How to derive this identityProving $ 2n choose n = 2^n frac 1 cdot 3 cdot 5 cdots (2n-1)n! $Spivak's Calculus (Chapter 2, Exercise 17)Help manipulating simple cubic equationHelp needed verifying a trigonometric identityHow to prove this identity (equality)?Does $L^2 = P^2$ implies that $L = P$?How much velocity can a canister of fuel give a spaceship?Why $e^kn^k<k^ke^frac n-k2^k$ whenever $n geq k^2 2^k ln 2 (1+o(1)) $?Is it possible to perform a computation to know if an object is at rest and its location given a force versus position diagram?How was this sum simplified to a ratio of Sinh functions?
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How to derive this identity
Proving $ 2n choose n = 2^n frac 1 cdot 3 cdot 5 cdots (2n-1)n! $Spivak's Calculus (Chapter 2, Exercise 17)Help manipulating simple cubic equationHelp needed verifying a trigonometric identityHow to prove this identity (equality)?Does $L^2 = P^2$ implies that $L = P$?How much velocity can a canister of fuel give a spaceship?Why $e^kn^k<k^ke^frac n-k2^k$ whenever $n geq k^2 2^k ln 2 (1+o(1)) $?Is it possible to perform a computation to know if an object is at rest and its location given a force versus position diagram?How was this sum simplified to a ratio of Sinh functions?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
$$ frac ab -x= frac c-xdb+left(fracb-dbright) left(fraca-cb-d - x right) $$
It is easy to check it by computing right hand side. It feels unnatural and a little magical. I could't derive it starting from LHS
This identity is used in the proof of Stolz Cesaro theorem (https://ru.m.wikipedia.org/wiki/Теорема_Штольца). It is in russian I understood with the help of google translate .
algebra-precalculus
asked 10 hours ago
MilanMilan
5371 gold badge5 silver badges15 bronze badges
$endgroup$
|
show 2 more comments
$begingroup$
$$ frac ab -x= frac c-xdb+left(fracb-dbright) left(fraca-cb-d - x right) $$
It is easy to check it by computing right hand side. It feels unnatural and a little magical. I could't derive it starting from LHS
This identity is used in the proof of Stolz Cesaro theorem (https://ru.m.wikipedia.org/wiki/Теорема_Штольца). It is in russian I understood with the help of google translate .
algebra-precalculus
asked 10 hours ago
MilanMilan
5371 gold badge5 silver badges15 bronze badges
$endgroup$
$begingroup$
If you know what you are aiming at then you can start by adding and subtracting $fracc-xdb$ ...
$endgroup$
– Martin R
10 hours ago
3
$begingroup$
There is a reason why the LHS is transformed like this. $b$ and $d$ do not fall from the sky. It seems you have to read the proof again more carefully.
$endgroup$
– callculus
10 hours ago
$begingroup$
Step by step reduce the LHS to the RHS. Turn that upside down and you derive the RHS to the LHS.
$endgroup$
– steven gregory
8 hours ago
$begingroup$
@callculus It is transformed like that so we can show it is less than epsilon. In the proof i have the identity does fall from the sky. If you have a more intuitive explenation please share .
$endgroup$
– Milan
8 hours ago
$begingroup$
Do you have a reference for the proof? If so, please include the link in your question in order to make it self contained.
$endgroup$
– Somos
7 hours ago
|
show 2 more comments
$begingroup$
$$ frac ab -x= frac c-xdb+left(fracb-dbright) left(fraca-cb-d - x right) $$
It is easy to check it by computing right hand side. It feels unnatural and a little magical. I could't derive it starting from LHS
This identity is used in the proof of Stolz Cesaro theorem (https://ru.m.wikipedia.org/wiki/Теорема_Штольца). It is in russian I understood with the help of google translate .
algebra-precalculus
asked 10 hours ago
MilanMilan
5371 gold badge5 silver badges15 bronze badges
$endgroup$
$$ frac ab -x= frac c-xdb+left(fracb-dbright) left(fraca-cb-d - x right) $$
It is easy to check it by computing right hand side. It feels unnatural and a little magical. I could't derive it starting from LHS
This identity is used in the proof of Stolz Cesaro theorem (https://ru.m.wikipedia.org/wiki/Теорема_Штольца). It is in russian I understood with the help of google translate .
