New answers tagged

0

You have gone astray in several ways here. What you are looking for is points where both $\frac {\partial f}{\partial x} = 0$ and $\frac {\partial f}{\partial y} = 0$, but you've gotten distracted by where $\frac {\partial f}{\partial x} = \frac {\partial f}{\partial y}$, which indeed is necessary, but is not sufficient. The set of equations you need to ...


1

No. $$ [\sin^2(x)\cos(x^2)(2x)] + \sin(x^2)(2\sin x \cos x) $$ can be rewritten as $$ 2\sin x \left(x\sin(x) \cos(x^2)+\sin(x^2)\cos(x)\right) $$


3

We get \begin{align*} &(\sin^2 x)(\cos(x^2)) (2x)+(\sin (x^2))(2\sin x\cos x)\\ &=2\sin x(x\sin x\cos(x^2)+\sin(x^2)\cos x) \end{align*} simply by factoring the $2\sin x$ from each term. This is a bit different than the answer you provided, but it is what the answer should be.


1

I get \begin{align*} \frac{d}{dx}\left[\sin^2(x)\sin(x^2)\right]&=2\sin(x)\cos(x)\sin\left(x^2\right)+\sin^2(x)\cos\left(x^2\right)(2x)\\ &=2\sin(x)\left[\cos(x)\sin\left(x^2\right)+x\sin(x)\cos\left(x^2\right)\right]. \end{align*} The term with $\cos\left(x^2\right)$ should still have a $\sin(x)$ multiplying it after factoring, because it was ...


1

The way your Poisson Bracket is written, it does not make sense that $f$ and $g$ are "vector valued". So I will assume scalar values, with vector inputs. $$f,g:\mathbb{R}^n\mapsto \mathbb{R}$$ Using index notation you have $$\begin{align} \frac{\partial f(x+Ay)}{\partial y_i} &= \frac{\partial f(x+Ay)}{\partial x_j}\frac{\partial (Ay)_j}{\partial y_i} ...


1

Theorem 25.1 in Rockafellar's Convex Analysis states that if the subdifferential of $f$ at $x$ is a singleton, then $f$ is differentiable at $x$. Let $u,v\in \partial f(x)$. Note that for any direction $y$, if $\alpha >0$ $$\langle y,u \rangle = \frac 1{\alpha} \langle y,\alpha u \rangle \leq \frac{f(x+\alpha y)-f(x)}{\alpha}$$ and if $\alpha <0$ $$\...


0

Here is a sketch of the missing algebra. Given the FOC in terms of the row vector $a^T$ $$\eqalign{ (x^Tx)a^T + \Big(\frac{\lambda}{\sqrt{a^Ta}}\Big) a^T = x^TR \\ \Bigg(x^Tx + \frac{\lambda}{\sqrt{a^Ta}}\Bigg) a^T = r^T \\ \Bigg(\chi^2 + \frac{\lambda}{\alpha}\Bigg) a^T = r^T \\ }$$ So $a$ is seen to be a scalar multiple of $r$. Square both sides and solve ...


1

An alternative approach is to use a Taylor expansion approach $$ f(X + \varepsilon M) = tr( Y (X + \varepsilon M)^{-1}) = tr( Y X^{-1} (I + \varepsilon M X^{-1})^{-1}) $$ and $$ (I + \varepsilon M X^{-1})^{-1} = I - \varepsilon M X^{-1} + O(\varepsilon^2) $$ which yields $$ f(X + \varepsilon M) = f(X) - \varepsilon tr( Y X^{-1} M X^{-1} ) + O(\varepsilon^2) $...


1

Apply the first part of the fundamental theorem of calculus $$\frac{d}{dt}\int_a^t f(x)dx=f(t)$$ where every $x$ in the integrand is replaced by $t$.


