Question about Taylor Series While loop.

2013 3 6
3 个回答
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I'm completely stuck on this While Loop using Taylor Series.
x = input('Input the angle in radians: ');
Cos_Estimate = 0;
k = 0;
Errrr = 1
while Errrr > .000001
if mod(k,2)
Sign = 1;
else
Sign = -1;
end
k = k + 2;
Cos_Estimate = Cos_Estimate + (x^k/(factorial(k)*Sign));
Errrr = abs(Cos_Estimate - cos(x));
end
err = abs(Cos_Estimate - cos(x));
fprintf('The estimated cosine value based on the Taylor Series is: %0.6f \n',Cos_Estimate)
fprintf('The actual cosine value is : %0.6f \n',cos(x))
fprintf('The estimation error is: %0.6f \n',err)
fprintf('The number of terms required was: \n',term)
Now I am getting NaN for my variable. I'm stuck.

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Azzi Abdelmalek
Azzi Abdelmalek 2013-3-6
编辑：Azzi Abdelmalek 2013-3-6

0 个投票

Try this
x = input('Input the angle in radians: ');
Cos_Estimate = 1;
Errrr=1
Sign = 1;
k = 0;
while Errrr > .000001 & k<60
Sign =-Sign;
k = k + 2;
Cos_Estimate = Cos_Estimate + (x^k/(factorial(k)*Sign))
Errrr = abs(Cos_Estimate - cos(x))
end
display(Cos_Estimate)

更多回答（2 个）

Babak
Babak 2013-3-6

1 个投票

add this line to the beginning of your code:
Errrr =1;

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Derryn
Derryn 2013-3-6
Added it, but now I get this as my output:
EDU>> Lab7_Prob3Script
Input the angle in radians: pi
The estimated cosine value based on the Taylor Series is: NaN
The actual cosine value is : -1.000000
The estimation error is: NaN
The number of terms required was:

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Azzi Abdelmalek
Azzi Abdelmalek 2013-3-6
编辑：Azzi Abdelmalek 2013-3-6

0 个投票

Your code never enter in the loop because Errrr is not defined

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Derryn
Derryn 2013-3-6
编辑：Derryn 2013-3-6
Added it, but now I get this as my output:
EDU>> Lab7_Prob3Script
Input the angle in radians: pi
The estimated cosine value based on the Taylor Series is: NaN
The actual cosine value is : -1.000000
The estimation error is: NaN
The number of terms required was:
Derryn
Derryn 2013-3-6
How do I fix my variable from being NaN?
This is the first time this problem has happened to me!
Matt Kindig
Matt Kindig 2013-3-6
编辑：Matt Kindig 2013-3-6
The issue is because the factorial() function, at very high values of k, exceeds the limit for double-precision numbers (specifically, you get a value of Inf). To demonstrate, run this at the command line:
>> factorial(160:180)
Notice that after the 11th column (i.e. factorial(170)) the factorial no longer is defined. When dividing by Inf, you get NaN in Cos_Estimate.
To fix this, the easiest way might be to add a second condition to your while loop, to terminate when k exceeds 170 or so.
Also, you don't define 'term' anywhere in your code, so the last line throws an error.
Matt Kindig
Matt Kindig 2013-3-6
编辑：Matt Kindig 2013-3-6
Also, what exactly is the point of the "estimation error" calculation (calculation of 'err')? By definition, won't this be equal to Errrr, because you have defined it that way? Also, won't 'err' always be less than 0.000001, because of the way you have defined your loop?
Derryn
Derryn 2013-3-6
Ah ok thanks. Does this look better?
x = input('Input the angle in radians: ');
Cos_Estimate = 0; k = 0;
count = 0;
Errrr = 1;
while Errrr > .000001
count = count + 1;
if mod(k,2)
Sign = 1;
else
Sign = -1;
end
k = k + 2;
Cos_Estimate = Cos_Estimate + (x^k/(factorial(k)*Sign));
Errrr = abs(Cos_Estimate - cos(x));
if k > 170
break
end
end
err = abs(Cos_Estimate - cos(x));
fprintf('The estimated cosine value based on the Taylor Series is: %0.6f \n',Cos_Estimate)
fprintf('The actual cosine value is : %0.6f \n',cos(x))
fprintf('The estimation error is: %0.6f \n',err)
fprintf('The number of terms required was: %0.0f \n',count)
I'm still getting the wrong value for Cos_Estimate

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