Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Make upper bounds for $\frac{1}{2^{n-k-1}}\;.\frac{1}{1-2^{\beta-(n-k)}}$. I want to make upper bounds for the following fraction respected to $\beta$ only (not $k$ or $n$).
\begin{align*}\frac{1}{2^{n-k-1}}\;.\frac{1}{1-2^{\beta-(n-k)}}
\end{align*}
where $\beta\simeq 0.7$, $0\leq k\leq n$, and $k, n\in\mathbb{N}... | For each integers $0\le k\le n$ put
$$f(n,k)=\frac{1}{2^{n-k-1}}\;.\frac{1}{1-2^{\beta-(n-k)}}=\frac 1{2^{n-k-1}-2^{\beta-1}}.$$
If $n=k$ then $f(n, k)$ is negative. If $n-k\ge 1$ then $f(n,k)\le \frac 1{1-2^{\beta-1}}$ and the equality is attained when $n-k=1$. Thus the required tight upper bound is $\frac 1{1-2^{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3373174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove hard inequality Let $a,b,c>0$, prove that: $$\frac{1}{(2a+b)^2}+\frac{1}{(2b+c)^2}+\frac{1}{(2c+a)^2}\geq\frac{1}{ab+bc+ca}$$
I tried to use the inequality $\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\geq\frac{(a+b+c)^2}{x+y+z} \forall x,y,z>0$ but that's all I can do. I think this inequality is too tight to use AM... | Let $x=2a+b,y=2b+c,z=2c+a$, then the inequality becomes:$$\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\ge\dfrac{27}{5x^2+5y^2+5z^2-2xy-2yz-2zx}$$ By using $a^2+b^2+c^2\ge ab+bc+ca$,
$$\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}=\dfrac{\left(xy\right)^2+\left(yz\right)^2+\left(zx\right)^2}{x^2y^2z^2}\ge\dfrac{x^2yz+xy^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3376262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Greatest integer function problem If $\alpha$ is real root of equation $$x^5 -x^3 +x-2$$ , then [$\alpha$^6 ] is equal to ? Here [.] represents greatest integer function . Any hint please!
| $x^5 - x^3 + x = \frac {x(x^6 + 1)}{x^2 + 1}\\
x(x^4 - x^2 +1) = \frac {x}{x^2+1}(x^6 + 1)\\
\frac {x}{x^2+1}(x^6 + 1) = 2$
We can conclude that $x> 0$
as $(x^6 + 1)> 0$ we need $\frac {x}{x^2+1} > 0$
$\frac {x}{x^2+1} < \frac 12$ for all x
$(x^6 + 1) > 4\\
\alpha^6 > 3$
But is it possible for $\alpha^6 > 4$?
$f(x) = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3376939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Calculate limit with or without the use of l'Hospital's rule or series expansions $$\lim_{x\to0^{+}}\left(\frac{\cos^2x}{x}-\frac{e^x}{\sin x}\right)$$
One can easily calculate this limit by using series expansions for the functions appearing inside the round brackets, yielding $-1$.
My question is: can anyone give ano... | $\begin{array}\\
\dfrac{\cos^2x}{x}-\dfrac{e^x}{\sin x}
&=\dfrac{1-\sin^2x}{x}-\dfrac{e^x}{x+O(x^3)}\\
&=\dfrac{1-(x+O(x^3))^2}{x}-\dfrac{1+x+O(x^2)}{x+O(x^3)}\\
&=\dfrac{1-x^2+O(x^3)}{x}-\dfrac{1+x+O(x^2)}{x}(1+O(x^2))\\
&=\dfrac{1-x^2+O(x^3)-(1+x+O(x^2)(1+O(x^2))}{x}\\
&=\dfrac{1-x^2+O(x^3)-(1+x+O(x^2)}{x}\\
&=\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3377349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Show whether $ x^3$ is $O(g(x))$ where $g(x)=x^2 $ I know that $x^3$ is $O(x^3)$ or any exponent higher than 3, but how do I show mathematically that $x^3$ is not $O(x^2)$?
| If you want to show that it's NOT the case that $x^3=O(x^2)$, then you are showing that $x^2=o(x^3)$, or in other words, that $x^2$ is "little-oh" of $x^3$, which communicates that $x^2$ is of a smaller order of growth than $x^3$.
As gimusi has noted, you could show $x^2=o(x^3)$ via limits as follows
$$\lim_{x\rightarr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3378304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
prove that ${3^{3n}} + 3^{2n} + 3^{n } + 1$ is divided by $4$. by induction I tried to take the 3 out but it is not helping me much.
| By induction?
Well if we assume $3^{3n} + 3^{2n} + 3^n + 1= 4K$ then
$3^{3(n+1)} +3^{2(n+1)} + 3^{n+1} + 1=$
$3^{3n+3} + 3^{2n + 2} + 3^{n+1} + 1=$
$3^{3n}*27 + 3^{2n}*9 + 3^n*3 + 1 =$
$[3^{3n} + 3^{2n} + 3^n+1] + 26*3^{3n} + 8*3^{2n} + 2*3^n =$
$4K + 4(6*3^{2n} + 2*3^{2n}) + 2(3^{3n} + 3^n) =$
$4[K+(6*3^{2n} + 2*3^{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3381956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
Markov Chain: Calculating Expectation Reach a Certain Set of States Suppose I have a Markov chain $Z_k$ with $6$ states, as depicted below:
The probability of moving from one node to a neighboring node is $1/2$. For example, the probability of moving from node $1$ to node $2$ is $1/2$ and the probability of moving fr... | As an alternative approach in the spirit of your initial attempt, you can condition on the number $2k$ of steps to reach $\{3,4,5\}$ from $1$:
\begin{align}
E(T_B)
&= \sum_{k=1}^\infty 2k\ P(\text{$2k$ steps}) \\
&= \sum_{k=1}^\infty 2k\ 2^k \left(\frac{1}{2}\right)^{2k} \\
&= \sum_{k=1}^\infty k \left(\frac{1}{2}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3386266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Question: particular formula for sums of squares and square of sums What is a general / particular formula (likely combinatorics or number theory) that describes a relationship between the sum of squares and square of sums? I remember once seeing a formula where “the ration between the sum of three numbers squared and ... | For the series of natural number :
$$1^2 + 2^2 + 3^2 +4^2 + ....+n^2 = \frac{n(n+1)(2n+1)}{6}$$
And
$$(1+2+3+4+...n)^2 = \left[\frac{n(n+1)}{2}\right]^2 = \frac{n^2(n+1)^2}{4}$$
So,
$$ \frac{1^2 + 2^2 + 3^2 +4^2 + ....+n^2}{(1+2+3+4+...n)^2} = \frac{n(n+1)(2n+1)}{6} \times \frac{4}{n^2(n+1)^2}$$
$$ = \frac23 \left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3386954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Does the fractional equation $\frac{1}{x-5} +\frac{1}{x+5}=\frac{2x+1}{x^2-25}$ have any solutions? We have a partial fraction equation:
$$\frac{1}{x-5} +\frac{1}{x+5}=\frac{2x+1}{x^2-25}$$
I multiplied the equation by the common denominator $(x+5)(x-5)$ and got $0=1$. Is this correct?
| Yes it is equivalent to
$$\frac{2x}{x^2-25}=\frac{2x+1}{x^2-25}\iff 2x=2x+1$$
which indeed has not solutions for $x\in \mathbb R$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3392171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving $\lim_{x\to 1}\sqrt{2x} = \sqrt 2 $ using $\varepsilon$-$\delta$
Prove that $$\lim_{x\to 1}\sqrt{2x} = \sqrt 2 $$ using $\varepsilon$-$\delta$.
What I've tried is
$$\left|\frac{(\sqrt{2x}-{\sqrt 2})(\sqrt {2x}+{\sqrt 2})}{\sqrt {2x}-{\sqrt 2}}\right|
=\left|\frac{2x-2}{\sqrt {2x}-{\sqrt 2}}\right|$$ and be... | You're aiming to make $|\sqrt{2x} - \sqrt{2}|$ very small. Rationalising the numerator is a good idea, but you seemed to rationalise $|\sqrt{2x} + \sqrt{2}|$ instead. Rather, we have
$$|\sqrt{2x} - \sqrt{2}| = \left|\frac{(\sqrt{2x} - \sqrt{2})(\sqrt{2x} + \sqrt{2})}{\sqrt{2x} + \sqrt{2}}\right| = \frac{|2x - 2|}{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3393010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Compute $\sum_{n=1}^{\infty} \frac{ H_{n/2}}{(2n+1)^3}$ How to prove that
$$S=\displaystyle \sum_{n=1}^{\infty} \frac{ H_{n/2}}{(2n+1)^3} \quad=\quad \frac{\pi^2G}{4}-\frac{21\zeta(3)\ln(2)}{8}+\frac{\pi^4}{64}+\frac{\Psi^{(3)}(\frac{1}{4})}{512}- \frac{\Psi^{(3)}(\frac{3}{4})}
{512}$$
This problem was proposed by @... | Another proof besides Cornel's one:
Let
\begin{align}
S&=\sum_{n=0}^\infty(-1)^n(H_{n/2}-H_n+\ln2)e^{ix(2n+1)}\\
&=\sum_{n=0}^\infty(-1)^n\int_0^1\frac{y^n \ dy}{1+y}e^{ix(2n+1)}\\
&=\int_0^1\frac{e^{ix}\ dy}{1+y}\sum_{n=0}^\infty\left(-ye^{2ix}\right)^n\\
&=\int_0^1\frac{e^{ix}\ dy}{(1+y)(1+e^{2ix}y)}\\
&=\frac{e^{ix}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3393844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
$ a_n = \frac{a_{n-1}(a_{n-1} + 1)}{a_{n-2}}.$ and $ T = 3.73205080..$? Consider the following sequence :
Let $a_1 = a_2 = 1.$
For integer $ n > 2 : $
$$a_n = \frac{a_{n-1}(a_{n-1} + 1)}{a_{n-2}}.$$
$$ T = \lim_{k \to \infty} \frac{a_k}{ a_{k - 1}}.$$
$$T = ??$$
What is the value of $T$ ?
Is there a closed form or inte... | EDIT: By already knowing the solution to your recurrence a priori, I am able to solve your problem. However, a more interesting question might be why your recurrence is equivalent to
$$a_n=5a_{n-1}-5a_{n-2}+a_{n-3}$$
and how to see this a priori.
Your recurrence is solved by the sequence (as can be checked by direct ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3396793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Limit of an inverse function Let $f:\mathbb R\to \mathbb R$ be an invertible function such that $$\lim_{x\to a} f(x)=b$$
for some $a,b\in \mathbb R$.
Does it follow that $$\lim_{x\to b}f^{-1}(x)= a,$$
where $f^{-1}$ denotes the inverse function of $f$?
