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My question is about the Taylor formula of $f(x,y)=sin xsin y$
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I have to solve the following exercise:
Find the secondary approach of $f(x,y)=sin xsin y$ near the point $(0,0)$. How accurate is the approach if $|x|leq 0,1$ and $|y|leq0,1$?
(I have to prove that $|E|leq frac86(0,1)^3$ )
$-$ I found :
the secondary approach is for $n=2$
beginalignf(x,y)= &f(0,0)+xf_x+yf_y+frac12(x^2f_xx+2xyf_xy+y^2f_yy)\&+ frac16(x^3f_xxx+3x^2yf_xxy +3y^2xf_xyy+y^3f_yyy)(Æ,n)endalign
$f(0,0)=0$
$f_x(0,0)=cos0sin0=0$, $f_y(0,0)=sin0cos0=0$
$f_xx(0,0)=-sin 0sin0=0$, $f_yy(0,0)=-sin0sin0=0$
$f_xy(0,0)=cos0cos0=1$
therefore $f(x,y)simeq 0+0+0+frac12(x^20+2xy+y^20)=xy$,
the error estimation is $|E(x,y)|=frac16(x^3f_xxx+dots+y^3f_yyy)(Æ,n)$
$-$How am I going to prove that $|E|leq frac86(0,1)^3?$ I don't know what I have to do. Are the things that I've done above correct?
taylor-expansion
edited Jul 25 at 11:56
Bernard
110k635103
asked Jul 25 at 11:29
Deppie3910
205
add a comment |Â
up vote
1
down vote
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I have to solve the following exercise:
Find the secondary approach of $f(x,y)=sin xsin y$ near the point $(0,0)$. How accurate is the approach if $|x|leq 0,1$ and $|y|leq0,1$?
(I have to prove that $|E|leq frac86(0,1)^3$ )
$-$ I found :
the secondary approach is for $n=2$
beginalignf(x,y)= &f(0,0)+xf_x+yf_y+frac12(x^2f_xx+2xyf_xy+y^2f_yy)\&+ frac16(x^3f_xxx+3x^2yf_xxy +3y^2xf_xyy+y^3f_yyy)(Æ,n)endalign
$f(0,0)=0$
$f_x(0,0)=cos0sin0=0$, $f_y(0,0)=sin0cos0=0$
$f_xx(0,0)=-sin 0sin0=0$, $f_yy(0,0)=-sin0sin0=0$
$f_xy(0,0)=cos0cos0=1$
therefore $f(x,y)simeq 0+0+0+frac12(x^20+2xy+y^20)=xy$,
the error estimation is $|E(x,y)|=frac16(x^3f_xxx+dots+y^3f_yyy)(Æ,n)$
$-$How am I going to prove that $|E|leq frac86(0,1)^3?$ I don't know what I have to do. Are the things that I've done above correct?
taylor-expansion
edited Jul 25 at 11:56
Bernard
110k635103
asked Jul 25 at 11:29
Deppie3910
205
1
You can use Lagrange's form of the remainder, as it looks for several variables. The derivatives that appear therein evaluated at unknown points are nevertheless sines and cosines, which have uniform bounds, $1$.
– user578878
Jul 25 at 11:32
1
This particular problem has the simplifying feature that the function is factorial, $f(x,y)=g_1(x)g_2(y)$. Therefore, you could deduce the bound from Lagrange's remainder for one variable applied to $sin(x)$ and $sin(y)$ separately.
– user578878
Jul 25 at 11:37
1
To continue on the path you delineate in your question (since, pace some comments above, the first question here is not to imagine other approaches), simply note that every third derivative involved in your last formula for $E$ is at most $1$, that the prefactors $x^iy^j$ involved are all bounded by $0.1^3$ and that the prefactor $frac16$ then yields exactly the bound $frac16cdot8cdot0.1^3$ you have been asked to show.
– Did
Jul 25 at 11:49
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have to solve the following exercise:
Find the secondary approach of $f(x,y)=sin xsin y$ near the point $(0,0)$. How accurate is the approach if $|x|leq 0,1$ and $|y|leq0,1$?
