Taylor expansion of $x^2ye^{1-xy}$ at the point $(1,1)$
Multi tool use
$begingroup$
I'm asked to find the second order Taylor expansion of $x^2ye^{1-xy}$ at the point $(1,1)$. I would like that you check my work, thank you.
So first I took all needed derivatives :
$$f_x=2xye^{1-xy}+x^2ye^{1-xy}(-y)=e^{1-xy}(2xy-x^2y^2)$$
$$f_y=x^2e^{1-xy}+x^2ye^{1-xy}(-x)=e^{1-xy}(x^2-x^3y)$$
$$f_{xx}=(-y)e^{1-xy}(2xy-x^2y^2)+e^{1-xy}(2y-2xy^2)=e^{1-xy}(x^2y^3-4xy^2+2y)$$
$$f_{yy}=(-x)e^{1-xy}(x^2-x^3y)+e^{1-xy}(-x^3)=e^{1-xy}(x^4y-2x^3)$$
$$f_{xy}=f_{yx}=(-x)e^{1-xy}(2xy-x^2y^2)+e^{1-xy}(2x-2x^2y)=e^{1-xy}(x^3y^2-4x^2y+2x)$$
and $f(1,1)=1, f_x(1,1)=1, f_y(1,1)=0, f_{xx}(1,1)=-1, f_{yy}(1,1)=-1, f_{xy}(1,1)=f_{yx}(1,1)=-1$
So now, I plug everything into the formula:
$$f(1,1)+f_x(1,1)(x-1)+f_y(1,1)+frac{1}{2!}[f_{xx}(1,1)(x-1)^2+2f_{xy}(x-1)(y-1)+f_{yy}(1,1)(y-1)^2]$$
$$=1+(x-1)+frac{1}{2!}[-(x-1)^2-2(x-1)(y-1)-(y-1)^2]$$
And I think that this should be the answer, do you agree with that ? Or did I forgot something/ made a mistake somewhere ? I checked the derivatives in WolframAlpha but still, I would like your feedback to be sure.
Thanks for your help !
real-analysis calculus analysis multivariable-calculus taylor-expansion
asked Jan 17 at 18:20
PoujhPoujh
609516
$endgroup$
add a comment |
$begingroup$
I'm asked to find the second order Taylor expansion of $x^2ye^{1-xy}$ at the point $(1,1)$. I would like that you check my work, thank you.
So first I took all needed derivatives :
$$f_x=2xye^{1-xy}+x^2ye^{1-xy}(-y)=e^{1-xy}(2xy-x^2y^2)$$
$$f_y=x^2e^{1-xy}+x^2ye^{1-xy}(-x)=e^{1-xy}(x^2-x^3y)$$
$$f_{xx}=(-y)e^{1-xy}(2xy-x^2y^2)+e^{1-xy}(2y-2xy^2)=e^{1-xy}(x^2y^3-4xy^2+2y)$$
$$f_{yy}=(-x)e^{1-xy}(x^2-x^3y)+e^{1-xy}(-x^3)=e^{1-xy}(x^4y-2x^3)$$
$$f_{xy}=f_{yx}=(-x)e^{1-xy}(2xy-x^2y^2)+e^{1-xy}(2x-2x^2y)=e^{1-xy}(x^3y^2-4x^2y+2x)$$
and $f(1,1)=1, f_x(1,1)=1, f_y(1,1)=0, f_{xx}(1,1)=-1, f_{yy}(1,1)=-1, f_{xy}(1,1)=f_{yx}(1,1)=-1$
So now, I plug everything into the formula:
$$f(1,1)+f_x(1,1)(x-1)+f_y(1,1)+frac{1}{2!}[f_{xx}(1,1)(x-1)^2+2f_{xy}(x-1)(y-1)+f_{yy}(1,1)(y-1)^2]$$
$$=1+(x-1)+frac{1}{2!}[-(x-1)^2-2(x-1)(y-1)-(y-1)^2]$$
And I think that this should be the answer, do you agree with that ? Or did I forgot something/ made a mistake somewhere ? I checked the derivatives in WolframAlpha but still, I would like your feedback to be sure.
Thanks for your help !
real-analysis calculus analysis multivariable-calculus taylor-expansion
asked Jan 17 at 18:20
PoujhPoujh
609516
$endgroup$
add a comment |
$begingroup$
I'm asked to find the second order Taylor expansion of $x^2ye^{1-xy}$ at the point $(1,1)$. I would like that you check my work, thank you.
