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Show that $\sqrt[3]{1+\sqrt{3}}$ isn't an element of the field $\mathbb{Q}(\sqrt{3} ,\sqrt[3]{2})$ Setting $\alpha = \sqrt[3]{2}$ and $a+b\alpha+c\alpha^2=\sqrt[3]{1+\sqrt{3}}$ for $a,b,c$ in $\mathbb{Q}(\sqrt{3})$ (The minimal polynomial for $\sqrt[3]{2}$ in $\mathbb{Q}(\sqrt{3})$ is $x^3-2=0$ since if it were reducible that would imply the existence of $\sqrt[3]{2}$ in $\mathbb{Q}(\sqrt{3})$ which is easily shown not to be true) . This comes down to proving that the equations $$ a^3+9ab^2=1 $$ and $$ 3b^3+3a^2b=1$$ have no solutions $a,b$ in $\mathbb{Q}(\sqrt{3})$. This is obviously extremely difficult to prove so there should be a better approach.
This is too long for a comment.
Considering the the irreducible polynomials for $\sqrt[3]{2}$ and $\sqrt[3]{1+\sqrt{3}}$ over $\mathbb{Q}(\sqrt{3})$ we see they are $x^3-(1+\sqrt{3})$ and $x^3-2$ respectively. If the two extensions in the comments are equal that would imply that there is an automorphism of the field $\mathbb{Q}(\sqrt{3})$ which fixes $\mathbb{Q}$ and sends $1+\sqrt{3}$ to $2$ but there is only one nontrivial automrphism of this field that fixes $\mathbb{Q}$ and it sends $\sqrt{3}$ to $-\sqrt{3}$. So the two extensions can't be isomorphic, let alone equal. Is this correct?
| Here is an alternative solution, which is essentially the same as Lubin's but with a more elementary presentation.
Lemma. Let $K$ be a subfield of $\mathbb C$ with $\sqrt[3]{2}\not\in K$ (whence it easily follows that $[K(\sqrt[3]{2}):K]=3$). Let $k\in K$. Then $k$
is a cube in $K(\sqrt[3]{2})$ iff one of $k,\frac{k}{2},\frac{k}{4}$ is already
a cube in $K$.
Proof of lemma. The "if" direction is obvious, so let us concentrate on the "only if" part : suppose that $k$ is a cube in $K(\sqrt[3]{2})$, so that
$k=(a+b\sqrt[3]{2}+c\sqrt[3]{4})^3$ for some $a,b,c\in K$. Putting $\theta=a+b\sqrt[3]{2}+c\sqrt[3]{4}$, one can compute that
$$\theta^2=(2ac + b^2)\sqrt[3]{4} + 2(ab + c^2)\sqrt[3]{2} + (a^2 + 4bc)$$
and
$$\theta^3=3(ab^2+a^2c+2bc^2)\sqrt[3]{4} + 3(a^2b+2b^2c+2ac^2)\sqrt[3]{2} + (a^3+2b^3+4c^3+12abc)$$
Since $\theta^3=k\in K$, we must have $z_1=ab^2+a^2c+2bc^2=0$ and $z_2=a^2b+2b^2c+2ac^2=0$. Now $az_1-bz_2=3c(a^3-2b^3)$, so either $a^3-2b^3=0$
(in which case $a=b=0$ because $\sqrt[3]{2}\not\in K$) or $c=0$. It easily follows that at most one of $a,b,c$ is nonzero and the lemma is proved.
Using the lemma above with $k=1+\sqrt{3}$ and $K={\mathbb Q}(k)$, we now have to
see if $k,\frac{k}{2},\frac{k}{4}$ are cubes in $K$ or not.
If $k=(a+b\sqrt{3})^3$ with $a,b\in{\mathbb Q}$, we deduce $a^3+9ab^2=1$ and
$3b^3+3a^2b=1$. We can write $a=\frac{u}{q},b=\frac{v}{q}$ where $u,v,q$ are integers. Then $u^3+9uv^2=q^3$ and $3u^3+3u^2b=q^3$. Reasoning modulo $5$, it can easily be checked that the only solution to this system in ${{\mathbb F}_5}^3$ is $(u,v,q)=(0,0,0)$. Returning to $\mathbb Z$, this means that each of $u,v,q$ must be divisible by $5$. But by infinite descent, this clearly implies that the only solution is $(u,v,q)=(0,0,0)$ in $\mathbb Z$ also, contradiction.
Similarly, in the two other cases we obtain a system where $(0,0,0)$ is the only solution modulo $5$. This finishes the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3326308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 2
} |
tangent inequality in triangle Let $a$, $b$ and $c$ be the measures of angles of a triangle (in radians).
It is asked to prove that
$$\tan^2\left(\dfrac{\pi-a}{4}\right)+\tan^2\left(\dfrac{\pi-b}{4}\right)+\tan^2\left(\dfrac{\pi-c}{4}\right) \ge 1$$
When does equality occur ?
My try :
Letting $u:= \tan\left(\dfrac{\pi-a}{4}\right)$ and $v:= \tan\left(\dfrac{\pi-b}{4}\right)$ the inequality reduces to proving
$$u^2+v^2+\dfrac{(1-uv)^2}{(u+v)^2} \ge 1\quad\quad (*)$$ ($$u,v\in (0,1)$$)
( using $a+b+c=\pi$ and the formula for $\tan(x+y)$ and the fact that $\tan\left(\dfrac{\pi}{2}-x\right)=\dfrac{1}{\tan x}$ )
I'm having trouble in proving that last inequality.
Any suggestions are welcome.
Thanks.
Edit : is the following reasoning to prove the inequality (*) sound ?
(*) is obvious when $u^2 + v^2 \ge 1$ so we only have to deal with the case $u^2 + v^2 \le 1$ which we assume true in what follows.
(*) $\iff (u^2 + v^2)(u + v)^2 \ge (u+v)^2+(1-uv)^2$
Setting $x:= u^2 + v^2$ and $a:=uv$ we get
(*) $\iff x^2+(2a-1)x \ge a^2+4a-1$
Now some calculus $(x^2+(2a-1)x)' = 2x + 2a-1$
the function $\phi:x\mapsto x^2+(2a-1)x$ then has a minimum at $\dfrac 12 - a$ which is $\phi\left(\dfrac 12 - a\right) = -a^2+a-\dfrac 14$
It then suffices to have $-a^2+a-\dfrac 14 \ge a^2+4a-1$
This last ineq is equivalent to
$8a^2+12a-3 \le 0$
which, in turn, is equivalent to
$a \in \left[\dfrac{-6-\sqrt{60}}{8}, \dfrac{-6+\sqrt{60}}{8}\right]$
Recall that we're working under the assumption $u^2 +v^2 \le 1$, that yields in particular that $uv \le \dfrac 12$.
since $ \dfrac{-6-\sqrt{60}}{8} \le 0 \le a := uv \le \dfrac 12 \le \dfrac{-6+\sqrt{60}}{8}$, we're done.
Thanks for taking time to check the correctness of the above proof.
| A different formulation:
Let
$\displaystyle \alpha = \frac{\pi - A}{4}$
$\displaystyle \beta = \frac{\pi - B}{4}$
$\displaystyle \gamma = \frac{\pi - C}{4}$
We need to find the minimum value of
$\displaystyle \tan^2 \alpha + \tan^2 \beta + \tan^2 \gamma$
subject to
$\displaystyle \alpha + \beta + \gamma = \frac{\pi}{2}$
Also $\displaystyle 0 \leq \alpha, \beta, \gamma \leq \frac{\pi}{4}$
Let
$\displaystyle f(\alpha, \beta, \gamma) = \tan^2 \alpha + \tan^2 \beta + \tan^2 \gamma + \lambda \left(\alpha + \beta + \gamma - \frac{\pi}{2} \right)$
Now set
$\displaystyle \frac{\partial f}{\partial \alpha} = \frac{\partial f}{\partial \beta} = \frac{\partial f}{\partial \gamma} = 0$
We immediately have
$\displaystyle \alpha = \beta = \gamma = \frac{\pi}{6}$
and finally the minimum value of $f$ is
$\displaystyle f \left( \frac{\pi}{6}, \frac{\pi}{6}, \frac{\pi}{6} \right) = 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3327429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
How to quickly compute the determinant of given matrix I need to find the determinant of given matrix :
$\begin{bmatrix}
1&0&0&0&0&2\\
0&1&0&0&2&0\\
0&0&1&2&0&0\\
0&0&2&1&0&0\\
0&2&0&0&1&0\\
2&0&0&0&0&1\\
\end{bmatrix}$
I know that It can be computed with the help of row operations;by applying
R1 $\to$ R1 + R2+R3+R4+R5+R6, and then subsequently simplifying using further row operations to find the determinant.
However my question is :
Is there another way to compute the determinant quickly ?,because the row operations are quite time taking and error prone.
I thought of using eigenvalues here but could not reach the solution
Can anyone help me to find this determinant quickly ?
Thank you
| By reordering both rows and columns you can see that the determinant is equal to that of a block diagonal matrix:
$$
\begin{vmatrix}
1&2&0&0&0&0\\
2&1&0&0&0&0\\
0&0&1&2&0&0\\
0&0&2&1&0&0\\
0&0&0&0&1&2\\
0&0&0&0&2&1\\
\end{vmatrix}
= \left(\begin{vmatrix}
1&2 \\ 2&1
\end{vmatrix}\right)^3
$$
Or with Schur complements: We have
$$
M = \begin{pmatrix}
A & B \\ C &D
\end{pmatrix} \text{ with } A = D = \begin{pmatrix}
1 & 0 & 0 \\ 0 & 1 & 0 \\ 0& 0 & 1
\end{pmatrix} \text{ and } B = C = \begin{pmatrix}
0 & 0 & 2 \\ 0 & 2 & 0 \\ 2& 0 & 0
\end{pmatrix}
$$
and therefore (with $I_3$ denoting the $3\times 3$ identity matrix):
$$
\det M = \det(A) \det(D - CA^{-1}B) = \det(I_3 - B^2) = \det(-3I_3) = -27 \, .
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3330252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Eliminating parameter $\beta$ from $x=\cos 3 \beta + \sin 3 \beta$, $y = \cos \beta - \sin \beta$
Based on the given parametric equations:
$$\begin{align}
x &=\cos 3 \beta + \sin 3 \beta \\
y &= \cos \beta \phantom{3}- \sin \beta
\end{align}$$
Eliminate the parameter $\beta$ to prove that $x-3y+2y^3=0$.
What I got so far:
$$\cos 3 \beta + \sin 3 \beta = ( \cos \beta - \sin \beta)(1+4\sin\beta\cos\beta)$$
Which trigonometric identity should I use to proceed?
| Making
$$
y = \frac 12\left(e^{i\beta}+e^{-i \beta}\right)-\frac{1}{2i}\left(e^{i\beta}-e^{-i\beta}\right)
$$
we have after powering and collecting
$$
y^3 = \frac 32\left(\cos\beta-\sin\beta\right)-\frac 12\left(\cos(3\beta)+\sin(3\beta)\right)
$$
then follows
$$
y^3 = \frac 32 y - \frac 12 x
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3333626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Please explain the formula for the sum of the cubes and the difference: $a^3 - b^3$ and $a^3 + b^3$? I have not yet fully mastered the formula for accelerated multiplication. I was able to understand everything except the last two. I would like to deal with them. Why have : $$ a^3 + b^3 = (a+b)(a^2 -ab + b^2)^* $$ and $$ a^3 - b^3 = (a-b)(a^2 + ab + b^2) $$ Can someone paint me a formula in detail. Why is this the way it works.
| There is a very general factorisation formula, once taught in high school:
$$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\dots+a^{n-k}b^k+\dots+ab^{n-2}+b^{n-1}).$$
(To memorise it, the second factor is the sum of all monomials in $a$ and $b$ of total degree $n-1$.)
The simplest proof consists in showing first by induction the particular case
$$1-x^n=(1-x)\cdot(1+x+x^2+\dots+x^n), $$
then, for the general case, we may suppose that $a\ne 0$ and we set $x=\frac ba$, so that
\begin{align}
a^n-b^n&=a^n(1-x^n)=a^n(1-x)(1+x+x^2+\dots+x^{n-1})\\
&=(a-ax)(a^{n-1}+a^{n-1}x+a^{n-1}x^2+\dots+a^{n-1}x^{n-1})
\end{align}
and observing that, for each $k$, $\;a^{n-1}x^k=a^{n-1-k}b^k$.
For the sum of powers, there is an analogous formula for $n$ odd, with an alternating sum:
$$a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+\dots+(-1)^{k-1}a^{n-k}b^k+\dots-ab^{n-2}+b^{n-1}).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3338889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Simple sum with $i = 2$ step I need to calculate the following sum:
$$S_{n} = \sum_{2 \leq i \leq n } (3i - 2)$$
(two more conditions for the above sum: $n$ is even and $i$ with step $2$ (not sure how to do multi-line))
Adding image of the task with multi-line visible:
I wrote down the first few terms and it looks like this:
$4, 10, 16, 22, ...$
I re-arranged a bit and obtained such sum:
$$S_{n} = \sum_{i = 0}^{n} 4+6i$$
and its terms are the same: $4, 10, 16, 22, ...$
Is the top boundary correct?
Proceeding with the solution:
$$S_{n} = \sum_{i = 0}^{n} 4+6i$$
$$S_{n} = 4 \sum_{i = 0}^{n} + 6 \sum_{i = 0}^{n} i$$
$$S_{n} = 4n + 4 + 6 \sum_{i = 0}^{n} i$$
$$S_{n} = 4n + 4 + 6 \frac{(1+n)n}{2}$$
$$S_{n} = 4n + 4 + 3 (n^2 + n)$$
$$S_{n} = 3n^2 + 7n + 4$$
Testing for $n = 3$:
$$S_{3} = 3(3^2) + 21 + 4 = 52 = \Bigg(4 + 10 + 16 + 22 \Bigg)$$
So the result seems correct. But is it for sure? Plugging $3$ into $n$ should take into account only first $3$ or first $4$ elements? (works fine for first $4$ elements, but seems kinda wrong for me)
|
We consider the original task
\begin{align*}
S_N=\sum_{{2\leq i\leq N}\atop{i\ \mathrm{with\ step\ } 2}}\left(3i-2\right)\qquad\qquad N\mathrm{\ even}
\end{align*}
We calculate for small $N=2,4,6$ (considering even $N$ only)
\begin{align*}
S_2&=\sum_{{2\leq i\leq 2}\atop{i\ \mathrm{with\ step\ } 2}}(3i-2)=3\cdot 2-2=4\\
S_4&=\sum_{{2\leq i\leq 4}\atop{i\ \mathrm{with\ step\ } 2}}(3i-2)=(3\cdot 2-2)+(3\cdot 4-2)=4+10=14\\
S_6&=\sum_{{2\leq i\leq 6}\atop{i\ \mathrm{with\ step\ } 2}}(3i-2)=(3\cdot 2-2)+(3\cdot 4-2)+(3\cdot 6-2)=4+10+16=30\\
\end{align*}
Generally we obtain for even $N$:
\begin{align*}
\color{blue}{S_N}&=\sum_{{2\leq i\leq N}\atop{{i\ \mathrm{with\ step\ }2}}}\left(3i-2\right)\\
&=\sum_{{i=2}\atop {i\ \mathrm{with\ step\ }2}}^N\left(3i-2\right)\\
&=\sum_{i=1}^{N/2}\left(3(2i)-2\right)\tag{1}\\
&=6\sum_{i=1}^{N/2}i-2\sum_{i=1}^{N/2}1\\
&=6\cdot\frac{1}{2}\cdot\frac{N}{2}\left(\frac{N}{2}+1\right)-2\cdot\frac{N}{2}\tag{2}\\
&\,\,\color{blue}{=\frac{3}{4}N^2+\frac{1}{2}N}
\end{align*}
which gives for small values of $N$: $S_2=4,S_4=14,S_6=30$ as expected.
Comment:
*
*In (1) we respect the step-width $2$ of $i$ by substituting $i$ with $2i$. We also have to set the lower limit to $1$ and the upper limit to $N/2$ as compensation.
*In (2) we use the finite geometric sum formula $\sum_{i=1}^ni=\frac{1}{2}n(n+1)$.
Hint: Your sum $S_n$ is not correct, since there are brackets missing and an index shift was not appropriately performed. But it can be easily corrected. We have with $n=\frac{N}{2}$
\begin{align*}
\sum_{{i=2}\atop{\mathrm{step\ width}\ 2}}^{N}(3i-2)&=\sum_{i=1}^{N/2}\color{blue}{(}3(2i)-2\color{blue}{)}\\
&=\sum_{i=1}^n\color{blue}{(}6i-2\color{blue}{)}\\
&=\sum_{i=0}^{\color{blue}{n-1}}\color{blue}{(}6i+4\color{blue}{)}\\
&=\ldots
\end{align*}
which gives for $n=1,2,3,\ldots$ the sequence $4,14,30,\ldots$ as it should be.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3339066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Integrating the real part of $(1-ix)/\sqrt{a^2-(1-ix)^2}$ What is the solution to the following integral (if there is any)?
$$I = \operatorname{Re}\int_0^{\infty}dx \frac{1-ix}{\sqrt{a^2-(1-ix)^2}},$$
where it is known that $a>1$.
My approach:
I'm not getting any progress with taking the integral before the real part, but I'm pretty sure I'm allowed to exchange integration and real part as I want. If so, then by introducing $b(x) = a^2+x^2-1$, we can rewrite the integral as
$$ I = \int_0^{\infty}dx \frac{\cos\left(\frac{1}{2}\arg\left(b+2ix\right)\right) -x\sin\left(\frac{1}{2}\arg\left(b+2ix\right)\right)}{\sqrt[4]{b^2+4x^2}}.$$
By using that $\sin(\frac{1}{2}\arcsin(x)) = \frac{1}{2}(\sqrt{x+1}-\sqrt{1-x})$, and $\cos(\frac{1}{2}\arccos(x)) = \frac{1}{\sqrt{2}}\sqrt{x+1}$, I believe that the integral can be rewritten
$$I = \int_0^{\infty}dx \frac{ \frac{1}{\sqrt{2}}\sqrt{\frac{b}{b^2+4x^2} + 1} - \frac{x}{2}\left[\sqrt{\frac{2x}{b^2+4x^2}+1} - \sqrt{1 - \frac{2x}{b^2+4x^2}}\right]}{\sqrt[4]{b^2+4x^2}} =
\int_0^{\infty}dx \frac{ \frac{1}{\sqrt{2}}\sqrt{b^2+4x^2+1} -\frac{x}{2}\left[\sqrt{b^2+4x^2+2x} - \sqrt{b^2+4x^2-2x}\right] }{(b^2+4x^2)^{3/4}}.$$
Substituting $b$ back in again and collecting powers of x:
$$ I = I_1 - I_2 + I_3,$$
where
$$ I_1 = \frac{1}{\sqrt{2}}\int_0^{\infty}dx \frac{\sqrt{x^4+x^2(3+2a^2)+a^2(a^2-1)}}{\left[x^4+x^2(2+2a^2)+a^2(a^2-1)+1\right]^{3/4}},$$
$$ I_2 = \int_0^{\infty}dx\frac{\frac{x}{2}\sqrt{x^4+x^2(2+2a^2)+2x+a^2(a^2-1)+1} }{\left[x^4+x^2(2+2a^2)+a^2(a^2-1)+1\right]^{3/4}},$$
$$I_3 = \int_0^{\infty}dx\frac{\frac{x}{2}\sqrt{x^4+x^2(2+2a^2)-2x+a^2(a^2-1)+1} }{\left[x^4+x^2(2+2a^2)+a^2(a^2-1)+1\right]^{3/4}},$$
but here my progress ends since I can't seem to take these integrals.
| Let $f(x) = (1 - i x)/\sqrt {a^2 - (1 - i x)^2}$ and assume that $\sqrt z$ is the principal value of the square root. Since $\lim_{x \to \infty} f(x) = -i$, $\int_0^\infty f(x) dx$ diverges. However,
$$\int_0^A \operatorname {Re}(f(x)) dx =
\operatorname {Re} \int_0^A f(x) dx$$
for $A \in \mathbb R$, because the integral on the rhs exists and $dx$ is real (that is, all $\Delta x_i$ in a Riemann sum are real). Then
$$\operatorname {Re} \int_0^A f(x) dx =
\operatorname {Re} \left(
-i \sqrt {(x + i)^2 + a^2} \, \bigg\rvert_{x = 0}^A \right) =
\operatorname {Im} \sqrt {(A + i)^2 + a^2}.$$
For $z$ in the upper half-plane,
$$\operatorname {Im} \sqrt z =
\frac {\operatorname {Im} z} {\sqrt {2 (|z| + \operatorname {Re} z)}}.$$
It follows that
$$\lim_{A \to \infty} \operatorname {Im} \sqrt {(A + i)^2 + a^2} = 1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3340048",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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What is the method to factor $x^3 + 1$? In the solution to a problem, it's stated that
We see that $x^3+1=(x+1)(x^2-x+1)$.
Why is this, and what method can I use for similar problems with different coefficients?
The full problem is
Find the remainder when $x^{81}+x^{48}+2x^{27}+x^6+3$ is divided by $x^3+1$.
| Since $(-1)^3+1=0,$ we know that $x^3+1$ is divisible by $x+1$, which gives the following way:
$$x^3+1=x^3+x^2-x^2-x+x+1=x^2(x+1)-x(x+1)+x+1=(x+1)(x^2-x+1).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3341005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
How would I go about computing this finite sum? How would I go about computing the sum
$$
\sum_{k=1}^{n} \dfrac{(-k^2+2k+1)2^k}{(k(k+1))^2}.
$$
I have tried partial fractions but have gotten stuck trying to find the coefficients. I decomposed it like this:
$$
\dfrac{2^k(-k^2+2k+1)}{(k(k+1))^2} = \frac{a_0}{k} + \frac{a_1}{k^2} + \frac{a_2}{k+1} + \frac{a_3}{(k+1)^2}
$$.
| Observe that
$$\dfrac{(-k^2+2k+1)2^k}{(k(k+1))^2}= \dfrac{-2^{k+1}}{(k+1)^2}+\dfrac{2^k}{k^2}.$$
Then take sum as $k$ varies from $1$ to $n$.
$$\sum_{k=1}^n \dfrac{-2^{k+1}}{(k+1)^2}+\dfrac{2^k}{k^2}=\sum_{k=1}^n \dfrac{-2^{k+1}}{(k+1)^2}+\sum_{k=1}^n \dfrac{2^k}{k^2}$$
This being an alternate sum of elements of same type, you are left with $$2-\dfrac{2^{n+1}}{(n+1)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3342041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Reducing fractions How can I get [if it's possible]:
$$ 2- \frac{1}{n+1} $$
from this:
$$2- \frac{1}{n} + \frac{1}{n(n+1)} $$
EDIT:
I started like this:
$$2- \frac{1}{n} + \frac{1}{n(n+1)}= 2-\frac{n+1}{n(n+1)}+\frac{1}{n(n+1)} =2-\frac{n+2}{n(n+1)} $$
How do I proceed from here?
|
I started like this: $$2- \frac{1}{n} + \frac{1}{n(n+1)}=
2-\frac{n+1}{n(n+1)}+\frac{1}{n(n+1)} =2-\frac{n+2}{n(n+1)} $$
The mistake happened here:
$$2-\frac{n+1}{n(n+1)}+\frac{1}{n(n+1)} =2-\frac{n+2}{n(n+1)}$$
because you forgot the minus sign. You can rewrite the left hand side as
$$2 +\frac{-n-1}{n(n+1)} + \frac{1}{n(n+1)}$$ and continue from there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3342762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is there any simpler way to find $\sin 2 y$ from $\cos(x+y)=\tfrac13$ and $\cos(x-y)=\tfrac15$? Is there any simpler way to find $\sin 2 y$ from $\cos(x+y)=\tfrac13$ and $\cos(x-y)=\tfrac15$? Note: $x$ and $y$ are obtuse angles.
My attempt that is not simple is as follows.
Expand both known constraints, so we have
\begin{align}
\cos x \cos y &=4/15\\
\sin x \sin y &=-1/15
\end{align}
Eliminate $x$ using $\sin^2 x +\cos^ 2 x=1$, we have
$$
225 \sin^4 y -210 \sin^2 y +1=0
$$
with its solution
$\sin^2 y = \frac{7\pm4\sqrt3}{15}$.
Then, $\cos^2 y = \frac{4(2\mp\sqrt3)}{15}$.
\begin{align}
\sin^2(2y) &= 4\cos^2 y\sin^2 y\\
&= 4 \times \frac{4(2\mp\sqrt3)}{15}\times \frac{7\pm4\sqrt3}{15} \\
\sin 2 y & = - \frac{4}{15}\sqrt{(2\mp\sqrt3)(7\pm4\sqrt3)}
\end{align}
$\sin 2y$ must be negative.
Edit
Thank you for your effort to answer my question. However, the existing answers seem to be more complicated than my attempt above.
