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If $a,b$ are positive real numbers, such that $a+b=1$, then prove that $(a+ \dfrac {1}{a})^3 +(b+ \dfrac {1}{b})^3 \ge \dfrac {125}{4}$
If $a,b$ are positive real numbers, such that $a+b=1$, then prove that $$\bigg(a+ \dfrac {1}{a}\bigg)^3 +\bigg(b+ \dfrac {1}{b}\bigg)^3 \ge \dfrac {125}{4}$$
I just learnt to prove If $a$ and $b$ are positive real numbers such that $a+b=1$, then prove that $\big(a+\frac{1}{a}\big)^2 +\big(b+\frac{1}{b}\big)^2 \ge\frac{25}{2}$ a few days ago, and it was just basic application of CS. But I find this one really difficult. I can not apply CS here directly on the numbers $a+\frac{1}{a}$ and $b+\frac{1}{b}$ because CS is for squares. So some manipulation is needed. Anything from hint to full answer will be appreciated.
|
An alternative approach: $h(x)=x+\frac{1}{x}$ is positive and log-convex over $(0,1)$, since
$$\frac{d^2}{dx^2}\log h(x)=\frac{(1+3x^2)+(x^2-x^4)}{(x+x^3)^2}>0.$$
In particular $h(x)^3$ is convex and the claim follows from Jensen inequality, as already remarked by Micheal Rozenberg.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Envelope of a family of curves
Show that the envelope of the family of curves $$\frac{x}{a}+\frac{y}{4-a}=1$$ is the parabola $$\sqrt{y}+\sqrt{x}=2$$
I know how we can get the envelope, but I could not get the required relation. I differentiated the family of curves w.r.t $a$. I got
$$a=\frac{1}{2}(x-y+4)$$
I then substituted by $a$ in the family of curves and got
$$-x^2+2xy-y^2+8y+8x-16=0$$
How to complete to get $\sqrt{x}+\sqrt{y}=2$?
|
Calculate $y$ from last equation:
$$
y = \frac{2x+8 \pm \sqrt{(2x+8)^2 - 4(16+x^2-8x)}}{2}
=
\frac{2x+8 \pm 8\sqrt{x}}{2} = x \pm 4\sqrt x + 4.
$$
Now if you want to get form $\sqrt x + \sqrt y = 2$, you need to assume that $0 \leq x,y \leq 4$, hence $y = x - 4\sqrt x + 4 = (2 - \sqrt x)^2$. Now taking square root of both sides yields desired result.
|
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"url": "https://math.stackexchange.com/questions/2495315",
"timestamp": "2023-03-29T00:00:00",
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|
Computing $\lim_{n\to\infty}\left(\sqrt{\frac{n}{(n+1)^2(n+2)}}t+\frac{n}{n+1}\right )^n$ I am struggling to find the limit of $(\sqrt{\frac{n}{(n+1)^2(n+2)}}t+\frac{n}{n+1} )^n$ as $n$ goes to $\infty$.
I know that the limit of $(1+\frac{t}{n})^n$ as $n$ goes to $\infty$ equals $e^t$ and that my limit should approach $e^{t-1}$.
However, I am having a hard time using and manipulating the known $(1+\frac{t}{n})^n$ limit to find the value that my limit approaches.
I tried $\lim_{n \to \infty} (\sqrt{\frac{n}{(n+1)^2(n+2)}}t+\frac{n}{n+1} )^n=(\lim_{n \to \infty} (\sqrt{\frac{n}{(n+1)^2(n+2)}}t+\lim_{n \to \infty} \frac{n}{n+1})^n$ to avoid using $\lim_{n \to \infty}(1+\frac{t}{n})^n=e^t$ but I am unsure if it is correct.
If possible, I would prefer a very simple method that just uses properties of limits.
Any help is appreciated, thank you in advanced.
|
In the same spirit as other answers, you could even go beyond the limit.
Starting with $$A=\left(\sqrt{\frac{n}{(n+1)^2(n+2)}}t+\frac{n}{n+1}\right )^n$$
$$\log(A)=n \log\left(\sqrt{\frac{n}{(n+1)^2(n+2)}}t+\frac{n}{n+1}\right )$$ Now, using long division (or better, Taylor series)
$$\frac{n}{(n+1)^2(n+2)}=\frac 1 {n^2}-\frac{4}{n^3}+O\left(\frac{1}{n^4}\right)$$
$$\sqrt{\frac{n}{(n+1)^2(n+2)}}=\frac 1 {n}-\frac{2}{n^2}+O\left(\frac{1}{n^3}\right)$$ $$\frac{n}{n+1}=1-\frac 1 {n}+\frac{1}{n^2}+O\left(\frac{1}{n^3}\right)$$ making
$$\sqrt{\frac{n}{(n+1)^2(n+2)}}t+\frac{n}{n+1}=1+\frac{t-1}{n}+\frac{1-2 t}{n^2}+O\left(\frac{1}{n^3}\right)$$
$$\log\left(\sqrt{\frac{n}{(n+1)^2(n+2)}}t+\frac{n}{n+1}\right )=\frac{t-1}{n}+\frac{-\frac{t^2}{2}-t+\frac{1}{2}}{n^2}+O\left(\frac{1}{n^3}\right)$$ $$\log(A)=(t-1)+\frac{-\frac{t^2}{2}-t+\frac{1}{2}}{n}+O\left(\frac{1}{n^2}\right)$$ $$A=e^{\log(A)}=e^{t-1}\left(1-\frac{t^2+2t-1 }{2n}\right)+O\left(\frac{1}{n^2}\right)$$ which shows the limit and how it is approached.
|
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|
Given the ratios of distances from three corners of a rectangle, find the coordinates of a point with said ratios.
I think the picture can explain it better than words, but I'm wondering how to figure this out. Given three ratios of distances from corners, not lengths (in the picture I set the base, $base=1$, to be the distance from the top left corner, while the other two corners are of lengths $\alpha\cdot$$base$ and $\beta\cdot$$base$) and given the height $H$ and width $W$ of a rectangle, what are the coordinates of a point with said ratios? I'm sure the Apollonian Theorem comes into play, but I can't quite figure it out. Thanks!
|
Segment $1$ and segment $\alpha$ create a circle of possibilities called the Circle of Apollonius, with endpoints of its diameter at $\frac{W}{1+\alpha}$ and $\frac{W}{1-\alpha}$. $\beta$ and $H$ work similarly. You can then intersect these circles to find two solutions.
To find points where we can be via $\alpha$ and $W$, we can use the distance formula from the two points $(0,0)$ and $(W,0)$:
$$
\begin{align}
\alpha\sqrt{ \left(x^2 + y^2\right)} &= \sqrt{\left(x-W\right)^2 + y^2}\\
\alpha^2 \left(x^2 + y^2\right) &= \left(x-W\right)^2 + y^2\\
\alpha^2x^2 + \alpha^2y^2 &= x^2 - 2Wx + W^2 + y^2 \\
\left(\alpha^2-1\right)x^2 + \left(\alpha^2-1\right)y^2 + 2Wx &= W^2\\
x^2 + y^2 + \frac{2W}{\alpha^2-1}x &= \frac{W^2}{\alpha^2-1}\\
x^2 + \frac{2W}{\alpha^2-1}x + \left(\frac{W^2}{\alpha^2-1}\right)^2 + y^2 &= \frac{W^2}{\alpha^2-1} + \left(\frac{W}{\alpha^2-1}\right)^2\\
\left(x + \frac{W}{\alpha^2-1}\right)^2 + y^2 + &= \frac{\left(\alpha^2-1\right)W^2}{\left(\alpha^2-1\right)^2} + \frac{W^2}{\left(\alpha^2-1\right)^2}\\
\left(x - \frac{W}{1-\alpha^2}\right)^2 + y^2 &= \frac{\alpha^2W^2}{\left(\alpha^2-1\right)^2}\\
\left(x - \frac{W}{1-\alpha^2}\right)^2 + y^2 &= \left(\frac{\alpha W}{\alpha^2-1}\right)^2\\
\end{align}$$
So the candidate points form a circle centered at $A = \left(\frac{W}{1-\alpha^2}, 0\right)$ with radius $a = \pm\frac{\alpha W}{\alpha^2-1}$. We can do the same thing with $H$ and $\beta$ to get a second circle centered at $B = \left(0, \frac{H}{1-\beta^2}\right)$ with radius $b = \pm\frac{\beta H}{\beta^2-1}$.
Now, let's find the intersections of these circles.
First, we need the distance between the centers:
$$d=\sqrt{\left(\frac{W}{1-\alpha^2}\right)^2 + \left(\frac{H}{1-\beta^2}\right)^2}$$
Now, the intersection. I'll use circle $a$ as the first circle. At this point the calculations are getting a little too nasty, and I haven't found anything nice after here, so you'll just have to do the math by plugging previous results in:
The distance from $A$ to the segment between the two solution points is $u = \frac{d^2 + a^2 - b^2}{2d}$. The distance from the line between $A$ and $B$ to the solution points is $v = \sqrt{u^2 - a^2}$. A unit vector pointing from $A$ to $B$ is $U = \frac{B-A}{d}$. Then $V$ is a unit vector perpendicular to $U$ - just switch the coordinates and flip one's sign. Finally, we can find the (up to) two solution points:
$$P = A + uU \pm vV$$
|
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|
max and min of $f(x,y)=\frac{xy+y^2}{x^2+y^2}$ The function is symmetric because $f(x,y)=f(-x,-y)$. It's defined in $\mathbb{R}^2-\{{(0,0)}\}$.If I calculate the gradient , it's null in the origin. Have I to prove $f$ is limited in $\mathbb{R}^2$?
|
By the arithmetic mean-geometric mean inequality
$$
|xy| \leq \tfrac{1}{2}(x^2 + y^2)
$$
so
$$
\left|\frac{xy + y^2}{x^2 + y^2}\right| \leq \frac{|xy|}{x^2 + y^2} + \frac{|y|^2}{x^2 + y^2} \leq \frac{1}{2} + 1 = \frac{3}{2}
$$
Since $|f(x,y)| \leq \frac{3}{2}$, we know
$$
-\frac{3}{2} \leq f(x,y) \leq \frac{3}{2}
$$
and $f$ is bounded both above and below.
|
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|
Solving a Linear First Order Differential Equation
$u_x+x^2y^4u_y=0 , u(1,y)=cos(2y) \\
\frac{dy}{dx}= x^2y^4 \implies \int y^{-4} dy = \int x^2 dx \implies -\frac{y^{-3}}{3}= \frac{x^3}{3} +C \\
C= -1/3(x^3+y^{-3}) \\
u(x,y)=f(C)=f(-1/3(x^3+y^{-3})) \\
\text{Given auxiliary condition: } u(1,y)= f(\frac{-1}{3}(1+y^{-3}))= \cos(2y).$
Solving for the general form of $f$ is where I get stuck.
I am unsure if I can simplify this problem by using the fact that $C$ is a constant. In other words can is the following a valid step:
$C= -1/3(x^3+y^{-3}) \implies C=(x^3+y^{-3})$
Doing so would mean I would then proceed with $f(1+y^{-3})= \cos(2y)$ and thus solve this form of $f$.
|
$$u_x+x^2y^4u_y=0 \tag 1$$
With the method of characteristics :
$$\frac{dx}{1}=\frac{dy}{x^2y^4}=\frac{du}{0}$$
A first family of characteristic curves comes from necesserarily $du=0 \quad\implies\quad u=c_1$
A second family of characteristic curves comes from $\frac{dx}{1}=\frac{dy}{x^2y^4}$ which is a separable ODE easy to solve : $x^3+\frac{1}{y^3}=c_2$
The general solution of Eq.1 expressed on implicit form is :
$$\Phi\left(\left(x^3+\frac{1}{y^3}\right)\;,\:u\right)=0$$
where $\Phi(X,Y)$ is any differentiable function of two variables.
Or, equivalent, on explicit form :
$$u=F\left(x^3+\frac{1}{y^3}\right) \tag 2$$
where $F(X)$ is any differentiable function.
Then we have to determine which function $F(X)$ satisfies the condition $u(1,y)=\cos(2y)$
$$F\left(1+\frac{1}{y^3}\right)=\cos(2y)$$
In particular $X=1+\frac{1}{y^3}\quad\to\quad y=(X-1)^{-1/3}\quad\to\quad F(X)= \cos(2(X-1)^{-1/3})$
So F(X) is determined. Putting it into Eq.(2) where $X=x^3+\frac{1}{y^3}$ leads to :
$$u=\cos\left(2\left(x^3+\frac{1}{y^3}-1\right)^{-1/3}\right)$$
|
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|
Prove That $ \sqrt{\frac xy + \frac yz}+ \sqrt{\frac yz + \frac zx}+ \sqrt{\frac zx + \frac xy}\ge 3 $ for x, y, z $\gt 0$, prove that: $$ \sqrt{\frac xy + \frac yz}+ \sqrt{\frac yz + \frac zx}+ \sqrt{\frac zx + \frac xy}\ge 3 $$
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Also, we can use Minkowski and AM-GM
$$\sum_{cyc}\sqrt{\frac{x}{y}+\frac{y}{z}}\geq\sqrt{\left(\sum_{cyc}\sqrt{\frac{x}{y}}\right)^2+\left(\sum_{cyc}\sqrt{\frac{y}{z}}\right)^2}\geq\sqrt{3^2+3^2}=3\sqrt2\geq3.$$
|
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|
Infinite series for $e$... How do you prove that $e=\sum_{n=0}^{\infty}\frac{1}{n!}$? Here I am assuming $e:=\lim_{n\to\infty}(1+\frac{1}{n})^n$. Do you have any good PDF file or booklet available online on this? I do not like how my analysis text handles this...
|
First prove by the ratio test that the series $\displaystyle \sum_{k=0}^{\infty} \frac{1}{k!}$ converges and denote the sum by $S$.
Then note that
$$\begin{align*}
\left( 1 + \frac{1}{n} \right)^n & = \sum_{k=0}^n \binom{n}{k} \cdot 1^{n-k} \cdot \left( \frac{1}{n} \right)^k = \sum_{k=0}^n \frac{n (n-1) \ldots (n-k+1)}{k!} \cdot \frac{1}{n^k} \\[1ex]
& = \sum_{k=0}^n \frac{1}{k!} \cdot \left( 1 - \frac{1}{n} \right)\left(1-\frac{2}{n}\right) \ldots \left( 1 - \frac{k-1}{n} \right) = \sum_{k=0}^n \frac{1}{k!} \cdot P_n^{(k)}
\end{align*}$$
where
$$P_n^{(k)} = \left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right) \ldots \left(1-\frac{k-1}{n}\right).$$
We can see that for each $k \in \mathbb{N}$ we have $0 \leqslant P_n^{(k)} \leqslant 1$ and $\displaystyle \lim_{n \to \infty} P_n^{(k)} = 1$.
Fix $\varepsilon > 0$ and let $K$ be so large that $\displaystyle \sum_{k=0}^K \frac{1}{k!} \geqslant (1-\varepsilon)S$. Now let $N$ be so large that whenever $n \geqslant N$, for $k = 0, 1, \ldots, K$ we have $P_n^{(k)} > 1-\varepsilon$.
So for $n \geqslant N$
$$\begin{align*}
(1-\varepsilon)^2 S \leqslant (1-\varepsilon) \sum_{k=0}^K \frac{1}{k!} \leqslant \sum_{k=0}^K \frac{1}{k!} P_n^{(k)} \leqslant \left(1+\frac{1}{n}\right)^n \leqslant \sum_{k=0}^n \frac{1}{k!} \leqslant S.
\end{align*}$$
Therefore
$$e = \lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n = S = \sum_{k=0}^{\infty} \frac{1}{k!}.$$
|
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"language": "en",
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|
Find the exact value of $\cos\frac{2\pi}{5}$ by solving equation There's this concept in the topic of complex numbers which I don't really understand much (and was devastated upon realising it'll appear in the topic often) - trigonometry!
I'm super lost to be honest.
We are asked to express $\cos3\theta$ and $\cos2\theta$ in terms of $\cos\theta$.
Then show that $\cos3\theta=\cos2\theta$ can be written as $4z^3-2z^2-3z+1=0$ where $z=\cos\theta$.
Lastly, by solving the equation above for $z$, we are to find out the value of $\cos\frac{2\pi}{5}$.
I attempted the question with just subbing in $z=\cos\theta$ into the equation and I got: $$4\cos^3\theta-3\cos\theta=\cos2\theta$$ and don't really know where to go next.
|
We have
\begin{eqnarray*}
\cos( 2 \theta) =2 \cos^2(\theta)-1 \\
\cos( 4 \theta) =4 \cos^3(\theta)-3 \cos( \theta)
\end{eqnarray*}
Let $z=\cos( \theta) $ then $\cos(3 \theta) = \cos(2 \theta)$ becomes $4z^3-2z^2-3z+1=0$. This will factorise $(z-1)(4z^2+2z-1)=0$, so $\color{red}{\cos(\frac{2 \pi}{5} )=\frac{-1+\sqrt{5}}{4}}$.
|
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|
Proving $(1+a)^n\ge 1+na+\frac{n(n-1)}{2}a^{2}+\frac{n(n-1)(n-2)}{6}a^3$ for all $n\in\mathbb N$ and all $a\ge -1.$ I was asked to prove the the following without induction. Could someone please verify whether my proof is right? Thank you in advance.
For any real number $a\ge -1$ and every natural number n, the statement, $$(1+a)^n\ge 1+na+\frac{n(n-1)}{2}a^{2}+\frac{n(n-1)(n-2)}{6}a^3,$$ holds.
Proof. From the binomial theorem we can see that, $$(1+a)^n\ge 1+na+\frac{n(n-1)}{2}a^{2}+\frac{n(n-1)(n-2)}{6}a^3,$$ becomes, $$\sum^{n}_{k=0}\binom{n}{k}a^{k}\ge\sum^{n}_{k=0}\binom{n}{k}a^{k}-\sum^{n}_{k=4}\binom{n}{k}a^{k}.$$ Also, if $a\lt b$, let us define $\sum^{a}_{i=b}f(i)=0$, by convention. Then for n=1 we have,\begin{align}\sum^{1}_{k=0}\binom{n}{k}a^{k}&\ge\sum^{1}_{k=0}\binom{n}{k}a^{k}-\sum^{1}_{k=4}\binom{n}{k}a^{k}\\ \sum^{1}_{k=0}\binom{n}{k}a^{k}&\ge\sum^{1}_{k=0}\binom{n}{k}a^{k}-0,\end{align} which is true, since both remaining sums are equal to each other. It is similarly the case for $n=2$ and $n=3$. More generally, \begin{align}\sum^{n}_{k=0}\binom{n}{k}a^{k}&\ge\sum^{n}_{k=0}\binom{n}{k}a^{k}-\sum^{n}_{k=4}\binom{n}{k}a^{k}\\ 1+\binom{n}{1}a^{1}+...+\binom{n}{k}a^{k}+...+\binom{n}{n}a^{n}&\ge1+\binom{n}{1}a^{1}+\binom{n}{2}a^{2}+\binom{n}{3}a^{3}\\ \binom{n}{4}a^4+\binom{n}{5}a^5+...+\binom{n}{k}a^{k}+...+\binom{n}{n}a^{n}&\ge0.\end{align}
To determine whether this last statement is true, we first consider $a\ge0$, and we see that it is trivially true. Next we consider the case where $-1\le a\lt0.$ We let $a=-b$, so $0\lt b\le 1$ and notice that, $$\binom{n}{4}a^4+\binom{n}{5}a^5+...+\binom{n}{k}a^{k}+...+\binom{n}{n}a^{n}$$ then becomes, $$\binom{n}{4}b^4-\binom{n}{5}b^5+...+(-1)^{k}\binom{n}{k}b^{k}+...+(-1)^{n}\binom{n}{n}b^{n}.$$
Also from the binomial theorem we can see that,
\begin{align}
(1+(-1))^{n}-(1+(-1))^{3}+(1+(-1))^{1}\\
=\sum^{n}_{k=0}\binom{n}{k}1^{n-k}(-1)^{k}-\sum^{3}_{k=0}\binom{n}{k}1^{n-k}(-1)^{k}+\sum^{1}_{k=0}\binom{n}{k}1^{n-k}(-1)^{k}\\
0=1-\binom{n}{1}+\binom{n}{4}-\binom{n}{5}+\binom{n}{6}+...+(-1)^{k}\binom{n}{k}+...+(-1)^{n}\binom{n}{n}\\
n-1=\binom{n}{4}-\binom{n}{5}+\binom{n}{6}+...+(-1)^{k}\binom{n}{k}+...+(-1)^{n}\binom{n}{n}\ge0,
\end{align}
and by noticing that $b^4\ge b^5\ge b^6\ge ...\ge b^k$, we can see that each binomial term $\binom{n}{k}$ is multiplied by a factor of $b^k$, making each term smaller than the term before.
Thus,
\begin{align}
\binom{n}{4}b^4-\binom{n}{5}b^5+...+(-1)^n\binom{n}{n}b^n\ge0\\
\binom{n}{4}a^4+\binom{n}{5}a^5+...+\binom{n}{n}a^n\ge0,\tag{by substitution}
\end{align} as desired.
Working backward, \begin{align}\binom{n}{4}a^4+\binom{n}{5}a^5+...+\binom{n}{k}a^{k}+...+\binom{n}{n}a^{n}&\ge0\\ 1+\binom{n}{1}a^{1}+...+\binom{n}{k}a^{k}+...+\binom{n}{n}a^{n}&\ge1+\binom{n}{1}a^{1}+\binom{n}{2}a^{2}+\binom{n}{3}a^{3}\\ \sum^{n}_{k=0}\binom{n}{k}a^{k}&\ge\sum^{n}_{k=0}\binom{n}{k}a^{k}-\sum^{n}_{k=4}\binom{n}{k}a^{k}\\ (1+a)^n&\ge 1+na+\frac{n(n-1)}{2}a^{2}+\frac{n(n-1)(n-2)}{6}a^3.\ \ \ \blacksquare\end{align}
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This is not an answer to the question about your proof. I am offering an alternative proof,using Taylor's Theorem. Note that, for integers $n$ and $k$ with $n> k\geq 0$,
$$(1+a)^n-\sum_{r=0}^k\,\binom{n}{r}\,a^r=(k+1)\binom{n}{k+1}\,\int_{0}^a\,(1+x)^{n-k-1}\,(a-x)^k\,\text{d}x\,.$$
Your question is a particular case where $k=3$, which leads to
$$(1+a)^n-\sum_{r=0}^3\,\binom{n}{r}\,a^r=4\binom{n}{4}\,\int_{0}^a\,(1+x)^{n-4}\,(a-x)^3\,\text{d}x\geq 0\,,$$
and the equality holds iff $a=0$. You can also use the derivative form:
$$(1+a)^n-\sum_{r=0}^3\,\binom{n}{r}\,a^r=\binom{n}{k+1}\,(1+\zeta)^{n-k-1}\,a^4\geq0\,,$$
where $\zeta$ is a number between $0$ and $a$.
|
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|
Can the difference of 2 undefined limits be defined? Is this limit defined or undefined?
$$\lim\limits_{x \to 0+} \left(\sqrt{\frac{1}{x}+2}-\sqrt{\frac{1}{x}}\right)$$
When I apply the rule of difference of limits, it's undefined. But, when I manipulate it, it gives me zero. And the graph of the function indicates it's defined on the right side.
By multiplying by $\frac{\sqrt{\frac{1}{x}+2}+\sqrt{\frac{1}{x}}}{\sqrt{\frac{1}{x}+2}+\sqrt{\frac{1}{x}}}$:
$$\lim\limits_{x \to 0+} \frac{\left( \sqrt{\frac{1}{x}+2}-\sqrt{\frac{1}{x}} \, \right) \left(\sqrt{\frac{1}{x}+2}+\sqrt{\frac{1}{x}} \, \right)}{\sqrt{\frac{1}{x}+2}+\sqrt{\frac{1}{x}}}$$
$$=\lim\limits_{x \to 0+} \frac{\frac{1}{x}+2-\frac{1}{x}}{\sqrt{\frac{1}{x}+2}+\sqrt{\frac{1}{x}}}$$
$$=\lim\limits_{x \to 0+} \frac{2}{\sqrt{\frac{1}{x}+2}+\sqrt{\frac{1}{x}}}$$
Then, we multiply by $\frac{\sqrt{x}}{\sqrt{x}}$:
$$=\lim\limits_{x \to 0} \frac{2\sqrt{x}}{\sqrt{1+2x}+1}$$
And, we substitute:
$$=\frac{2\sqrt{0}}{\sqrt{1+2\times0}+1} = 0$$
So, is this limit defined or not? and what's my error, if any?
|
You write $x\to 0+$ but it's often written as $x\to 0^+$. You asked if your limit is defined or undefined. I'll answer that with another two ways.
$$\lim_{x\to 0^+} \left(\sqrt{\frac{1}{x}+2}-\sqrt{\frac{1}{x}}\right)=$$
$$=\lim_{x\to 0^+}\left(\frac{\sqrt{1+2x}-1}{x}\cdot \sqrt{x}\right)=$$
Use the definition of a derivative and a limit multiplication rule/law.
$$=(\sqrt{1+2x})\bigg|_{x=0^+}\cdot 0=0,$$
because the derivative is a real number.
Another way is using Newton's generalized binomial theorem, also see Binomial series, which converges when $|x|<1$ but it can diverge when $|x|\ge 1$, see the Binomial series link for more information. $$\lim_{x\to 0^+}\frac{\sqrt{1+2x}-1}{x}=$$
$$=\lim_{x\to 0^+}\frac{(1+2x)^{\frac{1}{2}}-1}{x}=$$
$$=\lim_{x\to 0^+}\frac{1+\frac{1}{2}2x+o(x)-1}{x}=$$
$$=\lim_{x\to 0^+}\left(1+\frac{o(x)}{x}\right)=1+0=1$$
|
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|
Prove that $a_n=(1-\frac{1}{n})^n$ is monotonically increasing sequence I try to solve it Bernoulli inequality but it too complicated, am I missing something easier?
My try-
$$\frac{a_n}{a_{n+1}}=\frac{(1-\frac{1}{n})^n}{(1-\frac{1}{n+1})^{n+1}}\\=(\frac{1}{1-\frac{1}{n+1}})(\frac{\frac{n-1}{n}}{\frac{n}{n+1}})^n\\=(\frac{1}{1-\frac{1}{n+1}})(1-\frac{1}{n^2})^n<1\\\iff (1-\frac{1}{n^2})^n<1-\frac{1}{n+1}$$
here is where that I want to use Bernoulli...
|
With Bernoulli, but rewriting to make it simpler:
\begin{align}
&\Bigl(1-\frac1{n+1}\Bigr)^{n+1}>\Bigl(1-\frac1{n}\Bigr)^{n}\iff\Bigl(\frac n{n+1}\Bigr)^{n+1}>\Bigl(\frac{n-1}{n}\Bigr)^{n}\\
\iff&\Bigl(\frac{n^2}{n^2-1}\Bigr)^{n}>\frac{n+1}n=1+\frac1n\iff\Bigl(1+\frac{1}{n^2-1}\Bigr)^{n}>1+\frac1n.
\end{align}
Now, by Bernoulli's inequality,
$$\Bigl(1+\frac{1}{n^2-1}\Bigr)^{n}>1\frac{n}{n^2-1}>1+\frac{n}{n^2}=1+\frac1n.$$
|
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|
inequality with parameter - designate param that we get R for which param of A set of solutions is R?
$$-3 < {x^2 - ax - 2\over x^2 - x +1} < 2$$
hm
my trying:
$x^2-x+1$ is always + so I can multiply by $x^2-x+1$ instead of $(x^2-x+1)^2$
result should be set of $(-1 ;2 $ )
|
We need
$$x^2-ax-2<2x^2-2x+2$$ or
$$x^2+(a-2)x+4>0,$$ for which we need
$$(a-2)^2-4\cdot1\cdot4<0.$$
Also, we need $$x^2-ax-2+3x^2-3x+3>0$$ or
$$4x^2-(a+3)x+1>0,$$ for which we need $$(a+3)^2-4\cdot4\cdot1<0.$$
Now, solve this system.
I got $$-2<a<1.$$
|
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|
A proof regarding floor I have been trying to prove that:
$$x \geqslant 1 \implies \text{floor}(x) \geqslant \frac{x}{2}$$
I know this claim is true because I have compared the graph of $x/2$ and $\text{floor}(x)$. However, I just can't seem to get the right intuition has to how to go about proving this. I'm not asking for the proof. I'm asking for a hint, or maybe something that will push me in the right direction.
|
Either $\lfloor x \rfloor = 2k$ or $\lfloor x \rfloor = 2k+1$ for some integer $k \ge 0$.
CASE $1. \quad$ $x = 2k + \epsilon$ where $0 \le \epsilon < 1$
\begin{align}
\lfloor x \rfloor \ge \dfrac x2
&\iff 2k \ge k + \dfrac{\epsilon}{2} \\
&\iff k \ge \dfrac{\epsilon}{2} \\
&\iff \text{$(k=\epsilon = 0)$ or $(k \ge 1)$} \\
&\iff \text{$(x=0)$ or
$(x = 2k + \epsilon$ where $k \ge 1$ and $0 \le \epsilon < 1)$}
\end{align}
CASE $2. \quad$ $x = 2k + 1 + \epsilon$ where $0 \le \epsilon < 1$
Note that $0 \le \epsilon < 1
\iff -\dfrac 12 \le \dfrac{\epsilon}{2} - \dfrac 12 < 0$
\begin{align}
\lfloor x \rfloor \ge \dfrac x2
&\iff 2k + 1 \ge k + \dfrac 12 + \dfrac{\epsilon}{2} \\
&\iff k \ge \dfrac{\epsilon}{2} - \dfrac 12 \\
&\iff \text{$(x = 2k + 1 + \epsilon$
where $k \ge 0$ and $0 \le \epsilon < 1)$}
\end{align}
So $ \lfloor x \rfloor \ge \dfrac x2$ iff $x=0$ or $ \lfloor x \rfloor \ge 1$
|
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|
Prime factorization of integers Find the prime factorization of the following integers:
$e) 2^{30} -1$
Click here for the solutions
I used the formulas:
$a^2-1^2=(a-1)(a+1)$
$a^3+b^3=(a+b)(a^2-ab+b^2)$
$a^3-b^3=(a-b)(a^2+ab+b^2)$
And I became:
$2^{30} -1$
$=(2^{15}-1)(2^{15}+1)$
$=(2^5-1)(2^{10}+2^5+1)(2^5+1)(2^{10}-2^5+1)$
$=31*1057*32*993$
$=2^5*3*7*31*151*331$
What did I do wrong?
|
a^n - b^n = (a - b)(a^(n-1) + a^(n-2)b + a^(n-3)b^2 + ... + ab^(n-2) + b^(n-1)) - just prove it.
So, a^n - 1 = (a - 1)(a^(n-1) + a^(n-2) + ... + a + 1).
May be it can help you
|
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|
Solving Geometric Progression. Question
$\text{While solving recurrence relation ,}$
$$T(n)=T(\sqrt{n})+n$$
I got stuck in following Geometric series.
$$T(n)=n^{\frac{1}{2^{0}}}+n^{\frac{1}{2^{1}}}+n^{\frac{1}{2^{2}}}+...n^{\frac{1}{2^{k-1}}}$$
where $k=\log \log n$
Please help me out.
|
We must specify the numbers $n$ for which the recurrence holds (it cannot hold for $n=1$).
Suppose there is a real number $B$ and a function $T:[\sqrt{2},\infty)\rightarrow\mathbb{R}$ such that the following two properties hold:
1) $|T(x)|\leq B$ for all $x \in [\sqrt{2},2)$.
2) $T(x) = T(\sqrt{x}) + x$ for all $x \geq 2$.
Lemma
a) All real numbers $x \geq 2$ can be written $x=z^{2^k}$ for some real number $z \in [\sqrt{2},2)$ and some integer $k \in \{1, 2, 3, \ldots\}$.
b) If $z \in [\sqrt{2},2)$ and $n \in \{1, 2, 3, ...\}$, we have the equality:
$$ \boxed{T(z^{2^n}) = T(z) + z^{2^1} + z^{2^2} + ... + z^{2^n}} $$
c) The function $T$ satisfies
$$ \boxed{-B + x \leq T(x) \leq B + x + \left[\frac{\sqrt{2}}{\sqrt{2}-1}\right]\sqrt{x} \quad, \forall x \geq 4}$$
and so
$$ \lim_{x\rightarrow\infty} \frac{T(x)}{x} = 1 $$
Proof
Proof of (a): Fix $x \geq 2$. Take repeated square roots until the result first decreases below 2, that is, until we find a positive integer $k$ such that $x^{(1/2)^k}<2$ but $x^{(1/2)^{k-1}}\geq 2$. Define $z=x^{(1/2)^k}$ and note that $\sqrt{2}\leq z < 2$. $\Box$
Proof of (b): Fix $z \in [\sqrt{2},2)$. Then $z^2\geq 2$ and from the recurrence relation we get
\begin{align}
T(z^{2^1}) &= T(z) + z^{2^1} \\
T(z^{2^2}) &= T(z^{2^1}) + z^{2^2} = T(z) + z^{2^1} + z^{2^2} \\
T(z^{2^3}) &= T(z^{2^2}) + z^{2^3} = T(z) + z^{2^1} + z^{2^2}+ z^{2^3}
\end{align}
And so on. So for any $z \in [\sqrt{2},2]$ and any positive integer $n$ we have:
$$T(z^{2^n}) = T(z) + z^{2^1} + z^{2^2} + ... + z^{2^n} $$
This proves part (b). $\Box$
Proof of (c): Fix $x \geq 4$. Then $\sqrt{x}\geq 2$ and from part (a) we know $\sqrt{x}=z^{2^k}$ for some real number $z \in [\sqrt{2},2)$ and some integer $k \in \{1, 2, 3, ...\}$. Hence, $x =z^{2^{k+1}}$. Define $n=k+1$, so that $x=z^{2^n}$, and note that $n\geq 2$. We know from part (b) that
$$T(z^{2^n}) = T(z) + z^{2^1} + z^{2^2} + ... + z^{2^n} \quad (Eq. 1)$$
Thus:
\begin{align}
T(z) + z^{2^n} &\leq T(z^{2^n}) \quad \mbox{[from (Eq. 1)]} \\
&= T(z) + z^{2^n} + (z^{2^1} + ... + z^{2^{n-1}}) \quad \mbox{[from (Eq. 1)]}\\
&\leq T(z) + z^{2^n} + \sum_{i=2}^{2^{n-1}} z^i \quad \mbox{[since this fills in the missing $z^i$ terms]}\\
&= T(z) + z^{2^n} + z^2\frac{(z^{2^{n-1}-1}-1)}{z-1} \\
&\leq T(z) + z^{2^n} + \left[\frac{z}{z-1} \right]z^{2^{n-1}}\\
&\leq T(z) + z^{2^n} +\left[\sup_{y\in [\sqrt{2},2]}\frac{y}{y-1}\right] z^{2^{n-1}} \\
&= T(z) + z^{2^n} +\left[\frac{\sqrt{2}}{\sqrt{2}-1}\right]\sqrt{z^{2^n}}
\end{align}
Since $-B\leq T(z) \leq B$ we obtain
$$ -B + z^{2^n} \leq T(z^{2^n}) \leq B + z^{2^n} + \left[\frac{\sqrt{2}}{\sqrt{2}-1}\right]\sqrt{z^{2^n}} $$
Substituting $x=z^{2^n}$ gives the result. $\Box$
|
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|
A special improper integral I need to evaluate the following integral:
$$\int_{0}^\infty dx \left(\frac{x^2}{2x^2+1}\right)^a\left(\frac{x^2}{2x^2+t_y^2}\right)^b\left(\frac{x^2}{2x^2+t_z^2}\right)^c\frac{1}{x^3}$$
where, $a$, $b$, $c$, $t_y$ and $t_z$ are real.
I have succeeded to perform this integral only for two special cases:
1. $t_y$ = $t_z$ =1 and 2. $t_z$ = $t_y$
I shall be very thankful if someone could suggest a way to perform the integral above for $t_z \neq t_y$. Looking forward to hearing.
|
$$\int_0^\infty\left(\dfrac{x^2}{2x^2+1}\right)^a\left(\dfrac{x^2}{2x^2+t_y^2}\right)^b\left(\dfrac{x^2}{2x^2+t_z^2}\right)^c\dfrac{1}{x^3}~dx$$
$$=-2^{-a-b-c}\int_0^\infty\left(\dfrac{1}{1+\dfrac{1}{2x^2}}\right)^a\left(\dfrac{1}{1+\dfrac{t_y^2}{2x^2}}\right)^b\left(\dfrac{1}{1+\dfrac{t_z^2}{2x^2}}\right)^c~d\left(\dfrac{1}{2x^2}\right)$$
$$=2^{-a-b-c}\int_0^\infty\left(1+x\right)^{-a}\left(1+t_y^2x\right)^{-b}\left(1+t_z^2x\right)^{-c}~dx$$
$$=2^{-a-b-c}\int_0^1\left(1+\dfrac{u}{1-u}\right)^{-a}\left(1+\dfrac{t_y^2u}{1-u}\right)^{-b}\left(1+\dfrac{t_z^2u}{1-u}\right)^{-c}~d\left(\dfrac{u}{1-u}\right)$$
$$=2^{-a-b-c}\int_0^1(1-u)^{a+b+c-2}(1+(t_y^2-1)u)^{-b}(1+(t_z^2-1)u)^{-c}~du$$
$$=\dfrac{2^{-a-b-c}F_1(1,b,c,a+b+c;1-t_y^2,1-t_z^2)}{a+b+c-1}~\text{when}~0<t_y,t_z<\sqrt2$$
(according to http://en.wikipedia.org/wiki/Appell_series#Integral_representations)
$$=2^{-a-b-c}\int_0^1u^{a+b+c-2}((t_y^2-1)+(2-t_y^2)u)^{-b}((t_z^2-1)+(2-t_z^2)u)^{-c}~du$$
$$=\dfrac{2^{-a-b-c}}{(t_y^2-1)^b(t_z^2-1)^c}\int_0^1u^{a+b+c-2}\left(1-\dfrac{(t_y^2-2)u}{t_y^2-1}\right)^{-b}\left(1-\dfrac{(t_z^2-2)u}{t_z^2-1}\right)^{-c}~du$$
$$=\dfrac{2^{-a-b-c}}{(a+b+c-1)(t_y^2-1)^b(t_z^2-1)^c}F_1\left(a+b+c-1,b,c,a+b+c;\dfrac{t_y^2-2}{t_y^2-1},\dfrac{t_z^2-2}{t_z^2-1}\right)~\text{when}~t_y,t_z\geq\sqrt2$$
(according to http://en.wikipedia.org/wiki/Appell_series#Integral_representations)
|
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|
Ordinary Generating Function for the number of solution :$ x_1 + x_2 + \cdot\cdot\cdot + x_k = n$ Let $a_n$ be the number of solutions of the equation:
$$x_1 + x_2 + \cdot\cdot\cdot + x_k = n$$
where $x_i$ is the positive odd integer.
Hereto I would like to find the ordinary generating function for the sequence $a_n$
ODG has is a function satisfies the below :
$$G(a_n;n) = \sum_0^\infty a_nx^n$$
But what is the typical algorithm to find the ordinary generating function that uniquely determines its coefficient that corresponds to the regarding combinatorial situation?
Anyone can guide me where to start from?
|
The generating function is
$$
\begin{align}
&\left(\sum_{j=0}^\infty x^{2j+1}\right)^k\tag1\\
&=\left(\frac{x}{1-x^2}\right)^k\tag2\\
&=x^k\sum_{j=0}^\infty(-1)^j\binom{-k}{j}x^{2j}\tag3\\
&=\sum_{j=0}^\infty\binom{k+j-1}{j}x^{2j+k}\tag4\\
&=\sum_{n=k}^\infty\binom{\frac{n+k-2}2}{\frac{n-k}2}x^n\ [2|(n-k)]\tag5
\end{align}
$$
Explanation:
$(1)$: the exponent of $x$ chosen for each factor
$\phantom{(1)\text{: }}$represents the value chosen for each summand
$(2)$: sum the geometric series
$(3)$: apply the Binomial Theorem
$(4)$: $\binom{-n}{k}=(-1)^k\binom{n+k-1}{k}$ see this answer
$(5)$: $n=2j+k\implies j=\frac{n-k}2\ [2|(n-k)]$ where $[\cdots]$ are Iverson brackets
In step $(1)$, the exponent of $x$ chosen for each factor represents the value chosen for each summand. For example, if $k=4$,
$$
\scriptsize\left(\color{#CCC}{x^1+x^3+}x^5\color{#CCC}{+x^7+\cdots}\right)\left(\color{#CCC}{x^1+}x^3\color{#CCC}{+x^5+x^7+\cdots}\right)\left(\color{#CCC}{x^1+x^3+}x^5\color{#CCC}{+x^7+\cdots}\right)\left(\color{#CCC}{x^1+x^3+x^5+}x^7\color{#CCC}{+\cdots}\right)
$$
represents the sum $5+3+5+7=20$, and
$$
\scriptsize\left(\color{#CCC}{x^1+x^3+}x^5\color{#CCC}{+x^7+\cdots}\right)
\left(x^1\color{#CCC}{+x^3+x^5+x^7+\cdots}\right)
\left(\color{#CCC}{x^1+x^3+x^5+}x^7\color{#CCC}{+\cdots}\right)\left(\color{#CCC}{x^1+x^3+x^5+}x^7\color{#CCC}{+\cdots}\right)
$$
represents the sum $5+1+7+7=20$.
|
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Coefficient of $x^{n-2}$ in $(x-1)(x-2)(x-3)\dotsm(x-n)$ Question
Find the coefficient of $x^{n-2}$ in the expression $$(x-1)(x-2)(x-3)\dots(x-n)~~.$$
My approach
The coefficient of $x^n$ is $1$. The coefficient of $x^{n-1}$ is $- \frac{n(n+1)}{2}$
But I cannot proceed from here.
I would appreciate any help.
|
Taking the long way:
\begin{align}
f_{n}(x) &= (x-1)(x-2)\cdots(x-n) \\
f_{1}(x) &= x-1 \\
f_{2}(x) &= x^2 - 3x + 2 \\
f_{3}(x) &= x^3 - 6x^2 + 11x - 6\\
f_{4}(x) &= x^4 - 10x^3 + 35x^2 = 50x + 24 \\
f_{5}(x) &= x^5 - 15x^4 + 85x^3 - 215x^2 + 274x - 120.
\end{align}
From here it is determined that $[x^n] \, f_{n}(x) = 1$, $[x^{n-1}] \, f_{n}(x) = -\binom{n+1}{2} = s(n+1,n)$ and $[x^{n-2}] \, f_{n}(x) \in \{ 2, 11, 35, 85, 175, \cdots \}$. This pattern follows the (signed) Stirling numbers of the first kind, $s(n+2,n)$ or
$$[x^{n-2}] \, f_{n}(x) = \frac{(n-1) (n) (n+1) (3 n + 2)}{4!} = s(n+1, n-1).$$
One may compare the values obtained to A000914.
|
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|
Values of $a,b,c$ such that $\lim_{x \to 0}\frac{x(a+b-\cos x)-c\sin x}{x^5}=1$ Find the values of $a,b,c$ such that $$\lim_{x \to 0}\frac{x(a+b-\cos x)-c\sin x}{x^5}=1$$
Here's what I have got so far
Using L'Hospital's rule,
$$\lim_{x \to 0}\frac{a+b-\cos x+x\sin x-c\cos x}{5x^4}=1$$
So,$a+b-1=0$
Again,
$$\lim_{x \to 0}\frac{\sin x+\sin x+x\cos x+c\sin x}{20x^3}=1$$
But the solution is given as $a=120,b=60,c=180$
There is no way that $a+b=1$
|
Since $$\lim_{x \to 0}\frac{x(a+b-\cos x)-c\sin x}{x^5}=1$$ is given, we need
$$(x(a+b-\cos x)-c\sin x)'_{x=0}=0$$ or
$$(a+b-\cos{x}+x\sin{x}-c\cos{x})_{x=0}=0$$ or
$$a+b-c-1=0.$$
Thus, $$\lim_{x \to 0}\frac{a+b-(1+c)\cos{x}+x\sin{x}}{5x^4}=1.$$
Hence, we need $$(a+b-(1+c)\cos{x}+x\sin{x})'_{x=0}=0$$ or
$$((1+c)\sin{x}+\sin{x}+x\cos{x})_{x=0}=0,$$ which is true, which says
$$\lim_{x \to 0}\frac{(2+c)\sin{x}+x\cos{x}}{20x^3}=1$$ and from here
$$((2+c)\sin{x}+x\cos{x})'_{x=0}=0$$ or
$$\lim_{x\rightarrow0}\frac{(3+c)\cos{x}-x\sin{x}}{60x^2}=1,$$ where $c=-3$ and we get $-\frac{1}{60}=1$, which is wrong.
|
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|
Is there a matrix that is NOT I, but also has eigenvalue of 1? I am looking for a matrix that is not I, but also has eigenvalue of 1. Are there any? Can someone please show me an example and how it has eigenvalue 1.
|
Here is a simple example: set
$D = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}; \tag 1$
we have
$D \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}, \tag 2$
and
$D \begin{pmatrix} -1 \\ 1 \end{pmatrix} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{pmatrix} -1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \end{pmatrix} = -\begin{pmatrix} -1 \\ 1 \end{pmatrix}; \tag 3$
we see from (2) and (3) that the eigenvalues of $D$ are $\pm 1$. We can also find the eigenvalues of $D$ by evaluating the characteristic polynomial $p_D(\mu)$:
$p_D(\mu) = \det(D - \mu I) = \det \left ( \begin{bmatrix} -\mu & 1 \\ 1 & -\mu \end{bmatrix} \right ) = \mu^2 -1; \tag 4$
the roots of
$p_D(\mu) = \mu^2 - 1 \tag 5$
are $\mu = \pm 1$, in agreement with (2)-(3).
|
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|
Find, with proof, all integers $k$ satisfying the equation Find, with proof, all integers satisfying the equation
\begin{equation}
\frac{k-15}{2000} + \frac{k - 12}{2003} + \frac{k - 9}{2006} + \frac{k - 6}{2009} + \frac{k - 3}{2012} = \frac{k - 2000}{15} + \frac{k - 2003}{12} + \frac{k - 2006}{9} + \frac{k - 2009}{6} + \frac{k - 2012}{3}.
\end{equation}
This is a problem from a math competition, and here is the solution that is provided:
I can see how $k = 2015$ can be found via inspection; however, the value of $a$ is not as intuitive to me. Could someone either provide an alternate explanation, or clarify how to obtain the value of $a$?
Additionally, how is the inequality $\frac{1}{3} - \frac{5}{2000} > 0$ achieved?
|
They are just moving all the terms including $k$ to the right and all the ones that are constants to the left to make $b$. The term $\frac {k-2000}{15}$ becomes $\frac 1{15}k-\frac {2000}{15}$ and the second half goes to the right. The $\frac 1{15}$ in the expression for $a$ comes from the first term on the right, the $\frac 1{12}$ from the second, and so on. All the negative ones came from the left side.
|
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|
Common tangent of two ellipses Find the common tangent to: $4(x-4)^2 +25y^2 = 100$ and $4(x+1)^2 +y^2 = 4$.
I have found the derivatives of the above two equations:
$\dfrac{dy}{dx}=\dfrac{16-4x}{25y}$ and $\dfrac{dy}{dx}=\dfrac{-(4x+4)}{y}$
What do I do next?
|
Attempt:
1)$\dfrac{(x-4)^2}{5^2} + \dfrac{y^2}{2^2}=1;$
2) $(x+1)^2 +\dfrac{y^2}{2^2} =1;$
Draw them.
1) Major axis $5$, minor axis $2.$
Centred at $(4,0).$
2) Major axis $2$, minor axis $1.$
Centred at $(-1,0).$
The only common tangents :
1) At $(4,2)$ for ellipse $1.$
2) At $(-1,2)$ for ellipse $2.$
$y=2$;
Check your $dy/dx$ at these points.
Can you find the other common tangent?
|
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|
How can I calculate the limit $\lim\limits_{x \to 7} \frac{\sqrt{x+2} - \sqrt[3]{x+20}}{\sqrt[4]{x+9} - 2}$ without L'Hospital's rule? I have a problem with calculation of the limit:
$$\lim\limits_{x \to 7} \frac{\sqrt{x+2} - \sqrt[3]{x+20}}{\sqrt[4]{x+9} - 2}$$
Is there a way to calculate it? How can I do it?
|
Let $f(x)=\sqrt{x+2}-\sqrt[3]{x+20}$ and $g(x)=\sqrt[4]{x+9}-2$, then
\begin{align*}
\lim_{x\rightarrow 7}\dfrac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}&=\lim_{x\rightarrow 7}\dfrac{f(x)-f(7)}{x-7}\lim_{x\rightarrow 7}\dfrac{x-7}{g(x)-g(7)}\\
&=f'(7)\times\dfrac{1}{g'(7)}.
\end{align*}
Alternatively, if we let
\begin{align*}
\dfrac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}&=\dfrac{\dfrac{u^{1/2}-v^{1/2}}{u^{1/3}+u^{1/6}v^{1/6}+v^{1/3}}}{\dfrac{x-7}{(x+9)^{1/4}+2}\dfrac{1}{(x+9)^{1/2}+4}},
\end{align*}
where $u=(x+2)^{3}$, $v=(x+20)^{2}$, now $u^{1/2}-v^{1/2}=\dfrac{u-v}{u^{1/2}+v^{1/2}}$, where $u-v=(x-7)(x^{2}+12x+56)$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Error in $\int\limits_0^{\infty}dx\,\frac {\log^2 x}{a^2+x^2}$ This is my work for solving the improper integral$$I=\int\limits_0^{\infty}dx\,\frac {\log^2x}{x^2+a^2}$$I feel like I did everything write, but when I substitute values into $a$, it doesn’t match up with Wolfram Alpha.
First substitute $x=\frac {a^2}u$ so$$\begin{align*}I & =-\int\limits_{\infty}^0du\,\frac {2\log^2a-\log^2 u}{x^2+a^2}\\ & =\int\limits_0^{\infty}du\,\frac {2\log^2a}{a^2+x^2}-\int\limits_0^{\infty}du\,\frac {\log^2 u}{a^2+x^2}\end{align*}$$Hence$$\begin{align*}I & =\int\limits_0^{\infty}du\,\frac {\log^2a}{a^2+x^2}\\ & =\frac {\log^2a}{a^2}\int\limits_0^{\pi/2}dt\,\frac {a\sec^2t}{1+\tan^2t}\\ & =\frac {\pi\log^2a}{2a}\end{align*}$$However, when $a=e$ Wolfram Alpha evaluates the integral numerically as$$I\approx 2.00369$$however the input that I arrived at evaluates numerically$$\frac {\pi}{2e}\approx0.5778$$Where did I go wrong? And how would you go about solving this integral?
|
Here is one way to evaluate your integral (there are no doubt other ways).
