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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