Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Basic Number Theory Question involving quadratic equations and squares Question : Let $m,n\in \mathbb{N}$ such that $2m^2 + m = 2n^2 + n$, then prove that $m - n$ and $2m + 2n + 1$
are perfect squares.
$\begin{align}2m^2-2n^2 &=n-m\\
-2(n-m)(m+n) &= n-m \\\end{align}$
If $m-n \neq 0
$\begin{align}\Rightarrow 2(m+n) &... | $$ 2m^2 + m = 2 n^2 + n $$
$$ 2m^2 - 2n^2 + m - n = 0 $$
$$ 2(m^2-n^2) +(m-n) = 0 $$
$$ 2(m+n)(m-n) + (m-n) = 0 $$
$$ ( 2m+2n )(m-n) + 1 (m-n) = 0 $$
$$ (2m+2n+1)(m-n) = 0 $$
With integers, the left factor is odd so nonzero, thus $m=n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3706998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving $|a+1|+|2a+5|=5$ How to solve this equation?
$$|a+1|+|2a+5|=5$$
Here is my solution but it seems that my answers are wrong.
$$|a+1|+|2a+5|=5$$
Case 1:
$$a+1+2a+5=5$$
$$a = \frac{1}{3}$$
Case 2:
$$a+1-2a-5=5$$
$$a=-11$$
| Squaring both sides gives
$$(a+1)^2+|(a+1)(2a+5)|+(2a+5)^2=5^2,$$
or
$$|(a+1)(2a+5)|=25-(a^2+2a+1)-(4a^2+20a+25),$$
which gives
$$|(a+1)(2a+5)|=-5a^2-22a-1.$$
Squaring again, we obtain
$$(a+1)^2(2a+5)^2=(5a^2+22a+1)^2,$$
or
$$(2a^2+7a+5)^2=(5a^2+22a+1)^2,$$
or
$$(2a^2+7a+5)^2-(5a^2+22a+1)^2=0.$$
Factorising then gives
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3709128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Binomial Expansion $(1+kx)^4 (1+x)^n$
Question
In the expansion of $(1+kx)^4 (1+x)^n$, the coefficient of $x$ is $13$ and the coefficient of $x^2$ is $74$.
Find the possible values of $k$ and $n$.
My solution
$$
\begin{split}
(1+kx)^4 &= 1+4kx+6k^2 x^2+4k^3 x^3+k^4 x^4\\
(1+x)^n &= 1+ \binom{n}{1}x+\binom{n}{2} x^... | The second equation gives
$$n(n-1) + 8nk + 12k^2 = 148$$
(by multiplying the equation by $2$ and noting $\binom n2 = \frac{n(n-1)}2.$
We have $n+4k=13$. Write $n = 13-4k$, and substitute this into the above equation, to obtain
$$(13-4k)(12-4k) + 8(13-4k)k + 12k^2 = 148,$$
which is a quadratic in $k$. You can now solve ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3710413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate: $\int \frac{dx}{(x^2+x+1)^2},$ without using any kind of substitutions. I know there exists a reduction formula for the integral: $$\int\frac1{(ax^2+bx+c)^n}\,dx.$$ But this uses substitutions. So for the integral in question, let:
\begin{align}
I&=\int\frac{dx}{(x^2+x+1)^2}\\\\
&=\int\frac{dx}{\{(x-\omega)(x... | If your condition is integration without substitutions, using partial fractions is a proper approach but not easy in this case because computing the arbitrary constants $A, B, C, D$ is a bit laborious.
$$\frac{1}{(x-\omega)^2(x-\omega^2)^2}=\frac{A}{(x-\omega)}+\frac{B}{(x-\omega)^2}+\frac{C}{(x-\omega^2)}+\frac{D}{(x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3710992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Sum of Squares for $a^2+b^2+c^2+d^2+abcd+1\ge ab+bc+cd+da + ac+bd$ $\color{red}{\textrm{Update}}$
To clarify: I hope to see a (simple) SOS solution similar to the one of the following 3-variable problem, no other assumptions e.g. $a\ge b\ge c\ge d$.
Let us see an example. In the link below, by letting $y = \mathrm{mid}... | Since $$1+abcd\geq2\sqrt{abcd}$$ and after replacing $a$ at $a^2$, $b$ at $b^2$, $c$ at $c^2$ and $d$ at $d^2$, we need to prove that
$$a^4+b^4+c^4+d^4+2abcd\geq a^2b^2+a^2c^2+a^2d^2+b^2c^2+b^2d^2+c^2d^2$$ for still non-negative variables, which is true by SOS!
Indeed, let $a\geq b\geq c\geq d$.
Thus, $$a^4+b^4+c^4+d^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3711967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Determine all pairs of positive integers $(a,b)$ such that $a^2+b^2+ab$ is a perfect square. Consideration via mod $4$ shows that $a,b=0$ mod $4$ or one of between $a,b$ is $=1$ mod $4$ while the other is $=0 $ mod $4$.
Considering $(a+b)^2,a^2+an+b^2,(a+b-1)^2$ as we can deduce that $a,b>2$.
| Characterization of all solutions (non-trivial) to the Diophantine equation $$a^2+ab+b^2=x^2\quad a,b,x\in\mathbb{Z}$$ using Algebraic Number Theory
$\mathrm{WLOG}$ assume that $\gcd(a,b)=1$.We know that $\mathbb{Z}[\omega]$ is a $\mathrm{UFD}$ where $\omega=\frac{1+\sqrt{-3}}{2}$. Now we can factorize $$a^2+ab+b^2=(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3715194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
If $\mathrm{M,N}$ are $3\times 2, 2 \times 3$ matrices such that $\mathrm{MN}=$ is given. Then $\mathrm{det(NM)}$ is?
If $\mathrm{M,N}$ are $3\times 2, 2 \times 3$ matrices such that $\mathrm{MN}=\pmatrix{8& 2 & -2\\2& 5& 4\\-2& 4&5}$, then $\mathrm{det(NM)}$ is?
($\mathrm{NM}$ is invertible.)
$\mathrm{det(MN)}$ must... | One way to proceed, not knowing the relevant theorem( new one for me as well) is to guess that, being symmetric, this is a Gram matrix. Look for integer rectangles...
$$
\left(
\begin{array}{rr}
2 & 2 \\
2 & -1 \\
1 & -2
\end{array}
\right)
\left(
\begin{array}{rrr}
2 & 2 & 1 \\
2 & -1 & -2
\end{array}
\right) =
\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3717101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Deriving Law of Cosines from Law of Sines
How to eliminate $\alpha$ from the Law of Sines of plane trigonometry
$$ \dfrac{a}{\sin \alpha}= \dfrac{c}{\sin \gamma} =\dfrac{b}{\sin (\gamma+\alpha)} =2R $$
in order to arrive at the Law of Cosines
$$ c^2= a^2+b^2-2 a b \cos \gamma \;?$$
Starting to isolate $\alpha$
$$ \alp... | First, let's get the following relationship.
From $$\frac{a}{\sin\alpha}=\frac{c}{\sin\gamma}=\frac{b}{\sin(\alpha+\gamma)}$$
using componendo and dividendo,
$$\frac{a\cos\gamma+c\cos\alpha}{\sin\alpha\cos\gamma+\cos\alpha\sin\gamma}=\frac{b}{\sin(\alpha+\gamma)}$$
$$a\cos\gamma+c\cos\alpha=b$$
However, these are unnec... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3718421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
If $abc=1$, then how do you prove $\frac{b-1}{bc+1}+\frac{c-1}{ac+1}+\frac{a-1}{ab+1} \geq 0$? If $abc=1$, then how do you prove $\frac{b-1}{bc+1}+\frac{c-1}{ac+1}+\frac{a-1}{ab+1} \geq 0$?
I tried substitution on the bottom (for example $\frac{b-1}{\frac{1}{a}+1}$), but I then a very similar term. What should I do?
I ... | We have$:$
$$\frac{b-1}{bc+1}+\frac{c-1}{ac+1}+\frac{a-1}{ab+1}$$
$$={\sum \frac {2\,b{c}^{2} \left( ab-1 \right) ^{2}+b{c}^{2} \left( a-1
\right) ^{2}}{ 3\left( ab+1 \right) \left( bc+1 \right) \left( ac+1
\right) }} -\frac{f(a,b,c) }{{ 3\left( ab+1
\right) \left( bc+1 \right) \left( ac+1 \right) }}\geq 0$$
whe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3718759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
When will an ellipse ‘fall’ into a parabola?
Consider the parabola $y=x^2$ and an ellipse which ‘rests’ on it, given by the equation $$\frac{x^2}{a^2} +\frac{(y-h)^2}{b^2}=1$$The goal is to find all ordered pairs $(a,b)$ for which the ellipse doesn’t fall to the origin, namely it touches the parabola at two distinct p... | I'd look at the problem in a different way. Suppose we are given $b = h > 0$ such that the ellipse touches the parabola's vertex. What is the largest $a$, say $a^*$, such that the ellipse has no other intersection points with the parabola? For $a > a^*$, the ellipse cannot have $h = b$, thus such an $(a,b)$ pair can... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3721190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prove $x^4 + x^2 +1$ is always greater than $x^3 + x$ Let's say P is equal to $x^4 + x^2 +1$ and $Q$ is equal to $x^3 + x$.
For $x <0$, $P$ is positive and $Q$ is negative. Hence, in this region, $P>Q$.
For $x=0$, $P>Q$.
Also, for $x = 1$, $P>Q$.
For $x > 1$, I factored out $P$ as $x^2(x^2+1) + 1$ and $Q$ as $x(x^2+1)$... | Proving that $x^4 + x^2 +1$ is always above $x^3 + x$ si the same as proving that
$$f(x)=x^4-x^3+x^2-x+1$$ is always positive. We have
$$f'(x)=4 x^3-3 x^2+2 x-1$$ has only one real root; using the hyprbolic method, we find that its real root is given by
$$x_*=\frac{1}{12} \left(3+2 \sqrt{15} \sinh \left(\frac{1}{3} \si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3721404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 7
} |
Consider $\dot{x}=4x^{2}-16$. I am solving the ODE above, it is a question from Strogatz Nonlinear dynamics and chaos, chapter 2 question 2.2.1.
