Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How to factorise $x^4 - 3x^3 + 2$, so as to compute the limit of a quotient? Question:
Find the limit: $$\lim_{x \to 1}\frac{x^4 - 3x^3 + 2}{x^3 -5x^2+3x+1}$$
The denominator can be simplified to: $$(x-1)(x^2+x)$$
However, I am unable to factor the numerator in a proper manner (so that $(x-1)$ will cancel out)
I know... | Here are the steps
$$\lim_{x \to 1}\frac{x^4 - 3x^3 + 2}{x^3 -5x^2+3x+1}$$
$$= \lim_{x \to 1}\frac{(x-1)(x^3- 2x^2-2x-2)}{(x-1)(x^2-4x-1)} $$
$$= \lim_{x \to 1}\frac{x^3- 2x^2-2x-2}{x^2-4x-1} $$
$$= \frac{1- 2-2-2}{1-4-1} = \frac{1- 6}{1-5} = \frac{5}{4} $$
| {
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"url": "https://math.stackexchange.com/questions/1163127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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$y''+y'^{2}+y=0$ equation solution How would you solve this differential equation $y''+y'^{2}+y=0$? I can't apply the ansatz method (or more formally apply the characteristic polynomial method). Thanks
| $$
y'' + y'^2 + y = 0
$$
we can rewrite as this (prove it)
$$
\frac{1}{2}\frac{d}{dy}p^2 + p^2 + y = \frac{d}{dy}p^2 + 2p^2 + 2y = 0
$$
where $p = y'$.
$$
p^2 = \mathrm{e}^{-2y}\left(-\int 2y\mathrm{e}^{2y}dy + C \right)
$$
thus solution in terms of $p$
$$
p = \sqrt{\frac{1}{2} -y + C\mathrm{e}^{-2y}} = y'
$$
or
$$
\i... | {
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"timestamp": "2023-03-29T00:00:00",
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Permutations with odd-length cycles I need to find - as a homework problem - the exponential generating function for the number of permutations of $n$ consisting of an even number of odd-length cycles. I can retrieve the exponential generating function for the number of permutations with just odd-length cycles, by
$$
E... | There are two possible interpretations here, the first, permutations
consisting of an even number of odd cycles and some even cycles and
second, permutations consisting of an even number of odd cycles only.
First interpretation.
Observe that the generating function of permutations with odd cycles
marked is
$$G(z... | {
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"url": "https://math.stackexchange.com/questions/1163681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Simplifying help: If ${n \choose 3} + {n+3-1 \choose 3} = (n)_3$, compute $n$. If ${n \choose 3} + {n+3-1 \choose 3} = (n)_3$, compute $n$.
So far, I have:
$\frac{n!}{3! (n-3)!} + \frac{(n+2)!}{3! (n-1)!}$
Then I simplified to:
$\frac{1}{6} (n-2)(n-1)(n) + \frac{1}{6} (n)(n+1)(n+2)$
Is this correct, and if so, what is ... | You are nearly ready.
Substitute $(n)_3 = n \cdot (n-1) \cdot (n-2)$, and you will see both sides are divisable by n. Thus, converting the result to a polinomial form, you will get an at most 2-grade equation, which will be soon very easy to solve.
$$\frac{1}{6}(n-2)(n-1)n + \frac{1}{6}n(n+1)(n+2) = n(n-1)(n-2)$$
Multi... | {
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"timestamp": "2023-03-29T00:00:00",
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Solve: $-3(6t^3-1)^6 -3t[6(6t^3-1)^5(18t^2)]$ Solve: $-3(6t^3-1)^6 + -3t[6(6t^3-1)^5(18t^2)]$
I don't know how to multiply the two equations and then add them.
| Let $A = 6t^3-1$.
\begin{align} -3(6t^3-1)^6 + -3t[6(6t^3-1)^5(18t^2)] &= -3A^6-3t(108A^5t^2) \\
&= -3A^6-324A^5t^3\\
&=-3A^5(A+108t^3)\\
&=-3(6t^3-1)^5[(6t^3-1)+108t^3]\\
&=-3(6t^3-1)^5(114t^3-1)\\
\end{align}
So $t=\sqrt[3]{\frac{1}{6}}, \sqrt[3]{\frac{1}{114}}.$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Prove by mathematical induction that $\forall n \in \mathbb{N} : \sum_{k=1}^{2n} \frac{(-1)^{k+1}}{k} = \sum_{k=n+1}^{2n} \frac{1}{k} $ Prove by mathematical induction that:
$$\forall n \in \mathbb{N} : \sum_{k=1}^{2n} \frac{(-1)^{k+1}}{k} = \sum_{k=n+1}^{2n} \frac{1}{k} $$
Step 1: Show that the statement is true for $... | When you go from $n=p$ to $n=p+1$, what you want to show is
$$\sum_{k=1}^{2(p+1)} \frac{(-1)^{k+1}}{k} = \sum_{k={\color{red}{(p+1)}}+1}^{2(p+1)} \frac{1}{k}. $$
The crucial part I think you're overlooking is in red. To get the desired result, you have to pull the first term out of your sum $\sum_{k=p+1}^{2p+1} \frac{1... | {
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"timestamp": "2023-03-29T00:00:00",
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Finding square root of $-5-12i$ by formula and by De Moivre's Theorem I was trying to obtain the square root of $-5-12i$ by the formula for square root (given below) and also by De Moivre's theorem and verify that both give the same result. But the two results are somehow not matching for this complex number. I am writ... | There is a way without using $\arctan$.
Let the square root be $a+bi$. Then, $a^2+2abi-b^2 = -5-12i$, so $a^2-b^2 = -5 \text{(eq. 1)}$ and $2abi = -12i \ \text{(eq. 2)}; ab = -6 \ \text{(eq. 3)}$.
From equation $3$, $a = -\frac{6}{b}$. Substituting into equation $1$, $(-\frac{6}{b})^2-b^2+5=0$, and so:
$$\frac{36}{b^2}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $x^3-y^3 = (x-y)(x^2+xy+y^2)$ without expand the right side? I can prove that $x^3-y^3 = (x-y)(x^2+xy+y^2)$ by expanding the right side.
*
*$x^3-y^3 = (x-y)x^2 + (x-y)(xy) + (x-y)y^2$
*$\implies x^3 - x^2y + x^2y -xy^2 + xy^2 - y^3$
*$\implies x^3 - y^3$
I was wondering what are other ways to prove... | You could do the enclidian division of $X^3 - Y^3$ by $X-Y$ in the ring $A[X]$, where $A = \mathbf{Z}[Y]$, has $X-Y$ has unit leading coefficient.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1172119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 12,
"answer_id": 8
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Inequality between real numbers $a^ab^bc^c<(abc)^{\frac{a+b+c}{3}}$ let $a,b$ and $c$ positive reals. Shows that $a^ab^bc^c<(abc)^{\frac{a+b+c}{3}}$
How prove this inequality. Thanks
a suggestion please to prove this inequality
| Let's transform $a^ab^bc^c\geq(abc)^{(a+b+c)/3}$ by taking $\log$:
$$
a^ab^bc^c\geq(abc)^{(a+b+c)/3}\iff a\log a+b\log b+c\log c\geq\frac{a+b+c}{3}\log(abc)\\
\iff\frac{a}{a+b+c}\log a+\frac{b}{a+b+c}\log b+\frac{c}{a+b+c}\log c\geq\log(\sqrt[3]{abc}).
$$
Let's prove the last inequality above. With $f(x)=-\log x=\log(1... | {
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"timestamp": "2023-03-29T00:00:00",
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Which of following inequalities hold in interval 0 to pi/2
i tried using calculator and i got 1,2,4 correct .But i am not sure about how to prove them
| For C,
I recognize the right side
as the first 3 terms
of the binomial theorem expansion
of
$\sqrt{1+x}$.
If it is squared,
we get
$\begin{array}\\
(1+\frac{x}{2}-\frac{x^2}{8})^2
&=1+\frac{x^2}{4}+\frac{x^4}{64}
+2\frac{x}{2}-2\frac{x^2}{8}-2\frac{x^3}{16}\\
&=1+x-\frac{x^3}{8}+\frac{x^4}{64}\\
&=1+x-\frac{x^3}{8}(1-\... | {
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"url": "https://math.stackexchange.com/questions/1174120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve $16x^{-3}=-2$
Solve $16x^{-3}=-2$.
My working:
\begin{align}
16x^{-3}&=-2\\
\frac{1}{16x^{3}}&=-2\\
\frac{16x^3}{16x^3}&=-32x^3\\
1&=-32x^{3}\\
-32x^{3}&=1\\
-32x&=\sqrt[3]{1}\\
-32x&=1\\
x&=\frac{-1}{32}
\end{align}
Is this right? What have I done wrong?
| $$16x^{-3} = -2$$
$$\frac{16}{x^3} = -2$$
$$-8 = x^3$$
$$x = -2$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Critical numbers of the function: $x\sqrt{5-x}$ Let f(x) = $$\displaystyle f(x) = x\sqrt{5-x} $$
On the interval: [-6,4]
Critical numbers are the the values of x in the domain of f for which f'(x) = 0 or f'(x) is undefined.
Derivative of the function:
$$ \frac{1}{2} \cdot x (5-x)^{\frac{-1}{2}} \cdot -1$$
$$ \frac {\fr... | $\dfrac{d(x\sqrt{5-x})}{dx}=\dfrac{10-3x}{2\sqrt{5-x}}$
$10-3x=0$ imply the critical number is $x=\dfrac{10}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1175845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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chain rule derivative This is an derivatives Problem.
find the tangent to $y=\sqrt{x^2-x+5\;}\;$ at $x=5$.
$y = \boxed{\;?\;}$
What I did was first find y by plugin x into the equation. The answer is $y = 5$. Then I found the derivative of both $x^2-x+5$ and $\sqrt{x^2-x+5\;}\;$ the answers for those are $2x-1$ and $... | You are on a right track. You have found the derivative and now just plug $x=5$ in it and you have:
$$f'(x) = \frac{2x-1}{2\sqrt{x^2-x+5}}=\frac{10-1}{2\sqrt{25-5+2}} = \frac{9}{10}$$
This gives you the slope of the tangent at the point $(5,5)$ so you have $y=\frac 9{10}x + b$. But note that $(5,5)$ is a point that lie... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 1
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Finding the interval after substitution Given this problem
$$8\cdot 3^{\sqrt{x}+\sqrt[4]{x}}+9^{\sqrt[4]{x}+1}\geq 9^{\sqrt{x}}$$
$$8\cdot 3^{\sqrt{x}+\sqrt[4]{x}}+3^{2\sqrt[4]{x}+2}\geq 3^{2\sqrt{x}}\\8\cdot 3^{\sqrt{x}+\sqrt[4]{x}-2\sqrt{x}}+3^{2\sqrt[4]{x}+2-2\sqrt{x}}\geq 1\\8\cdot 3^{\sqrt[4]{x}-\sqrt{x}}+3^{2\s... | Your work seems correct: the only place equality takes place is at $x=16$. Substituting that into your original equation checks, with $6561$ on both sides.
