Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Can a triangle ABC be made if $\frac{\cos A}{2}=\frac{\cos B}{3}=\frac{\cos C}{7}$
I would like to know the simplest approach to find out whether a triangle ABC will be made if $$\frac{\cos A}{2}=\frac{\cos B}{3}=\frac{\cos C}{7}$$
The counterpart questions for sine and tangent can be handled as follows:
*
*If $\df... | Hint use may use the identity
$$\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C=1$$
| {
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"url": "https://math.stackexchange.com/questions/3834491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 2
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Understanding convergence of $a_n := \frac{2n^3 +n^2 +3}{n^3-4}$ We proved by definition that the sequence $a_n := \frac{2n^3 +n^2 +3}{n^3-4}$ converges.
Let $\epsilon > 0$. Choose $N := \lceil \frac{24}{\epsilon} \rceil + 2 $.
Then for all $n \geq N$ it holds, that
$$|a_n - 2| = \big |\frac{2n^3+n^2+3}{n^3-4} - \frac{... | I like to do these by dividing out the highest power of the variable. You get $\dfrac{2n^3+n^2+3}{n^3-4}=\dfrac{2+1/n+3/n^3}{1-4/n^3}\to2/1=2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Induction proof of a known harmonic sum I want to prove that $$\frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2^n} \leq 1$$ only by induction!
I check for the first one, $\frac12 \leq 1 $ correct.
Then I assume for $n=k$ : $$\frac12 + \frac14 + \dots + \frac{1}{2^k} \leq 1$$
And Try and prove for $n=k+1$
$$\frac12 + \fr... | Hint : Prove the following stronger hypothesis induction :
$$\frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2^n} = 1 - \frac{1}{2^n}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Maximizing $\frac{y+1}{x+2}$ when $(x-3)^2 + (y-3)^2 = 6$ Suppose $x$ and $y$ are real numbers such that $(x-3)^2 + (y-3)^2 = 6.$ Than, maximize $\frac{y+1}{x+2}.$
I do in fact realize that this is a double post, but it's a 5 year old question and I don't feel as if it is appropriate to bump it. I did as the original ... | Going from what you have (and avoiding calculus):
$(k^2 + 1)x^2 + (4k^2 -4k-3)x + (k^2 - 16k + 19)= 0$
If we plug this into the quadratic formula:
$x = \frac {-b \pm \sqrt {b^2 - 4ac}}{2a}$
When $b^2 - 4ac < 0$ the formula breaks.
Setting $b^2 - 4ac = 0$ will give us the extreme values of $k.$
$(4k^2 - 8k - 6)^2 - 4(k... | {
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Calculate the value of the following limit So I have given the limit:
$$\lim_{x\to0} \frac{2\sin\left ( e^{-\frac{x^2}{2}} -\cos x \right)}{(\arctan(\sinh(x^2))^2}$$
I have been struggling for hours with it. Since i got the undefined form when i put $x=0$ i tried out with L'Hopital method and I come to this point:
$$\l... | We have that
$$\frac{2\sin\left ( e^{\frac{-x^2}2}-\cos x \right )}{(\arctan(\sinh(x^2))^2}
=$$
$$=\left(\frac{\sinh(x^2)}{\arctan(\sinh(x^2)}\right)^2\cdot\left(\frac{x^2}{\sinh(x^2)}\right)^2\cdot \frac{\sin\left ( e^{\frac{-x^2}2}-\cos x \right )}{e^{\frac{-x^2}2}-\cos x }\cdot2\cdot\frac{e^{\frac{-x^2}2}-\cos x }{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3838421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $|ax^2 + bx + c| \le 1$ for all $x$ in $[0,1]$ then evaluate $|a|$, $|b|$ and $|c|$
Solve the following:
If $|ax^2 + bx + c| \le 1$ for all $x$ in $[0,1]$, then
i) $|a| \le 8$
ii) $|b| \ge 8$
iii) $|c| \le 1$
iv) $|a| + |b| + |c| \le 17$
Solution from my textbook:
Put $x = 0, 1 \text{ and } \frac{1} {2}$ to get:
... | Using triangle inequality to simplify the other answer, we have for example:
$$|b+3c|=|a+2b+4c-(a+b+c)|\le|a+2b+4c|+|a+b+c|\le4+1=5$$
$$|b|=|b+3c-3c|\le |b+3c|+|3c|\le5+3=8$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Determining whether a subset is a subspace The problems are as follows:
Determine whether the following subsets of $\mathbb{R}^3$ are subspaces of $\mathbb{R}^3$.
*
*$A = \{(u^2, v^2, w^2) \,|\, u, v, w \in \mathbb{R} \}$,
*$B = \left\{(a, b, c) \,|\,
\begin{pmatrix}
a & b & c\\
1 & 2 & 0\\
0 & 1 & 2
\end{pmatrix} ... | You proof seems fine to me but for case $A$ we should observe that in general
$$(u_1^2, v_1^2, w_1^2) + (u_2^2, v_2^2, w_2^2)\neq ((u_1+u_2)^2, (v_1+v_2)^2, (w_1+w_2)^2)$$
and you can easily find by yourself an example where equality doesn't hold which is a necessary step to complete the proof.
| {
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Find the area bounded by the curve $x^4+y^4=x^2+y^2$ I am stuck with this problem which deals with evaluating an Area
The problem reads :
Find the area bounded by the curve $x^4+y^4=x^2+y^2$.
I tried factorizing the expression and expressing $y$ in terms of $x$, not able to proceed with that idea. Someone please help... | It is possible to do this without using polar coordinates- but not easy. I would start by noting that only even powers of x and y are involved so the graph of this is symmetric about the x and y axes. That means that we can calculate the area for x and y positive and multiply by 4.
There is a little problem in that y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3840121",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Prove that the diophantine equation $(xz+1)(yz+1)=az^{k}+1$ has infinitely many solutions in positive integers. Given two positive integers $a$ and $k>3$ : From experimental data, it appears the diophantine equation
$(xz+1)(yz+1)=az^{k}+1$
has infinitely many solutions in positive integers $x,y, z$.
To motivate the que... | Getting there. Here is $k=4.$ a family of solutions to
$$ a z^4 + 1 = (xz+1)(yz+1) $$
is parametrized by integer $t$ with
$$ y=at $$
$$ x = a^4 t^5 - at $$
$$ z = a^2 t^3 $$
Both sides of the equation are
$$ a^9 t^{12} + 1 $$
=======================================
For that matter, we can take care of all $k \n... | {
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Closed Form Solution for a Recurrence Relation We start out with: $a_n = 3a_{n-1} - 7, a_0 = 2$. Is the following valid?
$$a_{n-1} = 3(3a_{n-2}-7)-7 \\ a_{n-1} = 3^2a_{n-2}-(3\cdot-7) - 7 \\ \vdots \\ a_k = 3^ka_{n-k} -(3^{k-1}\cdot-7) \cdots -7\\ \text{Let $k = n$} \\ a_n = 3^ka_0 - (3^{n-1}\cdot-7)-\cdots-7$$
I don't... | As Greg Martin has already pointed out, your closed form cannot be right, because it doesn’t satisfy the recurrence. You could also see whether it generates the right values for $a_1,a_2$, and $a_3$, say, and find that it doesn’t.
There is a slightly better way to organize this sort of ‘unwinding’ of a simple recurrenc... | {
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$\frac{a^3}{b+c}+\frac{b^3}{a+c}+\frac{c^3}{a+b}\geq \frac{3}{2}$ for $a,b,c\in\mathbb{R}^+$ with $abc=1$ Suppose that $a,b,c$ are positive reals such that $abc=1$. Prove that $$\frac{a^3}{b+c}+\frac{b^3}{a+c}+\frac{c^3}{a+b}\geq \frac{3}{2}.$$
Hint: Use Titu's lemma.
My approach: I am trying to use Titu's lemma direct... | Also, by Holder and AM-GM $$\sum_{cyc}\frac{a^3}{b+c}\geq\frac{(a+b+c)^3}{3\sum\limits_{cyc}(b+c)}=\frac{1}{6}(a+b+c)^2\geq\frac{9}{6}=\frac{3}{2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find coefficient of $x^3y^4z^5$ in polynomial $(x + y + z)^8(x + y + z + 1)^8$
Find coefficient of $x^3y^4z^5$ in polynomial $(x + y + z)^8(x + y + z + 1)^8$
It is pretty easy to see that our goal is to choose from each multiplier in this polynomial $x,y,z,1$ in certain amounts. Since sum of powers equals $12$ and we... | Objective: To find coefficient of $\ x^3y^4z^5\ $ in $\ (x + y + z)^8(x + y + z + 1)^8$
First expand $$(1+(x+y+z))^8=\sum_{r=0}^{8} {8 \choose r}(x+y+z)^r$$
Now multiplying $(1+(x+y+z))^8$ with $(x+y+z)^8$ gives us,
$$(x+y+z)^8(1+(x+y+z))^8=\sum_{r=0}^{8} {8 \choose r}(x+y+z)^{r+8} \label{1} \tag{1}$$
Now using multino... | {
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"timestamp": "2023-03-29T00:00:00",
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How to construct a solution to a homogeneous system of linear equations? given the system of linear equations
$\begin{array}{rrrrr}x_{1} & + & x_{2} & + & 2 x_{3} & = & 2 \\ 2 x_{1} & + & 5 x_{2} & - & x_{3} & = & 2 \\ 3 x_{1} & + & 6 x_{2} & + & x_{3} & = & 4\end{array}$
I found the solutions
$x=\left(\begin{array}{c}... | Since both $x$ and $y$ are solutions of the original system, $x-y$ is a solution of the homogenous system.
| {
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How to solve $\sqrt{2} + \sqrt{3} - \frac{m}{n} > \frac{1}{cn^4}, \forall n \ge 1$? If $\frac{m}{n}$ is an irreducible fraction such that $1 > \sqrt{2} + \sqrt{3} - \frac{m}{n} > 0$, then there exists a constant $c > 0$ so that $$\sqrt{2} + \sqrt{3} - \frac{m}{n} > \frac{1}{cn^4}, \forall n \ge 1$$
From the first inequ... | It is an old result of Liouville :https://en.m.wikipedia.org/wiki/Diophantine_approximation
It can be applied because your number $\sqrt{2}+\sqrt{3}$ is algebraic of degree $4$.
| {
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Remainder of dividing polynomial of the $n$th degree
What is the remainder when dividing the polynomial
$$P(x)=x^n+x^{n-1}+\cdots+x+1$$ with the polynomial
$$x^3-x$$ if $n$ is a natural odd number?
So, what I know so far is:
$$P(x)=Q(x)D(x)+R(x)$$
In this case I'll call $Q(x) = x^3-x$
$$Q(x) = 0 \iff x=\pm1$$
So from... | Your approach is fine but you've missed the root $x=0$ of $x^3-x$.