Edit: When I consider the $\epsilon,\delta$-definition of the lim... | If $x=a$, consider the sequence $(a_{n})=(a+\frac{b-a}{2}\frac{1}{n})_{n\geq 1}$ and the sequence $(b_{n})(b+\frac{b-a}{2}\frac{1}{n})_{n\geq1}$. We define $f$ as follows
$$f(x)=\begin{cases}b+\frac{b-a}{2}\frac{1}{2n-1}&\text{ if }x=a+\frac{b-a}{2}\frac{1}{n},\ n>0.\\ b+\frac{b-a}{2}\frac{1}{2n}&\text{ if }x=b+\frac{b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3397326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 2,
"answer_id": 1
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What are the integer solutions of the equation $(x^2 -1)(y^2 -1)=2(7xy-24)$ Can you help with this one? I've been trying fruitlessly for hours
$$(x^2 -1)(y^2 -1)=2(7xy-24)$$
| Let me elaborate on my hint given above in the comment section. I will focus on non-negative integer solutions because it is easy to see that if $(x,y)$ is a solution, then so is $(-x,-y)$.
Observe that the given equation can be written as :
\begin{align*}
(xy-7)^2 & =x^2+y^2\\
(xy-6)^2+13 & =(x+y)^2\\
(xy-6)^2-(x+y)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3400660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Group with simple relations I'm working on the following algebra problem:
Prove that if elements $a$ and $b$ of some group satisfy the relations
1) $a^5 = b^3 = 1$ and
2) $b^{-1}ab = a^2$
then $a = 1$.
Here is an example of what I've tried:
If $a^5 = 1$, then using relation 2) , multiplying on the right by $a^3$ on... | If $b^{-1}ab=a^2$, $b^3 =1$ and $a^5=1$, then
$$\begin{align*}
b^{-2}ab^2 &= b^{-1}(b^{-1}ab)b\\
&= b^{-1}a^2b\\
&= (b^{-1}ab)^2\\
&= (a^2)^2 = a^4 = a^{-1}.
\end{align*}$$
So therefore,
$$\begin{align*}
a&= b^{-3}ab^3\\
&= b^{-1}(b^{-2}ab^2)b\\
&= b^{-1}(a^{-1})b\\
&= (b^{-1}ab)^{-1}\\
&= (a^2)^{-1}\\
&= a^3.
\end{ali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3407718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How can I solve the following inequality? I have the inequality:
$lg((x^3-x-1)^2) < 2 lg(x^3+x-1)$
And I'm not sure how should I go about solving it. I wrote it like this:
$2lg(x^3-x-1) < 2lg(x^3+x-1)$
$lg(x^3-x-1) < lg(x^3 + x - 1)$ (*)
Here I have the conditions:
$x^3-x-1 > 0$
$x^3+x-1 > 0$
At first this stumpe... | The domain it's $$x^3-x-1\neq0$$ and $$x^3+x-1>0.$$
Now, since $$\ln(x^3-x-1)^2=2\ln|x^3-x-1|,$$ we need to solve
$$|x^3-x-1|<x^3+x-1$$ or
$$-x^3-x+1<x^3-x-1<x^3+x-1.$$
The left inequality gives $x>1$ and the right inequality gives $x>0,$ which gives the answer:
$$(1,+\infty)\setminus\{x|x^3-x-1=0\}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3410334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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In the range $0\leq x \lt 2\pi$ the equation has how many solutions $\sin^8 {x}+\cos^6 {x}=1$ In the range $0\leq x \lt 2\pi$ the equation has how many solutions
$$\sin^8 {x}+\cos^6 {x}=1$$
What i did
$\cos^6 {x}=1-\sin^8 {x}=(1-\sin^4 {x})(1+\sin^4 {x})=(1-\sin^2 {x})(1+\sin^2 {x})(1+\sin^4 {x})$
$\cos^4 {x}=(1+\sin^2... | One trick is letting $t=\sin^2(x)$, so that the equation reduces to a polynomial $t^4+(1-t)^3 - 1 = t^4 - t^3 + 3t^2 - 3t = t(t^3-t^2+3t-3) = t(t-1)(t^2+3)=0$
Thus either $t=0,1$ so that $\sin(x) = 0,1,-1$. In $[0, 2\pi)$, we clearly have solutions $x=0, \pi, \frac{\pi}{2}, \frac{3\pi}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3410740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Determining $n$ from its divisors Let $d_i$ be the $i^{th}$ smallest divisor of $n$.
Suppose that $n$ has $3$ prime factors and that:
$$n = d_{13}+d_{14}+d_{15}$$
$$d_{15}+1 = (d_5+1)^3$$
Is it possible to uniquely determine $n$?
I tried doing some stuff with writing $n=p^\alpha q^\beta r^\gamma$ and working modulo dif... | Well, $n<3d_{15}$ and $d_{15}\mid n$, so either $n=d_{15}$ or $n=2d_{15}$. However, in that first case, $d_{14}+d_{13}$ would be zero and that's impossible. Hence,
$$d_{15}=\frac n2.$$
It follows that $n<4d_{14}$ and because $d_{14}\mid n$ and $d_{14}<n/2$, we must have $d_{14}\in\{n/3,n/4\}$. But if $d_{14}=n/4$, then... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3415873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Prove convergence of $\sum_{n=1}^{\infty}\frac{\sin^3\left(\frac{\pi n}{n+5}\right)}{\sqrt{n^2+n}-n}$
I need to prove convergence of the series.
$$
\sum_{n=1}^{\infty}a_n,\ \
a_n=\frac{\sin^3\left(\frac{\pi n}{n+5}\right)}{\sqrt{n^2+n}-n}
$$
Well, at first I noticed that $\forall n\in\mathbb{N}\ a_n>0$:
$$
\left.
... | We have that
$$\sin^3\left(\frac{\pi n}{n+5}\right)=\sin^3\left(\frac{\pi n+\pi 5-\pi 5}{n+5}\right)=\sin^3\left(\pi-\frac{\pi 5}{n+5}\right)=$$
$$=\sin^3\left(\frac{\pi 5}{n+5}\right) \sim \left(\frac{\pi 5}{n+5}\right)^3$$
and
$$\frac{1}{\sqrt{n^2+n}-n}=\frac{1}{\sqrt{n^2+n}-n}\frac{\sqrt{n^2+n}+n}{\sqrt{n^2+n}+n}=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3416389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$|{\sqrt{a^2+b^2}-\sqrt{a^2+c^2}}|\le|b-c|$ where $a,b,c\in\mathbb{R}$ Show that $|{\sqrt{a^2+b^2}-\sqrt{a^2+c^2}}|\le|b-c|$ where $a,b,c\in\mathbb{R}$
I'd like to get an hint on how to get started. What I thought to do so far is dividing to cases to get rid of the absolute value. $(++, +-, -+, --)$
but it looks messy.... | For all $a,b,c \in \mathbb{R}$,
\begin{align}
0 &\leq (b-c)^2\\
2bc &\leq b^2 + c^2\\
2a^2bc &\leq a^2b^2 + a^2c^2\\
a^4 + 2a^2bc + b^2c^2 &\leq a^4 + a^2b^2 + a^2c^2 + b^2c^2\\
a^2 + bc&\leq \sqrt{a^4 + a^2b^2 + a^2c^2 + b^2c^2} \label{1}\tag{1}\\
2a^2 + 2bc&\leq 2\sqrt{a^4 + a^2b^2 + a^2c^2 + b^2c^2}\\
2a^2 -2\sqrt{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3416600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find all positive integers m,n and primes $p\geq5$ such that $m(4m^2+m+12)=3(p^n-1)$ I did it like this. We can manipulate the equation to come to
$\frac{(m^2+3)(4m+1)}{3}=p^n$
Now 3 divides either $(m^2+3) or (4m+1) $
If we assume 3 divides $(m^2+3)$ and it is equal to $(4m+1)$ (my intuition) .we get
$\frac{m^2+3}{3... | This is a variation on John Omielan's answer, starting with the equation in the form $(m^2+3)(4m+1)=3p^n$. As John notes, each factor $m^2+3$ and $4m+1$ must contain a power of the prime $p$, since $m^2+3$ and $4m+1$ are both greater than $3$ if $m\ge1$.
Now using properties of the $\gcd$ relation, we see that
$$\begin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3418575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Eigenvalues and Eigenvectors of a 3 by 3 matrix I want to find eigenvalues and eigenvectors of this matrix:
$$
\begin{bmatrix}
2 & 3 & 0 \\
3 & 6 & 1 \\
0 & 1 & 6 \\
\end{bmatrix}
$$
So eigenvalues:
$$
\begin{vmatrix}
2-\lambda_1 & 3 & 0 \\
3 & 6-\lambda_2 & 1 \\
0 & 1 & 6-\lambda_3... | The strategy is that you define one of the variables $x$, $y$, or $z$ as independent variable. Then two other variables will be dependent on it. Let's choose $z$ to be the independent. In this case your eigenvector corresponding to $\lambda_1=3-\sqrt{7}$ is $(\dfrac{(2+\sqrt{7})(4+\sqrt{7})}{3}z,-(3+\sqrt{7})z,z)$.
Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3421849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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$\frac{a^2+b^2+1}{ab+1}=k$. Find $k$ $a$ and $b$ are positive integers such that
$$\frac{a^2+b^2+1}{ab+1}=k$$
where $k$ is positive integer. Find all values $k$
My work:
$a$ is the root of the equation
$$x^2-kbx+b^2+1-k=0$$
Let $c$ is another root of this equation. Then
$c=kb-a$ and $ac=b^2+1-k$
| Note that, mod $ab+1$, $0=b^2(a^2+b^2+1)=b^4+b^2+1$. Therefore, $b^4+b^2+1 > 0$ is divisible by $ab+1$, thus $b(b^3+b) \geq ab$. So if $b \neq 0$, then this can be written as $k \leq b^2+1$ which entails (if $a > 0$), $c \geq 0$.
