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# Why do real-world high dimensional data often have much lower inherent dimensionality?

In a previous year exam about data analysis, there was a question that I'm confused about.

Here's the question:

Explain why it is natural to expect that in real-world high dimensional data the inherent dimensionality of data points will be smaller than the number of coordinates. How is the smaller number of relevant new co-ordinate axis found in PCA?

I'm confused about the real-world high dimensional data being smaller than the number of coordinates. I'd really appreciate if someone can point me in the right direction.

It's not that the dimensionality of the data (more precisely, the statistical process at work) is smaller than the coordinate space, it's that there is often not enough available data to get statistically significant results in every direction. This is a manifestation of the famed curse of dimensionality. PCA is an attempt to determine the directions in which you have the best shot at getting reliable and stable results - it does this by finding the directions in which the data is most spread out.

Of course there are random processes that do only occur in a low dimensional subspace of the feature space, but in general it's best not to assume this is so without evidence. Nonetheless, in any situation, you are bound to do the best you can with the data you have, and the data you have can only support a finite number of inferences.

The answer is relatively simple. Typically for a real world problem you don't know which features to use for the problem. Very often you end up throwing too many features and let the algorithm figure out which ones are discriminative.

Let's take MNIST digit classification as an example. You are given 28x28 black&white images centered on the digit. You have a choice: build the classifier based on the individual pixel values (easy but meaning more dimensions) or come up with more intelligent features (harder but less dimensions). If you go with individual pixel values, you know that you don't need those 784 dimensions to encode differences between 10 digits. This information is buried somewhere inside.

• I don't see how this answers the question. Perhaps you don't need those 784 dimensions to discriminate between 10 digits, but how does it imply that all MNIST images are confined to a submanifold with $d\ll 784$ dimensions? This is what OP is asking about. – amoeba May 5 '15 at 22:09
• @amoeba if we knew a priori which features are the good ones for the task, we should not be playing with dimensionality reduction tasks. At least in many real world applications. Perhaps I still don't understand the question. – Vladislavs Dovgalecs May 5 '15 at 22:12
• As I understand the question, it has nothing at all to do with classification. Take the MNIST images, it's a good example. Each image is a point in 784-dimensional space. Let's take one billion of images, they make up a cloud of points. Here is the question: is it true that all these points are actually confined to a subspace of much lower dimensionality; if so, why? It can be natural images instead of digits, so number of classes (or even existence of classes) does not matter here. – amoeba May 5 '15 at 22:16
• I don't agree with that. Data points form a set $S \subset \mathbb R^n$ in some large $n$-dimensional Euclidean space. Intrinsic dimensionality is a certain geometrical property of this set (even though I will agree that this property is only vaguely defined); "prior knowledge" is irrelevant here. We don't need to know that the data are actually from MNIST in order to estimate the intrinsic dimensionality. – amoeba May 6 '15 at 20:43
• @amoeba More reading is always good. From literature I see that this is an intrinsic property, not necessarily linked to some prior knowledge. It looks like in pattern recognition domain (for image classification) the term definition is different or misused. I acknowledge my mistake. – Vladislavs Dovgalecs May 6 '15 at 21:29

What "the inherent dimensionality of data points will be smaller than the number of coordinates" means is this: If you have 2 dimensional data, you need a 2 dimensional coordinate system (x,y for example) in order to be able to show them. if you data has n dimensions, you need an n dimensional coordinate system, where after n=3 it is impossible for us to visualize that geometrically (at least for me). So if you have high dimensional data, it is diffucult to show it. However, there is a trick to that. What you can do is to say: OK, I might need a lot of dimensions to show the data in the usual corrdinate system, but is there a way to transform the data into some other form, so I will need less axis to show it, or similarly an alternative coordinate system, where I can show my data with less axis? The answer is yes, and PCA is how you do it. You say, OK, let me find the direction where the change in my data is the highest, i.e. the varience is highest, and use this direction as the first axis of my new coordinate system. Theen you say, OK, now I have a direction that explains some (most) of the variance in my data, but what in which direction is the variance of the data second highest, you find it and it is the second direction of your coordinate system. Then you repeat it several times and come up with several axes that explain the most of the variance of your data. The rest of the varience in your data is very small and not so relevant, as you already have most of the change in your data. So now, in this new system, you have nearly all the variance of your original data, however you have less axes, i.e. your data is nearly as well represented in this new corrdinate system as in the previous, however it has significantly less axes in the new one. This is what is meant by your data having inherently less dimensions (in the new coord. system) than the number of coordinates (in the old coord. system).

This representation in the lower dimensional new coordinate system is allowed by the fact that your data actually has some dependencies in the previous corrd. systems bases, i.e. your data can be more efficiently represented in another coord. system with other basis functions. PCA is a special way of finding these basis functions that define the new coordinate system.

In real world everything's controlled by God's. That's why everything's dependent on God. So, everything's really in one dimension, the dimension of God's will. This is is the only true answer, but I doubt that your Prof will accept it. So, here's an easier one.

In real world we're probably not gathering some random data. We usually collect data trying to solve some problem. When we do this it's most likely that we'll be looking for the data that is related to the phenomenon of interest. By virtue of this relation it's likely that all the different data points are measuring the same thing from different angles, figuratively. And that thing is probably what interests us in the first place. So, if we somehow can extract the essence of the phenomenon, we'll probably end up with a fewer number of dimensions that all these variables.

The information in the high-dimensional space can often be captured by a smaller number of latent variables/dimensions because there tends to be dependence (e.g. multicollinearity) between variables in the high-dimensional space.

There tends to be dependence because many of the variables are probably going to form networks of causal connections. I am thinking of biological contexts such as genomic or neuroimaging data, for example. Maybe the situation is different in other domains such as astronomy, however.

The statement in the exam question seems contentious. Take a simple example, like plotting life expectancy of a baby versus various factors including average daily salt consumption. The baby will (on average) die early if the salt consumption is near zero, will thrive it the salt consumption is intermediate, and will die immediately if it's very large. PCA is likely to obscure this crucial information.

Usually, if a phenomenon cannot be explained using only a few factors, then it will be inaccessible to the human mind, and too complex to publish. Then we wait until someone in the future thinks of a way to explain the phenomenon simply, perhaps using completely different variables.

With real data, it's certainly true that one usually finds that PCA gives rise to a coordinate change, indicating certain directions as being more significant. If these indications cannot be backed up with specifically designed experiments or with existing theory or with previous results, then the research may be discontinued. The overall result may be a bias in the assumptions made by examiners as to the nature of the scientific enterprise.


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# Tag Info

1

This has been asked many times already. Volatility always refers to a model. And unless stated otherwise this model is the Black-Scholes model. In this model the volatility is the standard deviation of the log-returns divided by the square-root of time:  \log(\frac{S_{t}}{S_0}) = (r - \frac{1}{2}\sigma^2)t + \sigma W_t \sim \mathcal{N}\left( (r - ...

1

Look at the B-S parameters for the dynamics of the stock. $\frac{dS}{S} = \mu dt + \sigma dt$ $\sigma$ is independent of strike in the B-S model, which means all derivatives priced assuming these dynamics should have the same volatility. This clearly is not the case given the existence of smile and skew. You can't assume the BS model produces the "fair" ...

1

One reason is that implied volatility measures the relative value of the option as the price of an option depends on various parameters. As everyone has its own pricing model, it's insane to quote all parameters. This little simple IV tells you everything you'd need to know for valuation.

2

The most common use for implied volatility in valuation is for asseing options or option like postions. A volatile instrument is likley to activate or put an option postion in the money just on the basis of its volatility rather than any fundamental change in the intrinsic or fair market value of the underlying. This needs to be taken into account when ...

1

A pithy way to put it is "implied volatility is the wrong number to put in the wrong formula to get the right price." That is, implied volatility is by definition the parameter $\sigma$ to plug into the Black-Scholes option pricing formula to get the market price of a vanilla option. This is called "volatility," but in reality it isn't the same as the ...

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No. Implied volatility isn't a historical measure of standard deviation. Implied volatility is used to relate a market price to some model, be that Black-Scholes or something more sophisticated. Another way to phrase it, implied vol is that single vol input into a model, such that the model reproduces the market prices. Different models will have ...

Top 50 recent answers are included

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# 2.5 Motion equations for constant acceleration in one dimension  (Page 7/8)

 Page 7 / 8

A manned rocket accelerates at a rate of ${\text{20 m/s}}^{2}$ during launch. How long does it take the rocket to reach a velocity of 400 m/s?

To answer this, choose an equation that allows you to solve for time $t$ , given only $a$ , ${v}_{0}$ , and $v$ .

$v\phantom{\rule{0.15em}{0ex}}=\phantom{\rule{0.15em}{0ex}}{v}_{0}+\text{at}$

Rearrange to solve for $t$ .

$t=\frac{v-v{}_{0}\text{}}{a}=\frac{\text{400 m/s}-\text{0 m/s}}{{\text{20 m/s}}^{2}}=\text{20 s}$

## Section summary

• To simplify calculations we take acceleration to be constant, so that $\stackrel{-}{a}=a$ at all times.
• We also take initial time to be zero.
• Initial position and velocity are given a subscript 0; final values have no subscript. Thus,
$\left(\begin{array}{lll}\Delta t& =& t\\ \Delta x& =& x-{x}_{0}\\ \Delta v& =& v-{v}_{0}\end{array}}$
• The following kinematic equations for motion with constant $a$ are useful:
$x={x}_{0}+\stackrel{-}{v}t$
$\stackrel{-}{v}=\frac{{v}_{0}+v}{2}$
$v={v}_{0}+\text{at}$
$x={x}_{0}+{v}_{0}t+\frac{1}{2}{\text{at}}^{2}$
${v}^{2}={v}_{0}^{2}+2a\left(x-{x}_{0}\right)$
• In vertical motion, $y$ is substituted for $x$ .

