New answers tagged

2

Just take the remainder after dividing it by $7$, that is $$ 7 - x \mod 7$$


1

The syntax is mostly correct. The only improvement I see is to write the $k$-th derivative as $f^{(k)}$, not $f^k$. This also applies when $k$ is a known number. There is an alternative notation with apostrophes. So you should write $f^{(0)}(x)$ or $f(x)$, $f^{(1)}(x)$ or $f'(x)$, $f^{(2)}(x)$ or $f''(x)$ and so forth. The second one is generally only used ...


1

Yes, you should add $\Delta t \to 0$ to be explicit, but in general, $o(\Delta t)$ is understood to hold for $\Delta t \to 0$, so I think it is fine not to add it, but there is no harm in writing it. To both notations you give: They talk about different things. The first one is the definition of $y$ being differentiable at $t$, iff $$ y(t + \Delta t) = y(...


6

No. A unit argument can show it. Assuming $f$ is dimensionless and $x$ has units of length, $f'$ has dimensions of inverse length and $f''$ has dimensions of inverse length$^2$, so you can't compare them. The numeric comparison will depend on the units of measure of $x$. For a specific example, take $f(x)=\sin(2x)$ then $\max f'=2, \max f''=4$


1

Simple counterexample: consider the function $$f(x)=x^2$$ on the interval $(0,\frac{1}{2})$.


0

This is related to Buchstab function. Let $\Phi(x,y)$ be the number of positive integers $n\leq x$ which do not have any prime divisor $p< y$. Then it is known that for any fixed $U>1$ $$ \Phi(x,y)=\frac{x w(u)}{\log y} - \frac y{\log y} +O\left(\frac x{(\log x)^2} \right) $$ where $y=x^{\frac 1u}$ with $1\leq u\leq U$ , $y\geq 2$, and $w(u)$ ...


0

The distance function $d: (Y \times Y, d_{max}) \to \Bbb R$ is continuous. Hence, for all $n \ge n_0$, we have for all $x \in X$, $$d(f(x), f_n(x)) = d(\lim_{m \to \infty} f_m(x), f_n(x)) = d( \lim_{m \to \infty} (f_m(x), f_n(x))) = \lim_{m \to \infty} d(f_m(x), f_n(x)) \le \epsilon$$


4

Your mistake $$f'(0)=2 \quad \text{and} \quad f''(0)=4$$ Another way to proceed: if you know the series expansion of both $e^x$ and $\sin(2x)$: $$e^x=1+x+\frac {x^2}{2}+\frac {x^3}6+\cdots$$ $$\sin(2x)=2x-\frac {4x^3}3+\cdots$$ Then $$e^x\sin(2x)=\left(1+x+\frac {x^2}{2}+\frac {x^3}6+\cdots\right)\left(2x-\frac {4x^3}3+\cdots\right)$$ $$=2x-\frac {4x^3}3+...


1

Based on your work above $f''(0) = 4$ check this. Where did the $x^1$ term go? $f(x) = \frac {0}{0!} + \frac {2}{1!}x + \frac {4}{2!}x^2 + \frac {-2}{3!} x^3 \cdots$ I would say $3^{rd}$ degree polynomial, and not "grade." But that would be correct, the highest exponent of $x = 3.$ And, the highest derivative you would need is 3. What about the ...


1

You miscalculated the first and second order terms, remember that $sin(0) = 0$ and $ cos(0) = 1$.


1

In the case $$f(x_1)>f(x_1)\Leftrightarrow g(x_1)>g(x_2)$$ for all $x_1,x_2\in \mathbb{R}$, we would say that $g$ is a monotonic transformation of $f$. This is because $g$ can be written as a composition $g=h\circ f$ where $h$ is a strictly increasing function.


1

A different, although perhaps slightly more complicated, approach than the other answers: Find the vertex of this parabola and determine which values of $k$ put the vertex below the $x$-axis. Combine that with the fact you've already found: $k < 0$. Completing the square is one method of finding the vertex: $$ kx^2 - 2x + k = k\left(x^2 - \frac{2}{k}...


4

First off, we require $k<0$ then we require that $f$ has no roots so that it is always negative. That is, we need its discriminant to be $< 0$. i.e: $$4 - 4k^2 < 0 \iff 1- k^2 < 0 \iff k^2 > 1$$ since we know that $k<0$ then this becomes $k < -1$.