algebra-precalculus
algebra-precalculus
asked 10 hours ago
MilanMilan
5371 gold badge5 silver badges15 bronze badges
asked 10 hours ago
MilanMilan
5371 gold badge5 silver badges15 bronze badges
edited 7 hours ago
Milan
asked 10 hours ago
MilanMilan
5371 gold badge5 silver badges15 bronze badges
asked 10 hours ago
MilanMilan
5371 gold badge5 silver badges15 bronze badges
asked 10 hours ago
MilanMilan
5371 gold badge5 silver badges15 bronze badges
5371 gold badge5 silver badges15 bronze badges
$begingroup$
If you know what you are aiming at then you can start by adding and subtracting $fracc-xdb$ ...
$endgroup$
– Martin R
10 hours ago
3
$begingroup$
There is a reason why the LHS is transformed like this. $b$ and $d$ do not fall from the sky. It seems you have to read the proof again more carefully.
$endgroup$
– callculus
10 hours ago
$begingroup$
Step by step reduce the LHS to the RHS. Turn that upside down and you derive the RHS to the LHS.
$endgroup$
– steven gregory
8 hours ago
$begingroup$
@callculus It is transformed like that so we can show it is less than epsilon. In the proof i have the identity does fall from the sky. If you have a more intuitive explenation please share .
$endgroup$
– Milan
8 hours ago
$begingroup$
Do you have a reference for the proof? If so, please include the link in your question in order to make it self contained.
$endgroup$
– Somos
7 hours ago
|
show 2 more comments
$begingroup$
If you know what you are aiming at then you can start by adding and subtracting $fracc-xdb$ ...
$endgroup$
– Martin R
10 hours ago
3
$begingroup$
There is a reason why the LHS is transformed like this. $b$ and $d$ do not fall from the sky. It seems you have to read the proof again more carefully.
$endgroup$
– callculus
10 hours ago
$begingroup$
Step by step reduce the LHS to the RHS. Turn that upside down and you derive the RHS to the LHS.
$endgroup$
– steven gregory
8 hours ago
$begingroup$
@callculus It is transformed like that so we can show it is less than epsilon. In the proof i have the identity does fall from the sky. If you have a more intuitive explenation please share .
$endgroup$
– Milan
8 hours ago
$begingroup$
Do you have a reference for the proof? If so, please include the link in your question in order to make it self contained.
$endgroup$
– Somos
7 hours ago
$begingroup$
If you know what you are aiming at then you can start by adding and subtracting $fracc-xdb$ ...
$endgroup$
– Martin R
10 hours ago
$begingroup$
If you know what you are aiming at then you can start by adding and subtracting $fracc-xdb$ ...
$endgroup$
– Martin R
10 hours ago
3
3
$begingroup$
There is a reason why the LHS is transformed like this. $b$ and $d$ do not fall from the sky. It seems you have to read the proof again more carefully.
$endgroup$
– callculus
10 hours ago
$begingroup$
There is a reason why the LHS is transformed like this. $b$ and $d$ do not fall from the sky. It seems you have to read the proof again more carefully.
$endgroup$
– callculus
10 hours ago
$begingroup$
Step by step reduce the LHS to the RHS. Turn that upside down and you derive the RHS to the LHS.
$endgroup$
– steven gregory
8 hours ago
$begingroup$
Step by step reduce the LHS to the RHS. Turn that upside down and you derive the RHS to the LHS.
$endgroup$
– steven gregory
8 hours ago
$begingroup$
@callculus It is transformed like that so we can show it is less than epsilon. In the proof i have the identity does fall from the sky. If you have a more intuitive explenation please share .
$endgroup$
– Milan
8 hours ago
$begingroup$
@callculus It is transformed like that so we can show it is less than epsilon. In the proof i have the identity does fall from the sky. If you have a more intuitive explenation please share .
$endgroup$
– Milan
8 hours ago
$begingroup$
Do you have a reference for the proof? If so, please include the link in your question in order to make it self contained.
$endgroup$
– Somos
7 hours ago
$begingroup$
Do you have a reference for the proof? If so, please include the link in your question in order to make it self contained.
$endgroup$
– Somos
7 hours ago
|
show 2 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Consider the task of cooking up a straight line by combining two other straight lines.
Given a straight line $y=-x+a$
You want to represent this as a sum of two other straight lines.
