# Tag Info

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One way to define $x(z)^{y(z)}$ is to use the universal cover of $\mathbb{C}^\times = \mathbb{C} \setminus \{0\}$, which can be defined by the map : $$\mathbb{C} \rightarrow \mathbb{C}^\times, t \mapsto z=\exp(t).$$ When $\mathrm{Re}(t) \to -\infty$ then $z \to 0$. Hence the two functions $z \mapsto x(z)$ and $z \mapsto y(z)$ on $\mathbb{C}^\times$ define ...

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It is worth graphing these functions (and playing with the parameters). For $x>1$ they are both monotone a similar rate of growth to an exponential. The difference is over the range $[0,1]$, where $I_{\frac{1}{4}}$ is more like $x^{\frac{1}{4}}$ and $I_{\frac{-1}{4}}$ is more like $\frac{1}{x^{\frac{1}{4}}}$. But the main point I want to make is that our ...

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You can take the taylor series around x=0 which is the sum from 0 to infinity of $(-x^4+x^2)^/k!$ take as many terms as you want for accuracy and integrate the polynomial. You'll get $x+x^3/3-x^5/10-5x^7/42+x^9/216+41x^{11}/3120+O(x^13)$ Integrate this over a suitably large domain which graphically looks like from -2 to 2, improve the number of terms in the ...

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In general, $~\displaystyle\int_0^\infty\exp\Big(-\sqrt[N]x\Big)~dx~=~N!~,~$ so even a relatively simple looking expression like $\displaystyle\int_0^\infty\exp\Big(-x^4\Big)~dx~=~\Big(\tfrac14\Big)!~=~\Gamma\bigg(1+\frac14\bigg)~=~\Gamma\bigg(\frac54\bigg)~$ cannot be expressed in terms of elementary functions, let alone a slightly more complex one, ...

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Note: $(e^x-1)(e^x+1)=(e^x)^2+e^x-e^x-1=e^{2x}−1$ Also note that the denominator $e^x-1\ne0\implies e^x\ne1\implies x\ne0$ $$\frac{e^{2x}−1}{e^x−1}=\frac{(e^x-1)(e^x+1)}{e^x−1}=e^x+1 \text{ where } x\ne0$$

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Substitute $t = e^x \implies e^{2x} = (e^x)^2 = t^2$, which translates the expression to $\dfrac{t^2 - 1}{t-1} = \dfrac{(t-1)(t+1)}{(t-1)}$. Then what is the result when replacing $t = e^x$?

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One simple way to look at this is to compare $e^x$ and $e^{-x}$. The first is "growing" and the second is "decaying." It all depends on the sign of $x$. Check it out for yourself: https://www.wolframalpha.com/input/?i=plot+e%5Ex https://www.wolframalpha.com/input/?i=plot+e%5E-x They are the same graph except flipped around the y-axis. Different fields ...

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I'm no mathematician, and there may be more concrete definitions, but this is how I think of a function as exponential "growth" and "decay." Theory If an exponential function is "skyrocketing" (for lack of better terminology) and heads towards $\pm\infty$, then it's "growing" (you can think of it as "absolute" growth, and disregard the sign). If an ...

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Note that we cannot have a solution with $x<0$ because then LHS>0 but RHS<0 (where I using the standard abbreviations for left-hand side (here $2^x$) and right-hand side (here $4x$). Notice that $2^0>4\cdot0$, but $2^1<4\cdot1$ and $2^5>4\cdot5$, so we expect one solution between 0 and 1 and another between 1 and 5, although so far we have ...

