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  1. #1 Polynomial 
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    I need some help in a homework. My teacher find a polynomial f(x) with 'n' terms where the square of f(x) has also 'n' terms.

    I had to submit this before 2 months.


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    Forum Sophomore NimaRahnemoon's Avatar
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    Errrr... Trick question?

    Just to test him out tell him (x+ix)^2 and tell him that it is 2x^2+2ix^2. It really isn't... In math i^2 is -1. That's what he will probably say... and in reply say (((-1)^(.5))((-1)^(.5)))^2 is ((-1*-1)^(.5))^2 which is 1. He'll sit there puzzled while you laugh at him, and if he is anything like my teacher he'll be pissed at you for proving him wrong, which is always fun.

    If you don't know what I'm talking about, try squaring (x+ix) yourself and you might understand.

    But for reals, there isn't anyway that I can think of doing this. The thing I pointed out doesn't work in real mathamatics, again cuz i^2 is actually considered -1.

    Edit: I almost forgot if he says 2x^2 + 2ix^2 can combine into (2+2i)x^2, which is one term. Tell him it isn't proper etiquette to combine imaginary and real numbers like that, and therefore it should be two terms.

    Bothering teachers is fun.


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    Forum Professor wallaby's Avatar
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    geez you must be fun to teach.
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    Forum Radioactive Isotope mitchellmckain's Avatar
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    Quote Originally Posted by anandsatya
    I need some help in a homework. My teacher find a polynomial f(x) with 'n' terms where the square of f(x) has also 'n' terms.

    I had to submit this before 2 months.
    I think there is only one possible answer to this question. That is, there are many polynomials which satisfy this condition but they all have one obvious thing in common.

    Let me give you a hint.

    take a general polynomial of three terms

    f(x) = ax^2 +bx +c

    and square it

    f(x)^2 = a^2x^4 + 2abx^3 + b^2x^2 + 2bcx + c^2

    Well the only way any of these terms could be zero is if a, b or c is zero
    if a is zero (but not b or c) then f has 2 terms and f^2 has 3 terms
    the same thing if b or c is zero instead of a.

    but what if ......

    well, you you need to figure this out for yourself. It is your homework.

    Quote Originally Posted by Nima Rahnemoon
    Just to test him out tell him (x+ix)^2 and tell him that it is 2x^2+2ix^2. It really isn't... In math i^2 is -1. That's what he will probably say... and in reply say (((-1)^(.5))((-1)^(.5)))^2 is ((-1*-1)^(.5))^2 which is 1. He'll sit there puzzled while you laugh at him, and if he is anything like my teacher he'll be pissed at you for proving him wrong, which is always fun.

    If you don't know what I'm talking about, try squaring (x+ix) yourself and you might understand.

    But for reals, there isn't anyway that I can think of doing this. The thing I pointed out doesn't work in real mathamatics, again cuz i^2 is actually considered -1.

    Edit: I almost forgot if he says 2x^2 + 2ix^2 can combine into (2+2i)x^2, which is one term. Tell him it isn't proper etiquette to combine imaginary and real numbers like that, and therefore it should be two terms.

    Bothering teachers is fun.
    Your teacher is extra dumb if he buys any of this.
    (x+ix)^2 = 2ix^2
    Big deal? What does that prove?

    And I don't know what,
    "(((-1)^(.5))((-1)^(.5)))^2 = ((-1*-1)^(.5))^2 = 1"
    is supposed to prove since this is the same as
    (i * i)^2 = (-1)^2 = 1
    perfectly consistent with i^2 = -1.

    So I think Nima is either easily amused or he has conned himself. I certainly wonder if he really knows how to multiply polynomials or complex numbers.

    (1+i)^2 = 2i
    besides the usual polynomial method of multiplying this out we can use the exponential form
    1+i = sqrt(2) exp(i pi/4)
    which when squared gives you 2 exp(i pi/2)
    but exp(i pi/2) = i
    so (1+i)^2 = 2i

    Squaring a complex number doubles its angle on the complex plane.
    1+i has an angle of 45 degrees or pi/4 on the complex plane
    so its square has an angle of 90 degrees on the complex plane which means that it points along the imaginary axis.