Since both sides of your inequality are continuous for $x\ge 0$, the solution set for your inequality is either $[0,16]$ or $[16,\infty)$. A quick calculator check ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1177025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A fair 6-sided die is rolled 6 times, what's the probability the outcome has exactly 2 or 3 elements? A fair $6$-sided die is rolled $6$ times independently. For any outcome, this is the set of numbers that showed up at least once in the different rolls. For example, the outcome is $(2,3,3,3,5,5)$, the element set is $... | Exactly two numbers appear:
There are $\binom{6}{2}$ ways for two of the six numbers to appear. On each of the six rolls of the die, there are two possible outcomes, giving $\binom{6}{2} \cdot 2^6$ possible outcomes in which two of the six numbers appear. However, of these $2^6$ sequences, there are two in which the ... | {
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Given the $ x+y+z =3$ Prove that $ x^2+y^2+z^2 \geq 3xyz$ Given the $ x+y+z =3$ and x, y and z are positive numbers. How to prove that $ x^2+y^2+z^2 \geq 3xyz$.
I tried many methods but I failed.
I did the AM-HM in-equality, but failed.
$\frac{\frac{x}{yz} + \frac{y}{zx} + \frac{z}{xy}}{3} \geq \frac{3}{\frac{xy}{z}+... | use AM-GM inequality
$$3=x+y+z\ge 3\sqrt[3]{xyz}\Longrightarrow xyz\le 1$$
and
$$x^2+y^2+z^2\ge 3\sqrt[3]{x^2y^2z^2}=3\sqrt[3]{(xyz)^2\cdot 1}\ge\sqrt[3]{(xyz)^2\cdot(xyz)}= 3\sqrt[3]{(xyz)^3}= 3xyz$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1179131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Proving infinity limit of a multivariable function I have a function $f(x,y,z): \mathbb{R}^3\rightarrow \mathbb{R}$ which is $f(x,y,z)=3x^2+z^2+y^2-2xy+14$.
I'm trying to show that $f(x,y,z)\rightarrow \infty$ when $||(x,y,z)|| \rightarrow \infty$.
(Or formally: $\forall_{M>0}\exists_{R>0}\forall_{(x,y,z) \in \mathbb{R... | Solved! Given $M>0$, setting $R=\sqrt{2M}$, then for all vectors $(x,y,z) \in \mathbb{R}^3$ for which $||(x,y,z)||>R$, it holds that:
$f(x,y,z)=3x2+z2+y2−2xy+14=$
$0.5(x^2+y^2+z^2)+0.5(x^2+z^2+(2x-y)^2+28)$
$\geq 0.5(x^2+y^2+z^2) > M$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1184124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Evaluation of $\int\frac{1}{x^4-5x^2+16}dx$
Evaluation of $\displaystyle \int\frac{1}{x^4-5x^2+16}\,dx$
$\bf{My\; Try::}$ Given $$\displaystyle \int\frac{1}{x^4-5x^2+16}dx = \frac{1}{8}\int\frac{\left(x^2+4\right)-\left(x^2-4\right)}{x^4-5x^2+16}\,dx$$
So We get $$\displaystyle = \frac{1}{8}\int\frac{x^2+4}{x^4-5x^2+... | You could notice that $x^4-5x^2+16$ can be factored as $(x-a)(x-b)(x+a)(x+b)$ where $a$ and $b$ are complex numbers. Then partial fraction decomposition would lead to quite simple terms and the result for integral will just be the sum of four logarithms.
However, it is more likely that the recombination of the results ... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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"answer_id": 2
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Quadratic Residues $\pmod {2^n}$ I'd imagine this is a duplicate question, but I can't find it:
How many quadratic residues are there $\pmod{2^n}$.
I tried small $n$: $n=1: 2, n=2:2, n=3: 3, n=4: 4, n=5: $not 5: 0, 1, 4, 9, 16, 25
No pattern :/
| There is a basic solution that only uses modular arithmetic.
Let $p(n)$ be the number of quadratic residues modulo $2^n$. We know that $p(1) = p(2) = 2$. Now suppose $n \geq 3$. We will proof 3 small lemmas.
Lemma 1. It suffices to consider the residues of the numbers $0,1,...,2^{n-2}$
Proof. Indeed, the following rela... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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evaluate the double integral $\int_0^4 \int_{\sqrt{y}}^2 \sqrt{x^2+y}\, dxdy$ evaluate the double integral
$\int_0^4 \int_{\sqrt{y}}^2 \sqrt{x^2+y}\, dxdy$
Hi all, could someone give me a hint on this question?
I've actually tried converting to polar coordinates but i cant seem to get the limits. But if polar coordinat... | Here is a polar treatment.
First get rid of the square root by $z=\sqrt{y}$ with $dy=2zdz$: $$I=\int_0^4 \int_{\sqrt{y}}^2 \sqrt{x^2+y}\, dxdy=2\int_0^2 \int_{z}^2 \sqrt{x^2+z^2}z\, dxdz$$
Now draw a graph to see the integration regions:
Therefore with $z=r\sin \theta$, $x=r \cos \theta$ and $dxdz=r dr d\theta$ we ha... | {
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"url": "https://math.stackexchange.com/questions/1187943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Show that $f(n) = -\sqrt{n} + n\log\left(1 + \frac{1}{\sqrt{n}}\right), f:\mathbb{N}\rightarrow \mathbb{R}$ is a decreasing function. The way I tried to approach this question was to consider it as a continuous $f: \mathbb{R} \rightarrow \mathbb{R}$. I then tried to show that $\frac{d}{dx}f < 0$. So
\begin{align*}
\fra... | Hint: To complete what you had begun, compute the second derivative and study the function $f'$. Deduce its sign from its table of variations.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove: If $a$, $b$, and $c$ are consecutive integers such that $a< b < c $ then $a^3 + b^3 \neq c^3$.
Prove: If $a$, $b$, and $c$ are consecutive integers such that $a< b < c $ then $a^3 + b^3 \neq c^3$.
My Attempt: I start with direct proof.
Let $a,b,c$ be consecutive integers and $a< b < c $, there exists a ... | hint:If $k^3 = 3k^2+9k+7 \to k(k^2-3k-9) = 7 \to k\mid 7 \to k = \pm 1, \pm 7$. Can you continue?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1191907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Locus of a midpoint Let $Γ_1$ be a circle of radius $4$, and let $Γ_2$ be a circle of radius $14$. The distance between the centers of $Γ_1$ and $Γ_2$ is $25$. Let $A$ be a variable point on $Γ_1$, let $B$ be a variable point on $Γ_2$, and let $M$ be the midpoint of $AB$. Let $S$ be the set of all possible locations of... | This is most likely analytic geometry, so I'd say we can put:
$$\Gamma_1\;:\;\; x^2+y^2=16\;\;;\;\;\;\Gamma_2\;:\;\; (x-25)^2+y^2=196$$
Take now for example
$$\begin{cases}A=\left(x\,,\,\sqrt{16-x^2}\right)\in\Gamma_1&,\;-4\le x\le4\\{}\\B'=\left(11\,,\,0\right)\in\Gamma_2\end{cases}$$
The midpoint of $\;AB'\;$ is give... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1192134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Taylor series of $x/(x^2-4x+5)$ I'm supposed to find the Taylor series of this function (I can choose to center it at any A I want):
$$f(x)= x/(x^2-4x+5)$$
When I derivate, it only gets more and more confusing. How can I make any sense out of this?
| i think you can do a taylor series about a general center $x = a.$ of course there is more work to do. i am going to try to find a series about $x = a.$ make a change of variable $x = a + h,$ then we have
$$\frac{x}{x^2 - 4x + 5} = \frac{a + h}{(a+h)^2 - 4(a+h) + 5} = \frac{a + h}{\left(a^2 - 4a + 5 + 2h(a-2)+h^2\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1193204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is there another way to solve this integral? My way to solve this integral. I wonder is there another way to solve it as it's very long for me.
$$\int_{0}^{\pi}\frac{1-\sin (x)}{\sin (x)+1}dx$$
Let $$u=\tan (\frac{x}{2})$$
$$du=\frac{1}{2}\sec ^2(\frac{x}{2})dx $$
By Weierstrass Substitution
$$\sin (x)=\frac{2u}{u^2+1... | Yet another way. Using $u=\tan(x)$, we get
$$
\begin{align}
\int\frac{1-\sin(x)}{1+\sin(x)}\,\mathrm{d}x
&=\int\lower{2pt}{\frac{1-\frac{u}{\sqrt{1+u^2}}}{1+\frac{u}{\sqrt{1+u^2}}}\frac{\mathrm{d}u}{1+u^2}}\\
&=\int\left(1-\raise{2pt}{\frac{u}{\sqrt{1+u^2}}}\right)^{\!\!2}\,\mathrm{d}u\\
&=\int\left(2-\raise{2pt}{\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1194139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 9,
"answer_id": 5
} |
How many positive solutions are there (Positive 3 tuples)? I want to find how many positive solutions for the Diophantine equation $4x + 2y + 5z = 100$
I found a particular solution $(x,y,z) = (50,-50,0)$ then I found a general solution (basis) $s(-2,-1,2) + t(1,-2,0)$ for the homogeneous equation $4x + 2y +5z = 0$ and... | Since $5z = 100 - 4x - 2y$, $z$ must be even. Since $x, y, z > 0$, there are only nine possible values of $z$, namely, $z = 2, 4, 6, 8, 10, 12, 14, 16, 18$. By substituting these values for $z$, we can reduce the problem to determining the number of solutions of the equations
\begin{align*}
4x + 2y & = 10\\
4x + 2y &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1194915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Various evaluations of the series $\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}$ I recently ran into this series:
$$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}$$
Of course this is just a special case of the Beta Dirichlet Function , for $s=3$.
I had given the following solution:
$$\begin{aligned}
1-\frac{1}{3^3}+\frac... | I do not know how much this could help you; so forgive me if I am out off topic.
