# Tag Info

8

Actually, the truth is slightly more complicated than that. The problem is that if I tell you that $\cos x = t$, I have not yet given you enough informations for you to reconstruct what $x$ is. For example, $x=\frac{\pi}{2}$ and $x=-\frac{\pi}{2}$ both satisfy the equation $\cos x = 0$. In general, therefore, it is not possible to define an inverse of the ...

6

By definition functions have domains, proving injectivity without considering the domain is something devoided of sense, the domain of a function is part of its essence. In practice there can be times in which whatever the domain is, the proofs will look the same (if the functions are in fact injective), but they are necessarily different because the ...

5

For $x>1$, i.e. $\log x>0$, we can write $x=e^z$, and $z>0$. The inequality becomes $$z \leq e^{\frac{z}{2}}-e^{-\frac{z}{2}}.$$ This is trivially true, for example by using the definition $$e^u = \sum_{k=0}^\infty \frac{u^k}{k!}.$$ The case $0<x<1$ can be treated in a similar fashion. I know that this proof is not calculus-free, but it ...

4

Define $t:=\log x$. We have to show that for each $t\neq 0$, $$\frac{e^t-1}t\geqslant e^{t/2},$$ and, after having multiplied on both sides by $e^{-t/2}$, we are reduced to show that for each positive $t$, $$e^t-e^{-t}\geqslant 2t.$$ This inequality is easier to handle. We indeed recognize classical hyperbolic sine. If we know the power series expansion ...

4

What you can say is that $$x^4 +1 = 2\sqrt{2x - 1}\tag{1}$$ $$x^4 + 1 = 2x\tag{2}$$ share one solution: when $x = 1$, both equations are satisfied. But each equation has a second solution not shared by the other. Real solutions to $(1)$ are $x = 1,\;x\approx 0.68682$. Real solutions to $(2)$ are $\;x = 1,\;\text{ and }\;x \approx 0.54369.$ ...

4

You have found the solution for the first derivative, the idea is then to use this solution to bootstrap yourself to the next derivative. First of all, you may assume that $f(0) =0$, since you may always translate the function vertically without affecting zeroes of the derivatives. Thus the question can be thought of equivalently as: If ...