Question
\begin{equation}
\dot{x}=4x^{2}-16
\end{equation}
Answer
\begin{equation}
\frac{\dot{x}}{4x^{2}-16} = 1\\
\frac{\dot{x}}{x^{2}-4} = 4\\
{{dx\over dt}\over x^2-4}=4 \... | $$I=\int \frac{dx}{x^{2}-4}$$
Partial fraction decomposition gives us:
$$I=\int \left ( \frac A {x-2}-\dfrac B {x+2} \right)dx$$
$$I=\int \frac{x(A-B)+2(A+B)}{x^{2}-4}dx$$
$$\implies A=B=\dfrac 14$$
$$I=\dfrac 14\int \left ( \frac 1 {x-2}-\dfrac 1 {x+2} \right)dx$$
Then integrate with $\ln $ function .
$$I=\dfrac 14 \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3723410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Inequality 6 deg For $a,b,c\ge 0$ Prove that $$4(a^2+b^2+c^2)^3\ge 3(a^3+b^3+c^3+3abc)^2$$
My attempt: $$LHS-RHS=12(a-b)^2(b-c)^2(c-a)^2+2(ab+bc+ca)\sum_{sym} a^2(a-b)(a-c)$$
$$+\left(\sum_{sym} a(a-b)(a-c)\right)^2+14\left(\sum_{sym} a^3b^3+3a^2b^2c^2-abc\sum_{sym} ab(a+b)\right)+\sum_{sym} c^2(a-b)^2[2(a+b)^2+(a-b)^2... | For example $$4(a^2+b^2+c^2)^3-(a^3+b^3+c^3+3abc)^2=$$
$$=\sum_{cyc}(a^6+12a^4b^2+12a^4c^2-6a^3b^3-18a^4bc-a^2b^2c^2)\geq0,$$ which is true by Muirhead.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3724654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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The quotient of a quotient group by another quotient group Let $G$, $M$, and $N$ be the cyclic groups given by
$$
G \colon= \left\langle a \colon a^{12} = e \right\rangle = \left\{ e = a^0, a, a^2, \ldots, a^{11} \right\},
$$
$$
M \colon= \left\langle a^2 \right\rangle = \left\{ e, a^2, a^4, a^6, a^8, a^{10} \ri... | Theorem: Let N,K be normal subgroups of G with $N\subseteq K\subseteq G$, then (1) N is a normal subgroup of K,(2) $K/N$ is a normal subgroup of $G/N$, and (3) $(G/N)/(K/N)\cong G/K$.
Proof:
(1): Since N is a normal subgroup of G, $N=gNg^{-1}$ for all $g\in G$. In particular, $N=kNk^{-1}$ for all $k\in K$ since $K\subs... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3733959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Added angle formula to solve this indefinite integral $\int\frac{2\cos x-\sin x}{3\sin x+5\cos x }\,dx$ Starting from this very nice question Integrate $\int\frac{2\cos x-\sin x}{3\sin x+5\cos x }\,dx$ and the relative answers, I would to understand because this integral $$\int\frac{2\cos x-\sin x}{3\sin x+5\cos x }\,d... | Assuming you have
$$
\int\frac{a\cos x+b\sin x}{c\cos x+d\sin x}\,dx
$$
(with $ad-bc\ne0$, to avoid trivial cases) you can indeed write the denominator as $k\cos(x+\varphi)$ and do the substitution $y=x+\varphi$, so the numerator becomes
$$
a\cos\varphi\cos y-a\sin\varphi\sin y+b\cos\varphi\sin y-b\sin\varphi\cos y
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3734680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Finding All Solutions For $\sin(x) = x^2$ Hello everyone how can I find the count of the solution for $\sin(x) = x^2$?
I know there is a one solution in $x = 0$ and for the other solutions I tried to find the extreme point of the function: $y = x^2 - \sin(x)$ and $y'$ is:
$y' = 2x -\cos(x)$ but I don't know how to solv... | Just for the fun of it !
There is no explicit solution for the zero of function $$f(x)=2x -\cos(x)=0$$ If you need it, use Newton method which will converge quite fast as shown in the table
$$\left(
\begin{array}{cc}
n & x_n \\
0 & 0.000000 \\
1 & 0.500000 \\
2 & 0.450627 \\
3 & 0.450184
\end{array}
\right)$$ Anoth... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3735053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Sum of geometric series when exponent is $2n$, not $n$? I have a probability below which denotes the chance of catching a fish.
$$P = \left(\frac{1}{4}\right) + \left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^2 + \left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^4 + \dots $$
I can find a generalized form of $P$ by assu... | Notice, you must take common ratio $r=\left(\frac{3}{4}\right)^2$
$$\therefore \ \ P = \frac{a}{1-r} = \frac{\frac{1}{4}}{1-\left(\frac{3}{4}\right)^2} = \color{blue}{\frac47}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3738481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find coordinates of point in $\mathbb{R^4}$. There is a problem here in Q. $5$ on the last page. It states to find coordinates of point $p$.
Taking point $a=(3,2,5,1), \ b=(3,4,7,1), \ c= (5,8,9,3)$.
Also, $b$ has two coordinates in common with $a$, and $p$ lies on the same line as $a,b$.
So, those two coordinates of... | Using the property of inner product directly is neater.
You can still use the information that $a, b, p$ is collinear.
$$\frac{x-2}{4-2}=\frac{y-5}{7-5}$$
$$x-2=y-5$$
$$y-x=3$$
Along with what you found
$$x+y=17$$
We have $y=10$, $x=7$.
Hence $p=(3,7,10,1).$
Remark: the picture also illustrated that the two circles in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3739053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to simplify $\cos(\frac{1}{3}\cos^{-1}(x))$ and $\sin(\frac{1}{3}\cos^{-1}(x))$? How to simplify this trigonometric expression
$$
\cos \left(\frac{1}{3} \cos ^{-1}(b)\right)
$$
I want to write this equation as
$$
\cos \left(\frac{1}{3} \cos ^{-1}(b)\right) = \cos(\cos^{-1}(B)) = B
$$
Where $B$ is written in terms o... | Let $y=\frac13\cos^{-1}(b)$ then $\cos(3y)=b$. Which gives $4\cos^3y-3\cos y=b$. Hence your $B$ is a solution of $
4x^3-3x-b=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3739741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to prove $(n + \frac{1}{2})\log(1+\frac{1}{n})$ is increasing in $n$? Is $(n + \frac{1}{2})\log(1+\frac{1}{n})$ increasing in $n$?
I attempted to differentiate $p_n=(n + \frac{1}{2})\log(1+\frac{1}{n})$ with respect to $n$.
$p_n^{'} = \log(1+\frac{1}{n}) + \frac{(n+\frac{1}{2})}{(1+\frac{1}{n})}(-\frac{1}{n^2}) = \... | The derivative of the function being
$$f(n)=\log \left(1+\frac{1}{n}\right)-\frac{n+\frac{1}{2}}{\left(1+\frac{1}{n}\right)
n^2}$$
$f(1)=\log (2)-\frac{3}{4} <0$ and when $n$ is large
$$f(n)=-\frac{1}{6 n^3}+O\left(\frac{1}{n^4}\right)$$
Morover
$$f'(n)=\frac{1}{2 n^2 (n+1)^2}$$ make that the function never changes ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3740793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Problem with Summation of series Question: What is the value of $$\frac{1}{3^2+1}+\frac{1}{4^2+2}+\frac{1}{5^2+3} ...$$ up to infinite terms?
Answer: $\frac{13}{36}$
My Approach:
I first find out the general term ($T_n$)$${T_n}=\frac{1}{(n+2)^2+n}=\frac{1}{n^2+5n+4}=\frac{1}{(n+4)(n+1)}=\frac{1}{3}\left(\frac{1}{n+1}-\... | You were so close of solving it!
The final step is of the form
$$\frac{1}{3}\sum_{n=1}^\infty{\frac{1}{n+1}-\frac{1}{n+4}}$$
Writing down the series
$$\require{cancel} \frac{1}{3}\left[\left(\frac{1}{2}-\cancel{\frac{1}{5}}\right)+\left(\frac{1}{3}-\frac{1}{6}\right)+\left(\frac{1}{4}-\frac{1}{7})\right)+\left(\cancel{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3741476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
How prove this $\sum_{i=n+2}^{+\infty}\frac{1}{i^2}>\frac{2n+5}{2(n+2)^2}$ let $n$ be postive integer,show that
$$\sum_{i=n+2}^{+\infty}\dfrac{1}{i^2}>\dfrac{2n+5}{2(n+2)^2}\tag{1}$$
I know
$$\sum_{i=n+2}^{+\infty}\dfrac{1}{i^2}>\int_{n+2}^{+\infty}\dfrac{1}{x^2}dx=\dfrac{1}{n+2}$$
But $$\dfrac{1}{n+2}-\dfrac{2n+5}{2(n... | There is a rather elegant solution, which uses the convexity of the function. Since $f(x) = \frac{1}{x^2}$ is convex, let us consider an area element under the curve versus the area of the trapezoid formed by joining $(n+2,f(n+2))$ and $(n+3,f(n+3)$ by a straight line
$$dA < \frac{1}{2}\left(\frac{1}{(n+2)^2} + \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3743211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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how can I integrate $\int \dfrac{2x+3}{x^3-9x}dx$? How can I integrate this $$\int \dfrac{2x+3}{x^3-9x}dx?$$
My work
$$\dfrac{2x+3}{x^3-9x}=\dfrac{2x+3}{x(x+3)(x-3)}$$
$$=\dfrac{A}{x-3}+\dfrac{B}{x}=\dfrac{C}{x+3}$$
after solving, I got $A=1/2$, $B=-1/3$, $C=-1/6$
partial fractions decompositions:
$$\dfrac{2x+3}{x^3-9x... | Your working is correct. Use absolute values for logarithms in final answer
For substitution, let $x=3\sec\theta\implies dx=3\sec\theta\tan\theta d\ \theta $
$$\int \dfrac{2x+3}{x^3-9x}dx=\int \frac{2\sec\theta+3}{3\sec\theta(9\tan^2\theta)}3\sec\theta\tan\theta d\ \theta $$
$$=\frac19\int \frac{2\sec\theta+3}{\tan\th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3744234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Problematic solution related to finding the function range
The Question: Find the range of the following function
$$y=\dfrac{x}{2}+\dfrac{8}{x}$$
Solution $-1.$ (the solution given to me)
By Cauchy inequality,
$$\dfrac{x}{2}+\dfrac{8}{x}≥2\sqrt{ \dfrac{x}{2}× \dfrac{8}{x}}=4$$ where $x>0$ $$\dfrac{x}{2}+\dfrac{8}{x... |
*
*Question $-1$ :Do you find the solution $-1$ perfect?
No, I don't.
Firstly, the inequality in solution $-1$ is correct.
If $x\lt 0$, then by the inequality of arithmetic and geometric means, we have
$$\frac x2+\frac 8x=-\bigg(\frac{-x}{2}+\frac{8}{-x}\bigg)\le -2\sqrt{\frac{-x}{2}\times\frac{8}{-x}}=-4$$
So, th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3745060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Integrate $\int_0^1 \frac{x^2-1}{\left(x^4+x^3+x^2+x+1\right)\ln{x}} \mathop{dx}$ Insane integral $$\int_0^1 \frac{x^2-1}{\left(x^4+x^3+x^2+x+1\right)\ln x} \mathop{dx}$$
I know $x^4+x^3+x^2+x+1=\frac{x^5-1}{x-1}$ but does it help? I think $u=\ln{x}$ might be necessary some point.
