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16

This is easily checked, by definition of $e^x$. For $x > 0$ we have $$ e^x = 1 + x + \sum_{k = 2}^\infty\frac{x^k}{k!}>x$$ and for $x \leq 0$ we have $$x \leq 0 < e^x.$$


8

This is a pretty neat problem actually so I'm only giving you a little hint. Start by rewriting as $$\sum_{k=0}^{n-1}\frac{1}{k!}(k-n)^{k}e^{-(k-n)}=\sum_{k=0}^{n-1}\frac{1}{k!}\frac{d^{k}}{dx^{k}}e^{x(k-n)}|_{x=-1}=\sum_{k=0}^{n-1}\frac{1}{2\pi i}\int_{C}\frac{e^{z(k-n)}}{(z+1)^{k+1}}dz$$ Since $e^{z(k-n)}$ is analytic in all of $\mathbb{C}$ I applied ...


7

By Bernoulli's inequality, $$ \left(1+\frac xn\right)^n \ge 1+x \qquad\text{if $n > |x|$.} $$ Letting $n\to\infty$ yields $e^x \ge 1+x > x$. (I like this argument because it doesn't need to treat $x<0$ specially.)


6

If $a\not=0$, let $k=ah$ and note that $h\to0$ iff $k\to0$, which gives $$\lim_{h\to0}{(b^a)^h-1\over h}=a\lim_{h\to0}{b^{ah}-1\over ah}=a\lim_{k\to0}{b^k-1\over k}$$ If $a=0$, then $(b^a)^h-1=0$, so the limit is obviously $0$.


6

Your work is correct. Notice also that we can differentiate this function by simpler way: $$y=a^x=e^{\ln(a) x}$$ so using the chain rule $$\frac{dy}{dx}=\ln(a)e^{\ln(a) x}=\ln(a) a^x$$


5

You seem to have the right idea. $$1\text{ e+11}=1\cdot10^{+11}$$ You should know that $10$ raised to any positive integer is a $1$ with that many $0$s behind it. So $$10^{+11}=100000000000$$


5

You have to apply the chain rule: if $f(x)$ is a differentiable function the the derivative of $e^{f(x)}$ is $f'(x)e^{f(x)}$.


5

this holds because for $-1\lt u \lt 1$, $-u-u^2/2-...=\ln(1-u)$ thus $$e^{-u-u^2/2-...}=e^{\ln(1-u)}=1-u$$


4

One can notice that your formula is the expected number of $[0,1]$-uniformly distributed random variables that are needed for their sum to exceed $n$. (See http://mathworld.wolfram.com/UniformSumDistribution.html.) Moreover, the question http://mathoverflow.net/questions/141368/error-term-for-renewal-function discusses the behaviour of the error term ...


4

Try $\begin{cases} x=3\cos^3\theta\\ y=3\sin^3\theta\\ \end{cases}$ or, for rational solutions, $\begin{cases} x=3\left({{2t}\over{t^2+1}}\right)^3\\ y=3\left({{t^2-1}\over{t^2+1}}\right)^3\\ \end{cases}$


4

The function $f(x)=3^{x-1}+5^{x-1}$ increases monotonically for all $x\in\mathbb{R}$. Hence, if $f(x_0)=34$ for some $x_0\in\mathbb{R}$, then $f(x)<34$ for all $x<x_0$ and $f(x)>34$ for all $x>x_0$.


4

Yes, you did apply well this property of logarithm: $ln(ab)=ln(a) + ln(b)$, and the fact that $ln(x)$ is the inverse of $e^x$


3

You have to use the chain rule here. Writing $f(x) = e^x$ and $g(x) = -2x$ we have $h(x) := f(g(x)) = e^{-2x}$, hence by the chain rule $$ h'(x) = f'(g(x))g'(x) $$ Now $f'(x) = e^x$, hence $f'(g(x)) = e^{-2x}$, and $g'(x) = -2$, this gives $$ h'(x) = f'(g(x))g'(x) = e^{-2x} \cdot (-2) = -2e^{-2x} $$


3

I must say that at first I was quite skeptical that such function could exist, but it turned out that one can prove its existence. I have written down the proof but I will give here only sketch it (you should not have any problems computing it for yourself, nice exercise) Let us define $b_1=2\pi \rm i m_1$ and $\displaystyle b_n=2\pi \rm i m_n+ \log ...


3

For $|x|<1,$ using infinite Geometric Series, $$(1-x)^{-1}=1+x+x^2+\cdots$$ Integrate either sides to find $$-\ln(1-x)=\sum_{r=1}^\infty\frac{x^r}r$$ Now $$\ln y=z\implies y=e^z$$ If $z$ is finite real, real $y>0$


2

We have $$\begin{align} \frac{e^{-ita}-e^{-itb}}{it}e^{itx} &= \frac{e^{it(x-a)}-e^{it(x-b)}}{it}\\ &= \frac{\cos\left( t(x-a)\right) - \cos \left(t(x-b)\right)}{it} + \frac{\sin \left(t(x-a)\right) - \sin \left(t(x-b)\right)}{t} \end{align}$$ by the addition theorem for the exponential function and Euler's formula $e^{iz} = \cos z + i\sin z$. The ...


