# Tag Info

11

Forget that $n$ is an integer. For $x > 0$ we have $$\lim_{y\to\infty} \left(1+\frac{x}{y}\right)^y = \lim_{y\to\infty} \left(1+\frac{1}{\frac{y}{x}}\right)^{x\cdot \frac{y}{x}}$$ If $y \to \infty$ then $z:= \frac{y}{x} \to \infty$ as well so change of variables gives that this is equal to $$\lim_{z\to\infty} \left(1+\frac1z\right)^{xz}$$ which is the ...

6

$$\sqrt[n]{9^n+7^n}=9\sqrt[n]{1+\left(\frac{7}{9}\right)^n}\rightarrow9.$$

6

We consider the equation $$3^x+4^x=2^x+5^x~~~~(1)$$ Use Lagranges Mean Value Theorem (LMVT) for the function $f(t)=t^x$ for two intervals $(2,3)$ and $(4,5)$. So $$\frac{3^x-2^x}{3-2}=xt_1^{x-1}, ~~~t_1 \in (2,3)~~~~(2)$$ and $$\frac{5^x-4^x}{5-4}=xt_2^{x-1}, ~~~t_2 \in (4,5)~~~~(3).$$ By equating (2) and (3), we get (1) and $$xt_1^{x-1} = xt_2^{x-1}, ~~~... 6 Let x=0. Thus,$$b+c=0$$Let x=\frac{\pi}{2}. Thus,$$a+c\cdot e^{\frac{\pi}{2}}=0.$$Now, substitute x=-\frac{\pi}{2} and solve this system. We obtain:$$-a+c\cdot e^{-\frac{\pi}{2}}=0.$$Thus,$$c\left(e^{\frac{\pi}{2}}+e^{\frac{\pi}{2}}\right)=0$$or$$c=0,$$which gives a=0 and b=0. 5 For positives a, b and c such that a and c are different from 1, we obtain:$$c^{\log_cb}=b=\left(c^{\log_ca}\right)^{\log_ab}=c^{\log_ab\log_ca}.$$We used$$(a^x)^y=a^{xy}.$$5 Using the result that$$ \mathrm{PV}\sum_{k\in\mathbb{Z}}\frac1{x+k}=\pi\cot{\pi x} $$we get$$ \begin{align} \sum_{k=1}^\infty\frac1{1+k^2\pi^2} &=\frac{i}{2\pi}\mathrm{PV}\sum_{k\in\mathbb{Z}}\frac1{i/\pi+k}-\frac12\\ &=\frac{i}{2\pi}\pi\cot\left(\pi\frac{i}\pi\right)-\frac12\\[3pt] &=\frac12\coth(1)-\frac12\\[3pt] &=\frac1{e^2-1} \end{...

5

Raise both sides to the $n$th power: $$\left(\frac{n+2}{n-2}\right)^n < (n+2)^2,$$ then multiply through by $(n-2)^2/(n+2)^2$: $$\left(\frac{n+2}{n-2}\right)^{n-2} = \left(1+\frac{4}{n-2}\right)^{n-2} < (n-2)^2.$$ The left side is bounded by $e^4$, so as long as $n-2 > e^2$, the inequality is guaranteed to be satisfied. This proves it true for ...

5

Take$$\begin{array}{rccc}f\colon&\mathbb R&\longrightarrow&\mathbb R\\&x&\mapsto&\begin{cases}x+1&\text{ if }x\in\mathbb Z\\x&\text{ otherwise.}\end{cases}\end{array}$$Then $(\forall x\in\mathbb R):f(x+1)>f(x)$. However, $f$ is not monotonically increasing. It is more natural to deduce that your function is monotonically ...

4

Let $f(t)=t^x.$ $x>1$ or $x<0$. Since $f$ is a convex non-linear function and $(5,2)\succ(4,3),$ by Karamata we obtain: $$f(5)+f(2)>f(4)+f(3)$$ or $$5^x+2^x>4^x+3^x,$$ which says that in this case our equation has no roots. $0<x<1.$ Here, $f$ is a concave function and by Karamata again we obtain: $$5^x+2^x<4^x+3^x,$$ which says that ...

