Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Equation with different bases (exponential) I seem to be stuck with the following equation right here:
$$2^x + 2^{x+1} = 3^{x+2} + 3^{x+3}$$
| Note that $2^x+2^{x+1}=2^x+2\cdot 2^x=3\cdot 2^x$.
Similarly, $3^{x+2}+3^{x+3}=9\cdot 4\cdot 3^x$.
So our equation can be rewritten as $3\cdot 2^x=9\cdot 4\cdot 3^x$, or equivalently $(3/2)^x=1/12$, which can be solved using logarithms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1558897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
If $f(n+2)-5f(n+1)+6f(n)=0$ and $f(0)=2$ and $f(1)=5\;,$ Then $f(n)=$
$(1)\;\;$ If $f(n+1)-f(n) = 3$ and $f(0) = 1\;,$ Then $f(n) = \;,$ Where $n\in \mathbb{W}$
$(2)\;\;$ If $f(n+2)-5f(n+1)+6f(n)=0$ and $f(0)=2$ and $f(1)=5\;,$ Then $f(n)=$
Where $n\in \mathbb{W}$
$\bf{My\; Try\;, (1)::}$ Let $f(n)=a_{n}\;,$ Then Our... | Let the generating function of the sequence $a$ defined by
$$a_{n + 2} - 5a_{n + 1} + 6a_n= 0$$
be
$$g(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4 + \dots$$
Then
$$-5xg(x) = -5a_0x -5a_1x^2 - 5a_2x^3 - 5a_3x^4 - \dots$$
and
$$6x^2g(x) = 6a_0x^2 + 6a_1x^3 + 6a_2x^4 + 6a_3x^5 + \dots$$
Adding all three up,
$$\begin{alig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1560592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Derivative of $10^x\cdot\log_{10}(x)$
Derive $10^x\cdot\log_{10}(x)$
$$10^x\cdot \ln(10)\cdot \log_{10}(x)+\frac{1}{x\cdot \ln(10)}\cdot 10^x$$
But WolframAlpha gives another solution. Where am I wrong?
| Use the facts
$$(a^x)' = a^x \ln a, \quad (\log_a(x))' = \frac{1}{x\ln a}$$
for $a > 0$ and $a \neq 1$.
Then use the multiplication formula for differentiation to get
\begin{align}
& (10^x \log_{10}x)' \\
= & (10^x)'\log_{10}x + 10^x (\log_{10}x)' \\
= & 10^x \ln (10) \log_{10}x + 10^x \frac{1}{x\ln 10}.
\end{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1562999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 2
} |
Convergence of $\frac{1}{5} + \frac{1}{9} + \frac{1}{13} + ... = \sum_{i=1}^{\infty} \frac{1}{1+4i}.$ I've been working on an approximation for a problem in my numerical methods course, and I seemed to have run into the series
$$\frac{1}{5} + \frac{1}{9} + \frac{1}{13} + ... = \sum_{i=1}^{\infty} \frac{1}{1+4i}.$$
I'm ... | An elementary argument, though not the most simple or quick: let $a_n = (1+4n)^{-1}$. Then let $b_n = (4+4n)^{-1} = 4^{-1} (1+n)^{-1}$, so that we clearly have for all positive integer $n$, $$a_n > b_n.$$ Now note that $$S_n = \sum_{k=1}^{n-1} b_k = \frac{1}{4} \sum_{k=1}^{n-1} \frac{1}{k+1} = \frac{1}{4} \left( -1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1566947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Polynomial $P(x)$ contains only terms of odd degree.When $P(x)$ is divided by $(x-3)$ ,the remainder is $6$. Polynomial $P(x)$ contains only terms of odd degree.When $P(x)$ is divided by $(x-3)$ ,the remainder is $6$.If $P(x)$ is divided by $(x^2-9)$ then remainder is $g(x).$Find the value of $g(2).$
As $P(x)$ is a po... | Since it is an odd degree polynomial
P(–a)= –p(a)
Let P(x)=(x^–9)Q(x)+ax+b
(where Q is quotient and (ax+b) = g(x) = remainder)
Given P(3)=6
Therefore P(-3) = -6
Solve these two equations to get value of a and b
We get a=2 and b=0
G(x) =ax+b
G(2)= 2(2)+0 =4
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1567477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
On proving $\lim \frac{n^2}{n^2+n+1} = 1$
Prove that $\lim \frac{n^2}{n^2+n+1} = 1$
Let $\varepsilon > 0$ and let $N = \frac{1}{\varepsilon}.$ Then $n > N$ implies $n > \frac{1}{\varepsilon} \implies \frac{1}{n} < \varepsilon.$
But $\displaystyle \frac{1}{n} = \frac{n+1}{n(n+1)} = \frac{n+1}{n^2+n} > \frac{n+1}{n^2+n... | Your stream of thought is right; just notice that $1/\varepsilon$ is not necessarily an integer, which matters if we are talking about a sequence, which is a map defined on a subset of the set of all integers. Instead, you may write $N := \lceil 1/\varepsilon \rceil + 1$, by which you can ensure the choice of $N$ to be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1568875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find $x,y,z>0$ such that $x+y+z=1$ and $x^2+y^2+z^2$ is minimal How can I find $3$ positive numbers that have a sum of $1$ and the sum of their squares is minimum?
So far I have:
$$x+y+z=1 \qquad \implies \qquad z=1-(x+y)$$
So,
$$f(x,y)=xyz=xy(1-x-y)$$
But I'm stuck from here. Hints?
| Lagrange multipliers? So $f(x,y,z,\lambda) = x^2 + y^2 + z^2 - \lambda(x+y+z-1)$.
So critical points obey: $2x - \lambda = 0, 2y - \lambda = 0, 2z - \lambda = 0, x+y+z-1 = 0$. So $x = y = z = \frac{1}{3}$, which gives sum of squares $\frac{1}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1572128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 10,
"answer_id": 6
} |
Series expansion of integral Consider the function $I(y)=\int_0^\infty e^{-\sqrt{x^2+y^2}} \mathrm{d} x$. I'd like to see the leading order term of $I(y)$ about $y=0$, so I expand the integrand:
$$
e^{-\sqrt{x^2+y^2}}=e^{-x}-e^{-x}\frac{1}{2x}y^2+e^{-x}\frac{1+x}{8x^3}y^4+\dots
$$
However, only the integral of the $0^\... | Let $I(y)$ be the integral of interest given by
$$\begin{align}
I(y)&=\int_0^\infty e^{-\sqrt{x^2+y^2}}\,dx\\\\
&=\int_0^{|y|} e^{-\sqrt{x^2+y^2}}\,dx+\int_{|y|}^\infty e^{-\sqrt{x^2+y^2}}\,dx \tag 1
\end{align}$$
For the first integral on the right-hand side of $(1)$, we note that $\sqrt{x^2+y^2}\le \sqrt{2}|y|$. Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1576249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Divisibility of $2^n-n^2$ by 7 How many positive integers $n<10^4$ are there such that $2^n - n^2$ is divisible by 7?
| $10^4$ is small enough that it is possible to write a computer program to find the answer, but it is also possible to solve explicitly as follows:
We note that the possible remainders of $n^2$ modulo $7$ are $0, 1, 2$ and $4$. Since $2^n$ is never divisible by $7$, the possibilities that we have for which $2^n-n^2$ is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1576496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $x$ and $y$ are divisible by $3$ when $x^2+6u^2=2y^2+3v^2$
a) Solve $x^2\equiv 2y^2\pmod{3}$
b) Use part a to prove that $x$ and $y$ are divisible by $3$ when $x^2+6u^2=2y^2+3v^2$
My attempt:
$x^2\equiv 2y^2\pmod{3}\iff $$(y^{-1}x)^2 \equiv 2\pmod{3}$ but $\left(\frac{2}{3}\right)=-1$ since $3 \equiv 3\pm... | Once you have $x \equiv y \equiv 0 \pmod 3, $ you get a little extra, since $x^2 \equiv y^2 \equiv 0 \pmod 9. $ This says something about the other two letters, $ 6 u^2 \equiv 3 v^2 \pmod 9. $ Then $ 2 u^2 \equiv v^2 \pmod 3. $ Finally, this part of the conclusion is $u \equiv v \equiv 0 \pmod 3. $
Stated carefully... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1578537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
how to prove that a solution exists for a fourth order ode Does there exists a solution of the initial value problem
$$(x^2 − 4)\frac {d^4y}{dx^4} + 2x \frac {d^2y}{dx^2} + (\sin x)y(x) = 0$$
where $y(0) = 0, y'(0) = 1, y''(0) = 1, y'''(0) = −1$
So far I have $y_1=y$, $y_2= \frac {dy}{dx}$ and so on til $y_n = \frac {d... | I do not guarantee that no mistakes have been made.
Let
\begin{equation*}
y_{n}=\partial _{x}^{n-1}y
\end{equation*}
Then
\begin{eqnarray*}
\partial _{x}y_{1} &=&y_{2} \\
\partial _{x}y_{2} &=&y_{3} \\
\partial _{x}y_{3} &=&y_{4} \\
\partial _{x}y_{4} &=&y_{5}
\end{eqnarray*}
and
\begin{eqnarray*}
(x^{2}-4)\partial _{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1579598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Compute $\lim_{n\to\infty}\int_0^1\frac{x\sin{nx}}{1+x^2n^6}dx$ $\lim_{n\to\infty}\int_0^1\frac{x\sin{nx}}{1+x^2n^6}dx$
My Work:
By the comparison principle: $$\lim_{n\to\infty}\int_0^1\frac{x\sin{nx}}{1+x^2n^6}dx\le\lim_{n\to\infty}\int_0^1\frac x{1+x^2n^6}dx$$ I think that the way to compute this integral has to do ... | Hint: For $n\ge1$, $$\left|\frac{x\sin(nx)}{1+x^2n^6}\right|\le x$$ so Dominated Convergence should apply if the domain of integration is $[0,1]$.
