# Tag Info

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=yto 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. LetG(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$zW(z)$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$Q(10,2)=30\Leftrightarrow10^{1-\alpha}\cdot10^\alpha\cdot2^\beta=30\Leftrightarrow10\cdot 2^\beta=30\Leftrightarrow$... 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$$2 For any integer k, the numerator is 1 - 1 = 0. If k is also a multiple of 2N then the bottom is also 1-1 = 0. This is indeterminate. (It seems their N/2's should be 2N's.) Whenever k is not a multiple of 2N, the exponential on the bottom is not equal to one, so the expression on the bottom is nonzero, this whole expression is zero since ... 0 We need to prove 2 things: if a function is of exponential order then there exist constants..., if there exist constants... then the function is of exponential order. If there are constants a,M and t_0>0 such that |f(t)|\leq Me^{at} for every t>t_0, then in particular t>0, because t>t_0>0. Therefore the function is of ... 1 Here is one approach. Let b > 1. If you compute the derivative of the function b^x, you find that the answer is just b^x multiplied by an (annoying) constant. There is a value of b for which this constant is equal to 1. That's nice! With this special value of b, the derivative of b^x is just b^x, the same thing we started with. ... 2 These two familiar sums are the Taylor series for e^x about 0. To get e itself, you evaluate this series at x=1. Derivation: The nth term of the Taylor series of a function f about a is$$ \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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