2

The exponential function is a convex function, since the second derivative equals the function, so it is non-negative. This implies that the graphics of $f(x)=\exp(x)$ lies above the tangent line in $x=0$, i.e.: $$\forall \in\mathbb{R},\quad e^x \geq x+1.$$


2

No, for first one the pattern is $-i,-1,i,1,...$ and for second one the pattern is $i,-1,-i,1,..$


2

Using $(b^a)^x = b^{ax}$ and the chain rule, we get \begin{align} \log(b^a) &= \left.\frac{\mathrm d}{\mathrm dx} b^{ax}\right|_{x=0} = \left.\left(\frac{\mathrm d}{\mathrm dy} b^y\right)\right|_{y=0} \cdot\left.\left( \frac{\mathrm d}{\mathrm dx} ax\right)\right|_{x=0} = \log(b)\cdot a. \end{align}


2

Hint: $\displaystyle \log(b^a) = \frac{d(b^{ax})}{dx}(0)$ and using the chain rule $$\frac{d}{dx}b^{ax} = \frac{d}{dx}(b^{x})^a = a(b^x)^{a-1}\frac{d}{dx}b^{x}$$


2

Let's work with $A=M_n(\mathbb{C})$. Consider $x\in A$, then $\|x\|<1$ for some subordinate matrix norm iff $\;\mathrm{Spec}(x)\subset D^{\circ}$ where $D^{\circ}=\lbrace z\in\mathbb{C}\text{ s.t. }|z|<1\rbrace$ is the open unit disc. Indeed, if $\|\cdot\|$ is subordinate to some norm $|\cdot|$ on $\mathbb{C}^n$, then for any nonzero ...


2

It is correct!!! If you want to solve for $t$, it is as followed: $$\ln{(z)}=\ln{(a)}-bt \Rightarrow bt=\ln{(a)}-\ln{(z)}=\ln{(\frac{a}{z})} \Rightarrow t=\frac{1}{b} \ln{(\frac{a}{z})}, \ \text{ where } z,a>0$$ Like @Alex R. says in the comment, from the relation $$z=a e^{-bt}$$ if $\displaystyle{z=0 \Rightarrow a=0 \text{ and } b,t \text{ are ...


2

It is just a short notation to make good use of limited display and storage space. The used string in base 10 representation, called exponential or scientific notation, and the encoded rational number are related by $$ (\underbrace{\pm d.ddd\cdots d}_{\mbox{mantissa }m} \,\,\, \mbox{E}\underbrace{\pm dd\cdots d}_{\mbox{exponent }k})_{10} = m \cdot 10^k $$ ...


2

A function like $x\mathrm{e}^x+10\mathrm{e}^x$ does not have an $(x,y)$ intercept. The graph $y=x\mathrm{e}^x+10\mathrm{e}^x$ can have an $x$-intercept, where it crosses the $x$-axis; and it can have a $y$-intercept, where it crosses the $y$-axis. Consider the equation $y=x\mathrm{e}^x+10\mathrm{e}^x$. There is a common factor of $\mathrm{e}^x$, and so ...


2

One standard way to find a parametrization of $F(x,y)=0$ is to set $y=tx$. In your case it gives the following parametrization of non-trivial branch: $$\Large x=t^{\frac{1}{t-1}},\quad y=t^{\frac{t}{t-1}},$$ for $t>0$, $t\neq 1$.


1

Hint: It seems to me that one way could be using the Rolle's Theorem for function $$f(x)=3^{x-1}+5^{x-1}-34$$ to get a contradiction.


1

By definition of logarithms, $x_t=\ln {\dfrac{C_{t+1}}{C_t}}$


1

It may be helpful to note that $$\frac{d}{dt}\int^{\infty}_{t}f(u)du=-f(t)$$


1

Some other important formulas regarding $\sin(x)$ that you didn't mention are the infinite product $$\sin(x) = x \prod_{k=1}^{\infty}\Big( 1 - \frac{x^2}{\pi^2 k^2} \Big)$$ and the partial fractions decomposition $$\frac{1}{\sin(x)^2} = \sum_{k=-\infty}^{\infty} \frac{1}{(x-\pi k)^2}, \; \; x \notin \pi \cdot \mathbb{Z},$$ although I guess the latter only ...


1

A rigorous way to define the sine function is to consider it as the solution to the IVP: $$ \begin{cases} y^{\prime \prime} + y = 0\\ y(0) = 0 \\ y^{\prime}(0) = 1 \end{cases} $$



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