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Calculus Books

  1. Jan 4, 2008 #1
    Does anyone here have any recomendations for Calculus books for someone just starting out?
     
  2. jcsd
  3. Jan 4, 2008 #2
    Calculus Made Easy

    Calculus by Stewart
     
  4. Jan 4, 2008 #3
    I would recommend Schaum's Outline: Beginning Calculus. You only need some algebra and basic trig to understand it
     
  5. Jan 4, 2008 #4

    HallsofIvy

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    The difficulty with Schaum's Outline is that it is strong on calculation and formulas, rather weak on theory. Unfortunately, it is especially easy to focus on formulas in calculus and not realize that you are not learning the theory.
     
  6. Jan 4, 2008 #5

    mathwonk

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    lectures on freshman calculus by cruse and granberg.
     
  7. Jan 4, 2008 #6

    stewartcs

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  8. Jan 4, 2008 #7
    What about Calculus for Dummies? I heard that it was a really good book, but I guess it depends on the reader. For those of you who may have read it, is there much assumed knowledge?
     
  9. Jan 4, 2008 #8

    mathwonk

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    if the shoe fits wear it, otherwise get a book for an intelligent person.
     
  10. Jan 4, 2008 #9

    stewartcs

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    You might be hard pressed to get someone here to actually admit that they read Calculus for Dummies! :rolleyes:
     
  11. Jan 4, 2008 #10

    symbolipoint

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    I strongly agree with HallsOfIvy's opinion about Shaum's Ouline books:

    After using about three of them, I found they did not really help enough. A good, old, standard Calculus book such as authored by Sallas & Hill, or Larson & Hostetler, or Anton would be much better.
     
  12. Jan 4, 2008 #11
    stewartcs said:

    You might be hard pressed to get someone here to actually admit that they read Calculus for Dummies!


    Lol, good point, but I've never done Calculus in my life, and I just realized that I'm going to need it for the major I want. I'm in my last year of High School, and I'm starting to regret that I didn't take Calculus. I'm okay in math, but I've never been exactly brilliant in it. I just need a good book that doesn't have much assumed knowledge. Also, I want a book with workable problems and things of that nature.

    Is it possible to teach yourself Calculus if you have to right materials? Or am I asking for a miracle?

    Thanks, I appreciate everyone's advice so far!
     
  13. Jan 4, 2008 #12

    mathwonk

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    go to the library and choose for yourself the book that appeals to you. but just because you are a novice does not mean you are a dummy. choose a book for the intelligent layperson or beginner. a dummy cannot learn calculus.
     
  14. Jan 4, 2008 #13

    mathwonk

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    calc 1

    here is a free beginning differential calculus course.

    Day One, 2200 sp 2000 Real Numbers
    Real numbers are in one one correspondence with points on the line, and are used especially to measure lengths. A real number is represented symbolically as a finite or infinite decimal, with the finite decimals being the ones ending in all zeroes. The finite decimals have another expansion ending in all 9’s, (except for the number zero = 0.000000..... which has only that one expansion).
    Whole numbers or integers, are those real numbers corresponding to decimals with all zeroes after the decimal point. Rational numbers, or quotients of whole numbers, are a larger subclass of real numbers, corresponding to decimals whose expansions eventually repeat after some point (e.g. 3.4762989898......., where the repeating part is the ......98.....). A real number whose decimal expansion never repeats nor terminates in all zeroes (a form of repeating) is an “irrational” real number, since it cannot be written as a quotient of two whole numbers. (One can see by looking carefully at the result of the division process that dividing two whole numbers always results in an eventually repeating decimal. As a student I had trouble proving this, but try a few examples.)
    Since calculators contain only a finite number of decimal places, they cannot represent most real numbers, and thus the answers they give to most questions involving real numbers are almost always (slightly) wrong. Since real numbers are so complicated to describe explicitly, it is not easy even to perform familiar operations on them such as adding them and multiplying them. Some form of limit process is necessary even for these simple operations to be done correctly.
    But it is easy to add and multiply finite decimals, and every real number can be represented as closely as desired by a sequence of approximations by finite decimals. Thus one way to define sums of two positive real numbers is to approximate the summands by finite decimals, add the approximations, and then define the real sum as the unique number which is approximated by the sequence of approximate sums.
    I.e. add the integer parts of the two summands and take this as the first approximation to the sum, then add the approximations which also include the “tenths” parts and take this as the second approximation. Then add the two approximations which include the hundredths parts and call this the third approximation. This process never stops but it gives an infinite sequence of approximations to the answer, and these approximations determine the answer.
    I.e. the answer can be approximated in this way to any degree of accuracy (which is better than a calculator can do) and we can thus describe the actual sum theoretically as “the smallest real number which is not smaller than any of the approximating finite sums. Every problem in this course is solved by some such “limit”, i.e. infinite approximation, process.
     
