Mixing
gradient for sum on 100 variables analytically
$begingroup$
I need to minimize a function with x in $R^{100}$ and $a_i$ is a given vector.
The function itself is:
$sum_{i=1}^{500} log(1- a_i^t x) - sum_{i=1}^{100} log(1-x_i^2) $
The first thing I thought about was to separate the sums:
$sum_{i=1}^{500} log(1- a_i^t x)$
Then remove the sum and do derivative of $log(1- a_i^t x)$ then add the sum back to it. Same thing is done for the other sum.
At the end I got:
$sum_{i=1}^{500} frac {- a_i^t }{ln10(1 - a_i^tx)} - sum_{i=1}^{100} frac{-2x_i}{ln10(1-x_i^2)} $
You can see the whole opration on the picture I attached.
How I got this derivative
My questions is that now I calculated the derivative with respect to x which is in $R^{100}$
I researched the gradient and watched a few videos about it , it looks something like :
grad = $$
begin{matrix}
df(x,y,z)/dx \
df(x,y,z)/dy \
df(x,y,z)/dz \
end{matrix}
$$
However, with the work I previously did I am very confused now. How do I get the gradient in this case?
multivariable-calculus gradient-descent
asked Feb 2 at 9:40
Sara KatSara Kat
203
$endgroup$
add a comment |
$begingroup$
I need to minimize a function with x in $R^{100}$ and $a_i$ is a given vector.
The function itself is:
$sum_{i=1}^{500} log(1- a_i^t x) - sum_{i=1}^{100} log(1-x_i^2) $
The first thing I thought about was to separate the sums:
$sum_{i=1}^{500} log(1- a_i^t x)$
Then remove the sum and do derivative of $log(1- a_i^t x)$ then add the sum back to it. Same thing is done for the other sum.
At the end I got:
$sum_{i=1}^{500} frac {- a_i^t }{ln10(1 - a_i^tx)} - sum_{i=1}^{100} frac{-2x_i}{ln10(1-x_i^2)} $
You can see the whole opration on the picture I attached.
How I got this derivative
My questions is that now I calculated the derivative with respect to x which is in $R^{100}$
I researched the gradient and watched a few videos about it , it looks something like :
grad = $$
begin{matrix}
df(x,y,z)/dx \
df(x,y,z)/dy \
df(x,y,z)/dz \
end{matrix}
$$
However, with the work I previously did I am very confused now. How do I get the gradient in this case?
multivariable-calculus gradient-descent
asked Feb 2 at 9:40
Sara KatSara Kat
203
$endgroup$
$begingroup$
$x$ is a vector. This is not Calc I, where you just differentiate with respect to it. You need to differentiate with respect to each of the $x_j$ that make up $x$ instead.
$endgroup$
– Paul Sinclair
Feb 2 at 20:18
add a comment |
$begingroup$
I need to minimize a function with x in $R^{100}$ and $a_i$ is a given vector.
The function itself is:
$sum_{i=1}^{500} log(1- a_i^t x) - sum_{i=1}^{100} log(1-x_i^2) $
The first thing I thought about was to separate the sums:
$sum_{i=1}^{500} log(1- a_i^t x)$
Then remove the sum and do derivative of $log(1- a_i^t x)$ then add the sum back to it. Same thing is done for the other sum.
At the end I got:
$sum_{i=1}^{500} frac {- a_i^t }{ln10(1 - a_i^tx)} - sum_{i=1}^{100} frac{-2x_i}{ln10(1-x_i^2)} $
You can see the whole opration on the picture I attached.
How I got this derivative
My questions is that now I calculated the derivative with respect to x which is in $R^{100}$
I researched the gradient and watched a few videos about it , it looks something like :
grad = $$
begin{matrix}
df(x,y,z)/dx \
df(x,y,z)/dy \
df(x,y,z)/dz \
end{matrix}
$$
However, with the work I previously did I am very confused now. How do I get the gradient in this case?
multivariable-calculus gradient-descent
asked Feb 2 at 9:40
Sara KatSara Kat
203
$endgroup$
I need to minimize a function with x in $R^{100}$ and $a_i$ is a given vector.
The function itself is:
$sum_{i=1}^{500} log(1- a_i^t x) - sum_{i=1}^{100} log(1-x_i^2) $
The first thing I thought about was to separate the sums:
$sum_{i=1}^{500} log(1- a_i^t x)$
Then remove the sum and do derivative of $log(1- a_i^t x)$ then add the sum back to it. Same thing is done for the other sum.
At the end I got:
$sum_{i=1}^{500} frac {- a_i^t }{ln10(1 - a_i^tx)} - sum_{i=1}^{100} frac{-2x_i}{ln10(1-x_i^2)} $
You can see the whole opration on the picture I attached.
How I got this derivative
My questions is that now I calculated the derivative with respect to x which is in $R^{100}$
I researched the gradient and watched a few videos about it , it looks something like :
grad = $$
begin{matrix}
df(x,y,z)/dx \
df(x,y,z)/dy \
df(x,y,z)/dz \
end{matrix}
$$
However, with the work I previously did I am very confused now. How do I get the gradient in this case?
multivariable-calculus gradient-descent
multivariable-calculus gradient-descent
asked Feb 2 at 9:40
Sara KatSara Kat
203
asked Feb 2 at 9:40
Sara KatSara Kat
203
edited Feb 2 at 20:42
Sara Kat
asked Feb 2 at 9:40
Sara KatSara Kat
203
asked Feb 2 at 9:40
Sara KatSara Kat
203
asked Feb 2 at 9:40
Sara KatSara Kat
203
203
$begingroup$
$x$ is a vector. This is not Calc I, where you just differentiate with respect to it. You need to differentiate with respect to each of the $x_j$ that make up $x$ instead.
$endgroup$
– Paul Sinclair
Feb 2 at 20:18
add a comment |
$begingroup$
$x$ is a vector. This is not Calc I, where you just differentiate with respect to it. You need to differentiate with respect to each of the $x_j$ that make up $x$ instead.
$endgroup$
– Paul Sinclair
Feb 2 at 20:18
$begingroup$
$x$ is a vector. This is not Calc I, where you just differentiate with respect to it. You need to differentiate with respect to each of the $x_j$ that make up $x$ instead.
$endgroup$
– Paul Sinclair
Feb 2 at 20:18
$begingroup$
$x$ is a vector. This is not Calc I, where you just differentiate with respect to it. You need to differentiate with respect to each of the $x_j$ that make up $x$ instead.
$endgroup$
– Paul Sinclair
Feb 2 at 20:18
add a comment |
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$begingroup$
$x$ is a vector. This is not Calc I, where you just differentiate with respect to it. You need to differentiate with respect to each of the $x_j$ that make up $x$ instead.
$endgroup$
– Paul Sinclair
Feb 2 at 20:18