(I have to prove that $|E|leq frac86(0,1)^3$ )
$-$ I found :
the secondary approach is for $n=2$
beginalignf(x,y)= &f(0,0)+xf_x+yf_y+frac12(x^2f_xx+2xyf_xy+y^2f_yy)\&+ frac16(x^3f_xxx+3x^2yf_xxy +3y^2xf_xyy+y^3f_yyy)(Æ,n)endalign
$f(0,0)=0$
$f_x(0,0)=cos0sin0=0$, $f_y(0,0)=sin0cos0=0$
$f_xx(0,0)=-sin 0sin0=0$, $f_yy(0,0)=-sin0sin0=0$
$f_xy(0,0)=cos0cos0=1$
therefore $f(x,y)simeq 0+0+0+frac12(x^20+2xy+y^20)=xy$,
the error estimation is $|E(x,y)|=frac16(x^3f_xxx+dots+y^3f_yyy)(Æ,n)$
$-$How am I going to prove that $|E|leq frac86(0,1)^3?$ I don't know what I have to do. Are the things that I've done above correct?
taylor-expansion
edited Jul 25 at 11:56
Bernard
110k635103
asked Jul 25 at 11:29
Deppie3910
205
I have to solve the following exercise:
Find the secondary approach of $f(x,y)=sin xsin y$ near the point $(0,0)$. How accurate is the approach if $|x|leq 0,1$ and $|y|leq0,1$?
(I have to prove that $|E|leq frac86(0,1)^3$ )
$-$ I found :
the secondary approach is for $n=2$
beginalignf(x,y)= &f(0,0)+xf_x+yf_y+frac12(x^2f_xx+2xyf_xy+y^2f_yy)\&+ frac16(x^3f_xxx+3x^2yf_xxy +3y^2xf_xyy+y^3f_yyy)(Æ,n)endalign
$f(0,0)=0$
$f_x(0,0)=cos0sin0=0$, $f_y(0,0)=sin0cos0=0$
$f_xx(0,0)=-sin 0sin0=0$, $f_yy(0,0)=-sin0sin0=0$
$f_xy(0,0)=cos0cos0=1$
therefore $f(x,y)simeq 0+0+0+frac12(x^20+2xy+y^20)=xy$,
the error estimation is $|E(x,y)|=frac16(x^3f_xxx+dots+y^3f_yyy)(Æ,n)$
$-$How am I going to prove that $|E|leq frac86(0,1)^3?$ I don't know what I have to do. Are the things that I've done above correct?
taylor-expansion
edited Jul 25 at 11:56
Bernard
110k635103
asked Jul 25 at 11:29
Deppie3910
205
edited Jul 25 at 11:56
Bernard
110k635103
edited Jul 25 at 11:56
Bernard
110k635103
edited Jul 25 at 11:56
Bernard
110k635103
110k635103
asked Jul 25 at 11:29
Deppie3910
205
asked Jul 25 at 11:29
Deppie3910
205
asked Jul 25 at 11:29
Deppie3910
205
205
1
You can use Lagrange's form of the remainder, as it looks for several variables. The derivatives that appear therein evaluated at unknown points are nevertheless sines and cosines, which have uniform bounds, $1$.
– user578878
Jul 25 at 11:32
1
This particular problem has the simplifying feature that the function is factorial, $f(x,y)=g_1(x)g_2(y)$. Therefore, you could deduce the bound from Lagrange's remainder for one variable applied to $sin(x)$ and $sin(y)$ separately.
– user578878
Jul 25 at 11:37
1
To continue on the path you delineate in your question (since, pace some comments above, the first question here is not to imagine other approaches), simply note that every third derivative involved in your last formula for $E$ is at most $1$, that the prefactors $x^iy^j$ involved are all bounded by $0.1^3$ and that the prefactor $frac16$ then yields exactly the bound $frac16cdot8cdot0.1^3$ you have been asked to show.