If the Domain of Integration is $\boldsymbol{[0,\infty)}$
Integration by Parts gives
$$
\begin{align}
\int_0^\infty\frac{x\sin(nx)}{1+x^2n^6}\,\mathrm{d}x
&=-\frac1n\int_0^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1580090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
distinct roots of the equation $A\sin^3 x+B\cos^3 x+C =0$
The number of distinct roots of the equation $A\sin^3 x+B\cos^3 x+C =0$
no two of which differ by $2\pi$ is, Where $A,B,C\in \mathbb{R}$
$\bf{(a)}\;\;\;\;\;\; 3\;\;\;\;\;\; (b)\;\;\;\;\;\; 4\;\;\;\;\;\; (c)\;\;\;\;\;\; 6\;\;\;\;\;\; (d)\;\;\;\;\;\; infinite$
$... | Here, I found a restriction for the constants $A$, $B$, $C$ that wanted to share. We know by the well-known inequalities $|\sin x| \le 1$ and $|\cos x| \le 1$ that
$$-(|A|+|B|) \le A\sin^3 x+B\cos^3 x \le (|A|+|B|)$$
So, if we wish the equation to have a root we must have
$$-(|A|+|B|) \le -C \le (|A|+|B|)$$
Or equivale... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1580975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
The chart-problem; problem solving In how many ways can we construct a $6\times 6$ chart with only $1$ and $-1$ such that in every row and column, the product is always positive?
| There are $\color{red}{2^{25}}$ ways of filling a $6\times 6$ grid with $0$ or $1$ in such a way that along every row or column there is an even number of $1$s. We may first fill a $5\times 5$ subgrid in any way we like, then complete those rows and columns in the only way they fit our constraints. The only square we s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1582132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
I would like to calculate $\lim_ {n \to \infty} {\frac{n+\lfloor \sqrt{n} \rfloor^2}{n-\lfloor \sqrt{n} \rfloor}}$ I would like to calculate the following limit: $$\lim_ {n \to \infty} {\frac{n+\lfloor \sqrt{n} \rfloor^2}{n-\lfloor \sqrt{n} \rfloor}}$$
where $\lfloor x \rfloor$ is floor of $x$ and $x ∈ R$.
Now I know t... | You may observe that, as $n \to \infty$,
$$
\begin{align}
{\frac{n+\lfloor \sqrt{n} \rfloor^2}{n-\lfloor \sqrt{n} \rfloor}}&={\frac{2n+(\lfloor \sqrt{n} \rfloor-\sqrt{n})(\lfloor \sqrt{n} \rfloor+\sqrt{n})}{n-\lfloor \sqrt{n} \rfloor}}\\\\
&={\frac{2+(\lfloor \sqrt{n} \rfloor-\sqrt{n})(\lfloor \sqrt{n} \rfloor+\sqrt{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1582388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\lim \limits_{n\rightarrow\infty}\sin^2(\pi\sqrt{n^2 + n})$ I'm struggling to find the limit at infinity of : $$\lim \limits_{n\rightarrow\infty}\sin^2(\pi\sqrt{n^2 + n}), n\in\Bbb N$$
I know it is $1$ but I don't understand why this is wrong :
$\sin^2(\pi\sqrt{n^2 + n}) = \sin^2(\pi*n\sqrt{1 + \frac{1}{n}})$... | First of all, notice that the function $f (\theta) = \sin^2 \theta$ is periodic with a period $\pi$. Then, for any integer $n$ we should have $\sin^2 (\pi \sqrt {n^2 + n}) = \sin^2 \left (\pi \left(\sqrt {n^2 + n} - n\right)\right)$. Since $$\lim_{n \to \infty} \left(\sqrt {n^2 + n} - n\right) = \frac {1} {2},$$ it bec... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1586034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Proving $\sqrt{2a + 2\sqrt{a^2 - b}}=\sqrt{a-\sqrt{b}} + \sqrt{a+\sqrt{b}}$, where $a\geq 0,\ b\geq 0$. Who can help me to prove this equation?
$$\sqrt{2a + 2\sqrt{a^2 - b}}=\sqrt{a-\sqrt{b}} + \sqrt{a+\sqrt{b}}$$ Where, $$a\geq 0,\ b\geq 0$$
| Notice, here is the correct proof $$LHS=\sqrt{2a+2\sqrt{a^2-b}}$$
$$=\sqrt{(a-\sqrt{b})+(a+\sqrt b)+2\sqrt{a^2-(\sqrt b)^2}}$$
$$=\sqrt{\underbrace{\left(\sqrt{a-\sqrt{b}}\right)^2}_{A^2}+\underbrace{\left(\sqrt{a+\sqrt b}\right)^2}_{B^2}+2\underbrace{\sqrt{(a-\sqrt b)}}_{A}\underbrace{\sqrt{(a+\sqrt b)}}_{B}}$$
using ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1586309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Inverse Laplace problem using partition fraction Hello I am solving inverse Laplace transform using partial fraction. The question is:
$$ X(s) = \frac{10(s+1)}{s(s^2+4s+8)} => \frac{10(s+1)}{s((s+2)^2+4)} $$
$$ \frac {C1} {s} + \frac{C2}{((s+2)+4)} + \frac{C3}{((s+2)^2+4)} $$
for finding C1
$$ \frac{10(s+1)}{s((s+2)^2... | Notice, for factorization of denominator you should find the roots of quadratic equation: $s^2+4s+8=0$ as $$s=\frac{-4\pm\sqrt{4^2-4(1)(8)}}{2(1)}=-2\pm 2i$$
$$\implies s^2+4s+8=(s+2+2i)(s+2-2i)$$
Now, one should have the following partial fractions $$X(s)=\frac{10(s+1)}{s(s^2+4s+8)}=\frac{10(s+1)}{s(s+2+2i)(s+2-2i)}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1586387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to integrate $\int\limits_0^1 \left(-1\right)^{^{\left\lfloor\frac{1}{x}\right\rfloor}} dx$? As my title says, I need help integrating with floor functions,
$$\int\limits_0^1 \left(-1\right)^{^{\left\lfloor\frac{1}{x}\right\rfloor}} dx$$
What does this even mean exactly? How would approach this?
| Essentially, you have this:
$$\int_0^1 \left(-1\right)^{^{\left\lfloor\frac{1}{x}\right\rfloor}} dx=\int_\frac{1}{2}^1\left(-1\right)^1dx+\int_\frac{1}{3}^\frac{1}{2}\left(-1\right)^2dx+\int_\frac{1}{4}^\frac{1}{3}\left(-1\right)^3dx+\int_\frac{1}{5}^\frac{1}{4}\left(-1\right)^4dx+\ldots$$
$$=\left(-1+\frac{1}{2}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1587037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
} |
Find one basis of kernel, one basis of image of the linear transformation and it's defect Find one basis of kernel, one basis of image of the linear transformation
$T:\mathcal{M}_{2\times 2}(\mathbb{R})\rightarrow \mathcal{M}_{2\times 3}(\mathbb{R})$ and defect of $T$ where $T(X)=X \begin{bmatrix}
0 & 0... | The space of $2\times 2$ matrices has as basis the elementary matrices $e_{11},e_{12},e_{21},e_{22}$ with a $1$ in the $ij$ position and $0$ elsewhere, so you can calculate the matrix of $T$ relative to this basis and with the corresponding basis of the space of $2\times 3$ matrices. Note that $e_{ij}$ has trace $0$ if... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1587959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Inequality exercise (olympiad) For positive $a$, $b$, $c$ such that $abc=1$. Show that $$(ab+bc+ca)(a+b+c)+6\geq 5(a+b+c).$$
From the LHS, using AM-GM, we see that $(ab+bc+ca)(a+b+c)+6\geq 3(abc)^{2/3}3(abc)^{1/3}+6=15$. But clearly, 15 can be less than $5(a+b+c)$. So this technique won't work. I fail to see some other... | Lemma: for any postive $x,y,z>0$,and $xyz=1$ we have
$$(xy+yz+xz)^2+3\ge 4xyz(x+y+z)$$
proof:since Use Schur inequality we have $$a^3+b^3+c^3+3abc\ge ab(a+b)+bc(b+c)+ac(a+c)\ge 2((ab)^{3/2}+(bc)^{3/2}+(ac)^{3/2})$$
take $a'^2=a^3,b'^2=b^3,c'^2=c^3$ and Assmue that $a'b'c'=1$so we have
$$a'^2+b'^2+c'^2+3\ge 2(a'b'+b'c'... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
An infinite nested radical problem From this link, problem 36, I found that
$$\sqrt{4+\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4+\sqrt{4-...}}}}}}=2\left(\cos{\dfrac{4\pi}{19}}+\cos{\dfrac{6\pi}{19}}+\cos{\dfrac{10\pi}{19}}\right).$$
The signs : + + - + + - + + - ... .
How to prove it?
Furthermore, how to represent $\sqrt{7+2\sqr... | I. Solution
You wish to find the exact value of $x,y$ for,
$$x=\sqrt{7+2\sqrt{7-2\sqrt{7-2\sqrt{7+2\sqrt{7-2\sqrt{7-...}}}}}}\tag1$$
$$y=\sqrt{4+\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4+\sqrt{4-...}}}}}}\tag2$$
and specify that both have period $3$. So all you do is substitute at the correct point,
$$x=\sqrt{7+2\sqrt{7-2\sqrt{7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Prove that $\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ac}{1+c} \geq 3.$
Let $a,b,c$ be positive real numbers such that $abc = 1$. Prove that $$\dfrac{1+ab}{1+a}+\dfrac{1+bc}{1+b}+\dfrac{1+ac}{1+c} \geq 3.$$
This looks symmetric, so should I prove it for just $a \leq b \leq c$ and then the other cases follow?
| Since $1=abc$, we have
\begin{align}
\dfrac{1+ab}{1+a} + \dfrac{1+bc}{1+b} + \dfrac{1+ca}{1+c} & = \dfrac{abc+ab}{1+a} + \dfrac{abc+bc}{1+b} + \dfrac{abc+ca}{1+c}\\
& = ab \dfrac{1+c}{1+a} + bc \dfrac{1+a}{1+b} + ca \dfrac{1+b}{1+c}\\
& \geq 3\sqrt[3]{ ab \dfrac{\color{red}{1+c}}{\color{blue}{1+a}} \cdot bc \dfrac{\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1589172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Let $k = 2008^2 + 2^{2008}$. What is the last digit of $k^2 + 2^k$? Let $k = 2008^2 + 2^{2008}$. What is the last digit of $k^2 + 2^k.$
I thought of this
$$2008^2+2^{2008}\pmod{10} ≡ {-2}^2+{2^4}^{502}\pmod{10} ≡ 4+{-4}^{502}\pmod{10} ≡ 4+6^{251} \pmod{10}$$
but I still cannot prove it. Maybe there is a clever solution... | $$2008^2=2000^2+2*2000*8+8^2=>4 (mod 10)$$
$2^m=6 (mod 10) $ (m = $2008$)
$$4+6=0 (mod 10)$$
$$ k=0 (mod 10)$$
$$ k^2=0 (mod 10)$$
So, it is obvious that last digit of $2^k$ is $2^4=6 (mod 10)$ and last digit of $k^2$ is $0$.