    Oh and "proper etiquette" is to combine all real and imaginary parts and treat them as a single coefficient. So x+ix = (1+i)x is a polynomial with a single term.
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  6. #5  
    Forum Sophomore NimaRahnemoon's Avatar
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    Mhmm, I get most of it. I thought that x+ix is considered two terms since one is real and the other is imaginary.

    Well here is what I said in better format and with explanation, because that is really messy, and isn't very easy to understand.

    So we start of with
    (x+ix)
    The square of this is simply a²+2ab+b² where a=x and b=ix
    a²= x², 2ab= 2ix², b²= (ix)² or i²x². Here is what it is put together.
    (x+ix)²= x²+2ix²+i²x²
    So far so good?

    Now what is i²... based on the other topic we realize that i²= ±1

    Here is the approach
    i²= sqrt(-1)sqrt(-1)=(-1)^½(-1)^½= (-1*-1)^½= 1^½= 1 (by definition the sqrt of a number is positive, unless the situation says it can be negative)
    There might be some confusion on this (-1)^½(-1)^½=(-1*-1)^½, this is true becase they have the same degree...

    Then there is the proof that i² is -1... I am pretty sure we all know why...

    Given this, let's go back to the equation:
    (x+ix)²= x²+2ix²+i²x²
    so we replace the i² with ±1

    Since it is ±1, we get two equations... I, for the purpose of the original question, am only interested in the positive side.

    (x+ix)²= x²+2ix²+(1)(x²)= 2x²+2ix²
    In middle school I was taught that 2x² and 2ix² are considered two terms, because one is imaginary and one is real. So I consider this to be the answer to the original question.

    Mitchell, you are absolutely right that (x+i)² is 2i, however this depends on what you see i² as being.

    Once again, I still think that this question has no answer... but this is the best answer I can think of...

    Crap, time flies when you are doing math, I'm late for class.

    Edit: I am wrong... deepest apologies... i² = -1 and only -1, check the other topic named "i²" to see why.
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    Forum Radioactive Isotope mitchellmckain's Avatar
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    Quote Originally Posted by Nima Rahnemoon
    Now what is i²... based on the other topic we realize that i²= ±1

    Here is the approach
    i²= sqrt(-1)sqrt(-1)=(-1)^½(-1)^½= (-1*-1)^½= 1^½= 1 (by definition the sqrt of a number is positive, unless the situation says it can be negative)
    There might be some confusion on this (-1)^½(-1)^½=(-1*-1)^½, this is true becase they have the same degree...

    Then there is the proof that i² is -1... I am pretty sure we all know why...


    Edit: I am wrong... deepest apologies... i² = -1 and only -1, check the other topic named "i²" to see why.
    Yes you are wrong.
    However, (-1)^½(-1)^½=(-1*-1)^½ is much more interesting and that is not what you said before. It exploits the fact that the inverse of the square function is not the single valued square root function. In fact since you are squaring i you have to extended the square and square root functions to the complex plane and now you must use complex analysis and the concept of branching functions. In which case square root is a complex function with two branches, mapping every point on the complex plane to two different points on the complex plane and so (-1)^½ = ±i and (1)^½ = ±1 and so (-1)^½(-1)^½=(-1*-1)^½ is true only if you choose the right combination of branches. Anyway the problem with (-1)^½(-1)^½=(-1*-1)^½ is not square of i which is most definitely -1, but with the square root function.

    Quote Originally Posted by Nima Rahnemoon
    In middle school I was taught that 2x² and 2ix² are considered two terms, because one is imaginary and one is real. So I consider this to be the answer to the original question.
    Well you were taught wrong, but perhaps there is a confusion of terminology here. A polynomial is simply defined as a mathematical expression that is a sum of a number of terms. But that does not mean that 3x + x^3 +2 - 4x is a polynomial with 4 terms. To count the number of terms it is presumed that you put the polynomial into standard form. So the real question is what is the standard form of a polynomial and the answer is
    f(x) = a x^n + b x^(n-1) + ... + c x^2 + d x + e
    Which means that if you have two terms with the same power of x you must add their coefficients together.
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  8. #7  
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    "Multiplying a complex number by i simply rotates it 90 degrees around 0, so 1i = i lies on the imaginary axis of the complex plane, a distance of 1 from zero. The number ii = i^2, then, is the number 1 rotated 180 degrees about zero, which puts it at -1. The number -i = -1i = i^3 is 1 rotated 270 degrees, and i^4 = 1."

