Estimate of the remainder term of a complex exponential
$begingroup$
Let $f(xi)=e^{pi i lambda^{-1}|xi|^2}$ and
$$
f_N(xi)=sum_{j=0}^Nfrac{(pi ilambda^{-1}|xi|^2)^j}{j!}.
$$
Show that there exists $C>0$ such that for all $lambda >1$ and $xiinmathbf{R}^d$:
$$
|f(xi)-f_N(xi)|leq Cfrac{|xi|^{2(N+1)}}{lambda^{N+1}}.
$$
This question is motivated by an attempt to understand a step in Wolff's notes on the stationary phase method. If one only needs the estimate uniformly for $xi$ near zero, then I can see that why this is true since the higher order term of $|xi|^{2m}$ ($m>N$) is dominated by $|xi|^{2N+2}$. But I don't see how this is true for all $xiinmathbf{R}^d$.
complex-analysis inequality taylor-expansion
asked Jan 1 at 22:52
user587192user587192
1,825315
$endgroup$
add a comment |
$begingroup$
Let $f(xi)=e^{pi i lambda^{-1}|xi|^2}$ and
$$
f_N(xi)=sum_{j=0}^Nfrac{(pi ilambda^{-1}|xi|^2)^j}{j!}.
$$
Show that there exists $C>0$ such that for all $lambda >1$ and $xiinmathbf{R}^d$:
$$
|f(xi)-f_N(xi)|leq Cfrac{|xi|^{2(N+1)}}{lambda^{N+1}}.
$$
This question is motivated by an attempt to understand a step in Wolff's notes on the stationary phase method. If one only needs the estimate uniformly for $xi$ near zero, then I can see that why this is true since the higher order term of $|xi|^{2m}$ ($m>N$) is dominated by $|xi|^{2N+2}$. But I don't see how this is true for all $xiinmathbf{R}^d$.
complex-analysis inequality taylor-expansion
asked Jan 1 at 22:52
user587192user587192
1,825315
$endgroup$
add a comment |
$begingroup$
Let $f(xi)=e^{pi i lambda^{-1}|xi|^2}$ and
$$
f_N(xi)=sum_{j=0}^Nfrac{(pi ilambda^{-1}|xi|^2)^j}{j!}.
$$
Show that there exists $C>0$ such that for all $lambda >1$ and $xiinmathbf{R}^d$:
$$
|f(xi)-f_N(xi)|leq Cfrac{|xi|^{2(N+1)}}{lambda^{N+1}}.
$$
This question is motivated by an attempt to understand a step in Wolff's notes on the stationary phase method. If one only needs the estimate uniformly for $xi$ near zero, then I can see that why this is true since the higher order term of $|xi|^{2m}$ ($m>N$) is dominated by $|xi|^{2N+2}$. But I don't see how this is true for all $xiinmathbf{R}^d$.
complex-analysis inequality taylor-expansion
asked Jan 1 at 22:52
user587192user587192
1,825315
$endgroup$
Let $f(xi)=e^{pi i lambda^{-1}|xi|^2}$ and
$$
f_N(xi)=sum_{j=0}^Nfrac{(pi ilambda^{-1}|xi|^2)^j}{j!}.
$$
Show that there exists $C>0$ such that for all $lambda >1$ and $xiinmathbf{R}^d$:
$$
|f(xi)-f_N(xi)|leq Cfrac{|xi|^{2(N+1)}}{lambda^{N+1}}.
$$
This question is motivated by an attempt to understand a step in Wolff's notes on the stationary phase method. If one only needs the estimate uniformly for $xi$ near zero, then I can see that why this is true since the higher order term of $|xi|^{2m}$ ($m>N$) is dominated by $|xi|^{2N+2}$. But I don't see how this is true for all $xiinmathbf{R}^d$.
