Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Different ways of computing $\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}$
Possible Duplicate:
On the sequence $x_{n+1} = \sqrt{c+x_n}$
I am wondering how many different solutions one can get to the following question:
Calculate $\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}$
Post your favorite solution please.
| Consider the sequence $$x_n = \sqrt{1 + \sqrt{1 + \sqrt{1 + \cdots \text{ n times}}}}$$ i.e. $$x_{n+1} = \sqrt{1+x_n}.$$
For instance, $x_1 = \sqrt{1}$, $x_2 = \sqrt{1+\sqrt{1}}$, $x_3 = \sqrt{1+\sqrt{1+\sqrt{1}}}$.
We will prove that the sequence is monotonically increasing and is bounded above by $\phi$ where $\phi =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/156131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Putting ${n \choose 0} + {n \choose 5} + {n \choose 10} + \cdots + {n \choose 5k} + \cdots$ in a closed form As the title says, I'm trying to transform $\displaystyle{n \choose 0} + {n \choose 5} + {n \choose 10} + \cdots + {n \choose 5k} + \cdots$ into a closed form. My work:
$\displaystyle\left(1 + \exp\frac{2i\pi}{5... | As you know, $\sum_{j=0}^n {n \choose j} = 2^n$, and if $\omega$ is a primitive $5$'th root of unity
$\sum_{j=0}^n {n \choose j} \omega^j = (1 + \omega)^n$. Now $\sum_{i=0}^4 \omega^{ij} = 5$ if $j$ is divisible by $5$ and $0$ otherwise. So
$$\sum_{k} {n \choose {5k}} = \frac{1}{5} \sum_{i=0}^4 \sum_{j=0}^n {n \choos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/156635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Limit finding of an indeterminate form: $\lim\limits_{x\to0} \frac{x^3}{\tan^3(2x)}$ Here is the limit I'm trying to find out:
$$\lim_{x\rightarrow 0} \frac{x^3}{\tan^3(2x)}$$
Since it is an indeterminate form, I simply applied l'Hopital's Rule and I ended up with:
$$\lim_{x\rightarrow 0} \frac{x^3}{\tan^3(2x)} = \lim_... | Another idea using $\,\,\displaystyle{\frac{\sin x}{x}\underset{x\to 0}{\longrightarrow}1\,,\,\cos kx\underset{x\to 0}\longrightarrow 1\,\,(k=\text{a constant})\,\,,\,\sin 2x=2\sin x\cos x}$:
$$\frac{x^3}{\tan^3 2x}=\frac{x^3}{\frac{\sin^32x}{\cos^32x}}=\cos^32x\frac{x^3}{\left(2\sin x\cos x\right)^3}=\frac{1}{8}\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/157301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 4
} |
What other substitutions could I use to evaluate this integral? Consider the integral
$$ \int x^2\sqrt{2 + x} \, dx$$
I need to find the value of this integral, yet all its (seemingly) possible substitutions don't allow me to cancel appropriate terms. Here are three substitutions and their outcomes, all of which cover ... | You could even do this problem by parts actually.
Let $u = x^2$ and $dv = \sqrt{x + 2}$. So $du = x$ and $v = \dfrac{2}{3}(x+2)^{3/2}$. So
$$
\int x^2\sqrt{x + 2} dx = \dfrac{2}{3}x^2(x+2)^{3/2} - \dfrac{2}{3}\int x (x+2)^{3/2}
$$
Now let $s = x$ and $dt = (x+2)^{3/2}$. So $ds = 1$ and $t = \dfrac{2}{5}(x+2)^{5/2}$ The... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/157548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\frac{{a}^{2}}{b-1}+\frac{{b}^{2}}{a-1}\geq8$ I need to prove that for any real number $a>1$ and $b>1$ the following inequality is true:
$$\frac{{a}^{2}}{b-1}+\frac{{b}^{2}}{a-1}\geq8$$
| If you don't mind I give the details of @Vasea Igor's answer.
Define $f(a,b):=\frac{a^2}{b-1}+\frac{b^2}{a-1}$, where $a,b>1$. Then
$$
\frac{\partial}{\partial a}f(a,b)=\frac{2a(a-1)^2-(b-1)b^2}{(b-1)(a-1)^2}=\frac{2a^3-4a^2+2a-b^3+b^2}{(b-1)(a-1)^2},
$$
$$
\frac{\partial}{\partial b}f(a,b)=-\frac{a^2}{(b-1)^2}+\frac{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/157688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
"answer_id": 4
} |
Limit of $ \sum_{n=m+1}^{\infty} \frac{m}{n\sqrt{n^2-m^2}},m\rightarrow\infty$ I would like to show that $$ \sum_{n=m+1}^{\infty} \frac{m}{n\sqrt{n^2-m^2}}\rightarrow_{m\rightarrow \infty}\frac{\pi}{2}$$
Using integrals:
$$ m\int_{m+1}^{\infty} \frac{\mathrm dx}{x \sqrt{x^2-m^2}} \leq \sum_{n=m+1}^{\infty} \frac{m}{n\s... | The initial inequality appears to be set up correctly so the problem may lie with your evaluation of that integral. The result comes much quicker via Riemann Sums:
$$\sum_{n=m+1}^{\infty} \frac{m}{n\sqrt{n^2-m^2}}= \frac{1}{m} \sum_{n=m+1}^{\infty} \frac{m}{n\sqrt{(n/m)^2-1}} \to \int^{\infty}_1 \frac{1}{x\sqrt{x^2-1} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/159065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
calculate generally the determinant of $A = a_{ij} = \begin{cases}a & i \neq j \\ 1 & i=j \end{cases}$ calculate generally the determinant of $A = a_{ij} = \begin{cases}a & i \neq j \\ 1 & i=j \end{cases} = \begin{pmatrix}
1 & a & a & · & a \\
· & · & · & · \\
a & a & a & · & 1 \\
\end{pmatrix}$
Any hints?
| First add all collumns to the first one and so $ \displaystyle{ C_1 \to \left((n-1)a+1, (n-1)a+1 , \cdots , (n-1)a+1 \right)^T }$
now the determinant $\det A=[(n-1)a+1] \cdot$
\begin{pmatrix}
1 & a & a & \cdots & a \\
1& 1& a & \cdots &a \\
· & · & · & · \\
1 & a & a & \cdots & 1 \\
\end{pmatrix}
subtracking now from ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/159546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
How to solve this quadratic congruence equation How to solve $x^2+x+1\equiv 0\pmod {11}$ ?
I know that in some equations like $ax\equiv b\pmod d$ if $(a,d)=1$ then the equation has one and only one solution $x\equiv ba^{\phi(d)-1}\pmod d$. Any help will be appreciated. ;)
| You use the quadratic formula!
No, really. But you need to interpret the terms correctly: rather than "dividing" by $2a$ (here, $2$) you need to multiply by a number $r$ such that $2r\equiv 1\pmod{11}$ (namely, $r=6$). And rather than trying to find a square root, here $\sqrt{b^2-4ac} = \sqrt{1-4} = \sqrt{-3}$, you wan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/160385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 4,
"answer_id": 1
} |
Integral as a limit of a sum How do I get
$$
\int_a^b \frac{1}{x}dx = \ln\left(\frac{b}{a}\right)$$
as a limit of sum. The constant width partition of the interval $(a,b)$ doesn't seem to work.
| The constant width partition works fine. No need for fancy partitions. To wit:
Let $x_k = a+\frac{k}{N}(b-a)$, with $k = 0,...,N$. We have, of course, $x_0=a$, $x_N = b$.
Using the Taylor series expansion of $x \mapsto \ln x$, and the mean value theorem, we obtain the bounds, for $x \geq y$:
$$ \ln y + \frac{x-y}{y} - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/161659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Two algebra questions I have two questions that I need help with:
1) How many single digit even natural number solutions are there for the equation $A+B+C+D = 24$ such that $A+B > C+D$
A)20 B)11 C)16 D)24
2) Three positive real numbers $x,y,z$ are such that $x+y+Z = 1$. which of the following inequalities best discribe... | Brian's answer is fine. If we are allowed to use cauchy-schwarz inequality
then $$xy+yz+zx\le \sqrt{x^2+y^2+z^2}. \sqrt{x^2+y^2+z^2}=x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+zx)=1-2(xy+yz+zx)$$
Hence $xy+yz+zx\le \frac{1}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/162949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Why do the endpoints of the Maclaurin series for arcsin converge? The series $$\sum_{n=0}^\infty {{-\frac {1} 2} \choose n} \frac{(-1)^n}{2n+1}$$ is an endpoint for the Maclaurin series for arcsin(x). (The other endpoint is just the negative of this one.) I played around with this a bit and turned it into three (potent... | Here's another approach.
We write the shorthand $m!$ for $\Gamma(m+1)$, where $\Gamma$ is the gamma function.
Let $a_n = \displaystyle\frac{(-1)^n}{2n+1} {-\frac{1}{2} \choose n}$.
Then
$$\begin{eqnarray*}
\frac{a_{n+1}}{a_n} &=& -\frac{2n+1}{2n+3} \,
\frac{{-\frac{1}{2}\choose n+1}}{{-\frac{1}{2}\choose n}} \\
&=& ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/164889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
All integer solutions for $x^4-y^4=15$ I'm trying to find all the integer solutions for $x^4-y^4=15$.
I know that the options are $x^2-y^2=5, x^2+y^2=3$, or $x^2-y^2=1, x^2+y^2=15$, or $x^2-y^2=15, x^2+y^2=1$, and the last one $x^2-y^2=3, x^2+y^2=5$.
Only the last one is valid. $x^2+y^2=15$ is not solvable since the p... | $x^4-y^4=15\implies (x^2-y^2)(x^2+y^2)=15$. Factorization of $15=1.15$ or $3.5$. Case $1$, $(x^2-y^2)=1\implies (x+y)=1$ and $(x-y)=1$ which have only integer solutions $0,1$ but then $(x^2+y^2)\neq15$, therefore, factorization is $3.5$ which implies $(x^2+y^2)=5$ and $(x^2-y^2)=3$ which gives $x^2=4$ and $y^2=1\implie... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/166070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Recurrence relation $C_n = n + 1 + \dfrac{2}{n}\sum\limits_{k=0}^{n-1}C_k$. A Discrete Mathematics book from which I'm self-studying ("Discrete Mathematics and Its Applications", by Kenneth Rosen) asks me to do the following:
Given the following recurrence relation:
$$C_n = n + 1 + \frac{2}{n}\sum_{k=0}^{n-1}C_k$$
The ... | You can get rid of the sum on the RHS by subtracting:
For $C_{n+1}$:
\begin{align*}
C_{n+1} &= n + 2 + \dfrac{2}{n+1}\sum_{k=0}^{n}C_k \\
(n + 1) C_{n+1} &= (n + 1)(n + 2) + 2 \sum_{k=0}^{n}C_k \tag{1}
\end{align*}
For $C_n$:
\begin{align*}
C_n &= n + 1 + \dfrac{2}{n}\sum_{k=0}^{n-1}C_k \\
n C_n &= n(n + 1) + 2 \sum_{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/168214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Proving :$\frac{1}{2ab^2+1}+\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}\ge1$ Let $a,b,c>0$ be real numbers such that $a+b+c=3$,how to prove that? :
$$\frac{1}{2ab^2+1}+\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}\ge1$$
| By AM>GM
$$
\frac{a+b+c}{3}=1 \Rightarrow abc\le 1 \\
\Rightarrow \frac{1}{2ab^2+1} = \frac{1}{2abc\frac{b}{c}+1}\ge\frac{1}{2\frac{b}{c}+1}=\frac{c}{2b+c} \\
S =
\frac{1}{2ab^2+1}+\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}\ge\frac{c}{2b+c}+\frac{a}{2c+a}+\frac{b}{2a+b}
$$
$$
\begin{align}
3S-3 & \ge \frac{3c-2b-c}{2b+c}+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/168704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
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Turn fractions into $\mathbb Z_7$ elements I had to perform a division between two polinomials $2x^2+3x+4$ and $3x+4$, my book suggests to do this operation without worrying about the modulo. So my result is $(3x+4)(\frac{2}{3}x+\frac{1}{9})+\frac{32}{9}$. Unfortunately my book fails to explain how should I perform the... | Deal first with $1/3$. In the modulo $7$ world, this is a number $x$ such that $3x\equiv 1\pmod{7}$.