## Problems&Exercises

An Olympic-class sprinter starts a race with an acceleration of $4\text{.}{\text{50 m/s}}^{2}$ . (a) What is her speed 2.40 s later? (b) Sketch a graph of her position vs. time for this period.

(a) $\text{10}\text{.}8\phantom{\rule{0.25em}{0ex}}\text{m/s}$

(b)

A well-thrown ball is caught in a well-padded mitt. If the deceleration of the ball is $2\text{.}\text{10}×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}$ , and 1.85 ms $\left(\text{1 ms}={\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}\text{s}\right)$ elapses from the time the ball first touches the mitt until it stops, what was the initial velocity of the ball?

38.9 m/s (about 87 miles per hour)

A bullet in a gun is accelerated from the firing chamber to the end of the barrel at an average rate of $6\text{.20}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}$ for $8\text{.}\text{10}×{\text{10}}^{-4}\phantom{\rule{0.25em}{0ex}}\text{s}$ . What is its muzzle velocity (that is, its final velocity)?

(a) A light-rail commuter train accelerates at a rate of $1\text{.}{\text{35 m/s}}^{2}$ . How long does it take to reach its top speed of 80.0 km/h, starting from rest? (b) The same train ordinarily decelerates at a rate of $1\text{.}{\text{65 m/s}}^{2}$ . How long does it take to come to a stop from its top speed? (c) In emergencies the train can decelerate more rapidly, coming to rest from 80.0 km/h in 8.30 s. What is its emergency deceleration in ${\text{m/s}}^{2}$ ?

(a) $\text{16}\text{.}\text{5 s}$

(b) $\text{13}\text{.}\text{5 s}$

(c) $-2\text{.}{\text{68 m/s}}^{2}$

While entering a freeway, a car accelerates from rest at a rate of $2\text{.}{\text{40 m/s}}^{2}$ for 12.0 s. (a) Draw a sketch of the situation. (b) List the knowns in this problem. (c) How far does the car travel in those 12.0 s? To solve this part, first identify the unknown, and then discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, check your units, and discuss whether the answer is reasonable. (d) What is the car’s final velocity? Solve for this unknown in the same manner as in part (c), showing all steps explicitly.

At the end of a race, a runner decelerates from a velocity of 9.00 m/s at a rate of $2\text{.}{\text{00 m/s}}^{2}$ . (a) How far does she travel in the next 5.00 s? (b) What is her final velocity? (c) Evaluate the result. Does it make sense?

(a) $\text{20}\text{.}\text{0 m}$

(b) $-1\text{.}\text{00 m/s}$

(c) This result does not really make sense. If the runner starts at 9.00 m/s and decelerates at $2\text{.}{\text{00 m/s}}^{2}$ , then she will have stopped after 4.50 s. If she continues to decelerate, she will be running backwards.

Professional Application:

Blood is accelerated from rest to 30.0 cm/s in a distance of 1.80 cm by the left ventricle of the heart. (a) Make a sketch of the situation. (b) List the knowns in this problem. (c) How long does the acceleration take? To solve this part, first identify the unknown, and then discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking your units. (d) Is the answer reasonable when compared with the time for a heartbeat?

full meaning of GPS system
how to prove that Newton's law of universal gravitation F = GmM ______ R²
sir dose it apply to the human system
prove that the centrimental force Fc= M1V² _________ r
prove that centripetal force Fc = MV² ______ r
Kaka
how lesers can transmit information
griffts bridge derivative
below me
please explain; when a glass rod is rubbed with silk, it becomes positive and the silk becomes negative- yet both attracts dust. does dust have third types of charge that is attracted to both positive and negative
what is a conductor
Timothy
hello
Timothy
below me
why below you
Timothy
no....I said below me ...... nothing below .....ok?
dust particles contains both positive and negative charge particles
Mbutene
corona charge can verify
Stephen
when pressure increases the temperature remain what?
what is frequency
define precision briefly
CT scanners do not detect details smaller than about 0.5 mm. Is this limitation due to the wavelength of x rays? Explain.
hope this helps
what's critical angle
The Critical Angle Derivation So the critical angle is defined as the angle of incidence that provides an angle of refraction of 90-degrees. Make particular note that the critical angle is an angle of incidence value. For the water-air boundary, the critical angle is 48.6-degrees.
okay whatever
Chidalu
pls who can give the definition of relative density?
Temiloluwa
the ratio of the density of a substance to the density of a standard, usually water for a liquid or solid, and air for a gas.
Chidalu
What is momentum
mass ×velocity
Chidalu
it is the product of mass ×velocity of an object
Chidalu
how do I highlight a sentence]p? I select the sentence but get options like copy or web search but no highlight. tks. src
then you can edit your work anyway you want
Wat is the relationship between Instataneous velocity
Instantaneous velocity is defined as the rate of change of position for a time interval which is almost equal to zero
Astronomy


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Latex in Blogger: Another Alternative

Last year, I shared about writing latex in Blogger using Watchmath. Sadly, the server has been offline for a very long time.  I have also shared  about a Free Online Latex generator at Codecogs.com, a good alternative.

Codecogs.com has a WYSIWYG environment. You just have to choose the math symbol and it will automatically generate the Latex code. To embed the code in Blogger, just copy the HTML code to your post (be sure that you paste in the HTML  window).

Aside from embedding code in HTML, Codecogs.com also provides embed code s for XML, phpBB, Tiddlywiki, and WordPress.

Quarter in Review – The Most Popular Posts

It’s almost the end of the first quarter of 2012. It’s a very busy week, so no time to write this week. Anyway, in case you missed reading some articles, here are the most popular posts for this quarter.

The Most Popular Posts

A US President’s Proof of the Pythagorean Theorem

James Garfield, the 20th president of the United States, came up with an original proof of the Pythagorean Theorem in 1876 when he was still a Congressman. His proof was published in New England Journal of Education.

Recall that the Pythagorean Theorem states that given a right triangle with sides $a$, $b$, and hypotenuse $c$, the following equation is always satisified:

$a^2 + b^2 = c^2$.

President Garfield’s proof is quite simple. We can do this in three steps:

1. Find the area of figure above using the trapezoid
2. Find the area of the same figure using the three triangles
3. Equate the results in 1 and 2. » Read more
1 2 3 10


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# IEEE Journal on Selected Areas in Communications

## Filter Results

Displaying Results 1 - 19 of 19
• ### Fundamental design issues for the future Internet

Publication Year: 1995, Page(s):1176 - 1188
Cited by:  Papers (594)  |  Patents (9)
| | PDF (1544 KB)

The Internet has been a startling and dramatic success. Originally designed to link together a small group of researchers, the Internet is now used by many millions of people. However, multimedia applications, with their novel traffic characteristics and service requirements, pose an interesting challenge to the technical foundations of the Internet. We address some of the fundamental architectura... View full abstract»

• ### An algorithm for combined cell-site selection and power control to maximize cellular spread spectrum capacity

Publication Year: 1995, Page(s):1332 - 1340
Cited by:  Papers (295)  |  Patents (21)
| | PDF (796 KB)

There is much current interest in spread spectrum wireless mobile communications and in particular the issue of spread spectrum wireless capacity. We characterize spread spectrum cellular capacity and provide a combined power control, cell-site selection algorithm that enables this capacity to be achieved. The algorithm adapts users' transmitter power levels and switches them between cell-sites, a... View full abstract»

• ### Optimal buffer sharing

Publication Year: 1995, Page(s):1229 - 1240
Cited by:  Papers (51)  |  Patents (4)
| | PDF (916 KB)

We address the problem of designing optimal buffer management policies in shared memory switches when packets already accepted in the switch can be dropped (pushed-out). Our goal is to maximize the overall throughput, or equivalently to minimize the overall loss probability in the system. For a system with two output ports, we prove that the optimal policy is of push-out with threshold type (POT).... View full abstract»

• ### A protocol for random multiple access of packets with mixed priorities in wireless networks

Publication Year: 1995, Page(s):1324 - 1331
Cited by:  Papers (10)
| | PDF (588 KB)

Random multiple access (RMA) protocols comprise an important class for the accessing/signaling stage in multimedia ATM wireless networks. Within the RMA class, the ALOHA algorithm is most widely used at this point in time, while, as is well known, the latter algorithm becomes unstable, as the user population increases. In this paper, we consider the part-and-try algorithm, instead, for a system wi... View full abstract»

• ### A new feedback congestion control policy for long propagation delays

Publication Year: 1995, Page(s):1284 - 1295
Cited by:  Papers (23)  |  Patents (2)
| | PDF (1076 KB)

This paper presents a new feedback congestion control mechanism for the flow control of the best-effort available bit rate (ABR) traffic in ATM networks. This new mechanism belongs to the class of feedback control schemes that ensure no data losses and operate based on simple stop' and start' signals. A novelty presented by this paper is a methodology which, for a given set of desired properties... View full abstract»

• ### Optimal flow control of a stochastic fluid-flow system

Publication Year: 1995, Page(s):1219 - 1228
Cited by:  Papers (10)
| | PDF (760 KB)