4

This may not be exactly the type of reference you're seeking, but G. H. Hardy's book Orders of Infinity may be of interest. Wikipedia's definitions of "degree" are defined by a function's behavior near infinity, and therefore do not bound the number of real roots for an arbitrary function. If $b > 0$ and $c$ are real numbers, the function $$f(x) = c + ... 4 \Rightarrow \frac{f(x+y)}{x+y} -\frac{f(x-y)}{x-y} =4xy=(x+y)^2-(x-y)^2 or, \frac{f(x+y)}{x+y}-(x+y)^2=\frac{f(x-y)}{x-y}-(x-y)^2 make the substitutions, x\to \frac{x+y}{2} and y \to \frac{x-y}{2} Then the above becomes \frac{f(x)}{x}-(x)^2=\frac{f(y)}{y}-(y)^2=k(say) There you have f(x)=x^3+kx 4 Under the assumption that \epsilon is constant across a, then no additional assumptions are needed. Suppose that x_a is the fixed point of f_a, and choose e>0. Then there is a \delta such that for all b within \delta of b, you have e\epsilon\geq \vert f_b(x_a)-f_a(x_a)\vert=\vert f_b(x_a)-x_a\vert. Now, repeatedly apply f_b to ... 3 A function f:\ A\to B is tantamount to a subset G_f\subset A\times B having certain properties. In most cases A and B are clearly specified in advance, and one can then start right away to investigate whether f is injective or not. Very often A and/or B have to be surmised from the context. While the exact envisaged range B (e.g., {\mathbb ... 3 Part 1)$$f(x)=ax^n$$So, let n=2k. Then,$$f(x)=ax^n=a(x^2)^k$$Obviously, x^2 is an even function ((-x)^2=x^2). So,$$f(x)=ax^n=a(x^2)^k$$is an even function. Part 2) Let f(x),g(x) be two even functions. Then,$$f(x)=f(-x),\qquad g(x)=g(-x)$$Adding the two gives$$f(x)+g(x)=h(x)$$Using the above relation then gives ... 2 I'd like to offer another approach to building \tilde h. First step: h is C^1 and non-zero on K, hence there exists \epsilon>0 such that h(x)\ne0 whenever dist(x,K)\le\epsilon. Second step: let's take a function$$\phi(x) = \begin{cases}c\exp\left(-\frac{1}{1-|x|^2}\right), &|x|<1,\\0,&\text{otherwise.}\end{cases}$$where c is ... 2 The only functions which have this property are the constant functions. Why? Suppose f(x) \geq f(m) for all x. Then x=m is the global minimum of f. Likewise, if f(x) \leq f(M) for all x. Then x=M is the global maximum of f. So if f(M)=f(m), we have f(m) \leq f(x) \leq f(M)=f(m) for all x. This forces f(x)=f(m)=f(M) for all x so ... 2 Let the periodic function be f(x). One way to approach this would be to set$$ g(s):=\int_0^s(f(x)-\inf f(x))\,\text dx $$For example, consider f(x)=\sin x. The infimum of \sin is -1. So we would have$$g(s)=\int_0^s(\sin(x)-(-1))\,\text dx=\left.x-\cos x\right|_0^s=s+1-\cos s$$The result: 2 It is correct.$$t=\dfrac{-30\pm \sqrt{30^2-4(-5)(60-h)}}{2(-5)}t=\dfrac{-30\pm \sqrt{900+20(60-h)}}{-10}t=\dfrac{-30\pm \sqrt{900+1200-20h}}{-10}t=\dfrac{-30\pm \sqrt{2100-20h}}{-10}t=\dfrac{-30}{-10}\pm \dfrac{\sqrt{2100-20h}}{-10}t=3 \mp \dfrac{\sqrt{2100-20h}}{10}t=3 \mp \dfrac{\sqrt{2100-20h}}{\sqrt{100}}t= 3 \mp ...

2

HINT : \begin{align}f'(x) &= 3x^2 + 2ax + b \\ &= 3(x + \frac{a}{3})^2 + b - \frac{a^2}{3}\\ &\ge b - \frac{a^2}{3}\end{align} Now, $f'(x) > 0 \implies f(x)$ is an increasing function (an unrelated but good question to think about : does $f(x)$ increasing $\implies f'(x) > 0$?). What can you now say about the probability for which ...

2

We can view $\bar{x}_a$ as a minimizer of the continuous function $x\mapsto \|f_a(x)-f_a(f_a(x))\|$. If $f$ is jointly continuous as a function of $\mathbb{R}^n\times\mathbb{R}$, then the argmin correspondence that maps $a$ to the set of minimizers of this functions is upper hemi-continuous by Berge's maximum theorem (one has to show that locally all ...

2

You can use the contraction mapping estimates directly. You have the estimate $\|\bar{x}_a - f_a^{(k)}(x_0)\| \le {(1-\epsilon)^k \over \epsilon} \|f_a(x_0) - x_0\|$, so we can see that if we let $B = \sup_{a \in B(\hat{a},1)} \|f_a(x_0) - x_0\|$, then $\|\bar{x}_a - f_a^{(k)}(x_0)\| \le {(1-\epsilon)^k \over \epsilon} B$ for all $a \in B(\hat{a},1)$. So, ...