How do we find such an $x$? Well $7$ is small, so try all the possibilities for $x$. These are $1$ (well, not really!), $2$, $3$, $4$, $5$, and $6$. Try $x=2$. Does it work? Let's check. We have $(3)(2)\equiv 6\pmod{7}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/169568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Pythagorean Theorem for imaginary numbers If we let one leg be real-valued and the other leg equal $bi$ then the Pythagorean Theorem changes to $a^2-b^2=c^2$ which results in some kooky numbers.
For what reason does this not make sense? Does the Theorem only work on real numbers? Why not imaginary?
| In the vein of @AndréNicolas' comment that "ideas like this can be useful" ...
Heron's formula for the area, $W$, of a (non-obtuse) triangle in terms of the lengths of its sides is equivalent to the Pythagorean Theorem for Right-Corner Tetrahedra:
$$W^2 = X^2 + Y^2 + Z^2$$
where $W$ is the "hypotenuse-face" opposite th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/169680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
} |
Proof of $\int_0^\infty \frac{\sin x}{\sqrt{x}}dx=\sqrt{\frac{\pi}{2}}$ Numerically it seems to be true that
$$
\int_0^\infty \frac{\sin x}{\sqrt{x}}dx=\sqrt{\frac{\pi}{2}}.
$$
Any ideas how to prove this?
| Here is an another approach, which I show only a heuristic calculation:
$$\begin{align*}
\int_{0}^{\infty} \frac{\sin x}{\sqrt{x}} \; dx
&= \int_{0}^{\infty} \left( \frac{1}{\Gamma\left(\frac{1}{2}\right)} \int_{0}^{\infty} t^{-1/2} e^{-xt} \; dt \right) \sin x \; dx \\
&\stackrel{\ast}{=} \frac{1}{\Gamma\left(\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/171970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 8,
"answer_id": 4
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Evaluating $\int \frac{l\sin x+m\cos x}{(a\sin x+b\cos x)^2}dx$ How do I integrate this expression:
$$\int \frac{l\sin x+m\cos x}{(a\sin x+b\cos x)^2}dx$$.I got this in a book.I do not know how to evaluate integrals of this type.
| Use the famous equation
$$ a\cos x + b\sin x = \pm \sqrt{a^2 + b^2}\cos\left(x - \arctan(b/a)\right)$$
and let $\alpha = \arctan(b/a)$
to get
$$
\begin{aligned}
\mathcal{J}&=\int\dfrac{\ell \sin x + m\cos x}{(a^2 + b^2)\cos^2\left(x - \alpha\right)}\,\mathrm{d}x\\
&=\frac{1}{a^2 + b^2}\int\dfrac{\ell\sin(x - \alpha+\al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/172698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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About the Legendre differential equation Consider the Legendre differential equation $$ (1-x^2) y'' - 2xy' + n(n+1)y = 0 $$ Then its solution is given by $$ y = c_1 P_n (x) + \text{an infinite series} $$ In fact $y = c_1 P_n (x) + c_2 Q_n (x) $ where $P_n$ is Legendre polynomials and $Q_n $ is Legendre function of the ... | The infinite series you are talking about stems from the recurrence relation that can be derived when solving the Legendre differential equation: plugging in
$$y = \sum_{j=0}^{\infty}a_jx^j $$
into
$$[(1-x^2)y']' + n(n+1)y = 0$$
you get
$$\frac{a_{k+2}}{a_k} = \frac{k(k+1) - n(n+1)}{(k+1)(k+2)}$$
Obviously, this leads ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/176357",
"timestamp": "2023-03-29T00:00:00",
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Determine whether the following series converges or diverges I'm stuck on this problem:
$$1 + {1\cdot2\over1\cdot3} + {1\cdot2\cdot3\over1\cdot3\cdot5}+ {1\cdot2\cdot3\cdot4\over 1\cdot3\cdot5\cdot7} +\cdots$$
I've simplified the numerators $n!$ but can't figure out how to represent the denominators.
How do I go about ... | $$ 1 + {1\cdot2\over1\cdot 2\cdot3}2^1{(1)} + {1\cdot2\cdot3\over1\cdot2\cdot3\cdot4\cdot5}2^2(1\cdot2)+ {1\cdot2\cdot3\cdot4\over 1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7}2^3(1\cdot2\cdot3) +\cdots
= \sum_{n=1}^{\infty}{n!(n-1)!2^{n-1} \over (2n-1)!}$$
It converges by ratio test
$$ \lim_{n\rightarrow \infty} {(n+1)!n!2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/176743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Quadratic equation with absolute value Prepping for the GMAT, I came across the following question:
What is the product of all solutions of:
$$x^2 - 4x + 6 = 3 - |x - 1|?$$
First, I set up two equations, ie:
$$x^2 - 4x + 6 = 3 - (x - 1),$$
and
$$x^2 - 4x + 6 = 3 - (-1) \times (x - 1).$$
These factor down to $3$ solu... | The solutions to
$$x^2-4x+6 = 3-(x-1)$$
are only valid when $x-1 \ge 0$, i.e. when $x \ge 1$. Likewise, the solutions to
$$x^2-4x+6 = 3+(x-1)$$
are only valid when $x-1 \le 0$, i.e. when $x \le 1$.
Solving the first equation gives $x=1,2$, both of which are valid. Solving the second gives $x=1,4$. Notice that in this l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/177825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How to integrate $\frac{1}{\sqrt{1+x^2}}$ using substitution? How you integrate $\frac{1}{\sqrt{1+x^2}}$ using following substitution? $1+x^2=t$ $\Rightarrow$ $x=\sqrt{t-1} \Rightarrow dx = \frac{dt}{2\sqrt{t-1}}dt$... Now I'm stuck. I don't know how to proceed using substitution rule.
| $$A=\int\frac{1}{\sqrt[]{1+x^2}}$$
Let, $x = \tan\theta$
$dx = \sec^{2}\theta{d\theta}$
substitute, $x$, $dx$
$$A=\int\left(\frac{1}{\sec\theta}\right){\sec^{2}\theta{d\theta}}$$
$$A=\int{\sec\theta{d\theta}}$$
$$A=\int{\sec\theta\left(\frac{\sec\theta + \tan\theta}{\sec\theta + \tan\theta}\right){d\theta}}$$
$$A=\int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/179129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 1
} |
Evaluating $\int_0^1{\frac{1}{(x+3)^2}}\ln\left(\frac{x+1}{x+3}\right)dx$ using $\frac{dy}{dx}=\frac{2}{(x+3)^2}$ where $y=\frac{x+1}{x+3}$ Find derivative of $$y= \frac{ax+b}{cx+d}$$
I found it to be $$\frac{dy}{dx}=\frac{a}{cx+d}-\frac{c(ax+b)}{(cx+d)^2}$$
Use it to evaluate:
$$\int_0^1{\frac{1}{(x+3)^2}}\ln\left(\fr... | The solution I have now is:
$$\frac{1}{2}\int_\frac{1}{3}^\frac{1}{2}\ln(y)dy=\frac{1}{2}\left[y(ln(y)-1)\right]_\frac{1}{3}^\frac{1}{2}$$
$$=\frac{1}{2}\left[\frac{1}{2}\ln(\frac{1}{2})-\frac{1}{2}-\frac{1}{3}\ln(\frac{1}{3})+\frac{1}{3}\right]\\$$
$$=\frac{1}{2}\ln\left(\frac{1}{2}\right)-\frac{1}{4}-\frac{1}{3}\ln\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/179475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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No closed form for the partial sum of ${n\choose k}$ for $k \le K$? In Concrete Mathematics, the authors state that there is no closed form for
$$\sum_{k\le K}{n\choose k}.$$
This is stated shortly after the statement of (5.17) in section 5.1 (2nd edition of the book).
How do they know this is true?
| Proof of the answer by nczksv.
$$
\begin{array}{l}
f = \dfrac{(1+x)^n}{x^K \cdot (1-x)}
= \left(\dfrac{1+x}{x}\right)^n \cdot \dfrac{x^{n-K}}{1-x}\\
= \left(4^n+1\right)^n \cdot 4^n \cdot \dfrac{x^{n-K}}{4^n - 1}\\
= \dfrac{\left(4^n+1\right)^n \cdot 4^n }{(4^n - 1)(4^n)^{(n-K)}}
\end{array}
$$
Let $A = 4^n - 1$, mod ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
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Use the identity $\cos(A-B) -\cos(A+B) = 2\sin(A)\sin(B)$ to prove another identity and evaluate a sum Use the identity $$\cos(A-B) -\cos(A+B) = 2\sin(A)\sin(B)$$ to prove that:
$$2\sin(j\theta)\sin(\frac{1}{2}\theta)=\cos((j-\frac{1}{2})(\theta))-\cos((j+\frac{1}{2})\theta).$$
This seemed almost too easy, so I am wond... | Your proof of the second identity is correct. If $\sin \left( \frac{1}{2}\theta \right) \neq 0$ (what does this mean in terms of $\theta$?) this identity is equivalent to
$$\begin{equation*}
\sin \left( j\theta \right) =\frac{\cos \left( \left( j-\frac{1}{2}\right)
\theta \right) -\cos \left( \left( j+\frac{1}{2}\right... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality $a+b+c \geqslant abc +2$ Assuming $a,b,c \in (0, \infty)$, we need to prove that:
$$a+b+c \geqslant a b c+2 \quad \text{if} \quad ab+bc+ca=3$$
Can you give me an idea, please? This inequality seem to be known, but I didn't manage to solve it.
| Solve for $c$ from the constraint:
$$
c = \frac{3- a b}{a+b}
$$
and substitute back into the inequality:
$$
a + b + \frac{3-a b}{a+b} \geqslant 2 + \frac{a b(3-a b)}{a+b} \implies \frac{3 + a^2 b^2 + a^2 + b^2 -2 a b -2a -2b+3}{a+b} \geqslant 0
$$
Multiply both sides by $(a+b)$:
$$
3 + a^2 b^2 + a^2 + b^2 -2 a b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/181121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 4,
"answer_id": 3
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Congruence inequality Given $n>2$, by calculation or otherwise deduce that $5^{2^{n-3}} \neq -1 \pmod {2^n}$
Note:The problem arose when I tried to deduce $\langle5\rangle \cap \langle2^n-1\rangle=\{1\}$ in the group $\mathbb{Z}^{*}_{2^n}$, I have showed $\operatorname{ord}(5)=2^{n-2}$
| We show by induction that if $n \ge 3$, then $5^{2^{n-3}}\equiv 1+2^{n-1}\pmod{2^n}$. And it is clear that $1+2^{n-1}\not\equiv -1 \pmod{2^n}$ if $n \ge 3$.