Rewriting a little the expression $$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}x^n=\frac{1}{8} \,\Phi \left(-x,3,\frac{1}{2}\right)$$ where appears the Lerch transcendent function. Now, using $x=1$, we can get the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1195285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
} |
Find solution of $(1-x^2)y''-xy'+p^2y=0, p \in \mathbb{R}$ The following differential equation is given:
$$(1-x^2)y''-xy'+p^2y=0, p \in \mathbb{R}$$
*
*Find the general solution of the differential equation at the interval $(-1,1)$ (with the method of power series).
*Are there solutions of the differential equation... | The Equation can be rewritten as( D^2-(x/(1-x^2))×D+x^2/1-x^2)y=0.
This equation can easily be solved using one integral method .
As P+XQ=0 then solution will assume type of y=xv.
substitute dy/dx by xdv/dx+v ,d^2y/dx^2 by xd^2/dx^2+2dv/dx and y by xv
This equation will be converted to linear diffrential equation with ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1197701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove that $f(x) = \sqrt{x^2 + x}$ where $x \in [0, +\infty)$ is uniformly continuous Prove that $f(x) = \sqrt{x^2 + x}$ where $x \in [0, +\infty)$ is uniformly continuous.
So lets take:
${\mid \sqrt{x^2+x} - \sqrt{y^2+y} \mid}^2 \, \leqslant \,\, {\mid \sqrt{x^2+x} - \sqrt{y^2+y} \mid}{\mid \sqrt{x^2+x} + \sqrt{y^2+y... | For $\;x\in [1,\infty)\;$:
$$\left(\sqrt{x^2+x}\right)'=\frac{2x+1}{2\sqrt{x^2+x}}\le\frac{2x+1}{2x}=1+\frac1{2x}\stackrel{\text{Why?}}\le\frac32$$
Thus, having a bounded derivative makes $\;\sqrt{x^2+x}\;$ uniformly continuous in $\;[1,\infty]\;$, and being continuous in the bounded, closed interval $\;[0,1]\;$ it is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1198508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find the equation of tangent line to the curve $x=\cos(t) + \cos(2t)$, $y= \sin(t) + \sin(2t)$ at the point $(-1,1)$. I get the equation for the slope as $\frac{\cos(t) + 2\cos(2t)}{-\sin(t) -2\sin(2t)}$ but I'm unsure how to solve for the value of $t$. I know I need to sub in $-1$ and $1$ for $x$ and $y$ but the equat... | Given is curve:
$$\begin{gathered}
x(t) = \cos (t) + \cos (2t) \hfill \\
y(t) = \sin (t) + \sin (2t) \hfill \\
\end{gathered}$$
$$\begin{gathered}
\varphi (t) = \left( {\begin{array}{*{20}{c}}
{x(t)} \\
{y(t)}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{\cos (t) + \cos (2t)} \\
{\sin (t) +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1199259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
How come, in this problem, the maximum product is always achieved using only $2$s and $3$s? Consider the following problem.
Given a number $N$, write it as a sum $n = n_1 + n_2 + \cdots + n_k$, such that the product $p = n_1 \times n_2 \times \cdots \times n_k$ is maximized.
For example, $11$ can be written as $2 + 3 +... | For every sequence of natural numbers $\{a_1,\dots,a_n\}$ such that $N=\sum\limits_{k=1}^{n}a_k$:
*
*If $a_k=4$, then you can get an equivalent value by decomposing it into $2+2$ since $2\cdot2=4$
*If $a_k=5$, then you can get a better value by decomposing it into $2+3$ since $2\cdot3=6>5$
*If $a_k=6$, then you ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1200745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Combinatorial graph theory proof Why does $\binom{1/2}{n+1} * (-1)^n * 2^{2*n+1}$ equal $1/(n+1) * \binom{2n}{n}$? I came across this in an exercise in Graph Theory and Its Applications by Gross and Yellen. I haven't been able to solve it, can anyone give any insight?
| We know that,
$$\binom{x}k=\frac{x^{\underline k}}{k!}$$
where $x^{\underline k}$ is the falling factorial function defined as $x^{\underline k}=x(x-1)(x-2)\cdots (x-k+1)$
Also, $x!!$ is the double factorial function defined as,
$$x!!=\begin{cases}x(x-2)(x-4)\cdots 4\cdot 2\cdot 1~,~\textrm{when }x\textrm{ is even}\\x(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1204781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proof : Limit of a sequence
Prove from the definition of the limit of a sequence that $$\lim_{n\to\infty} \frac{2n^2+\cos(n)} {n^2+1} = 2 $$
(that is, for a given $\epsilon > 0$, find an explicit $N_\epsilon$)
Please explain.
I choose $n_0$ such that ${n_0} > \sqrt {\frac{3}{\varepsilon }} $. Then how can I say that... | Let $\epsilon > 0$ be given. Choose an $n_0 \in \mathbb{N}$ such that $n_0 > \sqrt{\frac 1{\epsilon}}$. Then for all $n \in \mathbb{N}$ such that $n \ge n_0$, we have
\begin{align}
\left| {\frac{{2{n^2} + \cos n}}
{{{n^2} + 1}} -2 } \right|&=\left| {\frac{{2{n^2} + \cos n}}
{{{n^2} + 1}} - \frac{2(n^2+1)}{n^2+1}} \righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1207083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Combinatorial sum inequality Prove the following inequality:
$$ \forall k\in\left\{4n+5:n\in\mathbb{N}\right\},\qquad\sum_{m=0}^{\frac{k-1}{2}}{\left( -1 \right) }^{m}\binom{k}{2m}2^{2m}\neq 1. $$
I'm particularly interested in a combinatorial or number-theoretical solution.
| We just have to compute:
$$\sum_{m=0}^{2n+2}\binom{4n+5}{2m}(2i)^{2m}=\frac{1}{2}\left(\sum_{k=0}^{4n+5}\binom{4n+5}{k}(2i)^k+\sum_{k=0}^{4n+5}\binom{4n+5}{k}(-2i)^k\right) $$
that is:
$$\sum_{m=0}^{2n+2}\binom{4n+5}{2m}(2i)^{2m}=\frac{(1+2i)^{4n+5}+(1-2i)^{4n+5}}{2}.$$
If we set:
$$ A_n = \frac{(1+2i)^{n}+(1-2i)^{n}}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1209899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Differential equation : $y' = (x+1)/(xy+x)$ So, I have the following differential equation to solve : $$y' = \frac{x+1}{xy+x}$$
After several steps, I get here :
$t^2 + 2t = 2x + 2ln(x) + c$
How do I isolate $t ?$
thank you!
By the way, $ t=f(x)=y$
| Start by dividing through by $x$:
$$
y'=\frac{1+\frac{1}{x}}{y+1} \\
y'(y+1)=1+\frac{1}{x} \\
\int y'(y+1)dx=\int 1+\frac{1}{x}dx \\
\frac{y^2}{2}+2y=2x+2\ln(x)+c~\text{for constant $c$} \\
y^2+2y+1=2x+2\ln(x)+2c+1 \\
(y+1)^2=2x+2\ln(x)+2c+1 \\
y+1=\pm \sqrt{2x+2\ln(x)+2c+1} \\
y=\pm \sqrt{2x+2\ln(x)+2c+1}-1.~_{\square... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1210662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Lowering powers of $\cos^2x \sin^4x$ First, I will be straight up, this is a homework question.
I need to write $\cos^2 x \sin^4 x$ in terms of cosine to the first power. I know that $\sin^4x$ =
$$ \frac{3-4\cos 2x+\cos 4x}{8}$$
from there I go:
$$ \frac{1+\cos 2x}{2} \cdot \frac{3-4\cos 2x+\cos 4x}{8}$$
$$ \frac{3(1 ... | I found:
$$cos^2(x)sin^4(x)=\frac{1}{32}(-cos(2x)-2cos(4x)+cos(6x)+x)$$
Verify the indenty:
$$32cos(x)^2sin(x)^4=2-cos(2x)-2cos(4x)+cos(6x)$$
$$6-2cos(2x)-4-4cos(4x)+2cos(4x)+cos(2x)+cos(6x)=2-cos(2x)-2cos(4x)+cos(6x)$$
$$\frac{1}{32}(-cos(2x)-2cos(4x)+cos(6x)+x)=\frac{1}{32}(-cos(2x)-2cos(4x)+cos(6x)+x)$$
The left han... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1216162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If $|\sin^2 x+ 17 - x^2|=|16-x^2|+2 \sin ^2 x + \cos ^2x$ then $x$ lies in what interval If $|\sin^2 x+ 17 - x^2|=|16-x^2|+2 \sin ^2 x + \cos ^2x~$ then $x$ lies in what interval?
Hints Please? I dont know how to approach this problem.
| Observe that $|\sin^2x+1+16-x^2|=|16-x^2|+\sin^2x+1$ $(*)$
Set $y=\sin^2x+1$ and $z=16-x^2$. Then $(*)$ reduce to
$|z+y|=|z|+y$. Taking square of both sides we get $y(z-|z|)=0$
Hence, since $\sin^2x+1\neq0$ we have $16-x^2=|16-x^2|$ and this is true only for $ -4\leq x\leq 4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1216955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proving $\frac{1}{\sqrt{2}}=1-\frac{1}{2^2}-\frac{1}{2!2^4}-\frac{3!!}{3!2^6}-\frac{5!!}{4!2^8}-\cdots$ How can I prove
$$\frac{1}{\sqrt{2}}=1-\frac{1}{2^2}-\frac{1}{2!2^4}-\frac{3!!}{3!2^6}-\frac{5!!}{4!2^8}-\cdots$$
I wanted to prove it by using the Taylor series of $\sqrt{2}$, but I couldnt do.
