calculus - Limit contradiction in L'Hopitals Rule and Special Trig limits

While in Calculus the other day I stumbled upon a contradiction in L'Hopitals Rule vs. Special Trig Limits.

The problem looks like this

$$
\lim_{x\to 0} \frac{\tan(x)−x}{x^3}
$$

Using L'Hopitals rule (because the limit = 0/0)
$$
\lim_{x\to 0} \frac{\tan(x)−x}{x^3} = \lim_{x\to 0} \frac{\frac{d}{dx}(\tan(x)−x)}{\frac{d}{dx}(x^3)} = \lim_{x\to 0} \frac{\sec(x)^2−1}{3x^2}
$$
Again using L'Hopitals rule (because the limit = 0/0)
$$
\lim_{x\to 0} \frac{\sec(x)^2−1}{3x^2} = \lim_{x\to 0} \frac{\frac{d}{dx}\sec(x)^2−1}{\frac{d}{dx}3x^2} = \lim_{x\to 0} \frac{2\sec(x)^2\tan(x)}{6x} = \lim_{x\to 0} \frac{\sec(x)^2\tan(x)}{3x}
$$
Again using L'Hopitals rule (because the limit = 0/0)

$$
\lim_{x\to 0} \frac{\sec(x)^2\tan(x)}{3x} = \lim_{x\to 0} \frac{\frac{d}{dx}\sec(x)^2\tan(x)}{\frac{d}{dx}3x} = \lim_{x\to 0} \frac{4\sec(x)^2\tan(x)^2 + 2\sec(x)^4}{6} = \frac{1}{3}
$$
Now, using special trig limits...
$$
\lim_{x\to 0} \frac{\tan(x)−x}{x^3} = \lim_{x\to 0} \frac{\tan(x)}{x^3}-\frac{x}{x^3} = \lim_{x\to 0} \frac{\tan(x)}{x}*\frac{1}{x^2}-\frac{1}{x^2}
$$
With knowledge of $\tan(x)$ trig limit
$$
\lim_{x\to 0} \frac{\tan(x)}{x} = 1

$$
You can now simplify.
$$
(\lim_{x\to 0} \frac{\tan(x)}{x}*\frac{1}{x^2}-\frac{1}{x^2}) = (\lim_{x\to 0} \frac{\tan(x)}{x}*\lim_{x\to 0}\frac{1}{x^2}-\lim_{x\to 0}\frac{1}{x^2}) = (1 * \lim_{x\to 0}\frac{1}{x^2}-\lim_{x\to 0}\frac{1}{x^2})
$$
Now the end result
$$
\lim_{x\to 0}\frac{1}{x^2}-\lim_{x\to 0}\frac{1}{x^2} = \lim_{x\to 0}\frac{1}{x^2}-\frac{1}{x^2} = 0
$$
1/3 does not = 0. Therefore there is a discrepancy through special trig limits and L'Hopitals rule. There is the same anomaly with

$$
\lim_{x\to0}\frac{\sin(x)-x}{x^3} = (\text{through L'Hopitals}) -\frac{1}{6} \text{ or (through special trig limits) } 0
$$
On a calculator in graph or table mode as you approach 0 it seems to be 1/3.
But as it turns out, if you get very precise, the limit actually seems to be approaching 0.

This happens the same with the graph, it seems to be parabolically approaching 1/3 but if you zoom in to an extreme you see it actually approaches zero.

This original work is my own as published on Sunday, October 9, 2016 at 9:51pm. I claim all knowledge credits and fallacies that may come with this discrepancy. But overall please, prove or disprove this, or at least explain why this occurs. Thank you for your time.

(I will try to get a picture of the graph and table)

sequences and series - Is $sum_{i=0}^infty (i) + c = sum_{k=0}^infty (2k) + c + sum_{j=0}^infty (2j+1) + c$?

Basically, the question is, "is the sum of all positive numbers equal to the sum of all positive even numbers and odd numbers?" (which is obviously yes) but with a twist: for every number, there is a constant $c$ which is also an integer.

$$\sum_{i=0}^\infty (i) + c = \sum_{k=0}^\infty (2k) + c + \sum_{j=0}^\infty (2j+1) + c$$

It really feels like this equation is true, as there should be an equal amount of $c$'s in both sides but I am not a mathematician at all, just equipped with high school maths, I wanted a proper explanation for this from you guys! Couldn't find this question when googling, sorry in advance if there is one.