Take now, for a given $k$, $(a,b)$ a solution in non-negative integers with $a \geq b$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3425706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
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:
$$572\equiv 2\pmod {10} \\
572^2 \equiv 2^2 \equiv 4\pmod{10} \\
572^3 \equiv 2^3 \equiv 8\pmod{10} \\
572^4 \equiv 2^4 \equiv 6\pmod{10} \\
572^5 \equiv 2^5 \equiv 2\pmod{10} \\
572^6 \... | Applying $\ xy\bmod xz\, =\, x(y\bmod z)\ = $ mod Distributive Law to factor out $\,x = 2\,$ below
$\ \ \ \ \begin{align}(2a)^{\large 2+4N}\!\bmod 10 &\,=\, 2(2\, \color{#c00}2^{\large\color{#c00} 4N} a^{\large 2} \color{#c00}a^{\large\color{#c00} 4N}\bmod 5)\\[.2em]
&\,=\, 2(2a^{\large 2}\bmod 5)\ \ {\rm by}\ \ 5\nmi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3428727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Linear Algebra: Determine if the set of polynomials span P2 The polynomial given is { ${1 - 2x + x^2, 2 - 3x, x^2 + 1, 2x^2 + x}$}
What I do not understand is, how am I going to solve the augmented matrix if it only spans to P2 (meaning the highest power will be 2?) but I am given 4 equations.
| It's clear that the system $\{x^2,x,1\}$ generates $P2$, since
$$P2 = \{ax^2+bx+c\ |\ a,b,c\in\Bbb{R}\} = span(x^2,x,1).$$
So if we can show that the system
$$\begin{align}&\text{I. } &1 - &2x &+ x^2 \\ &\text{II. } &2 - &3x & \\ &\text{III. } &1\ \ \ &\ \ + &x^2 \\ &\text{IV. } & &x+ &2x^2\end{align}$$
generates $\{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3432587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Let $f(x)=(x+1)(x+2)(x+3)(x+4)+5$ where $x\in[-6,6]$. If the range of the function is [a,b] where $a,b\in N$, then find the value of (a+b) Let $f(x)=(x+1)(x+2)(x+3)(x+4)+5$ where $x\in[-6,6]$. If the range of the function is [a,b] where $a,b\in N$, then find the value of (a+b)
My attempt is as follows:-
Rewrite $f(x)=g... | The range will be f(X)$\in$[4,5045]
It is simple approach will be solution to quadratic equation as in article. https://www.mathsdiscussion.com/solution-of-quadratic-equation/
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3432833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Please help how to work the inequation (x-1)/(x-5)<0 I am studying for college finals and I cannot solve the inequation (x-1)/(x-5)<0 using the method the professor taught us, which I have to use in the exam.
This is another inequation using said method:
I noticed that the inequation I posted has a different symbol th... | Remember if $\frac ab < 0$ then $a$ and $b$ are "different signs".
So if $\frac {x-1}{x-5} < 0$ then either
1) $x - 1 > 0$ and $x -5 < 0$
OR
2) $x-1 < 0$ and $x-5 > 0$.
In case 1) we have $x - 1> 0$ so $x > 1$ and $x-5 < 0$ so $x < 5$. So $x$ is between $1$ and $5$ or $1 < x < 5$
In case 2) we have $x -1 < 0$ and $x<... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3433425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Calculate cumulative probability for X~U(0,1) I'm trying to calculate
$$ P\left( |X - \mu_X| \geq k \sigma_X \right) $$
for $X\sim$uniform(0,1). I've calculated that $E[X]=1/2$, Var$[X]=1/12$, know that
$$ f_X(x) = 1, 0 \leq x \leq 1 $$ and calculated that
\begin{equation}
P(X \leq x) = F_X(x) = \begin{cases} 0 & x<0... | you have done well but there are some points you are missing, first of all:
\begin{equation}
P(X \leq x) = F_X(x) = \begin{cases} 0 & x<0,\\ x & 0\leq x < 1\\ 1 & x \geq1 \end{cases}.
\end{equation}
and another one is in the last step of calculations, where:
\begin{equation}
\begin{aligned}
P\left(\left| X-\frac{1}{2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3437268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the interpretation of the difference? We have the linear maps \begin{equation*}f:\begin{pmatrix}x_1 \\ x_2 \\ x_3\end{pmatrix} \mapsto \begin{pmatrix}x_1+x_2 \\ x_3 \\ x_1+x_2\end{pmatrix} \ \text{ and } \ h:\begin{pmatrix}x_1 \\ x_2 \\ x_3\end{pmatrix} \mapsto \begin{pmatrix}x_1+ x_2 \\ x_3 \\ x_1-x_2\end{pmat... | Yes it is correct, the interpretation is that $v$ belongs to the nullspace of $f$ while by $h$ vectors $v$ and $w$ are mapped into two linearly independent vectors.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3437989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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How do I prove that $\sum_{i=3}^{n} \frac{i-2}{\binom{i}{2}} < \frac n 4$ for all natural $n > 3$? How do I prove:
$$\frac{3-2}{\binom 3 2} + \frac {4 - 2}{\binom 4 2} + \dots + \frac{n-2}{\binom n 2} < \frac n 4$$
I tested this sum on a variety of $n$ from $2$ to $100$ and they all seem to be less than $\frac n 4$, ho... | $\displaystyle \sum\limits_{i=3}^n (i-2){\binom i 2}^{-1} = 2\sum\limits_{i=3}^n \frac{i-2}{i(i-1)}=2H_{n-1}-4+\frac{4}{n}$
From the assumption $\displaystyle~2H_{n-1}-4+\frac{4}{n}<\frac{n}{4}~$ we conclude for $~n\to n+1~$ :
$\displaystyle 2H_n-4+\frac{4}{n+1}= \left(2H_{n-1}-4+\frac{4}{n}\right) + \left(\frac{4}{n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3438956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Why does this process map every fraction to the golden ratio? Start with any positive fraction $\frac{a}{b}$.
First add the denominator to the numerator: $$\frac{a}{b} \rightarrow \frac{a+b}{b}$$
Then add the (new) numerator to the denominator:
$$\frac{a+b}{b} \rightarrow \frac{a+b}{a+2b}$$
So $\frac{2}{5} \rightarrow ... | Let $f$ be the map that takes $a/b$ to $(a+b)/(a+2b)$. We can prove inductively that the $n$th iteration of this process gives
$$f^n(a/b) = \frac{F_{n}a + F_{n+1}b}{F_{n+1}a + F_{n+2}b},$$
where $F_n$ is the $n$th Fibonacci number. Since $b$ is always non-zero, asymptotically, this ratio approaches
$$\lim_{n\rightarrow... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3440647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "40",
"answer_count": 5,
"answer_id": 2
} |
Find $\lim\limits_{n \to \infty} \sqrt[3]{n^3+2n^2+1}-\sqrt[3]{n^3-1}$. I have to find the limit:
$$\lim\limits_{n \to \infty} \sqrt[3]{n^3+2n^2+1}-\sqrt[3]{n^3-1}$$
I tried multiplying with the conjugate of the formula:
$$(a-b)(a^2+ab+b^2)=a^3-b^3$$
So I got:
$$\lim\limits_{n \to \infty} \dfrac{n^3+2n^2+1-n^3+1}{\sqrt... | It may be easier with binomial expansion:
$n\left(1+\dfrac2n+\dfrac1{n^3}\right)^{1/3}-n\left(1-\dfrac1{n^3}\right)^{1/3}$
$=n\left[1+\dfrac13\left(\dfrac2n+\dfrac1{n^3}\right)+o\left(\dfrac1{n^2}\right)
-\left(1-\dfrac13\dfrac1{n^3}+o\left(\dfrac1{n^6}\right)\right)\right]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3442861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Prove $4^n+5^n+6^n$ is divisible by 15 Prove by induction:
$4^n+5^n+6^n$ is divisible by 15 for positive odd integers
For $n=2k-1,n≥1$ (odd integer)
$4^{2k-1}+5^{2k-1}+6^{2k-1}=15N$
To prove $n=2k+1$, (consecutive odd integer)
$4^{2k+1}+5^{2k+1}+6^{2k+1}=(4)4^{2k}+(5)5^{2k}+(6)6^{2k}$,
How do I substitute the statem... | You could go mod $3$ and mod $5$ and conclude, an alternate proof is by induction : of course $4^n + 5^n + 6^n$ is divisible by $15$ when $n=1$. However, note that if $n \geq 3$:
$$
4^n + 5^n + 6^n - 4^{n-2} - 5^{n-2} - 6^{n-2} \\=\color{blue}{(4^n - 4^{n-2})} + \color{green}{(5^n - 5^{n-2})} + \color{red}{(6^n - 6^{n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3444343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
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... | $2\cos^2(x)+\sin(x)=1$
Using $2\cos^2(x)-1=\cos(2x)$ we have $\cos(2x)=-\sin(x)=\cos(\pi/2+x).$
Hence $2x=\pm(\pi/2+x)+2n\pi$, $n\in\mathbb{Z}$ and $x=\pi/2+2n\pi$ or $x=-\pi/6+2n\pi/3.$ The second expression can be re-written as $-\pi/6\pm2\pi/3+2k\pi$ or $-\pi/6+2k\pi$ giving the three solutions
\begin{align}
x&=\pi... | {
"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": 3
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Find the $n^{th}$ derivative of $y=\dfrac {x^n}{x-1}$. Find the $n^{th}$ derivative of $y=\dfrac {x^n}{x-1}$.
My Attempt:
$$y=\dfrac {x^n}{x-1}$$
$$y=x^n\cdot(x-1)^{-1}$$
Differentiating both sides,
$$y_{1}=x^n\cdot(-1)\cdot(x-1)^{-2}+(x-1)^{-1}\cdot n\cdot x^{(n-1)}$$
$$y_{1}=x^n\cdot(-1)\cdot(x-1)^{-2}+(x-1)^{-1}\cdo... | $f(x) = \frac{x^n}{x-1} = \frac{x^n-1+1}{x-1} = x^{n-1} + x^{n-2} + \ldots + x + 1 + \frac{1}{x-1} \Rightarrow$
$f^{(n)}(x) = \left(x^{n-1} + x^{n-1} + \ldots + x + 1\right)^{(n)} + \left(\frac{1}{x-1}\right)^{(n)} = 0 + (-1)^n\frac{n!}{(x-1)^{n+1}} = (-1)^n\frac{n!}{(x-1)^{n+1}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3446661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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why $\lim_{x\to+∞}\dfrac{(1+x)^{3/2}}{x^{1/2}} -x\\$ \begin{align}
& \lim_{x\to+∞}\frac{(1+x)^{3/2}}{x^{1/2}} -x\\[8pt]
= {} &\lim_{x\to+∞} \dfrac{(1+(1/x))^{3/2}-1}{1/x} \\[8pt]
= {} & \lim_{x\to+∞} \dfrac{\frac{3}{2}\cdot\frac{1}{x}}{\frac{1}{x}}
\end{align}
Detailed picture with clearer specifications
| The manipulations may be:
$${(1+x)^\frac32 \over \sqrt x }-x = {(1+x)^\frac32\over x^\frac12 } - {x^\frac32\over x^\frac12}$$
$$= {x^\frac32\over x^\frac12}\left(\left(1+\frac 1x\right)^\frac32 - 1\right)$$
Using Newton’s general binomial theorem we have $\left(1+\frac 1x\right)^\frac32 = 1 + \binom{3/2}1 \frac 1x + \b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3448617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $\frac{1}{MA^2}+\frac{1}{MB^2}+\frac{1}{MC^2}+\frac{1}{MD^2}\geq2$ for $M$ inside square $ABCD$ with side length $2$
Show that $$\frac{1}{MA^2}+\frac{1}{MB^2}+\frac{1}{MC^2}+\frac{1}{MD^2}\geq2$$ for any point $M$ inside of the square $ABCD$ whose side length is $2$.
