Derivation of Padé approximant to exponential function: unclear step in Gautschi
$begingroup$
On pages 363 to 365 in “Numerical Analysis” by Gautschi (2012) is a derivation of the Padé approximant of the exponential function.
I am stuck on a step in the beginning of the differentiation below (I typed this verbatim from the book; the relevant statement is at the bottom):
Let $v(t) = t^n (1-t)^m$.
By Leibniz's rule, one finds:
begin{align}
v^{(r)}(t) &= sum^{r}_{k=0} binom{r}{k} [t^n]^{(k)} [(1-t)^m]^{(r-k)} \
&= sum^r_{k=0} binom{r}{k}binom{n}{k} k! t^{n-k} binom{m}{r-k}(r-k)!(1-t)^{m-r+k}(-1)^{r-k};
end{align}
hence, in particular,
begin{align}
v^{(r)}(0) &= (-1)^{r-n} binom{m}{r-n}r! quad text{if} quad rgeq n; quad v^{(r)}(0)=0 quad text{if} quad r<n;\
v^{(r)}(1) &= (-1)^m binom{n}{r-m} r! quad text{if} quad rgeq m; quad v^{(r)}(1) = 0 quad text{if} quad r < m.
end{align}
Given any integer $qgeq 0$, repeated integration by parts yields
begin{equation}
int^1_0 mathrm{e}^{tz} v(t) mathrm{d}t = mathrm{e}^z sum^q_{r=0} (-1)^r frac{v^{(r)}(1)}{z^{r+1}} - sum^q_{r=0} (-1)^r frac{v^{(r)}(0)}{z^{r+1}} + frac{(-1)^{q+1}}{z^{q+1}} int^1_0 mathrm{e}^{tz} v^{(q+1)}(t)mathrm{d}t.
end{equation}
The following statement I struggle with:
Putting here $q = n+m$, so that $v^{(q+1)}(t) equiv 0$, ...
So if I understand the thought behind this correctly, then the integral on the RHS is supposed to vanish by choosing $q$ appropriately. But I don't see how this is supposed to happen? The $equiv$ sign I find extra confusing. Does the author intend to define it to vanish? That doesn't quite make sense to me, so it's probably not it.
What's going on here?
integration ordinary-differential-equations exponential-function pade-approximation
asked Jan 16 at 0:33
mSSMmSSM
15118
$endgroup$
add a comment |
$begingroup$
On pages 363 to 365 in “Numerical Analysis” by Gautschi (2012) is a derivation of the Padé approximant of the exponential function.
I am stuck on a step in the beginning of the differentiation below (I typed this verbatim from the book; the relevant statement is at the bottom):
Let $v(t) = t^n (1-t)^m$.
By Leibniz's rule, one finds:
begin{align}
v^{(r)}(t) &= sum^{r}_{k=0} binom{r}{k} [t^n]^{(k)} [(1-t)^m]^{(r-k)} \
&= sum^r_{k=0} binom{r}{k}binom{n}{k} k! t^{n-k} binom{m}{r-k}(r-k)!(1-t)^{m-r+k}(-1)^{r-k};
end{align}
hence, in particular,
begin{align}
v^{(r)}(0) &= (-1)^{r-n} binom{m}{r-n}r! quad text{if} quad rgeq n; quad v^{(r)}(0)=0 quad text{if} quad r<n;\
v^{(r)}(1) &= (-1)^m binom{n}{r-m} r! quad text{if} quad rgeq m; quad v^{(r)}(1) = 0 quad text{if} quad r < m.
end{align}
Given any integer $qgeq 0$, repeated integration by parts yields
begin{equation}
int^1_0 mathrm{e}^{tz} v(t) mathrm{d}t = mathrm{e}^z sum^q_{r=0} (-1)^r frac{v^{(r)}(1)}{z^{r+1}} - sum^q_{r=0} (-1)^r frac{v^{(r)}(0)}{z^{r+1}} + frac{(-1)^{q+1}}{z^{q+1}} int^1_0 mathrm{e}^{tz} v^{(q+1)}(t)mathrm{d}t.
end{equation}
The following statement I struggle with:
Putting here $q = n+m$, so that $v^{(q+1)}(t) equiv 0$, ...
