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Evaluate integral for $\int \sin^2 (x+\frac{\pi}{6}) dx$ Can someone walk me through how to evaluate the integral $$\int \sin^2 (x+\frac{\pi}{6}) dx?$$
I get as far as
$$\int \frac {1 - \cos(x + \frac{\pi}{6})}2dx,$$
but I am not sure how to proceed.
| Careful with formula
$$
\sin^2\left(t\right)=\frac{1-\cos\left(2t\right)}{2}
$$
Then using linearity
$$
\int \frac{1-\cos\left(2x+\frac{\pi}{3}\right)}{2}\text{d}x=\int \frac{1}{2}-\frac{1}{2}\int \cos\left(2x+\frac{\pi}{3}\right)
$$
Hence
$$
\int \frac{1-\displaystyle \cos\left(2x+\frac{\pi}{3}\right)}{2}\text{d}x=\frac{x}{2}-\frac{1}{4}\sin\left(2x+\frac{\pi}{3}\right)+K$$
| {
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"url": "https://math.stackexchange.com/questions/2787969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Prove that $\sqrt{11}-1$ is irrational by contradiction I am working on an assignment in discrete structures and I am blocked trying to prove that $\sqrt{11}-1$ is an irrational number using proof by contradiction and prime factorization.
I am perfectly fine doing it with only $\sqrt{11}$, but I am completely thrown off by the $-1$ when it comes to the prime factorization part.
My current solution looks like this :
$$ \sqrt {11} -1 = \frac {a}{b}$$
$$ \sqrt {11} = \frac {a}{b} + 1$$
$$ \sqrt {11} = \frac {a+b}{b}$$
$$ 11 = \left(\frac {a+b}{b}\right)^2$$
$$ 11 = \frac {(a+b)^2}{b^2}$$
$$ 11 = \frac {a^2 + 2ab + b^2}{b^2}$$
$$ 11 b^2 = a^2 + 2ab +b ^2$$
$$ 10b^2 = a^2 + 2ab $$
At that point, is it acceptable to conclude that a² is a multiple of 11 even though we have a trailing $2ab$?
The required method is then to conclude using prime factorization that $a = 11k$ and replace all that in the formula above to also prove $b$, however, I am again stuck with the ending $2ab$.
Would it instead be correct to prove that $\sqrt{11}$ is rational using the usual method and that, by extension, $\sqrt{11} - 1$ is also rational?
Thank you
| From $10b^2=a^2+2ab$ you can see that $2\mid a^2$, hence $a=2m$ (for some $m$), from which you can conclude that $b^2=2m^2+2mb-4b^2$, which implies $2\mid b$, contradicting the (tacit) assumption $\gcd(a,b)=1$.
This is an interesting variant on the standard proof of irrationality in that it only invokes the implication $2\mid n^2\implies 2\mid n$ instead of the more general implication $p\mid n^2\implies p\mid n$ (with $p$ prime). Indeed, it gives an easy proof for the irrationality of $\sqrt{4k+3}$, whether $4k+3$ is prime or not:
$$\sqrt{4k+3}-1={a\over b}\implies(4k+3)b^2=a^2+2ab+b^2\implies a^2=2((2k+1)b^2-ab)\\
\implies a=2m\\
\implies2m^2=(2k+1)b^2-2mb\\
\implies b^2=2(m^2-kb^2+mb)\\
\implies 2\mid b
$$
which contradicts the assumption $\gcd(a,b)=1$. Thus $\sqrt{4k+3}-1$ is irrational, hence $\sqrt{4k+3}$ is irrational.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2789537",
"timestamp": "2023-03-29T00:00:00",
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Unclear calculus problem I don't know if that's called calculus or precalculus in the Anglosphere, so please pardon me if I got the terminology incorrect.
I've been given the second derivative of $f(x): f''\left(x\right)=\frac{2x-1}{\sqrt{-x^2+x+6}} $.
I was asked to find $f'(x)$, if it is known that the slopes of all tangent lines to the graph of $f(x)$ in the range of $-2<x<3$ are not smaller than -1.
I integrated the second derivative:
$$f'\left(x\right)=\int \:\frac{2x-1}{\sqrt{-x^2+x+6}}dx\:=\:-2\sqrt{-x^2+x+6}\:+\:C$$
So the task at hand is essentially to find the value of C. The problem, however, is that I did not understand how to do that. The only thing I managed to conclude from the given information (about the slopes of the tangent lines) is that $f'\left(x\right)\ge -1$, and I don't know how to use an inequality to find C.
If someone can make it clearer, I'll be extremely glad :)
P.S. The final answer (according to the textbook) is $f'\left(x\right)=\:-2\sqrt{-x^2+x+6}\:+\:4$.
| Just continue
So
For $-2 < x < 3$ we have $-2\sqrt{-x^2+x+6} + C \ge -1$
$2\sqrt{-x^2 + x + 6} \le 1 + C$
So $ C \ge \max(2\sqrt{-x^2 + x + 6})_{x\in (-2,3)} - 1$.
So must find $\max(2\sqrt{-x^2 + x + 6})_{x\in (-2,3)}$. As $-x^2 +x+6$ must be non-negative we may assume $\max(2\sqrt{-x^2 + x + 6})_{x\in (-2,3)}$ where $-x^2 + x + 6$ does (assuming we get a non-negative value).
So solving for $(-x^2 + x + 6)' = -2x + 1 = 0$ whe have at $x = \frac 12$ is an extrema (and as $(-x^2 + x + 6)'' = -2 < 0$ it is a maximum).
So $C \ge 2\sqrt{-\frac 14 + \frac 12 + 6} - 1$
$= 2\sqrt{6\frac 14}-1 = 2\sqrt {\frac {25}4} - 1=5-1 = 4$.
So
$f'(x) = -2\sqrt{-x^2+x+6} + C$ where $C \ge 4$. I don't see anything in the problem that suggest $C$ can't be larger than $4$. However if it said there were one or more tangent lines where the slope is $-1$ but none where it is less, then $C$ must equal $4$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A tricky algebraic inequality This is an old inequality but I haven't seen a satisfactory solution yet and am hoping someone here can provide one. There are a couple of brute force solutions but they provide no insight into the inequality and I'd be surprised if there isn't a trick to it:
$$x\frac{(y+z)^2}{(1+yz)^2} + y\frac{(x+z)^2}{(1+xz)^2} + z\frac{(x+y)^2}{(1+xy)^2} \ge \frac{3\sqrt 3}{4}$$
for $xy+yz+zx = 1$, all positive.
My attempt was to try to find a lower bound of the left side in terms of symmetric quantities like $u=x+y+z$ and $w=xyz$, however I haven't had much success despite multiple attempts. For example, the left hand side above can be bounded from below by
$$\sum x\frac{(y+z)^2}{(1+yz)^2} \ge\frac{\left(\sum x(y+z)^2\right)^3}{\left(\sum (y+z)(1-y^2z^2)\right)^2} = \frac{\left(u+3w\right)^3}{\left(u+2u^2w+w\right)^2}$$
but the above has a minimum just a tad less than $3\sqrt 3/4$.
I have also tried a cotangent substitution and breaking the symmetry (i.e. assuming $x\ge y\ge z$) but I didn't get far.
I don't know the origin of the inequality but it's supposed to be competition level so I suspect it has a nice and tricky solution and not just a brute force one. So, I am hoping someone here will find it.
| We need to prove that
$$\sum_{cyc}\frac{x(y+z)^2}{(xy+xz+2yz)^2}\geq\frac{3}4\sqrt{\frac{3}{xy+xz+yz}}$$ or
$$\sum_{cyc}\frac{\frac{1}{x}\left(\frac{1}{y}+\frac{1}{z}\right)^2}{\left(\frac{1}{xy}+\frac{1}{xz}+\frac{2}{yz}\right)^2}\geq\frac{3}{4}\sqrt{\frac{3}{\frac{1}{xy}+\frac{1}{xz}+\frac{1}{yz}}}$$ or
$$\sum_{cyc}\frac{x(y+z)^2}{(y+z+2x)^2}\geq\frac{3}{4}\sqrt{\frac{3xyz}{x+y+z}}.$$
Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3.$
Thus, we need to prove that
$$\frac{u(-11w^6+756u^3w^3-126uv^2w^3+972u^4v^2-576u^2v^4)}{(w^3+9uv^2+54u^3)^2}\geq\frac{1}{4}\sqrt{\frac{w^3}{u}}$$ or $f(v^2)\geq0$, where
$$f(v^2)=4u(-11w^6+756u^3w^3-126uv^2w^3+972u^4v^2-576u^2v^4)-$$
$$-(w^3+9uv^2+54u^3)^2\sqrt{\frac{w^3}{u}}.$$
Now, it's obvious that $f''(v^2)<0$, which says that $f$ is a concave function.
But the concave function gets a minimal value for an extreme value of $v^2$,
which happens for equality case of two variables.
Since $f(v^2)\geq0$ is a homogeneous inequality it's enough to assume that $y=z=1$
and we need to prove that
$$\frac{4x}{(2+2x)^2}+\frac{2(x+1)^2}{(x+3)^2}\geq\frac{3}{4}\sqrt{\frac{3x}{x+2}}$$ or
$$(x-1)^2(37x^7+346x^6+1339x^5+2700x^4+2891x^3+1466x^2+309x+128)\geq0.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2793183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Probability that $z$ is EVEN satisfying the equation $x + y + z = 10$ is Question
Three randomly chosen non negative integers $x, y \text{ and } z$ are found to satisfy the equation $x + y + z = 10$. Then the probability that $z$ is even, is"?
My Approach
Calculating Sample space -:
Number of possible solution for $x + y + z = 10$
$$=\binom{10+3-1}{10}=12 \times 3=66$$
Possible outcome for $z$ to be even =$6(0,2,4,6,8,10)$
Hence the required probability$$=\frac{6}{66}=\frac{1}{11}$$
But the answer is $\frac{6}{11}$
Am I missing something?
| $x$ and $y$ need to have the same parity, so that we can predict that the probability will be close to $\frac12$.
For a given $z$, there are $11-z$ combinations of $x,y$, which are all of the same parity or of the opposite parity.
Hence
$$p=\frac{11+9+7+5+3+1}{11+9+7+5+3+1+10+8+6+4+2}=\frac{36}{66}.$$
Using summation formulas,
$$p=\frac{\left(\dfrac{12}2\right)^2}{\dfrac{11\cdot12}2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2795440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Express generic "penthagorean" triplet If asked to express the generic pythagorean triplets satisfying
$a^2+b^2=c^2$ you would answer
$$
a = k(r^2 - s^2), b = 2krs, c = k(r^2+s^2)
$$
with $k,r,s \in \Bbb N$ and $r>s$ and $r \ne s \mod 2$ and $\gcd(r,s) = 1$. This generates every pythagorean triplet exactly once as the pair $(r,s)$ takes on its allowed values.
I'm looking to derive an analogous formula to generate every "penthagorean" triplet, that is, every instance where the sum of two pentagonal numbers is a pentagonal number. That is, find all $(a,b,c)$ satisfying
$$
\frac{3a^2-a}2+\frac{3b^2-b}2=\frac{3c^2-c}2
$$
Deriving the formula for pythagorean numbers, using only elementary number theory such as considerations in mods $2$ and $4$ is so straightforward that I thought the analogous formula would be easy to obtain, but I am getting stuck.
| from pages 124,125 of Magnus. If we have integers $a,b,c,d$ with $ad-bc=1$ and $a+b+c+d \equiv 0 \pmod 2,$ and we take
$$
\left(
\begin{array}{ccc}
\frac{1}{2} \left( a^2 + b^2 + c^2 + d^2 \right)&ab+cd&\frac{1}{2} \left( a^2 - b^2 + c^2 - d^2 \right) \\
ac+bd&ad+bc&ac-bd \\
\frac{1}{2} \left( a^2 + b^2 - c^2 - d^2 \right)&ab-cd&\frac{1}{2} \left( a^2 - b^2 - c^2 + d^2 \right) \\
\end{array}
\right)
\left(
\begin{array}{c}
x \\
y\\
z \\
\end{array}
\right) =
\left(
\begin{array}{c}
u \\
v \\
w \\
\end{array}
\right)
$$
THEN
$$ u^2 - v^2 - w^2 = x^2 - y^2 - z^2 $$
At a minimum, the modular group takes a solution to your problem to another. Oh, multiply by 12 and complete the square,
$$ (6p-1)^2 + (6q-1)^2 = 1 + (6r-1)^2 $$
so that $6r-1$ becomes either $x$ or $u$
The observations in Magnus (1974) go back to Fricke and Klein (1897)
I cannot recall whether this gives all solutions to $x^2 - y^2 - z^2 = -1.$ Worth experimenting. The 3 by 3 matrix has determinant 1 and should preserve $\gcd(x,y,z).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2795850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do I find out the value of $p$ for $\sin p + \cos p = 0$? I tried doing this by putting $$\frac{ \sin p}{\sqrt2} +\frac{ \cos p}{\sqrt2} = 0 $$ which implies $$ \sin(p + π/4) = \sin 0 $$ which implies $p+ π/4 = nπ $. Now according to the question I'm solving $p = 2πt/T$. I need to get the relation $t = 3T/8$. How do I get this? Am I going the right way?
| $$\sin x + \cos x = 0$$
$$\Rightarrow \sin^2 x + 2 \sin x \cos x + \cos^2 x = 0$$
$$\Rightarrow \sin(2x) + 1 = 0$$
$$\Rightarrow \sin(2x) = -1$$
Since we know that $\sin x = -1$ when $x = \frac{3\pi}{2} + 2\pi n$, then $\sin(2x) = -1$ when $x = \frac{3\pi}{4} + \pi n $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2796511",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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Let $S = \{1,2,3,4,5,6\}$ be a sample space of equally likely outcomes Let $S = \{1,2,3,4,5,6\}$ be a sample space of equally likely outcomes such that $P(\{s\}) = \frac{1}{6}$, $\forall s \in S$ let X = $I_{(0,9,3,1)}, Y = I_{(2,4,4,2)}$ and $W = X + Y$
Compute $P(W = 1)$
My attempt:
$$W(1) = X(1) + Y(1) = I_{(0,9,3,1)}(1) + I_{(2,4,4,2)}(1) = 1 + 0 = 1$$
Since $P(\{s\}) = 1/6$, then $P(1) = 1$.
Would this be correct?
| Not entirely:
For one you are correct in saying:
$$
W(1) = 1
$$
But notice also that
\begin{align}
W(2) &= X(2) +Y(2) = 0 + 1 = 1, \\
W(3) &= X(3) +Y(3) = 1 + 0 = 1, \\
W(4) &= X(4) +Y(4) = 0 + 1 = 1, \\
W(5) &= X(5) +Y(5) = 0 + 0 = 0, \\
W(6) &= X(6) +Y(6) = 0 + 0 = 0.
\end{align}
So we get that
\begin{align}
P(W = 1) &= P \bigl(\{1, 2, 3, 4 \} \bigr) \\
&= P\bigl(\{1\}) + P\bigl(\{2\}) + P\bigl(\{3\}) + P\bigl(\{4\}) \\
&= \frac 16 + \frac 16 +\frac 16 +\frac 16 = \frac 46.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2801495",
"timestamp": "2023-03-29T00:00:00",
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Is there any formula to find $\sin^{-1}a +\sin^{-1}b$ We known that$$\tan^{-1} a +\ tan^{-1}b=\tan^{-1}\left(\frac{a+b}{1-ab}\right).$$
Now we derive the above formula. Let$$
\tan^{-1}a=\theta _1 \implies \tan\theta_1=a,\\
\tan^{-1}b=\theta _2 \implies \tan\theta_2=b,\\
\theta _1+\theta _2 = \tan^{-1}a+ \tan^{-1}b,\\
\tan(\theta _1+\theta _2) = \tan(\tan^{-1}a+\tan^{-1}b)·
\tan(\theta _1+\theta _2)\\
=\frac{ \tan(\tan^{-1}a)+\tan( \tan^{-1}b)}{1- \tan(\tan^{-1}a)· \tan(\tan^{-1}b)},\\
(\theta _1+\theta _2) =\tan^{-1}\left(\frac {a+b}{1-ab}\right).
$$
Therefore, $\tan^{-1} a +\ tan^{-1}b=\tan^{-1}\left(\dfrac{a+b}{1-ab}\right)$.
For all $\theta _1$ and $\theta _2$ such that $0 \le \theta _1,\theta _2 \lt 90$, in the similar way I try to find $\sin^{-1}a +\sin^{-1}b$. Let$$
\sin^{-1}a=\theta_1, \quad \sin^{-1}b=\theta_2,\\
\sin(\theta_1+\theta_2)=\sin(\sin^{-1}a +\sin^{-1}b),\\
\sin(\theta_1+\theta_2)=\sin(\sin^{-1}a)·\cos(sin^{-1}b) +\sin(\sin^{-1}b)·\cos(sin^{-1}a),\\
\theta_1+\theta_2=\sin(\sin^{-1}a)·\cos(sin^{-1}b) +\sin(\sin^{-1}b)·\cos(sin^{-1}a),\\
\sin^{-1}a+\sin^{-1}b=\sin(\sin^{-1}a).cos(sin^{-1}b) +\sin(\sin^{-1}b).cos(sin^{-1}a),\\
\sin^{-1}a+\sin^{-1}b=\sin(a)·\cos(\sin^{-1}b) +\sin(b)·\cos(\sin^{-1}a).$$
It cannot be write in the form only using $a$ and $b$.
$$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A·\tan B}.$$ It contains only $\tan$ terms. But $\sin(A+B)=\sin A·\cos B+\cos A·\sin B$, which contains both $\sin$ and $\cos$ terms.
Is there any formula for $\sin^{-1}a+\sin^{-1}b$?
| Assume that $\arcsin(a)+\arcsin(b)\in\left[-\frac\pi2,\frac\pi2\right]$. Then, since$$\sin\left(\arcsin(a)+\arcsin(b)\right)=a\sqrt{1-b^2}+b\sqrt{1-a^2},$$we have$$\arcsin(a)+\arcsin(b)=\arcsin\left(a\sqrt{1-b^2}+b\sqrt{1-a^2}\right).$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Integral using the Substitution y = $\frac{1}{2}(\sqrt{a}x^2+\frac{\sqrt{b}}{x^2})$ Given that $\int_{-\infty}^{\infty} e^{(-\frac{1}{2}x^2)}dx = \sqrt{2\pi}$,
Find $I(a,b) = \int_{-\infty}^{\infty} e^{-\frac{1}{2}(ax^2+\frac{b}{x^2})}dx$.
The question suggests using the substitution: $$y = \frac{1}{2}(\sqrt{a}x-\frac{\sqrt{b}}{x})$$ if careful attention is paid to the limits. I'm not sure how the limits of integral changes as I can't see why the limits wouldn't remain the same using the substitution.
| The key here is to note that this integral is improper at zero, so before performing the substitution notice the integrand is even so we may rewrite the integral as
$$I(a,b)=2\int_{0+}^\infty e^{-\frac{1}{2}(ax^2+\frac{b}{x^2})}dx$$
Let $y=\sqrt{a}x-\frac{\sqrt{b}}{x}$. Note the bounds become $-\infty$ to $\infty$, since as $x\rightarrow 0^+$, $y\rightarrow -\infty$. Also notice $y^2=ax^2+\frac{b}{x^2}-2\sqrt{ab}$, and
$$dy=\sqrt{a}+\frac{\sqrt{b}}{x^2}=\frac{\sqrt{y^2+4\sqrt{ab}}-y}{2\sqrt{b}}\sqrt{y^2+4\sqrt{ab}}\ dx$$ so the integral is
$$I(a,b)=2\int_{-\infty}^\infty e^{-\frac{1}{2}(y^2+2\sqrt{ab})}\frac{2\sqrt{b}}{\sqrt{y^2+4\sqrt{ab}}}\frac{1}{\sqrt{y^2+4\sqrt{ab}}-y}dy$$
Combining the positive and negative sides of the axis gives
$$I(a,b)=2\int_{0}^\infty e^{-\frac{1}{2}(y^2+2\sqrt{ab})}\frac{2\sqrt{b}}{\sqrt{y^2+4\sqrt{ab}}}\left(\frac{1}{\sqrt{y^2+4\sqrt{ab}}-y}+\frac{1}{\sqrt{y^2+4\sqrt{ab}}+y}\right)dy$$
$$=\frac{2}{\sqrt{a}}\int_0^\infty e^{-\frac{1}{2}(y^2+2\sqrt{ab})}dy$$
And from there you can use your identity.
| {
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"url": "https://math.stackexchange.com/questions/2802706",
"timestamp": "2023-03-29T00:00:00",
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Find a fraction $\frac{m}{n}$ which satisfies the given condition Find a fraction such that all of $\frac{m}{n}$, $\frac{m+1}{n+1}$, $\frac{m+2}{n+2}$, $\frac{m+3}{n+3}$, $\frac{m+4}{n+4}$, $\frac{m+5}{n+5}$ are reducible by cancellation. Condition: $m≠n$.
What I tried was... I wrote $$\frac{m}{n}=k$$ Then, I replaced $m$ in the fractions as $nk$. And after a bit of simple manipulation, I obtained $$k+\frac{1-k}{n+1},$$ $$k+\frac{2-k}{n+2},$$ $$k+\frac{3-k}{n+3},$$ $$k+\frac{4-k}{n+4},$$ $$k+\frac{5-k}{n+5}$$ Now I do not know how to proceed any further.
| If $m=30a$ and $n=30b$, then
$${m\over n}={30a\over30b}\\{m+2\over n+2}={2(15a+2)\over2(15b+1)}\\{m+3\over n+3}={3(10a+1)\over3(10b+1)}\\{m+4\over n+4}={2(15a+2)\over2(15b+2)}\\\text{and}\\{m+5\over n+5}={5(6a+1)\over5(6b+1)}$$ are all reducible. It remains to choose $a$ and $b$ so that
$$30a+1\over30b+1$$
is reducible. If we try $b=1$, we see we need $31\mid30a+1$, or $a\equiv1$ mod $31$, so $a=32$ does the trick. Thus $m/n=960/30$ has the desired property.
| {
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Number of solutions to the congruence $x^2\equiv 121\pmod {1800}$ I'm trying to find the number of solutions to this congruence:
$$x^2\equiv 121\pmod {1800}$$
I thought about writing it as a system of congruences. As $1800=3^2 \cdot 5^2 \cdot 2^3$, we get:
$x^2\equiv 121\pmod {5^2} \;,\; x^2\equiv 121\pmod {3^2} \;,\; x^2\equiv 121\pmod {2^3}$
$\Downarrow$
$ x^2\equiv 21\pmod {5^2} \;,\; x^2\equiv 4\pmod {3^2} \;,\; x^2\equiv 1\pmod {2^3} $
Now I'm pretty stuck with how to solve each of the above quadratic congruences, is there a fast way to do it?
Thanks
| Hint:
We need $\left(\dfrac x{11}\right)^2\equiv1\pmod{3^2,5^2,2^3}$
Now we can prove $y^2\equiv1\pmod{p^n}$ has exactly two solutions for prime $p\ge3$ and integer $n\ge1$
Finally we can apply Chinese Remainder Theorem to find the number of in-congruent solutions to be $$4\cdot2^2$$
See also: Number of solutions of $x^2=1$ in $\mathbb{Z}/n\mathbb{Z}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Convex sum order If I have a strictly convex function $f(x)$ with $f''(x)>0$
and if I know that for some $a\le b \le c$ and $x \le y \le z$ I have
$$a+b+c = x+y+z$$
$$f(a)+f(b)+f(c)=f(x)+f(y)+f(z)$$
can I conclude that at least one of the following must be true about the order?
$$ a \le x \le y \le b \le c \le z$$
$$ x \le a \le b \le y \le z \le c$$
Perhaps an equivalent question is to consider coincident centroids of two triangles with vertices on the convex curve, looking something like this diagram with pairs of vertices on the inside of the order
though I realise that I cannot push this analogy too far as it would not be necessarily true if say the curve was a circle and the triangles equilateral
| For simplicity, let us assume that all the points are different, i.e., $a < b < c$, $x < y < z$ and the all the points $a,b,c$ are different from $x,y,z$.
(Maybe some arguments can be transferred to the general case).
Now, the order of the points can be described by a permutation of $abcxyz$ which preserves the order of $abc$ and $xyz$.
By symmetry, we can assume w.l.o.g. that the permutation starts with $a$.
Let us consider permutations starting with $ab$.
*
*$abc$ can only be completed to $abcxyz$ and this cannot satisfy $a + b + c = x + y + z$.
*$abx$ can be completed to
*
*
*$abxycz$: this cannot satisfy $a + b + c = x + y + z$.
*
*
*$abxcyz$: this cannot satisfy $a + b + c = x + y + z$.
*
*
*$abxyzc$: by your triangly analogy, the entire triangle $xyz$ lies below the segment $bc$, hence the centroids of $abc$ and $xyz$ cannot coincide.
These arguments show that the permutation has to start with $ax$. By similar arguments we can conclude that the permutation has to end with $zc$ or $cz$.
Hence, four possibilities remain:
*
*$ax by cz$
*$ax yb cz$
*$ax by zc$
*$ax yb zc$
The first case cannot satisfy $a + b + c = x + y + z$. The second case is what you would like to have. Therefore, it remains to rule out the last two cases. By symmetry, it is enough to rule out $axbyzc$.
The strict convexity gives
\begin{align}
f(x) &< f(a) \frac{b-x}{b-a} + f(b) \frac{x-a}{b-a}, \\
f(y) &< f(b) \frac{c-y}{c-b} + f(c) \frac{y-b}{c-b}, \\
f(z) &< f(b) \frac{c-z}{c-b} + f(c) \frac{z-b}{c-b}.
\end{align}
Adding these inequalities and using $f(a) + f(b) + f(c) = f(x) + f(y) + f(z)$ gives
\begin{equation*}
f(a) + f(b) + f(c)
<
f(a) \frac{b-x}{b-a} + f(b) \Big[ \frac{x-a}{b-a} + \frac{2c-z-y}{c-b}\Big] + f(c) \frac{z+y-2b}{c-b}.
\end{equation*}
Rearranging terms and using $a + b + c = x + y + z$
gives
\begin{equation*}
0
<
(x-a)
\Bigg[
\frac{f(b) - f(a)}{b-a}
-
\frac{f(c) - f(b)}{c-b}
\Bigg].
\end{equation*}
This, however, is a contradiction to the strict convexity which yields
\begin{equation*}
\frac{f(b) - f(a)}{b-a}
<
\frac{f(c) - f(b)}{c-b}.
\end{equation*}
This rules out the case $axbyzc$.
Hence, if $a$ is the smallest point, the permutation has to be $axybcz$, i.e.,
$a < x < y < b < c < z$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2808323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Someone's Inequality ? $(a + b)^2 \le 2(a^2 + b^2) $ For real $a, b$ then $(a + b)^2 \le 2(a^2 + b^2) $
This fairly trivial inequality crops up a lot in my reading on (Lebesgue) integration, is it named after someone ? It extends rather obviously for positive reals to $a^2 + b^2 \le (a + b)^2 \le 2(a^2 + b^2) $.
Proof (if you need it):
$0 \le (a - b)^2 = a^2 + b^2 -2ab \implies 2ab \le a^2 + b^2$
$(a + b)^2 = a^2 + b^2 + 2ab$ which by previous $ \le 2(a^2 + b^2) $.
Application:
If $f, g$ are positive functions then $(f + g)^2$ is integrable $\iff$ $f^2, g^2$ are integrable since $f^2 + g^2 \le (f + g)^2 \le 2(f^2 + g^2) $ pointwise.
| This follows from the Cauchy–Schwarz inequality
$$
|\langle \mathbf{u},\mathbf{v}\rangle| ^2 \leq \langle \mathbf{u},\mathbf{u}\rangle \cdot \langle \mathbf{v},\mathbf{v}\rangle,
$$
with
$
\mathbf{u} = (1,1)$
and
$
\mathbf{v} = (a,b)
$
in $\mathbb R^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2808947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Spivak's Calculus: Finding the formula for the sum of a series. In Spivak's Calculus, Chapter 2, the second problem of the set asks you to find a formula for the following series:
$$\sum_{i=1}^n(2i-1)$$ and $$\sum_{i=1}^n(2i-1)^2$$
Now, for the former, this was fairly straightforward: it sums to $i^2$ and I have a proof that I'm happy with. However, for the latter, I cannot identify a pattern in terms of $n$ to begin building a proof. The series proceeds $1^2 + 3^2 + 5^2 + 7^2 + \cdots + (2n-1)^2$, so the sum for the first few indices would be $1, 10, 35,$ and $84.$ Not only have I failed to come up with an expression of this in terms of $n$, but I'm not even sure what I should be considering to lead me to such a formula.
| $\sum_\limits{k=1}^n (2k - 1)^2 = \sum_\limits{k=1}^n 4k^2 - \sum_\limits{k=1}^n 4k + \sum_\limits{k=1}^n 1 = 4\left(\sum_\limits{k=1}^n k^2\right) - 2n(n+1) + n$
You need a way to find $\sum_\limits{k=1}^n k^2$
There are lots of other ways to derive this... but this one is nice.
