Mixing
Taylor's expansion of a norm
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I want to Taylor expand (2nd order) the function, $f(x)=e^{{|x|}^2}$ (where $x$ is $= (x,y)in R^2$) and I also want to include the 3rd order integral remainder.
I know that for $e^{x^2}$ I have: $f(x) = e^{x^2} + 2xe^{x^2}(x-x_0) + (2xe^{x^2} + 4x^2e^{x^2})(x-x_0)^2/2$. How does this change when I have the norm and how would I make this expression tidier/simpler.
Also for the integral remainder: I have $E_3(x)=1/6int_{x_0}^x(x-t)^3f^4(t)dt $
Again is this correct and how does it change with the norm?
real-analysis
asked Jan 23 at 22:25
JackJack
1268
$endgroup$
add a comment |
$begingroup$
I want to Taylor expand (2nd order) the function, $f(x)=e^{{|x|}^2}$ (where $x$ is $= (x,y)in R^2$) and I also want to include the 3rd order integral remainder.
I know that for $e^{x^2}$ I have: $f(x) = e^{x^2} + 2xe^{x^2}(x-x_0) + (2xe^{x^2} + 4x^2e^{x^2})(x-x_0)^2/2$. How does this change when I have the norm and how would I make this expression tidier/simpler.
Also for the integral remainder: I have $E_3(x)=1/6int_{x_0}^x(x-t)^3f^4(t)dt $
Again is this correct and how does it change with the norm?
real-analysis
asked Jan 23 at 22:25
JackJack
1268
$endgroup$
1
$begingroup$
taylor expand around what point?
$endgroup$
– Calvin Khor
Jan 23 at 22:31
$begingroup$
any arbitrary x_0.
$endgroup$
– Jack
Jan 23 at 22:32
$begingroup$
The strategy is to consider the line segment $[x_0,x]subsetmathbb R^n$ (here $n=2$) parameterised by $gamma(t) = x_0 + t(x-x_0)$ and then taylor expand the 1D function $ f(gamma(t))$ at $t=0$.
$endgroup$
– Calvin Khor
Jan 23 at 22:39
$begingroup$
After Taylor expanding you should get a polynomial in $x$ with constant coefficients, which your expression is not. Can you find the mistake?
$endgroup$
– lcv
Jan 23 at 23:33
add a comment |
$begingroup$
I want to Taylor expand (2nd order) the function, $f(x)=e^{{|x|}^2}$ (where $x$ is $= (x,y)in R^2$) and I also want to include the 3rd order integral remainder.
I know that for $e^{x^2}$ I have: $f(x) = e^{x^2} + 2xe^{x^2}(x-x_0) + (2xe^{x^2} + 4x^2e^{x^2})(x-x_0)^2/2$. How does this change when I have the norm and how would I make this expression tidier/simpler.
Also for the integral remainder: I have $E_3(x)=1/6int_{x_0}^x(x-t)^3f^4(t)dt $
Again is this correct and how does it change with the norm?
real-analysis
asked Jan 23 at 22:25
JackJack
1268
$endgroup$
I want to Taylor expand (2nd order) the function, $f(x)=e^{{|x|}^2}$ (where $x$ is $= (x,y)in R^2$) and I also want to include the 3rd order integral remainder.
I know that for $e^{x^2}$ I have: $f(x) = e^{x^2} + 2xe^{x^2}(x-x_0) + (2xe^{x^2} + 4x^2e^{x^2})(x-x_0)^2/2$. How does this change when I have the norm and how would I make this expression tidier/simpler.
Also for the integral remainder: I have $E_3(x)=1/6int_{x_0}^x(x-t)^3f^4(t)dt $
Again is this correct and how does it change with the norm?
real-analysis
real-analysis
asked Jan 23 at 22:25
JackJack
1268
asked Jan 23 at 22:25
JackJack
1268
asked Jan 23 at 22:25
JackJack
1268
asked Jan 23 at 22:25
JackJack
1268
asked Jan 23 at 22:25
JackJack
1268
1268
1
$begingroup$
taylor expand around what point?
