Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Roots in equation In the equation
$\sqrt{x-2} - \sqrt{6x-11} + \sqrt{x+3} =0$
I got the roots of $x$ being $6$ and $7\sqrt{3}$.
Considering the graph shows only $6$ as being a valid solution, how should I go as figuring this out in the equation itself?
| Let $x=7\sqrt 3$ into $$\sqrt{x-2} - \sqrt{6x-11} + \sqrt{x+3} $$ and you get $$ \sqrt{x-2} - \sqrt{6x-11} + \sqrt{x+3} \approx -0.7869 \ne 0$$
Thus $x=7\sqrt 3$ is not a solution.
| {
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"url": "https://math.stackexchange.com/questions/3376256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Prove that $(5^{\frac 12} + 2)^{\frac 13}- (5^{\frac 12} - 2)^{\frac 13}$ is an integer and find its value I had proceed this question by taking
$$x =(5^{\frac 12} + 2)^{\frac 13}- (5^{\frac 12} - 2)^{\frac 13}$$
Then
$$x + (5^{\frac 12} -2)^{\frac 13} =(5^{\frac 12} + 2)^{\frac 13}$$
And then cubing both sides and the... | Let $a=\sqrt[3]{\sqrt{5}+2}$, $b=\sqrt[3]{\sqrt{5}-2}$ and $x=a-b$. Then,
$$x^3=\left(a-b\right)^3=a^3-3a^2b+3ab^2-b^3=a^3-b^3-3ab\left(a-b\right)\\=\sqrt{5}+2-\left(\sqrt{5}-2\right)-3x\sqrt[3]{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}=4-3x$$
Then the answer is the real root of the equation $x^3+3x-4=0$,
$$x^3-3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3376554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Solve Euler Project #9 only mathematically - Pythagorean triplet The "Euler Project" problem 9 (https://projecteuler.net/problem=9) asks to solve:
$a^2$ + $b^2$ = $c^2$
a + b + c = 1000
I find answers solving it with brute-force and programmatically, but is there a way to solve the problem ONLY mathematically? Can some... | The triplets are all of the form
$a=u(n^2-m^2),
b=2umn,
c=u(n^2+m^2)
$
with $n > m$
so
$a+b+c
=u(2n^2+2mn)
=2un(n+m)
$.
We must have
$n > m$.
Therefore
$500
=un(n+m)
$.
If
$500 = rst
$
with
$s < t$
then
$u = r,
n = s,
n+m = t
$
so
$m = t-n
=t-s
$.
We must have
$n > m$
so
$s > t-s$
or
$s < t < 2s$.
Playing around a bit,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3378407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Nesbitt by Nesbitt The title not says I'm Nesbitt but just only says it's a refinement of Nesbitt's inequality by itself so we have :
Let $a,b,c>0$ and $a\geq b \geq c$ then we have :
$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}}+\frac{\sqrt{b}}{\sqrt{a}+\sqrt{c}}+\frac{\sqrt{c}}{... | We need to prove that: $$\sum_{cyc}\frac{a^2}{b^2+c^2}\geq\sum_{cyc}\frac{a}{b+c},$$where a, b and c are positives.
Indeed, $$\sum_{cyc}\frac{a^2}{b^2+c^2}-\sum_{cyc}\frac{a}{b+c}=\sum_{cyc}\frac{ab(a-b)-ac(c-a)}{(b^2+c^2)(b+c)}=$$
$$=\sum_{cyc}(a-b)\left(\frac{ab}{(b^2+c^2)(b+c)}-\frac{ab}{(c^2+a^2)(c+a)}\right)=$$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3380408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Does $\lim_{(x,y) \to (0,0)} \frac{x y \sin^2 y}{x^2 y}$ exists?
$\displaystyle\lim_{(x,y) \to (0,0)} \frac{x y \sin^2 y}{x^2 y}$
Along $x =y$, it becomes
\begin{equation}
\lim_{(x,y) \to (0,0)}\frac{y^2 \sin^2 y}{y^3} = 0.
\end{equation}
But along $x = y^3$,
\begin{equation}
\lim_{(x,y) \to (0,0)}\frac{y^4 \sin^2 y}... | You are right indeed
$$\frac{x y \sin^2 y}{x^2 y}=\frac{\sin^2y}{y^2 }\frac{y}{x}$$
and
*
*$\frac{\sin^2y}{y^2 }\to 1$
but
*
*$\frac{y}{x}$
has not limit.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3382524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Solve in $\mathbb N^{2}$ the following equation : $5^{2x}-3\cdot2^{2y}+5^{x}2^{y-1}-2^{y-1}-2\cdot5^{x}+1=0$ Question :
Solve for natural number the equation :
$5^{2x}-3\cdot2^{2y}+5^{x}2^{y-1}-2^{y-1}-2\cdot5^{x}+1=0$
My try :
Let : $X=5^{x}$ and $Y=2^{y}$ so above equation
equivalent :
$2X^{2}+(Y-4)X-6Y^{2}-Y+2... | $5^{2x}-3\cdot 2^{2y}+5^{x}2^{y-1}-2^{y-1}-2\cdot5^{x}+1=0$
$(5^{2x}-2\cdot 5^x + 1)+ (5^{x}2^{y-1}-2^{y-1})-3\cdot 2^{2y}=0$
$(5^x-1)^2 + 2^{y-1}(5^x - 1)-3\cdot 2^{2y}=0$
If we solve for $5^x-1$ we get
$5^x-1 =\frac {-2^{y-1} \pm\sqrt{2^{2y-2} + 12*2^{2y}}}{2}=$
$\frac {-2^{y-1} \pm\sqrt{2^{2y-2} + 48*2^{2y-2}}}{2}=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3383436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Calculate the value of $\frac{AN}{AD}$ and $\frac{ME}{MP}$.
$M, N$ and $P$ are respectively the midpoint of $BC$, the centroid of $\triangle ABC$ and a point on $CA$ such that $NP \parallel BC$. A line passing through point $B$ intersects $AM$ and $PM$ respectively at $D$ and $E$ such that $BD = 5DE$. Calculate the va... | I will go an other way, it is the simplest way to check.
We try to apply the theorems of Ceva and/or Menelaus. The situation is as follows.
Let $Q$ be the mid point of $AC$.
Let $Z$ be the intersection of the parallels
*
*to $BC$ through $A$,
*to $AB$ through $C$,
so that $ABCZ$ is a parallelogram. $Q$ is the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3385651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to find the basis (and cartesian equation(s)) of a sum of two vectorial subspaces? I have the following vectorial subspaces :
U = span $\left(\begin{pmatrix} 2 \\ 0 \\ 1 \\ -2 \end{pmatrix},\begin{pmatrix} 3 \\ 6 \\ 9 \\ -12 \end{pmatrix} \right)$ and V = span $\left(\begin{pmatrix} 0 \\ 2 \\ 1 \\ 0 \end{pmatrix},\... | We have obtained that a basis is given by the following vectors
$$\begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix}= a\begin{pmatrix} 2 \\ 0 \\ 1 \\ -2 \end{pmatrix} +b\begin{pmatrix} 3 \\ 6 \\ 9 \\ -12 \end{pmatrix}+ c\begin{pmatrix} 0 \\ 2 \\ 1 \\ 0 \end{pmatrix}$$
to obtain the cartesian equation we need to eli... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3393333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Integrate $\sin^4(x)$ Consider the integral
$$\int \sin^4(x)dx$$
Now I could separate the $\sin^4(x)$ into two $\sin^2(x)$ terms and the use power reducing formula
$$\int \sin^2(x)\sin^2(x)dx $$
$$\int \frac{1-\cos(2x)}{2}*\frac{1-\cos(2x)}{2}dx $$
$$\int \frac{(1-\cos(2x))^2}{4} dx$$
$$\int\frac{1}{4}-\frac{2\cos(2x)... | First of all, I think you mistyped the last line of your integral, changing a 2 for an 8. Besides, you can use the linearity of the integral to write $$ \int(1-2\cos(2x)+\cos^{2}(2x))dx = \int 1 dx - 2\int \cos(2x)dx+\int \cos^{2}(2x)dx. $$ Now, note that $$\cos^{2}(2x) + \sin^{2}(2x) = 1$$ and $$ \cos^{2}(2x)-\sin^{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3393747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Calculate $\lim_{x\to \pi/4}\cot(x)^{\cot(4*x)}$ without L'Hôpital's rule How can I calculate limit
$$\lim_{x\to \pi/4}\cot(x)^{\cot(4*x)}$$
without using L'Hôpital's rule?
What I have tried so far:
I tried to use the fact that $\lim_{\alpha\to 0}(1 + \alpha)^{1/\alpha} = e$ and do the following:
$$\lim_{x\to \pi/4}\co... | We substitute $c = \cot(x)$ and we get
$$
\lim\limits_{x\to\pi/4} \cot(x)^{\cot(4x)}
=\lim\limits_{c\to 1}\; c^{\frac{c^4-6c^2+1}{4c^3-4c}}
$$
because
$$
\cot(4x) = \frac{\cot^4x-6\cot^2x+1}{4\cot^3x-4\cot x}
$$
Then we have
$$
\lim\limits_{c\to 1} \;c^{\frac{c^4-6c^2+1}{4c^3-4c}}
=\lim\limits_{c\to 1} \;\exp\left(\ln(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3394636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Showing $\lim_{x\to 0}\frac{\sin(x^2\sin\frac{1}{x})}{x}=0$ $$\lim_{x\to 0}\frac{\sin(x^2\sin\frac{1}{x})}{x}=\lim_{x\to 0 }\frac{x^2\sin\frac{1}{x}}{x}=\lim_{x\to 0} x\sin\frac{1}{x}=0$$
Is this solution right?
Thank you very much!
| $$\lim_{x\to 0}|\frac{\sin(x^2\sin\frac{1}{x})}{x}|=$$
$$\lim_{x\to 0 }|\frac{\sin(x^2\sin\frac{1}{x})}{x^2} (x)|\le $$
$$\lim_{x\to 0 }|\frac{(x^2\sin\frac{1}{x})}{x^2} (x)| =$$
$$\lim_{x\to 0}| x\sin(\frac{1}{x})|=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3395587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Number of possible polynomials
Let $a,b,c,d$ be four integers (not necessarily distinct) in the set $\{1,2,3,4,5\}$. Find the number of polynomials of the form $x^4+ ax^3 + bx^2 + cx +d$ which is divisible by $x+1$.
My Try:
Let $f(x) = x^4+ ax^3 + bx^2 + cx +d$, then $f(-1) = 0$. Thus $1+ (b+d) = c+a$. On counting ca... | The counting can be a bit simplified using your intermediate result as follows:
*
*You have $a-b+c-d = 1 \Leftrightarrow (a-1) + (5-b) + (c-1) + (5-d) = 9$
So, your question is equivalent to counting the number of integer solutions of
$$a' + b' + c' + d' = 9 \mbox{ with } a',b',c',d' \in \{0,1,2,3,4\}$$
Now, let th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3400404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
quick way to count Suppose we have 10 sticks with length 1-10, respectively. Pick three from them, how many triangles can we form?
