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Thread: Who does A Level Maths and can help me with my coursework??

  1. #1
    Senior Member 8-destiny-8 is a splendid one to behold 8-destiny-8 is a splendid one to behold 8-destiny-8 is a splendid one to behold 8-destiny-8 is a splendid one to behold 8-destiny-8 is a splendid one to behold 8-destiny-8 is a splendid one to behold 8-destiny-8 is a splendid one to behold 8-destiny-8's Avatar
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    Default Who does A Level Maths and can help me with my coursework??

    Hi, I'm kind of desperate, I really want to finish my coursework and get it out of the way but I can't because I'm stuck on something.
    I am on the last part of the Change of Sign method and all I have to do now is to show an instance where the method does not work.. I don't really know how to do that, so could somebody please help me? I'd be extremely grateful

  2. #2
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    Well...I'm a math teacher. Can you explain this issue more thoroughly?

    Edit: simple google search gave me this.

    http://www.coursework.info/AS_and_A_...od_L67876.html
    Last edited by Capernicus; 08-06-2009 at 08:35 AM.


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  3. #3
    Senior Member 8-destiny-8 is a splendid one to behold 8-destiny-8 is a splendid one to behold 8-destiny-8 is a splendid one to behold 8-destiny-8 is a splendid one to behold 8-destiny-8 is a splendid one to behold 8-destiny-8 is a splendid one to behold 8-destiny-8 is a splendid one to behold 8-destiny-8's Avatar
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    Thank you very much for the link. I searched on google too, but I didn't want to use the example given on the website I found because that would be cheating.. I just need someone to teach me what to do..
    What I'm stuck on is this.. I have to present an example where the Change of Sign method for finding the root of an equation fails to work, ie does not find all the roots of the equation, finds a root other than that expected or finds a false root. I don't know how to show this exactly nor can I come up with a suitable equation, so I have no idea what to do. Sorry if I am still too vague, but I would really appreciate your help.

  4. #4
    4: [Classified brah] Eris has a reputation beyond repute Eris has a reputation beyond repute Eris has a reputation beyond repute Eris has a reputation beyond repute Eris has a reputation beyond repute Eris has a reputation beyond repute Eris has a reputation beyond repute Eris has a reputation beyond repute Eris has a reputation beyond repute Eris has a reputation beyond repute Eris has a reputation beyond repute Eris's Avatar
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    Quote Originally Posted by 8-destiny-8 View Post
    Thank you very much for the link. I searched on google too, but I didn't want to use the example given on the website I found because that would be cheating.. I just need someone to teach me what to do..
    What I'm stuck on is this.. I have to present an example where the Change of Sign method for finding the root of an equation fails to work, ie does not find all the roots of the equation, finds a root other than that expected or finds a false root. I don't know how to show this exactly nor can I come up with a suitable equation, so I have no idea what to do. Sorry if I am still too vague, but I would really appreciate your help.
    Can you elaborate on this method. I have never heard of it before. I have studied ridiculous amounts of maths though, so I should be able to help you if I know what you're speaking of.



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    The change of sign method relies on a function F(x) that will change sign when it passes through zero. So, any function that does not cross the axes would be an example of when it doesn't work.

    Edit: To give you an example the function y= 1/x would be an instance where it does not work because it does not pass through zero and therefore the function will never equal to zero. It has no root.
    Last edited by OminousCloud; 08-06-2009 at 09:00 AM.


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  6. #6
    Senior Member 8-destiny-8 is a splendid one to behold 8-destiny-8 is a splendid one to behold 8-destiny-8 is a splendid one to behold 8-destiny-8 is a splendid one to behold 8-destiny-8 is a splendid one to behold 8-destiny-8 is a splendid one to behold 8-destiny-8 is a splendid one to behold 8-destiny-8's Avatar
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    Hi, thank you for offering your help I copied and pasted this from a website, I hope this helps:

    When solving any complex equation we:


    Rearrange the equation into the form f(x) = 0
    Sketch the graph of this function (using a calculator or by plotting points.)

    The solutions to our equation are where this graph crosses the x-axis. These solutions are called the roots of the equation.

    To locate where a root is we have to find limits between which the solution lies. i.e. Find an x-value that has a positive value for f(x), and an x-value that has a negative value for f(x).

    If the line is continuous then at some point in between these x-values the graph, (and value for f(x)), must = 0. This is a solution to our equation:


    i.e. If f(a) < 0, and f(b) > 0, then the root lies in the interval a < x < b.

    This idea is called a change of sign.

    We will not be able to find the root exactly, but we will be able to 'home in' on the root until we have it to the desired degree of accuracy.

