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2

If you insist on wanting to show that your sequence is Cauchy without using convergence: Suppose $m \geq n \geq N$. In that case $$|e^{-m} - e^{-n}| = e^{-n} - e^{-m} = e^{-n} (1 - e^{n - m}) \leq e^{-n} \leq e^{-N}.$$ Now for $\varepsilon > 0$, you can pick $N$ such that $e^{-N} \leq \varepsilon$.


2

Hint: If $m \ge n$, then $ 0 < e^{-m} \le e^{-n}$, and so $0 \le e^{-n}-e^{-m} < e^{-n}$. Similarly, if $n \ge m$, then $0 < e^{-n} \le e^{-m}$, and so $-e^{-m} < e^{-n}-e^{-m} \le 0$.


1

It converges to $0$! So it is in particular a Cauchy sequence (every convergent sequence is Cauchy).


1

We have $E(X_1+\cdots +X_n)=E(X_1)+\cdots +E(X_n)$, and, by independence, $\text{Var}(X_1+\cdots+X_n)=\text{Var}(X_1)+\cdots +\text{Var}(X_n)$. The $X_i$ have exponential distribution parameter $\lambda=\frac{1}{2}$. It is a standard fact that the mean of such an exponentially distributed random variable is $\frac{1}{\lambda}$, and the variance is ...


5

The answer is simple: $$(a^b)^c = a^{bc}$$ is an equality which only holds for real numbers with $a>0$ and does not necesarily hold on complex numbers.


1

Yep. As I found many years ago (1996 or so), this is what you need to solve to determine the equation of a catenary of a given length through two given points. Here is what I came up with to solve this problem (copied with editing from the document I wrote): Solving $r = \sinh(A)/A$ for $A$. Write the equation $r = \sinh(A)/A$ as $A r - \sinh(A) = 0$ ...


1

This function is exponential by the 2nd definition, but not by the first. The second definition ("a function with $a^x$ in it") is mathematically vague and makes little sense: think for example of $\sin (a^x)$ - or even $\log(a^x)$ Strictly speaking, the exponential function is one: $f(x)= e^x$ One can extend that to functions of the form $f(x)=a ...


4

I'ts not "exponential" in the sense of the derivative being proportional to the value, no. It does, however, have "exponential growth", in the sense that there's a constant $C$ with $$ |f(x)| \ge C u^x $$ for large enough $x$ and for some $u > 1$. In computer science, such functions are sometimes sloppily called 'exponential', even though they could ...


4

This kind of equations, which mix polynomial and trigonometric or hyperbolic terms do not show analytical solutions (beside the trivial $x=0$) and only numerical methods should be used. If you want me to elaborate on this topic, just post. Please notice that we can write the equation in a more simple form changing variable $ax=y$ to get $$\sinh(y)=c y$$ ...


2

Without any of the usual theoretical tools, you’ll need to do a bit of pattern-recognition. Notice the following pattern: $$\begin{align*} a_0=0&\\ a_1=1&=2\cdot0+1\\ a_2=1&=2\cdot1-1\\ a_3=3&=2\cdot1+1\\ a_4=5&=2\cdot3-1\\ a_5=11&=2\cdot5+1\\ a_6=21&=2\cdot11-1 \end{align*}$$ This suggests the first-order recurrence ...


1

One way to do this is to use generating functions. Let $G(x)=\sum_{n=0}^{\infty}a_nx^n$. We have the relation : $a_n=a_{n-1}+2a_{n-2}$. Multiply both sides by $x^n$ and summing from $n=2$ to $\infty$ we get: $G(x)-a_0-a_1x=x(G(x)-a_0)+2x^2G(x)$. Then we get: $G(x)(1-x-2x^2)=a_0-a_0x+a_1x=x$ (since $a_0=0,a_1=1$). So, ...