Letting the slope of one line to $-m$ forces the slope of other line to $-(1-m)$.
You can get that by comparing like terms in :
$$colorred-x+a = colorblue-mx + c - m'x+c'$$
Also constant term is $a-c$.
So above equation becomes
$$colorred-x+a = colorblue-mx + c + (1-m)left(dfraca-c1-m-xright)$$
Replacing $m$ by $fracdb$ and $a$ by $fracab$ gives your eqn.
answered 9 hours ago
ganeshie8ganeshie8
4,7391 gold badge11 silver badges19 bronze badges
$endgroup$
1
$begingroup$
that's an interesting way to concretize and visualize why such a transformation (+1)
$endgroup$
– G Cab
8 hours ago
add a comment |
$begingroup$
Starting from the LHS
$$fracab-x$$
we can subtract and add $fracc-xdb$ to form
$$Big(fracab-xBig) - fracc-xdb + fracc-xdb$$
which rearranges to
$$fracc-xdb+Big(fracab-x-fracc-xdbBig)$$
or
$$fracc-xdb+Big(fraca-c+xdb-xBig)$$
which is equivalent to
$$fracc-xdb+Big(fraca-bx-c+dxbBig)$$
where $left(fracb-dbright) left(fraca-cb-d - x right) = Big(fraca-bx-c+dxbBig)$ and therefore we have the RHS
$$fracc-xdb+left(fracb-dbright) left(fraca-cb-d - x right)$$
answered 8 hours ago
Axion004Axion004
1,4176 silver badges17 bronze badges
$endgroup$
add a comment |
$begingroup$
Let
$$
Z=:fracleft(c-xdright)b+left(fracb-dbright):left(fraca-cb-d:-:x:right):$$
Let
$$ K=fracleft(c-xdright)b$$
Let
$$ N=left(fracb-dbright):left(fraca-cb-d:-:x:right):$$
$$Z=K+N$$
We will rewrite $left(fraca-cb-d:-:x:right):$ as:
$$left(fraca-cb-d:-:x:right):=left(fraca-cb-d:-:x:fracleft(b-dright)left(b-dright):right):$$
$$N:=left(fracb-dbright)left(fraca-cb-d:-:x:fracleft(b-dright)left(b-dright):right):=left(fracb-dbright)left(fraca-c:-xleft(b-dright)b-d:right)$$
$$N:=left(fraca-c:-xleft(b-dright)b:right)$$
$$N:=left(fraca-c:-xb+xdb:right)$$
Using $Z=K+N$ and substituting the last expression for $N$ we get:
$$z:=fracleft(c-xdright)b+left(fraca-c:-xb+xdb:right) $$
$$z:=left(fraca:-xbb:right)$$
$$ z:=fraca:b-x$$
answered 9 hours ago
NoChanceNoChance
4,4782 gold badges13 silver badges22 bronze badges
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider the task of cooking up a straight line by combining two other straight lines.
Given a straight line $y=-x+a$
You want to represent this as a sum of two other straight lines.
Letting the slope of one line to $-m$ forces the slope of other line to $-(1-m)$.
You can get that by comparing like terms in :
$$colorred-x+a = colorblue-mx + c - m'x+c'$$
Also constant term is $a-c$.
So above equation becomes
$$colorred-x+a = colorblue-mx + c + (1-m)left(dfraca-c1-m-xright)$$
Replacing $m$ by $fracdb$ and $a$ by $fracab$ gives your eqn.
answered 9 hours ago
ganeshie8ganeshie8
4,7391 gold badge11 silver badges19 bronze badges
$endgroup$
1
$begingroup$
that's an interesting way to concretize and visualize why such a transformation (+1)
$endgroup$
– G Cab
8 hours ago
add a comment |
$begingroup$
Consider the task of cooking up a straight line by combining two other straight lines.
Given a straight line $y=-x+a$
You want to represent this as a sum of two other straight lines.
Letting the slope of one line to $-m$ forces the slope of other line to $-(1-m)$.
You can get that by comparing like terms in :
$$colorred-x+a = colorblue-mx + c - m'x+c'$$
Also constant term is $a-c$.