  7. #7
    LUCKY DUCK Capernicus has a reputation beyond repute Capernicus has a reputation beyond repute Capernicus has a reputation beyond repute Capernicus has a reputation beyond repute Capernicus has a reputation beyond repute Capernicus has a reputation beyond repute Capernicus has a reputation beyond repute Capernicus has a reputation beyond repute Capernicus has a reputation beyond repute Capernicus has a reputation beyond repute Capernicus has a reputation beyond repute Capernicus's Avatar
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    Not quite true Mini. See, the test requires the function to have both positive and negative values. So, merely any function, for the sake of argument let's day f (x) = ln |x|, does not necessarily have both pos and neg values, thereby not passing this test. I see where you were going with it though.

    I think this would be best achieved with a piecewise function.

    f (x) = { x if x =/= 0
    ______{ 1 if x = 0

    See, this function technically has no zeros but, simply plugging in values getting infinitely closer to 0 (from both sides) would give you this pos/neg approaching values.

    Hope that helps.
    Last edited by Capernicus; 08-06-2009 at 09:10 AM.


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  8. #8
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    Quote Originally Posted by Capernicus View Post
    Not quite true Mini. See, the test requires the function to have both positive and negative values. So, merely any function, for the sake of argument let's day f (x) = ln |x|, does not necessarily have both pos and neg values, thereby not passing this test. I see where you were going with it though.

    I think this would be best achieved with a piecewise function.

    f (x) = { x if x =/= 0
    ______{ 1 if x = 0

    See, this function technically has no zeros but, simply plugging in values getting infinitely closer to 0 (from both sides) would give you this pos/neg approaching values.

    Hope that helps.
    What? =/


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  9. #9
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    Quote Originally Posted by Capernicus View Post
    Not quite true Mini. See, the test requires the function to have both positive and negative values. So, merely any function, for the sake of argument let's day f (x) = ln |x|, does not necessarily have both pos and neg values, thereby not passing this test. I see where you were going with it though.

    I think this would be best achieved with a piecewise function.

    f (x) = { x if x =/= 0
    ______{ 1 if x = 0

    See, this function technically has no zeros but, simply plugging in values getting infinitely closer to 0 (from both sides) would give you this pos/neg approaching values.

    Hope that helps.
    I think the method assumes a nice function though.

    One problem with the method is that it does not work with functions that merely "skip" off the x axis. Like x^2. It has a double root at 0, but the change of sign method won't be able to tell you where.

    The method seems sound though, it is a pretty straightforward application of the mean value theorem.



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  10. #10
    Senior Member 8-destiny-8 is a splendid one to behold 8-destiny-8 is a splendid one to behold 8-destiny-8 is a splendid one to behold 8-destiny-8 is a splendid one to behold 8-destiny-8 is a splendid one to behold 8-destiny-8 is a splendid one to behold 8-destiny-8 is a splendid one to behold 8-destiny-8's Avatar
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    Sorry, I don't understand, I haven't been taught those types of functions yet.

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    Quote Originally Posted by 8-destiny-8 View Post
    Sorry, I don't understand, I haven't been taught those types of functions yet.
    Simple version: You wanted an example where it doesn't work. Try using it to find the root of

    y(x) = x*x



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  12. #12
    Senior Member 8-destiny-8 is a splendid one to behold 8-destiny-8 is a splendid one to behold 8-destiny-8 is a splendid one to behold 8-destiny-8 is a splendid one to behold 8-destiny-8 is a splendid one to behold 8-destiny-8 is a splendid one to behold 8-destiny-8 is a splendid one to behold 8-destiny-8's Avatar
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    Quote Originally Posted by Eris View Post
    Simple version: You wanted an example where it doesn't work. Try using it to find the root of

    y(x) = x*x
    Sorry, that was a reply to Capernicus. I understand what you said, it helped a lot, thank you!

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    Okay, maybe this will clarify.

    I considered your f (x) = 1/x example and it doesn't work. See, the change of sign test would seem to indicate that the root (y = 0) is located at x = 0, yes? This is because of the negative values to the left and positive values to the right. But as one tests values whose absolute values become increasingly small ( from 0.1 > 0.01 > 0.001), the y-values get increasingly large. So the change of sign test would not indicate that this is a root.

    Eris: does it really assume a nice function? Hm...

    destiny: you don't know about piecewise functions? o.o

    Well, the world is truly going to implode: Eris is being more understandable than me.


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    Senior Member 8-destiny-8 is a splendid one to behold 8-destiny-8 is a splendid one to behold 8-destiny-8 is a splendid one to behold 8-destiny-8 is a splendid one to behold 8-destiny-8 is a splendid one to behold 8-destiny-8 is a splendid one to behold 8-destiny-8 is a splendid one to behold 8-destiny-8's Avatar
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    No, I don't know what piecewise functions are. I haven't been taught about them yet.