We get $R(0)=P(0)=1$, $R(1)=P(1)=n+1$, $R(-1)=P(-1)=0$.
The quadratic polynomial that interpolates this data is $\frac{n-1}{2} x^2+\frac{n+1}{2}x+1$.
The same approach works when $n$ is even (the interpolation data is different because $P(-1)=1$) and we... | {
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Triples satisfying three equations with six solutions What are all triples $(x, y, z)$ such that all three of the following equations are satisfied:$$x(y + z - 5) = 7,$$$$y(x + z - 5) = 7,$$$$x^2 + y^2 = 50.$$According to Wolfram Alpha we know that $(-7,-1,5)$, $(-5,-5,43/5)$, $(-1,-7,5)$, $(1,7,5)$, $(5,5,7/5)$, $(7,1... | By subtracting the first two equations, we have $$(z-5)(x-y)=0$$
Let's consider two cases:
*
*If $z=5$, then we have $xy=7$ and $x^2+y^2=50$.
$$x^2+\frac{49}{x^2}=50$$
$$x^4-50x^2+49=0$$
$$(x^2-49)(x^2-1)=0$$
Solving this would give us $4$ solutions, corresponding to when $x=\pm 7, \pm 1$. Given $z$ and $x$, $y$ is ... | {
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a recursion in roots of the polynomial let $P(x)$ is a polynomial which satisfies property $\psi$ where property $\psi$ is given by
whenever r is a root of $P(x) = 0$ then $r^2 - 4$ is also a root of the given equation.
i) if $P(x)$ is a quadratic polynomial of the form $x^2 + ax + b$ then find all the possible equat... | Part i):
Let $r, s$ be the two roots of $P$.
By assumption, $r^2 - 4$ and $s^2 - 4$ all belong to the set $\{r, s\}$. Thus there are several possibilities:
*
*$r^2 - 4 = r$, $s^2 - 4 = s$.
In this case, $r$ and $s$ are the two roots of $x^2 - x - 4$, and we have $P(x) = x^2 - x - 4$.
*$r^2 - 4 = r$, $s^2 - 4 = r$.
... | {
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prove $\sum_{cyc}\frac{a^2}{a+2b^2}\ge 1$
prove $$\sum_{cyc}\frac{a^2}{a+2b^2}\ge 1$$ holds for all positives $a,b,c$ when $\sqrt{a}+\sqrt{b}+\sqrt{c}=3$ or $ab+bc+ca=3$
Background Taking $\sqrt{a}+\sqrt{b}+\sqrt{c}=3$ This was left as an exercise to the reader in the book 'Secrets in inequalities'.This comes under ... | For $ab+bc+ca=3.$ We need to prove
$$ a+b+c \geqslant 1 + \frac{2}{3}\sum (ab)^{2/3}.$$
From condition we get
$$a+b+c \geqslant \sqrt{3(ab+bc+ca)}=3.$$
Now, using the AM-GM inequality we have
$$\sum (ab)^{2/3} = \sum \sqrt[3]{1 \cdot ab \cdot ab} \leqslant \sum \frac{1+2ab}{3} = 1+\frac{2}{3}(ab+bc+ca)=3.$$
So
$$1 + \f... | {
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Multiple proofs of $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}2$ Here is my question:
Let $a,b,c\in\mathbb{R^+}$, prove that $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}2$$
Here is my solution:
From C-S inequality, we get $$\sum_{cyc}(a+b)\sum_{cyc}\frac{1}{a+b}\geq3^2,$$
which is equivalent to ... | One answer using AM-GM.
From AM-GM, we obtain two inequalities:
*
*$a^{\frac{3}2}+b^{\frac{3}2}+b^{\frac{3}2}\geq3a^\frac{1}2b$
*$a^{\frac{3}2}+c^{\frac{3}2}+c^{\frac{3}2}\geq3a^\frac{1}2c$
Add them up and we get:
$$2(a^{\frac{3}2}+b^{\frac{3}2}+c^{\frac{3}2})\geq3a^\frac{1}2(b+c)\iff\frac{a}{b+c}\geq\frac{a^\frac{3... | {
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Why is -8 $\equiv$ 6 mod 7? I read in a book that $-8 \equiv 6 \bmod 7$ which means that $-8$ and $6$ leave the same remainder when divided by $7.$
The remainder when $-8$ is divided by $7$ is $-1.$ But when $6$ is divided by $7,$ isn't the remainder $6$?
I recognise that we can write $7\cdot1 - 1=6$, so from here it s... | The definition is $a\equiv b\mod n$ iff $n$ divides $a-b$.
Here $-8\equiv 6\mod 7$ since $7$ divides $-8-6 = -14$.
Another definition is that $a\equiv b\mod n$ iff both leave the same remainder modulo $n$. But note that the remainder is between $0$ and $n-1$.
Here $-8$ and $6$ have the same remainder $6$ modulo $7$, s... | {
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Remainder Theorem: $f(x) = x^3+1$ divided by $\operatorname {d}(x)= x$ Take for example $\operatorname {f}(x) = x^3+1$ divided by $\operatorname {d}(x)= x-1$ $$\frac {x^3+1}{x-1}$$
The Remainder Theorem tells us that the remainder will be $$\operatorname {f}(zero-of-\operatorname {d}(x))$$
In this case the remainder sh... | $$\frac{x^3+1}{x-1}=x^2+x+1+\frac{2}{x-1}$$
Here $f(x)=x^3+1$, so the remainder when $f(x)$ is divided by the linear factor $x-1$ is a constant 2 as seen above and which is nothing but $f(1)=2$.
If the divisor is quadratic, the remainder is linear function of $x$ or a constant.
| {
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$n$ is odd if and only if there exists an $a \in \mathbb{Z}$ such that $n^4=16a+1$ Write a formal proof.
Suppose that $n \in ℕ$. Prove $n$ is odd if and only if there exists $a \in \mathbb{Z}$ such that $n^4 = 16a + 1$.
There exists $k\in \mathbb{Z}$ such that $n=2k+1$. So I've used this formula for odd numbers
$$n^4=(... | If $n$ is odd then there exists an integer $k$ so that $n$ can be written as $n=2k +1$. And with respect to that $k, n^4 = (2k+1)^4 = 16k^4 + 4*8k^3 + 6*4k^2 + 4k + 1=16k^4 + 32k^2 + 24k^2 + 4k +1$.
And $n^4= 16(k^4 + 2k^3 + \frac {3k^2 + k}2) + 1= 16(k^4 + 2k^3 + \frac {k(3k+1)}2) + 1$.
So if we can prove $\frac {k(... | {
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How can you calculate the rank of an nxn matrix with the given conditions? Let $A=(a_{i,j})$ a square matrix whose elements are:
*
*$0$ if $i=j$.
*$1$ if $j>i$.
*$-1$ if $j<i$.
Is there a simple way to find its rank?
| that was neat. Same rank as the matrix with entry $1$ when $j=i+1,$ then $-1$ when $j=i-1,$ otherwise $0$
Congruence:
$$ \tiny
\left(
\begin{array}{rrrrr}
1&0&0&0&0 \\
0&1&0&0&0 \\
0&-1&1&0&0 \\
0&0&-1&1&0 \\
0&0&0&-1&1 \\
\end{array}
\right)
\left(
\begin{array}{rrrrr}
0&1&1&1&1\\
-1&0&1&1&1 \\
-1&-1&0&1&1 \\
-1&-1&-1... | {
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Let $x,y,z$ are the lengths of sides of a triangle such that $x+y+z=2$. Find the range of $xy+yz+xz-xyz$ .
Let $x,y,z$ are the lengths of sides of a triangle such that $x+y+z=2$. Find the range of $xy+yz+xz-xyz$ .
What I Tried: I have tried by doing $(x+y+z)(x+y+z)=2*2 = 4.$
Also I got $2(xy+yz+xz)+(x^2+y^2+z^2)=4$.... | Given any three positive numbers $x,y,z$, it can form the sides of a non-degenerate triangle if and only if we can find three positive numbers $u, v, w$ such that
$$x = v +w,\quad y = u + w\quad\text{ and }\quad z = u +v$$
(This is known as the Ravi's substituion).
In terms of them, the condition $x + y + z = 2$ is equ... | {
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Find $\lim_{x \to 3} \frac{x^{15} - 3^{15} - 15 \cdot 3^{14}(x-3)}{(x-3)^2}$ Please help me find:
$$
\lim_{x \to 3} \frac{x^{15} - 3^{15} - 15 \cdot 3^{14}(x-3)}{(x-3)^2}
$$
I cannot use L'Hospital's rule.
I tried to eliminate $x-3$, but I have no idea what to do next.
$$
\lim_{x \to 3} \frac{x^{15} - 3^{15} - 15 \cd... | Now, $$\frac{x^{14}+3x^{13}+...+3^{13}x-14\cdot3^{14}}{x-3}=\frac{x^{14}-3^{14}}{x-3}+3\cdot\frac{x^{13}-3^{13}}{x-3}+...+3^{13}\rightarrow$$
$$=14\cdot3^{13}+3\cdot13\cdot3^{12}+...+3^{13}=(14+13+...+1)3^{13}=105\cdot3^{13}=35\cdot3^{14}.$$
I used the following:
$$a^n-b^n=(a-b)\left(a^{n-1}+a^{n-2}b+...+ab^{n-2}+b^{n-... | {
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Evaluating the challenging sum $\sum _{k=1}^{\infty }\frac{H_{2k}}{k^3\:4^k}\binom{2k}{k}$. I managed to evaluate the sum, my approach can be found $\underline{\operatorname{below as an answer}}$, I'd truly appreciate if any of you could share new methods to evaluate this series, thank you.
The following are the short ... | First part.
Evaluating $$\int _0^1\frac{\operatorname{Li}_3\left(x\right)}{x\sqrt{1-x^2}}\:dx$$
$$\int _0^1\frac{\operatorname{Li}_3\left(x\right)}{x\sqrt{1-x^2}}\:dx=\frac{1}{2}\int _0^1\ln ^2\left(t\right)\left(\int _0^1\frac{1}{\left(1-tx\right)\sqrt{1-x^2}}\:dx\right)dt$$
$$=\frac{\pi }{4}\int _0^1\frac{\ln ^2\le... | {
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$\sqrt{1-x^2}$ is not differentiable at $x = 1$. Please give me an easier proof if exists.
Prove that $\sqrt{1-x^2}$ is not differentiable at $x = 1$.
My proof is here:
Let $0 < h \leq 2$.
