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19

$y = 2 e^{x^2/2}$ satisfies the d.e. $y' = x y$ with $y(0) = 2$, while $z = e^{x} + e^{-x} = 2 \cosh(x)$ satisfies $z' = \tanh(x) z$ with $z(0)=2$. Since $x \ge \tanh(x) $ for $x \ge 0$, Gronwall's inequality does the rest.


14

As @flawr points out, the L.H.S. is $\displaystyle \cosh x = \frac{e^x+e^{-x}}{2}$ has an infinite product representation: $$\cosh x = \prod\limits_{n=1}^{\infty} \left(1+\frac{4x^2}{\pi^2(2n-1)^2}\right) \le \exp \sum\limits_{n=1}^{\infty}\frac{4x^2}{\pi^2(2n-1)^2} = e^{x^2/2}$$


14

$$\Longleftrightarrow 2^x(2^{x+1}+1)=2^3(2^{x+1}+1)$$


12

$$\large x^{\log(a)} = (e^{\log(x)})^{\log(a)} = e^{\log(x)\log(a)} = a^{\log(x)}$$ This only works if $x$ and $a$ are both positive real numbers.


8

In $2004$, using series compression, Brothers proposed $$e=\sum_{n=0}^{\infty}\frac{2n+2}{(2n+1)!}$$ which converges extremely fast. Using $10$ terms, you get $$e\approx \frac{69439789852104840011}{25545471085854720000}\approx 2.71828182845905$$ which has an error of $9 \times 10^{-22}$ while, for the same number of terms, the classical expansion would ...


7

The key here is to transform this into a quadratic equation. $$2^{2x+1} - 2^{x+4} = 2^3 - 2^x$$ can be rewritten as $$2 \cdot 2^{2x} - 2^4 \cdot 2^x = 2^3 -2^x.$$ Temporarily we will express $2^x=u$ and we find that the equation turns into $$2u^2 - 16u = 8 - u.$$ Finally writing this as $$2u^2 - 15u - 8 = 0$$ we can factor or use the quadratic equation to ...


7

This equation cannot be solved using “traditional” algebraic manipulations. In this case, one would use the Lambert W function: $$W(x): x = W(x)\cdot e^{W(x)}$$ or in other words, it is the solution of the equation $x = w e^w$. With this knowledge, we can try to substitute $y:=\frac{1}{x}$: $$\Rightarrow 0 = e^y-\frac{1}{y} \Rightarrow \frac{1}{y} = e^y ...


6

Note that by the Mean Value Theorem, we have $$ \frac{\tanh(x)}{x}=\mathrm{sech}^2(\xi)\tag{1} $$ for some $\xi$ between $0$ and $x$. Therefore, for $x\ne0$, $$ 0\lt\frac{\tanh(x)}{x}\lt1\tag{2} $$ Thus, for $x\ne0$, $$ \begin{align} x\frac{\mathrm{d}}{\mathrm{d}x}\left(e^{x-x^2/2}+e^{-x-x^2/2}\right) &=(x-x^2)e^{x-x^2/2}-(x+x^2)e^{-x-x^2/2}\\ ...


6

The function $f(x) = xe^x$ does not have an inverse that can be expressed in terms of any finite algebraic combination of the usual functions. The two branches that form its inverse are known as the Lambert W function(s). The equation $$ y = xe^x $$ can be solved for $y$ using a numerical approach, if an approximate solution is desired.


5

First, since both sides are even functions, it is sufficient to prove that $e^x+e^{-x} \leq 2e^{x^2/2}$ for all $x \geq 0$. Then we can change the problem to the equivalent problem of proving that for $x \geq 0$ $$ u(x) \equiv \frac{1}{e^x+e^{-x}} \geq \frac{1}{2}e^{-x^2/2} \equiv v(x)$$ We start by observing that $u(0) = v(0).$ Now in the immortal words of ...


5

I think this is basically a different way to state exactly what André Nicolas said in a comment, but observe that the largest power of a given prime $p$ found as a factor in a number less than $n$ is explicitly given by $$ \lfloor\log_p(n)\rfloor=\left\lfloor\frac{\ln n}{\ln p}\right\rfloor $$ so we can write $LCM(n)=LCM(1,2,...,n)$ explicitly as $$ ...


4

Hint: Make the substitution $u=8^x$. Now the equation becomes $u^2+u-20=0$.


4

$$\overline{e^z}=\overline{e^xe^{iy}}=e^x(\overline{\cos y+i\sin y})=e^x(\cos y-i\sin y)=e^x(\cos(-y)+i\sin(-y))=e^{x-iy}=e^{\overline{z}}$$


4

You won't get a closed-form solution, but numerical methods can be used.


4

A better formula to use would be $$y=a\left(1+\frac{r}{k}\right)^{kt},$$ where $k$ is the number of times the interest is compounded per year. So, plugging in your information gives $$y=3750\left(1+\frac{.06}{12}\right)^{132}=\$7243.55$$


3

Remember that $e^{i\theta} = \cos\theta+ i\sin\theta$ and so $e^{2\pi i} = 1$. So you can mark these four points on the unit circle, and you'll see that they are at angles, respectively, $-3\pi/4$, $-\pi/4$, $\pi/4$, and $3\pi/4$. Thus, these are the vertices of a square, and the four complex numbers add up to $0$.


