Hello,
I'm trying to use my calculator to find an approximation of a solution to \(\displaystyle sinx=x^2-3x+1\) to four decimal places.
The iterative step of Newton's Method: \(\displaystyle x_{n+1}=x_{n}-\dfrac{f(x_{n})}{f'(x_{n})}\)
Also, to use Newton's Method, the function must = 0.
\(\displaystyle sinx=x^2-3x+1\)
\(\displaystyle f(x)=sinx-x^2+3x-1\)
Now calculate f'(x)
\(\displaystyle f'(x)=cosx-2x+3\)
To choose our \(\displaystyle x_{0}\), we need to find where f(x)=0, by doing so I picked random values and plugged it into f(x) to see where there's a zero in between.
f(0) = -1
f(1) = 1.8
So there has to be a 0 between 0 and 1 so I will pick \(\displaystyle x_{0}=0.5\)
I also discovered where there is another 0.
f(2) = 1.9
f(3) = -0.85
So there has to be a 0 between 2 and 3 so I will pick \(\displaystyle x_{0}=2.5\)
So now I'm going to solve for the first part where there's a 0 between 0 and 1, and I want to find the first iteration (\(\displaystyle x_{1}\)) by plugging in my first guess of 0.5 into the f(x) andf'(x) that's used in \(\displaystyle x_{n+1}=x_{n}-
\dfrac{f(x_{n})}{f'(x_{n})}\) to get \(\displaystyle x_{n+1}=0.5-\dfrac{sin(0.5)-(0.5)^2+3(0.5)-1}{cos(0.5)-2(0.5)+3}\)
My answer for the first iteration is 0.268881345
My question is, how do I use my calculator to get the next iteration? I'm using a TI-30IIS scientific Calculator and this is my input:
First I type 0.5 and click enter which will be stored as my "ANS" (answer).
Note: ANS = 0.5
Then type the equation in: \(\displaystyle ANS-(sin(ANS)-ANS^2+3*ANS-1)/(cos(ANS)-2*ANS+3)\)
This equals \(\displaystyle 0.268881345\). Now how do I calculate the next iteration? (i.e. \(\displaystyle x_{1}\))
Thank you.
I'm trying to use my calculator to find an approximation of a solution to \(\displaystyle sinx=x^2-3x+1\) to four decimal places.
The iterative step of Newton's Method: \(\displaystyle x_{n+1}=x_{n}-\dfrac{f(x_{n})}{f'(x_{n})}\)
Also, to use Newton's Method, the function must = 0.
\(\displaystyle sinx=x^2-3x+1\)
\(\displaystyle f(x)=sinx-x^2+3x-1\)
Now calculate f'(x)
\(\displaystyle f'(x)=cosx-2x+3\)
To choose our \(\displaystyle x_{0}\), we need to find where f(x)=0, by doing so I picked random values and plugged it into f(x) to see where there's a zero in between.
f(0) = -1
f(1) = 1.8
So there has to be a 0 between 0 and 1 so I will pick \(\displaystyle x_{0}=0.5\)
I also discovered where there is another 0.
f(2) = 1.9
f(3) = -0.85
So there has to be a 0 between 2 and 3 so I will pick \(\displaystyle x_{0}=2.5\)
So now I'm going to solve for the first part where there's a 0 between 0 and 1, and I want to find the first iteration (\(\displaystyle x_{1}\)) by plugging in my first guess of 0.5 into the f(x) andf'(x) that's used in \(\displaystyle x_{n+1}=x_{n}-
\dfrac{f(x_{n})}{f'(x_{n})}\) to get \(\displaystyle x_{n+1}=0.5-\dfrac{sin(0.5)-(0.5)^2+3(0.5)-1}{cos(0.5)-2(0.5)+3}\)
My answer for the first iteration is 0.268881345
My question is, how do I use my calculator to get the next iteration? I'm using a TI-30IIS scientific Calculator and this is my input:
First I type 0.5 and click enter which will be stored as my "ANS" (answer).
Note: ANS = 0.5
Then type the equation in: \(\displaystyle ANS-(sin(ANS)-ANS^2+3*ANS-1)/(cos(ANS)-2*ANS+3)\)
This equals \(\displaystyle 0.268881345\). Now how do I calculate the next iteration? (i.e. \(\displaystyle x_{1}\))
Thank you.
Last edited:
