Using Newton's Method to Approximate Solutions to Equations

Key Questions

• How do you use Newton's Method to approximate the value of cube root?

  The Newton-Raphson method approximates the roots of a function. So, we need a function whose root is the cube root we're trying to calculate.

  Let's say we're trying to find the cube root of #3#. And let's say that #x# is the cube root of #3#. Therefore,

  #x^3 = 3#

  For the Newton-Raphson method to be able to work its magic, we need to set this equation to zero.

  #x^3 - 3 = 0#

  Now we will recall the iterative equation for Newton-Raphson.

  #x_(n+1) = x_n - (f(x_n))/(f'(x_n))#

  Substituting for #f(x) = x^3 - 3# gives us:

  #x_(n+1) = x_n - ((x_n)^3 - 3)/(3*(x_n)^2)#

  Now, we pick an arbitrary number, (the closer it actually is to #root3(3)# the better) for #x_0#. Let's use #x_0 = 0.5#. Then we substitute each previous number for #x_n# back into the equation to get a closer and closer approximation to a solution of #x^3 - 3 = 0#.

  #x_(1) = 0.5 - ((0.5)^3 - 3)/(3*(0.5)^2) = 4.33333 bar 3#
  #x_(2) = x_1 - ((x_1)^3 - 3)/(3*(x_1)^2) approx 2.94214333#
  #x_(3) = x_2 - ((x_2)^3 - 3)/(3*(x_2)^2) approx 2.07695292#
  #x_(4) = x_3 - ((x_3)^3 - 3)/(3*(x_3)^2) approx 1.61645303#
  #x_(5) = x_4 - ((x_4)^3 - 3)/(3*(x_4)^2) approx 1.46034889#
  #x_(6) = x_5 - ((x_5)^3 - 3)/(3*(x_5)^2) approx 1.44247296#
  #x_(7) = x_6 - ((x_6)^3 - 3)/(3*(x_6)^2) approx 1.4422496#
  #x_(8) = x_7 - ((x_7)^3 - 3)/(3*(x_7)^2) approx 1.44224957#

  You can see that with only 8 iterations, we've obtained an approximation of #root 3(3)# which is correct to 8 decimal places!

  You can apply this same logic to whatever cube root you'd like to find, just use #x^3 - a = 0# as your equation instead, where #a# is the number whose cube root you're looking for.

  Jacob F. · 18 · Aug 22 2014
• What is Newton's Method?

  Newton's Method is a mathematical tool often used in numerical analysis, which serves to approximate the zeroes or roots of a function (that is, all #x: f(x)=0#).

  The method is constructed as follows: given a function #f(x)# defined over the domain of real numbers #x#, and the derivative of said function (#f'(x)#), one begins with an estimate or "guess" as to where the function's root might lie. For example, suppose one is presented with the function #f(x) = x^2 +x -2.5#. This is similar to another function #g(x) = x^2 + x - 2#, whose roots are #x=1# and #x=-2#. Thus, thanks to this similarity, one might use #x=1# or #x=-2# as guesses to start Newton's Method with f(x).

  (Alternately, if a graphical representation is available but the exact root is not listed, an acceptable approximation might be the nearest whole number to the root).

  Whatever method used, we declare this initial guess to be #x_0#. We arrive at a better approximation, #x_1#, by employing the Method: #x_1 = x_0 - f(x_0)/(f'(x_0))#. Essentially, by utilizing the derivative, one is able to increment closer to the actual value. In the above example, #f(x) = x^2 + x - 2.5#, if we assume #x_0 = 1#, then #x_1 = 1 - f(1)/(f'(1)) = 1 - (-.5)/(3) = 7/6 or approx 1.16667#.