    So i^2 is only -1. Making i^2 positive 1 would be wrong, because it doesn't make sense graphically. This is the reason I think I am wrong

    If you don't understand what I am saying and need a visual, this site has a picture that shows what I'm talking about:

    http://www.clarku.edu/~djoyce/complex/mult.html

    However, (-1)^½(-1)^½=(-1*-1)^½ is much more interesting and that is not what you said before.
    Btw, I did say that i^2 can be 1 in my first post, but I didn't format it neatly, so it was hard to read.

    (((-1)^(.5))((-1)^(.5)))^2 is ((-1*-1)^(.5))^2 which is 1
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    Forum Radioactive Isotope mitchellmckain's Avatar
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    Quote Originally Posted by Nima Rahnemoon

    "Multiplying a complex number by i simply rotates it 90 degrees around 0, so 1i = i lies on the imaginary axis of the complex plane, a distance of 1 from zero. The number ii = i^2, then, is the number 1 rotated 180 degrees about zero, which puts it at -1. The number -i = -1i = i^3 is 1 rotated 270 degrees, and i^4 = 1."
    Of course I understand this. I did almost the same thing in my first post. But instead of say that multiplying by i rotates by 90 degrees, the point I made was that squaring a complex nubmer doubles its angle in the complex plane. i is at 90 degrees so squaring it gives you -1 at 180 degrees. Likewise 1+i is at 45 degrees so squaring it gives you 2i which is at 90 degrees.

    Quote Originally Posted by Nima Rahnemoon
    However, (-1)^½(-1)^½=(-1*-1)^½ is much more interesting and that is not what you said before.
    Btw, I did say that i^2 can be 1 in my first post, but I didn't format it neatly, so it was hard to read.

    (((-1)^(.5))((-1)^(.5)))^2 is ((-1*-1)^(.5))^2 which is 1
    No the problem for me was not the formatting but the fact that you were squaring both sides so I did not see what you were getting at.

    So I was recognizing that there was a real puzzle associated with (-1)^½(-1)^½=(-1*-1)^½. It was only your conclusion that i²= ±1 that was wrong. The real problem is in the square root function.

    Here I can explain the problem of branching using these angles.

    Just like squaring a complex number doubles its angle in the complex plane, the square root of a complex number halves its angle in the complex plane. So use this to consider the square root of 1. Is the angle 0 or is it 360 degrees? If you use 0 then half that is still zero and you get 1 again, but if you use 360 degrees (which is supposedly the same thing) then half of that is 180 degrees which is -1. Get it? How about -1? The angle is 180 degrees, and half of that is 90 degrees which is i, just like you said. But you can also think of -1 as being at the angle 540 degrees and half of that is 270 degrees which is -i. If you add another 360 it just brings you back to your first answer. For example, -1 is also at 900 degrees, and half of that is 450 degrees which is i again.
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  10. #9  
    Forum Sophomore NimaRahnemoon's Avatar
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    Quote Originally Posted by mitchellmckain
    Just like squaring a complex number doubles its angle in the complex plane, the square root of a complex number halves its angle in the complex plane. So use this to consider the square root of 1. Is the angle 0 or is it 360 degrees? If you use 0 then half that is still zero and you get 1 again, but if you use 360 degrees (which is supposedly the same thing) then half of that is 180 degrees which is -1. Get it? How about -1? The angle is 180 degrees, and half of that is 90 degrees which is i, just like you said. But you can also think of -1 as being at the angle 540 degrees and half of that is 270 degrees which is -i. If you add another 360 it just brings you back to your first answer. For example, -1 is also at 900 degrees, and half of that is 450 degrees which is i again.
    Wow, that is awesome. I tested it till i to the five and it works. what would ln|i| be? Graphically... Or ln|-1| ... ln(1).... which ever is better for explanation.

