Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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The right way of proving LHS = RHS I'm currently brushing up on my math before heading back to school for my bachelor's and am practicing on some questions.
Given the following equation :
$$\frac{x^2 - 3x}{x^2 - 9} = 1-\frac{3x-9}{x^2-9}.$$
and that I have to prove that the LHS = RHS; How would I go about doing so?
Wha... | First, to show the desired equality, we have: $$\frac{x^2-3x}{x^2-9}=\frac{x^2-9-3x+9}{x^2-9}=\frac{x^2-9}{x^2-9}-\frac{3x-9}{x^2-9}=1-\frac{3x-9}{x^2-9}.$$
Now, assuming that you are trying to find the integral of $\dfrac{x^2-3x}{x^2-9}$, we have:
\begin{align}
\int{\dfrac{x^2-3x}{x^2-9}dx}&=\int\left({1-\dfrac{3x-9}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1533333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Indefinite integrals with absolute values Which is the right way to solve indefinite integrals which contain absolute values? For example if I have
$\int |2x+3| e^x dx$
Can I consider the sign function and integrate separetly? I mean doing:
$ Sign(2x+3) \int (2x+3)e^x dx$
Or maybe I should use the definition of absolu... | $\int|2x+3|e^x~dx$
$=\text{sgn}(2x+3)\int(2x+3)e^x~dx$
$=\text{sgn}(2x+3)\int_{-\frac{3}{2}}^x(2x+3)e^x~dx+C$
$=\text{sgn}(2x+3)\int_{-\frac{3}{2}}^x(2x+3)~d(e^x)+C$
$=\text{sgn}(2x+3)[(2x+3)e^x]_{-\frac{3}{2}}^x-\text{sgn}(2x+3)\int_{-\frac{3}{2}}^xe^x~d(2x+3)+C$
$=\text{sgn}(2x+3)(2x+3)e^x-2~\text{sgn}(2x+3)\int_{-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1533992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $ \int \frac{\sin x+\cos x}{\sin^4x+\cos^4x}\,\text{d}x$. $$
\int \frac{\sin x+\cos x}{\sin^4x+\cos^4x}\,\text{d}x.
$$
What I have tried:
I tried writing denominator as
$ \sin^4x+\cos^4x = 1-2\sin^2x\cos^2x $ and $ 2\sin^2x\cos^2x = \frac{1}{2}\sin^2(2x) $
so the integral becomes,
$$
\int\frac{\sin x+\cos ... | I suggest that you instead split the integral as
$$
\int\frac{\sin x+\cos x}{\sin^4x+\cos^4x}\,dx=\int\frac{\sin x}{(1-\cos^2x)^2+\cos^4x}\,dx+\int\frac{\cos x}{\sin^4x+(1-\sin^2x)^2}\,dx,
$$
and then let $u=\cos x$ and $u=\sin x$ in the respective integral. You will get a (somewhat nasty) integral of a rational functi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1534821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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What is the sum of the series $x^{n-1} + 2x^{n-2} + 3x^{n-3} + 4x^{n-4} + \cdots + nx^{n-n}$? $x^{n-1} + 2x^{n-2} + 3x^{n-3} + 4x^{n-4} + \cdots + nx^{n-n}$
I'm not sure if this can even be summed. Any help is appreciated.
| It is
$S=x^{n-1}+2x^{n-2}+3x^{n-3}+\ldots+(n-2)\cdot x^2+(n-1)\cdot x^1+n\cdot x^0 \quad (1)$
$x\cdot S=x^{n}+2x^{n-1}+3x^{n-2}+\ldots+(n-2)\cdot x^3+(n-1)\cdot x^2+n\cdot x^1 \quad (2)$
Subtracting (1) from (2):
$(x-1)S=\color{blue}{x^n+x^{n-1}+x^{n-2}+x^{n-3}+\ldots + x^2+x}-n$
The formula for the blue expression is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1540463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Limit of a fraction with a square root
Find $$\lim_{x \to 2} \frac{4-x^2}{3-\sqrt{x^2+5}}$$ (without L'Hopital)
Where is the following wrong? (The limit is 6.)
\begin{align}\lim_{x \to 2} \frac{4-x^2}{3-\sqrt{x^2+5}}& =\lim_{x \to 2} \frac{(2-x)(2+x)}{\sqrt{9-x^2-5}}= \\
& =\lim_{x \to 2} \frac{(2-x)(2+x)}{\sqrt{-x^2... | $$\lim_{x\to 2}\frac{4-x^2}{3-\sqrt{x^2+5}}=\lim_{x\to 2}\frac{\left(4-x^2\right)\left(3+\sqrt{x^2+5}\right)}{\left(3-\sqrt{x^2+5}\right)\left(3+\sqrt{x^2+5}\right)}$$
$$=\lim_{x\to 2}\frac{\left(4-x^2\right)\left(3+\sqrt{x^2+5}\right)}{3^2-\left(\sqrt{x^2+5}\right)^2}=\lim_{x\to 2}\frac{\left(4-x^2\right)\left(3+\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1541799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Prove that $\lim_{x\to 0} \frac{a^x-1}{x}=\ln a$ Without using L'Hopital's rule, prove that
$$\lim_{x\to 0} \frac{a^x-1}{x}=\ln a$$
| Don't know if that is rigorous enough for you, but this is one way using $e^t = \sum_{k=0}^{\infty} \frac{t^k}{k!}$, $e^{\ln x} = x$ and some results about absolute convergent series:
\begin{align*}
\frac{a^x - 1}{x} & = (e^{x \cdot \ln a} -1)/x = \left(\sum_{k=0}^{\infty} \frac{(x \cdot \ln a)^k}{k!} - 1 \right)/x\\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1542483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
System of congruence relations Solve the system of congruence relations:
$2x+3y\equiv 1\pmod {11}$
$x+4y\equiv 4\pmod {11}$
Could someone give a hint how to solve this system.
I know that Chinese remainder theorem can't be used because modulo numbers ($11$) are not relatively prime.
| Write your system in matrix form: $\left( \begin{matrix} 2 & 3 \\ 1 & 4 \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right) = \left( \begin{matrix} 1 \\ 4 \end{matrix} \right) \pmod {11}$. The matrix has determinant $5$ which is invertible $\mod 11$, therefore the matrix itself is invertible and its ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1542611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding $\binom n0+\binom n3+\binom n6+\cdots $ How to get
$$\binom n0 + \binom n3 + \binom n6 + \cdots$$
MY ATTEMPT
$$(1+\omega)^n = \binom n0 + \binom n1 \omega^1 + \binom n2 \omega^2 + \cdots$$
$$(1+\omega^2)^n = \binom n0 + \binom n1 \omega^2 + \binom n2 \omega^4 + \cdots $$
$$(1 + 1)^n = 2^n = \binom n0 + \binom... | You have
$$(1+\omega)^n+(1+\omega^2)^n+2^n=3\left(\binom n0+\binom n3+\binom n6+\cdots\right)$$
Now note that
$$(1+\omega)^n+(1+\omega^2)^n=(-\omega^2)^n+(-\omega)^n=(-1)^n(\omega^n+\omega^{2n})$$
This is equal to $(-1)^n\cdot 2$ if $n\equiv 0\pmod 3$ or $(-1)^n\cdot (-1)=(-1)^{n+1}$ if $n\not\equiv 0\pmod 3.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1547522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Can you solve $y'+x+e^y=0$ by series expansion?
Find an approximate solution by series expansion of $y(x)$ around $x =
0$ up to fourth order in $x$ given the inital conditions $y(0)=0$
Let
$$
y=\sum_0^{\infty}a_nx^n \implies y'=\sum_0^{\infty}a_n nx^{n-1} \\
\implies \sum_0^{\infty}a_n nx^{n-1}+x+e^{\sum_0^{\infty}... | Here is another way of solving the equation:
$$y'(x)+x+e^{y(x)}=0\Longleftrightarrow$$
$$\frac{\text{d}y(x)}{\text{d}x}+x+e^{y(x)}=0\Longleftrightarrow$$
Let $y(x)=\ln(v(x))$, which gives $\frac{\text{d}y(x)}{\text{d}x}=\frac{\frac{\text{d}v(x)}{\text{d}x}}{v(x)}$:
$$x+v(x)+\frac{\frac{\text{d}v(x)}{\text{d}x}}{v(x)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1547602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Use Complex Integrals/ Residue to evaluate $\int_0^\infty \frac{dx}{(x+1)^3 + 1}$
Use Complex Integrals/ Residue to evaluate $\int_0^\infty \frac{dx}{(x+1)^3 + 1}$
I'm not sure how to do this integration. It looks like partial fractions but I'm unsure.
| We have
$$I = \int_0^{\infty} \dfrac{dx}{(1+x)^3+1} = \int_1^{\infty} \dfrac{dx}{x^3+1} = \int_1^0 \dfrac{-dx/x^2}{1/x^3+1} = \int_0^1 \dfrac{xdx}{1+x^3}$$
Hence,
$$I = \int_0^1\dfrac{x+1}{3(x^2-x+1)}dx - \int_0^1\dfrac{dx}{3(x+1)} = \dfrac{\pi}{3\sqrt{3}}-\dfrac{\log2}3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1547915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding $\sqrt{(14+6\sqrt 5)^3}+\sqrt{(14-6\sqrt 5)^3}$ Find $$\sqrt{(14+6\sqrt{5})^3}+ \sqrt{(14-6\sqrt{5})^3}$$
A.$72$
B.$144$
C.$64\sqrt{5}$
D.$32\sqrt{5}$
How to cancel out the square root?
| Note that $$14\pm 6\sqrt{5}=14\pm 2\sqrt{45}=9+5\pm 2\sqrt{9\times 5}=(\sqrt{9}\pm\sqrt{5})^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1548512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Given the conditions above, find when $x$, $y$, $z$ satisfy below: $ (x^2-1)(y+1)=\dfrac{z^2+1}{y-1}$ Let $x,y,z \in \mathbb{Z^+}$ and $x \neq y \neq z$.
Given the conditions above, find when $x$, $y$, $z$ satisfy below:
$$ (x^2-1)(y+1)=\frac{z^2+1}{y-1}\,.$$
What I did was I factored the numerator to
$$(x+1)(x-1)(y+1)... | The only solution is $x = y = z = 0$.
Suppose there were a solution with x, y, and z not all zero. Choose the maximum $k$ with $2^k | gcf(x, y, z)$. So $x = 2^k m$, $y = 2^k n$, and $z = 2^k p$, and $m$, $n$, and $p$ are not all even.
$(xy)^2 = x^2 + y^2 + z^2$
$(2^k m 2^k n)^2 = (2^k m)^2 + (2^k n)^2 + (2^k p)^2$
$(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1552477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Integrate $\int{ \frac{r}{(h^2 + r^2 - 2^{1/2}hr)^{1/2}} dr }$ This integral comes from a physics book when calculating potential difference between vertex and center of cone. I'm not good at integration. Please help.
update:
This is the required answer.