| We have to prove:
$$\frac{1}{\sqrt{2}}=\frac{3}{4}-\sum_{n\geq 1}\frac{(2n-1)!!}{(n+1)!2^{2n+2}}=\frac{3}{4}-\frac{1}{4}\sum_{n\geq 1}\frac{(2n)!}{n!(n+1)!8^{n}}$$
that follows from:
$$ \sum_{n\geq 1}\frac{(2n)!}{n!(n+1)!}z^n = \frac{1-\sqrt{1-4z}-2z}{2z}.\tag{1}$$
$(1)$ is just a minor variation of the generating func... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1217847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
find taylor series to fourth term I'm wondering if there is faster method than just calculating derivatives with finding taylor series up to 4 term of function $\displaystyle f(x)=\frac{(1+x^4)}{(1+2x)^3(1-2x)^2}$
| Since the Taylor series of $g(x)=\frac{x^4}{(1+2x)^3(1-2x)^2}$ in a neighbourhood of the origin is given by $x^4+o(x^4)$, it is enough to compute the Taylor series of:
$$\begin{eqnarray*} h(x)&=&\frac{1}{(1+2x)^3(1-2x)^2}\\&=&\left(1-6 x+24 x^2-80 x^3+240 x^4+o(x^4)\right)\left(1+4 x+12 x^2+32 x^3+80 x^4+o(x^4)\right)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1219248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the best polynomial approximation of the piecewise function Find the best approximation of the function
$$ f(x)= \begin{cases}
1 - x \quad\text{ for } 0 \le x \le 1 \\
1 + x \quad\text{ for } -1 \le x \le 0 \\
\end{cases}$$
in the interval $[-1,~ 1]$ by a polynomial $p(x)$ of degree $\le 4$, meaning that the int... | $\newcommand{\Er}{{\rm Error}}$
Define the problem and solve the differential calculus:
$$\begin{align}
%
\Er(A, B, C, D, E)
%
&= \int_{-1}^{1} (Ax^4 + Bx^3 + Cx^2 + Dx + E - f(x))^2 ~dx
%
\\ \\ &= \int_{-1}^{0} (Ax^4 + Bx^3 + Cx^2 + Dx + E - 1 - x)^2 ~dx \\
\\ &+ \int_{0}^{1} (Ax^4 + Bx^3 + Cx^2 + Dx + E - 1 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1220082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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diophantine equation $x^3+x^2-16=2^y$ Solve in integers: $x^3+x^2-16=2^y$.
my attempt:
of course $y\ge 0$, then $2^y\ge 1$, so $x\ge 1$.
for $y=0,1,2,3$ there is no good $x$.
so $y\ge 4$ and we have equation $x^2(x+1)=16(2^z+1)$, where $z=y-4\ge 0$.
what now?
| This is only a solution for even values of $x$:
For $x$ even, $x^2$ is even and $x+1$ is odd. So $\gcd(16, x+1) = 1$. So $16 \mid x^2$. So $4 \mid x$. So $x = 4k$ for some $k$.
$16k^2(4k+1)=16(2^z+1)$
$k^2(4k+1)=2^z+1$
If $k$ is even then the RHS is even and we have a contradiction. So $k$ must be odd. So $k = 2t+1$ fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1220402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
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The greatest common divisor Problem : Confirm the following properties of the greatest common divisor:
If $\gcd(a,b) = 1$, and $c \mid (a+b)$, then $\gcd(a,c) = \gcd(b,c) = 1$.
Is this right? This is my answer:
*
*$\gcd(a,b) = 1$ => There exist integers $x$ and $y$ such that: $ax + by = 1$.
*$c \mid (a+b)$ => The... | I think your reasoning is right, and I can provide a shorter proof: $\gcd(a,a+b)=\gcd(a,b)=1$ and similarly $\gcd(b,a+b)=\gcd(b,a)=1$. Then $c\mid a+b$ implies $\gcd(a,c) = \gcd(b,c) = 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1221190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Prove that a subset is normal Let $H$ be the subset of $GL(2, \mathbb R)$ consisting of all matrices of the form
$$ \left[
\begin{array}{cc}
x & 0\\
0 & x\\
\end{array}
\right] $$
where $x \neq 0$
Prove that H is normal.
UPDATE
So if I show $\left[
\begin{array}{cc}
h & 0\\
0 & h\\
\end{array}
\right]$ ... | You are indeed on the wrong track. Sounds like you have shown $H$ to be a subgroup. As for proving $H$ is a normal subgroup, we pick any arbitrary element $\left[\begin{array}{cc}
a & b\\
c & d\\
\end{array}\right] \in \text{GL}_2(\mathbb{R})$ and prove that $\left[\begin{array}{cc}
a & b\\
c & d\\
\end{array}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1221858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to deduce that $1\cdot 1 + 2\cdot 1 + 2\cdot 2 + 3\cdot 1+3\cdot 2+3\cdot 3 +...+(n\cdot n) = n(n+1)(n+2)(3n+1)/24$ I know how to reason $$1\cdot2 + 2\cdot3 + 3\cdot4 + n(n-1) = \frac{1}{3}n(n-1)(n+1)$$
However, I'm stuck on proving $$1\cdot1 + (2\cdot1 + 2\cdot2) + (3\cdot1+3\cdot2+3\cdot3) + \cdots +(n\cdot 1+...... | Base case:
Prove true for n=1
$ 1 * 1 = \frac{1(1 + 1)(1 + 2)(3(1) + 1)}{24} $
Inductive step:
Assume for n = k, prove for n = k+1. If we let the property be equal to P(n)
$ P(k+1) = P(k) + ((k+1) * 1 + ... +(k+1)*(k+1)) $
$ P(k+1) = \frac{k(k + 1)(k + 2)(3k + 1)}{24} + ((k+1) * 1 + ... +(k+1)*(k+1)) $
From here you sh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1225311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Solving second order differential equation in a hurry During last test I had to solve second order ODE: $y'' + 3y = e^x + 1$. I managed to write the general solution of homogenous equation (with RHS replaced by zero):
$$y = c_1 \cos \left(\sqrt{3} x\right) + c_2 \sin \left(\sqrt{3} x\right).$$
That wasn't hard. But how... | try the method undetermined coefficients; try a particular solution of the form $$ y = ae^x + b, y' = ae^x, y'' = ae^x. $$ sub this in $$y'' + 3y = e^x + 1 \to ae^x + 3(ae^x + b)=e^x + 1 \to a = \frac 14, b = \frac 13 $$
therefore, a particular solution is $$y_p = \frac 14 e^x + \frac 13, y_g = Ce^{-3x}+y_p $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1225626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proving $\lim_{x\to c}x^3=c^3$ for any $c\in\mathbb R$ using $\epsilon$-$\delta$ definition
$\lim_{x\to 3}x^3=c^3$ for any $c\in\mathbb R$
Let $\epsilon>0$. Then
$$|x^3-c^3|=|x-c||x^2+xc+c^2|.$$
Let $\delta=\min\{1,\epsilon/x^2+xc+c^2\}$.
Then if $0<|x-c|<\delta$ and therefore since $|x-c|<\epsilon/(x^2+xc+c^2)$,
$$|... | Given $\epsilon>0$, we need $\delta>0$ such that if $0<|x-c|<\delta$, then $|x^3-c^3|<\epsilon$. Now,
$$
|x^3-c^3| = |x-c||x^2+xc+c^2|.
$$
If $|x-c|<1$, then we have that $-1<x-c<1$ or simply $c-1<x<c+1$ so that
$$
x^2+xc+c^2<(c+1)^2+(c+1)(c)+c^2=(c^2+2c+1)+(c^2+c)+c^2=3c^2+3c+1,
$$
and so
$$
|x^3-c^3| = |x-c||x^2+xc+c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1227691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Bound for the sum of a finite sequence Consider ${\bf c} = (a,b) \in \mathbb{R}^2$ with $0< \|{\bf c}\| < 1.$ Let $n \in \mathbb{N} $ and define
\begin{align*}
F_{n}(k) & := \frac{ [a + x_{n}(k)]^2}{ [a + x_{n}(k)]^2 + [b + y_{n}(k)]^2}
\end{align*}
where $\omega := 2\pi/n,$ $x_{n}(k) := \cos(2 k \pi/n)$ and $y_{n}(k... | One idea is the following:
Let $z_k=x_n(k)+iy_n(k)=e^{ik\omega}$ and $c=a+ib=re^{i\theta}$ with $0<r<1$. Then, $F_n(k)=\cos^2\theta_k,\ 0\le k\le n-1$ where $\theta_k=\angle (z_k+c)$ so that $F_n(k)=\pi+\frac{\pi}{n}\sum_{k=0}^{n-1}\cos 2\theta_k$.
I think this can be helpful in finding some bounds.
Edit: One can note... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1229977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
multivariable limit of $\frac{x^2-y^2}{\sqrt{x^2+y^2}}$ Calculate multivariable limit of $$\lim_{(x,y) \rightarrow (0,0)}\frac{x^2-y^2}{\sqrt{x^2+y^2}}$$
How to do that? I was trying to figure out any transformations e.g. multiplying by denominator but I do not think it gives me any solution.
| $$\lim_{(x,y) \rightarrow (0,0)}\frac{x^2-y^2}{\sqrt{x^2+y^2}}$$
Transforming $x=r\cos\theta, y=r\sin\theta$.
$$\lim_{(x,y) \rightarrow (0,0)}\frac{x^2-y^2}{\sqrt{x^2+y^2}}=\lim_{r \rightarrow 0}\frac{r^2\cos^2\theta-r^2\sin^2\theta}{\sqrt{r^2\cos^2\theta+r^2\sin^2\theta}}=\lim_{r \rightarrow 0}\frac{r^2(\cos^2\theta-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1232753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
The Galois closure of $\mathbb{F}_2(x,y) / \mathbb{F}_2(x)$ Consider the field extension $\mathbb{F}_2(x,y) / \mathbb{F}_2(x)$ where $y$ is a root of the polynomial $g(T) = f(x,T) \in \mathbb{F}_2(x)[T]$, with
$$
\begin{array}{l}
f(x,y) = x^{12} + x^{10}y^2 + x^{10}y + x^{10} + x^9y^2 + x^9y + x^8y^4 + x^8y^2 + x^8 ... | Let $p(X,Y) = (X^{16}+X)(Y^2+Y)+(Y^{16}+Y)(X^2+X)$.
If $p(x,y) = p(x,z) = 0$ then $p(x,y+z) = 0$ so for a given value of $x$, the roots of the corresponding polynomial form a $\Bbb F_2$-subspace, usually $4$-dimensional.
Notice that for any $x \in \overline{\Bbb F_2}$,
$p(x,Y)$ has a double root $\iff$ $0$ is a doubl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1233228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to validate the basis of the space of solutions of a system of linear equations Problem
Solve the following nonhomogeneous system of linear equations:
$$
\begin{aligned}
x_{1} + 2 x_{2} + 3 x_{3} + x_{4} &= 4,\\
2 x_{1} + 2 x_{2} + 3 x_{3} + x_{4} &= 5,\\
3 x_{1} + 3 x_{2} + 4 x_{3} + x_{4} &= 6
\end{aligned}$$
Ple... | I'm not sure if the teacher's comment was for specifically this question, but you're right. The solution to this system lies on a line and not a plane as the teacher's comment suggests.
