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To prove the statement, simply calculate XD = DX entrywise. One gets xijdj =dixij for every i ≠ j. Your proof is wrong and incomplete. While the diagonal entries of D are distinct, one of them may be zero. So, one of the "eigenvector" you produce may be the zero vector. In addition, you said " X = DXD−1 hence X is diagonal".
28. The symbol. dy dx. means the derivative of y with respect to x. If y = f(x) is a function of x, then the symbol is defined as. dy dx =limh→0 f(x + h) − f(x) h. and this is is (again) called the derivative of y or the derivative of f. Note that it again is a function of x in this case. Note that we do not here define this as dy divided ...
In my AI textbook there is this paragraph, without any explanation. The sigmoid function is defined as follows $$\sigma (x) = \frac{1}{1+e^{-x}}.$$ This function is easy to differentiate
In our case f(x) = xTAx and σ(t) = x + th, f ′ (x; h) = lim t → 0 (x + th)TA(x + th) − xTAx t f ′ (x; h) = xTAh + hTAx. Since A is symmetric and we have the follwing: f ′ (x; h) = xTAh + hTAx = xTAh + xTATh f ′ (x; h) = xT(A + AT)h. So the differential/gradient is simply 2xTA. f ′ (x; h) = 2xTAh. Share. Cite.
Actually in MathTrain's definition, dx d does describe a mapping, namely inv ∘ d dx ∘ inv where inv: (x ↦ f(x)) ↦ (x ↦ 1 f (x)). And while inv is not a linear function, applying it twice "cancels" its nonlinearity out (well, except for the factor 0, but that could be defined by continuity), so dx d is even linear, hence an operator.
As a somewhat simple-minded start, $\begin{align} &\int_0^{s^{-1}(1-s(0))} \left(s(0)[s^{-1}(1-s(0))]+\int_0^{\theta} (1-s^{-1}(1-s(x)))dx\right) d \theta +\int_{s ...
3. The integral which you describe has no closed form which is to say that it cannot be expressed in elementary functions. For example, you can express ∫x2dx in elementary functions such as x3 3 + C. However, the indefinite integral from (−∞, ∞) does exist and it is π−−√ so explicitly: ∫∞ −∞e−x2 = π−−√. Note the ...
and then I tried substituting: t = sinxcosx and got ∫ tdt 2(1 − 2t2)√1 − 4t2. Another way would maybe be to make two integrals: ∫ 1 sin4x + cos4xdx = ∫ 1 (1 − √2sinxcosx)(1 + √2sinxcosx) dx = 1 2∫ 1 1 − √2sinxcosxdx + 1 2∫ 1 1 + √2sinxcosxdx. ... and again I tried t = tanx 2 (4th degree polynomial) and t = √ ...
Given the straight line boundary off center, there is no advantage using the polar coordinates. Instead, integrate as follows, A = ∫5 − 4∫√25 − y2 5 − y 3 xydxdy. where the lower and upper limits -4 and 5 are the y -intersections of the line with the circle. The x -integration leads to. A = 1 2∫5 − 4y(25 − y2 − (5 − y)2 9 ...
The $\Sigma$ sign is a sigma and stands for "sum". In an integral you take the limit as $\delta x$ goes to zero. So we replace the sigma with another type of s: $\int$. And the $\delta$ gets changed to a d. So it is now written: $\int f (x) dx $. and it is the "integral of f (x) with respect to x". But the dx doesn't mean anything on it's own.