So above equation becomes
$$colorred-x+a = colorblue-mx + c + (1-m)left(dfraca-c1-m-xright)$$
Replacing $m$ by $fracdb$ and $a$ by $fracab$ gives your eqn.
answered 9 hours ago
ganeshie8ganeshie8
4,7391 gold badge11 silver badges19 bronze badges
$endgroup$
1
$begingroup$
that's an interesting way to concretize and visualize why such a transformation (+1)
$endgroup$
– G Cab
8 hours ago
add a comment |
$begingroup$
Consider the task of cooking up a straight line by combining two other straight lines.
Given a straight line $y=-x+a$
You want to represent this as a sum of two other straight lines.
Letting the slope of one line to $-m$ forces the slope of other line to $-(1-m)$.
You can get that by comparing like terms in :
$$colorred-x+a = colorblue-mx + c - m'x+c'$$
Also constant term is $a-c$.
So above equation becomes
$$colorred-x+a = colorblue-mx + c + (1-m)left(dfraca-c1-m-xright)$$
Replacing $m$ by $fracdb$ and $a$ by $fracab$ gives your eqn.
answered 9 hours ago
ganeshie8ganeshie8
4,7391 gold badge11 silver badges19 bronze badges
$endgroup$
Consider the task of cooking up a straight line by combining two other straight lines.
Given a straight line $y=-x+a$
You want to represent this as a sum of two other straight lines.
Letting the slope of one line to $-m$ forces the slope of other line to $-(1-m)$.
You can get that by comparing like terms in :
$$colorred-x+a = colorblue-mx + c - m'x+c'$$
Also constant term is $a-c$.
So above equation becomes
$$colorred-x+a = colorblue-mx + c + (1-m)left(dfraca-c1-m-xright)$$
Replacing $m$ by $fracdb$ and $a$ by $fracab$ gives your eqn.
answered 9 hours ago
ganeshie8ganeshie8
4,7391 gold badge11 silver badges19 bronze badges
edited 9 hours ago
answered 9 hours ago
ganeshie8ganeshie8
4,7391 gold badge11 silver badges19 bronze badges
answered 9 hours ago
ganeshie8ganeshie8
4,7391 gold badge11 silver badges19 bronze badges
answered 9 hours ago
ganeshie8ganeshie8
4,7391 gold badge11 silver badges19 bronze badges
4,7391 gold badge11 silver badges19 bronze badges
1
$begingroup$
that's an interesting way to concretize and visualize why such a transformation (+1)
$endgroup$
– G Cab
8 hours ago
add a comment |
1
$begingroup$
that's an interesting way to concretize and visualize why such a transformation (+1)
$endgroup$
– G Cab
8 hours ago
1
1
$begingroup$
that's an interesting way to concretize and visualize why such a transformation (+1)
$endgroup$
– G Cab
8 hours ago
$begingroup$
that's an interesting way to concretize and visualize why such a transformation (+1)
$endgroup$
– G Cab
8 hours ago
add a comment |
$begingroup$
Starting from the LHS
$$fracab-x$$
we can subtract and add $fracc-xdb$ to form
$$Big(fracab-xBig) - fracc-xdb + fracc-xdb$$
which rearranges to
$$fracc-xdb+Big(fracab-x-fracc-xdbBig)$$
or
$$fracc-xdb+Big(fraca-c+xdb-xBig)$$
which is equivalent to
$$fracc-xdb+Big(fraca-bx-c+dxbBig)$$
where $left(fracb-dbright) left(fraca-cb-d - x right) = Big(fraca-bx-c+dxbBig)$ and therefore we have the RHS
$$fracc-xdb+left(fracb-dbright) left(fraca-cb-d - x right)$$
answered 8 hours ago
Axion004Axion004
1,4176 silver badges17 bronze badges
$endgroup$
add a comment |
$begingroup$
Starting from the LHS
$$fracab-x$$
we can subtract and add $fracc-xdb$ to form
