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Is it possible to learn pre-algebra to pre-calc in 9 months?

  1. Oct 29, 2015 #1
    I am currently in year 9 (9 grade for those in US) and I have a really rusty and a weak math background. I have 2 months of summer holidays coming up. I should be done with pre algebra in mid December. During my summer holidays I have more than 50 hours a week avalible for study and I was just is it possible to learn start learning or finishing learning pre calculus till July next year? if your answer is yes please suggest some resources that I can use.
     
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  3. Oct 29, 2015 #2

    micromass

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    Depends on the person. Some can do it, some can't.
     
  4. Oct 29, 2015 #3

    Vanadium 50

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    I read War and Peace in twenty minutes.
    It's about Russia.
     
  5. Oct 29, 2015 #4

    Krylov

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    You really required twenty minutes to figure that out?
     
  6. Oct 29, 2015 #5
    How does that concern me?
     
  7. Oct 29, 2015 #6
    Nine months is plenty of time if you study properly. You're still young so don't feel the need to rush.
    But can you make minute rice in 58 seconds?
     
  8. Oct 29, 2015 #7

    Vanadium 50

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    I see that absolutely everyone missed the point. The point is that while it may be possible to get through that much material, but that's a different thing than mastering that much material.

    You are talking about getting through four years of work in one summer. That's maybe 2000 hours of work - work guided by a professional - which you propose doing in no more than 600 hours: 50 hours/week x 12 weeks. So your proposal is for some who is "really rusty and [has] a weak math background" to learn without guidance 3x faster than someone with guidance.

    Does this sound reasonable to you? And even if you were to get through the material 3x faster, do you think you will have mastered it?
     
  9. Oct 29, 2015 #8

    berkeman

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    (its a metaphor...) :smile:
     
  10. Oct 29, 2015 #9

    symbolipoint

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    NO!

    You still need to study Intermediate Algebra for a year and Trigonometry which in "high school" is often incorporated into a "mathematical analysis" course, the equivalent of either a review of Intermediate Algebra or some increased level of advancement algebra. You might also need a Geometry course (another year) with proofs.
     
  11. Oct 29, 2015 #10

    Student100

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    I wouldn't worry as much about how far you can advance during your summer vacation, but how much of the above you can take care of. Priority number one should be rectifying your already rusty and weak background, any advanced study you manage to squeeze in is just icing on the cake.

    From the sounds of it you'll eventually take the classes formally at any rate, build a solid foundation.
     
  12. Oct 30, 2015 #11
    Why do you need to learn so much in such a short time? Is there any reason for it?
    I think it might be theoretically possible if you get a professional tutor but I think you need more time to understand and remember all the concepts.
    Unless you really must do this, I would not try to do this. You have got plenty of time and you should build your foundations and practice simpler calculations until you have them in your blood add we say ☺ no need to hurry!
     
  13. Oct 30, 2015 #12

    mathwonk

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    I don't think it is impossible for you to learn precalculus in 9 months. but we don't know for sure. all you can do is try. a book many people here often recommend for elementary math is Basic Mathematics by Lang. You might try that:

    https://www.amazon.com/Basic-Mathem...218834&sr=1-1&keywords=lang+basic+mathematics

    I think you should try, in algebra, to learn about linear and quadratic polynomials and the quadratic formula, and the "root/factor" theorem that x=a is a root of a polynomial f(x) if and only if (x-a) is a factor of f. In basic geometry you should know the concepts of congruence and similarity and the fundamental criteria for detecting these relations for triangles, as well as the basic Pythagorean theorem. In trig, learn about the circular functions sine and cosine, and their basic laws like the addition formulas and the law of cosines (generalized Pythagorean theorem), One should also learn about exponential functions and their basic rules like how exponentiating turns addition into multiplication, and the opposite for log functions. One should also practice some graphing of simple equations and understand the idea of "solving by graphing".

    I think Lang's book covers all this in under 500 pages.

    As a possibly more detailed alternative, and maybe more fun, I like the books Elementary Algebra and Geometry, by Harold Jacobs, but they don't cover trig I believe.

    As others have said, don't worry too much about how far you are getting, but just how much you are understanding.
     
