Can this equation be simplified further?
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I'm trying to simplify the following equation:
$y = dfrac{1-2exp(-x)cos(x)+exp(-2x)}{1+2exp(-x)sin(x)-exp(-2x)}$
I suspect that a simpler form using complex exponents exists, but I can't find it.
For context, this equation describes the effective conductivity due to the skin effect of a flat conductor as a function of its thickness. I just removed some scale factors for simplicity. The underlying differential equation gives rise to expressions of the form $exp(pm(1+i)x)$, which is where the $sin(x)$ and $cos(x)$ came from.
trigonometry complex-numbers
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I'm trying to simplify the following equation:
$y = dfrac{1-2exp(-x)cos(x)+exp(-2x)}{1+2exp(-x)sin(x)-exp(-2x)}$
I suspect that a simpler form using complex exponents exists, but I can't find it.
For context, this equation describes the effective conductivity due to the skin effect of a flat conductor as a function of its thickness. I just removed some scale factors for simplicity. The underlying differential equation gives rise to expressions of the form $exp(pm(1+i)x)$, which is where the $sin(x)$ and $cos(x)$ came from.
trigonometry complex-numbers
New contributor
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add a comment |
$begingroup$
I'm trying to simplify the following equation:
$y = dfrac{1-2exp(-x)cos(x)+exp(-2x)}{1+2exp(-x)sin(x)-exp(-2x)}$
I suspect that a simpler form using complex exponents exists, but I can't find it.
For context, this equation describes the effective conductivity due to the skin effect of a flat conductor as a function of its thickness. I just removed some scale factors for simplicity. The underlying differential equation gives rise to expressions of the form $exp(pm(1+i)x)$, which is where the $sin(x)$ and $cos(x)$ came from.
trigonometry complex-numbers
New contributor
$endgroup$
I'm trying to simplify the following equation:
$y = dfrac{1-2exp(-x)cos(x)+exp(-2x)}{1+2exp(-x)sin(x)-exp(-2x)}$
I suspect that a simpler form using complex exponents exists, but I can't find it.
For context, this equation describes the effective conductivity due to the skin effect of a flat conductor as a function of its thickness. I just removed some scale factors for simplicity. The underlying differential equation gives rise to expressions of the form $exp(pm(1+i)x)$, which is where the $sin(x)$ and $cos(x)$ came from.
trigonometry complex-numbers
trigonometry complex-numbers
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asked 2 hours ago
Maarten BaertMaarten Baert
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2 Answers
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$begingroup$
$$y=frac{1-2e^{-x}cos(x)+e^{-2x}}{1+2e^{-x}sin(x)-e^{-2x}}cdotfrac{e^x}{e^x}=frac{e^x-2cos(x)+e^{-x}}{e^x+2sin(x)-e^{-x}}cdotfrac{frac{1}{2}}{frac{1}{2}}$$ $$=frac{frac{e^x+e^{-x}}{2}-cos(x)}{frac{e^x-e^{-x}}{2}+sin(x)}=frac{cosh(x)-cos(x)}{sinh(x)+sin(x)}$$
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Nice! Would it be possible to rewrite this using $tan$ or $tanh$? Unfortunately $sinh(x)$ and $cosh(x)$ cause numerical issues (overflow) for large values of $x$.
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– Maarten Baert
1 hour ago
1
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If you want you can divide the top and bottom by $cosh(x)$ to get a $tanh(x)$ but this makes both the numerator and denominator more complicated.
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– coreyman317
1 hour ago
add a comment |
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After coreyman317's answer and your comment about large values of $x$, you could notice that for $x >24$
$$frac{cosh(x)-cos(x)}{sinh(x)+sin(x)} sim coth(x)$$ for an error $ < 10^{-10}$
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Your Answer
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2 Answers
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$begingroup$
$$y=frac{1-2e^{-x}cos(x)+e^{-2x}}{1+2e^{-x}sin(x)-e^{-2x}}cdotfrac{e^x}{e^x}=frac{e^x-2cos(x)+e^{-x}}{e^x+2sin(x)-e^{-x}}cdotfrac{frac{1}{2}}{frac{1}{2}}$$ $$=frac{frac{e^x+e^{-x}}{2}-cos(x)}{frac{e^x-e^{-x}}{2}+sin(x)}=frac{cosh(x)-cos(x)}{sinh(x)+sin(x)}$$
$endgroup$
$begingroup$
Nice! Would it be possible to rewrite this using $tan$ or $tanh$? Unfortunately $sinh(x)$ and $cosh(x)$ cause numerical issues (overflow) for large values of $x$.
$endgroup$
– Maarten Baert
1 hour ago
1
$begingroup$
If you want you can divide the top and bottom by $cosh(x)$ to get a $tanh(x)$ but this makes both the numerator and denominator more complicated.
$endgroup$
– coreyman317
1 hour ago
add a comment |
$begingroup$
$$y=frac{1-2e^{-x}cos(x)+e^{-2x}}{1+2e^{-x}sin(x)-e^{-2x}}cdotfrac{e^x}{e^x}=frac{e^x-2cos(x)+e^{-x}}{e^x+2sin(x)-e^{-x}}cdotfrac{frac{1}{2}}{frac{1}{2}}$$ $$=frac{frac{e^x+e^{-x}}{2}-cos(x)}{frac{e^x-e^{-x}}{2}+sin(x)}=frac{cosh(x)-cos(x)}{sinh(x)+sin(x)}$$
$endgroup$
$begingroup$
Nice! Would it be possible to rewrite this using $tan$ or $tanh$? Unfortunately $sinh(x)$ and $cosh(x)$ cause numerical issues (overflow) for large values of $x$.
