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Find the second degree Taylor polynomial of $f(x,y)=e^-x^2-y^2cos(xy)$ at $x_0=0$, $y_0=0$.
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up vote
2
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I would like to either confirm my solution is correct, or find the error in it. I used the following MATLAB code to try to check my answer, but the solution it gave differs from mine.
f = exp(-(x^2+y^2))cos(xy);
taylor(f,[x,y],[0,0])
ans =
x^4/2 + (x^2*y^2)/2 - x^2 + y^4/2 - y^2 + 1
My solution:
The second degree Taylor polynomial at the point $(a,b)$ is given by
$$p_2(x,y)=f(a,b)+Df(a,b)left[beginarrayc
x-a\y-b
endarrayright]+frac12[x-ambox y-b]Hf(a,b)left[beginarrayc
x-a\y-b
endarrayright]$$
where $Df(a,b)$ is the Jacobian matrix and $Hf(a,b)$ is the Hessian matrix (i.e. first and second order partial derivatives, respectively). Thus, we begin by computing the partial derivatives:
beginequation*
beginsplit
fracpartial fpartial x(x,y)=-e^-x^2-y^2(ymbox sin(xy)+2xmbox cos(xy))\
fracpartial fpartial y(x,y)=-e^-x^2-y^2(xmbox sin(xy)+2ymbox cos(xy))\
fracpartial^2 fpartial x^2(x,y)=e^-x^2-y^2((4x^2-y^2-2)mboxcos(xy)+4xymbox sin(xy))\
fracpartial^2 fpartial y^2(x,y)=e^-x^2-y^2(4xymbox sin(xy)-(x^2-4y^2+2)mboxcos(xy))\
fracpartial^2 fpartial xpartial y(x,y)=fracpartial^2 fpartial ypartial x(x,y)=e^-x^2-y^2((2x^2+2y^2-1)mboxsin(xy)+3xymbox cos(xy))
endsplit
endequation*
Then, at the point $(x_0,y_0)=(0,0)$,
beginequation*
beginsplit
f(0,0)=1\
fracpartial fpartial x(0,0)=fracpartial fpartial y(0,0)=0\
fracpartial^2 fpartial x^2(0,0)=fracpartial^2 fpartial y^2(0,0)=-2\
fracpartial^2 fpartial xpartial y(0,0)=fracpartial^2 fpartial ypartial x(0,0)=0
endsplit
endequation*
Therefore, the second degree Taylor polynomial at the point $(0,0)$ is
beginequation*
beginsplit
p_2(x,y)=f(0,0)+Df(0,0)left[beginarrayc
x-0\y-0
endarrayright]+frac12[x-0mbox y-0]Hf(0,0)left[
beginarrayc
x-0\y-0
endarrayright]\
=1+[0mbox 0]left[
beginarrayc
x\y
endarrayright]+frac12[xmbox y]beginbmatrix
-2&0\0&-2
endbmatrixleft[
beginarrayc
x\y
endarrayright]=1-x^2-y^2
endsplit
endequation*
multivariable-calculus proof-verification matlab
asked Jul 25 at 1:29
Atsina
513113
add a comment |Â
up vote
2
down vote
favorite
I would like to either confirm my solution is correct, or find the error in it. I used the following MATLAB code to try to check my answer, but the solution it gave differs from mine.
