# Tag Info

27

Note that applying function to a matrix is meant in sense of series, that is if $$\phi(z) = c_0 + c_1 z + c_2 z^2 + \dots$$ then $$\phi(\mathbf A) = c_0 \mathbf I + c_1 \mathbf A + c_2 \mathbf A^2 + \dots.$$ Observe that $$\phi(z) = \frac{e^z - 1}{z} = 1 + \frac{z}{2} + \frac{z^2}{6} + \dots = \sum_{k = 1}^\infty \frac{z^{k-1}}{k!}.$$ Also, to your ...

27

You can use the fact that, $\max\{x,y\}=\frac{x+y+|x-y|}{2}$. Therefore, $\max\{x,3x\}=\frac{x+3x+|x-3x|}{2} = \frac{4x+|-2x|}{2} =2x+|x|$.

19

$f^{-1}(A)$ is the preimage of the set $A$, it exists even if $f$ is not invertible. For $f: X\to Y$, it is defined as $$f^{-1}(A) = \{x\in X: f(x)\in A\}.$$ The notation is slightly confusing, I would say, but one gets used to it. It isn't really that bad of a notation, since, if $f$ is actually invertible and its inverse is $g$, then $g(A) = f^{-1}(A)$ ...

18

$$f(x)+f(-x)=\log\left[\left(x+\sqrt{x^2+1}\right)\cdot\left(-x+\sqrt{x^2+1}\right)\right]=\log(x^2+1-x^2)=0.$$

12

I will probably put more emphasis on those transformation properties of the $\max(\cdot)$ function which is reasonably "obvious" and easy to remember/use. For example, $\max(a+b,a+c) = a + \max(b,c)$. $\max(ab,ac) = a \max(b,c)$ when $a > 0$. $\max(a,-a) = |a|$. This may help a student to get familiar with such "tools" for attacking similar problems. ...

10

$f^{-1}$ here does not mean the inverse of $f$. It is a horrifically bad overuse of notation and can be very misleading. $f^{-1}(A)$ where $A$ is a set is the pre-image, i.e. $$f^{-1}(A) = \{x\in X: f(x) \in A\}.$$ The pre-image always exists. If $f$ were invertible, then this would coincide with what you think; however the pre-image takes care of the ...

8

$$f(-x) = \ln\left(-x + \sqrt{x^2 + 1}\right) = \ln\left( \left(-x + \sqrt{x^2 + 1}\right) \frac{x +\sqrt{x^2 + 1}}{x +\sqrt{x^2 + 1}} \right) \\ = \ln\left(\frac{1}{x +\sqrt{x^2 + 1}}\right) = -f(x)$$

8

Note that since $$\exp(A)=I+A+\frac1{2!}A^2+\frac1{3!}A^3+\cdots$$ then $A$ and $\exp(A)$ commute. Thus also $\exp(A)-I$ and $A^{-1}$ commute.

6

max{x,3x} + min{x,3x} = 4x max{x,3x} - min{x,3x} = |2x| ---------------------------- 2max{x,3x} = 4x + 2|x| max{x,3x} = 2x + |x| As a bonus, min{x,3x} = 2x - |x|

6

Just to summarize, there is a pairing here: $$x+y=1\;\;\;\Rightarrow\;\;\;f(x)+f(y)=1$$ This is readily verified by direct computation. In this case that means we can combine the given terms in pairs, so the answer is 5.

4

One has $$\cos(0x)=1,\quad \cos(1x)=\cos x,\qquad\cos(2x)=2\cos^2x-1\ .$$ This shows that for $0\leq n\leq 2$ the functions $x\mapsto \cos(nx)$ can be written "as polynomials in $\cos x\>$". Furthermore, it follows from the addition theorem for $\cos$ that $$\cos\bigl((n+1)x\bigr)+\cos\bigl((n-1)x\bigr)=2\cos(nx)\cos x\ ,$$ since the $\sin$ terms cancel. ...

4

I like the question! It is not immediately obvious to me, however, that it suffices to match your form to the given function at three points. Perhaps that is the case, but if so it requires a separate argument. Alternatively, having discovered the final form you could verify it directly. I would address the original problem this way: Your function, $f$, ...

4

HINT: Don't work hard, work smart. Recall, or prove, that ${}^C({}^BA)\sim{}^{C\times B}A$. Now use the facts that $\mathcal P(X)\sim{}^X2$ and that $\Bbb{Z^+}\sim\Bbb{Z^+\times Z^+}$.

4

Try writing out the series for $exp$ and simplify the equation before you evaluate! You wouldn't have to worry about dividing!

4

Hint: consider $f(x+k\cdot 346)$ for the first few values of $k$.

3

The argument below is a parity argument. Let us suppose that each player makes a tick mark in her notebook for every drawn game she is involved in. Note that a drawn game results in $2$ tick marks, so the total number of tick marks is even. Each player makes $0$ to $17$ tick marks. If all the players have different numbers of draws, then the total ...

3

Hint: use $$(a-b)*(a+b) = a^2 - b^2$$ and $$\log\frac{1}{a} = - \log(a)$$ properties.

