Sheen1101
Member
+12|7046
I have a calculus problem that I don't understand how to do and need help......

Use Newton's method to approximate the indicated root of the equation correct to six decimal places.
The positive root of 2cos(x) = x4
mcminty
Moderating your content for the Australian Govt.
+879|7148|Sydney, Australia

Sheen1101 wrote:

I have a calculus problem that I don't understand how to do and need help......

Use Newton's method to approximate the indicated root of the equation correct to six decimal places.
The positive root of 2cos(x) = x4
Welll...

Newton's method uses the formula x1 = x0 - f(x)/f'(x), where x0 is an initial guess of what the root is. You then sub x1 into the formula to get x2, and so on.

I would suggest using f(x) = 2cos(x) - x4


From wikipedia, which might help you:

onsider the problem of finding the positive number x with cos(x) = x3. We can rephrase that as finding the zero of f(x) = cos(x) − x3. We have f'(x) = −sin(x) − 3x2. Since cos(x) ≤ 1 for all x and x3 > 1 for x>1, we know that our zero lies between 0 and 1. We try a starting value of x0 = 0.5. (Note that a starting value of 0 will lead to an undefined result, showing the importance of using a starting point that is close to the zero.)
Sheen1101
Member
+12|7046

mcminty wrote:

Sheen1101 wrote:

I have a calculus problem that I don't understand how to do and need help......

Use Newton's method to approximate the indicated root of the equation correct to six decimal places.
The positive root of 2cos(x) = x4
Welll...

Newton's method uses the formula x1 = x0 - f(x)/f'(x), where x0 is an initial guess of what the root is. You then sub x1 into the formula to get x2, and so on.

I would suggest using f(x) = 2cos(x) - x4


From wikipedia, which might help you:

onsider the problem of finding the positive number x with cos(x) = x3. We can rephrase that as finding the zero of f(x) = cos(x) − x3. We have f'(x) = −sin(x) − 3x2. Since cos(x) ≤ 1 for all x and x3 > 1 for x>1, we know that our zero lies between 0 and 1. We try a starting value of x0 = 0.5. (Note that a starting value of 0 will lead to an undefined result, showing the importance of using a starting point that is close to the zero.)
k thanks

Board footer

Privacy Policy - © 2025 Jeff Minard