0

$a_n = a_{n-1} + 2a_{n-2} \to x^2 - x - 2 = 0 \to (x-2)(x+1) = 0 \to x = 2, -1 \to a_n = A2^n + B(-1)^n$. $a_0 = 0, a_1 = 1\to A+B=0, 2A-B=1 \to A = \dfrac{1}{3}, B = -\dfrac{1}{3} \to a_n = \dfrac{2^n -(-1)^n}{3}$. Thus: $a_2 = \dfrac{2^2 - (-1)^2}{3} = 1, a_3 = 3, a_4 = 5, a_5 = 11$.


4

All your differential equations after the first one are implied by the first equation. If $y'=y+1$ then differentiation of it gives $y''=y'$ and further differentiation gives the successive equations. So it is really just about solving the first equation with the given initial value, and this you can do.


0

The solution to $$ y' = y + 1,\ y(0)=1$$ is $y = 2e^x -1$ which can be found from your method of choice for first-order non-homogeneous ODEs. Now let's see what the derivatives look like $$y' = 2e^x$$ $$y'' = 2e^x$$ $$y''' = 2e^x$$ We can clearly see $y^{(k+1)} = y^{(k)}\ \forall k\in\mathbb{Z},k\geq1 $. Therefor all $n$th-derivatives of $y$ are equal ...


1

You can rewrite your equation as $$-x\ln 10 e^{-x\ln 10}=-\frac{\ln 10 }{10},$$ and then use Lambert W function to solve the obtained equation: $-x\ln 10 = W\left(-\frac{\ln 10 }{10}\right)$. As you can read on wiki, for some values of $z$ $W(z)$ is multiply defined, hence one one branch you obtain $\forall z>0 W(z\ln z) = \ln z$, which in our case ...


18

Well you have that $$y'=y''=y^{(3)}\cdots$$ only function that is a derivative of itself is $$ae^{x}$$ for some $a$ so $$y'=ae^x$$ and $$y=y'-1=ae^x-1$$ And since $y(0)=1$ than $a-1=1$ so $a=2$


0

That's integral linearity : $$\int_a^b c\ f(t)dt = c\int_a^b f(t)dt$$ Here with $c = -1$. For example, $$\int_a^b c\ dt = c\int_a^b dt = c\ (b-a)$$


1

There are several mistakes. First, $A=\frac{10}{10^\alpha1^\beta}=10^{1-\alpha}$ (this was correct!). But now, you use this for the second assignment, and I think you mixed up $\alpha$ and $\beta$. You should have had $Q(10,2)=30$ $\Leftrightarrow$ $10^{1-\alpha}\cdot10^\alpha\cdot2^\beta=30$ $\Leftrightarrow$ $10\cdot 2^\beta=30$ $\Leftrightarrow$ ...


2

We have that: $$ g(t) = \int_{\mathbb{R}}\frac{\sin(\pi x)}{\pi x}e^{-itx}\,dx = \mathbb{1}_{(-\pi,\pi)}(t)+\frac{1}{2}\mathbb{1}_{\{\pi\}\cup\{-\pi\}}(t).\tag{1}$$ To prove such an identity, we may compute the inverse Fourier transform of $\mathbb{1}_{(-\pi,\pi)}$, or notice that: $$ g(t) = 2\int_{0}^{+\infty}\frac{\sin(\pi x)}{\pi x}\cos(tx)\,dx = ...


3

Notice that the last digits of powers of $7$ run in the repeating sequence $7,9,3,1$. Thus, what you really need to know is what $7^7$ is congruent to modulo $4$, not modulo $10$, so as to tell where in the sequence $7^{7^7}$ falls. $7^k\bmod 4$ is easily computed from $k$; how?


4

Note that $7^2 \equiv 49 \equiv -1 \bmod 10$, and so $7^4 \equiv 1 \bmod 10$. Hence if $a=4q+r$ then $$7^a \equiv 7^{4q+r} \equiv (7^4)^q \cdot 7^r \equiv 7^r \bmod 4$$ This reduces your problem to finding $7^7 \bmod 4$... but this should be easy since $7 \equiv -1 \bmod 4$.