The result holds when $n=3$. Now we do the induction step. Suppose that we know that for a certain $k$, we have $5^{2^{k-3}}\equiv 1+2^{k-1}\pmod{2^k}$. We show... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/182270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Writing 1/3 as a sum of other numbers Is it possible to write $0.3333(3)=\frac{1}{3}$ as sum of $\frac{1}{4} + \cdots + \cdots\frac{1}{512} + \cdots$ so that denominator is a power of $2$ and always different?
As far as I can prove, it should be an infinite series, but I can be wrong.
In case if it can't be written usi... | More generally we can do this for any fraction $\frac{m}{n}$. We know that the sum of the infinite geometric series
$$ S=a+ar+ar^2+ar^3+\dots=\frac{a}{1-r} $$
if $ |r| < 1 $. So we could take $r=-\frac{1}{n-1}$ and $a=\frac{m}{n-1}$ giving
$$ S=\frac{\frac{m}{n-1}}{1+\frac{1}{n-1}}=\frac{\frac{m}{n-1}}{\frac{n}{n-1}}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/186417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Showing that $ \left( \frac{a+2b}{a+2c}\right)^3+\left( \frac{b+2c}{b+2a}\right)^3+\left( \frac{c+2a}{c+2b}\right)^3 \geq3 $ I would like to show that:
$$ \left( \frac{a+2b}{a+2c}\right)^3+\left( \frac{b+2c}{b+2a}\right)^3+\left( \frac{c+2a}{c+2b}\right)^3 \geq3 $$
Using AM-GM inequality:
$$ \left( \frac{a+2b}{a+2c}\ri... | Or maybe just do it by Holder's Inequality?
$$\left(\sum_{cyc} \left(\frac{a+2b}{a+2c} \right)^3 \right) \left(\sum_{cyc}(a+2b)(a+2c)\right)\left(\sum_{cyc}(a+2c)\right)^2 \ge \left(\sum_{cyc}(a+2b)\right)^4$$
As $\displaystyle \sum_{cyc}(a+2c) = 3(a+b+c) = \sum_{cyc}(a+2b)$, and $$\displaystyle \sum_{cyc}(a+2b)(a+2c) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/188449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Prove that $ \frac{1}{2}\cot^{-1}\frac{2\sqrt[3]{4}+1}{\sqrt{3}}+\frac{1}{3}\tan^{-1}\frac{\sqrt[3]{4}+1}{\sqrt{3}}=\dfrac{\pi}{6}. $ Prove that $$ \frac{1}{2}\cot^{-1}\frac{2\sqrt[3]{4}+1}{\sqrt{3}}+\frac{1}{3}\tan^{-1}\frac{\sqrt[3]{4}+1}{\sqrt{3}}=\dfrac{\pi}{6}. $$
| Let $y^3=4$ and $x^2=3$
So, we need to prove $ \frac{1}{2}\cot^{-1}\frac{2y+1}{x}+\frac{1}{3}\tan^{-1}\frac{y+1}{x}=\dfrac{\pi}{6}. $
or $3\cot^{-1}\frac{2y+1}{x}+2\tan^{-1}\frac{y+1}{x}=\pi$
or $3(\frac{\pi}{2}-\tan^{-1}\frac{2y+1}{x})+2\tan^{-1}\frac{y+1}{x}=\pi$
as $\tan^{-1}a+\cot^{-1}a=\frac{\pi}{2}$
or we need t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/193128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Inequality. $\sqrt{\frac{11a}{5a+6b}}+\sqrt{\frac{11b}{5b+6c}}+\sqrt{\frac{11c}{5c+6a}} \leq 3$
Let $a,b,c$ be positive numbers . Prove the following inequality:
$$\sqrt{\frac{11a}{5a+6b}}+\sqrt{\frac{11b}{5b+6c}}+\sqrt{\frac{11c}{5c+6a}} \leq 3.$$
What I tried:
I used Cauchy-Schwarz in the following form
$\sqrt{Ax}+... | We can prove the more general inequality for certain values of the ratio between b and a:
$$\sqrt{\frac{(a+b)x}{ax+by}}+\sqrt{\frac{(a+b)y}{ay+bz}}+\sqrt{\frac{(a+b)z}{az+bx}} \leq 3$$
if 0.8152 < b/a < 1.2267
Re-working the LHS and applying Cauchy-Schwarz we get:
$$(\sqrt{az + bx}\sqrt{\frac{(a + b) x}{(a z + b x... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 0
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FFT processor which can only be used once Given an 8 point FFT processor, which can be used only once, compute the DFTs of the sequence.
$$x_1(n)=[1,8,6,7,4,2,3,1]$$
$$x_2(n)=[1,4,3,2,8,7,6,1]$$
| Let $y(n)=x_1(n)+jx_2(n)$.
$$\therefore y(n)=[(1+1j), (8+4j), (6+3j), (7+2j), (4+8j), (2+7j), (3+6j), (1+1j)]$$
Now we can calculate the DFT of $y(n)$ by using the single 8 point FFT processor.
$$\begin{align}
\therefore Y(k) & = \sum_{n=0}^{N-1}{y(n)e^{-{2\pi{}jnk\over N}}} \\
& =\begin{bmatrix}
(32 + 32j) \\
\left(-6... | {
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"url": "https://math.stackexchange.com/questions/194034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Auxiliary trigonometric identities [Identidades Trigonométricas Auxiliares] I need to show two auxiliary trigonometric identities:
1) $\sec^2x = \tan ^2x + 1 (\cos x \neq 0)$
2) $\csc^2x = \cot^2x +1 (\sin x \neq 0)$
How could I do it?
[Original Portuguese]
Identidades Trigonométricas Auxiliares
Sendo preciso mostrar ... | Starting from $\sin^2 x + \cos^2 x = 1$, you get, dividing by $\cos^2 x$ (if this is $\ne 0$)
\[ \frac{\sin^2x}{\cos^2 x} + \frac{\cos^2 x}{\cos^2 x} = \frac 1{\cos^2 x} \]
which gives
\[ \tan^2 x + 1 = \mathrm{sec}^2 x \]
Dividing the first identity by $\sin^2 x$ we get
\[ \frac{\sin^2x}{\sin^2 x} + \frac{\cos^2 x}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/194371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proofs using linear congruences We have just covered solving linear congruences, and I am confused about how to use them in proofs. I understand that the linear congruence $$cx \equiv b \pmod m$$ has a unique solution $\bmod m$ if $\gcd(c,m) = 1$, but a general approach to problems escapes me.
Sample Questions
Prove t... | Hint $\rm(1)\ \ \ x^2 \equiv n,\,\ y^2\equiv -1\:\Rightarrow\: -n\equiv x^2y^2\equiv (xy)^2.\ $ But $\rm\:mod\ 65\!:\ {-}1 \equiv 64\equiv (\_ )^2.$
$\rm(2)\ \ mod\ 4\!:\ x^2\!+\!y^2\!+\!z^2 \equiv\, 3\:\Rightarrow\:x,y,z\:$ odd, by $\rm\:odd^2\equiv 1,\ even^2\equiv 0.\:$ Therefore we deduce $\rm\phantom{(2)\ \ } mo... | {
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"source": "stackexchange",
"question_score": "5",
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How to show that that $\sum_{n=1}^{\infty}\left( \frac{1}{3n-1} + \frac{1}{3n-2}- \frac{2}{3n}\right)= \ln\left(3\right)$? $$
\mbox{How to show that that}\qquad
\sum_{n = 1}^{\infty}\left({1 \over 3n - 1} + {1 \over 3n - 2} - {2 \over 3n}\right)
=
\ln\left(3\right)\ {\large ?}
$$
$$
\mbox{or}\quad
1 + \frac{1}{2} -\fra... | Let's try one line proof using Riemann sums. So, our limit may be written as
$$\lim_{n\to\infty}\sum_{k=1}^{n} \left(\frac1{3k-2}+\frac1{3k-1}+\frac1{3k}-\frac1k\right)=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{2n}\frac1{1+\frac{k}{n}}=\int_0^2\frac{1}{1+x}=\ln3.$$
Q.E.D.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/197595",
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"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 1
} |
By using properties of determinants show that determinant is equal to $(1+a^2+b^2)^3$ $$\begin{vmatrix}1+a^2-b^2&2ab&-2b\\
2ab&1-a^2+b^2&2a\\
2b&-2a&1-a^2-b^2\end{vmatrix}=(1+a^2+b^2)^3$$
I have been trying to solve the above determinant. But unfortunately my answer is always coming as:
$$1+3a^2+3a^4+a^6+3a^2b^4+3b^2+4... | Assuming $a,b\in\mathbb{R}$ (and if not, proving it for real $a,b$ should allow you it to extend it to complex $a,b$), each row vector has the norm $(1+a^2+b^2)$ and the row vectors are all orthogonal so the matrix divided by $(1+a^2+b^2)$ is an orthogonal matrix and the absolute value of its determinant is 1 which imm... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to represent $(43.3)_7$ in base-8? I am trying to represent $(43.3)_7$ in base-8.
But in only one integer digit by truncating the rest and using the numerical unsigned representation.
| $43.3_7 = 4*7 + 3 + \frac 37=$
$4(2*3 + 1) + 3 + \frac 37*\frac 88=$
$8*3 + (4+3) + \frac{\frac {24}7}8=$
$3*8 + 7 + \frac 38 + \frac 37\frac 18=$
$3*8 + 7*8^0 + 3*8^{-1} + \frac {24}7*8^{-2}=$
$3*8 + 7*8^0 + 3*8^{-1} + 3*8^{-2}+ \frac 37*8^{-2} =$
.... via induction....