$\frac{\sqrt{1-(1-h)^2}-\sqrt{1-1^2}}{-h} = \frac{\sqrt{2h-h^2}}{-h} = \frac{2h-h^2}{-h\sqrt{2h-h^2}} = \frac{h-2}{\sqrt{2h-h^2}... | The derivative of $x\mapsto\sqrt{1-x^2}$ exists only for $-1<x<1$:
$$f'(x)=-\frac{x}{\sqrt{1-x^2}}$$
Since the function is continuous on $[-1,1]$, we can compute the limits of the derivative. Then
\begin{align}
\lim_{x\to1^+}f'(x)&=i\infty\\
\lim_{x\to 1^-}f'(x)&=-\infty
\end{align}
so the function is not differentiabl... | {
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Find the internal angle of the triangle between $\overrightarrow a \& \overrightarrow b $. If $\overrightarrow a = \hat j + \sqrt 3 \hat k;\overrightarrow b = - \hat j + \sqrt 3 \hat k;\overrightarrow c = 2\sqrt 3 \hat k$ form a triangle then find the internal angle of the triangle between $\overrightarrow a \& \o... | By your work $$\cos\theta=-\frac{2}{2\cdot2}-\frac{1}{2}.$$
Let $\vec{a}=\vec{AB},$ $\vec{b}=\vec{BC}$ and $\vec{c}=\vec{AC}.$
Thus, $$\cos\theta=\cos\measuredangle ABC=\cos\left(180^{\circ}-\measuredangle(\vec{a},\vec{b})\right)=-\frac{1}{2}.$$$$
| {
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Is this proof regarding odd perfect numbers valid? (Edited in response to a comment.)
Here are my:
QUESTIONS
(1) Is this proof regarding odd perfect numbers valid, particularly the part where it says
$$\dfrac{2n^2}{D(n^2)} \neq (q + 1)?$$
(2) If the proof is incorrect, how can the argument be mended to produce a val... | (1)
Your proof looks invalid to me.
You are saying "Since $2n^2$ is even and $D(n^2) = 2n^2 - \sigma(n^2)$ is always odd (since $n^2$ is a square), then $\dfrac{2n^2}{D(n^2)}$ cannot be an integer".
This is not true since $\frac{\text{even}}{\text{odd}}$ can be an integer. For example, $\frac{6}{3}=2$.
Since $D(n^2)$ ... | {
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Prove that $\alpha^n<\sum_{k=1}^n {2n \choose n+k}k^2<\beta^n$, when $0<\alpha<4<\beta$ are constants
Let $\alpha,\beta$ constants such that $0<\alpha<4<\beta$. Prove that $\exists n_0.\forall n>n_0$ $$\alpha^n<\sum_{k=1}^n {2n \choose n+k}k^2<\beta^n$$
My attempt:
$$\sum_{k=1}^n {2n \choose n+k}k^2=\sum_{k=1}^n \fr... | \begin{align*}
\sum_{k=1}^n\binom{2n}{n+k}k^2
&=\frac12\sum_{k=-n}^{n}\binom{2n}{n+k}k^2
\\&=\frac12\sum_{k=0}^{2n}\binom{2n}{k}(k-n)^2
\\&=\frac18\left.\frac{d^2}{dx^2}\sum_{k=0}^{2n}\binom{2n}{k}e^{2(k-n)x}\right|_{x=0}
\\&=\frac18\left.\frac{d^2}{dx^2}(e^x+e^{-x})^{2n}\right|_{x=0}
\\&=2^{2n-3}\left.\frac{d^2}{dx^2}... | {
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How to define $\delta$ to prove $\lim\limits_{x \to 9} \sqrt{x-5} = 2$ When tackling this problem, parting from the assumption that $\lvert \sqrt{x-5} - 2\rvert \lt \epsilon$, I arrived through arithmetical manipulation at $$\frac{\lvert x-9\rvert}{\lvert \sqrt{x-5} + 2\rvert} \lt \epsilon,$$ which has the $\lvert x-9\... | Assuming $|x-9|<\delta$ we have
$$
\left| \sqrt{x-5} - 2 \right|
= \left| \frac{(\sqrt{x-5} - 2)(\sqrt{x-5} + 2)}{\sqrt{x-5} + 2} \right|
= \left| \frac{x-9}{\sqrt{x-5} + 2} \right|
< \frac{\delta}{\sqrt{x-5} + 2}
.
$$
Now note that $\frac{1}{\sqrt{x-5} + 2} < \frac{1}{2}$ as long as $x-5>0.$ The latter can be achieved... | {
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Let $n$ be an integer. If the tens digit of $n^2$ is 7, what is the units digit of $n^2$?
Let $n$ be an integer. If the tens digit of $n^2$ is 7, what is the units digit of $n^2$?
So $n^2 \equiv 7 \pmod{100}$? If this is the case then this can be written as $n^2 = 100k +7$, where $k \in \Bbb Z.$
Here one can see that... | Hint You are looking for adigit $k$ such that
$$n^2 \equiv 70+ k \pmod{100}$$
By the Chinese Remainder Theorem this is equivalent to
$$n^2 \equiv 2+ k \pmod{4}\\
n^2 \equiv k-5\pmod{25}
$$
The quadratic residues modulo $4$ are $0,1$, therefore $k \in \{ 2,3, 6,7 \}$. You have now to figure for which of those $k-5$ is ... | {
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Hard limit $\lim_{x\to\infty}\Bigg((x(x+1)(x+2))^{\frac{1}{3}}-\Big(2(x+1)-\Big(x^x(x+1)^{x+1}(x+2)^{x+2}\Big)^{\frac{1}{3x+3}}\Big)\Bigg)=0$ Prove that :
$$\lim_{x\to\infty}\Bigg((x(x+1)(x+2))^{\frac{1}{3}}-\Big(2(x+1)-\Big(x^x(x+1)^{x+1}(x+2)^{x+2}\Big)^{\frac{1}{3x+3}}\Big)\Bigg)=0$$
I can prove that :
$$\lim_{x\to\... | It suffices to note that as $x\to +\infty$,
$$\left(1+\frac{a}{x}\right)^{m\frac{x+b}{x+c}}=
\exp\left(m\frac{1+b/x}{1+c/x}\log\left(1+\frac{a}{x}\right)\right)=1+\frac{ma}{x}+o(1/x).$$
Then,
\begin{align*}(x(x+1)(x+2))^{\frac{1}{3}}&=x\left(1+\frac{1}{x}\right)^{\frac{1}{3}}\left(1+\frac{2}{x}\right)^{\frac{1}{3}}\\
&... | {
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Prove that $\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ... + \frac{1}{\sqrt{n}} \leq 3\sqrt{n+1} - 3$ Prove that $\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ... + \frac{1}{\sqrt{n}} \leq 3\sqrt{n+1} - 3$ for every natural $n$.
I've already tried to write it like this:
$$\frac{\sq... | By induction we have
*
*base case: $n=1 \implies 1 \le 3\sqrt 2-3$
*inductive step: we assume true
$$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + … + \frac{1}{\sqrt{n}} \leq 3\sqrt{n+1} - 3 \tag 1$$
and we need to prove that
$$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + … + \fra... | {
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Why I cannot get the same answer if I do substitution with $x=a \cos\theta$ for $\int \sqrt{a^2-x^2}dx$ compared with substitution $x=a \sin\theta$? Why I cannot get the same answer if I do substitution with $x=a \cos\theta$ for $\int \sqrt{a^2-x^2}dx$ compared with substitution $x=a \sin\theta$?
$\int \sqrt{a^2-x^2}dx... | Method$\#1:$
If $\theta=\arccos\dfrac xa=\dfrac\pi2-\arcsin\dfrac xa,$
$\cos\theta=\dfrac xa,0\le\theta\le\pi$
$$dx=-a\sin\theta\ d\theta\text{ and }\sqrt{a^2-x^2}=a\sin\theta$$
$$\int\sqrt{a^2-x^2}\ dx=-\int a^2\sin^2\theta\ d\theta=\dfrac{a^2}2\int(\cos2\theta-1)\ d\theta=\dfrac{a^2\sin2\theta}4-\dfrac{a^2\theta}2$$
... | {
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How can I calculate Slope and Y-intercept in Multiple Regression? What is the formula for Slope and Y-intercept in Multiple Linear Regression? We can easily find Slope and Y-intercept of Linear Regression meaning the data having only one Independent Variable?
Is there any general way to calculate it?
| Suppose you have the following regression function: $ y_i = \beta_{0} + \beta_{1} x_{i1} + \cdots + \beta_{p} x_{ip} + \varepsilon_i$, where $\varepsilon_i$ is the random part (white noise). Here you have $p+1$ parameters. To estimate the the parameters $b_0,b_1,\ldots, b_p$ we need the following matrix and vectors.
$\... | {
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Prove by induction $\left(1-\frac{1}{2}\right)\ldots\left(1-\frac{1}{2^{n}}\right)\geq\frac{1}{4}+\frac{1}{2^{n+1}}$ Prove by induction $\left(1-\frac{1}{2}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{8}\right)\ldots\left(1-\frac{1}{2^{n}}\right)\geq\frac{1}{4}+\frac{1}{2^{n+1}}$
What would be the best way to solv... | To prove a statement by induction, we must first prove the statement works for $n=1$, then extend that statement by supposing that the given statement is true for $n$, then proving it maintains truth for $n+1$. I am also assuming $n$ is an positive integer.
The logic behind that is if the statement works for $n=1$, the... | {
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The diagram shows an equilateral triangle $ADE$ inside a square $ABCD$ . What is the value of $\frac{[\Delta ADE]}{[\Delta DEC]}$ .
The diagram shows an equilateral triangle $ADE$ inside a square $ABCD$ . What is the value of $\frac{[\Delta ADE]}{[\Delta DEC]}$ .
What I Tried: Here is the diagram :-
You can see I ma... |
By Area of triangle using trigonometry
Area (DEC) $ = \frac12 (CD)(DE) \sin (30^\circ) = \frac 14 x^2$
| {
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Prove the identity $\frac{1}{1+x^{1}}+\cdots+\frac{2^{n}}{1+x^{2^{n}}}=\frac{1}{x-1}+\frac{2^{n+1}}{1+x^{2^{n+1}}}$ Show that for all non-negatives integers $n$, it is true that
$$\frac{1}{1+x^{1}}+\frac{2}{1+x^{2}}+\frac{4}{1+x^{4}}+\frac{8}{1+x^{8}}+\cdots+\frac{2^{n}}{1+x^{2^{n}}}=\frac{1}{x-1}+\frac{2^{n+1}}{1+x^{2... | A proof by induction is elementary and not difficult. Let $$S_n(x) = \sum_{k=0}^n \frac{2^k}{1 + x^{2^k}}, \quad P_n(x) = \frac{1}{x-1} + \frac{2^{n+1}}{\color{red}{1 - x^{2^{n+1}}}}.$$ Note that you have a typographical error for $P_n(x)$; the correct expression is shown in red text. The claim is that for all nonne... | {
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By using the properties of definite integrals, evaluate $\int_0^{\pi}\frac{x}{1+\sin x}dx$
By using the properties of definite integrals, evaluate $\int_0^{\pi}\frac{x}{1+\sin x}dx$.
