<Text-field style="_cstyle294" layout="Heading 1"><Equation executable="false" style="_cstyle294" input-equation="" display="LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRicvJSVzaXplR1EjMTJGJy8lK2JhY2tncm91bmRHUS5bMjU1LDI1NSwyNTVdRicvJSxtYXRodmFyaWFudEdRJ25vcm1hbEYn">LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYjLUkjbWlHRiQ2I1EhRic=</Equation><Font bold="true">Newton's Amazing Method: Estimate Pi to How Many Places?</Font></Text-field>Note: You may notice differences between this Maple worksheet and the equivalent Mathematica notebook. These differences were introduced to preserve the content of these modules and were necessary because of major functional differences between Maple and Mathematica.
<Text-field style="Heading 1" layout="Heading 1">Introduction</Text-field>OBJECTIVE: Use Maple to implement Newton's Method.To how many decimal places can you estimate LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVElJnBpO0YnLyUnaXRhbGljR1EmZmFsc2VGJy8lLG1hdGh2YXJpYW50R1Enbm9ybWFsRidGMg== or LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkmbXNxcnRHRiQ2Iy1JI21uR0YkNiRRIjJGJy8lLG1hdGh2YXJpYW50R1Enbm9ybWFsRidGMg==? In this module, Isaac Newton and Maple team up to help you in this endeavor. Estimates to thousands of decimal places are at your fingertips when you have such an awesome pair to help you. You will also encounter some of the hazards in using Newton's Method.
<Text-field style="_cstyle257" layout="Heading 1"><Font size="18">Technology Guidelines</Font></Text-field>
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<Text-field style="Heading 1" layout="Heading 1">Part I: Using Newton's Method to Estimate <Equation executable="false" style="2D Comment" input-equation="" display="LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVElJnBpO0YnLyUnaXRhbGljR1EmZmFsc2VGJy8lLG1hdGh2YXJpYW50R1Enbm9ybWFsRidGMg==">LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVElJnBpO0YnLyUnaXRhbGljR1EmZmFsc2VGJy8lLG1hdGh2YXJpYW50R1Enbm9ybWFsRidGMg==</Equation></Text-field>
<Text-field style="Heading 1" layout="Heading 1">Setting Up The Problem</Text-field>For those of us who enjoy mathematical challenges, estimating irrational numbers like LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVElJnBpO0YnLyUnaXRhbGljR1EmZmFsc2VGJy8lLG1hdGh2YXJpYW50R1Enbm9ybWFsRidGMg== , e and LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkmbXNxcnRHRiQ2Iy1JI21uR0YkNiRRIjJGJy8lLG1hdGh2YXJpYW50R1Enbm9ybWFsRidGMg== to a large number of decimal places is an exercise of enjoyment.To estimate Pi, we will use Newton's method to estimate the 0 of the function 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, for 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LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEieEYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy9GM1Enbm9ybWFsRic=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, since the exact value of this root is LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVElJnBpO0YnLyUnaXRhbGljR1EmZmFsc2VGJy8lLG1hdGh2YXJpYW50R1Enbm9ybWFsRidGMg==. First we define the function f(x).restart;
NumericEventHandler(division_by_zero=exception):
f(x):=tan(x);
We start by plotting LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2Ji1GLDYlUSR0YW5GJy8lJ2l0YWxpY0dRJmZhbHNlRicvJSxtYXRodmFyaWFudEdRJ25vcm1hbEYnLUkjbW9HRiQ2LVEwJkFwcGx5RnVuY3Rpb247RidGNy8lJmZlbmNlR0Y2LyUqc2VwYXJhdG9yR0Y2LyUpc3RyZXRjaHlHRjYvJSpzeW1tZXRyaWNHRjYvJShsYXJnZW9wR0Y2LyUubW92YWJsZWxpbWl0c0dGNi8lJ2FjY2VudEdGNi8lJ2xzcGFjZUdRJjAuMGVtRicvJSdyc3BhY2VHRk4tSShtZmVuY2VkR0YkNiQtRiM2JC1GLDYlUSJ4RicvRjVRJXRydWVGJy9GOFEnaXRhbGljRidGN0Y3RjdGK0Y3 over the specified domain to determine the approximate location of an x-value that makes the value of the function equal to 0.
plot(f(x), x=Pi/2..3*Pi/2, -2..2, color=blue, thickness=2, labels=["x","f(x)"]);
The plot window is cut down by specifying the plot range to -2..2, so that we can see the root clearly.