  15. Jan 4, 2008 #14

    mathwonk

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    calc 2

    Day 2, 2200, jan.13, 2000. Least upper bounds Recall how to describe the real number defined by an infinite decimal in terms of the sequence of finite decimals in its expansion. We said that the real number represented by an infinite decimal, is the smallest real number which is not smaller than any of its finite decimal approximations. This is the first example of the important concept of “least upper bound”. We define an “upper bound” of a set S of real numbers as a number L which is not smaller than any number in S, and a "least upper bound" as an upper bound such that no smaller number is an upper bound. I.e. L is a least upper bound (lub) for the set of numbers in S if and only if: 1) L is an upper bound for numbers in S, i.e. for every number x in S, x ≤ L, and 2) No number smaller than L is an upper bound for all numbers in S, i.e. for any number M < L, there is some number x in S with x > M. Ex: find the l.u.b. of the slopes of all lines joining points of form (1,1) and (x,x2) where x < 1. Solution: By the two point formula, the slope of the line joining (1,1) and (x,x2), is (x^2-1)/(x-1) = x+1, if x < 1. Thus the least number larger than all numbers of form 1+x with x < 1, is 2. I.e. property 1) holds since if x < 1, then 1+x < 2. And property 2) holds since x < 1 implies 1+x < 2. Then if M is any number less than 2, there is an x with x < 1, and 1+x > M. I.e. M < 2 so M-1 < 1, so just take x greater than M-1, and less than 1, i.e. M-1 < x < 1. Then M < 1+x < 2, as desired. Thus the slopes of the secant lines to the curve y = x^2, joining points (1,1) ans (x,x^2) for x<1, have form 1+x, and the smallest number larger than all these slopes shopuld be the slope of the tangent line to the curve y = x^2 at the point (1,1). Namely this slope should be 2. I.e. from looking at the graph of the curve y = x^2, near the point (1,1), we see that the tangent line at (1,1) should have slope larger than the slopes of any of the secant lines joining points (x,x^2), and (1,1), but the slope of these secants becomes arbitrarily near the slopes of the tangent line. Thus the tangent line's slope should be the smallest number larger than the slope of every such secant line, namely 2. Similarly, if we take the point (a,a^2) on the curve y = x^2, the secant lines joining points (a,a^2) and (x,x^2) for x < 2, have slope a+x where x < a. Thus the lub of these slopes is 2a. Hence the slope of the tangent line to y = x^2 at (a,a^2) should be 2a.

    Exercise: Show the slope of the tanegnt line to th curve y = x^3 at (1,1) is 3, and at (a,a^3) is 3a^2.
     
