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10

We can prove a stronger statement: the equation above has $n - 1$ real positive roots and a negative real one, and there are no other roots. Let $$g(x) = \sum_{i = 1}^n \frac1{x-a_i} - \frac1x,\qquad a_i \in (0, +\infty).$$ Note that $g(x)$ is defined in $\mathbb R \setminus \{0, a_1, \ldots, a_n\}$ and it's also continuous. Without loss of generality, ...


5

Good question, this follows because the function $log$ is strictly increasing. This means that whenever $x > y$, $log(x)$ is greater than $log(y)$. So in this case, if you can find the value x, such that $f(x)$ is maximised. Then that same value $x$, must be the maximum of $log(f(x))$. I hope that makes sense (and apologies for the formatting, I'm ...


5

The modified Bessel functions of the first kind fulfill the recurrence relation: $$ I_n(\alpha)=\frac{\alpha}{2n}\left(I_{n-1}(\alpha)-I_{n+1}(\alpha)\right)\tag{1}$$ due to the integral representation: $$ I_n(\alpha) = \frac{1}{\pi}\int_{0}^{\pi}\cos(nx)\,e^{\alpha\cos x}\,dx \tag{2}$$ and the cosine addition formulas. A straightforward consequence of $(1)$ ...


5

I concur with Arthur that your work looks OK minus a few technicalities, but those are rather a big deal in real analysis aren't they? Thus, I would write up your proof as follows: Given $\epsilon>0$, we need $\delta>0$ such that if $|x-1|<\delta$ then $|(x^3+5x^2-2)-4)|<\epsilon$. Now, $$ |x^3+5x^2-6|=|(x-1)(x^2+6x+6)|=|x-1||x^2+6x+6|. $$ If $|...


5

Method 1. As a first remark, Stirling's approximation will do the job: $$ a_n = \frac{n!^2 4^{n}}{(2n)!} = \frac{4^n}{\binom{2n}{n}} \operatorname*{\sim}_{n\to\infty} \sqrt{\pi n} $$ and the series trivially diverges (the general term does not even converge to zero). Method 2. A first simpler idea (than Stirling's): recognize a Binomial coefficient. For ...


4

You're on the right track, but you gave up too much in the numerator. Instead, observe that $$ \frac{2\sqrt{n}+1}{n^2+n+1}\leq \frac{2\sqrt{n}+1}{n^2}\leq \frac{3\sqrt{n}}{n^2}=\frac{3}{n^{\frac{3}{2}}}$$ and $\sum_{n=1}^{\infty}\frac{3}{n^{\frac{3}{2}}}$ converges.


4

Hint: $$\dfrac{d}{dt} \sum_{n=0}^\infty t^n = \sum_{n=1}^\infty n t^{n-1}$$ for $|t|<1$.


4

No, you are confusing things and you are not doing what you are required to do: you are not required to integrate the derivative of $\arctan$, but $\arctan$ itself. Start by finding the Taylor series of $\arctan x$ around $0$: $$(\arctan x)' = \frac 1 {1 + x^2} = \sum _{n = 0} ^\infty (-x^2)^n \implies \arctan x = \int \sum _{n = 0} ^\infty (-x^2)^n \ \Bbb ...


3

$p$ is a constant, so $\frac{1}{p+1}$ is a constant, so $\frac{x^{p+1}}{p+1}=\frac{1}{p+1}x^{p+1}$. Just because there's a quotient doesn't mean we need to think of it as a quotient.


3

The Taylor series (for $|x|\leq 1$) is $$\tan^{-1}(x)=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}x^{2n+1}$$ So $$\tan^{-1}(2x)=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}2^{2n+1}x^{2n+1}$$ Integrating $$\int \tan^{-1}(2x)\,dx=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)(2n+2)}2^{2n+1}x^{2n+2}$$ Using the bounds $$\int_0^{\frac 1{10}} \tan^{-1}(2x)\,dx=\frac{1}{100}-\frac{1}{...


