Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
solve the equation in $\Bbb C $ \begin{array}{l}{\text {Solve in } \mathbb{C}}: \\ {x^{2}+\left(\frac{x}{x+1}\right)^{2}=3} \\ {\text { my try: }} \\ {x^{2}(x+1)^{2}+x^{2}=3(x+1)^{2}} \\ {x^{2}\left(x^{2}+2 x+1\right)-3\left(x^{2}+2 x+1\right)+x^{2}=0} \\ ({x^{2}-3 )\left(x^{2}+2 x+1\right)+x^{2}=0} \\ {x^{4}+2 x^{3}-x^{2}-6 x-3=0} \\ {\text { Now what should I do? }}\end{array}
| An easier solution than expanding like you did:
$$x^2+\left(\frac{x}{x+1}\right)^2=3$$
$$\left(x-\frac{x}{x+1}\right)^2+\frac{2x^2}{x+1}=3$$
Put $\frac{x^2}{x+1}=t$. Then, $t^2+2t=3$. So, $t=1$ or $t=-3$.
This gives that either $x^2-x-1=0$ or $x^2+3x+3=0$ which can be solved easily.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3276986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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Show that $\sum_{j=0}^{2m} x^{2^nj} $ has at least $n+1$ distinct factors Show that
$\sum_{j=0}^{2m} x^{2^nj}
$
has at least $n+1$
distinct factors.
This is a generalization
(with $x=2$ and $m=1$)
of this problem
that I saw on quora:
Show that
$2^{2^{n+1}}+2^{2^n}+1
$
has at least $n+1$
distinct divisors.
Here is my solution.
I use
$x^{2^n}-1
=(x-1)\prod_{k=0}^{n-1}(x^{2^k}+1)
$,
$(x-1)\sum_{j=0}^{m-1} x^{j}
=x^m-1
$
and
$(x+1)\sum_{j=0}^{2m} (-1)^jx^{j}
=x^{2m+1}+1
$.
Let
$a_m(x)
=\sum_{j=0}^{m-1} x^{j}
$
and
$b_m(x)
=\sum_{j=0}^{m-1} (-1)^jx^{j}
=a_m(-x)
$.
Then
$(x-1)a_m(x)
=x^m-1
$
and
$(x+1)b_{2m+1}(x)
=x^{2m+1}+1
$.
$\begin{array}\\
a_{2m}(x^{2^n})
&=\sum_{j=0}^{2m} x^{2^nj}\\
&=\dfrac{(x^{2^n})^{2m+1}-1}{x^{2^n}-1}\\
&=\dfrac{(x^{2m+1})^{2^n}-1}{x^{2^n}-1}\\
&=\dfrac{(x^{2m+1}-1)\prod_{k=0}^{n-1}((x^{2m+1})^{2^k}+1)}{(x-1)\prod_{k=0}^{n-1}(x^{2^k}+1)}\\
&=\dfrac{a_{2m}(x)\prod_{k=0}^{n-1}((x^{2^k})^{2m+1}+1)}{\prod_{k=0}^{n-1}(x^{2^k}+1)}\\
&=\dfrac{a_{2m}(x)\prod_{k=0}^{n-1}((x^{2^k}+1)b_{2m}(x^{2^k}))}{\prod_{k=0}^{n-1}(x^{2^k}+1)}\\
&=a_{2m}(x)\prod_{k=0}^{n-1}b_{2m}(x^{2^k})\\
\end{array}
$
| If $n>0$, then
$$
\sum_{j=0}^{2m} x^{2^nj} = \frac{x^{2^n(2m+1)}-1}{x^{2^n}-1} = \left[\frac{x^{2^{n-1}(2m+1)}+1}{x^{2^{n-1}}+1}\right]\left[\frac{x^{2^{n-1}(2m+1)}-1}{x^{2^{n-1}}-1}\right] = \left[\sum_{j=0}^{2m} \left(-1\right)^j x^{2^{n-1}j}\right] \sum_{j=0}^{2m} x^{2^{n-1}j}.
$$
By induction,
$$
\sum_{j=0}^{2m} x^{2^nj} = \left(\sum_{j=0}^{2m}x^j\right)\prod_{k=0}^{n-1}\left[\sum_{j=0}^{2m} \left(-1\right)^j x^{2^{k}j}\right].
$$
I'm reasonably certain these polynomial factors are irreducible, so this gives the desired $n+1$ distinct factors.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3277189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Evaluating $\lim\limits_{x\to 2} \frac{\sqrt[3]{x} - \sqrt[3]{2}}{x^3 - 8}$ without l'Hospital's rule
How to evaluate the following limit?
$$
\lim\limits_{x\to 2} \dfrac{\sqrt[3]{x} - \sqrt[3]{2}}{x^3 - 8}
$$
I factored the denominator into $(x-2)(x^2+2x+4)$, but I couldn't go on from there.
| Just another way.
$$\lim\limits_{x\to 2} \dfrac{\sqrt[3]{x} - \sqrt[3]{2}}{x^3 - 8}=\lim\limits_{t\to 0}\frac{\sqrt[3]{t+2}-\sqrt[3]{2}}{t \left(t^2+6 t+12\right)} $$ Now, using the binomial expansion or Taylor series
$$\sqrt[3]{t+2}=\sqrt[3]{2}+\frac{t}{3\ 2^{2/3}}-\frac{t^2}{18\ 2^{2/3}}+O\left(t^3\right)$$
$$\frac{\sqrt[3]{t+2}-\sqrt[3]{2}}{t \left(t^2+6 t+12\right)}=\frac{\frac{t}{3\ 2^{2/3}}-\frac{t^2}{18\ 2^{2/3}}+O\left(t^3\right) }{t \left(t^2+6 t+12\right)}=\frac{\frac{1}{3\ 2^{2/3}}-\frac{t}{18\ 2^{2/3}}+O\left(t^2\right) }{12+6 t+t^2}\to \frac{1}{36\ 2^{2/3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3279312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\lfloor N/ \lfloor \sqrt{N} \rfloor^2\rfloor = 1$ for $N > 8$
Prove that $\lfloor N/ \lfloor \sqrt{N} \rfloor^2\rfloor = 1$ for $N > 8$.
This is simple to check by listing and the only exceptions are for $N = 2,3,8$. I am looking for a simple proof if possible. This problem arises from solving polynomials with constraints on the range of coefficients.
| Suppose $N = n^2 + k$ such that $0 \leq k < 2n+1$ so that we have $n^2 \leq N < (n+1)^2$. Then $\lfloor \sqrt{N} \rfloor = n$.
Now, we can simplify as below:
$$
\Bigg\lfloor \frac{N}{{\lfloor \sqrt{N} \rfloor}^2} \Bigg\rfloor
= \bigg\lfloor \frac{n^2 + k}{n^2} \bigg\rfloor
= 1 + \bigg\lfloor \frac{k}{n^2} \bigg\rfloor
$$
Since $k < 2n + 1$, we also have
$$ \frac{k}{n^2} < \frac{2n + 1}{n^2} = \frac{2}{n} + \frac{1}{n^2} $$
Thus, $1 + \Big\lfloor \frac{k}{n^2} \Big\rfloor = 1 + \Big\lfloor \frac{2}{n} \Big\rfloor$ and for this to be $1$, we need $\Big\lfloor \frac{2}{n} \Big\rfloor = 0$ i.e. $n \geq 3$ or equivalently $N \geq 9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3279469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Finding out which vectors are linearly dependent We have following vectors: $(1,2,1,3),(2,-1,3,-1),(3,1,4,2),(1,-3,1,-2)$. I'm trying to find out which of those are linearly dependent and remove them. What I did was to create a matrix: $\begin{pmatrix}
1 & 2 & 3 & 1\\
2 & -1 & 1 & -3\\
1 & 3 & 4 & 1\\
3 & -1 & 2 & -2
\end{pmatrix}$ and by solving the matrix I found out, that the last row will get removed, so they are linearly dependend. However, how do I find out which two vectors are linearly dependent? Thanks
| Just reduce it to row echelon form, keeping track of what you do. Start with $$
\begin{pmatrix}
1 & 2 & 3 & 1 &R_1\\
2 & -1 & 1 & -3 &R_2\\
1 & 3 & 4 & 1 &R_3\\
3 & -1 & 2 & -2 &R_4
\end{pmatrix}$$
Then
\begin{pmatrix}
1 & 2 & 3 & 1 &R_1\\
0 & -5 & -5 & -5 &R_2-2R_1\\
0 & 1 & 1 & 0 &R_3 -R_1\\
0 & -7 & -7 & -5 &R_4-3R_1
\end{pmatrix}
and continue in this fashion.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Algorithm for coefficients in an ascending linear sum I want to create a monotonically increasing sequence where each element of the sequence is the linear sum of a given finite set of positive irrational numbers. As a concrete example, consider the set $\{\sqrt{2}, \sqrt{3}, \sqrt{5}\}$ and the linear sum $s = \beta_1 \sqrt{2} + \beta_2 \sqrt{3} + \beta_3 \sqrt{5}$ . The sequence I want is:
$$
0 \times\sqrt{2} + 0 \times \sqrt{3} + 0 \times \sqrt{5} = 0\\
1 \times\sqrt{2} + 0 \times \sqrt{3} + 0 \times \sqrt{5} \approx 1.41\\
0 \times\sqrt{2} + 1 \times \sqrt{3} + 0 \times \sqrt{5} \approx 1.73\\
0 \times\sqrt{2} + 0 \times \sqrt{3} + 1 \times \sqrt{5} \approx 2.24\\
2 \times\sqrt{2} + 0 \times \sqrt{3} + 0 \times \sqrt{5} \approx 2.82\\
1 \times\sqrt{2} + 1 \times \sqrt{3} + 0 \times \sqrt{5} \approx 3.15\\
0 \times\sqrt{2} + 2 \times \sqrt{3} + 0 \times \sqrt{5} \approx 3.46\\
1 \times\sqrt{2} + 0 \times \sqrt{3} + 1 \times \sqrt{5} \approx 3.65\\
0 \times\sqrt{2} + 1 \times \sqrt{3} + 1 \times \sqrt{5} \approx 3.97\\
0 \times\sqrt{2} + 0 \times \sqrt{3} + 2 \times \sqrt{5} \approx 4.47\\
2 \times\sqrt{2} + 1 \times \sqrt{3} + 0 \times \sqrt{5} \approx 4.56\\
1 \times\sqrt{2} + 2 \times \sqrt{3} + 0 \times \sqrt{5} \approx 4.88,
$$
that particular sequence having been derived by trial and error!
Given a set of positive irrationals $\{n_1, n_2, \ldots , n_j\}$, is there a simple algorithm for generating successive integer $\beta_1, \beta_2, \ldots, \beta_j$, so that the sum $\beta_1 n_1 + \beta_2 n_2 + \ldots + \beta_j n_j$ forms a complete ascending sequence; that is, a sequence that doesn't skip any allowable sum.
I'm neither a mathematician nor a computer scientist but on the face of it, this problem seems to have some similar elements to the Knapsack Problem. If that's so, there might not be a guaranteed perfect algorithm other than an exhaustive search (over a limited range), but there might nonetheless be a very good (approximate) algorithm ... and I'd like to know what it is!
| Build a list of linear sums, starting from the coefficients $(0,0,0)\to0$. For every irrational, scan the list backwards to find the smallest element such that if you add that irrational to it, the sum exceeds the bottom of the list. This gives you $j$ candidates and you pick the smallest.
E.g.
Initially
$$0\sqrt2+0\sqrt3+0\sqrt5$$
and the next candidates are
$$\color{green}1\sqrt2+0\sqrt3+0\sqrt5, 0\sqrt2+\color{green}1\sqrt3+0\sqrt5,0\sqrt2+0\sqrt3+\color{green}1\sqrt5.$$
Obviously, the new list is
$$0\sqrt2+0\sqrt3+0\sqrt5\\\color{green}1\sqrt2+0\sqrt3+0\sqrt5$$
The next candidates are
$$\color{green}2\sqrt2+0\sqrt3+0\sqrt5,0\sqrt2+\color{green}1\sqrt3+0\sqrt5,0\sqrt2+0\sqrt3+\color{green}1\sqrt5.$$
The list becomes
$$0\sqrt2+0\sqrt3+0\sqrt5\\1\sqrt2+0\sqrt3+0\sqrt5\\0\sqrt2+\color{green}1\sqrt3+0\sqrt5$$
and so on.
In fact the implementation is easy: you keep a list of $j$ candidates, each being a tuple of multiplicities. Repeatedly, you pick the smallest sum and increment the multiplicity of the corresponding irrational.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3281597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
How can we check if a system of equations accepts an integer solution? I have the following system:
$$
x_1 + y_1 + z_1 = 0
$$
$$
x_2 + y_2 + z_2 = 0
$$
$$
x_1^2+y_1^2+z_1^2 = x_2^2 + y_2^2 + z_2^2
$$
$$
x_1x_2 + y_1y_2 + z_1z_2 = 0
$$
Geometrically, this is equivalent to find an orthogonal basis composed only of integer values, for a plane whose equation in the 3D space is $ x+y+z=0 $
I tried finding a solution using a python script by testing with numbers up to 50, but couldn't find any. I also tried solving it geometrically. I'm staring to think that there is not an integer solution, but I'm not sure how it could be proven. Thanks in advance for sharing your thoughts.
| Let $x_1=a$, $y_1=b$, $z_1=c$, $x_2=x$, $y_2=y$ and $z_2=z$.
Thus, $$a+b+c=x+y+z=0,$$ $$a^2+b^2+c^2=x^2+y^2+z^2$$ and $$ax+by+cz=0.$$
Hence, $$ax+by+(a+b)(x+y)=0,$$ which gives
$$(2a+b)x+(a+2b)y=0$$ and since
$$a^2+b^2+(a+b)^2=x^2+y^2+(x+y)^2,$$ we obtain:
$$a^2+ab+b^2=x^2+xy+y^2,$$ which gives
$$(a^2+ab+b^2)(2a+b)^2=y^2((a+2b)^2-(a+2b)(2a+b)+(2a+b)^2)$$ or
$$(a^2+ab+b^2)(2a+b)^2=3y^2(a^2+ab+b^2).$$
Can you end it now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3281683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Rationalize the denominator of $\frac{\sqrt{7\sqrt{3}+4\sqrt{5}}}{\sqrt{7\sqrt{3}-4\sqrt{5}}}$ I have to rationalize the denominator of $A =
\frac{\sqrt{7\sqrt{3}+4\sqrt{5}}}{\sqrt{7\sqrt{3}-4\sqrt{5}}}$. I multiplied the fraction by $\frac{\sqrt{7\sqrt{3}-4\sqrt{5}}}{\sqrt{7\sqrt{3}-4\sqrt{5}}}$, and $A=\frac{\sqrt{67}}{7\sqrt{3}-4\sqrt{5}}$. Then I multiplied by $\frac{7\sqrt{3}+4\sqrt{5}}{7\sqrt{3}+4\sqrt{5}}$, and $A=\frac{7\sqrt{201}+4\sqrt{335}}{67}$. Can someone tell me if there is a better solution and whether I am right?
| Various others have suggested mulrltiplying by $\sqrt{7\sqrt{3}\color{blue}{+}4\sqrt{5}}$ in the first step. The reason this is preferable is because you get a product with at most one square root in the denominator, which simplifies the computation; thus
$\dfrac{7\sqrt{3}+4\sqrt{5}}{\color{blue}{\sqrt{67}}}$
versus your expression
$\dfrac{\sqrt{67}}{7\sqrt{3}-4\sqrt{5}}$
The suggested alternative has a hidden advantage. Though not in this case, sometimes the remaining square root has a squared quantity in the denominator allowing a step to be saved. For instance, compare
$\dfrac{1}{\sqrt{2\sqrt{21}-4\sqrt{5}}}=\dfrac{\sqrt{2\sqrt{21}+4\sqrt{5}}}{\sqrt{4}}=\dfrac{\sqrt{2\sqrt{21}+4\sqrt{5}}}{2}$
with
$\dfrac{1}{\sqrt{2\sqrt{21}-4\sqrt{5}}}=\dfrac{\sqrt{2\sqrt{21}-4\sqrt{5}}}{2\sqrt{21}-4\sqrt{5}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3282708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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If $z + \frac{1}{z} = r$, where $r$ be a real number and $|r| < 2$. Find $|z|$. Let $r$ be a real number, $|r| < 2,$ and let $z$ be a complex number such that
$$z + \frac{1}{z} = r.$$Find $|z|.$ I am actually stuck on this problem. Any help will be appreciated. Thank you
| Multiply the equation by $z$, subtract $rz$, you get
$$
z^2 -rz + 1 = 0.
$$
Can you take it from here?
Hint: $ax^2 + bx + c = 0$ has solutions $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Edit
There are a bunch of heavy answers now. In my opinion the following solution is much more elementary. Just note that that $|z| = z \bar z$ for any complex number. Then use the quadratic formula to obtain $z = \frac{r - \sqrt{r^2 - 4}}{2}$ and $\bar z = \frac{r + \sqrt{r^2 - 4}}{2}$. Then
$$
z \bar z = \left( \frac{r - \sqrt{r^2 - 4}}{2}\right)\left( \frac{r + \sqrt{r^2 - 4}}{2}\right) = \frac{r^2 - r^2 + 4}{4} = 1.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3284501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
} |
Stability equilibrium point of non linear system with nonpositive eigenvalues. We have the following non linear system:\begin{equation}
\dot{X}=f(X),
\end{equation}
and lets suppose that $f(X_0)=0$. We also have that $\lambda_1,\dots,\lambda_n$ are the eigenvales of $D_f(X_0)$ and there are $Re(\lambda_1),\dots,Re(\lambda_k)=0$ for some $k<n$ and $Re(\lambda_{k+1}),\dots,Re(\lambda_n)<0$ .
What can we say about the stability of the system? Can we say is stable because it is not unstable?
| Here is a very simple example, a system on $\Bbb R^2$ which has a single equilibrium, a center, at $(0,0)$; the system is however unstable.
We let
$\dot x = y + x^3, \tag 1$
$\dot y = -x + y^3; \tag 2$
then
$\dot x = \dot y = 0 \Longrightarrow y = -x^3, \; x = y^3, \tag 3$
whence
$x = y^3 = (-x^3)^3 = -x^9, \tag 4$
$x^9 + x = 0, \tag 5$
$x(x^8 + 1) = 0; \tag 6$
only real solution is evidently
$x = 0, \tag 7$
since $x^8 + 1 > 0$ for all real $x$; thus, the only equilibrium of (1)-(2) is
$(x, y) = (x, -x^3) = (0, 0); \tag 8$
the Jacobian matrix of (1)-(2) is
$J(x, y) = \begin{bmatrix} 3x^2 & 1 \\ -1 & 3y^2 \end{bmatrix}, \tag 9$
whence
$J(0, 0) = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}; \tag{10}$
the characteristic polynomial of $J(0,0)$ is
$\det(J(0,0) - \lambda I) = \det \left (\begin{bmatrix} -\lambda & 1 \\ -1 & -\lambda \end{bmatrix} \right ) = \lambda^2 + 1; \tag{11}$
thus the eigenvalues of $J(0, 0)$ are $\pm i$; $(0,0)$ is a center; but with
$r^2 = x^2 + y^2, \tag{12}$
we have
$\dfrac{d}{dt}(r^2) = 2r\dot r = 2x \dot x + 2y\dot y$
$= 2xy + 2x^4 - 2yx + 2y^4 = 2(x^4 + y^4) > 0, \forall (x, y) \ne (0, 0); \tag{13}$
thus $r^2$, and hence $r$, is strictly monotonically increasing for all $r \ne 0$; the equilibrium point $(0,0)$, albeit a center, is unstable.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3287238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Finding all elements in $x\in U_{143}$ such that $x^2=1 \pmod{143}$
Find all elements in $x\in U_{143}$ such that $x^2=1 \pmod{143}$
I am kind of stuck here. I know that $143=11*13$, and perhaps looking at $U_{11},U_{13}$ will help but I am unable to find a solution so far.
| So we have $13\mid (x-1)(x+1)$ so $x=13k\pm 1$ and similary $x=11l\pm 1$
*
*if $13k+1 = 11l+1\implies 11\mid k\implies k=11s \implies x= 143s+1\implies \boxed{x=1}$
*if $13k-1 = 11l-1\implies 11\mid k\implies k=11s \implies $$x= 143s-1\implies \boxed{ x=142}$
*if $13k+1 = 11l-1\implies 11\mid 13k+2\implies 11\mid 2(k+1) \implies $ $k=11s-1\implies x=143s-12 \implies $ $
\boxed{x= 131}$
*if $13k-1 = 11l+1\implies 11\mid 13k-2\implies 11\mid 2(k-1) \implies $ $k=11s+1\implies x=143s+12 \implies \boxed{x= 12}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3292157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For positive integer numbers, so $\min\{\frac{c}{a},\frac{c}{b}\}+\lfloor\frac{c}{a}\rfloor\lfloor\frac{c}{b}\rfloor\geqq c\lfloor\frac{c}{ab}\rfloor$
Given three positive integer numbers $a, b, c$. Prove that
$$\min\left \{ \frac{c}{a}, \frac{c}{b} \right \}+ \left \lfloor \frac{c}{a} \right \rfloor\left \lfloor \frac{c}{b} \right \rfloor\geqq c\left \lfloor \frac{c}{ab} \right \rfloor$$
I think midway calculations are not easy formulas, I can't find what kind of formula transformation to solve! So I need to the help! Thanks for all the nice comments and interests from MathematicsSE-ees!
| The inequality requested to be proven is
$$\min\left( \frac{c}{a}, \frac{c}{b} \right) + \left\lfloor \frac{c}{a} \right\rfloor \left\lfloor \frac{c}{b} \right\rfloor \geqq c\left\lfloor \frac{c}{ab} \right\rfloor \tag{1}\label{eq1}$$
Since it's symmetric in $a$ and $b$, WLOG let $b \ge a$, so
$$\min\left(\frac{c}{a},\frac{c}{b}\right) = \frac{c}{b} \tag{2}\label{eq2}$$
Let
$$c = kab + d, \; k \ge 0, \; 0 \le d \lt ab \tag{3}\label{eq3}$$
The RHS of \eqref{eq1} becomes
$$c\left \lfloor \frac{c}{ab} \right \rfloor = (kab + d)(k) = abk^2 + kd \tag{4}\label{eq4}$$
Consider the case of $c \lt b$. This means $\left\lfloor \frac{c}{a} \right\rfloor \left\lfloor \frac{c}{b} \right\rfloor = 0$ and, since $a \ge 1$, also $c\left\lfloor \frac{c}{ab} \right\rfloor = 0$. Since $\frac{c}{b} \gt 0$, this means \eqref{eq1} holds with a strict inequality.
Otherwise, with $c \ge b$ (so $\frac{c}{a}$ and $\frac{c}{b}$ are each $\ge 1$, allowing the inequality below to subtract $1$ from each value and still have them be non-negative numbers), using \eqref{eq2}, the LHS of \eqref{eq1} becomes
\begin{align}
\frac{c}{b} + \left\lfloor \frac{c}{a} \right\rfloor \left\lfloor \frac{c}{b} \right\rfloor & = ka + \frac{d}{b} + \left\lfloor kb + \frac{d}{a} \right\rfloor \left\lfloor ka + \frac{d}{b} \right\rfloor \\
& \gt ka + \frac{d}{b} + \left(kb + \frac{d}{a} - 1\right)\left(ka + \frac{d}{b} - 1\right) \\
& = ka + \frac{d}{b} + abk^2 + kd - kb + kd + \frac{d^2}{ab} - \frac{d}{a} - ka - \frac{d}{b} + 1 \\
& = abk^2 + kd + \left(-kb + kd + \frac{d^2}{ab} - \frac{d}{a} + 1\right) \\
& = abk^2 + kd + \left(k(d-b) + \frac{d}{a}\left(\frac{d}{b} - 1\right) + 1\right) \tag{5}\label{eq5}
\end{align}
If $d \ge b$, then $k(d-b) + \frac{d}{a}\left(\frac{d}{b} - 1\right) + 1 \ge 1$, so the RHS of \eqref{eq5} is greater than that of \eqref{eq4}, meaning \eqref{eq1} holds with the strict inequality.
Next, consider the case where $d \lt b$. Let
$$d = ma + n, \; m \ge 0, \; 0 \le n \lt a \tag{6}\label{eq6}$$
Now, starting from the RHS of the first line of \eqref{eq5}, this gives
\begin{align}
ka + \frac{d}{b} + \left\lfloor kb + \frac{d}{a} \right\rfloor \left\lfloor ka + \frac{d}{b} \right\rfloor & = ka + \frac{d}{b} + (kb + m)(ka) \\
& \ge ka + (kb + m)(ka) \\
& = ka + abk^2 + mka \\
& = abk^2 + k(ma + a) \\
& \ge abk^2 + k(ma + n) \\
& = abk^2 + kd \tag{7}\label{eq7}
\end{align}
Once again, this shows that \eqref{eq1} holds. Note the only possibility for the first inequality in \eqref{eq7} to be an equals is if $d = 0$ since it's ignoring the $\frac{d}{b}$ term. Also, the only possibility of the second equality to be an equals is if $k = 0$ since $a \gt n$. Thus, both can be equals only if $k = d = 0$, but then this gives $c = 0$ in \eqref{eq3}, but the problem says $c \gt 0$. Thus, once again, it's a strict inequality in \eqref{eq1}.
As all possible cases have been considered, this shows \eqref{eq1} is always true, with it actually always having the LHS $\gt$ RHS.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3292489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Let $f (x$) = $\sqrt{−x^2 + 20x + 400} + \sqrt {x^2 − 20x}$. How many elements in the range of $f$ are integers?
Let $f (x$) = $\sqrt{−x^2 + 20x + 400} + \sqrt {x^2 − 20x}$. How many elements in the range of $f$ are integers?
I first let $y= x^2 -20x +100$.
Then substitute it in the function -------> $f(x) = \sqrt{-y+500}+\sqrt{y-100}$.