So the answer is $6+0=6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1589634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Inequality olympiad For all positive numbers $a,b,c$, prove that
$$\frac{a^3}{b^2-bc+c^2}+\frac{b^3}{a^2-ac+c^2}+\frac{c^3}{a^2-ab+b^2}\geq 3 \frac{(ab+bc+ac)}{a+b+c}$$
Note that both side are homogeneous of degree 1, so I think it is safe to assume $a+b+c=1$ but this does not go very far.
Any ideas/hint?
Thanks
| By Cauchy-Schwarz inequality:
$$\sum_{\text{cyc}}\dfrac{a^3}{b^2-bc+c^2}\left(\sum_{\text{cyc}}a\left(b^2-bc+c^2\right)\right)\ge \left(a^2+b^2+c^2\right)^2$$
In fact, you can prove the following stronger inequality:
$$\frac{\left(a^2+b^2+c^2\right)^2}{\sum_{\text{cyc}}a\left(b^2-bc+c^2\right)}\ge a+b+c\ge3\dfrac{ab+bc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
The number of real roots of the equation $1+a_1x+a_2x^2+.....+a_nx^n=0$ where $|x|<\frac{1}{3}$ and $|a_i|<2 \forall i=1,2,3....,n$ is The number of real roots of the equation $1+a_1x+a_2x^2+.....+a_nx^n=0$ where $|x|<\frac{1}{3}$ and $|a_i|<2 \forall i=1,2,3....,n$ is
$(a)n$ if $n$ is even
$(b)1$ if $n$ is odd.
$(c)... | Clarifying something the original poster said: When the degree of the polynomial is $n$, there are $n$ complex roots (counting multiplicity). Hence there are at most $n$ real roots.
Motivation: Noting how the range of $x$ and $a_i$ are restricted, we want to show that all the terms of $a_ix^i$ combined cannot "overpowe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\frac{1}{\sin 8^\circ}+\frac{1}{\sin 16^\circ}+....+\frac{1}{\sin 4096^\circ}+\frac{1}{\sin 8192^\circ}=\frac{1}{\sin \alpha}$,find $\alpha$ Let $\frac{1}{\sin 8^\circ}+\frac{1}{\sin 16^\circ}+\frac{1}{\sin 32^\circ}+....+\frac{1}{\sin 4096^\circ}+\frac{1}{\sin 8192^\circ}=\frac{1}{\sin \alpha}$ where $\alpha\in(0,90^... | HINT:
For $\sin A\ne0\iff A\ne m\pi$ where $m$ is any integer,
$$\cot A-\cot2A=\dfrac{\sin(2A-A)}{\sin2A\sin A}=\csc2A$$
Do you recognize the Telescoping Series?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Solve the functional equation $f(xy)=e^{xy-x-y} \big( e^yf(x)+e^xf(y) \big)$ If $f: \Bbb R ^+ \to \Bbb R$ satisfies $$f(xy)=e^{xy-x-y} \big( e^yf(x)+e^xf(y) \big)$$ for all $x, y \ge 0$ and if $f'(1)=e$, determine $f(x)$.
I'm a beginner. Can someone give me some hints for this sum?
I'm getting $f(1)=0$. Is that right?... | \begin{align*}
f(xy)&=e^{xy-x-y} \left( e^yf(x)+e^xf(y) \right) \\
f(1) &= e^0 f(1) + e^0 f(1) \\
f(1) &= 0
\end{align*}
If $x>0$, set $y = \frac{x+h}{x}$, then $xy=x+h$.
\begin{align*}
f(xy)&= f(x+h) \\
&= e^{x\frac{x+h}{x}-x-\frac{x+h}{x}} \left( e^\frac{x+h}{x} f(x)+e^xf(\frac{x+h}{x}) \right) \\
&= e^{h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Which quantity is greater, $\frac{x^2}{y+\frac1y}$ or $\frac{y^2}{x+\frac1x}$? $x \gt y$, $ xy \neq 0$
A= $ x^2\over {y+{1\over y}}$
B= $ y^2\over {x+{1\over x}}$
Options:
1) Quantity A is greater.
2) Quantity B is greater.
3) The two quantities are equal.
4) The relationship cannot be determined from the given informa... | Suppose I compute the difference $A-B$ and see if it is always of one sign.
$$\begin{align}A - B &=\frac{x^2}{y+1/y} - \frac{y^2}{x+1/x}\\
&=\frac{yx^2}{y^2+1} - \frac{xy^2}{x^2+1}\\
&=\frac{xy\left(x(x^2+1)-y(y^2+1)\right)}{(x^2+1)(y^2+1)}\\
&=\frac{xy(x-y)(x^2+xy+y^2+1)}{(x^2+1)(y^2+1)}\\
&=\frac{xy(x-y)\left((x+y/2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"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?
| Induction on k.
Base Case: 1=1
Induction Hypothesis: $(1+2+{...}+k)^2$=$1^3 + {...} + k^3$
Based on the hypothesis:
$(1+2+{...}+k+(k+1))^2$
=$(1+2+{...}+k)^2$+2$(1+2+{...}+k)(k+1)$+$(k+1)^2$
=$1^3 + {...} + k^3$+$k(k+1)^2$+$(k+1)^2$
=$1^3 + {...} + k^3 + (k+1)^3$
Done.
| {
"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": 0
} |
Trigonometry with Quadratic Equations If $\tan A$ and $\tan B$ are the roots of $x^2+px+q=0$, then prove that
$$\sin^2(A+B)+p \sin(A+B) \cos(A+B) + q \cos^2(A+B) = q$$
I tried the question but with $q$ other terms came associated.
| Using the sum and product formulae we have,
$q=tanAtanB, $ $-p=tanA+tanB$
And,
$tan(A+B)=\frac{tanA+tanB}{1-tanAtanB} \Rightarrow tan(A+B)=\frac{-p}{1-q}$
Now,
$\frac{p^{2}-p^{2}}{1-q}=0 \Rightarrow (\frac{p}{1-q})^{2}(1-q)+p(\frac{-p}{1-q})=0 \Rightarrow [tan(A+B)]^{2}(1-q)+p[tan(A+B)]=0 \Rightarrow [\frac{sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
How do I find the roots of this polynomial of degree $4$? I am studying for finals and in the review packet is shown this problem:
$$P(x)=2x^4 + 5x^3 + 5x^2 + 20x - 12$$
I don't know what to do, I have already tried looking in the textbook and Khan academy.
| Simple factorisation solves your problem.
$$P(x)=2x^4+5x^3+5x^2+20x-12$$ $$=2x^4+5x^3-3x^2+8x^2+20x-12$$
$$=x^2(2x^2+5x-3)+4(2x^2+5x-3)$$ $$=(x^2+4)(2x^2+5x-3)$$ $$=(x^2+4)(2x-1)(x+3)$$
Hence the $4$ roots are $x=\frac{1}{2}$,$x=-3$,$x=2i$ and $x=-2i$.
Hope this helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Is there a formula for computing the number of arrangements from a set of N elements groupped in K groups? Eg: N=3, K=2
There will be two groups in each solution. We need to calculate the number of such possible solutions. Consider the set S={1,2,3}. The possible solutions are:
{1} {2,3}
{1} {3,2}
{2} {1,3}
{2} {3,1}
{... | We assume that the example for $N=3$, $K=2$ is correct, meaning that the internal order of the elements in individual groups matters.
There are $N!$ ways of lining up our $N$ objects. That produces $N-1$ interobject "gaps." We choose $K-1$ of them to put a separator into.
That can be done in $\binom{N-1}{K-1}$ ways, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Simultaneous Equations (Stuck on the algebra)
Question: Solve the following simultaneous equations for real values of x and y
$$
\left\{
\begin{array}{l}
9^{2x+y} - 9^x \times 3^y = 6 \\
\log_{x+1}(y+3) + \log_{x+1}(y+x+4) = 3
\end{array}
\right.
$$
What I have attempted; for the first equation
$$
9^{2... | Writing the last equation
$$ \log_{x+1}(4-2x) + \log_{x+1}(5-x) = 3 $$
as
$$ \log_{x+1}[(4-2x)(5-x)] = 3 $$
you can get both term as exponents of $x+1$, so that
$$ (4-2x)(5-x) = (x+1)^3 $$
This equation can be easily solved, leading to $x=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
The value of series $\sum_{n=1}^{\infty}\frac{n}{2^n}$ The value of series
$\sum_{n=1}^{\infty}\frac{n}{2^n}$
I try to write some terms,but of no use. Is there any general method to approach such questions.
Thanks
| Notice that
\begin{align*}
\sum_{n=1}^\infty \frac{n}{2^n}
&= \frac{1}{2} \sum_{n=1}^\infty \frac{n}{2^{n-1}}
= \frac{1}{2} \sum_{n=0}^\infty \frac{n+1}{2^n}
= \frac{1}{2} \left( \sum_{n=0}^\infty \frac{n}{2^n}
+ \sum_{n=0}^\infty \frac{1}{2^n} \right) \\
&= \frac{1}{2} \sum_{n=1}^\infty \frac{n}{2^n} + 1,
\end{ali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
How do I show $f(x+2)-f(x)>2 \forall x$? For the function
$f(x)=x\cos{\frac{1}{x}}$, $x\geq1$,
How do I show that $f(x+2)-f(x)>2 \forall x$?
