134

There are two things I should point out. One is that the arithmetic mean doesn't properly measure annual growth rate. The second is why. The correct calculation for average annual growth is geometric mean. Let $r_1,r_2,r_3,\ldots,r_n$ be the yearly growth of a particular investment/portfolio/whatever. Then if you invest $P$ into this investment, after $n$ ...


43

After a 50% loss you need a 100% gain to break even. In that scenario the arithmetic average return is 25% and the geometric average return is 0%. It is more important to maximize geometric rather than arithmetic average return -- and this is intimately connected with the concept of risk-adjusted return and mean-variance optimization. Given a set of ...


27

This looks like the difference between an arithmetic mean and a geometric mean. Each year, if you invest $\$1000$ at the start and cash out at the end of the year, you want to use the arithmetic mean pointing at Mr Hare But if instead you invest $\$1000$ at the start of year $1$ and keep all the money invested until the end of year $5$ then you want to ...


26

From a friend who doesn't want to use math.se... Write the inequality as $$ \frac{1}{1+x} \left(1+x\right)^{1+\frac{1}{x}} + \frac{x}{1+x} \left(1+\frac{1}{x}\right)^{1+x} \leq 4.$$ Let $f(x) =(1+x)^{1+\frac{1}{x}}$. If $f$ is concave, then we are done since this means that $$4=f(1)=f((1- \alpha)x + \alpha y)) \geq (1- \alpha) f(x) + \alpha f(y) = \...


18

Write your approximation $$a^{c-bc} \cdot (a+1)^{cb}=a^c\left(1+\frac 1a\right)^{bc}$$ and your approximation is $$\left(1+\frac 1a\right)^{b}\approx 1+\frac ba$$ Which is the first two terms of the binomial expansion. It will be reasonably accurate when $\frac ba \ll 1$ The next term is $\frac {b(b-1)}{2a^2}$


14

This is a somewhat subtle concept. Say you have two fund managers, Alice and Bob. Alice has return exactly 5% per year and Bob has return either 0% or 10% per year with equal probability. Then clearly both fund managers will have the same average return (in expectation) if you define "average return" as just the arithmetic mean of your percent returns ...


13

This starts with a description of an answer, and then drills down to details. This question is more interesting than many. The original question's answer seems to be "no". One can construct, inductively, a sequence of intervals $I_k=[a_k,a_{k+1}]$ (with $a_k$ increasing to $1$), and values for $f$ on $I_k$, and a sequence of $n_k\to\infty$, so that the ...


13

$13/9$ is a convergent of the continued fraction for $e^{1/e}$, so it is expected to be unusually close to $e^{1/e}$. The continued fraction looks like $$ e^{1/e} = 1+\frac{1}{\displaystyle 2+\frac{1}{\displaystyle 4+\frac{1}{55+\ddots}}} $$ and since $55$ is exceptionally large, the approximation $$ 1+\frac{1}{\displaystyle 2+\frac{1}{\displaystyle 4}} = \...


12

One may write \begin{align} \sum_{n = 1}^{ \infty}\frac{n^2}{n!}&=\sum_{n = 1}^{ \infty}\frac{n(n-1)+n}{n!} \\\\&=\sum_{n = 1}^{ \infty}\frac{n(n-1)}{n!}+\sum_{n = 1}^{ \infty}\frac{n}{n!} \\\\&=\sum_{n = 2}^{ \infty}\frac{1}{(n-2)!}+\sum_{n = 1}^{ \infty}\frac{1}{(n-1)!} \end{align}Can you take it from here?


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 ...


10

Your second computation is equal to your first one: $$1 + \frac{2+\frac{3+\frac{4+\frac{5+ \frac{\cdots}{\cdots}}{5}}{4}}{3}}{2} = 1 + \frac{2}{2} + \frac{\frac{3}{3} + \frac{\frac{4}{4} + \frac{\frac{5}{5} + \frac{\frac{6}{6} + \frac{\cdots}{\cdots}}{5}}{4}}{3}}{2} = 2 + \frac{1+\frac{1+\frac{1+\frac{1+ \frac{\cdots}{\cdots}}{5}}{4}}{3}}{2} $$ I hope this ...