I could manage to prove this using analy... |
COMMENT:
An experimental approach;
In a Pythagorean triple (a, b,c) it can be seen that $c^2<2.5 ab$.Now considering the sketch we have:
$MA^2<2.5(S_{AHME} =AE \times AH)$
$MB^2<2.5(S_{HBGM} =AE \times BH)$
$MC^2<2.5(S_{EMFC} =AH \times EC)$
$MD^2<2.5 (S_{MGDF}=BH \times EC)$
$\frac{1}{MA^2}+\frac{1}{MB^2}+\frac{1}{M... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3449457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Show function is continuous at $(0,0)$ Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ be a function with
$$\space f(x, y) = \begin{equation}
\begin{cases}
\dfrac{y^2\log(1+x^2y^2)}{\sqrt{x^4+y^4}}, & (x, y) \neq (0, 0)\\
0, & (x, y) = 0
\end{cases}\end{equation}$$
Show that f is continious.
I've already shown that f is ... | Let $|x|,|y| <1$.
$\log (1+x^2y^2) \le x^2y^2 \le x^4 +y^4 \le (x^2+y^2);$
$\epsilon >0$ be given.
Choose $\delta =\epsilon^{1/2}$.
Used
AM-GM: $x^4+y^4 \ge 2x^2y^2$; and
$\log (1+z) \le z$ for $z \ge -1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3452216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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How to simplify $(a-b)^2(y-x)+(b-a)(x-y)^2-(b-a)(x-y)$? How to simplify the following?:
$(a-b)^2(y-x)+(b-a)(x-y)^2-(b-a)(x-y)$
I have performed the following, but I think what I have arrived at is not the final step:
$(b-a)((b-a)(y-x)+(x-y)^2-(x-y))$
$ (b-a)((b-a)(y-x)+(y-x)^2+(y-x))$
$ (b-a)((y-x)((b-a)+(y-x)+1)))$
$ ... | Adjust the signs and binomials right away:
$$(a-b)^2(y-x)+(b-a)(x-y)^2-(b-a)(x-y)=\\
-(b-a)^2(x-y)+(b-a)(x-y)^2-(b-a)(x-y)=\\
(b-a)(x-y)[-(b-a)+(x-y)-1]=\\
(b-a)(x-y)[x-y+a-b-1]=\\
(a-b)(x-y)(y-x-a+b+1).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3454440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Proving that $8^x+4^x\geq 5^x+6^x$ for $x\geq 0$. I want to prove that $$8^x+4^x\geq 6^x+5^x$$ for all $x\geq 0$. How can I do this?
My attempt:
I try by AM-GM: $$8^x+4^x\geq 2\sqrt{8^x4^x}=2(\sqrt{32})^x.$$
However, $\sqrt{32}\approx 5.5$ so I am not sure if $$2(\sqrt{32})^x\geq 5^x+6^x$$ is true.
Also, I try to compu... | Note that $8^x=(6^x)^{\log_6(8)}$ so that \begin{equation}\tag 1\label 14^x+8^x-6^x-5^x\geq 8^x-6^x-5^x\to\infty\end{equation} as $x\to\infty$.
I will show that the only non-negative solution to \begin{equation}\tag 2\label 28^x+4^x=6^x+5^x\end{equation} is $x=0$. Note that $$8^x+4^x=6^x+5^x\iff 8^x-6^x=5^x-4^x.$$
By ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3454535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 3
} |
Probability of k heads being the longest run in a row in n flips? EDIT: I miswrote the question I was trying to originally convey. Below is the correct question.
How can I use the Bernoulli distribution calculate the probability of the longest run of heads:
$0$ heads in a row in $5$ flips being the longest run
$1$ head... | Maximum run of zero heads: The only way this can occur is if all five flips are tails. For a fair coin, this has probability
$$\left(\frac{1}{2}\right)^5 = \frac{1}{32}$$
of occurring.
Maximum run of one head: For this to occur, there must either be one, two, or three heads in the sequence, no two of which are conse... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3456137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Let $f(x,y)=\frac{x^2+y^2} x$ In each point of the circle $x^2+y^2-2y=0$ calculate the directional derivative
Let $f(x,y)=\frac{x^2+y^2} x$
a) In each point of the circle $x^2+y^2-2y=0$, what is the value of the directional derivative with respect to a vector $(a,b) \in R^2$
My approach:
the circle is $x^2+(y-1)^2=1$... | We have that
$$x^2+y^2-2y=0 \iff x^2+(y-1)^2=1$$
which can be parametrized by
$$(x,y)=(\cos \theta, \sin \theta +1) \implies \hat v=(-\sin \theta, \cos \theta), \,\theta\in[0,2\pi)$$
and therefore
$$\nabla f(x,y) =\left(\frac{x^2-y^2}{x^2}, \frac{2y}{x}\right)=\left(\frac{-2\sin^2\theta-2\sin \theta}{\cos^2 \theta}, \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3456249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Solve without L'Hopital's rule: $\lim_{x\to0}\frac{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\tan(2x)}$ Solve without L'Hopital's rule:
$$\displaystyle\lim_{x\to0}{\frac{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\tan(2x)}}$$
My work:
$\displaystyle\lim_{x\to0}{\frac{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\tan(2x)}}=\... | Using Taylor series from the strating point.
Use the standard series of $\cosh(t)$ and make $t=3x^2$ gives
$$\cosh(x^3)=1+\frac{9 x^4}{2}+O\left(x^{8}\right)$$ Using the binomial expansion or Taylor series again
$$\sqrt{\cosh(x^3) }=1+\frac{9 x^4}{4}+O\left(x^{8}\right)$$
Use the standard series of $e^t$ and make $t=4x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3457730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Suppose two cubic polynomials f (x) and g(x) satisfy the following: f (2) = g(4); f (4) = g(8); f (8) = g(16); f (16) = g(32) + 64. Suppose two cubic polynomials $f (x)$ and $g(x)$ satisfy the following: If $f(2) = g(4)$; $f (4) = g(8)$;
$f (8) = g(16)$; $f (16) = g(32) + 64$. What is the value of $g(128) − f (64)$?
Th... | Let $F(x)=g(2x)-f(x) $. We have $F(2)=0,F(4)=0,F(8)=0$. So
\begin{eqnarray*}
F(x) = \lambda (x-2)(x-4)(x-8).
\end{eqnarray*}
Using $F(16)=-64$ gives
\begin{eqnarray*}
F(x) = -\frac{(x-2)(x-4)(x-8)}{3 \times 7} .
\end{eqnarray*}
So
\begin{eqnarray*}
F(64) = -\frac{62 \times 60 \times 56}{3 \times 7} =- 8 \times 20 \time... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3460032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\dfrac{dy}{dx}$ if $y=\sin^{-1}\bigg[\sqrt{x-ax}-\sqrt{a-ax}\bigg]$
Derivative of $y=\sin^{-1}\bigg[\sqrt{x-ax}-\sqrt{a-ax}\bigg]$
Put $\sqrt{x}=\sin\alpha,\;\sqrt{a}=\sin\beta$
$$
y=\sin^{-1}\bigg[\sqrt{x-ax}-\sqrt{a-ax}\bigg]=\sin^{-1}\bigg[\sqrt{x(1-a)}-\sqrt{a(1-x)}\bigg]\\
=\sin^{-1}\bigg[\sin\alpha|\cos\b... | Using
$$ (\arcsin u)'=\frac1{\sqrt{1-u^2}}u' $$
one has
\begin{eqnarray}
\frac{dy}{dx}&=&\frac{1}{\sqrt{1-(\sqrt{x-ax}-\sqrt{a-ax})^2}}(\sqrt{x-ax}-\sqrt{a-ax})'\\
&=&\frac1{\sqrt{1-(x-ax)-(a-ax)+2\sqrt{a(1-a)x(1-x)}}}\bigg[\frac{\sqrt{1-a}}{2\sqrt x}+\frac{\sqrt a}{2\sqrt{1-x}}\bigg]\\
&=&\frac{1}{\sqrt{(1-a)(1-x)+ax+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3461824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Proving that a list of perfect square numbers is complete Well, I have a number $n$ that is given by:
$$n=1+12x^2\left(1+x\right)\tag1$$
I want to find $x\in\mathbb{Z}$ such that $n$ is a perfect square.
I found the following solutions:
$$\left(x,n\right)=\left\{\left(-1,1^2\right),\left(0,1^2\right),\left(1,5^2\right... | COMMENT:
n is odd, let $n=(2m+1)^2$ ; m∈Z, then:
$4m(m+1)=12x^2(1+x)$ ⇒ $m(m+1)=3 x^2 (1+x)$
Suppose $m=3t$, then:
$t(3t+1)=x^2(1+x)$
⇒ $n=(2m+1)^2=(6t+1)^2$
For example $n=31^2= (6\times 5 +1)^2$ or $n=55^2=(6\times 9+1)^2$
Therefore there may be more solutions. Now let $m+1=3t$ then:
$(3t-1)3t=3x^2(1+x)$ ⇒ $(3t-1) t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3463339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Error in finding the derivative of the general quadratic function $f(x)=ax^2+bx+c$ Where is the error in this 'proof' that the derivative of the quadratic function is equal to $x(2a+1)$? Note: I use $h$ to denote a small number.
$$f(x)=ax^2+bx+c\\f'(x)=\lim\limits_{h \to 0} \frac{a(x+h)^2+b(x+h)+c-(ax^2+bx+c)}{h}\\=\li... | You wrote $$\lim\limits_{h \to 0}\frac{ax^2+2ahx+ah^2+bx+\color{red}{hx}+c-ax^2-bx-c}{h}$$ instead of $$\lim\limits_{h \to 0}\frac{ax^2+2ahx+ah^2+bx+\color{blue}{bh}+c-ax^2-bx-c}{h}.$$
Simplifying, it becomes
$$\lim\limits_{h \to 0}\frac{ax^2+2ahx+ah^2+bx+bh+c-ax^2-bx-c}{h}=\lim_{h\to0}\frac{h(2ax+b)}{h}=2ax+b$$
as de... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3464817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve $yy^\prime+x=\sqrt{x^2+y^2}$
Solve $yy^\prime+x=\sqrt{x^2+y^2}$
Tried dividing by $y$ to get $$y^\prime+\frac{x}{y}=\frac{\sqrt{x^2+y^2}}{y}$$
$$(y^\prime)^2+2\frac{x}{y}+\frac{x^2}{y^2}=\frac{x^2+y^2}{y^2}$$
$$(y^\prime)^2+2\frac{x}{y}=1$$
$$y^\prime=\sqrt{1-2\frac{x}{y}}$$
Tried using $v=\frac{y}{x}$
$$y^\pri... | Another way:
$$yy^\prime+x=\sqrt{x^2+y^2}$$
$$ydy+xdx=\sqrt{x^2+y^2}dx$$
$$\frac 12 (dy^2+dx^2)=\sqrt{x^2+y^2}dx$$
$$\frac 12 \frac {d(x^2+y^2)}{\sqrt{x^2+y^2}}=dx$$
Integrate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3466125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Limit $\lim\limits_{n\to\infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+2}+\frac{n}{n^2+3}+\cdots+\frac{n}{n^2+n}\right)$ $\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2}+\dfrac{n}{n^2+3}\cdots\cdots+\dfrac{n}{n^2+n}\right)$
Can we write it as following
$E=\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}\right)... | That is not legitimate, because the number of terms inside the limit are growing as $n$ tends to infinity.