– Did
Jul 25 at 11:49
add a comment |Â
1
You can use Lagrange's form of the remainder, as it looks for several variables. The derivatives that appear therein evaluated at unknown points are nevertheless sines and cosines, which have uniform bounds, $1$.
– user578878
Jul 25 at 11:32
1
This particular problem has the simplifying feature that the function is factorial, $f(x,y)=g_1(x)g_2(y)$. Therefore, you could deduce the bound from Lagrange's remainder for one variable applied to $sin(x)$ and $sin(y)$ separately.
– user578878
Jul 25 at 11:37
1
To continue on the path you delineate in your question (since, pace some comments above, the first question here is not to imagine other approaches), simply note that every third derivative involved in your last formula for $E$ is at most $1$, that the prefactors $x^iy^j$ involved are all bounded by $0.1^3$ and that the prefactor $frac16$ then yields exactly the bound $frac16cdot8cdot0.1^3$ you have been asked to show.
– Did
Jul 25 at 11:49
1
1
You can use Lagrange's form of the remainder, as it looks for several variables. The derivatives that appear therein evaluated at unknown points are nevertheless sines and cosines, which have uniform bounds, $1$.
– user578878
Jul 25 at 11:32
You can use Lagrange's form of the remainder, as it looks for several variables. The derivatives that appear therein evaluated at unknown points are nevertheless sines and cosines, which have uniform bounds, $1$.
– user578878
Jul 25 at 11:32
1
1
This particular problem has the simplifying feature that the function is factorial, $f(x,y)=g_1(x)g_2(y)$. Therefore, you could deduce the bound from Lagrange's remainder for one variable applied to $sin(x)$ and $sin(y)$ separately.
– user578878
Jul 25 at 11:37
This particular problem has the simplifying feature that the function is factorial, $f(x,y)=g_1(x)g_2(y)$. Therefore, you could deduce the bound from Lagrange's remainder for one variable applied to $sin(x)$ and $sin(y)$ separately.
– user578878
Jul 25 at 11:37
1
1
To continue on the path you delineate in your question (since, pace some comments above, the first question here is not to imagine other approaches), simply note that every third derivative involved in your last formula for $E$ is at most $1$, that the prefactors $x^iy^j$ involved are all bounded by $0.1^3$ and that the prefactor $frac16$ then yields exactly the bound $frac16cdot8cdot0.1^3$ you have been asked to show.
– Did
Jul 25 at 11:49
To continue on the path you delineate in your question (since, pace some comments above, the first question here is not to imagine other approaches), simply note that every third derivative involved in your last formula for $E$ is at most $1$, that the prefactors $x^iy^j$ involved are all bounded by $0.1^3$ and that the prefactor $frac16$ then yields exactly the bound $frac16cdot8cdot0.1^3$ you have been asked to show.
– Did
Jul 25 at 11:49
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
I suppose that in $0,1$, the comma is a decimal comma.
Observe that all partial derivatives at any order are the products of a sine or a cosine in $x$ and a sine or a cosine in $y$, up to a $-$ sign. So the absolute value of these derivatives are bounded by $1$, and by the triangle inequality,
$$|E(x,y)|=frac16|(x^3f_xxx+dots+y^3f_yyy)(Æ,n)|le frac16|(|x|^3 +3x^2|y|+3|x|y^2+|y|^3)$$
which is no more than $;dfrac16(0.1+0.1)^3=dfrac8610^-3;$ on $;[-0.1,0.1]times[-0.1,0.1]$.
answered Jul 25 at 12:09
Bernard
110k635103
Thank you very much!!
– Deppie3910
Jul 25 at 15:30
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
I suppose that in $0,1$, the comma is a decimal comma.
Observe that all partial derivatives at any order are the products of a sine or a cosine in $x$ and a sine or a cosine in $y$, up to a $-$ sign. So the absolute value of these derivatives are bounded by $1$, and by the triangle inequality,
$$|E(x,y)|=frac16|(x^3f_xxx+dots+y^3f_yyy)(Æ,n)|le frac16|(|x|^3 +3x^2|y|+3|x|y^2+|y|^3)$$
which is no more than $;dfrac16(0.1+0.1)^3=dfrac8610^-3;$ on $;[-0.1,0.1]times[-0.1,0.1]$.
answered Jul 25 at 12:09
Bernard
110k635103
Thank you very much!!