The second integral is in ln form. I have reached there by using... | Let me write your integral as $$ \int \frac{x}{\sqrt{x^2-\sqrt{2}ax+a^2}}dx$$
Then first consider the change of variable $u={x^2-\sqrt{2}ax+a^2} $, the $du=(2x-\sqrt{2}a )dx$, thus the integral becomes: $$ \int \frac{x}{\sqrt{x^2-\sqrt{2}ax+a^2}}dx = \frac{1}{2} \int \frac{2x-\sqrt{2}a+\sqrt{2}a}{\sqrt{x^2-\sqrt{2}a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1554458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Convergence of $\sum_{n = 1}^\infty 1/n^2$. I know that $\sum_{n=1}^\infty 1/n$ diverges whereas $\sum_{n=1}^\infty 1/n^2$ converges.
Intuitively, I do not see the difference. If $n \to \infty$, the denominators in both fractions will be so big that the fraction approaches zero. So why doesn't both the series converge ... | To see that $\displaystyle\sum_{n=1}^\infty \frac{1}{n}$ diverges, you can group the terms,
$$\begin{alignat*}{2}
\frac{1}{2} + \frac{1}{3} &> \frac{1}{2}+\frac{1}{2} &= 1\\
\frac{1}{4} + \dots +\frac{1}{7} &> \frac{1}{4} + \dots + \frac{1}{4} &= 1\\
&\;\;\vdots&\vdots\;\;\;\,\\
\frac{1}{2^n}+\dots+\frac{1}{2^{n+1}-1} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1557826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A fence of $y$ ft is $x$ ft from a wall, find shortest ladder using trigonometry
A fence $y$ ft high is $x$ ft from a wall. Find the length of the
shortest ladder that will rest with one end on the ground and the
other end on the wall.
This is a classic problem that is expected to be solved with trigonometry but ... | Your last assumption is the problem:
$$
\tan\theta = \frac{\sin\theta}{\cos\theta} = (\frac{y}{x})^{1/3} \\
\frac{\sin^2\theta}{\cos^2\theta} = (\frac{y}{x})^{2/3} \\
x^{2/3}\sin^2\theta = y^{2/3}\cos^2\theta \\
\sin^2\theta + \cos^2\theta = \frac{y^{2/3}\cos^2\theta}{x^{2/3}} + \cos^2\theta \\
\cos^2\theta(\frac{y^{2/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1558907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Evaluate the integral $\int \sqrt{(x-a)(b-x)}$ I'm trying to figure out how to evaluate the following integral:
$$\int \sqrt{(x-a)(b-x)} \, dx $$
I've tried various trig substitutions, but can't seem to get anywhere. This is an exercise in Apostol's Calculus Volume 1 (Section 6.22, Exercise 46). The solution provided ... | Notice, $$\int \sqrt{(x-a)(b-x)}\ dx$$ $$=\int\sqrt{(a+b)x-x^2-ab}\ dx$$
$$=\int \sqrt{\left(\frac{a+b}{2}\right)^2-\left(x-\frac{a+b}{2}\right)^2-ab}\ dx$$
$$=\int \sqrt{\left(\frac{a+b}{2}\right)^2-\left(x-\frac{a+b}{2}\right)^2-ab}\ dx$$
$$=\int \sqrt{\left(\frac{\sqrt{a^2+b^2}}{2}\right)^2-\left(x-\frac{a+b}{2}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1559121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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If $a,b$ are the roots of the equation $x^2-2x+3=0$ obtain the equation whose roots are $a^3-3a^2+5a-2$, $b^3-b^2+b+5$ I have been trying this using sum of roots and product of roots but it gets too lengthy. So I found the roots of the given equation which are imaginary and tried to replace the values in the two given ... | We have
$$ a^3-3a^2+5a-2=(a-1)(a^2-2a+3)+1=1$$
and
$$b^3-b^2+b+5=(b+1)(b^2-2b+3)+2=2. $$
So the desired polynomial is
$$ (X-1)(X-2)=X^2-3X+2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1559812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Find the Values of n and k for which the determinant of the Matrix M(n,k) is Singular I have been stuck on this problem for a couple of days, I don't want the answer, but I would appreciate some help in finding it! Thanks in advance!
Consider a Symmetric Square Matrix $M(n,k)$ such that there is $n$ lines and $n$ colum... | Let me demonstrate the process for $M(4, 7)$. By adding rows $2-4$ to the first row, we move from $M(4,7)$ to the matrix
$$ B = \begin{pmatrix} 4 & 4 & 4 & 4 \\ -1 & 7 & -1 & -1 \\ -1 & -1 & 7 & -1 \\ -1 & -1 & -1 & 7 \end{pmatrix} $$
and we have $\det M(4,7) = \det B$. Multiplying the first row by $\frac{1}{4}$ we mov... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1560450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Power series representation of $\frac{1+x}{1-x}$ explanation? I don't really get how any of this works. I tried looking at this power function example already to figure it out. But I got lost when this part comes up. $$\sum_{n=0}^{\infty}x^n=x^0+\sum_{n=1}^{\infty}x^n=1+\sum_{n=1}^{\infty}x^n$$
Why is this entire ste... | You have a sum of two geometric series.
\begin{align}
\frac 1 {1-x} & = 1 + x + x^2 + x^3 + \cdots \\[10pt]
\frac x {1-x} & = x + x^2 + x^3 + x^4 + \cdots
\end{align}
You're looking for the sum of the two above.
\begin{align}
\sum_{n=0}^\infty x^n & = x^0 + x^1 + x^2 + x^3 + x^4 + \cdots \\[10pt]
& = x^0 + \Big( x^1 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1561199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Evaluation of $\int_{0}^{1} \frac{1}{x+\sqrt{1-x^2}} \space dx$ Evaluate
$$\int_{0}^{1} \frac{1}{x+\sqrt{1-x^2}} \space dx$$
My main concern is finding the indefinite integral as once i have that the rest is fairly straight forward. Please give a detailed answer with reference to why you made each substitution (what in... | Since no one used it, let me consider the case of the antiderivative $$I=\int\frac{\cos x}{\sin x + \cos x} \space dx$$ Now, use the tangent half-angle substitution (Weierstrass substitution) $t=\tan(\frac x 2)$.
We so obtain $$I=\int \frac{2 \left(t^2-1\right)}{t^4-2 t^3-2 t-1}\space dt$$ But $$t^4-2 t^3-2 t-1=(t^2+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1561559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove for all reals: $a^2+b^2+c^2\geq 2(a+b+c)-3$ Problem:
Prove this for all reals $a,b,c$:
$$a^2+b^2+c^2\geq 2(a+b+c)-3$$
Attempt:
I am trying to work backwards.
$$a^2+b^2+c^2\geq 2(a+b+c)-3$$
$$(a+b+c)^2\geq 2(a+b+c)-3 + 2(ab+bc+ca)$$
$$(a+b+c)^2-2(a+b+c)\geq 2(ab+bc+ca)-3 $$
$$(a+b+c)(a+b+c-2)\geq 2(ab+bc+ca)-3 ... | it is equivalent to $$(a-1)^2+(b-1)^2+(c-1)^2\geq 0%$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1561923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solve $y^3-3y-\sqrt{2}=0$ using trigonometry This is a part of a larger question.
I had to show that for $4x^3-3x-\cos 3\alpha=0$ one of the solutions is $\cos \alpha$ and then find the other two solutions. Here they are:
$$4x^3-3x-\cos 3\alpha = (x-\cos \alpha)(2x+\cos \alpha + \sqrt{3} \sin \alpha)(2x+\cos \alpha - \... | Recall the triplication formula:
$$
\cos3\alpha=4\cos^3\alpha-3\cos\alpha
$$
so, if your equation is $y^3-3y=\sqrt{2}$, you can first set $y=at$, so
$$
a^3t^3-3at=\sqrt{2}
$$
and you'd like that $a^3/3a=4/3$, so you can take $a=2$: $8t^3-6t=\sqrt{2}$; setting $t=\cos\alpha$, we get
$$
\cos3\alpha=\frac{\sqrt{2}}{2}
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1562655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Derivative of $10^x\cdot\log_{10}(x)$
Derive $10^x\cdot\log_{10}(x)$
$$10^x\cdot \ln(10)\cdot \log_{10}(x)+\frac{1}{x\cdot \ln(10)}\cdot 10^x$$
But WolframAlpha gives another solution. Where am I wrong?
| $$\frac{\text{d}}{\text{d}x}\left(10^x\cdot\log_{10}(x)\right)=\frac{\text{d}}{\text{d}x}\left(\frac{10^x\ln(x)}{\ln(10)}\right)=$$
$$\frac{\frac{\text{d}}{\text{d}x}\left(10^x\ln(x)\right)}{\ln(10)}=\frac{\ln(x)\frac{\text{d}}{\text{d}x}(10^x)+10^x\frac{\text{d}}{\text{d}x}(\ln(x))}{\ln(10)}=$$
$$\frac{10^x\ln(10)\ln(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1562999",
"timestamp": "2023-03-29T00:00:00",
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} |
Find all irreducible polynomials of degrees 1,2 and 4 over $\mathbb{F_2}$. Find all irreducible polynomials of degrees 1,2 and 4 over $\mathbb{F_2}$.
Attempt: Suppose $f(x) = x^4 + a_3x^3 + a_2x^2 + a_1x + a_0 \in \mathbb{F_2}[x]$. Then since $\mathbb{F_2} =${$0,1$}, then we have either $0$ or $1$ for each $a_i$. Then... | Degree $1$, clearly $x$ and $x+1$.
Degree $2$, notice the last coefficient must be one, so there are only two options, $x^2+x+1$ and $x^2+1$. Clearly only $x^2+x+1$ is irreducible.
Degree $4$. There are $8$ polynomials to consider, again, because the last coefficient is $1$, now notice a polynomial is divisible by $x+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1563558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
$a^2-b^2$ will be a perfect square if the ratio of $a$ and $b$ is
$a^2-b^2$ will be a perfect square if the ratio of $a$ and $b$ is
Options are: A) $1:2$, B) $3:4$, C) $5:4$, D) $4:5$, E) None of these
So my first observation was that $|a|$ has to be greater than $|b|$. So only options that suffice this condition a... | A) $1^2-2^2=-3$,
B) $3^2-4^2=-7$,
C) $5^2-4^2=9$,
D) $4^2-5^2=-9$.
This rules out A), B) and D).
C) implies $(5k)^2-(4k)^2=(3k)^2$, so that the property holds for all $(a,b)=k(5,4)$.