<Text-field style="Heading 1" layout="Heading 1">Applying Newton's Method</Text-field>We will now write a Maple procedure which implements Newton's Method. It takes as input a function f, the variable x, and a starting guess for the zero. At each iteration, it prints out the iteration number n and the calculated value of x for that iteration. When two consecutive calculated values of x are equal to the precision specified by Digits, the procedure terminates and returns the value of the root.Newton := proc(f, x, start_value)
local formula, current_value, next_value, residual, n:
formula := x - f / diff(f,x):
current_value := evalf(start_value):
residual := 1:
n := 0:
while residual <> 0 do
next_value := evalf(eval(formula, x = current_value),Digits+2):
residual := evalf(next_value - current_value):
printf("n = %a %a = %a \134n", n, x, current_value):
n := n + 1:
current_value := next_value:
od:
end:
We set the precision at 100 digits for the estimate of the zero. After we go through this once you will want to come back and change this value.
Digits := 100:
Using the graph of the function, we specify a starting value of LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2Ji1GLDYlUSJ4RicvJSdpdGFsaWNHUSV0cnVlRicvJSxtYXRodmFyaWFudEdRJ2l0YWxpY0YnLUkjbW9HRiQ2LVEiPUYnL0Y4USdub3JtYWxGJy8lJmZlbmNlR1EmZmFsc2VGJy8lKnNlcGFyYXRvckdGQi8lKXN0cmV0Y2h5R0ZCLyUqc3ltbWV0cmljR0ZCLyUobGFyZ2VvcEdGQi8lLm1vdmFibGVsaW1pdHNHRkIvJSdhY2NlbnRHRkIvJSdsc3BhY2VHUSwwLjI3Nzc3NzhlbUYnLyUncnNwYWNlR0ZRLUkjbW5HRiQ2JFEiM0YnRj5GPkYrRj4= for the iterations.
our_estimate := Newton(tan(x),x,3);
<Text-field style="Heading 1" layout="Heading 1">Testing The Results</Text-field>How do we know that Newton's method is working correctly and that our estimate of LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVElJnBpO0YnLyUnaXRhbGljR1EmZmFsc2VGJy8lLG1hdGh2YXJpYW50R1Enbm9ybWFsRidGMg== is correct? We can compare it with Maple's estimate of LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVElJnBpO0YnLyUnaXRhbGljR1EmZmFsc2VGJy8lLG1hdGh2YXJpYW50R1Enbm9ybWFsRidGMg== to the same number of digits.evalf( Pi - our_estimate );If Maple's estimate of Pi is correct to the specified number of digits (which it is), then so is ours. Not bad! What is truly amazing is the rate at which Newton's method converges to the zero. Six iterations for 100 digits of accuracy - Wow! And, because it takes so few iterations, it is also very fast. But wait, it gets even better. Go back up and change Digits to 1000 or even 10000. See what happens. As you increase Digits, you may want to change your initial guess to be the value returned by your previous run of Newton (i.e. 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) to avoid recomputing the same initial iterations.
<Text-field style="Heading 1" layout="Heading 1">You Try It - Part I</Text-field>
<Text-field style="Heading 1" layout="Heading 1">More About <Equation executable="false" style="2D Comment" input-equation="" display="LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVElJnBpO0YnLyUnaXRhbGljR1EmZmFsc2VGJy8lLG1hdGh2YXJpYW50R1Enbm9ybWFsRidGMg==">LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVElJnBpO0YnLyUnaXRhbGljR1EmZmFsc2VGJy8lLG1hdGh2YXJpYW50R1Enbm9ybWFsRidGMg==</Equation></Text-field>Do some research to find out what the current best estimate of LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVElJnBpO0YnLyUnaXRhbGljR1EmZmFsc2VGJy8lLG1hdGh2YXJpYW50R1Enbm9ybWFsRidGMg== is, and what method was used to compute it. Research the history of the efforts that mathematicians have put into this problem.