By the way, I am confused in deciding which the correct pair among $(2\mp\sqrt3)(7\pm4\sqrt3)$ is.
| $90<x,y<180^\circ$
$180<x+y<360\implies x+y=360-\arccos(1/3)$
If $x-y>0,x-y=\arccos(1/5)$
$\sin2y=\sin(360-\arccos(1/3)-\arccos(1/5))=-\sin(\arccos(1/3)+\arccos(1/5))$
Now $\arccos(1/3)+\arccos(1/5)=\arcsin(2\sqrt2/3)+\arccos(2\sqrt6/5)$
Use Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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} |
Prove that $\int_0^1 \frac{x^2}{\sqrt{x^4+1}} \, dx=\frac{\sqrt{2}}{2}-\frac{\pi ^{3/2}}{\Gamma \left(\frac{1}{4}\right)^2}$ How to show
$$\int_0^1 \frac{x^2}{\sqrt{x^4+1}} \, dx=\frac{\sqrt{2}}{2}-\frac{\pi ^{3/2}}{\Gamma \left(\frac{1}{4}\right)^2}$$
I tried hypergeometric expansion, yielding $\, _2F_1\left(\frac{1}{2},\frac{3}{4};\frac{7}{4};-1\right)$. Can this be evaluated analytically? Any help will be appreciated.
| \begin{align}
\int\limits_{0}^{1} \dfrac{x^{2}}{\sqrt{1+x^{4}}}\,\mathrm{d}x &= \dfrac{1}{4}\int\limits_{0}^{1} \dfrac{1}{x^{1/4}\sqrt{1+x}}\,\mathrm{d}x \\
&= \dfrac{1}{4}\int\limits_{1}^{2} \dfrac{1}{\left(x-1\right)^{1/4}\sqrt{x}}\,\mathrm{d}x \\
&= \dfrac{1}{4}\int\limits_{1/2}^{1} \dfrac{1 -\color{red}{x} + \color{red}{x}}{\left(1-x\right)^{1/4}x^{5/4}}\,\mathrm{d}x \label{subst to 1/x} \tag{1} \\
&= \color{blue}{\dfrac{1}{4}\int\limits_{1/2}^{1} \dfrac{\left(1-x\right)^{3/4}}{x^{5/4}}\,\mathrm{d}x} + \dfrac{1}{4}\int\limits_{1/2}^{1} \dfrac{1}{\left(1-x\right)^{1/4}x^{1/4}}\,\mathrm{d}x \\
&\overset{\mathrm{IBP}}{=} \color{blue}{\left.-\dfrac{\left(1-x\right)^{3/4}}{x^{1/4}}\right\vert_{1/2}^{1}-\dfrac{3}{4}\int\limits_{1/2}^{1} \dfrac{1}{\left(1-x\right)^{1/4}x^{1/4}}\,\mathrm{d}x} + \dfrac{1}{4}\int\limits_{1/2}^{1} \dfrac{1}{\left(1-x\right)^{1/4}x^{1/4}}\,\mathrm{d}x \\
&= \dfrac{1}{\sqrt{2}}-\dfrac{1}{2}\int\limits_{1/2}^{1} \dfrac{1}{\left(1-x\right)^{1/4}x^{1/4}}\,\mathrm{d}x \\
&= \dfrac{1}{\sqrt{2}}-\dfrac{1}{4}\operatorname{B}\left(\dfrac{3}{4},\dfrac{3}{4}\right)
\end{align}
At \eqref{subst to 1/x} substitution $x \to x^{-1}$ was made.
| {
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"source": "stackexchange",
"question_score": "6",
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The sum of infinite series $(1/2)(1/5)^2 + (2/3)(1/5)^3 + (3/4)(1/5)^4..................$ Initially, I broke this series as $1/2 = 1-1/2, 2/3 = 1-1/3, 3/4 = 1-1/4$, then I got two series as it is
$$= (1/5)^2 + (1/5)^3 + (1/5)^4 +.......-[1/2(1/5)^2 + 1/3(1/5)^3 + 1/4(1/5)^4 +.....]$$
After solving the first part, we reached at this point
$$= 1/20 - [1/2(1/5)^2 + 1/3(1/5)^3 + 1/4(1/5)^4 +.....]$$
Please help someone in solving this question. I am very grateful to you.
| We know the GP progression formula for when common ratio less than $1$ which is pretty simple to derive and is $\displaystyle \sum_{k=1}^\infty r^{k} = \dfrac{1}{1-r} $ from there we can write the following,
\begin{align} \int \sum\limits_{k=2}^\infty r^{k-1} dr = \int \frac{1}{1-r} dr \end{align}
After integration we get:
\begin{align} \sum\limits_{k=2}^\infty \frac{r^{k}}{k}=-\ln(1-r)\end{align}
Now start from $k=2$ instead of $k=1$ and let $r=\dfrac{1}{5}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Algebraic Manipulations to solve for q If $p-q=\sqrt3$ and $\sqrt{\sqrt{3}-\sqrt{p}}=q$, find $q$.
Some work
$$\bigg(\sqrt{\sqrt{3}+\sqrt{p}}
\bigg)\bigg(\sqrt{\sqrt{3}-\sqrt{p}}\bigg)=\sqrt{3-p}$$
$$\sqrt{\sqrt{3}+\sqrt{p}}=\frac{\sqrt{3-p}}{q}$$
$$q^2\sqrt3+q^2\sqrt{p}=3-p$$
$$(\sqrt3+\sqrt{p})q^2+p-3=0$$
I can't find any way to use the first fact.
| Square the second equation and rearrange
\begin{eqnarray*}
\sqrt{3}-q^2 = \sqrt{p}.
\end{eqnarray*}
Square this and subtract $q$ and we can use the second equation to get
\begin{eqnarray*}
q^4 -2 q^2 \sqrt{3}-q +3 -\sqrt{3}=0.
\end{eqnarray*}
Now this can be factorised !
\begin{eqnarray*}
(q^2-q - \sqrt{3})(q^2+q +1 -\sqrt{3})=0.
\end{eqnarray*}
Should be a doddle from here ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3353394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all $x\in \mathbb{Z}$ with $x \equiv 2 \mod 4$ , $x \equiv 8\mod9$ and $x \equiv 1\mod5$
Find all $x\in \mathbb{Z}$ with $x \equiv 2 \mod 4$ , $x \equiv 8\mod9$ and $x \equiv 1\mod5$
My attempt:
$\gcd\left(9,5,4\right) = 1 = 9r+5s+4\times0$
$9 = 1\times5+4$
$5 = 1\times4+1$
so $1 = 5-(9-5) = 2\times5 - 1\times9$
thus $r=-1$ and $s = 2$
so $x = 2\times5\times8 -1\times9\times1=80 - 9=71\mod 180$
But the problem is that $71\equiv8\mod9$ and $71\equiv1\mod5$, but $71 \ne 2 \mod 4$. What is wrong?
| Hint $\ \overbrace{x = \color{#0a0}{-1\!+\!9}(a\!+\!5(b\!+\!4c))}^{\text{by iterated division}}\,$ so $\overbrace{\color{#90f}{\bmod 5}\Rightarrow a\equiv 3}^{\large\color{#90f}{x\ \equiv\ 1}};\ $ $\overbrace{\color{#c00}{\bmod 4}\Rightarrow b\equiv 0}^{\large\color{#c00}{x\ \equiv\ 2}},\,$ so $\ \bbox[5px,border:1px solid #c00]{x= 26\!+\!180c}$
Remark $ $ What went wrong in your method is you didn't correctly apply the CRT formula (or you incorrectly tried to generalize the formula for two congruences). Applying the linked formula yields
$\!\begin{align} x\,&\equiv\, \color{#c00}{2\pmod{4}}\ \ \ \ {\rm and}\, \ \ \ x\equiv \color{#0a0}{-1\pmod{9}}\ \ \ \ \ {\rm and}\,\ \ \ \ \ x\equiv \color{#90f}{1\pmod{5}}\\[.5em]
\iff x\ &\equiv\, \color{#c00}2(9\cdot 5)\overbrace{((9\cdot 5)^{-1}\!\color{#c00}{\bmod 4)}}^{\large\! 1/45\ \equiv\ \color{#c00}{1/1}\ } \color{#0a0}{-1} (4\cdot 5)\overbrace{((4\cdot 5)^{-1}\!\color{#0a0}{\bmod 9)}}^{\large 1/20\ \equiv\ 10/2\ \equiv\ \color{#0a0}{5/1}} + \color{#90f}1(4\cdot 9)\overbrace{((4\cdot 9)^{-1}\!\color{#90f}{\bmod 5)}}^{\large\! 1/36\ \equiv\ \color{#90f}{1/1}}\\[.5em]
&\equiv\, \color{#c00}2(9\cdot 5)\,(\color{#c00}{1/1}) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \color{#0a0}{-1} (4\cdot 5)\,(\color{#0a0}{5/1}) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \:\! + \color{#90f}1(4\cdot 9)\,(\color{#90f}{1/1})\\[.5em]
&\equiv\ \color{#c00}{90}\ \ - \ \ \color{#0a0}{100}\ \ +\ \ \color{#90f}{36}\ \bbox[5px,border:1px solid #c00]{\equiv\, 26\pmod{\!180}}
\end{align}$
See this answer for an intuitive explanation of the genesis of the above CRT formula (which will help you to correctly remember it and efficiently apply it).
| {
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"source": "stackexchange",
"question_score": "2",
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How to apply Lagrange Multipliers with modulus ??
I had the question $$x^2+y^2+xy=1$$ and I had to find maximum value of $$\vert xy(x^2+y^2) \vert$$
I tried using Lagrange multipliers by forming
$x^3+3xy^2 = \lambda (2y+x)$
And $y^3+3yx^2 = \lambda (2x+y)$
Now I subtract it to get $(x-y)^3 = \lambda (y-x)$
And as $\lambda ≠ 0 , x=y.$ This gives$ x = {1 \over \sqrt{3}}$ and the expression gives max as $\frac{2}{9}$ but the answer is $2$ . What have I done wrong.
I know it's modulus , but with negative too, there will only be a change in the sign of $\lambda$ which will not affect any working.
Also Lagrange Multipliers is something I read as an extra piece of info online, so maybe I might be missing something trivial here.
Also is there any requirement on $\lambda$ to be positive ??
| Hint.
Following the fact
$$
\max |f| = \max(-f,f) \ \ \text{and}\ \ \min |f| = \min(-f,f)
$$
With $f = x y(x^2+y^2)$ the lagrangian can be established as
$$
L(x,y,\lambda) = s f+\lambda(x^2+y^2+x y -1)
$$
where $s=\pm$
The stationary points are given by
$$
\nabla L = 0 = \left\{
\begin{array}{rcl}
2 s y x^2+\lambda (2 x+y)+s y \left(x^2+y^2\right)&=&0 \\
2 s x y^2+\lambda (x+2 y)+s x \left(x^2+y^2\right)&=&0 \\
x^2+y x+y^2-1 &=&0\\
\end{array}
\right.
$$
and solving we have
$$
\left[
\begin{array}{ccc}
s f & x & y \\
-2 s & -1 & 1 \\
-2 s & 1 & -1 \\
\frac{2 s}{9} & -\frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{3}} \\
\frac{2 s}{9} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\
\end{array}
\right]
$$
NOTE
Due to its homogeneity the problem, using $y = \mu x$, can be transformed into one without restrictions as
$$
f(\mu) = \frac{\mu(1+\mu^2)}{(1+\mu+\mu^2)^2}
$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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$ \frac{x_1}{1+x_1^2} + \frac{x_2}{1+x_1^2+x_2^2} +...+\frac{x_n}{1+x_1^2+x_2^2+...x_n^2} \le \sqrt{n}$ for $x_i > 0$ If $ x_1, x_2 , x_3........x_n $ are n positive reals prove that
$$ \frac{x_1}{1+x_1^2} + \frac{x_2}{1+x_1^2+x_2^2} +...+\frac{x_n}{1+x_1^2+x_2^2+...x_n^2} \le \sqrt{n}$$
So this is an IMO 2001 proposed question.It was in the section of Cauchy Schwartz inequalities in my book. It seems that there is a Cauchy hidden because of the root n but then we need to find two series whose terms are those of the LHS and the product of sum of whose squares gives n . I'm at a loss in finding that
Thanks for any help
| Let us call the individual terms of the inequality as $A_i$, then by AM-RMS inequality we have
$$\frac{\sum A_i}{n} \le \sqrt{\frac{\sum A_i^2}{n}},~\mbox{where} ~A_i=\frac{x_i}{1+x_1^2+x_2^2+x_3^2+...x_i^2}.~~~~(1)$$
So it would suffice to prove that $\sum A_i^2 \le 1.$
Note that for $i\ge 2$, $$ A_i^2=\frac{x_1^2}{(1+x_1^2+x_3^2+x_3^2+...x_i^2)^2}\le \frac{x_i^2}{(1+x_1^2+x_3^2+x_3^2+...x_{i-1}^2)~ (1+x_1^2+x_3^2+x_3^2+...x_{i}^2)}.$$
$$=\frac{1}{1+x_1^2+x_2^2+x_3^2+...+x_{i-1}^2}-\frac{1}{1+x_1^2+x_2^2+x_3^2+...+x_{i}^2}.$$
We also have $$\frac{x_1^2}{1+x_1^2} \le 1- \frac{1}{1+x_1^2}.$$
Summing all these we get $$\sum_{i=1}^{n} A_i^2 \le 1-\frac{1}{1+x_1^2+x_2^2+x_3^2+...+x_{i}^2}<1.$$
Finally from (1) it follows that $$\sum_{i=1}^{n} A_i < \sqrt{n}$$
| {
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"source": "stackexchange",
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Proving $n^2(n^2-1)(n^2+1)=60\lambda$ such that $\lambda\in\mathbb{Z}^{+}$ I'm supposed to prove that the product of three successive natural numbers the middle of which is the square of a natural number is divisible by $60$. Here's my attempt.
My Attempt:
$$\text{P}=(n^2-1)n^2(n^2+1)=n(n-1)(n+1)[n(n^2+1)]$$
It is now enough to prove that $n(n^2+1)$ is divisible by $10$. But for $n=4$, $4(17)\ne10\lambda$ but for $n=4$, $\text{P}$ is $4080=60\cdot68$ which means apart from just being a multiple of $6$, $n(n-1)(n+1)$ is actually helping $n(n^2+1)$ with a $5$ to sustain divisibility by $60$.
How to tackle this? Thanks
| Note that $n^5-n=n(n^2-1)(n^2+1)$ is divisible by $5$ by little Fermat. $n^3-n$ is likewise divisible by $3$ and either $n^2$ or $n^2-1$ is divisible by $4$.
| {
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"source": "stackexchange",
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$\alpha$, $\beta$, $\gamma$ are the root of $x^3-x^2+px-1=0$. $(\alpha^3+1)(\beta^3+1)(\gamma^3+1)=2019$. $\alpha$, $\beta$, $\gamma$ are the root of $x^3-x^2+px-1=0$. $(\alpha^3+1)(\beta^3+1)(\gamma^3+1)=2019$. What is the product of all possible value of $p$?
Note that $p$ could be a complex number.
I tried some basic Vieta's formulas, couldn't find easy way to simplify...
| $$x^3-x^2+px-1=0 \Rightarrow (x^3-1)^3-(x^2-px)^3=0 \Rightarrow x^9+(3p-.4)x^6+(p^3-3p+3)-1=0.$$
Let us transform this equation by $y=x^3+1 \rightarrow x=(y-1)^{1/3}$. Then
we get a cubic Eq. for $y$ as
$$y^3+y^2(3p-7)+(p^3-9p+14)y-p^3+6p-9=0,$$ $y_1, y_2, y_3$ are its roots.
Then $$y_1 y_2 y_3=p^3-6p+9=2019 \Rightarrow p^3-6p-2010=0.$$
The roots of $p$ are $p_1,p_2,p_3$ and their product is: $p_1p_2p_3=2010$
| {
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"url": "https://math.stackexchange.com/questions/3359868",
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"source": "stackexchange",
"question_score": "1",
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Find $\int \frac{\sin2x+2\tan x}{\cos^6x+6\cos^2x+4}dx$
Find $\displaystyle\int \frac{\sin2x+2\tan x}{\cos^6x+6\cos^2x+4}dx$
My approach is as follow
$\cos^2x=t$;
$\sin2x\ dx=-dt$
Therefore,
\begin{align}
\int \frac{\sin2x+2\tan x}{\cos^6x+6\cos^2x+4}dx&=\int \frac{\sin2x+\frac{2\tan x\cos^2x}{\cos^2x}}{\cos^6x+6\cos^2x+4}dx\\
&=\int \frac{\sin2x+\frac{\sin2x}{\cos^2x}}{\cos^6x+6\cos^2x+4}dx\\
&=\int \frac{-(1+\frac{1}{t})}{t^3+6t+4}dt
\end{align}
I am not able to proceed from here
| You are almost there.
Note that\begin{align}\frac{1+\frac1t}{t^3+6t+4}&=\frac{t+1}{t(t^3+6t+4)}\\&=\frac14\left(\frac1t-\frac{t^2+2}{t^3+6t+4}\right).\end{align}Now, use the fact that $(t^3+6t+4)'=3(t^2+2)$.
| {
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Finding the Taylor Series Expansion using Binomial Series, then obtaining a subsequent Expansion. Hi all doing some prep work for a course and really struggling to wrap my head around some of the revision questions. Some help would be really appreciated. There are two halfs, part one is as follows:
Consider the function:
$$ f(x) =\frac{1}{1-x} $$
Find the Taylor series expansion by expanding as a binomial series.
My solution is using $(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + ...$
In this case, $n = -1$ and $x = -1$.
So by substituting I end up with $$ 1 + x + x^2 + x^3 + ... $$
So the Taylor Series Expansion in this case is:
$$\sum_{n=0}^\infty x^n $$
Ok so far, now the second half.
Using your result for f (x) obtain the Taylor series for:
$$ g(x) =\frac{x^2}{1+x^2} $$
From what I understand reading, I need to convert f(x) into g(x) then apply whatever mathematical steps made to the terms in the series, which I did:
$$ f(x) =\frac{1}{1-x} * \frac{1-x}{1} = \frac{1}{1}$$
$$ \frac{1}{1} * \frac {1}{x^2+1} = \frac{1}{x^2+1}$$
$$ \frac{1}{x^2+1} * x^2 = \frac{x^2}{x^2+1} = g(x)$$
I'll leave out the individual steps, but applying these to the terms of the series, I get:
$$ \frac{x^2}{x^2+1} - \frac{x^3}{x^2+1} + \frac{x^3}{x^2+1} + \frac{x^4}{x^2+1} + \frac{x^4}{x^2+1}$$
Which is as far as I'm comfortable going, since this really doesn't feel right at all for a solution. I'd really appreciate if someone could point me in the right direction or let me know if I'm made some catastrophic error. I'm really not sure how to write the final terms in sigma notation either. Feel free to give me a slap on the wrist if I've made any silly errors.
|
We obtain from $f(x)=\frac{1}{1-x}=\sum_{n=0}^\infty x^n$
\begin{align*}
\color{blue}{x^2f(-x^2)}&=x^2\sum_{n=0}^\infty(-x^2)^n\\
&=x^2\sum_{n=0}^\infty(-1)^nx^{2n}\\
&=\sum_{n=0}^\infty (-1)^nx^{2n+2}\\
&\,\,\color{blue}{=\sum_{n=1}^\infty(-1)^{n-1}x^{2n}}
\end{align*}
In the last line we shift the index by $1$ in order to have $x^{2n}$.
| {
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Find all natural numbers $n$ such that $n+1$ divides $3n+11$ Following the example of my teacher:
Find all natural numbers $n$ such that $n-2$ divides $n+5$.
$$n+5 = n-2+7$$
As $n-2|n-2$, $n-2$ will divide $n+5$ if and only if $n-2|7$. Yet, $7$ has the divisors $-7$, $-1$, $1$, $7$ hence the equations:
*
*$n-2=-7 \Leftrightarrow n = -5 $
*$n-2=-1 \Leftrightarrow n = 1 $
*$n-2=7 \Leftrightarrow n = 3 $
*$n-2=-1 \Leftrightarrow n = 9 $
So $S = \{ 1, 3, 9 \}$
I decompose $n+1$ the exact same way:
$$n+1 = 3n+11 - 2(n+5)$$
But I'd get stuck as $2(n+5)$ since only $-2$, $-1$, $1$ and $2$ are divisors, which don't satisify the equation as I'd hoped:
*
*$3n+11 = -2 \Leftrightarrow n =\frac{-13}{3}$
*$3n+11 = -1 \Leftrightarrow n = -4$
*$3n+11 = 1 \Leftrightarrow n =\frac{-10}{3}$
*$3n+11 = 1 \Leftrightarrow n = -3$
Any clues?
| When the example has
$$ n + 5 = n-2 + 7 $$
it might have been more clearly written
$$ n + 5 = 1\cdot(n-2) + 7 $$
to highlight the quotient, $1$, and remainder, $7$, of the division $\frac{n+5}{n-2}$.
For your problem,
$$ 3n + 11 = 3 \cdot (n+1) + 8 \text{.} $$
That is, the quotient is $3$ and the remainder is $8$. So you want to inspect the choices of $n$ such that $(n+1) \mid 8$.
| {
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What is $\liminf_{n\to\infty} ( \frac{p_{n}}{p_{n}+p_{n+1}})$ Let $p_{n}$ a prime number and $p_{n+1}$ is the next prime.
How to calculate $\liminf_{n\to\infty} ( \frac{p_{n}}{p_{n}+p_{n+1}})$
Edited :
here is my attempt :
$\frac{p_{n}}{p_{n}+p_{n+1}}\approx \frac {n ln(n)}{p_{n}+p_{n+1}} \frac{p_{n}}{n ln(n)} $
| In fact, your lim inf is a lim. First, observe that
$$
(n+1)\ln(n+1) = n \ln(n+1) + \ln(n+1) = \\
n[\ln(n+1) - \ln(n) + \ln(n)] + \ln(n+1) = \\
n \ln(n) + n \ln\left(\frac{n+1}{n}\right) + \ln(n+1)
$$
Thus (by the prime number theorem), we have
$$
\lim_{n \to \infty} \frac{p_n}{p_{n+1}} =
\lim_{n \to \infty} \frac{p_n}{p_{n+1}} \cdot \frac{n \ln(n)}{n \ln(n)} \cdot \frac{(n+1)\ln(n+1)}{(n+1)\ln(n+1)}\\
=
\lim_{n \to \infty} \frac{p_n}{n\ln(n)} \cdot \frac{(n+1)\ln(n+1)}{p_{n+1}} \cdot \frac{n\ln(n)}{(n+1)\ln(n+1)}\\
=
\lim_{n \to \infty} \frac{n\ln(n)}{(n+1)\ln(n+1)}\\
= \left[\lim_{n \to \infty} \frac{(n+1)\ln(n+1)}{n\ln(n)}\right]^{-1}\\
= \left[\lim_{n \to \infty}1 + \frac{\ln[(n+1)/n]}{\ln(n+1)} + \frac{\frac{\ln(n+1)}{\ln(n)}}{n }\right]^{-1} = 1^{-1} = 1.
$$
it follows that
$$
\lim_{n \to \infty} \frac{p_n}{p_n + p_{n+1}} =
\lim_{n \to \infty} \frac{1}{1 + \frac{p_{n+1}}{p_n}} = \frac{1}{1+1} = \frac 12.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3364297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Prove inequalities $\frac 34 \le I(a,b) \le 1$ Given the expression,
$$ I(a,b) = \frac{a^2}{(1+a)(a+b)} + \frac{b^2}{(1+b)(a+b)} + \frac{1}{(1+a)(1+b)}$$
where $a\ge 0$ and $b\ge 0$, prove the following inequalities:
$$\frac 34 \le I(a,b) \le 1$$
I had trouble figuring it out. Tried some inequity techniques I am aware of and had no lack so far. I am not sure if it is wise to go brute force by examining the derivatives of $I(a,b)$, which I feel would be pretty messy due to its dual dimension.
Appreciate if anyone could offer a viable approach for the proof.
| Putting everything over a common denominator shows that
$$I(a,b) = \frac{a^2+b^2+a^2b+ab^2+a+b}{a^2+b^2+a^2b+ab^2+a+b+2ab} = \frac{(1+a)(1+b)(a+b)-2ab}{(1+a)(1+b)(a+b)} \leq 1$$
since $a,b\geq 0$. Simplifying this expression, we get
$$I(a,b) = 1 - \frac{2ab}{(1+a)(1+b)(a+b)}$$
Now take a look at the second term and consider it rewritten in the following way:
$$\frac{2ab}{(1+a)(1+b)(a+b)} = \frac{\frac{ab}{4}}{\frac{a^2+b^2+a^2b+ab^2+a+b+ab+ab}{8}}$$
then by AM-GM inequality
$$\frac{a^2+b^2+a^2b+ab^2+a+b+ab+ab}{8} \geq (a^8b^8)^{\frac{1}{8}} = ab$$
$$\implies \frac{2ab}{(1+a)(1+b)(a+b)} \leq \frac{\frac{ab}{4}}{ab} = \frac{1}{4}$$
which means that
$$I(a,b) \geq 1 -\frac{1}{4} = \frac{3}{4}$$
| {
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"source": "stackexchange",
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"answer_count": 3,
"answer_id": 0
} |
Determinant of matrix $[a_{ij}]$ where $a_{ij}=ij$ if $i\ne j$ and $a_{ij}=1+ij$ if $i=j$ Let $A=[a_{ij}]$ be a square matrix of order $n$ whose entries are given as follows.
For $1\leq i,j\leq n$ we have $a_{ij}=ij$ if $i\neq j$ and $a_{ij}=1+ij$ if $i=j$.
I have to evaluate the determinant.
