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estimate roots newtons linearization
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|dw:1353960049993:dw|
yes
If you define a function f(x)= x^2-5 it will have a zero when x is sqrt(5) You can use Newton's method to find that zero (x value) \[ x_{n+1}= x_n - \frac{f(x_n)}{f'(x_n)} \]
can you find f'(x) ?
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but i need the estimation of root5 i think ur r thinking other one
Maybe you mean a taylor series approximation? f(x) ~ f(a) + f'(a)(x-a)
hmm yup
I would pick a=4 because you know the square root of 4
exactly the same
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ohk
here f(x)= \(x^{\frac{1}{2}}\) \[ f'(x)= \frac{1}{2}x^{-1/2} \] evaluate this at x= a = 4
|dw:1353961871007:dw|
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