To check that your answer is correct, substitute the general form back into the system. Here's an example with the first equation:
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1236440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to integrate $x\ln(x+1)$? I am trying to compute $\int x\ln (x+1)\, dx$. I tried integrating by parts and ended up with:
$$\int x\ln(x+1)\,dx = \frac{1}{2}x^2\ln(x+1) - \frac{1}{2}\int\frac{x^2}{x+1}\,dx$$ but I'm stuck here.
| $$\int \dfrac{x^2}{x+1}dx = \int \dfrac{x^2-1}{x+1}dx + \int \dfrac{dx}{x+1} = \int (x-1) dx + \int \dfrac{dx}{x+1} = \dfrac{x^2}2 - x + \ln(x+1)+ c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1236753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Find the function that equals to $1-x^3+x^6-x^9+ \cdots$
Find the function that equals to $1-x^3+x^6-x^9+ \cdots$ for all $|x| < 1$
I know that $\frac{1}{1+x} = 1-x+x^2-x^3+...$ But I couldn't find the pattern here
| Hint: if $|f(x)| < 1$ then $$\frac{1}{1+f(x)} = 1 - f(x) + f(x)^2 - f(x)^3 + \dotsb$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1236846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove by mathematical induction: $\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n^2}>1$ Could anybody help me by checking this solution and maybe giving me a cleaner one.
Prove by mathematical induction:
$$\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n^2}>1; n\geq2$$.
So after I check special cases for $n=2,3$, I have to prove... | If you accept a proof without induction, here is one:
\begin{align*}
\frac 1n+\Bigl(\frac 1{n+1}+\dots+ \frac1{n^2}\Bigr)>\frac 1n +(n^2-n)\cdot\frac 1{n^2}=\frac 1n+1-\frac1n=1.
\end{align*}
This computation is valid if $n^2>n$, i.e. if n>1.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1239518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 1
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How do you find the factorial of a decimal or negative number and what does it show us? I know that you can find the factorial of positive integers where n!= n(n-1)...2 x 1. However, what if you want to find the factorial of a negative integer or a decimal? I tried to do it on my calculator and it gave an answer howeve... | As already noted by @ncmathsadist, the Gamma Function $$\Gamma(z)=(z-1)!$$ can be extended to a meromorphic function defined on the complex plane without the non-positive integers.
Here's an instructive example how to work with negative factorials presented in section $3.6$ of $A=B$ by M. Petkovsek, H. Wilf and D. Zei... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1242602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Find the values of $\sin 69^{\circ},\sin 18^{\circ} , \tan 23^{\circ}$
Calculate $\sin 69^{\circ},\sin 18^{\circ} , \tan 23^{\circ}$. accurate upto two decimal places or in surds .
$\begin{align}\sin 69^{\circ}&=\sin (60+9)^{\circ}\\~\\
&=\sin (60^{\circ})\cos (9^{\circ})+\cos (60^{\circ})\sin (9^{\circ})\\~\\
&=\dfr... | I do not know if you know multiple angle trig formulas.
Let $ A = 18 ^\circ. $ In a right angled triangle if acute angles are $ 2A= 36 ^\circ, \,3A= 54^\circ$,
$ \sin 2 A = \cos 3A $
$ 2 \sin A \cos A = 4 \cos^3 A -3 \cos A $
simplifying and solving for $ \sin A $ gives you
$$ \sin 18^\circ =\dfrac{\sqrt{5}-1} {4}. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1243415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Solve for $x:1 + \tan^2(x) = 8\sin^2(x)$ I have a tricky problem , I tried several methods and I can't seem to get a definite answer.
$1 + \tan^2(x) = 8\sin^2(x), x \in [\frac{\pi}{6} , \frac{\pi}{2}]$
I got to $8\cos^4(x)-8\cos^2(x)+1=0$ and found that $\cos^2(x) = \frac{1}{4}[2-\sqrt{2}]$ but that is not too useful... | Use the identity $$1+\tan^2(x) = \sec^2(x)$$ Then multiply your equation through by $\cos^2(x)$ to get $$\begin{align*}1 = 8\sin^2(x)\cos^2(x) \\ = 2(2\sin(x)\cos(x))^2 \\ = 2(\sin(2x))^2 \\ \implies \frac{1}{2} = (\sin(2x))^2 \\ \implies \pm \frac{\sqrt{2}}{2} = \sin(2x)\end{align*}$$ Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1248525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
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$a^2 = 2b^3 = 3c^5$ Find the smallest value of $abc$. We have following equation:
$a^2 = 2b^3 = 3c^5$
Where $a, b, c$ are natural numbers.
Find the smallest possible value of product $abc$.
| Since $a^2=2b^3$, this means $a = \sqrt{2b} \cdot b$. Hence, $b=2l^2$. This gives us $a=4l^3$.
We also need $a^2 = 3c^5 \implies 16l^6 = 3c^5 \implies c = \left(\dfrac{2^4l}3\right)^{1/5} \cdot l$. This means $l = 3\cdot 2 \cdot m^5$.
Hence, we have $(a,b,c) = (864m^{15},72m^{10},12m^6)$. The minimum is obtained when $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1249775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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In how many ways can $1000000$ be expressed as a product of five distinct positive integers? I'm trying to solve the following problem:
"In how many ways can the number $1000000$ be expressed as a product of five distinct positive integers?"
Here is my attempt:
Since $1000000 = 2^6 \cdot 5^6$, each of its divisors has ... | The number in question is the coefficient of $x^6 y^6 z^5$ in the product
$$\prod_{0\le i,j\le 6}(1+x^i y^j z).$$
Here $z$ counts the number of factors (we want $5$); $x$ and $y$ tag the powers of $2$ and $5$ respectively; we want both of those to be $6$. Distinctness is ensured because, for each factor $2^i 5^j$, we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1251256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 4,
"answer_id": 3
} |
How to express sum as triple summation I am trying to express the following sequences as summations:
$$
1+2^2+3^2+4^4+5^4+6^4+7^4
$$
and
$$
1+(2+3)^2 + (4+5+6+7)^4
$$
as summations. I think they will likely be triple summations, so something like this:
$$
\sum\sum\sum n^k
$$
but I can't think of what to put on the top... | To me,
it looks like
at each successive iteration,
the number of terms doubles
((1), (2, 3), (4, 5, 6, 7))
and the exponent
doubles
(1, 2, 4).
To get the elements in the terms,
since
$1+2+4+...+2^n
=2^{n+1}-1
$,
if stage 1 is just $(1)$,
the terms added at stage $n$
are from
$2^{n-1}$
to
$2^n-1$
and the exponent
is $2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1253397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Factorize Trigonometric Equation: $ 3\sin(x)^2 - 2\sin(x)\cos(x) - \cos(x)^2 = 0 $ I have a problem with the following trigonometric equation:
$$ 3\sin(x)^2 - 2\sin(x)\cos(x) - \cos(x)^2 = 0 $$
It's from the book Engineering Mathematics 7th edition by Stroud.
The book is giving the answer, but I can't seem to be able t... | Think of it like this:
Take $a=\sin x$ and $b=\cos x$. Then you have,
$$3a^2-2ab-b^2=3a^2-3ab+ab-b^2=3a(a-b)+b(a-b)=(a-b)(3a+b)$$
Substitute everything back and you have,
$$3\sin^2 x-2\sin x\cos x-\cos^2 x=(\sin x-\cos x)(3\sin x+\cos x)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1254334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Solving $a_n=5a(n/3)-6a(n/9)+2log_3n$ using domain transformation $a_n=5a(n/3)-6a(n/9)+2log_3n$, For $n\ge9$ and n is a power of 3.
$a_3=1$, and $a_1=0$
Transforming the first two terms is straightforward, but I'm not sure what to do with the log term. Should I rewrite it somehow? The fact that it isn't attached to the... | Here is a closely related recurrence that has the same complexity as
the one in the OP and admits an exact solution for all $n.$ This
computation resembles the following MSE
link, the difference
being that this one does not depend on the digits of $n.$
Suppose we start by solving t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1257998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How can I simplify $\sqrt{3^2 + 3^2\tan^2\theta}$? $$\sqrt{3^2 + 3^2\tan^2\theta}$$
$$ = (3)(3\tan\theta) = 9\tan\theta $$
I've simplified it like this but I'm not sure if that's correct.
| $$\begin{align} \sqrt{3^2 + 3^2\tan^2\theta} &= \sqrt{9(1+\tan^2 \theta)} \\
& = \sqrt 9 \cdot \sqrt {1 + \tan^2 \theta} \\
& = 3\sqrt{\sec^2\theta} \\
&= 3|\sec \theta| \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1259681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Proof that $(n^7-n^3)(n^5+n^3)+n^{21}-n^{13}$ is a multiple of $3$. I proved that $$(n^7-n^3)(n^5+n^3)+n^{21}-n^{13}$$ is a multiple of $3$ through the use of Little Fermat's theorem but i want to know if there exist other proofs(maybe for induction). How can I demonstrate it?
This my proof:
$$n^3(n^4-1)(n^5+n^3)+n^{13... | Another approach
we see that $n(n-1)(n+1)$ divides both $n^7-n^3=n^3(n-1)(n+1)(n^2+1)$ and $n^{21}-n^{13}=n^{13}(n-1)(n+1)(n^2+1)(n^4+1)$
and we know that $(n-1)n(n+1)$ is a multiple of $3$ because the product of three consecutive integers is divisible by $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1261045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
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A way to split this determinant as a product of two? The question asks if the det of the following 3*3 matrix is divisible by $$ x^4,x^3,x^2,x $$
$$
\begin{bmatrix}
a^2+x^2 & ab & ac \\
ab & b^2+x^2 & bc \\
ac & bc & c^2+x^2 \\
\end{bmatrix}
$$
I just shovelled through using the standard method and got $$ x^4(b^2+c^2+... | The matrix can be written as
$$A = x^2I_3 + \begin{bmatrix}a\\b\\c\end{bmatrix} \begin{bmatrix}a & b & c\end{bmatrix}$$
Now by Sylvester determinant theorem, we have
\begin{align}
\det(A) & = x^6 \det\left(I_3 + \dfrac1{x^2}\begin{bmatrix}a\\b\\c\end{bmatrix} \begin{bmatrix}a & b & c\end{bmatrix}\right) = x^6 \left(1+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1261497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Help with solving mathematical induction problem I need help with the following:
Use mathematical induction to prove that for every $n\in N$,
$$
\sum_{k=1}^n\frac{1}{\cos kx \cos(k+1)x}=\frac{\tan(n+1)x-\tan x}{\sin x}
$$
For $n=1$, the statement is true.
