Strange question about exponentsBasic math questionQuestion about the graph of the square root functionHint to show $tanh(z)=fracsinh(2x)+isin(2y)cosh(2x)+cos(2y)$?Algebra Fraction Problem - VariableA bowler has averaged 164 for the first 21 games of a 90-game season.Find $limlimits_xto 0 fracsqrt 1+x - sqrt 1-xsqrt[3] 1+x - sqrt[3] 1-x$Question about Properties of ExponentsEvaluating $lim _tto inftyfrac1-fracsqrttsqrtt+12-fracsqrt4t:+:1sqrtt+2$Help with function question $f(2x + 1) = 2f(x) + 1$How to solve this simple Algebraic equation? (Wolfram & SymboLab both stumped…)
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Strange question about exponents
Basic math questionQuestion about the graph of the square root functionHint to show $tanh(z)=fracsinh(2x)+isin(2y)cosh(2x)+cos(2y)$?Algebra Fraction Problem - VariableA bowler has averaged 164 for the first 21 games of a 90-game season.Find $limlimits_xto 0 fracsqrt 1+x - sqrt 1-xsqrt[3] 1+x - sqrt[3] 1-x$Question about Properties of ExponentsEvaluating $lim _tto inftyfrac1-fracsqrttsqrtt+12-fracsqrt4t:+:1sqrtt+2$Help with function question $f(2x + 1) = 2f(x) + 1$How to solve this simple Algebraic equation? (Wolfram & SymboLab both stumped…)
$begingroup$
I've been stumped for a while trying to figure out the flaw in this logic. Consider some positive, real x. Then
$$e^ix = e^2pi ifracx2pi = (e^2pi i)^fracx2pi = 1^fracx2pi = 1$$
I've felt like an idiot for hours. What am I missing?
algebra-precalculus
asked Mar 13 at 7:46
louie mcconnelllouie mcconnell
1,4881824
$endgroup$
add a comment |
$begingroup$
I've been stumped for a while trying to figure out the flaw in this logic. Consider some positive, real x. Then
$$e^ix = e^2pi ifracx2pi = (e^2pi i)^fracx2pi = 1^fracx2pi = 1$$
I've felt like an idiot for hours. What am I missing?
algebra-precalculus
asked Mar 13 at 7:46
louie mcconnelllouie mcconnell
1,4881824
$endgroup$
$begingroup$
Perhaps an analogy would be something like $(-1)^x = (-1)^2fracx2 = left[(-1)^2right]^fracx2 = 1^fracx2 = 1.$ I've simply written essentially the same string of equalities you've written using $e^pi i$ in place of $e^i.$
$endgroup$
– Dave L. Renfro
Mar 13 at 9:06
add a comment |
$begingroup$
I've been stumped for a while trying to figure out the flaw in this logic. Consider some positive, real x. Then
$$e^ix = e^2pi ifracx2pi = (e^2pi i)^fracx2pi = 1^fracx2pi = 1$$
I've felt like an idiot for hours. What am I missing?
algebra-precalculus
asked Mar 13 at 7:46
louie mcconnelllouie mcconnell
1,4881824
$endgroup$
I've been stumped for a while trying to figure out the flaw in this logic. Consider some positive, real x. Then
$$e^ix = e^2pi ifracx2pi = (e^2pi i)^fracx2pi = 1^fracx2pi = 1$$
I've felt like an idiot for hours. What am I missing?