-1

You can find that $k \ne 0$, otherwise the function would be always non-negative. Thus, $$ x(x - \frac2k) = -1 $$ and solving the related inequality provides the range of values for $k$. $$ \begin{cases} x < -1 \\ x < \frac{k+2}{k} \end{cases} \implies k < -1. $$


1

The fundamental fact is that, intuitively, $sin(X) \approx X$ when $X$ is small. Here you replace $X$ with $\frac{sin(x) + tan(x)}{2}$ and $\frac{sin(x) - tan(x)}{2}$ and that's how you get rid of the outter $sin$. Formally, you can say that $sin(x) \sim_{0} x$ meaning that $sin(x) = \epsilon(x) x$ or equivalently $x = \epsilon'(x)sin(x)$ where $\epsilon(x)$...


1

For small angles($ \theta$ $\rightarrow$ $ 0$), we can say $sin \theta=\theta$..(i) For $ \theta$, $\rightarrow$ $ 0$, $sin\theta \rightarrow0$ $tan\theta \rightarrow0$ $sin\theta + tan\theta \rightarrow0$ Using ..(i), $sin\frac{(sin\theta + tan\theta)}{2}=\frac{(sin\theta + tan\theta)}{2}$


2

We already have plenty of slick answers through creative telescoping, so I will go for the overkill. By crude estimations we have $\sum_{r\geq 3}\frac{4}{r^2+3}\leq\frac{\pi}{2}$, hence: $$ \sum_{r\geq 3}\arctan\left(\frac{4}{r^2+3}\right) = \text{Arg}\prod_{r\geq 3}\frac{r^2+3+4i}{r^2}\tag{1}=\text{Arg}\prod_{r\geq 3}\left(1+\frac{(2+i)^2}{r^2}\right)$$ ...


3

$$ \begin{align} \sum_{r=1}^\infty\arctan\left(\frac4{r^2+3}\right) &=\sum_{r=1}^\infty\left[\arctan\left(\frac2{r-1}\right)-\arctan\left(\frac2{r+1}\right)\right]\\ &=\lim_{n\to\infty}\left[\sum_{r=1}^n\arctan\left(\frac2{r-1}\right)-\sum_{r=3}^{n+2}\arctan\left(\frac2{r-1}\right)\right]\\ &=\lim_{n\to\infty}\left[\frac\pi2+\arctan(2)-\arctan\...


3

$$\sum_{r=1}^N\arctan\left(\frac{r+1}2\right)-\sum_{r=1}^N\arctan\left(\frac{r-1}2\right)=\sum_{r=1}^N\arctan\left(\frac{r+1}2\right)-\sum_{r=-1}^{N-2}\arctan\left(\frac{r+1}2\right)$$ $$=\sum_{r=1}^N\arctan\left(\frac{r+1}2\right)-\sum_{r=1}^N\arctan\left(\frac{r+1}2\right)-\arctan\left(0\right)-\arctan\left(\frac{1}2\right)+\arctan\left(\frac{N}2\right)+\...


4

One may observe that summing $$ u_{r+1}-u_{r-1} $$ may be simplified as telescoping sum: $$ \sum_{r=1}^N\left(u_{r+1}-u_{r-1}\right)=\sum_{r=1}^N\left(u_{r+1}-u_{r}\right)+\sum_{r=1}^N\left(u_{r}-u_{r-1}\right)=u_{N+1}+u_N-u_1-u_0. $$ From the identity $$ \arctan \left(\frac{4}{r^2+3} \right)=\arctan \left(\frac{r+1}2 \right)-\arctan \left(\frac{r-1}2 ...


1

We may compute the whole Taylor series from the Weierstrass product: $$ \cos(x)=\prod_{n\geq 0}\left(1-\frac{4x^2}{(2n+1)^2\pi^2}\right) \tag{1}$$ and the Taylor series of $\log(1-x)$, leading to: $$ \log\cos(x) = -\sum_{n\geq 0}\sum_{m\geq 1}\frac{4^m x^{2m}}{m\pi^{2m}(2n+1)^{2m}}=-\sum_{m\geq 1}\frac{(4^m-1)\,\zeta(2m)}{m\,\pi^{2m}}\,x^{2m}.\tag{2}$$ Since ...


1

For the derivatives one get: $$f'(x_0)=-\tan(x)$$ $$f''(x_0)=-\frac{1}{\cos^2(x)}$$ (Use the chainrule with $u=\log(x), v=\cos(x)$) Then you get: $$T_{f,0,2}(x)=\sum\limits_{j=1}^2 \frac{f^{(j)}(0)}{n!}\cdot x^n$$ $$T_{f,0,2}(x)=0-\tan(0)(x-0)-\frac{1}{2}x^2$$ $$T_{f,0,2}(x)=-\frac{1}{2}x^2$$


1

Then answer is no, the statement is not true. Take for example the cube root $x \mapsto x^\frac{1}{3}$. This map is uniformly continuous on $\mathbb{R}$. But its derivative and itself are unbounded.


