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18

$$\begin{align} \ln(1+e^x)-\frac x2 &= \ln(1+e^x)-\ln(e^{\frac x2}) \\ &= \ln\left((1+e^x)e^{\frac {-x}2}\right) \\ &=\ln(e^{\frac {-x}2} +e^{\frac x2}) \\ \end{align}$$ From this it should be obvious that the function is indeed even.


11

In a lot of cases, the domain and codomain change the function itself. The domain is clearly important, since a change of the domain changes what you're allowed to "plug into" the function. For example, if I have a function $f(x) = 2x$, it's not a function yet since I need to define the domain first. For example, if the domain is: $\{0\}$ then the ...


7

You can solve the equations for $E$ and $O$ in terms of $f$: $$\begin{align*} E(x)&=\frac12\big(f(x)+f(-x)\big)\\ O(x)&=\frac12\big(f(x)-f(-x)\big) \end{align*}$$ for each $x\in\Bbb R$. Thus, $E$ and $O$ are completely determined by the function $f$.


6

The domain and the codomain are intrinsic parts of a function. Consider the following examples: $$ f:\mathbb R\to\mathbb R,\ \ \ g:\mathbb R\to[0,\infty),\ \ \ h:[0,\infty)\to[0,\infty),\ \ \ j:[0,\infty)\to\mathbb R,\ \ \ k:\mathbb Z\to\mathbb N, $$ where $$ f(x)=x^2,\ \ \ \ g(x)=x^2,\ \ \ h(x)=x^2,\ \ \ \ j(x)=x^2\ \ \ k(x)=x^2. $$ We have that $h$ is ...


5

$$y=\frac{x+3}{x+1}=\frac{(x+1)+2}{x+1}=1+\frac2{x+1}$$ $$y=\frac2x \rightarrow y=\frac2{x+1}\rightarrow y= 1+\frac2{x+1}$$


5

Hint. One may recall that, for all $x \in \mathbb{R}$, $$ -1\leq \cos (x) \leq 1. $$ Then do you think it is possible to find a real number $x$ such that for example $$\cos(x)=2\,?$$


4

Another way of treating this is that a continuous function $ \ f(x) \ $ can be "separated" into "even" and "odd" components, $$ f_e(x) \ = \ \frac{f(x) \ + \ f(-x)}{2} \ \ \ \text{and} \ \ \ f_o(x) \ = \ \frac{f(x) \ - \ f(-x)}{2} \ \ . $$ Here, we have $$ f_o(x) \ = \ \frac{[ \ \log(1+e^x) \ - \ \frac{x}{2} \ ] \ - \ [ \ \log(1+e^{-x}) \ - \ ...


4

The last implication is an error. $$(x_1-1)^2=(x_2-1)^2 \Rightarrow |x_1-1|=|x_2-1|$$ So, either $x_1=x_2$, or $x_1=2-x_2$.


4

We have a counterexample. $f(0)=0=f(2)$, though $0\neq2$.


4

I would like to write an addendum to Travis' answer. By Lagrange's inversion formula, the solution $x$ of $x^3-x+z=0$ can be written as: $$ x = \sum_{k\geq 0}\binom{3k}{k}\frac{z^{2k+1}}{2k+1} $$ under the assumption $|z|< \frac{2}{\sqrt{27}} $ that makes such a series convergent. It follows that the inverse function $g(y)$ of $f(x)=x^3+x$ can be ...


4

GCF means greatest common factor, but usually you will meet it as GCD, menaing greatest common divisor. On the other hand LCM means least common multiplier. Anyway to solve the problem, one can use a well-know identity $LCM(a,b) \cdot GCD(a,b) = a\cdot b$. So just substitute $a=24$ and the given values to get the answer.


4

This holds in a linear well order (the other answers show this is needed). For suppose $L = \{x \in A: f(x) < x \}$ is non-empty. Then this has a minimal element $m \in L$. But then $f(m) < m$ and so $f(f(m)) < f(m)$ as $f$ is a morphism. This shows that $f(m) \in L$ and $f(m) < m$ so this contradicts minimality. So $L$ is empty. Done.


4

Notice that, by definition of an inverse function: $f(g(x))=x$ Then: $\frac{d}{dx} f(g(x))=1$ $f'(g(x)) \cdot g'(x)=1$ (by the Chain Rule) And, as @Alex G. pointed out: $g'(x)=\frac{1}{f'(g(x))}$ So, then it should follow immediately that: $g'(\frac{π}{4}+1)=\frac{1}{f'(g(1+\frac{π}{4}))}$ And then just do the calculations... EDIT: Sure enough, ...