So if I understand the thought behind this correctly, then the integral on the RHS is supposed to vanish by choosing $q$ appropriately. But I don't see how this is supposed to happen? The $equiv$ sign I find extra confusing. Does the author intend to define it to vanish? That doesn't quite make sense to me, so it's probably not it.
What's going on here?
integration ordinary-differential-equations exponential-function pade-approximation
asked Jan 16 at 0:33
mSSMmSSM
15118
$endgroup$
1
$begingroup$
You have a polynomial of degree $q$, so its $q^{rm th}$ derivative is a constant, the $(q+1)^{rm st}$ is thus zero. Was that the question?
$endgroup$
– LutzL
Jan 16 at 0:46
$begingroup$
Yes, that was indeed the problem. I was staring to much at the expansion of the Leibniz rule, that I didn't see the obvious result... :(
$endgroup$
– mSSM
Jan 19 at 0:51
add a comment |
$begingroup$
On pages 363 to 365 in “Numerical Analysis” by Gautschi (2012) is a derivation of the Padé approximant of the exponential function.
I am stuck on a step in the beginning of the differentiation below (I typed this verbatim from the book; the relevant statement is at the bottom):
Let $v(t) = t^n (1-t)^m$.
By Leibniz's rule, one finds:
begin{align}
v^{(r)}(t) &= sum^{r}_{k=0} binom{r}{k} [t^n]^{(k)} [(1-t)^m]^{(r-k)} \
&= sum^r_{k=0} binom{r}{k}binom{n}{k} k! t^{n-k} binom{m}{r-k}(r-k)!(1-t)^{m-r+k}(-1)^{r-k};
end{align}
hence, in particular,
begin{align}
v^{(r)}(0) &= (-1)^{r-n} binom{m}{r-n}r! quad text{if} quad rgeq n; quad v^{(r)}(0)=0 quad text{if} quad r<n;\
v^{(r)}(1) &= (-1)^m binom{n}{r-m} r! quad text{if} quad rgeq m; quad v^{(r)}(1) = 0 quad text{if} quad r < m.
end{align}
Given any integer $qgeq 0$, repeated integration by parts yields
begin{equation}
int^1_0 mathrm{e}^{tz} v(t) mathrm{d}t = mathrm{e}^z sum^q_{r=0} (-1)^r frac{v^{(r)}(1)}{z^{r+1}} - sum^q_{r=0} (-1)^r frac{v^{(r)}(0)}{z^{r+1}} + frac{(-1)^{q+1}}{z^{q+1}} int^1_0 mathrm{e}^{tz} v^{(q+1)}(t)mathrm{d}t.
end{equation}
The following statement I struggle with:
Putting here $q = n+m$, so that $v^{(q+1)}(t) equiv 0$, ...
So if I understand the thought behind this correctly, then the integral on the RHS is supposed to vanish by choosing $q$ appropriately. But I don't see how this is supposed to happen? The $equiv$ sign I find extra confusing. Does the author intend to define it to vanish? That doesn't quite make sense to me, so it's probably not it.
What's going on here?
integration ordinary-differential-equations exponential-function pade-approximation
asked Jan 16 at 0:33
mSSMmSSM
15118
$endgroup$
On pages 363 to 365 in “Numerical Analysis” by Gautschi (2012) is a derivation of the Padé approximant of the exponential function.
I am stuck on a step in the beginning of the differentiation below (I typed this verbatim from the book; the relevant statement is at the bottom):
Let $v(t) = t^n (1-t)^m$.
