Filter löschen
Filter löschen

Loop structure for fixed point.

2 Ansichten (letzte 30 Tage)
형석 김
형석 김 am 16 Apr. 2021
Kommentiert: 형석 김 am 17 Apr. 2021
Hello, I have problem with Loop structure for fixed point.
This is the problem, and i made code like this.
It only caculates 1 time and just stops. Which mean, it's not repeating my code, xb=x; and x=sqrt(1.8*x+2.5)
if i manually repeat xb=x; and x=sqrt(1.8*x+2.5) the answer is correct, however it's not repeating those code.
What's wrong about my loop and How to fix it?
  5 Kommentare
Rik
Rik am 17 Apr. 2021
How large is that difference? And what is the code you're currently using? And what results are you expecting?
형석 김
형석 김 am 17 Apr. 2021
Bearbeitet: 형석 김 am 17 Apr. 2021
@Rik The value of x i got is 2.7892, and the correct answer when I calculated manually is 2.7424. I've also calculated error when x=2.7892 manually, but error sllightly exceeds 0.05, which is (2.9333-2.7892)/2.78792=0.05166.

Melden Sie sich an, um zu kommentieren.

Akzeptierte Antwort

Walter Roberson
Walter Roberson am 17 Apr. 2021
and the correct answer when I calculated manually is 2.7424
No it is not. The correct answer is 2.71934053986603
format long g
roots([-1 1.8 2.5])
You tried the transform
-x^2 + 1.8*x + 2.5 == 0
implies 1.8*x + 2.5 == x^2
implies sqrt(1.8*x + 2.5) == sqrt(x^2)
and you deduced from that that
x == sqrt(1.8*x + 2.5)
but that is not correct. sqrt(x^2) is not x: it is abs(x) . So
abs(x) == sqrt(1.8*x + 2.5)
and you should be working both branches of that.
I suggest that you instead
-x^2 + 1.8*x + 2.5 == 0
implies 1.8*x == x^2 - 2.5
implies x == (x^2 - 2.5)/1.8

Weitere Antworten (0)

Kategorien

Mehr zu Loops and Conditional Statements finden Sie in Help Center und File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by