Answer

Note that the sum: $$1+2+3+… $$ diverges, so they sum to $\infty$ on both sides.

algebra precalculus - Show (via Complex Numbers): $frac{cosalphacosbeta}{cos^2theta}+frac{sinalphasinbeta}{sin^2theta}+1=0$ under given conditions

$\alpha$ and $\beta$ do not differ by an even multiple of $\pi$. If $\theta$ satisfies $$\frac{\cos\alpha}{\cos\theta}+ \frac{\sin\alpha}{\sin\theta}=\frac{\cos\beta}{\cos\theta}+\frac{\sin\beta}{\sin\theta}=1$$ then show that $$\frac{\cos\alpha\cos\beta}{\cos^2\theta}+\frac{\sin\alpha\sin\beta}{\sin^2\theta}+1=0$$
I wish to solve this problem using some elegant method, preferably complex numbers.

I've tried using the fact that $\alpha$ and $\beta$ satisfy an equation of the form $\cos x/\cos\theta + \sin x/\sin\theta = 1$, and got the required result. See my solution here: https://www.pdf-archive.com/2017/07/01/solution

I'm guessing there's an easier way to go about it. Thanks in advance!

Answer

Well, I'm not sure I can do that in a very elegant way, but it might be shorter. I'm using addition theorems, and the identity $$\sin x -\sin y=2\sin\frac{x-y}{2}\,\cos\frac{x+y}{2}$$ following immediately from them.
Multiplying the given equations by $\sin\theta\,\cos\theta,$ we get
$$\sin(\alpha+\theta)=\sin\theta\,\cos\theta=\sin(\beta+\theta),$$

but $$0=\sin(\alpha+\theta)-\sin(\beta+\theta)=2\sin\frac{\alpha-\beta}{2}\,\cos\left(\frac{\alpha+\beta}{2}+\theta\right).$$ The first factor is $\neq0$ by assumption, so $$\cos\left(\frac{\alpha+\beta}{2}+\theta\right)=0.$$ Multiplying by $2\sin\left(\frac{\alpha+\beta}{2}-\theta\right)$ and using the above identity, you get $\sin(\alpha+\beta)-\sin2\theta=0$, and this means (using $\sin2\theta=2\sin\theta\,\cos\theta$ and dividing by $\sin\theta\,\cos\theta$)
$$\frac{\sin\alpha\,\cos\beta}{\sin\theta\,\cos\theta}+\frac{\sin\beta\,\cos\alpha}{\sin\theta\,\cos\theta}=2.$$ Now you have
$$\left(\frac{\cos\alpha}{\cos\theta}+ \frac{\sin\alpha}{\sin\theta}\right)\,\left(\frac{\cos\beta}{\cos\theta}+\frac{\sin\beta}{\sin\theta}\right)-\left(\frac{\sin\alpha\,\cos\beta}{\sin\theta\,\cos\theta}+\frac{\sin\beta\,\cos\alpha}{\sin\theta\,\cos\theta}\right)=1\cdot1-2=-1,$$ and that gives your required result after simplifying.

trigonometry - Prove this trigonometric identity in quadrilateral

If $\alpha,\beta,\gamma,\delta$ are angles in quadrilateral different from $90^\circ$, prove the following:

$$ \frac{\tan\alpha+\tan\beta+\tan\gamma+\tan\delta}{\tan\alpha\tan\beta\tan\gamma\tan\delta}=\cot\alpha+\cot\beta+\cot\gamma+\cot\delta $$

I tried different transformations with using $\alpha+\beta+\gamma+\delta=2\pi$ in equation above, but no success. Am I missing some not-so-well-known formula?

Answer

It follows directly from $\tan(\alpha + \beta + \gamma + \delta) = 0$ and the sum angle formula for $\tan$ (see here: Tangent sum using symmetric polynomials)

Using that formula we get (from numerator = 0) that

$$ \tan \alpha + \tan \beta + \tan \gamma + \tan \delta = $$

$$\tan \alpha\tan \beta\tan \gamma+ \tan \alpha\tan \beta\tan \delta + \tan \alpha\tan \gamma\tan \delta + \tan \beta\tan \gamma\tan \delta$$

divididing by $ \tan \alpha\tan \beta\tan \gamma\tan \delta$ gives the result.

calculus - Evaluating a parametric integral

I need some help to evaluate the following integral.

$$\int_{0}^{\infty}{\mathrm{d}x \over x^{\alpha}\left(x + 1\right)}$$ where $\alpha \in \left(0,1\right)$

I've tried many ways (the best one seems to be developing by Taylor series) but actually I have no solution.

Some ideas?
Thank you.

Answer

I think we will need some complex analysis here.