    Last edited: Oct 30, 2015
  14. Oct 30, 2015 #13

    mathwonk

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    here is a small introduction to trig, just to show you how easy it is:

    Elements of trigonometry:

    Maybe I will try to write a short trig book. I hated trig in high school and never learned it until decades later. It seemed like a bunch of meaningless complicated formulas. Then I eventually found out how simple it is. Basically it is a way to measure angles. If you put the eraser of your pencil on the table in front of you and raise the pointed end up off the table, the pencil makes a certain angle with the table. How to measure that angle?

    One way is to measure how high off the table is the point. Of course that depends both on the angle and the length of the pencil. So use a pencil of length one. Then the height of the point off the table is called the "sine" of the angle. (If you don't have a "length one" pencil, then the height divided by the length, is the sine.)

    The sine is a vertical measure. The other way to measure the angle is a horizontal measure. Shine a light straight down on the pencil and measure how long the shadow is on the table. If the pencil has length one, that shadow length is the "cosine" of the angle. (If the pencil has any other length than one, the cosine is the length of the shadow on the table divided by the pencil length, as before.)

    There is one basic fact about sine and cosine that rules all others, the "pythagorean theorem". I.e. because the vertical distance from the point to the table forms a right angle with the shadow on the table, and the shaft of the pencil is opposite that right angle, we get the relation sine^2 + cosine^2 = 1^2 = 1, true for every angle.

    See if you agree that when the pencil is vertical, sine = 1, and cosine = 0. Thus sine(90degrees) = 1 and cosine(90degrees) = 0. Also sine(0degrees) = 0, while cosine(0degrees) = 1, (when the pencil is lying flat on the table.)


    That's about all there is to trig, except for a couple of tricky formulas that tell you how sine and cosine change when you double the angle, and that are hard to remember. (But there is a cute trick for remembering those too.) There are also some other angle measures that are not really new, but they are taught in school, like "tangent" = sine/cosine. There is also cotangent = cosine/sine, secant = 1/cosine, and cosecant = 1/sine, but you don't need them as much.

    If you are familiar with the unit circle in the (x,y) plane, then an angle is represented just by a point on the circle. I.e. the angle is the one made by the positive x axis and the radius from the center of the circle to the point on the circumference. The angle vertex is at the center. Then the x and y coordinates of that point on the circle are exactly the cosine and sine of the angle.

    This is the right way to do trig, using the unit circle. The size of the angle is measured, in "radians" rather than degrees, by the length of the circular arc cut off by the angle. Thus in radian measure a full 360 degree angle is 2pi radians, since 2pi = the length of a circle of radius one.

    Thus 90 degrees = pi/2 radians. Hence in radians, which are always used in calculus, sine(pi/2) = 1, and cosine(pi/2) = 0, while cosine(0) = 1, and sine(0) = 0.

    Homework:
    The other useful facts are that sine(pi/4) = sqrt(2)/2, and sine(pi/6) = 1/2, and sine(pi/3) = sqrt(3)/2. I did these in my head so see if they are correct, using pythagoras on a "30-60-90" triangle, and a "45-45-90" triangle. And see if you can use the pythagorean formula above to deduce that tan^2 + 1 = sec^2.


    Suggestion: if this is too opaque now, learn from one of those other longer sources and then come back and read this again. This is really about all there is to it, honest.


    Question: if an angle can be precisely measured just by its radian measure, why do we use two numbers, sine and cosine?

    Answer: it's easier to compute sine and cosine, because they are lengths of straight line segments, whereas radian measure is the length of a circular arc.

    In fact that is the whole reason for the existence of the subject of trig, to relate "linear" measure, to circular measure.
     
  15. Oct 30, 2015 #14

    mathwonk

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    hers a bit more trig:

    The cute trick is the fact that if you use complex numbers, then sine and cosine are a special case of the exponential function! And the laws of exponents are simpler than those for sine and cosine.

    Lets just treat this as a magic trick that works. If you have an angle of t radians, i.e. that an angle with vertex at the center of a unit radius circle, that cuts off an arc of length t on the circle, then cos(t) + i.sin(t) = e^(it), where e is a certain beautiful positive real number between 2 and 3, and i is a new "complex" number with i^2 = -1.

    Think of the numbers on the x axis as the usual "real" numbers, whose squares are all positive, and then i is a number on the y axis, at unit distance from the origin. Real numbers make angle zero (with the x axis) and i makes angle pi/2 (I started to say 90 degrees, as old habits die hard).