I just wrote the matrix but don't know how to proceed further.
| Let the determinant be $D_n$. Then
$$
D_n = \det
\begin{bmatrix}
1 + 1^2 & 1 \times 2 & 1\times 3 & \cdots & 1\times n\\
2\times 1 & 1 + 2^2 & 2\times 3 & \cdots & 2 \times n \\
3\times 1 & 3 \times 2 & 1 + 3^2 & \cdots & 3 \times n \\
\vdots & \vdots &\vdots & \ddots & \vdots\\
n\times 1 & n \times 2 & n\times 3 & \cdots & 1 + n^2
\end{bmatrix} =
\det
\begin{bmatrix}
1 + 1^2 & 1 \times 2 & 1\times 3 & \cdots & 0\\
2\times 1 & 1 + 2^2 & 2\times 3 & \cdots & 0 \\
3\times 1 & 3 \times 2 & 1 + 3^2 & \cdots & 0\\
\vdots & \vdots &\vdots & \ddots & \vdots\\
n\times 1 & n \times 2 & n\times 3 & \cdots & 1
\end{bmatrix} +\det
\begin{bmatrix}
1 + 1^2 & 1 \times 2 & 1\times 3 & \cdots & 1\times n\\
2\times 1 & 1 + 2^2 & 2\times 3 & \cdots & 2 \times n \\
3\times 1 & 3 \times 2 & 1 + 3^2 & \cdots & 3 \times n \\
\vdots & \vdots &\vdots & \ddots & \vdots\\
n\times 1 & n \times 2 & n\times 3 & \cdots & n^2
\end{bmatrix} = D_{n-1} + n \det
\begin{bmatrix}
1 + 1^2 & 1 \times 2 & 1\times 3 & \cdots & 1\\
2\times 1 & 1 + 2^2 & 2\times 3 & \cdots & 2 \\
3\times 1 & 3 \times 2 & 1 + 3^2 & \cdots & 3 \\
\vdots & \vdots &\vdots & \ddots & \vdots\\
n\times 1 & n \times 2 & n\times 3 & \cdots & n
\end{bmatrix} = D_{n-1} + n \det
\begin{bmatrix}
1 & 0& 0& \cdots & 1\\
0 & 1 &0 & \cdots & 2 \\
0& 0 & 1 & \cdots & 3 \\
\vdots & \vdots &\vdots & \ddots & \vdots\\
0&0&0 & \cdots & n
\end{bmatrix} = D_{n-1} + n^2,
$$
i.e.
$$
D_n = D_{n-1} + n^2.
$$
Therefore
$$
D_n = D_2 + (3^2 + 4^2 + \dots + n^2) = 1 + (1^2+2^2 + \cdots + n^2) = 1 + \frac {n(n+1)(2n+1)}6.
$$
Another trial
preparation
$\newcommand \bm \boldsymbol $
$ \newcommand \trans {^{\mathsf T}}$
For $$\bm A\in \mathrm M_m, \bm D \in \mathrm M_n, \bm B \in \mathrm M_{m,n}, \bm C \in \mathrm M_{n,m},$$ let $\bm M$ be a block matrix
$$
\bm M = \begin {bmatrix}
\bm A & \bm B \\ \bm C & \bm D
\end{bmatrix},
$$
1. if $\bm A$ is invertible, then $$\det \bm M = \det \bm A \det (\bm D - \bm C\bm A^{-1}\bm B); $$
2. if $\bm D$ is invertible, then $$\det \bm M = \det \bm D \det (\bm A - \bm B\bm D^{-1}\bm C); $$
3. Specifically when both $\bm A, \bm D$ are invertible,
$$
\det \bm D \det (\bm A - \bm B\bm D^{-1}\bm C)= \det \bm A \det (\bm D - \bm C\bm A^{-1}\bm B).
$$
Proof. 1. Take the determinant on the equation
$$
\begin{bmatrix} \bm I_m & \bm O\\ -\bm C\bm A^{-1} & \bm I_n \end{bmatrix}\bm M = \begin{bmatrix}\bm A & \bm B \\ \bm O & \bm D - \bm C\bm A^{-1} \bm B \end{bmatrix}.
$$
2. Similar argument. 3. Put 1, 2 together.
Now for the question. Let that matrix be $\bm A$. Then $$\bm A - \bm I_n = [jk]_{n \times n} = \bm c\bm c \trans $$ where $$
\bm c = \begin{bmatrix} 1 & 2 & 3 & \cdots & n\end{bmatrix}\trans.
$$
So by the formula we mentioned,
$$
\det \bm A =- \det (-\bm I_1)\det (\bm I_n - \bm c (-\bm I_1)^{-1} \bm c \trans) \stackrel {\bigstar}= -\det (\bm I_n) \det (-\bm I_1 - \bm c \trans \bm I_n^{-1} \bm c) = 1 + \bm {c}\trans \bm c = 1 + (1^2 + 2^2 + \cdots + n^2) = 1 + \frac {n (n+1)(2n+1)}6.
$$
| {
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"url": "https://math.stackexchange.com/questions/3367056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Using AM-GM inequality prove that $(1+\sqrt{2})(1+\sqrt{3})(1+\sqrt{5}) \lt 8\sqrt{30}$. It is trivial to prove that $(1+\sqrt{2})(1+\sqrt{3})(1+\sqrt{5}) \le
8\sqrt{30}$ using numeric methods. For example by multiplying
$(1+\sqrt{2}) \le 3 $
$(1+\sqrt{3}) \le 3 $
$(1+\sqrt{5}) \le 4 $
We get:
$(1+\sqrt{2})(1+\sqrt{3})(1+\sqrt{5}) < 36$
while $8\sqrt{30} \gt 40 $ because $\sqrt{30} \gt 5 $ . However for this
particular problem it is asked a solution using the AM-GM inequality and I am not able to find one. Can anyone help me?
| We use the inequality $$\frac{a+b}{2}\le \sqrt{\frac{a^2+b^2}{2}}$$ so we get
$$\frac{1+\sqrt{2}}{2}\times\frac{1+\sqrt{3}}{2}\times\frac{1+\sqrt{5}}{2}\le \sqrt{30}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3368413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
In $\triangle ABC$, $AD$ and $BE$ are medians, and $AD \perp BE$. If $AC=14$ and $BC=22$, find $AB$.
In $\triangle ABC$, $AD$ and $BE$ are medians, and $AD \perp BE$. If $AC=14$ and $BC=22$, find $AB$.
I'm using Apollonius's theorem to find the medians. Instead, I found
$$BE^2-AD^2=72$$
Not sure how to proceed.
| Let $G$ be the centroid of $\triangle ABC$. Suppose that $3x=|AD|$ and $3y=|BE|$. Then by Pythagoras
$$7^2=\left(\frac{1}{2}|AC|\right)^2=|AD|^2=|AG|^2+|GD|^2=(2x)^2+y^2$$
and
$$11^2=\left(\frac{1}{2}|BC|\right)^2=|BE|^2=|BG|^2+|GE|^2=(2y)^2+x^2.$$
Thus
$$170=7^2+11^2=(4x^2+y^2)+(4y^2+x^2)=5(x^2+y^2).$$
Therefore,
$$|AB|^2=|AG|^2+|BG|^2=(2x)^2+(2y)^2=4(x^2+y^2)=\frac{4}{5}(170)=136.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
First order linear differential equation (possible error in book) Consider following differential equation in $$\frac{dx}{(3x^2+y^2)/x}=\frac{dy}{(3y^2+x^2)/y}=\frac{dz}{-(x^2+y^2)/z}$$
$\require{enclose}$
Taking multipliers $x, y, 4z$ we get $$xdx+ydy+4zdz=0 \\ \enclose{box}{x^2+y^2+4z^2=C_1}$$
I have this function correct.
Book writes: for taking first two fraction $xdx-ydy=0$ and hence $$\bbox[5px,border:2px solid red]{x^2-y^2=C_2}$$
I am sure this is the correct solution.
We would have on subtracting numerator and denominator $$\frac{xdx-ydy}{2x^2-2y^2}$$
for which denominator isn't 0 so how can we assert $xdy-ydx=0$
I am not missing anything am I?
Possible Solution
$$\frac{xdx-ydy}{2x^2-2y^2}=\frac{xdx+ydy}{4x^2+4y^2} \\ \frac{d(x^2-y^2)}{4(x^2-y^2)}=\frac{d(x^2+y^2)}{8(x^2+y^2)} \\ \enclose{box}{\frac{(x^2-y^2)^2}{x^2+y^2}=C_2}$$
| You are right, two charecteristic equations are :
$$x^2+y^2+4z^2=C_1$$
$$\frac{(x^2-y^2)^2}{x^2+y^2}=C_2$$
From that, the general solution of the PDE expressed on the form of implicit equation $C_1=F(C_2)$ with arbitrary function $F$ :
$$x^2+y^2+4z^2=F\left(\frac{(x^2-y^2)^2}{x^2+y^2}\right)$$
On explicit form :
$$z(x,y)=\pm\frac12\sqrt{-x^2-y^2+F\left(\frac{(x^2-y^2)^2}{x^2+y^2}\right)}$$
NOTE :
$xdx-ydy=0$ is not correct in the general case thus $x^2-y^2=C_3$ is not a correct characteristic equation.
NOTE :
The problem is simplified with the change of variables
$\quad\begin{cases} X=x^2\\ Y=y^2\\ Z=z^2 \end{cases}$ :
$$\frac{dX}{3X+Y}=\frac{dY}{3Y+X}=-\frac{dZ}{X+Y}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $|x_n-x_{n+1}|=\frac{1}{2^{n-1}}$ using Mathematical induction $x_1=1$ $x_2=2$
$x_n=\frac{1}{2}(x_{n-2}+x_{n-1})$ for n $\gt$ 2. We have to prove that $|x_n-x_{n+1}|=\frac{1}{2^{n-1}}$
What I tried :
For n=1, $|x_1-x_{2}|=1 =\frac{1}{2^{0}}$
Let for n=k assumption be true. Hence $|x_k-x_{k+1}|=\frac{1}{2^{k-1}}$
For $n=k+1$
$|x_{k+1}-x_{k+2}|=\frac{1}{2}|x_{k-1}+x_{k}-x_{k}-x_{k+1}|= \frac{1}{2}|(x_{k-1}-x_{k})+(x_{k}-x_{k+1})| \le \frac{1}{2}|\frac{1}{2^{k-2}}+\frac{1}{2^{k-1}}|$
I'm stuck here, what do i do now? Can anyone help please?
| Using triangle inequality will probably not help you, as you're supposed to prove an exact equality, not an inequality. Instead, just substitute $x_{k+2} = \frac{1}{2}(x_k + x_{k+1})$. Then,
\begin{align*}
|x_{k+1} - x_{k+2}| &= \left|x_{k+1} - \frac{1}{2}(x_k + x_{k+1})\right| \\
&= \left|\frac{1}{2}x_k - x_{k+1}\right| \\
&= \frac{1}{2} \cdot \frac{1}{2^{k -1}} = \frac{1}{2^k},
\end{align*}
as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3372905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 1
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Given three positive numbers $a,b,c$. Prove that $\sum\limits_{sym}\frac{a+b}{c}\geqq 2\sqrt{(\sum\limits_{sym}a)(\sum\limits_{sym}\frac{a}{bc}})$ .
(A problem due to Mr. Le Khanh Sy). Given three positive numbers $a, b, c$. Prove that
$$\sum\limits_{sym}\frac{a+ b}{c}\geqq 2\sqrt{(\sum\limits_{sym}a)(\sum\limits_{sym}\frac{a}{bc}})$$
I'm eagerly interested in learning one method which assumes $c\not\equiv {\rm mid}(\!a, b, c\!)$. But if $c\equiv {\rm mid}(\!a, b, c\!)$:
$$2\sqrt{(\!\frac{a}{bc}+ \frac{b}{ca}+ \frac{c}{ab}\!)(\!a+ b+ c\!)}\leqq c(\!\frac{a}{bc}+ \frac{b}{ca}+ \frac{c}{ab}\!)+ \frac{a+ b+ c}{c}= \frac{a+ b}{c}+ \frac{a}{b}+ \frac{b}{a}+ 1+ \frac{c^{2}}{ab}$$
We need to prove
$$\begin{align} \frac{a+ b}{c}+ \frac{a}{b}+ \frac{b}{a}+ 1+ \frac{c^{2}}{ab}\leqq \frac{a+ b}{c}+ \frac{b+ c}{a} & + \frac{c+ a}{b}\Leftrightarrow 1+ \frac{c^{2}}{ab}\leqq \frac{c}{a}+ \frac{c}{b}\Leftrightarrow \\ & \Leftrightarrow (\frac{c}{a}- 1)(\frac{c}{b}- 1)\leqq 0\Leftrightarrow \frac{(c- a)(c- b)}{ab}\leqq 0 \end{align}$$
Who can teach me what would we do if $c\not\equiv {\rm mid}(\!a, b, c\!)$ ? I am goin' to set a bounty, thank u so much
| Need to prove$:$ $$\displaystyle \left( {\dfrac {a+b}{c}}+{\dfrac {b+c}{a}}+{\dfrac {a+c}{b}} \right) ^{2
}\geqslant 4 \left( a+b+c \right) \left( {\dfrac {a}{bc}}+{\dfrac {b}{ac}}+{
\dfrac {c}{ab}} \right) $$
Or $$\displaystyle \,{\frac { 2\left( ab-2\,ac+bc \right) ^{2}}{ \left( {a}^{2}+{c}^{2} \right) {b}^{2}}}+{\frac {2 \left( a-c \right) ^{2} \left[2\,b{a}^{2}+abc+2\,b{c}^{2}-ac(a+c) \right]^{2}}{ \left( {a}^{2}+{ c}^{2} \right) \left( {a}^{2}+{b}^{2}+{c}^{2} \right) {c}^{2}{a}^{2}} }+$$
$$+{\dfrac { \left( a-b \right) ^{2} \left( b-c \right) ^{2} \left( a-c
\right) ^{2} \left( {a}^{2}+4\,ab+4\,ac+{b}^{2}+4\,bc+{c}^{2}
\right) }{ \left( {a}^{2}+{b}^{2}+{c}^{2} \right) {c}^{2}{a}^{2}{b}^{
2}}}\geqslant 0$$
Which is true.
Text for the SOS's expression: Click here.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Line integral in parameterized curve $a,b\gt 0$ and $n \in \mathbb Z$ \ {$0$}. Parameterized curve $C: x=a\cos nt, y= b\sin nt$
where $0 \le t \le2\pi$ and n is how many times that goes around origo.
I want to calculate line integral:
$\omega(C) = \frac{1}{2\pi}\oint_C \frac{x\,dy-y\, dx}{x^2+y^2}$
I have calculated:
$dx=-a\sin( nt)n$
$dy = b\cos(nt)n$
then after that I put the values in "right" places and:
$\frac{a\,b\,n}{(a^2+b^2)(\cos^2(nt))+b^2}$
Then next step is integration and the limit values are $[0,2\pi]$.
$\omega(C) = \frac{1}{2\pi}\int_0^{2\pi}\frac{a\,b\,n}{(a^2+b^2)(\cos^2(nt))+b^2} dt$
I get some random answer for that so I know that is wrong. I think the problem is in the integration but I don't know where is the problem.
Answer what I got:
$\frac{2\tan^{-1}(\frac{b\tan(2n\pi}{a})\vert a b\vert - ab(mod(4n-1,2)-4n-1)\pi}{4\vert ab \vert\pi}$
| So you were able to find
$$
\frac{1}{2\pi}\oint_C \frac{x\,dy-y\, dx}{x^2+y^2}
= \dfrac{2\tan^{-1}\left(\frac{b\tan(2n\pi)}{a}\right)\vert a b\vert - ab(\mod(4n-1,2)-4n-1)\pi}{4\vert ab \vert\pi}
$$
That looks pretty complicated, but:
*
*$\tan(2n\pi) = 0$ for any integer $n$. Therefore
$$
\tan^{-1}\left(\frac{b}{a}\tan(2n\pi)\right) = 0
$$
*$4n-1$ is odd, so $(4n-1) \bmod 2 = 1$. Therefore
$$
-ab((4n-1)\bmod 2 -(4n+1))\pi = 4\pi nab
$$
So the expression simplifies to:
$$
\frac{1}{2\pi}\oint_C \frac{x\,dy-y\, dx}{x^2+y^2} = \frac{4\pi n ab}{4|ab|\pi} = \frac{ab}{|ab|} n = \pm n
$$
where the $+$ is taken when $a$ and $b$ have the same sign, and $-$ otherwise.
But you can approach this another way. Notice that $C$ is the same ellipse $C_0$ traversed $n$ times. So
$$
\frac{1}{2\pi}\oint_C \frac{x\,dy-y\, dx}{x^2+y^2}
=\frac{n}{2\pi}\oint_{C_0} \frac{x\,dy-y\, dx}{x^2+y^2}
$$
Also, setting $P=\frac{-y}{x^2+y^2}$ and $Q = \frac{x}{x^2+y^2}$, we see that $P_y = Q_x$ away from the origin. The ellipse $C_0$ is homologous to either the unit circle $S^1$, or its reverse, depending on whether $ab>0$ or $ab < 0$. So by Green's Theorem,
$$
\oint_{C_0} \frac{x\,dy-y\, dx}{x^2+y^2} = \pm \oint_{S^1} \frac{x\,dy-y\, dx}{x^2+y^2}
$$
If you parametrize the unit circle with $x=\cos t$ and $y=\sin t$, then you'll see
$$
\oint_{S^1} \frac{x\,dy-y\, dx}{x^2+y^2}
= \int_0^{2\pi} 1 \,dt = 2\pi
$$
and this matches your previous computation.
| {
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Let $ a \in \mathbb{N}^\ast $, prove that $ \frac{1}{2a} - \frac{1}{2a^3} < \sqrt{a^2+1} - a < \frac{1}{2a} $ I have the following problem to solve. It's about convergent sequences.
Let $ a \in \mathbb{N}^\ast $, prove that:
$$ \frac{1}{2a} - \frac{1}{2a^3} < \sqrt{a^2+1} - a < \frac{1}{2a} $$
To solve it, I first solve:
$$ 0 < \sqrt{a^{2}+1} - a - \frac{1}{2a} + \frac{1}{2a^3} $$
To do so, I compute the result for $ a = 1 $ which is $ \sqrt{2} - \frac{11}{8} > 0$. Then I was thinking of studying the variation of the sequence with it's derivative but it makes weird equations:
$$ U_a = \sqrt{a^{2}+1} - a - \frac{1}{2a} + \frac{1}{2a^3} $$
$$ U_a' = \frac{a}{\sqrt{a^2 + 1}} - 1 + \frac{1}{2a^2} - \frac{3}{2a^4} $$
$$ U_a' = 0 \Leftrightarrow \frac{2a^5 \sqrt{a^2+1} + 2a^4}{7a^2+9} - 1 = 0 $$
I'm lost from here, someone can help me ?
If it helps, we are studying limited expansions.
Thank you.
| Note that $$\sqrt{a^2+1}<\sqrt{a^2+1+\frac{1}{4a^2}}=\sqrt{\left(a+\frac1{2a}\right)^2}=a+\frac1{2a}.$$
Thus
$$\sqrt{a^2+1}-a<\frac{1}{2a}.$$
On the other hand
$$\sqrt{a^2+1}+a<2a+\frac{1}{2a}.$$
Therefore
$$\sqrt{a^2+1}-a=\frac{1}{\sqrt{a^2+1}+a}>\frac{1}{2a+\frac{1}{2a}}.$$
However
$$\frac{1}{2a+\frac{1}{2a}}=\frac{1}{2a\left(1+\frac{1}{4a^2}\right)}>\frac{1-\frac{1}{(4a^2)^2}}{2a\left(1+\frac{1}{4a^2}\right)}=\frac{1}{2a}\left(1-\frac{1}{4a^2}\right).$$
Hence we actually have a stronger inequality
$$\sqrt{a^2+1}-a>\frac{1}{2a}-\frac{1}{8a^3}>\frac{1}{2a}-\frac{1}{2a^3}.$$
| {
"language": "en",
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What is probability to get two balls of different colors, if one of them is blue?
There are $12$ red, $8$ green and $10$ blue balls in a box. Two balls are taken out at random. What is probability to get two balls of different colors, if one of them is blue?
We can use the formula of Conditional probability. Let $B=\{\text{got blue ball}\}$, $A = \{\text{two balls of different colors are taken}\}$.
$$P(A \mid B) = \frac{P(AB)}{P(B)}$$
$$P(B) = \frac{C^1_{10} \cdot C^1_{29}}{C^2_{30}}$$
$$P(AB)= \frac{C^1_{10} \cdot C^1_{20}}{C^2_{30}}$$
And answer is $20/29$. But it's wrong, and I don't understand why. Right is $40/49$.
| You computed the probability of obtaining at least one blue ball incorrectly. For the favorable cases, you can have either two blue balls, which occurs with probability
$$\Pr(\text{two blue}) = \frac{\dbinom{10}{2}}{\dbinom{30}{2}}$$
or one blue ball and one ball of a different color, which you correctly calculated is
$$\Pr(\text{exactly one blue}) = \frac{\dbinom{10}{1}\dbinom{20}{1}}{\dbinom{30}{2}}$$
so the probability that at least one of the balls is blue is
$$\Pr(\text{at least one blue}) = \frac{\dbinom{10}{2} + \dbinom{10}{1}\dbinom{20}{1}}{\dbinom{30}{2}}$$
Thus, the probability that two different balls of different colors are obtained if at least one blue ball is selected is
\begin{align*}
\Pr(\text{two balls of different colors} \mid \text{at least one blue selected}) & = \frac{\frac{\binom{10}{1}\binom{20}{1}}{\binom{30}{2}}}{\frac{\binom{10}{2} + \binom{10}{1}\binom{20}{1}}{\binom{30}{2}}}\\
& = \frac{\dbinom{10}{1}\dbinom{20}{1}}{\dbinom{10}{2} + \dbinom{10}{1}\dbinom{20}{1}}
\end{align*}
In your attempt to calculate $\Pr(\text{at least one blue})$, you selected a blue ball and one of the remaining balls. Doing so counts each selection with $2$ blue balls twice, once for each way you could designate one of the blue balls as the blue ball you have chosen. Observe that
$$\color{red}{\binom{2}{1}}\binom{10}{2} + \binom{10}{1}\binom{20}{1} = \color{red}{\binom{10}{1}\binom{29}{1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3376815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Demonstrate that $(x-(\frac{x^2-2}{2*x})\bigr)^2 \gt 2$ for the given conditions. I am stuck on a proof where I need to demonstrate that $(x-(\frac{x^2-2}{2*x})\bigr)^2 \gt 2$.
The proof provides me with the information that $x^2\gt 2$ and $x>0$.
I've taken the following steps to simplify the algebra...to the point where I arrive at:
$\frac{(x^2+2)^2}{(2*x)^2} = \frac {x^4+4*x^2+4}{4*x^2}$
I simplified this further to produce: $1 + \frac {1}{4}*x^2 + \frac{1}{x^2}$
In looking at this simplification, I see that I need to demonstrate (remembering the conditions), $\frac {1}{4}*x^2 + \frac{1}{x^2} \gt 1$
Now, $\frac{1}{4}*x^2 \gt \frac{1}{2}$ and $\frac{1}{x^2}\lt \frac{1}{2}$ but I cannot figure out how to conclude that the addition of the two outputs something greater than 1. I imagine that I somehow need to show that the first term $\frac{1}{4}*x^2$ is somehow bigger than $0.5$ than $\frac{1}{x^2}$ is less than $0.5$.
In attempt to address this, I performed the following operation:
$\frac{\frac{1}{4}*x^2}{\frac{1}{x^2}} = \frac{1}{4}*x^4$
I know that $x^4$ must be greater than $4$...therefore $\frac{\frac{1}{4}*x^2}{\frac{1}{x^2}} \gt 1$.
Have I therefore successfully demonstrated that the addition of the two fractions:
$\frac {1}{4}*x^2 + \frac{1}{x^2} \gt 1$
and can conclude that:
$(x-(\frac{x^2-2}{2*x})\bigr)^2 \gt 2$ ?
Edit: While the answers below also provide the solution to my question (albeit using a different strategy from the very beginning), I have figured out how to solve my question by finishing the procedure that I pursued above
Specifically, revisiting $\frac {1}{4}*x^2 + \frac{1}{x^2} \gt 1$, let's subtract $1$ from either side and then multiply through by $4x^2$ to produce:
$x^4-4x^2+4 \gt 0$ (multiplying by $4x^2$ will not change the direction of the inequality because $x\gt0$ )
We can factor this to $(x^2-2)^2$ and because $x \gt 0$, this number will always be greater than $0$.
| It is equivalent to $$\frac{(x^2-2)^2}{4x^2}>0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the value of the periodic continued fraction with given terms
Find the value of the periodic continued fraction with the terms
$1, 3, 4, 3, 2, 3, 4, 3, 2, 3, 4, 3, 2, . . .$
We see that it starts to be periodic after $1$, i.e, $3,4,3,2$ then $3,4,3,2$ etc...
I know that $x= \frac{A_{k+1}}{B_{k+1}}$ = $\frac{A_{k-1}+x.A_K}{B_{k-1}+x.B_K}$ where $q_{k+1}=x$
I have $$x={3+\cfrac{1}{4+\cfrac{1}{3+\cfrac{1}{2+\cfrac{1}{x}}}}}$$ for the periodic part. If I use my formula to compute the right-hand side, I will end up with a quadratic in $x$. Solving for $x$ gives me the following quadratic formula: $5x^2-14x-7=0$, then I did $1+\frac1x$ to find the whole continued fraction, and I got the same quadratic equation. Is this correct?
| if we take $u=x+1,$ we find that u > 2 has purely periodic c.f. $(2,3,4,3).$
Next
$$
\left(
\begin{array}{cc}
2 & 1 \\
1 & 0 \\
\end{array}
\right)
\left(
\begin{array}{cc}
3 & 1 \\
1 & 0 \\
\end{array}
\right)
\left(
\begin{array}{cc}
4 & 1 \\
1 & 0 \\
\end{array}
\right)
\left(
\begin{array}{cc}
3 & 1 \\
1 & 0 \\
\end{array}
\right) =
\left(
\begin{array}{cc}
7 & 2 \\
3 & 1 \\
\end{array}
\right)
\left(
\begin{array}{cc}
13 & 4 \\
3 & 1 \\
\end{array}
\right) =
\left(
\begin{array}{cc}
97 & 30 \\
42 & 13 \\
\end{array}
\right)
$$
Then
$$ u = \frac{97u+30}{42u + 13} $$
gives
$$ 42u^2 + 13 u = 97 u + 30 $$
or
$$ 7 u^2 - 14 u - 5 = 0. $$
Since $u > 0,$
$$ u = \frac{7 + 2 \sqrt{21}}{7} = 1 + \frac{ 2 \sqrt{21}}{7} $$
and $x=u-1$ is
$$ x = \frac{ 2 \sqrt{21}}{7} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3382551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Tricky epsilon-delta proof $$\lim_{x\to -1}\frac{x-1}{x^2-x+1}=-\frac{2}{3}$$
What I've got so far is that:
$$\forall\epsilon>0,\exists\delta>0\text{ s.t. }0<|x-(-1)|< \delta\implies\left|\frac{x-1}{x^2-x+1}-(-\frac{2}{3})\right|<\epsilon\\
\forall\epsilon>0,\exists\delta>0\text{ s.t. }0<| x+1|< \delta\implies\left|\frac{(x+1)(2x-1)}{3x^2-3x+3}\right|<\epsilon$$
How do I go about finding a value for delta from here? Thanks.
| Take the case where $x\in [-2,0]$ (i.e. $\delta < 1$). Then we have the following inequality
$$\left|\frac{(x+1)(2x-1)}{3(x^2-x+1)}\right| < \frac{5}{3}|x+1|$$
by maximizing the numerator and minimizing the denominator. So set $\delta = \min(\frac{3}{5}\epsilon,1)$ and the proof step for the limit follows in both cases.