Suppose that the statement is true for $n=m\in N$, and prove th... | this is a telescoping series. you have $$\frac{\sin (k+1)x}{cos(k+1)x} - \frac{\sin kx}{\cos kx}=\frac{\sin( k+1)x\cos kx - \sin kx\cos(k+1)x}{cos(k+1)\cos kx} = \frac{\sin x}{cos(k+1)\cos kx} \tag 1$$
adding the equations for $k = 1, 2, \cdots, n$ gives you $$\frac{\sin (n+1)x}{cos(n+1)x} - \frac{\sin x}{\cos x} = \si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1262046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to evaluate the $\lim\limits_{x\to 0}\frac {2\sin(x)-\arctan(x)-x\cos(x^2)}{x^5}$, using power series? How to evaluate the $\displaystyle\lim\limits_{x\to 0}\frac {2\sin(x)-\arctan(x)-x\cos(x^2)}{x^5}$, using power series?
It made sense to first try and build the numerator using power series that are commonly used... | Since the denominator has degree $5$, you just need to stop at $x^5$ in the power series expansions. So
\begin{align}
2\sin x&=2x-\frac{x^3}{3}+\frac{x^5}{60}+o(x^5)\\
-\arctan x&=-x+\frac{x^3}{3}-\frac{x^5}{5}+o(x^5)\\
-x\cos(x^2)&=-x+\frac{x^5}{2}+o(x^5)
\end{align}
Summing up we get
$$
\left(\frac{1}{60}-\frac{1}{5}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1265265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve the following recurrence relation: $S(1) = 2$; $S(n) = 2S(n-1)+n2^n, n \ge 2$ Solve the following recurrence relation:
$$\begin{align}
S(1) &= 2 \\
S(n) &= 2S(n-1) + n 2^n, n \ge 2
\end{align}$$
I tried expanding the relation, but could not figure out what the closed relation is:
+-------+----------------------... | Given
$$
S(n)=2S(n-1)+n2^n\tag{1}
$$
Consider $T(n)=2^{-n}S(n)$. $T(1)=1$ and, multiplying $(1)$ by $2^{-n}$, we have the recurrence
$$
T(n)=T(n-1)+n\tag{2}
$$
Therefore, $T(n)=\frac{n^2+n}2$ and thus,
$$
S(n)=(n^2+n)\,2^{n-1}\tag{3}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1267101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Finding $\int_{0}^{\infty }\frac{1}{1+x^4}dx$ finding $$\int_{0}^{\infty }\frac{1}{1+x^4}dx$$
My attempt is:
let $x=\sqrt{u}$
$dx=\frac{1}{2\sqrt{u}}$
$$\int_{0}^{\infty }\frac{1}{2\sqrt{u}(1+u^2)}du$$
here I stopped because I don't know how to complete this solution. Any help please.
| $$\int_{0}^{\infty }\frac{1}{1+x^4}dx=$$
$$\int_{0}^{\infty }\left(\frac{\sqrt{2}x-2}{4(-x^2+\sqrt{2}x-1)}+\frac{\sqrt{2}x+2}{4(x^2+\sqrt{2}x+1)}\right)dx$$
The antiderivative of $\left(\frac{\sqrt{2}x-2}{4(-x^2+\sqrt{2}x-1)}+\frac{\sqrt{2}x+2}{4(x^2+\sqrt{2}x+1)}\right) $ is:
$$\frac{1}{4\sqrt{2}}(-\log(x^2-\sqrt{2}x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1268550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Evaluate $\int_0^1 \int_\sqrt{y}^1 \int_0^{x^2+y^2} dz dx dy$.
Evaluate $\int_0^1 \int_\sqrt{y}^1 \int_0^{x^2+y^2} dz dx dy$.
Attempt:
$$ \int_0^1 \int_\sqrt{y}^1 \int_0^{x^2+y^2} dz dx dy
= \int_0^1 \int_\sqrt{y}^1 x^2 + y^2 dx dy
= 1/3 + 1/3 - 2/15 - 2/7 = \frac{26}{105}.$$
However, the solution should be $26/35$ ... | $$ \int_0^1 \int_\sqrt{y}^1 \int_0^{x^2+y^2} dz dx dy
= \int_0^1 \int_\sqrt{y}^1 ( x^2 + y^2) dx dy
$$
$$\int_\sqrt{y}^1 ( x^2 + y^2) dx=\left(\frac{1}{3}x^3+y^2x\right)_{x=\sqrt{y}}^{x=1} =\frac{1}{3}+y^2-\frac{1}{3}y^{3/2}-y^{5/2}$$
$$
\int_0^1(\frac{1}{3}+y^2-\frac{1}{3}y^{3/2}-y^{5/2})dy=\left( \frac{1}{3}y +\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1268849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Continued product in $\sin$ series Find the value of the product $$(\sin 1°)(\sin 3°)(\sin 5°)\ldots(\sin 89°)$$ I tried multiplying and dividing by $2$ and then combining and then converting into cosine, but doesn't work out.
| $$\prod_{i=1}^{45}\sin(2i-1)^\circ=\prod_{j=1}^{45}\cos(2j-1)^\circ$$
Let $\cos45x=\cos45^\circ$
$\implies45x=360^\circ n\pm45^\circ\iff x=8^\circ n\pm1^\circ$ where $n$ is any integer
Considering the '+' sign,
$1\le8n+1\le90\iff0\le n\le11 \ \ \ \ (1)$
$91\le8n+1\le180\iff12\le n\le22, \ \ \ \ (2)$
$n=12\implies\cos(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1270286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$f(x)$ is a polynomial satisfying $2 + f(x)f(y)=f(x)+f(y)+f(xy)$, find $f(f(2)$), given $f(2)=5.$
If f(x) is a polynomial satisfying $2 + f(x)f(y)=f(x)+f(y)+f(xy)$, find $f(f(2))$, given $f(2)=5.$
ATTEMPT:- $f(f(2))=f(5)$, We can find $f(0)$,$f(1)$ and $f(1/2)$ to be $1,2$ and $5/4$ respectively.
we can change the f... | Substitute $y = 2$ to get $2+f(x)f(2) = f(x)+f(2)+f(2x)$, which simplifies to $f(2x)-1 = 4(f(x)-1)$ for all reals $x$.
Now, suppose that $x_0 \neq 0$ is a zero of $f(x)-1$. Then $f(2x_0)-1 = 4(f(x_0)-1) = 4 \cdot 0 = 0$, and so, $2x_0$ is also a zero of $f(x)-1$. Continue this process indefinitely to get that $2^kx_0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1270658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Let $a,b$ be relative integers such that $2a+3b$ is divisible by $11$. Prove that $a^2-5b^2$ is also divisible by $11$. The divisibility for $11$ of $a^2 - 5b^2$ can be easily verified; in fact: $$a \equiv \frac {-3}{2}b \pmod {11}$$ therefore $$\frac {9}{4}\cdot b^2 - 5b^2 = 11(-\frac{b^2}{4}) \equiv 0 \pmod {11}.$$
T... | $11\mid 3(2a+4b)(2a-4b)-11a^2+11\cdot 2b^2=a^2-5b^2$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A probability question - Dice Throw Sum of three dice (six faced) throws is $15$. What is the probability that first throw was $4$?
The way I thought of solving this was...
- Given - sum of second and third throw is $11$
- Probability of getting first throw = $4$ is $1$ out of $6$, that is $\frac{1}{6}$
Is this correct... | Joffan pretty much pointed how to find the answer.
$P(D_{1}=4\mid Sum=15) = \frac {ways\ to\ get\ 11\ with\ 2\ dices}{ways\ to\ get\ 15\ with\ 3\ dices}$ (Note that getting 11 with 2 dices is like getting 4 with the first one).
$ways\ to\ get\ 11\ with\ 2\ dices = 2$ (both $(5,6)$ and $(6,5)$)
$ways\ to\ get\ 15\ with... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1273086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given $f(1)=10,f(2)=20,f(3)=30$ find $f(12)+f(-8)$ for a 4-th degree monic polynomial If $f(x)=x^4+ax^3+bx^2+cx+d$.
Given $f(1)=10,f(2)=20,f(3)=30$ find $f(12)+f(-8)$.
This problem has troubled me a lot.The more I try to solve it,it becomes lengthier.
My problem is that there are four unknowns and only three equations.... | Let $g(x) = f(x) - 10x$, we have
$$g(1) = g(2) = g(3) = 0\quad\implies\quad (x-1)(x-2)(x-3) \;\;\text{ divides }\;\;g(x)$$
As a result, $f(x)$ has the form
$$f(x) = 10x + (x-t)(x-1)(x-2)(x-3)$$
for suitably chosen constant $t$. Notice
$$\begin{align}
(12-1)(12-2)(12-3) &= 990\\
(-8-2)(-8-1)(-8-3) &= -990
\end{align}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1274133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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How prove $\frac{a^2}{(a+b)^2}+\frac{b^2}{(b+c)^2}+\frac{c^2}{(c+a)^2} \ge \frac{3}{4}+\frac{(a-b)(b-c)(a-c)}{(a+b+c)^3-3abc} $? Let $a \ge b \ge c >0$ . How can I prove
$$\frac{a^2}{(a+b)^2}+\frac{b^2}{(b+c)^2}+\frac{c^2}{(c+a)^2} \ge \frac{3}{4}+\frac{(a-b)(b-c)(a-c)}{(a+b+c)^3-3abc}. $$
Maybe a simple way?
| I think a simplest way here is a full expanding:
$$\sum\limits_{cyc}(5a^7b^2+a^7c^2+13a^6b^3+9a^6c^3+18a^5b^4+18a^5c^4+2a^7bc+15a^6b^2c+11a^6c^2b+$$
$$+11a^5b^3c+15a^5c^3b+16a^4b^4c-6a^5b^2c^2-50a^4b^3c^2-50a^4c^3b^2-28a^2b^2c^2)\geq0,$$
which is obviously true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1274242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Is $\frac{a^2+b^2}{2}=c^2$ possible? I am looking for an integer solution to the equation:
$$\frac{a^2+b^2}{2}=c^2(a\neq b\neq c)$$
That is a square number that is the mean of two other square numbers, is this possible? And if so please can you give me an example?
| We know that $(x-y)^2+(x+y)^2=2(x^2+y^2).$
So, if $a=x+y$ and $b=x-y$, then $c^2=x^2+y^2.$
Now see Formulas for generating Pythagorean triples.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving $9^n - 8n - 1$ is divisible by $8^2$ for $n\ge 0$? My textbook provided the following proof:
*Base case: When $n=0, 9^n-8n-1=0=64\cdot0$, so $64\mid\left(9^n-8n-1\right)$.
