# Tag Info

15

You may write, as $N \to +\infty$, \begin{align} e^{3/2}\prod_{n=2}^{N}e\left(1-\dfrac{1}{n^2}\right)^{n^2}&=e^{3/2}\times\prod_{n=2}^{N}e\times\prod_{n=2}^{N}\left(1-\dfrac{1}{n^2}\right)^{n^2}\\\\ &=e^{3/2}e^{N-1}\times\prod_{n=2}^{N}\dfrac{(n-1)^{n^2}}{n^{n^2}}\dfrac{(n+1)^{n^2}}{n^{n^2}}\\\\ ... 6 You can rewrite the product as\sum_{n=2}^{\infty} \left [1+n^2 \log{\left (1-\frac1{n^2} \right )} \right ] $$Now,$$n^2 \log{\left (1-\frac1{n^2} \right )} = -\int_0^1 \frac{du}{n^2-u} $$So the sum is equal to$$-\int_0^1 du \, u \sum_{n=2}^{\infty} \frac1{n^2-u} $$Using the fact that$$\sum_{n=-\infty}^{\infty} \frac1{n^2-u} = -\frac{\pi ...

4

$$2^x \cdot 4^{1-x}= 8^{-x}$$ $$\implies 2^x\cdot (2^2)^{1-x}=(2^3)^{-x}$$ $$\implies2^x\cdot 2^{2-2x}=2^{-3x}$$ $$\implies 2^{x+2-2x}=2^{-3x}$$ $$\implies 2^{2-x}=2^{-3x}$$ comparing both side we have, $$2-x=-3x$$ $$\implies -2x=2$$ $$\implies x=-1$$

-2

If you are referring to $2^x * 4^{1-x} = 8^{-x}$ then here are the steps $2^x * 4^{1-x} = 8^{-x}$ $2^x * 4^1 4^{-x} = 8^{-x}$ $2^x * 4\frac{1}{4^{x}} = \frac{1}{8^{x}}$ $4=\frac{4^x}{8^x 2^x}$ $4={\frac{1}{4}}^x$ $log_{\frac{1}{4}}(4)=x$ $x=-1$ Hope this helps!

2

Using $2^x\cdot 4^{1-x}=8^{-x}$ from the comments, write everything as a power of two: $2^x\cdot (2^2)^{1-x}=(2^3)^{-x}$, and then use the rules of exponents. You need these: $(a^b)^c=a^{bc}$ and $a^ma^n=a^{m+n}$. Once you have simplified the expression to something like $2^a=2^b$, you can take log base 2 to get $a=b$, and then solve for $x$.

1

Seems to be a case for the Lambert W-function: \begin{align} e^{x-1}(x+1) &=2 \iff \\ e^{x+1}(x+1) &=2e^2 \iff \\ f(x+1) &= 2e^2 \Rightarrow \\ x + 1 &= f^{-1}(2e^2) \Rightarrow \\ x &= f^{-1}(2e^2) - 1 \end{align} where $f(x) = x e^x$ and $f^{-1} = W$, which is that Lambert W-function. We need the branch with positive real numbers, ...

1

The Lambert W function is defined as $z=W(z)e^{W(z)}$. We have $$(x+1)e^{x-1}=2 \tag 1$$ Multiplying both side of $(1)$ by $e^2$ reveals that \begin{align} (x+1)e^{x+1}&=2e^2\\\\ &\implies x+1=W(2e^2)\\\\ &\implies x=W(2e^2)-1\\\\ &\implies x=2-1\\\\ &\implies x=1 \end{align} NOTE: For the more general case for which ...

1

Set, for simplicity, $x-1=t$, so the equation becomes $e^t(t+2)=2$. Consider the function $$f(t)=e^t(t+2)-2$$ which is defined for all real $t$. Next $$\lim_{t\to-\infty}f(t)=-2, \qquad \lim_{t\to-\infty}f(t)=\infty$$ (prove it). So we see that at least one solution exist, because by the intermediate value term, $f$ must assume the value $0$. Let's have ...