So first I took all needed derivatives :
$$f_x=2xye^{1-xy}+x^2ye^{1-xy}(-y)=e^{1-xy}(2xy-x^2y^2)$$
$$f_y=x^2e^{1-xy}+x^2ye^{1-xy}(-x)=e^{1-xy}(x^2-x^3y)$$
$$f_{xx}=(-y)e^{1-xy}(2xy-x^2y^2)+e^{1-xy}(2y-2xy^2)=e^{1-xy}(x^2y^3-4xy^2+2y)$$
$$f_{yy}=(-x)e^{1-xy}(x^2-x^3y)+e^{1-xy}(-x^3)=e^{1-xy}(x^4y-2x^3)$$
$$f_{xy}=f_{yx}=(-x)e^{1-xy}(2xy-x^2y^2)+e^{1-xy}(2x-2x^2y)=e^{1-xy}(x^3y^2-4x^2y+2x)$$
and $f(1,1)=1, f_x(1,1)=1, f_y(1,1)=0, f_{xx}(1,1)=-1, f_{yy}(1,1)=-1, f_{xy}(1,1)=f_{yx}(1,1)=-1$
So now, I plug everything into the formula:
$$f(1,1)+f_x(1,1)(x-1)+f_y(1,1)+frac{1}{2!}[f_{xx}(1,1)(x-1)^2+2f_{xy}(x-1)(y-1)+f_{yy}(1,1)(y-1)^2]$$
$$=1+(x-1)+frac{1}{2!}[-(x-1)^2-2(x-1)(y-1)-(y-1)^2]$$
And I think that this should be the answer, do you agree with that ? Or did I forgot something/ made a mistake somewhere ? I checked the derivatives in WolframAlpha but still, I would like your feedback to be sure.
Thanks for your help !
real-analysis calculus analysis multivariable-calculus taylor-expansion
asked Jan 17 at 18:20
PoujhPoujh
609516
$endgroup$
I'm asked to find the second order Taylor expansion of $x^2ye^{1-xy}$ at the point $(1,1)$. I would like that you check my work, thank you.
So first I took all needed derivatives :
$$f_x=2xye^{1-xy}+x^2ye^{1-xy}(-y)=e^{1-xy}(2xy-x^2y^2)$$
$$f_y=x^2e^{1-xy}+x^2ye^{1-xy}(-x)=e^{1-xy}(x^2-x^3y)$$
$$f_{xx}=(-y)e^{1-xy}(2xy-x^2y^2)+e^{1-xy}(2y-2xy^2)=e^{1-xy}(x^2y^3-4xy^2+2y)$$
$$f_{yy}=(-x)e^{1-xy}(x^2-x^3y)+e^{1-xy}(-x^3)=e^{1-xy}(x^4y-2x^3)$$
$$f_{xy}=f_{yx}=(-x)e^{1-xy}(2xy-x^2y^2)+e^{1-xy}(2x-2x^2y)=e^{1-xy}(x^3y^2-4x^2y+2x)$$
and $f(1,1)=1, f_x(1,1)=1, f_y(1,1)=0, f_{xx}(1,1)=-1, f_{yy}(1,1)=-1, f_{xy}(1,1)=f_{yx}(1,1)=-1$
So now, I plug everything into the formula:
$$f(1,1)+f_x(1,1)(x-1)+f_y(1,1)+frac{1}{2!}[f_{xx}(1,1)(x-1)^2+2f_{xy}(x-1)(y-1)+f_{yy}(1,1)(y-1)^2]$$
$$=1+(x-1)+frac{1}{2!}[-(x-1)^2-2(x-1)(y-1)-(y-1)^2]$$
And I think that this should be the answer, do you agree with that ? Or did I forgot something/ made a mistake somewhere ? I checked the derivatives in WolframAlpha but still, I would like your feedback to be sure.
Thanks for your help !
real-analysis calculus analysis multivariable-calculus taylor-expansion
real-analysis calculus analysis multivariable-calculus taylor-expansion
asked Jan 17 at 18:20
PoujhPoujh
609516
asked Jan 17 at 18:20
PoujhPoujh
609516
asked Jan 17 at 18:20
PoujhPoujh
609516
asked Jan 17 at 18:20
PoujhPoujh
609516
asked Jan 17 at 18:20
PoujhPoujh
609516
609516
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
One can also use formal algebraic operations with taylor polynomials taken modulo the ideal generated by monomials of degree $3$ in the factors $s=(x-1)$ and $t=(y-1)$. I will denote this ideal by $mathcal O$.