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Lambert solution $$2^x=4x \\ e^{x\log 2} = 4x \\ 1=4xe^{-x\log 2} \\ \frac{-\log 2}{4}=(-x\log 2)e^{-x\log 2} \\ W\left(\frac{-\log 2}{4}\right) = -x\log 2 \\ \frac{-1}{\log 2}W\left(\frac{-\log 2}{4}\right) = x$$ Using the two real branches of $W$, we get two real solutions: $$\frac{-1}{\log 2}W_0\left(\frac{-\log 2}{4}\right) = 0.3099069\dots \\ \frac{-... 1 You can algebraically rearrange this equation for hours and still get nowhere, because we cannot use algebra to solve it. Frustrating, isn't it? In short, you are better off letting a computer solve this for you via the Lambert W-Function (Wolfram Alpha will do this). Alternatively, you can graph the two sides of the equation and find the approximate ... 2 The first derivative equation is$$k_1(x+1)e^x-k_2e^{k_2x}-k_3=0$$and the second one,$$k_1(x+2)e^x-k_2^2e^{k_2x}=0.$$The solution(s) of the latter can be formulated in terms of the Lambert function W. https://en.wikipedia.org/wiki/Lambert_W_function#Examples The roots of the second derivative correspond to the extrema of the first derivative, which ... 1 Answer: e^{i m \phi} is the same thing as e^{i x} when x = m \phi. Note that i m \phi stands for the product i \, m \, \phi and not the imaginary part of \phi here. This should be interpreted as a family of periodic functions e^{i m \phi} and e^{-i m \phi} indexed by all non-negative integers m = 0, 1, 2, \dots. 8 If such a function exists, then you may add any function whose integral between 0 and 1 is zero. Now you should find one particular solution. For this, a constant (with respect to x) function suffices. 1 Note that if u=3^x, then u^2=3^{2x}. So in terms of u we have u^2-u\ge2 or u^2-u-2\ge0. You can factor the quadratic polynomial u^2-u-2 to get a solution for u, then take logarithms to get it in terms of x. 3 solutions you found are not correct. Correct solutions are y\le -1 and y\ge 2 but first solution gives no value of x becuase y is always positive. So now 3^x\ge 2 gives x\ge log_32 which is the final answer. Hope this helps ! 1 When you have  -1 > y or  y > 2, replace y with 3^x, it gives you (remembering that e > 0):$$e^{x\ln(3)}>2$$Which yields to:$$x > \dfrac{\ln(2)}{\ln(3)}1 You know that the basic exponential growth/decay equation is $$A=A_0e^{rt}$$ You are told that when t=120 that $$A=\tfrac{1}{2}A_0e^{120r}$$ which you solved correctly for r. Now you wish to know the value of t which causes $$0.60A_0=A_0e^{-0.005762265\,t}$$ so ... 1 The Taylor expansion of \exp\sin x around zero is 1+x+x^{2}/2+O(x^{4}). Therefore, the error is \begin{align*} \left|\exp\sin x-(1+x+x^{2}+x^{3})\right| & =\left|-x^{2}/2+O(x^{3})\right|\\ & \leq|x^{2}|/2+|O(x^{3})|\\ & \approx|x^{2}|/2 & \text{for }|x|\text{ small}. \end{align*} 0 As you noted the Taylor series for f(x) = e^x is f(x) \approx 1 + x + \frac{x^2}{2} + \frac{x^3}{6}. Now plug in \sin x and use the fact that for values close to x we have \sin x \approx x. So we have:e^{\sin x} \approx 1 + x + \frac{x^2}{2} + \frac{x^3}{6}$$Now the error of the first approximation is around \frac{x^2}{2} + \frac{5x^3}{6}. ... 0 Convergence is guaranteed for any t=-e^{-x}, as$$\lim_{n\to\infty}\left(1+\frac tn\right)^n=\lim_{m\to\infty}\left(1+\frac 1m\right)^{mt}=\left(\lim_{m\to\infty}\left(1+\frac 1m\right)^m\right)^t=e^t.$$No property of f_n is required. 0 It also works using L'Hopital's rule observing$$\lim_{n\rightarrow\infty}\left(1-\frac{e^{-x}}{n}\right)^{n}=e^{\lim_{n\rightarrow\infty}n\log\left(1-\frac{e^{-x}}{n}\right)} $$and so$$\lim_{n\rightarrow\infty}n\log\left(1-\frac{e^{-x}}{n}\right)=-e^{-x}\lim_{n\rightarrow\infty}\frac{\log\left(1-\frac{e^{-x}}{n}\right)}{-\frac{e^{-x}}{n}} =-e^{-x}\...