Which means, $100\leq{y}\leq{500}$. I then tried to find values of $y$ that would make both radicals disappear. I found four values of $y$ that made $f(x)$ integers; $100, 500, 356,$ and $244$. I also checked their discriminants and all of them were greater than $0$, which means each value of $y$ means two solutions for $x$, meaning, there are $8$ elements in the range of $f$ are integers. But the correct answer is $9$ and I can't seem to find the last one.
| One way to solve this problem is to differentiate the function. $$\frac{d}{dx}f(x) = \frac{-x+10}{\sqrt{-x^2+20x+400}} + \frac{x-10}{\sqrt{x^2-20x}}$$. This equals $0$ at $x = 10 \pm 10\sqrt{3}$. The domain of $f(x)$ is $[10-10\sqrt{5}, 0]$ $\cup$ $[20, 10+10\sqrt{5}]$. This means the six values to plug into the function are $10-10\sqrt{5}, 10-10\sqrt{3}, 0, 20, 10+10\sqrt{3}, 10+10\sqrt{5}$.
Plugging in these values finds $f(x) = 20, 20\sqrt{2}, 20, 20, 20\sqrt{2}, 20$, respectively. The minimum of these is $20$, the maximum is $20\sqrt{2}$. This means the range of the function is $[20, 20\sqrt{2}]$. Since $20\sqrt{2}$ ~ $28.28$, there are $9$ integers in the range.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3292837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the Center of Mass of a Solid Find the Center of Mass of a Solid of constant density $\rho$ bounded below by the disk $R: x^2 + y^2 \leq 4$ in the plane $z=0$ and above by the paraboloid $z = 4 - x^2 - y^2.$
(Hint: By symmetry, $\overline{x} = \overline{y} = 0,$ so you only have to compute $\overline{z}$.)
Can anyone help me set up this integral? I am having trouble figuring out where to start with this problem. Thanks
| First calculate the volume of the solid using polar coordinates
$$V = \iint_{x^2+y^2 \le 4} \int_{z = 0}^{4-x^2-y^2} dzdxdy = \iint_{x^2+y^2 \le 4} (4-x^2-y^2)\,dxdy = 2\pi \int_{r=0}^2 r(4-r^2)\,dr = 8\pi$$
Then we have
\begin{align}
z_{\text{cm}} &= \frac1V\iint_{x^2+y^2 \le 4} \int_{z = 0}^{4-x^2-y^2} z\,dzdxdy \\
&= \frac1{2V} \iint_{x^2+y^2 \le 4} (4-x^2-y^2)^2\,dxdy\\
&= \frac\pi{V} \int_{r=0}^2 r(4-r^2)^2\,dr\\
&= \frac18 \cdot \frac{32}3\\
&= \frac43
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Find triple summation rel in a closed form $S=\sum_{n=1}^{\infty}\sum_{m=1}^{n}\sum_{k=1}^{m}\frac{1}{(n+1)(k+1)(m+1)nmk}$
Evaluate $\displaystyle S=\sum_{n=1}^{\infty}\sum_{m=1}^{n}\sum_{k=1}^{m}\frac{1}{(n+1)(k+1)(m+1)nmk}$
My attempt :
Let $$A=\sum_{k=1}^{m}\frac{1}{k(k+1)}
=\sum_{k=1}^{m}\left( \frac1{k}-\frac1{k+1} \right) = \frac{m}{m+1}$$
and a second sum :
$$B=\sum_{m=1}^{n}\frac{1}{(m+1)^{2}}$$
from here how I can complete ??
| Continuing from the calculation above we have to sum:
$$S=\sum_{n=1}^\infty\frac{1}{n(n+1)}\sum_{m=1}^n\frac{1}{(m+1)^2}=\sum_{n=1}^{\infty}\frac{H_2(n+1)-1}{n(n+1)}\\=\sum_{n=1}^{\infty}[\frac{H_2(n+1)}{n}-\frac{H_2(n+1)}{n+1}]-\sum_{n=1}^{\infty}\frac{1}{n(n+1)}$$
Using the identity $H_2(n+1)=H_2(n)+\frac{1}{(n+1)^2}$
$$S=\sum_{n=1}^{\infty}[\frac{H_2(n)}{n}-\frac{H_2(n+1)}{n+1}]-1+\sum_{n=1}^{\infty}\frac{1}{n(n+1)^2}$$
And since the first sum telescopes to $1$ and the identity $\frac{1}{n(n+1)^2}=\frac{1}{n(n+1)}-\frac{1}{(n+1)^2}$ holds we obtain that:
$$S=\sum_{n=1}^{\infty}\frac{1}{n(n+1)^2}=\sum_{n=1}^{\infty}\Big[\frac{1}{n(n+1)}-\frac{1}{(n+1)^2}\Big]=1-(\zeta(2)-1)=2-\frac{\pi^2}{6}$$
| {
"language": "en",
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Proof: If $x^n = y^n$ and n is even, then $x=y$ or $x=-y$. This is a problem from Spivak's Calculus 4th ed., Chapter 1, Problem 6(d)
Proof: If $x^n = y^n$ and n is even, then $x=y$ or $x=-y$.
I tried to prove it in the following way but I'm not sure if the proof makes sense, and the author uses another method.
Proof: Let $n=k+1$, $k$ is odd such that $x^{k+1}=y^{k+1}$. This would only be possible if $x^k=y^k$ or $x^k=-y^k$. This is because, if $x^k=y^k$ and $k$ is odd, then $x=y$ (I proved this on a previous exercise). Therefore, $x^k\cdot x = y^k \cdot x \Rightarrow x^{k+1}=y^{k+1}$.
Similarly, $x^k=-y^k \Rightarrow x=-y \Rightarrow x^{k+1}=y^{k+1}$. Thus, I have proved that $x^{k+1}=y^{k+1}$ is possible only if $x^k=y^k$ or $x^k=-y^k$. As I have already proved, $x^k=-y^k \Rightarrow x=-y$ and $x^k=y^k \Rightarrow x=y $
$\therefore x^n=y^n$ and $n$ is even $\Rightarrow x=y$ or $x=-y$
| Let $n=2k$ be even and let $x^n =y^n$
We get $$x^{2k}=y^{2k}$$
Thus $$x^{2k}-y^{2k}=0$$
$$(x^2)^k-(y^2)^k=0$$
Factoring we get
$$(x^2-y^2)(x^{2k-2}+...+y^{2k-2})=0$$
That implies $$ x^2-y^2=0$$
Thus $$x=\pm y$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the minimum of a sum of certain multiplication of unit vectors Let $g_1,\dots,g_n\in\mathbb{R}^n$, and denote by $g_{ij}$ the $j$-th coordinate of $g_i$.
1) Assume that $\|g_i\|=1$ and that $g_{ii}=0$ for all $i$. I need to find:
$\displaystyle{\min_{g_1,\dots,g_n}} \sum_{i=1}^n\sum_{j\neq i} g_{ij}\cdot g_{ji}$,
and for which vectors $g_1,\dots,g_n$ this minimum is achieved?
2) Adding on the above assumptions, we also assume that $\displaystyle{\sum_{i=1}^ng_i =0}$. Now what is the minimum and for which vectors this minimum is achieved?
My try on the problem: for the first question, I think the minimum is $-2\sqrt{n-1}$ which is achieved for example by $g_1 =\begin{pmatrix} 0 \\ 1/\sqrt{n-1}\\ \vdots \\ 1/\sqrt{n-1} \end{pmatrix}, g_2=\dots =g_n = \begin{pmatrix} -1 \\ 0\\ \vdots \\ 0 \end{pmatrix}$. But couldn't find a proof that this is actually the minimum.
For the second problem, my guess is that the minimum should be zero. Another way I tried to look at the problem is by Hadamard product. Denote by $\circ$ the Hadamard product of matrices, let $G$ be a matrix with rows $g_i$, then the sum would be equal to:
$\begin{pmatrix} 1 \dots 1 \end{pmatrix} (G \circ G^\top) \begin{pmatrix} 1 \\ \vdots \\ 1 \end{pmatrix}$
so it may be enough that the extra condition of the second question makes $G\circ G^\top$ positive semi definite. Showing that would prove that the minimum is zero.
| Both answers are $-n$.
Lemma $\,$ For any $G\in\mathcal{M_{n\times n}}(\mathbb{R})$
$$\text{tr} (G^TG)+ \text{tr} (G^2) \ge 0$$
And the equality holds iff $G^T=-G$.
Proof $\,\,$ Write $G=(g_{ij})_{n\times n}$. Then we have
\begin{align}
(G^TG)_{ii} &= \sum_{j=1}^n g^2_{ji}\\
(G^2)_{ii} &= \sum_{j=1}^n g_{ij} g_{ji}\\
(GG^T)_{ii} &= \sum_{j=1}^n g^2_{ij}
\end{align}
Via the definition of trace, we have
\begin{align}
\text{tr} (G^TG)+ \text{tr} (G^2)
&=
\frac{1}{2} \text{tr}(G^TG) + \text{tr}(G^2) + \frac{1}{2}\text{tr}(GG^T)
\\&=
\sum_{i=1}^n \sum_{j=1}^n \left(\frac{1}{2}g^2_{ji} + g_{ij} g_{ji} +\frac{1}{2}g^2_{ij}\right)
\\ &= \frac{1}{2} \sum_{j=1}^n \sum_{i=1}^n (g_{ij} + g_{ji} )^2
\\ &\ge 0
\end{align}
And the equality holds iff $g_{ij}=-g_{ji}$ , i.e. $G^T=-G$.
Q.E.D.
Then we come back to the original problem. Put $G=(g_{ij})_{n\times n}$.
Note that $g_{ii}=0$ and $\|g_i\|=1$, then from the definition of trace we know
$$\sum_{i=1}^n\sum_{j\neq i} g_{ij}\cdot g_{ji}= \text{tr}(G^2)$$
$$\text{tr}(G^TG)=\sum_{i=1}^n \|g_i\| =n$$
Thus via the lemma we have
$$\sum_{i=1}^n\sum_{j\neq i} g_{ij}\cdot g_{ji} = \text{tr}(G^2)\ge -n$$
For $n$ being odd, define $g_{ij}$ by
$$
g_{ij} =
\begin{cases}
(-1)^{i+j}\frac{1}{\sqrt{n-1}} & \mbox{if } i<j \\
-g_{ji} & \mbox{if } i>j\\
0 & \mbox{if } i=j
\end{cases}
$$
For example when $n=3$ we define $G$ by
$\begin{pmatrix}
0 & -\frac{1}{\sqrt 2} & +\frac{1}{\sqrt 2} \\
+\frac{1}{\sqrt 2} & 0 & -\frac{1}{\sqrt 2} \\
-\frac{1}{\sqrt 2} & +\frac{1}{\sqrt 2} & 0
\end{pmatrix}$.
For $n \equiv 0 \mod 4$, we define $G$ by
$$
g_{ij} =
\begin{cases}
(-1)^{i+j} \sqrt\frac{2}{n} & \mbox{if } i\le \frac{n}{2} \mbox{ and } j > \frac{n}{2} \\
-g_{ji} & \mbox{if } i > \frac{n}{2} \mbox{ and } j \le \frac{n}{2}\\
0 & \mbox{other cases}
\end{cases}
$$
For $n \equiv 2 \mod 4$, we define $G$ by
$$
g_{ij} =
\begin{cases}
(-1)^{i+j} \sqrt\frac{2}{{n-2}} & \mbox{if } i\le \frac{n}{2} \mbox{ , } j > \frac{n}{2} \mbox{ and } i > j-\frac{n}{2} \\
-g_{{j-\frac{n}{2}},{i+\frac{n}{2}}} & \mbox{if } i\le \frac{n}{2} \mbox{ , } j > \frac{n}{2} \mbox{ and } i< j-\frac{n}{2} \\
-g_{ji} & \mbox{if } i > \frac{n}{2} \mbox{ and } j \le \frac{n}{2}\\
0 & \mbox{other cases}
\end{cases}
$$
For example when $n=4$ we define $G$ by $\begin{pmatrix}
0 & 0 & +\frac{1}{\sqrt 2} & -\frac{1}{\sqrt 2} \\
0 & 0 & -\frac{1}{\sqrt 2} & +\frac{1}{\sqrt 2} \\
-\frac{1}{\sqrt 2} & +\frac{1}{\sqrt 2} & 0 & 0 \\
+\frac{1}{\sqrt 2} & -\frac{1}{\sqrt 2} & 0 & 0 \\
\end{pmatrix}$,
and when $n=6$ we define $G$ by
$\begin{pmatrix}
0 & 0 & 0 & 0 & +\frac{1}{\sqrt 2} & -\frac{1}{\sqrt 2} \\
0 & 0 & 0 & -\frac{1}{\sqrt 2} & 0 & +\frac{1}{\sqrt 2} \\
0 & 0 & 0 & +\frac{1}{\sqrt 2} & -\frac{1}{\sqrt 2} & 0 \\
0 & +\frac{1}{\sqrt 2} & -\frac{1}{\sqrt 2} & 0 & 0 & 0 \\
-\frac{1}{\sqrt 2} & 0 & +\frac{1}{\sqrt 2} & 0 & 0 & 0 \\
+\frac{1}{\sqrt 2} & -\frac{1}{\sqrt 2} & 0 & 0 & 0 & 0 \\
\end{pmatrix}$.
Note that the construction satisfies both 1 and 2. Thus for $n\ge 3$, the minimums are always $-n$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is there a more elegant way to prove that $\sum_{n=1}^{\infty}{\frac{n^3+2n^2+1}{n^4-10n-7}}$ diverges?
Is there a more elegant way to prove that
$\sum_{n=1}^{\infty}{\frac{n^3+2n^2+1}{n^4-10n-7}}$ diverges?
Here's my (to my mind) messy proof using the comparison test:$$\frac{n^3+2n^2+1}{n^4-10n-7}\geq \frac{n^3}{n^4-10n-7}=\frac{1}{n-\frac{10}{n^2}-\frac{7}{n^3}}\geq \frac{1}{n} \Longleftrightarrow n\geq n-\frac{10}{n^2}-\frac{7}{n^3} \geq 0$$
since the harmonic series diverges, the aforementioned series diverges. Is there a more elegant way? Maybe a neater diverging series to compare it to? I can't use any tests beside the quotient, root and comparsion test.
Especially these rational polynomials really bother me!
| For $n>17$, $0<n^4-10n-7<n^4$ and $n^3+2n^2+1>n^3,$ so $$\frac{n^3+2n^2+1}{n^4-10n-7}>\frac{n^3}{n^4}=\frac{1}{n}$$ for $n>17.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $xyz$ if $x-\frac{1}{x}=y$, and $y-\frac{1}{y}=z$, and $z-\frac{1}{z}=x$ If
$$x-\frac{1}{x}=y, \qquad y-\frac{1}{y}=z, \qquad z-\frac{1}{z}=x$$
find the value of $xyz$.
This is how far I proceeded:
$x+y+z=z-1/z+x-1/x+y-1/y=>1/x+1/y+1/z=0
=>xy+yz+zx=0$
Also from question,
$x^2-1=xy,y^2-1=yz,z^2-1=zx$.
Adding $x^2+y^2+z^2-3=xy+yz+zx=0 =>x^2+y^2+z^2=3$ .
I am stuck here please help.
This image gives some hint, but I am unable to understand it.
| Given: $x-\frac1x=y,y-\frac1y=z,z-\frac1z=x$, you found correctly:
$$x-\frac1x+y-\frac1y+z-\frac1z=y+z+x \Rightarrow \frac1x+\frac1y+\frac1z=0 \Rightarrow xy+yz+zx=0 \ \ (1)\\
x^2-1+y^2-1+z^2-1=xy+yz+zx \Rightarrow x^2+y^2+z^2=3 \ \ (2)$$
Square each and add them all:
$$x^2+\frac1{x^2}-2+y^2+\frac1{y^2}-2+z^2+\frac1{z^2}-2=y^2+z^2+x^2 \Rightarrow \frac1{x^2}+\frac1{y^2}+\frac1{z^2}=6 \Rightarrow \\
x^2y^2+y^2z^2+z^2x^2=6x^2y^2z^2 \ \ (3)$$
Square $(1)$:
$$\underbrace{x^2y^2+y^2z^2+z^2x^2}_{6x^2y^2z^2}+2xyz(x+y+z)=0 \Rightarrow x+y+z=-3xyz \ \ (4)$$
Square $(4)$:
$$\underbrace{x^2+y^2+z^2}_{3}+2(\underbrace{xy+yz+zx}_{0})=9x^2y^2z^2 \Rightarrow 3=9x^2y^2z^2 \Rightarrow xyz=\pm \frac1{\sqrt{3}}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Value of $\lim\limits_{x\to-\infty}{(4x^2-x)^{1/2} +2x}$ The value of the $$\lim\limits_{x\to-\infty}{(4x^2-x)^{1/2} +2x}$$ is?
The answer given is $1/4$.
I rationalized and got $$\lim\limits_{x\to-\infty} \frac {-x}{|x|[(4- \frac {1}{x})^{1/2}-2]}$$ how to proceed further?
| A more elementary solution :
$$\lim_{x\to -\infty} \big( \sqrt{4x^2-x} +2x \big) =\lim_{x\to -\infty} \frac{-x}{\sqrt{4x^2-x} -2x} =\lim_{x\to \infty} \frac{x}{\sqrt{4x^2+x}+2x} =\lim_{x\to\infty} \frac{1}{\sqrt{4+\frac{1}{x}}+2} = \frac{1}{\sqrt{4}+2}=\frac{1}{4}$$
where I used that
$$\lim_{x\to -\infty} f(x)=\lim_{x\to \infty} f(-x)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3310615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve the equation $4 \sin x \cos (2x) \sin (3x) = 1$ for $0^{\circ} < x < 180^{\circ}$ I have tried using identities but none of them worked for me.
Eg: $4 \sin x(1 - 2 \sin^2 x)(3 \sin x - 4 \sin^3 x) = 1$
| Hint:
Make a partial linearisation: $\;2\sin x\sin 3x=\cos(3x-x)-\cos(3x+x)$, so
\begin{align}
4 \sin x \cos (2x) \sin (3x)&=2\cos 2x(\cos 2x-\cos 4x)=2\cos^22x-2\cos 2x\cos 4x\\
&=1+\cos 4x-2\cos 2x\cos 4x,
\end{align}
and after simplification, the equation becomes
$$\cos 4x(1-2\cos 2x)=0.$$
Can you solve this one?
| {
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Prove that a certain series grows as $O(\log n)$ The problem below is an exercise that came up during self-study. I cannot even easily convince myself of its truth.
Problem: There is a sequence of reals $1 = a_1 \leq a_2 \leq \ldots \leq a_n$ satisfying $a_{i+1}-a_i \leq \sqrt{a_i}$ for all $1 \leq i < n$. Prove that $$\sum_{i=1}^{n-1} \frac{a_{i+1}-a_i}{a_i} \leq O(\log n).$$
As a first step, letting $f(n)$ denote the LHS in the desired inequality, we can write $$f(n) = \sum_{i=1}^{n-1} \frac{1}{\sqrt{a_i}}.$$ My next thought is to show that this is less than a series that we know grows as $O(\log n)$, maybe the harmonic series, but I am stuck at this point.
| Note that $$a_{k+1} \le a_k+\sqrt{a_k} < a_k + \sqrt{a_k}+\tfrac{1}{4} = \left(\sqrt{a_k}+\tfrac{1}{2}\right)^2,$$ i.e. $\sqrt{a_{k+1}} < \sqrt{a_k}+\tfrac{1}{2}$ for all $k$. Hence, we can use induction to show that $\sqrt{a_n} \le \tfrac{n+1}{2}$ for all $n$.
Then for all $n$, we have:
\begin{align*}
\sum_{k = 1}^{n-1}\dfrac{a_{k+1}-a_k}{a_k} &\le \sum_{k = 1}^{n-1}\dfrac{a_{k+1}-a_k}{\left(\sqrt{a_{k+1}}-\tfrac{1}{2}\right)^2} & \text{since} \ \sqrt{a_{k+1}} < \sqrt{a_k}+\tfrac{1}{2}
\\
&= \sum_{k = 1}^{n-1}\int_{a_k}^{a_{k+1}}\dfrac{1}{\left(\sqrt{a_{k+1}}-\tfrac{1}{2}\right)^2}\,dx
\\
&\le \sum_{k = 1}^{n-1}\int_{a_k}^{a_{k+1}}\dfrac{1}{\left(\sqrt{x}-\tfrac{1}{2}\right)^2}\,dx & \text{since} \ \dfrac{1}{(\sqrt{x}-\tfrac{1}{2})^2} \ \text{is decreasing}
\\
&= \int_{1}^{a_n}\dfrac{1}{\left(\sqrt{x}-\tfrac{1}{2}\right)^2}\,dx
\\
&= \left[2\log\left(2\sqrt{x}-1\right) - \dfrac{2}{2\sqrt{x}-1}\right]_{1}^{a_n}
\\
&= 2\log\left(2\sqrt{a_n}-1\right) - \dfrac{2}{2\sqrt{a_n}-1} + 2
\\
&\le 2\log\left(2\sqrt{a_n}-1\right)+2
\\
&\le 2 \log n + 2. & \text{since} \ \sqrt{a_n} \le \tfrac{n+1}{2}
\end{align*}
Therefore, $\displaystyle\sum_{k = 1}^{n-1}\dfrac{a_{k+1}-a_k}{a_k} = O(\log n)$ as desired.
| {
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"timestamp": "2023-03-29T00:00:00",
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Help and verification on $1$-$1$ functions and their images $a) \space f(x) = 3x-7$ $E=\mathbb{R}$
The image of $f$ on $E$ is $\mathbb{R}$ correct?
$b) f(x) = e^{(\frac{1}{x})}$ with domain $(0, \infty)$
The image of $f$ on $E$ is $(0,\infty)$ correct?
c) I can verify that the following function is $1-1$, but I cannot find the image of the following function:
$f(x) = \tan(x) \space E= (\frac{\pi}{2},\frac{3\pi}{2})$
d) $f(x) = x^2 + 2x - 5 \space E = (-\infty,-6]$
$\frac{dy}{dx} = 2x + 2$
$f'(x) \lt 0$ on $(-\infty, -6]$ thus $f$ is decreasing on the interval. So $f$ is $1 - 1$
The image of $f$ is $[19, \infty)$
Is my proof of $1-1$ correct? what about the image?
$e) f(x) = \frac{x}{x^2+1} \space E = [-1,1]$
$\frac{dy}{dx} \space = \space \frac{x^2 + 1 - 2x^2}{(x^2+1)^2}$
$ = \frac{1-x^2}{(x^2+1)^2}$
The image of $f$ is $-\frac{1}{2},\frac{1}{2}$ is any of this correct? H0w do I show its $1 - 1$? Plugging in the domain gives 0.
f) $f(x) = 3x - \vert x \vert + \vert x - 2 \vert E = \mathbb{R}$
How do I show the image and prove that it is $1-1$?
|
$f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = 3x - 7$
You are correct that the range is $\mathbb{R}$.
You can show the function is injective (one-to-one) by showing that if $f(x_1) = f(x_2)$, then $x_1 = x_2$. Alternatively, you can show that the function is strictly increasing.
$f: (0, \infty) \to \mathbb{R}$ defined by $f(x) = e^{1/x}$
As Gerry Myerson indicated in the comments, your range is wrong.
\begin{align*}
\lim_{x \to \infty} f(x) & = \lim_{x \to \infty} e^{1/x} = \lim_{t \to 0^+} e^t = 1\\
\lim_{x \to 0^+} f(x) & = \lim_{x \to 0^+} e^{1/x} = \infty
\end{align*}
Since $f$ is continuous on $(0, \infty)$, the range of the function is $(1, \infty)$.
You can show the function is injective by demonstrating that it is strictly decreasing on its domain.
$f: \left(\dfrac{\pi}{2}, \dfrac{3\pi}{2}\right) \to \mathbb{R}$ defined by $f(x) = \tan x$
You are correct that $f$ is injective.
As for the range, one way to define the tangent function is as the point where the line containing the terminal side of the angle intersects the line $x = 1$.
As $x$ increases from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$, $f(x)$ increases from $-\infty$ to $\infty$. Since $f$ is periodic with period $\pi$, it has the same range in the interval $\left(\frac{\pi}{2}, \frac{3\pi}{2}\right)$. The graph of the tangent function on the domain
$$\bigcup_{k \in \mathbb{Z}} \left(-\frac{\pi}{2} + k\pi, \frac{\pi}{2} + k\pi\right)$$
is shown below.
$f: (-\infty, -6] \to \mathbb{R}$ defined by $f(x) = x^2 + 2x - 5$
Your solution is correct.
$f: [-1, 1] \to \mathbb{R}$ defined by $f(x) = \dfrac{x}{x^2 + 1}$
You correctly calculated the derivative. Since the function is differentiable, it is continuous. Moreover, notice that
$$f'(x) = \frac{1 - x^2}{(x^2 + 1)^2} > 0$$
in the open interval $(-1, 1)$. Hence, it is strictly increasing on the closed interval $[-1, 1]$. Therefore, the function is injective. Since $f(-1) = -1/2$ and $f(1) = 1/2$, the function has range $[-1/2, 1/2]$.
$f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = 3x - |x| + |x - 2|$
Observe that $|x| = x$ when $x \geq 0$ and that $|x| = -x$ when $x < 0$. Also, observe that $|x - 2| = x - 2$ when $x \geq 2$ and that $|x - 2| = -x + 2$ when $x < 2$. Hence,
\begin{align*}
f(x) & = \begin{cases}
3x + x - x + 2 & \text{if $x < 0$}\\
3x - x - x + 2 & \text{if $0 \leq x < 2$}\\
3x - x + x - 2 & \text{if $x \geq 2$}
\end{cases}\\
& = \begin{cases}
3x + 2 & \text{if $x < 0$}\\
x + 2 & \text{if $0 \leq x < 2$}\\
3x - 2 & \text{if $x \geq 2$}
\end{cases}
\end{align*}
To show that the function is injective, show that it is strictly increasing. As for finding the image, try graphing the piecewise function.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the following equation: $\sin x \cos x = \frac{1}{2}$ I am required to solve the following equation:
$$\sin x \cos x = \frac{1}{2}$$
My attempt:
Rewriting $\cos x$
$$\sin x \sqrt{1 - \sin^2 x} = \frac{1}{2}$$
Squaring both sides
$$\bigl(\sin x \sqrt{1 - \sin^2 x}\bigr)^2 = \bigl(\frac{1}{2}\bigr)^2$$
$$\sin^2 x (1 - \sin^2 x) = \frac{1}{4}$$
Expanding left side and multiplying both sides by 4
$$\sin^2 x - \sin^4 x = \frac{1}{4}$$
$$4\sin^2 x - 4\sin^4 x = 1$$
$$4\sin^2 x - 4\sin^4 x -1 = 0$$
Reordering left side
$$- 4\sin^4 x + 4\sin^2 x -1 = 0$$
$$4\sin^4 x - 4\sin^2 x + 1 = 0$$
Expression above can be factored as
$$(2\sin^2 x - 1)(2\sin^2 x - 1) = 0$$
$$(2\sin^2 x - 1)^2 = 0$$
It follows that
$$2\sin^2 x - 1 = 0 $$
$$\sin^2 x = \frac{1}{2} $$
$$\sin x = ± \frac{1}{\sqrt{2}} $$
So the resulting angles are: $45^{\circ},135^{\circ},225^{\circ},315^{\circ}$
Is my solution correct?