3

$$F=\sin^2(x+\alpha)+\sin^2(x+\beta)-2\cos(\alpha-\beta)\sin(x+\alpha)\sin(x+\beta)$$ $$=1-\{\underbrace{\cos^2(x+\alpha)-\sin^2(x+\beta)}\}-\cos(\alpha-\beta)\{\underbrace{2\sin(x+\alpha)\sin(x+\beta)}\}$$ Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$ and Werner Formula$(2\sin A\sin B=\cdots),$ $$D=1-\cos(2x+\alpha+\beta)\cos(\...


1

Note that axiomatic set theory is usually formalized in a logical language where there are no set-valued expressions other than variables. Whenever something that looks like a set-valued expression seems to appear in a formula, it is actually to be understood as an abbreviation of a more complex logical formula that intuitively claim that some variable $x$ ...


0

Why not $$F(x) = \{y \mid \{\{x\},\{x,y\}\} \in F\}\;?$$ This $y$ is guaranteed to exists and be unique by the definition of $F$ being a function.


0

Domain of $(f \circ g)(x) $ is A and co domain is C.


0

$(f\circ g)(x) = f(g(x))$, so first you need $g(x)$ to be defined, and then you need $g(x)$ to be in the domain of $f$. So the domain is $\{x \in B: g(x) \in A\} = g^{-1}(A)$. You could take the codomain to be $B$ since that's the codomain of $f$. The range is $f(g(g^{-1}(A)))$.


3

$$\text {Let }\quad L=\sin^2 (x+a)+\sin^2(x+b).$$ $$\text {Let }\quad R=-2\cos (a-b)\sin (x+a) \sin (x+b).$$ $$\text {We have }\quad L=\frac {1}{2}(1-\cos (2x+2a) +\frac {1}{2}(1-\cos (2x+2b).$$ Now with $y=2x+a+b$ and $z=a-b$ we have $$\cos (2x+2a)=\cos (y+z)=\cos y \cos z-\sin y \sin z. $$ $$\cos (2x+2b)=\cos (y-z)=\cos y \cos z +\sin y \sin z .$$ $$\...


15

Taking @SiongthyeGoh's comment, in the above picture, $AE=1$, $\angle BAE=x$, $\angle BAC=\beta$, and $\angle BAD=\alpha$. It follows that $DE=\sin(x+\alpha)$ and $CE=\sin(x+\beta)$. Applying the cosine theorem to $\triangle DCE$, the given expression is equal to $DC^2$, hence is a constant.


8

Let $$E=\sin^2(x+\alpha)+\sin^2(x+\beta)-\color {blue}{2\cos(\alpha-\beta)\sin(x+\alpha)}\sin(x+\beta)$$ Now use $2\cos A \sin B = \sin (A+B) - \sin (A-B)$. $$E=\sin^2(x+\alpha)+\sin^2(x+\beta)-\left[\sin\left(\alpha-\beta +x+\alpha\right)-\sin\left(\alpha-\beta -x-\alpha\right)\right]\sin(x+\beta)$$ $$E=\sin^2(x+\alpha)+\sin^2(x+\beta)-\left[\sin\left(...


0

Since it is quadratic then you can represent it as $$ ax^2 + bx + c = 0 $$ and this must be true for both of your data points, i.e., $$ 0 = a(0)^2 + b(0) + c $$ and $$ 0 = a(6)^2+ b(6) + c $$ The first data point's equation leads to $c=0$. Then the second gives $ 36a + 6b = 0 $ or $6a+b=0$. Substituting $b=-6a$ gives you your equation $$ y = ax^2-6ax $$ ...


2

The quadratic functions which vanish at $x=0$ and $x=6$ are one-parameter family, and their equation, as you found it, is: $$y=ax(x-6).$$ There's a maximum only if $a<0$. If $a>0$, the function has a minimum. In any case the extremum is obtained at the midsum of the roots, i.e. at $x=3$. Hence the maximum, if $a<0$, is $$y(3)=-9a.$$


0

Assume that $Z_{f}\in\text{im }f$, so that some $a\in A$ exists with $f\left(a\right)=Z_{f}$. For $a$ there are two possibilities: $a\in Z_{f}$ or $a\notin Z_{f}$. If $a\in Z_{f}$ then $a\notin f\left(a\right)=Z_{f}$ and a contradiction is found. If $a\notin Z_{f}$ then $a\in f\left(a\right)=Z_{f}$ and a contradiction is found.