$$Big(fracab-xBig) - fracc-xdb + fracc-xdb$$
which rearranges to
$$fracc-xdb+Big(fracab-x-fracc-xdbBig)$$
or
$$fracc-xdb+Big(fraca-c+xdb-xBig)$$
which is equivalent to
$$fracc-xdb+Big(fraca-bx-c+dxbBig)$$
where $left(fracb-dbright) left(fraca-cb-d - x right) = Big(fraca-bx-c+dxbBig)$ and therefore we have the RHS
$$fracc-xdb+left(fracb-dbright) left(fraca-cb-d - x right)$$
answered 8 hours ago
Axion004Axion004
1,4176 silver badges17 bronze badges
$endgroup$
add a comment |
$begingroup$
Starting from the LHS
$$fracab-x$$
we can subtract and add $fracc-xdb$ to form
$$Big(fracab-xBig) - fracc-xdb + fracc-xdb$$
which rearranges to
$$fracc-xdb+Big(fracab-x-fracc-xdbBig)$$
or
$$fracc-xdb+Big(fraca-c+xdb-xBig)$$
which is equivalent to
$$fracc-xdb+Big(fraca-bx-c+dxbBig)$$
where $left(fracb-dbright) left(fraca-cb-d - x right) = Big(fraca-bx-c+dxbBig)$ and therefore we have the RHS
$$fracc-xdb+left(fracb-dbright) left(fraca-cb-d - x right)$$
answered 8 hours ago
Axion004Axion004
1,4176 silver badges17 bronze badges
$endgroup$
Starting from the LHS
$$fracab-x$$
we can subtract and add $fracc-xdb$ to form
$$Big(fracab-xBig) - fracc-xdb + fracc-xdb$$
which rearranges to
$$fracc-xdb+Big(fracab-x-fracc-xdbBig)$$
or
$$fracc-xdb+Big(fraca-c+xdb-xBig)$$
which is equivalent to
$$fracc-xdb+Big(fraca-bx-c+dxbBig)$$
where $left(fracb-dbright) left(fraca-cb-d - x right) = Big(fraca-bx-c+dxbBig)$ and therefore we have the RHS
$$fracc-xdb+left(fracb-dbright) left(fraca-cb-d - x right)$$
answered 8 hours ago
Axion004Axion004
1,4176 silver badges17 bronze badges
answered 8 hours ago
Axion004Axion004
1,4176 silver badges17 bronze badges
answered 8 hours ago
Axion004Axion004
1,4176 silver badges17 bronze badges
answered 8 hours ago
Axion004Axion004
1,4176 silver badges17 bronze badges
1,4176 silver badges17 bronze badges
add a comment |
add a comment |
$begingroup$
Let
$$
Z=:fracleft(c-xdright)b+left(fracb-dbright):left(fraca-cb-d:-:x:right):$$
Let
$$ K=fracleft(c-xdright)b$$
Let
$$ N=left(fracb-dbright):left(fraca-cb-d:-:x:right):$$
$$Z=K+N$$
We will rewrite $left(fraca-cb-d:-:x:right):$ as:
$$left(fraca-cb-d:-:x:right):=left(fraca-cb-d:-:x:fracleft(b-dright)left(b-dright):right):$$
$$N:=left(fracb-dbright)left(fraca-cb-d:-:x:fracleft(b-dright)left(b-dright):right):=left(fracb-dbright)left(fraca-c:-xleft(b-dright)b-d:right)$$
$$N:=left(fraca-c:-xleft(b-dright)b:right)$$
$$N:=left(fraca-c:-xb+xdb:right)$$
Using $Z=K+N$ and substituting the last expression for $N$ we get:
$$z:=fracleft(c-xdright)b+left(fraca-c:-xb+xdb:right) $$
$$z:=left(fraca:-xbb:right)$$
$$ z:=fraca:b-x$$
answered 9 hours ago
NoChanceNoChance
4,4782 gold badges13 silver badges22 bronze badges
$endgroup$
add a comment |
$begingroup$
Let
$$
Z=:fracleft(c-xdright)b+left(fracb-dbright):left(fraca-cb-d:-:x:right):$$
Let
$$ K=fracleft(c-xdright)b$$
Let
$$ N=left(fracb-dbright):left(fraca-cb-d:-:x:right):$$
$$Z=K+N$$
We will rewrite $left(fraca-cb-d:-:x:right):$ as:
$$left(fraca-cb-d:-:x:right):=left(fraca-cb-d:-:x:fracleft(b-dright)left(b-dright):right):$$
$$N:=left(fracb-dbright)left(fraca-cb-d:-:x:fracleft(b-dright)left(b-dright):right):=left(fracb-dbright)left(fraca-c:-xleft(b-dright)b-d:right)$$
$$N:=left(fraca-c:-xleft(b-dright)b:right)$$