Rather, one sees that
\begin{align*}
\sum_{k=1}^{n}\dfrac{n}{n^{2}+n}\leq\sum_{k=1}^{n}\dfrac{n}{n^{2}+k}\leq\sum_{k=1}^{n}\dfrac{n}{n^{2}+1},
\end{align*}
the left and right-sided both tend to $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3467697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 2
} |
evaluate $\lim_{n\to\infty}\left(\frac{n^3-n^2+2n+3}{n^3+2n^2-1}\right)^{\left(\frac{8n^3-3n^2+2n-2}{10n^2+3n-1}\right)}$ $$
\lim_{n\to\infty}\left(\frac{n^3-n^2+2n+3}{n^3+2n^2-1}\right)^{\left(\frac{8n^3-3n^2+2n-2}{10n^2+3n-1}\right)}
$$
I have tried the following:
$$
\lim_{n\to\infty}\left(\frac{n^3-n^2+2n+3}{n^3+2n^... | Write$$f(n):=\frac{n^3-n^2+2n+3}{n^3+2n_2-1}=\frac{1-\frac1n+o\left(\frac1n\right)}{1+\frac2n+o\left(\frac1n\right)}=1-\frac3n+o\left(\frac1n\right)$$and$$g(n):=\frac{8n^3-3n^2+2n-2}{10n^2+3n-1}=\frac45n+o(1)$$so$$\lim_{n\to\infty}f(n)^{g(n)}=\exp\left(-3\cdot\frac45\right)=\exp\left(-\frac{12}{5}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3468124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding the eigenvalues of a $3 \times 3$ matrix without the determinant I am trying to find the eigenvalues of
$$A = \begin{pmatrix} 0 & -2 & -3 \\ -1 & 1 & -1 \\ 2 & 2 & 5\end{pmatrix}$$
Without using the determinant. I first tried doing something like I did here Finding the eigenvalues of $\begin{pmatrix} a & b \\ ... | You have to look at the factors you're dividing the rows by to make diagonal elements $1$: their product turns out to be $(+/-)$ the characteristic polynomial, so its roots will be the eigenvalues. You say you want to do this "without using the determinant", but your row reduction procedure (keeping track of those fac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3468409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Show $x^2+y^2=9z+3$ has no integer solutions
Show $x^2+y^2=9z+3$ has no integer solutions
So I know that $x^2+y^2=3(3z+1)$
And since $3\mid (9z+3)$ and $3\not\mid (3z+1)$ then $9z+3$ cannot be a square since it has a prime divisor which has a power less then $2$.
So does know this isn't a pythagorean triple $(x,y,\sq... | You've shown that $3|9x + 3$. You've also shown that $9\not \mid 9x+3$.
SO that means $3|x^2 + y^2$ so but $9\not \mid x^2 + y^2$.
If $x$ and $y$ are divisible by $3$ then $x^2 + y^2$ is divisible by $9$. So that's out. If one of $x$ or $y$ is divisible by $3$ but the other isn't then $x^2 + y^2$ is not divisible by... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3468562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Would this be a legal move in a Matrix? Here I have a matrix:
\begin{pmatrix}
b^2c^2 & bc & b+c \\
c^2a^2 & ca & c+a \\
a^2b^2 & ab & a+b \\
\end{pmatrix}
I was wondering can I do the following operation on this:
$C2 \to C2 \times C2$
So it would give me the following:
\begin{pmatrix}
b^2c^2 & b^2c^... | Since this will (for most $a$ and $b$) change the value of the determinant, lets use co-factor multiplication by your matrix $M$ to prove this, which shouldn't be too bad given it is a $3x3$ matrix. I will denote each $2x2$ co factor as $A_i$ for simplicity.
$$det(M) = (b+c)det(A_1) - (c+a)det(A_2) + (a+b)det(A_3)$$
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3472138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\int_0^\infty \frac{\sin^n x}{x^m}dx$ could be expressed via $\pi$ or $\log$ I want to show some results first (they were computed by MMA)
$$
\int_0^\infty \frac{\sin^5 x}{x^3} dx =\frac{5}{32}{\color{Red}\pi} \quad
\int_0^\infty \frac{\sin^5 x}{x^5} dx =\frac{115}{384} {\color{Red}\pi} \\
\int_0^\infty \frac{\sin^5 ... | The aim of this answer is to give the explicit expression for the value of the integrals. Essentially it is a development of the previous answer. Notice that for convenience I changed $m$ to $m+1$.
We are going to prove:
For all $0\le m<n$:
$$
\int_0^\infty\frac{\sin^n x}{x^{m+1}}dx
=\frac{(-1)^{\left\lfloor\frac{n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
$\begin{array}{|l}x^2-y^2=7 \\ x^2+xy+y^2=37\end{array}$ Any suggestions how to solve: $$\begin{array}{|l}x^2-y^2=7 \\ x^2+xy+y^2=37\end{array}$$
I can simplify the system and get a homogeneous polynomial of degree $2$, but I think it must have an easier way.
| Another possible approach consists in considering the change of variables $x = r\cos(\theta)$ and $y = r\sin(\theta)$, from whence we get
\begin{align*}
\begin{cases}
r^{2}\cos(2\theta) = 7\\\\
r^{2}(2 + \sin(2\theta)) = 74
\end{cases} \Longleftrightarrow r^{2} = \frac{7}{\cos(2\theta)} = \frac{74}{2+\sin(2\theta)}
\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Find the next year of the year $2016$ written as $n^{m-1}(n^{m}-1)$ See that : $2016=2^{6-1}(2^{6}-1)$
Find the next year of the year $2016$ written as $n^{m-1}(n^{m}-1)$ where $n,m>1$ natural numbers
I don't have any ideas to start in this problem
My be we need the factorization of $2016=2^{5}3^{2}.7$
But I need know... | Since $n^{m-1}(n^{m}-1)$ increases quite fast , we can try inputting some values of $m$
For $m=2$ , we get $n(n^2-1) \gt 2016$ , which is true for $n \ge 13. $ So the smallest value is at $n = 13$ , which gives $13 \times 168 = \color{#2dd}{2184 }.$
For $m=3$ , we get $n^2(n^3-1) \gt 2016$ , which is true for $n ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Normal to the curve $y=x^2$ forming the shortest chord
Find the normal to the curve $y=x^2$ forming the shortest chord
Normal at point $P(x,y)=(t,t^2)$ is $x+2ty=t+2t^3$, which meets the point $Q(m,m^2)$
$$
m+2tm^2=t+2t^3\implies m-t=-2t(m^2-t^2)=-2t(m-t)(m+t)\\
t+m=\frac{-1}{2t}\implies m=-t-\frac{1}{2t}\\
T=\Delta^... | $\left(\frac{4t^2+1}{2t}\right)^2\frac{4t^2+1}{4t^2}=\frac{(4t^2+1)^3}{(2t)^4}$ and not $\left(\frac{4t^2+1}{2t}\right)^3$. With this correction it should continue like user207119's answer to the linked question.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3477867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Finding the asymptotes of $(y-a)^2(x^2-a^2)=x^4+a^4$. Problem: Finding the asymptotes of $(y-a)^2(x^2-a^2)=x^4+a^4$.
My efforts: $(y-a)^2(x^2-a^2)=x^4+a^4\implies (y-a)^2=\dfrac{x^4+a^4}{x^2-a^2}\implies x=\pm a$ are vertical asymptotes. What are the remaining asymptotes?
| This doesn't satisfy
the usual definition
of asymptote but ...
If
$(y-a)^2(x^2-a^2)=x^4+a^4
$
then
$\begin{array}\\
(y-a)^2
&=\dfrac{x^4+a^4}{x^2-a^2}\\
&=\dfrac{x^4-a^2x^2+a^2x^2+a^4}{x^2-a^2}\\
&=\dfrac{x^2(x^2-a^2)+a^2(x^2+a^2)}{x^2-a^2}\\
&=x^2+\dfrac{a^2(x^2-a^2+2a^2)}{x^2-a^2}\\
&=x^2+a^2+\dfrac{2a^4}{x^2-a^2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3478895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Limit of $\sin\left(\sqrt {x+2}\right)-\sin\left(\sqrt{x+4}\right)$ as $x\to\infty$ I wish to find the limit of $$\sin\left(\sqrt {x+2}\right)-\sin\left(\sqrt{x+4}\right)$$ as $x\to\infty$. I think that this limit does not exist.
| Since
$$
\sin p - \sin q = 2\cos \left( {\frac{{p + q}}
{2}} \right)\sin \left( {\frac{{p - q}}
{2}} \right)
$$
you have that
$$
\begin{gathered}
\left| {\sin p - \sin q} \right| = 2\left| {\cos \left( {\frac{{p + q}}
{2}} \right)\sin \left( {\frac{{p - q}}
{2}} \right)} \right| = \hfill \\
\hfill \\
= 2\left|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3482503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Finding mixed probability density function. Please check my answer. Given random variable $X$ with cumulative distributive function
$$
F_X(x)=
\begin{cases}
0&x<0
\\
\dfrac{1}{4}x^2&0\leq x<1
\\
\dfrac{1}{2}&1\leq x<2
\\
\dfrac{1}{3}x&2\leq x<3
\\
1&x\geq 3
\end{cases}.
$$
Find the probability density function of $X$.
... | As you have correctly observed, $F_X$ is not continuous and thus $X$ cannot have an (absolutely) continuous distribtion. So there is no (measurable) positive function $f$ such that
$$F_X(x) = \int_{-\infty}^x f(t) dt, x \in \mathbb{R}$$
and thus no 'density' exists.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3483619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Square and rectangle The following figure shows an infinite checkered mesh, made up of $1$ cm squares, with the vertices of the highlighted squares.
