{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Hw_Sol - Solutions to selected exercises Use the Bisection...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Solutions to selected exercises Use the Bisection method to find solutions accurate to within 10 2 for x 3 7 x 2 + 14 x 6 = 0 on [0,1]. Solution: Let f ( x ) = x 3 7 x 2 + 14 x 6 = 0. Note that f (0) = 6 < 0 and f (1) = 2 > 0, therefore, based on the Intermediate Value Theorem, since f is continuous, there is p (0 , 1) such that f ( p ) = 0. Let a 0 = 0, b 0 = 1, with f ( a 0 ) < 0, f ( b 0 ) > 0. Let p 0 = a 0 + b 0 a 0 2 = 0 . 5, and we have f ( p 0 ) = 0 . 6250 < 0 (the same sign as f ( a 0 ), therefore a 1 = p 0 = 0 . 5, b 1 = b 0 = 1 and repeat: p 1 = 0 . 75, ... This yields the following results for p n and f ( p n ): n p n f ( p n ) 0 0.5 -0.6250000 1 0.75000000 +0.9843750 2 0.62500000 +0.2597656 3 0.56250000 -0.1618652 4 0.59375000 +0.0540466 5 0.57812500 -0.0526237 6 0.58593750 +0.0010313 Use the theorem from the course to find a bound for the number of iterations needed to achieve an approximation with accuracy 10 3 to the solution of x 3 x 1 = 0 lying in the interval [1 , 4]. Solution: Let’s first verify that f has a zero in the interval [1 , 4]: f (1) = 2 < 0, f (4) = 64 > 0, therefore, since f is continuous, by the Intermediate Value Theorem, f has a zero in [1 , 4]. By the theorem from the course, we impose: | p n p | ≤ b a 2 n = 3 2 n 10 3 , then 3 · 10 3 2 n n log 10 (3 · 10 3 ) log 10 (2) 11 . 55 Use algebraic manipulation to show that each of the following functions has a fixed point at p precisely when f ( p ) = 0, where f ( x ) = x 4 +2 x 2 x 3.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}