If you don't want to use the multiple choices, let
$$a^2-b^2=c^2.$$
You can factor as
$$(a+b)(a-b)=c^2=pq$$
and solve with
$$a=\frac{p+q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1565481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show $\sin(\frac{\pi}{3})=\frac{1}{2}\sqrt{3}$ I have to show that
$$\sin\left(\frac{\pi}{3}\right)=\frac{1}{2}\sqrt{3}$$
and
$$\cos\left(\frac{\pi}{3}\right)=\frac{1}{2}$$
Should I use the exponential function?
| Not to confuse you. But..
$$x=\dfrac{\pi}{3}$$
$$3x=\pi$$
$$\sin 3x=\sin \pi$$
$$3\sin x-4\sin^3x=0$$
$$\sin x=0 \text{ or } \sin x = \dfrac{\sqrt{3}}{2} \text{ or } \sin x = \dfrac{-\sqrt{3}}{2}$$
$$\text{as } 0<\dfrac{\pi}{3}<\dfrac{\pi}{2} \text{, } \sin \dfrac{\pi}{3} = \dfrac{\sqrt{3}}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1567824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 1
} |
Find all functions $f$ such that $f(x-f(y)) = f(f(x)) - f(y) - 1$ Find all functions $f : \mathbb{Z} \to \mathbb{Z}$ such that $f(x-f(y)) = f(f(x)) - f(y) - 1$.
So far, I've managed to prove that if $f$ is linear, then either $f(x) = x + 1$ or $f(x) = -1$ must be true. I did this by plugging in $x=0$ to the above equa... | Let $f\colon \Bbb Z\to \Bbb Z$ be a any function function with $$\tag0f(x-f(y)) = f(f(x)) - f(y) - 1$$
for all $x,y\in\Bbb Z$.
Letting $y=f(x)$ we find $$f(x-f(f(x)))=-1. $$
So for $a=-f(f(0))$ we have $f(a)=-1$. Then with $y=a$, $$\tag1f(x+1)=f(f(x)) $$
Then $(0)$ becomes
$$\tag2 f(x-f(y))=f(x+1)-f(y)-1 $$
Or with $g(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1568106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
} |
How to see this inequality Consider
$$
\frac{dx}{dt}=\sqrt{x^2+1}+t^2,\qquad x\in\mathbb{R}.
$$
Then I do not see why it is
$$
\lvert\frac{dx}{dt}\rvert\leq 2\lvert x\rvert+t^2.
$$
| The steps that come to mind are
$$\begin{array}{}
\frac{dx}{dt}=\sqrt{x^2+1}+t^2 \\
|\frac{dx}{dt}|=|\sqrt{x^2+1}+t^2| \\
|\frac{dx}{dt}|\le|\sqrt{x^2+1}|+|t^2| & \qquad \text{Triangle Inequality} \\
|\frac{dx}{dt}|\le \sqrt{x^2+1} + t^2 \\
\end{array}$$
If we had $\sqrt{x^2+1} \le 2|x|$, then we could get your final ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1572133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $n^{n^{n^{n}}}- n^{n^{n}}$ is divisible by $1989$ Problem
Let $n$ be a positive integer with $n \geq 3$. Show that $$n^{n^{n^{n}}}- n^{n^{n}}$$ is divisible by $1989$.
I don't really know where to begin with this question. Maybe I could do some case work on $n$ being even or odd, but I am not sure if that ... | $1989$ factors as $3^2\cdot 13\cdot 17$, so it is enough to show that $n\uparrow\uparrow 4 - n\uparrow\uparrow 3 \equiv 0 \pmod a$ for $a=9,13,17$.
ANALYSIS. Suppose $n$ is coprime to $a$ -- otherwise the congruence clearly holds. Then,
$$ n\uparrow\uparrow 4 - n\uparrow\uparrow 3 = n^{n\uparrow\uparrow 2} (n^{n\uparro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1572802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 1
} |
Simplify $2\cos(6\pi/7) + 2\cos(2\pi/7) + 2\cos(4\pi/7) + 1$ . I am trying to simplify
$$2\cos(6\pi/7) + 2\cos(2\pi/7) + 2\cos(4\pi/7) + 1$$
However if I plug this in the calculator the answer is zero. Is there a way to keep on simplifying without the calculator?
I know the identity $2\cos(\theta) = (e^{i\theta} +e^{... | HINT:
Your sum equals
$$\sum_{k=0}^6 \cos\frac{2 k \pi}{7}$$
because $\cos (\theta) = \cos (2\pi - \theta)$, so for instance $\cos\frac{4 \pi}{7} = \cos\frac{10 \pi}{7}$. Now this sum equals the real part of
$$\sum_{k=0}^6 (\cos\frac{2 k \pi}{7}+ i \sin \frac{2 k \pi}{7}) = \sum_{k=0}^6 (\cos\frac{2 \pi}{7}+ i \sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1574377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Calculus - limit of a function: $\lim\limits_{x \to {\pi \over 3}} {\sin (x-{\pi \over 3})\over {1 - 2\cos x}}$ How do you compute the following limit without using the l'Hopital rule?
If you were allowed to use it, it becomes easy and the result is $\sqrt{3}\over 3$ but without it, I am not sure how to proceed. $$\li... | $\frac {\sin x\cos \frac {\pi}{3}- \cos x\sin\frac {\pi}{3}}{1+2\cos x}\\
\frac {\frac 12\sin x- \frac {\sqrt 3}{2} \cos x}{1-2\cos x}\\
\frac {\sin x- \sqrt 3\cos x}{2-4\cos x}\\
\frac {\sin^2 x- 3\cos^2 x}{(2-4\cos x)(\sin x + \sqrt 3 \cos x)}\\
\frac {\sin^2 x + \cos^2 x- 4\cos^2 x}{(2-4\cos x)(\sin x + \sqrt 3 \cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1575541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
On $p^2 + nq^2 = z^2,\;p^2 - nq^2 = t^2$ and the "congruent number problem" (Much revised for brevity.) An integer $n$ is a congruent number if there are rationals $a,b,c$ such that,
$$a^2+b^2 = c^2\\
\tfrac{1}{2}ab = n$$
or, alternatively, the elliptic curve,
$$y = x^3-n^2x = x(x-n)(x+n)\tag1$$
is solvable in the rati... | Probably for the system.
$$\left\{\begin{aligned}&x^2+qy^2=a^2\\&x^2-qy^2=b^2\end{aligned}\right.$$
Write a simple formula as these numbers just to find.
$$x=p^4+s^4$$
$$y=2ps\sqrt{\frac{(p^2+s^2)(p^2-s^2)}{q}}$$
$$a=p^4+2p^2s^2-s^4$$
$$b=s^4+2p^2s^2-p^4$$
Or so.
$$x=p^4+6p^2s^2+s^4$$
$$y=2(p^2-s^2)\sqrt{\frac{2ps(p^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1576861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the length of AB In the diagram 4 circles of equal radius stand in a row in such a way that each circle touches the next one. $P$ is a point on the circumference of the first circle. The center of the fourth circle is point $Q$. The line $PQ$ goes through the centers of all four circles. $PC$ is a tangent on the f... | Let $\angle CPQ=\alpha$. We have : $\sin \alpha=\dfrac{CQ}{QP}=\dfrac{7}{7^2}= \dfrac{1}{7}$.
The distance from PC tho the center of the second circle is $d=3\cdot 7 \cdot \sin \alpha=3$ so the chord $AB$ is $AB=2\sqrt{7^2-3^2}=4\sqrt{10}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1579850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Integral $\int_0^\infty\frac{\tanh^2(x)}{x^2}dx$ It appears that
$$\int_0^\infty\frac{\tanh^2(x)}{x^2}dx\stackrel{\color{gray}?}=\frac{14\,\zeta(3)}{\pi^2}.\tag1$$
(so far I have about $1000$ decimal digits to confirm that).
After changing variable $x=-\tfrac12\ln z$, it takes an equivalent form
$$\int_0^1\frac{(1-z)^2... | $$\mathcal{A}(m,n)=\int_{0}^{\infty}\frac{\tanh(z)^m}{z^n}\text{d}z\qquad m+1>n>1$$
We have
$$\begin{aligned}
&\mathcal{A}\left(1,p+1\right)
=\frac{(2^{p+1}-1)}{\pi^p\cos\frac{p\pi}{2} } \zeta(p+1)\\
&\mathcal{A}\left(2,p+1\right)=
\frac{(p+1)(2^{p+2}-1)}{\pi^{p+1}\sin\frac{p\pi}{2} } \zeta(p+2)\\
&\mathcal{A}\left(3,p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1582943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "41",
"answer_count": 8,
"answer_id": 1
} |
What are all the concordant forms $n$ such that $a^2+b^2 = c^2,\,a^2+nb^2=d^2$ for $n<1000$? Part I. The list of congruent numbers $n<10^4$ such that the system,
$$a^2-nb^2 = c^2$$
$$a^2+nb^2 = d^2$$
has a solution in the positive integers is known (A003273)
$$n = 5, 6, 7, 13, 14, 15, 20, 21, 22, 23, 24, 28, 29, 30, 31... | While individ in his answer has shown there is an infinite number of concordant forms $n$, it can also be shown there is an infinite number of the special case $n = \pm N^2$. Given the system,
$$a^2+b^2 = c^2\tag1$$
$$a^2+nb^2 = d^2\tag2$$
First, let $a,\,b,\,c = p^2-q^2,\,2pq,\,p^2+q^2$ to satisfy $(1)$. The second be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1584485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Find dimension and a basis of a subspaces $U+V$, $U \cap V$ in terms of the parameter $\alpha$ Let $U=span((1,1,1,2),(1,2,2,\alpha))$ and $V=span((1,3,4,\alpha+2),(1,4,\alpha,\alpha+1))$ are the subspaces of $\mathbb{R^4}$. ($\alpha\in\mathbb{R}$).
Find dimension and one basis of $U+V$, $U \cap V$ in terms of $\alpha$.... | First, note that $\DeclareMathOperator{rref}{rref}$
\begin{align*}
\rref
\begin{bmatrix}
1&1&1&2\\
1&2&2&\alpha
\end{bmatrix}
&=
\begin{bmatrix}
1&0&0&4-\alpha\\
0&1&1&\alpha-2
\end{bmatrix}
&
\rref
\begin{bmatrix}
1&3&4&2+\alpha\\
1&4&\alpha&1+\alpha
\end{bmatrix}
&=
\begin{bmatrix}
1&0&16-3\,\alpha&\alpha+5\\
0&1&\al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1584810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How do you evaluate this summation: $S=\sum\limits_{r=0}^{15} (-1)^r \frac{\binom{15}{r}}{\binom{r+3}{r}}$ Find S:
$$S=\sum_{r=0}^{15} (-1)^r \frac{\binom{15}{r}}{\binom{r+3}{r}}$$
My attempt:
I tried writing the summation as:
$$S=3!(15!)\sum_{r=0}^{15} (-1)^r \frac{1}{(15-r)!(r+3)!}$$
and tried to convert it to a tel... | $$S=\sum_{r=0}^{15} (-1)^r \frac{\binom{15}{r}}{\binom{r+3}{r}} = \sum_{r=0}^{15} (-1)^r \frac{15!}{r! (15-r)!} \frac{3!\cdot r!}{(r+3)!}$$
$$= 15!\cdot3!\cdot\sum_{m=3}^{18} (-1)^{m+1} \frac{1}{(18-m)!} \frac{1}{(m)!}$$
$$= \frac{3!}{16 \cdot 17 \cdot 18} \sum_{m=3}^{18} (-1)^{m+1} \binom{18}{m}$$
$$= \frac{3!}{16 \cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1585842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Nesbitt's Inequality $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}$ I'm reading a book which focus in inequality.
