Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Problem in trigonometry integral $$\int \frac{1}{(9-x^2)^{\frac{3}{2}}}dx$$
Let $x=3\sin (u)$
$dx=3\cos (u)$
So, $(9-x^2)^{\frac{3}{2}}=(9-9\sin ^2(u))^{\frac{3}{2}}$
$=27\cos ^3(u)$
$u=\sin ^-1(\frac{x}{3})$
My problem is how to substitute u into the integral to become $3\int \frac{\sec ^2(u)}{27}$du?
| You are pretty much there.
$\int \frac{1}{(9-x^2)^{\frac{3}{2}}}dx = \int \frac{3\cos u}{27 cos^3 u}du = \frac{1}{9}\int \sec^2 u \; du$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1199299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is this 5th root in the set of natural numbers?
Is $$\sqrt[5]{x(x+1)(x^4 + x^2 + 1)} \in \mathbb{N}$$ for some $x$?
I am not asking for all $x$, but just for some natural number $x$?
I don't believe so, but I may be wrong?
Suppose $x=1$, you get $\sqrt[5]{6}$ which isnt natural.
$x=2$, you get $\sqrt[5]{126}$
$x=3$... | Note $(x+1)(x^4+x^2+1)=x^5+x^4+x^3+x^2+x+1$. But for $x\in\mathbb N, x>0$ we have
$$
x^5 < x^5+x^4+x^3+x^2+x+1 < x^5+5x^4+10x^3+10x^2+5x+1 = (x+1)^5
$$
so that $x^5+x^4+x^3+x^2+x+1$ is not a perfect fifth power, since it it strictly between two consecutive fifth powers.
Next, note that $x$ and $x^5+x^4+x^3+x^2+x+1$ ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1201196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A quick way to prove the inequality $\frac{\sqrt x+\sqrt y}{2}\le \sqrt{\frac{x+y}{2}}$ Can anyone suggest a quick way to prove this inequality?
$$\frac{\sqrt x+\sqrt y}{2}\le \sqrt{\frac{x+y}{2}}$$
| Both sides are positive so you may square the inequality ot obtain $$\begin{align*}\left(\frac{\sqrt{x}+\sqrt{y}}{2}\right)^2\le\left(\sqrt{\frac{x+y}{2}}\right)^2 &\iff \frac{x+2\sqrt{xy}+y}{4}\le \frac{x+y}{2} \\&\iff \frac{2\sqrt{xy}}{4}\le\frac{x+y}{2}-\frac{x+y}{4} \\&\iff \sqrt{xy}\le \frac{x+y}{2}\\[0.2cm]& \iff... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1203291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Prove by induction that $n^3 + 11n$ is divisible by $6$ for every positive integer $n$. Prove by induction that $n^3 + 11n$ is divisible by $6$ for every positive integer $n$.
I've started by letting $P(n) = n^3+11n$
$P(1)=12$ (divisible by 6, so $P(1)$ is true.)
Assume $P(k)=k^3+11k$ is divisible by 6.
$P(k+1)=(k+1)^... | For any positive integer $n$, let $S(n)$ denote the statement
$$
S(n) : 6\mid (n^3+11n).
$$
Base step: For $n=1, S(1)$ gives $1^3+11(1) = 12 = 2\cdot 6$. Thus, $S(1)$ holds.
Inductive step: Let $k\geq 1$ be fixed, and suppose that $S(k)$ holds; in particular, let $\ell$ be an integer with $6\ell = k^3+11k$. Then
\begin... | {
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How to simplify recurrence relation? I'm having trouble seeing how
$$5(2^{n-1} + 5\cdot 3^{n-1}) - 6(2^{n-2} + 5\cdot3^{n-2})$$
simplifies to:
$$2^{n-2}\cdot (10 - 6) + 3^{n-2} \cdot (75 - 30)$$
How can I simplify the above expression to get to the second version?
| hint: $$5\cdot 2^{n-1} = 5\cdot 2\cdot 2^{n-2} = 10\cdot 2^{n-2}, 5\cdot 3^{n-1} = 5\cdot 3\cdot 3^{n-2} = 15\cdot 3^{n-2}$$, and combine like terms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1204510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof that real multiplication distributes over addition using Dedekind cuts? Proving $\forall x,y,z\in \mathbb R,\:x(y+z)=xy+xz$
There is a very concise proof of this in ProofWiki using Cauchy sequences, but I was wondering whether the same would be possible using Dedekind cuts.
| There is, but as with most of the checks of the field axioms using Dedekind Cuts, it isn't pretty.
First recall that multiplication is given by, for $A,B$ positive
\begin{equation}
A\cdot B =\{a\cdot b \mid a\in A \wedge 0<a \wedge b \in B \wedge 0<b \} \cup \{ q\in \mathbb{Q}\mid q\leq 0\}.
\end{equation}
More general... | {
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Problem with nonhomogeneous recurrence relations I studying Discrete maths during this semester and I need your help.
I have been trying to solve one non-homogeneous recurrence relation and read many-many guides how to do this, but I haven't found the way.
However, it's here:
$A_n = -3A_{n-1} +4A_{n-2}+ 3.2^n$
Anyways... | Let $B_n = A_n - A_{n-1}$. Then
$$
B_n = 3.2^n - 4B_{n-1}.
$$
Let $C_n = B_n/(-4)^n$. Then
$$
C_n = (-0.8)^n + C_{n-1}.
$$
Therefore
$$
C_n = \sum_{k=1}^n (-0.8)^k + C_0 = -\tfrac{4}{9} (1-(-0.8)^n) + C_0 = C_0-\tfrac{4}{9} + \tfrac{4}{9} (-0.8)^n.
$$
This implies that
$$
B_n = (B_0-\tfrac{4}{9})(-4)^n + \tfrac{4}{9} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1206174",
"timestamp": "2023-03-29T00:00:00",
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Roots to the quartic equation, $(x+1)^2+(x+2)^3+(x+3)^4=2$ Solving with Mathematica gives me the four roots, $$x=-4,-2,\dfrac{-7\pm\sqrt5}{2}$$ Is there some trick to solving this that doesn't involve expanding and/or factoring by grouping?
| Hint: Write
$$(x + 3 - 2)^2 + (x+3 - 1)^3 + (x+3)^4 = 2$$
Let $y = x+3$ then
$$p(y) = y^2 - 4y + 4 + y^3 - 3y^2 + 3y -1 + y^4 = 2 \implies p(y) =y^4 + y^3 -2y^2 - y +1 = 0 $$
Notice that $1$ and $-1$ are roots of $p(y)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Maximum value of function given minimum value Suppose there is a function $f(x)=\frac{x^2-2x+b}{x^2+2x+b}$ (the problem doesn't specify, but I am assuming $b$ is a real) that has a minimum value of $\frac{1}{2}$. What is the maximum value of $f(x)$?
My first instinct was to divide out everything, getting that $f(x)=1-... | Consider $$y=\frac{x^2-2x+b}{x^2+2x+b}$$
Or $$(y-1)x^2+2(y+1)x+b(y-1)=0$$
In this quadratic equation in $x$, for those values of $y$ which exist in range, there should exist real value(s) of $x$.
Hence, Discriminant $\geq 0 $.
$$4(y+1)^2-4b(y-1)^2\geq0$$
$$(1-b)y^2+2y(1+b)+(1-b)\geq0$$
Now this is a quadratic in $y$ wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1209748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Problem in Differential Equation (How to proceed?) $$y\frac{dy}{dx}=\sqrt{1-y^2}$$, y=0 when x=0
$$\frac{\frac{dy}{dx}y}{\sqrt{1-y^2}}=1$$
$$\int\frac{\frac{dy}{dx}y}{\sqrt{1-y^2}}dx=\int 1dx$$
$$-\sqrt{1-y^2}=x+c_1$$
$$1-y^2=(x+c_1)^2$$
$$y^2=1-(x+c_1)^2$$
$$y=\pm \sqrt{1-(x+c_1)^2}$$
When x=0, y=0
$$0=\pm \sqrt{1-c^2... | $$\int_0^y \frac{y\, dy}{\sqrt{1-y^2}} = \int_0^x dx \to \sqrt{1-y^2} - 1=x$$ solving it gives you $$y = \pm \sqrt{2x-x^2}$$ is the circle or radius $1$ centered at $(1,0).$
$\bf note:$ use the definite integral whenever possible.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Mathematical Induction Proof Question dealing with integers How would you use mathematical induction to prove that
$1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + \dots + n \cdot (n + 1) \cdot (n + 2) = \frac{n(n + 1)(n + 2)(n + 3)}{4}$
I tried proving the base case of $n = 1$ but the left half is much larger than the right ... | You want to prove
$$\underbrace{1 \times 2 \times 3 + 2 \times 3 \times 4 + \dots + (n + 1)(n + 2)(n+3)}_a = \underbrace{\frac{(n + 1)(n + 2)(n + 3)(n+4)}{4}}_b$$
and you are allowed to assume
$$\underbrace{1 \times 2 \times 3 + 2 \times 3 \times 4 + \dots + n(n + 1)(n + 2)}_c = \underbrace{\frac{n(n + 1)(n + 2)(n + 3)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1212364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\sqrt{2011} + \sqrt{2013} + \sqrt{2015} + \sqrt{2017} < 4 \sqrt{2014}$ How to prove that $\sqrt{2011} + \sqrt{2013} + \sqrt{2015} + \sqrt{2017} < 4\sqrt{2014}$ without using calculator?
| Hint:
$$\sqrt{k-r}+\sqrt{k+r}<2\sqrt{k},\ \ \ \forall k\ge 0, k\ge |r|, r\neq 0$$
Since:
$$\iff 2k+2\sqrt{k^2-r^2}<4k\iff 2\sqrt{k^2-r^2}<2k\iff k^2-r^2<k^2\iff -r^2<0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1213378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Finding a limit with two independent variables: $\lim_{(x,y)\to (0,0)}\frac{x^2y^2}{x^2+y^2}$ I must find the following limit: $$\lim_{(x,y)\to (0,0)}\frac{x^2y^2}{x^2+y^2}$$ Substituting $y=mx$ and $y=x^2$, I have found the limit to be $0$ both times, as $x \to 0$. I have thus assumed that the above limit is $0$, and ... | Given $\varepsilon>0$, take $\delta=\sqrt{\varepsilon}$. Then,
\begin{align*}
0<\sqrt{x^2+y^2}<\delta\quad&\Rightarrow \quad x^2<\delta^2\text{ and }y^2<\delta^2\\
&\Rightarrow \quad\frac{1}{x^2}+\frac{1}{y^2}>\frac{2}{\delta^2}\\
&\Rightarrow \quad \left|\frac{x^2y^2}{x^2+y^2}-0\right|=\frac{1}{\frac{1}{y^2}+\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1213448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Denesting a square root: $\sqrt{7 + \sqrt{14}}$ Write:
$$\sqrt{7 + \sqrt{14}} = a + b\sqrt{c}$$
Form. $$7 + \sqrt{14} = a^2 + 2ab\sqrt{c} + b^2c$$
$a^2 + b^2c = 7$ and $2ab = 1$, and $c = 14$
But that doesn't seem right as $a, b,$ wont be integers?