| $\cos{\frac{1}{x}}=1-\frac{1}{x^22!}+\frac{1}{x^44!}-....$
$x\cos{\frac{1}{x}}=x-\frac{1}{x2!}+\frac{1}{x^34!}-....$
So $(x+2)\cos{\frac{1}{x+2}}=(x+2)-\frac{1}{(x+2)2!}+\frac{1}{(x+2)^34!}-....$
So $f(x+2)-f(x)=2+\{\frac{1}{2x}-\frac{1}{2(x+2)}\}+...$
$=2+\frac{1}{x(x+2)}+...>2$
NOTE: As the number of terms in the sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Evaluate the integral $\int_{-1}^1 \frac{1}{x^2-2x\cos\alpha+1}\mathrm dx,\alpha\in(0,\pi)$ For the equation $x^2-2x\cos\alpha+1=0$ solutions are
$$x_1=\cos\alpha-\sqrt{\cos^2\alpha-1},x_2=\cos\alpha+\sqrt{\cos^2\alpha-1}\Rightarrow$$
$$\int_{-1}^1 \frac{1}{x^2-2x\cos\alpha+1}\mathrm dx=\int_{-1}^1 \frac{1}{(x-\cos\alp... | If we let $t = \cos \alpha$, we have $$\frac{1}{x^2 - 2t x + 1} = \frac{1}{(x-t)^2 + (1-t^2)} = \frac{1}{1-t^2} \cdot \frac{1}{\frac{(x-t)^2}{1-t^2} + 1}.$$ This suggests using the substitution $$u = \frac{x-t}{\sqrt{1-t^2}}, \quad du = \frac{1}{\sqrt{1-t^2}} \, dx.$$ This results in the indefinite integral $$\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Help finding the determinant of a 4x4 matrix? Sorry for the lack of notation but the work should be easy to follow if you know what you are doing. Okay my problem is that the book says it can be done by expanding across any column or row but the only way to get what the book does in their practice example is to choose ... | Steve explained where you made a mistake in your calculations. And Patrick explained how you can save computations by judiciously choosing the rows/ columns you expand along. Just for fun, I'll explain a different way of evaluating the determinant. I'm just going to use the relationship between the elementary row/ co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 3
} |
Integral $\int\left(\frac{1}{x^4+x^2+1}\right)dx$ Someone can halp me to solve this integral:
$$\int\left(\frac{1}{x^4+x^2+1}\right)$$
solution$$\frac{1}{4}\ln\left(\frac{x^2+x+1}{x^2-x+1}\right)+\frac{1}{2\sqrt3}\arctan\frac {x^2-1}{x\sqrt3}$$
I don't manage using partial fraction because $${x^4+x^2+1}$$ has $\Delta\... | Hint :
$$x^4+x^2+1=(x^2+1)^2−x^2=(x^2−x+1)(x^2+x+1)$$
So $$∫ \frac {1}{x^4+x^2+1}dx=\int\frac {1}{(x^2−x+1)(x^2+x+1)}dx$$
$$=\int\frac {1-x}{2(x^2−x+1)}+\frac {1+x}{2(x^2+x+1)}dx$$
$$=-\int\frac {x-1}{2(x^2−x+1)}dx+\int\frac {x+1}{2(x^2+x+1)}dx$$
$$=-\int\frac {2x-2}{4(x^2−x+1)}dx+\int\frac {2x+2}{4(x^2+x+1)}dx$$
$$=-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Which of the following matrices has the same determinant as matrix $G$ Which of the following matrices has the same determinant as matrix:
$G$ = $ \begin{bmatrix}
1 & 2 & 3 \\
-2 & -3 & 2 \\
1 & 2 & 1
\end{bmatrix}$
I could just calculate each individual determinant; however I am prep... | Note that the matrix $B$ is obtained from $G$ by adding $\DeclareMathOperator{Row}{Row}\Row_1(G)$ to $\Row_3(G)$. Elementary row operations of this form leave the determinant unchanged. Thus $\det B=\det G$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding product of odd/even integers This is a question that came to mind and I was trying multiple ways from using combinatorial methods, graphical approaches, etc. The question I have is, how can I find the product of all the odd or even numbers less than some positive integer $n$.
For example:
If $n=6$, then the pr... | Case for Odd Product:
Since you want the product less than $n$, you are looking for the product of,
$$ 1\cdot 3\cdot 5\cdot 7\cdot 9\cdot \ldots \cdot (n-1) $$
Now, notice that you don't want the even numbers, so just divide by them,
$$\frac {1\cdot 2\cdot 3\cdot 4\cdot 5\cdot \ldots \cdot n} {2\cdot 4\cdot 6\cdot 8\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
The greatest integer less than or equal to the number $R=(8+3\sqrt{7})^{20}$
Given $$R=(8+3\sqrt{7})^{20}, $$ if $\lfloor R \rfloor$ is Greatest integer less than or equal to $R$, then which of the following option(s) is/are true?
*
*$\lfloor R \rfloor$ is an even number
*$\lfloor R \rfloor$ is an odd number
*$R-\... | Building on Anomaly's answer, let $a_n = (8 + 3\sqrt{7})^n + (8 - 3 \sqrt{7})^n$. Prove that $a_n$ can be defined by induction by $a_0 = 2$, $a_1 = 16$, $a_n = 16a_{n-1} - a_{n-2}$. This will help in obtaining information about $a_{20}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Proving $x = y$ or $x = -y$ when $x^n = y^n$ and $n$ is even I'm currently going through Spivak's calculus, and after a lot of effort, i still can't seem to be able to figure this one out.
The problem states that you need to prove that $x = y$ or $x = -y$ if $x^n = y^n$
I tried to use the formula derived earlier for $x... | Let $n=2p$. For convenience let us denote $y=a$. From the algebraic identities
\begin{eqnarray}
x^{2p}-a^{2p} &=&(x-a)\sum_{k=0}^{2p-1}a^{k}x^{2p-1-k}, \tag{1} \\
\sum_{k=0}^{2p-1}a^{k}x^{2p-1-k} &=&(x+a)\sum_{k=0}^{p-1}a^{2k}x^{2p-2-2k},\tag{2}
\end{eqnarray}
we conclude that
\begin{equation}
x^{2p}-a^{2p}=(x-a)(x+a)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Let $x,y \ge 0 $.Find the maximum value of $\cfrac{(3x+4y)^2}{x^2+y^2}$ Let $x,y \ge 0 $.Find the maximum value of $\cfrac{(3x+4y)^2}{x^2+y^2}$
This exercise is from the chapter of my book regarding the Cauchy-Schwarz inequality but I don't know if I've applied it correctly .
The problem asks for the maximum value of t... | I think ,you can consider as taking $x=\frac{4y}{3}$;
$\left(\left(\sqrt{R}\right)^2 +(\frac{4}{3})^2\right) (x^2+y^2) \ge (3x+4y)^2$
which R=25 for maximum value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
prove that $x_{n+1}=\frac{x_n(x_n^2+15)}{3x_n^2+5}$ is cubic order of convergence near $x_0=\sqrt{5}$ To solve the equation
$$x^2-5=0$$
There exitsts a iteration method
$$x_{n+1}=\frac{x_n(x_n^2+15)}{3x_n^2+5}$$
I know that it is cubic convergence but I don't know how to prove it.
I have tried the following
$$
\begi... | In general, Halley's method for the square root reads as
$$
x_{n+1}=x_n·\frac{x_n^2+3a}{3x_n^2+a}
$$
The difference to the square root develops then as
$$
x_{n+1}-\sqrt{a}=\frac{x_n^3+3x_n\sqrt a^2 - 3x_n^2\sqrt a-\sqrt a^3}{3x_n^2+a}
=\frac{(x_n-\sqrt a)^3}{3x_n^2+a}
$$
which gives you the third order convergence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1604421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Some equations from Russian maths book. Could you please help me with solving these equations. I would like to solve them in the most sneaky way. All of the exercises in this book can be solved in some clever way which I can't often find.
$$
\frac{(x-1)(x-2)(x-3)(x-4)}{(x+1)(x+2)(x+3)(x+4)} = 1
$$
$$
\frac{6}{(x+1)(x+2... | First question:
Let $y=(x-2)(x-3)$. So it's equivalent to
$$\frac{y(y-2)}{(y+10x)(y+10x-2)}=1$$
$$y^2-2y=y^2+20xy+100x^2-2y-20x=0$$
$$20x=100x^2+20xy$$
For $x\neq 0$,
$$1=5x+(x-2)(x-3)$$
$$x^2=-5$$
$$x=\pm\sqrt{5} i$$
Otherwise $x=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1604713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
If squaring a number means multiplying that number with itself then shouldn't taking square root of a number mean to divide a number by itself? If squaring a number means multiplying that number with itself then shouldn't taking square root of a number mean to divide a number by itself?
For example the square of $2$ ... | Assuming $x > 0$
Algebraically:
$$
\begin{matrix}
x \cdot x = x^2 & \rightarrow & \sqrt{x \cdot x} = \sqrt{x^2} & \rightarrow & \sqrt{x} \cdot \sqrt{x} = x \\
\downarrow & & & & \downarrow \\
x^2 \div x = x & \rightarrow & \sqrt{x^2 \div x} = \sqrt{x} & \rightarrow & x \div \sqrt{x} = \sqrt{x}
\end{matrix}
$$
Visual... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1605183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "119",
"answer_count": 25,
"answer_id": 16
} |
How to solve $\lim _{x\to \sqrt{2}}\left(\frac{e^{x^2}+e^2(1-x^2)}{[\ln(x^2-3\sqrt{2}x+5)]^2}\right)$? I have a problem with this limit, i have no idea how to compute it. Can you explain the method and the steps used(without Hopital if is possible)? Thanks
$$\lim_{x\to \sqrt{2}}\left(\frac{e^{x^2}+e^2(1-x^2)}{[\ln(x^2-... | $$\lim _{x\to \sqrt{2}}\left(\frac{e^{x^2}+e^2(1-x^2)}{[\ln(x^2-3\sqrt{2}x+5)]^2}\right)$$
$$=e^2\lim _{x\to \sqrt{2}}\dfrac{e^{x^2-2}-1-(x^2-2)}{x^2-3\sqrt2 x+4}\cdot\dfrac1{\left(\lim _{x\to \sqrt{2}}\dfrac{\ln(1+x^2-3\sqrt2 x+4)}{x^2-3\sqrt2 x+4}\right)^2}$$
Now,
$$\lim _{x\to \sqrt{2}}\dfrac{e^{x^2-2}-1-(x^2-2)}{(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1605491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Express a general vector $(x, y, z)$ in $R^3$ as a linear combination of $(1, 2, 1),(1, 0, -1),(1, -2, 1)$ $a + b + c = x$
$2a + 0 - 2c = y$
$a – b + c = z$
Then, $(a + b + c, 2a - 2c, a - b + c) = (x, y, z)$, but this looks horrendous unless they mean I should simplify this somehow or something entirely different. Am ... | hint: $a+c=\dfrac{x+z}{2}, a-c = \dfrac{y}{2}\Rightarrow 2a = \dfrac{x+y+z}{2}\Rightarrow a = \dfrac{x+y+z}{4}$, and similarly $b = \dfrac{x-z}{2}$. Can you solve for $c$ ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1606191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\lim_{x\to 0}\frac{\sin(\ln(x+1))-\ln(\sin(x)+1)}{\sin^4\frac{x}{2}}$ Compute the following limit: $$L=\lim_{x\to 0}\frac{\sin(\ln(x+1))-\ln(\sin(x)+1)}{\sin^4\left(\frac{x}{2}\right)}$$
| It's easiest to use Taylor expansions about $x = 0$. First the denominator to leading order, so we can tell how many terms are needed:
$$
\sin\left(\frac{x}{2}\right)^4 = \frac{x^4}{16} + O(x^6)
$$
So we need expansions to this order also for the numerator:
$$
\sin(\ln(x+1)) = x-\frac{x^2}{2}+\frac{x^3}{6}-\frac{x^5}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1606931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Proving the inequality $ \frac {x+y}{x^2+y^2}\leq \frac 12 \left(\frac {1}{x}+\frac{1}{y}\right)$ Let $x$ and $y$ be two positive numbers:
Prove that $$ \left( \frac {x+y}{x^2+y^2}\right) \leq \frac 12 \left(\frac {1}{x}+\frac{1}{y}\right).$$
I answered this one by squaring the two expressions. And therefore finding ... | An alternative approach:
$\left( \frac {x+y}{x^2+y^2}\right) \leq \frac 12 \left(\frac {1}{x}+\frac{1}{y}\right) \Longleftrightarrow 4x^2y + 4xy^2 \leq (x^2 + y^2)(2y + 2x) \Longleftrightarrow x^2y + y^2x \leq x^3 + y^3$
We assume WLOG $x \leq y \Longrightarrow \exists c\in \mathbb{R}_{≥0}$ such that $x+c = y$
Hence... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1611226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
The sequence $\phi_n= 1+\frac{1}{2^4}+\frac{1}{3^4}+\ldots+\frac{1}{n^4}$ is bounded above . A sequence $(\phi_n)$ is defined as follows : $$\phi_n= 1+\frac{1}{2^4}+\frac{1}{3^4}+\ldots+\frac{1}{n^4}$$
Show that the sequence is convergent.