I'm stuck in this question. Let $a,b,c$ be positive real numbers. (Nesbitt's inequality) Prove the inequality $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}$$
So the first step of so... | HINT: set
$$b+c=x$$
$$c+a=y$$
$$a+b=z$$
adding we get
$$a+b+c=\frac{x+y+z}{2}$$ and we can compute $$a+x=\frac{x+y+z}{2}$$ thus $$a=\frac{-x+y+z}{2}$$ etc
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1589751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Better proof for $\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} \equiv \frac{1+\cos x}{\sin x}$ It's required to prove that
$$\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} \equiv \frac{1+\cos x}{\sin x}$$
I managed to go about out it two ways:
*
*Show it is equivalent to $\mathsf{true}$:
$$\frac{1+\cos x + \sin x}{1... | if
$$
\frac{a}{b}=\frac{c}{d}=k
$$
then
$$
\frac{a+c}{b+d} = \frac{kb+kd}{b+d} =k =\frac{a}{b}
$$
since $$1-\cos^2 x =(1+\cos x)(1-\cos x) =\sin^2 x$$ we have
$$
\frac{1+\cos x}{\sin x} = \frac{\sin x}{1-\cos x} =\frac{1+\cos x +\sin x}{1 -\cos x +\sin x}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
} |
Prove that $xy+yz+zx+\frac{x^2y^2}{z^2}+\frac{y^2z^2}{x^2}+\frac{z^2x^2}{y^2}\ge 2x^2+2y^2+2z^2$ for every $x,y,z$ strictly positive I checked the inequality for $y=z$. In this particular case, after some simplifications, the inequality becomes:
$$
2x^3+y^3\ge 3x^2y,
$$
which is true, according to the arithmetic-geomet... | Let $\frac{xy}{z}=a^2$, $\frac{yz}{x}=b^2$, and $\frac{zx}{y}=c^2$.
Now, we have $y=ab$, $z=bc$, and $x=ca$.
It now suffices to prove $$a^4+b^4+c^4+a^2bc+b^2ca+c^2ab \ge 2(a^2b^2+b^2c^2+c^2a^2)$$
From Schur's Inequality, we have $$a^2(a-b)(a-c)+b^2(b-c)(b-a)+c^2(c-a)(c-b)=(a^4+b^4+c^4)-(a^3b+a^3c+b^3c+b^3a+c^3a+c^3b)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Convergence of sequence of ratio of consecutive terms. I would like to prove that the following sequence of ratios $(x_n)$ where $x_n = c_n/c_{n-1}$ converges to a finite limit $L$:
$$\lim_{n\to\infty} x_n = \lim_{n\to\infty}\frac{c_n}{c_{n-1}} = L$$
Where $c_n$ is defined by the recurrence relation:
$c_n = c_{n-1} + \... | It is clear that $c_n\rightarrow\infty$, and $x_n\ge 1$ for all $n$. Note that
$$ x_n = \frac{c_n}{c_{n-1}} = \frac{c_{n-1} + \lfloor\frac{c_{n-4}}{2}\rfloor}{c_{n-1}} = 1 + \frac{c_{n-2}}{c_{n-1}}\frac{c_{n-3}}{c_{n-2}}\frac{c_{n-4}}{c_{n-3}}\frac{\lfloor\frac{c_{n-4}}{2}\rfloor}{c_{n-4}} = 1 + \frac{1}{x_{n-1}x_{n-2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Integral $\int_0^{1/2}\arcsin x\cdot\ln^3x\,dx$ It's a follow-up to my previous question.
Can we find an anti-derivative
$$\int\arcsin x\cdot\ln^3x\,dx$$
or, at least, evaluate the definite integral
$$\int_0^{1/2}\arcsin x\cdot\ln^3x\,dx$$
in a closed form (ideally, as a combination of elementary functions and polyloga... | The indefinite integral can be expressed in terms of hypergeometrics.
We start with integration by parts with $dv=\sin^{-1}(x)$ and $u=\log^3x$.
$$\begin{align}
I&=\int \arcsin x \cdot \ln^3x\,dx\\
&=\ln^3x\cdot\left(\sqrt{1-x^2}+x\arcsin x\right)-\int \left(\sqrt{1-x^2}+x\arcsin x\right)\cdot \frac{3\ln^2x}{x}\,dx
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Uniform convergence of $\sum_{n=0}^{\infty} \frac{\int_{\sin nx}^{\sin(n+1)x}\sin t^2dt \int_{nx}^{\infty}\frac{dt}{\sqrt{t^4+1}}}{1+n^3x^2}$ I am preparing for the exam. Please help me to solve the following problem:
Given a series
$$\sum_{n=0}^{\infty} \frac{\int_{\sin nx}^{\sin(n+1)x}\sin t^2dt \int_{nx}^{\infty}\f... | First
$$
\begin{align}
\int_0^\infty\frac{\mathrm{d}t}{\sqrt{t^4+1}}
&=\frac14\int_0^\infty\frac{t^{-3/4}\,\mathrm{d}t}{(t+1)^{1/2}}\\
&=\frac14\frac{\Gamma\left(\frac14\right)\Gamma\left(\frac14\right)}{\Gamma\left(\frac12\right)}\\
&\le\int_0^1\mathrm{d}t+\int_1^\infty\frac{\mathrm{d}t}{t^2}\\[9pt]
&=1+1=2
\end{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Prove $(1+2+...+k)^2 = 1^3 + ... + k^3$ using induction I need to prove that
$$(1+2+{...}+k)^2 = 1^3 + {...} + k^3$$
using induction.
So the base case holds for $0$ because $0 = 0$ (and also for $1$: $1^2 = 1^3 = 1$)
I can't prove it for $k+1$ no matter what I try! Can you give me a hint?
| We have $(a+b)^2=a^2+b^2 + 2ab.$ Thus $$(1+2+...+(k+1))^2=(1+2+...+k)^2+(k+1)^2+ 2(1+2+...+k)(k+1)=1^3+...+k^3+(k+1)^2 + 2(1+2+...+k)(k+1)$$
Using $1+...+k=\dfrac{k(k+1)}{2}$ we get $$(1+2+...+k+1)^2=1^3+...+k^3+ (k+1)^2 + k(k+1)^2=1^3+...+k^3+ (k+1)^2(k+1)=1^3+...+(k+1)^3$$ and the inductive step is proven.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1592512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Find the real ordered pairs $(x,y)$ satisfying $x^2+x=y^4+y^3+y^2+y.$ Find the real ordered pairs $(x,y)$ satisfying $x^2+x=y^4+y^3+y^2+y.$
$x^2+x=y^4+y^3+y^2+y$
$4x^2+4x=4y^4+4y^3+4y^2+4y$
$4x^2+4x+1=4y^4+4y^3+4y^2+4y+1$
$(2x+1)^2=4y^4+4y^3+4y^2+4y+1$
I am stuck here.Is there a general method to solve such type of eq... | HINT: $$x^2+x=y^4+y^3+y^2+y \Rightarrow \left(x+\frac{1}{2}\right)^2=y^3(y+1)+\left(y+\frac{1}{2}\right)^2$$
$$\Rightarrow \left(x+\frac{1}{2}\right)^2-\left(y+\frac{1}{2}\right)^2=y^3(y+1)$$
$$\Rightarrow (x+y+1)(x-y)=y^3(y+1)$$
Now consider the $4$ cases where $x$ is even or odd and $y$ is even or odd and use divisib... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\sum_{n=1}^{\infty}(\frac{2}{9})^na_n=\frac{1}{3}$ Let $a_n(n\geq1)$ be the sequence of numbers defined by the recurrence relation$\hspace{1cm}a_1=1,\hspace{1cm}a_n=a_{n-1}a_1+a_{n-2}a_2+...+a_2a_{n-2}+a_1a_{n-1}$
Prove that $\sum_{n=1}^{\infty}(\frac{2}{9})^na_n=\frac{1}{3}$
Let $\alpha=\sum_{n=1}^{\infty... | Let $f(x)$ be $\sum_{i=1}^\infty a_ix^i$. Note that $$f(x) \cdot f(x) = \sum_{i=1}^\infty a_ix^i \cdot \sum_{i=1} a_ix^i = \sum_{i=2}^\infty \sum_{j=1}^{n-1} a_ja_{n-j} x^i = \sum_{i=2}^\infty a_ix^i = f(x)-a_1x=f(x)-x$$
Now we have $f(x)^2-f(x)+x=0$, or $f(x)=\frac{1 \pm \sqrt{1-4x}}{2}$.
Since $f(0)=0$, we have $f(x)... | {
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Find all real numbers $x,y > 1$ such that $\frac{x^2}{y-1}+\frac{y^2}{x-1} = 8$
Find all real numbers $x,y > 1$ such that $$\frac{x^2}{y-1}+\frac{y^2}{x-1} = 8$$
Attempt
We can rewrite this as $x^2(x-1)+y^2(y-1) = 8(x-1)(y-1)$. Then I get a multivariate cubic, which I find hard to find all solutions to.
| Note $a=x-1$ and $b=y-1$ and apply $r^2+s^2\geqslant2rs$. The solution is $x=y=2$.
| {
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Evaluate the following integral I got stuck with the integral below. I have tried to make it look like the derivative of arctan.
$$\int \frac{2-x}{x^2-x+1}\,dx$$
Thank you!
| $$\int \frac{2-x}{x^2-x+1}\ dx=\frac{1}{2}\int \frac{3-(2x-1)}{x^2-x+1}dx\\=\frac{1}{2}\int \frac{3}{x^2-x+1}\ dx-\frac 12\int \frac{
2x-1}{x^2-x+1}\ dx\\=\frac{3}{2}\int \frac{1}{\left(x-\frac 12\right)^2+\frac{3}{4}} dx-\frac 12\int \frac{x^2-x+1}{x^2-x+1}dx$$
Substitute $u:=x^2-x+1$ and $du=(2x-1)dx$
Substitute $\va... | {
"language": "en",
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Area of the shadow of a regular polygon inscribed in a sphere. Consider the situation given below:
Let a regular polygon be inscribed in a sphere such that its circumcentre is at a distance $r$ from the centre of the sphere of radius $R$. A point source of light is kept at the centre of the sphere. How can we calculate... | Let $\theta = \cos^{-1}\frac{r}{R}$ and $\phi = \frac{2\pi}{n}$ where
$n$ is the number of sides of the regular polygon.