5

If I am understanding correctly you would like to know if there is a suitable function $f$ such that $e^{a+b+c}=f(a)+f(b)+f(c)$ for all values of $a,b,c$. Notice such a function would satisfy $f(0)+f(0)+f(0)=e^0=1$. So $f(0)=\frac{1}{3}$. From here we can determine the function uniquely, since we must have $e^x=f(x)+f(0)+f(0)=f(x)+\frac{2}{3}$. So the ...


1

No. It is multiplicative, not additive. $$ \exp(A + B + C) = \exp(A) \exp(B) \exp(C) $$ Counter example: $$ \exp(x+x+x) = f(x) + f(x) + f(x) \iff f(x) = 1/3 \exp(3x) \\ e = \exp(1 + 0 + 0) =^! 1/3 \exp(3) + 2/3 \exp(0) > 7 $$


1

For $\epsilon>0$ we have $\left(1-\frac{\epsilon}{n}\right)^{n}\rightarrow e^{-\epsilon}<1$ and $\left(1+\frac{\epsilon}{n}\right)^{n}\rightarrow e^{\epsilon}>1$. So $\left(1-\frac{nx_{n}}{n}\right)^{n}=\left(1-x_{n}\right)^{n}\rightarrow1$ tells us that eventually $-\epsilon<nx_{n}<\epsilon$.


4

For $n$ be sufficiently large, we have $(1-x_n)^n >0$ (since it converges to 1). We can then say that $n \log(1-x_n)$ tends to zero. Necessarily $\log(1-x_n)$ tends to zero. So $x_n$ tends to zero and we have that : $ \quad \log(1+x) \underset{0}{\sim} x$ Then $nx_n \to 0$.


0

By the very definition (you may be required to provide justifications): $$f'(x_0)=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=\lim_{x\to x_0}\frac1{x-x_0}\sum_{n=1}^\infty\frac{x^n-x_0^n}{n!}=$$ $$=\lim_{x\to ...


2

Hint:$\displaystyle\sum_{n=0}^{\infty}\dfrac{x^n}{n!}=e^x $ ($e^x$ Taylor series)


0

HINT You have already the "final" exprssion : $f(n) = (31 \times 5^n -3)/4$. What you have to do is to prove - by induction - that it holds, i.e. that it satisfies the recurrence relation for $f(n)$. (i) base case : $n=0$ For $n=0$ the recursive definition of $f(n)$ has : $f(0)=7$. We have to check that it matches with the closed expression : ...


3

By the first condition, we have that $f(a/b) = f(1/b)^a$ and $f(1/b)^b = f(1)$ for any $a/b \in \mathbb{Q}$. Let $f(1) = ce^{i\theta}$. Then $f(1/b) = c^{1/b}e^{i(\theta + 2\pi k)/b}$ where $k$ is an integer depending on $b$. It follows $f(a/b) = e^{a/b}e^{i(\theta + 2\pi k)a/b}$. By continuity of $f$ and density of rationals, it follows that $k$ is locally ...


1

Since $(x^{-k})' =(-k)x^{-k-1} $, he first terms of $f^{(n)}(x)$ are $f'(x) =(-2x^{-3})e^{-x^{-2}} =\frac{-2}{x^{3}}e^{-x^{-2}} $ and $f''(x) =e^{-x^{-2}}(6x^{-4}+(-2x^{-3})^2) =e^{-x^{-2}}(6x^{-4}+4x^{-6}) =e^{-x^{-2}}\frac{6x^2+4}{x^6} $. Suppose $f^{(n)}(x) =\frac{p_n(x)}{x^{cn}}e^{-x^{-2}} =p_n(x)x^{-cn}e^{-x^{-2}} $. Since $(p_n(x)x^{-cn})' ...


1

You have to relabel the elements: use a bijection $\phi$ from $\mathbb N$ to $\mathbb N^2$ and define $B_n:=A_{\phi(n)}$. The assumption gives that $\mathbb P(\limsup_n B_n)=0$ hence $\mathbb P(\mbox{number of } n \mbox{ such that } B_{n} \mbox { occurs} < \infty) =1$.


