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ksaimouli
 4 years ago
estimate roots newtons linearization
ksaimouli
 4 years ago
estimate roots newtons linearization

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ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1353960049993:dw

phi
 4 years ago
Best ResponseYou've already chosen the best response.1Is this what you mean http://en.wikipedia.org/wiki/Newton's_method

phi
 4 years ago
Best ResponseYou've already chosen the best response.1If you define a function f(x)= x^25 it will have a zero when x is sqrt(5) You can use Newton's method to find that zero (x value) \[ x_{n+1}= x_n  \frac{f(x_n)}{f'(x_n)} \]

ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.0but i need the estimation of root5 i think ur r thinking other one

phi
 4 years ago
Best ResponseYou've already chosen the best response.1Maybe you mean a taylor series approximation? f(x) ~ f(a) + f'(a)(xa)

phi
 4 years ago
Best ResponseYou've already chosen the best response.1I would pick a=4 because you know the square root of 4

phi
 4 years ago
Best ResponseYou've already chosen the best response.1here f(x)= \(x^{\frac{1}{2}}\) \[ f'(x)= \frac{1}{2}x^{1/2} \] evaluate this at x= a = 4

ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1353961871007:dw

phi
 4 years ago
Best ResponseYou've already chosen the best response.1yes, so your approximate function is f(x) = 2 + (1/4)*(x4) reasonable for x values near 4

ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.0how did u get 2 u said 4

phi
 4 years ago
Best ResponseYou've already chosen the best response.1you start with f(x) ~ f(a) + f'(a) (xa) f(x)= sqrt(x) with a=4, f(a) is sqrt(4)

ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1353962194869:dw

phi
 4 years ago
Best ResponseYou've already chosen the best response.1the question asks for sqrt(5) , so plug in 5 for x into your estimation equation

phi
 4 years ago
Best ResponseYou've already chosen the best response.1yes, and 2.25^2 is 5.0625 which is pretty close to 5

ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.0got u one small doubt what if it is root31 shall i pick root36 or root25

phi
 4 years ago
Best ResponseYou've already chosen the best response.1pick the closer one. 31 is in the middle, so either end will not be very accurate.

ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.0root30 but this is not perfect square

phi
 4 years ago
Best ResponseYou've already chosen the best response.130 is not a perfect square. so you want to pick a nearby "a" where you know the root. 25 is the closest square. 36 is almost as close. Using 25 is probably better by a little bit that using 36

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0not to butt in but newton's method in this case is identical to mechanics rule

phi
 4 years ago
Best ResponseYou've already chosen the best response.1I never heard of Professor Mechanic

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you can use \[x_{n+1}=\frac{1}{2}\left (x_n+\frac{5}{x_n}\right)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0mechanic predated newton in fact i think mechanic was babylonian, maybe circa 3000 BC it is an old method for finding roots, but if you do the algebra with newtons method for square roots it ends up being the same. the recursion to find \(\sqrt{a}\) is \[x_{n+1}=\frac{1}{2}\left (x_n+\frac{a}{x_n}\right)\]

phi
 4 years ago
Best ResponseYou've already chosen the best response.1es, for f(x)= x^2 a and f'(x)= 2x newton's x f(x)/f'(x) is x  (x^2a)/2x = (2x^2 x^2 +a)/2x which becomes mechanic with a few more steps
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