    Sorry for deviating from the topic of the thread....

    Edit: Nevermind, I realized that the absolute value of 'i' is still 'i', but 'i' isn't really a negative number. So is ln(i) possible?
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    Forum Radioactive Isotope mitchellmckain's Avatar
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    Quote Originally Posted by Nima Rahnemoon

    What would ln|i| be? Graphically... Or ln|-1| ... ln(1).... which ever is better for explanation.

    Sorry for deviating from the topic of the thread....

    Edit: Nevermind, I realized that the absolute value of 'i' is still 'i', but 'i' isn't really a negative number. So is ln(i) possible?
    Nah! You didn't deviate it I did. But then I already answered the orignial question as much as it should be.

    well |z| where z is a complex variable is usually defined as the square root of (z times its complex conjugate), so |i| = sqrt[(i)(-i)] = 1. Graphically |z| is the distance to the origin on the complex plane. Anyway since |z| is always real, ln|z| is just the usual real function ln(x) of this distance to the to the origin on the complex plane. The distance is always positive, so no problem exists, although it goes to negative infinity at the origin.

    So your Edit is wrong, you had it right the first time, ln|i| = ln(1) = 0

    The complex function ln(z) is defined using the exponential form of complex number z = |z| exp(i a) = |z| (cos a + i sin a) where a is the angle in the complex plane, so ln(z) = ln|z| + i a.

    So without the ||, ln(i) = i pi/2. So the ln(z) function gives a convenient way of expressing the angle in the complex plane: a = Im(ln(z)), where Im(z) means the imaginary part of z without the i.

    Anyway considering the branching character of ln(z), we see that it is has an infinite number of branches. So i pi/2 is only the value for ln(i) on one branch. Actually ln(i) = i pi/2, -3i pi/2, 5i pi/2, -7i pi/2, 9i pi/2, ....
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  12. #11  
    Forum Sophomore NimaRahnemoon's Avatar
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    Ohh, that's cool.

    I thought absolute value is sqrt(x²)

    |i|= sqrt(i²), and I believed we agree that i² is equal to -1, so |i|= sqrt(-1)=i

    So your Edit is wrong, you had it right the first time, ln|i| = ln(1) = 0
    Then the other definition, the distance from x to 0 is the |x|. So, the distance from i to 0 on the complex axis is i. right? But I do agree that the absolute value of a number should be a real number. So I guess that sqrt(x²) applies to real numbers only and sqrt[x(conjugatex)] is for imaginary, but then how does that work graphically, because the distance from i to 0 is i.

    z = |z| exp(i a) = |z| (cos a + i sin a) where a is the angle in the complex plane, so ln(z) = ln|z| + i a.
    Oooo, good stuff! Arg, dam teachers don't teach didly squat. Did you learn from classes or from books? So far classes haven't done me any good and frankly books usually make things more complicated than they really are.
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    Forum Radioactive Isotope mitchellmckain's Avatar
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    Quote Originally Posted by Nima Rahnemoon
    Then the other definition, the distance from x to 0 is the |x|. So, the distance from i to 0 on the complex axis is i. right? But I do agree that the absolute value of a number should be a real number. So I guess that sqrt(x²) applies to real numbers only and sqrt[x(conjugatex)] is for imaginary, but then how does that work graphically, because the distance from i to 0 is i.

    z = |z| exp(i a) = |z| (cos a + i sin a) where a is the angle in the complex plane, so ln(z) = ln|z| + i a.
    Oooo, good stuff! Arg, dam teachers don't teach didly squat. Did you learn from classes or from books? So far classes haven't done me any good and frankly books usually make things more complicated than they really are.
    No by distance on the complex plane I mean treating it just like an x-y plane so the distance from i to 0 is 1 not i.

    Remember I said z = |z| exp(i a) = |z| (cos a + i sin a)
    so for i remember that a is pi/2 or 90 degrees so cos a = 0 and sin a = 1
    thus you have i = |i| (0+i), which means |i| must be 1 not i.

    I got it in a college class I think, which means a book (text for the class), but teacher was pretty good as I remember. But these days you can get it off the internet (try for example, google complex logarithm).
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