2

One easy solution is $x = 1$ which you can find by inspection. $e^{1-1} = e^0 = 1$ but $1+1 = 2$. So $e^{1-1}(1+1) = 2$. The solution(s) are not going to be solvable using elementary techniques. In fact the solution(s) are $$x = W_n(2e^2) - 1$$ where $W_n$ is the analytical continuation of the product log function. Note: $W_n(2e^2) -1 = 1$ Otherwise, ...

1

\begin{align*} F_Y(y) &= \Pr[Y \le y] = \sum_{n=1}^\infty \Pr[Y \le y \mid N = n]\Pr[N = n] \\ &= \sum_{n=1}^\infty \int_{t=0}^y \frac{\lambda^n t^{n-1} e^{-\lambda t}}{\Gamma(n)} \, dt \cdot (1-p)^{n-1} p \\ &= \int_{t=0}^y \lambda p e^{-\lambda t} \sum_{n=1}^\infty \frac{(\lambda t (1-p))^{n-1}}{(n-1)!} \, dt \\ &= \int_{t=0}^y \lambda p ... 0 I don't think you can. Look at something like v(t) = e^{-t}\sin(g(t) t). Then v'(t) = e^{-t}(\sin(g(t)t) + g(t)\sin(g(t)t). If you choose g(t) so that it grows fast enough, your original function will be bounded from above by exponential decay, but you can make the derivative grow as fast as you like. Edit: This reply doesn't look at the positivity ... 1 The answer is no. Take v(t) = e^{-t} (1+\sin(e^x) ) v'(t) = \cos(e^t)-e^{-t} (1+\sin(e^t)) and v'(t) doesn't converge to 0 1 No. I will show a non-negative function which does the job (it's easy enough to turn it into a positive one). Take v(x)=e^{-x}\cos^2(e^x). Clearly we have v(x)\leq v(0)e^{-x}. The derivative, which is -e^x\sin(2e^x), is not bounded. 3 Use the Inverse function theorem. The jacobian of f is e^x\ne0. This gives local invertibility. To show that f is not a injective on \mathbb{R}^2, consider f(x,y) and f(x,y+2\,k\,\pi), k\in\mathbb{Z}. 1 Suppose: A_i and B_i for each i = 1, 2, \ldots represent times for A and B to complete their i^{\rm th} race, respectively. Once the i^{\rm th} race is complete, the (i+1)^{\rm th} race is begun immediately, without regard to the status of the other racer (i.e., the two processes are independent). Let N_A(t) and N_B(t) represent the total ... 1 Denote by F the event B>A (B wins before A ) since the events are independent and denote S the event A>B\Bbb{P}(B>A, n \text{ times}) = \Bbb{P} (FF \ldots FS) = \bigg(\frac{\alpha}{\alpha + \beta}\bigg)^n \frac{\beta}{\alpha + \beta} $$since \Bbb{P}(A>B) = P(A \geq B) = 1 - P(B>A) 4 There is no closed solution. However note that:$$e^x = \frac{1}{x}$$has one real root. To see this you can consider it as a function , differentiate , determine the range etc. Therefore we have a root. This root is a famous constant denoted as \Omega. Its approximate value is 0.5671. Otherwise, this root can be expressed via Lambert W.That is if ... 2 This can be rewritten as$$ xe^x = 1$$And here, there is no "simple" answer, you need to introduce the Lambert W function. 2 2 \ln(x)= \ln(x^{2}+x-3) so \ln(x^{2})=\ln(x^{2}+x-3) so x^{2}=x^{2}+x-3 so x=3. Added later : Notice that we must have x>0 and x^{2}+x-3>0, so 3 is an acceptable value!. 5 You can't exponentiate both sides just yet (well, you can, but I'd rather not), let's see what you can do instead using 2 \ln x = \ln x^2 giving us$$\ln x^2 = \ln (x^2 + x -3).$$Now you can raise e to the power of each side (exponentiate each side) and get x^2 = x^2 + x - 3 which is solvable and gives x = 3. Let's see if this works:$$e^{2 \ln 3 ...