If $\epsilon > \frac{5}{3}$:
$$|x+1|<1\implies \left|\frac{(x+1)(2x-1)}{3(x^2-x+1)}\right| < \frac{5}{3}|x+1| < \frac{5}{3} < \epsilon$$
If $\epsilon \leq \frac{5}{3}$:
$$|x+1|<\frac{3}{5}\epsilon \implies \left|\frac{(x+1)(2x-1)}{3(x^2-x+1)}\right| < \frac{5}{3}|x+1| < \epsilon$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3385093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\tan(\frac{x}{2}) = \frac{\sin x}{1 + \cos x} $ using the quadratic formula I am trying to prove the fact that $\tan \frac{x}{2} = \frac{1-\cos x}{\sin x}$ or alternatively $\tan \frac{x}{2} = \frac{1- \cos x}{\sin x}$. (I understand that it can be proved using the half-angle identities of $\sin$ and $\cos$ but I want to understand how to get to the solution from this specific method of derivation.)
\begin{align*}
\tan(2x) &= \frac{2\tan(x)}{1-\tan(x)^2} \\
\tan(x) &= \frac{2\tan(\frac{x}{2})}{1-\tan(\frac{x}{2})^2} \\
\end{align*}
I now let $A=\tan x$ and $B=\tan \frac{x}{2}$
\begin{align*}
A\cdot(1-B^2) &= 2B\\
AB^2+2B-A &= 0 \\
\end{align*}
Now I solve for B using the quadratic formula.
\begin{align*}
B &= \frac{-2\pm \sqrt{4+4A^2}}{2A} \\
B &= \frac{-1\pm \sqrt{1+A^2}}{A} \\
\tan(\frac{x}{2}) &= \frac{-1\pm \sqrt{1+\tan(x)^2}}{\tan(x)}\\
\tan(\frac{x}{2}) &= \frac{-1\pm \sqrt{(\sec x)^2}}{\tan(x)}\\
\tan(\frac{x}{2}) &= \frac{-1\pm |\sec x|}{\tan(x)}
\end{align*}
I am confused as to how to continue at this point (firstly, not sure how to deal with the absolute value, and secondly not sure how to deal with the plus-minus).
Any help is greatly appreciated, as I feel like I do not fully understand how to manipulate absolute values and the meaning of the plus-minus.
| The quadratic formula alone won't help you obviate the $\pm$ sign. It's better to note $\sin x=\frac{2t}{1+t^2}$ ($t$ is more common notation than $B$) while $\cos x=\frac{1-t^2}{1+t^2}$, so $\frac{\sin x}{1+\cos x}=t$. Alternatively, $$\frac{\sin x}{1+\cos x}=\frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos^2\frac{x}{2}}=t.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The polynomial $x^{2k}+1+(x+1)^{2k}$ is not divisible by $x^2+x+1$. Find the value of $k \in \mathbb N$.
The polynomial $x^{2k}+1+(x+1)^{2k}$ is not divisible by $x^2+x+1$. Find the value of $k\in \mathbb{N}$.
I tried finding out the roots of $x^2+x+1$ which were $\dfrac{-1±\sqrt{3}i}{2}$ but in vain. I got no result other than making the polynomial more complicated.
Here's what I got : $$\left(\frac{-1±\sqrt{3}i}{2}\right)^{2k}+1+\left(\frac{-1±\sqrt{3}i}{2}+1\right)^{2k}.$$
Now, I don't know what to do next.
Any help would be appreciated.
| Using the remainder theoram, putting $x^2 + x + 1 = 0$ in the polynomial $P(x) =x^{2k} + 1 + (x + 1)^{2k}$ we will get the remaimder.
$$
P(x) = 1 + x^{2k} + (x^2 + 2x + 1)^k = 1 + x^{2k} + x^k
$$
We can also prove that if $x^2 + x + 1 = 0$, $x^3 = 1$.
We can now clearly see that for $k$ is not equal to $3q$, $P(x)=0$
But given it should not be divisible by $x^2 + x + 1$, Hence $k$ must be $3q$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all real values of the parameter a for which the equation $x^4+2ax^3+x^2+2ax+1=0$ has Find all real values of the parameter a for which the equation $x^4+2ax^3+x^2+2ax+1=0$ has
1) exactly two distinct negative roots
2) at least two distinct negative roots
I tried to factorize it but didn't get any breakthrough.
| Hint: Try $x^4+2ax^3+x^2+2ax+1=(x^2+bx+1)(x^2+cx+1)$.
| {
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Prove that the inequality $\frac{n^3}{3} < 3n-3$ applies to $n = 2$ and for all other $n \in \mathbb{N}$ the inequality does not hold I have been given the following task:
Prove that the following inequality applies to $n = 2$ and for all other $n \in \mathbb{N}$ the inequality does not hold:
$$\frac{n^3}{3} < 3n-3$$
My idea was to prove the statement by using induction.
For $n=1$ it follows:
$\frac{1^3}{3} = \frac{1}{3} \nless 0 = 3 * 1-3$
For $n=2$ it follows:
$\frac{2^3}{3} = \frac{8}{3} < 3 = 3 * 2-3$
For $n=3$ it follows:
$\frac{3^3}{3} = \frac{27}{3} = 9 \nless 6 = 3 * 3-3$
I now assume that the inequality does not hold for $n \in \mathbb{N} \backslash \{2\}.$
n $\rightarrow$ n+1
$\frac{(n+1)^3}{3} = \frac{n^3+3n^2+3n+1}{3} = \frac{n^3}{3}+ \frac{3n^2+3n+1}{3} < 3n-3 + \frac{3n^2+3n+1}{3}=\frac{3(3n-3)+3n^2+3n+1}{3}=\frac{3n^2+12n-8}{3}$
But how do I continue from this step or is this even the wrong approach?
| Let consider the reverse $\frac{n^3}{3} > 3n-3$ then we can prove the induction step as follows
$$\frac{(n+1)^3}{3} =\frac{n^3}{3}+\frac{3n^2+3n+1}{3}> 3n-3+\frac{3n^2+3n+1}{3}>3n-3+3=3(n+1)-3$$
we have used that for $n\ge 2$
$$\frac{3n^2+3n+1}{3}>3 \iff3n^2+3n-8>0$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find all $x\in \mathbb R$ such that $\sqrt[3]{x+2}+\sqrt[3]{x+3}+\sqrt[3]{x+4}=0$ The question is as the title says:
Find all $x\in \mathbb R$ such that $\sqrt[3]{x+2}+\sqrt[3]{x+3}+\sqrt[3]{x+4}=0$.
I struggle to even start this question.
By inspection, I see that $x$ must be negative. Playing around yields $x=-3$ as a solution, though I do not know how to prove that there are no other solutions.
Upon differentiation, I obtain:
\begin{align}
\frac{d}{dx} (\sqrt[3]{x+2}+\sqrt[3]{x+3}+\sqrt[3]{x+4})&= \frac{1}{3}(\sqrt[3]{(x+2)^{-2}}+\sqrt[3]{(x+3)^{-2}}+\sqrt[3]{({x+4})^{-2}})\\
\end{align}
which is always positive for all real values of $x$, implying that the function defined as $f(x)=\sqrt[3]{x+2} + \sqrt[3]{x+3} + \sqrt[3]{x+4}$ is strictly increasing.
Is there a better way to solve this equation?
| $y:=x+3$;
$f(y)=(y-1)^{1/3}+y^{1/3}+(y+1)^{1/3}=0$;
By inspection $y=0$ is a solution.
0) $y^{1/3}$ is an odd, increasing function.
1) $y \ge 1$ is ruled out, since $f(y)>0$
2) $y \le -1$ is ruled out, since $f(y)<0$.
Remains $1< y <1$;
3) $f(y)> 0$ for $0<y<1$; since $-1< (y-1) <0$, $1<(y+1) <2$:
4) Similarly $f(y) <0$ for $-1<y <0$;
It follows $y=0$ is the only solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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Series using partial sums... Determine if the following series converge by studying the partial sums. If it converges, compute its value.
$\sum_{n=1}^{\infty} \frac{2n+3}{(n+1)(n+2)}$.
We use partial fractions so
\begin{equation*}
\begin{split}
\frac{2n+3}{(n+1)(n+2)} &= \frac{A}{n+1}+\frac{B}{n+2} \\
\Longleftrightarrow2n+3 &= A(n+2)+B(n+1) \\
&= An+2A+Bn+B \\
&= (A+B)n+2A+B.
\end{split}
\end{equation*}
Equating terms tells us that $A+B = 2$ and $2A+B = 3$. We can find $A,B$ and $C$ by solving
\begin{equation*}
\begin{bmatrix}
1 & 1 \\
2 & 1
\end{bmatrix}\begin{bmatrix}
A \\
B
\end{bmatrix} = \begin{bmatrix}
2 \\
3
\end{bmatrix}.
\end{equation*}
Doing so yields $A = B = 1$. This means the partial sum is
\begin{equation*}
S_k = \sum_{n=1}^{k} \frac{2n+3}{(n+1)(n+2)} = \sum_{n=1}^{k} \left(\frac{1}{n+1}+\frac{1}{n+2}\right) = \left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{3}+\frac{1}{4}\right)+\ldots+\left(\frac{1}{k+1}+\frac{1}{k+2}\right).
\end{equation*}
Is this correct?
| Your method is correct, I just put a simpler approach if you are interested. We see that $$\frac{2n+3}{(n+1)(n+2)}>\frac{2n}{n^2+3n+2}>\frac{2n} {6n^2}=\frac{1}{3n} $$
But $\sum{\frac{1}{n}}$ diverges.....
| {
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How to get a closed-form for this sum? I'm trying to calculate the $[x^n]F(x)$ from a generating function $F(x) = \frac{1}{(1-2x^2)^2}$ and I came across with this expression involving a sum:
$$2^n\sum_{k=0}^n k^k (n-k)^{n-k}$$
What I did so far was calculate the $[x^n]\frac{1}{1-2x^2}$, witch is $(2n)^n$ if I'm not wrong, and then making a convolution $\frac{1}{1-2x^2} . \frac{1}{1-2x^2}$ to find $[x^n]$.
My question is, how can I obtain a closed-form for the sum above?
| It would seem that based on the series
$$\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$$
that
$$\frac{1}{1- 2 \, x^2} = \sum_{n=0}^{\infty} 2^{n} \, x^{2n} = \sum_{n=0}^{\infty} 2^{n/2} \, \left(\frac{1+(-1)^{n}}{2}\right) \, x^n.$$
From this then:
\begin{align}
\frac{1}{(1- 2 x^2)^2} &= \sum_{r=0}^{\infty} \sum_{s=0}^{\infty} 2^{(r+s)/2} \, \left(\frac{1+(-1)^{r}}{2}\right)\left(\frac{1+(-1)^{s}}{2}\right) \, x^{r+s} \\
&= \sum_{n=0}^{\infty} \sum_{s=0}^{n} 2^{n/2 -2} \, (1+(-1)^{n-s})(1+(-1)^{s}) \, x^{n} \\
&= \sum_{n=0}^{\infty} 2^{n/2 - 2} \, (1 + (-1)^n) \, \sum_{s=0}^{n} (1 + (-1)^{s}) \, x^n \\
&= \sum_{n=0}^{\infty} 2^{n/2 -3} \, [2 (n+1) (1 + (-1)^n) + (1 + (-1)^n)^2] \, x^n \\
&= \sum_{n=0}^{\infty} \left( \frac{2^{n/2} \, (n + 2) \, (1 + (-1)^{n})}{4} \right) \, x^n.
\end{align}
Alternatively using
$$\frac{1}{(1-x)^2} = \sum_{n=0}^{\infty} (n+1) \, x^{n}$$
then
\begin{align}
\frac{1}{(1- 2 x^2)^2} &= \sum_{n=0}^{\infty} 2^{n} \, (n+1) \, x^{2n} \\
&= \sum_{n=0}^{\infty} \left(\frac{2^{n/2} \, (n+2) \, (1 + (-1)^{n})}{4} \right) \, x^n.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all the value of $(x, y, z)$ with $x, y, z \in \Bbb R$ $$x^2 + 4 = y^3 + 4x - z^3$$
$$y^2 + 4 = z^3 + 4y - x^3$$
$$z^2 + 4 = x^3 + 4z - y^3$$
What's the proper method to solve this question? I've tried any method but still not get the answer.
| $$x^2 + 4 = y^3 + 4x - z^3 \implies x^2 -4x +4 = y^3-z^3 \implies (x-2)^2 = y^3-z^3 \quad\text{ (1.)}$$
$$y^2 + 4 = z^3 + 4y - x^3 \implies y^2 -4y +4 = z^3 -x^3 \implies (y-2)^2 = z^3-z^3 \quad\text{ (2.)}$$
$$z^2 + 4 = x^3 + 4z - y^3 \implies z^2 -4z +4 = x^3-y^3 \implies (z-2)^2 = x^3-y^3 \quad\text{ (3.)}$$
Adding all three equations:
$$(x-2)^2 + (y-2)^2 + (z-2)^2 = 0$$
$$x-2 = 0 \,\,,\,\, y-2 = 0\,\,,\,z-2=0$$
$$\boxed{x=y=z=2}$$
| {
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$x_1$ and $x_2$ are the solutiton of $\frac{2\cdot \sin(x) \cdot \cos(2x)}{\cos (x) \cdot \sin (2x)} - 5 \tan(x) + 5 = 0$ $x_1$ and $x_2$ are the solutiton of $\frac{2\cdot \sin(x) \cdot \cos(2x)}{\cos (x) \cdot \sin (2x)} - 5 \tan(x) + 5 = 0$
then, $\tan(x_1 + x_2) = ....$
i can do it by doing it
$\dfrac{2\cdot \sin(x) \cdot (\cos^2x - \sin^2x)}{\cos (x) \cdot 2 \sin x \cos x} - 5 \tan(x) + 5 = 0$
leads to $(\sin x - \cos x)(\sin x + 6 \cos x) = 0$
but it's complicated, do you know the less complicated way to solve it?
| The equation is $ \frac{2\tan x}{\tan 2x} - 5 \tan x + 5 = 0$, as one sees from rearranging the first term. The double angle formula for $\tan$ gives $\frac{2 \tan x }{\tan 2x} = 1 - \tan^2 x$ and therefore $(1 - \tan^2 x) - 5 \tan x + 5 = 0$, which simplifies to $\tan^2 x + 5 \tan x - 6 = 0$. This finally gives $\tan x = -6$ or $\tan x = +1$.
Finally, using the addition formula, we get $\tan(x_1 + x_2) = \frac{-6 + 1}{1 - (-6)\cdot(+1)} = \frac{-5}{7}$.
| {
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Prove $\sin^3A-\cos^3A=\left(\sin^2A-\cos^2A\right)(1-2\sin^2A\cos^2A)$ Prove $\sin^3A-\cos^3A=\left(\sin^2A-\cos^2A\right)(1-2\sin^2A\cos^2A)$
My attempt is as follows:
Taking LHS:
$$\left(\sin A-\cos A\right)(1+\sin A\cos A)$$
$$\left(\sin^2A-\cos^2A\right)\frac{\left(1+\sin A\cos A\right)}{\left(\sin A+\cos A\right)}$$
$$\left(\sin^2A-\cos^2A\right)\frac{(\left(\sin A+\cos A\right)^2-\sin A\cos A)}{\sin A+\cos A}$$
$$\left(\sin^2A-\cos^2A\right)\left(\sin A+\cos A-\frac{\sin A\cos A}{\sin A+\cos A}\right)$$
I was not getting any breakthroughs from here.
So I tried RHS:
$$(\sin A-\cos A)(\sin A+\cos A)(1-2\sin^2A\cos^2A)$$
$$(\sin A-\cos A)(\sin A+\cos A)((\sin^2A+\cos^2A)^2-2\sin^2A\cos^2A)$$
$$(\sin A-\cos A)(\sin A+\cos A)(\sin^4A+\cos^4A)$$
Even from here I was not getting breakthroughs, what am i missing?
Please help me.
| With $x=\pi$ the LHS is equal to $1$ while the RHS is equal to $-1$ therefore the identity is false.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Inverse trigonometry proof How would I go about proving
$$\arccos x = \arctan \frac{\sqrt {c-x^2}}{x} $$
Where $c$ is a constant and $0< x ≤1$
| Suppose $\theta = \arccos x$
then $\cos \theta = x$
if $0<x\le 1$ then $0\le \theta < \frac {\pi}{2}$
$\sin \theta = \sqrt {1-x^2}$ and $\tan\theta = \frac {\sin\theta}{\cos\theta} = \frac {\sqrt {1-x^2}}{x}$
$\theta = \arccos x = \arctan \frac {\sqrt {1-x^2}}{x}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3404845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $\lim_{n \to \infty} \left(\frac{3n-2}{3n+1}\right)^{2n} = \frac{1}{e^2}$?
It's known that $\lim_{n \to \infty} \left(1 + \frac{x}{n} \right)^n = e^x$.
Using the above statement, prove $\lim_{n \to \infty} \left(\frac{3n-2}{3n+1}\right)^{2n} = \frac{1}{e^2}$.
My attempt
Obviously, we want to reach a statement such as $$\lim_{n \to \infty} \left(1 + \frac{-2}{n}\right)^n \quad \text{ or } \quad \lim_{n \to \infty} \left(1 + \frac{-1}{n}\right)^n \cdot \lim_{n \to \infty} \left(1 + \frac{-1}{n}\right)^n$$
in order to be able to apply the above condition. However, I was unable to achieve this. The furthest I've got was the following:
\begin{align}
\left(\frac{n-2}{3n+1}\right)^{2n} &= \left( \frac{9n^2 - 12n + 4}{9n^2 + 6n + 1} \right)^n\\
&= \left(1 + \frac{-18n+3}{9n^2+6n+1}\right)^n\\
f(n)&= \left(1 + \frac{-2 + \frac{3}{n}}{n+\frac{2}{3} + \frac{1}{9n}} \right)^n
\end{align}
It seems quite obvious that $\lim_{n \to \infty} \left(f(n)\right) = \left(1 + \frac{-2}{n + \frac{2}{3}}\right)^n$, however, this is not exactly equal to the statement given above. Are you able to ignore the constant and apply the condition regardness? If so, why? How would you go about solving this problem?
| Observe that \begin{eqnarray*}
(\frac{3n-2}{3n+1})^{2n} & = & \left(\frac{3n+1-3}{3n+1}\right)^{2n}\\
& = & \left\{ \left(1+\frac{(-3)}{3n+1}\right)^{-1}\left(1+\frac{(-3)}{3n+1}\right)^{3n+1}\right\} ^{\frac{2}{3}}.
\end{eqnarray*}
We assume the fact without proof: For any $x\in\mathbb{R}$, $\lim_{n\rightarrow\infty}(1+\frac{x}{n})^{n}=e^{x}$.
Let $y_{n}=\left(1+\frac{(-3)}{n}\right)^{n}$. Then,
\begin{eqnarray*}
(\frac{3n-2}{3n+1})^{2n} & = & \left\{ \left(1+\frac{(-3)}{3n+1}\right)^{-1}y_{3n+1}\right\} ^{\frac{2}{3}}.
\end{eqnarray*}
Since $y_{n}\rightarrow e^{-3}$ and $(y_{3n+1})_{n}$ is a subsequence
of $(y_{n})_{n}$, we have $y_{3n+1}\rightarrow e^{-3}$ too as $n\rightarrow\infty$.
Note that $\left(1+\frac{(-3)}{3n+1}\right)^{-1}\rightarrow1$ as
$n\rightarrow\infty$. Therefore
\begin{eqnarray*}
\lim_{n\rightarrow\infty}(\frac{3n-2}{3n+1})^{2n} & = & \left\{ \lim_{n\rightarrow\infty}\left(1+\frac{(-3)}{3n+1}\right)^{-1}y_{3n+1}\right\} ^{\frac{2}{3}}\\
& = & \left\{ e^{-3}\right\} ^{\frac{2}{3}}\\
& = & e^{-2}.
\end{eqnarray*}
In the above, we have used the fact that $x\mapsto x^{\frac{2}{3}}$ is
continuous.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3405914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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How can we attain 3 solutions when using Vieta's Substitution? Consider $x^3+px=q$. Substituting $x:=w-\frac{p}{3w}$ will reduce it to $w^3-\frac{p^3}{27w^3}-q=0$. Multiplying with $w^3$ will result in $(w^3)^2-qw^3-\frac{1}{27}p^3=0$. Using the quadratic formula, we get$$w^3=-\frac{q\pm \sqrt{q^2+\frac{4}{27}p^3}}{2}$$
So this would imply that if $q^2<-\frac{4p^3}{27}$, there would be no solution in the real numbers. But shouldn't there always be a realy solution?
Cf: http://mathworld.wolfram.com/VietasSubstitution.html
| When $q^2<-\frac{4p^3}{27}$, $w^3$ becomes a complex number and there are three distinctive complex values for $w$ once the cubit root is taken on $w^3$.
The solution to the cubic equation $x^3+px+q$ is instead,
$$x=w-\frac{p}{3w}$$
As a result, there are three distinctive roots for $x$ evaluated from the three $w$’s. However, the roots result in real value due to the cancellation of the imaginary parts of $w$ and $ -\frac{p}{3w}$.
Edit: The solutions to $x^3-\lambda x +2=0$, asked in comments.
Three cases:
1) If $\lambda=3$, i.e. $q^2+\frac{4p^3}{27}=0$, there are three real roots, one double and one single.
$$x_1= -2,\>\>\>\>\>x_{2,3}=1$$
2) If $\lambda<3$, there are three mixed roots, one real and two complex. The real root is given by
$$x_1= \left( -1+\sqrt{1-\frac{\lambda^3}{27}} \right)^{1/3}
+ \left( -1-\sqrt{1-\frac{\lambda^3}{27}}\right)^{1/3}$$
3) If $\lambda>3$, there are three distinctive real roots. Normally they are given by their trigonometric forms as follows,
$$x_1=2\sqrt{\frac {\lambda}{3}} \cos\theta$$
$$x_2=2\sqrt{\frac {\lambda}{3}} \cos\left( \theta-\frac{2\pi}{3}\right)$$
$$x_3=2\sqrt{\frac {\lambda}{3}} \cos\left( \theta-\frac{4\pi}{3}\right)$$
where,
$$\theta= \frac13 \cos^{-1}\left( -\frac{3}{\lambda}\sqrt{\frac{3}{\lambda}}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3407536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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$\lim_{x\to 0} \frac{x\sin(3x)}{1-\cos(6x)}$ $$\lim_{x\to 0} \frac{x\sin(3x)}{1-\cos(6x)}$$
I tried the following but it doesn't seem to work...
$$= \lim_{x\to 0} \frac{x}{2} \cdot \frac{\sin(3x)}{3x}\cdot\frac{6x}{1-\cos(6x)}$$
$$= 0$$
But the result of this limit is $\frac{1}{6}$.
Am I missing something or did I make any glaring mistakes?
| We can also make the substitution $x \mapsto \frac{u}{3}$:
$$\frac{1}{3} \frac{u \sin u}{1 - \cos 2u} = \frac{1}{3} \frac{u \sin u}{1 - \cos^2 u + \sin^2 u} = \frac{1}{3} \frac{u \sin u}{2 \sin^2 u} = \frac{1}{6}\frac{u}{\sin u}.$$
and since $\frac{u}{\sin u} \to 1$ as $u \to 0$:
$$\lim_{u \to 0} \frac{1}{3} \frac{u \sin u}{1- \cos 2u} = \frac{1}{6} \times 1 = \frac{1}{6}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3408633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
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Why does $\sqrt{x\sqrt{x}}$ equal $x^{3/4}$ instead of $x^{1/4}$? When I try to solve the equation, I get $x^{1/4}$ since the two roots share the same index and exponent, which means that I ought to be able to multiply across.
Instead, the solution in the textbook is as follows.
$$\left (xx^{1/2} \right )^{1/2}=\left ( x^{3/2} \right )^{1/2}=x^{3/4}$$
Where did $x^{3/2}$ come from? I can't figure out how I get a 3 from any of that. And why is my original intuition incorrect?
| The notation is $\sqrt{x\sqrt x}= m$
Then $m^2 = x\sqrt x$.
So $(m^2)^2 = (x\sqrt x)^2 = x^2*(\sqrt x)^2 = x^2*x = x^3$
So $m^4 = x^3$ and so
$m = \sqrt[4]{x^3} = x^{\frac 34}$
....
The book did it this way:
$\sqrt{x\sqrt x} = (x\cdot x^{\frac 12})^{\frac 12}=$
$(x^1\cdot x^{\frac 12})^{\frac 12} =$
$(x^{1+ \frac 12})^{\frac 12} =$
$(x^{\frac 32})^{\frac 12} =$ (this is where the $\frac 32$ came from. $1+\frac 12 = \frac 32$).
$x^{\frac 32\cdot \frac 12} = x^{\frac 34}$.
......
Note: $\sqrt{\sqrt x} = x^{\frac 14}$ is where your intuitions comes from. But $\sqrt{x \sqrt x}$ has an extra $x$ in it. But there is more than one way to do things: $\sqrt{x \sqrt{x}}=\sqrt{\sqrt{x^2}\sqrt x} = \sqrt{\sqrt{x^2*x}} = \sqrt{\sqrt{x^3}} = (x^3)^{\frac 14} = x^{\frac 34}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3409229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
In the range $0\leq x \lt 2\pi$ the equation has how many solutions $\sin^8 {x}+\cos^6 {x}=1$ In the range $0\leq x \lt 2\pi$ the equation has how many solutions
$$\sin^8 {x}+\cos^6 {x}=1$$
What i did
$\cos^6 {x}=1-\sin^8 {x}=(1-\sin^4 {x})(1+\sin^4 {x})=(1-\sin^2 {x})(1+\sin^2 {x})(1+\sin^4 {x})$
$\cos^4 {x}=(1+\sin^2 {x})(1+\sin^4 {x}) , \cos^2{x}=0$
$(1-\sin^2{x})^2=(1+\sin^2 {x})(1+\sin^4 {x})$
$-3\sin^2{x}=\sin^6{x}$
Which is not possilbe
Is there a trick or something to solve this equation or to know how many solutions are there ?
| Hint: Write your equation in the form $$- \left( \sin \left( x \right) \right) ^{2} \left( \cos \left( x
\right) \right) ^{2} \left( \left( \cos \left( x \right) \right) ^
{4}-3\, \left( \cos \left( x \right) \right) ^{2}+4 \right)
=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3410740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
If $x$ and $y$ are real number such that $x^2+2xy-y^2=6$, then find the minimum value of $(x^2+y^2)^2$ If $x$ and $y$ are real number such that $x^2+2xy-y^2=6$, then find the maximum value of $(x^2+y^2)^2$
My attempt is as follows:
$$(x-y)^2\ge 0$$
$$x^2+y^2\ge 2xy$$
$$2(x^2+y^2)\ge x^2+y^2+2xy$$
\begin{equation}
2(x^2+y^2)\ge (x+y)^2\tag{1}
\end{equation}
Solving the given equation:
$$x^2+2xy-y^2=6$$
$$(x+y)^2=2y^2+6$$
So putting the value of $(x+y)^2$ in equation $1$
$$2(x^2+y^2)\ge 2y^2+6$$
$$x^2+y^2\ge y^2+3$$
\begin{equation}
x^2\ge 3\tag{2}
\end{equation}
So $x\in \left(-\infty,-\sqrt{3}\right) \cup \left(\sqrt{3},\infty\right)$
But how to proceed from here?
| Play with linear combinations of $x^2+y^2$ and the expression from the given equation. This way we find for $a>1$:
$$a(x^2+y^2) - 6 = (a-1)x^2-2xy+(a+1)y^2=(\sqrt{a-1}x-\sqrt{a+1}y)^2+(2\sqrt{a^2-1}-2)xy$$
We get rid of the last term if we let $a=\sqrt 2$:
$$\tag1\sqrt 2(x^2+y^2) -6 = \left(\sqrt{\sqrt 2-1}\,x-\sqrt{\sqrt 2+1}\,y\right)^2\ge 0.$$
This gives us the lower bound $x^2+y^2\ge 3\sqrt 2$ (and so $(x^2+y^2)^2\ge18$).