Note that a square and rectangle with vertices in the mesh are highlighted. The square has exactly $5 $ mesh points inside and the rectangle has exactly $2$ mesh points in... | For part b, the square with vertices at $(1,0)$, $(0,10)$, $(10,11)$, and $(11,1)$ and its mirror image (across the line $y=x$) does the trick: each contains the $100$ points $(i,j)$ with $1\le i,j\le10$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3484461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding the Fourier series of $\frac{4-2\cos x}{5-4\cos x}$ I am required to find the Fourier Series of the following function, on $[-\pi,\pi]$:
$$f(x)=\frac{4-2\cos(x)}{5-4\cos(x)}$$
This question seemed simple, at first. I managed to compute $a_0$ (which wasn't too easy) and noticed that $b_n=0$ for all $n\in\mathbb{... | Here is a noncomplex way to do the integral. Multiply top and bottom by $5+4\cos x$:
$$\frac{20+6\cos x +8\cos^2x}{25-16\cos^2 x} = -\frac{1}{2} +\frac{6\cos x}{9+16\sin^2 x} +\frac{65}{2}\frac{1}{9+16\sin^2 x}$$
$$I = \int_{-\pi}^\pi -\frac{1}{2} + \frac{6\cos x}{9+16\sin^2 x} +\frac{65}{2}\frac{1}{9+16\sin^2 x} \:dx ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3485809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How to evaluate $\int_{0}^{1}\frac{1-x}{1+x}\cdot \frac{x^n}{\sqrt{x^4-2x^2+1}}\mathrm dx$ I am trying to calculate this integral, but I find it is very challenging
$$\int_{0}^{1}\frac{1-x}{1+x}\cdot \frac{x^n}{\sqrt{x^4-2x^2+1}}\mathrm dx$$
but somehow I have managed to local its closed form to be $$(-1)^n\left(\frac{... | My solution is similar to that of @omegadot but I think it may be a little cleaner. For $n\in\mathbb{Z}^+$ let
$$C_n=(-1)^n\int_0^1\frac{x^{n-1}}{x+1}dx$$
Then
$$C_1=-\int_0^1\frac{dx}{x+1}=\left.-\ln(x+1)\right|_0^1=-\ln2$$
$$C_{n+1}-C_n=(-1)^{n+1}\int_0^1\frac{x^n+x^{n-1}}{x+1}dx=(-1)^{n+1}\int_0^1x^{n-1}dx=\frac{(-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3486268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Finding the inverse function of: $y=3x^2-6x+1$ I am trying to find the inverse function, $f^{-1}(x)$ of this function, $y=3x^2-6x+1$
Normally the teacher told us to interchange the $x$ and $y$
$x=3y^2-6y+1$ and... now transpose the formula to make $y$ the subject. But this look not possible. Is there another alternati... | $y=3x^2-6x+1=3(x^2-2x)+1$;
$y=3(x-1)^2 -3+1=$
$3(x-1)^2-2$;
A parabola opening upwards, vertex $(1,-2)$;
Local inverse for:
1) $x\gt 1$:
$(1/3)(y+2)=(x-1)^2$;
$x= 1+\sqrt{(1/3)(y+2)}$;
Switch:
$y=1+\sqrt{(1/3)(x+2)}$, $x \gt -2$.
Local inverse for
2) $x \lt 1$;
$-\sqrt{(1/3)(y+2)}=x-1$;
Switch:
$y=1-\sqrt{(1/3)(x+2)}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3487759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find polynomials $P(x)P(x-3) = P(x^2)$
Find all polynomials $P \in \Bbb R[x]$ such that $P(x)P(x-3) = P(x^2) \quad \forall x \in \Bbb R$
I have a solution but I'm not sure about that. Please check it for me.
It is easy to see that $P(x) = 0, P(x) = 1$ satisfied.
Consider $P(x) \neq c$ :
We have $P(x+3)P(x) = P((x+3)... | You point out that if $a$ is a root then $a^{2}$ is a root and $a+3$ is a root.
You don't need to divide into three cases to determine that means either infinitely or zero roots.
By induction $a + 3k; k \in \mathbb N$ are roots. If $j\ne k$ then $a+3j \ne a + 3k$ so if there is one root $a$ there are at least counta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3488111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
How to do partial fractions with a denominator to the power of a variable? How can I take the following sum and simplify it with partial fractions? $$\sum^{\infty}_{k=1}\frac{k-1}{2^{k+1}}$$
I know the denominator can be rewritten as $(2^k)(2)$, but how do I deal with the $2^k$ when doing partial fractions?
Usually,... | Here's a recurrence for
these power sums.
$\begin{array}\\
s_m(x)
&=\sum_{n=0}^{\infty} n^mx^n\\
&=\sum_{n=1}^{\infty} n^mx^n\\
&=x\sum_{n=1}^{\infty} n^mx^{n-1}\\
&=x\sum_{n=0}^{\infty} (n+1)^mx^{n}\\
&=x\sum_{n=0}^{\infty} x^{n}\sum_{k=0}^m \binom{m}{k}n^k\\
&=x\sum_{k=0}^m \binom{m}{k}\sum_{n=0}^{\infty} x^{n}n^k\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3489328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Finding the remainder when $5^{55}+3^{55}$ is divided by $16$
Find the remainder when $5^{55}+3^{55}$ is divided by $16$.
What I try
$a^{n}+b^{n}$ is divided by $a+b$ when $n\in $ set of odd natural number.
So $5^{55}+3^{55}$ is divided by $5+3=8$
But did not know how to solve original problem
Help me please
| Continuing your method:
$$5^{55}+3^{55}=(5+3)(5^{54}-5^{53}\cdot 3+5^{52}\cdot 3^2-\cdots +5^2\cdot 3^{52}-5\cdot 3^{53}+3^{54}) \Rightarrow \\
\small 5^{54}-5^{53}\cdot 3+5^{52}\cdot 3^2-\cdots +5^2\cdot 3^{52}-5\cdot 3^{53}+3^{54}\equiv 1-1+1-\cdots+1-1+1\equiv 1\pmod{2}$$
Hence:
$$5^{55}+3^{55}\equiv 8 \pmod{16}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3490156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 4
} |
Limit $\lim\limits_{x \to +\infty} \sqrt[n]{(x+a_1)(x+a_2)...(x+a_n)} - x$ =? What is the limit of $ f(x) = \sqrt[n]{(x+a_1)(x+a_2)...(x+a_n)} - x$ when $x$ goes to plus infinity? The number $n$ is fixed and $a_i$ are some constants.
All I know is the answer:
$(a_1 + a_2 + ... + a_n) / n$
I figured out that I can limi... | Notice that $$(x+a_1)(x+a_2)\cdot (x+a_n)=x^n+ b x^{n-1}+\mathcal{O}(x^{n-2})$$
Where $b=a_1+a_2+\dots +a_n$, and
$$\sqrt[n]{x^n+bx^{n-1}+\mathcal{O}(x^{n-2})}=\left(x^n+bx^{n-1}+\mathcal{O}(x^{n-2})\right)^{1/n}=x \left(1+bx^{-1} +\mathcal{O}(x^{-2})\right)^{1/n}$$
Now, we can use the first order Taylor approximation... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3491441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Integral $\int_0^e \left(\operatorname{W}(x)^{2}x-\frac{6x}{8}-\frac{3\operatorname{W}(x)}{8}+\frac{3}{8}\right)\,dx=0$ Hi I was playing with the Lambert function when I wondering myself about that :
Prove that :
$$\int_0^e \left(\operatorname{W}(x)^2 x-\frac{6x}{8}-\frac{3\operatorname{W}(x)}{8}+\frac{3}{8} \right... | By the definition of the Lambert $W$ function we have, taking $W(x)=u\Rightarrow x=ue^u$, that $$\int_0^e W(x)^2 x \, dx = \int_0^1 u^3 e^{2u} (u+1) \, du = \frac{3}{8} (e^2-1)$$ and $$-\frac{3}{8} \int_0^e W(x) \, dx = \int_0^1 ue^u (u+1) \, du = -\frac{3}{8} (e-1)$$ and trivially $$\int_0^e \left(\frac{3}{8}-\frac{6}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3494169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Limit of sum $x^3+x^5+x^7+x^9+...)$ I am asked to give the limit of:
$$ x^3+x^5+x^7+x^9+... \quad x\in(-1,1)$$
So I do the following:
The sum of the first $n$ terms will be equal to:
$$x^3+x^5+x^7+...+x^{3+2(n-1)}$$
I factor out $x^3$, I get:
$$x^3(1+x^2+x^4+..+x^{2(n-1)})$$
I also factor out $x^2$, I get:
$$x^3 x^2(1/... | The problem is that when you factor out $x^2$, the remaining powers should still differ by two, because multiplying powers is additive. What you really wanted to do is write $y=x^2$, and it becomes
$$\frac{x^3}{1-y}$$
and we can simply substitute $x^2$ for $y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3494339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Can I apply elementary row operation then find eigen values of a matrix? Suppose if a matrix is given as
$$ \begin{bmatrix}
4 & 6\\
2 & 9
\end{bmatrix}$$
We have to find its eigenvalues and eigenvectors.
Can we first apply elementary row operation . Then find eigenvalues.
Is their any relation on the matrix if it is ... | First compute the eigenvalues as follows:
$$
\begin{align}
\det(\lambda I-A)&=\det\left(
\begin{bmatrix}
\lambda & 0 \\
0 & \lambda
\end{bmatrix}
-
\begin{bmatrix}
4 & 6 \\
2 & 9
\end{bmatrix}
\right)
= \det \left(
\begin{bmatrix}
\lambda-4 & -6 \\
-2 & \lambda-9
\end{bmatrix}
\right) \\
&=
(\lambda-4)(\lambda-9)-(-2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3498199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
$\begin{cases} x^4+x^2y^2+y^4=21 \\ x^2+xy+y^2=3 \end{cases}$
I should solve the following system: $$\begin{cases} x^4+x^2y^2+y^4=21 \\
x^2+xy+y^2=3 \end{cases}$$
by reducing the system to a system of second degree.
What can I look for in such situations? What is the way to solve this kind of systems? The only th... | Multiply the first by $x^2-y^2$ and the second by $x-y$.$$x^6-y^6=21(x^2-y^2),\\x^3-y^3=3(x-y)$$
Then take the ratio
$$x^3+y^3=7(x+y).$$
Adding the two above,
$$2x^3=10x+4y$$
and using $2y=x^3-5x$,
$$4x^2+2x(x^3-5x)+(x^3-5x)^2-12=x^6-8x^4+19x^2-12=0.$$
By trial and error, $x^2=1$ are solutions and we factor as
$$(x^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3500677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Solving for the closed form of recurrence relations using characteristic polynomial I know how to find the closed form of some recurrence relations such as
those that are similar to the Fibonacci Sequence. I am not sure how to solve a recurrence relation using the characteristic polynomial when there is a constant invo... | There is always the matrix approach:
$$
\begin{pmatrix}
a_{n} \\ 1
\end{pmatrix}
=
\begin{pmatrix}
3 & -1 \\
0 & 1
\end{pmatrix}
\begin{pmatrix}
a_{n-1} \\ 1
\end{pmatrix}
$$
The characteristic polynomial of that matrix is $x^2-4x+3=(x - 3)(x - 1)$ and so $a_n=\alpha 3^n + \beta 1^n$. The coefficients are determined b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3501413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solving logarithm leaving in terms of $p$ and $q$ I would like to check the steps if Part a) is done correctly.
