New answers tagged

1

Just do a case split. If $y<0$, then $y-\sqrt{y^2-4}<0$ and $y+\sqrt{y^2-4}<0$ (note that $|y|>\sqrt{y^2-4}>0$). If $y>0$, then $y+\sqrt{y^2-4}>0$ and $y-\sqrt{y^2-4}$.


1

I am not sure if this might be helpful, but I bumped into the following example while answering this question Squeeze theorem on expression with minus. There, help was required for the following limit $$\lim_{x \rightarrow \infty }(-6e^{11x} + 9\sin(x) + 3e^{8x} )^{8/4x}\ .$$ The function $f(x)=(-6e^{11x} + 9\sin(x) + 3e^{8x} )^{8/4x}$ is not defined for ...


7

Another example is $$\lim_{x\to 0}\frac1{x^2}$$ which does not exist in $\mathbb R$, but does exist in the extended reals $[-\infty,\infty]$. It also exists in the extended complex numbers (Riemann sphere, roughly $\hat{\mathbb C}=\mathbb C\cup\{\infty\}$). Note that $$\lim_{x\to 0}\frac1{x}$$ does not exist in the reals or the extended reals, but does ...


4

For example the sequence $a_n = \bigg(1+\frac{1}{n}\bigg)^n$ has all $a_n \in \mathbb{Q}$ but $$\lim_{n\rightarrow \infty} \bigg(1+\frac{1}{n}\bigg)^n =e \notin \mathbb{Q} $$ If however, $a_n \in \mathbb{R}$ then $(\lim_{n\rightarrow \infty} a_n) \in \mathbb{R}$ (if it exists) since $\mathbb{R}$is complete.


4

Yes for example a sequence of rational numbers might converge to $\sqrt{2}$ which is not rational. . So the sequence does not converge in $\Bbb Q$ but does converge in $\Bbb R$. If your sequence is in $\Bbb R$ though, then since $\Bbb R$ is complete, if the sequence is Cauchy then it does converge somewhere. So if it doesn't converge in $\Bbb R$ it won't ...


0

Check for maxima/minima of the polynomial, and find its zeros. You know it is continuous, so your function is also (unless $f(x) = 0$). Splicing all this together will give you the range.


0

Write it as follows: $${ f(x)=2x^{ 2 }+4x+3 }=2{ \left( x+1 \right) }^{ 2 }+1$$ so the range of the function is : $$1\le x<+\infty $$


0

Change of variables : $\begin{cases} s=x+y \\ t=x-y\\ \end{cases} \quad \to \quad$ $f(x,y) = g(s,t) $ $$\frac{\partial f}{\partial x} = \frac{\partial g}{\partial s} + \frac{\partial g}{\partial t}$$ $$\frac{\partial f}{\partial y} = \frac{\partial g}{\partial s} - \frac{\partial g}{\partial t}$$ $\left(\frac{\partial f}{\partial x} \right)_{(a,b)} = ...


2

Another way: For $|x|<1$, we have: $$(x^3+x^4+x^5+...)^3=x^9(1+x+x^2+...)^3=x^9(1-x)^{-1}.$$ Now $(1-x)^{-3}$ is half of the second derivative of $(1-x)^{-1}.$ The second derivative of $(1-x)^{-1}=(1+x+x^2+...)$ is $(1.2+2.3 x+3.4 x^2+4.5 x^3+...). $ Half the co-efficient of $x^3$ in this, which is $(1/2).4.5=10,$ is therefore the co-efficient of ...


0

By trigonometric identity $$\mp\sin x=\sin\pi\cos x\pm\sin x\cos\pi=\sin(x\pm\pi)$$ Thus, $\;\sin(2\theta+\pi)=-\sin\theta\;$ , and from here that minus sign there. Acclaration to comment: $$-\pi<\theta<-\frac\pi2\stackrel{\cdot2}\implies-2\pi<2\theta<-\pi\stackrel{+\pi}\implies -\pi<\pi+2\theta<0$$


6

$$(x^3+x^4+x^5+x^6+\cdots)^3=x^9(1+x+x^2+\cdots)^3=x^9\left(\dfrac1{1-x}\right)^3=x^9(1-x)^{-3}$$ Now, we need the coefficient of $x^3$ in $(1-x)^{-3}$ Now the $r+1,(r\ge0)$th term of $(1-x)^{-3}$ is $$\dfrac{-3(-4)(-5)\cdots(-r)(-r-1)(-r-2)}{1\cdot2\cdot3\cdot r}(-x)^r=\binom{r+2}2x^r$$