Is the inequality $(1)$ sharp, i.e., can equality hold? Clearly, equality holds iff $x=\sqrt{\sqrt 2+1}\,t$ and $y=\sqrt{\sqrt 2-1}\,t$ for some $t$. But does there actually exist $t$ such that the original condition is met? We compute
$$ x^2+2xy-y^2=(\sqrt2+1)t^2+2t^2+(\sqrt2-1)t^2=\sqrt 2t^2$$
and this is indeed $=6$ for suitable $t$ (namely $t=\sqrt{3\sqrt 2}$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3411531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Computing the nth derivative using the binomial theorem. Let n be an element of the set of natural numbers
Let $F(x)=x^2(1+x)^n$ and write $F^n$ for the nth derivative of the function $F$.
Compute $F^n$ by applying the Binomial Theorem to $(1+x)^n$.
I don't understand the step where I need to find the derivative of these terms:
= $x^2 + {n \choose 1}x^3 + {n \choose 2} x^4 + .... + {n \choose n-1} x^{n+1} + x^{n+2} $
Even though I do understand how to show that for the first 3 terms the nth derivative is 0, but for the last two I have no clue??
Really appreciate any hints!!
| Write it in decreasing powers of $x$ and use subscript $n$:
$$F_n(x)=x^2(1+x)^n=x^2(x+1)^n=\\
\color{red}{x^{n+2}}+\color{green}{nx^{n+1}}+\color{blue}{\frac{n(n-1)}{2}x^{n}}+\frac{n(n-1)(n-2)}{6}x^{n-1}+\cdots+nx^3+x^2$$
Note that for $f(x)=x^n$:
$$f^{(0)}(x)=f(x)=x^n\\
f^{(k)}(x)=\begin{cases}n(n-1)(n-2)\cdots(n-k+1)x^{n-k},k\le n\\ 0,k>n\end{cases}$$
Hence:
$$F_0(x)=x^2 \Rightarrow F_0^{(0)}(x)=F_0(x)=x^2;\\
F_n^{(n)}(x)=\color{red}{(n+2)(n+1)n\cdots3\cdot x^2}+\color{green}{n\cdot (n+1)n\cdots1\cdot x}+\\
\color{blue}{\frac{n(n-1)}{2}\cdot n(n-1)\cdots 1}=\\
\frac12(n+2)(n+1)\cdot n!\cdot x^2+n(n+1)\cdot n!\cdot x+\frac{n(n-1)}{2}\cdot n!=\\
n!\left(\frac12(n+2)(n+1)x^2+n(n+1)x+\frac{n(n-1)}{2}\right).$$
Verify:
$$F_1(x)=x^2(x+1)=x^3+x^2 \Rightarrow F_1^{(1)}(x)=1!(3x^2+2x)=3x^2+2x;\\
F_2(x)=x^2(x+1)^2=x^4+2x^3+x^2 \Rightarrow \\
F_2^{(2)}(x)=2!(6x^2+6x+1)=12x^2+12x+2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3413696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving an inequality using AM-GM
Let $ a,b,c > 0$ such that $ abc = 1$. Prove that
$$\frac {ab}{a^2 + b^2 + \sqrt {c}} + \frac {bc}{b^2 + c^2 + \sqrt {a}} + \frac {ca}{c^2 + a^2 + \sqrt {b}}\le 1.$$
$$\frac{ab}{a^2+b^2+\sqrt{c}}=\frac{1}{\frac{a}{b}+\frac{b}{c}+\sqrt{c^3}}$$
(AM-GM): $$\frac{a}{b}+\frac{b}{a}+\sqrt{c^3}\geq 3\sqrt{c} $$
$$\frac{1}{\frac{a}{b}+\frac{b}{a}+\sqrt{c^3}}\leq \frac{1}{3\sqrt{c}}$$
Analogous: $$\frac{1}{\frac{b}{c}+\frac{c}{b}+\sqrt{a^3}} \leq \frac{1}{3\sqrt a}$$
$$\frac{1}{\frac{a}{c}+\frac{c}{a}+\sqrt{b^3}}\leq \frac{1}{3\sqrt b}$$
$$LHS \leq \frac{1}{3}+\frac{1}{3}+\frac{1}{3} =1$$
Am I right?
| By AM-GM and C-S we obtain:
$$\sum_{cyc}\frac{ab}{a^2+b^2+\sqrt{c}}\leq\sum_{cyc}\frac{ab}{2ab+\sqrt{c^2ab}}=\sum_{cyc}\frac{\sqrt{ab}}{2\sqrt{ab}+c}=$$
$$=\frac{3}{2}-\sum_{cyc}\left(\frac{1}{2}-\frac{\sqrt{ab}}{2\sqrt{ab}+c}\right)=\frac{3}{2}-\sum_{cyc}\frac{c}{2(2\sqrt{ab}+c)}\leq$$
$$\leq\frac{3}{2}-\frac{\left(\sum\limits_{cyc}\sqrt{c}\right)^2}{2\sum\limits_{cyc}(2\sqrt{ab}+c)}=\frac{3}{2}-\frac{1}{2}=1.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Prove that $x$ is an integer if $x^4-x$ and $x^3-x$ are integers. Given that $x^4-x$ and $x^3-x$ are integers.
Prove that $x$ is an integer.
Expanding, adding and subtracting the given expressions I can conclude that the followings are integers:$x(x-1)(x+1)$, $x^3(x-1)$, $x(x-1)(x^2+x+1)$ are integers.
Could you help me solve the problem?
| I will assume that $x$ is a priori living in some integral commutative ring, e.g. let's say $x$ is a complex number. It wouldn't be as easy (or as true) if $x$ were in some non-commutative or non-integral ring.
Write $a = x^4 - x$ and $b = x^3 - x$, which are integers by assumption.
We have
(1) $x^4 - x - a = 0$,
(2) $x^3 - x - b = 0$.
Subtracting $x$ times (2) from (1), we get
(3) $x^2 + (b - 1)x - a = 0$.
Subtracting $(x - (b - 1))$ times (3) from (2), we get
(4) $(b^2 - 2b + a)x = a(b - 1) + b$.
Now there are two possibilities.
Firstly, if $b^2 - 2b + a = 0$, then we must also have $a(b - 1) + b = 0$.
Substituting $a = 2b - b^2$, we get $b^3 - 3b^2 + b = 0$. But $b$ is an integer, so it must follow that $b = 0$.
Hence $x^3 - x = 0$, or $x = 0, \pm 1$, which are all integers.
Secondly, if $b^2 - 2b + a \neq 0$, then we may divide out and get $x = \frac{a(b - 1) + b}{b^2 - 2b + a}$. This means that $x$ is a rational number.
But $x$ is also a root of the polynomial $P(X) = X^3 - X - b\in\mathbb{Z}[X]$, so $x$ must be an integer, since $\mathbb{Z}$ is integrally closed.
Alternatively, a down-to-earth proof may work like this: write $x = \frac{p}{q}$ with $p, q$ integers and $\gcd(p, q) = 1$.
We then have $x^3 - x = \frac{p^3 - pq^2}{q^3}$, which is assumed to be an integer. This means $q\mid p^3 - pq^2$, or $q\mid p^3$.
But $q$ is prime to $p$, so the only possibility is $q = 1$.
Hence $x = p$ is an integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3420991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Finding the value of $c$ to make $-\dfrac{1}{2}x^2+x+c$ a perfect square trinomial
I should find the value of $c$ to make: $$-\dfrac{1}{2}x^2+x+c$$ a
perfect square trinomial.
I really messed up: $-\dfrac{1}{2}x^2+x+c=x-\dfrac{1}{2}x^2+c$, but it seems like $-\dfrac{1}{2}x^2$ isn't $a^2\boldsymbol{-2ab}+b^2$. Can you help me?
| We need $B^2-4AC=0$ that is
$$1+2c=0 \implies c=-\frac12$$
indeed we obtain
$$-\frac12(x^2-2x+1)=\left(\frac i{\sqrt2}(x-1)\right)^2$$
Therefore there is no $c$ to have a perfect square on reals.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3424744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is it possible to represent the natural number "1" as the sum of p-series in this way? My argument: $$1=(\frac{1}{2})^2+(\frac{1}{3})^2+\cdots+(\frac{1}{2})^3+(\frac{1}{3})^3+\cdots=\sum_{k=2}^\infty (\frac{1}{k})^2+\sum_{k=2}^\infty (\frac{1}{k})^3+\cdots .$$
Explanation) First, for any natural number $n\geq2$, the following holds:
$$\sum_{k=1}^\infty (\frac{1}{n})^k=(\frac{1}{n})+(\frac{1}{n})^2+(\frac{1}{n})^3+\cdots= \frac{\frac{1}{n}}{1-\frac{1}{n}}=\frac{1}{n-1}.$$ (the infinite geometric series.)
So, we obtain
$$1=(\frac{1}{2})+(\frac{1}{2})^2+\cdots=\sum_{k=1}^\infty (\frac{1}{2})^k,$$
$$\frac{1}{2}=(\frac{1}{3})+(\frac{1}{3})^2+\cdots=\sum_{k=1}^\infty (\frac{1}{3})^k,$$
$$\frac{1}{3}=(\frac{1}{4})+(\frac{1}{4})^2+\cdots=\sum_{k=1}^\infty (\frac{1}{4})^k,$$
and so on.
From above equalities,
$$1=(\frac{1}{2})+(\frac{1}{2})^2+\cdots$$
$$=((\frac{1}{3})+(\frac{1}{3})^2+\cdots)+(\frac{1}{2})^2+(\frac{1}{2})^3+\cdots$$
$$=((\frac{1}{4})+(\frac{1}{4})^2+\cdots)+(\frac{1}{3})^2+(\frac{1}{3})^3+\cdots+(\frac{1}{2})^2+(\frac{1}{2})^3+\cdots$$
$$=\cdots .$$
If $p\ge2$, then p-series absolutely converges. Hence we can change the order of terms in series as follows:
$$1=(\frac{1}{2})^2+(\frac{1}{3})^2+(\frac{1}{4})^2+\cdots$$
$$+(\frac{1}{2})^3+(\frac{1}{3})^3+(\frac{1}{4})^3+\cdots$$
$$+(\frac{1}{2})^4+(\frac{1}{3})^4+(\frac{1}{4})^4+\cdots$$
$$=\sum_{k=2}^\infty (\frac{1}{k})^2+\sum_{k=2}^\infty (\frac{1}{k})^3+\sum_{k=2}^\infty (\frac{1}{k})^4+\cdots.$$
Thus, "1" becomes the sum of p-series(exactly from $k=2$ to infinity).
Is this explanation correct?
| So what you are saying is
$1
=\sum_{m=2}^{\infty} \sum_{k=2}^{\infty} \dfrac1{k^m}
$.
Let's check.
$\begin{array}\\
\sum_{m=2}^{\infty} \sum_{k=2}^{\infty} \dfrac1{k^m}
&=\sum_{k=2}^{\infty} \sum_{m=2}^{\infty} \dfrac1{k^m}
\qquad\text{(reverse order of summation)}\\
&=\sum_{k=2}^{\infty}\dfrac{1/k^2}{1-1/k}
\qquad\text{(just a geometric series)}\\
&=\sum_{k=2}^{\infty}\dfrac{1}{k^2-k}
\qquad\text{(multiply num and dec by }k^2)\\
&=\sum_{k=2}^{\infty}\dfrac{1}{k(k-1)}
\qquad\text{(rewrite)}\\
&=\sum_{k=2}^{\infty}(\dfrac1{k-1}-\dfrac1{k})
\qquad\text{(now we can telescope)}\\
&=1\\
\end{array}
$
Yup.
| {
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Simple continued fraction of $\sqrt{d}$ with period of shortest length $3$ This is the problem:
Does there exist positive integer $d$ ( which is not a perfect square ) such that the length of the least period in the simple continued fraction of $\sqrt{d}$ is $3$?
Consider the following theorem
Theorem : If the positive integer $d$ is not a perfect square, the simple continued fraction of $\sqrt{d}$ has the form $\sqrt{d} = [a_0;\overline{a_1,a_2,\cdots,a_{r-1},2a_o}]$ with $a_o = \lfloor d \rfloor$. Here $r$ denotes the length of the least period in the expansion of $\sqrt{d}$. Where $\lfloor x \rfloor$ denotes the greatest integer function/ floor function of $x$.
We want to solve for non-square $d$ where$\sqrt{d} = [a_0;\overline{a_1,a_2,2a_o}]$, and $a_o = \lfloor d \rfloor$. Since $d$ is a positive integer, $a_o = \lfloor d \rfloor \ge 1$, and $a_1 , a_2$ are positive integers by definition. Please note that the converse of the above theorem is not true, for example, consider $[1;\overline{1,1,2}] = \sqrt{10}/2$ and $[0;\overline{1,1,0}] = \sqrt{2}/2$. I calculated first few continued fractions for $\sqrt{d}$, $$\begin{array}{c|c|c} \sqrt{d} & \text{Continued fraction} & r\\ \hline
√2 & [1;\bar{2}] & 1 \\
√3 & [1;\overline{1,2}] & 2 \\
√5 & [2;\bar{4}] & 1\\
√6 & [2;\overline{2,4}] & 2\\
√7 & [2;\overline{1,1,1,4}] & 4\\
√8 & [2;\overline{1,4}] & 2\\
√10 & [3;\bar{6}] & 1\\
√11 & [3;\overline{3,6}] & 2\\
√12 & [3;\overline{2,6}] & 2\\
√13 & [3;\overline{1,1,1,1,6}] & 5\\
√14 & [3;\overline{1,2,1,6}] & 4\\
√15 & [3;\overline{1,6}] & 2\\
√17 & [4;\bar{8}] & 1\\
√18 & [4;\overline{4,8}] & 2\\
√19 & [4;\overline{2,1,3,1,2,8}] & 6 \\
√20 & [4;\overline{2,8}] & 2\\
√21 & [4;\overline{1,1,2,1,1,8}] & 6\\
√22 & [4;\overline{1,2,4,2,1,8}] & 6\\
√23 & [4;\overline{1,3,1,8}] & 4\\
√24 & [4;\overline{1,8}] & 2\\ \end{array}$$
As we can see for $1< d \le 24, r \ne 3$. Also, on a side note, observe that there does not exist two consecutive intergers $d$ and $d+1$ such that both $\sqrt{d}$ and $\sqrt{d+1}$ have $r=1$, moreover there are infinitely $\sqrt{d}$ such that the length of there least period is $1$ or $2$, $\sqrt{n^2+1} = [n;\overline{2n}]$, $\sqrt{n^2+2} = [n;\overline{n,2n}]$ and
$\sqrt{n^2-1} = [n-1;\overline{1,2(n-1)}]$ , where $n \in \mathbb{N}$ .Even for $r=4$, we have $\sqrt{n^2-2} = [n-1; \overline{1,n-2,1,2(n-1)}]$, $n>2$. Now i have a hunch that no such $d$ exists for which $\sqrt{d}$ have $r=3$. Any hints on how to prove this ? In general does there exist a number $m$ such that $r\ne m $ for any $\sqrt{d}$ ?
| Just working numerically, $41$ is the least example, with $$\sqrt {41}=[6; \overline {2,2,12}]$$
here is a tabulated list of the periods of $\sqrt d$.
OEIS provides a list of $d$ for which the period is $3$, and that link provides a way of generating infinitely many examples.
| {
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Find the number of integers between 1 and 25 so that the expression is divided by 6 Find the number of integers so that the $n^2+3n+2$ is divided by 6 when $1 \le n \le 25$.
I start by $6\mid n^2+3n+2 6|(n+1)(n+2) \implies 6k=(n+1)(n+2) $ when $ k \in Z$
we can see that for $n=$ $1,2,4,5,7,8,10,11....$ will work, but I am not sure how to continue from here .
(some HINT can be very appreciated ).
Additionally I would love see a combinatorics approach for this .
Thank you guys.
| Being divisible by $6$ means being even and divisible by $3$.
*
*Modulo $2$, $n^2+3n+2\equiv n^2+n\equiv n+n \text{ (by Fermat) }\equiv 0$. So it is always even.
*Modulo $3$, $n^2+3n+2\equiv n^2+2$, and it is congruent to $0$ if and only if $n^2\equiv 1\mod 3$, which happens only if $n\equiv \pm 1$, in other words, if and only if $n$ is not divisible by $3$.
So we simply can take the list of integers between $1$ and $25$ and remove the $8$ multiples of $3$ from this list. There remains $17$ integers.
| {
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Complexity of Quicksort I am currently struggling with a problem on time complexity of the Quicksort algorithm in the average case.
I calculated the average time complexity of the algorithm to be
$T_n \quad = \quad \frac{1}{n} \cdot \sum_{k=1}^n \left( T_{k-1} + T_{n-k} + n - 1 \right)$
Which can also be written as
$T_n \quad = \quad n - 1 + \frac{2}{n} \cdot \sum_{k=1}^n T_{k-1}$
So far so well; now I have to calculate the term:
$n\cdot T_n - (n-1)\cdot T_{n-1}$
and solve the outcoming linear recursion of first order.
My ansatz is to insert the upper equation into this term, which gives me
$n\cdot T_n - (n-1)\cdot T_{n-1} \quad = \quad n^2 - n + 2 \cdot \sum_{k=1}^n T_{k-1} - \left[ (n-1)(n-2) + 2 \cdot \sum_{k=1}^{n-1} T_{k-1} \right]$
Whereof I get
$n\cdot T_n - (n-1)\cdot T_{n-1} \quad = \quad 2 \cdot \left[ (n-1) + T_{n-1} \right]$
Now, I am stuck. Because if I solve the equation for $T_n$, I obtain
$T_n \quad = \quad \frac{1}{n} \cdot \left[ (n+1) T_{n-1} + 2(n-1) \right]$
This is not linear though, meaning I made a mistake somewhere, or there is another way to solve this.
Does anyone have an idea how to carry on or how to solve this?
Sincerely,
Octavius
| I finally managed to get it down. I start from the above mentioned term
$n \cdot T_n \quad = \quad (n+1) \cdot T_{n-1} + 2(n-1)$
Since I only want to know magnitudes and asymptotic behavior I simply discard the $-2$ term, remaining with
$n \cdot T_n \quad = \quad (n+1) \cdot T_{n-1} + 2n$
which is equivalent to
$\frac{T_n}{n+1} \quad = \quad \frac{2}{n+1} + \frac{T_{n-1}}{n}$ .
Basically I am done now, because all I have to do is recursively insert the equation:
$\frac{T_n}{n+1} \quad = \quad \frac{2}{n+1} + \frac{2}{n} + \frac{T_{n-2}}{n-1} \quad = \quad \frac{2}{n+1} + \frac{2}{n} + \frac{2}{n-1} + \frac{T_{n-3}}{n-2} \quad = \quad \text{etc.}$
In a more compact form:
$\frac{T_n}{n+1} \quad = \quad 2 \cdot \sum_{j=2}^{n+1} \frac{1}{j}$
Which is the $(n+1)$-th partial sum of the harmonic series, meaning
$\sum_{j=2}^{n+1} \frac{1}{j} \quad \in \quad O(\log n)$
In conclusion:
$ T_n \quad \in \quad O(n \cdot \log n)$
That is what I wanted to find as a solution in agreement with relevant literature solutions.
| {
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How to prove $n \ge 2 {\left ( \sum_{k=1}^n \frac{1}{k} \right)}^2$ for all $n \ge 2$? I'm trying to prove that $$n \ge 2 {\left ( \sum_{k=1}^n \frac{1}{k} \right)}^2$$ for all $n \ge 2$. I try induction on $n$ as follows:
My attempt:
The inequality holds for $n=2$. Let it holds for $n$. Our goal is to show that $$2 {\left ( \sum_{k=1}^n \frac{1}{k} \right)}^2+1 \ge 2 {\left ( \sum_{k=1}^{n+1} \frac{1}{k} \right)}^2$$
This is equivalent to $$1 \ge \frac{1}{n+1} \left ( \frac{1}{n+1}+ \sum_{k=1}^n \frac{1}{k} \right) =\frac{1}{(n+1)^2} + \frac{1}{n+1} \sum_{k=1}^n \frac{1}{k}$$
I'm unable to approximate the sum $\sum_{k=1}^n \frac{1}{k}$. Could you please shed me some light on the last step?
| Note that the statement itself tells you how to approximate $\sum_{k=1}^{n} \frac{1}{k}$, so with wishful thinking we hope it is sufficient (which need not be the case).
Change the statement to showing that $\sqrt{\frac{n}{2}} \geq \sum_{k=1}^{n} \frac{1}{k}$.
Then, the induction step requires us to show that $\sum_{k=1}^{n+1} \frac{1}{k} \leq \sqrt{\frac{n}{2}} + \frac{1}{n+1} \leq \sqrt{ \frac{n+1}{2}}$
Since $\sqrt{ \frac{n+1}{2}} - \sqrt{ \frac{n}{2}} = \frac{\frac{1}{2}}{\sqrt{ \frac{n+1}{2}} + \sqrt{ \frac{n}{2}}} \geq \frac{1}{2*\sqrt{2(n+1)}} \geq \frac{1}{n+1}$ when $n+1 \geq 8$.
So, start the induction at $n= 7$, and check the initial base cases.
| {
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Show that $\frac{a}{c} + \frac{b}{d} +\frac{c}{a} + \frac{d}{b}\le-12$ Let $a$,$b$,$c$ and $d$ be non-zero, pairwise different real numbers such that $ \frac{a}{b} +\frac{b}{c} +\frac{c}{d} + \frac{d}{a}=4$ and $ac=bd$ .
Show that $\frac{a}{c} + \frac{b}{d} +\frac{c}{a} + \frac{d}{b}\le-12$ and that $-12$ is the maximum.
I simplified the inequality to prove: $a^2+b^2+c^2+d^2\le -12ac$
But I am not sure what to do next.
Hints and solutions would be appreciated.
Taken from the 2018 Pan African Math Olympiad http://pamo-official.org/problemes/PAMO_2018_Problems_En.pdf
| Let $u = \frac{a}{b},\,v= \frac{b}{c},$ then $\frac{c}{d}=\frac{1}{u},\,\frac{d}{a}=\frac{1}{v}$ we get $u+v+\frac{1}{u}+\frac{1}{v}=4,$ and
$$P = \frac{a}{c} + \frac{b}{d} + \frac{c}{a} + \frac{d}{b} = \left(u+\frac{1}{u}\right)\left(v+\frac{1}{v}\right).$$
Now see $u,v$ can't satisfied $uv > 0.$ Indeed if $u > 0,\,v>0$ then
$$4 = u+\frac{1}{u}+v+\frac{1}{v} \geqslant 2 + 2 =4.$$
Equality occur when $u=v=1$ (contradiction the distinct numbers).
If $u<0,\,v<0$ replace $(u,v)$ by $(-x,-y)$ with $x>0,\,y>0$ then
$$4 = -\left(x+y+\frac 1 x + \frac 1 y\right)< 0 . $$
Thefore $uv < 0,$ now
$$P = -12 + \frac{(u+1)^2(v+1)^2}{uv}-2\left(u+v+\frac{1}{u}+\frac{1}{v}-4\right) \leqslant -12.$$
See here or here
| {
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Help with algebra, rearrange equations I have:
$$
r=\frac{1-x^2-y^2}{(1-x)^2+y^2} \tag 1
$$
And want to write $(1)$ as:
$$
\Big (x-\frac{r}{1+r}\Big )^2+y^2=\Big (\frac{1}{1+r}\Big )^2 \tag 2
$$
First method, starting from $(1)$:
$$
r=\frac{1-x^2-y^2}{x^2-2x+1+y^2}\iff
$$
$$
r(x^2-2x+1+y^2)=1-x^2-y^2 \iff
$$
$$
rx^2-r2x+r+ry^2=1-x^2-y^2 \iff
$$
$$
rx^2+x^2-r2x+r+ry^2+y^2=1 \iff
$$
$$
x^2(r+1)-r2x+r+y^2(r+1)=1 \iff
$$
$$
x^2-x\frac{2r}{r+1}+\frac{r}{r+1}+y^2=\frac{1}{r+1} \tag 3
$$
Completing the square of $x^2-x\frac{2r}{r+1}$:
$$
x^2-x\frac{2r}{r+1}+\Big (\frac{r}{1+r} \Big )^2-\Big (\frac{r}{1+r} \Big )^2=\Big (x-\frac{r}{1+r}\Big )^2-\Big (\frac{r}{1+r} \Big )^2
$$
Inserting in $(3)$ gives:
$$
\Big (x-\frac{r}{1+r}\Big )^2-\Big (\frac{r}{1+r} \Big )^2+\frac{r}{r+1}+y^2=\frac{1}{r+1} \tag 4
$$
I'm stuck here, what is next?
