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  1. #61
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    Originally Posted by Movjam View Post
    Interesting, thanks for the response mate.

    Anyone care to delve in a little into abstract algebra and analysis? Currently reading a book called "A Brief Guide to the Great Equations" by Robert P. Crease (great book btw, would recommend) which mentions Euler being a pioner of analysis in maths with respect to all of his work with infinite summations and his equation of e^(ipi) + 1 = 0. So something along proof and derivations of equations?
    As an introductory text, and I have a bunch, look at (an earlier edition of) Joseph Gallian's Contemporary Abstract Algebra. It will always be on my shelf. A second ood alternative is Dummit and Foote's. As far as Analysis, the standard I believe is Rudin's Principle's of Mathematical Analysis. For a slightly softer introduction (but still rigorous), Bartle's Real Analysis is fine book that is easy to read.

    The two subjects are staples for higher mathematics, and that is why most uni's require their graduate students to be familiar with them. In a nutshell, algebra is the study of mathematical structures and their influences, implications, and generalizations. A more advanced treatment of sets and their operations, but far more general then what an average US highschool student would consider "algebra." As an example: consider the set of Integers {...-2,-1,0,1,2,...}. This set is naturally endowed with addition (primary) and multiplication (secondary).. The secondary operation "distributes" over the primary. One wishes to generalize the properties of this kind of set by saying it is a Ring, and more specifically what they call an Integral Domain. Another example: the rational numbers... this is the set of Integers together with their multiplicative inverses. One generalizes this to a (countable) Field which is a set that contains objects which have additive and multiplicative inverses, indentities and inherits all features from the Ring associated to it.

    Introdictory Analysis is essentially a rehash of The Calculus, going far deeper, seeing and proving its foundations. In fact, many call Introductory Analysis "Advanced Calculus." One inportant thing to note is that Real Analysis is fundamentally different from "Real Algebra" for the simple fact that one cannot create the real numbers as an "algebraic evolution" from that of the rationals. The notion of limit is required. But even so, these two subjects interact heavily. The Reals form a Field, and hence can be studied algebraically as well. The set of all real polynomials forms a Ring. The set of all 1-1 continuous functions forms a Group (a set with only one operation instead of two, but still maintains inverses). The set of all continuous functions forms a Monoid (a set with an operation, which need not have inverses). The "operation" or "multiplication" in the latter two is actually function composition. To say that a set has a binary operation means that when you perform the operation on two things of this type, you get another one of the same type.
    Last edited by MiscMathematician; 08-23-2010 at 07:57 AM.
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  2. #62
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    Halp!



    Supposed to find the limit as x approaches 3+, 3-, and 3. I understand how limits work but I'm confused by the system of equations. If I plug 3.001 into the first equation, then x is greater than 3. WHAT MEAN?
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  3. #63
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    Originally Posted by Errorist View Post
    Halp!



    Supposed to find the limit as x approaches 3+, 3-, and 3. I understand how limits work but I'm confused by the system of equations. If I plug 3.001 into the first equation, then x is greater than 3. WHAT MEAN?
    3+ is what u approach from the right, so just plug in 3 for the equation where x>3 to get the limit.
    3- use the equation for x<3
    and for 3 just use the equation where x=3





    for you brahs that know about the fourier series, i have a question
    if f(x)=sin^2(x), how do you show that the fourier series is 1/2-cos(2x)/2. i know that the 2 are equivalent, but going through and solving for the coefficients i keep getting 1/2, but not with the other part of the answer.
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    Hypotheses Non Fingo Errorist's Avatar
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    Originally Posted by charity4thepoor View Post
    3+ is what u approach from the right, so just plug in 3 for the equation where x>3 to get the limit.
    3- use the equation for x<3
    and for 3 just use the equation where x=3
    Ok, how do I plug something into 0 when x=3 though? The answer isn't 0 and the limit does exist.
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  5. #65
    Jags Are Bake charity4thepoor's Avatar
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    Originally Posted by Errorist View Post
    Ok, how do I plug something into 0 when x=3 though? The answer isn't 0 and the limit does exist.
    f(3)=0
    f(3-)=3
    f(3+)=3


    in order for the limit to exist lim(3-)=lim(3+)
    so the limit does exist but f(3)=0

    i dunno what to do it that isn't it brah
    Last edited by charity4thepoor; 09-05-2010 at 06:06 PM.
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    Ok, I put 3 in as the limit for all answers and it says they're all correct now, which makes sense. The limits are both 3 as x approaches 3- and 3+, so that makes sense that the limit as x approaches 3, is 3 since it's the same on both sides. I still don't understand how that works for the function 0, x=3 though.
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  7. #67
    Jags Are Bake charity4thepoor's Avatar
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    Originally Posted by Errorist View Post
    Ok, I put 3 in as the limit for all answers and it says they're all correct now, which makes sense. The limits are both 3 as x approaches 3- and 3+, so that makes sense that the limit as x approaches 3, is 3 since it's the same on both sides. I still don't understand how that works for the function 0, x=3 though.
    the limit exists if lim(x-)=lim(x+) and if that does exist then the limit is the number both approach. the point at x=3 and f(3)=0 is just to throw you off and make you think since you are just learning it
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    Originally Posted by turkey_server View Post
    No. Take a = 1. Then b = 3, and b > a. Take a = -1. Then b = -3, and b < a.
    uh dude.