My attempt:
(Using the property $\int_0^{2a}f(x)dx=\int_0^a(f(x)+f(2a-x))dx$)
$$\int_0^{2\frac{\pi}{2}}\frac{x}{1+\sin x}dx=\int_0^{\fr... | Alternatively, I like to try to change sine to cosine and then tackle it by the properties of odd and even functions.
First of all, I use the substitution $y=\dfrac{\pi}{2}-x,$ then
$$
\begin{aligned}
I &=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\frac{\pi}{2}-y}{1+\cos y} d y \\
&=\frac{\pi}{2} \int_{-\frac{\pi}{2}... | {
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Recurring Sequence with Exponent Working with recurring sequences and generating functions, I'm generally lost on solving a general expression of $a_n$ for any $n$ when the next part of the sequence, that is $a_{n+1}$, is in the form of an exponent, such that $a_n = a_{n-1} +k^{n-1}$, where k is some constant. I have ... | We use ordinary generating functions. Let $A(x) = \sum_{i=0}^n a_n x^n$. Then we have (summing from $n=1$)
\begin{align}
a_n &= 2a_{n-1} + 5^{n-1},\\
\sum_{n=1}^\infty a_nx^n &= \sum_{n=1}^\infty 2a_{n-1} x^n + \sum_{n=1}^\infty 5^{n-1} x^n,\\
A(x) - a_0 &= 2x\sum_{n=1}^\infty a_{n-1} x^{n-1} + x\sum_{n=1}^\infty 5^{n-... | {
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$16$ people, $8$ men, $8$ women, divide the group to $8$ couples, what is the chance for exactly $3$ male couples? We have a group of $16$ people, $8$ men, $8$ women.
We divide them to $8$ couples.
Let $X$ Be the number of couples make of $2$ men.
Calculate: $P(X = 3)$
I am not sure how to approach this.
I thought tha... | The number of ways to make $n$ couples from $2n$ people is
$$\frac{(2n)!}{n!\cdot 2^n}$$
This is derived by starting like you did, multiplying
$$\begin{align}
&\quad\binom{2n}{2} \cdot \binom{2n-2}{2} \cdot \binom{2n-4}{2} \cdots \binom{4}{2} \cdot \binom{2}{2} \\
\\
=& \quad\frac{2n(2n-1)}{2} \cdot \frac{(2n-2)(2n-3)}... | {
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A convergence test for series I need to prove the following result: let $\left( a_{n}\right) \subset
\left( 0,\infty \right) $ such that
\begin{equation*}
\frac{a_{n+1}}{a_{n}}=1-\frac{1}{n}-\frac{x_{n}}{n\ln n},
\quad
(X)
\end{equation*}
where $x_{n}\geq x>1$. Prove that
\begin{equation*}
\sum_{n\geq 1}a_{n}
\end{equa... | Here is a method which can be applied to both cases : let's do the first one, the other one will be similar. We have $a_n > 0$ such that
$$\frac{a_{n+1}}{a_n} = 1 - \frac{1}{n} - \frac{x_n}{n \ln (n)}$$
with $x_n \geq x > 1$ for a real number $x$. Let $y=(1+x)/2$. Consider $$u_n = \frac{1}{n \ln^{y}(n)}$$
One has
$$\fr... | {
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Contour integration with pole I wanted to integrate
$$\oint \frac{\sqrt{z+5}}{z^5}$$
around a circular contour radius $1$ center $0$.
So the function has pole at 0. How can I proceed from here?
| Cauchy's integral formula is
$$f^{(n)} (z_0) = \frac{n!}{2\pi i} \oint \frac{f(z)}{(z-z_0)^{n+1}} dz$$
so
$$ \oint \frac{\sqrt{z+5}}{z^5} dz = \frac{2\pi i}{4!} \,\left.\frac{d^4}{dz^4}(z+5)^{1/2}\right|_{z=0}$$
$$ \oint \frac{\sqrt{z+5}}{z^5} dz = -\frac{2\pi i}{4!} \, \left. \frac{15}{16(5+z)^{7/2}}\right|_{z=0}$$
$$... | {
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Find the flux integral of the vector field $F=(0,0,-1)$ through the cone $z=\sqrt{x^2+y^2}$, $x^2+y^2 \leq 1$
A uniform fluid flowing vertically downward (heavy rain) is described by the vector field $F=(0,0,-1)$. Find the total flow through the cone $z=\sqrt{x^2+y^2}$, $x^2+y^2 \leq 1$.
b)Now consider $F=(-\frac{\s... | Please note your normal vector is pointing upward. As per question, You need normal vector pointing downward through the cone surface. So the signs of your final answer will change otherwise it looks fine.
From cross product, you get $(-\frac{x}{\sqrt{x^2+y^2}}, -\frac{y}{\sqrt{x^2+y^2}}, 1)$.
Please see the third coor... | {
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Prove $\sin x\ge x-\frac{x^3}{3!}$
Prove $$\sin x\ge x-\frac{x^3}{3!}$$ for $x\ge 0$.
I know that a calculus/continuity proof exists, but I am curious if this can be proved without that. Here is a sketch I have made.;
we will use for $x\ge 0$;
$$\sin x\le x\le \tan x$$
proof of this see here
I got a weaker bound :... | Well, from the celebrated triplication formula, we get $$\sin3x=3\,\sin x-4\,\sin^3x.$$ This, of course, means that $$3^{k}\sin x/3^k-3^{k+1}\sin x/3^{k+1}=-4\cdot3^{k}\,\sin^3 x/3^{k+1}.$$ Summing the telescopic series and observing $3^{k}\sin x/3^k\to x$ as $k\to\infty$, we get
$$\sin x-x= -4\,\sum^\infty_{k=0}3^{k}\... | {
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A circle of radius 1 is inscribed inside a regular octagon (a polygon with eight sides of length b). Calculate the octagon’s perimeter and its area.
A circle of radius 1 is inscribed inside a regular octagon (a polygon with eight sides of length b). Calculate the octagon’s perimeter and its area.
Hint: Split the octag... | To add onto @JoshuaWang, tan($22.5^o$)=tan$\frac{\frac π4}{2}$
Then using the half angle formula for tangent which can be found from the sine and cosine half angle formulas:
$tan^2\frac π8$= $\frac{1-cos(π/4)}{1+cos(π/4)}$=$\frac{1-\frac{1}{\sqrt 2}}{1+\frac{1}{\sqrt 2}}$= $\frac{\frac{\sqrt2}{\sqrt2}-\frac{1}{\sqrt 2}... | {
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Problem of minimum in physics.
It is necessary to go from a point $A(0,0)$ to a point $B(a,b)$ walking from A to $P(x,0)$ with a speed $v_1$ and then until the point B with a speed $v_2$.
Find where is the point in which it is necessary to abandon x axis in order to have the minimum time to complete the path.
I've cal... | You are correct that
$$ \frac{dt}{dx} = \frac{1}{v_1} - \frac{a-x}{v_2 \sqrt{(a-x)^2+b^2}} \tag1$$
and that $\frac{dt}{dx} = 0$ implies
$$ v_2 \sqrt{(a-x)^2+b^2} = v_1 (a-x), \tag2$$
which (I think--I did not check your arithmetic here) implies
$$ x^2(v_2^2-v_1^2)+x(-2av_2^2+2av_1^2)+(a^2v_2^2+b^2v_2^2-v_1^2a^2) = 0. \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3908495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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How many 5-digit numbers such that the sum of their digits is a multiple of 4? My textbook has the same problem, except for 5-digit numbers the sum of whose digits is a multiple of 5. The general approach is:
Select the first 4 digits. Number of possible ways of doing this = $9\times 10 \times 10 \times 10$ since the t... | Denote by $a_i(n)$ $\>(0\leq i\leq3)$ the number of $n$-digit decimal strings having a digit sum leaving remainder $i$ modulo $4$, and collect the $a_i(n)$ into $a(n):=\bigl(a_0(n), a_1(n), a_2(n), a_3(n)\bigr)$, the latter considered as a column vector. We then have $$a(1)=(2,3,2,2)\ .$$ For the latter digits the mult... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3909479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Show that $(\sqrt{3}+1)^{2n}+(\sqrt{3}-1)^{2n}$ and $(\sqrt{3}+1)^{2n+1}+(1-\sqrt{3})^{2n+1}$.........? Show that $(\sqrt{3}+1)^{2n}+(\sqrt{3}-1)^{2n}$ and $(\sqrt{3}+1)^{2n+1}+(1-\sqrt{3})^{2n+1}$ are both divisible by $2^{n+1}$. Is this the highest power of $2$ dividing either of the numbers?
I am not well versed wit... | From binomial expansion it follows that
$$[(2+\sqrt{3})^n+(2-\sqrt{3})^n]$$ $$=[2^n+{n \choose 2} 2^{n-2} 3+{n \choose 4}2^{n-4} 3^2+...]=2K, K \in N$$
And $$[(2+\sqrt{3})^n-(2-\sqrt{3})^n]$$ $$=2\sqrt{3}[2^{n-1}+{n \choose 1} 2^{n-1}+{n \choose 3}2^{n-3} 3+...]=2\sqrt{3}L, L \in N$$
$$S_1=(1+\sqrt{3})^{2n}+(1-\sqrt{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3910795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving that, if $f(k)=\prod_{i=1}^ka_i+\sum_{b=1}^{k-1}(1-a_{k-b})\prod_{i=1}^ba_{k-b+i}$, then $f(k+1)=f(k)\cdot a_{k+1}+(1 - a_k)a_{k+1}$
Given that $$f(k) = \prod_{i=1}^k a_i + \sum_{b=1}^{k-1} (1-a_{k-b}) \prod_{i=1}^b a_{k-b+i}$$ for all $k$ where $(a_1, a_2, a_3, \ldots )$ are random constants, prove that: $$f(... | We can considerably simplify $f(k)$ which makes the proof easy.
We obtain
\begin{align*}
\color{blue}{f(k)}&=\prod_{i=1}^k a_i+\sum_{b=1}^{k-1}\left(1-a_{k-b}\right)\prod_{i=1}^b a_{k-b+i}\\
&=\prod_{i=1}^k a_i+\sum_{b=1}^{k-1}\left(1-a_{b}\right)\prod_{i=1}^{k-b} a_{b+i}\tag{1}\\
&=\prod_{i=1}^k a_i+\sum_{b=1}^{k-1}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3915158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Equation of Circle touching a straight line and passing through the centroid of a triangle and a particular point If the equation $3{x^2}y + 2x{y^2} - 18xy = 0$ represent the sides of a triangle whose centroid is G, then the equation of the circle which touches the line $x-y+1=0$ and passing through G and $(1,1)$ is | ${x^2} + {y^2} - x - y = 0$
B: ${x^2} + {y^2} - 9x - y + 8 = 0$
C: ${x^2} + {y^2} - 4x + 2y = 0$
D: ${x^2} + {y^2} + 9x + y + 8 = 0$
I got the answer but had the choice not been provided how would I find the solution.
