# Tag Info

2

To have any solution you need the quadratic equation $$p y^2+2=y$$ to have a positive solution, since $y=3^x$ is always positive. To have a unique solution you need there to be only one positive solution. This will happen when either there is only one solution which happens to be positive, or when there is a positive and a nonpositive solution. In standard ...

3

First limit is easy. Let $$f(n) = \left(1 + \frac{1}{n!}\right)^{n}$$ We will use the fact that the sequence $a_{n} = (1 + (1/n))^{n}$ is strictly increasing for $n \geq 3$ and tends to a positive limit (when $n \to \infty$) denoted by $e$. Also $2 < e < 3$. These facts can be proven (and given in many textbooks) without any standard theory of ...

0

Consider the first expression $$A=\left(1 + \frac{1}{n!}\right)^n$$ So $$\log(A)=n\log\left(1 + \frac{1}{n!}\right)\approx \frac n{n!}=\frac 1{(n-1)!}$$ So, $\log(A)$ goes to $0$ and then $A$ to $1$. Consider the second expression $$B_n=\left(1 + \frac{1}{n!}\right)^{n^n}$$ So $$\log(B)={n^n}\log\left(1 + \frac{1}{n!}\right)\approx \frac {n^n}{n!}$$ Now, ...

3

HINT: $$\lim_{n \to \infty} \left(1 + \frac{1}{n!}\right)^n=\lim_{n \to \infty} \left(\left(1 + \frac{1}{n!}\right)^{n!}\right)^{n/n!}$$ and the inner limit is $e$, hence the final one is 1. The second case is similar.

5

Let $r=s/t$ where $s\in\mathbb{Z}$ and $r\in\mathbb{N}$. Assume $e^r=p/q$ where $p,q\in\mathbb{N}$. Then $$pqt^nJ_n(r)= p^2 t^n A_n\left(\frac{s}{t}\right)+ q^2 t^n B_n\left(\frac{s}{t}\right)\in\mathbb{Z} \tag{1}$$ Note that $$0<J_n(x)\leq\frac{x^{2n}}{n!}\int_{-x}^xe^tdt=\frac{2 x^{2n}\sinh x}{n!}$$ So $$0<pqt^nJ_n(r)\leq \frac{2 pq t^n ... 1 I'm not sure what you are allowed to assume. Do you know that e is transcendental and can you use that? If so consider e^{\frac{p}{q}}=l where l\in \mathbb{Q}. What happens when you look at the polynomial x^p-l^q ? Looking at the first part of your question I'm assuming it's unlikely you are allowed to use the fact that e  is transcendental ... 1 Recall the definition of exponential function of one variable in terms of power series:$$ e^x := \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots $$Similarly, we can define the matrix exponent of n\times n matrix A:$$ e^A : = I + A + \frac{1}{2!}A^2 + \frac{1}{3!}A^3 + \dots = \sum_{n=0}^{\infty}\frac{A^n}{n!}, ...

1

A slight refinement: $$0 < f(k) - e^{-k} = \sum_{k=2}^{\infty} e^{-kn^2} < \sum_{k=2}^{\infty} e^{-kn} = \frac{e^{-2k}}{1-e^{-k}},$$ which is bounded by, say, $2e^{-2k}$ for $1-e^{-k}>1/2$, i.e. $k>\log{2}$. Hence, in the Poincaré-type asymptotics, $$f(k) = e^{-k} + O(e^{-2k}).$$ Aside: The $(\pi/k)^{1/2}$ bound is useful when $k$ is ...

1

For $k\ge 1,$ $$\sum_{n=1}^{\infty}e^{-kn^2} \le \sum_{n=1}^{\infty}e^{-kn} = \frac{e^{-k}}{1-e^{-k}}\le \frac{1}{1-e^{-1}}\cdot e^{-k}.$$

2

There is no way to express in elementary functions. In fact, go to http://mathworld.wolfram.com/JacobiThetaFunctions.html and you will find $$f(k)=\frac12(-1+\vartheta_3(0,e^{-k})).$$

3

I think other answers given here assume the existence of a nice function $e^{x}$ and this makes the proof considerably simpler. However I believe that it is better to approach the problem of solving $f'(x) = f(x)$ without knowing anything about $e^{x}$. When we go down this path our final result is the following: There exists a unique function ...

-1

I guess exponential is too crazy for that kind of stuff - it's growing - well, exponentially - so basically, always growing more than ever before - not just getting bigger - but getting more bigger than it ever got bigger before. All I'm saying is that the derivative always grows, i.e. that the second derivative is positive. Problem is, all derivatives are ...