Choose the coordinate system so that the vertices of the regular polygon are located at
$$\vec{v}_k = (R\sin\theta\cos(k\phi), R\sin\theta\sin(k\phi), R\cos\theta) \quad\text{ for } k = 0, 1, \ldots,... | {
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Prove that if $a$ and $b$ are nonnegative real numbers, then $(a^7+b^7)(a^2+b^2) \ge (a^5+b^5)(a^4+b^4)$ Prove that if $a$ and $b$ are nonnegative real numbers, then $(a^7+b^7)(a^2+b^2) \ge (a^5+b^5)(a^4+b^4)$
My try
My book gives as a hint to move everything to the left hand side of the inequality and then factor an... | You can simplify considerably by multiplying out both sides $$a^9+a^7b^2+a^2b^7+b^9\ge a^9+a^5b^4+a^4b^5+b^9$$
Cancelling the common terms from each side and dividing through by $a^2b^2$ gives $$a^5+b^5\ge a^3b^2+a^2b^3$$
Now move everything to the LHS and factorise.
Note if $a$ or $b$ is zero, equality is obvious.
| {
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Prove that $a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq 2(ab+bc+ca).$
Let $a,b,c$ be three nonnegative real numbers. Prove that $$a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq 2(ab+bc+ca).$$
It seems that the inequality $a^2+b^2+c^2 \geq ab+bc+ca$ will be of use here. If I use that then I will get $a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2}... | Let $a=x^3$, $b=y^3$, $c=z^3$, then it can be rewritten as:
$$
x^6+y^6+z^6+3 x^2 y^2 z^2-2 \left(x^3 y^3+x^3 z^3+y^3 z^3\right)\geq 0
$$
Use the following notations:
$$S_{3}:=xyz\qquad S_2:=xy+yz+xz\qquad S_1=x+y+z$$
Then:
$$
x^6+y^6+z^6=S_1^6-6 S_2 S_1^4+6 S_3 S_1^3+9 S_2^2 S_1^2-12 S_2 S_3 S_1-2 S_2^3+3 S_3^2
$$
$$
x... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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prove that $(x-y)^3+(y-z)^3+(z-x)^3 > 0.$
If $x<y<z$ are real numbers, prove that $$(x-y)^3+(y-z)^3+(z-x)^3 > 0.$$
Attempt
We have that $(x-y)^3+(y-z)^3+(z-x)^3=-3 x^2 y+3 x^2 z+3 x y^2-3 x z^2-3 y^2 z+3 y z^2$. Grouping yields $3(x^2z+xy^2+yz^2)-3(x^2y+xz^2+y^2z)$. Then I was thinking of using rearrangement but then... | Notice that the polynomial
$$S = (x-y)^3 + (y-z)^3 + (z-x)^3$$
is zero whenever two of the variables are equal so we must have
$$S = C(y-x) (z-x) (z-y)$$
for some constant $C$. A direct computation gives $C=3$ and the result follows since all factors are positive when $x < y < z$.
| {
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Prove $\sqrt{n}+\sqrt{n+k}$ is irrational For what $k\in\mathbb N$, $\sqrt{n}+\sqrt{n+k}$ is irrational? ($\forall n\in\mathbb N$)
| This result is not true in general for example, suppose that $n$ is a square $n=l^2$ and $k=0$ $\sqrt{l^2+0}+\sqrt{l^2}$ is rational, suppose that $n=9, k=16$, $\sqrt{16+9}+\sqrt{9}$ is rational.
The question is given $n$ for what values of $k$, $\sqrt{n+k}+\sqrt{n}$ is irrrational?
Proposition
Suppose that $k,n\in N$ ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that if $a,b,$ and $c$ are positive real numbers then $(a+b)(a+c) \geq 2 \sqrt{abc(a+b+c)}.$
Prove that if $a,b,$ and $c$ are positive real numbers then $$(a+b)(a+c) \geq 2 \sqrt{abc(a+b+c)}.$$
This looks like a simple question. We can apply AM-GM twice to get $(a+b)(a+c) \geq 4a\sqrt{bc}$. Then how do I use th... | We have
$$(a+b)(a+c) = a(a+b+c) +bc.$$
We can now apply AM-GM.
| {
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Prove the identity $\binom{2n+1}{0} + \binom{2n+1}{1} + \cdots + \binom{2n+1}{n} = 4^n$ I've worked out a proof, but I was wondering about alternate, possibly more elegant ways to prove the statement. This is my (hopefully correct) proof:
Starting from the identity $2^m = \sum_{k=0}^m \binom{m}{k}$ (easily derived from... | Applying what you mention on (in this context nicer) odd $2n+1$ instead of even $2n$ we find:$$2\times4^n=2^{2n+1}=\sum_{k=0}^{2n+1}\binom{2n+1}{k}=2\sum_{k=0}^{n}\binom{2n+1}{k}$$
Now divide by $2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to evaluate this Integral $\int { {\sqrt{5^2+K^2}}dK \over {\sqrt{10^2+K^2}K}} $ While working on an Exact Differential Equation, I encounter the following Integral.
$$\int { {\sqrt{5^2+K^2}} \over {K\sqrt{10^2+K^2}}} dK$$
I have tried substitution and all the other elementary methods, but the Integral simply refu... | This is not a trivial one $$I=\int\frac{\sqrt{x^2+25}}{x \sqrt{x^2+100}}\,dx$$ Let us try using $$\frac{\sqrt{x^2+25}}{ \sqrt{x^2+100}}=u^2 \implies x=\frac{5 \sqrt{1-4 u^4}}{\sqrt{u^4-1}}\implies dx= \frac{30 u^3}{\sqrt{1-4 u^4} \left(u^4-1\right)^{3/2}}du$$ So, $$I=-\int\frac{6 u^5}{4 u^8-5 u^4+1}\,du$$ Now, since th... | {
"language": "en",
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"source": "stackexchange",
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Proving sum of product forms a pattern in n * nnnnnn.... I am consider a problem regarding numbers which are, in decimal, one digit repeated - for instance, $88888888$ is such a number. In particular, I am looking at the following problem:
The sum of the digits of the number $$8\cdot \underbrace{88\ldots 88}_{n\text{ ... | $a*aaaa....aaa = a^2*1111111....111$ (m a's and m 1's)
Three cases to consider:
$a^2$ has one digit. (i.e. $a = 1,2,3$)
$a^2$ has two digits and the sum of the digits is less than 10. (i.e. $a = 4,5,6, 9$)
$a^2$ has two digits and the sum of the digits is 10 or more. (i.e. $a = 7,8$)
====
If $a^2 = b$ has one digit,$b$... | {
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Second derivative of $x^3+y^3=1$ using implicit differentiation I need to find the $D_x^2y$ of $x^3+y^3=1$ using implicit differentiation
So,
$$
x^3 + y^3 =1 \\
3x^2+3y^2 \cdot D_xy = 0 \\
3y^2 \cdot D_xy= -3x^2 \\
D_xy = - {x^2 \over y^2}
$$
Now I need to find the $D_x^2y$.
I am pretty sure that means the second d... | If $Dx$ is the first derivative and $Dx^2$ is the second derivative, than your first derivative is correct. For the second derivative we have:
$$
\frac{d}{dx}y'=\frac{d}{dx}\left(\frac{-x^2}{y^2}\right)$$
that, using fraction rule and chain rule for $y$, becomes:
$$\frac{-2xy^2+2x^2y(y')}{y^4}$$
substituting $y'=-x^2/y... | {
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"timestamp": "2023-03-29T00:00:00",
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Find the value of $\frac{i}{4-\pi}\int_{|z|=4}\frac{1}{z\cos{z}}dz$ Find the value of $$\frac{i}{4-\pi}\int_{|z|=4}\frac{1}{z\cos{z}}dz$$.
My attempt:
The integrand has singularities at $z=0, \frac{\pi}{2}, \frac{-\pi}{2}$, so
$$\frac{i}{4-\pi}\int_{|z|=4}\frac{1}{z\cos{z}}dz=\frac{i}{4-\pi}2\pi i ~Res_{z=z_k}\phi(z)=... | The computation of the residues is messed up. You must consider the function as a whole, but you tossed away $1/z$ and $\cos z$ where it mattered. The rule is: $${\rm Res}\left(\frac{f(z)}{g(z)},a\right) = \frac{f(a)}{g'(a)},$$if $a$ is a simple pole of $g$. Let's save the $i/(4-\pi)$ for later. We have: $$\int_{|z|=4}... | {
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"timestamp": "2023-03-29T00:00:00",
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Find the least value of $4\csc^{2} x+9\sin^{2} x$
Find the least value of $4\csc^{2} x+9\sin^{2} x$
$a.)\ 14 \ \ \ \ \ \ \ \ \ b.)\ 10 \\
c.)\ 11 \ \ \ \ \ \ \ \ \ \color{green}{d.)\ 12} $
$4\csc^{2} x+9\sin^{2} x \\
= \dfrac{4}{\sin^{2} x} +9\sin^{2} x \\
= \dfrac{4+9\sin^{4} x}{\sin^{2} x} \\
= 13 \ \ \ \ \ \ \ \... | By AM-GM inequality
$$4\csc^{2} x+9\sin^{2} x \geq 2 \sqrt{36\csc^{2} x\sin^{2} x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Find the minimum value of $\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta$
Find the minimum value of
$\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta$
$a.)\ 1 \ \ \ \ \ \ \ \ \ \ \ \ b.)\ 3 \\
c.)\ 5 \ \ \ \ \ \ \ \ \ \... | Hint:
We can use the Pythagorean identities $\color{blue}{\sin^2\theta+\cos^2\theta=1}$, $\color{blue}{\sec^2 \theta=\tan^2 \theta+1}$ and $\color{blue}{\csc^2\theta=\cot^2 \theta+1}$, giving us
\begin{align}\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta&=3+2\tan^2\thet... | {
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"timestamp": "2023-03-29T00:00:00",
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Convergence of $\int_\Bbb{R^2} \frac{\log(x^2+y^2)}{x^2+y^2}$ I tried to figure out if $\int_\Bbb{R^2} \frac{\log(x^2+y^2)}{x^2+y^2}$ converges
I think I should split it into two integrals $$\lim_{k\to\infty}\int_{B_\frac{1}{k}} \frac{\log(x^2+y^2)}{x^2+y^2} + \lim_{r\to\infty}\int_{B_r} \frac{\log(x^2+y^2)}{x^2+y^2}$... | You can form the following chain of inequalities
$$ \int_4^\infty \frac{\log r^2}{r} dr > \int_4^\infty \frac{1}{r} dr = \lim_{b \to \infty} \log b - \log 4 = \infty.$$
So your integral diverges at $\infty$. In fact, it also diverges at $0$.
| {
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$f(6)=144$ and $f(n+3) = f(n+2)\{f(n+1)+f(n)\}$, Then $f(7) =$? Given that $f(6)=144$ and $f(n+3) = f(n+2) \cdot\Big(f(n+1)+f(n)\Big)$ $[$For $n = 1,2,3,4]$
Then find the value of $f(7)$.
The solution is not unique but all of them are positive integers.
