How to derive relation between relative and log deviations?
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I ran into the following approximation:
$$frac{x_t - x_0}{x_0} approx log(x_t/x_0) + frac{1}{2}log(x_t/x_0)^2 $$
which is supposedly a second-order approximation around a fixed value $x_0$. How does this come about?
taylor-expansion
asked Dec 13 '18 at 15:47
JanedoJanedo
1
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add a comment |
$begingroup$
I ran into the following approximation:
$$frac{x_t - x_0}{x_0} approx log(x_t/x_0) + frac{1}{2}log(x_t/x_0)^2 $$
which is supposedly a second-order approximation around a fixed value $x_0$. How does this come about?
taylor-expansion
asked Dec 13 '18 at 15:47
JanedoJanedo
1
$endgroup$
add a comment |
$begingroup$
I ran into the following approximation:
$$frac{x_t - x_0}{x_0} approx log(x_t/x_0) + frac{1}{2}log(x_t/x_0)^2 $$
which is supposedly a second-order approximation around a fixed value $x_0$. How does this come about?
taylor-expansion
asked Dec 13 '18 at 15:47
JanedoJanedo
1
$endgroup$
I ran into the following approximation:
$$frac{x_t - x_0}{x_0} approx log(x_t/x_0) + frac{1}{2}log(x_t/x_0)^2 $$
which is supposedly a second-order approximation around a fixed value $x_0$. How does this come about?
taylor-expansion
taylor-expansion
asked Dec 13 '18 at 15:47
JanedoJanedo
1
asked Dec 13 '18 at 15:47
JanedoJanedo
1
asked Dec 13 '18 at 15:47
JanedoJanedo
1
asked Dec 13 '18 at 15:47
JanedoJanedo
1
asked Dec 13 '18 at 15:47
JanedoJanedo
1
1
add a comment |
add a comment |
1 Answer
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$begingroup$
Taylor expand $ln(x/x_0)$ around $x_0$
$$
lnleft(frac{x}{x_0}right) = frac{x - x_0}{x_0} - frac{(x - x_0)^2}{2x_0^2} + frac{(x - x_0)^3}{3x_0^3} + cdots
$$
Similarly
$$
ln^2left(frac{x}{x_0}right) = frac{(x - x_0)^2}{x_0^2} - frac{(x - x_0)^3}{x_0^3} + cdots
$$
So that
$$
lnleft(frac{x}{x_0}right) + frac{1}{2}ln^2left(frac{x}{x_0}right) = frac{x-x_0}{x_0} - frac{(x - x_0)^3}{6x_0^3} + cdots
$$
answered Dec 13 '18 at 15:57
caveraccaverac
14.8k31130
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1 Answer
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1 Answer
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$begingroup$
Taylor expand $ln(x/x_0)$ around $x_0$
$$
lnleft(frac{x}{x_0}right) = frac{x - x_0}{x_0} - frac{(x - x_0)^2}{2x_0^2} + frac{(x - x_0)^3}{3x_0^3} + cdots
$$
Similarly
$$
ln^2left(frac{x}{x_0}right) = frac{(x - x_0)^2}{x_0^2} - frac{(x - x_0)^3}{x_0^3} + cdots
$$
So that
$$
lnleft(frac{x}{x_0}right) + frac{1}{2}ln^2left(frac{x}{x_0}right) = frac{x-x_0}{x_0} - frac{(x - x_0)^3}{6x_0^3} + cdots
$$
answered Dec 13 '18 at 15:57
caveraccaverac
14.8k31130
$endgroup$
add a comment |
$begingroup$
Taylor expand $ln(x/x_0)$ around $x_0$
$$
lnleft(frac{x}{x_0}right) = frac{x - x_0}{x_0} - frac{(x - x_0)^2}{2x_0^2} + frac{(x - x_0)^3}{3x_0^3} + cdots
$$
Similarly
$$
ln^2left(frac{x}{x_0}right) = frac{(x - x_0)^2}{x_0^2} - frac{(x - x_0)^3}{x_0^3} + cdots
$$
So that
$$
lnleft(frac{x}{x_0}right) + frac{1}{2}ln^2left(frac{x}{x_0}right) = frac{x-x_0}{x_0} - frac{(x - x_0)^3}{6x_0^3} + cdots
$$
answered Dec 13 '18 at 15:57
caveraccaverac
14.8k31130
$endgroup$
add a comment |
$begingroup$
Taylor expand $ln(x/x_0)$ around $x_0$
$$
lnleft(frac{x}{x_0}right) = frac{x - x_0}{x_0} - frac{(x - x_0)^2}{2x_0^2} + frac{(x - x_0)^3}{3x_0^3} + cdots
$$
Similarly
$$
ln^2left(frac{x}{x_0}right) = frac{(x - x_0)^2}{x_0^2} - frac{(x - x_0)^3}{x_0^3} + cdots
$$
So that
$$
lnleft(frac{x}{x_0}right) + frac{1}{2}ln^2left(frac{x}{x_0}right) = frac{x-x_0}{x_0} - frac{(x - x_0)^3}{6x_0^3} + cdots
$$
answered Dec 13 '18 at 15:57
caveraccaverac
14.8k31130
$endgroup$
Taylor expand $ln(x/x_0)$ around $x_0$
$$
lnleft(frac{x}{x_0}right) = frac{x - x_0}{x_0} - frac{(x - x_0)^2}{2x_0^2} + frac{(x - x_0)^3}{3x_0^3} + cdots
$$
Similarly
$$
ln^2left(frac{x}{x_0}right) = frac{(x - x_0)^2}{x_0^2} - frac{(x - x_0)^3}{x_0^3} + cdots
$$
So that
$$
lnleft(frac{x}{x_0}right) + frac{1}{2}ln^2left(frac{x}{x_0}right) = frac{x-x_0}{x_0} - frac{(x - x_0)^3}{6x_0^3} + cdots
$$
answered Dec 13 '18 at 15:57
caveraccaverac
14.8k31130
answered Dec 13 '18 at 15:57
caveraccaverac
14.8k31130
answered Dec 13 '18 at 15:57
caveraccaverac
14.8k31130
answered Dec 13 '18 at 15:57
caveraccaverac
14.8k31130
14.8k31130
add a comment |
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