Why $|e^{itx}| = 1$?
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I'm studying some demonstrations of properties of characteristic function in which I have to use that $|e^{itx}| = 1$ but I don't understand it at all. Could you give a clue to demonstrate it?
probability-theory characteristic-functions
asked Dec 29 '18 at 12:19
Miguel AnguitaMiguel Anguita
505
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add a comment |
$begingroup$
I'm studying some demonstrations of properties of characteristic function in which I have to use that $|e^{itx}| = 1$ but I don't understand it at all. Could you give a clue to demonstrate it?
probability-theory characteristic-functions
asked Dec 29 '18 at 12:19
Miguel AnguitaMiguel Anguita
505
$endgroup$
7
$begingroup$
... because $cos (tx)^2+sin(tx)^2=1$
$endgroup$
– Fakemistake
Dec 29 '18 at 12:21
add a comment |
$begingroup$
I'm studying some demonstrations of properties of characteristic function in which I have to use that $|e^{itx}| = 1$ but I don't understand it at all. Could you give a clue to demonstrate it?
probability-theory characteristic-functions
asked Dec 29 '18 at 12:19
Miguel AnguitaMiguel Anguita
505
$endgroup$
I'm studying some demonstrations of properties of characteristic function in which I have to use that $|e^{itx}| = 1$ but I don't understand it at all. Could you give a clue to demonstrate it?
probability-theory characteristic-functions
probability-theory characteristic-functions
asked Dec 29 '18 at 12:19
Miguel AnguitaMiguel Anguita
505
asked Dec 29 '18 at 12:19
Miguel AnguitaMiguel Anguita
505
asked Dec 29 '18 at 12:19
Miguel AnguitaMiguel Anguita
505
asked Dec 29 '18 at 12:19
Miguel AnguitaMiguel Anguita
505
asked Dec 29 '18 at 12:19
Miguel AnguitaMiguel Anguita
505
505
7
$begingroup$
... because $cos (tx)^2+sin(tx)^2=1$
$endgroup$
– Fakemistake
Dec 29 '18 at 12:21
add a comment |
7
$begingroup$
... because $cos (tx)^2+sin(tx)^2=1$
$endgroup$
– Fakemistake
Dec 29 '18 at 12:21
7
7
$begingroup$
... because $cos (tx)^2+sin(tx)^2=1$
$endgroup$
– Fakemistake
Dec 29 '18 at 12:21
$begingroup$
... because $cos (tx)^2+sin(tx)^2=1$
$endgroup$
– Fakemistake
Dec 29 '18 at 12:21
add a comment |
2 Answers
2
active
oldest
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$begingroup$
A key property of the exponential map is
$$e^ze^{z^prime} = e^{z+z^prime}$$
This can be proven using Cauchy product.
Based on that, you get for $y in mathbb R$
$$vert e^{iy}vert^2= e^{iy} overline{e^{iy}}=e^{iy}e^{-iy}=e^{iy-iy}=1$$
answered Dec 29 '18 at 12:28
mathcounterexamples.netmathcounterexamples.net
27k22158
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$begingroup$
Thank you very much to you and all who answered!
$endgroup$
– Miguel Anguita
Dec 29 '18 at 12:53
add a comment |
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Hint: $$e^{itx}=cos(tx)+isin(tx)$$
answered Dec 29 '18 at 12:23
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
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2 Answers
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active
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2 Answers
2
active
oldest
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active
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votes
$begingroup$
A key property of the exponential map is
$$e^ze^{z^prime} = e^{z+z^prime}$$
This can be proven using Cauchy product.
Based on that, you get for $y in mathbb R$
$$vert e^{iy}vert^2= e^{iy} overline{e^{iy}}=e^{iy}e^{-iy}=e^{iy-iy}=1$$
answered Dec 29 '18 at 12:28
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
$begingroup$
Thank you very much to you and all who answered!
$endgroup$
– Miguel Anguita
Dec 29 '18 at 12:53
add a comment |
$begingroup$
A key property of the exponential map is
$$e^ze^{z^prime} = e^{z+z^prime}$$
This can be proven using Cauchy product.
Based on that, you get for $y in mathbb R$
$$vert e^{iy}vert^2= e^{iy} overline{e^{iy}}=e^{iy}e^{-iy}=e^{iy-iy}=1$$
answered Dec 29 '18 at 12:28
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
$begingroup$
Thank you very much to you and all who answered!
$endgroup$
– Miguel Anguita
Dec 29 '18 at 12:53
add a comment |
$begingroup$
A key property of the exponential map is
$$e^ze^{z^prime} = e^{z+z^prime}$$
This can be proven using Cauchy product.
Based on that, you get for $y in mathbb R$
$$vert e^{iy}vert^2= e^{iy} overline{e^{iy}}=e^{iy}e^{-iy}=e^{iy-iy}=1$$
answered Dec 29 '18 at 12:28
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
A key property of the exponential map is
$$e^ze^{z^prime} = e^{z+z^prime}$$
This can be proven using Cauchy product.
Based on that, you get for $y in mathbb R$
$$vert e^{iy}vert^2= e^{iy} overline{e^{iy}}=e^{iy}e^{-iy}=e^{iy-iy}=1$$
answered Dec 29 '18 at 12:28
mathcounterexamples.netmathcounterexamples.net
27k22158
answered Dec 29 '18 at 12:28
mathcounterexamples.netmathcounterexamples.net
27k22158
answered Dec 29 '18 at 12:28
mathcounterexamples.netmathcounterexamples.net
27k22158
answered Dec 29 '18 at 12:28
mathcounterexamples.netmathcounterexamples.net
27k22158
27k22158
$begingroup$
Thank you very much to you and all who answered!
$endgroup$
– Miguel Anguita
Dec 29 '18 at 12:53
add a comment |
$begingroup$
Thank you very much to you and all who answered!
$endgroup$
– Miguel Anguita
Dec 29 '18 at 12:53
$begingroup$
Thank you very much to you and all who answered!
$endgroup$
– Miguel Anguita
Dec 29 '18 at 12:53
$begingroup$
Thank you very much to you and all who answered!
$endgroup$
– Miguel Anguita
Dec 29 '18 at 12:53
add a comment |
$begingroup$
Hint: $$e^{itx}=cos(tx)+isin(tx)$$
answered Dec 29 '18 at 12:23
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
$endgroup$
add a comment |
$begingroup$
Hint: $$e^{itx}=cos(tx)+isin(tx)$$
answered Dec 29 '18 at 12:23
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
$endgroup$
add a comment |
$begingroup$
Hint: $$e^{itx}=cos(tx)+isin(tx)$$
answered Dec 29 '18 at 12:23
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
$endgroup$
Hint: $$e^{itx}=cos(tx)+isin(tx)$$
answered Dec 29 '18 at 12:23
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
answered Dec 29 '18 at 12:23
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
answered Dec 29 '18 at 12:23
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
answered Dec 29 '18 at 12:23
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
77.8k42866
add a comment |
add a comment |
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$begingroup$
... because $cos (tx)^2+sin(tx)^2=1$
$endgroup$
– Fakemistake
Dec 29 '18 at 12:21