I can't find a way out.
| Letting $a=f(1)$, $b=f(2)$, $c=f(3)$, $d=f(4)$, $e=f(5)$, $144=f(6)$, and $g=f(7)$ for notational simplicity, we have
$$\begin{align}
d&=c(b+a)\\
e&=d(c+b)\\
144&=e(d+c)\\
g&=144(e+d)\\
\end{align}$$
If I've done the algebra correctly, we can eliminate $d$ and $e$ leaving
$$144=c^2(a+b+1)(a+b)(b+c)$$
and
$$g=144c(a+b+1... | {
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Prove that $\frac{2^{122}+1}{5}$ is a composite number As in the title. It's very easy to show that $5|2^{122}+1$, but what should I do next to show that $\frac{2^{122}+1}{5}$ is a composite number? I'm looking for hints.
| $$1+2^{122}=\left(1+2^{61}\right)^2-\left(2^{31}\right)^2$$
$$=\left(1+2^{61}+2^{31}\right)\left(1+2^{61}-2^{31}\right)$$
Both factors are larger than $5$, also $$2^{122}\equiv \left(2^4\right)^{30}\cdot 2^2\equiv (1)^{30}\cdot 2^2\equiv -1\pmod{5},$$ so your result follows.
More generally, the following is called the ... | {
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"url": "https://math.stackexchange.com/questions/1622756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Compute $\lim_{n \to +\infty} n^{-\frac12 \left(1+\frac{1}{n}\right)} \left(1^1 \cdot 2^2 \cdot 3^3 \cdots n^n \right)^{\frac{1}{n^2}}$
How to compute
$$\displaystyle \lim_{n \to +\infty} n^{-\dfrac12 \left(1+\dfrac{1}{n}\right)} \left(1^1\cdot 2^2 \cdot 3^3 \cdots n^n \right)^{\dfrac{1}{n^2}}$$
I'm interested in ... | Another approach, considering $$A_n= n^{-\dfrac12 \left(1+\dfrac{1}{n}\right)} \left(\prod_{i=1}^n i^i \right)^{\dfrac{1}{n^2}}=n^{-\dfrac12 \left(1+\dfrac{1}{n}\right)} H(n)^{\dfrac{1}{n^2}}$$ where appears the hyperfactorial function. Taking logarithms $$\log(A_n)={-\dfrac12 \left(1+\dfrac{1}{n}\right)}\log(n)+\dfrac... | {
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"url": "https://math.stackexchange.com/questions/1624690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Integral $\int_0^{\pi/2} \sin(ax)\cos(x)\,dx$ I have to evaluate an integral $I(a) = \sin(ax)\cos(x)$ from $0$ to $\pi/2$.The variable of $a$ is not is greater than $1$:
$$\int_0^{\pi/2} \sin(ax)\cos(x)\,dx$$
I attempted to change the function to $[\sin(ax+x)+\sin(ax-x)]/2$ and then integrate, but I am left with (-)co... | Your approach is good. Integrate the following identity term-wise:
$$
\sin(a x) \cos(x) = \frac{1}{2} \sin\left(\left(1+a\right)x\right) - \frac{1}{2} \sin\left(\left(1-a\right) x\right)
$$
getting
$$
\int_0^{\pi/2} \sin(a x) \cos(x) \mathrm{d}x = \frac{1}{2} \int_0^{\pi/2} \sin\left(\left(1+a\right)x\right)\math... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Result of matrix $A^{2016}$ I want to find the result of $A^{2016}$ but I cannot find any pattern except for the zeros in the middle row and column.
$$A=\begin{bmatrix}1 & 0 & {-2}\\0 & 0 & {0}\\3 & 0 & {-4}\end{bmatrix}$$
| Note that $A$ has characteristic polynomial $p(\lambda)=-\lambda^3-3\lambda^2-2\lambda$, which yields $\lambda_1=-2$, $\lambda_2=-1$ and $\lambda_3=0$. Since each eigenvalue is different and there are 3, then $A$ is diagonalizable.
It is easy to see that $A=PDP^{-1}$ with
$P=\left(\begin{array}{ccc}2&1&0\\0&0&1\\3&1&0... | {
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Proving Trig Identities (Complex Numbers)
Question: Prove that if $z = \cos (\theta) + i\sin(\theta)$, then
$$ z^n + {1\over z^n} = 2\cos(n\theta) $$
Hence prove that $\cos^6(\theta)$ $=$ $\frac 1{32}$$(\cos(6\theta)$ + $6\cos(4\theta)$ + $15\cos(2\theta)$ + $10$)
I learnt to prove the first part in another post l... | By De Moivre's formula if $z = \cos (\theta) + i\sin(\theta)$ then $z^n = \cos (n\theta) + i\sin(n\theta)$.
So $$z^n = \cos (n\theta) + i\sin(n\theta)$$
and
$$z^{-n} = \cos (-n\theta) + i\sin(-n\theta)=\cos (n\theta) - i\sin(n\theta)$$
So $$z^n+z^{-n}=2\cos (n\theta)$$
Thus when $n=1$ ,
$$z+z^{-1}=2\cos (\theta) \Ri... | {
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"answer_id": 1
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Solution to the equation $\sqrt{x^2 - 2x + 1} - 5 = 0$ I had this equation on my exam :
$$\sqrt{x^2-2x+1} - 5 = 0$$
My friends said the the solution could be :
$$|x-1| = -5$$
So the solution is nothing!
But I say the solution is:
$$x^2-2x+1 = 25 $$
so $$x = 6\ |\ x = -4$$
My Question here is which solution is right... | It is $$\sqrt{x^2-2x+1} - 5 = 0 \\ \Rightarrow \sqrt{(x-1)^2} - 5 = 0 \\ \Rightarrow |x-1|-5=0 \\ \Rightarrow |x-1|=5 \\ \Rightarrow x-1=5 \text{ or } x-1=-5 \\ \Rightarrow x=6 \text{ or } x=-4$$ So, your solution is correct!!
$$$$
If the equation is $\sqrt{x^2-2x+1} + 5 =0$ then we have the following:
$$\sqrt{x^2-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1630525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Discrete Time Markov Chain question Let $\{X_n : n \ge 0 \}$ be a Markov chain with state space $ \{0, 1, 2, 3\} $ and transition matrix
$$P=\begin{pmatrix}
\frac{1}{4} & 0 & \frac{1}{2} & \frac{1}{4}\\
0 & \frac{1}{5} & 0 & \frac{4}{5}\\
0 & 1 & 0 & 0\\
\frac{1}{3} & \frac{1}{3} & 0 & \frac{1}{3}\\
\end{pmatrix}
$$
A... | First of all, notice that $Z_{n+1} = 2 \Leftrightarrow X_{n+1} = 2$ and $Z_n = 0 \Leftrightarrow X_n \in \{0,1\}$. So
$$P(Z_{n+1} = 2 \mid Z_n = 0, Z_{n-1} = 2) = P(X_{n+1} = 2 \mid X_n \in \{0,1\}, X_{n-1} = 2).$$
Write this quantity as
$$P(X_{n+1} = 2 \mid \{X_n = 0\} \cup \{X_n = 1\}, X_{n-1} = 2).$$
The strong Mark... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1633487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
A series related to $\pi\approx 2\sqrt{1+\sqrt{2}}$ This question follows a suggestion by Tito Piezas in Is there an integral or series for $\frac{\pi}{3}-1-\frac{1}{15\sqrt{2}}$?
Q: Is there a series by Ramanujan that justifies the approximation $\pi\approx2\sqrt{1+\sqrt{2}}?$
| The answer needs some context.
Part I. One may ask why,
$$\frac{1}{2\pi\sqrt{2}} - \frac{1103}{99^2} \approx 10^{-9}\tag1$$
is such a good approximation? In fact, the convergents of the continued fraction of $\displaystyle\frac{1}{2\pi\sqrt{2}}$ start as,
$$0,\;\frac{1}{8},\;\color{brown}{\frac{1}{3^2}},\;\frac{8}{71}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1633782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Bounded sequence $a_n=\sqrt{4+2 \sqrt{4+\cdots+2 \sqrt{4+2 \sqrt{4+2 \sqrt{4+4}}}}}$ Let
$$a_n=\sqrt{4+2 \sqrt{4+\cdots+2 \sqrt{4+2 \sqrt{4+2 \sqrt{4+4}}}}}$$
the sign $\sqrt{}$ occurs $n$ times.
a) Prove, that $a_n< \sqrt{5}+1$ for all $n$.
b) Find $\lim_{n\rightarrow \infty } a_n$
Author O.Kukush
| For the case (b) when $n \to \infty$ we can write the following,
\begin{equation}
a_n^2 = 4 + 2a_n \\ a_n^2 - 2a_n - 4 = 0 \\ a_n = \cfrac{2+\sqrt{20}}{2} = 1 + \sqrt{5}
\end{equation}
So, it may be stated that for all other $n$, $a_n < 1 + \sqrt{5} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1634548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Limit of the sequence defined by a recurrence Given a recurrence formula for an arithmetic sequence, $$U_{n} = \frac{1}{2+U_{n-1}}$$
Show that$$\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2+ ...}}}} = (SomeGivenValue)$$
How can we solve questions like this?
| The method is to make a substitution $U_n=\frac{T_n}{T_{n+1}}$ and you would get much more tangible
$$T_{n+1}^2-2T_nT_{n+1}-T_n^2=0$$
$$(T_{n+1}-T_n-\sqrt{2}T_n)(T_{n+1}-T_n+\sqrt{2}T_n)=0$$
$$T_{n+1}-T_n-\sqrt{2}T_n=0$$
$$T_{n+1}-T_n+\sqrt{2}T_n=0$$
These two you solve classically assuming $T_n=a^n$ and when you subst... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1637100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Probability an integer chosen at random from 1 to 1000 is divisible by 3,5, or 7 A number is chosen at random from the first 1,000 positive integers. What is the probability that it's divisible by 3,5, or 7?
So I started off by breaking the problem up and having:
divisible by 3: p(a)
divisible by 5: p(b)
divisible by ... | Use inclusion/exclusion principle:
*
*Include the amount of numbers divisible by $3$, which is $\Big\lfloor\frac{1000}{3}\Big\rfloor=333$
*Include the amount of numbers divisible by $5$, which is $\Big\lfloor\frac{1000}{5}\Big\rfloor=200$
*Include the amount of numbers divisible by $7$, which is $\Big\lfloor\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1637495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to simplify $\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}$ The answer is 2. But I want to learn how to simplify this expression without the use of calculator.
| More generally, if $a + b \sqrt{2}$ (with $a$, $b$ integers) has a cube root in $\mathbb Z[\sqrt{2}]$, the norm $N(a + b \sqrt{2}) = a^2 - 2 b^2 \sqrt{2}$ must be the cube of an integer, say $m^3$, and then that cube root would be
$x + y \sqrt{2}$ with $x^2 - 2 y^2 = m$. From $(x+y \sqrt{2})^3 = a + b \sqrt{2}$ we get... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1639957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
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How do I finish this trig integral $\int_0^{\pi/4}\frac{\sin^2 \theta}{\cos \theta}d\theta$? I got up to the part where it's $$\frac{9}{125}\int_0^{\large \frac{\pi}{4}}\frac{\sin^2\theta}{\cos\theta}\,\,d\theta$$
but I can't figure out how to finish it off.
