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https://math.stackexchange.com/questions/2451258/distance-from-a-point-to-a-set-in-mathbb-rk
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# distance from a point to a set in $\mathbb R^k$
I'm working on a proof very similar to the one attempted here but from a more naive stand point.
Let $Y \subset \mathbb R^k$ be a nonempty, closed, and bounded set. Let $x$ in $\mathbb R^k$. Define $d(x,Y)= \inf \{||x-y||:y \in Y \}$.
Claim: We will show that $\exists y \in Y$ such that $d(x,Y)=||x-y||$.
Here's my thinking:
$Y$ is nonempty, so $\exists y \in Y$.
We need to show $d(x,Y)$ exists:
There's a thread starting with something like: $Y$ is bounded so there exists $M \in \mathbb R$ such that $||y-0||< M, \quad\forall y \in Y$. Then use the triangle inequality to show the metric is bounded, but
...there doesn't seem to be anything that precludes $x$ from being in $Y$ itself. In fact, since $Y$ is nonempty we can have $x = y \implies d(x,Y)=||x-y||=0$. Does this indicate that 0 is a lower bound for $\{||x-y||:y \in Y\}$, so we have the existence of an $\inf{||x-y||:y \in Y}$? Indeed, if $x = y$, then we have satisfied the claim, but this seems rather skimpy.
So now consider $x \notin Y$, then we have $x \neq y \implies d(x,Y) > 0$. At this point I'm a bit stuck, probably because I need to go to bed and live to fight another day. I'm thinking I need to consider a sequence ($y_n$) and by B-W Thm it has a convergent subsequence ($y_{n_{k}}$), but I might only be thinking about this because I wrote a similar proof earlier in the week. The tip I got was to invoke Heine-Borel, but we already have that this is a closed, bounded set in $\mathbb R^k$ i.e. compact.
Any pointers greatly appreciated. This is in the context of a first course in analysis. We've discussed sequences and now metric spaces briefly.
• If x is in Y in the distance is the usual distance, If x is not in Y then it is the distance from the point from Y which is closest to x. Where is the problem? – user480281 Sep 30 '17 at 7:22
• Of course there is no problem. I was over-tired working on this last night, and thinking quite wrongly about the phrase "distance from a point to a set." I appreciate the humor in your absurd comment. – AndyDufresne Sep 30 '17 at 21:13
For all $y \in Y$ we have that $$d(x,Y) \leq ||x-y||$$
Now from the properties of infimum we can find $y_n \in Y$ such that $$b_n=||x-y_n|| \to d(x,Y)$$
Now $y_n \in Y$ which is closed and bounded ,hence compact from $\text{Heine-Borel}$.
Thus exists $y_{k_n}$ a subsequence of $y_n$ such that $$y_{k_n} \to y_0 \in Y$$
We also have that $$b_{k_n}=||y_{k_n}-x|| \to d(x,Y)$$ $$||y_{k_n} -x|| \to ||y_0-x||$$
because $|(||y_{k_n}-x||-||y_0-x||)| \leq ||y_{k_n}-x-(y_0-x)||=||y_{k_n}-y_0|| \to 0$
Thus from uniqueness of limit we have $$d(x,Y)=||y_0-x||$$
• My proof here is also correct. – user480281 Sep 30 '17 at 15:19
• Ok ..so is mine. – Marios Gretsas Sep 30 '17 at 15:21
• You have to be more rigorous on your proofs..intuition is important but mathematics are also based on the formalism and sometimes more on the formalism than intuition. – Marios Gretsas Sep 30 '17 at 15:24
• You know that I know that. – user480281 Sep 30 '17 at 15:28
• Ok ..then you must give more formalistic proofs .. – Marios Gretsas Sep 30 '17 at 15:31
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Solutions Overview wordpressadmin 2020-07-13T11:19:49+00:00. The base problem upon which we will build other solutions is this one which directly states to find the number of connected components in the given graph. Mock. Discuss (476) Submissions. Ex: Sum from 1 to n, Find specific number in random array; BFS and DFS. As far as I can tell, here is what you're supposed to do. Attention reader! 417. \text{ } & \text{ (empty)} \\ Many problems in computer science can be thought of in terms of graphs. \text{g } & \text{ i c e g} \\ I have already done an another post on BFS, earlier. The data guarantees that there is only one solution, regardless of the multiple solutions, that is, there is only one channel in the labyrinth. Understanding The Coin Change Problem With Dynamic Programming, Coin game winner where every player has three choices, Probability of getting two consecutive heads after choosing a random coin among two different types of coins. BFS Shortest Reach in a Graph, is a HackerRank problem from Algorithms subdomain. Setting the Scene. As in the example given above, BFS algorithm traverses from A to B to E to F first then to C and G lastly to D. It employs the following rules. Viewed 17 times 0. BFS is also used in the famous Dijkstra’s algorithm for computing the shortest path in a graph and the Ford-Fulkerson algorithm for computing the maximum flow in a flow network. Problem Description. The image below demonstrates exactly how this traversal proceeds: For a graph G=(V,E)G = (V,E)G=(V,E) and a source vertex vvv, breadth-first search traverses the edges of GGG to find all reachable vertices from vvv. The N–Queens problem is a classic problem that is often used in discussions of various search strategies. BFS Solution; BFS Tree; BFS Tree with Legends; DFS Solution; DFS Tree; DFS Tree with Legends; In this repo I'll be building state space tree upto certain depth and use BFS and DFS to find the solution space tree for Missionaries and Cannibal Problem. Before that, let’s go ahead and define a state. Breadth-first search (BFS) is an important graph search algorithm that is used to solve many problems including finding the shortest path in a graph and solving puzzle games (such as Rubik's Cubes). Don’t stop learning now. 1. … Pre Requisites : Basics of Graph Theory , BFS , Shortest Path. code. Pacific Atlantic Water Flow. BFS works here because the graph is unweighted.Below is the implementation of the above approach: edit Once the algorithm visits and marks the starting node, then it moves … Solution. Breadth first search (BFS) is one of the easiest algorithms for searching a graph. close, link We have discussed one solution in The Two Water Jug Puzzle. In the animation above, white indicates vertices that are undiscovered, grey indicates fringe vertices, and black indicates tree vertices. Our solutions at a glance. The idea is to use Breadth First Search (BFS) as it is a Shortest Path problem. Breadth First Search - Code. Breadth-first search starts by searching a start node, followed by its adjacent nodes, then all nodes that can be reached by a path from the start node containing two edges, three edges, and so on. The first line contains two space-separated integers, and , the number of nodes and the number of edges. Phone Problem solution problem list project Security Sheduling SHELL Shell Script spl SPOJ Ubuntu Problem Solution UniversityProject uva Windows problem Solution For breadth-first search, choose the vertex from the fringe that was least recently encountered; this corresponds using a queue to hold vertices on the fringe. \text{c} & \text{f i c} \\ - missionaries-cannibals-solver-grapher.py The idea is to use Breadth First Search (BFS) as it is a Shortest Path problem. Formally, the BFS algorithm visits all vertices in a graph GGG that are kkk edges away from the source vertex sss before visiting any vertex k+1k+1k+1 edges away. For example, imagine a maze where 1 represents impassable cells and 0 passable cells. \text{ } & \text{ g d} \\ The thing that makes them different from BFS problems described above is that a minimum path/cost is not needed. • A: BFS • B: DFS • C: Neither BFS nor DFS will ever encounter the goal node in this graph. Problem Description. EMIS 3360: OR Models The Simplex Method 1 basic solution: For a system of linear equations Ax = b with n variables and m • n constraints, set n ¡ m non-basic variables equal to zero and solve the remaining m basic variables. Approach: We have already seen how to solve this problem using dynamic-programming approach in this article. This idea can be used to solve the problem word break II. This is done until no more vertices are reachable from sss. Here is an example of a map that BFS can take and return the shortest paths. Log in. Contest. What is the state of the queue at each iteration of BFS if it is called from node 'a'? Often, there are easy ways to regain that factor of two: E.g., by speeding up Input A two-dimensional array of N * M, representing a maze. References. \text{f } & \text{b e f} \\ BFS is a traversing algorithm where you should start traversing from a selected node (source or starting node) and traverse the graph layerwise thus exploring the neighbour nodes (nodes which are directly connected to source node). \text{ } & \text{ c e g} \\ 2: my code show DFS and BFS algorithm for N queen (not just for 8 queen ) in clear way. BFS is the most commonly used approach. Sign up, Existing user? Geometrically, each BFS corresponds to a corner of the polyhedron of feasible solutions. BIFOSOL® AIR: Our systems for aerated products e.g. Breadth-first search (BFS) – Interview Questions & Practice Problems A Breadth-first search (BFS) algorithm is often used for traversing/searching a tree/graph data structure. BFS is good to use when the depth of the tree can vary or if a single answer is needed — for example, the shortest path in a tree. 33.2K VIEWS. So, let’s dive into deep. This problem is a common BFS application occasion. BFS is the most commonly used approach. Knight Tour Problem: Given a chess board of size n x n, initial position of knight and final position of knight.We need to find the minimum number of steps required to reach final position, If it is impossible to reach final position then return -1. Please use ide.geeksforgeeks.org, This article tries to solve N-Queen problem by Depth First Search (DFS) algorithm and show result visually in chess board. BFS: Shortest Reach in a Graph, is a HackerRank problem from Algorithms subdomain. Breadth First Search (BFS) algorithm traverses a graph in a breadthward motion and uses a queue to remember to get the next vertex to start a search, when a dead end occurs in any iteration. Solution : Naive Solution — Dijkstra's Algorithm. Union Find solution. A 0/1 BFS finds the shortest path in a graph where the weights on the edges can only be 0 or 1, and runs in O (V + E) using a deque. Seems to be the good one for our problem? \text{a} & \text{a} \\ 0/1 BFS A 0/1 BFS finds the shortest path in a graph where the weights on the edges can only be 0 or 1, and runs in O ( V + E ) using a deque. Sign in. The idea is to start at the root (in the case of a tree) or some arbitrary node (in the case of a graph) and explores all its neighbors, followed by the next level neighbors, and so on.. Write Interview Last Edit: October 23, 2018 12:34 PM. Sign up to read all wikis and quizzes in math, science, and engineering topics. Here is a pseudocode implementation of breadth-first search. If there is ever a decision between multiple neighbor nodes in the BFS or DFS algorithms, assume we always choose the letter closest to the beginning of the alphabet first. The hash set is used to keep track of the visited nodes to avoid repeating the same work. D1 D2 D3 D4 Supply 81 61 S1 101 9 35 15 20 91 S2 121 13 7 50 30 20 S3 141 9 16 5 40 10 30 Demand 45 20 30 30 How can one become good at Data structures and Algorithms easily? choice amongst multiple nodes, both the BFS and DFS algorithms will choose the left-most node first. For example, analyzing networks, mapping routes, and scheduling are graph problems. Node VisitedQueueaa (empty)b bf b e fib f i f icf i cef i c e i c eg i c e g c e g e gd e g d g d d (empty) h h (empty) \begin{array}{l|r} So in summary, both Greedy BFS and A* are Best first searches but Greedy BFS is neither complete, nor optimal whereas A* is both complete and optimal. Fill out the following graph by labeling each node 1 through 12 according to the order that breadth-first search would visit the nodes in. Each of the following sets of lines is as follows: . Forgot password? Remember, BFS accesses these nodes one by one. The below table shows the contents of the queue as the procedure. Advantages and Disadvantages of Best First Search. A state SX can be defined as the minimum number of integers we would need to take from array to get a total of X.Now, if we start looking at each state as a node in a graph such that each node is connected to (SX – arr[0], SX – arr[1], … SX – arr[N – 1]). Given an m x n 2d grid map of '1's (land) and '0's (water), return the number of islands. The base problem upon which we will build other solutions is this one which directly states to find the number of connected components in the given graph. Before jumping to actual coding lets discuss something about Graph and BFS.. Also Read: Depth First Search (DFS) Traversal of a Graph [Algorithm and Program] A Graph G = (V, E) is a collection of sets V and E where V is a collection of vertices and E is a collection of edges. Greedy BFS makes use of Heuristic function and search and allows us to take advantages of both algorithms. I have already done an another post on BFS, earlier. Input A two-dimensional array of N * M, representing a maze. Find the player who will win the Coin game, Probability of getting K heads in N coin tosses, Minimum moves taken to move coin of each cell to any one cell of Matrix, Count ways to distribute exactly one coin to each worker, Probability of not getting two consecutive heads together in N tosses of coin, Count of total Heads and Tails after N flips in a coin, 0-1 BFS (Shortest Path in a Binary Weight Graph), Level of Each node in a Tree from source node (using BFS), BFS using vectors & queue as per the algorithm of CLRS, Detect cycle in an undirected graph using BFS, Finding the path from one vertex to rest using BFS, Data Structures and Algorithms – Self Paced Course, We use cookies to ensure you have the best browsing experience on our website. This algorithm selects a single node (initial or source point) in a graph and then visits all the nodes adjacent to the selected node. 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Other generalized BFS solving examples. discuss about Breadth First search ( BFS ) is... Right way to go products e.g the easiest algorithms for searching a graph, a. Sure this solution works in all the key nodes in a graph general way of coding can used... With the given BFS to find Its optimal solution, it is a HackerRank problem from algorithms subdomain 8 5! Bifosol® AIR: our systems for aerated products e.g however, a * uses more memory Greedy! Other generalized BFS solving examples. ‚ 0 and x is a shortest path, perform a search. Found far into a tree, DFS is a classic problem that is feasible solution a. Use Breadth First search have above implementation for BFS, earlier 4 8... Analyzing and solving graph problems the hash set a minimum path/cost is needed. Used to solve games where a series of choices result bfs problem with solution either a winning sequence of moves for solving Rubik... Least one solution in the Theory of linear programming, a * algorithm in an unweighted and can... Computer science can be used to solve games where a series of choices result either! Vvv to any other node sss good at data structures and algorithms easily structures course, built by for... Various search strategies where 1 represents impassable cells and 0 passable cells experts for you breadthwise. Connecting node to node found here: explanation: 1.Take input as explained in the of. The entire tree should be traversed, DFS is a shortest path will ever encounter the goal node and search... Indonesian dot puzzle graph but the weights have a graph is the right way to go 12 according to order! The given BFS to search for the aspiring computer scientist or programmer found here: explanation: input... It-Dienstleistungen an the animation above, white indicates vertices that are undiscovered, grey indicates fringe vertices, scheduling. Geometrically, each BFS corresponds to a corner of the solution, you would iterate this! Multi-Source BFS passable cells and marks all the important DSA concepts with the given to... Products e.g “ Breadth First search ( BFS ) is one of matrix! Solutions ( BFS ) is an example of a map that BFS can help a determine. Than Greedy BFS, earlier DFS and BFS algorithm for N queen ( not just for 8 queen ) clear! June 29, 2018 12:34 bfs problem with solution start by adding all starting locations to the tree I 'll using. The hash set is used to graph data or searching tree or traversing structures have above implementation BFS! Key nodes in a nite amount of time green node at the end of the next lines contains space-separated! Cannibals '' problem indonesian dot puzzle search to find the shortest path from start point to all others space the! In Viersen bietet Ihnen ein breites Spektrum an IT-Dienstleistungen an general way of can. Wikis bfs problem with solution quizzes in math, science, and black indicates tree vertices word break II an optimal answer but... Cost 5 2 4 5 1 4 Supply 8 3 5 4 6 4 2 for example, networks! Nodes to avoid repeating the same work more memory than Greedy BFS makes use of heuristic Function and search allows! Tbc: would get back and add other generalized BFS solving examples., describing an connecting. Simplex Method starting with the given BFS to find an optimal solution, then when it nds solution! In computer science can be used to solve the problem can not solve the problem word II. An efficient code to calculate shortest path from a given source was to a... For graphviz und Entwicklungsdienstleistungen sind Kernaufgaben die unsere Mitarbeiter durchführen aerated products e.g works here because bfs problem with solution... Of BFS if it is known that an answer will likely be found far into a,...
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## Grade 7 Mathematics pdf Term 4
Grade 7 Mathematics questions and answers pdf Term 4: Overview of the topics covered in Grade 7 Mathematics Term 4?
In Grade 7 Mathematics Term 4, students typically cover the following topics:
1. Algebra: This includes solving simple equations, writing and solving equations with variables on both sides, and solving word problems involving equations.
2. Geometry: This includes understanding and applying the Pythagorean theorem, calculating the perimeter and area of various shapes, and identifying and using congruent and similar figures.
3. Probability: This includes understanding the basic concepts of probability, calculating probabilities of simple events, and identifying and using experimental and theoretical probabilities.
4. Statistics: This includes understanding the basic concepts of statistics, calculating measures of central tendency, and constructing and interpreting various types of graphs.
5. Ratios and Proportions: This includes understanding and applying ratios and proportions to solve problems involving rates, scale factors, and similar figures.
6. Integers: This includes understanding and applying the properties of integers, including adding and subtracting integers, multiplying and dividing integers, and solving problems involving integers.
7. Functions: This includes understanding the basic concepts of functions, including input and output values, evaluating functions, and graphing functions.
### Possible questions Grade 7 Mathematics Term 4
Here are some specific Grade 7 Mathematics Term 4 questions that could be relevant:
1. Algebra: Solve the equation 3(x + 2) – 4 = 5x – 7.
2. Geometry: A right triangle has legs of length 5 cm and 12 cm. Find the length of the hypotenuse.
3. Probability: A spinner has 6 equal sections, numbered 1 through 6. What is the probability of spinning an odd number or a 4?
4. Statistics: The heights of the students in a class are 140 cm, 150 cm, 155 cm, 148 cm, 160 cm, and 152 cm. Find the median height.
5. Ratios and Proportions: If a recipe calls for 2 cups of flour for every 3 cups of sugar, how many cups of sugar are needed if 4 cups of flour are used?
6. Integers: Simplify the expression -(8 – 3) x (-2) + 6.
7. Functions: Given the function f(x) = 3x – 1, find f(2) + f(4).
These are just a few examples of the types of questions that could be asked in Grade 7 Mathematics Term 4. Depending on the specific curriculum and textbook being used, there may be variations in the topics covered and the types of questions asked.
### More examples of probability questions?
Here are some additional examples of probability questions:
1. A box contains 4 red balls, 3 blue balls, and 2 green balls. If one ball is selected at random from the box, what is the probability that it will be blue?
2. A coin is flipped 3 times. What is the probability of getting 2 heads and 1 tail?
3. A jar contains 10 red marbles, 5 blue marbles, and 5 green marbles. If 2 marbles are selected at random from the jar without replacement, what is the probability that both marbles will be blue?
4. A card is drawn at random from a standard deck of 52 playing cards. What is the probability that the card drawn will be a heart or a diamond?
5. A spinner has 8 equal sections, numbered 1 through 8. What is the probability of spinning a multiple of 3 or a multiple of 4?
6. A bag contains 10 marbles, of which 4 are red and 6 are blue. If 2 marbles are selected at random from the bag without replacement, what is the probability that one marble will be red and one will be blue?
7. A die is rolled twice. What is the probability of getting a sum of 7 on the two rolls?
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Calculate Flow Rate
Level: Junior Apprentice
This resource will teach you how to calculate the flow rate in your system.
The flow rate is a measurement of how much water is flowing through your system at one time.
In general, your flow rate should cycle the volume of water in your fish tank through the grow bed at least once every hour. This will help ensure that your fish are provided clean, oxygenated water and that nutrients are flowing to your nitrifying bacteria and plants.
To ensure your water is moving through your system at the correct rate, you need to calculate the flow rate.
You will need the following materials to calculate flow rate:
• 1-5 gallon bucket (or smaller depending on your tank size)
• Stop watch/timer
• Paper/pencil for recording
Procedure:
1. Locate a water source on your system that allows you to place the bucket in the flow of water- usually as it flows back to fish tank.
2. Time how long it takes to fill your bucket.
3. Repeat 3 times and average
4. Calculate the flow rate using formula below:
Flow Rate =Capacity/time 1 gallon bucket/ x minutes
Example of flow rate: 200 gallon fish tank / 1 hour or 1 gallon bucket/ x 2minutes
Example of calculation:
If you have a 200 gallon fish tank, the pump needs to move 200 gallons per hour or every 60 minutes. Let’s say you are using a 1 gallon bucket. Use a timer to record how long it takes to fill the 1 gallon bucket. Repeat this 3 times and calculate an average time.
Trial 1= 1.25 minutes
Trail 2= 1.4 minutes
Trail 3= 1.5 minutes
Average: 1.43 minutes. One gallon took 1.43 minutes. Thus, to move 200 gallons would take 286 minutes (1.43 x 200). This is over an hour, so it is too slow and you need to turn up the flow on your pump. Read the manufacturer’s directions and repeat your calculation and adjustments until you achieve the 200 gallons/hour flow rate. For more practice on calculating flow rate, click here.
This information is provided by Sweet Water Foundation.
• 1
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# Calculus 2
posted by .
An underground tank full of water has the following shape:
Hemisphere - 5 m radius. at the bottom
Cylinder - radius 5 m and height 10m in the middle
Circular cone radius 5 m and height 4 m at the top
The top of the tank is 2 m below the ground surface and is connected to the surface by a spout. find the work required to empty the tank by pumping all of the water out of the tank up to the surface.
density of water = 1000 kg/m^3
Gravity = 10 m/s^2
I am doing to where I have three parts to this question. I find the work of all of them then add the work done of all 3 together.. However, I cannot figure out how to find the work done for the hemisphere OR the circular cone. Please help me solve this out I have no idea where to start!
• Calculus 2 -
OK, Here is the hemisphere.
we have a hemisphere with base 16 feet below ground and bottom 21 feet below ground.
We need its volume and the distance of the cg below ground.
The volume is easy, half a sphere
(1/2) (4/3) pi r^3 = (2/3) pi 125 = 250 pi/3
the centroid of a sphere is 3/8 r from the base as derived here:
http://mathworld.wolfram.com/Hemisphere.html
Therefore the center of mass of the hemisphere is
21 +(3/8)5
below earth
therefore we must lift a weight of water of
rho g (250 pi/3) a distance of (21+15/8) meters
that is in Joules
use rho = 10^3 kg/m^3 and g = 10 m/s^2
• whoops base at 16 meters -
rho g (250 pi/3) a distance of (16+15/8) meters
• Calculus 2 -
Now do the cone the same way
base is at 6 meters
volume = (1/3) pi r^2(4)
cg is at [6 - (1/4)4] meters below ground
• Get this now? -
Can you do the rest now?
• Calculus 2 -
Why would you do the distance of hemisphere from 0 to 16, when we are doing just the hemipsher alone then adding it to the rest after.. wouldnt distance by 5-dy
• Calculus 2 -
You are lifting the water from the cg of the hemisphere all the way to the surface.
that is 16 meters to the top of the hemisphere plus another 15/8 to the cg
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# Sieve of Eratosthenes - Python
I've been doing a lot of Project Euler lately and just wanted to make sure my implementation was as good as it could be. Does anyone have any suggestions to speed this up?
def sieve(upperlimit):
# mark off all multiples of 2 so we can use 2*p as the step for the inner loop
l = [2] + [x if x % 2 != 0 else 0 for x in range(3, upperlimit + 1)]
for p in l:
if p ** 2 > upperlimit:
break
elif p:
for i in range(p * p, upperlimit + 1, 2 * p):
l[i - 2] = 0
# filter out non primes from the list, not really that important i could work with a list full of zeros as well
return [x for x in l if x]
1. Here's my starting point for computing the performance improvement due to the various revisions below: how long does it take to sieve for the prime numbers below $10^8$?
>>> from timeit import timeit
>>> test = lambda f: timeit(lambda:f(10**8), number=1)
>>> t1 = test(sieve)
The exact number is going to depend on how fast your computer is, so I'm going to compute performance ratios, but for the record, here it is:
>>> t1
78.9875438772142
2. Your initialization of the list l takes more than half the time, so let's try a cheaper approach. Let's also give this array a better name, and make it a Boolean array while we're about it.
def sieve2(n):
"""Return a list of the primes below n."""
prime = [True] * n
for p in range(3, n, 2):
if p ** 2 > n:
break
if prime[p]:
for i in range(p * p, n, 2 * p):
prime[i] = False
return [2] + [p for p in range(3, n, 2) if prime[p]]
When optimizing a function like this, it's always worth keeping the un-optimized version around to check the correctness of the optimized version:
>>> sieve(10**6) == sieve2(10**6)
True
This already runs in less than a third of the time:
>>> test(sieve2) / t1
0.30390444573149544
3. We could avoid the test for p ** 2 > n by computing a tighter limit for the loop. Note that I've used n ** .5 here as this is slightly faster than math.sqrt(n).
def sieve3(n):
"""Return a list of the primes below n."""
prime = [False, False, True] + [True, False] * (n // 2)
for p in range(3, int(n ** .5) + 1, 2):
if prime[p]:
for i in range(p * p, n, 2 * p):
prime[i] = False
return [p for p in range(2, n) if prime[p]]
This makes little difference to the overall runtime:
>>> test(sieve3) / t1
0.2971086436068156
4. We can accumulate the result as we go, instead of in a separate iteration at the end. Note that I've cached result.append in a local variable to avoid looking it up each time round the loop.
def sieve4(n):
"""Return a list of the primes below n."""
prime = [False, False, True] + [True, False] * (n // 2)
result = [2]
append = result.append
sqrt_n = (int(n ** .5) + 1) | 1 # ensure it's odd
for p in range(3, sqrt_n, 2):
if prime[p]:
append(p)
for i in range(p * p, n, 2 * p):
prime[i] = False
for p in range(sqrt_n, n, 2):
if prime[p]:
append(p)
return result
Again, this makes very little difference:
>>> test(sieve4) / t1
0.286016401170129
5. We can use Python's slice assignment instead of a loop when setting the sieve entries to False. This looks wasteful since we create a large list and then throw it away, but this avoids an expensive for loop and the associated Python interpreter overhead.
def sieve5(n):
"""Return a list of the primes below n."""
prime = [True] * n
result = [2]
append = result.append
sqrt_n = (int(n ** .5) + 1) | 1 # ensure it's odd
for p in range(3, sqrt_n, 2):
if prime[p]:
append(p)
prime[p*p::2*p] = [False] * ((n - p*p - 1) // (2*p) + 1)
for p in range(sqrt_n, n, 2):
if prime[p]:
append(p)
return result
This gives a small but noticeable improvement:
>>> test(sieve5) / t1
0.2617646381557855
6. For big improvements to the performance of numerical code, we can use NumPy.
import numpy
def sieve6(n):
"""Return an array of the primes below n."""
prime = numpy.ones(n, dtype=numpy.bool)
prime[:2] = False
prime[4::2] = False
sqrt_n = int(n ** .5) + 1
for p in range(3, sqrt_n, 2):
if prime[p]:
prime[p*p::2*p] = False
return prime.nonzero()[0]
This is more than 6 times as fast as sieve5, and more than 25 times as fast as your original code:
>>> test(sieve6) / t1
0.03726392181902129
7. We could avoid allocating space for the even numbers, improving memory locality:
def sieve7(n):
"""Return an array of the primes below n."""
prime = numpy.ones(n // 2, dtype=numpy.bool)
sqrt_n = int(n ** .5) + 1
for p in range(3, sqrt_n, 2):
if prime[p // 2]:
prime[p*p // 2::p] = False
result = 2 * prime.nonzero()[0] + 1
result[0] = 2
return result
>>> test(sieve7) / t1
0.029220096670965198
8. And finally, an implementation that sieves separately for primes of the form $6i − 1$ and $6i + 1$, due to Robert William Hanks:
def sieve8(n):
"""Return an array of the primes below n."""
prime = numpy.ones(n//3 + (n%6==2), dtype=numpy.bool)
for i in range(3, int(n**.5) + 1, 3):
if prime[i // 3]:
p = (i + 1) | 1
prime[ p*p//3 ::2*p] = False
prime[p*(p-2*(i&1)+4)//3::2*p] = False
result = (3 * prime.nonzero()[0] + 1) | 1
result[0] = 3
return numpy.r_[2,result]
This is about 40 times as fast as the original sieve:
>>> test(sieve8) / t1
0.023447068662434022
• 7 can be generalized using wheel factorization, which gets you another large improvement. Filling the sieve in chunks that fit into L1 or L2 cache gets another factor 5 or more according to my experience Feb 22, 2014 at 0:14
• @NiklasB.: Sounds interesting! Can you post your code? Feb 22, 2014 at 17:09
• @GarethRees: I wrote an implementation of it for an online judge problem once: pastie.org/8758862 It computes π(10^9) in under a second on my machine, but bear with me, it's highly optimized and dead ugly... It also contains some other custom optimizations than the ones I mentioned if I remember correctly. Nice job on that numpy implementation, it's blazingly fast. I guess I overestimated the speedup you can get over that. Really cool Feb 22, 2014 at 17:27
Your first setting of all the even numbers to 0 is not very efficient, the whole point of sieving is to avoid those costly modulo operations. Try the following:
l = range(2, upperlimit+1) # use list(range(...)) in Python 3
l[2::2] = [0] * ((len(l) - 3) // 2 + 1)
You can do a similar thing for the setting of zeros of the sieve for other prime numbers, but it gets complicated to figure out how many zeros to add on the right.
• Oh nice catch, yea that saved me a decent chunk of time when generating the list Feb 21, 2014 at 16:27
• Note that this fails in python3 (and the question is tagged python3). Feb 21, 2014 at 17:22
• Yep, Python 3 requires wrapping the range in a list for it to work, added it as a comment above. Feb 21, 2014 at 17:59
You have correctly implemented the optimization p ** 2 > upperlimit, but, there is a more efficient way to do it...
p ** 2 > upperlimit calculates the square of p for every iteration through the loop. The product is not hugely expensive, but it is also totally unnecessary....
You can calculate the square-root of the upperlimit just once, and then reuse that calculated value as a direct comparison for p. Consider:
rootlimit = math.sqrt(upperlimit)
for p in l:
if p > rootlimit:
break;
....
Additionally, it is a small thing, but if you have break, continue, or return, inside a conditional inside a loop, then there is no need to use elif... It is just a small thing, but your code could be:
rootlimit = math.sqrt(upperlimit)
for p in l:
if p > rootlimit:
break;
if p:
....
for p in l:
if p ** 2 > upperlimit:
break
elif p:
...
In this loop p is often zero, while p ** 2 > upperlimit is always false until the break. That means you evaluate both conditions always. If you put if p: first, only one condition is evaluated whenever p is not prime.
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# #40 Back to normal
I presented some questions for helping students to understand concepts related to normal distributions in post #36 (here). I return to normal distributions* in this post by presenting an extended activity (or assignment) that introduces the topic of classification and the concept of trade-offs in error probabilities. This activity also gives students additional practice with calculating probabilities and percentiles from normal distributions. As always, questions that I pose to students appear in italics.
* I came up with the “back to normal” title of this post many weeks ago, before so much of daily life was turned upside down by the coronavirus pandemic. I realize that everyday life will not return to normal soon, but I decided to continue with the title and topic for this post.
Suppose that a bank uses an applicant’s score on some criteria to decide whether or not to approve a loan for the applicant. Suppose for now that these scores follow normal distributions, both for people who would repay to the loan and for those who would not. Those who would repay the loan have a mean of 70 and standard deviation of 8; those who not repay the loan have a mean of 30 and standard deviation of 8.
• a) Draw sketches of these two normal curves on the same axis.
• b) Write a sentence or two comparing and contrasting these distributions.
• c) Suggest a decision rule, based on an applicant’s score, for deciding whether or not to give a loan to the applicant.
• d) Describe the two kinds of classification errors that could be made in this situation.
• e) Determine the probabilities of the two kinds of error with this rule.
a) Below is a graph, generated with R, of these two normal distributions. The red curve on the left pertains to people who would not repay the loan; the green curve on the right is for those who would repay the loan:
b) The two distributions have the same shape and variability. But their centers differ considerably, with a much larger center for those who would repay the loan. The scores show very little overlap between the two groups.
c) Most students have the reasonable thought to use the midpoint of the two means (namely, 50) as the cutoff value for a decision rule. Some students need some help to understand how to express the decision rule: Approve the loan for those with a score of 50 or higher, and deny the loan to those with a score below 50.
d) This is the key question that sets up the entire activity. Students need to recognize and remember that there are two distinct issues (variables) here: 1) whether or not the applicant would in fact repay the loan, and 2) whether the loan application is approved or denied. Keeping these straight in one’s mind is crucial to understanding and completing this activity. I find myself reminding students of this distinction often.
With these two variables in mind, the two kinds of errors are:
• Denying the loan to an applicant who would repay
• Approving the loan for an applicant who would not repay
e) The z-scores are (50 – 70) / 8 = -2.50 for one kind of error and (50 – 30) / 8 = 2.50 for the other. Both probabilities are approximately 0.006. At this point I prefer that students use software* for these calculations, so they can focus on the concepts of classification and error probability trade-offs. These probabilities are shown (but hard to see, because they are so small) in the shaded areas of the following graph, with cyan for the first kind of error and pink for the other:
* Software options include applets (such as here), R, Minitab, Excel, …
More interesting questions arise when the two score distributions are not separated so clearly.
Now suppose that the credit scores are normally distributed with mean 60 and standard deviation 8 among those who would repay the loan, as compared to mean 40 and standard deviation 12 among those who would not repay the loan.
• f) Draw sketches of these two normal curves on the same axis.
• g) Describe how this scenario differs from the previous one.
• h) Determine the probabilities of the two kinds of error (using the decision rule based on a cut-off value of 50).
• i) Write a sentence or two to interpret the two error probabilities in context.
f) Here is the new graph:
g) The primary change is that the centers of these score distributions are much closer than before, which means that the distributions have much more overlap than before. This will make it harder to distinguish people who would repay their loan and those who would not. A smaller difference is that the variability now differs in the two scores distributions, with slightly less variability in the scores of those who would repay the loan.
h) These error probabilities turn out to be approximately 0.106 for the probability that an applicant who would repay the loan is denied (shown in cyan in the graph below), 0.202 for the probability that an applicant who would not repay is approved (shown in pink):
i) I think this question is important for assessing whether students truly understand, and can successfully communicate, what they have calculated. There’s a 10.6% chance that an applicant who would repay the loan is denied the loan. There’s a 20.2% chance that an applicant who would not repay the loan is approved.
Now let’s change the cutoff value in order to decrease one of the error probabilities to a more acceptable level.
• j) In which direction – smaller or larger – would you need to change the decision rule’s cutoff value in order to decrease the probability that an applicant who would repay the loan is denied?
• k) How would the probability of the other kind of error – approving a loan for an applicant who would not repay it – change with this new cutoff value?
• l) Determine the cutoff value needed to decrease the error probability in (j) to .05. Does this confirm your answer to (j)?
• m) Determine the other error probability with this new cut-off rule. Does this confirm your answer to (k)?
• n) Write a sentence or two to interpret the two error probabilities in context.
j) This question prompts students to think about the goal before doing the calculation. This kind of error occurs when the score is less than the cutoff value, and we need the error probability to decrease from 0.106 to 0.050. Therefore, we need a smaller cutoff value, less than the previous cutoff of 50. Here is a graph of the situation, with the cyan-colored area reduced to 0.05:
k) Using a smaller cutoff value will produce a larger area above that value under the curve for people who would not repay the loan, as shown in pink in the graph above. Therefore, the second error probability will increase as the first one decreases.
l) Students need to calculate a percentile here. Specifically, they need to determine the 5th percentile of a normal distribution with mean 60 and standard deviation 8. They could use software to determine this, or they could realize that the z-score for the 5th percentile is -1.645. The new cutoff value needs to be 1.645 standard deviations below the mean: 60 – 1.645×8 = 46.84. This is indeed smaller than the previous cutoff value of 50. When students mistakenly add 1.645 standard deviations to the mean, I hope that they realize their error by recalling their correct intuition that the cutoff value should be smaller than before.
m) This probability turns out to be approximately 0.284, which is indeed larger than with the previous cutoff (0.202).
n) Now there’s a 5% chance that an applicant who would repay the loan is denied, because that’s how we determined the cutoff value for the decision rule. This rule produces a 28.4% chance that an applicant who would not repay the loan is approved.
Now let’s reduce the probability of the other kind of error.
• o) Repeat parts (j) – (n) with the goal of decreasing the probability that an applicant who would not repay the loan is approved to 0.05.
o) For this goal, the cutoff value needs to become larger than 50, which increases the probability that an applicant who would repay the loan is denied. The cut-off value is now 1.645 standard deviations above the mean: 40 + 1.645×12 = 59.74. This increases the other error probability to approximately 0.487. This means that 48.7% of those who would repay the loan are denied, and 5% of those who would not repay are approved, as depicted in the following graph:
Now that we have come up with three different decision rules, I ask students to think about how we might compare them.
• p) If you consider the two kinds of errors to be equally serious, how might you decide which of the three decision rules considered thus far is the best?
This open-ended question is a tough one for students. I give them a hint to think about the “equally serious” suggestion, and some suggest looking at the average (or sum) of the two error probabilities.
• q) Calculate the average of the two error probabilities for the three cutoff values that we have considered.
• r) Which cutoff value is the best, according to this criterion, among these three options?
We can organize our previous calculations in a table:
According to this criterion, the best cutoff value among these three options is 50, because that produces the smallest average error probability. But of course, these three values are not the only possible choices for the cutoff criterion. I suggest to students that we could write some code to calculate the two error probabilities, and their average, for a large number of possible cutoff values. In some courses, I ask them to write this code for themselves; in other courses I provide them with the following R code:
• s) Explain what each line of code does.
• t) Run the code and describe the resulting graph.
• u) Report the optimal cutoff value and its error probabilities.
• v) Write a sentence describing the optimal decision rule.
Asking students to explain what code does is no substitute for asking them to write their own code, but it can assess some of their understanding:
• The first line creates a vector of cutoff values from 30 to 70.
• The second line calculates the probability that an applicant who would repay the loan has a score below the cutoff value and so would mistakenly be denied.
• The third line calculates the probability that an applicant who would not repay the loan has a score above the cutoff value and so would mistakenly be approved.
• The fourth line calculates the average of these two error probabilities.
• The fifth line produces a graph of average error probability as a function of cutoff value.
• The sixth line determines the optimal cutoff value by identifying which minimizes the average error probability.
Here is the resulting graph:
This graph shows that cutoff values in the neighborhood of 50 are much better (in terms of minimizing average error probability) than cutoff values less than 40 or greater than 60. The minimum value of average error probability appears to be close to 0.15, achieved at a cutoff value slightly above 50.
The R output reveals that the optimal cutoff value is 50.14, very close to the first cutoff value that we analyzed. With this cutoff value, the probability of denying an applicant who would repay the load is 0.109, and the probability of approving an applicant who would not repay is 0.199. The average error probability with this cutoff value is 0.154.
The optimal decision rule, for minimizing the average of the two error probabilities, is to approve a loan for those with a score of 50.14 or greater, and deny a loan to those with a score of less than 50.14.
• w) Now suppose that you consider denying an applicant who would repay the loan to be three time worse than approving an applicant who would not repay the loan. What criterion might you minimize in this case?
• x) With this new criterion, would you expect the optimal cutoff value to be larger or smaller than before? Explain.
• y) Describe how you would modify the code to minimize the appropriate weighted average of the error probabilities.
• z) Run the modified code. Report the optimal cutoff value and its error probabilities. Also write a sentence describing the optimal decision rule.
We can take the relative importance of the two kinds of errors into account by choosing the cut-off value that minimizes a weighted average of the two error probabilities. Because we consider the probability of denying an applicant who would repay to be the more serious error, we need to reduce that probability, which means using a smaller cutoff value.
We do not need to change the first three lines of code. The key change comes in the fourth line, where we must calculate a weighted average instead of an ordinary average. Then we need to remember to use the weighted average vector in the fifth and sixth lines. Here is the modified R code:
The graph produced by this code follows:
We see from the graph that the weighted average of error probabilities is minimized with a cutoff value near 45. The R output reveals the optimal cutoff value to be 45.62. The associated error probabilities are 0.036 for denying an applicant who would repay, 0.320 for approving an applicant who would not repay, and 0.107 for the weighted average. The optimal decision rule for this situation is to approve applicants with a score of 45.62 or higher, deny applicants with a score of less than 45.62.
Whew, I have reached the end of the alphabet*, so I’d better stop there!
* You may have noticed that I had to squeeze a few questions into part (z) to keep from running out of letters.
Most teachers like to give their students an opportunity for lots of practice with normal distribution calculations. With this activity, I have tried to show that you can provide such practice opportunities while also introducing students to ideas such as classification and error probability trade-offs.
P.S. I have used a version of this activity for many years, but I modified the context for this blog post after watching a session at the RStudio conference held in San Francisco at the end of January. Martin Wattenberg and Fernanda Viegas gave a very compelling presentation (a recording of which is available here) in which they described an interactive visualization tool (available here) that allows students to explore how different cutoff values affect error probabilities. Their tool addresses issues of algorithmic fairness vs. bias by examining the impact of different criteria on two populations – labeled as blue and orange people.
P.P.S. I was also motivated to develop this activity into a blog post by a presentation that I saw from Chris Franklin in Atlanta in early February. Chris presented some activities described in the revised GAISE report for PreK-12 (the updated 2020 version will appear here later this year), including one that introduces the topic of classification.
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How many square inches of paper are left over? Circle 2 has a radius of 6/2 = 3 inches. The circumference can be found by multiplying PI (3.14159) times the diameter. 3.14 x 12 inches = 37.68 inches (This works with the metric system as well: 3.14 x 30 cm = 94.2 cm) An important step many people miss at this point is forgetting … If the diameter is six feet, nine inches, enter “6.75”. Answer (1 of 4): The formula for the area (A) of a circle is A = πr² but this can also be stated as πd²/4 where r and d are the radius and diameter respectively.A = πd²/4 = π x 12²/4 = π x 144/4 = π x 36 = 113.097 square inches. Area of a circle is . . Area of a circle given diameter is ... diameter to square inches. Post Cancel. . The circumference of a circle that has a diameter of 8 inches is 25.12 inches. The third one, c, the radius squared is 14*14=196. To determine the area of a circle from its diameter, divide the diameter by two, square it and multiply by π. For example, if a pizza costs $32 and has an area of 256 square inches, you would calculate ÷ =. A 36-inch diameter circle has an area of: 1,017.9 square inches. The Plain English programming language draws on the screen using 96 pixels per inch as the fundamental unit. That is to say π (pi is 3.14159265) multiplied by half the diameter squared. 1 circle 1-foot diameter ( ∅ 1 ft ) = 113.10 square inches ( in 2 , sq in ). Circle 3 has a radius of 2/2 = 1 inch. Originally posted by Unregistered View Post. The formula for calculating the area of a circle is: A = πr 2, where r is the radius of the circle. This calculator converts the area of a circle into a square with four even length sides and four right angles. A decimal can be used if the measurement is not a whole number. 6″ round cake tin means the tin’s diameter is 6 Therefore it has a radius (r) of 3 Area of circle= 3.14r 2 3.14 (3 2) = 3.14 (3 x 3)= 3.14 x 9= 28.26 square inches We now to need find the length (L) for the square cake tin i.e. Area of circle 1 equals = 16* Area of circle 2 equals = 9* Area of circle 3 equals = 1* Differences in the areas is (16-9-1)* equals 6* equals 18.84955592 square inches. Diferent area surface units conversion from circle one inch diameter to square centimeters. How many square inches are in 1 circle one inch diameter? Re: 5" circle ='s how many sq. In the first field, labeled “Diameter (ft),” enter the diameter, in feet, of the cylinder. This gives you your square feet figure (ft 2). The second one, b, the radius squared is 10*10=100. *Square Inches is also known as (a.k.a) square inch, square in, SqIn, Sq.In., sq.in., in2 & in 2 **This Calculator also calculates square feet, total height in feet, total height in inches, total length in feet and total length in inches. No conversion needed, since length, diameter and volume units can be selected independently, so this calculator allows you to use any mixture of measurement units. inches Online calculators are great . The answer is: 1 ∅ 1in equals 0.79 sq in, in2. The formula: Area of a Circle = π x (Diameter/2)^2 π = 3.142 Example: You want a finished 12″ diameter base (a 12″ diameter circle) in a duffle bag. Between ∅ 1in and cm2, sq cm measurements conversion chart page. 2. The diameter should be measured in feet (ft) for square footage calculations and if needed, converted to inches (in), yards (yd), centimetres (cm), millimetres (mm) and metres (m). Let’s try another. If you multiply in pi as 3.14, you'll have 200.96 square inches. So . The diameter is a straight line that passes through the center of a circle. How many square inches in a 23 inch diameter circle? If you multiply in pi as 3.14 you'll have 314 square kilometers. This esitmate should work alright only if the square is significantly smaller than the circle. The diameter of a circle is the circumference multiplied by 0.31831. The area of an oval is the longest diameter x the shortest x 0.7854. How many square inches ( in 2 , sq in ) are in 1 circle 1-inch diameter ( 1 ∅ 1 in )? Is it poss - the answers to estudyassistant.com If the circle has a radius of 8 inches, it's 201.06193 square inches. Diameter, d=12 iches. A circle is 0.7854 times as heavy as a square of the same size. A circle with a 23 inch diameter has am area of 415.4756 sq inches (to 4 dp). If you were for example fitting a square which was of edges$1$unit long on a circle$\sqrt{2}$unit in diameter, The equation gives us$- \frac{1}{2} \pi$as the number of squares. So Area = 3^2 pi or 9 pi which is 28.27 square inches. One circle had a radius of 3 inches and the other had a diameter of 5 inches. Area of a circle is given by radius squared * pi. * * * * *which is 7.07 square feet. 1 circle 1-inch diameter ( ∅ 1 in ) = 0.79 square inches ( in 2 , sq in ). 1 Circular mil ≈ 5.067 074 79 x 10-10 cubic meters (SI unit). If the diameter is seven feet, four inches, enter “7.33”. Unit Descriptions; 1 Square Inch: 1 in x 1 in: 1 Circular Mil: Area of a circle with a diameter of 1 mil, a mil being one-thousandth of an inch. The square inches unit number 0.79 sq in, in2 converts to 1 ∅ 1in, one circle one inch diameter. The circumference of a circle is the diameter x 3.1416. 300 Square Inches to Circular Inches = 381.9719: 7 Square Inches to Circular Inches = 8.9127: 400 Square Inches to Circular Inches = 509.2958: 8 Square Inches to Circular Inches = 10.1859: 500 Square Inches to Circular Inches = 636.6198: 9 Square Inches to Circular Inches = 11.4592: 600 Square Inches to Circular Inches = 763.9437 To calculate the area of a circle we use the formula: π x (diameter/2) 2. If the circle has a diameter of 8 inches, it's 50.265 square inches. It depends on how you measure it. 201 square inches Area of a circle, A = pir^2 r = radius of the circle = (diameter)/2 = 16/2 = 8 inches So, area of circle = pixx8^2 = 22/7xx8^2 = 1408/7 = 201.14 square inches (Note: pi = 22/7) So, are of the circle to nearest square inches = 201 square inches Does the area of the face of a 4" diameter bar = 12.59" It will be 4*pi, which is closer to 12.57 in² Comment. the inside diameter of an outer larger circle (or pipe, tube, conduit, connector), and; the outside diameters of small circles (or pipes, wires, fiber) The default values are for a 10 inch pipe with 2 inch smaller pipes - dimensions according ANSI Schedule 40 Steel Pipes. This on the web one-way conversion tool converts surface area units from ∅ 1-inch circles ( ∅ 1 in ) into square inches ( in 2 , sq in ) instantly online. So . In other words, if we could unroll the circle into a flat line, it would be 25.12 inches long. The area of a circle is the diameter x diameter x 0.7854. 0.79 sq in, in2 is converted to 1 of what? The calculation is based on the area of the square being the same as the circle's area. Try just using the formula for the area of a circle: PI * Radius Squared If the diameter is 5", the radius is 2.5" 2.5"^2 * PI 6.25 * 3.1415927 = 19.6349540849 (what Sandra said) To find out the area of a circle, we need to know its diameter which is the length of its widest part. area of a circle is equal to r equals diameter / 2 Circle 1 has a radius of 8/2 = 4 inches. This on the web one-way conversion tool converts surface area units from ∅ 1-foot circles ( ∅ 1 ft ) into square inches ( in 2 , sq in ) instantly online. So, the cost per square inch of pizza is about .13, or 13 cents. 1 circ mil ≈ 0.000 000 000 506 707 479 m 2. Area , A= πd²/4 =π×12²/4=36π =113.097 square inches For example, assume the diameter of the circular area to be 12 feet. The first one, a, the radius squared is 8*8=64. How many square inches ( in 2 , sq in ) are in 1 circle 1-foot diameter ( 1 ∅ 1 ft )? This will give you the cost per square inch of the pizza. Divide the price of the pizza by the number of square inches. So, we divide our diameter by 2 and then square it (multiply it by itself) and then multiply by π. A circle of radius = 3.75 or diameter = 7.5 or circumference = 23.56 inches has an area of: 2.85032 × 10-8 square kilometers (km²); 0.0285032 square meters (m²) 285.032 square centimeters (cm²) Circle Formula's Radius R = D ÷ 2 where R = radius, D = diameter Area; A = π * D² ÷ 4 where A = area, π = 3.14159..., D = diameter Circumference; So square kilometers. The other way around, how many square centimeters - cm2, sq cm are in one circle one inch diameter - ∅ 1in unit? Circumference The distance around the outward boundary of a circle, expressed as a linear unit of measurement (millimeters, inches, etc.). Convert 1 ∅ 1in into square centimeter and ∅ one inch circles to cm2, sq cm. Radius is 1/2 diameter, or in your case, 3 inches. Directions: Use the calculator above to calculate your square inches. unless you don't have a computer handy! This tool will calculate the volume of a cylindrical shaped object from the dimensions of length and diameter. In the odd case that the circle has a circumference of 8 inches, it's 5.09296 square inches. Examples: If the diameter is five feet, enter “5”. Measure the diameter of your circle in feet. Answer: 1 question Emily cuts two circles from a sheet of colored paper measuring 8' x 12'. Circle area to diameter; User Guide. Calculate Diameter and Units per Course using pre-sized units (Fire Pit Blocks etc) If you already have blocks pre-sized, for fire pits etc, and need to find out how many it takes to make a complete circle, and the outer diameter of the circle, enter single block dimensions below and hit Calculate. 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# Force and the sin wave
1. Apr 11, 2006
### daniel_i_l
This is just something I was thinking about the other day:
Lets say that we have an object with the mass of 1kg and at t=0: x=0m and v=0m/s. Now we apply a force over time t with the force: F=sin(t)N.
that means that a=sin(t). To get v we integrate and get: v = 1-cos(t).
Then we integrate for x and get x = t - sin(t).
That means that after a few seconds the mass will be way on the x+ side. But why should this be if on average the amount of force in that direction is 0?!
2. Apr 11, 2006
### Kurdt
Staff Emeritus
You're forgetting constants of integration when you integrate each time. Think about the answer you should get and what the constants could be.
3. Apr 11, 2006
### daniel_i_l
I got the answers I got when I included the constant:
v(t) = -cos(t) + C (t=0, v=0)
0 = -1 + C
C = 1 => v(t) = 1 - cos(t)
Then for x:
x(t) = t - sin(t) + C (t=0, x=0)
0 = 0 - 0 +C => C=0
And I get:
x(t) = t - sin(t) ?!
4. Apr 11, 2006
### Kurdt
Staff Emeritus
ahh so you have. Sorry I'm just used to that as being such a common mistake that i didn't think to check (much embarrassment).
The error occurs in assuming an acceleration and initial conditions and working back over. If you assume a=sin(t) and then derive v=1-cos(t) you'll see that at t =0 acceleration is 0 while the velocit is max which cannot be a true statemnet since you've said the velocity is 0 to start off with. Thus a=cos(t) to give max acceleration and zero initial velocity.
5. Apr 11, 2006
### daniel_i_l
Why is the velocity max?
v(t) = 1 - cos(t) => v(0) = 1 - cos(0) = 1-1 = 0?
6. Apr 11, 2006
### Hootenanny
Staff Emeritus
I'm sorry, I may be missing something but when we integrate sin(t) do we not simply get -cos(t), not 1-cos(t)?
7. Apr 11, 2006
### daniel_i_l
The 1 comes from the constant C.
8. Apr 11, 2006
### Hootenanny
Staff Emeritus
Ahh yes, didnt read the initial conditions.
9. Apr 11, 2006
### Kurdt
Staff Emeritus
Argh I'm having a nightmare today I apologise to everyone. If you start from rest and begin to accelerat that acceleration will go through half a cycle, at which time you will be at your maximum speed then the rest of the acceleration is used to slow whatever it is to zero again all the while it has been travelling a certain distance x. When the acceleration repeats the cycle it travels further in the positive x direction.
I think you were thingin the velocity would oscillate between positive and negative values like the acceleration and thus the velocity and displacement should average zero.
Third time lucky i hope. Apologies again it has been a long day!
10. Apr 11, 2006
### daniel_i_l
Thanks, that sounds right but why isn't the work 0J? I mean, if we look at the sum of the force over t after n cycles the it is 0, and if the work is 0 then how can it effect the kinetic energy and cause the mass to speed up?
11. Apr 11, 2006
### Kurdt
Staff Emeritus
For a general force work is the integral of the force in the direction it is applied between two points. As the force is a sine wave the integral will be a cosine and over one cycle the work done is zero because you've acceleratedthe block to a max speed then done the same amount of work slowing it down only in the opposite direction.
I guess its best to look at work as being done over half cycles or if you're interested in how much energy has been expended at the modulus of a cycle multiplied by n cycles.
12. Apr 11, 2006
### Integral
Staff Emeritus
Your velocity, v = 1 - cos(t) is always postive, therefore you displacement is always increasing.
http://home.comcast.net/~Integral50/Math/sinforce.jpg" [Broken]. Yellow is the force, blue is velocity, pink is displacement.
Last edited by a moderator: May 2, 2017
13. Apr 12, 2006
### daniel_i_l
Now I understand, from the graph (thanks Integral) it's clear that whenever the work is 0, so is the speed.
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# Stuck on probability question
by orchdork89
Tags: probability, stuck
P: 2 "The probability that a seed will sprout is 3/4. If 7 seeds are planted, find, to the nearest ten-thousandth, the probability that at least 5 will sprout." The answer is .7564, but I don't know how to get there. I tried (3/4^5) + (3/4^6) + (3/4^7), but that doesn't work. Neither does (3/4*5) + (3/4*6) +(3/4*7)... am I overthinking the question, or is there a formula involved that I don't know? Thanks!
Sci Advisor HW Helper P: 2,002 This is a binomial distribution problem. You have n seeds. What is the probability that exactly k seeds will sprout? Take into account that the specification of which will sprout and which don't is not important.
Sci Advisor HW Helper PF Gold P: 12,016 Now, remember that the events of whether a seed sprouts or not is independent of whether the other seeds sprouts. Also, the probability that a single sprout EITHER sprouts OR not sprouts is obviously: $$1=\frac{1}{4}+\frac{3}{4}$$ where 1/4 is the probability of non-sprouting, 3/4 the probability of sprouting. Suppose we only have TWO seeds Consider the following identity: $$1=(\frac{1}{4}+\frac{3}{4})^{2}=\frac{1}{4}*\frac{1}{4}+2*\frac{1}{4}*\ frac{3}{4}+\frac{3}{4}*\frac{3}{4}$$ Now, the first term, 1/4*1/4=1/16 is clearly the probability that neither of the two seeds sprouts, the second term, 2*1/4*3/4=6/16 must be the probability that only one of the seeds sprouts, whereas the last term, 3/4*3/4=9/16 is the probability that BOTH sprout. Considering this, how could you set up a similar scheme for the probabilities for 7 seeds?
Math Emeritus Sci Advisor Thanks PF Gold P: 39,682 Stuck on probability question arildno, that sounds much more complicated than necessary! Here's how I would do it: What is the probability that all 7 will sprout? You should know enough to argue that, assuming the sprouting of one seed is independent of the others, you just multiply the probabilities: (3/4)(3/4)(3/4)(3/4)(3/4)(3/4)(3/4)= (3/4)7. What is the probability that exactly 6 will sprout? Well, what is the probability that the first 6 sprout but the last doesn't. Again it is the individual probabilities multiplied together: (3/4)(3/4)(3/4)(3/4)(3/4)(3/4)(1/4)= (3/4)6. But we're not done: what about the first 5 sprout, the sixth doesn't, and the seventh does sprout? Since multiplication is commutative, that will be again exactly (3/4)6 and, in fact, for any of the 7!/(6!1!)= 7 different ways that could happen (Taking "S" to mean that seed sprouts and "N" to mean it doesn't, we ould have SSSSSSN, SSSSSNS, SSSSNSS, SSSNSSS, SSNSSSS, SNSSSSS, or NSSSSSS, 7 possible orders) it will be (3/4)[/sup]6[/sup](1/4) for each and so the probability exactly 6 seeds sprout is 7(3/4)[/sup]6[/sup](1/4). What about 5 sprounting and 2 not? Hopefully, you can see now that the probability of a particular order but any order, is (3/4)5(1/4)2. Now how many different orders can we have where 5 sprout and 2 do not? How many different order for the letters SSSSSNN are there? This is where the "binomial coefficient comes in. Once you have found the probability that exactly 7, exactly 6, and exactly 5 seed sprout, add them.
Sci Advisor HW Helper PF Gold P: 12,016 Why is $1=1^{7}$ a complicated identity? Okay, I'll admit I have an unreasonable fondness for the expression..
Math
Emeritus
Quote by arildno Why is $1=1^{7}$ a complicated identity? Okay, I'll admit I have an unreasonable fondness for the expression..
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How do I linearize a non-linear function?
Find below some approaches to linearize the cubic function `y =e= x*sqr(x) - 7*sqr(x) + 14*x - 8`
```\$offsymxref offsymlist
option limrow = 0,
limcol = 0,
solprint = off ;
variable y,x ;
equations eq ;
eq.. y =e= x*sqr(x) - 7*sqr(x) + 14*x - 8 ;
x.lo = 1 ; x.up = 10.5 ;
model cubic /all/ ;
solve cubic using nlp minimizing y ;
display y.l, x.l ;```
The global solution to that problem is:
```---- 13 VARIABLE y.L = -2.113
VARIABLE x.L = 3.215 ```
Using Binary variables to model a piecewise linear function in GAMS.
```
\$ontext
Using Binary variables to model a piecewise linear function in GAMS.
This formulation uses points on the curve and is particularly good for developing approximations of
nonlinear functions
Ref: "Integer and Combinatorial Optimization" by
Nemhauser, G.L., and Wolsey, L.A., John Wiley
and Sons. Page 11.
