# Tag Info

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### Does a function $f(x)$ exist such that $f(x+1)-f(x)=\frac{1}{x}$. And if so what is it.

Yes. Let $L(z)$ denote the logarithm of the gamma function. Then $L(x + 1) - L(x) = \log(x)$. So the derivative $L'(x)$ (i.e. $\Gamma'(x) / \Gamma(x)$) satisfies the desired functional equation.

### Functions which are undefined at a point

Well if we have a function $f:X\to Y$, then we cannot talk about $f(a)$ if $a\notin X$, because we haven't defined what that means, and there is no way to "guess" what it could be, as we ...

### Help understanding example of not a function

A function is defined as having exactly one output for each input in the domain. The expression $\pm \sqrt{\frac 13x}$ does not provide a unique output for every input $x$, so it does not represent a ...
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### Let $f:(0,1)\to \mathbb R$ be a twice differentiable function. Which of the following is/are FALSE?

For a) $f(x)=\sqrt x$ is a counter-example. For b) $f(x)=x^{3/2}$ is a counter-example. For c) $f(x)=3x-x^{2}$ is a counter-example. d) is true: Just integrate twice.
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### Solve $\arcsin x=\arccos x, x\in[-1,1]$

Actually, for each $x\in[-1,1]$, $\sin(\arccos x)=\sqrt{1-x^2}$. Obviously, if $x<0$, the equation $x=\sqrt{1-x^2}$ has no solutions. And, if $x\geqslant0$,\begin{align}x=\sqrt{1-x^2}&\iff x^2=...
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### If $\lim_{x\to\infty} f(x)$ is finite, find $\lim_{x\to\infty} xf'(x)$

Suppose $\lim_{x \to \infty} xf'(x)=c>0$. Then there exists $x_0$ such that $f(x)-f(x_0)=\int_{x_0}^{x} f'(t)dt>\int_{x_0}^{x} \frac c {2t}dt=\frac c 2[\ln x -\ln x_0]$ for all $x >x_0$. ...
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### Differentiable top-k function

Below is a simple differentiable top-k for PyTorch I wrote. See also the code here. The idea is to pick some sigmoid function, e.g. the simple $\sigma(x) = \frac{1}{1 + e^{-x}}$, and express your ...

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### Empty set - short questions regarding Empty set.

The empty set is just as valid of an element of a set as any other potential element and exists as an element and increases the size just as any other element's inclusion would have caused. Here, I ...
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### Prove that if $A$ and $B$ are denumerable sets then $A\cup B$ is denumerable

First note that in your two cases you used the function $h$ in one of them where I think you intended $g$. Next note that in fact $h$ is not injective if $A$ and $B$ are not disjoint. I would suggest ...
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1 vote
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### Solution verification: Some preimage and image proofs of a map

Right, so I've writtne most of my feedback about your proof in the comments above. This is going to focus on how I would phrase my own arguments. I'm going to actually prove the parts where it says &...
1 vote

### Proving the restriction of a function is the composition of that function with the natural inclusion.

$i:S\hookrightarrow X$ is defined by $i(s)=s,\,\forall s\in S$. If $f:X\to Y$, then the restriction $f\restriction _S: S\to Y$ is by definition $f\restriction _S(s)= f(s),\,\forall s\in S$. Now just ...
1 vote
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### Intuitive understanding for $f(x)=f'(x)$ in $(a,b)$ when $f(a)=f(b)=0$

The Question asks about the Intuition, hence I will give Pictorial Intuitive Heuristics about this. When $f(x)$ has Multiple Crossings (with Multiple Maxima & Minima Points) about the x-axis, we ...
1 vote
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### For what values ​of $k$, $f(x)=\lvert x^{2}+(k-1) \lvert x \rvert -k \rvert$ is non-differentiable at five points?

$$f(x)=|x^2-(k-1)|x|-k|$$ will have 5 points of non differentiabilty if the quadratic $x^2-(k-1)x-k=0$ has both roots distinct and positive. For this we need $(k-1)^2+4k=(k+1)^2>0$, which always ...
1 vote

### For what values ​of $k$, $f(x)=\lvert x^{2}+(k-1) \lvert x \rvert -k \rvert$ is non-differentiable at five points?

$$f(x)=|(|x|+k)(|x|-1)|$$ For $5$ points of non-differentiability, both the roots of $(x+k)(x-1)$ should lie on the positive $x-$axis. Thus $k\lt0$. At $k=-1$, we'll have a repeated root, thus we won'...
1 vote

### For what values ​of $k$, $f(x)=\lvert x^{2}+(k-1) \lvert x \rvert -k \rvert$ is non-differentiable at five points?

If $x>0$, then $|x|=x$ and $$x^2+(k-1)|x|-k = x^2+(k-1)x-k = (x+k)(x-1)$$ If $x<0$, then $|x|=-x$ and $$x^2+(k-1)|x|-k = x^2-(k-1)x-k = (x-k)(x+1)$$ Now if $f$ is a real function ...
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### Proving that $\cos{x} + \{x\}$ is not periodic?

Define $f(x):= \{x\};\ g(x) := \cos(x).$ Suppose $f+g$ has period $\alpha.$ Then, $$f(-\alpha) + g(-\alpha) = f(0) + g(0) = 1 = f(\alpha) + g(\alpha).\quad \text{This gives:}$$  \{-\alpha\} + \...

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