By the way the original problem was:
$$\int_0^{0.6}\frac{x^2... | Use the trick to add and remove a $\cos^2\theta$ term in the numerator of the integral:
$$\int_0^{\pi/4}\frac{\sin^2\theta + \cos^2\theta - \cos^2\theta}{\cos\theta}\ \text{d}\theta = \int_0^{\pi/4}\frac{1}{\cos\theta} - \cos\theta\ \text{d}\theta$$
Which now you can split.
Remembering that $$\frac{1}{\cos\theta} = \te... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1646897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the integral $\int \frac{1+x}{\sqrt{1-x^2}}\mathrm dx$ The integral can be represented as
$$
\int \frac{1+x}{\sqrt{1-x^2}}\mathrm dx=
\int \left(\frac{1+x}{1-x}\right)^{1/2}\mathrm dx
$$
Substitution $$t=\frac{1+x}{1-x}\Rightarrow x=\frac{t-1}{t+1}\Rightarrow dx=\frac{2}{(t+1)^2}dt\Rightarrow \int\limits \left(\f... | No substitutions:
$$
\int\left(\frac{1}{\sqrt{1-x^2}}-\frac{-x}{\sqrt{1-x^2}}\right)\,dx
=\arcsin x-\sqrt{1-x^2}+c
$$
You can also do that way; continue with $u=\sqrt{t}$, so $t=u^2$ and $dt=2u\,du$; so you get
$$
\int\frac{4u^2}{(u^2+1)^2}\,du=
\int 2u\cdot\frac{2u}{(u^2+1)^2}\,du
$$
Noticing that $2u$ is the derivati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1647349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Evaluation of $\lim_{n\rightarrow \infty}\sum_{k=1}^n\sin \left(\frac{n}{n^2+k^2}\right)$
Evaluation of $\displaystyle \lim_{n\rightarrow \infty}\sin \left(\frac{n}{n^2+1}\right)+\sin \left(\frac{n}{n^2+2^2}\right)+\cdots+\sin \left(\frac{n}{n^2+n^2}\right)$
$\bf{My Try::}$ We can write the Sum as $$\lim_{n\rightarro... | We can use Riemann sums to evaluate the limit. We have
$$\sum_{k=1}^n\sin\left(\frac{n}{n^2+k^2}\right)=\sum_{k=1}^n\,n\sin\left(\frac{1/n}{1+(k/n)^2}\right)\frac1n\to \int_0^1\frac{1}{1+x^2}\,dx=\pi/4$$
since we have $$\left(\frac{1}{1+(k/n)^2}\right)-\frac{1}{6n^2}\left(\frac{1}{1+(k/n)^2}\right)^3\le n\sin\left(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1647807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
Calculate $\frac{1}{\sin(x)} +\frac{1}{\cos(x)}$ if $\sin(x)+\cos(x)=\frac{7}{5}$ If
\begin{equation}
\sin(x) + \cos(x) = \frac{7}{5},
\end{equation}
then what's the value of
\begin{equation}
\frac{1}{\sin(x)} + \frac{1}{\cos(x)}\text{?}
\end{equation}
Meaning the value of $\sin(x)$, $\cos(x)$ (the denominator) wi... | Notice, $$\frac{1}{\sin x}+\frac{1}{\cos x}$$
$$=\frac{\sin x+\cos x}{\sin x\cos x}$$
$$=2\cdot \frac{\sin x+\cos x}{2\sin x\cos x}$$
$$=2\cdot \frac{\sin x+\cos x}{(\sin x+\cos x)^2-1}$$
setting the value of $\sin x+\cos x$,
$$=2\cdot \frac{\frac 75}{\left(\frac{7}{5}\right)^2-1}$$
$$=\frac{35}{12}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1649606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 10,
"answer_id": 7
} |
Diophantine equation: $\frac1x=\frac{a}{x+y}-\frac1y$ For how many different natural values of $a$ the Diophantine equation: $\frac1x=\frac{a}{x+y}-\frac1y$ has natural roots?
I rearranged the equation as: $xy+x^2+y^2=(a-1)xy$ , hence I said we must have: $(a-1)>2$
I tried another idea: reform the equation to: $x^2+(2... | Rearrange it to
$$\frac{a}{x+y} = \frac{1}{x} + \frac{1}{y} = \frac{y+x}{xy}.\tag{1}$$
Multiply $(1)$ with $x+y$ to obtain
$$a = \frac{(x+y)^2}{xy}.\tag{2}$$
So the question is how many natural numbers are values of $\frac{(x+y)^2}{xy}$ with $x,y\in \mathbb{N}$. Suppose $x,y\in \mathbb{N}$ with $\frac{(x+y)^2}{xy}\in \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1650688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Coefficents of cubic polynomial and its least root Let $x^3-(m+n+1)x^2+(m+n-3+mn)x-(m-1)(n-1)=0$,
be a cubic polynomial with positive roots, where $m,n \ge2$ are natural nos. For fixed $m+n$, say $15$, it turns out that least root of the polynomial will be smallest in case of $m=2,n=13$ i.e the case in which differenc... | Some Remarks:
There is an explicit formula to solve cubic equations here . So you can always use it and analyze the roots in a brute-force way. I will follow a somewhat simpler method. I will assume that :
in the cases you are interested in you always have three (not necessarily distinct) real roots... otherwise the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1656617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
If $x,y,z>0$ and $x+y+z=1$ Then prove that $xy(x+y)^2+yz(y+z)^2+zx(z+x)^2\geq 4xyz$
If $x,y,z$ are positive real number and $x+y+z=1\,$ Then prove that
$xy(x+y)^2+yz(y+z)^2+zx(z+x)^2\geq 4xyz$
Let $$f(x,y,z)=xy(x+y)^2+yz(y+z)^2+zx(z+x)^2$$
Then $$\frac{f(x,y,z)}{xyz} = \frac{(x+y)^2}{z}+\frac{(y+z)^2}{x}+\frac{(z+x)^... | Brilliant answer from @deepsea but it took me a while to understand it. For anyone else struggling here are some baby steps:
$$
\frac{f(x,y,z)}{xyz}=\frac{(x+y)^2}{z}+\frac{(y+z)^2}{x}+\frac{(z+x)^2}{y}
$$
$$
=\sum_{cyc}\frac{(x+y)^2}{z}=\sum_{cyc}\frac{(y+z)^2}{x}=\sum_{cyc}\frac{(1-x)^2}{x}\ \text{, because}\ 1-x=y+z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1657914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Value of the product: $ \sqrt{2} \sqrt{2 - \sqrt{2}} \sqrt{2 - \sqrt{2 - \sqrt{2}}} \sqrt{2 - \sqrt{2 - \sqrt{2-\sqrt{2}}}} \cdots $ =? Let the recursive sequence
$$ a_0 = 0, \qquad a_{n+1} = \sqrt{2-a_n},\,\,n\in\mathbb N.
$$
T
Can we find the value of the product
$$
\prod_{n=1}^{\infty}{a_n}?
$$
Well, from here I don... | Another approach is the following one: if we assume $ a_n = 2\cos(\theta_n) $ it follows that
$$ \cos(\theta_{n+1})=\sqrt{\frac{1-\cos\theta_n}{2}} = \sin\left(\frac{\theta_n}{2}\right) = \cos\left(\frac{\pi-\theta_n}{2}\right)\tag{1} $$
from which we have $\theta_{n+1}=\frac{\pi-\theta_n}{2}$ and, by induction:
$$ \th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1658073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 2,
"answer_id": 1
} |
$\int\frac{a^2\sin^2x+b^2\cos^2x}{a^4\sin^2x+b^4\cos^2x}dx$ $\int\frac{a^2\sin^2x+b^2\cos^2x}{a^4\sin^2x+b^4\cos^2x}dx$
My Try:
$\int\frac{a^2\sin^2x+b^2\cos^2x}{a^4\sin^2x+b^4\cos^2x}dx$
$\int\frac{a^2\tan^2x+b^2}{a^4\tan^2x+b^4}dx$
$\int\frac{1}{b^2}\frac{\frac{a^2}{b^2}\tan^2x+1}{\frac{a^4}{b^4}\tan^2x+1}dx$
I do n... | The integrand can be written as $\frac{1}{a^2+b^2}\left[1+\frac{2a^2b^2}{\left(a^4+b^4\right)+\left(b^4-a^4\right)\cos 2\theta}\right]$. It is easy from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1658195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Basic Homogenous Differential Equation Solve the following differential equation
$$\frac{dy}{dx} = \frac{2x^2y+x^3}{y^3+2x^3}$$
I have identified that the differential equation is homogeneous because $f(rx,ry)$ = $r^3f(x,y)$ . I then used the substitution $u = y/x$ and $\frac{dy}{dx} = x\frac{du}{dx} + u$ reducing the ... | On substituting $u=\frac{y}{x}$ in the differential equation $$\frac{dy}{dx} = \frac{2x^2y+x^3}{y^3+2x^3}$$
we get
$$x\frac{du}{dx} + u = \frac{2u+1}{u^3+2}$$
or $$x\frac{du}{dx} = \frac{2u+1-u^4-2u}{u^3+2}$$
or $$\frac{u^3+2}{1-u^4}du = \frac{dx}{x}$$
or $$-\frac{1}{4}\cdot \frac{-4u^3}{1-u^4}du + \frac{2}{(1-u^2)(1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1660529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
integrate $\int \frac{dx}{1+cos^2x}$
$$\int \frac{dx}{1+\cos^2x}$$
I used $\cos x=\frac{1-v^2}{1+v^2}$ and $dx=\frac{2dv}{1+v^2}$
and got $$2\int \frac{dv}{v^4-v^2+1}=2\int \frac{dv}{(v^2-\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}=\frac{4}{\sqrt{3}}arctan(\frac{2v^2-1}{\sqrt{3}})+c$$
How to continue?
| Note that $\cos^2(x)=\frac{1+\cos(2x)}{2}$. Then,
$$\int \frac{1}{1+\cos^2(x)}\,dx=\int \frac{2}{3+\cos(2x)}\,dx$$
Enforcing the substitution $x=u/2$ yields
$$ \int \frac{2}{3+\cos(2x)}\,dx=\int \frac{1}{3+\cos(u)}\,du \tag 1$$
Now, making the Weierstrass Substitution in $(1)$, as in the OP, we find
$$\int \frac{1}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1660682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
integrate $\int_0^{\frac{\pi}{4}}\frac{dx}{2+\tan x}$
$$\int^{\frac{\pi}{4}}_{0}\frac{dx}{2+\tan x}$$
$v=\tan(\frac{x}{2})$
$\tan x=\frac{2v}{1-v^2}$
$dx=\frac{2\,dv}{1+v^2}$
$$\int^{\frac{\pi}{4}}_0 \frac{dx}{2+\tan x}=\int^{\frac{\pi}{8}}_0 \frac{\frac{2\,dv}{1+v^2}}{2+\frac{2v}{1-v^2}}=\int^{\frac{\pi}{8}}_0 \frac... | I have done the sum u just look at it.