For Part b), how do I continue from below? I seem to stuck for $\log_{10}(5)$…
Here is the problem:
Given that $p = \log_{10} 2$ and $q = \log_{10} 7$, express the following in terms of $p$ and $q$.
a) $\log... | Part a is correct and $\log_{10} 5+\log_{10} 2=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3504307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Given that $\piThe left side of the expression reduces to $2\sin x$ and the right side reduces $2+2\sin x$
Their addition gives $2+4\sin x$ . The correct answer is 2
I realize that this has got to do with the fact that $\pi<x<3\pi/2$ , in which case $\sin x$ will be -ve.
But then the expression should be $2-4\sin x$.... | $$\sqrt{4\sin^4x+\sin^2(2x)}+4\cos^2\left(\frac\pi4-\frac x2\right)$$
$$=\sqrt{4\sin^4x+4\sin^2x\cdot\cos^2x}+2\left(\cos\left(\frac\pi2-x\right)+1\right)$$
$$=2\vert \sin x \vert \sqrt{\sin^2x+\cos^2x}+2\sin x+2$$
$$=-2\sin x+2\sin x+2=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3505174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
We roll a fair die until a $5$ appears. What is the expected value of the minimum value rolled?
Question: Given a fair dice, we roll until we get a $5.$ What is the expected value of the minimum value rolled?
Answer is $\frac{137}{60}.$
There is a similar question asked in MSE but I do not understand the method used ... | First, $X$ is not the minimum value rolled before obtaining a $5$, it is the minimum value rolled up to and including the first roll that comes up $5$, so that $X=5$ is possible.
The event $X=5$ means that $5$ comes up before any of $1$, $2$, $3$, or $4$ (we don't care about $6$). Since each of the five numbers is equa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3505861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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Proving the inequality $\prod_{n=1}^\infty \left( 1+\frac1{n^2+\ln n} \right) < \frac72$ I am struggling to prove
$$\prod_{n=1}^k \left( 1+\frac1{n^2+\ln n} \right) < \frac72$$
For all $k \geq 1$. Clearly, this is true for $k=1$, so it suffices to show
$$\prod_{n=1}^\infty \left( 1+\frac1{n^2+\ln n} \right) < \frac72$... | I have not full solution to this problem.
I want to prove by induction that:
$$ \prod_{n=1}^{k} \left(1 + \frac{1}{n^2 + \ln{n}}\right) < \frac{7}{2} \frac{(k+1)^2 + \ln(k+1)}{(k+2)^2 + \ln(k+2)} < \frac{7}{2}$$
It's obvious for $k=1$.
Suppose we proved for $k$:
$$ \prod_{n=1}^{k + 1} \left(1 + \frac{1}{n^2 + \ln{n}}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3506406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Find the radius of the inscribed circle of a right triangle. Find the radius of the inscribed circle of a right triangle. The triangle's height is $\sqrt{6} + \sqrt{2}$ while the bisector of the right angle is 4.
This seemed like a generic similar triangles problem but it isn't that simple when I tried to solve it. Any... |
Find the radius $r$ of the inscribed circle of a right triangle
$ABC$
given its height $|CD|=h_c=\sqrt6+\sqrt2$
and bisector $|CE|=4$.
Let $I$ be the center of the inscribed circle
and $A_t,B_t,C_t$ its touching points
\begin{align}
\triangle CED,\ \triangle IEC_t,\triangle CB_tI:\quad
\frac{h_c}{\beta_c}
&=
\frac{r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3506942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 2
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$\int\sqrt{x^2\sqrt{x^3\sqrt{x^4\sqrt{x^5\sqrt{x^6\sqrt{x^7\sqrt{x^8\ldots}}}}}}}\,dx$ I was attempting to solve an MIT integration bee problem (1) when I misread the integral and wrote (2) instead.
$$\int\sqrt{x\cdot \sqrt[3]{x\cdot \sqrt[4]{x\cdot\sqrt[5]{x\ldots } }}}\,dx\tag{1}$$
$$\int\sqrt{x^2\sqrt{x^3\sqrt{x^4\... | The binomial theoem gives$$(1-x)^{-2}=\sum_{n\ge0}(n+1)(-1)^nx^n,$$so the exponent is$$-1+(1-1/2)^{-2}=3.$$So the integral is $\int x^3dx=\frac14x^4+C$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3507554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
An urn of 4 balls with 2 colors. Pick 2 balls and place them back 4 times. What's the probability of picking 2 balls of the same color twice in a row? An urn of 4 balls with 2 colors. Pick 2 balls and place them back 4 times. What's the probability of picking 2 balls of the same color twice in a row?
So the probability... | Assumptions:
*
*There are two balls of one color and two balls of a second color
*Suppose the colors are Green and Red. A successful outcome occurs in any of these cases when you choose two green followed by two green, two green followed by two red, two red followed by two green, or two red followed by two red.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3509877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
if $mx+3|x+4|-2=0$ has no solutions, solve for $m$
If $mx+3|x+4|-2=0$ has no solutions, which of the following value could be $m$?
(A)5
(B)$-\frac{1}{2}$
(C)-3
(D)-6
(E)$\frac{10}{3}$
my attempt:
$$mx-2=-3|x+4| \\ m^2x^2-4mx+4=9x^2+72x+144 \\ (9-m^2)x^2+(4m+72)x+140=0$$
because the equation has no solutions, therefor... | 1) $y_1=3|x-(-4)| \ge 0$ ;
2) $y_2= Mx +2$, where $M=-m$
$M=3=-m$ no intersection (Why?)
No intersection for $3 \ge M >1/2$ (Why?);
$3 \ge -m >1/2$, or
$-3 \le m < -1/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3510944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Linear differential equations, integrating factor Solve the following differential equation:
$$ dr+(2r \cot\theta+\sin 2\theta)d\theta=0$$
I have tried like this:
$$ \frac{dr}{d\theta}+2r\cot\theta=-\sin{2\theta}$$
\begin{align}
I.F. &=e^{\int{2\cot\theta d\theta}}\\
& =e^{-2\log\sin\theta}\\
& =\frac 1{\sin^2\theta}\\... | You messed up with your multiplication. Multiplying the integral factor to both sides should yield:
$$\dfrac{d}{d\theta}\left(\dfrac{r}{\sin^2\theta}\right)=\left(\dfrac{1}{\sin\theta}\right)^2\dfrac{dr}{d\theta}+2\cot\theta\left(\dfrac{1}{\sin\theta}\right)^2 r=-\sin2\theta\left(\dfrac{1}{\sin\theta}\right)^2=-2\cot\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3511335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Inverse of $I + A$ I am trying to solve the following exercise in Artin, without breaking into cases for even and odd $k$.
A square matrix $A$ is called nilpotent if $A^k = 0$ for some $k > 0$. Prove that if $A$ is nilpotent, then $I + A$ is invertible.
Here's my attempt.
I claim that the inverse of $I + A$ is
$$... | We know that, if $|x|<1$, then
$$\dfrac{1}{1+x} = 1 - x + x^2 - x^3 + x^4 -x^5 + \cdots$$
For just any square matrix, $\mathbf A$, it is not true that
$$\dfrac{1}{1+\mathbf A} =
1 - \mathbf A + \mathbf A^2 - \mathbf A^3 + \mathbf A^4 -\mathbf A^5 + \cdots$$
But, since $\mathbf A$ is nilpotent, this becomes the fin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3511926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Why is $(1+\frac{1}{n})^n < (1+\frac{1}{m})^{m+1}$? Why is it that
$$
\left(1+\frac{1}{n}\right)^n < \left(1+\frac{1}{m}\right)^{m+1},
$$
for any natural numbers $m, n$?
I have tried expanding using the binomial series and splitting into cases. I understand why it is trivially true when $m=n$ but I am not sure if the... | Both sequences
$$
a_n=\left(1+\frac{1}{n}\right)^n, \quad
b_n=\left(1+\frac{1}{n}\right)^{n+1}
$$
are monotonic. In particular, $a_n$ is increasing and $b_n$ is decreasing.
Hence, if $m,n\in \mathbb N$ and $k=$max$\{m,n\}$, then
$$
a_n\le a_k\le b_k\le b_m.
$$
Why are these sequences monotonic?
We have
$$
\frac{a_{n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3512177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
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Find locus of $\Delta ABC$ centroid with orthocentre at origin and side slopes 2, 3 and 5 Let $ABC$ be a triangle with slopes of the sides $AB$, $BC$, $CA$ are $2,3,5$ respectively. Given origin is the orthocentre of the triangle $ABC$. Then find the locus of the centroid of the triangle $ABC$.
Since sides have slo... | Respectively, the vertexes A, B and C are on the three altitude lines you obtained $3y+x=0$, $5y+x=0$ and $2y+x=0$. So, let their coordinates be $A(a,-\frac a3)$, $B(b,-\frac b5)$ and $C(c,-\frac c2)$. Then, use them to match the three side slopes
$$\frac{-\frac b5 + \frac a3 }{b-a}=2,\>\>\>\>\>\>\>
\frac{-\frac b5 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3512736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Birthday problem-Probability exactly $2$ triples and $4$ pairs if $20$ people in room Say there are 20 people in a room. What is the probability there are exactly 2 triples and 4 pairs. Is my answer shown below correct?
Assume 365 days in the year.
$P= \dfrac{\binom{365}{2}\binom{363}{4}\binom{20}{3}\binom{17}{3}\binom... | You have solved the problem correctly.
There are $\binom{365}{2}$ ways to choose the two days for the triples. There are $\binom{20}{3}$ ways to choose which three people share the earlier of these two birthdays and $\binom{17}{3}$ ways to choose which three of the remaining people share the later of these two birth... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3513720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How to multiply $\sqrt{5x^2+2x+1}$ with $\frac{1}{x}$? I believe the answer says, $\sqrt{5+\frac{2}{x}+\frac{1}{x^2}}$, but I don't see how they obtained it. I'm quite rusty with radicals.
| Just a slight variation to the approaches already posted.