| (edit for a little glitch in writing)
so, basically you want to solve
$$
\int_{0}^{1} \frac{(x^2-1)(x-1)}{(x^5-1)\ln x} \mathrm{d}x
$$
let
$$
I(s) = \int_{0}^{1} \frac{(x^2-1)(x-1)x^s}{(x^5-1)\ln x} \mathrm{d}x
$$
take derivative of which
$$
\begin{aligned}
I'(s) & = \int_{0}^{1} \frac{(x^2-1)(x-1)x^s}{x^5-1} \mathrm{d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3745320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 1
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Find value of $\dfrac{(1+\tan^2\frac{5\pi}{12})({1-\tan^2\frac{11\pi}{12}})}{\tan\frac{\pi}{12}\tan\frac{17\pi}{12}}$ My attempt :
$$\dfrac{\left(1+\tan^2\dfrac{5\pi}{12}\right)\left(1-\tan^2\dfrac{\pi}{12}\right)}{\tan\dfrac{\pi}{12}\tan\dfrac{5\pi}{12}}$$
Change into variable form
$$\dfrac{(1+a^2)(1-b^2)}{ab}$$
$$\df... | We can write
$$ F=\frac{(1+\tan^2(5\pi/12))(1-\tan^2(\pi/12))}{\tan(5\pi/12) \tan(\pi/12)}$$
$$\implies F=\frac{2}{\frac{2\tan(\pi/12)}{1-\tan^2(\pi/12)}}\frac{\sec^2(5\pi/12)}{\tan(5\pi/12)}.$$
$$F=2 \cot (\pi/6) \frac{\csc^2(\pi/12)}{\cot(\pi/12)}=4 \sqrt{3} \csc(\pi/6)=8\sqrt{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3745781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A concentric circles related problem. Given a circle of radius $AC= a$ with center in $C(C_x,0)$ and with $C_x>0$. Given an angle $\beta$ (between the points B C F) and a smaller circle of radius $BC =\displaystyle \frac{a+b}{2}$ , also with center in $C$ where the measure $\displaystyle b = \frac{a(1-\sin \beta)}{1+\s... | Let's define $\theta:=\frac12(90^\circ-\beta)$, so that $\beta=90^\circ-2\theta$, as this gives
$$b = a\frac{1-\sin\beta}{1+\sin\beta}=a\frac{1-\cos2\theta}{1+\cos2\theta}=a\frac{2\sin^2\theta}{2\cos^2\theta}=a\tan^2\theta \tag{1}$$
Then, defining $c:=C_x$ to reduce visual clutter, the target condition becomes
$$c=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3747422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Help with the last step in solving $\lim_{x\to0}\frac{(1+\sin x +\sin^2 x)^{1/x}-(1+\sin x)^{1/x}}x$ I managed to solve part of this limit but can't get the final step right. Here's the limit:
$$
\lim_{x\to0} {
\frac
{
\left(
1+\sin{x}+\sin^2{x}
\right)
^{1/x}
-
\left(
1+\sin{x}
\right)
^{1/x}
}... | You should note that the expression in numerator is of the form $A-B$ with both $A, B$ tending to $e$. We can write $$A-B=B\cdot\frac{\exp(\log A-\log B) - 1}{\log A-\log B} \cdot(\log A - \log B) $$ Since the middle factor tends to $1$ the desired limit is equal to the limit of $$e\cdot\frac{\log A-\log B} {x} =e\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3748637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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How find the maximum of the complex number inequality
Let $z_{i}(i=1,2,\cdots,n)$ be complex numbers such that $|z_{i}|\le 1,z_{1}+z_{2}+\cdots+z_{n}=0$. Define
$$f_{n}(z_{1},z_{2},\cdots,z_{n})=|z^3_{1}+z^3_{2}+\cdots+z^3_{n}|$$
*
*If $n=4$, find the maximum of $f_{n}(z_{1},z_{2},\cdots,z_{n})$.
*For any positive ... | Here's the asymptotic answer: $n - \max f_n$ converges to zero! Justification will be broken up into mod $3$ cases.
(Sorry, I don't have a more exact answer and this was too long to write as a comment. )
Suppose $n=3k$, then $z_{k} = \exp(k \cdot 2\pi i/3)$ gives $f_n = n$.
Suppose $n=3k+1$, the intuition is to onl... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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Let $f(x)$ be a polynomial of degree $8$ such that $f(r)=\frac1r$, for $r=1,2,3,\ldots,9$. Find $\frac1{f(10)}$. Question:
Let $f(x)$ be a polynomial of degree $8$ such that $f(r)=\frac1r$, for $r=1,2,3,\ldots,9$. Find $\frac1{f(10)}$.
My Approach: We know that $f(r)=\frac1r$, which implies that $$rf(r)-1=0$$
Using t... | You are right that $rf(r)-1$ is a polynomial of degree $9$ with zeros at $1, 2, \ldots, 9$. But that determines the polynomial only up to a constant factor, i.e.
$$
rf(r)-1=C(r-1)(r-2)(r-3)(r-4)(r-5)(r-6)(r-7)(r-8)(r-9)
$$
for some constant $C$ (which can be determined by setting $r=0$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3754308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 0
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Calculate the determinant of $n^\text{th}$ order
Calculate the determinant of $n^\text{th}$ order:
$$
\begin{vmatrix}
1 + a_1 & 1 + a_1^2 & \dots & 1 + a_1^n \\
1 + a_2 & 1 + a_2^2 & \dots & 1 + a_2^n \\
\vdots & \vdots & \ddots & \vdots \\
1 + a_n & 1 + a_n^2 & \dots & 1 + a_n^n \\
\end{vmatrix}
$$
So whe... | When after experimentation we find that the determinant is a product of some factors, it's a good idea to see if the matrix itself can be factored into a product of matrices. If the matrix $A = BC$, where the determinants of $B$ and $C$ are easily calculated, then we can recover $\det(A)$ using the identity $\det (BC) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3755483",
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"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
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Evaluating $\lim\limits_{x\to \infty}\left(\frac{20^x-1}{19x}\right)^{\frac{1}{x}}$ Evaluate the limit $\lim\limits_{x\to \infty}\left(\dfrac{20^x-1}{19x}\right)^{\frac{1}{x}}$.
My Attempt
$$\lim_{x\to \infty}\left(\frac{20^x-1}{19x}\right)^{\frac{1}{x}}=\lim_{x\to \infty}\left(\frac{(1+19)^x-1}{19x}\right)^{\frac{1}{x... | You can calculate the limit directly using the standard limits $\lim_{x\to \infty} x^{\frac 1x}= 1$ and $\lim_{x\to \infty} a^{\frac 1x}= 1$ for any $a>0$ as follows:
Note that
$$\frac{20}{(19x)^{\frac 1x}} =\frac{(20^x)^{\frac 1x}}{(19x)^{\frac 1x}} > \frac{(20^x-1)^{\frac 1x}}{(19x)^{\frac 1x}} >\frac{(20^x-\frac 12... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3756148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Proving $\sqrt{x^2-xz+z^2} + \sqrt{y^2-yz+z^2} \geq \sqrt{x^2+xy+y^2}$ algebraically
The question is to prove that for any positive real numbers $x$, $y$ and $z$,
$$\sqrt{x^2-xz+z^2} + \sqrt{y^2-yz+z^2} \geq \sqrt{x^2+xy+y^2}$$
So I decided to do some squaring on both sides and expanding:
$$\sqrt{x^2-xz+z^2} + \sqrt{... | Let there be a triangle $ABC$ with vertices as $A(x,0), B(z/2,z\sqrt{3}/2), C(-y/2,y\sqrt{3}/2),$ then $$AB=\sqrt{x^2+z^2-xz},~ BC=\sqrt{y^2+z^2-yz}, ~CA=\sqrt{x^2+y^2+xy}.$$
Finally, the triangular inequality $$AB+BC \ge CA.$$
proves the result. The equality holds when $A,B,C$ are collinear.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3756829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
A tough definite integral using contour integration For some reason, I am guessing that for any fixed $s_1,s_2>0$ and $\varepsilon >0$ being small, we have
\begin{align}&\quad-\int_0^\infty \frac{1}{2\pi} \log\left(\frac{(x-s_1)^2+s^2_2}{(x+s_1)^2+s^2_2}\right)\frac{4x\sin\varepsilon}{x^4-2x^2\cos\varepsilon +1}\,dx\\&... | As suggested, a contour integration technique can be used to evaluate this integral. Notice first that the integrand is an even function of $x$, then
\begin{align}
I&=- \frac{1}{2\pi}\int_0^\infty \log\left(\frac{(x-s_1)^2+s^2_2}{(x+s_1)^2+s^2_2}\right)\frac{4x\sin\varepsilon}{x^4-2x^2\cos\varepsilon +1}\,dx\\
&=- \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3758148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove that $\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2} > \frac{3n}{2n+1}$ for all $n \geq 2$ by induction
Prove that $\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2} > \frac{3n}{2n+1}$ for all $n \geq 2$ by induction
I tried to add $\frac{1}{(k+1)^2}$ to both sides of this inequality assuming it's ... | Now, $$\frac{3k}{2k+1}+\frac{1}{(k+1)^2}-\frac{3(k+1)}{2k+3}=\frac{k(k+2)}{(k+1)^2(2k+1)(2k+3)}>0$$ and by induction we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3758676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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$2^{3^{n}}+1=(2+1)\left(2^{2}-2+1\right)\left(2^{2 \cdot 3}-2^{3}+1\right) \cdots\left(2^{2 \cdot 3^{n-1}}-2^{3^{n-1}}+1\right)$ I was reading a solution in which author used this identity without giving any hint that how it comes ...{maybe it is obvious}
$2^{3^{n}}+1=(2+1)\left(2^{2}-2+1\right)\left(2^{2 \cdot 3}-2^{3... | HInt:
Use $$(x+1)(x^2-x+1)=x^3+1$$, successively.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3758765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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How many unordered pairs of positive integers $(a,b)$ are there such that $\operatorname{lcm}(a,b) = 126000$?
How many unordered pairs of positive integers $(a,b)$ are there such that $\operatorname{lcm}(a,b) = 126000$?
Attempt:
Let $h= \gcd(A,B)$ so $A=hr$ and $B=hp$, and $$phr=\operatorname{lcm}(A,B)=3^2\cdot 7\cdo... | I can't really follow what's going on in your attempt - it doesn't help that you use some letters with multiple inconsistent meanings and seem to start off with HCF/LCM swapped.
However, writing $A=2^a3^b5^c7^d$ and $B=2^e3^f5^g7^h$ you have $\max(a,e)=4$, etc. This means there are $2\times 5-1=9$ choices for $(a,e)$: ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3760396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Solving $\sqrt{1+\sqrt{2x-x^2}} + \sqrt{1-\sqrt{2x-x^2}} = \sqrt{4-2x}$ Can we find the solutions for this equation?
$$\sqrt{1+\sqrt{2x-x^2}} + \sqrt{1-\sqrt{2x-x^2}} = \sqrt{4-2x}, \quad x \in \mathbb{R}$$
I tried to amplify the second square root in the $LHS$ with the conjugate and then use AM-GM in order to find whe... | The domain gives $0\leq x\leq2$.
Now, $$\sqrt{1+\sqrt{2x-x^2}} + \sqrt{1-\sqrt{2x-x^2}} = \sqrt{\left(\sqrt{1+\sqrt{2x-x^2}} + \sqrt{1-\sqrt{2x-x^2}}\right)^2}=$$
$$=\sqrt{2+2\sqrt{1-2x+x^2}}=\sqrt{2+2|x-1|}.$$
Thus, it's enough to solve $$1+|x-1|=2-x.$$
Can you end it now?
