Suppose we have a function f(x1,x2)=x21+3x2+25𝑓(𝑥1,𝑥2)=𝑥12+3𝑥2+25 which we want to minimize the given function using the gradient descent algorithm. We initialize (x1,x2)=(0,0)(𝑥1,𝑥2)=(0,0). What will be the value of x1𝑥1 after ten updates in the gradient descent process?(Let η𝜂 be 1) 0 -3 −4.5 −3
Question
Suppose we have a function f(x1,x2)=x21+3x2+25𝑓(𝑥1,𝑥2)=𝑥12+3𝑥2+25 which we want to minimize the given function using the gradient descent algorithm. We initialize (x1,x2)=(0,0)(𝑥1,𝑥2)=(0,0). What will be the value of x1𝑥1 after ten updates in the gradient descent process?(Let η𝜂 be 1) 0 -3 −4.5 −3
Solution 1
The gradient descent algorithm is an iterative optimization algorithm for finding the minimum of a function. Here's how you can apply it to the function f(x1, x2) = x1^2 + 3x2 + 25.
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Initialize the values of x1 and x2 to 0.
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Compute the gradient of the function, which is the vector of its partial derivatives. The partial derivative of f with respect to x1 is 2x1, and the partial derivative with respect to x2 is 3.
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Update the values of x1 and x2 by subtracting the gradient times the learning rate (η) from the current values. Since η is 1, the update rule is x1 = x1 - 2x1 and x2 = x2 - 3.
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Repeat steps 2 and 3 for ten iterations.
Let's compute the value of x1 after ten updates:
- After the first update, x1 = 0 - 2*0 = 0.
- After the second update, x1 = 0 - 2*0 = 0.
- ...
- After the tenth update, x1 = 0 - 2*0 = 0.
So, the value of x1 after ten updates in the gradient descent process is 0.
Solution 2
The gradient descent algorithm is an iterative optimization algorithm for finding the minimum of a function. Here's how you can apply it to the function f(x1,x2)=x1^2+3x2+25.
-
Initialize the values of x1 and x2 to 0.
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Compute the gradient of the function, which is the vector of its partial derivatives. The partial derivative of f with respect to x1 is 2x1, and with respect to x2 is 3.
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Update the values of x1 and x2 by subtracting the gradient times the learning rate (η) from the current values. The learning rate is given as 1. So, the update equations are: x1 = x1 - η * (2x1) = x1 - 2x1 x2 = x2 - η * 3 = x2 - 3
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Repeat steps 2 and 3 for ten iterations.
Let's compute the value of x1 after ten updates:
- After the first update: x1 = 0 - 2*0 = 0
- After the second update: x1 = 0 - 2*0 = 0
- ...
As you can see, the value of x1 remains 0 after each update because the gradient (2x1) is also 0. So, after ten updates, the value of x1 is still 0.
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