Proving that $-frac12[log(1-e^{2ix})+log(1-e^{-2ix})]=-frac12log(2-2cos2x)$
$begingroup$
I was looking at this answer and I was confused to see the equality
$$-frac12left[;logleft(1-e^{2ix}right)+logleft(1-e^{-2ix}right);right]=-frac12log(2-2cos2x)$$
Which I am unable to prove.
I have tried
$$
S=-frac12log(1-e^{2ix})-frac12log(1-e^{-2ix})=-frac12logfrac{1-e^{2ix}}{1-e^{-2ix}}
$$
Then with $u=e^{ix}$:
$$frac{1-u^2}{1-frac1{u^2}}=frac{1-u^2}{1-frac1{u^2}}frac{u^2}{u^2}=-u^2frac{u^2-1}{u^2-1}=-u^2$$
So
$$S=-frac12log(-e^{2ix})=-frac{ipi}2-ix$$
Which I'm fairly certain is wrong. I'm sure there is something realy simple I'm missing here, could I have some help? Thanks.
algebra-precalculus proof-explanation
edited Jan 13 at 1:15
Blue
49.7k870158
asked Jan 13 at 1:03
clathratusclathratus
5,1441439
$endgroup$
add a comment |
$begingroup$
I was looking at this answer and I was confused to see the equality
$$-frac12left[;logleft(1-e^{2ix}right)+logleft(1-e^{-2ix}right);right]=-frac12log(2-2cos2x)$$
Which I am unable to prove.
I have tried
$$
S=-frac12log(1-e^{2ix})-frac12log(1-e^{-2ix})=-frac12logfrac{1-e^{2ix}}{1-e^{-2ix}}
$$
Then with $u=e^{ix}$:
$$frac{1-u^2}{1-frac1{u^2}}=frac{1-u^2}{1-frac1{u^2}}frac{u^2}{u^2}=-u^2frac{u^2-1}{u^2-1}=-u^2$$
So
$$S=-frac12log(-e^{2ix})=-frac{ipi}2-ix$$
Which I'm fairly certain is wrong. I'm sure there is something realy simple I'm missing here, could I have some help? Thanks.
algebra-precalculus proof-explanation
edited Jan 13 at 1:15
Blue
49.7k870158
asked Jan 13 at 1:03
clathratusclathratus
5,1441439
$endgroup$
add a comment |
$begingroup$
I was looking at this answer and I was confused to see the equality
$$-frac12left[;logleft(1-e^{2ix}right)+logleft(1-e^{-2ix}right);right]=-frac12log(2-2cos2x)$$
Which I am unable to prove.
I have tried
$$
S=-frac12log(1-e^{2ix})-frac12log(1-e^{-2ix})=-frac12logfrac{1-e^{2ix}}{1-e^{-2ix}}
$$
Then with $u=e^{ix}$:
$$frac{1-u^2}{1-frac1{u^2}}=frac{1-u^2}{1-frac1{u^2}}frac{u^2}{u^2}=-u^2frac{u^2-1}{u^2-1}=-u^2$$
So
$$S=-frac12log(-e^{2ix})=-frac{ipi}2-ix$$
Which I'm fairly certain is wrong. I'm sure there is something realy simple I'm missing here, could I have some help? Thanks.
algebra-precalculus proof-explanation
edited Jan 13 at 1:15
Blue
49.7k870158
asked Jan 13 at 1:03
clathratusclathratus
5,1441439
$endgroup$
I was looking at this answer and I was confused to see the equality
$$-frac12left[;logleft(1-e^{2ix}right)+logleft(1-e^{-2ix}right);right]=-frac12log(2-2cos2x)$$
Which I am unable to prove.
I have tried
$$
S=-frac12log(1-e^{2ix})-frac12log(1-e^{-2ix})=-frac12logfrac{1-e^{2ix}}{1-e^{-2ix}}
$$
Then with $u=e^{ix}$:
$$frac{1-u^2}{1-frac1{u^2}}=frac{1-u^2}{1-frac1{u^2}}frac{u^2}{u^2}=-u^2frac{u^2-1}{u^2-1}=-u^2$$
So
$$S=-frac12log(-e^{2ix})=-frac{ipi}2-ix$$
Which I'm fairly certain is wrong. I'm sure there is something realy simple I'm missing here, could I have some help? Thanks.
