1
vote
0answers
53 views

How to prove that $b^{x+y} = b^x b^y$ using this approach?

Fix $b>1$. If $m$, $n$, $p$, $q$ are integers, $n > 0$, $q > 0$, $r = m/n = p/q$, then I can prove that $(b^m)^{1/n} = (b^p)^{1/q}$. Hence it makes sense to define $b^r = (b^m)^{1/n}$. I ...
3
votes
4answers
73 views

Existence of solution in $x,y \in (a,b)$ of $ \bigg(\dfrac { a+b}2\bigg)^{x+y}=a^xb^y$

Let $a<b$ be positive real numbers , then is it true that there exist $x,y \in (a,b)$ such that $ \bigg(\dfrac { a+b}2\bigg)^{x+y}=a^xb^y$ ?
6
votes
2answers
222 views

Evaluating a limit. What makes the equality right?

I'm reading a proof of a limit calculation. The limit is: $$\lim\limits_{x\to 0}\left(\frac{a^x+b^x}{2}\right)^\frac{1}{x}$$ where $a,b>0$. The aother claims that: $$\lim\limits_{x\to ...
0
votes
2answers
46 views

When $\ln(1+y) = y + o(y)$?

I was reading a proof which utilize the fact that: $\ln(1+y) = y + o(y)$ http://math.stackexchange.com/a/842557/160028 I'm not so sure what is the meaning of $\ln(1+y) = y + o(y)$. When is it ...
5
votes
3answers
125 views

Prove $\lim_{n\to\infty} \frac{(2^{2^n}+1)(2^{2^n}+3)(2^{2^n}+5)\cdots (2^{2^n+1}+1)}{(2^{2^n})(2^{2^n}+2)(2^{2^n}+4)\cdots (2^{2^n+1})}=\sqrt{2}$

Question: Prove or disprove $$I=\lim_{n\to\infty} \frac{(2^{2^n}+1)(2^{2^n}+3)(2^{2^n}+5)\cdots (2^{2^n+1}+1)}{(2^{2^n})(2^{2^n}+2)(2^{2^n}+4)\cdots (2^{2^n+1})}=\sqrt{2}$$ I know ...
4
votes
0answers
68 views

Definition of $0^z$, $z$ complex

Usually we define the function $a^b$ over the complexes using the exponential function, as $e^{b\log a}$. This function has some issues with multivalued-ness, but it still more or less satisfies ...
1
vote
1answer
32 views

Transforming a power tower to a product

It is possible to write the product of a sequence of terms $a_i$ as a function of the sum of a sequence of functions of these terms: $$\prod_i a_i=f\left(\sum_i g(a_i)\right)$$ where $f=\exp$ and ...
6
votes
2answers
159 views

Calculating limit in parts. Why possible?

Let $f$, continuous function, differentiable at $x=1$ and $f(1)>0$. Consider the following equation: $$\lim \limits_{x\to 1} ...
2
votes
4answers
140 views

Proving exponential is growing faster than polynomial

Let $P(x)$, a polynomial which isn't the zero-polynomial. I want to prove the following limits $$\lim \limits_{x\to\infty} \left|P(x)\right|e^{-x} = 0$$ $$\lim \limits_{x\to-\infty} ...
2
votes
2answers
95 views

Is $\lim_{x\to 0+} \frac{\ln(x)}{\ln(x)} = \frac{-\infty}{-\infty} = 1$

$$\lim_{x\to 0^+} \sin(x)^\frac{1}{\ln(x)} = ... = \exp \left(\lim_{x\to 0^+} \frac{\ln(\frac{\sin x}{x}) + \ln(x)}{\ln(x)}\right)$$ Now, from continuity we can evaluate each term separately. ...
1
vote
1answer
26 views

Could somebody validate my proof for the limit of $a^{x_n}$ when $x_n \to c$?

So, here is the clear formulation of the problem: let $(x_n) $ be a convergent sequence of positive numbers, with $x_n \to c$. I want to prove that the sequence $(y_n) $, with $y_n=a^{x_n} $, ...
3
votes
1answer
71 views

On the equation $\exp(a x+b)=\ln(x)$

I am confronted with: $$\exp(a x+b)=\ln(x)$$ for $a,b$ reals and $a<0$, $b>0$. I need the (unique) solution for $x$. My first target is (if it exists) an analytic solution in terms of ...
13
votes
5answers
437 views

Value of $(-1)^x$ for $x$ irrational

I was working on an analysis problem when this question arose in one my proofs. I think it may be either $-1$ or $1$, but it seems like there can only be an arbitrary way to assign this. So is there ...
0
votes
1answer
31 views

About definition of $a^b$ with $a \in \Bbb{R} \wedge b \in \Bbb{N}$

-- let $a,c \in \mathbb{R}$, and $b \in \Bbb{N}$, with $\Bbb{R}$ is a complete ordered field, $c \triangleq a^b$ if $c=\begin{cases} 1, & \mbox{if } a\neq 0 \wedge b=0\\ 0, & \mbox{if } a=0 ...
2
votes
0answers
45 views

Exponential equations solving methods?