I counted one by one and got 50. Is there a quick way? Any help would be appreciated.
| Call the three sticks $a < b < c$. Pick $c$ first. The other two must sum to more than $c$. Pick $a$ second. So we have $$\sum_{c=1}^{10} \sum_{a=1}^{c-1} \sum_{b=a+1}^{c-1} [a+b > c]$$
Working from the inside out you should be able to tighten the bound on $b$ to remove that Iverson bracket and replace the inner sum wi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3400992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Maximum value of $a+b+c$ in an inequality Given that $a$, $b$ and $c$ are real positive numbers, find the maximum possible value of $a+b+c$, if
$$a^2+b^2+c^2+ab+ac+bc\le1.$$
From the AM-GM theorem, I have
$$a^2+b^2+c^2+ab+ac+bc\geq 6\sqrt[6]{a^4b^4c^4} = 6\sqrt[3]{a^2b^2c^2} \\
6\sqrt[3]{a^2b^2c^2} \le1 \\
a^2b^2c^2 \l... | $$a+b+c=\sqrt{\sum_{cyc}(a^2+2ab)}\leq\sqrt{\sum_{cyc}\left(a^2+\frac{1}{2}a^2+\frac{3}{2}ab\right)}\leq\sqrt{\frac{3}{2}}.$$
The equality occurs for $a=b=c$ and $\sum\limits_{cyc}(a^2+ab)=1,$ which says that we got a maximal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3401308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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How to Solve This Exponential Limit without Derivate / L'Hôspital's Rule can someone teach me how can I solve this limit without using the L'Hopital's Rule?
$$\lim_{x\to 0} \left( \frac{2+x^{2}}{2-x^{2}} \right)^{\frac{1}{x^2}}$$
Thanks in advance.
| $$
\frac{2+x^2}{2-x^2}=1+\frac{2x^2}{2-x^2}=1+\frac{x^2}{1-\frac{x^2}{2}}
$$
and $y=\frac{x^2}{1-\frac{x^2}{2}}\to 0$, as $x\to 0$. Hence
$$
\left(1+\frac{x^2}{1-\frac{x^2}{2}}\right)^{\frac{1-\frac{x^2}{2}}{x^2}}=(1+y)^{1/y}\to e,
$$
as $x\to 0$.
Finally
$$
\left(\frac{2+x^2}{2-x^2}\right)^{1/x^2}=\left(\left(1+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3401448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Minimum number of $k$-partitions of a set of size $n$ to enumerate all $n \choose k$ combinations Given a set $\mathcal{S}$ of size $n$, let a $k$-partition of $\mathcal{S}$ be a grouping into $k$ disjoint classes $$(S_1 ,S_2,...,S_k)$$
Where the $S_i$ do not necessarily contain the same number of elements.
Let $C(S_1... | Rather than $N$ I'm going to use the OEIS convention $T(n, k)$ (for triangle, I think).
Misha Lavrov's answer gives the bound $$T(n, k) \ge \frac{\log n - \log(k-1)}{\log k - \log(k-1)}$$
Another, easier, lower bound uses the auxiliary function A152072 which gives the largest possible size of the Cartesian product $S_1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3403318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Is $g(x_1, x_2) = (\alpha - x_1)^2 + (\max \{\alpha, x_1\} + \beta - x_2)^2$ convex?
Let $\alpha \geq 0$ and $\beta \geq 0$. Can we prove or disprove the following function is convex on $x_2 \geq x_1 \geq 0$?
$$
g(x_1, x_2) = (\alpha - x_1)^2 + (\max \{\alpha, x_1\} + \beta - x_2)^2
$$
My Approach: It is clear that... | As you already pointed out, $(\max \{\alpha, x_1\} + \beta - x_2)^2$ might descrease as $x_1$ increases when $x_1 + \beta - x_2 < 0$. So, to show a function is non-convex, one only needs to find a counter example.
Try the following three points (find a large enough $n > 2$ so that $\alpha - \frac{1}{n}\beta > 0$, and l... | {
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"url": "https://math.stackexchange.com/questions/3403587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Proving the injectivity of function I have a subset $B = \{(x,y) : x^2 + y^2 \leq 1 \} \subseteq \mathbb{R}^2$.
I am trying to prove that function
$g(x,y) = (\frac{x^2 + y^2}{x^2 + y^2 + 1}x, \frac{x^2 + y^2}{x^2 + y^2 + 1}y)$
is an injective function $g: \mathbb{R}^2 \rightarrow B$.
I attempted to prove this by direc... | We have $g(10,0)=\frac{1000}{101}(1,0)\not\in B$. So, $g$ does not map the plane to the disc $B$. Nevertheless, $g$ is an injective self-map of the plane.
To prove injectivity, try to pass to the polar coordinates $x=r\cos\theta$, $y=r\sin\theta$. Then $$g(x,y)=\frac{r^3}{r^2+1}(\cos\theta,\sin\theta).$$
Assume that $g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3404526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Following Fibonacci Prove that $ F_ {n + 2} = \sqrt{\frac{F_n {F_ {n + 1} ^ 2}(3 {F_n} +4 {F_ {n + 1}}) + 1 }{{F_n} ^ 2 + {F_ {n + 1}} ^ 2}} $?
I discovered this property from an attempt to solve the following problem:
Defining $ A_n = \sqrt{{F_n} ^ 2 + {F_{n + 2}} ^ 2}, $ the numbers $ A_n, $ $ A_ {n + 1} $ and $ A_{... | To prove
$F_ {n + 2} = \sqrt{\frac{F_n {F_ {n + 1} ^ 2}(3 {F_n} +4 {F_ {n + 1}}) + 1 }{{F_n} ^ 2 + {F_ {n + 1}} ^ 2}}
$,
I would write it as
$F^2_{n + 2}(F_n ^ 2 + F_ {n + 1}^ 2)
= F_n F_ {n + 1} ^ 2(3 F_n +4 F_ {n + 1}) + 1
$
or
$(F_{n+1}+F_n)^2(F_n ^ 2 + F_ {n + 1}^ 2)
= F_n F_ {n + 1} ^ 2(3 F_n +4 F_ {n + 1}) + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3407394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Find all rational triplets $(a,b,c)$ that are roots of the equation $x^3+ax^2+bx+c=0.$
Find all rational triplets $(a,b,c)$ that are roots of the equation $$x^3+ax^2+bx+c=0$$
My work so far:
By Vieta's formulas we have
$a+b+c=-a\;(1),$
$ab+ac+bc=b\;(2),$ and
$abc=-c\;(3).$
From $(3),$ either $c=0$ or $ab=-1$. If $... | Two years too late
$$(x^3+ax^2+bx+c)-(x-a)(x-b)(x-c)=$$ $$c(a b +1)- (c (a+b)+b(a-1))x+ (2 a+b+c)x^2=0\tag 1$$ So $c=-2a-b$ replaced in $(1)$ leads to
$$-(2 a+b) (a b+1)+[2 a (a+b)+b (b+1)]x=0 \tag 2$$
If $\color{red}{b=-2a}$, then $a(a-1)=0$, that is to say $a=0$ or $a=1$.
If $a=0$, then $b=0$ and $c=0$
If $a=1$, then... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3407974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Two points on a curve that have a common tangent line I have a curve $y = x^3 - x^4$ with an oblique common tangent line across the top of the curve. What is the equation of the tangent?
Have spent hours on this but getting no-where.
| Generally, $y=x^3-x^4=ax+b$ have 4 roots for $x$.
$$x^3-x^4 = ax+b \Rightarrow (x-x_1)(x-x_2)(x-x_3)(x-x_4) = 0 $$
If $ax+b$ is a tangent line on 2 points of the curve, that means
$$x^3-x^4 = ax+b \Rightarrow (x-x_1)^2(x-x_2)^2 = 0 $$
Then we can have
$$
x^4 - x^3 +0\cdot x^2 + ax + b = 0
\Rightarrow
$$
$$
x^4 - 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3408898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove that $\sqrt[n]{n} < (1 + \frac{1}{\sqrt{n}})^2$ for all n in the naturals. I need to prove that $\sqrt[n]{n} < (1 + \frac{1}{\sqrt{n}})^2$ for all n in the naturals.
I started by using Bernoulli's inequality:
$(1+\frac{2}{\sqrt{n}}) < (1 + \frac{1}{\sqrt{n}})^2$
I can say that:
$(1+\frac{2}{\sqrt{n}}) = (1+\frac... | $$\begin{align}
\left(1+{1\over\sqrt n}\right)^{2n}
&=\left(1+{2\over\sqrt n}+{1\over n}\right)^n\\
&\gt\left(1+{2\over\sqrt n} \right)^n\\
&\ge1+{n\choose1}{2\over\sqrt n}+{n\choose2}\left(2\over\sqrt n\right)^2\\
&\gt1+0+2(n-1)\\
&=2n-1\\
&\ge n
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3409831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 5
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$\int \frac{1}{(x^2-4)^2}dx$ Calculate:
$$\int \frac{1}{(x^2-4)^2}dx.$$
I tried Partial Fractions method first I write:
$$\frac{1}{(x^2-4)^2}=\frac{A}{X-2}+\frac{Bx+C}{(x-2)^2}+\frac{D}{x+2}+\frac{Ex+F}{(x+2)^2}.$$
We have:
$$A(x-2)(x+2)^2+(Bx+C)(x+2)^2+D(x+2)(x-2)^2+(Ex+F)(x-2)^2=1.$$
$$(A+B+D+E)x^3+(4A-2A+4B+C-4D+2D... | Here is a compact approach with the substitution $x=2\cosh t$,
$$I= \int \frac{1}{(x^2-4)^2}dx$$
$$=\frac{1}{8} \int \text{csch}^3 tdt
= -\frac{1}{8} \int \text{csch} t \>d(\coth t)$$
$$= -\frac{1}{8} \left(\text{csch} t \coth t + \int \text{csch} t \coth^2 tdt \right)$$
$$= -\frac{1}{8} \left(\text{csch} t \coth t + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3409950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Find the minimum of :$P=a+b+c-ab-bc-ca$ soluLet $a,b,c$ be positive real numbers and $a+b+c+abc=4$.
We can rewrite the first equation as $a+b+c=4-abc.$ Then,
\begin{align*}
P&=a+b+c-ab-bc-ca\\&=(4-abc)-ab-bc-ca\\&=4-abc-ab-bc-ca-(a+b+c+1)+(a+b+c+1)\\&=4-(abc+ab+bc+ca+a+b+c+1)+a+b+c+1\\&=4-(a+1)(b+1)(c+1)+a+b+c+1\\&=5+a... | We may suppose that $a\le b\le c$.
Supposing that $a\gt 1$ gives $4=a+b+c+abc\gt 4$, so we have $a\le 1$.
Supposing that $c\lt 1$ gives $4=a+b+c+abc\lt 4$, so we have $c\ge 1$.
Now using $$b=\frac{4-a-c}{1+ac}$$
we have
$$\begin{align}P&=a+b+c-ab-bc-ca
\\\\&=a+c-ac+(1-a-c)b
\\\\&=a+c-ac+(1-a-c)\cdot \frac{4-a-c}{1+ac}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3410321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\left(1+\sin\frac{\pi}{5}+i\cos\frac{\pi}{5}\right)^5+i\left(1+\sin\frac{\pi}{5}-i\cos\frac{\pi}{5}\right)^5$
Evaluate
$$\left(1+\sin\frac{\pi}{5}+i\cos\frac{\pi}{5}\right)^5+i\left(1+\sin\frac{\pi}{5}-i\cos\frac{\pi}{5}\right)^5$$
I did this by $$\left(1+\sin\frac{\pi}{5}+i\cos\frac{\pi}{5}\right)^5=\lef... | Say $z = \sin\frac{\pi}{5}+i\cos\frac{\pi}{5}$.