Write the numbers in a triangle
$\begin {array}{} &&&1\\&&2&&2\\&3&&3&&3\\&&&\vdots&&\ddots\\n&&n&\cdots\end{array}$
The sum of each row is $k^2.$ And, the sum of the whole trianglular array is $\sum_\limits{k=1}^n k^2$
Take this triangle and rotate it 60 degrees clockwise and 60 degrees counter clockwise and sum the 3 triangles together.
$\begin {array}{} &&&1\\&&2&&2\\&3&&3&&3\end{array}+\begin {array}{} &&&n\\&&n-1&&n\\&n-2&&n-1&&n\\\end{array}+\begin {array}{} &&&n\\&&n&&n-1\\&n&&n-1&&n-2\end{array}$
And we get:
$\begin {array}{} &&&2n+1\\&&2n+1&&2n+1\\&2n+1&&2n+1&&2n+1\\&&&\vdots&&\ddots\end{array}$
What remains how many elements are in the triangular array? $\sum_\limits{k=1}^n k$
$3\sum_\limits{k=1}^n k^2 = (2n+1)\sum_\limits{k=1}^n k\\
\sum_\limits{k=1}^n k^2 = \frac {n(n+1)(2n+1)}{6}$
Another way to do it is:
$\sum_\limits{k=1}^n (k+1)^3 - k^3 = (n+1)^3-1\\
\sum_\limits{k=1}^n (3k^2 + 3k + 1)\\
\sum_\limits{k=1}^n 3k^2 + \sum_\limits{k=1}^n 3k + \sum_\limits{k=1}^n 1\\
\sum_\limits{k=1}^n 3k^2 + 3\frac {n(n+1)}{2} + n = n^3 + 3n^2 + 3n \\
3\sum_\limits{k=1}^n k^2 = n^3 + \frac 32 n^2 + \frac 12 n\\
\sum_\limits{k=1}^n k^2 = \frac 16 (n)(n+1)(2n+1)$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solution of the functional equation $f(x) + f(y) =f\left(x\sqrt{1-y^2 }+y\sqrt{1-x^2 }\right) $
If
$$f(x) + f(y) =f\left(x\sqrt{1-y^2 }+y\sqrt{1-x^2 }\right)\text,$$
prove that
$$f(4x^3 -3x) + 3f(x) =0\text.$$
I started by substituting $x = y$, in the expression and I get
$$2f(x)=f\left(2x\sqrt{1-x^2}\right)\text.$$
Then, I also substitute $y = \sqrt{(1-x^2)}$ and I get
$$f(x) +f\left(x\sqrt{1-x^2}\right) = 0\text.$$
How shall I proceed further and solve this problem?
| I'll assume the arguments of $f$ are always between $-1$ and $1$. Then
$$2f(x)=f(2x\sqrt{1-x^2})$$
and
$$3f(x)=f(x)+f(2x\sqrt{1-x^2})=f(y)$$
where
$$
\begin{split}
y &= x\sqrt{1-4x^2(1-x^2)}+\sqrt{1-x^2}2x\sqrt{1-x^2}\\
&= x\sqrt{1-4x^2+4x^4}+2x\left(1-x^2\right)\\
&= x\left(1-2x^2\right)+2x\left(1-x^2\right)=3x-4x^3.
\end{split}
$$
But $f(y)+f(-y)=f(0)=f(0)+f(0)$ and so $3f(x)=f(-y)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2811680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
integral of $ \int (x^3+1)^{1/3} / x^2 dx $ What do you think about
$$ \int \frac {(x^3+1)^{1/3}}{x^2}\, dx $$
how do you compute this ? Is it possible to use the Euler substitution?
In fact, I don't know if the integral on my sheet is $$ \int \frac {(x^3+1)^{1/3}} {x^2} \, dx $$
or
$$ \int \frac{x^2 + 1}{x \sqrt {x^4 - x^2 + 1} } \frac {(x^3+1)^{1/3}} {x^2} \,dx $$
would the second expression make more sense? I thought that the second one is a little bit over complicated.
|
Hint : Divide numerator & denominator by $x$ & write the integrand
as $I= \int\frac {(1+\frac1 {x^3})^{1/3}}x dx $
The whole calculus :
\begin{align}
I =& \int \frac{(x^3 +1)^{1/3} }{x^2} dx \\
=& \int \frac{(1+ x^{-3})^{1/3} }{x} dx
\end{align}
let $y= x^{-3} \implies dy = \frac{-1}{3} x^{-4} dx $
$$ I = \frac{-1}{3} \int \frac {(1+y)^{1/3}}ydy $$
let $ z^3 = 1+y \implies dz =3z^2dy$
\begin{align}
I &= \int \frac {z^3}{1-z^3}dz \\
&= \int \frac{z^3 + 1 - 1}{1-z^3}dz \\
&= \int \frac{z^3 - 1}{1-z^3}dz + \int \frac{ 1 }{1-z^3}dz \\
&= - \int dz + \int \frac{ 1 }{1-z^3}dz \\
however \\
\int \frac{ 1 }{z^3-1}dz &= \int \frac{1}{ (z-1)(z^2+z+1) } dz \\
&= \frac{1}{3} \int \frac{1}{z-1} dz - \frac{1}{3} \int \frac{z+2}{z^2 + z + 1} dz \\
&=\frac{1}{3} \int \frac{1}{z-1} dz - \frac{1}{6} \int \frac{2z+1}{z^2 + z + 1} dz - \frac{1}{2} \int \frac{1}{z^2 + z + 1} dz \\
&=\frac{1}{3} \int \frac{1}{z-1} dz - \frac{1}{6} \int \frac{2z+1}{z^2 + z + 1} dz - \frac{1}{2} \int \frac{1}{ (z+ \frac{1}{2})^2 + (\frac{\sqrt 3}{2})^2 } dz \\
&=\frac{1}{3} \ln( z-1 ) - \frac{1}{6} \ln(z^2 + z + 1) - \frac{1}{\sqrt 3} \arctan ( \frac{2z + 1}{ \sqrt 3} ) + cst \\
so \\
I &= -z + \frac{1}{3} \ln( z-1 ) - \frac{1}{6} \ln(z^2 + z + 1) - \frac{1}{\sqrt 3} \arctan ( \frac{2z + 1}{ \sqrt 3} ) + cst \\
i.e. \\
I &= -\sqrt[3]{1 + x^{-3} } + \frac{1}{3} \ln( \sqrt[3]{1 + x^{-3} } -1 ) - \frac{1}{6} \ln(\sqrt[3]{1 + x^{-3} } ^2 + \sqrt[3]{1 + x^{-3} } + 1) \\ &- \frac{1}{\sqrt 3} \arctan ( \frac{2\sqrt[3]{1 + x^{-3} } + 1}{ \sqrt 3} ) + cst \\
\end{align}
$$ \int\frac {(1+\frac1 {x^3})^{1/3}}x dx = -\sqrt[3]{1 + x^{-3} } +
\frac{1}{3} \ln( \sqrt[3]{1 + x^{-3} } -1 ) \\ - \frac{1}{6}
\ln(\sqrt[3]{1 + x^{-3} } ^2 + \sqrt[3]{1 + x^{-3} } + 1) -
\frac{1}{\sqrt 3} \arctan ( \frac{2\sqrt[3]{1 + x^{-3} } + 1}{ \sqrt
3} ) + cst $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2812681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the value of a sum using Riemann sum theorem
Question: Find the value of $\sum_{i=1}^{n}(\frac{1}{n-i})^{c}$ for large $n$.
\begin{align}
\sum_{i=1}^{n}(\frac{1}{n-i})^{c}
& = \sum_{i=1}^{n}(\frac{1}{n})^{c}(\frac{1}{1-\frac{i}{n}})^{c}
\\ & = \frac{n}{n} \times \sum_{i=1}^{n}(\frac{1}{n})^{c}(\frac{1}{1-\frac{i}{n}})^{c}
\\ & = n(\frac{1}{n})^{c} \sum_{i=1}^{n}\frac{1}{n}(\frac{1}{1-\frac{i}{n}})^{c} \qquad(1)
\end{align}
Let $f(x) = (\frac{1}{1-x})^{c}$, by using Riemann-sum theorem, we have
\begin{align}
\lim_{n\rightarrow \infty}\sum_{i=1}^{n}\frac{1}{n}(\frac{1}{1-\frac{i}{n}})^{c}
& = \int_{0}^{1} (\frac{1}{1-x})^{c} = A \qquad(2)
\end{align}
By using $(1)$ and $(2)$, for sufficently large $n$, we have
$$\bbox[5px,border:2px solid #C0A000]{\sum_{i=1}^{n}(\frac{1}{n-i})^{c} = A\times n(\frac{1}{n})^{c}}$$
The presented proof has a problem, $f(x)$ is not defined in the closed interval $[0,1]$. How can I solve this problem?
Definition (Riemann-sum theorem) Let $f(x)$ be a function dened on a closed interval $[a, b]$. Then, we have
$$\lim_{n\rightarrow \infty}\sum_{i=1}^{n}f\Big(a +(\frac{b - a}{n})i\Big)\frac{1}{n}=\int_{a}^{b}f(x)dx$$
| \begin{align}\label{eq:7777}
& \frac{2}{\sqrt{n-i} + \sqrt{n-i+1}} \leq \frac{1}{\sqrt{n-i}} \leq \frac{2}{\sqrt{n-i} + \sqrt{n-i-1}} \nonumber\\
& \qquad \Rightarrow 2(\sqrt{n-i+1} - \sqrt{n-i}) \leq \frac{1}{\sqrt{n-i}} \leq 2(\sqrt{n-i} - \sqrt{n-i-1}) \nonumber\\
& \qquad \Rightarrow 2 \sum_{i=1}^{n-1}(\sqrt{n-i+1} - \sqrt{n-i}) \leq \sum_{i=1}^{n-1} \frac{1}{\sqrt{n-i}} \leq 2 \sum_{i=1}^{n-1}(\sqrt{n-i} - \sqrt{n-i-1}) \nonumber\\
& \qquad \Rightarrow 2 (\sqrt{n}-1) \leq \sum_{i=1}^{n-1} \frac{1}{\sqrt{n-i}} \leq 2 \sqrt{n-1}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that if ${x_1, x_2, x_3}$ are roots of ${x^3 + px + q = 0}$ then ${x_1^3+x_2^3 + x_3^3 = 3x_1x_2x_3}$ How to prove that ${x_1^3+x_2^3 + x_3^3 = 3x_1x_2x_3}$ holds in case ${x_1, x_2, x_3}$ are roots of the polynomial?
I've tried the following approach:
If $x_1$, $x_2$ and $x_3$ are roots then
$$(x-x_1)(x-x_2)(x-x_3) = x^3+px+q = 0$$
Now find the coefficient near the powers of $x$:
$$
x^3 - (x_1 + x_2 + x_3)x^2 + (x_1x_2 + x_1x_3 + x_2x_3)x - x_1x_2x_3 = x^3+px+q
$$
That means that I can write a system of equations:
$$
\begin{cases}
-(x_1 + x_2 + x_3) = 0 \\
x_1x_2 + x_1x_3 + x_2x_3 = p \\
- x_1x_2x_3 = q
\end{cases}
$$
At this point I got stuck. I've tried to raise $x_1 + x_2 + x_3$ to 3 power and expand the terms, but that didn't give me any insights. It feels like I have to play with the system of equations in some way but not sure what exact.
| Rewrite $x_1,x_2,x_3$ with $a,b,c$. From first Vieta formula we have $$a+b+c=0$$
so $a+b=-c$ and so on...
Now $$a^3+b^3+c^3= (a+b)(a^2-ab+b^2)+c^3 = c(\underbrace{-a^2+ab-b^2+c^2}_I)$$
Since
$$I = -a^2+ab-b^2+c^2 = a(b-a)+(c-b)(c+b) = $$ $$a(b-a)-a(c-b) = a(2b-a-c)=a(2b+b)=3ab$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Simplifying nested radicals with higher-order radicals I've seen that $$\sin1^{\circ}=\frac{1}{2i}\sqrt[3]{\frac{1}{4}\sqrt{8+\sqrt{3}+\sqrt{15}+\sqrt{10-2\sqrt{5}}}+\frac{i}{4}\sqrt{8-\sqrt{3}-\sqrt{15}-\sqrt{10-2\sqrt{5}}}}-\frac{1}{2i}\sqrt[3]{\frac{1}{4}\sqrt{8+\sqrt{3}+\sqrt{15}+\sqrt{10-2\sqrt{5}}}-\frac{i}{4}\sqrt{8-\sqrt{3}-\sqrt{15}-\sqrt{10-2\sqrt{5}}}}.$$
But then someone was able to simplify this neat, but long, expression with higher-order radicals, and they said they used De Moivre's theorem: $$\sin1^{\circ}=\frac{1}{2i}\sqrt[30]{\frac{\sqrt{3}}{2}+\frac{i}{2}}-\frac{1}{2i}\sqrt[30]{\frac{\sqrt{3}}{2}-\frac{i}{2}}.$$
I have been looking at this for a while now, and I cannot see how they were able to successfully do this. I am very impressed by the result and would like to use a similar technique to simplify nested radicals in the future.
Edit: It seems like the person who originally used De Moivre's theorem did not use it to directly simplify the longer radical expression, but rather found $\sin1^{\circ}$ by the method I figured out in my answer to this question. I do think there is limited value to writing the exact value of, say, $\sin1^{\circ}$ out, but which way do you think is better, the longer combination of square and cube roots, or the compact thirtieth-root?
| $\sin \theta = \dfrac{e^{i\theta}-e^{-i\theta}}{2i}$
$\begin{align*}\sin 1^\circ & = \sin\left(\dfrac{\pi}{180} \right) \\ & = \dfrac{e^{i\tfrac{\pi}{180}}-e^{-i\tfrac{\pi}{180}}}{2i} \\ & =\dfrac{\left(e^{i\tfrac{\pi}{6}}\right)^{1/30}-\left(e^{-i\tfrac{\pi}{6}}\right)^{1/30}}{2i} \\ & = \dfrac{\sqrt[30]{\cos \left( \dfrac{\pi}{6} \right) + i \sin \left( \dfrac{\pi}{6} \right)} - \sqrt[30]{\cos\left( \dfrac{\pi}{6} \right) - i \sin \left( \dfrac{\pi}{6} \right) } }{2i} \\ & = \dfrac{\sqrt[30]{\dfrac{\sqrt{3}}{2} + \dfrac{i}{2}} - \sqrt[30]{\dfrac{\sqrt{3}}{2} - \dfrac{i}{2}} }{2i}\end{align*}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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I am stuck with some details when graphing this function
Let $f(x) = \frac{x^3 + 1}{x}. $ Plot the function using derivatives criteria.
First of all, $f$ is indeterminate at $x=0.$ So I would have to deal with 2 intervals, separately: $(- \infty, 0)$, and $(0, \infty)$.
I would also have to consider:
$\lim_{\to \infty} f(x) = \lim_{\to \infty} \frac{1}{x} + x^2 = 0 + \infty = \infty$
$\lim_{\to -\infty} f(x) = \lim_{\to -\infty} \frac{1}{x} + x^2 = 0 + (-\infty)^2 = \infty$
If $x\neq 0, \frac{x^3 + 1}{x} = 0 \implies x^3 + 1 = 0,$ so the only root in $\mathbb{R}$ is $x=-1.$
If $x>0, \lim_{x \to 0} \frac{x^3 + 1}{x} = ( \lim_{x \to 0} \frac{1}{x} ) ( \lim_{x \to 0} x^3 + 1) = (\infty)(1) = \infty $
And if $x <0, \lim_{x \to 0} \frac{x^3 + 1}{x} = ( \lim_{x \to 0} \frac{1}{x} ) ( \lim_{x \to 0} x^3 + 1) = (-\infty)(1) = -\infty $
Then $f '(x) = \frac{2x^3 -1}{x^2}$
It should happen that, at $(\infty, 0)$, $f$ is decreasing. And $x<0 \implies 2x^3 -1 < 0 \implies \frac{2x^3 -1}{x^2} < 0.$ So $f'(x) < 0 \implies f$ is decreasing at $(-\infty, 0)$
and it should also happen that, at $(0, \infty)$, the function is increasing.
But, if $x<0$, it doesn't neccessarily happen that $2x^3 -1 > 0,$ So I am stuck here. However if i could assume that $2x^3 -1 > 0, $ then $f '(x) = \frac{2x^3 -1}{x^2} > 0,$ so $f$ would be increasing.
To find critical points, $f '(x) = \frac{2x^3 -1}{x^2} = 0 \implies_{x \neq 0} x = \frac{1}{\sqrt[3]{2}},$ or $x = -\frac{1}{\sqrt[3]{2}} ,$ and $f''(x) = \frac{2x^3+2}{x^3},$ so $f''(\frac{1}{\sqrt[3]{2}}) = 4(\frac{1}{2} + 1) = 6 >0,$ that would mean that $f$ has a local minimum. So my last question is why doesn't $x = -\frac{1}{\sqrt[3]{2}}
$, the other critical point appears in the graph?
| A qualitative approach.
The function can be written as $x^2+\dfrac1x$, and is the sum of a parabola and an equilateral hyperbola.
The hyperbola has a vertical asymptote at $x=0$ and the horizontal asymptote $y=0$. Hence for small $|x|$ it is dominant, and for large $|x|$, it is neglectible and the function gets closer and closer to the parabola.
In the negatives, both functions are decreasing and the sum is decreasing (to $-\infty$).
In the positives we sum increasing and decreasing and we cannot conclude about the variation. The first derivative is $2x-\dfrac1{x^2}$, which has a single positive root, hence the function has a single minimum.
This is confirmed by a plot:
| {
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Does $4(x^3+2x^2+x)^3(3x^2+4x+1) = 4x^3(3x+1)(x+1)^7$? Please provide proof
The answer in the back of the book is different to both my calculations and also the online calculator I crosschecked my answer with...
The Question:
"Differentiate with respect to $x$:"
$
(x^3+2x^2+x)^4
$
My Answer: $4(x^3+2x^2+x)^3(3x^2+4x+1)$
The Book's Answer: $4x^3(3x+1)(x+1)^7$
| $ \frac{d}{dx}(x^3+2x^2+x)^4 = 4(x^3+2x^2+x)^3\cdot(\frac{d}{dx}(x^3+2x^2+x))$ by the chain rule, then we get: $ \frac{d}{dx}(x^3+2x^2+x)^4 = 4(x^3+2x^2+x)^3\cdot(3x^2+4x+1) = 4x^3(x^2+2x+1)^3\cdot(3x^2+4x+1) $.
Now the rest should just be algebra: (note that $(x+1)^2=(x^2+2x+1)$)
$ 4x^3((x+1)^2)^3(3x^2+4x+1) = 4x^3(x+1)^6(3x^2+4x+1) $,
then show that $ (x+1)(3x+1)= (3x^2 +4x + 1) $, and insert in the above equation:
$ \frac{d}{dx}(x^3+2x^2+x)^4 = 4x^3(x+1)^7(3x+1) $, and you're done.
| {
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"timestamp": "2023-03-29T00:00:00",
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Intersection spheroid -plane I have a spheroid $S$ with $a$ is the equatorial radius, and $b$ is the polar radius, and ($a>b$)
I would get the intersection between $S$ and a plane $P: ux+vy+wz+d=0$
Then calculating the semi-axis major and minor and the center of this intersection.
There is always an intersection (resulting an ellipse, a circle, or a point), so no empty intersection.
My try:
From $P$, I get $x = (d-vy-wz)/u$, then I substituted this $x$ into the equation of $S$, I got a general equation of conic in the form $Ay^2+Byz+Cz^2+Dy+Ez+F=0$
Comparing that to my intersection's equation, I get:
$A=bv^2+bu^2$
$B=2vwb^2$
$C=bw^2+au^2$
$D=2dvb^2$
$E=2dwb^2$
$F=b^2d^2-a^2b^2u^2$
I calculated $θ$ then I rotated in order to remove yz-term.
then the new coefficients are:
$A'= Acos^2(θ) + Bcos(θ)sin(θ) + Csin^2(θ)$
$B'= 0$
$C'= Asin^2(θ) - Bcos(θ)sin(θ) + Ccos^2(θ)$
$D'= Dcos(θ)+ Esin(θ)$
$E'= -Dsin(θ) + Ecos(θ)$
$F'= F$
calculating the semi-axis major and minor of the intersection $a'$ and $b'$
$a'=sqrt((-4F'A'C' + C'D'^2 + A'E'^2)/(4A'C'^2))$
$b'=sqrt((-4F'A'C' + C'D'^2 + A'E'^2)/(4A'^2C'))$
The coordinates of the intersection's center $C$ are $y'_c$ and $z'_c$:
$y'_c= -D'/2A'$
$z'_c= -E'/2C'$
rotating back by $θ$ to get the coordinates $y_c$ and $z_c$, I got:
$y_c = y'_ccos(θ) - z'_csin(θ)$
$z_c = y'_csin(θ) + z'_ccos(θ)$
My question:
I compared these results with geogebra, I found that these semi-axis major and minor $a'$ and $b'$ are of the intersection projected onto the yOz plane, not of the intersection itself in 3D, also the same problem with the coordinates of the center $C$, that it must be given in 3D coordinates!
Any help please
| Equation of spheroid:
$$\frac{x^2+y^2}{a^2}+\frac{z^2}{b^2}=1 \tag{1}$$
Let $(X,Y,Z)$ be the centre of the section, then equation of plane section is
$$\frac{Xx+Yy}{a^2}+\frac{Zz}{b^2}=\frac{X^2+Y^2}{a^2}+\frac{Z^2}{b^2} \tag{2}$$
Comparing coefficients,
$$-\frac{(u,v,w)}{d}=
\frac{\left( \dfrac{X}{a^2}, \dfrac{Y}{a^2}, \dfrac{Z}{b^2} \right)}
{\dfrac{X^2+Y^2}{a^2}+\dfrac{Z^2}{b^2}}
\implies
(X,Y,Z)=-
\frac{d\left( \dfrac{u}{a^2}, \dfrac{v}{a^2}, \dfrac{w}{b^2} \right)}{\dfrac{u^2+v^2}{a^2}+\dfrac{w^2}{b^2}}$$
$(1)-(2)\times 2$,
$$\frac{(x-X)^2+(y-Y)^2}{a^2}+\frac{(z-Z)^2}{b^2}=1-\frac{X^2+Y^2}{a^2}-\frac{Z^2}{b^2}$$
which is a re-scaled spheroid of $(1)$ translated by $(X,Y,Z)$.
Intersection occurs only when
$$\lambda^2=1-\dfrac{X^2+Y^2}{a^2}-\dfrac{Z^2}{b^2} \ge 0$$
By symmetry, putting
*
*$z=Z$ gives the vertices on the major axis;
*$(x,y)=k(X,Y)$ gives the vertices on the minor axis.
With an aid of Mathematica,
*
*the semi-major axis is $$a\sqrt{1-\frac{X^2+Y^2}{a^2}-\frac{Z^2}{b^2}}$$
*the semi-minor axis is $$
\frac
{ab\sqrt{
\left( 1-\dfrac{X^2+Y^2}{a^2}-\dfrac{Z^2}{b^2} \right)
\left( \dfrac{X^2+Y^2}{a^4}+\dfrac{Z^2}{b^4} \right)}}
{\sqrt{\dfrac{X^2+Y^2}{a^2}+\dfrac{Z^2}{b^2}}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve $\cos x +\cos y - \cos(x+y)=\frac 3 2$ Solve $\cos x +\cos y - \cos(x+y)=\frac 3 2$ where $x,y\in [0,\pi]$.
I am trying to solve this but I am stuck. I know that $x=y=\pi/3$ is a solution but how do I show this is the only one? I think there are no others! Hints would be appreciated
| For $x,y \in [0,\pi]$
\begin{align}
\cos x &= u \\
\cos y &= v \\
\sin x &= \sqrt{1-u^2} \\
& \ge 0 \\
\sin y &=\sqrt{1-v^2} \\
& \ge 0 \\
u+v-uv+\sqrt{1-u^2} \sqrt{1-v^2} &= \frac{3}{2} \\
\sqrt{1-u^2} \sqrt{1-v^2} &= \frac{3}{2}+uv-u-v
\end{align}
By AM $\ge$ GM,
$$\frac{1-u^2+1-v^2}{2} \ge \frac{3}{2}+uv-u-v \tag{1}$$
equality holds if $1-u^2=1-v^2$ and rearranging $(1)$ gives
$$0 \ge \frac{(u+v-1)^2}{2} \tag{2}$$
But $\dfrac{(u+v-1)^2}{2}$ is non-negative which forces the equality to hold.
Therefore,
$$u+v=1 \land u^2=v^2$$
$$u+v=1 \land (u+v)(u-v)=0$$
$$u=v=\frac{1}{2} \implies x=y=\frac{\pi}{3}$$
Further points to be noticed
Considering the quadratic in $u$ of
$$(1-u^2)(1-v^2)=\left( \frac{3}{2}+uv-u-v \right)^2$$
Discriminant $$\Delta_u =4v^4-4v^3-3v^2+4v-1=(v^2-1)(2v-1)^2$$
*
*$v^2>1 \implies \Delta_u>0$ but $v$ is beyond the range of cosine
*$v=1 \implies \Delta_u=0$ but gives $0=\frac{1}{4}$, so no solutions
*$v=-1 \implies \Delta_u=0$ but gives $u=\frac{5}{4}$ which is beyond the range of cosine
*$v=\frac{1}{2} \implies \Delta_u=0$ that gives $u=\frac{1}{2}$
*$\Delta_u<0$ elsewhere
| {
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Prove by induction that $n^4-4n^2$ is divisible by 3 for all integers $n\geq1$. For the induction case, we should show that $(n+1)^4-4(n+1)^2$ is also divisible by 3 assuming that 3 divides $n^4-4n^2$. So,
$$ \begin{align} (n+1)^4-4(n+1)&=(n^4+4n^3+6n^2+4n+1)-4(n^2+2n+1)
\\ &=n^4+4n^3+2n^2-4n-3
\\ &=n^4+2n^2+(-6n^2+6n^2)+4n^3-4n-3
\\ &=(n^4-4n^2) + (4n^3+6n^2-4n)-3
\end{align}$$
Now $(n^4-4n^2)$ is divisible by 3, and $-3$ is divisible by 3. Now I am stuck on what to do to the remaining expression.
So, how to show that $4n^3+6n^2-4n$ should be divisible by 3? Or is there a better way to prove the statement in the title? Thank you!
| Here is a way to prove it without induction. Notice that $$n^4-4n^2=n^2(n-2)(n+2)$$
If $3\not\mid n$ then $3\mid (n+1)-3$, that is $3\mid n-2$, or $3\mid (n+2)$.
| {
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Choose $y\in\mathbb{R}$ so that $\sin y=\frac{u}{\sqrt{u^2+v^2}}$ and $\cos y=\frac{v}{\sqrt{u^2+v^2}}$. I know the identity $\sin^2y+\cos^2y=1$. Also, I notice that
$$\left(\frac{u}{\sqrt{u^2+v^2}}\right)^2+\left(\frac{v}{\sqrt{u^2+v^2}}\right)^2=1$$
without $u,v$ vanishing simultaneously. But I am not sure whether $\exists y\in\mathbb{R}$ s.t. $\sin y=\frac{u}{\sqrt{u^2+v^2}}$ and $\cos y=\frac{v}{\sqrt{u^2+v^2}}$. Does anyone have an idea? Thanks a lot.
| The identity $$\left(\frac{u}{\sqrt{u^2+v^2}}\right)^2+\left(\frac{v}{\sqrt{u^2+v^2}}\right)^2=1$$
means that the point $\left(\frac{u}{\sqrt{u^2+v^2}},\frac{v}{\sqrt{u^2+v^2}}\right)$ is on the unit circle $x^2+y^2=1$. By definition, there exists an angle $y$ (actually, an infinity of angles) such that $\cos(y)=\frac{u}{\sqrt{u^2+v^2}}$ and $\sin(y)=\frac{v}{\sqrt{u^2+v^2}}$.