I got $0\leq x\leq1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3763208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Computing $2 \binom{n}{0} + 2^2 \frac{\binom{n}{1}}{2} + 2^3 \frac{\binom{n}{2}}{3} + \cdots + 2^{n+1} \frac{\binom{n}{n}}{n+1}$ How can I compute the sum $2 \binom{n}{0} + 2^2 \frac{\binom{n}{1}}{2} +
2^3 \frac{\binom{n}{2}}{3} + \cdots + 2^{n+1} \frac{\binom{n}{n}}{n+1}$? I think I should expand $(1+ \sqrt{2})^n$ or... | The expression looks like it has something to do with binomial expansion of $\left(x+1\right)^n=\sum_{k=0}^{n}{\binom{n}{k}x^k}$ evaluated at $x=2$ but with each term being integrated. So we need to integrate both sides with respect to $x$ to get $\frac{\left(x+1\right)^{n+1}}{n+1}+c=\sum_{k=0}^{n}{\binom{n}{k}\frac{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3764054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $|a + b| = |a| + |b| \iff a\overline{b} \ge 0$ I'm reading a complex analysis book. In this book, the author establishes the following statement
If $a,b \in \mathbb{C}$, then $|a + b| = |a| + |b| \iff \left(a\overline{b}\in \mathbb{R}\right) \wedge \left(a\overline{b}\ge 0\right)$
This statement didn't s... | This could be useful:
$$ |a + b|^2 = |a|^2 +|b|^2 + 2 \mathrm{Re} (a\bar{b}) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3765616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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How to evaluate the following limit: $\lim_{x\to 0}\frac{12^x-4^x}{9^x-3^x}$? How can I compute this limit
$$\lim_{x\to 0}\dfrac{12^x-4^x}{9^x-3^x}\text{?}$$
My solution is here:
$$\lim_{x\to 0}\dfrac{12^x-4^x}{9^x-3^x}=\dfrac{1-1}{1-1} = \dfrac{0}{0}$$
I used L'H$\hat{\mathrm{o}}$pital's rule:
\begin{align*}
\lim_{x\t... | As an alternative, you can use
$$a^{x} = e^{x \ln(a)} = 1 + x\ln(a) + \frac{x^2\ln^{2}(a)}{2!} + \mathcal{O}(x^{3})$$
therefore
\begin{align}
\frac{a^{x} - b^{x}}{c^{x} - d^{x}} &= \frac{x\,(\ln(a) - \ln(b)) + \frac{x^2}{2} \, (\ln^{2}(a) - \ln^{2}(b)) + \mathcal{O}(x^{3})}{x\,(\ln(c) - \ln(d)) + \frac{x^2}{2} \, (\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3767178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 1
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Use Cauchy formula to solve $\int _0^{2\pi} \frac{dt}{a\cos t+ b\sin t +c} $ given $\sqrt{(a^2+b^2)}=1Use Cauchy formula to solve $$\int _0^{2\pi} \frac{dt}{a \cos t+ b \sin t +c} $$ Given $\sqrt{(a^2+b^2)}=1<c$.
I tried a variable substitution, but nothing elegant. Can anyone solve it using Cauchy formula?
Edit: there... | This can be done without taking the help of integration limits too.
Evaluate $$\int \frac{dt}{a \cos t+ b \sin t +c} $$
,where $\sqrt{(a^2+b^2)}=1<c$.
I don't know what Cauchy's formula you are talking about. But here is a nice way to do it.
Since $\sin t=\frac{2\tan(\frac x2)}{1+\tan^2(\frac x2)},\cos t=\frac{1-\tan... | {
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"source": "stackexchange",
"question_score": "1",
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Positive integer solutions to $\frac{1}{a} + \frac{1}{b} = \frac{c}{d}$ I was looking at the equation $$\frac{1}{a}+\frac{1}{b} = \frac{c}{d}\,,$$ where $c$ and $d$ are positive integers such that $\gcd(c,d) = 1$.
I was trying to find positive integer solutions to this equation for $a, b$, given any $c$ and $d$ that s... | The first equation is equivalent to (ac-d)(bc-d) = d^2. From there, you can find all the ways that two numbers multiply to d^2. If you have two numbers e and f that multiply to be d^2, then you can solve ac-d = e, bc-d = f individually. This stems from Simon's Favorite Factoring Trick that another person mentioned.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3768155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Find the complete solution set of the equation ${\sin ^{ -1}}( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} ) = \frac{\pi }{4} + {\sin ^{ - 1}}x$ is The complete solution set of the equation ${\sin ^{ - 1}}\left( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} \right) = \frac{\pi }{4} + {\sin ^{ - 1}}x$ is
(A)[-1,0]
(B)... | The other answers pointed out some very creative ways of solving your problem, but I just wanted to complete your algebraic manipulation for you.
$\left( {\left( {\frac{{x\sqrt {1 - {x^2}} + 1 - {x^2}}}{{\sqrt 2 }}} \right) - x\sqrt
{\left( {\frac{{1 - 2x\sqrt {1 - {x^2}} }}{2}} \right)} } \right) = \frac{1}{{\sqrt 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3771166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Evaluate trigonometric integral $ \int_{0}^{\pi / 2} \frac{x^{3} \cos x }{3 \sin x-\sin 3 x}dx $ Evaluate: $$
\int_{0}^{\pi / 2} \frac{x^{3} \cos x d x}{3 \sin x-\sin 3 x}
$$
Here I can see that the denominator nicely converts into $4\sin^{3}{x}$ so I basically get $$
\int_{0}^{\pi / 2}\left(\frac{x}{\sin x}\right)^{3}... | \begin{aligned}
& \int_{0}^{\frac{x}{2}} \frac{x^{2} \cos x}{3 \sin x-\sin 3 x} d x \\
=& \int_{0}^{\frac{\pi}{2}} \frac{x^{3} \cos x}{3 \sin x-3 \sin x+4 \sin ^{3} x} \\
=& \frac{1}{4} \int_{0}^{\frac{\pi}{2}} \frac{x^{3} \cos x}{\sin ^{3} x} d x \\
=&-\frac{1}{8} \int_{0}^{\frac{\pi}{2}} x^{3} d\left(\frac{1}{\sin ^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3771273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Find $ED$ in the triangle $ABC$
$AB = 6$, $AB \| EG$, $BC = 8$, $AC = 10$
I only have this:
From here, we know $BF =FG, AF=DA, GC=DC, OG=OF=OD$, also $$\frac{6}{GE}=\frac{8}{GC}=\frac{10}{EC}$$
And, we can see $$\frac{OG}{OC} = \frac{GC}{EC} \implies OG = \frac{4}{5}OE$$
From the same $$\frac{4}{5}EC=CG=DC=EC - ED \im... | The radius $r$ of circle inscribed in right $\Delta ABC$ with legs $AB=6$, $BC=8$ and hypotenuse $AC=10$ is given by $$r=\frac{\text{Area}}{\text{semi-perimeter}}=\frac{\frac12(6)(8)}{\frac{1}{2}(6+8+10)}=2\implies OG=OF=GB=r=2$$
$$CG=CB-GB=8-2=6\implies CD=CG=6$$
In similar right triangles $\Delta EGC\sim \Delta ABC$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3771780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Convergence or divergence of $\sum^{\infty}_{n=1}\frac{2^n\cdot n^5}{n!}$ using integral test
Finding convergence or Divergence of series $$\sum^{\infty}_{n=1}\frac{2^n\cdot n^5}{n!}$$ using integral test.
What i try::
$$\frac{2}{1}+\frac{4\cdot 32}{2!}+\frac{8\cdot 3^5}{3!}+\sum^{\infty}_{n=4}\frac{2^n\cdot n^5}{n!}... | I don't know "you have to use integral test " or ...? but I suggest ratio test
$$\left|\frac{a_{n+1}}{a_n}\right|=
\\\dfrac{\dfrac{2^{n+1}\cdot (n+1)^5}{(n+1)!}}{\dfrac{2^n\cdot n^5}{n!}}\\=\frac{2(n+1)^5}{(n+1)n^5}\\=\frac{2(n+1)^4}{n^5}\to 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3771866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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Does $-6(x-\frac{1}{2})^2$ = $-\frac{3}{2}(2x-1)^2$? I'm confused about a question I posted this morning.
I am trying to understand if $-6(x-\frac{1}{2})^2$ can be rewritten as $-\frac{3}{2}(2x-1)^2$?
I tried multiplying out the expression $-6(x-\frac{1}{2})^2$ to a polynomial form $36x^2-36x+9$ but that didn't take me... | We have
$$-6\left(x-\frac{1}{2}\right)^2=-\frac 6{\color{blue}4}\cdot \color{blue}4\cdot\left(x-\frac{1}{2}\right)^2=-\frac 3{\color{blue}2}\cdot \color{blue}{2^2}\cdot\left(x-\frac{1}{2}\right)^2=\\=-\frac 32\left(\color{blue}2\cdot x-\color{blue}2\cdot\frac{1}{2}\right)^2=-\frac 32(2x-1)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3772654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Conditional Probability Denominator doubt Let X be a rv such that $P(X = 2) = 1/4$ and its CDF is given by
$$F_{X}(x) =
\begin{cases}
0,& x< -3\\
\frac{3}{4}(x+3),& -3\leq x <2 \\
3/4,& 2 \leq x <4\\
\frac{3}{64} x^2,& 4 \leq x < \frac{8}{\sqrt{3}} \\
1,& x \geq \frac{8}{\sqrt{3}}
\end{cases}
$$
$x=2... | Having the CDF no integral is needed.
You can solve the problem in this way
$$\mathbb{P}[Z<3|X\geq 2]=\frac{F_X(3)-F_X(2^-)}{1-F_X(2^-)}=\frac{\frac{3}{4}-[\frac{x+3}{10}]_{x=2}}{1-[\frac{x+3}{10}]_{x=2}}=\frac{1}{2}$$
The current CDF is wrong. It cannot be $\frac{3}{4}(x+3)$ when $-3 \leq x <2$ because it is $F(2^-)>1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3773077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Derivatives of $ \frac{1}{r} $ and Dirac delta function I am trying to understand the formula
\begin{equation}
\nabla^2\left(\frac{1}{|{\bf r}-{\bf r}'|}\right) = - 4 \pi \delta(\bf{r}-\bf{r}'), \qquad\qquad {\rm (I)}
\end{equation}
where ${\bf r}=(x,y,z)$. This is something heavily used in electrostatics and the steps... | A1. If you are not familiar with distribution theory, we might consider an alternative approach using the idea of approximate Dirac delta function. Indeed, define
$$ f_{\epsilon}(\mathbf{x}) = \frac{1}{\sqrt{\|\mathbf{x}\|^2+\epsilon^2}}=\frac{1}{\sqrt{x^2+y^2+z^2+\epsilon^2}}. $$
Then its Laplacian is
$$ \Delta f_{\ep... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3774483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to transform $z$ into $\hat z$ Assume that we have vector $a = \begin{bmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{bmatrix} \in \mathbb R^n$, $b = \begin{bmatrix} b_1 \\ b_2 \\ \vdots \\ b_n \end{bmatrix} \in \mathbb R^n$, $c = \begin{bmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{bmatrix} \in \mathbb R^n$.
Then, the vecto... | I believe the Kronecker product can be pretty much useful. You could take.
Approach 1: Take:
$$z_{1} = \begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \\0 \\ 0\end{pmatrix} \quad z_{2} = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 1 \\ 0 \\ 0 \end{pmatrix} \quad z_{3}=\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 1 \\1 \end{pmatrix} $$
so that:
$$\hat{z} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3776655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Prove that $b^2-4ac$ can not be a perfect square Given $a$,$b$,$c$ are odd integers Prove that $b^2-4ac$ can not be a perfect square.
