Let's solve this step by step. The main concept here is Nesterov Accelerated Gradient (NAG). The trick is that NAG calculates the gradient at a look-ahead position, not at the current weight.

Step 1: Write down the given values

We need to calculate:

$\boxed{{\displaystyle {\mathrm{\nabla }}_{w}L\text{ at the look-ahead position}}}$$\boxed{\nabla_w L \text{ at the look-ahead position}}$

Step 2: Understand the NAG formula

NAG uses two equations:

Velocity update:

$u_t=\beta u_{t-1}+\mathrm{\nabla }L(w_t-\beta u_{t-1})$$u_t=\beta u_{t-1}+\nabla L(w_t-\beta u_{t-1})$

Weight update:

$w_{t+1}=w_t-\eta u_t$$w_{t+1}=w_t-\eta u_t$

Notice something important:

NAG first looks ahead using:

$w_t-\beta u_{t-1}$$w_t-\beta u_{t-1}$

Then calculates the gradient at that position.

Step 3: Calculate the look-ahead weight

We are given:

$w_9=2.5,u_8=0.8$$w_9=2.5,\quad u_8=0.8$

Using:

$w_{10}=w_9-\beta u_8$$w_{10}=w_9-\beta u_8$

Substitute:

$=2.5-(0.9\times 0.8)$$=2.5-(0.9\times0.8)$

$=2.5-0.72$$=2.5-0.72$

$\boxed{{\displaystyle w_{10}=1.78}}$$\boxed{w_{10}=1.78}$

Step 4: Calculate the look-ahead bias

Similarly:

$b_{10}=b_9-\beta u_8$$b_{10}=b_9-\beta u_8$

Substitute:

$=0-(0.9\times 0.8)$$=0-(0.9\times0.8)$

$\boxed{{\displaystyle b_{10}=-0.72}}$$\boxed{b_{10}=-0.72}$

Now we have our look-ahead position:

• Weight = 1.78

• Bias = -0.72

Step 5: Calculate the sigmoid output

Remember the sigmoid formula:

$\hat{y}=\frac{1}{1+e^{-z}}$$\hat y=\frac{1}{1+e^{-z}}$

Where:

$z=wx+b$$z=wx+b$

Substitute the look-ahead values:

$z=(1.78\times 0.5)+(-0.72)$$z=(1.78\times0.5)+(-0.72)$

$z=0.89-0.72$$z=0.89-0.72$

$\boxed{{\displaystyle z=0.17}}$$\boxed{z=0.17}$

Now apply sigmoid:

$\hat{y}=\frac{1}{1+e^{-0.17}}$$\hat y=\frac{1}{1+e^{-0.17}}$

$\boxed{{\displaystyle \hat{y}\approx 0.5424}}$$\boxed{\hat y\approx0.5424}$

Step 6: Calculate the gradient

This is the most important step.

For a sigmoid neuron with the given MSE loss:

$\boxed{{\displaystyle \frac{\partial L}{\partial w}=(\hat{y}-y)\hat{y}(1-\hat{y})x}}$$\boxed{ \frac{\partial L}{\partial w} =(\hat y-y)\hat y(1-\hat y)x }$

We know:

• $\hat{y}=0.5424$$\hat y=0.5424$

• $y=1$$y=1$

• $x=0.5$$x=0.5$

Substitute:

$=(0.5424-1)(0.5424)(1-0.5424)(0.5)$$=(0.5424-1)(0.5424)(1-0.5424)(0.5)$

Solve each part:

$=(-0.4576)(0.5424)(0.4576)(0.5)$$=( -0.4576)(0.5424)(0.4576)(0.5)$

$\boxed{{\displaystyle \frac{\partial L}{\partial w}\approx -0.0569}}$$\boxed{\frac{\partial L}{\partial w}\approx-0.0569}$

Final answer: −0.057 ≈ −0.056

What to remember for the exam

The entire question follows just these 4 steps:

Important exam trick:

Don't confuse these two:

• Normal Gradient Descent: Calculate gradient at the current weight.

• Nesterov (NAG): Calculate gradient at the look-ahead weight.

And one more thing: The question asks for the gradient of the look-ahead value, NOT the next updated weight. Therefore, you don't need to calculate the final velocity or update the weight using learning rate $0.1$$0.1$.