Here after step 2 I have used the method where u have to do like this
Af(x) +Bf'(x) =numerator. and here A and B are constant.
Here f(x)=denominator =2cosx +sinx
Hence f'(x) =-2sinx +cosx
So we get A = 2/5 B= 1/5
Thanks for asking such a nice question.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1662272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 5
} |
Finding sine of an angle in degrees without $\pi$ The following is found using a combination of: (a) a polygon with an infinite number of sides is a circle, (b) the perimeter of that polygon is the circumference of the circle that it becomes (of course), (c) the sine theorem, and (d) the ratio between the circumference... | As noted in a comment, there are equations you can write for the
trigonometric functions of rational portions of a right angle
(and therefore for the sine of any rational number of degrees)
without using $\pi$ or any trigonometric functions.
To actually use these equations may prove somewhat cumbersome, however.
Instea... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1663296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Evaluating the integral $\int \sqrt{1 + \frac{1}{x^2}} dx$
$$\int \sqrt{1 + \frac{1}{x^2}} dx$$
This is from the problem calculating the arc length of $y=\log{x}$.
I tried $x = \sinh{t}$ or $\frac{1}{x} = \tan{t}$ but all failed.
| If $x>0$,
$$\begin{align*}
I&=\int \sqrt{1 + \frac1{x^2}} \, dx \\
&= \int \frac{\sqrt{x^2+1}}{x} \, dx\\
&= -\frac12 \int \frac{(1+t^2)^2}{t^2(1-t^2)}\,dt \\
&= \int \frac{(1+u^2)^2}{u(1-u^2)^2}\,du
\end{align*}$$
by either substituting
$$t=\sqrt{1+x^2}-x \implies x=\frac{1-t^2}{2t} \implies dx = -\frac{1+t^2}{2t^2}\,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1663637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
How is Faulhaber's formula derived? I have been wanting to understand how to find the sum of this series.
$$1^p + 2^p + 3^p +{\dots} + n^p$$
I am familiar with Gauss' diagonalised adding trick for the sum of the first $n$ natural numbers.
I can prove the formulas for
$$\begin{align}
\sum_{1}^{n} k^2 &= \frac{n(n+1)(... | The following interesting derivation is from Aigner's "A Course in Enumeration" (Springer, 2007).
Remember that if we have the following exponential generating functions:
$\begin{align}
\widehat{A}(z)
&= \sum_{n \ge 0} a_n \frac{z^n}{n!} \\
\widehat{B}(z)
&= \sum_{n \ge 0} b_n \frac{z^n}{n!}
\end{align}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1667258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 0
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Show that $(n + 1)a^n < \frac{b^{n + 1} - a^{n + 1}}{(b-a)} < (n + 1)b^n$ $(b-a)(b^n + b^{n - 1}a + b^{n - 2}a^2 + \ldots + a^n)$
$= (b^{n + 1} - ab^n) + (ab^n - a^2b^{n - 1}) + (a^2b^{n - 1} - a^3b^{n - 2}) + \ldots + (a^nb - a^{n + 1})$
$= b^{n + 1} - a^{n + 1}$, so
$(b^n + b^{n - 1}a + b^{n - 2}a^2 + \ldots + a^n) =... | Hint
We have that
$$\dfrac{b^{n+1}-a^{n+1}}{b-a}=b^n+b^{n-1}a+b^{n-2}a^2+\cdots +a^2b^{n-2}+ba^{n-1}+a^n$$
Is not $a^n< b^ia^{n-i}< b^n, \forall i\in \{0,\dots, n\}?$ Are not there $n+1$ summands?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1667450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Solving recurrence relation? Consider the recurrence relation $a_1=8,a_n=6n^2+2n+a_{n−1}$. Let $a_{99}=K\times10^4$ .The value of $K$ is______ .
My attempt :
$a_n=6n^2+2n+a_{n−1}$
$=6n^2+2n+6(n−1)^2+2(n−1)+a_{n−2}$
$=6n^2+2n+6(n−1)^2+2(n−1)+6(n−2)^2+2(n−2)+......+a_1$
$=6n^2+2n+6(n−1)^2+2(n−1)+6(n−2)^2+2(n−2)+......+... | The idea is to find a closed form solution. We start with
$$
a_1 = 8 \\
6n^2+2n = a_n -a_{n−1}
$$
which is an inhomogenous linear recurrence relation and look for the previous
sequence elements:
$$
6(n-1)^2 + 2(n-1) = 6n^2 -12 n + 6 + 2n - 2 = 6 n^2 - 10n + 4
= a_{n-1} - a_{n-2}
$$
Substraction gives
$$
12n - 4 = a_n ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1668618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Distributing k objects among m people where each person gets i objects such that aIn an examination, the score in each of the four subjects ( say ) - A ,B,C & D can be between integers 0 and 10. Then , how many are there such that the student can secure a total of 21 ?
My attempt - My initial attempt was to find out t... | Let $a$, $b$, $c$, $d$ denote the student's scores in subjects $A$, $B$, $C$, and $D$, respectively. Then
$$a + b + c + d = 21 \tag{1}$$
is an equation in the non-negative integers subject to the constraints $a, b, c, d \leq 10$.
A particular solution of equation 1 corresponds to the insertion of three addition sig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1672852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
AB-BA so that you get another matrix I want to find out what the matrices are (a) and (b) so that $ab-ba = $
$$
\begin {pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\
\end {pmatrix} $$
This is the outcome of $ab-ba$ is. Ive been struggling with this for the past hour or so. All matric... | I like $$ A = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \\
B = \begin{pmatrix} 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{pmatrix} \text{.} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1675188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $d$ when $(\sin x)^7 = a \sin 7x + b \sin 5x + c \sin 3x + d \sin x$ There exist constants $a$, $b$, $c$, and $d$ such that
$(\sin x)^7 = a \sin 7x + b \sin 5x + c \sin 3x + d \sin x$
for all angles $x$. Find $d$.
| Interesting answers .Let me add another, just for filing a somewhat more "theoretical" one, using Chebyshev polynomials $U_n(x)$ of the second kind defined by
$$U_0(x)=1\\U_1(x)=2x\\U_{n+1}(x)=2x\space U_n(x)-U_{n-1}(x)$$
and for which one has the formula
$$U_n(\cos x)=\frac{\sin (n+1)x}{\sin x}\qquad (*)$$
We use the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1675940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Interval of convergence of $\sum_{n=1}^{\infty} \frac{k(k+1)(k+2)\cdot \cdot \cdot (k + n - 1)x^n}{n!}$
Given the series
$$ \sum_{n=1}^{\infty} \frac{k(k+1)(k+2)\cdot \cdot \cdot (k + n - 1)x^n}{n!} \quad \quad k \geq 1 $$
Find the interval of convergence.
I started by applying the Ratio test
$$
\lim_{n\to \infty}... | Note that for $k\ge1$, we have
$$
\frac{k(k+1)\cdots(k+n-1)}{n!}=\frac k1\frac{k+1}2\cdots\frac{k+n-1}{n}\ge1
$$
Thus, for $|x|=1$, the terms do not go to $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1676848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
integrate $\iint\limits_D \frac{2a}{\sqrt{4a^2-x^2-y^2}}dxdy, D: (x-a)^2+y^2 \le a^2$ $$I=\iint\limits_D \frac{2a}{\sqrt{4a^2-x^2-y^2}}dxdy, D: (x-a)^2+y^2 \le a^2(a>0)$$
The difficulty is to find a proper and easy way to solve this double integrals.
If do it like this, $ 0\le x \le 2a, -\sqrt{2ax-x^2} \le y \le \sqrt{... | Let $x=r\cos(\theta),y=r\sin(\theta)$. We know that $0\le x\le 2a,-a\le y\le a$, hence $0\le r\le 2a$ and $\displaystyle -\frac{\pi}{2}\le \theta\le \frac{\pi}{2}$. The Jacobian is $r$, thus \begin{align*}\iint\limits_D \frac{2a}{\sqrt{4a^2-x^2-y^2}}\text{d}x\text{d}y&=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1677123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Factorize $(x^2+y^2+z^2)(x+y+z)(x+y-z)(y+z-x)(z+x-y)-8x^2y^2z^2$ I am unable to factorize this over $\mathbb{Z}:$
$$(x^2+y^2+z^2)(x+y+z)(x+y-z)(y+z-x)(z+x-y)-8x^2y^2z^2$$
Since, this from an exercise of a book (E. J. Barbeau, polynomials) it must have a neat factorization.
I tried to guess some factors by putting $x-y... |
We write the expression as polynomial in $z^2$. This way we obtain
\begin{align*}
P(x,y,z)&=(x^2+y^2+z^2)(x+y+z)(x+y-z)(y+z-x)(z+x-y)-8x^2y^2z^2\\
&=-(x^2+y^2+z^2)\left((x+y)^2-z^2\right)\left((x-y)^2-z^2\right)-8x^2y^2z^2\\
&=-z^6+(x^2+y^2)z^4+(x^2-y^2)^2z^2-(x^2-y^2)^2(x^2+y^2)
\end{align*}
Analyzing the last... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1678933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
evaluate $\sum^{\infty}_{n=2} \frac{3}{10^n}$
evaluate $$\sum^{\infty}_{n=2} \frac{3}{10^n}$$
I know I can factor out $$\sum^{\infty}_{n=2} \frac{3}{10^n}=3\sum^{\infty}_{n=2} \frac{1}{10^n}$$
And I know that the sequence converges $${{\large \frac{1}{10^{n+1}}}\over{\large \frac{1}{10^n}}}=\frac{1}{10}<1$$
But how d... | You know that $\sum_{n\ge 1}q^n$ converges (you have $q=\frac{1}{10}$, similarily it works for all $q\in(-1,1)$), so let $$S=\sum_{n\ge 1}q^n=q+q^2+q^3+\dots$$ then $$qS=q^2+q^3+\dots = S-q$$ so$$S=\frac{q}{1-q}$$
Hence
$$\sum_{n=2}^{\infty}\frac{3}{10^n}=3\left(\sum_{n=1}^{\infty}\left(\frac{1}{10}\right)^n-\frac{1}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1680556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 3
} |
Maximum value of $\frac{3x^2+9x+17}{x^2+2x+9}$ If $x$ is real, the maximum value of $\frac{3x^2+9x+17}{x^2+2x+9}$ is?
Is it necessary that this function will attain maximum when the denominator will be minimum?
| Note that by division $\dfrac{3x^2+9x+17}{x^2+2x+9}=3+\dfrac{3x-10}{x^2+2x+9}$, so if the first expression has a maximum (or minimum) value it will occur when $\dfrac{3x-10}{x^2+2x+9}$ has its maximum (or minimum) value. The advantage of using $\dfrac{3x-10}{x^2+2x+9}$ is that the derivative is algebraically simpler.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1681057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
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.