My try:Let $a=2k_1+1,b=2n+1,c=2k_2+1;n,k_1,k_2 \in I$
$b^2-4ac=(2n+1)^2-4(2k_1+1)(2k_2+1)$
$\implies b^2-4ac=4n^2+4n+1-16k_1k_2-8k_2-8k_1-4 $
| Suppose $b^2 - 4ac$ were a perfect square. Then the quadratic $ax^2 + bx + c$ has a rational root by the quadratic formula. By the rational roots theorem, the root can be written as $\frac{p}{q}$ with $p$ a factor of $c$, $q$ a factor of $a$. In particular, $p$ and $q$ are both odd. Plugging in, we have
$a (\frac{p}{q}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3777972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Proving $(1+a^2)(1+b^2)(1+c^2)\geq8 $ I tried this question in two ways-
Suppose a, b, c are three positive real numbers verifying $ab+bc+ca = 3$. Prove that $$ (1+a^2)(1+b^2)(1+c^2)\geq8 $$
Approach 1:
$$\prod_{cyc} {(1+a^2)}= \left({a^2\over2}+1+{a^2\over2}\right)\left({b^2\over2}+{b^2\over2}+1\right)\left(1+{c^2\... | Let $a=\sqrt3\tan\alpha$, $b=\sqrt3\tan\beta$ and $c=\sqrt3\tan\gamma,$ where $\{\alpha,\beta,\gamma\}\subset\left(0,\frac{\pi}{2}\right)$.
Thus, $$\alpha+\beta+\gamma=\frac{\pi}{2}$$ and we need to prove that:
$$\ln\left((1+a^2)(1+b^2)(1+c^2)\right)\geq3\ln2$$ or
$$\ln(1+a^2)-\ln2+\ln(1+b^2)-\ln2+\ln(1+c^2)-\ln2\geq0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3779630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Bernoulli's Inequality for $-1 \leq x\leq 0$ My original goal was to prove that
$$\lim_{x\to 0}\frac{e^x-1}{x}=1$$
using the squeeze theorem as we haven't seen differentiability yet and thus I cannot use arguments such as Taylor series nor Bernoulli's theorem, nor can I use induction. For that I wanted to find a lower ... | You can use this supplement to Bernoulli's inequality:
If $0<x<1$, we have
$$(1-x)^n<1-nx+\frac{n(n-1)}2x^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3780633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Counting the number of integers with given restrictions Question: Consider the numbers $1$ through $99,999$ in their ordinary decimal representations. How many contain exactly one of each of the digits $2, 3, 4, 5$?
Answer: $720$.
Attempt at deriving the answer:
We have two cases: four digit numbers and five digit num... | To answer the question you asked: In the case where the non-$x$ value is in the third position, you missed permuting the fourth and fifth digits of the number, so that term should be $6\cdot 2\cdot 6\cdot 2$ (rather than $6\cdot 2\cdot 6$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3780987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Minimal polynomial of $\alpha + \beta$ over $\mathbb{Q}$ Let $\alpha,\beta \in \mathbb{C}$ such that $\alpha^3+\alpha+1=0$ and $\beta^2+\beta-3=0$ . Find the minimal polynomial of $\alpha+\beta$ over $\mathbb{Q}$.
I was trying the usual trick for this kind of problems, that is
Let $\gamma = \alpha+\beta$, and therefore... | Note that
*
*The polynomial $a^3+a+1$ has three distinct roots, exactly one of which is real.$\\[4pt]$
*The polynomial $b^2+b-3$ has two distinct roots, both of which are real.
By pairing one of the non-real roots of $a^3+a+1$ with each of the roots of $b^2+b-3$, it follows that the minimal polynomial of $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3786857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
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The value of $\int_0^1\ (\prod_{r=1}^n x+r).(\sum_{k=1}^n \frac{1}{x+k}).dx $ is $$\int_0^1\ (\prod_{r=1}^n x+r).(\sum_{k=1}^n \frac{1}{x+k}).dx $$
My working:
$$\int_0^1\ [(x+1)(x+2)(x+3)...(x+n)]\times[\frac{1}{x+1}\ + \frac{1}{x+2}\ + \frac{1}{x+3}\ +\ ...\ + \frac{1}{x+n}].dx \\$$
$\int_0^1[(x+1)(x+2)(x+3)...(x+n)]... | We have
$$P(x)= \prod\limits_{k=1}^{n}\left(x+k\right)$$
From which
$$P'(x)=\sum\limits_{i=1}^{n}\prod\limits_{k=1,k\ne i}^{n}\left(x+k\right)$$
then
$$\frac{P'(x)}{P(x)}=\sum\limits_{i=1}^{n}\frac{1}{x+i}$$
As a result
$$\int\limits_0^1 \left(\prod\limits_{r=1}^n (x+r)\right)\cdot \left(\sum_{k=1}^n \frac{1}{x+k}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3786984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Proving $\cos a-\cos b-\cos c\geq -\frac{3}{2}$, where $a+b+c=2\pi$ and $a,b,c>0$ I need help proving the following trig inequality:
$$\cos a-\cos b-\cos c\geq -\frac{3}{2}$$ where $a+b+c=2\pi$ and $a,b,c>0$.
I've found that equality occurs when $a=\frac{4\pi}{3},b=c=\frac{\pi}{3},$ but I'm not sure how to prove it i... | We need to prove that $$\cos{a}-2\cos\frac{b+c}{2}\cos\frac{b-c}{2}+\frac{3}{2}\geq0$$ or
$$2\cos^2\frac{a}{2}-1+2\cos\frac{a}{2}\cos\frac{b-c}{2}+\frac{3}{2}\geq0$$ or
$$4\cos^2\frac{a}{2}+4\cos\frac{a}{2}\cos\frac{b-c}{2}+1\geq0$$ or
$$\left(2\cos\frac{a}{2}+\cos\frac{b-c}{2}\right)^2+\sin^2\frac{b-c}{2}\geq0,$$ whic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3789885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Find the minimum value of $x_1^2+x_2^2+x_3^2+x_4^2$ subject to $x_1+x_2+x_3+x_4=a$ and $x_1-x_2+x_3-x_4=b$. Question: Find the minimum value of $x_1^2+x_2^2+x_3^2+x_4^2$ subject to $x_1+x_2+x_3+x_4=a$ and $x_1-x_2+x_3-x_4=b$.
My attempt: It can be easily seen that $x_1+x_3=\frac{a+b}{2}$ and $x_2+x_4=\frac{a-b}{2}$. Fu... | Why not use a Lagrangian and find an optimal value for a constrained optimization problem?
That is,
$$
\begin{array}{cl}
\min_{x} & x^T x \\
\text{subject to} & v_1^T x = a, v_2^T x = b
\end{array}
$$
where $x = [\begin{array}{cccc} x_1 & x_2 & x_3 & x_4 \end{array}]^T$, $v_1 = [\begin{array}{cccc} 1 & 1 & 1 & 1 \end{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3790537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Given $U_n=\int_0^\frac{\pi}{2} x\sin^n x dx$, find $\frac{100U_{10}-1}{U_8}$
If
$$U_n=\int_0^\frac{\pi}{2} x\sin^n x dx$$
Find $\frac{100U_{10}-1}{U_8}$
Answer: $90$
My Attempt:
I tried applying Integration By Parts, and when that failed, I tried the substitution $x\rightarrow \frac{\pi}{2} -x$ , only to establish a... | Given,$$u_{n}=\int^{\frac{\pi}{2}}_{0} x \sin^{n}x\ dx=\int^{\frac{\pi}{2}}_{0} (x \cdot \sin x) \sin^{n-1}x\ dx$$
Using Integration by parts:
$$u_{n}=\left[\sin^{n-1}x(-x \cdot \cos x + \sin x)\right]^{\frac{\pi}{2}}_{0}-\int^{\frac{\pi}{2}}_{0} (-x \cdot \cos x + \sin x) (n-1)\sin^{n-2}x \cos x\ dx$$
$$\Rightarrow u_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3790662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why does $-8^{\frac{1}{3}}$ have $2$, $e^{\frac{\pi}{3}}$ and $e^{\frac{5\pi}{3}}$? Use DeMoivre’s theorem to find $-8^{\frac{1}{3}}$. Express your answer in complex form.
Select one:
a. –2
b. – 2, 2 cis ($\pi$/3)
c. – 2, 2 cis ($\pi$/3), 2 cis (5$\pi$/3)
d. 2, 2 cis ($\pi$/3), 2 cis (5$\pi$/3)
e. None of these
The c... | Since $-8=8e^{i({\pi+2n\pi})}$ we have $(-8)^{\frac{1}{3}}=2e^{i\frac{\pi+2n\pi}{3}}$ for $n\in\mathbb Z$ (and you want the cases when $n=0,1$ and $2$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3791784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A continuous function $f:\left[-\frac{\pi}{4},\frac{\pi}{4}\right]\to[-1,1]$ and differentiable on $\left(-\frac{\pi}{4},\frac{\pi}{4}\right)$.
$\blacksquare~$ Problem: Suppose a continuous function $f:\left[-\frac{\pi}{4},\frac{\pi}{4}\right]\to[-1,1]$ and differentiable on $\left(-\frac{\pi}{4},\frac{\pi}{4}\right)... | Your solution is fine, you may just simpify a bit just one step:
You can write from the begging
$\left\vert \frac{g\left(\frac{\pi}{4}\right) - g\left(-\frac{\pi}{4} \right) }{\frac{\pi}{2}}\right\vert = \vert g'(x_0)\vert \quad \text{for some } x_0 \in \left(- \frac{\pi}{4}, \frac{\pi}{4} \right)$ so you can get imme... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3793797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
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The equation $2x^2+ax+(b+3)=0$ has real roots. Find the minimum value of $a^2+b^2$ Question:
The equation $2x^2+ax+(b+3)=0$ has real roots. Find the minimum value of $a^2+b^2$
My working:
$D=a^2-8(b+3)\geqq0$
$a^2\geqq8(b+3)=8b+24$
Add $b^2$ to both sides
$a^2+b^2\geqq b^2+8b+24=(b+4)^2+8\geqq 8$
$\therefore {(a^2+b^2)... | The minimum value for
$ a^2 + b^2 = 8(b+3) + b^2 $, because $ a^2 \geqq 8(b+3) $.
The value of $b$ that minimizes the quadratic $ 8(b+3) + b^2 $ is:
$ 8 + 2b =0 \Rightarrow b = -4$
From $ a^2 \geqq 8(b+3) \Rightarrow b \geqq -3$
Therefore, the minimum value of $ a^2 + b^2 $ is when $ b = -3 $ and $ a = 0 $
$ \Rightarro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3794277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Solving $\sqrt{9-x^2} > x^2 + 1$ without graphic calculator for the exact form
Is there any way to solve this inequality without using a graphic calculator to get the exact form?
$$\sqrt{9-x^2} > x^2 + 1$$
I've tried completing the square but I end up with
$$\frac{3 - \sqrt{41}}{2} < x^2 < \frac{3 + \sqrt{41}}{2}$$
w... | $\sqrt{9-x^{2}} > x^{2} + 1
\\\textsf{because both the sides are positive, you can square both sides}
\\ 9-x^{2} > (x^2 + 1)^{2}
\\ 9-x^{2} > x^{4} + 2x^{2} + 1
\\ x^{4} + 3x^{2} - 9 < 0
\\ \textsf{assume x^2 = t}
\\ t^2 +3t - 9 < 0
\\ t \in \left(\frac{-3 - \sqrt{41}}{2},\frac{-3 + \sqrt{41}}{2}\right)
\\ x^{2} \in \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3797646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Proving $\frac{a^3+b^3+c^3}{3}-abc\ge \frac{3}{4}\sqrt{(a-b)^2(b-c)^2(c-a)^2}$ For $a,b,c\ge 0$ Prove that $$\frac{a^3+b^3+c^3}{3}-abc\ge \frac{3}{4}\sqrt{(a-b)^2(b-c)^2(c-a)^2}$$
My Attempt WLOG $b=\text{mid} \{a,b,c\},$
$$\left(\dfrac{a^3+b^3+c^3}{3}-abc\right)^2-\dfrac{9}{16}(a-b)^2(b-c)^2(c-a)^2$$
\begin{align*} &=... | This is SOS's proof
Since $$\left(\dfrac{a^3+b^3+c^3}{3}-abc\right)^2-\dfrac{9}{16}(a-b)^2(b-c)^2(c-a)^2$$
$$=\sum \Big[{\dfrac {1}{140}}\, \left( 11a+11b-70c \right) ^{2}+{\dfrac {1944
}{35}}\,ab+{\dfrac {999}{140}}\, \left( a-b \right) ^{2}\Big](a-b)^4 \geqslant 0$$
So we are done.