| Set $r=\sqrt{7+\sqrt{14}}$; then $r^2=7+\sqrt{14}$ and so
$$
14=r^4-14r^2+49
$$
or
$$
r^4-14r^2+35=0
$$
The polynomial $X^4-14X^2+35=0$ is irreducible over the rational numbers by Eisenstein's criterion (with $7$), so the degree of $r$ over the rationals is $4$. A number of the form $a+b\sqrt{c}$ with rational $a,b,c$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1214527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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If $a,b,c$ are integers such that $4a^3+2b^3+c^3=6abc$, is $a=b=c=0$? If $a,b,c$ are integers such that $4a^3+2b^3+c^3=6abc$ , then is it true that $a=b=c=0$ ? I was thinking of infinite descent but can't actually proceed , please help. Thanks in advance
| We may assume that $a$, $b$, and $c$ are relatively prime. If not, we can divide out by the cube of the common factor. Observe that $$c^3=2(3abc-2a^3-b^3),$$ so $c$ is even. Let $c=2n$.
Then, the equation becomes $4a^3+2b^3+8n^3=12abn$. Therefore,
$$
2b^3=4(3abn-a^3-2n^3).
$$
Hence, $b$ is even. Let $b=2m$.
Then, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1217320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$x^3-3x^2+(a^2+2)x-a^2$ has 3 roots $x_1,x_2,x_3$ such that $\sin \tfrac{2\pi x_1}{3}+\sin \tfrac{2\pi x_3}{3}=2\sin \tfrac{2\pi x_2}{3}$. Find $a$. $x^3-3x^2+(a^2+2)x-a^2$ has 3 roots $x_1,x_2,x_3$ such that
$\sin \dfrac{2\pi x_1}{3}+\sin \dfrac{2\pi x_3}{3}=2\sin \dfrac{2\pi x_2}{3}$.
Find $a$
(Bulgari 1998)
| Observe
\begin{align*}x^3-3x^2+(a^2+2)x-a^2&=x(x^3-3x+2)+a^2(x-1)\\ &=[x(x-2)+a^2](x-1)\\ &=(x^2-2x+a^2)(x-1)\end{align*}
And the roots of $x^3-3x^2+(a^2+2)x-a^2$ are $1+\sqrt{1-a^2},\;1-\sqrt{1-a^2}\;\text{and}\; 1.$
If we set $a=1$ or $a=-1$, then the three roots are equal to $1$ and the equality holds.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1218672",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Calculation of real root values of $x$ in $\sqrt{x+1}-\sqrt{x-1}=\sqrt{4x-1}.$
Calculation of x real root values from $ y(x)=\sqrt{x+1}-\sqrt{x-1}-\sqrt{4x-1} $
$\bf{My\; Solution::}$ Here domain of equation is $\displaystyle x\geq 1$. So squaring both sides we get
$\displaystyle (x+1)+(x-1)-2\sqrt{x^2-1}=(4x-1)$.
$... | For $x\ge1$,
$$l(x):=\sqrt{x+1}-\sqrt{x-1}=\dfrac2{\sqrt{x+1}+\sqrt{x-1}}\le\sqrt2$$ and
$$r(x):=\sqrt{4x-1}\ge\sqrt3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1220800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
What is $2^{7!}\bmod{2987}$ Find the remainder when $2^{7!}$ is divided by $2987$.
I tried to factorise $2987$ to make it simple but it was in vain.
| $2^{\large 7!}\!\pmod{\color{brown}{29}87}\,$ can be computed using only simple mental arithmetic.
Note that $\ \color{brown}{29}\!\mid\! 87,\,$ i.e. $\, 29\cdot 3 = 87\ $ so $\ 2987 = 29(100\!+\!3) = 29\cdot 103$
Since $\, p = 29,103\,$ are prime, we can use little Fermat to reduce exponents, i.e.
$(1)\qquad\qquad\ \... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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"answer_id": 1
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How to factorize $x^4+2x^2-x+2$? look at this:
$$x^4+2x^2-x+2$$
How to factorize it?
It should be changed to be in the form of standard factorization formulas.
| $$f(x)=x^4+2x^2-x+1=(x^4-x^3+x^2)+(x^3 -x^2+x)+(2x^2-2x+2)$$
$$f(x)=x^2(x^2-x+1)+x(x^2-x+1)+2(x^2-x+2)=(x^2-x+1)(x^2+x+2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1222443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that the greatest common factor of $m+n$ and $m^2+n^2$ is 1 or 2 if $m$ and $n$ are relatively prime. Prove that the greatest common factor of $m+n$ and $m^2+n^2$ is $1$ or $2$ if $m$ and $n$ are relatively prime natural numbers.
Can anyone give a step-by-step answer for this?
| Let $d\lvert m+n$ and $d\lvert m^2+n^2$.
Then $d\lvert(m^2+n^2)-(m-n)(m+n)$, so $d\lvert 2n^2$;
$\;$and $d\lvert(m^2+n^2)+(m-n)(m+n)$, so $d\lvert 2m^2$.
Since $\gcd(2m^2, 2n^2)=2\gcd(m^2,n^2)=2\cdot1=2$,
$\hspace{.3 in}d\lvert2$ so $d=1$ or $d=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1225647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Number of subsets of length 7 I have the following summation:
$$\sum\limits_{k=7}^{n} {k-1\choose 6} $$
and apparently it counts the number of subsets of {1, 2, . . . , n} having size 7.
Why is this?
| It is because of Pascal's recurrence. The recurrence is that $\binom{i}{j}+\binom{i}{j-1} = \binom{i+1}{j}$ provided that $0< j\leq i$.
\begin{align*}
\binom{6}{6}+\binom{7}{6} &= \binom{7}{7} + \binom{7}{6} \\
&= \binom{8}{7} \\
\binom{6}{6}+\binom{7}{6} + \binom{8}{6} &= \binom{8}{7} + \bin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1229396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The height of two triangles My textbook says the height of the following triangles ($BC$) is $37.5$
Because
$$
\widehat{B_1}=\widehat{B_2}=30 \Rightarrow BD=50 \Rightarrow DC=25 \Rightarrow AC=75
$$
and since in right triangles the side opposite the 30 degrees angle is half the length of the hypotenuse, the height... | Using their approach:
$BC^2 + 75^2 = 4BC^2$
$3BC^2 = 3^2 \times 25^2$
$BC = \sqrt{3} \times 25$
SO you may conclude that your book had an error in calculating $\sqrt3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1229649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int \frac{dx}{1+\sin x+\cos x}$ Evaluate $$\int \frac{1}{1+\sin x+\cos x}\:dx$$
I tried several ways but all of them didn't work
I tried to use Integration-By-Parts method but it's going to give me a more complicated integral
I also tried u-substitution but all of my choices of u didn't work
Any suggestions?... | It is easy to recall the half-angle identities $$\begin{align*} \sin \frac{x}{2} &= \sqrt{\frac{1-\cos x}{2}}, \\ \cos \frac{x}{2} &= \sqrt{\frac{1+\cos x}{2}}, \end{align*}$$ from which the tangent half-angle identity can be written $$\tan^2 \frac{x}{2} = \frac{1-\cos x}{1+\cos x} = \frac{\sin^2 x}{(1+\cos x)^2},$$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1230164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
Finding Surface area of a curve rotated around the x axis I need to calculate the surface area obtained by rotating $\sin\pi x$, $0\le x \le 1$ about the x-axis. So the surface area equation i think i have to use is:
$$A = \int_0^1 2\pi y \sqrt{1+(dy/dx)^2} \, dx$$
so what I did so far is
$$A = \int_0^1 2\pi \sin{\pi ... | You're fine as far as $\tan{\theta} = \pi \cos{\pi x}$, so $\sec^2{\theta} \, d\theta = -\pi^2 \sin{\pi x}$. The limits require some more care: when $x=0$, $\tan{\theta} = \pi$, so we have the rather weird $\arctan{\pi}=P$, say, as the lower limit. $x=1$ gives $\theta=-P$ as the upper limit. Hence, switching the limits... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1230744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is there another way to solve $\int \frac{x}{\sqrt{2x-1}}dx$? $$\int \frac{x}{\sqrt{2x-1}}dx$$
Let $u=2x-1$
$du=2dx$
$$=\frac{1}{2}\int \frac{u+1}{2\sqrt{u}}du$$
$$=\frac{1}{2}\int (\frac{\sqrt{u}}{2}+\frac{1}{2\sqrt{u}})du$$
$$=\frac{1}{4}\int \sqrt{u}du+\frac{1}{4}\int\frac{1}{\sqrt{u}}du$$
$$=\frac{u^{\frac{3}{2}}}{... | While the approaches already posted work well, here is an approach that provides a straight forward way of deriving the result.
