calculus - Evaluating a limit with Taylor Series

I would like to find the following limit using Taylor Series:

$$\lim_{x\to0}\frac{6\sinh x-6x-x^3}{x^4(6+x^2)\sinh x}.$$

Now my question is the following: How do I know exactly how to approximate the numerator and the denominator? We're clearly going to need at least the first three terms of $\sinh x = x + \frac{x^3}{6} + \frac{x^5}{120}+...$ in the numerator to get something nonzero.

Could you please show me what a rigorous application of Taylors theorem would look like in this case?

(This is not homework - I'm preparing for an exam)

Thank you.

Answer

You ask for rigorous application? What about this:

$$\lim_{x\to0}\frac{6\sinh x-6x-x^3}{x^4(6+x^2)\sinh x}=\lim_{x\to0}\frac{6\left(x + \frac{x^3}{6} + \frac{x^5}{120}+o(x^5)\right)-6x-x^3}{x^4(6+x^2)( x+o(x))}\\=\lim_{x\to0}\frac{\frac{x^5}{20}+o(x^5)}{6x^5+o(x^5)}=\frac1{120}\quad?$$

limits - Is the function $f(n,m)=n^2+m^2$ in $O(nm)$

I am trying to solve a problem related to big o with two variables. I'll write the exercise below.

Prove or disprove the statement: $n^2+m^2=O(nm)$

Just in case, I'll write one definition needed:

Let $f,g:\mathbb N^k \to \mathbb N$, we say $f \in O(g)$ if and only if there are some numbers $c \in \mathbb N$ and $\overrightarrow{n_0} \in \mathbb N^k$ such that for all $\overrightarrow{n} >\overrightarrow{n_0}$,$$f(\overrightarrow{n}) \leq cg(\overrightarrow{n})$$

where $\overrightarrow{n_1} > \overrightarrow{n_2}$ if and only if $n_{1_i} > n_{2_i}$ for all $i=1,...,k$.

I think that this statement is false but I couldn't write a formal proof of this. I would really appreciate any suggestions. Thanks in advance.

calculus - Evaluating $int {sin theta + cos theta over sqrt{sin 2theta} }dtheta$

$$\int {\sin \theta + \cos \theta \over \sqrt{\sin 2\theta} }d\theta$$

---

After simplifying I get,

$${1\over \sqrt{2}}\int {\tan \theta + 1 \over \sqrt{\tan \theta}} d\theta$$

Substituting $u = \tan \theta$

$${1\over \sqrt{2}}\int {u + 1\over \sqrt{u}(1 + u^2)} du$$

Substituting $t^2 = u$

$${\sqrt{2}}\int {t^2 + 1\over (1 + t^4)} dt = \sqrt{2} \int {1 + 1/t^2 \over(t - 1/t)^2 + 2 }dt$$

Substituting $z = t - 1/t$

$$\sqrt{2} \int {1\over z^2 + 2 }dz = \arctan\left(\tan \theta - 1\over \sqrt{2\tan \theta} \right) + C = {\arcsin(2\sqrt{\sin2\theta} (\sin\theta - \cos \theta))\over 2} + C$$

Which is far from the right answer of $\arcsin(\sin \theta - \cos \theta) + C$.

Where did I go wrong ?

Edit :

I would like to know where I am wrong rather than knowing how to solve the problem.

Answer

Your factor before $\arctan$ should be $1$ rather than $1/2$; except for that, your antiderivative is correct. The error is in the conversion to $\arcsin$. If you use $\arctan \alpha = \arcsin(\alpha/\sqrt{\alpha^2+1})$, you will get the right answer.

Update: Now that the factor 1/2 has been edited away, things are fine up to that point. And actually the last step is almost OK too! (I dismissed it a bit too quickly before.)