Let
$$I = \int^\infty_0 \frac{\ln^2 x}{a^2 + x^2} \, dx, \quad a > 0.$$
Setting $x = \dfrac{a}{u}$ then $dx = -\dfrac{a}{u^2} \, du$ and we have
\begin{align*}
I &= a \int^\infty_0 \frac{\ln^2 \left (\frac{a}{u} \right )}{\frac{a^2}{u^2} + a^2} \cdot \frac{du}{u^2}\\
&= \frac{1}{a} \int^\infty_0 \frac{\ln^2 \left (\frac{a}{u} \right )}{1 + u^2} \, du\\
&= \frac{1}{a} \int^\infty_0 \frac{\left [\ln a - \ln u \right ]^2}{1 + u^2} \, du\\
&= \frac{\ln^2 a}{a} \int^\infty_0 \frac{du}{1 + u^2} - \frac{2 \ln a}{a} \int^\infty_0 \frac{\ln u}{1 + u^2} \, du + \frac{1}{a} \int^\infty_0 \frac{\ln^2 u}{1 + u^2} \, du\\
&= \frac{\ln^2 a}{a} I_1 - \frac{2 \ln a}{a} I_2 + \frac{1}{a} I_3.
\end{align*}
Now evaluating each of these integrals. The first is trival. We have
$$I_1 = \int^\infty_0 \frac{du}{1 + u^2} = \left [\tan^{-1} u \right]^\infty_0 = \frac{\pi}{2}.$$
For the second, setting $u = \dfrac{1}{y}, du = -\dfrac{1}{y^2} \, dy$. Thus
$$I_2 = \int^\infty_0 \frac{\ln u}{1 + u^2} \, du = \int^\infty_0 \frac{\ln \left (\frac{1}{y} \right )}{1 + \frac{1}{y^2}} \cdot \frac{dy}{y^2} = - \int^\infty_0 \frac{\ln y}{1 + y^2},$$
or $I_2 = 0$.
For the third integral we begin by dividing the interval of integration as follows
$$I_3 = \int^\infty_0 \frac{\ln^2 u}{1 + u^2} \, du = \int^1_0 \frac{\ln^2 u}{1 + u^2} \, du + \int^\infty_1 \frac{\ln^2 u}{1 + u^2} \, du.$$
If in the rightmost integral we set $u = \dfrac{1}{y}, du = -\dfrac{1}{y^2} \, dy$, then
\begin{align*}
I_3 &= \int^1_0 \frac{\ln^2 u}{1 + u^2} \, du + \int^1_0 \frac{\ln^2 \left (\frac{1}{y} \right )}{1 + \frac{1}{y^2}} \cdot \frac{dy}{y^2}\\
&= \int^1_0 \frac{\ln^2 u}{1 + u^2} \, du + \int^1_0 \frac{\ln^2 y}{1 + y^2} \, dy\\
&= 2 \int^1_0 \frac{\ln^2 u}{1 + u^2} \, du.
\end{align*}
Noting the term $1/(1 + u^2)$ can be written as the sum of a geometric series, namely
$$\frac{1}{1 + u^2} = \sum^\infty_{n = 0} (-1)^n u^{2n}, \quad |u| < 1$$
replacing the term in the integral with this sum, after interchanging the integral sign with the summation we have
$$I_3 = 2 \sum^\infty_{n = 0} (-1)^n \int^1_0 u^{2n} \ln^2 u \, du.$$
After performing integration by parts twice, we are left with
$$I_3 = 4 \sum^\infty_{n = 0} \frac{(-1)^n}{(2n + 1)^3}.$$
From the definition for the Dirichlet beta function $\beta (s)$, namely
$$\beta (s) = \sum^\infty_{n = 0} \frac{(-1)^n}{(2n + 1)^s},$$
we recognise our sum as corresponding to $\beta (3)$. Thus
$$I_3 = 4 \beta (3).$$
The value for $\beta (3)$ can be readily found (see here for example). It is
$$\beta (3) = \frac{\pi^3}{32},$$
giving
$$I_3 = 4 \cdot \frac{\pi^3}{32} = \frac{\pi^3}{8}.$$
So we finally have
$$I = \frac{\ln^2 a}{a} \cdot \frac{\pi}{2} + \frac{1}{a} \cdot \frac{\pi^3}{8},$$
or
$$\int^\infty_0 \frac{\ln^2 x}{a^2 + x^2} \, dx = \frac{\pi}{2a} \left (\ln^2 a + \frac{\pi^2}{4} \right ).$$
|
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|
Expectation of the Minimum of 2 random variables Let Z = min{|X|,|Y|}, with $X , Y \sim Normal(0,1)$.
Then, show that $E(Z) = \frac{2(\sqrt{2} - 1)}{\sqrt{\pi}}$
So far I got
$F_{Z}(z) = 1-[1 - F_{|X|}(z)][1-F_{|Y|}(z)]$
$F_{Z}(z) = 1-4F_{X}(-z)F_{Y}(-z)$
$F_{Z}(z) = 1-\frac{4}{2\pi}\int_{-\infty}^{-z}\int_{-\infty}^{-z}e^{-\frac{x^{2}+y^{2}}{2}}dxdy$
But I do not know where to go from here.
|
We know that $F_{|X|}(x)=Pr(-x\leq{X}\leq{x})=F_{X}(x)-F_{X}(-x)$
Hence, $f_{|X|}(x)=f_{X}(x)+f_{X}(-x)=2f_{X}(x)$
Therefore,
$f_{|X|}(x)=\sqrt{\frac{2}{\pi}}e^{\frac{-x^{2}}{2}}$ and similarly,
$f_{|Y|}(y)=\sqrt{\frac{2}{\pi}}e^{\frac{-y^{2}}{2}}$
Then, by symmetry of Normal curve, we can simplify to:
$E[min(|X|,|Y|)] = \frac{2}{\pi}(\int_{x=0}^{\infty}\int_{y=0}^{x}ye^{\frac{-x^{2}}{2}}e^{\frac{-y^{2}}{2}}dydx + \int_{y=0}^{\infty}\int_{x=0}^{y}xe^{\frac{-x^{2}}{2}}e^{\frac{-y^{2}}{2}}dxdy)$
$E[min(|X|,|Y|)] = \frac{4}{\pi}\int_{x=0}^{\infty}\int_{y=0}^{x}ye^{\frac{-x^{2}}{2}}e^{\frac{-y^{2}}{2}}dydx$
$E[min(|X|,|Y|)] = \frac{4}{\pi}\int_{x=0}^{\infty}e^{\frac{-x^{2}}{2}}(1-e^{\frac{-x^{2}}{2}})dx$
$E[min(|X|,|Y|)] = \frac{4}{\pi}\int_{x=0}^{\infty}(e^{\frac{-x^{2}}{2}}-e^{-x^{2}})dx$
$E[min(|X|,|Y|)] = \frac{4}{\pi}(\frac{1}{2})\int_{0}^{\infty}(\sqrt{2\pi}\frac{1}{\sqrt{2\pi}}e^{\frac{-x^{2}}{2}}-\sqrt{\pi}\frac{1}{\sqrt{\pi}}e^{-x^{2}})dx$
Since, we have 2 pdf's:
$Normal(0,1) : \frac{1}{\sqrt{2\pi}}e^{\frac{-x^{2}}{2}}$
$Normal(0,\frac{1}{2}) : \frac{1}{\sqrt{\pi}}e^{-x^{2}}$
Thus,
$E[min(|X|,|Y|)] = \frac{2}{\pi}(\sqrt{2\pi}-\sqrt{\pi})$
$E[min(|X|,|Y|)] = \frac{2({\sqrt{2}-1})}{\sqrt{\pi}}$
Hence, proved.
|
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|
Find $x$ such that $2^x+3^x-4^x+6^x-9^x=1$ The question:
Find values of $x$ such that $2^x+3^x-4^x+6^x-9^x=1$, $\forall x \in \mathbb R$.
Notice the numbers $4$, $6$ and $9$ can be expressed as powers of $2$ and/or $3$. Hence let $a = 2^x$ and $b=3^x$.
\begin{align}
1 & = 2^x+3^x-4^x+6^x-9^x \\
& = 2^x + 3^x - (2^2)^x + (2\cdot3)^x-(3^2)^x\\
& = 2^x + 3^x - (2^x)^2 + 2^x\cdot3^x-(3^x)^2 \\
& = a+b-a^2+ab-b^2 \\
0 & = a^2-ab+b^2-a-b+1
\end{align}
\begin{align}
0 & = a^2-ab+b^2-a-b+1 \\
& = 2a^2-2ab+2b^2-2a-2b+2 \\
& = (a^2-2ab+b^2)+(a^2-2a+1)+(b^2-2b+1) \\
& = (a-b)^2 + (a-1)^2 + (b-1)^2
\end{align}
This is where I am stuck. I am convinced that this factorisation could help solve the question, but I don't know how. Also, once we find values for $x$, we must prove that there are no further values of $x$. Could someone complete the question?
|
If the sum of a finite number of non-negative expressions is $0$, each of them has to be zero.
In other words, when $a,b,c\geq0$
$a+b+c=0 \implies a=b=c=0$
You have done the hard part by showing that it can be written as the sum of three squares.
This means that $a=b$, $a=1$ and $b=1$. What does that tell you about $x$?
|
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|
The ± sign in square root As I was doing a SAT question when I came across this question:
$\sqrt {x-a} = x-4$
If $a=2$,what is the solution set of the equation?
Options
*
*{$3,6$}
*{$2$}
*{$3$}
*{$6$} Correct Answer
I evaluated the equation and got $0=(x-3)(x-6)$If you put those number in the equation, you should get:
For 3:
$\sqrt {3-2} = 3-4$
Since $\sqrt {1} = ±1$
$±1 = -1$
For 6:
$\sqrt {6-2} = 6-4$
Since $\sqrt {4} = ±2$
$±2 = 2$
For the answer, they(SAT) evaluated $\sqrt {1}$ as $\sqrt {1} = -1$ and $\sqrt {4}$ as $\sqrt {4} = 2$ Why is it that $\sqrt {1}$ is equal to $-1$ and not $1$ and why $\sqrt {4}$ is equal to $2$ and not $-2$ Why isn't the solution set {$3,6$} a correct answer?
|
You factored it correctly, but your mistake is considering both roots of $x$. When we say $\sqrt{x}$, we refer to the principal root of $x$; that is, the $positive$ square root of $x$ (issue in red, corrections in blue):
For 3:
$\sqrt {3-2} = 3-4$
$\sqrt {1} = \color{red}{±}1$
$\color{blue}{\sqrt {1} = -1}$
$\color{blue}{1 = -1 \, \, \, \, \, \, \, \text{false; not in solution set}}$
For 6:
$\sqrt {6-2} = 6-4$
$\sqrt {4} = \color{red}{±}2$
$\color{blue}{\sqrt {4} = 2}$
$\color{blue}{2 = 2 \, \, \, \, \, \, \, \text{true; in solution set}}$
The extraneous solution $x=3$ was brought about when you squared the radical, because squaring removes the restriction that $\sqrt{u} \geq 0$.
|
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|
Finding the modulo class of $n \mod 7$ from that of $n^2 + n^3 \equiv 0\pmod 7$ The remainders of $n\mod 7$ will have the set of equivalence classes with residue:
$\{0,1,2,3,4,5,6\}$.
The remainders of $n^2\mod 7$ will have the set of equivalence classes with residue: $\{0,1,4,2\}$.
The remainders of $n^3\mod 7$ will have the set of equivalence classes with residue: $\{0,1,6\}$.
Regarding the $n^2 + n^3 \equiv 0\pmod 7$, I have taken the following approach:
Simply pairwise add the two sets given for $n^2$ and $n^3$. The pairwise combinations that add up to $7$ and hence lead to $ \equiv 0\pmod 7$ are :
(i) $1,6$
So, $n^2$ should have residue $1$; while $n^3$ should have residue $6$.
For, $n^2$ the values of n that satisfy are: $1, 6$
For, $n^3$ the values of n that satisfy are: $3, 5, 6$
So, the value of $n = 6k$ $ \forall k \in \mathbb {Z}$.
I am unable to have a theoretical basis for such pair-wise addition, and request the same; even if I am correct.
|
Since $7$ is prime and $n^2+n^3\equiv0\pmod{7}$ you have that either $n^2\equiv0\pmod{7}$ or $n+1\equiv0\pmod{7}$. Thus
$$
n\equiv0\pmod{7}
\qquad\text{or}\qquad
n\equiv6\pmod{7}
$$
|
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|
Find $\frac {\alpha}{\beta} + \frac {\beta}{\alpha}$ if $\alpha^2+3 \alpha+1=\beta^2+3\beta+1=0$ The question:
Let $\alpha$ and $\beta$ be $2$ distinct real numbers which such that $\alpha^2+3 \alpha+1=\beta^2+3\beta+1=0$. Find the value of $\frac {\alpha}{\beta} + \frac {\beta}{\alpha}$.
This problem is seems to be related to Vieta's Theorem, but so far I have not used it. This is my working out:
\begin{align}
\frac {\alpha}{\beta} + \frac {\beta}{\alpha} & = \frac {\alpha^2}{\alpha \beta} + \frac {\beta^2}{\alpha \beta} \\
& = \frac {\alpha^2+\beta^2}{\alpha \beta}
\end{align}
Vieta's Theorem:
Let $p(x)=ax^2+bx+c$ be a quadratic polynomial with zeros $\alpha$,$~\beta$. Then
$$\frac {-b}{a}=\alpha + \beta \\
\frac{c}{a} = \alpha \cdot \beta $$
Well, it is clear that the question wants us to find $\alpha^2 + \beta^2$ and $\alpha \beta$.
\begin{align}
\alpha^2+\beta^2 & = \alpha^2+2\alpha \beta + \beta^2 -2\alpha\beta \\
& = (\alpha+\beta)^2-2(\alpha\beta)
\end{align}
We have:
\begin{align}
\alpha^2 + 3\alpha + 1 & = \beta^2 + 3\beta + 1 \\
0 & =\alpha^2 - \beta^2+3\alpha-3\beta \\
& = (\alpha+\beta)(\alpha-\beta)+3(\alpha-\beta) \\
& = (\alpha-\beta)(\alpha+\beta+3) \\
\therefore~ \alpha + \beta & = -3\tag{reject $\alpha=\beta$}
\end{align}
This is where I am stuck because I am unable to find $\alpha\beta$ by algebraic manipulation. I have thought about trying to find $\alpha$ and $\beta$ through the quadratic formula, but it seems quite tedious, so it is a last resort. Is there a method to finish this question off?
|
$$\alpha^2+3 \alpha+1=\beta^2+3\beta+1=0$$
Implies that $\alpha$ and $\beta$ are the roots of $x^2 + 3x + 1$, thus:
$$x^2 + 3x + 1 = (x - \alpha)(x - \beta)$$
Expand $(x - \alpha)(x - \beta)$ to find $-\alpha - \beta = 3$ and $\alpha\beta = 1$.
|
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|
find $x$ where $x^3$ $\equiv$ $1\mod p$, given a primitive root modulo of $p$ How to efficiently find $x$ where $x^3$ $\equiv$ $1\mod p$, given a primitive root of $p$ in the range of $[0,p-1]$?
$p$ is a prime.
Is is true that it only has three roots?
So if one of the primitive roots is given, how to efficient found all such x?
|
COMMENT (posted because is long for write as a comment).- There are three roots if and only if $p=3m+1$ and only a root if and only if $p=3m+2$.
Since $x^{p-1}-1=0$, if $x^3-1=0$ has three roots then
$\dfrac{x^{p-1}-1}{x^3-1}\in\mathbb F_p$ and if only one root (obviously equal to $1$) then $\dfrac{x^{p-1}-1}{x^3-1}\notin\mathbb F_p$.
Examples.-In $\mathbb F_{13}$ one has $\dfrac{x^{12}-1}{x^3-1}=(x^6+1)(x^3+1)$ and $x=1,3,9$
In $\mathbb F_{11}$ one has $\dfrac{x^{10}-1}{x^3-1}=x^7+x^4+x+\dfrac{1}{x^2+x+1}\notin\mathbb F_{11}[x]$ and just $x=1$
|
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|
probability of rolling at least N of C when rolling X dice I have been looking this up but am struggling to find a straightforward answer.
I am still confused about the process.
So basically what is the probability of rolling at least three of a particular number say X when rolling five dice.
I believe the way to go is work out the probability of rolling 5 of the same number then add it too the probability of 4 of the same number and so on.
so 1/6^5 + 1/6^4 + 1/6^3 + 1/6^2.
Is this correct or am I leaving something out?
I originally phrased my question wrong asking the probability of getting x of the same number.
A good example of what I mean is
The probability of rolling at least three 2's when rolling five dice
|
First choose one to be shown $3$ times. There are $6\choose{1}$ ways to choose this die. Then use the standard binomial to get the probability of that selected die showing up exactly $3$ times.
$${6\choose{1}}\cdot{5\choose{3}}\cdot\frac{1}{6}^3\cdot\frac{5}{6}^2 \approx .193$$
And for the same number being shown exactly $4$ times:
$${6\choose{1}}\cdot{5\choose{4}}\cdot\frac{1}{6}^4\cdot\frac{5}{6}^1 \approx .0193$$
And for the same number being shown exactly $5$ times:
$${6\choose{1}}\cdot{5\choose{5}}\cdot\frac{1}{6}^5\cdot\frac{5}{6}^0 \approx 7.713\cdot10^{-4}$$
Then summing these $3$ results we get $p \approx 0.213$
For a fixed value:
We take away the $6\choose{1}$'s since we are no longer choosing $1$ of the $6$ die to be shown at least $3$ times. We have
$${5\choose{3}}\cdot\frac{1}{6}^3\cdot\frac{5}{6}^2 + {5\choose{4}}\cdot\frac{1}{6}^4\cdot\frac{5}{6}^1 + {5\choose{5}}\cdot\frac{1}{6}^5\cdot\frac{5}{6}^0 \approx 0.03549$$
|
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|
Coefficient of $x^{50}$ in the polynomial. What is the coefficient of $x^{50}$ in the polynomial
$[x+\binom{50}{0}][x+3\binom{50}{1}][x+5\binom{50}{2}].....[x+(2n+1)\binom{50}{50}]$
The solution given in one of the books was ,
We have ,
$[x+\binom{50}{0}][x+3\binom{50}{1}][x+5\binom{50}{2}].....[x+(2n+1)\binom{50}{50}]$
$=x^{51} +x^{50}[\binom{50}{0}+ 3\binom{50}{1} + 5\binom{50}{2} .... 101\binom{50}{50}$]
And from here they have found out the coefficient of $x^{50}$
However I am not able to understand, how is the coefficient of $x^{50}$ the sum of all of the binomial coefficients in the polynomial? Is there a general method to determine the coefficients of the second or the third highest degree variable of any polynomial using binomial theorem?
|
To get one $x^{50}$ term, you take the $x$ term (coefficient: one) from fifty sets of parentheses and a single constant term from the single remaining set of parentheses. Adding all of these up gives you the total coefficient of $x^{50}$.
As a simple analogy, consider the $x^2$ term in $(x+1)(x+3)(x+5)$.
You take the $x$ term from $(x+1)$, multiply it by the $x$ term from $(x+3)$ and finally by the constant term from $(x+5)$. That gives you $(x)(x)(5) = 5x^2$. Rotating through the possible combinations here, you get $5x^2 + 3x^2 + x^2 = 9x^2$. The same method is at play here.
|
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|
Prove that the roots of $ax^2+bx+c=0$, $a\neq 0$ and $a,b,c\in R$ will be real if $a(a+7b+49c)+c(a-b+c)<0$
Prove that the roots of $ax^2+bx+c=0$, $a\neq 0$ and $a,b,c\in \mathbb{R}$ will be real if $$a(a+7b+49c)+c(a-b+c)<0$$
My Attempt:
Given
\begin{align}
a(a+7b+49c)+c(a-b+c) &< 0 \\
49a \left( \dfrac {a}{49} + \dfrac {b}{7} + c \right)+c(a-b+c) &< 0 \\
49a \cdot f\left( \dfrac {1}{7} \right) + f(0)\cdot f(-1) &< 0
\end{align}
|
$$a(a+7b+49c)+c(a-b+c)\lt 0$$
is equivalent to
$$ac\lt \frac{-a^2-7ab+bc-c^2}{50}$$
So, we get
$$\begin{align}b^2-4ac&\gt b^2-\frac{4}{50}(-a^2-7ab+bc-c^2)\\\\&=\frac{1}{50}(50b^2+4a^2+28ab-4bc+4c^2)\\\\&=\frac{1}{50}\left(4\left(a+\frac{7b}{2}\right)^2+(b-2c)^2\right)\ge 0\end{align}$$
|
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|
Find a limit of a function W/OUT l'Hopital's rule. I've got an expression: $\lim_{x\to 0}$ $\frac {log(6-\frac 5{cosx})}{\sin^2 x}$
The question is: how to find limit without l'Hopital's rule?
|
Another way using Taylor series
$$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^5\right)$$
$$\frac 1 {\cos(x)}=1+\frac{x^2}{2}+\frac{5 x^4}{24}+O\left(x^5\right)$$
$$6-\frac 5 {\cos(x)}=1-\frac{5 x^2}{2}-\frac{25 x^4}{24}+O\left(x^5\right)$$
$$\log\left(6-\frac 5 {\cos(x)} \right)=-\frac{5 x^2}{2}-\frac{25 x^4}{6}+O\left(x^5\right)$$
$$\sin(x)=x-\frac{x^3}{6}+O\left(x^5\right)$$
$$\sin^2(x)=x^2-\frac{x^4}{3}+O\left(x^6\right)$$
$$\frac{\log\left(6-\frac 5 {\cos(x)} \right) } {\sin^2(x) }=\frac{-\frac{5 x^2}{2}-\frac{25 x^4}{6}+O\left(x^5\right) }{x^2-\frac{x^4}{3}+O\left(x^6\right) }=-\frac{5}{2}-5 x^2+O\left(x^3\right)$$ which shows the limit and also how it is approached.
|
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|
Prove $\tan(\frac{\alpha}{2})\tan(\frac{\beta}{2})=\frac{1}{5}$ Given $2\sin(\alpha)+2\sin(\beta)=3\sin(\alpha+\beta)$,
prove that $\tan(\frac{\alpha}{2})\tan(\frac{\beta}{2})=\frac{1}{5}$
Also we know that all the expressions are different from zero and defined. Including the expressions we received during the solution.
Tried to play with it, didn't seem to work for me.
|
Using the formulas of half-angle substitution, also called Weierstrass substitution formulas,
$$\cos(u)=\dfrac{1-t^2}{1+t^2} , \ \sin(u)=\dfrac{2t}{1+t^2} \ \ \ \text{where} \ \ \ t:=\tan(\tfrac{u}{2}),$$
one has to prove, setting $a=\tan(\alpha/2)$ and $b=\tan(\beta/2)$, that :
$$2\dfrac{2a}{1+a^2}+2\dfrac{2b}{1+b^2}=3 \dfrac{2a}{1+a^2}\dfrac{1-b^2}{1+b^2} +3 \dfrac{2b}{1+b^2}\dfrac{1-a^2}{1+a^2} \implies ab=\dfrac15$$
(we have used $\sin(\alpha+\beta)=\sin(\alpha)\cos(\beta)+\sin(\beta)\cos(\alpha)$).
The first expression, once we have multiplied both sides by $(1+a^2)(1+b^2)$ and cancelled by 2, gives:
$$2a+2ab^2+2b+2ba^2 = 3a-3a^2b+3b-3a^2b$$
$$\iff 5ab(a+b)=a+b$$
giving indeed, under the supplementary assumption that $a+b \neq 0$
$$ab=\tfrac15.$$
Remark: the overall advantage of formulas (*) is that they convert trigonometrical issues into fully algebraic formulations.
|
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|
How many solutions does this system of equations have? I have $2$ equations :
$$-\frac{a}{2}x-\frac{b}{2}+cx+d=x+2$$
$$-2ax-2b+cx+d=2x+1$$
with $a, b, c, x \neq 0$
We have to find all possible solutions of $a, b, c, d$ that make the equations true for all $x$.
I found one solution
$a=-\frac{2}{3}$,
$b=\frac{2}{3}$,
$c=\frac{2}{3}$,
$d=\frac{7}{3}$
But, I wonder if there are any other solutions to this system of equations?
|
Write the equations as:
$$
\begin{cases}
\begin{align}
(a-2c+2)x+b-2d+4 = 0 \\
(2a-c+2)x+2b-d+1 = 0
\end{align}
\end{cases}
$$
A polynomial is identical $0$ (i.e. for all $x$) iff all its coefficients are $0\,$, which gives the system to solve:
$$
\begin{cases}
\begin{align}
a - 2c + 2 = 0 \\
2a - c + 2 = 0 \\[7px]
b - 2d + 4 = 0 \\
2b -d + 1 = 0
\end{align}
\end{cases}
$$
The system has the unique solution as posted $\;a=-\frac{2}{3}$, $b=\frac{2}{3}$, $c=\frac{2}{3}$, $d=\frac{7}{3}\,$.
|
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|
Ratio of parts of an intersected segment in a rectangle
In rectangle $ABCD$, points $E$ and $F$ lie on sides $BC$ and $CD$ respectively. Point $F$ is the midpoint of $CD$ and $BE=\frac13BC$. Segments $AC$ and $FE$ intersect at point $P$. What is the ratio of $AP$ to $PC$? Express your answer as a common fraction.