    Every point in the x,y plane represents a complex number, and to multiply them you multiply their lengths and add their angles. So i^2 has length 1^2 = 1, and angle pi/2 + pi/2 = pi = 180 degrees. So i^2 is at -1 on the x axis!

    Just as you can take powers and exponents of real numbers you can also take complex exponents of real numbers like e, and then for any real number t, the complex number e^(it) is the same as the complex number cos(t) + i.sin(t). I.e. it has x coordinate cos(t) and y coordinate (or "i coordinate") sin(t).


    Now here is the cool part, just as e^(a+b) = e^a.e^b, when a,b are real, also e^i(s+t) = e^is.e^it. that gives us relations for cos(s+t) and sin(s+t) and the same functions of s and t.

    E.g. suppose we want to know what cos(2t) is,

    Well, e^i(2t) = e^i(t+t) = e^it.e^it. Now substitute the cos and sin formula in and multiply out.

    I.e. e^i(2t) = cos(2t) + i.sin(2t), and e^it = cos(t)+i.sin(t),

    so from e^i(2t) = e^it.e^it, we get (purely mechanically, as in Saxon!),

    that cos(2t) + i.sin(2t) = (cos(t)+i.sin(t)).(cos(t)+isin(t))

    = cos^2(t) + i^2 sin^2(t) + 2i.cos(t)sin(t), I hope.


    Since i^2 = -1, that gives

    cos(2t) + i.sin(2t) = [cos^2(t)-sin^2(t)] + i.[2cos(t)sin(t)].

    equating real and imaginary (i.e. x and y) parts, gives both formulas at once!

    cos(2t) = cos^2(t) - sin^2(t), and sin(2t) = 2cos(t)sin(t),

    the so - called "double angle formulas for cos and sin.



    Homework: check similarly the "addition laws" for cos and sin:

    cos(s+t) = cos(s)cos(t) - sin(s)sin(t),

    and sin(s+t) = cos(s)sin(t) + sin(s)cos(t).


    OK this was tedious, but it all follows from just remembering two formulas:

    e^it = cos(t) + i.sin(t), and e^(a+b) = e^a.e^b, (the addition law for exponentials)

    and most people find those easier to remember than the addition formulas for sin and cos.
     
  16. Oct 30, 2015 #15

    mathwonk

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    Ah yes, there is also the law of cosines, which actually occurs in Book II of Euclid as Prop. II.12 and Prop.II.13 I think, (and law of sines). So one already knows basic trig if one has had a classical course in Euclidean geometry, I will discuss this after dinner.

    The law of cosines is the pythagorean theorem for non right triangles, and follows from the right triangle case.

    Ok here's the law of cosines, from euclid. Prop. II.12. Hmmm... the data limit here is exceeded even by a one page pdf file. Here is the word file but without the pictures. the original pdf file is week 2 day three in this link (epsilon camp notes):

    http://alpha.math.uga.edu/~roy/camp2011/10.pdf

    teaser: there is a point of view in which the law of cosines becomes just the familiar rule (a-b)^2 = a^2 - 2ab + b^2.

    Law of Cosines
    Pythagoras says the square on the side opposite a right angle “equals” the two squares on the sides containing the angle. If the angle is acute, the square on the side opposite it is smaller than the two squares on the sides containing it, and if obtuse the square it is greater. The law of cosines tells exactly how much less or how much greater; in particular it says the discrepancy is twice the area of a certain rectangle. These are propositions 12-13, Book II of Euclid.

    Prop. II.12 (Law of cosines, obtuse case): Let ABC be a triangle on the base BC, with obtuse angle at C, and vertex at A. Drop a perpendicular from A to the line extending base BC, meeting it at X, outside segment BC.
    Then (AB)^2 = (AC)^2 + (BC)^2 + 2 (BC)(CX).

    Proof: By Pythagoras applied to right triangle AXB, we have (AB)^2 = (AX)^2 + (BX)^2. From IV.4, this equals (AX)^2 + (CX)^2 + (BC)^2 + 2 (BC)(CX). By Pythagoras applied to triangle AXC, this equals (AC)^2 + (BC)^2 + 2 (BC)(CX). QED.