$\big($We have $-iz = \cos\frac{\pi}{5}-i\sin\frac{\pi}{5} = \cos\frac{-\pi}{5}+i\sin\frac{-\pi}{5}$, so $\boxed{z^5 = -i}$ by De'Moivre formula.$\big) $
Notice that $\bar{z}={1\over z}$.
Then your expression is \begin{eqnarray}w &=& (1+z)^5+i(1+\bar{z})^5\\
&=& (1+z)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3413692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Evaluate the following limit probably by Riemann Sum: $\lim_{n\to\infty} \sum_{k=1}^{n} \sin\left(\frac{(2k-1)a}{n^2}\right)$ I've tried solving the following limit and I think that it might be possible to transform the sequence into a Riemann Sum.
$$\lim_{n\to\infty} \sum_{k=1}^{n} \sin\left(\frac{(2k-1)a}{n^2}\right)... | Try using complex numbers:
$$\sum\limits_{k=1}^{n} \sin\left(\frac{(2k-1)a}{n^2}\right)=\Im\left(\sum\limits_{k=1}^{n}z^{2k-1}\right)=
\Im\left(\frac{z (z^{2n} - 1)}{z^2 - 1}\right)= ...$$
where $z=e^{i\cdot \frac{a}{n^2}}$
$$...=\Im\left(\frac{\left(\cos{\frac{a}{n^2}} + i\sin{\frac{a}{n^2}}\right)\left(\cos{\frac{2a}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3418821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Understand roots of $(z+i)^2=(\sqrt3+i)^3$ I'm trying to solve the equation
$$(z+i)^2=(\sqrt3+i)^3$$ but I don't know how to extract
the roots
$$(z+i)^2=(\sqrt3+i)^3 \rightarrow (z+i)^2=8i \rightarrow z^2+(2i)z-(8i+1)=0$$
$z_{1,2}=-i \pm \sqrt{8i}$.
According to my book the solutions are $2+i$ and $-2-3i$
and I think t... | You have $z_{1,2} = -i + K$ where $K^2 = 8i$.
So find $K$ where $K^2 = 8i$.....
Let $K = a+bi$ so $K^2 = (a^2 -b^2) + 2abi = 0 + 8i$ so $a^2-b^2 =0$ and $2ab=8$
So solve for $a$ and $b$.....
$a^2 - b^2 =0$ means $a=\pm b$ and $2ab = 8$ means $ab=4$ means $a$ and $b$ are both the same sign. So $a=b$ and $ab =a^2=4$ so ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3419620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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How to solve a difficult system Problem: solve in $\mathbb R^{3}$ this system
$$
\begin{cases}y(1+x+x^2+x^3)=\dfrac{z}{16}\\ y^2x(1+x+2x^2+x^3+x^4)=\dfrac{2z+17}{16}\\
y^3x^3(1+x+x^2+x^3)=\dfrac{z}{16}\\
y^4x^6=1
\end{cases}
$$
Wolfram alpha gives $(x,y,z)=\left(4,\dfrac{1}{8},170\right)$.
I don't know how to solve i... | The fourth equation is redundant. From the first and the third equation (by equating them and noting that $1+x+x^2+x^3=0$ iff $x=-1$, but $x=-1$ doesn't yield a real solution), you can already deduce that $y^2x^3=1$. If you set $x=t^2$ and $y=1/{t^3}$, then the first and the second equations are now $$\frac{z}{16}=t^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3420374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Writing polynomial using powers of (x-a) I have a question related to writing a polynomial using powers of binomial of form $(x-a).$
I found an example: polynomial $P(x) = x^4 + 2x^3-3x^2-4x+1$ can be written as
$ (x+1)^4-2(x+1)^3-3(x+1)^2+4(x+1)+1$ using powers of $(x+1)$ and Horner's Method. How do we obtain this r... | This is similar to changing base of a number system.
So you can use long division.
$$P(x) = x^4 + 2x^3-3x^2-4x+1$$
To express it as powers of $(x+1)$, divide $P(x)$ by the new base $(x+1)$
$$\begin{align}
x^4 + 2x^3-3x^2-4x+1 &= (x+1)(x^3+x^2-4x) + \color{blue}{1} \\
x^3+x^2-4x &= (x+1)(x^2-4)+ \color{blue}{4}\\
x^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3420690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find integers $x$ and $y$ such that $8^x-9^y=431$ Find integers $x$ and $y$ such that $8^x-9^y=431$
My working:
By taking mod 9 and 16, I got $x$ odd and $y$ even.
Also $8^x>431\implies x\ge 3$
For $x=3$ I got $y=2$
| $$ 8^{y+d} - 9^y = 431 $$
$$ 8^d = \frac{431 + 9^y}{8^y} $$
$$ 8^d = (\frac{9}{8})^y+\frac{431}{8^y} $$
$$ 8^d \sim (\frac{9}{8})^y $$
$$ 3d\cdot ln(2) \sim y \cdot(2ln(3)-3ln(2)) $$
$$ d \sim y \cdot(\frac{2\cdot ln(3)-3\cdot ln(2)}{3\cdot ln(2)}) $$
$$ d \sim y \cdot 0.0566416671474374543024926292985... $$
$$ \frac{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3421712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Heptadecagon Derivation I am currently very interested in the derivation of the constructability of the 17-gon by Carl Friedrich Gauß.
Has someone got an easy explanation for the solution of
$$x^{17} - 1=0?$$
That was the equation he solved with which he showed
\begin{align}\cos \frac{360^\circ}{17}&=\frac{1}{16}\left... | This is an elementary proof.
Let $\varphi=\frac\pi{17}$,
$$S=-\sum_{n=1}^8(-1)^n\cos(n\varphi)$$
Multiplication by $2\cos(\varphi/2)$ gives:
\begin{align}
2S\cos\left(\frac\varphi 2\right)
&=-\sum_{n=1}^8(-1)^n\left(\cos\left(\frac{2n-1}2\varphi\right)-\cos\left(\frac{2n+1}2\varphi\right)\right)\\
&=\cos\left(\frac 12\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3427186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find all $n≥1$ natural numbers such that : $n^{2}=1+(n-1)!$ Problem :
Find all $n≥1$ natural numbers such that : $n^{2}=1+(n-1)!$
My try :
$n=1$ we find : $1=1+1$ $×$
$n=2$ we find : $4=1+1$ $×$
$n=3$ we find : $9=1+2$ $×$
$n=4$ we find : $16=1+6$ $×$
$n=5$ we find : $25=1+24$ $√$
Now how I prove $n=5$ only the ... | $n^2 = 1 + (n-1)!$
$n^2 -1 = (n-1)!$
$(n-1)(n+1) = (n-1)!$ (If we assume $n>1$)
$n+1 = \frac {(n-1)!}{n-1} = (n-2)!$.
$n-2 + 3 = (n-2)!$ (If we assume $n > 2$)
$1 + \frac {3}{n-2} = \frac {(n-2)!}{n-2} = (n-3)!$ is an integer.
So $n-2|3$. But $3$ is prime. So either $n-2=1$ or $n-2 = 3$. But we assume $n > 2$ so $n=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3427795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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Find the units digit of $572^{42}$ The idea of this exercise is that you use the modulus to get the right answer.
What I did was:
$$572\equiv 2\pmod {10} \\
572^2 \equiv 2^2 \equiv 4\pmod{10} \\
572^3 \equiv 2^3 \equiv 8\pmod{10} \\
572^4 \equiv 2^4 \equiv 6\pmod{10} \\
572^5 \equiv 2^5 \equiv 2\pmod{10} \\
572^6 \... | Since you found $572 \equiv 2\pmod{10} $ you have to check only what is $$2^{42} \equiv ?\pmod{10} $$
Since $2^5 \equiv 2$ we have $$2^{42} = 2^2\cdot (\color{red}{2^5})^8 \equiv 2^2\cdot \color{red}{2}^8 \equiv 2^{10} \equiv (\color{blue}{2^5})^2 \equiv \color{blue}{2}^2\equiv 4 \pmod {10}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3428727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
If a variable chord of the hyperbola subtend a right angle at the centre, find the radius of the circle it is tangent to
If a variable line $x\cos\alpha+y\sin\alpha=p$ which is a chord of the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$, $(b>a)$ subtend a right angle at the centre of hyperbola,then prove that it al... | The line equation seems a bit of a distraction. Here's how I'd solve the problem.
Suppose $H$ and $K$ determine a chord of the hyperbola that subtends a right angle at its center (the origin), $O$. We can write
$$H = (h \cos\phi, h \sin\phi) \qquad K = (k \sin\phi,-k \cos\phi)$$
where $h:=|OH|$, $k:=|OK|$, and $\phi$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3429601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Let $f(x)=(x+1)(x+2)(x+3)(x+4)+5$ where $x\in[-6,6]$. If the range of the function is [a,b] where $a,b\in N$, then find the value of (a+b) Let $f(x)=(x+1)(x+2)(x+3)(x+4)+5$ where $x\in[-6,6]$. If the range of the function is [a,b] where $a,b\in N$, then find the value of (a+b)
My attempt is as follows:-
Rewrite $f(x)=g... | The function is symmetric about the line $x = -2.5$
We could write
$f(x) = ((x+2.5) + 1.5)((x+2.5)+0.5)((x+2.5)-0.5)((x+2.5) - 1.5) + 5$
Which equals
$f(x) = ((x+2.5)^2 - 1.5^2)((x+2.5)^2-0.5^2) + 5$
Now lets do the substituion $u = (x+2.5)^2$
$(u - 2.25)(u-0.25) + 5$
Which has its minimum when $u = \frac {5}{4}$
$(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3432833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Prove convergence of $n \cdot \left(\sqrt{1+ \frac{1}{n}} -1\right)$ I am working on some old analysis exams and i got stuck on this exercise :
Using the epsilon definition show that $a_{n} = n \cdot \left(\sqrt{1+ \frac{1}{n}} -1\right)$ converges and determine its limit.