Assume for example that $u>0$ and $v>0$ (*), so that $y$ is in the first quadrant. Then, $y=\arcsin(\frac{v}{\sqrt{u^2+v^2}})+2k\pi$ and $x=\arccos(\frac{u}{\sqrt{u^2+v^2}})+2k\pi$, where $k$ is an integer.
(*) For other possible signs of $u$ and $v$, we obtain similar formulas.
| {
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Finding the mean of all 9-digit numbers formed from four $4$s and five $5$s
I need to find the mean of the numbers ($9$-digit) formed of four $4$s and five $5$s .
MY WORK:
In order to find the sum I do the following and find sum of digits :
$$25\times5!=3000$$
$$16\times4!=384$$
So, sum of all possible numbers :
$$3384(1+10+10^2+ ... + 10^8)$$
$$=3384\times10^7$$
For finding the amount of numbers formed, I do :
$$\frac{9!}{4!5!}$$
$$=126$$
This, the mean is :$$\frac{3384\times10^7}{126}$$
I don't know if I'm correct or not...
| For the reader with an interest in generating functions we have
from first principles the closed form
$${9\choose 4}^{-1}
\left. \frac{\partial}{\partial w} [z^4]
\prod_{q=0}^8 (z \times w^{4 \times 10^q} +
w^{5 \times 10^q}) \right|_{w=1.}$$
We may treat the derivative first and obtain
$${9\choose 4}^{-1}
\left. [z^4]
\prod_{q=0}^8 (z \times w^{4 \times 10^q} +
w^{5 \times 10^q}) \\ \times
\sum_{q=0}^8 \frac{4\times 10^q z \times w^{4 \times 10^q - 1} +
5 \times 10^q \times w^{5 \times 10^q-1}}{z \times w^{4 \times 10^q} +
w^{5 \times 10^q}}
\right|_{w=1.}$$
Evaluating at $w=1$ yields
$${9\choose 4}^{-1}
[z^4]
\prod_{q=0}^8 (z + 1) \times
\sum_{q=0}^8 \frac{4\times 10^q z +
5 \times 10^q}{z + 1}
\\ = {9\choose 4}^{-1}
[z^4] (1+z)^8
\sum_{q=0}^8 (4\times 10^q z + 5 \times 10^q)
\\ = {9\choose 4}^{-1}
[z^4] (1+z)^8
(444.444.444 z + 555.555.555)
\\ = {9\choose 4}^{-1} 111.111.111
[z^4] (1+z)^8 (4z + 5)
\\ = {9\choose 4}^{-1} 111.111.111
\left(4{8\choose 3} + 5 {8\choose 4}\right).$$
We have confirmed the result by the replies that were
first to appear, namely
$$\bbox[5px,border:2px solid #00A000]{506172839.}$$
| {
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"question_score": "3",
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Does hypergeometric distribution apply in this case? A box contains 4 red balls, 5 blue balls and 2 green balls.
Six balls are drawn without replacement, find the chance that exactly one of the first two is blue AND exactly two of the last four are blue.
The probability that I came up with is:
$\frac {{\binom {2}{1}}\binom {4}{2}}{\binom {6}{3}}$
I feel like this is not correct, as the probability came out to be 60%
I'm kind of wondering if hypergeometric distribution applies in this problem if so how do I properly set up the equation?
Thank you!
|
A box contains 4 red balls, 5 blue balls and 2 green balls.
Six balls are drawn without replacement, find the chance that exactly one of the first two is blue AND exactly two of the last four are blue.
The total ways to draw groups of $2$ and $4$ balls from a heap of $4+5+2$ (ie $11$) is $\binom {11}2\binom{11-2}{4}$. This is also written as $\binom{11}{2,4,5}$ (which is a multinomial coefficient).
The ways to draw the first two as exactly $1$ from $5$ blue and one from $4+2$ (ie $6$) not blues, and the last four as exactly $2$ from $5-1$ remaining blue and $2$ from $6-1$ remaining not blues is $\binom{5}{1}\binom {6}{1}\binom{5-1}{2}\binom{6-1}{2}$ or $\binom{5}{1,2,2}\binom{6}{1,2,3}$.
The probability is therefore $$\dfrac{\dbinom{5}{1,2,2}\dbinom{6}{1,2,3}}{\dbinom{11}{2,4,5}}$$
Alternatively we may seek the probability for placing the five blue balls in: one from the two first positions, two from the last four, and leaving two among the five unselected, when selecting an arrangement from the eleven balls. $$\dfrac{\dbinom 21\dbinom 4 2\dbinom 52}{\dbinom{11}{5}}$$
These are, of course, equal.
| {
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"timestamp": "2023-03-29T00:00:00",
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Short way to evaluate $\int_{0}^{1}\ln(\frac{a-x^2}{a+x^2})\cdot\frac{dx}{x^2\sqrt{1-x^2}}$ I would like to evaluate this integral, $$I=\int_{0}^{1}\ln\left(\frac{a-x^2}{a+x^2}\right)\cdot\frac{\mathrm dx}{x^2\sqrt{1-x^2}}$$
An approach:
1. Using integration by parts
$u=\ln\left(\frac{a-x^2}{a+x^2}\right)$, $du=\frac{x^2+a}{x^2-a}\left[\frac{2x}{a+x^2}+\frac{2x(a-x^2)}{a+x^2}\right]dx$
$dv=\frac{dx}{x^2\sqrt{1-x^2}}$, $v=-\frac{\sqrt{1-x^2}}{x}$
$$I=2\int_{0}^{1}\frac{\sqrt{1-x^2}}{x^2+a}dx-2\int_{0}^{1}\frac{\sqrt{1-x^2}}{x^2-a}dx=2I_1-2I_2$$
Integral $I_1$: Make a substitution of: $x=\sin(v)$, $v=\arcsin(x)$, $dx=\cos(v)dv$
$$I_1=\int_{0}^{\pi/2}\frac{\cos^2(v)}{a+\sin^2(v)}dv$$
Using trig to rewrite: $$I_1=\int_{0}^{\pi/2}\sec^2(v)\frac{dv}{[1+\tan^2(v)][(a+1)\tan^2(v)+a]}$$
Make another substitiution of: $v=\tan(y)$, $dy=\frac{1}{\sec^2(v)}dv$ and using partial fraction decomp:
$$I_1=(a+1)\int_{0}^{\infty}\frac{dy}{a+(a+1)y^2}-\frac{\pi}{2}=(a+1)I_3-\frac{\pi}{2}$$
With this substitution: $t=\frac{\sqrt{a+1}}{\sqrt{a}}y$, $dy=\frac{\sqrt{a}}{\sqrt{a+1}}dt$
$$I_3=\frac{1}{\sqrt{a(a+1)}}\cdot \frac{\pi}{2}$$
$$I_1=\frac{\pi}{2}\left(\frac{\sqrt{a+1}}{\sqrt{a}}-1\right)$$
In the same manner, $$I_2=\frac{\pi}{2}\left(\frac{\sqrt{a-1}}{\sqrt{a}}-1\right)$$
Finally, $$I=\pi\left(\frac{\sqrt{a-1}-\sqrt{a+1}}{\sqrt{a}}\right)$$
I am looking for another short approach of evaluating integral $I$
| Assuming $a>1$, the original integral equals
$$ \frac{1}{2}\int_{0}^{1}\log\left(\frac{a-x}{a+x}\right)\frac{dx}{x\sqrt{x(1-x)}}\,dx=-\int_{0}^{1}\frac{\text{arctanh}(x/a)}{x\sqrt{x(1-x)}}\,dx \tag{1}$$
or
$$ -\sum_{n\geq 0}\int_{0}^{1}\frac{x^{2n-1/2}}{(2n+1)a^{2n+1}\sqrt{1-x}}\,dx=-\sum_{n\geq 0}\frac{\pi\binom{4n}{2n}}{(2n+1)16^n a^{2n+1}}\tag{2}$$
hence it is enough to recall the generating function for Catalan numbers to recover the wanted identity.
It is worth noticing that the Maclaurin series of the squared arcsine allows to compute the similar integral
$$ \int_{0}^{1}\log\left(\frac{a^2+x^2}{a^2-x^2}\right)\frac{dx}{\color{red}{x}\sqrt{1-x^2}} $$
or the hypergeometric function $\phantom{}_4 F_3\left(\tfrac{1}{2},\tfrac{1}{2},1,1;\tfrac{3}{4},\tfrac{5}{4},\tfrac{3}{2};z\right)$ in terms of $\arcsin^2$ and $\text{arcsinh}^2$, in a very similar and efficient way.
| {
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What is going wrong in this log expansion? I am getting a weird result here:
Let $p_1 = q_1 + \Delta$ and Let $p_2 = q_2 - \Delta$
I use the expansion $\log(1-x) = -x -x^2/2 -x^3/3 - ...$ in the third step. This expansion is valid for $x<1$ so there should be no problem with that.
\begin{align*}
D(\vec{p}||\vec{q}) &= p_1\log{\frac{p_1}{q_1}} + p_2\log{\frac{p_2}{q_2}}\\
&= p_1\log{\frac{q_1 + \Delta}{q_1}} + p_2\log{\frac{q_2 - \Delta}{q_2}}\\
&= p_1(\frac{\Delta}{q_1} - \frac{\Delta^2}{q_1^2} + \frac{\Delta^3}{q_1^3} - ...) + p_2(-\frac{\Delta}{q_2} - \frac{\Delta^2}{q_2^2} - \frac{\Delta^3}{q_2^3} - ...)\\
&= (q_1 + \Delta)(\frac{\Delta}{q_1} - \frac{\Delta^2}{q_1^2} + \frac{\Delta^3}{q_1^3} - ...) + (q_2 - \Delta)(-\frac{\Delta}{q_2} - \frac{\Delta^2}{q_2^2} - \frac{\Delta^3}{q_2^3} - ...)\\
&=0
\end{align*}
The last equality follows from comparing terms of powers of $\Delta$ (am I not allowed to do that?).
| There are at least two errors:
$$p_1\log{\frac{q_1 + \Delta}{q_1}} = p_1\left(\frac{\Delta}{q_1} - \frac{\Delta^2}{\color{red}{2}q_1^2} - \frac{\Delta^3}{\color{red}{3}q_1^3} - \cdots\right)$$
Similar for the second term.
And even if the low order terms cancel, the result is not zero. Using Maple I get (if I did not make mistakes)
$$D(\vec{p}||\vec{q}) = \frac{1}{2}\left(\frac{1}{q_1}+\frac{1}{q_2}\right) \Delta^2 + O(\Delta^3)$$
| {
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Range of $(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2$ is
If $a_{1},a_{2},a_{3},a_{4}\in \mathbb{R}$ and $a_{1}+a_{2}+a_{3}+a_{4}=0$ and $a^2_{1}+a^2_{2}+a^2_{3}+a^2_{4}=1.$
Then Range of $$E =(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2$$ is
Try:
From $$(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2$$
$$=2(a^2_{1}+a^2_{2}+a^2_{3}+a^2_{4})+2(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{4}+a_{4}a_{1})$$
$$=2-2(a_{1}+a_{3})(a_{2}+a_{4})=2+2(a_{1}+a_{3})^2\geq 2$$
and equality hold when $\displaystyle a_{1}=-a_{3}$ and $a_{2}=-a_{4}$
Could some help me how to find its upper bound, Thanks
| You can square the given equation:
$$a_{1}+a_{2}+a_{3}+a_{4}=0 \Rightarrow \\
a_1^2+a_2^2+a_3^2+a_4^2+2(a_1a_2+a_1a_3+a_1a_4+a_2a_3+a_2a_4+a_3a_4)=0 \Rightarrow \\
2(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{4}+a_{4}a_{1})=-1-2a_1a_3-2a_2a_4.$$
Hence:
$$(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2=\\
=2(a^2_{1}+a^2_{2}+a^2_{3}+a^2_{4})\color{red}{-}2(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{4}+a_{4}a_{1})=\\
2-(-1-2a_1a_3-2a_2a_4)=3+2a_1a_3+2a_2a_4\le 3+(a_1^2+a_3^2)+(a_2^2+a_4^2)=4.$$
the equality occurs for $a_1=a_3, a_2=a_4$.
| {
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Need hint for equation system I stumbled upon the following in an old math book and could use a hint:
The sum of the volumes of two cubes is 35, whereas the sum of the
surface area of the two cubes is 78. Calculate the lengths of the
sides of the two cubes.
So, if we let $x$ be the length of the side of one cube, and $y$ be the length of the side of the other, we must have:
$$x^3 + y^3 = 35$$
$$6x^2 + 6y^2 = 78$$
i.e.:
$$x^3 + y^3 = 35$$
$$x^2 + y^2 = 13$$
How to proceed from there? If I try the obvious things like isolating x or y and plug into the other one I get a mess. Is there some type of substitution required?
| $35=x^3+y^3=(x+y)(x^2-xy+y^2)=$
$=(x+y)(13-xy)$
But
$(x+y)^2=x^2+y^2+2xy=13+2xy$ and so
$xy=\frac{1}{2}[(x+y)^2-13]$ and it can be replaced to first equation to get that
$35=(x+y)(13-\frac{1}{2}((x+y)^2-13))$
and if you define $z:=x+y$ you must resolve
$70=z(39-z^2)$
(You can observe that the solutions of the equation are $z=2$,$z=5$ and $z=-7$)
So the form of the solutions of your equations is $(x,z-x)$ where $z$ is the solution of that particular equation.
Now you can calulate
$x^2+(z-x)^2=13$
and you have that the unique solutions of your sistem are of the form $(x,z-x)$ such that verify the equation:
$ 2x^2-2zx+(z^2-13)=0$
| {
"language": "en",
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"question_score": "1",
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} |
Does this pattern continue $\lfloor\sqrt{44}\rfloor=6, \lfloor\sqrt{4444}\rfloor=66,\dots$? By observing the following I have a feeling that the pattern continues.
$$\lfloor \sqrt{44} \rfloor=6$$
$$\lfloor \sqrt{4444} \rfloor=66$$
$$\lfloor \sqrt{444444} \rfloor=666$$
$$\lfloor \sqrt{44444444} \rfloor=6666$$
But I'm unable to prove it. Your help will be appreciated.
| This can be proved a bit more simple.
The general term can be written as $4 \frac{10^{2n} - 1}{9}$. Taking square root will give you $\frac{2}{3} \sqrt{10^{2n} - 1}$. As $\frac{2}{3} \approx 0.66666666666$ and $\sqrt{10^{2n} - 1} \approx \sqrt{10^{2n}} = 10^n$, this explains why the result is $6;66;666;...$ etc. To prove rigorously, observe that: $66...66 = \frac{2}{3} 10^n - \frac{2}{3} < \frac{2}{3} \sqrt{10^{2n} - 1} < \frac{2}{3} 10^n = 66...66.67$.
| {
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"source": "stackexchange",
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Does the equation $q^2\, n(n+1) = p^2\, m(m+1)$ have solutions only if $m(m+1) = q^2\, k(k+1)$, $kI have the following equation
\begin{equation}
q^2\, n(n+1) = p^2\, m(m+1), \qquad(1)
\end{equation}
where $m,n,p,q \in \mathbb{N}$ and $p>q>2$ and $\text{gcd}(p,q)=1$.
I could find numerically solutions only when $m(m+1) = q^2\, k(k+1)$ and $n(n+1) = p^2\, k(k+1)$, where $k<m$, $k \in \mathbb{N}$. For example:
*
*$p=35$, $q=6$, $m=8$, $n=49$, so
$$m(m+1)=8\cdot 9=6^2 \cdot (1 \cdot 2)$$
$$n(n+1)=49 \cdot 50=35^2 \cdot (1 \cdot 2)$$
*$p=99$, $q=10$, $m=24$, $n=242$, so
$$m(m+1)=24\cdot 25=10^2 \cdot (2 \cdot 3)$$
$$n(n+1)=242\cdot 243 = 99^2 \cdot (2 \cdot 3)$$
Is it possible to prove (or disprove) that the equation (1) has solution only if $m(m+1) = q^2\, k(k+1)$ and $n(n+1) = p^2\, k(k+1)$, $k<m$ ?
Edit:
From eq. (1) and the given conditions it follows that $q^2 | m(m+1)$ and $p^2 | n(n+1)$. Therefore
\begin{align}
m(m+1) &= q^2 d, \qquad (2a)\\
n(n+1) &= p^2 d, \qquad (2b)\\
\end{align}
where $d = \text{gcd}(m(m+1),\, n(n+1))$. Since $m(m+1)$ and $n(n+1)$ are even numbers, it follows from (2) that $d$ is even non-square number. Numerically I could find solutions to the eq. (1) only if $d = k(k+1)$, where $k<m$, $k \in \mathbb{N}$.
| First of all, a number $N$ is of the form $m(m+1)$ if and only if $4N+1$ is a perfect square.
Since $p$ and $q$ are coprime, we have $q^2|m(m+1)$, so $$\frac{m(m+1)}{q^2}$$ is a positive integer and we have $$S:=\frac{m(m+1)}{q^2}=\frac{n(n+1)}{p^2}$$
Using the criterion above, we get that $$4Sq^2+1$$ and $$4Sp^2+1$$ must be a perfect square, so the $y$-values of the pairs $(x/y)$ solving the Pell-equation $$x^2-4Sy^2=1$$ include $y=p$ and $y=q$. If the fundamental solution is $(a/b)$ it is easy to see that every $y$ must be a multiple of $b$. But $p$ and $q$ are coprime, hence $b=1$. This means that $4S+1$ is a perfect square, so $S=k(k+1)$ and of course $k<m$, completing the proof.
| {
"language": "en",
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Prove $ \min \left(a+b+\frac1a+\frac1b \right) = 3\sqrt{2}\:$ given $a^2+b^2=1$ Prove that
$$ \min\left(a+b+\frac1a+\frac1b\right) = 3\sqrt{2}$$
Given $$a^2+b^2=1 \quad(a,b \in \mathbb R^+)$$
Without using calculus.
$\mathbf {My Attempt}$
I tried the AM-GM, but this gives $\min = 4 $.
I used Cauchy-Schwarz to get $\quad (a+b)^2 \le 2(a^2+b^2) = 2\quad \Rightarrow\quad a+b\le \sqrt{2}$
But using Titu's Lemma I get $\quad \frac1a+\frac1b \ge \frac{4}{a+b}\quad \Rightarrow\quad \frac1a+\frac1b \ge 2\sqrt{2}$
I'm stuck here, any hint?
| Without Calculus:
Multiplying both sides of $a+b+\dfrac 1a + \dfrac 1b \ge 3 \sqrt 2$ by $ab$ we get
$$a^2b+b^2a+a+b \ge 3 \sqrt2 ab$$
which factors as
$$(a+b)(ab+1) \ge 3 \sqrt2 ab$$
and squaring both sides yields
$$(a+b)^2(ab+1)^2 \ge 18(ab)^2$$
But $(a+b)^2 = 1+2ab$, and substituting $x = ab$ get
$$(1+2x)(x+1)^2\ge18x^2$$
Furthermore, we know $x \in\left [0, \dfrac 12 \right ]$ because $a^2+b^2 \ge 2ab$, so $\dfrac 12 \ge ab$.
Thus it remains to show that $f(x) = (1+2x)(x+1)^2 - 18x^2 \ge 0$ on $\left[0, \dfrac 12 \right]$. But expanding $f(x)$ and using the rational root theorem we can factor it as $f(x) = 2(x-1/2)(x^2-6x-1)$. Now $(x- 1/2)$ is non-positive on the required interval, and you can find the roots of $x^2-6x-1$ using the quadratic formula to show that it is negative on that interval.
Done!
| {
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"answer_id": 2
} |
Find the Determinant when $p$ and $q$ are roots of $x^2-2x+5=0$ If $p$ and $q$ are roots of $x^2-2x+5=0$ Then find value of
$$\Delta=\begin{vmatrix}
1 & 1+p^2+q^2 & 1+p^3+q^3\\
1+p^2+q^2& 1+p^4+q^4 & 1+p^5+q^5\\
1+p^3+q^3 & 1+p^5+q^5 & 1+p^6+q^6
\end{vmatrix}$$
My try: I tried to express the determinant as product of two determinants but in vain.
Secondly i tried taking $$p=\sqrt{5}\frac{1+2i}{\sqrt{5}}=\sqrt{5}(\cos t+i\sin t)$$ and $q$ its conjugate, but still its lengthy
| Use Viète's formulas:
$$p+q=2\qquad pq=5$$
$$p^2+q^2=(p+q)^2-2pq=-6$$
$$p^3+q^3=(p+q)(p^2+q^2-pq)=-22$$
$$p^4+q^4=(p^2+q^2)^2-2(pq)^2=-14$$
$$p^5+q^5=(p+q)(p^4+q^4)-pq(p^3+q^3)=82$$
$$p^6+q^6=(p^3+q^3)^2-2(pq)^3=234$$
from which we may easily work out the determinant as
$$\begin{vmatrix}
1 & 1-6 & 1-22\\
1-6& 1-14 & 1+82\\
1-22 & 1+82 & 1+234\end{vmatrix}=7344$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Number of ways of forming 10 student committee from 5 classes of 30 students each There are 5 classes with 30 students each. How many ways can a committee of 10 students be formed if each class has to have at least one student on the committee?
I figured that we first have to choose 5 people from each class, so there are $10^5$ options. There remain total of 29*5=145 students to choose from, and we can fill remaining 5 spots any way we want, i.e., $\binom{145}{5}$ options. Thus, I thought answer would be $$10^5 \dot\ \binom{145}{5} $$.
Apparently this is the wrong answer. I can see how the inclusion/exclusion principle also leads to the right answer. That is,
$$ {150 \choose 10}- 5{120 \choose 10} + {5 \choose 2}{90 \choose 10} - {5 \choose 3}{60 \choose 10} + {5 \choose 4}{30 \choose 10} $$
I don't see though what I overlooked and why my answer is different?
Any advice on when to simply use inclusion/exclusion and when to use other methods?
| Here is how it works without inclusion-exclusion.
There are 7 ways of partitioning 10 into 5 positive parts. The number may be checked here https://oeis.org/A008284
The problem is similar to a poker game hands problem but here we have 5 colors and 30 values.
For each kind of "hand" we have to do the specific calculus.
$case \ 10=6+1+1+1+1$ :
there are $\binom{5}{1}$ choices of the long color. For each one, there are $\binom{30}{6}$ choices of the values. For each other short color there are ${\binom{30}{1}}$ choices, for a total of :
$\binom{5}{1} \binom{30}{6}{\binom{30}{1}}^4$
$case \ 10=5+2+1+1+1$ :
there are $\binom{5}{1}$ choices of the long color. There are $\binom{4}{1}$ choices for the color of a pair. For the long color, there are $\binom{30}{5}$ choices of the values. For the pair there are ${\binom{30}{2}}$ choices; the total is for a total of:
$\binom{5}{1} \binom{4}{1} \binom{30}{5} \binom{30}{2} {\binom{30}{1}}^3$
Let's write for the sake of comparing the complete formula:
$$\binom{5}{1} \binom{30}{6}{\binom{30}{1}}^4 + \binom{5}{1} \binom{4}{1} \binom{30}{5} \binom{30}{2} {\binom{30}{1}}^3 + \binom {5}{1} \binom{4}{1}\binom{30}{4}\binom{ 30}{3}\binom{30}{1}^3+ \binom{5}{1} \binom{4}{2} \binom{ 30}{4} \binom{ 30}{2}^2 \binom{30}{1}^2 + \binom{5}{1} \binom{4}{2} \binom{ 30}{2} \binom{ 30}{3}^2 \binom{30}{1}^2 + \binom{5}{1} \binom{4}{1} \binom{ 30}{3} \binom{ 30}{1} \binom{30}{2}^3 + \binom{30}{2}^5 $$
After computing all seven cases, we get the required 645666069796875. I think, given the similarity with poker where there are involved a lot of money, someone would have invented in the last centuries a shorter way of computing hands.
The inclusion-exclusion is possible since we can manipulate that "at least one" and the calculus is shorter since the classes are equal.
| {
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"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
} |
Minimum value of $\frac{b+1}{a+b-2}$
If $a^2 + b^2= 1 $ and $u$ is the minimum value of the $\dfrac{b+1}{a+b-2}$, then find the value of $u^2$.
Attempt:
Then I tried this way: Let $a= bk$ for some real $k$.
Then I got $f(b)$ in terms of b and k which is minmum when $b = \dfrac{2-k}{2(k+1)}$ ... then again I got an equation in $k$ which didn't simplify.
Please suggest an efficient way to solve it.
| Try with $b=\cos 2x$ and $a= \sin 2x$.
\begin{eqnarray}{b+1\over a+b-2}&=& {2\cos^2 x\over -\cos^2x+2\sin x \cos x -3\sin^2x}\\
&=& {2\over -1+2\tan x -3\tan^2x}\\
&=& {2\over -1+2t -3t^2}
\end{eqnarray}
where $t= \tan x $. So the expression will take a minimum when quadratic function $g(t)=-3t^2+2t-1$ will take a maximum. Note that $g(t)<0$ for all $t \in \mathbb{R}$
So $$ u= {2\over -{2\over 3}} = -3\implies ....$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
a tough sum of binomial coefficients
Find the sum: $$\sum_{i=0}^{2}\sum_{j=0}^{2}\binom{2}{i}\binom{2}{j}\binom{2}{k-i-j}\binom{4}{k-l+i+j},\space\space 0\leq k,l\leq 6$$
I know to find $\sum_{i=0}^{2}\binom{2}{i}\binom{2}{2-i}$, I need to find the coefficient of $x^2$ of $(1+x)^4$ (which is $\binom{4}{2}$). But I failed to use that trick here. Any help appreciated!
| Here is a variant which could be seen as generalisation of OP's example. We use the coefficient of operator $[z^k]$ to denote the coefficient of $z^n$ in a series. This way we can write for instance
\begin{align*}
[z^k](1+z)^n=\binom{n}{k}
\end{align*}
and we also use Iverson brackets which are defined as
\begin{align*}
[[P(z)]]=\begin{cases}
1&\qquad P(z) \ \text{ true}\\
0&\qquad P(z) \ \text{ false}
\end{cases}
\end{align*}
We obtain for $0\leq k,l\leq 6$:
\begin{align*}
\color{blue}{\sum_{i=0}^2}&\color{blue}{\sum_{j=0}^2\binom{2}{i}\binom{2}{j}\binom{2}{k-i-j}\binom{4}{k-l+i+j}}\\
&=\sum_{i=0}^2\binom{2}{i}\sum_{j=0}^2\binom{2}{j}[z^{k-i-j}](1+z)^2[u^{k-l+i+j}](1+u)^4\tag{1}\\
&=[z^k][u^{k-l}](1+z)^2(1+u)^4\sum_{i=0}^2\binom{2}{i}\left(\frac{z}{u}\right)^i\sum_{j=0}^2\binom{2}{j}\left(\frac{z}{u}\right)^j\tag{2}\\
&=[z^k][u^{k-l}](1+z)^2(1+u)^4\left(1+\frac{z}{u}\right)^4\tag{3}\\
&=[u^{k-l}]\left([z^k]+2[z^{k-1}]+[z^{k-2}]\right)\left(1+\frac{z}{u}\right)^4(1+u)^4\tag{4}\\
&=[u^{k-l}]\left(\binom{4}{k}u^{-k}+2\binom{4}{k-1}[[k\geq 1]]u^{1-k}\right.\\
&\qquad\qquad\quad\left.+\binom{4}{k-2}[[k\geq 2]]u^{2-k}\right)(1+u)^4\tag{5}\\
&=\left(\binom{4}{k}[u^{2k-l}]+2\binom{4}{k-1}[[k\geq 1]][u^{2k-l-1}]\right.\\
&\qquad\qquad\quad\left.+\binom{4}{k-2}[[k\geq 2]][u^{2k-l-2}]\right)(1+u)^4\\
&\,\,\color{blue}{=\binom{4}{k}\binom{4}{2k-l}[[2k\geq l]]+2\binom{4}{k-1}\binom{4}{2k-l-1}[[k\geq 1]][[2k\geq l+1]]}\\
&\qquad\qquad\quad\color{blue}{+\binom{4}{k-2}\binom{4}{2k-l-2}[[k\geq 2]][[2k\geq l+2]]}
\end{align*}
Comment:
*
*In (1) we apply the coefficient of operator twice.