$$N:=left(fraca-c:-xb+xdb:right)$$
Using $Z=K+N$ and substituting the last expression for $N$ we get:
$$z:=fracleft(c-xdright)b+left(fraca-c:-xb+xdb:right) $$
$$z:=left(fraca:-xbb:right)$$
$$ z:=fraca:b-x$$
answered 9 hours ago
NoChanceNoChance
4,4782 gold badges13 silver badges22 bronze badges
$endgroup$
add a comment |
$begingroup$
Let
$$
Z=:fracleft(c-xdright)b+left(fracb-dbright):left(fraca-cb-d:-:x:right):$$
Let
$$ K=fracleft(c-xdright)b$$
Let
$$ N=left(fracb-dbright):left(fraca-cb-d:-:x:right):$$
$$Z=K+N$$
We will rewrite $left(fraca-cb-d:-:x:right):$ as:
$$left(fraca-cb-d:-:x:right):=left(fraca-cb-d:-:x:fracleft(b-dright)left(b-dright):right):$$
$$N:=left(fracb-dbright)left(fraca-cb-d:-:x:fracleft(b-dright)left(b-dright):right):=left(fracb-dbright)left(fraca-c:-xleft(b-dright)b-d:right)$$
$$N:=left(fraca-c:-xleft(b-dright)b:right)$$
$$N:=left(fraca-c:-xb+xdb:right)$$
Using $Z=K+N$ and substituting the last expression for $N$ we get:
$$z:=fracleft(c-xdright)b+left(fraca-c:-xb+xdb:right) $$
$$z:=left(fraca:-xbb:right)$$
$$ z:=fraca:b-x$$
answered 9 hours ago
NoChanceNoChance
4,4782 gold badges13 silver badges22 bronze badges
$endgroup$
Let
$$
Z=:fracleft(c-xdright)b+left(fracb-dbright):left(fraca-cb-d:-:x:right):$$
Let
$$ K=fracleft(c-xdright)b$$
Let
$$ N=left(fracb-dbright):left(fraca-cb-d:-:x:right):$$
$$Z=K+N$$
We will rewrite $left(fraca-cb-d:-:x:right):$ as:
$$left(fraca-cb-d:-:x:right):=left(fraca-cb-d:-:x:fracleft(b-dright)left(b-dright):right):$$
$$N:=left(fracb-dbright)left(fraca-cb-d:-:x:fracleft(b-dright)left(b-dright):right):=left(fracb-dbright)left(fraca-c:-xleft(b-dright)b-d:right)$$
$$N:=left(fraca-c:-xleft(b-dright)b:right)$$
$$N:=left(fraca-c:-xb+xdb:right)$$
Using $Z=K+N$ and substituting the last expression for $N$ we get:
$$z:=fracleft(c-xdright)b+left(fraca-c:-xb+xdb:right) $$
$$z:=left(fraca:-xbb:right)$$
$$ z:=fraca:b-x$$
answered 9 hours ago
NoChanceNoChance
4,4782 gold badges13 silver badges22 bronze badges
edited 7 hours ago
answered 9 hours ago
NoChanceNoChance
4,4782 gold badges13 silver badges22 bronze badges
answered 9 hours ago
NoChanceNoChance
4,4782 gold badges13 silver badges22 bronze badges
answered 9 hours ago
NoChanceNoChance
4,4782 gold badges13 silver badges22 bronze badges
4,4782 gold badges13 silver badges22 bronze badges
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$begingroup$
If you know what you are aiming at then you can start by adding and subtracting $fracc-xdb$ ...
$endgroup$
– Martin R
10 hours ago
3
$begingroup$
There is a reason why the LHS is transformed like this. $b$ and $d$ do not fall from the sky. It seems you have to read the proof again more carefully.
$endgroup$
– callculus
10 hours ago
$begingroup$
Step by step reduce the LHS to the RHS. Turn that upside down and you derive the RHS to the LHS.
$endgroup$
– steven gregory
8 hours ago
$begingroup$
@callculus It is transformed like that so we can show it is less than epsilon. In the proof i have the identity does fall from the sky. If you have a more intuitive explenation please share .
$endgroup$
– Milan
8 hours ago
$begingroup$
Do you have a reference for the proof? If so, please include the link in your question in order to make it self contained.
$endgroup$
– Somos
7 hours ago