Knowing that the limit is 1/2, I know need ... | You started well. You need to show that $ \left|\frac{1}{\sqrt{1+ \frac{1}{n}} + 1} - \frac{1}{2} \right|<\epsilon$ for $n>N$ where $N$ is some number.
$$ \left|\frac{1}{\sqrt{1+ \frac{1}{n}} + 1} - \frac{1}{2}\right|=\left|\frac{\sqrt n}{\sqrt{n+1} +\sqrt n} - \frac{1}{2} \right|=\left|\frac{2\sqrt n -\sqrt{n+1} -\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3432944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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How to compute $\sum_{n=1}^\infty\frac{(-1)^nH_{n/2}}{n^4}$? How to prove
$$\sum_{n=1}^\infty\frac{(-1)^nH_{n/2}}{n^4}=\frac18\zeta(2)\zeta(3)-\frac{25}{32}\zeta(5)?$$
I came across this series while I was working on a nice integral $\int_0^1\frac{\ln(1+x)\operatorname{Li}_3(-x)}{x}dx$ and because I managed to calcu... | Different approach
$$S=\sum_{n=1}^\infty\frac{(-1)^nH_{n/2}}{n^4}=-H_{1/2}+\sum_{n=2}^\infty\frac{(-1)^nH_{n/2}}{n^4},\quad H_{1/2}=2\ln2-2$$
use the fact that
$$\sum_{n=2}^\infty f(n)=\sum_{n=1}^\infty f(2n)+\sum_{n=1}^\infty f(2n+1)$$
$$\Longrightarrow S=2-2\ln2+\frac1{16}\sum_{n=1}^\infty\frac{H_{n}}{n^4}-\sum_{n=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3433136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given a polynomial f(x)=x(x+1)(x+2)(x+3)+1 Given a polynomial $f(x)=x(x+1)(x+2)(x+3)+1$ and $p$ an odd prime, prove that it exists a number $n$ such that $p$ divides $f(n)$ if and only if it exists an integer $m$ such that $p$ divides $m^2-5$.
| You have that
$$f(x) = x(x + 1)(x + 2)(x + 3) + 1 \tag{1}\label{eq1A}$$
Since $p$ is an odd prime, then any $n$ with $p \mid f(n)$ is true iff only $p \mid 16f(n)$. Multiply $f(n)$ by $16$ and distribute the powers of $2$ to each factor with $n$ in it to get
$$16f(n) = (2n)(2n + 2)(2n + 4)(2n + 6) + 16 \tag{2}\label{eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3433261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Convergence of $\left( \frac{1}{1} \right)^2+\left( \frac{1}{2}+\frac{1}{3} \right)^2+\cdots$ Does the following series converge? If yes, what is its value in simplest form?
$$\left( \frac{1}{1} \right)^2+\left( \frac{1}{2}+\frac{1}{3} \right)^2+\left( \frac{1}{4}+\frac{1}{5}+\frac{1}{6} \right)^2+\left( \frac{1}{7}+\f... | Notice, that:
$$\left(\frac{1}{1}\right)^2+\left(\frac{1}{2} + \frac{1}{3}\right)^2 + \dots<\left(\frac{1}{1}\right)^2+\left(\frac{2}{2}\right)^2+\left(\frac{3}{4}\right)^2+\left(\frac{4}{7}\right)^2=2+\sum_{n=2}^{\infty}\left(\frac{n}{\frac{n(n-1)}{2}+1}\right)^2<2+\sum_{n=2}^{\infty}\left(\frac{2}{n-1}\right)^2$$
Sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3436804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
} |
Solve for positive real solutions of cyclic equations $x+y^2+2xy=9$, $y+z^2+2yz=47$, $z+x^2+2xz=16$
Solve over positive reals $$x+y^2+2xy=9$$
$$y+z^2+2yz=47$$
$$z+x^2+2xz=16$$
With standard manipulation we get that $x+y+z=8$. Thus $x=8-y-z$ and we have two equations,
$$(8-y-z)^2+y^2+2(8-y-z)y=9$$
$$y+z^2+2yz=47$$
Whi... | $\begin{cases}
x+y^2+2xy=9\\
y+z^2+2yz=47\\
z+x^2+2xz=16
\end{cases} \implies
\begin{cases}
(2 x + 2 y)^2 - (2 x - 1)^2 = 35\\
(2 y + 2 z)^2 - (2 y - 1)^2 = 187\\
(2 z + 2 x)^2 - (2 z - 1)^2 = 63
\end{cases} \implies
\begin{cases}
x=1\\
y=2\\
z=5
\end{cases}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3437528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Convergence of a Cauchy Product I have the following series obtained via the Cauchy Product of the alternating harmonic series with itself
$\left( \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \right ) \cdot \left( \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \right ) = \sum_{n=1}^{\infty} \sum_{k=1}^{n} \frac{(-1)^k}{k} \cdot \frac{(-... | Since
$$\sum_{n=1}^\infty \frac{(-1)^n}{n} = \sum_{n=0}^\infty \frac{(-1)^{n+1}}{n+1},$$
we have, using the Cauchy product formula,
$$\begin{align}\left(\sum_{n=1}^\infty \frac{(-1)^n}{n}\right)\left(\sum_{n=1}^\infty \frac{(-1)^n}{n}\right) &= \left(\sum_{n=0}^\infty \frac{(-1)^{n+1}}{n+1}\right)\left(\sum_{n=0}^\inf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3438301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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How do I prove that $\sum_{i=3}^{n} \frac{i-2}{\binom{i}{2}} < \frac n 4$ for all natural $n > 3$? How do I prove:
$$\frac{3-2}{\binom 3 2} + \frac {4 - 2}{\binom 4 2} + \dots + \frac{n-2}{\binom n 2} < \frac n 4$$
I tested this sum on a variety of $n$ from $2$ to $100$ and they all seem to be less than $\frac n 4$, ho... | For any $n \ge 8$, we have $$\dfrac{n-2}{\binom{n}{2}} = \dfrac{n-2}{\tfrac{n(n-1)}{2}} = \dfrac{2}{n} \cdot \dfrac{n-2}{n-1} \le \dfrac{2}{n} \le \dfrac{1}{4},$$
and thus, $$\sum_{k = 3}^{n-1}\dfrac{k-2}{\binom{k}{2}} < \dfrac{n-1}{4} \implies \sum_{k = 3}^{n}\dfrac{k-2}{\binom{k}{2}} < \dfrac{n}{4}.$$
So if you can p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3438956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Understanding derivation of Catalan Numbers I encountered a derivation for finding the nth Catalan Number $C_n$ using generating functions in the book Diskrete Strukturen, Band 1 (Steger 2007, p.178).
Given is the recursive function $C_0 := 1$, $C_n := \sum_{k=1}^n C_{k-1}C{n-k} \;\;\; (\forall n \geq 1)$.
The first pa... | For the third equality,
\begin{align*}
-\frac{1}{2}\frac{\frac{1}{2} (\frac{1}{2} - 1) (\frac{1}{2} - 2) \ldots(\frac{1}{2} - n)}{ (n+1)!}(-4)^{n+1}&=
(-1)^n\cdot\frac{1}{2}\cdot\left(\frac{1}{2}\right)^{n+1}\frac{1(2-1)(4-1)\cdots(2n-1)}{(n+1)!}(-1)^n4^{2n+2}\\
&= 2^{-n-2}4^{2n+2}\frac{1\cdot 3\cdot 5\cdots 2n-1}{(n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3440813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve the equation $\cos^{-1}\frac{x^2-1}{x^2+1}+\tan^{-1}\frac{2x}{x^2-1}=\frac{2\pi}{3}$ $\cos^{-1}\dfrac{x^2-1}{x^2+1}+\tan^{-1}\dfrac{2x}{x^2-1}=\dfrac{2\pi}{3}$
Let's first find the domain
$$-1<=\dfrac{x^2-1}{x^2+1}<=1$$
$$\dfrac{x^2-1}{x^2+1}>=-1 \text { and } \dfrac{x^2-1}{x^2+1}<=1$$
$$\dfrac{x^2-1+x^2+1}{x^2+1... | Let $x=\tan\dfrac t2$ for $t\in(-\pi,\pi)$. The equation reads
$$\tan^{-1}(-\tan t)+\cos^{-1}(-\cos t)=\frac{2\pi}3$$
or
$$\tan^{-1}(\tan t)+\cos^{-1}(\cos t)=\frac{\pi}3.$$
By lazily checking on a plot, we see that the function has a null slope in the negatives, where there are no solutions.
In the positives, the slo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3442053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Prove that f(2000x)=2000f(x) Given function $f:\mathbb{R}\rightarrow \mathbb{R}$ such that $f(x+y+2xy)=f(x)+f(y)+2f(xy)$. Prove that $f(2000x)=2000f(x)$.
Letting $x=y=1$ yields $f(4)=4f(1)$, which means that $f(n)=nf(1)$. But I cannot prove this by induction.
And is it possible to prove that this function is monotonou... | We first prove by induction that we do in fact have that $f(n) = nf(1)$ for all integers $n$.
Taking $x = y = 0$ in the functional equation, we get that $f(0) = 4f(0)$, and so $f(0) = 0$.
Taking $y = -1$ in the functional equation gives us that
$$
f(-x - 1) = f(x) + f(-1) + 2f(-x) \label{a}\tag{1}
$$
for all real num... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3443441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
An Analytic Inequality $$\frac 12\ln(AC-B^2)\leq\frac{1}{2\pi}\int_{0}^{2\pi}\ln(A\sin^2\theta+2B\sin\theta\cos\theta+C\cos^2\theta)d\theta(A,C>0)$$
I see this inequality from a book about Honeycomb structure.It can be proved by dividing $[0,2\pi]$.
$$|AC-B^2|^{\frac n2}\leq\prod_{l=1}^{n}\left[A\sin^2\frac{\pi(2l-1)}{... | The integral can be evaluated in closed form. The expression under the logarithm is $$A\frac{1-\cos 2\theta}{2}+B\sin 2\theta+C\frac{1+\cos 2\theta}{2}=\frac{A+C}{2}+\frac{C-A}{2}\cos 2\theta+B\sin 2\theta,$$ which is equal to $A'+B'\cos(2\theta-\phi)$ with $$A'=\frac{A+C}{2},\qquad B'=\sqrt{B^2+\Big(\frac{A-C}{2}\Big)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3444725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prove that $(x + \sqrt[3]{abc})^3 \le (x + a)(x + b)(x + c) \le ( x + \frac{a + b + c}{3})^3$ Let $x,$ $a,$ $b,$ $c$ be nonnegative real numbers. Prove that
$$(x + \sqrt[3]{abc})^3 \le (x + a)(x + b)(x + c) \le \left( x + \frac{a + b + c}{3} \right)^3.$$
I know that this problem is a RHS-AM-GM-HM problem, but I am unsu... | Using AM
$$\left( x + \frac{a + b + c}{3} \right)^3=
\left(\frac{(x+a) + (x+b) + (x+c)}{3} \right)^3\overset{AM}{\geq} \\
\left(\sqrt[3]{(x+a)(x+b)(x+c)}\right)^3=
(x+a)(x+b)(x+c)$$
Using HUYGEN’S INEQUALITY (which is easy to prove using AM-GM)
$$(x+a)(x+b)(x+c)=x^3\left(1+\frac{a}{x}\right)\left(1+\frac{b}{x}\right)\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3445233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find $\lim_{x\to 0}\frac{(a^x-1)^3}{\sin(x\log a)\log(1+x^2\log a^2)}$
If $$f(x)= \frac{(a^x-1)^3}{\sin(x\log a)\log(1+x^2\log a^2)}$$ is continuous at $x=0$ then find $f(0)$
$$
f(0)=\lim_{x\to0}\frac{(a^x-1)^3}{\sin(x\log a)\log(1+x^2\log a^2)}\\
=\lim_{x\to0}\big[\frac{a^x-1}{x}\big]^3.\frac{x\log a}{\sin(x\log a)}... | We can use that
$$\frac{(a^x-1)^3}{\sin(x\log a)\log(1+x^2\log (a^2))}=\frac{(a^x-1)^3}{x^3}\frac{x^3}{\sin(x\log a)\log(1+x^2\log (a^2))}$$
and by standard limits
$$\frac{(a^x-1)^3}{x^3} \to \log^3 a$$
$$\frac{x^3}{\sin(x\log a)\log(1+x^2\log (a^2))}=\frac{x\log a}{\sin(x\log a)}\frac{x^2\log (a^2)}{\log(1+x^2\log (a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3445428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solve $2 \cos^2 x+ \sin x=1$ for all possible $x$ $2\cos^2 x+\sin x=1$
$\Rightarrow 2(1-\sin^2 x)+\sin x=1$
$\Rightarrow 2-2 \sin^2 x+\sin x=1$
$\Rightarrow 0=2 \sin^2 x- \sin x-1$
And so:
$0 = (2 \sin x+1)(\sin x-1)$
So we have to find the solutions of each of these factors separately:
$2 \sin x+1=0$
$\Rightarrow \sin... | Yes your solution is very nice and correct, as a slightly different alternative
$$2\cos^2(x)+\sin (x)=1 \iff 2(1-\sin x)(1+\sin x)+\sin x-1=0 $$
$$\iff (\sin x-1)(-2-2\sin x)+(\sin x-1)=0 \iff (\sin x-1)(-1-2\sin x)=0$$
which indeed leads to the same solutions, or also from here by $t=\sin x$
$$2-2 \sin^2 x+\sin x=1 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3445884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
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How do we prove this inequality? Suppose $a,b,c > 0$. Prove that
$$\frac{a^2}{b^2} +\frac{b^2}{c^2} + \frac{c^2}{a^2} \geq \frac ab + \frac bc + \frac ca.$$
I've tried multiplying everything by the denominator and then I tried to use the rearrangement inequality, but it didn't yield the result I was looking for. I coul... | After clearing the denominators we have to prove that
$$a^4c^2+b^4a^2+b^2c^4\geq abc(a^2c+ab^2+bc^2)$$
Now use that $$x^2+y^2+z^2\geq xy+yz+zx$$
This is $$(a^2c)^2+(bc^2)^2+(ab^2)^2\geq a^2cab^2+a^2cbc^2+ab^2bc^2=abc(a^2b+ac^2+b^2c)$$
Since$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=\frac{a^2c+ab^2+a^2b}{abc}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3446925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Compute $\int\frac{du}{\sqrt{a^2-u^2}}$ Please, can you explain why isn't the $$\int\frac{du}{\sqrt{a^2-u^2}} = \int\frac{du}{a\sqrt{1-(u/a)^2}}=\frac1a\sin^{-1}\bigg(\frac{u}{a}\bigg)+c$$ if $a > 0$ is a positive constant.