This paper is concerned with the optimal flow control of an ATM switching element in a broadband-integrated services digital network. We model the switching element as a stochastic fluid-flow system with a finite buffer, a constant output rate, and Markov-modulated fluid input. There is a cost of holding fluid and a reward for admitting the fluid to the buffer. We study the optimal flow control po... View full abstract»

• ### A framework for uplink power control in cellular radio systems

Publication Year: 1995, Page(s):1341 - 1347
Cited by:  Papers (1496)  |  Patents (20)
| | PDF (584 KB)

In cellular wireless communication systems, transmitted power is regulated to provide each user an acceptable connection by limiting the interference caused by other users. Several models have been considered including: (1) fixed base station assignment where the assignment of users to base stations is fixed, (2) minimum power assignment where a user is iteratively assigned to the base station at ... View full abstract»

• ### Architecting noncooperative networks

Publication Year: 1995, Page(s):1241 - 1251
Cited by:  Papers (109)  |  Patents (1)
| | PDF (1116 KB)

In noncooperative networks users make control decisions that optimize their individual performance measure. Focusing on routing, two methodologies for architecting noncooperative networks are devised, that improve the overall network performance. These methodologies are motivated by problem settings arising in the provisioning and the run time phases of the network. For either phase, Nash equilibr... View full abstract»

• ### Pricing congestible network resources

Publication Year: 1995, Page(s):1141 - 1149
Cited by:  Papers (304)  |  Patents (11)
| | PDF (764 KB)

We describe the basic economic theory of pricing a congestible resource such as an FTP server, a router, a Web site, etc. In particular, we examine the implications of “congestion pricing” as a way to encourage efficient use of network resources. We explore the implications of flat pricing and congestion pricing for capacity expansion in centrally planned, competitive, and monopolistic... View full abstract»

• ### On the message and time complexity of protocols for reliable broadcasts/multicasts in networks with omission failures

Publication Year: 1995, Page(s):1296 - 1308
Cited by:  Papers (6)
| | PDF (1240 KB)

This paper presents a fundamental study on the message and time complexity of reliable broadcast/multicast protocols in point-to-point networks subject to omission failures. We assume a weakly synchronous model in which there is a known upper bound on the delay in delivering a message from one process to another. A faulty process may omit sending or receiving messages; this characterizes a common ... View full abstract»

• ### Network programming methods for loss networks

Publication Year: 1995, Page(s):1189 - 1198
Cited by:  Papers (12)
| | PDF (900 KB)

This paper describes how some of the insights available from the stochastic analysis of dynamic routing may be incorporated into the classical mathematical programming approach to the design of networks. In particular, we present the results of a number of numerical investigations into network architectures for circuit-switched communication networks. Our investigations use recent theoretical resu... View full abstract»

• ### Stochastic control of handoffs in cellular networks

Publication Year: 1995, Page(s):1348 - 1362
Cited by:  Papers (13)  |  Patents (2)
| | PDF (1136 KB)

A dynamic programming formulation is used to obtain an optimal strategy for the handoff problem in cellular radio systems. The formulation includes the modeling of the underlying randomness in received signal strengths as well as the movements of the mobile. The cost function is designed such that there is a cost associated with switching and a reward for improving the quality of the call. The opt... View full abstract»

• ### Feedback-based flow control of B-ISDN/ATM networks

Publication Year: 1995, Page(s):1252 - 1266
Cited by:  Papers (16)
| | PDF (1184 KB)

We consider a system comprising of a single bottleneck switch/node that is fed by N independent Markov-modulated fluid sources. There is a fixed propagation delay incurred by the traffic between these sources and the switch. We assume that the switch sends periodic feedback in the form of a single congestion indicator bit. This feedback also incurs a fixed propagation delay in reaching the sources... View full abstract»

• ### Connection establishment in high-speed networks

Publication Year: 1995, Page(s):1150 - 1161
Cited by:  Papers (23)  |  Patents (12)
| | PDF (1232 KB)

The evolving view of connection establishment for connection-oriented services in high-speed networks such as ATM involves a contract negotiation process between a user agent and a network agent. The first stage consists of separate roles for the user and the network. The user agent must characterize the information streams that will be transmitted and the performance parameters that define the de... View full abstract»

• ### The effect of bandwidth allocation policies on delay in unidirectional bus networks

Publication Year: 1995, Page(s):1309 - 1323
Cited by:  Papers (5)
| | PDF (1224 KB)

We consider the problem of allocating bandwidth fairly to each node in a shared, unidirectional bus network. We focus on the pi persistent protocol, since these are open loop policies designed to operate well in high speed networks, which have a very large bandwidth-delay product and feedback in the upstream direction is not available in a timely manner. First, we introduce an improvem... View full abstract»

• ### A new degree of freedom in ATM network dimensioning: optimizing the logical configuration

Publication Year: 1995, Page(s):1199 - 1206
Cited by:  Papers (34)  |  Patents (6)
| | PDF (696 KB)

A mathematical model is presented that provides a well-defined formulation of the logical configuration problem of ATM networks (the carriers of future B-ISDN) with the objective of maximizing the total expected network revenue, given the physical network parameters and the traffic requirements of each virtual subnetwork. A two-phase solution procedure is developed in which the decision variables ... View full abstract»

• ### Adaptive algorithms for feedback-based flow control in high-speed, wide-area ATM networks

Publication Year: 1995, Page(s):1267 - 1283
Cited by:  Papers (65)
| | PDF (1140 KB)

This paper gives a class of flow control algorithms for the adaptive allocation of bandwidths to virtual connections (VC) in high-speed, wide-area ATM networks. The feedback rate to the source from the network is parsimonious, with each feedback bit indicating whether the buffer at a distant switch is above or below a threshold. The service discipline at the switch is first-come-first-served. The ... View full abstract»

• ### Billing users and pricing for TCP

Publication Year: 1995, Page(s):1162 - 1175
Cited by:  Papers (42)  |  Patents (86)
| | PDF (944 KB)

This paper presents a system for billing users for their TCP traffic. This is achieved by postponing the establishment of connections while the user is contacted, verifying in a secure way that they are prepared to pay. By presenting the user with cost and price information, the system can be used for cost recovery and to encourage efficient use of network resources. The system requires no changes... View full abstract»

• ### Analysis of controlled multiplexing systems via numerical stochastic control methods

Publication Year: 1995, Page(s):1207 - 1218
Cited by:  Papers (1)
| | PDF (1104 KB)

The paper demonstrates the usefulness of numerical methods of stochastic control theory for the design, analysis and control of multiplexing type systems and networks, as well as ATM type systems. The sources are of the Markov-modulated type, although the final results hold for other types (e.g. low-order AR schemes). Control problems arise when we wish to control cell loss due to buffer overflows... View full abstract»

## Aims & Scope

IEEE Journal on Selected Areas in Communications focuses on all telecommunications, including telephone, telegraphy, facsimile, and point-to-point television, by electromagnetic propagation.

Full Aims & Scope

## Meet Our Editors

Editor-in-Chief
Muriel Médard
MIT

Cambridge, MA 02139

617-253-3617

medard@mit.edu


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The average of five positive numbers is 308. The average of first two numbers is 482.5 and the average of last two

### Question Asked by a Student from EXXAMM.com Team

Q 2262745635.     The average of five positive numbers is 308. The average of
first two numbers is 482.5 and the average of last two
numbers is 258.5. What is the third number?
IBPS-CLERK 2017 Mock Prelims
A

224

B

58

C

121

D

Cannot be determined

E

None of these

#### HINT

(Provided By a Student and Checked/Corrected by EXXAMM.com Team)

#### Access free resources including

• 100% free video lectures with detailed notes and examples
• Previous Year Papers
• Mock Tests
• Practices question categorized in topics and 4 levels with detailed solutions
• Syllabus & Pattern Analysis


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Gaussian linear experts in the protocol of online regression incorporate information about the Gaussian nature of the noise in the outcomes. Each expert $\theta \in \mathbb{R}^n$ predicts the following probability distribution over the outcomes $y\in\mathbb{R}$:
$\displaystyle{\xi^\theta(y) = \frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(\theta'x_t-y)^2}{2\sigma^2}}, y\in \mathbb{R}}$
at each step, where the variance $\sigma^2$ is the same for all the experts, and is assumed to be known. The natural loss function for these experts is the logarithmic loss function $\lambda(\xi^\theta,y) = -\ln \xi^\theta(y)$.


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# The vapour pressure of n-hexane and n-heptane are 46 and 13mm Hg.If the vapour pressure of a solution of these two liquids is 38mmHg,the mole fraction of n-hexane in solution is

$\begin{array}{1 1}0.24\\0.56\\0.66\\0.76\end{array}$

$38=x(46)+(1-x)13$
$\Rightarrow x=(38-13)/(46-13)=25/33=0.76$


text:
# Transformation over time interval of a non-autonomous differential equation

Consider a differential equation $$\dot{x} = v(t,x)$$. This differential equation is equivalent to autonomous differential equation $$\dot{y} = w(y)$$, with $$y = (\alpha,x)$$ and $$w(y) = (1,v(t,x))$$. By rectification theorem, we are always guaranteed a unique solution in a neighbourhood.

In Arnold's ODE book - chapter 2, section 4. He defines transformation over the time interval and gives a corollary. Which I couldn't prove? This image has the definition and corollary.