$$\begin{align}
\int \frac{x}{\sqrt{2x-1}} dx &= \frac12 \int \frac{2x}{\sqrt{2x-1}} \,\,dx\\\\
&=\frac12 \int \frac{2x-1+1}{\sqrt{2x-1}} \,\,dx\\\\
&=\frac12 \int \sqrt{2x-1}\,\,\,dx+\frac12\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1231507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How to prove this inequality? $(a+b+c=1)$ Show that if $a,b,c$ are positive reals and $a+b+c=1$, then the following must hold:
$$\frac{2(a^3+b^3+c^3)}{abc}+3 \geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$
What I have tried is using $abc \leq \frac{1}{27}$ $(a+b+c \geq 3\sqrt[3]{abc}) $ and multiplying everything by $abc$,... | if $a,b,c>0$ then we get
$$\frac{2(a^3+b^3+c^3)}{abc}+3\geq \frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}$$
this is equivalent to
$$\frac{2(a^3+b^3+c^3)}{abc}\geq \frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}$$
and this
$$2(a^3+b^3+c^3)\geq bc(b+c)+ac(a+c)+ab(a+b)$$
and now note that
$$a^3+b^3=(a+b)(a^2-ab+b^2)\geq ab(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1233153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Is $\mathbb{Q}(\sqrt{2},\sqrt[3]{5})=\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})$? I understand that $\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})\subseteq\mathbb{Q}(\sqrt{2},\sqrt[3]{5})$
But I am struggling to algebraically show that $\sqrt{2}$,$\sqrt[3]{5}\in\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})$ to conclude that $\mathbb{Q}(\sqrt{2},\sqrt[3]{... | Let $x=\sqrt2+\sqrt[3]5$ for simplicity. By manually expanding we get
\begin{array}{rrrrrrr}
1 = & 1 \\
x = & & 1\cdot 2^{1/2} & + 1\cdot 5^{1/3} \\
x^2 = & 2 & & & + 2\cdot 2^{1/2}5^{1/3} & +1\cdot 5^{2/3} \\
x^3 = & 5 & + 2\cdot 2^{1/2} & + 6\cdot 5^{1/3} & & & +3\cdot2^{1/2}5^{2/3} \\
x^4 = & 4 & + 20\cdot 2^{1/2} &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1234930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
Find the closed-form for $\sum_{i=0}^n(-1)^i(\frac{1}{2})^i$ I start with simplifying:
$$\sum_{i=0}^n(-1)^i(\frac{1}{2})^i=\sum_{i=0}^n(-\frac{1}{2})^i$$
then:
$$S = 1 + (-\frac{1}{2}) + (-\frac{1}{2})^2 + ... +(-\frac{1}{2})^n$$
$$(-\frac{1}{2})S = (-\frac{1}{2}) + (-\frac{1}{2})^2 + ... +(-\frac{1}{2})^n+(-\frac{1}{2... | Here's another approach using the formula for finite geometric series
\begin{align*}
\sum_{i=0}^nx^n=\frac{1-x^{n+1}}{1-x}\tag{1}
\end{align*}
We obtain
\begin{align*}
\sum_{i=0}^n(-1)^i\left(\frac{1}{2}\right)^i&=\sum_{i=0}^n\left(-\frac{1}{2}\right)^i\\
&=\frac{1-\left(-\frac{1}{2}\right)^{n+1}}{1-\left(-\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1242383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Algebraic Manipulation What is the best method to get the LHS equal to RHS?
$\frac{n(n+1)(n+2)}{3} + (n+1)(n+2) = \frac{(n+1)(n+2)(n+3)}{3}$
Thank you.
| Your equation is :
$$ \frac{n(n+1)(n+2)}{3} + (n+1)(n+2) = \frac{(n+1)(n+2)(n+3)}{3} \tag{$\star$}$$
Notice that the product $(n+1)(n+2)$ is common to all the terms in $(\star)$. If $n \in \mathbb{Z} \smallsetminus \left\{-2,-1 \right\}$, then $(n+1)(n+2) \neq 0$ and you can $\color{blue}{\text{simplify}}$ by $(n+1)(n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1243683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Determinant of a 5 × 5 matrix I have a little problem with a determinant.
Let $A = (a_{ij}) \in \mathbb{R}^{(n, n)}, n \ge 4$ with
$$a_{ij} =
\begin{cases}
x \quad \mbox{for } \,i = 2, \,\, j \ge 4,\\
d \quad \mbox{for } \,i \ge j, \\
0 \quad \mbox{else.}
\end{cases}$$
So for example, if we choose $n = 5$, the ... |
My first idea was to split this matrix into a product of a triangular matrix $T$ and a rest matrix $R$ so that $A=T⋅R$.
That's very much a way to do it. The technique is called LU Decomposition. It produces a lower and upper triangular matrix, allowing trivial determinate calculations. For this reason, you actually o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1244967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Prove equality of two numbers written in complex polar form. Show that these two numbers are equal:
$$
z_1=\frac{e^{\tfrac{2\pi i}{9}}-e^{\tfrac{5\pi i}{9}}}{1-e^{\tfrac{7\pi i}{9}}}
$$
and
$$z_2=\frac{e^{\tfrac{\pi i}{9}}-e^{\tfrac{3\pi i}{9}}}{1-e^{\tfrac{4\pi i}{9}}}=\frac{1}{e^{\tfrac{\pi i}{9}}+e^{\tfrac{-\pi i}{... | Both complex numbers are of the form:$$\frac{e^{ia}-e^{ib}}{1-e^{i(a+b)}}=\frac{e^{ia}-e^{ib}}{1-e^{i(a+b)}}\times\frac{1+e^{-i(a+b)}}{1+e^{-i(a+b)}}=\frac{(e^{ia}-e^{-ia})-(e^{ib}-e^{-ib})}{-e^{i(a+b)}+e^{-i(a+b)}}\tag{1}$$We know that:$$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$$$\therefore2i\sin(x)=e^{ix}-e^{-ix}\tag{2}$$U... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1245471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If $ \cos(\theta) = - \frac{2}{3} $ and $ 450^{\circ} < \theta < 540^{\circ} $, find... If $ \cos(\theta) = - \frac{2}{3} $ and $ 450^{\circ} < \theta < 540^{\circ} $, find:
*
*The exact value of $ \cos \! \left( \frac{1}{2} \theta \right) $.
*The exact value of $ \tan(2 \theta) $.
What I’ve tried:
*
*I took t... | You were given that $\cos\theta = -\dfrac{2}{3}$, with $450^\circ < \theta < 540^\circ$.
Recall that
$$\cos\left(\frac{\theta}{2}\right) = \pm\sqrt{\frac{1 + \cos\theta}{2}}$$
where the sign is determined by the measure of angle $\frac{\theta}{2}$. Since $450^\circ < \theta < 540^\circ$, we may conclude that $225^\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1247745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If there must be at least one person in each table, in how many ways can 6 people be seated in 3 tables?
If there must be at least one person in each table, in how many ways can 6 people be seated in 3 tables?
I know there are three possible ways to split the set of people P into three distinct groups ($1+1+4,1+2+3,2... | We consider the 3 cases mentioned above:
1) If they are split 4/1/1, there are $\dbinom{6}{4}$ ways to select the 4 people sitting together, and $3!$ ways to arrange them at their table, so this gives $\dbinom{6}{4}(3!)=15(6)=90$ possibilities.
2) If they are split 3/2/1, there are $\dbinom{6}{3}$ ways to select the 3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1248388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Finding $\int \frac{1+\sin x \cos x}{1-5\sin^2 x}dx$
Find $\int \frac{1+\sin x \cos x}{1-5\sin^2 x}dx$
I used a bit of trig identities to get: $\int \frac {2+\sin (2x)}{-4+\cos(2x)}dx$ and using the substitution: $t= \tan (2x)$ I got to a long partial fractions calculation which doesn't seem right.
Any hints on how t... | The denominator of your second integral should be $-3+5\cos 2x$, because
from the identities $\sin x\cos x=\frac{\sin 2x}{2}$ and $\sin ^{2}x=\frac{
1-\cos 2x}{2}$ we obtain
\begin{equation*}
\frac{1+\sin x\cos x}{1-5\sin ^{2}x}=\frac{2+\sin 2x}{-3+5\cos 2x}.
\end{equation*}
To evaluate
\begin{equation*}
\int \frac{2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1250020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to prove that $ \sin \angle{GAB}+\sin \angle{GBC}+\sin \angle{GCA} \le \frac{3}{2} $ for a triangle $ABC$ with centroid $G$? Let $ G $ be the centroid of $ \triangle ABC $ , such that $ \measuredangle{GAB}=x,\measuredangle{GBC}=y,\measuredangle{GCA}=z $.
How do I prove that :
$$ \sin x +\sin y +\sin z\le \frac{3... | Let $AA'$, $BB'$ and $CC'$ be medians of the triangle.
Thus, from $\Delta ABA'$ we obtain:
$$\frac{\frac{a}{2}}{\sin{x}}=\frac{AA'}{\sin\beta}$$ or
$$\sin{x}=\frac{a\sin\beta}{2AA'}$$ or
$$\sin{x}=\frac{a\cdot\frac{2S}{ac}}{\sqrt{2b^2+2c^2-a^2}}$$ or
$$\sin{x}=\frac{2S}{c\sqrt{2b^2+2c^2-a^2}}.$$
Thus, it remains to pro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1250884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 2,
"answer_id": 1
} |
Evaluate the limit $\displaystyle\lim_{x \to 0}\frac{(e-\left(1 + x\right)^{1/x})}{\tan x}$. How to evaluate the following limit
$$\displaystyle\lim_{x\to 0} \dfrac{e-\left(1 + x\right)^{1/x}}{\tan x}$$
I have tried to solve it using L-Hospital's Rule, but it creates utter mess.
Thanks for your generous help in advance... | This is more a comment than an answer since previous answers show how to obtain the limit.
In the same spirit as abel's anwer, let me use a few more terms (keeping abel's notation) $$\ln y = \frac 1 x\ln(1 + x) =1-\frac{x}{2}+\frac{x^2}{3}+O\left(x^3\right) $$ $$y=e-\frac{e x}{2}+\frac{11 e x^2}{24}+O\left(x^3\right)$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1252232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
A real $2 \times 2 $ matrix $M$ such that $M^2 = \tiny \begin{pmatrix} -1&0 \\ 0&-1-\epsilon \\ \end{pmatrix}$ , then :
A real $2 \times 2 $ matrix $M$ such that $$M^2 = \begin{pmatrix} -1&0 \\ 0&-1-\epsilon \\ \end{pmatrix}$$
(a) exists for all $\epsilon > 0$.
(b) does not exist for any $\epsilon > 0$.
(c) exists... | no matter if $\epsilon \neq 0$ is , there is no real matrix $M.$
for $\epsilon = 0, M = \pmatrix{0&b\\-\frac 1b&0}, b\neq 0$
pick a matrix $$M = \pmatrix{a&b\\c&d}, M^2 =\pmatrix{a^2 + bc&(a+d)b\\(a+d)c&bc+d^2}=\pmatrix{-1&0\\0&-1-\epsilon} $$ we have the constraints $$a^2 + bc = -1, bc+d^2 = -1-\epsilon \to a^2 - d^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1252848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Show that $\arctan x+\arctan y = \pi+\arctan\frac{x+y}{1-xy}$, if $xy\gt 1$ Show that $\arctan x+\arctan y = \pi+\arctan\frac{x+y}{1-xy}$, if $xy\gt 1$
I am stuck at understanding why the constraint $xy\gt 1$. Here is my work so far
let $\arctan x =a\implies x=\tan a$
let $\arctan y =b\implies y=\tan b$
therefore $\fr... | The reason is that $\frac{\pi}{2} < |a+b| < \pi$. $|a+b| < \pi$ is easy to see.