| Let $v = \dfrac ua \implies u = a\cdot v$, and $\mathrm du = a\cdot\mathrm dv$.
Substitute in original expression,
$$\require{cancel}\begin{align}\int\dfrac{\mathrm du}{\sqrt{a^2 - u^2}} &\equiv \int\dfrac{a\cdot\mathrm dv}{\sqrt{a^2 - a^2v^2}}\\ &= \int\dfrac{a\cdot\mathrm dv}{a\sqrt{1 - v^2}}\\ &= \int\dfrac{\cancel{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3447178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Prove that for all $n ≥ 1$, $\sum_{i=1}^{n}\frac{1}{i^2}≤2 - \frac{1}{n}$
Prove that for all $n ≥ 1 $
$$\sum_{i=1}^{n}\frac{1}{i^2}≤2 - \frac{1}{n}$$
My attempt:
By induction.
Base case: $n = 1$
$$1 ≤ 2 - 1 = 1$$
Induction step:
Suppose it is true for $n$, i.e
$$\sum_{i=1}^{n}\frac{1}{i^2}≤2 - \frac{1}{n}$$
Adding ... | Proving that
$$\sum_{k=1}^{n} \frac{1}{k^2} <2-\frac{1}{n}.$$
Notice that $$\frac{1}{k^2} <\frac{1}{k(k-1)}=\frac{1}{k-1}- \frac{1}{k},~~~ k \ge 2.$$Starting the telescopic summing from $k=2$ we get
$$\sum_{k=2}^n \frac{1}{k^2} \le 1-\frac{1}{n} \implies \sum_{k=1}^{n} \frac{1}{k^2} <2-\frac{1}{n}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3451379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Smallest number of points required to estimate value of integral Find the smallest number of points required to estimate the value of $$\int_{0}^{1} sin(x^2\pi)dx$$ with error less than $10^{-8}$ using the Composite Simpson method. In the estimation n, use the absolute maximum of the corresponding derivative of $(1+x^2... | It might be easier than a direct computation of the sequence of derivative expressions to find the coefficient of $s^4$ in $f(x+s)=\sin(\pi(x+s)^2)$.
\begin{align}
\sin(\pi x^2 + \pi(2x+s)s)
&=\sin(\pi x^2)\cos(\pi(2x+s)s)+\cos(\pi x^2)\sin(\pi(2x+s)s)
\\
&=\sin(\pi x^2)\left[1-\frac12\pi^2(2x+s)^2s^2+\frac1{24}\pi^4(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3451907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that $\frac{(m_a)^4+(m_b)^4+(m_c)^4}{a^4+b^4+c^4} = \frac{9}{16}$ Let $ m_a, m_b $ and $ m_c $ be the medians relative to the $ a, b, c $ sides of a triangle, show that:
$$\frac{(m_a)^4+(m_b)^4+(m_c)^4}{a^4+b^4+c^4} = \frac{9}{16}$$
What i tried:
ust use Stewart’s theorem. We have $m=n=a/2,d=m_a$ so
$$\frac{a^3}{... | $$\begin{align}
m_a^4+m_b^4+m_c^4&=\frac1{16}\sum_\text{cyc.}(2b^2+2c^2-a^2)^2\\
&=\frac1{16}\sum_\text{cyc.}4b^4+4c^4+a^4+8b^2c^2-4a^2b^ 2-4c^2a^2\\
&=\frac1{16}(9a^4+9b^4+9c^4).
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3453618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Use DeMoivre's Theorem to prove $ \cos 5x = 16 \cos^5x - 20\cos^3x+5\cos x$ I need to prove the following equalities using DMT:
$ \cos 5x = 16 \cos^5x$$ - 20\cos^3x$$+5\cos x$
and
$ \sin 5x = 16 \sin^5x$$ - 20\sin^3x$$+5\sin x$
Can someone help me with this question?
(Attempt: $\cos5x+i\sin5x=(\cos x+i\sin x)^5$)
| By DeMoivre's Theorem:
$$\begin{equation}\begin{aligned}
cos\,5x &= cos\,5x + i\,sin\,5x \\
&= (cos\,x + i\,sin\,x)^5 \\
&= cos^5\,x + 10\,cos^3\,sin^2\,x\,i + 5\,cos\,x\,sin^4\,x\,i^4\\
&= cos^5\,x - 10\,cos^3 x\,sin^2\,x + 5\,cos\,x\,sin^4\,x \\
&= cos^5\,x - 10\,cos^3\,x (1 - cos^2\,x) + 5\,cos\,x (1 - cos^2\,x)^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3454051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Knowing that $a,b,c\ge0$, prove that $\sum_\text{cyc}a^2 \cdot \left[\sum_\text{cyc}\frac{1}{(b - c)^2}\right] \ge\frac{11 + 5\sqrt5}{2}$.
Knowing that $a$, $b$ and $c$ are non-negatives, prove that
a/ $$(bc + ca + ab) \cdot \left[\frac{1}{(b - c)^2} + \frac{1}{(c - a)^2} + \frac{1}{(a - b)^2}\right] \ge 4$$
b/ $$(a^2... | One may consider the Buffalo Way because there are terms like $(a-b)$ which get simplified by the Buffalo substitution:
a) By cyclicity of the inequality we may assume $a\le b, c$. Hence, there are $x,y\geq 0$ such that $b=a+x$ and $c=a+y$. In fact, by well-definedness of your expression, we know that $x>0,y>0$ and $x\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3454633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the determinant of A find the determinant of A using only determinant properties
$$
A=\begin{bmatrix}
1 & 3 & -4 \\
-2 & 1 & 2 \\
-9 & 15 & 0 \\
\end{bmatrix}
$$
| Row reduction does not change the determinant:
$$
\begin{bmatrix} 1 & 3 & -4 \\ -2 & 1 & 2 \\ -9 & 15 & 0 \\ \end{bmatrix}
\to
\begin{bmatrix} 1 & 3 & -4 \\ 0 & 7 & -6 \\ 0 & 42 & -36 \\ \end{bmatrix}
\to
\begin{bmatrix} 1 & 3 & -4 \\ 0 & 7 & -6 \\ 0 & 0 & 0\\ \end{bmatrix}
$$
The determinant is zero because the last m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3454786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Mathematical Induction Involving The Floor Function I need to prove the identity $\lfloor \sqrt{n}+\sqrt{n+1}\rfloor=\lfloor\sqrt{4n+2}\rfloor$ for all natural numbers $n$.
I wanted to use mathematical induction. The identity is true for $n=1$. Then, I assume $$\lfloor \sqrt{k}+\sqrt{k+1}\rfloor=\lfloor\sqrt{4k+2}\rf... | The identity is true for $n = 0$, so consider $n \ge 1$. Also, let
$$m = \sqrt{n} + \sqrt{n + 1} \tag{1}\label{eq1A}$$
With $m \gt 0 \; \to \; m = \sqrt{m^2}$, we get
$$\begin{equation}\begin{aligned}
m & = \sqrt{\left(\sqrt{n} + \sqrt{n + 1}\right)^2} \\
& = \sqrt{n + 2\sqrt{n(n+1)} + n + 1} \\
& = \sqrt{2n + 1 + 2\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3456333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
matrix representations of complex, dual and split-complex numbers For complex, dual and split-complex numbers there are matrix representations:
$$a+b \cdot i \equiv a\begin{pmatrix} 1 & 0\\0 & 1 \\
\end{pmatrix} +b\begin{pmatrix}0 & -1\\1 & 0 \\
\end{pmatrix}, i^2=-1,$$
$$a+b \cdot \epsilon \equiv a\begin{pmatr... | In general, any matrix $B$ that you choose as the imaginary unit will have a minimal polynomial $p(x) $ such that $p(B) =0$. Any first or second degree minimal polynomial is possible. The resulting ring is $\mathbb R[x] /(p(x)) $. You can get rings that are not isomorphic to any of the three you mention this way, for e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3457471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to remove square roots from denominator in $\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}$? I have this math problem: Remove all square roots in the denominator of
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = ?$
The obvious solution is to multiply the fraction with $\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}-\sqrt{y}}$.
$\f... | Multiply denominator and numerator with $\sqrt x+\sqrt y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3458074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find the limit of $\sqrt[n]{n^2 + n}$ To find the limit I got the $\sqrt[3n]{n^2+n}$
Particularly, $\sqrt[3n]{n^2+n} \ge 1 \rightarrow \sqrt[3n]{n^2+n} = 1 + d_n$ where $d_n\ge 0$.
According to the Bernoulli's rule
$\sqrt{n^2+n} = (1+d_n)^n \ge d_n\cdot n \rightarrow d_n \le \frac{\sqrt{n^2+n}}{n}$
The $\frac{\sqrt{n^... | If your limit exists, say it is equal to $L$. Then
$$
\begin{split}
\ln L &= \ln \left( \lim_{n \to \infty} \left(n^2+n\right)^{1/n} \right) \\
&= \lim_{n \to \infty} \ln \left( \left(n^2+n\right)^{1/n} \right) \\
&= \lim_{n \to \infty} \frac{\ln \left(n^2+n\right)}{n} \\
&= \lim_{n \to \infty} \left[... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3460379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 0
} |
How to find the overlapping area of a semicircle and a rectangle I am being asked to calculate the pink area that is overlapping between the semicircle and the rectangle. I was only given the radius of the circle (5), the equation of a circle $(x^2+y^2 = r^2).$ I have found the intersection points between the circle an... | There are several ways you could approach this problem, but I think this would be the most straight-forward:
*
*Split up the pink region into two: $0\le x\le 4,$ and $4\le x\le 5.$ For the first region, you have a rectangle of area $3\cdot 4=12.$
*For the other region, you can perform an integral. You have that $5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3462889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
If $ab+bc+ca\ge1$, prove that $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{\sqrt{3}}{abc}$ The following problem is from CHKMO 2018 Problem 1:
If $ab+bc+ca\ge1$, prove that $$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{\sqrt{3}}{abc}$$
I tried to use Cauchy–Schwarz inequality, by try multiplying different... | Just as what was done in Post 1, we can substitute;
$$a=\frac{1}{x}$$
$$b=\frac{1}{y}$$
$$c=\frac{1}{z}.$$
Then, we have that;
\begin{align*}
& \text{ } \text{ } \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } ab+bc+ca \geq 1\\
&\implies \frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\geq 1\\
&\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3465571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
$(x^2+1)(y^2+1)=z^2+1$
Prove that $$(x^2+1)(y^2+1)=z^2+1$$ has infinitely many solutions when $x,y,z \in \Bbb N$
I couldn't even find one solution for this equation.
