derivatives – How is the notation in this Hessian Matrix calculated?

I’m sorry I’m not sure how to word this question. I’ve returned to school after a long break where I was working full time. In school I did calculus, but most of it seems to have left me.

I have started schooling again in data science and I am trying to compute a Hessian Matrix for a simple function. The function is:

$$f(x_1, x_2) = (x_1-1)^2 + 100(x_1^2-x_2)^2$$

I have calculated the gradient vector by taking the first order derivative

$$Delta f(x_1, x_2) = begin{bmatrix}
2(x_1-1) + 400x_1(x_1^2-x_2) \
-200(x_1^2-x_2)
end{bmatrix}$$

In attempting to calculate the Hessian Matrix I am confused by the notation of entry 1,2 and 2,1:

$$ Delta^2f(x_1, x_2) = begin{bmatrix}
frac{delta^2f}{delta x_1^2} & frac{delta^2f}{delta x_2 delta x_1}\
frac{delta^2f}{delta x_1 delta x_2} & frac{delta^2f}{delta x_2^2} \
end{bmatrix}$$

For entry (1,1) and (2,2), I just retake the derivative of the above gradient vector

$$ frac{delta^2f}{delta x_1^2} = frac{delta f}{delta x_1} (2(x_1-1) + 400x_1(x_1^2-x_2)) = 1200x^2-400x_2+2 $$

and

$$ frac{delta^2f}{delta x_2^2} = frac{delta f}{delta x_2} ( -200x_1^2+200x_2 ) = 200 $$

Therefore the matrix as it stands is:

$$ Delta^2f(x_1, x_2) = begin{bmatrix}
1200x^2-400x_2+2 & frac{delta^2f}{delta x_2 delta x_1}\
frac{delta^2f}{delta x_1 delta x_2} & 200 \
end{bmatrix}$$

How would I go about calculating $$ frac{delta^2f}{delta x_2 delta x_1} and frac{delta^2f}{delta x_1 delta x_2} $$?

Thank you for your time

calculus and analysis – Derivative with respect to a variable is leaving unevaluated derivatives of functions

I have the following

Inactive(Log)(Sqrt(
 E^(-0.96 t - 0.96 Conjugate(t))
   Conjugate(
   Cosh((sqr t)/2) - (1.76 Sinh((sqr t)/2))/sqr) (Cosh((sqr t)/2) - (
    1.76 Sinh((sqr t)/2))/sqr)))

Removing inactivate, upon taking the derivative w.r.t t, I end up with

E^(0.96 t + 
    0.96 Conjugate(t)) (E^(-0.96 t - 0.96 Conjugate(t))
       Conjugate(
       Cosh((sqr t)/2) - (1.76 Sinh((sqr t)/2))/
        sqr) (-0.88 Cosh((sqr t)/2) + 1/2 sqr Sinh((sqr t)/2)) + 
     E^(-0.96 t - 0.96 Conjugate(t))
       Conjugate(
       Cosh((sqr t)/2) - (1.76 Sinh((sqr t)/2))/
        sqr) (Cosh((sqr t)/2) - (1.76 Sinh((sqr t)/2))/sqr) (-0.96 - 
        0.96 Derivative(1)(Conjugate)(t)) + 
     E^(-0.96 t - 
       0.96 Conjugate(t)) (Cosh((sqr t)/2) - (1.76 Sinh((sqr t)/2))/
        sqr) (-0.88 Cosh((sqr t)/2) + 
        1/2 sqr Sinh((sqr t)/2)) Derivative(1)(Conjugate)(
       Cosh((sqr t)/2) - (1.76 Sinh((sqr t)/2))/sqr)))/(4 Conjugate(
    Cosh((sqr t)/2) - (1.76 Sinh((sqr t)/2))/sqr) (Cosh((sqr t)/2) - (
     1.76 Sinh((sqr t)/2))/sqr)

This gives me 54.1743 (-0.0204715 - 0.0204715 Derivative(1)(Conjugate)(0.177417) + 0.00461473 (-0.96 - 0.96 Derivative(1)(Conjugate)(1)))
when a value for t is passed in.
As you can see, the issue is the unevaluated D(conjugate(t),t), which prevents me from obtaining a plottable value. I have attempted to make use of Inactivate and activate, as other posts with a similar issue have mentioned, but this doesn’t seem to be addressing the issue.

As I see it, I need all derivatives w.r.t t to be evaluated before the replace all comes into play for any value of t, which I thought inactivate and activate would allow. But this does not seem to be the case, as I just get the value of t is not a variable, which makes sense given its trying then to take the derivative with respect to a number. What am I missing here?