3

If you assume $\forall n\geq 1$, then I gather any primitive of $f(x) =\frac{e^x-1}{x}$ should do. The rationale behind it is to find a power series $F$ with $F(x) = \sum_{n=1}^\infty \frac{f^{(n)}(0)}{n!} x^n$ (and non-zero radius of convergence) satisfying what you want. That is, $$ F(x) = \sum_{n=1}^\infty \frac{x^n}{n\cdot n!}. $$ Deriving this, you get ...


3

Look at the polygonal representation of two spaces. Now removing a disc from the middle, the rest of the space will deformation retract into the boundary, which is nothing but wedge of two circle. (Just draw the picture of polygonal presentation, you can actually see what is happening.)


3

HINT: First write $$\left|\frac{x-1}{x^2+1}+1\right|=\left|\frac{x(x+1)}{x^2+1}\right|\le |x|\,|x+1|$$ Second, restrict $x$, say for example $-3/2\le x\le -1/2$. Can you finish now?


3

From $f^{-1}(x)=\frac{1}{f(x)} \tag 1$ by replacing $x: = f(1)$ we get: $1 = \frac{1}{f(f(1))}$ so $f(f(1))= 1$ By applying $f$ to (1) we get $x=f(\frac{1}{f(x)})$, so $1=f(\frac{1}{f(1)})$. Therefore $f(f(1))=f(\frac{1}{f(1)})$ and, using injectivity, $f(1)=\frac{1}{f(1)}$ then $f(1)=1$


3

A function is a rule. You consider some input values, consider the rule, and then find an output value. But this isn't a very rigorous definition. We require a further degree of precision necessary for good mathematics: Consider your favorite function, $$f(x)=x^2$$ A casual observer may consider this to be a parabola defined on the real line, but is that ...


3

If "better" means better suited for certain purposes then yes. Fourier Series, for example, allow approximation which are extremely well suited for application in physics in general, also for numerical approximation (fast Fourier transform) or for analysing partial differential equations (since they map differential operators to multiplicatin operators). It ...


3

Utterly impossible: $$y?1=y\cdot 1 = y,$$ $$y' = \frac{d(y?1)}{dx}=\frac{dy}{dx}\cdot\frac{d1}{dx} = 0$$


3

I am also quite new to this topic, so I am not sure whether it is true, but here my ideas: $u_n = 2u_{n-1} + u_{n-2}$, since we can form a word of length $n$ (not ending in $z$) by either taking a word of length $n-1$ and adding either $x$ or $y$ or by taking a word of length $n-2$ and adding $zx$. $u_n = u_{n-1} + 2u_{n-2}$, since we can fill $n$ spaces ...


2

strictly monotone increasing. Any such function is continuous almost everywhere. But what do you mean by singularities? It could have jump discontinuities, in fact, it could have infinitely many. As pointed out by Merlinsbeard, it is in fact almost everywhere differentiable (which is even stronger)


2

You can complete the square so that you have $y = (x-1)^2 - 1$. Now, for given $y \geq -1$, we have $x_1, x_2$ which map to $y$, namely $x_1 = \sqrt{y + 1} + 1$ and $x_2 = 1 - \sqrt{y+1}$ and so it is not injective.


2

From $a^2=b^2$ you can deduce that either $a=b$ or $a=-b$. So it's incorrect to deduce from $(x_1-1)^2=(x_2-1)^2$ that $x_1-1=x_2-1$.


2

Hint: Consider what possible values $f(0)$ could have, using the fact that $$ f^4(x) = \int\limits_0^x t^2 f^3(t) dt. $$ Use this to find $C$.


2

$f(x)$ is a contraction, so it has a unique fixed point by the contraction mapping theorem, that is a point where $f(x)=x$. So there is exactly one point where $y=f(x)=x$.


2

Suppose that $f(x)=E_1(x)+O_1(x)=E_2(x)+O_2(x)$, with the $E_i$ even and the $O_i$ odd ($i=1,2$). Then $$ E_1(x)-E_2(x)=O_2(x)-O_1(x)$$ The left-hand side is even, and the right-hand side is odd. But the only function which is both even and odd is the zero function.


2

Schaum's outlines are a great resource and contain lots of exercises to work through. For the "Mathematics 1" categories listed, you could probably find those all in the beginning and advanced calculus outlines. For the "Mathematics 2" categories, Schaum's Outline of Differential Equations will cover part of it. For the rest, I'd recommend Applied Partial ...


2

For $x,y \in I$ we have $$|\alpha f(x)+\beta g(x) - \alpha f(y) - \beta g(y)|\leq \alpha |f(x)-f(y)| +\beta |g(x)-g(y)|$$ and the result follows by assumption.


2

If $T:X\to Y$ is a linear operator with graph $G$, then note that $G$ is a vector subspace of $X\oplus Y$. Using the fact that $X$ is finite dimensional try to convince yourself that $G$ is finite dimensional and hence closed. Now, use the Closed Graph Theorem.



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