3

You seem to be thinking of ${\mathrm{d}\over\mathrm{dx}}$ as one of the functions in the chain rule. The chain rule (in one variable) applies to the composition of two real-valued functions of a real variable. ${\mathrm{d}\over\mathrm{dx}}$ is not such a function. It takes differentiable functions of a real variable to functions of a real variable. The ...


2

The basic thing you need is $f_u$ and $f_v.$ I'll describe how to get the first. The second is like unto it. So, we have that $$f_u=\frac{f_x}{u_x}+\frac{f_y}{u_y}.$$ Similarly, obtain $f_v.$ Then the gradient is $(f_u,f_v).$ Thus, at the point $(x,y)=(-1,2),$ the gradient is given by $\nabla_{(-1,2)}=(f_u(-1,2),f_v(-1,2)),$ so that in the direction $(3,-4),...


3

This is false as stated (probably it's true with some reasonable additional hypothesis, maybe regarding monotonicity of $f'$). The Idea: We take $f(x)=\int_0^x f'(t)\,dt$, where $f'>0$ is constructed so that $f'(n)\to\infty$, but $f'$ is very small except at points very close to a positive integer, so that $\int_0^\infty f'\le1$. Then $f(x)\le1$, so the ...


2

Because $f$ is differentiable at $0$ (note that we do not even need that $f$ is differentiable or even continuous elsewhere), we have that $\lim_x\to 0{f(x)-f(0)}{x}$ exists, hence with $\epsilon:=1$ there exists $\delta>0$ such that $$\left|\frac{f(x)-f(0)}{x}-f'(0)\right|<1 $$ for $|x|<\delta$. In paticular, $f(x)$ is between two lines with ...


4

In order for $f$ to be differentiable in $0$, it has to be continuous in $0$. Therefore $$\lim_{n\rightarrow\infty}f(x_{n})=f(0)$$ if $\lim_{n\rightarrow\infty}x_{n}=0$.


0

Hint An easier solution is $$f(x)=e^{ax}\sin (bx+c)=\Im\{ e^{ax+jbx+jc}\}$$therefore $${d^n f(x)\over dx^n}=\Im\{{d^n\over dx^n} e^{(a+jb)x+jc}\}=\Im\{e^{jc}{d^n\over dx^n}e^{(a+jb)x}\}=\Im \{e^{jc}(a+jb)^ne^{(a+jb)x}\}$$


0

The function $x\mapsto|x|$ is differentiable over $\mathbb{R}\setminus\{0\}$, so there is no problem in applying the chain rule: for $x\ne0$, $$ \frac{d}{dx}\log\lvert x\rvert=\frac{1}{\lvert x\rvert}\frac{\lvert x\rvert}{x}=\frac{1}{x} $$


0

First things first, no expression will work at $x=0$, since $\log|x|$ is not defined there. Also, $|x|$ is differentiable everywhere except at $x=0$, so things should work fin elsewhere. It's often easier to work with $|x|$ as a piecewise-defined function, so: If $x>0, |x|=x$, you can apply the chain rule there and get $f'(x)=\frac{1}{|x|}=\frac{1}{x}$ ...


0

For non-zero x, we could still use the formula $\frac{d}{dx}\log f(x)=\frac{f'(x)}{f(x)}$ Since $f'(x)=\begin{cases}1&x>0\\-1&x<0\end{cases}$ for $f(x)=|x|$ so $\frac{d}{dx}\log |x|=\begin{cases}\frac{1}{|x|}&x>0\\\frac{-1}{|x|}&x<0\end{cases}=\frac{1}{x}$ given $x\neq 0$


4

Since $\log(|x|)$ is not defined at $x=0$, we may investigate the differentiability for $x\not=0$. We distinguish two cases depending on the sign of $x$. For $x>0$, we know that $|x|=x$ and $$\frac{d}{dx}\left(\log(|x|)\right)=\frac{d}{dx}\left(\log(x)\right)=\frac{1}{x}.$$ On the other hand, for $x<0$, we have that $|x|=-x$ and therefore $$\frac{d}{...


