Linearization using Taylor series
$begingroup$
I'm trying to understand this approximation:
$x sqrt{bigg(1+alphaBig(frac{y}{x}-1Big)bigg)bigg/bigg(1+alphaBig(frac{x}{y}-1Big)bigg)} approx xbigg(1+frac{alpha}{2}Big(frac{y}{x}-1-frac{x}{y}+1Big)bigg)$
and this one:
$Largefrac{(x^3 (1-alpha)+y^3alpha)Big(frac{(1-alpha)}{x}+frac{alpha}{y}Big)}{(x(1-alpha)+yalpha)^2}approx 1+alphafrac{x}{y}bigg(frac{y^2}{x^2}-1bigg)^2$
According to the brief explanation, we derive the approximation using Taylor series linearization.
I'm familiar with Taylor expansion of $f(xpm ah)$, but not with linearization/approximation using Taylor.
I suspect there's some derivation involved, but I'm not too sure.
Can someone help?
taylor-expansion linearization
asked Dec 18 '18 at 20:17
JIM BOYJIM BOY
356
$endgroup$
add a comment |
$begingroup$
I'm trying to understand this approximation:
$x sqrt{bigg(1+alphaBig(frac{y}{x}-1Big)bigg)bigg/bigg(1+alphaBig(frac{x}{y}-1Big)bigg)} approx xbigg(1+frac{alpha}{2}Big(frac{y}{x}-1-frac{x}{y}+1Big)bigg)$
and this one:
$Largefrac{(x^3 (1-alpha)+y^3alpha)Big(frac{(1-alpha)}{x}+frac{alpha}{y}Big)}{(x(1-alpha)+yalpha)^2}approx 1+alphafrac{x}{y}bigg(frac{y^2}{x^2}-1bigg)^2$
According to the brief explanation, we derive the approximation using Taylor series linearization.
I'm familiar with Taylor expansion of $f(xpm ah)$, but not with linearization/approximation using Taylor.
I suspect there's some derivation involved, but I'm not too sure.
Can someone help?
taylor-expansion linearization
asked Dec 18 '18 at 20:17
JIM BOYJIM BOY
356
$endgroup$
$begingroup$
If $f(x) = f(a) + f'(a)(x - a) + f''(a)/2! (x-a)^2 dots $ is the Taylor series of $f(x)$, then the linearization is just the first two terms: $L(x) = f(a) + f'(a)(x-a)$. Note that this is really just the tangent line to $f(x)$ at $x = a$.
$endgroup$
– JavaMan
Dec 18 '18 at 20:20
$begingroup$
I'm still confused, we have three variables, $x$ $y$ and $alpha$. How should I derivate the equation? With respect to $x$, $y$ or $alpha$?
$endgroup$
– JIM BOY
Dec 18 '18 at 20:35
$begingroup$
I believe $alpha$ is the variable that is infinitesimal. $x$ and $y$ can just be seen as some coefficients.
$endgroup$
– MoonKnight
Dec 18 '18 at 21:15
add a comment |
$begingroup$
I'm trying to understand this approximation:
$x sqrt{bigg(1+alphaBig(frac{y}{x}-1Big)bigg)bigg/bigg(1+alphaBig(frac{x}{y}-1Big)bigg)} approx xbigg(1+frac{alpha}{2}Big(frac{y}{x}-1-frac{x}{y}+1Big)bigg)$
and this one:
$Largefrac{(x^3 (1-alpha)+y^3alpha)Big(frac{(1-alpha)}{x}+frac{alpha}{y}Big)}{(x(1-alpha)+yalpha)^2}approx 1+alphafrac{x}{y}bigg(frac{y^2}{x^2}-1bigg)^2$
According to the brief explanation, we derive the approximation using Taylor series linearization.
I'm familiar with Taylor expansion of $f(xpm ah)$, but not with linearization/approximation using Taylor.
I suspect there's some derivation involved, but I'm not too sure.
Can someone help?
taylor-expansion linearization
asked Dec 18 '18 at 20:17
JIM BOYJIM BOY
356
$endgroup$
I'm trying to understand this approximation:
$x sqrt{bigg(1+alphaBig(frac{y}{x}-1Big)bigg)bigg/bigg(1+alphaBig(frac{x}{y}-1Big)bigg)} approx xbigg(1+frac{alpha}{2}Big(frac{y}{x}-1-frac{x}{y}+1Big)bigg)$
and this one:
$Largefrac{(x^3 (1-alpha)+y^3alpha)Big(frac{(1-alpha)}{x}+frac{alpha}{y}Big)}{(x(1-alpha)+yalpha)^2}approx 1+alphafrac{x}{y}bigg(frac{y^2}{x^2}-1bigg)^2$
According to the brief explanation, we derive the approximation using Taylor series linearization.
