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Need to get A+ on the next math midterm

  1. Oct 18, 2011 #1
    Hello everybody. After getting back my last math midterm I feel like a damn orangutan who's one wooden box short of getting that dangling banana. My question is, what are your strategies for acing your math midterms? I understand most of the material and I do study fastidiously but somehow that I manage to screw up on some questions by way of really dumb mistakes. As well, there's always the dreaded "not covered in lecture or mention on the homework assignments and practice midterms" questions that involve some novel application of the concepts we learned on the midterms too. So what are some of your innovative study and reviewing strategies? As well, what are some of the secrets to flawless performance on test? Oh yeah, mynext midterm's probably gonna be on Chain rule, directional derivatives, gradient, Max/Min and Lagrange's Method.

    Thanks all for their wisdom.
    Peace.
     
  2. jcsd
  3. Oct 18, 2011 #2

    chiro

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    I currently am doing a subject on DE's (differential equations) and he was responding to the mid session test which apparently a significant number of people didn't pass.

    He said that the most important thing is to understand the ideas presented in lectures. If you understand those you can translate those problems to other problems and even formulate your own results if this is the case.

    To me, this is what mathematics is all about. It's about knowing the real core of what you are doing and using that to create new methods or to solve new problems.

    If you know the why, then back it up with the how by doing as much work as you need to.

    In terms of flawless performance, I can't really give any advice on that, but a common sense solution tells me to first understand the ideas and then get some practice in applying them. I don't think there is any safe way to get full marks, but if you understand what you are doing, you should get a decent score.

    Ask yourself "What's the point of all this?" and take it from there.
     
  4. Oct 18, 2011 #3
    Haha yeah... the point is to understand, I mean its about what you learn not the marks after all. But at the same time, it's gotta look good on that transcript for the employers.

    I've made it my personal goal to get an A+ in math and I'm working towards that. It's not that the material itself is difficult, in fact it's quite simple but damn I gotta somehow convince my prof that I know >90% of the material somehow.
     
  5. Oct 18, 2011 #4
    Do tons of practice problems. Find problems in your book that don't look familiar and try to solve them without using any references.
     
  6. Oct 19, 2011 #5
    Once you are familiar with an approach, do as kylem suggested. I would also suggest once you are familiar with the problems in question, try to find another way to solve them on your own. It kind of helps with the understanding, etc...
     
  7. Oct 19, 2011 #6
    Learn the material inside out. Ex, suppose the current topic is taxonomy in a microbiology course, find a microbiology textbook, look up taxanomy in the index and learn the entire concept. Then compare your notes to the prof's notes and take it from there
     
  8. Oct 19, 2011 #7

    mathwonk

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    when i needed an A in a graduate real analysis class as a senior i did several things. I "chose " a teacher i had taken before and knew he favored asking straightforward questions like: prove the radon nikodym theorem, or some such.

    then i went to every class to be sure i got all the material covered there and heard any tips about what we should certainly know.

    then i re copied all the notes and basically memorized the proof of every theorem proved in class.

    then i bought at least three additional books, Munroe, Halmos, Riesz - Nagy, (there was no official text for the class) and read those as well, for fuller understanding and coverage of material not mentioned in class.

    then i went to the house library and researched copies of old exams by this professor from previous years and worked all of them in detail.

    then i took the exam and basically got 100% (they did not return the exams). at least one theorem asked there was not presented in class but was in the book of halmos i had read. still some brilliant kid finished and walked out almost 30 minutes before me. but i got my A.

    In a class with a more demanding professor who asks new material and solutions of original proofs and problems, you need even more understanding and ability to create proofs. Then you must attempt to prove the theorems yourself, not just memorize the ones in the books and course, and branch out and try to think of ways to use the ideas in the course to do other things. Work as many hard exercises as possible. Talk to people who understand the topic and ask for their insight. explain the material to others. Study with another student in the class at about your level of ability or a little higher. Then try to relax right beforehand.
     
    Last edited: Oct 19, 2011
  9. Oct 19, 2011 #8

    mathwonk

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    By the way the basic facts in that subject you are studying are these:

    1) know the definition of the total derivative (differential) as a linear transformation.