For the best constant, assume $c=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3798764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Given that $a,b,c$ satisfy the equation $x^3-2007 x +2002=0$, then find $\frac{a-1}{a+1}+\frac{b-1}{b+1} +\frac{c-1}{c+1}$
Given that $a,b,c$ satisfy the equation $x^3-2007 x +2002=0$, then find $\frac{a-1}{a+1}+\frac{b-1}{b+1} +\frac{c-1}{c+1}$
The concept of transformation of roots can be applied here.
So replace
$... | If $x$ is any of $a,b,c$ then plugging in $x=\frac{1+y}{1-y}$ gives the new equation $$y^3 - 2y^2 + \frac{999}{1002} \,y+\frac{1}{1002}=0 \, .$$ Finding the objective then amounts to finding the sum $A+B+C$ for the 3 solutions of this equation which is just the negative of the coefficient of the $y^2$ term in the gener... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3799807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the minimum value of $f(x) = x + \frac{x}{x^2 + 1} + \frac{x(x + 4)}{x^2 + 2} + \frac{2(x + 2)}{x(x^2 + 2)}$ Find the minimum value of
$$f(x) = x + \frac{x}{x^2 + 1} + \frac{x(x + 4)}{x^2 + 2} + \frac{2(x + 2)}{x(x^2 + 2)}$$ for $x > 0.$
I'm not sure how to start this problem. I think AM-GM is the best way, but I'... | We have
$$f'(x) = \frac{(x^4+x^2+2)(x^3+2x^2+2x+2)(x^3-2x^2+2x-2)}{x^2(x^2+1)^2(x^2+2)^2}.$$
Because $x>0,$ so
$$f'(x) = 0 \Rightarrow x = x_0 = \frac{\sqrt[3]{17+3\sqrt{33}}}{3}-\frac{2}{3\sqrt[3]{17+3\sqrt{33}}}+\frac{2}{3} \approx 1.5437$$
Therefore
$$f(x) \geqslant f(x_0) = 5.$$
So $f(x)_{\min} = 5,$ equality occu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3804362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that$(1-\omega+\omega^2)(1-\omega^2+\omega^4)(1-\omega^4+\omega^8)…$ to 2n factors$=2^{2n}$ Prove that $(1-\omega+\omega^2)(1-\omega^2+\omega^4)(1-\omega^4+\omega^8)…$ to 2n factors$=2^{2n}$ where $\omega$ is the cube root of unity
My attempt:
$(1+\omega^n+\omega^{2n})=0$
$\Rightarrow (1-\omega^n+\omega^{2n})=-2\... | Since $$a^3+b^3=(a+b)(a^2-ab+b^2),$$ we obtain: $$\prod_{k=1}^{2n}(1-w^{2^{k-1}}+w^{2^{k}})=\prod_{k=1}^{2n}\frac{\left(1+w^{3\cdot2^{k-1}}\right)}{1+w^{2^{k-1}}}=\frac{2^{2n}}{((1+w)(1+w^2))^n}=2^{2n}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3805820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Sides $\frac{|b - c|}{\sqrt{(b^2 + 1)(c^2 + 1)}}, \frac{|c - a|}{\sqrt{(c^2 + 1)(a^2 + 1)}}, \frac{|a - b|}{\sqrt{(a^2 + 1)(b^2 + 1)}}$ of a triangle.
Prove that for all pairwise distinct $a, b, c \in \mathbb R$, $$\frac{|b - c|}{\sqrt{b^2 + 1}\sqrt{c^2 + 1}}, \frac{|c - a|}{\sqrt{c^2 + 1}\sqrt{a^2 + 1}}, \frac{|a - b... | Hint.
Given the three sides $l_1,l_2,l_3$ we habe
$$
\cos\theta_1 = \frac{l_2^2+l_3^2-l_1^2}{2l_2l_3}
$$
now making
$$
\cases{
l_1^2 = \frac{(b-c)^2}{\left(b^2+1\right) \left(c^2+1\right)}\\
l_2^2 = \frac{(c-a)^2}{\left(a^2+1\right) \left(c^2+1\right)}\\
l_3^2 = \frac{(a-b)^2}{\left(a^2+1\right) \left(b^2+1\right)}
}
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3807438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Limit of an integral $\lim_{r\to 0^+} \int_0^1 \left(\frac{f(x)r}{x^2+r^2}\right )~dx$ Suppose $f$ is a real-valued continuous function on the unit interval $[0,1]$. How can we compute $$\lim_{r\to 0^+} \int_0^1 \left(\dfrac{f(x)r}{x^2+r^2} \right)~dx?$$
If we let $\min f(x)=m$ and $\max f(x)=M$, then we have $$m\tan^{... | Define $M$ such that $|f(x)|<M$ for $x\in [0,1]$. First, note that for all fixed $\gamma>0$ and all $r>0$ we have
$$r\int_\gamma^1\frac{f(x)}{x^2+r^2}dx\leq rM\int_\gamma^1 \frac{1}{x^2+r^2}dx<rM\int_\gamma^1 \frac{1}{x^2}dx=rM\left(\frac{1}{\gamma}-1\right)$$
$$r\int_\gamma^1\frac{f(x)}{x^2+r^2}dx\geq -rM\int_\gamma^1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3814285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Number of real roots $x^8-x^5+x^2-x+1=0$ Find the number of real roots of $x^8-x^5+x^2-x+1$.
My attempt: $f(x)$ has $4$ sign changes and $f(-x)$ has no sign changes, so the possibility of having real roots is $4+0=4$. Since this is a polynomial of degree $8$ it should have $8-4=4$ imaginary roots.
It is quite impossibl... | If $x\geq1$ so
$$x^8-x^5+x^2-x+1=(x^8-x^5)+(x^2-x)+1>0.$$
If $0\le x<1$ so
$$x^8-x^5+x^2-x+1=1-x+x^2(1-x^3)+x^8>0.$$
If $x<0$ so after replacing $x$ on $-x$ we obtain:
$$x^8+x^5+x^2+x+1$$ has no positive roots by the Descarte's rule.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3816206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
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Generating set for $SL_2(\mathbb{C})$ Let
\begin{equation}
A =
\begin{pmatrix}
\lambda & 0\\
0 & \frac{1}{\lambda}\\
\end{pmatrix}
\end{equation}
be a matrix in $SL_2(\mathbb{C}$). I am trying to prove that $SL_2(\mathbb{C})$ is generated by the matrices
\begin{equation}
\begin{pmatrix}
1 & z\\
0 & 1\\
\end{pmatrix},
... | How about this:
$A = \begin{pmatrix}
1 & -\lambda \\
0 & 1
\end{pmatrix} \cdot \begin{pmatrix}
1 & 0\\
\frac{1}{\lambda} & 1\\
\end{pmatrix} \cdot \begin{pmatrix}
1 & 0\\
-1 & 1\\
\end{pmatrix} \cdot \begin{pmatrix}
1 & 1\\
0 & 1\\
\end{pmatrix} \cdot \begin{pmatrix}
1 & 0\\
\lambda - 1 & 1\\
\end{pmatrix}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3816757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Evaluating the limit: $\lim_{x\rightarrow\infty}\frac{(ax+b)^{n+2}}{(ex+d)^n}-\frac{a^{n+2}x^2}{e^n}-\frac{x(b^{n+2}e^n-a^{n+2}d^n)}{e^{2n}}$ Can someone please evaluate this limit for me, I have been breaking my head for the past 2-3 days..... Any help would be appreciated.
$$\lim_{x\rightarrow\infty}\left \{ \frac{(a... | Set $p=b^{n+2}e^n-a^{n+2}d^n$
\begin{align}
&\lim_{x\to\infty}\left\{\frac{(ax+b)^{n+2}}{(ex+d)^n}-\frac{a^{n+2}x^2}{e^n}-\frac {x(b^{n+2}e^n-a^{n+2}d^n)}{e^{2n}}\right\}=\\
&\qquad\lim_{x\to\infty}\left\{\frac{(ax+b)^{n+2}}{(ex+d)^n}-\frac{a^{n+2}x^2}{e ^n}-\frac{px}{e^{2n}}\right\}=\\
&\qquad\lim_{x\to\infty}\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3817328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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combinatorial identity another solution? I would appreciate if somebody could help me with the following problem:
I need to show that,
$$\sum^{33}_{k=0}\binom{100}{3k}=\sum^{49}_{k=0}4^k$$
My proof is the following:
$$(1+x)^{100} = a_0 + a_1x + a_2x^2 + \cdots + a_{100}x^{100}$$
Let
$A = a_0 + a_3 + a_6 + \cdots + a_{9... | My argument uses more or less a discrete Fourier transform in a standard way to pick off every third coefficient. Let $\zeta_3 = \exp(2\pi i/3)$ be a primitive 3rd root of unity. Note that $1^k + \zeta_3^k + \zeta_3^{2k} = 3\delta_{3 \mid k}$. Letting $p(x) := (1+x)^{100}$, we see
$$\begin{align*}
\frac{p(1)+p(\zeta_3)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3819424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Finding $k$ such that $\frac{3x^2+kx-44}{x^2-121}$ simplifies to $\frac{3x-4}{x-11}$
If the rational expression $\frac{3x^2+kx-44}{x^2-121}$, where $k$ is an element of $\mathbb{W}$, simplifies to $\frac{3x-4}{x-11}$, then the value of $k$ must be?
my work:
$$\frac{3x^2+kx-44}{x^2-121}= \frac{3x-4}{x-11}$$
so the ans... | hint
If $$\frac{3x^2+kx-44}{x^2-121}=\frac{3x-4}{x-11}$$
Then, with $ x=1$, we get
$$\frac{3+k-44}{1-121}=\frac{1}{10}$$
and
$$k=-12+41=29$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3820408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can we prove that $\operatorname{Tr}(ABC) = \operatorname{Tr}(CBA)$? I would like to verify the claim:
$$\operatorname{Tr}(ABC) = \operatorname{Tr}(CBA)$$
I tried verifying through an example:
Given the following $3$ different matrices:
\begin{align}
A & = \begin{pmatrix}
1 & 2 & 3 & 4 \\
5 & 6 & 7 & 8 \\
9 & 10 & 11... | In fact, $\mathrm{Tr}\,(ABC)=\mathrm{Tr}\,(CBA)$ cannot hold for all $A$, $B$, $C$ so long as the ring of scalars is nontrivial. For let
\begin{align}
A &= \begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix} \\[2ex]
B &= \begin{pmatrix}0 & 0\\ 1 & 0\end{pmatrix} \\[2ex]
C &= \begin{pmatrix}1 & 0\\ 0 & 0\end{pmatrix} \\[2ex]
\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3824594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Floor function bounding From CMC:
What is the sum of the square of the real numbers $x$ for which $x^2 - 20\lfloor x\rfloor + 19 = 0$?