Second method, starting from $(2)$:
$$
\Big (x-\frac{r}{1+r}\Big )^2+y^2=\Big (\frac{1}{1+r}\Big )^2 \iff
$$
$$
\Big (\frac{x(r+1)-r}{1+r}\Big )^2+y^2=\Big (\frac{1}{1+r}\Big )^2 \iff
$$
$$
(x(r+1)-r)^2+y^2(1+r)^2=1 \tag 5
$$
Expand $(x(r+1)-r)^2$:
$$
(x(r+1)-r)^2=x^2(1+r)^2-x2r(1+r)+r^2
$$
Inserting in $(5)$ gives:
$$
x^2(1+r)^2-x2r(1+r)+r^2+y^2(1+r)^2=1 \iff
$$
$$
x^2(1+2r+r^2)-2rx-2xr^2+r^2+y^2(1+2r+r^2)=1\iff
$$
$$
x^2+2rx^2+r^2x^2-2rx-2xr^2+r^2+y^2+2ry^2+r^2y^2=1\iff
$$
Collect $r^2$ and $r$:
$$
r^2(x^2-2x+1+y^2)+r(2x^2-2x+2y^2) = 1-x^2-y^2 \iff
$$
$$
r^2((x-1)^2+y^2)+r(2x(x-1)+2y^2) = 1-x^2-y^2 \tag 6
$$
I'm stuck here, I don't know how to go from $(6)$ to $(1)$.
| Your equation $$\left(x-\frac{r}{1+r}\right)^2+y^2-\frac{1}{(1+r)^2}=0$$ is equivalent to
$$rx^2+ry^2-2xr+x^2+y^2+r-1$$.
| {
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Find $g$(3) if $g(x)g(y)=g(x)+g(y)+g(xy)-2$ and $g(2)=5$
If $g(x)$ is a polynomial function satisfying $g(x)g(y)=g(x)+g(y)+g(xy)-2$ for all real $x$ and $y$ and $g(2)=5$, then find $g$(3)
Putting $x=1,y=2$
$$
5g(1)=g(1)+5+5-2\implies \boxed{g(1)=2}
$$
And I think I can evaluate for $g(1/2)$ also.
Solution given in my reference is $g(3)=10$, but I think I am stuck with this.
| The point is that $g$ is a polynomial.
Thus write $g(x) = a_n x^n + \dotsc + a_0$, and you have an equality:
$$\sum_{i, j = 0}^n a_i a_j x^i y^j = (\sum_{i = 0}^n a_i (x^iy^i + x^i + y^i)) - 2.$$
Since this is true for all $x, y$, it means that it is an equality as polynomials in $x, y$.
By comparing coefficients, we see that $a_i a_j$ must be zero whenever $i > j > 0$ (because there is no $x^i y^j$ term on the right).
Therefore $g$ is of the form $ax^n + b$, and our equation becomes:
$$a^2x^ny^n + abx^n + aby^n + b^2 = ax^ny^n + b + ax^n + b + ay^n + b - 2,$$
from which we deduce that $a = 0$ or $a = 1$.
If $a = 0$, then $b^2 = 3b - 2$, so that $b = 1$ or $b = 2$: this contradicts $g(2) = 5$.
The only possibility is then $a = 1$. Comparing coefficients before $x^n$, we have $ab = a$, which tells us $b = 1$.
It turns out that any polynomial $g$ of the form $x^n + 1$ will satisfy that equation for all $x, y$.
From $g(2) = 5$ we then conclude that $n = 2$, hence $g(3) = 3^2 + 1 = 10$.
| {
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Why does this process map every fraction to the golden ratio? Start with any positive fraction $\frac{a}{b}$.
First add the denominator to the numerator: $$\frac{a}{b} \rightarrow \frac{a+b}{b}$$
Then add the (new) numerator to the denominator:
$$\frac{a+b}{b} \rightarrow \frac{a+b}{a+2b}$$
So $\frac{2}{5} \rightarrow \frac{7}{5} \rightarrow \frac{7}{12}$.
Repeating this process appears to map every fraction to
$\phi$ and $\frac{1}{\phi}$:
$$
\begin{array}{ccccccccccc}
\frac{2}{5} & \frac{7}{5} & \frac{7}{12} & \frac{19}{12} & \frac{19}{31} & \frac{50}{31} & \frac{50}{81} & \frac{131}{81} & \frac{131}{212} & \frac{343}{212} & \frac{343}{555} \\
0.4 & 1.40 & 0.583 & 1.58 & 0.613 & 1.61 & 0.617 & 1.62 & 0.618 & 1.62 & 0.618 \\
\end{array}
$$
Another example:
$$
\begin{array}{ccccccccccc}
\frac{11}{7} & \frac{18}{7} & \frac{18}{25} & \frac{43}{25} & \frac{43}{68} & \frac{111}{68} & \frac{111}{179} & \frac{290}{179} & \frac{290}{469} & \frac{759}{469} & \frac{759}{1228} \\
1.57143 & 2.57 & 0.720 & 1.72 & 0.632 & 1.63 & 0.620 & 1.62 & 0.618 & 1.62 & 0.618 \\
\end{array}
$$
Q. Why?
| You are re-stating the Fibonacci sequence, and by the theory of linear recurrences, the terms are quasi-proportional to the powers of the largest root of the characteristic equation
$$\phi^2-\phi-1=0.$$
Hence, the ratio of successive terms quickly tends to $\phi$.
| {
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Find $\lim\limits_{n \to \infty} \sqrt[3]{n^3+2n^2+1}-\sqrt[3]{n^3-1}$. I have to find the limit:
$$\lim\limits_{n \to \infty} \sqrt[3]{n^3+2n^2+1}-\sqrt[3]{n^3-1}$$
I tried multiplying with the conjugate of the formula:
$$(a-b)(a^2+ab+b^2)=a^3-b^3$$
So I got:
$$\lim\limits_{n \to \infty} \dfrac{n^3+2n^2+1-n^3+1}{\sqrt[3]{(n^3+2n^2+1)^2} + \sqrt[3]{(n^3+2n^2+1)(n^3-1)} + \sqrt[3]{(n^3-1)^2}}$$
$$\lim\limits_{n \to \infty} \dfrac{2n^2+2}{\sqrt[3]{n^6+4n^5+4n^4+2n^3+4n^2+1} + \sqrt[3]{n^6+2n^5-2n^2-1} + \sqrt[3]{n^6-2n^3+1}}$$
And I saw that we can factor $n^2$ in the denominator and if we do the same in the numerator, we'd get that the limit is equal to $2$. The problem is that my textbook claims that this limit is actually $\dfrac{2}{3}$. I don't see why should I have a $3$ in the denominator since the coefficient of $n^2$ would be $1$ if I would carry out the factoring to detail. So, what did I do wrong?
| Since
$(n+1)^3
=n^3+3n^2+$lower order terms,
$(n+2/3)^3
=n^3+2n^2+$lower order terms.
Therefore the left cube root
is about $n+2/3$.
Since the right cube root
is just about $n$,
the difference is about $2/3$.
| {
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Prove formula from Pascal's triangle I found this in my Norwegian mathematics book. I have solved a)-d) but am stuck at e). Nice if you have feedback on what I have attempted to do.
a) Where are the numbers of the form $\binom{k}{l}$ in Pascal's triangle? Where are the numbers of the form $\binom{k}{2}$? In general, if i is a fixed number and k varies, where will you find the numbers $\binom{k}{i}$?
My answer: I find it on the third diagonal line from the left. If i is fixed, you find it on the ith diagonal line from the left
b) Check through some examples that the following equality seems to hold:
$\sum_{k=1}^n \binom{k}{i} = \binom {n+1}{i+1}$
Can you explain it through Pascals triangle?
My answer:
$\sum_{k=2}^4\binom{k}{2}=\binom{5}{3}=10$
In Pascal's triangle this is the sum all from the third diagonal line from the left up to k=4.
The result is $\binom {n+1}{i+1}$
c) Prove the formula b) by induction on n.
My answer:
The equation holds for n. Assuming that the equation also holds for n+1.
$\sum_ {k=1}^{n+1}\binom{n+1}{i+1}+\binom{n+1}{i}$
$=\frac{(n+1)!}{(i+1)!(n+1)!}+\frac{(n+1)!}{i!(n+1-i)!}$
$=\frac{(n+1)!(n+1-i)}!{(i+1)!(n+1-i)!}+\frac{(n+1)!(i+1)}{(i+1)!(n+1-i)!}$
$=\frac{(n+1)!(n+1-i+i+1)}{(i+1)!(n+1-i)!}$
$=\frac{(n+2)!}{(i+1)!((n+2)-(i+1))!}$
$=\binom{n+2}{i+1}$
This is correct according to Pascal's triangle.
d) Show that the formula can be rewritten to:
$\sum_{k=1}^n k(k-1)(k-2)\dots (k-i+1)=\frac{(n+1)n(n-1)\dots(n-i+1)}{i+1}$
My thoughts: Multiply by (n-i)! / (k-1)! on both sides.
$\sum_{k=1}^n \binom{k}{i} = \binom {n+1}{i+1}$
$\sum_{k=i}^n\frac{k!}{i!(k-i)!}\dot (k-i)!=\sum_{k=1}^nk(k-1)(k-2)\dots(k-i+1)$
$\binom {n+1}{i+1}\dot (n-i)!=\frac{(n+1)!}{(i+1)!(n-i)!}\dot (n-1)!$
$=\frac{(n+1)n(n-1)\dots(n-i+1)}{i+1}$
e) Show how the formula in d) can be used to find expressions for the sums $\sum_{k=1}^nk^2$, $\sum_{k=1}^nk^3$ etc. Hint: Start with the formula
$\sum_{k=1}^nk=\frac{n(n+1)}{2}$
And work your way up the potentials.
My thoughts: I just know how to get the formula $\sum_{k=1}^nk=\frac{n(n+1)}{2}$
| I have the solution for $\sum_{k=1}^nk^2$
$k^2=k(k-1)+k$ so that $\sum_{k=1}^nk^2=\sum_{k=1}^nk(k-1)+\sum_{k=1}^nk$
$\sum_{k=1}^n k(k-1)(k-2)\dots (k-i+1)=\frac{(n+1)n(n-1)\dots(n-i+1)}{i+1}$
$\sum_{k=1}^n k(k-1)=\frac{(n+1)n}{2+1}=\frac{(n+1)n(n-1)}{3}$
$\sum_{k=1}^nk^2=\sum_{k=1}^nk(k-1)+\frac{n(n+1)}{2}=\frac{(n+1)n(n-1)}{3}+\frac{n(n+1)}{2}=\frac{2(n+1)n(n-1)}{3*2}+\frac{3n(n+1)}{2*3}=\frac{n(n+1)(2(n-1)+3))}6=\frac{n(n+1)(2n+1)}6$
The solution for $\sum_{k=1}^nk^3$
$\sum_{k=1}^n k(k-1)(k-2))=\sum_{k=1}^nk^3-\sum_{k=1}^n3k^2+2\sum_{k=1}^n3k=\frac{(n+1)n(n-1)(n-2)}{4}$
$\sum_{k=1}^nk^3 -3\frac{n(n+1)(2n+1)}6 + 2\frac{n(n+1)}{2}=\frac{(n+1)n(n-1)(n-2)}{4}$
$\sum_{k=1}^nk^3=\frac{(n+1)n(n-1)(n-2)}{4}+3\frac{n(n+1)(2n+1)}6-2\frac{n(n+1)}{2}$
$\frac{2(n+1)n(n-1)(n-2)}{8}+4\frac{n(n+1)(2n+1)}8-8\frac{n(n+1)}{8}$
$=\frac{n^2(n+1)^2}{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3443611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove $4^n+5^n+6^n$ is divisible by 15 Prove by induction:
$4^n+5^n+6^n$ is divisible by 15 for positive odd integers
For $n=2k-1,n≥1$ (odd integer)
$4^{2k-1}+5^{2k-1}+6^{2k-1}=15N$
To prove $n=2k+1$, (consecutive odd integer)
$4^{2k+1}+5^{2k+1}+6^{2k+1}=(4)4^{2k}+(5)5^{2k}+(6)6^{2k}$,
How do I substitute the statement where $n=2k-1$ to the above, to factor out 15 in order to prove divisibility? Would it be easier to assume $n=k$ is odd and prove $n=k+2$ is divisible by 15?
| As you suggested, it's notationally simpler to suppose $4^k+5^k+6^k$ is divisible by $15$ and consider
$$4^{k+2}+5^{k+2}+6^{k+2} = 16\cdot 4^k + 25\cdot 5^k + 36\cdot 6^k.$$
Subtracting the original expression, we get $15\cdot 4^k + 24\cdot 5^k + 35\cdot 6^k$. The first term is divisible by $15$. Now note that
$$24\cdot 5^k +35\cdot 6^k = 15\cdot 8\cdot 5^{k-1} + 15\cdot 14\cdot 6^{k-1}$$
is likewise divisible by $15$. Thus, $4^{k+2}+5^{k+2}+6^{k+2}$ is indeed divisible by $15$.
Query: Where did we use that $k$ is odd? Well, obviously to start the induction. But where else?
| {
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"url": "https://math.stackexchange.com/questions/3444343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solve for $x,y$ $\begin{cases}\sin^2x+\sin^2y=\frac{3}{4}\\x+y=\frac{5\pi}{12}\end{cases}$
Designate $x,y\in\left(0,\frac{\pi}{2}\right)$ which fulfills
$\begin{cases}\sin^2x+\sin^2y=\frac{3}{4}\\x+y=\frac{5\pi}{12}\end{cases}$
My proof:
$\cos2x=1-2\sin^2x\Rightarrow \sin^2x=\frac{1}{2}-\frac{1}{2}\cos2x$
$\sin^2x+\sin^2y=\frac{3}{4}\Rightarrow1-\frac{1}{2}\left(\cos2x+\cos2y\right)=\frac{3}{4}\Rightarrow \cos2x+\cos2y=\frac{1}{2}\\ \cos x+\cos y=2\cos\frac{x+y}{2}\cos\frac{x-y}{2}\\\cos2x+\cos2y=\frac{1}{2}\Rightarrow \cos(x+y)\cos(x-y)=\frac{1}{4}\\\cos\left(\frac{5\pi}{12}\right)\cos\left(2x-\frac{5\pi}{12}\right)=\frac{1}{4}\\\cos2x=2\cos^2x-1\Rightarrow\cos\left(\frac{10\pi}{12}\right)=2\cos^2\left(\frac{5\pi}{12}\right)-1\Rightarrow\cos\left(\frac{5\pi}{12}\right)=\sqrt{\frac{-\frac{\sqrt{3}}{2}+1}{2}}=\frac{\sqrt{2-\sqrt{3}}}{2}\\\cos\left(2x-\frac{5\pi}{12}\right)=\frac{1}{4}\cdot\frac{2}{\sqrt{2-\sqrt{3}}}=\frac{1}{2\sqrt{2-\sqrt{3}}}$
I don't know what to do next,I don't know if it makes sense what I wrote
| I think this is a question about mental arihmetic, which $x=\pi/6, y= \pi/4$.
Or proceed from third step, $\cos2x+\cos2y=2\cos(x+y)(\cos(x-y))=1/2$
hence $2\cos(5\pi/12)(\cos(x-y))=1/2$
then using calculator, or as follows, you compute $\cos(x-y)=\dfrac{\sqrt{2}(\sqrt{3}+1)}{4}$ (note that $\cos(5\pi/12)$=...
Either calculator, or by remember that $\cos(x-y)=\cos(\pi/12)$,
then solving equation gives the answer
$\\ \\\\\\\\\\ \\\ \\\ \\\ $
Or by direct computation, gives $1-\cos(2y)^2 \dfrac{(\sqrt{3}-1)^4}{4}=\cos(2y)^2+2\cos(2y)+1$, hence $\cos(2y)=0$, giving answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3447480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Is there a better way to solve this equation? I came across this equation:
$x + \dfrac{3x}{\sqrt{x^2 - 9}} = \dfrac{35}{4}$
Wolfram Alpha found 2 roots: $x=5$ and $x=\dfrac{15}{4}$, which "coincidentally" add up to $\dfrac{35}{4}$. So I'm thinking there should be a better way to solve it than the naïve way of bringing the fractions together and then squaring. Is there any?
| As $x>0$
WLOG $x=3\sec t, 0< t<\dfrac\pi2\implies \sin t,\cos t>0$
$$\dfrac{35}4=3\sec t+\dfrac{3\sec t}{3\tan t} \iff\sec t+\csc t=\dfrac{35}{12}$$
Method$\#1:$
Let $\sin t+\cos t=u$
$$\dfrac{35}{12}=\dfrac{2u}{u^2-1}$$
$$\iff0=35u^2-24u-35=7u(5u-7)+5(5u-7)=(5u-7)(7u+5)$$
As $u>0, u=\sin t+\cos t=\dfrac75$
Now use $(\sin t+\cos t)^2+(\cos t-\sin t)^2=2$
$\cos t-\sin t=\pm\sqrt{2-\left(\dfrac75\right)^2}=\pm\dfrac15$
$2\cos t=\cos t-\sin t+\cos t+\sin t=\dfrac{7\pm1}5$
Method$\#2:$
As $\sec^2t+\csc^2t=\sec^2t\csc^2t$
Let $u=\sec t\csc t=\dfrac2{\sin2t}$
Squaring we get $$\left(\dfrac{35}{12}\right)^2=u^2+2u\iff(u+1)^2=\left(\dfrac{37}{12}\right)^2\implies u=\dfrac{25}{12}\text{ as }u>0$$
$\sin2t=\dfrac2u=\dfrac{24}{25}\implies\cos2t=\pm\dfrac7{25}$
$x=\dfrac3{\cos t}=+\dfrac6{\sqrt{2(1+\cos2t)}}=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3449031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Find $\underset{x\rightarrow 0}\lim{\frac{1-\cos{x}\sqrt{\cos{2x}}}{x\sin{x}}}$
Find
$$\underset{x\rightarrow 0}\lim{\frac{1-\cos{x}\sqrt{\cos{2x}}}{x\sin{x}}}$$
My work.$$\underset{x\rightarrow 0}\lim\frac{1}{x\sin{x}}=\frac{\underset{x\rightarrow0}\lim{\;\frac{\sin{x}}{x}}}{\underset{x\rightarrow 0}\lim{\;x\sin{x}}}=\underset{x\rightarrow 0}\lim\frac{1}{x^2}$$
$$\underset{x\rightarrow 0}\lim{\frac{\cos{x}}{x\sin{x}}}=\underset{x\rightarrow 0}\lim{\frac{\sin{2x}}{2x\sin^2{x}}}=\underset{x\rightarrow 0}\lim{\frac{\sin{2x}}{2x}}\cdot\underset{x\rightarrow 0}\lim{\frac{1}{\sin^2{x}}}=\frac{1}{x^2}$$
$$\underset{x\rightarrow 0}\lim{\sqrt{\cos{2x}}}=\underset{x\rightarrow 0}\lim{\sqrt{1-2\sin^2{x}}}=\underset{x\rightarrow 0}\lim{\sqrt{1-2x^2}}$$
L' Hopital's rule:
$\underset{x\rightarrow 0}\lim{\frac{1-\cos{x}\sqrt{\cos{2x}}}{x\sin{x}}}=\underset{x\rightarrow 0}\lim{\frac{1-\sqrt{1-2x^2}}
{x^2}}=\underset{x\rightarrow 0}\lim{\frac{-4x}{x^3\sqrt{1-2x^2}}}$
What should I do next?
| As $x\to 0$, $x\sin x\sim x^2$. Also
$$\cos x=1-\frac{x^2}2+O(x^4),$$
$$\cos2x=1-2x^2+O(x^4),$$
$$\sqrt{\cos2x}=1-x^2+O(x^4),$$
$$(\cos x)(\sqrt{\cos2x})=1-\frac{3x^2}2+O(x^4),$$
$$1-(\cos x)(\sqrt{\cos2x})\sim\frac{3x^2}2.$$
Therefore
$$\lim_{x\to0}\frac{1-(\cos x)(\sqrt{\cos2x})}{x\sin x}=\frac32.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3450548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Limit of this sequence $\lim_{n\to \infty}\frac{ \sqrt{2n+2} - \sqrt{2n-2}}{\sqrt{3n+1} - \sqrt{3n}}$ I am trying to calculate the limit of this sequence :
$$\lim_{n\to \infty}\frac{ \sqrt{2n+2} - \sqrt{2n-2}}{\sqrt{3n+1} - \sqrt{3n}}$$
I tried two methods and the two methods leaded me to infinity or 4/0.
Anything would be helpful , thanks.
| We can use that
$$ \frac{ \sqrt{2n+2} - \sqrt{2n-2}}{\sqrt{3n+1} - \sqrt{3n}}=$$
$$=\frac{ \sqrt{2n+2} - \sqrt{2n-2}}{\sqrt{3n+1} - \sqrt{3n}}
\cdot \frac{\sqrt{2n+2} + \sqrt{2n-2}}{ \sqrt{2n+2} + \sqrt{2n-2}}
\cdot\frac{\sqrt{3n+1} + \sqrt{3n}}{ \sqrt{3n+1} + \sqrt{3n}}=$$
$$=4\cdot \frac{\sqrt{3n+1} + \sqrt{3n}}{ \sqrt{2n+2} + \sqrt{2n-2}}=4\cdot \frac{\sqrt{3+\frac1n} + \sqrt{3}}{ \sqrt{2+\frac2n} + \sqrt{2-\frac2n}}\to 4\frac{2\sqrt 3}{2\sqrt 2}=2\sqrt 6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3452352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculation of the limit $\lim_{x\to 0} (\cos x)^{1/x^2}$ without De l'Hospital/Landau's symbols/asymptotic comparison I have calculate this limit
$$\lim_{x\to 0}\ (\cos x)^{1/x^2}$$
with these steps. I have considered that:
$$(\cos x)^{1/x^2}=(\cos x -1+1)^{1/x^2}=\left(\frac{1}{\frac{1}{\cos x -1}}+1\right)^{\frac{1}{x^2}}$$
I remember that $1/(\cos x -1)$ when $x\to 0$ the limit is $\infty$. Hence
$$\left(\frac{1}{\frac{1}{\cos x -1}}+1\right)^{\frac{1}{x^2}}=\left[\left(\frac{1}{\frac{1}{\cos x -1}}+1\right)^{\frac{1}{x^2}}\right]^{\frac{\frac{1}{\cos x -1}}{\frac{1}{\cos x -1}}}=\left[\left(\frac{1}{\frac{1}{\cos x -1}}+1\right)^{\frac{1}{\cos x-1}}\right]^{\frac{\frac{1}{x^2}}{\frac{1}{\cos x -1}}} \tag{1}$$
But if I take
$$p=\frac{1}{\cos x -1}\xrightarrow{x\to 0}p\to \infty$$
therefore I consider the
$$\lim_{p\to \infty}\left(1+\frac 1p\right)^p=e$$
Consequently for the $(1)$,
$$\left(\frac{1}{\frac{1}{\cos x -1}}+1\right)^{\frac{1}{\cos x-1}}\xrightarrow{p\to \infty} e$$
and the exponent
$$\lim_{x\to 0}\frac{\frac{1}{x^2}}{\frac{1}{-(-\cos x +1)}}=-\frac 12\tag{2}$$
At the end $\displaystyle \lim_{x\to 0}\ (\cos x)^{1/x^2}=e^{-\frac 12}$.
I have followed this strategy in my classroom with my students. Is there a shorter solution to the exercise than the one I have given?
| For all $x \in \mathbb{R}$ holds
$$1 - \frac{x^2}2\le \cos x \le 1 - \frac{x^2}2 + \frac{x^4}{24}$$
so
$$\left(1 - \frac{x^2}2\right)^{1/x^2}\le (\cos x)^{1/x^2} \le \left(1 - \frac{x^2}2 + \frac{x^4}{24}\right)^{1/x^2}$$
Both bounds are easily seen to converge to $e^{-\frac12}$ when $x \to 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3453794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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Prove that $W(n,\sum_{k=1}^{n}k^{2})=2(n-1)$ Let $F_{a}(n)$ be the digit sum of $n$ in base $a$,
define $W(a,b)=F_{a}(a^{\lceil\frac{\log{b}}{\log{a}} \rceil}-b)$,
prove that $\displaystyle\ W(n,\sum_{k=1}^{n}k^{2})=2(n-1)$ if $n−1 \in 6\mathbb{N} \pm 1$.
| Let
\begin{align}
&a=n&
&b=\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}6
\end{align}
We have
$$\frac{a^3}3\leq b\leq\frac{a^3}3+a^2$$
from which
$$2<\frac{\log(b)}{\log(a)}<3$$ hence $\lceil\log(b)/\log(a)\rceil=3$ and $W(a,b)=F_a(a^3-b)$.
Moreover, if $b=d_0+d_1a+d_2a^2$ with $0\leq d_i\leq a-1$ then
$$a^3-b=(a-d_0)+(a-d_1-1)a+(a-d_2-1)a^2$$
Since $a\nmid b$, we have $d_0\neq 0$, hence
\begin{align}
W(a,b)
&=(a-d_0)+(a-d_1-1)+(a-d_2-1)\\
&=3a-(d_0+d_1+d_2)-2\\
&=3n-2-F_n(b)
\end{align}
and it remains to show $F_n(b)=n$.
If $n=6q$, then $b=q+(3q)n+(2q)n^2$, hence $F_n(b)=q+3q+2q=6q=n$.
If $n=6q+2$, then $b=(1+3q)+qn+(1+2q)n^2$, hence $F_n(b)=(1+3q)+q+(1+2q)=2+6q=n$.
| {
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"url": "https://math.stackexchange.com/questions/3456752",
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"source": "stackexchange",
"question_score": "4",
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How many seating arrangements of $m$ people in a row of $n$ seats are there if $k$ of the people must sit together? Suppose there are five people $A, B, C, D, E$ to be seated in a row of eight seats $S_1, \ldots, S_8$.
(1) How many possibilities are there if $A$ and $B$ are to sit next to each other?
(2) How many possibilities are there if $A$ and $B$ are not to sit next to each other?
My Attempt:
First Approach:
There are $8$ options of a seat for $A$. However, if $A$ is to be seated in $S_1$ or $S_8$, then $B$ can only be seated in $S_2$ or $S_7$, respectively. On the other hand, if $A$ is to be seated in any one of $S_2$ through $S_7$, then $B$ has two options of seat in each case.
