Choose the calculator you like. And with his help find out how many books has stephen king written.
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https://www.gradesaver.com/textbooks/math/precalculus/precalculus-6th-edition/chapter-4-inverse-exponential-and-logarithmic-functions-chapter-4-test-prep-review-exercises-page-492/55
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Precalculus (6th Edition)
$\dfrac {10^{0.25}-7}{2}$
$\log \left( 2x+7\right) =0.25\Rightarrow 2x+7=10^{0.25}\Rightarrow x=\dfrac {10^{0.25}-7}{2}$
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https://physics.stackexchange.com/questions/214983/effect-of-redshift-on-energy-conservation?rq=1
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# Effect of redshift on energy conservation [duplicate]
Light coming from galaxies that are going away from us is redshifted. Since the energy of a photon is purely dependent on its frequency one may conclude that the energy of these photons decreases. The same light coming from the same star in the same galaxy will be seen to a planet in that galaxy as it "actually" is. So dependent on the relative motion the energy will be seen, differently. How does this not conflict with the conservation of energy? Or the "total" energy of the universe depend on the frame?
I was thinking that the light coming to us is located on a larger range since its period is larger. So the light carries the same energy for both observers. Even if its average energy is smaller total energies are same.
However, I could not convince myself and I think I am making wrong interpretations.
• Even in Newtonian physics energy depends on relative motion. If I am stationary I measure my own kinetic energy to be zero. However if you are moving relative to me at velocity $\mathbf{v}$, in your frame I am moving past you at velocity $-\mathbf{v}$ and therefore you measure my kinetic energy to be $Mv^2/2$. The conservation of energy simply states that in a given inertial frame, energy does not appear from nowhere. Different inertial observers do not have to measure the same energy, however. Oct 27, 2015 at 23:01
• You are right, but somehow I did not think about this issue like this before. Probably, I am making the mistake by trying to keep the physical quantities constant even if I change the frame. However, I am still confused about energy conservation. Oct 27, 2015 at 23:06
• If we assume no acceleration and we ignore spreading and we ignore time dilation, then it's really pretty simple. If the object is moving towards you the same amount of energy is spread out over a smaller distance over the time that it takes the light to reach you. If it's moving away from you, the energy is spread out over a larger distance. The red or blue shift, (we can assume equal number of photons), is essential to the conservation of energy. I obviously simplified the question quite a bit, to just 2 objects, but I think that's the gist of the answer. Oct 27, 2015 at 23:33
• The total energy of the universe has not been determined observationally, it's therefor not possible to say if energy is actually conserved on the scale of the universe or not. I may well not be. More likely, in my personal opinion, is that total energy is conserved and that we merely don't understand how photons actually interact with the physical vacuum on the scale of the entire universe. Oct 27, 2015 at 23:37
• It's curious to me that observer-specific effects are also present in quantum physics, since reference frames are in general a relativistic concept, by definition. However if a hydrogen atom A, moving at a high speed away from another hydrogen atom B, emits a photon at the frequency of the alpha Lyman line (by means of an electron jumping from an excited state in the second orbital down to the first orbital), that quantum of energy of the emitted photon cannot excite an electron in atom B to jump from the first to the second orbital -- the energy in the reference frame of B is insufficient. Oct 19, 2021 at 21:58
I prefer to look at it based on certain Laws and observations.
1. The First and second Law of Thermodynamics are true Energy is conserved and the universe is moving towards increasing entropy.
2. The Hubble constant is accurate and the universe is expanding at an exponential rate given by $a(t) = e^{Ht}$, where the constant $H$ is the Hubble expansion rate and $t$ is time. As in all FLRW spaces, $a(t)$, the scale factor, describes the expansion of physical spatial distances.
3. Redshift is the displacement of spectral lines toward longer wavelengths (the red end of the spectrum) in radiation from distant galaxies and celestial objects. This is interpreted as a Doppler shift that is proportional to the velocity of recession and thus to distance.
4. Blueshift a shift toward shorter wavelengths of the spectral lines of a celestial object, caused by the motion of the object toward the observer.
You say -
Light coming from galaxies that are going away from us is redshifted. Since the energy of a photon is purely dependent on its frequency one may conclude that the energy of these photons decreases. The same light coming from the same star in the same galaxy will be seen to a planet in that galaxy as it "actually" is. So dependent on the relative motion the energy will be seen, differently. How does this not conflict with the conservation of energy?
Ok let's say that is true we are experiencing a redshift but at the same time let's consider another observer who is moving at the same speed as our recession speed but towards the photon emitter source he is experiencing a blueshift. Each photon will have redshift and a corresponding blueshift depending on the relative frame of refences{one moving towards and one moving away}. The following picture makes it clearer.
"Dopplerfrequenz" by Charly Whisky 18:20, 27 January 2007 (yyy) - Own work. Licensed under CC BY-SA 3.0 via Commons.
An animation illustrating how the Doppler effect causes a car engine or siren to sound higher in pitch when it is approaching than when it is receding. The pink circles represent sound waves.
The energy from each siren waves doesn't change but depending where you are you will here a higher or lower pitch.
Now on a cosmologic scale the with an exponentially expanding universe with local Hubble horizons and trying to decide if the universe is a closed or open system leads to a cluster of different methods to try and verify the Conservation of energy of the universe. But a good way to think about this is to imagine open local systems exchanging matter and energy with each other across the universe and the sum of their energy diferences will be zero thereby conserving energy.
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The OEIS mourns the passing of Jim Simons and is grateful to the Simons Foundation for its support of research in many branches of science, including the OEIS.
The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A320285 Semiprimes followed by successive gaps 4, 6, 9. 1
8203, 9703, 18163, 35823, 72687, 72847, 75759, 95695, 97959, 132879, 177159, 194127, 198763, 201099, 210379, 223807, 226887, 228043, 299227, 306283, 344779, 347527, 351399, 360763, 403467, 407107, 454143, 487927, 506467, 514927, 516487, 532803, 537367, 538903, 546847, 556707, 562819 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS 4, 6, 9 are the first 3 semiprimes (A001358). Are there semiprimes followed by gaps {4, 6, 9, 10} = the first 4 semiprimes? Answer: No, one of them would be divisible by 4. - Giovanni Resta, Oct 23 2018 Semiprimes s such that the first semiprime after s equals s+4, the next one equals s+10, and the next one equals s+19. - Harvey P. Dale, Sep 25 2022 LINKS Harvey P. Dale, Table of n, a(n) for n = 1..4000 MATHEMATICA spQ[n_] := Plus @@ Last /@ FactorInteger[n] == 2; Select[ Range[10^6/2] 2 + 1, AllTrue[# + {0, 4, 10, 19}, spQ] && Count[ Range[ #+1, #+18], x_ /; spQ@ x] == 2 &] (* Giovanni Resta, Oct 23 2018 *) SequencePosition[If[PrimeOmega[#]==2, 1, 0]&/@Range[600000], {1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1}][[All, 1]] (* Harvey P. Dale, Sep 25 2022 *) PROG (PARI) next_semiprime(n) = for(x=n, oo, if(bigomega(x)==2, return(x))) is(n) = if(bigomega(n)!=2, return(0)); my(v=[n, next_semiprime(n+1), next_semiprime(next_semiprime(n+1)+1), next_semiprime(next_semiprime(next_semiprime(n+1)+1)+1)]); v[2]-v[1]==4 && v[3]-v[2]==6 && v[4]-v[3]==9 \\ Felix Fröhlich, Oct 23 2018 CROSSREFS Cf. A001358. Sequence in context: A168346 A331357 A045060 * A168471 A031844 A210008 Adjacent sequences: A320282 A320283 A320284 * A320286 A320287 A320288 KEYWORD nonn AUTHOR Zak Seidov, Oct 09 2018 STATUS approved
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Last modified May 28 01:12 EDT 2024. Contains 372900 sequences. (Running on oeis4.)
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Simplify (ab^2)(5a^2b^3)(3a^3)
Dec 18, 2014
Best Answer
#1
+20981
+10
(a·b²)(5·a²·b³)(3·a³) = (5·3)(a·a²·a³)(b²·b³) = (15)(a6)(b5) = 15a6b5
(add exponents when you multiply numbers with the same base)
Dec 18, 2014
1+0 Answers
#1
+20981
+10
Best Answer
(a·b²)(5·a²·b³)(3·a³) = (5·3)(a·a²·a³)(b²·b³) = (15)(a6)(b5) = 15a6b5
(add exponents when you multiply numbers with the same base)
geno3141 Dec 18, 2014
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https://schoollearningcommons.info/question/what-is-the-difference-between-largest-and-smallest-number-of-four-digits-24981892-54/
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## what is the difference between largest and smallest number of four digits ?
Question
what is the difference between largest and smallest number of four digits ?
in progress 0
1 month 2021-08-12T04:41:44+00:00 2 Answers 0 views 0
4086
Solution :
The greatest 4-digit number formed with the given digits is 8764.
The smallest 4-digit number formed with the given digits is 4678.
∴ The difference between the two numbers
8764-4678 = 4086.
Step-by-step explanation:
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https://en.m.wikipedia.org/wiki/Differential_of_a_function
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# Differential of a function
In calculus, the differential represents the principal part of the change in a function y = f(x) with respect to changes in the independent variable. The differential dy is defined by
${\displaystyle dy=f'(x)\,dx,}$
where ${\displaystyle f'(x)}$ is the derivative of f with respect to x, and dx is an additional real variable (so that dy is a function of x and dx). The notation is such that the equation
${\displaystyle dy={\frac {dy}{dx}}\,dx}$
holds, where the derivative is represented in the Leibniz notation dy/dx, and this is consistent with regarding the derivative as the quotient of the differentials. One also writes
${\displaystyle df(x)=f'(x)\,dx.}$
The precise meaning of the variables dy and dx depends on the context of the application and the required level of mathematical rigor. The domain of these variables may take on a particular geometrical significance if the differential is regarded as a particular differential form, or analytical significance if the differential is regarded as a linear approximation to the increment of a function. Traditionally, the variables dx and dy are considered to be very small (infinitesimal), and this interpretation is made rigorous in non-standard analysis.
## History and usage
The differential was first introduced via an intuitive or heuristic definition by Isaac Newton and furthered by Gottfried Leibniz, who thought of the differential dy as an infinitely small (or infinitesimal) change in the value y of the function, corresponding to an infinitely small change dx in the function's argument x. For that reason, the instantaneous rate of change of y with respect to x, which is the value of the derivative of the function, is denoted by the fraction
${\displaystyle {\frac {dy}{dx}}}$
in what is called the Leibniz notation for derivatives. The quotient dy/dx is not infinitely small; rather it is a real number.
The use of infinitesimals in this form was widely criticized, for instance by the famous pamphlet The Analyst by Bishop Berkeley. Augustin-Louis Cauchy (1823) defined the differential without appeal to the atomism of Leibniz's infinitesimals.[1][2] Instead, Cauchy, following d'Alembert, inverted the logical order of Leibniz and his successors: the derivative itself became the fundamental object, defined as a limit of difference quotients, and the differentials were then defined in terms of it. That is, one was free to define the differential dy by an expression
${\displaystyle dy=f'(x)\,dx}$
in which dy and dx are simply new variables taking finite real values,[3] not fixed infinitesimals as they had been for Leibniz.[4]
According to Boyer (1959, p. 12), Cauchy's approach was a significant logical improvement over the infinitesimal approach of Leibniz because, instead of invoking the metaphysical notion of infinitesimals, the quantities dy and dx could now be manipulated in exactly the same manner as any other real quantities in a meaningful way. Cauchy's overall conceptual approach to differentials remains the standard one in modern analytical treatments,[5] although the final word on rigor, a fully modern notion of the limit, was ultimately due to Karl Weierstrass.[6]
In physical treatments, such as those applied to the theory of thermodynamics, the infinitesimal view still prevails. Courant & John (1999, p. 184) reconcile the physical use of infinitesimal differentials with the mathematical impossibility of them as follows. The differentials represent finite non-zero values that are smaller than the degree of accuracy required for the particular purpose for which they are intended. Thus "physical infinitesimals" need not appeal to a corresponding mathematical infinitesimal in order to have a precise sense.
Following twentieth-century developments in mathematical analysis and differential geometry, it became clear that the notion of the differential of a function could be extended in a variety of ways. In real analysis, it is more desirable to deal directly with the differential as the principal part of the increment of a function. This leads directly to the notion that the differential of a function at a point is a linear functional of an increment Δx. This approach allows the differential (as a linear map) to be developed for a variety of more sophisticated spaces, ultimately giving rise to such notions as the Fréchet or Gateaux derivative. Likewise, in differential geometry, the differential of a function at a point is a linear function of a tangent vector (an "infinitely small displacement"), which exhibits it as a kind of one-form: the exterior derivative of the function. In non-standard calculus, differentials are regarded as infinitesimals, which can themselves be put on a rigorous footing (see differential (infinitesimal)).
## Definition
The differential of a function ƒ(x) at a point x0.
The differential is defined in modern treatments of differential calculus as follows.[7] The differential of a function f(x) of a single real variable x is the function df of two independent real variables x and Δx given by
${\displaystyle df(x,\Delta x){\stackrel {\mathrm {def} }{=}}f'(x)\,\Delta x.}$
One or both of the arguments may be suppressed, i.e., one may see df(x) or simply df. If y = f(x), the differential may also be written as dy. Since dx(x, Δx) = Δx it is conventional to write dx = Δx, so that the following equality holds:
${\displaystyle df(x)=f'(x)\,dx}$
This notion of differential is broadly applicable when a linear approximation to a function is sought, in which the value of the increment Δx is small enough. More precisely, if f is a differentiable function at x, then the difference in y-values
${\displaystyle \Delta y{\stackrel {\rm {def}}{=}}f(x+\Delta x)-f(x)}$
satisfies
${\displaystyle \Delta y=f'(x)\,\Delta x+\varepsilon =df(x)+\varepsilon \,}$
where the error ε in the approximation satisfies εx → 0 as Δx → 0. In other words, one has the approximate identity
${\displaystyle \Delta y\approx dy}$
in which the error can be made as small as desired relative to Δx by constraining Δx to be sufficiently small; that is to say,
${\displaystyle {\frac {\Delta y-dy}{\Delta x}}\to 0}$
as Δx → 0. For this reason, the differential of a function is known as the principal (linear) part in the increment of a function: the differential is a linear function of the increment Δx, and although the error ε may be nonlinear, it tends to zero rapidly as Δx tends to zero.
## Differentials in several variables
Operator / Function ${\displaystyle f(x)}$ ${\displaystyle f(x,y,u(x,y),v(x,y))}$
Differential 1: ${\displaystyle df\,{\overset {\underset {\mathrm {def} }{}}{=}}\,f'_{x}\,dx}$ 2: ${\displaystyle d_{x}f\,{\overset {\underset {\mathrm {def} }{}}{=}}\,f'_{x}\,dx}$
Partial derivative ${\displaystyle f'_{x}\,{\overset {\underset {\mathrm {(1)} }{}}{=}}\,{\frac {df}{dx}}}$ ${\displaystyle f'_{x}\,{\overset {\underset {\mathrm {(2)} }{}}{=}}\,{\frac {d_{x}f}{dx}}={\frac {\partial f}{\partial x}}}$
Total derivative ${\displaystyle {\frac {df}{dx}}\,{\overset {\underset {\mathrm {(1)} }{}}{=}}\,f'_{x}}$ ${\displaystyle {\frac {df}{dx}}\,{\overset {\underset {\mathrm {(3)} }{}}{=}}\,f'_{x}+f'_{u}{\frac {du}{dx}}+f'_{v}{\frac {dv}{dx}};(f'_{y}{\frac {dy}{dx}}=0)}$
Following Goursat (1904, I, §15), for functions of more than one independent variable,
${\displaystyle y=f(x_{1},\dots ,x_{n}),}$
the partial differential of y with respect to any one of the variables x1 is the principal part of the change in y resulting from a change dx1 in that one variable. The partial differential is therefore
${\displaystyle {\frac {\partial y}{\partial x_{1}}}dx_{1}}$
involving the partial derivative of y with respect to x1. The sum of the partial differentials with respect to all of the independent variables is the total differential
${\displaystyle dy={\frac {\partial y}{\partial x_{1}}}dx_{1}+\cdots +{\frac {\partial y}{\partial x_{n}}}dx_{n},}$
which is the principal part of the change in y resulting from changes in the independent variables xi.
More precisely, in the context of multivariable calculus, following Courant (1937b), if f is a differentiable function, then by the definition of differentiability, the increment
{\displaystyle {\begin{aligned}\Delta y&{}{\stackrel {\mathrm {def} }{=}}f(x_{1}+\Delta x_{1},\dots ,x_{n}+\Delta x_{n})-f(x_{1},\dots ,x_{n})\\&{}={\frac {\partial y}{\partial x_{1}}}\Delta x_{1}+\cdots +{\frac {\partial y}{\partial x_{n}}}\Delta x_{n}+\varepsilon _{1}\Delta x_{1}+\cdots +\varepsilon _{n}\Delta x_{n}\end{aligned}}}
where the error terms ε i tend to zero as the increments Δxi jointly tend to zero. The total differential is then rigorously defined as
${\displaystyle dy={\frac {\partial y}{\partial x_{1}}}\Delta x_{1}+\cdots +{\frac {\partial y}{\partial x_{n}}}\Delta x_{n}.}$
Since, with this definition,
${\displaystyle dx_{i}(\Delta x_{1},\dots ,\Delta x_{n})=\Delta x_{i},}$
one has
${\displaystyle dy={\frac {\partial y}{\partial x_{1}}}\,dx_{1}+\cdots +{\frac {\partial y}{\partial x_{n}}}\,dx_{n}.}$
As in the case of one variable, the approximate identity holds
${\displaystyle dy\approx \Delta y}$
in which the total error can be made as small as desired relative to ${\displaystyle {\sqrt {\Delta x_{1}^{2}+\cdots +\Delta x_{n}^{2}}}}$ by confining attention to sufficiently small increments.
### Application of the total differential to error estimation
In measurement, the total differential is used in estimating the error Δf of a function f based on the errors Δx, Δy, ... of the parameters x, y, …. Assuming that the interval is short enough for the change to be approximately linear:
Δf(x) = f'(x) × Δx
and that all variables are independent, then for all variables,
${\displaystyle \Delta f=f_{x}\Delta x+f_{y}\Delta y+\cdots }$
This is because the derivative fx with respect to the particular parameter x gives the sensitivity of the function f to a change in x, in particular the error Δx. As they are assumed to be independent, the analysis describes the worst-case scenario. The absolute values of the component errors are used, because after simple computation, the derivative may have a negative sign. From this principle the error rules of summation, multiplication etc. are derived, e.g.:
Let f(a, b) = a × b;
Δf = faΔa + fbΔb; evaluating the derivatives
Δf = bΔa + aΔb; dividing by f, which is a × b
Δf/f = Δa/a + Δb/b
That is to say, in multiplication, the total relative error is the sum of the relative errors of the parameters.
To illustrate how this depends on the function considered, consider the case where the function is f(a, b) = a ln b instead. Then, it can be computed that the error estimate is
Δf/f = Δa/a + Δb/(b ln b)
with an extra 'ln b' factor not found in the case of a simple product. This additional factor tends to make the error smaller, as ln b is not as large as a bare b.
## Higher-order differentials
Higher-order differentials of a function y = f(x) of a single variable x can be defined via:[8]
${\displaystyle d^{2}y=d(dy)=d(f'(x)dx)=(df'(x))dx=f''(x)\,(dx)^{2},}$
and, in general,
${\displaystyle d^{n}y=f^{(n)}(x)\,(dx)^{n}.}$
Informally, this motivates Leibniz's notation for higher-order derivatives
${\displaystyle f^{(n)}(x)={\frac {d^{n}f}{dx^{n}}}.}$
When the independent variable x itself is permitted to depend on other variables, then the expression becomes more complicated, as it must include also higher order differentials in x itself. Thus, for instance,
{\displaystyle {\begin{aligned}d^{2}y&=f''(x)\,(dx)^{2}+f'(x)d^{2}x\\d^{3}y&=f'''(x)\,(dx)^{3}+3f''(x)dx\,d^{2}x+f'(x)d^{3}x\end{aligned}}}
and so forth.
Similar considerations apply to defining higher order differentials of functions of several variables. For example, if f is a function of two variables x and y, then
${\displaystyle d^{n}f=\sum _{k=0}^{n}{\binom {n}{k}}{\frac {\partial ^{n}f}{\partial x^{k}\partial y^{n-k}}}(dx)^{k}(dy)^{n-k},}$
where ${\textstyle {\binom {n}{k}}}$ is a binomial coefficient. In more variables, an analogous expression holds, but with an appropriate multinomial expansion rather than binomial expansion.[9]
Higher order differentials in several variables also become more complicated when the independent variables are themselves allowed to depend on other variables. For instance, for a function f of x and y which are allowed to depend on auxiliary variables, one has
${\displaystyle d^{2}f=\left({\frac {\partial ^{2}f}{\partial x^{2}}}(dx)^{2}+2{\frac {\partial ^{2}f}{\partial x\partial y}}dx\,dy+{\frac {\partial ^{2}f}{\partial y^{2}}}(dy)^{2}\right)+{\frac {\partial f}{\partial x}}d^{2}x+{\frac {\partial f}{\partial y}}d^{2}y.}$
Because of this notational infelicity, the use of higher order differentials was roundly criticized by Hadamard 1935, who concluded:
Enfin, que signifie ou que représente l'égalité
${\displaystyle d^{2}z=r\,dx^{2}+2s\,dx\,dy+t\,dy^{2}\,?}$
A mon avis, rien du tout.