10

Note that by the binomial theorem $$\left(\frac{19}{13}\right)^{31}=\left(1+\frac{6}{13}\right)^{31}>\binom{31}{3}\frac{6^3}{13^3}=\frac{31\cdot 30\cdot 29\cdot 36}{13^3}>\frac{26^4}{13^3}=16\cdot 13>13^2$$ which implies that $19^{31}>13^{33}$. Bonus question. By using a similar approach we can show the stronger inequality $19^{31}>13^{34}$. ...


10

Fractional powers of negative numbers are not uniquely defined, and the "general rule" $(a^m)^n=a^{m\times n}$ does not always work when $m$ and $n$ are not integers.


10

J.W. Tanner has communicated the main point and provided some links to questions that provide more details. I'd like to try to tell the (mostly) whole story in one place. Recall that the standard definition of $a^b$ for $a \in \mathbb{R}_{>0}$, $b\in \mathbb{R}$ is $$a^b := e^{b\ln(a)}$$ Where the exponential function can be defined in several ways-- ...


9

Let $x$ be a percentage. If the return in a good year is $(1+x)$ and the return in a bad year is $(1-x)$, the the total return is given by $$(1+x)(1-x)=1-x^2$$ What this tells us is that negative returns have a much greater impact on investments than positive returns at the same rate. In the example given, even though Mr. Hare was able to achieve a $56%$ ...


9

Hint: Rewrite the expression as $$\frac{1}{7}e^{-2x^2}\left(1-4x^2\right) = \frac{1-4x^2}{7e^{2x^2}}$$ Now, notice the growth of the numerator and denominator. Which grows more quickly?


8

Here's three ways of doing it: Taylor You can use Taylor, i.e. $e^{x} \sim 1 + x$, so $$\frac{1 - e^{-x}}{e^{x} - 1} \sim \frac{1 - (1-x)}{1+x-1} = \frac{x}{x} =1$$ Factorizing Another (but equivalent) way is factor $e^{x}$ from the denominator $$\displaystyle \lim_{x\to0}{1 - e^{-x}\over e^x - 1} = \displaystyle \lim_{x\to0}{1 - e^{-x}\over e^{x}(1 - e^{...


8

You can still convert it into a polynomial, since $$2^{5x} - 8 = 2\cdot 2^{2x}$$ converts to $$y^5-8=2y^2$$ if you introduce $y=2^x$.


8

If $t = \log_2(3^x-1) = \log_3(2^x+1)$, we have $2^t = 3^x - 1$ and $3^t = 2^x + 1$. Thus $3^t + 2^t = 3^x + 2^x$. It's easy to see that $3^x + 2^x$ is an increasing function of $x$, therefore we must have $t=x$. Now with $t=x$ the equation becomes $3^x - 2^x = 1$. Dividing by $2^x$, write it as $(3/2)^x - 1 = (1/2)^x$. Now the left side is an increasing ...


8

$$\ln(e^{2x}-3e^x)\ne\dfrac{\ln e^{2x}}{\ln 3e^x}$$ Let $u=e^x$, this transforms the equation to a quadratic: $u^2-3u-4=0$ which can be used to obtain values for $u$ and eventually $x$.


7

Hint: $$ \sum_{k=1}^\infty 2^{-k}(e^{-k}-e^{-k-1}) = \sum_{k=1}^{\infty} (2e)^{-k} \left( 1- e^{-1}\right)= \left(1 - \frac{1}{e}\right) \sum_{k=1}^{\infty} \frac{1}{(2e)^k}. $$ Can you finish the problem from here?