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I assume that $y_1$ means $y'$ and $y_2$ means $y''$. So differentiating we get $y'=-2y\frac{1}{\sqrt{1-x^2}}$. Differentiating again we get $y''=-2xy\frac{1}{(1-x^2)^{3/2}}+4y\frac{1}{1-x^2}$. Hence $$(1-x^2)y''-xy'=-2xy\frac{1}{\sqrt{1-x^2}}+4y+2xy\frac{1}{\sqrt{1-x^2}}=4y$$ So the equation holds with $\lambda=4$. Note that this does not agree with the ...

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Use that the power series of the exponential function converges for any $\;z\in\Bbb C\;$ : $$\frac1{e^z-1}=\frac1{1+z+\frac{z^2}2+\ldots-1}=\frac1{z\left(1+\frac z2+\mathcal O(z^2)+\ldots\right)}=$$ $$=\frac1{z}\left(1-\frac z2+\frac{z^2}4-\ldots\right)=\frac1z-\frac12+\ldots$$ The above is already enough to know, for example, the residue at $\;z=0\;$ of ...

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On differentiating; $y'=$$y\frac{-2}{√(1-x^2)} which on squaring gives, (1-x^2)y^2=4y^2. Differentiate again and after cancellation of 2y' on both sides, you get (1-x^2)y''-xy'-4y=0 0$$\frac{1}{z}-\frac{1}{2}+\frac{z}{12}-\frac{z^3}{720}+\frac{z^5}{30240}-\frac{z^7}{1209600}+\frac{z^9}{4790016}-\frac{691z^{11}}{1307674368000}+O(z^{13})$$I think it is no simple expression,so this is the result used mathematica. 1 The Laurent series at 0 is defined with the help of the Bernoulli numbers. 1 If one knows the following Taylor series expansion, as u \to 0,$$ \log(1+u)=u+O(u^2) $$then one may write, for any fixed real number t, as n \to \infty,$$ \left( 1+\frac{t}{n} \right)^{n}=e^{\large n\log\left(1+\frac{t}{n}\right)}=e^{\large n\left(\frac{t}{n}+O\left(\frac{1}{n^2}\right)\right)}=e^{t+O\left(\frac{1}{n}\right)} $$which gives$$ \... 1$2^x + 4^x= 2\Rightarrow2^x (1 + 2^x ) = 2\Rightarrow 1 + 2^x = 2 ^{1 - x}\Rightarrow1 + 2^x = 2 ^{- x} \times 2$Now Set$ y = 2^x$; then we have$1 + y = y^{-1} \times 2 \Rightarrowy^2 + y -2 = 0$. Which has solutions$y = 1$and$y = -2$.$y = -2 $is unacceptable, because$y = 2^x$is a positive function. So$y = 2^x = 0$is ... 1 we have $$2^{2x}+2^x-2=0$$ and with $$2^x=t$$ you will get $$t^2+t-2=0$$ a quadratic equation to solve. 3$a^x \cdot a^y=a^{x+y}$is the identity you are using$a^x+a^y=a^{x+y}$which is not correct 9 $$2^x+2^{2x}=2$$ Now put$2^x=t$$$t+t^{2}=2$$ $$t^{2}+t-2=0$$ $$(t-1)(t+2)=0$$ Thus$t=1$or$t=-22^x=1$or$2^x=-2$Since$2^x>0 $for all real$x$,$2^x=1=2^0$Therefore$x=0$1 The first limit diverges to$\infty$. You can see this by looking at the graph. Also, for the second limit, when$x \to 2^+$, it diverges to$-\infty$, when$x \to 2^-$, it diverges to$\infty$. The left and right limits are not the same, so the second limit does not exist. 11 Because$2\arctan\left(\frac{e^x-1}{e^x+1}\right)=2\left(\arctan(e^x)-\frac\pi4\right)=2\arctan(e^x)+C'$. The results differ by a constant. 1 It isn't true. Note that for$R=0$the left side is$0$and the right side$1/4$. Since both sides are continuous, that inequality persists for some$R > 0$. 