12 votes

derivative of the max function

No, $g(y)$ is not differentiable in general. For example, let $h: \mathbb{R} \to [0, 1]$ be a smooth function with $h(x) = 0$ when $x \leq 0$ and $h(x) = 1$ when $x \geq 1$. (Such a function exists. ...
David Gao's user avatar
  • 4,890
12 votes

Interpreting a notation in calculus of variations (differentiating with respect to a derivative)

Consider a functional $J[y]$ defined by: $$J[y] = \int_a^b F(x, y, y') dx \tag{1}$$ I think it will be helpful to remind you right away that your notation implies that you have already agreed to ...
hft's user avatar
  • 556
9 votes

Are differentials on their own in stochastic calculus just an abuse of notation?

Differentials in stochastic calculus have very precise interpretations. The stochastic differential equation in differential form $$dX_t = \mu (t, X_t) dt + \sigma (t, X_t) dB_t$$ translates precisely ...
Jose Avilez's user avatar
  • 12.9k
6 votes
Accepted

How to calculate $\int(yy'' + (y')^2)\,dx$?

Even if you couldn't have noticed the $vu' + uv'$, you could still do through integration Integrating through by-parts: \begin{align*} \int yy'' + (y')^2 \, dx &= \int yy'' \, dx + \int (y')^2 \, ...
OpateItZOpatoOpate's user avatar
5 votes
Accepted

$|f(x)-f(y)| < |g(x) - g(y)|$ when $|f'| < |g'|$

Fix $x<y$. By Cauchy’s mean-value theorem (which is completely elementary, and is simply a “clever” application of the usual mean-value theorem), there is a number $\xi\in (x,y)$ such that \begin{...
peek-a-boo's user avatar
  • 55.9k
5 votes
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Finding an increasing function on $\mathbb{R}^+$ such that $f(0)=0$, $\lim_{x\to\infty}f(x)=1$, and $f\circ\left[(f')^{-1}\right]=\frac1{1+x}$

Let $\phi(x)=1/(1+x)$ for $x\in(0,\infty)$ we have $\phi^{-1}(y)=\frac1y-1$ for $y\in(0,1)$. Now $f\circ((f')^{-1})=\phi$ means $(f')^{-1}=f^{-1}\circ\phi$. Take inverse $f'=\phi^{-1}\circ f$. So you ...
Liding Yao's user avatar
  • 1,831
3 votes

Complex Integral ML Lemma

You missed the fact that the $ML$ is just that; an inequality. Anyway, your approach does now allow you to compute that integral, since you were not able to prove that $\displaystyle\lim_{R\to\infty}\...
José Carlos Santos's user avatar
3 votes
Accepted

Complex Integral ML Lemma

What you've shown is that $$\limsup_{R\to\infty}\left\lvert\int_{\Delta_R}\frac{ze^{iz}}{z^2+1}\,\mathrm{d}z\right\rvert\leq\pi,$$ not that the limit of the integral is $\pi$. You will need a better ...
Lorago's user avatar
  • 9,584
3 votes

How is $|z|^2$ not analytic?

Let me summarize the points already made in comments: The definition of "analytic at $p$" is "differentiable in some neighborhood of $p$". I.e., it is not equivalent to "...
Paul Sinclair's user avatar
3 votes

Interpreting a notation in calculus of variations (differentiating with respect to a derivative)

The "functional" is a way to convert functions to real numbers : It maps a function (of 1 variable or 2 variables or more variables) to a real number. Here the functional is $J[y]$ to ...
Prem's user avatar
  • 9,880
2 votes

Advanced calculus book recommendations.

I am reading "Analysis on Manifolds" by James R. Munkres and I was reading "Calculus on Manifolds" by Michael Spivak. First I tried to finish "Calculus on Manifolds" by ...
佐武五郎's user avatar
2 votes

Why is $\frac{d^n}{dx^n}(y(x))$ the notation for the $n$th derivative of $y(x)$, instead of $\frac{d^n}{d^nx}(y(x))$?

This Leibnizian notation makes a lot of sense. Let's start with the first derivative $f'(x)=\frac{dy}{dx}$. Here the differential $dx$ is the independent variable and the differential $dy$ is the ...
Mikhail Katz's user avatar
  • 42.6k
2 votes
Accepted

Is this function differentiable? (Error in the notes?)

This function $f$ is only defined on $[0,\infty)\times\Bbb R$. If we define differentiability of $f$ at $(0,0)$ (which is only on the boundary) by mimicking the usual definition for points interior to ...
Anne Bauval's user avatar
  • 35.4k
2 votes

derivative of the max function

For $x,y \in [-1,1]$, define $f(x,y) = xy$. Then, $g(y) = \max_x f(x,y) = \max (f(-1,y), f(1,y)) =\max (-y, y) = |y|$ which is not differentiable.
chi's user avatar
  • 2,163
2 votes
Accepted

What is $\int (y'(x))^2 dx$?