$3*8 + 7*8^0 + 3*8^{-1} + 3*8^{-2} + 3*8^{-3} + ... | {
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"source": "stackexchange",
"question_score": "1",
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Inequality. $2(x^2+y^2+z^2)^2 \geq 3(x^2y^2+y^2z^2+z^2x^2)+3(x^3y+y^3z+z^3x)$ How do I prove that :
$$2(x^2+y^2+z^2)^2 \geq 3(x^2y^2+y^2z^2+z^2x^2)+3(x^3y+y^3z+z^3x).$$
It seems to be easy , but I have no idea:)
thanks :)
| We start the same way as uforoboa: Expanding the left hand side, you are trying to show that
$$2x^4 + 2y^4 + 2z^4 + 4(x^2y^2+y^2z^2+z^2x^2) \geq 3(x^2y^2+y^2z^2+z^2x^2)+3(x^3y+y^3z+z^3x)$$
This equivalent to
$$2x^4 + 2y^4 + 2z^4 +x^2y^2+y^2z^2+z^2x^2 \geq 3(x^3y+y^3z+z^3x)$$
By AM-GM one has $x^3y \leq 1/2(x^4 + x^2y^... | {
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Prove/disprove $\sum x^2y^2 \ge \sum x^3y$ Prove $x^2y^2+y^2z^2+z^2x^2 \ge$ or $\le x^3y+y^3z+z^3x$ where $x,y,z$ are real numbers.
Actually, I have reached here from this problem:
Inequality. $2(x^2+y^2+z^2)^2 \geq 3(x^2y^2+y^2z^2+z^2x^2)+3(x^3y+y^3z+z^3x)$
| Define $f(x,y,z)=x^2y^2+y^2z^2+z^2x^2-x^3y-y^3z-z^3x$. Then $f(1,1,2)=-2$ and $f(1,1,-1)=4$. Therefore, neither the inequality nor its reverse is true.
For positive numbers, we also have $f(3,2,1)=-16$ and $f(4,6,1)=24$. So even for all positive numbers the inequality is not true.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Showing that $\lim_{k\rightarrow 0}\int_0^1\frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}} = \int_0^1\frac{dx}{\sqrt{(1-x^2)}}$ So I'm trying to show that:
$$\lim_{k\rightarrow 0}\int_0^1\frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}} = \int_0^1\frac{dx}{\sqrt{(1-x^2)}}$$
I guess this boils down to a solid understanding of uniform convergence... | What you want to show is that
$$\frac{1}{\sqrt{(1-x^2)(1-k^2x^2)}}\to \frac{1}{\sqrt{1-x^2}}$$
on uniformly $[0,1-\epsilon]$ for all $\epsilon>0$ and that
$$\lim\limits_{\epsilon\to 0}\int_0^{1-\epsilon}\frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}}\to \int_0^{1}\frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}}$$ uniformly in $k$. The first ... | {
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"url": "https://math.stackexchange.com/questions/202594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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How do we solve the equation? How do we solve the following equation in the set of real numbers?
$$(x+1)\cdot \sqrt{x+2} + (x+6)\cdot \sqrt{x+7}=(x+3)\cdot (x+4).$$
I wrote the given equation has the form
\begin{equation*}
(x+1)(\sqrt{x + 2} - 2) + (x + 6)(\sqrt{x+7} - 3) = (x-2)(x+4)
\end{equation*}
This equation i... | We have
$$\dfrac{x+1}{\sqrt{x+2}+2}<\dfrac{x+2}{\sqrt{x+2}+2}\leqslant \dfrac{x+2}{2}$$
and
$$\dfrac{x+6}{\sqrt{x+7}+3}<\dfrac{x+6}{3}\leqslant \dfrac{x+6}{2}.$$
Therefore
$$\dfrac{x+1}{\sqrt{x+2}+2}+\dfrac{x+6}{\sqrt{x+7}+3}<\dfrac{x+2}{2}+\dfrac{x+6}{2}=x+4$$
Thus the equation
\begin{equation*}
\dfrac{x+1}{\sqrt{x+2... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Calculate a sum involving nth root of unity Calculate
$$1+2\epsilon+3\epsilon^{2}+\cdots+n\epsilon^{n-1}$$
Where $\epsilon$ is nth root of unity.
There is a hint that says: multiply by $(1-\epsilon)$
Doing this multiplication I get:
$$1+\epsilon+\epsilon^{2}+\epsilon^{3}+\cdots+\epsilon^{n-1}-n$$
The answer is: $-\frac... | Of course, when $\epsilon=1$,
$$
\begin{align}
&1+2\epsilon+3\epsilon^2+4\epsilon^3+\dots+n\epsilon^{n-1}\\
&=1+2+2+4+\dots+n\\
&=\frac{n(n+1)}{2}
\end{align}
$$
When $\epsilon\ne1$
$$
\left.
\begin{align}
1+\epsilon+\epsilon^2+\epsilon^3+\dots+\epsilon^{n-1}&=\frac{1-\epsilon^n}{1-\epsilon}\\
\epsilon+\epsilon^2+\epsi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/206523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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prove $\frac{1}{ n+1}+\frac{1}{ n+2}+\cdots+\frac{1}{2n}<\frac{25}{36}$ by mathematical induction How to prove
$$\frac{1}{ n+1}+\frac{1}{ n+2}+\cdots+\frac{1}{2n}<\frac{25}{36}$$
by Mathematical induction，n$\ge $1
| I didn't prove it by mathematical induction.
Set $a_{n}=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}$, then by calculation, $a_{n+1}-a_{n}=\frac{1}{2n+1}-\frac{1}{2n+2}$, $a_{n}$ is strictly monotonically increasing.
So we can obtain that
$$\begin{align*}
a_{n}&=a_{1}+\sum_{k=1}^{n-1}(a_{k+1}-a_{k})\\
&=\frac{1}{2}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/207675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 8,
"answer_id": 6
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Special Differential Equation I ended up with a differential equation that looks like this:
$$\frac{d^2y}{dx^2} + \frac 1 x \frac{dy}{dx} - \frac{ay}{x^2} + \left(b -\frac c x - e x \right )y = 0.$$
I tried with Mathematica. But could not get the sensible answer. May you help me out how to solve it or give me some re... | $\dfrac{d^2y}{dx^2}+\dfrac{1}{x}\dfrac{dy}{dx}-\dfrac{ay}{x^2}+\left(b-\dfrac{c}{x}-ex\right)y=0$
$\dfrac{d^2y}{dx^2}+\dfrac{1}{x}\dfrac{dy}{dx}-\left(ex-b+\dfrac{c}{x}+\dfrac{a}{x^2}\right)y=0$
Let $y=\dfrac{u}{\sqrt{x}}$ ,
Then $\dfrac{dy}{dx}=\dfrac{1}{\sqrt{x}}\dfrac{du}{dx}-\dfrac{u}{2x\sqrt{x}}$
$\dfrac{d^2y}{dx^... | {
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"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
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expanding $(a+b+c+...)^x$ where $x$ is natural neatly Suppose that $a,b,c,d,...$ are unknown variables.
One wishes to expand $(a+b+c..)^x$ where $x$ is natural number ina neat manner (for e.g. using combination, sigma etc.).
What would be some way?
Also, what would be the number of terms that would allow writing neatly... | One can expand $(x_1+x_2+x_3+\dots +x_m)^n$ by using the multinomial theorem. It states that
$$(x_1+x_2+x_3+\dots +x_m)^n=\sum_{k_1+k_2+\dots+k_m=n}\binom{n}{k_1,k_2,k_3,\dots,k_m}\prod_{t=1}^mx_t^{k_t}$$
$$=\sum_{k_1+k_2+\dots+k_m=n}\binom{n}{k_1,k_2,k_3,\dots,k_m}x_1^{k_1}x_2^{k_2}x_3^{k_3}\cdots x_m^{k_m}$$
Where $... | {
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"source": "stackexchange",
"question_score": "1",
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Finding the limit $ \lim_{x\to 0} \frac{1-\cos x \cos(2x)}{x^2}$ I cannot find the following limit:
$$ \lim_{x\to 0} \frac{1-\cos x \cos(2x)}{x^2} \, . $$
Please, help me.
| Here is a proof using Tools 1-3 below and, I think, nothing else.
Tool 1: $2\cos(a)\cos(b)=\cos(a+b)+\cos(a-b)$.
Tool 2: $1-\cos(a)=2\sin^2(\frac{a}2)$.
Tool 3: Let $u(z)=\frac{\sin(z)}z$, then $u(z)\to1$ when $z\to0$. (You said you knew that.)
Using Tools 1 and 2, one gets
$$
\frac{1-\cos x\cos 2x}{x^2}\stackrel{\text... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/214444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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How I convert Decimal Using negetive Base 2? I dont have any idea how i convert deciaml using negetive base 2?
i googling today long time but couldn't find any solution.
can any tell me process?
i need to learn it .
| One can use the standard algorithm for converting from one base to another. Divide by the target base $b$ to get an integer quotient and a remainder in the chosen range of digits, generally but not always $\{0,\dots,|b|-1\}$; replace the original number by the quotient and repeat; continue until you get a quotient of $... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Finding a closed formula for $1\cdot2\cdot3\cdots k +\dots + n(n+1)(n+2)\cdots(k+n-1)$ Considering the following formulae:
(i) $1+2+3+..+n = n(n+1)/2$
(ii) $1\cdot2+2\cdot3+3\cdot4+...+n(n+1) = n(n+1)(n+2)/3$
(iii) $1\cdot2\cdot3+2\cdot3\cdot4+...+n(n+1)(n+2) = n(n+1)(n+2)(n+3)/4$
Find and prove a 'closed formula' for ... | The pattern looks pretty clear: you have
$$\begin{align*}
&\sum_{i=1}^ni=\frac12n(n+1)\\
&\sum_{i=1}^ni(i+1)=\frac13n(n+1)(n+2)\\
&\sum_{i=1}^ni(i+1)(i+2)=\frac14n(n+1)(n+2)(n+3)\;,
\end{align*}\tag{1}$$
where the righthand sides are closed formulas for the lefthand sides. Now you want
$$\sum_{i=1}^ni(i+1)(i+2)\dots(i... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
} |
What are good ways to find the limits of this function? Compute: $\displaystyle\lim_{x\rightarrow 2}\frac{\sqrt{x+1}-\sqrt{ 1-x}}{x}$
| I assume you are interested in $$\lim_{x \to 0} \dfrac{\sqrt{1+x} - \sqrt{1-x}}x$$
Multiply and divide by $\sqrt{1+x} + \sqrt{1-x}$ to get
$$\dfrac{\sqrt{1+x} - \sqrt{1-x}}x \times \dfrac{\sqrt{1+x} + \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}} = \dfrac{(1+x)-(1-x)}{x(\sqrt{1+x} + \sqrt{1-x})} = \dfrac{2x}{x(\sqrt{1+x} + \sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/220448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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As shown in the figure: Prove that $a^2+b^2=c^2$ Geometry: Buildings in the triangle
Other triangles with the same property:
$1.$ 12 18 6 12 30 102
$2.$ 15 30 15 15 15 90
$3.$ 24 30 54 24 6 42
$4.$ 30 10 40 30 20 50 (proposed problem in sense of clockwise)
$5.$ 36 12 6 12 18 96
$6.$ 36... | Denote the length of the lower side (the hypotenuse of the big triangle) by $h$. Then the side containing $c$ is of length $h\sin 40^\circ=h\cos 50^\circ$ (Do you see why?)