complex-analysis inequality taylor-expansion
complex-analysis inequality taylor-expansion
asked Jan 1 at 22:52
user587192user587192
1,825315
asked Jan 1 at 22:52
user587192user587192
1,825315
edited Jan 2 at 1:18
user587192
asked Jan 1 at 22:52
user587192user587192
1,825315
asked Jan 1 at 22:52
user587192user587192
1,825315
asked Jan 1 at 22:52
user587192user587192
1,825315
1,825315
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can show this by using the integral form of the remainder in Taylor's theorem:
$$
e^{ix} = sum_{j=0}^Nfrac{(ix)^j}{j!} + frac{i^{N+1}}{N!}int_0^x(x-s)^Ne^{is},ds.
$$
From this it follows that
$$
bigg|e^{ix}-sum_{j=0}^Nfrac{(ix)^j}{j!}bigg| le frac{|x|^{N+1}}{(N+1)!}.
$$
Now substituting $x = pilambda^{-1}|xi|^2$ shows that the desired inequality holds for all $xiinmathbf R^d$ and all $lambda > 0$.
answered Jan 1 at 23:22
AOrtizAOrtiz
10.5k21441
$endgroup$
1
$begingroup$
@user587192: The $i$ is crucial. Otherwise, by the reverse triangle inequality, we would have that $|e^z|$ is majorized by a polynomial of degree $N+1$ on $mathbf C$, which is certainly not true since the exponential function restricted to the real axis grows faster than any polynomial as $mathrm{Re}(z)to+infty$.
$endgroup$
– AOrtiz
Jan 2 at 1:15
$begingroup$
Thanks again! I accidentally deleted my previous comment: "Is it crucial that one has the imaginary number $i$ in the exponent of $e^{ix}$? Your argument seems to suggest that the estimate would work if $i$ is not there".
$endgroup$
– user587192
Jan 2 at 1:21
1
$begingroup$
@user587192 to estimate the integral you need the fact that $|e^{is}| =1$.
$endgroup$
– Jacky Chong
Jan 2 at 1:47
$begingroup$
@JackyChong: Indeed, one nice property of $e^{ix}$ is that all the derivatives are bounded, while it is not true for $e^x$. I didn't realize that when I posted the first comment, which is now deleted.
$endgroup$
– user587192
Jan 2 at 1:59
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can show this by using the integral form of the remainder in Taylor's theorem:
$$
e^{ix} = sum_{j=0}^Nfrac{(ix)^j}{j!} + frac{i^{N+1}}{N!}int_0^x(x-s)^Ne^{is},ds.
$$
From this it follows that
$$
bigg|e^{ix}-sum_{j=0}^Nfrac{(ix)^j}{j!}bigg| le frac{|x|^{N+1}}{(N+1)!}.
$$
Now substituting $x = pilambda^{-1}|xi|^2$ shows that the desired inequality holds for all $xiinmathbf R^d$ and all $lambda > 0$.
answered Jan 1 at 23:22
AOrtizAOrtiz
10.5k21441
$endgroup$
1
$begingroup$
@user587192: The $i$ is crucial. Otherwise, by the reverse triangle inequality, we would have that $|e^z|$ is majorized by a polynomial of degree $N+1$ on $mathbf C$, which is certainly not true since the exponential function restricted to the real axis grows faster than any polynomial as $mathrm{Re}(z)to+infty$.
$endgroup$
– AOrtiz
Jan 2 at 1:15
$begingroup$
Thanks again! I accidentally deleted my previous comment: "Is it crucial that one has the imaginary number $i$ in the exponent of $e^{ix}$? Your argument seems to suggest that the estimate would work if $i$ is not there".
$endgroup$
– user587192
Jan 2 at 1:21
1
$begingroup$
@user587192 to estimate the integral you need the fact that $|e^{is}| =1$.
$endgroup$
– Jacky Chong
Jan 2 at 1:47
$begingroup$
@JackyChong: Indeed, one nice property of $e^{ix}$ is that all the derivatives are bounded, while it is not true for $e^x$. I didn't realize that when I posted the first comment, which is now deleted.