*In (2) we use the linearity of the coefficient of operator and apply the rule $[z^{p-q}]A(x)=[z^p]z^qA(z)$.
*In (3) we apply the binomial theorem twice.
*In (4) we expand $(1+z)^2$ and select the coefficient of $[z^k]$.
\begin{align*}
[z^k]&(1+z)^2\left(1+\frac{z}{u}\right)^4\\
&=[z^k](1+2z+z^2)\left(1+\frac{z}{u}\right)^4\\
&=\left([z^k]+2[z^{k-1}]+[z^{k-2}]\right)\left(1+\frac{z}{u}\right)^4\\
&=\left([z^k]+2[z^{k-1}]+[z^{k-2}]\right)\sum_{j=0}^4\binom{4}{j}\left(\frac{z}{u}\right)^j\\
&=[z^k]\sum_{j=0}^4\binom{4}{j}\left(\frac{z}{u}\right)^j
+2[z^{k-1}]\sum_{j=0}^4\binom{4}{j}\left(\frac{z}{u}\right)^j
+[z^{k-2}]\sum_{j=0}^4\binom{4}{j}\left(\frac{z}{u}\right)^j\\
\end{align*}
*In (5) we select the coefficients of $[z^{k-a}]$ in $\left(1+\frac{z}{u}\right)^4$ with $0\leq a\leq 2$. We use Iverson brackets to set terms to zero if the lower part of binomial coefficients is less than zero. We do a similar job with $[u^{k-l}]$ in the following lines.
| {
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find all the points on the line $y = 1 - x$ which are $2$ units from $(1, -1)$ I am really struggling with this one. I'm teaching my self pre-calc out of a book and it isn't showing me how to do this. I've been all over the internet and could only find a few examples. I only know how to solve quadratic equations by converting them to vertex form and would like to stick with this method until it really sinks in. What am I doing wrong?
1.) Distance formula $\sqrt{(x-1)^2 + (-1 -1 + x)^2}=2$
2.) remove sqrt, $(x - 1)(x - 1) + (x - 2)(x - 2) = 4$
3.) multiply, $x^2 - 2x +1 + x^2 -4x +4 = 4$
4.) combine, $2x^2 -6x +5 = 4$
5.) general form, $2x^2 -6x +1$
6.) convert to vertex form (find the square), $2(x^2 - 3x + 1.5^2)-2(1.5)^2+1$
7.) Vertex form, $2(x-1.5)^2 -3.5$
8.) Solve for x, $x-1.5 = \pm\sqrt{1.75}$
9.) $x = 1.5 - 1.32$ and $x = 1.5 + 1.32$
10.) $x = 0.18$ and $2.82$
When I plug these two $x$ values back into the vertex form of the quadratic equation, I'm getting $y = 0.02$ for both $x$ values. These points are not on the line. Can someone tell me what I'm doing wrong please?
| Another, possibly easier way to tackle this is to find the intersection points between a circle centered at $(1,-1)$ with radius $2$, and $y=1-x.$
The equation for the circle would be $$(x-1)^2+(y+1)^2=4$$ The equation of the line is $$y = 1-x$$
We can plug in $1-x$ for $y$ in the equation for the circle to get $$(x-1)^2+(2-x)^2=4$$
Expanding,
$$x^2-2x+1+4-4x+x^2=4$$
or
$$2x^2-6x+1=0$$
If you use the quadratic equation, you'll get $\frac32\pm\frac{\sqrt 7}2$. Plugging back into to either equation (preferably the linear equation), you'll get the coordinates as $(\frac32\pm\frac{\sqrt 7}2,-\frac12\pm\frac{\sqrt 7}2)$
| {
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Find $\log _{24}48$ if $\log_{12}36=k$ Find $\log _{24}48$ if $\log_{12}36=k$
My method:
We have $$\frac{\log 36}{\log 12}=k$$ $\implies$
$$\frac{\log 12+\log 3}{\log 12}=k$$ $\implies$
$$\frac{\log3}{2\log 2+\log 3}=k-1$$ So
$$\log 3=(k-1)t \tag{1}$$
$$2\log 2+\log 3=t$$ $\implies$
$$\log 2=\frac{(2-k)t}{2} \tag{2}$$
Now $$\log _{24}48=\frac{\log 48}{\log 24}=\frac{4\log 2+\log 3}{3\log 2+\log 3}=\frac{2(2-k)+k-1}{3\left(\frac{2-k}{2}\right)+k-1}=\frac{6-2k}{4-k}$$
is there any other approach?
| With $z = \log_2(3)$:
$$L = \log_{24}(48) = \frac{\ln(48)}{\ln(24)} = \frac{\ln(2^4\times 3)}{\ln(2^3 \times 3)} = \frac{4\ln(2) + \ln(3)}{3\ln(2) + \ln(3)} = \frac{4+z}{3 + z}$$
$$K = \log_{12}{(36)} = \frac{\ln(36)}{\ln(12)} = \frac{\ln(2^2\times 3^2)}{\ln(2^2 \times 3)} = \frac{2\ln(2) + 2\ln(3)}{2\ln(2) + \ln(3)} = \frac{2+2z}{2 + z}$$
Just solve for $z$ and plug:
$$z = \frac{2K-2}{K-2}$$
$$L = \frac{2K - 6}{K-4}$$
| {
"language": "en",
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"source": "stackexchange",
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Inverse matrix of matrix (all rows equal) plus identity matrix Let $A$ be a matrix where all rows are equal, for example,
$$A=\left[\begin{array}{ccc}
a_{1} & a_{2} & a_{3} \\
a_{1} & a_{2} & a_{3} \\
a_{1} & a_{2} & a_{3}
\end{array}\right]$$
Then what is the inverse of the matrix $B=I+A$, where $I$ is the identity matrix? For example,
$$B=\left[\begin{array}{ccc}
a_{1}+1 & a_{2} & a_{3} \\
a_{1} & a_{2}+1 & a_{3} \\
a_{1} & a_{2} & a_{3}+1
\end{array}\right]$$
I have a conjecture, which computation has so far confirmed:
$$B^{-1}=I-\frac{A}{\mbox{tr}(A)+1}$$
Why is this true?
| The inverse of $B$:
$$B^{-1}=\frac{\text{adj}{(B)}}{\det(B)}.$$
The determinant of $B$:
$$\det(B)=\begin{vmatrix}a_{1}+1 & a_{2} & a_{3} \\
a_{1} & a_{2}+1 & a_{3} \\
a_{1} & a_{2} & a_{3}+1\end{vmatrix}=
\begin{vmatrix}a_{1}+1 & a_{2} & a_{3} \\
-1 & 1 & 0 \\
-1 & 0 & 1\end{vmatrix}=a_1+a_2+a_3+1.$$
The adjugate of $B$:
$$\text{adj}(B)=\text{C}^T=\\
\begin{pmatrix}
(a_2+1)(a_3+1)-a_2a_3 & -a_2(a_3+1)+a_2a_3 & a_2a_3-a_3(a_2+1) \\
-a_1(a_3+1)+a_1a_3 & (a_1+1)(a_3+1)-a_1a_3 & -a_3(a_1+1)+a_1a_3 \\
a_1a_2-a_1(a_2+1) & -a_2(a_1+1)+a_1a_2 & (a_1+1)(a_2+1)-a_1a_2
\end{pmatrix}=\\
\begin{pmatrix}
a_1+a_2+a_3+1-a_1 & -a_2 & -a_3 \\
-a_1 & a_1+a_2+a_3+1-a_2 & -a_3 \\
-a_1 & -a_2 & a_1+a_2+a_3+1-a_3
\end{pmatrix}=\\
\begin{pmatrix}
a_1+a_2+a_3+1 & 0 & 0 \\
0 & a_1+a_2+a_3+1 & 0 \\
0 & 0 & a_1+a_2+a_3+1
\end{pmatrix}-
\begin{pmatrix}
a_1 & a_2 & a_3 \\
a_1 & a_2 & a_3 \\
a_1 & a_2 & a_3
\end{pmatrix}.
$$
Note: $C^T$ is the transpose of the cofactor matrix of $B$.
Hence, the result follows.
| {
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"url": "https://math.stackexchange.com/questions/2858266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Prove: $\sin\frac{\pi}{20}+\cos\frac{\pi}{20}+\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}=\frac{\sqrt2}{2}$
Prove: $$\sin\frac{\pi}{20}+\cos\frac{\pi}{20}+\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}=\frac{\sqrt2}{2}$$
ok, what I saw instantly is that:
$$\sin\frac{\pi}{20}+\sin\frac{3\pi}{20}=2\sin\frac{2\pi}{20}\cos\frac{\pi}{20}$$
and that,
$$\cos\frac{\pi}{20}-\cos\frac{3\pi}{20}=-2\sin\frac{2\pi}{20}\sin\frac{\pi}{20}$$
So,
$$2\sin\frac{2\pi}{20}(\cos\frac{\pi}{20}-\sin\frac{\pi}{20})=\frac{\sqrt2}{2}=\sin\frac{5\pi}{20}$$
Unfortunately, I can't find a way to continue this, any ideas or different ways of proof?
*Taken out of the TAU entry exams (no solutions are offered)
| So, you have
$$2\sin\frac{2\pi}{20}(\cos\frac{\pi}{20}+\sin\frac{\pi}{20})=2\sin\frac{2\pi}{20}\left(\frac{\cos^2\frac{\pi}{20}-\sin^2\frac{\pi}{20}}{\cos\frac{\pi}{20}-\sin\frac{\pi}{20}}\right)= \frac{\sqrt2}{2},$$
or
$$ = \frac{2 \sin\frac{2\pi}{20}\cos\frac{2\pi}{20}}{\cos\frac{\pi}{20}-\sin\frac{\pi}{20}} = \frac{\sin\frac{\pi}{5}}{\cos\frac{\pi}{20}-\sin\frac{\pi}{20}}=\frac{\sqrt2}{2}.$$
As $\sin\frac{\pi}{20}= \sin\left(\frac{\pi}{4}-\frac{\pi}{5}\right) = \frac{\sqrt{2}}{2}(\cos\frac{\pi}{5}-\sin\frac{\pi}{5})$ and $\cos\frac{\pi}{20}= \frac{\sqrt{2}}{2}(\cos\frac{\pi}{5}+\sin\frac{\pi}{5})$, we have
$$ \frac{\sin\frac{\pi}{5}}{\frac{\sqrt{2}}{2}2 \sin\frac{\pi}{5}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2858498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Is there a simpler way to determine m, n, p, such that the following holds for all reals? I am given the following equation and I am asked to find $m$, $n$ and $p$, so that the equation holds for all reals:
$$ \sin^4x + \cos^4x + m(\sin^6x + \cos^6x) + n(\sin^8x + \cos^8x) + p(\sin^{10}x + \cos^{10}x) = 1, \space \forall x \in \mathbb R $$
I have managed to solve it by, in advance, calculating the following power reduction formulas, and then applying them to the equation:
$$
\sin^4x + \cos^4x = 1 - \frac12\sin^2{2x} \\
\sin^6x + \cos^6x = 1 - \frac34\sin^2{2x} \\
\sin^8x + \cos^8x = 1 - \sin^2{2x} + \frac18\sin^4{2x} \\
\sin^{10}x + \cos^{10}x = 1 - \frac54\sin^2{2x} + \frac5{16}\sin^4{2x}
$$
As a result, I managed to simplify it to the following, which is only in terms of powers of $\sin{2x}$:
$$ \left( 1+m+n+p \right) - \left( \frac12 + \frac{3m}4 + n + \frac{5p}4 \right) \sin^2{2x} + \left( \frac{n}8 + \frac{5p}{16} \right) \sin^4{2x} = 1, \space \forall x \in \mathbb R $$
I then came to the conclusion that the only possible way for which this can be true is if:
$1+m+n+p=1$, $ \frac12 + \frac{3m}4 + n + \frac{5p}4 = 0 $ and $ \frac{n}8 + \frac{5p}{16} = 0 $.
By solving the system of equations below I arrive at the solutions $m=6$, $n=-10$, $p=4$.
$$
\left\{
\begin{aligned}
1 + m + n + p = 1 \\
\frac12 + \frac{3m}4 + n + \frac{5p}4 = 0 \\
\frac{n}8 + \frac{5p}{16} = 0
\end{aligned}
\right.
$$
My question is if my reasoning is correct and if there exists any simpler way to solve the problem.
| Consider the polynomial
$$P(t):=-1+t^2+(1-t)^2+mt^3+m(1-t)^3+nt^4+n(1-t)^4+pt^5+p(1-t)^5\,.$$
If $t=\sin^2(x)$ is a root for every $x\in\mathbb{R}$, then $P\equiv 0$ identically. In particular,
$$0=[t^4]\,P(t)=2n+5p\,,$$
$$0=[t^2]\,P(t)=2+3m+6n+10p\,,$$
and
$$0=[t^0]\,P(t)=P(0)=m+n+p\,,$$
where $[t^k]\,P(t)$ denotes the coefficient of the term $t^k$ in $P(t)$ for every $k=0,1,2,\ldots$. (Here, the coefficients are found via the Binomial Theorem.) It follows immediately that $m=6$, $n=-10$, and $p=4$.
We still need to show that $P\equiv 0$ is indeed the case. First, $\deg(P)\le4$ can be easily seen. Therefore, we need to check that $[t]\,P(t)=0$ and $[t^3]\,P(t)=0$, but
$$[t]\,P(t)=-2-3m-4n-5p=-\big([t^2]\,P(t)\big)+\big([t^4]\,P(t)\big)=0$$
and
$$[t^3]\,P(t)=-4n-10p=-2\,\big([t^4]\,P(t)\big)=0\,.$$
Thus, $P$ is indeed the zero polynomial.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2858672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
$\lim_{n \to \infty} \sum_{k=1}^{\left\lfloor\frac{n}{2}\right\rfloor}{\binom{n-k}{k}\frac{1}{2^{n-k}}}$?
Consider the following limit:
$$\lim_{n \to \infty} \sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor}{\binom{n-k}{k}\frac{1}{2^{n-k}}}.$$
I can find the limit numerically, but is it possible to compute it analytically?
The answer is $\frac{2}{3}$.
| Let $a_n$ denote the expression in the limit. Using the identity $n! = \int_{0}^{\infty} x^n e^{-x} \, dx$, we have
\begin{align*}
a_n
&= \sum_{k=0}^{\lfloor n/2\rfloor} \frac{1}{k!(n-2k)!2^{n-k}} \int_{0}^{\infty} x^{n-k} e^{-x} \, dx \\
&= \int_{0}^{\infty} \underbrace{ \sum_{k=0}^{\lfloor n/2\rfloor} \frac{(x/2)^k}{k!} \frac{(x/2)^{n-2k}}{(n-2k)!} }_{=(\diamond)} e^{-x} \, dx.
\end{align*}
The inner sum $(\diamond)$ can be analyzed using the idea of Cauchy product. Indeed, write
$$ (\diamond) = \sum_{\substack{j,k \geq 0 \\ j+2k = n}} \frac{(x/2)^k}{k!} \frac{(x/2)^j}{j!}. $$
Then the generating function for $(a_n)$ is given by
\begin{align*}
\sum_{n=0}^{\infty} a_n z^n
&= \int_{0}^{\infty} \sum_{n=0}^{\infty} \sum_{\substack{j,k \geq 0 \\ j+2k = n}} \frac{(z^2 x/2)^k}{k!} \frac{(zx/2)^j}{j!} e^{-x} \, dx \\
&= \int_{0}^{\infty} \exp\left( \frac{z^2x}{2} + \frac{zx}{2} - x \right) \, dx \\
&= \frac{1}{1 - \frac{z}{2} - \frac{z^2}{2}} \\
&= \frac{2}{3}\cdot\frac{1}{1-z} + \frac{1}{3}\cdot\frac{1}{1+\frac{z}{2}} \\
&= \sum_{n=0}^{\infty} \frac{2 + \left(-\frac{1}{2}\right)^n}{3} z^n.
\end{align*}
Therefore
$$ \lim_{n\to\infty} a_n = \lim_{n\to\infty} \frac{2 + \left(-\frac{1}{2}\right)^n}{3} = \frac{2}{3}. $$
As a sanity check, the following numerical computation confirms the formula $a_n = \frac{2 + \left(-\frac{1}{2}\right)^n}{3}$ derived from above.
$\hspace{2em}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\langle 3 \rangle$ in $\textbf{Z}[\omega]$ ramifies, not splits, right? Potentially dumb question here, please bear with me, I want to make sure I have not overlooked some important subtlety.
As you know, $$\omega = \frac{-1 + \sqrt{-3}}{2}$$ is a complex cubit root of unity, that is, $\omega^3 = 1$. This slightly obscures the fact that $1 + 2 \omega = \sqrt{-3}$, and since $(\sqrt{-3})^2 = -3$, it follows that $\langle 3 \rangle = \langle \sqrt{-3} \rangle^2$. Meaning that $\langle 3 \rangle$ ramifies, it doesn't split. This even though $$\left( \frac{3 - \sqrt{-3}}{2} \right) \left( \frac{3 + \sqrt{-3}}{2} \right) = 3.$$
That's because $$\left( \frac{-1 + \sqrt{-3}}{2} \right) \sqrt{-3} = \frac{-\sqrt{-3} - 3}{2},$$ or $\omega(1 + 2 \omega) = \omega + 2 \omega^2$. Since $\omega$ is a unit, the ideals $\langle 1 + 2 \omega \rangle$ and $\langle \omega + 2 \omega^2 \rangle$ are in fact the same.
Are these calculations correct? Have I drawn the right conclusion?
| It seems fine, although I must admit I had trouble understanding what you want.
Anyway, first to conclude that $\langle 3 \rangle $ ramifies you need to prove that $\left \langle \sqrt{-3} \right \rangle$ is a prime ideal. This shouldn't be too hard to see, as it's norm is $3$ a prime number.
Then you can establish the following relation
$$\left \langle \sqrt{-3} \right \rangle = \left \langle \frac{3 - \sqrt{-3}}{2} \right \rangle = \left \langle \frac{3 + \sqrt{-3}}{2} \right \rangle$$
to conclude that the both ways lead to the same prime ideal factorization of $\langle 3 \rangle$. As you have already noticed we have that $\left \langle \sqrt{-3} \right \rangle = \left \langle \frac{3 + \sqrt{-3}}{2} \right \rangle$, as $\left( \frac{1 - \sqrt{-3}}{2} \right) \sqrt{-3} = \frac{3 + \sqrt{-3} }{2}$
Similarly $\left \langle \sqrt{-3} \right \rangle = \left \langle \frac{3 - \sqrt{-3}}{2} \right \rangle$, as $\left( \frac{-1 - \sqrt{-3}}{2} \right) \sqrt{-3} = \frac{3 - \sqrt{-3} }{2}$ and the left-most factor is $\omega^2$, another unit.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2861748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Determining the last column so that the resulting matrix is an orthogonal matrix
Determine the last column so that the resulting matrix is an orthogonal
matrix $$\begin{bmatrix} \dfrac{1}{\sqrt{2}} & \dfrac{1}{\sqrt{6}} & ? \\ \dfrac{1}{\sqrt{2}} & -\dfrac{1}{\sqrt{6}} & ? \\ 0 & \dfrac{2}{\sqrt{6}} & ? \end{bmatrix}$$
Can anyone please provide hints to solve this?
| Let $[ x \ y \ z]$ be the vector you are looking for, then it must satisfy
$$x \frac{1}{\sqrt{2}} +y \frac{1}{\sqrt{2}}+z (0)= 0$$
which is $y = - x$ and
$$x \frac{1}{\sqrt{6}} -y \frac{1}{\sqrt{6}}+z \frac{2}{\sqrt{6}} = 0$$
which is $z = \frac{1}{2}y - \frac{1}{2}x$. Using $y = -x$ in $z = \frac{1}{2}y - \frac{1}{2}x$, we get
$$z = -x$$
So the vector you are looking for has the form $[x, \ -x, \ -x]$
Choose $x = \frac{1}{\sqrt{3}}$ (for example), we get $[\frac{1}{\sqrt{3}}, \ -\frac{1}{\sqrt{3}}, \ -\frac{1}{\sqrt{3}}]$, which is a possible last column.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2861966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
How to determine if a set of five $2\times2$ matrices is independent $$S=\bigg\{\left[\begin{matrix}1&2\\2&1\end{matrix}\right], \left[\begin{matrix}2&1\\-1&2\end{matrix}\right], \left[\begin{matrix}0&1\\1&2\end{matrix}\right],\left[\begin{matrix}1&0\\1&1\end{matrix}\right],
\left[\begin{matrix}1&4\\0&3\end{matrix}\right]\bigg\}$$
How can I determine if a set of five $2\times2$ matrices are independent?
| As has been pointed out, four matrices form a basis for the $2\times2$ matrices (the easiest would be
$$
\left[\begin{matrix}1&0\\0&0\end{matrix}\right], \left[\begin{matrix}0&1\\0&0\end{matrix}\right], \left[\begin{matrix}0&0\\1&0\end{matrix}\right], \left[\begin{matrix}0&0\\0&1\end{matrix}\right]
$$) so five matrices cannot be linearly dependent.
In your case the dependence is
$$
\left[\begin{matrix}1&2\\2&1\end{matrix}\right] + \left[\begin{matrix}2&1\\-1&2\end{matrix}\right] + \left[\begin{matrix}0&1\\1&2\end{matrix}\right] - 2\left[\begin{matrix}1&0\\1&1\end{matrix}\right] -
\left[\begin{matrix}1&4\\0&3\end{matrix}\right] =
\left[\begin{matrix}0&0\\0&0\end{matrix}\right].
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2862389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
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State the possible Cartesian equations of $p_2$. Given a plane $p_1$ to have the equation $x+2y-3z=0$ with points $B = (0,3,2), M = (1.5,3,2.5)$. Another plane $p_2$ has a perpendicular distance of $0.5$ with $p_1$. State the possible Cartesian equations of $p_2$.
I think I used a wrong way but got the correct answer, I let $Q$ be a point $(x,y,z)$ on $p_2$, and I know $BQ = (x,y-3,z-2)$ then i use length of projection formula to get $$|x+2y-3z| = \frac{2}{\sqrt{14}}$$
Then I let $x = \frac{2}{\sqrt{14}}, y = 0, z=0$ to get a point $Q$.
| The plane $p_2$ is parallel to $p_1$ so it is of the form $x+2y-3z=c$.
The plane $p_1$ passes through the origin so the distance of $p_1$ and $p_2$ is equal to the distance of the origin and $p_2$.
Distance of a plane $Ax+By+Cz = D$ from the origin is given by $\frac{|D|}{\sqrt{A^2+B^2+C^2}}$ so in this case we have
$$0.5 = d(0, p_2) = \frac{|c|}{\sqrt{1^2+2^2+3^2}} = \frac{|c|}{\sqrt{14}}$$
so $c = \pm \frac{\sqrt{14}}2$. Therefore $p_2$ is given by
$x+2y-3z=\pm \frac{\sqrt{14}}2$.
| {
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"url": "https://math.stackexchange.com/questions/2863362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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how to prove the below statement If $$\frac{1}{a+w}+\frac{1}{b+w}+\frac{1}{c+w}+\frac{1}{d+w}=\frac2w$$ where $a,b,c,d$ are real numbers and $w$ is a non real complex cube root of unity then prove that $$\frac{1}{a+w^2}+\frac{1}{b+w^2}+\frac{1}{c+w^2}+\frac{1}{d+w^2}=2w$$
I have no idea as how to start the problem. Please help me.
| Hint:
$ω^2$ is the conjugate of $ω$, so
$$\frac{1}{a+ω^2}+\frac{1}{b+ω^2}+\frac{1}{c+ω^2}+\frac{1}{d+ω^2}=\overline{\,\frac{1}{a+ω}+\frac{1}{b+ω}+\frac{1}{c+ω}+\frac{1}{d+ω}\,}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2864323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve: $2^x\Bigl(2^x-1\Bigl) + 2^{x-1}\Bigl(2^{x-1} -1 \Bigl) + .... + 2^{x-99}\Bigl(2^{x-99} - 1\Bigl) = 0$ The question says to find the value of $x$ if, $$2^x\Bigl(2^x-1\Bigl) + 2^{x-1}\Bigl(2^{x-1} -1 \Bigl) + .... + 2^{x-99}\Bigl(2^{x-99} - 1 \Bigl)= 0$$
My approach:
I rewrote the expression as,
$$2^x\Bigl(2^x-1\Bigl) + \frac{2^x}{2}\Bigl(\frac{2^x}{2} -1 \Bigl) + .... + \frac{2^x}{2^{99}} \Bigl(\frac{2^x}{2^{99}} - 1 \Bigl)= 0$$
I then took $\bigl(2^x\bigl)$ common and wrote it as,
$$2^x \Biggl[ \Bigl(2^x - 1\Bigl) + \frac{1}{2^1}\Bigl(2^x -2^1\Bigl) + \frac{1}{2^2}\Bigl(2^x - 2^2\Bigl) + \;\ldots + \frac{1}{2^{99}} \Bigl(2^x - 2^{99}\Bigl)\Biggl] = 0$$
After further simplification I got,
$$\frac{2^x}{2^{99}} \Biggl[ \Bigl(2^x\cdot2^{99} - 2^{99}\Bigl) + \Bigl(2^x \cdot 2^{98} - 2^{99}\Bigl) + \ldots + \bigl(2^x -2^{99}\bigl)\Biggl] = 0$$
Taking $-2^{99}$ common I got,
$$-2^x \Biggl[ \Bigl( 2^{x+99} + 2^{x+98} + \ldots + 2^{x+2} + 2^{x+1} + 2^x \Bigl)\Biggl]= 0$$
Now the inside can be expressed as $$\sum ^ {n= 99} _{n=1} a_n$$ Where $a_n$ are the terms of the GP.
Thus we can see that either $$-2^x= 0$$ Or, $$\sum ^ {n= 99} _{n=1} a_n = 0$$
Since the first condition is not poossible, thus, $$\sum ^ {n= 99} _{n=1} a_n = 0$$
So,
$$2^{x + 99} \Biggl(\frac{1-\frac{1}{2^{100}}}{1-\frac{1}{2}} \Biggl) = 0$$
Either way once I solve this, I am not getting an answer that is even in the options. The answers are all in the form of logarithmic expressions.
Any help would be appreciated. We have to find the value of $x$.
| Hint.
$$
2^{2x}\sum_{k=0}^{99}2^{-2k} =2^x\sum_{k=0}^{99}2^{-k}
$$
hence
$$
2^x = \frac{\sum_{k=0}^{99}2^{-k}}{\sum_{k=0}^{99}2^{-2k}} = 1.5\to x = 0.5849625007211562
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2865206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How do I solve this logarithm problem? I'm trying to solve this problem:
If $\log_{27}(a)=b$, find $\log_{\sqrt[6]{a}}\sqrt{3}$
However, I'm unable to see any connection in those given information. How can I solve this logarithm?
| Given $\log_{27}(a) = b$ then $a = 27^b$. Then:
$$\sqrt[6]{a} = (27^b)^\frac{1}{6} = (3^{3 \cdot \frac{1}{6}})^b = 3^\frac{b}{2}$$
$$\log_{\sqrt[6]{a}}x = \log_{3^\frac{b}{2}}x = \frac{\log_{3}x}{\log_{3}3^\frac{b}{2}} = \frac{2}{b}\log_{3}x$$
Since $x = \sqrt{3}$ then
$$\frac{2}{b}\log_{3}x = \frac{2}{b}\cdot\frac{1}{2} = \frac{1}{b}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2865831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given $t = \tan \frac{\theta}{2}$, show $\sin \theta = \frac{2t}{1+t^2}$ Given $t = \tan \frac{\theta}{2}$, show $\sin \theta = \frac{2t}{1+t^2}$
There are a few ways to approach it, one of the way i encountered is that using the $\tan2\theta$ formula, we get $$\tan\theta = \frac{2t}{1-t^2}$$
By trigonometry, we know the ratio of the triangle, that is the opposite to adjacent is $2t : 1-t^2$, and hence it follows the hypotenuse is $\sqrt{(2t)^2+(1-t^2)^2} = \sqrt{(t^2+1)^2} = (t^2+1)$
My question is, can the above square rooted answer be $-(t^2+1)$ also? Why do we reject the negative answer?
| I think the second "solution" comes from squaring then unsquaring. This process can often add solutions to an equation. Using a much simpler example:
$$
\begin{align}
& &x-1 &= 2
\\ &\Rightarrow &(x-1)^2 &= 4
\\ &\Rightarrow &x^2-2x+1 &= 4
\\ &\Rightarrow &x^2-2x-3 &= 0
\\ &\Rightarrow &(x-3)(x+1) &= 0
\\ &\Rightarrow & x &=3,-1
\end{align}
$$
Whenever you square an equation in order to solve it, you should check all the "solutions" you come out with.