Here $$\Phi(t_0,x_0;t)$$ is the unique solution for the intial value problem $$\dot{x} = v(t,x)$$ and with $$x(t_0)= x_0$$

• A hint for 3): when you go directly from time $t_0$ to $t$ the result is the same as when you go first from $t_0$ to $s$ and then from $s$ to $t$ (look at Fig. 70). – user539887 Mar 7 at 7:41


text:
Complex Numbers Bundle｜8 Self-Grading Math Google Forms｜Editable! complex_conjugate online. 3. Sum of the angles in a triangle is 180 degree worksheet. Write down the complex conjugates of: i) 5 + 3j ii) iii) iv) v) 4 vi) Where appropriate put both the original complex number and its conjugate into a + jb form and check your results. Proving triangle congruence worksheet. 45 min. Properties of parallelogram worksheet. Now, take the complex conjugate of the last wave-function you wrote - I assume for simplicity A = real, $$Ψ^*_k(x,t) = A \exp\bigl(i\bigl[-kx - ħk^2\frac{-t}{2m}\bigr]\bigr)$$ You see what we got? The questions in this maze focus only and only on dividing a Binomial complex number by a Bino, Complex Numbers Test Sets: Worksheet – 01. Some of the worksheets for this concept are Conjugate of complex numbers 1, Complex conjugate root theorem, 2 5 and 2 6b complex zeros and the fundamental theorem of, Complex zeros and the fundamental theorem of algebra, Irrational and imaginary root theorems, 1pf1 complex analysis, Complex numbers in polar form demoivres theorem, Complex numbers and the complex exponential. Some of the worksheets for this concept are Conjugating verbs to go work, Conjugate acid base pairs name chem work 19 2, How to conjugate french verbs present tense, How to conjugate portuguese verbs, Conjugate of complex numbers 1, Rationalizing imaginary denominators, Dividing radical. Dividing Complex Numbers. The complex conjugate root theorem tells us that complex roots are always found in pairs. So if z =a +bi, its complex conjugate, z , is defined by z =a −bi Any complex number a+bi has a complex conjugate a −bi and from Activity 5 it can be seen that ()a +bi ()a−bi is a real number. Returns the complex conjugate of a complex number in x + yi or x + yj text format. (ii) (10 - 5i)/ (6 + 2i) The document prints on standard paper but can be trimmed to fit a composition notebook (that's how I use it). Found worksheet you are looking for? Worksheet will open in a new window. Write the expression as a complex number in standard form. Evaluate. Become adept at adding, subtracting, multiplying, and dividing complex numbers with this free worksheet. Division - Dividing complex numbers is just as simpler as writing complex numbers in fraction form and then resolving them. 1. Students will practice adding, subtracting, multiplying, and dividing complex numbers with this coloring activity. And the simplest reason or the most basic place where this is useful is when you multiply any complex number times its conjugate, you're going to get a real number. Some of the worksheets displayed are Conjugating verbs to go work, Conjugate acid base pairs name chem work 19 2, How to conjugate french verbs present tense, How to conjugate portuguese verbs, Conjugate of complex numbers 1, Rationalizing imaginary denominators, Dividing radical. Scalar multiplication is involved in some of the questions, Included in this Product:2 similar versions for the price of 1! 14 differentiated worksheets covering all Complex Number content, for the new AQA further maths A-Level. For the complex number , identify the real number and the imaginary number.2. In particular, it is helpful for them to understand why the Upgrade your homework and classwork options with assignments that motivate your kiddos to battle all the way to the end of the page!Herowork is designed to be engaging repeated skill practice with a twis, Complex Roots, Imaginary Numbers, Conjugates, & Operations Notes, Algebra 2: i and Complex Conjugates Mini Formative Assessment, 30% off! Note that you are not asked to write the complex conjugates in the form a + jb, but simply to state the complex conjugate of … Version 1 has all answers on the outside of the, Have you seen the preview? (i) [ (5 + 9i) + (2 − 4i)] whole bar For any complex number w= c+dithe number c−diis called its complex conjugate. Use as a formative assessment after teaching this concept. Tutorial 1 In the following tutorial we further explain the complex conjugate root theorem. Students simpli, Complex Numbers Triples ActivityWith this triples matching activity, students will practice simplifying, adding, subtracting, multiplying, and dividing complex numbers. Conjugate To Be - Displaying top 8 worksheets found for this concept.. The the time goes toward the past and the particle moves in opposite direction (back to the source). When b=0, z is real, when a=0, we say that z is pure imaginary. For division, students must be able to rationalize the denominator, which includes multiplying by the conjugate. These complex numbers are a pair of complex conjugates. D p 9A TlPlW nrGi0g ih 0t zsM r2eLsteurgvWeed 4.8 P.3.6 Rationalizing Denominators & Conjugates STANDARD A.REI.B.4 AI/AII. Write the conjugate of each. On slide 1, the student adds or subtracts two complex numbers. There are 20 problems total, separated into two columns. Modulus of a Complex Number. The complex conjugate of a complex number is the number with equal real part and imaginary part equal in magnitude, but the complex value is opposite in sign. Summary : complex_conjugate function calculates conjugate of a complex number online. So for =ඹ+ම then ̅=ඹ−ම Just like with dealing with surds, we can also rationalist the denominator, when dealing with complex numbers. Use the method of completing the square to transform any quadratic equation in x into an equation of the form (x–p) 2 =q that has the same solutions. ©m w260 H1z4x qK ju 5t 3ab 7S Xokf vtIw SaLr lev 4L3L TC B.P n wAClnl F CrDiEguh8t 8s9 vr2e RsYePr RvkePdb.W I SMsa 7dTe l kw 7i 3t Ohm zI an2f ViDnbidt7e N 7A zlpgTeFb6r5a L e2z. Let's look at an example: 4 - 7 i and 4 + 7 i . ©H z2 C0U1x2w VK4uNtvaL WSQoTf xtyw haDr6e 1 IL mLhC t. L Y uA 7l 4l 9 UrUi3gKh8t Gsz crXeOs0eurtv YeJd D.9 W yM Haxdde s 9wxi pt8h6 PI3n XfYitngi 1t neq EA LlAg Pe5b NryaZ s2 K.H Worksheet by Kuta Software LLC Algebra 2: Complex Numbers ENTIRE UNIT Mini Formative Assessments, Conjugates and Quotients of Complex Numbers Guided Notes and HW, Quadratics Revisited - Complex Numbers Algebra 2 Unit 1 Bundle, Divide Complex Numbers using Conjugates Guided Notes, Algebra 2 - Conjugates of Complex Numbers, Complex Conjugates Theorem Doodle Graphic Organizer, Multiply A Complex Number and its Conjugate: 24 Task Cards, Complex Conjugate - Find the Conjugate, Moduli, and Quotients of Complex Numbers, Math Operations with Complex Numbers Mystery Drawing, Complex Numbers No Prep FUN Notes Doodle Pages also for Distance Learning, Algebra 2 FUN Notes Doodle Bundle also for Distance Learning Packets, Simplifying Complex Numbers Color Worksheet, Simplifying Complex Numbers Coloring Activity, Operations With Complex Numbers Cut & Paste Activity {Algebra 2}, Scavenger Hunt {School/Home/Stations} - Dividing Complex Numbers, Complex Numbers Digital Drag & Drop Activity - Distance Learning, Algebra 2 Curriculum - Texas TEKS Aligned, Complex Numbers Review or Quiz x 2｜2 Similar Versions!｜Math Google Forms. And I want to emphasize. Example: Displaying top 8 worksheets found for - Conjugate. Given a complex number z = a + b i {\displaystyle z=a+bi} (where a and b are real numbers), the complex conjugate of z {\displaystyle z} , often denoted as z ¯ {\displaystyle {\overline {z}}} , is equal to a − b i . Displaying top 8 worksheets found for - Conjugate To Be. W, Teach your students the techniques to determine tricky limits.Two techniques that are often found on the AP Exam are using the conjugate to rationalize a numerator and simplifying complex fractions. Write the problem in fractional form. A way todo thisisto utilizethe fact that(A+B)(A−B)=A2−B2 in order to eliminatesquare roots via squaring. It's the perfect, Your Algebra students will love learning about Complex Numbers with this fun format.Topics Include:Adding and subtractingMultiplying Dividing (rationalizing)Finding conjugatesFinding the absolute value (modulus)Graphing complex numbers. Simplify a … Change your answers in Q1 to exact x + yi form. 5. Power of a Complex Number. Then plot all eight complex numbers in the same complex plane. b. Hopefully yours will find it enjoyable and helpful as well :-) In some classes I ran this activity in class while in others I assigned it at home as a homework. Some of the worksheets for this concept are How to conjugate french verbs present tense, Conjugating verbs to go work, Conjugate acid base pairs name chem work 19 2, Conjugate of complex numbers 1, Bronsted, How to conjugate portuguese verbs, How to conjugate french verbs imperfect tense, … In general, the complex conjugate of a complex number is simply that number with the sign of the imaginary part reversed, that is, if z = x + i y, then the complex conjugate is z ∗ = x − i y. Then plot all eight complex numbers in the same complex plane. Adding and subtracting complex numbers. Students love to doodle while learning and it has been shown th, 25 carefully thought out problems on simplifying complex numbers. Simplify a bi a bi 2. $\overline{z}$ = 25 and p + q = 7 where $\overline{z}$ is the complex conjugate of z. ©f i2 N0O12F EKunt la i ZS3onf MtMwtaQrUeC 0LWLoCX.o F hA jl jln DrDiag ght sc fr 1ersve1r2vte od P.a G XMXaCdde 9 9waiht5hB 1I2nAfUizn ZibtMeV fA Sl Agesb 7rfa G G2D.Z Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 2 Name_____ Operations with Complex Numbers Date_____ Period____ Simplify. 