I'm not sure where to start, do I need to use the fact that $FCE$ is a right triangle, or just the ratios, any help would be appreciated.
|
$A = (0,0)$
$B = (x,0)$
$C = (x,-y)$
$D = (0,-y)$
$E = (x,-y/3)$
$F = (x/2,-y)$
$P = (tx,-ty)$
$P = \lambda E + (1-\lambda)F$
$tx = \lambda x + (1-\lambda)x/2 = \lambda x + x/2 - \lambda x/2 = (\lambda + 1/2 - \lambda/2) x $
$t = \frac{(\lambda + 1)}{2}$
$-\frac{(\lambda + 1)}{2}y = \lambda (-y/3) + (1-\lambda)(-y) = \lambda (-y/3) + -y + \lambda y = (\frac{2 \lambda}{3} - 1) y$
$-\frac{(\lambda + 1)}{2} = (\frac{2 \lambda}{3} - 1) \text{ iff } \lambda = \frac{3}{7} \text{ iff } t = \frac{5}{7} \text{ iff } 1-t = \frac{2}{7}$
$\text{Length of } AP = \sqrt{ (\frac{5}{7}x)^2 + (\frac{5}{7}y)^2}$
$\text{Length of } BC = \sqrt{ (\frac{2}{7}x)^2 + (\frac{2}{7}y)^2}$
ANS: $\frac{Len(AP)}{Len(PC)} = \frac {\frac{5}{7}} {\frac{2}{7}} = \frac {5}{2} $
Note: From $[\lambda = \frac{3}{7}]$ and $[1 - \lambda = \frac{4}{7}]$ we also get $\frac{Len(FP)}{Len(PE)} = \frac{3}{4}$ 'free of charge'.
|
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|
Convergence of $\sum_{n=0}^{+ \infty} \frac{n^3 - 5n^2 + \pi}{2n^5 - \sqrt{3}}$
Does $$\sum_{n=0}^{+\infty} \frac{n^3 - 5n^2 + \pi}{2n^5 - \sqrt{3}}$$
converge?
My attempt:
$$\forall n \in \mathbb{N} \setminus\{0\}:\frac{n^3-5n^2 + \pi }{2n^5 -\sqrt{3}} = \frac{1}{n^2}\frac{1-5/n + \pi/n^2}{2- \sqrt{3}/n^5}$$
And because
$$\frac{1-5/n + \pi/n^2}{2- \sqrt{3}/n^5} \to \frac{1}{2}$$
there exists $n_0$ such that $\forall n > n_0: \left\vert \dfrac{1-5/n + \pi/n^2}{2- \sqrt{3}/n^5} - \dfrac{1}{2}\right\vert < 1$
Hence, for $n \geq n_0$:
$$ \frac{1-5/n + \pi/n^2}{2- \sqrt{3}/n^5} < \frac{3}{2}$$
Thus: $$\frac{n^3-5n^2 + \pi }{2n^5 -\sqrt{3}} < \frac{3}{2}\frac{1}{n^2}$$
and for $n_1$ sufficiently large, the sequence of terms is positive (as both the numerator and the denumerator converge to $+ \infty$, so for $n$ large enough the numerator and denumerator are positive, and hence the quotient is positive).
Let $N := \max\{n_0,n_1\}$
Then, because $\sum\dfrac{1}{n^2}$ converges, we conclude using the comparison test that the series
$$\sum_{n > N}^{+ \infty} \frac{n^3-5n^2 + \pi }{2n^5 -\sqrt{3}}$$
converges, and hence, the given series converges.
Is this correct? (I tried to explain every single step). Do you have any comments? Is there an easier approach?
|
your general term is equivalent to $1/n^2$, so it converges.
|
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|
Find the general term of the sequence, starting with n=1 Find the general term of the sequence, starting with n=1, determine whether the sequence converges, and if so find its limit.
$$\frac{3}{2^2 - 1^2}, \frac{4}{3^2 - 2^2} , \frac{5}{4^2 - 3^2}, \cdots$$
Can you help me with this,I know how to solve the problem with n= 0 but is that different with n= 1?
|
Observe the series first.
$$ \frac{1\color{green}{+2}}{(1\color{green}{+1})^2-(1)^2}, \frac{2\color{green}{+2}}{(2\color{green}{+1})^2-(2)^2}, \frac{3\color{green}{+2}}{(3\color{green}{+1})^2-(3)^2}, ....
\text{upto} \frac{n\color{green}{+2}}{(n\color{green}{+1})^2-(n)^2} $$
So now,
General Term or $ n^{th} $ term can be written as:
$$ \frac{(n+2)}{(n+1)^2-n^2} $$
$$ \frac{(n+2)}{(\require{cancel} \cancel{n^2}+2n+1-
\cancel{n^2}) }$$
$$ = \frac{(n+2)}{(2n+1)}$$
|
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|
$25! \pmod {78125}$ $25! \pmod{78125}$ is a problem I'm working on.
Since $78125$ looked very divisible by $5$, I checked, and found that $78125 = 5^7$.
Then I thought, if there are seven factors 5 in $25!$, then $25! \equiv 0 \mod 78125$, but I only found $25$ to be divisible by 5 six times, so I don't think that got me anywhere.
Am I even on the right track here?
Thanks in advance for any help!
|
If $\frac{25!}{5^6}\equiv k \pmod{5}$, then $5\mid\frac{25!}{5^6}-k$, i.e. $5^7\mid 25!-k\cdot 5^6$ so $25!\equiv k\cdot 5^6 \pmod{5^7}$. What remains is to find $k$.
For that, note $\frac{25!}{5^6}=(1\cdot 2 \cdot 3 \cdot 4) \cdot (6
\cdot 7 \cdot 8 \cdot 9) \cdot (11 \cdot 12 \cdot 13 \cdot 14) \cdot (16 \cdot 17 \cdot 18 \cdot 19) \cdot (21 \cdot 22 \cdot 23 \cdot 24)\cdot(\frac55 \cdot \frac{10}{5} \cdot \frac{15}{5} \cdot \frac{20}{5})\equiv (4!)^6\equiv (-1)^6=1 \pmod 5$ , so $k=1$.
That gives us the answer: $25!\equiv 5^6 \pmod {5^7}$.
|
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|
Prove: $(\frac{1+i\sqrt{7}}{2})^4+(\frac{1-i\sqrt{7}}{2})^4=1$ $$\left(\frac{1+i\sqrt{7}}{2}\right)^4+\left(\frac{1-i\sqrt{7}}{2}\right)^4=1$$
I tried moving the left exponent to the RHS to then make difference of squares exp. $(x^2)^2$. Didn't get the same on both sides though. Any help?
|
Yet another way: let $a=\frac{1+i\sqrt{7}}{2}$ and $b=\frac{1-i\sqrt{7}}{2}$ then $a+b=1, ab=2$ so $a,b$ are the roots of the quadratic $x^2-x+2=0\,$.
Therefore $a^2=a-2\,$, then successively:
*
*$a^3 = a\cdot a^2 = a \cdot(a-2) = a^2-2a = (a-2)-2a=-a-2$
*$a^4= a \cdot a^3 = a(-a-2) = -a^2-2a=-(a-2)-2a=-3a+2$
The same holds for $\,b\,$ so $\;a^4+b^4=(-3a+2)+(-3b+2)=-3(a+b)+4=-3\cdot 1+4=1\,$.
|
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|
$\lim_{x\to 0} \left(\frac{1}{1-\cos x}-\frac{2}{x^2}\right)$ Find the limits :
$$\lim_{x\to 0} \left(\frac{1}{1-\cos x}-\frac{2}{x^2}\right)$$
My Try :
$$\lim_{x\to 0} \left(\frac{1}{1-\cos x}-\frac{2}{x^2}\right)=\lim_{x\to 0}\frac{x^2-2(1-\cos x)}{x^2(1-\cos x)}$$
Now what do I do ?
|
Using series expansion:
$$\cos x = 1 - \frac{x^2}{2}+\frac{x^4}{24}+ O(x^6)$$
$$\frac{1}{1-\cos x}=\frac{1}{x^2}\frac{2}{1 -\frac{1}{12}x^2 + O(x^4)}$$
$$\frac{1}{1-\cos x}-\frac{2}{x^2}=\frac{2}{x^2}\left(\frac{ \frac{1}{12}x^2 + O(x^4) }{1 -\frac{1}{12}x^2 + O(x^4)}\right) \to \frac16$$
|
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|
Equation with integers $x$, $y$ If $x$, $y$ positive integers ($x<y$), how can I solve the equation $x+y=14\sqrt{xy-48}$ ?
|
Write $x^2 - 194xy + y^2$ in matrix form:
$$
\pmatrix{x & y}
\pmatrix{ 1 & -97 \\ -97 & 1}
\pmatrix{x \\ y}
$$
This matrix has eigenvectors $(1,\pm 1)$, which leads us to consider $u=x+y$ and $v=x-y$.
Then $x^2 - 194xy + y^2 + 9408=0$ becomes $49 v^2 - 48 u^2+ 9408=0$.
Since $9408 =2^6×3×7^2$, this implies that $u=7u_1$ and $v=12v_1$ and so $3v_1^2-u_1^2+4=0$, a Pell equation.
|
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|
Proving whether the series $\sum_{n=1}^\infty \frac{(-1)^n}{n-(-1)^n}$ converges. I've updated my proof to be complete now, edited for proof-verification!
We know that for the partial sums with even an uneven terms, the following holds:
$S_{2N}=\sum_{n=1}^{2N} \frac{(-1)^n}{n-(-1)^n} = -\frac{1}{2} + \frac{1}{1} -\frac{1}{4} +\frac{1}{3} -\frac{1}{6} +\frac{1}{5}-\dots+\frac{1}{2N-1}$
$= \frac{1}{2\times1} +\frac{1}{4\times3} + \frac{1}{6\times5}+\dots+\frac{1}{2N(2N-1)} = \sum_{n=1}^{2N} \frac{1}{2n(2n-1)}$
We may rewrite the series in pairs as we know it will have an even amount of terms.
$S_{2N+1} = \sum_{n=1}^{2N+1} \frac{(-1)^n}{n-(-1)^n} = -\frac{1}{2} + \frac{1}{1} -\frac{1}{4} +\frac{1}{3}-\dots+\frac{1}{2N-1} -\frac{1}{2N+2}$
$=\frac{1}{2\times1} +\frac{1}{4\times3} + \frac{1}{6\times5}+\dots+\frac{1}{2N(2N-1)} - \frac{1}{2N+2} = \sum_{n=1}^{2N} \frac{1}{2n(2n-1)}-\frac{1}{2N+2}$
As $n\in\mathbb{N}$, we know that $n\geq1$ so:
$n\geq1 \iff 3n\geq3 \iff 3n^2\geq3n \iff 3n^2-3n\geq0 \iff 4n^2-2n \geq n^2+n$
So: $2n(2n-1)\geq n(n+1) \iff \frac{1}{2n(2n-1)}\leq \frac{1}{n(n+1)}$ for all $n\geq1$.
As the series of the latter sequence converges, we can conclude, by the comparison, test that the series $\sum_{n=1}^{2N} \frac{1}{2n(2n-1)}$ converges.
Suppose it converges to $s$, then we know $^{\lim S_{2N}}_{N\to\infty} = s$ and thus $\lim_{N\to\infty}[S_{2N+1}] = s - (\lim_{N\to\infty}[\frac{1}{2N+2}]) = s-0 = s.$ As the partial sums ending with even and uneven terms both converge to the same limit, the series $\sum_{n=1}^\infty \frac{(-1)^n}{n-(-1)^n}$ converges. $\tag*{$\Box$}$
|
\begin{align*}
\sum_{k=2}^{N}\dfrac{1}{2k(2k-1)}\leq\sum_{k=2}^{N}\dfrac{1}{2k(2k-(k/2))}=\dfrac{1}{3}\sum_{k=2}^{N}\dfrac{1}{k^{2}}<\dfrac{1}{3}\sum_{k=2}^{\infty}\dfrac{1}{k^{2}}<\infty,
\end{align*}
so $\{S_{2N}\}$ is convergent, so is $\{S_{2N+1}\}$ because $\lim_{N}(S_{2N+1}-S_{2N})=0$ (so they have the same limit). Then $\{S_{N}\}$ is convergent.
|
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|
Derivative of $f\left(x\right)=\arctan\left(\sqrt{\frac{1+x}{1-x}}\right)$ I am trying to find the derivative of $f\left(x\right)=\arctan\left(\sqrt{\frac{1+x}{1-x}}\right)$ by only using the formula $\arctan\left(u\left(x\right)\right)'=\frac{u'\left(x\right)}{u\left(x\right)^2+1}$. I don't honestly understand this formula, here are however my calculations. Is this the right way to find the derivative? And how do I proceed?
$$\LARGE
\frac{\frac{1}{2\sqrt{\frac{1+x}{1-x}}}\cdot \frac{\left(1-x\right)-\left(1+x\right)}{x^2-2x+1}}{\frac{x+1}{1-x}+1}
$$
I also appreciate if you can explain what that "formula" exactly is.
|
For your convenience, you could first write
$$\frac{1+x}{1-x}=\frac2{1-x}-1\implies\left(\frac{1+x}{1-x}\right)'=\frac2{(1-x)^2}$$
Thus, with the aid of the chain rule:
$$\left(\arctan\sqrt\frac{1+x}{1-x}\right)'=\frac1{1+\frac{1+x}{1-x}}\cdot\frac1{2\sqrt\frac{1+x}{1-x}}\cdot\frac2{(1-x)^2}=$$
$$=\frac{1-x}2\cdot\frac{\sqrt{1-x}}{2\sqrt{1+x}}\cdot\frac2{(1-x)^2}=\frac1{2\sqrt{1-x^2}}$$
|
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|
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f(xy - 1) + f(x)f(y) = 2xy - 1$ Using induction, I proved that $f(x) = x$ and $f(x) = -x^2$ work, but only for rational numbers.
How can I prove them for all real numbers?
|
So we assume $$f(xy - 1) + f(x)f(y) = 2xy - 1\tag1$$ for all real $x,y$. The RHS is not constant, so constants aren't solutions. Substituting $y=0$ in (1) gives
$$f(-1)+f(x)f(0)=-1,$$ and as $f(x)$ is not a constant, we get
$$f(0)=0,\quad f(-1)=-1\tag2.$$
Letting $y=1$, we obtain $$f(x-1)+f(x)f(1)=2x-1\tag3.$$ Replacing $x$ by $xy$ in (3) gives $f(xy-1)+f(xy)f(1)=2xy-1$, and together with (1), this implies
$$f(x)f(y)=f(xy)f(1)\tag4.$$ But as the OP pointed out already in a comment, $f(1)$ can have only two possible values: from (3) with $x=1$, we get $f(0)+f(1)^2=1$, i.e. $f(1)=\pm1$.
Case $f(1)=1$: From (3), we get $$f(x-1)=2x-1-f(x)\tag5,$$ from (4) with $y=-1$ we conclude $$f(-x)=f(x)f(-1)f(1)=-f(x)\tag{odd}.$$ Replacing $x$ by $-x$ in (5) gives $f(-x-1)=-2x-1-f(-x)$, using (odd), we arrive at
$$f(x+1)=2x+1-f(x)\tag6.$$
From (1) with $y=x$, we have $$f(x^2-1)+f(x)^2=2x^2-1\tag{7}.$$ But $x^2-1=(x-1)(x+1)$, so (using (5) and (6))
$$f(x^2-1)+f(x)^2=[2x-1-f(x)][2x+1-f(x)]+f(x)^2=2x^2-1,$$ and after some simplification, $$[f(x)-x]^2=0,$$ i.e. $f(x)=x$.
Case $f(1)=-1$: Here, (4) with $y=-1$ gives
$$f(-x)=f(x)\tag{even},$$ and instead of (5) and (6), we get in the same way
$$f(x-1)=2x-1+f(x)$$ and $$f(x+1)=-2x-1+f(x).$$ Since in this case
$f(x^2-1)=-f(x-1)f(x+1)$, (7) becomes
$$-[2x-1+f(x)][-2x-1+f(x)]+f(x)^2=2x^2-1,$$ simplifying to
$f(x)=-x^2$.
|
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|
Find the minimum of expression: $\frac{2-x}{3+x}+\frac{2-y}{3+y}+\frac{2-z}{3+z}$ If $x+y+z=1$ and $x,y,z$ are positive numbers, Find the minimum of expression:
$$\frac{2-x}{3+x}+\frac{2-y}{3+y}+\frac{2-z}{3+z}$$
My solution:
$$\left[\frac{2-x}{3+x}+\frac{2-y}{3+y}+\frac{2-z}{3+z} \right]_{min}\Rightarrow \left[\frac 5{x+3}+\frac 5{y+3}+\frac 5{z+3}-3\right]_{min}\Rightarrow \left[ \frac 1{x+3}+\frac 1{y+3}+\frac 1{z+3}\right]_{min}$$
And we have $a+b+c≥3\sqrt[3]{abc}$
Which that $\left[ a+b+c\right]_{min}\Rightarrow a=b=c$
and from here, we get $$\frac 1{x+3}=\frac1{y+3}=\frac 1{z+3} \Rightarrow x=y=z$$
Finally, It must be $x=y=z=\frac 13$
and $$\left[\frac{2-x}{3+x}+\frac{2-y}{3+y}+\frac{2-z}{3+z} \right]_{min}=\frac 32$$
Am I correct?
Note: I'm sorry for wrong mathematical symbols.
|
For $x=y=z=\frac{1}{3}$ we get a value $\frac{3}{2}.$
We'll prove that it's a minimal value.
Indeed, let $x=\frac{a}{3}$, $y=\frac{b}{3}$ and $z=\frac{c}{3}.$
Hence, $a+b+c=3$ and
$$\sum_{cyc}\frac{2-x}{3+x}-\frac{3}{2}=\sum_{cyc}\left(\frac{2-\frac{a}{3}}{3+\frac{a}{3}}-\frac{1}{2}\right)=\frac{3}{2}\sum_{cyc}\frac{1-a}{9+a}=$$
$$=\frac{3}{2}\sum_{cyc}\left(\frac{1-a}{9+a}+\frac{a-1}{10}\right)=\frac{3}{20}\sum_{cyc}(a-1)\left(1-\frac{10}{9+a}\right)=\frac{3}{20}\sum_{cyc}\frac{(a-1)^2}{9+a}\geq0.$$
Id est, your answer is right.
We can end your idea by the following way.
By C-S $$\sum_{cyc}\frac{2-x}{3+x}=\sum_{cyc}\frac{5-3-x}{3+x}=5\sum_{cyc}\frac{1}{3+x}-3=$$
$$=\frac{1}{2}\sum_{cyc}(3+x)\sum_{cyc}\frac{1}{3+x}-3\geq\frac{1}{2}\cdot9-3=\frac{3}{2}.$$
Also, we can use Rearrangement.
Indeed, triples $(1-3x,1-3y,1-3z)$ and $\left(\frac{1}{3+x},\frac{1}{3+y},\frac{1}{3+z}\right)$ are the same ordered.
Thus, by Chebyshov's inequality we obtain:
$$\sum_{cyc}\frac{2-x}{3+x}-\frac{3}{2}=\sum_{cyc}\left(\frac{2-x}{3+x}-\frac{1}{2}\right)=\frac{1}{2}\sum_{cyc}\frac{1-3x}{3+x}\geq$$
$$\geq\frac{1}{6}\sum_{cyc}(1-3x)\sum_{cyc}\frac{1}{3+x}=0.$$
|
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"url": "https://math.stackexchange.com/questions/2578231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
2nd solution of $\cos x \cos 2x\cos 3x= \frac 1 4 $
$\cos x \cos 2x\cos 3x= \dfrac 1 4 $
Attempt explained:
$(2\cos x \cos 3x)\cos 2x = \frac1 2 $
$(\cos 4x +\cos 2x )\cos 2x = \frac 1 2 \\\cos ^2y + \cos y (2\cos^2y- 1)= \frac1 2 \\ $
(Let, y = 2x)
$\implies 4\cos^3 y+2\cos^2y- 2\cos y-1=0$
I solved this equation using Rational Root Theorem and got $y = \frac 1 2$
$\implies x= m\pi \pm \dfrac\pi3 \forall x\in \mathbb {Z}$
Using Remainder theorem, the other solution is $\cos^2 y = \dfrac 1 2 $
$\implies x = \dfrac n2\pi \pm\dfrac \pi 8 $
But answer key states: $ x= m\pi \pm \dfrac\pi3or x =(2n+1)\dfrac \pi 8$
Why don't I get the second solution correct?
|
How does $\cos^2y=\dfrac12$ imply $$\dfrac{n\pi}2\pm\dfrac\pi6?$$
In fact $\cos^2y=\dfrac12\iff\cos2y=0$
$\implies2y=m\pi+\dfrac\pi2,4y=\pi(2m+1)$ where $m$ is any integer $\ \ \ \ (1)$
Again, $\cos^2y=\cos^2\dfrac\pi4\iff\sin^2y=\sin^2\dfrac\pi4$
Using Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $
$y=n\pi\pm\dfrac\pi4,4y=\pi(4n\pm1)$ where $n$ is any integer $\ \ \ \ (2)$
Set some values of $m,n$ to identify $\{2m+1\},\{4n\pm1\}$ to be esentially the same set
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2578575",
"timestamp": "2023-03-29T00:00:00",
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|
Is there a criterion for showing that a two variable continuous function $f:\Bbb S^1\times\Bbb S^1\to \Bbb R^2$ pass from the origin? Is there a criterion for showing that a symmetric ($f(x,y)=f(y,x)$) two variable continuous function $f:\Bbb S^1\times\Bbb S^1\to \Bbb R^2$ pass from the origin? i.e.
$$\exists\,(x,y)\in \Bbb S^1\times\Bbb S^1,\,x\neq y\quad s.t. \quad f(x,y)=(0,0).$$
where we know that $f$ pass from all four regions and $f(x,x)=(0,0)$ for all $x\in\Bbb S^1$. I tried to define a real value function from $f$ and use from intermediate value theorem but I couldn't find a helpful map. Is it sufficient to show ${\rm Im} f$ is simply connected?
|
Let $\alpha(\theta)=\left\lgroup
\begin{array}{c}
\cos \theta
\\
\sin \theta
\end{array}
\right\rgroup \in \mathbb{S}^1$ and $\beta(\varphi)=\left\lgroup
\begin{array}{c}
\cos \varphi
\\
\sin \varphi
\end{array}
\right\rgroup\in\mathbb{S}^1$ parametrizations of $\mathbb{S}^1$ whit $0\leq \theta\leq 2\cdot \pi$ and $0\leq \varphi\leq 2\cdot \pi$.
By Poincaré-Miranda theorem if there are $\theta^\prime,\theta^{\prime\prime}\in [0,2\pi]$ and $\varphi^\prime,\varphi^{\prime\prime}\in [0,2\pi]$ such that
$$
f_1
\left(
\left\lgroup
\begin{array}{c}
\cos \theta^\prime
\\
\sin \theta^{\prime}
\end{array}
\right\rgroup
,
\left\lgroup
\begin{array}{c}
\cos \varphi
\\
\sin \varphi
\end{array}
\right\rgroup
\right)
<0,
\quad
f_1
\left(
\left\lgroup
\begin{array}{c}
\cos \theta^{\prime\prime}
\\
\sin \theta^{\prime\prime}
\end{array}
\right\rgroup
,
\left\lgroup
\begin{array}{c}
\cos \varphi
\\
\sin \varphi
\end{array}
\right\rgroup
\right)
>0,\quad \forall \varphi \in [\varphi^{\prime},\varphi^{\prime\prime}]
$$
$$
f_2
\left(
\left\lgroup
\begin{array}{c}
\cos \theta
\\
\sin \theta
\end{array}
\right\rgroup
,
\left\lgroup
\begin{array}{c}
\cos \varphi^{\prime}
\\
\sin \varphi^{\prime}
\end{array}
\right\rgroup
\right)
<0,
\quad
f_2
\left(
\left\lgroup
\begin{array}{c}
\cos \theta
\\
\sin \theta
\end{array}
\right\rgroup
,
\left\lgroup
\begin{array}{c}
\cos \varphi^{\prime\prime}
\\
\sin \varphi^{\prime\prime}
\end{array}
\right\rgroup
\right)
>0,\quad \forall \theta \in [\theta^{\prime},\theta^{\prime\prime}]
$$
then there are $\theta^\ast$ and $\varphi^\ast$ such that
$$
f_1
\left(
\left\lgroup
\begin{array}{c}
\cos \theta^{\ast}
\\
\sin \theta^{\ast}
\end{array}
\right\rgroup
,
\left\lgroup
\begin{array}{c}
\cos \varphi^{\ast}
\\
\sin \varphi^{\ast}
\end{array}
\right\rgroup
\right)
=0\;\;
\mbox{ and }
\;\;
f_2
\left(
\left\lgroup
\begin{array}{c}
\cos \theta^{\ast}
\\
\sin \theta^{\ast}
\end{array}
\right\rgroup
,
\left\lgroup
\begin{array}{c}
\cos \varphi^{\ast}
\\
\sin \varphi^{\ast}
\end{array}
\right\rgroup
\right)
=0,
$$
The Poincaré–Miranda theorem is a generalization of the intermediate value theorem.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integral of x² from 0 to b - using archimedes sum of squares - Apostol's I'm reading Apostol's Calculus book and in the first chapter is presented the way archimedes found the sum of the square and how it can be used to calculate the integral of $x²$. But I'm not able to follow some steps of the proof.
from the book:
We subdivided the base in $n$ parts each with length $\ \frac{b}{n} $, a typical point corresponds to $\frac{kb}{n}$ where $k$ takes values from $k = 1, 2, 3, ..., n$
We can construct rectangles from for each $k th$ point:
$Base = \frac{b}{n}$
$Height = (\frac{kb}{n})^2$
$Area = Base * Height = \frac{b}{n} . (\frac{kb}{n})^2$
$Area = \frac{b^3}{n^3}.k^2 $
If we sum all the rectangles, we get a bit more than the area under the curve $x^2$
$S_{big} = \frac{b^3}{n^3}.(1² + 2² + 3² + ... + (n-1)² + n²)$
If we can construct smaller rectangles, using $n-1$ points, we we get a bit less than the area under the curve $x^2$.