6

From the OP, the coefficient of $x^{12}$ in $(x^3 + x^4 + x^5 + x^6 + \cdots)^3$ is equal to that of $x^3$ in $(1+x+x^2+x^3)^3$. This is equivalent to asking the number of ways to pick one $x^i$ below in each row so that the product of the $x^i$ picked in each row is of the form $kx^3$ for some number $k$. $$ \require{enclose} \bbox[border:2px solid red] { ...


2

Suppose you calculated $f$ from $1$ to $5$ as you did. Now suppose you know the value of $f(i)$ with $1 \leq i \leq n$ and you have got only $2$ values. Can you calculate $f(n+1)$? Of course: if it is even it is equal to $f((n+1)/2)$ which is $x$ or $y$, if it is odd it is $f((n+1)-5)$ which is $x$ or $y$ (note that $n+1-5=n-4>0$ so it is a good value for ...


8

It's basically the number of ways you can write $12$ as a sum of three integers all greater than or equal to $3$, where order matters. So $12=3+3+6$, $12=3+4+5$, $12=4+4+4$. The first can be rearranged three ways, the second can be rearranged six ways and the last can only be arranged one way. So yes, I believe the answer is $10$.


2

Since $a > b > 1$, and since $\log(x)$ is an increasing function of $x$, then $$\log(a) > \log(b) > 0.$$ Multiplying these two inequalities together: $$a > b > 1$$ and $$\log(a) > \log(b) > 0$$ gives you $$a\log(a) > b\log(b).$$ By a property of logarithms ($x\log(x) = \log(x^x)$), this implies $$\log(a^a) > \log(b^b).$$ Again, ...


1

Alternatively, you could make a table using the recurrence you've found with $f(x+1)=f(x)-f(x-1)$, and, after completing for $f(17),f(18),f(19)$, see that a recurrence arises. \begin{array} {|r|r|} \hline x &16 &17 &18 &19 &20 &21 &22 &23 \\ \hline f(x) &20 &-16 &-36 &-20 &16 &36 &20 &-16 \\ ...


0

For $f(x) = f(x+1) + f(x-1)$, one can get (as was shown in the previous answer) the following characteristic equation $\lambda^2-\lambda +1 = 0$, which yields $\lambda_{1,2} = \frac{1\pm i\sqrt{3}}{2} = e^{\pm i \frac{\pi}{3}}$. Then the overall for the function is defined as $f(x) = \alpha e^{ ix \frac{\pi}{3}} + \beta e^{ -ix \frac{\pi}{3}}$, and using ...


4

For the originally posted question, $f(x)=f(x+1)-f(x-1)$ Rearranging, we have the recurrence relation $f(x+1)=f(x)+f(x-1)$ Looking at its characteristic polynomial, it is of the form $\lambda^2=\lambda+1$ Solving for $\lambda$, we get $\lambda = \frac{1+\sqrt{5}}{2}$ and $\frac{1-\sqrt{5}}{2}$ as the solutions. This implies that $f(x)$ will be of the ...


2

If the question is as stated, the answer will have millions of digits because it is a Fibonacci-like series. If the question is $f(x)=f(x+1)+f(x-1)$, then it repeats every six terms.


11

If $a\gt b$ and $a\gt1$ and $b\gt1$ and $a,b\in\mathbb{R}$ then this implies:$$a^b\gt b^b\tag{1}$$ Also, if $a\gt b$ then this implies:$$a^a\gt a^b\tag{2}$$ Now, making use of result (1) in result (2) yields:$$a^a\gt a^b\gt b^b$$


10

You're asking to prove that the map $f(x):=x^x=\exp(x\log x)$ is strictly increasing. But since $x\mapsto x$ and $x\mapsto\log x$ are both strictly increasing and the first one is $>1$ in $\Bbb R_{>1}$, their product is such (in $\Bbb R_{>1}$). Then the exponential is strictly increasing too, so $f$ is the composition of strictly increasing maps, ...