Notice that when $x>0, y>0$, $a+b$ reaches minimum when $xy=1$. Also, $\tan^{-1} \frac{1}{x} = \cot^{-1} x$, and $\tan^{-1} x + \tan^{-1} \frac{1}{x} = \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$.
Similarly, when $x<0, y<0$, we should have $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1255968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
Compute $\lim_\limits{n\to\infty}a_n$ where $a_{n+2}=\sqrt{a_n.a_{n+1}}$ I managed to show that the limit exists, but I don't know how to compute it.
EDIT:
There are initial terms: $a_1=1$ and $a_2=2$.
| We have $a_2\gt a_1\gt0$ and $a_{n+2}^2=a_{n+1}a_n$.
So,$$\begin{align}
&a_3^2=a_2a_1\\
\implies&a_3^2\gt a_1^2\\
\implies &a_3\gt a_1
\end{align}$$
and $$\begin{align}
&a_3^2=a_2a_1\\
\implies &a_3^2\lt a_2^2\\
\implies &a_3\lt a_2
\end{align}$$
So, we have $$\color{red}{a_2\gt a_3\gt a_1\gt 0}$$
Similarly, $$\begin{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1260341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Evaluating trigonometric limit: $\lim_{x \to 0} \frac{ x\tan 2x - 2x \tan x}{(1-\cos 2x)^2}$
Evaluate $\lim_{x \to 0} \cfrac{ x\tan 2x - 2x \tan x}{(1-\cos 2x)^2} $
This is what I've tried yet:
$$\begin{align} & \cfrac{x(\tan 2x - 2\tan x)}{4\sin^4 x} \\
=&\cfrac{x\left\{\left(\frac{2\tan x}{1-\tan^2 x} \right) - 2\... | Your solution looks good! Give yourself a pat on your back.
Another way is to expand the trigonometric polynomials using Taylor series, i.e.,
\begin{align}
\dfrac{x(\tan(2x)-2\tan(x))}{(1-\cos(2x))^2} & = x \cdot \dfrac{2x+ \dfrac{(2x)^3}3+ \mathcal{O}(x^5) - 2x - 2\cdot\dfrac{x^3}3 + \mathcal{O}(x^5)}{\left(1-\left(1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1263968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
How to prove that $\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\leq\frac{3\sqrt{3}}{4}$? Let $a,b,c>0: (a+b)(b+c)(c+a)=ab+bc+ca$. How to prove that $$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\leq\frac{3\sqrt{3}}{4}$$
| Let $x=\frac{a}{\sqrt{(a+b)(a+c)}}$ $y=\frac{b}{\sqrt{(b+a)(b+c)}}$ and $z=\frac{c}{\sqrt{(c+a)(c+b)}}$.
Then we have $x^{2}+y^{2}+z^{2}+2xyz=1$. Note that for angles $A$, $B$ and $C$ in some acute triangle:$\cos^{2}A+\cos^{2}B+\cos^{2}C+2\cos A\cos B\cos C=1$. Hence $x$,$y$ and $z$ can be replaced by $\cos A$, $\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1264369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 0
} |
find coefficient of $x^{50}$ Let $f(x)=\frac{1}{(1+x)(1+x^2)(1+x^4)}$then find the coefficient of term $x^{50}$ in $(f(x))^3$.I think that we can set $$(f(x))^3=\frac{a}{(1+x)^3}+\frac{b}{(1+x^2)^3}+\frac{c}{(1+x^4)^3}$$ and find a,b and c then use Taylor seri .
| It won't work like that: $f(x)^3 = \dfrac{1}{(1+x)^3 (1+x^2)^3 (1+x^4)^3}$, and then the partial-fraction expansion will include terms in $1/(1+x)$, $1/(1+x)^2$, ...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1267677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Induction Clarification I had this problem: Is it always necessary to go from n to (n + 1) or from (n - 1) to n in the inductive hypothesis? Is the "direction" always important?
Here is my solution to one such proof, which turns out to be totally wrong. Is it really so?
$\underline{\textrm{BS}}.$ for n = 1
$\binom{2... | Yes, this is wrong. The idea of induction is like a row of dominos. If the first falls, then the one behind must fall to. But what you prove is that if the second falls then the first falls. But you can't prove that the second falls.
There is a variant of induction that can be used:
*
*Prove $P(1)$.
*Prove that if ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1267938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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According to Buckingham Theorem the rank of $A$ should be $2$ A physical system is described by a law of the form $f(E,P,A)=0$, where $E,P,A$ represent, respectively, energy, pressure and surface area. Find an equivalent physical law that relates suitable dimensionless quantities.
That' what I have tried so far:
1st st... | You don't need to do $A^n$.
You can see the 1st and 2nd row are multiple of each other, so one of them can be eliminated to $0$. That makes the rank $2$.
Alternatively, switch the 1st row and last row to make it
$$\pmatrix{2&-1&2\\-2&-2&0\\1&1&0}$$
Then divide the second row by $-2$, and subtract from the last row the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1268776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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compute improper integrals using integration by parts Compute
\begin{equation*}
\int_0^\infty \frac{\sin^4(x)}{x^2}~dx\text{ and }\int_0^\infty \frac{\sin (ax) \cos (bx)}{x}~dx.
\end{equation*}
For the first integral I tried letting $u = \sin ^4 x$ and $dv= \frac{1}{x^2}~dx$, which simplified to $\int_0 ^\infty \frac{... | For the first one, we have
\begin{align}
\int_0^{\infty} \dfrac{\sin^4(x)}{x^2}dx & = \int_0^{\infty} \dfrac{\sin^2(x)(1-\cos^2(x))}{x^2}dx = \int_0^{\infty} \dfrac{\sin^2(x)}{x^2}dx - \int_0^{\infty} \dfrac{\sin^2(x)\cos^2(x)}{x^2}dx\\
& = \int_0^{\infty} \text{sinc}^2(x) dx - \int_0^{\infty} \dfrac{\sin^2(2x)}{(2x)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1270106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Determining all the positive integers $n$ such that $n^4+n^3+n^2+n+1$ is a perfect square. I successfully thought of bounding our expression examining consecutive squares that attain values close to it, and this led to the solution I'll post as an answer, which was the one reported. However, before that, I had briefly ... | For all positive $n$ we have $$\require\cancel \left(n^2+\frac{n}{2}\right)^2=\cancel{n^4+n^3}+\frac{n^2}{4}<\cancel{n^4+n^3}+n^2+n+1 \\ \frac{n^2}{4}<n^2+n+1 $$ and on the other hand $$\require\cancel \left(n^2+\frac{n+2}{2}\right)^2=n^4+n^3+2n^2+\frac{n^2+4n+4}{4}>n^4+n^3+n^2+n+1 \\ \cancel{n^4+n^3}+\frac{9}{4}n^2\ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1270599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
For which values of $a,b,c$ is the matrix $A$ invertible? $A=\begin{pmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{pmatrix}$
$$\Rightarrow\det(A)=\begin{vmatrix}b&c\\b^2&c^2\end{vmatrix}-\begin{vmatrix}a&c\\a^2&c^2\end{vmatrix}+\begin{vmatrix}a&b\\a^2&b^2\end{vmatrix}\\=ab^2-a^2b-ac^2+a^2c+bc^2-b^2c\\=a^2(c-b)+b^2(a-c)+c^2(b-a)... | You can rewrite the determinant as $(b-a)(c-a)(c-b)$, so $\det(A)$ is nonzero precisely when $a,b,c$ are all distinct (note that one of $a,b,c$ can be zero).
Your matrix is the $3\times 3$ case of the Vandermonde matrix, whose determinant is generally a similar product of differences.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1270699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 4
} |
How do you add two series together How do you add the series
$$\frac{1}{2}\left(\sum_{n=0}^{\infty}\frac{2^{n}}{(z-3)^{n+1}} + \sum_{n=0}^{\infty}\frac{(z-3)^{n}}{4^{n+1}}\right)$$
?
is this right?
$$\begin{aligned}
&\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{2^{n}}{(z-3)^{n+1}} + \frac{(z-3)^{n}}{4^{n+1}}\right)\\
=\... | Well, you're doing fine so far. You can do a bit better by noting that $$4^{n+1}=\left(2^2\right)^{n+1}=2^{2(n+1)}=2^{2n+2}$$ and that $$4^{n+1}\cdot(z-3)^{n+1}=\bigl(4(z-3)\bigr)^{n+1}=(4z-12)^{n+1}.$$
You could also distribute the $\frac12$ through, if you like.
I do have to wonder what you gain from combining these ... | {
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"question_score": "2",
"answer_count": 3,
"answer_id": 0
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The value of ${\sum_{k=0}^{20}}(-1)^k\binom{30}{k}\binom{30}{k+10}$ $\newcommand{\b}[1]{\left(#1\right)}
\newcommand{\c}[1]{{}^{30}{\mathbb C}_{#1}}
\newcommand{\r}[1]{\frac1{x^{#1}}}$
The value of $$\sum_{k=0}^{20}(-1)^k\binom{30}{k}\binom{30}{k+10}$$
It is also the coefficient of $x^{10}$ in:
$$\b{\c0\r0-\c1\r1+\c2\r... | Vandermonde's Identity yields
$$
\begin{align}
\sum_{k=0}^{20}\binom{30}{k}\binom{30}{k+10}
&=\sum_{k=0}^{20}\binom{30}{k}\binom{30}{20-k}\\[6pt]
&=\binom{60}{20}
\end{align}
$$
In the product,
$$
\begin{align}
\left(x-\frac1x\right)^{30}
&=\sum_{k=0}^{30}(-1)^k\binom{30}{k}x^{30-k}x^{-k}\\
&=\sum_{k=0}^{30}(-1)^k\bin... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Vandermonde determinant and linearly independent Let $a_1,a_2,a_3,b_1,b_2,b_3,b_4,b_5,b_6\in \mathbb{C}$ such that $a_i\not=a_j$ for all $i\not=j.$
If $$\begin{vmatrix}
a_1 & a_2& a_3 & b_1 \\
a_1^2 & a_2^{2} & a_3^{2} & b_2\\
a_1^3 & a_2^{3} & a_3^{3} & b_3\\
a_1^4 & a_2^{4} & a_3^{4} & b_4\\
\end{vmatrix} =0,$$
... | Suppose that none of $a_i$ is zero. Otherwise the proof is obvoius.