Now let's find the length of segment from the rightmost bottom vertex to the one at the bottom of $b$ (denote it by $l_1$): Using the law of sines,... | {
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"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 2
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The quadratic form $x^2 + ny^2$ via prime factors Elementary algebra shows that the product of two numbers in the form $x^2 + ny^2$ again has the same form, since if $p = (a^2 + nb^2)$ and $q = (c^2 + nd^2)$,
$$pq = (a^2 + nb^2)(c^2 + nd^2) = (ac \pm nbd)^2 + n(ad \mp bc)^2$$
My question is: Assuming that a number $z$... | Alright, the answer to the actual question asked is yes, as follows. I am taking a prime $p$ with $\gcd(p,n) = 1.$ Then I am demanding $u^2 + n v^2 = p.$ Next, I am taking some number, composite or prime, call it $Q,$ and demand about $pQ$ rather than $Q$ itself,
$$ pQ = r^2 + n s^2. $$
First, we get
$$ \left( \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/229201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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Prove this property for an arbitrary circle Prove that in an arbitrary circle, the point on the circle closest to the origin must lie on the extended line connecting the circle's centre and the origin.
|
Let the equation of the circle be $(x-a)^2+(y-b)^2=r^2$
So, any point on the circle can be $P(r\cos \theta+a,r\sin \theta+b)$
If the distance$(d)$ from the point $(r\cos \theta+a,r\sin \theta+b)$ from the origin,
$d^2=(r\cos \theta+a)^2+(r\sin \theta+b)^2=r^2+a^2+b^2+2r(a\cos \theta+b\sin \theta)=r^2+a^2+b^2+2r\sqrt{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/231319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
Integral of cosec squared ($\operatorname{cosec}^2x$, $\csc^2x$) According to my sheet of standard integrals,
$\int \csc^2x \, dx = -\cot x + C$.
I am interested in a proof for the integral of $\operatorname{cosec}^2x$ that does not require differentiating $\cot x$. (I already know how to prove it using differentiation... | Question: $\displaystyle\int{\csc^2(x)dx} = $
Using substitution: $u = \dfrac{1}{\sin^2(x)}$
$\dfrac{du}{dx} = \dfrac{-2\cos(x)}{\sin^3(x)} \quad \longrightarrow du = \left(\dfrac{-2\cos(x)}{\sin(x)}\right)\times \left(\dfrac{1}{\sin^2(x)}\right)dx$
$\therefore -\dfrac{1}{2}\tan(x)du = \dfrac{1}{\sin^2(x)} dx \qquad (... | {
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"source": "stackexchange",
"question_score": "4",
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How do you compute the limit with multiple variables without fail? $$\lim\limits_{(x,y)\rightarrow (0,0)} \dfrac{(x^3-y^3) \sin (2x^2+3y^2)}{x^2+2y^2}$$
I don't know what to do doesn't it give always $0$?
Whether $x=0$ $x=y$ or $y=0$ or $x=y^2$ it always give $0$ since it goes to $(0,0)$
| $\left | \dfrac{(x^3-y^3) \sin (2x^2+3y^2)}{x^2+2y^2} \right |\leq \dfrac{|x^3-y^3| (2x^2+3y^2)}{x^2+2y^2} \leq \dfrac{|x^3-y^3| (2x^2+4y^2)}{x^2+2y^2}\leq 2|x^3-y^3|$
Where we used the inequality $|\sin (x)| \leq |x|$
ADDED Let $f(x,y)= x^3-y^3$, now knowing that $ \lim _{ (x,y) \rightarrow (0,0)} f(x,y)= 0$
It mean... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Taylor Polynomial of $\arctan$ of given Degree and Error
Replace the following function by its taylor polynomial of the given
grade, and approximate the error in the given interval:
$$f(x) = \arctan(x) \textrm{ by } T_3(f,x,0) \textrm{ in } |x| \le\frac{1}{10}$$
My solution and thoughts
We only need the first three... | Your work looks good to me, except for a mistake in the last step (probably a typo). Other than that, it's pretty much exactly the way I would hope one of my students would solve this problem.
The mistake is that you lost the $99$ in the final step, so that at the end you should have $$
\left|R_3(x)\right| \le \frac{9... | {
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"source": "stackexchange",
"question_score": "2",
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How to multiply polynominals I can't figure out how to multiply these polynominals $$(5x^2+3x^4-7x^3+5x+8)(2x^2-4x+9-6x^2+7x)$$
I tried multiplying like this
$$(5x^2+3x^4-7x^3+5x+8)(2x^2-4x+9-6x^2+7x)$$
$$3x^4-7x^3+(5x^2)(2x^2)(-6x^2)+(5x)(7x)(-4x)+(8)(9)$$
$$3x^4-7x^3-60x^2-140x+72$$
It says the answer is $$-12x^6-19x... | To multiply two polynomials, distribute each term in the left set of parentheses over the entire collection of terms in the right set of parentheses. Your example would start out like:
$$
\begin{align*}
& (\color{red}{5x^2+3x^4-7x^3+5x+8})(2x^2-4x+9-6x^2+7x)\\
= & \color{red}{5x^2}(2x^2-4x+9-6x^2+7x)\\
& \color{red}{+ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/253522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Limit of $\left(\frac{n^{2}+8n-1}{n^{2}-4n-5}\right)^{n}$, is the following true?
Using the previous question*, arithmetics of bordes, and the sandwich theorem, find the limits of the following sequenes:
b. $\left(\dfrac{n^{2}+8n-1}{n^{2}-4n-5}\right)^{n}$
(*) the previous question is to prove that $\lim_{n\to \inf... | Note that for any $\epsilon > 0$,
$$1 + \dfrac{8-\epsilon}{n} \le 1 + \dfrac{8}{n} - \dfrac{1}{n^2} \le 1 + \dfrac{8}{n}$$
for sufficiently large $n$. Therefore
$$ \eqalign{e^{8-\epsilon} &= \lim_{n \to \infty} \left(1 + \dfrac{8-\epsilon}{n}\right)^n \le \liminf_{n \to \infty} \left(1 + \dfrac{8}{n} - \dfrac{1}{n^2}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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5 linear equations in 5 unknowns I need an example of 5 linearly independent equations with 5 variables. How can I write such a equation set. As an example:
0 0 0 0 1 | -4
0 0 1 -1 0 | 3
1 0 0 0 1 | 2
... | You can choose any $\hat{x}_{1} , \ldots , \hat{x}_{5}$, and use the system
$$
\begin{bmatrix}
1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
x_{1} \\
x_{2} \\
x_{3} \\
x_{4} \\
x_{5}
\end{bmatrix}
=
\begin{bmatrix}
\hat{x}_{1} \\
\hat... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Proving square root of a square is the same as absolute value Lets say I have a function defined as $f(x) = \sqrt {x^2}$. Common knowledge of square roots tells you to simplify to $f(x) = x$ (we'll call that $g(x)$) which may be the same problem, but it isn't the same equation. For example, say I put $-1$ into them:
$\... | When $x < 0$, $|x| = -x > 0$. $-x \ne x$ (unless $x=0$) and $(-x)^2 = x^2$. There are two "square roots" of any positive number $y$, i.e. numbers whose square is $y$, and the positive one is called $\sqrt{y}$. So $\sqrt{x^2} = -x = |x|$ when $x < 0$, and $\sqrt{x^2} = x = |x|$ when $x \ge 0$.
| {
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"source": "stackexchange",
"question_score": "20",
"answer_count": 3,
"answer_id": 0
} |
Find the upper and lower limits of $xf(x)$, as $x\rightarrow \infty$ Define $$f(x)=\int_{x}^{x+1}\sin(t^2)dt$$
Find the upper and lower limits $xf(x)$, as $x\rightarrow \infty$.
I find the answer as $+1, -1$ since $|\sin(x)| \le 1$. (Of course I calculated that function)
Is that right or did I miss something?
=======... | $$
\begin{align}
f(x)
&=\int_x^{x+1}\sin(t^2)\,\mathrm{d}t\\
&=-\int_x^{x+1}\frac1{2t}\,\mathrm{d}\cos(t^2)\\
&=\frac{\cos(x^2)}{2x}-\frac{\cos((x+1)^2)}{2(x+1)}
-\int_x^{x+1}\cos(t^2)\frac1{2t^2}\,\mathrm{d}t\\
&=\frac{\cos(x^2)}{2x}-\frac{\cos((x+1)^2)}{2(x+1)}+O\left(\frac1{x^2}\right)\tag{1}
\end{align}
$$
So
$$
\b... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Finding the sum of all solutions $2x + 3y = n$ has exactly $2011$ non-negative integral solutions. Determine the SUM of the possible values of $n$.
| Let us verify for $n=m[2,3]+k$ where $0\le k<[2,3]$
If $k=0\implies 2x+3y=6m\implies 3(2m-y)=2x,$ so $y$ must even $=2z$(say),
So, $x=3(m-z)\implies 0\le z\le m$ so $z$(hence $y$) can assume $m+1$ values in non-negative integers.
If $k=1\implies 2x+3y=6m+1\implies 3(y-2m)+1=2x,$ so $y$ must odd $=2z+1$(say), where $z... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help in proving $ \left( 1 + \frac{1}{n} \right)^{n} \leq \sum\limits_{k=0}^{n} \frac{1}{k!} < 3 $. I am trying to prove this statement for all $ n \geq 1 $ using induction:
$$
\left( 1 + \frac{1}{n} \right)^{n} \leq \sum_{k=0}^{n} \frac{1}{k!} < 3.
$$
I said:
*
*Base case $ n = 1 $:
$$
\left( 1 + \frac{1}{1} \rig... | For all $ n \in \mathbb{N} $, we have
\begin{align}
\left( 1 + \frac{1}{n} \right)^{n}
&= \sum_{k=0}^{n} \binom{n}{k} \left( \frac{1}{n} \right)^{k} \quad (\text{By the Binomial Theorem.}) \\
&= \sum_{k=0}^{n} \frac{n!}{k!(n - k)!} \cdot \frac{1}{n^{k}} \quad (\text{By the definition of the binomial coeffic... | {
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"source": "stackexchange",
"question_score": "6",
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Evaluate $\int {2x\over x^2-1}dx$ My friend evaluated this before he went to bed: $$\int {2x\over x^2-1}dx$$
The answer was $\log(x^2-1)$.
I just can't figure out how that works. I know that $\int \frac1x dx = \log|x|$, so what just happened to $2x$?
| $\frac{2x}{x^2-1}=\frac{(x+1)+(x-1)}{x^2-1}=\frac1{x-1}+\frac1{x+1}$
Alternatively, using Partial Fraction Decomposition, let $\frac{2x}{x^2-1}=\frac{(x+1)+(x-1)}{x^2-1}=\frac A{x+1}+\frac B{x-1}$ where $A,B$ are arbitrary constants.