– Deppie3910
Jul 25 at 15:30
add a comment |Â
up vote
2
down vote
accepted
I suppose that in $0,1$, the comma is a decimal comma.
Observe that all partial derivatives at any order are the products of a sine or a cosine in $x$ and a sine or a cosine in $y$, up to a $-$ sign. So the absolute value of these derivatives are bounded by $1$, and by the triangle inequality,
$$|E(x,y)|=frac16|(x^3f_xxx+dots+y^3f_yyy)(Æ,n)|le frac16|(|x|^3 +3x^2|y|+3|x|y^2+|y|^3)$$
which is no more than $;dfrac16(0.1+0.1)^3=dfrac8610^-3;$ on $;[-0.1,0.1]times[-0.1,0.1]$.
answered Jul 25 at 12:09
Bernard
110k635103
Thank you very much!!
– Deppie3910
Jul 25 at 15:30
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
I suppose that in $0,1$, the comma is a decimal comma.
Observe that all partial derivatives at any order are the products of a sine or a cosine in $x$ and a sine or a cosine in $y$, up to a $-$ sign. So the absolute value of these derivatives are bounded by $1$, and by the triangle inequality,
$$|E(x,y)|=frac16|(x^3f_xxx+dots+y^3f_yyy)(Æ,n)|le frac16|(|x|^3 +3x^2|y|+3|x|y^2+|y|^3)$$
which is no more than $;dfrac16(0.1+0.1)^3=dfrac8610^-3;$ on $;[-0.1,0.1]times[-0.1,0.1]$.
answered Jul 25 at 12:09
Bernard
110k635103
I suppose that in $0,1$, the comma is a decimal comma.
Observe that all partial derivatives at any order are the products of a sine or a cosine in $x$ and a sine or a cosine in $y$, up to a $-$ sign. So the absolute value of these derivatives are bounded by $1$, and by the triangle inequality,
$$|E(x,y)|=frac16|(x^3f_xxx+dots+y^3f_yyy)(Æ,n)|le frac16|(|x|^3 +3x^2|y|+3|x|y^2+|y|^3)$$
which is no more than $;dfrac16(0.1+0.1)^3=dfrac8610^-3;$ on $;[-0.1,0.1]times[-0.1,0.1]$.
answered Jul 25 at 12:09
Bernard
110k635103
answered Jul 25 at 12:09
Bernard
110k635103
answered Jul 25 at 12:09
Bernard
110k635103
answered Jul 25 at 12:09
Bernard
110k635103
110k635103
Thank you very much!!
– Deppie3910
Jul 25 at 15:30
add a comment |Â
Thank you very much!!
– Deppie3910
Jul 25 at 15:30
Thank you very much!!
– Deppie3910
Jul 25 at 15:30
Thank you very much!!
– Deppie3910
Jul 25 at 15:30
add a comment |Â
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1
You can use Lagrange's form of the remainder, as it looks for several variables. The derivatives that appear therein evaluated at unknown points are nevertheless sines and cosines, which have uniform bounds, $1$.
– user578878
Jul 25 at 11:32
1
This particular problem has the simplifying feature that the function is factorial, $f(x,y)=g_1(x)g_2(y)$. Therefore, you could deduce the bound from Lagrange's remainder for one variable applied to $sin(x)$ and $sin(y)$ separately.
– user578878
Jul 25 at 11:37
1
To continue on the path you delineate in your question (since, pace some comments above, the first question here is not to imagine other approaches), simply note that every third derivative involved in your last formula for $E$ is at most $1$, that the prefactors $x^iy^j$ involved are all bounded by $0.1^3$ and that the prefactor $frac16$ then yields exactly the bound $frac16cdot8cdot0.1^3$ you have been asked to show.
– Did
Jul 25 at 11:49