The reason why I am asking is, the author of the book used different method, and the end result he got was:
$$\sin2x = 1$$
So $2x = \sin^{-1}(1) = 90^{\circ},450$, and thus $x = 45^{\circ},225^{\circ}$
| There's a useful identity you can use.
$2\sin(x)\cos(x)=\sin(2x)$
Now, rewrite it as $\frac{\sin(2x)}{2}=\sin{x}\cos{x}$.
Swap $\frac{\sin(2x)}{2}$ with your LHS and you have:
$\frac{\sin(2x)}{2}=\frac{1}{2}$ as the following.
We are now left with $\sin(2x)=1$.
| {
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"answer_id": 5
} |
Solving $(xy)y'= x^2+3y^2$ I am having a very frustrating time with the back book that says my answer is way off but to me everything looks fine:
\begin{align*}
(xy)y'&= x^2+3y^2\\
y' &= \frac{x^2}{xy} + \frac{3y^2}{xy}\\
y' &= \frac{x}{y} + \frac{3y}{x}\\
y' &= \frac{1}{v} + 3v\\
y' &= \frac{1 + 3v^2}{v}\\
v+\frac{dv}{dx}x &= \frac{1+3v^2}{v}\\
\frac{dv}{dx}x&= \frac{1+3v^2-v^2}{v}\\
\frac{dv}{dx}x &= \frac{1+2v^2}{v}\\
\int \frac{v}{2v^2+1}\,dv &= \int\frac{1}{x}\,dx\\
u &= 2v^2+1\\
du &= 4v\,dv\\
dv &= \frac{1}{4v}\,du\\
\int \frac{v}{u} \frac{1}{4v}\,du &= \int \frac{1}{x} \,dx\\
\int \frac{1}{4u}\,du &= \ln|x| + c\\
\frac{1}{4} \int \frac{1}{u}\,du &= \ln|x| +c\\
\frac{1}{4} \ln|2v^2 + 1| &= \ln |x| + c\\
\ln|2v^2 + 1|&= 4\ln|x|+c\\
2v^2 + 1 &= e^{4\ln|x|}e^c\\
2v^2 + 1 &= Cx^4\\
2v^2 &= Cx^4\\
v^2 &= Cx^4\\
\frac{y}{x} &= \sqrt{Cx^4}\\
y &= x\sqrt{Cx^4}
\end{align*}
However the book says the answer is $x^2 + 2y^2 = Cx^6.$ I am fairly sure there are no mistakes.
| As you set
$$v=\frac{y}{x}$$
you need to substitute it back into
$$2v^2+1=Cx^4$$
which forms
$$2\Big(\frac{y}{x}\Big)^2 + 1 = Cx^4 \implies x^2+2y^2 = Cx^6$$
| {
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Finding the point inside a triangle to minimize the total distance to three vertices Let $\mathbf a, \mathbf b, \mathbf c$ be three vectors. We want to minimize the function
$$
f(\mathbf r)=|\mathbf r-\mathbf a|+|\mathbf r-\mathbf b|+|\mathbf r-\mathbf c|.
$$
Now as usual we evaluate the gradient:
$$
\nabla f=\frac{\mathbf r-\mathbf a}{|\mathbf r-\mathbf a|}+\frac{\mathbf r-\mathbf b}{|\mathbf r-\mathbf b|}+\frac{\mathbf r-\mathbf c}{|\mathbf r-\mathbf c|}=0.
$$
The sum of three unit vectors is zero, so the angle between any two of them is $\frac{2}{3}\pi$.
Question: can I find an expression of $\mathbf r$ in terms of $\mathbf a, \mathbf b, \mathbf c$? Please avoid writing the vectors into their components. I am looking forward to seeing a beautiful solution.
| A beautiful solution can indeed be derived, with all the symmetry preserved. It is suitably more involved, as expected.
Start with the zero gradient condition
$$
\frac{\mathbf r-\mathbf A}{|\mathbf r-\mathbf A|}+\frac{\mathbf r-\mathbf B}{|\mathbf r-\mathbf B|}+\frac{\mathbf r-\mathbf C}{|\mathbf r-\mathbf C|}=0
$$
which leads to
$$\mathbf r =\frac{yz\mathbf A + zx\mathbf B + xy \mathbf C}{xy+yz+zx}
$$
where the scalers are $x=|\mathbf r-\mathbf A|$, $y=|\mathbf r-\mathbf B|$ and $z=|\mathbf r-\mathbf C|$. Because of the angles all being $2\pi/3$ between them, they satisfy, according to the cosine rule,
$$
a^2=y^2+z^2+yz\\
b^2=z^2+x^2+zx\\
c^2=x^2+y^2+xy
$$
where $a$, $b$ and $c$ are the side lengths of the triangle, opposite of vertices A, B and C, respectively. Specifically, $a=|\mathbf B-\mathbf C|$, $b=|\mathbf C-\mathbf A|$ and $c=|\mathbf A-\mathbf B|$.
The joint equations above can be worked out, albeit in lengthy and tedious steps. Nonetheless, it produces the following solution
$$
\mathbf r =\frac{k_a\mathbf A + k_b\mathbf B + k_c \mathbf C}{(8/\sqrt{3})(a^2+b^2+c^2)K+32K^2}
$$
with the coefficients
$$
k_a=\left(a^2+ 4K/\sqrt{3}\right)^2 -(b^2-c^2)^2\\
k_b=\left(b^2+ 4K/\sqrt{3}\right)^2 -(c^2-a^2)^2\\
k_c=\left(c^2+ 4K/\sqrt{3}\right)^2 -(a^2-b^2)^2
$$
and $K$ representing the area of the triangle, i.e.
$$
K=\frac{1}{4}\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}.
$$
| {
"language": "en",
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"source": "stackexchange",
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Limit of several variable function using Taylor expansion I need to find the following limit:
$$ \lim_{(x,y) \to (0,0)} \frac{x^2+y^2}{1-\cos x\cos y}$$
My approach to find the limit is to first plug in the point, this doesn't work as the function is not continuous at that point. My next step is to notice that there are functions of $\cos$ of $x$ and of $y$ separately which means that I can Taylor expand and this is a very close approximation to the original function and we can therefore treat it as the actual function. So I Taylor expand both $\cos x$ and $\cos y$ to two terms to get the new limit:
$$ \lim_{(x,y) \to (0,0)} \frac{x^2+y^2}{1-((1-\frac{x^2}{2})(1-\frac{y^2}{2}))}$$
This simplifies to :
$$ \lim_{(x,y) \to (0,0)} \frac{x^2+y^2}{\frac{x^2+y^2}{2}-\frac{x^2y^2}{4} }$$
$$ \lim_{(x,y) \to (0,0)} \frac{x^2+y^2}{x^2+y^2} \frac{1}{\frac{1}{2}-\frac{x^2y^2}{4(x^2+y^2)} }$$
the term
$$ -\frac{x^2y^2}{4(x^2+y^2)} $$ goes to zero and so the limit goes to $2$. Can someone check my work please, also I Taylor expanded both $\cos x$ and $\cos y$ to the same amount of terms, does it matter to how many terms I expand is there some unwritten rule about this since I notice that it affects the outcome. Also let me know if there is an alternative approach to solve this limit and how it is justified in order for me to get a better understanding of how to approach several variable limits. Thank you for your time and help :)
| Another Taylor-approach without polar coordinates could be as follows:
Having in mind that
*
*$\frac{1}{2}\left(\cos u + \cos v \right) = \cos \frac{u+v}{2}\cdot \cos \frac{u-v}{2}$
you may apply the coordinate transformation
*
*$(\star): x= \frac{u+v}{2}, y= \frac{u-v}{2}$
So, plugging this into the given expression you get
\begin{eqnarray*}\frac{x^2+y^2}{1-\cos x\cos y}
& \stackrel{(*)}{=} & \frac{u^2+v^2}{2-(\cos u + \cos v)} \\
& \stackrel{Taylor}{=} & \frac{u^2+v^2}{2-(1-\frac{u^2}{2} + o(u^2)+ 1-\frac{v^2}{2} + o(v^2))}\\
& = & \frac{1}{\frac{1}{2} + \frac{o(u^2)+o(v^2)}{u^2+v^2}} \\
& \stackrel{(\star\star): (u,v)\to (0,0)}{\longrightarrow} & \frac{1}{\frac{1}{2} + 0} = 2
\end{eqnarray*}
The last step $(\star\star)$ is justified as follows:
$$ \left|\frac{o(u^2)+o(v^2)}{u^2+v^2} \right|\leq \left|\frac{o(u^2)}{u^2} \right| + \left|\frac{o(v^2)}{v^2} \right| \stackrel{(u,v)\to (0,0)}{\rightarrow} 0
$$
| {
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Proof: $\binom{x+y+n-1}{n} = \sum_{k=0}^{n} \binom{x+n-k-1}{n-k} \binom{y+k-1}{k}$ I wanted to prove following equation
$\binom{x+y+n-1}{n} = \sum_{k=0}^{n} \binom{x+n-k-1}{n-k} \binom{y+k-1}{k}$
Using Vandermonde's identity
$\binom{a+b}{t} = \sum_{k=0}^{t} \binom{a}{t-k} \binom{b}{k}$
whereas $a=x+n-k-1$ and $b=y+k-1$ but it doesn't add up.
Where is my mistake? How do i proof it?
| We have for our sum
$$\sum_{k=0}^n {x+n-k-1\choose n-k} {y+k-1\choose k}
\\ = [z^n] (1+z)^{x+n-1}
\sum_{k=0}^n {y+k-1\choose k} z^k (1+z)^{-k}.$$
The coefficient extractor controls the range of the sum:
$$[z^n] (1+z)^{x+n-1}
\sum_{k\ge 0} {y+k-1\choose k} z^k (1+z)^{-k}
\\ = [z^n] (1+z)^{x+n-1}
\frac{1}{(1-z/(1+z))^y}
\\ = [z^n] (1+z)^{x+n-1} (1+z)^y =
[z^n] (1+z)^{x+y+n-1} \\ = {x+y+n-1\choose n}$$
as claimed.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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What are the solutions for $ n(n+1)=p^2$ for n belongs to $N$ The following is my approach:
$ n^2+n =p^2$
$ n^2+n+\frac{1}{4} = p^2 + \frac{1}{4}$
$ (n+\frac{1}{2})^2 = p^2+\frac{1}{4}$
$ (n+\frac{1}{2}-p) (n+\frac{1}{2}+p) = \frac{1}{4}$
I am not able to proceed further from here. Any suggestion on what to do next?
| Note that $\gcd(n, n+1) = \gcd(n, 1) = 1$ do eithet $n=p^2, n+1 = 1$ or $n+1 = p^2, n = 1$; both of which are impossible.
| {
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Let a and b be natural numbers such that $2a - b, a - 2b$ and $a + b$ are all distinct squares. What is the smallest possible value of $b$? Let a and b be natural numbers such that $2a - b, a - 2b$ and $a + b$ are all distinct squares.
What is the smallest possible value of $b$?
Let, $2a-b=k^2, a-2b=p^2, a+b=q^2$.
$k^2=p^2+q^2$ after adding any of the two equations.
How to proceed further?
| If you subtract the last two you get $q^2-p^2=3b$. If you add the first and last you get $k^2+q^2=3a$. No primitive Pythagorean triangle has legs that differ by a multiple of $3$, so we need a triangle that has a common factor of $3$. The smallest such is $9-12-15$ and we find
$$3b=144-81=63\\b=21\\3a=225+144=369\\a=123$$
This is the smallest $b$ because the difference of the two legs must be at least $3$. If the shorter leg is $c$ we have $(c+3)^2-c^2=6c+9$ and $b$ will grow with the shorter leg.
| {
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Limit with polar coordinates $ \lim_{(x,y) \to (0,0)} \frac{x^3y}{x^4+y^2} = 0$ I need to prove that
$$ \lim_{(x,y) \to (0,0)} \frac{x^3y}{x^4+y^2} = 0$$
I'm using polar coordinates
$$\lim_{\rho \to 0} \frac{\rho^4\cos^3(\theta)\sin(\theta)}{\rho^2(\rho^2\cos^4(\theta)+\sin^2(\theta))} = \lim_{\rho \to 0} \frac{\rho^2\cos^3(\theta)\sin(\theta)}{\rho^2\cos^4(\theta)+\sin^2(\theta)} $$
From that point I can't find a function $g(\rho)\to 0$ for $\rho\to0$.
$$|\frac{\rho^2\cos^3(\theta)\sin(\theta)}{\rho^2\cos^4(\theta)+\sin^2(\theta)}|\leq\frac{\rho^2}{\rho^2\cos^4(\theta)+\sin^2(\theta)}$$
This is where I get stuck. Thank you in advance for your help!
| The proof with polar is messy, but I will finish it, and then provide an alternative approach.
Polar Approach:
The key is to consider the cases when $\sin^2(\theta)<\rho$ and $\sin^2(\theta)\ge\rho$.
When $\sin^2(\theta)<\rho$ we have
$$\frac{\rho^2|\cos^3(\theta)\sin(\theta)|}{\rho^2\cos^4(\theta)+\sin^2(\theta)}<\frac{\rho^3|\cos^3(\theta)|}{\rho^2\cos^4(\theta)}=\frac\rho{\sqrt{1-\sin^2(\theta)}}<\frac\rho{\sqrt{1-\rho}}\to0$$
When $\sin^2(\theta)\ge\rho$ we have
$$\frac{\rho^2|\cos^3(\theta)\sin(\theta)|}{\rho^2\cos^4(\theta)+\sin^2(\theta)}<\frac{\rho^2}{\sin^2(\theta)}\le\frac{\rho^2}\rho=\rho\to0$$
Hence the limit is $0$, independent of $\theta$.
Alternative Approach:
Polar relies on $\rho^2=x^2+y^2$, which, in this case, results in a fairly messy expression. Alternatively, we can let $\rho^2=x^4+y^2$ to match our denominator. We thus have the inequalities $\rho^2\ge x^4$ and $\rho^2\ge y^2$, or $|x|\le\sqrt\rho$ and $|y|\le\rho$ (since $\rho>0$). Then we have
$$\frac{|x^3y|}{x^4+y^2}\le\frac{(\sqrt\rho)^3\rho}{\rho^2}=\sqrt\rho\to0$$
which is a much cleaner proof.
| {
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Prove inequality with logarithms:$\log_{a}{\frac{a+b}{2}} + \log_{b}{\frac{a+b}{2}} \ge 2$ Let $ a, b \in (1, \infty)$. Prove that:
$$\log_{a}{(\frac{a+b}{2})} + \log_{b}{(\frac{a+b}{2})} \ge 2$$
I tried switching the bases on each of the logaritm but I got stuck:
$$\frac{\log_{\frac{a+b}{2}}{(ab)}}{\log_{\frac{a+b}{2}}{(a)}\log_{\frac{a+b}{2}}{(b)}}$$
| Let's set $f(x)=\dfrac 1{\ln(x)}$
We have $f''(x)=\dfrac{2+\ln(x)}{x^2\ln(x)^3}>0$ on $(1,+\infty)$, thus $f$ is convex.
The result is just an application of the convexity inequality
$$f\left(\frac{a+b}2\right)\le \frac 12f(a)+\frac 12f(b)\iff \frac 1{\ln(a)}+\frac 1{\ln(b)}\ge \frac 2{\ln(\frac{a+b}2)}$$
| {
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Integral involving nested Log I've been trying to solve this integral for a few days.
$$\int_0^{\infty}\left(\frac{1}{n}\left(t+n\right)\ln\left(\frac{t+n}{t}\right)-\ln\left(\frac{1}{n}\left(t+n\right)\ln\left(\frac{t+n}{t}\right)\right)-1\right)dt$$
For $n\gt0$.
I'm able to solve most of the integral until I got stuck trying to solve
$$\int\log\left(\log\left(\frac{t+n}{t}\right)\right)dt$$
Edit:
We first see that with a substitution we take $n$ out of the problem. Thus the integral we want to solve has a value of $0.38033\dots$ @Yuriy S has helped find an alternate form for integral. I want to contribute another alternate form that can be derived from Yuriy's form.
$$I_1=-\frac{1}{4}+\int_{0}^{\infty}\left(-\frac{e^s-1}{2e^s}+\ln\left(e^s-1\right)-\ln\left(s\right)\right)\frac{e^s}{\left(e^s-1\right)^2}ds$$
Another Update:
I discovered that
$$\begin{align}
I_1+\frac14&=-\int_x^\infty\frac{1}{t(e^t-1)}dt-\left(-\frac1x-\frac{\ln{x}}{2}+\sum_{n=2}^\infty\frac{B_n}{n!(n-1)}x^{n-1}\right)
\\&=\sum_{n=1}^\infty \text{Ei}(-xn)-\left(-\frac1x-\frac{\ln{x}}{2}+\sum_{n=2}^\infty\frac{B_n}{n!(n-1)}x^{n-1}\right)
\end{align}$$ for $0\lt x\lt 2\pi$. Here the non-integral part on the rhs is the series expansion of the integral part at $x=0$.
| We can actually find a term by term expansion of the $f(s)=s+(e^s-1)\left(\ln(e^s-1)-\ln s-1\right)$ used in YuriyS's answer. If we rearrange $f(s)$, we get $$f(x) = (x+1-e^x) + (e^x-1)(\ln(e^x-1)-\ln(x))$$
If we look at $x+1-e^x$, this has a known Taylor series (which converges for all real) of $$-\sum_{n=2}^\infty \frac{x^n}{n!}$$
We also know that $$e^x-1 = \sum_{n=1}^\infty \frac{x^n}{n!}$$ which again converges for all real.
$\ln(e^x-1)-\ln(x)$ is a bit more tricky. If we differentiate it, we get $$\frac{e^x}{e^x-1}-\frac{1}{x} = 1+\frac{1}{e^x-1}-\frac{1}{x}$$ Here we can use the fact that $\frac{x}{e^x-1} = \sum_{n=0}^\infty \frac{B_n}{n!} x^n$ where $B_n$ are the Bernoulli numbers. If we divide by $x$ and add $1-\frac{1}{x}$, we get $$1+\frac{1}{e^x-1}-\frac{1}{x} = \frac{1}{2}+\sum_{n=2}^{\infty}\frac{B_n}{n!} x^{n-1}$$ Integrating, we then have that $$\ln(e^x-1)-\ln(x) = \frac{x}{2} + \sum_{n=2}^\infty\frac{B_n}{n! \cdot n}x^n$$
We now have that $$f(x) = -\sum_{n=2}^\infty \frac{x^n}{n!} + \sum_{n=1}^\infty \frac{x^n}{n!} \cdot \left(\frac{x}{2} + \sum_{n=2}^\infty\frac{B_n}{n! \cdot n}x^n\right) = -\sum_{n=2}^\infty \frac{x^n}{n!} + \frac{x}{2} \sum_{n=1}^\infty \frac{x^n}{n!} + \sum_{n=1}^\infty \frac{x^n}{n!} \cdot \sum_{m=2}^\infty\frac{B_m}{m! \cdot m}x^m$$
$$f(x) = -\sum_{n=2}^\infty \frac{x^n}{n!} + \sum_{n=2}^\infty \frac{x^{n}}{2(n-1)!} + \sum_{n=3}^\infty \cdot \sum_{m=2}^{n-1}\frac{B_m}{(n-m)!m! \cdot m}x^n$$
Finally, we get the closed form for $a_n$ in $f(x) = \sum_{n=3}^\infty a_n x^n$ as $$a_n = \frac{n-2}{2(n!)}+\sum_{m=2}^{n-1}\frac{B_m}{(n-m)!m! \cdot m} = \sum_{m=2}^{n-1}\left(\frac{B_m}{(n-m)!m! \cdot m}+\frac{1}{2(n!)}\right)$$
Using the same $I_1 = \int_0^\infty f(s) \frac{e^s ds}{(e^s-1)^3}$ as YuriyS, we now want to find $$I_1 = \sum_{n=3}^\infty a_n \frac{n!}{2} (\zeta(n-1)-\zeta(n)) = \sum_{n=2}^\infty \left(a_{n+1}\frac{(n+1)!}{2}-a_{n}\frac{n!}{2}\right)\zeta(n)$$ I am not exactly sure what to do from here, but at least it is in the form of an infinite series instead of an integral.
Edit: As YuriyS mentioned in the comments, $a_n n!$ can be stated neatly as $b_n=\frac{n-2}{2}+\sum_{m=2}^{n-1}\frac{B_m \binom{n}{m}}{m}$. This means that $$I_1 = \sum_{n=3}^\infty \frac{b_n}{2} (\zeta(n-1)-\zeta(n)) = \sum_{n=2}^\infty \left(\frac{b_{n+1}}{2}-\frac{b_n}{2}\right)\zeta(n)$$
Edit 2: Unfortunately, these series will diverge, as mentioned in the comments, making it impossible for use in calculating $I_1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that $n^4-6n^3+11n^2-6n$ is divisible by $4$ for every integer $n$. So I am trying to refresh my memory when it comes to modular arithmetic but I am having some difficulties. The problem as states in the title is:
Show that $n^{4}-6n^{3}+11n^{2}-6n$ is divisible by $4$ för every
integer $n$.
So what I first did was reformulating the problem in terms of congruences, in the following way:
$$4\vert (n^{4}-6n^{3}+11n^{2}-6n) \iff 4\vert (n^{4}+11n^{2}-(6n^{3}+6n))\\ \hspace{6cm} \iff n^{4}+11n^{2}\equiv 6n^{3}+6n\pmod 4$$
Now, one initial question I have is if it is correct that the only cases I need to check is:
$$n^{4}+11n^{2}\equiv 0\equiv 6n^{3}+6n\pmod 4 \\ n^{4}+11n^{2}\equiv 1\equiv 6n^{3}+6n\pmod 4\\ n^{4}+11n^{2}\equiv 2\equiv 6n^{3}+6n\pmod 4\\n^{4}+11n^{2}\equiv 3\equiv 6n^{3}+6n\pmod 4,$$
since $0,1,2,3$ are the possible remainders when considering division by $4$?
If this is true, it still seams like a really tedious approach to solving the problem. So I am wondering if there is some other way of solving it?
Thanks in advance.
| Continuing your idea (which isn't quite what I would have done) you can reduce the coefficients.
$n^{4}-6n^{3}+11n^{2}-6n \equiv 0 \pmod 4 \iff$
$n^4 + 11n^2 \equiv 6n^3 + 6n \pmod 4\iff$
$n^4 - n^2 \equiv 2n^3 + 2n\pmod 4\iff$
$n^2(n-1)(n+1) \equiv 2n(n^2 + 1)\mod 4\iff$
The check if $n\equiv 0\pmod 4$ then LHS and RHS are both $\equiv 0\pmod 4$.
If $n\equiv 3\equiv -1 \pmod 4$ or $n \equiv 1\pmod 4$ then $n\equiv \pm 1\pmod 4$ and $n\mp 1 \equiv 0\pmod 4$ so LHS is $\equiv 0 \pmod 4$. And $n^2 +1\equiv 2\pmod 4$ an $2n \equiv \pm 2$ so RHS is $\equiv \pm 4 \equiv 0 \pmod 4$.
And if $n \equiv 2 \pmod 4$ then $n^2 \equiv 4 \equiv 0\pmod 4$ so LHS $\equiv 0\pmod 4$. meanwhile $2n\equiv 4\equiv 0 \pmod 4$. So RHS $\equiv 0 \pmod 4$.
So in all cases: $n^2(n+1)(n-1) \equiv 2n(n^2+1)\equiv 0\pmod 4$.
......
But I wouldn't have done it that way.
Perhaps the must informative, but most needlessly calculating way would be to factor.
$n^4 -6n^2 + 11n^2 -6n = n(n^3 - 6n^2 + 11n -6)=$
$n(n-1)(n^2 -5n^2 + 6)=n(n-1)(n-2)(n-3)$
Note: $n-3, n-2,n-1, n$ are four consecutive numbers! So ONE of them is divisible by $4$. So the product must be divisible by $4$.
That is an important lesson! And an important strategy to learn.
But ... factoring is hard.
.......
If I were to do it my way, I'd reduce
$n^4 -6n^3 +11n^2 -6n \equiv n^4 + 2n^3 -n^2+2n\pmod 4$.
For $n\equiv 0,1,2,-1\pmod n$ we have $n^2 \equiv0,1,0,1\pmod n$ and $n^3\equiv 0,1,0,-1\pmod n$ and $n^4 \equiv 0,1,0,1\equiv n^2$ we'd have
$2n \equiv 0,2,0,-2\equiv 0,2,0,2$ and $2n^3 \equiv 0,2,0,2\equiv 2n$
So we have $n^4 + 2n^3 - n^2 + 2n \equiv n^2 +2n - n^2+ 2n\equiv 4n\equiv 0*n\equiv 0\pmod 4$.
.....
Which hints at Bill Dubuque's parity argument.
If $n$ is even than $n^4, n^3, n^2, 6n$ are all divisible by $4$.
If $n$ is odd then $n\equiv \pm 1$ and $n^4-6n^3 + 11n^2 -6n\equiv 1\mp 6+11 \mp 6=12 \mp 12\equiv 0\mp 0 \equiv 0\pmod 4$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
$\lim\limits_{n \to \infty} \cos^2\left(\pi \sqrt[3]{n^3+n^2+2n}\right) $, where $n \in \mathbb{N}$ $$
\lim\limits_{n \to \infty} \cos^2\left(\pi \sqrt[3]{n^3+n^2+2n}\right)
$$
where $n \in \mathbb{N}$.
In this question, what I thought was, since $n \to \infty$ and $\cos ^2x$ is periodic , all I need is actually the fractional part of this. And it easy to say that $n+1> \sqrt[3]{n^3+n^2+2n}> n$.