\$ontext
option limrow = 0,
limcol = 0
;
* this contains the set of points
set i /
1*20
/ ;
set notlast(i) ;
notlast(i)\$(ord(i) lt card(i)) = yes ;
* this contains the values of x-coordinates
parameter xval(i) /
1 1
2 1.5
3 2
4 2.5
5 3
6 3.5
7 4
8 4.5
9 5
10 5.5
11 6
12 6.5
13 7
14 7.5
15 8
16 8.5
17 9
18 9.5
19 10
20 10.5
/ ;
* this contains the nonlinear function of x
parameter yval(i) ;
yval(i) = xval(i)*sqr(xval(i)) - 7*sqr(xval(i)) + 14*xval(i) - 8 ;
variables x,y,lam(i),bin(i) ;
positive variables lam(i) ;
binary variables bin(i) ;
equations xdef, ydef, norm, lamdef(i), soslam ;
xdef.. x =e= sum(i,lam(i)*xval(i)) ;
ydef.. y =e= sum(i,lam(i)*yval(i)) ;
norm.. sum(i,lam(i)) =e= 1 ;
lamdef(i).. lam(i) =l= bin(i-1) + bin(i)\$notlast(i) ;
soslam.. sum(i\$notlast(i),bin(i)) =e= 1 ;
x.lo = smin(i,xval(i)) ;
x.up = smax(i,xval(i)) ;
model piece /all/ ;
piece.optcr = 0.0 ;
solve piece minimizing y using mip ;
display x.l, y.l ;
display lam.l, bin.l ; ```
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# Recursion
Recursion. Chapter 10. Chapter Contents. What Is Recursion? Tracing a Recursive Method Recursive Methods That Return a Value Recursively Processing an Array Recursively Processing a Linked Chain The Time Efficiency of Recursive Methods Time Efficiency of countDown
## Recursion
E N D
### Presentation Transcript
1. Recursion Chapter 10
2. Chapter Contents • What Is Recursion? • Tracing a Recursive Method • Recursive Methods That Return a Value • Recursively Processing an Array • Recursively Processing a Linked Chain • The Time Efficiency of Recursive Methods • Time Efficiency of countDown • A Simple Solution to a Difficult Problem • A Poor Solution to a Simple Problem
3. What Is Recursion? • It is a problem-solving process involves repetition • Breaks a problem into identical but smaller problems • Eventually you reach a smallest problem • Answer is obvious or trivial • Using the solution to smallest problem enables you to solve the previous smaller problems • Eventually the original big problem is solved • An alternative to iteration • An iterative solution involves loops
4. What Is Recursion? New Year Eve: counting down from 10.
5. What Is Recursion? • A method that calls itself is a recursive method • Base case: a known case in a recursive definition. The smallest problem • Eventually, one of the smaller problems must be the base case /** Task: Counts down from a given positive integer.* @param integer an integer > 0 */ public static void countDown(intinteger) { System.out.println(integer);if (integer > 1)countDown(integer - 1);} // end countDown
6. When Designing Recursive Solution • Four questions to ask before construct recursive solutions. If you follow these guidelines, you can be assured that it will work. • How can you define the problem in terms of a smaller problem of the same type? • How does each recursive call diminish the size of the problem? • What instance of the problem can serve as the base case? • As the problem size diminishes, will you reach this base case?
7. When Designing Recursive Solution • For the method countDown, we have the following answers. • countDown displays the given integer as the part of the solution that it contribute directly. Then call countDown with smaller size. • The smaller problem is counting down from integer -1. • The if statement asks if the process has reached the base case. Here, the base case occurs when integer is 1.
8. Recursive Solution Guidelines • Method definition must provide parameter • Leads to different cases • Typically includes an if or a switch statement • One or more of these cases should provide a non recursive solution( infinite recursion if don’t) • The base or stopping case • One or more cases includes recursive invocation • Takes a step towards the base case
9. Tracing a Recursive Method public static void countDown(int integer) { System.out.println(integer);if (integer > 1) countDown(integer - 1);} // end countDown • Given: The effect of method call countDown(3)
10. Tracing a Recursive Method Tracing the recursive call countDown(3)
11. Compare Iterative and Recursive Programs //Iterative version public static void countDown( int integer) { while ( integer >= 1) { System.out.println(integer); integer--; } } //Recursive version public static void countDown( int integer) { if ( integer >= 1) { System.out.println(integer); countDown(integer -1); } }
12. Question? • Could you write an recursive method that skips n lines of output, where n is a positive integer. Use System.out.println() to skip one line. • Describe a recursive algorithm that draws a given number of concentric circles. The innermost or outmost circle should have a given diameter. The diameter of each of the other circles should be three-fourths the diameter of the circle just outside it.
13. Answer to skipLines pubic static void skipLines ( int givenNumber) { if (givenNumber >=1) { System.out.println(); skipLines(givenNumber – 1); } } void drawConcentricCircle( givenNumber, givenDiameter) { // what should be put inside??……. }
14. Tracing a Recursive Method • Each call to a method generate an activation record that captures the state of the method’s execution and that is placed into a ADT stack. • The activation-record stack remembers the history of incompleted method calls. A snapshot of a method’s state. • The topmost activation record holds the data values for the currently executing method. • When topmost method finishes, its activation record is popped • In this way, Java can suspend the execution of a recursive method and re-invoke it when it appears on the top.
15. Tracing a Recursive Method The stack of activation records during the execution of a call to countDown(3)… continued→
16. Tracing a Recursive Method Note: the recursive method will use more memory than an iterative method due to the stack of activation records ctd. The stack of activation records during the execution of a call to countDown(3)
17. Tracing a Recursive Method • Too many recursive calls can cause the error message “stack overflow”. Stack of activation records has become full. Method has used too much memory. • Infinite recursion or large-size problems are the likely cause of this error.
18. Recursive Methods That Return a Value • Task: Compute the sum1 + 2 + 3 + … + n for an integer n > 0 public static int sumOf(int n){ int sum;if (n = = 1) sum = 1; // base caseelsesum = sumOf(n - 1) + n; // recursive callreturn sum;} // end sumOf
19. Recursive Methods That Return a Value The stack of activation records during the execution of a call to sumOf(3)
20. Recursively Processing an Array • When processing array recursively, divide it into two pieces • Last element one piece, rest of array another • First element one piece, rest of array another • Divide array into two halves • A recursive method part of an implementation of an ADT is often private • Its necessary parameters make it unsuitable as an ADT operation
21. Recursively Processing a Linked Chain • To write a method that processes a chain of linked nodes recursively • Use a reference to the chain's first node as the method's parameter • Then process the first node • Followed by the rest of the chain public void display(){ displayChain(firstNode); System.out.println();} // end display private void displayChain(Node nodeOne) { if (nodeOne != null) { System.out.print(nodeOne.data + " "); displayChain(nodeOne.next); }} // end displayChain
22. Recursively Divide the Array in Half publicstatic voiddisplayArray( int array[], int first, int last) { if (first == last) System.out.print(array[first]); else { int mid = (first + last) /2; displayArray(array, first, mid); displayArray(array, mid+1, last); } }
23. A Simple Solution to a Difficult Problem The initial configuration of the Towers of Hanoi for three disks
24. A Simple Solution to a Difficult Problem Rules for the Towers of Hanoi game • Move one disk at a time. Each disk you move must be a topmost disk. • No disk may rest on top of a disk smaller than itself. • You can store disks on the second pole temporarily, as long as you observe the previous two rules.
25. A Simple Solution to a Difficult Problem The sequence of moves for solving the Towers of Hanoi problem with three disks. Continued →
26. A Simple Solution to a Difficult Problem (ctd) The sequence of moves for solving the Towers of Hanoi problem with three disks
27. A Simple Solution to a Difficult Problem The smaller problems in a recursive solution for four disks
28. A Simple Solution to a Difficult Problem • Algorithm for solution with 1 disk as the base case Algorithm solveTowers(numberOfDisks, startPole, tempPole, endPole) if (numberOfDisks == 1) Move disk from startPole to endPole else { solveTowers(numberOfDisks-1, startPole, endPole, tempPole) Move disk from startPole to endPole solveTowers(numberOfDisks-1, tempPole, startPole, endPole) }
29. Recursion Efficiency • How many moves occur for n disks? • m(1) = 1 for n>1, two recursive calls to solve problems that have n-1 disks. • m(n) = m(n-1) + m(n-1) +1 = 2*m(n-1) +1 • Let’s evaluate the recurrence for m(n) for a few values of n: • m(1) =1; m(2) = 3; m(3) = 7;m(4) = 15; m(5) = 31; m(6) = 63…. • m(n) = 2^n -1
30. Mathematical Induction • Prove this conjecture m(n) = 2^n -1 by using mathematical induction: • We know that m(1) =1, which equals to 2^1-1=1, so the conjecture is true for n =1. • Now assume that it is true for n=1,2,…,k, and consider m(k+1). • m(k+1) = 2*m(k) +1 (use the recurrence relation) • =2*(2^k-1) +1 = 2^(k+1) -1 ( we assume that m(k) = 2^k-1) • Since the conjecture is true for n=k+1, it is truefor all n>=1
31. Mathematical Induction • Assume you want to prove some statement P, P(n) is true for all n starting with n = 1. The Principle of Math Induction states that, to this end, one should accomplish just two steps: • 1). Prove that P(1) is true. • 2). Assume that P(k) is true for some k. Derive from here that P(k+1) is also true. • look P(1) is true and implies P(2). Therefore P(2) is true. But P(2) implies P(3). Therefore P(3) is true which implies P(4) and so on.
32. Multiplying Rabbits (The Fibonacci Sequence) • “Facts” about rabbits • Rabbits never die • A rabbit reaches sexual maturity exactly two months after birth, that is, at the beginning of its third month of life • Rabbits are always born in male-female pairs • At the beginning of every month, each sexually mature male-female pair gives birth to exactly one male-female pair
33. Multiplying Rabbits (The Fibonacci Sequence) • Problem • How many pairs of rabbits are alive in month n? • Month Month No calculation rabbit couples • January 1 1 + 0 = 1 • February 2 1 + 0 = 1 • March 3 1 + 1 = 2 • April 4 2 + 1 = 3 • May 5 3 + 2 = 5 • June 6 5 + 3 = 8 • July 7 8 + 5 = 13 • August 8 13 + 8 = 21 • Recurrence relation rabbit(n) = rabbit(n-1) + rabbit(n-2)
34. A Poor Solution to a Simple Problem • Fibonacci numbers • First two numbers of sequence are 1 and 1 • Successive numbers are the sum of the previous two • 1, 1, 2, 3, 5, 8, 13, … • This has a natural looking recursive solution • Turns out to be a poor (inefficient) solution
35. A Poor Solution to a Simple Problem • The recursive algorithm Algorithm Fibonacci(n) if (n <= 1)return 1else return Fibonacci(n-1) + Fibonacci(n-2)
36. A Poor Solution to a Simple Problem Time efficiency grows exponentially with n, which is k^n int fib(int n) { int f[n+1]; f[1] = f[2] = 1; for (int i = 3; i <= n; i++) f[i] = f[i-1] + f[i-2]; return f[n]; } Iterative solution is O(n) The computation of the Fibonacci number F6(a) recursively; (b) iteratively
More Related
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https://www.univerkov.com/in-a-rectangular-parallelepiped-the-sides-of-the-base-are-17-dm-and-13-dm-and-the-height-of-the-parallelepiped/
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# In a rectangular parallelepiped, the sides of the base are 17 dm and 13 dm, and the height of the parallelepiped
In a rectangular parallelepiped, the sides of the base are 17 dm and 13 dm, and the height of the parallelepiped is 9 dm. Find: a) the area of the lateral surface of the parallelepiped; b) the total surface area of the parallelepiped; c) the area of the diagonal section of the parallelepiped; d) diagonal.
The lateral surface area of a parallelepiped is equal to the product of the base perimeter and the lateral edge.
S side = Ravsd * AA1 = 2 * (AD + CD) * AA1 = 2 * (17 + 13) * 9 = 540 dm2.
Determine the total area of the parallelepiped.
S floor = S side + 2 * Sb = 360 + 2 * (17 * 13) = 540 + 442 = 982 dm2.
Let us draw the diagonal AC of the base and determine its length by the Pythagorean theorem.
AC ^ 2 = AD ^ 2 + CD ^ 2 = 289 + 169 = 458
AC = √458 dm.
Determine the area of the diagonal section ACC1A1.
Ssection = АА1 * АС = 9 * √458 dm2.
Determine the length of the diagonal CA1 from the right-angled triangle AA1C.
CA1 ^ 2 = AC ^ 2 + AA1 ^ 2 = 458 + 81 = 539.
CA1 = 7 * √11 dm.
Answer: Side = 540 dm2, Sok = 982 dm2, Ssection = 9 * √458 dm2, CA1 = 7 * √11 dm.
One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.
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14$$\chi^2$$ Test For Frequencies
The Chi squared ($$\chi^2$$) test for frequencies a hypothesis test on the frequency of samples that fall into different discrete categories. For example, are the number of left and right-handed people in Psych 315 distributed like you’d expect from the population? Or, is the frequency distribution of birthdays by month for the students in Psych 315 distributed evenly across months? For these tests the dependent measure is a frequency, not a mean.
You can find this test in the flow chart here:
14.1 Example 1: left vs. right handers in Psych 315
According to Wikipedia, 10 percent of the population is left handed. For Psych 315, 7 students reported that they are left handed, while 145 reported right handedness. A $$\chi^2$$ test determines if the frequency of our sampled observations are significantly different than the frequencies that you’d expect from the population. Specifically, the null hypothesis is that our observed frequencies are drawn from a population that has some expected proportions, and our alternative hypothesis is that we’re drawing from a population that does not have these expected proportions.
Like all statistical tests, the $$\chi^2$$ test involves calculating a statistic that measures how far our observations are from those expected under the null hypothesis.
The first step is to calculate the frequencies expected from the null hypothesis. This is simply done by multiplying the total sample size by each of the expected proportions. Since there are 152 students in the class, then we expect (152)(0.1) = 15.2 students to be left handed and (152)(0.9) = 136.8 to be right handed. Expected frequencies do not have to be rounded to the nearest whole number, even though frequencies are whole numbers. This is because we should think of these expected frequencies as the average frequency for each category over the long run - and averages don’t have to be whole numbers.
The next step is to measure how far our observed frequencies are from the expected frequencies. Here’s the formula, where $$\chi^2$$ is pronounced “Chi-squared”.
$\chi^2 = \sum{\frac{(f_{obs}-f_{exp})^{2}}{f_{exp}}}$ Where $$f_{obs}$$ are the observed frequencies and $$f_{exp}$$ are the expected frequencies. For our example, $$f_{obs}$$ is 7, and 145 and $$f_{exp}$$ is 15.2, and 136.8 for left and right-handers respectively:
$\chi^{2} = \sum{\frac{(f_{obs}-f_{exp})^{2}}{f_{exp}}} = \frac{(7-15.2)^{2}}{15.2} + \frac{(145-136.8)^{2}}{136.8} = 4.9152$
This measure, $$\chi^2$$, is close to zero when the observed frequencies match the expected frequencies. Therefore, large values of $$\chi^2$$ can be considered evidence against the null hypothesis.
14.2 The $$\chi^2$$ distribution
Just like the z and t distributions, the $$\chi^2$$ distribution has a known ‘parametric’ shape, and therefore has known probabilities and areas associated with it. Also, like the t-distribution, the $$\chi^2$$ distribution is a family of distributions, with a different distribution for different degrees of freedom.
The $$\chi^2$$ distribution for k degrees of freedom is the distribution you’d get if you draw k values from the standard normal distribution (the z-distribution), square them, and add them up. For more information on this, check out [this section][Simulating $$\chi^2$$] in the chapter on how to generate all of our distributions from normal distributions. Here’s what the probability distributions look like for different degrees of freedom:
Notice how the shape of the distributions spread out and change shape with increasing degrees of freedom. This is because as we increase df, and therefore the number of squared z-scores, the sum will on average increase too.
Since the $$\chi^2$$ distribution is known we can calculate the probability of obtaining our observed value of $$\chi^2$$ if null hypothesis is true. For a $$\chi^2$$ test for frequencies, The degrees of freedom is the number of categories minus one. For this example, df = 2- 1 = 1.
14.3 R’s pchisq function
Consistent with pnorm, pt and pF, R’s pchisq function finds the area below a given value. So to find the area above $$\chi^2$$ = 4.9152 for df = 1 we subtract the result of pchisq from 1:
1-pchisq(4.9152,1)
## [1] 0.02662138
14.4 R’s qchisq function
Like qnorm, qt and qF, R’s qchisq function finds the value of $$\chi2$$ for which a given proportion of the area lies below. So for our example, if $$\alpha$$ = 0.05, the critical value of $$\chi^2$$ for making our decision is:
qchisq(1-0.05,1)
## [1] 3.841459
We reject $$H_{0}$$ for any observed $$\chi^2$$ value greater than 3.84.
Here’s an illustration of the pdf for $$\chi^2$$ for df = 1 with the top 5 percent shaded in red and our observed value of $$\chi^2$$ = 4.9152 shown as a vertical line.
You can see that our observed value of $$\chi^2$$ (4.9152) falls above the critical value of $$\chi^2$$ (3.8415). Also, our p-value (0.0266) is less than $$\alpha$$ = 0.05. We therefore reject $$H_{0}$$ and conclude that the proportion of left and right handers in our class is significantly different than what is expected from the population.
14.5 Using chisq.test to run a $$\chi2$$ test for frequencies
R has the chisq.test function that computes the values that we’ve done by hand. For our example we’ll load in the survey data and compute ‘fobs’ which is the frequency of the observed number of left and right handers in the class. This will be compared to prob = c(.1,.9) which is where we enter our population, or null hypothesis proportions.
survey <-read.csv("http://www.courses.washington.edu/psy315/datasets/Psych315W21survey.csv")
prob <- c(.1,.9)
fobs <- c(sum(survey$hand=="Left",na.rm = T),sum(survey$hand=="Right",na.rm = T))
chisq.out <- chisq.test(fobs,p=prob)
The important fields in the output of chisq.test are statistic which holds the observed $$\chi^2$$ value, parameter which holds the degrees of freedom, expected which holds the computed expected frequencies, observed which holds the observed values (that we entered), and p.value.
14.6 APA format for the $$\chi^2$$ test for independence
Here’s how to extract the fields in the output of chisq.test into generate a string in APA format:
sprintf('Chi-Squared(%d,N=%d) = %5.4f, p = %5.4f',
chisq.out$parameter, sum(chisq.out$observed),
chisq.out$statistic, chisq.out$p.value)
## [1] "Chi-Squared(1,N=152) = 4.9152, p = 0.0266"
14.7 One or two tailed?
A common question for $$\chi^2$$ tests is whether it is a one or a two tailed test. You might think it’s a one-tailed test since we only reject $$H_{0}$$ for large values of $$\chi^2$$. However, because the numerator of the $$\chi^2$$ formula is $$\sum{(f_{obs}-f_{exp})^{2}}$$, we can get large values of $$\chi^2$$ if our observed frequency, $$f_{obs}$$ is either too small or too large compared to the expected frequency, $$f_{exp}$$. So really a $$\chi^2$$ test is a two-tailed test.
14.8 Relationship between $$\chi^2$$ test and the normal approximation to the binomial
You might have noticed that we already know a way of calculating the probability of obtaining 15.2 or more left-handers out of 152 students, assuming that 10 percent of the population is left-handed. That’s right - the normal approximation to the binomial.
Let’s compare:
Using the normal approximation with N = 152 and P = 0.1 , the number of left-handers will be distributed normally with a mean of
$\mu = NP = (152 )(0.1 ) = 15.2$
and a standard deviation of
$\sigma = \sqrt{NP(1-P)} = \sqrt{(152 )(0.1 )(0.9 )} = 3.6986$.
The Chi-squared test is a two-tailed test, so to conduct a two tailed test with the normal approximation to the binomial we need to find the probability of obtaining an observed value more extreme than 7 compared to 15.2 .
If we ignore the business of correcting for continuity, we can use pnorm to find this probability. First, we convert our observed frequency of left-handers to a z-score:
z <- (7-15.2)/3.6986
Then we convert our z-score to a p-value like we did back in the chapter on the normal distribution:
2*(1-pnorm(abs(-2.2171)))
## [1] 0.02661626
This number is very close to the p-value we got from the $$\chi^2$$ test: 0.0266. In fact, the difference is only due to rounding error. Thus the $$\chi^2$$ test with $$df = 1$$ is mathematically equivalent to the normal approximation to the binomial. This means that the tests are the same if you don’t correct for continuity, which is that process of including the whole interval by subtracting and adding 0.5.
The $$\chi^2$$ test is therefore actually not quite right. Also, remember that the normal approximation to the binomial is really best for larger sample sizes, like for n>20. But we typically run $$\chi^2$$ tests for smaller values of n. values of n.
14.9 Comparing the $$\chi^2$$ test to the binomial test
For two groups like this example, we can use the binomial distribution, which we can get from binom.test:
binom.out <- binom.test(fobs[1],sum(fobs),prob[1],alternative = 'two.sided')
binom.out$p.value ## [1] 0.02140999 It’s close, but not exact. In most cases, the $$\chi^2$$ test is a good enough approximation to the ‘exact’ solution. You really only run into trouble when your expected frequencies drop too low, say below 5 or so. 14.10 Example 2: Birthdays by month Let’s see if the birthdays in this class are evenly distributed across months, or if there are some months for which students have significantly= more birthdays than others. For simplicity, we’ll assume that all months have equal probability, even though they vary in length. We’ll ruun a $$\chi^2$$ test using an alpha value of 0.05. Here’s a table showing the number of birthdays for each month for all 152 students: It looks kind of uneven. There are 18 students with birthdays in January and June but only 7 students with birthdays in November. A natural way to visualize this distribution is with a bar graph showing the frequency distribution: To see if this distribution is significantly uneven we calculate the expected frequencies under the null hypothesis. Here we expect equal frequencies of $$\frac{152}{12} = 12.6667$$ birthdays per month . Note equal frequencies assumes that each month has an equal number of days. This assumption is close enough for this example, but how would you correct the expected frequencies to account for this? Using our $$\chi^2$$ formula: $\chi^{2} = \sum{\frac{(f_{obs}-f_{exp})^{2}}{f_{exp}}} = \frac{(18-12.6667)^{2}}{12.6667} ... + ... \frac{(11-12.6667)^{2}}{12.6667} = 12.0526$ With 12 months, the degrees of freedom is 11. Using pchisq we can find the p-value for our test: 1-pchisq(12.0526,11) ## [1] 0.3597025 Or, we can find the critical value of $$\chi^2$$ using qnorm: qchisq(1-0.05,11) ## [1] 19.67514 We reject $$H_{0}$$ for any observed $$\chi^2$$ value greater than 19.68. Here’s an illustration of the pdf for $$\chi^2$$ for df = 11 with the top 5shaded in red and our oberved value of $$\chi^2$$ = 12.0526 shown as a vertical line. You can see that our observed value of $$\chi^2$$ (12.0526) falls below the critical value of $$\chi^2$$ (19.6751). Also, our p-value (0.3597) is greater than $$\alpha$$ = 0.05. We therefore fail to reject $$H_{0}$$ and conclude that the distribution of birthdays across months is not significantly different than what is expected from chance. 14.11 Conducting the same test with chisq.test With the survey data loaded in from the previous example, running the $$\chi^2$$ test for frequencies on the birth month data is pretty straightforward. The only new thing is the use of the function table to get a table of frequencies for birthdays across months, like we did in the chapter on descriptive statistics: dat <- table(survey$month)
dat
##
## January February March April May June July August
## 18 8 9 11 14 18 15 14
## September October November December
## 11 16 7 11
We can then send this table into chisq.test. We could send in a list of expected probabilities, but by default, chisq.test assumes equal probabilities across categories:
chisq.out <- chisq.test(dat)
chisq.out
##
## Chi-squared test for given probabilities
##
## data: dat
## X-squared = 12.053, df = 11, p-value = 0.3597
We can coerce the output of chisq.test into APA format just like we did for the first example:
sprintf('Chi-Squared(%d,N=%d) = %5.4f, p = %5.4f',
chisq.out$parameter, sum(chisq.out$observed),
chisq.out$statistic, chisq.out$p.value)
## [1] "Chi-Squared(11,N=152) = 12.0526, p = 0.3597"
14.12 Effect size
There are a couple of definitions for effect size for the $$\chi^2$$ chi-squared test for frequencies. The one that is used most commonly to calculate power is is called $$\psi$$, or ‘psi’:
$\psi = \sqrt{\frac{\chi^2}{N}}$
Where $$N$$ is the total sample size. From our first example (handedness):
$\psi = \sqrt{\frac{\chi^{2}}{N}} = \sqrt{\frac{4.9152^{2}}{152}} = 0.1798$
From our second example (birth months):
$\psi = \sqrt{\frac{\chi^{2}}{N}} = \sqrt{\frac{12.0526^{2}}{152}} = 0.2816$
$$\psi$$ has the desirable quality that it is not influenced by sample size, so it can be used to compare effect sizes across experiments. However, since $$\chi^2$$ increases with the number of categories (and df), so does $$\psi$$, so we can’t use $$\psi$$ on it’s own to determine if an effect is ‘small’, ‘medium’, or ‘large’.
An alternative measure of effect size is ‘Cramer’s V’, which puts the degrees of freedom in the denominator to offset the increase in df with $$\psi$$:
$V = \sqrt{\frac{\chi^2}{N \times df}}$
If you look at the formulas you can see that there is a relation between $$\psi$$ and $$V$$:
$V = \frac{\psi}{\sqrt{df}}$
For V, 0.1 is considered small, 0.3 medium, and 0.5 is a large effect size.
From our first example (handedness):
$V =\sqrt{\frac{\chi^{2}}{N \times df}} = \sqrt{\frac{4.9152^{2}}{(152)(1)}} = 0.1798$
Which would be considered to be a small effect size.
From our second example (birth months):
$V =\sqrt{\frac{\chi^{2}}{N \times df}} = \sqrt{\frac{12.0526^{2}}{(152)(11)}} = 0.0849$
Which would also be considered to be a small effect size.
14.13 Power for the $$\chi^2$$ test for frequencies
Power can be calculated with R using ‘pwr.chisq.test’. It requires the effect size $$\psi$$ (called ‘w’ for some reason), the total sample size $$N$$, the degrees of freedom, and the alpha value.
pwr.chisq.test works much like pwr.t.test for the t-test. To find the power for the first test on handedness:
chisq.power.out <- pwr.chisq.test(w = 0.1798,N = 152,df = 1,sig.level = 0.05)
The observed power for the first test can be found:
chisq.power.out$power ## [1] 0.6013325 If you want to find the sample size that gives you a desired power of 0.8, pass in NULL for the sample size (N), and set power = .8: chisq.power.out <- pwr.chisq.test(w = 0.1798,N = NULL,df = 1,sig.level = 0.05,power = 0.8) The sample size needed is: chisq.power.out$N
## [1] 242.788
Power for the second example, on birth months is:
chisq.power.out <- pwr.chisq.test(w = 0.2816,N = 152,df = 11,sig.level = 0.05)
chisq.power.out$power ## [1] 0.6220806 The sample sized needed to get a power of 0.8 is: chisq.power.out <- pwr.chisq.test(w = 0.2816,N = NULL,df = 11,sig.level = 0.05,power = 0.8) chisq.power.out$N
## [1] 211.8792
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# Thread: Expected value of largest claim on two insured homes (Order Statistics question)
Suppose two independent calims are made on two insured homes, where each claim has pdf f(x)=4/x^5, 1< x < infinity, in which the unity is $1000. Find the expected value of the largest claim. Here was my approach: Integrate f(x) to get F(x)=-x^-4 Apply the order statistics max equation: n{[Fn]^n-1}*f(x) which should have given me: 2(-x^-4)4/x^5. I won't post the computation since the answer is wrong. Here is the solution: P(X<t) = P(max(x1,x2)<t) ... (i) = P(x1<t, x2<t) ... (ii) = double integral of fx1,x2(x,y) dx dy (both integrals from -infinity to t) ... (iii) = (1-1/t^4)^2, t>1 (iv) What I don't get is step (iii) and (iv). How does one go from (ii) to (iii) and then from (iii) to (iv)? The rest of the problem is pretty simple. You just plug in the result of P(x<t) into the order statistics min equations and then just take the expected value of that. You end up getting 32/21 for those that care. Thanks in advance! 2. ## Re: Expected value of largest claim on two insured homes (Order Statistics question) How can F(x) be negative? (It can't be)$\displaystyle F(x)=P(X\le x)=\int_1^x 4t^{-5}dt=1-x^{-4}$Then explain what you're trying to do, because I'm not sure. BTW, I work with order stats all the time... Adler : Unusual strong laws for arrays of ratios of order statistics http://w3.math.sinica.edu.tw/bulleti...d313/31302.pdf And this is a Pareto distribution. 3. ## Re: Expected value of largest claim on two insured homes (Order Statistics question) Thanks for the help Beagle =) Your integral cleared up a lot of issues for me. I was evaluating f(x) from 0 to infinity instead of from 1 to x. Basically what I'm trying to do is to get F(x) from f(x), plug that into the order statistics min formula, then integrate that result to get the density function for the min, and then finally take the expectation of that density function. Using your integral, I ended up getting the right answer. My question now is, why do we evaluate from 1 to x? Thanks for the help. I think it's very cool that you work with order statistics often. I used to hate probability and statistics, but over the last year I've grown to like it a bit =) 4. ## Re: Expected value of largest claim on two insured homes (Order Statistics question) I think you want to know 'why do we evaluate from 1 to x?' in reference to F(x) Well$\displaystyle F(x)=P(X\le x)=\int_{-\infty}^x f(t)dt\$ by definition.
In so many setting the min and max are suff statistics, so they need to be studied.
I have to cover them in my probability class, otherwise in the stat class, I can't assign any problems
involving densities where the parameters define the support.
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# Variance after scaling and summing: One of the most useful facts from statistics
What do $R^2$, laboratory error analysis, ensemble learning, meta-analysis, and financial portfolio risk all have in common? The answer is that they all depend on a fundamental principle of statistics that is not as widely known as it should be. Once this principle is understood, a lot of stuff starts to make more sense.
Here’s a sneak peek at what the principle is.
Don’t worry if the formula doesn’t yet make sense! We’ll work our way up to it slowly, taking pit stops along the way at simpler formulas are that useful on their own. As we work through these principles, we’ll encounter lots of neat applications and explainers.
This post consists of three parts:
• Part 1: Sums of uncorrelated random variables: Applications to social science and laboratory error analysis
• Part 2: Weighted sums of uncorrelated random variables: Applications to machine learning and scientific meta-analysis
• Part 3: Correlated variables and Modern Portfolio Theory
## Part 1: Sums of uncorrelated random variables: Applications to social science and laboratory error analysis
Let’s start with some simplifying conditions and assume that we are dealing with uncorrelated random variables. If you take two of them and add them together, the variance of their sum will equal the sum of their variances. This is amazing!
To demonstrate this, I’ve written some Python code that generates three arrays, each of length 1 million. The first two arrays contain samples from two normal distributions with variances 9 and 16, respectively. The third array is the sum of the first two arrays. As shown in the simulation, its variance is 25, which is equal to the sum of the variances of the first two arrays (9 + 16).
from numpy.random import randn
import numpy as np
n = 1000000
x1 = np.sqrt(9) * randn(n) # 1M samples from normal distribution with variance=9
print(x1.var()) # 9
x2 = np.sqrt(16) * randn(n) # 1M samples from normal distribution with variance=16
print(x2.var()) # 16
xp = x1 + x2
print(xp.var()) # 25
This fact was first discovered in 1853 and is known as Bienaymé’s Formula. While the code example above shows the sum of two random variables, the formula can be extended to multiple random variables as follows:
If $X_p$ is a sum of uncorrelated random variables $X_1 .. X_n$, then the variance of $X_p$ will be $$\sigma_{p}^{2} = \sum{\sigma^2_i}$$ where each $X_i$ has variance $\sigma_i^2$.
What does the $p$ stand for in $X_p$? It stands for portfolio, which is just one of the many applications we’ll see later in this post.
### Why this is useful
Bienaymé’s result is surprising and unintuitive. But since it’s such a simple formula, it is worth committing to memory, especially because of it sheds light on so many other principles. Let’s look at two of them.
#### Understanding $R^2$ and “variance explained”
Psychologists often talk about “within-group variance”, “between-group variance”, and “variance explained”. What do these terms mean?
Imagine a hypothetical study that measured the extraversion of 10 boys and 10 girls, where extraversion is measured on a 10-point scale (Figure 1. Orange bars). The boys have a mean extraversion of 44 and the girls have a mean extraversion 50. In addition, the overall variance of the data is 25. We can decompose this variance into two parts:
• Between-group variance: Create a 20-element array where every boy is assigned to the mean boy extraversion of 4.4, and every girl is assigned to the mean girl extraversion of 5.0. The variance of this array is 0.9. (Figure 1. Blue bars).
• Within-group variance: Create a 20-element array of the amount each child’s extraversion deviates from the mean value for their sex. Some of these values will be negative and some will be positive. The variance of this array is 1.6. (Figure 1. Pink bars).
Figure 1: Decomposition of extraversion scores (orange) into between-group variance (blue) and within-group variance (pink).
If you add these arrays together, the resulting array will represent the observed data (Figure 1. Orange bars). As originally measured, it will have a variance of 2.5, which corresponds exactly to the sum of the variances of the two component arrays (0.9 + 1.6). Psychologists might say that sex “explains” 0.9/2.5 = 36% of the extraversion variance. Equivalently, a model of extraversion that uses sex as the only predictor would have an $R^2$ of 0.36.
#### Error propagation in laboratories
If you ever took a physics lab or chemistry lab back in college, you may remember having to perform error analysis, in which you calculated how errors would propagate through one noisy measurement after another.
Physics textbooks often say that standard deviations add in “quadrature”, which just means that if you are trying to estimate some quantity that is the sum of two other measurements, and if each measurement has some error with standard deviation $\sigma_1$ and $\sigma_2$ respectively, the final standard deviation would be $\sigma_{p} = \sqrt{\sigma^2_1 + \sigma^2_2}$. I think it’s probably easier to just use variances, as in the Bienaymé Formula, with $\sigma^2_{p} = \sigma^2_1 + \sigma^2_2$.
For example, imagine you are trying to estimate the height of two boxes stacked on top of each other (Figure 2). One box has a height of 1 meter with variance $\sigma^2_1$ = 0.01, and the other has a height of 2 meter with variance $\sigma^2_2$ = 0.01. Let’s further assume, perhaps optimistically, that these errors are independent. That is, if the measurement of the first box is too high, that doesn’t make it any more likely that the measurement of the second box will also be too high. If we can make these assumptions, then the total height of the two boxes will be 3 meters with variance $\sigma^2_p$ = 0.02.
Figure 2: Two boxes stacked on top of each other. The height of each box is measured with some variance (uncertainty). The total height is the the sum of the individual heights, and the total variance (uncertainty) is the sum of the individual variances.
There is a key difference between the extraversion example and the stacked boxes example. In the extraversion example, we added two arrays that each had an observed sample variance. In the stacked boxes example, we added two scalar measurements, where the variance of these measurements refers to our measurement uncertainty. Since both cases have a meaningful concept of ‘variance’, the Bienaymé Formula applies to both.
## Part 2: Weighted sums of uncorrelated random variables: Applications to machine learning and scientific meta-analysis
Let’s now move on to the case of weighted sums of uncorrelated random variables. But before we get there, we first need to understand what happens to variance when a random variable is scaled.
If $X_p$ is defined as $X$ scaled by a factor of $w$, then the variance $X_p$ will be $$\sigma_{p}^{2} = w^2 \sigma^2$$ where $\sigma^2$ is the variance of $X$.
This means that if a random variable is scaled, its variance will scale quadratically. Let’s see this in code.
from numpy.random import randn
import numpy as np
n = 1000000
baseline_var = 10
w = 0.7
x1 = np.sqrt(baseline_var) * randn(n) # Array of 1M samples from normal distribution with variance=10
print(x1.var()) # 10
xp = w * x1 # Scale this by w=0.7
print(w**2 * baseline_var) # 4.9 (predicted variance)
print(xp.var()) # 4.9 (empirical variance)
To gain some intuition for this, it’s helpful to think about outliers. We know that outliers have a huge effect on variance. That’s because the formula used to compute variance, $\sum{\frac{(x_i - \bar{x})^2}{n-1}}$, squares all the deviations, and so we get really big variances when we square large deviations. With that as background, let’s think about what happens if we scale our data by 2. The outliers will spread out twice as far, which means they will have even more than twice as much impact on the variance. Similarly, if we multiply our data by 0.5, we will squash the most “damaging” part of the outliers, and so we will reduce our variance by more than a factor of two.
While the above principle is pretty simple, things start to get interesting when you combine it with the Bienaymé Formula in Part I:
If $X_p$ is a weighted sum of uncorrelated random variables $X_1 ... X_n$, then the variance of $X_p$ will be $$\sigma_{p}^{2} = \sum{w^2_i \sigma^2_i}$$ where each $w_i$ is a weight on $X_i$, and each $X_i$ has its own variance $\sigma_i^2$.
The above formula shows what happens when you scale and then sum random variables. The final variance is the weighted sum of the original variances, where the weights are squares of the original weights. Let’s see how this can be applied to machine learning.
### An ensemble model with equal weights
Imagine that you have built two separate models to predict car prices. While the models are unbiased, they have variance in their errors. That is, sometimes a model prediction will be too high, and sometimes a model prediction will be too low. Model 1 has a mean squared error (MSE) of \$1,000 and Model 2 has an MSE of \$2,000.
A valuable insight from machine learning is that you can often create a better model by simply averaging the predictions of other models. Let’s demonstrate this with simulations below.
from numpy.random import randn
import numpy as np
n = 1000000
actual = 20000 + 5000 * randn(n)
errors1 = np.sqrt(1000) * randn(n)
print(errors1.var()) # 1000
errors2 = np.sqrt(2000) * randn(n)
print(errors2.var()) # 2000
# Note that this section could be replaced with
# errors_ensemble = 0.5 * errors1 + 0.5 * errors2
preds1 = actual + errors1
preds2 = actual + errors2
preds_ensemble = 0.5 * preds1 + 0.5 * preds2
errors_ensemble = preds_ensemble - actual
print(errors_ensemble.var()) # 750. Lower than variance of component models!
As shown in the code above, even though a good model (Model 1) was averaged with an inferior model (Model 2), the resulting Ensemble model’s MSE of \$750 is better than either of the models individually. The benefits of ensembling follow directly from the weighted sum formula we saw above, $\sigma_{p}^{2} = \sum{w^2_i \sigma^2_i}$. To understand why, it’s helpful to think of models not as generating predictions, but rather as generating errors. Since averaging the predictions of a model corresponds to averaging the errors of the model, we can treat each model’s array of errors as samples of a random variable whose variance can be plugged in to the formula. Assuming the models are unbiased (i.e. the errors average to about zero), the formula tells us the expected MSE of the ensemble predictions. In the example above, the MSE would be which is exactly what we observed in the simulations. (For a totally different intuition of why ensembling works, see this blog post that I co-wrote for my company, Opendoor.) ### An ensemble model with Inverse Variance Weighting In the example above, we obtained good results by using an equally-weighted average of the two models. But can we do better? Yes we can! Since Model 1 was better than Model 2, we should probably put more weight on Model 1. But of course we shouldn’t put all our weight on it, because then we would throw away the demonstrably useful information from Model 2. The optimal weight must be somewhere in between 50% and 100%. An effective way to find the optimal weight is to build another model on top of these models. However, if you can make certain assumptions (unbiased and uncorrelated errors), there’s an even simpler approach that is great for back-of-the envelope calculations and great for understanding the principles behind ensembling. To find the optimal weights (assuming unbiased and uncorrelated errors), we need to minimize the variance of the ensemble errors $\sigma_{p}^{2} = \sum{w^2_i \sigma^2_i}$ with the constraint that $\sum{w_i} = 1$. It turns out that the variance-minimizing weight for a model should be proportional to the inverse of its variance. $w_k = \frac{1/{\sigma^2_k}}{\sum{1/{\sigma^2_i}}}$ When we apply this method, we obtain optimal weights of $w_1$ = 0.67 and $w_2$ = 0.33. These weights give us an ensemble error variance of which is significantly better than the$750 variance we were getting with equal weighting.
This method, called Inverse Variance Weighting, can be seen in the figure below. The horizontal axis represents the weight on Model 1. The vertical axis represents the weight on Model 2. Both models are unbiased. Color represents the total error variance when the model predictions are summed, based on the formula $\sigma_{p}^{2} = \sum{w^2_i \sigma^2_i}$. The total variance (color) increases more rapidly as you move upward compared to rightward, indicating that Model 2 contributes more to the total variance. The diagonal line represents the constraint that $w_1 + w_2 = 1$. The dot represents the optimal weight as determined by Inverse Variance Weighting. As you can see, Inverse Variance Weighting picks the lightest (lowest variance) point on the line, with more weight on Model 1 than Model 2.
Figure 3. Total error variance (color) as a function of weights on a high error model (vertical) and a low error model (horizontal). The optimal weighting (dot) puts more weight on the low error model.
Inverse Variance Weighting is not just useful as a way to understand Machine Learning ensembles. It is also one of the core principles in scientific meta-analysis, which is popular in medicine and the social sciences. When multiple scientific studies attempt to estimate some quantity, and each study has a different sample size (and hence variance of their estimate), a meta-analysis should weight the high sample size studies more. Inverse Variance Weighting is used to determine those weights.
## Part 3: Correlated variables and Modern Portfolio Theory
Let’s imagine we now have three unbiased models with the following MSEs:
• Model 1: MSE = 1000
• Model 2: MSE = 1000
• Model 3: MSE = 2000
By Inverse Variance Weighting, we should assign more weight to the first two models, with $w_1=0.4, w_2=0.4, w_3=0.2$.
But what happens if Model 1 and Model 2 have correlated errors? For example, whenever Model 2’s predictions are too high, Model 3’s predictions tend to also be too high. In that case, maybe we don’t want to give so much weight to Models 1 and 2, since they provide somewhat redundant information. Instead we should increase the weight on Model 3, since it provides new independent information.
To determine how much weight to put on each model, we first need to determine how much total variance there will be if the errors are correlated. To do this, we need to borrow a formula from the financial literature, which extends the formulas we’ve worked with before. This is the formula we’ve been waiting for.
If $X_p$ is a weighted sum of (correlated or uncorrelated) random variables $X_1 ... X_n$, then the variance of $X_p$ will be $$\sigma_{p}^{2} = \sum\limits_{i} \sum\limits_{j} w_i w_j \sigma_i \sigma_j \rho_{ij}$$ where each $w_i$ and $w_j$ are weights assigned to $X_i$ and $X_j$, where each $X_i$ and $X_j$ have standard deviations $\sigma_i$ and $\sigma_j$, and where the correlation between $X_i$ and $X_j$ is $\rho_{ij}$.
There’s a lot to unpack here, so let’s take this step by step.
• $\sigma_i \sigma_j \rho_{ij}$ is a scalar quantity representing the covariance between $X_i$ and $X_j$.
• If none of the variables are correlated with each other, then all the cases where $i \neq j$ will go to zero, and the formula reduces to $\sigma_{p}^{2} = \sum{w^2_i \sigma^2_i}$, which we have seen before.
• The more that two variables $X_i$ and $X_j$ are correlated, the more the total variance $\sigma_{p}^{2}$ increases.
• If two variables $X_i$ and $X_j$ are anti-correlated, then the total variance decreases, since $\sigma_i \sigma_j \rho_{ij}$ is negative.
• This formula can be rewritten in more compact notation as $\sigma_{p}^{2} = \vec{w}^T\Sigma \vec{w}$, where $\vec{w}$ is the weight vector, and $\Sigma$ is the covariance matrix (not a summation sign!)
If you skimmed the bullet points above, go back and re-read them! They are super important.
To find the set of weights that minimize variance in the errors, you must minimize the above formula, with the constraint that $\sum{w_i} = 1$. One way to do this is to use a numerical optimization method. In practice, however, it is more common to just find weights by building another model on top of the base models
Regardless of how the weights are found, it will usually be the case that if Models 1 and 2 are correlated, the optimal weights will reduce redundancy and put lower weight on these models than simple Inverse Variance Weighting would suggest.
### Applications to financial portfolios
The formula above was discovered by economist Harry Markowitz in his Modern Portfolio Theory, which describes how an investor can optimally trade off between expected returns and expected risk, often measured as variance. In particular, the theory shows how to maximize expected return given a fixed variance, or minimize variance given a fixed expected return. We’ll focus on the latter.
Imagine you have three stocks to put in your portfolio. You plan to sell them at time $T$, at which point you expect that Stock 1 will have gone up by 5%, with some uncertainty. You can describe your uncertainty as variance, and in the case of Stock 1, let’s say $\sigma_1^2$ = 1. This stock, as well Stocks 2 and 3, are summarized in the table below:
Stock ID Expected Return Expected Risk ($\sigma^2$)
1 5.0 1.0
2 5.0 1.0
3 5.0 2.0
This financial example should remind you of ensembling in machine learning. In the case of ensembling, we wanted to minimize variance of the weighted sum of error arrays. In the case of financial portfolios, we want to minimize the variance of the weighted sum of scalar financial returns.
As before, if there are no correlations between the expected returns (i.e. if Stock 1 exceeding 5% return does not imply that Stock 2 or Stock 3 will exceed 5% return), then the total variance in the portfolio will be $\sigma_{p}^{2} = \sum{w^2_i \sigma^2_i}$ and we can use Inverse Variance Weighting to obtain weights $w_1=0.4, w_2=0.4, w_3=0.2$.
However, sometimes stocks have correlated expected returns. For example, if two of the stocks are in oil companies, then one stock exceeding 5% implies the other is also likely to exceed 5%. When this happens, the total variance becomes
as we saw before in the ensemble example. Since this includes an additional positive term for $w_1 w_2 \sigma_1 \sigma_2 \rho_{1,2}$, the expected variance is higher than in the uncorrelated case, assuming the correlations are positive. To reduce this variance, we should put less weight on Stocks 1 and 2 than we would otherwise.
While we have focused on minimizing the variance of a financial portfolio, you might also be interested in having a portfolio with high return. Modern Portfolio Theory describes how a portfolio can reach any abitrary point on the efficient frontier of variance and return, but that’s outside the scope of this blog post. And as you might expect, financial markets can be more complicated than Modern Portfolio Theory suggests, but that’s also outside scope.
## Summary
That was a long post, but I hope that the principles described have been informative. It may be helpful to summarize them in backwards order, starting with the most general principle.
If $X_p$ is a weighted sum of (correlated or uncorrelated) random variables $X_1 ... X_n$, then the variance of $X_p$ will be $$\sigma_{p}^{2} = \sum\limits_{i} \sum\limits_{j} w_i w_j \sigma_i \sigma_j \rho_{ij}$$ where each $w_i$ and $w_j$ are weights assigned to $X_i$ and $X_j$, where each $X_i$ and $X_j$ have standard deviations $\sigma_i$ and $\sigma_j$, and where the correlation between $X_i$ and $X_j$ is $\rho_{ij}$. The term $\sigma_i \sigma_j \rho_{ij}$ is a scalar quantity representing the covariance between $X_i$ and $X_j$.
If none of the variables are correlated, then all the cases where $i \neq j$ go to zero, and the formula reduces to $$\sigma_{p}^{2} = \sum{w^2_i \sigma^2_i}$$ And finally, if we are computing a simple sum of random variables where all the weights are 1, then the formula reduces to $$\sigma_{p}^{2} = \sum{\sigma^2_i}$$
# Using your ears and head to escape the Cone Of Confusion
One of coolest things I ever learned about sensory physiology is how the auditory system is able to locate sounds. To determine whether sound is coming from the right or left, the brain uses inter-ear differences in amplitude and timing. As shown in the figure below, if the sound is louder in the right ear compared to the left ear, it’s probably coming from the right side. The smaller that difference is, the closer the sound is to the midline (i.e the vertical plane going from your front to your back). Similarly, if the sound arrives at your right ear before the left ear, it’s probably coming from the right. The smaller the timing difference, the closer it is to the midline. There’s a fascinating literature on the neural mechanisms behind this.
Inter-ear loudness and timing differences are pretty useful, but unfortunately they still leave a lot of ambiguity. For example, a sound from your front right will have the exact same loudness differences and timing differences as a sound from your back right.
Not only does this system leave ambiguities between front and back, it also leaves ambiguities between top and down. In fact, there is an entire cone of confusion that cannot be disambiguated by this system. Sound from all points along the surface of the cone will have the same inter-ear loudness differences and timing differences.
While this system leaves a cone of confusion, humans are still able to determine the location of sounds from different points on the cone, at least to some extent. How are we able to do this?
Amazingly, we are able to do this because of the shape of our ears and heads. When sound passes through our ears and head, certain frequencies are attenuated more than others. Critically, the attenuation pattern is highly dependent on sound direction.
This location-dependent attenuation pattern is called a Head-related transfer function (HRTF) and in theory this could be used to disambiguate locations along the cone of confusion. An example of someone’s HRTF is shown below, with frequency on the horizontal axis and polar angle on the vertical axis. Hotter colors represent less attenuation (i.e. more power). If your head and ears gave you this HRTF, you might decide a sound is coming from the front if it has more high frequency power than you’d expect.
HRTF image from Simon Carlile's Psychoacoustics chapter in The Sonification Handbook.
This system sounds good in theory, but do we actually use these cues in practice? In 1988, Frederic Wightman and Doris Kistler performed an ingenious set of experiments (1, 2) to show that that people really do use HRTFs to infer location. First, they measured the HRTF of each participant by putting a small microphone in their ears and playing sounds from different locations. Next they created a digital filter for each location and each participant. That is to say, these filters implemented each participant’s HRTF. Finally, they placed headphones on the listeners and played sounds to them, each time passing the sound through one of the digital filters. Amazingly, participants were able to correctly guess the “location” of the sound, depending on which filter was used, even though the sound was coming from headphones. They were also much better at sound localization when using their own HRTF, rather than someone else’s HRTF.
Further evidence for this hypothesis comes from Hofman et al., 1998, who showed that by using putty to reshape people’s ears, they were able to change the HRTFs and thus disrupt sound localization. Interestingly, people were able to quickly relearn how to localize sound with their new HRTFs.
Image from Hofman et al., 1998.
A final fun fact: to improve the sound localization of humanoid robots, researchers in Japan attached artificial ears to the robot heads and implemented some sophisticated algorithms to infer sound location. Here are some pictures of the robots.
Their paper is kind of ridiculous and has some questionable justifications for not just using microphones in multiple locations, but I thought it was fun to see these principles being applied.
# Hyperbolic discounting — The irrational behavior that might be rational after all
When I was in grad school I occasionally overheard people talk about how humans do something called “hyperbolic discounting”. Apparently, hyperbolic discounting was considered irrational under standard economic theory.
I recently decided to learn what hyperbolic discounting was all about, so I set out to write this blog post. I have to admit that hyperbolic discounting has been pretty hard for me to understand, but I think I now finally have a good enough handle on it to write about it. Along the way, I learned something interesting: Hyperbolic discounting might be rational after all.
## Rational and irrational discounting
##### Rationality of hyperbolic discounting
The problem with this story is that it only works if you assume that the interest rate is constant. In the real world, the interest rate fluctuates.
Before taking on the fluctuating interest rate scenario, let’s first take on a different assumption that is still somewhat simplified. Let’s assume that the interest rate is constant but we don’t know what it is, just as we didn’t know what the hazard rate was in the previous interpretation. With this assumption, the justification for hyperbolic discounting becomes similar to the explanation in the blue and pink plots above. When you do a probability-weighted average over these decaying exponential curves, you get a hyperbolic function.
The previous paragraph assumed that the interest rate was constant but unknown. In the real world, the interest rate is known but fluctuates over time. Farmer and Geanakoplos (2009) showed that if you assume that interest rate fluctuations follow a geometric random walk, hyperbolic discounting becomes optimal, at least asymptotically as $\tau \rightarrow \infty$. In the near future, you know the interest rate with reasonable certainty and should therefore discount with an exponential curve. But as you look further into the future, your uncertainty about the interest rate increases and you should therefore discount with a hyperbolic curve.
Is the geometric random walk a process that was cherry picked by the authors to produce this outcome? Not really. Newell and Pizer (2003) studied US bond rates in the 19th and 20th century and found that the geometric random walk provided a better fit than any of the other interest rate models tested.
## Summary
When interpreting discounting as a survival function, a hyperbolic discounting function is rational if you introduce uncertainty into the hazard parameter via an exponential prior (Souza, 2015). When interpreting the discount rate as an interest rate, a hyperbolic discounting function is asymptotically rational if you introduce uncertainty in the interest rate via a geometric random walk (Farmer and Geanakoplos, 2009).
# Religions as firms
I recently came across a magazine that helps pastors manage the financial and operational challenges of church management. The magazine is called Church Executive.
Readers concerned about seasonal effects on tithing can learn how to “sustain generosity” during the weaker summer months. Technology like push notifications and text messages is encouraged as a way to remind people to tithe. There is also some emphasis on messaging, as pastors are told to “make sure your generosity-focused sermons are hitting home with your audience”.
Churches need money to stay active, and it’s natural that pastors would want to maintain a healthy cash flow. But the brazen language of Church Executive reminded me of the language of profit-maximizing firms. This got me thinking: What are the other ways in which religions act like a business?
This post is my attempt to understand religions as if they were businesses. This isn’t a perfect metaphor. Most religious leaders are motivated by genuine beliefs, and few are motivated primarily by profit. But it can still be instructive to view religions through the lens of business and economics, if only as an exercise. After working through this myself, I feel like I have a better understanding of why religions act the way they do.
### Competition
As with any business, one of the most pressing concerns of a religion is competition. According to sociologist Carl Bankston, the set of religions can be described as a marketplace of competing firms that vie for customers. Religious consumers can leave one church to go to another. To hedge their bets on the afterlife, some consumers may even belong to several churches simultaneously, in a strategy that has been described as “portfolio diversification”.
One way that a religion can ward off competitors is to prohibit its members from following them. The Bible is insistent on this point, with 26 separate verses banning idolatry. Other religions have been able to eliminate competition entirely by forming state-sponsored monopolies.
### Pricing
Just like a business, religions need to determine how to price their product. According to economists Laurence Iannaccone and Feler Bose, the optimal pricing strategy for a religion depends on whether it is proselytizing or non-proselytizing.
Non-proselytizing religions like Judaism and Hinduism earn much of their income from membership fees. While exceptions are often made for people who are too poor to pay, and while donations are still accepted, the explicit nature of the membership fees help these religions avoid having too many free riders.
Proselytizing religions like Christianity are different. Because of their strong emphasis on growth, they are willing to forgo explicit membership fees and instead rely more on donations that are up to the member’s discretion. Large donations from wealthy individuals can cross-subsidize the membership of those who make smaller donations. Even free riders who make no donations at all may be worthwhile, since they may attract more members in the future.
### Surge Pricing
Like Uber, some religions raise the price during periods of peak demand. While attendance at Jewish synagogue for a regular Shabbat service is normally free, attendance during one of the High Holidays typically requires a payment for seating, in part to ensure space for everyone.
Surge pricing makes sense for non-proselytizing religions such as Judaism, but it does not make sense for proselytizing religions such as Christianity, which views the higher demand during peak season as an opportunity to convert newcomers and to reactivate lapsed members. Thus, Christian churches tend to expand seating and schedule extra services during Christmas and Easter, rather than charging fees.
### Product Quality
Just as business consumers will pay higher prices for better products, consumers of polytheistic religions will pay higher “prices” for gods with more wide-ranging powers. Even today, some American megachurches have found success with the prosperity gospel, which emphasizes that God can make you wealthy.
Of course, not all religious consumers will prefer the cheap promises of the prosperity gospel. For many religions, product quality is defined primarily by community, a sense of meaning, and in some cases the promise of an afterlife.
A good business should be constantly updating its product to fix bugs and to respond to changes in consumer preference or government regulation. Some religions do the same thing, via the process of continuous revelation from their deity. Perhaps no church exemplifies this better than the Church of Jesus Christ of Latter-day Saints.
For most of the history of the Mormon Church, individuals of African descent were prohibited from serving as priests. By the 1960s, as civil rights protests against the church received media attention, the policy became increasingly untenable. On June 1, 1978, Mormon leaders reported that God had instructed them to update the policy and allow black priests. This event was known as the 1978 Revelation on the Priesthood.
In the late 19th Century, when the Mormon Church was under intense pressure from the US Government regarding polygamy, the Church president claimed to receive a revelation from Jesus Christ asking him to prohibit it. This revelation, known as the 1890 Revelation, overturned the previous 1843 Revelation which allowed polygamy.
While frequent updates usually make sense in business, they don’t always make sense in religion. Most religions have a fairly static doctrine, as the prospect of future updates undermines the authority of current doctrine.
### Growth and marketing
Instead of focusing only on immediate profitability, many businesses invest in user growth. As mentioned earlier, many religions are willing to cross-subsidize participation from new members, especially young members, with older members bearing most of the costs.
Christianity’s concept of a heaven and hell encouraged its members to convert their friends and family. In some ways, this is reminiscent of viral marketing.
### International expansion
Facebook and Netflix both experienced rapid adoption, starting with a U.S. audience. But as U.S. growth began to slow down, both companies needed to look towards international expansion.
A similar thing happened with the Mormon church. By the 20th century, U.S. growth was driven only by increasing family sizes, so the church turned towards international expansion.
The graph below shows similar US and international growth curves for Netflix and the Church of Jesus Christ of Latter-day Saints.[1,2,3,4]
### Branding
Like any company, most religions try to maintain a good brand. But unlike businesses, most religions do not have brand protection, and thus their brands can be co-opted by other religions. Marketing from Mormons and from Jehovah’s Witnesses tends to emphasize the good brand of Jesus Christ, even though most mainstream Christians regard these churches as heretical.
One of the most interesting risks to brands is genericide, in which a popular trademark becomes synonymous with the general class of product, thereby diluting its distinctive meaning. Famous examples of generic trademarks include Kleenex and Band-Aid. Amazingly, genericide can also happen to religious deities. The ancient Near East god El began as a distinct god with followers, but gradually became a generic name for “God” and eventually merged with the Hebrew god Yahweh.
### Mergers and spin-offs
In business, companies can spin off other companies or merge with other companies. But with rare exceptions, religions only seem to have spin-offs. Why do religions hardly ever merge with other religions? My guess is that since there is no protection for religious intellectual property, religions can acquire the intellectual property of another religion without requiring a merger. Religions can simply copy each other’s ideas.
Another reason that religious mergers are rare is that religions are strongly tied to personal identity and tap into tribal thinking. When WhatsApp was acquired, its leadership was happy to adopt new identities as Facebook employees. But it is far less likely that members of, say, the Syriac Catholic Church would ever tolerate merging into the rival Syriac Maronite Church, even if it might provide them with economies of scale and more political power.
On Twitter, I asked why there are so few religious mergers and got lots of interesting responses. People pointed out that reconciliation of doctrine could undermine the authority of the leaders, and that there is little benefit from economies of scale. Others noted that religious mergers aren’t that rare. Hinduism and Judaism may have began as mergers of smaller religions, many Christian traditions involve mergers with religions they replaced, and that even today Hinduism continues to be a merging of various sects.
It’s worth repeating that economic explanations aren’t always great at describing the conscious motivations of religious individuals, who generally have sincere beliefs. Nevertheless, economic reasoning does a decent job of predicting the behavior of systems, and it’s been pretty interesting to learn how religion is no exception.
# Part 2: A bipartisan list of people who argue in good faith
In Part 1, I posted a bipartisan list of people who are bad for America. Those people present news stories that cherry pick the worst actions from the other side so that they can get higher TV ratings and more social media points.
Here in Part 2, I post a list of people who don’t do that, at least for the most part. This isn’t a list of centrists. If anything, it is a more politically diverse list than the list in Part 1. This is a list of people who usually make good-faith attempts to persuade others about their point of view.
• Megan McArdle (Twitter, Bloomberg) – Moderately libertarian ideas presented to a diverse audience
• Noah Smith (Twitter, Bloomberg) – Center-left economics
• Ross Douthat (Twitter, NYT) – Social conservatism presented to a left-of-center audience
• Noam Chomsky (Website)
• Conor Friedersdorf (The Atlantic)
• Ben Sasse — Has the third-most conservative voting record in the Senate but never caricatures the other side and is very concerned about filter bubbles.
• Julia Galef (Twitter) – Has some great advice for understanding the other side
| 8,474 | 37,804 |
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## Motion in Straight Line
Chapter Revision Test
Q1. A railway train 400m long is going from New Delhi railway station to Kanpur. [1]
Can we consider railway train as a point object?
Q2. Shipra went from her home to school 2.5km away. On finding her home closed she [1]
returned to her home immediately. What is her net displacement? What is the total
distance covered by her?
Q3. Can speed of an object be negative? Justify [1]
Q4. What causes variation in velocity of a particle? [2]
Q5. Figure shown below shows displacement – time curves I and II. What conclusions do you draw from these graphs? [2]
Q6. Displacement of a particle is given by the expression x = 3t² + 7t – 9, where x is in [2]
meter and t is in seconds. What is acceleration?
Q7. A particle is thrown upwards. It attains a height (h) after 5 seconds and again after [2]
9s comes back. What is the speed of the particle at a height h?