Induction step: Suppose that $n\in\mathbb N$ and $64\mid\left(9^n-8n-1\right)$. Then there is some integer $k$ such that $9^n-8n-1=0=64... | Use induction.
You know that the assertion is true for $n = 1.$
Assume that the assertion is true for $n = N.$
Then, the problem reduces to showing that
$$(64) ~\text{divides}~ \{[9^{N+2} - 8(N+1)] - [9^{N+1} - 8N]\}$$
$$ ~=~ [9^{N+1}(9 - 1)] - 8 = 8 \times [9^{N+1} - 1].$$
So, the problem reduces to showing that
$$(8)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1279028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that $\sum_{k = 0}^{4} (1+x)^k = \sum_{k=1}^5 \binom{5}{k}x^{k-1}$ Question:
Show that:
$$\sum_{k = 0}^{4} (1+x)^k = \sum_{k=1}^5 \binom{5}{k}x^{k-1}$$
then go on to prove the general case that:
$$\sum_{k = 0}^{n-1} (1+x)^k = \sum_{k=1}^n \binom{n}{k}x^{k-1}$$
Attempted solution:
It might be doable to first prove... | Using $a^n-1=(a-1)(\sum_{k=0}^{n-1}a^k)\implies\sum_{k=0}^{n-1}a^k=\frac{a^n-1}{a-1}$ for the first equality below (with $a=x+1$), we have
$$
\sum_{k=0}^{n-1}(x+1)^k=\frac{(x+1)^n-1}{x+1-1}=\frac{1}{x}(-1+(x+1)^n)=\frac{1}{x}\left(-1+\sum_{k=0}^n\binom{n}{k}x^k\right)=\frac{1}{x}\sum_{k=1}^n\binom{n}{k}x^k
$$
which sim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1284853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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How would I make a graph for $\sqrt{x+1}-3$ I have to make a table for $\sqrt{x+1}-3$ and I can't figure out how to find the $x$ values. I know that I have to get the middle section to equal perfect squares, which are $0,\ 1,\ 4,\ 9$ but I don't know how. Please help!
Is it $2,\ 3,\ 6$ and $12$?
The problem is when I... | To obtain integer values for $y = \sqrt{x + 1} - 3$, $x + 1$ must be a perfect square. The first five perfect squares are $0, 1, 4, 9, 16$. Thus, we must set $x + 1$ equal to $0, 1, 4, 9, 16$.
If $x + 1 = 0$, then $x = -1$. If $x = -1$, then $y = \sqrt{-1 + 1} - 3 = \sqrt{0} - 3 = -3$. Hence, the point $(-1, -3)$ i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1286411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Inequality for sides and height of right angle triangle
Someone recently posed the question to me for the above, is c+h or a+b greater, without originally the x and y lengths. I used this method: (mainly pythagorus)
$a^2+b^2=c^2=(x+y)^2=x^2+y^2+2xy$
$a^2=x^2+h^2$ and $b^2=y^2+h^2$
therefore $x^2+h^2+y^2+h^2=x^2+y^2+2x... | I think we get $$a+b<c+h$$
Squaring we obtain
$$a^2+b^2+2ab<c^2+h^2+2hc$$
thus we have
$$2ab<h^2+2hc$$
$$2hc<h^2+2hc$$
because $ab=ch$ (area formulas) and we get $$h^2>0,$$ which is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1286755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help with an Inverse Trigonometry Integral 2 Evaluate
$$\int^{1/{\sqrt{3}}}_{-1/{\sqrt{3}}} \frac{x^4}{1-x^4}\cos^{-1}\frac{2x}{1+x^2} \mathrm{d}x$$
The Solution: We try to eliminate $\cos^{-1}\frac{2x}{1+x^2}$ by using the relation $$\pi - cos^{-1}(a) = cos^{-1}(a)$$
Consider the first 3 steps:$$$$
$$I=\int^{1/{\sq... | Let $I$ be given by
$$I=\int_{-\sqrt{3}/3}^{\sqrt{3}/3}\frac{x^4}{1-x^4} \arccos\left(\frac{2x}{1+x^2}\right)dx$$
Upon substituting $x\to -x$ we find that
$$\begin{align}
I&=\int_{-\sqrt{3}/3}^{\sqrt{3}/3}\frac{x^4}{1-x^4} \arccos\left(\frac{-2x}{1+x^2}\right)dx\\\\
&=\int_{-\sqrt{3}/3}^{\sqrt{3}/3}\frac{x^4}{1-x^4} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1288621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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solving $a = \sqrt{b + x} + \sqrt{c + x}$ for $x$ I'm trying to solve a very simple looking square root equation but nothing seems to work. The equation has this form (solve for $x$):
$$
a = \sqrt{b + x} + \sqrt{c + x}
$$
Squaring both sides obviously doesn't help since it will still give me a square root. Rearranging ... | To solve
$$
a = \sqrt{b + x} + \sqrt{c + x}
$$
you must square twice, since you have two square roots.
First, for $\sqrt{b + x}$ and $\sqrt{c + x}$ to both exist, it must be $x\ge-b$ and $x\ge-c$, so
$$
x\ge\max(-b,-c).
$$
Then:
$$
a - \sqrt{b + x} = \sqrt{c + x}.\tag{1}
$$
Squaring once we find
\begin{align}
& a^2 + b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1288718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Formal power series coefficient problem Find the coefficient of:
$[x^{33}](x+x^3)(1+5x^6)^{-13}(1-8x^9)^{-37}$
I have figured out that I need to use this identity:
$(1-x)^{-k} = \sum\limits_{i>=0} \binom {n+k-1} {k-1} x^n $
But I have no clue how to proceed with this, I have been stuck with this for hours please help... | $$
\begin{align}
[x^{33}](x+x^3)(1+5x^6)^{-13}(1-8x^9)^{-37}
&=[x^{32}](1+5x^6)^{-13}(1-8x^9)^{-37}\\
&+[x^{30}](1+5x^6)^{-13}(1-8x^9)^{-37}\\[6pt]
&=[x^{30}](1+5x^6)^{-13}(1-8x^9)^{-37}\\[6pt]
&=[x^{10}](1+5x^2)^{-13}(1-8x^3)^{-37}
\end{align}
$$
The binomial theorem gives
$$
\begin{align}
(1+5x^2)^{-13}
&=\sum_{j=0}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1289014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $\alpha$ lies between $0$ and $4$.
Let $a,b,c$ be the length of the sides of the triangle $ABC$ .
Given $(a+b+c)(b+c-a)=\alpha bc$.Then Prove that the value of $\alpha$ lies in between $0$ and $4$.
$\begin{align}(a+b+c)(b+c-a)&=\alpha bc\\
\implies \alpha&=\dfrac{b^2+c^2-a^2}{bc}+2\\
\alpha&=2\cos A+2\... | As given $(a+b+c)(b+c-a)=\alpha bc$ then we can simply as follows
$$(b+c)^2-a^2=\alpha bc \implies \alpha=\frac{b^2+c^2+2bc-a^2}{bc}=\frac{b^2+c^2-a^2}{bc}+2=2+2\cos A$$ Now, for a triangle to exist, we have a condition for angle $A$ as $0<A<\pi$ thus we get $-1<\cos A<1$ hence, we get
$$(2-2)<\alpha<(2+2) \implies 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1291780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Need help with taylor series.
Evaluate the limit
$$\lim\limits_{x \to 1} \frac{1-x + \ln x}{1+ \cos πx}$$
The limit im trying to get is $-\frac{1}{π^2}$ as I've solved from l'Hopitals rule.
Now I need to solve the limit by using Taylor Series and this is what i did so far
$$\begin{align*}
f(x) &= 1-x + \ln x = 1 -x... | You do not need the Tailor expansion but only the first derivative, the rule of L'Hôpital gives us:
$$
\lim\limits_{x \to 1} \frac{1-x + \ln x}{1+ \cos πx}=\lim\limits_{x \to 1} \frac{\frac{1}{x}-1}{-\pi\,\sin\pi x}=\lim\limits_{x \to 1} \dfrac{\dfrac{1}{x^2}}{-\pi^2\cos\pi x} = -\dfrac{1}{\pi^2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1292372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Finding the equation of a curve where the gradient is $ax + b$ at all points. The gradient of a curve is $ax + b$ at all points, where $a$ and $b$ are constants. Find the equation of the curve given that it passes through the points $(0,4)$ and $(1,3)$ and that the tangent at $(1,3)$ is parallel to the $x$-axis.
My wo... | Well, the tangent line at $(1,3)$ is parallel to the $x$-axis, meaning that the gradient when $x=3$ is $0.$
On the other hand, we know that the gradient is always of the form $ax+b$ at any given $x.$ In particular, the gradient when $x=3$ is $3a+b.$
Now you have two linear equations with variables $a$ and $b,$ which wi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1293066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all 4 digits numbers that $ABCD=(CD)^2$ Please help me to solve following problem:
Find all 4 digits numbers such that $ABCD=(CD)^2$.(any of $A,B,C,D$ is a digit!)
I know one of solutions is $5776=(76)^2$.
| We need $D^2\equiv D\pmod{10}$ hence $D(D-1)$ must be a multiple of $10$. This implies that $D\in\{0,1,5,6\}$.
Next, the tens digit of $(CD)^2=(10\cdot C+D)^2=100\cdot C^2+20\cdot C\cdot D+D^2$ is determined by the ones digit of $2\cdot C\cdot D$ and the tens digit of $D^2$.
*
*For $D=0$ we need $2\cdot 0\cdot C\equ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What mistake have I made when trying to evaluate the limit $\lim \limits _ {n \to \infty}n - \sqrt{n+a} \sqrt{n+b}$? Suppose $a$ and $b$ are positive constants.
$$\lim \limits _ {n \to \infty}n - \sqrt{n+a} \sqrt{n+b} = ?$$
What I did first:
I rearranged $\sqrt{n+a} \sqrt{n+b} = n \sqrt{1+ \frac{a}{n}} \sqrt{1+ \frac{b... | It is true that both $a/n$ and $b/n$ tend to $0$ and $n\to\infty$, however the factor of $n$ in that term is approaching $\infty$ at the same time. So, analyzing this way, the second term gives the form $\infty\cdot 1$ and so the entire expression gives the indeterminate form $\infty-\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1297035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 0
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Hexagon Numbering Problem
So in the above hexagon figure, I have to arrange 1 to 7, inclusive, into the circles such that the three dark red triangles have the same sum. How many distinct arrangements can there be?
| First find which numbers will be common for all shaded triangles (center circle) so the remaining numbers can be ordered in three pairs with the same sum.