1

$\frac d {dx} \log|x|=\frac 1 x$ if $x \neq0$ but the derivative at $0$ does not even makes sense since the function is not defined at $0$.


1

What you have done is correct. The only possible solution is $f(x)=\frac 1 {g(x)} {\int_0^{x} yg'(y)dy}$ for $x>0$. But now $f(x)\to 0$ as $ x\to 0$ at least when $g$ is a nice function so the given IVP has no solution. [ $|f(x)| \leq \frac {x\int_0^{x} g'(y)dy} {|g(x)|}= x \to 0$ if $g'$ is positive, for example].


1

First, note that the title in the excerpt is incorrect. It should be "$n$th derivative of $e^{ax}\sin(bx+c)$". (1) Here we introducing new quantities $r$ and $\alpha$, so we may define them however we want. Essentially, this amounts to writing the pair $(a, b)$ in polar coordinates, as $(r, \alpha)_{\textrm{polar}}$. More explicitly, any such $r$ satisfies ...


1

$$y=~e^{ax}\sin(bx+c)~$$ Differentiating with respect to $~x~$, we have $$y_1=\frac{dy}{dx}=~a~e^{ax}\sin(bx+c)~+~b~e^{ax}\cos(bx+c)~$$ $$\implies y_1=~e^{ax}~\{~a~\sin(bx+c)~+~b~\cos(bx+c)\}~\tag1$$ For computation of higher-order derivatives it is convenient to express the constants $~a~$ and $~b~$ in terms of the constants $~r~$ and $~\alpha~$ defined by ...


1

Differentiate $f\circ \gamma$ (I use primes instead of overdots): $$(f\circ \gamma)'=[f( \gamma(t))]'=f'( \gamma)\gamma'(t)=(f'\circ \gamma)\gamma'.$$ Thus, we have that $$|(f\circ \gamma)'|=|f'( \gamma)||\gamma'(t)|=(r\circ\gamma)|\gamma'|=(r\circ\gamma)\gamma',$$ if $\gamma'\ge 0.$ Also, we have that $$\frac{(f\circ \gamma)'}{|(f\circ \gamma)'|}=\frac{(f'...


2

This is more or less how the proof of the power rule goes. The binomial theorem is handy to generalise the idea to arbitrary integral indices, but the proof of a specific case is much more clear to follow at first: \begin{align*} y + \delta y &= (x+\delta x)^3\\ &= x^3 + 3x^2\delta x+3x\delta x^2 + \delta x^3\\ \implies \delta y &= \...


1

Let $f(\mathbf{x})=\mathbf{x}^\top\mathbf{x}$. If write $\mathbf{x}=\mathbf{x}_0+\epsilon \mathbf{y}$ in the neighborhood of a point $\mathbf{x}_0$ then we may write \begin{align} f(\mathbf{x}) &=\mathbf{x}^\top\mathbf{x}\\ &=(\mathbf{x}_0+\epsilon \mathbf{y})^\top (\mathbf{x}_0+\epsilon \mathbf{y})\\ &=\mathbf{x}_0^\top \mathbf{x}_0+(2\mathbf{x}...


3

Consider the trace/Frobenius product (denoted by a colon) $$A:B = {\rm Tr}(A^TB)$$ As long as the matrices $(A,B)$ have same number of rows/columns, they can have any shape: tall-and-thin, square, short-and-fat. And of course, they can be row or column vectors in which case one recovers the ordinary dot-product $$a:b = {\rm Tr}(a^Tb) = a\cdot b$$ The ...


0

If this is related to the differential operator $d$ from differential geometry, then a property of $d$ is that $d(dx) = 0$ and $d(uv) = (du)v + u(dv)$.