1

$$B(f,T)=\frac{2hf^3}{c^2}\frac{1}{e^\frac{hf}{kT}-1}$$ $$\frac{\partial B}{\partial T}=\frac{2hf^3}{c^2}\frac{\partial}{\partial T}\left(\frac{1}{e^\frac{hf}{kT}-1}\right)=$$ $$=-\frac{2hf^3}{c^2}\left(\frac{1}{e^\frac{hf}{kT}-1}\right)^2\frac{\partial}{\partial T}\left(e^\frac{hf}{kT}-1\right)=$$ ...

2

My suggestion is to look at $\log B$. Take the derivate of the logged function wrt to $T$ and multiply by $B(T)$.

3

HINT : $$111x=\log_2(u)\Rightarrow x=\frac{\log_2(u)}{111}.$$

2

Hint: The product of the roots of the new cubic is $2^{111(x_1 + x_2 + x_3)}$, where $x_1, x_2, x_3$ are the roots. This is 2005 AIME I #8.

1

Hint: $$\sum_{k\geq1}\frac{\left(-\lambda t\right)^{k}}{k!}=\sum_{k\geq0}\frac{\left(-\lambda t\right)^{k}}{k!}-1=e^{-\lambda t}-1$$ and $$\frac{\lambda te^{-\lambda t}-\lambda t}{t}=\lambda e^{-\lambda t}-\lambda\rightarrow0.$$ as $t\rightarrow 0.$

0

To find the expected value of something, you multiply each of its values by the probability of those values, and add them all up, which in this case means you integrate. $$E(3X^2 - 6X + 2) = \int_0^{\infty} (3x^2 - 6x + 2)f(x)dx$$ where f(x) is the pdf of the exponential distribution, so $$E(3X^2 - 6X + 2) = \int_0^{\infty} (3x^2 - 6x + ... 2 As the exponential function is strictly growing, the equation 2^x=y has a single real solution in x (for y>0), which is called the logarithm (in base 2), denoted as \log_2(y).$$2^{\log_2(y)}=y.$$The values of 2^x for increasing integers x are 1,2,4,8,16,32\cdots, and for negative integers, \dfrac12,\dfrac14,\dfrac18\dfrac1{16}\cdots, ... 0 The property you employed to conclude x=3 from 2^x = 2^3 is called the injective property of the function f(x)=2^x. A function is said to be injective (or 1-1) if$$f(a)=f(b) \implies a=b.$$In the case of f(x) = n^x for a fixed positive n (except for n=1), the function is injective, so you can make the same conclusion. Thus 3^x = 27 \implies ... 4$$x = \log_2 y \iff y = 2^x$$The inverse is called the base two logarithm. In your case 2^x = 8 \iff x = \log_2 8 = \log_2 2^3 = 3. In general the inverse for a^x, where a> 1 is the base a logarithm. So y = a^x \iff x = \log_a y. 0 Those cases are easy to solve if you have something like the example you just gave. Look for how does logarithm and exponential functions work. In other cases like 2^x=\pi that is more complicated to computate manually. By computate manually I mean by giving an explicit value in decimal notation. So look some info about exponential and logarithmic ... 1 Applying the Binomial theorem, \begin{equation*} \lim_{n\to\infty}(1+\frac{x}{n})^n = \lim_{n\to\infty}1+x+\frac{(n-1)x^2}{2n}+\frac{(n-2)(n-1)x^3}{6n^2}+\frac{(n-3)(n-2)(n-1)x^4}{24n^3}+\frac{(n-4)(n-3)(n-2)(n-1)x^5}{120n^4}+... \\ =1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+... \\ =\sum_{n\geq 0}\frac{x^n}{n!}. \end{equation*} Note that ... 1 To some extent, you have in fact given the answer yourself, if you have (literally) tried everything then you have proven it is impossible to integrate in terms of elementary functions. This is indeed the case. This statement is similar to the impossibility of solving a polynomial of degree \geq 5 in terms of radicals; also the impossibility of "squaring ... 1 Just throwing it out there for you to see, I also like this proof:$$y=\ln xe^y=x$$after differentiating,$$e^y \frac{dy}{dx}=1 \begin{align} \frac{dy}{dx}&=\frac{1}{e^y}\\ &= \frac{1}{e^{\ln x}}\\ &= \frac{1}{x} \end{align} $$of course, that assumes you already know the derivative of e^x and the chain rule 4 Define$$e=\lim_{h\to 0} \left(1+h\right)^{1/h}.$$Then change variables h\mapsto h/x giving$$e=\lim_{h/x\to 0} \left(1+\frac{h}{x}\right)^{\frac{x}{h}}=\lim_{h\to 0} \left(1+\frac{h}{x}\right)^{\frac{x}{h}},$$where the limit in the second equality follows since h approaches 0 as h/x does. Since x is constant w.r.t. h, we can simplify by ... 11 If you can use the chain rule and the fact that the derivative of e^x is e^x and the fact that \ln(x) is differentiable, then we have:$$\frac{\mathrm{d} }{\mathrm{d} x} x = 1\frac{\mathrm{d} }{\mathrm{d} x} e^{\ln(x)} = e^{\ln(x)} \frac{\mathrm{d} }{\mathrm{d} x} \ln(x) = 1e^{\ln(x)} \frac{\mathrm{d} }{\mathrm{d} x} \ln(x) = 1x ...

