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Need help writing 'math' from English

  1. Aug 14, 2014 #1
    I need to write, "(k)1/2= a finite term recursive expression of 'k' where 'k' is the set of all Reals where 'k' is greater than 0 and 'k' for all complex values" in math.

  2. jcsd
  3. Aug 15, 2014 #2

    Simon Bridge

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    Basic setbuilding notation:

    ...where do you get stuck?
    Personally I'm having trouble following the English description of what k is supposed to be.

    By "k is the set of... " do you mean that k is a member of the set, or is the label assigned to the set?
    Considering that the set of reals is a subset of complex numbers, are you saying that k is any complex number that is not a real number 0 or less?
    Last edited: Aug 15, 2014
  4. Aug 15, 2014 #3


    Staff: Mentor

    I too am confused by what you wrote as the description for k. The first part is pretty clear; namely, {k ##\in## R | k > 0}, but this part --
    -- seems to be incomplete.
  5. Aug 15, 2014 #4
    I didn't know that, and as such the answer to your last question is yes.

    'k' is the variable in the radical for a general form function (if this wording is awkward please let me know), as in a general form where √k=f(k) that is exact for all Complex values (since the Reals are a subset of complex) for k>0 .

    Considering these conditions, is this right?
    $$\{k\in \mathbb{C}\mid k>0\}$$

    Is there a global way to stating recursivity and finite termed expressions or do we just keep these statements in English?

    Performing derivations has never been an issue but the technical vernacular has always been troublesome and often proves itself more difficult a task than the work itself, hence concentrating more effort on this area to gain competency (in the meantime, please pardon my lack of refinement).
    Last edited: Aug 15, 2014
  6. Aug 15, 2014 #5
    I can understand your confusion, I made the mistake of not realizing that the set of all Reals is a subset of all Complex values. This is likely an elementary case so it is very good to have corrected.

    Considering this new insight, does my last post make more sense?
  7. Aug 15, 2014 #6


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    The Complex numbers have no natural order. So it makes little sense to write "k>0" when k is complex.

    I think that what you are trying to get at is that √k is well defined when k is both real and positive.

    For complex k there are always two solutions for x2=k except when k=0.

    For k positive and real, one solution is positive and one is negative. √k is taken to denote the positive one.

    For k negative and real, the two solutions are complex conjugates whose real part is zero. One is equal to 0+√(-k) i. The other is equal to 0-√(-k)i. Neither one is preferred.

    For k complex with a non-zero imaginary part the two solutions will be 180 degrees apart on the complex plane. Neither one is preferred.
    Last edited: Aug 15, 2014
  8. Aug 15, 2014 #7
    Okay so how about this,

    $$\{k\in \mathbb{C}\mid k>0-\infty i\}$$
    or should I split it up since complex numbers have no natural order and therefor the inequality is inappropriate?

    I was also thinking something along these lines as well,
    $$\{k\in \mathbb{C}\mid k\} \cup \{k\in \mathbb{R}\mid k>0\} $$ but this isn't correct either since it includes the whole set of complex numbers which includes both positive and negative reals. Is there a symbol in the notation to replace $$\cup$$ with something that represents 'except where' and then we just change the inequality to 'k< or equal to 0'?
  9. Aug 15, 2014 #8
    My apologies Simon, I forgot to answer this part of your post.

    By "k is the set of..." I was trying to imply at what values ('the set') is the variable k from f(k) true for √k.
  10. Aug 15, 2014 #9

    Simon Bridge

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    ##\mathbb C \cup \{x\in\mathbb R:x>0\}## is just the set of complex numbers.
    Remember the reals are a subset of complex numbers.
    The union of a set and it's subset is itself.

    Similarly, the intersection of a set and it's subset is the subset.

    You want to exclude the subset.

    k is a complex number that is not a negative real number.
    (Is k allowed to be zero? Proceeding as if k cannot be zero either.)

    i.e. The real part of k can be negative, but only if the imaginary part is not zero.

    Something like:$$k\in\mathbb C : k\notin \{x\in\mathbb R: x\leq 0\}$$... ie. k is a member of the relative compliment of complex numbers and negative real numbers.$$k\in\mathbb C \backslash \{x\in\mathbb R: x\leq 0\}$$
    How about extracting the definition of a complex number by introducing two arbitrary reals:$$k=x+iy: x,y\in \mathbb R \land \forall (x,y):y=0, x>0$$ ... there's probably a better way of doing that...

    Are there numbers that are not in the set of complex numbers that you don't want k to be?
    If not, then the representation is very simple.

    Aside: If ##x,y\in\mathbb R## and ##i=\sqrt{-1}## then ##z=x+iy\in\mathbb C##.
    Since y=0 is a real number, it follows that ##z=x\in\mathbb C##,
    i.e. all real numbers are members of the set of complex numbers.

    Notes: we have all been using "blackboard bold" for special sets.
    Some people advise against it i.e.
    Krantz, S., Handbook of Typography for the Mathematical Sciences, Chapman & Hall/CRC, Boca Raton, Florida, 2001, p. 35.
    Last edited: Aug 15, 2014
  11. Aug 16, 2014 #10


    Staff: Mentor

    I would write the set this way:
    {z ##\in## C | Re(z) > 0}. This gives you the right half of the complex plane.
  12. Aug 16, 2014 #11

    Simon Bridge

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    <puzzled> which set would you write that way?