Because this sequence is monotonic, proving it is bounded above will be sufficie... | The following approach will also give a reasonable bound:
$$1+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}+\frac{1}{5^4}+\frac{1}{6^4}+\frac{1}{7^4}+\frac{1}{8^4}+\frac{1}{9^4}+\ldots+\frac{1}{n^4} \\ < 1+\frac{1}{2^4}+\frac{1}{2^4}+\frac{1}{4^4}+\frac{1}{4^4}+\frac{1}{4^4}+\frac{1}{4^4}+\frac{1}{8^4}+\frac{1}{8^4}+\ldots... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1612011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Positive correlation with the sequence $\sqrt{ij}/2-\min(i,j)$ Is there a sequence of positive real numbers $x_1,\ldots,x_n$ for which
$$
\sum_{1\leq i,j\leq n}\left[\frac{\sqrt{ij}}{2}-\min(i,j)\right]x_ix_j> 0?
$$
| Alternative solution:
There exists such $n$ and a sequence of positive real numbers $x_1, x_2, \cdots, x_n$.
Let $A$ denote the matrix whose $(i,j)$-th entry is $\frac{\sqrt{ij}}{2}$.
Then, $A = uu^\mathsf{T}$ where
$u = \frac{1}{\sqrt{2}}[\sqrt{1}, \sqrt{2}, \cdots, \sqrt{n}]^\mathsf{T}$.
Let $B$ denote the matrix who... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1614111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the matrix of linear transformation with respect to different basis Let $f:P_2[x]\rightarrow M_{2\times 2}(\mathbb{R})$ is a linear transformation defined as $f(a+bx+cx^2)= \begin{bmatrix}
b+c & a \\
b & c \\
\end{bmatrix}$
Find the matrix of $f$ with respect to basis $ ({\be... | Your question is fuzzy: $f$ has one matrix with respect to the two given bases. So there is only one case.
Let's call the matrix we want to find $M_f$ and write the coefficients of the two given vectors down:
The vector $a + bx + cx^2$ with respect to the basis $(1,x,x^2)$ is $(a,b,c)$.
The vector $\begin{bmatrix}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1615093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solving this limit $\lim_{x\to 3}\frac{x^2-9-3+\sqrt{x+6}}{x^2-9}$. The question is $\lim_\limits{x\to 3}\frac{x^2-9-3+\sqrt{x+6}}{x^2-9}$.
I hope you guys understand why I have written the numerator like that. So my progress is nothing but $1+\frac{\sqrt{x+6}-3}{x^2-9}$.
Now how do I rationalize the numerator?
It is g... | Set $\sqrt{x+6}-3=y\implies x=(y+3)^2-6=y^2+6y+3$
$$\implies\lim_{x\to3}\dfrac{\sqrt{x+6}-3}{x^2-9}$$
$$=\lim_{y\to0}\dfrac y{(y^2+6y+3)^2-9}$$
$$=\lim_{y\to0}\dfrac y{(y^2+6y)^2+6(y^2+6y)}$$
As $y\to0,y\ne0$
So, cancel out $y$ from N & D to get
$$\lim_{y\to0}\dfrac1{y(y+6)^2+6(y+6)}=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1615527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Solving integral using trig substitution $\tan(x/2)=t$ I have problems with solving the following integral:
$$ \int{{\sin x - \cos x}\over {\sin x + \cos x}} \, dx$$
Could anybody please help me to find the solution and show me the method how it can be solved? I already tried to solve similar ones but I get always stuc... | \begin{align}
\tan \frac x 2 & = t \\[8pt]
\frac x 2 & = \arctan t \\[8pt]
x & = 2\arctan t \\[8pt]
\sin x & = \sin(2\arctan t) = \sin(2 \, \bullet) = 2 \sin(\bullet)\cos(\bullet) \\
& = 2\sin(\arctan t)\cos(\arctan t), \\[8pt]
\cos x & = \cos(2\arctan t) = \cos(2\,\bullet) = \cos^2(\bullet) - \sin^2(\bullet) \\
& = \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
Let $f:[1,\infty)\to R$ be a monotonic and differentiable function and $f(1)=1.$If $N$ is the number of solutions of $f(f(x))=\frac{1}{x^2-2x+2}$. Let $f:[1,\infty)\to R$ be a monotonic and differentiable function and $f(1)=1.$If $N$ is the number of solutions of $f(f(x))=\frac{1}{x^2-2x+2}$.Find $N.$
$\frac{1}{x^2-2x... | Hint: If $f(x)$ is monotonic, then $f(f(x))$ is (possibly non-strictly) increasing, so…
Full solution: Clearly if $f$ is increasing, so is $f\circ f$. Furthermore, if $f$ is decreasing, then $x \leq y$ implies $f(x) \geq f(y)$, which in turn implies $f(f(x)) \leq f(f(y))$. Either way, $f \circ f$ is increasing.
Sinc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to show $\frac{1}{2\sqrt{2} + \sqrt{3}} = \frac{2\sqrt{2} - \sqrt{3}}{5}$? Show that:
$$ \dfrac{1}{2\sqrt2+\sqrt3}=\dfrac{2\sqrt2-\sqrt3}{5}$$
So I multiplied everything by $\sqrt3$
Then I got
$$\frac{\sqrt{3}}{2\sqrt{2}+3}$$
Then multiply it by $\sqrt2$ to obtain
$$\frac{\sqrt{2}\sqrt{3}}{2 \cdot 3+3}$$
Which is ... | $$(2\sqrt2-\sqrt3)(2\sqrt2+\sqrt3)=(2\sqrt2)^2-(\sqrt3)^2=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1618339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Complex number - wrong result at the end I need to solve this:
$$ \frac{i^4+3}{i-1}$$
On my book the result should be: $-2-2i$ but I get: $-1-2i$ and I do not understand where the error is.
My steps:
$$ \frac{i^4+3}{i-1} = \frac{i^4+3}{i-1} \cdot \frac{-1-i}{-1-i}$$
$$ \frac{(i^4+3)(-1-i) + (-1-i)(i-1)}{(i-1)(-1-i)}$$
... | Following your line of algebraic manipulation,
\begin{align}
\frac{i^4+3}{i-1} &= \frac{i^4+3}{i-1} \cdot \frac{-1-i}{-1-i} \\
&= \frac{(i^4+3)(-1-i)}{(i-1)(-1-i)} \\
&= \frac{(1+3)(-1-i)}{-i-i^{2}+1+i} \\
\end{align}
I'll leave to you to finish the details. I think the mistake was i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1619360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Find all $p$'s such that $p^4 + p^3 + p^2 + p +1$ is a perfect square.
Let $p$ be a prime number. Find all $p$'s such that $p^4 + p^3 + p^2 + p +1$ is a perfect square.
I tried rewriting the expression as $p^4 + p^3 + p^2 + p +1 = x^2 \iff (p^2 + p)(p^2 + 1)=(x - 1)(x + 1)$. Then I think I need to bound this using $... | Note that $p^2=q^4+q^3+q^2+q+1$ has solutions only for $p=11$ and $q=3$.
Indeed we can write $$\left(q^2+\frac{q}{2}\right)^2={q^4+q^3}+\frac{q^2}{4}<{q^4+q^3}+q^2+q+1 \\ \frac{q^2}{4}<q^2+q+1 $$ and on the other hand $$ \left(q^2+\frac{q+2}{2}\right)^2=q^4+q^3+2q^2+\frac{q^2+4q+4}{4}>q^4+q^3+q^2+q+1 \\ {q^4+q^3}+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1619971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
What is the probability that two numbers between 1 and 10 picked at random sum to a number greater than 5? We have the numbers $1$ through $10$ in a box, we pick one at random, write it down and put it back in the box. We pick another of those numbers at random and write it down again. If we add the two numbers, what i... | You could solve this with generating functions. The generating function for this situation, equivalent to rolling a fair 10-sided die twice, is:
$$(x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + x^9 + x^{10})^2$$
which expands to
$$x^2 + 2 x^3 + 3 x^4 + 4 x^5 + 5 x^6 + 6 x^7 + 7 x^8 + 8 x^9 +
9 x^{10} + 10 x^{11} + 9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1623793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
Solutions to $x+y+z-2 = (x-y)(y-z)(z-x)$ Show that the equation $$x+y+z-2 = (x-y)(y-z)(z-x)$$ has infinite solutions $(x,y,z)$ with $x, y,z$ distinct integers.
In my attempt to solve the problem only found solutions form $x=y, z=2-2x$.