That is: Finally, what is meant, or represented, by the equality [...]? In my opinion, nothing at all. In spite of this skepticism, higher order differentials did emerge as an important tool in analysis.[10]
In these contexts, the nth order differential of the function f applied to an increment Δx is defined by
${\displaystyle d^{n}f(x,\Delta x)=\left.{\frac {d^{n}}{dt^{n}}}f(x+t\Delta x)\right|_{t=0}}$
or an equivalent expression, such as
${\displaystyle \lim _{t\to 0}{\frac {\Delta _{t\Delta x}^{n}f}{t^{n}}}}$
where ${\displaystyle \Delta _{t\Delta x}^{n}f}$ is an nth forward difference with increment tΔx.
This definition makes sense as well if f is a function of several variables (for simplicity taken here as a vector argument). Then the nth differential defined in this way is a homogeneous function of degree n in the vector increment Δx. Furthermore, the Taylor series of f at the point x is given by
${\displaystyle f(x+\Delta x)\sim f(x)+df(x,\Delta x)+{\frac {1}{2}}d^{2}f(x,\Delta x)+\cdots +{\frac {1}{n!}}d^{n}f(x,\Delta x)+\cdots }$
The higher order Gateaux derivative generalizes these considerations to infinite dimensional spaces.
## Properties
A number of properties of the differential follow in a straightforward manner from the corresponding properties of the derivative, partial derivative, and total derivative. These include:[11]
• Linearity: For constants a and b and differentiable functions f and g,
${\displaystyle d(af+bg)=a\,df+b\,dg.}$
${\displaystyle d(fg)=f\,dg+g\,df.}$
An operation d with these two properties is known in abstract algebra as a derivation. They imply the Power rule
${\displaystyle d(f^{n})=nf^{n-1}df}$
In addition, various forms of the chain rule hold, in increasing level of generality:[12]
• If y = f(u) is a differentiable function of the variable u and u = g(x) is a differentiable function of x, then
${\displaystyle dy=f'(u)\,du=f'(g(x))g'(x)\,dx.}$
{\displaystyle {\begin{aligned}dy&={\frac {dy}{dt}}dt\\&={\frac {\partial y}{\partial x_{1}}}dx_{1}+\cdots +{\frac {\partial y}{\partial x_{n}}}dx_{n}\\&={\frac {\partial y}{\partial x_{1}}}{\frac {dx_{1}}{dt}}\,dt+\cdots +{\frac {\partial y}{\partial x_{n}}}{\frac {dx_{n}}{dt}}\,dt.\end{aligned}}}
Heuristically, the chain rule for several variables can itself be understood by dividing through both sides of this equation by the infinitely small quantity dt.
• More general analogous expressions hold, in which the intermediate variables xi depend on more than one variable.
## General formulation
A consistent notion of differential can be developed for a function f : RnRm between two Euclidean spaces. Let xx ∈ Rn be a pair of Euclidean vectors. The increment in the function f is
${\displaystyle \Delta f=f(\mathbf {x} +\Delta \mathbf {x} )-f(\mathbf {x} ).}$
If there exists an m × n matrix A such that
${\displaystyle \Delta f=A\Delta \mathbf {x} +\|\Delta \mathbf {x} \|{\boldsymbol {\varepsilon }}}$
in which the vector ε → 0 as Δx → 0, then f is by definition differentiable at the point x. The matrix A is sometimes known as the Jacobian matrix, and the linear transformation that associates to the increment Δx ∈ Rn the vector AΔx ∈ Rm is, in this general setting, known as the differential df(x) of f at the point x. This is precisely the Fréchet derivative, and the same construction can be made to work for a function between any Banach spaces.
Another fruitful point of view is to define the differential directly as a kind of directional derivative:
${\displaystyle df(\mathbf {x} ,\mathbf {h} )=\lim _{t\to 0}{\frac {f(\mathbf {x} +t\mathbf {h} )-f(\mathbf {x} )}{t}}=\left.{\frac {d}{dt}}f(\mathbf {x} +t\mathbf {h} )\right|_{t=0},}$
which is the approach already taken for defining higher order differentials (and is most nearly the definition set forth by Cauchy). If t represents time and x position, then h represents a velocity instead of a displacement as we have heretofore regarded it. This yields yet another refinement of the notion of differential: that it should be a linear function of a kinematic velocity. The set of all velocities through a given point of space is known as the tangent space, and so df gives a linear function on the tangent space: a differential form. With this interpretation, the differential of f is known as the exterior derivative, and has broad application in differential geometry because the notion of velocities and the tangent space makes sense on any differentiable manifold. If, in addition, the output value of f also represents a position (in a Euclidean space), then a dimensional analysis confirms that the output value of df must be a velocity. If one treats the differential in this manner, then it is known as the pushforward since it "pushes" velocities from a source space into velocities in a target space.
## Other approaches
Although the notion of having an infinitesimal increment dx is not well-defined in modern mathematical analysis, a variety of techniques exist for defining the infinitesimal differential so that the differential of a function can be handled in a manner that does not clash with the Leibniz notation. These include:
## Examples and applications
Differentials may be effectively used in numerical analysis to study the propagation of experimental errors in a calculation, and thus the overall numerical stability of a problem (Courant 1937a). Suppose that the variable x represents the outcome of an experiment and y is the result of a numerical computation applied to x. The question is to what extent errors in the measurement of x influence the outcome of the computation of y. If the x is known to within Δx of its true value, then Taylor's theorem gives the following estimate on the error Δy in the computation of y:
${\displaystyle \Delta y=f'(x)\Delta x+{\frac {(\Delta x)^{2}}{2}}f''(\xi )}$
where ξ = x + θΔx for some 0 < θ < 1. If Δx is small, then the second order term is negligible, so that Δy is, for practical purposes, well-approximated by dy = f'(xx.
The differential is often useful to rewrite a differential equation
${\displaystyle {\frac {dy}{dx}}=g(x)}$
in the form
${\displaystyle dy=g(x)\,dx,}$
in particular when one wants to separate the variables.
## Notes
1. ^ For a detailed historical account of the differential, see Boyer 1959, especially page 275 for Cauchy's contribution on the subject. An abbreviated account appears in Kline 1972, Chapter 40.
2. ^ Cauchy explicitly denied the possibility of actual infinitesimal and infinite quantities (Boyer 1959, pp. 273–275), and took the radically different point of view that "a variable quantity becomes infinitely small when its numerical value decreases indefinitely in such a way as to converge to zero" (Cauchy 1823, p. 12; translation from Boyer 1959, p. 273).
3. ^ Boyer 1959, p. 275
4. ^ Boyer 1959, p. 12: "The differentials as thus defined are only new variables, and not fixed infinitesimals..."
5. ^ Courant 1937a, II, §9: "Here we remark merely in passing that it is possible to use this approximate representation of the increment Δy by the linear expression hf(x) to construct a logically satisfactory definition of a "differential", as was done by Cauchy in particular."
6. ^ Boyer 1959, p. 284
7. ^ See, for instance, the influential treatises of Courant 1937a, Kline 1977, Goursat 1904, and Hardy 1905. Tertiary sources for this definition include also Tolstov 2001 and Itô 1993, §106.
8. ^ Cauchy 1823. See also, for instance, Goursat 1904, I, §14.
9. ^ Goursat 1904, I, §14
10. ^ In particular to infinite dimensional holomorphy (Hille & Phillips 1974) and numerical analysis via the calculus of finite differences.
11. ^ Goursat 1904, I, §17
12. ^ Goursat 1904, I, §§14,16
13. ^
14. ^
15. ^ See Robinson 1996 and Keisler 1986.
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# Challenge
A student was asked to find the volume of the bowl-shaped figure generated when the curve y = x2 from x = 0 to x = 2 is revolved around the y-axis. She used the disk method and found the volume to be $\displaystyle \int_{0}^{4}{\pi ydy}=8\pi$. To check her work she used the method of cylindrical shells and found the same answer: $\displaystyle\int_{0}^{2}{2\pi x\left( 4-{{x}^{2}} \right)dx}=8\pi$.
The second part of the question asked for the value of x for which the bowl would be half full. So she first solved the equation $\displaystyle \int_{0}^{k}{\pi ydy=\frac{1}{2}\cdot 8\pi }$ and found $k=2\sqrt{2}$. This is a y-value so the corresponding x-value is $\sqrt{2\sqrt{2}}\approx 1.682$. She again checked her work by shells by solving $\displaystyle \int_{0}^{k}{2\pi x\left( 4-{{x}^{2}} \right)dx=\frac{1}{2}\cdot 8\pi }$ and found that $k=\sqrt{-2\left( \sqrt{2}-2 \right)}\approx 1.082$. (This is the x-value.)
Both computations are correct. Can you explain to her why her answers are different? Use the comment box below to share your explanations. I will post mine in a week or so.
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## Show a list of "Not-Sold Critical Itens" per client
I am developing a visual that shows what additional products from my portfolio I should offer to each client.
The reasoning behind it is something like this:
• I choose a channel/segment of clients (columns present in my dimension table dim_client)
• I list all the materials sold for this clients
• I check what products are bought by more than a specified X% of the total clients (this itens are now called "SUPERMIX")
• After that, I check every client that are part of this channel/segment, and list what itens from the Supermix they are NOT buying.
So far how I've done (Refer to the image below):
1. I put a filter for channel and segment acting in the page, as well as a date of sales
2. I created the following measures:
---------------------------------------------------------------------------------------
----------------------------------------------------------------------------------------
----------------------------------------------------------------------------------------
N_Clients_Percentage_Of_Total =
//here we check the % of clients that buy each item
VAR _all_clients =
CALCULATE(
ALLSELECTED(dim_Products)
)
--------------------------------------------------------------------------------------------
Trigger_Perc_Client =
//Here we check if the percentage of clients that buy the item is greater than our trigger
VAR _trigger_select = Trigger_0_100[Trigger_0_100_Valor]
RETURN IF([N_Clients_Percentage_Of_Total]>=_trigger_select,1,0)
----------------------------------------------------------------------------------------------------
Trigger_Perc_Client_ALL =
//basically we just put a "ALLSELECTED" to be able to use this measure=1 as filter in the visual
CALCULATE([Trigger_Perc_Cliente],ALLSELECTED(dim_Clientes))
------------------------------------------------------------------------------------------------
• Now I create the left table, with the PK_Material column, the [N_clients_Buy] and the [N_Clients_Percentage_Of_Total] measure columns. The "Star" appears based on the result of the [Trigger_Perc_Client]. In this example, every item that is bought by more than 50% of the clients is flagged.
So far so good! I created my "SuperMix" porfolio 😀
• The right table is basically a matrix with "PK_Client" and "PK_Material" and the [Client_Buys_N_Itens] measure. By using the [Trigger_Perc_Client_ALL] = 1 as a filter, I see only the "supermix" products, which is great. So I can see which supermix products each client already buy.
But finally, what I really want is to see what supermix product each client DOES NOT buy. I thought that by selecting the "Show itens with no data" in the visual would show all the itens for all the custmores, but with a "N_Itens" =0 for the non-sold itens. But whenever I try it the program starts running forever and never finishes.
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# Commutative Associative and Distributive Laws
In ordinary scalar algebra, the additive and multiplicative operations obey the commutative, associative, and distributive laws as follows:
Commutative law of addition: Commutative law of multiplication: Associative law of addition; Associative law of multiplication: Distributive law:
a 4 b = b 4 a ah = ha (<t + b)+c = a+{h + c) (ab)c = a(bc) a(b 4 c) = ab 4 ac
These have been referred to during the discussion of the similarly named laws applicable to the union and intersection of sets. Most, but not all, of these laws also apply to matrix operations—the significant exception being the commutative law of multiplication.
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Copyright © University of Cambridge. All rights reserved.
## 'I'm Eight' printed from http://nrich.maths.org/
### Show menu
Rachel from West Flegg Middle School has made decisions about things such as what numbers she wanted to use and what sort of mathematics she used. And she look for patterns! She says:
Hi, I'm Rachel. I am nearly eleven, so I thought I would write about "I'm Eleven''. In my investigation of different ways to find eleven, I will be using addition, subtraction, fractions, decimals and timesing. Some sums will use all and some will use some, but whatever, I will make eleven.
1 (10 + 78) $\div$ 8 = 11 2 (0.8 $\times$ 10) +3 = 11 3 0.11 $\times$100 = 11 4 ((11/12 of 72) $\div$ 11) +5 = 11 5 50 - 39 = 11 6 (3 $\times$ 12) - (100 $\div$ 4)= 11
I also, apart from these sums, found 2 sets of patterns. Here they are:
Pattern 1
(4 $\times$3) - 1 = 11 (5 $\times$3) - 4 = 11 (6 $\times$3) - 7 = 11 (7 $\times$3) - 10 = 11
Pattern 2
(4 $\times$11)-(3 $\times$11) = 11 (5 $\times$11)-(4 $\times$11) = 11 (6 $\times$11)-(5 $\times$11) = 11 (7 $\times$11)-(6 $\times$11) = 11
7 (132 $\div$ 10) - 2.2 = 11 8 52 - 41 = 11 9 (77 $\div$ 11) + 4 = 11 10 (11 $\times$ 11) $\div$ 11= 11 11 249.15 $\div$ 22.65 = 11 12 3$^2$ + 2 = 11
In 2015 we had a number of solutions sent in from The Spinney School.
Here is a glimpse of them, but all the pupils' work can be see in this file Spinney School.doc .
Thank you all at the Spinney for your good work and thinking through the whole idea of using what you know to get your total.
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# RS Aggarwal Class 10 Solutions Chapter 11 - Arithmetic Progressions Ex 11A (11.1)
## RS Aggarwal Class 10 Chapter 11 - Arithmetic Progressions Ex 11A (11.1) Solutions Free PDF
Q.1: Find the 6th term from the end of the AP 17, 14, 11, …….., (-40) .
Sol:
Here a = 17, d = (14 – 17) = – 3, I = -40
And n = 6
Now, $n^ {th}$ Â term from the end = [I – (n – 1) d]
= [- 40 – (6 – 1) (-3)]
= [- 40 + 5 x 3]
= – 40+15
= – 25
Hence, the 6th term from the end is   – 25.
Q2: Is 184 a term of the AP 3, 7, 11, 15, ……. ?
Sol:
The given AP is 3, 7, 11, 15, …….
Here, a = 3 and d = 7 – 3 = 4
Let nth term of the given AP be 184.
Then,
$a_{n} = 184$
$\Rightarrow 3 + (n – 1) \times 4 = 184$
$\Rightarrow 4n – 1 = 184$
$\Rightarrow 4n = 185$
$\Rightarrow n = \frac{184}{4} = 46 \frac{1}{4}$
But the number of terms cannot be a fraction.
Hence, 184 is not a term of the given AP.
Q.3: Is – 150 a term of the AP 11, 8, 5, 2, …… ?
Sol:
The given AP is 11, 8, 5, 2, ……
Here, a = 11 and d = 8 – 11 = – 3
Let nth term of the given AP be 150.
Then,
$a_{n} = 150$
$\Rightarrow 11 + (n – 1) \times (-3) = 150$
$\Rightarrow – 3n + 14 = – 150$
$\Rightarrow – 3n = – 164$
$\Rightarrow n = \frac{164}{3} = 54 \frac{2}{3}$
But the number of terms cannot be a fraction.
Hence, -150 is not a term of the given AP.
Q.4: Which term of the AP 121, 117, 113, ….. is its first negative term?
Sol:
The given AP is 121, 117, 113, …..
Here, a = 121 and d= 117 – 121 = -4
Let the nth term of the given AP be the first negative term. Then,
$a_{n} < 0$
$\Rightarrow 121 + (n – 1) \times (-4) < 0$
$\Rightarrow 125 – 4n < 0$
$\Rightarrow -4n < -125$
$\Rightarrow n > \frac{125}{4} = 31\frac{1}{4}$
$Therefore, n = 32$
Hence, the 32nd term is the first negative term of the given AP.
Q.5: Which term of the AP is $20, 19\frac{1}{4}, 18\frac{1}{2}, 17\frac{3}{4}, …….$ is its negative term?
Sol:
The given AP is $20, 19\frac{1}{4}, 18\frac{1}{2}, 17\frac{3}{4}, …….$
Here, a = 20 and d = $19\frac{1}{4} – 20 = \frac{77}{4} – 20 = -\frac{3}{4}$
Let the nth term of the given AP be the first negative term. Then,
$a_{n} < 0$
$\Rightarrow 20 + (n – 1) \times \left ( -\frac{3}{4} \right ) Â < 0$
$\Rightarrow 20 + \frac{3}{4} – \frac{3}{4}n < 0$
$\Rightarrow \frac{83}{4} – \frac{3}{4}n < 0$
$\Rightarrow -\frac{3}{4}n < – \frac{83}{4}$
$\Rightarrow n > \frac{83}{3} = 27 \frac{2}{3}$
$Therefore, n = 28$
Hence, the 28th term is the first negative term of the given AP.
Q.6: The 7th term of an AP is -4 and its 13th term is -16. Find the AP.
Sol:
In the given AP, let the first term = a, and common difference = d
Then, $T_ {n}$ = a + (n – 1) d
$\Rightarrow$ Â Â $T_ {7}$ = a + (7 – 1)d, and $T_{13}$ Â = a + (13 – 1)d
$\Rightarrow$ Â Â $T_ {7}$ = a + 6d, $T_ {13}$ = a + 12d
Now, $T_ {7}$ = -4
$\Rightarrow$ a + 6d = -4 – – – (1)
$T_{13}$ = -16
$\Rightarrow$ Â Â a + 12d = -16 – – – (2)
Subtracting (1) from (2), we get
$\Rightarrow$ 6d = -12 d = -2
Putting d = -2 in (1), we get
a + 6 (-2) = -4
$\Rightarrow$ a – 12 = -4
$\Rightarrow$ Â Â a = 8
Thus, a = 8, and d = -2
So the required AP is 8, 6, 4, 2, 0   …
Q.7: The 4th term of an AP is 0. Prove that its 25th term is triple its 11th term.
Sol:
In the given AP, let the first term = a, and common difference = d
Then, $T_ {n}$ = a + (n – 1) d
Now, $T_ {4}$ = a + (4 – 1) d
$\Rightarrow$ a + 3 d = 0……..(i)
$\Rightarrow$ a = -3d
Again, $T_ {11}$ = a + (11 – 1)d = Â a + 10d
= -3d + 10 d = 7d             [Using (i) ]
Also, $T_ {25}$ = a + (25 – 1)d =  a + 24d = -3d + 24d = 21d                       [Using (i) ]
i.e. $T_{25} = 3 \times 7d = Â 3 \times T_{11}$
Hence, 25th term is triple its 11th term.
Q.8: The 8th term of an AP is zero. Prove that its 38th term is triple its 18th term.
Sol:
Let a be the first term and d be the common difference of the AP. Then,
$a_{8} = 0$
$\Rightarrow a + (8 – 1)d = 0$
$\Rightarrow a + 7 d = 0$
$\Rightarrow a = -7d$ …………..(i)
Now,
$\Rightarrow \frac{a_{38}}{a_{18}} = \frac{a + (38 – 1)d}{a + (18 – 1)d}$
$\Rightarrow \frac{a_{38}}{a_{18}} = \frac{ – 7d + 37 d} {- 7d + 17d}$
$\Rightarrow \frac{a_{38}}{a_{18}} = \frac{30d } {10 d} = 3$
$a_{38} = 3 \times a_{18}$
Hence, the 38th term of the AP is triple its 18th term.
Q.9: The 4th term of an AP is 11. The sum of the 5th and 7th term of this AP is 34. Find its common difference.
Sol:
Let a be the first term and d be the common difference of the AP. Then,
$a_{4} = 11$
$\Rightarrow a + (4 – 1)d = 11$
$\Rightarrow a + 3 d = 11$        …………(i)
Now,
$a_{5} + a_{7} = 34$ Â Â Â Â Â Â Â Â Â Â Â Â (Given)
$\Rightarrow (a + 4d) + (a + 6d) = 34$
$\Rightarrow 2a + 10d = 34$
$\Rightarrow a + 5d = 17$            ……….(ii)
From (i) and (ii) we get
11 – 3d + 5d = 17
$\Rightarrow 2d = 17 – 11 = 6$
$\Rightarrow d = 3$
Hence, the common difference of the AP is 3.
Q.10: The 9th term of an AP is -32 and the sum of its 11th and 13th term is -94. Find the common difference of the AP.
Sol:
Let a be the first term and d be the common difference of the AP. Then,
$a_{4} = 11$
$\Rightarrow a + (9 – 1)d = – 32$
$\Rightarrow a + 8 d = -32$        …………(i)
Now,
$a_{11} + a_{13} = -94$ Â Â Â Â Â Â Â Â Â Â Â Â (Given)
$\Rightarrow (a + 10d) + (a + 12d) = -94$
$\Rightarrow 2a + 22d = -94$
$\Rightarrow a + 11d = -47$            ……….(ii)
From (i) and (ii) we get
– 32 – 8d + 11d = -47
$\Rightarrow 3d = -47 + 32 = -15$
$\Rightarrow d = -5$
Hence, the common difference of the AP is -5.
Q.11: Determine the nth term of the AP whose 7th term is -1 and 16th term is 17.
Sol:
Let a be the first term and d be the common difference of the AP. Then,
$a_{ 7 }$ = -1
→ a + (7 – 1) d = -1       $\left [ a_{ n } = a + (n – 1)d \right ]$
→ a + 6d = -1      … (1)
Also,
$a_{ 16 }$ = 17
→ a + 15d = 17             …(2)
From (1) and (2), we get
-1 – 6d + 15d = 17
→ 9d = 17 + 1 = 18
→ d = 2
Putting d = 2 in (1), we get
a + 6 x 2 = -1
→ a = -1 – 12 = -13
Therefore, $a_{ n } = a + (n – 1)d$
= -13 + (n – 1) x 2
= 2n – 15
Hence, the nth term of the AP is (2n – 15)
Q.12: If 4 times the 4th term of an AP is equal to 18 times its 18th term then find its 22nd term.