algebra-precalculus
algebra-precalculus
asked Mar 13 at 7:46
louie mcconnelllouie mcconnell
1,4881824
asked Mar 13 at 7:46
louie mcconnelllouie mcconnell
1,4881824
asked Mar 13 at 7:46
louie mcconnelllouie mcconnell
1,4881824
asked Mar 13 at 7:46
louie mcconnelllouie mcconnell
1,4881824
asked Mar 13 at 7:46
louie mcconnelllouie mcconnell
1,4881824
1,4881824
$begingroup$
Perhaps an analogy would be something like $(-1)^x = (-1)^2fracx2 = left[(-1)^2right]^fracx2 = 1^fracx2 = 1.$ I've simply written essentially the same string of equalities you've written using $e^pi i$ in place of $e^i.$
$endgroup$
– Dave L. Renfro
Mar 13 at 9:06
add a comment |
$begingroup$
Perhaps an analogy would be something like $(-1)^x = (-1)^2fracx2 = left[(-1)^2right]^fracx2 = 1^fracx2 = 1.$ I've simply written essentially the same string of equalities you've written using $e^pi i$ in place of $e^i.$
$endgroup$
– Dave L. Renfro
Mar 13 at 9:06
$begingroup$
Perhaps an analogy would be something like $(-1)^x = (-1)^2fracx2 = left[(-1)^2right]^fracx2 = 1^fracx2 = 1.$ I've simply written essentially the same string of equalities you've written using $e^pi i$ in place of $e^i.$
$endgroup$
– Dave L. Renfro
Mar 13 at 9:06
$begingroup$
Perhaps an analogy would be something like $(-1)^x = (-1)^2fracx2 = left[(-1)^2right]^fracx2 = 1^fracx2 = 1.$ I've simply written essentially the same string of equalities you've written using $e^pi i$ in place of $e^i.$
$endgroup$
– Dave L. Renfro
Mar 13 at 9:06
add a comment |
1 Answer
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$begingroup$
$z to z^frac x 2pi$ is a mutivalued function and just one of its values for $z=e^2pi i$ is $e^2pi ifrac x 2pi$. So you cannot conclude that $e^ix=1$.
To understand exactly what is happening you have to get familiar with logarithms in the complex plane.
answered Mar 13 at 7:51
Kavi Rama MurthyKavi Rama Murthy
68.1k53068
$endgroup$
add a comment |
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1 Answer
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$begingroup$
$z to z^frac x 2pi$ is a mutivalued function and just one of its values for $z=e^2pi i$ is $e^2pi ifrac x 2pi$. So you cannot conclude that $e^ix=1$.
To understand exactly what is happening you have to get familiar with logarithms in the complex plane.
answered Mar 13 at 7:51
Kavi Rama MurthyKavi Rama Murthy
68.1k53068
$endgroup$
add a comment |
$begingroup$
$z to z^frac x 2pi$ is a mutivalued function and just one of its values for $z=e^2pi i$ is $e^2pi ifrac x 2pi$. So you cannot conclude that $e^ix=1$.
To understand exactly what is happening you have to get familiar with logarithms in the complex plane.
answered Mar 13 at 7:51
Kavi Rama MurthyKavi Rama Murthy
68.1k53068
$endgroup$
add a comment |
$begingroup$
$z to z^frac x 2pi$ is a mutivalued function and just one of its values for $z=e^2pi i$ is $e^2pi ifrac x 2pi$. So you cannot conclude that $e^ix=1$.
To understand exactly what is happening you have to get familiar with logarithms in the complex plane.
answered Mar 13 at 7:51
Kavi Rama MurthyKavi Rama Murthy
68.1k53068
$endgroup$
$z to z^frac x 2pi$ is a mutivalued function and just one of its values for $z=e^2pi i$ is $e^2pi ifrac x 2pi$. So you cannot conclude that $e^ix=1$.
To understand exactly what is happening you have to get familiar with logarithms in the complex plane.
answered Mar 13 at 7:51
Kavi Rama MurthyKavi Rama Murthy
68.1k53068
answered Mar 13 at 7:51
Kavi Rama MurthyKavi Rama Murthy
68.1k53068
answered Mar 13 at 7:51
Kavi Rama MurthyKavi Rama Murthy
68.1k53068
answered Mar 13 at 7:51
Kavi Rama MurthyKavi Rama Murthy
68.1k53068
68.1k53068
add a comment |
add a comment |
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$begingroup$
Perhaps an analogy would be something like $(-1)^x = (-1)^2fracx2 = left[(-1)^2right]^fracx2 = 1^fracx2 = 1.$ I've simply written essentially the same string of equalities you've written using $e^pi i$ in place of $e^i.$
$endgroup$
– Dave L. Renfro
Mar 13 at 9:06