29 min. F Worksheet by Kuta Software LLC Math for College Readiness Name_____ Date_____ Period____ ©2 A2G0f1t4 r pK MuStEad YS5oHfXt Lw ia br8eW QLFLrC x. Question 2. It has the same real component x, but the imaginary component is negated. There are 20 problems total, separated into two columns. Dividing Radicals. Plus model problems explained step by step We learn the theorem and illustrate how it can be used for finding a polynomial's zeros. ... Conjugate of a Complex Number. Conjugate of complex number. The level of complexity includes rationalizing the denominator by using the conjugate with monomial over monomial and binomial over monomial division. Description : Writing z = a + ib where a and b are real is called algebraic form of a complex number z : a is the real part of z; b is the imaginary part of z. 4. The complex conjugate zeros, or roots, theorem, for polynomials, enables us to find a polynomial's complex zeros in pairs. a. Find the quadratic equation that has 5 - 3i as one of its roots. Example: Conjugate of 7 – 5i = 7 + 5i. Write the conjugate of each. Are you getting the free resources, updates, and special offers we send out every week in our teacher newsletter? Write the expression as a complex number in standard form 5. Multiply the numerator and the denominator by the conjugate of the denominator. complex_conjugate online. Explain why complex conjugates do not have an “i” term Challenge Problems 1. For example, the complex conjugate of $$3 + 4i$$ is $$3 − 4i$$. Complex Numbers in Polar Form. nth Root of a Complex Number. Types of angles worksheet. Algebra II Complex Conjugate Worksheet Name: _ Find all zeros (real and imaginary) of each polynomial function using the Square Roots of a Complex Number. Each sheet contains a short summary in addition to the differentiated questions. 28 min. 2. Write the conjugate of each. a) 6 i b) 11 i c) 24 i 4. EXAMPLE 6 Find 8 −1 . Polynomials Complex Conjugate Root Theorem Worksheet 2 Answer each of the following without using a calculator and using the boxes provided for your answers. Also included in: Maze - BUNDLE IMAGINARY NUMBERS (Complex Numbers). Rez = Rez z = z; Imz = 0 Imz = Imz z = z; Rez = 0 z +z = 2Rez z = z z z = 2iImz zz = jzj2 jzj = jzj 1=z = z=jzj2; z ̸= 0 argz = argz; z ̸= 0 z = 1=z; jzj = 1 z1 +z2 = z1 +z2 z1z2 = z1z2 z1=z2 = z1=z2; z2 ̸= 0 z1 +:::+zn = z1 +:::+zn z1::: zn = z1::: zn 2. REF: 010714b 2 ANS: 3 REF: 089325siii 3 ANS: 2 REF: 088620siii 4 ANS: 4 If 1 i is one solution, the other is 1 i. To download/print, click on pop-out icon or print icon to worksheet to print or download. ... Sal discusses the usefulness of the conjugate of the complex number is this video. This lesson covers how to find a conjugate of a complex number as well as how to simplify expressions including division by a complex number. Complementary and supplementary worksheet. Our comprehensive worksheet numbers are free to download, easy to use, and very flexible. The real complex numbers lie on the x–axis, which is then called the real axis, while the imaginary numbers lie on the 7 + 2i 3 – 5i The complex conjugate of 3 – 5i is 3 + 5i. “Relearning” h, Heroes are not made overnight, they are developed with practice. Guided notes and practice problems on dividing complex numbers. Great for use in the classroom when first learning the topic, or as homework or revision sheets. This resource works well as independent prac, In this activity, students will practice operations with complex numbers. 2. Conjugate - Displaying top 8 worksheets found for this concept.. Discussion: In Class Exercise – 02. Then plot all eight complex numbers in the same complex plane.4. Students simpli You can see in the graph of f(x) = x2 + 1 below that f has no real zeros. Write the expression as a complex number in standard form. Some of the worksheets for this concept are Conjugate of complex numbers 1, Complex conjugate root theorem, 2 5 and 2 6b complex zeros and the fundamental theorem of, Complex zeros and the fundamental theorem of algebra, Irrational and imaginary root theorems, 1pf1 complex analysis, Complex numbers in polar form demoivres theorem, Complex numbers and … (x) 0 x 08 x 5 g = x 4 − 4 2 − 2 − 2 i is a zero − 4 + 3 The following diagram explains complex conjugate pairs. These complex worksheet numbers are randomly created and never repeated, so you have an infinite inventory of quality complex worksheet numbers to use in class or at home. Not only does, Students will simplify 18 algebraic expressions with complex numbers/imaginary numbers including adding, subtracting, multiplying and dividing complex numbers (includes rationalizing the denominator by multiplying by the conjugate) (Algebra 2 Curriculum) Worksheet Q1: Find the complex conjugate of − 7 − and the sum of this number with its complex conjugate. Quick and easy warm-up or exit ticket. A complex conjugate is formed by changing the sign between two terms in a complex number. There is a variety of problems from basic operations, rationalizing, conjugating, distributing, and negative radicals. Show all of your working. The answer to a puzzle will appear at the bottom of each worksheet once the solutions are found.Complex Numbe, This maze is part of : Maze - BUNDLE COMPLEX NUMBERSThis product is a good review of checking for understanding how to Divide Complex Numbers and Rationalizing the Denominator - Binomial/Binomial ONLY. Area and perimeter worksheets. First, he shows both graphically and algebraically adding a complex number and its conjugate. In other words if we find, or are given, one complex root, then we can state that its complex conjugate is also a root. Multiply the next few complex number pairs and try to recognize a pattern 21) 3 2 3 2 ii 22) ii 4 5 4 5 23) ii7 3 7 3 24) Look carefully at questions 21-23. 1. Notation: w= c+ di, w¯ = c−di. 15 min. Find the conjugate of a complex number; use conjugates to find moduli and quotients of complex numbers. Complex Conjugates. Summary : complex_conjugate function calculates conjugate of a complex number online. Then plot all eight complex numbers in the same complex plane. Free worksheet(pdf) and answer key on Dividing Complex Numbers. Students will simplify 20 dividing radical expressions problems WITHOUT variables in this independent practice riddles worksheet. When b=0, z is real, when a=0, we say that z is pure imaginary. Also included in: Complex Numbers Bundle｜8 Self-Grading Math Google Forms｜Editable! Enjoy these free printable sheets focusing on the complex and imaginary numbers typically covered unit in algebra 2 each worksheet has model problems worked out step by step practice problems as well as challenge questions at the sheets end. There are two versions of the puzzle included. Conjugate - Displaying top 8 worksheets found for this concept.. 2. Multiply the next few complex number pairs and try to recognize a pattern 21) 3 2 3 2 ii 22) ii 4 5 4 5 23) ii7 3 7 3 24) Look carefully at questions 21-23. Multiplying by the Conjugate Sometimes it is useful to eliminate square roots from a fractional expression. Complex numbers is vital in high school math. A.REI.B.4: Complex Conjugate Root Theorem Answer Section 1 ANS: 3 The product of the roots equals . The complex conjugate of a complex number is the number with equal real part and imaginary part equal in magnitude, but the complex value is opposite in sign. We also work through some typical exam style questions. These notes go along step-by-step with a presentation sold separately. The real part of the resultant number = 5 and the imaginary part of the resultant number = 6i. Scroll down the page for more examples and solutions for dividing complex numbers. G Worksheet by Kuta Software LLC Math for College Readiness Name_____ Date_____ Period____ This right here is the conjugate. Maze - Operations on Complex Numbers - Division of Binomial by Binomial, Maze - BUNDLE IMAGINARY NUMBERS (Complex Numbers), Learning Limits - 5: Rationalizing a Numerator and Complex Fractions, Herowork - Conjugates - Wilma Rudolph Mystery Pic & Weekday Joke. (iii) 3i... (2) If z = x + iy , find the following in rectangular form. Use the conjugate root theorem and the … Some of the worksheets for this concept are How to conjugate french verbs present tense, Conjugating verbs to go work, Conjugate acid base pairs name chem work 19 2, Conjugate of complex numbers 1, Bronsted, How to conjugate portuguese verbs, How to conjugate french verbs imperfect tense, Rationalizing imaginary denominators. When you access the file you, This five problem assessment helps you to quickly determine students' understanding of the term "i" within complex numbers (and that it's equal to square root of negative 1) as well as i squared and how to write complex conjugates. Write the conjugate of each. For example, the complex conjugate of 3 + 4 i is 3 − 4 i. There are 20 problems total, separated into two columns. Complementary and supplementary word problems worksheet. 4. 2. Complex number any number that can be written in the form where … Simplifying Expressions with Complex Numbers. Introduction to Complex Numbers ... i2 = –1 To divide complex numbers, multiply the numerator and denominator by the complex conjugate of the complex number in the denominator of the fraction. Use basic rules like combining the like terms together, FOIL method, multiplying the top and bottom by the complex conjugate of the denominator to simplify complex-valued expressions. Sum of two complex numbers a + bi and c + di is given as: (a + bi) + (c + di) = (a + c) + (b + d)i. Complex conjugation negates the imaginary component, so as a transformation of the plane C all points are reflected in the real axis (that is, points above and below the real axis are exchanged). I implement. Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. Subset, Universal Set, Power Set & Intervals. Write your answers in polar form. Perform operations like addition, subtraction and multiplication on complex numbers, write the complex numbers in standard form, identify the real and imaginary parts, find the conjugate, graph complex numbers, rationalize the denominator, find the absolute value, modulus, and argument in this collection of printable complex number worksheets. 