    -1 - b = 2

    2 + 1 = 3

    yeah cause 3 = -3...no
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    Registered User Muckle_Ewe's Avatar
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    Originally Posted by Spaghettii View Post
    uh dude.

    -1 - b = 2

    2 + 1 = 3

    yeah cause 3 = -3...no
    Wat?

    b = 3a

    This is satisfied by both a=1, b=3 and a=-1, b=-3. but in one case a<b and in the other a>b.
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  10. #70
    Registered User kbm12345's Avatar
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    Could someone he;lp me answer this please:

    b. A property of transposes is, for any two (compatible) matrices X
    and Y, Y'X' = (XY)'.
    Prove that (A')-1 = (A-1)' (for any invertible matrix A).

    A to the power of -1
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  11. #71
    ▶⛅💩🔗   🔊📶 99%🔌🔋3:14 MiscMathematician's Avatar
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    Originally Posted by kbm12345 View Post
    Could someone he;lp me answer this please:

    b. A property of transposes is, for any two (compatible) matrices X
    and Y, Y'X' = (XY)'.
    Prove that (A')-1 = (A-1)' (for any invertible matrix A).

    A to the power of -1
    Use ^(-1) to denote powers from now on.

    I = I' = (AA^(-1))' = A'(A^(-1))'

    Thus A is invertible <=> A' is invertible.

    We multiply on the left by (A')^(-1) and we're done.
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    Hypotheses Non Fingo Errorist's Avatar
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    Reps for not just answers, but explanation of this stuff so I can actually understand wtf is going on. It's labor day weekend and I'm pretty sure I'm not gonna be able to get a hold of my instructor, but this online hw is due 1am tomorrow. I have waded through hours of videos and none of them have shown problems like these, nor do I know how to enter these into wolframalpha.

    1.)
    If possible, choose k so that the following function is continuous on any interval:


    2.)
    Let

    Find each point of discontinuity of f, and for each give the value of the point of discontinuity and evaluate the indicated one-sided limits. If you have more than one point, give them in numerical order, from smallest to largest.

    3.)
    Let

    Find:


    4.)
    Let f be defined by

    Find (in terms of m) -1+ and -1-
    Find the value of m so that
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  13. #73
    ▶⛅💩🔗   🔊📶 99%🔌🔋3:14 MiscMathematician's Avatar
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    Originally Posted by Errorist View Post
    Reps for not just answers, but explanation of this stuff so I can actually understand wtf is going on. It's labor day weekend and I'm pretty sure I'm not gonna be able to get a hold of my instructor, but this online hw is due 1am tomorrow. I have waded through hours of videos and none of them have shown problems like these, nor do I know how to enter these into wolframalpha.

    1.)
    If possible, choose k so that the following function is continuous on any interval:


    Let k=20. The function is continuous at 2 then, and everywhere else.
    2.)
    Let

    Find each point of discontinuity of f, and for each give the value of the point of discontinuity and evaluate the indicated one-sided limits. If you have more than one point, give them in numerical order, from smallest to largest.

    Since both functions are continuous, the only (potential) point of discontinuity will be on the bounds of the piecewise definition. Plug in x=5. If the values agree, it is everywhere continuous, if they do not, there is a single discontinuity.


    3.)
    Let

    Find:


    Sorry, should be 2-(9+6h+h^2) (too lazy to retype the tex)

    4.)
    Let f be defined by

    Find (in terms of m) -1+ and -1-
    Find the value of m so that
    [/quote]


    Plug in -1 into the first and the second. You will get the two values it is asking for (in general the + sign means use the function for which x > that value, - for <). Set these two values equal to each other and solve for m.
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    sup misc math bros. my research is in dynamical systems, time series, and machine learning so I use some math
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    Originally Posted by MiscMathematician View Post


    Let k=20. The function is continuous at 2 then, and everywhere else.
    That's what I thought before. I factored this in my head the first time and forgot to cancel out the (x-2) so I got lost.