f = exp(-(x^2+y^2))cos(xy);
taylor(f,[x,y],[0,0])
ans =
x^4/2 + (x^2*y^2)/2 - x^2 + y^4/2 - y^2 + 1
My solution:
The second degree Taylor polynomial at the point $(a,b)$ is given by
$$p_2(x,y)=f(a,b)+Df(a,b)left[beginarrayc
x-a\y-b
endarrayright]+frac12[x-ambox y-b]Hf(a,b)left[beginarrayc
x-a\y-b
endarrayright]$$
where $Df(a,b)$ is the Jacobian matrix and $Hf(a,b)$ is the Hessian matrix (i.e. first and second order partial derivatives, respectively). Thus, we begin by computing the partial derivatives:
beginequation*
beginsplit
fracpartial fpartial x(x,y)=-e^-x^2-y^2(ymbox sin(xy)+2xmbox cos(xy))\
fracpartial fpartial y(x,y)=-e^-x^2-y^2(xmbox sin(xy)+2ymbox cos(xy))\
fracpartial^2 fpartial x^2(x,y)=e^-x^2-y^2((4x^2-y^2-2)mboxcos(xy)+4xymbox sin(xy))\
fracpartial^2 fpartial y^2(x,y)=e^-x^2-y^2(4xymbox sin(xy)-(x^2-4y^2+2)mboxcos(xy))\
fracpartial^2 fpartial xpartial y(x,y)=fracpartial^2 fpartial ypartial x(x,y)=e^-x^2-y^2((2x^2+2y^2-1)mboxsin(xy)+3xymbox cos(xy))
endsplit
endequation*
Then, at the point $(x_0,y_0)=(0,0)$,
beginequation*
beginsplit
f(0,0)=1\
fracpartial fpartial x(0,0)=fracpartial fpartial y(0,0)=0\
fracpartial^2 fpartial x^2(0,0)=fracpartial^2 fpartial y^2(0,0)=-2\
fracpartial^2 fpartial xpartial y(0,0)=fracpartial^2 fpartial ypartial x(0,0)=0
endsplit
endequation*
Therefore, the second degree Taylor polynomial at the point $(0,0)$ is
beginequation*
beginsplit
p_2(x,y)=f(0,0)+Df(0,0)left[beginarrayc
x-0\y-0
endarrayright]+frac12[x-0mbox y-0]Hf(0,0)left[
beginarrayc
x-0\y-0
endarrayright]\
=1+[0mbox 0]left[
beginarrayc
x\y
endarrayright]+frac12[xmbox y]beginbmatrix
-2&0\0&-2
endbmatrixleft[
beginarrayc
x\y
endarrayright]=1-x^2-y^2
endsplit
endequation*
multivariable-calculus proof-verification matlab
asked Jul 25 at 1:29
Atsina
513113
What’s your question?
– amd
Jul 25 at 2:19
@amd "I would like to either confirm my solution is correct, or find the error in it."
– Atsina
Jul 25 at 2:23
The Matlab output that you've got is the fourth-degree Taylor approximation, not second-degree. Either add'Order',2to the arguments or drop the higher-order terms yourself.
– amd
Jul 25 at 6:05
@amd Thank you!
– Atsina
Jul 25 at 15:26
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I would like to either confirm my solution is correct, or find the error in it. I used the following MATLAB code to try to check my answer, but the solution it gave differs from mine.
f = exp(-(x^2+y^2))cos(xy);
taylor(f,[x,y],[0,0])
ans =
x^4/2 + (x^2*y^2)/2 - x^2 + y^4/2 - y^2 + 1
My solution:
The second degree Taylor polynomial at the point $(a,b)$ is given by
$$p_2(x,y)=f(a,b)+Df(a,b)left[beginarrayc
x-a\y-b
endarrayright]+frac12[x-ambox y-b]Hf(a,b)left[beginarrayc
x-a\y-b
endarrayright]$$
where $Df(a,b)$ is the Jacobian matrix and $Hf(a,b)$ is the Hessian matrix (i.e. first and second order partial derivatives, respectively). Thus, we begin by computing the partial derivatives:
beginequation*
beginsplit
fracpartial fpartial x(x,y)=-e^-x^2-y^2(ymbox sin(xy)+2xmbox cos(xy))\
fracpartial fpartial y(x,y)=-e^-x^2-y^2(xmbox sin(xy)+2ymbox cos(xy))\
fracpartial^2 fpartial x^2(x,y)=e^-x^2-y^2((4x^2-y^2-2)mboxcos(xy)+4xymbox sin(xy))\
