Mistake in the computation via partial fractions
$begingroup$
This is a computation made in Titchmash's Introduction to Zeta functions. I was trying to reverse the computation. However, I kept missing factors.
Consider $frac{1- xyz^2}{(1-z)(1-xz)(1-yz)(1-xyz)}$. This is the trick used to compute $frac{zeta(s)zeta(s-a)zeta(s-b)zeta(s-a-b)}{zeta(2s-a-b)}$ where $zeta$ is the Riemann zeta function. I always miss $1-xyz^2$ in partial fractions.
$$frac{1}{(1-z)(1-xz)}=frac{1}{1-x}(frac{1}{1-z}-frac{x}{1-xz})=f$$
$$frac{1}{(1-yz)(1-xyz)}=frac{1}{1-x}(frac{1}{1-yz}-frac{x}{1-xyz})=g$$
So $(1-xyz^2)fg$ is the original expression. It suffices to consider $fg$. There will be 4 terms with quadratic denominators. Consider $(1-x)^2fg=(frac{1}{1-z}-frac{x}{1-xz})(frac{1}{1-yz}-frac{x}{1-xyz})$ instead.
There are only 2 terms containing $frac{1}{1-z}$.
$$frac{1}{(1-z)(1-yz)}=(frac{1}{1-z}-frac{y}{1-yz})frac{1}{1-y}$$
And
$$-frac{x}{(1-z)(1-xyz)}=-frac{x}{1-xy}(frac{1}{1-z}-frac{xy}{1-xyz})$$
Now combine $frac{1}{1-z}$ coefficients. $frac{1}{1-y}-frac{x}{1-xy}=frac{1-x}{(1-y)(1-xy)}$.
However I do not see removal of $(1-xy z^2)$(i.e. $frac{1}{1-xy z^2}$ did not show up anywhere in the computation.)
The final step in the book gives $frac{1- xyz^2}{(1-z)(1-xz)(1-yz)(1-xyz)}=frac{1}{(1-x)(1-y)}(frac{1}{1-z}-frac{x}{1-xz}-frac{y}{1-yz}+frac{xy}{1-xyz})$.(If one starts with partial fraction between $frac{1}{(1-z)(1-xz)}$ and $frac{1}{(1-yz)(1-xyz)}$, then the desired result can be achieved.
$textbf{Q:}$ Where is the mistake in above computation?(Say I only care to trace error from $frac{1}{1-z}$ coefficients. The computation of those coefficients should be independent of each other.)
real-analysis calculus complex-analysis algebra-precalculus partial-fractions
edited Dec 28 '18 at 17:11
Larry
2,55031131
asked Dec 28 '18 at 16:11
user45765user45765
2,7282724
$endgroup$
add a comment |
$begingroup$
This is a computation made in Titchmash's Introduction to Zeta functions. I was trying to reverse the computation. However, I kept missing factors.
Consider $frac{1- xyz^2}{(1-z)(1-xz)(1-yz)(1-xyz)}$. This is the trick used to compute $frac{zeta(s)zeta(s-a)zeta(s-b)zeta(s-a-b)}{zeta(2s-a-b)}$ where $zeta$ is the Riemann zeta function. I always miss $1-xyz^2$ in partial fractions.
$$frac{1}{(1-z)(1-xz)}=frac{1}{1-x}(frac{1}{1-z}-frac{x}{1-xz})=f$$
$$frac{1}{(1-yz)(1-xyz)}=frac{1}{1-x}(frac{1}{1-yz}-frac{x}{1-xyz})=g$$
So $(1-xyz^2)fg$ is the original expression. It suffices to consider $fg$. There will be 4 terms with quadratic denominators. Consider $(1-x)^2fg=(frac{1}{1-z}-frac{x}{1-xz})(frac{1}{1-yz}-frac{x}{1-xyz})$ instead.
There are only 2 terms containing $frac{1}{1-z}$.
$$frac{1}{(1-z)(1-yz)}=(frac{1}{1-z}-frac{y}{1-yz})frac{1}{1-y}$$
And
$$-frac{x}{(1-z)(1-xyz)}=-frac{x}{1-xy}(frac{1}{1-z}-frac{xy}{1-xyz})$$
Now combine $frac{1}{1-z}$ coefficients. $frac{1}{1-y}-frac{x}{1-xy}=frac{1-x}{(1-y)(1-xy)}$.
However I do not see removal of $(1-xy z^2)$(i.e. $frac{1}{1-xy z^2}$ did not show up anywhere in the computation.)
