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Can this equation be simplified further?
The Next CEO of Stack OverflowHow can this trig equation be simplified?Can this be simplified?How can this equation be simplified this way? Transmission line: ZinTaking the integral of a strange function.Evaluating Complex ExpressionsHow should trigonometric expressions be simplified?How can we show $cos^6x+sin^6x=1-3sin^2x cos^2x$?Roots of complex quadratic polynomialCan $Asin^2t + Bsin tcos t + Csin t + Dcos t + E = 0$ be solved algebraically?How was this equation simplified?
$begingroup$
I'm trying to simplify the following equation:
$y = dfrac1-2exp(-x)cos(x)+exp(-2x)1+2exp(-x)sin(x)-exp(-2x)$
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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I'm trying to simplify the following equation:
$y = dfrac1-2exp(-x)cos(x)+exp(-2x)1+2exp(-x)sin(x)-exp(-2x)$
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
$endgroup$
add a comment |
$begingroup$
I'm trying to simplify the following equation:
$y = dfrac1-2exp(-x)cos(x)+exp(-2x)1+2exp(-x)sin(x)-exp(-2x)$
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:
$y = dfrac1-2exp(-x)cos(x)+exp(-2x)1+2exp(-x)sin(x)-exp(-2x)$
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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New contributor
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asked 2 hours ago
Maarten BaertMaarten Baert
132
132
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2 Answers
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$begingroup$
$$y=frac1-2e^-xcos(x)+e^-2x1+2e^-xsin(x)-e^-2xcdotfrace^xe^x=frace^x-2cos(x)+e^-xe^x+2sin(x)-e^-xcdotfracfrac12frac12$$ $$=fracfrace^x+e^-x2-cos(x)frace^x-e^-x2+sin(x)=fraccosh(x)-cos(x)sinh(x)+sin(x)$$
$endgroup$
$begingroup$
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 |
$begingroup$
After coreyman317's answer and your comment about large values of $x$, you could notice that for $x >24$
$$fraccosh(x)-cos(x)sinh(x)+sin(x) sim coth(x)$$ for an error $ < 10^-10$
$endgroup$
add a comment |
Your Answer
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2 Answers
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2 Answers
2
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$begingroup$
$$y=frac1-2e^-xcos(x)+e^-2x1+2e^-xsin(x)-e^-2xcdotfrace^xe^x=frace^x-2cos(x)+e^-xe^x+2sin(x)-e^-xcdotfracfrac12frac12$$ $$=fracfrace^x+e^-x2-cos(x)frace^x-e^-x2+sin(x)=fraccosh(x)-cos(x)sinh(x)+sin(x)$$
$endgroup$
$begingroup$
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 |
$begingroup$
$$y=frac1-2e^-xcos(x)+e^-2x1+2e^-xsin(x)-e^-2xcdotfrace^xe^x=frace^x-2cos(x)+e^-xe^x+2sin(x)-e^-xcdotfracfrac12frac12$$ $$=fracfrace^x+e^-x2-cos(x)frace^x-e^-x2+sin(x)=fraccosh(x)-cos(x)sinh(x)+sin(x)$$
$endgroup$
$begingroup$
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 |
$begingroup$
$$y=frac1-2e^-xcos(x)+e^-2x1+2e^-xsin(x)-e^-2xcdotfrace^xe^x=frace^x-2cos(x)+e^-xe^x+2sin(x)-e^-xcdotfracfrac12frac12$$ $$=fracfrace^x+e^-x2-cos(x)frace^x-e^-x2+sin(x)=fraccosh(x)-cos(x)sinh(x)+sin(x)$$
$endgroup$
$$y=frac1-2e^-xcos(x)+e^-2x1+2e^-xsin(x)-e^-2xcdotfrace^xe^x=frace^x-2cos(x)+e^-xe^x+2sin(x)-e^-xcdotfracfrac12frac12$$ $$=fracfrace^x+e^-x2-cos(x)frace^x-e^-x2+sin(x)=fraccosh(x)-cos(x)sinh(x)+sin(x)$$
answered 2 hours ago
coreyman317coreyman317
1,069420
1,069420
$begingroup$
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 $x >24$
$$fraccosh(x)-cos(x)sinh(x)+sin(x) sim coth(x)$$ 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 $x >24$
$$fraccosh(x)-cos(x)sinh(x)+sin(x) sim coth(x)$$ 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 $x >24$
$$fraccosh(x)-cos(x)sinh(x)+sin(x) sim coth(x)$$ for an error $ < 10^-10$
$endgroup$
After coreyman317's answer and your comment about large values of $x$, you could notice that for $x >24$
$$fraccosh(x)-cos(x)sinh(x)+sin(x) sim coth(x)$$ for an error $ < 10^-10$
answered 1 hour 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.
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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