So, $2x=(A+B)x+B-A$
Comparing the constant terms in either of the identity, $B-A=0\im... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/266684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 0
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Compute the series Compute the series
$$1)\space\sum_{n=1}^{\infty}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n+1}\right)\frac{1}{n(n+1)}$$
$$2)\space\sum_{n=1}^{\infty}\left(1-\frac{1}{2}+\frac{1}{3}-\cdots(-1)^{n}\frac{1}{n+1}\right)\frac{1}{n(n+1)}$$
| For the first part, recognize that $\frac {1}{n(n+1)} = \frac {1}{n} - \frac {1}{n+1}$. So the sum can we written as
$$ \sum_{n=1}^{\infty} ( 1 + \frac {1}{2} + \frac {1}{3}+ \ldots + \frac {1}{n+1} ) ( \frac {1}{n} - \frac {1}{n+1}) $$
Break this up into the separate terms of the right parenthesis, and we find that c... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Simplify these products of permutations Can someone please explain how to get these answers? Everytime I think I understand the method, I end up getting a completely different answer to the one provided. How exactly do you go about simplifying these?
[Note: Read from left to right]
$$(i)\;\;\; (1,2,3).(2,4)(1,3,5) = (1... | To find, for example, $(1,2,3)\cdot(2,4)(1,3,5)$ you need to apply it on every $1\leq i\leq 5$:
$$\begin{array}{l}
[1](1,2,3)\cdot(2,4)(1,3,5)=[2](2,4)(1,3,5)=4\\
[2](1,2,3)\cdot(2,4)(1,3,5)=[3](2,4)(1,3,5)=5\\
[3](1,2,3)\cdot(2,4)(1,3,5)=[1](2,4)(1,3,5)=3\\
[4](1,2,3)\cdot(2,4)(1,3,5)=[4](2,4)(1,3,5)=2\\
[5](1,2,3)\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/269231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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The area of a triangle determined by the bisectors. How can I calculate the area of a triangle determined by the interior bisectors? What I want to say it is represented in the following picture:
$AQ$ is the bisector of the angle $\angle BAC$, $BR$ -bisector for $\angle ABC$ and $CP$ -bisector for $\angle ACB$. Now,... | We'll derive the equation using the fact:
$$A_{PQR}=A_{ABC}-A_{PBR}-A_{RCQ}-A_{QAP}, \quad (I)$$
Using the angle bisector theorem we get:
$$BP=\frac{ac}{a+b},\quad (1)$$
$$BR=\frac{ac}{b+c}, \quad (2)$$
$$CR=\frac{ab}{b+c},\quad (3)$$
$$CQ=\frac{ab}{a+c},\quad (4)$$
$$AQ=\frac{bc}{a+c},\quad (5)$$
and
$$AP=\frac{bc}{a+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/269496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Solve this inequality $\prod_{i=1}^{50} \frac {2i-1}{2i} < \frac {1}{10}$ Prove that
$ \frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot \frac{7}{8}\cdot \frac{9}{10}\cdot \frac{11}{12}\cdot \frac{13}{14}...\cdot \frac{91}{92}\cdot \frac{93}{94}\cdot \frac{95}{96}\frac{97}{98}\cdot \frac{99}{100} <\frac{1}{10}$
| The product of the first 31 fractions is $0.100923...$ while the product of the first 32 fractions is $0.099346...$, so it looks like you only need the first 32 of the 50 fractions to get below $1/10$. Using more fractions (all less than 1) will only make that smaller.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/274753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
calculate $\lim_{x\rightarrow\frac{\pi}{4}}\frac{\cos2x}{\cos x-\sin x}$ $$\lim_{x\rightarrow\frac{\pi}{4}}\frac{\cos(2x)}{\cos(x)-\sin(x)}=\lim_{x\rightarrow\frac{\pi}{4}}\frac{2\cos^{2}(x)-1}{\cos(x)-\sqrt{1-\cos^{2}(x)}}$$
$$t=\cos(x)$$
$$\lim_{x\rightarrow\frac{\pi}{4}}\frac{2t^{2}-1}{t-\sqrt{1-t^{2}}}=\lim_{x\righ... | $$ \lim_{x\to\frac{\pi}{4}}\frac{\cos2x}{\cos x-\sin x}=
\lim_{x\to\frac{\pi}{4}}\frac{\cos^2 x-\sin^2 x}{\cos x-\sin x}=$$
$$ \lim_{x\rightarrow\frac{\pi}{4}}\frac{(\cos x- \sin x)(\cos x+\sin x)}{\cos x-\sin x}=$$
$$= \lim_{x\rightarrow\frac{\pi}{4}}{\cos x+\sin x}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/275215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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Is this possible If $r_1, r_2, t_1,$ and $t_2$ are real numbers and if $\left|r_{1}\right|<\left|r_{2}\right|$, $\left|t_{1}\right|<\left|t_{2}\right|$
and
$$
\left|\left(r_{1}-r_{2}\right)\left(t_{1}-t_{2}\right)\right|>\left|\left(r_{1}+r_{2}\right)\left(t_{1}+t_{2}\right)-2\left(r_{1}r_{2}+t_{1}t_{2}\right)\right|,... | The second inequality $$|r_{1}+r_{2}-(t_{1}+t_{2})|>||r_{1}|-|t_{1}||+||r_{2}|-|t_{2}||$$holds if $r_{1}> t_{1}>0$, $r_{2}> 0> t_{2}$, then we have $$|r_{1}+r_{2}-(t_{1}+t_{2})|=|r_{1}-t_{1}|+|r_{2}-t_{2}|>||r_{1}|-|t_{1}||+||r_{2}|-|t_{2}||$$
By assumption $r_{1}<r_{2}$, $0<t_{1}<-t_{2}$, then the left hand of the fir... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/276768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Optimization of $2x+3y+z$ under the constraint $x^2+ y^2+ z^2= 1$
Let $x, y$ and $z$ be real numbers such that $x^2+ y^2+ z^2= 1$. Find the
maximum and minimum values of $2x + 3y + z$ .
How can I able to solve the problem? Thanks for your help.
| Using Lagrange Multiplier, $$f(x,y,z,\lambda)=2x+3y+z-\lambda(x^2+y^2+z^2-1)$$
$$\frac{\delta f}{\delta x}=2-2\lambda x$$ etc,
For the extreme values of $f, \frac{\delta f}{\delta x}=0$ etc,
$\implies x=\frac1{\lambda},y=\frac3{2\lambda},z=\frac1{2\lambda}\implies 2x+3y+z=\frac7{\lambda}$
But $x^2+y^2+z^2=1\implies \l... | {
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Evaluate $\int_0^1\left(\frac{1}{\ln x} + \frac{1}{1-x}\right)^2 \mathrm dx$ Evaluate
$$\int_0^1\left(\frac{1}{\ln x} + \frac{1}{1-x}\right)^2 \mathrm dx$$
| New answer (Jun 6, 2022). About 9 and half years since my first solution, it suddenly dawned on me that I can compute this integral in another way:
Note that, for $0 < x < 1$,
$$ \frac{1}{1-x} + \frac{1}{\log x} = \int_{0}^{1} \frac{1-x^s}{1-x} \, \mathrm{d}s. $$
Plugging this to OP's integral and interchanging the ord... | {
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Two different solutions to integral Given the very simple integral
\begin{equation}
\int -\frac{1}{2x} dx
\end{equation}
The obvious solution is
\begin{equation}
\int -\frac{1}{2x} dx = -\frac{1}{2} \int \frac{1}{x} dx = -\frac{1}{2} \ln{|x|} + C
\end{equation}
However, by the following integration rule
\begin{equation... | Both of the results is Ok. Note that there is no need that two constants $C$ in the first result and $C$ in the second one are the same. Here, we have $C_1=C_2\times0.5\ln|2|$.
| {
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Prove $\frac{a}{bc}+ \frac{b}{ca}+ \frac{c}{ab} \ge 1$ Let $a,b,c$ be positive real numbers such that $\dfrac{1}{bc}+ \dfrac{1}{ca}+ \dfrac{1}{ab} \ge 1$. Prove that $\dfrac{a}{bc}+ \dfrac{b}{ca}+ \dfrac{c}{ab} \ge 1$.
| multiply abc
$a+b+c\geq abc $
$a^2+b^2+c^2\geq abc $
$a^2+b^2+c^2-(a+b+c)$
$⇔(a-\frac12)^2+(b-\frac12)^2+(c-\frac12)^2\geq \frac34$
therefore
$a^2+b^2+c^2\geq \frac34+abc$
| {
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Fibonacci Proof
Prove that: $$F_1F_2+F_2F_3+F_3F_4+\cdots+F_{2n-1}F_{2n}=F_{2n}^2$$
I set it up so:
$$F^2(2k) + F(2k+1)F(2k+2) = F^2(2k+2)$$
I've tried:
$$F(2k)^2 + F(2k+1)*F(2k+2) = F(2k+2)*F(2k)+F(2k+2)*F(2k+1)$$
$$F(2k)^2 = F(2k+2)*F(2k)$$
$$F(2k+1)*F(2k) - F(2k)*F(2k-1) = F(2k)*F(2k)+F(2k)*F(2k+1)$$
$$-F(2k)*F(2k... | $$F_{r+2}^2-F_r^2=(F_{r+2}-F_r)(F_{r+2}+F_r)=F_{r+1}(F_{r+2}+F_r)$$
So, $$F_{r+2}^2-F_r^2=F_{r+2}F_{r+1}+F_{r+1}F_r$$
We can utilize this fact to prove the required proposition without using induction.