2 Multiplying throughout by$4e^R$to get $$2(e^{2R})-2\ge e^{2R}\iff e^{2R}\ge2\iff2R\ge\ln2$$ 0 The equation is of the form a^x +b^x= c^x. In a more general setting , have a look at Fermat's Last Theorem Wiki Fermat's Last Theorem , it states that the equation will admit solution for positive integers a,b,c only for x=1 and x=2 ( Pythagora's theorem) ; for x>2, no three positive integers a,b,c will satisfy the equation 2 Hint we can write it as$(\frac{1}{(1+\frac{1}{n})})^{2n}=\frac{1}{e^2}$1$\lim_\limits{n\to \infty}\frac{n^{2n}}{(n+1)^{2n}}$divide top and bottom by$n^{2n}\lim_\limits{n\to \infty}\frac{1}{(1+\frac 1n)^{2n}} = \frac 1{e^2}$1 $$\lim _{ n\rightarrow \infty } \frac { n^{ 2n } }{ (n+1)^{ 2n } } =\lim _{ n\rightarrow \infty }{ { \left( \frac { 1 }{ { \left( 1+\frac { 1 }{ n } \right) }^{ n } } \right) }^{ 2 } } =\frac { 1 }{ { e }^{ 2 } }$$ 0 This is why everyone should learn GEMS: Groupings, exponenents, multiplication, subtraction/addition instead of PEMDAS. The expression in the absolute value is a grouping. 2 $$\sum_{s\geq 1}\frac{s}{(s-1)!}\lambda^{s-1} = \sum_{n\geq 0}\frac{n+1}{n!}\lambda^{n}=\frac{d}{d\lambda}\left(\lambda e^{\lambda}\right)=\color{red}{(\lambda+1)\,e^{\lambda}}.$$ 2 Hint $$\sum_{s=1}^\infty \frac{\lambda^{s-1}}{(s-1)!}s=\frac{\mathrm d }{\mathrm d \lambda}\sum_{s=1}^\infty \frac{\lambda ^s}{(s-1)!}$$ 1 $$\frac{\lambda^{s-1}}{(s-1)!}s=\frac{\lambda^{s-1}}{(s-1)!}(s-1)+\frac{\lambda^{s-1}}{(s-1)!}$$ Can you go on? 2 Your approach to evaluate the limit$\lim\limits_{x \to 0}\dfrac{e^{x} - 1}{x}$fails because you are trying to use quotient rule for limits which works only when the limit of denominator is non-zero. The justification given in$[1]$is completely wrong although it does look like it is intuitively correct. The result $$\lim_{x \to 0}e^{x} = \lim_{x \to 0} 1 ... 9 An "infinite sum" is not a sum. An infinite sum is the limit of a sequence :$$\sum_{n=0}^\infty=\lim_{N\to\infty} \sum_{n=0}^N.$$An the limit of a sequence of rational numbers is not necessarily rational. 3 Take any known irrational number, x. You can always represent x by the sum of an infinite number of rational numbers. For example, take the decimal representation of x: x = 5.1938527\ldots and then each term in the sum could form one of the decimal digits: x =5 + \frac{1}{10} + \frac{9}{100} + \frac{3}{1000} + \cdots Therefore the sum of an ... 4 The "tail" when we truncate the usual series for e just after the term \frac{1}{9!} is$$\frac{1}{10!}+\frac{1}{11!}+\frac{1}{12!}+\cdots.$$This sum is less than the sum of the infinite geometric series$$\frac{1}{10!}\left(1+\frac{1}{11}+\frac{1}{11^2}+\cdots\right),$$which is \frac{11}{10\cdot 10!}. This is less than our target error. So we can ... 1 I have a promising idea but still have to work the details. There is a class of integrals that connect \pi and e that come from:$$ \int_{0}^{+\infty}\frac{\cos(x)}{1+x^2}=\frac{\pi}{2e}\tag{1} $$Now we may apply integration by parts multiple times, reaching:$$ \frac{\pi}{e} = \int_{0}^{+\infty}\frac{p(x)(1-\cos x)}{(1+x^2)^k}\,dx \tag{2}$$with$p(x)\$...

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