Instead of the longer $mathcal O(s^3,s^2t, st^2, t^3)$.
Then:
$$
begin{aligned}
&x^2yexp(1-xy) +mathcal O
\
%&=
%((x-1)+1)^2((y-1)+1)exp(1-((x-1)+1)((y-1)+1))
%+mathcal O
%\
&=(1+s)^2(1+t)exp(1-(1+s)(1+t))
+mathcal O
\
&=(1+s)^2(1+t)exp(-s-t-st)
+mathcal O
\
&=(1+t+2s+2st+s^2)left(1-frac 1{1!}(s+t+st)+frac 1{2!}(s+t)^2right)
+mathcal O
\
&=(1+t+2s+2st+s^2)
-(1+t+2s)frac 1{1!}(s+t+st)
+1cdotfrac 1{2!}(s+t)^2
+mathcal O
\
&=(1+t+2s+2st+s^2)
\
&qquad-(s+st+2s^2)-(t+t^2+2st)-st
\
&qquadqquad+frac 1{2!}(s^2+2st+t^2)
+mathcal O
\
&=
1+s-frac 12s^2-st-frac 12t^2+mathcal O .
end{aligned}
$$
All computations were included. (Many copy+paste issues.)
Computer check:
sage: var('x,y,s,t');
sage: taylor( (1+s)^2 * (1+t) * exp(1-(1+s)*(1+t)), (s,0), (t,0), 2 )
-1/2*s^2 - s*t - 1/2*t^2 + s + 1
sage: taylor( x^2 * y * exp( 1-x*y ), (x,1), (y,1), 2 )
-1/2*(x - 1)^2 - (x - 1)*(y - 1) - 1/2*(y - 1)^2 + x
(The $s,t$-version is better, for some reason, sage wanted to simplify the $(x-1)+1$, corresponding to $s+1$, and printed that $x$. Else, everything is the same. Well, the order of the terms is not the one we want, but...)
answered Jan 17 at 19:06
dan_fuleadan_fulea
6,5781312
$endgroup$
$begingroup$
So just to be sure, you make a substitution for $(x-1)$ and $(y-1)$ and arrange the exponent so that we can compute everything in terms of $s=(x-1)$ and $t=(y-1)$. You then use the well known Taylor series expansion of $e^x$ with those substitution variables so that we can calculate the Taylor expansion as if we had only one variable (this variable being a "mix" of t and s), is that correct ?
$endgroup$
– Poujh
Jan 17 at 19:15
1
$begingroup$
@Poujh Yes, all algebraic operations like addition, multiplication and composition (with care about the point to expand around, so $exp$ was expanded around $0$, since $-s-t-st=1-xy$ is "around zero") are compatible with building the Taylor polynomial modulo some fixed degree. Building the Taylor polynomial from the definition works of course, but against time, e.g. in an exam, it is safer to avoid derivatives (of derivatives) and compute them in the relevant points, for my taste.
$endgroup$
– dan_fulea
Jan 17 at 19:20
add a comment |
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$begingroup$
One can also use formal algebraic operations with taylor polynomials taken modulo the ideal generated by monomials of degree $3$ in the factors $s=(x-1)$ and $t=(y-1)$. I will denote this ideal by $mathcal O$.
Instead of the longer $mathcal O(s^3,s^2t, st^2, t^3)$.
Then:
$$
begin{aligned}
&x^2yexp(1-xy) +mathcal O
\
%&=
%((x-1)+1)^2((y-1)+1)exp(1-((x-1)+1)((y-1)+1))
%+mathcal O
%\
&=(1+s)^2(1+t)exp(1-(1+s)(1+t))
+mathcal O
\
&=(1+s)^2(1+t)exp(-s-t-st)
+mathcal O
\
&=(1+t+2s+2st+s^2)left(1-frac 1{1!}(s+t+st)+frac 1{2!}(s+t)^2right)
+mathcal O
\
&=(1+t+2s+2st+s^2)
-(1+t+2s)frac 1{1!}(s+t+st)
+1cdotfrac 1{2!}(s+t)^2
+mathcal O
\
&=(1+t+2s+2st+s^2)
\
&qquad-(s+st+2s^2)-(t+t^2+2st)-st
\
&qquadqquad+frac 1{2!}(s^2+2st+t^2)
+mathcal O
\
&=
1+s-frac 12s^2-st-frac 12t^2+mathcal O .
end{aligned}
$$
All computations were included. (Many copy+paste issues.)