Let $x = a_1, y = a_2, z = a_3$.
$$
\begin{vmatrix}
x^k & y^k & z^k & b_k\\
x^{k+1} & y^{k+1} & z^{k+1} & b_{k+1}\\
x^{k+2} & y^{k+2} & z^{k+2} & b_{k+2}\\
x^{k+3} & y^{k+3} & z^{k+3} & b_{k+3}
\end{vmatrix} = x^k y^k z^k
\begin{vmatrix}
1 & 1 & 1 & b... | {
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"answer_count": 1,
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Find numbers $a, b, c$ given that $a+b+c=12$, $a^2+b^2+c^2=50$, and $a^3+b^3+c^3=168$
Let $a+b+c=12$, $a^2+b^2+c^2=50$, and $a^3+b^3+c^3=168$. Find $a,b,c$
Suppose $a, b, c$ are roots of $P(x)$.
$$P(x) = k(x - a)(x - b)(x - c)$$
But then I get $(k = 1)$
$$P(x) = x^3 - 12x^2 + x(ab + ac + bc) - abc$$
Cant go further.... | \begin{eqnarray}ab+bc+ca=\frac{1}{2}((a+b+c)^2-(a^2+b^2+c^2))=47\end{eqnarray}
\begin{align*}a^3+b^3+c^3-3abc&=(a+b+c)(a^2+b^2+c^2-(ab+bc+ca))\\ abc&=44\end{align*}
So $a, b, c$ are roots of the polynomial $x^3-12x^2+47x-44=0$.
| {
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Integrate $\int \frac{x^5 dx}{\sqrt{1+x^3}}$ I took $1+x^3$ as $t^2$ . I also split $x^5$ as $x^2 .x^3$ . Then I subsituted the differentiated value in in $x^2$ . I put $x^3$ as $1- t^2$ . I am getting the last step as $2/9[\sqrt{1+x^3}x^3 ]$ but this is the wrong answer , i should get $2/9[\sqrt{1+x^3}(x^3 +2)]$.
Plea... | It looks like you've only made a minor mistake, but since I couldn't tell exactly where it happened, I have provided a step-by-step solution starting with the substitution you chose.
If $t^2=1+x^3$, then $2t\,dt=3x^2\,dx$, so
\begin{align*}
\int \frac{x^5\,dx}{\sqrt{1+x^3}}&=\int \frac{x^3\cdot 3x^2\,dx}{3\sqrt{1+x^3}}... | {
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"answer_id": 0
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Quadratic solutions puzzle The equation $x^2+ax+b=0$, where $a\neq b$, has solutions $x=a$ and $x=b$. How many such equations are there?
I'm getting $1$ equation as I can only find $a=b=0$ as an equation, which is not allowed.
$$x=\frac{±\sqrt{a^2-4 b}-a}2$$
$x=a$ or $b$ so these are the equations
$$a=\frac{\sqrt{a^2-4... | Let's take $a=\frac{\sqrt{a^2-4 b}-a}2$ for example:
$$a=\frac{\sqrt{a^2-4 b}-a}2$$
$$2a=\sqrt{a^2-4 b}-a$$
$$3a=\sqrt{a^2-4 b}$$
$$9a^2=a^2-4 b$$
$$8a^2=-4 b$$
$$b=-2 a^2$$
Now you can take that $b$ and substitute in $b=\frac{\sqrt{a^2-4 b}-a}2$ to find a.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Given $a,b,c,d>0$ and $a^2+b^2+c^2+d^2=1$, prove $a+b+c+d\ge a^3+b^3+c^3+d^3+ab+ac+ad+bc+bd+cd$
Given $a,b,c,d>0$ and $a^2+b^2+c^2+d^2=1$, prove $$a+b+c+d\ge a^3+b^3+c^3+d^3+ab+ac+ad+bc+bd+cd$$
The inequality can be written in the condensed form
$$\sum\limits_{Sym}a\ge\sum\limits_{Sym}a^3+\sum\limits_{Sym}ab$$
I was... | Let $a+b+c+d=4u$, $ab+ac+bc+ad+bd+cd=6v^2$ and $abc+abd+acd+bcd=4w^3$.
Hence, $16u^2-12v^2=1$ and our inequality is equivalent to $3v^6-4uv^2w^3+w^6\geq0$.
By Roll's theorem there are $x>0$, $y>0$ and $z>0$, for which
$x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
After this substitution we need to prove that
$\sum\limits... | {
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"question_score": "5",
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find the solutions to the equation $4\sin^2\theta + 1 = 6\sin\theta$ in the interval $0^\circ \leq \theta < 360^\circ$ Find the solutions to the following equation for $0^\circ \leq θ < 360^\circ$:
$$4\sin^2 θ + 1 = 6\sin θ$$
My work:
$$4\sin^2\theta - 6\sin\theta + 1 = 0$$
Factor
$$\sin\theta= \frac{1}{4}(3+ \sqrt{5})... | The domain of arcsin is $[-1,1]$, the co-domain is $[-\pi/2, \pi/2]$
So you have only the solution $a = arcsin((3-\sqrt5)/4)$. That's because you cannot take arcsin from $(3+\sqrt5)/4$, as it's not in the domain.
Then ... given that your $sin$ value $(3-\sqrt5)/4$ is in $(0, 1)$, you conclude that $0 < a < \pi/2$. Fro... | {
"language": "en",
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"source": "stackexchange",
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how to evaluate the product $\prod _{n=2}^\infty (1+ \frac{1}{n^2}+\frac{1}{n^4}+\frac{1}{n^6}+\cdots )$? Evaluating the infinite product of $\prod _{n=2}^\infty (1+ \frac{1}{n^2}+\frac{1}{n^4}+\frac{1}{n^6}+\cdots )$. Please Help.
| The product is
$$\prod_{n=2}^{\infty} \frac1{1-\frac1{n^2}} = \prod_{n=2}^{\infty} \frac{n^2}{n^2-1} = \prod_{n=2}^{\infty} \frac{n^2}{(n-1)(n+1)} = \frac{2 \cdot 2}{1 \cdot 3}\cdot \frac{3 \cdot 3}{2 \cdot 4}\cdot \frac{4 \cdot 4}{3 \cdot 5} \cdots = 2$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Closed forms for definite integrals involving error functions I have been working for a while with these kinds of integrals
$$\int_0^\infty dx\,\text{erfc}\left(c +i x\right)\exp \left(-\frac{1}{2}d^2x^2+i cx\right)$$
$$\int_\Lambda^\infty dx\,\frac{1}{x}\text{erfc}\left(c +i x\right) \exp \left(-\frac{1}{2}d^2x^2+i cx... | If $u =\operatorname{erfc}(c+ix)$ then
$$du = \dfrac {-2} {\sqrt\pi} e^{-(c+ix)^2} i \, dx = \dfrac {-2} {\sqrt\pi} e^{-(c^2-2icx - x^2)^2} i \, dx. \tag 1$$
We need to work with
$$
\exp\left( \frac {-1}2 d^2x^2 + icx\right).
$$
The exponent is
$$
\frac{-d^2} 2 \left( x^2 - \frac{2icx}{d^2} - \frac{c^2}{d^4} \right) - ... | {
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"source": "stackexchange",
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Find the 2x2 matrix given 2 equations. Find the $2 \times 2$ matrix ${A}$ such that ${A}^2 = {A}$ and
${A} \begin{pmatrix} 7 \\ -1 \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \end{pmatrix}.$
I have tried to express A as a matrix with variables a,b,c, and d, but it gets too messy with the equations. Any help is appreciated... | Hint: we have
$$A\begin{pmatrix} 7\\-1\end{pmatrix}=\begin{pmatrix} 6\\2\end{pmatrix}$$
And therefore
$$\begin{align}A\begin{pmatrix} 6\\2\end{pmatrix}
&=A^2\begin{pmatrix} 7\\-1\end{pmatrix}\\
&=A\begin{pmatrix} 7\\-1\end{pmatrix}\\
&=\begin{pmatrix} 6\\2\end{pmatrix}
\end{align}$$
So if $A=\begin{pmatrix} a&b\\c&d\en... | {
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Another combined limit I've tried to get rid of those logarithms, but still, no result has came.
$$\lim_{x\to 0 x \gt 0} \frac{\ln(x+ \sqrt{x^2+1})}{\ln{(\cos{x})}}$$
Please help
| Using $\lim\limits_{x\to0}\frac{\log(1+x)}x=1$ and $\lim\limits_{x\to0}\frac{\sin(x)}x=1$,
$$
\begin{align}
&\lim_{x\to0^+}\frac{\log\left(x+\sqrt{x^2+1}\right)}{\log(\cos(x))}\\
&=\lim_{x\to0^+}\frac{\log\left(1+x+\sqrt{x^2+1}-1\right)}{\log(1+\cos(x)-1)}\\
&=\lim_{x\to0^+}\frac{\log\left(1+x+\frac{x^2}{\sqrt{x^2+1}+1... | {
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"answer_count": 6,
"answer_id": 4
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show that $\frac{1}{F_{1}}+\frac{2}{F_{2}}+\cdots+\frac{n}{F_{n}}<13$ Let $F_{n}$ is Fibonacci number,ie.($F_{n}=F_{n-1}+F_{n-2},F_{1}=F_{2}=1$)
show that
$$\dfrac{1}{F_{1}}+\dfrac{2}{F_{2}}+\cdots+\dfrac{n}{F_{n}}<13$$
if we use
Closed-form expression
$$F_{n}=\dfrac{1}{\sqrt{5}}\left(\left(\dfrac{1+\sqrt{5}}{2}\right... | Clearly: $F_n > F_{n-1}$, for $n \geq 3$
Thus, $F_n = F_{n-1} + F_{n-2} > 2*F_{n-2} > 2^2*F_{n-4}>...>2^{(n-3)/2}*F_3 = 2^{(n-1)/2}$
Now, your inequality can be written as: $\sum\limits_{i=3}^n \frac{i}{F_i} < 10$
The $LHS < \sum\limits_{i=3}^n \frac{i}{2^{(i-1)/2}} = 2^{1/2}*\sum\limits_{i=3}^n \frac{i}{2^{i/2}} < 2^{... | {
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"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 4
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Similar matrix for two projections Let there be two matrices $A = \frac{1}{2} \begin{pmatrix}
1 & 1\\
1 & 1\\
\end{pmatrix}$ and $B = \frac{1}{4} \begin{pmatrix}
2 & 4\\
1 & 2\\ \end{pmatrix}$ over the field $\mathbb{F}=\mathbb{Q}$. Find a invertible matrix $S$, s.t. $S^{-1} A S ... | An example is $S= \left( \begin{array}{cc} 1 & 0\\ 0&2 \end{array} \right)$.
