Please dont copy paste answers from a program, show work please. 1.Use Newton\'s
ID: 2841824 • Letter: P
Question
Please dont copy paste answers from a program, show work please.
1.Use Newton's method to find all roots of the equation correct to six decimal places.
(smaller value)
(larger value)
2.Use Newton's method to find all roots of the equation correct to six decimal places.
100100
Use Newton's method to find all roots of the equation correct to six decimal places. Use Newton's method to find all roots of the equation correct to six decimal places. Use Newton's method to approximate the given number correct to eight decimal places.
Explanation / Answer
dyoann only gave one solution to the first problem. I believe the third problem was the 100th root of 100.
The Newton method uses iteration xn+1 = xn - f(xn)/f'(xn)
1. (x-3)^2 = ln(x)
(x-3)^2 - ln(x) = 0
f(x) = (x-3)^2 - ln(x)
The derivative is 2(x-3) - 1/x
The second derivative is 2 + 1/x^2 > 0 for all positive x, so the function will only have at most 2 zeros.
f(2) = (-1)^2 - ln(2) = 1 - ln(2) > 0
f(3) = 0^2 - ln(3) < 0
f(4) = 1^2 - ln(3) = 1 - ln(3) is still < 0
However, f(5) = 2^2 - ln(5) = 4 - ln(5) > 0 (recall that ln e = 1, or ln 2.7 is approximately one, so ln 8.3 is approximately 2 to get reasonable first guesses)
We shall use 2 and 4 as our first guesses.
Solutions are 2.130352 and 4.197724
2. e^x = 4 - 5x
e^x - 4 + 5x = 0
The derivative is always positive, so there is only one solution.
f(x) = e^x - 4 + 5x
f'(x) = e^x + 5
f(0) = -4
f(1) = e - 4 + 5 = e + 1 is approximately 3.7 > 0
Thus, the solution is in [0,1]
We see in the table below that the solution is 0.477570 (remember to round up).
Let's start at 1/2
3. For the 100th root of 100, this means x^100 = 100, or x^100 - 100 = 0
The derivative is 100x^99
sqrt(10) is about 3.16, so x^25 is about 3.16
e is about 2.71, so 3.16 is about e^1.2
Thus, let's start at 1 + 1.2/25 = 1.048
The solution is 1.047129