Put $r=2n-2,2n-4,\cdots,4,2,0$ to get,
$$F_{2n}^2-F_{2n-2}^2=F_{2n}F_{2n-1}+F_{2n-1}F_{2n-2}$$
$$F_{2n-2}^2-F_{2n-4... | {
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Review on double integral Forgot how to do these :
$$\displaystyle\int_0^1\int_0^1\frac{\text{d}x\text{d}y}{1-x^2y^2}$$
| Consider the integral over $x$:
$$\int_0^1 dx \; \frac{1}{1-x^2 y^2} $$
Make the substitution $x = \sin{\theta}/y$, $dx = \cos{\theta}/y \: d \theta$:
$$\begin{align} &= \frac{1}{y} \int_0^{\arcsin{y}} d \theta \sec{\theta} \\ &= \frac{1}{y} [\log{(\sec{\theta} + \tan{\theta})]_{0}^{\arcsin{y}}} \\ &= \frac{1}{2 y} \l... | {
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"answer_count": 3,
"answer_id": 0
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Approaches to integrate $\int_0^1 \frac{x}{\sqrt{a+bx+cx^2}} dx$ I search for a good approach (or a general solution technique) to integrate
$$\int_0^1 \frac{x}{\sqrt{a+bx+cx^2}} dx$$
My knowledge on integration ends with rational functions.
| $\displaystyle \int\frac{x}{\sqrt{a+bx+cx^2}}dx = \frac{1}{2c}\int\frac{\left(2cx+b\right)-b}{\sqrt{a+bx+cx^2}}dx$
$ \displaystyle = \frac{1}{2c}\int\frac{2cx+b}{\sqrt{a+bx+cx^2}}dx-\frac{b}{2c}\int\frac{1}{\sqrt{a+bx+cx^2}}dx$
for First one put $a+bx+cx^2 = t^2$ and $\left(2cx+b \right)dx = 2tdt$
$ \displaystyle = \fr... | {
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A definite integral with hyperbolic cosines I want to show that $$ \int_{0}^{\infty} \frac{\cosh (ax) \cosh (bx)}{\cosh (\pi x)} \ dx = \frac{\cos ( \frac{a}{2} ) \cos ( \frac{b}{2})} {\cos (a) + \cos (b)} \ , \ |a|+|b| < \pi.$$
I thought one approach would be to integrate the appropriate function around a rectangle w... | I have a way of showing this without contour integration in the complex plane. There is a bit of a trick involved and, frankly, Mathematica misleads. It should be noted that the condition $|a|+|b| < \pi$ is needed for the integral to converge. Basically, rewrite the $\cosh$'s as exponentials:
$$\begin{align} \int_{0... | {
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Asymptotic expansion of $ I_n = \int_0^{\pi/4} \tan(x)^n \mathrm dx $ I'm trying to compute the asymptotic expansion of
$$ I_n = \int_0^{\pi/4} \tan(x)^n \mathrm dx $$
Here is what I've done:
Change of variable $$ t= \tan x $$
$$ I_n = \int_0^1 \frac{t^n \mathrm dt}{1+t^2} = \int_0^1 \frac{(1-t)^n \mathrm dt}{t^2-2t+2}... | May be this can give you an asymptotics. Consider even case
$$
I_{2k}=\int\limits_{0}^{1}\frac{t^{2k}}{t^2+1}dt=\int\limits_{0}^{1}(-1)^k\left(\sum\limits_{j=0}^{k-1}(-1)^{j+1}t^{2j}+\frac{1}{t^2+1}\right)dt=\\ (-1)^k\left(-\sum\limits_{j=0}^{k-1}\frac{(-1)^j}{2j+1}+\frac{\pi}{4}\right)=\sum\limits_{j=k}^\infty\frac{(... | {
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determinant $s=n+1$ Need help
$A=\begin{vmatrix}
s&s&s &\cdots & s&s\\
s&1&s &\cdots & s&s\\
s&s&2 &\cdots & s&s\\\vdots & \vdots&\vdots&\ddots&\vdots&\vdots&\\
s&s&s &\cdots & n-1&s\\
s&s&s &\cdots & s&n\\
\end{vmatrix}$
a) calculate $\det A$ when $s =n+1$
Prove that in that case is invertible
b) It may happened that ... | Adding a multiple of one row to another row in a square matrix does not change the determinant (link). Hence, by subtracting the first row from all the others, we find
$$A=\begin{pmatrix}
s&s&s &\cdots & s&s\\
s&1&s &\cdots & s&s\\
s&s&2 &\cdots & s&s\\
\vdots & \vdots&\vdots&\ddots&\vdots&\vdots&\\
s&s&s &\cdots & n-... | {
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Calculus integration problem: $\int \sin^5 (x) \cos^2 (x)\,dx$ What's the integration of $$\int \sin^5 (x) \cos^2 (x)\,dx?$$
| $$ \int \sin^5 (x) \cos^2(x) dx $$
$$= \int(\sin^2(x))^2 \cos^2(x) \sin(x) dx$$
$$=-\int(1 - \cos^2(x))^2 cos^2(x) (-sin(x) dx) $$
Let $u = \cos(x)$
$\implies du = -\sin(x) dx$
$$= -\int(1 - u^2)² u² (du)$$
$$= -\int(1 - 2u^2 + u^4) u^2 du $$
$$= -\int(u^2 - 2u^4+ u^6) du$$
$$= -\left(\frac{u^3}{3} - \frac{2u^5}{5... | {
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General expression for exponentiating a matrix? $A= \begin{pmatrix}
\cos(x) & \sin(x) \\
-\sin(x) & \cos(x) \\
\end{pmatrix}$
I found that
$A^2 = \begin{pmatrix}
\cos^2(x)-\sin^2(x) & 2\cos(x)\sin(x) \\
-2\cos(x)\sin(x) & \cos^2(x)-\sin^2(x) \\
\end{pmatrix}$
$A^2 = \beg... | In general, following formula holds: (Proof of this formula uses induction.)
$$A^n =\begin{pmatrix}
\cos (nx) & -\sin(nx) \\
\sin(nx) & \cos(nx)
\end{pmatrix}$$
In fact, matrix $\begin{pmatrix}
\cos (x) & -\sin(x) \\
\sin(x) & \cos(x)
\end{pmatrix}$ behave similar to $\cos x + i \sin x$.
| {
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complex power series Sorry for taking this from another question, but the second part was never answered, and I'm not sure how to get there. From: Prove the following equation of complex power series.
Show that for $|z| \lt 1$ with $z \in \Bbb C$, we have
$$ \sum_{k=0}^\infty \frac{{z^2}^k}{1-{z^2}^{k+1}} = \frac{z}{1-... | For the first identity, you're trying to prove that
$$
0 = \frac{z}{1-z} - \sum_{k=0}^\infty \frac{{z^2}^k}{1-{z^2}^{k+1}} = \lim_{K\to\infty} \bigg( \frac{z}{1-z} - \sum_{k=0}^K\frac{{z^2}^k}{1-{z^2}^{k+1}} \bigg)
$$
for $|z|<1$. If you work out the exact value of the expression on the right-hand side for small values... | {
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Problem with simple integral I'm trying to solve this simple integral:
$$\frac12 \int \frac{x^2}{\sqrt{x + 1}} dx$$
Here's what I have done so far:
*
*$\displaystyle t = \sqrt{x + 1} \Leftrightarrow x = t^2 - 1 \Rightarrow dx = 2t dt$
*$\displaystyle \frac12 \int \frac{x^2}{\sqrt{x + 1}} dx = \int \frac{t (t^2 - 1)... | $$\begin{align}\frac{x^2}{\sqrt{x+1}} &= \frac{x^2-1}{\sqrt{x+1}} + \frac{1}{\sqrt{x+1}}\\&=(x-1)\sqrt{x+1} + \frac{1}{\sqrt{x+1}}\\&=(x+1)^{3/2} - 2\sqrt{x+1} + \frac{1}{\sqrt{x+1}}\end{align}$$
I think you can integrate each of these terms.
| {
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Prove that $3^{2^{k}} - 1$ is divisible by $2^{k+2}$? Prove that $3^{2^{k}} - 1$ is divisible by $2^{k+2}$ for $k \ge 1$?
Suppose $p,q,r$ are positive integers satisfying $\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}<1$, prove that $\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}\le \dfrac{41}{42} $ ?
| For the second question, assume without loss of generality that $p\le q\le r$. Let $$d=1-\left(\frac1p+\frac1q+\frac1r\right)\;;$$ $d$ is rational, so let $d=\dfrac{m}n$ in lowest terms. If $$1>\frac1p+\frac1q+\frac1r>\frac{41}{42}\;,$$ then $d<\dfrac1{42}$, so $n>42$. Clearly $n\mid pqr$, so $pqr>42$, and since $4^3=6... | {
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Finding an explicit solution of a differential equation
Find an explicit solution of an initial value problem $$\frac{dy}{dx}=\frac{2x}{1+2y}, \; y(2)=0$$
Attempt: I have no problem finding the general solution which is
$y+y^2=x^2+C$
Then, I find the implict solution which I think is correct, but I am not sure
$y+... | Complete the square on the right hand side i.e.
\begin{align}
y+y^2 & = x^2-4\\
y^2 + y + \dfrac14 & = x^2-4 + \dfrac14\\
\left( y + \dfrac12 \right)^2 & = x^2-\dfrac{15}4\\
y + \dfrac12 & = \pm \sqrt{x^2 - \dfrac{15}4}\\
y & = - \dfrac12 \pm \dfrac{\sqrt{4x^2-15}}2
\end{align}
Further, we need $y(2) = 0$ and hence $y ... | {
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A log improper integral Evaluate :
$$\int_0^{\frac{\pi}{2}}\ln ^2\left(\cos ^2x\right)\text{d}x$$
I found it can be simplified to
$$\int_0^{\frac{\pi}{2}}4\ln ^2\left(\cos x\right)\text{d}x$$
I found the exact value in the table of integrals:
$$2\pi\left(\ln ^22+\frac{\pi ^2}{12}\right)$$
Anyone knows how to evaluate t... | First of all, substituting $x\to\frac\pi2-x$ yields
$$\int_0^{\frac\pi2} \ln(\cos^2(x)) \, dx = \int_0^{\frac\pi2} \ln(\sin^2(x)) \, dx = 2 \int_0^{\frac\pi2} \ln(\sin(x)) \, dx = -\pi\ln(2)$$
where the log-sine integral can be computed using the identity
$$\ln(\sin(x)) = -\ln(2) - \sum_{k=1}^\infty \frac{\cos(2kx)}k$$... | {
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The Least Natural number $n$ which has $18$ divisors The Least Natural number $n$ which has $18$ divisors
My Try:: $18 = 2\times 3 \times 3 = 3^2 \times 2$
Now How can I solve it
Thanks
| The smallest prime power $n=p^k$ with 18 divisors would need $k=17$ and hence be $2^{17}$.
If $n=p^kq^m$ with two primes $p,q$ and $k\ge m\ge1$, then it has $(k+1)(m+1)$ divisors, which allows $k=8,m=1$ or $k=5,m=2$ and suggests $n=2^83$ or $n=2^53^2$. (Of course, it is best to take the smallest possible primes and use... | {
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Simplification of $\sqrt{(1-\cos\alpha \cos\beta)^2-\sin^2\alpha \sin^2\beta}$ Simplify the expression $$\sqrt{(1-\cos\alpha \cos\beta)^2-\sin^2\alpha \sin^2\beta}$$
I have done this way : $(1-\cos\alpha \cos\beta)^2 = 1-2\cos\alpha \cos\beta +\cos^2\alpha \cos^2\beta$
Please guide further....
| $$\begin{align}
&\phantom{=\;}( 1 - \cos\alpha \cos\beta)^2 - \sin^2\alpha \sin^2\beta \\
&= ( 1 - \cos\alpha \cos\beta)^2 - (\sin\alpha \sin\beta)^2 &(1) \\
&= ( 1 - \cos\alpha \cos\beta - \sin\alpha \sin\beta)( 1 - \cos\alpha \cos\beta + \sin\alpha \sin\beta) &(2) \\
&= \left( 1 - (\cos\alpha \cos\beta + \sin\alpha \... | {
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Verifying Trigonometric Identities: $2\cos^2x-1 = \frac{1-\tan^2x}{1+\tan^2x}$ Verify that $$ 2\cos^2x-1 = \frac{1-\tan^2x}{1+\tan^2x}$$
| It's generally easier to turn more-complicated stuff into less-complicated stuff; and one common suggestion is always "first convert to sine and cosine":
$$\begin{align}
\frac{1-\tan^2\theta}{1+\tan^2\theta}
&= \frac{1-\frac{\sin^2\theta}{\cos^2\theta}}{1+\frac{\sin^2\theta}{\cos^2\theta}} \\
&=\frac{\left(\cos^2\thet... | {
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The number of real roots of $(x+3)^4 + (x+5)^4 = 16$ I was solving some problems and I came across this question:
Q: The number of real roots of $(x+3)^4 + (x+5)^4 = 16$ is
(a) 0 (b) 2
(c) 4 (d) none of these
Solution: put $y = x + (3+5... | What you need is the expansion of $(a+b)^4$. This is a special case of the very important Binomial Theorem. We have
$$(a+b)^4=a^4+\binom{4}{1}a^3b+\binom{4}{2}a^2b^2+\binom{4}{3}ab^3+b^4.$$
This simplifies to $a^4+4a^3b+6a^2b^2+4ab^3+b^4$.
