Revisting the initial value problem of \[ (z)\\ (z)\\ = F(z;z_0) (z_0)\\ (z_0) \\ \] The question is to solve this subject to the initial condition of when power is launched only into the mode a at z₀ = 0. This means $(0)=0$ and $(0)\neq 0$. \[ F(z;z_0)= {\beta_c}e^{i \delta (z+z_0)} & }{\beta_c}sin\beta_c(z-z_0)e^{i \delta (z+z_0)}\\ }{\beta_c}sin\beta_c(z-z_0)e^{-i \delta (z+z_0)} &{\beta_c}e^{-i \delta (z+z_0)} \\ \] Subject to our initial conditions then \[ (z)\\ (z)\\ = F(z;0) (0)\\ 0 \\ \] Carrying the matrix multiplication step by step \[ (z)\\ (z)\\ = {\beta_c}e^{i \delta (z)} & }{\beta_c}sin\beta_c(z)e^{i \delta (z)}\\ }{\beta_c}sin\beta_c(z)e^{-i \delta (z)} &{\beta_c}e^{-i \delta (z)} \\ (0)\\ 0 \\ \] \[ (z)\\ (z)\\ = (0) \beta_c cos \beta_c(z)-i\delta sin \beta_c(z)}{\beta_c}e^{i \delta (z)} & 0\\ (0) }{\beta_c}sin\beta_c(z)e^{-i \delta (z)} & 0\\ \] This leaves us then with \[(z)=(0)\left( cos \beta_c z -{\beta_c} sin \beta_c z \right) e^{i \delta z} \] \[(z)=(0)\left(}{\beta_c} sin \beta_c z \right) e^{-i \delta z} \] Looking back at my slides I wrote that $(z)=(0)\left(}{\beta_c} sin \beta_c z \right) e^{-i \delta z} $, which given my initial conditions would be wrong since $(0)=0$. Thankfully, this error doesn’t propagate through the rest of the presentation as that error only shows up on slide 8 and the rest of the figures I made are still accurate. I want to point out another error I made in the paper in regards to figure 1 here. The error is that in the figure in the paper, I wrote the same error for the power, corresponding to the blue line in figure one. I wrote ${P_b(o)}$, when in fact it should be ${P_a(o)}$, It was corrected for the presentation and here.

Coupled-mode theory is concerned with coupling spatial modes of differing polarizations, distributions, or both. To understand codirectional coupling it is useful to have an understand of background material that builds to codirectional coupling. First, consider coupling normal modes in a single waveguide that is affected by a perturbation. Such case is single-waveguide mode coupling. The perturbation in question is spatially dependent and is represented as ΔP(r), a perturbing polarization. Consider the following Maxwell’s equations \[ \nabla \times E=i \omega \mu_0 H\] \[\nabla \times H=-i \omega \epsilon E-i \omega \Delta P\] Consider two sets of fields (E₁, H₁) and (E₂, H₂), they satisfy the Lorentz reciprocity theorem give by ∇ ⋅ (E₁×H₂*+E₂*×H₁) = −iω(E₁⋅ΔP₂*−E₂*⋅ΔP₁) For ΔP₁ = ΔP and ΔP₂ = 0 and integrating over the result for the cross section of the waveguide in question, we get \[ {\partial z} A_{\nu}(z)e^{i\left(-\right)} z=i \omega e^{-i z} ^{\infty} ^{\infty} E_{\mu}^{*} \cdot \Delta P dxdy \] Evoking orthonormality, we can get the coupled-mode equation \[ \pm }{\partial z} =i \omega e^{-i z} ^{\infty} ^{\infty} E_{\nu}^{*} \cdot \Delta P dxdy \] The plus sign indicates forward propagating modes when Bν > 0 and the minus sign indicates a backward propagating mode with Bν < 0 Many applications are concerned with the coupling between two modes. This coupling between two modes can be within the same waveguide or can be coupled between two parallel waveguides. For a system where we are interested in coupling two modes for either the parallel waveguides case or within the same waveguide, the two modes are described by two amplitudes A and B. The coupled equations are given by \[\pm {\partial z}=i A+ i B e^{i(\beta_b-\beta_a)z} \] \[\pm {\partial z}=i B + i A e^{i(\beta_a-\beta_b)z} \] for the two modes we want to solve. The coupling coefficients are given as part of a matrix C = [Cνμ] and $=[_{\nu \mu}]$ They are given as \[C_{\nu \mu}=^{\infty} ^{\infty} \left( E_{\nu}^{*} \times H_{\mu} + E_{\mu} \times H_\nu \right) \cdot dxdy=c_{\mu \nu}^{*}\] \[ _{\nu \mu}=\omega ^{\infty} ^{\infty} E_{\nu}^{*} \cdot \Delta \cdot E_{\mu} dxdy\] where Δϵ is the perturbation applied to the system. We then simplify the math further by removing the self coupling terms in equations (5) and (6) by expressing our normal mode coefficients by \[A(z)=(z)e^{\pm i ^{z} (z)dz} \] \[B(z)=(z)e^{\pm i ^{z} (z)dz}\] For cases of interest, perturbation will either be independent of z or be a periodic funciton of z. This further reduces our coupling coefficients to \[ \pm }{\partial z}=i e^{i 2 \delta z} \] \[ \pm }{\partial z}=i e^{-i 2 \delta z}\] The parameter of 2δ is the phase mismatch between the two coupled modes. CODIRECTIONAL COUPLING Codirectional coupling is when the coupling of two propagating modes are in the same direction, over some length l, where βa and βb are both greater than zero. Coupling equations to be used are \[}{\partial z}=i e^{i 2 \delta z} \] \[ }{\partial z}=i e^{-i 2 \delta z}\] The general solution of this system is solved as an initial value problem in matrix form is given as $$ (z)\\ (z)\\ = F(z;z_0) (z_0)\\ (z_0)\\ $$ Where F(z; z₀) is the forward coupling matrix and it relates field amplitudes at an intial value of z₀ to those at z. For the most simple case when power is launched into only mode a at z = 0 , giving $(0)=0$. With z = 0 \[(z)=(0)\left( cos \beta_c z -{\beta_c} sin \beta_c z \right) e^{i \delta z} \] \[(z)=(0)\left(}{\beta_c} sin \beta_c z \right) e^{-i \delta z} \] where $\beta_c=\left( +\delta^2\right)^{2}$ Then looking at the power of the two modes when they are completely phase matched, that is when δ = 0 \[{P_a(0)}=|(z)}{(0)}|^2=cos^2 \beta_c z \] \[{P_a(0)}=|(z)}{(0)}|^2=sin^2 \beta_c z\] where κab = κba* Define the coupling efficiency for a length l as $\eta=|^2}{\beta_c ^2} sin^2 \beta_c z$ as well as the coupling length $l_c={2 \beta_c}$. I have plotted the phase matching condition of δ = 0 in figure 1. It is seen that we can only have complete power transfer when we are phase matched between the two coupled modes.