19 min. Answer key included. For division, students must be able to rationalize the denominator, which includes multiplying, Students will practice adding, subtracting, multiplying, and dividing complex numbers with this coloring activity. Conjugate Showing top 8 worksheets in the category - Conjugate . My kids loved this activity. Algebra 2: Complex Numbers ENTIRE UNIT Mini Formative Assessments, Also included in: Quadratics Revisited - Complex Numbers Algebra 2 Unit 1 Bundle, Also included in: Algebra 2 Credit Recovery, Also included in: Algebra 1 Mystery Drawings Volume 1, Also included in: Algebra 2 FUN Notes Doodle Bundle also for Distance Learning Packets, Also included in: Algebra 2 Activities Growing Bundle, Also included in: Algebra 2 Curriculum - Texas TEKS Aligned. Complex Numbers Test It covers the following questions: 1. For example, the complex conjugate of $$3 + 4i$$ is $$3 − 4i$$. The following notation is used for the real and imaginary parts of a complex number z. (x) x x 3 x x f = 5 4 + 6 3 − 1 2 − 6 + 8 is a zero − 2 2. About This Quiz & Worksheet About This Quiz & Worksheet Test your ability to solve complex conjugate problems and find the product of these mathematical expressions in this … 2. They need to do their Herowork! A complex number is a number that can be expressed in the form a + bi, where a and b are real numbers,... Complex Conjugates. Some of the worksheets for this concept are Conjugating verbs to go work, Conjugate acid base pairs name chem work 19 2, How to conjugate french verbs present tense, How to conjugate portuguese verbs, Conjugate of complex numbers 1, Rationalizing imaginary denominators, Dividing radical. These complex numbers are known as complex conjugates. Explain why complex conjugates do not have an “i” term Challenge Problems 1. Equal, Conjugate, Opposite and Reciprocal Complex Numbers. If = + then ̅(pronounced zed bar), is given by = − , and this is called the complex conjugate of z. Divide Complex Numbers Dividing by Complex Numbers For the complex number , identify the real number and the imaginary number. But 7 minus 5i is also the conjugate of 7 plus 5i, for obvious reasons. If you solve the corresponding equation 0 = x2 + 1, you find that x = ,which has no real solutions. 1. Find all the complex numbers of the form z = p + qi , where p and q are real numbers such that z. Word document can be edited. In other cases, I used these scavenger hunts as sta, This drag & drop activity for use with Google Slides is great for distance learning or one-to-one classrooms!This activity contains 2 slides over Complex Numbers Operations. It covers the following questions: Scroll down the page for more examples and solutions. The file it's editable and in the form of a google slide. Dealing with complex numbers number, identify the real number and the number! Th, 25 carefully thought out problems on dividing complex numbers and imaginary terms 24 scaffolded questions start. Simplify 20 dividing radical expressions problems without variables in this independent practice riddles worksheet doodle... Following notation is used for Finding a polynomial 's zeros number, identify the real and terms! Order to eliminatesquare roots via squaring version 1 has all answers on the outside of,! Of 3 + 5i document reader options Product:2 similar versions for the complex numbers cis to exact x yi... A+B ) ( A−B ) =A2−B2 in complex conjugate worksheet to eliminatesquare roots via squaring %... Fit a composition Notebook ( that 's how i use it ) total, separated into columns. Contains a short summary in addition to the differentiated questions will learn practice. Send out every week in our teacher newsletter love to doodle while learning and has. Number any number that has each of these arguments for any complex number z a variety problems. Past and the imaginary part... Plotting complex numbers parts are added to imaginary terms are added imaginary... W¯ = c−di ” in the following notation is used for the complex Theorem!, multiplying, and special offers we send out every week in our teacher newsletter − 7 − the! X, but the imaginary part of the resultant number = 6i special we! = c−di one of its roots 24 i 4 2 4 i, identify the real number the...: Maze - Bundle imaginary numbers ( complex numbers use conjugates to find moduli quotients! And operations with complex numbers in the form of a complex number of numbers!: this resource is for teaching students complex roots, imaginary numbers, conjugates and operations with complex Bundle｜8... Number c−diis called its complex conjugate of a complex number, identify the number... Worksheets found for this concept + 7 i and 4 + 7 i and +. Journals in my classroom for almost 20 years roots from a fractional expression to come up the... Numbers in the form of a complex number that can be used for Finding a polynomial 's zeros have! ) =A2−B2 in order to eliminatesquare roots via squaring Bundle, also included in this practice... Standard form number in standard form 20 years, but the imaginary number displaying top 8 worksheets for. Also the conjugate with monomial over monomial division ( A+B ) ( A−B ) in..., moduli, and negative radicals the Theorem and illustrate how it can be written in the tutorial! The product of the form z = a + bi is: a –.! If a complex number any number that can be used for Finding polynomial! Become adept at complex conjugate worksheet, subtracting, multiplying, and negative radicals 7 2i... Numbers to find a complex number is a lesson for Algebra 2 Notebook Guided notes Bundle, included! Sal discusses the usefulness of the imaginary part... Plotting complex numbers this. Where p complex conjugate worksheet q are real numbers such that z is real, a=0. Nrgi0G ih 0t zsM r2eLsteurgvWeed 4.8 P.3.6 rationalizing Denominators & conjugates complex numbers in the same complex.. 3 − 4 i B ) 7 i lesson of my limit series conjugates do not an! Classroom for almost 20 years Denominators & conjugates complex numbers + 4 i direction back! Will simplify 20 dividing radical expressions problems without variables in this Product:2 similar versions for the complex root! Must be able to rationalize the denominator by using the conjugate is obtained by changing the sign of angles! The square roots of negative numbers complex number and the denominator, which includes multiplying by the conjugate Sometimes is. Following questions: 1 4.8 P.3.6 rationalizing Denominators & conjugates complex numbers it. 24 unique task cards ask student to come up with the conjugate numbers ) from Math at... Or revision sheets = 5 and the denominator, when a=0, we will using! As a formative assessment after teaching this concept, w¯ = c−di this activity - 7 i and +! Are free to download, easy to use, and quotients of complex worksheet! Multiplying, and dividing complex numbers in the graph of f ( ). Such that z is real, when a=0, we say that z is real, when,... B. conjugate to be while learning and it has the same complex plane a complex conjugate as homework or sheets! Differentiated questions special triangles to find a complex number that has 5 - 3i as one its.: find the quadratic equation that has 5 - 3i as one of its.. ( worksheet ) complex numbers in the classroom when first learning the topic, or as or. Eight complex numbers of the conjugate of a Google slide known as the Argand diagram or plane! Eliminate square roots from a fractional expression summary in addition to the:. The questions, included in: Algebra 2 Notebook Guided notes Bundle, also included in: 2. Using a calculator and using the conjugate with monomial over monomial division easy to,. Q1 to exact x + yi form imaginary component is negated evaluate an expression we say z..., easy to use, and dividing complex numbers a pair of complex conjugates complex_conjugate function calculates conjugate a... A lesson for Algebra 2 Notebook Guided notes complex conjugate worksheet practice problems on dividing complex numbers easy and with... Plotting complex numbers + 4 i questions, included in: Algebra 2 Notebook notes. ” h, Heroes are not made overnight, they are developed with practice 's zeros includes multiplying the... Product:2 similar versions for the price of 1 used Math journals in classroom... 0 = x2 + 1 below that f has no real zeros a variety of problems from basic operations rationalizing. Bundle｜8 Self-Grading Math Google Forms｜Editable is included.I have used Math journals in my classroom for almost 20 years the equation... This resource is for teaching students complex roots, imaginary numbers, conjugates and operations with complex numbers this. Each question corresponds to a matching Answer that gets colored in to form a symmetrical design when a=0 we... Up with the conjugate, moduli, and special offers we send out week... Symmetrical design this activity, or as homework or revision sheets use in the fifth lesson of limit... Separated into two columns 2 i 3 x, but the imaginary number find quadratic. Algebraically adding a complex number, identify the real number and the moves... The fifth lesson of my limit series independent practice riddles worksheet: 3 the product the. Of my limit series square roots of negative numbers - Bundle imaginary numbers, conjugates operations..., for obvious reasons: Maze - Bundle imaginary numbers to evaluate an expression − 7 and. For Algebra 2 Notebook Guided notes and practice both “ tricks ” in the fifth lesson my!, moduli, and very flexible parts are added together and imaginary part of the of! Reader options multiply the numerator and the imaginary part imaginary terms are added to imaginary terms are added and! Bundle｜8 Self-Grading Math Google Forms｜Editable Power Set & Intervals be able to rationalize the,. To the differentiated questions Heroes are not made overnight, they are developed with practice Power Set &.... Fact that ( A+B ) ( A−B ) =A2−B2 in order to eliminatesquare roots squaring! And practice both “ tricks ” in the form z = p qi! The corresponding equation 0 = x2 + 1, the student adds or subtracts two complex numbers: 4 7..., he shows both graphically and algebraically adding a complex number has no real zeros tricks! My classroom for almost 20 years “ i ” term Challenge problems 1 of. 8 – Finding conjugates of complex number online example, the complex conjugate root Theorem tells us complex! Classroom for almost 20 years imaginary number.2 sign between two terms in a complex number: complex_conjugate function conjugate! Download or print using the properties of conjugate numbers to evaluate an.... Bithen summary: complex_conjugate function calculates conjugate of the roots equals with monomial over monomial.... 6 i B ) 7 i and 4 + 7 i, which includes multiplying by the of! In Opposite direction ( back to the differentiated questions after teaching this concept, but the part. Polynomials complex conjugate of a complex number 10 4 i is 3 + 4i\ ) marketplace teachers! Bundle｜8 Self-Grading Math Google Forms｜Editable operations with complex numbers variety of problems from basic operations, rationalizing,,... You complex conjugate worksheet the free resources, updates, and negative radicals “ tricks ” the... The browser document reader options back to the differentiated questions so is its conjugate... Has the same complex plane is its complex conjugate and it has been shown th, 25 carefully thought problems... All the complex number in standard form numbers dividing by complex numbers dividing complex. 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complex conjugate worksheet 2021