My steps are elaborated below
$3{x^2}y + 2x{y^2} - 18xy = 0 \Rightarrow xy\left( {3x + 2y - 18} \r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3917805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove inverse trigonometric equation $2\tan^{-1}2=\pi-\cos^{-1}\frac{3}{5}$ The question is:
Prove that: $2tan^{-1}2=\pi-cos^{-1}\frac{3}{5}$ Hint: use the fact that $tan(\pi-x)=-tanx$
I did the question without using the Hint, but I don't know how to do it using the hint.
Quick working out of what I've done:
\begin{... | Hint:
Using Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$
$$2\tan^{-1}2=\pi+\tan^{-1}\dfrac{2\cdot2}{1-2^2}=\pi+\tan^{-1}\left(?\right)$$
Now using Principal values $\tan^{-1}(-y)=-\tan^{-1}y$
Again, if $\tan^{-1}\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3918968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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What is the relationship between $\frac{1}{3}x_1+\frac{1}{3}x_2+\frac{1}{3}x_3$ and $x_1^{\frac{1}{3}}x_2^{\frac{1}{3}}x_3^{\frac{1}{3}}$? Is $\frac{1}{3}x_1+\frac{1}{3}x_2+\frac{1}{3}x_3$ > or < or = $x_1^{\frac{1}{3}}x_2^{\frac{1}{3}}x_3^{\frac{1}{3}}$, given $x_1$, $x_2$ and $x_3$ are all positive?
I know there exis... | As you said, $x_1$,$x_2$,$x_3$ are positive reals. Furthermore, by AM-GM inequality $\frac{x_1 + x_2 + x_3}{3} \geq (x_1)^\frac{1}{3}(x_2)^\frac{1}{3}(x_3)^\frac{1}{3}$
so the proof is using just AM-GM inequality.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find scale, rotation, and translation that maps three points to three points in 3D Given three source points $\mathbf{P}_1 = [x_1\ y_1\ z_1]^T,$
$\mathbf{P}_2 = [x_2\ y_2\ z_2]^T,$ $\mathbf{P}_3 = [x_3\ y_3\ z_3]^T,$ and
three destination points $\mathbf{P}'_1 = [x'_1\ y'_1\ z'_1]^T,$
$\mathbf{P}'_2 = [x'_2\ y'_2\ z'_2... | The assumption is that this can be accomplished with a similarity transformation.
We use the following sequence of transformations to compose the desired transformation:
*
*Translate $\mathbf{P}_0$ to the origin;
*Orient $\triangle \mathbf{P}_0 \mathbf{P}_1 \mathbf{P}_2$
to align with $\triangle \mathbf{P}'_0 \mathb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3922403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is this a correct way to show that $\sum_{n \geq 0} \frac{n^3}{n!}=5e$ Is this a correct way to show that $\sum_{n \geq 0} \frac{n^3}{n!}=5e$ ?
$$S_3 = \sum_{n \geq 0} \frac{n^3}{n!}=\sum_{n \geq 1} \frac{n^2}{(n-1)!} \implies$$
$$S'_3=S_3-e=\sum_{n \geq 1} \frac{n^2-1^2}{(n-1)!}=\sum_{n \geq 2} \frac{n+1}{(n-2)!} \imp... | Seems good to me. An alternate way - which is slightly neater - would be to first write $n^3$ in a convenient way:
\begin{align} n^3 &= n^2(n-1) + n^2 \\
&= n(n-1)(n-2) + 2n(n-1) + n^2 \\
&= n(n-1)(n-2) + 2n(n-1) + n(n-1) + n.\end{align}
Then, since all the sums involved converge, \begin{align} \sum_{n \ge 0} \frac{n^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3925523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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On the convergence of improper integrals Question
Determine the value of $r$ for which the integral $$\int^{\infty}_1 (\frac r {x + 1} - \frac {3x} {2x^2 + r})\ \mathrm {d}x$$ is convergent and evaluate the integral for that value of $r$.
My working
It is trivial to see that the integral is divergent for $r = 0\ $, so ... | I have finally managed to answer the question :)
It is trivial to see that the integral is divergent for $r = 0\ $, so let $r \neq 0$.
If $r \neq 0$, note that we can always choose $a$ large enough such that $$\int^{\infty}_1 (\frac r {x + 1} - \frac {3x} {2x^2 + r})\ \mathrm {d}x = \int^a_1 (\frac r {x + 1} - \frac {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3925912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
To derive quantity $k=\frac{2-p^2}{p+2}$ for positive rational $p$ such that $p^2>2$ and $(p+k)^2>2$ I am reading Baby Rudin and stuck at one small point.
$A$ is set of all positive rationals $p$ such that $p^2<2$ and $B$ is set of all positive rationals $p$ such that $p^2>2$. We wanted to show $A$ contains no largest ... | We get $$(p+k)^2 >2\implies p+k\gt\sqrt 2\implies k\gt \sqrt 2-p$$
Multiplying $\sqrt 2-p$ by $\dfrac{\sqrt 2+p}{\sqrt 2+p}\ (=1)$ gives $\dfrac{2-p^2}{\sqrt 2+p}$, and so we have
$$k\gt \dfrac{2-p^2}{\sqrt 2+p}$$
It follows from $2\gt \sqrt 2$ and $2-p^2\lt 0$ that $\dfrac{2-p^2}{2+p}\gt\dfrac{2-p^2}{\sqrt 2+p}$.
Ther... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3926373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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How to prove the inequality $abc(a+b+c)^2≤(a^3+b^3+c^3)(ab+bc+ca)$? I need to prove something like that:
For $a,b,c>0$ prove: $abc(a+b+c)^2≤(a^3+b^3+c^3)(ab+bc+ca)$.
I know that $3abc≤(a^3+b^3+c^3)$, but then I derived $3(ab+bc+ca) ≤ (a+b+c)^2$, I can't move on.
Can anyone help me?
| Hint: Using CS inequality,
$$(a^3+b^3+c^3)\left(\frac1a+\frac1b+\frac1c\right)\geqslant (a+b+c)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3926463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why $\int{\frac{dr}{r\sqrt{c^{2}r^4 -1}}} = -\frac{1}{2} \arcsin\left(\frac{1}{cr^2}\right)$? I must solve the following integral, where $c$ is a constant $$\int{\frac{dr}{r\sqrt{c^{2}r^4 -1}}}$$ When I compute and do the calculations, I obtain that $$\int{\frac{dr}{r\sqrt{c^{2}r^4 -1}}} = \frac{1}{2} \arctan \left(\sq... | In fact
$$ \tan\bigg(\cos^{-1}\bigg(\frac{1}{cr^2}\bigg)\bigg)=\frac{\sin\bigg(\cos^{-1}\bigg(\frac{1}{cr^2}\bigg)\bigg)}{\cos\bigg(\cos^{-1}\bigg(\frac{1}{cr^2}\bigg)\bigg)}=\frac{\sqrt{1-\frac1{c^2r^4}}}{\frac{1}{cr^2}}=\sqrt{c^2r^4-1}. $$
Since $\frac{1}{cr^2}>0$, $\cos^{-1}\bigg(\frac{1}{cr^2}\bigg)\in(0,\pi/2)$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3928172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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What is a simple formula that can create the patterns $+1 +2 +3 +4$, or $+1 +3 +5 +7$, or $+1 +4 + 7 +10$, etc. Originally I was looking for some simpler more intuitive way to appreciate the value of squaring something. I looked for how much it was increasing by every time I add $+1$ to the size of what was being squar... | A polynomial of degree $n$ will have the $n$th finite difference equal to $n!$ times the leading coefficient of the polynomial. In your example, you have
\begin{array}{|c|c|c|c|}
\hline
x & x^2 & \text{1st difference} & \text{2nd difference} \\
\hline
1 & 1 & & \\
2 & 4 & 3 & \\
3 & 9 & 5 & 2 \\
4 & 16 & 7 & 2 \\
5 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3934864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Number Theory : Solve the system of congruence $28x+17y\equiv 18 \pmod{41}$ and $31x+11y\equiv 35\pmod{41}$ Number Theory : Solve the system of congruence
(1) $28x+17y\equiv 18 \pmod{41}$
(2) $31x+11y\equiv 35\pmod{41}$
Attempt :
we know that equation (1) and (2) are in the same$\pmod{41}$. so we can use Modular a... | Alternatively, use linear algebra. The system is
$$
\begin{pmatrix} 28 & 17 \\ 31 & 11 \end{pmatrix}
\begin{pmatrix} x \\ y\end{pmatrix}
=
\begin{pmatrix} 18 \\ 35 \end{pmatrix}
$$
The inverse of the matrix is
$$
-\frac{1}{219}\begin{pmatrix} \hphantom-11 & -17 \\ -31 & \hphantom-28 \end{pmatrix}
$$
It remains to f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3935020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 1
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Simplifying $\log_x 25 - \left(\frac{\log_5 x}{4} - 2\log_5 x\right)$ My teacher gave me the following expression to change into a single expression:
$$\log_x 25 - \left(\frac{\log_5 x}{4} - 2\log_5 x\right)$$
This is what I got after working it out:
$$\log_x 25 - \log_5 \left(\frac{x}{4x^2}\right)$$
| $$\begin{align}
& \log_{x}25 - \left(\dfrac{\log_{5}x}{4} - 2\log_{5}x\right) \\
= & \log_{x}5^2 + \dfrac{7}{4}\log_{5}x \\
= & 2\log_{x}5 + \dfrac{7}{4}\log_{5}x \\
= & \dfrac{2}{\log_{5}x} + \dfrac{7}{4}\log_{5}x \\
= & \dfrac{7\log_{5}x + 8}{4\log_{5}x} \\
= & \dfrac{\log_{5}x^7 + \log_{5}5^8}{\log_{5}x^4} \\
= & \d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3936543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Determine the number of three-digit numbers that are divisible by 3 Determine the number of three-digit numbers that are divisible by 3 and that you can form with 3 different numbers among 1, 2, 3, 5, 6, 8
I found 5 numbers or 5 different ways with a sum that is divisible by 3. And each of them has 6 combinations becau... | $$\begin{align}2,5,8&\equiv -1\pmod 3\\3,6&\equiv 0\pmod 3\\1&\equiv 1\pmod 3\end{align}$$
There are $3\times 2\times 1=6$ ways to select one number from each group and $1$ way to select three numbers from a single group, for a total of $7$ combinations.