  16. Jan 4, 2008 #15

    mathwonk

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    calc. #3

    Day 3, 2200, jan.18, 2000. Area of circle as a least upper bound
    Recall the definition of the “least upper bound” of a set S of real numbers, is “the smallest real number not smaller than any number in the set S”. Then 1 is the least upper bound of the infinite set of real numbers of form {.9, .99, .999, .9999, ...........}, since 1 is laregr than all these numnbers, and yet these numbers eventually become larger than any number less than 1.
    Indeed this is what it means to say the infinite decimal .9999999..... represents (i.e. equals) the number 1, since the real number represented by an infinite decimal is the least upper bound of the set of all finite decimals obtained by truncating the infinite decimal at all finite stages.
    E.g. the real number represented by the infinite decimal .333333....., is by definition the least upper bound of the sequence of finite decimals {.3, .33, .333, .3333, .......}, and equals 1/3.
    Recall that area is measured in “square units” and that the area of a plane region is intuitively the number of unit squares that will fit inside the region. We make sense of the area of regions such as rectangles, parallelograms, triangles, and polygons, by cutting up a certain number of unit squares and reassembling them to fit inside the region.
    Then consider a region such as a circle, where no amount of cutting and reassembling can ever make a finite number of unit squares fit perfectly inside the region. One way to make sense of the concept of area here is to approximate the area of the circle by the areas of an infinite number of simpler regions for which we understand area better, such as inscribed polygons. We define the area of a circle as the least upper bound of the areas of all (convex) polygons which can be inscribed in the circle.
    We will use the Pythagorean “distance formula” to measure the length of a line segment and we define the length of a circle to be the least upper bound of the lengths (i.e. perimeters) of all polygons which can be inscribed in the circle. We define the number π to be the ratio π = C/2r of the circumference to twice the radius of any circle, (accepting that this ratio is the same for all circles). We argue that the area of a circle of radius r is πr2 as follows: the area of the circle is the lub of the areas of all inscribed polygons. Looking only at regular polygons, i.e. ones with all sides the same length, seems acceptable. Then we see by dividing the polygon up into triangles that the area of the polygon is half the product of the perimeter of the polygon times the common height of each triangle, (where that common height is the distance from the center of the polygon to the center of any one edge of the polygon). We agree the lub of this product of increasing factors, is the product of the lub’s of the factors. Thus the area of the circle is 1/2 the product of the lub of the heights of the triangles times the lub of the perimeters of the polygons. But the lub of the heights of the triangles is the radius of the circle, and the lub of the perimeters of the polygons is the circumference of the circle. Thus the lub of the areas of the inscribed polygons is 1/2 the product of the radius of the circle times its circumference, i.e. A = (1/2)(2πr)(r) = πr2.
     
  17. Jan 4, 2008 #16

    mathwonk

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    calc #4

    Day 4, 2200, 1/20/2000. Tangent slopes as limits
    We try finding the slope of a tangent line to a curve, by approximating it by slopes of secant lines. It is easier to compute the slope of a secant line than the slope of a tangent line, because we have two points to work with for a secant line, but only one point to work with for a tangent line. A tangent line to a curve at a point, is a line that passes through the point and has the “same direction” as the curve, near that point, whatever that means. Then we ask how to approximate a tangent line to a curve at a given point p.
    It seems that it can be done by choosing another point q also on the curve, and close to p, and constructing the secant line joining those two points. I.e. if q is a point on the curve and q is near p, then the secant line joining q to p should be a good approximation to the tangent line of the curve at p. In particular the slope of the secant line joining q to p, should be a good approximation to the slope of the tangent line at p. Since we know the tangent line to our curve at p should pass through p, the only other thing we need to compute that tangent line is its slope. So we take an infinite number of points q getting closer and closer to p, and take the slopes of all those secant lines. Then we say the slope of the tangent line is whatever number those secant slopes are approximating, provided such a number exists. I.e. the slope of the tangent line at p is the “limit”, as the point q approaches p, of the slopes of secant lines joining p to q.
    We have thus described roughly the slope of the tangent line to a curve. The problem of computing that slope remains. Then we asked what number is being approximated by x2, as x is taken closer and closer to 3. We agreed that as x gets closer and closer to 3, (but not equal to it), that x2 gets closer to 9 (even though it does not quite equal it).

    I.e. as we take numbers closer and closer to 3, and square them, we are getting better and better approximations to the number 9. Given any number, we can consider a sequence of approximations to that number. Conversely, given a sequence of approximations, we can ask what number they are approxiamtions to. That number being aproximated, is called the limit of the sequence of approximations. So the limit of the numbers x^2, as x approaches 3, is 9.
    Then we ask what number is being approximated by the numbers (x^2-9)/(x-3) as x approaches 3. This one is not as easy, because we can not set x=3 and evaluate. If we look at approximations for numbers x near 3, say x = 3.1, we get (9.61 - 9)/(3.1 - 3) = .61/.1 = 6.1. If x = 3.01, we get (9.0601 - 9)/(3.01 - 3) = (.0601)/.01 = 6.01. I will go out on a limb here and guess the number being approximated is 6. To actually compute this, simplify the expression
    (x^2-9)/(x-3) by factoring it, getting [(x+3)(x-3)]/(x-3. Then since our approximations are computed by setting x equal to numbers near 3, but not equal to 3, the non zero factors
    (x-3)/(x-3) cancel, so we can use the simpler expression x+3 instead, whose limit as x approaches 3, is clearly 6.
    In the same way, as stated earlier, the limit of the approximations (x^2-a^2)/(x-a), i.e. of x+a, as x approaches a, is 2a.
     
    Last edited: Jan 4, 2008
  18. Jan 4, 2008 #17

    mathwonk

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    oops some of the symbols are not coming across. so i stop here.
     
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