    2) know the definition of the directional derivatiove, and know that that the directional derivative can exist in every direction while the function still may not have a total derivative,

    3) know that if the total derivative does exist, then the directional derivative in every direction does also, and i fact the directional derivative in the direction v is just the value of the total derivative (as linear map) evaluated at v.

    4) know that if the directional derivatives exists in the axis directions (partial derivatives) and exist on a whole open neighborhood of the point, and if the partial derivatives are continuous at the point, then also the total derivative does exist there and is represented by the matrix of partial derivatives.

    The moral is the total derivative is the conceptual version of the derivative and the partial derivatives are a computational tool enabling you to calculate the total derivative.

    5) The chain rule says that the total derivative of a composition of two differentiable maps is the composition of their total derivatives, as linear maps, and its matrix of partials is the matrix product of the matrices of partials of the two components mappings.

    6) The gradient is nothing but the derivative in the special case of a real valued function, hence is represented by a one rowed matrix of partials.

    7) In case a differentiable real valued function has a local extremum when restricted to a curve or line in it domain, the derivative will equal zero in that direction. If there is a local extremum for the function itelf at the point then all partials and all directional derivatives will be zero there. and thus the gradient matrix or gradient vector, will equal zero there.

    8) A continuous function defined on a closed bounded set will have a global maximum and minimum there, either on the boundary of the set, or at an interior point where the gradient is either zero or does not exist.

    9) E.g. to find the max of function defined on a disc, one checks the points inside the disc where he gradient is zero, and then one restricts the function to the boundary circle, reparametrizing it by polar coordinates and again finds the points where he one variable derivative is zero.

    10) Another way to approach the previous problem of examining the values on the boundary, is to find the gradient of the original function and look for points of the boundary where that graident is perpendicular to the boundary, i.e. parallel to the radius of the boundary circle.

    11) More generally the explanation for the trick mentioned in the previous no.#10, is called La Grange's method. It is the trivial observation that the derivative of the restriction of the map, is the restriction of the derivative. Thus the derivative of the restriction of f to the boundary circle is the restriction to the boundary circle of the gradient. Now the gradient acts by dot product, hence its restriction to the boundary circle is zero precisely when it is perpendicular to the boundary circle. In particular the normal, or perpendicular, vector to the surface or curve defined by g=0, is the gradient of g, since the derivative of the g must be zero when restricted to a set where g is constant.

    Repeat" basic fact: the gradient of g is perpendicular to the set where g is constant.

    12) By the previous reasoning, the method of lagrange is often taught in the following mindless way: to maximize the restriction of a function f to the surface or curve defined by g = 0, find the points of g=0 where the gradient of f is perpendicular to the surface g=0. I.e. find the points of g=0 where the gradient of f is parallel to the gradient of g.

    I.e. solve g=0, and gradg = c.gradf, for some scalar c, called the "lagrange multiplier". This gives a name, the multiplier, to the least meaningful element of the whole procedure, namely the unimportant scalar c. It is very difficult to give good example questions for this subject, and most of them are solvable without finding the scalar c.


    That's about all I recall at the moment about your question. good luck.
     
  10. Oct 19, 2011 #9

    mathwonk

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    e.g. to maximize a linear function like f(x,y) = x+y on an ellipse, find the points of the ellipse where the gradient (1,1) of f is perpendicular to the boundary of the ellipse, i.e. to the tangent line.

    If the ellipse is defined by say x^2 + 4y^2 = 1, then this is g=0 where g(x,y) = x^2 + 4y^2 -1, and gradg = (2x,8y), so look for points (x,y) on the ellipse where (x,4y) is parallel to (1,1), i.e. where x = 4y.

    Substituting x = 4y in the equation of the ellipse gives 16y^2 + 4y^2 = 1, or 20y^2 = 1, or y^2 = 1/20.

    Please check this or see where i screwed this up. this is the sort of trivial question they ask in mickey mouse books like hass weir and thomas.