3

Notice that for any $n \in \mathbb N$, we have that: \begin{align*} (2n)! &= \color{red}{(2n)}(2n - 1)\color{red}{(2n - 2)}(2n - 3)\color{red}{(2n - 4)}(2n - 5) \cdots \color{red}{(4)}(3)\color{red}{(2)}(1) \\ &\leq \color{red}{(2n)}(2n)\color{red}{(2n - 2)}(2n - 2)\color{red}{(2n - 4)}(2n - 4) \cdots \color{red}{(4)}(4)\color{red}{(2)}(2) \\ &= ...


3

Here is an intuitive explanation. The second order approximation to $f$ near $x_0 \in \mathbb R^2$ is $f(x) \approx f(x_0) + \nabla f(x_0)^T (x - x_0) + \frac12 (x - x_0)^T \nabla^2 f(x_0) (x - x_0)$. If $x_0$ is a local extremum for $f$, then $\nabla f(x_0) = 0$, so the behavior of $f$ near $x_0$ is determined by the quadratic term. If $\nabla^2 f(x_0)$ ...


3

For any $c \in \Bbb C$, $f(x) := x^c$ is defined as $\exp(c \log x)$, where, typically, we take $\log$ to be the principal branch of logarithm. This function is not differentiable at any point in $(-\infty, 0]$ (though, it is differentiable everywhere else). So, we have to assume that $x \in \Bbb C \setminus (-\infty, 0]$. If this is the case, $$f'(x) =c \...


3

I will draw the areas considered for integrating $\int_0^2 2-\sqrt{4-x^2}dx$ because it is easier to draw. The left one is the left rectangle method because it uses the point at the left side of the interval. The second one is the right rectangle method because it uses the point at the right of the interval. The third is the trapezoid method. The ...


2

For second part , $$y=\ln \ln x$$ $$\frac{dy}{dx}=\frac{d(\ln \ln x)}{d(\ln x)}.\frac{d(\ln x)}{d(x)}$$ $$\frac{dy}{dx}=\frac{1}{\ln x}.\frac{1}{x}$$


2

Let $$g(x):=f(x)-\frac{x^3}{3}$$ for all $x\in\mathbb{R}$. Then, $g$ satisfies Cauchy's functional equation $$g(x+y)=g(x)+g(y)$$ for all $x,y\in\mathbb{R}$ with the additional property that $g$ is continuous at $0$ (since $g'(0)=f'(0)=-1$). Thus, there exists $k\in\mathbb{R}$ such that $$g(x)=kx$$ for all $x\in\mathbb{R}$. Finally, as $k=g'(0)=-1$, we get ...


2

It appears that the integral when $n=2$ can be represented in terms of elliptic integrals: $$ I(2)=\frac{\pi}{2}-\frac{1}{\sqrt{6}}\left(\Pi\left(\frac23\mid\frac13\right)-K\left(\frac13\right)\right). $$ Here the arguments of elliptic functions follow Mathematica conventions: that is, $$ K(m)=\int^{\pi/2}_{0}\frac{d\theta}{\sqrt{1-m\sin^2\theta}} $$ and ...


2

Let $\epsilon>0$ be given. We have to find a number $\delta>0$ such that $\left|\frac{x^2-5x}{x^2+2}+1\right|<\epsilon$ whenever $|x-2|<\delta$. But, as Andre notes, $$ \left|\frac{x^2-5x}{x^2+2}+1\right|=\left|\frac{2x^2-5x+2}{x^2+2}\right|=|x-2|\left|\frac{2x-1}{x^2+1}\right|. $$ We find a positive constant $C$ such that $\left|\frac{2x-1}{x^2+...


2

Let $f:D \subseteq \Bbb R^2 \to R$ be a twice differentiable function. Then we'd call $f$ a surface. The directional derivative $D_vf(x,y)$ is the derivative of $f$ at $(x,y)$ in the direction of the unit vector $v$. This gives the scalar slope at $(x,y)$ in that direction. The second directional derivative ${D_v}^2f(x,y)$ is the second derivative of $...