7

Note that the derivative of $x^x$ is $(\ln(x)+1)x^x>0$(see here) if $x>e^{-1}$, implying $x^x$ is a strictly increasing function if $x>e^{-1}$. This implies your claim is true not only when $a,b>1$ but also $a,b>e^{-1}$


3

Try this sequence of inequalities: $$b^b < b^a < a^a $$


0

When dealing with logarithms, vertical shifts, and horizontal stretches/compressions are considered to be one and the same. $$\log(a\cdot x)=\log(x)+c$$ We don't say it doesn't have a vertical shift, it does, but it also has a horizontal stretch/compression that is exactly related to the vertical shifting. To understand why this is so, return to the ...


0

Just computing the boundary points where $\frac{1/2}{\cos x}=\pm 1$ doesn't give you much information about what to do with those boundary points. Instead, it is easier to think: $\frac{1/2}{\cos x}$ is in $[-1,1]$ exactly when $\cos x \le -\frac12$ or $\cos x \ge \frac 12$. At this point stop doing symbolic algebra and instead draw a graph of the cosine ...


0

First, I want to note that this is not a full solution, it just makes finding $a,b$ with $\frac 1a+\frac 1b=\frac pq$ a little easier. All solutions to $$a+b|ab$$ are $(a,b)=(\alpha\gamma(\alpha+\beta),\beta\gamma(\alpha+\beta))$ for some $\alpha,\beta,\gamma$. So if $$\frac 1a+\frac 1b=\frac 1n$$ then substitution yields $n=\alpha\beta\gamma$. But we wish ...


1

By Gronwall's inequality, $f$ is identically zero.


0

HINT: What does the second derivative of this function tells us about how many maximum points we can find on its domain?


1

When in doubt, use parentheses. I would codify your observations thus: Parentheses may be omitted only under these circumstances: When the function symbol provides its own bracketing. For example, you can write $\sqrt{1 + x}$. Of course when you don't have TeX like we do here, you may have to write something like sqrt(1 + x), even though in this ...


0

If you want $\cos^{-1}(x)$ to be a function, then it should only have one result for any given element in the range $[-1, 1]$. This is because in order for a rule to be considered a function, it must have only one unique output for any given input. It is true that there are an infinite number of angles whose cosine is equal to $\frac{1}{2}$. However, we ...


2

In general, reducing complexity of notation is always good. That is, unless it introduces ambiguity into your statement. As such, I would advise omitting parentheses, unless complexity of the expression dictates a necessity. For instance, many authors use $ Tx$ to denote $ T(x)$, where $T$ is usually a linear transformation. But sometimes you might have a ...


0

Just because $A \subset B$ doesn't necessarily mean that $\forall b \in B \exists a \in A \colon b = f(a)$; for instance: $f \colon \mathbb N \rightarrow \mathbb N*, f(x) = 2 \times x + 1$. $ \mathbb N* \subset \mathbb N$. The function is injective, but there exists $b$ such that $ \nexists a \in A \colon b = f(a)$, for instance $b = 0$. That's the problem ...


1

You're correct, in that cos(x) is not a one-to-one function; there are infinite values of x for every value of y. This will reflect on the inverse cos function as well. Indeed, sometimes this needs to be utilized to get the correct answer to a problem. However, the arccos you'll find in a calculator is by definition restricted to the range of 0 < x < ...


0

You can consider $\arccos: [-1,1]\to [0,\pi]$ or $\arccos: [-1,1]\to [r,r+\pi]$ for any $r>0.$ Once you fixed the target there is only one value for $\arccos x.$ But if you don't fix the target you have infinite values. In some situations (for example, if you are working with angles in the second quadrant) you must need an answer that belongs to $[\pi/2, ...


2

Just argue like this: $B\subset A$ gives an injection $i:B \rightarrow A$ (the standard inclusion: every $b \in B$ is sent to itself). By hp, you are given $f: A \rightarrow B$ injective, so that by Cantor-Berstein you know $A$ and $B$ are in bijection.


1

Well, your friend is wrong. $f(a)$ maps to exactly one $c$ and $g(c)$ maps to exactly one $b \in A$ but there is no reason in any of the pluperfect hells to assume that $b$ is the same as the original $a$. Anyway you comment isn't true. Let $f: (0,1) \rightarrow (0,1/2)$ $f(x) = \frac 1 4 x$. There is no $x \in (0,1)$ such that $f(x) = 1/3$. Of course ...