1

Suppose $Z_f$ is in the image of $f$. Then $Z_f = f(a)$ for some $a \in A$. Use the definition of $Z_f$, and the assumption that $Z_f = f(a)$, to prove that neither $a \in Z_f$ nor $a \not \in Z_f$, which is a contradiction since it must be true that either $a \in Z_f$ or $a \not \in Z_f$.


1

In general $$x\in f^{-1}(X)\iff f(x)\in X\tag1$$ Observe that $x\in X\implies f(x)\in f(X)\implies x\in f^{-1}(f(X))$ so it is always true that: $$X\subseteq f^{-1}(f(X))\tag2$$ Essential is the information $f^{-1}(f(X))\subseteq X$ for every $X\subseteq A$. Applying this on $X=\{a\}$ where $a\in A$ we get $f^{-1}\left(f\left(\left\{ a\right\} \right)\...


3

The brakes reduce it's velocity by $10$ m/s in every second. So the acceleration is $-10 \, \text{m/s}^2$. So $$y(t)=v-10\,\text{ms}^{-2}t$$ $$x(t)=100\,\text{ms}^{-1}t+\frac{1}{2}(-10\,\text{ms}^{-2})t^2$$


0

$f$ is a decreasing function. So $$f(f(x))>f(f(-x))$$ if and only if $$f(x)<f(-x)$$ if and only if $$x>-x$$ which is true for all positive $x$. In a comment to another answer you appear to say that the inequality to solve is actually $$f(f(f(x)))>f(f(-x))$$ In that case the same logic tells you this is true if and only if $$ \begin{align} f(x)...


0

if we have $f(x)=30-2x-x^3$ then we get $f(-x)=30+2x+x^3$ and you must solve $$30-2(30-2x-x^3)-(30-2x-x^3)^3>30-2(30+2x+x^3)-(30+2x+x^3)$$ can you simplify this?


1

Choose any $a_1, a_2 \in A$ such that $F(a_1) = F(a_2)$. We want to show that $a_1 = a_2$. Indeed, since $F^{-1}$ is a well-defined function, we know that it must map equal inputs to the same output. Thus, it follows that $F^{-1}(F(a_1)) = F^{-1}(F(a_2))$ so that $a_1 = a_2$, as desired. $~~\blacksquare$


1

The question as stated is a little confusing. I will assume that you are trying to prove the following: If $f : A \to B$ possesses an inverse, then that inverse is one-to-one. The proof is as follows: Suppose $f^{-1}(x)=f^{-1}(y)$ for $x, y \in B$. Then $f(f^{-1}(x)) = f(f^{-1}(y))$, which is precisely the statement that $x=y$.


-1

If you can assume the top statement (above "Prove") : $F^{-1}$ is one-to-one iff $F^{-1}$ is a bijection iff $F$ is a bijection iff $F$ is one-to-one. (All functions are onto.)


0

A lot of the time, mathematical notation is decided by historical reasons rather than logical ones. This is true with function application. These all mean the same thing, unless it's obvious from context that they don't. For example, the notation $fx$ is common when $f$ is a linear transformation. Indeed, $Ax$ is common shorthand for $A(x)$; they mean the ...


1

$\forall x \in (0,1), f(x) = f(x^2) = f(x^4) = ... = f(x^{2^n})= \displaystyle \lim_{ n \to \infty}f(x^{2^n}) = f(0) = C$


0

The function $$f(x) = \left ( \frac{2+\cos x}{3} \right) ^{x^4}$$ has this property. This follows from the fact that $\int_0^{2\pi} [(2+\cos x)/3]^n\, dx \sim 1/\sqrt n$ as $n\to \infty$ (by Laplace's method).


1

You know that you can use $\frac 0 0$ and $\frac \infty \infty$ in L'Hôpital's rule. Thus, if you have $0\cdot\infty$ where $f(k)\rightarrow0$ and $g(k)\rightarrow\infty$, you can rewrite $\lim_{k\rightarrow c}{f(k)\cdot g(k)}$ as $\lim_{k\rightarrow c}{\frac {f(k)}{\frac 1 {g(k)}}}$. Now the problem is no longer in the form $0\cdot \infty$ but $\frac 0 0$, ...


2

Because the question is not in indeterminate form ie it's not 0/0


7

l'Hopital's rule only works for the limits "$0/0$" and "$\infty/\infty$". To compute this limit try this: $$\frac{\sin x+\cos x}x=\frac{\sin x}x+\frac{\cos x}x$$



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