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Working on a Little Personal Project

  1. Apr 3, 2013 #1
    I enjoy learning things and then working on my own personal studies. I'm loving calculus right now and am working with it as much as possible in between my other classes. The problem I have set out for myself right now is to find a limit that represents ∏. Before I go on can someone please explain to me how to use the math font? When I try to use it all I get is a bunch of words and symbols I don't know how to use it. Thanks.
  2. jcsd
  3. Apr 4, 2013 #2
    Hi, I'm learning the latex language as well. Here are a few starting points based on a limited breadth of knowledge I have on it so far:

    First off, in order for the equations to display, you must have the symbols within the itex or tex brackets. That means:




    Click quote on my post so you can see what I typed in order to make the word display like this.

    For limits, I haven't found the exact button in the latex reference, but by clicking the quote button on other members' posts, I've found the notation to be:

    [itex]\lim_{h\rightarrow 0 } \frac{f(x+h)-f(x)}{h}[/itex]


    [tex]\lim_{h\rightarrow 0 } \frac{f(x+h)-f(x)}{h}[/tex]

    So you type \lim_{stuff} ...just make sure you tack on the [itex] notation on each end.

    for exponentials, you can find the language in the latex reference table, its just a^{stuff} or 1^{2} or 20^{50} etc
    Last edited: Apr 4, 2013
  4. Apr 4, 2013 #3

    [tex]/lim_{rightarrow0 } \frac{ln(x+h)-ln(x)}{h}[/tex]
  5. Apr 4, 2013 #4
    [itex]\lim_rightarrow 0{f(x+h)-f(x)/h}[itex]
    Last edited by a moderator: Apr 4, 2013
  6. Apr 4, 2013 #5
    Thanks for the post. I will work on it but I have been up too long to work with it now. So from what it looks like it's just memorization? How do I know where to put spaces, brackets, slashes, ect? Thanks.
  7. Apr 4, 2013 #6


    Staff: Mentor

    Like this:
    $$\lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$$
    Right-click the expression above to see how I did it.
  8. Apr 4, 2013 #7

    This would have been correct, the only two reasons the limit didn't display is because A) you typed a forward slash before lim_{} instead of a backslash, which is my fault because that's what I wrote. And secondly, there should be a backslash before the word rightarrow It should have been:

    \lim_{\rightarrow 0 } instead of: /lim_{\rightarrow0 }

    Even without signifying a number before the rightarrow, the notation will still display..
    [tex]\lim_{\rightarrow 0 } \frac{ln(x+h)-ln(x)}{h}[/tex]

    but when you add a number there, you would put it just before the backslash and it will display like this:

    [tex]\lim_{h\rightarrow 0 } \frac{ln(x+h)-ln(x)}{h}[/tex]

    P.S. I didn't realize you could right click on the expressions to get the notation as Mark has stated. Much easier than clicking the quote button to see how expressions were written.
  9. Apr 4, 2013 #8


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    Staff Emeritus
    Science Advisor

  10. Apr 4, 2013 #9


    Staff: Mentor

    I always use \to instead of \rightarrow, since it's much quicker to type.
    The other form is \Rightarrow (renders as ##\Rightarrow##) which you can use to show that one statement implies another, like so:

    (x - 2)2 = 4 ##\Rightarrow## x = 0 or x = 4.
  11. Apr 4, 2013 #10
    Ok thank you everyone very much! I'm going to work on the font these upcoming days.
    \lim_{rightarrow 0 } /frac{ln(x+h)-ln(x)}{h} ...Just a practice limit. I will watch the tutorial as well over and over until I can get most of the little details memorized like spaces and what not, then I will work on the initial question I asked. Thanks.
  12. Apr 4, 2013 #11
    well that limit didn't work -_-
  13. Apr 5, 2013 #12
    Have a look at

    [tex]\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{1}{n^2}[/tex](Basel's problem, beautiful proof by Euler of its limit), and
    [tex]\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{1}{n^4}[/tex]

    Or perhaps more generally
    [tex]\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{1}{n^{2m}}[/tex]
  14. Apr 5, 2013 #13
    Thanks! The limit on all those is zero right? (As long as the value of m on the last problem is positive and greater than zero.)
  15. Apr 5, 2013 #14
    [tex]\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{n}{2n+3}[/tex]
  16. Apr 5, 2013 #15
    What????What did I do wrong?
  17. Apr 5, 2013 #16
    No, the limit of [tex]\lim_{n \rightarrow \infty} \frac{1}{n^{2m}}[/tex] goes to 0, not the sum. For example:

    [tex]\lim_{n \rightarrow \infty} \sum_0^{n}\frac{1}{n^2}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}...= \frac{\pi ^2}{6}[/tex]
  18. Apr 5, 2013 #17
    I'm sorry I don't know what I was thinking when I wrote that. That limit is not what I would have expected though. Is it just the result of a proof?
  19. Apr 6, 2013 #18
    Yes, as is everything in maths I suppose (it's not an axiom, if that's what you mean).
  20. Apr 6, 2013 #19
    Just out of curiosity how would I find that proof? I think its interesting that something apparently non trigonometric has a sum involving pi. Although I know a great deal of problems can be turned into trigonometric ones, i'm just a little curious about this one. Thanks.
  21. Apr 6, 2013 #20
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