By Leibniz's rule, one finds:
begin{align}
v^{(r)}(t) &= sum^{r}_{k=0} binom{r}{k} [t^n]^{(k)} [(1-t)^m]^{(r-k)} \
&= sum^r_{k=0} binom{r}{k}binom{n}{k} k! t^{n-k} binom{m}{r-k}(r-k)!(1-t)^{m-r+k}(-1)^{r-k};
end{align}
hence, in particular,
begin{align}
v^{(r)}(0) &= (-1)^{r-n} binom{m}{r-n}r! quad text{if} quad rgeq n; quad v^{(r)}(0)=0 quad text{if} quad r<n;\
v^{(r)}(1) &= (-1)^m binom{n}{r-m} r! quad text{if} quad rgeq m; quad v^{(r)}(1) = 0 quad text{if} quad r < m.
end{align}
Given any integer $qgeq 0$, repeated integration by parts yields
begin{equation}
int^1_0 mathrm{e}^{tz} v(t) mathrm{d}t = mathrm{e}^z sum^q_{r=0} (-1)^r frac{v^{(r)}(1)}{z^{r+1}} - sum^q_{r=0} (-1)^r frac{v^{(r)}(0)}{z^{r+1}} + frac{(-1)^{q+1}}{z^{q+1}} int^1_0 mathrm{e}^{tz} v^{(q+1)}(t)mathrm{d}t.
end{equation}
The following statement I struggle with:
Putting here $q = n+m$, so that $v^{(q+1)}(t) equiv 0$, ...
So if I understand the thought behind this correctly, then the integral on the RHS is supposed to vanish by choosing $q$ appropriately. But I don't see how this is supposed to happen? The $equiv$ sign I find extra confusing. Does the author intend to define it to vanish? That doesn't quite make sense to me, so it's probably not it.
What's going on here?
integration ordinary-differential-equations exponential-function pade-approximation
integration ordinary-differential-equations exponential-function pade-approximation
asked Jan 16 at 0:33
mSSMmSSM
15118
asked Jan 16 at 0:33
mSSMmSSM
15118
asked Jan 16 at 0:33
mSSMmSSM
15118
asked Jan 16 at 0:33
mSSMmSSM
15118
asked Jan 16 at 0:33
mSSMmSSM
15118
15118
1
$begingroup$
You have a polynomial of degree $q$, so its $q^{rm th}$ derivative is a constant, the $(q+1)^{rm st}$ is thus zero. Was that the question?
$endgroup$
– LutzL
Jan 16 at 0:46
$begingroup$
Yes, that was indeed the problem. I was staring to much at the expansion of the Leibniz rule, that I didn't see the obvious result... :(
$endgroup$
– mSSM
Jan 19 at 0:51
add a comment |
1
$begingroup$
You have a polynomial of degree $q$, so its $q^{rm th}$ derivative is a constant, the $(q+1)^{rm st}$ is thus zero. Was that the question?
$endgroup$
– LutzL
Jan 16 at 0:46
$begingroup$
Yes, that was indeed the problem. I was staring to much at the expansion of the Leibniz rule, that I didn't see the obvious result... :(
$endgroup$
– mSSM
Jan 19 at 0:51
1
1
$begingroup$
You have a polynomial of degree $q$, so its $q^{rm th}$ derivative is a constant, the $(q+1)^{rm st}$ is thus zero. Was that the question?
$endgroup$
– LutzL
Jan 16 at 0:46
$begingroup$
You have a polynomial of degree $q$, so its $q^{rm th}$ derivative is a constant, the $(q+1)^{rm st}$ is thus zero. Was that the question?
$endgroup$
– LutzL
Jan 16 at 0:46
$begingroup$
Yes, that was indeed the problem. I was staring to much at the expansion of the Leibniz rule, that I didn't see the obvious result... :(
$endgroup$
– mSSM
Jan 19 at 0:51
$begingroup$
Yes, that was indeed the problem. I was staring to much at the expansion of the Leibniz rule, that I didn't see the obvious result... :(
$endgroup$
– mSSM
Jan 19 at 0:51
add a comment |
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$begingroup$
You have a polynomial of degree $q$, so its $q^{rm th}$ derivative is a constant, the $(q+1)^{rm st}$ is thus zero. Was that the question?
$endgroup$
– LutzL
Jan 16 at 0:46
$begingroup$
Yes, that was indeed the problem. I was staring to much at the expansion of the Leibniz rule, that I didn't see the obvious result... :(
$endgroup$
– mSSM
Jan 19 at 0:51