New answers tagged

0

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 ...


1

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 ...


0

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 ...


2

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, ...


2

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


3

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 $t = e^x$?


0

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 ...


1

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 ...


0

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 ...


0

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 \begin{equation} A=A_0e^{rt} \end{equation} You are told that when $t=120$ that \begin{equation} A=\tfrac{1}{2}A_0e^{120r} \end{equation} which you solved correctly for $r$. Now you wish to know the value of $t$ which causes \begin{equation} 0.60A_0=A_0e^{-0.005762265\,t} \end{equation} 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}\...


1

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 ...


0

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 ...


1

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

$2^x + 4^x= 2$ $\Rightarrow$ $2^x (1 + 2^x ) = 2$ $\Rightarrow $ $1 + 2^x = 2 ^{1 - x}$ $\Rightarrow$ $1 + 2^x = 2 ^{- x} \times 2$ Now Set $ y = 2^x$; then we have $1 + y = y^{-1} \times 2 $ $\Rightarrow$ $y^2 + y -2 = 0$. 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 $t=-2$ $2^x=1$ 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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