fracpartial^2 fpartial y^2(x,y)=e^-x^2-y^2(4xymbox sin(xy)-(x^2-4y^2+2)mboxcos(xy))\
fracpartial^2 fpartial xpartial y(x,y)=fracpartial^2 fpartial ypartial x(x,y)=e^-x^2-y^2((2x^2+2y^2-1)mboxsin(xy)+3xymbox cos(xy))
endsplit
endequation*
Then, at the point $(x_0,y_0)=(0,0)$,
beginequation*
beginsplit
f(0,0)=1\
fracpartial fpartial x(0,0)=fracpartial fpartial y(0,0)=0\
fracpartial^2 fpartial x^2(0,0)=fracpartial^2 fpartial y^2(0,0)=-2\
fracpartial^2 fpartial xpartial y(0,0)=fracpartial^2 fpartial ypartial x(0,0)=0
endsplit
endequation*
Therefore, the second degree Taylor polynomial at the point $(0,0)$ is
beginequation*
beginsplit
p_2(x,y)=f(0,0)+Df(0,0)left[beginarrayc
x-0\y-0
endarrayright]+frac12[x-0mbox y-0]Hf(0,0)left[
beginarrayc
x-0\y-0
endarrayright]\
=1+[0mbox 0]left[
beginarrayc
x\y
endarrayright]+frac12[xmbox y]beginbmatrix
-2&0\0&-2
endbmatrixleft[
beginarrayc
x\y
endarrayright]=1-x^2-y^2
endsplit
endequation*
multivariable-calculus proof-verification matlab
asked Jul 25 at 1:29
Atsina
513113
I would like to either confirm my solution is correct, or find the error in it. I used the following MATLAB code to try to check my answer, but the solution it gave differs from mine.
f = exp(-(x^2+y^2))cos(xy);
taylor(f,[x,y],[0,0])
ans =
x^4/2 + (x^2*y^2)/2 - x^2 + y^4/2 - y^2 + 1
My solution:
The second degree Taylor polynomial at the point $(a,b)$ is given by
$$p_2(x,y)=f(a,b)+Df(a,b)left[beginarrayc
x-a\y-b
endarrayright]+frac12[x-ambox y-b]Hf(a,b)left[beginarrayc
x-a\y-b
endarrayright]$$
where $Df(a,b)$ is the Jacobian matrix and $Hf(a,b)$ is the Hessian matrix (i.e. first and second order partial derivatives, respectively). Thus, we begin by computing the partial derivatives:
beginequation*
beginsplit
fracpartial fpartial x(x,y)=-e^-x^2-y^2(ymbox sin(xy)+2xmbox cos(xy))\
fracpartial fpartial y(x,y)=-e^-x^2-y^2(xmbox sin(xy)+2ymbox cos(xy))\
fracpartial^2 fpartial x^2(x,y)=e^-x^2-y^2((4x^2-y^2-2)mboxcos(xy)+4xymbox sin(xy))\
fracpartial^2 fpartial y^2(x,y)=e^-x^2-y^2(4xymbox sin(xy)-(x^2-4y^2+2)mboxcos(xy))\
fracpartial^2 fpartial xpartial y(x,y)=fracpartial^2 fpartial ypartial x(x,y)=e^-x^2-y^2((2x^2+2y^2-1)mboxsin(xy)+3xymbox cos(xy))
endsplit
endequation*
Then, at the point $(x_0,y_0)=(0,0)$,
beginequation*
beginsplit
f(0,0)=1\
fracpartial fpartial x(0,0)=fracpartial fpartial y(0,0)=0\
fracpartial^2 fpartial x^2(0,0)=fracpartial^2 fpartial y^2(0,0)=-2\
fracpartial^2 fpartial xpartial y(0,0)=fracpartial^2 fpartial ypartial x(0,0)=0
endsplit
endequation*
Therefore, the second degree Taylor polynomial at the point $(0,0)$ is
beginequation*
beginsplit
p_2(x,y)=f(0,0)+Df(0,0)left[beginarrayc
x-0\y-0
endarrayright]+frac12[x-0mbox y-0]Hf(0,0)left[
beginarrayc
x-0\y-0
endarrayright]\
=1+[0mbox 0]left[
beginarrayc
x\y
endarrayright]+frac12[xmbox y]beginbmatrix
-2&0\0&-2
endbmatrixleft[
beginarrayc
x\y
endarrayright]=1-x^2-y^2
endsplit
endequation*
multivariable-calculus proof-verification matlab
asked Jul 25 at 1:29
Atsina
513113
asked Jul 25 at 1:29
Atsina
513113
asked Jul 25 at 1:29
Atsina
513113
asked Jul 25 at 1:29
Atsina
513113
513113
What’s your question?