It is true that you cannot do this integral without knowing what $y$ is. But we can simplify it: $$ \int (y'(x))^2 dx = \int y'(x)y'(x) dx=y(x)y'(x)-\int y''(x)y(x)dx $$ There isn't much we can do ...
Masd's user avatar
  • 631
2 votes

Example of a function that is differentiable only at points $x$ where $x$ is an integer

Try $f(x) = 1_\mathbb{Q}(x) \sin^2 {2 \pi x}$.
copper.hat's user avatar
  • 173k
2 votes

Does $U'>0$ and $U''<0$ imply $U'''\geq 0$?

No, not necessarily. Take $f:(0,\pi/2)\to\mathbb{R}$ given by $f(x)=\sin(x)$. Then, $f'(x)=\cos(x)>0$ and $f''(x)=-\sin(x)<0$ on $(0,\pi/2)$ but $f'''(x)=-\cos(x)<0$ on $(0,\pi/2)$.
Julio Puerta's user avatar
  • 4,461
2 votes

Intuitive difference between complex differentiability and $R^2 \to R^2$ real differentiability

One intuition is: Complex Differentiability = Real Differentiability + Local Angle Preserving (as long as the derivative $\neq 0$). This "local angle preserving" is the main point behind the ...
balddraz's user avatar
  • 7,620
1 vote

derivative of the max function

$f(x,y) = y \sin x$. Then $g(y) = |y|$.
copper.hat's user avatar
  • 173k
1 vote

Finding the point where the tangent to the curve becomes parallel to the $x$-axis

As you said, find the point where the derivative is $0$. The derivative of $x-\ln x\;$ is $1-\frac1x,\;$ which is $0$ when $x=1$. When $x=1,\; y=x-\ln x=1.\;$ So the point you want is $(1,1)$.
J. W. Tanner's user avatar
  • 60.7k
1 vote

Classifying Singularities of a Rational Function

You can actually compute the Laurent series for the whole function: Note: $$\frac{1}{z-3} = -\frac{1}{3\left(1-\frac{z}{3}\right)} = -\frac{1}{3}\sum_{j=0}^{\infty}\frac{z^j}{3^j} \quad |z|<3 $$ $$\...
Bertrand87's user avatar
  • 2,181
1 vote
Accepted

Classifying Singularities of a Rational Function

As $$\lim_{z\to 0} \dfrac{z+2}{z-3}=-\dfrac{2}{3},$$ which is neither $0$ nor $\infty$, this factor does not affect the classification of $z=0$. And $\cos(1/z)$ has a essential singularity at $z=0$, ...
Julio Puerta's user avatar
  • 4,461
1 vote
Accepted

Can you have a function that is differentiable, yet the jacobian matrix is not continuous?

$f(x)=x^2 \cos ({1\over x})$ does the job : $f'(x)=\sin ({1\over x})+2x \cos ({1\over x})$ if $x=\not = 0$ and $f'(0)=0$. Note that $f'({1\over \pi/2+2n\pi})$ goes to $1$ as $n\to \infty$.
Thomas's user avatar
  • 7,505
1 vote
Accepted

Derivative of trace involving product and function of matrix

$ \def\o{{\tt1}} \def\bR#1{\big(#1\big)} \def\BR#1{\Big[#1\Big]} \def\LR#1{\left(#1\right)} \def\op#1{\operatorname{#1}} \def\trace#1{\op{Tr}\LR{#1}} \def\frob#1{\left\| #1 \right\|_F} \def\q{\quad} \...
greg's user avatar
  • 36k
1 vote

Derivative of trace involving product and function of matrix

You shouldn't use $\operatorname{tr}(A^T B)$. Use $⟨A∣B⟩$ instead (Frobenius inner product / tensordot). Note that $g:ℝ^{m×n}→ℝ,Z↦⟨A∣Z⟩$ is a linear function, so ...
Hyperplane's user avatar
  • 11.7k
1 vote

Prove that a function doesn't have a horizontal asymptote

hint Assume, it has. So, $$\exists a\in R :\lim_{x\to \infty} (f(x)-a)=0$$ $\implies$ $$\lim_{x\to \infty}(f(x+1)-f(x))=0$$ $\implies$ $$\lim_{x\to\infty} f'(C_x)=0$$ with MVT and $x<C_x<x+1$, ...
hamam_Abdallah's user avatar
1 vote

Interpreting a notation in calculus of variations (differentiating with respect to a derivative)

$F$ is a function from $\mathbb R^3$ to $\mathbb R$. If you label points in $\mathbb R^3$ as $(x,y,z)$, then $\frac{\partial F}{\partial x}$, $\frac{\partial F}{\partial y}$ and $\frac{\partial F}{\...
ultralegend5385's user avatar
1 vote

derivative of the natural logarithm

By definition, $[\ln{u}]′=\frac{du}{u}$. With $f(x)=\ln{x}$ you have derivative $\frac{dx}{x}=\frac{1}{x}$ but with $𝑓(𝑥)=𝑥^2$ and $u=𝑥^2$ you have derivative $\frac{du}{u}=\frac{2x}{x^2}=\frac{2}{...
RobinSparrow's user avatar
  • 1,298
1 vote

Approximation of a non differentiable function

If the function $f$ has partial derivatives at $(a,b)$, then the function $g(x)=f(x,b)$ is differentiable at $x=a$ and the function $h(y)=f(a,y)$ is differentiable at $y=b$. Thus, restricting to these ...
Ted Shifrin's user avatar

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