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1

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


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 ...


-3

$$ \log(a^b) = \int_1^{a^b} \frac{dx}x. $$ Suppose $w^{\,b}=x$. Then $bw^{b-1}\,dw= dx$, so $b\dfrac{dw}{w} = \dfrac{dx}x$. As $x$ goes from $1$ to $a^b$, then $w$ goes from $1$ to $a$. Hence, $$ \log(a^b) = \int_1^{a^b} \frac{dx}x = \int_1^a b \frac{dw}w = b\log a. $$


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}


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 $0$.


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 ...


0

The answers are all helpful so far, but the original question was on the right track: there is indeed an error function that results. The reason is that the original integral, which Henry correctly pointed out is over t, not $\tau$, is from 0 to $\infty$, and not from $-\infty$ to $\infty$. This changes everything. In what follows, I took the liberty of ...


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.


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 $x>x_0$.


0

This is called index shift. Introduce $m=n+N$. Since $n$ runs from $-N$ to $N$, the new index $m$ runs from $0$ to $2N$. So, $$\sum_{n=-N}^{n=N} e^{ 2 \pi i n t} = \sum_{m=0}^{m=2N} e^{ 2 \pi i (m-N) t} = \sum_{m=0}^{m=2N} e^{ 2 \pi i m t}e^{ -2 \pi i N t}$$ Here we see the common factor $e^{ -2 \pi i N t}$ and move it outside to get $$e^{ -2 \pi i N ...


0

If you have two particular points, $(x_1, y_1)$, and $(x_2, y_2)$, then you have two equations and two unknowns: $$A=\frac{y_1}{1-e^{-x_1/B}}$$ $$B=-x_2\left(\ln\left(\frac{y_2-A}{-A}\right)\right)^{-1}$$ Since $x_1$, $y_1$, $x_2$, and $y_2$ are constants, this is like solving an a regular two equation to unknown algebra problem, albeit a nasty one.


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 $e^x$


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 ...


4

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


0

I'm assuming you want some kind of closed form for $N(\alpha,\beta).$ For that, set $\beta(x-x_1)=t.$ Then, $\beta \,\mathrm{d}x=\mathrm{d}t.$ It follows that $$N(\alpha,\beta)=\beta^{-\alpha-1} \int_0^{\beta(x_2-x_1)} t^{\alpha} e^{-t}\, \mathrm{d}t$$ It is known that $$\gamma(s,x)= \int_0^{x} t^{s-1} e^{-t}\, \mathrm{d}t$$ where $\gamma(s,x)$ is the ...


0

The Lambert W-function will do the trick. We start with $s^x = x+1$, and want to have something of the form $f(x)e^{f(x)} = W^{-1}\bigl(f(x)\bigr)$ \begin{align*} s^x &= x+1\\ \iff \frac{s^{x+1}}s &= x+1\\ \iff e^{(x+1)\log s} &= s(x+1)\\ \iff (x+1)e^{-(x+1)\log s} &= \frac 1s\\ \iff -\log s\cdot (x+1)\cdot e^{-(x+1)\log s} &= ...


0

The first function, as in your comment, models $P(x)=A(1+0.63)^x$ for $P(x)=y, A=1298$ and where the percentage is an increase of $63$%. The second function can be considered similarly, $P(x)=f(x)=A(1+q)^x=2(1+(-0.35))^x=2(1-0.35)^x=2(0.65)^x,$ which is a decrease of $35$%. Note that, counter to your comment, $1-0.63=0.37$ which is not present in either ...


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

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 ...


0

You can use $A=Pe^{rt}$ if you want. But then we have to find $r$. In one year, the price goes from $1$ to $1.06$, so $1.06=e^r$, giving $r=\ln(1.06)$. Now continue. Alternately, use the equation $A=P(1.06)^t$. Let $q$ be the amount of time it takes for the price to quintuple. Put $A=5P$. Then we are solving the equation $(1.06)^q=5$. Take the logarithm of ...


0

You don't actually care what the current cost is. You are asked to find the $t$ that corresponds to $\frac AP=5$, which is when the price of anything has been multiplied by $5$. Your equation $A=Pe^{0.06t}$ gives an annual rate higher than $6\%$-plug in $t=1$ to get $\frac AP\approx 1.0618$ for a annual rate of $6.18\%$


0

$\mathcal{L}^{-1}_{s\to t}\left\{e^{-\left(\frac{b}{b+s}\right)^k}\right\}$ $=\mathcal{L}^{-1}_{s\to t}\left\{\sum\limits_{n=0}^\infty\dfrac{(-1)^nb^{kn}}{n!(b+s)^{kn}}\right\}$ $=\mathcal{L}^{-1}_{s\to t}\left\{1+\sum\limits_{n=1}^\infty\dfrac{(-1)^nb^{kn}}{n!(b+s)^{kn}}\right\}$ ...


0

For the special case $k=1$ you should have the answer $$ {{\rm e}^{-bt}} \left( i\sqrt {{\frac {b}{t}}} {\rm I_{1} \left(\,2\,\sqrt {-bt}\right)}+\delta \left( t \right) \right) ,$$ where $I_n(x)$ is the modified Bessel function of the first kind and $\delta(x)$ is the dirac delta function.