Suppose
$A = \sqrt{5x^2 + 2x + 1}$
$B = \frac{1}{x}$
If $AB = C$, then ${(AB)}^2 = C^2$ or $A^2B^2 = C^2$
$A^2$ = $5x^2 + 2x + 1$
$B^2$ = $\frac{1}{x^2}$
So $C^2 = 5 + \frac{2}{x} + \frac{1}{x^2}$
And to return to C: $$\sqrt{C^2} = \pm\sqrt{5 + \frac{2}{x} + \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3516409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Why does't quadratic formula work to factor polynomial when $a \ne 1$? $$2x^2 + 3x + 1$$
applying quadratic formula:
$$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$$
$$a=2, b=3, c=1$$
$$x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot1}}{2\cdot2}$$
$$x = \frac{-3 \pm \sqrt{9-8}}{4}$$
$$x = \frac{1}{4}[-3 + 1],~~~x=\frac{1}{4}[-3-1... | On the quadratic formula:
$$ax^2 +bx+c=0$$
$$r_i =\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$
$$a\cdot(x-r_1)(x-r_2)$$
First, the "$a\cdot$" term in the factored result "$a\cdot(x-r_1)(x-r_2)$" is an often overlooked term when using the quadratic formula to factor a second order polynomial, because usually a=1, or the "$a\cdot$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3516494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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If $\cos^{-1}(\frac{x}{a})+\cos^{-1}(\frac{y}{b})=\theta$, then $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{2xy} {ab}\cos(\theta)$ If $\cos^{-1}(\frac{x}{a})+\cos^{-1}(\frac{y}{b})=\theta$, prove
that $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{2xy}{ab}\cos(\theta)=\sin^2(\theta)$
My trial:Let:
$\cos^{-1}(\frac{x}{a})=\alpha$, ... | Using
\begin{eqnarray*}
\cos(a+b) = \cos(a) \cos(b)- \sin(a) \sin(b).
\end{eqnarray*}
We have
\begin{eqnarray*}
\frac{xy}{ab} - \sqrt{ \left( 1- \frac{x^2}{a^2} \right) \left( 1- \frac{y^2}{b^2} \right) } = \cos( \theta).
\end{eqnarray*}
Now move the $\frac{xy}{ab}$ to RHS, Square and rearrange.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3517512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find the $n$-th power of a $3{\times}3$ matrix using the Cayley-Hamilton theorem. I need to find $A^n$ of the matrix $A=\begin{pmatrix}
2&0 & 2\\
0& 2 & 1\\
0& 0 & 3
\end{pmatrix}$ using Cayley-Hamilton theorem.
I found the characteristic polynomial $P(A)=(2-A)^2(3-A)$ from which I got $A^3=7A^2-16A+12$. How to c... | We can compute $A^2$ directly:
$$
A^2 = \left(
\begin{array}{ccc}
2 & 0 & 2 \\
0 & 2 & 1 \\
0 & 0 & 3 \\
\end{array}
\right)\left(
\begin{array}{ccc}
2 & 0 & 2 \\
0 & 2 & 1 \\
0 & 0 & 3 \\
\end{array}
\right)
=
\left(
\begin{array}{ccc}
4 & 0 & 10 \\
0 & 4 & 5 \\
0 & 0 & 9 \\
\end{array}
\right).
$$
From the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3517795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Find the remainder when $p(x)$ is divided by $x^2 -1$ Polynomial $p(x)$ leaves a remainder of $4$ when divided by $x-1$ and a remainder of $-2$ when divided by $x+1$.
Find the remainder when $p(x)$ is divided by $x^2 -1$ .
According to Remainder Theorem, when a polynomial $p(x)$ is divided by $(ax+b)$, the remainder i... | $3x+1 = 3(x-1) + 4\\3x+1 = 3(x+1)-2$
So your answer $3x+1$ is correct.
| {
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"timestamp": "2023-03-29T00:00:00",
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Does $\iint_D \frac{x^2}{x^2+y^2} dx dy $ converge on $D= \left\{ (x, y) : x^2+y^2\leq ax \right\} $ ? If yes, what value does it converge to? Powers equal to $2$ entice the polar substitution.
$D= \left\{ (x, y) : x^2+y^2\leq ax \right\}$, so $0 \leq r \leq a \cos \phi.$ For the domain to make sense, we need either $\... | I get the same result as your first one. But I don't see the need for special functions, or introducing the $\epsilon$ you use.
Consider the region $x^2+y^2\leq ax$. This is the same as $$\left(x-\frac{a}{2}\right)^2+y^2\leq\frac{a^2}{4}$$
So the region is a disc centered at $\left(\frac{a}{2},0\right)$ with its bounda... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove $f(x)=x^3+x$ is injective How can algebraically prove $f(x)=x^{3}+x$ is injective.
I can get to $a^2+b^2+ab=-1$ but I can't go any further.
| It suffices to note that $f(x)$ is a strictly increasing function on the reals.
That said, if you do want to go that route, we have
$$\begin{align}
f(a) = f(b)
&\iff
a^3+a = b^3 +b
\\&\iff
(a^3-b^3)+(a-b) = 0
\\&\iff
(a-b)(a^2+ab+b^2)+(a-b) = 0
\\&\iff
(a-b)(a^2+ab+b^2+1) = 0
\\&\iff
a = b \,\,\,\text{ or }\,\,\, a^2+... | {
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"timestamp": "2023-03-29T00:00:00",
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How to make a generating function to solve this problem? How many ways are there to put 8 identical balls into 3 boxes so that no box has more than 4 balls in it? But Box 3 can only have up to 2 balls inside it.
I started the problem by trying to write out the polynomials. For the first two boxes I thought it would be ... | Your generating function approach to giving the answer $6$ as the coefficient of $x^8$ in the expansion of $(1+x+x^2)(1+x+x^2+x^3+x^4)^2$ is correct and the possibilities are:
4 + 4 + 0
3 + 4 + 1
4 + 3 + 1
2 + 4 + 2
3 + 3 + 2
4 + 2 + 2
Your method assumes the balls are indistinguishable but the boxes are distinguisha... | {
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"timestamp": "2023-03-29T00:00:00",
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What's the generating function for $\sum_{n=1}^\infty\frac{\overline{H}_n}{n^2}x^n\ ?$ Is there closed form for
$$\sum_{n=1}^\infty\frac{\overline{H}_n}{n^2}x^n\ ?$$
where $\overline{H}_n=\sum_{k=1}^n\frac{(-1)^{k-1}}{k}$ is the alternating harmonic number.
My approach,
In this paper page $95$ Eq $(5)$ we have
$$\sum... | This is a long comment to https://math.stackexchange.com/a/3523732/198592 which just provides my result for comparison.
Let $\overline{H}_n=\sum_{k=1}^{n}(-1)^{k+1}\frac{1}{k}$ be the alternating harmonic sum and define the generating function of order $q=0,1,2,...$ as
$$g_{q}(x) = \sum_{n=1}^\infty\frac{\overline{H}_n... | {
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"timestamp": "2023-03-29T00:00:00",
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The real numbers $w, x, y, z$ satisfy the equalities $w+x+y+z=0$ and $w^2+x^2+y^2+z^2=1$. Prove that $-1\le wx+xy+yz+zw\le0$ The real numbers $w, x, y, z$ satisfy the equalities $w+x+y+z=0$ and $w^2+x^2+y^2+z^2=1$. Prove that $-1\le wx+xy+yz+zw\le0$
Supposing $wx+xy+yz+zw>0$. I've proven via contradiction that $wx+xy+y... | We have $a^2+b^2 \geq 2ab$ for any two real numbers $a$ and $b$. So
$$
\begin{aligned}
0=(x+y+z+w)^2 &= x^2+y^2+z^2+w^2+2(xy+yz+zw+wx)+2(xz+wy) \\
&=1+2(xy+yz+zw+wx)+2(xz+wy) \\
&\leq 1+2(xy+yz+zw+wx)+(x^2+z^2+y^2+w^2) \\
&=2+2(xy+yz+zw+wx)
\end{aligned}
$$
It follows that
$$-1 \leq xy+yz+zx+zw$$
Equality is achieved ... | {
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"timestamp": "2023-03-29T00:00:00",
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For △ABC, prove $\frac a{h_a} + \frac b{h_b} + \frac c{h_c} \ge 2 (\tan\frac{\alpha}2+ \tan\frac{\beta}2 + \tan\frac{\gamma}2)$
Given $\triangle ABC$, (using the main parameters and notation), prove that $$ \frac{a}{h_a} + \frac{b}{h_b} + \frac{c}{h_c} \ge 2 \cdot \left(\tan\frac{\alpha}{2} + \tan\frac{\beta}{2} + \ta... | Use the sine rule and the area formuli of the triangle $\frac12 h_a a = \frac12 bc\sin\alpha$, $\frac12 h_b b = \frac12 ca\sin\beta$, $\frac12 h_c c = \frac12 ab\sin\gamma$ to express
$$\frac{a}{h_a} + \frac{b}{h_b} + \frac{c}{h_c} = \frac{\sin^2\alpha +\sin^2\beta+\sin^2\gamma}{\sin\alpha \sin\beta\sin\gamma}\tag 1$$
... | {
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"timestamp": "2023-03-29T00:00:00",
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Second order IVP differential equation So the given material and equations I have are
$$(1):x=C_1\cos(t)+C_2\sin(t)$$
$$x \left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}$$
$$x'\left(\frac{\pi}{3}\right)=0$$
First I sub in the $\frac{\pi}{3}$ and the $\frac{\sqrt{3}}{2}$ into my first equation and get
$$\frac{\sqrt{3}}{2... | $$0=-\frac{\sqrt{3}}{2}C_1+\frac{1}{2}C_2$$
$$ \implies \sqrt{3}C_1=C_2$$
Plug that in first equation.
$$\frac{\sqrt{3}}{2}=\frac{1}{2}C_1+\frac{\sqrt{3}}{2}C_2$$
$$\frac{\sqrt{3}}{2}=\frac{1}{2}C_1+\frac{{3}}{2}C_1$$
$$C_1=\frac{\sqrt{3}}{4}$$
$$ \implies C_2=\sqrt{3}C_1=\frac {3} 4$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $P=\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}$ when $a+b+c=0$
Let $a,b,c$ such that $$a + b + c =0$$ and
$$P=\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}$$ is defined.
Find the value of $P$.
This is a very queer problem.
| Elegant approach
Think of $p(a, b, c)$ as a function $p : X\to\mathbb{R}$, where $X = \{(a, b, c)\in\mathbb{R}^3 : a+b+c=0\}$.
If $p$ is a constant function, i.e. if $p(a, b, c)\equiv P$ for some constant $P$, then in particular, $p(a, b, c) = P$ for any values of $a, b, c$ we might care to choose, so long as they sat... | {
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"timestamp": "2023-03-29T00:00:00",
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In how many different ways can apples be placed in the box if at least there is one apple of each color?
Suppose there are $4$ red apples, $5$ green apples, and $6$ yellow apples, $9$ of them will be put into a box. In how many different ways can apples be placed in the box if at least there is one apple of each color... | The generating function for the number of ways to arrange $n$ apples is
$$
\begin{align}
&\left(x+x^2+x^3+x^4\right)\left(x+x^2+x^3+x^4+x^5\right)\left(x+x^2+x^3+x^4+x^5+x^6\right)\\
&=\frac{x-x^5}{1-x}\frac{x-x^6}{1-x}\frac{x-x^7}{1-x}\\
&=\left(x^3-x^7-x^8-x^9+x^{12}+x^{13}+x^{14}-x^{18}\right)\sum_{k=0}^\infty\binom... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3530271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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