Computer check:
sage: var('x,y,s,t');
sage: taylor( (1+s)^2 * (1+t) * exp(1-(1+s)*(1+t)), (s,0), (t,0), 2 )
-1/2*s^2 - s*t - 1/2*t^2 + s + 1
sage: taylor( x^2 * y * exp( 1-x*y ), (x,1), (y,1), 2 )
-1/2*(x - 1)^2 - (x - 1)*(y - 1) - 1/2*(y - 1)^2 + x
(The $s,t$-version is better, for some reason, sage wanted to simplify the $(x-1)+1$, corresponding to $s+1$, and printed that $x$. Else, everything is the same. Well, the order of the terms is not the one we want, but...)
answered Jan 17 at 19:06
dan_fuleadan_fulea
6,5781312
$endgroup$
$begingroup$
So just to be sure, you make a substitution for $(x-1)$ and $(y-1)$ and arrange the exponent so that we can compute everything in terms of $s=(x-1)$ and $t=(y-1)$. You then use the well known Taylor series expansion of $e^x$ with those substitution variables so that we can calculate the Taylor expansion as if we had only one variable (this variable being a "mix" of t and s), is that correct ?
$endgroup$
– Poujh
Jan 17 at 19:15
1
$begingroup$
@Poujh Yes, all algebraic operations like addition, multiplication and composition (with care about the point to expand around, so $exp$ was expanded around $0$, since $-s-t-st=1-xy$ is "around zero") are compatible with building the Taylor polynomial modulo some fixed degree. Building the Taylor polynomial from the definition works of course, but against time, e.g. in an exam, it is safer to avoid derivatives (of derivatives) and compute them in the relevant points, for my taste.
$endgroup$
– dan_fulea
Jan 17 at 19:20
add a comment |
$begingroup$
One can also use formal algebraic operations with taylor polynomials taken modulo the ideal generated by monomials of degree $3$ in the factors $s=(x-1)$ and $t=(y-1)$. I will denote this ideal by $mathcal O$.
Instead of the longer $mathcal O(s^3,s^2t, st^2, t^3)$.
Then:
$$
begin{aligned}
&x^2yexp(1-xy) +mathcal O
\
%&=
%((x-1)+1)^2((y-1)+1)exp(1-((x-1)+1)((y-1)+1))
%+mathcal O
%\
&=(1+s)^2(1+t)exp(1-(1+s)(1+t))
+mathcal O
\
&=(1+s)^2(1+t)exp(-s-t-st)
+mathcal O
\
&=(1+t+2s+2st+s^2)left(1-frac 1{1!}(s+t+st)+frac 1{2!}(s+t)^2right)
+mathcal O
\
&=(1+t+2s+2st+s^2)
-(1+t+2s)frac 1{1!}(s+t+st)
+1cdotfrac 1{2!}(s+t)^2
+mathcal O
\
&=(1+t+2s+2st+s^2)
\
&qquad-(s+st+2s^2)-(t+t^2+2st)-st
\
&qquadqquad+frac 1{2!}(s^2+2st+t^2)
+mathcal O
\
&=
1+s-frac 12s^2-st-frac 12t^2+mathcal O .
end{aligned}
$$
All computations were included. (Many copy+paste issues.)
Computer check:
sage: var('x,y,s,t');
sage: taylor( (1+s)^2 * (1+t) * exp(1-(1+s)*(1+t)), (s,0), (t,0), 2 )
-1/2*s^2 - s*t - 1/2*t^2 + s + 1
sage: taylor( x^2 * y * exp( 1-x*y ), (x,1), (y,1), 2 )
-1/2*(x - 1)^2 - (x - 1)*(y - 1) - 1/2*(y - 1)^2 + x
(The $s,t$-version is better, for some reason, sage wanted to simplify the $(x-1)+1$, corresponding to $s+1$, and printed that $x$. Else, everything is the same. Well, the order of the terms is not the one we want, but...)
answered Jan 17 at 19:06
dan_fuleadan_fulea
6,5781312
$endgroup$
$begingroup$
So just to be sure, you make a substitution for $(x-1)$ and $(y-1)$ and arrange the exponent so that we can compute everything in terms of $s=(x-1)$ and $t=(y-1)$. You then use the well known Taylor series expansion of $e^x$ with those substitution variables so that we can calculate the Taylor expansion as if we had only one variable (this variable being a "mix" of t and s), is that correct ?