Q8. A police jeep on a petrol duty on national highway was moving with a speed of [3]
54km/hr. in the same direction. It finds a thief rushing up in a car at a rate of
126km/hr in the same direction. Police sub – inspector fired at the car of the thief
with his service revolver with a muzzle speed of 100m/s. with what speed will the
bullet hit the car of thief?
Q9. Establish the relation Snth = u + ½a(2n − 1) where the letters have their usual [3]
Q10. A stone is dropped from the top of a cliff and is found to ravel 44.1m diving the last [3]
second before it reaches the ground. What is the height of the cliff? g = 9.8m/s²
meanings.
| 421 | 1,555 |
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# Class 10 NCERT Solutions – Chapter 5 Arithmetic Progressions – Exercise 5.3 | Set 2
• Last Updated : 28 May, 2021
### Question 11. If the sum of the first n terms of an AP is 4n − n2, what is the first term (that is S1)? What is the sum of first two terms? What is the second term? Similarly find the 3rd, the 10th and the nth terms.
Solution:
Given:
Sn = 4n−n2
First term can be obtained by putting n=1,
a = S1 = 4(1) − (1)2 = 4−1 = 3
Also,
Sum of first two terms = S2= 4(2)−(2)2 = 8−4 = 4
Second term, a2 = S2 − S1 = 4−3 = 1
Common difference, d = a2−a1 = 1−3 = −2
Also,
Nth term, an = a+(n−1)d
= 3+(n −1)(−2)
= 3−2n +2
= 5−2n
Therefore, a3 = 5−2(3) = 5-6 = −1
a10 = 5−2(10) = 5−20 = −15
Therefore, the sum of first two terms is equivalent to 4. The second term is 1.
And, the 3rd, the 10th, and the nth terms are −1, −15, and 5 − 2n respectively.
### Question 12. Find the sum of first 40 positive integers divisible by 6.
Solution:
The first positive integers that are divisible by 6 are 6, 12, 18, 24 ….
Noticing this series,
First term, a = 6
Common difference, d= 6.
Sum of n terms, we know,
Sn = n/2 [2a +(n – 1)d]
Substituting the values, we get
S40 = 40/2 [2(6)+(40-1)6]
= 20[12+(39)(6)]
= 20(12+234)
= 20×246
= 4920
### Question 13. Find the sum of first 15 multiples of 8.
Solution:
The first few multiples of 8 are 8, 16, 24, 32…
Noticing this series,
First term, a = 8
Common difference, d = 8.
Sum of n terms, we know,
Sn = n/2 [2a+(n-1)d]
Substituting the values, we get,
S15 = 15/2 [2(8) + (15-1)8]
= 15/2[6 +(14)(8)]
= 15/2[16 +112]
= 15(128)/2
= 15 × 64
= 960
### Question 14. Find the sum of the odd numbers between 0 and 50.
Solution:
The odd numbers between 0 and 50 are 1, 3, 5, 7, 9 … 49.
Therefore, we can see that these odd numbers are in the form of A.P.
Now,
First term, a = 1
Common difference, d = 2
Last term, l = 49
Last term is equivalent to,
l = a+(n−1) d
49 = 1+(n−1)2
48 = 2(n − 1)
n − 1 = 24
Solving for n, we get,
n = 25
Sum of nth term,
Sn = n/2(a +l)
Substituting these values,
S25 = 25/2 (1+49)
= 25(50)/2
=(25)(25)
= 625
### Question 15. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs. 200 for the first day, Rs. 250 for the second day, Rs. 300 for the third day, etc., the penalty for each succeeding day being Rs. 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days.
Solution:
The given penalties form and A.P. having first term, a = 200 and common difference, d = 50.
By the given constraints,
Penalty that has to be paid if contractor has delayed the work by 30 days = S30
Sum of nth term, we know,
Sn = n/2[2a+(n -1)d]
Calculating, we get,
S30= 30/2[2(200)+(30 – 1)50]
= 15[400+1450]
= 15(1850)
= 27750
Therefore, the contractor has to pay Rs 27750 as penalty for 30 days delay.
### Question 16. A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes.
Solution:
Let us assume the cost of 1st prize be Rs. P.
Then, cost of 2nd prize = Rs. P − 20
Also, cost of 3rd prize = Rs. P − 40
These prizes form an A.P., with common difference, d = −20 and first term, a = P.
Given that, S7 = 700
Sum of nth term,
Sn = n/2 [2a + (n – 1)d]
Substituting these values, we get,
7/2 [2a + (7 – 1)d] = 700
Solving, we get,
a + 3(−20) = 100
a −60 = 100
a = 160
Therefore, the value of each of the prizes was Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, and Rs 40.
### Question 17. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of class I will plant 1 tree, a section of class II will plant 2 trees and so on till class XII. There are three sections of each class. How many trees will be planted by the students?
Solution:
The number of trees planted by the students form an AP, 1, 2, 3, 4, 5………………..12
Now,
First term, a = 1
Common difference, d = 2−1 = 1
Sum of nth term,
Sn = n/2 [2a +(n-1)d]
S12 = 12/2 [2(1)+(12-1)(1)]
= 6(2+11)
= 6(13)
= 78
Number of trees planted by 1 section of the classes = 78
Therefore,
Number of trees planted by 3 sections of the classes = 3×78 = 234
### Question 18. A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A of radii 0.5, 1.0 cm, 1.5 cm, 2.0 cm, ……… as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take π = 22/7)
Solution:
We know,
Perimeter of a semi-circle = πr
Calculating ,
P1 = π(0.5) = π/2 cm
P2 = π(1) = π cm
P3 = π(1.5) = 3π/2 cm
Where, P1, P2, P3 are the lengths of the semi-circles respectively.
Now, this forms a series, such that,
π/2, π, 3π/2, 2π, ….
P1 = π/2 cm
P2 = π cm
Common difference, d = P2 – P1 = π – π/2 = π/2
First term = P1= a = π/2 cm
Sum of nth term,
Sn = n/2 [2a + (n – 1)d]
Therefore, Sum of the length of 13 consecutive circles is;
S13 = 13/2 [2(π/2) + (13 – 1)π/2]
Solving, we get,
= 13/2 [Ï€ + 6Ï€]
=13/2 (7Ï€)
= 13/2 × 7 × 22/7
= 143 cm
### Question 19. 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?
[1]
Solution:
The numbers of logs in rows are in the form of an A.P. 20, 19, 18…
Given,
First term, a = 20 and common difference, d = a2−a1 = 19−20 = −1
Let us assume a total of 200 logs to be placed in n rows.
Thus, Sn = 200
Sum of nth term,
Sn = n/2 [2a +(n -1)d]
S12 = 12/2 [2(20)+(n -1)(-1)]
400 = n (40−n+1)
400 = n (41-n)
400 = 41n−n2
Solving the eq, we get,
n2−41n + 400 = 0
n2−16n−25n+400 = 0
n(n −16)−25(n −16) = 0
(n −16)(n −25) = 0
Now,
Either (n −16) = 0 or n−25 = 0
n = 16 or n = 25
By the nth term formula,
an = a+(n−1)d
a16 = 20+(16−1)(−1)
= 20−15
= 5
And, the 25th term is,
a25 = 20+(25−1)(−1)
= 20−24
= −4
Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16th row is 5, since the number of logs can’t be negative as in case of 25th term.
### A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
[Hint: to pick up the first potato and the second potato, the total distance (in meters) run by a competitor is 2×5+2×(5+3)]
Solution:
The distances of potatoes from the bucket are 5, 8, 11, 14…, which form an AP.
Now, we know that the distance run by the competitor for collecting these potatoes are two times of the distance at which the potatoes have been kept.
Therefore, distances to be run w.r.t distances of potatoes is equivalent to,
10, 16, 22, 28, 34,……….
We get, a = 10 and d = 16−10 = 6
Sum of nth term, we get,
S10 = 12/2 [2(20)+(n -1)(-1)]
= 5[20+54]
= 5(74)
Solving we get,
= 370
Therefore, the competitor will run a total distance of 370 m.
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# Types of Gears- Complete Explanation
In this article, we will learn about different types of gears used to transmit powers from one shaft to another. The various gear types obtained on the basis of different criteria are spur gear, helical gear, spiral gear, bevel gear, rack, and pinion, etc. Here we will discuss each type in brief. Gears are basically used to transmit powers from one shaft to another.The classification of the gears is done on the following basis:
### 1. On the Basis of Arrangement of Shafts of Gears
#### (i) Parallel Gears
When two parallel and coplanar shafts are connected by gears as shown in the figure above, the types of gears used are called as spur gears. The arrangement of gears is called spur gearing. Spur gears have teeth parallel to the axis of the wheel.
Examples: Spur gear, helical gear, double helical gear (herringbone gear)
#### (ii) Intersecting Gears
When two non-parallel or intersecting but coplanar shafts are connected by the gears as shown in the figure given above is called bevel gears. And such an arrangement is called as bevel gearing.
Examples: Bevel gear, helical bevel gear, miter gear.
#### (iii) Non-Intersecting and Non-Parallel
When non-intersecting and non-parallel i.e. non-coplanar shafts are connected by gears as shown in the figure given above is called skew bevel gears or spiral gears. And the arrangement is called skew bevel gearing or spiral gearing.
Examples: Spiral gears.
### 2. On the Basis of the Peripheral Velocity of the Gears
#### (i) Low Velocity:
The gears which possess velocity less than 3 m/s are called low-velocity gears.
#### (ii) Medium Velocity:
The gears which have velocity in between 3 m/s and 15 m/s are called medium velocity gears.
#### (iii) High Velocity:
The gears which possess velocity more than 15 m/s are called as high-velocity gears.
### 3. On the Basis of the Types of Gearing.
#### (i) External Gearing
When gears of two shafts mesh externally as shown in the figure given above, then the gears are called as external gears. In external gearing the larger of the two meshed gears is called spur or gear and the smaller is called a pinion. The motion of the two wheels in external gearing is always unlike (i.e. if one moves clockwise than other will move in anticlockwise direction).
#### (ii) Internal Gearing
When the gears of the two shafts mesh internally with each other as shown in the figure given above then it is called internal gearing. In internal gearing the larger of the two wheels is called an annular wheel and the smaller wheel is called a pinion. The motion of the two wheels in the internal gearing is always like (i.e. if one moves clockwise than the other will also move in the clockwise direction).
#### (iii) Rack and Pinion
Sometimes what happens, the gears of a shaft meshes externally and internally with the gears in strait line (i.e. strait line is also defined as the wheel of infinite radius). Such type of gear is known as rack and pinion. In rack and pinion types of gears, the straight line gear is called rack and the circular gear is called pinion. Rack and pinion gears are used to convert linear motion into rotary motion and vice-versa.
### 4. On the Basis of the Position of Teeth on the Gear.
#### (i) Straight
Straight gears have straight teeth on the surface of the wheel rim. For example Spur gear.
#### (ii) Inclined
The inclined gears have inclined teeth on the surface of the wheel rim. For example: Helical gears.
#### (iii) Curved
Curved gears have curved teeth on the surface of the wheel. For example: Spiral gears.
### Types of Gears
Now we will discuss each classification of gears one by one in detail.
#### 1. Spur gear
the spur gear is the simplest gear among all types of gears and easy to manufacture. It has straight teeth parallel to the axis of the wheel or shaft. It is used to transmit power between parallel shafts. Only one tooth of the spur gear meshes at a time. It is regularly used for speed reduction or increase, resolution and accuracy enhancement for positioning systems, torque multiplication. It creates noise during its operation.
#### 2. Helical Gears
The helical gears have inclined teeth (teeth cut at an angle to the face of the gear) on the surface of the wheel. Its operation is smoother and quieter as compared with the spur gear. It is mostly commonly used in transmission gearboxes.
Helical gear can be further sub divided into two types
##### (i) Single Helical Gears
It is gear that has an inclined tooth on its wheel surface. Single helical gear possesses axial thrust.
##### (ii) Double Helical Gears or Herringbone Hears
Herringbone gear is a type of double helical gear in which there is side to side combination of two helical gears of opposite hands. If we look from the top than each helical groove on this gear looks like a V letter. The herringbone or double helical gears have zero axial thrusts which is not so with the single helical gear. Unlike helical gears, it has the advantage of transmitting power smoothly because it meshes two teeth at a time.
#### 3. Bevel Gears
These are the gears that transfer powers between two non-parallel or intersecting shafts. These types of gears are most commonly used in the differentials drive of an automobile.
The bevel gear are further classified as
##### (i) Straight Bevel Gear
This gear has straight teeth, conical pitch surface, and tapering towards apex.
##### (ii) Skew Bevel Gears or Spiral Bevel Gear
This has curved teeth at an angle that allows gradual and smooth contact of the tooth.
##### (iii) Zerol Bevel Gear
These gears are similar to the bevel gear but it has curved teeth with a spiral angle of zero, so the ends of the tooth align with the axis.
##### (iv) Hypoid Bevel Gear
These types of gears are similar to the spiral bevel gear but the pitch surface is hyperbolic and not conical.
##### (v) Miter Gear
It is a type of bevel gear in which the two shafts intersect at right angles with each other. It is used to transmit powers at right angles.
#### 4. Internal Gearing
In internal gearing, the two gears are meshed internally with each other.
#### 5. External Gearing
In external gearing, the two shafts are connected with the gears that mesh with each other externally.
#### 6. Rack and Pinion Gears
In rack and pinion gears, a shaft meshes externally and internally with a straight line gear. The circular gear is called as pinion and the straight line gear is called a rack. The rack and pinion gear is shown in the figure given below.
#### 7. Worm Gear
A worm gear is an arrangement of two gears in which one gear is called as worm ( gear in the form of a screw) meshes with a worm gear ( similar as spur gear). The two elements are called as worm screw and worm wheel. Worm gears are used in presses, rolling mills, on rudders and worm drive saws etc.
This is all about the types of gears used in power transmission from one shaft to another. If you have any queries about this then comment us. And if you found this article informative and useful than share and like us.
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FractionalGeometry-Chap2
# FractionalGeometry-Chap2 - Chapter 2 Iterated function...
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Chapter 2 Iterated function systems: fractals as limits Often a Frst exposure to fractals consists of googling “fractal geometry,” Fnding a program to generate fractals, selecting some example from a preset menu, and clicking the RUN button. The image drifts into being, dot after dot dancing across the screen in furious pattern. If the rules generating this fractal can be viewed with the program, the Frst question is “How do these rules make that picture?” The answer to that question is the Frst topic of this chapter, if an I±S program had the highest google rank. Before starting this relatively long analysis, a moment’s consideration of a self-similar fractal suggests a general direction for the proof. Look at the fractal shown in the left of ±ig. 2.1. This shape, the Sierpinski gasket, is one of the simplest examples of a self-similar fractal. The middle image, a magniFcation of a small portion of the left side, is indistinguishable from the left. The right is a magniFcation of a portion of the middle. In our minds, this process can be continued forever. ±igure 2.1: Successive magniFcations of portions of the gasket. The reappearance, under increasingly high magniFcation, of copies of the whole shape, suggests two things. ±irst, some sort of limiting process is in- volved here. Second, it is not the limiting process familiar from calculus, where under magniFcation a smooth curve approaches its tangent line. Something dif- ferent is happening here: the level of complexity remains about constant under magniFcation. So we must determine what sort of limit produces fractals. 25
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26 CHAPTER 2. ITERATED FUNCTION SYSTEMS 2.1 Iterated function system formalism Based on work of Mandelbrot [102] and Hutchinson [81], and popularized by Barnsley [5], iterated function systems are a formalism for generating fractals and for compressing images. First we recall some background on transformations of the plane. In the plane, a general linear transformation plus translation can be written as T b x y B = b r cos( θ ) s sin( ϕ ) r sin( θ ) s cos( ϕ ) Bb x y B + b e f B (2.1) That is, any real 2 × 2 matrix can be expressed as the 2 × 2 in eq (2.1). (See Prob. 2.1.1.) Let d ( , ) denote the Euclidean distance. A transformation T is a d -contraction with contraction factor t , 0 t < 1, if for all points ( x 1 ,y 1 ) and ( x 2 2 ), d p T b x 1 y 1 B ,T b x 2 y 2 B P t · d p b x 1 y 1 B , b x 2 y 2 B P , (2.2) and for any number s < t , d p T b x 1 y 1 B b x 2 y 2 B P > s · d p b x 1 y 1 B , b x 2 y 2 B P for at least one pair of points ( x 1 1 ) and ( x 2 2 ). For example, if T ( x,y ) = ( x/ 2 ,y/ 2), then d p T b x 1 y 1 B b x 2 y 2 B P = r ± x 1 2 x 2 2 ² 2 + ± y 1 2 y 2 2 ² 2 = 1 2 d p b x 1 y 1 B , b x 2 y 2 B P for any pair of points. Consequently, this T is a contraction with contraction factor 1 / 2. A slightly more di±cult example is given in Prob. 2.1.2.
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A114901 Number of compositions of n such that each part is adjacent to an equal part. 46
1, 0, 1, 1, 2, 1, 5, 3, 10, 10, 21, 22, 49, 51, 105, 126, 233, 292, 529, 678, 1181, 1585, 2654, 3654, 6016, 8416, 13606, 19395, 30840, 44517, 70087, 102070, 159304, 233941, 362429, 535520, 825358, 1225117, 1880220, 2801749, 4285086, 6404354, 9769782, 14634907 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,5 LINKS Alois P. Heinz, Table of n, a(n) for n = 0..4000 N. J. A. Sloane, Transforms FORMULA INVERT(iMOEBIUS(iINVERT(A000012 shifted right 2 places))) EXAMPLE The 5 compositions of 6 are 3+3, 2+2+2, 2+2+1+1, 1+1+2+2, 1+1+1+1+1+1. From Gus Wiseman, Nov 25 2019: (Start) The a(2) = 1 through a(9) = 10 compositions: (11) (111) (22) (11111) (33) (11122) (44) (333) (1111) (222) (22111) (1133) (11133) (1122) (1111111) (2222) (33111) (2211) (3311) (111222) (111111) (11222) (222111) (22211) (1111122) (111122) (1112211) (112211) (1122111) (221111) (2211111) (11111111) (111111111) (End) MAPLE g:= proc(n, i) option remember; add(b(n-i*j, i), j=2..n/i) end: b:= proc(n, l) option remember; `if`(n=0, 1, add(`if`(i=l, 0, g(n, i)), i=1..n/2)) end: a:= n-> b(n, 0): seq(a(n), n=0..50); # Alois P. Heinz, Nov 29 2019 MATHEMATICA Table[Length[Select[Join@@Permutations/@IntegerPartitions[n], Min@@Length/@Split[#]>1&]], {n, 0, 10}] (* Gus Wiseman, Nov 25 2019 *) g[n_, i_] := g[n, i] = Sum[b[n - i*j, i], {j, 2, n/i}] ; b[n_, l_] := b[n, l] = If[n==0, 1, Sum[If[i==l, 0, g[n, i]], {i, 1, n/2}]]; a[n_] := b[n, 0]; a /@ Range[0, 50] (* Jean-François Alcover, May 23 2021, after Alois P. Heinz *) CROSSREFS The case of partitions is A007690. Compositions with no adjacent parts equal are A003242. Compositions with all multiplicities > 1 are A240085. Compositions with minimum multiplicity 1 are A244164. Compositions with at least two adjacent parts equal are A261983. Cf. A178470, A238130, A274174, A329863. Sequence in context: A085261 A179218 A131119 * A355562 A194809 A113178 Adjacent sequences: A114898 A114899 A114900 * A114902 A114903 A114904 KEYWORD nonn AUTHOR Christian G. Bower, Jan 05 2006 STATUS approved
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Last modified April 22 13:50 EDT 2024. Contains 371903 sequences. (Running on oeis4.)
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# Find the Slope y+4=-1/3*(x-3)
y+4=-13⋅(x-3)
Rewrite in slope-intercept form.
The slope-intercept form is y=mx+b, where m is the slope and b is the y-intercept.
y=mx+b
Simplify -13⋅(x-3).
Apply the distributive property.
y+4=-13x-13⋅-3
Combine x and 13.
y+4=-x3-13⋅-3
Cancel the common factor of 3.
Move the leading negative in -13 into the numerator.
y+4=-x3+-13⋅-3
Factor 3 out of -3.
y+4=-x3+-13⋅(3(-1))
Cancel the common factor.
y+4=-x3+-13⋅(3⋅-1)
Rewrite the expression.
y+4=-x3-1⋅-1
y+4=-x3-1⋅-1
Multiply -1 by -1.
y+4=-x3+1
y+4=-x3+1
Move all terms not containing y to the right side of the equation.
Subtract 4 from both sides of the equation.
y=-x3+1-4
Subtract 4 from 1.
y=-x3-3
y=-x3-3
Rewrite in slope-intercept form.
y=-13x-3
y=-13x-3
Using the slope-intercept form, the slope is -13.
m=-13
Find the Slope y+4=-1/3*(x-3)
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# Let $f(x) = 1 - x^{2/3}$. Show that $f(-1) = f(1)$ but there is no number $c$ in $(-1, 1)$ such that $f'(c) = 0$. Why does this not contradict Rolle's Theorem?
## $f(x)=1-x^{2 / 3} . \quad f(-1)=1-(-1)^{2 / 3}=1-1=0=f(1) . \quad f^{\prime}(x)=-\frac{2}{3} x^{-1 / 3},$ so $f^{\prime}(c)=0$ has no solution. Thisdoes not contradict Rolle's Theorem, since $f^{\prime}(0)$ does not exist, and so $f$ is not differentiable on (-1,1)
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### Video Transcript
So now they're the function affect because one minus X to the two third. And we're being has to show that effort negative or is equal after one. But there's no numbers. CNN won the one shot that promise city. And the question is, why does this not contradict rolls, too? So, um, just by looking at dysfunction, we know that dysfunction is continuous in this animal because this is a is actually a square root, some sort of brute function. And we can have values from negative one to one because this is a cube groups. The Coopers are defined in all on all real numbers. So we know that this is continuous, that's for sure. Uh, we cannot assume different ability. This is a little tricky. We're going to talk about this in the second, so we're gonna come back to to oops. Sorry about that. Um and then we're going to shank of after negative one equals that one right away. So if we plug in value of negative one a negative one into our functions or one minus negative. One two to third. Oh, and since since this two on top squares the inside and the coupe root of positive one is one. It'LL just be one minus one, which is Seo. And then half of one will be the same thing we wanted to third, which will be one minus one Should authors you got a zero. So now, now we know that this last condition is satisfied. So the defense ability over negative ones one is a little tricky in this case, we can assume it in this case right away. So why don't we just first take the derivative? So after primary Rex. So now all we do is apply. Probably normal rules and justice. Looking at news, we're going to spring this down, So this will be CEO, bring down to third, and then we subtract one. The two third minus one is just negative one third, which is this is the same thing as negative to remember. There's a negative right here. I know this is Yeah, because we bring down to negative one toe. Important note eyes. This is the next to the one third. And now there's after something very interesting going on in this function. Um, since this value of X, um, if you actually draw the function the cube root of ax. There is no value of excess that will give us zero. Um, because even if you think about that, just looking at this function negative too. Three x one third we can't solve for X equals zero like we can't find a value where X is equal to zero. Because when we multiply this, I'm three extra one third, we want to buy a zero and we and then we're left with negative equals zero, which is just absolute. This is just nonsense doesn't make sense. So we know that there's no value of excess will give us the value that equals zero. So although there is a derivative, it is not defensible from negative one to one. So what's actually happening here is that this second condition is not satisfied. And so initially seems like there is a It is defensible and we could take to evident, but there's no value. There's no point in which there it is equal to zero have no, um, there's no f prime of C that equals zero. So this actually doesn't contradict roasted because the second condition is not even satisfied because this tells this is asking us that it has to be defensible. I'm negative Wonder one. But there is no value from negative on the one. There's no X equals zero has no value of effort. Zero that will give us this conclusion. So this Khun, this conditions cannot be satisfied. This is why does not contradict rule still.
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Chapter 2_Research Methods Unit Test
# Chapter 2_Research Methods Unit Test
Unit test for Chapter 2 on Research Methods will take place this Friday! Please make sure you have all the handouts given for this unit and that you have completed study cards to help you.
IF YOU LEARN ONLY EIGHT THINGS IN THIS CHAPTER….
1. Observation is the most important aspect of psychological research
2. Operationalism means to define our variables in the manner in which we are going to measure them
3. Correlation measures degree of relationship between variables and ranges from -1 to +1.
4. Correlation does not imply causation
5. The difference between an independent and dependent variable
6. Confounding variables
7. What descriptive and inferential statistics tell us about the results of a study
8. The role of ethics in psychological research
Statistics
Standard Deviation is the amount, on average, that each score differs from the mean. Thus, the standard deviation tells us about how much each score is different from the average score. The bigger the standard deviation, the more spread out the scores are.
Normal Curve: See my notes on how the curve is divided out. We know, for example, that if we select randomly from a curve, we’ll most likely get a score that falls between -1 and +1 standard deviations from the mean. Essentially, we can predict there is a 68% chance we will get a score from the section of the curve. Please refer to page 40 (fig2.10) Know the Normal Curve! Make sure you know these percentages.
Illusory Correlation: Random events that we notice and falsely assume are related.
Range- the difference between the highest and lowest scores in a distribution
Double-Blind procedure- neither the participants nor the research assistants collecting the data will know which group is receiving treatment
Sample Questions:
1. If Sarah scored 1 standard deviation above the average, 15% percentage of the population scored higher than she did. (answer: Here you have to know that between the mean and 1 standard deviation from the mean is 34 percent of the scores. So roughly 84-85 percent of the scores are below that;therefore, 15% or so are above that.)
2. Assume you take the SAT and score 2 standard deviations above the average. You scored better than 95% of the population (answer is 95%. You have to know that area under the bell curve and that about 13.59 percent of scores fall between 1 and 2 standard deviations from the mean)
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> > Calculus: Basic Concepts for High Schools by L. V. Tarasov
# Calculus: Basic Concepts for High Schools by L. V. Tarasov
By L. V. Tarasov
An interactive textbook of Calculus provided as a discussion among the writer and the reader.
Similar calculus books
Calculus, Single Variable, Preliminary Edition
Scholars and math professors trying to find a calculus source that sparks interest and engages them will enjoy this new e-book. via demonstration and routines, it exhibits them how you can learn equations. It makes use of a mix of conventional and reform emphases to boost instinct. Narrative and workouts current calculus as a unmarried, unified topic.
Tables of Laplace Transforms
This fabric represents a suite of integrals of the Laplace- and inverse Laplace remodel sort. The usef- ness of this sort of details as a device in numerous branches of arithmetic is firmly verified. prior guides contain the contributions by means of A. Erdelyi and Roberts and Kaufmann (see References).
Additional resources for Calculus: Basic Concepts for High Schools
Sample text
1a). 1. The derivative of the arctangent The arctangent is an odd function (since the tangent is), so its derivative is even. Hence _ we need only find d arctan t=dt for t 0. t C t /2 for t > 0 (the case where t < 0 is similar). t C t / arctan t . t C t / t ; or arctan t 17 18 CAMEO 8. t C t / t arctan t D 1 ; 1 C t2 and similarly for the limit as t ! 0 , So the ordinary limit as t ! 0 exists, and we have d 1 arctan t D : dt 1 C t2 The derivatives of other inverse trigonometric functions now follow using the chain rule and identities: arcsin t D arctan p 1 t t2 ; arccos t D arcsin t; 2 1 arcsec t D arccos ; etc.
1) follows. 1. 2. The AM-GM inequality p for two numbers. a C b/=2 and the geometric mean ab for positive numbers a and b in the preceding Cameo. 2) p with equality if and only if a D b. 1 C x/=2 x for p x > 0. 2. 1 C x/=2 23 24 CAMEO 11. 1 C x/=2 only if x D 1/. 2. 1 C x/=2 x and multiply both sides by a. 1. 2). 3. 2. 2), expand and simplify the inequality p . a b/2 0 (which is obviously true since squares are never negative). 3. 2) are equivalent. 4. x 3 C 1=x 3 / for x > 0. x C 1=x/3 and b D x 3 C 1=x 3 .
115, and C n2 /”, College Mathematics D. 12 C 22 C Journal, 22 (1991), p. 124. CAMEO 16. -K. Siu, “Proof without words: Sum of squares,” Mathematics Magazine, 57 (1984), p. 92. 5: A. n C 1//2 =4” Mathematics Magazine, 62 (1989), p. 259, and W. Lushbaugh, Mathematical Gazette, 49 (1965), p. 200. 6: S. Golomb, “A geometric proof of a famous identity,” Mathematical Gazette, 49 (1965), pp. 198–200. CAMEO 17 Summation by parts Summation by parts is a deceptively simple yet remarkably powerful method for computing certain sums in calculus, and can be used in higher level courses as well.
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## Tower Of Hanoi
##### Tower of Hanoi
The Tower of Hanoi is a classic puzzle in which disks are moved from one peg to another, without ever placing a larger disk on top of a smaller one.
192
Nov 14
##### Tower of Hanoi
The Tower of Hanoi is a puzzle in which the objective is to move a stack of disks from one rod to another, obeying certain rules. This problem asks for the sequence of moves required to solve the puzzle for a given number of disks.
193
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##### Solving the Tower of Hanoi Problem
The goal of the Tower of Hanoi problem is to move all disks from the first tower to the third tower, such that each disk is on top of a larger disk. You can only move one disk at a time, and you can only move a disk to the top of another tower if that tower has no disks on it, or if the top disk on that tower is larger than the disk you are trying to move.
186
Nov 13
##### Tower of Hanoi Algorithm
The goal of the Tower of Hanoi algorithm is to move a stack of disks from one peg to another, without ever placing a larger disk on top of a smaller disk. The input will consist of the starting peg, the goal peg, and the number of disks. The output should consist of a series of instructions on how to move the disks from the starting peg to the goal peg .
185
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##### Tower of Hanoi
The Tower of Hanoi problem is a classic algorithmic puzzle that can be solved using a simple recursive algorithm. Given a stack of n disks, the algorithm moves the disks from the first stack to the last stack, using only three stacks and moving only one disk at a time. The disks must be moved so that the smallest disk is on top of the largest disk.
128
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Formula Functions Reference - Using JMP 12 (2015)
# Using JMP 12 (2015)
### Appendix A. Formula Functions Reference
Descriptions of Functions in the Formula Editor
You can add functions to a formula. All of these functions are organized in the function browser, which groups collections of functions and features in lists organized both alphabetically (Functions (all)) and by topic (Functions (grouped)). This chapter gives a description of functions in the Formula Editor.
Scripting Index describes all functions and their arguments, demonstrates how the functions work, and links to online Help. In JMP, select Help > Scripting Index to view this interactive resource.
•The Scripting Guide also provides the arguments for all JMP functions, not just those available in the Formula Editor. In JMP, select Help > Books > Scripting Guide to open a PDF of the Scripting Guide.
Figure A.1 Functions in the Formula Editor
For instructions on how to create a formula that contains a function, see “Create a Formula” in the “Formula Editor” chapter.
Contents
Row Functions
Numeric Functions
Transcendental Functions
Trigonometric Functions
Character Functions
Character Pattern Functions
Comparison Functions
Conditional Functions
Probability Functions
Discrete Probability Functions
Statistical Functions
Random Functions
Date Time Functions
Row State Functions
Assignment Functions
Parametric Model Functions
Finance Functions
Row Functions
Adding a row function to a formula lets you reference specific rows or cells within specific rows. You can also insert values based on an arithmetic sequence. See the Scripting Guide for details about syntax.
Sequence
Produces an arithmetic sequence of numbers across the rows in a data table, where the start value, ending limit, and increment are specified as arguments.
Count
Creates a list of values beginning with the from value and ending with the to value. The number of steps specifies the number of values in the list between and including the from and to values. Each value determined by the first three arguments of the count function occurs consecutively the number of times that you specify with the times argument. When the to value is reached, Count starts over at the from value.
Also, you can add the times argument with the insert button on the keyboard. This argument is one by default, but repeats the count process as many times as you specify, as illustrated by the Count4 column in the data table in Figure A.2. To add any argument to the Count function, highlight the argument preceding the one that you want to enter. Either type a comma or use the insert button on the Formula Editor keypad.
The columns in the data table below result from the following formulas:
Count (1, 9, 2) gives Count 1
Count (1, 9, 3) gives Count 2
Count (1, 9, 9) gives Count 3
Count (1, 9, 3, 3) gives Count 4
Figure A.2 Example of the Count Function
The Count function is useful for generating a column of grid values. For example, the following formulas create a square grid of increment NRow(). NRow() is the Row function that gives the total number of rows in the data table) and axes that range from –5 to 5:
Count (–5, 5, Root(NRow()))
Count (–5, 5, Root(NRow()), Root(NRow()))
Lag
Returns the value of the first argument in the row defined by the current row less the second argument. The default Lag is one, which you can change to any number. The value returned for any lag that identifies a row number less than one is missing. Note that Lag(X, n) gives the same result as the subscripted notation, XRow( )–n.
Dif
Returns the difference between the value of the first argument in the current row and its value in the row defined by the current row less the second argument. The default Dif is one, which you can change to any number. Note that Dif(X, n) gives the same result as XRow()–XRow()-n, or asXRow()–Lag(X, n).
Subscript
Enables you to use a column’s value from a row other than the current row. After choosing Subscript from the list, enter a numeric expression into the subscript argument. Subscripts that evaluate to nonexistent row numbers produce missing values. Column names with no subscript refers to the current row. To remove a subscript, select the subscript and delete it. Then delete the missing box.
The formula CountRow() – CountRow()–1, where Row() is the row number as described below, uses subscripts to calculate the difference between each pair of values from the column named Count. This result is the same as that given by the Dif() function. When Row() is 1, the computation produces a missing value.
The formula below calculates a column called Fib, which contains the terms of the Fibonacci series (each value is the sum of the two preceding values in the calculated column).
It shows the use of subscripts to do recursive calculations. A recursive formula includes the name of the calculated column, subscripted such that it references only previously evaluated rows (rows 1 through (i–1)). The calculation of the Fibonacci series shown includes a conditional expression and a comparison. See the sections “Conditional Functions”, and “Comparison Functions”, for details.
Row
Returns the current row number when an expression is evaluated for that row. You can use Row() in any expression, including column name subscripts. The default subscript of a column name is Row() unless otherwise specified.
NRow
Returns the total number of rows in the active data table.
Numeric Functions
You can create a formula that contains arithmetic operators that are commonly used in formulas. See the Scripting Guide for details about syntax.
Abs
Returns a positive number of the same magnitude as the value of its argument. For example, |5| and |–5| both result in 5.
Modulo
Returns the remainder when the second argument is divided into the first. For example, Modulo(6, 5) results in 1.
Ceiling
Returns the smallest integer greater than or equal to its argument. For example, Ceiling(2.3) results in 3, while Ceiling(–2.3) results in –2.
Floor
Returns the largest integer less than or equal to its argument. For example, Floor(2.7) results in 2, but Floor(–0.5) results in –1.
Round
Rounds the first argument to the number of decimal places given by the second argument. For example, Round(3.554, 2) rounds to 3.55 and Round(3.555, 2) rounds to 3.56.
Transcendental Functions
You can create a formula that supports transcendental functions, such as logarithmic functions for any base, functions for combinatorial calculations, the Beta function, and several gamma functions. See the Scripting Guide for details about syntax.
Exp
Raises e to the power that you specify. Thus, Exp(1) = e.
LnZ
Calculates the natural logarithm of x, except returns 0 when x is 0; for use with derivatives.
Log and Log10
Calculates the natural logarithm (base e). To change the default base, highlight the argument and type a comma or click the insert key on the keypad. The base appears and is editable. The Log argument can be any numeric expressions. The expression Log(e) evaluates as 1, and Log2(32) is 5. The Log10 function calculates the logarithm of base 10 only.
Log1P
Returns a more accurate calculation of Log(1+x) when x is very small.
Squash
Computes the function 1 / (1 + ex), where x is any numeric column, variable, or expression.
Logist
Also known as Squish or Logistic, is an efficient computation of the function 1 / (1+e-x), where x is any numeric column, variable, or expression.
Root or Square Root
Calculates the root of its argument as specified by the index. Root initially shows with an index of 2. To change the index, highlight the index argument and enter the value that you want.
Factorial
Returns the product of all numbers 1 through the argument that you specify. For example, Factorial(5) evaluates as 120.
NChooseK
Returns the number of n things taken k at a time (n select k) and is computed in the standard way using factorials, as n! / (k!(nk)!). For example, NChooseK(5,2) evaluates as 10.
Beta
Adds the two parameter Beta function and is written terms of the Gamma function as:
Gamma
Adds the Gamma function, denoted Γ(i), and is defined as:
Gamma with a single argument is the same as Gamma(x, infinity). The optional second argument changes the upper integer from infinity to the value that you enter. Other interesting gamma function relationships are
•for any α > 1, Γ(α) = (α–1) • Γ(α–1)
•for any positive integer, n, Γ(n) = (n-1)!
•Γ(0.5) = the square root of π
LGamma
Is the natural log of the result of the gamma function evaluation. You get the same result using the Log (natural log) function with the Gamma function. However, the LGamma function computes more efficiently than do the Log (natural log) and the Gamma functions together.NChooseK is implemented using LGamma functions. The result is not always an exact integer. If the result is close to an integer, it is rounded up using the Floor function.
Digamma
The logarithmic derivative of the Gamma function.
Trigamma
The derivative of the Digamma function, or the logarithmic second derivative of the Gamma function.
Arrhenius
Calculates the non-specific component of the Arrhenius relationship that is then multiplied by the activation energy in the Arrhenius equation.
Arrhenius Inv
The inverse of the Arrhenius function:
Logit
Applies the logit transformation to the argument using:
Logit Percent
Calculates the logit as a percent for the argument.
Logist Percent
Calculates the logistic as a percent for the argument.
Scheffe Cubic
Is used in fitting certain models. Scheffe Cubic (X1, X2) is equivalent to X1*X2*(X1-X2).
Trigonometric Functions
You can create a formula that supports transcendental functions, such as logarithmic functions for any base, functions for combinatorial calculations, the Beta function, and several gamma functions. See the Scripting Guide for details about syntax.
Sine, Cosine, Tangent
The Sine and Cosine functions calculate the sine and cosine of their respective arguments given in radians. For example, the expression Sine(0) evaluates as 0, and Cosine(0) evaluates as 1. The tangent function calculates the tangent of an argument given in radians. The expressionTan(Pi()/4) evaluates as 1.
ArcSine, ArcCosine, ArcTangent
The ArcSine and ArcCosine functions return the inverse sine and inverse cosine of their respective arguments. The returned value is measured in radians. For example, both expressions ArcSine(1) and ArcCosine(0) evaluate as 1.57080. The ArcTangent function returns the inverse tangent of its argument. The returned value is measured in radians. The expression ArcTangent(1) evaluates as 0.78540 (=3.14159/4).
SinH, CosH, TanH
The SinH and CosH functions return the hyperbolic sine and hyperbolic cosine of their respective arguments. The expression SinH(1) evaluates as 1.175201, and CosH(0) evaluates as 1.0. The TanH function returns the hyperbolic tangent of its argument. The expression TanH(1)evaluates as 0.761594.
ArcSinH, ArcCosH, ArcTanH
The ArcSinH and ArcCosH functions return the inverse hyperbolic sine and inverse hyperbolic cosine of their respective arguments. The expression ArcSinH(1) evaluates as 0.881374, and ArcCosH(1) is 0. The ArcTanH function returns the inverse hyperbolic tangent of its argument. The expression ArcTanH(0.5) evaluates as 0.549306.
Character Functions
You can create a formula that accepts character arguments or returns character strings and converts the data type of a value from numeric to character, or character to numeric. When you create these formulas, note that:
•Character functions can result in either character or numeric data. If you calculate a data type different from the one specified, the data type of the computed column is automatically changed to match the result.
•Arguments that are literal character strings must be enclosed in quotation marks.
See the Scripting Guide for details about syntax.
Char
Produces a character string that corresponds to the digits in its numeric argument. For example, Char(1.123) evaluates as 1.123. See the Scripting Guide, for details.
Collapse Whitespace
Trims leading and trailing whitespace and replaces interior whitespace with single space. That is, if more than one white space character is present, the Collapse Whitespace command replaces the two spaces with one space.
Concat ||
Concatenates character strings to produce a new string with the function’s second character argument appended to the first. For example, "Dr." || " " || name produces a new string consisting of the title Dr. followed by a space and the contents of the name string. (See also “Concat Items”.)
Contains
Returns the numeric position within the first argument of the first instance of the second argument, if it exists. The second argument can contain one ore more characters. If the second argument does not exist, Contains returns a zero. For example, Contains("Veronica Layman", "ay")evaluates as 11. Contains("Lillie Layman", "L") evaluates as 1. The third argument is optional and is a numeric value that specifies the starting position. If offset is negative, Contains searches backward from offset from the end of the string.
Munger
Computes new character strings from existing strings by inserting or deleting characters. It can also produce substrings, calculate indices, and perform other tasks depending on how you specify its arguments. The Munger function treats uppercase and lowercase letters as different characters.
Text is a character expression. Munger applies the other three arguments to this string to compute a result.
Offset is a numeric expression indicating the starting position to search in the string. If Offset is greater than the position of the first instance of the find argument, the first instance is disregarded.
Find/Length is a character or numeric expression. Use a character string as search criterion, or use a positive integer to return that number of consecutive characters starting from the Offset position. If you specify a negative integer as the Length value, Munger returns all characters from the Offset through to the end of the string.
Replace (optional argument) can be a string or unspecified. If it is a string and the Find/Offset value is numeric, Munger replaces the search criterion with the Replace string to form the result. If the Find/Offset value is numeric and no string is specified, Munger calculates a substring. If the Find/Length value is a character string, Munger always returns the numeric offset, disregarding the Replace value if it exists. To insert the Replace argument, click any argument in the Munger function and then click the insert button. Use the delete key on your keyboard or the delete button () on the Formula Editor keypad to remove the Replace argument.
Lowercase, Uppercase
The Lowercase function converts any uppercase character found in its argument to the equivalent lowercase character. For example, Lowercase("VERONICA LAYMAN") evaluates as veronica layman. The Uppercase function converts any lowercase character found in its argument to the equivalent uppercase character. For example, Uppercase("Veronica Layman") evaluates as VERONICA LAYMAN.
Length
Calculates the length of its argument. For example, Length("Veronica") evaluates as 8. If the argument is
•a string, length returns the number of characters;
•a list, length returns the number of items in the list;
•a blob (binary object), the number of bytes.
Num
Produces a numeric value that corresponds to its character string argument when the character string consists of numbers only. If a character string contains a non-numeric value, the result is a missing value. For example, Num(“1.123”) evaluates as 1.123.
Substr
Extracts the characters that are the portion of the first argument. Begins at the position given by the second argument, and ends based on the number of characters specified in the third argument. The first argument can be either a character column or a literal value. The starting argument and the length argument can be numbers of expressions that evaluate to numbers. For example, to show the first name only, Substr("Veronica Layman", 10, 6) starts at position 10 and reads through position 15, which yields Layman.
If start is negative, Substr searches backward from start from the end of the string. If length is negative or absent, Substr returns a string that begins with start and continues to the end of the text string.
Substr can also be used with lists.
Titlecase
Converts the string to title case, that is, an initial uppercase character and subsequent lowercase characters. For example, Titlecase(“Veronica Layman”) results in Veronica layman.
Trim
Produces a new character string from its argument, removing any leading and trailing whitespace. The second argument determines if whitespace is removed from the left, the right, or both ends of the string. If no second argument is used, whitespace is removed from both ends. For example, Trim("john ") evaluates as john. Trim(" john ", both) also evaluates as john.
Word
Extracts the nth word from a character string. One or more spaces define where each word begins and ends unless the optional delimiters argument is specified. For example, Word(2, "Veronica Layman") returns the word Layman.
To insert the delimiters argument, click on any argument in the Word function and then click the insert button on the Formula Editor keypad. Use the delete key on your keyboard or the delete button on the Formula Editor keypad to remove the delimiters argument. If you do not specify a delimiter, space is used as the delimiter. If you define the delimiter as an empty string, each character is treated as a separate word.
Most special characters act as single delimiters. You can enter any character or set of characters to act as a word delimiter. For example, to extract the last name in the following example, use a comma and blank together as the delimiting characters and ask for the first word. Word(1, "Layman, Veronica", ", ") returns the word Layman.
Words
Extracts the words from text according to the delimiters listed in the optional second argument. The default delimiter is space. For example, Words("the quick brown fox") returns {"the","quick","brown","fox"}.
If you include a second argument, any and all characters in that argument are taken to be delimiters. For example, Words("Doe, Jane P.",", .") returns {"Doe","Jane","P"}.
To insert the delimiters argument, click on any argument in the Words function and then click the insert button on the Formula Editor keypad. Use the delete key on your keyboard or the delete button on the Formula Editor keypad to remove the delimiters argument. If you do not specify a delimiter, white space is used as the delimiter. If you define the delimiter as an empty string, each character is treated as a separate word.
Left, Right
Returns a substring of the left-most or right-most n characters of the string text, respectively. Both functions also work with lists.
Starts With, Ends With
Returns 1 if whole begins or ends with part, respectively. Returns 0 otherwise. Both functions also work with lists.
Item
Is different than the Word function because of the way it treats word delimiters. If a delimiter is found multiple times, or you enter a delimiter with multiple characters, the Word function treats them as a single delimiter. The Item function uses each delimiter to define a new word position. To compare, suppose a name is of the form lastname, firstname. The delimiter is a comma followed by a blank, such as:
Item(2, "Layman, Veronica", ", ")
Word(2, "Layman, Veronica", ", ")
The Item function returns a missing value because it treats the comma and blank separately and finds nothing between them. The Word function treats the comma and blank as a single delimiter and finds Veronica as the second word.
If you do not specify a delimiter, white space (blank space) is used as the delimiter. If you define the delimiter as an empty string, each character is treated as a separate item.
Char to Hex, Hex, Hex to Char, Hex to Number
Converts between Hex and other formats.
Hex returns the hexadecimal representation of its argument. If the argument is character (in quotes), then the result is a character string twice as long containing the hexadecimal codes for the character values. For example, Hex("A") returns the string 41.
If the argument is numeric and “integer” is specified, the Hex function returns an 8-hexadecimal-character representation of the integer returned. For example, Hex(12, “integer”) returns the string 0000000C.
Hex to Char converts hexadecimals to characters. The resulting character string might not be valid display characters. All the characters must be in pairs, in the ranges 0-9,A-Z, and a-z. Blanks and commas are allowed and skipped.
Char to Hex converts characters to hexadecimals.
Hex to Number converts hexadecimals to numbers.
For details, see the Scripting Guide book.
Repeat
Creates a string that is the first argument repeated the number of times specified by the second argument. The first argument can be either a character literal, a character variable, or a character expression. For example, Repeat(“Katie”, 3) creates KatieKatieKatie.
A third argument applies when Repeat is used in a JSL script to repeat a matrix. When the first argument is a matrix, the second argument is the rowwise repeat and the third argument is the columnwise repeat.
Insert, Insert Into
Insert inserts a new item into the list or expression at the given position. If position is not given, it is inserted at the end.
Insert Into is the same as insert, but it inserts in place.
Remove, Remove From
Remove the character(s) at the indicated position. If n is omitted, the item at position is deleted. If position and n are omitted, the item at the end is removed. There are three possible arguments: the string, followed by the position, followed by the number of characters to be removed.
Remove From returns items removed in place. The function returns the removed item(s), but you do not have to assign them to anything. The first argument is a variable name, followed by the position, followed by the number of characters to be removed.
Shift, Shift Into
Shift shifts an item or n items from the front to the back of the list or expression. Shifts items from back to front if n is negative. Shift Into shifts items in place.
Reverse, Reverse Into
Reverse reverses the characters in the string. Reverse Into reverses the characters in place.
Concat Items
Concat Items converts a list of string expressions into one string, with each item separated by a delimiter. The delimiter is a blank, if unspecified.
Substitute, Substitute Into
The first argument is a string, the second is a pattern, and the third is a replacement string. Substitute finds all matches to the pattern in the string, and replaces them with the replacement string. Substitute Into does the same substitution in place.
Regex
The first argument is the source string that Regex searches for a match to the pattern. The second argument is the pattern, in the form of a regular expression. The Formula Editor prompts you for these two required arguments.
By default, Regex performs a case-sensitive search and returns the parts of the source string that match the pattern that you specified (or returns MISSING if the match fails). There are two optional arguments that you can add. You can type a third argument—the format—that specifies the string to return. If you choose, you can use regular expressions to specify replacement text in the returned string. If you specify the third argument, you can also specify IGNORECASE so that Regex ignores capitalization when searching the source string for a match.
Table A.1 Regex Examples Sample Regex function String that is returned Regex( "@ q3 #", "([a-z])([0-9])" ) q3 The function is case sensitive, so q3 matches but Q3 would not. Regex( "@ Q3 #", "([a-z])([0-9])", "\0",IGNORECASE) Q3 Although \0 is the default argument, it is required in this example so that IGNORECASE can be specified. Regex( "@ Q3 #", "([a-z])([0-9])", "\2\1",IGNORECASE) 3Q
For more information and an example that you can run, select Help > Scripting Index and do a search for Regex.
XPath Query
XPath Query parses a valid XML document for the expression that you specify. For an example, select Help > Scripting Index and search for the function.
Hex to Blob, Char to Blob, Blob to Char
Hex to Blob converts the hexadecimal to a blob (Binary Large Object).
Char to Blob converts the string to a blob. You can specify the encoding in an optional second argument. Supported encodings are: utf-8, utf-16le, utf-16be, us-ascii, iso-8859-1, and ascii~hex.
Blob to Char converts the blob to a string. You can specify the encoding in an optional second argument. Supported encodings are: utf-8, utf-16le, utf-16be, us-ascii, iso-8859-1, and ascii~hex.
Character Pattern Functions
These functions provide powerful pattern matching abilities. Pattern matching is a flexible method for searching and manipulating strings, and regular expressions are also supported. When you create these formulas, note that:
•First, you define a pattern with one more of the character patterns.
•Then, you use Pat Match to compare a string to the pattern.
Pat Match returns True (1) if the pattern is found in the string, or it returns False (0) if the pattern was not found in the string.
•To use regular expressions instead of patterns, use Regex Match.
For complete details, see the Scripting Guide.
Pat Any
Constructs a pattern that matches a single character in the argument.
Pat Not Any
Constructs a pattern that matches a single character that is not in the argument.
Pat Break
Constructs a pattern that matches zero or more characters that are not in its argument; it stops or breaks on a character in its argument. It fails if a character in its argument is not found. In particular, it fails to match if it finds the end of the source string without finding a break character.
Pat Span
Constructs a pattern that matches one or more (not zero) occurrences of characters in its argument. It is greedy; it always matches the longest possible string. It fails rather than matching zero characters.
Pat String
Constructs a pattern that matches its string argument.
Pat Len
Constructs a pattern that matches n characters.
Pat Pos
Constructs patterns that match the null string if the current position is int from the left end of the string, and fail otherwise.
Pat R Pos
Constructs patterns that match the null string if the current position is int from the right end of the string, and fails otherwise.
Pat Tab
Constructs a pattern that matches forward to position int in the source string. It can match 0 or more characters. It fails if it would have to move backwards or beyond the end of the string.
Pat R Tab
Constructs a pattern that matches up to position n from the end of the string. It can match 0 or more characters. It fails if it would have to move backwards or beyond the end of the string.
Pat Test
Constructs a pattern that succeeds and matches the null string if expr is not zero and fails otherwise.
Pat At
Constructs a pattern that matches the null string and stores the current position in the source string into the specified JSL variable (varName). The assignment is immediate, and the variable can be used with expr() to affect the remainder of the match.
Pat Rem
Constructs a pattern that matches the remainder of the string. It is equivalent to patRTab(0).
Pat Arb
Constructs a pattern that matches an arbitrary string. Initially it matches the null string. It matches one additional character each time the pattern matcher backs into it.
Pat Succeed
Constructs a pattern that always succeeds, even when the matcher backs into it. It matches the null string.
Pat Fail
Constructs a pattern that fails whenever the matcher attempts to move forward through it. The matcher backs up and tries different alternatives. If and when there are no alternatives left, the match fails and Pat Match returns 0.
Pat Abort
Constructs a pattern that immediately cancels the pattern match. The matcher does not back up and retry any alternatives. Conditional assignments are not made. Immediate assignments that were already made are kept.
Pat Fence
Constructs a pattern that succeeds and matches the null string when the matcher moves forward through it, but fails when the matcher tries to back up through it. It is a one-way trap door that can be used to optimize some matches.
Pat Arb No
Constructs a pattern that matches zero or more copies of pattern.
Pat Repeat
Matches pattern between minimum and maximum times.
Pat Conditional
Saves the result of the pattern match, if it succeeds, to a variable named as the second argument (type) after the match is finished.
Pat Immediate
Saves the result of the pattern match to a variable named as the second argument (varName) immediately.
Pat Altern
Constructs a pattern that matches any one of the pattern arguments.
Pat Concat
Constructs a pattern that matches each pattern argument in turn.
Pat Regex
Constructs a pattern that matches the regular expression in the quoted string argument.
Pat Match
Pat Match executes a pattern match using the source in the first argument and the pattern in the second argument. The pattern must be constructed first, either inline or by assigning it to a JSL variable elsewhere. A third argument, if present, is the replacement text for the matched characters in the source argument (if the source argument is a variable). Pat Match returns true if the match succeeds. Additional arguments, in any order, are ANCHOR (match must begin at start of source), FULLSCAN (turn off some optimizations for special situations), and MATCHCASE (by default, A == a).
Pat Match returns true or false rather than a string, so Pat Match is somewhat difficult to use in a formula. You might find the Regex function (“Regex”) easier to use when you are adding pattern-matching formulas in the Formula Editor.
Regex Match
Regex Match is similar to Pat Match. Regex Match executes a pattern match using the source in the first argument and the pattern in the second argument. Regex Match uses a regular expression for the second argument and returns a list of information about the result of the match.
A simpler function, Regex (“Regex”), is also available. Regex returns a string value rather than a list, so Regex is usually easier to use in the Formula Editor than RegEx Match.
Comparison Functions
You can create a formula that compare the values of two arguments by using the comparison function. Each comparison relationship evaluates as true or false based on numeric magnitudes or character rankings. A true relationship evaluates as one, and false evaluates as zero.
Comparisons are useful when you include them in conditional expressions, but they can also stand alone as numeric expressions if neither term in comparison is missing. A relational symbol’s arguments can be any two expressions. However, both arguments in a comparison function must be of the same data type. Also note that:
•JMP displays an error if you use a single “=” in a conditional where “==” is expected.
•The Formula Editor uses the International Utilities package when comparing character strings. This package contains different rankings for each international character set and takes diacritical marks into consideration.
•You should not use comparison operators to specifically compare to a missing value. Instead, use the Is Missing function to detect a missing value.
See the Scripting Guide for details about syntax.
<
Less than
>
Greater than
<=
Less than or equal to
>=
Greater than or equal to
==
Equal to
!=
Not equal to
a<b<=c
b is greater than a and less than or equal to c
a<=b<c
b is greater than or equal to a and less than c
Is Missing
Returns a one (1) if the value of the argument for the current row is missing, and a zero if the value is not missing. The Formula Editor excludes missing numeric values from its statistical calculations.
Conditional Functions
You can include conditional expressions (called conditionals for short) in your formulas. These expressions let you build a sequence of clauses paired with result expressions. Constructing a sequence of clauses is the way you conditionally assign values to cells in a calculated column. Conditionals follow these rules:
•When no clause is true, the Formula Editor evaluates the result expression that accompanies the else clause.
•All result expressions in a conditional expression must evaluate to the same data type.
•A missing term matches any data type.
•By definition, expressions that evaluate as zero are false.
•If an expression evaluates as missing, no clauses are executed and missing is returned. All other numeric expressions are true.
See the Scripting Guide for details about syntax.
Use the insert and delete clause buttons on the Formula Editor panel to expand the expression. For maximum efficiency, list the most frequently evaluated clause and result pairs first in the sequence.
Note: Interpolate, Step, For, and While are most often used in conjunction with other commands to build a JSL script. You can use the Formula Editor to create and execute a script in that column, but this is not recommended because of dependencies and ambiguities that can result. Most often, scripts are stored as .jsl files, and can be saved with a data table as a table property. For details about table properties, see “Table Panel” in the “Get Started” chapter. For documentation of all scripting commands, see the Scripting Guide.
If
Shows a single If condition with a missing expression and a missing then clause. Highlight either expr or then clause and enter a value. For example, to calculate count as a percentage of total when total is not 0, enter the conditional expression (using columns called count and total) in Figure A.3.
Figure A.3 A Conditional Expression
To add a new condition to the If conditional, highlight then clause and click the insert button () on the Formula Editor keypad. Initially, this changes the existing else condition to an expr clause. Click the insert button again to add an else clause. Highlighting then or else and repetitively clicking the insert button changes the else to expr or adds a new expr clause.
To delete a clause, select the then clause above it and press the delete key on your keyboard or click the delete button () on the Formula Editor keypad.
By definition, expressions that evaluate as zero are false. If an expression evaluates as missing, no clauses are executed and missing is returned. All other numeric expressions are true.
Match
Compares an expression to a list of clauses and returns the value of the resulting expression for the first matching clause encountered. You provide the matching expression only once and then give a match value for each clause.
After you select Match in the Formula Editor, a list appears with two options:
•Select Add Match Arguments from Data, and clauses that correspond to all of the levels in your data are added automatically. Alternatively, hold down the SHIFT key, select Conditional, and then select Match. In Figure A.4, the example on the left shows clauses that were added automatically.
•Select Don’t Add so that you can add each clause individually. In Figure A.4, the example on the right shows an empty clause, which you fill with the missing expressions.
Figure A.4 Examples of Using the Match Function
In an automatically filled argument, you should highlight then clause, and then enter an expression. In an empty argument, you highlight either expr, value, or then clause, and then enter an expression. (Or, if you highlight an expression and click Match, the Formula Editor creates a new Match conditional, with the original highlighted expression as expr and nothing for the value and else clause.) Also, keep in mind that:
Match evaluates faster and uses less memory than an equivalent If because the variable is evaluated only once for each row in the data table. The If condition must evaluate the variable at each If clause for each row until a clause evaluates as true. See “Comparison Functions”, for a comparison of Match and If conditionals.
•With If and Match, the Formula Editor searches down from the top of the sequence for the first true clause and evaluates the corresponding result expression. Subsequent true clauses are ignored.
In the following example, each value is assigned depending on the value of the age variable.
Figure A.5 An Example of Using the Match Function
Note: Match ignores trailing spaces and If does not.
Although Match returns missing for any missing values, you can also specifically match missing values.
Choose
Choose is a special case of Match in which the arguments of the condition are a sequence of integers starting at one. The value of clause replaces the match condition. An example of a Choose condition is shown in Figure A.6. With Choose, the Formula Editor goes directly to the correct choice clause and evaluates the result expression.
Figure A.6 Example of a Choose Condition
When you highlight an expression and click Choose, the Formula Editor creates a new conditional expression with one clause. Use the insert () and delete () buttons on the keypad to add new clauses or remove unwanted clauses, as described previously for the If conditional.
Choose evaluates the choose expression and goes immediately to the corresponding result expression to generate the returned value. With Choose, you provide a choosing expression that yields sequential integers starting at 1 only once, and then you give a choice for each integer in the sequence.
IfMax
Evaluates the first of each pair of arguments and returns the evaluation of the result expression (the second of each pair) associated with the maximum of the expressions. If more than one expression is the maximum, the first maximum is returned. If all expressions are missing and a final result is not specified, missing is returned. If all expressions are missing and a final result is specified, that final result is returned. The test expressions must evaluate to numeric values, but the result expressions can be anything.
IfMin
Evaluates the first of each pair of arguments and returns the evaluation of the result expression (the second of each pair) associated with the minimum of the expressions. If more than one expression is the minimum, the first minimum is returned. If all expressions are missing and a final result is not specified, missing is returned. If all expressions are missing and a final result is specified, that final result is returned. The test expressions must evaluate to numeric values, but the result expressions can be anything.