$S_{small} = \frac{b^3}{n^3}.(1² + 2² + 3² + ... + (n-1)²)$
So the real area under the curve $x^2$ is between the two areas:
$S_{small} < A < S_{big}$
After a bit of algebra we get that:
$S_{big} = \frac{b^3}{n^3} . (\frac{n^3}{3} + \frac{n²}{2} + \frac{n}{6})$
$S_{small} = \frac{b^3}{n^3} . (\frac{n^3}{3} - \frac{n²}{2} + \frac{n}{6})$
To prove that $A$ is $\frac{b^3}{3}$ he uses this inequalities:
$1² + 2² + 3² + ... + (n-1)² < \frac{n³}{3} < 1² + 2² + 3² + ... + n²$
But I don't understand where the $\frac{n³}{3}$ came from. And can't follow the proof
EDIT:
After taking the average of the two expression:
$\frac{(\frac{n^3}{3} + \frac{n²}{2} + \frac{n}{6}) + (\frac{n^3}{3} - \frac{n²}{2} + \frac{n}{6})}{2} = \frac{n^3}{3} + \frac{n}{6}$
So I understan where $\frac{n³}{3}$ came from, but why is $\frac{n}{6}$ is thrown away?
|
You may want to read this. It really is not an easy or intuitive inequality to find.
Alternatively, you can consider this:
Eventually, you would want $n$ to approach $+\infty$.
So let's consider
\begin{align}
S_{\text{big}} &= \frac{b^3}{n^3} \cdot \left(\frac{n^3}{3} + \frac{n²}{2} + \frac{n}{6}\right)\\
&=b^3\left(\frac{1}{3} + \frac{1}{2n} + \frac{1}{6n^2}\right)\\
\lim_{n\to\infty}S_{\text{big}} &=b^3\left(\frac{1}{3} +0+0\right)\\
&=\frac{b^3}{3}
\end{align}
Similarly, you will have
$$\lim_{n\to\infty}S_{\text{small}} =\frac{b^3}{3}$$
too.
So you can actually just squeeze here already.
|
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"timestamp": "2023-03-29T00:00:00",
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|
how to solve this equation using logarithms for the equation :
$ 3^{x^2} + 3^x = 90 $
my solution was :
$3^{x^2} + 3 ^x = 3^{2^2} + 3^2$
so $x=2$
but i want to know if there is any solution by using logarithms ?
when using wolframAlpha th solution was $ x= -2.02356 $
or $x = 2 $
but How ?
|
If we divide both sides by $3$ first, we get $$3^{x^2-1} + 3^{x-1} = 30$$
Then, we can write it as $$(3^{x-1})^{x+1}+3^{x-1} = 3^{x-1}(3^{x-1})^{x}+3^{x-1}= 30$$
Now, if we bracket $3^{x-1}$, we get $$3^{x-1}(3^{x^2-x}+1)=30$$
From here, if there exists an integer solution, $30$ must be factorized as $3 \cdot 10$ so that $3^{x-1}$ is an integer (Notice that in other factorizations of $30$, there is no term $3^k$ where $k \in \mathbb{Z^+}$). So we can try $3^{x-1} = 3 \implies x=2$ and we see that it is a solution of this equation since $3^{x^2-x}+1 = 10$ when $x=2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2582559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
How can I solve the following problem with floor functions?
Ideally I'd like to find an analytical solution, though I doubt one exists. If an analytical doesn't exist, any insight into tackling this problem numerically would be greatly appreciated. I can't think of anything other than naive brute-force.
Given $c$, find $x$ such that $y$1$=$$y$2$=$$d$.
$$y_1 = (x+h)-\lfloor x + h \rfloor$$
$$y_2 = \frac{(-x \cdot s)}{(x+s)} + h - \left \lfloor \frac{(-x \cdot s)}{(x+s)}+h \right \rfloor$$
where,
$$c \in \mathbb{Z^+}$$
$$s=\sqrt c $$
$$d=s-\lfloor s \rfloor$$
$$h=2 \cdot d- \lfloor 2 \cdot d \rfloor$$
For example,
Given $c=77$,
$$s=\sqrt c \approx 8.775$$
$$d=s-\lfloor s \rfloor \approx 0.775$$
$$h=2 \cdot d- \lfloor 2 \cdot d \rfloor \approx 0.550$$
Dot notation is just scalar multiplication.
Here's what this problem looks like visually.
where the black line is $y_1$, the red line is $y_2$, the blue line is $d$ and the red arrow is pointing to the $x$ that satisfies the condition. In this case $x \approx 2.225$.
So what we have is essentially two sawtooth waves, $y_1$ with a period of $1$ and $y_2$ with a period that decreases non-linearly as a function of $x$ and I'm looking for the $x$ at which both lines equal $d$.
Since $x$ has a period of $1$, we know that the intersection will have to occur at some $x$ with fractional part $1-d$.
Brute-forcing won't be sufficient since this needs to work for extremely large $c$. So, without an analytical solution I'm trying to find some way to attack this problem numerically but can't find any way to narrow scope.
|
Define $\{x\} = x - [x]$ for $x \in \mathbb{R}$, then$$
d = \{\sqrt{c}\}, \quad h = \{2d\}.
$$
For $c \in \mathbb{N}_+$, suppose $m \in \mathbb{N}_+$ satisfies $m^2 \leqslant c < (m + 1)^2$. Note that $\displaystyle c \neq \left(m + \frac{1}{2}\right)^2$ since $c \in \mathbb{N}_+$.
Case 1: $\displaystyle m^2 \leqslant c < \left(m + \frac{1}{2}\right)^2$. In this case,$$
d = \{\sqrt{c}\} = \sqrt{c} - m \in \left(0, \frac{1}{2}\right), \quad h = \{2d\} = 2d.$$
Because$$
\{x + h\} = d, \quad \left\{-\frac{\sqrt{c} x}{x + \sqrt{c}} + h\right\} = d,
$$
there exist $n_1, n_2 \in \mathbb{Z}$ such that$$
d = x + h - n_1, \quad d = -\frac{\sqrt{c} x}{x + \sqrt{c}} + h - n_2,
$$
which imply $x = n_1 + d - h = n_1 - d$ and\begin{align*}
n_2 &= h - d - \frac{\sqrt{c} x}{x + \sqrt{c}} = d - \frac{\sqrt{c} (n_1 - d)}{n_1 - d + \sqrt{c}}\\
&= (\sqrt{c} - m) - \frac{\sqrt{c} (n_1 - (\sqrt{c} - m))}{n_1 - (\sqrt{c} - m) + \sqrt{c}}\\
&= \sqrt{c} - m - \frac{\sqrt{c} (n_1 + m - \sqrt{c})}{n_1 + m} = \frac{c}{n_1 + m} - m.
\end{align*}
Note that $c, n_1, n_2$ are all integers, so each solution $x$ one-one corresponds to an integer $n_1$ such that $n_1 + m \neq 0$ and $n_1 + m | c$. Hence all solutions are$$
x = n_1 - d = (n_1 + m) - (m + d) = k - \sqrt{c},
$$
where the integer $k$ divides $c$.
Case 2: $\displaystyle \left(m + \frac{1}{2}\right)^2 < c < (m + 1)^2$. In this case,$$
d = \{\sqrt{c}\} = \sqrt{c} - m \in \left(\frac{1}{2}, 1\right), \quad h = \{2d\} = 2d - 1.$$
Because$$
\{x + h\} = d, \quad \left\{-\frac{\sqrt{c} x}{x + \sqrt{c}} + h\right\} = d,
$$
there exist $n_1, n_2 \in \mathbb{Z}$ such that$$
d = x + h - n_1, \quad d = -\frac{\sqrt{c} x}{x + \sqrt{c}} + h - n_2,
$$
which imply $x = n_1 + d - h = n_1 - d + 1$ and\begin{align*}
n_2 &= h - d - \frac{\sqrt{c} x}{x + \sqrt{c}} = d - 1 - \frac{\sqrt{c} (n_1 - d + 1)}{n_1 - d + 1 + \sqrt{c}}\\
&= (\sqrt{c} - m) - 1 - \frac{\sqrt{c} (n_1 - (\sqrt{c} - m) + 1)}{n_1 - (\sqrt{c} - m) + 1 + \sqrt{c}}\\
&= \sqrt{c} - m - 1 - \frac{\sqrt{c} (n_1 + m + 1 - \sqrt{c})}{n_1 + m + 1} = \frac{c}{n_1 + m + 1} - m - 1.
\end{align*}
Note that $c, n_1, n_2$ are all integers, so each solution $x$ one-one corresponds to an integer $n_1$ such that $n_1 + m + 1\neq 0$ and $n_1 + m + 1 | c$. Hence all solutions are$$
x = n_1 - d + 1 = (n_1 + m + 1) - (m + d) = k - \sqrt{c},
$$
where the integer $k$ divides $c$.
|
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|
Taylor limits with sine I'm having troubles calculating these two limits (I prefer to write a sigle question including both of them, instead of two different ones).
This one
$$\\ \lim_{x\rightarrow 0} \frac{ x-\sin^2(\sqrt x)-\sin^2(x)} {x^2} $$ I tried expanding with Taylors at different orders but the square root in the sine argument gave problems with the o() grades, leading for example to things like this $$ o(\sqrt x^3) $$ that does not seem easy to manage. I tried rewriting the limit using $$ \sin^2(x) + \cos^2(x) = 1 $$, obtaining
$$\\ \lim_{x\rightarrow 0} \frac{ x-1+\cos^2(\sqrt x)-\sin^2(x)} {x^2} $$
since the McLaurin expasion for the cosine looked a little better to me, but I keep obtaining a wrong result.
The second limit in this one instead
$$\\ \lim_{x\rightarrow 0} \frac{ [\frac{1}{(1-x)} +e^x]^2 -4e^{2x} -2x^2} {x^3} $$
and I happily wrote the McLaurin expansion stopping at the second order, that are
$$(1-x)^{-1} = 1+x+\frac{x^2}{2} + o(x^2)$$
$$e^x = 1+x+\frac{x^2}{2} +o(x^2)$$
$$e^{2x} = 1 + 2x+2x^2+o(x^2)$$
doing the calculation leads to this
$$\\ \lim_{x\rightarrow 0}\frac{\frac{-4}{3}x^3-2x^2}{x^3}$$
$$\\ \lim_{x\rightarrow 0}\frac{-4}{3} - \frac{2}{x}$$ and here I have problems understating what is going on. I don't know if x tends to 0 from right or left, so I can't evaluate the result (that would be, $$ +\infty$$ or $$ -\infty$$ ). Does this mean I did some wrong calculation? Do I need to take some more orders in the McLaurin expansion?
|
If you're uncertain about $\sqrt{x}$, substitute $\sqrt{x}=t$, so the limit becomes
$$
\lim_{t\to0^+}\frac{t^2-\sin^2(t)-\sin^2(t^2)}{t^4}
$$
Of course we just need Taylor up to degree $4$:
\begin{align}
\sin^2(t)&=\left(t-\frac{t^3}{6}+o(t^3)\right)^2=t^2-\frac{t^4}{3}+o(t^4)\\[4px]
\sin^2(t^2)&=(t^2+o(t^2))^2=t^4+o(t^4)
\end{align}
Thus your limit is
$$
\lim_{t\to0^+}\frac{t^2-t^2+t^4/3-t^4+o(t^4)}{t^4}=-\frac{2}{3}
$$
For the second limit you should not stop at degree $2$ in the numerator, because the denominator has degree $3$.
You have $$\left(\frac{1}{1-x}+e^x\right)^{\!2}=\left(2+2x+\frac{3}{2}x^2+\frac{7}{6}x^3+o(x^3)\right)^{\!2}=4+4x^2+8x+6x^2+\frac{14}{3}x^3+6x^3+o(x^3)$$ so the numerator becomes $$4+8x+10x^2+\frac{32}{3}x^3-4\left(1+2x+\frac{(2x)^2}{2}+\frac{(2x)^3}{6}\right)-2x^2+o(x^3)=\frac{16}{3}x^3+o(x^3)$$
|
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|
Finding The Basis Of An Intersection Of Two Subspaces
Let $$U=\operatorname{Span}\{v_1,v_2,v_3\}, V=\operatorname{Span}\{v_4,v_5,v_6\}$$
Where $$v_1=(1,28,2,39),v_2=(2,28,2,39),v_3=(-1,28,2,39)\\v_4=(0,8,0,11),v_5=(0,31,1,43),v_6=(0,-3,0,-4)$$
Find a basis for $U\cap V$
I have forgot the algorithm for finding intersection of subspaces, but in the exercise I was given an algorithm that I have yet met and would like to understand, how and why it works.
The first step is to find an homogeneous system s.t the subspace is the solution set (Null space).
To do so for $U$ we look at $$\left(\begin{array}{ccc|c}
1 & 2 & -1 &x\\
28 & 28& 28&y\\
2 & 2 & 2& z\\
39& 39&39& w \\
\end{array}\right)\sim \left(\begin{array}{ccc|c}
1 & 2 & -1 &x\\
1 & 1& 1& \frac{y}{28}\\
0 & 0 & 0& \frac{z}{2}-\frac{y}{28}\\
0& 0&0& \frac{w}{39}-\frac{y}{28} \\
\end{array}\right)$$
So the matrix that $U$ is her solution set is $$ \begin{pmatrix} 0&-\frac{1}{28}&\frac{1}{2}&0\\ 0&-\frac{1}{28}&0&\frac{1}{39} \end{pmatrix}$$
Doing the same with $V$ we get
$$\left(\begin{array}{ccc|c}
0 & 0 & 0 &x\\
8& 31& -3&y\\
0 & 1 & 0& z\\
11& 43&-4& w \\
\end{array}\right)\sim \left(\begin{array}{ccc|c}
8 & 31 & -3 &y\\
0 & 1 & 0& z\\
0 & 0& 1& 8w-11y-3z\\
0& 0&0& x \\
\end{array}\right)$$
So the matrix that $V$ is her solution set is $$ \begin{pmatrix} 1&0&0&0\end{pmatrix}$$
To find the intersection we solution for $$\left(\begin{array}{cccc|c}
0 & -\frac{1}{28} & \frac{1}{2} &0 &0\\
0 & -\frac{1}{28}& &\frac{1}{39}& 0\\
1 & 0 & 0& 0 &0\\
\end{array}\right) $$
Which is $$ \begin{pmatrix} 0&28&2&39\end{pmatrix}$$
a. Why does putting the vectors in columns and $x,y,z,w$ in the $b$ vector give us a system which the vectors are their solutions?
b. Why looking at the null space of the vectors (why we put the vectors in horizontal and not in vertical) that we found give us the basis of intersection?
C. Is there another way to find the basis of intersection?
|
a. Why does putting the vectors in columns and $x,y,z,w$ in the $b$ vector give us a system which the vectors are their solutions?
Call $u=\begin{bmatrix} 1 & 2 & -1\\
28 & 28& 28\\
2 & 2 & 2\\
39& 39&39\end{bmatrix}$ and $\xi=\begin{bmatrix}x\\y\\z\\w\end{bmatrix}$.
Then $(x,y,z,w)\in U$ if and only if $\operatorname{rank}(u|\xi)=\operatorname{rank}(u)$.
The matrix $u$ is simplified by row operations which are performed with a multiplication on the left by a invertible matrix $a$, thus
$$a=\begin{bmatrix}
1&0&0&0\\0&\frac 1{28}&0&0\\0&-\frac 1{28}&\frac 12&0\\0&-\frac 1{28}&0&\frac 1{39}
\end{bmatrix}\quad
au=\begin{bmatrix} 1 & 2 & -1\\
1 & 1& 1\\
0 & 0 & 0\\
0& 0&0
\end{bmatrix}\quad a\xi=\begin{bmatrix}x\\\frac y{28}\\\frac z2-\frac y{28}\\\frac w{39}-\frac y{28}\end{bmatrix}$$
But $\operatorname{rank}(u|\xi)=\operatorname{rank}(u)$ if and only if $\operatorname{rank}(au|a\xi)=\operatorname{rank}(au)$.
This is equivalent to the system
$$\left\{\begin{array}{c}
\frac z2-\frac y{28}=0\\\frac w{39}-\frac y{28}=0
\end{array}\right.\iff\begin{bmatrix} 0&-\frac{1}{28}&\frac{1}{2}&0\\ 0&-\frac{1}{28}&0&\frac{1}{39}\end{bmatrix}\begin{bmatrix}x\\y\\z\\w\end{bmatrix}=\begin{bmatrix}0\\0\\0\\0\end{bmatrix}$$
hence $U$ is the kernel of your set solution
\begin{bmatrix} 0&-\frac{1}{28}&\frac{1}{2}&0\\ 0&-\frac{1}{28}&0&\frac{1}{39}\end{bmatrix}
Concluding: you put the generators of $U$ and $x,y,z,w$ in column in order to have
$$(x,y,z,w)\in U\iff\operatorname{rank}(u|\xi)=\operatorname{rank}(u)$$
and perform row operations; the kernel of the set solution gives your subspace $U$.
Thus we write our subspaces $U,V$ as kernel of two matrix.
b. Why looking at the null space of the vectors (why we put the
vectors in horizontal and not in vertical) that we found give us the
basis of intersection?
For matrix $A,B$ with the same number of columns, we have
$$\ker(A)\cap\ker(B)=\ker\begin{pmatrix}A\\B\end{pmatrix}$$
hence from
$$U=\ker
\begin{bmatrix} 0&-\frac{1}{28}&\frac{1}{2}&0\\ 0&-\frac{1}{28}&0&\frac{1}{39}\end{bmatrix}\quad V=\ker\begin{bmatrix}1&0&0&0\end{bmatrix}$$
we get
$$U\cap V=\ker\begin{bmatrix}0&-\frac{1}{28}&\frac{1}{2}&0\\ 0&-\frac{1}{28}&0&\frac{1}{39}\\1&0&0&0\end{bmatrix}$$
|
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|
Given matrix times vector, find the inverse of the matrix times the same vector. Given $AV= \begin{bmatrix}8\\8 \\ 8 \\8\end{bmatrix}$, where A is a matrix and V is a vector, find $A^{-1}V$.
My thought are the following:
Step 1:
Times $A^{-1}$ for both sides.
$A^{-1}AV=A^{-1} \begin{bmatrix}8\\8 \\ 8 \\8\end{bmatrix}$
$V=8A^{-1} \begin{bmatrix}1\\1 \\ 1 \\1\end{bmatrix}$
$\frac{1}{8} V=A^{-1} \begin{bmatrix}1\\1 \\ 1 \\1\end{bmatrix}$
Step 2:
Can I assume $AV=\lambda V=8*\begin{bmatrix}1\\1 \\ 1 \\1\end{bmatrix}$ here? Hence, $V=\begin{bmatrix}1\\1 \\ 1 \\1\end{bmatrix}$? Is this a correct step?
Step 3:
If step 2 is correct, then
$\frac{1}{8}V = \frac{1}{8} \begin{bmatrix}1\\1 \\ 1 \\1\end{bmatrix}=A^{-1} \begin{bmatrix}1\\1 \\ 1 \\1\end{bmatrix}=A^{-1} V=\begin{bmatrix} \frac{1}{8} \\ \frac{1}{8} \\ \frac{1}{8} \\\frac{1}{8}\end{bmatrix}$
Does the above inference sounds right?
|
What if I add extra information regarding the vector $V$ as the following,
Question: Given $AV= \begin{bmatrix}8\\8 \\ 8 \\8\end{bmatrix}$, where A is a matrix and ${V=\begin{bmatrix}1\\1 \\ 1 \\1\end{bmatrix}}$ , find $A^{-1}V$.
Times $A^{-1}$ for both sides.
$A^{-1}AV=A^{-1} \begin{bmatrix}8\\8 \\ 8 \\8\end{bmatrix}=A^{-1} * 8 * V$
$V=A^{-1} * 8 * V$
$\frac{1}{8}V = A^{-1} V=\begin{bmatrix} \frac{1}{8} \\ \frac{1}{8} \\ \frac{1}{8} \\\frac{1}{8}\end{bmatrix}$
Add information about $V$ seems become a trivial question?
Explaination:I did not use the assumption that V is the eigenvector of the matrix A here. Just factor out 8 from $[8,8,8,8]^T$ to become $V$.
|
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|
Explicit form of the function $F\left(p\right)=\int_{0}^{\, \pi/2}\ln\left(1+p\sin^2\left(t\right)\right)\text{d}t$ Let $x \in \mathbb{R}^{+}$, I wonder how to prove that
$$
F\left(x\right)=\int_{0}^{\, \pi/2}\ln\left(1+x\sin^2\left(t\right)\right)\text{d}t=\pi \ln\left(\frac{1}{2}\left(1+\sqrt{1+x}\right)\right)
$$
It would beautifully show that
$$
\int_{0}^{\, \pi/2}\ln\left(1+4\sin^2\left(t\right)\right)\text{d}t=\pi \ln\left(\varphi\right)
$$
I've no idea how to proceed, except maybe that
$$
F\left(x\right)=\pi\ln\left(\sqrt{1+x}\right)-x\int_{0}^{\, \pi/2}\frac{\sin\left(2t\right)}{1+x\sin^2\left(t\right)}\text{d}t
$$
Then I guess the second will be the missing part, but how to compute it easily ?
|
Consider
$$ F'(x) =\int_0^{\pi/2} \frac{\sin^2 t}{1+x\sin^2t}dt = \int_0^{\pi/2} \frac{1}{\csc^2t + x}dt $$
Let $u = \cot t$
$$ \begin{align}
F'(x) &= \int_0^\infty \frac{1}{(u^2+1)(u^2+1+x)}du \\
&= \frac{1}{x}\int_0^\infty \left(\frac{1}{u^2+1} - \frac{1}{u^2+1+x} \right) du \\
&= \frac{1}{x}\left(\frac{\pi}{2} - \frac{\pi}{2\sqrt{1+x}}\right) \\
&= \frac{\pi}{2\sqrt{1+x}(\sqrt{1+x}+1)}
\end{align} $$
Since $F(0)=0$, we have
$$ \begin{align}
F(x) &= \frac{\pi}{2}\int_0^x \frac{1}{\sqrt{1+t}(\sqrt{1+t}+1)}dt \\
&= \pi \int_1^{\sqrt{1+x}} \frac{1}{1+u}du \\
&= \pi\ln \left(\frac{1 + \sqrt{1+x}}{2}\right)
\end{align} $$
|
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|
Sum of all Fibonacci numbers $1+1+2+3+5+8+\cdots = -1$? I just found the sum of all Fibonacci numbers and I don't know if its right or not.
The Fibonacci sequence goes like this : $1,1,2,3,5,8,13,\dots$ and so on
So the Fibonacci series is this $1+1+2+3+5+8+13+\dots$
Let $1+1+2+3+5+8+\dots=x$
$$\begin{align}
1 + 1 + 2 + 3 + 5 + \dots &= x\\
1 + 1 + 2 + 3 + \dots &= x\\
1 + 2 + 3 + 5 + 8 + \dots &= 2x \text{ (shifting and adding)}
\end{align}$$
We in fact get the same sequence. But the new sequence is one less than the original sequence. So the new sequence is $x-1$.
But $x-1=2x$ which implies that $x=-1$.
So $1+1+2+3+5+8+\dots=x$ which means...
$1+1+2+3+5+8+13+21+\dots=-1$
Is this right or wrong? Can someone please tell? Thanks...
|
You're assuming that the limit of the sum of the first $n$ Fibonacci numbers exists as $n \to \infty$, which it doesn't. Which is to say that in order to apply your method, the series must be convergent but the series diverges, so your method is wrong. It is, however, related to the power series of the Fibonacci numbers, that is to say $$
1+z+2z^2+3z^3+5z^4+8z^5+... = \sum_{n=0}^\infty F_{n+1}z^n=\frac{1}{1-z-z^2}
$$
One can see that when we put $z=1$, the value is $-1$, as you assert in your answer.
However, this series is only convergent when $|z|<\dfrac{1}{\phi}$, where $\phi$ is the golden ratio, as should follow from Binet's Formula. See here for a derivation of the upper identity, and more information.
|
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|
$\exp(z^2)=i$ solution check Find all complex solutions to the equation $\exp(z^2)=i$.
Attempt
We have $z^2 = x^2-y^2+2ixy$ where $z=x+iy$. Then
$$exp(x^2-y^2)=1 \text{ and } 2xy=\frac{\pi}{2}$$
So $x^2=y^2$ and $xy = \frac{\pi}{4}$ Thus $x = y$ and so $x=y= \pm \frac{\sqrt{\pi}}{2}$
I just don't feel very confident about this for some reason.
|
If $\exp(z^2) = i$ then $z^2 = \log i = \ln|i| + i\arg i = 0 + i(\frac{\pi}{2} + 2\pi n) = \frac{4n+1}{2}\pi i$ where $n$ is an integer. These are the numbers $\ldots, -\frac{7\pi i}{2}, -\frac{3 \pi i}{2}, \frac{\pi i}{2}, \frac{5 \pi i}{2}, \frac{9\pi i}{2}, \ldots$. Taking square roots (using that $\sqrt{i} = \pm\frac{1+i}{\sqrt{2}}$)
we get $$z = \pm \frac{1+i}{2} \sqrt{(4n+1)\pi}$$
for $n \geqslant 0$ and
$$z = \pm\frac{-1+i}{2} \sqrt{(-4n-1)\pi}$$
for $n < 0$.
|
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|
$f$ is differentiable at x=1, with $f(1)>0$. Find $\lim \limits_{n\to \infty}\left( \frac{f\left(1+\frac{1}{n}\right)}{f(1)} \right)^\frac{1}{n} $ $f$ is differentiable at $x=1$, with $f(1)>0$. Find $\lim \limits_{n\to \infty}\left( \frac{f\left(1+\frac{1}{n}\right)}{f(1)} \right)^\frac{1}{n}$.
Here's what I got so far:
$$\left( \frac{f(1+\frac{1}{n})}{f(1)} \right)^\frac{1}{n} = \exp\left(\frac{1}{n}\left(\log(f\left(1+\frac{1}{n}\right) - \log f(1)\right)\right) = \\
= \exp\left(\frac{1}{n}\left(\log f\left(1+\frac{1}{n}\right) - \log f(1)\right) \cdot \frac{\frac{1}{n}}{\frac{1}{n}}\right)$$
Now, if $f$ is differentiable at $x=1$, then so it $\log(f)$. Therefore, as $n\rightarrow\infty$:
$$\lim \limits_{n\to \infty}\left(\frac{f\left(1+\frac{1}{n}\right)}{f(1)} \right)^\frac{1}{n} = \exp\left(\lim \limits_{n\to \infty}\frac{1}{n^2} \cdot(\log f)'(1)\right) = e^0 = 1$$
I'm not too sure about that fact that the derivative cancels out, as I could have gotten that zero exponent in a much simpler way. Am I missing something?
|
Just a suggestion (if I may) to make life easier and to get more than the limit.