There are $3!=6$ permutations of each combination, giving a total... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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show that $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ if $abc \geq 1 $ We want to show that $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ if $abc \geq 1 $
I've tried to use AM-GM directly on the LHS but that obviously failed.
Simplifyi... | $$\begin{align}\frac{a}{b} + \frac{b}{c} + \frac{c}{a}
&\geq \frac{(abc)^{1/3}}{a} + \frac{(abc)^{1/3}}{b} + \frac{(abc)^{1/3}}{c}\\
&\geq \frac{1}{a} + \frac{1}{b} + \frac{1}{c}
\end{align}$$
The second inequality is obvious while the first one comes from Karamata's Majorization Inequality with the convexity of $f(x)... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integrating $\int{\frac{x\,dx}{x^2 + 3x -4}}$ I saw this "Beat the Integral" problem and wanted to be sure I was approaching it correctly.
The integral is $$\int{\frac{x}{x^2 + 3x -4}dx}$$
So I decide I want to decompose this, because it's a degree-2 polynomial in the denominator and a degree-1 in the numerator, so tha... | $$I=\int\frac{x}{x^2+3x+4}dx=\frac12\int\frac{2x+3}{x^2+3x+4}-\frac{3}{x^2+3x+4}$$
$$I=\frac12\ln|x^2+3x+4|-\frac32\int\frac{1}{(x+3/2)^2+7/4}dx$$
then you can let $u=x+3/2,\,du=dx$ so you have $u^2+(\sqrt{7}/2)^2=\frac74\left((2u/\sqrt{7})^2+1\right)$ now let $v=2u/\sqrt{7},\,dv=2/\sqrt{7}du$ now you have an integral ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}{ {{1}\over{2\pi}} e^{{{-1}\over{2}}(4x^2-2xy+3y^2)} ~\mathrm dx \mathrm dy}$ How to calculate that integral? I have no idea what substitution to use. I think polar coordinates will be needed here.
$$\displaystyle \int_{-\infty}^{\infty} \int_{-\infty}^{\infty... | You can even do it with $(x,y)$. Complete the square first
$$\int{ \frac{1}{2\pi}} e^{{{-1}\over{2}}(4x^2-2xy+3y^2)} dx=\frac{e^{-\frac{11 y^2}{8}} \text{erf}\left(\frac{4 x-y}{2 \sqrt{2}}\right)}{4
\sqrt{2 \pi }}$$
$$\int_{-\infty}^\infty{ \frac{1}{2\pi}} e^{{{-1}\over{2}}(4x^2-2xy+3y^2)} dx=\frac{e^{-\frac{11 y^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3946976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove $\frac{\ln(1+x)}{1+x}=x-\left(1+\frac{1}{2}\right)x^2+\left(1+\frac{1}{2}+\frac{1}{3}\right)x^3-\ldots$ I have to prove that :
$$\frac{\ln(1+x)}{1+x}=x-\left(1+\frac{1}{2}\right)x^2+\left(1+\frac{1}{2}+\frac{1}{3}\right)x^3-\ldots$$
I already know that the Maclaurin series of $\ln(1+x)$ is $x-\frac{1}{2}x^2+\frac... | You must multiply the series
$$\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}+O\left(x^6\right)$$
$$\frac 1{1+x}=1-x+x^2-x^3+x^4-x^5+O\left(x^6\right)$$
$$\frac{\log(1+x)}{1+x}=\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}+O\left(x^6\right) \right) \times$$ $$\left(1-x+x^2-x^3+x^4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3947763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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(Multiplicative) inverse of $\alpha = (\sqrt[3]{7})^2 + 3\sqrt[3]{7} + 1 \in\mathbb Q[\sqrt[3]{7}]$ Set $\mathbb Q[\sqrt[3]{7}] = \{F(\sqrt[3]{7}) \mid F ∈ Q[x]\}$ is a field (with the usual addition and the usual multiplication).
Calculate the (multiplicative) inverse of
$$\alpha = (\sqrt[3]{7})^2 + 3\sqrt[3]{7} + 1 \... | You actually have two possibilities:
*
*Either, denoting $x=\sqrt[3]7$, you try to find a linearcombination $ax^2+bx+c$ such that
$\;(ax^2+bx+)(x^2+3x+1)=1$, which leads to solving the linear system
$$\begin{cases}a+3b+c=0\\7a+b+3c=0\\21a+7b+c=1\end{cases},$$
which can be solved finding the reduced row echelon form ... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\frac{1}{2}\cdot\frac{3}{4}\cdot...\cdot\frac{2n-1}{2n}<\frac{1}{\sqrt{3n}}$ for all $n$. Prove for all $n$: $\frac{1}{2}\cdot\frac{3}{4}\cdot...\cdot\frac{2n-1}{2n}<\frac{1}{\sqrt{3n}}$.
Using induction, I tried the brain-dead method and went straight for $$\frac{2n+1}{2n+2}\cdot\frac{1}{\sqrt{3n}}<\frac{1}{\sq... | A First Approach
$$
\begin{align}
n\prod_{k=1}^n\left(\frac{2k-1}{2k}\right)^2
&=\frac14\prod_{k=2}^n\left(\frac{2k-1}{2k}\right)^2\frac{k}{k-1}\tag{1a}\\
&=\frac14\prod_{k=2}^n\frac{2k-1}{2k}\frac{2k-1}{2k-2}\tag{1b}\\
&=\frac14\prod_{k=2}^n\frac{\color{#C00}{k-1/2}}{\color{#090}{k}}\frac{\color{#75F}{k-1/2}}{\color{#... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3953464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Assuming the expression converges, determine the largest integer $n\le 9,000,000$ for which $\sqrt {n+\sqrt {n+\sqrt {n+\cdots}}}$ is rational. What I have done is that I supposed $y=\sqrt {n+\sqrt {n+\sqrt {n+\cdots}}}$.
So, $y^2-n=y$, which is $y^2-y+\frac{1}{4}=n+\frac{1}{4}$. So, $(y-\frac{1}{2})^2=n+{1\over 4}=\fr... | Let $m=2k+1$. Varying $k$ gives all possible values of $n$ for $y$ to be rational:
$$m^2=4k^2+4k+1=4(k^2+k)+1\implies n=k^2+k=k(k+1)\le9\cdot10^6$$
It is easily seen that the solution corresponds to $k=2999$, i.e. $n=8997000$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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partial fractions when the fraction cannot be decomposed I am trying to find partial fractions of $\frac {1}{(x^2+1)^2}$. All the coefficients I get are zeros except the coefficient for the constant term which is 1, leaving me with the fraction I started with, so it seems like the fraction cannot be decomposed. How can... | In the real-valued partial fraction method, your term
$$
\frac{1}{(x^2+1)^2}
$$
is the best you can do. Evaluation of integrals
$$
\int\frac{dx}{(ax^2+bx+c)^n}
$$
is done as follows: For $n=1$, complete the square and it is an arctangent. For $n>1$ there is a reduction formula from integration by parts:
\begin{align... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Finding probability of all outcomes occurring in n repetitions Suppose you have numbers 1, 2, 3, 4. You can pick each of these numbers with probability 1/4. After n iid selections, what is the probability that you have seen all 4 numbers? Express this as a function of n.
For example, if n=5, and you draw the numbers 1,... | You may just count systematically as follows:
Number of sequences of length $n$ containing only $1$ digit: $$\color{blue}{\Rightarrow 4}$$
Number of sequences of length $n$ containing exactly $2$ digits:
*
*Choose two of the digits: $\color{blue}{\binom 42}$
*Number of sequences of length $n$ containing at least one... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3960242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve the equation $\sqrt{45x^2-30x+1}=7+6x-9x^2$ Solve the equation $$\sqrt{45x^2-30x+1}=7+6x-9x^2.$$
So we have $\sqrt{45x^2-30x+1}=7+6x-9x^2\iff \begin{cases}7+6x-9x^2\ge0\\45x^2-30x+1=(7+6x-9x^2)^2\end{cases}.$ The inequality gives $x\in\left[\dfrac{1-2\sqrt{2}}{3};\dfrac{1+2\sqrt{2}}{3}\right].$ I am not sure how ... | Let $u=9x^2-6x$ and rewrite the equation as
$$\sqrt{5u+1}=7-u$$
Squaring both sides gives $5u+1=49-14u+u^2$, or
$$u^2-19u+48=(u-16)(u-3)=0$$
We see that $u=16$ is not a solution, since $\sqrt{81}\not=-9$. This leaves us with $u=3$, which is a valid solution, since $\sqrt{16}=4$, and from this we have
$$9x^2-6x=3\implie... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $a ( n ) = b( n + 2)$
let $a(n)$ denotes the number of ways of expressing the positive integer $n$ as an ordered sum of 1's and 2's. Let $b(n)$ denote the number of ways of expressing n as an ordered sum of integers greater than 1. prove that $a(n) = b(n+2)$. for $n=1,2,3...$
My approach:
$a(1) = 1$ (only... | Hint: Observe that the sequence $1, 2, 3, 5, 8, \ldots $ looks like the Fibonacci sequence, offset by 1
Hint: Show that $a_n$ is indeed the Fibonacci sequence (offset by 1), because $ a_{n+2} = a_{n+1 } + a_{n} $, with initial conditions $ a_1 = 1, a_2 = 2$.
Hint: Show that $b_n, n \geq 2$ is indeed the Fibonacci seque... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find $\lim_{x \to \infty}{\frac{(x+1)^{\frac{2}{3}}-(x-1)^{\frac{2}{3}}}{(x+2)^{\frac{2}{3}}-(x-2)^{\frac{2}{3}}}}$ As stated in the title.
My attempt. Dividing through $(x-2)^{\frac{2}{3}}$.
$$L=\lim_{x \to \infty}{\frac{(x+1)^{\frac{2}{3}}-(x-1)^{\frac{2}{3}}}{(x+2)^{\frac{2}{3}}-(x-2)^{\frac{2}{3}}}}=\lim_{x \to \in... | Here is how I would do it without using L'hopital, or series expansion :
Using the identity $ a^{3}-b^{3}=\left(a-b\right)\left(a^{2}+ab+b^{2}\right) $, we have : $$ \left(x+1\right)^{2}-\left(x-1\right)^{2}=\left(\left(x+1\right)^{\frac{2}{3}}-\left(x-1\right)^{\frac{2}{3}}\right)\left(\left(x+1\right)^{\frac{4}{3}}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3963376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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The $2$nd, the $1$st and the $3$rd term of an arithmetic progression form a geometric progression An arithmetic progression is given with a common difference $\ne0.$ The $2nd$, the $1st$ and the $3rd$ term of the ap form a geometric progression. Find the common ratio.