INTRODUCTION Hedrick Casimir first introduced the idea of the vacuum field in 1948. Interest started when Casimir and Polder earlier in 1948 described the interaction between q perfectly conducting plate and an atom by using the Van Der Waals-London forces, then correcting it for retardation effects in the limit of large distances. Then the next issue was to show the effects between the interaction between two perfectly conducting plates using this Van Der Waals retarded interaction. It is useful to understand the properties of the paper that lead to the Casimir effect. Casimir considers a cubic cavity of a volume L³ bounded by perfectly conducting square plate with side L be parallel to the XY face and investigates the situation where plate is at a small distance from XY face and situation where the distance between is a distance ${2}$ [1]. Looking at the resonant frequencies of this system, $ {2} \sum \hbar \omega $, the sum being over all the resonant frequencies, notice the problem that this goes to infinity and is not physical. The differences between the frequencies is well defined for the two situations. Labeling the situations of small interaction and long interaction, differences are \[{2}(\sum \hbar \omega )_I -{2}(\sum \hbar \omega)_{II}\] and the value given is the interaction between the plate and the XY face. Cavity will have dimensions \[0 \leq X \leq L, 0 \leq Y \leq L,0 \leq Z \leq a \] with the wavenumbers given by \[ k_i={L}, k_z={a} \] where i=x,y and a is the small distance to be used between plate and XY face. The total K is given by $K==$. For every kx, ky, kz, there is two be two standing waves unless ni is zero, which case there is one standing wave. kx, ky can be regarded as continuous variables for very large L, then in polar coordinates \[ {2} \sum \hbar \omega = {\pi^2 2} ^{\infty}^{\infty}{a^2}+x^2} x dx\] where the notation of (0)1 means the n=0 term is to be multiplied by ${2}$. When a becomes large, the summation can be turned into an integral. Now the interaction energy between plates for the situations above is \[{2} \sum \hbar \omega = {\pi^2 2} ^{\infty}^{\infty}{a^2}+x^2} x dx-^{\infty} ^{\infty} x dx({\pi}) \] For sake of brevity, the result is given for a force per cm^2 \[F={240 a^4} \] Casimir concluded that there exists a force between the two plates which is independent of material and interpreted as a zero point pressure of electromagnetic waves, this result is now known as the Casimir effect. This result is hard to understand without a knowledge of quantum field theory or quantum electrodynamics. A brief anecdote on QFT will be given to try to understand the Casimir effect and the vacuum field. Quantum Field Theory states that all the fundamental fields are quantized at each and every point in space. A field can be seen as space being filled with interconnected vibrating balls and springs. Vibrations in a field then propagate and are governed by the appropriate wave equation for the field in question. The second quantization of quantum field theory or canonical quantization requires that each ball and spring combination be quantized. The field at each point in space is a simple harmonic oscillator and its quantization puts a quantum harmonic oscillator at each point. The vacuum we assume to be empty is not so empty after all. As it would be, the vacuum has all the properties a particle may have such as spin, energy but on average these cancel out. If these properties cancel shouldn’t the vacuum be empty? The exception is the vacuum energy. Since the vacuum field’s quantization is a quantum oscillator, we know that the ground state of a quantum harmonic oscillator is $E_0={2} \hbar \omega$. For the vacuum field, this is called the zero point energy. This is the same energy Casimir uses when calculating the force between the two plates. [2]