$endgroup$
– Poujh
Jan 17 at 19:15
1
$begingroup$
@Poujh Yes, all algebraic operations like addition, multiplication and composition (with care about the point to expand around, so $exp$ was expanded around $0$, since $-s-t-st=1-xy$ is "around zero") are compatible with building the Taylor polynomial modulo some fixed degree. Building the Taylor polynomial from the definition works of course, but against time, e.g. in an exam, it is safer to avoid derivatives (of derivatives) and compute them in the relevant points, for my taste.
$endgroup$
– dan_fulea
Jan 17 at 19:20
add a comment |
$begingroup$
One can also use formal algebraic operations with taylor polynomials taken modulo the ideal generated by monomials of degree $3$ in the factors $s=(x-1)$ and $t=(y-1)$. I will denote this ideal by $mathcal O$.
Instead of the longer $mathcal O(s^3,s^2t, st^2, t^3)$.
Then:
$$
begin{aligned}
&x^2yexp(1-xy) +mathcal O
\
%&=
%((x-1)+1)^2((y-1)+1)exp(1-((x-1)+1)((y-1)+1))
%+mathcal O
%\
&=(1+s)^2(1+t)exp(1-(1+s)(1+t))
+mathcal O
\
&=(1+s)^2(1+t)exp(-s-t-st)
+mathcal O
\
&=(1+t+2s+2st+s^2)left(1-frac 1{1!}(s+t+st)+frac 1{2!}(s+t)^2right)
+mathcal O
\
&=(1+t+2s+2st+s^2)
-(1+t+2s)frac 1{1!}(s+t+st)
+1cdotfrac 1{2!}(s+t)^2
+mathcal O
\
&=(1+t+2s+2st+s^2)
\
&qquad-(s+st+2s^2)-(t+t^2+2st)-st
\
&qquadqquad+frac 1{2!}(s^2+2st+t^2)
+mathcal O
\
&=
1+s-frac 12s^2-st-frac 12t^2+mathcal O .
end{aligned}
$$
All computations were included. (Many copy+paste issues.)
Computer check:
sage: var('x,y,s,t');
sage: taylor( (1+s)^2 * (1+t) * exp(1-(1+s)*(1+t)), (s,0), (t,0), 2 )
-1/2*s^2 - s*t - 1/2*t^2 + s + 1
sage: taylor( x^2 * y * exp( 1-x*y ), (x,1), (y,1), 2 )
-1/2*(x - 1)^2 - (x - 1)*(y - 1) - 1/2*(y - 1)^2 + x
(The $s,t$-version is better, for some reason, sage wanted to simplify the $(x-1)+1$, corresponding to $s+1$, and printed that $x$. Else, everything is the same. Well, the order of the terms is not the one we want, but...)
answered Jan 17 at 19:06
dan_fuleadan_fulea
6,5781312
$endgroup$
One can also use formal algebraic operations with taylor polynomials taken modulo the ideal generated by monomials of degree $3$ in the factors $s=(x-1)$ and $t=(y-1)$. I will denote this ideal by $mathcal O$.
Instead of the longer $mathcal O(s^3,s^2t, st^2, t^3)$.
Then:
$$
begin{aligned}
&x^2yexp(1-xy) +mathcal O
\
%&=
%((x-1)+1)^2((y-1)+1)exp(1-((x-1)+1)((y-1)+1))
%+mathcal O
%\
&=(1+s)^2(1+t)exp(1-(1+s)(1+t))
+mathcal O
\
&=(1+s)^2(1+t)exp(-s-t-st)
+mathcal O
\
&=(1+t+2s+2st+s^2)left(1-frac 1{1!}(s+t+st)+frac 1{2!}(s+t)^2right)
+mathcal O
\
&=(1+t+2s+2st+s^2)
-(1+t+2s)frac 1{1!}(s+t+st)
+1cdotfrac 1{2!}(s+t)^2
+mathcal O
\
&=(1+t+2s+2st+s^2)
\
&qquad-(s+st+2s^2)-(t+t^2+2st)-st
\
&qquadqquad+frac 1{2!}(s^2+2st+t^2)
+mathcal O
\
&=
1+s-frac 12s^2-st-frac 12t^2+mathcal O .
end{aligned}
$$
All computations were included. (Many copy+paste issues.)
