1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How 3=2 is true?,help me to find out the mistakes

  1. Feb 4, 2013 #1
    m^2+m^2+m^2+.........up to m term =m^2*m=m^3
    Which is identity .So we use derivative for this identity.
    Now, d/dm(m^2+m^2+m^2+.....up to m term)=d/dm(m^3)
    Or 2m+2m+2m+........up to m term=3m^2
    Or 2m*m=3m^2
    Or 2m^2=3m^2 but it is not possible.So there should be some mistakes.But I'm not getting it.
    What I suspect are as follows:
    In L.H.S we differentiated first,then add all terms.In R.S we added all then differentiated it.Is the mistake for maintaining different orders of operation?
    I suspect another reason like while differentiating R.S, number of terms is considered as variable and multiplied with m^2 that leads to m^3,but as it is no. of terms if I assume it as constant then, m*d/dm(m^2)=m*2m=2m^2,which matches with L.S. Is this okay?or something else?
    Thanks for your help.
     
    Last edited: Feb 4, 2013
  2. jcsd
  3. Feb 4, 2013 #2
    Here is my thought:
    you have m-terms of m^2. That is, m*m^2. Taking derivative, d\dm (m*m^2)=m^2+m*2m=3m^2.
    The problem with d/dm(m^2+m^2+m^2+.....up to m term) is that m-term is also a function of m. So, technically, you need to differentiate the function (m-term)
     
  4. Feb 4, 2013 #3

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    You are not allowed to take a derivative of a sum like ##S(m) = \sum_{i=1}^m a_i## with respect to m. Think about it: if S(m) had a derivative, it would be
    [tex] S'(m) = \lim_{h \to 0} \frac{\sum_{i=1}^{m+h} a_i - \sum_{i=1}^m a_i}{h}
    = \lim_{h \to 0} \frac{\sum_{i=m}^{m+h} a_i}{h}.[/tex] If you take, for example, h = 0.001, what on earth could you possibly mean by ##\sum_{i=m}^{m + 0.001} a_i?## It makes no sense.
     
  5. Feb 4, 2013 #4
    Thanks to all for your help.
    I couldn't understand this line,
    ##\sum_{i=1}^{m+h} a_i - \sum_{i=1}^m a_i=\sum_{i=m}^{m+h} a_i##
    ##\sum_{i=m}^{m+h} a_i=a_m+a_{m+1}+a_{m+2}+...+a_{m+h}## (assuming h is some integer)But terms upto ##a_m## should be canceled by ##\sum_{i=1}^{i=m}a_i## then what remains is ##a_{m+1}+a_{m+2}+...+a_{m+h}## which is ##\sum_{i=m+1}^{i=m+h}a_i## .Now in this case as h→0, after subtraction we'll get a term ##a_{m+h}## that fails to exist.
    Another thing I couldn't understand is:
    [tex]S(m)=\sum^{i=1}_{i=m}a_i[/tex],Whether a is constant or function of m?
    However ,it seems I got your point that as number of terms can't be a fraction so we can't get derivative,like ##a_{m+0.001}## can't exist.Am I right?
     
  6. Feb 4, 2013 #5

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Whew! Glad to see that's settled, because it is well known that 3 = 2 only for a large value of 2. :smile:
     
  7. Feb 4, 2013 #6
    R: How 3=2 is true?,help me to find out the mistakes

    Your left hand term is only defined for integer values of m, so d/dm makes no sense.
     
  8. Feb 4, 2013 #7

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Yes, order has been restored. We can all sleep easy tonight! :wink:
     
  9. Feb 4, 2013 #8

    Mark44

    Staff: Mentor

    Or a very small value of 3...
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: How 3=2 is true?,help me to find out the mistakes
  1. Please help me out (Replies: 2)

Loading...