Once $A$ and $B$ have been seated, there are $6$ seats left for $C$, and then $5$ seats left for $D$, and finally $4$ seats left for $E$.
In this way, the total number of seating arrangements are
$$ 2 \times 1 \times 6 \times 5 \times 4 + 6 \times 2 \times 6 \times 5 \times 4 = 240 + 1440 = 1680. $$
Is this solution correct?
Second Approach:
Now let us treat $A, B$ as one entity. Then we have four entities to be accommodated in seven spots, for which there are $7 \times 6 \times 5 \times 4 = 840$ ways. And, each one of these $840$ arrangements, the members $A$ and $B$ of the block $A, B$ can be arranged within the block in two distinct ways. Therefore there are $ 840 \times 2 = 1680$ possible seating arrangements.
Is my answer correct? If so, then are both of my approaches also correct? If not, then where are the problems.
More generally, I have the following question:
Let $k, m, n$ be any natural numbers such that $k \leq m \leq n$. Then how many ways are there in total of seating $m$ people $P_1 \ldots, P_m$ in $n$ seats $S_1, \ldots, S_n$ such that some $k$ of these people insist on sitting next to each other?
My Attempt Using the Second Approach:
Let us treat that $k$ people as one block. Then there are
$$
\begin{align}
& (n-k+1) \underbrace{(n-k) (n-k-1) \ldots }_{(m-k) \mbox{ factors}} \\
&= (n-k+1) \big[ \ (n-k) (n-k-1) \ldots \big( (n-k) - (m-k-1) \big) \ \big] \\
&= (n-k+1)(n-k) \ldots (n-m+1) \\
&= ^{n-k+1}P_{n-m} .
\end{align}
$$
Finally, corresponding to each of the above arrangements with the $k$ people considered as one block, there are $k!$ ways of arranging the $k$ people within the block amongst themselves.
Hence there are a total of
$$ k! \ ^{n-k+1}P_{n-m} $$
ways of seating $m$ people in $n$ seats with $k$ people seated next to each other.
| Both of your solutions to the first problem are correct. However, your general formula is not. Notice that in the first problem, $n = 8$, $m = 5$, and $k = 2$, so your formula gives
$$2!P(8 - 2 + 1, 8 - 5) = 2!P(7, 3) = 2! \cdot 7 \cdot 6 \cdot 5 = 2 \cdot 210 = 420$$
Let's see what went wrong.
We wish to seat $m$ people, $k$ of whom must sit consecutively, in $n$ seats. Since the block takes up $k$ of the $n$ places, it must begin in one of the first $n - (k - 1) = n - k + 1$ positions. Once the block has been placed, there are $n - k$ seats left for the remaining $m - k$ people. They can be arranged in those seats in $P(n - k, m - k)$ ways. The people within the block can be arranged in $k!$ ways, which gives us the formula
$$(n - k + 1)P(n - k, m - k)k!$$
As a sanity check, let's try our formula when $n = 8$, $k = 2$, and $m = 5$. It gives
$$(8 - 2 + 1)P(8 - 2, 5 - 2)2! = 7 \cdot P(6, 3) \cdot 2! = 7 \cdot 6 \cdot 5 \cdot 4 \cdot 2 = 1680$$
which agrees with the answer you obtained in your example.
Observe that
\begin{align*}
(n - k + 1)P(n - k, m - k) & = (n - k + 1) \cdot \frac{(n - k)!}{[(n - k) - (m - k)]!}\\
& = \frac{(n - k + 1)(n - k)!}{(n - m)!}\\
& = \frac{(n - k + 1)!}{(n - m)!}\\
& = \frac{(n - k + 1)!}{[(n - k + 1) - (m - k + 1)]!}\\
& = P(n - k + 1, m - k + 1)
\end{align*}
so we could write our formula in the form
$$P(n - k + 1, m - k + 1)k!$$
As a sanity check, note that if $n = 8$, $k = 2$, and $m = 5$, then
$$P(8 - 2 + 1, 5 - 2 + 1)2! = P(7, 4)2! = 7 \cdot 6 \cdot 5 \cdot 4 \cdot 2 = 1680$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3457074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Asymptotic evaluations of a limit: $\lim_{x\to 0}\frac{\ln(1+x^3\cos 3x)}{2x^3(1-\sqrt{1+\cos 3x)}}$ I have this simple limit
$$\lim_{x\to 0}\frac{\ln(1+x^3\cos 3x)}{2x^3(1-\sqrt{1+\cos 3x})}$$
I have solved this with these steps:
$$\frac{\ln(1+x^3\cos 3x)}{2x^3(1-\sqrt{1+\cos 3x})}=\frac{\ln(1+x^3\cos 3x)(1+\sqrt{1+\cos 3x})}{2x^3(1-\sqrt{1+\cos 3x})\cdot(1-\sqrt{1+\cos 3x})}$$
$$=-\frac 12 \frac{\ln(1+x^3\cos 3x)}{x^3\cos 3x}\cdot (1+\sqrt{1+\cos 3x})\stackrel{x\to 0}{=}-\frac{1+\sqrt 2}{2}$$
My actual problem is to use the asymptotic evaluations: if I use $$\ln(1+x^3\cos 3x)\sim (1+x^3\cos 3x)$$
and
$$\sqrt[n]{1+\psi(x)}-1\sim \frac{\psi(x)}{n}$$
where $\psi(x)=\cos 3x$, I obtain
$$\frac{\ln(1+x^3\cos 3x)}{2x^3(1-\sqrt{1+\cos 3x})}\asymp\frac{(1+x^3\cos 3x)}{2x^3\left(-\frac 12\cos 3x\right)}$$
but is $x\to 0$ this limit diverges. I do not often use asymptotic evaluations but, it may be my fatigue, but I do not find the error at the moment. Thank you for your help.
| The fact is that, $\ln(1+x^{3}\cos 3x)\approx x^{3}\cos 3x$, without the additional $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3457599",
"timestamp": "2023-03-29T00:00:00",
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Solve without L'Hopital's rule: $\lim_{x\to0}\frac{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\tan(2x)}$ Solve without L'Hopital's rule:
$$\displaystyle\lim_{x\to0}{\frac{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\tan(2x)}}$$
My work:
$\displaystyle\lim_{x\to0}{\frac{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\tan(2x)}}=\displaystyle\lim_{x\to0}{\frac{\cos{(2x)}\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\sin{(2x)}}}$
$\displaystyle\lim_{x\to0}{\frac{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\sin{2x}}}=\displaystyle\lim_{x\to0}{\frac{\cosh(3x^2)\cdot e^{8x^3}-1}{x^2\sin{(2x)}}}\cdot\displaystyle\lim_{x\to0}{\frac{1}{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}+1}}$
All of my attempts failed.
$$\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}+1\;\;\text{is continuous &decreasing }$$
Source: Matematička analiza 1, 2. kolokvij
| You need to make use of the following standard limits $$\lim_{t\to 0}\frac{e^t-1}{t}=1,\lim_{t\to 0}\frac{\tan t} {t} =1,\lim_{t\to 0}\frac{\sinh t} {t} =1,\lim_{t\to a} \frac{t^n-a^n} {t-a} =na^{n-1}$$ Using the second limit above the limit in question is equal to the limit of $$\frac{\sqrt{\cosh(3x^2)}e^{4x^3}-1}{2x^3}$$ Adding and subtracting $e^{4x^3}$ in numerator we can split the above fraction as $$e^{4x^3}\cdot\frac{\cosh^{1/2}(3x^2)-1}{2x^3}+\frac{e^{4x^3}-1}{2x^3}$$ and this can be further rewritten as $$e^{4x^3}\cdot\frac{\cosh^{1/2}(3x^2)-1}{\cosh(3x^2)-1}\cdot\frac{\cosh(3x^2)-1}{2x^3}+2\cdot\frac{e^{4x^3}-1}{4x^3}$$ The limit of above expression is equal to that of $$1\cdot\frac{1}{2}\cdot\frac{\cosh(3x^2)-1}{2x^3}+2$$ via the standard limits mentioned at the start of the answer. Next we can use the identity $$\cosh t =1+2\sinh^2(t/2)$$ to rewrite the first fraction as $$\frac{\sinh^2(3x^2/2)}{(3x^2/2)^2}\cdot \frac{(3x^2/2)^2}{2x^3}$$ which tends to $1^2\cdot 0=0$. The desired limit is thus $2$.
A combination of algebraic manipulation and standard limits is sufficient to solve most limit problems encountered in a typical calculus course.
| {
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Solve intial value problem using power series $xy''+y'+2y=0$ with y(1) =2, y'(1) =4.
What I tried:
\begin{equation}
xy'' +y'+2y = 0
\end{equation}
Let $y=\sum_{k=0}^{\infty}c_kx^k$, $y^\prime=\sum_{k=1}^{\infty}kc_kx^{k-1}$, $y^{\prime\prime}=\sum_{k=2}^{\infty}k(k-1)c_kx^{k-2}$
Then
\begin{equation}
x\sum_{k=2}^{\infty}k(k-1)c_kx^{k-2}+\sum_{k=1}^{\infty}kc_{k+1}x^{k-1}+2\sum_{k=0}^{\infty}c_kx^k=0
\end{equation}
\begin{equation}
\sum_{k=2}^{\infty}k(k+1)c_kx^{k-1}+\sum_{k=1}^{\infty}kc_kx^{k-1}+\sum_{k=0}^{\infty}2c_kx^k=0
\end{equation}
\begin{equation}
\sum_{k=0}^{\infty}k(k+1)c_{k+1}x^{k}+\sum_{k=0}^{\infty}(k+1)c_{k+1}x^{k}+\sum_{k=0}^{\infty}2kc_kx^{k}=0
\end{equation}
\begin{equation}
\sum_{k=0}^{\infty}[k(k+1)c_{k+1}+(k+1)c_{k+1}+2kc_k]x^k=0
\end{equation}
Therefore
\begin{equation}
k(k+1)c_{k+1}+(k+1)c_{k+1}+2kc_k=0 \text{ for }k\ge0
\end{equation}
So
\begin{equation}
c_{k+1}=\frac{-2k}{(k+1)^2}c_k
\end{equation}
This is how much I have tried. The answer is given to be:
$y=2+4(x-1)-4(x-1)^2+\frac{4}{5}(x-1)^3-\frac{1}{3}(x-1)^4+\frac{2}{15}(x-1)^5+.... $
How do I obtain this form. Pls help.
Question link: http://imgur.com/a/gTdCYbk
| $$
\begin {cases}
xy''+y'+2y=0, \\
y(1) =2, y'(1) =4
\end{cases}
$$
Hint
There are some mistakes in your calculations. And you should use this serie instead since you are given initial conditions at $x=1$ and not at $x=0$:
$$y=\sum_{k=0}^{\infty}c_k(x-1)^k,$$
$$ y'(x)=\sum_{k=0}^{\infty}(k+1)c_{k+1}(x-1)^{k}$$
$$ y''(x)=\sum_{k=1}^{\infty}k(k+1)c_{k+1}(x-1)^{k-1}$$
And you have to use the initial conditions. $y(1)=2$ means that $c_0=2$ and $y'(1)=4$ means that $c_1=4$.
Be carrefull because the equation becomes
$$xy''+y'+2y=0, $$
$$(x-1)y''+y''+y'+2y=0$$
I got this:
$$(k+1)^2c_{k+1}+2c_k+(k+1)(k+2)c_{k+2}=0$$
And also that
$$2c_2+c_1+2c_0=0 \implies c_2=-4$$
For $c_3$ I got $4/3$ and not $4/5$ as stated in your book.
And my serie looks like this :
$$y=2+4(x-1)-4(x-1)^2+\frac{4}{3}(x-1)^3-\frac{1}{3}(x-1)^4++....$$
| {
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"source": "stackexchange",
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Show that $\sum_{k=0}^\infty \frac{e^{-x}(k-x)^2x^k}{k!} = x$ Show that $$\sum_{k=0}^\infty \frac{e^{-x}(k-x)^2x^k}{k!} = x$$
I guess I can use Maclaurin expansion of $e^x = \sum_{k=0}^\infty \frac{x^k}{k!}$.
| Consider instead
$$\begin{align*}
\sum_{k=0}^\infty (k(k-1)-2kx+x^2)\frac{x^k}{k!}
&= \sum_{k=2}^\infty k(k-1)\frac{x^k}{k!} - 2\sum_{k=1}^\infty k\frac{x^{k+1}}{k!} + \sum_{k=0}^\infty \frac{x^{k+2}}{k!}
\\ &= \sum_{k=2}^\infty \frac{x^k}{(k-2)!} - 2\sum_{k=1}^\infty \frac{x^{k+1}}{(k-1)!} + \sum_{k=0}^\infty \frac{x^{k+2}}{k!} \\ &= 0
\end{align*}$$
Then adding $\displaystyle\sum_{k=0}^\infty k\frac{x^k}{k!}$ to both sides gives
$$\sum_{k=0}^\infty (k-x)^2\frac{x^k}{k!} = \sum_{k=0}^\infty k\frac{x^k}{k!} = \sum_{k=0}^\infty \frac{x^{k+1}}{k!} = xe^x$$
which gives the intended result by multiplying both sides by $e^{-x}$
| {
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Find the limit of $\sqrt[n]{n^2 + n}$ To find the limit I got the $\sqrt[3n]{n^2+n}$
Particularly, $\sqrt[3n]{n^2+n} \ge 1 \rightarrow \sqrt[3n]{n^2+n} = 1 + d_n$ where $d_n\ge 0$.
According to the Bernoulli's rule
$\sqrt{n^2+n} = (1+d_n)^n \ge d_n\cdot n \rightarrow d_n \le \frac{\sqrt{n^2+n}}{n}$
The $\frac{\sqrt{n^2+n}}{n} \rightarrow 1$, so $\lim d_n=1 $
So, $\lim\sqrt[n]{n^2+n} = \lim (1+d_n)^3 = \lim(1+3d_n^2+3d_n+d_n^3) =8$
However, $\sqrt[n]{n^2+n}$ tends to $1$. Where is the problem of my solution ? Can you give me a hint of how can I solve it with Bernoulli's rule ?
| Hint: This is pretty straightforward if you accept that $\sqrt[n]n\to1$. For, $\sqrt[n]{n^2+n}=\sqrt[n]n\cdot \sqrt[n]{n+1}$. Note $\sqrt[n]{n+1}\to1$ easily.
| {
"language": "en",
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Polynomial with integer coefficients: given values at point find minimal value for a point A polynomial $P(x)$ with integer coefficients satisfies the following: $P(5) = 25, P(7) = 49, P(9) = 81$. How do I find the minimal possible value of $|P(10)|$.
| We can write $p(x)=q(x)(x-5)(x-7)(x-9)+x^2$.
Now $p(10)=q(10)(10-5)(10-7)(10-9)+100=q(10) \cdot 5 \cdot 3 \cdot 1 +100=15 \cdot q(10)+100$
So we want $q(10)$ to be "as close as possible" to $-\frac{100}{15}=-6,67..$, i.e. we want $q(10)=-7$ and then $|p(10)|=|-15 \cdot 7 +100|=|-105+100|=5$. So for example take $q(x)=x-17$ and then we get the minimal value for $|p(10)|=5$.
EDIT: unicity of form for representing $p(x)$
Suppose $g(x) \neq x^2$ such that $g(5)=25, g(7)=49, g(9)=81$. Then $g(x)-x^2$ has at least three roots, namely $5,7,9$ so $g(x)-x^2=r(x)(x-5)(x-7)(x-9)$ or in other words $g(x)=r(x)(x-5)(x-7)(x-9)+x^2$. Now we have two different representations for $p(x)$, i.e. $p(x)=q(x)(x-5)(x-7)(x-9)+x^2$ and $p(x)=h(x)(x-5)(x-7)(x-9)+g(x)$ , but we can substitute $g(x)=r(x)(x-5)(x-7)(x-9)+x^2$ inside the first representation for $p(x)$ and get
$$p(x)=h(x)(x-5)(x-7)(x-9)+r(x)(x-5)(x-7)(x-9)+x^2=\\=[h(x)+r(x)](x-5)(x-7)(x-9)+x^2=q(x)(x-5)(x-7)(x-9)+x^2$$
where $q(x)=r(x)+h(x)$. So we can assume WLOG the representation $p(x)=q(x)(x-5)(x-7)(x-9)+x^2$.
| {
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Evaluate the following integral: $\int_{0}^{\pi/4} \frac{\sin\left(x\right)+\cos\left(x\right)}{9+ 16\sin\left(2x\right)} \, dx$ This is a question for the 2006 MIT Integration Bee that went unanswered by the contestants. I am not sure how to solve it either. I was only able to use the double angle formula to simplify the integral: $\sin\left({2x}\right) = 2 \sin\left(x\right) \cos\left(x\right)$
The final answer given was: $\frac{1}{20} \ln \left({3}\right)$
$$\int_{0}^{\frac{\pi}{4}} \dfrac{\sin\left(x\right)+\cos\left(x\right)}{9+ 16\sin\left(2x\right)} \, dx = $$
$$\int_{0}^{\frac{\pi}{4}} \dfrac{\sin\left(x\right)+\cos\left(x\right)}{9+ 32 \sin\left(x\right) \cos\left(x\right)} \, dx $$
| $$I=\int_{0}^{\pi/4}\frac{\sin x +\cos x}{9+16 \sin 2x} dx.$$
Use $\sin 2x=1-(\sin x- \cos x)^2$ and re-write
$$I=\int_{0}^{\pi/4} \frac{\sin x +\cos x}{25-16(\sin x -\cos x)^2} dx,$$
Now use $\sin x -\cos x=t \implies (\cos x+ \sin x ) dx=dt$, then
$$I=\frac{1}{16}\int_{-1}^{0} \frac{dt}{25/16-t^2}=\left .\frac{1}{16} \frac{4}{10}\log\frac{5/4+t}{5/4-t} \right|_{-1}^{0}= \frac{1}{40} \log 9=\frac{1}{20}\log3.$$
| {
"language": "en",
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Proving bijectivity of $f:[0,\infty)\to[0,2)$, where $f(x)=\frac{1}{x}+1$ for $x>1$, and $f(x)=x$ for $x\le 1$
Let $f: [0, \infty) \rightarrow [0,2)$ be defined by $$f(x) = \left\{
\begin{array}{ll}
\frac{1}{x} + 1, & x > 1 \\
x, & x\le 1 \\
\end{array}
\right.$$
The question asks to prove that the function is bijective.
I have to prove it is injective and surjective.
Injectivity:
if $x,y<1$, then $x=y$
if $x,y>1$ then, $\frac{1}{x}+1=\frac{1}{y}+1$, then $x=y$
I get this bit, but I don't get how to prove the surjective part of the proof.
Please could you help?
| Injectivity: Suppose $f(x) = f(y)$.
There are four cases: $0 \le x \le 1$ and $0\le y \le 1$ and $f(x) = x = f(y) = y$. Thus $x = y$.
$0 \le x \le 1$ and $y > 1$ and $f(x) = x = f(y) = \frac 1y +1$. So $x = \frac 1y + 1$. And $y= \frac 1{x-1}$. But if $y > 1$ then $0 < \frac 1y < 1$ and $1 < \frac 1y + 1< 2$ which contradicts $0\le x \le 1$. (Likewise $0\le x=\frac 1y + 1 \le 1\implies -1 \le \frac 1y \le 0\implies y < 0$ also a contradiction).
$x> 1$ and $0 \le y \le 1$. .... also gives a contradiction... Too similar to the above to bear discussing.
[Note: we could have avoided these contradictory options by noting that $0\le w \le 1 \implies 0\le w=f(w) \le 1$ and $w > 1 \implies 0<\frac 1w < 1\implies 1 < \frac 1w + 1 = f(w) < 2$. So $f(w) \in [0,1] \iff w \in [0,1]$ and $f(w) \in (1,2) \iff w\in (0,1)$.]
Fourth and final case $x > 1$ and $y > 1$. Then $\frac 1x + 1 = \frac 1y + 1$ so $\frac 1x = \frac 1y$ so $x=y$.
So $f(x)=f(y)\implies x=y$.
Surjectivity:
Let $k \in [0, 2)$. If $1 < k < 2$ and if $2>\frac 1x + 1 = k >1$ then $1> \frac 1x = k- 1 > 0$ and $x =\frac 1{k-1} > 1$.
So if $1 < k < 2$ then $\frac 1{k-1} >1$ so $f(\frac 1{k-1}) = k$.
If if $0\le k \le 1$ then $f(k) = k$.
Either way for all $k\in [0,\infty)$ there is $x$ either equal to $k$ if $k \le 1$ or equal to $\frac 1{k-1}$ if $k > 1$ so that $f(x) = k$.
| {
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Proof explanation of $\sum_{k=0}^{n-1}(n+k)(n-k) = \frac{1}{6}n(n+1)(4n-1), n \geq 1$ I don't understand the proof of
$$\sum_{k=0}^{n-1}(n+k)(n-k) = \frac{1}{6}n(n+1)(4n-1), n \geq 1$$
via induction.
$$\text{Base case}: n=1 \Rightarrow \sum_{k=0}^0(1+k)(1-k)=1=\frac{1}{6}(1+1)(4-1) = \frac{6}{6}$$
$$\text{Inductive step:} \sum_{k=0}^n (n+1+k)(n+1-k) = \sum_{k=0}^n[(n+k)(n-k)+(n-k)+(n+k)+1]$$
$$= \sum_{k=0}^n (n+k)(n-k)+\sum_{k=0}^n1+2n \cdot \sum_{k=0}^n1$$
$$=\sum_{k=0}^{n-1}(n+k)(n-k)+(n+n)(n-n)+(n+1)+2n\cdot(n+1)$$
$$= \text{(IH)} \frac{1}{6}n(n+1)(4n-1)+(n+1)+2n(n+1)$$
$$ = \frac{1}{6} (n+1)(n(4n-1)+6+12n)$$
$$= \frac{1}{6}(n+1) (4n^2+11n+6)$$
$$= \frac{1}{6}(4n+3)(n+2)$$
I don't understand the first 3 lines of the inductive step.
Why do we get
$$\sum_{k=0}^n[(n+k)(n-k)+(n-k)+(n+k)+1]$$
and
$$= \sum_{k=0}^n (n+k)(n-k)+\sum_{k=0}^n1+2n \cdot \sum_{k=0}^n1$$
$$=\sum_{k=0}^{n-1}(n+k)(n-k)+(n+n)(n-n)+(n+1)+2n\cdot(n+1)$$
| Here
$$\sum_{k=0}^n \left[(n+1+k)(n+1-k)\right] = \sum_{k=0}^n\left[(n+k)(n-k)+(n-k)+(n+k)+1\right]$$
we are using that
$$(n+1+k)(n+1-k)=((n+k)+1)((n-k)-1)=$$
$$=(n+k)(n-k)+(n-k)+(n+k)+1$$
and then since $(n+k)+(n-k)=2n$
$$\sum_{k=0}^n\left[(n+k)(n-k)+(n-k)+(n+k)+1\right]=$$
$$=\sum_{k=0}^n \left[(n+k)(n-k)\right]+$$
$$+\sum_{k=0}^n 2n+$$
$$+\sum_{k=0}^n 1$$
and for the last step, we have used that
$$\sum_{k=0}^n \left[(n+k)(n-k)\right]=\sum_{k=0}^{n-1}\left[(n+k)(n-k)\right]+(n+n)(n-n)$$
| {
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Limit $\lim\limits_{n\to\infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+2}+\frac{n}{n^2+3}+\cdots+\frac{n}{n^2+n}\right)$ $\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2}+\dfrac{n}{n^2+3}\cdots\cdots+\dfrac{n}{n^2+n}\right)$
Can we write it as following
$E=\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}\right)+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+2}\right)+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+3}\right)\cdots\cdots+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+n}\right)\tag{1}$
Let's see what happens:-
$$\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}\right)$$
$$\lim\limits_{n\to\infty}\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n^2}}=0$$
In the same way for further terms, we will get $0$
Let's also confirm for general term
$$\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+n}\right)$$
$$\lim\limits_{n\to\infty}\left(\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n}}\right)=0$$
So the whole expression $E$ will be zero
But actual answer is $1$
Let's see what happens if we evaluate the original expression $OE=\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2}+\dfrac{n}{n^2+3}\cdots\cdots+\dfrac{n}{n^2+n}\right)$
$OE=\lim\limits_{n\to\infty}\left(\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n^2}}+\dfrac{\dfrac{1}{n}}{1+\dfrac{2}{n^2}}+\dfrac{\dfrac{1}{n}}{1+\dfrac{3}{n^2}}\cdots\cdots+\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n}}\right)$
Now we can easily see that each term inside the bracket is tending to $0$, so can we say that sum of all terms upto infinity as well tends to zero?
I think we cannot because the quantity is not exactly zero, it is tending to zero, so when we add the values tending to zero upto infinity, we may not get zero.
But I got the following counter thought:-
$\lim\limits_{x\to0}\dfrac{(1+x)^\frac{1}{3}-1}{x}$
As we know $(1+x)^n=1+nx+\dfrac{n(n-1)}{2}x^2+\dfrac{n(n-1)(n-2)}{6}x^3\cdots\cdots\infty$ where $|x|<1$
$\lim\limits_{x\to0}\dfrac{\left(1+\dfrac{1}{3}x-\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{1}{2}x^2+\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{5}{3}\cdot\dfrac{1}{6}x^3\cdots\cdots\right)-1}{x}$
$\lim\limits_{x\to0}\dfrac{1}{3}-\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{1}{2}x+\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{5}{3}\cdot\dfrac{1}{6}x^2\cdots\cdots$
Now here also all the terms except $\dfrac{1}{3}$ are tending to $0$. So here also we can say that the whole quantity may not turn out to be zero as we are adding all terms upto infinity.
But surprisingly $\dfrac{1}{3}$ is the correct answer.