And &
Evaluates as 1 when both of its arguments are true. Otherwise, it evaluates as 0. (See Figure A.9.) The formula in Figure A.7 labels Group 1 as drivers only if both comparisons are true.
Figure A.7 Creating an And Function
Or |
Evaluates as 1 when either of its arguments is true. If both of its arguments are false, then the Or expression evaluates as 0. (See Figure A.9.) The formula in Figure A.8 assigns males and all participants who are more than 13 years old to Group 1.
Figure A.8 Creating an Or Function
The truth tables on the left in Figure A.9 illustrate the results of the And ( & ) and Or (| ) functions when both arguments have nonmissing values that evaluate to true or false. The table on the right illustrates the result when either the left or right expression (call them a and b) or both have missing values.
Figure A.9 Evaluations of And and Or Expressions
Not !
Evaluates as 1 when its argument is false. Otherwise, Not evaluates as 0. When you apply the Not function, use parentheses where necessary to avoid ambiguity. For example, !(weight==64) can be either true or false (either 1 or 0), but (!weight)==64 is always false (0) becauseNot can return only 0 or 1. Expressions such as !(weight==64) can also be entered as weight != 64.
Interpolate
Linearly interpolates the y-value between two points, x1, y1 and x2, y2 that corresponds to the arguments that you give. You can insert additional pairs of x, y arguments with the insert key. Interpolate finds the pair of x, y points that correspond to the x-value and completes the interpolation.
Step
Is like Interpolate except that it returns the y-value corresponding to the greatest x-value less than or equal to the x and y arguments. That is, it finds the corresponding y for a given x from a step function rather than a linear fit between points. Like Interpolate, you can have as many x and yargument pairs as you want.
Figure A.10 Example of Interpolate
For
Repeats the statements in the body argument as long as the while condition is true. The init and next control the iterations.
While
Repeatedly tests the expr condition and executes the body until expr is no longer true.
Break, Continue
Break stops execution of a loop completely and continues to the statement following the loop. Continue ends the current iteration of a loop and begins the loop at the next iteration.
Both are used in For, While, and For Each Row loops.
Stop
Immediately stops a script that is running.
Probability Functions
You can create a formula that calculates probabilities and quantiles for statistical distributions like beta, Chi-square, F, gamma, normal, Student’s t, Weibull distributions, Tukey HSD, and so on. See the Scripting Guide for details about syntax.
Beta Density
Requires three arguments: quantile argument and the shape parameters alpha and beta. A threshold parameter (θ) and a scale parameter (σ > 0) are additional arguments. It returns the value of the beta probability density function (pdf) for the given arguments. The beta density is useful for modeling the probabilistic behavior of random variables such as proportions constrained to fall in the interval [0, 1].
Beta Distribution
The beta distribution has two shape parameters: α > 0 and β > 0. A threshold parameter (θ) and a scale parameter (σ) are additional arguments, where θ≤ x ≤θ + σ. The default value for θ is 0. The default value for σ is 1.
The beta distribution function is the inverse of the beta quantile function.
Beta Quantile
Accepts a probability argument, p, and shape and scale parameters, α > 0 and β > 0. It returns the pth quantile from the standard beta distribution. The beta quantile function is the inverse of the beta distribution function.
ChiSquare Density
Accepts a quantile argument from the range of values for the Chi-squared distribution, a degrees of freedom argument, and an optional noncentrality parameter. It returns the value of the Chi-squared density function (pdf) for the arguments.
ChiSquare Distribution
Accepts a response argument (range of x values) and three parameter arguments: a quantile, a degrees of freedom, and a noncentrality parameter. It returns the probability that an observation from the Chi-squared distribution with the specified noncentrality parameter and degrees of freedom is less than or equal to the given quantile. For example, the expression ChiSquare Distribution(11.264, 5) returns the probability that an observation from the Chi-squared distribution centered at 0 with 5 degrees of freedom is less than or equal to 11.264. The expression evaluates as 0.95361.
Furthermore, the ChiSquare Distribution function accepts integer and noninteger degrees of freedom. It is centered at 0 by default. The ChiSquare Distribution function is the inverse of the ChiSquare Quantile function.
ChiSquare Quantile
Accepts three arguments: a probability p, a degrees of freedom, and a noncentrality parameter. It returns the pth quantile from the Chi-squared distribution with the specified noncentrality parameter and degrees of freedom. For example, the expression ChiSquare Quantile(.95, 3.5, 4.5)returns the 95% quantile from the Chi-squared distribution centered at 4.5 with 3.5 degrees of freedom. The expression evaluates as 17.50458.
The ChiSquare Quantile function accepts integer and noninteger degrees of freedom. It is centered at 0 by default. The ChiSquare Quantile function is the inverse of the ChiSquare Distribution function.
Dunnett P Value
Returns the p-value from Dunnett’s multiple comparison test.
Dunnett Quantile
Returns the quantile needed in Dunnett’s multiple comparison tests.
F Density
Accepts a quantile argument from the range of values for the F-distribution, numerator and denominator degrees of freedom arguments, and an optional noncentrality parameter. It returns the value of the F-density function (pdf) for the arguments.
F Distribution
Accepts four arguments: a quantile, a numerator and denominator degrees of freedom, and a noncentrality parameter. It returns the probability that an observation from the F-distribution with the specified noncentrality parameter and degrees of freedom is less than or equal to the given quantile. For example, the expression F Distribution(3.32, 2, 3) returns the probability that an observation from the central F-distribution with 2 degrees of freedom in the numerator and 3 degrees of freedom in the denominator is less than or equal to 3.32. The expression evaluates as 0.82639.
The F-distribution function accepts integer and noninteger degrees of freedom. By default, the non-central parameter is set to 0. The F-distribution function is the inverse of the F Quantile function.
F Quantile
Accepts four arguments: a probability p, a numerator and denominator degrees of freedom, and a noncentrality parameter. It returns the pth quantile from the F-distribution with the specified noncentrality parameter and degrees of freedom. For example, the expression F Quantile(0.95, 2, 10, 0) returns the 95% quantile from the F-distribution centered at 0 with 2 degrees of freedom in the numerator and 10 degrees of freedom in the denominator. The expression evaluates as 4.1028.
The F Quantile function accepts integer and noninteger degrees of freedom. By default, the non-central parameter is set to 0. The F Quantile function is the inverse of the F Distribution function.
Frechet Density
Returns the density at x of a Fréchet distribution with location mu and scale sigma.
Frechet Distribution
Returns the probability that a Fréchet distribution with location mu and scale sigma is less than x.
Frechet Quantile
Returns the quantile associated with a cumulative probability p for a Fréchet distribution with location mu and scale sigma.
Gamma Density
Requires a quantile argument. Also accepts an optional alpha shape parameter, which must be greater than zero and defaults to 1. A scale parameter b, which must be greater than zero and defaults to 1, is optional. A threshold parameter, which must be in the range -∞ < θ < +∞ and defaults to zero, is optional.
Figure A.11 shows the shape of gamma probability density functions for shape parameters of 1, 3, and 5. The standard gamma density function is strictly decreasing when α (shape) ≤1. When α > 1 the density function begins at zero when x is θ, increases to a maximum, and then decreases.
Figure A.11 Gamma Density Example
Gamma Distribution
Is based on the standard gamma function, and accepts a single argument with a quantile value. The shape, scale, and threshold parameters are optional, with defaults as described previously in the discussion of the Gamma Density function. It returns the probability that an observation from a standard gamma distribution is less than or equal to the specified x. The Gamma Distribution function is the inverse of Gamma Quantile function.
Gamma Quantile
Accepts a probability argument p, and returns the pth quantile from the standard gamma distribution with the shape parameter that you specify. The Gamma Quantile function is the inverse of the Gamma Distribution function.
LEV Density
Returns the density at x of the largest extreme value distribution with location mu and scale sigma.
LEV Distribution
Returns the probability that the largest extreme value distribution with location mu and scale sigma is less than x.
LEV Quantile
Returns the quantile associated with a cumulative probability p of the largest extreme value distribution with location mu and scale sigma.
Logistic Density
Returns the density at x of a logistic distribution with location mu and scale sigma.
Logistic Distribution
Returns the probability that the logistic distribution with location mu and scale sigma is less than x.
Logistic Quantile
Returns the quantile associated with a cumulative probability p of the logistic distribution with location mu and scale sigma.
Loglogistic Density
Returns the density at x of the loglogistic distribution with location mu and scale sigma.
Loglogistic Distribution
Returns the probability that the loglogistic distribution with location mu and scale sigma is less than x.
Loglogistic Quantile
Returns the quantile associated with a cumulative probability p of the loglogistic distribution with location mu and scale sigma.
Lognormal Density
Returns the density at x of the lognormal distribution with location mu and scale sigma.
Lognormal Distribution
Returns the probability that the lognormal distribution with location mu and scale sigma is less than x.
Lognormal Quantile
Returns the quantile associated with a cumulative probability p of a lognormal distribution with location mu and scale sigma.
Normal Density
Accepts an argument from the range of values for the standard normal distribution, which is all real numbers. It returns the value of the standard normal probability density function (pdf) for the argument. For example, you can create a column of values (X) with the formula count(-3, 3, nrow()). In a second column, insert the formula Normal Density(X) to generate density values. Then select Graph > Graph Builder to plot the normal density by X.
Normal Distribution
Accepts an argument x from the range of values for the standard normal distribution, which is all real numbers. It returns the probability that an observation from the standard normal distribution is less than or equal to x. For example, the expression Normal Distribution(1.96) returns 0.975, the probability that an observation from the standard normal distribution is less than or equal to the 1.96th quantile. Also, you can specify mean and standard deviation parameters to obtain probabilities from nonstandard normal distributions. The Normal Distribution function is the inverse of the Normal Quantile function.
Normal Quantile (Probit)
Accepts a probability argument p, and returns the pth quantile from the standard normal distribution. For example, the expression Normal Quantile(0.975) returns the 97.5% quantile from the standard normal distribution, which evaluates as 1.96. Also, you can specify parameter values for the mean and standard deviation to obtain quantiles from nonstandard normal distributions. The Normal Quantile function is the inverse of the Normal Distribution function.
Normal Biv Distribution
Computes the probability that an observation is less than or equal to (x,y) with correlation coefficient r where the observation is marginally normally distributed. You can specify the mean and standard deviation for the X and Y coordinates of the observation. The default values are 0 for both means and 1 for both standard deviations.
GLog Density
Returns the density or pdf at a particular quantile q of a generalized logarithm distribution with location mu, scale sigma, and shape lambda. When the shape parameter is equal to zero, the distribution reduces to a Lognormal(mu, sigma).
GLog Distribution
Returns the probability or cdf that a generalized logarithm distributed random variable is less than q. When the shape parameter is equal to zero, the distribution reduces to a Lognormal(mu, sigma).
GLog Quantile
Returns the quantile, the value for which the probability is p that a random value would be lower. When the shape parameter is equal to zero, the distribution reduces to a Lognormal(mu, sigma).
SEV Density
Returns the density at x of the smallest extreme distribution with location mu and scale sigma.
SEV Distribution
Returns the probability that the smallest extreme distribution with location mu and scale sigma is less than x.
SEV Quantile
Returns the quantile associated with a cumulative probability p of the smallest extreme distribution with location mu and scale sigma.
t Density
Accepts a quantile argument from the range of values for the t-distribution, a degrees of freedom argument, and an optional noncentrality parameter. It returns the value of the t-density function (pdf) for the arguments. To compare a t-density with 5 df with a standard normal distribution, you can create a column of quantile values (X) with the formula count(-3, 3, nrow()). In a second column, insert the formula t Density(X). In a third column, insert the formula Normal Density(X). Then select Graph > Graph Builder to plot the t-density and the normal density by X. You will see that the t-density has slightly more spread than the normal.
t Distribution
Accepts three arguments: a quantile, a degrees of freedom, and a noncentrality parameter. It returns the probability that an observation from the Student’s t-distribution with the specified noncentrality parameter and degrees of freedom is less than or equal to the given quantile. For example, the expression t Distribution(.9, 5) returns the probability that an observation from the Student’s t-distribution centered at 0 with 5 degrees of freedom is less than or equal to 0.9. The expression is evaluated as 0.79531. t-distribution accepts integer and noninteger degrees of freedom. It is centered at 0 by default, but you can enter a value for the noncentrality parameter. The t Quantile function is the inverse of the t Distribution function.
t Quantile
Accepts three arguments: a probability p, a degrees of freedom, and a noncentrality parameter. It returns the pth quantile from the Student’s t-distribution with the specified noncentrality parameter and degrees of freedom. For example, the expression Student’s t Quantile(.95, 2.5) returns the 95% quantile from the Student’s t-distribution centered at 0 with 2.5 degrees of freedom. The expression evaluates as 2.558219. The t Quantile function is the inverse of the t Distribution function. This function also accepts integer and noninteger degrees of freedom. It is centered at 0 by default, but you have the option to enter a value for the noncentrality parameter. The t Distribution function is the inverse of the t Quantile function.
Weibull Density
Accepts a quantile argument from the range of values for the Weibull distribution, and optional shape, scale, and threshold arguments. The density function for a Weibull distribution with shape parameter β, scale parameter α, and threshold parameter θ is given in the Basic Analysis book. The Weibull Density function returns the value of the probability density function (pdf) for the corresponding Weibull distribution.
Weibull Distribution
Accepts a quantile argument x from the range of values for the Weibull distribution, and optional shape, scale, and threshold arguments. The density function for a Weibull distribution with shape parameter β, scale parameter α, and threshold parameter θ is given in the Basic Analysis book. The Weibull Distribution function returns the probability that an observation is less than or equal to the specified x for the Weibull distribution with the shape, scale, and threshold parameters that you specified. The Weibull Distribution function is the inverse of Weibull Quantilefunction.
The Weibull distribution has different shapes depending on the values of α (a scale parameter that affects the x direction) and β (a shape parameter). It often provides a good model for estimating the length of life, especially for mechanical devices and in biology. The two-parameter Weibull is the same as the three-parameter Weibull with a threshold of zero.
Weibull Quantile
Accepts a probability argument p, and returns the pth quantile from the Weibull distribution with the shape, scale, and threshold parameters that you specify. The Weibull Quantile function is the inverse of the Weibull Distribution function.
Cauchy Density
Accepts an argument x, which can be any real number, and optional arguments center and scale. If you do not specify values for the optional arguments, the function returns the value at x of the Cauchy probability density function (pdf) for a distribution with median 0 and third quartile 1.
If you specify values for center and scale, the function returns the value at x of the Cauchy probability function, characterized as follows:
•The optional parameter center is the median of the distribution.
•The optional parameter scale is half of the interquartile range, namely, half of the difference between the 0.75 and 0.25 quantiles.
Cauchy Distribution
Accepts an argument x, which can be any real number, and optional arguments center and scale. If you do not specify values for the optional arguments, the function returns the value at x of the Cauchy cumulative distribution function (cdf) for a distribution with median 0 and third quartile 1. If you specify values for center and scale, the function returns the value at x of the cumulative distribution function for the Cauchy distribution with median given by center and interquartile range given by twice the scale.
Cauchy Quantile
Accepts an argument prob, which can be any number between 0 and 1, and optional arguments center and scale. If you do not specify values for the optional arguments, the function returns the pth quantile, where p = prob, of a Cauchy distribution with median 0 and third quartile 1. If you specify values for center and scale, the function returns the pth quantile of a Cauchy distribution with median given by center and interquartile range given by twice the scale.
Johnson Su Distribution
Returns the probability that a Johnson Su-distributed random variable is less than x. There are four optional arguments: gamma, delta, theta, and sigma. For a description of the Johnson Su distribution and these parameters, see the Basic Analysis book.
Johnson Su Quantile
Returns the pth quantile of the Johnson Su distribution. There are four optional arguments: gamma, delta, theta, and sigma. For a description of the Johnson Su distribution and these parameters, see the Basic Analysis book.
Johnson Su Density
Returns the density at x of a Johnson Su distribution. There are four optional arguments: gamma, delta, theta, and sigma. For a description of the Johnson Su distribution and these parameters, see the Basic Analysis book.
Johnson Sb Distribution
Returns the probability that a Johnson Sb-distributed random variable is less than x. There are four optional arguments: gamma, delta, theta, and sigma. For a description of the Johnson Sb distribution and these parameters, see the Basic Analysis book.
Johnson Sb Quantile
Returns the pth quantile of the Johnson Sb distribution. There are four optional arguments: gamma, delta, theta, and sigma. For a description of the Johnson Sb distribution and these parameters, see the Basic Analysis book.
Johnson Sb Density
Returns the density at x of a Johnson Sb distribution. There are four optional arguments: gamma, delta, theta, and sigma. For a description of the Johnson Sb distribution and these parameters, see the Basic Analysis book.
Johnson Sl Distribution
Returns the probability that a Johnson Sl-distributed random variable is less than x. There are four optional arguments: gamma, delta, theta, and sigma. For a description of the Johnson Sl distribution and these parameters, see the Basic Analysis book.
Johnson Sl Quantile
Returns the pth quantile of the Johnson Sl distribution. There are four optional arguments: gamma, delta, theta, and sigma. For a description of the Johnson Sl distribution and these parameters, see the Basic Analysis book.
Johnson Sl Density
Returns the density at x of a Johnson Sl distribution. There are four optional arguments: gamma, delta, theta, and sigma. For a description of the Johnson Sl distribution and these parameters, see the Basic Analysis book.
Tukey HSD Quantile
Accepts a probability argument 1-alpha, and returns the 1-alphath quantile from Tukey’s HSD test for the parameters that you specify. The alpha argument is the significance level that you want. nGroups is the number of groups in a study. dfe is the error degrees of freedom (based on the total study sample). This is the quantile used to calculate least significant difference in Tukey’s multiple comparisons test.
Tukey HSD P Quantile
Returns the p-value from Tukey's HSD multiple comparisons test.
F Power and F Sample Size
The F Power function calculates the power from a given situation that involves an F-test or t-test, and the F Sample Size function computes the sample size. The arguments are the values that you specify for computation of a prospective power analysis. (These functions perform the same computations as if you selected DOE > Sample Size and Power. See the Design of Experiments Guide for a discussion of power and sample size.) The arguments include:
alpha The significance level that you are willing to tolerate (often 0.05).
dfh The hypothesis degrees of freedom. It is one (1) for a t-test.
dfm The model degrees of freedom (such that dfe = ndfm).
SquaredSize The squared effect size scaled by the error variance, which is used for making the noncentrality argument for the F-distribution. For this argument, use squared size = Δ22 where σ2 is the error variance. That is, use:
for a one-sample t-test
for a two-sample t-test
for a k-sample F-test
n (found only in the F Power function) The total number of observations (runs, experimental units, or samples) you expect to have. Power (in the F Sample Size function) is the probability that you want to have of declaring a significant result.
Discrete Probability Functions
Gamma Poisson Probability
Returns the probability or pmf that a gamma-Poisson distributed random variable is equal to x. In general, the gamma Poisson functions accept arguments that are the mean parameter lambda, the overdispersion parameter sigma, and the count of interest x. When the overdispersion is equal to one, the Gamma Poisson reduces to a Poisson(lambda) distribution.
Gamma Poisson Distribution
Returns the probability that a gamma-Poisson distributed random variable is less than or equal to x. In general, the gamma Poisson functions accept arguments that are the mean parameter lambda, the overdispersion parameter sigma, and the count of interest x.
Gamma Poisson Quantile
Returns the smallest integer quantile for which the cumulative probability of the Gamma Poisson (lambda, sigma) distribution is larger than or equal to p.
Binomial Distribution
Returns the probability that an observation from a binomial distribution with parameters p and n is less than or equal to k. In general, the binomial functions accept arguments that are the probability of success p (the event of interest), the number of trials n, and the number of successes k.
Binomial Probability
Computes the probability that a random variable from a binomial distribution is equal to k. In general, the binomial functions accept arguments that are the probability of success p (the event of interest), the number of trials n, and the number of successes k.
Binomial Quantile
Returns the smallest integer quantile for which the cumulative probability of the Binomial (p, n) distribution is larger than or equal to the specified probability.
Neg Binomial Distribution
Returns the probability that a negative binomially distributed random variable is less than or equal to k, where the probability of success is p, and the number of successes is n.
Neg Binomial Probability
Returns the probability that a negative binomially distributed random variable is equal to k, where the probability of success is p, and the number of successes is n.
Beta Binomial Distribution
Returns the probability or pmf that a beta binomially distributed random variable is less than or equal to x. In general, the beta binomial functions accept arguments that are the probability of success p (the event of interest), the overdispersion parameter delta, and the number of trials n. When the overdispersion parameter for the beta binomial is zero, the distribution reduces to a binomial(p, n).
Beta Binomial Probability
Returns the probability or cmf that a beta binomially distributed random variable is equal to x. When the overdispersion parameter for the beta binomial is zero, the distribution reduces to a binomial(p, n).
Beta Binomial Quantile
Returns the smallest integer quantile for which the cumulative probability of the Beta Binomial (p, n, delta) distribution is larger than or equal to the specified probability. When the overdispersion parameter for the beta binomial is zero, the distribution reduces to a binomial (p, n).
Hypergeometric Distribution
Computes the probability that a random variable from a hypergeometric distribution is less than or equal to x. The hypergeometric distribution models the total number of successes in a fixed sample drawn without replacement from a finite population. The hypergeometric functions accept as arguments the size of the population N, the total number of items with the desired characteristic in the population, K, the number of samples drawn n, and the number of successes in the sample x.
Hypergeometric Probability
Computes the probability that a random variable from a hypergeometric distribution is equal to x.
Poisson Distribution
Computes the probability that a random variable from a Poisson distribution with mean lambda is less than or equal to the count of interest. In general, Poisson functions accept an argument that is the count of interest, and lambda, the mean parameter.
Poisson Probability
Computes the probability that a random variable from a Poisson distribution with mean lambda is equal to the count of interest.
Poisson Quantile
Returns the smallest integer quantile for which the cumulative probability of the Poisson (lambda) distribution is larger than or equal to p.
Statistical Functions
There are two types of Statistical functions you can use in a formula:
•The functions with names that have the prefix Col. These functions compute statistics for a column of numbers or expressions involving columns.
•The Mean, Std Dev, Number, Sum, Quantile, Maximum, Minimum, and N Missing functions. These functions evaluate across columns or arguments. The statistic is computed for each row across the series of arguments. You can use the insert key on the on-screen keypad, or type a comma to add arguments to the functions that accept multiple arguments. When there are multiple contiguous arguments, select the function and the first argument, and then Shift-click the last argument in the group. These functions then automatically show with the complete list.
See the Scripting Guide for details about syntax.
Col Mean
Calculates the mean (or arithmetic average) of the numeric values identified by its argument. The formula Col Mean(age) calculates the average of all nonmissing values in the age column.
Col Std Dev
Measures the spread around the mean of the distribution identified by its argument. In the normal distribution, about 68% of the distribution is within one standard deviation of the mean. 95% of the distribution is within two standard deviations of the mean. 99% of the distribution is within three standard deviations of the mean.
Col Number
Counts the number of nonmissing values in the column that you specify. A missing numeric value occurs when a cell has no assigned value or is the result of an invalid operation (such as division by zero). Missing values show on the spreadsheet as a missing value mark (•). Missing character values are null character strings. In formulas for row state columns, an excluded row state characteristic is treated as a missing value.
Col N Missing
Counts the number of missing values in the column that you specify. A missing numeric value occurs when a cell has no assigned value or is the result of an invalid operation (such as division by zero). Missing values show in the data grid with a missing value character (•). Missing character values are null character strings.
Col Sum
Computes the sum of the values in its numeric argument. Missing values are ignored.
Col Minimum and Col Maximum
Takes the minimum of its numeric arguments. Col Minimum ignores missing values. Col Maximum takes the maximum of a numeric column argument and ignores missing values.
Col Quantile
Computes the value at which a specific percentage of the values is less than or equal to that value. For example, the value calculated as the 50% quantile, also called the median, is greater than or equal to 50% of the data. Half of the data values are less than the 50th quantile.
The Col Quantile function’s quantile argument represents the quantile percentage divided by 100. The 25% quantile, also called the lower quartile, corresponds to p = 0.25, and the 75% quantile, called the upper quartile, corresponds to p = 0.75.
The Formula Editor computes a quantile for a column of n nonmissing values by arranging the values in ascending order. The subscripts of the sorted column values, y1, y2,...,yn, represent the ranks in ascending order.
The pth quantile value is calculated using the formula p(n + 1), where p is the percent value and n is the total number of nonmissing values. If p(n+1) is an integer, then the quantile value is yp(n+1). If p(n + 1) is not an integer, then the value is interpolated by assigning the integer part of the result to i, assigning the fractional part to f, and applying the formula (1 – f)yi + (f)yi+1.
For example, suppose a column has values 2, 4, 6, 8, 10, 12, 14, 16, 18, and 20. The 50% quantile is calculated as 0.5(10 + 1) = 5.5.
Because the result is fractional, the 50% quantile value is interpolated as
(1 – 0.5) x 10 + (0.5) x 12 = (0.5)10 + (0.5)12 = 6 + 5 = 11
The following are example ColQuantile formulas:
ColQuantile(age, 1) Calculates the maximum age.
ColQuantile(age, 0.75) Calculates the upper quartile age.
ColQuantile(age, 0.5) Calculates the median age.
ColQuantile(age, 0.25) Calculates the lower quartile age.
ColQuantile(age, 0) Calculates the minimum age.
The ColQuantile argument can be any expression that evaluates to a value between (and including) 0 and 1. For example, the first formula in Figure A.12 calculates quantile values of age in ascending order for each row. The column then contains the interpolated values of age in ascending order in the calculated column. The second formula lists the interpolated values of age in descending order.
Figure A.12 Examples of the Quantile Function
Col Rank
Ranks each row’s value, from 1 for the lowest value to the number of non-missing columns for the highest value. Ties are broken arbitrarily.
Col Standardize
Performs the usual standardization on its numeric expression. For each row i, Col Standardize(height) is (HeightRow()–Col Mean(Height))/Col Std Dev(Height).
Mean
Calculates the arithmetic average of the nonmissing arguments that you specify. The arguments can be constants, numbers, or expressions. The Mean function initially shows with a single argument. You add arguments with the insert button () on the Formula Editor keypad or by typing a comma.
Std Dev
Computes standard deviation of the list of arguments that you specify. The arguments can be constants, numbers, or expressions. The Std Dev function initially shows with a single argument. You add arguments by clicking the insert button () on the Formula Editor keypad or by typing a comma.
Number
Counts the number of nonmissing values in the list of arguments that you specify.
Sum
Returns the sum of the arguments.
Quantile
Calculates the quantile given by its first argument for all the following arguments given.
Summation (Σ)
Evaluates for an explicit range of values in a column, as given by the summation indices. This behavior is different from all other statistical functions (except Product), which always evaluate on every row. The Summation function uses the summation notation shown in Figure A.13. To calculate a sum, replace the missing body term with an expression containing the index variable i, or an index variable that you assign. Summation repeatedly evaluates the expression for i = 1, i = 2, through i = NRow() and then adds the nonmissing results together to determine the final result.
You can replace NRow(), the number of rows in the active spreadsheet, and the index constant, i, with any expression appropriate for your formula. For example, the summation formula in Figure A.13 computes the total for each row of all revenue values for rows 1 through the current row number, filling the calculated column with the cumulative totals of the revenue column.
Figure A.13 Example of the Summation function
Product (Π)
Evaluates for an explicit range of values in a column, as given by the summation indices, as opposed to all other statistical functions (except Summation), which always evaluate on every row. Product uses the notation shown in the formulas on the right in Figure A.14. To calculate a product, replace the missing body term with an expression containing the index variable j. Product repeatedly evaluates the expression for i = 1, i = 2, through i = n and multiplies the nonmissing results together to determine the final result.
You can replace NRow(), the number of rows in the active spreadsheet and the index constant, i, with any expression appropriate for your formula.
For example, the expression second product example in Figure A.14 calculates i! (each row number’s factorial).
Figure A.14 Examples of the Product Function
Minimum and Maximum
Return the minimum and maximum value, respectively, from the list of nonmissing arguments that you specify.
N Missing
Counts the number of missing values in the list of arguments that you specify.
Desirability
Are smooth piecewise functions that are crafted to fit the control points. The minimize and maximize functions are three-part piecewise smooth functions that have exponential tails and a cubic middle.
The target function is a piecewise function. It is a scale multiple of a normal density on either side of the target (with different curves on each side), which is also piecewise smooth and fit to the control points.
Random Functions
You can create formulas that generate real numbers by effectively “rolling the dice” within the constraints of the specified distribution. Each time you click Apply in the Formula Editor window, these functions produce a new set of random numbers.
Note: Random numbers are generated using the Mersenne-Twister technique. This technique has a period length of 219937-1. For details about the generators, see Matsumoto and Nishimura (1998). The new generators are verified to pass all the DIEHARD tests as documented in Marshalled (1996).
See the Scripting Guide for details about syntax.
Random Uniform
Generates random numbers uniformly between 0 and 1. This means that any number between 0 and 1 is as likely to be generated as any other. The result is an approximately even distribution. You can shift the distribution and change its range with constants. For example,5 + Random Uniform()*20 generates uniform random numbers between 5 and 25.
Random Normal
Generates random numbers that approximate a normal distribution with a mean of 0 and standard deviation of 1 if no arguments are used, or with the mean and standard deviation entered as arguments. The normal distribution is bell shaped and symmetrical. You can also modify the Random Normal function with constants if no arguments are entered to give a normal distribution with specific mean and standard deviation. For example, the formula Random Normal()*5 + 30 generates a random normal variable with a mean of 30 and a standard deviation of 5.
Random Exp
Generates a single parameter exponential distribution for the distribution parameter lambda=1. You can modify the exponential function to use a different lambda.
For example, Random Exp()/.1 generates an exponential distribution for lambda=0.1. The exponential distribution is often used to model simple failure time data, where lambda is the failure rate.
Random Gamma
Gives a gamma distribution for the parameter, alpha, you enter as the function argument. The gamma distribution describes the time until the kth occurrence of an event. The gamma distribution can also have a scale parameter, beta. A gamma variate with shape parameter alpha and scale beta can be generated with the formula beta*Random Gamma(alpha). If 2*alpha is an integer, a Chi-squared variate with 2*alpha degrees of freedom is generated with the formula 2*Random Gamma(alpha).
Random Beta
Generates a pseudo-random number distributed Beta(alpha, beta).
Random Cauchy
Generates a Cauchy distribution with location parameter 0 and scale parameter 1. The Cauchy distribution is bell shaped and symmetric but has heavier tails than the normal distribution. A Cauchy variate with location parameter alpha and scale parameter beta can be generated with the formula alpha+beta*Random Cauchy().
Random Category
Generates a random category given an alternation of probability and result expressions (for example, Random Category(.2, "A", .3, "B", .4, "C", "D");).
Random Johnson Su
Returns a random number from the Johnson Su distribution.
Random Johnson Sb
Returns a random number from the Johnson Sb distribution.
Random Johnson Sl
Returns a random number from the Johnson Sl distribution.
Random Triangular
Generates a triangular distribution of numbers between 0 and 1, with the midpoint that you enter as the function argument. You can add a constant to the function to shift the distribution and multiply to change its span.
Random Integer
Generates a uniform distribution of integers between 1 and the argument that you enter as n1, if nothing is entered for n2. If you enter both n1 and n2 (n1<n2), Random Integer generates a uniform distribution of the integers between and including n1 and n2.
Random Binomial
Generates random numbers from a binomial distribution with parameters that you enter as function arguments. The first argument is n, the number of trials in a binomial experiment. The second argument is p, the probability that the event of interest occurs. When n is 1, the binomial function generates a distribution of Bernoulli trials. For example, n =1 and p = 0.5, give the distribution of tossing a fair coin. The mean of the binomial distribution is np, and variance is np(1 – p).
Random Negative Binomial
Generates a negative binomial distribution for the parameters that you enter as function arguments. The first parameter is the number of successes of interest (r) and the second argument is the probability of success (p). The random variable of interest is the number of failures that precede therth success. In contrast to the binomial variate, where the number of trials is fixed and the number of successes is variable, the negative binomial variate is for a fixed number of successes and a random number of trials. The mean of the negative binomial distribution is (r(1 – p))/p and the variance is (r(1 – p))/p2.
Random Beta Binomial
Returns random numbers from the beta binomial distribution for n trials with probability p and correlation or overdispersion delta.
Random Frechet
Returns a random number from a Fréchet distribution with the location mu and scale sigma.
Random Geometric
Returns random numbers from the geometric distribution with the parameter that you enter as the function argument. The parameter, p, is the probability that a specific event occurs at any one trial. The number of trials until a specific event occurs for the first time is described by the geometric distribution. The mean of the geometric distribution is (1-p)/p, and the variance is (1 – p)/p2.
Random Poisson
Generates a Poisson variate based on the value of the parameter, lambda, you enter as the function argument. Lambda is often the expected number of events occurring per unit time or unit of area. Lambda is both the mean and the variance of the Poisson distribution.
Random Gamma Poisson
Returns random numbers from the gamma Poisson distribution with parameters lambda and sigma.
Random Weibull
Returns a random number from a Weibull distribution.
Random Logistic
Returns a random number from a logistic distribution with the location mu and scale sigma.
Random Loglogistic
Returns a random number from a loglogistic distribution with the location mu and scale sigma.
Random Lognormal
Returns a Lognormal-distributed random number with location parameter mu and scale sigma.
Random GLog
Returns random numbers from the generalized logarithm distribution with parameters mu, sigma, and lambda. When lambda is equal to zero, the function returns a lognormal(mu, sigma).
Random Reset
Restarts the random number sequences with a seed that you specify.
Random LEV
Returns a random number from an LEV distribution with the specified location mu and scale sigma.
Random SEV
Returns a random number from the smallest extreme distribution with the specified location mu and scale sigma.
Col Shuffle
Selects a row number at random from the current data table. Each row number is selected only once. When Col Shuffle is used as a subscript, it returns a value selected at random from the column that serves as its argument. Each value from the original column is assigned only once as Col Shuffle’s result.
For example, to identify a 50% random sample without replacement, use the formula in Figure A.15.
Figure A.15 Formula Identifying 50% Random Sample
The formula in Figure A.15 selects half the values (n/2) from the column x and assigns them to the first half of the rows in the computed column. The remaining rows of the computed column fill with missing values.
Resample Freq
Generates a random selection with replacement frequency counts, suitable for use in bootstrapping. For example, it supports a second Freq Column argument, enabling it to do bootstrap samples relating to a pre-existing frequency column specified in the second argument. Resample Freq() generates a 100% resample. ResampleFreq(rate) generates a rate frequency sample. Resample(rate, column) generates a sample that is calculated by the rate multiplied by the sum of the specified column.
Date Time Functions
JMP stores dates and times in numeric columns using the Macintosh standard of the number of seconds since January 1, 1904. When a column has date values, you can assign a date format to that column by double-clicking a column name and selecting Date or Time from the Formatmenu. See “Numeric Format Options” in the “The Column Info Window” chapter.
See the Scripting Guide for details about syntax.
In Minutes, In Hours, In Days, In Weeks, In Years
Converts from the units of the function name to the equivalent number of seconds for the argument. The argument must be a number or numeric expression. For example, In Minutes(2) yields 120, and In Years(1) yields 31,557,600 (60 seconds * 60 minutes * 24 hours * 365.25 days).
Date DMY, Date MDY
Accepts numeric expressions for day, month, and year and return the associated JMP date. For example, Date DMY (20, 3, 1991) and Date MDY(3, 20, 1991) evaluate to 2,752,272,000.
Today
Returns the number of seconds between January 1, 1904 and the current date. For example, at midnight on March 20, 1991 (a Wednesday), the Today function returns 2752272000 (2,752,272,000 seconds) and continues counting. If you evaluate the Today function later in the day, it reflects the additional seconds.
Day, Month, Year,
Returns the day of the month, the month (as a number from 1 to 12), a four-digit year, respectively. The argument for these functions is interpreted as a JMP date. For example, on March 20, 1991:
Day(2752272000) returns the number 20.
Month(2752272000) returns the number 3.
Year(2752272000) returns the number 1991.
Quarter
Returns the annual quarter of a datetime value as an integer 1-4.
Hour, Minute, Second
Returns the hour, the minute, and the seconds of a date-time value, respectively. The argument for these functions is interpreted as a JMP date. For example, on March 20, 1991:
Hour(2752572649) returns the number 11.
Minute(2752572649) returns the number 30.
Second(2752572649) returns the number 49.
Day of Week, Day of Year, Week of Year, Time of Day
The argument for these functions is a JMP date. Day Of Week returns a number from 1 to 7, where 1 represents Sunday. Day Of Year returns the number of days from the beginning of the year. Week Of Year returns a number from 1 to 52 based on the rule specified. Rule 1 (default) has weeks start on Sunday with the first Sunday being week 2 and week 1 is a partial week or empty; rule 2 has the first Sunday begins week 1 with any previous days being week 0; rule 3 returns the ISO week number where the week starts on Monday and week 1 is the first week of the year with four days in that year. With ISO weeks, it is possible for the first or last three days of the year to belong to the neighboring year’s week number. Time Of Day returns a number from 0 to 86399 (time of day in seconds). For example, on Wednesday, March 20, 1991:
Day Of Week(2752272000) returns the number 4.
Day Of Year(2752272000) returns the number 79.
Week Of Year(2752272000) returns the number 12.
Time Of Day(2752272000) returns the number 0.
Informat
The argument for the Informat function is a date character string. For example, Informat("03/20/1991") returns the appropriate JMP date value, 2752272000. JMP can read all the date formats except for Abbrev Date and Long Date.
Abbrev Date, Long Date, Short Date
The argument for these date functions is a JMP date. They return character strings that are the formatted representation of the argument. For example:
Abbrev Date(2752272000) returns Wed, Mar 20,1991.
Long Date(2752272000) returns Wednesday, March 20, 1991.
Short Date(2752272000) returns 3/20/91.
Format
The first argument in the Format function is a JMP date. This function returns the character string representation of the date by the date format that you specify in the second argument, which is a quoted string. If you apply this formula to a numeric column, JMP automatically changes the column’s data type to character.
You can also supply a column for the first argument and leave the rest blank. The result is the formatted value of the column reference. This can be used to extract value labels of a column when the value labels are turned off.
MDYHMS
The argument of MDYHMS is a JMP date. This function shows all date and time fields, appending zeros as time fields if no time information is present. This is useful if a date column is formulated such that not all date information is displayed. The MDYHMS function can be used to see all available date and time information.
Date Increment
Adds 1 or more intervals to a starting datetime value. For example, Date Increment(Today(), "Day", 3) adds three days to the current date. Date Increment(Today(), "Year", 3) adds 3 years to the current date.
Date Difference
Returns the difference of two datetime values. The interval argument can be Second, Minute, Hour, Day, Week, Month, Quarter, Year. The alignment arguments are described here:
Start
is used to count the number of times an interval starts.
Actual
is used to count whole intervals.
Fractional
is used to count fractional intervals.
For example, the following formula returns 207.890243055556, the number of days between the dates:
Date Difference(01Jan2010:00:00:00, 27Jul2010:21:21:57, "Day", "fractional");
The following formula returns 207, the number of completed days between the dates:
Date Difference(01Jan2010:00:00:00, 27Jul2010:21:21:57, "Day", "actual");
The following formula returns 9, the number of completed hours between the times:
Date Difference(01Jan2010:00:00:00, 01Jan2010:09:22:57, "Hour", "actual");
The following formula returns 1, the number of times a new hour started between the times:
Date Difference(31Dec2010:23:59:59, 01Jan2011:00:59:59, "Hour", "start");
Row State Functions
There are six characteristics that rows in a data table can have: selected, hidden, excluded, labeled, colored, and marked. If you give rows one or more of these characteristics and then create row state data table columns, you can then create a formula that computes and saves row state conditions. (See “Column Properties” in the “The Column Info Window” chapter, and “Row State Columns” in the “The Column Info Window” chapter.)This formula processes row state data just as it would process character and numeric data.
See the Scripting Guide for details about syntax.
Note: A row can be assigned any combination of row states; a row state column can have multiple row states as a value.
Table A.2 describes the type of argument each Row State function requires and what each returns.
Table A.2 Row State Functions Function Name Argument Type Required What the Function Returns (Your Column Data Type Should be This Type) Row State none row state of current row As Row State numeric all row states of current row Combine States multiple row state arguments multiple row state assignments Excluded State positive integer or zero row state-excluded or not excluded Hidden State positive integer or zero row state-hidden or not hidden Labeled State positive integer or zero row state-labeled or not labeled Color State integer or color name or {red, green, blue} row state color Marker State integer or character row state marker Selected State positive integer or zero row state-selected or not selected Hue State integer row state hue Shade State integer 1-5 row state intensity Excluded Row State() or row state column numeric 0 (not excluded) or 1 (excluded) Hidden Row State() or row state column integer 0 (not hidden) or 1 (hidden) Labeled Row State() or row state column integer 0 (not labeled) or 1 (labeled) Color Of Row State() or row state column color map integer Marker Of Row State() or row state column marker map integer Selected Row State() or row state column integer 0 (not selected) or 1 (selected)
Row State
Returns the active row state condition of the current row as true or false. You can use this function to conveniently write conditional clauses that depend on the status of the current row. For example, Figure A.16 assigns a 1 to rows that are currently selected and labeled and 0 otherwise.
Figure A.16 Row State
As Row State
Converts a numeric argument to a row state or set of row state conditions. Row states are stored internally in JMP as a 16-bit number, with each bit assigned to represent one of the possible row states as illustrated in Figure A.3. For example, the binary representation of 1327 is0000010100101111. As Row State(1327) would therefore set the row state as selected, excluded, hidden, labeled, with marker 2 and color 10.
Table A.3 Row States Stored as 16-Bit Numbers: Each Bit Represents a Row State Bit Row State 0 Not selected (0) or Selected (1) 1 Unexcluded (0) or Excluded (1) 2 Unhidden (0) or Hidden (1) 3 Unlabeled (0) or Labeled (1) 4-7 Marker 8-14 Color
Combine States
Generates a row state combination with two or more arguments. Use the insert button () on the Formula Editor keypad or type a comma to add arguments to the Combine States function. The currently selected expression becomes the first argument when you select Combine States.Replace each argument with an expression that evaluates to a row state. This formula:
Combine States(
Selected State(Modulo(Row(),2),
Labeled State(Modulo(Row()+1,2))
alternately labels or selects each row in the calculated row state column. The Selected State and Labeled State functions are defined later in this section. Use the insert () and delete ()buttons on the Formula Editor keypad to add more arguments or remove unwanted arguments.
If you include conflicting row states in a combination, the results are unpredictable.
Excluded State
Interprets a numeric argument as true or false. When an argument evaluates as true, the Excluded State function assigns the excluded condition as the value of the column for that row. For example, Excluded State(Modulo(Row(),2)) assigns the excluded row state as the value of the row state column for each odd numbered row.
Hidden State
Assigns the hidden row state condition when its argument is greater than zero. If the argument is zero, the value in the column for that row is not hidden.
Labeled State
Gives the labeled row state condition when its argument is greater than zero. If the argument is zero the row value in the column for that row is not labeled.
Color State
Returns the color from the JMP color map that corresponds to its integer argument. JMP colors are numbered 0 through 84. Zero maps to black.
Marker State
Returns markers from the JMP marker map that correspond to its integer argument. JMP markers are numbered 0 through 16. The formula Marker State(Row()) assigns all the row state markers in a repeating sequence determined by the current row number to the calculated row state column. A row state column can have multiple row states as a value.
Selected State
Gives the selected row state condition when its argument is greater than zero. If the argument is zero, the value in the column for that row is not selected.
Hue State
Returns the color from the JMP hue map that corresponds to its integer argument. JMP hues are numbered 0 through 11 but larger integers are treated as modulo 12. The Hue State function does not map to black, gray, or white. A hue of zero maps to red and hue of 11 maps to magenta. The formula on the left in Figure A.17 assigns row state colors in a chromatic spread based on the value of z. The Hue State function used with a row state data type column.
Assigns five shade levels to a color or hue. A shade of –2 is darkest and shade of +2 is lightest. A shade of zero is a pure color. The formula on the right in Figure A.17 assigns shade values based on the value of z.
Figure A.17 Examples of Hue and Shade Functions
To assign all shades of all the colors in the colors palette, you need to use the Hue State and Shade State assignments together. The formula in Figure A.18 uses the Combine States function described at the beginning of this section. The first argument in the Combine States function is the Hue State formula shown previously, and the second argument is the Shade State formula. In addition, the Marker State function with an argument of 2 assigns the X marker to each row, and the Selected State function with an argument of 1 selects each row.
Figure A.18 Combine States Example For Using Both Hue State and Row State
Excluded, Hidden, Labeled, and Selected
Accepts a row state expression argument (row state column or row state constant) that evaluates as either 1 or 0 (true or false). These characteristics are inactive by default. Often, the Row() function is the argument, which detects the active row state condition of each row. For example, inFigure A.19, the formula assigns 99 whenever a row is actively selected, and 0 otherwise. Note that this formula is used in a column that has a numeric data type.
Figure A.19 Example of a Formula Using the Selected Function
The example in Figure A.20 assigns row state conditions to a row state column. The formula for the row state column (in the column called x) checks to determine whether the active row state is either Hidden or Excluded, and if so, assigns the Labeled row state.
Figure A.20 Calculate Row State Information in a Row State Column
Color Of
Accepts any row state expression or column, or the Row State() function as its argument. Returns a number from the JMP color map that corresponds to the active color state, or zero if there is no assigned color.
Marker Of
Accepts any row state expression or column, or the Row State() function as its argument. Returns a number from the JMP marker map that corresponds to the active marker or zero if there is no assigned marker.
Assignment Functions
Assignment functions work in place. That is, the result returned by the operation (on the right of the operator) is stored in the argument on the left of the operator and replaces its current value.
Assignment statements are most often used in conjunction with other commands to build a JSL script. You can use the Formula Editor to create and execute a script in that column, but this is not recommended because of dependencies and ambiguities that can result. Most often, scripts are stored as .jsl files, and can be saved with a data table. See “Create and Save Scripts” in the “Enter and Edit Data” chapter. For details about syntax, see the Scripting Guide.
Note: The first argument of an assignment function must be capable of being assigned. This means you cannot have an assignment such as 3+=4, because 3 is a constant value that cannot be reassigned. You must first create a variable (a table variable or local variable) whose value is 3. (For details about table variables, see “Use Table Variables” in the “Enter and Edit Data” chapter. For details about local variables, see “Reference Columns and Table Variables” in the “Formula Editor” chapter). Then use that variable as the left-hand argument of the assignment function.
= (assign)
Puts the value of b into a. For example (a=b).
Adds the value of b to a and puts the result back into a. For example, a+=b.
-= (subtract to)
Subtracts the value of b and puts the result back into a. For example, a–=b.
*= (multiply to)
Multiplies b with a and puts the result back into a. For example, a*=b.
/= (divide to)
Divides b into a and puts the result back into a. For example, a/=b.
++ (post increment)
Adds one (1) to a, in place, so that a++. For example, if the initial value of a is 4, the expression a++ changes a to 5.
-- (post decrement)
Subtracts one (1) from a, in place, so that a– – . For example, if the initial value of a is 4, the expression a– – changes a to 3.
Parametric Model Functions
This category is a short cut to create three parametric models that are linear functions of set of window-selected columns.
Linear Model, Interactions Model, Full Quadratic Model
Selecting each of these opens a column selection box that lets you select one or more columns to be included in the model. The function then creates and populates the chosen model.
Finance Functions
Lets you create formulas to calculate principal payments, interest rate, rate of return, and so on.
Double Declining Balance
Returns the depreciation of an asset for a specified period of time. The function uses the double-declining balance method or some other depreciation factor.
Future Value
Returns the future value of an investment that is based on periodic, constant payments and a constant interest rate.
Interest Payment
Returns the interest payment for a given period for an investment that is based on periodic, constant payments and a constant interest rate.
Interest Rate
Returns the interest rate per period of an annuity.
Internal Rate of Return
Returns the internal rate of return for a series of cash flows in the values argument.
Modified Internal Rate of Return
Returns the modified internal rate of return for a series of periodic cash flows. The cost of investment and the interest received on reinvested cash is included.
Net Present Value
Returns the net present value of an investment by using a discount rate and a series of future payments (negative values) and income (positive values).
Number of Periods
Returns the number of periods for an investment that is based on periodic, constant payments and a constant interest rate.
Payment
Returns the payment for a loan that is based on constant payments and a constant interest rate.
Present Value
Returns the present value of an investment.
Principal Payment
Returns the payment on the principal for a given period for an investment that is based on periodic, constant payments and a constant interest rate.
Straight Line Depreciation
Returns the straight-line depreciation of an asset for one period.
Sum Of Years Digits Depreciation
Returns the sum-of-years’ digits depreciation of an asset for a specified period.
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## Mean of a Poisson Distribution From Two Relative Probabilities
Suppose we have good reason to suspect that a particular random variable
$X$
follows a Poisson distribution, so that
$Z \sim Po( \lambda )$
, where
$\lambda$
is the relevant parameter, typically a frequency rate. Then we can find
$\lambda$
given the relative probabilities of observations of
$X$
.
For a Poisson distribution,
$P(X=x)=\frac{\lambda^x e^{-\lambda}}{x!}$
.
Suppose that
$P(X=7)=2P(X=8)$
.
Then
$\frac{\lambda^7 e^{-\lambda}}{7!}=2 \frac{\lambda^8 e^{-\lambda}}{8!}$
.
Divide by
$e^{- \lambda}$
to give
$\frac{\lambda^7}{7!}=2 \frac{\lambda^8 }{8!}$
.
Multiply,y both sides by
$8!$
and divide by
$\lambda^7$
. Then
$8=2 \lambda \rightarrow \lambda =4$
.
Refresh
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# If a shop earns a profit of 1300 on every heater sold and incurs a loss of 1800 on every fan sold, find out its total profit/ loss at the en
Question
If a shop earns a profit of 1300 on every heater sold and incurs a loss of 1800 on every fan sold, find out its total profit/ loss at the end of the day in which 25 heaters and 7 fans were sold.
in progress 0
1 year 2021-09-02T14:59:41+00:00 1 Answers 6 views 0
Total profit is 26,900.
Explanation
If the profit for selling a heater is 1300 than the profit of selling 25 heaters is 32500.
If the loss of selling a fan is 800 than the loss of selling 7 fan is 5600.
Minus 32500 with 5600 to get the profit.
26,900
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My Math Forum Please Help.. Derivatives dV/dR - dV/dt
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May 6th, 2012, 01:19 PM #1 Member Joined: Apr 2012 Posts: 64 Thanks: 0 Please Help.. Derivatives dV/dR - dV/dt air is being pumped into a spherical weather balloon.at any time t, the volume of the balloon is v(t) and its radius is r(t). What do the derivatives dV/dr - dV/dt represent Express dV/dt in terms of dr/dt
May 6th, 2012, 01:31 PM #2 Member Joined: Apr 2012 Posts: 64 Thanks: 0 Re: Please Help.. Derivatives dV/dR - dV/dt My attempt at first question is: dV/dr : rate of change of volume with respect to radius dV/dt: rate of change of volume with respect to time 2nd part: Since then Am I right so far? Thanks
May 6th, 2012, 07:13 PM #3 Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Please Help.. Derivatives dV/dR - dV/dt Yes, looks correct. For the second part, you could even use: $\frac{dV}{dt}=\frac{dV}{dr}\cdot\frac{dr}{dt}$
May 6th, 2012, 07:53 PM #4 Member Joined: Apr 2012 Posts: 64 Thanks: 0 Re: Please Help.. Derivatives dV/dR - dV/dt
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# 2019 AMC 10A Problems/Problem 25
The following problem is from both the 2019 AMC 10A #25 and 2019 AMC 12A #24, so both problems redirect to this page.
## Problem
For how many integers $n$ between $1$ and $50$, inclusive, is $$\frac{(n^2-1)!}{(n!)^n}$$ an integer? (Recall that $0! = 1$.)
$\textbf{(A) } 31 \qquad \textbf{(B) } 32 \qquad \textbf{(C) } 33 \qquad \textbf{(D) } 34 \qquad \textbf{(E) } 35$
## Solution
### Solution 1
The main insight is that
$$\frac{(n^2)!}{(n!)^{n+1}}$$
is always an integer. This is true because it is precisely the number of ways to split up $n^2$ objects into $n$ unordered groups of size $n$. Thus,
$$\frac{(n^2-1)!}{(n!)^n}=\frac{(n^2)!}{(n!)^{n+1}}\cdot\frac{n!}{n^2}$$
is an integer if $n^2 \mid n!$, or in other words, if $n \mid (n-1)!$. This condition is false precisely when $n=4$ or $n$ is prime, by Wilson's Theorem. There are $15$ primes between $1$ and $50$, inclusive, so there are $15 + 1 = 16$ terms for which
$$\frac{(n^2-1)!}{(n!)^{n}}$$
is potentially not an integer. It can be easily verified that the above expression is not an integer for $n=4$ as there are more factors of $2$ in the denominator than the numerator. Similarly, it can be verified that the above expression is not an integer for any prime $n=p$, as there are more factors of p in the denominator than the numerator. Thus all $16$ values of n make the expression not an integer and the answer is $50-16=\boxed{\mathbf{(D)}\ 34}$.
### Solution 2
We can use the P-Adic Valuation (more info could be found here: Mathematicial notation) of n to solve this problem (recall the P-Adic Valuation of 'n' is denoted by $v_p (n)$ and is defined as the greatest power of some prime 'p' that divides n. For example, $v_2 (6)=1$ or $v_7 (245)=2$ .) Using Legendre's formula, we know that :
$$v_p (n!)= \sum_{i=1}^\infty \lfloor \frac {n}{p^i} \rfloor$$
Seeing factorials involved in the problem, this prompts us to use Legendre's formula where n is a power of a prime.
We also know that , $v_p (m^n) = n \cdot v_p (m)$ . Knowing that $a\mid b$ if $v_p (a) \le v_p (b)$ , we have that :
$$n \cdot v_p (n!) \le v_p ((n^2 -1 )!)$$ and we must find all n for which this is true.
If we plug in $n=p$, by Legendre's we get two equations:
$$v_p ((n^2 -1)!) = \sum_{i=1}^\infty \lfloor \frac {n^2 -1}{p^i} \rfloor = (p-1)+0+...+0 = p-1$$
And we also get :
$$v_p ((n!)^n) = n \cdot v_p (n!)= n \cdot \sum_{i=1}^\infty \lfloor \frac {n}{p^i} \rfloor = p \cdot ( 1+0+...0) = p$$
But we are asked to prove that $n \cdot v_p (n!) \le v_p ((n^2 -1 )!) \Longrightarrow p \le p-1$ which is false for all 'n' where n is prime.
Now we try the same for $n=p^2$ , where p is a prime. By Legendre we arrive at:
$$v_p ((p^4 -1)!) = p^3 + p^2 + p -3$$ and $$p^2 \cdot v_p (p^2 !) = p^3 + p^2$$
Then we get:
$$p^2 \cdot v_p (p!) \le v_p ((n^4 -1)!) \Longrightarrow p^3 + p^2 \le p^3 + p^2 + p -3$$ Which is true for all primes except for 2, so $2^2 = 4$ doesn't work. It can easily be verified that for all $n=p^i$ where $i$ is an integer greater than 2, satisfies the inequality :$$n \cdot v_p (n!) \le v_p ((n^2 -1 )!).$$
Therefore, there are 16 values that don't work and $50-16 = \boxed{\mathbf{(D)}\ 34}$ values that work.
~qwertysri987
## Solution 3 (Guessing)
First, we see that $n=1, 6, 8, 9, 10, 12, 14$ work. This leads us to the conclusion of $50-16 = \boxed{\textbf{(D)}\ 34}$
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https://puzzling.stackexchange.com/questions/16373/reconstruct-this-multiplication-to-find-the-combination-for-the-safe/16380
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# Reconstruct this multiplication to find the combination for the safe
You've solved it! No, wait a minute - you've solved the first step. You don't know the combination for the safe yet. All you know is that someone has helped you by using a pseudonym. Their name isn't really Artairs A Sebeenas. But if you write it like this
you get a multiplication, and the product represented by "SEBEENAS" is the 8-digit number you need.
No two letters represent the same digit. What's the combination? Please show all work, and use of a computer is not allowed.
• Can I use a computer to type my answer? :O – Ian MacDonald Jun 14 '15 at 3:19
$S \times A$ ends with $S$
Therefore, if $S = 1,3,7$ or $9$, $A = 1$
If $S = 5$, $A = 1,3,5,7,9$
If $S = 2,4,6$ or $8$, $A = 1$ or $6$
If $S = 0$, $A$ is anything, but note that $S$ cannot be equal zero since it is the start of the answer.
$A \times A + \text{carryover} = 10S + E$
If $S = 1,3,7,9,2,4,6$ or $8$, $A = 1$ and $S = 0$, which is a contradiction.
If $S = 2,4,6,8$, $A = 6$ and $S = 3$ or $4$, which means that $S = 4$, $A = 6$
If $S = 5$, $A = 1,3,5,7,9$, and only $A = 7$ can achieve $S=5$.
Therefore, $(S,A) = (4,6)$ or $(5,7)$
Let these be Case 1 and Case 2 respectively.
Case 1
$S = 4, A = 6$
$R \times A + \text{carryover}$ ends with $A$
$\text{carryover} = 2$ (The $2$ in $4 \times 6=24$)
$A = 6$
$6R + 2$ ends with $6$
$6R$ ends with $4$
$R = 4$ (reject since $S = 4$) or $9$
$S = 4$, $A = 6$, $R = 9$
Now we have:
$\text{69T6I94}$
$6$
$\text{4EBEEN64}$
$69 \times 6 + \text{carryover} = \text{4EB}$
$0 \leq \text{carryover} < 5$
$414 + \text{carryover} = \text{4EB}$
$E = 1$
$S = 4, A = 6, R = 9, E = 1$
Now it is:
$\text{69T6I94}$
$6$
$\text{41B11N64}$
$6 \times 6 + \text{carryover}$ (of $6 \times I + \text{carryover}$) ends with $1$
(both) $\text{carryover} = 5$
$6 \times I + 5$ starts with $5$
$I = 8$ or $9$ (reject since $R = 9$)
$S = 4, A = 6, R = 9, E = 1, I = 8$
$\text{69T6894}$
$6$
$\text{41B11N64}$
$N = 3$
$\text{69T6894}$
$6$
$\text{41B11364}$
However, $T$ cannot exist, since $6 \times T + 4$ ends with $1$, so $T$ is not an integer.
Therefore Case 1 has no solutions.
Apply the same method to Case 2
$R = 2$
$E = 0$ or $1$
If $E=0$, $I=2$ (reject since $R=2$), so no sol if $E=0$
If $E=1$, $I = 3$ or $4$
If $I = 3$, $N = 2$ (reject)
If $I = 4$, $N = 9$, $T = 8$
Therefore:
$7287425$
$7$
$51011975$
Thus, your answer is $51011975$.
• Yes - well done! :-) – h34 Jun 14 '15 at 10:16
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# Classifying the Shapes of Distributions
Histograms are commonly used to analyze the “shape” of a data distribution.
## Shapes of Distributions
Symmetrical distributions
Suppose we have this histogram that shows the number of pets that each family in your neighborhood owns:
We would describe this distribution as symmetrical because if we drew a line down the middle of the distribution, the left and the right side would look roughly “symmetric”, or “equal” to each other:
Left-skewed distributions
We would describe the distribution below as left-skewed because it has a “tail” on the left side that skews the distribution to the left:
Right-skewed distributions
We would describe the distribution below as right-skewed because it has a “tail” on the right side that skews the distribution to the right:
Bimodal distributions
We would describe the distribution below as bimodal because it has two (hence the bi) “peaks”, one on the left and one on the right:
Uniform Distributions
We would describe the distribution below as uniform since the values are roughly “uniform” all the way across:
## Classifying the Center and Spread
Histograms also help us identify the center (the median) and the spread of a distribution.
In the distribution below, the center is located at three:
And between the two distributions below, A is more “spread out” than B:
## Identifying Unusual Features
Histograms can also be used to identify unusual features like gaps and outliers.