Sol:
Let a be the first term and d be the common difference of the AP. Then,
4 x $a_{ 4 } = 18 \times a_{ 18 }$ Â Â Â (Given)
→ 4 (a + 3d) = 18 (a + 17d)    $\left [ a_{ n } = a + (n – 1)d \right ]$
→ 2 (a + 3d) = 9 (a + 17d)
→ 2a + 6d = 9a + 153d
→ 7a = -147d
→ a = -21d
→ a + 21d = 0
→ a + (22 – 1)d = 0
→ $a_{ 22 }$ = 0
Hence, the 22nd term of the AP is 0.
Q.13: If 10 times the 10th term of an AP is equal to 15 times the 15th term, show that its 25th term is zero.
Sol:
Let a be the first term and d be the common difference of the AP. Then,
10 x $a_{ 10 } = 15 \times a_{ 15 }$ (Given)
→ 10 (a + 9d) = 15 (a + 14d)  $\left [ a_{ n } = a + (n – 1)d \right ]$
→ 2 (a + 9d) = 3(a + 14d)
→ 2a + 18d = 3a + 42d
→ a = -24d
→ a + 24d = 0
→ a + (25 – 1) d = 0
→ $a_{ 25 }$ = 0
Hence, the 25th term of the AP is 0.
Q.14: Find the common difference of an AP whose first term is 5 and the sum of its first four terms is half the sum of the next four terms.
Sol:
Let the common difference of the AP be d.
First term, a = 5
Now,
$a_{ 1 } + a_{ 2 } + a_{ 3 } + a_{ 4 } = \frac{ 1 }{ 2 } (a_{ 5 } + a_{ 6 } + a_{ 7 } + a_{ 8 })$Â Â Â Â Â Â Â Â Â Â Â (Given)
→ a + (a + d) + (a + 2d) + (a + 3d) = $\frac{ 1 }{ 2 }$ [ (a + 4d) + ( a + 5d) + (a + 6d) + (a + 7d)]
 $\left [ a_{ n } = a + (n – 1)d \right ]$
→ 4a + 6d = $\frac{ 1 }{ 2 }$ (4a + 22d)
→ 8a + 12d = 4a + 22d
→ 22d – 12d = 8a – 4a
→ 10d = 4a
→ d = $\frac{ 2 }{ 5 }a$
→ d = $\frac{ 2 }{ 5 } \times 5 = 2$  (a = 5)
Hence, the common difference of the AP is 2.
Q.15: The sum of the 2nd and the 7th terms of an AP is 30. If its 15th term is 1 less than twice its 8th term, find the AP.
Sol:
Let a be the first term and d be the common difference of the AP. Then,
$a_{ 2 } + a_{ 7 } = 30$
Therefore, (a + d) + (a + 6d) = 30Â Â Â Â $\left [ a_{ n } = a + (n – 1)d \right ]$
→ 2a + 7d = 30             ….(1)
Also,
$a_{ 15 } = 2a_{ 8 } – 1$Â Â Â Â Â Â Â Â Â (Given)
→ a + 14d = 2 (a + 7d) – 1
→ a + 14d = 2a + 14d – 1
→ -a = -1
→ a = 1
Putting a = 1 in (1), we get
2 x 1 + 7d = 30
→ 7d = 30 – 2 = 28
→ d = 4
So,
$a_{ 2 }$ = a + d = 1 + 4 = 5
$a_{ 3 }$ = a +2d = 1 + 2 x 4 = 9,…
Hence, the AP is 1, 5, 9, 13,…
Q.16: For what value of n, the nth terms of the arithmetic progressions 63, 65, 67,… and 3, 10, 17,.. are equal?
Sol:
Let the nth term of the given progressions be $t_{ n }$ and $T_{ n }$ ,respectively.
The first AP is 63, 65, 67,…
Let its first term be a and the common difference be d.
Then a = 63, and d = (65 – 63) = 2
So, its $n^{th}$ term is given by
$T_{ n }$ = A + (n -1) D
→ 3 + (n -1) x 7
→ 7n – 4
Now, $t_{ n }$ = $T_{ n }$
→ 61 + 2n = 7n – 4
→ 65 = 5n
→ n = 13
Hence, the 13th terms of the AP’s are the same.
Q.17: The 17th term of an AP is 5 times more than twice its 8th term. If the 11th term of the AP is 43, find its nth term.
Sol:
Let a be the first term and d be the common difference of the AP. Then,
$a_{ 17 } = 2a_{ 8 } + 5$Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (Given)
Therefore, a + 16d = 2 (a + 7d) + 5Â Â $\left [ a_{ n } = a + (n – 1)d \right ]$
→ a + 16d = 2a + 14d + 5
→ a – 2d = -5       …(1)
Also,
$a_{ 11 }$ = 43Â Â Â Â Â Â Â Â Â Â Â Â Â Â (Given)
→ a + 10d = 43             …(2)
From (1) and (2), we get
-5 + 2d + 10d = 43
→ 12d = 43 + 5 = 48
→ d = 4
Putting d = 4 in (1), we get
a – 2 x 4 = -5
→ d = 4
Therefore, $a_{ n } = a + (n – 1)d$
= 3 + (n – 1) x 4
= 4n – 1
Hence, the nth term of the AP is (4n – 1)
#### Practise This Question
On combining two aqueous solutions of soluble ionic compounds, a precipitate is formed when
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Relativistic Energy
# Relativistic Energy
## What is Relativistic energy
Relativistic energy is the total energy of a particle as observed by an observer in a reference frame in which the particle is moving at a relativistic speed. It is a consequence of the theory of relativity, which states that the laws of physics are the same for all observers, regardless of their relative motion.
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The relativistic energy of a particle with mass m and velocity v is given by the following equation:
E = mc^2 / √(1-v^2/c^2)
where:
E is the relativistic energy of the particle m is the rest mass of the particle v is the velocity of the particle c is the speed of light in a vacuum (approximately 3.00 x 10^8 m/s)
This equation is known as the mass-energy equivalence formula and shows that the energy of a particle increases as its velocity increases, approaching infinity as the velocity approaches the speed of light.
In classical physics, the total energy of a particle is equal to its kinetic energy (the energy it possesses due to its motion) plus its potential energy (the energy it possesses due to its position or configuration). However, in the theory of relativity, the concept of kinetic energy is modified to include the rest mass energy of the particle, and the total energy of a particle is given by the mass-energy equivalence formula above.
## Relativistic energy Derivation
The relativistic energy of a particle with mass m and velocity v is given by the following equation:
E = mc^2 / √(1-v^2/c^2)
where:
E is the relativistic energy of the particle m is the mass of the particle v is the velocity of the particle c is the speed of light in a vacuum (approximately 3.00 x 10^8 m/s)
This equation is known as the mass-energy equivalence formula and is a consequence of the theory of relativity. It states that the total energy of a particle is equal to its rest mass energy (mc^2) plus its kinetic energy (Ek). In the case of a particle moving at a velocity v, the kinetic energy can be expressed as:
Ek = mc^2 * (√(1-v^2/c^2) – 1)
Substituting this expression for Ek into the mass-energy equivalence formula gives us:
E = mc^2 + mc^2 * (√(1-v^2/c^2) – 1)
Which simplifies to:
E = mc^2 / √(1-v^2/c^2)
This equation shows that the energy of a particle increases as its velocity increases, approaching infinity as the velocity approaches the speed of light.
## Application of Relativistic energy
The concept of relativistic energy is important in a number of fields, including particle physics, astronomy, and engineering. Some applications of relativistic energy include:
1. Particle accelerators: Relativistic energy is used to calculate the energy of particles moving at speeds close to the speed of light in particle accelerators.
2. Cosmic rays: Cosmic rays are high-energy particles that originate from outer space and can be detected on Earth. Relativistic energy is used to calculate the energy of these particles.
3. Nuclear power: In nuclear power plants, the energy released by nuclear reactions is calculated using the mass-energy equivalence formula.
4. Relativistic rocket propulsion: Theoretical designs for rocket engines that would use the mass-energy equivalence formula to achieve relativistic velocities have been proposed.
5. High-energy physics: Relativistic energy is important in the study of high-energy particle collisions, such as those that occur at the Large Hadron Collider.
6. Astrophysics: Relativistic energy is used to calculate the energy of particles and objects in space, such as black holes and neutron stars.
7. Medical imaging: Relativistic energy is used in PET (Positron Emission Tomography) scanners, which use high-energy particles to create images of the inside of the human body.
## Related content
Difference Between Mass and Weight Differences & Comparisons Articles in Physics Important Topic of Physics: Reynolds Number Distance Speed Time Formula Refractive Index Formula Mass Formula Electric Current Formula Electric Power Formula Resistivity Formula Weight Formula
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# If the tens digit x and the units digit y of a positive
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If the tens digit x and the units digit y of a positive [#permalink] 26 Feb 2006, 09:43
If the tens digit x and the units digit y of a positive integer n are
reversed, the resulting integer is 9 more than n. What is y in terms of x ?
(A) 10 - x
(B) 9 - x
(C) x + 9
(D) x - 1
(E) x + 1
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E.
From first statement:
x*10+y=n
When x and y are reversed:
y*10+x=n+9
Solving both equations:
10y+x=10x+y+9
9y=9x+9
y=x+1
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Before reverse n = 10x + y
After reverse number becomes 10y+x --> this is 9 more than n
10y+x-9 = 10x+y
9y = 9x+9
y = x+1
Ans E
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The product of the units digit, the tens digit, and the 8 14 Oct 2008, 14:46
The product of the units digit, the tens digit, and the 2 17 May 2008, 01:57
The product of the units digit, the tens digit, and the 0 16 Nov 2014, 08:52
The product of the units digit, the tens digit, and the 4 08 Feb 2008, 02:43
if the tens digit x and the units digit y of a positive 6 18 Oct 2006, 15:07
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Search found 172 matches
Mon Mar 28, 2011 7:31 pm
Forum: Combinatorics
Topic: Probability: coin toss
Replies: 7
Views: 3179
Re: Probability: coin toss
A valid sequence is a sequence of tosses which doesn't contain a pair of consecutive heads. Lets define three sequences : i) $T_1=2,\ T_2=3,\ T_{i+2}=T_{i+1}+T_i$ i) $t_1=1,\ t_2=2,\ t_{i+2}=t_{i+1}+t_i$ i) $h_1=1,\ h_2=1,\ h_{i+2}=h_{i+1}+h_i$ Notice that if the coin is tossed for $k$ times, the nu...
Sun Mar 27, 2011 6:47 pm
Forum: Geometry
Topic: Game of pool? (own)
Replies: 6
Views: 2933
Re: Game of pool? (own)
I will explain my solution later (Sorry... actually I don't have enough time to write it right now )
I just want to be sure about the solution. Is it
?
Mon Feb 21, 2011 6:49 pm
Forum: Secondary Level
Topic: Circle is symetric about diameter
Replies: 11
Views: 4754
Re: Circle is symetric about diameter
the reflection of A and B are ACP and BCQ. Actually the term 'reflection' is almost same as physics as Moon vai said. So the reflection of a man will be a man, a tree will be a tree, a line will be a line, a point will be a point. Probably you didn't mean reflection here. Please attach a photo (you...
Thu Feb 17, 2011 8:39 am
Forum: Secondary Level
Topic: Circle is symetric about diameter
Replies: 11
Views: 4754
Easy to prove but interesting :
$A$ and $B$ are two points on a circle and $C$ is a point on diameter $MN$ such that $\angle ACN=\angle BCN$
Prove that $AC=BC$
Wed Feb 16, 2011 7:48 pm
Forum: Computer Science
Topic: number of common divisors
Replies: 4
Views: 2385
Re: number of common divisors
আমি জানি না এইটা কাজ করবে কিনা। তবে চেষ্টা করে দেখা যায়। একটা সংখ্যা যদি a,b কে করে তাইলে তাদের gcd কেউ ভাগ করবে। input দেওয়া সংখ্যা ২ টার gcd বের করে তার উৎপাদক সংখ্যা বের করা যায়। এক্ষেত্রে আরকি একটা সংখ্যার prime power factorization করা লাগতিসে, আগে যেখানে ২ টা সংখ্যার করা লাগতো। gcd বের করতে ...
Sat Feb 12, 2011 6:59 am
Forum: International Olympiad in Informatics (IOI)
Replies: 21
Views: 11202
2010 e HSC dise ba tar por HSC dibe emon sobai allowed.
btw, chorom boka ta ke?
Wed Feb 09, 2011 9:23 pm
Forum: International Olympiad in Informatics (IOI)
Replies: 21
Views: 11202
হুম, তবে আশার (বা দুরাশার) কথা হইল বেশি লোকজন উৎসাহী না
HandaramTheGreat wrote:সবকিছু শেষ হতে আনুমানিক ক'টা বাজতে পারে?
গতবার ৪.৩০ টার দিকে result দেয়া হইছিল মনে হয়...
Wed Feb 09, 2011 1:42 pm
Forum: International Olympiad in Informatics (IOI)
Topic: Algorithms
Replies: 9
Views: 5499
Algorithms
Can anyone suggest some useful algorithms? For example, I have just learned Dijkstra's algorithm.
Important: Please don't think that, Zzzz probably knows this 'easy and wellknown' algorithm, so I need not to suggest it.
Wed Feb 09, 2011 7:03 am
Forum: International Olympiad in Informatics (IOI)
Replies: 21
Views: 11202
If $P$ divides $(P-1)!+1$, prove that $P$ is a prime.
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# AP Calculus AB : Relationship between differentiability and continuity
## Example Questions
### Example Question #1 : Relationship Between Differentiability And Continuity
The function is differentiable at the point . List which of the following statements must be true about :
1) The limit exists.
_________________________________________________________
2)
_________________________________________________________
3)
_________________________________________________________
4)
_________________________________________________________
5)
1,3,4,and 5
1, 2, and 4
1, and 5
All must be true.
1, 3, and 5
1, 2, and 4
Explanation:
1) If a function is differentiable, then by definition of differentiability the limit defined by,
exists. Therefore (1) is required by definition of differentiability. _______________________________________________________________
2) If a function is differentiable at a point then it must also be continuous at that point. (This is not conversely true).
For a function to be continuous at a point we must have:
Therefore (2) and (4) are required.
-----------------------------------------------------------------------------------------
3)
This is not required, the left side of the equation is the definition of a derivative at a point for a function . The derivative at a point does not have to equal to the function value at that point, it is equal to the slope at that point. Therefore 3 does not have to be true.
However, we can note that it is possible for a function and its' derivative to be equal for a given point. Sine and cosine, for instance will intersect periodically. Another example would be the exponential function which has itself as its' derivative
______________________________________________________________
4) See 2
_______________________________________________________________
5)
Again, the function does not have to approach the same limit as its' derivative. It is possible for a function to behave in this manner, such as in the case of sine and its' derivative cosine, which will both have the same limit at points where they intersect.
### Example Question #1 : Concept Of The Derivative
When the limit fails to exist,
The function is not continuous at .
The function is not defined at .
None of the above necessarily
The function is not differentiable at .
The function is not differentiable at .
Explanation:
By definition of differentiability, when the limit exists. When exists, we say the function is 'differentiable at '.
### Example Question #3 : Concept Of The Derivative
Which of the following functions is differentiable at , but not continuous there?
They are all differentiable and continuous at
They are all differentiable and continuous at
Explanation:
All of the functions are differentiable at . If you examine the graph of each of the functions, they are all defined at , and do not have a corner, cusp, or a jump there; they are all smooth and connected (Not necessarily everywhere, just at ). Additionally it is not possible to have a function that is differentiable at a point, but not continuous at that same point; differentiablity implies continuity.
### Example Question #4 : Concept Of The Derivative
For which of the following functions does a limit exist at , but not a y-value?
Explanation:
To answer the question, we must find an equation which satisfies two criteria:
(1) it must have limits on either side of that approach the same value and (2) it must have a hole at .
Each of the possible answers provide situations which demonstrate each combination of (1) and (2). That is to say, some of the equations include both a limit and a y-value at neither, or,in the case of the piecewise function, a y-value and a limit that does not exist.
In the function, , the numerator factors to
while the denominator factors to . As a result, the graph of this
function resembles that for , but with a hole at . Therefore, the limit
at exists, even though the y-value is undefined at .
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## Start your FREE trial today!
866-387-9139 mathseeds@3plearning.com
Back to Articles
SEP 27, 2018
Mathseeds fully covers first grade math skills—students have fun doing the self‑paced online lessons, which are aligned to Common Core and state standards. Free trial.
First grade math is a significant year for young learners, who will start to build on the basic mathematical knowledge learned in kindergarten. By first grade, students should be able to identify basic shapes, describe measurable attributes such as length, width, and height, and understand the concepts of addition ("putting together") and subtraction ("taking away").
There are some big differences between kindergarten and first grade math standards. Students at a first grade level are expected to move beyond a mere understanding of numbers and know how to independently reason with numbers.
In first grade, children are introduced to time, money, and the meaning of numbers greater than those they can count. First graders begin solving simple addition and subtraction problems. They learn to skip count by 2s, 5s, and 10s—a skill that will help them later when solving math equations. They also begin to work with 2- and 3‑dimensional geometric shapes.
First grade math students are expected to be able to measure length, tell time, and represent and interpret data. Additionally, they should be able to use place value concepts to add and subtract, and know how to represent and solve simple addition and subtraction problems.
The Mathseeds program fully covers grades K–3 math skills, featuring 50 lessons per grade level.
Lessons 51–100 feature Mango the monkey and friends as they introduce and explore different math concepts. Children learn to count to 100, order numbers, and identify ordinal numbers. They develop an understanding of place value including regrouping, and also practice their subtraction skills.
Students learn how to add and subtract to 10, and then within 100. Strategies include counting on, counting back, near doubles, and using number fact families. They learn how to skip count by 2s, 5s, and 10s, as well as the early multiplication and division skills of grouping and sharing.
Later, students learn to identify bills and coins, and use addition to find amounts of money. They explore fractions, focusing on wholes, halves, and fourths. Students continue to investigate the features of 2D shapes and 3D objects. They follow simple directions to a particular location and learn to read clocks to the half‑hour. They also work with early chance concepts, tally charts, and simple picture graphs.
## An Overview of 1st Grade Math Lessons in Mathseeds
### Lesson Number and Name
51 Addition to 10 with Two and Three groups
52 Sorting and Grouping 2D Shapes
53 Subtraction 1
54 O'clock
55 Near and Far
56 Subtraction 2
57 Position 1
58 Subtraction on a Number Line
59 Area
60 Counting 20-30
61 Wholes and Halves
62 Sorting and Grouping 3D Objects
63 Ordinal Numbers
64 Money
66 Halves and Quarters
67 Counting 30-40
### Lesson Number and Name
68 Find the Difference 1
69 Putting Shapes Together
70 O'clock & Half Past
71 Sharing 1
72 Doubles to Double 10
73 Mass
74 Grouping
75 Counting 40-50
76 The Equal Sign
77 Skip Counting by 2s & 5s
78 Position 2
79 Counting by 10s
80 Data 1
81 Counting 50-70
82 Chance 1
83 Money 2
84 Measuring Length
### Lesson Number and Name
85 Find the Difference 2
86 Counting 70-100
87 Half Past and Digital Time
89 Capacity 2
90 Skip Counting
91 Near Doubles to 20
92 Change from \$20
93 Number Fact Families
94 Position 3
96 Bridging to Ten
97 Data 2
99 3D Objects
100 Subtracting Unknown Numbers
## 1st Grade Math and Common Core Standards
The Mathseeds lessons provide comprehensive coverage of K–3 Math Common Core state standards, Texas TEKS, and other state standards.
The correlation charts here clearly show the scope and range of the Mathseeds lessons across all K–2 math domains, standards, content, and skills (grade 3 charts coming soon).
Teachers can use these charts to identify the ways in which Mathseeds lessons meet the learning objectives and standards for your first grade math students.
## 1st Grade Math Worksheets and Lesson Plans
Mathseeds features a comprehensive range of teacher resources, including first grade worksheets, lesson plans, and assessment material. There is a set of teacher resources to match each of the 200 lessons in the program. View some sample worksheets here.
The Teacher Toolkit in Mathseeds also features math posters to brighten up the classroom, digital big books for each grade level, and much more.
Are you new to Mathseeds? Start a free trial here of the award-winning online math program for students in grades K–3.
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# CLASS NOTES
```CLASS NOTES
SYNTHETIC DIVISION
When factoring or evaluating polynomials we often find that it is convenient to divide a
polynomial by a linear (first degree) binomial of the form x − k where k is a real number. In
certain problems we must use trial-and-error to find a particular value of k. The resulting
process of repeated divisions can be tedious and time-consuming. The use of synthetic division
(instead of long division) can save a lot of time and effort.
It must be emphasized that synthetic division may only be used when dividing a polynomial
by a binomial of the form x − k .
bg
Illustration: Suppose we wish to divide the polynomial P x = 5x 3 − 6 x 2 − 28 x − 2 by x − 3 .
The k-value is 3 (the number subtracted from x in the binomial x − 3 ). P x is a polynomial of
degree 3, and the four coefficients of P x are 5, − 6, − 28 and − 2 .
bg
bg
The division may now be expressed as shown below. Note that there is a blank space above the
horizontal line.
3
g
5
−6
− 28
−2
We “bring down” (copy) the leading coefficient below the line in the same column.
3
g
5
B
−6
− 28
−2
5
We now multiply 5 (the number below the line in the first column) by the k-value (3) and write
the result in the next column (second column) above the line.
3
g
5
−6
− 28
−2
15
5
Then we add the two numbers that are above the line in the second column and write the sum
below the line.
3
g
5
−6
− 28
−2
15
5
9
We now multiply 9 (the number below the line in the second column) by the k-value (3) and
write the result in the next column (third column) above the line.
Page 1 of 10
3
g
5
−6
− 28
15
27
5
−2
9
Then we add the two numbers that are above the line in the third column and write the sum
below the line.