So we have the ap: $a_1,a_1+d,a_1+2d$ and the gp: $... | The three terms of a geometric series $a_0, a_1= a_0 + d, a_2 = a_0+2d$ or if it is easier $a_0= a_1-d, a_1, a_2= a_1 + d$.
The three terms of a geometric series is $b_0, b_1=b_0\cdot r, b_2=b_0\cdot r^2$ or if it is easier $b_0=\frac {b_1}r, b_1, b_2= b_1\cdot r$.
We are told that $a_1, a_1 -d, a_1 +d$ form a geometri... | {
"language": "en",
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"source": "stackexchange",
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How to do integral $\int\frac1{x^3+x+1}dx$? I've been stuck by this integral. I only knew that roots of the cubic equation are complex.
Please help me with this. $$\int\frac{dx}{x^3+x+1}$$
| Since $x^3+x+1=(x-a)((x-b)^2+c^2)$ with$$\begin{align}a&:=\sqrt[3]{\frac{\sqrt{93}-9}{18}}-\sqrt[3]{\frac{2/3}{\sqrt{93}-9}},\\b&:=\frac{1}{\sqrt[3]{12(\sqrt{93}-9)}}-\sqrt[3]{\frac{\sqrt{93}-9}{144}},\\c&:=\sqrt{3}\left(\frac{1}{\sqrt[3]{12(\sqrt{93}-9)}}+\sqrt[3]{\frac{\sqrt{93}-9}{144}}\right),\end{align}$$the integ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3967108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given $f(x) = x^2 + \ln (1 + \frac{1}{x})$, prove that $\forall x > 0, x \in \mathbb{R}: f(x) \geq \frac{1 + 2\ln(2)}{4}$ Given $f(x) = x^2 + \ln (1 + \frac{1}{x})$, prove that $\forall x\in(0,+\infty): f(x) \geq \frac{1 + 2\ln(2)}4$.
That would be same as proving that $x^2 + \ln (\frac{x + 1}{x}) - \frac{1 + \ln(4)}{4... | You are almost there.
Since $f(0) = f(+\infty) = +\infty$ and there is a unique solution $x_0 > 0$ such that $f'(x_0)=0$ (because $2x^2(1+x)$ is an increasing function of $x$) we only need to show that $f(x_0) > \frac{1+2 \ln 2}{4}$.
Since $x_0 \in \left(\frac 12, 1\right)$, then
$$f(x_0) = x_0^2 + \ln \left( 1+\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3968778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given that $a = \sqrt[3]4 + \sqrt[3]2 + 1$, find $\frac{3}{a} + \frac{3}{a^2} + \frac{1}{a^3}$.
Given that $a = \sqrt[3]4 + \sqrt[3]2 + 1$, find $\frac{3}{a} + \frac{3}{a^2} + \frac{1}{a^3}$.
What I Tried: I only figured out that:- $$\rightarrow a = 2^\frac{2}{3} + 2^\frac{1}{3} + 2^\frac{0}{3}$$ Yet this does not he... | Hint: As $$a=\frac{(\sqrt[3]{4}+\sqrt[3]{2}+1)(\sqrt[3]{2}-1)}{\sqrt[3]{2}-1}=\frac{1}{\sqrt[3]{2}-1} $$ $${\left(\frac{1}{a}+1\right)}^3=2$$ can you proceed ....
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplify, $\sqrt[3]\frac{(\sqrt{a-1} - \sqrt{a})^5}{(\sqrt{a-1} + \sqrt{a})} + \sqrt[3]\frac{(\sqrt{a-1} + \sqrt{a})^5}{(\sqrt{a} - \sqrt{a-1})}$.
Simplify, $\sqrt[3]\frac{(\sqrt{a-1} - \sqrt{a})^5}{(\sqrt{a-1} + \sqrt{a})} + \sqrt[3]\frac{(\sqrt{a-1}
+ \sqrt{a})^5}{(\sqrt{a} - \sqrt{a-1})}$.
What I Tried: I thought... | \begin{eqnarray*}
\sqrt[3]\frac{(\sqrt{a-1} - \sqrt{a})^5}{(\sqrt{a-1} + \sqrt{a})} + \sqrt[3]\frac{(\sqrt{a-1}
+ \sqrt{a})^5}{(\sqrt{a} - \sqrt{a-1})} = \frac{-(\sqrt{a-1} -\sqrt{a})^2 + (\sqrt{a-1} -\sqrt{a})^2}{\sqrt[3]{(\sqrt{a} - \sqrt{a-1})(\sqrt{a} + \sqrt{a-1})}} \\
= \frac{-(a-1+a-2 \sqrt{a(a-1)})+ a-1+a+2 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3970199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Consider the function $f(x) = \frac{1}{3^x + \sqrt{3}}.$ Find :- $\sqrt{3}[f(-5) + f(-4) + ... + f(3) + f(4) + f(5) + f(6)]$.
Consider the function $f(x) = \frac{1}{3^x + \sqrt{3}}.$ Find :-
$\sqrt{3}[f(-5) + f(-4) + ... + f(3) + f(4) + f(5) + f(6)]$.
What I Tried: I checked similar questions and answers in the Art o... | Let $\sqrt3=a$
$$\dfrac{f(x)}a=\dfrac1{a^{2x-1}+1}$$
If $\dfrac{a^{2x-1}}{1+a^{2x-1}}=f(px+q)=\dfrac1{1+a^{2(px+q)-1}}$
$\implies 1-2x=2(px+q)-1$
$\implies p=-1,q=1$
$$\implies f(x)+f(1-x)=a$$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3970350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Explanation for $\int_{\frac{\pi}{2}+(j-1)\pi}^{\frac{\pi}{2}+j\pi} \frac{|\cos(t)|}{\frac{\pi}{2}+j\pi} \,dt = \frac{2}{\frac{\pi}{2}+j\pi} $ I don't see how / why one can rewrite the integral as following:
$$\int_{\frac{\pi}{2}+(j-1)\pi}^{\frac{\pi}{2}+j\pi} \frac{|\cos(t)|}{\frac{\pi}{2}+j\pi} \,dt = \frac{2}{\fra... | We can write
\begin{align*}
\int_{\frac{\pi}{2}+(j-1)\pi}^{\frac{\pi}{2}+j\pi} \frac{|\cos(t)|}{\frac{\pi}{2}+j\pi} \ dt & = \frac{1}{\frac{\pi}{2}+j\pi} \int_{\frac{-\pi}{2}+j\pi}^{\frac{\pi}{2}+j\pi}|\cos(t)| \ dt\\
&\overset{\color{blue}{u=t-j\pi} }{=} \frac{1}{\frac{\pi}{2}+j\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3972672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Evaluate $\int_{2}^{7} \frac{x}{1-\sqrt{2+x}} d x$ We have the following integral:
$$
\int_{2}^{7} \frac{x}{1-\sqrt{2+x}}\, dx
$$
And this is my solution, which seems to be wrong, and I am failing to see where exactly I failed at:
We have $u=1-\sqrt{2+x}, x=u^2-2u-1, dx=-2\sqrt{2+x}\, du$, and we know that $x\geq -2$ a... | The first thing I would like to point out is that since you want to do a substitution, it is really extra (unnecessary) work writing $dx$ in terms of $x$ (where the aim of substitution is to "get rid" of the initial variable and simplify the integral in the process).
With that out of the way, let $u = 1 - \sqrt {2 + x}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Let $a,b,c\in\mathbb{Z}$, $1Let $a,b,c\in\mathbb{Z}$, $1<a<10$, $c$ is a prime number and $f(x)=ax^2+bx+c$. If $f(f(1))=f(f(2))=f(f(3))$, find $f'(f(1))+f(f'(2))+f'(f(3))$
My attempt:
\begin{align*}
f'(x)&=2ax+b\\
(f(f(x)))'&=f'(f(x))f'(x)\\
f'(f(x))&=\frac{(f(f(x)))'}{f'(x)}\\
\end{align*}
| Okay here's the outline of how you solve this.
first, you write down the system of equations
$$
0=f(f(2))-f(f(1))=(3a+b)*(b+5a^2+3ab+2ac) \\
0=f(f(3))-f(f(2))=(5a+b)*(b+13a^2+5ab+2ac)
$$
and this leaves you with four possibilities:
(a) $(3a+b=0)$ and $(5a+b=0)$ which is impossible since $a \ne 0$
(b) $(3a+b=0)$ and $(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3975895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to solve this equation $(x-1)(x+2) + 4(x-1)\sqrt{\dfrac{x+2}{x-1}} = 12$? $(x-1)(x+2) + 4(x-1)\sqrt{\dfrac{x+2}{x-1}} = 12$
$Domain: x\in (-\infty;-2]\cup(1;+\infty)$
$x = 2\longrightarrow$ Done
How to prove $x = \dfrac{-1-3\sqrt{17}}{2}$ is the last solution? With raise both side by power of two?
Any better way su... | Assuming $x\neq 1$ you can rewrite your equation as
$$(x-1)(x+2) + 4\sqrt{(x-1)(x+2)} = 12$$
Then call $z = \sqrt{(x+1)(x-2)}$ and solve the quadratic associated equation, and then you pullback to $x$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find n where $(2n+1)2^{4n+5} = 3 \pmod{7}$ For $n$ normal number, the book solved it like this:
If $n$ can be divided by $3$ (which is $n = 3k$) then $n = 21L + 9$.
If $n$ can't be divided by 3(Which is either $n = 3k +1$ or $n = 3k + 2$) then $n = 21L + 1$ or $n = 21L + 2$ .
But I didn't solve it like this.
My logic i... | This is somewhere between a comment and an answer.
The simplest way to solve this is that the value of $(2n+1)2^{4n+5}$ is precisely a function of $n \mod 42 = 6 \times 7$. More precisely, $2n+1 \mod 7$ is precisely a function of $n \mod 7$ while $2^{4n+5} \mod 7$ is precisely a function of $n \mod 6$. For each $i \in ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3984075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Maximum value of modulus in exponential form I am trying to find the maximum value of
$\left|e^{i\theta}-2\right|+\left|e^{i\theta}+2\right|\mbox{ for }0\le\theta\le2\pi$.