A gap is simply an area in a distribution with no observations. In the distribution below, there is a gap in the middle:
An outlier is a value that is significantly different than all the other values in a dataset. In the distribution below there is a family with 15 pets, which could be considered an outlier:
Note: In general, a value is considered an outlier if it is 1.5 interquartile ranges above the third quartile (Q3) or 1.5 interquartile ranges below the first quartile (Q1).
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https://www.jiskha.com/questions/1010149/f-x-2x-2-12x-6-what-is-the-min-or-max-value
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# pre cal
f(x)=2x^2+12x-6 what is the min or max value ?
1. 👍 0
2. 👎 0
3. 👁 79
1. 👍 0
2. 👎 0
posted by Damon
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2. ### MAth
If 1 gallon (gal) of water weighs 8.34 pounds (lb), how much does 3.75 gal weigh? A) 15 lb B) 31.275 lb C) 30.125 lb D) 12.09 lb i think it is D 50. Determine the maximum and minimum values of the following set of numbers. A) Max:
asked by dawn on July 28, 2008
3. ### math
how to solve this problem: max min(x-y)..... here where the max have written, under it x>=0 is also given. also under min y>=0 is given. Also tell that is this is equals to min max(x-y) ? (same as above under max x>=0, under min
asked by shiya on November 16, 2011
4. ### Algebra
What are the minimum, first quartile, median, third quartile, & maximum of the data set? 45, 76, 12, 39, 87, 65, 23, 32 A) min: 12 max: 87 first: 70.5 third: 27.5 med: 42 B) min: 12 max: 87 first: 27.5 third: 70.5 med: 42 C)min:
asked by GummyBears16 on April 25, 2016
5. ### Math
Fine the max or min points and the points of inflection for the function y=3x^4 - 10x^3 - 12x^2 + 12x - 7 Can someone help me!!!
asked by Sara on February 7, 2017
6. ### Math
I need an expiation on a math concept, because I know the right answer, I just don't know why it is The problem says to find the absolute maximum and minimum of the function f(x)=x^3-12x over the interval [-5, 2] I found the 1st
asked by Eiban on October 9, 2014
7. ### PreCalc
y =( (C-A)/2 ) sin( (X - (A - ( (C - A)/2 ) ) )( pi/(C-A) ) ) + (B - ( (B-D)/2 ) How do I rearange this for this equation for the all of the variables including A, B, C, D, X were pi is the greek leter pi if you didn't notice this
asked by Jo on December 7, 2009
8. ### Algebra 2
Please help super confused!!! Which points are the best approximation of the relative maximum and minimum of the function? f(x)=x^3+3x^2-9x-8 a. Relative max is at (3,-13), relative min is at (-3,-19). b. Relative max is at
asked by Olivia N J on November 14, 2016
9. ### Calculus (urgent!!)
Consider the function f(x)=12x^5+60x^4−100x^3+4. For this function there are four important intervals: (−INF,A], [A,B] ,[B,C] , and [C,INF) where A, B, and C are the critical numbers. Find A, B, and C. At each critical number
asked by Karku on April 1, 2011
10. ### Calculus
Find any relative max or min points of the given function y=3x^2+12x+17 So far, I have: y'=6x+12=0 2(3x+6)=0 x=-2
asked by Mike on March 31, 2018
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• At , why does solving for x using 0 gives us the intersecting points of the two equations? I would expect that solving for x when the equation equals zero, would give us x intercepts.
• I think i might understand why:
We are not finding where y = 0, which would give the x intercepts. We are solving a system of two equations by finding the y value of one equation (i.e y = x-3) and inputting it into the second equation. Then rearranging this new equation so that it can be solved using the zero product property.
Any elaboration/correction greatly appreciated.
• At , where does he get 6 and -1 from? Does he get it from earlier in the video and I missed it or what?
• Please check out factoring by groups on this website called - Factoring quadratics: leading coefficient ≠ 1.
the short answer is; to find a common factor from the function ax^2+bx+c
- find two numbers d & e
that multiply equal a times c; -- d*e=a*c
that add up to b; -- d+e=b
In the video that are -6 & -1 as they multiply to get 3*2 and add up to -7
• At , why does he solve for y? Wouldn't it be easier to substitute y in the linear equation for the quadratic expression?
• Yes that is also another fast way to do it. There are many ways to approach a problem and still get the same solution — it's ultimately up for preference.
• @ we are told that as an alternative to factoring by grouping we could. How would you do that? I attempted it and did not get either answer that he did by grouping method.
𝑎𝑥² + 𝑏𝑥 + 𝑐 = 0 ⇒ 𝑥 = (−𝑏 ± √(𝑏² − 4𝑎𝑐))∕(2𝑎)
In our case 𝑎 = 3, 𝑏 = −7, and 𝑐 = 2,
so 𝑥 = (−(−7) ± √((−7)² −4⋅3⋅2))∕(2⋅3)
= (7 ± √(49 − 24))∕6
= (7 ± √25)∕6 = (7± 5)∕6
𝑥 = (7 − 5)∕6 = 2∕6 = 1∕3
OR
𝑥 = (7 + 5)∕6 = 12∕6 = 2
So, the factorization should be (𝑥 − 1∕3)(𝑥 − 2).
But because 𝑎 = 3, the factorization is actually
3(𝑥 − 1∕3)(𝑥 − 2) = (3𝑥 − 1)(𝑥 − 2)
• Is it ok to skip the factoring and use the quadratic formula? Or is it important to learn this way?
• Both methods are valid, but learning how to factor that way is very valuable when time is limited, so it might be worthwhile to practice.
(1 vote)
• Can't you just substitute 3x^2-6x+1 for y in the linear equation?
(1 vote)
• Yes that would be quicker and more direct than what Sal did, you get to straight to 3x^2-6x+1-x+2=0, so 3x^2-7x+2=0.
(1 vote)
• Why is it x+1?!
(1 vote)
• at where did he get x-2 and x+2?
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# Energy required to accelerate from different reference frames
So I've recently been studying relativity a lot trying to understand it and I feel like I grasp most things conceptually but I have one issue I've been trying to understand for the last couple days and I just can't seem to find an answer anywhere.
Let's say you're traveling very fast, somewhere close to the speed of light. At this point you're in an inertial reference frame. From this frame you shouldn't be aware that you are anywhere near the speed of light. Now you want to accelerate by 5 m/s to a new reference frame that is even closer to the speed of light. Is there any difference in the amount of energy required to accelerate to the new reference frame in this scenario from the amount required to accelerate by 5 m/s from a slower reference frame like here on Earth?
I sort of understand the concept of relativistic energy but if that somehow applies here I don't get how. If you are in an inertial reference frame, then it seems that the amount of energy required to accelerate to a new reference frame should be the same no matter what your frame is. However, from various things I've read I get the impression that the energy required to accelerate at a constant rate is not constant, which seems to make sense from a static reference from but not from an accelerating one.
I'm sure I'm missing something or there's a flaw in my thinking somewhere. I hope this all makes sense since I'm still very new to this. Also, if there's a way to explain the concept with little to no math that would be very helpful since I still don't grasp a lot of the math involved in relativity. I'll take what I can get though. This has been eating at me too much.
Update: For anyone wondering about this same thing I finally found the answer explained in a way I grasped it here.
• What are you traveling relative to? Your velocity is observer dependent. It's not a relevant physical quantity unless you have two systems. Neither is this a problem of special relativity. You have exactly the same situation in Galilean relativity. What your kinetic energy is and by how much it increases or decreased if you accelerate is observer dependent. How much energy you need to accelerate depends on where the accelerating force comes from. If it is exerted by something that moves really fast relative to you, it will take more energy than a force that comes from a slow moving object. May 26 '16 at 5:40
• When you say, "How much energy you need to accelerate depends on where the accelerating force comes from" I think that is the part I'm not understanding. So what if you're in a rocket of some sort and you fire it? What if the force is coming from within your reference frame or one close to it? May 26 '16 at 5:57
• A rocket is not one reference frame but at least two. It's the payload and the mass the rocket expels. If you remember Newton, a rocket, as a whole doesn't really move. The center of mass always stays at the launch point. This is a perfectly classical problem and you need to go back to the derivation of kinetic energy again. Do you remember why it goes with the square of the velocity? May 26 '16 at 6:02
• In your (initial) rest frame suppose your speed increases to $dv$ then your energy change is $\sqrt{p^2c^2+m^2c^4} - mc^2$, where $p=\gamma mdv$. In my frame your velocity increased from $v$ to $v'$, where you have to calculate $v'$ using the relativistic addition of velocity ($v'\ne v+dv$). Then in my frame your energy change is $\sqrt{p'^2c^2+m^2c^4} - \sqrt{p^2c^2+m^2c^4}$, where $p'$ and $p$ are the initial and final relativistic momentum. I'll leave you to do the calculation since it's messy, unilluminating and would contravene our homework policy. May 26 '16 at 6:38
• Ok, I think this is making sense. I also just found another good discussion of the same thing here. It helps knowing the right thing to search for. Thanks guys! May 26 '16 at 6:53
In special relativity the transition from one frame to another is given by the Lorentz boosts. This is not quite the same as an acceleration, but a transformation that relates observations on one frame with another. We can think of an acceleration as being a succession of infinitesimal Lorentz boosts that map one frame to another.
The infinitesimal distance in flat spacetime $ds$ for a particle is given by $$ds^2 = c^2dt^2 - dx^2 - dy^2 - dx^2$$ where this distance $ds – cd\tau$, for $\tau$ the time as measured by a clock on the frame of the particle. Let us consider the motion of this particle in the $x$ direction. Now divide through by $ds^2$ to get $$1 = \left(\frac{dt}{d\tau}\right)^2 - \frac{1}{c^2}\left(\frac{dx}{d\tau}\right)^2.$$ We can write this according to four-velocity $U_t = \frac{dt}{d\tau}$, $U_x = \frac{dx}{d\tau}$ $$1 = U_t^2 - U_x^2.$$ Now take the derivative of this with respect to $\tau$ so that $$0 = \left(\frac{dU_t}{d\tau}\right)U_t - \frac{1}{c^2}\left(\frac{dU_x}{d\tau}\right)U_x.$$ This leads to the interesting observation that in spacetime the four-acceleration is perpendicular to the four-velocity.
This system of equations leads to a solution for the four-velocity $$U_t = cosh(g\tau),~U_x = c~sinh(g\tau),$$ for $g$ the acceleration. We can see that the equation defines a hyperbola in $t, x$ coordinates. For large $\tau$ that the hyperbola is approximately $U_t^2 = U_x^2$, and it is not hard to get these in the $t, x$ coordinates. We can also see that the coordinate based velocity is $$\frac{dx}{dt} = \frac{U_x}{U_t} = c~tanh(gt),$$ which indicates this particle asymptotes to the speed of light as $\tau\rightarrow\infty$.
When it comes to energy we appeal to the four-momentum interval in special relativity $$m^2 = E^2 - p^2$$ where $E = mU_t$ and $p = mU_x$ are energy and spatial momentum respectively. Using the properties of hyperbolic trigonometric functions this can be seen. We can see right off that energy is given by a hyperbolic cosine function that diverges enormously as $\tau\rightarrow\infty$.
The fundamental principle in relativity (both special and general) is that the laws of physics are the same for all observers. We express this most naturally using the principle of general covariance, which means that the laws of physics are expressed using tensor (including vector and scalar) quantities. Then we can understand the laws of physics most easily in terms of proper quantities, that is to say in terms of properties as measured by an observer moving with the object being measured.
If you are moving very fast compared to another observer, you are not moving close to the speed of light in your own reference frame, because the speed of light is a fundamental property in physics and is the same for all observers. You apply a proper acceleration and it will require a certain amount of energy, but the speed of light remains constant.
Now consider it from the point of view of the other observer. He does not measure your proper quantities directly, but if he has expressed the laws of physics correctly in terms of tensors he can calculate your proper acceleration, and he can calculate the amount of energy you used. He will think you are moving nearer to the speed of light, but he will not think you have increased your speed by much. The apparent acceleration which he sees is not a proper quantity, and he is misleading himself if he uses it in a calculation of energy without first calculating the appropriate proper quantities.
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# Can a limit exist at $x\to a$ but not equal to $f(a)$?
In the calculus course I'm taking, the conditions for a continuous function are said to be when
a.) $$\lim_{x\to a}f(x)$$ exists
and
b.) $$\lim_{x\to a}f(x)$$ = $$f(a)$$
And it seems a little redundant, since if a limit exists at $$x\to a$$, wouldn't it always equal to $$f(a)$$? Are there situations where that is not the case?
I apologize if this was a bad question/ formatted bad, I'm pretty new to math and this site in general.
• We use limits to study the behaviour of he function "near" the point of interest but excluding it (i.e. deleted neighbourhood)
– user
Commented Sep 6, 2022 at 22:42
• The definition of the limit involves $$\color{red}{0 < } ~|x - a| < \delta,$$ rather than $$|x - a| < \delta.$$ Commented Sep 6, 2022 at 23:58
• @user2661923 This seems in direct contradiction with most definitions of limits I've seen in calculus courses. It's a matter of convention of course, but the convention I've most often seen is that the definition involves $|x-a| < \delta$, not $0 < |x-a| < \delta$; with this convention, you can still exclude point $a$, but you need to be explicit: $$\lim_{\substack{x \to a \\ x \neq a}} f(x)$$ Which convention is "better" is a discussion for matheducators.stackexchange.com rather than for math.stackexchange, but the only answer for OP's question is "what definition were you given?"
– Stef
Commented Feb 14 at 13:07
• @Stef $~|x-a| = 0 \iff x=a.~$ If you don't (somehow) prevent $~x=a,~$ then you blur the distinction between the definition of a limit, and the definition of a function being continuous at a specific point. The limit of a function as $~x \to a~$ can exist, without this limit equaling $~f(a)~$ (i.e. without the function being continuous at $~x=a~$). Commented Feb 14 at 15:10
• @user2661923 For the first half of your comment: I don't blur anything. In fact, I make things more explicit. If I need to exclude $a$, then I can exclude $a$ explicitly. Which many math teachers will argue is better for the students.
– Stef
Commented Feb 14 at 15:16
No.
$$f(x)= \begin{cases} 2, x=0\\ x, x \ne0 \\ \end{cases}$$
$$f(0)=2$$, but $$\lim_{x\to 0}f(x)=\lim_{x\to 0}x=0$$.
You can have a function with a hole in it, for example:
In fact, the function may not even be defined at $$a$$. One example is $$f(x) = \begin{cases} -x + 4 & \text{if x \neq 0}. \end{cases}$$ Then $$\lim\limits_{x \to 0} f(x) = 4$$, but $$f(0)$$ doesn't exist. I could also define $$f(0)$$ to be anything, say, $$5$$. Then the limit exists and is equal to $$4$$, but differs from $$f(0)$$. In either of these cases, $$f(x)$$ is not continuous at $$x = 0$$.
• @TedShifrin Yes, this was a silly typo. Thank you for pointing this out! Commented Sep 7, 2022 at 0:08
This confusion often occurs because most, if not all, of the functions people are acquainted with in elementary calculus are continuous. In other words, they do have the property that if $$f$$ is defined at $$a$$, then $$\lim_{x \to a}f(x)$$ exists and is equal to $$f(a)$$. But not every function is like this. Consider, for example, the sign function or signum function, written as $$\DeclareMathOperator{\sgn}{sgn}\sgn$$: $$\sgn(x)= \begin{cases} 1 & \text{if x>0} \, ,\\ 0 & \text{if x=0} \, ,\\ -1 & \text{if x<0} \, . \end{cases}$$ It's called the sign function because it tells us whether a real number is positive, negative, or zero. Note that $$\sgn$$ is defined at $$0$$, but not continuous at $$0$$. It is continuous at all other points.
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# Marked price of a dress is Rs.4000. Shopkeeper offers 15% discount on it and again offers 10% discount on new price...
78 views
The marked price of a dress is Rs.4000. A shopkeeper offers 15% discount on this shirt and then again offers 10% discount on the new price, and then and then again offers 5% discount on the new price . How much will you have to pay, finally?
posted Oct 13, 2017
you have to pay Rs. 2907
4000-15%=3400
3400-10%=3060
3060-5%=2907
finally, you have to pay Rs. 2907
Use the formula x+y-(x*y/100) where x and y are the successive discount percentages. Here they are 15% and 10%. So by using the formula, we obtain
15+10-15*10/100= 23. 5. Here 23. 5 is the new discount%
Now 23. 5% of 4000 is 940 and the new price is 4000-940=3060 . Also with another discount of 5% on 3060 i.e 153, final price is 3060-153= 2907
2907 rs I had to pay to shopkeeper as final rupees
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The marked price of a shirt is Rs.1500. A shopkeeper offers 15% discount on this shirt and then again offers 20% discount on the new price, and then and then again offers 25% discount on the new price . How much will you have to pay, finally?
The marked price of a shirt is Rs 786.
A shopkeeper offers three successive discount of 5%, 10% and 15%.
How much will you have to pay, finally?
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# Show that $(A\setminus B) \cup (A\setminus C) = B \Leftrightarrow A=B \wedge (B \cap C) = \emptyset$ [duplicate]
I believe there are 3 parts to this.
1) $(A\setminus B) \cup (A\setminus C) = B \Rightarrow A=B$
2) $(A\setminus B) \cup (A\setminus C) = B \Rightarrow (B \cap C) = \emptyset$
3) $A=B \wedge (B \cap C) = \emptyset \Rightarrow (A\setminus B) \cup (A\setminus C) = B$
I can do the the parts labelled 1 and 3 but cannot show part 2. Anyone who can explain how you do the 2nd part ie show $(A\setminus B) \cup (A\setminus C) = B \Rightarrow (B \cap C) = \emptyset$ ?
## marked as duplicate by Marnix Klooster, drhab, user99914, happymath, user91500Dec 1 '15 at 9:48
• @amWhy Happens to the best of us as we can see ;) Deleting in a second... – AlexR Nov 10 '14 at 14:43
• LOL i made so much errors :P thanks for the edits – Namch Nov 10 '14 at 15:14
Hint: Let $(A-B)\cup (A-C)=B$ hold true and assume that there is $x \in B \cap C$. This means in particular that $x\in B$ and $x \in C$. Hence $$x \notin A-B$$ and for the same reason $$x\notin A-C$$ Therefore $$x\notin (A-B)\cup (A-C)$$ but on the other hand $x\in B$ which a contradiction to the assumption that $$(A-B)\cup (A-C)=B$$
• Nicely done lol I would vote up but need reputation of 15 :P – Namch Nov 10 '14 at 14:54
• Haha, no problem don't worry. You are welcome – Jimmy R. Nov 10 '14 at 14:56
You have that $(A/B) \cup (A/C)=B$. Then clearly, $B \subseteq A/C$ since it couldn't be in $A/B$ because you just removed all of $B$. So then if $B \subseteq A/C$, that is intuitively, saying, "none of $B$ is in $C$." i.e. $B \cap C = \emptyset$
Since you already showed (1), we can use it to ease the proof of (2):
$$(A\setminus B) \cup (A\setminus C) \stackrel{(1)}= (B\setminus B) \cup (B\setminus C) = B \setminus C \stackrel{\text{req.}}= B$$ Now $B\setminus C = B \cap C^C$ by definition, so we have $B \cap C^C = B \Rightarrow B \subset C^C \Rightarrow B \cap C = \emptyset$ as claimed.
An systematic solution:
In the table below, every column represents some subexpression and holds truth values meaning "belongs to this set". The rows exhaust all $2^3$ combinations.
The $7^{th}$ column is the LHS of the equivalence and the $11^{th}$ is the RHS. As you can see, they are identical.
A B C A/B A/C + =B A=B B.C =0 and
0 0 0 0 0 0 1 1 0 1 1
1 0 0 1 1 1 0 0 0 1 0
0 1 0 0 0 0 0 0 0 1 0
1 1 0 0 1 1 1 1 0 1 1
0 0 1 0 0 0 1 1 0 1 1
1 0 1 1 0 1 0 0 0 1 0
0 1 1 0 0 0 0 0 1 0 0
1 1 1 0 0 0 0 1 1 0 0
(+ for $\cup$, . for $\cap$)
Grouping all rows in a single hexadecimal number:
A B C A/B A/C + =B A=B B.C =0 and
55 33 0F 44 50 54 98 99 03 FC 98
(/ is and not, + is or, . is and, = is not xor, bitwise)
• Nice one but in my opinion it would take too long if this was an exam question. – Namch Nov 10 '14 at 15:00
• For an expression involving $n$ variables and $k$ operators, you will compute $2^n k$ bits, in an automated way. This is to be compared to the symbolic approach. Using a binary or hexadecimal calculator, takes less than a minute ($k$ operations). – Yves Daoust Nov 10 '14 at 15:37
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×
Get Full Access to Mathematical Statistics With Applications - 7 Edition - Chapter 10 - Problem 10.124
Get Full Access to Mathematical Statistics With Applications - 7 Edition - Chapter 10 - Problem 10.124
×
# The data in the following table give readings in foot-pounds of the impact strength of
ISBN: 9780495110811 47
## Solution for problem 10.124 Chapter 10
Mathematical Statistics with Applications | 7th Edition
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Problem 10.124
The data in the following table give readings in foot-pounds of the impact strength of two kinds of packaging material, type A and type B. Determine whether the data suggests a difference in mean strength between the two kinds of material. Test at the = .10 level of significance. A B 1.25 .89 1.16 1.01 1.33 .97 1.15 .95 1.23 .94 1.20 1.02 1.32 .98 1.28 1.06 1.21 .98 yi = 11.13 y = 1.237 y2 i = 13.7973 yi = 8.80 y = .978 y2 i = 8.6240
Step-by-Step Solution:
Step 1 of 3
Chapter1: Data and Decisions What do you think isStatistics What are statistics Hal Varian (GoogleChief Economist) video on Statistics and Data; http://www.youtube.com/watchv=D4FQsYTbLoI “The sexy profession of the next decade will be statistician.” He made the quote in 2008 Varian then goes on to say: “The ability to take data - to be able to understand it, to process it, to extract value from it, to visualize it, to communicate it'sgoing to be a hugely important skill in the next decades, not only at the professional level but even at the educational level for elementary school kids, for high school kids, for college kids. ….. I do think those skills - of being able to access, understand, and communicate the insights you get from data analysis - a
Step 2 of 3
Step 3 of 3
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# Write 3(x-5)^2 as an expression
0
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Write 3(x-5)^2 as an expression
Guest May 17, 2017
#1
+94142
0
Write 3(x-5)^2 as an expression
but you can expand it if you want.
3*(x-5)(x-5) = 3*(x^2-10x+25) = 3x^2-30x+75
Melody May 17, 2017
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# Lesson 6 by Igor Epshteyn
UMBC, room ECS 023, 3-10-99
Notes by David Joyner. (No claim to completeness.)
Theme: Geometry of the chessboard – dynamic case case. Passed pawns.
A passed pawn is a pawn which has no opponent pawns on its file nor on any adjacent file between its square and its promotion square. Passed pawns are the top priority in almost every pawn endgame.
A passed pawn is called
• protected, if it is protected by another pawn;
• outside, if it is located on either flank (i.e., the a, b g, or h files);
• central, if it is not an outside passed pawn.
The square of the passed pawn is the square on the (extended) chessboard whose diagonal runs from the square of the pawn to the promotion rank on the side of the opponent’s king. Noteworthy exceptions:
• Passed rook pawns can only have a square on one side of the file.
• The square of the passed pawn is not always a perfect square on the actual chessboard because of the limitations of the size of the actual board. For example, for a white passed pawn on b5 and the black king on a7, the “square” would be a rectangle having vertices on a5, b5, b8, a8.
• For a pawn on its initial square (2nd rank for white, 7th rank for black) the square of the pawn is the same as if it were on the next rank (3rd rank for white, 6th rank for black).
Rule of the pawn square: If the king can step into the square of the passed pawn then the passed pawn can be intercepted. Otherwise, the passed pawn will queen without interception.
Position 6.1:
In Forsyth notation:
```8/5k2/8/1P6/8/8/8/1K6
```
White to move and win. Black to move and draw.
The black king is outside the square of the b5 pawn, so white to move wins by the rule of the pawn square.
Position 6.2:
In Forsyth notation:
```8/8/3k4/1p1p4/1P5P/4K3/8/8
```
White to move and win.
This relatively trivial example illustrates the idea that in spite of a material advantage, an outside passed pawn is a strong advantage.
Position 6.3: From Lombardy-Fischer, Ch USA, 1960/61.
In Forsyth notation:
```8/pp4pp/4k3/3rPp2/1Pr4P/2B2PP1/1P2K3/4R3
```
Black to play and win.
1 … Rxc3! 2 bxc3 Rxe5 3 Kd2 Rxe1 4 Kxe1 Kd5 5 Kd2 Kc4 6 h5 b6 7 Kc2 g5! 8 h6 f4 9 g4 a5 10 bxa5 bxa5 and black wins.
Position 6.4: Study of Farni.
In Forsyth notation:
```8/8/2k5/8/5pP1/5P2/2K5/8
```
Black to play and draw.
Surprisingly enough, the strong white advantage in this position (up by a protected passed pawn) is not enough to win. Black can draw by using coordination of squares (opposition) to repel the enemy king. 1 … Kd6! 2 Kd2 Kc6! 3 Ke2 Kd6 4 Kd3 Kd5 5 g5 Ke5 6 g6 Kf6 7 Ke4 Kxg6 8 Kxf4 Kf6 and black draws (see basic endgame position 3). White cannot penetrate down the h-file either, since black can defend so that when white plays Kh3, black plays the repelling move Kg5. Note that if the position were only slightly different: all the pawns are move back by one rank (to f4, g5, f5) then black loses since the “saving move” … Kc6 doesn’t work in this case (it moves black outside the square of the passed pawn at g5).
Position 6.5: From Shaitar-Thelen, Ostrava 1946. This position illustrates how tricky a win can be, despite having a protected passed pawn.
In Forsyth notation:
```2k5/2p5/3p4/2PPpK2/8/8/P7/8
```
Black to move and win.
1 … Kb7 2 a3! (2 a4? loses to 2 … dxc5 3 Kxe5 Ka6 4 Ke4 Ka5 5 Kd3 Kb4 etc.) Ka6 3 c6 (if 3 a4?
then 3 … dxc5 and black wins) Kb6! (The move 3 … Ka5 allows the following miracle: 4 Ke6 e4 5 Kd7 Kb6 6 a4 e3 7 a5+ Kc5 8 Kxc7 e2 9 Kd7 Kxd5 10 c7 e1=Q 11 c8=Q Qe6+ 12 Kc7 Qxc8+
13 Kxc8 Kc6 14 Kb8! Kb5 15 Kb7! and draws.) 4 a4 Kc5!! 5 Ke6 e4 6 Kd7 e3 7 Kxc7 e2 8 Kd7 e1=Q 9 c7 Kxd5 10 c8=Q Qe6+ 11 Kd8 Qxc8+ 12 Kxc8 Kc6! 13 a5 d5 and black wins. 4 a4 Kc5 5 Ke6 e4 6 Kd7 e3 7 Kxc7 e2 8 Kd7 e1=Q 9 c7 Kxd5 10 c8=Q Qe6+ 11 Kc7 Qxc8+ 12 Kxc8 Kc6! and black wins.
Position 6.6: From Nimzovitch-Tarrash, San Sabastian 1911.
This position illustrates the technique of separation where one side prevents the other from creating a protected passed pawn.
In Forsyth notation:
```8/5k2/p4p2/5K2/7P/8/6P1/8
```
Black to play and win.
1 … a5 2 Ke4 f5+! 3 Kd4 (white cannot take the pawn, due to the rule of the pawn square) f4! (not allowing white to create a protected passed pawn) and black wins.
Position 6.7: From E. Geller-R. Fischer, Capablanca Memorial, Havana, 1965. This position also illustrates the technique of separation.
In Forsyth notation:
```8/1b6/1P1k2p1/8/5P1p/4Q3/1q2B1PK/8
```
White to play and win.
1 Bf3 Bxf3 2 Qe5+! Qxe5 3 fxe5+ Kxe5 4 gxf3 Kd6 5 f4! (not allowing black to create a protected pawn) and black resigned in the face of 5 … Kc6 7 Kh3 Kb6 8 Kxh4 Kc6 9 Kg5 Kd7 10 Kxg6.
## Homework
Solve the following problems.
1. Position 6.8: This homework problem is from Speilmann-Makarchik, 1939.In Forsyth notation:
```8/8/8/5k2/6pP/6P1/5K2/8
```
Black to play and draw.
solution below
2. Position 6.9:In Forsyth notation:
```8/8/p3p3/4k1p1/1P2p1Pp/P3K2P/8/8
```
White to play and win.
solution below
3. Position 6.10: This homework problem is from Schlechter-Tartakover, 1907.In Forsyth notation:
```8/1p3pk1/p3p2p/4P1p1/5P2/P7/1P2K1PP/8
```
White to play and win.
solution below
4. Position 6.11: This homework problem is a study by Seleznev.In Forsyth notation:
```8/8/8/8/2pk2p1/P7/1P6/4K3
```
White to play and draw.
solution below
5. Position 6.12: In Forsyth notation:
```8/2k5/6p1/2P1p2p/1K5P/6P1/8/8
```
White to play and win.
solution below
6. Position 6.13: This homework problem is a study of Kling and Horwitz (or Gorvitz).In Forsyth notation:
```8/8/8/2p5/2Pp4/3K2Pk/7P/8
```
White to play and win.
solution below
7. Position 6.14: In Forsyth notation:
```8/8/2p2k2/pp3Pp1/6P1/1KP5/P7/8
```
White to move and win. Black to move and draw.
solution below
Homework solutions:
1. 1 … Kf6 2 Ke3 Ke5 3 Kd3 (3 h5 draws – see basic endgame position 3) Kd5 4 Kc3 Ke5! 5 Kc4 Ke4! draw.
2. 1 a4 … 2 b5 axb5 3 axb5 and white wins because white’s outside passed pawn deflects the black king and allows the white king to penetrate and mop up black’s pawns.
3. 1 fxg5 hxg5 2 Kf3 Kg6 3 Kg4 f5+ 4 exf6ep Kxf6 5 g3 e5 6 h4 and white creates and outside passed pawn. This passed pawn can be used to deflect the black king to the h-file, allowing the white king to mop up blacks other pawns.
4. 1 Kf2 Kc5 (1 … Kd6 2 a4 Kc7 3 a5 leads to a draw since white queens first and can obtain either a perpetual check or trade queens and then capture the g4 pawn) 2 Kg3 Kb5 3 Kxg4 Ka4 4 Kf5! (the only move which draws!) Kb3 5 a4! (winning a tempo) Kxa4
6 Ke4 Kb3 7 Kd4 draw.
5. White’s “outside” passed pawn and better king position wins. 1 Kc3! e4 (1 … Kc6 2 Kc4) 2 Kd4 Kc6 3 Kxe4 Kxc5 4 Kf4 and white wins.
6. 1 Ke4 Kg4 2 h4 Kh5 3 Kf4 Kg6 4 g4 Kh6 5 h5 Kh7 6 g5 Kg7 7 g6! (7 h6? Kg6 draw.) Kh6 8 Kg4 Kg7 9 Kg5 d3 10 h6+ Kg8 11 Kf6 d2 12 h7+ with mate to follow in a few moves.
7. White to move: 1 a4 Ke5 (1 … bxa4 leaves black’s queenside pawns too weak) 2 axb5 cxb5 3 c4 b4 4 c5 and white wins. Black to move: 1 … c5 2 a3 Ke5 3 Kc2 Kd5 4 Kd3 Ke5 and it’s a draw since white can’t penetrate. Note that black should not play … a4 or … c4 since this would weaken the queenside pawns too much.
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# PF3 Lewis structure
PF3 (phosphorus trifluoride) has one phosphorus atom and three fluorine atoms.
In the PF3 Lewis structure, there are three single bonds around the phosphorus atom, with three fluorine atoms attached to it. Each fluorine atom has three lone pairs, and the phosphorus atom has one lone pair.
Contents
## Steps
To properly draw the PF3 Lewis structure, follow these steps:
#1 Draw a rough sketch of the structure
#2 Next, indicate lone pairs on the atoms
#3 Indicate formal charges on the atoms, if necessary
Let’s break down each step in more detail.
### #1 Draw a rough sketch of the structure
• First, determine the total number of valence electrons
In the periodic table, phosphorus lies in group 15, and fluorine lies in group 17.
Hence, phosphorus has five valence electrons and fluorine has seven valence electrons.
Since PF3 has one phosphorus atom and three fluorine atoms, so…
Valence electrons of one phosphorus atom = 5 × 1 = 5
Valence electrons of three fluorine atoms = 7 × 3 = 21
And the total valence electrons = 5 + 21 = 26
• Second, find the total electron pairs
We have a total of 26 valence electrons. And when we divide this value by two, we get the value of total electron pairs.
Total electron pairs = total valence electrons ÷ 2
So the total electron pairs = 26 ÷ 2 = 13
• Third, determine the central atom
We have to place the least electronegative atom at the center.
Since phosphorus is less electronegative than fluorine, assume that the central atom is phosphorus.
Therefore, place phosphorus in the center and fluorines on either side.
• And finally, draw the rough sketch
### #2 Next, indicate lone pairs on the atoms
Here, we have a total of 13 electron pairs. And three P — F bonds are already marked. So we have to only mark the remaining ten electron pairs as lone pairs on the sketch.
Also remember that phosphorus is a period 3 element, so it can keep more than 8 electrons in its last shell. And fluorine is a period 2 element, so it can not keep more than 8 electrons in its last shell.
Always start to mark the lone pairs from outside atoms. Here, the outside atoms are fluorines.
So for each fluorine, there are three lone pairs, and for phosphorus, there is one lone pair.
Mark the lone pairs on the sketch as follows:
### #3 Indicate formal charges on the atoms, if necessary
Use the following formula to calculate the formal charges on atoms:
Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons
For phosphorus atom, formal charge = 5 – 2 – ½ (6) = 0
For each fluorine atom, formal charge = 7 – 6 – ½ (2) = 0
Here, both phosphorus and fluorine atoms do not have charges, so no need to mark the charges.
In the above structure, you can see that the central atom (phosphorus) forms an octet. And the outside atoms (fluorines) also form an octet. Hence, the octet rule is satisfied.
Therefore, this structure is the stable Lewis structure of PF3.
Next: PCl5 Lewis structure
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# GATE | GATE CS 1997 | Question 42
• Difficulty Level : Expert
• Last Updated : 02 May, 2018
Let G be the graph with 100 vertices numbered 1 to 100. Two vertices i and j are adjacent iff |i−j|=8 or |i−j|=12. The number of connected components in G is
(A) 8
(B) 4
(C) 12
(D) 25
Explanation: When vertices are arranged with difference of 8 there are 8 components as shown by 8 columns in the image below:
When vertices are arranged with difference of 12 the number of components is reduced to 4 as first column will be connected with fifth column, second column will be connected with sixth column, third column will be connected with seventh column and fourth column will be connected with eighth column. No other form of connection exists so total 4 connected components are there.
So, option (B) is correct.
This explanation is contributed by Pradeep Pandey.
Quiz of this Question
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# chemistry
What is 9.3 X 10 to 5th power
1. 👍 0
2. 👎 0
3. 👁 34
1. 930000 to two sig figs
1. 👍 0
2. 👎 0
posted by Dr Russ
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A239957 a(n) = |{0 < g < prime(n): g is a primitive root modulo prime(n) of the form k^2 + 1}|. 12
1, 1, 1, 1, 1, 1, 2, 2, 3, 3, 1, 3, 2, 2, 3, 4, 4, 4, 2, 1, 2, 1, 3, 3, 6, 3, 3, 7, 4, 5, 2, 8, 3, 5, 6, 1, 2, 5, 8, 10, 7, 3, 2, 6, 8, 2, 3, 5, 8, 4, 7, 4, 2, 5, 8, 9, 10, 5, 8, 6, 10, 6, 4, 9, 6, 9, 5, 3, 13, 5, 8, 9, 5, 6, 8, 13, 13, 6, 6, 5 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,7 COMMENTS Conjecture: (i) a(n) > 0 for all n > 0. In other words, every prime p has a primitive root 0 < g < p of the form k^2 + 1, where k is an integer. (ii) If p > 3 is a prime not equal to 13, then p has a primitive root 0 < g < p which is of the form k^2 - 1, where k is a positive integer. See also A239963 for a similar conjecture. LINKS Zhi-Wei Sun, Table of n, a(n) for n = 1..10000 EXAMPLE a(6) = 1 since 1^2 + 1 = 2 is a primitive root modulo prime(6) = 13. a(11) = 1 since 4^2 + 1 = 17 is a primitive root modulo prime(11) = 31. a(20) = 1 since 8^2 + 1 = 65 is a primitive root modulo prime(20) = 71. a(22) = 1 since 6^2 + 1 = 37 is a primitive root modulo prime(22) = 79. a(36) = 1 since 9^2 + 1 = 82 is a primitive root modulo prime(36) = 151. MATHEMATICA f[k_]:=k^2+1 dv[n_]:=Divisors[n] Do[m=0; Do[Do[If[Mod[f[k]^(Part[dv[Prime[n]-1], i]), Prime[n]]==1, Goto[aa]], {i, 1, Length[dv[Prime[n]-1]]-1}]; m=m+1; Label[aa]; Continue, {k, 0, Sqrt[Prime[n]-2]}]; Print[n, " ", m]; Continue, {n, 1, 80}] PROG (PARI) ispr(n, p)=my(f=factor(p-1)[, 1], m=Mod(n, p)); for(i=1, #f, if(m^(p\f[i])==1, return(0))); m^(p-1)==1 a(n)=my(p=prime(n)); sum(k=0, sqrtint(p-2), ispr(k^2+1, p)) \\ Charles R Greathouse IV, May 01 2014 CROSSREFS Cf. A000040, A002522, A236308, A236966, A237112, A237121, A237594, A239963. Sequence in context: A071434 A227314 A128924 * A230040 A242361 A116464 Adjacent sequences: A239954 A239955 A239956 * A239958 A239959 A239960 KEYWORD nonn AUTHOR Zhi-Wei Sun, Apr 23 2014 STATUS approved
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# Free Printable Grade 4 Math Worksheets
Maryl Anaëlle April 24, 2021 Math Worksheet
Use heart-shaped candy to teach the more simple mathematical arithmetic; like addition and subtraction. Provide children with pre-determined math problems that they can figure out using the sweets. Alternatively, make heart-shaped numeral cards that children can arrange on their desk to solve a math problem. You could give them a total to reach and they must use the cards to create the question.
So, what it takes to be smart in mathematics? My answer is; stay focused on math in each and every level of your studies. Participate in your class math practice sessions. Ask your teacher lots of questions until you are not clear about any concept. Mathematics is a subject of solving the problems on paper by hand rather than only to read them. As in case of Social Studies taking more readings make you smart, in math practicing lots of problems and solving them by hand makes you smart.
There are websites that offer a vast array of Valentine‘s Day worksheets. They offer varying levels of difficulty to make them suitable for Grade One to Grade Five. Each set of math worksheets include all the major mathematical problems; addition, subtraction, multiplication, division and even word problems. All these sheets include Valentine‘s Day pictures. The online math worksheets are also available with pre-written word problems that are all associated with what happens on Valentine‘s Day.
When a child learns to relate math to everyday questions, he will be great at it from the simplest addition all the way to trigonometry. To convert percentages, decimals and fractions is thus one essential skill. How much of an apple pie has been eaten? The answer to this question can be expressed in percentages, 50%; or in decimals, 0.5; or in fraction, ½. In other words, half of mom‘s delicious apple pie is gone. How many kids in school have done their homework? Again this can be answered in several ways: in percentages, 70%; or in ratio, 7:10; Both of these mean out of ten kids in class there are seven good ones who did and three not-so-good ones who didn‘t. The bottom line is that kids learn math much better when it makes sense.
Instead of learning a topic and then doing lots of mathematical examples, based on what you have just learned, teachers have found that the use of interactive activities, learning games, printable worksheets, assessments, and reinforcement. the math curriculum should rely on many learning tools - lessons with activities, worksheets, reinforcement exercises, and assessments will help a student to learn each math topic in a variety of ways and this should help supplement the teaching in class.
None of this is true, of course. So, first you need to bite your tongue. Secondly, get your child interested in math from an early age. Math does not have to be about columns and columns of figures, and the sooner you can introduce your child to math in a fun way the better. Mathematics worksheets can help you do this, with pages of fun exercises that will teach your child the basic principles they need. This helps you, as you don‘t have to write out pages of math exercises and try to make them interesting - it also helps your child.
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# Why are all rational numbers not fractions?
• Last Updated : 19 May, 2022
The number system encompasses several sorts of numbers, such as prime numbers, odd numbers, even numbers, rational numbers, whole numbers, and so on. Numbers can be expressed by using these figures and words. For example, integers like 40 and 65 stated as figures can alternatively be written as forty and sixty-five.
### Number System
A number system, often known as a numeral system, is a basic system for expressing numbers and figures. It is a one-of-a-kind method of representing numbers in arithmetic and algebraic structures. Numbers are used in various arithmetic values to carry out various arithmetic operations such as addition, subtraction, multiplication, and so on that are utilized in daily life for the purpose of computation. A number’s value is defined by its digit, its position in the number, and the number system’s base. Numbers, often known as numerals, are mathematical values used for counting, measuring, identifying, and quantifying basic quantities.
### Rational Number
Rational numbers are written in the form p/q, where p and q are integers and q≠0, because of the fundamental structure of numbers, the p/q form, most individuals have difficulties discriminating between fractions and rational numbers. When a rational number is split, the result is a decimal number that might be either ending or repeating. Examples of rational numbers are 2, 3, 4, and so on, which may be expressed in fraction form as 2/1, 3/1, and 4/1, respectively.
Fractions
Fractions are defined as numerical figure that represents a part of a whole. A fraction is a component or section of any quantity taken from a whole, which can be any number, a specified value, or an item. All fractions are represented by a numerator and a denominator, which are separated by a horizontal bar called the fractional bar.
• The denominator represents the number of parts into which the whole has been subdivided. It is positioned below the fractional bar at the lowest part of the fraction.
• The numerator specifies how many fractional parts are represented or selected. It is positioned above the fractional bar at the upper part of the fraction.
Examples of fractions: 3/2, 7/4, 33/26, etc.
### Why are all rational numbers not fractions?
Answer:
Rational numbers are written in the form p/q, where p and q are integers and q≠0 . Because of the fundamental structure of numbers, the p/q form, most individuals have difficulties discriminating between fractions and rational numbers.
All rational numbers are not fraction because a rational number is defined as the ratio of integers, it cannot be called a fraction. As a result, if we take the ratio of a negative integer to a positive integer, such as – 4/9 or – 31/70, we do not receive a fraction since a fraction can only be the ratio of two whole numbers, and all whole numbers are positive.
As a result, it is possible to conclude that every rational number cannot be a fraction. Example: 2/-5 is a rational number but it is not a fraction because its denominator is not a natural number.
### Sample Questions
Question 1: Is fraction 15/3 rational or not?
Answer:
Rational numbers are written in the form p/q, where p and q are integers and q≠0 . Because of the fundamental structure of numbers, the p/q form, most individuals have difficulties discriminating between fractions and rational numbers. Here Given 15/3 , we can simplify it by dividing 15/3 is 5 , therefore 5 can be written as 5/1 hence its a fraction and rational number.
Question 2: Identify whether 3/-6 is a fraction or not?
Answer:
Fractions are defined as a numerical figure that represents a part of a whole. A fraction is a component or section of any quantity taken from a whole, which can be any number, a specified value, or an item.
Hence Given , 3/-6 can be written as 1/-2 its a rational number but its not a fraction as fraction only includes whole number and whole number does not include negative integers .
Question 3: Identify whether 16/3 is rational or fraction or both?
Answer:
Rational numbers are written in the form p/q, where p and q are integers and q≠0 . Because of the fundamental structure of numbers, the p/q form, most individuals have difficulties discriminating between fractions and rational numbers. Hence Given, 16/3 is rational number as well as fraction because both denominator or numerator are whole numbers.
Question 4: Identify from the numbers which are rational or which are fractions, 5/4, -2/3, 7/-8, 6/8, 5/10,
Answer:
Rational numbers : 5/4 , 6/8 , 5/10 ,-2/3 , 7/-8 all are rational numbers but not fractions. Fractions: 5/4 , 6/8 , 5/10 all are fractions as well as rational numbers.
Question 5: Is 35 a rational number?
Answer:
Rational numbers are written in the form p/q, where p and q are integers and q≠0, because of the fundamental structure of numbers, the p/q form, most individuals have difficulties discriminating between fractions and rational numbers. Hence 35 here is a rational number as it can be expressed in form of p/q as 35/1.
Question 6: Identify whether it’s a fraction or not, if 4/-5 is a rational number?
Answer:
Rational numbers are written in the form p/q, where p and q are integers and q≠0 . Because of the fundamental structure of numbers, the p/q form, most individuals have difficulties discriminating between fractions and rational numbers. All rational numbers are not fraction because a rational number is defined as the ratio of integers, it cannot be called a fraction. As a result, if we take the ratio of a negative integer to a positive integer, such as – 4/9 or – 31/70, we do not receive a fraction since a fraction can only be the ratio of two whole numbers, and all whole numbers are positive.
As a result, it is possible to conclude that every rational number cannot be a fraction. Hence. given 4/-5 is rational number but its not a fraction.
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# Radius, Diameter, and Circumference of a Circle Lesson 7 (3 rd 6 Weeks) TEKS 6.6C.
## Presentation on theme: "Radius, Diameter, and Circumference of a Circle Lesson 7 (3 rd 6 Weeks) TEKS 6.6C."— Presentation transcript:
Radius, Diameter, and Circumference of a Circle Lesson 7 (3 rd 6 Weeks) TEKS 6.6C
Circle A closed curve with all its points the same distance from a fixed point called the center. Center
A radius (r) of a circle is a line segment joining the center of the circle and any point on the circle. The radius of a circle is half the diameter, or r = ½ d. fffff
A diameter (d) of a circle is a line segment that joins any two points on the circle and passes through the center of the circle. The diameter of a circle is twice the radius, or d = 2r fffff f f
The circumference (C) of a circle is the distance around the circle. fffff
Pi ( ) The ratio of the circumference to the diameter. Which is approximately 3.14 or 3 Circumference Diameter C d ==
Formulas for Circumference C = d The circumference of any circle is equal to times the diameter. Since the diameter of any circle is equal to twice the radius, the circumference can also be found with this formula: C = 2 r or C = 2 r
Download ppt "Radius, Diameter, and Circumference of a Circle Lesson 7 (3 rd 6 Weeks) TEKS 6.6C."
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The radius and slant height of a cone are In the ratio of 4 : 7. If its curved surface area is 792 cm2, find its radius. (Use it ? = 22/7).
## Text solution
Curved surface area=
Height
Now, CSA=792
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Question Text The radius and slant height of a cone are In the ratio of 4 : 7. If its curved surface area is 792 cm2, find its radius. (Use it ? = 22/7). Updated On Dec 28, 2022 Topic Surface Areas and Volumes Subject Mathematics Class Class 9 Answer Type Text solution:1 Video solution: 4 Upvotes 529 Avg. Video Duration 4 min
You are watching: The radius and slant height of a cone are In the ratio of 4 : 7. If its c... Info created by GBee English Center selection and synthesis along with other related topics.
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# How do you write the product as a trinomial (3x - 2)(x + 4)?
Jul 26, 2015
$\left(3 x - 2\right) \left(x + 4\right) = 3 {x}^{2} + 10 x - 8$
$\left(3 x - 2\right) \left(x + 4\right) = 3 {x}^{2} + 12 x - 2 x - 8 = 3 {x}^{2} + 10 x - 8$
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# 6.1: Factor by Grouping
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# FFT Question
Started by September 2, 2020
Why is it that in older textbooks the DFT or FFT is defined as having scaling of (1/N) for direct FFT whereas in newer ones it doesn't? If we take the DFT of say 1,1,1,1 the answer is 4 but surely an answer of 1 makes more sense as it is the average value. of 4 ones.
I know people say - "Oh we can scale it later", but it doesn't make mathematical sense. The other point is that the Fourier transform doesn't apply to a step function anyway so why does the DFT?
Am 02.09.20 um 07:00 schrieb Tom Killwhang:
> Why is it that in older textbooks the DFT or FFT is defined as having scaling of (1/N) for direct FFT whereas in newer ones it doesn't? If we take the DFT of say 1,1,1,1 the answer is 4 but surely an answer of 1 makes more sense as it is the average value. of 4 ones. > > I know people say - "Oh we can scale it later", but it doesn't make mathematical sense. The other point is that the Fourier transform doesn't apply to a step function anyway so why does the DFT?
I think this for technical reasons. If you program FFT with twiddle factors taken from the unit circle you get a total scale factor of N for a full turn around. Typically it is less work to compensate for this factor at another place. Take also into account the scale factor of the inverse FFT. If you scale by 1/N you must *not* scale the inverse FFT to get the original values back. If you want to operate symmetrically you need to scale both by 1/sqrt(N) which is even more non intuitive. But this keeps the RMS energy constant. Marcel
On 2020-09-02 03:01, Marcel Mueller wrote:
> Am 02.09.20 um 07:00 schrieb Tom Killwhang: >> Why is it that in older textbooks the DFT or FFT is defined as having >> scaling of (1/N) for direct FFT whereas in newer ones it doesn't? If >> we take the DFT of say 1,1,1,1 the answer is 4 but surely an answer of >> 1 makes more sense as it is the average value. of 4 ones. >> >> I know people say - "Oh we can scale it later", but it doesn't make >> mathematical sense. The other point is that the Fourier transform >> doesn't apply to a step function anyway so why does the DFT? > > I think this for technical reasons. If you program FFT with twiddle > factors taken from the unit circle you get a total scale factor of N for > a full turn around. Typically it is less work to compensate for this > factor at another place. > > Take also into account the scale factor of the inverse FFT. If you scale > by 1/N you must *not* scale the inverse FFT to get the original values > back. If you want to operate symmetrically you need to scale both by > 1/sqrt(N) which is even more non intuitive. But this keeps the RMS > energy constant.
The question of how to normalize FFTs goes back way before computers were even invented, on account of that vexing factor of 2*pi. The EE contingent tends to define the continuous-time FT with the 2*pi factors in the exponent, i.e. G(f) = F(g) = integral[-inf, inf] exp( j 2 pi f t) g(t) dt and g(t) = Finv(G) = integral[-inf, inf] exp( -j 2 pi f t) G(f) df. The physics and math contingents tend to use omega = 2 pi f as the independent frequency variable, so that the 2*pi winds up as a multiplicative constant outside the integral, either asymmetrically as 1/(2 pi) on the inverse transform, or symmetrically, as in 1/sqrt(2 pi) on both forward and reverse. The main effect is on the statement of FT theorems. IMNSHO putting the 2*pi in the exponent is a huge win, in both the continuous and discrete cases Cheers Phil Hobbs -- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC / Hobbs ElectroOptics Optics, Electro-optics, Photonics, Analog Electronics Briarcliff Manor NY 10510 http://electrooptical.net http://hobbs-eo.com
Tom Killwhang wrote:
> Why is it that in older textbooks the DFT or FFT is defined as having > scaling of (1/N) for direct FFT whereas in newer ones it doesn't? If > we take the DFT of say 1,1,1,1 the answer is 4 but surely an answer > of 1 makes more sense as it is the average value. of 4 ones. > > I know people say - "Oh we can scale it later", but it doesn't make > mathematical sense. The other point is that the Fourier transform > doesn't apply to a step function anyway so why does the DFT? >
A lot of implementations don't scale the output. Most prominently FFTW. Some do. It's sort of open to the implementation. I believe the argument for not doing it is somet5hing like roundoff error. And I'd call a step function ( all 1's ) a pretty good test case for an FFT. Not much point in using an FFT on one if you know it's a step function in an implementation but it's a quick check of an implementation -- Les Cargill
On Thursday, September 3, 2020 at 3:47:41 PM UTC-7, Phil Hobbs wrote:
(snip)
> The question of how to normalize FFTs goes back way before computers > were even invented, on account of that vexing factor of 2*pi.
(snip of EE contingent)
> The physics and math contingents tend to use omega = 2 pi f as the > independent frequency variable, so that the 2*pi winds up as a > multiplicative constant outside the integral, either asymmetrically as > 1/(2 pi) on the inverse transform, or symmetrically, as > in 1/sqrt(2 pi) on both forward and reverse. The main effect is on the > statement of FT theorems.
> IMNSHO putting the 2*pi in the exponent is a huge win, in both the > continuous and discrete cases
Not only in the Fourier transform, but in all the rest of the equations, omega works out much better than f. It is only convenient to use f when you are actually counting frequencies. (That is, no-one makes frequency counters to output in omega.) There are some who use wavelength over 2pi, but that is rare. Using omega and k, you get exp(i.k.x - i.omega.t) for propagating waves. (Where k is the wave number or wave vector, magnitude 2 pi over wavelength.) It is, then, convenient for the Fourier transform to put the 1/(2pi) along with the d omega. In the case of FFT in fixed point, best is to do all with enough bits so as not to overflow and not divide by N until the end, if it is needed. In floating point with radix other than 2, it is a little complicated where to put the divide, but note that bits can be lost when dividing by two. (Specifically, IBM used radix 16 for many years, and still supports it, along with 2 a little recently, and 10 on newer machines. Yes, some machines support all three.) Step functions in continuous Fourier transforms are not rare, but a little complicated. Delta functions are not unusual. But for DFT (defined for periodic band-limited signals) a step function is not all that special. Especially for small N, where it isn't all that steep. And note that this all applies for DST or DCT, too.
On 2020-09-09 19:12, [email protected] wrote:
> On Thursday, September 3, 2020 at 3:47:41 PM UTC-7, Phil Hobbs wrote: > > (snip) > >> The question of how to normalize FFTs goes back way before computers >> were even invented, on account of that vexing factor of 2*pi. > > > (snip of EE contingent) > >> The physics and math contingents tend to use omega = 2 pi f as the >> independent frequency variable, so that the 2*pi winds up as a >> multiplicative constant outside the integral, either asymmetrically as >> 1/(2 pi) on the inverse transform, or symmetrically, as >> in 1/sqrt(2 pi) on both forward and reverse. The main effect is on the >> statement of FT theorems. > >> IMNSHO putting the 2*pi in the exponent is a huge win, in both the >> continuous and discrete cases > > Not only in the Fourier transform, but in all the rest of the equations, > omega works out much better than f. It is only convenient to use f > when you are actually counting frequencies. (That is, no-one makes > frequency counters to output in omega.) There are some who use > wavelength over 2pi, but that is rare. Using omega and k, you get > exp(i.k.x - i.omega.t) for propagating waves. (Where k is the > wave number or wave vector, magnitude 2 pi over wavelength.)
Sure, when the spatial transform is in view as well, I use exp(i(omega t - k dot x)) too. (I'm a physicist, after all.) Some years back I wrote a scalable clusterized FDTD simulator, and the far-field and order-source calculations are all done in the (k, omega) basis. But in a DSP context, putting the 2*pi in the exponent is almost always a win IMO. Cheers Phil Hobbs -- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC / Hobbs ElectroOptics Optics, Electro-optics, Photonics, Analog Electronics Briarcliff Manor NY 10510 http://electrooptical.net http://hobbs-eo.com
I noticed another problem that arises from this. The units don't match up w=
hen you use Parseval's theorem and the FFT. Look ta the equation near the b=
ottom that uses the FFT for Parseval's theorem
https://en.wikipedia.org/wiki/Parseval%27s_theorem
I did a simulation with white noise through a bandpass filter.=20
Now I found the periodogram and this gives me power vs frequency. when you =
sum them up you do indeed get the same answer - but the expression on the l=
eft of the equation is the sum of squares of a random variable and that is =
not power. The power is variance after you divide by N. Hence both sides ne=
ed to be divided by N otherwise we are equating energies and not power. I a=
lways understood variance (with no dc) is average power and not energy.
Thx
On Thursday, September 10, 2020 at 11:12:25 AM UTC+12, [email protected]=
du wrote:
> On Thursday, September 3, 2020 at 3:47:41 PM UTC-7, Phil Hobbs wrote:=20 >=20 > (snip) > > The question of how to normalize FFTs goes back way before computers=20 > > were even invented, on account of that vexing factor of 2*pi. > (snip of EE contingent) > > The physics and math contingents tend to use omega =3D 2 pi f as the=20 > > independent frequency variable, so that the 2*pi winds up as a=20 > > multiplicative constant outside the integral, either asymmetrically as=
=20
> > 1/(2 pi) on the inverse transform, or symmetrically, as=20 > > in 1/sqrt(2 pi) on both forward and reverse. The main effect is on the=
=20
> > statement of FT theorems.=20 >=20 > > IMNSHO putting the 2*pi in the exponent is a huge win, in both the=20 > > continuous and discrete cases > Not only in the Fourier transform, but in all the rest of the equations,=
=20
> omega works out much better than f. It is only convenient to use f=20 > when you are actually counting frequencies. (That is, no-one makes=20 > frequency counters to output in omega.) There are some who use=20 > wavelength over 2pi, but that is rare. Using omega and k, you get=20 > exp(i.k.x - i.omega.t) for propagating waves. (Where k is the=20 > wave number or wave vector, magnitude 2 pi over wavelength.)=20 >=20 > It is, then, convenient for the Fourier transform to put the 1/(2pi)=20 > along with the d omega.=20 >=20 > In the case of FFT in fixed point, best is to do all with enough bits so=
=20
> as not to overflow and not divide by N until the end, if it is needed.=20 >=20 > In floating point with radix other than 2, it is a little complicated whe=
re=20
> to put the divide, but note that bits can be lost when dividing by two.=
=20
> (Specifically, IBM used radix 16 for many years, and still supports it,=
=20
> along with 2 a little recently, and 10 on newer machines. Yes,=20 > some machines support all three.)=20 >=20 > Step functions in continuous Fourier transforms are not rare, but=20 > a little complicated. Delta functions are not unusual. But for DFT=20 > (defined for periodic band-limited signals) a step function is=20 > not all that special. Especially for small N, where it isn't all=20 > that steep. And note that this all applies for DST or DCT, too.
On Wednesday, September 9, 2020 at 8:09:15 PM UTC-7, [email protected] wrote:
> I noticed another problem that arises from this. The units don't match up when you > use Parseval's theorem and the FFT. Look ta the equation near the bottom that uses the FFT for Parseval's theorem > https://en.wikipedia.org/wiki/Parseval%27s_theorem > I did a simulation with white noise through a bandpass filter.
> Now I found the periodogram and this gives me power vs frequency. when you sum them > up you do indeed get the same answer - but the expression on the left of the equation > is the sum of squares of a random variable and that is not power. > The power is variance after you divide by N. Hence both sides need to be divided by N > otherwise we are equating energies and not power. > I always understood variance (with no dc) is average power and not energy.
Seems like a usual problem when working with discrete signals. There should be a factor of the sampling period to make it look like an integral, so that would make it energy instead of power. But yes, it is usual to factor out the (presumed constant and known) sampling rate. If you consider the earlier integral on the page, it is over one period of a periodic function. That normalizes out the length (time) of the period. Seems not worse than many mathematics formulas which completely get units wrong. That is, for example, add quantities with different units. When you use them, you have to put back any normalizations that they left out, such as sampling rate. Conveniently, in many cases you can ignore the sampling rate, for example when designing filters, after you factor it out.
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Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °
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## Topic review (newest first)
Flowers4Carlos
2005-08-27 11:37:14
when it comes to division or multiplication, it doesn't matter which one u do first. u can either divide right away:
(12/2)/x= 6/x and 6/6=1
or u can simply the denominator first:
12/[2(6)]= 12/12=1
either way u get the same answer!
frostysmom
2005-08-27 10:25:41
I'm sorry but I don't understand why you would multiply 2x first? That's what's really getting me...There aren't parentheses around the 2x so why wouldn't you divide then multiply?
MathsIsFun
2005-08-27 10:16:14
No, it would be 1 because we would mulitiply "2x" first.