    Exercise: Prove:
    Prop. II.13: (Law of cosines, acute case): Let triangle ABC on base BC have an acute angle at C, and vertex A. Drop a perpendicular from A to base BC, and assume it meets the base at X, between B and C.
    Then (AB)^2 = (AC)^2 + (BC)^2 - 2 (BC)(CX).

    Remarks: What does this theorem have to do with cosines? If you recall the definition of the cosine of angle <C in the picture for the acute case above, cos(<C) = |XC|/|AC|, the ratio of the numerical lengths of the two sides. Hence cos(<C).(AC) = XC, an equality of segments. Substituting this into Euclid’s formula above gives us
    (AB)^2 = (AC)^2 + (BC)^2 - 2 (AC)(BC).cos(<C), and this is the usual law of cosines in trigonometry. It also works for the obtuse case, since the cosine of an obtuse angle is negative, so the minus signs cancel and give us the formula in II.12 above.
     
  17. Oct 30, 2015 #16

    mathwonk

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    a few last remarks:

    Law of sines:

    Take a triangle with base AB and vertex C, and side a opposite angle A, side b opposite angle B, and so on.

    Assume angles A and B are both acute to make it easy. Now drop a perpendicular from vertex C to side AB hitting it at point Q between A and B. Let the length of that perpendicular CQ be x.

    Now by the very definition of sin and cos, we have that sin(A) = x/b, and sin(B) = x/a. Hence x = b.sin(A) = a.sin(B), so sin(A)/a = sin(B)/b. voila! I don't use this much. It is also true that
    sin(A)/a = sin(C)/c.


    Coordinate geometry and trig
    There is a nice way to do geometry in the (x,y) plane, using coordinates that allows the algebraic tools of addition and multiplication to enhance the geometry. For each point (a1,a2) in the (x,y) plane, we imagine an arrow going from the origin (0,0) to the point (a1,a2). We call this arrow A. Let B be another arrow fro the origin (0,0) to the point (b1,b2). If we add the coordinates getting another point (a1+b1, a2+b2), we can ask how the arrow A+B from the origin (0,0) to (a1+b1, a2+b2) is related to the first two arrows. It turns out that the arrow A+B forms the diagonal of a parallelogram with sides A and B. The vertices of this parallelogram are the points (0,0), (a1,a2), (a1+b1, a2+b2), and (b1,b2).

    E.g. if A = (1,0), and B = (0,1), then A+B = (1,1) is the diagonal of the square with vertices (0,0), (1,0), (1,1), and (0,1).
    Consider the triangle with vertices (0,0), (a1,a2), and (a1+b1, a2+b2). The arrows A and A+B form two sides of this triangle with common vertex (0,0). The third side, which goes from the point (a1,a2) to the point (a1+b1, a2+b2), is parallel to the arrow B, and has the same length. Thus if the arrows A and B are perpendicular, the arrow A+B is the hypotenuse of a right triangle whose sides have the same lengths as the arrows A and B. If we denote length of an arrow by | |, then by Pythagoras we get |A+B|^2 = |A|^2 + |B|^2.
    By the same reasoning, if we consider the triangle with two sides A and B, its third side is parallel to the arrow A-B, and has the same length. Hence by Pythagoras, if A and B are perpendicular, then |A-B|^2 = |A|^2 + |B|^2.

    We will define a multiplication of arrows that captures this theorem and also the more general law of cosines.
    To multiply two arrows A = (a1,a2) and B = (b1,b2), we define their “dot product”as A.B = a1b1 + a2b2, which is a number, rather than an arrow. It is easy to check that this multiplication has some of the properties of usual multiplication, like commutativity and distributivity for addition, and so on, but the product of two non - zero arrows can be zero. E.g. (1,0).(0,1) = 1.0 + 0.1 = 0+0 = 0.

    Moreover, the product of an arrow with itself is exactly the square of its length, i.e. A.A = (a1)^2 + (a2)^2 = |A|^2, by Pythagoras.
    In fact the dot product of two arrows is zero exactly when the arrows are perpendicular.

    I.e. consider A and B as two sides of a triangle. Then the third side is parallel to the arrow A-B, and has the same length. Hence |A-B|^2 = (A-B).(A-B) = A.A – 2A.B + B.B =|A|^2 +|B|^2 – 2A.B. But if A and B are perpendicular, then by Pythagoras we must have |A-B|^2 = |A|^2 + |B|^2, so A.B must be zero.