There are solutions as required by the statement?
| Set $x=y+a$ and $z=y-b$. The equation then becomes
$$ 3y+a-b-2 = -ab(a+b) $$
which simplifies to
$$ 3y-2 = -a^2b - ab^2 - a + b $$
If we set $a=2$, then every $b$ that is a multiple of $3$ will lead to a solution for $y$.
In general an integral $y$ is possible exactly when $a+1\equiv b\pmod 3$ or $a\equiv b\equiv 1\pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1626320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Evaluating the following limit: $\lim _{x\to \frac{\pi }{4}}\left(\tan\left(2x\right)\tan\left(\frac{\pi }{4}-x\right)\right)$ I don't find the right identities for this
$$\lim _{x\to \frac{\pi }{4}}\left(\tan\left(2x\right)\tan\left(\frac{\pi }{4}-x\right)\right)$$
Someone can help me ?
Thanks.
| For $x\neq \frac{\pi}{4}+k\pi/2$ and $x\neq \pi/2+k\pi$ ($k\in\mathbb{Z}$), we have
$$\tan(2x)=\frac{2\tan(x)}{1-\tan^{2}(x)}$$
and
$$\tan\left(\frac{\pi}{4}-x\right)=\frac{\tan\left(\frac{\pi}{4}\right)-\tan(x)}{1+\tan\left(\frac{\pi}{4}\right)\tan(x)}$$
Combining the two expressions and using the fact that $\tan(\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1626834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
$X^4 - 4Y^4 = -Z^2$ has no solutions in non zero integers I am trying to prove that $X^4 - 4Y^4 = -Z^2$ has no solutions in non zero integers. I know there are similar questions on MS, but that minus signs before the $Z$ gives me a hard time.
For the moment, I converted the equation to the equivalent $X^4 + Z^2 = 4Y^2$... | For another way of showing your required conclusion, I give you the following:
Hint:
$$Z^2 = 4Y^4 - X^4 = (2Y^2 + X^2)(2Y^2 - X^2)$$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1628270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How many integer solutions are there to the equation $x_1 + x_2 + x_3 + x_4 = 12$ with restrictions on $x_1,x_2,x_3,x_4$? How many integer solutions are there to the equation $x_1 + x_2 + x_3 + x_4 = 12$ with $x_i > 0$ for each $i \in \{1, 2, 3, 4\}$? How many solutions with $x_1 > 1$, $x_2 > 1$, $x_3 > 3$, $x_4 \geq ... | I am not sure why you have four variables in the statement of your question and three variables in your answer. I will assume you meant to work with four variables.
How many integer solutions are there to the equation $x_1 + x_2 + x_3 + x_4 = 12$ with $x_i > 0$ for each $i \in \{1, 2, 3, 4\}$?
We wish to solve the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1628696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
If $ A=\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+.........+\frac{1}{100\sqrt{99}}\;,$ Then $\lfloor A \rfloor =$
If $\displaystyle A=\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+.........+\frac{1}{100\sqrt{99}}\;,$ Then $\lfloor A \rfloor =$
Where $\lfloor x \rfloor$ represent floor fu... | Hint. Observe that, for $x\in [k,k+1],\, k=1,2,3,\cdots$,
$$
\frac1{(k+2)\sqrt{k+1}} \leq\frac1{(x+1)\sqrt{x}}\leq \frac1{(k+1)\sqrt{k}}.
$$
Integrating both sides from $x=k$ to $x=k+1$ gives
$$
\frac1{(k+2)\sqrt{k+1}} \leq\int_k^{k+1}\frac1{(x+1)\sqrt{x}}dx\leq \frac1{(k+1)\sqrt{k}}
$$ then, summing from $k=1$ to $k=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1634645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
} |
the minimum value of $a^2+b^2-a-\frac{2b}{3}$ using known standard inequalities If $a,b$ are real numbers satisfying $a+2b=3,$ then the minimum value of $a^2+b^2-a-\frac{2b}{3}$
Here $a+2b=3\implies a=3-2b$
$a^2+b^2-a-\frac{2b}{3}=(3-2b)^2+b^2-(3-2b)-\frac{2b}{3}$
$=9+4b^2-12b+b^2-3+2b-\frac{2b}{3}$
I diiferentiated i... | Hint:
$$ a^2+b^2-a-\frac{2b}{3} = (a - \frac12)^2 + (b - \frac13)^2 - \frac{13}{36}$$
and the first $2$ terms at the RHS can be seen as the square of the distance from a point to the point $(\frac12, \frac13)$. So the problem becomes finding the distance from a certain point to a line.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1635651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find numbers whose sum of digits equals a value How do I find all of the numbers in a given range whose sum of digits equal to a given value?
For example:
Range : 100 - 9000
Value : 4
Result : 103, 112, 121, 130, 202, 211, 220, 301, 310, ..., 4000, ...
| Observe that any four digit number larger than $4000$ will have a digit sum greater than $4$. Thus, the problem reduces to finding the number of three or four digit numbers with digit sum $4$.
For three digit numbers, let $x_2$ denote the hundreds digit, $x_1$ denote the tens digit, and $x_0$ denote the units digit. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1635838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Simplifying the arc length integral for a segment of a circle (Calc 2) So I know the length L of the curve $y=\sqrt{R^{2}-x^{2}}$ from $x=0$ to $x=a$ where $|a| < R$ is given by:
$$L= \int_0^a \frac{R}{\sqrt{R^{2}-x^{2}}}dx $$
Now I must set up the arc length integral and simplify it so that it is in the form listed ab... | Assume $R>a$ and $a,R\in\mathbb{R^+}$:
$$\text{L}=\int_{0}^{a}\sqrt{1+\left(\frac{\partial}{\partial x}\left[\sqrt{R^2-x^2}\right]\right)^2}\space\text{d}x=\int_{0}^{a}\sqrt{1+\left(-\frac{x}{\sqrt{R^2-x^2}}\right)^2}\space\text{d}x=$$
$$\int_{0}^{a}\sqrt{1+\frac{x^2}{R^2-x^2}}\space\text{d}x=\int_{0}^{a}\sqrt{\frac{R^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1636248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Is this relationship already known? I like math because it's a puzzle to me, but am really not very good at it. But I figured out the relationship below myself. Just curious, is this already pretty common knowledge? Kind of proud of myself for figuring it out, but my son who's getting math minor had never heard of it. ... | This is another way of stating that $n^2 = 1 + 3 + 5 + ...$; the sum of the first n-1 odd numbers.
$a - b = 1$ means $a = b + 1$ means $a + b = 2b + 1$ means $a + b$ is the (b-1)-th odd number.
So $a^2 = b^2 + a + b$ is the same as saying $(b + 1)^2 = b^2 + (b + 1) + b = b^2 + 2b + 1$; which can be shown by expansion: ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1636688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Let $a$ be a root of the cubic $x^3-21x+35=0$. Prove that $a^2+2a-14$ is a root of the cubic.
Let $a$ be a root of the cubic $x^3-21x+35=0$. Prove that $a^2+2a-14$
is a root of the cubic.
My effort
Working backwards I let $P(x)$ be a polynomial with roots $a,a^2+2a-14$ and $r$.
Thus, $$P(x)=(x-a)(x-r)(x-(a^2+2a-14)... | It's quite a lot easier to simply plug in $x=a^2+2a-14$ into $P(x)$ and see \begin{align}
x^3-21x+35&=(a^2+2a-14)^3-21(a^2+2a-14)+35\\
&=(35-21a + a^3) (-69 - 9 a + 6 a^2 + a^3)\\
&=0\cdot (-69 - 9 a + 6 a^2 + a^3)\\
&=0
\end{align}
Note that factorizing isn't too hard since you already "know" that one factor will be $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1638609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Convergence of $a_{n+1}=\sqrt{2-a_n}$ I'm attempting to prove the convergence (or divergence, though I strongly suspect it converges) of a sequence defined as $a_{n+1}=\sqrt{2-a_n}$ with $a_1=\sqrt{2}$.
I cannot use the monotonic sequence theorem as the sequence is not monotonically increasing. In fact, the first few ... | You have surely proved that $a_n\le 2$ for all $n$.
Consider the sequences $b_n=a_{2n-1}$ and $c_n=a_{2n}$. The recursions are
$$
b_{n+1}=a_{2n+1}=\sqrt{2-a_{2n}}=\sqrt{2-\sqrt{2-a_{2n-1}}}=
\sqrt{2-\sqrt{2-b_n}}
$$
Let's show $(b_n)$ is decreasing:
\begin{gather}
b_{n+1}\le b_n\\
\sqrt{2-\sqrt{2-b_n}}\le b_n\\
2-\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1640379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 4
} |
Maximum value of $\lambda$ It is given that a,b,c are be of same sign and a,b,c are in Harmonic progression i.e. $\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$ and also
$\frac{a+b}{2a-b}+\frac{c+b}{2c-b}\geq \sqrt{\lambda \sqrt{\lambda \sqrt{\lambda.............\infty}}}$.
Find the greatest possible value of $\lambda$.
How sh... | Let $a=1/x,\; b=1/y, \;c=1/z.$ We have $y=(x+z)/2$. The LHS of the inequality is $$\frac {y+x}{2 y-x}+\frac {y+z}{2 y-z}=\frac { z/2+ 3 x/2}{z}+\frac { x/2+ 3 z/2}{x}=1+\frac {3}{2} (x/z+z/x).$$ Since $x$ and $z$ (which are $1/a$ and $1/c$) have the same sign, $x/z$ and $z/x$ are positive, so $$x/z+z/x=(\;(x/z)^{1/2}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1641182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show if $A^TA = I$ and $\det A = 1$ , then $A$ is a rotational matrix Show if $A^TA = I$ and $\det A = 1$ where
$ A =
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
$, then $A =\begin{bmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix}$.
attempt:
Suppose $ A^TA =\begin{bmatri... | If you had $A =\begin{bmatrix} \cos\theta & \sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$ or $A =\begin{bmatrix} \cos\theta & -\sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$ instead of $A =\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$, would it then be true that $A^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1642625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Divisors of $75600$ of the type of $4n+2$
Find the total no. of divisors of $75600$ which of the type of $4n+2$
where $n\in \mathbb{N}$ and
$75600=2^4 \cdot 3^3 \cdot 5^2 \cdot 7^1$
Now I think divisors of type $(4n+2)$ should be of type
$2^1\cdot 3^a\cdot 5^b\cdot 7^c$. Am I correct?
| First of all, you are right.