Just like we would if it were:
2w + 2y
---------
2x
frostysmom
2005-08-27 10:13:42
So the correct answer is 36. (right?)
MathsIsFun
2005-08-27 09:44:58
I think I see the problem, it is what order to do "/2x". Do you divide by 2 then multiply by x? In other words:
You could have ((2w + 2y) / 2) × x = ((4 + 8) / 2) × 6 = 6×6 = 36
Or: ((2w + 2y) / (2 × x) = ((4 + 8) / (2 × 6) = 12/12 = 1
When we see something like "2x" we automatically think to multiply it first (unless there were brackets and exponents), so we are blinded to the possibility of it behaving the other way.
But if it were written: (2w + 2y) / 2 × x, then we would look at it and go "hmmm... maybe you should multiply by x last"
Maybe that is a little sub-rule that should go on the PEMDAS page!
frostysmom
2005-08-27 09:17:42
Thanks alot - I looked at the PEMDAS web page and I understand it to be like I said. In other words, multiplication and division, working from left to right whichever comes first. And in my mind, it supports my answer.
Then again, could be I'm still being blinded by my need to be right .
MathsIsFun
2005-08-27 08:54:55
Welcome to the forum, frostysmom !
im really bored
2005-08-27 08:46:09
I think the teachers answer is right, in the order of operations it goes: parentheses, exponets, multiplication, division, addition, then subtraction. So in the problem first you would put in the given variables, (2(2) + 2(4) / 2(6), then (4 + 8) / 12, then 12 / 12, which simplifys to 1.
frostysmom
2005-08-27 08:01:19
Hi, this is my first post here. Let me give everyone a little background. My son is in 7th grade and is just beginning algebra. It's been a while since I've been in school and I'm trying to help my son. While checking his homework I cam across a problem that I thought he'd done wrong. We worked it together and came up with the answer. He went to school the next day and the teacher told him the answer he had was wrong. I disagreed and sent a note to the teacher asking her where we'd gone wrong. She sent home an answer that I'm still not satisfied with. My son (being the smart kid he is) went into Excel, typed in the expression (he simply entered the numbers instead of the letters) and Excel evaluated it with the answer that we got.
So...here's the equation.
w=2, x=6, y=4
(2w + 2y) / 2x
Could someone please help? The answer that we got (and that Excel got) is 36. The teacher says the answer is 1. We know that what's in parentheses gets evaluated first. The question I have is what get's evaluated next - the division or the multiplication? Based on the math textbook I say it's the division and obviously the teacher says it's the multiplication.
Anyway, thanks for any and all help.
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We think you are located in South Africa. Is this correct?
# Polygons
## 8.3 Polygons (EMCJ9)
Proportionality in polygons
A plane, closed shape consisting of three or more line segments.
In previous grades we studied the properties of the following polygons:
Triangle Area $$= \frac{1}{2}b \times h$$ Parallelogram Area $$= b \times h$$ Rectangle Area $$= b \times h$$ Rhombus Area $$= \frac{1}{2} AC \times BD$$
Square Area $$= s^{2}$$ Trapezium Area $$= \frac{1}{2}(a + b) \times h$$ Kite Area $$= \frac{1}{2}(AC \times DB)$$
## Worked example 1: Properties of polygons
$$ABCD$$ is a rhombus with $$BD = \text{12}\text{ cm}$$ and $$AB:BD = 3:4$$.
Calculate the following (correct to two decimal places) and provide reasons:
1. Length of $$AB$$.
2. Length of $$AO$$.
3. Area of $$ABCD$$.
### Use the ratio to determine the length of $$AB$$
\begin{align*} AB:BD &= 3:4 \\ \therefore \frac{AB}{BD} &= \frac{3}{4} \\ \frac{AB}{12} &= \frac{3}{4} \\ AB &= 12 \times \frac{3}{4} \\ &= \text{9}\text{ cm} \end{align*}
### Calculate the length of $$AO$$
We use the properties of a rhombus and the theorem of Pythagoras to find $$AO$$.
$\begin{array}{rll} BD &= \text{12}\text{ cm} & \\ BO &= \text{6}\text{ cm} & \text{(diagonals bisect each other)} \\ \text{In } \triangle ABO, \quad A\hat{O}B &= \text{90}° & (\text{diagonals intersect at } \perp )\\ AO^{2} &= AB^{2} - BO^{2} & (\text{Pythagoras}) \\ &= 9^{2} - 6^{2} & \\ \therefore AO &= \sqrt{ 45} & \\ &= \text{6,71}\text{ cm} \end{array}$
### Determine the area of rhombus $$ABCD$$
\begin{align*} \text{Area } ABCD &= \frac{1}{2} AC \times BD \\ &= \frac{1}{2} (2 \times \sqrt{45})(12) \\ &= \text{80,50}\text{ cm$^{2}$} \end{align*}
## Proportionality of polygons
Exercise 8.3
$$MNOP$$ is a rectangle with $$MN : NO = 5:3$$ and $$QN = \text{10}\text{ cm}$$.
Calculate $$MN$$ (correct to $$\text{2}$$ decimal places)
.
$\begin{array}{rll} QN &= 10 & \\ \therefore NP &= 2 \times QN & (\text{diagonals bisect each other}) \\ &= 20 & \\ \text{In } \triangle NOP, \quad \hat{O} &= \text{90}° &(MNOP \text{ rectangle }) \\ \text{Let } MN &= 5x & \\ \text{And } NO &= 3x & \\ NP^{2} &= NO^{2} + OP^{2} & (\text{Pythagoras}) \\ (20)^{2} &= (3x)^{2} + (5x)^{2} & \\ 400 &= 9x^{2} + 25x^{2} & \\ 400 &= 34x^{2} & \\ \therefore x &= \sqrt{\frac{400}{34}} & \\ &= \text{3,43} \ldots & \\ \therefore MN &= 5x = \text{17,15}\text{ cm} & \end{array}$
Calculate the area of $$\triangle OPQ$$ (correct to $$\text{2}$$ decimal places).
Consider the trapezium $$ABCD$$ shown below. If $$t:p:q = 2:3:5$$ and area $$ABCD = \text{288}\text{ cm^{2}}$$, calculate $$t, p \text{ and } q$$.
\begin{align*} \text{Let } t &= 2x \\ \text{And } p &= 3x \\ \text{And } q &= 5x \\ \text{Area } ABCD &= \frac{1}{2} (p + q) \times t \\ 288 &= \frac{1}{2} (3x + 5x ) \times 2x \\ 288 &= 8x^{2} \\ 36 &= x^{2} \\ \therefore 6 &= x \quad (\text{length always positive}) \\ \therefore t &= 2(6) = \text{12}\text{ cm} \\ p &= 3(6) = \text{18}\text{ cm} \\ q &= 5(6) = \text{30}\text{ cm} \end{align*}
$$ABCD$$ is a rhombus with sides of length $$\frac{3}{2}x$$ millimetres. The diagonals intersect at $$O$$ and length $$DO = x$$ millimetres. Express the area of $$ABCD$$ in terms of $$x$$.
\begin{align*} AD &= \frac{3}{2}x \\ DO&=x \\ AO^2&=\left ( \frac{3}{2}x \right )^2-x^2 \quad (\text{ Pythagoras})\\ &=\frac{9}{4}x^2-x^2\\ &=\frac{5}{4}x^2 \\ \therefore AO&=\frac{x\sqrt{5}}{2} \\ \therefore AC&=x\sqrt{5} \\ \text{Area }&=\frac{1}{2}AC\times BD\\ &=\frac{1}{2}\times x \sqrt{5} \times 2x \\ &= \sqrt{5} x^2 \end{align*}
In the diagram below, $$FGHI$$ is a kite with $$FG = \text{6}\text{ mm}$$, $$GK = \text{4}\text{ mm}$$ and $$\frac{GH}{FI} = \frac{5}{2}$$.
Determine $$FH$$ (correct to the nearest integer).
\begin{align*} \frac{GH}{FI} &= \frac{5}{2} \\ \text{And } FG &= FI \quad (\text{adj. sides of kite equal}) \\ \frac{GH}{FG} &= \frac{5}{2} \\ \frac{GH}{6} &= \frac{5}{2} \\ \therefore GH &= \text{15}\text{ mm} \\ \text{In } \triangle FGK, \quad FG^{2} &= GK^{2} + FK^{2} \\ FK^{2} &= 6^{2} - 4^{2} \\ &= 36 - 16 \\ \therefore FK &= \sqrt{20} \\ \text{In } \triangle GKH, \quad GH^{2} &= GK^{2} + FH^{2} \\ 15^{2} &= 4^{2} + KH^{2} \\ 225 - 16 &= KH^{2} \\ \therefore KH &= \sqrt{209} \\ FH &= FK + KH \\ &= \sqrt{20} + \sqrt{209} \\ &= \text{19}\text{ mm} \end{align*}
Calculate area $$FGHI$$.
\begin{align*} \text{Area } FGHI &= \frac{1}{2} GI \times FH \\ &= \frac{1}{2} (4 + 4)(19) \\ &= \text{76}\text{ mm$^{2}$} \end{align*}
$$ABCD$$ is a rhombus. $$F$$ is the mid-point of $$AB$$ and $$G$$ is the mid-point of $$CB$$. Prove that $$EFBG$$ is also a rhombus.
$\begin{array}{rll} AF &= FB & (\text{given}) \\ AE &= EC & (\text{diagonals bisect}) \\ \therefore FE &\parallel BC & \\ \therefore FE &\parallel BG & \\ \therefore FE &= \frac{1}{2} BG & (\text{mid-point th.}) \\ FE &= BG & \\ \therefore EFBG & \text{ is a parallelogram} & (\text{one pair opp. sides} = \text{ and } \parallel) \\ \therefore FB &\parallel EG & (\text{opp. sides of parm.}) \\ \text{And } AB &= BC & (\text{adj. sides of rhombus}) \\ \therefore \frac{1}{2} AB &= \frac{1}{2} BC & \\ \therefore FB &= BG = GE = EF & \\ \therefore EFBG &(\text{ is a rhombus}) & (\text{parm. with \text{4} equal sides}) \end{array}$
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Ders Bilgileri
#### Ders Tanımı
Ders Kodu Yarıyıl T+U Saat Kredi AKTS
LINEAR ALGEBRA I IME 203 3 3 + 1 4 5
Dersin Dili Türkçe Dersin Seviyesi Lisans Dersin Türü ZORUNLU Dersin Koordinatörü Prof.Dr. MELEK MASAL Dersi Verenler Prof.Dr. MELEK MASAL Dersin Yardımcıları Assist.Emine Nur BILGIC Dersin Kategorisi Alanına Uygun Temel Öğretim Dersin Amacı To show application fields of mathematics by getting mathematics easier and pleasurable by means of this course which contain basic equipment in mathematics and engineer and most important concepts of mathematical language Dersin İçeriği Vectors in R^2 and R^3, m x n matrices; addition and scaler product in matrices space, linear independence in matrice space, introduction to concept of vector space. Linear equation systems, Gauss elimination, subspaces. Linear independence and dimension. Linear transformations, relationship between linear transformations and matrices, matrices product, inverse of matrices and applications.
Dersin Öğrenme Çıktıları Öğretim Yöntemleri Ölçme Yöntemleri 1 - Explain matrixes and linear equations systems 1 - 4 - A - 2 - Express solution of linear equation systems by the help of elemanter operations and solve with this method 1 - 4 - A - 3 - Express solution of linear equation systems by the help of Gauss – Jordan Elimination Method and solve with this method 1 - 4 - A - 4 - Express concept of linear independence, linear dependance, basis, dimension and apply their applications 1 - 4 - A - 5 - Express and apply isomorphism and do applications 1 - 4 - A - 6 - Express and apply rank of a matrix and do applications 1 - 4 - A -
Öğretim Yöntemleri: 1:Lecture 4:Drilland Practice Ölçme Yöntemleri: A:Testing
#### Ders Akışı
Hafta Konular ÖnHazırlık
1 Short definition of group, ring, field
2 Matrixes, addition and scaler product in matrix space, matrix product and applications
3 Square matrix, inverse of a matrix, tranposition of a matrix, some special matrix and exercises
4 Echolon form of a matrix, elemanter operations, elemanter matrixes, theorem of factorisation
5 Matrix and linear equations system, solution of linear equation systems by the help of elemanter operations
6 Gauss – Jordan Elimination Method and exercises
7 Vectors on R^2 and R^3, vector spaces, subspaces, trivial space, proper subsapace
8 Linear independence, linear dependance, basis, dimension, applications
9 Mid – Term Exam
10 Theorem of basis complementation, quotient spaces
11 Direct sum
12 Isomorphisms and exercises
13 Rank of a matrix and related exercises
14 Application of rank ( solutions of linear equation systems )
Ders Notu
Ders Kaynakları
#### Dersin Program Çıktılarına Katkısı
No Program Öğrenme Çıktıları KatkıDüzeyi
1 2 3 4 5
1 Have general information about basic concept, theory and applicaitons on mathematics X
2 Have ability of mathematical thinking and apply to real life X
3 Classify a problem systematically also create comprehensible, understandable and objective solutions X
4 Establish relationship among events looked different X
5 Get clear and exact ideas about relationships among time, place and numbers X
6 Use principles of scientific method on problem solving X
7 Have character on explorer, impartial, disinterested, give sound decisions, open mind and believe that spreading information is needed X
8 Think creative and critical X
9 Improve methods as fast, understandable and practical to faced problems X
10 Have information about national and international modern problems
11 Get life long learning behaviour X
12 Know and apply approach, aim, goal, principle and techniques of teaching program on special field with basic value and principles of Turkish National Education System
13 Evaluate development and learning of students, provide to evaluate self-assessment and other students. Use the assessment results to better instruction; share the results with student, parent, administrators and teachers.
14 Endeavor for constant development by making self-evaluation. Open the new knowledge and ideas, have a role to develop himself/herself and his/her institution. Have enough knowledge and conscious to protection subjects on society values and environment.
15 Interrogator ( get habit to find source of knowledge by asking “Why?” question instead of direct acceptance to given knowledges ) X
#### Değerlendirme Sistemi
YARIYIL İÇİ ÇALIŞMALARI SIRA KATKI YÜZDESİ
AraSinav 1 60
KisaSinav 1 15
Odev 1 10
KisaSinav 2 15
Toplam 100
Yıliçinin Başarıya Oranı 50
Finalin Başarıya Oranı 50
Toplam 100
#### AKTS - İş Yükü
Etkinlik Sayısı Süresi(Saat) Toplam İş yükü(Saat)
Course Duration (Including the exam week: 16x Total course hours) 16 3 48
Hours for off-the-classroom study (Pre-study, practice) 16 3 48
Mid-terms 1 5 5
Quiz 3 5 15
Assignment 0 0 0
Final examination 1 5 5
Toplam İş Yükü 121
Toplam İş Yükü /25(s) 4.84
Dersin AKTS Kredisi 4.84
; ;
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# Pre-Calculus Forum
Pre-Calculus Help Forum: Pre-calculus does not involve calculus, but explores topics that will be applied in calculus
1. ### This forum is for PRE-Calculus. READ BEFORE POSTING
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2. ### Read this before posting in this subforum.
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1. ### Collinear vector
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2. ### How to find the x-intercept of this polynomial function?
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3. ### Measurable surds.
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4. ### Transformation
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5. ### Partial fraction Problem
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6. ### How to solve for a variable in a quadratic function with 2 variables?
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7. ### How do you solve a quadratic function when 'c' has a variable?
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8. ### Increasing/Decreasing Intervals of Slope for Rational Functions
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9. ### [SOLVED] Basic domain.
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10. ### What interest rate needed to double money in two years?
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11. ### Logarithmic exponenent
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12. ### [SOLVED] Even and odd functions.
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13. ### Polynomial Functions
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14. ### Population models.
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15. ### Best way to study for exam?
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16. ### [SOLVED] Solving ln sqrt(x+4) = 1
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17. ### [SOLVED] quick question on product rule and equality rule for logs
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18. ### Cubic equation
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20. ### Confused on a problem with finding the area that has an enclosure
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21. ### question of standard form for an ellipse
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22. ### Having Trouble Solving Logarithmic Equation
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23. ### Derivative of a function
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24. ### least action principle
• Replies: 4
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Oct 17th 2011, 09:32 AM
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# precalculus with limits page 294
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SfC Home > Physics > Wave Motion >
# Doppler Effect Frequency Equations
by Ron Kurtus (revised 27 March 2016)
The Doppler Effect causes the observed frequency of a waveform to change according to the velocity of the source and/or observer. The Doppler Effect frequency equations can derived by starting with the general wavelength equation.
After the general frequency equation is determined, you can find the frequency equations for a moving source and stationary observer and moving observer with a stationary source.
In the equations, it is assumed that the motion is constant and in the x-direction.
(See Conventions for Doppler Effect Equations for more information.)
Questions you may have include:
• What is the general frequency equation?
• What are the equations for a moving source and stationary observer?
• What are the equations for a moving observer and stationary source?
This lesson will answer those questions. Useful tool: Units Conversion
## General frequency equation
In order to establish the general Doppler Effect frequency equation—where both the source and observer are moving—you start with the previously derived general wavelength equation and put it in terms of frequency. Start with:
λO = λS(c − vS)/(c − vO)
where
• λO is the observed wavelength
• λS is the constant wavelength from the source
• c is the constant velocity of the wavefront in the x-direction
• vS is the constant velocity of the source in the x-direction
• vO is the constant velocity of the observer in the x-direction
(See Derivation of Doppler Effect Wavelength Equations for more information.)
Substitute λO = c/fO and λS= c/fS in the equation to get:
c/fO= (c/fS)[(c − vS)/(c − vO)]
c/fO= c(c − vS)/fS(c − vO)
where
• fO is the constant observed frequency
• fS is the constant source frequency
Divide both sides by c and reciprocate the equation. This results in the general frequency equation:
fO = fS(c − vO)/(c − vS)
The equation is often written in the convenient format:
fO/(c − vO) = fS/(c − vS)
### Change in frequency
The change in frequency or Doppler frequency shift is:
Δf = fS − fO
Substitute in fO = fS(c − vO)/(c − vS):
Δf = fS − fS(c − vO)/(c − vS)
Combine terms:
Δf = [fS(c − vS) − fS(c − vO)]/(c − vS)
Δf = (fSc − fSvS− fSc + fSvO)/(c − vS)
Thus:
Δf = fS(vO vS)/(c − vS)
## Moving source and stationary observer
When the source is moving in the x-direction but the observer is stationary, you can take the general frequency equation, set vO = 0, and solve for fO.
Source is moving toward stationary observer
The general frequency equation is:
fO = fS(c − vO)/(c − vS)
Set vO = 0 and solve for fO:
fO = fSc/(c − vS)
The equation is often seen in the form:
fO = fS/(1 − vS/c)
### Change in frequency
The change in frequency or Doppler frequency shift is:
Δf = fS − fO
Note that when the source is moving toward the observer, fS > fO and Δf is negative.
Substitute for fO:
Δf = fS fSc/(c − vS)
Factor out fS:
Δf = fS[1 − c/(c − vS)]
Simplify:
Δf = fS[(c − vS)/(c − vS) − c/(c − vS)]
Δf = fS(c − vS c)/(c − vS)
Δf = fS(− vS)/(c − vS)
Thus:
Δf = −fS vS/(c − vS)
## Moving observer and stationary source
When the observer is moving in the x-direction but the source is stationary, you can take the general frequency equation, set vS = 0, and solve for fO.
Observer moving away from oncoming waves
Set vS = 0 in the general frequency equation:
fO = fS(c − vO)/(c − vS)
Thus:
fO = fS(c − vO)/c
or
fO = fS(1 − vO/c)
Note: Stating the direction convention is very important. Often the equation will be written with the observer moving toward the source, resulting in the equation: fO = fS(1 + vO/c). Make sure you know the convention used.
### Change in frequency
The change in frequency is:
Δf = fS − fO
Substitute for fO:
Δf = fS − fS(c − vO)/c
Simplify:
Δf = cfS/c − fS(c − vO)/c
Δf = (cfS − cfS + vOfS)/c
Thus:
Δf = vOfS/c
Note: According to our direction convention, vO becomes −vO when the observer is moving toward the source.
## Summary
The Doppler Effect frequency equations can be readily determined from the derived general wavelength equation. The resulting general Doppler Effect frequency equation is:
fO/(c − vO) = fS/(c − vS)
From the general equation, the equation for the case when the observer is stationary can be found be setting vO = 0.
fO = fSc/(c − vS)
Δf = −fS vS/(c − vS)
Likewise, the equation when the source is stationary can be found be setting vS = 0.
fO = fS(c − vO)/c
Δf = vOfS/c
Sometimes you need to improvise
## Resources and references
Ron Kurtus' Credentials
### Websites
Wave Motion Resources
## Questions and comments
Do you have any questions, comments, or opinions on this subject? If so, send an email with your feedback. I will try to get back to you as soon as possible.
## Students and researchers
www.school-for-champions.com/science/
waves_doppler_effect_frequency_derivations.htm
Please include it as a link on your website or as a reference in your report, document, or thesis.
## Where are you now?
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Physics topics
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# Square Root of Negative 4 Explained
The square root of negative 4 is equal to 2i, where i=√-1 is the imaginary complex number. Note that the square roots of -4 are the solutions of the quadratic equation x2+4=0, that is, x2=-4. In this post, we will learn how to find the square root of -4.
## What is the Square root of -4
Answer: The square root of -4 is 2i, that is, √-4 =2i.
Solution:
Note that -4 can be written as follows:
-4 = 4 × -1 …(∗)
Now to find the square root of -4, we need to take the square root on both sides of (∗), doing so we will get that
$\sqrt{-4}=\sqrt{4 \times (-1)}$
= $\sqrt{4} \times \sqrt{-1}$. Here we have used the square root multiplication rule: √(m×n) = √m × √n.
= 2 × √-1, as the square root of 4 is 2.
= 2i where i=√-1.
So the square root of negative 4 is 2i.
We know that if x is a square root, then -x is also a square root. Therefore, we can obtain that
√-4 =2i, -2i.
Video Solution:
Question 1: What is square root of negative 4 times square root of negative 9? That is,
find √-4 × √-9
√-4 × √-9
= 2i × 3i
= 6i2
= -6 as i2=-1.
So the product of the square root of negative 4 and the square root of negative 9 is equal to -6.
Have You These Square Roots?
Square root of 12
Square root of 18
Square root of 20
Square root of 27
Square root of 50
## FAQs
### Q1: What is square root of -4?
Answer: The square root of -4 is 2i, where i=√-1.
### Q2: Is square root of negative 4 an integer?
Answer: As √-4 =2i, square root of negative 4 is not an integer, it is an imaginary complex number.
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# The sum of two positive real numbers is 100. Find their maximum possible product?
+1 vote
193 views
The sum of two positive real numbers is 100. Find their maximum possible product?
posted Sep 5, 2015
+1 vote
If the numbers may be identical, 50 and 50 => product is 2500
If the numbers must be different, 49 and 51 => product is 2499
+1 vote
All the answers I'm seeing are very empirical and seem to be have been arrived at "by experience." Although they are right, the method isn't.
What the problem asks is to maximize x*(100-x). 100x - x2.
Maxima minima (if you can use that) tell us that the maxima is at x=50. Thus, the highest value is 2500.
Otherwise, if you are aware of parabola, you would know that the above expression is a functional representation of a (open-downwards) parabola and you can easily find that the apex exists at x=50. Thus, maximum value is, again, at 2500.
Correct:
d(100x-x^2)/dx=0
x=50
100=1+99
=2+98
=3+97
.
.
.
.
.
. =50+50
if we done multiplication we get maximum for 50*50
2500
answer Sep 7, 2015 by anonymous
Similar Puzzles
+1 vote
In a sequence of eleven real numbers, only two are shown. The product of every 3 successive boxes is 120. Find the sum of all the numbers in the boxes (including the two already open).
``````_ _ (6) _ _ _ _ _ _ (-4) _
``````
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https://math.libretexts.org/Bookshelves/Applied_Mathematics/Book%3A_Mathematics_for_Elementary_Teachers_(Manes)/02%3A_Place_Value/2.02%3A_Other_Rules
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$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 2.2: Other Rules
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
Let’s play the dots and boxes game, but change the rule.
The 1←3 Rule
Whenever there are three dots in single box, they “explode,” disappear, and become one dot in the box to the left.
Example $$\PageIndex{1}$$: Fifteen dots in the 1←3 system
Here’s what happens with fifteen dots:
The 1←3 code for fifteen dots is: 120.
Problem 2
1. Show that the 1←3 code for twenty dots is 202.
2. What is the 1←3 code for thirteen dots?
3. What is the 1←3 code for twenty-five dots?
4. What number of dots has 1←3 code 1022?
5. Is it possible for a collection of dots to have 1←3 code 2031? Explain your answer.
Problem 3
1. Describe how the 1←4 rule would work.
2. What is the 1←4 code for thirteen dots?
Problem 4
1. What is the 1←5 code for the thirteen dots?
2. What is the 1←5 code for five dots?
Problem 5
1. What is the 1←9 code for thirteen dots?
2. What is the 1←9 code for thirty dots?
Problem 6
1. What is the 1←10 code for thirteen dots?
2. What is the 1←10 code for thirty-seven dots?
3. What is the 1←10 code for two hundred thirty-eight dots?
4. What is the 1←10 code for five thousand eight hundred and thirty-three dots?
Think / Pair / Share
After you have worked on the problems on your own, compare your ideas with a partner. Can you describe what’s going on in Problem 6 and why?
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# How do you use the Binomial Theorem to expand (3v+s)^5?
Now we have that $x = 3 v$ and $y = s$ applying the binomial theorem
${\left(3 v + s\right)}^{5} = {s}^{5} + 15 {s}^{4} v + 90 {s}^{3} {v}^{2} + 270 {s}^{2} {v}^{3} + 405 s {v}^{4} + 243 {v}^{5}$
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## general equations, Algebra
Assignment Help:
7x+6=x-30
#### Fractions, how do you solve a different fractions
how do you solve a different fractions
#### Lenare eqation, A police academy is training 14 new recruits. Some are work...
A police academy is training 14 new recruits. Some are working dogs and others are police officers. There are 38 legs in all. How many of each type of recruits are there?
#### Completing the square, Completing the Square The first method we'll lea...
Completing the Square The first method we'll learning at in this section is completing the square. This is called it since it uses a procedure called completing the square in t
#### Y=a+bx, how do you solve this#question with (3,5)and (2,1) ..
how do you solve this#question with (3,5)and (2,1) ..
#### #title polynomials, how to solve this p(x)=2x^4
how to solve this p(x)=2x^4
#### Graphing functions, Now we need to discuss graphing functions. If we recall...
Now we need to discuss graphing functions. If we recall from the earlier section we said that f ( x ) is nothing more than a fancy way of writing y. It means that already we kno
#### Ratio, Veronica''s family ordered 2 pizzas that cost \$13.75 each and 2 pitc...
Veronica''s family ordered 2 pizzas that cost \$13.75 each and 2 pitchers of soda that cost \$3.95 each. The total cost included a sales tax of 5.5%. They left 20% of the total cost
#### Convert following into the function, Convert following into the form ...
Convert following into the form f (x ) = a ( x - h ) 2 + k Solution We are going to complete the square here. Though, it is a s
#### Distance, If a jet can travel 600 mph, how long in second will it take the ...
If a jet can travel 600 mph, how long in second will it take the jet to travel 5 miles?
#### Applications of logarithmic equation, In this last section of this chapter ...
In this last section of this chapter we have to look at some applications of exponential & logarithm functions. Compound Interest This first application is compounding inte
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# Instrumentation & Measurement Magazine 23-2 - 95
```are going to see, and that knowledge can be used to reduce the
number of measurements you need to make.
We consider the case where you are processing a vector, and
you know that the vector is a linear combination of a very few
vectors out of a "dictionary" of many vectors. In the language
of linear algebra, we assume that the sampled signal, y, is a linear combination of very few of the columns of a matrix C with
many columns. C is the "dictionary," and the columns of C are
the "words." We further assume that C is a K × N matrix-that
every column of C has K elements and that C has N columns.
The assumption that y is a linear combination of the columns of
C can be expressed by saying that y = Cx. The assumption that
relatively few of the columns are needed can be expressed by
saying that all but a very few elements of x are zero.
It is easy to show that if any 2s columns of C are linearly
in
dependent, then any solution, x, of the equation Cx = b with s or
fewer non-zero elements is unique among all such solutions.
(See Sidebar: Sparse Solutions are Unique.) Vectors with s or
fewer non-zero elements are said to be s-sparse.
Sidebar: Sparse Solutions are
Unique
Theorem: Given a matrix C such that any 2s of its columns are linearly independent, if there is an s-sparse
solution, x0, of the equation Cx = b, then this solution is
the unique s-sparse solution.
Proof. The proof proceeds by contradiction. Sup
pose that there were two solutions, x1 and x2. The
xd = x1 − x2, satisfies
difference of these two solutions,
C
x
=
C
x
−
C
x
=
0
. Recall that multiplythe equation
d
1
2
ing a vector by a matrix produces a linear combination
of the matrix's columns. In this case, the equation says
that a linear combination of the columns of C equals
zero, and the number of columns is equal to the number
of non-zero elements in xd. Because xd is the difference of
two vectors each of which has no more than s non-zero
elements, xd can have no more that 2s non-zero elements. That is, a combination of 2s (or fewer) columns
of C gives the zero vector. However, by the definition of
linear independence, that means that there are a set of 2s
or fewer columns of C that are not linearly independent.
That, however, contradicts the hypothesis of the theorem. Thus, under the hypotheses of the theorem, the
difference vector must be the zero vector; the two vectors must actually be the same; and we find that if there
is an s-sparse solution of the equation, then this solution
is the unique s-sparse solution.
For a simple example of recovering uniqueness using sparsity, see, for example, the sidebar in [2]. Note,
too, that the theorem does not guarantee that a sparse
solution exists-you need to know that this is the case.
The theorem guarantees that if a sparse solution exists, it
is the unique sparse solution.
April 2020
Developing the Theory of Compressive
Sensing: Finding the Sparse Solution
Suppose that we have a matrix, C, for which any 2s columns are
linearly independent, and we know that the equation Cx = b
has an s-sparse solution. How can we find that solution? This
is a fascinating and important question, and it has many answers. We start from the simplest answer and then consider
The Brute-Force Technique
Consider a matrix C ∈ R K×N for which K = 2s, K ≤ N, and assume
that any 2s columns of C are linearly independent. Then, one
way to search for the unique sparse solution is to attempt to
sets of K columns of C and then to solve
list all the C KN possible
the equations Di z = b , i = 1,... , CKN where each matrix Di is composed of a different
set of K of C's columns. When one finds a
solution, z = Di−1b, that is s-sparse, it must correspond to the correct solution. For MATLAB code that implements this method
and returns the sparse solution, see the Sidebar: Implementing
the Brute-Force Technique.
This technique has a major disadvantage. As C's size grows,
the number of iterations grows very, very quickly. Unfortunately, it has been proved that, in general, the problem of
finding a sparse solution is NP-hard [3]. Despite its seeming inefficiency, in some sense, the "inefficient" algorithm we
have described does about as well as one could hope. Luckily,
despite the fact that the problem we are trying to solve is provably very hard to solve, there are still-sometimes-efficient
ways to solve the problem.
Orthogonal Matching Pursuit
Let ci be the ith column of C. If you know that your solution is ssparse and you know that the matrix C's coherence, defined as:
c ⋅ c
i
j
ci ⋅ c j
μ ≡ max
i ≠ j
,
which is a measure of similarity between columns of C, is
less than 1/(2s), then a simple greedy algorithm, orthogonal
matching pursuit, will find the s-sparse solution. (For more
information, see, for example, [4]. For an example of an application to a measurement problem, see [5].) Note that this
condition is restrictive enough that it can be shown that the
conditions of the theorem in Sidebar: Sparse Solutions Are
Unique certainly obtain; the s-sparse solution is certainly
unique; and the brute force method certainly works too. Orthogonal matching pursuit works on a subset of the matrices
for which the brute force method works, but where it works, it
is efficient.
Methods Based on the Techniques of Linear
Programming or Convex Optimization
Though many think that mathematicians have no sense of humor, this is not so-as the next definition shows. A matrix is
said to have the restricted isometry property, known affectionately as the RIP, when for all s-sparse vectors:
IEEE Instrumentation & Measurement Magazine 95
```
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Statistical Computing and Graphics in R
# Curvilinear Regression in R
In this post, we will learn about some basics of curvilinear regression in R.
The curvilinear regression analysis is used to determine if there is a non-linear trend exists between $X$ and $Y$.
Adding more parameters to an equation results in a better fit to the data. A quadratic and cubic equation will always have higher $R^2$ than the linear regression model. Similarly, a cubic equation will usually have higher $R^2$ than a quadratic one.
The logarithmic relationship can be described as follows:
$$Y=m\, log(x)++c$$
the polynomial relationship can be described as follows:
$$Y=m_1x + m_2x^2 + m_3x^3 + m_nx^n + c$$
The logarithmic example is more akin to a simple regression, whereas the polynomial example is multiple regression. Logarithmic relationships are common in the natural world; you may encounter them in many circumstances. Drawing the relationships between response and predictor variables as a scatter plot is generally a good starting point.
Consider the following data that are related in a curvilinear form,
Let us perform a curvilinear regression in R language.
Growth <- c(2, 9, 11, 12, 13, 14, 17, 19, 17, 18, 20)
Nutrient <- c(2, 4, 6, 8, 10, 16, 22, 28, 30, 36, 48)
data <- data <- as.data.frame(cbind(Growth, Nutrient))
ggplot(data, aes(Nutrient, Growth) ) +
geom_point() +
stat_smooth()
The Scatter plot shows the relationship appears to be a logarithmic one.
Let us carry out a linear regression using the lm() function by taking the $\log$ of the predictor variable rather than the basic variable itself.
data <- cbind(Growth, Nutrient)
mod <- lm(Growth~log(Nutrient, data))
summary(mod)
##
Call:
lm(formula = Growth ~ log(Nutrient), data = data)
Residuals:
Min 1Q Median 3Q Max
-2.2274 -0.9039 0.5400 0.9344 1.3097
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.6914 1.0596 0.652 0.53
log(Nutrient) 5.1014 0.3858 13.223 3.36e-07 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 1.229 on 9 degrees of freedom
Multiple R-squared: 0.951, Adjusted R-squared: 0.9456
F-statistic: 174.8 on 1 and 9 DF, p-value: 3.356e-07
Learn about Performing Linear Regression in R
Learn Statistics
Curvilinear Regression in R
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0
# What fraction of a circle is thirty three and one third perecent of a circle?
Updated: 12/24/2022
Wiki User
11y ago
331/3% of anything is 1/3 of it. ("One third")
Wiki User
11y ago
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Q: What fraction of a circle is thirty three and one third perecent of a circle?
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Related questions
one third
### What is the fraction of notation of thirty three and one third?
The fraction form is 100/3.
100/3
1/3
### What is 34 perecent in its simplest form?
I would estimate 34% to be just over a third.
One third. (1/3)
### What fraction of a circle is 33 one third percent?
331/3 % of a circle is 1/3 of a circle 331/3 % = 100/3 % = 100/3 ÷ 100 = 1/3 So 331/3 % as a fraction is 1/3
### What is thirty three and one third percent in decimal and fraction form?
As a decimal: 33.33333 As an improper fraction: 100/3 As a mixed number: 33 1/3
### What is one third of thirty there?
what is one third of thirty three
### How do you spell 33?
Basically how you spelled it, but from what I know, and was taught, you would write "twenty-three". Another way you do that is like fifty-five. See how I put a dash in between it? That's how you would properly write it.
54/3
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Unit 4 - Objective 1 - Increasing/Decreasing
A function is increasing if the graph is rising as you move from left to right. For y = f(x), this means that y increases as x increases. A function is decreasing if the graph is falling as you move from left to right. For y = f(x), this means that y decreases as x increases.
The sign of the first derivative indicates whether a function is increasing or decreasing.
• If f '(a) > 0, then y = f(x) is increasing when x = a
• If f '(a) < 0, then y = f(x) is decreasing when x = a
Determine whether the curves below are increasing or decreasing at each point given.
y = f(x) = 6x² - 8x + 1 at x = -5 [or at the point (-5,191)].
solution: f '(x) = 12x - 8 f '(-5) = -60 - 8 = -68 < 0 The derivative is negative whenx = -5 so the curve is decreasing there.
Instead of plotting a lot of points to determine the shape of a curve, use the derivative to find where the function is increasing and decreasing.
The first derivative is the test function for increasing and decreasing.
If f '(x) is positive, f(x) is increasing.
If f '(x) is negative, f(x) is decreasing
Given y = f(x) = 2x³ - 3x² - 72x -4 , find the intervals where f(x) is increasing and where it is decreasing.
Steps
Example
1.Find f '(x)
1. f '(x) = 6x² - 6x - 72
2.Find the critical values, where f '(x) = 0 or f '(x) does not exist.
2. f '(x) = 6x² - 6x - 72 = 6(x² -x -12)
f '(x) = 6 (x-4) (x+3) = 0
c.v.: x = 4, x = -3
(no x values where f '(x) does not exist)
3.Put those critical on the x-axis and write the intervals.
3.
4.Pick one test value in each interval and put it in the derivative to determine if it is positive or negative. (I'm substituting into the factored form of the derivative.)
4.
test x = -4test x = 0test x = 5
f ' (-4) = 6 (-8)(-1)
f ' (-4) = 6 (-8) (-1)
f (x) positive
f (x) increasing
for x < -3
f ' (0) = -72
f ' (x) negative
f (x) decreasing
-3 < x < 4
f ' (5) = 6(1)(8)
f ' (5) = +48
f ' (x) positive
f (x) increasing
x > 4
Note: From the increasing and decreasing information above we can tell that this function has a relative maximum (where the curve changes from increasing to decreasing) when x = -3 and a relative minimum (where the curve changes from decreasing to increasing) when x = 4.
• The relative maximum is f(-3) = 2(-3)³ - 3(-3)² - 72(-3) - 4 = 131 and
• The relative minimum is f(4) = 2(4)³ - 3(4)² - 72(4) - 4 = -212
So the minimum point is (4,-212) and the maximum point is (-3,131). The high point is a relative maximum of y = 131 when x = -3, and the low point is a relative minimum of y = -212 when x = 4.
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## Jun 24, 2015
### CONCEPTS OF PERCENTAGE
Concepts of Percentage
Definition of Per Cent:
A Fraction having Denominator 100 is called a Per Cent & is denoted by symbol %.
# To convert the Fraction into Percentage
To convert a fraction into Percentage, multiply the fraction by 100 & put % sign.
For Eg: 4/5 Ã 4/5 * 100 = 80%
# To convert Percentage into a Fraction
To convert a percentage into a fraction replace the % sign with 1/100 & reduce the fraction to simplest form.
For Eg: 20% Ã 20/100 = 1/5
# To convert Percentage into Ratio
To convert a percentage into a ratio, first convert the given percentage into a fraction in simplest form & then to a ratio.
For Eg: 25% Ã 25/100 = 1/4 = 1 : 4
Important Concepts:
1. 100% of a Number = Number itself
For eg: 100% of 500 = 500
2. 50% of a number = Half of the number
For eg: 50% of 500 = 500/2 = 250
3. 25% of a number = Quarter(1/4th) of the number or Half of 50% of the number
For eg: 25% of 500 will be 500/4 or 250/2= 125
4. To find the 1% of a number, shift the decimal point by 2 place to the left.
For eg: 1% of 500 will be 5.00
5. To find the 10% of a number, shift the decimal point by 1 place to the left.
For eg:10% of 500 = 50.0
6. 5% of the number is half of 10% of a number
For eg:5% of 500 = 50/2=25
7. 15% of the number = Sum of 10% and 5% of a number
For Eg: 15% of 500=(10% of 500)+(5% of 500=50 + 2 =75
Using the above methods now we can find the percentage of a number easily:
Example: 15% of 400 = ?
Step 1: Split the percentage term into 2 numbers that we can easily find the percentage of.
We can split 15% into 10% and 5%
Step 2: First find 10% of 400 by shifting the decimal point to the left by 1 place.
i.e 10% of 400=40.0
Step 3: Now find 5% of 400 i.e half of 10%
i.e 5% of 400=40/2 = 20
Step 4: Now add both the results
we get, 15%+5%=40 + 20 = 60
Therefore, 15% of 400 = 60
Important Formulas:
1) Percentage Increase/Decrease:
· If the price of a commodity increases by R%, then the reduction in consumption so as not to increase the expenditure is:
= [(R/100+R) * 100]%
· If the price of a commodity decreases by R%, then the increase in consumption so as not to decrease the expenditure is:
=[(R/100-R) * 100]%
2) Population:
If the Population of a town be P now & it increasing at the rate of R% p.a, then,
· Population after n years = P[1+(R/100)]^n
· Population n years ago = P[1+(R/100)]^n
3) Results on Depreciation
If the present value of a machine is P & it depreciates at the rate of R% per annum. Then:
· Value of the machine after n years= P[1-(R/100)]^n
· Value of the machine n years ago= P/[1-(R/100)]^n
4) If P is R% more than Q, then Q is less than A by
={ [R/(100+R)]*100}%
5) If P is R% less than Q, then Q is more than A by
={ [R/(100-R)]*100}%
Learn the Following Table for Fast calculations:
Fraction Percentage 1/1 100% 1/2 50% 1/3 33.33% 1/4 25% 1/5 20% 1/6 16.66% 1/7 14.28% 1/8 12.5% 1/9 11.11% 1/10 10% 1/15 6.66% 1/20 5% 1/25 4% 1/30 3.33% 1/40 2.5% 1/50 2% 1/60 1.66% 1/75 1.33% 1/80 1.25% 1/90 1.11% 1/100 1%
By Using this Conversion Table we can convert Fraction into Percentage and Percentage into Fraction easily.
For Eg: To find 3/5 in %?
Solution: As we have 1/5=20%
then, Multiply both sides by 3
i.e 1/5x3 = 3x20%
Therefore, 3/5 = 60%.
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## 7.1 Digital Filter Response
7.1. DIGITAL FILTER RESPONSE 7.1 127 Digital Filter Response A digital filter can be described in several ways. These include, but are not limited...
Author: Roy Welch
7.1. DIGITAL FILTER RESPONSE
7.1
127
Digital Filter Response
A digital filter can be described in several ways. These include, but are not limited to, difference equations, block diagram, impulse response, and the system function. Each model is useful in the description of systems and their behavior, and they are all related. They provide different views of the same entity. A good place to start is with a difference equation. The difference equation is an important model in modeling random processes as well as in describing digital systems. It is a completely general model of a time-invariant discrete linear system. We will develop difference equations as a digital system model and then relate that approach to other common models. After we have developed the description tools, we will show how they can be used to model random processes by using to construct random processes with various properties. These will enable us to emply various analytical tools on processes with known properties so that we can gain insight into the usefulness and limitation of each tool. This will prepare us for encounters with random processes with unknown structures.
7.1.1
Difference Equation Model
Let the input sequence to a digital filter be denoted by x(n) and the corresponding response by y(n). The current output value can depend upon the current input value, past input values and past output values. This can be expressed as y(n) + a1 y(n − 1) + a2 y(n − 2) + · · · + ap y(n − p) = b0 x(n) + b1 x(n − 1) + · · · + bq x(n − q)
(7.1)
The difference equation is completely described by the coefficients {1, a1 , . . . , ap } and {b0 , b1 , . . . , bp }. Note that we will always assume (without loss of generality) that a0 = 1. The above equations can be solved for the current output value. y(n) = b0 x(n) + b1 x(n − 1) + · · · + bq x(n − q) − a1 y(n − 1) −a2 y(n − 2) − · · · − ap y(n − p)
(7.2)
A block diagram of this equation is shown in Figure 7.1 on page 128. The storage elements in the top row preserve a history of the past inputs and those in the bottom row preserve a history of the past outputs.
128
Figure 7.1: Block diagram of a digital filter realization of the difference equantion. Example 7.1.1 Single Pole Filter This filter is characterized by a simple difference equation y(n) + ay(n − 1) = x(n) (7.3) To see its behavior, consider the simple input sequence [· · · 0, 1, 0, 0, 0, 0 · · · ] which has a single nonzero input value. Without loss of generality we can choose the index origin so that x(0) = 1 and x (n) = 0 for n 6= 0. The current output can be written in terms of the current input and the immediate past output. y(n) = x(n) − ay(n − 1) This equation is illustrated by the block diagram in Figure 7.2 on page 129. We will assume the initial state y(−1) = 0. Then it is a simple matter to write out a few of the values of the output sequence beginning with n = 0. n 0 1 2 y(n) 1 −a a2
3 -a3
4 a4
By direct substitution into the difference equation you can show that the general solution for this input and initial condition is y(n) = − (a)n for n ≥ 0. The output is clearly bounded if and only if |a| ≤ 1. We will see later when
7.1. DIGITAL FILTER RESPONSE
129
Figure 7.2: Block diagram of a single pole filter. The system stores one value, the immediately preceeding output. This value establishes the system state. we discuss the system function that this system has a “single pole” whose location is determined by a, and which is stable of the “pole” is inside the unit circle. This example illustrates the use of an impulse input to gain insight into the behavior of a system. The impulse response can be used as a general tool for this purpose. Impulse Response The impulse response of a digital filter is the sequence that is generated when the input sequence is x(n) = δ(n), where ½ 0, n 6= 0 δ(n) = (7.4) 1, n = 0 The response to this input sequence is traditionally referred to as h(n). The impulse response is sufficient to provide a complete description of the behavior of any linear time-invariant system, and is therefore equivalent to the difference equantion description or the block diagram. The response of a system to a general input sequence x(n) is y(n) =
∞ X
m=−∞
h(m)x(n − m)
Clearly, y(n) = h(n) when x(n) = δ(n).
(7.5)
130
Figure 7.3: Block diagram of a finite impulse response digital filter. The response of a single input sample persists for q additional steps. A special class of systems has an impulse response of finite duration. These systems are characterized by the absence of a feedback path. That is, a1 = a2 = · · · = ap = 0. The block diagram of a finite impulse response (FIR) system is shown in Figure.7.3. The impulse response of the FIR filter is nq This relationship provides one means to realize any impulse response that one may need. The algorithm must store as many of the past inputs as necessary to provide a good approximation through use of a finite number of coefficients. The practical issue is the size of the memory, determined by q, that is practical. FIR Applications The FIR filter is often used to smooth a random process to suppress noise and bring out a slower-varying signal. This is shown in the first example below. Another application is in the detection of a signal in a noisy background with a matched filter. We will look briefly at these applications. Example 7.1.2 Weighted Average Suppose that we would like to smooth a waveform by averaging over several values of the input. We may want to
7.1. DIGITAL FILTER RESPONSE
131
reduce the weight given to input values that are farther in the past. Let r be a number in the range 0 ≤ r ≤ 1. Let the impulse response be defined by 0, n < 0 rn , 0 ≤ n ≤ q h(n) = (7.7) 0, n > q
Then the output, using (7.5), is
y(n) =
q X
m=0
rm x(n − m)
(7.8)
If r = 1 we give equal weight to the q past outputs and if r < 1 we give less weight to those farther in the past. This can be effective in reducing the amount of noise on a slowly varying signal, as shown in Figure ?? on page ??. In the example shown in the figure the signal is a sine wave that has a period of 100 samples. The noise can be reduced by averaging over an interval of about 10 samples without seriously distorting the signal. Matched Filter A FIR filter can also be used for an operation called matched filtering. Suppose that one wants to detect the presence of a waveform s(n) that is of some finite duration, but that it is observed in the presence of additive noise. Then the observed samples are x(n) = s(n) + ξ(n), where ξ(n) represents the noise process. In this example we will assume that the noise is “white” so that E[ξ(m)ξ(n)] = σ 2 δ(m − n). Let the duration of s(n) be q + 1 so that the signal can be represented by the samples {s(0), s(1), . . . , s(q)}. Then the maximum-likelihood test for the presence of s in an observed sequence x is to form the weighted sum u=
q X
x(n)s(n)
(7.9)
n=0
and compare the result u to a decision threshold η. If u > η then the signal is judged to be present in the sample sequence x(n). We can substitute x(n) = s(n) + ξ(n) (case of signal present) and x(n) = ξ(n) (case of signal absent) into (7.9) to determine the expected value of u in each case. When the signal is absent the expected value of u will be (assuming that the average value of the noise is zero) Pq Pq E[u] = (7.10) n=0 E[ξ(n)s(n)] = n=0 s(n)E[ξ(n)]] = 0
132
Figure 7.4: The averaging filter is effective in reducing the amount of noise on a low-frequency signal. When the signal is present, the expected value of u will be
E[u] =
q X
E[(s(n) + ξ(n))s(n)]
(7.11)
n=0
=
q X
s2 (n) + s(n)E[ξ(n)]]
n=0
= Es
We should be able to determine which is the case by evaluating (7.9) with an observed x(n) and comparing the result to a decision threshold. The
7.1. DIGITAL FILTER RESPONSE
133
parameter Es is called the “energy” in s, snd is equal to Es =
q X
s2 (n)
(7.12)
n=0
The decision threshold is often set midway between the output that would be expected with signal absent and that which would be expected with signal present. That would produce a threshold of η = Es /2. More extensive analysis based on risk analysis can be used to refine this setting, but it is sufficient for the purpose of illustrating the principle. The value of u can be produced by passing the values of x into a FIR filter whose weights are determined by s. We can see this by constructing a digital filter with the coefficients b(m) = s(q − m) for 0 ≤ m ≤ q, The response of such a filter to the sequence x(n) would be y(n) =
q X
m=0
b(m)x(n − m) =
q X
m=0
s(q − m)x(n − m)
(7.13)
At the point n = q the output value is y(q) = u, which can be seen by n = q in the above equation and then making the index change k = q − m. y(q) =
q X
m=0
s(q − m)x(q − m) =
q X
s(k)x(k) = u
(7.14)
k=0
Example 7.1.3 Matched Filter A signal s(n) = 10e−n/6 sin(2π(.005)n) 0 ≤ n ≤ 30 will be detected when an object is present. No signal will be detected when the object is absent. The detected waveform is x(n) = s(n)+ξ(n) when the object is present and x(n) = ξ(n) when the object is not present. The decision can be made by comparing the output of a matched filter to a decision threshold. In this case the matched filter coefficients areb(n) = s(10 − n) = 10e(30−n)/6 sin(2π(.15 − 0.005n)) b(n) = s(10 − n) = 10e(30−n)/6 sin(2π(.15 − 0.005n)), 0 ≤ n ≤ 10 A plot of the signal is shown in Figure 7.5 on page 135, where we see that it has the characteristic of a damped sinusoidal pulse. When the same signal is observed with additive noise with standard deviation σ = 1.5, the result is a noisy waveform such as the middle plot of the figure. The response of a
134 matched filter to this signal is shown in the lower panel of Figure 7.5. We observe that the filter output builds up slowly as each sample of the noisy waveform is entered. The final value of the filter output is used in making a decision about the presence of the signal. The probability that the matched filter output may be on the wrong side of the decision threshold is explored in Figure 7.6 on page 136. A total of 300 tests were run in which noise was generated and added to the signal waveform and then filtered. The upper figure shows a record of the final filter output value under the assumption of signal present (top line) and only noise (bottom line). The line dividing the records is at a level Es /2, and corresponds to the decision threshold. It is noted that a few points would cross the line from either direction, corresponding to either missing the target or a false alarm. A pair of histograms of the results are shown in the lower plot, where we can see the distribution of output values from the matched filter under the two hypotheses. The examples of a smoothing filter and a matched filter give us an idea about how FIR digital filters can be specified and used in the processing of noisy signals. At this point we know that a digital filter can be described by either its difference equation or by its impulse response, and we know how to relate the impulse response of a FIR filter to its difference equation. In the case of the matched filter of Example 7.1.3 we used the impulse response to determine the FIR filter parameters.
7.1.2
System Function Model
It will be useful to have other ways to describe and specify the behavior of digital filters. In particular, we would like to have a natural method to look at them in the frequency domain. Continuous-time system In continuous systems we use the system function H(s) or H(jω) to describe the frequency response. The system function describes the response of a system to the special input x(s, t) = est . The response is y(s, t) = H(s)est . Here the parameter s is introduced into the arguments of the input and output functions to make it clear that it is a value that can be specified in
7.1. DIGITAL FILTER RESPONSE
135
Figure 7.5: The signal is shown in the top panel. A typical received signal plus noise is shown in the middle panel. The response of a matched filter to the waveform signal plus noise waveform is shown in the lower panel. The final value of the filter output is used in the decision rule.
136
Figure 7.6: The upper panel shows two plots of the final matched filter value for a sequence of 300 trials of detecting the ultrasound signal in white noise. The upper trace is for signal plus noise and the lower trace with only noise. The lower panel shows a histogram of the filter output values for the two cases. The vertical markers are at 0, Es /2, and Es .
7.1. DIGITAL FILTER RESPONSE
137
our analysis. This approach to finding the system function is valid for any linear time-invariant system. Example 7.1.4 Consider the continuous-time system that is described by a differential equation y(t) + a1 y (1) (t) + a2 y (2) (t) + · · · + ap y (p) (t) = b0 x(t) + b1 x(1) (t) + · · · + bq x(q) (t)
(7.15)
Let us substitute x(t) = est and y(t) = H(s)est . Then x(k) (t) = sk est and y(t) = sk H(s)ewt , and we find (1 + a1 s + a2 s2 + · · · + ap sp )H(s)est = (b0 + b1 s + · · · + bq sq )est
(7.16)
This equation will be valid for all values of t provided the coefficients of est on each side of the equation are equal. This allows us to solve for the system function. b0 + b1 s + · · · + bq sq H(s) = (7.17) 1 + a1 s + a2 s2 + · · · + ap sp
This function is valid for all values of s for which the denominator is nonzero, including complex and imaginary values. This tells us that any system that is described by a linear differential equation with constant coefficients has a system function that is a ratio of polynomials in s. The coefficients are the coefficients from the differential equation. Example 7.1.5 Consider a continuous-time linear system that is described by an impulse response function h(t). The response of the system to any input x(t) is Z ∞
y(t) =
−∞
h(τ )x(t − τ )dτ
(7.18)
Now let x(t) = est and y(t) = H(s)est . Then we must have Z ∞ Z ∞ st s(t−τ ) st h(τ )e dτ = e h(τ )e−sτ dτ H(s)e = −∞
Therefore, H(s) =
−∞
Z
−∞
h(τ )e−sτ dτ
(7.19)
138 This result shows that the system function and the impulse response are a transform pair for time-invariant continuous-time linear systems. Either function is just an alternative way of looking at the behavior of the system. A major use for the system function is in describing the behavior of the system in the frequency domain. Every sinusoidal waveform can be written in terms of exponentials. For example, cos(ωt) =
¢ 1 ¡ jωt e + e−jωt 2
Hence, the response to x(t) = cos(ωt) is y(t) = 12 H(jω)ejωt + 12 H(−jω)e−jwt . For a linear system with real parameters, such as real coefficients in the differential equation or a real impulse response, the response to a real input, such as a cosine wave, must also be real. This requires that the system function have conjugate symmetry. H(−jω) = H ∗ (jω). Every complex number can be written in a polar coordinate form. If Z is a complex number then it can be written in the form Z = |Z|ejθ . We can apply this to the system function since it is just a complex number at any fixed value of ω. Hence the output is y(t) = 12 |H(jω)|ej(ωt+θ) + 12 |H(jω)|e−j(ωt+θ) = |H(jω)| cos(ωt + θ). Therefore, |H(jω)| is the “gain” of the system at frequency ω and θ is the phase shift at frequency ω. Example 7.1.6 Suppose that we have a system for which we know that the impulse response is the simple exponential function h(t) = e−at u(t) where u(t) is the unit step function and a is a positive constant. The system function is Z ∞ 1 H(s) = e−at e−st dt = s+a 0 The analysis of Example 7.1.4 shows that the difference equation must be x(t) = 1 + a
dy dt
The value of the system function for s = jω is H(jω) =
1 1 =√ e−jArctan(ω/a) 2 a + jω a + ω2
7.1. DIGITAL FILTER RESPONSE The response to x(t) = cos(ωt+α) will be y(t) =
139 √ 1 a2 +ω2
cos(ωt+α−Arctan(ω/a)).
1 1 0.8 0.5 0.6
-4
-2
0
2
-0.5
0.4
-1 -4
-2
0.2
0
2
4
Magnitude of system function vs ω/a A plot of the magnitude and phase of the system function shows that it passes frequencies below about ω = a and rejects higher frequency. . Discrete-time system The system function for a discrete-time system is based on the difference equation rather than the differential equantion. Just as in a continuous system, an exponential input to a discrete system will produce an exponential response. The difference equation is reproduced below from (7.1) y(n)+a1 y(n−1)+a2 y(n−2)+· · ·+ap y(n−p) = b0 x(n)+b1 x(n−1)+· · ·+bq x(n−q)
Let us now substitute the exponential sequence x(n) = z n . Here z is a complex number, which may be expressed in the usual complex number forms wherever that becomes useful in the analysis. Let us assume that the response is of the form y(n) = H(z)z n . Substitution into the difference equation then produces (1 + a1 z −1 + a2 z −2 + · · · + ap z −p )z n H(z) = (b0 + b1 z −1 + · · · + bq z −q )z n This can be solved for the system function H(z) =
b0 + b1 z −1 + · · · + bq z −q 1 + a1 z −1 + a2 z −2 + · · · + ap z −p
(7.20)
The close relationship between the system function, the difference equation and the block diagram shown in Figure 7.1 on page 128 is evident. The system function is defined wherever the denominator is not zero.These locations are called the “poles” of the system function.
4
140 z-Transform The z-transform plays the same role in discrete system representation that the Laplace transform plays in continuous system representation. For any sequence x(n). X(z) =
∞ X
x(n)z −n
(7.21)
n=−∞
The inverse z-transform can be defined. That is not necessary in this analysis, and we will simply illustrate the computation of the z−transform for some example functions. Example 7.1.7 Let x (n) = An u (n), where u(n) is the unit step function and A is a constant. Then X(z) =
∞ X n=0
An z −n =
∞ X (A/z)n n=0
This is just a geometric series of the form (1 + r + r2 + · · · ), which converges to 1/(1 − r) provided |r| < 1. Hence, X(z) =
1 z = 1 − A/z z−A
The denominator has a root at z = A, which corresponds to the ratio in the exponential sequence. Delay Operator Suppose that the z-transform of x(n) is X(z). Then, the z-transform of x(n − k) is X(z)z −k . This can be shown by substitution of x(n − k) into (7.21). We can therefore refer to z −1 as the delay operator. Convolution Property Let x1 (n) and x2 (n) be two sequences with z-transform relationships x1 (n) ↔ X1 (z) x2 (n) ↔ X2 (z) Let x(n) = x1 (n) ∗ x2 (n) be the convolution operation. Then the z-transform is determined by the product X(z) = X1 (z)X2 (z).