I'm familiar with Taylor expansion of $f(xpm ah)$, but not with linearization/approximation using Taylor.
I suspect there's some derivation involved, but I'm not too sure.
Can someone help?
taylor-expansion linearization
taylor-expansion linearization
asked Dec 18 '18 at 20:17
JIM BOYJIM BOY
356
asked Dec 18 '18 at 20:17
JIM BOYJIM BOY
356
asked Dec 18 '18 at 20:17
JIM BOYJIM BOY
356
asked Dec 18 '18 at 20:17
JIM BOYJIM BOY
356
asked Dec 18 '18 at 20:17
JIM BOYJIM BOY
356
356
$begingroup$
If $f(x) = f(a) + f'(a)(x - a) + f''(a)/2! (x-a)^2 dots $ is the Taylor series of $f(x)$, then the linearization is just the first two terms: $L(x) = f(a) + f'(a)(x-a)$. Note that this is really just the tangent line to $f(x)$ at $x = a$.
$endgroup$
– JavaMan
Dec 18 '18 at 20:20
$begingroup$
I'm still confused, we have three variables, $x$ $y$ and $alpha$. How should I derivate the equation? With respect to $x$, $y$ or $alpha$?
$endgroup$
– JIM BOY
Dec 18 '18 at 20:35
$begingroup$
I believe $alpha$ is the variable that is infinitesimal. $x$ and $y$ can just be seen as some coefficients.
$endgroup$
– MoonKnight
Dec 18 '18 at 21:15
add a comment |
$begingroup$
If $f(x) = f(a) + f'(a)(x - a) + f''(a)/2! (x-a)^2 dots $ is the Taylor series of $f(x)$, then the linearization is just the first two terms: $L(x) = f(a) + f'(a)(x-a)$. Note that this is really just the tangent line to $f(x)$ at $x = a$.
$endgroup$
– JavaMan
Dec 18 '18 at 20:20
$begingroup$
I'm still confused, we have three variables, $x$ $y$ and $alpha$. How should I derivate the equation? With respect to $x$, $y$ or $alpha$?
$endgroup$
– JIM BOY
Dec 18 '18 at 20:35
$begingroup$
I believe $alpha$ is the variable that is infinitesimal. $x$ and $y$ can just be seen as some coefficients.
$endgroup$
– MoonKnight
Dec 18 '18 at 21:15
$begingroup$
If $f(x) = f(a) + f'(a)(x - a) + f''(a)/2! (x-a)^2 dots $ is the Taylor series of $f(x)$, then the linearization is just the first two terms: $L(x) = f(a) + f'(a)(x-a)$. Note that this is really just the tangent line to $f(x)$ at $x = a$.
$endgroup$
– JavaMan
Dec 18 '18 at 20:20
$begingroup$
If $f(x) = f(a) + f'(a)(x - a) + f''(a)/2! (x-a)^2 dots $ is the Taylor series of $f(x)$, then the linearization is just the first two terms: $L(x) = f(a) + f'(a)(x-a)$. Note that this is really just the tangent line to $f(x)$ at $x = a$.
$endgroup$
– JavaMan
Dec 18 '18 at 20:20
$begingroup$
I'm still confused, we have three variables, $x$ $y$ and $alpha$. How should I derivate the equation? With respect to $x$, $y$ or $alpha$?
$endgroup$
– JIM BOY
Dec 18 '18 at 20:35
$begingroup$
I'm still confused, we have three variables, $x$ $y$ and $alpha$. How should I derivate the equation? With respect to $x$, $y$ or $alpha$?
$endgroup$
– JIM BOY
Dec 18 '18 at 20:35
$begingroup$
I believe $alpha$ is the variable that is infinitesimal. $x$ and $y$ can just be seen as some coefficients.
$endgroup$
– MoonKnight
Dec 18 '18 at 21:15
$begingroup$
I believe $alpha$ is the variable that is infinitesimal. $x$ and $y$ can just be seen as some coefficients.
$endgroup$
– MoonKnight
Dec 18 '18 at 21:15
add a comment |
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$begingroup$
If $f(x) = f(a) + f'(a)(x - a) + f''(a)/2! (x-a)^2 dots $ is the Taylor series of $f(x)$, then the linearization is just the first two terms: $L(x) = f(a) + f'(a)(x-a)$. Note that this is really just the tangent line to $f(x)$ at $x = a$.
$endgroup$
– JavaMan
Dec 18 '18 at 20:20
$begingroup$
I'm still confused, we have three variables, $x$ $y$ and $alpha$. How should I derivate the equation? With respect to $x$, $y$ or $alpha$?
$endgroup$
– JIM BOY
Dec 18 '18 at 20:35
$begingroup$
I believe $alpha$ is the variable that is infinitesimal. $x$ and $y$ can just be seen as some coefficients.
$endgroup$
– MoonKnight
Dec 18 '18 at 21:15