1

You already deduced that $f(1) = 2$. From the $f(f(n))=3n$ condition, we have that $f(f(1))=f(2)=3$, hence $f(2)=3$. Similarly, we have that $f(3)=f(f(2))=3\cdot 2=6$, and $f(6)=f(f(3))=3\cdot 3=9$, and $f(9)=f(f(6))=3 \cdot 6=18$. So until now we have that $$f(1)=2\, , \; f(2)=3 \, , \; f(3)=6 \, ,\; f(6)=9 \, , \; f(9)=18 \, .$$ Since $f(n)< ... 1 The only property that needs to have a relation to be a function is that (as you've said) function can only have one range per domain. Roughly speaking it 'means' that in a part of time (domain) you can be just in one place (range). It doesn't need to have any formal rule for a given relation (as$y=f(x)$) to be a function, it needs just to fulfill the ... 1 Indeed, your function does have a rule--infinitely-many possible rules, in fact. For example, if$g(x)$is literally any function$\Bbb R\to\Bbb R,$we could say that$F$is the function from$\{1,2,3,4\}$to$\Bbb R$given by ... 1 When$A$is a set,$|A|$is intended to mean the cardinality of$A$--the number of elements of$A.$So,$A$is a set with$m$elements and$B$is a set with$n$elements. Now, to build a function$f:A \to B,$for each$a\in A$we must pick a specific$b\in B$to call "$f(a).$" Given any$a\in A,$how many ways are there for us to pick such a$b\in B$? For ... 1 The set of all functions$A\to B$is sometimes denoted$B^A$, exactly because when$A,B$are finite then there are$|B|^{|A|}$(in the question$n^m$) functions. This is so because for every element$a\in A$there are$|B|$possibilities for$f(a)$, and these$|A|$choices are entirely independent and together determine$f$. Numbers of independent choices ... 1 HINT: For every$a_i$in$A$, there are$n$elements in$B$to which it can map. Try to proceed from here. SPOILERS BELOW Let$A=\{a_1,a_2, \cdots,a_m\}$and$B=\{b_1,b_2, \cdots,b_n\}$There are$n$possible values for$f(a_1)$, namely,$b_1,b_2, \cdots,b_n$. Similarly, there are$n$possible values for each$f(a_i)$. Thus net number of ways to map ... 1 The inverse of a function is with respect to function composition, not with respect to point-wise addition. So, the inverse of$e^x$is not$x^{-1}$but rather$\ln (x)$. Can you now work out the second function? As for describing these functions, it's a bit unclear what they aim at. Perhaps to draw a sketch of the graph, or to say something qualitative ... 1 No. if you consider $$f(x, y) = \frac{x^2}{y}$$, this has no limit as$(x, y) \to (0, 0)$basically because$y$can go to zero really fast (like$x^3$, for instance) so in this case the limit is$\infty$, or they can both go to zero with the same "speed", so the limit is$0$. $$f(x, x^3) \to \infty$$ but $$f(x, x) \to 0$$. What the text is saying is ... 1 Let$K_0$be the field of rationals. For any$n\ge 0$, let$K_{n+1}$be the intersection of all the subfields of the reals that contain$f(K_n)$. Finally, let$K$be the union of all the$K_n$. Note that$K$is closed under$f$. For if$a\in K$, then$a\in K_n$for some$n$, and therefore$f(a)\in K_{n+1}$. For countability, note that each$K_n$is ... 1 Remember that the logarithmic expression$y = \log(f(x))$exists when$f(x) > 0$, which implies that$x - 3 > 0$. So the domain of$y = \log(x - 3) + 2$is$x > 3$. Since the range of the logarithmic function with any linear equation is$\mathbb{R}$, the range of the given function is$\mathbb{R}$as well. 1 As you say, this is essentially truncating Taylor's series after the linear term(s). By Lagrange's form of the remainder in one variable (see Wikipedia or your favorite calculus textbook): $$f(x) = f(x_0) + f'(x_0) (x - x_0) + \frac{f''(\xi)}{2!}(x - x_0)^2$$ where$x_0 \le \xi \le x$. The approximation you cite is valid as long as the second-order term ... 1 The Chain Rule in Newton's notation says If$h(x) = f(g(x))$, then$h'(x) = f'(g(x)) * g'(x)$Since we see$2 - 2x^3$come up in$h(x)$, that gives us a strong suggestion that we can use$g(x) = 2 - 2x^3$and, therefore,$f(x) = x^4 + \frac{1}{x}$What remains is to find$f'(x)$and$g'(x)\$, and then plug it into the formula.

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