    Originally Posted by MiscMathematician View Post
    Since both functions are continuous, the only (potential) point of discontinuity will be on the bounds of the piecewise definition. Plug in x=5. If the values agree, it is everywhere continuous, if they do not, there is a single discontinuity.
    I'm still not understanding this. If they're both continuous, then how can there be discontinuity? If I plug in 5, then I get 35 on one function and 25 on the other. Other than that, they're both infinite functions.

    Originally Posted by MiscMathematician View Post

    Sorry, should be 2-(9+6h+h^2) (too lazy to retype the tex)
    So, I worked this out before and got (6h+h^2)/h but it doesn't accept that as an answer, nor does it accept 6h+h, 6, or -6. Also when h->0 then the limit is -6, but I'm not seeing how it comes out as a negative.

    Originally Posted by MiscMathematician View Post
    Plug in -1 into the first and the second. You will get the two values it is asking for (in general the + sign means use the function for which x > that value, - for <). Set these two values equal to each other and solve for m.
    I'm still doing something wrong. I got -1.6 for x->-1+ and -4 for x->-1- with m=0. The problem is that when m=0, the limit of -1+ is 8 and the limit of -1- is -8.

    Sorry for asking so many dumb questions. It's been a few semesters since I took algebra.
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    Sup math brahs?

    Second year Applied and Discrete Mathematics major. Eager to learn more.
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    applied math with concentrations in differential equations/probability and statistics. sup guys?
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    [QUOTE=Errorist;543664583]That's what I thought before. I factored this in my head the first time and forgot to cancel out the (x-2) so I got lost.

    I'm still not understanding this. If they're both continuous, then how can there be discontinuity? If I plug in 5, then I get 35 on one function and 25 on the other. Other than that, they're both infinite functions.

    Pretend the function is a road and when you plug in your x, you get your current altitude. The road is all nice until x=5. But then you fall 10 meters and die. This discontinues your life.


    So, I worked this out before and got (6h+h^2)/h but it doesn't accept that as an answer, nor does it accept 6h+h, 6, or -6. Also when h->0 then the limit is -6, but I'm not seeing how it comes out as a negative.
    Cancel the h. And when you do the initial subtraction you should have -6h-h^2 from distributing the -.

    I'm still doing something wrong. I got -1.6 for x->-1+ and -4 for x->-1- with m=0. The problem is that when m=0, the limit of -1+ is 8 and the limit of -1- is -8.

    Sorry for asking so many dumb questions. It's been a few semesters since I took algebra.
    Its cool man. You shouldn't get just a number. You aren't plugging in anything for m, only for x (-1). You should get -8-2m for the first, and 8+5m for the second. Setting these equal you'll get m=-16/7
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    Sup math brahs, taking 3 upper level math courses this semester to finish my minor and trying to decide if I want to stay another semester and just get a double major in math. Depends how well I handle this semester since I have two proof based classes and have never taken a proof heavy class.
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    Originally Posted by MiscMathematician View Post
    Pretend the function is a road and when you plug in your x, you get your current altitude. The road is all nice until x=5. But then you fall 10 meters and die. This discontinues your life.
    Ok, that makes a little bit more sense, but there's an infinite amount of Xs where this occurs. With that, it also says "If you have more than one point, give them in numerical order, from smallest to largest. If you have extra boxes, fill each in with an x."

    So lets say the first point is 5. According to the left hand side of the limit will be 35 and the right hand side will be 25. How many and which of these discontinuities am I supposed to list if there's an infinite amount of them? It's asking for 3 different points. Should I just give it the 1 point with it's left and right hand side limits, then fill the rest of the boxes up with Xs?



    Originally Posted by MiscMathematician View Post
    Cancel the h. And when you do the initial subtraction you should have -6h-h^2 from distributing the -.
    Oh ok, I forgot to distribute the - to 6h+h^2 when I opened up the parentheses. Also, I was confused with which h to cancel out. Is the h being canceled or factored out? Because I didn't know I could cancel out 1 of each. I thought it was just one of them.

    Originally Posted by MiscMathematician View Post
    Its cool man. You shouldn't get just a number. You aren't plugging in anything for m, only for x (-1). You should get -8-2m for the first, and 8+5m for the second. Setting these equal you'll get m=-16/7
    Ah ****. I originally got 0=7m because I added 8 to both sides instead of subtracting. I need to stop trying to do **** in my head.
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    Originally Posted by Darkwatcher759 View Post
    sup misc math bros. my research is in dynamical systems, time series, and machine learning so I use some math
    Sweet I'm gonna be doing a lot of Dynamical systems in my masters. Any particular interests? I quite like dealing with bifurcation theory and stable/unstable manifolds although I would like to understand chaos better.

    Originally Posted by Errorist View Post
    Ok, that makes a little bit more sense, but there's an infinite amount of Xs where this occurs. With that, it also says "If you have more than one point, give them in numerical order, from smallest to largest. If you have extra boxes, fill each in with an x."