2

You said formula give scores in $[0.3,0.4]$, which makes me guess your your raw data is in $[-0.6,-0.2]$. Try: $n\over{n+{(e^{-score})}}$, try varying the values of $n$ (n=5,5.5,6,7 etc will give you output in range $[.7,.8]$ if your raw data is in the domain mentioned above.) $n\over{n+n^{-score}}$, try varying the values of $n$. (n=6,7,8 etc will give ...

2

I guess your current raw scores are in the neighborhood of $-0.7$ or so, which yields a scaled score of $$\frac{1}{1+e^{0.7}} \doteq 0.33$$ If you want, you can replace the $1$ in both the numerator and the denominator by $6$, and then you would get $$\frac{6}{6+e^{-score}} = \frac{6}{6+e^{0.7}} \doteq 0.75$$ But this is fairly ad hoc. I don't know ...

0

If we look for the maximum of the curvature of the curve $y=e^{ax}$, we have $$\kappa=\frac{y''}{(1+(y')^2)^{3/2}}=\frac{a^2 e^{a x}}{\left(1+a^2 e^{2 a x}\right)^{3/2}}$$ $$\frac{d\kappa}{dx}=\frac{a^3e^{ax}(1-2a^2e^{2ax})}{\left(1+a^2 e^{2 a x}\right)^{5/2}}$$ So, the maximum curvature is obtained for $$x=-\frac{\log(2a^2)}{2a}$$ for which ...

1

OK, let's try a plodding pedestrian approach: $$f'(x)-f(x) = 0$$ The idea of an exponential multiplier $m(x)$ is that we write $$0 = m(x)f'(x) + (-m(x))f(x) = m(x)f'(x)+m'(x)f(x) = \Big( m(x)f(x)\Big)'.$$ For this to work we would need $-m(x)=m'(x)$. We do not need all functions $m$ satisfying this, but only one. So there's no need to wonder whether ...

14

Observe that if $f(x)=f'(x)$ then $$\left(\frac{f(x)}{e^x}\right)'=\frac{f'(x)-f(x)}{e^x}=0$$ Hence $\dfrac{f(x)}{e^x}$ is constant...

4

Consider $g(x) = f(x)\exp(-x)$. Then we have $g'(x) = f'(x)\exp(-x)-f(x)\exp(-x) = 0$. Thus, $g\equiv c$ for some constant $c$. Hence $f(x) = c\exp(x)$.

3

You have $\dfrac{dy}{dx}=y$. Often one writes $\dfrac{dy} y = dx$ and then evaluates both sides of $\displaystyle\int\frac{dy} y = \int dx$, etc. However, for a question like this perhaps one should be more careful. If $f(x)\ne 0$ for all $x$, then one has $\dfrac{f'(x)}{f(x)}=1$ for all $x$. This implies $$\frac{d}{dx} \log |f(x)| = 1$$ for all $x$. ...

