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18

Well you have that $$y'=y''=y^{(3)}\cdots$$ only function that is a derivative of itself is $$ae^{x}$$ for some $a$ so $$y'=ae^x$$ and $$y=y'-1=ae^x-1$$ And since $y(0)=1$ than $a-1=1$ so $a=2$


12

My favorite paper about $x^x$ is The $x^x$ Spindle, which appeared in Mathematics Magazine back in 1996. The main idea is to visualize the fact that we can write it as $$x^x = e^{x (\ln(x)+2k\pi i)}.$$ Note that for each choice of $k$, we get a different branch of the logarithm. Given any real number $x$, most of these branches will be complex valued. ...


5

The denominator of the exponent tends to zero from above, no matter how $(x,y)$ tends to $(0,0)$. So the exponent grows without bound to $+\infty$. Hence the expression itself does as well. (Some may say the limit doesn't exist since it is infinite, but that's a matter of convention.) If the numerator were $-1$ instead of $1$, the exponent would tend to ...


5

The answer is simple: $$(a^b)^c = a^{bc}$$ is an equality which only holds for real numbers with $a>0$ and does not necesarily hold on complex numbers.


5

Unfortunately, as you note, the limit $$\lim_{h\rightarrow0}\frac{a^h-1}h$$ is not easily worked with (since it equals $\log(a)$). We can, however, gleam two bits of information from it. Firstly, if we define the function $$f(a)=\lim_{h\rightarrow 0}\frac{a^h-1}h$$ we can show that it is monotonic increasing; in particular, notice that if $b>a$ then, from ...


5

Let $u=t^4$ so $t=u^{1/4}$ and $dt=\frac 14 u^{-3/4}$ and then $$\int_0^\infty e^{-t^4}dt=\frac14\int_0^\infty u^{-3/4}e^{-u}du=\frac14\int_0^\infty u^{-3/4}e^{-u}du=\frac14\Gamma\left(\frac14\right)=\Gamma\left(\frac54\right)$$ using the equality $$\Gamma(x+1)=x\Gamma(x)$$


5

If I am understanding correctly you would like to know if there is a suitable function $f$ such that $e^{a+b+c}=f(a)+f(b)+f(c)$ for all values of $a,b,c$. Notice such a function would satisfy $f(0)+f(0)+f(0)=e^0=1$. So $f(0)=\frac{1}{3}$. From here we can determine the function uniquely, since we must have $e^x=f(x)+f(0)+f(0)=f(x)+\frac{2}{3}$. So the ...


4

I'ts not "exponential" in the sense of the derivative being proportional to the value, no. It does, however, have "exponential growth", in the sense that there's a constant $C$ with $$ |f(x)| \ge C u^x $$ for large enough $x$ and for some $u > 1$. In computer science, such functions are sometimes sloppily called 'exponential', even though they could ...


4

By the first condition, we have that $f(a/b) = f(1/b)^a$ and $f(1/b)^b = f(1)$ for any $a/b \in \mathbb{Q}$. Let $f(1) = ce^{i\theta}$. Then $f(1/b) = c^{1/b}e^{i(\theta + 2\pi k)/b}$ where $k$ is an integer depending on $b$. It follows $f(a/b) = e^{a/b}e^{i(\theta + 2\pi k)a/b}$. By continuity of $f$ and density of rationals, it follows that $k$ is locally ...


4

This kind of equations, which mix polynomial and trigonometric or hyperbolic terms do not show analytical solutions (beside the trivial $x=0$) and only numerical methods should be used. If you want me to elaborate on this topic, just post. Please notice that we can write the equation in a more simple form changing variable $ax=y$ to get $$\sinh(y)=c y$$ ...


4

You're using $\log$ incorrectly, even in the real sense. It's not true, for instance, that $\log\left(e+e^{-1}\right)=0$. Hint: Instead, multiply both sides of $8=e^{iz}+e^{-iz}$ by $e^{iz}$ and solve the resulting quadratic in $e^{iz}$.


4

Hint: given an arbitrary monic polynomial $p(x)$ of degree $N$, let $x_k$, $k=1,\cdots,N$ be its roots. Then $$p(x)=(x-x_1)\cdots(x-x_N).$$ The coefficient of $x^{N-1}$ is $-x_1-x_2-\cdots-x_N$, i.e. negative sum of roots. Your polynomial is of degree 2011 (but not monic, just transform it into a monic one, note $x^{2012}$ is cancelled). Thus the sum of ...


4

All your differential equations after the first one are implied by the first equation. If $y'=y+1$ then differentiation of it gives $y''=y'$ and further differentiation gives the successive equations. So it is really just about solving the first equation with the given initial value, and this you can do.


4

One option is to solve this by diagonalization. Here's another option: Note that $$ A^2 = -w^2I $$ Thus, we have $$ A^n = \begin{cases} (-1)^{n/2}w^{n}I& n \text{ is even}\\ (-1)^{(n-1)/2}w^{n-1}A& n \text{ is odd} \end{cases} $$ Now, expand the matrix exponential $$ \exp(At)=\sum^{\infty}_{n=0}\frac{A^nt^n}{n!}= ...