  Often, one may be able to find the root another way (by using a graphing calculator, for example), and an exam item or textbook problem may demand a certain degree of accuracy (such as within 1% of the actual value). In such a case, if #x_1# is not an accurate enough approximation, one performs the iteration again, as often as needed for the desired degree of accuracy. The formula to find the general #x_n#, then, is #x_n = x_(n-1) - f(x_(n-1))/(f'(x_(n-1)))#

  Psykolord1989 . · 2 · Nov 24 2014

• How do you use Newton's Method to approximate #root5(20) # ?
• How do you use Newton's Method to approximate the value of cube root?
• How do you use Newton's Method to approximate the root of the equation #x^4-2x^3+5x^2-6=0# on the interval #[1,2]# ?
• How do you use Newton's Method to approximate the positive root of the equation #sin(x)=x^2# ?
• If a rough approximation for ln(5) is 1.609 how do you use this approximation and differentials to approximate ln(128/25)?
• How do you use linear approximation to estimate #g(2.95)# and #g(3.05)# if you know that #g(3)=-5#?
• How do you use a linear approximation to estimate #g(0.9)# and #g(1.1)# if we know that #g(1)=3# and #g'(x)=sqrt(x^2+15)#?
• How do you use differentials to estimate the value of #cos(63)#?
• When do you use newton's method?
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• How do you use #f(x) = sin(x^2-2)# to evaluate #(f(3.0002)-f(3))/0.0002#?
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• How do you find the Linear Approximation at x=0 of #y=sqrt(3+3x)#?
• How do you estimate Δf = f (a + Δx) - f (a) using the Linear Approximation given #f(x) = x – 4x^2#, a = 1, Δx = -0.3?
• How do you use differentials and the function #f(x,y) = arctan(x*y^2)# to approximate the value of f(0.94, 1.17)?
• How do you find the linear approximation L to f at the designated point P. compare the error in approximating f by L at the specified point Q with the distance between P and Q given #f(x,y) = 1/sqrt(x^2+y^2)#, P(4,3) and Q(3.92, 3.01)?
• A differentiable function has the property that f(5) = 4 and f'(5) = 3. What is the estimate for f(4.8) using the local linear approximation for f at x=5?
• A pendulum of length L feet has a period of: #T= 3.14(L^.5)/2(2^.5)# seconds. If you increase the length of a 1-foot pendulum by .01 feet, how much does its period increase?
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• How do you use a linear approximation or differentials to estimate #1/1002#?
• How do you use a linear approximation or differentials to estimate #tan44º#?
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• How do you use linear approximation to estimate #root3( 64.1)#?
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• A hemispherical dome of radius 40 feet is to be given 7 coats of paint, each of which is 1/100 inch thick. How do you use linear approximation to estimate the volume of paint needed for the job?
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• If f(-2)=1 and f'(-2)=5, what is the approximate value of f(-2.01)?
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• Question #109a1
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• How do you use Newton's approximation method with #f(x) = x^2 - 2# to iteratively solve for the positive zero of #f(x)# accurately to within 4 decimal places using at least #6# iterations?
• Question #aae8e
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• If point of inflection of #y=e^(-x^2)# is at #y=1/e^a#, #a=..........#?
• Question #3a704
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• How do you use Newton's method to find the approximate solution to the equation #x^5+x^3+x=1#?
• How do you use Newton's method to find the approximate solution to the equation #2x^5+3x=2#?
• How do you use Newton's method to find the approximate solution to the equation #x^4=x+1,x>0#?
• How do you use Newton's method to find the approximate solution to the equation #x^4=x+1,x<0#?
• How do you use Newton's method to find the approximate solution to the equation #x^3-10x+4=0, x>1#?
• How do you use Newton's method to find the approximate solution to the equation #x^3-10x+4=0, 0<x<1#?
• How do you use Newton's method to find the approximate solution to the equation #x+sqrtx=1#?
• How do you use Newton's method to find the approximate solution to the equation #x+1/sqrtx=3#?
• How do you use Newton's method to find the approximate solution to the equation #e^x=1/x#?
• How do you use Newton's method to find the approximate solution to the equation #e^x+x=4#?
• How do you use Newton's method to find the approximate solution to the equation #tanx=e^x, 0<x<pi/2#?
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• Question #4cf41
• Question #eda23
• Question #c3851
• Question #2096b
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• Question #b399e
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• Question #8c423
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