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Loop structure for fixed point.

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형석 김
형석 김 2021 年 4 月 16 日
コメント済み: 형석 김 2021 年 4 月 17 日
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 件のコメント
Rik
Rik 2021 年 4 月 17 日
How large is that difference? And what is the code you're currently using? And what results are you expecting?
형석 김
형석 김 2021 年 4 月 17 日
編集済み: 형석 김 2021 年 4 月 17 日
@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.

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Walter Roberson
Walter Roberson 2021 年 4 月 17 日
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
  1 件のコメント
형석 김
형석 김 2021 年 4 月 17 日
Thanks for your advice.

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