– amd
Jul 25 at 2:19
@amd "I would like to either confirm my solution is correct, or find the error in it."
– Atsina
Jul 25 at 2:23
The Matlab output that you've got is the fourth-degree Taylor approximation, not second-degree. Either add'Order',2to the arguments or drop the higher-order terms yourself.
– amd
Jul 25 at 6:05
@amd Thank you!
– Atsina
Jul 25 at 15:26
add a comment |Â
What’s your question?
– amd
Jul 25 at 2:19
@amd "I would like to either confirm my solution is correct, or find the error in it."
– Atsina
Jul 25 at 2:23
The Matlab output that you've got is the fourth-degree Taylor approximation, not second-degree. Either add'Order',2to the arguments or drop the higher-order terms yourself.
– amd
Jul 25 at 6:05
@amd Thank you!
– Atsina
Jul 25 at 15:26
What’s your question?
– amd
Jul 25 at 2:19
What’s your question?
– amd
Jul 25 at 2:19
@amd "I would like to either confirm my solution is correct, or find the error in it."
– Atsina
Jul 25 at 2:23
@amd "I would like to either confirm my solution is correct, or find the error in it."
– Atsina
Jul 25 at 2:23
The Matlab output that you've got is the fourth-degree Taylor approximation, not second-degree. Either add
'Order',2 to the arguments or drop the higher-order terms yourself.– amd
Jul 25 at 6:05
The Matlab output that you've got is the fourth-degree Taylor approximation, not second-degree. Either add
'Order',2 to the arguments or drop the higher-order terms yourself.– amd
Jul 25 at 6:05
@amd Thank you!
– Atsina
Jul 25 at 15:26
@amd Thank you!
– Atsina
Jul 25 at 15:26
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
I leave your answer's critique to someone with less tired eyes. I will tell you this much. There is an easier way to calculate: just substitute into known Taylor series for the exponential and cosine,
$$f(x,y) = e^-x^2-y^2cos(xy) = (1-x^2-y^2+cdots)(1-frac12(xy)^2+ cdots )$$
But, $(xy)^2$ is order $4$ so the answer is merely,
$$f(x,y) = e^-x^2-y^2cos(xy) = 1-x^2-y^2+cdots $$
The above is the Taylor expansion of $f$ at $(0,0)$ to order $2$.
answered Jul 25 at 2:30
James S. Cook
12.8k22668
Oh...duh. Thank you.
– Atsina
Jul 25 at 2:34
Glad to help, also now that I look at your answer, glad we agree on the final result. Of course, your approach is logical it's just not as lazy as mine.
– James S. Cook
Jul 25 at 15:22
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
I leave your answer's critique to someone with less tired eyes. I will tell you this much. There is an easier way to calculate: just substitute into known Taylor series for the exponential and cosine,
$$f(x,y) = e^-x^2-y^2cos(xy) = (1-x^2-y^2+cdots)(1-frac12(xy)^2+ cdots )$$
But, $(xy)^2$ is order $4$ so the answer is merely,
$$f(x,y) = e^-x^2-y^2cos(xy) = 1-x^2-y^2+cdots $$
The above is the Taylor expansion of $f$ at $(0,0)$ to order $2$.
answered Jul 25 at 2:30
James S. Cook
12.8k22668
Oh...duh. Thank you.
– Atsina
Jul 25 at 2:34
Glad to help, also now that I look at your answer, glad we agree on the final result. Of course, your approach is logical it's just not as lazy as mine.