PROBLEM 1 Part C Plots done using mathematica, figure 1 using equation 8.25 excluding the ϕ dependence and using the appropriate l values for each mode.

Graphene has garnered a lot of interest for its unique electronic properties and its potential applications. Graphene is a two-dimensional form of carbon in which atoms are arranged in a honeycomb lattice. It is a zero-gap semiconductor, this allows the ability to dope graphene to have high concentrations by applying a voltage. This property of electric gating is also used to control conductivity and the dielectric constant of graphene. There is extensive research on the many properties of this monolayer of carbon but I will focus on conductivty, surface waves supported by graphene, and research into applications with optical modulators. There are two absorption properties that are involved in light-graphene interaction, interband and intraband absorption for small signals. This is described by the complex conductivity, \[\sigma_g=(\omega,\mu_c,\Gamma,T)+(\omega,\mu_c,\Gamma,T)\], where ω is the angular frequency of light, μc is the chemical potential, Γ is scattering rate, and T is the temperature [1].Whats interesting is that the chemical potential can be tuned with electrical gating, thus the conductivity can be controlled by the same process. Chemical potential also controls what absorption process is occurring.Interband corresponding to absorption from the valence band to conduction band and intraband absoprtion corresponding to absoprtion from a semiconductor like optical property to a metal like optical property [2]. For incoming light with energy ℏω, interband absoprtion dominates when $\mu_c < {2}$ and intraband dominates when $\mu_c > {2}$ and it is theoretically predicted that intraband absoprtion would be dominate when $\mu_c \approx {1.67}$ [2]. Zhaolin Lu et al investigates graphene conductivty at T = 300K with a scattering rate ℏΓ = 5meV. Figure 1 plots real and imaginary parts of conductivity vs chemical potential and the dielectric constant as a funciton of chemical potential. Notice the sensitivity of conductivity with little change in chemical potential. Here $=1-{i\omega\epsilon_0\Delta}$ where Δ = 0.7nm is the effective thickness of graphene. Notice the dip in in the magnitude of the effective dielectric constant, it is near zero at μt. Physically this is the marking of the transition from a dielectric type graphene to a metal type graphene. The transition chemical potential is μt = 0.515eV for this experiment. Theoretical value from [1] predicts μt = 0.479eV which has about 7 % error. Since conductivity can be tuned by electrical gating so too can the effective dielectric constant.

This paper describes the methods to calbrate LRO's Diviner Lunar Radiometer Experiment. Like many radiometers, Diviner is sensitive to instrument temperature changes along the orbit of LRO. Regularly executed calibration blocks include instrument pointings to space and towards internal blackbodies at a known temperature. Data from these blocks serve to determine current offsets and current DN to radiance conversion value. A ground calibration campaign served to determine conversion tables over temperature.

Single-molecule Förster Resonance Energy Transfer (smFRET) allows probing intermolecular interactions and conformational changes in biomacromolecules, and represents an invaluable tool for studying cellular processes at the molecular scale. smFRET experiments can detect the distance between two fluorescent labels (donor and acceptor) in the 3-10 nm range. In the commonly employed confocal geometry, molecules are free to diffuse in solution. When a molecule traverses the excitation volume, it emits a burst of photons, which can be detected by single-photon avalanche diode (SPAD) detectors. The intensities of donor and acceptor fluorescence can then be related to the distance between the two fluorophores. While recent years have seen a growing number of contributions proposing improvements or new techniques in smFRET data analysis, rarely have those publications been accompanied by software implementation. In particular, despite the widespread application of smFRET, no complete software package for smFRET burst analysis is freely available to date. In this paper, we introduce FRETBursts, an open source software for analysis of freely-diffusing smFRET data. FRETBursts allows executing all the fundamental steps of smFRET bursts analysis using state-of-the-art as well as novel techniques, while providing an open, robust and well-documented implementation. Therefore, FRETBursts represents an ideal platform for comparison and development of new methods in burst analysis. We employ modern software engineering principles in order to minimize bugs and facilitate long-term maintainability. Furthermore, we place a strong focus on reproducibility by relying on Jupyter notebooks for FRETBursts execution. Notebooks are executable documents capturing all the steps of the analysis (including data files, input parameters, and results) and can be easily shared to replicate complete smFRET analyzes. Notebooks allow beginners to execute complex workflows and advanced users to customize the analysis for their own needs. By bundling analysis description, code and results in a single document, FRETBursts allows to seamless share analysis workflows and results, encourages reproducibility and facilitates collaboration among researchers in the single-molecule community.

