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 $$(a^x)^y=a^{xy}.$$


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

Coming up with any sort of answer here will require a large number of assumptions. The most important factor is how the growth rate changes with population. The one solid value you've given is that the average lifespan is 100 years. I will make the simplifying assumption that this is from all causes, so includes the impact of unnatural deaths. That means ...


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

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


2

The induction step: 2): If the statement $4^{2n} > 15n$ is true for one particular $n=k$ then it must also be true for $n = k+1$. You want to assume that you know $4^{2k} > 15k$ for one particular $k$. You can assume this was given to you by God. You need to prove therefor that $4^{2(k+1)} > 15(k+1)$ and to prove that you will use $4^{2k} > ...


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.


1

Set $y=0$ and your identity becomes $$\exp(ax)=a\exp(x),$$ which is notoriously false. Still not convinced ? With $x=1$, $$\exp(a)=a\,e\ ???$$ A correct statement is $$(\exp(x+y))^a=\exp(a(x+y))=\exp(ax+ay)=\\\exp(ax)\exp(ay)=(\exp(x))^a(\exp(y))^a=(\exp(x)\exp(y))^a.$$


1

Consider the concave function $f(t) = \sqrt[t]{1-t}$ (shown below.) Using Jensen’s inequality, $$f(a) + f(b) \leqslant 2f(\tfrac12) = \tfrac12$$ —- To show $f$ is concave, it is enough to show $g=\log f$ is concave, as $t\mapsto e^t$ is convex and increasing. Perhaps the easiest way for this is to note the Taylor series for $x \in (0, 1)$ for $g= -\...


1

Start with $n = 0$ and observe that the inequality is true. Then, for $n = k$, assume that $4^{2k} > 15$. Now, let $n = k + 1$. (You must show that $4^{2(k+1)} > 15(k+1) = 15k + 15$.) To this end, $$ 4^{2(k+1)} = 4^{2k+2} = 4^{2}4^{2k} \geq 16(15k) = (15 + 1)15k = 15k + 15(15k) > 15k + 15.$$ (as you wanted to show)


1

Base case: N=1 gives you 16>15, which is true. Suppose $4^{2n}>15n$ is true for n. Then $4^{2(n+1)}=4^{2n}16>16(15n)>15n+15$ You can prove the last inequality by induction as well.


1

First, is the basis step true? If so, that should be in your question. It would show you expended some effort. Second, the induction step start with $4^{2n}>15n.$ You do some math on that until it looks like $4^{2(n+2)}> 15(n+1).$ Then you all done but mopping up.


1

For $n=1$ we obtain: $16>15,$ which is true. Now, by the assumption that $4^{2n}>15n$ is true we obtain: $$4^{2(n+1)}=16\cdot4^{2n}>16\cdot15n>15(n+1).$$ Can you end it now?


1

For a rigorous argument, let $\eta(x)$ be any measurable function such that $\eta(x)/x \to 0$ as $x\to\infty$. In particular, for each $\epsilon \in (0, 1)$, there exists $R > 0$ such that $|\eta(x)/x| < \epsilon$ whenever $x \geq R$. So, if the function $\delta(t)$ is implicitly defined by the relation $$ \int_{t}^{\infty} e^{-x+\eta(x)} \, \mathrm{d}...


1

near $+\infty$, $$e^{-x+o(x)}\sim e^{-x} $$ and $$\int_0^{+\infty}e^{-x}dx \text{ converges}$$ thus $$\int_t^{+\infty}e^{x+o(x)}dx \sim \int_t^{+\infty}e^{-x}dx$$ or $$\int_t^{+\infty}e^{x+o(x)}dx \sim e^{-t}\sim e^{-t+o(t)}.$$


1

$c(t) = \sum_{m=1}^n a_m e^{i w_m t}$ the $w_m$ are distinct and the $a_m$ are non-zero. The shifts of the functions $e^{i w_m t}$ generate a vector space of dimension $n$, the same as the vector space generated by the shifts of $c(t)$. Pick some $t_1,\ldots,t_n$ such that the $c(t+t_j)$ are linearly independent, we get some $B_{m,j}$ such that $$\forall ...


1

I borrowed Paul Sinclair's formula for the net rate of population growth, $$ 0.04 \left( 1 - \frac P{10^{10}} \right) $$ where $P$ is the population. I assumed that at time $t = 0$ years, the Earth ships off $2$ billion of its $7$ billion population to the first colony. At time $t = 1$ year, the colony has one year of growth (from $2$ billion) and the Earth ...


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