algebra-precalculus proof-explanation
algebra-precalculus proof-explanation
edited Jan 13 at 1:15
Blue
49.7k870158
asked Jan 13 at 1:03
clathratusclathratus
5,1441439
edited Jan 13 at 1:15
Blue
49.7k870158
asked Jan 13 at 1:03
clathratusclathratus
5,1441439
edited Jan 13 at 1:15
Blue
49.7k870158
edited Jan 13 at 1:15
Blue
49.7k870158
edited Jan 13 at 1:15
Blue
49.7k870158
49.7k870158
asked Jan 13 at 1:03
clathratusclathratus
5,1441439
asked Jan 13 at 1:03
clathratusclathratus
5,1441439
asked Jan 13 at 1:03
clathratusclathratus
5,1441439
5,1441439
add a comment |
add a comment |
1 Answer
1
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votes
$begingroup$
This step is wrong.
$$S=-frac12log(1-e^{2ix})-frac12log(1-e^{-2ix})=-frac12logfrac{1-e^{2ix}}{1-e^{-2ix}}$$
Correct way:
begin{align}
-frac12[log(1-e^{2ix})+log(1-e^{-2ix})]&=-frac12[log((1-e^{2ix})(1-e^{-2ix}))]\
&=-frac12[log(1-e^{-2ix}-e^{2ix}+1))]\
&=-frac12log(2-2cos2x)
end{align}
answered Jan 13 at 1:12
Thomas ShelbyThomas Shelby
4,7382727
$endgroup$
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
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active
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votes
$begingroup$
This step is wrong.
$$S=-frac12log(1-e^{2ix})-frac12log(1-e^{-2ix})=-frac12logfrac{1-e^{2ix}}{1-e^{-2ix}}$$
Correct way:
begin{align}
-frac12[log(1-e^{2ix})+log(1-e^{-2ix})]&=-frac12[log((1-e^{2ix})(1-e^{-2ix}))]\
&=-frac12[log(1-e^{-2ix}-e^{2ix}+1))]\
&=-frac12log(2-2cos2x)
end{align}
answered Jan 13 at 1:12
Thomas ShelbyThomas Shelby
4,7382727
$endgroup$
add a comment |
$begingroup$
This step is wrong.
$$S=-frac12log(1-e^{2ix})-frac12log(1-e^{-2ix})=-frac12logfrac{1-e^{2ix}}{1-e^{-2ix}}$$
Correct way:
begin{align}
-frac12[log(1-e^{2ix})+log(1-e^{-2ix})]&=-frac12[log((1-e^{2ix})(1-e^{-2ix}))]\
&=-frac12[log(1-e^{-2ix}-e^{2ix}+1))]\
&=-frac12log(2-2cos2x)
end{align}
answered Jan 13 at 1:12
Thomas ShelbyThomas Shelby
4,7382727
$endgroup$
add a comment |
$begingroup$
This step is wrong.
$$S=-frac12log(1-e^{2ix})-frac12log(1-e^{-2ix})=-frac12logfrac{1-e^{2ix}}{1-e^{-2ix}}$$
Correct way:
begin{align}
-frac12[log(1-e^{2ix})+log(1-e^{-2ix})]&=-frac12[log((1-e^{2ix})(1-e^{-2ix}))]\
&=-frac12[log(1-e^{-2ix}-e^{2ix}+1))]\
&=-frac12log(2-2cos2x)
end{align}
answered Jan 13 at 1:12
Thomas ShelbyThomas Shelby
4,7382727
$endgroup$
This step is wrong.
$$S=-frac12log(1-e^{2ix})-frac12log(1-e^{-2ix})=-frac12logfrac{1-e^{2ix}}{1-e^{-2ix}}$$
Correct way:
begin{align}
-frac12[log(1-e^{2ix})+log(1-e^{-2ix})]&=-frac12[log((1-e^{2ix})(1-e^{-2ix}))]\
&=-frac12[log(1-e^{-2ix}-e^{2ix}+1))]\
&=-frac12log(2-2cos2x)
end{align}
answered Jan 13 at 1:12
Thomas ShelbyThomas Shelby
4,7382727
answered Jan 13 at 1:12
Thomas ShelbyThomas Shelby
4,7382727
answered Jan 13 at 1:12
Thomas ShelbyThomas Shelby
4,7382727
answered Jan 13 at 1:12
Thomas ShelbyThomas Shelby
4,7382727
4,7382727
add a comment |
add a comment |
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