_A 5-minute video demonstration of this paper is available at this YouTube link._ PREAMBLE A variety of research on human cognition demonstrates that humans learn and communicate best when more than one processing system (e.g. visual, auditory, touch) is used. And, related research also shows that, no matter how technical the material, most humans also retain and process information best when they can put a narrative "story" to it. So, when considering the future of scholarly communication, we should be careful not to do blithely away with the linear narrative format that articles and books have followed for centuries: instead, we should enrich it. Much more than text is used to communicate in Science. Figures, which include images, diagrams, graphs, charts, and more, have enriched scholarly articles since the time of Galileo, and ever-growing volumes of data underpin most scientific papers. When scientists communicate face-to-face, as in talks or small discussions, these figures are often the focus of the conversation. In the best discussions, scientists have the ability to manipulate the figures, and to access underlying data, in real-time, so as to test out various what-if scenarios, and to explain findings more clearly. THIS SHORT ARTICLE EXPLAINS—AND SHOWS WITH DEMONSTRATIONS—HOW SCHOLARLY "PAPERS" CAN MORPH INTO LONG-LASTING RICH RECORDS OF SCIENTIFIC DISCOURSE, enriched with deep data and code linkages, interactive figures, audio, video, and commenting.

First,second,and fourth sound were successfully found and plotted. Graphs show a lack of steepness in decay as sound modes approach Tλ, this could be due to refilling liquid helium later than recommended leaving less medium for the sound modes to propogate through. Scattering factor,n, was found to be n=1.239 ± .007 and porosity,P, was found to be 0.46 ± 0.02 close to the theoretical value of ≈40% porosity.

