Thread: Calculus - PreCalc Review

  1. #1
    essence of digital xddxogm3's Avatar
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    Calculus - PreCalc Review

    I have a question.
    I'm doing a cube.
    (a+b)^3
    the book says it should be the following.
    (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3

    but one of my b variable is delta

    (1) unsure how to make delta on the post.
    so we will use this to resemble delta

    so the actual problem is this.

    (x+x)^3

    when cubing this, would i square and cube delta?

    x^3+3x^3+3x^3^2+^3x^3
    "Hence to fight and conquer in all your battles is not supreme excellence;
    supreme excellence consists in breaking the enemy's resistance without fighting."
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  2. #2
    Dump Truck Internet valis's Avatar
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    yes you do, dx is simply (x + a bit), treat it just like a normal variable

  3. #3
    Toaster Zach L.'s Avatar
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    It is a normal variable. What you end up finding out, though, is that the higher order terms go much more quickly to zero than the lower order terms, making the higher order terms negligible.

    That is, (dx)^2 goes much more quickly to 0 than dx.
    The word rap as it applies to music is the result of a peculiar phonological rule which has stripped the word of its initial voiceless velar stop.

  4. #4
    aoeuhtns
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    Erm, there's a huge difference between 'dx' and 'delta x', so it's best not to confuse the two. The latter is usually used as an actual quantity, the difference between two distinct x values. The former is a differential, a different beast that when treated like an actual variable can cause incorrectness, unless you happen to correctly treat it as an actual value.

    Quote Originally Posted by xviddivxoggmp3
    (x+x)^3

    when cubing this, would i square and cube delta?

    x^3+3x^3+3x^3^2+^3x^3
    Unless the smilies went out of control, that looks horribly wrong.

    You don't square and cube delta alone -- delta is just part of the variable name "deltax" or "Δx" (I wrote two versions because maybe the Unicode doesn't work for everybody.) The delta character is part of the variable name the same way that you can use multiple letters in variable names in C. In other words,

    (x + deltax)^3 = (x^2 + x*deltax + deltax^2)(x + deltax) =
    x^3 + 3*(x^2)*deltax + 3*x*((deltax)^2) + deltax^3

    And let's see if the Unicode works:
    (x + Δx)^3 = (x^2 + x*Δx + Δx^2)(x + Δx) =
    x^3 + 3*(x^2)*Δx + 3*x*((Δx)^2) + Δx^3.

    So, any time you see "Δx" or "Δy" etc, it always means the same thing as "(Δx)" or "(Δy)". So (Δx)^3 is just a slightly more verbose way of saying Δx^3 -- they mean the same thing, since Δx is a single variable.

    Generally speaking, delta represents "change in", so one might want to read Δx as "change in x." Though "delta x" works too. They're the same amount of syllables :-)

  5. #5
    Toaster Zach L.'s Avatar
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    Eh... most of the books I've encountered lately used 'little delta x' as the differential; that's what I was speaking of. But yes, you are correct, there is a difference between the differential and the 'change in'. My mistake.
    Last edited by Zach L.; 08-31-2005 at 06:05 AM.
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  6. #6
    essence of digital xddxogm3's Avatar
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    so if you have delta and x combined as in deltax, it can not be seperated for further calculations? that makes since. I just wanted to make sure i couldn't take the x away from the delta when performing these calculations. so if i was factoring i would have to factor out a deltax and not just the x in the deltax.
    "Hence to fight and conquer in all your battles is not supreme excellence;
    supreme excellence consists in breaking the enemy's resistance without fighting."
    Art of War Sun Tzu

  7. #7
    & the hat of GPL slaying Thantos's Avatar
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    When in doubt use one of the most powerful tools avialable to you: Let

    As in: Let y = Δx
    And then redo the equation.

  8. #8
    essence of digital xddxogm3's Avatar
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    another question about Δx
    I have a problem that i'm trying to work.

    lim [2(x+Δx) - 2x]/Δx = (2x+2Δx-2x)/Δx = 2Δx/Δx
    Δx->0

    The book answer to the limit is 2, but i get 0 due to i go from

    2Δx/Δx = Δx

    why is this incorrect?

    substituting y does the same thing for me.
    y = Δx
    lim [2(x+y) - 2x]/y = (2x+2y-2x)/y = 2y/y = 2y/y = y
    y->0
    Last edited by xviddivxoggmp3; 09-06-2005 at 08:57 PM.
    "Hence to fight and conquer in all your battles is not supreme excellence;
    supreme excellence consists in breaking the enemy's resistance without fighting."
    Art of War Sun Tzu

  9. #9
    Cat Lover
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    (2*3)/3 = 2.

    (2*y)/y = 2

    (2*Δx)/Δx = 2

  10. #10
    essence of digital xddxogm3's Avatar
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    ok, i'm blind.
    opps.
    is there a list of operations you can do with Δx
    similar to an order of precedence/operations list
    Last edited by xviddivxoggmp3; 09-06-2005 at 10:49 PM.
    "Hence to fight and conquer in all your battles is not supreme excellence;
    supreme excellence consists in breaking the enemy's resistance without fighting."
    Art of War Sun Tzu

  11. #11
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    delta x is just a number, like any other number. Say you have an int with initial value of 2, which is then changed to 7. Then delta x (or the change in x) is 5. Or say you define delta x to be 0.00001. Then you could say if y - x < delta x then, for the current task, y == x. Or say you do an experiment and get three values for x that are 7, 8, and 9. Then you could say that x is 8 +/- delta x where delta x is 1.

    This concept of changes in a given value is used a lot in mathematics and chemistry. For example, you can view standard deviation in statistics as a change in a given value, or a delta x. Or describe what happens to a given chemical bond between two different atoms as the temperature rises from room temperature to 1000 degrees in units of 100 degrees (that is 100 degrees is a delta x for temperature).
    You're only born perfect.

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