EDIT: A way to find this matrix is: Take $S= \left( \begin{array}{cc} a & 0\\ b&c \end{array} \right)$. It's very important to take that 0 for simplify your work. (Generally, when you want to find a matrix with some properties, try to take as ... | {
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"question_score": "2",
"answer_count": 1,
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How many possible words of this type can be formed? We are making $10$ letter words using the letters $A,C,G,T$.
How many possible words are there of the form $A...AC...CG...GT...T$
This is where all of the $A's$ go before the all of the $C's$ and all of the $C's$ go before all of the $G's$ and all of the $G's$ go bef... | This problem is equivalent to finding the number of ways to partition $10$ into four parts, in which the order of the partitioning matters. For example, $10 = 2 + 3 + 3 + 2$ is not the same as $10 = 3 + 2 + 2 + 3$. The first number represents the number of A's, the second number represents the number of C's, and so on.... | {
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Evaluate $7^{8^9}\mod 100$ I'm preparing myself for discrete math exam and here's one of the preparation problems:
Evaluate $$7^{8^9}\mod 100$$
Here's my solution:
$7^2\equiv49 \mod 100\implies (7^2)^2\equiv49^2=2401\equiv 1\mod 100$
So, as it turns out $7^4\equiv 1\mod 100$. It will be useful later. Now let's examine ... | $7^2\equiv -1\mod 50$, hence $7^4\equiv 1\mod 100$.
Thus $\,7^{8^9}=\bigl(7^8\bigr)^{8^8}=\bigl((7^4)^2\bigr)^{8^8}\equiv 1\mod 100$.
| {
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"source": "stackexchange",
"question_score": "2",
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Simplify the expression and find the minimum value I want to simplify the expression
\begin{equation}
A = \frac{\sqrt{1-\sqrt{1-x^2}}\Big[\sqrt{(1+x)^3} + \sqrt{(1-x)^3} \Big]}{2-\sqrt{1-\sqrt{1-x^2}}}
\end{equation}
and find the minimum value of $\displaystyle \sqrt{2}A+\frac{1}{x}$.
I can only simplify to this point... | Try this: $u = \sqrt{1+x}, v = \sqrt{1-x} \to \sqrt{1-x^2} = uv, 2 = u^2+v^2$, and consider $f(u,v) = ...$
| {
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"source": "stackexchange",
"question_score": "1",
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Let $X$ be a set of primes $p$ so that $5^{p^2}+1 \equiv 0 \pmod {p^2}$ Which of these sets is $X$ equal to? $5^{p^2}+1\equiv 0\pmod {p^2}$
$1.$ $\emptyset $
$2.$ {$3$}
$3.$ All primes of the form $4k+3$
$4.$ All primes except $2$ and $5$
$5.$ All primes
This one is pretty easy to get right through the process of elimi... | Clearly $p\neq 2$, because $5^{2^2}+1\equiv 1^{2^2}+1\equiv 2\not\equiv 0\pmod{\! 2^2}$.
$\ p\mid 5^{p^2}+1\,\Rightarrow\, p\mid 5^{2p^2}-1$ and by little Fermat $p\mid 5^{p-1}-1$.
See this lemma to get $\,p\mid 5^{(2p^2,\,p-1)}-1=5^2-1=2^3\cdot 3\,\Rightarrow\, p=3$
$5^{3^2}+1\equiv 125^3+1\equiv (-1)^3+1\equ... | {
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"source": "stackexchange",
"question_score": "3",
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Finding the area under $\frac{3}{2x+1}$ and $3x-2$ Find the area between these curves.
$$y=\dfrac{3}{2x+1},\qquad y=3x-2;\qquad x=2\quad \text{et} \quad y=0 $$
Indeed,
I calculate the integral of the blue function between $1$ and $2$. Then, I will calculate the area of the triangle between the yellow line and the x-a... | As mentioned in the comments, you are almost correct.
But your first integral should start at $2/3$, not $3/2$. So you should have
$$\int _{\frac{2}{3}}^1\left(3x-2\right)dx+\int _1^2 \frac{3}{2x+1} dx = \frac{3}{2}\ln \left(\frac{5}{3}\right) + \frac{1}{6}.$$
As a sanity check, this makes much more sense for a few rea... | {
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If $a_1,a_2,a_3$ are roots $x^3+7x^2-8x+3,$ find the polynomial with roots $a_1^2,a_2^2,a_3^2$ If $a_1,a_2,a_3$ are the roots of the cubic $x^3+7x^2-8x +3,$ find the cubic polynomial whose roots are:
$a_1^2,a_2^2,a_3^2$ and the polynomial whose roots are $\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}.$ Not really sure wh... | Suppose $x^3+7x^2-8x+3=(x-a_1)(x-a_2)(x-a_3)$. Using difference of squares,
$$
\begin{align*}
(x^2-a_1^2)(x^2-a_2^2)(x^2-a_3^2)&=\left[(x-a_1)(x-a_2)(x-a_3)\right]\left[(x+a_1)(x+a_2)(x+a_3)\right]\\
&= \left[x^3+7x^2-8x+3\right]\left[x^3-7x^2-8x-3\right]\\
&=x^6-65 x^4+22 x^2-9.
\end{align*}
$$
Therefore $(x-a_1^2)(x-... | {
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"source": "stackexchange",
"question_score": "2",
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Solve the non-linear system of equations For real $x,y,z>0$ solve the system of equation
\begin{cases}
\dfrac{1}{x}-3 y+4 z=5,\\
\dfrac{1}{y}-4 z+5 x=3,\\
\dfrac{1}{z}-5 x+3 y=4,
\end{cases}
It is easy to check out that
$$
x =\frac 1 5, y= \frac 1 3, z=\frac 1 4
$$
is a solution.
How I could prove that th... | Multiply the first equation by $5x$, the second by $3y$, the third by $4z$. Then add all three up. This gives you $25x + 9y + 16z = 12$. Now this is clearly a useful intermediate result. For example, you can substitute it into the original equations to eliminate one variable. This yields six new relations:
[1] $4/x -25... | {
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"question_score": "2",
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Decompose $\frac{x^2-2x+3}{(x-1)^2(x^2+4)}$, partial fraction braindead decompose $\frac{x^2-2x+3}{(x-1)^2(x^2+4)}$
the way my teacher wants us to solve is by substitution values for x,
I set it up like this:
(after setting the variables to the common denominator and getting rid of the denominator in the original equat... | Hint : Start with $$\frac{x^2-2x+3}{(x-1)^2(x^2+4)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{CX+D}{x^2+4}$$
Multiply with $(x-1)^2(x^2+4)$ and then insert the special values.
| {
"language": "en",
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"source": "stackexchange",
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Trigonometry Identity proving If $\sin(x-y) =\cos y$ prove that $\tan y = \frac{1+ \sin y}{\cos y} $.
Is there an error with the question? I don't seem to be able to get the answer. Should it be $\tan x$ instead of $\tan y = \frac{1+\sin y}{\cos y}$ ?
$$\tan x\times \cos y = 1 + \sin y $$
$$\frac{\sin x}{\cos x} \time... | HINT:
$$\cos y=\sin(x-y)=\sin x\cos y-\cos x\sin y$$
$$\iff\cos x\sin y=\cos y(\sin x-1)\implies\dfrac{\sin y}{\cos y}=\dfrac{\sin x-1}{\cos x}$$
Now $\cos^2x=(1-\sin x)(1+\sin x)\implies\dfrac{1-\sin x}{\cos x}=\dfrac{\cos x}{1+\sin x}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1302239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is this entity with operators correct? Let define the operators $A = \frac{1}{\sqrt{2}}(x+\partial_x)$ and $B = \frac{1}{\sqrt{2}}(x-\partial_x)$. I am suppossed to check the identity $AB-BA=1$ but I cannot proof it. Is the identity correct?
My attempt:
$$AB = \frac{x}{\sqrt{2}}\left(\frac{1}{\sqrt{2}}\left(x-\partial_... | The identity is correct. The problem is $\partial_x\cdot x$ is not equal to $1$. Think about it this way:
$$(\partial_x\cdot x)(f) =\partial_x(xf)=f+x\partial_x f$$
So the operator $\partial_x\cdot x=1+x\partial_x$. If you do this for both $AB$ and $BA$, you will get the required identity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1302411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
First order differential equation: did i solve this equation right So i'm trying to solve:
$$x^2\frac{dy}{dx} + 2xy = y^3$$
I'm given this differential equation, that Bernoulli equation:
$$\frac{dy}{dx} + p(x)y = q(x)y^{n} $$
I think i've solved it and got
$$ u = \frac{2}{5x} +Cx^4$$
I'm just not sure i am right i... | After correction, the solution to the initial differential equation is
$$y(x) = \pm \sqrt{ \frac{5x} {5Cx^5 + 2} }$$
Regarding the $u$ transformation: from the B.E. we know (1) $y' = q y^n - p y$. From the $u$ definition we know that (2) $u' = (1-n) y^{-n} y'$. Substitute $y'$ from (1) into (2) to recover the new ODE f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1302602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Here are two fractions, $\frac{2}{3}$,$\frac{7}{8}$, which of these fractions are closer to $\frac{3}{4}$? I've been throwing this question around my family. No one has a clue, therefore can someone help?
I'm pretty sure this will be easy to do
| The distance between real numbers $x$ and $y$ is $|x-y|$. So the distance from $\frac{2}{3}$ to $\frac{3}{4}$ is
\begin{align*}
\Big|\frac{2}{3}-\frac{3}{4}\Big|=\Big|\frac{8}{12}-\frac{9}{12}\Big|=\frac{1}{12}.
\end{align*}
And the distance from $\frac{7}{8}$ to $\frac{3}{4}$ is
\begin{align*}
\Big|\frac{7}{8}-\frac{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1302968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solution to $y'=y^2-4$ I recognize this as a separable differential equation and receive the expression:
$\frac{dy}{y^2-4}=dx$
The issue comes about when evaluating the left hand side integral:
$\frac{dy}{y^2-4}$
I attempt to do this integral through partial fraction decomposition using the following logic:
$\frac{1}{(... | $$
\begin{align}
x
&=\int\frac{\,\mathrm{d}y}{y^2-4}\\
&=\frac14\int\left(\frac1{y-2}-\frac1{y+2}\right)\,\mathrm{d}y\\
&=\frac14\log\left(\frac{y-2}{y+2}\right)+C
\end{align}
$$
Therefore,
$$
\frac{y-2}{y+2}=ke^{4x}
$$
or, solving for $y$,
$$
y=2\,\frac{1+ke^{4x}}{1-ke^{4x}}
$$
We get your form of the answer by lettin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1304633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Rolling two dice, what is the probability of getting 6 on either of them, but not both? Rolling two dice, what is the probability of getting 6 on one of them, but not both?
| Total outcomes: $6^2 = 36$.