CHAPTER 8:HYPOTHESIS TESTING FOR POPULATION PROPORTIONS Testing a claim we calculated a 95% confidence interval for the true population proportion of UCLA students who travelled outside the US. \[(0.26,0.44)\] A 95% confidence interval of 26% to 44% means that - We are 95% confident that the true population proportion of UCLA students who travelled outside the US is between 26% and 44%. - 95% of random samples of size n = 100 will produce confidence intervals that contain the true population proportion. &nbsp; - The true population proportion,p, may be outside the interval,but we would expect it to be somehwat close to $$ - In our random sample of 100 students we had found that 35 of them have at some point in their lives travelled outside the US,$$= 0.35. - It is difficult to decide how close is close enough, or how far is too far, and this decision should not be made subjectively. HYPOTHESIS TESTING - In Statistics when testing claims we use an objective method called hypothesis testing - Given a sample proportion,  , and sample size, n, we can test claims about the population proportion, p. - We call these claims hypotheses - Our starting point, the status quo, is called the null hypothesis and the alternative claim is called the alternative hypothesis. - If our null hypothesis was that p = 0.35 and our sample yields  = 0.35, then the data are consistent with the null hypothesis, and we have no reason to not believe this hypothesis. - This doesn’t prove the hypothesis but we can say that the data support it. If our null hypothesis was different than p=0.35, lets say p=30 and our sample yields $$, then the data are not consistent with the hypothesis and we need to mke choices as to whethere this inconsistency is large enough to not believe the hypothesis. - If the inconsistency is significant, we reject the null hypothesis Example for hypothesis testing for one proportion Research conducted a few years ago showed that 35% of UCLA students had travelled outside the US. UCLA has recently implemented a new study abroad program and results of a new survey show that out of the 100 randomly sampled students 42 have travelled abroad. Is there significant evidence to suggest that the proportion of students at UCLA who have travelled abroad has increased after the implementation of the study abroad program? - population proportion used to 0.35 - new sample proportion is 0.42 - Testing the claim that the population proportion is now greater than 0.35 . - But is this difference statistically significant, i.e. are the data inconsistent enough? - We do a formal hypothesis test to answer this question. SETTING UP HYPOTHESES - Null hypothesis, denoted by Ho, specifies a population model parameter of interest and proposes a value for that parameter (p). \[H_0: p=0.35\] - Alternative hypothesis, denoted by HA, is the claim we are testing for.\[H_A :p > 0.35\] - Even though we are testing for the alternative hypothesis, we check to see whether or not the null hypothesis is plausible. - If the null hypothesis is not plausible, we reject the null hypothesis and conclude that there is sufficient evidence to support the alternative. If the null hypothesis is plausible, we fail to reject the null hypothesis and conclude that there isn’t sufficient evidence to support the alternative. Why do we check if Ho is plausible and not if HA is plausible? The same logic used in jury trials is used in statistical tests of hypothesis - We begin by assuming that the null hypothesis is true. - Next we consider whether the data are consistent with this hypothesis. - If they are, all we can do is retain the hypothesis we started with. If they are not, then like a jury, we ask whether they are unlikely beyond a reasonable doubt. &nbsp; - Hypothesis testing is very much like a court trial - this doesnt prove the hypothesis but we can say that the data supports it - H_0:Defendant is innocent H_A:Defendant is guility - We then present the evidence-collect data - Then we judge the evidence - “Could these data plausibly have happened by chance if the null hypothesis were true? - If they were very unlikely to have occurred, then the evidence raises more than a reasonable doubt in our minds about the null hypothesis. - Ultimately we must make a decision. How unlikely is unlikely? If the evidence is not strong enough to reject the presumption of innocent, the jury returns with a verdict of “not guilty” - The jury does not say that the defendant is innocent. - All it says is that there is not enough evidence to convict, to reject innocence. - The defendant may, in fact, be innocent, but the jury has no way to be sure. - Said statistically, we fail to reject the null hypothesis. - We never declare the null hypothesis to be true, because we simply do not know whether it’s true or not. - Therefore we never“accept the null hypothesis” How do we determine if H_0 is plausible? - In Statistics we can quantify our level of doubt. - How unlikely is it to get a random sample of 100 students where 42 have travelled abroad if in