8
votes
Accepted
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.
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5
votes
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 ...
- 3,371
4
votes
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 ...
- 363k
4
votes
Accepted
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.
- 5,478
3
votes
Accepted
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=...
- 402k
3
votes
Accepted
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 $x >x_0$. ...
- 5,478
3
votes
Accepted
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 ...
- 4,040
3
votes
Proving that $\cos{x} + \{x\}$ is not periodic?
We can pick two or more convenient values of $x$ that lead to a contradiction. If we pick $x = 0$, then $\cos(0) + \{0\} = 1 = \cos{p} + \{p\}$.
If we use $x = \pi$, we have $-1 + \{\pi\} = -\cos{p} + ...
- 322
3
votes
Let $f:(0,1)\to \mathbb R$ be a twice differentiable function. Which of the following is/are FALSE?
There are a number of issues with your reasoning for $(a)$ and $(b)$.
So, a counterexample here is to consider something like:
$$f(x)= \sqrt{1-x}$$
This is bounded on $(0,1)$ but its derivative is ...
- 4,496
3
votes
Functions which are undefined at a point
After reading your question and interactions in comments, this is what I understood you to be interested in:
We have some function $f:X\to Y$ and a relation $y=f(x)$ where $x,y$ are variables. When $...
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3
votes
Accepted
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 ...
- 72.2k
3
votes
Accepted
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 ...
- 4,584
2
votes
Accepted
For $z ∈ \mathbb C$ if the minimum value of $(|z-3 \sqrt{2}|+ |z-p \sqrt{2}i|) $ is $5 \sqrt{2}$ .Then what is the value of $p$?
Use the triangle inequality: $$|z_1|+|z_2|\ge |z_1\pm z_2|$$
$$F=|z-3\sqrt{2}|+|z-p\sqrt{2} i|\ge |(z-3\sqrt{2})-(z-p\sqrt{2}i|=|3\sqrt{2}+p\sqrt{2}i|$$
The min valueof $F$ is $$\sqrt{18+2p^2}=5\sqrt{...
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2
votes
Help understanding example of not a function
When they say $y = \pm \sqrt{\frac{1}{3}x}$ they are giving coordinate pairs. So every $x \geq 0$ value except $x = 0$ generates a pair (two coordinate pairs) of values: $\left[\left(x, \sqrt{\frac{1}{...
- 5,992
2
votes
Accepted
How can I convert the following discrete function to continuous?
By geometric series: $$f(x)=\frac{1}{2}(3^x-1)$$For positive integer $x$. However, using the same formula for any real number $x$ will provide a continuous extension in a natural way.
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2
votes
How can I convert the following discrete function to continuous?
Your given Discrete function $f(x) = \sum_{n=0}^{x-1} 3^{n}$ which is a Summation over a Geometric Progression.
We can evaluate that to get :
$f(x) = \frac{3^{x}-1}{2}$
We can now let $x$ be a real ...
- 675
2
votes
Accepted
Proving without using the given condition
Indeed, $f(A) \subseteq B$ is unnecessary for the exercises as they are currently presented.
Consider the following set of exercises:
(a) $A \subseteq B \implies f^{-1}(A) \subseteq f^{-1}(B)$
(b) $f^{...
- 669
2
votes
Accepted
$y' = g(x,y)$ equation intrepetation
In $v=g(x,y)$, the function $g$ is just a function of 2 variables.
In $y'=g(x,y)$ the function value stands for a slope, one may think of the line $y=y_0+g(x_0,y_0)(x-x_0)$ at the point $(x_0,y_0)$. ...
- 112k
2
votes
Let $f(x)=2\arccos x+4\operatorname{ arccot } x-3x^2-2x+10, x\in[-1,1]$. If $[a,b]$ is the range of $f(x)$, find $4a-b$.
The first thing that is required to be found in the domain of the function which is $[-1,1]$ as we have to satisfy the domain of $\arccos(x)$
That done, we proceed to find the derivative,
$$f′(𝑥)=−\...
2
votes
Proof that function ln$\frac {(2-e^x)} {(3+2e^x)}$ is surjective...
Let $x \in \mathbb R$. Let $y = e^x$ then $y > 0.$ Then we have
$$ \frac{2-0}{3-2 \cdot 0} = \frac 2 3>\frac{2-y}{3+2y}$$
since the function $t \mapsto \frac{2-t}{3+2t}$ decreases on $t > \...
- 1,673
1
vote
Finding $\displaystyle \sum \limits _{n=1}^{\infty}\frac{(-1)^n (H_{2n}-H_{n})}{n(2)^n \binom{2n}{n}}$
Edit
After writing the answer below I found from suggestions that an analytical answer was already found here.
Reformulating the sum as an integral
A well-known method to compute integrals is to ...
- 121
1
vote
Accepted
How to simplify this piecewise defined integral function
In all cases, the upper endpoint is $\min\{u+\frac t2,j+\frac12\}$ and the lower endpoint is $\max\{j-\frac12,u-\frac t2\}$. And of course, the integrand is so simple that you could just evaluate the ...
- 64.9k
1
vote
Accepted
How to prove $f(x) = \frac{e^{x^{2}} \sqrt {\sin x}}{\cos x}$ is continuous on $[0,1]$ using limit theorems?
Hint:
$f$ is continuous and $f\ge 0$ implies $\sqrt{f}$ is also continuous.
$f, g$ are continuous implies $fg$ is also continuous
$f, g$ are continuous and $\forall x$ ,$ g(x)\neq 0$ then $\frac{f}...
- 9,278
1
vote
Accepted
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 &...
- 4,496
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 ...
- 2,618
1
vote
Accepted
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 ...
- 675
1
vote
Accepted
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 ...
- 39.1k
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'...
- 5,785
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 ...
- 5,249
1
vote
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\} + \...
- 13k
Only top scored, non community-wiki answers of a minimum length are eligible