But evaluation of $\sqrt[3]{n^3+n^2+2n} - n$ , is getting tricky. I'm sure there must be a short way to solve it. Can someone help me with it?
| $\lim _{n \to \infty} \cos^2(\pi \sqrt[3]{n^3+n^2+2n}) = ?
$
$(n+1/3)^3
=n^3+3(1/3)n^2+3(1/9)n+1/27
=n^3+n^2+n/3+1/27
$
so
$n^3+n^2+2n
\gt (n+1/3)^3
$.
$(n+1/3+a/n)^3
=n^3+n^2+n(3a+1/3)+O(1)
\gt n^3+n^2+2n
$
if
$3a+1/3 > 2$
or
$a > 5/9
$.
Choosing
$a = 2/3$,
$(n+1/3+2/(3n))^3
\lt n^3+n^2+2n
$
so that
$n+1/3
\lt \sqrt[3]{n^3+n^2+2n}
\lt n+1/3+2/(3n)
$
for all large enough $n$.
From this you can proceed
as some of the other answers.
| {
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"url": "https://math.stackexchange.com/questions/3321399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Question related with the Calculation.(Easy but calculation process complicated.) Though idea for solving of these problems surely simple, I tried these over and over and Whenever I did it, My answer came out differently for each attempts. :(
(The Calculation process might be little complicated.)
Please help me.
$Q-1)$
$a+b = \sqrt {2 \sqrt 3 - \sqrt 2}, a-b= \sqrt {3\sqrt2 - \sqrt 3}$
Find the $a^4 + b^4 + 3a^2b^2$
$Q-2)$
Let the given right triangle that its perimeter's length is $2a$ and length of the Hypotenuse is $x$
Then Find the range of the $x$
My attempt for the Q1
Since $(a+b)^2 = {2 \sqrt 3 - \sqrt 2}$ and $(a-b)^2 = {3\sqrt2 - \sqrt 3}$
$a^2 + b^2 = {1 \over 2}{(\sqrt3 + 2\sqrt 2)}, ab = {1 \over 4}({3 \sqrt3 - 4\sqrt 2})$
$(a^2 + b^2)^2 + a^2b^2 = {1 \over 16}{(103- 8\sqrt 6)}$ [$Not$ $10$]
It seems the the answer is incorrect I thought for the Q1. But the one who suggest this question said the answer like the below.
The Answer)
$Q1)$ $10$
$Q2)$ $2(\sqrt2 - 1)a \le x < a$
Thanks.
| Your answer to part $1$ is correct.
For part $2$, let the interior angle between the leg and hypotenuse of the right triangle be $\theta$. Clearly, $0<\theta<\pi/2$. The perimeter of the triangle$$x+x\cos\theta+x\sin\theta=2a\implies x=\frac{2a}{1+\cos\theta+\sin\theta}$$For the given range of $\theta$, you have $1<\sin\theta+\cos\theta=\sqrt2\sin\left(\frac\pi4+\theta\right)\le\sqrt2$, so you get$$\frac{2a}{1+\sqrt2}\le x<\frac{2a}{1+1}$$
| {
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"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Simplify $\sqrt{10 + \sqrt{24} + \sqrt{40} + \sqrt{60}} $ Simplify
$$ \sqrt{10 + \sqrt{24} + \sqrt{40} + \sqrt{60}} $$
Attempt:
$$ \sqrt{10 + \sqrt{24} + \sqrt{40} + \sqrt{60}} = \sqrt{10 + 2\sqrt{6} + 2\sqrt{10} + 2\sqrt{15}} = \sqrt{10 + 2(\sqrt{6} + \sqrt{10} + \sqrt{15})} $$
let $X =\sqrt{10 + \sqrt{24} + \sqrt{40} + \sqrt{60}}$, then
$$ X^{2} = 10 + 2(\sqrt{6} + \sqrt{10} + \sqrt{15}) $$
How to continue?
| From $6 = 2\times 3$, $10 = 2\times 5$, $15 = 3 \times 5$, observe that $10 + 2(\sqrt6 + 2\sqrt{10} + 2\sqrt{15}) = (\sqrt2+\sqrt3+\sqrt5)^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Rate of change with $a(t)=\frac{1}{t+4}$ on $[9,9+h]$ I am working on an exercise to find the rate of change between points $[9, 9+h]$ with the function $a(t)=\frac{1}{t+4}$.
The solution provided is $\frac{-1}{13(13+h)}$ whereas I arrive at $\frac{\frac{1}{h}}{h}$.
My working:
$a(t_1)$ = $\frac{1}{9+4}$ = $\frac{1}{13}.$
$x(t_2)$ = $\frac{1}{9+h+4}$ = $\frac{1}{13+h}.$
The rate of change is: $\frac{a(t_2)-a(t_1)}{t_2-t_1}.$
So: $\dfrac{\frac{1}{13+h}-\frac{1}{13}}{9+h-9}$ = $\dfrac{\frac{1}{13}+\frac{1}{h}-\frac{1}{13}}{h}$ = $\dfrac{\frac{1}{h}}{h}.$
Where did I go wrong and how can I arrive at $\frac{-1}{13(13+h)}$?
| You made the critical mistake of writing
$$\frac1{13+h}=\frac1{13}+\frac1h$$
when this is in fact completely wrong. Starting from the last correct step:
$$\frac{\frac1{13+h}-\frac1{13}}{9+h-9}=\frac{\frac{13-(13+h)}{13(13+h)}}h$$
$$=\frac{-h}{13h(13+h)}=\frac{-1}{13(13+h)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3323150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Integration of $ \int x^{2} \sqrt{2x-6} dx $
$$ \int x^{2} \sqrt{2x-6} dx = ?$$
My Attempt:
by partial integration
$$ \int x^{2} \sqrt{2x-6} dx = \frac{x^{2} (2x-6)^{3/2}}{3}- \frac{2}{3} \int x(2x-6)^{3/2}dx$$
continuing partial integration
$$ = \frac{x^{2} (2x-6)^{3/2}}{3}- \frac{2}{3} \left[ \frac{x(2x-6)^{5/2}}{5} - \int \frac{(2x-6)^{5/2}}{5} dx\right] $$
$$ = (x^{2}/3)(2x-6)^{3/2} - (2x/15)(2x-6)^{5/2} + (2/105)(2x-6)^{7/2} + C$$
Is this the correct and best/simplest answer?
Strangely, the multiple choices only include answers in the form:
$$ A(2x+6)^{7/2} + B(2x+6)^{5/2} + C(2x+6)^{3/2} + D$$
where $A,B,C,D$ are constants.
| Your answer is correct. Compare it with WA answer:
$$= (x^{2}/3)(2x-6)^{3/2} - (2x/15)(2x-6)^{5/2} + (2/105)(2x-6)^{7/2} + C=\\
(2x-6)^{3/2}\left(\frac{x^2}{3}-\frac{2x(2x-6)}{15}+\frac{2(2x-6)^2}{105}\right)+C=\\
(2x-6)^{3/2}\cdot \frac{35x^2-28x^2+84x+8x^2-48x+72}{105}+C=\\
(2x-6)^{3/2}\cdot \frac{5x^2+12x+24}{35}+C=\\
2\sqrt{2}(x-3)^{3/2}\cdot \frac{5x^2+12x+24}{35}+C.$$
I don't think it can be transformed to the expected multiple choices with $2x+6$ as the binomial is under square root. So, I assume there is a typo in the source.
You can avoid the hard methods and calculate as follows:
$$\int x^{2} \sqrt{2x-6} dx = \frac14\int (2x)^{2} \sqrt{2x-6} dx = \frac14\int (2x-6+6)^{2} \sqrt{2x-6} dx = \\
\frac14\int (2x-6)^{2.5}dx+\frac14\int 12(2x-6)^{1.5}dx+\frac14\int 36(2x-6)^{0.5}dx=\\
\frac1{28}(2x-6)^{3.5}+\frac35(2x-6)^{2.5}+3(2x-6)^{1.5}+C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3323783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Find 3×3 orthogonal matrix satisfying the following conditions. Let $A$ be 3×3 orthogonal real matrix.
For $\mathbf{u}$ $= \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \mathbf{v}$ $= \begin{pmatrix} 0 \\ 0 \\ \sqrt3 \end{pmatrix}, \mathbf{w}$ $= \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}$,
$Au=v, Aw=-w$ and $detA=-1$
Find $A$.
I noted that $u, v, w$ are linearly independent. And I write
$A\begin{pmatrix}
1 & 0 & 1 \\
1 & 0 & -1 \\
1 & \sqrt3 & 0
\end{pmatrix}=\begin{pmatrix}
0 & a & -1 \\
0 & b & 1 \\
\sqrt3 & c & 0
\end{pmatrix}.$
Then $A=\begin{pmatrix}
0 & a & -1 \\
0 & b & 1 \\
\sqrt3 & c & 0
\end{pmatrix}\begin{pmatrix}
\frac{1}{2} & \frac{1}{2} & 0 \\
-\frac{1}{2\sqrt3} & -\frac{1}{2\sqrt3} & \frac{1}{\sqrt3} \\
\frac{1}{2} & -\frac{1}{2} & 0
\end{pmatrix}$
Now, how can I find $a, b, c$??
Am I doing this right?
If there is a better solution, please help me.
| Hint
As $A$ is an orthogonal matrix it preserves the inner product. Therefore $\Vert Av\Vert = \sqrt 3$. Also $\langle Av, Aw\rangle =\langle v, w\rangle=0$.
Finally use the equation $\det A =-1$.
This provides 3 equations to find $a,b,c$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
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How to find the derivative of $x{^2/3} - (x - 1)^{1/3}$ by the limit calculation? I had to solve an exercise and was going to ask for help here, but as I typed, I spotted an error myself. I felt sorry to throw away all that MathJax stuff I typed for more than an hour, so I decided to post the question anyway, since it still can be useful to someone. If someone has any suggestions, corrections, I would be very grateful.
This is the task itself, which I'd like to quote, at least for search engines to index, if someone else has this problem.
Graph the curves in Exercises 39-48.
a. Where do the graphs appear to have vertical tangent lines?
b. Confirm your findings in part (a) with limit calculations …
The problem itself:
I can't solve exercise 45: $y = x^{2/3} - (x - 1)^{1/3}$.
I would be grateful if someone told me how to solve this exactly, but any help or hint will be greatly appreciated.
I need to find the derivative of this function, but by limit calculations. I know, the derivative should be $\frac{2}{3x^{1/3}} - \frac{1}{3(x- 1)^{2/3}}$, but I can't find it through the limit:
$$
\lim_{h \to 0} \frac{(x + h)^{2/3} - (x - 1 + h)^{1/3} - x^{2/3} + (x- 1)^{1/3}}{h}
$$
Tried approach 1:
I tried to split the fraction in four ones, multiply both the nominator and denominator by a power of nominator necessary to get rid of the cubic root, like:
$$
\frac{x + h}{h(x + h)^{1/3}}
$$
But this didn't help, since dangling 'h's were still remaining in the expressions and I needed to combine the four fractions into a single one again, thus multiplying the nominators by cubic roots again.
Tried approach 2:
This one actually worked, and I'll post it as an answer.
| Let's split the limit fraction in two ones:
$$
\lim_{h \to 0} \frac{(x + h)^{2/3} - (x - 1 + h)^{1/3} - x^{2/3} + (x- 1)^{1/3}}{h} = \\
\lim_{h \to 0} (\frac{(x + h)^{2/3} - x^{2/3}}{h} - \frac{(x- 1)^{1/3} - (x - 1 + h)^{1/3} }{h})
$$
For simplicity let's deal with the fractions without the limit notation:
$$
\frac{(x + h)^{2/3}- x^{2/3}}{h} \text{let's call it the left part} - \frac{(x- 1)^{1/3} - (x - 1 + h)^{1/3}}{h} \text{let's call it the right part}
$$
I multiplied both the left and right parts to make the nominators become differences of the cubes (to get rid of the cubic roots).
The left part turned to be this:
$$
\require{cancel}
\frac{(x + h)^2 - x^2}{h((x + h)^{4/3} + x^{2/3}(x + h)^{2/3} + x^{4/3})} = \\
\frac{\cancel{x^2} + 2\cancel{h}x + \cancelto{h}{h^2} - \cancel{x^2}}{\cancel{h}((x + h)^{4/3} + x^{2/3}(x + h)^{2/3} + x^{4/3})} = \\
\frac{2x + h}{(x + h)^{4/3} + x^{2/3}(x + h)^{2/3} + x^{4/3}}
$$
the right part:
$$
\require{cancel}
\frac{\cancel{x} - \cancel{1} - \cancel{x} + \cancel{1} - \cancelto{1}{h}}{\cancel{h}((x - 1)^{2/3} + (x - 1)^{1/3}(x - 1 + h)^{1/3} + (x - 1 + h)^{2/3})} = \\
\frac{- 1}{(x - 1)^{2/3} + (x - 1)^{1/3}(x - 1 + h)^{1/3} + (x - 1 + h)^{2/3}}
$$
Since the h tends to 0, the limit is:
$$
\require{cancel}
\lim_{h \to 0}(\frac{2x + \cancelto{0}{h}}{(x + \cancelto{0}{h})^{4/3} + x^{2/3}(x + \cancelto{0}{h})^{2/3} + x^{4/3}} - \frac{1}{(x - 1)^{2/3} + (x - 1)^{1/3}(x - 1 + \cancelto{0}{h})^{1/3} + (x - 1 + \cancelto{0}{h})^{2/3}}) = \\
\frac{2x}{x^{4/3} + x^{2/3}x^{2/3} + x^{4/3}} - \frac{1}{(x - 1)^{2/3} + (x - 1)^{1/3}(x - 1)^{1/3} + (x - 1)^{2/3}} = \\
\frac{2\cancel{x}}{3\cancelto{x^{1/3}}{x^{4/3}}} - \frac{1}{3(x - 1)^{2/3}} = \\
\frac{2}{3x^{1/3}} - \frac{1}{3(x - 1)^{2/3}}
$$
| {
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"url": "https://math.stackexchange.com/questions/3326436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Compute the inverse of a matrix with partially known values I have an odd problem. I have a matrix $A$ of size 16 per 16. As an example, we will consider the 4 by 4 matrix below.
$$A = \begin{pmatrix}50 & 3 & 10 & 2\\\ 3 & 60 & 7 & 1\\\ 10 & 7 & 55 & 4\\\ 2 & 1 & 4 & 45 \end{pmatrix}$$
I want to find the matrix $B$ such as:
$$B = \begin{pmatrix}b_{11} & b_{12} & b_{13} & 0\\\ b_{21} & b_{22} & 0 & 0\\\ b_{31} & 0 & b_{33} & b_{34}\\\ 0 & 0 & b_{43} & b_{44} \end{pmatrix}$$
where I forced some of the values $b_{ij}$ to $0$ and where $C = B^{-1}$ such that $C_{ij}$ is equal to (or as close as possible) to $A_{ij}$ if $B_{ij} \neq 0$.
In other words, if I inverse the matrix $B$, I want to get the same values as $A$ where in the position $(i, j)$ $B$ is not set to $0$.
Additionally, the matrix $A$ and the matrix $B$ are symmetric. Moreover, the diagonal values $A_{ii}$ are about 10 times larger than the off-diagonal values $A_{ij}$.
Is there a way to solve this problem?
N.B: If anyone has a better title for this post... I'm open to suggestions :)
| We have $7$ equations in $7$ variables $b_{11},b_{12},b_{13},b_{22},b_{23},b_{33},b_{34}$, which have, for example, a solution
$$
B=\begin{pmatrix}
\frac{27497}{1321025} & - \frac{1}{997} & - \frac{1}{265} & 0 \\\
- \frac{1}{997} & \frac{50}{2991} & 0 & 0 \\\
- \frac{1}{265} & 0 & \frac{136093}{7167985} & - \frac{4}{2459} \\\
0 & 0 & - \frac{4}{2459} & \frac{55}{2459}
\end{pmatrix},
$$
where $B^{-1}$ has the required form. In fact, we have
$$
B=\begin{pmatrix}
50 & 3 & 10 & \frac{8}{11} \\\
3 & 60 & \frac{3}{5} & \frac{12}{275} \\\
10 & \frac{3}{5} & 55 & 4 \\\
\frac{8}{11} & \frac{12}{275} & 4 & 45
\end{pmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3328911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to flip all fractions in the power series for $\ln(1 + x)$? I am trying to evaluate this using power series:
$$1 + \frac{2}{2} + \frac{3}{2^2} + \frac{4}{2^3} + \dots$$
By using the power series for $\ln(1 + x)$, I have recognized that dividing through by $x$ and setting $x = -2$ will get you this:
$$1 + \frac{2}{2} + \frac{2^2}{3} + \frac{2^3}{4} + ..$$
This seems so close, but I can't seem to figure out how to flip each fraction so that it matches. How can I do this?
If I am on the completely wrong path and this is a coincidence, please point me in the right direction.
| We have that
$$\sum_{n=1}^\infty \frac{n}{2^{n-1}}=1 + \frac{2}{2} + \frac{3}{2^2} + \frac{4}{2^3} + \dots$$
where
\begin{align*}
\sum_{n=1}^\infty \frac{n}{2^{n-1}}
= \sum_{n=0}^\infty \frac{n+1}{2^n}
= \left( \sum_{n=0}^\infty \frac{n}{2^n}
+ \sum_{n=0}^\infty \frac{1}{2^n} \right)
&= \sum_{n=1}^\infty \frac{n}{2^n} + 2\tag{1}
\end{align*}
then by the geometric series we have that
$$
f(x)=\sum_{n=0}^\infty x^n \;\; =\;\; \frac{1}{1-x}.
$$
provided $|x|<1$ . Thus
$$xf'(x)=\sum_{n=1}^{\infty} nx^n=\frac{x}{(1-x)^2}$$
where we let $x=\frac{1}{2}$ to find
$$\sum_{n=1}^\infty \frac{n}{2^n}=2\tag{2}$$
therefore by $(1)$ and $(2)$ we find that
$$\sum_{n=1}^\infty \frac{n}{2^{n-1}}=4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3329042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $a, b, c$ are the sidelengths of a triangle, show that $a^2b(a−b) +b^2c(b−c)+c^2a(c−a)\ge0$. After many days of work and some help with some helpful Math Stack Exchange community members, I have only one inequality homework question which remains unsolved:
If $a, b, c$ are the sidelengths of a triangle, show that $a^2b(a−b) +b^2c(b−c)+c^2a(c−a)\ge0$.
My attempt:
Let $a=y+z, b=z+x, c=x+y$. Then $x,y,z\ge0$.
But after substitute into the inequality, and expand, I still cannot use Muirhead.
This question is one of the starred question and I can't do it.
Can someone help me? Any help is appreciated!
| Let $c=\min\{a,b,c\}$.
Thus, we need to prove that
$$\frac{a^2}{c}+\frac{b^2}{a}+\frac{c^2}{b}\geq\frac{a^2b^2+a^2c^2+b^2c^2}{abc}$$ or
$$\frac{a^2}{b}+\frac{b^2}{a}-a-b+\frac{a^2}{c}-\frac{a^2}{b}+\frac{c^2}{b}-c\geq\frac{a^2b^2+a^2c^2+b^2c^2-abc(a+b+c)}{abc}$$ or
$$\frac{(a-b)^2(a+b)}{ab}+\frac{(c^2-a^2)(c-b)}{bc}\geq\frac{(ac-bc)^2+(bc-ab)(ac-ab)}{abc}$$ or
$$\frac{(a-b)^2(a+b)}{ab}+\frac{(c-a)(c-b)(a+c)}{bc}\geq\frac{(a-b)^2c}{ab}+\frac{(c-a)(c-b)}{c}.$$
Can you end it now?
| {
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"url": "https://math.stackexchange.com/questions/3331907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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A nice Combinatorial Identity I am trying to show that $\forall N\in\mathbb{N}$,
$$\sum\limits_{n=0}^{N}\sum\limits_{k=0}^{N}\frac{\left(-1\right)^{n+k}}{n+k+1}{N\choose n}{N\choose k}{N+n\choose n}{N+k\choose k}=\frac{1}{2N+1}$$
It's backed by numerical verifications, but I can't come up with a proof.
So far, I tried using the generating function of $\left(\frac{1}{2N+1}\right)_{N\in\mathbb{N}}$, which is $\frac{\arctan\left(\sqrt{x}\right)}{\sqrt{x}}$, by showing that the LHS has the same generating function, but this calculation doesn't seem to lead me anywhere...
Any suggestion ?
Edit: the comment of bof (below this question) actually leads to a very simple proof.
Indeed, from bof's comment we have that the LHS is equal to $$\int_{0}^{1}\left(\sum\limits_{k=0}^{N}(-1)^k{N\choose k}{N+k\choose k}x^k\right)^2dx$$
And we recognize here the shifted Legendre Polynomials $\widetilde{P_N}(x)=\displaystyle\sum\limits_{k=0}^{N}(-1)^k{N\choose k}{N+k\choose k}x^k$.
And we know that the shifted Legendre Polynomials form a family of orthogonal polynomials with respect to the inner product $\langle f|g\rangle=\displaystyle\int_{0}^{1}f(x)g(x)dx$, and that their squared norm with respect to this product is $\langle\widetilde{P_n}|\widetilde{P_n}\rangle=\frac{1}{2n+1}$;
so this basically provides the desired result immediately.
| We seek to verify that
$$\sum_{n=0}^N \sum_{k=0}^N
\frac{(-1)^{n+k}}{n+k+1}
{N\choose n} {N\choose k}
{N+n\choose n} {N+k\choose k}
= \frac{1}{2N+1}.$$
Now we have
$${N\choose k} {N+n\choose n}
= \frac{(N+n)!}{(N-k)! \times k! \times n!}
= {N+n\choose n+k} {n+k\choose k}.$$
We get for the LHS
$$\sum_{n=0}^N \sum_{k=0}^N
\frac{(-1)^{n+k}}{n+k+1} {N+n\choose n+k}
{N\choose n} {N+k\choose k} {n+k\choose k}
\\ = \sum_{n=0}^N \frac{1}{N+n+1} \sum_{k=0}^N
(-1)^{n+k} {N+n+1\choose n+k+1}
{N\choose n} {N+k\choose k} {n+k\choose k}
\\ = \sum_{n=0}^N \frac{1}{N+n+1} {N\choose n}
\sum_{k=0}^N
(-1)^{n+k} {N+n+1\choose N-k}
{N+k\choose k} {n+k\choose k}
\\ = \sum_{n=0}^N \frac{1}{N+n+1}
[z^N] (1+z)^{N+n+1}
{N\choose n}
\sum_{k=0}^N
(-1)^{n+k} z^k
{N+k\choose N} {n+k\choose n}.$$
Now the coefficient extractor controls the range and we continue with
$$\sum_{n=0}^N \frac{1}{N+n+1}
[z^N] (1+z)^{N+n+1}
{N\choose n}
\\ \times \sum_{k\ge 0}
(-1)^{n+k} z^k
{N+k\choose N}
\;\underset{w}{\mathrm{res}}\; \frac{1}{w^{n+1}} \frac{1}{(1-w)^{k+1}}
\\ = \sum_{n=0}^N \frac{1}{N+n+1}
[z^N] (1+z)^{N+n+1}
{N\choose n}
\;\underset{w}{\mathrm{res}}\; \frac{1}{w^{n+1}} \frac{1}{1-w}
\\ \times \sum_{k\ge 0}
(-1)^{n+k} z^k
{N+k\choose N} \frac{1}{(1-w)^{k}}
\\ = \sum_{n=0}^N \frac{(-1)^n}{N+n+1}
[z^N] (1+z)^{N+n+1}
{N\choose n}
\\ \times \;\underset{w}{\mathrm{res}}\; \frac{1}{w^{n+1}} \frac{1}{1-w}
\frac{1}{(1+z/(1-w))^{N+1}}
\\ = \sum_{n=0}^N \frac{(-1)^n}{N+n+1}
[z^N] (1+z)^{N+n+1}
{N\choose n}
\\ \times \;\underset{w}{\mathrm{res}}\; \frac{1}{w^{n+1}}
\frac{(1-w)^N}{(1-w+z)^{N+1}}
\\ = \sum_{n=0}^N \frac{(-1)^n}{N+n+1}
[z^N] (1+z)^{n}
{N\choose n}
\\ \times \;\underset{w}{\mathrm{res}}\; \frac{1}{w^{n+1}}
\frac{(1-w)^N}{(1-w/(1+z))^{N+1}}
\\ = \sum_{n=0}^N \frac{(-1)^n}{N+n+1}
[z^N] (1+z)^{n}
{N\choose n}
\\ \times \sum_{k=0}^n (-1)^k {N\choose k}
{n-k+N\choose N} \frac{1}{(1+z)^{n-k}}
\\ = \sum_{n=0}^N \frac{(-1)^n}{N+n+1}
{N\choose n}
\\ \times \sum_{k=0}^n (-1)^k {N\choose k}
{n-k+N\choose N} [z^N] (1+z)^k.$$
Now for the coefficient extractor to be non-zero we must have $k\ge N$
which happens just once, namely when $n=N$ and $k=N.$ We get
$$\frac{(-1)^N}{2N+1}
{N\choose N}
(-1)^N {N\choose N}
{N-N+N\choose N}.$$
This expression does indeed simplify to
$$\bbox[5px,border:2px solid #00A000]{
\frac{1}{2N+1}}$$
as claimed.
| {
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"url": "https://math.stackexchange.com/questions/3333597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 1,
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} |
Combinatorics generating functions of a series Hi I found that the generating function of a series $a_n$ is:
$$\frac{(1-x)(1+2x)}{(1+3x)(1-3x)}$$
I need to find a formula for $a_n$.
I tried some things and found that the generating function is equal to:
$$\frac{1}{3}\cdot (1+2x)\cdot( \frac{2}{1+3x} + \frac{1}{1-3x})$$
but I cant get any further than that.
| We start from the generating function:
$$f(x)=\frac{(1-x)(1+2x)}{(1+3x)(1-3x)} = (1-x)\frac{(1+2x)}{(1+3x)(1-3x)} $$
The denominator of the fraction has the zeroes $1/3,-1/3$. Their reciprocals are therefore $3,-3$.
