$$\underset{n\to \mathrm{\infty}}{lim}{(\frac{1}{3}+\frac{{n}^{1000}}{{2}^{n}})}^{n}$$calculate

djo57bgfrqn
2022-10-15
Answered

$$\underset{n\to \mathrm{\infty}}{lim}{(\frac{1}{3}+\frac{{n}^{1000}}{{2}^{n}})}^{n}$$calculate

You can still ask an expert for help

spornya1

Answered 2022-10-16
Author has **18** answers

If $f(n)=(\frac{1}{3}+\frac{{n}^{1000}}{{2}^{n}}{)}^{n}$,

$\begin{array}{}g(n)& =\mathrm{ln}f(n)\\ & =n\mathrm{ln}(\frac{1}{3}+\frac{{n}^{1000}}{{2}^{n}})\\ & =n\mathrm{ln}(\frac{1}{3}(1+\frac{3{n}^{1000}}{{2}^{n}}))\\ & =n(\mathrm{ln}(1/3)+n\mathrm{ln}(1+\frac{3{n}^{1000}}{{2}^{n}})\\ & =-n\mathrm{ln}(3)+n\mathrm{ln}(1+\frac{3{n}^{1000}}{{2}^{n}})\end{array}$

Since $\frac{3{n}^{1000}}{{2}^{n}}\to 0$ and $\mathrm{ln}(1+x)\sim x$ for small x, $n\mathrm{ln}(1+\frac{3{n}^{1000}}{{2}^{n}})\sim n\frac{3{n}^{1000}}{{2}^{n}}\to 0$ so $g(n)\sim -n\mathrm{ln}(3)=\mathrm{ln}({3}^{-n})$ and $f(n)\sim 1/{3}^{n}\to 0$.

Used ln, not exp.

$\begin{array}{}g(n)& =\mathrm{ln}f(n)\\ & =n\mathrm{ln}(\frac{1}{3}+\frac{{n}^{1000}}{{2}^{n}})\\ & =n\mathrm{ln}(\frac{1}{3}(1+\frac{3{n}^{1000}}{{2}^{n}}))\\ & =n(\mathrm{ln}(1/3)+n\mathrm{ln}(1+\frac{3{n}^{1000}}{{2}^{n}})\\ & =-n\mathrm{ln}(3)+n\mathrm{ln}(1+\frac{3{n}^{1000}}{{2}^{n}})\end{array}$

Since $\frac{3{n}^{1000}}{{2}^{n}}\to 0$ and $\mathrm{ln}(1+x)\sim x$ for small x, $n\mathrm{ln}(1+\frac{3{n}^{1000}}{{2}^{n}})\sim n\frac{3{n}^{1000}}{{2}^{n}}\to 0$ so $g(n)\sim -n\mathrm{ln}(3)=\mathrm{ln}({3}^{-n})$ and $f(n)\sim 1/{3}^{n}\to 0$.

Used ln, not exp.

asked 2022-11-24

the equation f(x)′−f(x)=0 holds for the exponential function on the complex plane.Now what i dont understand is this.

"let $f(x)={a}_{0}+{a}_{1}X+{a}_{2}{X}^{2}........$ f is a polynomial with infinite degree ".Why is that. I dont understand how he came to that conclusion?I mean Why define it that way?.MAybe he could solve the ODE on the real numbers and avoid this "out of nowhere" polynomial or is there a connection?

"let $f(x)={a}_{0}+{a}_{1}X+{a}_{2}{X}^{2}........$ f is a polynomial with infinite degree ".Why is that. I dont understand how he came to that conclusion?I mean Why define it that way?.MAybe he could solve the ODE on the real numbers and avoid this "out of nowhere" polynomial or is there a connection?

asked 2022-11-04

Given the vector space, $C(-\mathrm{\infty},\mathrm{\infty})$ as the set of all continuous functions that are always continuous, is the set of all exponential functions, $U=\{{a}^{x}\mid a\ge 1\}$, a subspace of the given vector space?**As far as I'm aware, proving a subspace of a given vector space only requires you to prove closure under addition and scalar multiplication, but I'm kind of at a loss as to how to do this with exponential functions (I'm sure it's way simpler than I'm making it).****My argument so far is that the set U is a subset of the set of all differentiable functions, which itself is a subset of $C(-\mathrm{\infty},\mathrm{\infty})$, but I doubt that argument would hold up on my test, given how we've tested for subspaces in class (with closure).**

asked 2022-10-20

Given the series definition of the exponential function, i.e. $\mathrm{exp}(x)={\displaystyle \sum _{k=0}^{\mathrm{\infty}}{\displaystyle \frac{{x}^{k}}{k!}}}$. Given that I have already proven that polynomials are continuous, does from this fact follow the continuity of the exponential function ?

asked 2022-11-11

the pythagoras tree is a fractal generated by squares. for each square, two new smaller squares are constructed and connected by their corners to the original square.

how do i create an exponential function to find out how many squares will be in the next sequence? since every iteration, the previous amount of squares added is doubled, how can i phrase this in function form? for example:

iteration 1: 0 + 1 = 1 square,

iteration 2: 1 + 1(2) = 3 squares,

iteration 3: 3 + 2(2) = 7 squares

how do i create an exponential function to find out how many squares will be in the next sequence? since every iteration, the previous amount of squares added is doubled, how can i phrase this in function form? for example:

iteration 1: 0 + 1 = 1 square,

iteration 2: 1 + 1(2) = 3 squares,

iteration 3: 3 + 2(2) = 7 squares

asked 2022-08-07

Simplify. Write the answer as a single term without negative exponents.

$\frac{7{n}^{-}7-7{m}^{-}7}{m+{n}^{8}}$

$\frac{7{n}^{-}7-7{m}^{-}7}{m+{n}^{8}}$

asked 2022-10-13

I'm trying to derive the following two inequalities for the exponential function, where $x,y,z\in \mathbb{R}$, t>0 and a>1

$$|y-z|{e}^{-\frac{(y-z{)}^{2}}{2at}}\le \phantom{\rule{thickmathspace}{0ex}}{C}_{a}{t}^{\frac{1}{2}}{e}^{-\frac{(y-z{)}^{2}}{4at}},\phantom{\rule{0ex}{0ex}}{t}^{-\frac{1}{a}}|y-z{|}^{1+\frac{2}{a}}{e}^{-\frac{(y-z{)}^{2}}{2at}}\le \phantom{\rule{thickmathspace}{0ex}}{C}_{a}{t}^{\frac{1}{2}}{e}^{-\frac{(y-z{)}^{2}}{4at}},$$

with some constant ${C}_{a}>0$

I tried to derive it by using derivatives but it didn't work for the purpose.

$$|y-z|{e}^{-\frac{(y-z{)}^{2}}{2at}}\le \phantom{\rule{thickmathspace}{0ex}}{C}_{a}{t}^{\frac{1}{2}}{e}^{-\frac{(y-z{)}^{2}}{4at}},\phantom{\rule{0ex}{0ex}}{t}^{-\frac{1}{a}}|y-z{|}^{1+\frac{2}{a}}{e}^{-\frac{(y-z{)}^{2}}{2at}}\le \phantom{\rule{thickmathspace}{0ex}}{C}_{a}{t}^{\frac{1}{2}}{e}^{-\frac{(y-z{)}^{2}}{4at}},$$

with some constant ${C}_{a}>0$

I tried to derive it by using derivatives but it didn't work for the purpose.

asked 2022-10-22

Is the exponential function integrable?

is there, in general, a difference between integrable and integrable in the Riemann sense?

is there, in general, a difference between integrable and integrable in the Riemann sense?