I am feeling very confused in these two things. Please help me.
| You can use the Squeeze Theorem, with
$$ n\cdot \frac{n}{n^2 + n} \leq a_n \leq n\cdot \frac{n}{n^2 + 1} $$
And as $ n \to \infty$, you get $ 1 \leq \lim_{n \to \infty} a_n \leq 1 $, So, using the Squeeze Theorem, $a_n \to 1$.
| {
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value of $k$ in binomial sum
If $\displaystyle (1-x)^{\frac{1}{2}}=a_{0}+a_{1}x+a_{2}x^2+\cdots \cdots +\infty.$ and $a_{0}+a_{1}+a_{2}+\cdots+a_{10}=\frac{\binom{20}{10}}{k^{10}}.$ Then $k$ is
what i try
$(1-x)^{\frac{1}{2}}=1-\frac{1}{2}x+\frac{1}{2}\bigg(\frac{1}{2}-1\bigg)\frac{x^2}{2!}+\frac{1}{2}\bigg(\frac{1}{2}-1\bigg)\bigg(\frac{1}{2}-2\bigg)\frac{x^3}{3!}+\cdots\cdots$
$\displaystyle a_{0}=1,a_{1}=-\frac{1}{2},a_{2}=-\frac{1}{2^2}\frac{1}{2!},a_{3}=\frac{1\cdot 3}{2^3}\frac{1}{3!}+\cdots$
How do i solve it Help me pleses
| The binomial expansion is
$$\sqrt{1-x}=\sum_{n=0}^\infty (-1)^n\binom{\frac{1}{2}}{n}x^n$$ and so you need to solve for $k$
$$\sum_{n=0}^{10} (-1)^n\binom{\frac{1}{2}}{n}=\frac{\binom{20}{10}}{k^{10}}$$ The result should be a rather small integer. Just use Excel and inspection.
| {
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Finding how many numbers in a given set contain a given binary pattern I came across a weird question recently in a competition, and now that the competition is complete I'm wondering how to solve similar problems (the actual competition was non-calculator and had values in place of $n_{10}$ and $k_{10}$.
Given $S = \{x_{10} | x_{10} \in \mathbb{N}, n_{10} < x_{10} < k_{10}\}$, I want to find how many $y_{10}$ exist such that $y_{10} \in S$ and $y_2$ contains the string "101"
I had no idea where to even begin this problem, so I brute-forced the solution.
What is the "proper" approach that won't take up ten minutes?
| A practical method for tests
I had a go at solving your problem by recurrence relations in 'test conditions'. This did work but I found it much easier to write out the solution by splitting up the interval into simple chunks. So for your example the working would be as follows
$\begin{vmatrix}1&1&0&0&0&1&1&1 \\1&1&0&0&0&0&0&0\\\end{vmatrix} 1 \text { number}$
$\begin{vmatrix}1&0&1&1&1&1&1&1 \\1&0&1&0&0&0&0&0\\\end{vmatrix} 2^5 \text { numbers} $
$\begin{vmatrix}1&0&0&1&1&1&1&1 \\1&0&0&0&0&0&0&0\\\end{vmatrix} 3\times4-1 $
Then we can ignore the two most significant bits
$\begin{vmatrix}1&1&1&1&1&1 \\1&1&0&0&0&0\\\end{vmatrix} 3\times4-1-4$
$\begin{vmatrix}1&0&1&1&1&1 \\1&0&1&0&0&0\\\end{vmatrix} 2^3 \text{ numbers}$
$\begin{vmatrix}1&0&0&1&1&1 \\1&0&0&1&0&0\\\end{vmatrix} 1 \text { number}$
Total=$60$.
This example is 'general' in the sense that it shows how one can deal with all three possibilities for the most significant bits:-
$\begin{vmatrix}1&1\\\end{vmatrix}$
$\begin{vmatrix}1&0&1\\\end{vmatrix}$
$\begin{vmatrix}1&0&0\\\end{vmatrix}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3475466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$ABC$ rectangle in $A$ or $C$ iff $\frac{\sin(\alpha)+\sin(\gamma)}{\sin(\beta)}=\cot\left(\frac{\beta}{2}\right)$ The triangle $ABC$ is rectangle in $A$ or $C$ if and only if $\frac{\sin(\alpha)+\sin(\gamma)}{\sin(\beta)}=\cot\left(\frac{\beta}{2}\right)$, where $\alpha$ is the angle in $A$, $\beta$ is the angle in $B$ and $\gamma$ is the angle in $C$.
I have already managed to prove the ‘only if’ direction by putting $\alpha=\frac{\pi}{2}$ respectively $\gamma=\frac{\pi}{2}$ and using trigonometric formulas. However, I’m stuck for the ‘if’ direction. How could I proceed ?
Thanks for your help !
| Rewrite
$$\frac{\sin\alpha+\sin\gamma}{\sin\beta}
=\frac{2\sin\frac{\alpha+\gamma}{2}\cos\frac{\alpha-\gamma}{2}}{2\sin\frac{\beta}{2}\cos\frac{\beta}{2}}
=\frac{\cos\frac{\alpha-\gamma}{2}}{\sin\frac{\beta}{2}}
=\cot\frac{\beta}{2}$$
or
$$\cos\frac{\alpha-\gamma}{2}=\cot\frac{\beta}{2}\sin\frac{\beta}{2}=\cos\frac{\beta}{2}$$
which leads to
$$\frac{\alpha-\gamma}{2} = \pm\frac{\beta}{2}$$
Thus, either $\alpha = \beta+\gamma = 90^\circ$, or $\gamma= \beta+\alpha = 90^\circ$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
$\int_0^\infty \frac{\sin^n x}{x^m}dx$ could be expressed via $\pi$ or $\log$ I want to show some results first (they were computed by MMA)
$$
\int_0^\infty \frac{\sin^5 x}{x^3} dx =\frac{5}{32}{\color{Red}\pi} \quad
\int_0^\infty \frac{\sin^5 x}{x^5} dx =\frac{115}{384} {\color{Red}\pi} \\
\int_0^\infty \frac{\sin^5 x}{x^2} dx =\frac{5}{16}\,{\color{Red}\log}\, \frac{27}{5} \quad
\int_0^\infty \frac{\sin^5 x}{x^4} dx =-\frac{5}{96}(27\,{\color{Red }\log } \,3-25\,{\color{Red}\log }\,5) \\
$$
and
$$
\int_0^\infty \frac{\sin^6 x}{x^4} dx =\frac{1}{8} {\color{Red}\pi} \quad
\int_0^\infty \frac{\sin^6 x}{x^6} dx =\frac{11}{40} {\color{Red}\pi} \\
\int_0^\infty \frac{\sin^6 x}{x^3} dx =\frac{3}{16}\,{\color{Red}\log}\, \frac{256}{27} \quad
\int_0^\infty \frac{\sin^6 x}{x^5} dx ={\color{Red}\log}\, \frac{3^\frac{27}{16}}{4} \\
$$
As we can see, in the integral $\displaystyle\int_0^\infty \frac{\sin^nx}{x^m}dx$
, if $n-m$ is even, the integral is expressed via $\pi$, and if $n-m$ is odd, the integral is expressed via $\log$. Amazing for me, it seems always true such as
$$\int_0^\infty \frac{\sin^8 x}{x^2} dx =\frac{5\pi}{32}$$ and
$$\int_0^\infty \frac{\sin^8 x}{x^3} dx =\frac{9}{8}\log\frac{4}{3}$$
Do we have a general method to compute $\displaystyle\int_0^\infty \frac{\sin^nx}{x^m}dx$ which implies these laws?
My attempt
This post tells us how to compute $\displaystyle\int_0^\infty \frac{\sin^n x}{x^n}dx \tag{*}$
We can compute some other $\displaystyle\int_0^\infty \frac{\sin^nx}{x^m}dx$ via $(*)$, such as via the formulas
$$\displaystyle\int_{0}^{\infty}\dfrac{\sin^3 x}{x}\,dx = \dfrac{3}{4}\int_{0}^{\infty}\dfrac{\sin x}{x}\,dx - \dfrac{1}{4}\int_{0}^{\infty}\dfrac{\sin 3x}{x}\,dx$$
and
$$\int_0^\infty \frac{\sin^2 (2x)}{x^2}dx=\int_0^\infty \frac{4\sin^2 x-4\sin^4 x}{x^2}$$
But it's complicated to compute the general cases. Could you please share some ideas of a possible method to show the law mentioned above?
| A somewhat recursive form of the integral can be derived as follows:
$$\int_0^\infty\frac{\sin^n(x)}{x^m}dx = \int_0^\infty\sin^n(x)\frac{1}{x^m}dx =$$
integrating per partes and assuming $n > m-1$ leads to
$$=\frac{n}{m-1}\int_0^\infty\frac{\sin^{n-1}(x)}{x^{m-1}}\cos(x)dx=\frac{n}{m-1}\int_0^\infty\frac{\sin^{n-2}(x)}{x^{m-2}}\frac{\sin(2x)}{2x}dx.$$
EDIT:
As mentioned also in one of the comments to OP, in the reference book Gradshteyn and Ryzhik (3.821.12),
this is further expanded to:
$$\int_0^\infty\frac{\sin^n(x)}{x^m}dx=\frac{n(n-1)}{(m-1)(m-2)}\int_0^\infty\frac{\sin^{n-2}(x)}{x^{m-2}}dx-\frac{n^2}{(m-1)(m-2)}\int_0^\infty\frac{\sin^n(x)}{x^{m-2}}dx,$$
but I am not sure how to obtain this.
Still, the answer by @SimplyBeautiful gives a better intuition why there is sometimes "$\pi$" in the expression and why "$\ln x$" in the rest.
| {
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"url": "https://math.stackexchange.com/questions/3476409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Show that $\lim\limits_{x\to 0} \frac{\sin x\sin^{-1}x-x^2}{x^6}=\frac1{18}$ Question: Show that $\lim\limits_{x\to 0} \dfrac{\sin x\sin^{-1}x-x^2}{x^6}=\dfrac{1}{18}$
My effort: $\lim\limits_{x\to 0} \dfrac{\sin x\sin^{-1}x-x^2}{x^6}=\lim\limits_{x\to 0} \dfrac{\dfrac{\sin x}{x} x\sin^{-1}x-x^2}{x^6}=\lim\limits_{x\to 0} \dfrac{\sin^{-1}x-x}{x^5}=\lim\limits_{x\to 0} \dfrac{\frac{1}{\sqrt{1-x^2}}-1}{5x^4}$ .
Is my approach correct?
| Note,
$$\sin x\sin^{-1}x=(x-\frac{x^3}6 +\frac{x^5}{120}+O(x^7)) (x+\frac{x^3}6 +\frac{3x^5}{40}+O(x^7))=x^2+\frac{x^6}{18}+O(x^8)$$
Thus,
$$\lim\limits_{x\to 0} \dfrac{\sin x\sin^{-1}x-x^2}{x^6}
=\lim\limits_{x\to 0}\dfrac1{x^6} \left(\frac{x^6}{18}+O(x^8)\right)=\dfrac{1}{18}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Show that $\lim_{x\to 0} \frac{\tan x\tan^{-1}x-x^2}{x^6}=\frac{2}{9}$ Show that $$\lim_{x\to 0} \frac{\tan x\tan^{-1}x-x^2}{x^6}=\frac{2}{9}.$$
Proceed:
$$\tan x\tan^{-1}x=(x+\dfrac{1}{3}x^3+\dfrac{2}{15}x^5+O(x^7)) (x-\frac{1}{3}x^3+\frac{1}{5}x^5+O(x^7))=x^2+\frac{2x^6}{9}+O(x^8)$$
Thus,
$$\lim\limits_{x\to 0} \dfrac{\tan x\tan^{-1}x-x^2}{x^6}
=\lim\limits_{x\to 0}\dfrac1{x^6} \left(\frac{2x^6}{9}+O(x^8)\right)=\dfrac{2}{9}$$
Where is the problem?
| $$\tan^{-1} x=x-x/^3/3-x^5/5+..., \tan x=x+x^3/3+2x^5/15+...$$
Then $$L=\lim_{x \rightarrow 0}\frac{\tan x \tan^{-1} x-x^2}{x^{6}} =\lim_{x \rightarrow 0} \frac{2x^6/9+x^8/45+...}{x^6}=\frac{2}{9.}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3477174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Limit of $\sin\left(\sqrt {x+2}\right)-\sin\left(\sqrt{x+4}\right)$ as $x\to\infty$ I wish to find the limit of $$\sin\left(\sqrt {x+2}\right)-\sin\left(\sqrt{x+4}\right)$$ as $x\to\infty$. I think that this limit does not exist.
| Hint: $\sin A -\sin B =2\cos (\frac {A+B} 2) \sin (\frac {A-B} 2)$ and hence $|\sin (\sqrt {x+2} -\sin (\sqrt {x+4}) |\leq 2 |\sin (\sqrt {x+2} - \sqrt {x+4})/2| \to 0$ since $\sqrt {x+2} - \sqrt {x+4} =\frac {-2} {\sqrt {x+2} +\sqrt {x+4}} \to 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3482503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Multiple of a number and sum of a set Lets say there is a number 30, and this sum of numbers
3 + 5 + 7 + 9 ...
Now I want to find where this sum and a multiple of 30 gets equal.
for 30 its 4 = 30 * 4 = 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21
I don't have any math experience, just wanted to know if this could be calculated by some formula.
I can imagine, that it must somehow be related to circles/sine-waves, where 30 is the length of the circle and the sum of this numbers must somehow match a multiple of it.
Maybe there is some kind of formula for this already, would like to know, thanks
PS: would be nice, if you could also put the right tags to this question
| We know that the sum of first $n$ odd numbers is $n^2$ . So the sum $$1+3+5+\cdots+(n+1) = n^2 \\ \implies 3+5+7+\cdots+(n+1)=(n+1)^2-1$$Since we want it to be a multiple of $30$ , We have the following equation:
$$(n+1)^2- 1= 30k \implies n^2+2n-30k = 0$$
Using quadratic formula , we get:
$$n = \dfrac{-2\pm\sqrt{4+120k}}2\implies n = -1+\sqrt{1+30k}$$
Since $n$ is an integer , it follows that $(1+30k)$ must be a perfect square of an odd number.
$$1+30k = (2x+1)^2 \implies k = \dfrac{2x(x+1)}{15}$$
Since $2$ is not divisible by $15$ , it follows that $x^2+x\equiv 0\mod 15$
and we get the possible values of $x \equiv 0,5,9,14 \mod 15$ $($The proof is left to the readers .$)$
Our previous equation becomes ,
$$n = -1 + 2x+1 \implies n = 2x$$
So for $x \equiv 0,5,9,14 \mod 15$ , you can get a possible solution $$(n,k) = \left(2x , \dfrac{2x(x+1)}{15}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3482807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
For which value of $k$ the equations $y=2x-5$, $y=x+2$ and $y=kx-12$ have common solution For which value of $k$ the equations $y=2x-5$, $y=x+2$ and $y=kx-12$ have common solution.
Let $G_{2x-5}$ intersects $G_{x+2}$ at point $N(x_N;y_N)$. We can get that $x_N=7$ and $y_N=9$. From here $N$ must lie on $G_{kx-12}$. Therefore, $k=\dfrac{20}{7}$.
Here are the graphs of $y=2x-5$, $y=x+2$ and $\dfrac{20}{7}x-12$. What's the problem with the drawing? Why $N$ does not lie on the third function?
After your help:
| The problem is that $k = 3$ is the proper value since $3(7) - 12 = 9$. Your $k$ value of $\frac{20}{7}$ is $\frac{1}{7}$ too small, which is why the red line is slightly to the right & below where it should be.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3483501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find maximum of $C=2(x+y+z)-xy-yz-xz$
Let $x,y,z\ge 0$ such that $x^2+y^2+z^2=3$. Find the maximum of $$C=2(x+y+z)-xy-yz-xz.$$
I tried Schur and AM-GM inequality but I really have no idea about this problem. It is not homogeneous so it's hard for me.
| Let $(x+y+z) = a$ . Then $$(x+y+z)^2 = a^2 \implies (xy+yz+zx)=\dfrac{a^2-3}{2}$$
So $$2(x+y+z)-xy-yz-xz = 2a -\dfrac{a^2-3}2$$
$$ C = \dfrac{-a^2+4a+3}{2}$$
The maximum of this quadratic is at $a = 2$ , for which the max becomes :
$$2(x+y+z)-xy-yz-xz = \dfrac 72$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3484263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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For $a$, $b$, $c$ the sides of a triangle, show $ 7(a+b+c)^3-9(a+b+c)\left(a^2+b^2+c^2\right)-108abc\ge0$
If $a$, $b$, and $c$ are the three sidelengths of an arbitrary triangle, prove that the following inequality is true, with equality for equilateral triangles.
$$ 7\left(a+b+c\right)^3-9\left(a+b+c\right)\left(a^2+b^2+c^2\right)-108abc\ge0 \tag{1}$$
In expanded form:
$$ 6\left(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b \right)-\left(a^3+b^3+c^3\right)-33abc\ge0 \tag{2}$$
This a part of an ongoing research in triangle geometry and related to solving a cubic equation.
| \begin{align}
7\left(a+b+c\right)^3-9\left(a+b+c\right)\left(a^2+b^2+c^2\right)-108abc
&\ge0
\tag{1}\label{1}
\end{align}
As @DeepSea suggested, we can replace
the expressions in terms of side lengths $a,b,c$
with equivalent in terms of semiperimeter $\rho=\tfrac12(a+b+c)$,
inradius $r$ and
circumradius $R$ of the triangle, knowing that
\begin{align}
a+b+c&=2\rho
\tag{2}\label{2}
,\\
a^2+b^2+c^2&=2(\rho^2-r^2-4rR)
\tag{3}\label{3}
,\\
abc&=4\rho\,r\,R
\tag{4}\label{4}
,
\end{align}
so \eqref{1} is becomes
\begin{align}
7(2\rho)^3-9(2\rho)\cdot2(\rho^2-r^2-4rR)-108\cdot4\rho\,r\,R
&\ge0
,\\
20\rho^3+36\rho\,r^2-288\rho\,r\,R
&\ge0
,\\
5\rho^2+9 r^2-72 rR
&\ge0
\tag{5}\label{5}
,\\
\end{align}
And ve can divide \eqref{5} by $R^2$
and consider new $\rho,r$ that correspond
to a scaled triangle with $R=1$:
\begin{align}
5\rho^2+9 r^2-72 r
&\ge0
\tag{6}\label{6}
.
\end{align}
Using
the left part of
Gerretsen's Inequality,
\begin{align}
r\,(16\,R-5\,r)&\le\rho^2
,
\end{align}
we can check if/when
\begin{align}
5\,r\,(16-5\,r)+9 r^2-72 r
&\ge0
\end{align}
instead of \eqref{6}, which simplifies to
\begin{align}
1-2\,r
&\ge0
,
\end{align}
which holds for $r\in[0,\tfrac12]$,
that is, for all valid triangles.
Hence, \eqref{1}.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
$\frac5x-\frac3y=-4$ and $\frac3{x^2}+\frac6{y^2}=\frac{11}3$
Solve the system:
$$\begin{array}{|l} \dfrac{5}{x}-\dfrac{3}{y}=-4 \\ \dfrac{3}{x^2}+\dfrac{6}{y^2}=\dfrac{11}{3} \end{array}$$
First, we have $x,y \ne 0$. If we multiply the first equation by $xy$ and the second by $3x^2y^2$, we get $5y-3x=-4xy$ and $9y^2+18x^2=11x^2y^2$. It doesn't seem I'm on the right track. What else can I try?
| HINT:
Let $\dfrac{1}{x}=a$, $\dfrac{1}{y}=b$ and solve the following system:
$$5a-3b=-4$$ and $$3a^2+6b^2=\frac{11}{3}.$$
$b=\dfrac{5a+4}{3}$ gives $$3a^2+\frac{2(5a+4)^2}{3}=\frac{11}{3}$$ or
$$59a^2+80a+21=0$$ or
$$(a+1)(59a+21)=0.$$
Can you end it now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3484812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Finding a formula for a triangle similar to Pascal's. I need to find a formula expressed in binomial coefficients of the following triangle:
$$D_n^k=\begin{cases}
n \quad\quad\quad\quad\quad\text{ if } n=k \text{ or } k=0 \\
D_{n-1}^k+D_{n-1}^{k-1} \text{ otherwise}
\end{cases}$$
This triangle is the same as the Pascal's triangle, except for the base value ($n$ instead of $1$).
| We derive a formula with the help of generating functions for
\begin{align*}
D_n^k&=D_{n-1}^{k}+D_{n-1}^{k-1}\qquad\ \ n\geq 1,k\geq 1\tag{1}\\
D_0^k&=k\qquad\qquad\qquad\qquad k\geq 1\\
D_n^0&=n\qquad\qquad\qquad\qquad n\geq 1
\end{align*}
We obtain
\begin{align*}
D(x,y)&=\sum_{n=1}^\infty\sum_{k=0}^nD_n^kx^ny^k\\
&=\sum_{n=1}^\infty D_n^0x^n+\sum_{n=1}^\infty D_n^nx^ny^n+\sum_{n=2}^\infty\sum_{k=1}^{n-1}D_n^kx^ny^k\tag{2}\\
&=\sum_{n=1}^\infty nx^n+\sum_{n=1}^\infty nx^ny^n+\sum_{n=2}^\infty\sum_{k=1}^{n-1}\left(D_{n-1}^k+D_{n-1}^{k-1}\right)x^ny^k\tag{3}\\
&=\frac{x}{(1-x)^2}+\frac{xy}{(1-xy)^2}+x\sum_{n=1}^\infty\sum_{k=1}^{n}\left(D_n^k+D_n^{k-1}\right)x^ny^k\tag{4}\\
&=\frac{x}{(1-x)^2}+\frac{xy}{(1-xy)^2}+x\sum_{n=1}^\infty\sum_{k=1}^{n}D_n^kx^ny^k\\
&\qquad+xy\sum_{n=1}^\infty\sum_{k=0}^{n-1}D_n^kx^ny^k\tag{5}\\
&=\frac{x}{(1-x)^2}+\frac{xy}{(1-xy)^2}+x\left(D(x,y)-\sum_{n=1}^\infty nx^n\right)\\
&\qquad+xy\left(D(x,y)-\sum_{n=1}^\infty nx^ny^n\right)\\
&=\frac{x}{1-x}+\frac{xy}{1-xy}+x(1+y)D(x,y)\\
\color{blue}{D(x,y)}&\color{blue}{=\frac{1}{1-x(1+y)}\left(\frac{x}{1-x}+\frac{xy}{1-xy}\right)}\tag{6}
\end{align*}
Note the relationship of $D(x,y)$ with a generating function for the binomial coefficients $\binom{n}{k}$. We have for $ n,k\geq 0$:
\begin{align*}
\frac{1}{1-x(1+y)}=\sum_{n=0}^\infty x^n(1+y)^n=\sum_{n=0}^\infty\sum_{k=0}^n\color{blue}{\binom{n}{k}}x^ny^k
\end{align*}
Comment:
*
*In (2) we split the sum in order to apply the recurrence relation (1).
*In (3) we apply the recurrence relation (1).
*In (4) we use $\sum_{n=1}^\infty nz^n=z\sum_{n=1}^\infty nz^{n-1}=z\frac{d}{dz}\frac{1}{1-z}=\frac{z}{(1-z)^2}$. We also shift the index $n$ by one.
*In (5) we split the double sum and shift the index $k$ by one at the right-most sum.
We derive a formula for $D_n^k$ from the generating function (6). It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series.
We obtain from (6) for $n\geq 1$ and $0\leq k\leq n$:
\begin{align*}
\color{blue}{D_n^k}&=[x^ny^k]\frac{1}{1-x(1+y)}\left(\frac{x}{1-x}+\frac{xy}{1-xy}\right)\\
&=[x^ny^k]\sum_{j=0}^\infty x^j(1+y)^j\sum_{q=1}^\infty x^q\\
&\qquad+[x^ny^k]\sum_{j=0}^\infty x^j(1+y)^j\sum_{q=1}^\infty x^qy^q\tag{7}\\
&=[y^k]\sum_{j=0}^n[x^{n-j}](1+y)^j\sum_{q=1}^\infty x^q\\
&\qquad+[y^k]\sum_{j=0}^n[x^{n-j}](1+y)^j\sum_{q=1}^\infty x^qy^q\tag{8}\\
&=[y^k]\sum_{j=0}^{n-1}(1+y)^j+[y^k]\sum_{j=0}^{n-1}(1+y)^jy^{n-j}\tag{9}\\
&=[y^k]\frac{(1+y)^n-1}{(1+y)-1}+[y^k]y^n\frac{\left(\frac{1+y}{y}\right)^n-1}{\left(\frac{1+y}{y}\right)-1}\tag{10}\\
&=[y^{k+1}]\left((1+y)^n-1\right)+[y^{k-1}]\left((1+y)^n-y^n\right)\tag{11}\\
&\,\,\color{blue}{=\binom{n}{k+1}+\binom{n}{k-1}}\tag{12}
\end{align*}
Note in (12) we use for integers $p,q$, with $p$ non-negative, the convention $\binom{p}{q}=0$ if $q<0$ or $p<q$. See for instance formula (5.1) in Concrete Mathematics by R.L. Graham, D.E. Knuth and O. Patashnik.
Comment:
*
*In (7) we do a geometric series expansion multiple times.
*In (8) we apply the rule $[x^p]x^qA(x)=[x^{p-q}]A(x)$. We also set the upper indices to $n$ since values $j>n$ do not contribute.
*In (9) we select the coefficient of $x^{n-j}$.
*In (10) we apply the finite geometric summation formula.
*In (11) we do some simplifications.
*In (12) we select the coefficients accordingly.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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what $a$ that makes $(a - \sqrt 2) x^2 + (a - \sqrt 2)x + a - 1$ always below $x$-axis it means$(a - \sqrt 2) < 0$ also $D < 0$
so we get $a < \sqrt 2$ and also ${(a - \sqrt 2)}^2 - 4(a - \sqrt 2)(a - 1) < 0$
Solve ${(a - \sqrt 2)}^2 - 4(a - \sqrt 2)(a - 1) < 0$
$(a - \sqrt 2)[ (a - \sqrt 2) - 4(a - 1)] < 0$
i get $\frac {4 - \sqrt 2}{3} < a < \sqrt 2$
but my answer not in the options. is there something wrong?
| Just be careful here:
$$(a - \sqrt 2)[ (a - \sqrt 2) - 4(a - 1)] < 0$$
$$(a - \sqrt 2)[ -3a +4- \sqrt 2] < 0$$
and then:
$$a<\frac{4-\sqrt{2}}{3} \quad \text{or} \quad a>\sqrt{2}$$
and after that you have to take the intesection with $a<\sqrt{2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
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