Therefore, you need to count the number of divisors of $3^3\cdot5^2\cdot7^1$:
*
*In each divisor, the factor $3$ can appear between $0$ and $3$ times, i.e., $4$ different combinations
*In each divisor, the factor $5$ can appear between $0$ and $2$ times, i.e., $3$ different combinations... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1644796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Is the sequence $\sqrt{1+\frac{1}{n^2}}$ increasing or decreasing? Is the sequence $\sqrt{1+\frac{1}{n^2}}$ increasing or decreasing?
I simplified it to $\frac{\sqrt{n^2+1}}{n}$, and I tried $a_{n+1}-a_n$ and $\frac{a_{n+1}}{a_n}$, but neither seem to work, how should I proceed?
| $\sqrt{1+\frac{1}{n^2}} > \sqrt{1+\frac{1}{(n+1)^2}}$
$\iff 1+\frac{1}{n^2} > 1+\frac{1}{(n+1)^2}$
$\iff \frac{1}{n^2} > \frac{1}{(n+1)^2}$
$\iff {n^2} < {(n+1)^2}$
$\iff {n^2} < n^2+2n+1$
$\iff 0< 2n+1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1646107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Prove that $a+\frac{1}{b}>2$ or $b+\frac{1}{a}>2$ for two strict positive numbers Another Olympiad Problem, let $x$ and $a$ and $b$ be strictly real positive numbers.
*
*Prove that $x$+$\frac{1}{x}$$>$$2$ (proven)
*Than conclude that $a$+$\frac{1}{b}$$>$$2$ or
$b$+$\frac{1}{a}$$>$$2$
For the second ques... | Your first statement $x + \frac 1x > 2$ is not fully correct, you have
$$
x + \frac 1x \ge 2
$$
for $x > 0$, with equality exactly for $x = 1$.
And as the example $a=b=1$ shows, you can only conclude that
$$
a+\frac{1}{b} \ge 2 \quad \text{ or } \quad b+\frac{1}{a} \ge 2 \, .
$$
which is your claim with $\ge$ instead... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1649513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
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... | Assume that,
$\sin x=a, \cos x=b $
Given that : $$\sin x+\cos x=\frac75$$
$$a+b=\frac75\tag 1$$
$$\sin^2 x+\cos^2 x=1$$
$$a^2 +b^2 =1$$
$$(a+b)^2-2ab =1$$
$$(7/5)^2-2ab =1$$
$$ab=12/25\tag 2$$
solving (1), (2), i get $a=3/5, b=4/5$
therefore,
$$\frac1{\sin x}+\frac1{\cos x}=\frac1a+\frac1b$$$$=\frac{1}{3/5}+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1649606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 10,
"answer_id": 3
} |
Solve equation $\frac{1}{x}+\frac{1}{y}=\frac{2}{101}$ in naturals My try was $$\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}=\frac{2}{101}\\x+y=2k,xy=101k\\x=2k-y\\y(2k-y)=101k\\2ky-y^2=101k\\y^2-2ky+101k=0\\y=k+\sqrt{k^2-101k}\\x=k-\sqrt{k^2-101k}$$
Now $\sqrt{k^2-101k}$ has to be either integer or rational,if it's an integ... | $$\dfrac1x=\dfrac{2y-101}{101y}\iff x=\dfrac{101y}{2y-101}$$
If $d$ divides $2y-101,101y$
$d$ must divide $2(101y)-101(2y-101)=101^2$
So, $2y-101$ must divide $101^2$ to make $x$ an integer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1650875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
integrate $\int \frac{dx}{x\sqrt{1-x}}$
$$\int \frac{dx}{x\sqrt{1-x}}$$
$$\int \frac{dx}{x\sqrt{1-x}}$$
$u=1-x$
$du=-dx$
$$-\int \frac{du}{(1-u)\sqrt{u}}$$
$a(1-u)+b\sqrt{u}=1\Rightarrow a-au+b\sqrt{u}=1$
$a=1\Rightarrow b\sqrt{u}-u=0\Rightarrow b=\sqrt{u}$
$$\int \frac{\sqrt{u}}{1-u}du-\int \frac{du}{\sqrt{u}}=\int ... | $x=\sin^2(\theta)$
$$
\begin{align}
\int\frac{\mathrm{d}x}{x\sqrt{1-x}}
&=\int\frac{2\,\mathrm{d}\theta}{\sin(\theta)}\\
&=-2\int\frac{\mathrm{d}\cos(\theta)}{1-\cos^2(\theta)}\\
&=-\int\left(\frac1{1-\cos(\theta)}+\frac1{1+\cos(\theta)}\right)\mathrm{d}\cos(\theta)\\
&=\log\left(\frac{1-\cos(\theta)}{1+\cos(\theta)}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1651932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
What is wrong with this infinite sum We know that:
https://www.youtube.com/watch?v=w-I6XTVZXww
$$S=1+2+3+4+\cdots = -\frac{1}{12}$$
So multiplying each terms in the left hand side by $2$ gives:
$$2S =2+4+6+8+\cdots = -\frac{1}{6}$$
This is the sum of the even numbers
Furthermore, we can add it to itself but shifting th... | Interpreted literally (i.e., using the usual sense of limits of infinite series), the first line,
$$1 + 2 + 3 + \cdots = -\frac{1}{12} ,$$
is simply false, as the series on the l.h.s. diverges.
What's true, for example, is that there's a natural way to extend the function $$Z(s) := \sum_{k = 1}^{\infty} k^{-s} ,$$ whic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1653416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Total number of $4$ digit numbers whose product of digits is $72$
Total number of $4$ digit numbers whose product of digits is $72$
$\bf{My\; Try::}$ Here the possible factor of $72 = 2^{3}\cdot 3^2$
Now here we divide $2^3\cdot 3^3$ into product of $4$ numbers
Like $1\cdot 1\cdot 8\cdot 9$
and $1\cdot 2\cdot 4\cdot ... |
Where have I made a mistake?
*
*You forgot to count the number of permutations of $[2,3,3,4]$
*You miscounted the number of permutations of $[2,2,3,6]$ as $\frac{4!}{3!}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1654265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Another $1=2$ proof So a friend shows me this :
$x^4= x^2+x^2+ \cdots +x^2 $ ( i.e. $x^2$ added $x^2$ times)
Now take the derivative of both side;
$4x^3 = 2x + 2x + \cdots + 2x $;
So $4x^3 = 2x^3 \cdots $(1)
And so dividing by $x^3$ gives $2=1 \cdots $(2).
I know we can't divide by 0 so that makes (2) false, but to... | *
*First of all, the statement
$x^2$ added $x^2$ times
makes sense only if $x^2$ is a positive integer. Else if $x^2$ is not a positive integer, then the statement is meaningless.
*Moreover, from $(1)$, we have that $4x^3=2x^3 \Rightarrow 4x^3-2x^3=0 \Rightarrow 2x^3=0 \Rightarrow x^3=0$
And hence division by... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1658284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\int\frac{3x^2+1}{(x^2-1)^3}dx=\frac{-x}{(x^2-1)^2}+c$ Prove that $\int\frac{3x^2+1}{(x^2-1)^3}dx=\frac{-x}{(x^2-1)^2}+c$
My Try:
$\int\frac{3x^2+1}{(x^2-1)^3}dx$
Put $x=\sec\theta$
$\int\frac{3\sec^2\theta+1}{\tan^6\theta}\sec\theta\tan\theta d\theta$
Converting to $\sin$ and $\cos$ we get
$=\int\frac{(3+... | Another way is, simply, to take the derivative of $$\frac{-x}{(x^2-1)^2}+c$$ and verify that this derivative is no other that $$\frac{3x^2+1}{(x^2-1)^3}$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1658367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Find the value of $\frac{1}{20} + \frac{1}{30} + \frac{1}{42} + \frac{1}{56} + \frac{1}{72} + \frac{1}{90}$ Find the value of $p+q$, where $p$ and $q$ are two positive integers such that $p$ and $q$ have no common factor larger than $1$ and
$$\frac{1}{20} + \frac{1}{30} + \frac{1}{42} + \frac{1}{56} + \frac{1}{72} + \... | We know that
$$
S_n=\frac{1}{1 \cdot 2}+ \frac{1}{2 \cdot 3}+ \cdots+ \frac{1}{(n-1)n}=1-\frac{1}{n}
$$
Then yours sum is exactly $$S_{10}-S_4=(1-\frac{1}{10})-(1-\frac{1}{4})=\frac{1}{4}-\frac{1}{10}=\frac{5-2}{20}=\frac{3}{20}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1659227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How would you find the kinematic range using the position function? Knowing that the range is:
$$ R = \frac{v^2\sin2\theta}g $$
Taking the integral of the velocity function we have:
$$ R(T) = (V_i \cos\theta T + x_i)X +\left(-\frac{1}2gT^2+V_i\sin\theta T+y_i\right)Y$$
So, I know it hits the ground at:
$$ \text{time i... | \begin{align*}
x &= ut\cos \theta \\
y &= ut\sin \theta -\frac{gt^{2}}{2} \\
&= x\tan \theta -\frac{gx^{2}}{2u^{2}\cos^{2} \theta} \\
&= x\tan \theta \left( 1-\frac{gx}{2u^{2} \sin \theta \cos \theta} \right) \\
&= x\tan \theta \left( 1-\frac{gx}{u^{2} \sin 2\theta} \right) \\
R &= \frac{u^{2} \sin 2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1659318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
determinant of sum of anticommuting matrices Let $A, B \in \mathcal{M}_n(\mathbb{C})$ such that $AB=-BA$ and $det(A)=0$ and $det(A+B)=0.$
Prove that $det(A^3+B^3)=0$.
I don't really know how to approach this question.
| You only need $AB=-BA$ and $\det(A+B)=0$, not $\det A=0$, to prove that $\det(A^3+B^3)=0$. The reason is that $A^3+B^3$ can be factored such that $A+B$ is a factor.