Allowed outcomes is the complementary of unallowed outcomes which can be split into "both dice < 6" ( case 1 ) "and" "not both dice = 6" ( case 2 ):
$6^2 - 5^2 - 1^2 = 36-25-1 = 10$. We can write this as a binary number matrix with the indices equal to the dice results:
$\left(\begin{array}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1305186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Prove bounds of a strictly increasing sequence using integrals to approximate For the strictly increasing sequence $ \ x_n = \frac1{1^2} + \frac1{2^2} + \frac1{3^2} +\cdots+\frac1{n^2},$ for $n\ge1$.
(a) Prove the sequence is bounded above by $2$; deduce that is has a limit $L\le2$.
(b) Prove that the "tail" sequence $... | For the upper bound, notice that
$$1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} + \frac{1}{49} + \ldots < 1 + \frac{1}{4} + \frac{1}{4} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \ldots = \\ 1 + \frac{1}{4} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \ld... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1305582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Changing-sided dice probability problem. Suppose you roll a fair $6$-sided dice, and that the number you roll is $m$.
If $m=1$, stop. Otherwise, roll an $m$-sided dice. The number you roll is $n$. If $n=1$, stop. Otherwise roll an $n$-sided dice... etc.
What is the probability it will take exactly $x$ rolls to roll a 1... | This looks like a classic Markov Chain problem.
There are 6 states (6,5,4,3,2,1), and you start at 6 and end at 1. The transition matrix is:
$P=\left(\begin{array}{cccccc}
\frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6}\\
0 & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
An estimate for dyadic numbers I would like to prove that for some positive $\delta<1/2$ we have the following inequalities
$$ |\frac{1}{2^{n+1}} - \frac{1}{2^{m+1}}| \leqslant \delta\left( |\frac{1}{2^{n}} - \frac{1}{2^{m+1}}| + |\frac{1}{2^{m}} - \frac{1}{2^{n+1}}|\right)\qquad (m,n\in \mathbb{N})$$
$$ |\frac{1}{2^{n... | you can take $\delta=\frac{1}{3}$.
for any $a,b$ we have
$$\left|a-b\right|=\frac{1}{3}\left|3a-3b\right|=\frac{1}{3}\left|\left(2a-b\right)+\left(a-2b\right)\right|$$
now using triangle inequality and noting that $\left|a-2b\right|=\left|2b-a\right|$ we get
$$\le\frac{1}{3}\left(\left|2a-b\right|+\left|2b-a\right|\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1307000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving $\cos x+\sin x-1=0$ How does one solve this equation?
$$\cos {x}+\sin {x}-1=0$$
I have no idea how to start it.
Can anyone give me some hints? Is there an identity for $\cos{x}+\sin{x}$?
Thanks in advance!
| As $\sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\left(\cos x+\sin x\right)$ by the angle addition formula we find that:
\begin{equation}
\begin{aligned}
\cos x+\sin x-1&=0\\
\implies\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)&=1\\
\implies\sin\left(x+\frac{\pi}{4}\right)&=\frac{\sqrt{2}}{2}\\
\implies x+\frac{\pi}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1307085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 0
} |
$\sin(2x)-\cos(2x)-\sin(x)+\cos(x)=0$ I would like to share a trigonometry question here. Wonder is there another way to solve it or not.
$\sin(2x)-\cos(2x)-\sin(x)+\cos(x)=0$
$2\cos(x)\sin(x)-(1-2\sin^2(x))-\sin(x)+\cos(x)=0$
$2\cos(x)\sin(x)-1+2\sin^2(x)-\sin(x)+\cos(x)=0$
$2\sin^2(x)+2\cos(x)\sin(x)-\sin(x)+\cos(x)-... | If $\sin2A-\sin2B=\cos2A-\cos2B$
using Prosthaphaeresis Formulas, $$2\sin(A-B)\cos(A+B)=-2\sin(A-B)\sin(A+B)$$
$$2\sin(A-B)[\cos(A+B)+\sin(A+B)]=0$$
$\sin(A-B)=0\implies A-B=n\pi$ where $n$ is any integer
$\cos(A+B)+\sin(A+B)=0\iff\tan(A+B)=-1=\tan\left(-\dfrac\pi4\right)$
$A+B=m\pi-\dfrac\pi4$ where $m$ is any intege... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1307551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evalute big determinant Today in exam I tried to evaluate this determinant but failed, only somehow "guessed" the answer I got here. Now in home I've managed to find something intuitive, just want to know whether the approach is correct, and is there more faster way exist. Given determinant
$$\det\begin{vmatrix}
1 & 2 ... | First we can subtract from each row the following row
$$D=
\begin{vmatrix}
1 & 2 & 3 & \ldots & n-2 & n-1 & n\\
2 & 3 & 4 & \ldots & n-1 & n & n\\
3 & 4 & 5 & \ldots & n & n & n \\
\ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\
n & n & n & \ldots & n & n & n
\end{vmatrix}
=
\begin{vmatrix}
-1&-1 &-1 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1307884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Double integral over trapezoid Compute $$\iint_D \frac{1}{(1+(x+2y)^2)^2} \,dx\,dy$$ where $D$ is given by $x \geq 0 , \, y \geq 0, \, 1 \leq x+2y \leq 2 \\$.
I am supposed to solve it with the help of contour lines. By drawing $D$ we get a trapezoid with the corners $(1,0), \, (2,0), \, (0, 1/2), \, (0,1)$. A contour... | The height of the trapezoid is incorrect. Consider the line $y=2x$ which is perpendicular to the contour lines. This line intersects the contour $f(x,y) = \lambda$ at $\left(\frac{\lambda}{5} , \frac{2\lambda}{5}\right)$. So, the height is given by
$$\sqrt{\left(\frac{\lambda}{5} - \frac{1}{5}\right)^2 + \left( \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1308506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Is there a proof show that : $\cos(z)$ and $\sin(z)$ are images of unbounded functions? if we knew that :cos and sin are bounded function $\mathbb{R}$ for any real
number $x$ .
let $z $ be a complex variable , Is there a proof show that :
$\cos(z)$ and $\sin(z)$ are images of unbounded functions ?
Any kind of help is... | In the spirit of giving an explicit computation accessible to good high-school students:
If one accepts that
$$
\left.
\begin{gathered}
\exp(iz) = \cos z + i\sin z, \\
\exp(z + w) = \exp(z) \exp(w),
\end{gathered}\right\}\quad\text{for all complex $z$ and $w$,}
\tag{1}
$$
then unboundedness of both $\cos$ and $\sin$ al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1309043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Finding $\lim\limits_{n\to\infty }\frac{1+\frac12+\frac13+\cdots+\frac1n}{1+\frac13+\frac15+\cdots+\frac1{2n+1}}$ I need to compute: $\displaystyle\lim_{n\rightarrow \infty }\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n}}{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots+\frac{1}{2n+1}}$.
My Attempt: $\dis... | With the Euler-MacLaurin formula, we have
numerator = $\ln(n)+O(1)$ and denominator = $\frac12 \ln(2n+1) + O(1)$, so the limit is
$\lim_{n \to \infty}\frac{\ln(n)+O(1)}{\frac12 \ln(2n+1) + O(1)}= 2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1309606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 4
} |
Prove that $\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\ge a+b+c$ How to prove that
\begin{equation*}\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\ge a+b+c,\ where \ a,b,c>0\end{equation*}
I tried the following:
\begin{equation*}abc(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})\ge a+b+c\end{equation*}
Using Chebyshev's inequality
... | Multiplying both sides of your inequality with $abc>0$, you get equivalently that:
$ \displaystyle (ab)^2 + (bc)^2 + (ca)^2 \geq abc (a+b+c) $
This holds by the basic inequality $ \displaystyle x^2 + y^2 + z^2 \geq xy +yz+ xz $, which holds for all $x,y,z $ real.
edit: The basic inequality holds for all real, thank's ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1309677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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simple 2 sides inequality
$$2<\frac{x}{x-1}\leq 3$$
Is the only way is to multiple both sides by $(x-1)^2$?
so we get $2x^2-4x+2<x^2-x $ and $x^2-x<3x^2-6x+3$ which is $-x^2+3x-2$ and $-2x^2+5x-3<0$ so the sloutions are:
$1<x\leq \frac{3}{2}$ and $1<x\leq 2$ so overall it is $1<x\leq\frac{3}{2}$
| Why quadratic? Linear is sufficient.
If $x > 1$, then $(x-1)$ is positive so we can multiply by that to get: $2(x-1) \lt x \le 3(x-1)$. Left-hand gives us $x < 2$ and right-hand gives us $x \ge \frac32$. Thus: $\frac32 \le x < 2$.
If $x < 1$, then we have to flip the signs and get $2(x-1) > x \ge 3(x-1)$. But there, t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1310939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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Find all Integral solutions to $x+y+z=3$, $x^3+y^3+z^3=3$. Suppose that $x^3+y^3+z^3=3$ and $x+y+z=3$.
What are all integral solutions of this equation?
I can only find $x=y=z=1$.
| Using identity: $(x+y+z)^3 = x^3+y^3+z^3 + 3(x+y)(y+z)(z+x)$, we have:
$3^3 = 3 + 3(x+y)(y+z)(z+x) \Rightarrow (x+y)(y+z)(z+x) = 8$. From this you should be able to deduce the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1312183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Finer asymptotic estimate of an integral I'm studying the asymptotic behaviour for large $n\in \mathbb N$ of $\displaystyle \int_1^\infty \frac{1}{1+t^{n+1}}$
Using the substitution $u=(n+1)\ln(t)$, $$\displaystyle \int_1^\infty \frac{1}{1+t^{n+1}}=\frac{1}{n+1}\int_0^\infty\exp(\frac{u}{n+1})\frac{1}{1+\exp(u)}du$$
An... | As Lucian remarked,
$$ \int_0^\infty \dfrac{dt}{1+t^{n+1}} = \dfrac{\pi}{n+1} \csc\left(\dfrac{\pi}{n+1}\right) $$
That is asymptotically
$$ 1+{\frac {{\pi }^{2}}{6{n}^{2}}}-{\frac {{\pi }^{2}}{3{n}^{3}}
}+{\frac {{\pi }^{2} \left( 7\,{\pi }^{2}+180 \right) }{360\,{n}^{4}}}
+ \ldots$$
Now we have to subtract the inte... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1313358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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For which values of $a$ does $||x^2-16|-7|-2a=0$ have 3 roots? Here is the equation:
$$||x^2-16|-7|-2a=0$$
Please help solve this.
