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Taylor polynomial of third degree and error estimation

  • Thread starter Telemachus
  • Start date
832
30
1. Homework Statement
It seems that I'm a little bit lost about this exercise. It says: Find the taylors polynomial of third degree centered at the origin for [tex]z=\cos y \sin x[/tex]. Estimate the error for: [tex]\Delta x=-0.15,\Delta y=0.2[/tex].

So, I did the first part (the easy one), the taylors polynomial for z at (0,0) looks like this:

[tex]f(x,y)=x+\displaystyle\frac{1}{3!}(-x^3+3xy^2)+R_4((x,y),(0,0))[/tex]

Then I've found the expression for the error:
[tex]R_4=\displaystyle\frac{1}{4!}(\cos c_2 \sin c_1 x^4+4\sin c_2 \cos c_1 x^3y+6\cos c_2 \sin c_1 x^2y^2+4\sin c_2 \cos c_1 xy^3-\cos c_2 \sin c_1 y^4)[/tex]

Now, how do I estimate the error? I have to say that the error will be something like [tex]R_4\leq{}k[/tex], I have to find "k". Should I consider [tex]c_1=0.15, c_2=0.2[/tex]? How do I proceed from there?

Bye and thanks for posting!
 
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Answers and Replies

32,921
4,625
1. Homework Statement
It seems that I'm a little bit lost about this exercise. It says: Find the taylors polynomial of third degree centered at the origin for [tex]z=\cos y \sin x[/tex]. Estimate the error for: [tex]\Delta x=-0.15,\Delta y=0.2[/tex].

So, I did the first part (the easy one), the taylors polynomial for z at (0,0) looks like this:

[tex]f(x,y)=x+\displaystyle\frac{1}{3!}(-x^3+3xy^2)+R_4((x,y),(0,0))[/tex]

Then I've found the expression for the error:
[tex]R_4=\displaystyle\frac{1}{4!}(\cos c_2 \sin c_1 x^4+4\sin c_2 \cos c_1 x^3y+6\cos c_2 \sin c_1 x^2y^2+4\sin c_2 \cos c_1 xy^3-\cos c_2 \sin c_1 y^4)[/tex]

Now, how do I estimate the error? I have to say that the error will be something like [tex]R_4\leq{}k[/tex], I have to find "k". Should I consider [tex]c_1=0.15, c_2=0.2[/tex]? How do I proceed from there?

Bye and thanks for posting!
All of your sine and cosine expressions have a maximum value of 1.

You know what f(0, 0) is, right? You're interested in approximating f(0 - 0.15, 0 + 0.2)
 
832
30
With f you reefer to the taylors polynomial? its zero anyway, z its zero at that point, and f is zero too, cause the error its zero at that point.

I've considered [tex]c1=x=-0.5,c2=y=0.2[/tex] and I get an error of 0.000143, which I think its a good approximation, I've evaluated the function at that point, and the polinomial, and it gives an error of 0.00002.

Is this right?

What I did know was that [tex]x\leq{}c_1\leq{}0[/tex] and [tex]0\leq{}c_2\leq{}y[/tex], but I didn't know wich value should I take for c_1 and c_2 on the interval. I've just made it equal, but I think I should looked for a maximum. It worked on this case, but I think this wouldn't work in other cases.

Bye there, thanks Mark.
 
32,921
4,625
With f you reefer to the taylors polynomial?
No, I'm not. f is the function you want to approximate by its Taylor polynomial.
its zero anyway, z its zero at that point, and f is zero too, cause the error its zero at that point.

I've considered [tex]c1=x=-0.5,c2=y=0.2[/tex] and I get an error of 0.000143, which I think its a good approximation, I've evaluated the function at that point, and the polinomial, and it gives an error of 0.00002.
No, because you should not be setting c1 to -.5 or c2 to .2. c1 and c2 are unknown numbers in the intervals [-.5, 0] and [0, .2] respectively.
Is this right?

What I did know was that [tex]x\leq{}c_1\leq{}0[/tex] and [tex]0\leq{}c_2\leq{}y[/tex], but I didn't know wich value should I take for c_1 and c_2 on the interval. I've just made it equal, but I think I should looked for a maximum. It worked on this case, but I think this wouldn't work in other cases.

Bye there, thanks Mark.
I would just replace the cosine and sine terms by 1, since they will never be larger than 1.
 
832
30
Thanks Mark!
 

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