0

You seem to be on the right path to show $$ \mathrm{d}\left( \frac{x \,\mathrm{d} x+y \,\mathrm{d}y}{\sqrt{x^2 + y^2}} \right) = \frac{\left(x^2+y^2\right) (x \,\mathrm{d}^2 x+y \,\mathrm{d}^2 y)+x^2 (\mathrm{d}y)^2+y^2 (\mathrm{d}x)^2-2 x y \,\mathrm{d}x \,\mathrm{d}y}{\left(x^2+y^2\right)^{3/2}} \text{.} $$ Then $$ \frac{(y \,\mathrm{d}x - x \,\...


2

I think you are assuming that $\sum_{i=1}^n\sqrt{(x-a_i)^2 + (y-b_i)^2} = R \space \Rightarrow \space \sum_{i=1}^n(x-a_i)^2 + (y-b_i)^2 = R^2$ which is incorrect (unless $n=1$).


3

In order to find $$d(xdx)$$ you apply the product rule of differentiation. $$d(xdx)= dxdx + xd^2x = (dx)^2 + xd^2x$$


3

As $x^Tx$ is a scalar, $\color{blue}x^Tx=(\color{blue}x^Tx)^T=x^T\color{blue}x.$


2

You can take the differential of the relationship $$\eqalign{ p &= (1+r)(pA+wl) \\ }$$ to obtain (assuming $p,r$ are constant) $$\eqalign{ 0 &= (1+r)(p\,dA + l\,dw+ w\,dl) \\ l\,dw &= -(p\,dA + w\,dl) \\ }$$ This indicates how $(A,l)$ would need to change to compensate for a change in $w$, in such a way that $p$ is held constant. For example, ...


0

You know that $f'(z)=$ $\left[\begin{array}{cc} u_x & -v_x \\ v_x & u_x \end{array}\right] $ then $f'(a,b)$ equal $\left[\begin{array}{cc} u_x & -v_x \\ v_x & u_x \end{array}\right] \left[\begin{array}{c} a\\ b \end{array}\right]=\left[\begin{array}{c} au_x -bv_x\\ av_x+bu_x \end{array}\right]$ The book consider $a+ib = \left[\begin{...


0

As to the title of your question, Logistic function: where does it come from?, I can provide an intuition for the logistic function which is the common interpretation from a machine learning perspective. It seems that the underlying question has already been answered above, but I thought this interpretation could help with your intuition about logistic ...


6

If $f$ is differentiable and $f(x_1) = f(x_2) = f(x_3) = 0$ with $x_1 < x_2 < x_3$ then you can apply the mean-value theorem (or Rolle's theorem) to both intervals $[x_1,x_2]$ and $[x_2, x_3]$. It follows that $f'$ has a root in each of the open intervals $(x_1, x_2)$ and $(x_2, x_3)$. That makes (at least) two roots of the derivative. In the same ...


2

I think you can find all ingredients in Ullrich's book on pages 4-6. Ullrich explains that $f$ is complex differentiable at $z$ iff there exists $a \in \mathbb C$ such that $f(z+h) = f(z) +a \cdot h + o(h)$. Then $f'(z) = a$. The map $L : \mathbb C \to \mathbb C, L(h) = a \cdot h = f'(z)\cdot h,$ is $\mathbb C$-linear and thus trivially also $\mathbb R$-...


3

Writing your limits with an $h$ makes both equations look like the same limiting procedure. But there is a subtle difference: in your first equation we have $h \in \mathbb{C}$, whereas the second equation is real and hence $h \in \mathbb{R}$. Consequently the first equation is a lot stronger than the second. Instead of approaching $0$ only from the left or ...


2

Subtract $x_t$ to the LHS and RHS of [A] : $$x_{t+1}-x_t=k x_t (1-x_t)- k \frac{1}{k}x_t $$ $$\underbrace{\dfrac{x_{t+1}-x_t}{1}}_{\text{Discrete derivative}}=k x_t(1 - L x_t) \ \ \text{with} \ \ L:=1+\frac{1}{k}$$ Or, better, under the form (thanks to @Yuriy S for this remark) : $$\underbrace{\dfrac{x_{t+1}-x_t}{\Delta t}}_{\text{Discrete derivative}}=k'...