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# Haplotype

A haplotype (from the Greek: ἁπλοῦς, haploûs, "onefold, single, simple") in genetics is a combination of alleles (DNA sequences) at adjacent locations (loci) on a chromosome that are inherited together. A haplotype may be one locus, several loci, or an entire chromosome depending on the number of recombination events that have occurred between a given set of loci, if any occurred.

A second meaning of the term haplotype is a set of single-nucleotide polymorphisms (SNPs) on a single chromosome of a chromosome pair that are associated statistically. It is thought that these associations, and the identification of a few alleles of a haplotype sequence, can unambiguously identify all other polymorphic sites in its region. Such information is very valuable for investigating the genetics of common diseases, and has been investigated for the human species by the International HapMap Project.[1][2]

Many genetic testing companies use the term 'haplotype' to refer to an individual collection of short tandem repeat (STR) allele mutations within a genetic segment, while using the term 'haplogroup' to refer to the SNP/unique-event polymorphism (UEP) mutations which represents the clade to which a collection of potential haplotypes belong (the term clade referring to a set of people sharing a common ancestor).[3]

## Haplotype resolution

An organism's genotype may not define its haplotype uniquely. For example, consider a diploid organism and two bi-allelic loci (such as SNPs) on the same chromosome. Assume the first locus has alleles A or T and the second locus G or C. Both loci, then, have three possible genotypes: (AA, AT, and TT) and (GG, GC, and CC), respectively. For a given individual, there are nine possible configurations (haplotypes) at these two loci (shown in the Punnett square below). For individuals who are homozygous at one or both loci, the haplotypes are unambiguous - meaning that there is not any differentiation of haplotype T1T2 vs haplotype T2T1; where T1 and T2 are labeled to show that they are the same locus, but labeled as such to show it doesn't matter which order you consider them in, the end result is two T loci. For individuals heterozygous at both loci, the gametic phase is ambiguous - in these cases, you don't know which haplotype you have, e.g., TA vs AT.

AA AT TT
GG AG AG AG TG TG TG
GC AG AC AG TC
or
AC TG
TG TC
CC AC AC AC TC TC TC

The only unequivocal method of resolving phase ambiguity is by sequencing. However, it is possible to estimate the probability of a particular haplotype when phase is ambiguous using a sample of individuals.

Given the genotypes for a number of individuals, the haplotypes can be inferred by haplotype resolution or haplotype phasing techniques. These methods work by applying the observation that certain haplotypes are common in certain genomic regions. Therefore, given a set of possible haplotype resolutions, these methods choose those that use fewer different haplotypes overall. The specifics of these methods vary - some are based on combinatorial approaches (e.g., parsimony), whereas others use likelihood functions based on different models and assumptions such as the Hardy-Weinberg principle, the coalescent theory model, or perfect phylogeny. The parameters in these models are then estimated using algorithms such as the expectation-maximization algorithm (EM), Markov chain Monte Carlo (MCMC), or hidden Markov models (HMM).

Microfluidic whole genome haplotyping is a technique for the physical separation of individual chromosomes from a metaphase cell followed by direct resolution of the haplotype for each allele.

## Y-DNA haplotypes from genealogical DNA tests

Unlike other chromosomes, Y chromosomes do not come in pairs. Every human male has only one copy of that chromosome. This means that there is not any chance variation of which copy is inherited, and also (for most of the chromosome) not any shuffling between copies by recombination; so, unlike autosomal haplotypes, there is effectively not any randomisation of the Y-chromosome haplotype between generations. A human male should largely share the same Y chromosome as his father, give or take a few mutations; thus Y chromosomes tend to pass largely intact from father to son, with a small but accumulating number of mutations that can serve to differentiate male lineages. In particular, the Y-DNA represented as the numbered results of a Y-DNA genealogical DNA test should match, except for mutations.

### UEP results (SNP results)

Unique-event polymorphisms (UEPs) such as SNPs represent haplogroups. STRs represent haplotypes. The results that comprise the full Y-DNA haplotype from the Y chromosome DNA test can be divided into two parts: the results for UEPs, sometimes loosely called the SNP results as most UEPs are single-nucleotide polymorphisms, and the results for microsatellite short tandem repeat sequences (Y-STRs).

The UEP results represent the inheritance of events it is believed can be assumed to have happened only once in all human history. These can be used to identify the individual's Y-DNA haplogroup, his place in the "family tree" of the whole of humanity. Different Y-DNA haplogroups identify genetic populations that are often distinctly associated with particular geographic regions; their appearance in more recent populations located in different regions represents the migrations tens of thousands of years ago of the direct patrilineal ancestors of current individuals.

### Y-STR haplotypes

Genetic results also include the Y-STR haplotype, the set of results from the Y-STR markers tested.

Unlike the UEPs, the Y-STRs mutate much more easily, which allows them to be used to distinguish recent genealogy. But it also means that, rather than the population of descendants of a genetic event all sharing the same result, the Y-STR haplotypes are likely to have spread apart, to form a cluster of more or less similar results. Typically, this cluster will have a definite most probable center, the modal haplotype (presumably similar to the haplotype of the original founding event), and also a haplotype diversity — the degree to which it has become spread out. The further in the past the defining event occurred, and the more that subsequent population growth occurred early, the greater the haplotype diversity will be for a particular number of descendants. However, if the haplotype diversity is smaller for a particular number of descendants, this may indicate a more recent common ancestor, or a recent population expansion.

It is important to note that, unlike for UEPs, two individuals with a similar Y-STR haplotype may not necessarily share a similar ancestry. Y-STR events are not unique. Instead, the clusters of Y-STR haplotype results inherited from different events and different histories tend to overlap.

In most cases, it is a long time since the haplogroups' defining events, so typically the cluster of Y-STR haplotype results associated with descendents of that event has become rather broad. These results will tend to significantly overlap the (similarly broad) clusters of Y-STR haplotypes associated with other haplogroups. This makes it impossible for researchers to predict with absolute certainty to which Y-DNA haplogroup a Y-STR haplotype would point. If the UEPs are not tested, the Y-STRs may be used only to predict probabilities for haplogroup ancestry, but not certainties.

A similar scenario exists in trying to evaluate whether shared surnames indicate shared genetic ancestry. A cluster of similar Y-STR haplotypes may indicate a shared common ancestor, with an identifiable modal haplotype, but only if the cluster is sufficiently distinct from what may have happened by chance from different individuals who historically adopted the same name independently. Many names were adopted from common occupations, for instance, or were associated with habitation of particular sites. More extensive haplotype typing is needed to establish genetic genealogy. Commercial DNA-testing companies now offer their customers testing of more numerous sets of markers to improve definition of their genetic ancestry. The number of sets of markers tested has increased from 12 during the early years to 111 more recently.

Establishing plausible relatedness between different surnames data-mined from a database is significantly more difficult. The researcher must establish that the very nearest member of the population in question, chosen purposely from the population for that reason, would be unlikely to match by accident. This is more than establishing that a randomly selected member of the population is unlikely to have such a close match by accident. Because of the difficulty, establishing relatedness between different surnames as in such a scenario is likely to be impossible, except in special cases where there is specific information to drastically limit the size of the population of candidates under consideration.

## Diversity

Haplotype diversity is a measure of the uniqueness of a particular haplotype in a given population. The haplotype diversity (H) is computed as:[4]
$H=\frac{N}{N-1}(1- \sum_{i}x_i^2)$
where $x_i$ is the (relative) haplotype frequency of each haplotype in the sample and $N$ is the sample size. Haplotype diversity is given for each sample.

## Software

• FAMHAP[5] — FAMHAP is a software for single-marker analysis and, in particular, joint analysis of unphased genotype data from tightly linked markers (haplotype analysis).
• FugueEM based haplotype estimation and association tests in unrelated and nuclear families.
• HPlus[6] — A software package for imputation and testing of haplotypes in association studies using a modified method that incorporates the expectation-maximization algorithm and a Bayesian method known as progressive ligation.
• HaploBlockFinder — A software package for analyses of haplotype block structure.
• Haploscribe[7] — Reconstruction of whole-chromosome haplotypes based on all genotyped positions in a nuclear family, including rare variants.
• Haploview[8] — Visualisation of linkage disequilibrium, haplotype estimation and haplotype tagging (Homepage).
• HelixTree — Haplotype analysis software - Haplotype Trend Regression (HTR), haplotypic association tests, and haplotype frequency estimation using both the expectation-maximization (EM) algorithm and composite haplotype method (CHM).
• PHASE — A software for haplotype reconstruction, and recombination rate estimation from population data.
• SHAPEIT[9] — SHAPEIT2 is a program for haplotype estimation of SNP genotypes in large cohorts across whole chromosome.
• SNPHAPEM based software for estimating haplotype frequencies from unphased genotypes.
• WHAP[10]haplotype based association analysis.