fact the true population proportion is 35%? - How unlikely is it to get a random sample of 100 students where 42 have travelled abroad if in fact the true population proportion is 35%? To answer this question we use the model proposed by the null hypothesis as a given and calculate the probability that the event we have witnessed could happen. Prob( Observed or more extreme outcome | H_0 true)=Prob($$ > 0.42 | p=0.35) - This probability quantifies exactly how suprised we are to see our results and is called the P-VALUE Calculating the p-value - First we calculate the test statistic(Z-score) - Test statistic used for hypothesis testing for proportions is a z-score \[ {SD}\] - Remember CLT: in calculating z-score we use the mean and SD by THE CLT, and observed value is \[ \approx N\left(mean=p,SD={n}}\right)\] P COMES FROM NULL HYPOTHESIS Check the Conditions for the CLT - Random and Independent: The sample is collected randomly and the trials are independent of each ot - Large Sample: - Sample has at least 10 successes, np ≥ 10 - at least 10 failures n(1 − p)≥10 - Large Population: If the sample is collected without replacement, then the population size is at least 10 times the sample size. Example of calculating p-value for study abroad stats We have that H₀ : p = 0.35 and H_A:p > 0.35 \[z-score=-p}{{n}}}={{100}}}=1.47\] and p-value\[Prob( > 0.42 \vert p=0.35=prob(z>1.47)=1-0.9292=0.0708\] Decision based on the p-value - When the data are consistent with the model from the null hypothesis, the p-value is high and we are unable to reject the null hypothesis. - In that case, we have to“retain” the null hypothesis. - We can’t claim to have proved it; instead we fail to reject the null hypothesis and conclude that the difference we observed between the null hypothesis (p = 0.35) and the observed outcome ($$ = 0.42) is due to natural sampling variability (or chance). &nbsp; - If the p-value is low enough, we reject the null hypothesis since what we observed would be very unlikely if in fact the null model was true. - We call such results statistically significant How low a p-value is low enough? - we compare the p value to a give α: if pvalue < α → Reject H_0 - There is sufficient evidence to suggest to H_A is plausible if p value > α → fail to reject H_0 - There is not sufficient evidence to suggest that HA is plausible. The difference we are seeing between the null model and the observed outcome is due to natural sampling variability. - When p-value is low, it indicates that obtaining the observed $$ or even a more extreme outcome is highly unlikely under the assumption that Ho is true, therefore we reject that assumption. α level is the complement of the confidence level. Remember: WHEN CONSTRUCTING CONFIDENCE INTERVALS IF A CONFIDENCE LEVEL IS NOT SPECIFIED USE 95 PERCENT CONFIDENCE - Since 1 − 0.95 = 0.05, if a α level is not specified use α= 0.05 Decision based on the p-value with a p-value of 0.0708, do we reject of fail to reject H_0? Since p value is greater than α than we fail to reject H_0 What does the conclusion of the hypothesis mean in context of the research question? Answer: The data do not provide convincing evidence to suggest that the true population proportion of UCLA students who have travelled outside the US has increased. Interpreting the p-value - A p-value is a conditional probability - the probability of the observed or a more extreme outcome given that the null hypothesis is true. Prob(observed or more extreme outcome | H_0 true) - the p value is NOT the probability that null hypothesis is true - It’s not even the conditional probability that the null hypothesis is true given the data The following is the correct interpretation of the p value? If in fact the true population proportion of UCLA students who have travelled outside the US is 0.35, the probability of getting a random sample where the sample proportion is 0.42 or higher would be 0.0708. - When testing for population proportions, there are three possible alternative hypotheses: - H_A: p<P_0 - H_A: p>p_0 - H_A:p ≠ p_0 - We decide on which alternative hypothesis to use based on we hypothesize (or what the wording of the question instructs as to hypothesize): - Smaller, less, decreased, fewer - Larger, greater, more, increased - Different, not equal to, changed - We DO NOT decide on which alternative hypothesis to use based on what the data suggest. - HA: parameter ≠ hypothesized value is known as a two-sided alternative because we are equally interested in deviations on either side of the null hypothesis value. - For two-sided alternatives, the p-value is the probability of deviating in either direction from the null hypothesis value. - The other two alternative hypotheses are called - A one-sided alternative focuses on deviations from the null hypothesis value in only one direction. - Thus, the p-value for one-sided alternatives is the probability of deviating only in the direction of the alternative away from the null hypothesis value. THERES MORE TO DO HERE BUT FOR NOW SKIP TO CHAPTER 9