The question is only meaningful on an interval where $\sin 2\theta \ge 0$, say $0 \le \theta \le \pi/2$, and the answer $A=\arcsin(\sin\theta-\cos\theta)$ is correct on that whole interval, as is easily verified by differentiation (and note also that $|\sin\theta-\cos\theta| \le 1$ on that interval).

Now it seems that your expression $B$ actually agrees with the expression $A$ on the subinterval $\pi/12 \le \theta \le 5\pi/12$, whereas $B=\text{const.}-A$ on the intervals $0\le \theta \le \pi/12$ and $\le 5\pi/12 \le \theta \le \pi/2$. I haven't really bothered to analyze why this happens, I was just lazy and looked at plots of $A \pm B$. But surely (as Barry Cipra pointed out) it has something to do with the properties of the inverse trig functions; for example $\arcsin(\sin x) = \pi-x$ if $\pi/2 \le x \le \pi$.

asymptotics - Help proving $sum_{nle x}{ln{n}}=xln{x}-x+O(ln{x})$

Just learning a bit about big O notation and have come across this exercise.

The notation used is $$\sum_{n\le x}{\ln{n}}=x\ln{x}-x+O(\ln{x})$$
and I am assuming that is equivalent to

$$\sum_{n=1}^{x}{\ln{n}}=x\ln{x}-x+O(\ln{x}).$$

I know that
$$\int_1^x{\ln{t}}\operatorname{dt}=x\ln{x}-x+1$$
so I feel like I am almost there, just not entirely sure how to make the connection, in particular exactly how the big O comes into this.

Edit: I should make it clear, I can see (I have done a quick sketch) that the summation is an 'overshoot' of the integral, and that the difference between the summation and the integral is bounded by something, but I'm not sure how to get that it is precisely $\ln{x}$.

Also I have no idea what tags to use, feel free to change them.

elementary number theory - Find the last digit of $3^{1006}$

The way I usually do is to observe the last digit of $3^1$, $3^2$,... and find the loop. Then we divide $1006$ by the loop and see what's the remainder. Is it the best way to solve this question? What if the base number is large? Like $33^{1006}$? Though we can break $33$ into $3 \times 11$, the exponent of $11$ is still hard to calculate.

Answer

You have $$3^2=9\equiv -1\pmod{10}.$$

And $1006=503\times 2$, so

$$3^{1006}=(3^2)^{503}\equiv (-1)^{503}\equiv -1\equiv 9\pmod{10}.$$

So the last digit is $9$.

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And for something like $11$, you can use the fact that $11\equiv 1\pmod {10}.$

sequences and series - How can one simplify $frac 36 + frac {3cdot 5}{6cdot9} + frac{3cdot5cdot7}{6cdot9cdot12}$ up to $infty$?

How can one simplify $\cfrac 36 + \cfrac {3\cdot 5}{6\cdot9} + \cfrac{3\cdot5\cdot7}{6\cdot9\cdot12}$ up to $\infty$?

I am new to the topic so can the community please guide me on the approach one needs to take while attempting such questions?

Things I am aware of:

1. Permutations and combinations




2. Factorials and some basic properties that revolve around it.




3. Some basic results of AP, GP, HP





4. Basics of summation.

Thanks for reading.

Answer

Let $$S=\cfrac 36 + \cfrac {3\cdot 5}{6\cdot9} + \cfrac{3\cdot5\cdot7}{6\cdot9\cdot12}\cdots \cdots \infty$$

Then $$\frac{S}{3} =\frac{1\cdot 3}{3\cdot 6}+\frac{1\cdot 3\cdot 5}{3\cdot 6 \cdot 9}+\frac{1\cdot 3\cdot 5\cdot 7}{3\cdot 6 \cdot 9\cdot 12}+\cdots\cdots$$

So $$1+\frac{1}{3}+\frac{S}{3} =1+\frac{1}{3}+\frac{1\cdot 3}{3\cdot 6}+\frac{1\cdot 3\cdot 5}{3\cdot 6 \cdot 9}+\frac{1\cdot 3\cdot 5\cdot 7}{3\cdot 6 \cdot 9\cdot 12}+\cdots\cdots$$