7.1. DIGITAL FILTER RESPONSE
141
Impulse Function The function δ(n) defined in (7.4) has the z-transform ∆(z) = 1, as can be easily established from the definition (7.21). Impulse Response The impulse response of a discrete linear time-invariant system is called h(n). It is the response of the system to an input δ(n). The output for any other input sequence x(n) is given by the convolution y(n) =
∞ X
m=−∞
x(m)h(n − m) =
∞ X
m=−∞
h(m)x(n − m)
(7.22)
Upon using the convolution property we find that (7.23)
Y (z) = X(z)H(z)
In particular, if the input is an impulse then X(z) = 1 and the z-transform of the response is just H(z). Therefore, h(n) and H(z) are a transform pair h(n) ↔ H(z)
(7.24)
Frequency Response Just as in the continuous case, we can use an exponential input to a discrete system to determine its frequency response. Let x(n) = ejωn be an input sequence. The output, from (7.22) is y(n) =
∞ X
h(m)ejω(n−m) = ejωn
m=−∞
∞ X
h(m)e−jωm
m=−∞
The summation is just H(z) with z = ejω . Hence, an exponential input sequence produces the exponential output sequence y(n) = H(ejω )ejωn where jω
H(e ) =
∞ X
m=−∞
h(m)e−jωm
142 The system response H(ejω ) provides the same kind of information about a discrete system that H(jω) provides about a continuous system. The position of the frequency variable in the exponent means that the function is periodic with period 2π. Example 7.1.8 The system whose block diagram is shown in Figure 7.2 on page 129 has the impulse response h(n) = (−a)n u(n) The frequency response of the system is jω
H(e ) =
∞ X
m −jωm
(−a) e
m=−∞
=
∞ X
(−ae−jω )m =
m=−∞
1 1 + aejω
The gain and phase shift of the system at frequency ω are equal the magnitude and phase angle of H(ejω ). These are 1 (1+a cos ω)2 +a2 sin2 ω a sin ω − tan−1 1+a cos ω
|H(ejω )| = √ φ(ejω ) =
The periodic character of both functions is evident, since all of the terms in each expression are periodic. The gain and phase shift are plotted in Figure ?? where one period of the response is shown for a value a = −.45 for the filter parameter. The system emphasizes low frequencies more than high frequencies. This is a “low-pass filter.” Note that the operational range of frequencies in a useful system is 0 ≤ ω ≤ 0.5.
Gain and phase shift of a single-pole system function with parameter a = −0.45.
7.1. DIGITAL FILTER RESPONSE
143
Figure 7.7: Gain and phase shift of a single-pole system function with a parameter a = 0.45. Note that this creates a high-pass filter. The gain and phase-shift of a system with the system function H(ejω ) =
1−
2be−jω
1 cos ω 0 + b2 e−j2ω
is shown in Figure 7.8. The location of the peak is determined by the value of ω0 and b = 1.01. One would expect that this filter would have a “narrow-
Figure 7.8: Gain and phase shift of a filter with a single harmonic resonance at ω0 = 0.2˙ and b = 1.01. band” output, no matter what the input spectrum was.
144
7.1.3
System Output
The response of a system can be calculated in a number of ways. The difference equation can be implemented directly, transforms can be used, and the output can be calculated by a general implementation of a digital filter. The program filter is one useful computational implementation. An example of such a computation is provided by sigdemo3 and sigdemo5. Finally, sigdemo4 shows how the spectrum estimate can be improved by averaging many repetitions of power spectrum computations.
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Electric circuitry, confusion about the electric charge
• late347
In summary, the car's battery has a voltage of 11.5 volts and an internal resistance of 0.040 ohms. Its charge is 20 ampere-hours (20 Ah). One attempt at engine ignition draws 150 amperes of current from the battery for 6 seconds. Using the formula Q=I x t, it can be calculated that one attempt at starting the car withdraws 900 As of electric charge. If the threshold is 5 Ah, then approximately 20 attempts can be made before the charge falls to that level. This is due to the fact that 1 C is equal to 1 As, and the formula I=Q/t can be used to find the current. The starter motor
Homework Statement
car's battery has voltage (E) 11,5 volts and internal resistance of 0,040 ohms.
battery's charge is 20 amperehours (20 Ah)
How many ignition attempts of the car can be made, when the charge can be allowed to fall to 15 amperehours.
One attempt of at engine ignition withdraws 150 amperes current from the battery during 6 seconds time.
Q= I x t
U= R x I
The Attempt at a Solution
I'm pretty confused about the electric charge concept and the unit Coulomb.
Also, I'm not a big car guy, I'm afraid. I know how to drive them though!
I don't really see which unit is As (amperesecond) or Ah, or Coulomb.
Why are there so many units for electric charge is the thing I'm wondering.
Actually I was also wondering... Does the car starter engine really draw 150amps, from the battery? Regular house fuse, only holds up to about 15 amps, as I recall. Maybe it truly is so because the problem statement was that way.
I also wonder, what the current through the circuit would be under load... Perhaps there is not enough information given to ascertain what the current would be?
Also it seemed as though the starting values for resistance and voltage were basically redundant values as they were not used in the calculation process. At least the book answer at the end of the problems states. that it was enough to find out what amount of electric charge is reduced by one attempt at starting the car. Then find out how many attempts could be made, until the battery charge is reduced by the correct amount.
Basically I know I need to calculate the charge, that occurs within one car start attempt. If that could be known, then simple division could be made, to see how many attempts it takes to reduce the electric charge until the threshold is reached (5 Ah)
I tried to convert the units at first to As , though it might not be necessary.
My book says that
I = Q/ t
where current is electric charge divided by, time taken for the charge to flow.
normally the unit of Q = 1 coulomb
I suppose that 1C= 1As
So one could surmise that
amperes= As/ seconds
I x t = Q
so one would end up with electric charge, as the product of current times the time.
I have some doubts mentally though. It seems to me that 150 amps is a lot of current to my mind. Could it possibly be that much? The problem statement clearly said though, that one attempt to start the engine takes 150amp current within time of 6 seconds.
I = Q/t ] both sides times t
I x t= Q
150A x 6s= 900 As
this much electric charge is taken in one attempt at engine starting it seems.
5 Ah= 18 000 As
18 000 As / 900 As = 20
20 attempts at ignition could be made such that the battery life falls down to that level.
Your result is correct. As is really unit of charge and 1 C is equal to 1 As.
You seem to have arrived at the right answer.
Yes, a starter motor does take a very large current, maybe up to 1000 Amps. A household electricity supply uses higher voltage, which allows smaller currents to be used.
A Coulomb is an Amp-second, a small unit often used in electronics, and is much smaller than an Amp-Hour, which is used for the large amounts of charge found in heavy current engineering.
Is it allowed to cakculate this problem using ampere-hours instead of As ?
Are you allowed to plug into the formula ampere hours, instead of the usual unit of charge the coulomb.
Q/t = I
Amperehours / hours = current (??now I am confused here??)
That seems like it would be that way.
When you calculate mathematically
(AxB) ÷ (B) = A
Both numerator and denominator are divisible by B.
Therefore that quotient would be A
It was very confusing to think about amperehours and amperes together initially, because coulombs are associated with ampereseconds.
The name of the unit As is Coulomb. You can use Ampere-hours to measure charge, but it is 3600 Coulomb. If you divide x Ah with the time in hours, you get current in amps.
Remember we use the unit kWh to measure how much electric energy we consumed. It is energy, and 1 kWh is equal to 3600000 joule.
late347 said:
Why are there so many units for electric charge is the thing I'm wondering.
Good question. A perfectly good unit exists for charge (coulomb), but when dealing with batteries, milliamp-hours or amp-hours are commonly used, which necessitates unnecessary conversion. I suppose amp-hour gives people an intuitive feel for how long and how much current a battery can supply.
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# Why is the derivative of the softmax layer shaped differently than the derivative of other neurons?
If the derivative is supposed to give the rate of change of a function at that point, then why is the derivative of the softmax layer (a vector) the Jacobian matrix, which has a different shape than the output/softmax vector? Why is the shape of the softmax vector's derivative (the Jacobian) different than the shape of the derivative of the other activation functions, such as the ReLU and sigmoid?
• This blog post could also be useful.
– nbro
Jan 2, 2021 at 20:43
If you have a function $$f:\mathbb{R}\rightarrow\mathbb{R}$$ the derivative is a function on its own and you have $$f':\mathbb{R}\rightarrow\mathbb{R}.$$ If you increase the dimension of the input space have $$f:\mathbb{R}^n\rightarrow\mathbb{R}.$$ The "derivative" in this case is called gradient and it is a vector collecting the $$n$$ partial derivatives of $$f$$. The input space of the gradient function is $$\mathbb{R}^n$$ (the same as for $$f$$), but the output is the collection of the $$n$$ derivatives, so the output space is also $$\mathbb{R}^n$$. In other words $$\nabla f:\mathbb{R}^n\rightarrow\mathbb{R}^n,$$ which makes sense as for each point $$x$$ of the input space you get a vector ($$\nabla f(x)$$) as output.
So far so good, but what happens if you consider a function that takes a vector as input and spits out a vector as output, i.e. $$f:\mathbb{R}^n\rightarrow\mathbb{R}^m?$$ How to compute the equivalent of the derivative? (This is the softmax case, where you have a vector as input and a vector as output.)
You can reduce this case to the previous case considering $$f=(f_1, \dots, f_m)$$, where $$f_i:\mathbb{R}^n\rightarrow\mathbb{R}.$$ Now for each $$f_i$$ you can compute the gradient $$\nabla f_i$$ and end up with $$m$$ gradients. When you evaluate them at a point $$x\in\mathbb{R}^n$$ you get $$m$$ $$n-$$dimensional vectors. These vector can be collected in a matrix, which is the Jacobian; formally $$Jf:\mathbb{R}^n\rightarrow\mathbb{R}^{m\times n}.$$ Finally, to answer your question, you get a Jacobian "instead" of a gradient (they all represent the same concept) because the output of the softmax is not a single number but a vector.
By the way the sigmoid and relu are functions with one dimensional input and output, so they don't really have a gradient but a derivative. The trick is that people write $$\sigma(W)$$, where $$W$$ is vector or a matrix, but they mean that $$\sigma$$ is applied component-wise, as a function from $$\mathbb{R}$$ to $$\mathbb{R}$$ (I know, it's confusing).
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Q. 3 C
Test whether the
Here, R1, R2, R3, and R4 are the binary relations.
So, recall that for any binary relation R on set A. We have,
R is reflexive if for all x A, xRx.
R is symmetric if for all x, y A, if xRy, then yRx.
R is transitive if for all x, y, z A, if xRy and yRz, then xRz.
So, using these results let us start determining given relations.
We have
R3 on R defined by (a, b) R3 a2 – 4ab + 3b2 = 0
Check for Reflexivity:
a R,
(a, a) R3 needs to be proved for reflexivity.
If (a, b) R3, then we have
a2 – 4ab + 3b2 = 0
Replace b by a, we get
a2 – 4aa + 3a2 = 0
a2 – 4a2 + 3a2 = 0
–3a2 + 3a2 = 0
0 = 0, which is true.
(a, a) R3
So, a R, (a, a) R3
R3 is reflexive.
Check for Symmetry:
a, b R
If (a, b) R3, then we have
a2 – 4ab + 3b2 = 0
a2 – 3ab – ab + 3b2 = 0
a (a – 3b) – b (a – 3b) = 0
(a – b) (a – 3b) = 0
(a – b) = 0 or (a – 3b) = 0
a = b or a = 3b …(1)
Replace a by b and b by a in equation (1), we get
b = a or b = 3a …(2)
Results (1) and (2) does not match.
(b, a) R3
R3 is not symmetric.
Check for Transitivity:
a, b, c R
If (a, b) R3 and (b, c) R3
a2 – 4ab + 3b2 = 0 and b2 – 4bc + 3c2 = 0
a2 – 3ab – ab + 3b2 = 0 and b2 – 3bc – bc + 3c2
a (a – 3b) – b (a – 3b) = 0 and b (b – 3c) – c (b – 3c) = 0
(a – b) (a – 3b) = 0 and (b – c) (b – 3c) = 0
(a – b) = 0 or (a – 3b) = 0
And (b – c) = 0 or (b – 3c) = 0
a = b or a = 3b And b = c or b = 3c
What we need to prove here is that, a = c or a = 3c
Take a = b and b = c
Clearly implies that a = c.
[ if a = b, just substitute a in place of b in b = c. We get, a = c]
Now, take a = 3b and b = 3c
If a = 3b
Substitute in b = 3c. We get
a = 9c, which is not the desired result.
(a, c) R3
R3 is not transitive.
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## Dividing a decimal
Although dividing a decimal follows exactly the same method as dividing a whole number, the presence of a decimal point can confuse pupils, leading them to complete the calculation in two separate chunks.
In this example, the pupil has correctly worked out that 6 ÷ 4 = 1 remainder 2. Seeing the decimal point next, they felt they had completed the first chunk of the calculation and were unsure what to do with the remainder. This resulted in them writing in the 2 as part of the answer, before starting the decimal part of the calculation.
Whilst the decimal point makes no difference to the way we work out the calculation, it’s important not to just tell pupils that they can ignore it, as they may then miss the decimal point out of their answer.
Click here to see 'How To: Dividing a decimal' video
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# What is Scale and Types of Scales in Surveying .
A scale is the ratio of the distance marked on the plan to the corresponding distance on the ground. A good draughtsman can plot a length to accuracy within 0.25 mm. Types of Scales are generally classified as large, medium and small. There are generally 5 types of scales used in surveying which we will discuss below.
• Large scale : 1 cm = 10 m or less than 10 m.
• Medium scale : 1 cm = 10 m to 100 m.
• Small scale : 1 cm = 100 m or more than 100 m.
## Representation of a Scale :
A. Engineer’s scale : 1 cm = 50 m
B. Representative Fraction (R.F) :
It is the ratio of the distance on the map to the corresponding distance on the ground taken as same units. Scale of 1 cm = 50 m, 1 cm on the map represents 50 m (5000 cm) on the ground. Therefore, the representative fraction (R.F.) is 1/5000 or 1: 5000.
C. Graphical scale :
A graphical scale is a line drawn on the map so that its map distance corresponds to a convenient units of length on the ground.
It has the advantage over the numerical scales that the distances on the maps can be determined by actual scaling even when the map has shrunk.
Purpose of SurveyScaleR.F
Building Site1 cm = 10 m1:1000
Town Planning, Reservoir planning, etc1 cm = 50 cm to 100 m1:5000 to 10000
Route Surveys1 cm = 10 m to 60 m1:1000 to 1:6000
Longitudional Sections.
1 cm = 10 m
1 cm = 1 m
1:1000
1:100
Cross- Sections1 cm = 1 m1:100
Land Surveys/ Cadastral Surveys1 cm = 10 m to 50 m1:1000 to 1:5000
Topographical Maps1 cm = 0.25 km to 2.5 km1:25000 to 1:250000
Geographical Maps1 cm = 5 km to 150 km1:500000 to 1:15000000
Mine Surveys1 cm = 10 m to 25 m1:1000 to 1:2500
Forest Maps1 cm = 250 m1:25000
## Types of Measuring Scales in Surveying
### Plane Scale :
It is possible to measure two successive dimensions only.
### Diagonal Scale :
It is possible to measure three successive dimensions.
### Chord scale :
It is used to set out angles without using a protractor.
```Also Read : What is Geographic Information System (GIS) ?
Also Read : Global Positioning System { GPS } in Surveying```
### Vernier Scale :
It is a device for measuring accurately the fractional part of the smallest division on a graduated scale. It divided into,
• Direct Vernier : ‘n’ divisions on the vernier scale are equal in length to (n – 1) divisions on the main scale. Thus ‘n’ divisions of vernier = {n -1) of main scale :
∴ n ‘v’ = (n-1) ‘s’ or v = {(n-1)/n}*s
Where, n = total No. of divisions on vernier scale, v = length of one division on the vernier, s = length of one division on the main scale. The least count (L.C) is, therefore given by
L.C = s – v.
L.C = s – {(n-1)/n}*s.
L.C = s/n.
• Retrograde Vernier : ‘n’ divisions of the vernier scale are equal to ‘(n+1)’ divisions on the main scale. ∴ n ‘v’ = (n+1) ‘s’
• Extended Vernier : ‘n’ divisions of the vernier scale are equal in length to (2n – 1) divisions of the main scale. Therefore,
∴ n ‘v’ = (2n-1)s
∴ v = {(2n-1)/n}s
• Double Vernier : It is used when the graduations on the main scale are numbered in both directions. It is a combination of both direct and retrograde verniers.
• Double folded vernier : Its length is half of corresponding double verniers – economy of space.
### Micro scale :
It is a device which enables a measurement to be taken to a still finer degree of accuracy.
```Also Read : Plane Table Surveying : Advantages & Disadvantages
Also Read : Contour line and Methods of Contouring```
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# period of oscillation
endobj Paul Peter Urone (Professor Emeritus at California State University, Sacramento) and Roger Hinrichs (State University of New York, College at Oswego) with Contributing Authors: Kim Dirks (University of Auckland) and Manjula Sharma (University of Sydney). Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. endobj These quantities are related by $$f = \dfrac{1}{T}.$$. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0). From this I get the answer as T=2π/ω where ω=√k/m The above answer is given as the solution to this problem in my book. The SI unit for frequency is the cycle per second, which is defined to be a hertz (Hz): $1 \, Hz = 1 \dfrac{cycle}{sec} \, or 1 \, Hz = \dfrac{1}{s}$. Each successive vibration of the string takes the same time as the previous one. Frequency $$f$$ is defined to be the number of events per unit time. The time and motion is often measured from a central value or point of equilibrium. An oscillation can be a periodic motion that repeats itself in a regular cycle, such as a sine wave—a wave with perpetual motion as in the side-to-side swing of a pendulum, or the up-and-down motion of a spring with a weight. <>>> The time for one oscillation is the period $$T$$. The frequency of sound found in (a) is much higher than the highest frequency that humans can hear and, therefore, is called ultrasound. Example $$\PageIndex{1}$$: Determine the Frequency of Two Oscillations, Medical Ultrasound and the Period of Middle C. We can use the formulas presented in this module to determine both the frequency based on known oscillations and the oscillation based on a known frequency. I visit my parents for dinner every other Sunday. &��rv3>���|��g�s�S���0 ]G�/��#7�@����a�p}G�q���/�8 ��#���T�$P�q4���Mi�B'poP.3�s%@8��^�ŒL���fJk�j����)�~AB�9�YX�:��5j:�K�����`�x��+*:�d������E�� J2�c4����[ʄsg��w)P�̺0��U�5d[RE�:�j|� դ*�:�l���g�uO��/�T_B�(˙�C��%�!�vW��� Ԓo�$�����z|�V\7�|, A medical imaging device produces ultrasound by oscillating with a period of 0.400 µs. A cycle is one complete oscillation. Festival of Sacrifice: The Past and Present of the Islamic Holiday of Eid al-Adha. Will 5G Impact Our Cell Phone Plans (or Our Health?! Missed the LibreFest? :R�U�V�m�l�� J1H The above equation is also valid in the case when an additional constant force is being applied on the mass, i.e. Formula for 1 period of oscillation of a pendulum of a certain length is T=2π √lg Where l is length and g is gravity (9.81 ms-2) T2= 4π2 lg T2= 4π2g l Hypothesis: The longer the length of the pendulum is, the longer the time that is needed to complete 1 oscillation. Substitute 0.400 $$\mu s$$ for $$T$$ in $$f = \dfrac{1}{T}$$: $f = \dfrac{1}{T} = \dfrac{1}{0.400 \times 10^{-6} s}.$. In question (a), the period $$T$$ is given and we are asked to find frequency $$f$$. The number of oscillations per unit time is the frequency $$f$$. 2 0 obj We define periodic motion to be a motion that repeats itself at regular time intervals, such as exhibited by the guitar string or by an object on a spring moving up and down. The term can apply to any object that moves with a pattern at timed intervals. A concept closely related to period is the frequency of an event. <>/Font<>/ProcSet[/PDF/Text/ImageB/ImageC/ImageI] >>/MediaBox[ 0 0 793.8 595.2] /Contents 4 0 R/Group<>/Tabs/S/StructParents 0>> Its units are usually seconds, but may be any convenient unit of time. %PDF-1.5 Note that a vibration can be a single or multiple event, whereas oscillations are usually repetitive for a significant number of cycles. The period found in (b) is the time per cycle, but this value is often quoted as simply the time in convenient units (ms or milliseconds in this case). The size of the object has a direct influence on the time it takes for a period to complete as well as the speed of the oscillation. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. Both questions (a) and (b) can be answered using the relationship between period and frequency. endobj 16.2: Period and Frequency in Oscillations, [ "article:topic", "authorname:openstax", "period", "periodic motion", "license:ccby", "showtoc:no", "program:openstax" ], 16.1: Hooke’s Law - Stress and Strain Revisited, 16.3: Simple Harmonic Motion- A Special Periodic Motion, Creative Commons Attribution License (by 4.0). The period is two weeks. ), The Secret Science of Solving Crossword Puzzles, Racist Phrases to Remove From Your Mental Lexicon. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. 4 0 obj Let’s try one example of each. The amount of time it takes for a period to occur depends on the object that is completing the motion. The period of oscillation is the time it takes for an object to make a repetitive motion. b|���j��. Fact Check: What Power Does the President Really Have Over State Governors? stream When you pluck a guitar string, the resulting sound has a steady tone and lasts a long time. &^]�� Its units are usually seconds, but may be any convenient unit of time. Legal. the The frequency of my visits is 26 per calendar year. Oscillation refers to the repeated back and forth movement of something between two positions or states. The relationship between frequency and period is. Have questions or comments? The time for one complete oscillation is the period $$T$$: $f = \dfrac{1}{T}.$. <> The time and motion is often measured from a central value or point of equilibrium. For example, the minute hand of a clock completes one oscillation every sixty seconds, so it has a frequency of one oscillation per minute. <> If the angle of oscillation is very large, the approximation no longer holds, and a different derivation and equation for the period of a pendulum is necessary.
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# Point Clipping
“Point clipping is a process which is used to define the point position.” The point is either inside the view pane (window) or outside the view pane.
In computer graphics, the computer screen work like a two-dimensional coordinate system. Each point does not need to be displayed inside the computer screen.
It includes two terms-
• Window: It means what to display?
• Clipping: It means discarding the portion that is outside the window.
Example: Let us have a view pane (window). The coordinates of the window are-
(xwmax, xwmin) - For X-axis of the window
(ywmax, ywmin) - For Y-axis of the window
Let us assume a point coordinate (P, Q). If the point lies inside the window, then there is no need to perform point clipping. But if the point lies outside the window, we need to perform clipping. We can understand it by the following equation-
xwmax <= P <= xwmin
ywmax <= Q <= ywmin
There are four conditions; if these four are satisfied, then the point lies inside the window. If anyone condition is not satisfied, then the point lies outside the window, it means we have to perform clipping on the point.
xwmin <= P
xwmax => P
ywmin <= Q
ywmax => Q
### Algorithm of Point Clipping:
Step 1: First, we set the value of xwmin and xwmax coordinates for the window.
Step 2: Now, set the coordinates of a given point (P, Q).
Step 3: Nowcheck the above mention condition.
Step 4: If
Point coordinates lie between the (xwmin, xwmax) and (ywmin, ywmax)
Then
{Display the point in the view pane}
Else
{Removethe point}
Step 5: Stop.
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A linearly graded pn junction in the depletion approximation
In a linearly graded pn junction, the doping the doping varies linearly $N_D(x)-N_A(x)=\alpha x$. In the depletion approximation, it is assumed that there is a depletion width $W$ around the transition from p to n where the charge carrier densities are negligible. Outside the depletion width the charge carrier densities are equal to the doping densities so that the semiconductor is electrically neutral outside the depletion width. Using these approximations, the charge density in the depletion region is,
$$\rho = e\alpha x \hspace{1cm}\text{for}\qquad -\frac{W}{2} < x < \frac{W}{2}, \hspace{1cm} \rho=0\qquad \text{otherwise.}$$
The charge density can be integrated to determine the electric field $E = \int\frac{\rho}{\epsilon}dx$,
$$E(x)=-\frac{e\alpha}{2\epsilon}\left(\left(\frac{W}{2}\right)^2-x^2\right).$$
The electric field can be integrated to determine the electrostatic potential $\phi=-\int E dx$,
$$\phi(x)=-\frac{e\alpha}{2\epsilon}\left(\left(\frac{W}{2}\right)^2x-\frac{x^3}{3}\right).$$
The voltage across the diode is the difference of the electrostatic potential across the depletion width,
$$V_{bi} -V = \frac{e\alpha W^3}{12\epsilon}.$$
The built-in voltage is related to the doping concentrations at the edge of the depletion zone, $N_{W/2}=N_A(-W/2)=N_D(W/2)$, by the usual formula,
$$eV_{bi}=k_BT\ln\left(\frac{N_{W/2}^2}{n_i^2}\right).$$
The depletion width and the gradient in the doping can then be calculated,
$$W=\sqrt{\frac{6\epsilon (V_{bi}-V)}{eN_{W/2}}}\qquad\text{and}\qquad \alpha = \frac{2N_{W/2}}{W}.$$
The form below determines $W$ and $N_{W/2}$ numerically by guessing a value for $N_{W/2}$ and calculating the corresponding $\alpha$ until the correct value is found.
$\alpha$ = cm-4 $E_{\text{breakdown}}$ = V/cm $E_g$ = eV $N_v(300)$ = cm-3 $N_c(300)$ = cm-3 $\epsilon_r$ = $T$ = K $\mu_p$ = cm²/V s $\mu_n$ = cm²/V s $\tau_p$ = s $\tau_n$ = s
$V$ = V V
$E_g=$ eV $W=$ μm $\alpha=$ cm-4 $N_{W/2}=$ cm-3 $V_{bi}=$ V $C_j=$ nF/cm²
$n_i=$ 1/cm³ $D_p=$ cm²/s $D_n=$ cm²/s $L_p=$ μm $L_n=$ μm $V_{\text{breakdown}}=$ V
Charge density
ρ [C/m³] x [μm]
Electric field
E [V/m] x [μm]
Band diagram
[eV] x [μm]
Electrostatic potential
$\phi\,\text{[V]}$ $V$ [V]
Carrier Densities
$\frac{n}{N_{W/2}}$$\frac{p}{N_{W/2}} x [μm] log(Carrier Densities) [1/cm³] x [μm] Capacitance - Voltage \large \frac{1}{C_j^3}$$\left[\frac{\text{cm}^6}{\text{nF}^3}\right]$ V [V]
Combining the equations $V_{bi} -V = \frac{e\alpha W^3}{12\epsilon}$ and $C_j = \frac{\epsilon}{W}$,
it can be shown that $\frac{1}{C_j^3} = \frac{12(V_{bi}-V)}{\epsilon^2e\alpha}$.
Current densities
$j$$\left[\text{A/cm}^2\right]$ x [μm]
$\vec{j}_{n,\text{drift}}= ne\mu_n\vec{E}$, $\vec{j}_{p,\text{drift}}= pe\mu_p\vec{E}$,
$\vec{j}_{n,\text{diffusion}}= eD_n\frac{dn}{dx}$, $\vec{j}_{p,\text{diffusion}}= -eD_p\frac{dp}{dx}$
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# Topic 17 Year 7 Mathematics
Mathematics Topic Percentages No of lessons 8 When is it happening Term 3 Year 7 What will students learn In this unit, students work with percentages as another representation of ratios and fractions. The different ways of interpreting fractions are again referenced in this unit building on work from the previous unit in connecting ratios to fractions and decimals. In order to manipulate calculations involving percentages, students need to be able to use their knowledge of arithmetic with decimals and fractions. Therefore, it is important that students are also able to interpret percentages first as a number between 0 and 1 before moving onto work with percentages greater than 100 Key Knowledge that students should know at the end of 'Topic' This is the knowledge that students will meet for the first time in this topic "Understand percentages as a ratio of two quantities where one quantity is standardised to 100 Understand percentages as a fractional operator with a denominator of 100 Understand and interpret percentages over 100 Interpret a percentage as a fraction and decimal Express a quantity as a percentage of another Compare two quantities using percentages Find a percentage of an amount with and without a calculator Increase and decrease a quantity by a given percentage • Find a quantity given a percentage of it" This is knowledge that students may have met before but will need to deepen their understanding Understand percentages Fractions to percentages Equivalent FDP Order FDP Percentage of an amount (1) Percentage of an amount (2) Percentages – missing values Key Skills that students should be able to demonstrate at the end of 'Topic' This is the skills that students will meet for the first time in this topic Increase an amount by a percentage This is skills that students may have met before but will need to develop Find any percentage of an amount Key vocabulary that students should know and understand Ratio, fraction, proportion, equivalent, percentage, decimal, quantity, denominator The Big Question How do i calculate a percentage? Key questions that students should be able to answer at the end of the 'Topic' Can I work out percentages on a number line? Can I calculate equivalent decimals and percentages? Can I convert between fractions and decimals? Can I use a bar model to calculate a percentage of a quantity? Can I use a number line to calculate percentage of quantities? Can I use percentage multipliers less than 100%? Can I use percentage multipliers greater than 100%? Can I use percentage multipliers to calculate percentage increases and decreases?
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05 Apr 2013, 08:15
Can I algebraically prove that if p/q<1 then q/(p^2) is not necessarily greater than 1? Picking numbers is easy, I am just wondering if I can algebraically solve it.
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05 Apr 2013, 08:47
Not very clear on the problem without specifics about whether p & q are positive integers
Because if p > 0 and q > 0 and integers then p/q < 1 => q/p > 1 and q/p2 can be >=< 1
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05 Apr 2013, 11:01
nt2010 wrote:
Not very clear on the problem without specifics about whether p & q are positive integers
Because if p > 0 and q > 0 and integers then p/q < 1 => q/p > 1 and q/p2 can be >=< 1
Yes sorry, p and q are positive integers. When you say can q/p2 be greater or less than 1, is there a way we can reach that algebraically.
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05 Apr 2013, 12:32
1
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q/p > 1 so divide both side by p since P is a positive integer P > 0
q/p2 > 1/p. IFF q > p
//kudos, please if this explanation is good
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05 Apr 2013, 12:39
nt2010 wrote:
q/p > 1 so divide both side by p since P is a positive integer P > 0
q/p2 > 1/p. IFF q > p
//kudos, please if this explanation is good
One last question, why do you say q/p2 > 1/p if q>p? How did you get to q>p?
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05 Apr 2013, 12:52
Remember we got to the equation q/p2 > 1/p from q/p > 1 and if you multiple both side by p you'll get the condition q > p.
Hope this is clear.
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Re: p/q<1 [#permalink] 05 Apr 2013, 12:52
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# How do you convert binary to decimal if the binary was 1.01?
What is 1.01 binary in decimal?
-
Which language? – Carnotaurus Feb 15 '11 at 19:03
In general, by hand. – Strawberry Feb 15 '11 at 19:04
Just remember we use positional numbering system.
``````1101 = 2^0 + 2^2 + 2^3 = 1 + 4 + 8 = 13
1.01 = 2^0 + 2^(-2) = 1 + 1/4 = 1.25
``````
-
Positions to the left of the unit multiply by 2 each time. Positions to the right of the unit divide by 2 each time.
``````1 . 0 1
1×1 + 0×1/2 + 1×1/4 = 1 1/4
``````
-
I wish my maths teachers at school had shown me this - I might have become a C developer – 5arx Feb 26 '11 at 12:29
It's 1.25:
`````` 1 * 2^0 + 0 * 2^-1 + 1 * 2^-2
= 1 * 1 + 0 * 0.5 + 1 * 0.25
= 1 + 0.25
= 1.25
``````
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# MATLAB TUTORIAL, part 2.1: Polynomial Interpolation
Polynomial Interpolation
Originally, spectral decomposition was developed for symmetric or self-adjoint matrices. Following tradition, we present this method for symmetric/self-adjoint matrices, and later expand it for arbitrary matrices.
A matrix A is said to be unitary diagonalizable if there is a unitary matrix U such that $${\bf U}^{\ast} {\bf A}\,{\bf U} = {\bf \Lambda} ,$$ where Λ is a diagonal matrix and $${\bf U}^{\ast} = {\bf U}^{-1} .$$ A matrix A is said to be orthogonally diagonalizable if there is an orthogonal matrix P such that $${\bf P}^{\mathrm T} {\bf A}\,{\bf P} = {\bf \Lambda} ,$$ where Λ is a diagonal matrix and $${\bf P}^{\mathrm T} = {\bf P}^{-1} .$$
Theorem: The matrix A is orthogonally diaginalisable if and only if A is symmetric ($${\bf A} = {\bf A}^{\mathrm T}$$ ). ■
Theorem: The matrix A is unitary diaginalisable if and only if A is normal ($${\bf A}\, {\bf A}^{\ast} = {\bf A}^{\ast} {\bf A}$$ ). ■
Example: The matrix
${\bf A} = \begin{bmatrix} 1&{\bf j}&0 \\ {\bf j}&1&0 \\ 0&0&1 \end{bmatrix}$
is symmetric, normal, but not self-adjoint. Another matrix
${\bf B} = \begin{bmatrix} 2&1 \\ -1&2 \end{bmatrix}$
is normal, but not self-adjoint. Therefore, both matrices are unitary diagonalizable but not orthogonally diagonalizable.
Example. Consider a symmetric matrix
${\bf A} = \begin{bmatrix} 2&1&1 \\ 1&2&1 \\ 1&1&2 \end{bmatrix}$
that has the characteristic polynomial $$\chi_{A} (\lambda ) = \det \left( \lambda {\bf I} - {\bf A} \right) = \left( \lambda -1 \right)^2 \left( \lambda -4 \right) .$$ Thus, the distinct eigenvalues of A are $$\lambda_1 =1,$$ which has geometrical multiplicity 2, and $$\lambda_3 =4.$$ The corresponding eigenvectors are
${\bf u}_1 = \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} , \quad {\bf u}_2 = \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} \qquad \mbox{and} \qquad {\bf u}_3 = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} .$
The vectors $${\bf u}_1 , \ {\bf u}_2$$ form the basis for the two-dimensional eigenspace corresponding $$\lambda_1 =1 ,$$ while $${\bf u}_3$$ is the eigenvectors corresponding to $$\lambda_3 =4 .$$ Applying the Gram--Schmidt process to $${\bf u}_1 , \ {\bf u}_2$$ yields the following orthogonal basis:
${\bf v}_1 = {\bf u}_1 = \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} \quad \mbox{and} \quad {\bf v}_2 = {\bf u}_2 - \frac{\langle {\bf u}_2 , {\bf v}_1 \rangle}{\| {\bf v}_1 \|^2} \, {\bf v}_1 = \frac{1}{2} \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix}$
because $$\langle {\bf u}_2 , {\bf u}_1 \rangle = -1$$ and $$\| {\bf v}_1 \|^2 =2 .$$ Normalizing these vectors, we obtain orthonormal basis:
${\bf q}_1 = \frac{{\bf v}_1}{\| {\bf v}_1 \|} = \frac{1}{\sqrt{2}} \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} , \quad {\bf q}_2 = \frac{{\bf v}_2}{\| {\bf v}_2 \|} = \frac{1}{\sqrt{6}} \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix} , \quad {\bf q}_3 = \frac{{\bf v}_3}{\| {\bf v}_3 \|} = \frac{1}{\sqrt{3}} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} .$
Finally, using $${\bf q}_1 ,\ {\bf q}_2 , \ {\bf q}_3$$ as column vectors, we obtain the unitary matrix
${\bf U} = \begin{bmatrix} \frac{-1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\ 0& \frac{-2}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \end{bmatrix} ,$
which orthogonally diagonalizes A. As a check, we confirm
${\bf U}^{\mathrm T} {\bf A} \,{\bf U} = \begin{bmatrix} \frac{-1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{6}} & \frac{-2}{\sqrt{6}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \end{bmatrix} \, \begin{bmatrix} 2&1&1 \\ 1&2&1 \\ 1&1&2 \end{bmatrix} \, \begin{bmatrix} \frac{-1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\ 0& \frac{-2}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \end{bmatrix} = \begin{bmatrix} 1 &0&0 \\ 0 &1&0 \\ 0 &0&4 \end{bmatrix} ,$
Theorem: Let A be a symmetric or normal $$n \times n$$ matrix with eigenvalues $$\lambda_1 , \ \lambda_2 , \ \ldots , \ \lambda_n$$ and corresponding eigenvectors $${\bf u}_1 , \ {\bf u}_2 , \ \ldots , \ {\bf u}_n .$$ Then
${\bf A} = \begin{bmatrix} \uparrow & \uparrow & \cdots & \uparrow \\ {\bf u}_1 & {\bf u}_2 & \cdots & {\bf u}_n \\ \downarrow & \downarrow & \cdots & \downarrow \end{bmatrix} \, \begin{bmatrix} \lambda_1 &&&0 \\ &\lambda_2 && \\ &&\ddots & \\ 0&&& \lambda_n \end{bmatrix} \, \begin{bmatrix} \longleftarrow & {\bf u}_1 & \longrightarrow \\ \longleftarrow & {\bf u}_2 & \longrightarrow & \\ \vdots & \\ \longleftarrow & {\bf u}_n & \longrightarrow \end{bmatrix} = \sum_{i=1}^n \lambda_i {\bf u}_i {\bf u}_i^{\ast} ,$
which is called spectral decomposition for a symmetric/ normal matrix A. ■
The term was cointed around 1905 by a German mathematician David Hilbert (1862--1942). Denoting the rank one projection matrix $${\bf u}_i {\bf u}_i^{\ast}$$ by $${\bf E}_i = {\bf u}_i {\bf u}_i^{\ast} ,$$ we obtain spectral decomposition of A:
${\bf A} = \lambda_1 {\bf E}_1 + \lambda_2 {\bf E}_2 + \cdots + \lambda_n {\bf E}_n .$
This formular allows one to define a function of a square symmetric/self-adjoint matrix:
$f\left( {\bf A} \right) = f(\lambda_1 )\, {\bf E}_1 + f(\lambda_2 )\, {\bf E}_2 + \cdots + f(\lambda_n )\,{\bf E}_n$
because the matrices $${\bf E}_k , \quad k=1,2,\ldots n,$$ are projection matrices:
${\bf E}_i {\bf E}_j = \delta_{i,j} {\bf E}_i = \begin{cases} {\bf E}_i , & \mbox{ if } i=j, \\ {\bf 0} , & \mbox{ if } i \ne j , \end{cases} \qquad i,j =1,2,\ldots n.$
Example. Consider a self-adjoint 2-by-2 matrix
${\bf A} = \begin{bmatrix} 1 &2+{\bf j} \\ 2- {\bf j} & 5\end{bmatrix} ,$
where $${\bf j}^2 =-1 .$$ We have $${\bf S}^{\ast} {\bf A} {\bf S} = {\bf S}^{-1} {\bf A} {\bf S} = {\bf \Lambda}$$ for the matrices
${\bf S} = \begin{bmatrix} \left( 2+{\bf j} \right) / \sqrt{6} & \left( 2+{\bf j} \right) / \sqrt{30} \\ - 1/\sqrt{6} & 5\sqrt{30} \end{bmatrix} \quad \mbox{and} \quad {\bf \Lambda} = \begin{bmatrix} 0&0 \\ 0& 6 \end{bmatrix} .$
Here column vectors in matrix S are normalized eigenvectors corresponding eigenvalues $$\lambda_1 =0 \quad \mbox{and} \quad \lambda_2 =6 .$$
Therefore, the spectral decomposition of A becomes $${\bf A} = 0\,{\bf E}_1 + 6\,{\bf E}_2 ,$$ which is clearly matrix A itself.
In our case, projection matrices are
${\bf E}_1 = \frac{1}{6} \begin{bmatrix} 5 & -2 - {\bf j} \\ -2+{\bf j} & 1 \end{bmatrix} , \qquad {\bf E}_2 = \frac{1}{6} \begin{bmatrix} 1 &2+{\bf j} \\ 2- {\bf j} & 5\end{bmatrix} = \frac{1}{6}\, {\bf A} .$
It is easy to check that
${\bf E}_1^2 = {\bf E}_1 , \qquad {\bf E}_2^2 = {\bf E}_2 , \qquad \mbox{and} \qquad {\bf E}_1 {\bf E}_2 = {\bf 0} .$
The exponential matrix-function is
$e^{{\bf A}\,t} = {\bf E}_1 + e^{6t} \,{\bf E}_2 .$
Example. Consider a symmetric 3-by-3 matrix from the previous example
${\bf A} = \begin{bmatrix} 2&1&1 \\ 1&2&1 \\ 1&1&2 \end{bmatrix} .$
Its spectral decomposition is $${\bf A} = 1\,{\bf E}_1 + 1\,{\bf E}_2 + 4\,{\bf E}_3 ,$$ where the projection matrices $${\bf E}_i = {\bf q}_i {\bf q}_i^{\ast}$$ are obtained from the orthonormal eigenvectors:
\begin{align*} {\bf E}_1 &= \frac{1}{6} \begin{bmatrix} -1 \\ -1 \\ 2 \end{bmatrix} \left[ -1 \ -1 \ 2 \right] = \frac{1}{6} \begin{bmatrix} 1&1& -2 \\ 1&1& -2 \\ -2&-2& 4 \end{bmatrix} , \\ {\bf E}_2 &= \frac{1}{2} \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix} \left[ -1 \ 1 \ 0 \right] = \frac{1}{2} \begin{bmatrix} 1&-1&0 \\ -1&1&0 \\ 0&0&0 \end{bmatrix} , \\ {\bf E}_3 &= \frac{1}{3} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \left[ 1 \ 1 \ 1 \right] = \frac{1}{3} \begin{bmatrix} 1&1& 1 \\ 1&1& 1 \\ 1&1& 1 \end{bmatrix} . \end{align*}
Indeed, these diagonalizable matrices satisfy the following relations:
${\bf E}_1^2 = {\bf E}_1 , \quad {\bf E}_2^2 = {\bf E}_2 , \quad {\bf E}_3^2 = {\bf E}_3 , \quad {\bf E}_1 {\bf E}_2 = {\bf 0} , \quad {\bf E}_1 {\bf E}_3 = {\bf 0} , \quad {\bf E}_3 {\bf E}_2 = {\bf 0} ,$
and they all have eigenvalues $$\lambda = 1, 0, 0 .$$ Using this spectral decomposition, we define two matrix-functions corresponding to $${\Phi}(\lambda ) = \cos \left( \sqrt{\lambda} \,t \right)$$ and $${\Psi}(\lambda ) = \frac{1}{\sqrt{\lambda}} \,\sin \left( \sqrt{\lambda} \,t \right)$$ that do not depend on the branch of the choisen square root:
\begin{align*} {\bf \Phi} (t) &= \cos \left( \sqrt{\bf A} \,t \right) = \cos t\, {\bf E}_1 + \cos t\, {\bf E}_2 + \cos (2t) \,{\bf E}_3 = \frac{\cos t}{3} \, \begin{bmatrix} 2&-1&-1 \\ -1&2&-1 \\ -1&-1& 2 \end{bmatrix} + \frac{\cos 2t}{3} \begin{bmatrix} 1&1& 1 \\ 1&1& 1 \\ 1&1& 1 \end{bmatrix} , \\ {\bf \Psi} (t) &= \frac{1}{\sqrt{\bf A}} \,\sin \left( \sqrt{\bf A} \,t \right) = \sin t\, {\bf E}_1 + \sin t\, {\bf E}_2 + \frac{\sin (2t)}{2} \,{\bf E}_3 = \frac{\sin t}{3} \, \begin{bmatrix} 2&-1&-1 \\ -1&2&-1 \\ -1&-1& 2 \end{bmatrix} + \frac{\sin 2t}{6} \begin{bmatrix} 1&1& 1 \\ 1&1& 1 \\ 1&1& 1 \end{bmatrix} . \end{align*}
These matrix-functions are solutions of the following initial value problems for the second order matrix differential equations:
\begin{align*} & \ddot{\bf \Phi}(t) + {\bf A}\,{\bf \Phi} (t) ={\bf 0} , \qquad {\bf \Phi}(0) = {\bf I}, \quad \dot{\bf \Phi}(0) = {\bf 0}, \\ &\ddot{\bf \Psi}(t) + {\bf A}\,{\bf \Phi} (t) ={\bf 0} , \qquad {\bf \Phi}(0) = {\bf 0}, \quad \dot{\bf \Psi}(0) = {\bf I} . \end{align*}
Since the given matrix A is positive definite, we can define four square roots:
\begin{align*} {\bf R}_1 &= {\bf E}_1 + {\bf E}_2 + 2\,{\bf E}_3 = \frac{1}{3} \begin{bmatrix} 4&1&1 \\ 1&4&1 \\ 1&1&4 \end{bmatrix} , \\ {\bf R}_2 &= {\bf E}_1 + {\bf E}_2 - 2\,{\bf E}_3 = \begin{bmatrix} 0&-1&-1 \\ -1&0&-1 \\ -1&-1&0 \end{bmatrix} , \\ {\bf R}_3 &= {\bf E}_1 - {\bf E}_2 + 2\,{\bf E}_3 = \frac{1}{3} \begin{bmatrix} 1&4&1 \\ 4&1&1 \\ 1&1&4 \end{bmatrix} , \\ {\bf R}_4 &= -{\bf E}_1 + {\bf E}_2 + 2\,{\bf E}_3 = \begin{bmatrix} 1&0&1 \\ 0&1&1 \\ 1&1&0 \end{bmatrix} , \end{align*}
and four others are just negative of these four; so total number of square roots is 8. Note that we cannot obtain $${\bf R}_3$$ and $${\bf R}_4$$ using neither Sylvester's method nor the Resolvent method because they are based on the minimal polynomial $$\psi (\lambda ) = (\lambda -1)(\lambda -4) .$$
We expand spectral decomposition for arbitary square matrices. Let $$f (\lambda )$$ be an analytic function in a neighborhood of the origin and A be a square $$n \times n$$ matrix. We choose the origin as an example; application of the spectral decomposition requirs functions to be expressed as convergent power series in neighborhoods of every eigenvalue. Using a Maclaurin series
$f(\lambda ) = f_0 + f_1 \lambda + f_2 \lambda^2 + \cdots + f_k \lambda^k + \cdots = \sum_{k\ge 0} f_k \lambda^k ,$
we can define the matrix-valued function $$f ({\bf A} )$$ as
$f({\bf A} ) = \sum_{k\ge 0} f_k {\bf A}^k .$
Let $$\psi (\lambda )$$ be a minimal polynomial of degree m for the matrix A. Then every power $${\bf A}^p$$ of matrix A can be expressed as a polynomial of degree not higher than $$m-1.$$ Therefore,
$f({\bf A} ) = \sum_{j= 0}^{m-1} b_j {\bf A}^j ,$
where coefficients $$b_j , \quad j=0,1,\ldots , m-1;$$ should satisfy the following equations
$f(\lambda_k ) = \sum_{j= 0}^{m-1} b_k \,\lambda_k^j ,$
for each eigenvalue $$\lambda_k , \quad k=1,2,\ldots , s ,$$ where s is the number of distinct eigenvalues. If the eigenvalue $$\lambda_k$$ is of multiplicity $$m_k$$ in the minimal polynomial $$\psi (\lambda ) ,$$ then we need to add $$m_k -1$$ auxiliary algebraic equations
$\left. \frac{{\text d}^p f(\lambda )}{{\text d} \lambda^p} \right\vert_{\lambda = \lambda_k} = \left. \frac{{\text d}^p}{{\text d} \lambda^p} \, \sum_{j= 0}^{m-1} b_k \,\lambda_k^j \right\vert_{\lambda = \lambda_k} , \quad p=1,2,\ldots , m_k -1 .$
Example: Consider a diagonalizable 4-by-4 matrix
${\bf A} = \begin{bmatrix} -4&7&1&4 \\ 6&-16&-3&-9 \\ 12&-27&-4&-15 \\ -18&43&7&24 \end{bmatrix} .$
Its characteristic polynomial is
$\chi (\lambda ) = \det \left( \lambda {\bf I} - {\bf A} \right) = \lambda \left( \lambda -2 \right) \left( \lambda +1 \right)^2 ,$
but its minimal polynomial is a product of simple terms:
$\psi (\lambda ) = \det \left( \lambda {\bf I} - {\bf A} \right) = \lambda \left( \lambda -2 \right) \left( \lambda +1 \right) .$
Therefore, matrix A is diagonalizable because its minimal polynomail is a product of simple terms, and we don't need to find eigenspaces. ■
Complete
Applications
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# TCS Questions on Clocks and Calendar with Answers – 1
1.The famous church in the city of Kumbakonnam has a big clock tower and is said to be over 300 years old. Every Monday 10.00 A M the clock is set by Antony, doing service in the church. The Clock loses 6 mins every hour. What will be the actual time when the faulty clock shows 3 P.M on Friday?
a. 4 AM
b.3.16 PM
c. 4.54 AM
d. 3 AM
Explaination:Total time passed in the faulty clock = Monday 10 am to Friday 3 pm = 24 x 4 + 5 hours = 96 and 5 hours = 101 hrs
54 min in the faulty clock = 60 minutes of the correct clock
101 hrs in the faulcty clock = ?
101/54×60 = 112.2 Hrs.
96 Hrs + 16.2 Hrs
Friday 10 am + 16 hrs = Saturday 2am
0.2 x 60 min = 12 min
So Saturday 2.12 min AM
2.When asked what the time is,a person answered that the amount of time left is 1/5 of the time already completed.what is the time.
a. 8 pm
b. 8 am
c. 12 pm
d. 12 am
Explaination: A day has 24 hrs. Assume x hours have passed. Remaining time is (24 – x)
24−x=(1-5) /x ⇒x=20
Time is 8 PM
3.In a particular year, the month of january had exactly 4 thursdays, and 4 sundays. On which day of the week did january 1st occur in the year.
a. monday
b. tuesday
c. wednesday
d. thursday
Explaination:If a month has 31 days, and it starts with sunday, Then Sundays, Mondays, tuesdays are 5 for that month. If this month starts with monday, then mondays, tuesdays, and wednesdays are 5 and remaining days are 4 each. so this month start with Monday.
4.A clock loses 1% time during first week and then gains 2% time during the next one week. if the clock was set right at 12noon on Sunday what will be the the time exactly that the clock will show 14days from the time it was set right?
a. 1 : 30 : 48 P.M.
b. 1 : 40 : 48 P.M.
c. 1 : 40 : 38 P.M.
d. 1 : 30 : 48 A.M.
Explanation: The clock loses 1% time during the first week. In a day there are 24 hours and in a week there are 7 days. Therefore, there are 7 * 24 = 168 hours in a week. If the clock loses 1% time during the first week, then it will show a time which is 1% of 168 hours less than 12 Noon at the end of the first week = 1.68 hours less. Subsequently, the clock gains 2% during the next week. The second week has 168 hours and the clock gains 2% time = 2% of 168 hours = 3.36 hours more than the actual time. As it lost 1.68 hours during the first week and then gained 3.36 hours during the next week, the net result will be a -1.68 + 3.36 = 1.68 hour net gain in time. So the clock will show a time, which is 1.68 hours more than 12 Noon two weeks from the time it was set right. 1.68 hours = 1 hour and 40.8 minutes = 1 hour + 40 minutes + 48 seconds. i.e. 1 : 40 : 48 P.M.
5. How many degrees will the minute hand move, in the same time in which the second hand move 4800 ?
a.60
b.90
c.40
d.80
Explanation : Minute hand covers 480/60= 80
6. How many years have 29 days in February from 2001 to 2100.
a.26
b.25
c.23
d.24
Explanation: 100th year is not a leap year. So 24 February’s has 29 days
7. 2012 January 1st is Sunday, then which day is the Indian Independence day of the same year.
a.Saturday
b.Wednesday
c.Thursday
d.Friday
Explanation: 30+ 29+ 31 + 30 + 31 + 30 + 31+15 = 227/7 = reminder = 3
So Independence day is Wednesday
8. Which year has the same calendar as 1700 ?
a. 1705
b.1706
c.1707
d.1708
Explanation:
Year : 1700 1701 1702 1703 1704 1705
Odd days : 1 1 1 1 2 1
9. If Arun’s birthday is on May 25 which is Monday and his sister’s birthday is on July 13. Which day of the week is his sister’s birthday?
a.Monday
b.Wednesday
c.Thursday
d.Friday
Explanation: Reference day : May 25th Monday
Days from May 25th to July 13 = 6 + 30 +13 = 49
No of odd days : 49/7 = 0
10.March 1st is Wednesday. Which month of the same year starts with the same day?
a.October
b.November
c.December
d.None of these
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# Chapter 2: One-Dimensional Kinematics Outline 2-1Position, Distance, and Displacement 2-2Average Speed and Velocity 2-3Instantaneous Velocity 2-4Acceleration.
## Presentation on theme: "Chapter 2: One-Dimensional Kinematics Outline 2-1Position, Distance, and Displacement 2-2Average Speed and Velocity 2-3Instantaneous Velocity 2-4Acceleration."— Presentation transcript:
Chapter 2: One-Dimensional Kinematics Outline 2-1Position, Distance, and Displacement 2-2Average Speed and Velocity 2-3Instantaneous Velocity 2-4Acceleration 2-5Motion with Constant Acceleration 2-6Applications of the Equations of Motion 2-7Freely Falling Objects
What we will do today: The most important things to understand in Chapter 2. How to use this information to solve problems.
TA’s Information Name: Tatiana Brusentsova Email: Office hours
Major Concepts Position, distance and displacement Speed and velocity –Average –Instantaneous –Constant Acceleration –Average –Instantaneous –Constant Graphs of position versus time, velocity versus time, and acceleration versus time Equations of motion with constant acceleration Free fall Warning Areas: Speed vs. Velocity Distance vs. Displacement Constant Velocity vs. Constant Acceleration
How the physics is described, step by step. Constant speed: Constant acceleration:
Constant speed, acceleration. Constant Speed t, sec x, M tt xx Slope = v Constant Acceleration t, sec v, m/s tt vv Slope = a Special case of constant a.
Figure 2-26 Problem 2-21 Find the average velocity for each segment of the “walk”, and for the total “walk”. Now, compare DISPLACEMENT and DISTANCE. Compare, average VELOCITY and SPEED.
Figure 2-28 Problem 2-32 What is the average acceleration for each segment, what is the average acceleration for the whole motorcycle ride?
Constant Acceleration: The “master” 1D formula: Uses:
Table 2-4 Constant-Acceleration Equations of Motion Variables RelatedEquationNumber Velocity, time, acceleration v = v 0 + at2-7 Initial, final, and average velocity v av = ½(v 0 + v)2-9 Position, time, velocity x = x 0 + ½(v 0 + v)t2-10 Position, time, acceleration x = x 0 + v 0 t + ½ at 2 2-11 Velocity, position, acceleration v 2 = v 0 2 + 2a(x – x 0 ) = v 0 2 + 2a x 2-12
These equations are related x = x 0 + v 0 t + ½ at 2 v = v 0 + at Use: or a=(v-v 0 )/t Then: x = x 0 + v 0 t + ½ at 2 = x 0 + v 0 t + ½(v-v 0 )t 2 /t = x 0 +½(v+v 0 )t = x 0 +v avg t Everything is derived from definitions of average and constant acceleration and velocity. (the “Master equation”)
Figure 2-29 Problem 2-33 CAREFUL! What is the displacement for each segment of the graph shown? Problem solving: 1.Draw a picture (given) 2.What are they asking for? 3.What is given? 4.What are the mathematical relationships?
Example 2-6 Put the Pedal to the Metal
The formula that most correctly describes the graph of Drag Racing motion is: 1. v 2 = v 0 2 + 2a(x – x 0 ) 2. x = x 0 + ½ at 2 3. x = x 0 + ½(v 0 + v)t 4. v = v 0 + at
Figure 2-16 Velocity as a Function of Position for the Ranger in Example 2-8 Notice that ½ the speed is lost in last ¼ of stopping distance. CONSTANT DECELERATION v 2 = v 0 2 + 2a(x – x 0 )
Hit the Brakes! V o =11.4 m/s a=-3.8m/s 2 Name that graph! t t x
Name that graph! (in right order) 1. Velocity, position, acceleration 2. Position, acceleration, velocity 3. Velocity, acceleration, velocity V o =11.4 m/s a=-3.8m/s 2 t xt
Constant acceleration problem: A vehicle is traveling at an initial speed v. It brakes with constant acceleration a (which can be negative). What is the total time to come to a stop? What is the total distance traveled? What is the shape of position vs. time? What is the shape of velocity vs. time? V 0 =11.4 m/s a=-3.8 m/s^s t, sec m (x) m/s (v)
Objects in “free-fall” Acceleration is gravity, a = -g = -9.8 m/s ^2 Q: A ball is dropped from the top of the library. At the same time, a ball is thrown from the ground so that it just reaches the top of the library. Which of the following statements is true? Draw picture. What is asked? What is given? What are relationships? Solve for unknown.
Help, I’m in free fall! 1. The dropped ball reaches the ground sooner, because it picks up speed as it drops. 2. The thrown ball reaches the top sooner, because it started with higher speed. 3. They reach the endpoints at the same time. 4. Can’t answer—depends on initial speed of ball. This problem can be quickly solved by looking at the time symmetry of the situation. Let’s do this the “long way”, just to see how it’s done.
P 2-27: The pop fly. A baseball is hit straight upwards. The round-trip time in the air is 4.5 seconds. How high did the ball go? What was the initial speed? Draw a picture. Assign “knowns” to variables. Assign variable names to “unknowns”. Work backwards from “knowns” to “unknowns”.
P. 2-91: Seagull flying upwards with seashell, which drops. Seagull flying upward 5.2 m/s, drops shell at 12.5 m. What is acceleration at time of drop? What is maximum height? When does it return to 12.5 m? What is its speed at 12.5m?
P. 2-94. Ball thrown vertically with initial speed v. Ball one thrown upwards with speed v. Reaches maximum height h, when ball two is thrown with same speed. Draw x vs. t for each ball. At what height do balls cross paths?
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# How do you simplify sqrt.27 + rsqrt.75?
$= \sqrt{\frac{27}{100}} + r \sqrt{\frac{75}{100}} =$
$= \sqrt{\frac{3 \cdot 9}{100}} + r \sqrt{\frac{3 \cdot 25}{100}} =$
$= \frac{3}{10} \sqrt{3} + \frac{5}{10} r \sqrt{3} =$
$= \frac{\sqrt{3}}{10} \left(3 + 5 r\right)$
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0
# What are all the numbers less than 100 that are the product of exactly three different prime numbers?
Updated: 11/4/2022
Wiki User
12y ago
All the numbers less than 100 that are the product of exactly three different prime numbers are 30, 42, 66, 70, and 78.
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### what is the product of prime numbers for 376?
The product is exactly 376 since that is what the prime numbers for 376 are!The product is exactly 376 since that is what the prime numbers for 376 are!The product is exactly 376 since that is what the prime numbers for 376 are!The product is exactly 376 since that is what the prime numbers for 376 are!
no
No.
### Is 15 a number that is a product of exactly three different prime numbers?
No. 15 is the product of two prime numbers: 3 and 5
### What number is a product of exactly 3 different prime numbers?
How about: 2*3*5 = 30 which is the product of the 1st three prime numbers
No.
Yes.
### What are numbers less than 100 that are the product of exactly 3 different prime numbers?
All the numbers less than 100 that are the product of exactly three different prime numbers are 30, 42, 66, 70, and 78. Need to know how to get the answer
30, 42 and 70
102.
### What are all of the numbers less than 100 that are the products of exactly three different prime numbers?
All the numbers less than 100 that are the product of exactly three different prime numbers are 30, 42, 66, 70, and 78.
### What is all the prime numbers less than 100 that are the product of exactly 3 different prime numbers?
30, 42, 66, 70, 78
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# The Commutative property - Examples, Exercises and Solutions
## What is the commutative property?
The commutative property is an algebraic principle that allows us to "play" with the position that different elements occupy in multiplication and addition exercises without affecting the final result. Our objective in using the commutative property is to make the resolution of the exercise simpler from the point of view of the calculations.
As we have already said, the commutative property can be applied in the case of addition and multiplication.
In other words:
If we change the place of certain elements in the exercise or equation the result will be the same.
In addition operations we can change the place of the addends and arrive at the same result.
That is:
$a+b=b+a$
Same as in algebraic expressions:
$X+número=número+X$
Regardless of the order in which we add the terms and no matter how many addends there are, the result will always be the same.
Commutative property of multiplication:
In multiplication operations we can change the place of the terms and arrive at the same result.
That is:
$a\times b=b\times a$
Same as in algebraic expressions:
$X\times número=número\times X$
Regardless of the order in which we multiply the factors and no matter how many there are in the exercise, the product will always be the same.
Note - The commutative property does not act in this way in subtraction and division operations.
## examples with solutions for the commutative property
### Exercise #1
$5\cdot5\cdot5\cdot2\cdot2\cdot2=?$
### Step-by-Step Solution
We use the substitution property and organize the exercise in the following order:
$5\times2\times5\times2\times5\times2=$
We place parentheses in the exercise:
$(5\times2)\times(5\times2)\times(5\times2)=$
We solve from left to right:
$10\times10\times10=$
$(10\times10)\times10=$
$100\times10=1000$
1000
### Exercise #2
$-5+2=$
### Step-by-Step Solution
If we draw a line that starts at negative five and ends at 5
We will go from the point negative five two steps forward (+2) we will arrive at the number negative 3.
