ucla math 131a homework solutions

ucla math 131a homework solutions

Topic: UCLA Math 131A Homework Solutions

1. Introduction

The course is an introduction to differential and partial differential equations. Specifically, the course covers fundamental solution of Laplace’s equation. And yet each section’s homework is given separately, and we are in produce many homeworks in this quarter. So I decide to record each homework’s solutions, and may be the midterm. And I believe may be some students next year will take this class and may need a reference. And that’s what I am doing right now. This may help following students concentrate on this class and get high grade. Also, may help next TA. I have been a TA of this class for nearly a year, so if I am still a TA next year, I will know what the students will think and what the TA need to do. And I think this is also a good preparation for my final (be a good TA). And that’s all. The source has been recorded in your account. Or you can just search UCLA Math with Professor Thomas. By the way, he used to be a super great mathematician. And I find he have a Wikipedia page. I will have a “click” in the hot summer day. By the way, it is very easy to check if a man is a great mathematician: 【”】If we can find his picture with a pipe in his mouth, then this man must be a great mathematician with no exception; And if a great mathematician with a pipe in his mouth, then his picture must be used in some advertisement for some sorts.【”】 Now let’s draw Euler with a pipe in his mouth. Ha! It’s funny! But don’t try to draw, because it’s non-Euler and of course, nonsense! Professor P. Kevin’s researsh field is in differential geometry, I believe in 90%. Well, you can check in yourself. And he likes hiking and he have his university page link in the GPS. He used to be a respected public figure I think. Ah! and his birthday in the year of 1991 is different between some guys. May be I will change something after the midterm. Anyway leave these junks, he is a really good professor, the best in the department. And in the midterm, his mean score of his three classes is 6x out of 90. Ha, I know x. See? It may easy to guess how many questions in the midterm.

2. Problem Set 1 Solutions

The first thing to establish within Problem Set 1 solutions, and more generally when solving separable differential equations, is recognizing that we can factor the derivatives in the way that they are presented in the problem above and assign them to differentials on opposite sides of the equation. These are two key and related skills. It allows us to apply the reason for assigning the differentials really on in the problem and later results in applying calculus operations to separate the variables. We can assign the variable of the two functions on either side of the differential equation, say if x is with dx, and we have something like g(x) dx equals to f(y) dy, then with antiderivatives, essentially we are integrating both sides directly and with respect to different variables, and then solve for y in terms of x, or vice versa. I intentionally offer an alternative strategy for solving what turns out to be a rather simple separable differential equation in order to highlight a common mistake. The correct intuition for each step of the way is not always obvious. I wanted to communicate these strategies by way of drawing connections to more general heuristics when solving separable differential equations. In writing these solutions to Problem Set 1, then, I aimed to weave these alternative methods and the related narrative with concise explanations, modeling a step-by-step approach for a variety of audiences. Every sentence and every step serve to set up, provide context for, or give motivation for the calculus operations, the solutions, and the process as a whole. By doing so, the continuous narrative and overall structure of the work not only tie together division of the separate variables but also show why this procedure is justified.

3. Problem Set 2 Solutions

Case 2: If Y = X – μ, then Y ∼ N(0,σ^2) – that is, the random variable Y has the Gaussian distribution with the standard parameter σ^2 and the graph is shifted to the left by μ. Finally, the parameter μ enables us to locate the center of the centuries-old bell curve. For instance, a smaller parameter μ shifts the curve left and a larger parameter shifts the curve right. Also, the variance σ^2 helps demonstrate the dispersion and the flatter the curve, the bigger the variation. These relations are used in applying integration method over the domain of random variables in performing different calculations such as computing the expected value. It reveals that μ is a measure of the location of the peak whereas σ^2 is a measure of the spread of the Gaussian distribution curve.

Case 1: If X ∼ N(μ,σ^2), then X follows the Gaussian distribution with the mean μ and the variance σ^2. Each value in the domain is stretched vertically by σ and the graph is shifted to the right by μ.

We can identify or parameterize the Gaussian distribution with two cases – shift the graph to the negative axis by μ and stretch or shrink the graph with the square root of the parameter σ.

Assume the standard Gaussian distribution X ∼ N(0,1), which means its mean is 0 and its parameter μ = 0.

X ∼ N(μ,σ^2) = (1/√(2πσ^2)) * e^(-1/2((x-μ)/σ)^2)

1. Recall that the Gaussian distribution of a random variable X and the parameters μ and σ^2.

Assuming t is a constant, the given problem can be rewritten as ∫e−txdμ = N for some unknown constant factor N. The given equation also indicates that the standards of the Gaussian distribution must be the square root of the parameter μ.

4. Problem Set 3 Solutions

The domain for f is R, and the range for f is [0, oo). For any y >= 0, by letting x such that f(x) = y, we have e^x = y, so x = ln(y). This gives evidence that f(x) = e^x is surjective. This completes the proof that f is bijective. Because f is bijective, there is a unique function g that is the inverse of f. We will call this function g(x) = ln(x). By the definition of the inverse function, g(f(x)) = x for all x. We have g(f(x)) = g(e^x) = ln(e^x) = x for all x >= 0. Since x = f(x) = e^x > 0, we see that g(x) = ln(x) is the inverse of f. Therefore, we have shown that f(x) = e^x has an inverse function g(x) = ln(x). Also, we know that f(x) = e^x and g(x) = ln(x) are both surjective and injective because they have inverse functions. This completes the whole proof. We have shown a function f is bijective if and only if it has an inverse function. In other words, for any function f, there exists an inverse function if and only if f is bijective.

5. Conclusion

I am not sure how much I improved from the previous exam but I have learned a lot by completing this homework. First of all, my undergraduate studies were focused on engineering. However, after I started to work in a financial institute, I had to work with lots of stocks and I wanted to improve my skills in math since my undergraduate studies did not cover pure maths. I had taken mat 125A with Dr. David Marker at my first semester. The class was hard but Dr. Marker was an inspiration in writing rigorous proof and his lecture was one of the most fruitful lecture that I had in my life. My proof techniques in higher math has improved a lot and after completing this homework, I am not only confident with my understanding of the basic concepts of real analysis but I am also looking forward to take quals and applying math to solve real world problems. I will actively participate in the email chains that professor opens up for questions. I will ask a lot of questions because I believe the first step of learning a subject is to ask a question every time when a statement or a concept is introduced. By doing so, I will also keep every other student in the class to be active in the forum and hopefully, we can also help each other. I will also visit my professor’s office hour regularly and I will act on any comments that I did not include in my solutions. I will be open to suggestions and critics and I believe that through the process of constantly improving my own understanding of math can help me to become a better mathematician every day. Last, I have learned that math is a subject to train our critical thinking. Every single definition and every single theorem has its own nuance and has its own rules. As we journey through this amazing subject, we train our mind to see through the foginess and we train our heart to be patient when looking for a solution. And for me, this is the most beautiful thing that I have always been looking forward to. Math is a place to create and a place to be free with one’s creation. Math is an exploration for the unknown and a long lasting intellectual journey that is going to be huge fun.