$endgroup$
– user587192
Jan 2 at 1:59
add a comment |
$begingroup$
You can show this by using the integral form of the remainder in Taylor's theorem:
$$
e^{ix} = sum_{j=0}^Nfrac{(ix)^j}{j!} + frac{i^{N+1}}{N!}int_0^x(x-s)^Ne^{is},ds.
$$
From this it follows that
$$
bigg|e^{ix}-sum_{j=0}^Nfrac{(ix)^j}{j!}bigg| le frac{|x|^{N+1}}{(N+1)!}.
$$
Now substituting $x = pilambda^{-1}|xi|^2$ shows that the desired inequality holds for all $xiinmathbf R^d$ and all $lambda > 0$.
answered Jan 1 at 23:22
AOrtizAOrtiz
10.5k21441
$endgroup$
1
$begingroup$
@user587192: The $i$ is crucial. Otherwise, by the reverse triangle inequality, we would have that $|e^z|$ is majorized by a polynomial of degree $N+1$ on $mathbf C$, which is certainly not true since the exponential function restricted to the real axis grows faster than any polynomial as $mathrm{Re}(z)to+infty$.
$endgroup$
– AOrtiz
Jan 2 at 1:15
$begingroup$
Thanks again! I accidentally deleted my previous comment: "Is it crucial that one has the imaginary number $i$ in the exponent of $e^{ix}$? Your argument seems to suggest that the estimate would work if $i$ is not there".
$endgroup$
– user587192
Jan 2 at 1:21
1
$begingroup$
@user587192 to estimate the integral you need the fact that $|e^{is}| =1$.
$endgroup$
– Jacky Chong
Jan 2 at 1:47
$begingroup$
@JackyChong: Indeed, one nice property of $e^{ix}$ is that all the derivatives are bounded, while it is not true for $e^x$. I didn't realize that when I posted the first comment, which is now deleted.
$endgroup$
– user587192
Jan 2 at 1:59
add a comment |
$begingroup$
You can show this by using the integral form of the remainder in Taylor's theorem:
$$
e^{ix} = sum_{j=0}^Nfrac{(ix)^j}{j!} + frac{i^{N+1}}{N!}int_0^x(x-s)^Ne^{is},ds.
$$
From this it follows that
$$
bigg|e^{ix}-sum_{j=0}^Nfrac{(ix)^j}{j!}bigg| le frac{|x|^{N+1}}{(N+1)!}.
$$
Now substituting $x = pilambda^{-1}|xi|^2$ shows that the desired inequality holds for all $xiinmathbf R^d$ and all $lambda > 0$.
answered Jan 1 at 23:22
AOrtizAOrtiz
10.5k21441
$endgroup$
You can show this by using the integral form of the remainder in Taylor's theorem:
$$
e^{ix} = sum_{j=0}^Nfrac{(ix)^j}{j!} + frac{i^{N+1}}{N!}int_0^x(x-s)^Ne^{is},ds.
$$
From this it follows that
$$
bigg|e^{ix}-sum_{j=0}^Nfrac{(ix)^j}{j!}bigg| le frac{|x|^{N+1}}{(N+1)!}.
$$
Now substituting $x = pilambda^{-1}|xi|^2$ shows that the desired inequality holds for all $xiinmathbf R^d$ and all $lambda > 0$.
answered Jan 1 at 23:22
AOrtizAOrtiz
10.5k21441
edited Jan 1 at 23:27
answered Jan 1 at 23:22
AOrtizAOrtiz
10.5k21441
answered Jan 1 at 23:22
AOrtizAOrtiz
10.5k21441
answered Jan 1 at 23:22
AOrtizAOrtiz
10.5k21441
10.5k21441
1
$begingroup$
@user587192: The $i$ is crucial. Otherwise, by the reverse triangle inequality, we would have that $|e^z|$ is majorized by a polynomial of degree $N+1$ on $mathbf C$, which is certainly not true since the exponential function restricted to the real axis grows faster than any polynomial as $mathrm{Re}(z)to+infty$.
$endgroup$
– AOrtiz
Jan 2 at 1:15
$begingroup$
Thanks again! I accidentally deleted my previous comment: "Is it crucial that one has the imaginary number $i$ in the exponent of $e^{ix}$? Your argument seems to suggest that the estimate would work if $i$ is not there".