The final step in the book gives $frac{1- xyz^2}{(1-z)(1-xz)(1-yz)(1-xyz)}=frac{1}{(1-x)(1-y)}(frac{1}{1-z}-frac{x}{1-xz}-frac{y}{1-yz}+frac{xy}{1-xyz})$.(If one starts with partial fraction between $frac{1}{(1-z)(1-xz)}$ and $frac{1}{(1-yz)(1-xyz)}$, then the desired result can be achieved.
$textbf{Q:}$ Where is the mistake in above computation?(Say I only care to trace error from $frac{1}{1-z}$ coefficients. The computation of those coefficients should be independent of each other.)
real-analysis calculus complex-analysis algebra-precalculus partial-fractions
edited Dec 28 '18 at 17:11
Larry
2,55031131
asked Dec 28 '18 at 16:11
user45765user45765
2,7282724
$endgroup$
$begingroup$
I don't follow: there isn't any $frac1{1-xyz^2}$ in the partial fraction.
$endgroup$
– Lord Shark the Unknown
Dec 28 '18 at 16:17
$begingroup$
@LordSharktheUnknown The partial fraction final answer should be removal of $1-xy z^2$ part.(If I believe the expression is unique, then it should coincide with the final step of the book.)
$endgroup$
– user45765
Dec 28 '18 at 16:18
$begingroup$
@LordSharktheUnknown I guess appropriate question is where have I missed the factor $frac{1}{1-xyz^2}$.
$endgroup$
– user45765
Dec 28 '18 at 16:19
$begingroup$
You seem to have got half-way through the calculation and then given up. What happens if you keep going?
$endgroup$
– Lord Shark the Unknown
Dec 28 '18 at 16:33
$begingroup$
@LordSharktheUnknown I did the computation yesterday for all 4 terms 2 times. However, I did not observe $frac{1}{1-xyz^2}$ showing up in any of them.(This terms has to show up if the answer in the book is correct and I checked it did show up in a different computation I did.) That is why I felt something very wrong going on in my computation.
$endgroup$
– user45765
Dec 28 '18 at 16:41
add a comment |
$begingroup$
This is a computation made in Titchmash's Introduction to Zeta functions. I was trying to reverse the computation. However, I kept missing factors.
Consider $frac{1- xyz^2}{(1-z)(1-xz)(1-yz)(1-xyz)}$. This is the trick used to compute $frac{zeta(s)zeta(s-a)zeta(s-b)zeta(s-a-b)}{zeta(2s-a-b)}$ where $zeta$ is the Riemann zeta function. I always miss $1-xyz^2$ in partial fractions.
$$frac{1}{(1-z)(1-xz)}=frac{1}{1-x}(frac{1}{1-z}-frac{x}{1-xz})=f$$
$$frac{1}{(1-yz)(1-xyz)}=frac{1}{1-x}(frac{1}{1-yz}-frac{x}{1-xyz})=g$$
So $(1-xyz^2)fg$ is the original expression. It suffices to consider $fg$. There will be 4 terms with quadratic denominators. Consider $(1-x)^2fg=(frac{1}{1-z}-frac{x}{1-xz})(frac{1}{1-yz}-frac{x}{1-xyz})$ instead.
There are only 2 terms containing $frac{1}{1-z}$.
$$frac{1}{(1-z)(1-yz)}=(frac{1}{1-z}-frac{y}{1-yz})frac{1}{1-y}$$
And
$$-frac{x}{(1-z)(1-xyz)}=-frac{x}{1-xy}(frac{1}{1-z}-frac{xy}{1-xyz})$$
Now combine $frac{1}{1-z}$ coefficients. $frac{1}{1-y}-frac{x}{1-xy}=frac{1-x}{(1-y)(1-xy)}$.
However I do not see removal of $(1-xy z^2)$(i.e. $frac{1}{1-xy z^2}$ did not show up anywhere in the computation.)
The final step in the book gives $frac{1- xyz^2}{(1-z)(1-xz)(1-yz)(1-xyz)}=frac{1}{(1-x)(1-y)}(frac{1}{1-z}-frac{x}{1-xz}-frac{y}{1-yz}+frac{xy}{1-xyz})$.(If one starts with partial fraction between $frac{1}{(1-z)(1-xz)}$ and $frac{1}{(1-yz)(1-xyz)}$, then the desired result can be achieved.
$textbf{Q:}$ Where is the mistake in above computation?(Say I only care to trace error from $frac{1}{1-z}$ coefficients. The computation of those coefficients should be independent of each other.)
real-analysis calculus complex-analysis algebra-precalculus partial-fractions
edited Dec 28 '18 at 17:11
Larry
2,55031131
asked Dec 28 '18 at 16:11
user45765user45765
2,7282724
$endgroup$
This is a computation made in Titchmash's Introduction to Zeta functions. I was trying to reverse the computation. However, I kept missing factors.