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    Quote Originally Posted by Capernicus View Post
    Eris: does it really assume a nice function? Hm...
    I think it is derived from the fact that given a function at two points (a, f(a)) and (b,f(b)), if it is nice, it passes through all points between f(a) and f(b) at some X coordinates we're not concerned with. Therefore, given the fact that f changes sign between a and b, there must be a point where it is 0 in between. This is if and only if it is nice, where in this context is smooth and with smooth derivatives (of all orders?)

    I think this comes from the mean value theorem, though it was so long ago I took elementary calculus I can't for the life of me remember.
    Last edited by Eris; 08-06-2009 at 09:45 AM.



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  16. #16
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    Quote Originally Posted by Eris View Post
    I think it is derived from the fact that given a function at two points (a, f(a)) and (b,f(b)), if it is nice, it passes through all points between f(a) and f(b) at some X coordinates we're not concerned with. Therefore, given the fact that f changes sign between a and b, there must be a point where it is 0 in between.

    I think this comes from the mean value theorem, though it was so long ago I took elementary calculus I can't for the life of me remember.
    No you are quite correct about the mean value theorem. However, if this Change of Sign method were indeed derived from that...then...shouldn't it be impossible to find a counterexample? @__@ It is 7 am and I haven't slept, so forgive me if I'm retarded!


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    Quote Originally Posted by Capernicus View Post
    No you are quite correct about the mean value theorem. However, if this Change of Sign method were indeed derived from that...then...shouldn't it be impossible to find a counterexample? @__@ It is 7 am and I haven't slept, so forgive me if I'm retarded!
    No, the counter examples exist where some of the MVT conditions or change of sign conditions aren't met. In this case, if any of these fail, the change of sign method may fail:

    1. Fully defined curve (a gap in the function's definition at the root would be kinda ninja)
    2. Smooth curve (piecewise defined function and/or derivative)
    3. Sign-change around the root (fails at x^2, x^4, x^6, ...)
    Last edited by Eris; 08-06-2009 at 09:45 AM.



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    Quote Originally Posted by Eris View Post
    No, the counter examples exist where the MVT conditions + change of sign conditions aren't met. In this case, if any of these fail, the change of sign method may fail:

    1. Fully defined curve (a gap in the function's definition at the root would be kinda ninja)
    2. Smooth curve (piecewise defined function and/or derivative)
    3. Sign-change around the root (fails at x^2, x^4, x^6, ...)
    That's it, I'm going to bed.


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    Is this actually under inequalities?


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    Quote Originally Posted by Rikumi View Post
    Is this actually under inequalities?
    It's some sort of protocalculus. I don't know what it sorts under though.

    Quote Originally Posted by 8-destiny-8 View Post
    Sorry, that was a reply to Capernicus. I understand what you said, it helped a lot, thank you!
    I came up with another case it might not work. If you have a plot like this, where it touches the x-axis twice, you might miss the fact that it's two roots, and not one. Though it really stems from the same problem that arose in the x^2 case. (Actually x^3-x^2 = x^2(x-1), so it's exactly the same, you catch the x-1 root, but not the x^2 root)
    Last edited by Eris; 08-06-2009 at 10:46 AM.



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    Eris you're a freaking genius! I only got a B in calculus -_-

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    Quote Originally Posted by Eris View Post
    I came up with another case it might not work. If you have a plot like this, where it intersects the x-axis twice, you might miss the fact that it's two roots, and not one. Though it really stems from the same problem that arose in the x^2 case. (Actually x^3-x^2 = x^2(x-1), so it's exactly the same, you catch the x-1 root, but not the x^2 root)
    So that's the case.

    I was wondering somewhere around it.


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    Quote Originally Posted by Shin Natsume View Post
    Eris you're a freaking genius! I only got a B in calculus -_-
    Heh, I have studied quite a lot of mathematics (you need that to study physics). From the top of my head, I've taken classes in Calculus, Linear Algebra, Complex Analysis, Multi-variable Calculus, Ordinary and Partial Differential Equations, Function spaces and Vector Calculus (this is a serious pain).

    It actually isn't -that- much, it's like one or one and a half years of full time college studies.



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    I SHOULD take linear algebra next semester but the only class available for it is an online class and I do NOT think thats a good idea. Aside from that, I took single var calculus part one and two ^_^

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    Quote Originally Posted by Shin Natsume View Post
    I SHOULD take linear algebra next semester but the only class available for it is an online class and I do NOT think thats a good idea. Aside from that, I took single var calculus part one and two ^_^
    That's a shame. Linear Algebra is really useful if you want to get into 3D programming. I remember making a small 3D space exploration program as I took the class (it really helped me get a feel for a lot of the stuff in Lin. Algebra)



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