In my example, using the first line to check both "solutions", obviously $x=-1$ is rubbish.
In your example, checking both "solutions" $\sin\theta = \frac{2t}{1+t^2}$ and $\sin\theta = \frac{2t}{-(1+t^2)}$ is slightly more difficult. By considering the graphs of $y=\tan\frac{\theta}{2}$ and $y=\sin\theta$ we can see that $$\sin\theta>0 \iff t=\tan\frac{\theta}{2}>0 $$
This is true for $\sin\theta = \frac{2t}{1+t^2}$ but not for $\sin\theta = \frac{2t}{-(1+t^2)}$, so we discard the second solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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Log inequalities Solve the inequality
$\log(5^{1/x} +5^3) < \log 6 + \log 5^{1+\frac{1}{2x}}$
I came up with an equation
$5^{\frac{4x-1}{2x}} + 5^{\frac{-2x+1}{2x}}-6<0$
Which I couldn't solve, i tried to make it quadratic but I couldn't
| Yes, since $\log$ function is monotonic increasing we have
$$\log(5^{1/x}+5^3)<\log(6\cdot 5^{1+1/2x})\iff 5^{1/x}+5^3<6\cdot 5^{1+1/2x}$$
that is
$$5^{1/x-1-1/2x}+5^{3-1-1/2x}<6$$
and then your result
$$5^{\frac{4x-1}{2x}}+5^{\frac{1-2x}{2x}}-6<0$$
then
$$5^{2-\frac{1}{2x}}+5^{\frac{1}{2x}-1}-6<0$$
$$25\cdot 5^{-\frac{1}{2x}}+\frac15 \cdot 5^{\frac{1}{2x}}-6<0$$
that is by $y=5^{\frac{1}{2x}}$
$$\frac{25}y+\frac y 5-6<0 \implies y^2-30y+125<0$$
$$y=\frac{30\pm\sqrt{900-500}}{2}=15\pm 10 \implies 5<y<25$$
that is, observing that exponential function is monotonic increasing
$$5<5^{\frac{1}{2x}}<25=5^2$$
$$1<\frac{1}{2x}<2 \iff \frac12 <2x <1 \iff \frac 14 < x < \frac 12$$
| {
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"timestamp": "2023-03-29T00:00:00",
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For $a> 0$, determine $\lim_{n\rightarrow \infty}\sum_{k=1}^{n}( a^\frac{k}{n^2} - 1)$? For $a> 0$, determine $\lim_{n\rightarrow \infty}\sum_{k=1}^{n}( a^\frac{k}{n^2} - 1)$
My attempts : $\sum_{k=1}^{n}( a^\frac{k}{n^2} - 1) = ( a^\frac{1}{n^2}- 1) + ( a^\frac{2}{n^2}- 1)+......... +(a^\frac{n}{n^2}- 1)=a^\frac{1}{n^2}+ a^\frac{2}{n^2}++.....+a^\frac{n}{n^2}- n $
ThereFore $\lim_{n\rightarrow \infty}\sum_{k=1}^{n}( a^\frac{k}{n^2} - 1)= a^{\lim_{n\rightarrow \infty}\frac{1}{n^2}}+ a^{\lim_{n\rightarrow \infty}\frac{2}{n^2}}++.....+a^{\lim_{n\rightarrow \infty}\frac{n}{n^2}} - \lim_{n\rightarrow \infty} n = \lim_{n\rightarrow \infty} (a^0 + a^0 +.....+a^0) -\lim_{n\rightarrow \infty} n = \lim_{n\rightarrow \infty}( 1+1.....+1) -\lim_{n\rightarrow \infty} n=\lim_{n\rightarrow \infty} n -\lim_{n\rightarrow \infty} n = 0 $
Hence $\lim_{n\rightarrow \infty}\sum_{k=1}^{n}( a^\frac{k}{n^2} - 1)= 0$
Is my answer is correct ??
Any hints / solution will be appreciated
thanks u..
| Intuition:
since $e^u = 1+u + o(u)$ when $u\to 0$ and
$0 \leq \frac{k}{n^2}\ln a \leq \frac{\ln a}{n} \xrightarrow[n\to\infty]{} 0$ for every $1\leq k\leq n$, we "expect" to have
$$
\sum_{k=1}^{n}( a^{\frac{k}{n^2}}-1)
= \sum_{k=1}^{n}( e^{\frac{k}{n^2}\ln a}-1) \approx \sum_{k=1}^{n} \frac{k}{n^2}\ln a = \ln a\cdot \frac{n(n+1)}{2n^2}\xrightarrow[n\to\infty]{} \frac{\ln a}{2} \tag{1}
$$
so we should try and make this rigorous. (Sanity check: this is consistent with $a=1$, for which we do get $0$.)
Side note. Easy lower bound: using the standard inequality (proven e.g. by convexity) $e^x \geq 1+x$ for all $x$, we have
$$
\sum_{k=1}^{n}( e^{\frac{k}{n^2}\ln a}-1) \geq \sum_{k=1}^{n}\frac{k}{n^2}\ln a = \frac{\ln a}{2}\cdot \frac{n(n+1)}{n^2} \tag{2}
$$
and as mentioned we have $\lim_{n\to\infty} \frac{\ln a}{2}\cdot \frac{n(n+1)}{n^2} = \frac{\ln a}{2}$. So we have that the limit is at least $\frac{\ln a}{2}$... That was the easy part.
Proof of (1).
We can write
$$\begin{align}
\sum_{k=1}^{n}( e^{\frac{k}{n^2}\ln a}-1)
&=\sum_{k=1}^{n}{\underbrace{\big(e^{\frac{\ln a}{n^2}}\big) }_{=\alpha}}^k - n = \alpha\cdot \frac{1-\alpha^{n}}{1-\alpha} - n
= e^{\frac{\ln a}{n^2}}\frac{1-e^{\frac{\ln a}{n}}}{1-e^{\frac{\ln a}{n^2}}} - n \\
&= \left(1+ o\left(\frac{1}{n}\right)\right)\cdot \frac{\frac{\ln a}{n}+\frac{\ln^2 a}{2n^2}+o(\frac{1}{n^2})}{\frac{\ln a}{n^2}+o(\frac{1}{n^2})} - n\\
&= \left(1+ o\left(\frac{1}{n}\right)\right)\cdot \frac{n+\frac{\ln a}{2}+o(1)}{1+o(1)} - n\\
&= n+\frac{\ln a}{2}+o(1) - n\\
&= \frac{\ln a}{2}+o(1) \xrightarrow[n\to\infty]{} \frac{\ln a}{2} \tag{3}
\end{align}$$
where we used the Taylor series of $e^u$ as $u\to 0$, $e^u = 1+u+\frac{u^2}{2}+o(u)$, and that of $\frac{1}{1+u} = 1-u+o(u)$. This last (3) gives us the expected limit, $\frac{\ln a}{2}$.
| {
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How to simplify $\int{\sqrt[4]{1-8{{x}^{2}}+8{{x}^{4}}-4x\sqrt{{{x}^{2}}-1}+8{{x}^{3}}\sqrt{{{x}^{2}}-1}}dx}$? I have been asked about the following integral:
$$\int{\sqrt[4]{1-8{{x}^{2}}+8{{x}^{4}}-4x\sqrt{{{x}^{2}}-1}+8{{x}^{3}}\sqrt{{{x}^{2}}-1}}dx}$$
I think this is a joke of bad taste. I have tried every elementary method of integration which i know, also i tried integrating using Maple but as i suspected, the integrad doesn't have an anti derivative. Any ideas?
| Hint:
Let $\text{arcsec}x=t, x=\sec t$
Using Principal values, $0\le t\le\pi,t\ne\dfrac\pi2$
$\sin t=\sqrt{\left(1-\dfrac1x\right)^2}=\dfrac{\sqrt{x^2-1}}{|x|}$ as $\sin t\ge0$
$\tan t=\sin t\sec t=?$
$$1-8x^2+8x^4-4x\sqrt{x^2-1}+8x^3\sqrt{x^2-1}=\dfrac{\cos^4t-8\cos^2t+8-4\sin t\cos^2t+8\sin t}{\cos^4t}$$
Now $\cos^4t-8\cos^2t+8-4\sin t\cos^2t+8\sin t$
$=(1-\sin^2t)^2-8(1-\sin^2t)+8-4\sin t(1-\sin^2t)+8\sin t$
$=\sin^4t+4\sin^3t+6\sin^2t+4\sin t+1$
$=(\sin t+1)^4$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
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"answer_id": 2
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Gauss elimination. Where did I go wrong?
Gaussian elimination with back sub:
So my starting matrix:
\begin{bmatrix}
1 & -1 & 1 & -1
\\2 & 1 & -3 & 4
\\2 & 0 & 2 & 2
\end{bmatrix}
multiply the 2nd and 3rd row by -1 * (first row):
\begin{bmatrix}
1 & -1 & 1 & -1
\\0 & 3 & -5 & 6
\\0 & 2 & 0 & 4
\end{bmatrix}
then add -1(third row) to the 2nd row->
\begin{bmatrix}
1 & -1 & 1 & -1
\\0 & 1 & -5 & 2
\\0 & 2 & 0 & 4
\end{bmatrix}
add -2(2nd row) to the third row ->
\begin{bmatrix}
1 & -1 & 1 & -1
\\0 & 1 & -5 & 2
\\0 & 0 & 10 & 0
\end{bmatrix}
But then this seems to have no solution because $10z = 0$.... ugh
EDIT
As I was writing this, it occurred to me that $z = 0$, $y = 2$, $x = 1$. Is that right?
| Yes, your answer is correct.
Check that $(1,2,0)$ is a solution and also since the rank is $3$, there is a unique solution.
$10z=0 \implies z=0$, substitute that to other equations, we easily get $y=2$ and then $x-y+0 = -1 \implies x-2=-1 \implies x=1$.
| {
"language": "en",
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"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
compute the summation $\sum_ {n=1}^\infty \frac{2n-1 }{2\cdot4\cdots(2n)}= \text{?}$ compute the summation
$$\sum_ {n=1}^\infty \frac{2n-1 }{2\cdot4\cdots(2n)}= \text{?}$$
My attempts : i take $a_n =\frac{2n-1}{2\cdot4\cdots(2n)}$
Now
\begin{align}
& = \frac{2n}{2\cdot4\cdot6\cdots2n} -\frac{1}{2\cdot4\cdot6\cdots2n} \\[10pt]
& =\sum_ {n=1}^\infty \left(\frac{2n}{2\cdot4\cdot6\cdots2n} -\frac{1}{2\cdot4\cdot6\cdots2n}\right) \\[10pt]
& =\sum_ {n=1}^\infty \left(\frac{1}{1\cdot2\cdot3\cdots(n-1)} -\frac{1}{2(1\cdot2\cdot3\cdots n}\right) \\[10pt]
& =\sum_ {n=1}^\infty \left(\frac{1}{(n-1)!} - \frac{1}{2}\sum_ {n=2}^\infty \frac{1}{n!}\right) \\[10pt]
& = e - \frac{1}{2} (e- 1)= \frac{1}{2}(e+1)
\end{align}
Is it correct ???
if not correct then any hints/solution will be appreciated..
thanks in advance
| It is wrong since
$$
2\times 4\times 6\times\dotsb (2n)=2^n n!\quad (n\geq 1).
$$
Write
$$
\sum_{n=1}^\infty\frac{2n-1}{2^n n!}=\sum_{n=1}^\infty\frac{1}{2^{n-1}(n-1)!}-\sum_{n=1}^\infty\frac{(1/2)^n}{n!}
=\sum_{n=0}^\infty\frac{(1/2)^n}{n!}-\sum_{n=1}^\infty \frac{(1/2)^n}{n!}=1.
$$
Note that we don't need to know the value of the intermediate sums.
| {
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"question_score": "3",
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Can we solve $A+D\sin^{2}x=B\sin x+C\cos x$ without having to solve a quartic polynomial? Suppose the following equation
$$
A+D\sin^{2}x=B\sin x+C\cos x,
$$
where $A,B,C,D\in\mathbb{R}$ are the real constants. Initially, I tried to find its solution from a simple substitution
\begin{align*}
A-B\sin x+D\sin^{2}x & =\pm C\sqrt{1-\sin^{2}x},
\end{align*}
that after $t=\sin x$ leads to the following quartic equation
$$
(A^{2}-C^{2})-2ABt+(B^{2}+2AD+C^{2})t^{2}-2BDt^{3}+D^{2}t^{4}=0.
$$
The Weierstrass substitution, where $t=\tan\frac{x}{2}$, and
$$
\sin x=2\sin\frac{x}{2}\cos\frac{x}{2}=\frac{2\sin\frac{x}{2}}{\cos\frac{x}{2}}\cos^{2}\frac{x}{2}=\frac{2\tan\frac{x}{2}}{\frac{1}{\cos^{2}\frac{x}{2}}}=\frac{2\tan\frac{x}{2}}{1+\tan^{2}\frac{x}{2}}=\frac{2t}{1+t^{2}},
$$
and
$$
\cos x=\cos^{2}\frac{x}{2}-\sin^{2}\frac{x}{2}=\left(1-\frac{\sin^{2}\frac{x}{2}}{\cos^{2}\frac{x}{2}}\right)\cos^{2}\frac{x}{2}=\frac{1-\tan^{2}\frac{x}{2}}{\frac{1}{\cos^{2}\frac{x}{2}}}=\frac{1-\tan^{2}\frac{x}{2}}{1+\tan^{2}\frac{x}{2}}=\frac{1-t^{2}}{1+t^{2}},
$$
leads to
$$
A+D\frac{4t^{2}}{(1+t^{2})^{2}}=B\frac{2t}{1+t^{2}}+C\frac{1-t^{2}}{1+t^{2}},
$$
which is also the quartic equation for $t$
$$
(A+C)t^{4}-2Bt^{3}+2(A+2D)t^{2}-2Bt+A-C=0,
$$
$$
(A+C)t^{4}-2Bt(t^{2}+1)+2(A+2D)t^{2}+A-C=0.
$$
Is there any better substitution avoiding the transformation of the trigonometric identity to the quartic equation for $t$?
Thanks for your help.
| If you set $X=\cos x$ and $Y=\sin x$, the equation becomes geometrically finding the intersection of $DY^2-CX-BY-A=0$ (a parabola when $D\ne0$), with the circle $X^2+Y^2=1$, which generally has four solutions. So, no: you can't reduce the resolvent equation from degree $4$ unless there is some particular relation between the coefficients.
If $D=0$ the problem becomes intersecting a line (provided one among $B$ and $C$ is nonzero) with a circle and indeed can be reduced to a quadratic equation.
It's no better if you use $A=A\cos^2x+A\sin^2x$, because the conic becomes
$$
AX^2+(A-D)Y^2+CX+BY=0
$$
which is either an ellipse, for $A(A-D)>0$, or a hyperbola, for $A(A-D)<0$.
Another way to see it is setting $t=\tan(x/2)$, which transforms the equation into
$$
A+\frac{4Dt^2}{(1+t^2)^2}=\frac{2Bt}{1+t^2}+\frac{C(1-t^2)}{1+t^2}
$$
which is what you did. As you see, this generally is a quartic:
$$
(A+C)t^4-2Bt^3+(2A+4D)t^2-2Bt-C=0
$$
One gets a symmetry when $B=0$, because in this case the parabola is symmetric with respect to the $X$-axis: as you see, the quartic becomes biquadratic.
| {
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Calculating $\int_{-\infty}^{\infty}\left( \frac{\cos{\left (x \right )}}{x^{4} + 1} \right)dx$ via the Residue Theorem? In the text, "Function Theory of One Complex Variable" by Robert E. Greene and Steven G. Krantz. I'm inquiring if my proof of $(1)$ is valid ?
$\text{Proposition} \, \, \, (1) $
$$\int_{-\infty}^{\infty}\left( \frac{\cos{\left (x \right )}}{x^{4} + 1} \right)dx=\frac{\sqrt{2}}{2 e^{\frac{\sqrt{2}}{2}}} \left(\pi \sin{\left (\frac{\sqrt{2}}{2} \right )} + \pi \cos{\left (\frac{\sqrt{2}}{2} \right )}\right) $$
The first step on our quest to prove $(1)$, is one must consider a Semicircular Contour $\gamma_{R}$, assuming $R > 1$ define
$$\gamma_{R}^{1}(t) = t + i0 \, \, \text{if} \, \, -R \leq t \leq R $$
$$\gamma_{R}^{2}(t) = Re^{it} \, \, \text{if} \, \, \, \, \, \, 0\leq t \leq \pi.$$
One can call these two curves taken together $\gamma_{R}$ or $\gamma$, after picking our $\gamma_{R}$ one can consider $$\oint_{\gamma _{R}}\frac{e^{iz}}{z^{4}+1}dz.$$
It's trivial that,
$$\oint_{\gamma_{R}}g(z)~ dz = \oint_{\gamma_{R}^{1}} g(z) dz+ \oint_{\gamma_{R}^{2}}g(z) ~dz.$$
It's imperative that
$$\displaystyle \oint_{\gamma_{R}^{1}} g(z) dz \rightarrow \lim_{R \rightarrow \infty }\int_{-R}^{R} \frac{e^{ix}}{1+x^{4}} \, \, \text{as}\, \, R \rightarrow \infty $$
Using the Estimation Lemma one be relived to see
$$\bigg |\oint_{\gamma_{R}^{2}} g(z) dz \bigg | \leq \big\{\text{length}(\gamma_{R}^{2}) \big\} \cdot \sup_{\gamma_{R}^{2}}|g(z)|\leq \pi \cdot \frac{1}{R^{4} - 1}. $$
Now it's safe to say that
$$\lim_{R \rightarrow \infty}\bigg | \oint_{\gamma_{R}^{2}}\frac{1}{1+z^{4}} dz \bigg| \rightarrow 0. $$
It's easy to note after all our struggle that
$$ \int_{-\infty}^{\infty} \frac{cos(x)}{1+x^{4}} = Re \int_{-\infty}^{\infty} \frac{e^{ix}}{1+x^{4}} = Re\bigg( \frac{\sqrt{2}}{2 e^{\frac{\sqrt{2}}{2}}} \left(\pi \sin{\left (\frac{\sqrt{2}}{2} \right )} + \pi \cos{\left (\frac{\sqrt{2}}{2} \right )}\right)\bigg) = \frac{\sqrt{2}}{2 e^{\frac{\sqrt{2}}{2}}} \left(\pi \sin{\left (\frac{\sqrt{2}}{2} \right )} + \pi \cos{\left (\frac{\sqrt{2}}{2} \right )}\right) $$
The finial leg of our conquest is to consider that
$$\oint_{\gamma _{R}}\frac{e^{iz}}{z^{4}+1}dz = 2 \pi i \sum_{j=1,2,3,4} Ind_{\gamma} \cdot \operatorname{Res_{f}(P_{j})}$$
It's easy to calculate that
$$\operatorname{Res_{z = \sqrt[4]{-1}}} \Big(\frac{e^{iz}}{z^{4}+1} \Big) = \frac{1}{4}\sqrt[4]{-1} \, \, -e^{(-1)^{3/4}}$$
$$\operatorname{Res_{z = \sqrt[4]{-1}}} \Big(\frac{e^{iz}}{z^{4}+1} \Big) = \frac{1}{4}\sqrt[4]{-1} \, \, e^{(-1)^{3/4}}$$
$$\operatorname{Res_{z = -(-1)^{3/4}}} \Big(\frac{e^{iz}}{z^{4}+1} \Big) = \frac{1}{4}\sqrt[4]{-1} \, \, e^{\sqrt[4]{-1}}$$
$$\operatorname{Res_{z = (-1)^{3/4}}} \Big(\frac{e^{iz}}{z^{4}+1} \Big) = \frac{1}{4}\sqrt[4]{-1} \, \, e^{\sqrt[4]{-1}}$$
Putting the pieces together we have that
$$\oint_{\gamma _{R}}\frac{e^{iz}}{z^{4}+1}dz = 2 \pi i + \frac{1}{4}\sqrt[4]{-1} \, \, -e^{(-1)^{3/4}} + \frac{1}{4}\sqrt[4]{-1} \, \, e^{(-1)^{3/4}} + \frac{1}{4}\sqrt[4]{-1} \, \, e^{\sqrt[4]{-1}} + \frac{1}{4}\sqrt[4]{-1} \, \, e^{\sqrt[4]{-1}}= \frac{\sqrt{2}}{2 e^{\frac{\sqrt{2}}{2}}} \left(\pi \sin{\left (\frac{\sqrt{2}}{2} \right )} + \pi \cos{\left (\frac{\sqrt{2}}{2} \right )}\right).$$
| Most of it is correct, but there's a problem near the end, when you apply the residue theorem. The polynomial $x^4+1$ has four roots: $\exp\left(\frac{\pi i}4\right)$, $\exp\left(\frac{3\pi i}4\right)$, $\exp\left(\frac{5\pi i}4\right)$, and $\exp\left(\frac{7\pi i}4\right)$. But only the first two matter, since they're the only ones which are in the semicircle bounded by the image of $\gamma_R$.
| {
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Inequality $(1+x^k)^{k+1}\geq (1+x^{k+1})^k$ Let $k$ be a positive integer and $x$ a positive real number. Prove that $(1+x^k)^{k+1}\geq (1+x^{k+1})^k$.
This looks similar to Bernoulli's inequality. If we write $X=x^k$, the inequality is equivalent to $(1+X)^{\frac{k+1}{k}}\geq 1+X^{\frac{k+1}{k}}$, but this is not exactly in the form where we can apply Bernoulli.
| Case 1. $x\le 1$.
Then $x^k\ge x^{k+1}$ and therefore
$$(1+x^k)^{k+1} = (1+x^k)(1+x^k)^k \ge (1+x^k)^k \ge (1+x^{k+1})^k.$$
Case 2. $x>1$.
We divide the inequality by $x^{k(k+1)}$ and get the following equivalent form
$$\left(1+\left(\frac 1x \right)^k\right)^{k+1} \ge \left(1+\left(\frac 1x \right)^{k+1}\right)^k.$$
This is just Case 1. for the number $\dfrac 1x$.
| {
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"source": "stackexchange",
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Convergence of $a_1=1,a_2=\sqrt{7},a_3=\sqrt{7\sqrt{7}}, a_4=\sqrt{7\sqrt{7\sqrt{7}}}, \dots$ Define a sequence $a_n$ as follows: $a_1=1,a_2=\sqrt{7},a_3=\sqrt{7\sqrt{7}}, a_4=\sqrt{7\sqrt{7\sqrt{7}}}, \dots$
Determine if it's convergent and find its limit.
The sequence satisfies $a_n=\sqrt{7a_{n-1}}$. If it's convergent with limit $a$, then $a^2=7a$, so either $a=0$ or $a=7$. We show that the sequence is monotonically increasing and bounded above by $7$ so its limit cannot be equal $0$, thus it is $7$.
Claim. $a_n < 7$ for all $n$.
Proof: Induction on $n$. The base is clear. Assume $a_{n-1} < 7$. This is equivalent to saying that $7a_{n-1} < 49$, which happens iff $\sqrt {7a_{n-1}} < 7$. Then $a_n=\sqrt{7a_{n-1}} < 7$.
It remains to show $a_n$ is monotonically increasing.
Consider
$a_n/a_{n-1}=\sqrt{7/a_{n-1}}$. We prove that this is greater than $1$. It will follow that $a_n > a_{n-1}$.
Since $a_n < 7$, $7/a_{n-1}> 1$, whence $a_n/a_{n-1} > 1$. Thus the sequence is increasing and bounded above, so it's convergent. It's limit cannot be zero, so it must be $7$.
Is this a correct proof?
| Here's a totally different approach.
Let $T(a)=\sqrt{7a}$. For $a \geq 2$,
$$
T(a) \geq \sqrt{14} > \sqrt{4} = 2
\qquad
\text{and}
\qquad
|T^{\prime}(a)| = \frac{\sqrt{7}}{2\sqrt{a}} \leq \frac{\sqrt{7}}{2\sqrt{2}} = \frac{\sqrt{14}}{4} < \frac{\sqrt{16}}{4} = 1.
$$
Therefore, $T$ is a contraction on $[2, \infty)$.
By the Banach fixed point theorem, $T$ has a fixed point in $X$.
You already proved that this fixed point has to be $a=7$ (since $a=0 \notin X$).
Conclude by noting that $T(a_1)=T(1)\geq2$.
| {
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Sum of n terms of this series $\frac{1}{1.3} + \frac{2}{1.3.5} +\frac {3}{1.3.5.7} + \frac{4}{1.3.5.7.9}........ n $ Terms.
I Know the answer to this problem but I couldn't find any proper way to actually solve this question. I thought the denominators were the product of n odd natural numbers. So I wrote the nth term as:
$\frac{2^{n+1}(n+1)!.n }{[2(n+1)]!}$
But I don't know if that's correct or even what to do next.?
| the general term is given by $$a_n=\frac{n}{1*3*5....(2n+1)}$$
$$a_n=\frac{\frac{2n+1-1}{2}}{1*3*5....(2n+1)}$$
$$a_n=\frac{1}{2}*({\frac{2n+1}{1*3*5....(2n+1)}-\frac{1}{1*3*5...(2n+1)})}$$
$$a_n=\frac{1}{2}*({\frac{1}{1*3*5....(2n-1)}-\frac{1}{1*3*5...(2n+1)})}$$
for $n=1$ $$\frac{1}{2}*({\frac{1}{1}-\frac{1}{1*3})}$$
for $n=2$ $$\frac{1}{2}*({\frac{1}{1*3}-\frac{1}{1*3*5})}$$
for $n=3$ $$\frac{1}{2}*({\frac{1}{1*3*5}-\frac{1}{1*3*5*7})}$$
thus $$S_n=\frac{1}{2}*(1-\frac{1}{1*3*5*7*....(2n+1)}) $$
$$\lim_{n\to\infty} S_n=\frac{1}{2}$$
| {
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Evaluate $\lim_{x \to \infty} [(x+2)\tan^{-1}(x+2) -x \tan^{-1}x]$ $\underset{x \to \infty}{\lim} [(x+2)\tan^{-1}(x+2) -x \tan^{-1}x]=?$
My Try :$[(x+2)\tan^{-1}(x+2) -x \tan^{-1}x] = x \tan^{-1} \frac {2}{1+2x+x^2} + 2. \tan^{-1}(x+2)$
$\underset{x \to \infty}{\lim} x \tan^{-1} \frac {2}{1+2x+x^2} =0$ [By manipulating L'hospital] and $\underset{x \to \infty}{\lim}2. \tan^{-1}(x+2) = \pi$
so $\underset{x \to \infty}{\lim} [(x+2)\tan^{-1}(x+2) -x \tan^{-1}x]= \pi$
Can anyone please correct me If I have gone wrong anywhere?
| Your solution is fine but it can be shortened. By Lagrange's theorem $(x+2)\arctan(x+2)-x\arctan(x)$ is twice the value of the derivative of $z\arctan z$ at some $z\in(x,x+2)$. Since $\frac{d}{dz}z\arctan(z) = \underbrace{\frac{z}{1+z^2}}_{\to 0}+\underbrace{\arctan(z)}_{\to \pi/2}$, the wanted limit is $\pi$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Algebra - Solving for three unknowns. Find all possible solutions of
$$2^x + 3^y = z^2.$$
My approach.