Edit: I will also say that there are, obviously, a few more variable passed in here that are not mentioned, but since they are not causing an issue with the derivation, I have imitted them from the code.

derivatives – Rolles mean value theorem

Let f: R→ R be such that f”'(x) exist. Suppose f (a) = f'(a) = f(b) = f'(b) = 0 for some a <b, then

(A) f(c)=0 for some c € R

(C) f(x) > 0 for all x

(B) f(x) is never zero

(D) f(x) <0 for all x

I know mean value theorem upto just ist derivative, but it’s a 3rd level derivative, so I don’t know where to start from…

Please help…

calculus – Different (Non-Equivalent) Derivatives for the Same Function

$f(x) = left(x^2(3-x)right)^frac{1}{3}$

  1. $f(x) = x^frac{2}{3}(3-x)^frac{1}{3} \
    implies f'(x) = frac{2}{3}x^{-frac{1}{3}}(3-x)^{-frac{1}{3}} – frac{1}{3}x^frac{2}{3}(3-x)^{-frac{2}{3}} $

This is essentially using the quotient-rule (and a little bit of chain-rule).

  1. $f(x) = (3x^2 – x^3)^frac{1}{3} \
    implies f'(x) = (3x^2 – x^3)^{-frac{2}{3}}(2x-x^2)$

This was essentially using the chain-rule.

Even if I expand out the brackets, and simplify, I am not able to get one into the form of the other. They seem, apparently, to be different answers altogether!

Have I made a serious blunder somewhere?

media – Is there an easy way to get the URLs for image derivatives?

I’m working with complex image styles, and I’d like to be able to load all the different derivatives quickly to check they’re being processed correctly.

But I don’t see how to get the links easily. The best I can come up with is opening the node in JSONAPI and, with the patch from https://www.drupal.org/project/drupal/issues/2825812, getting the URLs to derivatives from that.

It there something easier?

Back in the D6 days, Image module showed the links to all the sizes on the image node itself.

cv.complex variables – Smooth functions with vanishing normal derivatives

Let $B_1 := {z ∈ C : |z| le 1}$, and let $C_0(B_1,mathbb C)$ be the space of continuous complex-valued functions on $B_1$ equipped with the uniform convergence topology.

How to show that, the set of smooth functions $f: B_1 rightarrow mathbb C$ whose normal derivative vanishes along the boundary ∂$B_1$ are dense in $C_0(B_1,mathbb C)$?

I tried to use the Stone-Weierstrass theorem. Couldn’t make much progress. Kindly share your views. Thank you.

derivatives – Prove that f(x) is differentiable on (0, inf) and not differentiable at x=0

Assume that e^x is differentiable on R, prove that
f(x)= x/(1+e^(1/x)) , x is not 0
f(x)= 0, x=0
is differentiable on (0,inf). Is f differentiable at x=0?

What I did is i use definition of derivation
lim h->0 (f(h)-f(0))/h and then it became lim h->0 1/(1+e^(1/h)) which it doesn’t exist.
So I guess it is not differentiable at x=0.
But how to prove that f(x) is differentiable at (0, inf) ?
And are my steps correct?
Thank you in advance.

dg.differential geometry – Writing a Taylor series with covariant derivatives (connections)?

A connection of a vector bundle $E$ on a manifold $M$ is a map $d_E: Omega^0(E) to Omega^1(E)$ that extends uniquely to a map $d_E: Omega^p(E) to Omega^{p+1}(E)$ while satisfying
$$
d_E(omega otimes s) = domega otimes s + (-1)^p omega wedge d_Es.
$$

The curvature tensor is defined to be $d_E circ d_E.$

Now, in very elementary differential geometry, the curvature (Gaussian or others) can be motivated by second order terms in Taylor series expansion of a parameterisation.

This leads to the question: can we write a Taylor series expansion with $d_E$? That would really help understanding curvature.

a nonstandard differential equation – product of a function and her consecutive derivatives equal 1

For which $ninmathbb{N}$ there exists an interval $I$ and a function $fcolon Itomathbb{R}$, that is $n$ times differentiable and $f(x)cdot f'(x)cdotldotscdot f^{(n)}(x)=1$ for all $xin I$?


Own problem.

For $n=1$ it’s easy to see that $I=(0,infty)$ and $f(x)=2sqrt{x}$ work fine.

It seems to complicate a lot already for $n=2$:
https://www.wolframalpha.com/input/?i=f%28x%29*f%27%28x%29*f%27%27%28x%29%3D1 (I hardly understand this reply).

What tools could crack a problem like this?