    If A and B are sides of any triangle with third side parallel to and of same length as A-B, then again |A-B|^2 = A|^2 +|B|^2 – 2A.B. This looks exactly like the law of cosines except that we have 2A.B in place of 2|A||B|cos© where c is the angle between A and B. Thus in fact A.B must equal |A||B|cos©.

    If on the other hand we knew that A.B = |A||B|cos©, then we get the law of cosines by expanding the dot product |A-B|^2 = (A-B).(A-B) = A.A – 2A.B + B.B = |A|^2 +|B|^2 – 2A.B =|A|^2 +|B|^2 – 2|A||B|cos©.

    This allows us to remember the easier formula A.B = |A||B|cos©, and then to recover the more complicated law of cosines.

    It also gives a way to calculate cosines without a calculator.
    E.g. the angle between the arrows A and B has cosine equal to (A.B)/|A||B|. E.g. the angle between A = (1,0) and B = (1,sqrt(3)) has cosine equal to 1/2. Remember what angle that is?


    To summarize that mess, define a product of two arrows A,B in the plane, beginning at the origin,

    as: A.B = |A||B|cos©, where | | denotes length, and c is the angle between the two arrows.

    Also define an addition of arrows, by placing the tail of one arrow at the head of the other and taking the sum to go from the tail of the first to the head of the second arrow.

    Then the third side of the triangle with arrows A and B as two sides, is parallel to the arrow A-B, and of same length.

    Then the law of cosines becomes the usual rule for expanding a square:

    (A-B).(A-B) = A.A - 2A.B + B.B and since the angle between any vector and itself is zero, we get:

    |A-B||A-B|cos(0) = |A||A|cos(0) - 2 |A||B| cos© + |B||B|cos(0).

    Then since cos(0)m = 1, we get

    |A-B|^2 = |A|^2 - 2|A||B| cos© + |B|^2, which is exactly the usual law of cosines.


    Caveat: I have not proved here that this multiplication is distributive, For that I probably need the law of cosines!
    (sometime later)...
    Well, duh, I guess that's the whole point: the law of cosines is equivalent to saying this multiplication is distributive, i.e. is a multiplication!

    So this does not reprove the law of cosines, it just restates it in a more natural form.

    Then the three term principle says that since the explicit multiplication rule A.B = a1b1 + a2b2, is also distributive, and agrees with the first one on vectors that are equal, it must agree as well on all vectors.


    My point here is to try to show how elementary math is illuminated when viewed from a higher point of view. No advanced math is being done here, but we are seeing elementary math more clearly I hope, by exposing its structure. This is what is not obtained from plug and chug treatments. If anyone has a book on trig, Saxon or otherwise, it might be interesting to compare its treatment with the one given here.


    By the way, if anyone has a child who looks at the law of cosines

    |C|^2 = |A|^2 +|B|^2 – 2|A||B|cos©,

    and remarks that the right hand side looks kind of like what you get when you expand (A-B)^2, or asks whether C = A-B, then

    you have a potential mathematician - certainly a child who is thinking like one. Mathematics is about looking for patterns, and analogies.
     
  18. Nov 1, 2015 #17
    Another good book that covers topics similar to Lang is "Fundamentals of Freshman Mathematics" by Allendoerfer/Oakley.
     
  19. Nov 1, 2015 #18

    symbolipoint

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    Referring to the topic of this, Is possible to learn Pre-Algebra through Pre-Calculus in 9 months? NO. Not for most people, at least.

    A good, talented student who understands Basic Mathematics well (meaning Number Sense, simple common Geometry) can skip Pre-Algebra;
    go directly to Algebra 1, and if learns it well, may be able to study College Algebra & Trigonometry. Given possible 6 months for Algebra 1 and another 6 months for Pre-Calculus, that makes 1 year. College Algebra contains all of Intermediate Algebra, and goes further. Not easy; but if the student is talented and puts in the effort, maybe he can manage.

    The question is about if the student can develop this knowledge and skill in that one-year time.
     
  20. Nov 1, 2015 #19

    I am preparing for ioi and need to learn the maths for it so I can learn discrete math and then learn algorithms and data structures.
     
  21. Nov 1, 2015 #20
    I am more than happy to spend a year but what about algebra 2? U didn't mention that. Sorry I am from Australia and not used to the US system.
     
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