I can replace $e^{i\theta}$ with $a+ib$ and then proceed to get a function of $a$;
Namely;
$$\sqrt{5+4a}+\sqrt{5-4a}$$
I can then use calculus to fi... | Let $f(\theta) = \left|e^{i\theta}-2\right|+\left|e^{i\theta}+2\right|$. Note that $f(\theta)\ge0$. We have $$f^2(\theta) = |e^{i\theta}-2|^2 + |e^{i\theta}+2|^2 + 2|e^{2i\theta}-4| = (e^{i\theta}- 2)(e^{-i\theta}- 2) + (e^{i\theta}+2)(e^{-i\theta}+2) + 2|e^{2i\theta}-4| = 1 -4\cos(\theta) +4 +1+4\cos(\theta)+4 +2|e^{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3985577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Number of integers at most $x$ with exactly two distinct prime factors I wish to find an asymptotic for the number of integers not exceeding $x$ with exactly two distinct prime factors. Here is a starting point:
Throughout $p$ and $q$ are primes. We are interested in $\frac{1}{2}\#\{(p,q,\alpha,\beta): p^{\alpha}q^{\be... | The asymptotic is
$$\frac{x\log \log x}{\log x}\,.$$
More generally, a result of Landau (1900) is that
$$\pi_k(x) \sim \rho_k(x) \sim \sigma_k(x) \sim \frac{x(\log \log x)^{k-1}}{(k-1)!\log x}\,,$$
where
\begin{align}
\pi_k(x) &= \# \{ n \leqslant x : \omega(n) = \Omega(n) = k\}\,,\\
\rho_k(x) &= \#\{ n \leqslant x : \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3987005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Discussion on the prolongation of the solution I have the following
$$(PC)\begin{cases}y'=\dfrac{1}{x+1}y-\dfrac{4x}{x+1}y^2\\y(3)=\dfrac{2}{b} \end{cases}\text{ with }b\ne 0$$
Determine for which values of $b$, it exists at least one solution defined in $I=[0,4]$.
$\textbf{My attempt:}$ I noticed that the function $... | You found correctly that
$$
(1+x)z(x)=2(x^2+C)\implies 4·\frac{b}{2}=2·(9+C)
$$
The solution $y$ becomes singular where $z=\frac1y$ has a root. As $(x+1)$ has no root on $[0,4]$, it remains to consider the other side. If $C>0$ then there is no problem. For negative $C$ to avoid roots on the interval one needs $C<-16$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3987698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to determine h(x) in a polynomial partial fraction decompostion Im susposed to do a partial fractional division;
$$ \frac{-2x^2 + 8x - 9} {(x-1) (x-3)^2}$$ Now I used the formula and this is what I got;
$$\frac {A}{(x-1)} + \frac{B}{x-3} + \frac{C}{(x-3)^2}$$
Now I've put the 3 fractions together,and got this
$$ \f... | You made a mistake when you put three fractions together:
$$
\frac{A(x-3)^2+B(x-1)(x-3)+C(x-1)}{(x-1)(x-3)^\color{red}{2}}.
$$
Now you set
$$
A(x-3)^2+B(x-1)(x-3)+C(x-1)=-2x^2+8x-9
$$
to find $A,B,C$.
*
*Set $x=1$: $4A=-2+8-9$.
*Set $x=3$: $2C=-18+24-9$.
*Set $x=2$: $A-B+C=-8+16-9$.
Now it should be very straightf... | {
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"url": "https://math.stackexchange.com/questions/3990210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Euler's proof of Fermat's Last Theorem for cubes I'm working through Richard Freidburg's An Adventurer's Guide to Number Theory, and on p. 149, chapter 7, he presents Euler's proof of Fermat's Last Theorem for cubes.
He starts with $z^3 = x^3 + y^3$, where $x, y, z$ have no factors in common. This implies that $z$ is e... | Well, if $x^3$ and $y^3$ are congruent to $0$ mod $9$, then so is $z^3$, which contradicts "no common factors". If they are both $1$ (resp $-1$) then $z^3$ is $2$ (resp $-2$) which contradicts that "cube should be congruent to $0,1,-1$".
| {
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Using quadratic residue show that $2b^2+ 3$ cannot be a divisor of $a^2-2$
I have tried working with residue of $2$ modulo $p$ where $p$ is a divisor of $2b^2+3$ but couldn't stablish the result.
| You have the right idea. Consider there are integers $a$ and $b$ with $2b^2 + 3 \mid a^2 - 2$. With $2b^2 + 3$, it's always odd and $\gt 1$, so it has at least one prime factor and all of these prime factors are odd. Let $p$ be any of these prime factors. Since $p \mid 2b^2 + 3$, then $p \mid a^2 - 2$. Thus, as you sta... | {
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"answer_id": 0
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Finding the intervals of increase and decrease and concavity of $y=\frac{\sqrt{1+|x-2|}}{1+|x|}$ I have the function $f(x)=\frac{\sqrt{1+|x-2|}}{1+|x|}$, which I need to analyze. I have already proven that $\lim\limits_{x \to +-\infty}{f(x)}=0$, therefore the function has only a horizontal asymptote at $y=0$. Then I lo... | $$f(x)=\left\{
\begin{array}{ll}
\frac{\sqrt{3-x}}{1-x} & x<0 \\
\frac{\sqrt{3-x}}{x+1} & 0 \le x<2 \\
\frac{\sqrt{x-1}}{x+1} & x\geq 2 \\
\end{array}\right.
$$
$$
f'(x)=
\left \{
\begin{array}{ll}
\frac{5-x}{2 \sqrt{3-x} (x-1)^2} & x<0 \\
\frac{x-7}{2 \sqrt{3-x} (x+1)^2} & 0<x<2 \\
\frac{3-x}{2 \sqrt{x-1} (x+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3995882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Show that there is only one real root of the equation $1 + a^2 + ax - x^3$ I have to show that for every $a>0$ the equation $1 + a^2 + ax - x^3 = 0$, has exactly one solution. I have made the graph of the function $f(x) = 1 + a^2 + ax - x^3$ on desmos and I can clearly see that. But, how can I prove it explicity?
Thank... | Oh... for shucks and giggles.
All cubic have at least one real solution. So let $x=r$ be one real root.
Then $f(x)=(1+a^2) + ax -x^3 = (x-r)(Ax^2 + Bx + C) = Ax^3 + (B-Ar)x^2 + (C-Br)x -Cr$ for some
$A=-1; B+r=0; C-Br= a; -Cr=1+a^2$
So $B = -r; C = -\frac {1+a^2}r$ (assuming $r \ne 0$ but as $f(0) = 1+a^2 \ne 0$ we k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3996005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
How can I prove $\sqrt{3}+\sqrt{4}+...+\sqrt{n}<\frac{1}{4}n^2$ for all whole numbers $n \geq 3$? $\sqrt{3}+\sqrt{4}+...+\sqrt{n}<\frac{1}{4}n^2 ; n \geq 3$
Induction first step:
$n=3$
$\sqrt{3}<\frac{9}{4}$ which is true
Induction second step:
$n=k$
$\sqrt{3}+\sqrt{4}+...+\sqrt{k}<\frac{1}{4}k^2$
Induction third step:... | You have a statement $S(n)$. You attempt to show $S(3)$, and that $S(n)$ implies $S(n+1)$ for all $n \geq 3$. Unfortunately this works only for $n \geq 4$.
You solve the conundrum by showing both $S(3)$ and $S(4)$, and since the induction statement is true for $n \geq 4$, $S(n)$ is also true for all $n \geq 5$.
It happ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4001045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
A triangle $ABC$ with $c=60,l_c=10$ and $P=125$ A triangle $ABC$ is given with $c=60,l_c=10$ and $P=125$, where $AB=c$, $l_c$ is the angle bisector of $\measuredangle C$ and $P$ is the perimeter. Find $a$ and $b$.
Here are my thoughts:
We have $\begin{cases}\dfrac{m}{n}=\dfrac{b}{a}\\l_c^2=ab-mn\end{cases}.$ From the ... | Length of angle bisector is given by
$\displaystyle l_c^2 = \frac{ab}{(a+b)^2} ((a+b)^2 - c^2)$
$a+b = 125-60 = 65, c = 60$
Leads to $\displaystyle 10^2 = \frac{ab}{65^2}(65^2-60^2)$
i.e., $ab = 676 \implies a(65-a) = 676$
We get a quadratic $a^2 - 65a + 676 = 0 \implies a = 13, 52$
So $a = 13, b = 52$ or $a = 52, b = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4003400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Calculate $\displaystyle \lim_{x \to 3} \frac{\sqrt{19-x} - 2\sqrt[4]{13+x}}{\sqrt[3]{11-x} - x + 1}$ Calculate:
$$\displaystyle \lim_{x \to 3} \frac{\sqrt{19-x} - 2\sqrt[4]{13+x}}{\sqrt[3]{11-x} - x + 1}$$
The problem with that case is that the roots are in different powers so multiplication in nominator and denomin... | Here is a simple strategy. Find the limits of those radicals and subtract that limit from them.
Thus for numerator the term $\sqrt{19-x}\to 4$ and hence subtract and add $4$ to get $$\sqrt{16-x}-4+4$$ The other term in numerator is $2\sqrt[4]{13+x}$ which tends to $4$ and replace it with $$2\sqrt[4]{13+x}-4+4$$ and see... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4004897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Prove that $\left \lfloor \frac{1+\lfloor na+1/a\rfloor}{a} \right \rfloor=n$ If $a \geq \frac{1+\sqrt{5}}{2}$ and $n \in \Bbb W$, prove that
$$\left \lfloor \frac{1+\left\lfloor \frac{1+na^2}{a}\right\rfloor}{a} \right \rfloor=n.$$
I could prove only when $a$ is an integer, that is $a \geq 2$. If $a \in \Bbb Z$ we hav... | We need to prove
*
*$$ \frac{1+\lfloor \frac{1+na^2}{a}\rfloor}{a} \geqslant n $$
*$$ \frac{1+\lfloor \frac{1+na^2}{a}\rfloor}{a} < n+1 $$
*
*$$ 1+\left\lfloor \frac{1+na^2}{a}\right\rfloor > \frac{1+na^2}{a} > an$$
so dividing by $a$ we obtain what we wanted to.
*$$ \frac{1+\lfloor \frac{1+na^2}{a}\rfloor}{a}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4006683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove that the algebraic expression is greater than zero Prove that $x^8 - x^5 + x^2 -x + 1 >0$ for $x \in \mathbb{R}$.
It's from the (junior) high-school competition and the idea is that not everyone there knows calculus, so that I'm looking for more "basic" justification.
My idea was to use AM-GM: $x^8 + x^2 \geq 2\s... | Let's complete the alternate sign polynomial with $\begin{cases}a=x^{8}-x^{5}+x^{2}-x+1\\b=-x^{7}+x^{6}+x^{4}-x^{3}\end{cases}$
When $x\le 0$ then $(x^5+x)\le 0$ and since $(x^8+x^2+1)\ge 1$ we conclude easily that $a>0$.
So let's focus on the case $x>0$.
Notice that $a+b=\dfrac{x^9+1}{x+1}>0$
But since $b=(1-x)(x^6-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4007910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 5
} |
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.