3
g
5
−6
− 28
15
27
9
−1
5
−2
We repeat the previous two steps one last time to get the final result.
3
g
5
−6
− 28
−2
15
27
−3
9
−1
−5
5
b
g
The first three numbers below the line 5, 9 and − 1 are the coefficients of the quotient. The
last number below the line −5 is the remainder R. We can therefore rewrite the division as the
sum of a polynomial (the quotient) and a rational (the remainder divided by the divisor).
b g
5x 3 − 6 x 2 − 28 x − 2
−5
= 5x 2 + 9 x − 1 +
x−3
x−3
bg bg bg
bg
bg
bg
NOTE: We can see that P 3 = 5 3 − 6 3 − 28 3 − 2 = − 5 . I.e. the value of P k is equal
to the remainder, R, when we divide P x by x − k . More on this later (Remainder Theorem).
3
Example: Perform the division
2
3x 4 − 4 x 2 + 9 x − 6
and rewrite as a polynomial plus rational.
x+2
b g
NOTE: The divisor may be written x − −2
→ k = −2 . Furthermore, the numerator, which
is a polynomial of degree 4, has five terms. The five terms are 3x 4 + 0 x 3 − 4 x 2 + 9 x − 6 . We
use synthetic division to find the quotient and remainder.
−2
g3
0
−6
3
−6
−4
9
−6
12 − 16
14
−7
8
8
→
→ The quotient is 3x 3 − 6 x 2 + 8 x − 7 and the remainder is 8.
3x 4 − 4 x 2 + 9 x − 6
8
= 3x 3 − 6 x 2 + 8 x − 7 +
x+2
x+2
Page 2 of 10
Exercises: Perform each division and rewrite as a polynomial plus a rational.
x 3 + 7 x 2 + 15
1.
x+4
2x4 + 7x3 − x + 5
2.
x+3
4 x 4 − 7 x 3 + 9 x 2 − 3x − 8
3.
x −1
3x 5 − 4 x + 1
4.
x−2
−4
g1
7
0
15
− 4 − 12 48
1.
3 − 12 63
1
−3
g2
7
2.
g4
0
1 −3
8 − 19
4 −3
2
g3
0
6
0
6
→
3
3 −5
0 −4
3 6 12 24 44
89
−19
2x4 + 7x3 − x + 5
= 2 x 3 + x 2 − 3x + 8 +
x+3
x+3
−5
4 x 4 − 7 x 3 + 9 x 2 − 3x − 8
= 4 x 3 − 3x 2 + 6 x + 3 +
x −1
x −1
1
6 12 24 48 88
4.
→
9 − 3 −8
4 −3
x 3 + 7 x 2 + 15
63
= x 2 + 3x − 12 +
x+4
x+4
5
9 − 24
−7
3.
−1
−6 −3
2
1
→
→
3x 5 − 4 x + 1
89
= 3x 4 + 6 x 3 + 12 x 2 + 24 x + 44 +
x−2
x−2
REMAINDER THEOREM
bg
At the end of the previous illustration we saw that P 3 = −5 . Similarly, in the example we saw
that P −2 = 8 . In both cases the value of the polynomial at x = k is equal to the remainder R.
The same principle applies in each of the exercises above. This principle illustrates the
Remainder Theorem.
b g
Remainder Theorem:
bg
bg
If the polynomial P x is divided by x − k , then the remainder R is equal to P k .
bg
Pk =R
Page 3 of 10
One application of the Remainder Theorem is the evaluation of polynomials at given k-values.
bg
Illustration: Suppose we wish to evaluate the polynomial P x = 2 x 4 + x 3 − 3x 2 + 5x + 4 at
k = −7 . Typically a student will find P −7 by replacing each x in the equation for P by −7 .
b g
Pb −7g = 2b −7g + b −7g − 3b −7g + 5b −7g + 4 or Pb −7g = 4281 . Even with the help of
4
3
2
This gives
a calculator this can be a tedious and frustrating calculation, especially if the number of terms is
large. If the student has learned Synthetic Division and is comfortable with it, then he/she will
find that the Remainder Theorem is a faster and more reliable way to find P −7 .
b g
g2
−7
−3
1
5
4
− 14
91 − 616
4277
2 − 13
88 − 611
4281
→
b g
P −7 = 4281
bg
Exercises: In each of the following evaluate the polynomial P x at the given k-value.
bg
3. Pb x g = 6 x
bg
4. Pb x g = 4 x + 2 x
1. P x = 2 x 3 − 8 x 2 − 4 x + 5 at k = 6
4
2. P x = 5x 4 − 3x 2 + 2 x − 1 at k = −4
− x 3 + 9 x 2 − 7 at k = 7
5
4
− 5x 3 + 6 x 2 − 3x − 8 at k = −5
6
g2
−8 −4
1.
2
−4
12
24 120
4
20 125
g5
0
−3
bg
→
2
P 6 = 125
−1
− 20 80 − 308 1224
2.
5 − 20
7
g6
−1
−5
9
6
41 296
g4
2
− 20
b g
→
P −4 = 1223
→
P 7 = 14497
77 − 306 1223
−7
0
42 287 2072 14504
3.
4.
5
−5
bg
2072 14497
6
−3
−8
90 − 425 2095 − 10460
→
4 − 18 85 − 419 2092 − 10468
Page 4 of 10
b g
P −5 = −10468
FACTOR THEOREM
bg
Pb x g
= Qb x g +
bg
We have seen that if a polynomial P x is divided by x − k to get the quotient Q x and
remainder R, then we can write
bg
x−k
R
. And we know, by the Remainder
x−k
Pb x g
= Qb x g . This last
bg
x−k
Pb x g = b x − k g ⋅ Qb x g which means that x − k is a factor of Pb x g .
Theorem, that P k = R . It follows that if P k = 0 then R = 0 and
equation may be rewritten
Factor Theorem:
bg
bg
The polynomial x − k is a factor of polynomial P x if and only if P k = 0 .
b g
bg
Example: Given that P −3 = 0 rewrite the polynomial P x = 6 x 3 + 19 x 2 + 2 x − 3 as the
product of two factors.
bg
Solution: By the Factor Theorem we know that x + 3 is one factor of P x . To get the other
factor we can use synthetic division to divide P x by x + 3 (keeping in mind that k = −3 ).
−3
g6
6
bg
2 −3
19
− 18 − 3
3
1 −1
0
→
b g b gc
h
P x = x + 3 6x 2 + x − 1
bg
Exercises: In each of the following rewrite the polynomial P x as the product of two factors.
bg
Pb x g = 2 x + x
b g
1. P x = x 4 + 2 x 3 − 7 x 2 − 20 x − 12 and given P −2 = 0
2.
5
bg
− 30 x 3 − 35x 2 + 38 x + 24 and given P 4 = 0
4
−2
g1
2 − 7 − 20 − 12
−2
1.
1
4
g2
2
14
12
0 −7
−6
0
1 − 30 − 35
8
2.
0
9
36
38
→
24
24 − 44 − 24
6 − 11 − 6
b g b gc
h
P x = x + 2 x3 − 7x − 6
→
b g b gc
h
P x = x − 4 2 x 4 + 9 x 3 + 6 x 2 − 11x − 6
0
Page 5 of 10
FACTORING POLYNOMIALS OF HIGHER DEGREE (degree higher than 2)
bg
Terminology: A zero of a polynomial P is a number k such that P k = 0 (we worked with
“zeros” in the preceding section). A zero of P is called a root or solution of the polynomial
equation P x = 0 . Clearly the technical difference between “zeros” and “roots” is of little
importance to us. Furthermore, the term “zero” can be confusing to the student (a “zero” of the
polynomial is not necessarily equal to zero). Therefore, in the remainder of this discussion the
term “root” will be used exclusively.
bg
The Rational Roots Test:
bg
Suppose the polynomial P x has integer coefficients. If the rational number dn (in
lowest terms) is a root of P then the numerator n is a factor of the constant term of P
and the denominator d is a factor of the leading coefficient.
The Integer Roots Test:
bg
Suppose the polynomial P x has integer coefficients with leading coefficient 1. If the
integer k is a root of P then k is a factor of the constant term of polynomial P.
The Rational Roots Test and the Integer Roots Test can only give us possible roots of a given
polynomial. We must test these possible values (using synthetic division) to determine which, if
any, of the possibilities are actual roots. Once we have the actual root(s) then we can write the
polynomial in factored form.
bg
Example 1: Completely factor the polynomial P x = x 3 − 5x 2 − 2 x + 24 .
Solution: Polynomial P has integer coefficients with leading coefficient 1. We can therefore
apply the Integer Roots Test. Any possible integer root of P must be a factor of the constant
term 24. The possible integer roots are 1, 2, 3, etc. as well as their negatives. The complete list
can be written ±1, ± 2, ± 3, ± 4, ± 6, ± 12, ± 24 . This gives a total of 14 possible integer roots!
At most three of these can be actual roots. It is also possible that none of them are actual roots.
They must be tested one by one using synthetic division (trial-and-error).
bg
If we try k =1 we get a non-zero remainder which is, as we know, P 1 .
1
g1
− 5 − 2 − 24
1 −4
−6
→
bg
P 1 = −30
→
bg
P1 ≠0
1 − 4 − 6 − 30
Therefore k =1 is not a root of P. Similarly k = −1 and k = 2 are not roots of P.
However, as shown below, k = −2 is a root of P. It follows, by the Factor theorem, that x + 2 is
one of the factors of P.
Page 6 of 10
−2
g1
− 5 − 2 − 24
− 2 14
24
1 − 7 12
0
b g b gc
→
P x = x + 2 x 2 − 7 x + 12
h
We see that the other factor, in addition to x + 2 , is x 2 − 7 x + 12 . This second factor can itself be
factored using the techniques of High School algebra. Or it can be factored by continuing to test
the possible integer roots. The final, factored, form is P x = x + 2 x − 3 x − 4 .
b g b gb gb g
bg
Example 2: Completely factor the polynomial P x = x 5 − 26 x 3 + 12 x 2 + 97 x + 60 .
Solution: The possible integer roots are ±1, ± 2, ± 3, ± 4, ± 5, ± 6, ± 10, etc. (applying the
Integer Roots Test). We find that k = −1 is a root of P.
−1
g1
0 − 26 12
−1
1
97
60
1 − 1 − 25 37
60
b g b gc
→
25 − 37 − 60
h
P x = x + 1 x 4 − x 3 − 25x 2 + 37 x + 60
0
Working with the other (second) factor we test k = −1 a second time (roots can repeat!) and find
that it is also a root of the second factor.
−1
g1
− 1 − 25 37
−1
2
60
23 − 60
1 − 2 − 23 60
→
b g b gb gc
h
P x = x + 1 x + 1 x 3 − 2 x 2 − 23x + 60
0
Working with the third factor we find that k = 3 is a root.
3
g1
− 2 − 23
3
1
60
3 − 60
1 − 20
b g b g b x − 3gc x
→
P x = x +1
2
2
h
+ x − 20
0
As in Example 1, the quadratic factor x 2 + x − 20 can be factored by using the techniques of
High School algebra or by continuing to test the possible integer roots. The factored form is
b g b g b x − 3gb x + 5gb x − 4g .
P x = x +1
2
bg
Example 3: Completely factor the polynomial P x = 6 x 4 − 5x 3 + 26 x 2 − 25x − 20 .
Solution: Because the leading coefficient is not equal to 1, we apply the Rational Roots Test.
Suppose that dn is a rational root of P.
Then the possible values of the numerator n are ±1, ± 2, ± 4, ± 5, ± 10, ± 20 .
The possible values of the denominator d are ±1, ± 2, ± 3, ± 6 .
Page 7 of 10
We find the possible combinations (n divided by d ) of the above numbers. The possible rational
roots are ±1, ± 2, ± 4, ± 5, ± 10, ± 20, ± 21 , ± 25 , ± 13 , ± 23 , ± 43 , ± 53 , ± 10
, ± 20
, ± 16 , ± 56 .
3
3
We find that k = − 21 is a root of P.
−1
2
j6
− 5 26 − 25 − 20
−3
4 − 15
20
6 −8
30 − 40
0
b g e jc
→
h
P x = x + 21 6 x 3 − 8 x 2 + 30 x − 40
Working with the second factor we find that k = 43 is a root.
4
3
j6
−8
6
30 − 40
8
0
40
0
30
0
b g e je jc
→
h
P x = x + 21 x − 43 6 x 2 + 30
c
h
The quadratic factor 6 x 2 + 30 can be rewritten as 6 x 2 + 5 . This quadratic factor can not be
factored further. The final (simplified) factored form is
b g e je jc
h
P x = 6 x + 21 x − 43 x 2 + 5
bg b
gb
gc
h
which can be rewritten as P x = 2 x + 1 3x − 4 x 2 + 5 .
Exercises: Completely factor each of the following polynomials.
bg
Pb x g = 4 x
1. P x = x 4 − 27 x 2 − 14 x + 120
3.
4
bg
Pb x g = 3x + 10 x
2. P x = x 4 + x 3 − 15x 2 + 23x − 10
− 9 x 3 − 25x 2 + 36 x + 36
4.
5
4
− 11x 3 − 46 x 2 − 4 x + 24
Solutions:
1. The possible integer roots are ±1, ± 2, ± 3, ± 4, ± 5, ± 6, ± 10, etc. (Integer Roots Test). We
test these possibilities (using synthetic division) to find the actual roots of the polynomial.
2
g1
0 − 27 − 14
2
120
4 − 46 − 120
1 2 − 23 − 60
b g b gc
h
→ k = 2 is a root → P x = x − 2 x 3 + 2 x 2 − 23x − 60
0
We continue by testing for roots in the second factor.
−3
g1
2 − 23 − 60
−3
3
60
1 − 1 − 20
0
b g b gb gc
h
→ k = −3 is a root → P x = x − 2 x + 3 x 2 − x − 20
The third factor may be factored by the techniques of High school algebra.
b g b gb gb gb g
P x = x −2 x + 3 x + 4 x −5
Page 8 of 10
2. The possible integer roots are ±1, ± 2, ± 5, ± 10 (Integer Roots Test). We test these
possibilities to find the actual roots of the polynomial.
1
g1
1 − 15
23 − 10
2 − 13
1
1 2 − 13
b g b gc
10
h
→ k = 1 is a root → P x = x − 1 x 3 + 2 x 2 − 13x + 10
10
0
We continue testing for roots in the second factor. As mentioned earlier, it is important to
test the same root a second time. Roots can, and sometimes do, repeat.
1
g1
2 − 13
10
b g b gb gc
3 − 10
1
1 3 − 10
h
→ k = 1 is a root (again) → P x = x − 1 x − 1 x 2 + 3x − 10
0
The third factor may be factored by the techniques of High school algebra.
b g b g b x + 5gb x − 2g
P x = x −1
2
3. The Rational Roots Test allows us to find the possible rational roots:
±1, ± 2, ± 3, ± 4, ± 6, ± 9, ± 12, ± 18, ± 36, ± 21 , ± 23 , ± 92 , ± 41 , ± 43 , ± 94
We test these possibilities to find the actual roots of the polynomial.
g4
2
− 9 − 25
36
36
b g b gc
− 2 − 54 − 36
8
4 − 1 − 27 − 18
h
→ k = 2 is a root → P x = x − 2 4 x 3 − x 2 − 27 x − 18
0
We continue by testing for roots in the second factor.
−2
g4
− 1 − 27 − 18
−8
18
18
4 −9
−9
0
b g b gb gc
→ k = −3 is a root → P x = x − 2 x + 2 4 x 2 − 9 x − 9
h
The third factor may be factored by the techniques of High school algebra.
b g b gb gb gb
g
P x = x − 2 x + 2 x − 3 4x + 3
4. The possible rational roots are ±1, ± 2, ± 3, ± 4, ± 6, ± 8, ± 12, ± 24, ± 13 , ± 23 , ± 43 , ± 83 .
We test these possibilities to find the actual roots of the polynomial.
g
− 1 3 10 − 11 − 46 − 4
−3 −7
18
3 7 − 18 − 28
24
28 − 24
24
b g b gc
→ − 1 is a root → P x = x + 1 3x 4 + 7 x 3 − 18 x 2 − 28 x + 24
0
Page 9 of 10
h
We continue by testing for roots in the second factor.
2
g3
7 − 18 − 28
6
24
8 − 12
3 13
b g b gb gc
16 − 24
26
→ k = 2 is a root → P x = x + 1 x − 2 3x 3 + 13x 2 + 8 x − 12
h
0
We continue by testing for roots in the third factor.
−2
g3
8 − 12
13
− 6 − 14
12
−6
0
3
7
b g b gb gb gc
h
→ k = −2 is a root → P x = x + 1 x − 2 x + 2 3x 2 + 7 x − 6
The fourth factor may be factored by the techniques of High school algebra.
b g b gb gb gb
gb g
P x = x + 1 x − 2 x + 2 3x − 2 x + 3
Drill Exercises: Completely factor each of the following polynomials.
bg
3. Pb x g = x + 9 x − 6 x − 116 x + 168
5. Pb x g = x + 8 x + 5x − 74 x − 120
7. Pb x g = 2 x + x − 85x + 42
9. Pb x g = 20 x + 12 x − 33x − 5x + 6
11. Pb x g = 8 x − 20 x − 18 x + 81x − 54
1. P x = x 3 + 5x 2 − 12 x − 36
4
3
2
4
3
2
3
2
4
4
3
2
3
2
b g b gb gb g
3. Pb x g = b x + 7gb x + 6gb x − 2g
5. Pb x g = b x + 5gb x + 4gb x + 2gb x − 3g
7. Pb x g = b x + 7gb2 x − 1gb x − 6g
9. Pb x g = b2 x + 3gb2 x + 1gb5x − 2gb x − 1g
11. Pb x g = b x + 2gb2 x − 3g
1. P x = x + 6 x + 2 x − 3
2
3
bg
4. Pb x g = x − 6 x − 15x + 80 x + 84 x − 144
6. Pb x g = x + 6 x − 28 x − 110 x − 117 x − 40
8. Pb x g = 24 x − 62 x + 31x + 12
10. Pb x g = 9 x + 30 x − 92 x − 498 x − 685x − 300
12. Pb x g = 72 x + 90 x − 383x − 287 x + 254 x − 40
2. P x = x 3 − 6 x 2 − 7 x + 60
5
4
3
5
4
3
3
5
2
2
4
5
2
3
2
4
3
2
b g b gb gb g
4. Pb x g = b x + 3gb x + 2gb x − 1gb x − 4gb x − 6g
6. Pb x g = b x + 8gb x − 5gb x + 1g
8. Pb x g = b4 x + 1gb2 x − 3gb3x − 4g
10. Pb x g = b x + 3gb x + 1gb x − 4gb3x + 5g
12. Pb x g = b2 x + 5gb3x + 4gb4 x − 1gb3x − 1gb x − 2g
2. P x = x + 3 x − 4 x − 5
3
2
Page 10 of 10
```
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# A conjecture on 4-coloring maximal planar graphs
The question/task is to prove/disprove the conjecture below.
Let $$G$$ be a maximal planar graph with a 4-coloring $$f$$. Let $$(a,b,c,d)$$ be a cycle in $$G$$. Let $$S$$ be the collection of all $$a,c$$-paths in $$G$$ and all $$b,d$$-paths in $$G$$.
Conjecture: At least two members of $$S$$ are bicolored.
(i.e., there exist distinct paths $$Q_1,Q_2\in S$$ and colors $$i,j,k,l\in\{1,2,3,4\}$$ such that $$f(u)\in\{i,j\}$$ for every vertex $$u$$ on $$Q_1$$ and $$f(v)\in\{k,l\}$$ for every vertex $$v$$ on $$Q_2$$).
Definitions:-
$$G$$ is a maximal planar graph if it can be drawn on a plane such that no edges cross and boundary of every face is a triangle. A 4-colouring $$f$$ of $$G$$ is a function $$f:V(G)\to\{1,2,3,4\}$$ such that $$f$$ map endpoints of each edge to different 'colors' (i.e. $$f(u)\neq f(v)$$ for every edge $$uv$$ of $$G$$).
Notes:-
It is easy to see that the conjecture is true if the cycle $$(a,b,c.d)$$ is tricolored (or bicolored). The following is the crux of the conjecture.
Let $$G$$ be a planar graph with a 4-coloring $$f$$. Let $$(a,b,c,b)$$ be a cycle in $$G$$ such that boundary of each face inside the cycle is a triangle. Suppose that the cycle $$(a,b,c,d)$$ receives all four colors.
Conjecture 2: Then, there is a bicolored $$a,c$$-path or a bicolored $$b,d$$-path inside the cycle.
If Conjecture 2 is true, then the main conjecture above is true (apply Conjecture 2 to outside region of the cycle $$(a,b,c,d)$$ ).
Theorem A.1. Let $$G$$ be an a-graph with boundary cycle $$uxvy$$ for the exterior 4-face and let $$G$$ have a 4-coloring $$c$$. Suppose, without loss of generality, that $$c(x)=1$$, $$c(y)=1$$ or 2, $$c(u)=3$$, and $$c(v)=3$$ or 4. Then there is either a 1–2 path between $$x$$ and $$y$$ or a 3–4 path between $$u$$ and $$v$$.
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# Problem Solving Through Programming In C NPTEL Week 3 Solutions
This set of MCQ(multiple choice questions) focuses on the Problem Solving Through Programming In C NPTEL Week 3 Solutions.
NOTE: You can check your answer immediately by clicking show answer button. Problem Solving Through Programming In C NPTEL Week 3 assignment answers” contains 10 questions.
Now, start attempting the quiz.