$endgroup$
– Maarten Baert
1 hour ago
1
$begingroup$
If you want you can divide the top and bottom by $cosh(x)$ to get a $tanh(x)$ but this makes both the numerator and denominator more complicated.
$endgroup$
– coreyman317
1 hour ago
add a comment |
$begingroup$
$$y=frac{1-2e^{-x}cos(x)+e^{-2x}}{1+2e^{-x}sin(x)-e^{-2x}}cdotfrac{e^x}{e^x}=frac{e^x-2cos(x)+e^{-x}}{e^x+2sin(x)-e^{-x}}cdotfrac{frac{1}{2}}{frac{1}{2}}$$ $$=frac{frac{e^x+e^{-x}}{2}-cos(x)}{frac{e^x-e^{-x}}{2}+sin(x)}=frac{cosh(x)-cos(x)}{sinh(x)+sin(x)}$$
$endgroup$
$$y=frac{1-2e^{-x}cos(x)+e^{-2x}}{1+2e^{-x}sin(x)-e^{-2x}}cdotfrac{e^x}{e^x}=frac{e^x-2cos(x)+e^{-x}}{e^x+2sin(x)-e^{-x}}cdotfrac{frac{1}{2}}{frac{1}{2}}$$ $$=frac{frac{e^x+e^{-x}}{2}-cos(x)}{frac{e^x-e^{-x}}{2}+sin(x)}=frac{cosh(x)-cos(x)}{sinh(x)+sin(x)}$$
answered 2 hours ago
coreyman317coreyman317
1,059420
1,059420
$begingroup$
Nice! Would it be possible to rewrite this using $tan$ or $tanh$? Unfortunately $sinh(x)$ and $cosh(x)$ cause numerical issues (overflow) for large values of $x$.
$endgroup$
– Maarten Baert
1 hour ago
1
$begingroup$
If you want you can divide the top and bottom by $cosh(x)$ to get a $tanh(x)$ but this makes both the numerator and denominator more complicated.
$endgroup$
– coreyman317
1 hour ago
add a comment |
$begingroup$
Nice! Would it be possible to rewrite this using $tan$ or $tanh$? Unfortunately $sinh(x)$ and $cosh(x)$ cause numerical issues (overflow) for large values of $x$.
$endgroup$
– Maarten Baert
1 hour ago
1
$begingroup$
If you want you can divide the top and bottom by $cosh(x)$ to get a $tanh(x)$ but this makes both the numerator and denominator more complicated.
$endgroup$
– coreyman317
1 hour ago
$begingroup$
Nice! Would it be possible to rewrite this using $tan$ or $tanh$? Unfortunately $sinh(x)$ and $cosh(x)$ cause numerical issues (overflow) for large values of $x$.
$endgroup$
– Maarten Baert
1 hour ago
$begingroup$
Nice! Would it be possible to rewrite this using $tan$ or $tanh$? Unfortunately $sinh(x)$ and $cosh(x)$ cause numerical issues (overflow) for large values of $x$.
$endgroup$
– Maarten Baert
1 hour ago
1
1
$begingroup$
If you want you can divide the top and bottom by $cosh(x)$ to get a $tanh(x)$ but this makes both the numerator and denominator more complicated.
$endgroup$
– coreyman317
1 hour ago
$begingroup$
If you want you can divide the top and bottom by $cosh(x)$ to get a $tanh(x)$ but this makes both the numerator and denominator more complicated.
$endgroup$
– coreyman317
1 hour ago
add a comment |
$begingroup$
After coreyman317's answer and your comment about large values of $x$, you could notice that for $x >24$
$$frac{cosh(x)-cos(x)}{sinh(x)+sin(x)} sim coth(x)$$ for an error $ < 10^{-10}$
$endgroup$
add a comment |
$begingroup$
After coreyman317's answer and your comment about large values of $x$, you could notice that for $x >24$
$$frac{cosh(x)-cos(x)}{sinh(x)+sin(x)} sim coth(x)$$ for an error $ < 10^{-10}$
$endgroup$
add a comment |
$begingroup$
After coreyman317's answer and your comment about large values of $x$, you could notice that for $x >24$
$$frac{cosh(x)-cos(x)}{sinh(x)+sin(x)} sim coth(x)$$ for an error $ < 10^{-10}$
$endgroup$
After coreyman317's answer and your comment about large values of $x$, you could notice that for $x >24$
$$frac{cosh(x)-cos(x)}{sinh(x)+sin(x)} sim coth(x)$$ for an error $ < 10^{-10}$
answered 56 mins ago
Claude LeiboviciClaude Leibovici
125k1158135
125k1158135
add a comment |
add a comment |
Maarten Baert is a new contributor. Be nice, and check out our Code of Conduct.
Maarten Baert is a new contributor. Be nice, and check out our Code of Conduct.
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