Consider $$A=\left( \frac{f\left(1+\frac{1}{n}\right)}{f(1)} \right)^{n^a}\implies \log(A)=n^a \log\left( \frac{f\left(1+\frac{1}{n}\right)}{f(1)} \right)$$ Assuming that $f(.)$ is continuoulsy differentiable at $x=1$, Taylor series gives for infinitely large values of $n$
$$f\left(1+\frac{1}{n}\right)=f(1)+\frac{f'(1)}{n}+\frac{f''(1)}{2 n^2}+O\left(\frac{1}{n^3}\right)$$
$$\frac{f\left(1+\frac{1}{n}\right)}{f(1)}=1+\frac{f'(1)}{f(1) n}+\frac{f''(1)}{2 f(1)
n^2}+O\left(\frac{1}{n^3}\right)$$ Continuing with Taylor
$$\log\left( \frac{f\left(1+\frac{1}{n}\right)}{f(1)} \right)=\frac{f'(1)}{f(1) n}+\frac{f(1) f''(1)-f'(1)^2}{2 f(1)^2
n^2}+O\left(\frac{1}{n^3}\right)$$ If $\color{red}{a=1}$, then
$$\log(A)=\frac{f'(1)}{f(1)}+\frac{f(1) f''(1)-f'(1)^2}{2 f(1)^2
n}+O\left(\frac{1}{n^2}\right)$$ Taylor again using $A=e^{\log(A)}$
$$A=e^{\frac{f'(1)}{f(1)}}\left(1+\frac{f(1) f''(1)-f'(1)^2}{2 f(1)^2 n}+O\left(\frac{1}{n^2}\right)\right)$$ which shows the limit and also how it is approached.
If $\color{red}{a=-1}$, then
$$\log(A)=\frac{f'(1)}{f(1) n^2}+O\left(\frac{1}{n^3}\right)$$
$$A=1+\frac{f'(1)}{f(1) n^2}+O\left(\frac{1}{n^3}\right)$$ which shows the limit and also how it is approached.
|
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|
How to prove that $\frac{a+b}{1+a+b} \leq \frac{a}{1+a} + \frac{b}{1+b}$ for non-negative $a,b$? If $a, b$ are non-negative real numbers, prove that
$$
\frac{a+b}{1+a+b} \leq \frac{a}{1+a} + \frac{b}{1+b}
$$
I am trying to prove this result. To that end I added $ab$ to both denominator and numerator as we know
$$
\frac{a+b}{1+a+b} \leq \frac{a+b+ab}{1+a+b+ab}
$$
which gives me
$$
\frac{a}{1+a} + \frac{b}{(1+a)(1+b)}
$$
How can I reduce the second term further, and get the required result?
|
$$
\frac{a+b}{1+a+b} =\frac{a}{1+a+b} + \frac{b}{1+a+b}
$$
then prove
$$
\frac{a}{1+a+b} \leq \frac{a}{1+a}
$$
|
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|
Find $ \lim_{n\rightarrow \infty} \frac{1}{n^2} + \frac{2}{(n+1)^2}+\dots+\frac{n+1}{(2n)^2} $ I want to find the following limit:
$$ \lim_{n\rightarrow \infty} \frac{1}{n^2} + \frac{2}{(n+1)^2}+ \frac{3}{(n+2)^2}+\dots+\frac{n+1}{(2n)^2} $$
I guess it converges to 0 and I tried to prove it using the squeeze theorem but it doesn't seem to work.
Any ideas?
|
Note that
$$\frac{1}{n^2} + \frac{2}{(n+1)^2}+ \frac{3}{(n+2)^2}+\dots+\frac{n+1}{(2n)^2}=\sum_{k=n}^{2n}\frac{k-n+1}{k^2}
=\sum_{k=n}^{2n}\frac{1}{k}-(n-1)\sum_{k=n}^{2n}\frac{1}{k^2}\sim \log \left(\frac{2n}{n-1}\right)-\frac{n+1}{2n}\to\log 2-\frac12$$
indeed for Harmonic series and Euler–Maclaurin formula
$$\sum_{k=n}^{2n}\frac{1}{k}=\sum_{k=1}^{2n}\frac{1}{k}-\sum_{k=1}^{n-1}\frac{1}{k}\sim\log2n-\log (2n-1)=\log \left(\frac{2n}{n-1}\right)$$
$$\sum_{k=n}^{2n}\frac{1}{k^2}=\sum_{k=1}^{2n}\frac{1}{k^2}-\sum_{k=1}^{n-1}\frac{1}{k^2}\sim -\frac1{2n}+\frac{1}{n-1}=\frac{n+1}{2n(n-1)}$$
|
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|
Confusion about solving quadratics $x^2-8x=5$ I got into a little confusion, hopefully someone can clear it up for me.
So I'm solving $x^2-8x=5$ right, and here are my steps:
$$1: x^2-8x=5$$
$$2: x^2-8x+16=21$$
$$3: (x-4)^2=21$$
$$4: x-4=\pm\sqrt{21}$$
$$x=\pm\sqrt{21} +4$$
So I have a question, why is it plus or minus? I thought the square root is usually positive?
|
We have the equation
$$x^2 - 8x = 5$$
Add $16$ to both sides so we can factor the LHS in a nice way, so
\begin{align}
x^2 - 8x + 16 &= 5 + 16 \\
(x-4)^2 &= 21
\end{align}
Now take the square root on both sides (we can do this because both sides of the equation are positive terms)
$$ \sqrt{(x-4)^2} = \sqrt{21}$$
Notice that $x$ could be a positive or negative number and still make the equation true, we have to consider such cases, so
\begin{align}
x-4 &= \pm \sqrt{21} \\
x &= 4 \pm \sqrt{21}
\end{align}
will be the solutions to the quadratic equation.
|
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|
Infinite series summation
$$\sum^{\infty}_{n=0}\frac{1}{(4n+5)(4n+6)\cdots \cdots (4n+11)}$$
Try: $$\sum^{\infty}_{n=0}\frac{(4n+4)!}{(4n+11)!} = \frac{1}{6!}\sum^{\infty}_{n=0}\frac{(4n+4)!\cdot 6!}{(4n+4+6+1)!} = \frac{1}{6!}\sum^{\infty}_{n=0}\int^{1}_{0}x^{4n+4}(1-x)^6dx$$
$$ \frac{1}{6!}\int^{1}_{0}\left(\sum^{\infty}_{n=0}x^{4n}\right)x^4(1-x)^6dx = \frac{1}{6!}\int^{1}_{0}\frac{x^4(1-x)^6}{1-x^4}dx$$
could some help me to solve it thanks
|
The telescoping series, as answered by lab bhattacharjee, is the good way to go.
With regard to the integral you properly wrote, using long division and partial fraction decomposition, $$\frac{x^4(1-x)^6}{1-x^4}=-x^6+6 x^5-15 x^4+20 x^3-16 x^2+12x-16+\frac{4 x}{x^2+1}+\frac{16}{x+1}$$
|
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|
Definite integral of a rational fraction Can I find the value of $$\int_3^{\infty}\frac{x-1}{(x^2-2x-3)^2}dx$$ by just factoring the fraction?
I tried to wrote:
$$\frac{x-1}{(x^2-2x-3)^2}=\frac{x-1}{(x^2-2x+1-4)^2}=\frac{x-1}{[(x-1)^2-2^2]^2}=\frac{x-1}{(x+1)^2\cdot(x-3)^2}$$ but didn't work out. Any ideas?
|
write $$\frac{x-1}{(x^2-2x-3)^2}$$ as $$-1/8\, \left( x+1 \right) ^{-2}+1/8\, \left( x-3 \right) ^{-2}$$
use that $$\frac{(x-1)}{(x^2-2x-3)^2}=\frac{x-1}{(x+1)^2(x-3)^2}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x-3}+\frac{D}{(x-3)^2}$$
it is
$$\frac{x-1}{(x^2-2x-3)^2}=\frac{A(x+1)(x-3)^2+B(x-3)^2+C(x-3)(x+1)^2+D(x+1)^2}{(x+1)^2(x-3)^2}$$
|
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|
Using only the digits 2,3,9, how many six-digit numbers can be formed which are divisible by 6?
Using only the digits $2,3,9$, how many six-digit numbers can be formed which are divisible by $6$?
The options are:
(A) $41$
(B) $80$
(C) $81$
(D) $161$
The last digit must be $2$. But I faced problem when calculating the number of number which are divisible by $3$. Somebody please help me.
|
Let's find a divisability test for 6.
\begin{equation}
1 = 1 \mod 6\\
10 = 4 \mod 6\\
100 = 4*10 = 4 \mod 6,\\
\text{and so on for higher powers of 10}
\end{equation}
Thus, we find: a number X is divisable by 6 iff, cutting of the last digit, taking the sum of the other digits times 4 and adding the last digit the result is divisable by 6.
You are asked for a 6 digit number using only $2,3,9$. We are thus asked to find $a,b,c,d,e,f \in {2, 3, 9}$ such that $4 * (a + b + c + d + e) + f = 0 \mod 6$. As you noted, the last digit must be $2$, which you can conclude from the equation above quite easily by noticing that f must be even. So we conclude $4 * (a + b + c + d + e) = 4 \mod 6$ thus $a + b + c + d + e = {1,4} \mod 6$ and $a + b + c + d + e = 1 \mod 3$ follows. Since both $6$ and $9$ reduce modulo 6, either 2 or 5 higher digits must be equal to $2$, the rest can be chosen freely.
\begin{equation}
\binom{5}{2} * 2^3 + \binom{5}{5} * 2^0 = 81
\end{equation}
|
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|
If $a_{1}>2$and $a_{n+1}=a_{n}^{2}-2$ then Find $\sum_{n=1}^{\infty}$ $\frac{1}{a_{1}a_{2}......a_{n}}$ Question If $a_{1}>2$and $\left\{ a_{n}\right\} be$ a recurrsive
sequence defined by setting $a_{n+1}=a_{n}^{2}-2$ then Find
$\sum_{n=1}^{\infty}$$\frac{1}{a_{1}a_{2}......a_{n}}$
Book's Answer
I have mentioned my problem in the image.Any and all help will be appreciated
|
From $a_n^2-4 = a_1^2 \ldots a_{n-1}^2(a_1^2-4)$ one gets $\frac {a_n^2}{a_1^2 \ldots a_{n-1}^2}- \frac 4 {a_1^2 \ldots a_{n-1}^2}= (a_1^2-4)$ therefore $\frac {a_n}{a_1 \ldots a_{n-1}} = \sqrt {\frac 4 {a_1^2 \ldots a_{n-1}^2} + (a_1^2-4)} \tag 1$
Now, since $\lim a_n= +\infty$ it follows $\lim a_1^2 \ldots a_{n-1}^2 = +\infty$ therefore $\lim \frac {a_n}{a_1 \ldots a_{n-1}} = \sqrt {(a_1^2-4)}$
|
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|
Proving that $ (A + I)^n=(2^n-1)A + I $ Assume that $ A^2 = A$ , where $A$ is a square matrix.
Then prove $(A + I)^n=(2^n-1)A + I $
Could someone help ?
|
By indution if $(A+I)^n=(2^n-1)A+I$ and $A^2=A$, we have
\begin{eqnarray}
(A+I)^{n+1}&=&(A+I)^n(A+I)=((2^n-1)A+I)(A+I)\\
&=&(2^n-1)A^2+A+(2^n-1)A+I\\
&=&2(2^n-1)A+A+I\\
&=&(2^{n+1}-2)A+A+I\\
&=&(2^{n+1}-1)A+I
\end{eqnarray}
|
{
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"url": "https://math.stackexchange.com/questions/2606699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Form an equation whose roots are $(a-b)^2,(b-c)^2,(c-a)^2.$ If $a,b,c$ are the roots of the equation $x^3+x+1=0,$ Then the equation whose roots are $(a-b)^2,(b-c)^2,(c-a)^2.$
Try: $a+b+c=0,ab+bc+ca=1,abc=-1$
Now $(a-b)^2+(b-c)^2+(c-a)^2=2(a+b+c)^2-6(ab+bc+ca)=-6$
Could some help me to explain short way to calculate product,Thanks
|
There is the following formula.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, $$\prod_{cyc}(a-b)^2=27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)=27\left(-\frac{4}{27}-1\right)=-31.$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Calculate limit with L'Hopital's rule I want to calculate the limit $\displaystyle{\lim_{x\rightarrow 0}\frac{x^2\cos \left (\frac{1}{x}\right )}{\sin x}}$.
I have done the following:
It holds that $\lim_{x\rightarrow 0}\frac{x^2\cos \left (\frac{1}{x}\right )}{\sin x}=\frac{0}{0}$.
So, we can use L'Hopital's rule:
\begin{align*}\lim_{x\rightarrow 0}\frac{x^2\cos \left (\frac{1}{x}\right )}{\sin x}&=\lim_{x\rightarrow 0}\frac{x^2\cos \left (\frac{1}{x}\right )}{\sin x} \\ &=\lim_{x\rightarrow 0}\frac{\left (x^2\cos \left (\frac{1}{x}\right )\right )'}{\left (\sin x\right )'} =\lim_{x\rightarrow 0}\frac{2x\cdot \cos \left (\frac{1}{x}\right )+x^2\cdot \left (-\sin \left (\frac{1}{x}\right )\right )\cdot \left (\frac{1}{x}\right )'}{\cos x} \\ &=\lim_{x\rightarrow 0}\frac{2x\cdot \cos \left (\frac{1}{x}\right )-x^2\cdot \sin \left (\frac{1}{x}\right )\cdot \left (-\frac{1}{x^2}\right )}{\cos x} \\ & =\lim_{x\rightarrow 0}\frac{2x\cdot \cos \left (\frac{1}{x}\right )+\sin \left (\frac{1}{x}\right )}{\cos x}=\lim_{x\rightarrow 0}\left (2x\cdot \cos \left (\frac{1}{x}\right )+\sin \left (\frac{1}{x}\right )\right ) \\ & =\lim_{x\rightarrow 0}\left (2x\cdot \cos \left (\frac{1}{x}\right )\right )+\lim_{x\rightarrow 0}\left (\sin \left (\frac{1}{x}\right )\right )\end{align*}
We calculate the two limits separately
*
*$\lim_{x\rightarrow 0}\left (2x\cdot \cos \left (\frac{1}{x}\right )\right )$ :
\begin{equation*}\left |\cos \left (\frac{1}{x}\right )\right |\leq 1 \Rightarrow -1\leq \cos \left (\frac{1}{x}\right )\leq 1 \Rightarrow -2x\leq 2x\cdot \cos \left (\frac{1}{x}\right )\leq 2x\end{equation*} we consider the limit $x\rightarrow 0$ and we get \begin{equation*}\lim_{x\rightarrow 0} \left (2x\cdot \cos \left (\frac{1}{x}\right ) \right )=0\end{equation*}
*
*How can we calculate the limit $\lim_{x\rightarrow 0}\left (\sin \left (\frac{1}{x}\right )\right )$ ?
|
The limit $$\lim_{x\rightarrow 0}\left (\sin \left (\frac{1}{x}\right )\right )$$
does not exist.
Notice that as $x\to 0^+$, $1/x\to \infty$ .
Therefore sin(1/x) jumps up and down between $-1$ and $1$ infinitely many times.
Thus the limit $$\lim_{x\rightarrow 0}\left (\sin \left (\frac{1}{x}\right )\right )$$ does not exist.
You need to try a different approach to avoid this limit.
|
{
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"url": "https://math.stackexchange.com/questions/2614160",
"timestamp": "2023-03-29T00:00:00",
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|
Limits of power of fractions I'm having trouble solving these limits:
$$\lim_{x\to \infty} \left (\frac{3-x}{2x+5}\right )^{3x}\\\lim_{x\to \infty} \left (\frac{2x+1}{3x-4}\right )^{1-2x}\\\lim_{x\to \infty} \left (\frac{2-3x}{1-3x}\right )^{x+4}$$
Are they even defined?
|
Hint: Knowing that $$e^a =\lim_{x\to \infty}\left(1+\frac{a}{x}\right)^x$$
show by yourself that:
$$\lim_{x\to \infty} \left (\frac{3-x}{2x+5}\right )^{3x}~~~~~~~~DNE$$
$$\lim_{x\to \infty} \left (\frac{2x+1}{3x-4}\right )^{1-2x}= \lim_{x\to \infty} \left (\frac{3}{2}-\frac{11}{4x+2}\right )^{2x-1}=\infty$$
$$\lim_{x\to \infty} \left (\frac{2-3x}{1-3x}\right )^{x+4}=\lim_{x\to \infty} \left (1-\frac{1}{3x-1}\right )^{3x-1 \frac{x+4}{3x-1}} =e^{-\frac{1}{3}}$$
use similar idea as here Proving that $\lim_{n\to\infty}\left ( \frac{2n-1}{2n+3} \right )^n=e^{-2}$ without using de l‘Hôspital
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Integral involving multiple variables $$\int\dfrac{a+bx^2}{\sqrt{3ax+bx^3}}\mathrm dx$$
So this is what I did:
$u = 3ax + bx^3$
$\dfrac{\mathrm du}{\mathrm dx}= 3a + 3bx^2$
$\mathrm du = 3a + 3bx^2 dx$
= $\displaystyle\int 3\cdot \frac{1}{\sqrt{u}}\mathrm du$
= $3\cdot \frac{u^{\frac{1}{2}}}{\left(\frac{1}{2}\right)}+c$
= $3\cdot \frac{\left(3ax+bx^3\right)^{\frac{1}{2}}}{\left(\frac{1}{2}\right)}+c$
but this is incorrect.
The right answer is actually
$$\frac{2}{3}\sqrt{3ax+bx^3}+c$$
which is not equivalent to the above answer even in simplest form.
Any help?
|
$$\begin{align}I&=\int\dfrac{a+bx^2}{\sqrt{3ax+bx^3}}\mathrm dx\\&=\dfrac13\int\dfrac{3a+3bx^2}{\sqrt{3ax+bx^3}}\,\mathrm dx\\&=\dfrac13\int\dfrac{\mathrm d(3ax+bx^3)}{\sqrt{3ax+bx^3}}\qquad\text{Use substitution now, let }u=3ax+bx^3\\&=\dfrac13\int\dfrac{\mathrm du}{\sqrt u}\\&=\dfrac13\cdot 2\sqrt u+C\\&=\dfrac23\sqrt{3ax+bx^3}+C\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2620169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
How to prove $\lim \limits_{x \to 1^-} \sum\limits_{n=0}^\infty (-1)^nx^{n²} = \frac{1}{2} \ $? $\lim \limits_{x \to 1^-} \displaystyle \sum_{n=0}^\infty (-1)^nx^{n²} = \frac{1}{2}$
The power $n^2$ is problematic. Can we bring this back to the study of usual power series?
I do not really have any idea for the moment.
|
EDITED. Here is yet another answer based on my recent answer. Indeed, if $P$ is a non-constant polynomial with coefficients in $\mathbb{R}$ such that $P(n) \to +\infty$ as $n \to +\infty$, one immediately deduces from the result in the link that
$$ \lim_{x \uparrow 1^-} \sum_{n=0}^{\infty} (-1)^n x^{P(n)}
= \lim_{s \to 0^+} \sum_{n=0}^{\infty} (-1)^n e^{-P(n)s}
= \frac{1}{2}, $$
which entails OP's question as a special case with $P(n) = n^2$.
Here is an elementary derivation. First, let $g : (0,\infty) \times (0, 1) \to \mathbb{R}$ by
$$ g(a,x) = \frac{1 - x^{a}}{1 - x^{2a+2}}. $$
We make the following observations on $g$.
Observation. $g$ is increasing in $a$ and non-increasing in $x$.
Its proof is more of less calculus computations, so we leave it to the end. To see how this function is related to our problem, notice that
$$ f(x) = \sum_{n=0}^{\infty} (-1)^n x^{n^2} = \sum_{n=0}^{\infty} \left( x^{4n^2} - x^{4(n+1)^2} \right) g(4n+1, x). $$
We prove that liminf and limsup of $f(x)$ as $x \uparrow 1$ are both $\frac{1}{2}$.
Liminf. An immediate consequence is that $g(4n+1, x) \geq \lim_{r\uparrow 1}g(4n+1, r) = \frac{4n+1}{8n+4}$. So for each fixed $N \geq 1$, we can bound $f(x)$ below first by truncating first $N$ terms and then by utilizing the aforementioned lower bound of $g(4n+1, x)$:
\begin{align*}
f(x)
&\geq \sum_{n=N}^{\infty} \left( x^{4n^2} - x^{4(n+1)^2} \right) \frac{4n+1}{8n+4} \\
&\geq \frac{4N+1}{8N+4} \sum_{n=N}^{\infty} \left( x^{4n^2} - x^{4(n+1)^2} \right)
= \frac{4N+1}{8N+4} x^{4N^2}.
\end{align*}
So it follows that
$$ \liminf_{x\uparrow 1}f(x) \geq \frac{4N+1}{8N+1} \xrightarrow[\quad N\to\infty \quad]{} \frac{1}{2}. $$
Limsup. For the other direction, fix $\epsilon > 0$ and define $N = N(\epsilon, x) = \lfloor \epsilon / \log(1/x) \rfloor$. Then for $x$ close to $1$, the sum of first $N$ terms can be bounded by using $g(4n+1, x) \leq g(4N-3, x)$:
\begin{align*}
\sum_{n=0}^{N-1} \left( x^{4n^2} - x^{4(n+1)^2} \right) g(4n+1, x)
&\leq \sum_{n=0}^{N-1} \left( x^{4n^2} - x^{4(n+1)^2} \right) g(4N-3,x) \\
&\leq g(4N-3,x)
= \frac{1 - e^{(4N-3)\log x}}{1 - e^{(8N-4)\log x}} \\
&\to \frac{1-e^{-4\epsilon}}{1-e^{-8\epsilon}} \quad \text{as } N \to \infty.
\end{align*}
For the remaining terms, we may utilize $g(4n+1, x) \leq g(\infty,x) = 1$ to obtain
\begin{align*}
\sum_{n=N}^{\infty} \left( x^{4n^2} - x^{4(n+1)^2} \right) g(4n+1, x)
&\leq \sum_{n=N}^{\infty} \left( x^{4n^2} - x^{4(n+1)^2} \right) \\
&= x^{4N^2}
= e^{4N^2 \log x}
\to 0 \quad \text{as } N \to \infty.
\end{align*}
So it follows that
$$ \limsup_{x\uparrow 1}f(x)
\leq \frac{1-e^{-4\epsilon}}{1-e^{-8\epsilon}}
\xrightarrow[\quad \epsilon \downarrow 0 \quad]{} \frac{1}{2}. $$
Here is the proof of the observation:
*
*We notice that
$$ \frac{\partial g}{\partial a}(a,x) = \frac{x^a \log (1/x)}{(1-x^{2a+2})^2} \left(x^{2a+2}-2 x^{a+2}+1\right) > 0 $$
since $x^{2a+2}-2 x^{a+2}+1 = x^2(x^a - 1)^2 + (1-x^2) > 0$. So $g$ is increasing in $a$ for any $x \in (0, 1)$.
*Similarly, we find that
$$ \frac{\partial g}{\partial x}(a,x) = - \frac{x^{a-1}}{(1-x^{2a+2})^2} \left( (a+2)x^{2a+2} + a - (2a+2) x^{a+2} \right). $$
By the AM-GM inequality, we have
$$ \frac{a+2}{2a+2} \cdot x^{2a+2} + \frac{a}{2a+2} \cdot 1 \geq x^{a+2} $$
and hence $g$ is non-increasing in $x$ for any $a \in (0, \infty)$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the centre of a circle given the equation of a tangent to the circle and the x coordinate of the centre The line with equation $2x+y-5=0$ is a tangent to the circle with equation $(x-3)^2 + (y-p)^2=5$
Find the two possible values of $p$.
|
The equation
$$(x-3)^2+(5-2x-p)^2=5$$
or
$$x^2+9-6x+4x^2+q^2-4qx=5$$
where $q=5-p $
must have only one solution.
$$5x^2-2x (2q+3)+4=0$$
$$\delta'=(2q+3)^2-20=0$$
thus
$$p=5-q=5-\frac {-3\pm 2\sqrt{5}}{2} $$
$$=\frac {13}{2}\pm \sqrt {5} $$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Find a combinatorial proof for $\binom{n+1}{k} = \binom{n}{k} + \binom{n-1}{k-1} + ... + \binom{n-k}{0}$
Let $n$ and $k$ be integers with $n \geq k \geq 0$. Find a combinatorial proof for
$$\binom{n+1}{k} = \binom{n}{k} + \binom{n-1}{k-1} + \cdots + \binom{n-k}{0} .$$
My approach:
I was thinking to use the binomial formula as in $$2^n = \sum{\binom{n}{k}1^k1^{n-k}} .$$
I also tried to use Pascal's Identity $\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}$.
|
A combinatorial proof based upon lattice paths.
The number of lattice paths of length $n+1$ from $(0,0)$ to $Q_k=(n-k+1,k)$ consisting of $(1,0)$-steps and $(0,1)$-steps is $$\binom{n+1}{n-k+1}=\binom{n+1}{k}$$ since we have to choose precisely $n-k+1$ $(1,0)$-steps from a total of $n+1$ steps.
