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January 9th, 2019, 10:22 AM   #1
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Linear approximation

Using linear approximation calculate sqrt(99)?
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January 9th, 2019, 11:00 AM   #2
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Using linear approximation calculate sqrt(99)?
Let $\displaystyle y(x) = \sqrt{x}$. Expand this in a Taylor series (with x small compareed to a):
$\displaystyle y(a - x) \approx y(a) + \left . \dfrac{dy}{dx} \right | _{x = a}\cdot (a - x)$

So let a = 100 and x = 1.

-Dan
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January 9th, 2019, 12:55 PM   #3
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Using linear approximation calculate sqrt(99)?
I'm lazy: slightly less than 10
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January 9th, 2019, 02:10 PM   #4
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It is between 9 and 10 and approximation will change for different values of x and a
$\displaystyle y(x)\approx y(a)+y’(a)(x-a) $
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January 9th, 2019, 04:50 PM   #5
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Quote:
Originally Posted by topsquark View Post
Let $\displaystyle y(x) = \sqrt{x}$. Expand this in a Taylor series (with x small compareed to a):
$\displaystyle y(a - x) \approx y(a) + \left . \dfrac{dy}{dx} \right | _{x = a}\cdot (a - x)$

So let a = 100 and x = 1.

-Dan
Shouldn't that be
$\displaystyle y(a - x) \approx y(a) - \left . \dfrac{dy}{dx} \right | _{x = a}\cdot x$
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January 9th, 2019, 04:58 PM   #6
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Originally Posted by v8archie View Post
Shouldn't that be
$\displaystyle y(a - x) \approx y(a) - \left . \dfrac{dy}{dx} \right | _{x = a}\cdot x$


Thanks for the catch.

-Dan
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