0

First, let's discuss what this equation means. $$f''(x)g'(x) + g''(x)f(x) = 0$$ It is by itself a relation between the two functions. If we fix $f(x)$ and two initial (or boundary) conditions, we can use the ODE to find $g(x)$. On the other hand, if we fix $g(x)$ and two initial (or boundary) conditions, we can use the ODE to find $f(x)$. I doubt we can ...


1

Let $h=g'$. Then $\frac {h '} h=-\frac {f''} f$. So $log (h(x))=C-\int_0^{x} \frac {f''(t)} {f(t)}\, dt$. Take exponential and integrate again to write $g$ in terms of $f$.


1

The angle of rotation of the clock's minute hand is $2\pi$ for $60$ minutes. For one minute the angle of rotation is $\frac{2\pi}{60}=\frac{\pi}{30}$. So the tip of the hand travels $\frac{\pi}{30}r$ on the circle. This is a short distance, compared to the circumference of the circle. So the small arc of circle can be confused with a straight line of same ...


0

If we measure from the center of the clock, at $6:01$ the $x$ position of the hand is $r\cos \frac \pi {30}$ and at $6:02$ it is $r \cos \frac \pi{15}$. If the time is $t$ minutes past $6:01$ the linear approximation to the horizontal position is $r\cos \frac \pi {30}+t(r \cos \frac \pi{15}-r \cos \frac \pi{30})$. It sounds like you are asked for one value,...


1

There is no difference between the notations - they mean exactly the same thing. However, at different times you will find one more useful than the other. For example, when doing u-substitution with integrals, the $\frac{d}{dx}$ is helpful. The same thing is true when using the chain rule - it is often easier to keep track of what is happening with $\frac{d}{...


0

Using prime ' usually indicates that we take the total derivative w.r.t. to all variables, whereas $\frac{d}{dx}$ indicates that we take the total derivative with respect to the variable $x$. Note that this is different from the partial derivative $\frac{\partial }{\partial x}$. As an example, consider $f(x, y, z)=x^2+y^2+3z$ where $y=\sin(x)$. Then $f' = ...


0

There is no mathematical difference. They are different notations for the same thing. (The first was due to Newton, the second to Leibniz.)


0

To deal with the $x^3$, observe that you can apply the geometric series trick $$\frac{1}{8-x^3}=\frac{1}{8}\frac{1}{1-\frac{x^3}{8}}=\frac{1}{8}\Big(1+\frac{x^3}{8}+\frac{x^6}{64}+\frac{x^9}{512}+\dots\Big)$$ and manipulate the Maclaurin Expansion (a Taylor series expansion centered at $x=0$). The Maclaurin Expansion of $\sin(x)$ is $$\sin(x)=\sum_{k=0}^{\...


0

$0=f'(0)=a(a-1)(a+2)\implies a=0,1$ or $-2$. Meanwhile, $0\le f''(0)=2a\implies a\ge0$. Hence $a=0,1$.


0

Just wanted to cover up the $\implies$ case, where the other answers doesn't addressed, and the OP just explained vaguely "differentiating and using the chain rule gives that the required derivatives of $f$ vanish". As stated in a similar question, it is not just that simple. Let's show that if $f \circ \psi^{-1}\colon \widetilde{\mathbb{R}}\to \mathbb{R}$ ...


0

If you want to refer to Wikipedia, you have to identify the proper variables. Here, you have $z(y)=f(y)$ and $y(x,s)=x+s$. Hence $\frac{dz}{dy}= \frac{df}{dy}$ and $\frac{\partial y}{\partial s}= 1$. Which leads to $$\frac{\partial f(x+s)}{\partial s}= \frac{df}{dy}(x+s)= f^\prime(x+s).$$


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