3

Unfortunately, there are 2 common convention in matrix calculus: Jacobian (Numerator) and Gradient (Denominator). For a function $f:\mathbb R^n \to \mathbb R^m$, then Jacobian convention: $\frac{\partial f}{\partial x}$ is represented by the $m\times n$ matrix with entries $\big(\tfrac{\partial f}{\partial x}\big)_{ij} = \tfrac{\partial f_i}{\partial x_j}$ ...


2

$g$ is not convex: consider the points $0,1,N$ where $N$ is a positive integer $>1$. We have $1=\frac 1 N (N)+(1-\frac 1 N) 0$. If $g$ is convex then we would have $g(1) \leq \frac 1 N g(N)+(1-\frac 1 N)g(0)$. If you let $N \to \infty$ in this you get the contradiction that $e \leq 1$. $\log g$ is also not convex. It is concave. So $-\log\, g$ is convex....


2

This is what padding is meant to handle. Indeed, computing convolutions (or correlations, more accurately) near an image edge will always end up "going out of bounds" (since the value at output pixel $(i,j)$ depends on image values within $(i\pm K,j\pm K)$ for a square kernel of size $K$). The question is what to do when an out-of-bounds is hit. One ...


2

This seems to be a combination of very precise data (in the Excel sheet) and very vague questions. Here is one possible approach. If a customer has only $1000$ denomination notes then the delivery boy will on average have to give the customer $500$ in change (in some cases it will be more than this, in other cases less). Out of $40$ deliveries the delivery ...


1

You know that $h_\theta(x)=\theta_1x$. Thus the cost function is $$J(\theta_1)=\frac{1}{2m}\sum_{i=1}^m(h_\theta(x^i)-y^i)^2=\frac{1}{2m}\sum_{i=1}^m(\theta_1x^i -y^i)^2$$ Setting the first derivative equal to $0$. For the derivative we use the chain rule. $$J^{'}(\theta_1)=\frac{1}{m}\sum_{i=1}^m(\theta_1x^i -y^i)\cdot x^i=0$$ I omit the factor $\...


1

The order of the polynomial ($M$) is the highest exponent of $x$ that appears in the expression of the polynomial. For example, $x^2$ is a degree-2 polynomial and $1+3x+3x^2+x^3$ is a degree-3 polynomial


1

$$L(w) = \sum_{i=1}^n \log\left( 1+\exp(-y_iw^Tx_i)\right)$$ \begin{align} \frac{\partial L}{\partial w_j} &= \sum_{i=1}^n \frac{0+\exp(-y_iw^Tx_i)\frac{\partial }{\partial w_j}(-y_iw^Tx_i)}{\left( 1+\exp(-y_iw^Tx_i)\right)} \\ &= \sum_{i=1}^n \frac{\exp(-y_iw^Tx_i)}{\left( 1+\exp(-y_iw^Tx_i)\right)} \cdot \frac{\partial}{\partial w_j}(-y_iw^Tx_i)\\ ...


1

My doubt is what is $hθ(x)$ here? Is it sigmoid function i.e $\frac{1}{1+e^{-z}}$? $h_{\theta}(x)$ is your hypothesis function. Which is given as $$ h_{\theta}(x) = \frac{1}{1+e^{-\theta^{t}x}} $$ So if you sigmoid function $\sigma(z)$ is given as $$ \sigma(z) =\frac{1}{1+e^{-z}} $$ Another doubt is what is $xj$ term here refer to? Is it our input ...


1

Consider an arbitrary collection of $d+1$ points $x_1,\ldots,x_{d+1}$ and consider for each $i$ $\displaystyle t_i:=\max_{1\leq j \leq d+1} (x_j)_i$. There are at most $d$ distinct points on the boundary of $(-\infty,t_1]\times \ldots\times (-\infty,t_d]$. Label these points with $1$ and all the rest with $0$. It is not possible to find an element of $\...


1

Let $M = U\Sigma V^{*}.$ Then what is $M^*M$? Given that $\Sigma$ is diagonal, what can you say about the eigenvalues of $M^*M$? Hint: Eigenvalues are invariant under ...


1

There are many different types of RNNs with different governing equations. In general, they are written with the update rule $$ (a_t,h_t) = f_r(h_{t-1},x_{t-1}|\theta) $$ for which the rule you state is simply a "special case" with $f_r(h_t,x_t|\theta)=h_t + f(h_t,x_t|\theta)$ and ignoring the output-per-timestep $a_t$. The loop comes from iterating the ...


1

Your reasoning, for the most part, is correct. However, as the sigmoid function deals with real numbers in the interval [0, 1], the weights and biases you've chosen are not ideal. Suppose the input was an image of the number 3, in this case, the 3rd (4th if you're counting from 1) neuron in the third layer is > 0.99 while all the other neurons in that layer ...


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