2

If you can use the definition of $e$ as: $$e:=\lim_{n\rightarrow∞}\left(1+\frac{1}{n}\right)^n$$ and the slightly modified form: $\displaystyle e^x=\lim_{n\rightarrow∞}\left(1+\frac{x}n\right)^n$ then, by setting $h=\frac1{x}$ you can calculate the desired limit.

16

The simpler way is to use the inverse function theorem for derivatives: If $f$ is a bijection from an interval $I$ onto an interval $J=f(I)$, which has a derivative at $x\in I$,and if $f'(x)\neq 0$, then $f^{-1}\colon J\to I$ has a derivative at $y=f(x)$, and $$\bigl(f^{-1}\bigr)'(y)=\frac1{f'(x)}=\frac1{f'\bigl(f^{-1}(y)\bigr)}.$$ As $(\mathrm ... 5 If you accept that exponentiation is continuous, then certainly $$\left(\lim_{n\to\infty}\left(1+\frac1n\right)^n\right)^x = \lim_{n\to\infty}\left(1+\frac1n\right)^{nx}$$ But if$u=nx$, then by substitution we have $$\lim_{n\to\infty}\left(1+\frac1n\right)^{nx}=\lim_{u\to\infty}\left(1+\frac{x}{u}\right)^u$$ 5 Use that$e^t\geq 1+t$for all$t\in\mathbb{R}$and$x+y\geq 0$: $$e^{x+y-2}=\left(e^{\frac{x+y}{2}-1}\right)^2\geq\left(\frac{x+y}{2}\right)^2.$$ 0 Let $$f(a,b)=\int_0^{+\infty}e^{ibs^2}J_0(as)\,s\,ds.$$ Changing variable to$t=\sqrt{b}s$gives$f(a,b)=g(a/\sqrt{b})/b, where $$g(c)=\int_0^{+\infty}e^{it^2}tJ_0(ct)\,dt.$$ Integrating by parts, \begin{aligned} g(c)&=\bigl[e^{it^2}/(2i) J_0(ct)\bigr]_0^{+\infty}-\int_0^{+\infty} e^{it^2}/(2i)(-c J_1(ct))\,dt\\ ... 1e^x-e^y<ke^x-k<e^y\ln(e^x-k)<y<x$$Graph on an example : 2 To sketch it out : Let x = r\cos\theta and y=r\sin\theta. Upon considering both x, y are \geq 0 we get \theta \in [0,\frac{\pi}{2}] and r>0. Plug in the substitutions into your initial inequality to get \frac{e^2r^2}{4} \leq e^{r(\cos\theta+\sin\theta)} . Now, on [0,\frac{\pi}{2}] we get (pretty intuitively) that \min(\sin\theta ... 1 Make the change of variable y = \frac{1}{x^2}, then y \to +\infty , when x \to 0$$ \frac{e^{\frac{-1}{x^2}}}{x} = \sqrt{y}e^{-y} $$2 As it is, your limit goes to \infty, since 1/x^2 is very big whenever x approaches zero. A very different story would be if you had to compute$$ \lim_{x\to 0}\frac{e^{-1/x^2}}{x}. 