# Tag Info

1

It can be solved using more elementary tools. It $\mathrm{e}^{f(z)}$ is a non-constant polynomial, then it has a root (due to the Fundamental tTheorem of Algebra), which contradicts the fact that the exponential does not vanish.

5

Hint: Consider the type of the singularity of $e^{f(z)}$ in $\infty$. Different hint: Does $e^{f(z)}$ have any zeros? The first hint generalises to show that $e^{h(z)}$ for analytic $h$ never can have a pole in an isolated singularity of $h$.

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1

$e^0=1$, and this is the slope of $e^x$ at $x=0$. So...

3

Hint. Take the logarithm of your term and since $n$ is large, use a first order Taylor expansion. Multiplied by the exponent, you will find $0$ as the limit of the logarithm, then $1$ for the limit.

4

Hint: rewrite your limit as $$\lim_{n\to\infty}\left[\left(1-\frac{\sqrt {2t}}{n}\right)^{-n/2}\left(1+\frac{\sqrt {2t}}{n}\right)^{-n/2}\right]$$ and use the product rule.

4

Hints: $$(1)\; \left(1-\frac{2t}{n^2}\right)^{-n/2}=\left(\left[\left(1-\frac{2t}{n^2}\right)^{n^2}\right]^{-1/2}\right)^{1/n}$$ $$(2)\;\text{ For any function}\;\;f(n)\;\;s.t.\;\;\lim_{n\to\infty}f(n)=\infty\;,\;\;\lim_{n\to\infty}\left(1+\frac x{f(n)}\right)^{f(n)}=e^x$$ $$(3)\;\lim_{n\to\infty}a_n=a>0\implies\lim_{n\to\infty}\sqrt[n]{a_n}=1$$

0

Another interesting special case is if $b=\frac{1}{a}\ne 1$, i.e., equation changes to $a^x+a^{-x}=x$. We can then find that if $a=a^\star=e^{\frac{1}{2\sinh q}}=1.392877\dots$, or $a=a^{\star\star}=e^{-\frac{1}{2\sinh q}}=0.7179\dots$ where $q$ is solution to $\coth q = q$ (this can be found in literature to be $q=1.19967874\dots$) then there is exactly ...

1

Hint: The complex exponential function has period $2 \pi i$, and therefore $\exp(2\pi i k) = \exp(0) = 1\;$ for integer $k$.

3

Here you go limit proof of e.

-1

If you do not want to involve "e", just compute the value of your term for successive value of n. Starting from 1 to 10 (step of 1), you will successively have : 2.00000, 2.25000, 2.37037, 2.44141, 2.48832, 2.52163, 2.54650, 2.56578, 2.58117, 2.59374. Continue with steps of 10; you will then have 2.65330, 2.67432, 2.68506, 2.69159, 2.69597, 2.69912, 2.70148, ...

