Derivation on equation with summation symbolDerivatives - Show equalityHow do you read a partial differential equation?How many solutions for an equation with simple restrictionsCompute derivative with respect to a matrixPartial differentiating implicitlyDerivative of a vector by a matrixDerivation of derivative of multivariate Gaussian w.r.t. covariance matrixHow to derive the substituted Partial Differential EquationHow to anti-derivative the partial derivative of a multivariate function across all variables to end up with the original functionPartial derivative of a function within a function, lambda calculus
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Derivation on equation with summation symbol
Derivatives - Show equalityHow do you read a partial differential equation?How many solutions for an equation with simple restrictionsCompute derivative with respect to a matrixPartial differentiating implicitlyDerivative of a vector by a matrixDerivation of derivative of multivariate Gaussian w.r.t. covariance matrixHow to derive the substituted Partial Differential EquationHow to anti-derivative the partial derivative of a multivariate function across all variables to end up with the original functionPartial derivative of a function within a function, lambda calculus
$begingroup$
I have an equation where:
$J_t = frac12sum(haty-y)^2$ where $haty$ is a vector of $hat y_1,hat y_2,...hat y_k$.
If i were to differentiate $J_t$ in terms of $y_1$, is it right to assume that the $sum$ symbol doesn't affect the differentiation process?
Hence,
$fracpartial J_tpartial y_1 = (hat y_1 - y)$
I would like to know if the above is a correct way to differentiate?
discrete-mathematics partial-derivative
$endgroup$
add a comment |
$begingroup$
I have an equation where:
$J_t = frac12sum(haty-y)^2$ where $haty$ is a vector of $hat y_1,hat y_2,...hat y_k$.
If i were to differentiate $J_t$ in terms of $y_1$, is it right to assume that the $sum$ symbol doesn't affect the differentiation process?
Hence,
$fracpartial J_tpartial y_1 = (hat y_1 - y)$
I would like to know if the above is a correct way to differentiate?
discrete-mathematics partial-derivative
$endgroup$
1
$begingroup$
Yes, if you pull the first addendum it would be this: $frac12(haty_1-y)^2+sumfrac12(haty_2-y)^2$ the second part of the sum, as you can see, is a constant in $haty_1$.
$endgroup$
– Bertrand Wittgenstein's Ghost
Mar 30 at 4:11
add a comment |
$begingroup$
I have an equation where:
$J_t = frac12sum(haty-y)^2$ where $haty$ is a vector of $hat y_1,hat y_2,...hat y_k$.
If i were to differentiate $J_t$ in terms of $y_1$, is it right to assume that the $sum$ symbol doesn't affect the differentiation process?
Hence,
$fracpartial J_tpartial y_1 = (hat y_1 - y)$
I would like to know if the above is a correct way to differentiate?
discrete-mathematics partial-derivative
$endgroup$
I have an equation where:
$J_t = frac12sum(haty-y)^2$ where $haty$ is a vector of $hat y_1,hat y_2,...hat y_k$.
If i were to differentiate $J_t$ in terms of $y_1$, is it right to assume that the $sum$ symbol doesn't affect the differentiation process?
Hence,
$fracpartial J_tpartial y_1 = (hat y_1 - y)$
I would like to know if the above is a correct way to differentiate?
discrete-mathematics partial-derivative
discrete-mathematics partial-derivative
asked Mar 30 at 3:42
MaxxxMaxxx
1255
1255
1
$begingroup$
Yes, if you pull the first addendum it would be this: $frac12(haty_1-y)^2+sumfrac12(haty_2-y)^2$ the second part of the sum, as you can see, is a constant in $haty_1$.
$endgroup$
– Bertrand Wittgenstein's Ghost
Mar 30 at 4:11
add a comment |
1
$begingroup$
Yes, if you pull the first addendum it would be this: $frac12(haty_1-y)^2+sumfrac12(haty_2-y)^2$ the second part of the sum, as you can see, is a constant in $haty_1$.
$endgroup$
– Bertrand Wittgenstein's Ghost
Mar 30 at 4:11
1
1
$begingroup$
Yes, if you pull the first addendum it would be this: $frac12(haty_1-y)^2+sumfrac12(haty_2-y)^2$ the second part of the sum, as you can see, is a constant in $haty_1$.
$endgroup$
– Bertrand Wittgenstein's Ghost
Mar 30 at 4:11
$begingroup$
Yes, if you pull the first addendum it would be this: $frac12(haty_1-y)^2+sumfrac12(haty_2-y)^2$ the second part of the sum, as you can see, is a constant in $haty_1$.
$endgroup$
– Bertrand Wittgenstein's Ghost
Mar 30 at 4:11
add a comment |
0
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$begingroup$
Yes, if you pull the first addendum it would be this: $frac12(haty_1-y)^2+sumfrac12(haty_2-y)^2$ the second part of the sum, as you can see, is a constant in $haty_1$.
$endgroup$
– Bertrand Wittgenstein's Ghost
Mar 30 at 4:11