We observe that each path which goes from $(0,0)$ to $Q_k$ crosses the vertical line $x=n-k$. We can therefore partition all $\binom{n+1}{k}$ paths according to points $P_j=(n-k,j), 0\leq j\leq k$ which are followed by a horizontal step to $(n-k+1,j)$ and the vertical steps which lead to $Q_k$.
Since there are precisely $\binom{n-k+j}{j}$ paths from $(0,0)$ to $P_j$ the number of all paths from $(0,0)$ to $Q_r=(n-k+1,k)$ is
\begin{align*}
\color{blue}{\binom{n+1}{k}=\binom{n}{k}+\binom{n-1}{k-1}+\cdots+\binom{n-k}{0}}
\end{align*}
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Solving $ \frac{x^2}{x-1} + \sqrt{x-1} +\frac{\sqrt{x-1}}{x^2} = \frac{x-1}{x^2} + \frac{1}{\sqrt{x-1}} + \frac{x^2}{\sqrt{x-1}}$
Find all real values of $ x>1$ which satisfy$$ \frac{x^2}{x-1} + \sqrt{x-1} +\frac{\sqrt{x-1}}{x^2} = \frac{x-1}{x^2} + \frac{1}{\sqrt{x-1}} + \frac{x^2}{\sqrt{x-1}}$$
Well I first observe that each factor has its own inverse involve in the equation and rewriting
gives $$\left( \frac{x^2}{x-1}-\frac{x-1}{x^2} \right) + \left(\sqrt{x-1} -\frac{1}{\sqrt{x-1}} \right)+\left(\frac{\sqrt{x-1}}{x^2} -\frac{x^2}{\sqrt{x-1}}\right) =0 $$
However after expanding all the terms it does not look wise to do so. Does anyone has an idea
|
Let $\sqrt {x-1}=t \implies x= 1+t^2 \quad x^2=(1+t^2)^2$
$$\frac{(1+t^2)^2}{t^2} + t +\frac{t}{(1+t^2)^2} = \frac{t^2}{(1+t^2)^2} + \frac{1}{t} + \frac{(1+t^2)^2}{t}$$
$$(1+t^2)^4 + t^3(1+t^2)^2 +t^3 = t^4 + t(1+t^2)^2 + t(1+t^2)^4$$
$$(t - 1)(t^2 - t + 1) (t^2 + t + 1) (t^4 + 2 t^2 - t + 1)=0 \implies t=1$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2626411",
"timestamp": "2023-03-29T00:00:00",
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|
Simplify $\tan^{-1}\Big[\frac{\cos x-\sin x}{\cos x+\sin x}\Big]$, where $x<\pi$ Write in the simplist form
$$
\tan^{-1}\bigg[\frac{\cos x-\sin x}{\cos x+\sin x}\bigg],\quad x<\pi
$$
My Attempt:
$$
x<\pi\implies -x>-\pi\implies \frac{\pi}{4}-x>\frac{-3\pi}{4}
$$
$$
\frac{\cos x-\sin x}{\cos x+\sin x}=\frac{\cos^2 x-\sin^2 x}{(\cos x+\sin x)^2}=\frac{\cos 2x}{1+2\sin x\cos x}=\frac{\cos 2x}{1+\sin 2x}\\=\frac{\sin(\tfrac{\pi}{2}-2x)}{1+\cos(\tfrac{\pi}{2}-2x)}=\frac{2\sin(\tfrac{\pi}{4}-x)\cos(\tfrac{\pi}{4}-x)}{2\cos^2(\tfrac{\pi}{4}-x)}={\tan(\tfrac{\pi}{4}-x)}\\
\implies {\tan(\tfrac{\pi}{4}-x)}=\tan\big[\tan^{-1}\frac{\cos x-\sin x}{\cos x+\sin x}\big]\\
\implies \tan^{-1}\frac{\cos x-\sin x}{\cos x+\sin x}=n\pi+\frac{\pi}{4}-x
$$
If $-\pi/2<\frac{\pi}{4}-x<\pi/2$,
$$
\tan^{-1}\frac{\cos x-\sin x}{\cos x+\sin x}=\frac{\pi}{4}-x
$$
this is what is given as answer.
But, since we have further restricted the domain to $\frac{-\pi}{2}<\frac{\pi}{4}-x<\frac{\pi}{2}\implies\frac{-\pi}{4}<x<\frac{3\pi}{4}$ from $x<\pi$, how can it be same as the original function ?
|
The given expression is defined for $x\neq \frac{3\pi}{4}+k\pi$ with the given condition $x<\pi$.
When you simplify the expression by $\arctan$ you need to restrict the domain to $\frac{-\pi}{4}<x<\frac{3\pi}{4}$ which is not in contrast with the initial one.
Thus I don't see any kind of problem with your answer.
|
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|
Divergence of a Series $\sum_{n=1}^\infty (\frac{1}{n!})(\frac{n}{e})^n$ I'm trying to show that $$\sum_{n=1}^\infty \left(\frac{1}{n!}\right)\left(\frac{n}{e}\right)^n$$ diverges by the Corollary of Raabe's Test.
Corollary of Raabe's Test: Let $X=\{x_n\}$ be a sequence of nonzero real numbers and let $a=\lim_{n \to \infty} (n(1-\frac{|x_n+1|}{x_n}))$, whenever this limit exists. Then the series is absolutely convergent when $a>1$ and is divergent when $a<1$.
What I find:
$$a= \lim_{n \to \infty} (n(1-\frac{1}{e}(1+\frac{1}{n})^n))=\infty \cdot 0$$ (which is indeterminate)
$$\lim_{n \to \infty} (n(1-\frac{1}{e}(1+\frac{1}{n})^n))=\lim_{n \to \infty} \frac{1-\frac{1}{e}(1+\frac{1}{n})^n)}{\frac{1}{n}}$$
Applying L'hospital's rule, I get: $$\lim_{n \to \infty} \frac{-\frac{n}{e}(1+\frac{1}{n})^{n-1}(-\frac{1}{n^2})}{-\frac{1}{n^2}}=\lim_{n \to \infty} (-\frac{n}{e}(1+\frac{1}{n})^{n-1})$$
Help!!! I do not know what to do next.
|
The limit is $a=1/2$.
Start by the approximation
$$\ln \left( \frac 1e (1+x)^{1/x} \right) = -1 + \frac 1x \ln (1+x) = -1 + 1-\frac 12 x +o(x) = -\frac 12 x +o(x) $$
Plugging $x= \frac 1n$ you get
$$1- \frac 1e \left( 1+ \frac 1n \right)^n = 1- e^{\ln \left( \frac 1e (1+\frac 1n)^{n} \right)} = 1- e^{-\frac{1}{2n} + o( 1/n)}$$
Now, again, using the limit
$$\lim_{x \to 0} \frac{1-e^x}{x} = -1$$
we have
$$a= \lim_{n \to \infty} \frac{1- e^{-\frac{1}{2n} + o( 1/n)}}{1/n} =
\lim_{n \to \infty} \frac{1- e^{-\frac{1}{2n} + o( 1/n)}}{-\frac{1}{2n} + o( 1/n)} \cdot \frac{-\frac{1}{2n} + o( 1/n)}{1/n}
= -1 \cdot \left( -\frac{1}{2} \right) = \frac{1}{2}$$
|
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|
Find the natural solutions of $a^3-b^3=999$ I want to find the natural solutions of $a^3-b^3=999$.
I got $a^3-b^3=(a-b)\cdot(a^2+ab+b^2)$, so if we consider the equation in $\mathbb{Z}/3\mathbb{Z}$ we get
$$(a-b)\cdot(a^2+ab+b^2) \equiv0 \text{ mod }3$$
and because $\mathbb{Z}/3\mathbb{Z}$ is a domain, we get
$$a\equiv b \text{ mod } 3 \text{ or } a^2+ab+b^2\equiv0 \text{ mod } 3.$$
Besides, the prime factorization of $999=3^3\cdot37$, but I don't know how to go on.
I would appreciate any hints.
|
So $(a-b)(a^2 + ab + b^2) = 3^3*37$
So $(a-b)|3^3*37$.
So $a-b = 3^j*37^k$ where $j = 0,1,2,3$ and $k=0,1$
And $(a^2 + ab +b^2) = \frac {3^3*27}{a-b} = 3^{3-j}*37^{1-k}$
That's 8 possible systems of equation.
But there are some obvious things to note.
If $37|a-b$ then $a - b \ge 37$ and ... that just seems wrong.
That means $a \ge 37$ and $a^2 + ab + b^2 > 37^2$ but if $37|a+b$ then the very largest that $a^2 + ab + b^2$ can be is $\frac {999}{37} = 27$.
So $37\not \mid a-b$ and $37|a^2 + ab + b^2$.
So $a-b = 3^k; k\le 3; a^2 +ab + b^2 = 3^{3-k}*37$
And there are only 4 systems of equations.
Likewise if $27|a-b$ then $a \ge 27$ and $a^2 + ab + b^2 > 27^2 > 37$ so that's not possible.
So $a-b = 3^k; k\le 2; a^2 + ab + b^2 = 3^{3-k}*37$.
And there are only 3 systems of equations.
$k = 0,1,2$.
If $k = 0$ we have $a= b+ 1$ and $(b+1)^2 + b(b+1) + b^2 = 999$
or $3b^2 + 3b + 1 = 999$ which isn't possible as LHS and RHS are not equivalent mod $3$.
If $k = 1$ we have $a = b+3$ and $(b + 3)^2 + b(b+3)+b^2 = 333$ and
$3b^2 + 9b + 9 = 333$ or
$b^2 + 3b - 108 = 0$ so $b = \frac {-3 +\sqrt {9+432}}2 = \frac {-3 + 21}2 = 9$ and $a = 12$
If $k = 2$ we have $a = b+9$ and $(b+9)^2 + b(b+9) + b^2 = 111$
$3b^2 + 27b + 81 = 111$
$b^2 + 9b + 27 = 37$
$b^2 + 9 - 10 = 0$
$(b+10)(b - 1) = 0$ so $b = 1$ and $a=10$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find all solutions of the equation $z^4+(-2+2\sqrt{2}i)z^3+(3-4\sqrt{2}i)z^2+(4+10\sqrt{2}i)z-10=0$ Find all solutions of the equation:
$z^4+(-2+2\sqrt{2}i)z^3+(3-4\sqrt{2}i)z^2+(4+10\sqrt{2}i)z-10=0$
I don't have any other idea than to guess them.
|
The peculiar form of the polynomial, with the same $\,\sqrt{2}\,$ present in the imaginary parts of the coefficients, only, suggests splitting it into two terms for closer inspection:
$$
\begin{align}
P(z) &= z^4+(-2+2\sqrt{2}i)z^3+(3-4\sqrt{2}i)z^2+(4+10\sqrt{2}i)z-10 \\
&= (z^4-2z^3+3z^2+4z-10) + i \,\sqrt{2}\,(2z^3 - 4 z^2+10z) \\
&= (z^4-2z^3+3z^2+4z-10) + i \,2\sqrt{2}\,z(z^2 - 2 z+5) \\
\end{align}
$$
The quadratic factor $\,z^2 - 2 z+5\,$ in the second term further suggests checking whether the first term maybe has it as a factor as well, and it turns out that it does, indeed:
$$
z^4-2z^3+3z^2+4z-10 = (z^2-2)(z^2 - 2 z+5)
$$
From here on, the polynomial factors, and the rest follows by routine calculations:
$$
P(z) = (z^2 + i\,2\,\sqrt{2}z- 2)(z^2-2z+5)
$$
Disclaimer: the shortcut above only works because the polynomial was specifically set up the way it was. Such "luck" virtually never happens in "real life" outside homework/contest problems.
|
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|
continued fraction of $\sqrt{41}$
Show that $\sqrt{41} = [6;\overline {2,2,12}]$
here's my try:
$$\sqrt{36}<\sqrt{41}<\sqrt{49}\implies6<\sqrt{41}<7\implies\lfloor\sqrt{41}\rfloor=6$$
$$\sqrt{41}=6+\sqrt{41}-6=6+\frac{1}{\frac{1}{\sqrt{41}-6}}$$
$$\frac{1}{\sqrt{41}-6}=\frac{\sqrt{41}+6}{41-36}=\frac{\sqrt{41}+6}{5}=\frac{12+\sqrt{41}-6}{5}=2+\frac{\sqrt{41}-4}{5}$$
So far,
$$\sqrt{41}=6+\frac{1}{2+\frac{\sqrt{41}-4}{5}}=6+\frac{1}{2+\frac{1}{\frac{5}{\sqrt{41}-4}}}$$
But, $$\frac{5}{\sqrt{41}-4}=\frac{5(\sqrt{41}+4)}{41-16}=\frac{\sqrt{41}+4}{5}=\frac{6+\sqrt{41}-2}{5}=\color{red}{1}+\frac{\sqrt{41}-1}{5}$$
It suppose to be $2$ and not $1$.
Where is the mistake? (I triple-checked and it seems fine to me)
|
$$\frac{6+\sqrt{41}-2}{5}=\frac{10+\sqrt{41}-6}{5}=2+\frac{\sqrt{41}-6}{5}$$
with $0<\sqrt{41}-6<5$
|
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|
Evaluate $\int_0^1 \frac{\mathrm{d}x}{\sqrt{1-x^2}}\frac{x }{1-k^2x^2}\log\left(\frac{1-x}{1+x}\right)$ I am trying to evaluate the following integral
$$I(k) = \int_0^1 \frac{\mathrm{d}x}{\sqrt{1-x^2}}\frac{x }{1-k^2x^2}\log\left(\frac{1-x}{1+x}\right)$$
with $0< k < 1$.
My attempt
By performing the substitution
$$y=\frac{1-x}{1+x} \Longleftrightarrow x=\frac{1-y}{1+y}$$
we have
$$ I(k) = \int_0^1 \mathrm{d}y \frac{\log y}{\sqrt{y}}\frac{y-1}{(k^2-1)y^2-2y(k^2+1)+(k^2-1)}$$
Now we can decompose
$$\frac{y-1}{(k^2-1)y^2-2y(k^2+1)+(k^2-1)}= \frac{1}{2k(1+k)}\frac{1}{y+a}- \frac{1}{2k(1-k)}\frac{1}{y+a^{-1}}$$
with $a=\frac{1-k}{1+k}$ runs from $0$ to $1$.
Therefore we can write
$$I(k) = \frac{1}{2k(1+k)}\color{blue}{\int_0^1 \mathrm{d}y \frac{\log y}{\sqrt{y}}\frac{1}{y+a}} - \frac{1}{2k(1-k)}\color{red}{\int_0^1 \mathrm{d}y \frac{\log y}{\sqrt{y}}\frac{1}{y+a^{-1}}}$$
So we have only to evaluate $\color{blue}{I_1}$ and $\color{red}{I_2}$.
Now we consider the following integral
$$ J(\sigma) = \int_0^1 \mathrm{d}x \frac{\log x}{\sqrt{x}}\frac{1}{x-\sigma^2}$$
so that $\color{blue}{I_1}=J(i\sqrt{a})$ and $\color{red}{I_2}=J(i/\sqrt{a})$.
By considering the map $x\mapsto x^2$ we can write
$$ J(\sigma)=4\int_0^1\mathrm{d}x \frac{\log x}{x^2-\sigma^2}= \frac{2}{\sigma}\left[\color{green}{\int_0^1 \frac{\log x}{x-\sigma}}-\color{green}{\int_0^1 \frac{\log x}{x+\sigma}}\right]$$
The problem then reduces to evaluate the $\color{green}{\text{green}}$ integrals. At this point I'm stuck. I think that it needs to be solved by using polylogarithms, but I don't really know how to use these functions.
Mathematica 11.0 says
$$J(\sigma)=4 \left(\frac{\Phi \left(\frac{1}{\sigma ^2},2,\frac{3}{2}\right)}{4 \sigma ^4}+\frac{1}{\sigma ^2}\right)$$
where $\Phi$ is the Lerch transcendent. I don't know if this result is true (numerical integration is somewhat problematic). However, if it is true, I don't know what to do next.
Any hint on how to proceed with the evaluation?
Thanks in advance!
|
A great approach on your part! Let me try another.
$$I(k) = \int_0^1 \frac{\mathrm{d}x}{\sqrt{1-x^2}}\frac{x }{1-k^2x^2}\log\left(\frac{1-x}{1+x}\right)=-2 \sum_{n=0}^\infty \frac{1}{2n+1} \int_0^1 \frac{x^{2n+2}}{\sqrt{1-x^2}}\frac{\mathrm{d}x }{1-k^2x^2} $$
So we really only need to solve the integral in the form:
$$Y(k,n+1)=\int_0^1 \frac{x^{2n+2}}{\sqrt{1-x^2}}\frac{\mathrm{d}x }{1-k^2x^2}$$
Note that:
$$Y(k,0)=\int_0^1 \frac{\mathrm{d}x }{\sqrt{1-x^2}(1-k^2x^2)}=\frac{\pi}{2 \sqrt{1-k^2}}$$
This can be seen directly from the antiderivative (check by differentiation):
$$ \int \frac{\mathrm{d}x }{\sqrt{1-x^2}(1-k^2x^2)} =\frac{1}{\sqrt{1-k^2}} \tan^{-1} \left( \frac{\sqrt{1-k^2} x}{\sqrt{1-x^2}} \right)$$
Let's try something for the general case:
$$Y(k,n+1)=\frac{1}{k^2}\int_0^1 \frac{x^{2n}(k^2x^2-1+1)}{\sqrt{1-x^2}}\frac{\mathrm{d}x }{1-k^2x^2}=$$
$$=-\frac{1}{k^2} \int_0^1 \frac{x^{2n} \mathrm{d}x}{\sqrt{1-x^2}}+\frac{1}{k^2} \int_0^1 \frac{x^{2n}}{\sqrt{1-x^2}}\frac{\mathrm{d}x }{1-k^2x^2}$$
From which we write a recurrence relation:
$$Y(k,n+1)=-\frac{1}{k^2} \int_0^1 \frac{x^{2n} \mathrm{d}x}{\sqrt{1-x^2}}+\frac{1}{k^2} Y(k,n)$$
Let us denote:
$$P(n)=\int_0^1 \frac{x^{2n} \mathrm{d}x}{\sqrt{1-x^2}}$$
Then:
$$Y(k,n+1)=-\frac{1}{k^2} \left( P(n)-Y(k,n) \right)$$
$P(n)$ is just Beta function and has the form:
$$P(n)=\frac{\sqrt{\pi}~ \Gamma \left(n+\frac{1}{2} \right)}{2 n!}$$
In other words, while this solution is by no means complete, we already obtained:
The series representation of the integral:
$$I(k)=-2 \sum_{n=0}^\infty \frac{1}{2n+1} Y(k,n+1) \tag{1}$$
The recurrence relation together with the initial condition, which allows us to compute all the terms:
$$Y(k,n+1)=\frac{1}{k^2} \left(Y(k,n)- \frac{\sqrt{\pi}~ \Gamma \left(n+\frac{1}{2} \right)}{2 n!} \right) \tag{2}$$
$$Y(k,0)=\frac{\pi}{2 \sqrt{1-k^2}} \tag{3}$$
I will check how this works numerically in a couple of days.
As can be seen from the integral, $Y(k,n)$ actually has a solution in terms of Gauss Hypergeometric function:
$$Y(k,n)=\frac{\sqrt{\pi}~ \Gamma \left(n+\frac{1}{2} \right)}{2 n!} {_2 F_1} \left(1, n+\frac{1}{2};n+1; k^2 \right)$$
But the recurrence relation is a more simple way to obtain the values without special software. $P(n)$ can be converted into the form of rational multiples of $\pi$.
Final remark: while this doesn't seem to lead us to a closed form, let me note that this approach works for a large class of integrals, as long as we can expand the non-algebraic function as a series.
Trying to derive a rational form for $P(n)$:
$$ \Gamma \left(n+\frac{1}{2} \right)=\left(n-1+\frac{1}{2} \right)\left(n-2+\frac{1}{2} \right) \cdots \left(1+\frac{1}{2} \right) \frac{1}{2} \Gamma \left(\frac{1}{2} \right)=$$
$$=\left(n-1+\frac{1}{2} \right)\left(n-2+\frac{1}{2} \right) \cdots \left(1+\frac{1}{2} \right) \frac{\sqrt{\pi}}{2}= $$
$$ =\left(2n-1 \right)\left(2n-3 \right) \cdots 3 \cdot \frac{\sqrt{\pi}}{2^n} $$
So:
$$P(n)=\frac{\pi}{2} \frac{1 \cdot 3 \cdots (2n-3)(2n-1) }{2^n~n!}$$
Separate case: $P(0)= \pi/2$.
Edit 2
I have checked the algorithm numerically, and it works, but the convergence is terrible, and a huge loss of significance is involved.
|
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|
What is $\lim_{x\to 2} \frac{\sqrt{x+2}-2}{x-2}$? I tried multiplying by the conjugate which gave:
$$\frac{x-2}{(x-2)\sqrt{x+2}+2x-4}$$
But i'm still gettting $\frac{0}{0}$. According to my textbook the answer should be $\frac{1}{4}$, but how do I get there?
|
Hint
$$\frac{\sqrt{x+2}-2}{x-2}\cdot \frac{\sqrt{x+2}+2}{\sqrt{x+2}+2}= \frac{\sqrt{x+2}^2-2^2}{(x-2)\sqrt{x+2}+2}$$
$$ \frac{x-2}{(x-2)\sqrt{x+2}+2}= \frac{1}{\sqrt{x+2}+2}$$
|
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|
Find the minimum value of $f(x,y,z)=\frac{x^2}{(x+y)(x+z)}+\frac{y^2}{(y+z)(y+x)}+\frac{z^2}{(z+x)(z+y)}$. Find the minimum value of $f(x,y,z)=\frac{x^2}{(x+y)(x+z)}+\frac{y^2}{(y+z)(y+x)}+\frac{z^2}{(z+x)(z+y)}$ for all notnegative value of $x,y,z$.
I think that minimum value is $\frac{3}{4}$ when $x=y=z$ but I have no prove.
|
By C-S $$\sum_{cyc}\frac{x^2}{(x+y)(x+z)}\geq\frac{(x+y+z)^2}{\sum\limits_{cyc}(x+y)(x+z)}=\frac{\sum\limits_{cyc}(x^2+2xy)}{\sum\limits_{cyc}(x^2+3xy)}\geq\frac{3}{4},$$
where the last inequality it's $$\sum_{cyc}(x-y)^2\geq0.$$
The equality occurs for $x=y=z$, which says that we got a minimal value.
Another way:
We need to prove that:
$$4\sum_{cyc}x^2(y+z)\geq3\prod_{cyc}(x+y)$$ or
$$\sum_{cyc}(x^2y+x^2z-2xyz)\geq0$$ or
$$\sum_{cyc}z(x-y)^2\geq0.$$
|
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|
Limit of $\frac{x^2-\log(1+x^2)}{x^2\sin^2x}$ as $x$ goes to $0$ As plugging $0$ in $\frac{x^2-\log(1+x^2)}{x^2\sin^2x}$ makes the function becomes undetermined form of $\frac{0}{0}$. I tried applying L'Hospital's rule but it became messy and did not look helpful if I do further differentiation. So I tried finding the taylor polynomial of $\log(1+x^2)$, which is $x^2-\frac{1}{2}x^4+$... to see if I could bound the absolute value of the fraction above with something like $|\frac{x^2-x^2}{x^2\sin^2x}|$ and apply the comparison lemma, but it looks like that does not work too...
|
Note that by Taylor's series
*
*$\log(1+x^2)=x^2-\frac{x^4}{2}+o(x^4)$
*$\sin x^2=x^2+o(x^2)$
thus
$$\frac{x^2-\log(1+x^2)}{x^2\sin^2x}=\frac{x^2-x^2+\frac{x^4}{2}+o(x^4)}{x^4+o(x^4)}=\frac{\frac12+o(1)}{1+o(1)}\to \frac12$$
|
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|
Addition of $2$ Events
Let $X$ and $Y$ be independent, each uniformly distributed on $\{1, 2, ..., n\}$. Find $P(X + Y = k)$ for $2 \le k \le 2n$.
\begin{align}P(X + Y = k) &= \sum_{(x,y)\,:\,x+y=k} P(x, y) \\
&= \sum_{(x,y)\,:\,x+y=k} \frac{1}{n^2} \\
&= (k - 1)\frac{1}{n^2} \\
&= \frac{k-1}{n^2} \end{align}
When $k = 2: (1, 1)$
When $k = 3: (1, 2), (2, 1)$
When $k = 4: (1, 3), (3, 1), (2, 2)$
When $k = 5: (1, 4), (4, 1), (2, 3), (3, 2)$
$$\#(x, y) = k - 1$$
Textbook Answer:
$\frac{k-1}{n^2}\,\,\,$ for $\,\,\,2 \le k \le n+1$
$\frac{2n-k+1}{n^2}\,\,\,$ for $\,\,\,n+2 \le k \le 2n$
Why are there $2$ intervals being considered?