Using $BA=-AB$ and always normalising products to the form $A^pB^q$, i.e. all $A$ factors before all $B$ factors, the product
$$
(A+rB)(A+B)(A+sB)=A^3+(r+s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1660516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
integrate $\int_0^{\frac{\pi}{4}}\frac{dx}{2+\tan x}$
$$\int^{\frac{\pi}{4}}_{0}\frac{dx}{2+\tan x}$$
$v=\tan(\frac{x}{2})$
$\tan x=\frac{2v}{1-v^2}$
$dx=\frac{2\,dv}{1+v^2}$
$$\int^{\frac{\pi}{4}}_0 \frac{dx}{2+\tan x}=\int^{\frac{\pi}{8}}_0 \frac{\frac{2\,dv}{1+v^2}}{2+\frac{2v}{1-v^2}}=\int^{\frac{\pi}{8}}_0 \frac... | The integrand is
$$f(x):=\frac{\cos(x)}{2\cos(x)+\sin(x)}.$$
We can form a linear combination to let the derivative of the denominator appear at the numerator:
$$af(x)+b=\frac{a\cos(x)+b(2\cos(x)+\sin(x))}{2\cos(x)+\sin(x)}=\frac{-2\sin(x)+\cos(x)}{2\cos(x)+\sin(x)},$$ is obtained with
$$b=-2,a=5.$$
Then by integratin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1662272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 1
} |
Prove that $(x+y+z)^2(yz+zx+xy)^2 \leq 3(y^2+yz+z^2)(z^2+zx+x^2)(x^2+xy+y^2)$ This is Problem 6 of the 2007 Indian National Math Olympiad (INMO).
If $x, y, z$ are positive real numbers, prove that
$(x+y+z)^2(yz+zx+xy)^2 \leq 3(y^2+yz+z^2)(z^2+zx+x^2)(x^2+xy+y^2).$
My best idea was to expand this and simplify. Althoug... |
Let $AF=x\ , BF=y\ , CF=z$.
$F-$ Fermat point of $\triangle ABC\ $
So inequality we can rewrite as : $x+y+z\le 3R$ , which is obviously true)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1663294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Generating function to find the number of ways to put marbles in a basket Write a generating function for the number of ways to make a basket of $n$
marbles, if you need to use at least one orange marble , an even number of
yellow marbles, at most 2 green marbles, and any number of red marbles.
Because the question is ... | Completing Brian Scott's answer, you are after the coefficient of $z^n$ in the following. This yields to partial fractions and using geometric series and (generalized) binomial theorem:
$\begin{align}
[z^n] \left( \sum_{k \ge 1} z^k \right)
\cdot &\left( 1 + z + z^2 \right)
\cdot \left( \sum_{k \g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1664493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculate the limit $\lim_{x \to 2} \frac{x^2\sqrt{x+2}-8}{4-x^2}$ Calculate the limit
$$\lim_{x \to 2} \frac{x^2\sqrt{x+2}-8}{4-x^2}$$
I tried to factorise and to simplify, but I can't find anything good.
$$\lim_{x \to 2} \frac{\frac{x^2(x+2)-8\sqrt{x+2}}{\sqrt{x+2}}}{(4-x^2)}$$
| L'Hôpital's rule is not necessary. Here are the steps
$$\lim\limits_{x\to 2} \frac{x^2\sqrt{x+2}-8}{4-x^2}$$
$$=\lim\limits_{x\to 2} \frac{8-x^2\sqrt{x+2}}{(x-2)(x+2)}$$
Let $t=\sqrt{x+2}$, then
$$\lim\limits_{t\to 2} \frac{8-\left(t^2-2\right)^2 t}{\left(t^2-4\right)t^2}$$
$$=\lim\limits_{t\to 2} \frac{\left(-t^5+4t^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1666082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
How to prove that the roots of this equation are integers? Let there be an equation $a^2 + 4ab + b^2 - 121 = 0$ where I want to prove that a,b are integers. Then I want to find whether there are integer values of $b$ for which $a$ is also an integer. Let us consider the case for $a$ $\rightarrow$ $a = \dfrac{-4b \pm \... | From $a^2+b^2+4ab-121=0$ we get $(a+b)^2+2ab=121$Or $2=\frac{11+a+b}{a} \frac{11-a-b}{b}$Now case 1: when the factors on right hand side are integers and coprime. Since 2 has only 2 factors, one of these has to be 1 and the other 2 Taking $\frac{11+a+b}{a}=2$ and $\frac{11-a-b}{b}=1$ we get $a=11 , b=0$
Similarly $\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1667323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find $\int \frac {x^2}{x^3+1} dx$. What is my mistake? $\int \frac {x^2}{x^3+1} dx$
$ u = 3x+1, du=3x^2 dx$
$\int \frac{3 du}{u} $
Am I wrong something? Why the answer is $\int \frac{du}{3u}$ instead of $\int \frac{3 du}{u} $ ?
Thank you.
| If you set $u=x^3+1$ (so that $du = 3x^2dx$), then $x^2dx = \frac{du}{3}$ and
$$
\int \frac{x^2 dx}{x^3 +1} = \int \frac{\frac{du}{3}}{u} =\int \frac{du}{3u}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1667729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Prove $\frac {1 + a + a^2 + \ldots +a^{n - 1} }{n} < \frac {1 + a + a^2 + \ldots +a^{n - 1} + a^n}{n + 1}$ if $1 < a$
Prove $\frac {1 + a + a^2 + \ldots +a^{n - 1} }{n} < \frac {1 + a + a^2 + \ldots +a^{n - 1} + a^n}{n + 1}$ if $1 < a$
Tried induction. Not sure where my mistake is, but what I did doesn't seem to make... | Multiply both sides on $n(n-1)$ then you can see:
$$na^{n} > \sum_{i=0}^{n}a^{i}$$
Right side is geometric progression:
$$na^{n}>\frac{a^{n}-1}{a-1}$$
Now you can continue by yourself.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1669178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Solve $a_n=2 a_{n-1} - a_{n-2} + 2^n$ using generating function I'm preparing to an exam and trying to solve $a_n=2 a_{n-1} - a_{n-2} + 2^n$, where $a_0=0$ and $a_1=1$.
This is my approach:
Let $A(z)=\sum_{n \geq 0} a_{n+2} z^{n+2}$, then:
$$\sum a_{n+2} z^{n+2} = 2 \sum a_{n+1} z^{n+2} - \sum a_n z^{n+2} + \sum (2z)^... | You have to be more careful with your indexing. I prefer to always cast the equation so that it is defined for $n \geq 0$ for greater consistency, but of course it doesn't matter as long as you keep track of your indices.
Anyway, the correct equation would be
$$A(z)(z^2-2z+1) = z + z^2 \frac{4}{1-2z}$$
if you defined ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1672276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove that $6$ divides $n^3+11n$? How can i show that
$$6\mid (n^3+11n)$$
My thoughts:
I show that
$$2\mid (n^3+11n)$$
$$3\mid (n^3+11n)$$
And
$$n^3+11n=n\cdot (n^2+11)$$
And if $n=x\cdot 3$ for all $x \in \mathbb{N}$ then:
$$3\mid (n^3+11n)$$
And if not:
The cross sum of$$n^2+11$$
is multiple of 3.
Can this be right o... | $$
n^3+11n
= 12 \binom{n}{1} + 6 \binom{n}{2} + 6 \binom{n}{3}
= 6\left(2 \binom{n}{1} + \binom{n}{2} + \binom{n}{3}\right)
$$
This is obtained by using repeated differences and Newton's interpolation formula:
$$
\begin{array}{llll}
0 & 12 & 30 & 60 & 108 & \\
12 & 18 & 30 & 48 & \\
6 & 12 & 18 & \\
6 & 6 & \\
0 & \\
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1673037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 3
} |
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 $d$.
| Evaluate the equation for different $x$:
$$\begin{align}\\
\frac\pi6&\to -a+b+2c&+d&=\frac1{64}\\
\frac\pi4&\to -a-b+\ c&+d&=\frac8{64}\\
\frac\pi3&\to +a-b&+d&=\frac{27}{64}\\
\frac\pi2&\to -a+b-\ c&+d&=\frac{64}{64}\\
\end{align}$$
(after normalization of the coefficient of $d$).
Elimination by the combination $(1)+3... | {
"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": 0
} |
Find the limit as $(x,y,z) \to (0,0,0)$ of a specific function $f(x,y,z)$
find $$\lim_{(x,y,z) \to (0,0,0)} \frac{\sin(x^2+y^2+z^2)}{\sqrt{x^2+y^2+z^2}} $$ if exist
$\sin$ is a continuous function so $\lim_{(x,y,z) \to (0,0,0)} \sin(x^2+y^2+z^2)=0$.
$\sqrt{x}$ is not defined for $x<0$ over $\mathbb{R}$ I need to find... | HINT:
$$\frac{\sin(x^2+y^2+z^2)}{\sqrt{x^2+y^2+z^2}}=\frac{\sin(x^2+y^2+z^2)}{x^2+y^2+z^2}\cdot\sqrt{x^2+y^2+z^2}\;.$$
What do you know about $\lim\limits_{x\to 0}\dfrac{\sin x}x$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1676229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\lim_{x \to 0}{\frac{\sin(4x)}{\sin(3x)}}$ without L'Hospital I need to evaluate
$$\lim_{x \to 0}{\frac{\sin(4x)}{\sin(3x)}}$$
I solved this using L'Hospital's Theorem and I got 4/3
However, is there a way to do this without applying this theoerm?
| Since $$\sin t=t-\frac{t^3}{3!}+\frac{t^5}{5!}+\ldots=\sum_{k=0}^{\infty}\frac{(-1)^kt^{2k+1}}{(2k+1)!}$$
we have
\begin{align}
\frac{\sin(4x)}{\sin(3x)}&=\frac{4x-\frac{(4x)^3}{3!}+\frac{(4x)^5}{5!}+\ldots}{3x-\frac{(3x)^3}{3!}+\frac{(3x)^5}{5!}+\ldots}\\
&=\frac{4-\frac{4^3}{3!}x^2+\frac{4^5}{5!}x^4+\ldots}{3-\frac{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1677117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Solve the following trigonometric equation using product cancellation Say I want to solve this equation using product cancellation:
$$\cos(2x+\frac {\pi}{2})\tan 3x = 0$$
The solutions are
$$\cos(2x+\frac {\pi}{2}) = 0 \space \vee \tan 3x = 0$$
Therefore
$$\cos(2x+\frac {\pi}{2}) = 0 \space \rightarrow 2x + \frac {\pi}... | The problem is that while $$\lim_{x \to \frac{\pi}{2}} \left[ \cos\left( 2x+\frac{\pi}{2} \right) \right]=0$$
is true, $$\lim_{x \to \frac{\pi}{2}} \left[ \tan\left(3x\right) \right]=\infty$$
so we cannot not just say that the product is equal to zero. We can take the limit of the product,
$$\lim_{x \to \frac{\pi}{2}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1678696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
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.