Any ideas on how to prove it?
Thanks in advance.
| $$\boxed{\color{red}{(x^2+1)(4x^4+1)= (2x^3+x)^2+1}}$$
Some motivation for that "proof without words" $$z^2 = x^2+y^2+x^2y^2$$
Let $y=xt$ then we have $$z^2 = x^2(1+t^2+x^2t^2)$$
so if we put $t=2x$ we get $$z^2 = x^2(1+4x^2+4x^4)=x^2(2x^2+1)^2$$
So we have triples $(x,2x^2,2x^3+x)$ where $x$ is arbitrary integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3467276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
What is the analytic solution for $y' = -\frac{a}{y} + \frac{b}{y^2}$? For $y: \mathbb{R}_+ \to \mathbb{R}_+$, if $y'(x) = -\frac{a}{y} + \frac{b}{y^2}$, then $y(x) = ?$
Can anyone give an exact solution of this type?
Update: Yes, $a,b$ are some positive constants. By separating the variables and integrating over $x,... | Suppose $y'=-\frac ay+\frac b{y^2}$. Then,
$$dx=\frac{y^2}{b-ay}\,dy$$
Now integrate and obtain an implicit equation for $y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3467348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Question on drawing 4 tickets form 7 tickets A bag contains 7 tickets marked with the numbers 0, 1, 2, ..., 6 respectively. A ticket is drawn and replaced; find the chance that after 4 drawings the sum of the numbers drawn is 8.
| We can view $4$ drawings to be $4$ different boxes and the number on the ticket drawn at $i$-th drawing is the number of balls in the $i$-th box after the balls are thrown to the boxes where $1 \leq i \leq 4.$ Now since the numbers on the tickets are from $0,1, 2 , \cdots ,6$ so that means each box will contain at most... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3468106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Inequality Proof $(x^2+1)(y^2+1)(z^2+1)\leq...$ I want to show, that for positive numbers $x,y,z$ with $xy,yz,zx\geq1$, $(x^2+1)(y^2+1)(z^2+1)\leq\left(\left(\frac{x+y+z}{3}\right)^2+1\right)^3$.
Using the AM-QM inequality for $(x^2+1),(y^2+1)$ and $(z^2+1)$, I already got $(x^2+1)(y^2+1)(z^2+1)\leq\left(\frac{x^2+y^2+... | Let $z = \max (x,y,z)$. Notice that
*
*$z^2 \geq zx \geq 1 $ so $ z \geq 1$.
*$ (x+y+z)^2 \geq 3 (xy+yz+zx) \geq 9$ so $(x+y+z) \geq 3$.
*Thus $z(x+y+z) \geq 3$.
*$(x+y)^2 \geq 4xy \geq 4 $.
*Thus $(x+y)(x+y+4z) = (x+y)^2 + 4z(x+y) \geq 4 + 8 = 12$.
Now, onto the problem.
First, as per Carl's... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3471617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Prove that the equation $x^3+2y^3+4z^3=9w^3$ has no solution $(x,y,z,w)\neq (0,0,0,0)$
Prove that the equation $x^3+2y^3+4z^3=9w^3$ has no solution $(x,y,z,w)\neq (0,0,0,0)$
So reduced $\mod 2$ I have $x^3\equiv w^3$, so if $x$ is odd, $w$ is odd and if $x$ is even, $w$ is even.
But I'm not really sure what else I s... | Suppose there is an integer solution $(x, y, z, w)$ for your equation, then we have
$$ x^3+2y^2+4z^3\equiv 0 \pmod 9$$
However, $0^3\equiv 3^3\equiv 6^3\equiv 0 \pmod 9$, $1^3\equiv 4^3\equiv 7^3\equiv 1 \pmod 9$, $2^3\equiv 5^3\equiv 8^3\equiv -1 \pmod 9$. So this implies there exists $a, b, c \in \{-1,0,1\}$ such tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3472127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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If $\tan2\theta=\frac{b}{a-c}$, then $\cos 2\theta=\frac{a-c}{\sqrt{b^2+(a-c)^2}}$ and $\sin2\theta=\frac{b}{\sqrt{b^2+(a-c)^2}}$ I am studying general equation of the second degree. While studying that chapter I came across
$$\tan2\theta=\frac{b}{a-c} \tag{1}$$
Now from (1), the author computed
$$\cos2\theta=\frac... | $$\tan(2\theta) = \frac{b}{a-c}$$
So we have a right triangle where the legs of the right triangle are of lengths $b$ and $a-c$, which means the hypotenuse should be $\sqrt{b^2 + (a-c)^2}$. We can, thus, compute the sine and cosine of the concerned angle using $\frac{\text{adj.}}{\text{hyp.}}$ and $\frac{\text{opp.}}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3473720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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show that $[f]_{B_1}$ is congruent to $[f]_{B_2}$
Let $f$ be a bilinear from on $\mathbb{R}^2$ defined by $$f(u,v)=2u_1v_1-3u_1v_2+u_2v_2,\quad u=(u_1,u_2),v=(v_1,v_2)$$
$(i)$ Find the matrices $[f]_{B_1}$ and $[f]_{B_2}$ relative to the bases $B_1=\{(1,0),(1,1)\}$ and $B_2=\{(2,1),(1,-1)\}$, respectively.
$(ii)$ Henc... | The matrices are not symmetric because $f$ is not symmetric.
For (ii), it is perhaps helpful to think about $[f]_B$ where $B$ is the standard basis. We have
\begin{align}
\begin{pmatrix} 1 & 0 \end{pmatrix}[f]_B \begin{pmatrix} 1 \\ 0 \end{pmatrix} &= 2,&
\begin{pmatrix} 1 & 0 \end{pmatrix}[f]_B \begin{pmatrix} 1 \\ 1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3473826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve equation $(2+ \sqrt{5})^{\frac{x}{2}}+2 \cdot \left(\frac{7 - 3 \sqrt{5}}{2}\right)^{\frac{x}{4}}=5$ Find the value of $x$ given the equation,
$$(2+ \sqrt{5})^{\frac{x}{2}}+2 \cdot \left(\frac{7 - 3 \sqrt{5}}{2}\right)^{\frac{x}{4}}=5$$
I think they are powers of $ \frac {1} {\Phi} $ or $ \Phi $. In case I think ... | Recognize
$$\left(\frac{7 - 3 \sqrt{5}}{2}\right)^{x/4}
= \left(\frac{3 - \sqrt{5}}{2}\right)^{x/2}
= \left(\frac{\sqrt{5}-1}{2}\right)^{x}$$
$$(2+ \sqrt{5})\left(\frac{3 - \sqrt{5}}{2}\right) = \frac2{\sqrt5-1}$$
and rewrite the given equation as
$$\left(\frac2{\sqrt5-1}\right)^{x/2}
+2 \cdot \left(\frac{\sqrt{5}-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3474255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Generating Function & Sequence Find the generating functions of the sequences
2, 1, 2, 1, 2, 1, . . .
I get $\frac{1}{1+x} + \frac{1}{1-x} = \frac{2}{1-x^2}$
But the solution ends up with $\frac{2}{1-x^2} + \frac{x}{1-x^2} = \frac{2+x}{1-x^2}$.
The solutions starts with $\sum_{n\ge 0} (2)x^{2n}+\sum_{n\ge 0} (1) x^{2n... | As $\frac{1}{1-x}$ is the generating series for $1,1,1,\dots$ and $\frac{1}{1-x}$ is the generating series for $1,-1,1,-1,\dots$, the series for which you are calculating the generating function for is $2,0,2,0,\dots$.
But we can get the result you want by noticing that $2,1,2,1,\dots$ is actually $\frac{3}{2},\frac{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3474486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
$ \lim_{n \to \infty} (1+ \frac{3}{n^2+n^4})^n$ I've tried to solve the limit
$$ \lim_{n \to \infty} (1+ \frac{3}{n^2+n^4})^n$$
but I'm not sure.
$ (1+ \frac{3}{n^2+n^4})^n = \sqrt [n^3]{(1+ \frac{3}{n^2+n^4})^{n^4}} \sim \sqrt [n^3]{(1+ \frac{3}{n^4})^{n^4}} \sim \sqrt [n^3]{e^3 } \rightarrow 1$
Is it r... | It is more or less correct, but how do you know that$$\lim_{n\to\infty}\left(1+\frac3{n^2+n^4}\right)^{n^4}=e^3?$$
It seems to me that it is more natural to do it as follows:\begin{align}\lim_{n\to\infty}\left(1+\frac3{n^2+n^4}\right)^{n^3}&=\lim_{n\to\infty}\left(\left(1+\frac3{n^2+n^4}\right)^{n^2+n^4}\right)^{\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3477465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Simplify $\frac{d}{dx}\frac{x^2}{1+\sqrt{x^2+1}}$ Simplify $$\frac{d}{dx}\frac{x^2}{1+\sqrt{x^2+1}}$$
I have manage to arrive to $$\frac{x^3+2x+2x\sqrt{x^2+1}}{(x^2+2)\sqrt{x^2+1}+2x^2+2}$$
But Wolfram manage to simplify to $$\frac{x}{\sqrt{x^2+1}}$$
| Write $t:=\sqrt{x^2+1}$ so $\frac{x^2}{1+\sqrt{x^2+1}}=\frac{t^2-1}{1+t}=t-1$ has derivative $\frac{dt}{dx}=\frac{x}{\sqrt{x^2+1}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3477760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Show that $\sum_{n=1}^{\infty} \frac{(-1)^{n}}{\sqrt{n}}\sin(1 + \frac{x}{n})$ converges uniformly on $\mathbb{R}$ I am trying to show the followng:
Show that $\sum_{n=1}^{\infty} \frac{(-1)^{n}}{\sqrt{n}}\sin(1 + \frac{x}{n})$ converges uniformly on $\mathbb{R}$.
I have shown it for a compact subset of $\mathbb{R}$ ho... | The sum is not uniformly convergent on $\mathbb{R}$.