Time between pulses of SL was found to be 3.069 × 10−5 ± 0.01 seconds, width of an SL pulse was found to be 1.925 ∗ 10−6 ± 0.04 seconds and Sonoluminescence was observed.

CHAPTER 5 REVIEW A RANDOM PHENOMENON is a situation where we know what outcomes could happen but we dont know in what order they did or will happen But we can calculate the probability with which each of those outcomes could occur. people are bad at identifying random samples so we rely on coin flips or random number tables to simulate randomness. Say if you want to simulate randomnes using a random number table and simulate rolling a die 10 times. In this example we choose a random line say 30, recall that for a die there 6 possible outcomes 1-6, so in the table eliminate any numbers greater 6. Read left to right and you get a series of numbers. This is an example of simulating randomness. RANDOMNESS TO PROBABILITY Say you have an itunes library of 1000 songs and there are 5 iron maiden songs in it. what is the probability that when you hit shuffle you get an iron maiden song? ${1000}=0.005$ DEFINITIONS OF PROBABILITY A FREQUENTIST: The probability of an event is its long-run relative frequency. Theoretical: we may not be able to predict which song we get each time we hit shuffle, but we know in the long run of the library, 5 out of 1000 will be iron maiden songs. Empirical: Listen to 100 songs on shuffle and 4 of them are iron maiden songs. empirical probability is then 4/100=0.04 a BAYESIAN: we wont use much in this class Definitions of Probability for any random phenomenon, each attempt is called a TRIAL and each trial generates an outcome. OUTCOME Each time you hit shuffle, thats a trial The songs that plays as a result of the shuffle is an outcome A SAMPLE SPACE is the collection of all possible OUTCOMES of a trial Sample space would be the 1000 itunes songs a comination of outcomes is called an EVENT for example, playing 2 iron maiden songs in a row SAMPLE SPACE a couple has two kids, what is the sample space for the gender of the kids? Well there are 4 possibilities S=BB,BG,GB,GG The probability that both will be boys is 0.25. INDEPENDENCE when thinking about what occurs with combinations of outcomes, things are simplifies if the individual trials are independent - This means that the outcome of one event doesnt influence the outcome or change the next outcome. - If the itunes shuffle is really random then each of the songs played are independent example for coing toss trial: Each coin toss outcome: Heads or tails probability: P(h)=P(t)=0.5 each time we flip we dont know which side will come up but in the long run, the probabilities will be 0.50 for each side. sample space: tossed once: s=h,t tossed twice:s=hh,tt,ht,th LAW OF LARGE NUMBERS - The law of large numbers says that the long run relative frequency of repeated indpendent events gets closer and closer to the true relative frequency as the number of trails increases. - if a coin is tossed many times, the overall percentage will settile down to about 50% as the umber of tosses increases - this is why probability is defined as a long run relative frequency event IMPOSSIBILITY TO A CERTAINTY probabilities must be between 0 and 1. 0 means impossible and 1 is certainty. COMPLIMENT RULE Set of ourcomes that are not in the event A is called the compliment of A, denoted AC. The probability of an event occuring is 1-Probability(it doesnt occur). \[Prob(A^C)=1-Prob(A)\] DISJOINT EVENTS Events that have no outcomes in common are called disjoint( or mutually exclusive). Examples: The outcome of a coing toss cant be a head and a tail. so disjoint student cant fail and pass an exam:disjoint ADDITION RULE for two disjoint events A and B (think like heads and tails for coins), the probability that one OR the other occurs is the sum of the two probabilies. \[Prob(A or B)=P(A)+P(B)\] MULTIPLICATION RULE For two independent events A and B, the probability that both A AND B occur is the product of the individual events together. \[Prob(A and B)=Prob(A)Prob(B)\] EXAMPLES In a multiple choice exam there are 5 questions and 4 choices for each question (a, b, c, d). Nancy has not studied for the exam at all, and decided to randomly guess the answers. What is the probability that she gets the first question right? 0.25 What is the probability that she gets the second question right? 0.25 What is the probability that the first question she gets right is the 5th question? if she get just the fifth one right that means the first 4 were wrong and the last one was right. so that is (0.75)⁴ * 0.25 = 0.0791 What is the probability that she gets all the questions right? (0.25)⁵ = 0.0010 What is the probability that the first question she gets at least one question right? They are asking for at least once question right so using the compliment rule, which is to say 1-probability(none right)=1-probability(all wrong)=1 − (0.75)⁵ = 0.7627 INDEPENDENT VS DISJOINT can two events be independent and disjoint? Remember independence means that the outcome of an event doesnt influence the outcome of another event. Disjoint means no outcomes in common like not being able to have heads or tails. If we know that the outcome of a coin toss was head, we know its not tails. So whether or not the outcome is tails it DEPENDS on whether or not the outcomes was heads. so that would be called disjoint and dependent. - disjoint events cannot be independent - since we know that disjoints have no outcomes in common, then we know if one occured, the other didnt. - just because two events can occur at the same time doesnt mean they need to be dependent. - two students can get an A in a class but they may not have studied together etc - if you did study together and both got As then your grade could be dependent EXAMPLES A middle school estimates that 20% of its students miss one day of school per semester due to sickness, 13% miss two days, and 5% miss three or more days. What is the probability that a student chosen at random doesn’t miss any days of school due to sickness? Not missing any school is the compliment of missing all days. so prob(nomisses)=1 − (0.20 + 0.13 + 0.05)=0.62 What is the probability that a student chosen at random misses no more than one day? prob(atmost1day)=Prob(nomisses)+prob(1miss)=0.82 If a parent has two kids at this middle school, what is the probability that neither will miss any school? prob(neithermissany)=(prob(nomiss))²=0.3844 We used the multiplication rule to calculate the probability in the previous example. a) What must be true about these kids to make that approach valid? independence. b) Do you think this assumption is reasonable? yes. what is the probability that one kid misses some school and the other doesn’t miss any? Prob(one misses some school)=prob(1 kid misses some school and the other doesnt)+prob(1st kid doesnt miss any school and the 2nd does)=0.38*0.62+0.62*0.38)=0.4712 GENERAL ADDITION RULE If A and B are disjoint,\[Prob(A or B)=prob(A)+prob(B)\] If A and B are not disjoint(independent) \[ prob (A or B)=Prob(A)+prob(B)-prob(A and B)\] EXAMPLE In a class where everyone’s native language is English, 70% of students speak Spanish, 45% speak French as a second language and 20% speak both. Assume that there are no students who speak another second language. What is the probability that a randomly selected students speaks Spanish or French? This is independent because its students who speak spanish or french, those dont effect each other. by the general addition rule then, prob(SpanishorFrench)=prob(spanish)+prob(french)−prob(both)=.70 + .45 − .20 = 0.95 CONDITIONAL PROBABILITY Bayes theorem: \[Prob(A\vert B)={Prob(B)}\] Say you have a chart that has smokers and there sex amd the total people were 1691. if people who smoke are 421 and female are 234 of the smokers a question could be, what is the probability that a randomly chosen smoker is female? This follows what bayes theorem is saying. probabilities of A given B. In this example we have the probability of find smokers given they are female form. so \[prob(F\vert S)={421/1691}=234/421\] this is something we could have guessed but it could get trickier. BACK TO MISSING SCHOOL EXAMPLE What is the probability that a randomly selected student speaks French given that they speak Spanish? prob(F|S)=Prob(FandS)/prob(S)=.20/.70 = 0.29 Check professors notes for burger example. GENERAL MULTIPLICATION RULE If A and B are independent, \[Prob(A and B)=Prob(A)*Prob(B)\] If A and B do not need to be independent, \[Prob( A and B)=Prob(A\vert B)*Prob(B)\] ONLY ASSUME INDEPENDENCE IF THE QUESTION SPECFIES IT OR YOU CHECKED FOR YOURSELF CHECKING FOR INDEPENDENCE USING CONDITIONAL PROBABILITIES if Prob(A)=Prob(A|B) then the events are independent. INDEPENDENT VS DISJOINT-RECAP if A and B are disjoint: - Prob(A and B)=0 - A and B are not independent If A and B are not disjoint: - Prob(A and B)>0 - A and B may or not be independent