We use $\lfloor x\rfloor\le x<\lfloor x\rfloor+1$ and eventually get the bounds $1\le x\le19$ and $x\ge 18,x\le 2.$ Of course, it's possible for $x$ not to be an integer, so how do we... | From $\lfloor x\rfloor=(x^2+19)/20\gt0$, we see that we must have $x\gt0$, hence $x=\sqrt{20\lfloor x\rfloor-19}$ (i.e., the positive, not the negative, square root). It follows that $x^2-20\lfloor x\rfloor+19=0$ has a (unique) solution with $\lfloor x\rfloor=n\in\mathbb{Z}^+$ if and only if $n\le\sqrt{20n-19}\lt n+1$.... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the inverse of $f(x)=3x^3+1$ I get $\sqrt[3]{\frac{x-1}{3}}$ whereas textbook shows answer as $\sqrt{\frac{x-1}{3}}3$ I am to find the inverse of $f(x)=3x^3+1$. My working:
$$f(x)=3x^3+1$$
$$x=3y^3+1$$
$$x-1=3y^3$$
$$\frac{x-1}{3}=y^3$$
$$f^{-1}(x)=\sqrt[3]\frac{x-1}{3}$$
The solutions section of my textbook says ... | This will be a typo, a misplaced $3$
Your answer $f^{-1}(x)=\sqrt[3]{\dfrac{x-1}{3}}$ is correct, as is $\left(\frac{x-1}{3}\right)^{\frac13}$, though you may need to specify using the real cube root rather the principal cube root when $x<1$
If $\sqrt{\frac{x-1}{3}}3$ had really been intended then I suspect it would ha... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving for $x$ and $y$ when $(3x + y)(x + 3y)\sqrt{xy} = 14$ Solve for $x$ and $y$ when $$(3x + y)(x + 3y)\sqrt{xy} = 14$$ $$(x+y)(x^2 + 14xy + y^2) = 36.$$
I was thinking of squaring the first equation and moving on from there, but I think it'll be a bit too messy. Is there a better way to start this problem?
| From the first equation we have $xy>0$.
Thus, the second give $x+y>0,$ which says $x>0$ and $y>0$.
Let $y=t^2x,$ where $t>0$.
Thus, $$18(3t^2+1)(t^2+3)t=7(t^2+1)(t^4+14t^2+1)$$ or
$$(t^2-6t+1)(7t^4-12t^3+26t^2-12t+7)=0$$ or
$$t^2-6t+1=0,$$
which gives $y=(17-12\sqrt2)x$ or $y=(17+12\sqrt2)x.$
Let $y=(17-12\sqrt2)x.$
Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3827639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving for $x$ when $\sqrt{\sqrt{3} - \sqrt{\sqrt{3} + x}} = x$ Suppose $x$ is a real number such that $\sqrt{\sqrt{3} - \sqrt{\sqrt{3} + x}} = x.$ Find $x.$
I first squared to get rid of the first square root, which gave me $\sqrt{3} - \sqrt{\sqrt{3} + x} = x^2.$ However, I'm not sure how to move on from there. Can ... | Comment:
If you put $x=\sqrt{\sqrt{3} - \sqrt{\sqrt{3} + x}} $ under the third radical you get:
$$\sqrt{\sqrt{3} - \sqrt{\sqrt{3} + \sqrt{\sqrt{3} - \sqrt{\sqrt{3} + \sqrt{\sqrt{3} - \sqrt{\sqrt{3} + }} . . .}} }} = x$$
Suppose there is no negative sign then by squaring both sides you get:
$$X^2=\sqrt 3 +X$$
Which give... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving that $\left[ \mathbb{Q} \left( \sqrt[3]{4+\sqrt{5}} \right ) : \mathbb{Q} \right] = 6$ Let $\alpha = \sqrt[3]{4+\sqrt{5}}$. I would like to prove that $\left[ \mathbb{Q} \left( \alpha \right ) : \mathbb{Q} \right] = 6$. We have $\alpha^3 = 4 + \sqrt{5}$, and so $(\alpha^3 - 4)^2 = 5$, hence $\alpha$ is a root o... | As $\Bbb Q(\sqrt 5)$ is a subfield, its automorphism $\sqrt 5\to-\sqrt 5$ can be extended, which means that $\sqrt[3]{4-\sqrt 5}\in\Bbb Q(\sqrt[3]{4+\sqrt 5})$ and also
$$ \sqrt[3]{11}=\sqrt[3]{4-\sqrt 5}\sqrt[3]{4+\sqrt 5}\in\Bbb Q(\sqrt[3]{4+\sqrt 5}).$$ Now, $[\Bbb Q(\sqrt[3]{11}):\Bbb Q]$ is clearly $3$. With a sub... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3831073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Proving that $\frac{1}{(2k+1)^4} - 2 \sum_{n=k+1}^\infty \left(\frac{1}{(2n)^4} - \frac{1}{(2n+1)^4}\right)$ is non negative Prove that the sequence
$$s_k = \frac{1}{(2k+1)^4} - 2 \sum_{n=k+1}^\infty \left(\frac{1}{(2n)^4} - \frac{1}{(2n+1)^4}\right)$$ is non-negative.
I would appreciate an elementary proof. I tried us... | $f(x) = 1/x^4$ is strictly convex, so that
$$
\frac{1}{(2n)^4} < \frac 12 \left(\frac{1}{(2n-1)^4} + \frac{1}{(2n+1)^4} \right)
$$
or
$$
\frac{1}{(2n)^4} - \frac{1}{(2n+1)^4} < \frac 12 \left(\frac{1}{(2n-1)^4} - \frac{1}{(2n+1)^4} \right)
$$
It follows that
$$
2 \sum_{k=n+1}^\infty \left(\frac{1}{(2n)^4} - \frac{1... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Is $(x^2+3)^3-4(x-1)^3$ an irreducible polynomial in $Q[x]$? I was trying to solve the problem below
Let $K \subset \mathbb{C}$ be a splitting field of $f(x) = x^3 − 2$ over $\mathbb{Q}$. Find a complex number $z$, such that $K = \mathbb{Q}(z)$.
All the roots of the polynomial $f(x)$ are $2^{1/3}$, $2^{1/3}\omega$ an... | It is irreducible by Eisenstein shift, which I explain how to find below.
Note$\, \bmod\color{#c00}3\!:\ f(x) := (x^2\!+\!3)^3\!-4(x\!-\!1)^3 \equiv x^6-x^3+1 \equiv (x\!+\!1)^6 \ $ is a prime power.
So Eisenstein works on $\,g(x) = f(x\!-\!1) \equiv x^6\ $ by
$\,g(0) = f(-1) = 96 \not\equiv 0\pmod{\!3^2}$
Key Idea be... | {
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Find the supremum of $|z|^2+\text{Re}(\overline{z}w)$ subject to $|z|^2+|w|^2=1$. I am trying to find the operator norm of $$A=\begin{pmatrix}3 & 1 \\
1 & 1\end{pmatrix}\in M_n(\mathbb{C}).$$This is what I have so far: If $|z|^2+|w|^2=1$, then $$||A\begin{pmatrix}z\\w\end{pmatrix}||_2=\sqrt2\sqrt{1+4(|z|^2+\text{Re}(\o... | You can use Lagrange multipliers. Let$$f(x,y,z,t)=|x+yi|^2+\operatorname{Re}\bigl((x-yi)(z+ti)\bigr)=x^2+y^2+xz+yt$$and let$$g(x,y,z,t)=|x+yi|^2+|z+iy|^2=x^2+y^2+z^2+t^2.$$So, you should solve the system$$\left\{\begin{array}{l}\nabla f(x,y,z,t)=\lambda\nabla g(x,y,z,t)\\g(x,y,z,t)=1\end{array}\right.$$There are severa... | {
"language": "en",
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"source": "stackexchange",
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Compute the limit $\lim\limits_{t \to + \infty} \int_0^{+ \infty} \frac{ \mathrm d x}{e^x+ \sin tx} $ I have been working on this limit for days, but I am not getting it. The question is
Compute the limit $$\lim_{t \to + \infty} \int_0^{+ \infty} \frac{ \mathrm d x}{e^x+ \sin (tx)}$$
Note that the integral is well de... | A slight modification of OP's attempt will lead to a solution. Indeed, write $I(t)$ for the integral and note that
$$ I(t) = \int_{0}^{\infty} \frac{\mathrm{d}x}{e^x + \sin(tx)}
\stackrel{(y=tx)}= \frac{1}{t} \int_{0}^{\infty} \frac{\mathrm{d}y}{e^{y/t} + \sin y}. $$
Also, define $J(t)$ by
$$ J(t)
= \frac{1}{t} \sum_{k... | {
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"question_score": "17",
"answer_count": 4,
"answer_id": 1
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Find a mistake in a computation of a determinant. I recently have to solve one exercise in which I need to compute the determinant of the matrix
$$ A= \begin{pmatrix}
a & 1 & 1 & 1 \\
1 & a & 1 & 1 \\
1 & 1 & a & 1 \\
1 & 1 & 1 & a \\
\end{pmatrix} ,$$
where $a\in\mathbb{R}$. This is how I proceed using famous prop... | In the first step it seems there is a sign problem indeed
$$\det(A)= \begin{vmatrix}
a & 1 & 1 & 1 \\
1 & a & 1 & 1 \\
1 & 1 & a & 1 \\
1 & 1 & 1 & a \\
\end{vmatrix}=\begin{vmatrix}
a & 1 & 1 & 1 \\
1 & a & 1 & 1 \\
1 & 1 & a & 1 \\
0 & 0 & 1-a & a-1 \\
\end{vmatrix}=\begin{vmatrix}
a & 1 & 1 & 1 \\
1 & a & 1... | {
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"source": "stackexchange",
"question_score": "3",
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Prove that $\lim_{(x,y)\to(1,2)}\frac{x}{x+y}=\frac{1}{3}$ How to prove that $$\lim_{(x,y)\to(1,2)}\frac{x}{x+y}=\frac{1}{3}$$ with $\varepsilon-\delta$ limit definition?
I know so far that $\forall \varepsilon > 0 \ \exists \delta > 0: 0 < d((x, y), (1, 2)) < \delta \Rightarrow \Big|\frac{x}{x+y} - \frac{1}{3}\Big| <... | We can choose $\delta$ such that $3x + 3y > c$ for some positive $c$, so we don't have to worry about the denominator.
For the numerator, note that:
$$|2x-y| = |2x - 2 - y + 2| \le |2x-2|+|y-2|$$
| {
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"source": "stackexchange",
"question_score": "1",
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Where did I go wrong in solving this equation?
