Numerical Methods for Civil Engineers A (APM3715)

1-Secant Method
Given:
𝑥 = cos𝑥 𝑓(𝑥) = cos𝑥 − 𝑥



Formula:
𝑓(𝑥 )(𝑥 − 𝑥 )
𝑥 =𝑥 −
𝑓(𝑥 ) − 𝑓(𝑥 )



Initial values:
𝑥 = 0, 𝑥 =1



🔁 Iteration 1
( . )( )
𝑓(0) = cos(0) − 0 = 1 𝑓(1) = 0.5403 − 1 = −0.4597 𝑥 = 1 − 𝑥 = 1−
.
.
𝑥 = 1 − 0.3149 = 0.6851
.




🔁 Iteration 2
. ( . )
𝑓(0.6851) = 0.7742 − 0.6851 = 0.0891 𝑥 = 0.6851 − ( )
𝑥 = 0.6851 −
. .
. ( . ) .
𝑥 = 0.6851 − 𝑥 = 0.6851 + 0.0512 = 0.7363
. .




🔁 Iteration 3
. ( . . )
𝑓(0.7363) = 0.7410 − 0.7363 = 0.0047 𝑥 = 0.7363 − 𝑥 = 0.7363 −
. .
. ( . ) .
𝑥 = 0.7363 − 𝑥 = 0.7363 + 0.0028 = 0.7391
. .




🔁 Iteration 4
𝑓(0.7391) ≈ 0

, ✅ Final Answer:
𝑥 ≈ 0.7391




2 — Newton-Raphson
Given:
𝑓(𝑥) = cos𝑥 − 𝑥 𝑓′(𝑥) = −sin𝑥 − 1



Formula:
𝑓(𝑥 )
𝑥 =𝑥 −
𝑓′(𝑥 )



Initial value:
𝑥 = 0.5



🔁 Iteration 1
.
𝑓(0.5) = 0.8776 − 0.5 = 0.3776 𝑓′(0.5) = −0.4794 − 1 = −1.4794 𝑥 = 0.5 − 𝑥 =
.
0.5 + 0.2552 = 0.7552



🔁 Iteration 2
𝑓(0.7552) = 0.7280 − 0.7552 = −0.0272 𝑓′(0.7552) = −0.6850 − 1 = −1.6850 𝑥 =
.
0.7552 − 𝑥 = 0.7552 − 0.0161 = 0.7391
.




🔁 Iteration 3
𝑓(0.7391) ≈ 0