4

For $n$ be sufficiently large, we have $(1-x_n)^n >0$ (since it converges to 1). We can then say that $n \log(1-x_n)$ tends to zero. Necessarily $\log(1-x_n)$ tends to zero. So $x_n$ tends to zero and we have that : $ \quad \log(1+x) \underset{0}{\sim} x$ Then $nx_n \to 0$.


4

Note that $7^2 \equiv 49 \equiv -1 \bmod 10$, and so $7^4 \equiv 1 \bmod 10$. Hence if $a=4q+r$ then $$7^a \equiv 7^{4q+r} \equiv (7^4)^q \cdot 7^r \equiv 7^r \bmod 4$$ This reduces your problem to finding $7^7 \bmod 4$... but this should be easy since $7 \equiv -1 \bmod 4$.


3

Notice that the last digits of powers of $7$ run in the repeating sequence $7,9,3,1$. Thus, what you really need to know is what $7^7$ is congruent to modulo $4$, not modulo $10$, so as to tell where in the sequence $7^{7^7}$ falls. $7^k\bmod 4$ is easily computed from $k$; how?


3

You can express the solution using Lambert's W function, but in practice you'd find it numerically. Take logs on both sides to get $$ 10000 = x \log_{10}(x) $$ and use bisection or Newton-Raphson to approximate the solution.


3

As you already know, considering the function $$G_n(x)=\log\left(n+\frac{n-1}{x-1}\right),$$ defined on $(1,+\infty)$, $\hat\lambda_n$ solves the identity $$\hat\lambda_n M_n=G_n(\hat\lambda_n M_n).$$ As you noted, this identity has no analytical solution. However, the function $G_n$ decreases on $(1,+\infty)$ from $G_n(1)=+\infty$ to $G_n(+\infty)=\log(n)$ ...


3

HINT: Show that the derivative of $2^x-x$ is positive for $x>?$.


3

Let $$A=\pmatrix{0&9\cr-1&0\cr}\ .$$ You can easily check that $$A^2=-9I$$ and so the exponential series gives $$\eqalign{e^{tA} &=I+tA+\frac{1}{2!}t^2A^2+\frac{1}{3!}t^3A^3+\cdots\cr &=\Bigl(I-\frac{1}{2!}(9t^2I)+\frac{1}{4!}(9^2t^4I)+\cdots\Bigr)+\Bigl(tA-\frac{1}{3!}(9t^3A)+\frac{1}{5!}(9^2t^5A)+\cdots\Bigr)\cr ...


3

If you insist on wanting to show that your sequence is Cauchy without using convergence: Suppose $m \geq n \geq N$. In that case $$|e^{-m} - e^{-n}| = e^{-n} - e^{-m} = e^{-n} (1 - e^{n - m}) \leq e^{-n} \leq e^{-N}.$$ Now for $\varepsilon > 0$, you can pick $N$ such that $e^{-N} \leq \varepsilon$.


3

In Factorial Inequality problem it is shown that $$n!<\left(\frac n2\right)^n.$$ For $n=2000$ we have $$2000!<\left(\frac {2000}{2}\right)^{2000}=1000^{2000}.$$


3

Note that $x^2+y^2=r^2$. This is the equation of a circle of radius $r$. Saying that $(x,y)\to (0,0)$ here means that the radius of this circle is approaching zero. So we can transform this to a one variable limit: $$ \lim_{r\to 0} y=\lim_{r\to 0} e^{\frac{1}{r^2}}$$ Now note that $$\log y=\frac{1}{r^2}\rightarrow \infty,$$ as $ r\to 0$. So $y \to \infty$. ...


3

First, note that $\frac {d}{dx} 0^x = 0^x$, so you really want to know that there is a unique positive real number $a$ for which $\frac {d}{dx} a^x = a^x$. Suppose that $\frac{d}{dx}a^x=a^x$ and $\frac{d}{dx}b^x=b^x$, with $a$ and $b$ both positive. Then by the quotient rule, ...


2

Technically speaking, you can not evaluate a function on the "points" $+\infty$ and $-\infty$, nevertheless, for some functions, they are accumulation points of the domain, thus you can evaluate the limit. In your case you have: $$ \lim_{x\longrightarrow +\infty} e^{-x}=\lim_{x\longrightarrow +\infty} \frac{1}{e^{x}}=0 $$ This is a standard limit, but You ...


2

You should know two facts. $$\begin{split} a^x \cdot a^y &= a^{x+y}\\ (a\cdot b)^x &= a^x \cdot b^x \end{split}$$ Then you can easily obtain next two facts $$\begin{split} \left(a^b\right)^c &= \underbrace{a^b \cdot a^b \cdot ~\dots~ \cdot a^b}_{\text{c times}} = a^{b+b+\dots+b} &= a^{b\cdot c}\\ \frac{a^x}{a^y} &= a^x \cdot ...


2

An exact answer is only possible in terms of the Lambert $W$ function (Wikipedia link): $$\begin{align*} 0 &= 4e^{-2x}-3x\\\\ 3x &= 4e^{-2x}\\\\ 2xe^{2x}&=\tfrac{8}{3}\\\\ ye^y&=\tfrac{8}{3}\;\;(y=2x)\\\\ y&=W(\tfrac{8}{3})\\\\ x&=\tfrac{1}{2} W(\tfrac{8}{3}) \end{align*}$$



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