4

Hint: Use the expansion $$\coth x=\sum_{n=-\infty}^{\infty}\dfrac{x}{n^2\pi^2+x^2}$$

4

By AM-GM $$2\cos \left ( \frac{x^{2}+x}{6} \right )=2^{x}+2^{-x}\geq2\sqrt{2^x\cdot2^{-x}}=2.$$ The equality occurs for $2^x=2^{-x},$ which gives $x=0$. But, $$2\cos \left ( \frac{x^{2}+x}{6} \right )\geq2$$ gives $$2\cos \left ( \frac{x^{2}+x}{6} \right )=2,$$ which says that it's ewnough to check that $0$ is indeed the root. Can you end it now?

4

Not true. $f(a,0) = \frac{1}{\sqrt{a}} \to \infty$ as $a \to 0+$.

3

Starting with $$\left(\frac{x}{3}\right)^n e^{-(x/3)^n} = [\mathrm{const.}]$$ first note that the exponent of $e$ contains a negative sign. We can produce that on the outside by negating: $$-\left(\frac{x}{3}\right)^n e^{-(x/3)^n} = -[\mathrm{const.}]$$ and then we see by grouping, that $$\left[-\left(\frac{x}{3}\right)^n\right] e^{\left[-\left(\frac{x}{... 3 Since$$e^z=-1\iff z=\pi i+2n\pi i\text{ for some }n\in\mathbb Z,$$and since the number i has two square roots: \pm\frac1{\sqrt2}(1+i), then$$e^{z^2}=-1\iff z=\pm\frac{\sqrt{(2n+1)\pi}}{\sqrt2}(1+i)\text{ for some }n\in\mathbb Z.$$3 What you have found is the slope function for your original function at any point where that function is defined. This includes the point that you're given. Just use y-y_0=m(x-x_0) to form the equation of the tangent line. The point is given ((x_0,y_0)=(1,0)) and you know the slope at that point:$$ y-0=f'(1)(x-1). $$3 Hint: 2^x + 2^{-x} \geq 2 by AM-GM, and 2\cos(\frac{x^2+x}{6})\leq 2 3 There is a unique real solution x, but it cannot be expressed in terms of elementary functions. Of course you can use numerical methods to approximate the value of x, and it should be clear that x is very slightly larger than 1. With the help of a computer I quickly found that$$x\approx1.0030899874071590958.$$2 It is always true that$$\ln(\exp(g))=g$$no matter how complex an expression g may be. There is no need to take the logarithm of g, or transform products into sums, or any other such thing: the logarithm undoes the exponential, no more, no less. So$$f(t)= e^{-(\theta/t)^{\beta}-c\frac{\Gamma\Big(\alpha,\big(\frac{\theta}{t}\big)^\beta\Big)}{\Gamma(\...

2

If you have Tet(x), then $$H_n(x)=\text{Tet}(\text{Tet}^{−1}(x)+n)$$ You could visit https://math.eretrandre.org/tetrationforum/index for questions and details. Also, math.eretrandre.org/tetrationforum/showthread.php?tid=1017 for an implmentation of $\text{Tet}^{−1}(x)$ and $\text{Tet}(x)$. Kneser's 1949 paper used exactly that equation to generate the ...

2

HINT: \begin{align} &\hphantom{=.}\int \ln(e^x\sin^3x)dx \\ &= \int \ln(e^x)+\ln(\sin^3x)dx \\ &= \int (x+3\ln(\sin x))dx \\ &= \frac{x^2}{2}+3\int \ln(\sin x)dx \end{align}

2

$$(1+x)e^{ax/(1+x)}=b$$ $$e^{a-a/(1+x)}=\frac{b}{1+x}$$ $$\frac1be^a=\frac1{1+x}e^{a/(1+x)}$$ $$\frac{a}be^a=\frac{a}{1+x}e^{a/(1+x)}$$ $$\frac{a}{1+x}=W_k{\left(\frac{a}be^a\right)}$$ $$1+x=\frac{a}{W_k{\left(\frac{a}be^a\right)}}$$ $$x=\frac{a}{W_k{\left(\frac{a}be^a\right)}}-1$$ Where $W_k(z)$ is the $k$th branch of the Lambert W function.