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The first formula fails to find the probability after $n+1$, because:
When $n=2$, the sample space is: $(1,1),(1,2),(2,1),(2,2)$, while $2\le k \le 4$ and
$$k=2: (1,1); k=3: (1,2),(2,1); k=4: (2,2)$$
Note that when $k=4$, we are ignoring $(1,3),(3,1)$, because they are not in the sample space. So the two formulas must be applied: one until the middle, another after it.
|
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|
Definite Integral = $\int_0^{2\pi} \frac{\sin^2\theta}{(1-a\cos\theta)^3}\,d\theta$ for $0\le a<1$ I am trying to find an expression for the following Definite Integral for the range $0 \leq a <1$:
$$D = \int_0^{2\pi} \frac{\sin^2\theta}{(1-a\cos\theta)^3}\,d\theta.$$
which gives (Wolfram Alpha)
$$D= \left[
\frac{\sin \theta(\cos \theta - a)}{2(a^2-1)(a \cos\theta-1)^2}
+\frac{\tanh^{-1}\left( \frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}} \right)}{(a^2-1)^{3/2}}\right]_0^{2\pi}
.$$
which can be expressed as
$$D= \left[
\frac{(a^2-1)^{1/2}\sin \theta(\cos \theta - a)+2 (a \cos\theta-1)^2\tanh^{-1}\left( \frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}} \right)}{2(a^2-1)^{3/2}(a \cos\theta-1)^2}
\right]_0^{2\pi}
.$$
This expression involves discontinuities and complex numbers which is beyond my present abilities to handle.
|
We can use the standard formula $$\int_{0}^{2\pi}\frac{dx}{1-a\cos x} =\frac{2\pi}{\sqrt{1-a^2}},|a|<1\tag{1}$$ Note that $$\frac{d} {dx} \frac{\sin x} {1-a\cos x} =\frac{\cos x-a} {(1-a\cos x) ^2} =-\frac{1}{a}\frac{a^2-1+1-a\cos x} {(1-a\cos x) ^2}$$ and integrating the above with respect to $x$ in interval $[0,2\pi]$ we get $$0=\frac{1-a^2}{a}\int_{0}^{2\pi}\frac{dx}{(1-a\cos x) ^2}-\frac{1}{a}\int_{0}^{2\pi}\frac{dx}{1-a\cos x} $$ ie $$\int_{0}^{2\pi}\frac{dx}{(1-a\cos x) ^2}=\frac{2\pi}{(1-a^2)^{3/2}}\tag{2}$$ Next we can observe that $$\frac{d} {dx} \frac{\sin x} {(1-a\cos x) ^2}=\frac{\cos x-a-a\sin^2x}{(1-a\cos x) ^3}$$ and integrating the above we obtain $$0=-aI+\frac{1-a^2}{a}\cdot J-\frac{1}{a}\cdot\frac{2\pi}{(1-a^2)^{3/2}}$$ ie $$(1-a^2)J-a^{2}I=\frac{2\pi}{(1-a^2)^{3/2}}\tag{3}$$ where $I$ is the integral in question and $J=\int_{0}^{2\pi}(1-a\cos x) ^{-3}\,dx$. We need another relationship between $I, J$ to evaluate both of them. Note that if $t=\cos x$ then $$\frac{\sin^2x}{(1-a\cos x) ^3}=\frac{1-t^2}{(1-at)^3}=\frac{A} {1-at}+\frac{B}{(1-at)^{2}}+\frac{C}{(1-at)^3}\tag{4}$$ where $$A=-\frac{1}{a^2},B=\frac{2}{a^2},C=-\frac{1-a^2}{a^2}$$ and on integrating equation $(4)$ we get $$a^2I+(1-a^2)J=\frac{2\pi(1+a^2)}{(1-a^2)^{3/2}}\tag{5}$$ Subtracting equation $(3)$ from equation $(5)$ we get $$I=\frac{\pi} {(1-a^2)^{3/2}}$$ Comparing this elementary solution with the one offered by Jack D'Aurizio shows us the power of Feynman's technique.
Formula $(1)$ is an easy consequence of the following result $$\int_{0}^{\pi}\frac{dx}{a-b\cos x} =\frac{\pi} {\sqrt{a^2-b^2}},a>|b|\tag{6}$$ which is (not so) easily proved via the substitution $$(a-b\cos x) (a+b\cos y) =a^2-b^2$$ The same substitution $$(1-a\cos x) (1+a\cos y) =1-a^2$$ has been used directly in this answer to evaluate the integral in question with much less effort.
|
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|
U-substitution of 2x in trigonometric substitution Find
$$\int^{3\sqrt{3}/2}_0\frac{x^3}{(4x^2+9)^{3/2}} \,\mathrm{d}x.$$
The text says to use substitution of $u = 2x$. How did they get $u = 2x$ and not $u = x^3$?
|
When $u = 2x$, $du = 2dx$,
$$\int^{3\sqrt3/2}_0\frac{x^3}{(4x^2+9)^{3/2}}dx
= \int^{3\sqrt{3}}_0\frac{u^3/8}{(u^2+9)^{3/2}}\frac{du}{2},
$$
this allows us to do further substitution $u = 3 \tan t$ to get rid of the root sign in the denominator.
If you use $u = x^3$, $du = 3x^2 dx$,
$$\int^{3\sqrt3/2}_0\frac{x^3}{(4x^2+9)^{3/2}}dx
= \int^{81\sqrt{3}/8}_0\frac{\sqrt[3]{u}/3}{(u^{2/3}+9)^{3/2}}du,
$$
which is more complicated then the first equality as this involves the cubic root.
|
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|
Showing that $ 1 + 2 x + 3 x^2 + 4 x^3 + \cdots + x^{10} = (1 + x + x^2 + x^3 + x^4 + x^5)^2$ I was studying a polynomial and Wolfram|Alpha had the following alternate form:
$$P(x) = 1 + 2 x + 3 x^2 + 4 x^3 + 5 x^4 + 6 x^5 + 5 x^6 + 4 x^7 + 3 x^8 + 2 x^9 + x^{10} = (1 + x + x^2 + x^3 + x^4 + x^5)^2$$
Of course, we can verify this through expansion, but if I were a mathematician without access to CAS, how might I notice that this is the case?
I suppose what I'm asking is how one should "see" that $P$ can be simplified to $(1 + x + x^2 + x^3 + x^4 + x^5)^2$? Is it a multinomial thing (which seems a bit too complicated for someone to "notice"), or is there something simpler about the polynomial that one could use to factor it?
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It is easier to expand $$(1 + x + x^2 + x^3 + x^4 + x^5)^2$$
to get $$ 1 + 2 x + 3 x^2 + 4 x^3 + 5 x^4 + 6 x^5 + 5 x^6 + 4 x^7 + 3 x^8 + 2 x^9 + x^{10}$$
All we have to do is check the squares and twice the products to see if the coefficients are correct.
How do we see that P(x) is a perfect square?
When evaluated at different integer values of x, we get perfect squares so, that may be helpful to make an intelligent guess.
|
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|
Solve the equation in positive integers $k^m+m^n=kmn$
Solve the equation in positive integers $$k^m+m^n=kmn$$
My work so far:
1) Let $k=m$. Then
$$k^k+k^n=k^2n$$
If $k=1$ then $n=2$; if $k=2$ then $n=2$ or $n=3$
Let $k\ge 2$. We see that $n\ge2$.
$$k^{k-2}+k^{n-2}=n$$
2) Let $k=n$. Then $$k^m+m^k=k^2m$$. I find $m=2$, $k=2$ or $k=4$
3) Let $m=n$. Then $$k^m+m^k=km^2$$. Similarly.
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Suppose first that $k$ is the largest among $k$, $m$, and $n$. Then, $$k^m<k^m+m^n=kmn\leq k^3$$ implies that $k>1$ and $m<3$. If $m=1$, then we have $k+1=kn$, which is not possible (recalling that $k>1$). If $m=2$, then $k^2+2^n=2kn$ implies that
$$0\leq (k-n)^2=n^2-2^n\,,$$
yielding $n\in \{2,3,4\}$. Hence, in this case where $k$ is the largest, the solutions in the form $(k,m,n)$ are $(2,2,2)$, $(4,2,3)$, and $(4,2,4)$.
Now, suppose that $m$ is the largest of the three variables. Then,
$$m^n<k^m+m^n=kmn\leq m^3 $$
implies that $m>1$ and $n<3$. We start with the subcase $n=1$, which leads to
$$k^m+m=km\text{ or }k^m = (k-1)m\,.$$
From the equation above, $k-1$ divides $k^m$, which only happens when $k=2$.
Nonetheless, this leads to a contradiction $2^m=m$. Thus, $n=2$ is the only hope here. Plugging $n=2$ into the original equation, we get
$$k^m+m^2=2km\text{ or }0\leq (m-k)^2=k^2-k^m\,.$$
This proves that $k=1$ or $m=2$. If $k=1$, then $m=1<2=n$, which is absurd. Thus, $m=2$ and we obtain $(k,m,n)=(2,2,2)$.
Finally, we assume that $n$ is the largest. We note that
$$m^n<k^m+m^n=kmn\leq n^3\,.$$
The inequality above implies that $m<4$. There are a few easy-to-check scenarios listed below.
*
*$m=1$: this case leads to $(k,m,n)=(1,1,2)$.
*$m=2$: this case leads to $(n-k)^2=n^2-2^n$, which, as before, results in possible solutions in the form $(k,m,n)$ being $(2,2,2)$, $(2,2,3)$, and $(4,2,4)$ (recalling the constraint $k\leq n$).
*$m=3$: this case leads to $3^n<k^3+3^n=3kn\leq 3n^2$, or $3^{n-1}<n^2$. We have immediately that $n=2$, but this contradicts the assumption that $n\geq m=3$.
In summary, there are five solutions $(k,m,n)\in \left(\mathbb{Z}_{>0}\right)^3$ to $k^m+m^n=kmn$. All solutions in the form $(k,m,n)$ are $(1,1,2)$, $(2,2,2)$, $(2,2,3)$, $(4,2,3)$, and $(4,2,4)$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to calculate the value of $\sum\limits_{k=0}^{\infty}\frac{1}{(3k+1)\cdot(3k+2)\cdot(3k+3)}$? How do I calculate the value of the series $$\sum_{k=0}^{\infty}\frac{1}{(3k+1)\cdot(3k+2)\cdot(3k+3)}= \frac{1}{1\cdot2\cdot3}+\frac{1}{4\cdot5\cdot6}+\frac{1}{7\cdot8\cdot9}+\cdots?$$
|
You can calculate the partial sum using
$$
\frac{1}{(3k+1)(3k+2)(3k+3)}=\frac{1}{2}\frac{1}{3k+1}-\frac{1}{3k+2}+\frac{1}{2}\frac{1}{3k+3}
$$
|
{
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"url": "https://math.stackexchange.com/questions/2644864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Finding $\lim x_n$ when $\left( 1+\frac{1}{n}\right)^{n+x_n}=1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}$
Let $x_n$ be the unique solution of the equation $$\left( 1+\frac{1}{n}\right)^{n+x_n}=1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}$$
Find $\lim_{n \to \infty} x_n$
I think that the limit must be $\frac{1}{2}$, because $\left(1+\frac{1}{n}\right)^{n+\frac{1}{2}}$ is decreasing and convergent to $e$, while $1+\frac{1}{1!}+\dots+\frac{1}{n!}$ is increasing and convergent to $e$, so $$\left(1+\frac{1}{n}\right)^{n+\frac{1}{2}}>1+\frac{1}{1!}+\dots+\frac{1}{n!}$$
which means that $\frac{1}{2}>x_n$. I also know that for $a \in [0,\frac{1}{2}), \left(1+\frac{1}{n}\right)^{n+a}$ is eventually increasing, but I don't know how to get a lower bound for $x_n$ which goes to $\frac{1}{2}$
|
The limit is indeed $\frac{1}{2}$. Due to Taylor's formula with integral remainder,
$$ \sum_{k=0}^{n}\frac{1}{k!} = e-\int_{0}^{1}\frac{(1-t)^n}{n!}\,e^t\,dt=e\left(1+O\left(\tfrac{1}{(n+1)!}\right)\right)=\exp\left(1+O\left(\tfrac{1}{(n+1)!}\right)\right)\tag{1} $$
while
$$ \left(1+\frac{1}{n}\right)^{n+x}=\exp\left[(n+x)\left(\frac{1}{n}-\frac{1}{2n^2}+o\left(\frac{1}{n^2}\right)\right)\right]=\exp\left[1+\frac{x-\frac{1}{2}}{n}+o\left(\frac{1}{n}\right)\right]\tag{2} $$
so by equating the RHSs of $(1)$ and $(2)$ we get $\lim_{n\to +\infty}x_n=\frac{1}{2}$ as expected.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Equation of a circle cutting another given circle orthogonally What is the equation of the circle which touches the line $x+y=5$ at $(-2,7)$ and cut the circle $$x^2+y^2+4x-6y+9=0$$ orthogonally?
I tried to denote the center of circle as $(h,k)$ and radius as $r$. Orthonogallity implies
\begin{align*}
(h+2)^2+(k-3)^2&=r^2+4,\\
(h+2)^2+(k-7)^2&=r^2.
\end{align*}
Solving equation gives $k=\dfrac{11}{2}$. How to find $h,r$?
|
The equation of tangent to a circle is given by $T=0$
Let the equation of required circle be $x^2+y^2+2gx+2fy+c=0$ and the tangent to that circle is at point $(-2,7)$. Hence using above information the equation of tangent at that point is given by $$(g-2)x+(f+7)y-2g+7f+c=0$$
But the equation of tangent is given to be $x+y-5=0$
Hence we get $$\frac {g-2}{1}= \frac {f+7}{1}=\frac {2g-7f-c}{5}$$
Using this information we get $g=f+9$ and $c=-10f-17$
Using the condition of orthogonality you already got $f=\frac {-11}{2}$
Now you just have to substitute the values to get the equation of required circle.
|
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|
Given that $X,Y$ are independent $N(0,1)$ , show that $\frac{XY}{\sqrt{X^2+Y^2}},\frac{X^2-Y^2}{2\sqrt{X^2+Y^2}}$ are independent $N(0,\frac{1}{4})$ It is given that $X,Y \overset{\text{i.i.d.}}{\sim} N(0,1)$
Show that $\frac{XY}{\sqrt{X^2+Y^2}},\frac{X^2-Y^2}{2\sqrt{X^2+Y^2}} \overset{\text{i.i.d.}}{\sim} N(0,\frac{1}{4})$
I was thinking of making polar transformations $X=r \cos \theta, Y=r \sin \theta$
Then I am getting stuck at ranges of $\theta$
|
Setting $U=\dfrac{XY}{\sqrt{X^2+Y^2}}, V=\dfrac{X^2-Y^2}{2\sqrt{X^2+Y^2}}$ and using moment generating functions:
\begin{align}
M_{(U,V)}(u,v) & =\mathbb{E}\left[\mathrm{e}^{\langle\,(u,v)\,;\,(U,V)\,\rangle}\right] \\[10pt]
& =\iint_{\mathbb{R}^2}\exp\left(u\frac{xy}{\sqrt{x^2+y^2}}+v\frac{x^2-y^2}{2\sqrt{x^2+y^2}}\right)\cdot f_{(X,Y)}(x,y)\,\mathrm{d}x\,\mathrm{d}y.
\end{align}
You can see here that $$M_{(U,V)}(u,v)=\exp\left(\frac18(u^2+v^2)\right)=\exp\left(\frac12(u,v)
\begin{pmatrix}1/4 & 0\\ 0 & 1/4\end{pmatrix}
(u,v)^T\right)$$
so that $(U,V)$ is normal multivariate with $\mathbf{0}$ mean and diagonal covariance matrix, meaning $U, V$ are independant $\mathcal{N}(0,1/4)$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Calculus sine proof Suppose that $a, b, c$ are non-zero acute angles such that
$$\frac{\sin(a − b)}{\sin(a + b)} + \frac{\sin(b − c)}{\sin(b + c)} + \frac{\sin(c − a)}{\sin(c + a)}= 0$$
Prove that at least two of $a, b, c$ are equal.
I have no idea how to begin.
|
Let $\tan{a}=x$, $\tan{b}=y$ and $\tan{c}=z$.
Thus, $$0=\sum_{cyc}\frac{\sin(a-b)}{\sin(a+b)}=\sum_{cyc}\frac{x-y}{x+y}=\frac{\sum\limits_{cyc}(x-y)(x+z)(y+z)}{\prod\limits_{cyc}(x+y)}=$$
$$=\frac{\sum\limits_{cyc}(x-y)(z^2+xy+xz+yz)}{\prod\limits_{cyc}(x+y)}=\frac{\sum\limits_{cyc}(x-y)z^2}{\prod\limits_{cyc}(x+y)}=$$
$$=\frac{\sum\limits_{cyc}(x^2y-x^2z)}{\prod\limits_{cyc}(x+y)}=\frac{(x-y)(x-z)(y-z)}{(x+y)(x+z)(y+z)},$$
which gives $$(a-b)(a-c)(b-c)=0$$ and we are done!
|
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"timestamp": "2023-03-29T00:00:00",
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|
How is $x(2x+7)+3$ equal to $(2x+1)(x+3)$? For some reason, $x(2x+7)+3$ seems like it should be equal to $(2x+7)(x+3)$ instead of $(2x+1)(x+3)$. How does $(2x+1)$ factor out of here?
The original equation was $2x^2+7x+3$.
Proof of this: https://www.desmos.com/calculator/2xpcqznkio
Notice how $x(2x+7)+3$ and $(2x+1)(x+3)$ overlap, but $(2x+7)(x+3)$ does not!
|
Let $7 = 1 + 6$, then $$\begin{align} x(2x+7) + 3 &= x(2x + 1 + 6) + 3 \\ &= x(2x + 1) + 6x + 3 \\ &= x(2x + 1) + 3(2x + 1) \\ &= (2x+1)(x+3)\end{align}$$ as should be desired.
|
{
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"url": "https://math.stackexchange.com/questions/2650119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Is there any way to prove $ \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2} \leq \frac{\pi^2}{6} $ by induction since $ \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6} $ we have that for each $n\in \Bbb N$ , $ \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2} \leq \frac{\pi^2}{6} $
my problem is can we prove the statement
"for each $n\in \Bbb N$ , $ \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2} \leq \frac{\pi^2}{6} $"
only using induction ?(without using the fact that convergence of the above series)
any ideas, thanks!
|
This is not a full proof yet, just an idea I had.
From Leibniz series we know that for $k \geq 1$:
$$\frac{\pi}{4} \geq 1-\frac{1}{3}+\frac{1}{5}-\dots +\frac{1}{4k-3}-\frac{1}{4k-1}$$
Squaring we have:
$$\frac{\pi^2}{6} \geq \frac{8}{3} \left( 1-\frac{1}{3}+\frac{1}{5}-\dots +\frac{1}{4k-3}-\frac{1}{4k-1} \right)^2$$
If we could prove that for any $k \geq 1$:
$$\frac{8}{3} \left( 1-\frac{1}{3}+\frac{1}{5}-\dots +\frac{1}{4k-3}-\frac{1}{4k-1} \right)^2 \geq 1+\frac{1}{2^2}+\dots+\frac{1}{k^2}$$
We are finished.
The base case:
$$\frac{8}{3} \left( 1-\frac{1}{3}\right)^2=\frac{32}{27} >1$$
The induction hypothesis is:
$$\frac{8}{3} \left( 1-\frac{1}{3}+\frac{1}{5}-\dots +\frac{1}{4k-3}-\frac{1}{4k-1} \right)^2 \geq 1+\frac{1}{2^2}+\dots+\frac{1}{k^2} \tag{1}$$
We need to prove that from (1) follows:
$$\frac{8}{3} \left( 1-\frac{1}{3}+\frac{1}{5}-\dots +\frac{1}{4k+1}-\frac{1}{4k+3} \right)^2 \geq 1+\frac{1}{2^2}+\dots+\frac{1}{k^2}+\frac{1}{(k+1)^2} \tag{2}$$
If this is indeed true, I'll try to finish the proof. Or I offer anyone else who wants to try the opportunity to post their answer before me.
For a few values of $k$ I checked the inequality (1) holds, so it should be possible to prove it by induction.
|
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|
Determining $A+B$, given $\sin A + \sin B = \sqrt{\frac{3}{2}}$ and $\cos A - \cos B = \sqrt{\frac12}$. Different approaches give different answers. The question:
Determine $A + B$ if $A$ and $B$ are acute angles such that:
$$\sin A + \sin B = \sqrt{\frac{3}{2}}$$
$$\cos A - \cos B = \sqrt{\frac{1}{2}} $$
Here are the two solutions that I found:
I think the problem with the second solution may have to do with the assumption that:
$$\cos\left(A + \frac{\pi}{6}\right) = \cos\left(B - \frac{\pi}{6}\right)$$
is equivalent to
$$A + \frac{\pi}{6} = -B + \frac{\pi}{6}$$
But can't you say that $\cos N = \cos M$ is equivalent to $\pm N = \pm M$ for all values of $N$ and $M$ (because cosine is an even function)?
|
Using Complex Addition
$$
e^{ia}+e^{i(\pi-b)}=\sqrt{\frac12}+i\sqrt{\frac32}=\color{#C00}{\sqrt2}e^{i\color{#090}{\pi/3}}
$$
*
*$b+a=\frac\pi2$ since $\left|e^{ia}+e^{i(\pi-b)}\right|^2=\color{#C00}{2}\implies\overbrace{\color{#C00}{2}=2+2\cos((\pi-b)-a)}^{\text{Law of Cosines}}$
*$b-a=\frac\pi3$ since $\frac{a+(\pi-b)}2=\color{#090}{\frac\pi3}$
Therefore,
$$
a=\frac\pi{12}\text{ and }b=\frac{5\pi}{12}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2652449",
"timestamp": "2023-03-29T00:00:00",
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|
Understanding the formula for stereographic projection of a point. I was wondering about the equation of line I can write which can help me finding the coordinates of Point $P'$ in relation with coordinates of points on the sphere that is $P$.
Let the $P'(X,Y)$, then how can I find these following coordinates -
I tried equating the slope of $NP'$ with that of the slope of $PP'$.
but could only write $\frac{y}{x} = \frac{Y-y}{X-x}$.
I donot know how it is getting these coordinates -
$(X,Y) = (\frac{2ax}{a-z},\frac{2ay}{a-z})$
and $(x,y,z) = (\frac{2aX}{X^2+Y^2+a^2},\frac{2aY}{X^2+Y^2+a^2},\frac{X^2+Y^2-a^2}{X^2+Y^2+a^2})$?
Here we are following this notation that - points lying on the sphere are represented by $(x,yz)$ and points lying outside the sphere by $X,Y$
|
Let $S$ be the point of tangency of the sphere and the plane,
and let $Q$ lie on line $NS$ such that angle $\angle NQP$ is a right angle.
Observe that triangles $\triangle NQP$ and $\triangle NSP'$ are similar,
with $NQ = a - z$ and $NS = 2a.$
Therefore $$\frac{P'S}{PQ} = \frac{2a}{a-z}. \tag1$$
But we also have $P = (x,y,z)$ while $Q = (0,0,z),$
and $P' = (X,Y,-a)$ (in three dimensions) while $S = (0,0,-a).$
By proportions,
$$\frac Xx = \frac Yy = \frac{P'S}{PQ},$$
and therefore (using equation $(1)$ to substitute for $\frac{P'S}{PQ}$),
$$
X = \frac{2a}{a-z}x \qquad\text{and}\qquad Y = \frac{2a}{a-z} y.
$$
To transform coordinates in the other direction, observe that
triangles $\triangle NPS$ and $\triangle NSP'$ are similar, with
$$
\frac{PN}{NS} = \frac{NS}{P'N}.
$$
Then
$$
\frac xX = \frac{PQ}{P'S} = \frac{PN}{P'N} = \frac{(NS)^2}{(P'N)^2},
$$
so
$$
x = \frac{(NS)^2}{(P'N)^2}X.
$$
Seeing that $P'N$ is the hypotenuse of a right triangle with legs
$\sqrt{X^2 + Y^2}$ and $2a,$
so $$(P'N)^2 = X^2 + Y^2 + 4a^2,$$
and recalling that $NS = 2a,$ we have
$$
x = \frac{4a^2}{X^2 + Y^2 + 4a^2}X. \tag2
$$
Similarly,
$$
y = \frac{4a^2}{X^2 + Y^2 + 4a^2}Y. \tag3
$$
Using the fact that $z^2 = a^2 - x^2 - y^2,$
using equations $(2)$ and $(3)$ to substitute for $x$ and $y,$
and performing some algebraic manipulations, we find that
$$
z = \frac{a(X^2 + Y^2 - 4a^2)}{X^2 + Y^2 + 4a^2}.
$$
The formulas you were trying to justify for $(x,y,z)$ are then seen
not to be correct.
Those formulas, multiplied by $a,$ would give $(x,y,z)$ in terms of the
coordinates of $P$ projected onto the plane $z = 0$
(not $z = -a$); that may explain where those formulas came from
(though this does not explain the missing factor of $a$).
If you multiply each of the formulas
$\frac{2aX}{X^2+Y^2+a^2},$ $\frac{2aY}{X^2+Y^2+a^2},$
and $\frac{X^2+Y^2-a^2}{X^2+Y^2+a^2}$ by $a,$
substitute $\frac X2$ for $X,$ and substitute $\frac Y2$ for $Y,$
you will have correct formulas for the reverse projection from
the plane $z = -a.$
|
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|
Given $x, y$ are acute angles such that $\sin y= 3\cos(x+y)\sin x$. Find the maximum value of $\tan y$. Given $x, y$ are acute angles such that $\sin y= 3\cos(x+y)\sin x$. Find the maximum value of $\tan y$.
Attempt at a solution:
$$(\sin y = 3(\cos x \cos y - \sin x \sin y) \sin x) \frac{1}{\cos y}$$
$$\tan y = (3 \cos x - 3 \sin x \tan y) \sin x$$
$$\tan y + 3 \sin^2 x + \tan y = 3 \sin x \cos x$$
$$\tan y = \frac{3 \sin x \cos x}{1 + 3 \sin^2 x}$$
I have also tried substituting $0$, $30$, $45$, $60$, $90$ to the values of $x$.
|
Hint:
Since
$$
\begin{align}
\sin(y)
&=3\cos(x+y)\sin(x)\\
&=3\cos(x)\cos(y)\sin(x)-3\sin(x)\sin(y)\sin(x)
\end{align}
$$
we get
$$
\begin{align}
\tan(y)
&=\frac{3\sin(x)\cos(x)}{1+3\sin^2(x)}\\
&=\frac{3\tan(x)}{\sec^2(x)+3\tan^2(x)}\\[3pt]
&=\frac{3\tan(x)}{1+4\tan^2(x)}
\end{align}
$$
Then note that $1+4\tan^2(x)=4\tan(x)+(2\tan(x)-1)^2$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.