Use Newton's method to find a value of x that satisfies: e^x- (x^2/2) = 0
Use the starting value x0=0. Display answer in 4 significant figures
e^x - x^2/2=0
2e^x - x^2 = 0
let y = 2e^x - x^2
y' = 2e^x - 2x
Newton said
newx = x - y/y'
= x -(2e^x-x^2)/(2e^x - 2x)
which simplified to
(2xe^x - x^2 - 2e^x)/(2e^x - 2x)
make a chart with
x ---- newx , then replacing x with newx in the next line
0 -1
-1 -.9034
-.9034 -.9012
-.9012 -.901201
x = -.901201 correct to 4 decimals after 3 iterations, not bad
check:
LS = e^x - x^2/2
= .406081662 - .406081621
= .000000041
fantastic result!!!
Could u please show how u got -.9034 too? Thanks!!!
To use Newton's method to find a value of x that satisfies the equation e^x - (x^2/2) = 0, we need to follow these steps:
1. Write down the equation: e^x - (x^2/2) = 0
2. Find the derivative of the equation: f'(x) = e^x - x
3. Choose a starting value, x0. In this case, x0 = 0.
4. Compute the next value in the sequence using the formula:
x1 = x0 - (f(x0) / f'(x0))
5. Repeat step 4 until the desired level of accuracy is achieved.
Let's apply these steps to find the value of x that satisfies the equation e^x - (x^2/2) = 0:
Step 1: The equation is e^x - (x^2/2) = 0.
Step 2: The derivative of the equation is f'(x) = e^x - x.
Step 3: The starting value is x0 = 0.
Step 4: Plug the starting value x0 = 0 into the formula:
x1 = x0 - (e^x0 - (x0^2/2)) / (e^x0 - x0)
x1 = 0 - (e^0 - (0^2/2)) / (e^0 - 0)
= 0 - (1 - 0) / (1 - 0)
= -1
Step 5: Repeat Step 4 until the desired level of accuracy is achieved. Let's continue to find x2:
x2 = x1 - (e^x1 - (x1^2/2)) / (e^x1 - x1)
Inserting the values:
x2 = -1 - (e^(-1) - ((-1)^2/2)) / (e^(-1) - (-1))
Continue this process until you achieve the desired level of accuracy, and stop when the difference between two consecutive x values is small enough. In this case, we need to stop after four significant figures.
After computing the values of x1, x2, and so on, you will arrive at a value of x that satisfies the equation e^x - (x^2/2) = 0 with four significant figures.