Put $a=y$, $b=1$, then $a=y$, $b=-1$, and add. There is a pleasant amount of c... | {
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"source": "stackexchange",
"question_score": "2",
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Trig substitution integral I am trying to find $$\int{ \frac {5x + 1}{x^2 + 4} dx}$$
The best approach would be to split up the fraction. According to Wolfram Alpha, the answer is $\frac{5}{2}\ln\left(x^2 + 4\right) + \frac{1}{2}\displaystyle\arctan\left(\frac x2\right)$ which seems OK, but when I try the trig substit... |
$$\frac{5}{2}\ln({\frac{x^2 + 4}{4}}) + \frac{1}{2}\arctan({x/2}) + c_1 \neq \frac{5}{2}\ln(x^2 + 4) + \frac{1}{2}\arctan(x/2) + C$$
Note that $$\frac{5}{2}\ln\left({\frac{x^2 + 4}{4}}\right) = \frac 52 \ln\left(x^2 + 4\right) - \frac 52 \left (\ln 4\right) = \frac 52 \ln(x^2 + 4) + c_2$$
So put $c_1 + c_2 = C$. Then... | {
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"question_score": "6",
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Prove by Induction $\frac{a_n-\sqrt{A}}{a_n+\sqrt{A}} = \left[\frac{a_1-\sqrt{A}}{a_1+\sqrt{A}}\right]^{2^{n-1}} $ $a_1=\frac{1}{2}(a_0+\frac{A}{a_0})$;
$a_2=\frac{1}{2}(a_1+\frac{A}{a_1})$; and
$a_{n+1}=\frac{1}{2}(a_n+\frac{A}{a_n})$ for $n \geq 2$; where $a\gt 0$, $A\gt 0$.
Prove:
$$\frac{a_n-\sqrt{A}}{a_n+\sqrt{A}}... | Let us start:
\begin{align}
\frac{a_n -\sqrt{A}}{a_n+\sqrt{A}} &\underbrace{=}_{*}
\frac{\frac{1}{2}a_{n-1}+ \frac{1}{2}\frac{A}{a_{n-1}} -\sqrt{A}}{\frac{1}{2}a_{n-1}+ \frac{1}{2}\frac{A}{a_{n-1}} +\sqrt{A}} \underbrace{=}_{**}
\frac{\frac{1}{2}a_{n-1}^ 2+ \frac{1}{2}{A} -\sqrt{A}a_{n-1}}{\frac{1}{2}a_{n-1}^ 2+ \fra... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Definite integration evaluation of $\int_{0}^{\pi}\frac{x}{(a^2\cos^2x+b^2\sin^2x)^2}dx$ OK, so the question says evaluate the integral
$$\int_{0}^{\pi}\frac{x}{(a^2\cos^2x+b^2\sin^2x)^2}dx$$
What I do is use the property that $\int_a^bf(x)dx=\int_a^bf(b+a-x)dx$ and this gives me ($I$ is the value of the integral)
$$\f... | $$
\begin{aligned}
\because I&=\int_0^\pi \frac{x}{\left(a^2 \cos ^2 x+b^2 \sin ^2 x\right)^2} \\& \stackrel{x\mapsto\pi-x}{=} \int_0^\pi \frac{\pi-x}{\left(a^2 \cos ^2 x+b^2 \sin ^2 x\right)^2} d x \\
&=\pi \int_0^\pi \frac{d x}{\left(a^2 \cos ^2 x+b^2 \sin ^2 x\right)^2}-I \\
\therefore I&=\frac{\pi}{2} \int_0^\pi \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/312145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Help with the inequality $\sum_{k=1}^{1006} \sqrt{k \cdot (2014-k)}<506^2\pi$ This question's been solved, come and look if you want to check out some hardcore solutions
Here's an inequality that needs to be proven:
Prove that
$\sqrt{1\cdot 2013} + \sqrt{2\cdot 2012} + \sqrt{3\cdot 2011} + \dots + \sqrt{1006\cdot 1008}... | $$\sum_{k=0}^{n-1} \sqrt{k \cdot (2n-k)} = 2n \left(\sum_{k=0}^{n-1} \sqrt{\dfrac{k}{2n} \cdot \left(1-\dfrac{k}{2n} \right)} \right) = (2n)^2 \left(\sum_{k=0}^{n-1} \sqrt{\dfrac{k}{2n} \cdot \left(1-\dfrac{k}{2n} \right)} \cdot \dfrac1{2n}\right)$$
$$\underbrace{\left(\sum_{k=0}^{n-1} \sqrt{\dfrac{k}{2n} \cdot \left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/312263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find the standard matrix for a linear transformation If $T: \Bbb R^3→ \Bbb R^3$ is a linear transformation such that:
$$
T \Bigg (\begin{bmatrix}-2 \\ 3 \\ -4 \\ \end{bmatrix} \Bigg) = \begin{bmatrix} 5\\ 3 \\ 14 \\ \end{bmatrix}$$ $$T \Bigg (\begin{bmatrix} 3 \\ -2 \\ 3 \\ \end{bmatrix} \Bigg) = \begin{bmatrix}-4 \\ 6... | Note that if $$
\begin{pmatrix}-2 \\ 3 \\ -4 \\ \end{pmatrix} = -2\begin{pmatrix}1 \\ 0\\ 0 \\ \end{pmatrix} +3\begin{pmatrix}0 \\ 1\\ 0 \\ \end{pmatrix} -4\begin{pmatrix}0 \\ 0\\ 1 \\ \end{pmatrix}
$$ so $$
T\begin{pmatrix}-2 \\ 3 \\ -4 \\ \end{pmatrix} = -2\times T\begin{pmatrix}1 \\ 0\\ 0 \\ \end{pmatrix} +3\times ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/313798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 3,
"answer_id": 2
} |
How to solve following system of non-linear algebraic equations I have a system
$$
5x^2 - 5y^2 - 3x + 9y = 0,
$$
$$
5x^3 + 5y^3 - 13xy - 15x^2 - y^2 = 0.
$$
How to solve it?
| First, you can observe that if $y=0$, then $x=0$, which gives out a solution.
Let us assume $y\neq 0$ now.
Take $\dfrac{x}{y}=\lambda$.
Then the first equation is:
$$(5\lambda^2-5)y^2+(9-3\lambda)y=0$$
which is
$$(5\lambda^2-5)y+(9-3\lambda)=0$$
So
$$y = \dfrac{3\lambda-9}{5\lambda^2-5}$$
The second one is:
$$(5\lambda... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/314317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculus Bonus Problem solve algebraically
$$\lim_{x\rightarrow 0}\frac{\sqrt[3]{1+x^2}-\sqrt[4]{1-2x}}{x+x^2}$$
without using L'hopital's rule
This is a bonus question for my Math Intensive Major Cal 1 class, I would like to know how to solve it since it stumped me on the test, and after many hours of working at this ... | We will use the formula
$$
a-b=\frac{a^{12}-b^{12}}{a^{11}+a^{10}b+\ldots+ab^{10}+b^{11}}
$$
with
$$
a=\sqrt[3]{1+x^2}\quad\mbox{and}\quad b=\sqrt[4]{1-2x}.
$$
This gives
$$
\sqrt[3]{1+x^2}-\sqrt[4]{1-2x}=\frac{(1+x^2)^4-(1-2x)^3}{a^{11}+a^{10}b+\ldots+ab^{10}+b^{11}}=\frac{6x-8x^2+\ldots}{a^{11}+a^{10}b+\ldots+ab^{10... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/315539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Help understanding the solution to #83 Section 9.1 Calculus 9e, by Larson.
Find $a_n$ of $1, \; \frac{-1}{1\times 3}, \; \frac{1}{1\times3\times5}, \frac{-1}{1\times 3\times 5 \times 7}$
I solved it as $\dfrac{\left(-1\right)^{n-1}}{\left(2n-1\right)!}$
The solution guide says: $\dfrac{\left(-1\right)^{n-1}2^n n!}{\... | No different results here but a somewhat different way of working through it.
\begin{array}{|c|c|c|c|c|}
\hline
\mathbf{n}&1& 2 &3& 4 & 5\\
\hline
a_n&1&\frac{-1}{1\cdot3}&\frac{1}{1\cdot3\cdot5}&\frac{-1}{1\cdot3\cdot5\cdot7}&\frac{1}{1\cdot3\cdot5\cdot7\cdot9}\\
\end{array}
Terms like $1 \cdot 3 \cdot 5 \cdot 7$ are ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/316827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$x_{n+1} = \sqrt{3x_n}$ converges for $x_1 = 1,x_1 = 27$ Proof Prove $x_{n+1} = \sqrt{3x_n}$ converges for $x_1 = 1,x_1 = 27$. (Separate problems for $x_1 = 1$ and $x_1 = 27$.)
EDIT: Took out bad algebra.
| There's no need to prove this using induction as you can find the value that it converges to!
Begin by letting $x_1=a$ and finding the first few terms, as follows:
\begin{align*}
x_1 &= a \\
x_2 &= \sqrt{3}\cdot\sqrt{a} = \sqrt{3a} \\
x_3 &= \sqrt{3}\cdot\sqrt{\sqrt{3a}} = \sqrt{3\sqrt{3a}}\\
x_4 &= \sqrt{3}\cdot\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/317554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
Show that the curve $x^2+y^2-3=0$ has no rational points
Show that the curve $x^2+y^2-3=0$ has no rational points, that is, no points $(x,y)$ with $x,y\in \mathbb{Q}$.
Update: Thanks for all the input! I've done my best to incorporate your suggestions and write up the proof. My explanation of why $\gcd(a,b,q)=1$ is a... | It is actually sufficient to analyze the curve modulo 3 (or 2, even), where the question becomes whether $-1 \equiv 2$ is a square modulo 3. You can reduce the proof to this step directly by clearing denominators/divisibility arguments, but this curve is related to the quaternion algebra generated by the square roots o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/319553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 4,
"answer_id": 3
} |
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.