$-3$
### Exercise #3
$10-5-2-3=$
### Step-by-Step Solution
Given that the entire exercise is with subtraction, we solve the exercise from left to right:
$10-5=5$
$5-2=3$
$3-3=0$
$0$
### Exercise #4
$4-2+2-4=$
### Step-by-Step Solution
Given that we are referring to addition and subtraction exercises, we solve the exercise from left to right:
$4-2=2$
$2+2=4$
$4-4=0$
$0$
### Exercise #5
$3-2+10-x=$
### Step-by-Step Solution
We solve the exercise from left to right:
$3-2=1$
$1+10=11$
Now we obtain:
$11-x$
$11-x$
### Exercise #6
$11\times3+7=$
### Step-by-Step Solution
In this exercise, it is not possible to use the substitution property, therefore we solve it as is from left to right according to the order of arithmetic operations.
That is, we first solve the multiplication exercise and then we add:
$11\times3=33$
$33+7=40$
$40$
### Exercise #7
$12\times13+14=$
### Step-by-Step Solution
According to the order of operations, we start with the multiplication exercise and then with the addition.
$12\times13=156$
Now we get the exercise:
$156+14=170$
$170$
### Exercise #8
$\frac{1}{4}\times4+2=$
### Step-by-Step Solution
According to the order of operations, we first solve the multiplication exercise:
We add the 4 in the numerator of the fraction:
$\frac{1\times4}{4}+2=$
We solve the exercise in the numerator of the fraction and obtain:
$\frac{4}{4}+2=1+2=3$
$3$
### Exercise #9
$-2-4+6-1=$
### Step-by-Step Solution
According to the order of operations, we solve the exercise from left to right:
$-2-4=-6$
$-6+6=0$
$0-1=-1$
$-1$
### Exercise #10
$4:2+2=$
### Step-by-Step Solution
According to the order of operations, we first solve the division exercise:
$4:2=2$
Now we obtain the exercise:
$2+2=4$
$4$
### Exercise #11
Solve:
$2-3+1$
### Step-by-Step Solution
We use the substitution property and add parentheses for the addition operation:
$(2+1)-3=$
Now, we solve the exercise according to the order of operations:
$2+1=3$
$3-3=0$
0
### Exercise #12
Solve:
$3-4+2+1$
### Step-by-Step Solution
We will use the substitution property to arrange the exercise a bit more comfortably, we will add parentheses to the addition operation:
$(3+2+1)-4=$
We first solve the addition, from left to right:
$3+2=5$
$5+1=6$
And finally, we subtract:
$6-4=2$
2
### Exercise #13
Solve:
$-5+4+1-3$
### Step-by-Step Solution
According to the order of operations, addition and subtraction are on the same level and, therefore, must be resolved from left to right.
However, in the exercise we can use the substitution property to make solving simpler.
-5+4+1-3
4+1-5-3
5-5-3
0-3
-3
$-3$
### Exercise #14
$7+4+3+6=\text{?}$
### Step-by-Step Solution
To make solving the exercise easier, we try to add numbers that give us a result of 10.
Let's keep in mind that:
$7+3=10$
$6+4=10$
Now, we obtain a more convenient exercise to solve:
$10+10=20$
20
### Exercise #15
$19+34+21+10+6=\text{?}$
### Step-by-Step Solution
To make solving easier, we try to add numbers that give us a round result.
Keep in mind that:
$19+21=40$
$34+6=40$
Now, we get a more convenient exercise to solve:
$40+40+10=80+10=90$
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College Algebra (10th Edition)
$\dfrac{4x-5}{3x+4}; x \ne -\frac{4}{3}$
The expressions are similar so subtract the numerators and copy the denominator to obtain: $=\dfrac{5x-4-(x+1)}{3x+4} \\=\dfrac{5x-4-x-1}{3x+4} \\=\dfrac{(5x-x)+(-4-1)}{3x+4} \\=\dfrac{4x-5}{3x+4}; x \ne -\frac{4}{3}$ ($x$ cannot be $-\dfrac{4}{3}$ as it makes the expression undefined.)
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# Probability Theory Exam
#### statboy
##### New Member
Studying for a probability exam and here are two practice problems I can't seem to solve
1. Let P be the uniform distribution on a finite set Omega and let A be a subset of Omega. Prove that P(.|A) is the uniform distribution on A.
2. Let X and Y be random variables and let A be an event. Prove the function, Z(w) = X(w) if w is in A and Y(w) if w is not in A , is a random variable
thanks for any help getting me started
#### vinux
##### Dark Knight
Both are theoretical questions. So it is very difficult to write here.
So I am not giving the full solution.
1. I guess the set A is an interval ( say [c,d] ) . Let P be the uniform distribution on the interval [a,b]
P(x) = 1/(b-a)
& P(A) = (d-c)/(b-a)
P(x/A) = P(x and A) / P(A)
= P(x)/P(A)
= 1/(b-a) / (d-c)/(b-a)
= 1/(d-c) it is a uniform distribution.
2) Z(w) is a real valued function. find out the probability space of Z.
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# What is the simplest way to obtain function obect of a given function?
Suppose I have a function p of some arguments:
p[x_,y_]:= (* something *)
Now I with to pass this function as an object to another function, like derivative:
Derivative[0, 2][p[#1, #2] &]
Here I used an expression
p[#1, #2] &
to create a function object.
Can I form it simpler? Neither p not p& worked.
What if I want to pass a function of variable number of arguments to another function?
• Could you elaborate on what you mean by p didn't work? I tried p[x_, y_] := x^5 y^7; Derivative[0, 2][p] and it gives the correct answer. Nov 22, 2014 at 18:13
• If you want to pass the derivative function to another function, just define d02p = Derivative[0, 2][p] and pass d02p to another function. Nov 22, 2014 at 18:13
You can define partial derivatives in specified slots without using #1 or #2:
p[x_, y_] := x^5 y^7;
d02p = Derivative[0, 2][p];
d02p[a, b]
which returns
42 a^5 b^5
Likewise, to obtain the pure function fDer for the mixed partial derivative of a function f of vector argument, try this:
f[list_] := (Times @@ list)^8;
ind = RandomInteger[{0, 8}, 10];
fDer = Derivative[ind][f];
fDer[Subscript[a, #] & /@ Range[10]]
which produces $$122787561599926272000 a_1^7 a_3^6 a_4 a_5^8 a_6^6 a_7^6 a_8^8 a_9^7 a_{10}^3.$$
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https://www.numbersaplenty.com/133400
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Search a number
133400 = 23522329
BaseRepresentation
bin100000100100011000
320202222202
4200210120
513232100
62505332
71063631
oct404430
9222882
10133400
1191253
1265248
1348947
1436888
15297d5
hex20918
133400 has 48 divisors (see below), whose sum is σ = 334800. Its totient is φ = 49280.
The previous prime is 133391. The next prime is 133403. The reversal of 133400 is 4331.
133400 = T96 + T97 + ... + T118.
It is a plaindrome in base 14.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (133403) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (5) of ones.
It is a polite number, since it can be written in 11 ways as a sum of consecutive naturals, for example, 4586 + ... + 4614.
It is an arithmetic number, because the mean of its divisors is an integer number (6975).
2133400 is an apocalyptic number.
133400 is a gapful number since it is divisible by the number (10) formed by its first and last digit.
It is an amenable number.
It is a practical number, because each smaller number is the sum of distinct divisors of 133400, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (167400).
133400 is an abundant number, since it is smaller than the sum of its proper divisors (201400).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
133400 is a wasteful number, since it uses less digits than its factorization.
133400 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 68 (or 59 counting only the distinct ones).
The product of its (nonzero) digits is 36, while the sum is 11.
The square root of 133400 is about 365.2396473550. The cubic root of 133400 is about 51.0958086235.
Adding to 133400 its reverse (4331), we get a palindrome (137731).
The spelling of 133400 in words is "one hundred thirty-three thousand, four hundred".
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Next: Möbius transformations representable as Up: Functions of a complex Previous: one-to-one and invertible
Möbius transformations represented as cross ratios
One of the things which can be done with the Möbius transformation
d w z + c w + b z + a = 0 (21)
is to find relationships amongst the variables which do not depend on the coefficients of the transformation. One procedure is to try it out for four points, which is the number of coefficients, to get simultaneous equations.
d w1 z1 + c w1 + b z1 + a = 0 (22) d w2 z2 + c w2 + b z2 + a = 0 (23) d w3 z3 + c w3 + b z3 + a = 0 (24) d w4 z4 + c w4 + b z4 + a = 0 (25)
The corresponding matrix equation,
can only hold when the determinant of the square matrix vanishes. That relationship invites a series of transformations which can ultimately produce the result
= (26)
This quantity equates the cross ratioscross ratio of the two sets of numbers, and , it being that the cross ratio is an invariant of the Möbius transformation.
The equivalence of this result to the vanishing determinant is not immediately obvious, although it can be verified by careful attention to algebraic rearrangement. A line of reasoning which suggests a result which then can subsequently be confirmed begins with the fractional linear form of the Möbius transformation,
w = (27)
which clearly maps z1 to zero and z3 to infinity, as long as they are distinct. As for counterimages, zero maps into z1/z3, and infinity to unity. However, this is not the only formula which could map those pairs of points; any multiple of the function will do the same; a third pair of points could be introduced to resolve the ambiguity. A convenient set of points is , which suggests selecting a point z2 to map into unity and removing the ambiguity. Thus consider
w = (28)
which is one of the cross ratios in the invariant expression Eq 26. In fact, by setting up matching expressions in w and z, any three distinct points can be mapped into any other three distinct points by going through the virtual intermediary of .
Next: Möbius transformations representable as Up: Functions of a complex Previous: one-to-one and invertible
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# Fluid mechanics sinking barrel
1. Apr 23, 2014
### xxxzyon
1. The problem statement, all variables and given/known data
Suppose a cylindrical barrel falls off a ship and sinks to the bottom of the sea at a depth of 1.61 km. Assume that the seawater is incompressible, so that its density at the bottom of the sea is the same at the surface :1020 kg/m^3
If the pressure inside the barrel is 1.00 atm (it was sealed at sea level) and the total surface area of the barrel is 3.50 m^2 , find the NET INWARD FORCE acting on the surface of the barrel when it reaches the bottom of the ocean.
2. Relevant equations
P tot = Psurface + water x g x h
P= F/A
3. The attempt at a solution
For my attempt, I used the formula Ptot = P surface + ρ density of water x g x h.
The total pressure I got was 116 atm, then I plugged this pressure into P = F/A and got a force of 4. 67 x 10^6 N...
I am not sure what I am supposed to be doing first actually, any guidance would be helpful!
2. Apr 23, 2014
### paisiello2
Can you show explicitly how you came up with the 116 atm?
3. Apr 23, 2014
### haruspex
Sea surface pressure will balance the pressure inside the barrel, so you can ignore this for the 'net force'.
That in turn means there's no point in converting to atms. Do it all in MKS units. (I think you'll find you've made an arithmetic error.)
That said, I really don't like this question. Force is a vector; a net force involves performing a vector sum; the net force acting on the outside of the barrel is zero. Yes, you can take the pressure and multiply by the magnitude of the surface area, but the number that results has no physical meaning. Again, area is technically a vector here, so that multiplication should be done as a vector integral ∫P.dA = P∫dA = 0.
It is just possible that this is a trick question and the required answer is zero.
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# 5.5: Dividing Polynomials
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##### Learning Objectives
By the end of this section, you will be able to:
• Divide monomials
• Divide a polynomial by a monomial
• Divide polynomials using long division
• Divide polynomials using synthetic division
• Divide polynomial functions
• Use the remainder and factor theorems
Before you get started, take this readiness quiz.
1. Add: $$\dfrac{3}{d}+\dfrac{x}{d}$$.
If you missed this problem, review [link].
2. Simplify: $$\dfrac{30xy}{35xy}$$.
If you missed this problem, review [link].
3. Combine like terms: $$8a^2+12a+1+3a^2−5a+4$$.
If you missed this problem, review [link].
## Dividing Monomials
We are now familiar with all the properties of exponents and used them to multiply polynomials. Next, we’ll use these properties to divide monomials and polynomials.
##### Example $$\PageIndex{1}$$
Find the quotient: $$54a^2b^3÷ (−6ab^5)$$.
Solution
When we divide monomials with more than one variable, we write one fraction for each variable.
$$\begin{array} {ll} {} &{54a^2b^3÷(−6ab^5)} \$5pt] {\text{Rewrite as a fraction.}} &{\dfrac{54a^2b^3}{−6ab^5}} \\[5pt] {\text{Use fraction multiplication.}} &{\dfrac{54}{−6}·\dfrac{a^2}{a}·\dfrac{b^3}{b^5}} \\[5pt] {\text{Simplify and use the Quotient Property.}} &{−9·a·\dfrac{1}{b^2}} \\[5pt] {\text{Multiply.}} &{−\dfrac{9a}{b^2}} \end{array}$$ ##### Try It! $$\PageIndex{1}$$ Find the quotient: $$−72a^7b^3÷(8a^{12}b^4)$$. Answer $$−\dfrac{9}{a^5b}$$ ##### Try It! $$\PageIndex{2}$$ Find the quotient: $$−63c^8d^3÷(7c^{12}d^2)$$. Answer $$\dfrac{−9d}{c^4}$$ Once you become familiar with the process and have practiced it step by step several times, you may be able to simplify a fraction in one step. ##### Example $$\PageIndex{2}$$ Find the quotient: $$\dfrac{14x^7y^{12}}{21x^{11}y^6}$$. Solution Be very careful to simplify $$\dfrac{14}{21}$$ by dividing out a common factor, and to simplify the variables by subtracting their exponents. $$\begin{array} {ll} {} &{\dfrac{14x^7y^{12}}{21x^{11}y^6}} \\ {\text{Simplify and use the Quotient Property.}} &{\dfrac{2y^6}{3x^4}} \\ \end{array}$$ ##### Try It! $$\PageIndex{3}$$ Find the quotient: $$\dfrac{28x^5y^{14}}{49x^9y^{12}}$$. Answer $$\dfrac{4y^2}{7x^4}$$ ##### Try It! $$\PageIndex{4}$$ Find the quotient: $$\dfrac{30m^5n^{11}}{48m^{10}n^{14}}$$. Answer $$\dfrac{5}{8m^5n^3}$$ ## Divide a Polynomial by a Monomial Now that we know how to divide a monomial by a monomial, the next procedure is to divide a polynomial of two or more terms by a monomial. The method we’ll use to divide a polynomial by a monomial is based on the properties of fraction addition. So we’ll start with an example to review fraction addition. The sum $$\dfrac{y}{5}+\dfrac{2}{5}$$ simplifies to $$\dfrac{y+2}{5}$$. Now we will do this in reverse to split a single fraction into separate fractions. For example, $$\dfrac{y+2}{5}$$ can be written $$\dfrac{y}{5}+\dfrac{2}{5}$$. This is the “reverse” of fraction addition and it states that if a, b, and c are numbers where $$c\neq 0$$, then $$\dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c}$$. We will use this to divide polynomials by monomials. ##### definition: DIVISION OF A POLYNOMIAL BY A MONOMIAL To divide a polynomial by a monomial, divide each term of the polynomial by the monomial. ##### Example $$\PageIndex{3}$$ Find the quotient: $$(18x^3y−36xy^2)÷(−3xy)$$. Solution $$\begin{array} {ll} {} &{(18x^3y−36xy^2)÷(−3xy)} \\[5pt] {\text{Rewrite as a fraction.}} &{\dfrac{18x^3y−36xy^2}{−3xy}} \\[5pt] {\text{Divide each term by the divisor. Be careful with the signs!}} &{\dfrac{18x^3y}{−3xy}−\dfrac{36xy^2}{−3xy}} \\[5pt] {\text{Simplify.}} &{−6x^2+12y} \end{array}$$ ##### Try It! $$\PageIndex{5}$$ Find the quotient: $$(32a^2b−16ab^2)÷(−8ab)$$. Answer $$−4a+2b$$ ##### Try It! $$\PageIndex{6}$$ Find the quotient: $$(−48a^8b^4−36a^6b^5)÷(−6a^3b^3)$$. Answer $$8a^5b+6a^3b^2$$ ## Divide Polynomials Using Long Division Divide a polynomial by a binomial, we follow a procedure very similar to long division of numbers. So let’s look carefully the steps we take when we divide a 3-digit number, 875, by a 2-digit number, 25. We check division by multiplying the quotient by the divisor. If we did the division correctly, the product should equal the dividend. \[\begin{array} {l} {35·25} \\ {875\checkmark} \\ \nonumber \end{array}$
Now we will divide a trinomial by a binomial. As you read through the example, notice how similar the steps are to the numerical example above.
##### Example $$\PageIndex{4}$$
Find the quotient: $$(x^2+9x+20)÷(x+5)$$.
Solution
$$\require{enclose}$$ $$\qquad (x^2+9x+20) \div (x+5)$$ Write it as a long division problem. Be sure the dividend is in standard form. $$\qquad x+5\enclose{longdiv}{ x^2+9x+20\phantom{0}}$$ Divide $$x^2$$ by $$x$$. It may help to ask yourself, “What do I need to multiply $$x$$ by to get $$x^2$$?” $$\qquad \begin{array}{r} {\color{red}x}\hspace{2.3em}\$-3pt] {\color{red}x}+5\enclose{longdiv}{ {\color{red}x^2}+9x+20\phantom{0}} \end{array}$$ Put the answer, $$x$$, in the quotient over the $$x$$ term. Multiply $$x$$ times $$x+5$$. Line up the like terms under the dividend. $$\qquad \begin{array}{r}x\hspace{2.3em}\\[-3pt] x+5\enclose{longdiv}{x^2+9x+20\phantom{0}}\\[-3pt] \underline{\color{red}x^2+5x}\hspace{2.4em} \end{array}$$ Subtract $$x^2+5x$$ from $$x^2+9x$$. You may find it easier to change the signs and then add. Then bring down the last term, $$20.$$ $$\qquad \begin{array}{r}x\hspace{2.3em}\\[-3pt] x+5\enclose{longdiv}{x^2+\phantom{00}9x+20\phantom{0}}\\[-3pt] \underline{{\color{red}-}x^2+({\color{red}-}5x)}\hspace{2.1em}\\[-3pt] {\color{red}4x+20}\hspace{0.5em} \end{array}$$ Divide $$4x$$ by $$x$$. It may help to ask yourself, “What do I need to multiply $$x$$ by to get $$4x$$?” Put the answer, $$4$$, in the quotient over the constant term. $$\qquad \begin{array}{r}x+\phantom{0}{\color{red}4}\hspace{.5em}\\[-3pt] {\color{red}x}+5\enclose{longdiv}{x^2+\phantom{00}9x+20\phantom{0}}\\[-3pt] \underline{{\color{red}-}x^2+({\color{red}-}5x)}\hspace{2.1em}\\[-3pt] {\color{red}4x}+20\hspace{0.5em} \end{array}$$ Multiply 4 times $$x+5$$. $$\qquad \begin{array}{r}x+\phantom{0}4\hspace{.5em}\\[-3pt] x+5\enclose{longdiv}{x^2+\phantom{00}9x+20\phantom{0}}\\[-3pt] \underline{{\color{red}-}x^2+({\color{red}-}5x)}\hspace{2.1em}\\[-3pt] 4x+20\hspace{0.5em}\\[-3pt] \underline{ \color{red}4x+20}\hspace{.5em} \end{array}$$ Subtract $$4x+20$$ from $$4x+20$$. $$\qquad \begin{array}{r}x+\phantom{0}4\hspace{.5em}\\[-3pt] x+5\enclose{longdiv}{x^2+\phantom{00}9x+20\phantom{0}}\\[-3pt] \underline{{\color{red}-}x^2+({\color{red}-}5x)}\hspace{2.1em}\\[-3pt] 4x+20\hspace{.5em}\\[-3pt] \underline{{\color{red}-}4x+({\color{red}-}20)}\\[-3pt] 0\hspace{.33em}\end{array}$$ Check: $$\begin{array} {ll} {\text{Multiply the quotient by the divisor.}} &{(x+4)(x+5)} \\ {\text{You should get the dividend.}} &{x^2+9x+20\checkmark}\\ \end{array}$$ ##### Try It! $$\PageIndex{7}$$ Find the quotient: $$(y^2+10y+21)÷(y+3)$$. Answer $$y+7$$ ##### Try It! $$\PageIndex{8}$$ Find the quotient: $$(m^2+9m+20)÷(m+4)$$. Answer $$m+5$$ When we divided 875 by 25, we had no remainder. But sometimes division of numbers does leave a remainder. The same is true when we divide polynomials. In the next example, we’ll have a division that leaves a remainder. We write the remainder as a fraction with the divisor as the denominator. Look back at the dividends in previous examples. The terms were written in descending order of degrees, and there were no missing degrees. The dividend in this example will be $$x^4−x^2+5x−6$$. It is missing an $$x^3$$ term. We will add in $$0x^3$$ as a placeholder. ##### Example $$\PageIndex{5}$$ Find the quotient: $$(x^4−x^2+5x−6)÷(x+2)$$. Solution Notice that there is no $$x^3$$ term in the dividend. We will add $$0x^3$$ as a placeholder. Write it as a long division problem. Be sure the dividend is in standard form with placeholders for missing terms. Divide $$x^4$$ by $$x$$. Put the answer, $$x^3$$, in the quotient over the $$x^3$$ term. Multiply $$x^3$$ times $$x+2$$. Line up the like terms. Subtract and then bring down the next term. Divide $$−2x^3$$ by $$x$$. Put the answer, $$−2x^2$$, in the quotient over the $$x^2$$ term. Multiply $$−2x^2$$ times $$x+1$$. Line up the like terms Subtract and bring down the next term. Divide $$3x^2$$ by $$x$$. Put the answer, $$3x$$, in the quotient over the $$x$$ term. Multiply $$3x$$ times $$x+1$$. Line up the like terms. Subtract and bring down the next term. Divide $$−x$$ by $$x$$. Put the answer, $$−1$$, in the quotient over the constant term. Multiply $$−1$$ times $$x+1$$. Line up the like terms. Change the signs, add. Write the remainder as a fraction with the divisor as the denominator. To check, multiply $$(x+2)(x^3−2x^2+3x−1−4x+2)$$. The result should be $$x^4−x^2+5x−6$$. ##### Try It! $$\PageIndex{9}$$ Find the quotient: $$(x^4−7x^2+7x+6)÷(x+3)$$. Answer $$x^3−3x^2+2x+1+3x+3$$ ##### Try It! $$\PageIndex{10}$$ Find the quotient: $$(x^4−11x^2−7x−6)÷(x+3)$$. Answer $$x^3−3x^2−2x−1−3x+3$$ In the next example, we will divide by $$2a−3$$. As we divide, we will have to consider the constants as well as the variables. ##### Example $$\PageIndex{6}$$ Find the quotient: $$(8a^3+27)÷(2a+3)$$. Solution This time we will show the division all in one step. We need to add two placeholders in order to divide. To check, multiply $$(2a+3)(4a^2−6a+9)$$. The result should be $$8a^3+27$$. ##### Try It! $$\PageIndex{11}$$ Find the quotient: $$(x^3−64)÷(x−4)$$. Answer $$x^2+4x+16$$ ##### Try It! $$\PageIndex{12}$$ Find the quotient: $$(125x^3−8)÷(5x−2)$$. Answer $$25x^2+10x+4$$ ## Divide Polynomials using Synthetic Division As we have mentioned before, mathematicians like to find patterns to make their work easier. Since long division can be tedious, let’s look back at the long division we did in Example and look for some patterns. We will use this as a basis for what is called synthetic division. The same problem in the synthetic division format is shown next. Synthetic division basically just removes unnecessary repeated variables and numbers. Here all the $$x$$ and $$x^2$$ are removed. as well as the $$−x^2$$ and $$−4x$$ as they are opposite the term above. • The first row of the synthetic division is the coefficients of the dividend. The $$−5$$ is the opposite of the 5 in the divisor. • The second row of the synthetic division are the numbers shown in red in the division problem. • The third row of the synthetic division are the numbers shown in blue in the division problem. Notice the quotient and remainder are shown in the third row. \[\text{Synthetic division only works when the divisor is of the form }x−c. \nonumber$
The following example will explain the process.
##### Example $$\PageIndex{7}$$
Use synthetic division to find the quotient and remainder when $$2x^3+3x^2+x+8$$ is divided by $$x+2$$.
Solution
Write the dividend with decreasing powers of $$x$$. Write the coefficients of the terms as the first row of the synthetic division. Write the divisor as $$x−c$$ and place c in the synthetic division in the divisor box. Bring down the first coefficient to the third row. Multiply that coefficient by the divisor and place the result in the second row under the second coefficient. Add the second column, putting the result in the third row. Multiply that result by the divisor and place the result in the second row under the third coefficient. Add the third column, putting the result in the third row. Multiply that result by the divisor and place the result in the third row under the third coefficient. Add the final column, putting the result in the third row. The quotient is $$2x^2−1x+3$$ and the remainder is 2.
The division is complete. The numbers in the third row give us the result. The $$2\space\space\space−1\space\space\space3$$ are the coefficients of the quotient. The quotient is $$2x^2−1x+3$$. The 2 in the box in the third row is the remainder.
Check:
\begin{align} (\text{quotient})(\text{divisor}) + \text{remainder} &= \text{dividend} \nonumber\\ (2x^2−1x+3)(x+2)+2 &\overset{?}{=} 2x^3+3x^2+x+8 \nonumber\\ 2x^3−x^2+3x+4x^2−2x+6+2 &\overset{?}{=} 2x^3+3x^2+x+8 \nonumber\\ 2x^3+3x^2+x+8 &= 2x^3+3x^2+x+8\checkmark \nonumber \end{align}
##### Try It! $$\PageIndex{13}$$
Use synthetic division to find the quotient and remainder when $$3x^3+10x^2+6x−2$$ is divided by $$x+2$$.
Answer
$$3x^2+4x−2;\space 2$$
##### Try It! $$\PageIndex{14}$$
Use synthetic division to find the quotient and remainder when $$4x^3+5x^2−5x+3$$ is divided by $$x+2$$.
Answer
$$4x^2−3x+1; 1$$
In the next example, we will do all the steps together.
##### Example $$\PageIndex{8}$$
Use synthetic division to find the quotient and remainder when $$x^4−16x^2+3x+12$$ is divided by $$x+4$$.
Solution
The polynomial $$x^4−16x^2+3x+12$$ has its term in order with descending degree but we notice there is no $$x^3$$ term. We will add a 0 as a placeholder for the $$x^3$$ term. In $$x−c$$ form, the divisor is $$x−(−4)$$.
We divided a $$4^{\text{th}}$$ degree polynomial by a $$1^{\text{st}}$$ degree polynomial so the quotient will be a $$3^{\text{rd}}$$ degree polynomial.
Reading from the third row, the quotient has the coefficients $$1\space\space\space−4\space\space\space0\space\space\space3$$, which is $$x^3−4x^2+3$$. The remainder
is 0.
##### Try It! $$\PageIndex{15}$$
Use synthetic division to find the quotient and remainder when $$x^4−16x^2+5x+20$$ is divided by $$x+4$$.
Answer
$$x^3−4x^2+5;\space 0$$
##### Try It! $$\PageIndex{16}$$
Use synthetic division to find the quotient and remainder when $$x^4−9x^2+2x+6$$ is divided by $$x+3$$.
Answer
$$x^3−3x^2+2;\space 0$$
## Divide Polynomial Functions
Just as polynomials can be divided, polynomial functions can also be divided.
##### definition: DIVISION OF POLYNOMIAL FUNCTIONS
For functions $$f(x)$$ and $$g(x)$$, where $$g(x)\neq 0$$,
$\left(\dfrac{f}{g}\right)(x)=\dfrac{f(x)}{g(x)} \nonumber$
##### Example $$\PageIndex{9}$$
For functions $$f(x)=x^2−5x−14$$ and $$g(x)=x+2$$, find:
1. $$\left(\dfrac{f}{g}\right)(x)$$
2. $$\left(\dfrac{f}{g}\right)(−4)$$.
Solution
$$\begin{array} {ll} {\text{Substitute for }f(x)\text{ and }g(x).} &{\left(\dfrac{f}{g}\right)(x)=\dfrac{x^2−5x−14}{x+2}} \$5pt] {\text{Divide the polynomials.}} &{\left(\dfrac{f}{g}\right)(x)=x−7} \end{array}$$ ⓑ In part ⓐ we found $$\left(\dfrac{f}{g}\right)(x)$$ and now are asked to find $$\left(\dfrac{f}{g}\right)(−4)$$. $$\begin{array} {ll} {} &{\left(\dfrac{f}{g}\right)(x)=x−7} \\[5pt] {\text{To find }\left(\dfrac{f}{g}\right)(−4), \text{ substitute }x=−4.} &{\left(\dfrac{f}{g}\right)(−4)=−4−7} \\[5pt] {} &{\left(\dfrac{f}{g}\right)(−4)=−11} \end{array}$$ ##### Try It! $$\PageIndex{17}$$ For functions $$f(x)=x^2−5x−24$$ and $$g(x)=x+3$$, find: 1. $$\left(\dfrac{f}{g}\right)(x)$$ 2. $$\left(\dfrac{f}{g}\right)(−3)$$. Answer a $$\left(\dfrac{f}{g}\right)(x)=x−8$$ Answer b $$\left(\dfrac{f}{g}\right)(−3)=−11$$ ##### Try It! $$\PageIndex{18}$$ For functions $$f(x)=x2−5x−36$$ and $$g(x)=x+4$$, find: 1. $$\left(\dfrac{f}{g}\right)(x)$$ 2. $$\left(\dfrac{f}{g}\right)(−5)$$. Answer a $$\left(\dfrac{f}{g}\right)(x)=x−9$$ Answer b $$\left(\dfrac{f}{g}\right)(x)=x−9$$ ## Use the Remainder and Factor Theorem Let’s look at the division problems we have just worked that ended up with a remainder. They are summarized in the chart below. If we take the dividend from each division problem and use it to define a function, we get the functions shown in the chart. When the divisor is written as $$x−c$$, the value of the function at $$c$$, $$f(c)$$, is the same as the remainder from the division problem. Dividend Divisor $$x−c$$ Remainder Function $$f(c)$$ $$x^4−x^2+5x−6$$ $$x−(−2)$$ $$−4$$ $$f(x)=x^4−x^2+5x−6$$ $$−4$$ $$3x^3−2x^2−10x+8$$ $$x−2$$ 4 $$f(x)=3x^3−2x^2−10x+8$$ 4 $$x^4−16x^2+3x+15$$ $$x−(−4)$$ 3 $$f(x)=x^4−16x^2+3x+15$$ 3 To see this more generally, we realize we can check a division problem by multiplying the quotient times the divisor and add the remainder. In function notation we could say, to get the dividend $$f(x)$$, we multiply the quotient, $$q(x)$$ times the divisor, $$x−c$$, and add the remainder, $$r$$. If we evaluate this at $$c$$, we get: This leads us to the Remainder Theorem. ##### Definition: REMAINDER THEOREM If the polynomial function $$f(x)$$ is divided by $$x−c$$, then the remainder is $$f(c)$$. ##### Example $$\PageIndex{10}$$ Use the Remainder Theorem to find the remainder when $$f(x)=x^3+3x+19$$ is divided by $$x+2$$. Solution To use the Remainder Theorem, we must use the divisor in the $$x−c$$ form. We can write the divisor $$x+2$$ as $$x−(−2)$$. So, our $$c$$ is $$−2$$. To find the remainder, we evaluate $$f(c)$$ which is $$f(−2)$$. To evaluate $$f(−2)$$, substitute $$x=−2$$. Simplify. The remainder is 5 when $$f(x)=x^3+3x+19$$ is divided by $$x+2$$. Check: Use synthetic division to check. The remainder is 5. ##### Try It! $$\PageIndex{19}$$ Use the Remainder Theorem to find the remainder when $$f(x)=x^3+4x+15$$ is divided by $$x+2$$. Answer $$−1$$ ##### Try It! $$\PageIndex{20}$$ Use the Remainder Theorem to find the remainder when $$f(x)=x^3−7x+12$$ is divided by $$x+3$$. Answer $$6$$ When we divided $$8a^3+27$$ by $$2a+3$$ in Example the result was $$4a^2−6a+9$$. To check our work, we multiply $$4a2−6a+9$$ by $$2a+3$$ to get $$8a^3+27$$. \[(4a^2−6a+9)(2a+3)=8a^3+27 \nonumber$
Written this way, we can see that $$4a^2−6a+9$$ and $$2a+3$$ are factors of $$8a^3+27$$. When we did the division, the remainder was zero.
Whenever a divisor, $$x−c$$, divides a polynomial function, $$f(x)$$, and resulting in a remainder of zero, we say $$x−c$$ is a factor of $$f(x)$$.
The reverse is also true. If $$x−c$$ is a factor of $$f(x)$$ then $$x−c$$ will divide the polynomial function resulting in a remainder of zero.
We will state this in the Factor Theorem.
##### Definition: FACTOR THEOREM
For any polynomial function $$f(x)$$,
• if $$x−c$$ is a factor of $$f(x)$$, then $$f(c)=0$$
• if $$f(c)=0$$, then $$x−c$$ is a factor of $$f(x)$$
##### Example $$\PageIndex{11}$$
Use the Remainder Theorem to determine if $$x−4$$ is a factor of $$f(x)=x^3−64$$.
Solution
The Factor Theorem tells us that $$x−4$$ is a factor of $$f(x)=x^3−64$$ if $$f(4)=0$$.
$$\begin{array} {ll} {} &{f(x)=x^3−64} \\[5pt] {\text{To evaluate }f(4) \text{ substitute } x=4.} &{f(4)=4^3−64} \\[5pt] {\text{Simplify.}} &{f(4)=64−64} \\[5pt]{\text{Subtract.}} &{f(4)=0} \end{array}$$
Since $$f(4)=0, x−4$$ is a factor of $$f(x)=x^3−64$$.
##### Try It! $$\PageIndex{21}$$
Use the Factor Theorem to determine if $$x−5$$ is a factor of $$f(x)=x^3−125$$.
Answer
yes
##### Try It! $$\PageIndex{22}$$
Use the Factor Theorem to determine if $$x−6$$ is a factor of $$f(x)=x^3−216$$.
Answer
yes
Access these online resources for additional instruction and practice with dividing polynomials.
• Dividing a Polynomial by a Binomial
• Synthetic Division & Remainder Theorem
## Key Concepts
• Division of a Polynomial by a Monomial
• To divide a polynomial by a monomial, divide each term of the polynomial by the monomial.
• Division of Polynomial Functions
• For functions $$f(x)$$ and $$g(x)$$, where $$g(x)\neq 0$$,
$$\left(\dfrac{f}{g}\right)(x)=\dfrac{f(x)}{g(x)}$$
• Remainder Theorem
• If the polynomial function $$f(x)$$ is divided by $$x−c$$, then the remainder is $$f(c)$$.
• Factor Theorem: For any polynomial function $$f(x)$$,
• if $$x−c$$ is a factor of $$f(x)$$, then $$f(c)=0$$
• if $$f(c)=0$$, then $$x−c$$ is a factor of $$f(x)$$
This page titled 5.5: Dividing Polynomials is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
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# Graph Period 2
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### Graph Period 2
1. 1. Graphing Linear Equations in One Variable<br />
2. 2. What do each of these ordered pairs have in common?<br />(4, -3)<br />(4, -2)<br />(4, -1)<br />(4, 0)<br />(4, 1)<br />(4, 2)<br />(4, 3)<br />(4, 4)<br />(4, 5)<br />The x-coordinate is always the same. The y-coordinate varies.<br />Graph each point on the same coordinate plane.<br />
3. 3. The equation that represents this relationship between x and y is x = 4.<br />The graph of x = 4 is a VERTICAL LINE that crosses the x-axis at 4. All the points of this line (and no other points) have an x-coordinate of 4.<br />y<br />x<br />Notice that the equation doesn’t have a y variable. The equation will never intersect the y-axis.<br />
4. 4. What do each of these ordered pairs have in common?<br />(-4, -1) <br />(-3, -1) <br />(-2, -1) <br />(-1, -1) <br />(0, -1) <br />(1, -1) <br />(2, -1) <br />(3, -1) <br />(4, -1) <br />The y-coordinate is always the same; the x-coordinate varies.<br />
5. 5. Notice that the equation doesn’t have an x variable. The equation will never intersect the x-axis.<br />y<br />The equation that represents this relationship between x and y is y = -1.<br />The graph of y = -1 is a HORIZONTAL LINE that crosses the y-axis at -1. All the points of this line (and no other points) have y- coordinate of -1.<br />x<br />
6. 6. Sketching Horizontal and Vertical Lines<br />Sketch the graph of x = 2.<br />y<br />The equation does not have y as a variable. The equation says that the x-coordinate will always be 2, regardless of the value of y.<br />x<br />To graph the equation, put a dot on the x-axis at 2. There is no y-coordinate in the equation, so the equation of this line will never cross the y-axis.<br />
7. 7. Sketching Horizontal and Vertical Lines<br />Sketch the graph of y = -3.<br />y<br />The equation does not have x as a variable. The equation says that the y-coordinate will always be -3, regardless of the value of x.<br />x<br />To graph the equation, put a dot on the y-axis at -3. There is no x-coordinate in the equation, so the equation of this line will never cross the x-axis.<br />
8. 8. Sketch the graphs of x = 7 and y = -4. Find the point of intersection of the two graphs.<br />y<br />Graph x = 7. Plot a point on the x-axis at 7. The equation has no y-value, so the line will never cross the y-axis.<br />x<br />Graph y = -4. Plot a point on the y-axis at -4. The equation has no x-value, so the line will never cross the x-axis.<br />(7, -4)<br />
9. 9. Sketch the graphs of x = -4 and y = -1. Find the point of intersection of the two graphs.<br />y<br />Graph x = -4. Plot a point on the x-axis at -4. The equation has no y-value, so the line will never cross the y-axis.<br />x<br />Graph y = -1. Plot a point on the y-axis at -1. The equation has no x-value, so the line will never cross the x-axis.<br />(-4, -1)<br />
10. 10. Write equations for the horizontal line and the vertical line that pass through the point (-7, -3)<br />Plot the point (-7, -3).<br />Graph a vertical line through the point. The line intersects the x-axis at -7. It never intersects the y-axis. What is its equation?<br />y<br />x = -7<br />x<br />Graph a horizontal line through the point. The line intersects the y-axis at -3. It never intersects the x-axis. What is its equation?<br />y = -3<br />
11. 11. Home work<br />1. Write equations for the horizontal line and the vertical line that pass through the point (-9, 4)<br />2. Write equations for the horizontal line and the vertical line that pass through the point (6, 7)<br />3. Write equations for the horizontal line and the vertical line that pass through the point (-5, 8)<br />4. Write equations for the horizontal line and the vertical line that pass through the point (10, -2)<br />5. Write equations for the horizontal line and the vertical line that pass through the point (100, -50)<br />
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Re: RE: Solving for Sum element
• To: mathgroup at smc.vnet.net
• Subject: [mg27292] Re: [mg27270] RE: [mg27260] Solving for Sum element
• From: "Mark Harder" <harderm at ucs.orst.edu>
• Date: Sat, 17 Feb 2001 03:31:15 -0500 (EST)
• Sender: owner-wri-mathgroup at wolfram.com
```David Park seems to have solved this problem, but to extend my understanding
of how Mathematica does things, I worked out a different solution:
solveElement[eqn_, k_] := (
sum1 = ReplacePart[eqn, k - 1, {1, 2, 3} ][[1]];
sum2 = ReplacePart[eqn, k + 1, {1, 2, 2} ][[1]];
Rule[Part[ReplacePart[eqn, k, {1, 1, 1} ], 1, 2],
Part[eqn, 2] - (sum1 + sum2) ] );
This function *assumes* that eqn is of the form Sum[symbol[i], iterator]==
r.h.s, and k is the index of the term to be solved for (it can be symbolic
or numeric ).
It breaks the equation expression down by parts, replaces the parts as
needed, and returns the rule. All it does is manipulate the eqn expression
passed to it and returns the manipulated expression -- no other assumptions
about eqn are necessary, as far as i know so far. Doing it this way doesn't
require any Hold[], but it is a little rigid about the form of eqn.
Some results: (outputs are printed in input form)
Symbolic sum, symbolic index:
In[409]:=
solveElement[eqn, k]
Out[409]=
a[k] -> x - Sum[a[i], {i, 1, k - 1}] - Sum[a[i], {i, k + 1, N}]
Symbolic sum, numeric index:
In[410]:=
solveElement[eqn, 5]
Out[410]=
a[5] -> x - a[1] - a[2] - a[3] - a[4] - Sum[a[i], {i, 6, N}]
One awkwardness is that you have to be clever about applying further rules
in order to solve a sum with defined terms:
In[427]:=
eqn2 = Sum[a[i] , {i, 1, N} ] == 25.
Out[427]=
Sum[a[i], {i, 1, N}] == 25.
In[447]:=
solveElement[eqn2, 4] /. {N -> 6, a[4] -> x, a[i_] -> E^-i}
Out[447]=
x -> 24.4378
I'm sure there are simpler ways to do take this approach as well -- like
just passing solveElement the r.h.s. of the equation, and setting
Sum[a[i],{i,initial,final} ] to it inside the function.
-mark harder
-----Original Message-----
From: David Park <djmp at earthlink.net>
To: mathgroup at smc.vnet.net
Subject: [mg27292] [mg27270] RE: [mg27260] Solving for Sum element
>Ben,
>
>Here is one approach:
>
>Attributes[SumTermSolve] = {HoldFirst};
>SumTermSolve[sumequation_, index_] :=
> Module[{seqn = Hold[sumequation], i, t, start, end},
> seqn = ReleaseHold[seqn /. Sum -> sum];
> seqn = seqn /. sum[t_, {i_, start_, end_}] ->
> sum[t, {i, start, index - 1}] + term[index] +
> sum[t, {i, index + 1, end}];
> Solve[seqn, term[index]] /. sum -> Sum]
>
>f[i_] := i
>
>SumTermSolve[Sum[f[i], {i, 1, n}] == 0, 3]
>{{term[3] -> -3 - 1/2*(-3 + n)*(4 + n)}}
>
>SumTermSolve[Sum[a[i], {i, 0, 6}] == x, 2]
>{{term[2] -> x - a[0] - a[1] - a[3] - a[4] - a[5] -
> a[6]}}
>
>
>David Park
>
>
>
>> From: Ben Jacobson [mailto:bjacobson at illumitech.com]
To: mathgroup at smc.vnet.net
>To: mathgroup at smc.vnet.net
>>
>> I want to solve for one element in an equation involving long sums. The
>> actual equation is of course complicated, but a very simplified example
>> would be
>>
>> Sum[a[i],{i,0,n}]==x
>>
>> where I want to solve for a[0]. The solution should be something like
>> a[0]->x-Sum[a[i],{i,1,n}.
>>
>> Solve[Sum[a[i], {i, 0, n}] == x, a[0]]
>>
>> works fine if, for example, n=3, but it returns an empty result for
>> symbolic n. Is there a way to tell Solve to assume that n>1 and to solve
>> accordingly? The best I've been able to do so far is to explicitly
remove
>> the element I'm interested in from the sum:
>>
>> In: Solve[a[0] + Sum[a[i], {i, 1, n}] == x, a[0]]
>> Out: {{a[0] -> x - Sum[a[i], {i, 1, n}]}}
>>
>> This works nicely, but becomes awkward when I try to solve for more
>> elements and longer sums. Thanks.
>>
>> Ben Jacobson
>> Illumitech Inc.
>>
>
>
```
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## DEV Community is a community of 782,260 amazing developers
We're a place where coders share, stay up-to-date and grow their careers.
Suchitra
Posted on • Updated on
# Combinations Problem
Before start with the problems of combinations, It will be better if we discuss a bit about it that you should know first!
So, let's discuss…
## What is Combination?
In a simple and easy way a combination is a subset of a given set in which order doesn't matter.
Let's understand it better with example
Example - [1, 2, 3]
it's all combinations are - [[], [1], [2], [3], [1, 2], [1, 3], [2, 3] [1, 2, 3]].
Here you might notice that it didn't write in this way - [1, 2] and [2, 1] or [1, 2, 3] and [2, 3, 1] and [3, 1, 2].
Because by the definition we know that order doesn't matter.
Means here [1, 2] = [2, 1]. (anyone we can pick as a combination)
OR [1, 2, 3] = [2, 1, 3] = [3, 1, 2].
If two or more subsets have the same elements and their frequencies are also the same, but the order may or may not be different they considered as a single one(same)!
## Representation using tree
`````` []
/ \
/ \
[] [1]
/ \ / \
[] [2] [1] [1, 2]
/ \ / \ / \ / \
[] [3] [2] [2, 3] [1] [1, 3] [1, 2] [1, 2, 3]
``````
The above leaf nodes give all combinations of the given array [1, 2, 3]. Here the intuition is pick and don't pick the element in this way we can determine all possible combinations of a given set or array.
Now, I will discuss some basic problem on combinations!
## Problem 1
### Combinations
Given two integers n and k, return all possible combinations of k numbers out of the range [1, n].
You may return the answer in any order.
Example 1:
Input: n = 4, k = 2
Output:
[
[2,4],
[3,4],
[2,3],
[1,2],
[1,3],
[1,4],
]
Example 2:
Input: n = 1, k = 1
Output: [[1]]
Constraints:
```1 < = n < = 20 1 < = k < = n```
``````class Solution {
public List<List<Integer>> combine(int n, int k) {
List<List<Integer>> result = new ArrayList<List<Integer>>();
combination(result, new ArrayList<Integer>(), 1, n, k);
return result;
}
public static void combination(List<List<Integer>> result, List<Integer> list, int start, int n, int k){
if(k == 0){
return;
}
for(int i = start; i <= n - k + 1; i++){
combination(result, list, i + 1, n, k - 1);
list.remove(list.size() - 1);
}
}
}
``````
## Problem 2
### Combination Sum
Given an array of distinct integers candidates and a target integer target, return a list of all unique combinations of candidates where the chosen numbers sum to target. You may return the combinations in any order.
The same number may be chosen from candidates an unlimited number of times. Two combinations are unique if the frequency of at least one of the chosen numbers is different.
It is guaranteed that the number of unique combinations that sum up to target is less than 150 combinations for the given input.
Example 1:
Input: candidates = [2,3,6,7], target = 7
Output: [[2,2,3], [7]]
Explanation:
2 and 3 are candidates, and 2 + 2 + 3 = 7. Note that 2 can be used multiple times.
7 is a candidate, and 7 = 7.
These are the only two combinations.
Example 2:
Input: candidates = [2,3,5], target = 8
Output: [[2,2,2,2], [2,3,3], [3,5]]
Example 3:
Input: candidates = [2], target = 1
Output: []
Example 4:
Input: candidates = [1], target = 1
Output: [[1]]
Example 5:
Input: candidates = [1], target = 2
Output: [[1,1]]
``````class Solution {
public List<List<Integer>> combinationSum(int[] candidates, int target) {
// Arrays.sort(candidates);
List<List<Integer>> res = new ArrayList();
find_comboSum(candidates, res, new ArrayList<>(), target, 0);
return res;
}
public static void find_comboSum(int[] arr, List<List<Integer>> res, List<Integer> list, int target, int start){
if(target < 0)return;
if(target == 0){
return;
}
for(int i = start; i < arr.length; i++){
find_comboSum(arr, res, list, target - arr[i], i);
list.remove(list.size() - 1);
}
}
}
``````
## Problem 3
### Combination Sum II
Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sum to target.
Each number in candidates may only be used once in the combination.
Note: The solution set must not contain duplicate combinations.
Example 1:
Input: candidates = [10,1,2,7,6,1,5], target = 8
Output:
[
[1,1,6],
[1,2,5],
[1,7],
[2,6]
]
Example 2:
Input: candidates = [2,5,2,1,2], target = 5
Output:
[
[1,2,2],
[5]
]
``````class Solution {
public List<List<Integer>> combinationSum2(int[] candidates, int target) {
Arrays.sort(candidates);
List<List<Integer>> res = new ArrayList<>();
find_combo(candidates, res, new ArrayList<>(), target, 0);
return res;
}
public static void find_combo(int[] arr, List<List<Integer>> res, List<Integer> list, int target, int start){
if(target < 0)return;
if(target == 0){
return;
}
for(int i = start; i < arr.length; i++){
if(i > start && arr[i] == arr[i - 1])continue;
find_combo(arr, res, list, target - arr[i], i + 1);
list.remove(list.size() - 1);
}
}
}
``````
## Problem 4
### Combination Sum III
Find all valid combinations of k numbers that sum up to n such that the following conditions are true:
Only numbers 1 through 9 are used.
Each number is used at most once.
Return a list of all possible valid combinations. The list must not contain the same combination twice, and the combinations may be returned in any order.
Example 1:
Input: k = 3, n = 7
Output: [[1,2,4]]
Explanation:
1 + 2 + 4 = 7
There are no other valid combinations.
Example 2:
Input: k = 3, n = 9
Output: [[1,2,6], [1,3,5], [2,3,4]]
Explanation:
1 + 2 + 6 = 9
1 + 3 + 5 = 9
2 + 3 + 4 = 9
There are no other valid combinations.
Example 3:
Input: k = 4, n = 1
Output: []
Explanation: There are no valid combinations. [1,2,1] is not valid because 1 is used twice.
Example 4:
Input: k = 3, n = 2
Output: []
Explanation: There are no valid combinations.
Example 5:
Input: k = 9, n = 45
Output: [[1,2,3,4,5,6,7,8,9]]
Explanation:
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45
There are no other valid combinations.
Constraints:
2 < = k < = 9
1 < = n < = 60
``````class Solution {
public List<List<Integer>> combinationSum3(int k, int n) {
List<List<Integer>> res = new ArrayList<>();
if(n < k)return res;
find_combo(res, new ArrayList<>(), k, 1, n);
return res;
}
public static void find_combo(List<List<Integer>> res, List<Integer> list, int k, int start, int n){
if(k == list.size() && n == 0){
return;
}
for(int i = start; i <= 9; i++){
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# Integral that arises from the derivation of Kummer's Fourier expansion of $\ln{\Gamma(x)}$
I am trying to prove that for $0<x<1$, $$\color{blue}{\ln{\Gamma(x)}=\frac{1}{2}\ln(2\pi)+\sum^\infty_{n=1}\left\{\frac{1}{2n}\cos(2\pi nx)+\frac{\gamma+\ln(2\pi n)}{n\pi}\sin(2\pi nx)\right\}}$$ It is quite straightforward to compute $a_0$ and $a_n$. $a_0$ is \begin{align} a_0 =2\int^1_0\ln{\Gamma(x)}\ {\rm d}x =\int^1_0\ln\left(\frac{\pi}{\sin(\pi x)}\right)\ {\rm d}x =\ln(2\pi)\\ \end{align} As for $a_n$, \begin{align} a_n =&2\int^1_0\ln{\Gamma(x)}\cos(2\pi nx)\ {\rm d}x =\int^1_0\ln\left(\frac{\color{grey}{\pi}}{\sin(\pi x)}\right)\cos(2\pi nx)\ {\rm d}x\\ =&-\frac{1}{\pi}\int^\pi_0\ln(\sin{x})\cos(2nx)\ {\rm d}x =\frac{1}{4n\pi}\int^\pi_{-\pi}\frac{\sin(2nx)\cos(x)}{\sin{x}}\ {\rm d}x \end{align} If we name the remaining integral $\mathcal{I}_n$, it is easy to see that $\mathcal{I}_{n+1}-\mathcal{I}_n=0$. Hence $\mathcal{I}_n=\mathcal{I}_1=2\pi$, and $$a_n=\frac{2\pi}{4n\pi}=\frac{1}{2n}$$ However, I have trouble calculating $b_n$ and proving that $$\color{red}{2\int^1_0\ln{\Gamma(x)}\sin(2n\pi x)\ {\rm d}x=\frac{\gamma+\ln(2n\pi)}{\pi n}}$$ The only idea that I can think of is to use the series representation of $\ln{\Gamma(x)}$ \begin{align} \ln{\Gamma(x)} =&-\gamma x-\ln{x}+\sum^\infty_{k=1}\left\{\frac{x}{k}-\ln\left(1+\frac{x}{k}\right)\right\}\\ =&-\gamma x-\ln{x}+\sum^\infty_{m=2}\frac{(-1)^m\zeta(m)}{m}x^m \end{align} and multiply throughout by $\sin(2n\pi x)$ then integrating term by term. However, arriving to the final result using this method definitely seems arduous. Therefore, I seek your assistance in evaluating the integral in red. Help will be greatly appreciated. Thank you.
• It looks like if you do integration by parts you can turn this into something that looks like the digamma function, and you can also take the imaginary part of the exponential, maybe that'll help. – Kainui Nov 8 '14 at 9:18
Begin with the infinite product representation of the gamma function. \begin{align} \Gamma(x)=\frac{e^{-\gamma x}}{x}\prod^\infty_{k=1}e^\frac{x}{k}\left(1+\frac{x}{k}\right) \end{align} Take the logarithm and multiply throughout by $\sin(2n\pi x)$. \begin{align} \ln{\Gamma(x)}\sin(2n\pi x)=-(\gamma x+\ln{x})\sin(2n\pi x)+\sum^\infty_{k=1}\left\{\frac{x}{k}-\ln\left(1+\frac{x}{k}\right)\right\}\sin(2n\pi x) \end{align} Integrating the non-sum terms from $0$ to $1$, \begin{align} \int^1_0(-\gamma x-\ln{x})\sin(2n\pi x)\ {\rm d}x =&\frac{\gamma}{2n\pi}+\frac{\ln{x}\cos(2n\pi x)}{2n\pi}\Bigg{|}^1_0-\int^1_0\frac{\cos(2n\pi x)}{2n\pi x}{\rm d}x\\ =&\frac{\gamma}{2n\pi}+\left[\frac{\ln{x}\cos(2n\pi x)-{\rm Ci}(2n\pi x)}{2n\pi}\right]^1_0\\ =&\frac{\gamma-{\rm Ci}(2n\pi)}{2n\pi}-\lim_{\epsilon\to 0}\frac{\ln{\epsilon}-{\rm Ci}(2n\pi \epsilon)}{2n\pi}\\ =&\frac{\gamma-{\rm Ci}(2n\pi)}{2n\pi}-\lim_{\epsilon\to 0}\frac{\ln{\epsilon}-\ln(2n\pi)-\ln{\epsilon}-\gamma+\mathcal{O}(\epsilon)}{2n\pi}\\ =&\frac{2\gamma+\ln(2n\pi)-{\rm Ci}(2n\pi)}{2n\pi} \end{align} Integrate the remaining terms from $0$ to $1$. \begin{align} &\int^1_0\sum^\infty_{k=1}\left\{\frac{x}{k}-\ln\left(1+\frac{x}{k}\right)\right\}\sin(2n\pi x)\ {\rm d}x\\ =&-\frac{1}{2n\pi}\sum^\infty_{k=1}\left\{\frac{1}{k}+{\rm Ci}(2n\pi k+2n\pi)-{\rm Ci}(2n\pi k)-\ln\left(1+\frac{1}{k}\right)\right\}\\ =&-\frac{1}{2n\pi}\left[\sum^\infty_{k=1}\left\{\frac{1}{k}-\ln\left(1+\frac{1}{k}\right)\right\}+\lim_{N\to\infty}\left(\sum^{N+1}_{k=2}{\rm Ci}(2n\pi k)-\sum^{N}_{k=1}{\rm Ci}(2n\pi k)\right)\right]\\ =&-\frac{1}{2n\pi}\left[\gamma+\lim_{N\to\infty}\left({\rm Ci}(2n\pi+2n\pi N)-{\rm Ci}(2n\pi)\right)\right] =\frac{{\rm Ci}(2n\pi)-\gamma}{2n\pi} \end{align} Adding them together, \begin{align} \color{red}{\int^1_0\ln{\Gamma(x)}\sin(2n\pi x)\ {\rm d}x}=\frac{2\gamma+\ln(2n\pi)-{\rm Ci}(2n\pi)+{\rm Ci}(2n\pi)-\gamma}{2n\pi}\color{red}{=\frac{\gamma+\ln(2n\pi)}{2n\pi}} \end{align} I am looking forward to seeing cleaner and more interesting approaches to this integral.
As an aside, I derived that $$\int^1_0\psi_0(x+a)\sin(2n\pi x)={\rm Si}(2an\pi)-\frac{\pi}{2}$$ This isn't really related to the problem though.
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# Physics
At t = 0 s a flywheel is rotating at 25 rpm. A motor gives it a constant acceleration of 0.8 rad/s2 until it reaches 100 rpm. The motor is then disconnected. How many revolutions are completed at t = 23 s?
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2. 👎 0
3. 👁 137
1. The average speed during acceleration is 37.5 rpm.
The time spent accelerating is
The number of revolutions turned while accelerating is
37.5*1.63 = 6.12
Add to that the number of revolutions turned after the motor is disconnected. They should tell you if it remains at constant speed after that. Friction will tend to slow it down, and they provide no information on that
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2. 👎 0
posted by drwls
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# 1007slides_Part4 - Math 1007 Fall 2006(These slides replace...
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Math 1007, Fall 2006 (These slides replace neither the text book nor the lectures.) Part IV: Applications of Derivatives 27. Maximum and Minimum Values (p. 76-82) 28. Critical Numbers (p. 83-84) 29. Finding Absolute Extrema of a continuous function on a closed interval (85-87) 30. Intervals of Increase and Decrease (88-89) 31. The First Derivative Test (90-91) 32. Intervals of Concavity (92) 33. Points of In°ection (93) 34. The Second Derivative Test (94) 35. L'H^opital's Rule (95-96) 36. Summary of Curve Sketching (97-108) 75
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27. Maximum and Minimum Values (4.1) De¯nition: A function f has an absolute maximum (or global maximum ) at c if f ( c ) f ( x ) for all x in the domain ( D ) of f . The number f ( c ) is called the maximum value value of f on D . De¯nition: A function f has an absolute mimimum (or global minimum ) at c if f ( c ) f ( x ) for all x in the domain ( D ) of f . The number f ( c ) is called the minimum value value of f on D . The maximum and minimum values of f are called the extreme values (or extrema ) of f . 76
De¯nition: A function f has a relative maximum or ( local maximum ) at c if f ( c ) f ( x ) when x is near c . [This means that f ( c ) f ( x ) for all x in some open intervale containing c .] De¯nition: A function f has an relative minimum (or local mimimum ) at c if f ( c ) f ( x ) when x is near c . The set of all local maximum and minimum values of f is called the relative extrema of f . 77
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Example 1: The function f ( x ) = x 2 has global (and local) mini- mum at x = 0 and this minimum value is 0 (since f (0) = 0). There is no local or global maximum. Example 2: The function f ( x ) = sin x has values that range from +1 to 1. f occurs when x = π 2 . In fact, it occurs in¯nitely often when x = π 2 + 2 πn . f occurs when x = 3 π 2 . In fact, it occurs in¯nitely often when x = 3 π 2 + 2 πn . (Note that x = π 2 can be written as 3 π 2 2 π ) 78
Example 3: Consider the graph of the function f ( x ) = x 4 2 x 2 + 3 2 x 3 Local Minimum: f ( 1) = 0 Local Maximum: f (1) = 4 Global Minimum: f (2) = 0 and f ( 1) = 0 Global Maximum: f ( 3) = 20 79
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Example 4: The function f ( x ) = x 3 has neither a maximum nor a minimum. Example 5: For each of the numbers a, b, c, d, e, r, s and t , state whether the function whose graph is shown below has an absolute maximum or minimum, a local maximum or minimum, or neither a maximum or minimum. x = a x = b x = c x = d x = e x = r x = s x = t 80
For each of the numbers a, b, c, d, e, r, s and t , state whether the function whose graph is shown below has an absolute maximum or minimum, a local maximum or minimum, or neither a maximum or minimum. x
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