– James S. Cook
Jul 25 at 15:22
add a comment |Â
up vote
1
down vote
accepted
I leave your answer's critique to someone with less tired eyes. I will tell you this much. There is an easier way to calculate: just substitute into known Taylor series for the exponential and cosine,
$$f(x,y) = e^-x^2-y^2cos(xy) = (1-x^2-y^2+cdots)(1-frac12(xy)^2+ cdots )$$
But, $(xy)^2$ is order $4$ so the answer is merely,
$$f(x,y) = e^-x^2-y^2cos(xy) = 1-x^2-y^2+cdots $$
The above is the Taylor expansion of $f$ at $(0,0)$ to order $2$.
answered Jul 25 at 2:30
James S. Cook
12.8k22668
Oh...duh. Thank you.
– Atsina
Jul 25 at 2:34
Glad to help, also now that I look at your answer, glad we agree on the final result. Of course, your approach is logical it's just not as lazy as mine.
– James S. Cook
Jul 25 at 15:22
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I leave your answer's critique to someone with less tired eyes. I will tell you this much. There is an easier way to calculate: just substitute into known Taylor series for the exponential and cosine,
$$f(x,y) = e^-x^2-y^2cos(xy) = (1-x^2-y^2+cdots)(1-frac12(xy)^2+ cdots )$$
But, $(xy)^2$ is order $4$ so the answer is merely,
$$f(x,y) = e^-x^2-y^2cos(xy) = 1-x^2-y^2+cdots $$
The above is the Taylor expansion of $f$ at $(0,0)$ to order $2$.
answered Jul 25 at 2:30
James S. Cook
12.8k22668
I leave your answer's critique to someone with less tired eyes. I will tell you this much. There is an easier way to calculate: just substitute into known Taylor series for the exponential and cosine,
$$f(x,y) = e^-x^2-y^2cos(xy) = (1-x^2-y^2+cdots)(1-frac12(xy)^2+ cdots )$$
But, $(xy)^2$ is order $4$ so the answer is merely,
$$f(x,y) = e^-x^2-y^2cos(xy) = 1-x^2-y^2+cdots $$
The above is the Taylor expansion of $f$ at $(0,0)$ to order $2$.
answered Jul 25 at 2:30
James S. Cook
12.8k22668
answered Jul 25 at 2:30
James S. Cook
12.8k22668
answered Jul 25 at 2:30
James S. Cook
12.8k22668
answered Jul 25 at 2:30
James S. Cook
12.8k22668
12.8k22668
Oh...duh. Thank you.
– Atsina
Jul 25 at 2:34
Glad to help, also now that I look at your answer, glad we agree on the final result. Of course, your approach is logical it's just not as lazy as mine.
– James S. Cook
Jul 25 at 15:22
add a comment |Â
Oh...duh. Thank you.
– Atsina
Jul 25 at 2:34
Glad to help, also now that I look at your answer, glad we agree on the final result. Of course, your approach is logical it's just not as lazy as mine.
– James S. Cook
Jul 25 at 15:22
Oh...duh. Thank you.
– Atsina
Jul 25 at 2:34
Oh...duh. Thank you.
– Atsina
Jul 25 at 2:34
Glad to help, also now that I look at your answer, glad we agree on the final result. Of course, your approach is logical it's just not as lazy as mine.
– James S. Cook
Jul 25 at 15:22
Glad to help, also now that I look at your answer, glad we agree on the final result. Of course, your approach is logical it's just not as lazy as mine.
– James S. Cook
Jul 25 at 15:22
add a comment |Â
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What’s your question?
– amd
Jul 25 at 2:19
@amd "I would like to either confirm my solution is correct, or find the error in it."
– Atsina
Jul 25 at 2:23
The Matlab output that you've got is the fourth-degree Taylor approximation, not second-degree. Either add
'Order',2to the arguments or drop the higher-order terms yourself.– amd
Jul 25 at 6:05
@amd Thank you!
– Atsina
Jul 25 at 15:26