First I substituted $x = 0$, and got the solution, then for $y = 0$. And for $x > 0$ and $y > 0$ , I just know the Pythagorean triple and got the solution as $(4,2,5)$ and $(4,2,-5)$. Please help me to solve it properly.
| If $x,y>0$ then working mod $3$ we have $2^x\equiv 1$ so $x$ is even. Working mod $4$ gives $y$ even. So $(2^{x/2})^2+(3^{y/2})^2=z^2$, meaning $(2^{x/2},3^{y/2},|z|)$ form a pythagorean triple.
Any pythagorean triple can be written as $(a(b^2-c^2), 2abc, a(b^2+c^2))$ for some positive integers $a,b,c$ (where the first two terms can be in either order). So we must have $2abc$ is a power of $2$, and so $a(b^2-c^2)$ is a power of $3$. This means $a=c=1$, $b$ is a power of $2$, and $(b-1)(b+1)$ is a power of $3$, hence $b$ must be $2$. This means the only option for $x,y>0$ is $2^{x/2}=8, 3^{y/2}=3, |z|=5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2882635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
For what values of $a$ and $b$ does $\lim_{x\rightarrow \infty}(\sqrt{x^2+x+1}-ax-b)=1$? I have a question about limits tending to infinity. I need to find the constants $a$ and $b$ for which this limit takes the value 1. Please, help! Thank you!
$$\lim_{x\rightarrow \infty}(\sqrt{x^2+x+1}-ax-b)=1.$$
I've tried various things, including trial and error (it gave me the values $a=1$ and $b=\frac{1}{2}$), but it seems that I can't find a way to do this.
Thank you!!
| $\lim_{x\to \infty} \frac{\sqrt{x^2+x+1}}{x} = \lim_{x\to \infty} \frac{x\sqrt{1+{\frac{1}{x}}+\frac{1}{x^2}}}{x} = 1=a$
$\lim_{x\to \infty} \sqrt{x^2+x+1}-ax = \lim_{x\to \infty} \sqrt{x^2+x+1}-x= \lim_{x\to \infty} \frac{x+1}{\sqrt{x^2+x+1}+x} = \frac{1}{2}$
$\lim_{x\to \infty} \sqrt{x^2+x+1}-ax+b=1 \implies 1-b=\frac{1}{2} \implies b=\frac{1}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2883026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Evaluate $ \lim _{x \to 0} \left[{\frac{x^2}{\sin x \tan x}} \right]$ where $[\cdot]$ denotes the greatest integer function. Evaluate $$\lim _{x \to 0} \left[{\frac{x^2}{\sin x \tan x}} \right]$$ where $[\cdot]$ denotes the greatest integer function.
Can anyone give me a hint to proceed?
I know that $$\frac {\sin x}{x} < 1$$ for all $x \in (-\pi/2 ,\pi/2) \setminus \{0\}$ and $$\frac {\tan x}{x} > 1$$ for all $x \in (-\pi/2 ,\pi/2) \setminus \{0\}$. But will these two inequalities be helpful here?
| Can you just use the known expansions for $\sin x, \tan x$ for small $x$? Then have $$\frac{x^2}{\sin x \tan x} = \frac{x^2}{(x - \frac{x^3}{6} + \cdots)(x + \frac{x^3}{3} + \cdots)}.$$ The rhs only including terms up to $x^2$ can be written $$\frac{1}{(1 - x^2/6)(1 + x^2/3)}$$ and the denominator to order $x^2$ is $1 + x^2/6$ so that finally have $$\frac{x^2}{\sin x \tan x} = \frac{1}{1 + \frac{x^2}{6} + \cdots} = 1 - x^2/6$$ and the integer part is 0.
Alternatively write the original expression as $$\frac{x^2 \cos x}{\sin^2 x}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2885453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Difference of numbers in a unit interval $x,y,z\in \mathbb{R}$ are chosen at random from the unit interval $[0, 1]$.
What is the probability that
$$\max(x,y,z) - \min(x,y,z) \leq \frac{2}{3}$$
EDIT- Solutions not using calculus would be appreciated, as this problem appeared on a test where calculus was not meant to be used.
| Let $E$ be the event $\{ \max(x, y, z) - \min(x, y, z) \le \frac{2}{3}\}$. Our goal is to find $\Pr[E]$. Now let's split $E$ into 6 events $E_{xy}, E_{yx}, E_{xz}, E_{zx}, E_{yz}, E_{zy}$, where
$$E_{xy} = \{ (x = \max(x, y, z)) \land (y = \min(x, y, z)) \land (x - y \le 2/3)\},$$
and the rest are defined similarly. Note that $E$ is the union of these six events and the probability of intersection of any two of these events is $0$. Also, due to symmetry, all these 6 events have the same probability. Hence $\Pr[E] = 6\Pr[E_{xy}]$.
It remains to compute $\Pr[E_{xy}]$. Note that $E_{xy}$ happens iff $y\le x\le y + 2/3$ and $z\in[y, x]$. I.e.
\begin{align*}
\Pr[E_{xy}] &= \int\limits_0^1 dy \int\limits_y^{\min(y + \frac{2}{3}, 1)} dx \int\limits_y^x dz
= \int\limits_0^1 dy \int\limits_y^{\min(y + \frac{2}{3}, 1)} (x - y) dx \\
&= \int\limits_0^1 dy \cdot \frac{(x - y)^2}{2} \Bigg|_{y}^{\min(y + \frac{2}{3}, 1)}
= \int\limits_0^1 \frac{\left(\min(y + \frac{2}{3}, 1) - y\right)^2}{2} dy\\
&= \int\limits_0^{\frac{1}{3}} \frac{\left(\frac{2}{3}\right)^2}{2}dy + \int\limits_{\frac{1}{3}}^1 \frac{(1 - y)^2}{2}dy \\
&= \frac{2}{27} -\frac{(1 - y)^3}{6} \Bigg |_{\frac{1}{3}}^1 = \frac{10}{81}.
\end{align*}
And thus the answer is $ 6\cdot \frac{10}{81} = \frac{20}{27}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Why $\sum_{n=1}^{\infty}\frac{n\left(\sin x\right)^{n}}{2+n^{2}}$ is not uniformly convergent on $[0,\;\frac{\pi}{2})$? Why $\sum_{n=1}^{\infty}\frac{n\left(\sin x\right)^{n}}{2+n^{2}}$
is not uniformly convergent on $[0,\;\frac{\pi}{2})$?
I was thinking that we need to show partial sums
\begin{equation}
\left|S_{2n}\left(x\right)-S_{n}\left(x\right)\right|\nrightarrow0
\end{equation}
for some $x\in[0,\;\frac{\pi}{2})$.
Any hint?
Is this correct?
\begin{align}
\left|S_{2n}\left(x\right)-S_{n}\left(x\right)\right| & =\left|\frac{\left(n+1\right)\left(\sin x\right)^{n+1}}{2+\left(n+1\right)^{2}}+\cdots+\frac{2n\left(\sin x\right)^{2n}}{2+\left(2n\right)^{2}}\right|\\
& =\frac{\left(n+1\right)\left(\sin x\right)^{n+1}}{2+\left(n+1\right)^{2}}+\cdots+\frac{2n\left(\sin x\right)^{2n}}{2+\left(2n\right)^{2}}\nonumber \\
& \geq\frac{\left(n+1\right)\left(\sin x\right)^{n+1}}{2+\left(2n\right)^{2}}+\cdots+\frac{2n\left(\sin x\right)^{2n}}{2+\left(2n\right)^{2}}\nonumber \\
& \geq\frac{\left(n+1\right)\left(\sin x\right)^{n+1}}{2+\left(2n\right)^{2}}+\cdots+\frac{\left(n+1\right)\left(\sin x\right)^{2n}}{2+\left(2n\right)^{2}}\nonumber \\
& \geq\frac{\left(n+1\right)\left(\sin x\right)^{2n}}{2+\left(2n\right)^{2}}+\cdots+\frac{\left(n+1\right)\left(\sin x\right)^{2n}}{2+\left(2n\right)^{2}}\nonumber \\
& =\frac{n\left(n+1\right)\left(\sin x\right)^{2n}}{2+\left(2n\right)^{2}}\nonumber
\end{align}
then let $x=\frac{\pi}{2}-\frac{1}{n}$, so we have
\begin{align}
\left|S_{2n}\left(x\right)-S_{n}\left(x\right)\right| & \geq\frac{n\left(n+1\right)\left(\sin x\right)^{2n}}{2+\left(2n\right)^{2}}\\
& =\frac{n\left(n+1\right)\left(\sin\left(\frac{\pi}{2}-\frac{1}{n}\right)\right)^{2n}}{2+\left(2n\right)^{2}}\nonumber \\
& \rightarrow0.25,\:not\:0\nonumber
\end{align}
| First, your proof looks fine to me. Below I sketch a different kind of proof.
Let $f(x)$ be the sum of this series. Note that each summand increases on $[0,\pi/2).$ Hence so does $f(x).$ It follows that $\lim_{x\to \pi/2^-} f(x)=L$ for some $L\in (0,\infty].$
Suppose the series converges uniformly to $f$ on $[0,\pi/2).$ Because each summand is bounded on $[0,\pi/2),$ so is every partial sum, which implies by uniform convergence that $f$ is bounded on $[0,\pi/2).$ Thus $L<\infty.$
Let $N\in \mathbb N.$ Then
$$\tag 1 L = \lim_{x\to \pi/2^-} f(x) \ge \lim_{x\to \pi/2^-}\sum_{n=1}^{N}\frac{n(\sin x)^{n}}{2+n^{2}} = \sum_{n=1}^{N}\frac{n}{2+n^{2}}.$$
Since $\sum_{n=1}^{\infty}\dfrac{n}{2+n^{2}} = \infty,$ the right side of $(1)$ can be made arbitrarily large. Thus $L=\infty,$ contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2887128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Finding closed form for $\sum_{k=1}^n k2^{k-1}$ I am trying to use the perturbation method to find a closed form for:
$$
S_n = \sum_{k=1}^n k2^{k-1}
$$
This is what I’ve tried so far:
$$
S_n + (n+1)2^n = 1 + \sum_{k=2}^{n+1} k2^{k-1}
$$
$$
S_n + (n+1)2^n = 1 + \sum_{k=1}^{n} (k+1)2^{k}
$$
$$
S_n + (n+1)2^n = 1 + \sum_{k=0}^{n-1} k2^{k-1}
$$
$$
S_n + (n+1)2^n = 1 + 0 + n2^{n-1} + \sum_{k=1}^{n} k2^{k-1}
$$
But I don’t think it is right since it would result in a $S_n - S_n = ...$
Could you please help me?
Thank you very much.
| Note the mistake: $\sum_{k=1}^{n} (k+1)2^{k}\ne \sum_{k=0}^{n-1} k2^{k-1}$.
Here is the right way:
$$\begin{align}S_n + (n+1)2^n &= 1 + \sum_{k=1}^{n} (k+1)2^{k}=\\
&=1+\sum_{k=0}^{n-1} (k+2)2^{k+1}=\\
&=1+\color{blue}{\sum_{k=0}^{n-1} k\cdot 2^{k+1}}+\color{red}{\sum_{k=0}^{n-1}2^{k+2}}=\\
&=1+\color{blue}{\sum_{k=0}^n k\cdot 2^{k+1}-n\cdot 2^{n+1}}+\color{red}{4(2^n-1)}=\\
&=1+4\sum_{k=0}^nk\cdot 2^{k-1}-n\cdot 2^{n+1}+4(2^n-1)=\\
&=1+4S_n-n\cdot 2^{n+1}+4(2^n-1) \Rightarrow \\
3S_n&=(n+1)2^n-1+n\cdot 2^{n+1}-4(2^n-1)=\\
&=3n\cdot 2^n-3\cdot 2^n+3 \Rightarrow \\
S_n&=n\cdot 2^n-2^n+1.\end{align}$$
| {
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"url": "https://math.stackexchange.com/questions/2888273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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From $5$ apples, $4$ mangoes and $3$ bananas, in how many ways we can select at least two fruits of each variety?
From $5$ apples, $4$ mangoes and $3$ bananas, in how many ways we can select at least two fruits of each variety if fruits of same species are different?
My attempts:
$$\underbrace{\bigg({5\choose 2}+{5\choose 3}+{5\choose 4}+{5\choose 5}\bigg)}_{\text{appples}}\cdot\underbrace{\bigg({4\choose 2}+{4\choose 3}+{4\choose 4}\bigg)}_\text{mangoes}\cdot\underbrace{\bigg({3\choose 2}+{3\choose 3}\bigg)}_\text{bananas}=1144$$
But answer is: $2^{12}-4$. Please help.
| Your solution is correct. We can confirm it with another approach.
The number of subsets of a set with $n$ elements is $2^n$ since each element is either included in a subset or it is not.
Since there are $5$ apples, there are $2^5$ subsets of apples. Of these, $\binom{5}{0} + \binom{5}{1} = 1 + 5 = 6$ have fewer than two apples. Thus, $2^5 - 6 = 32 - 6 = 26$ have at least two apples.
Since there are four mangoes, there are $2^4$ subsets of mangoes. Of these, $\binom{4}{0} + \binom{4}{1} = 1 + 4 = 5$ have fewer than two mangoes. Thus, $2^4 - 5 = 16 - 5 = 11$ have at least two mangoes.
Since there are three bananas, there are $2^3$ subsets of bananas. Of these, $\binom{3}{0} + \binom{3}{1} = 1 + 3 = 4$ have fewer than two bananas. Thus, $2^3 - 4 = 8 - 4 = 4$ have at least two bananas.
Hence, the number of ways of selecting at least two fruits of each variety is
$$26 \cdot 11 \cdot 4 = 1144$$
as you found.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2891092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Summation problem: $f(x)=1+\sum_{n=1}^{\infty}\frac{x^n}{n}$ I want to evaluate this summation:
$$S=1+x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+...+...$$
where, $|x|<1$
Here it is my approach
$$S=1+\sum_{n=1}^{\infty}\frac{x^n}{n}=f(x)$$
$$f'(x)=1+x+x^2+x^3+...+...=\frac{1}{1-x}$$
$$f(x)=\int f'(x)dx=\int \frac{1}{1-x} dx=-\ln(1-x)+C$$
$$f(0)=1 \Longrightarrow C=1 $$
$$S=1-\ln(1-x)$$
And here is my problem:
I calculated this sum for only $|x|<1$.
Then, I checked in Wolfram Alpha and I saw that, this sum $f(x)$ converges for $x=-1$. But, this creating a contradiction with my solution. Because, the series $$f'(x)=1+x+x^2+x^3+...+...=\frac{1}{1-x}$$ for $x=-1$ doesn't converge, diverges. Therefore, there is a problem in my solution. But I can not find the source of the problem.
How can I prove the formula $f(x)=1-\ln(1-x)$ is also correct at the point $x=-1$ ?
Thank you very much.
| It's well-known that $$\ln(1+y)=y-\frac{y^2}{2}+\frac{y^3}{3}-\frac{y^4}{4}+\cdots,~~~\forall y \in (-1,1].\tag1$$
Thus, let $y=-x.$ we have $$\ln(1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}+\cdots,~~~\forall x \in [-1,1).\tag2$$
As result, $$S=1+x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\cdots=1-\left(-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}+\cdots\right)=1-\ln(1-x).$$
Notice that, $(1)$ demands $-1<y\leq 1$. then in $(2)$, we demand $-1<-x\leq 1,$ namely, $-1\leq x<1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2892467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Binomial expansion lower bound $A^n + B^n \le (A+B)^n$ for non-integer $n$ By the Binomial expansion for integer powers,
$$ (A+B)^n = \sum_{k=0}^n {n\choose{k}} A^{n-k} B^{k}$$
(I'm assuming $A,B\ge 0$) and so we get the easy estimate $A^n + B^n \le (A+B)^n$ for any positive integer $n$.
Now what happens when we take $n$ to be non-integer? Well if $0<n\le1$, then we have the reverse inequality,
$$(A+B)^n \le A^n + B^n ,\qquad {(0<n\le 1)} $$
What about if $n>1$ is a non-integer? Can we still say $A^n + B^n \le (A+B)^n$? I'm pretty sure the answer is yes, but I wanted to check with you guys. Here is my quick argument. Let $n = p + \alpha$ where $p \in \mathbb{N}$ and $0<\alpha<1$.
\begin{align} A^n+B^n &= A^\alpha A^p + B^\alpha B^p \\ &\le \text{max}(A^\alpha,B^\alpha) (A^p+B^p) \\ &\le \text{max}(A^\alpha,B^\alpha) (A+B)^p \\ &\le (A+B)^\alpha (A+B)^p \\ &= (A+B)^n\end{align}
| If $A=B$,
$A^n + B^n \le (A+B)^n$
is
$2A^n \le (2A)^n
=2^nA^n$
or
$2 \le 2^n$\which is true since
$n \gt 1$.
Assuming
$A \lt B$,
dividing by
$B^n$
and letting
$A/B = x$,
$A^n + B^n \le (A+B)^n$
becomes
$1+x^n \le (1+x)^n$
with
$0 < x < 1$.
Let
$f(x)
= (1+x)^n-1-x^n$.
$f(0) = 0$.
$f'(x)
=n(1+x)^{n-1}-nx^{n-1}
=n((1+x)^{n-1}-x^{n-1})
\gt 0$
since
$n-1 > 0$
so
$(1+x)^{n-1}
\gt x^{n-1}$.
Therefore
$f(x) > 0$
for $x > 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2894592",
"timestamp": "2023-03-29T00:00:00",
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Limit of $\lim\limits_{x \rightarrow \infty}{(x-\sqrt \frac{x^3+x}{x+1})}$ - calculation correct? I just want to know if this way of getting the solution is correct.
We calculate $\lim\limits_{x \rightarrow \infty} (x-\sqrt \frac{x^3+x}{x+1}) = \frac {1}{2}$.
\begin{align}
& \left(x-\sqrt \frac{x^3+x}{x+1}\,\right) \cdot \frac{x+\sqrt \frac{x^3+x}{x+1}}{x+\sqrt \frac{x^3+x}{x+1}} = \frac{x^2 - \frac{x^3+x}{x+1}}{x+\sqrt \frac{x^3+x}{x+1}} = \frac{\frac{x^2(x+1) - x^3+x}{x+1} }{x+\sqrt \frac{x^3+x}{x+1}} \\[10pt]
= {} & \frac{\frac{x^2+x}{x+1} }{x+\sqrt \frac{x^3+x}{x+1}} = \frac{\frac{x+1}{1+\frac{1}{x}} }{x+\sqrt \frac{x^3+x}{x+1}} = \frac{\frac{x+1}{1+\frac{1}{x}}}{x+\sqrt x^2 \sqrt \frac{x+\frac{1}{x}}{x+1}} \\
= {} & \frac{\frac{x+1}{1+\frac{1}{x}} }{x+\sqrt x^2 \sqrt \frac{1+\frac{1}{x^2}}{1+\frac{1}{x}}} = \frac{x+1}{x+ x \sqrt 1} = \frac{x+1}{2x} \to \frac {1}{2}
\end{align}
I used this for symplifying:
*
*${(x-\sqrt \frac{x^3+x}{x+1})} \cdot {{(x+\sqrt \frac{x^3+x}{x+1})}} = x^2 -\frac{x^3+x}{x+1}=> (a-b)(a+b) = a^2-b^2$.
*$\sqrt \frac{x^3+x}{x+1} = \sqrt {x^2\cdot\frac{({x+\frac{1}{x}})}{x+1}} = \sqrt x^2 \sqrt\frac{x+\frac{1}{x}}{x+1}$.
I didn't write everything formally correct but hopefully you'll get the idea. The problem I see is that sometimes I'm applying $\lim\limits_{x \rightarrow \infty}$ to just the lower part of a fraction, when I probably have to apply it to both parts of the fraction? I'm talking about this part especially: $\lim\limits_{x \rightarrow \infty}\frac{x+1}{1-\frac{1}{x}} \sim x+1$.
As always very grateful for any comments/help. Cheers.
| As an alternative we have by binomial expansion
$$\sqrt \frac{x^3+x}{x+1}=x\sqrt \frac{x+1/x}{x+1}=x\sqrt \frac{x^2+1}{x^2+x}=x\sqrt \frac{(x+1)^2-2x}{x(x+1)}=x\sqrt{\frac{x+1}{x}-\frac{2}{x+1}}=x\sqrt{1+\frac1x-\frac{2}{x+1}} =x\left( 1+\frac1{2x}-\frac{1}{(x+1)}+o\left(\frac1x\right)\right)=x+\frac12-\frac{x}{(x+1)}+o(1)$$
and therefore
$$x-\sqrt \frac{x^3+x}{x+1}=x-x-\frac12+\frac{x}{(x+1)}+o(1)=-\frac12+\frac{x}{(x+1)}+o(1)\to \frac12$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given $f(x) = \arcsin\left(\frac{2x}{1+x^2}\right)$, find $f'(1)$.
Let
$$
f(x) = \arcsin\left(\frac{2x}{1+x^2}\right).
$$
What is the value of $f'(1)$?
The function splits into
$$
f(x) = \begin{cases}
\phantom{\pi-\,}2\arctan(x), & \text{if $-1 \leq x \leq 1$},\\
\pi - 2\arctan(x), & \text{if $x > 1$}.
\end{cases}
$$
I differentiated the function corresponding to $1$ but my answer is coming to be $1$, but the book tells me the answer does not exist. How can we say the derivative doesn't exist?
| Notice that
\begin{align*}
f'(x) &= \frac{1}{\sqrt{1-\left(\frac{2x}{1+x^2}\right)^2}}\frac{d}{dx}\left(\frac{2x}{1+x^2}\right)
\\&= \frac{1}{\sqrt{\left(\frac{1+x^2}{1+x^2}\right)^2-\left(\frac{2x}{1+x^2}\right)^2}}\cdot\frac{(1+x^2)(2) - (2x)(2x)}{(1+x^2)^2}
\\&= \sqrt{\frac{(1+x^2)^2}{(1-x^2)^2}}\cdot \frac{2(1-x^2)}{(1+x^2)^2}
\end{align*}
(Skipping a bit of the algebra here). You would like to cancel the square and square root, and in the case of $(1+x^2)$, you can. But $1-x^2$ could be positive or negative, so $\sqrt{(1-x^2)^2} = |1-x^2|$, not $1-x^2$. Therefore
$$
f'(x) = \frac{2}{1+x^2}\cdot \frac{1-x^2}{|1-x^2|}
$$
As $x\to 1$, the first factor tends to $2$, unambiguously. But the second factor is $\pm 1$ depending on whether the numerator is positive or negative. That is,
$$
\lim_{x\to 1^-} f'(x) = 1\qquad\text{and}\qquad\lim_{x\to 1^+} f'(x) = -1
$$
It follows that $f$ cannot be differentiable at $1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the shortest distance from a point to curved surface
Find the shortest distance from a point $(0,0,0)$ to curved surface $x^2+2y^2-z^2=5.$
What I have done is,
Let a point in the curved surface be $(a,b,c)$, and $\begin{bmatrix} {a - 0} \\
{b - 0}\\
{c - 0}\\
\end{bmatrix}$ be vector v.
A vector w perpendicular to curved suface at (a,b,c) would be
$ \begin{bmatrix} {2a} \\
{4b}\\
{-2c} \\
\end{bmatrix} $.
Distance is $\frac{10}{2\sqrt{a^2+4b^2+c^2}}$ .
Then $\begin{bmatrix} {a - 0} \\
{b - 0}\\
{c - 0}\\
\end{bmatrix} \cdot \begin{bmatrix} {2a} \\
{4b}\\
{-2c} \\
\end{bmatrix} = 2a^2+4b^2-2c^2 = \sqrt{a^2+b^2+c^2}\cdot 2\sqrt{a^2+4b^2+c^2}\cdot \cos{0} =10$.
and $\begin{bmatrix} {a - 0} \\
{b - 0}\\
{c - 0}\\
\end{bmatrix} \times \begin{bmatrix} {2a} \\
{4b}\\
{-2c} \\
\end{bmatrix} = \begin{bmatrix} {-2bc-4bc} \\
{2ac+2ac}\\
{4ab-2ab} \\
\end{bmatrix} = \begin{bmatrix} {-6bc} \\
{4ac}\\
{2ab} \\
\end{bmatrix}= \sqrt{()} \cdot \sqrt{()} \cdot \sin{0}$ = 0
$\quad \to$ For this to be true, two of a,b, and c have to be zero.
So the answer is $\frac{10}{2\sqrt{10}}=\frac{\sqrt{10}}{2}$.
Is this correct? and is there a better way?
| Alternatively, you can set up an optimization (minimization) problem.
Let the point $(a,b,c)$ belong to the curve $x^2+2y^2-z^2=5$. The squared distance from the point to the origin is:
$$d^2(a,b,c)=(a-0)^2+(b-0)^2+(c-0)^2=a^2+b^2+c^2,$$
which needs to be minimized subject to the constraint: $a^2+2b^2-c^2=5$.
Using the Lagrange multiplier's method:
$$L(a,b,c,\lambda)=a^2+b^2+c^2+\lambda (5-a^2-2b^2-c^2)\\
\begin{cases}L_a=2a-2a\lambda =0\\
L_b=2b-4b\lambda=0 \\
L_c=2c-2c\lambda =0\\
L_{\lambda}=5-a^2-2b^2-c^2=0\end{cases}\Rightarrow \\
(a,b,c)=\left(0,\pm \sqrt{\frac 52},0\right) \Rightarrow d^2=\frac 52 \Rightarrow d=\sqrt{\frac 52} \ \text{(min)}\\
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2898128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Show that $10^n \gt 6n^2+n$ Show for all $n \in \mathbb{N}$ $(n\geq1):$ $10^n \gt 6n^2+n$
My solution:
Base case: For $n=1$
$10^1 \gt 6 \cdot 1^2+1$
Inductive hypothesis:
$10^n \gt 6n^2+n \Rightarrow 10^{n+1} \gt 6\cdot(n+1)^2+(n+1)$
Inductive step:
$10^{n+1} \gt 6\cdot(n+1)^2+(n+1)$
$\Rightarrow$ $10^{n+1} \gt 6(n^2+2n+1)+n+1$
$\Rightarrow$ $10^{n+1} \gt 6n^2+12n+6+n+1$
$\Rightarrow$ $10^{n+1} \gt 6n^2+13n+7$
$\Rightarrow$ $10^n \cdot 10^1 \gt 6n^2+13n+7$
$\Rightarrow$ $(6n^2+n)\cdot10 \gt 6n^2+13n+7$
$\Rightarrow$ $60n^2+10n \gt 6n^2+13n+7$
I am stuck at this point. What techniques or tricks are there to solve the rest?
It is of interest to me, because I am currently practicing a lot of exercises related to convergences, inequalities and mathematical induction.
Any hints guiding me to the right direction I much appreciate.
| $$60n^2+10n \gt 6n^2+13n+7 \iff 54n^2>3n+7$$
Note that $$ 54n^2 \ge 54n=3n+51n\ge 3n+51>3n+7$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2898521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
$3a4b \equiv 0\pmod{3}$, $3a4b \equiv 3\pmod{5}$
When the four digit number $3a4b$ is divided by $5$, the remainder is $3$. This number can also be divided by $3$ without remainder. Evaluate $a$ and $b$.
We have two conditions as illustrated below
$$3a4b \equiv 0\pmod{3} \tag{1}$$
$$3a4b \equiv 3\pmod{5} \tag{2}$$
Simpiflying
$$3000 + 100a +40 +b \equiv 0\pmod{3}$$
Which yields
$$4a+b \equiv 0\pmod{3}$$
Here we get that
$$a+b \equiv 0\pmod{3}$$
No clue whether or not it seems correct. Could you assist?
With my best wishes!
| $$4a+b \equiv 0\pmod{3}$$
This step is wrong
$$3000 + 100a +40 +b \equiv 0\pmod{3}$$
$$3000 + 99a+a +39+1 +b \equiv 0\pmod{3}$$
$$a+1 +b \equiv 0\pmod{3}$$
$$a +b \equiv 2\pmod{3}$$
Also given that
$$3000 + 100a +40 +b \equiv 3\pmod{5}$$
$$b \equiv 3\pmod{5}$$
Now what can you say about $b$ where $0\le b\le9$ and
$b \equiv 3\pmod{5}$
| {
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"url": "https://math.stackexchange.com/questions/2901672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How can I find the limit of the following sequence $\sin ^2 (\pi \sqrt{n^2 + n})$? How can I find the limit of the following sequence:
$$\sin ^2 (\pi \sqrt{n^2 + n})$$
I feel that I will use the identity $$\sin ^2 (\pi \sqrt{n^2 + n}) = \frac{1}{2}(1- \cos(2 \pi \sqrt{n^2 + n})), $$
But then what? how can I deal with the limit of $\cos (2 \pi \sqrt{n^2 + n})$? I know that $\cos (n\pi) = (-1)^n$, if $n$ is a positive integer but then what?
| Intuitively, $n^2+n$ is almost exactly halfway between $n^2$ and $(n+1)^2$, and so $\sqrt{n^2+n}$ is very close to a half-integer. This means $\sin(\pi\sqrt{n^2+n})$ should be very close to $\pm 1$, and in either case $\sin^2(\pi\sqrt{n^2+n})$ will be very nearly $1$.