If you put $\boxed{1 \Rightarrow 2+7=3+6=4+5}$
If you put $2$ there is no possible arrangament
If you put $3$ there is no possible arrangament
If you put $\boxed{4 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1298570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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how many possible acute triangles with perimeter given How many possible acute triangles exist with perimeter 18? All sides are positive integers. The triangle (7,7,4) is the same as (4,7,7). I need the work in a way that a geometry 9th grade student would be able to come up with.
| Let's assume that $c \geq a \land c \geq b$; this forms the first constraint:
$$
c \geq \frac{18}{3} \implies c \geq 6
$$
A valid triangle requires $a + b > c$; substituting the perimeter constraint gives us:
$$
a + b > c \implies (18 - c) > c \implies c < 9
$$
You can work out the values for $a$ and $b$ from the three... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Determinant of matrices without expanding Show that $$\begin{array}{|ccc|}
-2a & a + b & c + a \\
a + b & -2b & b + c \\
c + a & c + b & -2c
\end{array} = 4(a+b)(b+c)(c+a)\text{.}$$
I added the all rows but couldn't get it.
| Let $x=b+c,y=c+a,z=a+b$. We claim that
$$
\left|\begin{pmatrix}
x-y-z & z & y\\
z & y-z-x & x\\
y & x & z-x-y
\end{pmatrix}\right|=4xyz.
$$
When $x=0$, add column 1 to columns 2 and 3 to obtain
$$
\left|\begin{pmatrix}
-y-z & -y & -z\\
z & y & z\\
y & y & z
\end{pmatrix}\right|=0.
$$
Thus by symmetry, $xyz$ divides the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1299090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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taking the limit $\lim\limits_{n\rightarrow \infty} {\frac{(3^{n+1} + 4)(7^n-47)}{(7^{n+1}-47)(3^n +4)} }$ I need help with a guide on how i will deal with this kind of problem.. This a part of my solution in series convergence. I find it hard taking the limit of this: $$\lim_{n\rightarrow \infty} {\frac{(3^{n+1} + 4)(... | $$\lim\limits_{n\to \infty} {\frac{(3^{n+1} + 4)(7^n-47)}{(7^{n+1}-47)(3^n +4)} }=
\lim\limits_{n\to \infty} {\frac{\frac{3^{n+1} + 4}{3^n}\cdot \frac{7^n-47}{7^n}}{\frac{7^{n+1}-47}{7^n}\cdot\frac{3^n +4}{3^n}} }=
\lim\limits_{n\to \infty} {\frac{\left(3+\frac{4}{3^n}\right)\left( 1-\frac{47}{7^n}\right)}{\left(7-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1300448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to integrate $\int dx \frac{1}{\cosh^2 x +a^2}$ How to integrate that function
$$\int \frac{1}{\cosh^2 x +a^2}dx$$,
What I did was rewrite $$\cosh^2 x = ({\frac {{e}^x+{e}^{-x}}{2}})^2 $$
then $$\int \frac{1}{({\frac {{e}^x+{e}^{-x}}{2}})^2 +a^2}dx=$$ $$\frac{1}{4}\int \frac{1}{e^{2x}+e^{-2x}+4{a}^2 }dx$$,
Mult... | Expand of the comment from lab.
\begin{align}
&\int \frac{1}{\cosh^2 x +a^2} dx\\
=&\int \frac{\mathrm{sech} ^2x}{1 +a^2\mathrm{sech} ^2x} dx \\
=&\int \frac{1}{1 +a^2(1-\tanh^2x)} d\tanh x \qquad\text{let $y=\tanh x$}\\
=&\int \frac{d y}{(1+a^2)-a^2y^2}\\
=&\frac{1}{a^2}\int \frac{d y}{-y^2+\frac{1+a^2}{a^2}}\qquad \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1303704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I show this inequality: $-2 \le \cos \theta (\sin \theta +\sqrt{\sin ^2 \theta +3})\le 2$
Show that $$-2 \le \cos \theta ~ (\sin \theta +\sqrt{\sin ^2 \theta +3})\le 2$$ for all value of $\theta$.
Trial: I know that $0\le \sin^2 \theta \le1 $. So, I have $\sqrt3 \le \sqrt{\sin ^2 \theta +3} \le 2 $. After tha... | My Solution:: Given $$f(\theta) = \cos \theta \left(\sin \theta + \sqrt{\sin^2 \theta + 3}\right)$$
Now let $$y=\sin \theta \cdot \cos \theta +\cos \theta \cdot \sqrt{\sin^2 \theta + 3}$$
Now using the Cauchy-Schwarz inequality, we get
$$\left(\sin^2 \theta +\cos^2 \theta \right)\cdot \left\{\cos^2 \theta + \left(\sq... | {
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"url": "https://math.stackexchange.com/questions/1303772",
"timestamp": "2023-03-29T00:00:00",
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How to prove that this matrix is positive definite? Let $\mathbf{A}=\begin{pmatrix}a^2+b^2 & b^2 & b^2 & ... & b^2 \\ b^2 & a^2+b^2 & b^2 & ... & b^2\\ \vdots & b^2 & \ddots & & b^2 \\ b^2 & \dots & & & a^2+b^2 \end{pmatrix}$, where $a,b\ne 0$. How can I be sure that this matrix is positive definite?
Any help would ... | Notice that
$$x^TAx=\sum_{j=1}^n(a^2+b^2)x_j^2+2\sum_{i\lt j\leqslant n}b^2x_ix_j=b^2\left(\sum_{j=1}^nx_j\right)^2 +a^2\sum_{j=1}^nx_j^2\geqslant a^2\sum_{j=1}^nx_j^2 $$
which is positive unless $x=0$. Note that we only have to require that $a\neq 0$, $b$ can be any real number.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1305477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Prove sequence $\left(\frac{1}{6n^2+1}\right)$ converges to $0$ I am asked to verify that the sequence $\left(\frac{1}{6n^2+1}\right)$ converges to $0$:
$$\lim \frac{1}{6n^2+1}=0.$$
Here is my work:
$$\left|\frac{1}{6n^2+1}-0\right|<\epsilon$$
$\frac{1}{6n^2+1}<\epsilon$, since $\frac{1}{6n^2+1}$ is positive
$$\frac{... | Let $\epsilon > 0$. Let's sketch it first. $$\frac{1}{6n^2+1} < \frac{1}{6n^2} < \epsilon.$$ Look: $$\frac{1}{6n^2} < \epsilon \iff 1 < 6n^2\epsilon \iff \frac{1}{6\epsilon} < n^2 \iff n > \frac{1}{\sqrt{6\epsilon}}.$$
Now we begin. Let $\epsilon > 0$. Then exists $n_0 \in \Bbb N$ such that $n_0 > 1/\sqrt{6\epsilon}$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1305831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Roots of the equation $x^2 + px + q = 0$ If $\tan A$ and $\tan B$ are the roots of the equation $x^2 + px + q = 0 $ , show that $$\sin^2(A+B) + p \sin(A+B)\cos(A+B) + q \cos^2(A+B) = q$$
I tried using the result that $\tan A + \tan B = -p $ and $(\tan A)(\tan B) = q$ and tried substituting in the original equation but... | you have $$\tan A + \tan B = -p, \tan A \tan B = q \implies \tan(A+B) = \frac{\tan A + \tan B}{1- \tan A \tan B}=\frac p{q-1} $$ and $$\sin(A+B) = \pm\frac p{\sqrt{p^2 + (q-1)^2}},\quad \cos(A+B) = \pm \frac {q-1}{\sqrt{p^2 + (q-1)^2}} $$
now, $$\begin{align}\sin^2(A+B) & + p \sin(A+B)\cos(A+B) + q \cos^2(A+B)\\
& = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Showing $ \int_0^{2 \pi } \frac{dt}{a^2 \cos^2 t + b^2 \sin^2 t} = \frac{2 \pi}{ab}$ The question:
Let $\gamma$ be a contour such that $0 \in I(\gamma),$ where $I$ is the interior of the contour. Show that
$$\int_\gamma z^n \, \text{d}z = \begin{cases} 2\pi i & \text{if } n = -1 \\ 0 & \text{otherwise} \end{cases}$$
... | We assume that $a$ and $b$ are real-valued and positive-valued parameters.
Let $z=e^{it}$ so that $\cos t = \frac12(z+z^{-1})$, $\sin t=\frac1{2i}(z-z^{-1})$, $dt=dz/iz$ and $t$ goes from $0$ to $2\pi$. Then,
$$\begin{align}
\int_0^{2\pi} \frac{dt}{a^2\cos^2 t+b^2\sin^2t}&=\oint_C \frac{1}{a^2\frac14(z+z^{-1})^2-b^2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Evaluate $\lim\limits_{x\to+\infty}\frac{\left(x+\sqrt{x^2-1}\right)^n+\left(x-\sqrt{x^2-1}\right)^n}{x^n},n\in \mathbb{N}$ If we use the following
$$a^n-b^n=\left(a-b\right)\left(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1}\right)=u\times t$$
$$u=x+\sqrt{x^2-1}-x+\sqrt{x^2-1}=2\sqrt{x^2-1}=2x\sqrt{1-\frac{1}{x}}$$
Now, th... | $$\lim\limits_{x\to+\infty}\frac{(x+\sqrt{x^2-1})^n+(x-\sqrt{x^2-1})^n}{x^n}
$$
$$= \lim\limits_{x\to+\infty}(1+\sqrt{1-x^{-2}})^n+ \lim\limits_{x\to+\infty}(1-\sqrt{1-x^{-2}})^n = 2^n + 0 = 2^n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Finding distance from point to line which is perpendicular to another line Find the distance of the point $(1,1,1)$ from $x+y+z=1$ measured perpendicular to the line $\frac{x}{2}=\frac{y}{3}=\frac{z}{6}$
| The equation of the straight line perpendicular to the given line $\frac{x}{2}=\frac{y}{3}=\frac{z}{6}$ and passing through the point $P(1,1,1)$ is:
$$\frac{x-1}{a}=\frac{y-1}{b}=\frac{z-1}{c}$$
where, $2a+3b+6c=0$.
Clearly, $(a,b,c)=(3,2,-2)$ is a solution.
Let $$\frac{x-1}{3}=\frac{y-1}{2}=\frac{z-1}{-2}=t$$
Then, a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1315133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.