0

Your result is right, but the way you explained it is muddled. Because the series expansion$$\exp z=\sum_{n=0}^{\infty}\dfrac{1}{n!} z^n$$holds over the entire complex plane, it holds also under the replacement $z\leftarrow z-\mathrm i\pi.$ Hence$$-\exp z=\exp z\,\exp(-\mathrm i\pi)=\exp(z-\mathrm i\pi)=\sum_{n=0}^{\infty}\dfrac{1}{n!} (z-\mathrm ... 0 Let's back up a bit. Recall how compound interest works when there are n compounding periods per year:$$ A=P\left(1+{r\over n}\right)^{nt}. Now thinking of there being more and more and more compounding periods in a year (every month, every day, every hour, every minute, every second, etc.) so that we are taking the limit as n\to\infty in the above ... 0 As you have said, just put the points into the equation: \begin{align} 5 & = a^2 + b \\ -4 & = a^0 + b \end{align} The second equation directly gives you the first value you need, inserted into the first equation you can easily calculate a. 1 Find maximum of the function \frac{ln x}{x} 1 Consider the function f(x)=x/\ln(x). For x \geq e, this is an increasing function on the domain by taking its derivative. Since \pi > e, f(\pi) > f(e) and this gives the inequality. 2 Here is one approach. Half-life T = 4.51 \times 10^9 years Ratio \dfrac{\mbox{U-238}}{\mbox{Lead}} = 0.9 = \dfrac{9}{10} This means \dfrac{\mbox{U-238 atoms}}{\mbox{Original U-238 atoms}}= \dfrac{\dfrac{9}{10}}{0.9 + 1} = \dfrac{9}{19} = \dfrac{N}{N_0} \dfrac{N}{N_0} = \left(\dfrac{1}{2}\right)^{\dfrac{t}{T}} t = \dfrac{T \ln ... 1 Your intuition is correct, the ratio of mean to median of a random variable X with density of shape abe^{-ax}+cde^{-cx} is not always the same as the ratio of mean to median of an exponentially distributed random variable. (The latter ratio, as your post pointed out, is \frac{1}{\ln 2}.) To show this, it is enough to give an example. Let X have ... 5 Suppose w is not in the image of f. Since f(z) - w is entire and never zero, it can be written as f(z) - w = \exp(g(z)) for some entire function g. Notice that f(z) - w = -w only when z = 0. Apply the little Picard theorem to to g(z) to get a contradiction. 1 n^{kn}=2^{k\,n\log_2(n)}. Does that help? EDIT: Added later. This let's us more clearly see thatn^{kn}=2^{k\,n\log_2(n)}\ll2^{2^n}$$We can come upon any function between k\,n\log_2(n) and 2^n to use in that exponent to find an intermediate function. Like say, n^2. Is 2^{n^2} "interesting"? Added even later: Imagine an n\times n array ... 3 As an illustration example of this problem, I propose here the basis of a possible algorithm; this will use a second order Newton iteration method and, as usual, the key point will be to find a reasonable starting point. Assuming that Lambert function is available, let me consider two functions$$f(x) =a^x+b^x-x$$and$$g(x)=2c^x-x$$Expanding both ... 1 You can use your idea f(i)=f(i-1)*k just fine. If you want n values, (hence n-1 factors of k) you need k=\left(\frac {\max}{\min}\right)^{\frac 1{n-1}} Now f(1)=\min, f(n)=\max, f(i)=\min\cdot k^{i-1} 6 For a \ge e^{1/e} = 1.444667861\ldots, we have a^x \ge x for all x. So we require at least a < e^{1/e} and b < e^{1/e}. And for a > e^{1/2e} = 1.201943368\ldots, we have a^x > x/2 for all x. So we require at least a \le e^{1/2e} or b \le e^{1/2e}. You can check these figures by differentiating f(x) = a^x - x (resp. f(x) = ... 2 Notice if a=b=1, then you get trivial solution x = 2. I claim that if a,b are both \geq2, then there is NO solution!. To see this, Suppose there is a solution x such that$$ a^x + b^x = x $$. Notice since a and b are greater than 2, then the following is trivially true a^x > x  and b^x > x . This implies that a^x + b^x > 2x ... 2 There is no closed-form solution in terms of standard functions, except in special cases. If a and b are given, you might try numerical methods, such as Newton's. 1 Can you solve the differential equation? You should have a solution with two constants-the initial amount and a. The two pieces of data give you two equations in two unknowns to find these constants. Added: your solution needs a constant of integration. The solution should be x=c\cdot e^{-at}. Now plug in the data you are given:$$1000=c\cdot ...

0

Ok, I made a mistake above, and put the chain rule result in the exponent! Wrong! Here is the correct way, using FTC: $$\huge{\frac{d}{dx}\int_0^{9\ln x}e^t \,dt=(e^{9lnx})\frac{9}{x}=(e^{lnx})^9(\frac{9}{x})=\frac{x^9}{1}\frac{9}{x}=9x^8}$$

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No special estimation is needed. It is correct to conclude that $f_n(a_n)$ converges to $f(a)$ in this case, because for large enough $n$ (in fact for $n > |a|$), all the $f_n$ are monotonic on an interval containing $a$. The principle is if $f_n(x) \to f(x)$ for all $x$, the limit $f(x)$ is continuous at $a$, and there is a $\delta > 0$ such ...

0

HINT: $$\left(1 + \frac{1}{\frac {n}{x}} \right)^{\frac{n}{x}} = e\text{, as} \frac{n}{x} \to \infty$$ Raising both sides to $x$ will give you what you want. Then just apply Bernoulli's Inequality as you've mentioned.

1

If $e=\lim_{n\to\infty} \left(1+\frac1n\right)^n$, then $$e^x=\lim_{n\to\infty} \left(1+\frac1n\right)^{xn}$$ Let $n=\frac ux$ (which we can do so long as $x$ is constant): \begin{align} e^x&=\lim_{u\to\infty} \left(1+\frac xu\right)^{u}\\ &=\lim_{u\to\infty}\sum_{k=0}^u{u\choose k}\left(\frac xu\right)^k\\ ...

0

At $x=0$ we have $e^x=x+1$ and the derivative of $e^x$ is $e^x$ while the derivative of $x+1$ is just 1. For $x>0$ we have $e^x>1$ since $\log(e^x)=x>\log(1)=0$. So $e^x$ grows faster than $x+1$ as $x$ increases. For $x<0$ we have $e^x<1$ for the same reasons so $e^x$ goes down slower than $x+1$ as $x$ decreases.

3

Proof of the inequality using the convexity Let $f(x)=e^x$ then we have $f''(x)=e^x\ge0$ so $f$ is a convex function and the equation of its tangent line at the point $x=0$ is $y=x+1$. What's the position of this tangent relative to the curve of $f$?

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