1 This is going to be a very primitive attempt to answer this question. Yet I think it is worth posting since this kind of calculations appear many times in calculus. We take the log of the limit and then we Taylor expand the following function: \log\left(2 - e^{\frac{1}{k}}\right) = \log\left(1 - \sum\limits_{p=1}^\infty \frac{1}{p!} ... 0 A few hints that might help you out: If the product in your expression is P = e^{\log P}, then you can use the property of the logarithm \log (2-e^{\frac{1}{k}}) = \log 2 +\log (1-\frac{e^{\frac{1}{k}}}{2}). For large k the last expression is close to -\log 2, so you can use Taylor series for expansion. 1 Consider the more general integral \begin{align} I_{n} = \int_{0}^{\infty} t \, e^{-b t^{2}} \, J_{n}(a t) \, dt \end{align} for which \begin{align} I_{n} &= \sum_{k=0}^{\infty} \frac{(-1)^{k} \, \left(\frac{a}{2}\right)^{2k+n}}{k! \, \Gamma(k+n+1)} \cdot \int_{0}^{\infty} e^{- b t^{2}} \, t^{2k+n+1} \, dt \\ &= \frac{1}{2} \, \sum_{k=0}^{\infty} ... 0 Hint:\mathcal{L}\left(J_0(a\sqrt{x})\right) = \frac{1}{t} e^{-\frac{a^2}{4t}},\tag{1}\mathcal{L}\left(J_1(a\sqrt{x})\right) = \frac{|a|\sqrt{\pi}}{t^{3/2}}\left(I_0\left(e^{-\frac{a^2}{8t}}\right)-I_1\left(e^{-\frac{a^2}{8t}}\right)\right) e^{-\frac{a^2}{8t}}\tag{2}(1) just follows from writing J_0 as its Taylor series. The same technique ... 0 Since the function is even then \begin{align} I = \int_{ - \infty }^\infty {\exp \left( { - \frac{{2\pi ^2 }}{{x^2 }}} \right)dx} = 2\int_0^\infty {\exp \left( { - \frac{{2\pi ^2 }}{{x^2 }}} \right)dx} \end{align} Setting \begin{align} x = \frac{{\sqrt 2 }}{{\sqrt u }}\pi ,\,\,\,\text{i.e.,}\,\,\,u = \frac{{2\pi ^2 }}{{x^2 }} \Rightarrow du = - ... 1 The problem is intrinsically nonlinear since you want to minimizeSSQ(a,b,c)=\sum_{i=1}^N\big(ae^{bt_i}+c-y_i\big)^2$and the nonlinear regression will require good (or at least reasonable and consistent) estimates for the three parameters. But, suppose that you assign a value to$b$; then defining$z_i=e^{bt_i}$the problem turns to be linear$(y=az+c)\$ ...

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