Estimation of Roots (1 Viewer)

[FDownes]

For some reason, when I enter the following information in to my calculator, I can't seem to get the correct answer. Could someone possibly explain why this is? The question asks;

a) Show that there is a root to the equation sin x = x - 1/2 between x = 0.5 and x = 1.8.

[This is the part I'm having trouble with. According to the answers, the equation should be in the form f(x) = sin x - x + 1/2, and then you simply sub in 0.5 and 1.8 for x. For some reason I keep getting f(0.5) = 0.008726... when the answers say it should be 0.479, and P(1.8) = -1.2685... when the answers sa it should be -0.326.]

b) Taking x = 1.2 as a first approximation to this root, apply Newton's method once to find a closer approximation to this root. Give your answer correct to 2 decimal places.

 

[lyounamu]

For some reason, when I enter the following information in to my calculator, I can't seem to get the correct answer. Could someone possibly explain why this is? The question asks;

a) Show that there is a root to the equation sin x = x - 1/2 between x = 0.5 and x = 1.8.

[This is the part I'm having trouble with. According to the answers, the equation should be in the form f(x) = sin x - x + 1/2, and then you simply sub in 0.5 and 1.8 for x. For some reason I keep getting f(0.5) = 0.008726... when the answers say it should be 0.479, and P(1.8) = -1.2685... when the answers sa it should be -0.326.]

b) Taking x = 1.2 as a first approximation to this root, apply Newton's method once to find a closer approximation to this root. Give your answer correct to 2 decimal places.

Did you write that in radians or degrees?

It should be radians.

2nd question will be sweet as long as you understood that you have to use radians.

Use the formula: f(x2) = x1 - f'(x1)/f(x1)

 

[conics2008]

Did you write that in radians or degrees?

It should be radians.

2nd question will be sweet as long as you understood that you have to use radians.

Use the formula: f(x2) = x1 - f'(x1)/f(x1)[/quote]

Just fix this up lym,

 

[lyounamu]

Did you write that in radians or degrees?

It should be radians.

2nd question will be sweet as long as you understood that you have to use radians.

Use the formula: f(x2) = x1 - f'(x1)/f(x1)[/quote]

Just fix this up lym,

I don't get it. Mine was initially right.

 

[tacogym27101990]

I don't get it. Mine was initially right.

the formula's x₂=x₁- [f(x₁)/f'(x₁)]
thats all thats wrong

 

[lyounamu]

the formula's x₂=x₁- [f(x₁)/f'(x₁)]
thats all thats wrong

Was it the other way around?

Ok, I didn't look at maths textbook for a month...lol

 

[tacogym27101990]

yeah dont worry
it was the one 3 unit formula i used to never be able to remember
i just came up with a way to remember it
i just remember its the opposite way to the differentiation of a log
sounds stupid, but it works =]

 

[vds700]

i just remember it by the equation

y - y1 = m(x-x1)
0 - f(x1) = f'(x1)(x - x1)
then its easy to see that f'(x) is on the bottom

 

[lyounamu]

i just remember it by the equation

y - y1 = m(x-x1)
0 - f(x1) = f'(x1)(x - x1)
then its easy to see that f'(x) is on the bottom

Wow, that's how the formula comes? Awesome discovery~~~~

 

[FDownes]

Did you write that in radians or degrees?

It should be radians.

I realised this a few minutes after posting, but I was busy and couldn't get back to the computer. A stupid mistake to make, but an easy one. Thanks.

 

[vds700]

Wow, that's how the formula comes? Awesome discovery~~~~

haha not really its a simple proof, in the fitzpatrick 3u book

 

[lyounamu]

haha not really its a simple proof, in the fitzpatrick 3u book

ah, damn. Thanks by the way~