Resonant frequencies measured at (0.69,1.93,2.92,3.69,4.31,4.82) Hz. Q factors varied from the sweeps and from measurements of alpha largely but this can be due to the effect of high amplitudes on the higher frequencies and the natural error in trying to measure the time decaying exponential of a wave that has been turned off from the amplifier.

CHAPTER 5 5.1 a)2,6,4,2,7,4,0,6,5,0 b)T,T,T,T,H, T,T,T,H,T c)No I only got 2 heads 5.10 7/4, −10% cant be real probabilities for 7/4 is larger than one and we cant have negative probabilities. 5.15 a)1/16 b)4/16 = 1/4 c)6/16 = 3/8 d)4/16 = 1/4 e)1/16 5.21 104/1858 = 5.6% 5.22 26/1858 = 1.4% 5.23 Were looking at the total numbers for Democrat or Republican, and since this is an or questions we add (689 + 469)/1858 = 62.3% 5.24 (593 + 530/1858 = 60.4% 5.25 Following the guided excersiceP(Liberal)=530/1858 = 28.5% P(Democrat)=689/1858 = 37.1% They arent mutually exclusive because you can see that there are people who are both liberal and democrat. P(LibandDem)=(306)/1858 = 16.5% If we dont subtract away the probability of people being democrat and liberal from that equation, you will be over counting. P(LiborDem)=(530 + 689 − 306)/1858 = 49.1% 5.33 a)P(Oddorlessthan3)=3/6 + 2/6 − 1/6 = 4/6 = 66.7% b)P(Oddorlessthan2)=3/6 + 1/6 − 1/6 = 50% 5.40 a)15/50 = 30% b)35/50 = 70% c)30/50=60%$\\ d)20/50=40\%$ e)1 because they are compliments. 5.44 a)1 − 0.23 − 0.41 = 0.36 b)0.41 + 0.36 = 0.77 c)0.41 + 0.23 = 0.64 d)a and c are complementary.16 or more mistakes mean 16 up to 30. 15 or fewer all values from 0 up to 15. If you add these together you get the whole experiment. 5.47 a)306/530 = 57.7% b)104/593 = 17.53% c)The highest percent is with liberal democrats. 5.57 Follow the guided exercise

INTRODUCTION The purpose of the lab was to find the normal modes of rectangular and cylindrical geometries and using that data of these frequencies obtain the speed of sound. Before the lab was started the first 33 modes for the rectangular box were found theoretically up to the (4, 0, 0) mode to compare to our results and the first 17 modes of the cylindrical up to the (4, 2, 0). The physics used to to find the resonant frequencies of the box come from seperation of variables of the pressure wave equation and the frequencies for the cylindrical geometry were found seperating variables again but in cylindrical coordinates to the pressure wave equation and examining the zeroes of the deriviatives of the bessel function. For practial purposes, c was taken to be $345 {s}$ to approximate resonant frequencies.

PROBLEM 1 Define the function we will be using for the forward differencing as \[u=ge^{ikx_j}\] \[}{}=iku\] and the forward differencing equation as \[\Delta_x^{'}f_j=-f_j}{\Delta}\] where are function f will be our functuon u. Putting this is we get \[\Delta_x^{'}u=}-ge^{ikx_j}}{\Delta}\] define xj + 1 = xj + Δ then the above equation becomes \[\Delta_x^{'}u=-ge^{ikx_j}}{\Delta}\] Factoring we get \[\Delta_x^{'}u=(e^{ik\Delta}-1)}{\Delta}\] and taylor expanding we get \[\approx(1+ik\Delta+{2}-1)}{\Delta}\] Then factoring \[=(1+{2})}{\Delta}\]. then \[\Delta_x^{'}u=}{}(1+{2})\] So \[\Delta_x^{'}=(1+{2})}{}\]

LAB QUESTIONS Question 1 The graph shows a linear relationship between at bats and runs. Its not perfectly linear but you can see that as the number of bats goes up the number of runs seem to go up as well. Can you make an accurate prediction? Not exactly but you would want the number of at bats to go up as the graph indicates your runs will go up too.

LAB QUESTIONS Question 1 The Unit of observation in this lab is births. There were 22 variables recorded.