Now campare the right side series

Using Binomial expansion of $$(1-x)^{-n} = 1+nx+\frac{n(n+1)}{2}x^2+\frac{n(n+1)(n+2)}{6}x^3+.......$$

So we get $$nx=\frac{1}{3}$$ and $$\frac{nx(nx+x)}{2}=\frac{1}{3}\cdot \frac{3}{6}$$

We get $$\frac{1}{3}\left(\frac{1}{3}+x\right)=\frac{1}{3}\Rightarrow x=\frac{2}{3}$$

So we get $$n=\frac{1}{2}$$

So our series sum is $$(1-x)^{-n} = \left(1-\frac{2}{3}\right)^{-\frac{1}{2}} = \sqrt{3}$$

So $$\frac{4}{3}+\frac{S}{3}=\sqrt{3}$$

So $$S=3\sqrt{3}-4.$$

elementary number theory - How $gcd (s,t)=1$ implies $gcd(s-t,s+t) =1$?

Let $s$ and $t$ be two relatively prime odd numbers. We want to know the $d=\gcd(s-t,s+t).$ Let $d=2^lm$ thus $s-t=2^lmk_1$ and $s+t=2^lmk_2$ for $k_1$ and $k_2$ odd numbers and $\gcd(k_1, k_2)=1.$ Summing and subtracting both sides of two equality-es and diving by $2$ results in $s=2^lm \frac{(k_1+k_2)}{2}$ and $t=2^lm \frac{(k_2-k_1)}{2}$ in which $\frac{(k_2 \pm k_1)}{2}$ are integers as both $k_1$ and $k_2$ are odd numbers and because $s$ and $t$ are relatively prime implies $l=0$ which is not possible since both $s-t$ and $s+t$ are even so must $l \ge 1$ !! Contradiction?

A rewritten version of this same question, for the author's benefit:

Let $s$ and $t$ be two distinct relatively prime odd numbers.

We want to compute $d=\gcd(s-t,s+t).$ Note that $s-t$ and $s+t$ are both nonzero because $s$ and $t$ are distinct.

Let $d=2^\ell m$, where $m$ is odd. Then since $s-t$ and $s+t$ are both multiples of $d$, we have
\begin{align}
s-t&=2^\ell mk_1 \text{and}\\
s+t&=2^\ell mk_2
\end{align}
where $k_1$ and $k_2$ are both odd. Furthermore, $\gcd(k_1, k_2)=1,$ for if it were some other number $u > 1$, then $ud$ would divide both $s+t$ and $s-t$, contradicting the fact that $d$ is their greatest common divisor.
Finally, since both $s-t$ and $s+t$ are even and nonzero, we must have $\ell \ge 1$, which we'll use shortly.

Summing and subtracting both sides of these two equalities and dividing by $2$ gives
\begin{align}

s&=2^\ell m \frac{(k_1+k_2)}{2} \text{ and}\\
t& =2^\ell m \frac{(k_2-k_1)}{2}
\end{align}
Note that because $k_1$ and $k_2$ are both odd, their sum and difference are both even, hence
$\frac{(k_2 \pm k_1)}{2}$ are both integers.

Now $s$ and $t$ are relatively prime (given), so $\ell$ must be zero (otherwise $2^\ell$ would divide both, and the gcd could not be $1$).

On the other hand, we showed above that $\ell \ge 1$. That's a contradiction.

=====

Now there's definitely something screwy in the "proof" above, because Jyrki's example shows that the gcd is actually $2$. But you'll have a lot easier time finding the error now that the "proof" is actually written clearly and coherently.

Answer

$k_1$ and $k_2$ don't have to be odd numbers, only one of them has to be. If $s$ and $t$ are both odd, $s+t$ and $s-t$ are both even, so you know that $d$ is also even.