7

No. Suppose that $f \not\equiv 0$. We solve the ODE: $$f(x) = f'(x) \implies \frac{f'(x)}{f(x)} = 1 \implies \int \frac{f'(x)}{f(x)}\,{\rm d}x = \int \,{\rm d}x \implies \ln |f(x)| = x+c, \quad c \in \Bbb R$$With this, $|f(x)| = e^{x+c} = e^ce^x$. Call $e^c = A > 0$. Eliminating the absolute value, we have $f(x) = Ae^x$, with $A \in \Bbb R \setminus ... 0 You can compute the maximum of the curvature of the curve$y=e^x$. This can be done by the formula $$\kappa=\frac{y''}{(1+(y')^2)^{3/2}}=\frac{e^x}{(1+e^{2x})^{3/2}}.$$ A simple computation shows that this is maximal for$x=-\frac{\ln 2}{2}$(after edit). However, for the function$A\,e^{Bx}$where$A,B>0$, the result may be different and the computed ... 1 If you have this differential equation: $$f'(x)=2f(x)$$ You can separate the variables (divide both sides by$f(x)$) to get: $$\frac{f'(x)}{f(x)}=2$$ Notice that in this case you "ignore" the solution$f(x)=0$because your left hand side would otherwise be undefined. You need to remember that$f(x)=0$would also be a solution to this equation. You can ... 1 observe that $$\left(\frac{f(x)}{e^{2x}}\right)'=\left(\frac{f'(x)-2f(x)}{e^{2x}}\right)=0$$ Hence the function$\frac{f(x)}{e^{2x}}$is constant. 1 No, this is not enough. Obviously the object of the problem is to make you "discover" this point, probably because you have heard of logarithm before. You have$f'(x)=2f(x)$It means that$\forall x$such that$f(x) \neq 0$, you have$\dfrac{f'(x)}{f(x)}=2$But can you integrate$\dfrac{f'(x)}{f(x)}$? 10 $$\lim_{n\to\infty} \frac{3^n}{2^n+3^n}=\lim_{n\to \infty}\frac{1}{(\frac{2}{3})^n+1} =1.$$Since$\frac{2}{3}<1$, so$(\frac{2}{3})^n\to 0$as$n\to \infty$. 3 Write $$f(x)=x^{\sqrt x}$$ Then $$g(x)=\ln f(x)=\sqrt x\ln x=\frac{\ln x}{x^{-1/2}}$$ Now use l'Hopital to compute $$\lim_{x\to0^+}g(x)$$ Since$x\mapsto e^x$is continuous, $$\lim_{x\to 0^+}f(x)=e^{\lim_{x\to0^+}g(x)}$$ 2 Hint $$\sqrt x\ln x=\frac{\ln(x)}{\frac{1}{\sqrt x}}$$ 2 In fact,$\ln x < \sqrt x$for all$x>0.$Let$f(x) = \sqrt x - \ln x.$Then$f'(x) = 1/(2\sqrt x) - 1/x.$This tells us$f'<0$on$(0,4), f'(4)=0,$and$f'>0$on$(4,\infty).$Thus$f(4)$is the absolute minimum of$f$on$[1,\infty).$It is quite simple to check that$f(4) > 0,$hence$f> 0$on$(0,\infty).$5 Hint: The series representation for the exponential will work. Use the fact that if$x$is positive then $$e^x\gt \frac{x^3}{3!}.$$ 0 You're nearly there, but not quite. You have to take the additional step to say that, if$Z=X/c$, then $$\tag{1}F_Z(x) = F_X(cx)=1-e^{-\lambda(cx)}=1-e^{-(c\lambda)x}=F_Y(x)$$ where$Y\sim\operatorname{Exp}(c\lambda)$, and therefore$Z$and$Y$are identically distributed. More tersely, ... 1 The following recursive formula might help. Suppose$g_n(x)$is defined by $$g_n(x) = f_n(x)^{f_{n-1}(x)^{{\dots}^{f_1(x)}}}$$ That is,$g_n(x) = f_n(x)^{g_{n-1}(x)}$. Then$\ln g_n(x) = g_{n-1}(x) \ln f_n(x)$, and so, after implicit differentiation, we get $$g'_n(x) = g_n(x) \cdot \left( g'_{n-1}(x) \ln f_n(x) + g_{n-1}(x) \frac{f'_n(x)}{f_n(x)} \right)$$ ... -3 By Mathematica:$\cos ^{2^{x^{\cos (x)}}}(x) \left(\log (2) 2^{x^{\cos (x)}} x^{\cos (x)} \log (\cos (x)) \left(\frac{\cos (x)}{x}-\log (x) \sin (x)\right)-2^{x^{\cos (x)}} \tan (x)\right)0 For the first one, you probably just need to simplify it. Because it isn't an equation, you can't really "solve'' it for anything. Here's how you could simplify that expression: \begin{align*} e^{8\ln\left(b^{\frac{1}{4}}\right)} &= e^{\ln\left(b^{\frac{1}{4}}\right)^8} \\ &= e^{\ln\left(b^{\frac{8}{4}}\right)} \\ &= e^{\ln\left(b^{2}\right)} ... 0 The first one isn't equal to anything to solve so it's unknown, the other is $$\ln(6x-2)=5$$ $$6x-2=e^5$$ $$6x=e^5+2$$ $$x=\frac{e^5+2}{6}$$ 1 In a general manner, any equation which can write $$f(x)=A+Bx+C\log(D+Ex)=0$$ has solutions in terms of Lambert function. If, for any reason, you cannot use it, only numerical methods will find the root. Probably, the simplest should be Newton which, starting from a "reasonable" guessx_0$will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ ... 