Computer check:
sage: var('x,y,s,t');
sage: taylor( (1+s)^2 * (1+t) * exp(1-(1+s)*(1+t)), (s,0), (t,0), 2 )
-1/2*s^2 - s*t - 1/2*t^2 + s + 1
sage: taylor( x^2 * y * exp( 1-x*y ), (x,1), (y,1), 2 )
-1/2*(x - 1)^2 - (x - 1)*(y - 1) - 1/2*(y - 1)^2 + x
(The $s,t$-version is better, for some reason, sage wanted to simplify the $(x-1)+1$, corresponding to $s+1$, and printed that $x$. Else, everything is the same. Well, the order of the terms is not the one we want, but...)
answered Jan 17 at 19:06
dan_fuleadan_fulea
6,5781312
answered Jan 17 at 19:06
dan_fuleadan_fulea
6,5781312
answered Jan 17 at 19:06
dan_fuleadan_fulea
6,5781312
answered Jan 17 at 19:06
dan_fuleadan_fulea
6,5781312
6,5781312
$begingroup$
So just to be sure, you make a substitution for $(x-1)$ and $(y-1)$ and arrange the exponent so that we can compute everything in terms of $s=(x-1)$ and $t=(y-1)$. You then use the well known Taylor series expansion of $e^x$ with those substitution variables so that we can calculate the Taylor expansion as if we had only one variable (this variable being a "mix" of t and s), is that correct ?
$endgroup$
– Poujh
Jan 17 at 19:15
1
$begingroup$
@Poujh Yes, all algebraic operations like addition, multiplication and composition (with care about the point to expand around, so $exp$ was expanded around $0$, since $-s-t-st=1-xy$ is "around zero") are compatible with building the Taylor polynomial modulo some fixed degree. Building the Taylor polynomial from the definition works of course, but against time, e.g. in an exam, it is safer to avoid derivatives (of derivatives) and compute them in the relevant points, for my taste.
$endgroup$
– dan_fulea
Jan 17 at 19:20
add a comment |
$begingroup$
So just to be sure, you make a substitution for $(x-1)$ and $(y-1)$ and arrange the exponent so that we can compute everything in terms of $s=(x-1)$ and $t=(y-1)$. You then use the well known Taylor series expansion of $e^x$ with those substitution variables so that we can calculate the Taylor expansion as if we had only one variable (this variable being a "mix" of t and s), is that correct ?
$endgroup$
– Poujh
Jan 17 at 19:15
1
$begingroup$
@Poujh Yes, all algebraic operations like addition, multiplication and composition (with care about the point to expand around, so $exp$ was expanded around $0$, since $-s-t-st=1-xy$ is "around zero") are compatible with building the Taylor polynomial modulo some fixed degree. Building the Taylor polynomial from the definition works of course, but against time, e.g. in an exam, it is safer to avoid derivatives (of derivatives) and compute them in the relevant points, for my taste.
$endgroup$
– dan_fulea
Jan 17 at 19:20
$begingroup$
So just to be sure, you make a substitution for $(x-1)$ and $(y-1)$ and arrange the exponent so that we can compute everything in terms of $s=(x-1)$ and $t=(y-1)$. You then use the well known Taylor series expansion of $e^x$ with those substitution variables so that we can calculate the Taylor expansion as if we had only one variable (this variable being a "mix" of t and s), is that correct ?
$endgroup$
– Poujh
Jan 17 at 19:15
$begingroup$
So just to be sure, you make a substitution for $(x-1)$ and $(y-1)$ and arrange the exponent so that we can compute everything in terms of $s=(x-1)$ and $t=(y-1)$. You then use the well known Taylor series expansion of $e^x$ with those substitution variables so that we can calculate the Taylor expansion as if we had only one variable (this variable being a "mix" of t and s), is that correct ?
$endgroup$
– Poujh
Jan 17 at 19:15
1
1
$begingroup$
@Poujh Yes, all algebraic operations like addition, multiplication and composition (with care about the point to expand around, so $exp$ was expanded around $0$, since $-s-t-st=1-xy$ is "around zero") are compatible with building the Taylor polynomial modulo some fixed degree. Building the Taylor polynomial from the definition works of course, but against time, e.g. in an exam, it is safer to avoid derivatives (of derivatives) and compute them in the relevant points, for my taste.
$endgroup$
– dan_fulea
Jan 17 at 19:20
$begingroup$
@Poujh Yes, all algebraic operations like addition, multiplication and composition (with care about the point to expand around, so $exp$ was expanded around $0$, since $-s-t-st=1-xy$ is "around zero") are compatible with building the Taylor polynomial modulo some fixed degree. Building the Taylor polynomial from the definition works of course, but against time, e.g. in an exam, it is safer to avoid derivatives (of derivatives) and compute them in the relevant points, for my taste.
$endgroup$
– dan_fulea
Jan 17 at 19:20
add a comment |
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