7

Your first calculation can be written as $$1+\frac12(2+\frac13(3+\frac14(4+\dots)))$$ which expands to $$1+\frac{2}{2}+\frac{3}{2\cdot 3}+\frac{4}{2\cdot 3 \cdot 4} + \dots$$ Cancelling the repeated number in each term turns it into $$1+\frac11+\frac1{1\cdot2}+\frac{1}{1\cdot2\cdot 3} + \dots$$ $$=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\...


7

Your attempt is wrong, sorry: you cannot just use particular cases. And the case $a=-e$ is impossible, because $a^x$ is only defined for $a>0$. The answer should be in terms of $a$, and using a single value is not enough. Consider the function $f(x)=a^x-x-2$. Then $$ f'(x)=a^x\log a-1 $$ (with $\log$ being the natural logarithm). This doesn't vanish for $...


7

$$ \begin{align} \prod_{m=2}^{n-1}e^{\frac{\pi in}m} &=e^{\pi in(H_{n-1}-1)}\\ &=(-1)^ne^{\pi inH_{n-1}}\\[9pt] &=(-1)^{n-1}e^{\pi inH_n}\tag1 \end{align} $$ where $H_n$ is the $n^\text{th}$ Harmonic Number. $$ H_n\sim\log(n)+\gamma+\frac1{2n}-\frac1{12n^2}+\frac1{120n^4}-\frac1{252n^6}+\frac1{240n^8}-\frac1{132n^{10}}\tag2 $$ and $\gamma$ is the ...


7

The system can be written as $$\begin{cases} \log_{10}(2^x)=\log_{10}(11-3^y)\\ \log_{10}(35-(2^{x})^2)=\log_{10}((3^{y})^2) \end{cases}$$ Now if $u:=2^x<\sqrt{35}$ and $v:=3^y<11$, we can throw away the logarithm and solve with respect to $u$ and $v$: $$\begin{cases} u+v=11\\ u^2+v^2=35 \end{cases}$$ Can you take it from here? P.S. Yes, as you ...


7

The proof is not correct because it assumes implicitely that if a sequence $(a_n)_{n\in\mathbb N}$ converges to $1$, then $\lim_{n\to\infty}{a_n}^n=1$. This is not true. For instance,$$\lim_{n\to\infty} \left (1 + \frac1n \right )=1\text{ and }\lim_{n\to\infty}\left(1 + \frac1n\right)^n=e\neq1.$$


7

The answer to such a question depends crucially on what you are trying to relate to what. There are many "definitions" of the function $e^x$ and one can try to relate them to each other in intuitive way. Here is a (partial?) list: 1) The function $e^x$ is the unique solution of the ODE $f'(x)=f(x)$ with initial value $f(0)=1$. The high brow rephrasing is ...


7

If we square both sides we get $$e^{-x^2}\frac1{x^2}=y^2$$ $$x^2e^{x^2}=\frac1{y^2}$$ $$x^2=W\Big(\frac1{y^2}\Big)$$ $$x=\sqrt{W\Big(\frac1{y^2}\Big)}$$


7

Since the $LHS$ is the sum of the form $y+\frac{1}{y}$, where $y$ is positive, hence the minimum value of $LHS$ is $2$ (By AM-GM inequality), which is attained when $y=1$. Also, the maximum value of $RHS$ is $2$. Hence, solution is $x$ such that $e^x=1$ and $2sin(x^3)=2$. Since, the only solution of the former is 0 and which doesn't satisfy latter, hence no ...


6

As a general rule, if you see a lot of products and exponents it may clear things up to take logs. In your case, $$(a+b)^c ≈ a^{c-bc} \cdot (a+1)^{cb}$$ becomes $$c \log(a+b) \approx c(1-b) \log(a) + cb \log(a+1)$$ or just $$\log(a+b) \approx (1-b) \log(a) + b \log(a+1).$$ This is equivalent to doing a linear interpolation of $\log x$ between the ...


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