Take $\epsilon \in (0,\frac{1}{10})$. Suppose $N$ is large such that $|\sum_{N \le n \le 2N} \frac{(-1)^n}{\sqrt{n}}\sin(1+\frac{x}{n})| \le \epsilon$ for all $x \in \mathbb{R}$. Take $X = 2\pi\prod_{N \le n \le 2N} n$. Independent of the specific choice of $X$, we o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3482417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
} |
Find $\lim_{x\to0}\frac{1-\cos x\cos2x\cos3x}{x^2}$
Find $$\lim_{x\to0}\dfrac{1-\cos x\cos2x\cos3x}{x^2}$$
My attempt is as follows:-
$$\lim_{x\to0}\dfrac{1-\dfrac{1}{2}\left(2\cos x\cos2x\right)\cos3x}{x^2}$$
$$\lim_{x\to0}\dfrac{1-\dfrac{1}{2}\left(\cos3x+\cos x\right)\cos3x}{x^2}$$
$$\lim_{x\to0}\dfrac{1-\dfrac{1}... | My preferred way is using Taylor expansion of cosine
$$\cos{x}=1-{x^2\over 2}+o(x^3)$$
This yields
$$\begin{align}
\cos{x}\cos{2x}\cos{3x}=&\left(1-{x^2\over 2}+o(x^3)\right)\cdot\\
& \left(1-{4x^2\over 2}+o(x^3)\right)\cdot\\
&\left(1-{9x^2\over 2}+o(x^3)\right)=1-7x^2+o(x^2)
\end{align}$$
And so
$${1-\cos{x}\cos{2x}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3483609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
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Sum of coefficients of $x^i$ (Multinomial theorem application)
A polynomial in $x$ is defined by
$$a_0+a_1x+a_2x^2+ \cdots + a_{2n}x^{2n}=(x+2x^2+ \cdots +nx^n)^2.$$
Show that the sum of all $a_i$, for $i\in\{n+1,n+2, \ldots , 2n\}$, is
$$ \frac {n(n+1)(5n^2+5n+2)} {24}.$$
I don't know how to proceed. I know th... | Write $(x+2x^2+\cdots+nx^n)^2=(x+2x^2+\cdots+nx^n)(x+2x^2+\cdots+nx^n)$ and for each coefficient in the first factor find the sum of the coefficients in the second factor with which it will enter into the desired sum: The coefficient $k$ in the first factor is paired with $n+1-k$ through $n$ in the second factor, so it... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3484240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
$\frac{x}{y}-\frac{y}{x}=\frac56$ and $x^2-y^2=5$
Solve the system: $$\begin{array}{|l}
\dfrac{x}{y}-\dfrac{y}{x}=\dfrac{5}{6} \\ x^2-y^2=5 \end{array}$$
First, we have $x,y \ne 0$. Let's write the first equation as:
$$\dfrac{x}{y}-\dfrac{y}{x}=\dfrac{5}{6} \Leftrightarrow \dfrac{x^2-y^2}{xy}=\dfrac{5}{6}$$
We have ... | Rewrite $\dfrac{x}{y}-\dfrac{y}{x}=\dfrac{5}{6}$ as $6x^2-5xy-6y^2=0$ and then factorize,
$$(2x-3y)(3x+2y)=0$$
to have $x=\frac32y$ and $x=-\frac23 y$. Plug them into $x^2-y^2=5$ to obtain the real solutions $(3,2)$ and $(-3,-2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3484688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Point at infinity on quartic elliptic curve Elliptic curve defined by
$$E_1: y^2=7 x^4+x^3+x^2+x+3, P_1=(-1,3)$$
can be transformed to
$$E_2: v^2=u^3-\frac{250 u}{3}-\frac{1249}{27}$$
Substitutions used are:
$$\left(x\to \frac{15 u-9 v+217}{39 u+9 v+209},y\to \frac{9 \left(54 u^3+639 u^2-27 v^2+592 v-16501\right)}{(39 ... |
$$E_1(-1,3)\to E_2(0,1,0)$$
$$E_1(-1,-3)\to E_2(-\frac{71}{9},-\frac{296}{27})$$
$$E_1(-\frac{5413}{16069},-\frac{434267883}{258212761})\to E_2(-\frac{71}{9},\frac{296}{27})$$
$$E_1(1,-\sqrt{7},0)\to E_2(\frac{40}{3}-6 \sqrt{7},-81+26 \sqrt{7})$$
$$E_1(1,\sqrt{7},0)\to E_2(\frac{40}{3}+6 \sqrt{7},-81-26 \sqrt{7})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3484803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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If $p, q$ and $r$ are distinct roots of $x^3-x^2+x-2=0$, find the value of $p^3+q^3+r^3$. If $p, q$ and $r$ are distinct roots of $x^3-x^2+x-2=0$, find the value of $p^3+q^3+r^3$.
Here's what I have got,
By Vieta's rule
$p+q+r=1\text{. ...........}(1)$
$pq+qr+pr=1\text{. ...........}(2)$
$p... | $$(x^3-2)^3=(x^2-x)^3$$
$$(x^3)^3-8-3(x^3)^22+3(x^3)2^2=(x^3)^2-(x^3)-3x^2\cdot x(x^3-2)$$
Replace $x^3=y$ to find
$$y^3-8-3y^2\cdot2+3y\cdot2^2=y^2-y-3y(y-2)$$
$$\iff y^3-y^2(6+1-3)+\cdots=0$$ whose roots are $ p^3,q^3,r^3$
$$\implies p^3+q^3+r^3=\dfrac{6+1-3}1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3486173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Birationally transforming general curve of genus 1 to Weierstrass form
What are general rules to birationally transform general curve of any
degree of genus 1 to Weierstrass form, provided we have one rational
point?
Example of curve of degree 12:
$$x^9 y^3+9 x^9 y^2+27 x^9 y+27 x^9+9 x^8 y^3+81 x^8 y^2+243 x^8 y... | $$C: x^9 y^3+9 x^9 y^2+27 x^9 y+27 x^9+9 x^8 y^3+81 x^8 y^2+243 x^8 y+243 x^8+35 x^7 y^3+318 x^7 y^2+963 x^7 y+972 x^7+74 x^6 y^3+687 x^6 y^2+2124 x^6 y+2187 x^6+90 x^5 y^3+871 x^5 y^2+2799 x^5 y+2988 x^5+67 x^4 y^3+692 x^4 y^2+2358 x^4 y+2655 x^4+39 x^3 y^3+415 x^3 y^2+1466 x^3 y+1717 x^3+21 x^2 y^3+211 x^2 y^2+723 x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3488571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to prove this formula for the determinant of a $4 \times 4$ tridiagonal matrix? This following is a problem from B. S. Grewal's Higher Engineering Mathematics.
Show
$$\begin{vmatrix} 2\cos(\theta) & 1 & 0 & 0 \\ 1 & 2 \cos(\theta) & 1
& 0 \\ 0 & 1 & 2 \cos(\theta) & 1 \\ 0 & 0 & 1 & 2 \cos(\theta)
\end{vmatrix}... | By an expansion on the first column one gets
\begin{align}\begin{vmatrix} 2\cos(\theta) & 1 & 0 & 0 \\ 1 & 2 \cos(\theta) & 1 & 0 \\ 0 & 1 & 2 \cos(\theta) & 1 \\ 0 & 0 & 1 & 2 \cos(\theta) \end{vmatrix} &= 2 \cos(\theta) \begin{vmatrix} 2 \cos(\theta) & 1 & 0 \\ 1 & 2 \cos(\theta) & 1 \\ 0 & 1 & 2 \cos(\theta) \end{v... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3489325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
If $\alpha, \beta, \gamma, \delta$ are distinct roots of equation $x^4 + x^2 + 1 = 0$ then $\alpha^6 + \beta^6 + \gamma^6 + \delta^6$ is I tried to find discriminant first and got two roots $$\frac{-1 + \sqrt{3}i}{2}$$ and $$\frac{-1 - \sqrt{3}i}{2}$$
I tried taking $x^2 = t$ and solving equation for root $w$ and $w^2$... | $$x^4=-(x^2+1)\Rightarrow x^6=-x^4-x^2=x^2+1-x^2=1$$
So $$\sum x^6=\sum 1=1+1+1+1=4$$
Where $\displaystyle \sum x^6=\alpha^6+\beta^6+\gamma^6+\delta^6.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3494035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 1
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If $n>3$ is not prime, show that we can find positive integers $a,b,c$ such that $n=ab+bc+ca+1$.
If $n>3$ is not prime, show that we can find positive integers $a,b,c$ such that $n=ab+bc+ca+1$.
When $n=4$, pick $a=b=c=1$. When $n=6,$ pick $a=2$ and $b=c=1$. When $n=8$, pick $a=b=c=2$. I don't get the pattern here. I ... | Notice that $xy = (x-1)(y-1)+(y-1) + (x-1)+1.$ Choose $a=x-1,b=y-1, $ and $c=1$. All are positive as $x,y>1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3494531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Show that $\int\limits_0^x \exp\left(-\frac{t^2}2\right)dt = \frac{f(x)}{g(x)}$
Let $f(x) = x+\dfrac{x^3}{1\cdot 3} + \dfrac{x^5}{1\cdot 3 \cdot 5} + \dfrac{x^7}{1\cdot 3\cdot 5\cdot 7}+\dots$ and let $g(x) = 1+\dfrac{x^2}{2} + \dfrac{x^4}{2\cdot 4} + \dfrac{x^6}{2\cdot 4\cdot 6} + \dots$ Show that $\displaystyle\int_... | Since $f^\prime+xf=1$, integration factors tell us a constant $c$ exists for which $f=\exp-\frac{x^2}{2}\cdot\int_c^x\exp\frac{t^2}{2}dt$. Since $f(0)=0$, $c=0$. The desired claim then follows from $g=\exp\frac{x^2}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3495436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Complex power series in Stein's book Let $$ F(z)=\sum\limits_{n=1}^{\infty}d(n)z^{n}$$for $ \left|z\right|<1 $,where $d(n)$ denotes the number of divisors of $ n $.Obeserve that the radius of convergence of this series is 1.Verify the identity $$\sum\limits_{n=1}^{\infty}d(n)z^{n}=\sum\limits_{n=1}^{\infty}\frac{z^{n}}... | Let be $
z=z\left( r \right) = re^{2\pi i\frac{p}
{q}}
$
We will prove that
$$
\mathop {\lim }\limits_{r \to 1^ - } \left| {\left( {1 - r} \right)F\left( {z\left( r \right)} \right)} \right| = + \infty
$$
As first we write
$$
F(z) = \sum\limits_{n = 1}^{ + \infty } {\frac{{z^n }}
{{1 - z^n }} = \sum\limits_A {\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3498017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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If $z = \frac{(\sqrt{3} + i)^n}{(\sqrt{3}-i)^m}$, find the relation between $m$ and $n$ such that $z$ is a real number. I am given the following number $z$:
$$z = \dfrac{(\sqrt{3} + i)^n}{(\sqrt{3} - i)^m}$$
with $n, m \in \mathbb{N}$. I have to find a relation between the natural numbers $n$ and $m$ such that the numb... | The key insight is to recognize roots of unity in the expression.