## References

1. ^ The International HapMap Consortium (2003). "The International HapMap Project". Nature 426 (6968): 789–796. doi:10.1038/nature02168. PMID 14685227.
2. ^ The International HapMap Consortium (2005). "A haplotype map of the human genome". Nature 437 (7063): 1299–1320. doi:10.1038/nature04226. PMC 1880871. PMID 16255080.
3. ^ Facts & Genes. Volume 7, Issue 3
4. ^ Masatoshi Nei and Fumio Tajima, "DNA polymorphism detectable by restriction endonucleases", Genetics 97:145 (1981)
5. ^ Becker T., Knapp M. (2004). "Maximum-likelihood estimation of haplotype frequencies in nuclear families". Genetic Epidemiology 27 (1): 21–32. doi:10.1002/gepi.10323. PMID 15185400.
6. ^ Li S.S., Khalid N., Carlson C., Zhao L.P. (2003). "Estimating haplotype frequencies and standard errors for multiple single nucleotide polymorphisms". Biostatistics 4 (4): 513–522. doi:10.1093/biostatistics/4.4.513. PMID 14557108.
7. ^ Roach J.C., Glusman G., Hubley R., Montsaroff S.Z., Holloway A.K., Mauldin D.E., Srivastava D., Garg V., Pollard K.S., Galas D.J., Hood L., Smit A.F.A. (2011). "Chromosomal Haplotypes by Genetic Phasing of Human Families". American Journal of Human Genetics 89 (3): 382–397. doi:10.1016/j.ajhg.2011.07.023. PMID 21855840.
8. ^ Barrett J.C., Fry B., Maller J., Daly M.J. (2005). "Haploview: analysis and visualization of LD and haplotype maps". Bioinformatics 21 (2): 263–265. doi:10.1093/bioinformatics/bth457. PMID 15297300.
9. ^ Delaneau O, Zagury J-F, Marchini J (2013). "Improved whole chromosome phasing for disease and population genetic studies.". Nature Methods 10: 5–6. doi:10.1038/nmeth.2307.
10. ^ Purcell S., Daly M. J., Sham P. C. (2007). "WHAP: haplotype-based association analysis". Bioinformatics 23 (2): 255–256. doi:10.1093/bioinformatics/btl580. PMID 17118959.


text:
, 22.06.2019 21:00 christianskyy7074

# C

Idk bro sorry  lava is hotter then magma but im  not sure about it alll
Yes, it’s awesome
: )

the photo and full question are both in attachments. mark as brainliest. i need 4 more brainliest to advance in rank.

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C$$\times = 2 = = = = \frac{3}{2 } 6$$...


text:
# Raytracer: Lights?

This topic is 4171 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

Hallo, I'm working on a raytracer, but I am having some problems with adding lights to it. For some reasons all my objects get black, as if no light is hitting them. I have tryed moving the light around to make sure that the objects are not simply placed in front of the light. I have some ideas on why this happens. One problem is that "lightCoef" is allways less or equal to zero. Not sure why, but for some reason it is. I did try to make "lighCoef" posetive by doing AbsDot() insted of Dot(), that made the spheres visible, but they had no gradient on them, so the color was just as flat as befor I tryed to add lights. Any ideas on what is wrong? I have been trying to solve this the whole day and is now close to giving up. If this is not enugh code to see what is wrong I can post more, but I tryed to only post what is needed as it is often boring to read through to much code. Thanks a lot for your time:)
                float r, g, b;

// Check which if there is a light affecting the sphere
for (int currentLight = 0; currentLight <= thisScene.getAmountOfObjects(); currentLight++)
{
Vector3D sphereRayIntersection;
Point rayOrg = currentRay.getOrigin();

// Add the start point of the ray and the direction of the sphere together
// and mult. it with the distance to closest sphere
sphereRayIntersection = currentRay.getDirection();
sphereRayIntersection.setX(sphereRayIntersection.getX() + rayOrg.getX());
sphereRayIntersection.setY(sphereRayIntersection.getY() + rayOrg.getY());
sphereRayIntersection.setZ(sphereRayIntersection.getZ() + rayOrg.getZ());

sphereRayIntersection.scale(closesSphereDistance);

// Find the normal to the sphere at point of intersection
Sphere thisSphere = thisScene.getSphere(closestSphere);
Vector3D normal;

normal = sphereRayIntersection - thisSphere.getPosition();

normal.normalize();

// Get the light direction
Vector3D lightDir;

Sphere thisLight = thisScene.getLight(currentLight);
lightDir = thisLight.getPosition() - sphereRayIntersection;

lightDir.normalize();

// Find light coefficient
float lightCoef = Dot(lightDir,normal);
if (lightCoef < 0)
lightCoef = 0;

// Get the color of the light and how much it affects the sphere
Material thisMaterial = thisLight.getMaterial();

float lightR = thisMaterial.getR() * lightCoef;
float lightG = thisMaterial.getG() * lightCoef;
float lightB = thisMaterial.getB() * lightCoef;

// Add the effect of the light to the color of the sphere
Material sphereColor = thisSphere.getMaterial();

r = ((sphereColor.getR() * lightR) / 255 );
g = ((sphereColor.getG() * lightG) / 255);
b = ((sphereColor.getB() * lightB) / 255);

// Make sure that the color is not higher than 255
if (r > 255)
r = 255;
if (g > 255)
g = 255;
if (b > 255)
b = 255;

//std::cout << r << "," << g << "," << b << std::endl;
}

// Assign the color to the pixel
int colorR = r;
int colorG = g;
int colorB = b;

colorBuffer[arrayTracer + 0] = colorR;
colorBuffer[arrayTracer + 1] = colorG;
colorBuffer[arrayTracer + 2] = colorB;

[/source] [Edited by - h3ro on April 19, 2007 5:42:43 PM]

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Are you sure your calculation for getting the intersection point is generating the right result? It should be (origin + (direction * distance)) not ((origin + direction) * distance). Most likely this is causing your surface normal to be incorrect, which is yielding the bogus lighting. Also, note that scaling the normal by the sphere's radius is totally useless since the next step is to normalize it, which forces the length to 1 [wink]

As for posting code - try making your [ code ] tags lowercase, or use [ source ] instead (which is preferable for long bits of code).

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Thanks for you post, but it did not really solve my problem.

Here is the picture I have now:
http://img89.imageshack.us/img89/18/raytracedscenehz9.jpg

As you see something must be wrong:P

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Perhaps your light is simply way too bright? :) Try to divide the final result by 10 or so.

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Hi there!

sphereRayIntersection = currentRay.getDirection();sphereRayIntersection.setX(sphereRayIntersection.getX() + rayOrg.getX());sphereRayIntersection.setY(sphereRayIntersection.getY() + rayOrg.getY());sphereRayIntersection.setZ(sphereRayIntersection.getZ() + rayOrg.getZ());                    sphereRayIntersection.scale(closesSphereDistance);

If 'currentRay' represents a ray of the form 'origin + t*dir' and 'closesSphereDistance' is supposed to be the parameter 't' for the ray at which it intersects the sphere, then the point 'sphereRayIntersection' should be 'origin + t*direction', not '(origin + direction)*t', just as ApochPiQ mentioned. So the above code snippet should be changed to something like this:

sphereRayIntersection = currentRay.getDirection();sphereRayIntersection.scale(closesSphereDistance);sphereRayIntersection.setX(sphereRayIntersection.getX() + rayOrg.getX());sphereRayIntersection.setY(sphereRayIntersection.getY() + rayOrg.getY());sphereRayIntersection.setZ(sphereRayIntersection.getZ() + rayOrg.getZ());

When you changed that you should really debug whether the result can be right or not (best to test that with a single sphere centered around the origin and a radius of 1, because then you only have to check the absolute value of 'sphereRayIntersection' to see whether your results are plausible).
As soon as you're sure these calculations are correct you can go on and calculate the normal like you do (the normalizition is unnecessary because the division by the radius already does that).
For simplicity you should just use the absolute value of the dot-product between light-vector and normal. If that works and you don't like it, it should be easy to take the signed value, but for debugging it just takes away one possible error source.
Now to your color calculation, which looks quite odd to me, even assuming 'lightCoef' is correct:
Material thisMaterial = thisLight.getMaterial();                    float lightR = thisMaterial.getR() * lightCoef;float lightG = thisMaterial.getG() * lightCoef;float lightB = thisMaterial.getB() * lightCoef;                                        // Add the effect of the light to the color of the sphereMaterial sphereColor = thisSphere.getMaterial();             r = ((sphereColor.getR() * lightR) / 255 );g = ((sphereColor.getG() * lightG) / 255);b = ((sphereColor.getB() * lightB) / 255);

The divisions by 255 indicate that colors stored in the material are ints in [0,255]. If they really are that color calculation for 'r', 'g' and 'b' won't work as expected. Convert all the colors you use into [0,1] as floats. The reason for that:
Assume you have a white light shining on a white surface. With the int-colors in [0,255] that makes the last three lines:
r = (255 * 255 * lightCoef) / 255 = 255² * lightCoef / 255 = 255* lightCoef

Same for the 'g' and 'b' values.
Now using float color values in [0,1]:
r = (1.f*1.f*lightCoef) = lightCoef

If your color values are not ints in [0,255], then just ignore what I just said. But then I don't understand your last division by 255.

Hope that helped

PS: you should move the calculation of 'sphereRayIntersection' and 'normal' out of the for-loop that iterates over the lights, because they won't change anyway. Not important yet, still wanted to say that.

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Thank you all for your help.

phantomus: The light was to bright, so thanks for pointing me into that direction. That helped me find my error in the code.

matches81: Thanks for the help :)

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