calculus - Find the limit as x approaches 0

$$\lim_{x\to 0} \frac{\sin(3x)}{x+\tan(4x)}$$

I am having trouble with this one.

I know that $\lim_{x\to 0}\frac{\sin(x)}{x} = 1$
and same for the inverse

also $\lim_{x\to 0}\frac{\tan(x)}{x} = 1$

I am able to simplify $\sin(x)$ easily by multiplying by $\frac{3x}{3x}$

I am lost as to what to do next however. I seems like a simple problem (and it probably is) and I also know that the limit of the whole function as $x$ approaches $0$ will be $3/5$.

I am also not allowed to use L'Hospital's rule.

Thanks!

Answer

Write

$$\frac{\sin(3x)}{x+\tan(4x)} = \frac{\sin(3x)/x}{1+\tan(4x)/x} = \frac{3\cdot \sin(3x)/(3x)}{1+4\cdot \tan(4x)/(4x)}$$

See what to do now?

Understanding newton interpolating polynomial

I understand what a Newton Interpolating Polynomial is and what it accomplishes, but I'm not exactly sure how it accomplishes it.
I'm not looking for a proof of why it works, just a layman explanation of how it is calculated, preferably with an example.

A lot like this question but instead of Lagrange Interpolating Polynomial it's about Newton's Interpolating Polynomial.

Answer

Polynomial interpolation amounts to solving a system of linear equations. Let's explain what that means: if you have some polynomials $p_1,\dots,p_n$ and want to interpolate the data $(x_1,y_1),\dots,(x_n,y_n)$, the interpolation problem says: find numbers $c_1,\dots,c_n$ so that

$$c_1 p_1(x_1) + c_2 p_2(x_1) + \dots + c_n p_n(x_1) = y_1 \\
c_1 p_1(x_2) + c_2 p_2(x_2) + \dots + c_n p_n(x_2) = y_2 \\
\vdots \\
c_1 p_1(x_n) + c_2 p_2(x_n) + \dots + c_n p_n(x_n) = y_n.$$

Thus, we can pick and choose the polynomials $p$ that we want to use to find these $c$s. Good choices will make this system of linear equations easy to solve. In Lagrange interpolation, you choose polynomials $\ell_k$ which vanish at all but one node and are equal to $1$ at the remaining node. Then the system just becomes $c_1=y_1,c_2=y_2$, etc.

In Newton interpolation we make the system a little bit harder to solve. Specifically, we have $p_1$ which is constant, $p_2$ which vanishes at $x_1$, $p_3$ which vanishes at $x_1$ and $x_2$, and so forth. This idea is fairly similar to Lagrange interpolation: once we've chosen an interpolant for the first $k$ points, we want every change we make to leave those $k$ points alone.

With this choice, the system above becomes

\begin{align}
c_1 p_1(x_1)= y_1 \\
c_1 p_1(x_2) + c_2 p_2(x_2) = y_2 \\
c_1 p_1(x_3) + c_2 p_2(x_3) + c_3 p_3(x_3) = y_3 \\
\vdots \\
c_1 p_1(x_n) + \dots + c_n p_n(x_n) = y_n.

\end{align}

This system is triangular, whereas the Lagrange system is diagonal. But it is still pretty easy to solve: we read off $c_1=\frac{y_1}{p_1(x_1)}$, then we plug that into the second equation and solve for $c_2$, then we plug those results into the third equation and solve for $c_3$, etc.

The only remaining thing is to choose these polynomials. But that's not difficult: just pick the simplest polynomials that vanish at $x_1,\dots,x_{k-1}$, namely $p_1=1$, $p_k=(x-x_1)\dots(x-x_{k-1})$ for $k \geq 2$.

An example: you want to interpolate $(1,2),(3,4),(6,10)$. You have $p_1=1,p_2=x-1,p_3=(x-1)(x-3)$. Your first equation says:

$$c_1=2.$$

Your second equation says

$$c_1 + c_2 \cdot 2 = 4.$$

Plug in what $c_1$ is:

$$2 + 2c_2 = 4$$

so $c_2=1$. Your third equation says

$$c_1 + c_2 \cdot 5 + c_3 \cdot 15 = 10.$$

Plug in what you know:

$$2 + 5 + 15c_3 = 10$$

so $c_3=\frac{1}{5}$. Thus your interpolant is $2+(x-1)+\frac{1}{5}(x-1)(x-3)$.

calculus - Prove: If $x$ has the property that $0leq x0$, then$ x=0$.