    In the set described in post #1 (and clarified later) the real part is allowed to be negative but only if the imaginary part is non-zero. If you wanted to represent that on the complex plane, you'd need a hole along the entire negative real axis. Everything else is allowed.

    The set quoted is just the complex plane.

    Neither of these are well represented by the right half of the complex plane.
    However, the Re/Im notation is probably a better way to handle my k=x+iy version... $$\{k\in\mathbb C : \forall\; \Im(k)=0, \; \Re(k)>0\}$$ ... "forall" may not be right there.

    Would this work: ##\{k\in\mathbb C : \text{Arg}(k)\neq\pi\}##

    Aside: put a backslash in front of the "Re(z)" in LaTeX and you get ##\Re(z)## ;)
  13. Aug 16, 2014 #12


    Staff: Mentor

    Maybe I misunderstood, especially since it has taken a number of posts to figure out what mesa is trying to get. What I got out of a quick read of the thread was that he was looking for complex numbers for which the real part was positive (or maybe nonnegative).
  14. Aug 16, 2014 #13
    Okay, this makes sense. All that is needed is the $$\notin$$ notation to exclude negative Reals from the complex plane. Or for the second example the '\' in place of $$: k\notin$$
    Very nice!

    Once again, this makes sense as well, thank you for the clarification!

    This part I am a little confused on, what numbers are left that are outside the set of complex numbers since it also contains all Reals when bi=0?

    The history of 'blackboard bold' is interesting and it seems was originally intended for when 'hand writing' these things as 'bold' (hard to do otherwise) and was never intended for printed works in the first place (although one could argue it is aesthetically pleasing).

    On another note, in order to improve with these types of things do you have another (preferably more frugal) recommendation for mathematical typography?
  15. Aug 16, 2014 #14

    Simon Bridge

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    See http://en.wikipedia.org/wiki/Complement_(set_theory [Broken]) for the use of compliments in set theory.

    I think the most succinct way is: ##\{k\in\mathbb C : \text{Arg}(k)\neq \pi\}##
    ... although I think that includes k=0, since it's argument is undefined. Easy to fix.

    What is the context of this?

    I'm assuming it's not an arbitrary homework question because of how tricky the wording was to parse. It would have been quite tricky to guide you to the compliment or argument solution too.

    There's quaternions etc.

    You'll notice when you go from reals to complex numbers, you lose the property of being ordered?

    Quaternions (a 4-dimensional normed R-algebra) lose commutativity, but are associative.
    Octonions (an 8-dimensional normed R-algebra) lose associativity, but are alternative.
    Sedenions (a 16-dimensional R-algebra) lose alternativity, and are also not a normed R-algebra, because there are zero-divisors.

    http://en.wikipedia.org/wiki/Hurwitz%27s_theorem_(normed_division_algebras [Broken])

    Last edited by a moderator: May 6, 2017
  16. Aug 16, 2014 #15
    Ii is fast becoming apparent I require more study on this topic.

    I enjoy finding identities and exploring math in general. This particular function I thought was a neat result but lacking the proper mathematical vernacular makes it difficult to share with people who really know these things.

    I haven't studied these, but have heard of them. I will look through your links!
    Last edited by a moderator: May 6, 2017
  17. Aug 16, 2014 #16

    Simon Bridge

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    Neat result of what?

    ... quaternions have application in computer science for their ability to efficiently represent rotations.

    Basically, define basis elements i,j,k (not the cartesian unit vectors) so that ##i^2=j^2=k^2=ijk=-1##
    A quaternion is q=a+bi+cj+dk.
    ... you can similarly look up the others.

    However, the reason I brought it up is because sometimes an overarching set can be left implicit ... i.e. ##\{a:a<0\}## would usually be interpreted as the set of negative reals, and ##\{a:\Re(a)>0\}## would imply complex numbers are intended.
    So you could write ##\{k:\text{Arg}(k)\neq \pi \}## and be confident that people would realise you meant to define k over the complex plane.

    It's a language - so there are lots of ways to express yourself. The exact approach you choose depends on what you want to draw the readers attention to.
  18. Aug 16, 2014 #17
    I was looking at the golden ratio and trying to work a general form, after the task was completed (and some new insights gained) it appeared it may be possible to build a general form function for radicals (our f(k)=√k). The result I thought was 'neat'.

    I will look into this, although will likely have some questions.
  19. Aug 17, 2014 #18

    Simon Bridge

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    Oh right.

    You were thinking that k can be any complex not a negative real?
    Positive reals can have real square-roots fersure, but I don't see how excluding negative reals from k helps considering you are allowing any complex number otherwise.

    Normally the radical indicates a positive square root so that if ##y=x^2## then ##x\in\{\pm\sqrt{y}\}## ... that can mess things up. The result is that the negative reals are excluded, by definition, from the solution set.
    This what you are thinking of?

    Don't know what you mean by "general form function" in this context.
    Anyway - that would be for another thread ;)
  20. Aug 17, 2014 #19
    Not quite (and my apologies for the confusion), this isn't an attempt to define where √k is true but to show where a specific identity is true.

    In other words if you want to see √3 as a function of '3' and 'phi' then this function will give an exact result but it does not work for √-3≠f(-3,phi), although it will for √(-3)^(1/2)=f((-3)^(1/2),phi). Does that clear things up?

    I can't think of an example off the the of my head but, as you say, a discussion for another thread.
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