0

A straighforward way is to compute the derivative (in $x$) of the two members of the egality, noticing for $x=1$ that both are $0$. The result is all these derivatives are null function.


2

Hint 1: multiply the integrand by $1 = \frac{t}{t}$, then use parts. Hint 2: in parts, let $dv = t e^{t^2}dt$ always; an antiderivative is $\frac{1}{2} e^{t^2}$. Hint 3: do this twice.


0

Try proving that for $a<1$ the left side is greater (obvious for negative $a$ and positive where $9^a<3$). For $a>1$ check that the right side is greater (the derivative would tell you which side is steeper).


1

Possible solution outline: Set $f(a)= 6a^2+3-9^a$. Then, as you observed, $f(1) = 0$. To deduce that this is the only root, try to show the that $f(a)$ is monotonically decreasing, i.e. $f'(a) < 0$ for all $a$. To do this, find the point for the maximum of $f'(a)$ by solving $f''(a)=0$. $$f''(a) = 12-9^a\log^29 = 0\\\implies ...


0

$$e^{-x^2}=e^{-(x-1+1)^2}=\sum_{n=0}^\infty\frac{(-1)^n(x-1+1)^{2n}}{n!}=\sum_{n=0}^\infty\frac{(-1)^n\sum_{k=0}^\infty{2n\choose k}(x-1)^k}{n!}=\sum_{n=0}^\infty\frac{(-1)^n{2n\choose 0}}{n!}+\sum_{n=0}^\infty\frac{(-1)^n{2n\choose 1}(x-1)}{n!}+\sum_{n=0}^\infty\frac{(-1)^n{2n\choose 2}(x-1)^2}{n!}+\cdots$$


2

For any integer $k$, the numerator is $1 - 1 = 0$. If $k$ is also a multiple of $2N$ then the bottom is also $1-1 = 0$. This is indeterminate. (It seems their $N/2$'s should be $2N$'s.) Whenever $k$ is not a multiple of $2N$, the exponential on the bottom is not equal to one, so the expression on the bottom is nonzero, this whole expression is zero since ...


0

We need to prove 2 things: if a function is of exponential order then there exist constants..., if there exist constants... then the function is of exponential order. If there are constants $a,M$ and $t_0>0$ such that $|f(t)|\leq Me^{at}$ for every $t>t_0$, then in particular $t>0$, because $t>t_0>0$. Therefore the function is of ...


1

Here is one approach. Let $b > 1$. If you compute the derivative of the function $b^x$, you find that the answer is just $b^x$ multiplied by an (annoying) constant. There is a value of $b$ for which this constant is equal to $1$. That's nice! With this special value of $b$, the derivative of $b^x$ is just $b^x$, the same thing we started with. ...


2

These two familiar sums are the Taylor series for $e^x$ about $0$. To get $e$ itself, you evaluate this series at $x=1.$ Derivation: The $n$th term of the Taylor series of a function $f$ about $a$ is $$ \frac{f^{(n)}(a)}{n!} (x-a)^n.$$ But if $f(x) \triangleq e^x$, then $f'(x) = e^x$, and by an inductive argument, $f^{(n)}(x) = e^x$ for every positive ...


2

By definition, $$e=\lim_{n\to\infty}\left(1+\frac1n\right)^n.$$ Using the binomial theorem, the $k^{th}$ term of the development is $${\binom nk}\frac1{n^k}=\frac{n(n-1)(n-2)\dots(n-k+1)}{k!.n.n.n\dots n},$$ and $$\lim_{n\to\infty}{\binom nk}\frac1{n^k}=\frac1{k!}.$$ For example, ...