Therefore the explicit formula for $\frac{(1+2x)}{(1+3x)(1-3x)} $ has the form
$$
a_n = \alpha\cdot (3)^n + \beta \cdot(-3)^n
$$
We have
$$
a_0 = \frac{(1+2x)}{(1+3x)(1-3x)} \bigg|_{x=0} =1\\
a_1 = \frac d {dx}\frac{(1+2x)}{(1+3x)(1-3x)} \bigg|_{x=0} = 2$$
And therefore obtain the linear equation system
$$
1 = \alpha + \beta \\
2 = \alpha\cdot (3) + \beta \cdot(-3)
$$
And with it the explicit formula
$$\frac{(1+2x)}{(1+3x)(1-3x)} =
\sum_{n\ge 0}
5/6·3^n + 1/6·(-3)^n x^n
$$
Therefore, we have:
$$
f(x) =
(1-x)\frac{(1+2x)}{(1+3x)(1-3x)} = (1-x)
\sum_{n\ge 0}
5/6·3^n + 1/6·(-3)^n x^n
\\ =
(\sum_{n\ge 0}
5/6·3^n + 1/6·(-3)^n x^n) - (\sum_{n\ge 0}
5/6·3^n + 1/6·(-3)^n x^{n+1})
\\
=1+\sum_{n\ge 1} (5/6·3^n + 1/6·(-3)^n - (5/6·3^{n - 1} + 1/6·(-3)^{n - 1})) x^n
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3333949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Why is the $\mathbb{Q}(i, \sqrt{3}, \sqrt[3]{5}) $splitting of the polynomial $(x^3 - 5)(x^2 + 1)$ and not just $(x^3 - 5)$? I know that $\zeta = \sqrt[3]{5}$ satisfies the polynomial $x^3-5 \in \mathbb{Q}[x]$, and splits completely in $\mathbb{Q}(i, \alpha, \zeta)$, where $\alpha = \sqrt{3}$.
I found the roots $x_1 = \zeta$, $x_2 = \zeta/2 + \zeta i \alpha /2$ and $x_3 = \zeta/2 - \zeta i \alpha /2$
And I can write $\zeta = x_1$, $\alpha = (2x_1/x_2 - 1)^4/3$ and $i = (x_3-x_2)/(x_1\alpha)$, and since the polynomial $(x^3-5)$ already splits completely over $\mathbb{Q}(i, \alpha, \zeta)$, why do I need the extra term $(x^2+1)$?
| To find the splitting field of $x^3-5$ on $\mathbb{Q}$ you have to extend $\mathbb{Q}$ with the root of the polynomial. So the first extension is $\mathbb{Q}(\sqrt[3]{5})$ then the other two roots of the polinomial, $\frac{-1+i\sqrt{3}}{2}\sqrt[3]{5}, \frac{-1-i\sqrt{3}}{2}\sqrt[3]{5}$, aren't in this extension, infact there isn't complex number in $\mathbb{Q}(\sqrt[3]{5})$. So now let's extend with another roots, like $\frac{-1+i\sqrt{3}}{2}\sqrt[3]{5}$, and we obtein $\mathbb{Q}(\sqrt[3]{5}, \frac{-1+i\sqrt{3}}{2}\sqrt[3]{5})$ but $\frac{\frac{-1+i\sqrt{3}}{2}\sqrt[3]{5},}{\frac{\sqrt[3]{5}}{2}}=-1+i\sqrt{3}$, so $i\sqrt{3} \in \mathbb{Q}(\sqrt[3]{5}, i\sqrt[3]{5}+\frac{1}{2}\sqrt[3]{5},)$. Then we can conclude that $\mathbb{Q}(\sqrt[3]{5}, \frac{-1+i\sqrt{3}}{2}\sqrt[3]{5})=\mathbb{Q}(\sqrt[3]{5}, i\sqrt{3})$ and this is the the splitting field of $x^3-5$. But $i, \sqrt{3} \not \in \mathbb{Q}(\sqrt[3]{5}, i\sqrt{3})$.
If we consider the splitting field of $(x^3-5)(x^2+1)$ we nead to extend $\mathbb{Q}(\sqrt[3]{5}, i\sqrt{3})$ with a root of $x^2+1$ and we obtain $\mathbb{Q}(\sqrt[3]{5}, i\sqrt{3}, i)=\mathbb{Q}(\sqrt[3]{5}, \sqrt{3}, i)$. So how it's easly to check the splitting field of $(x^3-5)(x^2+1)$ contains the splitting field of $(x^3-5)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3336245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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Can there be two inequalities? $$x+a+\sqrt{x^2+a^2}>b$$
I can consider this inequality for $a,b,x>0$.
1) $\sqrt{x^2+a^2}>b-(x+a)$
$$x^2+a^2>b^2-2b(x+a)+(x+a)^2$$
$$0>b^2-2b(x+a)+2ax$$
$$x(2b-2a)>b^2-2ba$$
$$x>\frac{b^2-2ba}{(2b-2a)}$$
2) $(x+a)-b>-\sqrt{x^2+a^2}$
$$x^2+a^2<b^2-2b(x+a)+(x+a)^2$$
$$0<b^2-2b(x+a)+2ax$$
$$x(2b-2a)<b^2-2ba$$
$$x<\frac{b^2-2ba}{(2b-2a)}$$
I get two inequalities. What is Im doing wrong?
| HINT
I am going to deal with the case in which $b - x - a \geq 0$. Otherwise, any value of $x$ that satisfies $x \geq b - a$ is a solution.
\begin{align*}
& x + a + \sqrt{x^{2} + a^{2}} > b \Longleftrightarrow \sqrt{x^{2} + a^{2}} > b - x - a \Longleftrightarrow
\begin{cases}
x^{2} + a^{2} > (b - x - a)^{2}\\\\
b - x - a \geq 0
\end{cases}\\\\
& \begin{cases}
x^{2} + a^{2} > b^{2} + x^{2} + a^{2} - 2bx - 2ab + 2ax\\\\
b - x - a \geq 0
\end{cases} \Longleftrightarrow
\begin{cases}
b^{2} - 2bx - 2ab + 2ax < 0\\\\
b - x - a \geq 0
\end{cases}\\\\
& \begin{cases}
b^{2} + 2x(a - b) - 2ab < 0\\\\
b - x - a \geq 0
\end{cases} \Longleftrightarrow
\begin{cases}
2x(a-b) < 2ab - b^{2}\\\\
b - x - a \geq0
\end{cases}
\end{align*}
Then you consider the three possible situations: $a > b$, $a = b$ or $a < b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3336347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Then find the range of m. Given that there are real constants $a,b,c,d$ such that the identity
$$Mx^2+2xy+y^2=(ax+by)^2+(cx+dy)^2$$ holds for all $x,y$ in the set of real numbers, what is the range of $M$?
| Multiplying out the right hand gives us $$Mx^2+xy+y^2=(a^2+c^2)x^2+2xy(ab+cd)+(b^2+d^2)y^2$$ from which we see that we want to simultaneously solve $$a^2+c^2=M\quad \quad ab+cd=1\quad \quad b^2+d^2=1$$
Apply Cauchy Schwarz to the vectors $\vec v = (a,c)$ and $\vec u = (b,d)$. We deduce that $$(ab+cd)^2≤(a^2+c^2)(b^2+d^2)$$ which in our case yields $$1≤M$$
Now, if we have $M≥1$ we can simply take $$(a,b,c,d)=\left(\sqrt M,\, \frac 1{\sqrt M}, \,0,\, \sqrt {1-\frac 1M}\right)$$
(noting that $M≥1$ implies that the square root expression for $d$ does give a real number).
So we see that the final answer is $$\boxed {M≥1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3338302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integral of $\frac{1}{\sin^2(x)\cos^2(x)}$ $$ I = \int\frac{1}{\sin^2(x)\cos^2(x)}$$
I have tried the following:
$$\sin^2x = \sin x \cdot \sin x = \frac{1}{2}(1 - \cos2x)$$
$$\cos^2x = \cos x \cdot \cos x = \frac{1}{2}(1 + \cos2x)$$
The integral becomes:
$$I = 4\int\frac{1}{1 - \cos^2(2x)} = 4\int\frac{1}{\sin^2(2x)}$$
Substitute $u = 2x \implies du = 2dx$
$$I = 2\int\frac{1}{\sin^2u} = -2\cot(u)$$
Plugging back x I get:
$$I = -2\cot(2x)$$
I tried plugging in the integral into an integral-calculator and the answer was: $\tan(x) - \cot(x)$. Can you help me identify what I did wrong?
| Your result is correct. Here's another approach:
$$\sin2x=2\sin x\cos x\implies \sin^2x\cos^2x=\frac14\sin^22x$$
and since $\;(\cot x)'=-\csc^2x=-\frac1{\sin^2x}\;$ , we get
$$\int\frac{dx}{\sin^2x\cos^2x}=\frac42\int\frac{d(2x)}{\sin^22x}=-2\cot 2x+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3339030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Simplex Method gives multiple, unbounded solutions but Graphical Method gives unique soution I'm taking an undergraduate course on Linear Programming and we were asked to solve the following problem using the Simplex Method:$$\max:~Z=3x+2y\\\text{subject to}\begin{cases}x+y\le20\\0\le x\le15\\x+3y\le45\\-3x+5y\le60\\y\text{ unrestricted in sign}\end{cases}$$The standard form of the LPP is$$\max:~Z=3x+2m-2n\\\text{subject to}\begin{cases}x+m-n+a=20\\x+b=15\\x+3m-3n+c=45\\-3x+5m-5n+d=60\\x,m,n,a,b,c,d\ge0\end{cases}$$where $y=m-n$. The optimal Simplex tableau was obtained$$\begin{matrix}&&3&2&-2&0&0&0&0\\&\text{Basis}&x&m&n&a&b&c&d&\text{RHS}\\2&m&0&1&-1&1&0&0&-1&5\\0&b&0&0&0&-3&1&0&2&15\\0&c&0&0&0&-5&0&1&8&80\\3&x&1&0&0&0&0&0&1&15\\&\text{Deviations}&0&0&\color{red}0&-2&0&0&-1&Z=55\end{matrix}$$Since all deviations are negative, the stopping criteria is fulfilled. But the deviation corresponding to non-basic $n$ is $0$, so this must be a case of multiple optimal solutions. With $n$ as the entering variable the minimum ratio test fails, which means this is also a case of unbounded solutions.
On solving the same question using Graphical method, I got a unique solution $Z=55$ at $(15,5)$. What is the problem in the Simplex Method?
| The simplex method will produce the correct answer:
$$\max:~Z=3x+2y\\\text{subject to}\begin{cases}x+y\le20\\0\le x\le15\\x+3y\le45\\-3x+5y\le60\\y\text{ unrestricted in sign}\end{cases} \Rightarrow \begin{cases}x+y+s_1=20\\x+s_2=15\\x+3y+s_3=45\\-3x+5y+s_4=60\\y\text{ unrestricted in sign}\end{cases}\\
\begin{array}{cccccc|c}
x&y&s_1&s_2&s_3&s_4&C\\
\hline
1&1&1&0&0&0&20&s_1\\
\boxed1&0&0&1&0&0&15&s_2\\
1&3&0&0&1&0&45&s_3\\
-3&5&0&0&0&1&60&s_4\\
\hline
-3&-2&0&0&0&0&0\\
\end{array} \Rightarrow \\
\begin{array}{cccccc|c}
x&y&s_1&s_2&s_3&s_4&C\\
\hline
0&\boxed1&1&-1&0&0&5&s_1\\
1&0&0&1&0&0&15&x\\
0&3&0&-1&1&0&30&s_3\\
0&5&0&3&0&1&105&s_4\\
\hline
0&-2&0&3&0&0&45\\
\end{array} \Rightarrow \\
\begin{array}{cccccc|c}
x&y&s_1&s_2&s_3&s_4&C\\
\hline
0&1&1&-1&0&0&\color{red}5&y\\
1&0&0&1&0&0&\color{red}{15}&x\\
0&0&-3&2&1&0&15&s_3\\
0&0&-5&8&0&1&80&s_4\\
\hline
0&0&2&1&0&0&\color{red}{55}\\
\end{array}$$
Thus: $z(15,5)=55$ is the maximum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3339768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find all function $f(x)$ such that $f(x^2 + x + 3)f(3x + 1) = f(6x^3 + 7x^2 + 16x + 3), \forall x \in \mathbb R$.
Find all function $f(x)$ such that $$\large f(x^2 + x + 3)f(3x + 1) = f(6x^3 + 7x^2 + 16x + 3), \forall x \in \mathbb R$$
I have provided my solution but there isn't a strong claim for why $f(x)$ is in the form of $(ax + b)^n$ where $a, b \in \mathbb R, n \ge 1$ (as Luca Bressan suggested).
| It can be seen that $f(x) = (ax + b)^n$ $(a, b \in \mathbb R, n \ge 1)$. We have that $$[a(x^2 + x + 3) + b][a(3x + 1) + b] = a(6x^3 + 7x^2 + 16x + 3) + b, \forall x \in \mathbb R$$
$$\iff a^2(3x^3 + 4x^2 + 10x + 3) + ab(x^2 + 4x + 4) + b^2$$
$$ = a(6x^3 + 7x^2 + 16x + 3) + b, \forall x \in \mathbb R$$
$$\iff 3a^2x^3 + (4a^2 + ab)x^2 + (10a^2 + 4ab)x + (3a^2 + 4ab + b^2)$$
$$ = 6ax^3 + 7ax^2 + 16ax + (3a + b), \forall x \in \mathbb R$$
$$\implies \left\{ \begin{align} 3a^2 &= 6a\\ 4a^2 + ab &= 7a\\ 10a^2 + 4ab &= 16a\\ 3a^2 + 4ab + b^2 &= 3a + b\end{align} \right., \forall x \in \mathbb R$$
$$\iff (a, b) \in \{(0, 0), (0, 1), (2, -1)\} \iff f(x) \in \{0, 1, (2x - 1)^n\}, \forall x \in \mathbb R, n \ge 1$$
| {
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Find $k^{th}$ power of a square matrix I am trying to find the $A^{k}$, for all $k \geq 2$ of a matrix, \begin{pmatrix} a & b \\ 0 & 1 \end{pmatrix}
My approach:
$A^{2}=\begin{pmatrix} a^2 & ab+b \\ 0 & 1 \end{pmatrix}$
$A^{3}=\begin{pmatrix} a^3 & a^{2}b+ab+b \\ 0 & 1 \end{pmatrix}$
$A^{4}=\begin{pmatrix} a^4 & a^{3}b+a^{2}b+ab+b \\ 0 & 1 \end{pmatrix}$
$A^{5}=\begin{pmatrix} a^5 & a^{4}b+a^{3}b+a^{2}b+ab+b \\ 0 & 1 \end{pmatrix}$
Continuing this way, we obtain
$A^{k}=\begin{pmatrix} a^k & (a^{k-2}+a^{k-3}+a^{k-4}+.....+1)b \\ 0 & 1 \end{pmatrix}$
I am stuck here! I was wondering if you could give me some hints to move further. I appreciate your time.
| Not the best way, but you could also try diagonalisation. The caveat with diagonalisation is that for certain values of $a$ and $b$ (in particular, if $a = 1$ and $b \neq 0$), the matrix won't be diagonalisable. However, if we make the assumption that $a \neq 1$, then we should end the process with a perfectly valid expression for $A^n$ that will work for all $a \neq 1$, and by continuity, we can conclude that it works for $a = 1$ too (or take the formula, and prove by induction).
Also, I know you want hints, so I hid everything behind spoiler boxes.
The eigenvalues are $1$ and $a$, and we will assume they are different. We have,
$$A - I = \begin{pmatrix} a - 1 & b \\ 0 & 0 \end{pmatrix},$$
with eigenvector $(-b, a - 1)$. Also,
$$A - aI = \begin{pmatrix} 0 & b \\ 0 & 1 - a \end{pmatrix},$$
with eigenvector $(1, 0)$. So, let
$$P = \begin{pmatrix} 1 & -b \\ 0 & a - 1\end{pmatrix},$$
giving us
$$P^{-1} = \frac{1}{a - 1}\begin{pmatrix} a - 1 & b \\ 0 & 1\end{pmatrix}.$$
We should have
$$A = P\begin{pmatrix} a & 0 \\ 0 & 1\end{pmatrix}P^{-1},$$
so
$$\begin{align*} A^n &= P\begin{pmatrix} a^n & 0 \\ 0 & 1\end{pmatrix}P^{-1} \\ &= \frac{1}{a - 1}\begin{pmatrix} 1 & -b \\ 0 & a - 1\end{pmatrix}\begin{pmatrix} a^n & 0 \\ 0 & 1\end{pmatrix}\begin{pmatrix} a - 1 & b \\ 0 & 1\end{pmatrix} \\ &= \frac{1}{a - 1}\begin{pmatrix} 1 & -b \\ 0 & a - 1\end{pmatrix}\begin{pmatrix} a^{n+1} - a^n & ba^n \\ 0 & 1\end{pmatrix} \\ &= \frac{1}{a - 1}\begin{pmatrix} a^{n+1} - a^n & ba^n - b \\ 0 & a - 1\end{pmatrix} \\ &= \begin{pmatrix} a^n & b \frac{a^n - 1}{a - 1} \\ 0 & 1\end{pmatrix} \\ &= \begin{pmatrix} a^n & b(1 + a + a^2 + \ldots + a^{n-1}) \\ 0 & 1\end{pmatrix}.\end{align*}$$
That last formula must hold (at least) for $a \neq 1$, but due to the continuity of matrix powers, it must also hold at $a = 1$ too.
| {
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Commutative matrices of a square matrix Question: Find the all matrice that commute with $B=\begin{pmatrix} b & 1 & 0 \\ 0 & b & 1 \\ 0 & 0 & b \end{pmatrix}$.
My work:
Let $A=\begin{pmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{pmatrix}$.
Now from $AB=BA$ implies,
$\begin{pmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{pmatrix} . \begin{pmatrix} b & 1 & 0 \\ 0 & b & 1 \\ 0 & 0 & b \end{pmatrix}
=\begin{pmatrix} b & 1 & 0 \\ 0 & b & 1 \\ 0 & 0 & b \end{pmatrix}.\begin{pmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{pmatrix}$
Then,
$\begin{pmatrix} a_1b & a_1+bb_1 & b_1+c_1b \\ a_2b & a_2+bb_2 & b_2+c_2b \\ a_3b & a_3+bb_3 & b_3+c_3b\end{pmatrix} =\begin{pmatrix} ba_1+a_2 & bb_1+b_2 & c_1b+c_2 \\ ba_2+a_3 & bb_2+b_3 & c_2b+c_3 \\ ba_3 & bb_3 & bc_3\end{pmatrix}$
Then, solving this,
$ a_1b= ba_1+a_2 \implies a_2=0$
$a_1+bb_1=bb_1+b_2 \implies a_1=b_2$
$b_1+c_1b=c_1b+c_2 \implies b_1=c_2$
$a_2b=ba_2+a_3 \implies a_3=0$
$a_2+bb_2=bb_2+b_3 \implies a_2=b_3$
$b_2+c_2b=c_2b+c_3 \implies b_2=c_3$
$a_3b=ba_3$
$a_3+bb_3= bb_3 \implies a_3=0$
$b_3+c_3b= bc_3 \implies b_3=0$.
Then, I am not sure about following setting. Is it right or wrong?
$A=\begin{pmatrix} a_1 & b_1 & ? \\ 0 & a_1=b_2 & b_1 \\ 0 & 0 & c_3 \end{pmatrix}$.
I was wondering if you could help to resolve this issue. I appreciate your time.
| It is automatically true that $B$ commutes with any matrix of the form
$$ xI + y B + z B^2. $$ Note that it is not necessary to consider $B^3$ or $B^4,$ as these can be absorbed into the given expression.
The nontrivial theorem is that, when the minimal polynomial agrees with the characteristic polynomial, then the only matrices commuting with $B$ are those polynomial expressions in $B.$ That applies here, the condition is equivalent to this: each eigenvalue occurs in just one Jordan block.
In sum, you are still a little off.
| {
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"question_score": "2",
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Distribute $13$ identical balls in $6$ cells. Find the number of distributions such that at least $10$ balls will be in the first 3 cells together Let $13$ identical balls be distributed in $6$ cells. Find the number of distributions in which there are at least $10$ balls in the first $3$ cells together.
My attempt:
First I'll divide into cases: $A_{0},A_{1},A_{2},A_{3}$
$\sum_{i=0}^{3}A_{i}=$ all distributions possible.
This is a question where the order matters and reparations
is allowed. Given $n$ balls and $k$ cells, there are: $\binom{n+k-1}{n}$
ways to distribute $n$ balls to $k$ cells.
$A_{0}$ - there are $13$ balls in the first $3$ cells, and $0$ balls in the $3$ other cells thus $k_{1}=3,n_{1}=13\Longrightarrow$$\binom{n+k-1}{n}=\binom{15}{13}=105$ ways to distribute $13$ balls between $3$ cells. The distribution of balls in the first $3$ cells is disjointed to the distribution of balls in the other $3$ cells, therefore we'd apply the multiplication principle:
number of ways to distribute $0$ balls in $3$ cells:
$\binom{2}{0}=1\Longrightarrow A_{0}=\binom{15}{13}=105$
$A_{1}$ - there are $12$ balls in the first $3$ cells,
and $1$ ball in the $3$ other cells:
$k_{1}=3,n_{1}=12,k_{2}=3,n_{2}=1 \Longrightarrow\binom{n_{1}+k_{1}-1}{n_{1}}\cdot\binom{n_{2}+k_{2}-1}{n_{2}}=\binom{15}{12}\cdot\binom{3}{1}=455\cdot3=1365$
$A_{2}$- there are $11$ balls in the first $3$ cells,
and $2$ ball in the $3$ other cells:
$k_{1}=3,n_{1}=11,k_{2}=3,n_{2}=2$}$\Longrightarrow\binom{n_{1}+k_{1}-1}{n_{1}}\cdot\binom{n_{2}+k_{2}-1}{n_{2}}=\binom{15}{11}\cdot\binom{4}{2}=1365\cdot6=8190$}
$A_{3}$- there are $10$ balls in the first $3$ cells,
and $3$ ball in the $3$ other cells:
$k_{1}=3,n_{1}=11,k_{2}=3,n_{2}=3 \Longrightarrow\binom{n_{1}+k_{1}-1}{n_{1}}\cdot\binom{n_{2}+k_{2}-1}{n_{2}}=\binom{15}{10}\cdot\binom{5}{3}=3003\cdot10=30030$}
$\sum_{i=0}^{3}A_{i}=105+1365+8190+30030=39690$
I'm not sure that my answer is correct, it seems to me that my result is to big.
| You're very close, but have made a small mistake in calculating ${n_1 + k_1 - 1 \choose n_1}$: the top argument does not always equal $15$. You should find:
$$A_0 = {15 \choose 2} {2 \choose 2} = 105$$
$$A_1 = {14 \choose 2} {3 \choose 2} = 273$$
$$A_2 = {13 \choose 2} {4 \choose 2} = 468$$
$$A_3 = {12 \choose 2} {5 \choose 2} = 660$$
$$A_0 + A_1 + A_2 + A_3 = 105 + 273 + 468 + 660 = 1506$$
| {
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"source": "stackexchange",
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How can I solve $\int e^{2\theta} \sin(3\theta)\, d\theta$ with integration by parts? $\int e^{2\theta}\sin(3\theta)d\theta$ seems to be leading me in circles. The integral I get when I use integration by parts, $\int e^{2\theta}\cos(3\theta)d\theta$ just leads me back to $\int e^{2\theta}\sin(3\theta)d\theta$. I am not sure how to solve it.
My Steps:
$\int e^{2\theta}\sin(3\theta)d\theta$
Let $u = \sin(3\theta)$ and $dv=e^{2\theta}d\theta$
Then $du = 3\cos(3\theta)d\theta$ and $v = \frac{1}{2}e^{2\theta}$
\begin{align*}
\int e^{2\theta} \sin(3 \theta)d\theta &= \frac{1}{2} e^{2\theta}\sin(3\theta) - \int\frac{1}{2}e^{2\theta}3\cos(3\theta)d\theta\\
&=e^{2\theta}\sin(3\theta) - \frac{3}{2}\int e^{2\theta}\cos(3\theta)d\theta\\
\end{align*}
$\int e^{2\theta}\cos(3\theta)d\theta$
Let $u = \cos(3\theta)$ and $dv = e^{2\theta}d\theta$
Then $du = -3\sin(3\theta)d\theta$ and $v=\frac{1}{2}e^{2\theta}$
\begin{align*}
\int e^{2\theta}\cos(3\theta) &= \frac{1}{2}e^{2\theta}\cos(3\theta)-\int (\frac{1}{2}e^{2\theta}\cdot-3\sin(3\theta))d\theta\\
&=\frac{1}{2}e^{2\theta}\cos(3\theta)+ \frac{3}{2} \int e^{2\theta}\sin(3\theta)d\theta
\end{align*}
So you can see I just keep going in circles. How can I break out of this loop?
| As lab bhattacharjee answered, in case integration by parts is not mandatory, you can make life easier considering that what you need is the imaginary part of
$$I=\int e^{2\theta} e^{3i \theta}\,d\theta=\int e^{(2+3i)\theta}\,d\theta=\frac {e^{(2+3i)\theta}}{(2+3i)}=\frac{2-3i}{13}e^{(2+3i)\theta}$$
$$I=\frac{3}{13} e^{2 \theta } \sin (3 \theta )+\frac{2}{13} e^{2 \theta } \cos (3
\theta )+i \left(\frac{2}{13} e^{2 \theta } \sin (3 \theta )-\frac{3}{13} e^{2
\theta } \cos (3 \theta )\right)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\int \frac{6\sin x \cos^2 x+\sin 2x-23\sin x} {(\cos x - 1)^2(5-\sin^2x)} \mathrm{dx}$ I got this question $\int \frac{6\sin x \cos^2 x+\sin 2x-23\sin x} {(\cos x - 1)^2(5-\sin^2x)} \mathrm{dx}$ and I solved it as thus:
$$\int \frac{6\sin x \cos^2 x+\sin 2x-23\sin x} {(\cos x - 1)^2(5-\sin^2x)} \mathrm{dx}$$
$$=\int \frac{6\sin x \cos^2 x+2 \sin x\cos x-23\sin x} {(\cos x - 1)^2(5-\sin^2x)} \mathrm{dx}$$
$$=\int \frac{\sin x(6\cos^2x+2\cos x-23)} {(\cos x - 1)^2(5-(1-\cos^2x))}\mathrm{dx}$$
$$=\int \frac{\sin x(6\cos^2x+2\cos x-23)} {(\cos x - 1)^2(4+\cos^2x))}\mathrm{dx}\\$$
$$\mathrm{put\;} \color{red}{u=\cos x}\quad \mathrm{\frac{du} {dx}}=-\sin x\quad \mathrm{dx} =\frac{\mathrm{du} } {-\sin x} \\
\require{cancel}
\int \frac{\sin x(6u^2+2u-23)} {(u- 1)^2(u^2+4)}\frac{\mathrm{du}} {-\sin x} =$$
$$=\int \frac{\cancel{\sin x}(6u^2+2u-23)} {(u- 1)^2(u^2+4)}\frac{\mathrm{du}} {-\cancel{\sin x}}\\$$
$$=\int \frac{23-2u-6u^2 } {(u-1)^2(u^2+4)}\mathrm{du}$$
$$\color{red} {\mathrm{Resolving\;}} \frac{23-2u-6u^2 } {(u-1)^2(u^2+4)}\color{red} {\mathrm{\; into\; partial\; fractions\;we \;have:}} \\$$
$$\frac{4u-5} {u^2+4} - \frac{4} {u-1}+ \frac{3} {(u-1)^2} =\frac{4u} {u^2+4} - \frac{5 } {u^2+ 4} - \frac{4} {u-1}+ \frac{3} {(u-1)^2} \\$$
$$\int\ \frac{23-2u-6u^2 } {(u-1)^2(u^2+4)}=\int \left(\frac{4u} {u^2+4} - \frac{5 } {u^2+ 4} - \frac{4} {u-1}+ \frac{3} {(u-1)^2}\right) \mathrm{du}$$
$$2\ln \{u^2 +4\}-\frac{5}{2 }\tan^{-1} \left\{\frac{u}{2} \right\}-4\ln\{u-1\}-\frac{3}{u-1}
\\$$
$$\mathrm{But} \;u=\cos x\\$$
Edit
$$\underline{2\ln \left|\cos^2x +4\right|-\frac{5}{2 }\tan^{-1} \left(\frac{\cos x}{2} \right) -4\ln\left|\cos x-1\right|-\frac{3}{\cos x-1}+\mathrm{C}}$$
Is there anything wrong with these steps?