More formally, we have
$$\sin^2(\pi\sqrt{n^2+n})=\sin^2[\pi(\sqrt{n^2+n}-n)]
$$
Now, multiplying by $1$ in the form $\frac{\sqrt{n^2+n}+n}{\sqrt{n^2+n}+n}$ gives
$$
\sin^2[\pi(\sqrt{n^2+n}-n)]=\sin^2\left(\pi\frac{n}{\sqrt{n^2+n}+n}\right)
$$
Since $\lim_{n \to \infty} \frac{n}{\sqrt{n^2+n}+n}=\frac{1}{2}$ and $\sin^2$ is continuous, this sequence has limit $\sin^2 \frac{\pi}{2}=1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Sum the series $\frac{n}{1⋅2⋅3}+\frac{n-1}{2⋅3⋅4}+ \frac{n-2}{3⋅4⋅5}+...\text{(upto n terms)}$
$\frac{n}{1⋅2⋅3}+\frac{n-1}{2⋅3⋅4}+ \frac{n-2}{3⋅4⋅5}+...\text{(upto n terms)}$
Clearly, we can see that $$T_r=\frac{n-r+1}{r(r+1)(r+2)}$$
Now, somehow, we have to make this telescoping.
But we do not have $n$ as a factor of the denominator, so we have to multiply the denominator by $n$ and after our manipulation, multiply the whole sum by $n$. But that too did not work.
PLease help
| Let $$u_r= \frac 1 r - \frac 1{r+1}$$
Then
$$\begin{split}
u_r-u_{r+1} &= \frac 1 r - \frac 1{r+1} -\bigg( \frac 1 {r+1} - \frac 1{r+2}\bigg) \\
&= \frac 1 r - \frac 2{r+1} + \frac 1 {r+2} \\
&= \frac{ (r+2)(r+1) -2r(r+2) + r(r+1)}{r(r+1)(r+2)}\\
&= \frac 2 {r(r+1)(r+2)}\\
\end{split}$$
With that, the sum to compute is $$\begin{split}
S_n &= \frac 1 2 \sum_{r=1}^n(n-r+1)(u_r-u_{r+1})\\
&= \frac 1 2 \bigg (\sum_{r=1}^n(n-r+1)u_r-\sum_{r=2}^{n+1} (n-r+2)u_r\bigg)\\
&= \frac 1 2 \bigg ( nu_1-u_{n+1} -\sum_{r=2}^n u_r \bigg)\\
&= \frac 1 2 \bigg ( nu_1 -\sum_{r=2}^{n+1} u_r \bigg)\\
&= \frac 1 2 \bigg (nu_1 - \frac 1 2 + \frac 1 {n+2} \bigg) \\
&= \frac 1 2 \bigg (\frac {n-1}2 + \frac 1 {n+2} \bigg) \\
\end{split}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2903555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Induction. Am I missing something or is there a mistake in the question? $$\sum_{k=1}^n k*3^k=\frac {3(3^n(2n-1)+1)} 4 $$
So let f(n)= $\sum_{k=1}^n k*3^k $
and g(n)=$\frac {3(3^n(2n-1)+1)} 4$
By induction hypothesis, $f(n+1) = f(n) + (n+1)3^{n+1} \overset{\text{i.h.}}{=} g(n) + (n+1) 3^{n+1} = g(n+1).$
$$\frac{3(3^{n+1}(2n+1)+1)}4=(n+1)(3^{n+1})+\frac{3(3^n(2n-1)+1))}4 $$
I am stuck afterward, please help, thanks.
| To do the inductive step, observe that:
$(n+1)3^{n+1}+\frac{3(3^n(2n-1)+1)}4=\frac{4(n+1)3^{n+1}+3(3^n(2n-1)+1)}4=\frac{(6n+3)3^{n+1}+3}4=\frac{3(2n+1)3^{n+1}+3}4=\frac{3(3^{n+1}(2(n+1)-1)+1)}4$...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2904076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
} |
Graph complex inequality $|1/z|<1$, when $z \in \mathbb{C}$ Problem
Graph complex inequality $|1/z|<1$, when $z \in \mathbb{C}$
Attempt to solve
if i set $z=x+iy$ when $(x,y) \in \mathbb{R}, z \in \mathbb{C}$
$$ |\frac{1}{x+iy}|<1 $$
Trying to multiply denominator and nominator with complex conjugate.
$$ |\frac{x-iy}{(x+iy)(x-iy)}| <3 $$
$$ |\frac{x-iy}{x\cdot x + x(-iy) + iy \cdot x + iy (-iy)}| <3$$
$$ |\frac{x-iy}{x^2-xyi+xyi-(iy)^2}|<3 $$
Utilize the fact that $i^2=-1$
$$ |\frac{x-iy}{x^2+y^2}|<3 $$
$$ |\frac{x}{x^2+y^2}-\frac{iy}{x^2+y^2}|<3 $$
$$ \sqrt{(\frac{x}{x^2+y^2})^2+(-\frac{iy}{x^2+y^2})^2}<3 $$
$$ (\frac{x}{x^2+y^2})^2+(-\frac{iy}{x^2+y^2})^2<3^2 $$
$$ \frac{x^2}{x^4+2x^2y^2+y^4}+\frac{i^2y^2}{x^4+2x^2y^2+y^4}<3^2 $$
$$ $$
I tried to plot the inequality with wolfram alpha and i got area outside of circle in $(0,0)
Not quite sure if this is correct.
| Why not just write
$$ |1/z| < 1 \quad\iff\quad |z| > 1 \quad\iff\quad x^2+y^2 > 1. $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving equation with fraction I don't understand how to get from the second to the third step in this equation:
$ - \frac { \sqrt { 2 - x ^ { 2 } } - x \left( \frac { - x } { \sqrt { 2 - x ^ { 2 } } } \right) } { \left( \sqrt { 2 - x ^ { 2 } } \right) ^ { 2 } } = - \frac { \sqrt { 2 - x ^ { 2 } } + \frac { x ^ { 2 } } { \sqrt { 2 - x ^ { 2 } } } } { \left( \sqrt { 2 - x ^ { 2 } } \right) ^ { 2 } } = - \frac { \frac { 2 - x ^ { 2 } } { \sqrt { 2 - x ^ { 2 } } } + \frac { x ^ { 2 } } { \sqrt { 2 - x ^ { 2 } } } } { \left( \sqrt { 2 - x ^ { 2 } } \right) ^ { 2 } } = - \frac { 2 } { \left( \sqrt { 2 - x ^ { 2 } } \right) ^ { 3 } } $
Why can we just add $ 2 - x ^ { 2 } $ in the numerator?
Step 1 to step 2, as well as step 3 to step 4 is clear to me.
| Hint : Note that $2-x^2 = (\sqrt{2-x^2}) * (\sqrt{2-x^2})$ and $(\sqrt{2-x^2}) = \frac{2-x^2}{(\sqrt{2-x^2})}$
| {
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"url": "https://math.stackexchange.com/questions/2906794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Monotonicity of the function $(1+x)^{\frac{1}{x}}\left(1+\frac{1}{x}\right)^x$. Let $f(x)=(1+x)^{\frac{1}{x}}\left(1+\frac{1}{x}\right)^x, 0<x\leq 1.$
Prove that $f$ is strictly increasing and $e<f(x)\leq 4.$
In order to study the Monotonicity of $f$, let
$$g(x)=\log f(x)=\frac{1}{x}\log (1+x)+x\log \left(1+\frac{1}{x}\right).$$
And $f$ and $g$ has the same Monotonicity. By computation,
$$g'(x)=\frac{1}{x^2}\left(\frac{x}{1+x}-\log (1+x)\right)+\log \left(1+\frac{1}{x}\right)-\frac{1}{1+x}.$$
As we know $\frac{x}{1+x}-\log (1+x)\leq 0$ and $\log \left(1+\frac{1}{x}\right)-\frac{1}{1+x}\geq 0$. So it does not determine the sign of $g'(x)$.
If we compute the second derivative $g''(x)$, you will find it is also difficult to determine the sign of $g''(x)$.
Our main goal is to prove $$\frac{1}{x^2}\left(\frac{x}{1+x}-\log (1+x)\right)+\log \left(1+\frac{1}{x}\right)-\frac{1}{1+x}>0.$$
Is there some tricks to prove this result. Any help and hint will welcome.
| The right inequality.
We can use the TL method here.
We need to prove that $$(1+a)^b(1+b)^a\leq4$$
for $a>0$, $b>0$ such that $ab=1$, which is $$\frac{\ln(1+a)}{a}+\frac{\ln(1+b)}{b}\leq2\ln2.$$
But $$\sum_{cyc}\left(\ln2-\frac{\ln(1+a)}{a}\right)=\sum_{cyc}f(a),$$
where $f(a)=\ln2-\frac{\ln(1+a)}{a}-(\ln2-0.5)\ln a$.
Easy to show that $f(a)\geq0$ for all $0<a\leq11$.
But for $a>11$ we obtain: $$(1+a)^b(1+b)^a=(1+a)^{\frac{1}{a}}\left(1+\frac{1}{a}\right)^a<(1+11)^{\frac{1}{11}}e<4.$$
The left inequality.
We need to prove that:
$$\frac{1}{x}\ln(1+x)+x\ln\left(1+\frac{1}{x}\right)>1$$ or $g(x)>0$, where
$$g(x)=(1+x^2)\ln(1+x)-x^2\ln{x}-x.$$
Indeed, $$g'(x)=2x\ln(1+x)+\frac{1+x^2}{1+x}-2x\ln{x}-x-1=2x\left(\ln\left(1+\frac{1}{x}\right)-\frac{1}{1+x}\right)>0$$ because
$$\left(\ln\left(1+\frac{1}{x}\right)-\frac{1}{1+x}\right)'=-\frac{1}{x(1+x)^2}<0$$ and
$$\ln\left(1+\frac{1}{1}\right)-\frac{1}{1+1}>0.$$
Id est, $$g(x)>\lim_{x\rightarrow0^+}g(x)=0$$ and we proved that $e<f(x)\leq4.$
Now, we'll prove that $f$ increases on $(0,1].$
By your work we need to prove that
$$\frac{1}{x^2}\left(\frac{x}{1+x}-\ln(1+x)\right)+\ln\left(1+\frac{1}{x}\right)-\frac{1}{1+x}\geq0$$ for all $0<x\leq1$ or
$$(x^2-1)\ln(1+x)-x^2\ln{x}+\frac{(1-x)x}{1+x}\geq0$$ and since $$\ln{x}\leq\frac{2(x-1)}{1+x},$$ it's enough to prove that
$$-(1-x^2)\ln(1+x)+\frac{2x^2(1-x)}{1+x}+\frac{(1-x)x}{1+x}\geq0$$ or
$$\ln(1+x)\leq\frac{x(2x+1)}{(1+x)^2},$$ which is smooth.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $2^t = a^b \pm 1$, what are all the possible values of $t$? Let $t$ be a positive integer such that $2^t = a^b \pm 1$ for some integers $a$ and $b$, each greater than $1$. What are all the possible values of $t$?
The question is taken from here. I know the answer is $3$, so $a = 3$ and $b = 2$. But how can I prove that it's unique? Can you give me a hint, please? Thanks!
| The answer is $t=3$.
This answer proves that the following two claims are true :
Claim 1 : If $t,a,b$ are integers such that $t\ge 1,a\ge 2$ and $b\ge 2$, then $2^t=a^b+1$ has no solutions.
Claim 2 : If $t,a,b$ are integers such that $t\ge 1,a\ge 2$ and $b\ge 2$, then $2^t=a^b-1$ has the only one solution $(t,a,b)=(3,3,2)$.
Proof for Claim 1 : $a$ has to be odd, and $a^b\equiv -1\pmod 4$ implies that $a\equiv -1\pmod 4$ and that $b$ has to be odd. Since $b\ge 3$, we have
$$2^t=(a+1)(a^{b-1}-a^{b-2}+a^{b-3}-\cdots +1)\tag1$$
Here, note that
$$a^{b-1}-a^{b-2}+a^{b-3}-\cdots +1=\frac{a^b+1}{a+1}\ge\frac{a^2+1}{a+1}\gt 1$$
and that
$$a^{b-1}-a^{b-2}+a^{b-3}-\cdots +1\equiv 1\pmod 2$$
It follows from $(1)$ that $2^t$ has an odd divisor greater than $1$, which is impossible. $\quad\square$
Proof for Claim 2 : $a$ has to be odd.
If $b$ is even ($b=2m$), then $2^t=(a^m-1)(a^m+1)$. Since $a^m-1\ge 1$, there are non-negative integers $u,v\ (u\lt v)$ such that $a^m-1=2^u$ and $a^m+1=2^v$ from which $2=2^v-2^u$ follows. If $u=0$, then RHS is odd, which is impossible. If $u\ge 2$, then RHS is divisible by $4$, which is impossible. So, $u=1$ gives $a=3,m=1,v=2$.
If $b$ is odd, then since $b\ge 3$, we have $2^t=(a-1)(a^{b-1}+a^{b-2}+\cdots +1)$. We see that $a^{b-1}+a^{b-2}+\cdots +1$ is odd greater than $1$. It follows that $2^t$ has an odd divisor greater than $1$, which is impossible. $\quad\square$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Calculate $\int \frac{x^4+1}{x^{12}-1} dx$ So I found this problem:
Calculate $$\int \frac{x^4+1}{x^{12}-1} dx$$ where $x\in(1, +\infty)$
and I don't have any ideea how to solve it. I tried to write $$x^{12}-1=(x^6+1)(x+1)(x-1)(x^2+x+1)(x^2-x+1)$$ but with $x^4+1$ I had no idea what to do, an obvious thing would be to add some terms (an $x^5$ would be helpful) and then discard them, but again I couldn't do anything. What should I do?
| Use $$x^6+1=(x^2+1)(x^4-x^2+1)$$
$$I={\displaystyle\int}\dfrac{x^4+1}{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)\left(x^4-x^2+1\right)}\,\mathrm{d}x$$
Apply partial Factor decomposition you get
$$I={\displaystyle\int}\left(-\dfrac{x^2+1}{6\left(x^4-x^2+1\right)}-\dfrac{2x+1}{12\left(x^2+x+1\right)}+\dfrac{2x-1}{12\left(x^2-x+1\right)}-\dfrac{1}{3\left(x^2+1\right)}-\dfrac{1}{6\left(x+1\right)}+\dfrac{1}{6\left(x-1\right)}\right)\mathrm{d}x$$
$$I=-\class{steps-node}{\cssId{steps-node-1}{\dfrac{1}{6}}}{\displaystyle\int}\dfrac{x^2+1}{x^4-x^2+1}\,\mathrm{d}x-\class{steps-node}{\cssId{steps-node-2}{\dfrac{1}{12}}}{\displaystyle\int}\dfrac{2x+1}{x^2+x+1}\,\mathrm{d}x+\class{steps-node}{\cssId{steps-node-3}{\dfrac{1}{12}}}{\displaystyle\int}\dfrac{2x-1}{x^2-x+1}\,\mathrm{d}x-\class{steps-node}{\cssId{steps-node-4}{\dfrac{1}{3}}}{\displaystyle\int}\dfrac{1}{x^2+1}\,\mathrm{d}x-\class{steps-node}{\cssId{steps-node-5}{\dfrac{1}{6}}}{\displaystyle\int}\dfrac{1}{x+1}\,\mathrm{d}x+\class{steps-node}{\cssId{steps-node-6}{\dfrac{1}{6}}}{\displaystyle\int}\dfrac{1}{x-1}\,\mathrm{d}x$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Arithmetic sequence problem $\frac{x+4}{x-3},\frac{x+6}{2},\frac{4}{x-2}$ Choose such x that the following
$$\frac{x+4}{x-3},\frac{x+6}{2},\frac{4}{x-2}$$
forms finite arithmetical sequence.
If I use the equation: $$2a_2=a_3+a_1$$
I always get the wrong answer.
| Let
$$
a_1=\frac{x+4}{x-3}, \quad a_2=\frac{x+6}{2},\quad a_3=\frac{4}{x-2}
$$
If $ a_1 $, $ a_2 $ and $ a_3 $ are in arithmetic progression the possible values for $ x $ are
$$
4,\quad \frac{-4+\sqrt{72}}{2} \quad\mbox{and}\quad \frac{-4-\sqrt{72}}{2}
$$
We have $2a_2={a_3+a_1}$, that is,
$$
{x+6}=\frac{x+4}{x-3}+\frac{4}{x-2}
$$
\begin{align}
(x+6)(x-3)(x-2)=(x+4)(x-2)+4(x-3)
\\
\\
[(x+6)(x-3)-(x+4)](x-2)-4(x-3)=0
\\
\\
[x^2+2x-22](x-2)-4x+12=0
\\
\\
x^3-30x+56=0
\end{align}
Note that $x=4$ is a integer solution this equation.
$$
x^3-30x+56=(x-4)(x^2+4x-14)=0
$$
The other two roots are
$$
\frac{-4+\sqrt{72}}{2} \quad \frac{-4-\sqrt{72}}{2}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Determine the greatest common divisor of polynomials $x^2+1$ and $x^3+1$ in $\Bbb Q[X]$. Exercise: Determine a gcd of the polynomials $x^2+1$ and $x^3+1$ in $\Bbb Q[X]$.. Write the gcd as a combination of the given polynomials.
Is it correct that I keep using long division until the result is $0$, and then the previous result would be a gcd?
x^2+1 / x^3 + 1 \ x
x^3 + x
________-
1 - x
1 - x / x^2 + 1 \ -x-1
x^2 - x
________-
x + 1
x - 1
________-
2
2 / 1 - x \ -1/2x + 1/2
- x
______-
1
1
______-
0
Conclusion: A gcd is $2$?
I'm not sure what's meant with "Write the gcd as a combination of the given polynomials.", so it would be great if someone could point me in the right direction.
| Notice that $x^2+1$ is irreducible over $\mathbb{Q}$. Let $d = \gcd(x^2+1, x^3+1)$. Then $d \mid x^2+1$ so $d$ is a nonzero constant, or $d$ is an associate of $x^2+1$.
However, $x^2+1 \not\mid x^3+1$ so $d$ must be constant. We can choose $d = 1$ since $\gcd$ is usually supposed to be monic.
Finally, notice that
$$1=\left[\frac{1}{2}(x+1) \right](x^3+1) + \left[\frac{1}{2}(-x^2-x+1)\right] (x^2+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2910644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Please show detailed steps of double integration of absolute difference Please show detailed steps of integration
$$\int_0^2 \int_0^1 0.5|x-y| dxdy$$
$$=\frac{1}{2}\int_0^2\int_0^y(y-x)dxdy + \frac{1}{2}\int_0^2\int_y^1(x-y)dxdy$$
$$=\frac{1}{2}\int_0^2\{y\int_0^ydx - \int_0^yxdx\}dy + \frac{1}{2}\int_0^2\{\int_y^1xdx - y\int_y^1dx\}dy$$
$$=\frac{1}{2}\int_0^2\frac{1}{2}y^2dy + \frac{1}{2}\int_0^2\frac{1}{2} -y + \frac{1}{2}y^2dy$$
$$
=2/3 + 1/6 = 5/6
$$
I can't get the $\frac{2}{3}$ solution.
| The problem is with your integration limits. You integrate $y$ from $0$ to $2$ and $x$ from $y$ to $1$. But For say $y=1.5$ it's the wrong way around. If you integrate $x$ from $0$ to $1$ and then you split the $y$ integration from $0$ to $x$ and from $x$ to $2$, you get the right answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2911118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding value of $\lim_{n\rightarrow \infty}\frac{1+\frac{1}{2}+\frac{1}{3}+ \cdots +\frac{1}{n^3}}{\ln(n)}$
Find the value of
$$\lim_{n\rightarrow \infty}\frac{1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n^3}}{\ln(n)}$$
My Try: Using Stolz-Cesaro,
Let $\displaystyle a_{n} = 1+\frac{1}{2}+\frac{1}{3}+\cdots \cdots +\frac{1}{n}$
and $b_{n} = \ln(n)$
So $\displaystyle \frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}} = \lim_{n\rightarrow \infty}\frac{1}{{(n+1)^3}}\cdot \frac{1}{\ln\bigg(1+\frac{1}{n}\bigg)} = 0$
Please explain if what I have done above is right.
| Your $a_n$ should go up to $\frac1{n^3}$. Then $a_{n+1}-a_n=\frac{1}{(n+1)^3}+\frac{1}{(n+1)^3-1}\dots+\frac1{n^3+2}+\frac{1}{n^3+1}$.
I don't know whether you can arrive there by your method, but you can find the limit by using the asymptotics of harmonic numbers / definition of the euler $\gamma$ constant. Then the limit should be $3$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
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Proving an equation holds for $x \neq 0$ I would like to show that for $x \neq 0$,
$\dfrac{1}{x} = 1 + (1 - x) + (1 - x)^2 + (1 - x)^3/x.$
One way would be to just expand everything, but is there an easier way?
The sum of a geometric series is $1/(1 - y)$ Letting $y = 1 - x,$ we see
$1/x = \sum_{n = 0}^{\infty} (1 - x)^{n}$
So this takes care of the first terms, but then I just need to show
$\sum_{n = 3}^{\infty} (1 - x)^{n} = (1 - x)^{3}/x.$
Or, maybe I'm approaching this completely incorrectly.
| We have for $x\neq 0$
$$\dfrac{1}{x} = 1 + (1 - x) + (1 - x)^2 + (1 - x)^3/x$$
$$ (1 - x)^3 + x(1 - x)^2\color{red}{+ x(1 - x)-(1-x)}=0 $$
$$(1 - x)^3 + x(1 - x)^2\color{red}{+ (1 - x)(x-1)}=0$$
$$(1 - x)^3 \color{blue}{+ x(1 - x)^2- (1 - x)^2=0}$$
$$(1 - x)^3 \color{blue}{+ (1 - x)^2(x-1)}=0$$
$$(1 - x)^3 \color{blue}{- (1 - x)^3}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2914035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The limit of $\frac{n^3-3}{2n^2+n-1}$ I have to find the limit of the sequence above.
Firstly, I tried to multiply out $n^3$, as it has the largest exponent.
$$\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} =
\lim_{n\to\infty}\frac{n^3(1-\frac{3}{n^3})}{n^3(\frac{2}{n} + \frac{1}{n^2} - \frac{1}{n^3})} =
\lim_{n\to\infty}\frac{1-\frac{3}{n^3}}{\frac{2}{n} + \frac{1}{n^2} - \frac{1}{n^3}}$$
$$
\begin{align}
\lim_{n\to\infty}1-\frac{3}{n^3} = 1 \\[1ex]
\lim_{n\to\infty}\frac{2}{n} + \frac{1}{n^2} - \frac{1}{n^3} = 0 \\[1ex]
\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} = \frac{1}{0}
\end{align}
$$
Then, after realizing $\frac{1}{0}$ might not be a plausible limit, I tried to multiply out the variable with the largest exponent in both the dividend and the divisor.
$$\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} = \lim_{n\to\infty}\frac{n^3(1 - \frac{3}{n^3})}{n^2(2 + \frac{1}{n} - \frac{1}{n^2})} =
\lim_{n\to\infty}n\cdot\frac{1 - \frac{3}{n^3}}{2 + \frac{1}{n} - \frac{1}{n^2}}$$
$$
\begin{align}
\lim_{n\to\infty}1-\frac{3}{n^3} = 1 \\
\lim_{n\to\infty}2 + \frac{1}{n} - \frac{1}{n^2} = 2 \\
\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} = \frac{1}{2} \\
\lim_{n\to\infty}n = \infty
\end{align}
$$
So, my questions about this problem:
*
*Could $\frac{1}{0}$ be a valid limit?
*Does $\infty\cdot\frac{1}{2}$
equal to $\infty$?
*In conclusion, what is the limit of the sequence
above? $\infty?$
Thank you!
| *
*No, it's not valid, it is not defined. Consider two sequences: $(a_n)_{n\in\mathbb{N}} = \frac{1}{2^{-n}}$ and $(b_n)_{n\in\mathbb{N}} = \frac{1}{-2^{-n}}$. You could say that limit of both of them is $\frac{1}{0}$, since $\pm2^{-n} \to 0$, but does this make sense? Do these sequences have the same limit?
*Saying that $\infty\cdot\frac{1}{2} = \infty$ is not very accurate, but your intuition is good. In general, for any sequence $(a_n)_{n\in\mathbb{N}}$ divergent to $+\infty$ and for any $x>0$, you have $\lim_{n\to\infty} x\cdot a_n = +\infty$. What happens for $x < 0$?
*It is $+\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2914728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 4
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A double sum or a definite integral. I am trying to evaluate the following double sum
\begin{eqnarray*}
\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{(-1)^{n+m}}{n(3n+m)}.
\end{eqnarray*}
Using the integral trick
\begin{eqnarray*}
\frac{1}{3n+m} =\int_0^1 y^{3n+m-1} dy,
\end{eqnarray*}
the sum can be transformed into integral
\begin{eqnarray*}
\int_0^1 \frac{ \ln(1+y^3)}{1+y} dy.
\end{eqnarray*}
Now "half" of this is easy (IBP & rearrange)
\begin{eqnarray*}
\int_0^1 \frac{ \ln(1+y)}{1+y} dy = \frac{1}{2} (\ln 2)^2.
\end{eqnarray*}
So we are left with
\begin{eqnarray*}
\int_0^1 \frac{ \ln(1-y+y^2)}{1+y} dy = \int_0^1 \frac{ (1-2y)\ln(1+y)}{1-y+y^2} dy.
\end{eqnarray*}
Now, apart from the obvious IBP done above, this integral has me stumped.
An exact evaluation would be nice, barring that, an expression in terms of dilogrithmic values or something similar would be helpful. Any comments or answers, gratefully received.
