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Simple mathematical problem

  1. Feb 5, 2014 #1
    Is this equation equal to:

    (e^(hf/kT)) - 1


    e^( (hf/kT) - 1 )

    http://s29.postimg.org/le6iqy3rb/exp.png [Broken]
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Feb 5, 2014 #2
    The former.
  4. Feb 5, 2014 #3


    Staff: Mentor

    First off, what you wrote is NOT an equation. An equation always has an = symbol in it.

    The image in the link is [exp(hf/kT) - 1].

    What you have written is ambiguous, as what you probably meant is this:
    $$e^{\frac{hf}{kT} - 1}$$

    What you actually wrote, though, is this:
    $$e^{\frac{hf}{k}T - 1}$$

    The brackets - [] - around the entire expression are unnecessary.
    Last edited by a moderator: May 6, 2017
  5. Feb 5, 2014 #4


    User Avatar

    Staff: Mentor

    Why did you put the -1 in the exponential? The parenthesis limit the argument of exp to hf/kT.

    My guess is that Mechatron did not write that himself, but saw it in a book. It's most probably related to the Planck distribution (blackbody radiation). As economicsnerd said, the correct reading is
    e^{\beta h f} - 1 \mbox{ where } \beta = \frac{1}{kT}
    The additional bracket [] might be there because it is part of a greater equation.
  6. Feb 15, 2014 #5
    Im sorry to be off topic but I'm using a i device and i can't see (what i think to be) mathematical symbols that are in this thread... They appear as dollar signs and other randoms, i was wondering if maybe I'm short of additional download or setting adjustments. All help is highly appreciated
  7. Feb 15, 2014 #6


    Staff: Mentor

    The posted image, which doesn't have the -1 term, doesn't match the expressions in the first post. In the first post Mechatron asks about these expressions:
    (e^(hf/kT)) - 1


    e^( (hf/kT) - 1 )

    In any case, this is moot, as Mechatron has been banned from PF.

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