2

This is not true. Take, for instance, $A=\left[\begin{smallmatrix}2\pi i&0\\0&0\end{smallmatrix}\right]$. It is skew-Hermitian and $e^A=\left[\begin{smallmatrix}1&0\\0&1\end{smallmatrix}\right]$. However, $A$ is not of that form that you mentioned.

2

Taking the logarithms of the desired identities $y_j = a b^{x_j}$ gives a linear equation system \begin{align} 1 \cdot \log a + x_1 \cdot \log b = \log y_1 \\ 1 \cdot\log a + x_2 \cdot\log b = \log y_2 \end{align} which can be solved for $(\log a,\log b)$ with the usual methods. The determinant of the coefficient matrix is $x_2 - x_1$, so that a ...

2

If $y_1=ab^{x_1}$ and $y_2=ab^{x_2}$, then$$\frac{y_2}{y_1}=\frac{ab^{x_2}}{ab^{x_1}}=b^{x_2-x_1}.$$So, take $b=\left(\frac{y_2}{y_1}\right)^{1/(x_2-x_1)}$. And now $a=\frac{y_1}{b^{x_1}}$.

2

Yes, your solution is correct. Here's how you find the inverse of an exponential function: $$y=3-e^x\implies e^x=3-y.$$ A logarithmic function is the inverse of an exponential function by definition: $$y=a^x\Longleftrightarrow x=\log_a{y}.$$ Therefore: $$3-y=e^x \Longleftrightarrow x=\ln{(3-y)}.$$ The only thing you need to do now is change the names of ...

2

$$9^{n}<7^{n}+9^{n}<9^{n}+9^{n}=2\cdot 9^{n}$$ $$9<\bigg(7^{n}+9^{n}\bigg)^{\frac{1}{n}}<\bigg(2\cdot 9^{n}\bigg)^{\frac{1}{n}}$$ Using Squeeze Theorem $$\lim_{n\rightarrow \infty}\bigg(7^{n}+9^{n}\bigg)^{\frac{1}{n}}=9.$$

2

This can be solved with the Lambert W function. It is the inverse function of $\exp(x)x$. First, multiply the equation by $\frac{a}{b}$ so the coefficients of $x$ match, then migrate the coefficient of the exponential into the exponent, like so: $$\exp(ax+\ln\frac{a}{b})+ax+\frac{a}{b}c=0.$$ Next, subtract $\frac{a}{b}c$ and add $\ln\frac{a}{b}$ so we ...

2

In general $$\lim_{x\to \infty} \left( 1+\frac{a}{x} \right) ^{bx}=e^{ab}$$ and they are only based on the definition of the number $e$ $$e:=\lim_{n\to \infty} \left( 1+\frac{1}{n} \right)^n$$ Here is a simple proof: \begin{align} \lim_{x\to \infty} \left( 1+\frac{a}{x} \right) ^{bx} &= \lim_{x\to \infty} \left[ \left( 1+\frac{a}{x} \right)^{x/a} \... 2 Apply binomial theorem resp. the binomial series, then \left(1+\frac1n\right)^{nx}=1+x+\sum_{k=2}^{\infty}\frac{x(x-\frac1n)...(x-\frac{k-1}n)}{k!} $$and$$ \left(1+\frac xn\right)^{n}=1+x+\sum_{k=2}^{\infty}\frac{(1-\frac1n)...(1-\frac{k-1}n)}{k!}x^k $$Both expansions converge in their coefficients to the same limit \frac{x^k}{k!}, the complicated ... 2 Consider that for x>0, n and \frac nx both tend to infinity so that$$\lim_{n\to\infty}\left(1+\dfrac1n\right)^n=\lim_{n\to\infty}\left(1+\dfrac xn\right)^{n/x}=e.$$Then by continuity, you can raise to the x^{th} power inside the limit,$$\lim_{n\to\infty}\left(1+\dfrac1n\right)^{nx}=\lim_{n\to\infty}\left(1+\dfrac xn\right)^n=e^x. For ...

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