We have
$$
z = \frac{(\sqrt{3} + i)^n}{(\sqrt{3}-i)^m} = \frac{(2\omega)^n}{(2\omega^5)^m} = 2^{n-5m}\omega^{n-5m}
$$
where $\omega^6=-1$. Therefore, we need $n-5m \equiv 0 \bmod 6$, or $n+m \equiv 0 \bmod 6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3498784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Convergence of $\sum_{n=1}^{\infty} \frac{x^n}{n+1}$ Would anyone happen to know what $\sum_{n=1}^{\infty} \frac{x^n}{n+1}$ converges to? My method goes as
$$
f(x)
= \sum_{n=1}^{\infty} \frac{x^n}{n+1}
= \frac{1}{x} \sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1}
= \frac{1}{x} \sum_{n=1}^{\infty} \int_0^x t^ndt
= \frac{1}{x}... | $$\sum_{n=1} \frac{x^n}{n+1} = \frac{1}{x}\sum_{n=2} \frac{x^n}{n} = \frac{1}{x}\left(\sum_{n=1} \frac{x^n}{n} - x\right) = -1 - \frac{\log(1-x)}{x}$$
for $0<|x| < 1$ (the limit to $0$ exists, and equals $0$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3499009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Numbers 'poetics' 1 and 2019
One number is called 'poetic' when it can be represented of only one form how $2^a+2^b+2^c$ such that $a\ge b \ge c\ge 0.$ How many poetic numbers are between $1$ and $2019?$
Attempt: I separate in cases, I find that every $2^k$ with $k\ge2 , 2^k + 1$ with $k\ge1$ and every case with $a>... | As $2^{11}>2047>2020>2^{10}+2^9$, there are no poetic numbers from $2020$ to $2047$.
All poetic numbers from $1$ to $2047$ actually lie between $1$ and $2019$.
Integers in this range can be written as binary numbers of no more than $11$ digits.
A poetic number should have at most three $1$'s in the digits when it is ex... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\lim_{x \to 1}\frac{\sin(x^2-x)}{x^2-1}$ and $\lim_{x \to 1}\frac{\ln(x^2+x-1)}{\ln(x^3+x-1)}$ I found these limits and I was unable to solve them due to the occurring indeterminations
$$\lim_{x \to 1}\frac{\sin(x^2-x)}{x^2-1}$$
$$\lim_{x \to 1}\frac{\ln(x^2+x-1)}{\ln(x^3+x-1)}$$
Can someone help me, please... | $$\lim_{x \to 1}\frac{\ln(x^2+x-1)}{\ln(x^3+x-1)} = \lim_{x \to 1} \frac{(2x+1)\cdot(x^3+x-1)}{(3x^2+1)\cdot(x^2+x-1)} = \frac{3}{4}$$ (using L'Hospital)
| {
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Help with inequality problem
Given $a$ , $b$ , $c \ge 0$ show that
$$\frac{a^2}{(a+b)(a+c)} + \frac{b^2}{(b+a)(b+c)}+ \frac{c^2}{(c+a)(c+b)} \ge \frac{3}{4}.$$
I tried using Titu's lemma on it, resulting in
$$\frac{a^2}{(a+b)(a+c)}+\frac{b^2}{(b+a)(b+c)}+ \frac{c^2}{(c+a)(c+b)}\ge \frac{(a+b+c)^2}{a^2+b^2+c^2 + 3(... | By C-S $$\sum_{cyc}\frac{a^2}{(a+b)(a+c)}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(a+b)(a+c)}\geq\frac{3}{4},$$ where the last inequality it's
$$4\sum_{cyc}(a^2+2ab)\geq3\sum_{cyc}\left(a^2+3ab\right)$$ or
$$\sum_{cyc}(a^2-ab)\geq0$$ or
$$\sum_{cyc}(2a^2-2ab)\geq0$$ or
$$\sum_{cyc}(a^2+b^2-2ab)\geq0$$ or $$\sum_{cyc}(a-b... | {
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Find density function for a variable $Y$, if $Y=X^2$ and $f_X=30x^2(1-x)^2, 0 I have to find $f_Y$ if $f_X$ is given as $f_X=30x^2(1-x)^2, 0 <x < 1$ and $Y=X^2$. How do I do this?
Ok, then I get: $F_Y=P(x \le y^{\frac{1}{2}}) \Rightarrow F_X(y^{\frac{1}{2}})$
$F(y)=\int^{\sqrt{y}}_0 30x^2(1-x)^2dx=\left[ u'=30x^2 \Righ... | $$F_Y(y)=P(X^2\le y)=P(-\sqrt y\le X\le\sqrt y)=P(X\le\sqrt y)=F_X(\sqrt y)$$
Since
$$f_X(x)=\begin{cases}30x^2(1-x)^2&\text{for }0<x<1\\0&\text{otherwise}\end{cases}$$
the CDF for $X$ is
$$F_X(x)=\int_{-\infty}^xf_X(t)\,\mathrm dt=\begin{cases}0&\text{for }x<0\\10 x^3 - 15 x^4 + 6 x^5&\text{for }0\le x<1\\1&\text{for ... | {
"language": "en",
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"source": "stackexchange",
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integral representation for $\sum_{k=0}^{x}k^{p}$ How the following integral representation can be derived?
$$\sum_{k=0}^{x}k^{p}=\int_{0}^{x+1}B_{p}\left(t\right)dt=\frac{B_{p+1}\left(x+1\right)-B_{p+1}}{p+1}$$
I know Faulhaber's formula which is as follows:
$$\sum_{k=0}^{x}k^{p}=\frac{1}{p+1}\sum_{j=0}^{p}B_{j}{{p+1}... | We use the generating function
\begin{align*}
\frac{te^{tx}}{e^t-1}=\sum_{n\geq 0}B_n(x)\frac{t^n}{n!}\tag{1}
\end{align*}
of Bernoulli polynomials to derive the integral representation.
From (1) we can show for $n\geq 1$
\begin{align*}
B_{n}(x+1)-B_n(x)&=nx^{n-1}\tag{2}
\end{align*}
as follows:
We obtain
\begin{al... | {
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Find $\int_0^\infty\frac{x\log x}{(1+x^2)^2}dx$
Evaluate $$
\int_0^\infty\frac{x\log x}{(1+x^2)^2}dx
$$
$$
\int\frac{x}{(1+x^2)^2}dx=\frac{-1}{2(1+x^2)}\\
\int_0^\infty\frac{x\log x}{(1+x^2)^2}dx=
\bigg[\log x\frac{-1}{2(1+x^2)}\bigg]_0^\infty-\int_0^\infty\frac{1}{x}.\frac{-1}{2(1+x^2)}dx\\
=0+\int_0^\infty\frac{1}{... | Note that, with $x=\frac1t$,
$$\int_1^\infty\frac{x\log x}{(1+x^2)^2}dx
=- \int_0^1\frac{t\log t}{(1+t^2)^2}dt$$
Thus, the integral is evaluated to zero.
| {
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Find area of ellipse $5x^2 -6xy +5y^2=8$
Find the area of ellipse whose equation in the $xy$- plane is given by $5x^2 -6xy +5y^2=8$
My attempt : I know that area of ellipse $ = \pi a b$ ,where $a$ is semi-major axis and $b$ is semi minor axis
Now if we make matrix $\begin{bmatrix} 5 & -3 \\-3& 5\end{bmat... | Center of ellipses is $(0,0)$ and any point on the chord passing through center is $(r\cos{\theta} , r\sin{\theta} )$ . Let chord intersect ellipses at points $A$ and $B$ . Put the point in equation of ellipses we get
$$(5\cos^2{\theta})r^2-(6\cos{\theta} \sin{\theta} ) r^2 + (5\sin^2{\theta} )r^2 =8$$
$$(5-3\sin^2{\t... | {
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Solution verification:$\lim_{x\to 2}\frac{\ln(x-1)}{3^{x-2}-5^{-x+2}}$
Evaluate without L'Hospital:$$\lim_{x\to
2}\frac{\ln(x-1)}{3^{x-2}-5^{-x+2}}$$
My attempt:
I used: $$\lim_{f(x)\to 0}\frac{\ln(1+f(x))}{f(x)}=1\;\&\;\lim_{f(x)\to 0}\frac{a^{f(x)}-1}{f(x)}=\ln a$$
$$
\begin{split}
L &= \lim_{x\to 2} \frac{\ln(x-1... | This is fine. Here is an alternative approach using taylor series:
First substitute $x-2=y$ to simplify it. Let the required limit be $l$. Then
$$l = \lim_{y\to0}\left(\dfrac{\ln(1+y)}{3^y-5^{-y}}\right)$$
$$ = \lim_{y\to0}\left(\dfrac{y-\dfrac{y^2}{2}+\cdots}{(1+y\ln3+\cdots)-(1-y\ln5+\cdots)}\right)$$
$$=\dfrac{1}{\l... | {
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"source": "stackexchange",
"question_score": "9",
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Why does't quadratic formula work to factor polynomial when $a \ne 1$? $$2x^2 + 3x + 1$$
applying quadratic formula:
$$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$$
$$a=2, b=3, c=1$$
$$x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot1}}{2\cdot2}$$
$$x = \frac{-3 \pm \sqrt{9-8}}{4}$$
$$x = \frac{1}{4}[-3 + 1],~~~x=\frac{1}{4}[-3-1... | You are not "solving" $$2x^2+3x+1$$
but $$
2x^2+3x+1 \mathbf{=0}.
$$
For the specific values of $x$ you found, it is indeed true that
$$
2x^2+3x+1 = x^2+\frac{3}{2}x+\frac{1}{2}=0 \quad \text{ where }x\in \left\{- \frac{1}{2}, -1\right\}
$$
Since we solved for when this expression is equal to zero, we are free to mult... | {
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Prove that if $x$ and $y$ are both not $0$ Prove that if $x$ and $y$ are both not $0,$ then
$$x^4+x^3y+x^2y^2+xy^3+y^4>0$$
I know this seems fairly easy but I'm fairly new to calculus and need some help proving that this is true. Appreciate the help!
| Another way:
Note that $0\leq x^2(x+y)^2=x^4+2x^3y+x^2y^2$, and similarly $y^4+2xy^3+x^2y^2\geq 0$. Adding these and dividing by $2$, we get $\frac12x^4+x^3y+x^2y^2+xy^3+\frac12y^4\geq 0$. Then add $\frac{x^4+y^4}{2}$, which is positive since $x$ and $y$ are not both $0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $A^2=a_1^2+a_2^2-a_3^2-a_4^2$ for all integers $A$.
Prove that:
For every $A\in\mathbb Z$, there exist infinitely many $\{a_1,a_2,a_3,a_4\}\subset\mathbb Z$ given $a_m\neq a_n $ such that $$A^2=a_1^2+a_2^2-a_3^2-a_4^2$$
After many hours, I found a general formula satisfying the statement for all $A$ and ... | Note that
$$a_1^2+a_2^2-a_3^2-a_4^2=\underbrace{(a_1+a_3)(a_1-a_3)}_{=:M}+\underbrace{(a_2+a_4)(a_2-a_4)}_{=:N} $$
where $M$ and $N$ can be any integer that is either odd or a multiple of $4$. In particular, negatives are allowed so that for each number ($A^2$ or otherwise), we find infinitely many such $M,N$.
Concret... | {
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Evaluating $\lim_{n \to \infty}(\sin\sqrt{n+1} - \sin\sqrt{n}\;)$ $$\lim_{n \to \infty}(\sin\sqrt{n+1} - \sin\sqrt{n}\;)$$
I know that due to continuity of sine I can split it in:
$$\lim_{n \to \infty}(\sin\sqrt{n+1} - \sin\sqrt{n}\;) = \lim_{n \to \infty}\sin\sqrt{n+1} -\lim_{n \to \infty}\sin\sqrt{n}$$
I also know t... | Your line,
$$ \lim_{n \to \infty}(\sin\sqrt{n+1} - \sin\sqrt{n}\;) \\
\qquad \qquad = \lim_{n \to \infty}\sin\sqrt{n+1} -\lim_{n \to \infty}\sin\sqrt{n} \text{,} $$
is only valid if the two limits on the right exist. Neither limit on the right exists, so this is an invalid operation.
Generally, you can make progres... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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