2

This is not an answer but it is too long for a comment. If we consider the more general term $$a_n = \frac{(n!)^2 k^{n}}{(2n)!} $$ and using Stirling approximation, just as Clement C. did in his answer, we have $$a_n \operatorname*{\sim}_{n\to\infty} \left(\frac k 4 \right)^n \sqrt{\pi n}$$ and $a_n$ goes through a maximum value when $n=\frac 1{2 \log\left(\...


2

The geometric series that you got is for $\dfrac d {dx} \arctan(2x)$, not for $\arctan(2x)$. You need to take an antiderivative of that.


2

Suppose $\{a_n\}$ is the sequence of terms of this series. Then it is clear that $\lim \sup (|a_n|)^{1/n} = 1 $ so that the radius of convergence is $1$.


2

Taking the absolute value of your algebra gives $x^2$. Since the ratio test looks for when this is $<1$, we ask ourselves when is $|x^2|<1$. The answer is when $-1<x<1$, and if $|x|>1$ it does not converge. This suffices to find the radius; it is $1$. If you want the interval of convergence as well, then you need to test when $|x|=1$. ...


2

Note that for $0<|x-1|<1$ $$\begin{align} \left|x^3+5x^2-6\right|&=|x^2+6x+6||x-1|\\\\ &<22|x-1|\\\\ &<\epsilon \end{align}$$ whenever $|x-1|<\delta=\min\left(1,\frac{\epsilon}{22}\right)$


2

You need to obtain a power series for $\arctan 2x$ and integrate that. What you have done is to obtain a power series for its derivative and integrate that, which is clearly wrong.


2

Hint Define (and fill in details) $$f(x):=\frac{A}{x-1} + \frac{B}{x-2}=\frac{(A+B)x-2A-B}{(x-1)(x-2)}$$ so $$\lim_{x\to1^+}f(x)=(-A)\lim_{x\to1^+}\frac1{(x-1)(x-2)}=\infty$$ $$\lim_{x\to2^-}f(x)=B\lim_{x\to2^-}\frac1{(x-1)(x-2)}=-\infty$$ and now use the IVT and the fact $\;f(x)\;$ is continuous at $\;(1,2)\;$


2

Let $$f(x) = \frac{A}{x - 1} + \frac{B}{x - 2}.$$ Observe that $f$ is defined on $\mathbb R \setminus \{1,2\}$ and it's continuous since it's a sum of continuous functions. Now, $\lim\limits_{x \to 1^+} f(x) = +\infty$ $\lim\limits_{x \to 2^-} f(x) = -\infty$ Therefore, from the definition of limit and the intermediate value theorem, it follows that $f$...


2

$4b_n=-a_{n+1}-2a_n$, $4b_{n+1}=-a_{n+2}-2a_{n+1}$, $-a_{n+2}-2a_{n+1}=16a_n-6a_{n+1}-12a_n$, $a_{n+2}-4a_{n+1}+4a_n=0$. Do you know how to solve that kind of recurrence? Here's an approach. $a_{n+2}-4a_{n+1}+4a_n=(a_{n+2}-2a_{n+1})-2(a_{n+1}-2a_n)=c_{n+1}-2c_n$, where we are defining $c_n$ by $c_n=a_{n+1}-a_n$. Now we have to solve $c_{n+1}-2c_n=0$, and ...


2

Hint...substitute $$\frac 1r-\frac 1p=u\cdot \frac {\epsilon}{p}$$ and then you have a standard $\arcsin$-type integral


2

$$\int \frac{dr}{r^2} \frac{1}{\sqrt{-(\frac{1}{r} - \frac{1}{p})^2 + \frac{\epsilon^2}{p^2} }}=-\int \frac { d\left( \frac { 1 }{ r } -\frac { 1 }{ p } \right) }{ \sqrt { \frac { \epsilon ^{ 2 } }{ p^{ 2 } } -\left( \frac { 1 }{ r } -\frac { 1 }{ p } \right) ^{ 2 } } } =-\frac { \epsilon }{ p } \int { \frac { d\frac { p\left( \frac { 1 }{ r } -\frac { ...



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