Recent content by nizi

I appreciate your thought-provoking response very much. My question has been answered. What I was particularly concerned about is whether I can use the term "i.i.d." here, even thought ##x_n## has certain units, e.g. kg. This is because the units of ##\beta_0## and ##\beta_1## are different in...

This time I target the following two-class Bayesian logistic regression as statistical models. $$y_n \sim \mathrm{Bernoulli}(q_n)$$ $$q_n = \sigma (\beta_0 + x_n \beta_1)$$ where ##n## is the index of the data and ##\sigma## is the logistic function. Since I assume ##\beta_0 + x_n \beta_1##...

I apologize for the delay. I sincerely appreciate your kind remarks. ##\{ \frac { \partial }{ \partial x }, \frac { \partial }{ \partial y } \}## is the basis of the tangent space, which I studied in linear algebra. WWGD-san, like fresh_42-san, has given me another perspective of partial...

I apologize for the delay. I sincerely appreciate your kind remarks. I checked the link, and although I have heard of vector bundle, Lie group, etc., the level of mathematics is too high for me to decipher. I am sorry for the trouble you have gone to in teaching me this. I have dabbled in...

I apologize for the delay. I sincerely appreciate your kind remarks. Since ##g## is known, you are saying that I should differentiate by ##t## in advance before notating the equation. Indeed, my equation notation gives the impression that ##g##, like ##f##, is an unknown function to be obtained...

Thank you all for your kind responses. I am currently out of the country on a business trip, so please give me some time to respond. I apologize for the delay in responding to your question.

Thank you for your detailed response, fresh_42-san (I am Japanese, and in Japanese, -san is added after the name as an honorific title.). > ##\frac { \partial g} { \partial t}## would be misleading because everyone would search for the other variables I completely agree with you. By the way...

Some may say that ##\frac{ \partial g }{ \partial t }## is correct because it is a term in a partial differential equation, but since ##g## is a one variable function with ##t## only, I think ##\frac{ dg }{ dt }## is correct according to the original usage of the derivative and partial...

It's a clue to change the equation so that it keeps the equality. For example trying premultiplying the inverse of A...

I believe it may help you understand the material time derivative and spatial time derivative. The material time derivative of the spatial field is in the following by use of the chain rule. [FONT="Times New Roman"] \frac{Df\left(\textbf{x},t\right)}{Dt}...

\frac{dy}{dx} = 3 x^2 + 1 using implicit differentiaion \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} = \frac{1}{3 x^2 + 1} accordingly replacing x by y \frac{dy}{dx} = \frac{1}{3 y^2 + 1} i.e. \frac{dy}{dx} = \frac{1}{3 \left( {x^3 + x } \right)^2 + 1} then you have only to substituting x=2 .

may I suggest this? A(S) = \int\int_S dA = \int_u \int_v \left| { \frac{\partial \mathop r\limits^ \to}{\partial u} \times \frac{\partial \mathop r\limits^ \to}{\partial v} } \right| du dv now considering spherical coordinates, and representing the integral surface with respect to...

using the second equation, \left( { x \times u } \right) \times u = v \times u ① u \left( { x \cdot u } \right) - x \left( { u \cdot u } \right) = v \times u using the first equation and developping ② x = \frac{1}{\left| u \right|} u - \frac{1}{\left| u \right|^{2}} v \times u hence x is...

b)use the rationalization of the numerator. n \left( { \sqrt{1+n^{2}}-n } \right) = \frac{n \left( { 1+n^{2}-n^{2} } \right)}{\sqrt{1+n^{2}}+n} c)watch the given hint carefully! \left( { \frac{3}{2} } \right)^{2n} < \left( { \frac{2n+1}{n+1} } \right)^{2n} you'd better prove the given...

Neglecting the given attempt, I put y = x^{x^{x}} z = x^{x} and develop as follows. \ln y = \ln x^{x^{x}} = z \ln x \frac{y'}{y} = z' \ln x + z \frac{1}{x} here I calculate the differentiation of z z = x^{x} \ln z = x \ln x \frac{z'}{z} = \ln x + x...