1 Although the Lambert W function is a perfectly valid solution, as described in the other answers, it is impossible to compute with a standard calculator, and is beyond your current math level anyway (judging from your question). The equation$2^{n-3}=20/n$tells you that$20/n$is a multiple of$2$. Furthermore, it should be obvious that$n>3$, since ... 3 Suppose$n\cdot 2^n=160$. Since$160=2^5\cdot 5$, you know that$5$divides$n$, so$n=5m$. Then $$5m\cdot2^{5m}=160$$ becomes$m\cdot 2^{5m}=32$and so$m=1$. However, the general solution of$x\cdot 2^x=a$, for arbitrary$a$, cannot be determined “explicitly”, without using “higher level” functions such as Lambert's$W$. 3 You can modify your equation as such $$\ln(2)n e^{\ln(2)n} = 160\ln(2)$$ So that $$n = \frac{1}{\ln(2)} W(160\ln(2))=5$$ Where$W$is the Lambert W function. I do not know any identity on W that would lead the result 5 without numerically computing W (which is usually done using root-finding). 1 If$a$is small, a first order Taylor expansion built at$a=0$gives $$(x+a)^{x+a}-x^{x+2a}=-a\, x^x\, \big(\log (x)-1\big)+O\left(a^2\right)$$ Do you think that this is sufficient to prove that $$\lim_ {a \to 0} X_a = e$$ Another way is starting from zoli's answer $$\frac{(x+a)\log(x+a)-x\log(x)}{a}=2\log(x)$$ and expand the lhs as a Taylor series built at ... 0 Plug in 5 for t and evaluate. Add that to the 5000 cells you already have. 2 Taking the logarithm of both sides of $$(x+a)^{x+a}=x^{x+2a}$$ we get $$(x+a)\ln(x+a)=x\ln(x)+2a\ln(x)$$ or $$\frac{(x+a)\ln(x+a)-x\ln(x)}{a}=2\ln(x). \tag 1$$ The left hand side tends to$\frac{d(x\ln(x))}{dx}=\ln(x)+1$if$a$tends to zero. The right hand side does not depend on$a$. That is, $$\ln(x)+1=2\ln(x).$$ As a result, whatever the solution ... 1 Rewriting the equation to$(1+\frac{a}{x})^{x+a}=x^a$and taking$\ln$, we have$(x+a) \ln(1+\frac{a}{x})=a\ln x$. Assuming that$X_a$is bounded for$a$from some neighbrhood of$0$, we take the Taylor approximation for small$a$of the left hand side: $$(X_a+a) (\frac{a}{X_a}+o(a))=a\ln X_a$$ and $$a+o(a) = a\ln X_a$$ which yields$\ln X_a=1$. 4 This might not be rigorous, but note that one has $$(x+a)^x=x^{x+a}\Rightarrow x\ln(x+a)=(x+a)\ln x\Rightarrow \frac{\ln(x+a)}{x+a}=\frac{\ln x}{x}.$$ Then, let$f(x)=\frac{\ln x}{x}$. One has$f'(x)=0\iff x=e$, and considering the graph of$y=f(x)$should give you the answer. 0 $$\sum_{n=0}^\infty\frac{n^2}{n!} = \sum_{n=1}^\infty\frac{n}{(n-1)!}=\sum_{n=0}^\infty\frac{n+1}{n!}=\frac{d}{dx}xe^x|_{x=1}=2e$$ 5 You got a very good start. Now rewrite the numerator$n$as$(n-1)+1$. Our sum is equal to $$\sum_1^\infty \frac{n-1}{(n-1)!}+\sum_1^\infty \frac{1}{(n-1)!}.$$ By the method you already used, the first sum is$\sum_2^\infty \frac{1}{(n-2)!}$. This sum is$e$. More simply, the second sum is$e$. 2 Factorials in denominator are always appealing. So start writing $$S=\sum_{n=0}^\infty\frac{ n^2}{n!}x^n=\sum_{n=0}^\infty\frac{ n(n-1)+n}{n!}x^n=x^2\sum_{n=0}^\infty\frac{ n(n-1)}{n!}x^{n-2}+x\sum_{n=0}^\infty\frac{ n}{n!}x^{n-1}$$ Recognize that the first sum is the second derivative of$\sum_{n=0}^\infty\frac{ x^n}{n!}=e^x$and that the second sum is the ... 1 Least squares folution to overdetermined system$Ax = b$is given by $$x = \left(A^\mathsf{T} A\right)^{-1} A^\mathsf{T} b$$ There's no need to do a transform$z = e^x$, it is solvable as it is. Rewrite the overdetermined system $$e^{2x_1} a + e^{x_1} b + c = f_1\\ e^{2x_2} a + e^{x_2} b + c = f_2\\ \vdots\\ e^{2x_n} a + e^{x_n} b + c = f_n\\$$ in matrix ... 1 Hint We can (after reindexing and checking convergence) write the sum as the value of the series $$\sum_{k = 0}^{\infty} \frac{n + 1}{n!} x^n$$ at$x = 1$. 2 Notice that $$I = I(M,c) := \int_{-\infty}^{\infty} \frac{e^{-y^2/2}\sinh^2 (cy)}{\cosh(Mcy)} \, dy = \frac{1}{c} \int_{-\infty}^{\infty} \frac{e^{-y^2/2c^2}\sinh^2 y}{\cosh(My)} \, dy.$$ Then as$c \to \infty$and$M > 2\$, $$I \sim \frac{1}{c} \int_{-\infty}^{\infty} \frac{\sinh^2 y}{\cosh(My)} \, dy = \frac{\pi}{2Mc}(\sec(\pi/M) - 1).$$

7

Hint: $$\frac{2^{2x}}{5^{x-1}}=5\frac{2^{2x}}{5^x}=5\frac{4^{x}}{5^x}=5\left(\frac{4}{5}\right)^x$$

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