Strange question about exponentsBasic math questionQuestion about the graph of the square root functionHint...
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Strange question about exponents
Basic math questionQuestion about the graph of the square root functionHint to show $tanh(z)=frac{sinh(2x)+isin(2y)}{cosh(2x)+cos(2y)}$?Algebra Fraction Problem - VariableA bowler has averaged 164 for the first 21 games of a 90-game season.Find $limlimits_{xto 0} frac{sqrt{ 1+x} - sqrt{ 1-x}}{sqrt[3]{ 1+x} - sqrt[3]{ 1-x}}$Question about Properties of ExponentsEvaluating $lim _{tto infty}frac{1-frac{sqrt{t}}{sqrt{t+1}}}{2-frac{sqrt{4t:+:1}}{sqrt{t+2}}}$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
$$e^{ix} = e^{2pi ifrac{x}{2pi}} = (e^{2pi i})^{frac{x}{2pi}} = 1^{frac{x}{2pi}} = 1$$
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
$$e^{ix} = e^{2pi ifrac{x}{2pi}} = (e^{2pi i})^{frac{x}{2pi}} = 1^{frac{x}{2pi}} = 1$$
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 $(-1)^x = (-1)^{2frac{x}{2}} = left[(-1)^2right]^{frac{x}{2}} = 1^{frac{x}{2}} = 1.$ 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
$$e^{ix} = e^{2pi ifrac{x}{2pi}} = (e^{2pi i})^{frac{x}{2pi}} = 1^{frac{x}{2pi}} = 1$$
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
$$e^{ix} = e^{2pi ifrac{x}{2pi}} = (e^{2pi i})^{frac{x}{2pi}} = 1^{frac{x}{2pi}} = 1$$
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 $(-1)^x = (-1)^{2frac{x}{2}} = left[(-1)^2right]^{frac{x}{2}} = 1^{frac{x}{2}} = 1.$ 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 $(-1)^x = (-1)^{2frac{x}{2}} = left[(-1)^2right]^{frac{x}{2}} = 1^{frac{x}{2}} = 1.$ 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 $(-1)^x = (-1)^{2frac{x}{2}} = left[(-1)^2right]^{frac{x}{2}} = 1^{frac{x}{2}} = 1.$ 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 $(-1)^x = (-1)^{2frac{x}{2}} = left[(-1)^2right]^{frac{x}{2}} = 1^{frac{x}{2}} = 1.$ 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
1
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oldest
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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
68.9k53169
$endgroup$
add a comment |
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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
68.9k53169
$endgroup$
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
68.9k53169
$endgroup$
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
68.9k53169
$endgroup$
$z to z^{frac x {2pi}}$ 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
68.9k53169
answered Mar 13 at 7:51
Kavi Rama MurthyKavi Rama Murthy
68.9k53169
answered Mar 13 at 7:51
Kavi Rama MurthyKavi Rama Murthy
68.9k53169
answered Mar 13 at 7:51
Kavi Rama MurthyKavi Rama Murthy
68.9k53169
68.9k53169
add a comment |
add a comment |
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$begingroup$
Perhaps an analogy would be something like $(-1)^x = (-1)^{2frac{x}{2}} = left[(-1)^2right]^{frac{x}{2}} = 1^{frac{x}{2}} = 1.$ 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