### Problem Solving Through Programming In C NPTEL Week 3 assignment answers
Q1. Which of the following statement is correct?
a) Operator precedence determines which operator is performed first in an expression with more than one operator with different precedence. Associativity is used when two operators of same precedence appear in an expression.
b) Operator associativity determines which operator is performed first in an expression with more than one operator with different associativity. Precedence is used when two operators of same precedence appear in an expression.
c) Operator precedence and assocativity are same.
d) None of the above
Q2. Find the output of the following C code.
a) 67
b) -36
c) 66
d) -37
Q3. What is the output of the following C code?
a) 0
b) 3
c) 4
d) Compilation error
Q4. Find the output of the following C code.
a) IITKGP
b) IITD and IITM
c) IITKGP and IITM
d) IITM
Q5. What will be the output?
a) Condition is true
b) Condition is false
c) Error
d) No output possible
Problem Solving Through Programming In C NPTEL Week 3 assignment answers
Q6. What is the output of the following program?
a) Programming on C 0
b) NPTEL 0
c) NPTEL 3
d) Compilation error
Q7. What is the output of the C program given below
a) true
b) false
c) Compiler dependent
d) Compiler error
Problem Solving Through Programming In C NPTEL Week 3 assignment answers
Q8. What will be the output?
a) 0
b) 1
c) 10
d) 30
Problem Solving Through Programming In C NPTEL Week 3 assignment answers
Q9. What will be the output?
a) TRUE
b) FALSE
c) Syntax Error
d) Compilation Error
Q10. What is the output of the following C code?
a) 10
b) 11
c) 20
d) Compiler error
Problem Solving Through Programming In C NPTEL Week 3 assignment answers
### Problem Solving Through Programming In C NPTEL Week 3 assignment answers
Q1. What should be the value of ‘b’ such that the output of the program will be 20?
a) 1
b) 2
c) 3
d) 4
Q2. Find the output of the following C code
a) IITKGP
b) IITD and IITM
c) IITKGP and IITM
d) IITM
Q3. What is the output of the following program?
a) Programming on C 0
b) NPTEL 0
c) NPTEL 3
d) Compilation error
Q4. Find the output of the following C code
a) -42
b) 24
c) 15
d) -34
Q5. Which of the following statement is correct?
a) Operator precedence determines which operator is performed first in an expression with more than one operator with different precedence. Associativity is used when two operators of same precedence appear in an expression
b) Operator associativity determines which operator is performed first in an expression with more than one operator with different associativity. Precedence is used when two operators of same precedence appear in an expression
c) Operator precedence and associativity are same.
d) None of the above
Problem Solving Through Programming In C NPTEL Week 3 assignment answers
Q6. What is the output of the C program given below?
a) true
b) false
c) Compiler dependent
d) Compiler error
Q7. What is the output of the following program?
a) Value of c and d are 1 and 3 respectively
b) Value of c and d are 1 and 3.333333 respectively
c) Value of c and d are 1.000000 and 3.000000 respectively
d) Value of c and d are 1 and 3.000000 respectively
Answer: d) Value of c and d are 1 and 3.000000 respectively
Problem Solving Through Programming In C NPTEL Week 3 assignment answers
Q8. What will be the output?
a) TRUE
b) FALSE
c) Syntax Error
d) Compilation Error
Problem Solving Through Programming In C NPTEL Week 3 assignment answers
Q9. What will be the output?
a) 0
b) 1
c) 7
d) Compilation error
Q10. The precedence of arithmetic operators is (from highest to lowest)
a) %, *, /, +, –
b) %, +, /, *, –
c) +, -, %, *, /
d) %, +, -, *, /
Answer: a) %, *, /, +, –
Problem Solving Through Programming In C NPTEL Week 3 assignment answers
### Problem Solving Through Programming In C NPTEL Week 3 assignment answers
Q1. Find the output of the following C program
a) 120, 120
b) 120, 130
c) 130, 120
d)130, 130
Q2. What will be the output of the following program?
a) 1
b) 0
c) -1
d) 2
Q3. Find the output of the following C code.
a) x1=4, x2=3
b) x1=-5, x2=-4
c) x1=2.5, x2=4.2
d) Roots are imaginary
Q4. Find the output of the following code.
a) 6
b) 4
c) 10
d) 14
Q5. Which of the following statements are correct?
I. The ‘else’ block is executed when condition inside ‘if’ statement is false.
II. One ‘if’ statement can have multiple ‘else’ statement.
a) Only I
b) Only II
c) Both I and II
d) None of the above is correct
Problem Solving Through Programming In C NPTEL Week 3 assignment answers
Q6. C modulo division operator ‘%’ can be applied on
a) only float variables
b) only int variables
c) int and float combination
d) any data types in C
Q7. The output of the following program will be
a) C programming
b) Java
Python
c) C programming
Java
d) Compilation error
Python
Problem Solving Through Programming In C NPTEL Week 3 assignment answers
Q8. What will be the output?
a) 2
b) 8
c) 10
d) 12
Problem Solving Through Programming In C NPTEL Week 3 assignment answers
Q9. What will be the output?
a) Right
b) Wrong
c) 0
d) No output
Q10. What will be the output of the program?
a) The answer will be 15
b) The answer will be 0
c) The answer will be 1
d) Compilation error
Problem Solving Through Programming In C NPTEL Week 3 assignment answers
### Previous Course – Week 3 assignment answers
Q1. Which of the following statements is correct?
a) Operator precedence determines which operator is performed first in an expression with more than one operator with different precedence. Associativity is used when two operators of same precedence appear in an expression
b) Operator associativity determines which operator is performed first in an expression with more than one operator with different associativity. Precedence is used when two operators of same precedence appear in an expression
c) Operator precedence and associativity are same.
d) None of the above
Q2. What is the output of the following program?
``````#include<stdio.h>
int main()
{
int x=11, y=5, z;
float w;
z=x%y;
w=x/y;
printf("Value of z and w are %d and %f respectively", z, w);
return 0;
}``````
a) Value of z and w are 1 and 2 respectively
b) Value of z and w are 1 and 2.200000 respectively
c) Value of z and w are 1.000000 and 2.200000 respectively
d) Value of z and w are 1 and 2.000000 respectively
Q3. What will be the output?
``````#include<stdio.h>
int main()
{
int a=4, b=15, c=29;
if(c>b>a)
printf("TRUE");
else
printf("FALSE");
return 0;
}``````
a) TRUE
b) FALSE
c) Syntax Error
d) Compilation Error
Q4. What will be the output of following program?
``````#include<stdio.h>
int main()
{
int x=(10 || 0) && (10);
printf("x=%d", x);
}``````
a) x=60
b) x=70
c) x=0
d) x=1
Q5. What will be the output?
``````#include<stdio.h>
int main()
{
int x=17, y=1;
if(!(!x) && y)
printf("%d", x);
else
printf("%d", y);
return 0;
}``````
a) 17
b) 18
c) 1
d) 0
Problem Solving Through Programming In C NPTEL Week 3 solutions
Q6. What is the output of the following C code?
``````#include<stdio.h>
int main()
{
int x=7, y=3, z=5;
printf("%d\n", x-x/y*y%z);
return 0;
}``````
a) 7
b) 6
c) 5
d) 0
Q7. What will be the output?
``````#include<stdio.h>
int main()
{
int x;
x = (11+3)%5/2;
printf("%d\n", x);
return 0;
}``````
a) 0
b) 1
c) 2
d) Compilation error
Problem Solving Through Programming In C NPTEL Week 3 solutions
Q8. Find the output of the following C code
``````#include<stdio.h>
int main()
{
int a=3, b=2;
a=a=b=0
printf("%d, %d", a,b);
return 0;
}``````
a) 1,2
b) 0,0
c) 3,2
d) 1,0
Problem Solving Through Programming In C NPTEL Week 3 solutions
Q9. The output of the following program will be
``````#include<stdio.h>
int main()
{
int a=0, b=1, c=-1;
if(a)
printf("IITKGP\n");
if(b)
printf("IITM \n");
if(c)
printf("IITR \n");
return 0;
}``````
a) IITKGP
b) IITMIITR
c) IITM
IITR
d) IITKGP
IITR
Q10. What is the output of this C code?
``````#include<stdio.h>
int main()
{
int x = 10;
if(x>0)
printf("inside if\n");
else if(x>0)
printf("inside elseif\n");
return 0;
}``````
a) inside if
b) inside elseif
c) inside if
inside elseif
d) compile time error
Problem Solving Through Programming In C NPTEL Week 3 solutions
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Programming in Java NPTEL week 1 quiz answers
Nptel – Deep Learning assignment solutions
For discussion about any question, join the below comment section. And get the solution of your query. Also, try to share your thoughts about the topics covered in this particular quiz.
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dous
Posted on
# Transposing a Matrix in Java
Sometimes you may need to transpose a matrix. For example while working on linear algebra. The main logic is writing the elements in matrix’s row as a column.
I choose Java to do this however you can use any language. If even the syntax changes, the main logic always will be the same.
## Main method
``````Scanner sc = new Scanner(System.in);
System.out.println("Row number:");
int row = sc.nextInt();
System.out.println("Column number:");
int col = sc.nextInt();
int[][] arr = new int[row][col];
for(int i = 0; i < arr.length; i++){
for(int j = 0; j < arr[0].length; j++){
System.out.println("Enter the arr["+i+"]["+j+"] element:");
arr[i][j] = sc.nextInt();
}
}
System.out.println();
print2dArr(arr);
System.out.println();
print2dArr(tranpose(arr));
sc.close();
``````
## Transpose Method
``````static int[][] tranpose(int[][] arr){
int[][] ret = new int[arr.length][arr[0].length];
for(int i = 0; i < arr.length; i++){
for(int j = 0; j < arr[0].length; j++){
ret[j][i] = arr[i][j];
}
}
return ret;
}
``````
## Print Method
``````static void print2dArr(int[][] arr){
for(int i = 0; i < arr.length; i++){
for(int j = 0; j < arr[0].length; j++){
System.out.print(arr[i][j]+" ");
}
System.out.println();
}
}
``````
## Output
``````Row number:
3
Column number:
3
Enter the arr[0][0] element: 1
Enter the arr[0][1] element: 6
Enter the arr[0][2] element: 2
Enter the arr[1][0] element: 7
Enter the arr[1][1] element: 8
Enter the arr[1][2] element: 3
Enter the arr[2][0] element: 2
Enter the arr[2][1] element: 3
Enter the arr[2][2] element: 1
1 6 2
7 8 3
2 3 1
1 7 2
6 8 3
2 3 1
``````
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hw13.solutions
# The projection is y u2 y u3 1 14 1 y u1 u1 u1 u2
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Unformatted text preview: , u3 , u4 #6. The projection is y · u1 y · u2 3 5 u1 + u2 = − u1 + u2 = (6, 4, 1) u1 · u1 u2 · u2 2 2 #10. The projection is y · u2 y · u3 1 14 1 y · u1 u1 + ++ = u1 + u2 − u3 = (5, 2, 3, 6). u1 · u1 u2 · u2 u3 · u3 3 3 3 The perpendicular vector is (3, 4, 5, 6) − (5, 2, 3, 6) = (−2, 2, 2, 0). 1 #14. The closest point in the projection z · v1 z · v2 1 v1 + v2 = v1 + 0v2 = (1, 0, −1/2, −3/2). v1 · v1 v2 · v2 2 ˆ #16. The projection is ˆ = (−1, −5, −3, 9) and the distance is ||y − y || = √y |(| − 4,...
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• Aug 17th 2011, 04:15 AM
jonoe
a - ( (b) / (c+x^2) )
The line goes through (2, $2/3$) and the gradient of the curve is at its maximum at x= root6/3 (square root of 6 divided by 3)
Prove a = 1, b = 2, c = 2.
Can I please have a detailed solution to this? I've tried so many ways, been hours already. It was in the dydx and integration section.
• Aug 17th 2011, 05:06 AM
jonoe
Re: Help!
sorry c wasnt 3, c = 2
• Aug 17th 2011, 05:13 AM
Siron
Re: Help!
For example, if the curve goes throught the point $\left(2,\frac{2}{3}\right)$ that means:
$\frac{2}{3}=a-\frac{b}{c+4}$
Have you tried something this way? Also with the other given information? ...
• Aug 17th 2011, 05:16 AM
jonoe
Re: Help!
Yes I was aware it was a curve. I've tried something like that. Still ended up with 2 unknowns. and 3 in the other.
And the point where the gradient is its max. Stuck on how to get that equation. Or use that info. Would it be double derivative?
• Aug 17th 2011, 05:29 AM
Siron
Re: Help!
Yes, calculate the second derivative of the function, then let $f''(x)=0$ and enter $x=\frac{\sqrt{6}}{3}$ therefore you can calculate $c$.
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Thursday, July 18, 2024
Latest:
# The Significance of the Personal Month Number in Timing and Decision-Making in Relationships
### Introduction
In numerology, each person has a unique Personal Month number that reflects the energy and vibration of that particular month for them. This number can have a significant impact on timing and decision-making in relationships, as it can provide insight into the potential outcomes of certain actions and guide individuals towards making choices that align with their goals and values.
### Calculating the Personal Month Number
To calculate your Personal Month number, you first need to know your Life Path number. This is calculated by adding together the digits in your birth date and reducing the sum to a single digit (with the exception of Master numbers 11 and 22, which are not reduced). For example, if your birth date is June 7, 1995, you would calculate your Life Path number as follows:
6 + 7 + 1 + 9 + 9 + 5 = 37
3 + 7 = 10
1 + 0 = 1
So, in this case, the individual’s Life Path number is 1.
To calculate your Personal Month number, you then need to add the current month’s number to your Life Path number. For example, if it is currently March (the third month of the year), you would add 3 to your Life Path number. If your Life Path number is 1, your Personal Month number for March would be 4 (1 + 3 = 4).
### The Meaning of the Personal Month Number
Each Personal Month number has a unique meaning and energy associated with it. For example:
• Personal Month 1: This is a time of new beginnings and taking on new challenges. It is a good time to start a new project or make a major change in your life.
• Personal Month 2: This is a time of balance and cooperation. It is a good time to focus on partnerships and teamwork.
• Personal Month 3: This is a time of creativity and self-expression. It is a good time to focus on your personal goals and express yourself through art or writing.
• Personal Month 4: This is a time of structure and stability. It is a good time to focus on organization and practical matters.
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# Tagged Questions
114 views
### What is a solution for a TSP with multiple salesmen and no return but known vertices and end points?
I don't know if I worded that correctly and I am not sure if it is even a TSP problem but here is the scenario. I am designing and trying to optimize a route planner for a delivery service. I have ...
110 views
### Route between A and B with stations between
i´m obviously missing the forest through the trees ... i know about the traveling salesman problem, but is there any other algorithm/problem which better fits my needs/description? I need to describe ...
268 views
### Can someone please explain 2.5-opt heuristic (also known as 2h-opt) in basic terms?
I am trying to find information that accurately explains 2.5-opt, but I am coming up short. I have read Jon Bentleys "Fast Algorithms for Geometric Traveling Salesman Problems", as well as several ...
533 views
### Approximation algorithm for TSP variant, fixed start and end anywhere but starting point + multiple visits at each vertex ALLOWED
NOTE: Due to the fact that the trip does not end at the same place it started and also the fact that every point can be visited more than once as long as I still visit all of them, this is not really ...
460 views
### Testing a Hamiltonian Path finder implementation
I am implementing an algorithm which finds an optimal Hamiltonian path in a directed graph. I have implemented an algorithm which appears to work reasonably well, however I am not entirely sure if ...
316 views
### How to find the least cost to visit every vertex in an undirected, weighted graph at least once starting from a predetermined vertex?
The path taken does not have to end back at the predetermined vertex. Basically, the traveling salesman problem except that a vertex can be visited more than one time. EDIT: There will be up to a ...
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Assignment 2: Exploring Parabolas
Michelle Greene
In this assignment, we want to explore the graph of y=ax^2 for different values of a.
Let's first look at how the graph changes when a is different positive integers. We will use y=x^2 as our original graph to compare changes to.
From this illustration, we can see that as a increases, the parabola stretches, or gets thinner. The vertex, however, stays the same.
What happens when a gets closer and closer to 0? Let's take a look (again, let's start with a=1 for comparison):
Drawing from this illustration, we notice that as a decreases from 1 and gets closer and closer to zero, the parabola widens, or gets fatter. We should expect this, because if a=0, we just get the line y=x, which is just the x-axis.
So, now that we know what happens with different positive values of a, let's explore negative values of a. For the sake of comparison, let's graph the same values of a, both negative and positive (I have used the same color for both negative and postive values to help us better see the effect.)
From these graphs, we can easily see that the corresponding negative values of a causes the graph to just reflect about the x-axis. The vertex nor the shape changes; only the way the parabola opens changes.
For fun, and to perhaps better illustrate, click here to see an animation of what has been described above.
In conclusion, changing the value of a in this equation does not change where the vertex is located. As a increases, the parabola gets skinnier, as a decreases, the parabola flattens out, and if a is negative, it is reflected about the x-axis, or it opens downward instead of upward.
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# Two people A and B are 700 km apart, they start towards each other at an interval of 3 hrs, A leaves before B. The speeds of A and B are 55 km/h and 5
23 views
in Aptitude
closed
Two people A and B are 700 km apart, they start towards each other at an interval of 3 hrs, A leaves before B. The speeds of A and B are 55 km/h and 52 km/h respectively, How much time will they take to meet after A starts?
1. 5 hrs
2. 7 hrs
3. 8 hrs
4. 11 hrs
5. 12 hrs
by (30.0k points)
selected
Correct Answer - Option 3 : 8 hrs
Given:
D = 700 km
A starts 3 hrs before B
A's speed = 55 km/h
B's speed = 52 km/h
Concept:
Relative speed in opposite direction is the sum of individual speeds.
Formula used:
Speed = distance/time
Calculation:
A's distance in 3 hrs = 55 × 3 = 165 km
Remaining distance = 700 - 165 = 535 km
Relative speed = 55 + 52 = 107 km/h
Time = 535/107 = 5 hrs
Required time = 3 + 5 = 8 hrs
∴ The time taken is 8 hrs
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The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A260678 Numbers n>0 for which n+(17-n)^2 is not prime. 2
33, 34, 37, 42, 49, 50, 51, 53, 56, 58, 60, 65, 67, 68, 69, 71, 72, 75, 78, 82, 83, 84, 85, 86, 88, 91, 94, 95, 97, 100, 101, 102, 105, 106, 107, 110, 111, 113, 114, 116, 117, 118, 119, 122, 123, 124, 128, 129, 132, 133, 134, 135, 136, 139, 141, 143, 148, 151, 152, 153 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Motivated by the fact that n+(17-n)^2 = 1+16^2, 2+15^2, ..., 16+1^2, 17+0^2, 18+1^2, 19+2^2, ..., 32+15^2 are all prime. This has an explanation through Heegener numbers, similar to Euler's prime-generating polynomial, cf. A002837 and related crossrefs. LINKS Robert Israel, Table of n, a(n) for n = 1..10000 MAPLE remove(t -> isprime(t+(17-t)^2), [\$1..200]); # Robert Israel, May 02 2017 MATHEMATICA Select[Range[200], !PrimeQ[# + (17 - #)^2] &] (* Vincenzo Librandi, Nov 16 2015 *) PROG (PARI) for(n=1, 999, isprime(n+(17-n)^2)||print1(n", ")) (MAGMA) [n: n in [1..180] | not IsPrime(n+(17-n)^2)]; // Vincenzo Librandi, Nov 16 2015 CROSSREFS Cf. A260679 (n+(17-n)^2), A007635 (primes in that sequence = primes of the form n^2+n+17). Cf. A002837 (n^2-n+41 is prime), A005846 (primes of form n^2+n+41), A007634 (n^2+n+41 is composite), A097823 (n^2+n+41 is not squarefree). Sequence in context: A112817 A115394 A335072 * A329658 A140143 A041543 Adjacent sequences: A260675 A260676 A260677 * A260679 A260680 A260681 KEYWORD nonn,easy AUTHOR M. F. Hasler, Nov 15 2015 STATUS approved
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Last modified August 7 09:16 EDT 2020. Contains 336274 sequences. (Running on oeis4.)
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## Physical Mathematics and the Fine-Structure Constant
.
Abstract: Research into ancient physical structures, some having been known as the seven wonders of the ancient world, inspired new developments in the early history of mathematics. At the other end of this spectrum of inquiry the research is concerned with the minimum of observations from physical data as exemplified by Eddington’s Principle. Current discussions of the interplay between physics and mathematics revive some of this early history of mathematics and offer insight into the fine-structure constant. Arthur Eddington’s work leads to a new calculation of the inverse fine-structure constant giving the same approximate value as ancient geometry combined with the golden ratio structure of the hydrogen atom. The hyperbolic function suggested by Alfred Landé leads to another result, involving the Laplace limit of Kepler’s equation, with the same approximate value and related to the aforementioned results. The accuracy of these results are consistent with the standard reference. Relationships between the four fundamental coupling constants are also found.
.
Inverse fine-structure constant is a root of: x4 – 136x3 – 136x2 – 818x + 1 = 0. This equation gives a value for x = 137.035 999 168 ….
CODATA (2014) Inverse alpha = 137.035 999 160 (33). The other root of the equation is approximately 1/818 and 818 = (136+1/3)6 = (4×136)+(2×137). OSF Preprints Physical_Mathematics_and_the_Fine-Structure_Constant
.
## Fine-Structure Constant from Golden Ratio Geometry
.
After a brief review of the golden ratio in history and our previous exposition of the fine-structure constant and equations with the exponential function, the fine-structure constant is studied in the context of other research calculating the fine-structure constant from the golden ratio geometry of the hydrogen atom. This research is extended and the fine-structure constant is then calculated in powers of the golden ratio to an accuracy consistent with the most recent publications. The mathematical constants associated with the golden ratio are also involved in both the calculation of the fine-structure constant and the proton-electron mass ratio. These constants are included in symbolic geometry of historical relevance in the science of the ancients.