For which $a$ does the equation have 3 roots?
| Api, first, use the definition of absolute value.
$||x^2-16|-7|-2a=0$ implies that two cases:
1) $|x^2-16|-7=2a$
This equation gives us $a=\frac{x^2-23}{2} $ and $a=\frac{x^2-9}{2}$.
2) $|x^2-16|-7=-2a$
Again, this equation gives us $a=\frac{23-x^2}{2} $ and $a=\frac{x^2+9}{2}$.
We should have the following this case.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1315422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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$\frac{1}{{9\choose r}} -\frac{1}{{10\choose r}} = \frac{11}{6{11\choose r}}$. Is there a way to find $r$ without using algebra?
$$\frac{1}{\dbinom 9r} -\frac{1}{{\dbinom{10}r}} = \frac{11}{6\times \dbinom{11}r}$$
I guess directly applying algebra for this problem would be enough. But are there any simpler and pretti... | Well, you can use Pascal's rule, or whatever it's called:
$$ \binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1} $$
to rewrite the left-hand side as
$$ \frac{\binom{10}{r}-\binom{9}{r}}{\binom{10}{r}\binom{9}{r}} = \frac{\binom{9}{r-1}}{\binom{10}{r}\binom{9}{r}}. $$
Multiplying up gives
$$ \frac{11}{6}\binom{10}{r}\binom... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1316148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Coefficient of operator and how to do it This question stems from this
$$
\frac{1}{x+z}- \frac{1}{x} = \sum_{k=0}^\infty \frac{z^k}{k!}\frac{d^k}{dx^k}[\frac{1}{x}]
$$
Now, i need to find the Bell Polynomial of $\frac{1}{x}$,
$$
[z^n]\left(\frac{(-z)^k}{x^k(x+z)^k}\right) = [z^n]\sum_{n \geq k} Y^{\Delta}(n,k,x) z^n = ... |
Here's the calculation of $[z^n]$:
\begin{align*}
[z^n]\frac{(-z)^k}{x^k(x+z)^k}&=[z^n]\frac{(-z)^k}{x^{2k}\left(1+\frac{z}{x}\right)^k}\\
&=\frac{(-1)^k}{x^{2k}}[z^n]\frac{z^k}{\left(1+\frac{z}{x}\right)^k}\\
&=\frac{(-1)^k}{x^{2k}}[z^{n-k}]\frac{1}{\left(1+\frac{z}{k}\right)^k}\tag{1}\\
&=\frac{(-1)^k}{x^{2k}}[z^{n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1318783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove $ \lim\limits_{x\to\infty}y_{n}=\sqrt{x}$ if $y_{n}=\frac{1}{2}\left(y_{n-1}+\frac{x}{y_{n-1}}\right),n\in \mathbb{N},x>0,y_{0}>0$ Can someone say how to solve this problem? In solution, it says that it stars with $$\frac{y_{n}-\sqrt{x}}{y_{n}+\sqrt{x}}=\left(\frac{y_{n-1}-\sqrt{x}}{y_{n-1}+\sqrt{x}}\right)^2,n\g... | The desired formula comes directly from the definitions.
\begin{align}
\frac{y_{n} - \sqrt{x}}{y_{n} + \sqrt{x}} &= \dfrac{\dfrac{1}{2}\left(y_{n - 1} + \dfrac{x}{y_{n - 1}}\right) - \sqrt{x}}{\dfrac{1}{2}\left(y_{n - 1} + \dfrac{x}{y_{n - 1}}\right) + \sqrt{x}}\notag\\
&= \dfrac{y_{n - 1}^{2} + x - 2y_{n - 1}\sqrt{x}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1320326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Need a hint to evaluate $\lim_{x \to 0} {\sin(x)+\sin(3x)+\sin(5x) \over \tan(2x)+\tan(4x)+\tan(6x)}$ I know that $\sin A + \sin B + \sin C = 4\cos({A \over 2})\cos({B \over 2})\cos({C \over 2})$ when $A+B+C=\pi$. If ${x \to 0}$ then I have a half circle, right?
If it is right then I have
$\tan(2x) + \tan(4x) + \tan(6x... | This is a much simpler take on this question and it uses the following result $$\lim_{x\to 0}\sin x = 0\tag{1}$$ from which we get $$\lim_{x \to 0}\cos x = 1\tag{2}$$ using the relation $\sin^{2}x + \cos^{2}x = 1$. Further dividing $(1)$ by $(2)$ we get $$\lim_{x \to 0}\tan x = 0\tag{3}$$ We can see that the numerator ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1323196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Simplifying sum equation. (Solving max integer encoded by n bits) Probably a lack of understanding of basic algebra. I can't get my head around why this sum to N equation simplifies to this much simpler form.
$$
\sum_{i=0}^{n-2} 2^{-i+n-2} + 2^i = 2^n - 2
$$
Background
To give you some background I am trying to derive... | Applying the proof of the geometric series to this particular case:
$$
MaxInt(n) = S(n) = \sum_{i=0}^{n-2} 2^i = 2^0 + 2^1 ... + 2^{n-3} + 2^{n-2}
$$
Adding up each term in the series:
$$
2S(n) = (2^0 + 2^0) + (2^1 + 2^1) ... + (2^{n-3} + 2^{n-3}) + (2^{n-2} + 2^{n-2})
$$
Simplifying:
$$
2S(n) = 2^1 + 2^2 ... + (2 \cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1323931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
a limit problem? $$\lim_{x\to 0}{e^x+ \ln{1-x\over e }\over \tan x-x} $$
i tried doing l'hospital. but couldn't do ! could anyone help ?
| This screams "Taylor" to me. First, $\log\frac{1-x}e=\log (1-x)-\log e=\log(1-x)-1$. We have
$$
e^x=1+x+\frac{x^2}2+\frac{x^3}6+o(x^4),\ \ \log(1-x)=-x+\frac{x^2}2-\frac{x^3}3+o(x^4),
$$
$$\tan x=x-\frac{x^3}3+o(x^5).
$$
So
\begin{align}
{e^x+ \ln{1-x\over e }\over \tan x-x}&=\frac{1+x+\frac{x^2}2+\frac{x^3}6+o(x^4)+(-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1324019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Calculate $\lim_{x \to \infty} (\sqrt{x^2 + 3x} - \sqrt{x^2 + 1})$ without using L'Hospital's rule Question:
Calculate
$$\lim_{x \to \infty} (\sqrt{x^2 + 3x} - \sqrt{x^2 + 1})$$
without using L'Hospital's rule.
Attempted solution:
First we multiply with the conjugate expression:
$$\lim_{x \to \infty} (\sqrt{x^2 + 3x} -... | There's an error in the factorisation of the denominator. No need to factor $x$ in the numerator, as you have a theorem for the limit at $\infty$ of a rational function. You should obtain, if $x>0$:
$$\lim_{x \to+ \infty}\frac{3x-1}{x\Bigl(\sqrt{1 + \dfrac3x} + \sqrt{1 + \dfrac1{x^2}}\Bigr)} = \lim_{x \to +\infty} \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1324388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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Prime number of the form $A^{B^C}+D^{E^F}$ My Brother asked me what is the smallest prime number of the form $A^{B^C}+D^{E^F}$ where A,B,C are three distinct prime numbers, and D,E,F are 3 distinct primes that is Permutations of those 3 primes.I realize that we must arrange the exponents to be as smallest as possible i... | $A = 2$ OR $D = 2$
$A \neq D$, $B \neq E$, $C \neq F$
$a^n$ generally increases faster as $n \rightarrow \infty$ than as $a$ does.
If we say that $A = 2$, then $A^{B^C}$ must be even. Therefore, $D^{E^F}$ must be odd.
Therefore we have $2^{B^C} + D^{2^F}$ OR $2^{B^C} + D^{F^2}$
Let's try $2^{5^3}+3^{2^5}$. It is prime!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1325858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Integral ${\large\int}_0^\infty\frac{dx}{\sqrt[4]{7+\cosh x}}$ How to prove the following conjectured identity?
$$\int_0^\infty\frac{dx}{\sqrt[4]{7+\cosh x}}\stackrel{\color{#a0a0a0}?}=\frac{\sqrt[4]6}{3\sqrt\pi}\Gamma^2\big(\tfrac14\big)\tag1$$
It holds numerically with precision of at least $1000$ decimal digits.
Are... | I will follow @user15302's idea. In this answer, I showed that
$$ \int_{0}^{\infty} \frac{dx}{(a + \cosh x)^{s}} \, dx = \frac{1}{(a+1)^{s}} \int_{0}^{1} \frac{v^{s-1}}{\sqrt{(1-v)(1-bv)}} \, dv, $$
where $b = \frac{a-1}{a+1}$. Now let $I$ denote the Vladimir's integral and set $s = \frac{1}{4}$ and $a = 7$. Then we ha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1326557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 4,
"answer_id": 1
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Proving “The sum of $n$ consecutive cubes is equal to the square of the sum of the first $n$ numbers.” This site
states:
Example $\boldsymbol 3$. The sum of consecutive cubes. Prove this remarkable fact of arithmetic: $$1^3 +2^3 +3^3 +\ldots +n^3 =(1 +2 +3 +\ldots +n)^2.$$
“The sum of $n$ consecutive cubes is equa... | Consider the infinite "multiplication table" array $M$ defined by $M_{i,j}=ij$
$M =
\left[\begin{array}{r}
1 & 2 & 3 & 4 & 5 \\
2 & 4 & 6 & 8 & 10 \\
3 & 6 & 9 & 12 & 15 & \cdots\\
4 & 8 & 12 & 16 & 20 \\
5 & 10 & 15 & 20 & 25 \\
&& \vdots &&& \ddots
\end{array}\right]$
Let $L(n) =
\{M_{i,j} : 1 \le i \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1328798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 2
} |
List the elements of the field $K = \mathbb{Z}_2[x]/f(x)$ where $f(x)=x^5+x^4+1$ and is irreducible Since $\dim_{\mathbb{Z}_2} K = \deg f(x)=5$, $K$ has $2^5=32$ elements. So constructing the field $K$, I get:
\begin{array}{|c|c|c|}
\hline \text{polynomial} & \text{power of $x$} & \text{logarithm} \\\hline
0 & 0 & -\... | One problem is, that you need to list all polynomials $f(x)\in \mathbb{F}_2[x]$ of degree $n\le 4$. This gives $2^{4+1}=32$ polynomials. For degree $5$, and hence higher degree, you can "reduce" the degree of the polynomial by the rule $x^5:=-x^4-1=x^4+1$. For example, the polynomial $x^3+1$ is missing in your list.
Ed... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1329867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.