If this integral has been seen before a reference & advise on how I would search & find something similar in the future would be gratefully appreciated.
| This integral won't evaluate to anything pretty, but we can write it in terms of the dilogarithm $\operatorname {Li_2}(x)$. Factoring a $-2$ out of the numerator and completing the square in the denominator gives us:
$$\int_0^1 \frac{ (1-2y)\ln(1+y)}{1-y+y^2} dy = -2\int_0^1 \frac{ (y-\frac12)\ln(1+y)}{(y-\frac12)^2 +\frac34} dy.$$
Then apply the substitution $x=y-\frac12$.
$$\int_{-\frac 12}^{\frac 12} \frac{x \ln(x+ \frac32)}{x^2+ \frac 34}dx = \int_{-\frac 12}^{\frac 12} \overbrace{\ln(x+ \frac32)}^{u} \cdot \underbrace {\frac {x}{x^2+ \frac 34} dx }_{dv}$$ Integration by parts show the original integral is equivalent to $\displaystyle \int_{-\frac 12}^{\frac12} \frac{\ln(x^2+\frac34)}{x+\frac 32}dx$. Now we can introduce yet another substitution, say $z=x+\frac 32$. Now we get:
$$\int_1^2 \frac {\ln \bigl((z- \frac 32)^2 + \frac 34\big)}{z}dz$$
We can factor the sum of squares by using the fact that $a^2 +b^2 = (a+bi)(a-bi)$.
$$\int_1^2 \frac {\ln\big((z- \frac 32 + \frac {\sqrt 3}{2}i)(z- \frac 32 - \frac {\sqrt 3}{2}i)\big)}zdz$$
Using the log rules, you can rewrite this as the sum of two integrals:
$$\int_1^2 \frac {\ln(z- \frac 32 + \frac {\sqrt 3}{2}i)}zdz + \int_1^2 \frac {\ln (z- \frac 32 - \frac {\sqrt 3}{2}i)}zdz \tag{1}$$
Now lets look for a general solution to $\displaystyle \int_1^2 \frac {\ln (x+a)}{x}dx $, where $a$ is any constant.
$$\int_1^2 \frac {\ln (x+a)}{x}dx = \int_1^2 \frac {\ln\big(a(\frac xa +1)\big)}{x}dx = \ln(a) \int_1^2 \frac {dx}x+ \int_1^2 \frac {\ln(\frac xa +1)}{x} dx.$$ Apply the substitution $u = - \frac xa$. This changes it to:
$$ \ln(a)\ln(2)+ \int \frac {\ln(1-u)}{u} du = \Big[\ln(a)\ln(x)+ \operatorname {Li_2} (- \frac xa)+ C\Big]_1^2$$
$$ \int_1^2 \frac {\ln (x+a)}{x}dx =\ln(a)\ln(2)+ \operatorname {Li_2} (- \frac 2a)- \operatorname {Li_2} (-\frac 1a) \tag{2}$$
Now, putting (2) into (1) yields:
$$ \ln(- \frac 32 + \frac {\sqrt 3}{2}i)\ln(2)+ \operatorname {Li_2} (- \frac 2{- \frac 32 + \frac {\sqrt 3}{2}i})- \operatorname {Li_2} (-\frac 1{- \frac 32 + \frac {\sqrt 3}{2}i}) + \ln(- \frac 32 - \frac {\sqrt 3}{2}i)\ln(2)+ \operatorname {Li_2} ( \frac 2{ \frac 32 + \frac {\sqrt 3}{2}i})- \operatorname {Li_2} (\frac 1{ \frac 32 + \frac {\sqrt 3}{2}i}) $$
There's probably a way to simplify this, but I bet it's very tedious. The imaginary part should end up being 0.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
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Proving Altitudes of Triangle can never form a Triangle Prove that Altitudes of Triangle can never form a Triangle
My try: we have altitudes proportional to reciprocals of sides of given triangle
Let $a,b,c$ are sides we have
$$a+b \gt c $$
$$b+c \gt a$$ and $$c+a \gt b$$
Now if $\frac{1}{a}$, $\frac{1}{b}$ and $\frac{1}{c}$ are sides of another Triangle we need to show that
$\frac{1}{a}+\frac{1}{b}$ Can never be Greater than $\frac{1}{c}$
Any hint?
| The condition for a triangle is
$c(a+b) > ab$.
For the 3-4-5 triangle,
this is
$5(3+4) > 3\cdot 4$
which is true.
So the altitudes do form a triangle
in this case.
Note that you can write
the condition as
$\begin{array}\\
0
&\gt ab-c(a+b)\\
&= ab-c(a+b)+c^2-c^2\\
&= (a-c)(b-c)-c^2\\
\text{or}\\
c^2
&\gt (a-c)(b-c)\\
&= (c-a)(c-b)\\
\end{array}
$
This is true if
$c$ is the largest side
of the triangle,
so it is always true.
Therefore,
the altitudes can
always form a triangle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2916383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the limit by using the definition of derivative. Here is the problem.
Let $f$ be the function that has the value of $f(1)=1$ and $f'(1)=2$. Find the value of
$$ L = \lim_{x \to 1} {\frac{\arctan{\sqrt{f(x)}-\arctan{f(x)}}}{ \left (\arcsin{\sqrt{f(x)}}-\arcsin{f(x)}\right)^2}} $$
I have tried using
$$
L=\lim_{x\to 1}
\frac{1}{x-1}\frac{\frac{\arctan{\sqrt{f(x)}}-\arctan{\sqrt{f(1)}}}{x-1}-\frac{\arctan{{f(x)}}-\arctan{{f(1)}}}{x-1}}
{\left [\frac{\arcsin{\sqrt{f(x)}}-\arcsin{\sqrt{f(1)}}}{x-1}-\frac{\arcsin{{f(x)}}-\arcsin{{f(1)}}}{x-1} \right ]^2}
$$
and reduced that big chunks by using $f'(a)=\lim_{x \to a} \frac{f(x)-f(a)}{x-a}$ which I got
$$
\begin{split}
L&=\lim_{a\to 1}
\frac{1}{a-1}\frac{\Big [ \arctan\sqrt{f(x)} \Big ]'_{x=a} - \Big [ \arctan{f(x)} \Big ]'_{x=a}}
{\Big [ \arcsin\sqrt{f(x)} \Big ]'_{x=a} - \Big [ \arcsin{f(x)} \Big ]'_{x=a}}\\[2em]
&=\lim_{a\to 1}
\frac{1}{a-1} \frac{\frac{1}{1+f(a)}\frac{1}{2\sqrt{f(a)}}f'(a)-\frac{1}{1+(f(a))^2}f'(a)}
{\left [ \frac{1}{\sqrt{1-f(a)}}\frac{1}{2\sqrt{f(a)}}f'(a)-\frac{1}{\sqrt{1-(f(a))^2}}f'(a) \right ]^2}\\[2em]
&=\lim_{a\to 1}
\frac{1}{f'(a)\frac{a-1}{1-f(a)}} \frac{\frac{1}{1+f(a)}\frac{1}{2\sqrt{f(a)}}-\frac{1}{1+(f(a))^2}}
{\left [ \frac{1}{2\sqrt{f(a)}}-\frac{1}{\sqrt{1+f(a)}} \right ]^2}\\[2em]
&=\lim_{a\to 1}{-\frac{\frac{1}{1+f(a)}\frac{1}{2\sqrt{f(a)}}-\frac{1}{1+(f(a))^2}}
{\left [ \frac{1}{2\sqrt{f(a)}}-\frac{1}{\sqrt{1+f(a)}} \right ]^2}}\\[2em]
&=\boxed{(\sqrt{2}+1)^2}
\end{split}
$$
but the answer keys tell me that the answer of this problem is $L=\left( \frac{\sqrt{2}+1}{2}\right)^2$.
So, Can someone please explain to me what did I do wrong?
| Probably the derivation is wrong because arcsin x is not differentiable at $x=1$, therefore your first step is not allowed.
As an alternative, since $f(x)$ is continuos at $x=1$ we have that
$$L = \lim_{y \to 1^-} {\frac{\arctan{\sqrt{y}-\arctan{y}}}{ \left (\arcsin{\sqrt{y}}-\arcsin{y}\right)^2}}$$
then we can use that
*
*$\arcsin(x) = \arctan\left(\frac{x}{\sqrt{1 - x^2}}\right)$
*$\arctan(u) - \arctan(v) = \arctan\left(\frac{u - v}{1 +
uv}\right) $
to obtain
$$L = \lim_{y \to 1^-}
\frac{ \arctan\left( \frac{\sqrt y -y}{1+y\sqrt y}\right)}
{ \left[\arctan \left(\frac{\sqrt y \sqrt{1-y^2}-y\sqrt{1-y}} { \sqrt{(1-y)(1-y^2)}+y\sqrt y} \right)\right]^2 }=\frac{3+ 2\sqrt 2}{4}=\left( \frac{\sqrt{2}+1}{2}\right)^2$$
indeed by $y=1-u$ with $u\to 0^+$ by binomial expansion $\sqrt y = 1-\frac12 u +o(u)$ it easy to show that
*
*$\arctan\left( \frac{\sqrt y -y}{1+y\sqrt y}\right)\sim \frac 14 u$
*$\arctan \left(\frac{\sqrt y \sqrt{1-y^2}-y\sqrt{1-y}} {
\sqrt{(1-y)(1-y^2)}+y\sqrt y} \right)\sim (\sqrt 2 -1) \sqrt u$
and therefore
$$\frac{ \arctan\left( \frac{\sqrt y -y}{1+y\sqrt y}\right)}
{ \left[\arctan \left(\frac{\sqrt y \sqrt{1-y^2}-y\sqrt{1-y}} { \sqrt{(1-y)(1-y^2)}+y\sqrt y} \right)\right]^2 }\sim \frac{\frac 14 u}{(\sqrt 2 -1)^2 u}=\frac{3+ 2\sqrt 2}{4}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proof Verification $\sum_{k=0}^n \binom{n}{k} = 2^n$ (Spivak's Calculus) $$\sum_{k=0}^n \binom{n}{k} = 2^n$$ I'll use induction to solve prove this. Then
$$\sum_{k=0}^n \binom{n}{k} = 2^n = \binom{n}{0} + \binom{n}{1} + ... + \binom{n}{n - 1} + \binom{n}{n}$$
First prove with n = 1
$$\binom{1}{0} + \binom{1}{1} = 2^1$$
Since $$\binom{1}{0} = \binom{1}{1} = 1$$
it's true.
Now suppose that is true with $n$ if is true with $n + 1$
Then, multiply both sides by two
$$2(2^n) = 2(\binom{n}{0} + \binom{n}{1} + ... + \binom{n}{n - 1} + \binom{n}{n})$$
$$2^{n+1} = 2\binom{n}{0} + 2\binom{n}{1} + ... + 2\binom{n}{n - 1} + 2\binom{n}{n}$$
$$2\binom{n}{0} + 2\binom{n}{1} + ... + 2\binom{n}{n - 1} + 2\binom{n}{n} = \binom{n}{0} + \binom{n}{0} + \binom{n}{1} + \binom{n}{1} + ... + \binom{n}{n} + \binom{n}{n} $$
The first term have two equal term, then, you sum the last one with the first one of the next term, and you'll get this
$$\binom{n}{0} + \binom{n}{1} +... + \binom{n}{n-1} + \binom{n}{n}$$
If we use this equation (Already proved)
$$\binom{n}{k-1} + \binom{n}{k} = \binom{n+1}{k} $$
Of course, we'll have two term without sum, one $\binom{n}{0}$ and $\binom{n}{n}$
We can write these two term like this
$$\binom{n}{0} = \binom{n}{n} = \binom{n+1}{0} = \binom{n+1}{n+1}$$
Then, we get
$$2^{n+1} = \binom{n+1}{0} + \binom{n+1}{1} + ... + \binom {n+1}{n+1}$$
And it's already proved. Note: Just if we take $0! = 1$
I have to prove these too.
$\sum_{k}^n \binom{n}{m} = 2^{n-1}$ If $m$ is even. And
$\sum_{j}^n \binom{n}{j} = 2^{n-1}$ If $j$ is odd. Then, I just said that
If
$$\sum_{m}^n \binom{n}{m} + \sum_{j}^n \binom{n}{j} = \sum_{k=0}^n \binom{n}{k} $$
Then
$$2^{n - 1} + 2^{n - 1} = 2^n$$
Which is true, then, I already prove this. And I have a last one.
$$\sum_{i=0}^n (-1)^i\binom{n}{i} = 0$$
if n is odd. Then
$$\binom{n}{0} - \binom{n}{1} + ... + \binom{n}{n-1} - \binom{n}{n} = 0$$
And that's can be solve knowing that $$\binom{n}{k} = \binom{n}{n - k}$$
And if n is even
$$\binom{n}{0} - \binom{n}{1} + ... - \binom{n}{n-1} + \binom{n}{n} = 0$$
That means that every negative term if when n is odd, then, we can use our two last prove to prove it
If
$$\sum_{m}^n \binom{n}{m} - \sum_{j}^n \binom{n}{j} = 0$$
Then
$$2^{n-1} - 2^{n-1} = 0$$
Which is true.
And that's it, I want to know if my proves are fine and are rigorous too and what is the meaning of every combinatorics prove .
I want to know too better approaches to prove these (Or forms more intuitive)
| A better mathematician than you or I once struggled as mightily with this question as you have, and I told him to expand $(1+1)^n$ with the binomial expansion.
| {
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"timestamp": "2023-03-29T00:00:00",
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Question about Paul Erdös' proof on the Sylvester-Schur Theorem I'm going through the proof and I am not clear the sufficiency of Erdös's argument at one step.
Here's the set up to the argument:
(1) Let $\{x\}$ be the least integer greater or equal to $x$.
(2) Let $a_i$ be shorthand for $\left\{\dfrac{n}{2^i}\right\}$ so that $a_1 =\left\{\dfrac{n}{2}\right\}, a_2 =\left\{\dfrac{n}{2^2}\right\}, a_k =\left\{\dfrac{n}{2^k}\right\}$ and $a_1 \ge a_2 \ge a_3 \ge \dots a_k$
(3) $a_k \le 2a_{k+1}$ since $a_k < \dfrac{n}{2^k}+1 = \dfrac{2n}{2^{k+1}}+1 \le 2a_{k+1} + 1$
(4) If $m$ is the first exponent for which $\dfrac{n}{2^m} \le 1$, then $a_m = 1$
Now, here's the conclusion that Erdös makes:
$\prod\limits_{p_i \le n}p_i\prod\limits_{p_k \le \sqrt{n}}p_k\prod\limits_{p_l \le \sqrt[3]{n}}p_l \dots \le {2a_1 \choose a_1}{2a_2 \choose a_2}\dots{2a_m\choose a_m}$
The argument for this is based on the following observations (which all seem reasonable to me -- I am just not clear how they add up to the conclusion above)
*
*$1 < y \le n$ is completely covered by the intervals:
$$a_m < y\le 2a_m, a_{m-1} < y \le 2a_{m-1}, \dots, a_1 < y \le 2a_1$$
*
*The interval $1 < y \le \left\lfloor\sqrt[k]{n}\right\rfloor$ is completely covered by the intervals:
$$\left\lfloor\sqrt[k]{a_m}\right\rfloor < y\le \left\lfloor\sqrt[k]{(2a_m)}\right\rfloor, \left\lfloor\sqrt[k]{a_{m-1}}\right\rfloor < y\le \left\lfloor\sqrt[k]{(2a_{m-1})}\right\rfloor, \dots,$$
I am not clear how this shows for the inequality above: "the right side being a multiple of the left."
| From the observations, we can say that
$$\prod\limits_{p_i \le \sqrt[k]n}p_i\le \left(\prod_{\left\lfloor\sqrt[k]{a_1}\right\rfloor < p_i\le \left\lfloor\sqrt[k]{2a_1}\right\rfloor}p_i\right)\left(\prod_{\left\lfloor\sqrt[k]{a_2}\right\rfloor < p_i\le \left\lfloor\sqrt[k]{2a_2}\right\rfloor}p_i\right)\cdots \left(\prod_{\left\lfloor\sqrt[k]{a_1}\right\rfloor < p_i\le \left\lfloor\sqrt[k]{2a_m}\right\rfloor}p_i\right)$$
and that the right side is a multiple of the left.
So,
$$\begin{align}&\prod\limits_{p_i \le n}p_i\prod\limits_{p_k \le \sqrt{n}}p_k\prod\limits_{p_l \le \sqrt[3]{n}}p_l \dots
\\\\&\le \left(\prod_{\left\lfloor a_1\right\rfloor < p_i\le \left\lfloor 2a_1\right\rfloor}p_i\right)\left(\prod_{\left\lfloor a_2\right\rfloor < p_i\le \left\lfloor 2a_2\right\rfloor}p_i\right)\cdots \left(\prod_{\left\lfloor a_1\right\rfloor < p_i\le \left\lfloor 2a_m\right\rfloor}p_i\right)
\\\\&\qquad\times \left(\prod_{\left\lfloor\sqrt{a_1}\right\rfloor < p_k\le \left\lfloor\sqrt{2a_1}\right\rfloor}p_k\right)\left(\prod_{\left\lfloor\sqrt{a_2}\right\rfloor < p_k\le \left\lfloor\sqrt{2a_2}\right\rfloor}p_k\right)\cdots \left(\prod_{\left\lfloor\sqrt{a_1}\right\rfloor < p_k\le \left\lfloor\sqrt{2a_m}\right\rfloor}p_k\right)
\\\\&\qquad\times \left(\prod_{\left\lfloor\sqrt[3]{a_1}\right\rfloor < p_l\le \left\lfloor\sqrt[3]{2a_1}\right\rfloor}p_l\right)\left(\prod_{\left\lfloor\sqrt[3]{a_2}\right\rfloor < p_l\le \left\lfloor\sqrt[3]{2a_2}\right\rfloor}p_l\right)\cdots \left(\prod_{\left\lfloor\sqrt[3]{a_1}\right\rfloor < p_l\le \left\lfloor\sqrt[3]{2a_m}\right\rfloor}p_l\right)
\\\\&\qquad\times\cdots
\\\\&=\left(\prod_{\left\lfloor a_1\right\rfloor < p_i\le \left\lfloor 2a_1\right\rfloor}p_i\right)\left(\prod_{\left\lfloor\sqrt{a_1}\right\rfloor < p_k\le \left\lfloor\sqrt{2a_1}\right\rfloor}p_k\right)\left(\prod_{\left\lfloor\sqrt[3]{a_1}\right\rfloor < p_l\le \left\lfloor\sqrt[3]{2a_1}\right\rfloor}p_l\right)\cdots
\\\\&\qquad\times\left(\prod_{\left\lfloor a_2\right\rfloor < p_i\le \left\lfloor 2a_2\right\rfloor}p_i\right)\left(\prod_{\left\lfloor \sqrt{a_2}\right\rfloor < p_k\le \left\lfloor \sqrt{2a_2}\right\rfloor}p_k\right)\left(\prod_{\left\lfloor \sqrt[3]{a_2}\right\rfloor < p_l\le \left\lfloor \sqrt[3]{2a_2}\right\rfloor}p_l\right)\cdots
\\\\&\qquad\times\cdots
\\\\&\qquad\times \left(\prod_{\left\lfloor{a_m}\right\rfloor < p_i\le \left\lfloor{2a_m}\right\rfloor}p_i\right)\left(\prod_{\left\lfloor\sqrt{a_m}\right\rfloor < p_k\le \left\lfloor\sqrt{2a_m}\right\rfloor}p_k\right)\left(\prod_{\left\lfloor\sqrt[3]{a_m}\right\rfloor < p_l\le \left\lfloor\sqrt[3]{2a_m}\right\rfloor}p_l\right)\cdots
\end{align}$$
Since $\binom{2a_k}{a_k}$ is divisible by any prime $p$ such that
$$\sqrt[s]{a_k}\lt p\le \sqrt[s]{2a_k}$$
for any positive integer $s$, we see that
$$\left(\prod_{\left\lfloor a_k\right\rfloor < p_i\le \left\lfloor 2a_k\right\rfloor}p_i\right)\left(\prod_{\left\lfloor\sqrt{a_k}\right\rfloor < p_k\le \left\lfloor\sqrt{2a_k}\right\rfloor}p_k\right)\left(\prod_{\left\lfloor\sqrt[3]{a_k}\right\rfloor < p_l\le \left\lfloor\sqrt[3]{2a_k}\right\rfloor}p_l\right)\cdots$$
divides
$$\binom{2a_k}{a_k}$$
Therefore, we can say that
$$\begin{align}&\left(\prod_{\left\lfloor a_1\right\rfloor < p_i\le \left\lfloor 2a_1\right\rfloor}p_i\right)\left(\prod_{\left\lfloor\sqrt{a_1}\right\rfloor < p_k\le \left\lfloor\sqrt{2a_1}\right\rfloor}p_k\right)\left(\prod_{\left\lfloor\sqrt[3]{a_1}\right\rfloor < p_l\le \left\lfloor\sqrt[3]{2a_1}\right\rfloor}p_l\right)\cdots
\\\\&\qquad\times\left(\prod_{\left\lfloor a_2\right\rfloor < p_i\le \left\lfloor 2a_2\right\rfloor}p_i\right)\left(\prod_{\left\lfloor \sqrt{a_2}\right\rfloor < p_k\le \left\lfloor \sqrt{2a_2}\right\rfloor}p_k\right)\left(\prod_{\left\lfloor \sqrt[3]{a_2}\right\rfloor < p_l\le \left\lfloor \sqrt[3]{2a_2}\right\rfloor}p_l\right)\cdots
\\\\&\qquad\times\cdots
\\\\&\qquad\times \left(\prod_{\left\lfloor{a_m}\right\rfloor < p_i\le \left\lfloor{2a_m}\right\rfloor}p_i\right)\left(\prod_{\left\lfloor\sqrt{a_m}\right\rfloor < p_k\le \left\lfloor\sqrt{2a_m}\right\rfloor}p_k\right)\left(\prod_{\left\lfloor\sqrt[3]{a_m}\right\rfloor < p_l\le \left\lfloor\sqrt[3]{2a_m}\right\rfloor}p_l\right)\cdots\end{align}$$
divides
$${2a_1 \choose a_1}{2a_2 \choose a_2}\dots{2a_m\choose a_m}$$
It follows from these that
$${2a_1 \choose a_1}{2a_2 \choose a_2}\dots{2a_m\choose a_m}$$
is a multiple of
$$\prod\limits_{p_i \le n}p_i\prod\limits_{p_k \le \sqrt{n}}p_k\prod\limits_{p_l \le \sqrt[3]{n}}p_l \dots $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2919395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Prove formula $\operatorname{arctanh} x = \frac12\,\ln \left(\frac{1+x}{1-x}\right)$ Problem
Prove formula $\operatorname{arctanh} x = \frac{1}{2} \ln \left(\frac{1+x}{1-x}\right)$
Attempt to solve
To start off with definition of functions $\sinh(x)$ and $\cosh(x)$
$$ \cosh(x)=\frac{e^x+e^{-x}}{2} $$
$$ \sinh(x) = \frac{e^x-e^{-x}}{2} $$
Hyperbolic tangent is defined as:
$$ \tanh(x)=\frac{\sinh(x)}{\cosh(x)}=\frac{e^x-e^{-x}}{e^{x}+e^{-x}} $$
Notation $\text{arctanh}(x)$ means area tangent which is inverse of hyperbolic tangent.
$$ \operatorname{arctanh}(x)=\tanh^{-1}(x) $$
Trying to invert the $\tanh(x)$ we get:
$$\begin{align}
\frac{e^x-e^{-x}}{e^x+e^{-x}} &= y &\implies\\
e^x-e^{-x}&=y(e^x+e^{-x}) &\implies\\
e^x-e^{-x}&=ye^{x}+ye^{-x} &\implies\\
e^x(1-y)&=e^{-x}(1+y) &\implies\\
\ln(e^x(1-y)) &= \ln(e^{-x}(1+y)) &\implies\\
\ln(e^x)+\ln(1-y) &= \ln(e^{-x})+\ln(1+y) &\implies\\
x + \ln(1-y) &= -x + \ln(1+y) &\implies\\
2x &= \ln(1+y)-\ln(1-y)&\implies\\
x &= \frac{1}{2} \ln \frac{1+y}{1-y}
\end{align}
$$
By switching variables we get:
$$ \implies \operatorname{arctanh}(y) = \frac{1}{2} \ln \left(\frac{1+y}{1-y}\right) $$
| Looks fine to me. Well done!
Just a small thing, a matter of taste: I'd prefer to do this:
\begin{align*}
e^x(1-y) &= e^{-x}(1+y)\\
e^{2x} &=\dfrac{1+y}{1-y}\\
x &= \dfrac{1}{2}\ln\left(\dfrac{1+y}{1-y}\right).
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2922852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How i can reduce this polynomial in $\mathbb{Z}_2 [x]$? How i can reduce the polynomial $f(x) = x^9+x^8+x^6+x^5+x^4+x^3+1 \in \mathbb{Z}_2 [x]$. I've tried use the reduction modulo a prime, but that don't solve the problem totally.
| Using sage the factorization is easy:
sage: R.<x> = PolynomialRing(GF(2))
sage: f = x^9 + x^8 + x^6 + x^5 + x^4 + x^3 + 1
sage: f.factor()
(x^3 + x^2 + 1) * (x^3 + x + 1)^2
(Assuming that "reduce" means "factor" in the OP.)
Manually, as a human, we expect factors - if any - till degree four. So it is a good try to see if one of the factors
*
*$x$, $(x+1)$,
*then $(x^2+x+1)$, the only irreducible polynomial in degree two,
*then $(x^3+x+1)$, $(x^3+x^2+1)$, the only irreducible polynomials in degree three,
*and finally $(x^4 + x + 1)$, $(x^4 + x^3 + 1)$, $(x^4 + x^3 + x^2 + x + 1)$, the only irreducible polynomials in degree four
divides the given polynomial.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2923560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $\sum\limits_{n=1}^{\infty}\frac{1}{p(n)}\in\mathbb{Q}$, is $\sum\limits_{n=1}^{\infty}\frac{n}{p(n)}\in\mathbb{Q}$? Suppose $p(n)$ is a polynomial with rational coefficients and rational roots of degree at least $3$. If we know
$$\sum_{n=1}^{\infty}\frac{1}{p(n)}\in\mathbb{Q}$$
are we able to infer that
$$\sum_{n=1}^{\infty}\frac{n}{p(n)}\in\mathbb{Q}?$$
I've tried several approaches to proving (or disproving) this to include the following:
-Looking for counterexamples
-Generating functions
-Residues
-Partial fraction decomposition
but nothing has yielded any positive or negative results. Any tips, terms, papers, methods, or generally topics that I could look into would also be welcome.
Edit: As noted by Carl Schildkraut below, if this is true, then we would automatically know that $\zeta(2k+1)$ was irrational. Since this seems to greatly increase the potential difficulty, I offer the following modification in order to simplify it:
Suppose $p(n)$ is a polynomial with rational coefficients, rational roots, $\deg(P)\geq 3$, and every root has order $1$. If we know
$$\sum_{n=1}^{\infty}\frac{1}{p(n)}\in\mathbb{Q}$$
are we able to infer that
$$\sum_{n=1}^{\infty}\frac{n}{p(n)}\in\mathbb{Q}?$$
| $\newcommand\Q{\mathbf{Q}}$
I object a little bit to the "accepted" answer. It would require proving that $\sqrt{3} \pi - 3 \log(2)$ is not a rational number. I'm not sure this is so obvious.
Here is an alternate solution which is less random:
Let~$b > a$ be distinct non-zero rational numbers such that $2a$ and $2b$ are integers but $a$ and $b$ are not.
Note that $a+b$ and $a-b$ will both be integers. Let
$$R= \displaystyle{\frac{a+b}{a-b}},$$
and let
$$f(x) = \frac{1}{a b(a+b)} \cdot x(x+a)(x+b)(x+a+b).$$
Note that
$$ \sum_{n=1}^{\infty} \frac{1}{f(n)} = \sum_{n=1}^{\infty}
\frac{ab(a+b)}{n(n+a)(n+b)(n+a+b)}
=
\sum_{n=1}^{\infty} \frac{1}{n} - \frac{1}{n+a+b} + \frac{R}{n+a} - \frac{R}{n+b} $$
$$= \sum_{n=1}^{b + a } \frac{1}{n} + R \sum_{n=1}^{b-a} \frac{1}{n+a} \in \Q + \Q = \Q$$
is a telescoping sum and thus rational. On the other hand, if
$$F(a,b) = \frac{1}{R} \sum_{n=1}^{\infty} \frac{n}{f(n)},$$
then
$$F(a,b) = \frac{1}{R} \sum_{n=1}^{\infty}
\frac{n ab(a+b)}{n(n+a)(n+b)(n+a+b)} = \sum_{n=1}^{\infty} \frac{b}{n+b} - \frac{a}{n+a} - \frac{b-a}{n+a+b}$$
$$= \sum_{n=1}^{\infty} \frac{a}{n+b} + \frac{b-a}{n+b} - \frac{a}{n+a} - \frac{b-a}{n+a+b}$$
$$= \sum_{n=1}^{\infty} \frac{a}{n+b} - \frac{a}{n+a} + \frac{b-a}{n+b} - \frac{b-a}{n+a+b}$$
$$= \left(\sum_{n=1}^{b-a} - \frac{a}{n+a} \right) + (b-a) \sum_{n=1}^{\infty} \frac{1}{n+b} - \frac{1}{n+a+b}$$
$$= \left(\sum_{n=1}^{b-a} - \frac{a}{n+a} \right) + (b-a)\left( \sum_{n=0}^{\infty} \frac{1}{n+1/2} - \frac{1}{n+1}
\right)
- (b-a) \sum_{n=0}^{b-1/2} \frac{1}{n+1/2} + (b-a) \sum_{n=0}^{a+b-1} \frac{1}{n+1}$$
$$ \in \Q + 2(b-a) \log 2 + \Q + \Q = \Q + 2(b-a) \log 2.$$
This is similar to the other solution, except showing that $\log(2)$ is irrational is a direct consequence of the transcendence of $e$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2923680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 2,
"answer_id": 1
} |
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