$endgroup$
– Calvin Khor
Jan 23 at 22:31
$begingroup$
any arbitrary x_0.
$endgroup$
– Jack
Jan 23 at 22:32
$begingroup$
The strategy is to consider the line segment $[x_0,x]subsetmathbb R^n$ (here $n=2$) parameterised by $gamma(t) = x_0 + t(x-x_0)$ and then taylor expand the 1D function $ f(gamma(t))$ at $t=0$.
$endgroup$
– Calvin Khor
Jan 23 at 22:39
$begingroup$
After Taylor expanding you should get a polynomial in $x$ with constant coefficients, which your expression is not. Can you find the mistake?
$endgroup$
– lcv
Jan 23 at 23:33
add a comment |
1
$begingroup$
taylor expand around what point?
$endgroup$
– Calvin Khor
Jan 23 at 22:31
$begingroup$
any arbitrary x_0.
$endgroup$
– Jack
Jan 23 at 22:32
$begingroup$
The strategy is to consider the line segment $[x_0,x]subsetmathbb R^n$ (here $n=2$) parameterised by $gamma(t) = x_0 + t(x-x_0)$ and then taylor expand the 1D function $ f(gamma(t))$ at $t=0$.
$endgroup$
– Calvin Khor
Jan 23 at 22:39
$begingroup$
After Taylor expanding you should get a polynomial in $x$ with constant coefficients, which your expression is not. Can you find the mistake?
$endgroup$
– lcv
Jan 23 at 23:33
1
1
$begingroup$
taylor expand around what point?
$endgroup$
– Calvin Khor
Jan 23 at 22:31
$begingroup$
taylor expand around what point?
$endgroup$
– Calvin Khor
Jan 23 at 22:31
$begingroup$
any arbitrary x_0.
$endgroup$
– Jack
Jan 23 at 22:32
$begingroup$
any arbitrary x_0.
$endgroup$
– Jack
Jan 23 at 22:32
$begingroup$
The strategy is to consider the line segment $[x_0,x]subsetmathbb R^n$ (here $n=2$) parameterised by $gamma(t) = x_0 + t(x-x_0)$ and then taylor expand the 1D function $ f(gamma(t))$ at $t=0$.
$endgroup$
– Calvin Khor
Jan 23 at 22:39
$begingroup$
The strategy is to consider the line segment $[x_0,x]subsetmathbb R^n$ (here $n=2$) parameterised by $gamma(t) = x_0 + t(x-x_0)$ and then taylor expand the 1D function $ f(gamma(t))$ at $t=0$.
$endgroup$
– Calvin Khor
Jan 23 at 22:39
$begingroup$
After Taylor expanding you should get a polynomial in $x$ with constant coefficients, which your expression is not. Can you find the mistake?
$endgroup$
– lcv
Jan 23 at 23:33
$begingroup$
After Taylor expanding you should get a polynomial in $x$ with constant coefficients, which your expression is not. Can you find the mistake?
$endgroup$
– lcv
Jan 23 at 23:33
add a comment |
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1
$begingroup$
taylor expand around what point?
$endgroup$
– Calvin Khor
Jan 23 at 22:31
$begingroup$
any arbitrary x_0.
$endgroup$
– Jack
Jan 23 at 22:32
$begingroup$
The strategy is to consider the line segment $[x_0,x]subsetmathbb R^n$ (here $n=2$) parameterised by $gamma(t) = x_0 + t(x-x_0)$ and then taylor expand the 1D function $ f(gamma(t))$ at $t=0$.
$endgroup$
– Calvin Khor
Jan 23 at 22:39
$begingroup$
After Taylor expanding you should get a polynomial in $x$ with constant coefficients, which your expression is not. Can you find the mistake?
$endgroup$
– lcv
Jan 23 at 23:33