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Approximation formula proof

  1. May 4, 2014 #1
    Hi.
    Please help me prove the approximation formula below given in my book. This is not homework question.
    Thanks.

    attachment.php?attachmentid=69383&stc=1&d=1399188366.jpg
     

    Attached Files:

  2. jcsd
  3. May 4, 2014 #2

    mfb

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    You just neglect ##\tau_1## or ##\tau_2##, whatever is smaller. Note that a larger value will lead to a larger exponential as well (for positive t).
     
  4. May 5, 2014 #3
    Thank you. However, that doesn't solve the problem. What need to be proved is different.
     
  5. May 5, 2014 #4
    [tex]\frac{τ_1e^{-\frac{t}{τ_1}}-τ_2e^{-\frac{t}{τ_2}}}{τ_1-τ_2}=\frac{e^{-\frac{t}{τ_1}}-(τ_2/τ_1)e^{-\frac{t}{τ_2}}}{1-(τ_2/τ_1)}=e^{-\frac{t}{τ_1}}\left(\frac{1-(τ_2/τ_1)e^{-t(\frac{1}{τ_2}-\frac{1}{τ_1})}}{1-(τ_2/τ_1)}\right)=e^{-\frac{t}{τ_1}}\left(\frac{1-(τ_2/τ_1)e^{-t\frac{(τ_1-τ_2)}{τ_1τ_2}}}{1-(τ_2/τ_1)}\right)[/tex]

    If τ1>>τ2, then [itex]e^{-t\frac{(τ_1-τ_2)}{τ_1τ_2}}<1[/itex].

    From this, it follows that, in the numerator, [itex](τ_2/τ_1)e^{-t\frac{(τ_1-τ_2)}{τ_1τ_2}}<<1[/itex]

    Also, in the denominator, [itex](τ_2/τ_1)<<1[/itex]

    So the term in parenthesis approaches unity.

    Chet
     
  6. May 6, 2014 #5

    mfb

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    Why? It shows that the big is a correct approximation.
     
  7. May 6, 2014 #6
    I guess the analysis I did in #4 did not work for the OP? (It was just a more fleshed-out version of what mfb was saying).

    Chet
     
  8. May 7, 2014 #7
    Thank you all.
    I think you misread the question a bit. The expression on the right hand side of the equation is [tex]e^{-\frac{t}{τ}}[/tex] with [tex]τ = τ_1 + τ_2[/tex].
     
  9. May 7, 2014 #8
    You're right. But that doesn't matter much. The same general procedure could be used for this.

    Chet
     
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