3

For any function $f : \mathbb R \rightarrow \mathbb R$, the set $f^{-1}( a, \infty)$ always exists, and it is defined by $$\{ x: \in \mathbb R : f(x) > a \},$$ although it is not always measurable.

3

Hint: This is based on @columbus8myhw's hint. Consider the polynomial $$kf(k)-1$$ which has degree 99 is zero for all $k=1,2,3, \cdots, 99$. So you know the polynomial $kf(k)-1$ has the form $$C(k-1)(k-2)(k-3)\cdots(k-99)$$ and you can find $C$ when you put $k=0$. In general, its a good idea to find a polynomial that has as many zeros as its degree ...

3

For the first highlighted step, note that \begin{align*} \| x - a \| &\le \max(\|x - a\|, \|y - b\|) = \|(x,y) - (a,b)\| \\ \| y - b \| &\le \max(\|x - a\|, \|y - b\|) = \|(x,y) - (a,b)\| \\ \| a \| &\le \max(\|a\|, \|b\|) = \|(a,b)\| \\ \| b \| &\le \max(\|a\|, \|b\|) = \|(a,b)\|. \end{align*} Therefore, \begin{align*} \| B \| \| x - a \| \| ...

3

As a first thought, I would say $f^{-1}$ is always injective, since a function is invertible if and only if it is injective and surjective. $f^{-1}$ is certainly invertible since its inverse is $f$, thus it is also injective. For the bounded question, consider the function f(x) = \left\{ \begin{aligned} &\dfrac{1}{1+(\frac{1}{x})^2} &&: x ... 3 Hint: Let y be a solution to x=x^2+3x+1, then you get a quadratic equation in f(y) which you can solve for f(y). 3 You can, but why let your computer "run" to bigger numbers, if running to \sqrt{n} is enough? If n is not a prime then n admits a prime factor p\leq \sqrt{n}. 2 The trigonometric functions relate an angle to a length. More precisely, in a rectangle triangle, the sine, cosine and tangent of an angle are the ratio of two of the sides.\cos(\theta)=\frac xr,\ \sin(\theta)=\frac yr,\ \tan(\theta)=\frac yx.$$For convenience, we will now assume that r=1, as this doesn't change the proportions. ... 2 Without imagination...$$ \frac{x+1}{x^3-x} = \frac{x+1}{x(x^2-1)} = \frac{x+1}{x(x-1)(x+1)} = \frac{1}{x(x-1)} $$2 Note that f(-x) = -ax + b|x|+ c, thus breaking into symmetrical and anti symmetrical combinations one obtains$$ f(x) - f(-x) = 2ax\\ f(x) + f(-x) = 2b|x| + 2c $$Thus$$ a = \frac{f(1) - f(-1)}{2} = \frac{3 + 1}{2} = 2\\ c = f(0) = 0\\ b = \frac{f(1) + f(-1)}{2} - c = \frac{3 - 1}{2} = 1 $$Second variation of the same idea:$$ f(-x) = \max(-x, -3x) = ...

2

Maybe you can try using sequential continuity, equivalent to continuity in metric (gen. 1st countable) : Let $x_n \rightarrow x , y_n \rightarrow y$ , we want to show $x_ny_n \rightarrow xy$ : We have that $(x_n-x)(y_n-y) \rightarrow 0$ as $n \rightarrow \infty$ , since each of the terms on the left can be made small -enough. Now: x_ny_n -xy= x_ny_n ... 2 As in Nemo's reply. \begin{align} f(x,y,z,t)&=-2\left( \left( x+5 \right) ^{ 2 }+\left( y+10 \right) ^{ 2 }+\left( z+15 \right) ^{ 2 }+t^{ 2 } \right) +200\left( x+y+z+t \right) \\ &=-2\left( \left( x+5 \right) ^{ 2 }+\left( y+10 \right) ^{ 2 }+\left( z+15 \right) ^{ 2 }+t^{ 2 } \right) +200\cdot 150 \\ &=-2\left( \left( x+5 \right) ^{ 2 ... 2 I can propose this method. Let y = \frac{x^2}{2} + cx^{3/2}. $$y is strictly increasing (when c\ge0), so inverse function exists. And we have y\to\infty when x\to\infty. x^{3/2}=o(x^2), and you start correctly:$$ x\sim\sqrt{2y}. $$Next step:$$ x\sim\sqrt{2y}(1+z(y)), $$where z(y) = o(1). Put back:$$ y = y(1+z)^2 + c(2y)^{3/4}(1+z)^{3/2}, ... 2 Fromf(x) =\frac12x^2+cx^{3/2} =\frac12 x^{2}(1 +2cx^{-1/2}) $,$\sqrt{f(x)} =\sqrt{\frac12} x(1 +2cx^{-1/2})^{1/2} =\sqrt{\frac12} x(1 +cx^{-1/2}+O(1/x)) $. Therefore$x =\sqrt{2 f(x)}(1 +2cx^{-1/2})^{-1/2} =\sqrt{2 f(x)}(1 +O(x^{-1/2})) \approx\sqrt{2 f(x)} $. Putting this back in,$\begin{array}\\ x &\approx\sqrt{2 f(x)}(1 +2c\left(\sqrt{2 ...

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