3

Hint: The $x$-intercept is when $\frac43 e^{3x}+2e^{2x}-8e^{x} = 0$. Now set $y = e^{x}$, so your equation is $$\frac{4}{3}y^3+2y^2-8y = 0 $$ which means $$y\left(\frac{4}{3}y^2+2y-8\right) = 0. $$


3

take the $\log$ for both sides to get $$\log (x^{log(a)})=\log (a^{\log(x)})$$ $$\log (a){log(x)}=\log( a){\log(x)}$$ It is clear that the $a$,$x$ should be positive


3

From your equation you get: $(3^{2x}+2^{3x})/{136^x}=1/6\Rightarrow (9/136)^x+(4/136)^x=1/6$. The second member is constant while the first one is a function in $x$: $f(x)=(3^{2x}+2^{3x})/{136^x}$. The function is strictly decreasing, so the equation has at most one solution. Checking with Wolphram Alpha you get a numerical solution. You can actually state ...


3

The power series of the exponential function is defined on $\Bbb R$ so we can differentiate it term by term on $\Bbb R$ and we get $$\exp'(x)=\exp(x)$$ Moreover, we see easily that $\exp(x)>0$ for $x\ge0$ and using the Cauchy product we get $$\exp(x)\exp(y)=\exp(x+y),\quad \forall (x,y)\in\Bbb R^2$$ hence $$\exp(-x)\exp(x)=\exp(0)=1\implies ...


3

Using any logarithm $\log$, we have $$\log n = \log (d^m) = m \log d,$$ so $$m = \frac{\log n}{\log d} = \log_d n.$$


3

Factor out $2^{x+1}+1$ from both sides, you will get: $$2^x(2^{x+1}+1)=2^3(2^{x+1}+1)$$ One solution will be $2^{x+1}+1=0$ and second will be $2^x=2^3$. First equation does not have real solutions, but over complex number it have a solution $$x=-1+\frac{i\pi}{\ln2}+\frac{2i\pi k}{\ln2},k\in\mathbb{Z}$$ Second equation have only one real solution $x=3$ and ...


2

The solutions are counting multiplicity and include complex solutions. It's called the fundamental theorem of algebra. Low-level proofs are not easy to come by, however.


2

Hint. You may write $$ (1+\frac{\epsilon}{2n})^n=e^{n \log\left( 1+ \frac{\epsilon}{2n}\right)} $$ and use $$ (e^u)'=u'\times e^u $$ giving here $$ \left(n \log\left( 1+ \frac{\epsilon}{2n}\right) \right)'\times e^{n \log\left( 1+ \frac{\epsilon}{2n}\right)} $$ or $$ \left(\log\left( 1+ \frac{\epsilon}{2n}\right)- \frac{\epsilon}{2n}\frac{1}{\left(1+ ...


2

$\large{e^{\ln(3+x)} = e^9}$ means, $\large{e^{\ln_e(3+x)} = e^9}$ and it is a logarithmic property that that, $\large{x^{log_xy}=y}$, therefore $e$ on the LHS "cancels" out.


2

This is a duplicate. I copy below my answer Using Taylor expansion we have $$\sin t=t (1+o(1))\text{ when }t\to 0.$$ At this point if $x$ goes to infinity, then $$\left(1+sin \frac{1}{x}\right)^x \sim \left(1+\frac{1}{x}\right)^x\to e.$$


2

Look at the series of $2\cosh x$: $$2\cosh x = 2 + x^2+\frac{x^4}{12}+ \ldots= \sum_{k=0}^\infty 2\frac{x^{2k}}{(2k)!}$$ Thus, when $|x|<1$: $$e^x+e^{-x}-2-x^2=\sum_{k=2}^\infty 2\frac{x^{2k}}{(2k)!} < x^4\sum_{k=2}^\infty\frac{2}{(2k)!}= (2\cosh 1 -3)x^4$$ $$\approx 0.086\ x^4 <x^4/6$$


2

The user OohAah pointed me in the right direction. $$ |f(x)-p_3(x)| = |R_4(x)| $$ $$ |R_4(\phi x)| = (e^{\phi x}+e^{-\phi x}) {x^4 \over 24} $$ where $ 0 \le \phi \le 1 $. Since $ |x| \le 1 $ it follows that $$ \phi x \le 1 $$ Ergo $$ |R_4(x)| \le (e+e^{-1}){x^4 \over 24} \le {x^4 \over 6} $$ If my mathematical writing is sloppy, please do point it ...


2

The approach you mention is difficult, but possible. I have presented it in my blog post. The main steps are as follows: 1) Define $a^{b}$ (without using any logs or $e$) rigorously for $a > 0$ and any real $b$. 2) Show that $\lim_{a \to 0}\dfrac{x^{a} - 1}{a} = f(x)$ exists for all $x > 0$ and hence defines a function of $x$. This function is ...


2

Concerning the antiderivative $$\int\frac{e^{x}}{x}dx=\text{Ei}(x)$$ where appears the exponential integral function which would present problems around $x=0$. But, provided $b>0$ $$\int_0^b \frac{1-e^ {-x}}{x}dx=\gamma+\log (b)+\Gamma (0,b)$$



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