<Text-field style="Heading 1" layout="Heading 1">Check out <Equation executable="false" style="2D Comment" input-equation="" display="LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkmbXNxcnRHRiQ2Iy1JI21uR0YkNiRRIjJGJy8lLG1hdGh2YXJpYW50R1Enbm9ybWFsRidGMg==">LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkmbXNxcnRHRiQ2Iy1JI21uR0YkNiRRIjJGJy8lLG1hdGh2YXJpYW50R1Enbm9ybWFsRidGMg==</Equation></Text-field>Construct a function f so that the Newton procedure can be used to estimate LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkmbXNxcnRHRiQ2Iy1JI21uR0YkNiRRIjJGJy8lLG1hdGh2YXJpYW50R1Enbm9ybWFsRidGMg==. Begin by defining a function that is 0 when LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2Ji1GLDYlUSJ4RicvJSdpdGFsaWNHUSV0cnVlRicvJSxtYXRodmFyaWFudEdRJ2l0YWxpY0YnLUkjbW9HRiQ2LVEiPUYnL0Y4USdub3JtYWxGJy8lJmZlbmNlR1EmZmFsc2VGJy8lKnNlcGFyYXRvckdGQi8lKXN0cmV0Y2h5R0ZCLyUqc3ltbWV0cmljR0ZCLyUobGFyZ2VvcEdGQi8lLm1vdmFibGVsaW1pdHNHRkIvJSdhY2NlbnRHRkIvJSdsc3BhY2VHUSwwLjI3Nzc3NzhlbUYnLyUncnNwYWNlR0ZRLUkmbXNxcnRHRiQ2Iy1JI21uR0YkNiRRIjJGJ0Y+Rj5GK0Y+. It is suggested that you try a polynomial. Replace the polynomial given below with your function.f := x^3 - 5;Change the bounds over which you plot your function f if desired.plot(f, x=-1..3, -7..7, color=blue, labels=["x","f(x)"]);Change the precision to 50 digits.Digits:=50:
Use the graph above to pick an initial estimate of the zero.
our_estimate := Newton(x^3-5,x,2);
Compare your result to LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkmbXNxcnRHRiQ2Iy1JI21uR0YkNiRRIjJGJy8lLG1hdGh2YXJpYW50R1Enbm9ybWFsRidGMg==, and see if it has given the desired precision. It clearly does not if you haven't changed f.
evalf( sqrt(2) - our_estimate );
<Text-field style="Heading 1" layout="Heading 1">Part II: Hazards in Applying Newton's Method</Text-field>Let's select a polynomial that we can't factor and begin by looking at its graph.f:=x^5-5*x^3-2*x^2+1;
plot(f, x=-3..3, color=blue, labels=["x","f(x)"]);
From the graph we see that there is a zero after LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2Ji1GLDYlUSJ4RicvJSdpdGFsaWNHUSV0cnVlRicvJSxtYXRodmFyaWFudEdRJ2l0YWxpY0YnLUkjbW9HRiQ2LVEiPUYnL0Y4USdub3JtYWxGJy8lJmZlbmNlR1EmZmFsc2VGJy8lKnNlcGFyYXRvckdGQi8lKXN0cmV0Y2h5R0ZCLyUqc3ltbWV0cmljR0ZCLyUobGFyZ2VvcEdGQi8lLm1vdmFibGVsaW1pdHNHRkIvJSdhY2NlbnRHRkIvJSdsc3BhY2VHUSwwLjI3Nzc3NzhlbUYnLyUncnNwYWNlR0ZRLUkjbW5HRiQ2JFEiMkYnRj5GPkYrRj4=. Use this as our initial guess to Newton's method.This function has three zeros in the domain over which the plot extends. What happens when we specify a starting value of 2?We iterate until two consecutive calculated values of x are equal to the precision specified by Digits. The symbol n is used to count the number of iterations required to achieve the specified precision.
Digits:=20:
Newton(f,x,2);
So we have found the zero close to LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2Ji1GLDYlUSJ4RicvJSdpdGFsaWNHUSV0cnVlRicvJSxtYXRodmFyaWFudEdRJ2l0YWxpY0YnLUkjbW9HRiQ2LVEiPUYnL0Y4USdub3JtYWxGJy8lJmZlbmNlR1EmZmFsc2VGJy8lKnNlcGFyYXRvckdGQi8lKXN0cmV0Y2h5R0ZCLyUqc3ltbWV0cmljR0ZCLyUobGFyZ2VvcEdGQi8lLm1vdmFibGVsaW1pdHNHRkIvJSdhY2NlbnRHRkIvJSdsc3BhY2VHUSwwLjI3Nzc3NzhlbUYnLyUncnNwYWNlR0ZRLUkjbW5HRiQ2JFEiMkYnRj5GPkYrRj4=. What if we had instead made an initial guess of LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2Ji1GLDYlUSJ4RicvJSdpdGFsaWNHUSV0cnVlRicvJSxtYXRodmFyaWFudEdRJ2l0YWxpY0YnLUkjbW9HRiQ2LVEiPUYnL0Y4USdub3JtYWxGJy8lJmZlbmNlR1EmZmFsc2VGJy8lKnNlcGFyYXRvckdGQi8lKXN0cmV0Y2h5R0ZCLyUqc3ltbWV0cmljR0ZCLyUobGFyZ2VvcEdGQi8lLm1vdmFibGVsaW1pdHNHRkIvJSdhY2NlbnRHRkIvJSdsc3BhY2VHUSwwLjI3Nzc3NzhlbUYnLyUncnNwYWNlR0ZRLUkjbW5HRiQ2JFEkMS41RidGPkY+RitGPg==?