Do you have an idea or general method to solve the following equation?: $$a^{\alpha x}+b^{\beta x} = c^{\gamma x}+ d^{\delta x}$$ when $a,b,c,d$ aren't zero, and $\alpha, \beta, \gamma, \delta$ are ...
0
votes
2answers
117 views

Prove $(a^b)^c = a^{bc}$

I need to prove the exponent identity $(a^b)^c = a^{bc}$, where $a,b,c \in \mathbb{Z}$. Apparently this proof is elementary/trivial, but I can't think of how to prove it. I need it as a lemma for ...
2
votes
2answers
98 views

Definitive answer to existence of real exponents

Well, I've been searching through this fórum and I know this question has been answered many times. But the answers I see, are kinda circular (I think). Let's start by the natural case. Natural case ...
5
votes
2answers
114 views

Extending exponentiation to reals

I've been reading through a course on exponential functions, starting from integer-valued exponents to rational ones as in: $x^r$ from $r\in \Bbb{N}$ to $\Bbb{Z}$, and combining them to rigorously ...
3
votes
3answers
167 views

Why do real powers need positive bases? (and to handle them)

I'm studying Calculus and I've stumbled across a concept I have some difficulties in fully grasping. That is "real powers". I don't understand the theory behind it and I think I don't understand very ...
2
votes
0answers
433 views

Exponent p-value generated in Excel

Excel gave me a p-value of 1.44909E-09 Notice is does not say .09 but 09 This is confusing me, I am trying to analyze my data but am stuck at this point. If it were E-9 it could be ...
4
votes
1answer
91 views

Why is real exponentiation continuous in the base?

I know that real exponentiation is continuous in the exponent ($f(x)=a^x$ is continuous), but how do we know real exponentation is continuous in the base? What I mean is, if $r$ is an arbitrary real ...
0
votes
1answer
137 views

For what $ \alpha \in \mathbb R$ is $ |x|^\alpha $ differentiable in $x=0$?

I came across the following question: For what $ \alpha \in \mathbb R$ is $ |x|^\alpha $ differentiable in $x=0$? What I have tried: Since for $ \alpha = 1 $ is clearly non-differentiable in ...
3
votes
1answer
213 views

Summation of powers inequality

Can anyone provide a slick proof of the following? Let $0 < x \le 1$. Then $\displaystyle \sum_{k=0}^{n-1} x^k \ge \frac {1} {1 - (1 - 1/n)x}$.
21
votes
1answer
2k views

Infinite tetration, convergence radius

I got this problem from my teacher as a optional challenge. I am open about this being a given problem, however it is not homework. The problem is stated as follows. Assume we have an infinite ...
12
votes
6answers
1k views

Proving the inequality $e^{-2x}\leq 1-x$

How do I prove the inequality $e^{-2x}\leq1-x$ for $0\leq x\leq1/2$?
5
votes
2answers
214 views

Slope definition of $e$

A common definition of $e$ is given as $$e = \lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^{n}$$ which can be proven to be equivalent to $$e=\lim_{h\rightarrow 0}\ ...
5
votes
1answer
1k views

Prove variant of triangle inequality containing p-th power for 0 < p < 1

Sorry if this is a trivial question, but I am kind of stuck with proving the following inequality and have been searching for a while: $\rho \left( \sum\limits_i^n d_i \right) \leq \sum\limits_i^n ...
4
votes
3answers
391 views

Power function inequality

Let $x$ and $p$ be real numbers with $x \ge 1$ and $p \ge 2$ . Show that $(x - 1)(x + 1)^{p - 1} \ge x^p - 1$ . I recently discovered this result. I am sure it is known, but it is new to me. It is ...
5
votes
3answers
392 views

Proof of exponential from homomorphism property

I am going back through a bunch of calculus I learned in high school and proving the stuff that they just told us was true. Along the way, I found I had to prove that if $f(x+y)=f(x)f(y)$ then $f$ is ...
1
vote
1answer
84 views

Simple estimation $e^{a\sqrt{r}} > r$

I want to prove a simple theorem about contour integration via residues and I need the following estimation: $e^{a\sqrt{r}} > r$ for any real a > 0 and r >> 0. Is this true? If so, what is an ...
110
votes
15answers
8k views

Zero to the zero power - Is $0^0=1$?

Could someone provide me with good explanation of why $0^0 = 1$? My train of thought: $x > 0$ $0^x = 0^{x-0} = 0^x/0^0$, so $0^0 = 0^x/0^x = ?$ Possible answers: $0^0 * 0^x = 1 * 0^x$, so ...