$\alpha^{-1}\simeq\frac{360}{\phi^{2}}-\frac{2}{\phi^{3}}+\frac{\mathit{A^{2}}}{K\phi^{4}}-\frac{\mathit{A^{\mathrm{3}}}}{K^{2}\phi^{5}}+\frac{A^{4}}{K^{3}\phi^{7}}\simeq137.035\,999\,168&space;.$
$A=e^{\pi}-7\pi-1$
α is the fine-structure constant, φ is the Golden Ratio, A is the Golden Apex of the Great Pyramid and K is the polygon circumscribing constant. 2016 CODATA: 137.035 999 160 (33).
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# 1 grain per US teaspoon [gr/US tsp] in kilograms per cubic foot
## gr/US tsp to kg/ft³ unit converter of density
1 grain per US teaspoon [gr/tsp] = 0.37 kilogram per cubic foot [kg/ft³]
### grains per US teaspoon to kilograms per cubic foot density conversion cards
• 1
through
25
grains per US teaspoon
• 1 gr/tsp to kg/ft³ = 0.37 kg/ft³
• 2 gr/tsp to kg/ft³ = 0.74 kg/ft³
• 3 gr/tsp to kg/ft³ = 1.12 kg/ft³
• 4 gr/tsp to kg/ft³ = 1.49 kg/ft³
• 5 gr/tsp to kg/ft³ = 1.86 kg/ft³
• 6 gr/tsp to kg/ft³ = 2.23 kg/ft³
• 7 gr/tsp to kg/ft³ = 2.61 kg/ft³
• 8 gr/tsp to kg/ft³ = 2.98 kg/ft³
• 9 gr/tsp to kg/ft³ = 3.35 kg/ft³
• 10 gr/tsp to kg/ft³ = 3.72 kg/ft³
• 11 gr/tsp to kg/ft³ = 4.09 kg/ft³
• 12 gr/tsp to kg/ft³ = 4.47 kg/ft³
• 13 gr/tsp to kg/ft³ = 4.84 kg/ft³
• 14 gr/tsp to kg/ft³ = 5.21 kg/ft³
• 15 gr/tsp to kg/ft³ = 5.58 kg/ft³
• 16 gr/tsp to kg/ft³ = 5.96 kg/ft³
• 17 gr/tsp to kg/ft³ = 6.33 kg/ft³
• 18 gr/tsp to kg/ft³ = 6.7 kg/ft³
• 19 gr/tsp to kg/ft³ = 7.07 kg/ft³
• 20 gr/tsp to kg/ft³ = 7.45 kg/ft³
• 21 gr/tsp to kg/ft³ = 7.82 kg/ft³
• 22 gr/tsp to kg/ft³ = 8.19 kg/ft³
• 23 gr/tsp to kg/ft³ = 8.56 kg/ft³
• 24 gr/tsp to kg/ft³ = 8.93 kg/ft³
• 25 gr/tsp to kg/ft³ = 9.31 kg/ft³
• 26
through
50
grains per US teaspoon
• 26 gr/tsp to kg/ft³ = 9.68 kg/ft³
• 27 gr/tsp to kg/ft³ = 10.05 kg/ft³
• 28 gr/tsp to kg/ft³ = 10.42 kg/ft³
• 29 gr/tsp to kg/ft³ = 10.8 kg/ft³
• 30 gr/tsp to kg/ft³ = 11.17 kg/ft³
• 31 gr/tsp to kg/ft³ = 11.54 kg/ft³
• 32 gr/tsp to kg/ft³ = 11.91 kg/ft³
• 33 gr/tsp to kg/ft³ = 12.28 kg/ft³
• 34 gr/tsp to kg/ft³ = 12.66 kg/ft³
• 35 gr/tsp to kg/ft³ = 13.03 kg/ft³
• 36 gr/tsp to kg/ft³ = 13.4 kg/ft³
• 37 gr/tsp to kg/ft³ = 13.77 kg/ft³
• 38 gr/tsp to kg/ft³ = 14.15 kg/ft³
• 39 gr/tsp to kg/ft³ = 14.52 kg/ft³
• 40 gr/tsp to kg/ft³ = 14.89 kg/ft³
• 41 gr/tsp to kg/ft³ = 15.26 kg/ft³
• 42 gr/tsp to kg/ft³ = 15.64 kg/ft³
• 43 gr/tsp to kg/ft³ = 16.01 kg/ft³
• 44 gr/tsp to kg/ft³ = 16.38 kg/ft³
• 45 gr/tsp to kg/ft³ = 16.75 kg/ft³
• 46 gr/tsp to kg/ft³ = 17.12 kg/ft³
• 47 gr/tsp to kg/ft³ = 17.5 kg/ft³
• 48 gr/tsp to kg/ft³ = 17.87 kg/ft³
• 49 gr/tsp to kg/ft³ = 18.24 kg/ft³
• 50 gr/tsp to kg/ft³ = 18.61 kg/ft³
• 51
through
75
grains per US teaspoon
• 51 gr/tsp to kg/ft³ = 18.99 kg/ft³
• 52 gr/tsp to kg/ft³ = 19.36 kg/ft³
• 53 gr/tsp to kg/ft³ = 19.73 kg/ft³
• 54 gr/tsp to kg/ft³ = 20.1 kg/ft³
• 55 gr/tsp to kg/ft³ = 20.47 kg/ft³
• 56 gr/tsp to kg/ft³ = 20.85 kg/ft³
• 57 gr/tsp to kg/ft³ = 21.22 kg/ft³
• 58 gr/tsp to kg/ft³ = 21.59 kg/ft³
• 59 gr/tsp to kg/ft³ = 21.96 kg/ft³
• 60 gr/tsp to kg/ft³ = 22.34 kg/ft³
• 61 gr/tsp to kg/ft³ = 22.71 kg/ft³
• 62 gr/tsp to kg/ft³ = 23.08 kg/ft³
• 63 gr/tsp to kg/ft³ = 23.45 kg/ft³
• 64 gr/tsp to kg/ft³ = 23.83 kg/ft³
• 65 gr/tsp to kg/ft³ = 24.2 kg/ft³
• 66 gr/tsp to kg/ft³ = 24.57 kg/ft³
• 67 gr/tsp to kg/ft³ = 24.94 kg/ft³
• 68 gr/tsp to kg/ft³ = 25.31 kg/ft³
• 69 gr/tsp to kg/ft³ = 25.69 kg/ft³
• 70 gr/tsp to kg/ft³ = 26.06 kg/ft³
• 71 gr/tsp to kg/ft³ = 26.43 kg/ft³
• 72 gr/tsp to kg/ft³ = 26.8 kg/ft³
• 73 gr/tsp to kg/ft³ = 27.18 kg/ft³
• 74 gr/tsp to kg/ft³ = 27.55 kg/ft³
• 75 gr/tsp to kg/ft³ = 27.92 kg/ft³
• 76
through
100
grains per US teaspoon
• 76 gr/tsp to kg/ft³ = 28.29 kg/ft³
• 77 gr/tsp to kg/ft³ = 28.66 kg/ft³
• 78 gr/tsp to kg/ft³ = 29.04 kg/ft³
• 79 gr/tsp to kg/ft³ = 29.41 kg/ft³
• 80 gr/tsp to kg/ft³ = 29.78 kg/ft³
• 81 gr/tsp to kg/ft³ = 30.15 kg/ft³
• 82 gr/tsp to kg/ft³ = 30.53 kg/ft³
• 83 gr/tsp to kg/ft³ = 30.9 kg/ft³
• 84 gr/tsp to kg/ft³ = 31.27 kg/ft³
• 85 gr/tsp to kg/ft³ = 31.64 kg/ft³
• 86 gr/tsp to kg/ft³ = 32.02 kg/ft³
• 87 gr/tsp to kg/ft³ = 32.39 kg/ft³
• 88 gr/tsp to kg/ft³ = 32.76 kg/ft³
• 89 gr/tsp to kg/ft³ = 33.13 kg/ft³
• 90 gr/tsp to kg/ft³ = 33.5 kg/ft³
• 91 gr/tsp to kg/ft³ = 33.88 kg/ft³
• 92 gr/tsp to kg/ft³ = 34.25 kg/ft³
• 93 gr/tsp to kg/ft³ = 34.62 kg/ft³
• 94 gr/tsp to kg/ft³ = 34.99 kg/ft³
• 95 gr/tsp to kg/ft³ = 35.37 kg/ft³
• 96 gr/tsp to kg/ft³ = 35.74 kg/ft³
• 97 gr/tsp to kg/ft³ = 36.11 kg/ft³
• 98 gr/tsp to kg/ft³ = 36.48 kg/ft³
• 99 gr/tsp to kg/ft³ = 36.85 kg/ft³
• 100 gr/tsp to kg/ft³ = 37.23 kg/ft³
• gr/tsp stands for gr/US tsp
#### Foods, Nutrients and Calories
RASPBERRY DANISH DOEGNUTS, UPC: 072470000095 contain(s) 363 calories per 100 grams or ≈3.527 ounces [ price ]
DUBBLE BUBBLE GUMS, UPC: 059642126043 contain(s) 375 calories per 100 grams or ≈3.527 ounces [ price ]
#### Gravels, Substances and Oils
Substrate, Eco-Complete weighs 1 538 kg/m³ (96.0142 lb/ft³) with specific gravity of 1.538 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Coal, Bituminous broken weighs 833 kg/m³ (52.00249 lb/ft³) [ weight to volume | volume to weight | price | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-417C, liquid (R417C) with temperature in the range of -51.12°C (-60.016°F) to 68.34°C (155.012°F)
#### Weights and Measurements
A pound per square nanometer is a non-metric measurement unit of surface or areal density
The fuel consumption or fuel economy measurement is used to estimate gas mileage and associated fuel cost for a specific vehicle.
mg/yd³ to oz t/US tsp conversion table, mg/yd³ to oz t/US tsp unit converter or convert between all units of density measurement.
#### Calculators
Calculate area of a trapezoid, its median, perimeter and sides
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# A small limit problem
## Homework Statement
lim (x-->infinity) [(sqrt(cosX)-cosX)/(x^2)]
## Homework Equations
10x!
tiny-tim
Homework Helper
Welcome to PF!
Hi alexd! Welcome to PF!
(have a square-root: √ and an infinity: ∞ )
lim (x-->infinity) [(sqrt(cosX)-cosX)/(x^2)]
Hint: draw the graph …
what does it look like?
(erm … you don't mean limx -> 0, do you?)
Thanx!
I drew the graph, it seems to be going to zero when x->∞
and unfortunetally it is x->∞ and not x->0
but i'm a bit stuck with the limit...
any ideas on how to approach it (analytically)?
tiny-tim
Homework Helper
I drew the graph, it seems to be going to zero when x->∞
ok … so why is it going to 0? …
how big is the bottom getting? and how big the top?
ok.. I think I see where u're going with this...
so it's actually not defined when x->∞
The case for x->0 is much easier - the answer is 1/4...
is that what u meant?
tiny-tim
Homework Helper
The case for x->0 is much easier - the answer is 1/4...
is that what u meant?
Yes
ok.. I think I see where u're going with this...
so it's actually not defined when x->∞
Yes it is defined …
just answer my previous question: how do the top and bottom (separately) behave?
well...
the bottom obviously goes to ∞ when x->∞
and as for the top.. I have no idea...
that's where i got stuck the 1st time..
what do you do with cosx when x->∞?
the graph is attached
#### Attachments
• 1.JPG
11.8 KB · Views: 280
tiny-tim
Homework Helper
Hi alexd!
(i can't see your graph yet, but i assume it's upsy-downsy )
well...
the bottom obviously goes to ∞ when x->∞
Yup!
and as for the top.. I have no idea...
that's where i got stuck the 1st time..
what do you do with cosx when x->∞?
I think you're worrying too much about the fact that it obviously doesn't converge
but the important point is that it doesn't get out of hand either
|cosx| can't be greater than 1 …
so the top/bottom … ?
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In the above example, for instance, the class of … But di erent ordered … If x and y are real numbers and , it is false that .For example, is true, but is false. If two elements are related by some equivalence relation, we will say that they are equivalent (under that relation). Equivalence relations. The set [x] ˘as de ned in the proof of Theorem 1 is called the equivalence class, or simply class of x under ˘. Let be an integer. Modular addition and subtraction. Problem 3. Examples of Reflexive, Symmetric, and Transitive Equivalence Properties An Equivalence Relationship always satisfies three conditions: De nition 4. Proof. Example. if there is with . Solution: Relation $\geq$ is reflexive and transitive, but it is not symmetric. For example, if [a] = [2] and [b] = [3], then [2] [3] = [2 3] = [6] = [0]: 2.List all the possible equivalence relations on the set A = fa;bg. An equivalence relation on a set induces a partition on it. What about the relation ?For no real number x is it true that , so reflexivity never holds.. Modulo Challenge (Addition and Subtraction) Modular multiplication. Equality Relation For example, take a look at numbers $4$ and $1$; $4 \geq 1$ does not imply that $1 \geq 4$. It was a homework problem. Show that the less-than relation on the set of real numbers is not an equivalence relation. Let . Let ˘be an equivalence relation on X. Practice: Modular multiplication. The intersection of two equivalence relations on a nonempty set A is an equivalence relation. It provides a formal way for specifying whether or not two quantities are the same with respect to a given setting or an attribute. Problem 2. We write X= ˘= f[x] ˘jx 2Xg. This is the currently selected item. Then is an equivalence relation. An equivalence relation is a relation that is reflexive, symmetric, and transitive. This is false. The equivalence relation is a key mathematical concept that generalizes the notion of equality. A rational number is the same thing as a fraction a=b, a;b2Z and b6= 0, and hence speci ed by the pair ( a;b) 2 Z (Zf 0g). The relation is symmetric but not transitive. We say is equal to modulo if is a multiple of , i.e. Equality modulo is an equivalence relation. If we consider the equivalence relation as de ned in Example 5, we have two equiva-lence … Some more examples… It is true that if and , then .Thus, is transitive. The last examples above illustrate a very important property of equivalence classes, namely that an equivalence class may have many di erent names. Answer: Thinking of an equivalence relation R on A as a subset of A A, the fact that R is re exive means that (For organizational purposes, it may be helpful to write the relations as subsets of A A.) Equivalence relations A motivating example for equivalence relations is the problem of con-structing the rational numbers. First we'll show that equality modulo is reflexive. 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Help needed regarding excel sheet..
India
January 4, 2010 6:35am CST
hi. i need help regarding excel sheet. i am new to use excel sheet formulas. so please help me out. well, i have a situation where i need to two coloums and return the value in thrid for every row. Example: A B C 1 1 2 2 1 3 3 1 4 like this i have to get the values, so i need the formula so that i can add two coloums and return the value into third coloum... the value in the 3coloum 1st row third cell must return to 1st coloum 2nd row first cell. this proces has to continue for the entire row, plz help me out guys
1 person likes this
1 response
@yelambar (144)
• India
4 Jan 10
if i understood you correctly try this.. A B C 1 1 1 =A1+B1 2 =C1 1 =A2+B2 3 =C2 1 =a3+b3 and so on. You can just enter the formula once in cell C1, once done you can then drag the cell down, it will copy all the formula's, same goes for cell A2. Hope this helps, if you have more questions please let me know. :)
@yelambar (144)
• India
4 Jan 10
what a mess, formulae in cell c1 will be =a1+b1, c2 will be =a2+b2 and so on.. while cell a2 will be =C1, a3 will be =c2 and so on. am i makes sense?
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1. ## velocity
A toy rocket vertically in such a way that t seconds after lift-off, it is
h(t)= -1/2t^2 + 30t feet above ground.
a. How high is the rocket after 40 seconds?
b. What is the average velocity of the rocket over the first 40 seconds of flight (between
t
= 0 and
t = 40)?
c. What is the (instantaneous) velocity of the rocket at lift-off (t = 0)? What is the velocity
after 40 seconds?
2. Originally Posted by cclia
A toy rocket vertically in such a way that t seconds after lift-off, it is
[LEFT]h(t)= -1/2t^2 + 30t feet above ground.
a. How high is the rocket after 40 seconds?
do you understand how functions are defined? do you realize that h(t) gives the height at any time t? do you realize, for example, if i wanted to find the height after 2 seconds, this would simply be h(2)?
b. What is the average velocity of the rocket over the first 40 seconds of flight (between t= 0 and t = 40)?
there is a formula for this, it should not have been too hard to find in your text.
the average velocity between t = 0 and t = 40 is given by
$\displaystyle \frac {h(40) - h(0)}{40 - 0}$
c. What is the (instantaneous) velocity of the rocket at lift-off (t = 0)? What is the velocity after 40 seconds?
hint: the derivative gives us the formula for the instantaneous velocity
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#### Please solve RD Sharma class 12 chapter Higher Order Derivatives exercise 11.1 question 41 maths textbook solution
Proved
Hint:
You must know the derivative of cot inverse $x$
Given:
$y=\left(\cot ^{-1} x\right)^{2}, \text { prove } y_{2}\left(x^{2}+1\right)^{2}+2 x\left(x^{2}+1\right) y_{1}=2$
Solution:
$y=\left(\cot ^{-1} x\right)^{2}$
\begin{aligned} &\frac{d y}{d x}=2 \cot ^{-1} x\left(\frac{-1}{1+x^{2}}\right)\\ &\left(1+x^{2}\right) \frac{d y}{d x}=-2 \cot ^{-1} x\\ &\text { Again differentiating }\\ &\left(1+x^{2}\right)^{2} \frac{d^{2} y}{d x^{2}}+2 x \frac{d y}{d x}=-2\left(\frac{-1}{1+x^{2}}\right) \end{aligned}
\begin{aligned} &\left(1+x^{2}\right)^{2} \frac{d^{2} y}{d x^{2}}+2 x \frac{d y}{d x}=\left(\frac{2}{1+x^{2}}\right) \\ &\left(1+x^{2}\right)^{2} \frac{d^{2} y}{d x^{2}}+2 x\left(1+x^{2}\right) \frac{d y}{d x}=2 \\ &y_{2}\left(x^{2}+1\right)^{2}+2 x\left(x^{2}+1\right) y_{1}=2 \end{aligned}
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https://mindmatters.ai/t/pi-as-irrational-number/
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Mind Matters Natural and Artificial Intelligence News and Analysis
TagPi as irrational number
4. How Almost Any Number Can Encode the Library of Congress
That’s a weird, counterintuitive — but quite real — consequence of the concept of infinity in math
We are used to dealing with simple numbers, like ½ and 2. Most numbers are not that simple. Most numbers, like 0.847859028378490… go on forever and ever without repeating or showing any pattern. Note that such numbers, called irrational numbers, have an infinite number of digits. And there are a lot of them. The number 0.847859028378490… for example differs from the number 0.847859023378490… (See if you can spot the difference.) If two numbers differ only at the billionth decimal and are otherwise the same, they are different numbers. Because an irrational number is infinitely long — and we have seen in the first three posts that weird things happen with infinity — we’d expect something weird to happen with irrational Read More ›
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### Sample Problem
At a school, 60% of the students are boys. If there are 40 more boys than girls in the school, how many students are in the school?
students
#### Solution
First find the percentage difference:
The percentage of girls: 100% – 60% = 40%
The percentage difference: 60% – 40% = 20%
Then use the formula: Whole = Different amount ÷ Percentage difference
Difference amount: 40 students
Percentage of difference: 20%
Thus, whole = 40 ÷ 20% = 200 students
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# csrcoo garbles unsorted complex16 matrices
## csrcoo garbles unsorted complex16 matrices
While using csrcoo to convert an unsorted (rows and colums) complex16 (double precision complex) matrix from COO to CSR using the option to sort the columns in the output, I discovered the array of matrix values gets garbled during the sort. The same operation appears to work if the input is already sorted, or using any other data type.
I'm guessing there's something wrong in how it handles swapping/gathering the data in the value arrray. Can anyone confirm?
5 Beiträge / 0 neu
Nähere Informationen zur Compiler-Optimierung finden Sie in unserem Optimierungshinweis.
Would you please provide a simple test case, including a sparse matrix and a code snippet showing how you call mkl_zcsrcoo?
And I also need information on the version of MKL and your OS.
Thanks.
A small program demonstrating my problem is attached. I am running Linux and using MKL 11.0.
## Anlagen:
AnhangGröße
1.27 KB
I can confirm that the problem exists also on Windows 7, MKL 10.3.12 and 11.0.2, 32- or 64-bit. The example code below has the values of the "val" array contrived such that they are also in sort order when printed out.
```
program tcnv
integer,parameter :: n=8,nnz=12
integer :: irow(nnz)=(/ 7, 5, 2, 0, 4, 0, 1, 6, 5, 6, 5, 3 /)
integer :: icol(nnz)=(/ 7, 2, 2, 1, 4, 0, 1, 6, 7, 1, 5, 2 /)
!order C 7 4 2 6 1 3 B 9 A 8 5
complex*16 :: val(nnz)=(/(12,22),(7,17),(4,14),(2,12),(6,16),(1,11), &
(3,13),(11,21),(9,19),(10,20),(8,18),(5,15)/)
complex*16 :: acsr(nnz)
integer :: job(6)=(/2,0,0,0,nnz,0/)
integer :: rowptr(n+1),colind(nnz)
character*198 :: vers
!
call mkl_get_version_string(vers)
write(*,'(A)')vers
call mkl_zcsrcoo(job,n,acsr,colind,rowptr,nnz,val,irow,icol,info)
do i=1,nnz
write(*,'(1x,i3,2x,2F8.1)')i,acsr(i)
end do
end program tcnv
```
The output is not correct:
```
1 12.0 2.0
2 1.0 11.0
3 3.0 13.0
4 4.0 5.0
5 14.0 6.0
6 15.0 16.0
7 7.0 17.0
8 9.0 19.0
9 8.0 18.0
10 11.0 21.0
11 10.0 20.0
12 12.0 22.0
```
Change the array to COMPLEX*8 and mkl_zcsrcoo to mkl_ccsrcoo, and you get
```
1 1.0 11.0
2 2.0 12.0
3 3.0 13.0
4 4.0 14.0
5 5.0 15.0
6 6.0 16.0
7 7.0 17.0
8 8.0 18.0
9 9.0 19.0
10 10.0 20.0
11 11.0 21.0
12 12.0 22.0
```
This does look like a bug. This issue is now escalated to the MKL engineering team. Thanks for catching it!