Wolfram alpha made me extremely skeptical.
Are there alternative steps?
| Mathematica gives exactly the same answer for you $u$-integral as
$-(3/(-1 + u)) - 5/2 ArcTan[u/2] - 4 Log[-1 + u] + 2 Log[4 + u^2], u=\cos x$
There is indeed no discrepancy with Mathematica and your answer. Cheer up!
| {
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"question_score": "2",
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If $A+B+C=180$, then $\frac{\sin2A+\sin2B+\sin2C}{\cos A+\cos B+\cos C-1}=8\cos\frac A2 \cos\frac B2 \cos\frac C2$ Then $2A+2B+2C =360$
So $$\sin 2C=-\sin(2A+2B)$$
Putting that in the equation
$$\frac{2\sin(A+B)\sin(A-B)-2\sin(A+B)\cos(A+B)}{\cos A+\cos B-\cos(A+B)+1}$$
$$\frac{2\sin(A+B)[\sin(A-B)-\cos(A+B)]}{\cos A+\cos B-\cos(A+B)+1}$$
I don’t know how to proceed. Please help me continue
| In a triangle, $$R^2(\sin(2A)+\sin(2B)+\sin(2C))=2\Delta$$
is a consequence of $[ABC]=[OAB]+[OBC]+[OCA]$ with $O$ being the circumcenter.
Carnot's theorem gives $R(\cos A+\cos B+\cos C-1) = r$, hence by $abc=4R\Delta$ the claim is equivalent to
$$ (a+b+c)\Delta = 2abc\cos\tfrac{A}{2}\cos\tfrac{B}{2}\cos\tfrac{C}{2}$$
which is simple to prove by squaring both sides, then exploiting Heron's formula and
$$ \cos^2\tfrac{A}{2}=\tfrac{1}{2}\left(1+\cos A\right)=\frac{1}{2}\left(\frac{b^2+c^2-a^2}{2bc}+1\right)=\frac{(a+b+c)(-a+b+c)}{4bc}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3353872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Solve $\tan x =\sec 42^\circ +\sqrt{3}$ For the trigonometric equation,
$$\tan x =\sec 42^\circ+\sqrt{3}$$
Find the angle $x$, where $0<x<180^\circ$.
I tried to solve for an unknown angle $x$ in a geometry problem with a trigonometric approach. I ended up with the trig equation above. Without hesitation, I reached my calculator, entering the right-hand-side and arctan-ing it for $x$.
To my surprise, the angle $x$ comes out at exactly 72 degrees. I did not expect such a neat relationship. Then, I thought I should have solved the equation analytically for the whole-degree angle without the calculator. I spent a good amount of time already and was not able to derive it yet.
Either the equation is not as innocent as it looks, or a straightforward method just eludes me.
| $\begin{align} \cos(42°) &= \cos(60°-18°) \cr
&= \cos(60°)\cos(18°) + \sin(60°)\sin(18°) \cr
&= {1\over2} (\cos(18°) + \sqrt3 \sin(18°)) \cr
\sec(42°) &= \left({2 \over \cos(18°) + \sqrt3 \sin(18°)}\right)
\left({\cos(18°) - \sqrt3 \sin(18°) \over \cos(18°) - \sqrt3 \sin(18°)}\right) \cr
&= {2(\cos(18°) - \sqrt3 \sin(18°)) \over \cos^2(18°) - 3\sin^2(18°)} \cr
&= \left({2\sin(18°) \over 1 -4 \sin^2(18°)}\right) (\cot(18°) - \sqrt3) \cr
\end{align}$
Let $s=\sin(18°)$, using multiple angles formula
$\sin(90°) = \sin(5 \times 18°) = 16s^5 - 20s^3 + 5s = 1$
$16s^5 - 20s^3 + 5s - 1 = 0$
$(s-1)(4s^2+2s-1)^2 = 0$
Since $s≠1$, we get $4s^2+2s-1 = 0\quad → \large{2s \over 1-4s^2} = 1$
$\tan(x) = \sec(42°) + \sqrt3 = (\cot(18°) - \sqrt3) + \sqrt3 = \tan(72°)$
| {
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How to check if $ x^2 + x = y^2 - 1$ has positive integer solutions? How to check if $ x^2 + x = y^2 - 1$ has positive integer solutions?
I've tried using different modulos to simplify problem, but nothing has worked yet.
| As AmateurMathPirate said, we have $(2y)^{2} - (2x+1)^{2} = 3$
Thus $(2y - 2x - 1)(2y + 2x + 1) = 3$
Then $(2y - 2x - 1)|3$ and $(2y + 2x +1)|3$
So $(2y - 2x -1) = 1, -1, 3,$ or $-3$
i) if $(2y - 2x - 1) = 3 \implies (2y + 2x + 1) = 1 \implies 2y + 2x = 0 \implies y = -x \implies x\leq 0$ or $y\leq 0$ then it can't be
ii) if $(2y - 2x - 1) = -3 \implies (2y + 2x + 1) = -1 \implies 2y + 2x = 2 \implies x + y = 1$, but $x,y\geq 1$ then $x + y \geq 2$ then it can't be
iii) if $(2y - 2x -1) = 1 \implies (2y + 2x + 1) = 3$ but $x,y\geq 1$ then $2y + 2x + 1 \geq 5$ then it can't be
iv) if $(2y - 2x - 1) = -1 \implies 2y+ 2x + 1 = -3$ but $2x + 2y +1 >0$ then it can't be
So there is not exist that positive intengers
| {
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Is there a proof for Euler's "other" formula for pi? Some context: The webpage Pi Formulas on Wolfram.com has a list of some well-known formulas for $\pi$. However, there is a formula, as stated below, which I can't find a proof for.
$$\pi = \frac{2}{\prod_{n=2}^{\infty}( 1 + \frac{(-1)^{\frac{p_n-1}{2}}}{p_n})} =\frac{2}{\prod_{n=1}^{\infty}( 1 + \frac{\sin(\frac{\pi}{2}p_n)}{p_n})}$$
where $p_n$ is the $n$-th prime number. Wolfram references a book called The Joy of Pi by David Blatner. The book states the formula, but never proves it nor leaves a reference. Apart from these two sources, my efforts to find a proof have been in vain. I checked via Python whether this equality was correct or not, and it is correct (as far as Python can verify).
My question: Does anyone know where a proof for this equality is somewhere in the Mathematical literature? If so, I would appreciate the proof being shown in an answer.
Edit: After much deliberation, I have discovered that Dirichlet Series representation of the Lebinez formula was not the answer to this problem. The famous Lebinez formula for $\pi$ is
$$\sum_{k=0}^{\infty} \frac{(-1)^{k}}{2k+1} = \frac{\pi}{4}.$$
The Dirichlet Series representation is
$$\prod_{p \equiv 1 \text{ (mod }4)}\left(\frac{p}{p-1}\right)\prod_{p \equiv 3 \text{ (mod }4)}\left(\frac{p}{p+1}\right).$$
Whilst initially this looked like the answer, upon review, I realized my error.
By expanding out our original formula divided by 2, we get
$$\frac{1}{\prod_{n=2}^{\infty}(1+\frac{(-1)^{\frac{p_n-1}{2}}}{p_n})} = (1-\frac{1}{3})^{-1}(1+\frac{1}{5})^{-1}(1-\frac{1}{7})^{-1}(1-\frac{1}{11})^{-1}\cdots$$
$$=\prod_{p \equiv 3 \text{ (mod }4)}\left(\frac{p}{p-1}\right)\prod_{p \equiv 1 \text{ (mod }4)}\left(\frac{p}{p+1}\right).$$
Notice that
$$\frac{1}{1 + \frac{(-1)^{\frac{p-1}{2}}}{p}} = \begin{cases}
\frac{p}{p+1}, & \text{if } p \equiv 1 \text{ (mod }4) \\
\frac{p}{p-1}, & \text{if } p \equiv 3 \text{ (mod }4) \\
\end{cases}
\neq
\begin{cases}
\frac{p}{p-1}, & \text{if } p \equiv 1 \text{ (mod }4) \\
\frac{p}{p+1}, & \text{if } p \equiv 3 \text{ (mod }4) \\
\end{cases}.$$
While the two products are very similar, they are NOT the same. So the question remains, where's the proof? I would imagine that one could prove Euler's "other" formula for $\pi$ from Lebinez's formula for $\pi$.
| We have
\begin{eqnarray*}
P&=&\frac{1}{\prod_{n=2}^{\infty}\left( 1 + \frac{(-1)^{\frac{p_n-1}{2}}}{p_n}\right)} \\ &=& \left(1+\frac{1}{3} \right)^{-1} \left(1-\frac{1}{5} \right)^{-1}\left(1+\frac{1}{7} \right)^{-1}\left(1+\frac{1}{11} \right)^{-1} \cdots
\end{eqnarray*}
Now expand each of these terms geometrically and multiply out. We will get the reciprocal of every odd number exactly once and with a minus sign if it is congruent to $3$ mod $4$ and a plus sign if it is $1$ mod $4$.
So we have
\begin{eqnarray*}
P=1 -\frac{1}{3} +\frac{1}{5} -\frac{1}{7} +\frac{1}{9}-\frac{1}{11} +\frac{1}{13} \cdots = \frac{\pi}{4}.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3360259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\exists c>0$ s.t $\sum_{n\geq x}\frac{1}{n^2}\leq \frac{c}{x}$ $\mathbf{Question:}$ Prove that there exists a constant $c>0$ such that for all $x \in [1, \infty)$, $\displaystyle\sum_{n \geq x}\frac{1}{n^2} \leq \frac{c}{x}$
$\mathbf{Attempt:}$ Evidently,
$\displaystyle\bigg[\displaystyle\sum_{n =1}^\infty\frac{1}{n^2}-\sum_{j=1}^{\lfloor x \rfloor} \frac{1}{j^2}\bigg]{\lfloor x \rfloor}\leq x\displaystyle\sum_{n \geq x}\frac{1}{n^2} \leq \displaystyle\bigg[\sum_{n =1}^\infty\frac{1}{n^2}-\sum_{j=1}^{\lfloor x \rfloor} \frac{1}{j^2}\bigg]{\lceil x \rceil} $.
This inequality arises since,
$\displaystyle\bigg[\sum_{n =1}^\infty\frac{1}{n^2}-\sum_{j=1}^{\lfloor x \rfloor} \frac{1}{j^2}\bigg]=\displaystyle\sum_{n \geq x}\frac{1}{n^2}$.
Now, let $\lfloor x\rfloor=m$
$\displaystyle\lim_{m \to \infty}\bigg[\sum_{n =1}^\infty\frac{1}{n^2}-\sum_{j=1}^m \frac{1}{j^2}\bigg]m$
$=\lim_{{m \to \infty},{r\to \infty}}\bigg[\frac{m}{(m+1)^2}+\frac{m}{(m+2)^2}+\frac{m}{(m+3)^2}+...+\frac{m}{(m+r)^2}+...\bigg]=$
$\lim_{{h \to 0},{r\to \infty}}h\bigg[\frac{1}{(1+h)^2}+\frac{1}{(1+2h)^2}+\frac{1}{(1+3h)^2}+...+\frac{1}{(1+rh)^2}+...\bigg]=\displaystyle\int_{x=0}^\infty\frac{1}{(1+x)^2}dx=1$
We have, $\lceil x\rceil=m+1$ and $\displaystyle\lim_{m \to \infty}\bigg[\sum_{n =1}^\infty\frac{1}{n^2}-\sum_{j=1}^m \frac{1}{j^2}\bigg](m+1)=\lim_{m \to \infty}\bigg[\sum_{n =1}^\infty\frac{1}{n^2}-\sum_{j=1}^m \frac{1}{j^2}\bigg]m$
Thereby, we conclude, $\displaystyle \lim_{x\to \infty}x\sum_{n \geq x}\frac{1}{n^2}=1 $.
Therefore, from the definition of limit, $\forall \varepsilon>0$, $\exists G>0$ such that $\bigg|x\displaystyle\sum_{n \geq x}\frac{1}{n^2}-1\bigg|<\varepsilon$ $\forall x>G$. Choosing $\varepsilon =1$, we get $G=G_0$ such that $x\displaystyle\sum_{n \geq x}\frac{1}{n^2}<2$, for any $x> G_0$.
We set $\displaystyle c=\max\bigg\{2, \sup_{x\in [1,G_0]}x\sum_{n \geq x}\frac{1}{n^2} \bigg\}$
Is this procedure valid? Kindly verify.
| I think you would be better off starting with an integral from the start:
$$\frac 1{n^2} \le \int_{n-1}^{n} \frac 1{t^2} \, dt.$$
For any integer $k$ [edited to address complaint] integer $k \ge 2$ you have
$$\sum_{n=k}^\infty \frac 1{n^2} \le \int_{k-1}^\infty \frac 1{t^2} \, dt = \frac 1{k-1}$$
From here it is not too difficult.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3361638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Write the expression so that the variable is only presented once and the exponents are positive. There is this problem
$\frac{10x^{\frac{1}{3}}y^{-\frac{1}{4}}}{15x^{-\frac{1}{2}}y^{\frac{3}{4}}}$
and i know that the answer is
$\frac{2x^{\frac{5}{6}}}{3y}$
but i cant find the logic of how is solved, Where do the coefficients come from?
I understand that fractions are the result of multiplication of expressions, but I do not understand how all come together.
| Note that $x^{r}x^{q} = x^{r+q}$, so if $r = -q$, then $x^{-q}x^{q} = x^{-q+q} = x^{0} = 1$. Thus, you can "remove" the negative exponent factors by multiplying the numerator & denominator by the factor using the corresponding positive exponent, as shown below
$$\begin{equation}\begin{aligned}
\frac{10x^{\frac{1}{3}}y^{-\frac{1}{4}}}{15x^{-\frac{1}{2}}y^{\frac{3}{4}}} & = \frac{2(5)x^{\frac{1}{3}}y^{-\frac{1}{4}}}{3(5)x^{-\frac{1}{2}}y^{\frac{3}{4}}} \left(\frac{x^{\frac{1}{2}}y^{\frac{1}{4}}}{x^{\frac{1}{2}}y^{\frac{1}{4}}}\right) \\
& = \frac{2x^{\frac{1}{3} + \frac{1}{2}}y^{\frac{-1}{4} + \frac{1}{4}}}{3x^{\frac{-1}{2} + \frac{1}{2}}y^{\frac{3}{4} + \frac{1}{4}}} \\
& = \frac{2x^{\frac{1}{3} + \frac{1}{2}}}{3y^{\frac{3}{4} + \frac{1}{4}}} \\
& = \frac{2x^{\frac{5}{6}}}{3y}\end{aligned}\end{equation}\tag{1}\label{eq1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3361769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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} |
Prove that the sequence $C_{n+1} = 1 + \frac{C_n}{C_n + 1}$ is monotonically increasing by induction I have a sequence
$$C_{n+1} = 1 + \frac{C_n}{C_n + 1}$$
With a base case $C_1 = 3/2$ and want to prove that it's monotonically increasing by induction.
Whenever I try to prove $C_n < C_{n+1} \implies C_{n+1} < C_{n+2}$
where $$C_{n+2} = 1 + \frac{2C_n + 1}{3C_n + 2} $$
I get to the inequality $$ \frac{C_n}{C_n + 1} < \frac{2C_n + 1}{(C_n + 1)^2} $$
but don't know how to get to the point where that implies that $C_{n+1} < C_{n+2}$
Thank
| By induction, it is clear that $C_n>0$ for all $n$. Put $c=C_n$.
You want : $C_{n+1}\gt C_n$, which is equivalent to
$$
\begin{array}{lcl}
C_{n+1}\gt C_n & \Leftrightarrow & 1+\frac{C_n}{C_n+1} \gt C_n \\
& \Leftrightarrow & 1+\frac{c}{1+c} \gt c \\
& \Leftrightarrow & \frac{1+2c}{1+c} \gt c \\
& \Leftrightarrow & 1+2c \gt c+c^2 \\
& \Leftrightarrow & 0 \gt -1-c+c^2 \\
& \Leftrightarrow & 0 \gt \big(c-\frac{1}{2}\big)^2-\frac{5}{4}\\
& \Leftrightarrow & \frac{5}{4} \gt \big(c-\frac{1}{2}\big)^2\\
& \Leftrightarrow & \frac{-\sqrt{5}}{2} \lt c-\frac{1}{2} \lt \frac{\sqrt{5}}{2}\\
& \Leftrightarrow & \frac{1-\sqrt{5}}{2} \lt c \lt \frac{1+\sqrt{5}}{2}\\
\end{array}
$$
Put $\alpha=\frac{1-\sqrt{5}}{2}$ and $\beta=\frac{1+\sqrt{5}}{2}$.
So, the goal is now to show that $\alpha \lt C_n \lt \beta$ for every $n$.
Can you finish from here ? (use induction and $C_{n+1}=2-\frac{1}{1+C_n}$).
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Find coefficient of $x^2$ Let $C$ be the coefficient of $x^2$ in the expansion of the product
$$(1-x)(1+2x)(1-3x) \cdots (1+14x)(1-15x)$$
Find: $\lvert C \rvert $
| Multiply the monomials one by one and discard the terms of degree above $2$:
$$1-x\\1+2x\to1+ x-2 x^2\\
1 -3 x\to1 -2 x-5 x^2\\
1+ 4 x\to1+ 2 x-13 x^2\\
1 -5 x\to1 -3 x-23 x^2\\
1+ 6 x\to1+ 3 x-41 x^2\\
1 -7 x\to1 -4 x-62 x^2\\
1+ 8 x\to1+ 4 x-94 x^2\\
1 -9 x\to1 -5 x-130 x^2\\
1+ 10 x\to1+ 5 x-180 x^2\\
1 -11 x\to1 -6 x-235 x^2\\
1+ 12 x\to1+ 6 x-307 x^2\\
1 -13 x\to1 -7 x-385 x^2\\
1+ 14 x\to1+ 7 x-483 x^2\\
1 -15 x\to1 -8 x\color{green}{-588} x^2\\$$
Notice that the terms of the first degree follow a simple pattern and that the final result is a sum of pairwise products.
$$0-2\cdot1-3\cdot1-4\cdot2-5\cdot2-\cdots15\cdot7$$
$$(1+ax+bx^2)(1+cx)\to1+(a+c)x+(b+ac)x^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3366536",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Why does $r = \frac 23 - \frac 12 T$ rather than $r = \frac 23$? I have the following question. Using Gaussian elimination, I got an answer different from the book's. Can anyone please tell me what I'm doing wrong?
At The Crispy Critter's Head Shop and Patchouli Emporium, along with their dried up weeds, sunflower seeds and astrological postcards, they sell an herbal tea blend. By weight,
*
*Type I herbal tea is 30% peppermint, 40% rose hips and 30% chamomile.
*Type II herbal tea is 40% peppermint, 20% rose hips and 40% chamomile.
*Type III herbal tea is 35% peppermint, 30% rose hips and 35% chamomile.
How much of each Type of tea is needed to make 2 pounds of a new blend of tea that is equal parts peppermint, rose hips and chamomile?
Here are my steps:
Matrix:
$(E1) \frac{3}{10}p +\frac{4}{10}r + \frac{3}{10}c = \frac{2}{3}$
$(E2) \frac{4}{10}p +\frac{2}{10}r + \frac{4}{10}c = \frac{2}{3}$
$(E3) \frac{35}{100}p +\frac{3}{10}r + \frac{35}{100}c = \frac{2}{3}$
I now make the leading coefficient in E1 a 1:
$(new E1) p +\frac{4}{3}r + c = \frac{20}{9}$
I eliminate p from E2:
$(E1) -\frac{4}{10}p -\frac{16}{30}r - \frac{4}{10}c = -\frac{8}{9}$
$(E2) \frac{4}{10}p +\frac{2}{10}r + \frac{4}{10}c = \frac{2}{3}$
and get new E2:
$(new E2) r = \frac{2}{3}$
I eliminate p from E3:
$(E3) -\frac{35}{100}p -\frac{140}{300}r - \frac{35}{100}c = -\frac{7}{9}$
$(E3) \frac{35}{100}p +\frac{90}{300}r + \frac{35}{100}c = \frac{6}{9}$
and get new E3:
$(new E3) r = \frac{2}{3}$
So I now have the following matrix:
$(new E1) p +\frac{4}{3}r + c = \frac{20}{9}$
$(new E2) r = \frac{2}{3}$
$(new E3) r = \frac{2}{3}$
Now this leaves c as a free variable so I set it equal to T and I get
c = t,
$r = \frac{2}{3}$, and $p = \frac{4}{3} - T $
But the book says the answer is c = t,
$r = \frac{2}{3} - \frac{1}{2}T$, and $p = \frac{4}{3} - \frac{1}{2}T $
Any ideas what I did wrong?
| Your equations are the wrong way round, it should be
$$
\begin{eqnarray}
\frac{3}{10}a + \frac{4}{10}b + \frac{35}{100}c = 2/3\\
\frac{4}{10}a + \frac{2}{10}b + \frac{3}{10}c = 2/3\\
\frac{3}{10}a + \frac{4}{10}b + \frac{35}{100}c = 2/3 \\
\end{eqnarray}
$$
Where $a$ is tea type 1, $b$ is tea type 2 and $c$ is tea type 3.
This way you sum up the herbs in the different teas and get 2/3 pounds of each.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3366642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
What is the integral part of *M*, if *M*= $\sqrt{2012* \sqrt{2013* \sqrt{2014* \sqrt{... \sqrt{( 2012^2-2)* \sqrt{(2012^2-1)* \sqrt{2012^2}}}}}}}$ What is the integral part of M, if
M=
$\sqrt{2012* \sqrt{2013* \sqrt{2014* \sqrt{... \sqrt{( 2012^2-3)* \sqrt{(2012^2-1)* \sqrt{2012^2}}}}}}}$
I have no clue how to solve this, hints aswell as full solutions would be appreciated, taken from the 2012 IWYMIC https://chiuchang.org/wp-content/uploads/sites/2/2018/01/2012-IWYMIC-Individual.x17381.pdf
| For convenience let $$f(n)=\sqrt{n\sqrt{(n+1)\sqrt{(n+2)\cdots\sqrt{(n^2-2)\sqrt{(n^2-1)\sqrt{n^2}}}}}}$$
where $n$ is a positive integer. So basically,
$$f(n)=\prod_{k=n}^{n^2}k^{\frac1{2^{k-n+1}}}.$$
Let $n>1$. As $k> n$ for each index $k=n+1,n+2,\ldots,n^2-2,n^2-1$, we get
$$f(n)>n^{\frac{1}{2}+\frac1{2^2}+\ldots+\frac{1}{2^{n^2-n}}}\cdot (n^2)^{\frac{1}{2^{n^2-n+1}}}=n^{1-\frac{1}{2^{n^2-n}}}n^{\frac1{2^{n^2-n}}}=n.$$
For the upper bound, we can easily see that
$$\sqrt{n^2}<n^2+1.$$
So $$\sqrt{(n^2-1)\sqrt{n^2}}<\sqrt{(n^2-1)(n^2+1)}<\sqrt{n^4}=n^2.$$
We continue by induction to show that
$$\sqrt{k\sqrt{(k+1)\sqrt{(k+2)\cdots\sqrt{(n^2-2)\sqrt{(n^2-1)\sqrt{n^2}}}}}}<k+1$$
for all positive integers $k\leq n^2$. In particular this shows that
$$f(n)<n+1.$$
Therefore
$$n\leq f(n)<n+1$$
for every positive integer $n$, with equality on LHS iff $n=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3366914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Sum of the first natural numbers: how many and what are the most common methods to verify it? We know that Gauss has shown that the sum $S$ of the first $n$ natural numbers is given by the relation:
$$S=\frac{n(n+1)}{2} \tag{*}$$
The proof that I remember most frequently is as follows:
Let be $S=1+2+\dotsb+(n-1)+n \tag{1}$ We can write $S$ it also as: $\tag{2} S=n+(n-1)+\dotsb+2+1.$
By adding up member to member we get:
$\tag{3} 2S=\underbrace{(n+1)+(n+1)+\dotsb+2+1}_{n-\mathrm{times}}.$
Hence we obtain the $(^\ast)$.
