Please need solution 1. Censider 4he functiou S.FOx) =aca02) and glx)= -Xin(3x)now, Except Speei Fred, use the Seed Secantmethod for a pproximation.%3D(a) The third Cp) and fetrth fifth (Ps) approxmatiònfor the[9,3;04], roundedvalueZero of gx) on the intervalto Seven deeimats areandP5 =b Compute f(8) (rounded oft toand g() (sCientific notation, rounded off toSu deamâts)Sit decimaus)(e) Determine, Pn; F(Pn) and 8 (P.) wherePn is the 1ntereept betwren fed and gointerval [o.1;0.2] withinon theof 10-5. whât is the required numberinterationsa tolerancen ofコ;ギ(R) =Gve Give Pay F(pn) and g CPn) roundedSix decimais%3D%3Dn=

Answered: I. Censder he function S. ând glx) F)…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

See similar textbooks

Related questions

Question

Please need solution

1. Censider 4he functiou S.
FOx) =aca02) and glx)= -Xin(3x)
now, Except Speei Fred, use the Seed Secant
method for a pproximation.
%3D
(a) The third Cp) and fetrth fifth (Ps) approxmatiòn
for the
[9,3;04], rounded
value
Zero of gx) on the interval
to Seven deeimats are
and
P5 =
b Compute f(8) (rounded oft to
and g() (sCientific notation, rounded off to
Su deamâts)
Sit decimaus)
(e) Determine, Pn; F(Pn) and 8 (P.) where
Pn is the 1ntereept betwren fed and go
interval [o.1;0.2] within
on the
of 10-5. whât is the required number
interations
a tolerance
n of
コ;ギ(R) =
Gve Give Pay F(pn) and g CPn) rounded
Six decimais
%3D
%3D
n=

Expert Solution

Step 1

Let $f (x) = - x \log (3 x)$

1st iteration :

$x 0 = 0.3 a n d x 1 = 0.4 f (x 0) = f (0.3) = 0.0137272 a n d f (x 1) = f (0.4) = - 0.0316725 ∴ x 2 = x 0 - f (x 0) \cdot \frac{x 1 - x 0}{f (x 1) - f (x 0)} x 2 = 0.3 - 0.0137272 \cdot \frac{0.4 - 0.3}{- 0.0316725 - 0.0137272} x 2 = 0.3302364 ∴ f (x 2) = f (0.3302364) = - 0.3302364 \log (0.9907092) = 0.0013387$

2nd iteration :

$x 1 = 0.4 a n d x 2 = 0.3302364 f (x 1) = f (0.4) = - 0.0316725 a n d f (x 2) = f (0.3302364) = 0.0013387 ∴ x 3 = x 1 - f (x 1) \cdot \frac{x 2 - x 1}{f (x 2) - f (x 1)} x 3 = 0.4 - (- 0.0316725) \cdot \frac{0.3302364 - 0.4}{0.0013387 - (- 0.0316725)} x 3 = 0.3330655 ∴ f (x 3) = f (0.3330655) = - 0.3330655 \log (0.9991965) = 0.0001163$

3rd iteration :

$x 2 = 0.3302364 a n d x 3 = 0.3330655 f (x 2) = f (0.3302364) = 0.0013387 a n d f (x 3) = f (0.3330655) = 0.0001163 ∴ x 4 = x 2 - f (x 2) \cdot \frac{x 3 - x 2}{f (x 3) - f (x 2)} x 4 = 0.3302364 - 0.0013387 \cdot \frac{0.3330655 - 0.3302364}{0.0001163 - 0.0013387} x 4 = 0.3333346 ∴ f (x 4) = f (0.3333346) = - 0.3333346 \log (1.0000038) = - 0.0000005$

4th iteration :

$x 3 = 0.3330655 a n d x 4 = 0.3333346 f (x 3) = f (0.3330655) = 0.0001163 a n d f (x 4) = f (0.3333346) = - 0.0000005 ∴ x 5 = x 3 - f (x 3) \cdot \frac{x 4 - x 3}{f (x 4) - f (x 3)} x 5 = 0.3330655 - 0.0001163 \cdot \frac{0.3333346 - 0.3330655}{- 0.0000005 - 0.0001163} x 5 = 0.3333333 ∴ f (x 5) = f (0.3333333) = - 0.3333333 \log (0.9999999) = 0$

The approximate root of the equation $- x \log (3 x) = 0$ using the Secant method is 0.3333333