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Approximation by differential

  1. Jun 16, 2017 #1
    1. The problem statement, all variables and given/known data
    Approximate ##~\sqrt[4]{17}~## by use of differential

    2. Relevant equations
    Differential: ##~dy=f(x)~dx##

    3. The attempt at a solution
    $$y=\sqrt[4]{x},~~dy=\frac{1}{4}x^{-3/4}=\frac{1}{4\sqrt[4]{x^3}}$$
    $$\sqrt[4]{16}=2,~~dx=1,~~dy=\frac{1}{4\sqrt[4]{x^2}}\cdot 1=0.149$$
    $$\sqrt[4]{17}=2.031$$
    The error is too big
     
  2. jcsd
  3. Jun 17, 2017 #2

    Ray Vickson

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    Homework Helper

    Nevertheless, your result is correct.

    If you want better accuracy you need to take additional, higher order terms (2nd derivatives, maybe 3rd derivatives,etc).
     
  4. Jun 17, 2017 #3
    Thank you Ray
     
  5. Jun 17, 2017 #4

    Mark44

    Staff: Mentor

    Here is how I would write things.
    ##f(x + \Delta x) \approx f(x) + df \approx f(x) + f'(x) \Delta x = x^{1/4} + \frac 1 4 x^{-3/4} \Delta x##
    When ##x = 16## and ##\Delta x = 1##, we have
    ##\sqrt[4]{16 + 1} \approx 16^{1/4} + \frac 1 4 \frac 1 {16^{3/4}} \cdot 1 = 2 + \frac 1 32 = 2.03125##
    By calculator, ##\sqrt[4]{17} \approx 2.03054##. Since ##\Delta x = 1## is relatively large in comparison to x = 17, the approximation using differentials isn't all that accurate. If ##\Delta x## were smaller, the approximation would be better.
     
  6. Jun 17, 2017 #5
    Thank you Mark44
    By the way, how do i copy your names here, i write them again. when i pause the mouse on your name it becomes a pointer and there is no option to copy
     
  7. Jun 17, 2017 #6

    WWGD

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    Just hit the 'Reply' button on the lower right of the post. You may also hite the 'Quote' button , also on the lower right.
    BTW, small mistake: dy=f'(x)dx , not dy=f(x)dx , unless f(x)=f'(x).
     
  8. Jun 18, 2017 #7
    Thanks, but i mean i want to copy your name, WWGD, to here, instead of looking and typing it. i usually thank every one that answered my question
     
  9. Jun 20, 2017 #8

    WWGD

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    Maybe you can just use the 'Like' button as a means of thanking.
     
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