Consider $frac{1- xyz^2}{(1-z)(1-xz)(1-yz)(1-xyz)}$. This is the trick used to compute $frac{zeta(s)zeta(s-a)zeta(s-b)zeta(s-a-b)}{zeta(2s-a-b)}$ where $zeta$ is the Riemann zeta function. I always miss $1-xyz^2$ in partial fractions.
$$frac{1}{(1-z)(1-xz)}=frac{1}{1-x}(frac{1}{1-z}-frac{x}{1-xz})=f$$
$$frac{1}{(1-yz)(1-xyz)}=frac{1}{1-x}(frac{1}{1-yz}-frac{x}{1-xyz})=g$$
So $(1-xyz^2)fg$ is the original expression. It suffices to consider $fg$. There will be 4 terms with quadratic denominators. Consider $(1-x)^2fg=(frac{1}{1-z}-frac{x}{1-xz})(frac{1}{1-yz}-frac{x}{1-xyz})$ instead.
There are only 2 terms containing $frac{1}{1-z}$.
$$frac{1}{(1-z)(1-yz)}=(frac{1}{1-z}-frac{y}{1-yz})frac{1}{1-y}$$
And
$$-frac{x}{(1-z)(1-xyz)}=-frac{x}{1-xy}(frac{1}{1-z}-frac{xy}{1-xyz})$$
Now combine $frac{1}{1-z}$ coefficients. $frac{1}{1-y}-frac{x}{1-xy}=frac{1-x}{(1-y)(1-xy)}$.
However I do not see removal of $(1-xy z^2)$(i.e. $frac{1}{1-xy z^2}$ did not show up anywhere in the computation.)
The final step in the book gives $frac{1- xyz^2}{(1-z)(1-xz)(1-yz)(1-xyz)}=frac{1}{(1-x)(1-y)}(frac{1}{1-z}-frac{x}{1-xz}-frac{y}{1-yz}+frac{xy}{1-xyz})$.(If one starts with partial fraction between $frac{1}{(1-z)(1-xz)}$ and $frac{1}{(1-yz)(1-xyz)}$, then the desired result can be achieved.
$textbf{Q:}$ Where is the mistake in above computation?(Say I only care to trace error from $frac{1}{1-z}$ coefficients. The computation of those coefficients should be independent of each other.)
real-analysis calculus complex-analysis algebra-precalculus partial-fractions
real-analysis calculus complex-analysis algebra-precalculus partial-fractions
edited Dec 28 '18 at 17:11
Larry
2,55031131
asked Dec 28 '18 at 16:11
user45765user45765
2,7282724
edited Dec 28 '18 at 17:11
Larry
2,55031131
asked Dec 28 '18 at 16:11
user45765user45765
2,7282724
edited Dec 28 '18 at 17:11
Larry
2,55031131
edited Dec 28 '18 at 17:11
Larry
2,55031131
edited Dec 28 '18 at 17:11
Larry
2,55031131
2,55031131
asked Dec 28 '18 at 16:11
user45765user45765
2,7282724
asked Dec 28 '18 at 16:11
user45765user45765
2,7282724
asked Dec 28 '18 at 16:11
user45765user45765
2,7282724
2,7282724
$begingroup$
I don't follow: there isn't any $frac1{1-xyz^2}$ in the partial fraction.
$endgroup$
– Lord Shark the Unknown
Dec 28 '18 at 16:17
$begingroup$
@LordSharktheUnknown The partial fraction final answer should be removal of $1-xy z^2$ part.(If I believe the expression is unique, then it should coincide with the final step of the book.)
$endgroup$
– user45765
Dec 28 '18 at 16:18
$begingroup$
@LordSharktheUnknown I guess appropriate question is where have I missed the factor $frac{1}{1-xyz^2}$.
$endgroup$
– user45765
Dec 28 '18 at 16:19
$begingroup$
You seem to have got half-way through the calculation and then given up. What happens if you keep going?
$endgroup$
– Lord Shark the Unknown
Dec 28 '18 at 16:33
$begingroup$
@LordSharktheUnknown I did the computation yesterday for all 4 terms 2 times. However, I did not observe $frac{1}{1-xyz^2}$ showing up in any of them.(This terms has to show up if the answer in the book is correct and I checked it did show up in a different computation I did.) That is why I felt something very wrong going on in my computation.
$endgroup$
– user45765
Dec 28 '18 at 16:41
add a comment |
$begingroup$
I don't follow: there isn't any $frac1{1-xyz^2}$ in the partial fraction.