    Loomis and Sternberg give the following example:

    maximize the surface area A = 2(xy+yz+xz) of a rectangular parallelepiped with sides x,y,z, subject to fixed volume V = xyz.

    The method says the minimum will be a

    "critical point of the function (A-cV) for some scalar c, and setting the gradient of this function equal to zero, gives:

    2(y+z) - cyz = 0 = 2(x+z) - cxz = 2(x+y) - cxy. These imply x=y=z.

    Then the constraint xyz = V implies x=y=z = V^(1/3)."
     
    Last edited: Oct 19, 2011
  11. Oct 19, 2011 #10

    mathwonk

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    no response to a complete survey of the subject?
     
  12. Oct 20, 2011 #11
    Whoa there haha... thanks for the advice. Though you stress absolute devotion, I'm more of the lazy work less for more type . Oh well, guess it's time to crack open that textbook and do the one thing I look forward to more than getting ****... math problems!
     
    Last edited: Oct 20, 2011
  13. Oct 20, 2011 #12
    And uh, oh my Linear Algebra class is next semester... lol, but thank you for the insights regardless.
     
    Last edited: Oct 20, 2011
  14. Oct 20, 2011 #13
    what class are you summarizing mathwonk? I plan to revisit this page when i take whatever course you are talking about
     
  15. Oct 20, 2011 #14
    Advanced Calculus/Vector Calculus, which I'm currently taking.

    Thanks for taking the time to write all that out Wonk!
     
  16. Oct 20, 2011 #15
    Great post Mathwonk! Thanks for that. :)
     
  17. Oct 21, 2011 #16

    mathwonk

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    here are my notes from 20 years ago on the meaning and definition of the total differential. these are some of the numerous sets of notes i have never yet posted online.
     

    Attached Files:

  18. Oct 22, 2011 #17

    mathwonk

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    Has anyone read my pdf notes yet? Do they help? Herp, I gather from you remark that your class did not even define the total derivative, or at least not correctly.

    A homogeneous linear map is a function L such that L(x+y) = L(x) + L(y), and L(cx) = cL(x), for all vectors x,y and all scalars c.

    The main fact about the derivative is that it is a linear map. i.e.the derivative of f at p is the linear map taking x to f'(p).x, not the number f'(p).

    This linear map called df(p) is characterized as the best linear approximation to the map

    called ∆f(p), taking x to f(p+x)-f(p).




    This is not so important in one variable but in several variable it is crucial, since then the numbers representing the derivative form a gradient vector or a matrix, and this makes the chain rule rather complicated, if you do not understand the linear map version.
    This stuff is usually taught several times from several points of view. Presumably you are in the sophomore junior level first encounter with several variables.

    My notes were from my senior level math major non honors, advanced calculus course.

    Amazingly by that time some people have actually already had a "linear algebra" course, but not one in which linear maps were even mentioned.

    Linear algebra is a prerequisite for understanding calculus, especially several variable calc, not the other way around, but courses are taught in a time honored backwards way that ignores the insights of mathematics gained over the past 80 years.


    of course your remarks about hard work also explain clearly why you are not getting the grade you desire.
     
    Last edited: Oct 22, 2011
  19. Oct 22, 2011 #18

    mathwonk

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    posting here is so educational. you start out assuming people want useful advice. then you give it, and it is predictably that they should work hard, and you give a prescription for exactly how to go about it. then you find out some of them want to believe that there is some witchcraft that will make them knowledgeable by turning around three times and spitting on their ankles or some such, and you know you are talking to fences, or something with no ears. hellooooo???? I am wasting my time here.
     
  20. Oct 22, 2011 #19
    ^You are not wasting your time. I thought your notes on the derivative were helpful. In fact I wondered why you don't put all your notes online. Many of the notes off your webpage have helped me understand ideas in algebra.
     
  21. Oct 23, 2011 #20
    Mathwonk your notes weren't a waste of time nor do you waste time. It would be great if the world had more people like you who were passionate about their field of interest and wanted to share that passion with others.
     
    Last edited: Oct 23, 2011
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