    So lets say the first point is 5. According to the left hand side of the limit will be 35 and the right hand side will be 25. How many and which of these discontinuities am I supposed to list if there's an infinite amount of them? It's asking for 3 different points. Should I just give it the 1 point with it's left and right hand side limits, then fill the rest of the boxes up with Xs?
    Hey man could you maybe elaborate on where you said "but there's an infinite amount of Xs where this occurs". There is only one discontinuity in that function which is at x=5. Which others do you think there are?

    Originally Posted by Errorist View Post
    Oh ok, I forgot to distribute the - to 6h+h^2 when I opened up the parentheses. Also, I was confused with which h to cancel out. Is the h being canceled or factored out? Because I didn't know I could cancel out 1 of each. I thought it was just one of them..
    If you have a fraction of the form...

    then this can also be written as so everything gets divided by the denominator.
    Last edited by Muckle_Ewe; 09-06-2010 at 03:53 PM.
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    Originally Posted by Muckle_Ewe View Post
    Sweet I'm gonna be doing a lot of Dynamical systems in my masters. Any particular interests? I quite like dealing with bifurcation theory and stable/unstable manifolds although I would like to understand chaos better.
    Just gotten into it, but it may play a big role in my thesis. Im using the approach of a reconstructed phase space from a 1-dimensional time series with the time embedding approach. Then in phase space I can detect desired features of my time series (i.e. in phase space oscillations have a distinct). Im working on real-time algorithms though so its not so mathematical as stability theory and bifurcations. Chaos is interesting, when Ive read about it, it is a lot more enjoyable than most topics
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    Originally Posted by Muckle_Ewe View Post
    Hey man could you maybe elaborate on where you said "but there's an infinite amount of Xs where this occurs". There is only one discontinuity in that function which is at x=5. Which others do you think there are?
    It was asking for at least 3 points of discontinuity, except it was just to trick me. I put in just the 1 point with both of its one sided limits and kept the rest of the entry boxes blank and it said it was correct.

    Originally Posted by Muckle_Ewe View Post
    If you have a fraction of the form...

    then this can also be written as so everything gets divided by the denominator.
    Oh thanks, duh. I have this habit of either forgetting or over complicating the simple stuff after I've been introduced to harder stuff. I do this on quizzes and tests also. I'll be able to calculate the hard problems and will miss or flat out forget how to do the easiest problems. It's embarrassing.
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    K wtf.


    I got this for the derivative and I know it's right


    but I can't figure out f'(1).
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    Originally Posted by errorist View Post
    k wtf.


    i got this for the derivative and i know it's right


    but i can't figure out f'(1).
    [(2+1^2)(-2*1) - (2-1^2)(2*1)] / (2+1^2)^2
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    Originally Posted by MiscMathematician View Post
    [(2+1^2)(-2*1) - (2-1^2)(2*1)] / (2+1^2)^2
    Wtf, I tried this multiple times before and the answer kept coming out wrong. I don't know what I was doing wrong.
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    I understand HOW to use the rules in calculus, but I'm having difficulty knowing WHEN to use them. Is there an easy way of determining which rules need to be applied, like certain things I should look for that are a dead give away? For example, I thought I was supposed to use the quotient rule on a problem the other day because it was in fractional form, but I was instead supposed to use the power rule.
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    Sounds interesting. Too bad I have no f*cking clue what's going on this thread.
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    Originally Posted by Errorist View Post
    I understand HOW to use the rules in calculus, but I'm having difficulty knowing WHEN to use them. Is there an easy way of determining which rules need to be applied, like certain things I should look for that are a dead give away? For example, I thought I was supposed to use the quotient rule on a problem the other day because it was in fractional form, but I was instead supposed to use the power rule.
    Forget the quotient rule. Anytime you have a quotient, rewrite it as a negative power and apply your chain and product rules.
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    Originally Posted by Errorist View Post
    I understand HOW to use the rules in calculus, but I'm having difficulty knowing WHEN to use them. Is there an easy way of determining which rules need to be applied, like certain things I should look for that are a dead give away? For example, I thought I was supposed to use the quotient rule on a problem the other day because it was in fractional form, but I was instead supposed to use the power rule.
    I don't really understand your question completely and this advice i'm about to give may come as no surprise.

    But you need to do problems - lots of problems!

    It's great that you're using this thread for help. Use the advice and techniques people have given here and do example problems. The best way to learn anything is through practice and there are a lot of calculus examples on the internet.

    I can't quite remember if I've posted this before, but Paul's Online Notes is absolutely one of the best calculus resources on the internet.

    http://tutorial.math.lamar.edu/Classes/CalcI/CalcI.aspx
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