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# Recursive function causing a stack overflow
I am trying to write a simple sieve function to calculate prime numbers in clojure. I've seen this question about writing an efficient sieve function, but I am not to that point yet. Right now I am just trying to write a very simple (and slow) sieve. Here is what I have come up with:
``````(defn sieve [potentials primes]
(if-let [p (first potentials)]
(recur (filter #(not= (mod % p) 0) potentials) (conj primes p))
primes))
``````
For small ranges it works fine, but causes a stack overflow for large ranges:
``````user=> (sieve (range 2 30) [])
[2 3 5 7 11 13 17 19 23 29]
user=> (sieve (range 2 15000) [])
java.lang.StackOverflowError (NO_SOURCE_FILE:0)
``````
I thought that by using `recur` this would be a non-stack-consuming looping construct? What am I missing?
• +1 for having stack overflow in the title of your question – radman Jun 1 '10 at 1:12
• Funny; works for me. What version of Clojure are you using, with what JVM, on what platform? Can you run `(range 2 15000)` without overflow? – Dominic Cooney Jun 1 '10 at 1:17
• Ubuntu 9.10, Java 1.6.0_15, latest snapshot of Clojure 1.2.0 – dbyrne Jun 1 '10 at 1:30
• Yes, I get an overflow at 15000. Can you run one million without overflowing? – dbyrne Jun 1 '10 at 1:31
• The title should be "non recursive function causing stack overflow". – Petr Gladkikh Dec 29 '14 at 16:47
You're being hit by `filter`'s laziness. Change `(filter ...)` to `(doall (filter ...))` in your `recur` form and the problem should go away.
A more in-depth explanation:
The call to `filter` returns a lazy seq, which materialises actual elements of the filtered seq as required. As written, your code stacks `filter` upon `filter` upon `filter`..., adding one more level of `filter`ing at each iteration; at some point this blows up. The solution is to force the whole result at each iteration so that the next one will do its filtering on a fully realised seq and return a fully realised seq instead of adding an extra layer of lazy seq processing; that's what `doall` does.
• Thanks! This fixed my problem. Excellent explanation. – dbyrne Jun 1 '10 at 1:44
• any thoughts how to find this out? maybe something like macroexpand? – edbond Jun 5 '10 at 20:29
• Have a look at the stack trace, I'd say. A pile of `clojure.lang.LazySeq` method calls would be a good indication that the problem is laziness-related. – Michał Marczyk Jun 5 '10 at 22:23
• Brilliant answer. – Robert Grant Feb 25 '14 at 13:42
Algorithmically the problem is that you continue filtering when there's no more purpose to it. Stopping as early as possible achieves quadratic reduction in recursion depth (`sqrt(n)` vs. `n`):
``````(defn sieve [potentials primes]
(if-let [p (first potentials)]
(if (> (* p p) (last potentials))
(concat primes potentials)
(recur (filter (fn [n] (not= (mod n p) 0)) potentials)
(conj primes p)))
primes))
``````
Runs OK for 16,000 (performing just 30 iterations instead of 1862), and for 160,000 too, on ideone. Even runs 5% faster without the `doall`.
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# FMC Takeoff Ref Page
## Recommended Posts
What is the significance of the QRH ON/OFF option at LSK 6R on the FMC Takeoff Ref page? I also notice when QRH is ON, there sometimes is a slight difference (1 or 2 knots) between one or more of the listed V speeds (eg, v1=119 vs V1=120) -- I ignore the slight difference. If I shouldn't ignore it, what should I do about it?
Thanks,
Al
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QRH means the FMC calculates your v speeds based on the QRH (quick reference handbook). With it off, you have to manually enter them.
Also, you might want to look up what v speeds actually are...
##### Share on other sites
QRH means the FMC calculates your v speeds based on the QRH (quick reference handbook). With it off, you have to manually enter them.
Also, you might want to look up what v speeds actually are...
Thanks for the input.
I know very well what the v speeds are. Just wanted to understand how the difference comes about on that page. If one set is based on the QRH, how did the FMC come up with the other set?
Al
##### Share on other sites
What is the significance of the QRH ON/OFF option at LSK 6R on the FMC Takeoff Ref page?
See page 11.40.49 in the PMDG NGX FCOMv2.
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See page 11.40.49 in the PMDG NGX FCOMv2.
Thanks.
Al
##### Share on other sites
Thanks for the input.
I know very well what the v speeds are. Just wanted to understand how the difference comes about on that page. If one set is based on the QRH, how did the FMC come up with the other set?
Al
As Kyle will say, aviation is not as rigid as one would think. I believe a small 1 or 2 kt difference is probably due to rounding or conversions. It will not impact your takeoff, as by the time the wheels leave the ground you are past V1 and Vr anyhow. Now if it was a 4kt or higher difference, I would look into it further.
##### Share on other sites
As Kyle will say, aviation is not as rigid as one would think.
Thanks for the mention, and you're definitely right. The cool thing about aviation is that we can get so finely detailed. The problem with aviation is that there is just enough variability that concentrating on details that are too fine becomes counterproductive.
At the same time, related to this thread topic, I'm not entirely sure I understand the issue. When you turn the QRH option off, the speeds should disappear. When you turn it on, the computed speeds should reappear. There isn't a "second set" of numbers at all from my experience, or even in the FCOM. Sounds almost like the the OP entered speeds from TOPCAT, or some other tool and is then comparing them to the QRH speeds. The speeds entered on the actual LSK spots are either calculated, or "downselected" from the QRH calc speeds. Again, the FMC doesn't have two sets of data here. If the speeds don't match the QRH speeds (the smaller numbers followed by a > symbol), then they must have been entered by hand, based on some other calculation. If this is the case, and they're slightly off (V1 is expected, since the FMC only calculated balanced field V1, while TOPCAT calcs runway-specific V1), refer to Wes' post above.
In other words, the very first time you show up on the TAKEOFF REF page, you should either see absolutely nothing (QRH OFF) or speeds to the left of blanks (---, which is the QRH ON option). OP: Is this not what you're seeing?
##### Share on other sites
At the same time, related to this thread topic, I'm not entirely sure I understand the issue. When you turn the QRH option off, the speeds should disappear. When you turn it on, the computed speeds should reappear. There isn't a "second set" of numbers at all from my experience, or even in the FCOM.
On FCOM v2 pg 11.40.48 you will see a representation of the FMC Takeoff Ref pg with two columns of V speeds that differ slightly (by 1 Kt). This is exactly the situation I've run into, but don't know why. In this situation pushing the LSKs does not transfer the QRH values into the right hand column. I’m not concerned about the 1Kt difference between a given v speed, just curious how this comes about and why pushing the respective LSK does not update the value.
Thx,
Al
Edit: Looks like the two columns of V speeds may be related to Change of Performance Data After V Speed Entry, see pgs 11.40.49 and 11.40.50. It is not clear to me which column of speeds is the "latest", although I would guess the right column since I can't find a way to "load" the left column using the LSKs.
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1. ## Salt problem
A tank with capacity of 400 gal of water originally contains 200 gal of water with 100 lb of salt in solution. Water containing 1 lb of salt per gallon is entering at a rate of 3 gal/ min, and the mixture is allowed to flow out of the tank at a rate of 2 gal/ min. Let Q(t) lb. be the amount of salt in the tank, V(t) gal be the volume of water in the tank.
Find the amount of salt in the tank at any time prior to the instant when the solution begins to overflow. Your answer should be a function in terms of t.
Doesn't Q(T) = t +200?
Find the concentration (in pounds per gallon) of salt in the tank when it is on the point of overflowing. Round your answer to 3 decimal places.
Doesn't this just equal 200 +200 = 400?
2. Originally Posted by lord12
A tank with capacity of 400 gal of water originally contains 200 gal of water with 100 lb of salt in solution. Water containing 1 lb of salt per gallon is entering at a rate of 3 gal/ min, and the mixture is allowed to flow out of the tank at a rate of 2 gal/ min. Let Q(t) lb. be the amount of salt in the tank, V(t) gal be the volume of water in the tank.
Find the amount of salt in the tank at any time prior to the instant when the solution begins to overflow. Your answer should be a function in terms of t.
Doesn't Q(T) = t +200?
Find the concentration (in pounds per gallon) of salt in the tank when it is on the point of overflowing. Round your answer to 3 decimal places.
Doesn't this just equal 200 +200 = 400?
Well if you think about it, initially you have 100 lbs of salt in 200 gallons or 1/2 lb/gal. The concentration of the incoming salt solution is 1 lb/gal. So if you assume perfect (and instanteous) mixing, the concentration will change from 1/2 to 1 (assuming an infinite tank). Here we must set up a differential equation for the process.
First the tank in fill up $V =(3-2)t + 200$.
Next, the change in the amount of salt A is
$\frac{dA}{dt} = r_i c_i - r_o c_o$
where ri - rate in, ro - rate out, ci - concentration in and co - concentration out.
so
$\frac{dA}{dt} = 3 \cdot 1 - 2 \cdot \frac{A}{V} = 3 - \frac{2A}{t+200}$
the initial condition
$A(0) = 100$. A linear ODE for A. Can you solve it?
3. I solve the diffeq and get that A = t +200
4. Originally Posted by lord12
I solve the diffeq and get that A = t +200
$A = t + 200 - \frac{4,000,000}{(t+200)^2}$.
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# Explanation Of ‘for-end’ Loop In MatLab With Examples
Actually, for loop is a common in general programming languages avaliable. Also Matlab has for loop to obtain required algorithms. In here, we explain how to use for loop in Matlab with very basic examples.
## How To Use ‘for’ Loop In MatLab Programming?
To use for loop in Matlab, you can click on the given link to remember the operators that are used in Matlab.
You can obtain loops for different queries to obtain algoritms in Matlab, by increasing the looped value in each loop by using ‘for’ loop. Taka a look at the example below;
``````>> x = [5+3j 2+6j 5j 5 -4 0 -8j 12 -1+8j];
im = 0; real = 0;
for y = 1:9
if imag(x(y))~=0
%query for each elements of x
real=real+1;
else
real = real+1;
end
end
real
im
real =
3
im =
5
``````
If you understand above program example that obtained in Matlab command window, you will learn the logic of for-end in Matlab. Assume that we want to create a program that finds the number of elements that has imaginary value and that has not imaginary value separately, of a vector in Matlab.
Assume that we have vector that has 9 elements inside it, and some of these elements has imaginary number. To do it, we created two of variables named ‘im’ and ‘real’ after the vector. These variables have initial values of 0. In general, you need to give inital values to your variables that will be used in for-end loop in Matlab. These values will change inside for-end loop.
To obtain for loop, you need to create a vector that has elements in increasing direction. For example, we created the vector ‘y’ for for-end loop above, that has elements of 1 to 9 inside it. In each loop, ‘y’ will take values starting from 1 ending at 8. Once for-end loop reaches to 9, program will exit the for-end loop, and go on other codes after the for-end loop.
To find out the imaginary element inside vector ‘x’, we created a if-else query. We said that in each loop, ‘y’ will take different values. So, if we build a program that asks to the vector ‘x’, “Does your current y. element(which is the value of ‘y’ in current loop…) has imaginary section?”
We asked this question with if-else in Matlab by indexing the each element of vector ‘x’ in each loop, inside ‘imag()’ command. If the imaginary section is not zero that we stated in if command above, the value of ‘im’ will be increased 1(im=im+1). If there is no imaginary section(so it is ‘else’ in our code), the value of ‘real’ will be increased 1.
At the end of the loop, we want to see the values of these two vectors. As you see in the result at command window, 3 of elements has no imaginary section, so they are real, and the number that has imaginary section is 5.
So in each loop, the variable that assigned to for-end command will take a new values. You can obtain programmes for these new values in each loop, inside the for-end command in Matlab.
Another example about for-end loop in Matlab;
``````x = [1 5 6 3; 4 6 3 1; 7 6 5 6;];
for y = 1:3
%indexing the rows of x matrix
for z = 1:4
%indexing the columns of x matrix
D(z,y)=x(y,z)
end
end
D =
1 4 7
5 6 6
6 3 5
3 1 6``````
This example is about the priority of the for-end loops that coded inside themselves. In here, we want to convert a 3×4 matrix which is ‘x’ into a 4×3 matrix which is ‘D’ above. So in each loop, we need to index the matrix elements individually to assign these valus to a new matrix elements.
We created a for-end loops that intertwined to each other. For the outmost for-end loop, we created a vector variable ‘y’, which will index the row of individual element from ‘x’ to assing it to matrix ‘D’ as column.
In innermost for-end loop, we created the vector variable ‘z’ to index the column adress of each elements of matrix ‘x’ to assign them into rows of matrix ‘D’.
Inside the innermost for-end loop, we wrote the code that will exchange the elements of matrix ‘x’ into matrix ‘D’, changing their row and column adresses to column and row adresses in ‘D’.
The created code will start the taking the value of 1 for ‘y’, and taking the value of 1 for ‘z’. The inner for-end loop has the priority; which will increase the value of ‘z’ up to 4, when the value of ‘y’ is 1. When the inner loop complete, outer loop will to ‘y’ the value of 2. So, code will work in this manner up to ‘y’ reaches the value of 4.
As you see in the result, matrix ‘x’ inverted to matrix ‘D’ in command window in Matlab.
YOU CAN LEARN MatLab IN MECHANICAL BASE; Click And Start To Learn MatLab!
If you want further code examples about for-end loop in Matlab, state it in comments.
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0
# IN football One knee in bounds equal two feet?
Updated: 9/28/2023
Wiki User
13y ago
The answer is yes. There is a book written by John Madden entitled "one Knee equals two feet (and everything else you need to know about football)" that deals with this subject in depth. The book was published in 1986 but can still be found on Amazon.com or ebay. The NFL rules article 3 states "A forward pass is complete (by the offense) or intercepted (by the defense) if a player, who is inbounds:
(a) secures control of the ball in his hands or arms prior to the ball touching the ground; and
(b) touches the ground inbounds with both feet or with any part of his body other than his hands.
Wiki User
13y ago
Wiki User
12y ago
Yes, they both have to be inbounds inorder for it to be counted as a catch.
You must maintain possession of the ball too.
Wiki User
13y ago
True
Earn +20 pts
Q: IN football One knee in bounds equal two feet?
Submit
Still have questions?
Related questions
### One knee equals two feet but what if one knee is in bounds and one knee comes down out of bounds?
It is incomplete.
### Is a player out of bounds if the ball and his upper body are out of bounds before his knee hits the ground in bounds?
Rules vary from state to state but in this case it is perfectly legal. So along as the players feet are in bounds. In highschool football its normally if one foot is in bounds and the other is not out of bounds and the player has control of the ball then its a catch
### Who said one knee eguals two feet?
John Madden, it's a football term. One knee hitting the ground is the equivalent to two feet hitting the ground for purposes of a receiver making a catch in or out of bounds. It is also the title of Madden's book.
### At what times is someone in possession of the ball considered out of bounds if ones knee goes down in bounds but the ball hits out of bounds?
If his knee touches inbounds, he is considered tackled in the field of play, regardless if the the ball ends up out of bounds. The clock will continue to run
### What muscles contract when you kick a football?
the quadriceps will contract and the hamstring will relax .the quariceps straightens the knee and the hamstring extends the hip and bends the knee to kcik the football the quadriceps will contract and the hamstring will relax .the quariceps straightens the knee and the hamstring extends the hip and bends the knee to kcik the football the quadriceps will contract and the hamstring will relax .the quariceps straightens the knee and the hamstring extends the hip and bends the knee to kcik the football
extends the knee
no
knee
### You landed on your feet and hurt your knee What is wrong?
I know when I did that, I twisted the ligaments in my knee.
knee injury
Playing football
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# Power point pass now
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### Power point pass now
1. 1. How did you do on your assessment? These 5 simple questions can tell us a lot about you, so let’s check them out!
2. 2. How did you do on question number one <ul><li>Subtract: 7/8 – 1/4 = (a) 5/8 </li></ul><ul><li>How did we get this? </li></ul><ul><li>There are a couple of rules for adding and subtracting fractions: </li></ul><ul><li>What you do to the bottom you must do to the top. </li></ul>
3. 3. More about fractions <ul><li>When adding or subtracting, you should always look for the common denominator. </li></ul><ul><li>So lets take a second look: </li></ul><ul><li>7 /8 </li></ul><ul><li>- 1/4 which one is your common denominator? </li></ul><ul><li>(8) is your common denominator, because out of the two denominators (8) is the only denominator that can have a number go into it. </li></ul>
4. 4. Denominators <ul><li>What do we mean? Looking at your two denominators: (4) and (8), (8) is the largest. (4) can go into (8) two times, but (8) cannot go into (4) making (8) your common denominator. </li></ul><ul><li>1 × 2 =2 what you do to the bottom </li></ul><ul><li>4 × 2 =8 you do to the top </li></ul><ul><li>now you have 2/8 – 7/8 = 5/8 </li></ul>
5. 5. How did you do on question number two <ul><li>Question two is not much different than number one, except now we are adding. </li></ul><ul><li>You are still looking for the common denominator in all (3) fractions. So take some time to figure what the most common denominator will be for: </li></ul><ul><li>3/4 + 7/8 + 1/4 </li></ul>
6. 6. More common denominators <ul><li>What did you get? </li></ul><ul><li>These are our converted fractions: </li></ul><ul><li>6/8 +2/8+7/8 as you can see, (8) is the common denominator </li></ul><ul><li>You should have gotten 6/8 from ¾ </li></ul><ul><li>You should have gotten 2/8 from ¼ </li></ul><ul><li>And 7/8 will remain the same! Now lets add. </li></ul>
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## KY_Fan
I have tried to figure this out on my own forever now. I have been getting hint from the net but can't figure it out. So I figured I would admit defeat and see what you guys could show me. I need the three numbers entered into this program to add together, as of right now they aren't adding. The numbers are just showing up. EDIT: I AM A NOOB AT PROGRAMMING
``````def main():
a = input("Please select a number:")
b = input("And another one:")
d = a+b+c
print(d)
main()
``````
## pyTony 888
Which numbers? I only see you taking three string input and then concatenating them, assuming you are using python3 as indicated by form of your print statements. No use to define a function for only one print call.
## KY_Fan
I have to input a random number each time it asks (After I run the code). Then instead of giving me the sum of the three random numbers I put in, it just displays the three random numbers. Where I am knew to this stuff it is kind of difficult for me to explain and my professor is a loser and won't help!! lol
## pyTony 888
``````def add():
a = input("Please select a number: ")
b = input("And another one: ")
c = input("One more please: ")
return float(a) + float(b) + float(c)
``````
## KY_Fan
Wooo! Thanks for showing me the proper code. I just can't believe I took this online CS course and professor is horrible. He doesn't help at ALL and he only gives us a day or so per assignment. Considering this is an intro to CS class you would think we would have more time! I am going to try and reconnect with one of my high school teachers that knows how to code.
## sneekula 969
This is really a question of what version of Python you are using.
If you use Python2 then input() is for numeric input, but with Python3 you will get strings.
With Python27 this will work:
``````# Python27
def main():
a = input("Please select a number:")
b = input("And another one:")
d = a+b+c
print(d)
main()
'''possible result ...
And another one:2
6
'''
``````
## KY_Fan
Well I am using Python3. I guess that is why I was having so much trouble, I'm not sure what my version of Python my book is using either. I guess I need to read up some more..
Thanks for the help!
## sneekula 969
``````# find the version of Python you are using
import platform
print(platform.python_version())
``````
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# Chef and Rainbow Array, why is it giving 1st subtas correct and onother one is wrong
t=int(input())
for i in range(t):
(flag1,flag2)=(0,0)
n=int(input())
a=[int(x) for x in input().split()]
k=n//2
if a[k]==7:
u=a[:k]
if n&1:
v=a[k+1:]
else:
v=a[k:]
v.reverse()
for j in range(len(u)-1):
if u[j+1]==u[j] or u[j+1]==u[j]+1:
continue
else:
break
else:
flag1=1
for j in range(len(v)-1):
if v[j+1]==v[j] or v[j+1]==v[j]+1:
continue
else:
break
else:
flag2=1
if u==v and flag1==1 and flag2==1:
print(“yes”)
else:
print(“no”)
else:
print(“no”)
//
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# Career Center > Careers in Statistics
## What Do Statisticians Do?
The world is becoming quantitative. More and more professions, from the everyday to the exotic, depend on data and numerical reasoning.
Data are not just numbers, but numbers that carry information about a specific setting and need to be interpreted in that setting. With the growth in the use of data comes a growing demand for the services of statisticians, who are experts in the following:
• Producing trustworthy data
• Analyzing data to make their meaning clear
• Drawing practical conclusions from data
### Examples of Statistics Careers
#### Medicine
The search for improved medical treatments rests on careful experiments that compare promising new treatments with the current state of the art. Statisticians work with medical teams to design experiments and analyze the complex data they produce.
#### Environment
Studies of the environment require data on the abundance and location of plants and animals, on the spread of pollution from its sources, and on the possible effects of changes in human activities. The data are often incomplete or uncertain, but statisticians can help uncover their meaning.
#### Industry
The future of many industries and their employees depends on improvement in the quality of goods and services and the efficiency with which they are produced and delivered. Improvement should be based on data, rather than guesswork. More companies are installing elaborate systems to collect and act on data to better serve their customers.
#### Government Surveys
How many people are unemployed this month? What do we export to China, and what do we import? Are rates of violent crime increasing or decreasing? The government wants data on issues such as these to guide policy, and government statistics agencies provide them by surveys of households and businesses.