3

It doesn't hold that $f$ gives the bijection. Consider $A=(0,2),B=(0,1)$ and $f(x)=\dfrac{x}{3}.$ The problem in your proof: Each element of $B$ is an element of $A,$ but this doesn't mean that an element of $B$ has a preimage. Think of the above example.


0

Okay, resulting polynomial is always negative on ($-\infty, -2$], has at least one root in (-2,-1). the polynomial is positive at -1, 0 and 1. At 2, the polynomial is negative, so there is at least one root in (1,2). The polynomial is always positive on [$3, \infty$). So three roots are accounted for and two are not. The nature of these roots is ...


0

So you are looking for a function $f:\mathbb{N}\rightarrow\mathcal{P}(\mathbb{Q})$,where $\mathcal{P}(\mathbb{Q})$ is the power set of the rationals, rather than a function $f:\mathbb{N}\rightarrow\mathbb{Q}$. The way you are posing the problem causes some confusion. $\mathcal H(m,k)$ obtains its min value for $m=k=1$ and then $\mathcal H(1,1)=1$ and its ...


0

To answer how to perform convolution with step functions, I would say there are two common ways: directly with integration (which is feasible for this problem but not for many others), or with the Laplace transform. However, doing integration with step functions is tricky and worth explaining, in my opinion, so I'll answer for future readers: Direct ...


1

Jump discontinuity occurs when sided limits exist. It means that $$\lim_{x\to a^+}f(a)\in\mathbb{R} \quad\text{and}\quad \lim_{x\to a^-}f(a)\in\mathbb{R}$$ So clearly this is not a jump discontinuity. The "infinite discontinuity" is often called "essential discontinuity" so that it's not misleading. This discontinuity is essential as the limit from the ...


1

Since $p$ and $q$ are continuous and $p(x)\le f(x)\le q(x)$, any point $x_0$ such that $p(x_0)=q(x_0)$ is a good candidate. Now, do the condition $a_{n-1}\ne b_{n-1}$ and the fact that $p,q$ are monic polynomials of (identical) even degree say anything about $\deg(p-q)$?


2

Hint: The difference of polynomials p and q is a polynomial with odd degree. So it has a root. Now Squeeze theorem or its modification would be useful for proving your statement.


2

Your function is a) not a function and b) is not continuous. f(0,- 1/2) = {0, -1/2} and f(0, 1/2) = {1/2, 0} have two values so it is not a function. It can be made a function by setting: $f(x,y)=\begin{cases}\left(0,\frac{1}{2}+y\right) & \text{ if }(x,y)\in[-1,1]\times\left[-1,-\frac{1}{2}\right)\\(0,y) & \text{ if }(x,y)\in ...


2

According to your definition, $f(0,-1/2)$ is being mapped to two distinct points : $(0,0)$ and $(0,-1/2)$. So it's not quite a function.


0

In the one dimensional case, when you take a derivative, you find the slope of the function in a certain point p, so you can take a linear approximation of the function "not so far" from that point p. In n dimensions, what would happen if you consider the derivatives (and therefore the slopes) with respect to the other variables?


1

Define $f:\mathbb{R}\to S$ to be $$ f(a)=\begin{pmatrix} a&0\\ 0&a\\ \end{pmatrix}=M_a$$ Let $M_k$ be given for $k\in\mathbb{R}$. Then we know that $\exists k\in \mathbb{R}$ such that $f(k)=M_k$ by definition. So, the function is surjective.


2

Presumably you mean in the first order language of set theory, involving the single predicate "$\in$". You need to be able to talk about unordered pairs, ordered pairs using the definition $\{u,\{u,v\}\}$, notions derived from these, and more. To express "all functions $f$ from $D$ to $C$ have property $P(f)$", first we need to be able to express "$f$ is a ...


1

The elements of $\mathbb Z$ are isolated points and the function is $0$ almost everywhere. Between integers, continuity is granted as the function is constant. At integers, we need to check if the function value coincides with the limit, it if exists. The limit is indeed $0$, as the function value is $0$ in any neighborhood of radius $<1$. Then ...


0

Contexts where codomain does not enter into the definition of a map are those in which map composition does not need to be taken into consideration. An object defined this way is often more properly called "graph of a map" or "parameteric set", where such a set is the image set (as in the case of, e.g., parametric surfaces). On the other hand, in map ...



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