Let's modify the original question so that instead of asking for the gradient of the look-ahead value, it asks for the next weight and bias after the NAG update.

Revised question (exam format)

A sigmoid neuron has a single input $x=0.5$$x=0.5$ and the true output is $y=1$$y=1$. The initial weight and bias are $w_0=5$$w_0=5$ and $b_0=-5$$b_0=-5$.

The model uses Nesterov Accelerated Gradient Descent (NAG) with momentum parameter $\beta =0.9$$\beta=0.9$ and learning rate $\eta =0.1$$\eta=0.1$. The loss function is:

$L(w)=\frac{1}{2}(\hat{y}-y{)}^2$$L(w)=\frac{1}{2}(\hat y-y)^2$

After 10 iterations (starting from iteration zero), the parameter values at iteration $t=9$$t=9$ are:

$u_8=0.8,w_9=2.5,b_9=0$$u_8=0.8,\quad w_9=2.5,\quad b_9=0$

Calculate the next updated weight and bias after one NAG iteration.

Step-by-step solution

Step 1: Calculate look-ahead weight

NAG formula:

$w_{lookahead}=w_9-\beta u_8$$w_{lookahead}=w_9-\beta u_8$

$=2.5-(0.9\times 0.8)$$=2.5-(0.9\times0.8)$

$\boxed{{\displaystyle w_{lookahead}=1.78}}$$\boxed{w_{lookahead}=1.78}$

Step 2: Calculate look-ahead bias

$b_{lookahead}=b_9-\beta u_8$$b_{lookahead}=b_9-\beta u_8$

$=0-(0.9\times 0.8)$$=0-(0.9\times0.8)$

$\boxed{{\displaystyle b_{lookahead}=-0.72}}$$\boxed{b_{lookahead}=-0.72}$

Step 3: Calculate predicted output

First calculate:

$z=wx+b$$z=wx+b$

$z=(1.78\times 0.5)-0.72$$z=(1.78\times0.5)-0.72$

$z=0.17$$z=0.17$

Apply sigmoid:

$\hat{y}=\frac{1}{1+e^{-0.17}}$$\hat y=\frac{1}{1+e^{-0.17}}$

$\boxed{{\displaystyle \hat{y}=0.5424}}$$\boxed{\hat y=0.5424}$

Step 4: Calculate gradient

For sigmoid with MSE:

$\mathrm{\nabla }L=(\hat{y}-y)\hat{y}(1-\hat{y})x$$\nabla L=(\hat y-y)\hat y(1-\hat y)x$

Substitute:

$=(0.5424-1)(0.5424)(1-0.5424)(0.5)$$=(0.5424-1)(0.5424)(1-0.5424)(0.5)$

$\boxed{{\displaystyle \mathrm{\nabla }L=-0.0569}}$$\boxed{\nabla L=-0.0569}$

Step 5: Calculate new velocity

$u_9=\beta u_8+\mathrm{\nabla }L$$u_9=\beta u_8+\nabla L$

$=0.9(0.8)-0.0569$$=0.9(0.8)-0.0569$

$\boxed{{\displaystyle u_9=0.6631}}$$\boxed{u_9=0.6631}$

Step 6: Calculate updated weight

$w_{10}=w_9-\eta u_9$$w_{10}=w_9-\eta u_9$

$=2.5-(0.1\times 0.6631)$$=2.5-(0.1\times0.6631)$

$\boxed{{\displaystyle w_{10}=2.43369}}$$\boxed{w_{10}=2.43369}$

Step 7: Calculate updated bias

$b_{10}=b_9-\eta u_9$$b_{10}=b_9-\eta u_9$

$=0-(0.1\times 0.6631)$$=0-(0.1\times0.6631)$

$\boxed{{\displaystyle b_{10}=-0.06631}}$$\boxed{b_{10}=-0.06631}$

Final answer

Updated weight

2.43369

Updated bias

-0.06631

What to remember for the exam

One important correction: In the original question, the look-ahead bias was calculated using the same velocity $u_8$$u_8$, as shown in the provided solution. We have followed that convention here.

Also, the given answer $-0.056$$-0.056$ is rounded; using more accurate sigmoid calculations gives approximately $-0.0569$$-0.0569$.