Newton(f,x,1.5);
This time the iterations converged to a different value of x. Why did that happen? Can you tell precisely where the break would occur? Does it have anything to do with the derivative of the function? What if we had started at x = 0?Newton(f,x,0);This time Newton's method didn't work at all. Why did this happen? It might help to look at the graph again. How might we modify the procedure Newton so that this is avoided?plot(f,x=-3..3, color=blue, labels=["x","f(x)"]);Let's try an initial guess of LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2Jy1GLDYlUSJ4RicvJSdpdGFsaWNHUSV0cnVlRicvJSxtYXRodmFyaWFudEdRJ2l0YWxpY0YnLUkjbW9HRiQ2LVEiPUYnL0Y4USdub3JtYWxGJy8lJmZlbmNlR1EmZmFsc2VGJy8lKnNlcGFyYXRvckdGQi8lKXN0cmV0Y2h5R0ZCLyUqc3ltbWV0cmljR0ZCLyUobGFyZ2VvcEdGQi8lLm1vdmFibGVsaW1pdHNHRkIvJSdhY2NlbnRHRkIvJSdsc3BhY2VHUSwwLjI3Nzc3NzhlbUYnLyUncnNwYWNlR0ZRLUYjNiUtRjs2LVEqJnVtaW51czA7RidGPkZARkNGRUZHRklGS0ZNL0ZQUSwwLjIyMjIyMjJlbUYnL0ZTRlotSSNtbkdGJDYkUSQxLjVGJ0Y+Rj5GK0Y+RitGPg==.Newton(f,x,-1.5);Why does this converge to the zero to the right instead of the closer zero to the left?
<Text-field style="Heading 1" layout="Heading 1">You Try It - Part II</Text-field>Here is a cubic polynomial. Study its graph.f:=x^3-5*x^2-2*x+11;
plot(f,x=-2..5.5, color=blue,labels=["x","f(x)"]);You decide where you should begin your iterations in order to compute each of the zeros of this polynomial. By changing the initial value, you determine which starting values converge to which zero. We start with x = 0.
Newton(f,x,0);What happened? It seems to have completely skipped over a zero! Why did this happen?
<Text-field style="Heading 1" layout="Heading 1">Animation of Newton's Method</Text-field>This Maple procedure shows Newton's method in action.NewtonPlot := proc(f, x, start_value, xrange)
local p, p1, p2, formula, current_value, next_value, residual, n, MAX_FRAMES:
MAX_FRAMES := 20:
formula := x - f / diff(f,x):
current_value := evalf(start_value):
residual := 1:
n := 0:
p1 := plot(f,x=xrange, args[5..nargs], color=blue):
p:= NULL:
while residual <> 0 do
if n > MAX_FRAMES then break fi:
next_value := evalf(eval(formula, x = current_value),Digits+2):
residual := evalf(next_value - current_value):
if residual = 0 then break fi:
p2 := plots[pointplot]({[next_value,0],[next_value,eval(f,x=next_value)]},connect=true,color=green):
p := p, plots[display](p1,
p2, plot(eval(f,x=current_value)/(current_value-next_value)*(x-next_value),x=xrange)):
n := n + 1:
current_value := next_value:
od:
plots[display](p,insequence=true):
end:Digits:=20:Try different starting values (the third argument in NewtonPlot( )) so that Newton's method will converge on each of the three zeros. Also, try to find the break-over points where Newton's method will converge to one root if the starting value is on one side of the break-over point, and to another root if the starting value is on the other side.
To animate, click on the graph below to bring up the animation toolbox at the top of the worksheet. Press the play button (the large arrow) to play the animation. Change the speed by clicking on the double arrow buttons, or show one frame at a time by pressing the arrow followed by a vertical bar.
NewtonPlot(x^3-5*x^2-2*x+11,x,0,-2..7);NewtonPlot(tan(x),x,2.2,Pi/2..3*Pi/2,-2..2);