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Linear Approximation

  1. Aug 16, 2007 #1
    1. The problem statement, all variables and given/known data

    OK, I'm doing this linear approximation problem:

    Approximate [tex]\sqrt{4.1} - \sqrt{3.9}[/tex]

    2. Relevant equations

    f(a + h) ~ f(a) + hf`(a)

    3. The attempt at a solution

    This is what I have done so far:

    I approximated each square root separately...

    4.1 = 4 + h
    h = .1
    f(x) = [tex]\sqrt{x}[/tex]
    f`(x) = [tex]1/(2\sqrt{x}[/tex]

    then I got:

    [tex]\sqrt{x+h} ~ \sqrt{x} + h/(2\sqrt{x})[/tex]
    so that the approximation of [tex]\sqrt{4.1}[/tex] is 2 + (.1/4)

    I did the same thing for [tex]\sqrt{3.9}[/tex] and got

    2 - (.1/4)

    Then I just took them and subtracted

    2 + (.1/4) - 2 + (.1/4)
    and got .2/4 as the approximation. It seems like it's erroneous. Could someone help me out with setting up this problem if I did it wrong?
     
  2. jcsd
  3. Aug 16, 2007 #2
    All you steps are right. Why do you think it is erroneous?
     
  4. Aug 16, 2007 #3
    It seems to me that it is a large margin of error, I was just wondering why this is so.

    Or maybe it's not too large, I'm not sure...
     
  5. Aug 16, 2007 #4

    learningphysics

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    Homework Helper

    everything looks right to me.

    check your calculator and calculate [tex]\sqrt{4.1} - \sqrt{3.9}[/tex]

    it's very close to 0.2/4 = 0.05
     
  6. Aug 16, 2007 #5
    Thanks.
     
  7. Aug 16, 2007 #6
    It's an approximation. Without a calculator, it's not an easy task for many to calculate the square root of an arbitrary number, unless it's a "nice" number like a perfect square. In those cases, you can use this method to approximate square roots to find a number reasonably close to the right one.
     
  8. Aug 16, 2007 #7
    So let's say I was doing something like e^(-0.1) - ln (1.1)

    I can just go about doing it thus:

    h = -0.1

    so that:
    -0.1 = 0 + h and 1.1 = 1 - h



    for the first term:

    f(x) = e^(x)
    f`(x) = e^(x)

    but then how do I go about it from there...

    Could someone help me setup the problem?

    I guess this would work, wouldn't it:

    e^(0) - h/(e^(0))
    1 - 0.1 = 0.9

    Which seems reasonable...

    And then for the ln(1.1) it would be
    f(x) = ln(x)
    f`(x) = 1/x

    ln(1) - h/x
    which would give me
    0 - (-0.1/1)
    = 0.1

    That doesn't make any sense...

    What'd I do wrong.
     
  9. Aug 16, 2007 #8

    Integral

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    Staff Emeritus
    Science Advisor
    Gold Member

    What's the problem?
     
  10. Aug 16, 2007 #9
    EDIT:

    Should I use -0.1 for h or can i use 0.1.....
     
    Last edited: Aug 16, 2007
  11. Aug 16, 2007 #10
    When I do:

    e^0 - (h / e^0)
    = 1 + (0.1/1)
    = 1.1

    It should be 0.9 ....

    I have the correct approximation for -ln(1.1) = 0.1

    but I made a mistake

    Shouldn't it be 1 PLUS 0.1 to give me 1.1? Cuz h = NEGATIVE 0.1

    What's wrong here?
     
  12. Aug 16, 2007 #11
    Holy crap man, nevermind. I've been up for like 3 days straight so I'm buggin' out.

    Thanks for your insight, guys :)
     
  13. Aug 16, 2007 #12

    learningphysics

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    Homework Helper

    Although the numbers come out right, I'm wondering why you are dividing by e^(0) instead of multiplying by e^(0)
     
  14. Aug 16, 2007 #13

    learningphysics

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    Homework Helper

    I would advise that you approximate each term separately so that you don't get confused with the signs... Also, define h consistently...

    so I'd do this:

    let h = [tex]x_{actual}-x_{approximate}[/tex]

    So if I'm approximating e^(-0.1) by e^0...

    h = -0.1 - 0 = -0.1

    And then use this exact formula where a refers to the approximate x value:

    f(a + h) ~ f(a) + hf`(a) (don't switch from h to -h or anything like that... be consistent)

    So: f(0 + (-0.1)) ~ f(0) + (-0.1)f'(0)

    ie: e^(0 + (-0.1)) ~ e^(0) + (-0.1)(e^(0)) = 0.9

    So e^(-0.1) ~ 0.9

    Now approximate ln(1.1) as a separate problem, defining h again...

    It is important to be consistent... worry about signs at the end...
     
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