$endgroup$
– Lord Shark the Unknown
Dec 28 '18 at 16:17
$begingroup$
@LordSharktheUnknown The partial fraction final answer should be removal of $1-xy z^2$ part.(If I believe the expression is unique, then it should coincide with the final step of the book.)
$endgroup$
– user45765
Dec 28 '18 at 16:18
$begingroup$
@LordSharktheUnknown I guess appropriate question is where have I missed the factor $frac{1}{1-xyz^2}$.
$endgroup$
– user45765
Dec 28 '18 at 16:19
$begingroup$
You seem to have got half-way through the calculation and then given up. What happens if you keep going?
$endgroup$
– Lord Shark the Unknown
Dec 28 '18 at 16:33
$begingroup$
@LordSharktheUnknown I did the computation yesterday for all 4 terms 2 times. However, I did not observe $frac{1}{1-xyz^2}$ showing up in any of them.(This terms has to show up if the answer in the book is correct and I checked it did show up in a different computation I did.) That is why I felt something very wrong going on in my computation.
$endgroup$
– user45765
Dec 28 '18 at 16:41
$begingroup$
I don't follow: there isn't any $frac1{1-xyz^2}$ in the partial fraction.
$endgroup$
– Lord Shark the Unknown
Dec 28 '18 at 16:17
$begingroup$
I don't follow: there isn't any $frac1{1-xyz^2}$ in the partial fraction.
$endgroup$
– Lord Shark the Unknown
Dec 28 '18 at 16:17
$begingroup$
@LordSharktheUnknown The partial fraction final answer should be removal of $1-xy z^2$ part.(If I believe the expression is unique, then it should coincide with the final step of the book.)
$endgroup$
– user45765
Dec 28 '18 at 16:18
$begingroup$
@LordSharktheUnknown The partial fraction final answer should be removal of $1-xy z^2$ part.(If I believe the expression is unique, then it should coincide with the final step of the book.)
$endgroup$
– user45765
Dec 28 '18 at 16:18
$begingroup$
@LordSharktheUnknown I guess appropriate question is where have I missed the factor $frac{1}{1-xyz^2}$.
$endgroup$
– user45765
Dec 28 '18 at 16:19
$begingroup$
@LordSharktheUnknown I guess appropriate question is where have I missed the factor $frac{1}{1-xyz^2}$.
$endgroup$
– user45765
Dec 28 '18 at 16:19
$begingroup$
You seem to have got half-way through the calculation and then given up. What happens if you keep going?
$endgroup$
– Lord Shark the Unknown
Dec 28 '18 at 16:33
$begingroup$
You seem to have got half-way through the calculation and then given up. What happens if you keep going?
$endgroup$
– Lord Shark the Unknown
Dec 28 '18 at 16:33
$begingroup$
@LordSharktheUnknown I did the computation yesterday for all 4 terms 2 times. However, I did not observe $frac{1}{1-xyz^2}$ showing up in any of them.(This terms has to show up if the answer in the book is correct and I checked it did show up in a different computation I did.) That is why I felt something very wrong going on in my computation.
$endgroup$
– user45765
Dec 28 '18 at 16:41
$begingroup$
@LordSharktheUnknown I did the computation yesterday for all 4 terms 2 times. However, I did not observe $frac{1}{1-xyz^2}$ showing up in any of them.(This terms has to show up if the answer in the book is correct and I checked it did show up in a different computation I did.) That is why I felt something very wrong going on in my computation.
$endgroup$
– user45765
Dec 28 '18 at 16:41
add a comment |
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$begingroup$
I don't follow: there isn't any $frac1{1-xyz^2}$ in the partial fraction.
$endgroup$
– Lord Shark the Unknown
Dec 28 '18 at 16:17
$begingroup$
@LordSharktheUnknown The partial fraction final answer should be removal of $1-xy z^2$ part.(If I believe the expression is unique, then it should coincide with the final step of the book.)
$endgroup$
– user45765
Dec 28 '18 at 16:18
$begingroup$
@LordSharktheUnknown I guess appropriate question is where have I missed the factor $frac{1}{1-xyz^2}$.
$endgroup$
– user45765
Dec 28 '18 at 16:19
$begingroup$
You seem to have got half-way through the calculation and then given up. What happens if you keep going?
$endgroup$
– Lord Shark the Unknown
Dec 28 '18 at 16:33
$begingroup$
@LordSharktheUnknown I did the computation yesterday for all 4 terms 2 times. However, I did not observe $frac{1}{1-xyz^2}$ showing up in any of them.(This terms has to show up if the answer in the book is correct and I checked it did show up in a different computation I did.) That is why I felt something very wrong going on in my computation.
$endgroup$
– user45765
Dec 28 '18 at 16:41