How many other simple methods exist to calculate the sum of the first natural numbers?
| The expression of $S_n$ must be a quadratic polynomial (because the difference $S_n-S_{n-1}=n$ is a linear polynomial). As $S_0=0$, it is of the form $an^2+bn.$
By identification,
$$\begin{cases}S_1=a+b=1,\\S_2=4a+2b=3\end{cases}$$
and $$a=b=\dfrac12.$$
The Lagrangian polynomial by $(0,0)$, $(1,1)$, $(2,3)$ is $\dfrac{x^2}2+\dfrac x2$.
$$S(n)-S(n-1)=an^2+bn-a(n-1)^2-b(n-1)=a(2n-1)+b=n$$
so that
$$a=b=\dfrac12.$$
A posteriori:
$$S_n-S_{n-1}=\frac{n(n+1)}2-\frac{(n-1)n}2=n.$$
As an arithmetic progression is linear, the average term is the average of the extreme terms.
$$\frac{S_n}n=\frac{1+n}2$$ and $$S_n=\frac{n(n+1)}2.$$
Let the function defined by a geometric series
$$f(x):=\sum_{k=0}^n x^k=\frac{x^{n+1}-1}{x-1}.$$
We differentiate on $x$,
$$f'(x)=\sum_{k=0}^n k x^{k-1}=\frac{x^{n+1}-1}{x-1}=\frac{(n+1)x^n(x-1)-(x^{n+1}-1)}{(x-1)^2}.$$
Then with $x=1$, using L'Hospital,
$$f'(1)=\sum_{k=0}^n k =\lim_{x\to0}\frac{nx^{n+1}-(n+1)x^n+1}{(x-1)^2} =\lim_{x\to0}\frac{n(n+1)x^{n}-(n+1)nx^{n-1}}{2(x-1)}=\frac{n(n+1)}2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3371266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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Help with $\int \frac{\sqrt{x^2 + 1}}{x}\:dx$ I have been starting at the following integral for the past 2 hours and can not see where I have gone wrong. Can anyone please assist in isolating where I've made a mistake.
My sanity thanks you in advance,
\begin{equation*}
I = \int \frac{\sqrt{x^2 + 1}}{x}\:dx
\end{equation*}
Let $x = \tan(s)$:
\begin{equation*}
\frac{dx}{ds} = \sec^2(s) \rightarrow dx = \sec^2(s)\:ds
\end{equation*}
Thus,
\begin{align*}
I &= \int \frac{\sqrt{x^2 + 1}}{x}\:dx = \int \frac{\sqrt{\tan^2(s) + 1}}{\tan(s)}\cdot \sec^2(s)\:ds = \int \frac{\sec(s)}{\tan(s)}\sec^2(s)\:ds = \int \frac{\sec^3(s)}{\tan(s)}\:ds \\
&= \int \frac{\frac{1}{\cos^3(s)}}{\frac{\sin(s)}{\cos(s)}}\:ds = \int \frac{1}{\cos^3(s)} \cdot \frac{\cos(s)}{\sin(s)}\:ds = \int \frac{1}{\cos^2(s)\sin(s)}\:ds = \int \frac{1}{\left(1 - \sin^2(s)\right)\sin(s)}\:ds \\
&= \int \frac{1}{\left(1 + \sin(s)\right)\left(1 - \sin(s)\right)\sin(s)}\:ds
\end{align*}
Applying a Partial Fraction Decomposition we see:
\begin{equation*}
\frac{1}{\left(1 + \sin(s)\right)\left(1 - \sin(s)\right)\sin(s)} = \frac{1}{\sin(s)} - \frac{1}{2}\cdot \frac{1}{1 + \sin(s)} - \frac{1}{2}\cdot \frac{1}{1 - \sin(s)}
\end{equation*}
Thus,
\begin{align*}
I &= \int \frac{1}{\left(1 + \sin(s)\right)\left(1 + \sin(s)\right)\sin(s)}\:ds = \int \left[\frac{1}{\sin(s)} - \frac{1}{2}\cdot \frac{1}{1 + \sin(s)} - \frac{1}{2}\cdot \frac{1}{1 - \sin(s)} \right]\:ds \\
&= \int \frac{1}{\sin(s)}\:ds - \frac{1}{2} \int \frac{1}{1 + \sin(s)}\:ds - \frac{1}{2} \int \frac{1}{1 - \sin(s)}\:ds
\end{align*}
We now employ the Weierstrass Substitution $t = \tan\left(\frac{s}{2} \right)$:
\begin{equation*}
ds = \frac{2}{1 + t^2}\:dt, \quad \sin(s) = \frac{2t}{1 + t^2}
\end{equation*}
Thus,
\begin{align*}
I&= \int \frac{1}{\sin(s)}\:ds - \frac{1}{2} \int \frac{1}{1 + \sin(s)}\:ds - \frac{1}{2} \int \frac{1}{1 - \sin(s)}\:ds \\
&= \int \frac{1}{\frac{2t}{1 + t^2}} \cdot \frac{2}{1 + t^2}\:dt - \frac{1}{2} \int \frac{1}{1 + \frac{2t}{1 + t^2}}\cdot \frac{2}{1 + t^2}\:dt - \frac{1}{2} \int \frac{1}{1 - \frac{2t}{1 + t^2}}\cdot \frac{2}{1 + t^2}\:dt \\
&= \int \frac{1}{t}\:dt - \int \frac{1}{1 + t^2 + 2t}\:dt - \int \frac{1}{1 + t^2 - 2t}\:dt \\
&= \int \frac{1}{t}\:dt - \int \frac{1}{\left(t + 1\right)^2}\:dt - \int \frac{1}{\left(t - 1\right)^2}\:dt = \ln\left|t\right| - -\frac{1}{t + 1} - - \frac{1}{t - 1} + C \\
&= \ln\left|t\right| +\frac{1}{t + 1} + \frac{1}{t - 1} + C = \ln\left|t\right| +\frac{2t}{t^2 - 1} + C
\end{align*}
Where $C$ is the constant of integration. Here:
\begin{equation*}
t = \tan\left(\frac{s}{2} \right) = \tan\left(\frac{\arctan(x)}{2} \right) = \frac{\sqrt{x^2 + 1} - 1}{x}
\end{equation*}
Thus,
\begin{align*}
I&= \ln\left|t\right| +\frac{2t}{t^2 - 1} + C = \ln\left|\frac{\sqrt{x^2 + 1} - 1}{x}\right| +\frac{2\left(\frac{\sqrt{x^2 + 1} - 1}{x}\right)}{\left(\frac{\sqrt{x^2 + 1} - 1}{x} \right)^2 - 1} + C \\
&= \ln\left|\frac{\sqrt{x^2 + 1} - 1}{x}\right| + 2 \cdot \frac{\sqrt{x^2 + 1} - 1}{x} \cdot \frac{1}{\left(\frac{x^2 + 1 - 2\sqrt{x^2 + 1} + 1}{x^2} \right) - 1} + C \\
&= \ln\left|\frac{\sqrt{x^2 + 1} - 1}{x}\right| + 2 \cdot \frac{\sqrt{x^2 + 1} - 1}{x} \cdot \frac{x^2}{\left(x^2 + 1 - 2\sqrt{x^2 + 1} + 1\right) - x^2} + C \\
&= \ln\left|\frac{\sqrt{x^2 + 1} - 1}{x}\right| + 2 \cdot \frac{\sqrt{x^2 + 1} - 1}{x} \cdot \frac{x^2}{2\left(1 - \sqrt{x^2 + 1} \right)} + C = \ln\left|\frac{\sqrt{x^2 + 1} - 1}{x}\right| - x + C \\
\end{align*}
| In some cases trigonometric substitution should be avoided
$$\displaystyle I=\int\dfrac{\sqrt{1+x^2}}xdx=\int x\cdot\dfrac{\sqrt{1+x^2}}{x^2}dx$$
Let $\sqrt{1+x^2}=u\implies1+x^2=u^2,x\ dx=u \ du$
$$I=\int\dfrac u{u^2-1}udu=1+\dfrac1{u^2-1}=1+\dfrac{1+u+1-u}{(1+u)(1-u)}=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3371363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
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I have troubles with proving the induction step: $\frac{1}{2} * \frac{3}{4} * \frac{5}{6} * ... * \frac{(2k+1)}{2k+2} ≤ \frac{1}{\sqrt{2k+3}}$ Let's assume that the unequality is right if $n = k$, that is:
$\frac{1}{2} * \frac{3}{4} * \frac{5}{6} * ... * \frac{(2k-1)}{2k} ≤ \frac{1}{\sqrt{2k+1}}$
Let's prove that the unequality is right for $n = k + 1$ as well, that is:
$\frac{1}{2} * \frac{3}{4} * \frac{5}{6} * ... * \frac{(2k+1)}{2k+2} ≤ \frac{1}{\sqrt{2k+3}}$
I've spent a comparable amount of time trying to prove this point but looks like I can't grasp what to do.
| Note that$$\frac12\times\frac34\times\frac56\times\cdots\times\frac{2k-1}{2k}\times\frac{2k+1}{2k+2}\leqslant\frac1{\sqrt{2k+1}}\times\frac{2k+1}{2k+2},$$by the induction hypothesis. So, it would be great if you could prove that$$\frac1{\sqrt{2k+1}}\times\frac{2k+1}{2k+2}\leqslant\frac1{\sqrt{2k+3}}\tag1$$But $(1)$ is equivalent to$$\left(\frac{2k+1}{2k+2}\right)^2\leqslant\frac{2k+1}{2k+3},$$which in turn is equivalent to$$(2k+1)(2k+3)\leqslant(2k+2)^2.$$Can you take it from here?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int \cos^2x\sin^4x\mathrm{d}x$
Evaluate integral $$\int \cos^2x\sin^4x\mathrm{d}x.$$
Attempt. Setting $\tan x=t$, gives:
$$\int \cos^2x\sin^4x\mathrm{d}x =\int \frac{1}{1+t^2} \,\left(\frac{t^2}{1+t^2}\right)^2 \frac{\mathrm{d}t}{1+t^2}=\int \frac{t^4}{(1+t^2)^4} \mathrm{d}t,$$
which does not seem to be elementary.
Thank in advance for the help.
| Here is to integrate economically,
$$\cos^2x\sin^4 x = \frac 18 \sin^2 2x (1-\cos 2x)= \frac {1}{16}-\frac {1}{16}\cos 4x-\frac 18 \sin^22x\cos 2x$$
Thus,
$$\int \cos^2x\sin^4x\mathrm{d}x = \frac {x}{16}-\frac {1}{64}\sin4x-\frac{1}{48}\sin^32x +C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3372371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to write $\sum_{n=0}^{N-1} \binom{2n}{n}x^{n+1}$ in closed form or short form? How to write $\sum_{n=0}^{N-1} \binom{2n}{n}x^{n+1}$ in closed form or short form?
We can write the above sum as
$\binom{0}{0}x+\binom{2}{1}x^2+\binom{4}{2}x^3+\binom{6}{3}x^4+\cdots+\binom{2N-2}{N-1}x^N$.
There is a formula $(1+x)^n=\sum_{k=0}^{n} \binom{n}{k}x^k$.
But how to apply it?
Is there any short form or any function that would give short form ?
| We can transform using the duplication formula DLMF
\begin{align}
\sum_{n=0}^{N-1} \binom{2n}{n}x^{n+1}&=x\sum_{n=0}^{N-1} \frac{x^n}{n!}\frac{\Gamma(2n+1)}{\Gamma(n+1)}\\
&=\frac{x}{\sqrt{\pi}}\sum_{n=0}^{N-1} \frac{(4x)^n}{n!}\frac{\Gamma(n+1)\Gamma(n+1/2)}{\Gamma(n+1)}\\
&=\frac{x}{2\sqrt{\pi}}\sum_{n=0}^{\infty} \frac{(4x)^n}{n!}\Gamma(n+1/2)-\frac{x}{\sqrt{\pi}}\sum_{n=N}^{\infty} \frac{(4x)^n}{n!}\Gamma(n+1/2)\\
&=\frac{x}{\sqrt{1-4x}}-\frac{x}{\sqrt{\pi}}\sum_{n=N}^{\infty} \frac{(4x)^n}{n!}\Gamma(n+1/2)
\end{align}
from the Newton's generalized binomial theorem. Now, shifting the index in the sum,
\begin{align}
\sum_{n=N}^{\infty} \frac{(4x)^n}{n!}\Gamma(n+1/2)&=\left( 4x \right)^N\sum_{k=0}^{\infty} \frac{(4x)^k}{(N+k)!}\Gamma(N+k+1/2)\\
&=\left( 4x \right)^N\sum_{k=0}^{\infty} \frac{(4x)^k}{k!}\frac{\Gamma(k+1)\Gamma(k+N+1/2)}{\Gamma(k+N+1)}\\
&=\left( 4x \right)^N\frac{\Gamma(N+1)}{\Gamma(N+1/2)}\,_2F_1\left( 1,N+1/2;N+1;4x) \right)
\end{align}
from the definition of the hypergeometric function. Finally, as noted by @Sil in the comment
\begin{equation}
\sum_{n=0}^{N-1} \binom{2n}{n}x^{n+1}=\frac{x}{\sqrt{1-4x}}- \binom{2N}{N}x^{N+1}\,_2F_1\left( 1,N+1/2;N+1;4x) \right)
\end{equation}
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove the inequality is true Here is a question that I need to prove
Prove that for $a, b \geq 0$
$$a^8+b^8\geq a^3b^5+a^5b^3$$
So far I have managed to simplify to
$$(a^3-b^3)(a^5-b^5)\geq 0$$
| Note that $$(a^3-b^3)(a^5-b^5) = (a-b)(a^2+ab+b^2)(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4)$$
$$=(a-b)^2(a^2+ab+b^2)(a^4+a^3b+a^2b^2+ab^3+b^4)\ge 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3378873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Given three non-negative numbers $a,b,c$. Prove that $\frac{a+b+c}{k}\geqq\sum\limits_{cyc}\frac{a-b}{b+ k}$ for $k= constant$ so that $k> 0$ .
Given three non-negative numbers $a, b, c$. Prove that
$$\frac{a+ b+ c}{k}\geqq \frac{a- b}{b+ k}+ \frac{b- c}{c+ k}+ \frac{c- a}{a+ k}$$
for $k= constant$ so that $k> 0$ .
For $k= 2$, we can use
$$\frac{a+ b}{2}\geqq \frac{a- b}{b+ 2}$$
but for $k= constant$ ? And I think it would be really really easy if expands. Thanks for help that a lot
| Suppose that $a$ is the largest and $c$ the smallest we see that: $$\frac{a} {k} \ge \frac{a-b} {b+k} $$ and $$\frac{b} {k} \ge \frac{b-c} {c+k} $$ and as $c-a$ is not positive we deduce: $$\frac{c} {k} \ge \frac{c-a} {a+k} $$
Now add these 3 inequalities .
| {
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Simplify this expansion : $\omega=(28+(\frac{5290}{3})^{\frac{3}{2}})^{\frac{1}{3}}+(28-(\frac{5290}{3})^{\frac{3}{2}})^{\frac{1}{3}}$ Find a simple closed form of :
$\omega=(28+(\frac{5290}{3})^{\frac{3}{2}})^{\frac{1}{3}}+(28-(\frac{5290}{3})^{\frac{3}{2}})^{\frac{1}{3}}$
My try :
Let :
$A=(28+(\frac{5290}{3})^{\frac{3}{2}})^{\frac{1}{3}}$
And
$B=(28-(\frac{5290}{3})^{\frac{3}{2}})^{\frac{1}{3}}$
Now :
$A^{3}+B^{3}=56$
But how I can now find $A$ and $B$ ?
| Given
$$\omega=\sqrt[3]{28+(\frac{5290}{3})^{\frac{3}{2}}}+\sqrt[3]{28-(\frac{5290}{3})^{\frac{3}{2}}}$$
Let $A=\sqrt[3]{28+(\frac{5290}{3})^{\frac{3}{2}}}\Rightarrow A^3=28+(\frac{5290}{3})^{\frac{3}{2}}$
and $B=\sqrt[3]{28-(\frac{5290}{3})^{\frac{3}{2}}}\Rightarrow B^3=28-(\frac{5290}{3})^{\frac{3}{2}}$.
Notice that $A^3+B^3=56$.
Also, $AB=\sqrt[3]{\big(28+(\frac{5290}{3})^{\frac{3}{2}}\big)\big(28-(\frac{5290}{3})^{\frac{3}{2}}\big)}=\sqrt[3]{28^2-(\frac{5290}{3})^3}=-\frac23\sqrt[3]{18504483479}$.
So we can write
\begin{align}
\omega&=A+B\\
\omega^3&=A^3+3A^2B+3AB^2+B^3=A^3+B^3+3AB\cdot(A+B)\\
\omega^3&=56+3(-\frac23\sqrt[3]{18504483479})\ \omega\\
\omega^3&=56-2\sqrt[3]{18504483479}\ \omega\\
\end{align}
At the end of the day, you are solving the cubic equation
$$\omega^3+2\sqrt[3]{18504483479}\ \omega-56=0\\$$
There has got to be a typo somewhere.
| {
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"url": "https://math.stackexchange.com/questions/3379567",
"timestamp": "2023-03-29T00:00:00",
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Help solve recurrence : f(x) = 2f(x-1) + x While trying to analyze the time complexity of the heapify algorithm ,
I came up with the following recurrence for it's time complexity in terms of heap height:
T(h) = 2T(h-1) + h
Could you help me approach this recurrence?
| $$a_k = 2a_{k-1} + k$$
$$\implies \dfrac{a_k}{2^k} = \dfrac{a_{k-1}}{2^{k-1}} + \dfrac{k}{2^k}$$
$$\implies \dfrac{a_k}{2^k} - \dfrac{a_{k-1}}{2^{k-1}} = \dfrac{k}{2^k}$$
$$\implies \sum_{k=1}^{n} \left(\dfrac{a_k}{2^k} - \dfrac{a_{k-1}}{2^{k-1}}\right) = \sum_{k=1}^{n} \dfrac{k}{2^k}$$
$$ \implies \dfrac{a_n}{2^n} - \dfrac{a_0}{2^0} = \sum_{k=1}^{n} \dfrac{k}{2^k} $$
Note that the R.H.S. is an Arithmetic-Geometric Progression. Thus,
$$\sum_{k=1}^{n} \dfrac{k}{2^k} = \dfrac{1}{2^n}(2^{n+1} - n -2)$$
$$\implies \dfrac{a_n}{2^n} - \dfrac{a_0}{2^0}= \dfrac{1}{2^n}(2^{n+1} - n -2) $$
$$ \displaystyle \implies \boxed{a_n = 2^n(a_0 + 2) - n -2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3382051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality $a^2+4b^2<1$. Let $a,b$ be two strictly positive real numbers that satisfy $a^3+b^3=a-b$. How can I prove that $a^2+4b^2<1$?
So because $a^3+b^3$ is $>0$ I know that $a>b$.
I tried using $a^3+b^3=(a+b)(a^2-ab+b^2)$ but that didn‘t lead me anywhere...
| Note that $$\frac {a^3+b^3 } { a-b}=1 $$
We show that $$a^2 +4b^2 <\frac {a^3+b^3 } { a-b}$$
Since $a-b$ is positive we may multiply both sides by $a-b$ to get $$(a^2+4b^2)(a-b)< a^3+b^3$$
After simplification on both sides we get $$a^2-4ab+5b^2>0$$
That is the same as $$(a-2b)^2+b^2 >0$$ which is true.
| {
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"url": "https://math.stackexchange.com/questions/3386240",
"timestamp": "2023-03-29T00:00:00",
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If $f(2x-3) = 4x-2$, then what is $f(x)$? I have this statement:
If $f(2x-3) = 4x-2,$ the function $f(x)$ is ...?
My attempt was:
Move the function $3$ units to the left
$f(2x) = 4(x+3) -2 = 4x+10$
Divide $x$ by $2$
$f(x) = 2x+10$
Verifiy $f(x) = 2x+10 \to f(2x) = 4x+10 \to f(2x-3) = 4(x-3)+10 = \underbrace{4x-2}_{f(2x-3)}$
But according to the guide the correct answer is $2x+4$ and i don't know why. Thans in advance.
| Let $y$ equal to an expression that will make $2y-3=x$ so that we can have $f(x)$.
This can be done if $y=\frac{x}{2}+1.5$
Plug $y$ in:
$f(2y-3)=4y-2$ and substitute the value of $y$ in terms of $x$:
$f(2(\frac{x}{2}+1.5)-3)=f(x)=4(\frac{x}{2}+1.5)-2=2x+4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3387446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Polynomials that induce the zero function mod $n$
*
*Which polynomials induce the zero function mod $n$?
In particular:
*
*What is the polynomial of least degree that induces the zero function mod $n$?
*What is the monic polynomial of least degree that induces the zero function mod $n$?
These are not vacuous questions because of the following general result:
If $r$ is the maximum exponent in the prime factorization of $n$, then $x \mapsto x^{r+\lambda (n)}-x^r$ is the zero function mod $n$. [Wikipedia]
Here, $\lambda$ is the Carmichael function.
*
*When is $x^{r+\lambda (n)}-x^r$ the monic polynomial of least degree that induces the zero function mod $n$?
Fermat's theorem implies that $x^n-x$ is the answer for $n$ prime: all polynomials that induce the zero function mod $n$ are a multiple of $x^n-x$.
How can this be generalized to composite $n$?
Here are some other examples:
$$
\begin{array}{rll}
n & L_n: \text{least degree} & M_n: \text{least degree monic}
\\ 2 & x^2+x
\\ 3 & x^3-x
\\ 4 & 2(x^2+x) & x^4-x^2
\\ 5 & x^5-x
\\ 6 & 3(x^2+x) & x^3-x
\\ 7 & x^7-x
\\ 8 & 4(x^2+x) & x^4+2x^3+3x^2+2x = x(x+1)(x^2+x+2)
\\ 9 & 3(x^3-x) & x^8-x \quad (???)
\\ 10 & 5(x^2+x) & x^5-x
\\ 11 & x^{11}-x
\\ 12 & 6(x^2+x) & x^4+5x^2+6x = x(x+1)(x^2-x+6)
\\ 13 & x^{13}-x
\\ 14 & 7(x^2+x) & ???
\\ 15 & 5(x^3-x) & x^5-x
\\ 21 & 7(x^3-x) & ???
\\ 24 & 12(x^2+x) & x^4+2x^3+11x^2+10x = x(x+1)(x^2+x+10)
\end{array}
$$
It seems clear that $L_{2m}=m(x^2+x)$, because $x^2+x$ is always even.
More generally,
*
*Is $L_{pq} = qL_p$ and $M_{pq}=L_q$ for $p<q$ primes?
*If $n=pm$ and $p$ is smallest prime divisor of $n$, then is $L_{pm}=mL_p$?
Corrections and additions welcome.
Please collect partial results as answers.
| These papers characterize the polynomials that induce the zero function mod $n$:
*
*On polynomial functions (mod $n$) by Singmaster (1974)
*Polynomials and their residue systems by Kempner (1921)
Kempner's paper is closer to what I have in mind. I'll have to check it more closely.
I came across Kempner's paper in the paper A basis for residual polynomials in $n$ variables by Litzinger (1935).
| {
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"timestamp": "2023-03-29T00:00:00",
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Confused about the approach to this Delta Epsilon Proof I have recently started with Calculus and got into Delta Epsilon Proofs as I found it intriguing. After I went over some basic questions involving delta epsilon proofs, I have reached a question which I was not sure how to approach.
This is the questions:
Prove that $\lim_{x\to 2} x^4 = 16$
So, what I have basically started doing is the following:
Let $δ$ = 1, and by the definition of a limit I arrived that $|x-2|$ < $1$, so I concluded that $|x + 2| < 5$. Then, I have arrived that $|x-2||x+2||x-2||x+2| < 5|x+2| * 5|x-2|$.
But I don't really know what to do next, and if my approach is even right.
| Well, in the end we want for all $\epsilon > 0$ there is a $\delta$ based on the $\epsilon$ so that $|x-2|<\delta \implies |x^4 -16|< \epsilon$.
If we go from $|x-2| < \delta \implies..... |x^4 - 16| < h(\delta)$ for some function $h$ we can set $|x^4 - 16| < h(\delta) < \epsilon$ and solve for $\delta$ in terms of $\epsilon$.
So
$|x-2| < \delta = h^{-1}(\epsilon)$ for some $h$ we will discover later.
$2 - \delta < x < 2 + \delta$
$(2-\delta)^4 < x < (2+\delta)^4$ (if we assume $0< \delta <2$)
$16 - 32\delta + 24\delta^2 -8\delta^3 + \delta^4 < x^4 < 16 + 32\delta + 24\delta^2 +8\delta^3 + \delta^4$
$-32\delta + 24\delta^2 - 8\delta^3 +\delta^4 < x^4 -16 <32\delta + 24\delta^2 +8\delta^3 + \delta^4$
$-32\delta - 24\delta^2 - 8\delta^3-\delta^4 <-32\delta + 24\delta^2 - 8\delta^3 +\delta^4< x^4 -16 <32\delta + 24\delta^2 +8\delta^3 + \delta^4$
(and if we assume $0 < \delta < 1$) we have
$-32\delta - 24\delta -8\delta -\delta < 32\delta - 24\delta^2 - 8\delta^3-\delta^4 <-32\delta + 24\delta^2 - 8\delta^3 +\delta^4< x^4 -16 <32\delta + 24\delta^2 +8\delta^3 + \delta^4 < 32\delta + 24\delta +8\delta + \delta$
so
$-65\delta < x^4 -16 < 65\delta$
$|x^4 - 16 |< 65\delta$
And if we let $\delta = h(\epsilon) = \min (1, \frac {\epsilon}{65})$ then
$|x^4 - 16| < 65\delta \le \epsilon$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to solve $\sin(3x)+\cos(2x)={1\over2}$ I have a trouble to continuing this problem.
My work so far :
$$\sin(3x)+\cos(2x)={1\over2}$$
I try using this :
$$\sin(3x)+\sin(90^\circ-2x)={1\over2}$$
$$2\sin\left({3x+90^\circ-2x\over2}\right)\cos\left({3x-90^\circ+2x\over2}\right)={1\over2}$$
$$4\sin\left({x+90^\circ\over2}\right)\cos\left({5x-90^\circ\over2}\right)=1$$
How to solving this equation ?