$endgroup$
– user587192
Jan 2 at 1:21
1
$begingroup$
@user587192 to estimate the integral you need the fact that $|e^{is}| =1$.
$endgroup$
– Jacky Chong
Jan 2 at 1:47
$begingroup$
@JackyChong: Indeed, one nice property of $e^{ix}$ is that all the derivatives are bounded, while it is not true for $e^x$. I didn't realize that when I posted the first comment, which is now deleted.
$endgroup$
– user587192
Jan 2 at 1:59
add a comment |
1
$begingroup$
@user587192: The $i$ is crucial. Otherwise, by the reverse triangle inequality, we would have that $|e^z|$ is majorized by a polynomial of degree $N+1$ on $mathbf C$, which is certainly not true since the exponential function restricted to the real axis grows faster than any polynomial as $mathrm{Re}(z)to+infty$.
$endgroup$
– AOrtiz
Jan 2 at 1:15
$begingroup$
Thanks again! I accidentally deleted my previous comment: "Is it crucial that one has the imaginary number $i$ in the exponent of $e^{ix}$? Your argument seems to suggest that the estimate would work if $i$ is not there".
$endgroup$
– user587192
Jan 2 at 1:21
1
$begingroup$
@user587192 to estimate the integral you need the fact that $|e^{is}| =1$.
$endgroup$
– Jacky Chong
Jan 2 at 1:47
$begingroup$
@JackyChong: Indeed, one nice property of $e^{ix}$ is that all the derivatives are bounded, while it is not true for $e^x$. I didn't realize that when I posted the first comment, which is now deleted.
$endgroup$
– user587192
Jan 2 at 1:59
1
1
$begingroup$
@user587192: The $i$ is crucial. Otherwise, by the reverse triangle inequality, we would have that $|e^z|$ is majorized by a polynomial of degree $N+1$ on $mathbf C$, which is certainly not true since the exponential function restricted to the real axis grows faster than any polynomial as $mathrm{Re}(z)to+infty$.
$endgroup$
– AOrtiz
Jan 2 at 1:15
$begingroup$
@user587192: The $i$ is crucial. Otherwise, by the reverse triangle inequality, we would have that $|e^z|$ is majorized by a polynomial of degree $N+1$ on $mathbf C$, which is certainly not true since the exponential function restricted to the real axis grows faster than any polynomial as $mathrm{Re}(z)to+infty$.
$endgroup$
– AOrtiz
Jan 2 at 1:15
$begingroup$
Thanks again! I accidentally deleted my previous comment: "Is it crucial that one has the imaginary number $i$ in the exponent of $e^{ix}$? Your argument seems to suggest that the estimate would work if $i$ is not there".
$endgroup$
– user587192
Jan 2 at 1:21
$begingroup$
Thanks again! I accidentally deleted my previous comment: "Is it crucial that one has the imaginary number $i$ in the exponent of $e^{ix}$? Your argument seems to suggest that the estimate would work if $i$ is not there".
$endgroup$
– user587192
Jan 2 at 1:21
1
1
$begingroup$
@user587192 to estimate the integral you need the fact that $|e^{is}| =1$.
$endgroup$
– Jacky Chong
Jan 2 at 1:47
$begingroup$
@user587192 to estimate the integral you need the fact that $|e^{is}| =1$.
$endgroup$
– Jacky Chong
Jan 2 at 1:47
$begingroup$
@JackyChong: Indeed, one nice property of $e^{ix}$ is that all the derivatives are bounded, while it is not true for $e^x$. I didn't realize that when I posted the first comment, which is now deleted.
$endgroup$
– user587192
Jan 2 at 1:59
$begingroup$
@JackyChong: Indeed, one nice property of $e^{ix}$ is that all the derivatives are bounded, while it is not true for $e^x$. I didn't realize that when I posted the first comment, which is now deleted.
$endgroup$
– user587192
Jan 2 at 1:59
add a comment |
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