0

So you need to compute $$ \frac1{2\pi i}\int_{-i\infty}^{i\infty}e^{xs-\left(\tfrac{b}{b+s}\right)^k}ds=e^{-bx} \frac1{2\pi i}\int_{-i\infty}^{i\infty}e^{x(b+s)-\left(\tfrac{b}{b+s}\right)^k}ds. $$ For $x$ positive and $k$ positive integer this is equal to $$ e^{-bx}\operatorname{res}_{z=0}e^{xz-\left(\tfrac{b}{z}\right)^k}. $$ The residue is the ...


0

Compute the Taylor expansion of the RHS: $$\begin{align} f(u)&=e^{-u-u^2/2-u^3/3-...},\\ f'(u)&=(-1-u-u^2-...)\ f(u)=\frac{f(u)}{u-1}\text{ (geometric series)},\\ f''(u)&=\frac{f'(u)(u-1)-f(u)}{(u-1)^2}=0\text{ (!)} \end{align}$$ Hence $$f(0)=1, f'(0)=-1, f''(0)=f'''(0)=...=0\implies f(u)=1-u.$$


5

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


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 $y>0$


0

Just as you prove the convergence of the geometric sequence you can also prove the convergence to zero if you multiply by a polynomial factor. Consider a sequence $p(n)\cdot q^n$ with a polynomial $p$ (with positive coefficients) of degree $d$ and $0<q<1$. Then $q=\frac1{1+α}$ with $α>0$. Using binomial expansion and removing some positive terms in ...


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 $$ h'(x) = f'(g(x))g'(x) = e^{-2x} \cdot (-2) = -2e^{-2x} $$


5

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


1

No, you didn't distribute the $\frac{1}{2}$ in the exponent properly. It should be $$e^{\frac{1}{2} \ln |a|}(\cos (\frac{1}{2}\arg{a}) + i \sin (\frac{1}{2}\arg{a})).$$ Now you can see that the choice of arg is important.


1

Not quite. The values of (using your "$j$") notation $e^{j\pi k/2}$ cycle among $1$, $j$, $-1$, and $-j$ in order as $k$ moves through the integers $0,1,2,\ldots$. They cycle in reverse order as you move backwards through the negative integers, which gives you the values for the other expression you wrote. Said differently, you can write $$e^{\pm ...


2

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


0

In the comments you mentioned that a linear function will also work. This is the simplest kind of function you can use. There is only one linear function satisfying your requirements, and that is: $f(x)=100-x$


-1

This is a question that shows the bad guessing abilities of the human when it comes to exponential growth. No calculation is required here - the bottle will be half full after 59 minutes, since it doubles every minute it will be full at minute 60 which equals 1 hour. So the bottle will be half full at 11.59 am. As an alternative you can also calculate it: ...


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 ...


1

Theorem: let $p \in \mathbb{Z^+}$ and $\alpha \in (0,1)$ be constant. Then $a_n=n^p\alpha^n\to0.$ $\underline{\text{Proof}}$ Consider the series $\sum\limits_{n=1}^{\infty}a_n$. Then we have $$\frac{a_{n+1}}{a_n}=\frac{(n+1)^p\alpha^{n+1}}{n^p \alpha^n}=\left(1+\frac{1}{n}\right)^p\alpha \to \underbrace{\alpha<1}_{\text{by assumption}}.$$ ...


0

To $a):$ You get $P(X\le 1500)=F(1500)=1-e^{-\frac{3}{2}}.$ So $P(X\ge 1500)=1-F(1500)=e^{-\frac{3}{2}},$ which is the probability that a bulb lasts $1500$ hours or more. Now you need to consider a binomial distribution $Y\in B(10,e^{-\frac{3}{2}})$ and get $P(Y\ge 7).$ That is: $$P(Y\ge ...


0

Start by finding the prob that a bulb does not last 1500 hours. : p=$\int_0^{1500} \frac{1}{1000}*e^{\frac{-1}{1000}*x}$. Find $q=1-p.$ Prob that atleast 7 bulbs function function for 1500 or more= $\sum _{i=7 to 10}(^{10}C_{i})q^{i}p^{10-i}$ Prob that a bulb does not last more than 2000 hours, r= $\int_{0}^{2000} \frac{1}{1000}*e^{\frac{-1}{1000}*x}. $ ...


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 $$ (\underbrace{\pm d.ddd\cdots d}_{\mbox{mantissa }m} \,\,\, \mbox{E}\underbrace{\pm dd\cdots d}_{\mbox{exponent }k})_{10} = m \cdot 10^k $$ ...


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 $$10^{+11}=100000000000$$


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

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


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.: $$\forall \in\mathbb{R},\quad e^x \geq x+1.$$


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.)


0

If $x \leq 0$, then $e^x > 0 \geq x$. Suppose $x > 0$ and let $f(x) = e^x - x $. So $f'(x) = e^x - 1 > e^0 - 1 = 0$. Hence $f$ is increasing and since $f(0) > 0$, $f(x) = e^x - x > 0$, finally $e^x > x$. The inequality is strict.


1

If we look at the initial value for each function $f(x)=x$ and $g(x)=e^x$ we see that $f(0)=0 < g(0)=1$. Now for $x>0$ we have $f'(x)=1$ and $g'(x) = e^x$ which tells us that $f'(x) < g'(x)$ for all $x > 0$, and so $$x \le e^x$$ for all $x \ge 0$. For negative $x$ we know that $e^x$ is strictly positive, so this tells us that $$x \le e^x$$ for ...


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 $$x \leq 0 < e^x.$$


1

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



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