1

When you multiply $\exp(x)$ by $\exp(y)$ by that definition, you get $\exp(x+y)$. That is one of the exponent laws, and is why $\exp(x)=e^x$ for some number $e$. Then $e^1=\exp(1)$ which is your sum. $$\exp(x)\exp(y)=\sum_{n=0}^{\infty}\frac{x^n}{n!}\sum_{m=0}^{\infty}\frac{y^m}{m!}\\ =\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\frac{x^ny^m}{n!m!}\\ ...


1

Let $x \in \mathbb{R}^+$. $f:x\longrightarrow \exp(x) \in C^\infty(\mathbb{R}, \mathbb{R})$, hence we can write $\forall n \in \mathbb{N}$ : $$\left|f(x)-\sum_{k=0}^n f^{(k)}(0)\frac{x^k}{k!}\right| \leq \frac{|x|^{n+1}}{(n+1)!}I_{n+1} \text{ where } I_{n+1}=sup_{[0,x]}|f^{n+1}|$$ And $\forall t \in [O,x] f^{(n)}(t)=\exp(t)$ which means $I_{n+1}=\exp(x)$ ...


3

Note that $x^2+y^2=r^2$. This is the equation of a circle of radius $r$. Saying that $(x,y)\to (0,0)$ here means that the radius of this circle is approaching zero. So we can transform this to a one variable limit: $$ \lim_{r\to 0} y=\lim_{r\to 0} e^{\frac{1}{r^2}}$$ Now note that $$\log y=\frac{1}{r^2}\rightarrow \infty,$$ as $ r\to 0$. So $y \to \infty$. ...


2

As $(x,y)\rightarrow 0$, $x^2+y^2 \rightarrow 0$ while always remaining positive. Since $\lim_{t\to 0^+} \frac{1}{t} = +\infty$, we find that $\lim_{(x,y)\to 0} \frac{1}{x^2 + y^2} = +\infty$. Finally, since $\lim_{t\to +\infty} e^t = +\infty$, we string it all together to get: $$ \lim_{(x,y)\to (0,0)} e^{\frac{1}{x^2+y^2}} = e^{\lim_{(x,y)\to (0,0)} ...


5

The denominator of the exponent tends to zero from above, no matter how $(x,y)$ tends to $(0,0)$. So the exponent grows without bound to $+\infty$. Hence the expression itself does as well. (Some may say the limit doesn't exist since it is infinite, but that's a matter of convention.) If the numerator were $-1$ instead of $1$, the exponent would tend to ...


0

You look up the Cumulative Distribution Function of the distribution. $CDF(x)$ is the chance the repair takes less than $x$ time. To get the chance it takes more, subtract from $1$.


2

When $t$ is positive, $U(t)$ is 1, and $U(-t)$ is zero, so you get only the first term, which equals $e^{-a|t|} = e^{-at}$. Now do the same thing when $t$ is negative... (I'm assuming here that your book/prof has the definition that $U(x) = 1$ for $x \ge 0$ and is zero otherwise.) (You might reasonably be worrying about the case $t = 0$, but if your ...


0

It is mostly index manipulation $$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$$ $$e^x-1=\sum_{n=1}^\infty\frac{x^n}{n!}$$ $$\frac{e^x-1}{x}=\sum_{n=1}^\infty\frac{x^{n-1}}{n!} =1+\sum_{n=2}^\infty\frac{x^{n-1}}{n!} =1+\sum_{n=1}^\infty\frac{x^{n}}{(n+1)!}$$ $$\frac{e^x-1}{x}-1=\sum_{n=1}^\infty\frac{x^{n}}{(n+1)!}$$ Now use the triangle inequality: ...


0

Before expanding it is convenient to rearrange the expression (assuming $x\neq0$): $$ \left|\frac{e^{x}-1}{x}-1\right|=\left|\frac{e^{x}-1-x}{x}\right|=\left|\frac{\sum_{2}^{\infty}\left(x^{n}/n!\right)}{x}\right|=\left|\sum_{2}^{\infty}\frac{x^{n-1}}{n!}\right|\leq\sum_{1}^{\infty}\frac{\left|x\right|^{n}}{(n+1)!} $$ where the last follows from ...



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