#### Market Research
Are consumer tastes in television programs changing? What are promising locations for a new retail outlet? Market researchers use both government data and their own surveys to answer questions such as these. Statisticians design the elaborate surveys that gather data for both public and private use.
#### The Nature of Statistics
Statistics provides the reasoning and methods for producing and understanding data. Statisticians are specialists, but statistics demands they be generalists, too.
#### Mathematics and Computers Are Involved ...
Statistics uses mathematics, but it is not abstract or isolated. Statisticians work with people from other professions to solve practical problems. Statistics uses modern computing to organize and analyze data, and statisticians command specialized tools. But the emphasis is on the data to be understood and the problem to be solved, rather than on computing for its own sake.
#### ... But Understanding the Data Is Crucial
Statisticians must know more than statistics. A statistician who works in medicine or in a manufacturing plant or in market research must learn enough medicine or engineering or marketing to understand the data in their setting. Statisticians need the ability to work with other people, to listen, and to communicate.
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{[ promptMessage ]}
Bookmark it
{[ promptMessage ]}
assignment3
assignment3 - mentation 3 Let A ={h R,S i| R and S are...
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CSC544 Assignment #3 due Thursday 3/10 in class version 1.0 Problems 1. A Turing machine with a doubly infinite tape is similar to an ordinary Turing machines, but its tape is infinite to the left as well as to the right. The tape is initially filled with blanks except for the portion that con- tains the input. Computation is defined as usual except that the head never encounters an end to the tape as it moves leftward. Show that this type of Turing machine recognizes exactly the class of Turing-recognizable languages. 2. Show that the collection of decidable languages is closed under comple-
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Unformatted text preview: mentation. 3. Let A = {h R,S i| R and S are regular expressions and L ( R ) ⊆ L ( S ) } . Show that A is decidable. (Hint: regular languages are closed under complemen-tation, intersection, and union). 4. Prove that EQ DFA is decidable by testing the two D±As on all strings up to a certain size. Calculate a size that works. 5. A language is prefx-Free iF no member is a proper prefx oF another mem-ber. Let PF REX = {h R i| R is a regular expression where L ( R ) is prefx-Free } . Show hat PF REX is decidable. 1...
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https://www.imperial.ac.uk/aeronautics/study/ug/current-students/modules/h401/?module=AERO96013&year=20_21
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# Aeronautical Engineering (MEng)
## Introduction to Turbulence and Turbulence Modelling
### Module aims
The course aims at providing the student with a minimum background knowledge and physical understanding necessary for the critical assessment of turbulent models which the student is very likely to be confronted with in subsequent courses and in industry.
### Learning outcomes
On successfully completing this course unit, students will be able to:
· describe the basic characteristics of turbulent flows, as well as their practical consequences concerning drag, dispersion of momentum/material/heat, etc;
· distinguish the differences in the mechanics of important classes of turbulent flows such as boundary free flows and wall flows;
· operate Reynolds decompositions and work out turbulent mean flow, energy, dissipation, pressure properties;
· manipulate basic turbulent models and understand their strengths and limitations;
· operate multiscale/Fourier analysis of homogeneous turbulence and derive the Kolmogorov theory of small-scale turbulence;
· derive the log-law of the wall in various ways whilst understanding the assumptions made and their limitations;
· understand the range of applicability of the Taylor frozen turbulence hypothesis fundamental for the interpretation of hot wire anemometry measurements.
· understand and operate various decompositions, mean and fluctuations on the one hand, multiscale/Fourier decompositions of fluctuations on the other, a very widely transferable skill to all types of data and PDE analyses.
· use fundamental knowledge to investigate and assess new and emerging technologies
· apply mathematical and computer-based models for solving aerodynamic fluid-flow problems in engineering
### Module syllabus
Introduction: basic characteristics of turbulent flows: 3-D, vortex stretching, energy transfer, enstrophy, strain rates, randomness and statistics.
Reynolds decomposition: mean flow turbulence fluctuations and energetics. Pivotal experimental observation concerning independence of kinetic energy dissipation rate from Reynolds number. The importance of pressure. Brief discussion of practical consequences to do with drag and dispersion of momentum, material and heat.
Examples of simple turbulent flows: two-dimensional free shear flows (jets and wakes), two-dimensional channel flow, turbulent boundary layer, decaying grid flow. Introduce the concept of statistical homogeneity and its consequences. Distinction between wall flows and free shear layers.
The Kolmogorov theory of homogeneous turbulence: The universality theory, the -5/3 law for the energy spectrum and the limitations of universality.
Wall turbulence. The log law. Anisotropic energy spectra and their scalings. Eddy structure of turbulent boundary layers and its relation to spectra.
for numerical simulation
Measurement of turbulent flows. Constant temperature anemometry, Taylor’s frozen flowfield hypothesis, basic uncertainty quantification.
### Teaching methods
Lectures, tutorials
### Assessments
Examined Assessment
2 hour written examination in January (100%)
Non-Examined Assessment
1 x Progress test (peer marked)
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http://mymathforum.com/advanced-statistics/3760-urn-contains-azure-balls-c-carmine-balls-help-print.html
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My Math Forum (http://mymathforum.com/math-forums.php)
- - An urn contains a azure balls and c carmine balls.. (Help) (http://mymathforum.com/advanced-statistics/3760-urn-contains-azure-balls-c-carmine-balls-help.html)
byron123 September 4th, 2008 11:56 PM
An urn contains a azure balls and c carmine balls.. (Help)
Quote:
An urn contains a azure balls and c carmine balls, where a.c is different 0 (zero). Balls are removed at random and discarded until the first time that a ball (B, say) is removed having a different colour from its predecessor. The ball B is now replaced and the procedure restarted. This process continues until the last ball is drawn from the urn. Show that this last ball is equally likely to be azure or carmine.
Can anyone help me out as to how to approach & solve this question? I guess something with induction but I could really use some help..
Thanks!
RFurball September 5th, 2008 03:10 AM
Perhaps you could try a few examples
That's a good question and I don't know the exact approach but if I were to try to solve it I would run through a couple of examples on paper.
First I would go through all the combinations of 2 azure and 1 carmine balls and see how the probability works out.
Next I would go through all combinations of 3 and 2 balls or 3 and 1 ball.
From these examples you may be able to pick out the pattern for a and c balls.
cknapp September 10th, 2008 09:39 AM
Re: An urn contains a azure balls and c carmine balls.. (Help)
Hmm... this is a tricky question, but some things to think about that *may* help:
What is the probabilty that your first azure ball happens on the k-th selection? First carmine ball?
Then you have (a+c)+1-k balls remaining: How many azure balls do you have? How many carmine? (I.e. what is the probability of selecting an a or a c?)
You have to do something inductive, I would assume.
All times are GMT -8. The time now is 06:16 AM.
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#### What is 717 percent of 57?
How much is 717 percent of 57? Use the calculator below to calculate a percentage, either as a percentage of a number, such as 717% of 57 or the percentage of 2 numbers. Change the numbers to calculate different amounts. Simply type into the input boxes and the answer will update.
## 717% of 57 = 408.69
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Calculating seven hundred and seventeen of fifty-seven How to calculate 717% of 57? Simply divide the percent by 100 and multiply by the number. For example, 717 /100 x 57 = 408.69 or 7.17 x 57 = 408.69
#### How much is 717 percent of the following numbers?
717 percent of 57.01 = 40876.17 717 percent of 57.02 = 40883.34 717 percent of 57.03 = 40890.51 717 percent of 57.04 = 40897.68 717 percent of 57.05 = 40904.85 717 percent of 57.06 = 40912.02 717 percent of 57.07 = 40919.19 717 percent of 57.08 = 40926.36 717 percent of 57.09 = 40933.53 717 percent of 57.1 = 40940.7 717 percent of 57.11 = 40947.87 717 percent of 57.12 = 40955.04 717 percent of 57.13 = 40962.21 717 percent of 57.14 = 40969.38 717 percent of 57.15 = 40976.55 717 percent of 57.16 = 40983.72 717 percent of 57.17 = 40990.89 717 percent of 57.18 = 40998.06 717 percent of 57.19 = 41005.23 717 percent of 57.2 = 41012.4 717 percent of 57.21 = 41019.57 717 percent of 57.22 = 41026.74 717 percent of 57.23 = 41033.91 717 percent of 57.24 = 41041.08 717 percent of 57.25 = 41048.25
717 percent of 57.26 = 41055.42 717 percent of 57.27 = 41062.59 717 percent of 57.28 = 41069.76 717 percent of 57.29 = 41076.93 717 percent of 57.3 = 41084.1 717 percent of 57.31 = 41091.27 717 percent of 57.32 = 41098.44 717 percent of 57.33 = 41105.61 717 percent of 57.34 = 41112.78 717 percent of 57.35 = 41119.95 717 percent of 57.36 = 41127.12 717 percent of 57.37 = 41134.29 717 percent of 57.38 = 41141.46 717 percent of 57.39 = 41148.63 717 percent of 57.4 = 41155.8 717 percent of 57.41 = 41162.97 717 percent of 57.42 = 41170.14 717 percent of 57.43 = 41177.31 717 percent of 57.44 = 41184.48 717 percent of 57.45 = 41191.65 717 percent of 57.46 = 41198.82 717 percent of 57.47 = 41205.99 717 percent of 57.48 = 41213.16 717 percent of 57.49 = 41220.33 717 percent of 57.5 = 41227.5
717 percent of 57.51 = 41234.67 717 percent of 57.52 = 41241.84 717 percent of 57.53 = 41249.01 717 percent of 57.54 = 41256.18 717 percent of 57.55 = 41263.35 717 percent of 57.56 = 41270.52 717 percent of 57.57 = 41277.69 717 percent of 57.58 = 41284.86 717 percent of 57.59 = 41292.03 717 percent of 57.6 = 41299.2 717 percent of 57.61 = 41306.37 717 percent of 57.62 = 41313.54 717 percent of 57.63 = 41320.71 717 percent of 57.64 = 41327.88 717 percent of 57.65 = 41335.05 717 percent of 57.66 = 41342.22 717 percent of 57.67 = 41349.39 717 percent of 57.68 = 41356.56 717 percent of 57.69 = 41363.73 717 percent of 57.7 = 41370.9 717 percent of 57.71 = 41378.07 717 percent of 57.72 = 41385.24 717 percent of 57.73 = 41392.41 717 percent of 57.74 = 41399.58 717 percent of 57.75 = 41406.75
717 percent of 57.76 = 41413.92 717 percent of 57.77 = 41421.09 717 percent of 57.78 = 41428.26 717 percent of 57.79 = 41435.43 717 percent of 57.8 = 41442.6 717 percent of 57.81 = 41449.77 717 percent of 57.82 = 41456.94 717 percent of 57.83 = 41464.11 717 percent of 57.84 = 41471.28 717 percent of 57.85 = 41478.45 717 percent of 57.86 = 41485.62 717 percent of 57.87 = 41492.79 717 percent of 57.88 = 41499.96 717 percent of 57.89 = 41507.13 717 percent of 57.9 = 41514.3 717 percent of 57.91 = 41521.47 717 percent of 57.92 = 41528.64 717 percent of 57.93 = 41535.81 717 percent of 57.94 = 41542.98 717 percent of 57.95 = 41550.15 717 percent of 57.96 = 41557.32 717 percent of 57.97 = 41564.49 717 percent of 57.98 = 41571.66 717 percent of 57.99 = 41578.83 717 percent of 58 = 41586
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https://betterlesson.com/lesson/resource/2029367/warm-up-absolute-value-functions-day-2
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## Warm Up- Absolute Value Functions Day 2 - Section 1: Warm Up and Homework Check
Warm Up- Absolute Value Functions Day 2
# Absolute Value Functions: Transformations Day 2
Unit 2: Modeling with Functions
Lesson 21 of 24
## Big Idea: Students model sun reflection and miniature golf in this extension on the transformation of absolute value functions.
Print Lesson
2 teachers like this lesson
Standards:
Subject(s):
Math, Algebra, Algebra 2, master teacher project, absolute value functions, modeling absolute value functions, transformation of functions
50 minutes
### Amelia Jamison
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Hello everyone,
Some college friends and myself are building a standalone 30W wind turbine and I am having trouble powering my arduino due with the Vin pin with my boost converter. Our generator output at cut-in speed is about 2.3V and I am trying to use an Yeeco 2.0V - 24.0V boost converter to turn on the arduino at the cut-in.
However, when I hook up the converter to the arduino, the output voltage is ALWAYS at 3.6V no matter what the input voltage is and the current is very high (about a 1V -> 300mA ratio). But when I disconnect the arduino the output voltage is 10.5V (which is what I need) and input voltage is where it was originally?? When I use a DC voltage supply I have to increase the current source to about 1A before I faintly see the "ON" light on the arduino. Also when using DC supply the inout voltage always falls to 1.6V unless if I increase the current. But I have seen the Due's current draw at 9V is about 120mA and at 12V its only 55mA? I at first thought the converter could not supply enough current but it can supply up to 2A!! To me it seems like an elementary (relatively speaking) issue but I just CAN'T figure out why this is happening
It sounds like something is taking current, what else is connected to the Due?
Does this happen with nothing else connected to the Due?
Are you sure the output voltage from your boost converter is clean, have you looked at it n an oscilloscope?
This is the link to the converter.
http://www.yeecoconverter.com/index.php?main_page=product_info&products_id=145
Nothing else connected literally just connections from DC supply -> converter -> arduino. Even in this simple setup it is the same result. I have tried breadboarding all 3 to the same ground bus rather than +- potential pins still the same. I have a feeling there is something weird in the logic of the boost converter when arduino is plugged in?
Strange things can happen when you put a boost converter (external), a 5volt buck converter (Due), and a linear 3.3volt regulator (Due MCU) in series.
Both switching regulators probably have UVL (under voltage lockout). The Due has.
And likely high startup currents, that the generator might not be able to supply on startup.
I don't see this working without some sort of battery/charging system.
Leo..
Is there a way to disable those undervoltage lockouts in the Due? I know that it is unwise but we are kind of desperate
You could try extra capacitors on the output of your boost converter. Maybe a 470uF with a 0.1 uF ceramic across it.
bcm78:
Is there a way to disable those undervoltage lockouts in the Due?
No.
Converters have losses. Doubling output voltage more than doubles the required input current of that booster.
You stand a better chance by just connecting the generator to V-in (assuming voltages are within range of the Due). Maybe a fat capacitor (thousands of uF) can help bridge the startup current. Dunno.
Can't you use an Arduino that's less power-hungry?
8Mhz Arduinos can work from 1.8volt.
leo..
A few thoughts:
And there is a brownout detector. A brownout can trigger a Reset and by reading SUPC_SR register you can know the source of the Reset (page 287, Sam3x datasheet)
Hi,
How many volts is your turbine?
Can you post a diagram of your project please, so we can see how you are powering everything, including your wind turbine and batteries?
Thanks.. Tom...
bcm78:
Hello everyone,
Some college friends and myself are building a standalone 30W wind turbine and I am having trouble powering my arduino due with the Vin pin with my boost converter. Our generator output at cut-in speed is about 2.3V and I am trying to use an Yeeco 2.0V - 24.0V boost converter to turn on the arduino at the cut-in.
However, when I hook up the converter to the arduino, the output voltage is ALWAYS at 3.6V no matter what the input voltage is and the current is very high (about a 1V -> 300mA ratio). But when I disconnect the arduino the output voltage is 10.5V (which is what I need) and input voltage is where it was originally?? When I use a DC voltage supply I have to increase the current source to about 1A before I faintly see the "ON" light on the arduino. Also when using DC supply the inout voltage always falls to 1.6V unless if I increase the current. But I have seen the Due's current draw at 9V is about 120mA and at 12V its only 55mA? I at first thought the converter could not supply enough current but it can supply up to 2A!! To me it seems like an elementary (relatively speaking) issue but I just CAN'T figure out why this is happening
First, let me offer a suggestion: Power the Due via, either the USB port, or via the 5V pin. Thus you bypass the onboard switching supply.
Topic #428302.0 -- Where discussed is use of 5V Out to power board
Next, a wind turbine generator's raw output is likely to be full of ripple (at a frequency determined by the turbine rotation) speed, right? If so, that could be a reason for odd behavior. Or, is there some sort of regulator in the turbine. BTW: when you say "generator", are you referring to the wind turbine? Or is "generator" something else? And if something else, is that a solid (i.e. well regulated) 2.3V? And, as the Turbine picks up speed, that 2.3V will increase, right? What is the maximum voltage? I.e. the voltage range?
And finally, your description of the problem is a bit confusing to me. When you say "when I hook up the converter to the arduino, the output voltage is ALWAYS at 3.6V", what output? The output of the Yeeco converter? Or, the 3.3V output on the Due? And, I'm not sure what you mean by 1V -> 300mA ratio. When you disconnect the Arduino, it's the output of the Yeeco Converter that goes to 10.5V, right...when the input to the converter is at, what, 2.3V? And, why is "10.5V", what you need? The Arduino Due has, at the VIN input, a range of 7V to 12V. Why isn't 7V the needed voltage (though, I suppose, that may not be the voltage the Due's SMPS is most efficient)? "and input voltage is where it was originally"...wait, when did the input voltage change? Do you mean input voltage was originally at 2.3V, and, when you connected the Veeco, it changed to some other voltage? What was the voltage it changed to when the Veeco was connected?
When I use a DC voltage supply I have to increase the current source to about 1A before I faintly see the "ON" light on the arduino.
That is strange...I wonder if you damaged the DC-to-DC converter on the Due? Do you have another Due that you can try powering with your "DC voltage supply", to see if the same thing happens? Also, as Wawa pointed out, this could be due to interaction between the two switch mode converters. Maybe a linear regulator following the Veeco -- a low dropout regulator, to minimize power loss -- will filter out noise that could be affecting the Due's regulator. [or, bypass the Due's regulator, all together, as I mentioned, above.
Also when using DC supply the inout voltage always falls to 1.6V unless if I increase the current.
I'm not sure what you mean by the "inout voltage". Do you mean the output voltage on your "DC voltage supply"?
but I have seen the Due's current draw at 9V is about 120mA and at 12V its only 55mA?
I don't have any experience with the quiescent power requirements of a Due, but the Switch mode supply on the other side of the Vin pin will make efficient use of power, so when the voltage goes up (at Vin), the current demand goes down, for an overall power demand that is somewhat equivalent (efficiency changes, and other sources of power loss (such as the 3.3V linear regulator) will cause deviations from an even power demand).
Sorry if this sounds snarky -- I in no way mean to be mean (couldn't resist ) -- just trying to help you, and all participating, get to the bottom of this ;D
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# Question Video: Solving Word Problems on Unit Rate That Involve Decimals Mathematics • 5th Grade
A car uses one litre of petrol to travel 10 kilometres. How many litres of petrol will be used on a journey of 676.8 kilometres?
02:41
### Video Transcript
A car uses one litre of petrol to travel 10 kilometres. How many litres of petrol will be used on a journey of 676.8 kilometres?
Having read this question, we can see that one litre of petrol is used for every 10 kilometres that a car travels. We are asked to calculate how many litres of petrol will be used on a journey of 676.8 kilometres. To find the answer, we need to find how many lots of 10 kilometres there are in this long distance. In other words, we need to find the answer to 676.8 divided by 10. We know that 676.8 consists of six hundreds, seven tens which have a value of 70, six ones, and eight-tenths.
And when we’re thinking about the place value of a number, each column to the right of another number is worth 10 times less. So, for example, 10 times less than a hundred is a ten. 10 times less than a ten is a one, and so on. So, if we want to divide 676.8 divided by 10, what we really want to do is for each digit in that number to be worth 10 times less.
In other words, they all need to shift one place to the right. 10 times less than eight-tenths is eight hundredths. 10 times less than six ones is six-tenths. 10 times less than seven tens, is seven ones. And if we divide six hundreds by 10, we get six tens. 676.8 divided by 10 is 67.68.
If a car uses one litre of petrol to travel 10 kilometres, we can find the number of litres of petrol that will be used on a journey of 676.8 kilometres by dividing this number by 10. We managed to divide 676.8 by 10 by shifting the digits one place to the right. Our answer contains the same four digits; they’ve just moved. The number of litres of petrol that will be used on a journey of 676.8 kilometres is 67.68 litres.
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https://nl.mathworks.com/matlabcentral/cody/problems/107-count-from-0-to-n-m-in-base-n/solutions/486983
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Cody
# Problem 107. Count from 0 to N^M in base N.
Solution 486983
Submitted on 15 Aug 2014 by bainhome
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% M = 2; N = 2; y_correct = [0 0 1 1; 0 1 0 1]; assert(isequal(countInBaseN(M,N),y_correct))
2 Pass
%% M = 2; N = 3; y_correct = [ 0 0 0 1 1 1 2 2 2; 0 1 2 0 1 2 0 1 2]; assert(isequal(countInBaseN(M,N),y_correct))
3 Pass
%% M = 3; N = 4; y_correct = [ 0 0 0 0 0 1 0 0 2 0 0 3 0 1 0 0 1 1 0 1 2 0 1 3 0 2 0 0 2 1 0 2 2 0 2 3 0 3 0 0 3 1 0 3 2 0 3 3 1 0 0 1 0 1 1 0 2 1 0 3 1 1 0 1 1 1 1 1 2 1 1 3 1 2 0 1 2 1 1 2 2 1 2 3 1 3 0 1 3 1 1 3 2 1 3 3 2 0 0 2 0 1 2 0 2 2 0 3 2 1 0 2 1 1 2 1 2 2 1 3 2 2 0 2 2 1 2 2 2 2 2 3 2 3 0 2 3 1 2 3 2 2 3 3 3 0 0 3 0 1 3 0 2 3 0 3 3 1 0 3 1 1 3 1 2 3 1 3 3 2 0 3 2 1 3 2 2 3 2 3 3 3 0 3 3 1 3 3 2 3 3 3]; assert(isequal(countInBaseN(M,N),y_correct'))
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