MHB Approximating the value of a function

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The discussion revolves around using binary search to approximate the square root of 5 by analyzing the function f(x) = x^2 - 5 within the interval [2, 3]. The user calculated values at various points and arrived at 2.1875 as the approximation. A participant clarified that the maximum deviation from this approximation is 0.0625, confirming that it meets the accuracy requirement of 1/16. The conversation highlights the method of determining accuracy in root-finding using binary search. Understanding the bounds of the approximation is crucial for verifying the desired accuracy.
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Hello,

I have a problem which I did solve, but I ain't sure of something in it, something small to be honest.

The question is:

Use a binary search and the function f(x)=x^2-5 in the domain [2,3] to find approximation for the value of the square root of 5, with an accuracy of 1/16.

What I did, I tried finding where the function is equal to 0, i.e. x^2-5=0 meaning x is the square root of 5.

I calculated f(2) and f(3), and then f(2.5), f(2.25), f(2.125) and finally f(2.1875), which is the answer (2.1875 is the answer). My simple question is, how, or where, do I see that I am 1/16 accurate ?

Thank you !
 
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Hi Yankel,

You didn't mention the values for f, but either way, the actual root is either between 2.125 and 2.1875, or between 2.1875 and 2.125.
So the maximum deviation from 2.1875 is 0.0625, which is 1/16.
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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