The value of the expression $ax^2 + bx + 1$ are $1$ and $4$ when $x$ takes the values of $2$ and $3$ respectively. Find the value of the expression when $x$ takes the value of $4.$
Here is my first attempt at solving this question:
\begin{align}
2a^2 + 2b + 1 &= 1 \lefta... | $$ 4a + 2b + 1 = 1 \leftarrow\text{(1)} $$
$$ 9a + 3b + 1 = 4 \leftarrow\text{(2)} $$
$$ 4a + 2b = 0 $$
$$ 9a + 3b = 3 $$
$$a=1, b=-2$$
| {
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Prove the inequality $1\le\int_1^4 \frac{1}{1+\sqrt(x)} \,dx$ I want to prove the inequality:
$$1\le\int_1^4 \frac{1}{1+\sqrt{x}} \,dx$$
This is my attempt:
The domain is $(0,\infty)$ and the range is $[0,1]$
So, $\frac{1}{1+\sqrt{x}} \ge \frac{1}{1+\sqrt{9}}=\frac{1}{4}$ (used a value for x=9)
So, $1\ge\int_1^4 \frac{... | The idea is correct but you picked $x=9$ instead of $x=4$. We have for $x\in [1,4]$
$$\frac{1}{1+\sqrt{x}}\geq \frac{1}{1+\sqrt{4}}=\frac{1}{1+2}=\frac{1}{3}$$
Then
$$\int_1^4 \frac{1}{1+\sqrt{x}}dx\geq \int_1^4 \frac{1}{3}dx =\frac{3}{3}=1$$
| {
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"source": "stackexchange",
"question_score": "1",
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how to show that $\csc x - \csc\left(\frac{\pi}{3} + x \right) + \csc\left(\frac{\pi}{3} - x\right) = 3 \csc 3x$?
How to show that $\csc x - \csc\left(\frac{\pi}{3} + x \right) + \csc\left(\frac{\pi}{3} - x\right) = 3 \csc 3x$?
My attempt:
\begin{align}
LHS &= \csc x - \csc\left(\frac{\pi}{3} + x\right) + \csc\left(\... | $$\frac{1}{\sin{x}}-\frac{1}{\sin\left(\frac{\pi}{3}+x\right)}+\frac{1}{\sin\left(\frac{\pi}{3}-x\right)}=$$
$$=\frac{\sin\left(\frac{\pi}{3}-x\right)\sin\left(\frac{\pi}{3}+x\right)+\sin{x}\left(\sin\left(\frac{\pi}{3}+x\right)-\sin\left(\frac{\pi}{3}-x\right)\right)}{\sin{x}\sin\left(\frac{\pi}{3}+x\right)\sin\left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3840253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Limit of multiple absolute values Let $f(x) = \frac{x^2+2x-3|1-x^2|+1}{2|x+1|-|x^2+x|}$. Find $\lim_{x\rightarrow -1}f(x)$.
I broke up each absolute value into parts:
$|1-x^2| = \begin{cases}
1-x^2 & -1 \leq x\leq 1 \\
-(1-x^2) & x>1,x<-1
\end{cases}
$ ,
$|x+1| = \begin{cases}
x+1 & x\geq -1 \\... | As an alternative, we have that by $x=y-1$ with $y\to 0$
$$\frac{x^2+2x-3|1-x^2|+1}{2|x+1|-|x^2+x|}=\frac{y^2-3|2y-y^2|}{2|y|-|y(y-1)|}=\frac{|y|-3|2-y|}{2-|y-1|}\to \frac{0-6}{2-1}=-6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
The Finite Sum $\sum_{r=1}^{n}\frac{1}{(3r-2)(3r+2)}$ and failure to Telescope I'm interested in when the partial fraction method of trying to get a series to telescope fails, and have alighted upon the interesting example of,
$$\sum_{r=1}^{n}\frac{1}{(3r-2)(3r+2)}$$
The standard method of tackling this would be to use... | We can write the sum as
\begin{align*}
\sum_{r = 1}^{n}\dfrac{1}{(3r-2)(3r+2)} &= \dfrac{1}{4}\sum_{r = 1}^{n}\left[\dfrac{1}{(3r-2)} - \dfrac{1}{(3r+2)}\right]
\\
&= \dfrac{1}{4}\sum_{r = 1}^{n}\int_{0}^{1}(x^{3r-3}-x^{3r+1})\,dx
\\
&= \dfrac{1}{4}\int_{0}^{1}\sum_{r = 1}^{n}(x^{3r-3}-x^{3r+1})\,dx
\\
&= \dfrac{1}{4}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
$\lim_{n\to\infty}\left( \frac1{4\cdot 7}+\frac1{7\cdot 10}+\ldots+\frac1{(3n+1)(3n+4)} \right) $ I am having trouble finding the infinite sum
$$
\lim_{n\to\infty}\left( \frac1{4\cdot 7}+\frac1{7\cdot 10}+\ldots+\frac1{(3n+1)(3n+4)} \right).
$$
I know that
$$
\lim_{n\to\infty}\frac1{(3n+1)(3n+4)} =0,
$$
but I have no i... | $\dfrac{1}{(3n+1)(3n+4)}=\dfrac{1}{3}\bigg(\dfrac{1}{3n+1}-\dfrac{1}{3n+4}\bigg)$
$\dfrac{1}{4\cdot 7}+\dfrac{1}{7\cdot 10}=\dfrac{1}{3}\bigg(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}\bigg)$ and so on ...
EDIT
$\dfrac{1}{4\cdot 7}+\dfrac{1}{7\cdot 10}+...+\dfrac{1}{(3n+1)(3n+4)}=\dfrac{1}{3}\bigg(\dfrac{1}{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3843359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Find the minimum of $P = (a - b)(b - c)(c - a)$
Given $a, b, c$ are real numbers such that $a^2 + b^2 + c^2 = ab + bc + ca + 6$. Find the minimum of
$$P = (a - b)(b - c)(c - a)$$
My solution:
*
*We have:
$$a^2 + b^2 + c^2 = ab + bc + ca + 6$$
$$\implies 2a^2 + 2b^2 + 2c^2 = 2ab + 2bc + 2ca + 12$$
$$\implies (a - b... | WLOG $a\ge b\ge c$ and let $x=a-b,y=b-c,z=c-a$
We observe $x+y+z=0$ with $x,y\ge 0$.and as you have found $x^2+y^2+z^2=12$
Elimination of $z$ results in: ${(x+y)}^2=6+xy ...(1)$
since ${(x+y)}^2\ge 4xy$ which means $0\le xy\le 2$
let $xy=t$
since $0\le t\le 2$
Now $x^2y^2z^2=t^2(6+t)\le 6.2^2+2^3=32$....(using (1) and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3844805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 0
} |
Using complex numbers, can sine be greater than $1?$ I worked on a brain teaser that started with $\sin{(z)}=10.$ It then solved for $\cos{(z)}=\sqrt{-99}.$
I tried solving for $z,$ using $z=x+iy$ then $z=ae^{bi}$ but got nowhere. So is this even possible?
| Yes, it can.
Let us find such $z$ so that $\sin(z) = 10$.
Firstly, as a consequence of the Euler's formula:
$ e^{iz} = \cos(z) + i \cdot \sin(z)$,
we have the following two identities which hold for all complex numbers $z$:
$$
\sin(z) = \frac{e^{iz} - e^{-iz}}{2i} \\
\cos(z) = \frac{e^{iz} + e^{-iz}}{2}
$$
Then:
$$
\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3844941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
The result of $\int{\sin^3x}\,\mathrm{d}x$ $$\int{\sin^3x}\,\mathrm{d}x$$
I find that this integration is ambiguous since I could get the answer with different approaches. Are these answers are valid and true? Could someone tell me why and how? And also, is there any proof stating that these two method I use results th... | Since $$\cos^3(x)=\frac{3}{4}\cos(x)+\frac{1}{4}\cos(3x)$$
Your first integral becomes $$\int \sin^3(x)dx=\dfrac{1}{3}\cos^3x - \cos x + C$$
$$=\frac{1}{3}\big[\frac{3}{4}\cos(x)+\frac{1}{4}\cos(3x)\big]-\cos(x)+C$$
$$=\frac{1}{12}\cos(3x)-\frac{3}{4}\cos(x)+C$$
Note that the constant of integration are not necessaril... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3845996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to calculate the integral $\int_0^{\infty}\frac{x^{1/2}}{1-x^2}\sin(ax)\sin[a(1-x)] dx$ How can I calculate the following integral:
$$\int_0^{\infty}\frac{x^{1/2}}{1-x^2}\sin(ax)\sin[a(1-x)] dx$$
where $a>0$.
It seems like the integrand is well defined without any singularities, but I don't have any clue how to pro... | Following exactly what @Maxim suggested in the comment
$$\int_0^\infty f(x) dx = \operatorname {Re} \operatorname {v. \! p.} \int_0^\infty g(x) dx = \operatorname {Re} \left(\int_0^{i \infty} g(x) dx + \pi i \operatorname* {Res}_{x = 1} g(x) \right).$$
taking $x=iu$
\begin{align}
\int_0^{i\infty}g(x) dx &=\frac{1}{2}i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3848772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Efficiently enumerate the boolean matrix of sparsity k such that no rows and columns are all 0s. Is there a way to enumerate the boolean matrix of $k$ entries of $1$ with no rows and columns all being $0$s?
e.g.
$k=1$, $\begin{bmatrix}1\end{bmatrix}$
$k=2$, $\begin{bmatrix}1 & 1\end{bmatrix}$
$\begin{bmatrix}1 \\ 1\end... | Here's one possible scheme.
We'll fill in the $1$s of the up-to-$n$-by-$n$ matrix one at a time, but going in lexicographic order. The first $1$ can be in any of the positions $(1,1), (1,2), \dots, (1,n)$. If we just placed a $1$ in position $(i,j)$, the next $1$ can be in any of the positions $$(i,j+1), (i,j+2), \dots... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3849531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Use the sum of the first ten terms to approximate the summation $\sum_1^{\infty} \frac{8n}{(n+1)3^n}$ Use the sum of the first ten terms estimate the summation and then to estimate the error
$\sum_1^{\infty} \frac{8n}{(n+1)3^n}$
Okay, so i added up the first ten terms and got approximately 1.101237.
The answer key says... | This method leads to a conservative estimation for the error since
$$\frac{8n}{(n+1)3^n} < \frac{8n}{n3^n} = \frac{8}{3^n} \implies R_{10, exact} = \sum_{n=11}^{\infty} \frac{8n}{(n+1)3^n}< R_{10, approx} = 8\sum_{n=11}^{\infty} \frac{1}{3^n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3852434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
When is $k^4-24k+16$ a perfect square. When is the equation $k^4-24k+16$ perfect square.
(k is an integer.)
I got this equation as discriminant while solving an equation. I tried to solve it but couldn't i tried to write it in a form of a square but couldnt solve it.I bashed a bit and found 0 and 3 as a solution. Any ... | I think this is a correct solution. Suppose that $k^4-24k+16$ is a square.
*
*Suppose that $-24k+16\ge 0$. Then $k^4-24k+16\ge (k^2+1)^2=k^4+2k^2+1$, so $-24k+16\ge 2k^2+1$ or $2k^2+24k-15\le 0$ or $k^2+12k-7.5\le 0$. This means $$-6-\sqrt{36+7.5}\le k\le -6+\sqrt{36+7.5}.$$
Which means $-13\le k\le 0$ (remember that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3853263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"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.