Take a branch of $1/z^a$ defined in $\mathbb{C}\setminus [0,+\infty)$ and consider the integral
$$\int_{\gamma}\frac{dz}{z^a(z+1)}$$
where $\gamma$ is a close path composed by an arc of a inner circle of radius $01$ and two parallel segments over and under the segment $[r,R]$.
Then by the residue theorem

$$\int_{\gamma}\frac{dz}{z^a(z+1)}=2\pi i\mbox{Res}\left(\frac{1}{z^a(z+1) },-1\right)=2\pi i e^{-i\pi a}.$$
Now we take the limit as $R\to+\infty$ and $r\to 0^+$.
It is easy to see that the integrals along the arcs of the circles goes to $0$. Hence
$$\int_0^{+\infty}\frac{dx}{x^a(x+1)}-\int_0^{+\infty}\frac{dx}{x^ae^{2\pi ia}(x+1)}=2\pi i e^{-i\pi a}$$
which implies that
$$\int_0^{+\infty}\frac{dx}{x^a(x+1)}=\frac{2\pi i e^{-i\pi a}}{1-e^{-2i\pi a}}=\frac{\pi}{\sin (\pi a)}.$$

set theory - Help with intuition on Cardinal Arithmetic Problems

It happens a lot to me that when I find an intuitive model (picture) of a mathematical entity, the proofs left as exercises in books are very easy to solve. For example when dealing with filters and ultrafilters on sets (specially $\omega$) I just need to imagine the Hasse diagram of the Poset $\langle\mathcal{P}(\omega),\subseteq\rangle$ and most proofs and definitions come naturally.

I have been trying to find a model/picture for arithmetic with cardinals that helps me solve the problems that come in some books but I don't know if cardinal numbers are too big and I cannot picture them correctly or I just have to use different models/pictures for different problems. So far the two that have been working best are parallel lines (when comparing cardinals) and my intuition with injective and surjective functions. But those approaches only work quickly with easy problems (adding or multiplying finitely many cardinals or relating the cardinality of two specific sets). However, this techniques become lengthy when new concepts are introduced (like cofinality and exponentiation of cardinals). Moreover, I have not been able to solve many problems with just these two methods. I'll quote two of those problems and try to make this ideas more clear:

1. (This one is in Andras Hajnal & Peter Hamburguer's Set Theory Book) If $\kappa$ is an infinite cardinal number and $\kappa=\sum_{\lambda
   

For this one I just changed $\kappa^{cf(\kappa)}$ for $\prod_{\lambda

2. (A friend gave me this one but I have a feeling there is a typo somewhere) Suppose that $\alpha$ is a limit ordinal and that $\langle\kappa_\xi\rangle_{\xi<\alpha}$ is a strictly increasing sequence of cardinals such that $\kappa=\sum_{\xi<\alpha}\kappa_\xi$ and if $0<\lambda
   
   

Although not stated in the problem, every cardinal $\kappa_\xi$ must be strictly less than $\kappa$ (otherwise the sum is greater than $\kappa$). And given that for every infinite cardinal $cf(\kappa)=\min\{\lambda\in CN_\infty\mid\forall\xi\in\lambda(\kappa_\xi<\kappa)\wedge\kappa=\sum_{\xi<\lambda}\kappa_\xi\}$ we must have that $cf(\kappa)\leq\alpha$. Under the same principle one just have to prove that $cf(\kappa^\lambda)\leq\alpha$. There is also a trivial inequality in this problem: $\kappa^\lambda=(\sum_{\xi<\alpha}\kappa_\xi)^\lambda\geq\sum_{\xi

So the questions are: Do you use a single model/picture to help you solve this kind of problems about infinite sums, products and exponentiation of cardinals? Which one? and could you provide a hint on how to use such a model/picture to solve problems 1. and 2.?

Answer

I just realized that you are asking for other methods than constructing explicit functions, however, I figure direct attacks might be the most intuitive for problems like this.

In order to show that $\kappa^{cf(\kappa)}\leq\prod_{\lambda

Given $f\in \kappa^{cf(\kappa)}$, recursively build a sequence $g$ such that $g(\lambda)=\begin{cases}
f(i)+1 && \text{if $i$ is the least such that f(i) is in }[0,\kappa_\lambda) \text{ and i is not used previously} \\
0 && otherwise
\end{cases} $.

Note we eventually exhaust all f-values as $cf(\kappa)$ is regular.

If $f\neq h$, then let $i\in cf(\kappa)$ be the least point where they differ. There are two cases to consider one is there exists a least undefined $\lambda$ such that $f(i), h(i)\in [0,\kappa_\lambda)$ and another case being there exists a least undefined $\lambda$ such that $\kappa_\lambda$ separates $f(i)$ and $h(i)$. In either case, the sequences constructed from two functions are different. So you have an injection.

radicals - Proving that for each prime number $p$, the number $sqrt{p}$ is irrational

I'm a total beginner and any help with this proof would be much appreciated. Not even sure where to begin.

Prove that for each prime number $p$, the square root of p is irrational.