I'm going through Apostol's Calculus I introduction, and I'm trying to prove this, but I'm having a little trouble doing it. It's proposed as an exercise in section I 3.5: order axioms.

So, what I get this theorem says is that, if a real number is smaller than all positive reals and greater or equal to zero, then it must be zero. I think it has something to do with real numbers' density; With my limited knowledge in math, I intuit there's no successor to 0 in ℝ, because, if I said it was 0.1, then there would be 0.01 which is smaller, but then I could find 0.001, and then 0.0001, and I could find infinitely many smaller numbers 0.00000000000000000000000000... the thing is that I remembered about infinitesimals, and remembered that the definition of a positive infinitesimal is a number α, such that ∀n|n∈ℕ : [0<α<$\frac{1}{n}$]. So if 0 is a real number and α is too, then 0+α is real number too, and it would be the next number to 0 in real numbers (I'm not sure about this, this is one of the voids I have). so, coming back to the theorem, if x is smaller than all positive reals, then it's smaller than 0+α, the number next to 0, so if it's smaller than that, it couldn't be a member of ℝ+, because the first positive real would be 0+α (again, another assumption based on intuition, maybe this would make mathematicians pull their hair in despair) and x≠0+α. So, since x∉ℝ+, but the sufficient conditions also states that 0⩽x, too, then x would have to be 0, because it's not 0+α, its successor.

The problem is that I have proven formally the other theorems proposed in the book, using the order and body axioms given there, and not using, I don't know... weird stuff like I do in here, also assuming things I intuit, instead of formalism. I put the aforementioned reasoning, just so you know I have tried to understand what I have to prove, instead of just putting it on the internet for other people to solve it for me, without putting in any effort. So, is an approach using infinitesimals valid for proving this, if so, is the way I'm trying to see it correct? if not, then, how should I prove this?

Thanks in advance.

Answer

This is about real numbers; infinitesimals do not belong.

The proof of this theorem depends on what you have available, but basically it goes like this:

Suppose that $x\ne0$. Then $x>0$ by the trichotomy axiom. Now you can take $h=x/2$ and you have a contradiction, because $0

combinatorics - combinatorial proof for binomial identity

Consider the following binomial identity:
$$
\sum_{k=0}^n(-1)^k\binom{n}{k}g(k)=0
$$
for every polynomial $g(k)$ with degree less than $n$.

My Proof
Every polynomial $g(k)$ of degree $t$ can be represented in the following form

$$
g(k) = \sum_{l=0}^tc_l(k)_l,
$$
where
$$
(k)_l=k(k-1)\ldots(k-l+1),
$$
ans $c_l$ are some coefficients.

For every $l
\begin{multline}
\sum_{k=0}^n(-1)^k\binom{n}{k}(k)_l=
\sum_{k=l}^n(-1)^k\binom{n}{k}(k)_l=
\sum_{k=l}^n(-1)^k\frac{n!k!}{k!(n-k)!(k-l)!}=
\sum_{k=l}^n(-1)^k\frac{n!k!}{k!(n-k)!(k-l)!}=
\frac{n!}{(n-l)!}\sum_{k=l}^n(-1)^k\frac{(n-l)!}{(n-k)!(k-l)!}=\\
\frac{n!}{(n-l)!}\sum_{k=l}^n(-1)^k\binom{n-l}{n-k}=0,
\end{multline}
therefore,
$$

\sum_{k=0}^n(-1)^k\binom{n}{k}g(k)=0.
$$

Question
Do you know some other proofs of this identity? I'm most interested in combinatorial proof.

Answer

Let $n$ a positive integer and $E$ the real vector space of polynomials with degree $

Consider the linear map $\varphi:E\to E,P(X)\mapsto P(X)-P(X+1)$.

For every nonzero $P(X)\in E$, we observe that $\deg(\varphi(P(X))<\deg(P(X))$. This leads to $\varphi^n=0$ (of course $\varphi^k$ means $\varphi\circ\cdots\circ\varphi$ with $k$ times $\varphi$).

Now consider the map $\psi:E\to E,P(X)\mapsto P(X+1)$, so that $\varphi=id_E-\psi$.

Since $\psi$ and $id_E$ commute, we can apply Newton's binomial formula and get :

$$\sum_{k=0}^n{n\choose k}(-1)^k\psi^k=\left(id_E-\psi\right)^n=\varphi^n=0$$

So, for any $P(X)\in E$ :

$$\sum_{k=0}^n{n\choose k}(-1)^kP(X+k)=0$$Evaluating in $0$, we finally get :

$$\sum_{k=0}^n{n\choose k}(-1)^kP(k)=0$$

calculus - Prove inequality using Mean Value Theorem

How would you prove the following inequality using the Mean Value Theorem:
$$1+2\ln x\leq x^2$$ for $x>0$.

Answer

Let $f (t)=1+2\ln(t)-t^2$ for $t>0$.

for $x>0$ , $f $ is continuous at $[x,1] $ or $[1,x]$ and differentiable at $(x,1) $ or $(1,x) $ thus by MVT, exists $c$ strictly between $x $ and $1$ such that

$$f (x)-f (1)=(x-1)f'(c) $$
$$1+2\ln (x)-x^2=2(x-1)(1/c -c) $$
$$=2 (x-1)\frac {1-c^2}{c} $$

If $x>1$ then $1-c^2<0$ and

if $x <1$ then $1-c^2>0$

thus in all cases
$$f (x)\le 0.$$

real analysis - Calculate the limit $left(frac{1}{sqrt{n^2+1}}+frac{1}{sqrt{n^2+2}}+...+frac{1}{sqrt{n^2+n}}right)$

I have to calculate the following limit:
$$\lim_{n\to\infty} \left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+...+\frac{1}{\sqrt{n^2+n}}\right)$$

I believe the limit equals $1$, and I think I can prove it with the squeeze theorem, but I don't really know how.

Any help is appreciated, I'd like to receive some hints if possible.

Thanks!

Answer

For every $n>0$,
$$\frac{n}{\sqrt{n^2+n}}\le\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots+\frac{1}{\sqrt{n^2+n}}\le\frac{n}{\sqrt{n^2+1}}$$

Can you continue with Squeeze theorem?

elementary number theory - How to convert a diophantine equation into parametric form?

Let be $\space 7x+9y=5 \space$ a linear diophantine equation, in two variables. What are the integer solutions for $x$ and $y$?

I know that $7x+9y=5$ is a cartesian equation for a line in the plane. Then I thought, if one could define $x$ and $y$ in terms of the same parameter, it would be possible to know all the integer solutions. But I don't have clue on how can convert an diophantine equantion in the form $ax+by=c$ to its parametric form.

Can you give me some hints?Thanks.

Answer

$\frac 97=1+\frac27=1+\frac1{\frac72}=1+\frac1{3+\frac12}$

So, the last but one convergent is $1+\frac13=\frac43$

Using Convergent property of continued fraction, $7\cdot4-9\cdot3=1$

$7x+9y=5(7\cdot4-9\cdot3)\implies 7(x-20)=-9(y+15)\implies x-20=\frac{-9(y+15)}{7}$ which is an integer.

So, $7\mid(y+15)$ as $(7,9)=1$ $\implies \frac{x-20}{-9}=\frac{y+15}7=z$ for some integer $z$

So, $y=7z-15=7(z-3)+6=7w+6$ where $w=z-3$ is any integer.

So, $x=-9z+20=-9(w-3)+20-27=-(9w+7)$

---

Alternatively, by observation $7x+9y=5=14-9$

or $7(x-2)=-9(y+1)$

or, $\frac{x-2}{-9}=\frac{y+1}7$

$\frac{-9(y+1)}7=x-2$ which is an integer, so is $\frac{y+1}7$ as $(7,9)=1$

So, $\frac{x-2}{-9}=\frac{y+1}7=u$ where $u$ is any integer.

So, $x=-9u+2,y=7u-1$