Thanks for your help.
| Recall that:
$$\sin(a+b) + \sin(a-b) = 2\sin(a)\cos(b).$$
Let $a=x$ and $b=2x$. Then:
$$\sin(x+2x) + \sin(x-2x) = 2\sin(x)\cos(2x) \Rightarrow\\
\sin(3x) + \sin(-x) = 2\sin(x)\cos(2x) \Rightarrow\\
\sin(3x) - \sin(x) = 2\sin(x)\cos(2x) \Rightarrow\\
\sin(3x) = 2\sin(x)\cos(2x) + \sin(x).$$
At this point, your equation can be rewritten as follows:
$$\sin(3x)+\cos(2x)={1\over2} \Rightarrow\\
2\sin(x)\cos(2x) + \sin(x) + \cos(2x) = {1\over2}.$$
Now, recall that $\cos(2x) = 1 - 2\sin^2(x).$ Then:
$$2\sin(x)(1-2\sin^2(x)) + \sin(x) + (1-2\sin^2(x)) = {1\over2} \Rightarrow \\
2\sin(x)-4\sin^3(x)+\sin(x)+1-2\sin^2(x) = {1\over2} \Rightarrow \\
8\sin^3(x)+4\sin^2(x)-6\sin(x)-1 = 0.$$
To solve the last line, read the answer by lab bhattacharjee, which suggests you to use the Cardano's formula for finding the roots of third order polynomial, obtained with the substitution $s = \sin(x)$. Also, the Ruffini's rule is a good method for finding the roots of polynomials.
| {
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"timestamp": "2023-03-29T00:00:00",
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Does $x \cdot \sin \left( \frac{1}{x} \right) = 1$ has solutions? A look to the plot show to you that the function $f(x) = x \cdot \sin \left( \frac{1}{x} \right) - 1 $ has no zero near the origin. Wolframalpha software says that $ x \cdot \sin \left( \frac{1}{x} \right) = 1$ if $x \approx 5.16\cdot 10^{15}$ but I suspect this huge number is a wrong solution due to numerical problems. Is it true that for some $x$, far away from the origin, the equation $$ x \cdot \sin \left( \frac{1}{x} \right) = 1 $$ is satisfied? Can, far away, $f(x)= x \cdot \sin \left( \frac{1}{x} \right)$ exceeds 1?
| Consider substituting \frac{1}{x}:
$$
x \cdot \sin \left( \frac{1}{x} \right) = 1\\
\iff\\
\sin \left( \frac{1}{x} \right) = \frac{1}{x}\\
\iff\\
\sin (z) = z\\
$$
This is only satisfied if $z=0$, implying $x=\infty$, which is what wolfram was trying to say.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $x\equiv 1\pmod{5}, x\equiv 2\pmod{6}, x\equiv 3\pmod{7}$ Solve $x\equiv 1\pmod{5}, x\equiv 2\pmod{6}, x\equiv 3\pmod{7}$
First I can see $x=5t+1, t\in Z$. Then they insert this into the second equation, which is $5t+1\equiv 2(mod6)$, which leads to $t\equiv 5mod6$. Then they get $t=6u+5$, I'm confused how they got to $t=6u+5$ from $5t+1$. Also if anyone has any strategies to prove these type of problems where there's an $x$ that has different moduluses.
| x % 5 = 1 = -4
x % 6 = 2 = -4
x % 7 = 3 = -4
This means x % LCM(5, 6, 7) = -4
x % 210 = -4
Therefore x = 210t - 4 (-4, 206, ...)
| {
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"timestamp": "2023-03-29T00:00:00",
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An elementary inequality for 3 positive real numbers Let $a, b, c > 0$ and $n \ge 0$ . Show that
$$\frac{a^2}{b^2 + nab} + \frac{b^2}{c^2 + nbc} + \frac{c^2}{a^2 + nca} \ge \frac{(a+b+c)^2}{(n+1)(ab+bc+ca)} $$
My first attempt was direct computations in the hope that this will lead to some simplifications or a known inequality , but this was not the case.
Another idea is to use induction over n .
For $n = 0$ the inequality become simpler :
$$\frac{a^2}{b^2} + \frac{b^2}{c^2} + \frac{c^2}{a^2} \ge \frac{(a+b+c)^2}{ab+bc+ca} $$
But still direct computation or manipulating simpler inequalities (like those between arithmetic , geometric and harmonic means ) don't seems to work .
Any suggestions are welcome. Thank you !
| By Holder
$$\sum_{cyc}\frac{a^2}{b^2+nab}=\frac{\sum\limits_{cyc}\frac{a^2}{b(b+na)}\sum\limits_{cyc}ab\sum\limits_{cyc}(b+na)}{\sum\limits_{cyc}ab\sum\limits_{cyc}(b+na)}\geq$$
$$\geq\frac{(a+b+c)^3}{\sum\limits_{cyc}ab\sum\limits_{cyc}(b+na)}=\frac{(a+b+c)^2}{(n+1)(ab+ac+bc)}.$$
| {
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Convergent series??? Determine for each of the following series whether or not it is convergent. You must justify your answer.
1. $\sum_{n=1}^{\infty} \left(\left(n^3+1\right)^{\frac{1}{3}}-n\right)$.
Could I use $a^3-b^3 = (a-b)(a^2+ab+b^2))$? If so how?
Using the hint given we get that
\begin{equation*}
\begin{split}
\left(n^3+1\right)^{\frac{1}{3}}-n &= \frac{\left[\left(n^3+1\right)^{\frac{1}{3}}\right]^3-n^3}{\left[\left(n^3+1\right)^{\frac{1}{3}}\right]^2+\left(n^3+1\right)^{\frac{1}{3}}\cdot n+n^2} \\
&= \frac{n^3+1-n^3}{\left(n^3+1\right)^{\frac{2}{3}}+n\left(n^3+1\right)^{\frac{1}{3}}+n^2} \\
&= \frac{1}{\left(n^3+1\right)^{\frac{2}{3}}+n\left(n^3+1\right)^{\frac{1}{3}}+n^2} \\
&\leq \frac{1}{n^2}.
\end{split}
\end{equation*}
But we know that $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges ($p$-series with $p = 2 > 1$). Thus, by the comparison test $\sum_{n=1}^{\infty} \left(\left(n^3+1\right)^{\frac{1}{3}}-n\right)$ converges as well.
2. $\sum_{n=2}^{\infty} \frac{1}{n\ln{(n)}^3}$.
The function $f:[2,\infty) \mapsto \mathbb{R}$ defined by $f(x) := \frac{1}{x\ln{(x)}^3}$ is continuous, positive and monotonely decreasing by examining the derivative $f'(x) = -\frac{\ln{(x)}+3}{x^2(\ln{(x)})^4} < 0$ for all $x\geq 2$.
Since $f(n) = \frac{1}{n\ln{(n)}^3}$, to determine whether or not the series converges it suffices to see what happens to $\int_{2}^{\infty} f(x) \; dx$. Doing so, letting $u = \ln{(x)} \Rightarrow du = \frac{dx}{x}$ gives
\begin{equation*}
\begin{split}
\int_{2}^{t} \frac{1}{x\ln{(x)}^3} \; dx &= \lim_{t\to\infty} \int_{\ln{(2)}}^{t} \frac{1}{u^3} \; du \\
&= \lim_{t\to\infty} \left[-\frac{1}{2u^2}\right]_{\ln{(2)}}^{t} \\
&= \frac{1}{2\ln{(2)}^2}
\end{split}
\end{equation*}
which is finite. This integral converges so by the integral test $\sum_{n=2}^{\infty} \frac{1}{n\ln{(n)}^3}$ converges as well.
3. $\sum_{n=1}^{\infty} \frac{\sin{(\pi+n)}}{\sqrt{n}+2^n}$.
First we note that
\begin{equation*}
\frac{\sin{(\pi+n)}}{\sqrt{n}+2^n} = -\frac{\sin{(n)}}{\sqrt{n}+2^n}.
\end{equation*}
This series is absolutely convergent if and only if
\begin{equation*}
\sum_{n=1}^{\infty} \left|-\frac{\sin{(n)}}{\sqrt{n}+2^n}\right| = \sum_{n=1}^{\infty} \frac{\sin{(n)}}{\sqrt{n}+2^n}
\end{equation*}
converges. We'll prove that this series does converge by using the comparsion test.
Observe that for any $k\geq 1$,
\begin{equation*}
\frac{\sin{(n)}}{\sqrt{n}+2^n}\leq \frac{1}{\sqrt{n}+2^n}\leq \frac{1}{2^n}.
\end{equation*}
But we know that the series $\sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^n$ converges as it's just a geometric series with ratio $\frac{1}{2} < 1$. So the comparison test tells us that $\sum_{n=1}^{\infty} \frac{\sin{(\pi+n)}}{\sqrt{n}+2^n}$ converges, and thus the original series is absolutely convergent.
| For the first one,
by Bernoulli's inequality,
$(1+x/n)^n \ge 1+x
$
so
$(1+x)^{1/n} \le 1+x/n
$.
Therefore
$\begin{array}\\
(n^m+1)^{1/m}-n
&=n((1+n^{-m})^{1/m}-1)\\
&\le n((1+n^{-m}/m)-1)\\
&= n(n^{-m}/m)\\
&= n^{-m+1}/m\\
\end{array}\\
$
and the sum of this converges if
$m-1 > 1
$
or
$m > 2
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3396721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding the summation of a series Determine the following sum
$$\frac{5}{3 \cdot 6 }\cdot \frac{1}{4^2} + \frac{5\cdot 8}{3 \cdot 6 \cdot 9 }\cdot \frac{1}{4^3} + \frac{5 \cdot 8 \cdot 11}{3 \cdot 6 \cdot 9 \cdot 12}\cdot \frac{1}{4^4} + \frac{5\cdot8\cdot 11\cdot 14}{3 \cdot 6\cdot 9\cdot 12 \cdot 15}\cdot \frac{1}{4^5} + \dots \dots$$
I tried to use Generalised Binomial Theorem, but I am unable to find $x,y,r$
The Generalized Binomial Theorem
$$(x+y)^{r}=\sum _{k=0}^{\infty }{r \choose k}x^{r-k}y^{k}$$ where the ${r \choose k }$ denotes the falling factorial.
I notice that the factors in the numerator is increasing, hence I am unable to use it.
| Hint: $$\frac{5 \cdot 8 \cdot 11 \cdot 14}{3 \cdot 6 \cdot 9 \cdot 12 \cdot 15}=\frac{1}{3 \cdot 5!}(1+\frac{2}{3})(2+\frac{2}{3})(3+\frac{2}{3})(4+\frac{2}{3})=\frac{(-1)^4}{3 \cdot 5!}\prod_{k=0}^3{\left(-\frac{5}{3}-k\right)}$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3398754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How can I prove $5^{3^n}+1$ is divisible by $3^{n+1}$ The problem was to prove that $5^{3^n}+1$ is divisible by $3^{n+1}$ for all nonnegative integers.
I attempted to prove the claim by induction.
Claim: For all $ n \ge 0 $ $$ 3^{n+1} ~|~ 5^{3^n}+1 $$
Basis case:
$$ 3^{0+1} = 3 ~|~ 6 = 5^{3^0} + 1 ~ ~\checkmark$$
Inductive step: Assume $$ 3^{k+1} ~|~ 5^{3^k} + 1 $$
It follows by the definition of divides, $ 5^{3^k } + 1 = 3^{k+1}\cdot m $ for some integer $m$.
Then $$~~~~~ 5^{3^{k+1}} +1
\\ = 5^{3^k \cdot 3 }+1
\\ = 5^{3 \cdot 3^k}+1
\\ = \left( 5^3 \right)^{3^k} +1
\\= 125^{3^k} +1 $$ I have no idea how to proceed. Am I on the right track? If it is possible can I get a hint.
One thing I did notice is that it might be easier to prove $$ 3^{n} ~|~ 5^{3^n}+1 $$ This statement seems to be true, since the basis case is then $1|6$.
| Try using the fact that $x^3 +1 = (x+1)(x^2 -x+1).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3401740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Argument of a complex fraction, why different answers? I want to take the argument of the following complex fraction. Using the second method I get a different answer, why is that?
$$
G(\omega)=
\frac{1}{(1+2\omega i)^2}
\tag 1
$$
Method 1:
\begin{align}
\arg\frac{1}{(1+2\omega i)^2}
&=\arg1-\arg\Big((1+2\omega i)^2\Big) \tag 2\\
&=\arg1-\arg\Big((1+2\omega i)(1+2\omega i)\Big) \tag 3\\
&=\arg1-\arg(1+2\omega i)-\arg(1+2\omega i) \tag 4\\
&=\arctan\frac{0}{1}-\arctan\Big(\frac{2\omega}{1}\Big)-\arctan\Big(\frac{2\omega}{1}\Big) \tag 5\\
&=-2\arctan(2\omega) \tag 6
\\
\end{align}
Method 2:
Expand the denominator:
$$
(1+2\omega i)^2=1-4\omega^2+4\omega i
$$
So I have
\begin{align}
\arg \frac{1}{(1+2\omega i)^2}
&=\arg\frac{1}{1-4\omega^2+4\omega i} \tag 7\\
&=\arg1-\arg(1-4\omega^2+4\omega i)\tag 8\\
&=-\arctan\bigg (\frac{4\omega}{1-4\omega^2}\bigg) \tag 9
\end{align}
So $(6)$ is not equal to $(9)$, What is wrong with method 2?
| Note that$$\tan\bigl(2\arctan(2\omega)\bigr)=\frac{4\omega}{1-4\omega^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3404060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Show that $\lim_{(x,y)\to (0,0)}\frac{x^4y^3}{x^2+y^4} = 0$. Show that
\begin{equation*}
\lim_{(x,y)\to (0,0)}\frac{x^4y^3}{x^2+y^4} = 0.
\end{equation*}
Let's show this formally using an $\epsilon-\delta$ proof. For $(x,y)\neq (0,0)$, let $\epsilon > 0$. Then if $(x,y)\in \mathbb{R}^2$ and $|(x,y)| < \frac{\epsilon}{3}$, then
\begin{equation*}
|y|^3\leq x^2+y^4 < \epsilon^3,
\end{equation*}
so $|y| < \sqrt[3]{\epsilon}$. Thus,
\begin{equation*}
\begin{split}
\left|\frac{x^4y^3}{x^2+y^4}-0\right| &= \left|\frac{x^4y^3}{x^2+y^4}\right| \\
&= \frac{x^4|y|^3}{x^2+y^4} \\
&\leq \frac{x^4y^3}{x^4} \\
&= |y|^3 \\
&< \epsilon
\end{split}
\end{equation*}
and we are done.
Where have I gone wrong? Thanks.
| Let $|x|<1$, $|y| <1$, $(x,y) \not = (0,0)$;
$x^2+y^4 \ge x^4+y^4\ge 2x^2y^2$ (AM-GM);
$|\dfrac {x^4y^3}{x^2+y^4}|\le |\dfrac {x^4y^3}{2x^2y^2}|=$
$(1/2)x^2|y| \le (x^2+y^2)\cdot 1$;
Choose $ \delta= \epsilon^{1/2}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 3,
"answer_id": 1
} |
Find $\lim_{x\to ∞} (\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x})$ without L'Hopital or Taylor series.
$$\large \lim_{x\to ∞} (\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x})$$
My try is as follows:
$$\large \lim_{x\to ∞} (\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x})=$$$$ \lim_{x\to
∞}x\left(\sqrt[3]{1+\frac{3}{x}}-\sqrt{1\ -\frac{2}{x}}\right)$$$$=\lim_{x\to ∞}x\lim_{x\to
∞}\left(\sqrt[3]{1+\frac{3}{x}}-\sqrt{1\ -\frac{2}{x}}\right)$$
which is $∞×0$ , but clearly this zero is not exactly zero. I was thinking about generalized binomial
theorem, but seems it will make the limit difficult, so how this kind of limits can be solved without using Taylor series or L'Hopital's rule?
| Using high-school math:$$\lim_{x\to ∞} (\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x})=\lim_{x\to ∞} \frac{\sqrt[3]{(x^{3}+3x^{2})^2}-x^{2}+2x}{\sqrt[3]{x^{3}+3x^{2}}+\sqrt{x^{2}-2x}}=$$ $$ \lim_{x\to ∞} \frac{(x^{3}+3x^{2})^2-(x^{2}-2x)^3}{(\sqrt[3]{x^{3}+3x^{2}}+\sqrt{x^{2}-2x})(\sqrt[3]{(x^{3}+3x^{2})^4}+\sqrt[3]{(x^{3}+3x^{2})^2}(x^{2}-2x)+(x^2-2x)^2)}=$$ $$\lim_{x\to ∞} \frac{12x^5-3x^4+8x^3}{(\sqrt[3]{x^{3}+3x^{2}}+\sqrt{x^{2}-2x})(\sqrt[3]{(x^{3}+3x^{2})^4}+\sqrt[3]{(x^{3}+3x^{2})^2}(x^{2}-2x)+(x^2-2x)^2)}=$$ $$[\text{leaving the highest power}]=\lim_{x\to ∞} \frac{12x^5}{(x+x)(x^4+x^4+x^4)}=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3408325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Trouble figuring out a complex number equation I'm struggling to prove this question, an help is greatly appreciated!
If $a=cis(\pi/5)$, prove that:
$$a^7=-a^2$$
and
$$a^9=-a^4$$
| Let's think about it geometrically. $\frac{7 \pi}{5}$ and $\frac{2\pi}{5}$ differ by $\pi$, and so do $\frac{9 \pi}{5}$ and $\frac{4\pi}{5}$.
Geometrically, this means that in the complex plane, $e^{ i\frac{7 \pi}{5}}$ and $e^{i \frac{2\pi}{5}}$ point in opposite directions. Written algebraically, we have $e^{ i\frac{7 \pi}{5}} = - e^{i \frac{2\pi}{5}}$, or $a^7 = - a^2$.
Similarly, $e^{i \frac{9 \pi}{5}}$ and $e^{i \frac{4\pi}{5}}$ point in opposite directions, so $e^{i \frac{9 \pi}{5}} = - e^{i \frac{4\pi}{5}}$, or $a^9 = - a^4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3409143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the $1000$th digit after the decimal point of $\sqrt{n},$ where $n=\underbrace{11\dots1}_{1998 \text{ 1's}}$
Find the $1000$th digit after the decimal point of $\sqrt{n}$, where $n=\underbrace{11\dots1}_{1998 \text{ 1's}}$.
Obviously, $\underbrace{11\dots1}_{1998 \text{ 1's}}=\dfrac{1}{9}\left(9\cdot10^{1997}+9\cdot 10^{1996}+\dots+9\right),$ so we want to find $\left(\dfrac{10^{1998}-1}{9}\right).$ If only there was some way to convert this expansion into some closed form. I'm not sure if calculus would be useful. The problem asks for a single digit, so if we consider repeating digits, everything will be a lot easier. There seems to be a pattern in the decimal expansions of numbers consisting of only $1.$ For instance,
$$\sqrt{1}=1,$$
$$\sqrt{11}=3.3166247...$$
$$\sqrt{111}=10.5356537...$$
$$\sqrt{1111}=33.3316666...$$
$$\sqrt{11111}=105.408728...$$
$$\sqrt{111111}=333.333166...$$
$$\sqrt{1111111}=1054.09250...$$
Every term of the form $\displaystyle\sum_{n=0}^{2k+1}10^n$ has $k+1$ $3$'s at the beginning and $k+1$ 3's right after the decimal expansion, followed by one $1,$ and $2(k+1)$ $6$'s. Proving this would prove that the $1000$th digit is $1.$ This is the same as showing that $\sqrt{\left(\dfrac{10^{2m}-1}{9}\right)}=\dfrac{10^{m}-1}{3}+\dfrac{1}{3}-\dfrac{1}{6}\cdot 10^{-m}+\epsilon_m$ where $|\epsilon_m|<10^{-2m}.$
Edit: the previous question I asked was inspired by the current one, but the previous question seemed to have a rather unpleasant answer, so I changed it.
| A little experimentation shows that if $n$ is the integer with $2m$ digits, where $m$ is an integer, all of them $1,$ then $\sqrt{n}$ has $m$ $3$'s, followed by a decimal point, $m$ $3$'s, a $1,$ and $2m$ $6$'s. If this was true, that would imply that the required digit is $1.$ Just substitute $2m=1998$ to verify.
Let $x=\dfrac{10^m-\frac{1}{2}10^{-m}}{3},m\in\mathbb{N}.$ Then $x$ has $m$ $3$'s, followed by a decimal point, followed by $m$ $3$'s, followed by one $1,$ followed by infinite $6$'s. So we just need to show that $(x-10^{-m-2})^2<n$ and $(x+10^{-m-2})^2>n,$ where $n=\dfrac{10^{2m}-1}{9}.$ This will show that the $(m+1)$st digit, or $1000$th digit, is indeed $1.$
$$(x-10^{-m-2})^2=\left(\dfrac{10^m}{3}-\dfrac{47}{300}10^{-m}\right)^2\\
=\dfrac{10^{2m}}{9}-\dfrac{47}{450}+\dfrac{2209}{90\;000}10^{-2m}\\
=n-\dfrac{47}{450}+\dfrac{2209}{90\;000}10^{-2m}\\
<n$$.
Similarly,
$$(x+10^{-m-2})^2=\left(\dfrac{10^m}{3}+\dfrac{47}{300}10^{-m}\right)^2\\
=\dfrac{10^{2m}}{9}+\dfrac{47}{450}+\dfrac{2209}{90\;000}10^{-2m}\\
=n+\dfrac{47}{450}+\dfrac{2209}{90\;000}10^{-2m}\\
>n$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3411118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
what is the limit as $x$ approaches infinity of $x-\sqrt{x^2-x+2}$ $$
\lim_{x \to \infty} \left( x - \sqrt{x^2 - x +2 } \right)
$$
I've tried rationalizing the expression but after repeated applications of L'Hospital's rule, it doesn't feel like I'm getting anywhere.
| Note that
$$ \sqrt{x^2 - x + 2} = \sqrt{x^2 -x + \frac{1}{4} + 1.75} = \sqrt{(x - \frac{1}{2})^2 + 1.75}$$.
What is the limit of $ \sqrt{(x - \frac{1}{2})^2 + 1.75} - \sqrt{(x - \frac{1}{2})^2}$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3412245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 0
} |
Dimension of Null Space of a given linear transformation Let $A =\begin{bmatrix}2 & -1\\-1 & 2\end{bmatrix}$
Define Linear Transformation $T$ as $AX - XA$ where $X$ is a $2$ x $2$ matrix. Find dimension of Null space of $T$?
My approach: Let $X = \begin{bmatrix}a & b\\c & d\end{bmatrix}$ Then simplifying the expression for T.
$AX = \begin{bmatrix}2 & -1\\-1 & 2\end{bmatrix} \times \begin{bmatrix}a & b\\c & d\end{bmatrix} = \begin{bmatrix}2a -c & 2b - d\\-a + 2c & -b+2d\end{bmatrix}$
and
$XA = \begin{bmatrix}a & b\\c & d\end{bmatrix} \times \begin{bmatrix}2 & -1\\-1 & 2\end{bmatrix} = \begin{bmatrix}2a - b & -a+2b\\2c-d & -c+2d\end{bmatrix}$
So $AX - XA = \begin{bmatrix}2a -c -2a + b& 2b - d +a-2b\\-a + 2c -2c +d & -b+2d +c -2d\end{bmatrix}$
Simplying it will give
$AX -XA = \begin{bmatrix}b -c & a - d\\d-a & c-b\end{bmatrix}$
Now null space is when $TX = 0$
so comparing element by element in $AX-XA = 0$
$\begin{bmatrix}b -c & a - d\\d-a & c-b\end{bmatrix} = \begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}$
we get $b = c$ and $a = d$ for matrices which will belong to $T's$ null space.
So $X = \begin{bmatrix}a & b \\b & a\end{bmatrix}$ will be mapped to $0$. $X$ could be written as
$X = a\times\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix} + b \times\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}$.
So the Dimension of Null space should be $2$.
Am I correct? I faced this problem in math subject gre paper and linear algebra is my weakest topic so if anyone could help me with this I would be really thankful.
| This given solution seems fine. So, +1, endorsed!
Here's another way of doing these calculations which some may find somewhat easier; also I think it lends further insight into the problem:
We may generalize and write
$A = \begin{bmatrix} a & b \\ b & a \end{bmatrix} = aI + bP, \tag 1$
where
$P = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}; \tag 2$
then since $I$ commutes with any matrix $X$,
$IX = XI, \tag 3$
it is easy to see that
$T(X) = AX - XA = b(PX - XP); \tag 4$
this always vanishes for $b = 0$ and for arbitrary $b \ne 0$ provided
$PX - XP = 0 \Longrightarrow PX = XP; \tag 5$
with
$X = \begin{bmatrix} x_1 & x_2 \\ x_3 & x_4 \end{bmatrix}, \tag 6$
(5) becomes
$\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} x_1 & x_2 \\ x_3 & x_4 \end{bmatrix} = \begin{bmatrix} x_1 & x_2 \\ x_3 & x_4 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}; \tag 7$
that is,
$\begin{bmatrix} x_3 & x_4 \\ x_1 & x_2 \end{bmatrix} = \begin{bmatrix} x_2 & x_1 \\ x_4 & x_3 \end{bmatrix}, \tag 8$
from which
$x_1 = x_4, \; x_2 = x_3; \tag 9$
we thus see that
$X = \begin{bmatrix} x_1 & x_2 \\ x_2 & x_1 \end{bmatrix} = x_1I + x_2P; \tag{10}$
comparing this to (1), we see that $X$ is in the same general family as $A$, and that if $b \ne 0$
$\dim \ker T = 2; \tag{11}$
but
$\dim \ker T = 4 \tag{11}$
when $b = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3413014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integrate $\int_{0}^{\infty}{\frac{(x^2+4)\ln(x)}{x^4+16}}dx$
$$\int_{0}^{\infty}{\dfrac{\left(x^2+4\right)\ln\left(x\right)}{x^4+16}}dx$$
I tried indefinite integration and got struck at $$\int_{0}^{\infty}{\dfrac{\left(u^2+1\right)\ln\left(u\right)}{u^4+1}}du.$$
I tried by-parts now taking $\ln(u)$ as my first function, but it doesn't seem to help. How to solve it and also is there an easy way out to solve it, as my method isn't that good.
| $$I=\int_0^\infty\frac{(4+x^2)\ln x}{16+x^4}\ dx\overset{x=2y}{=}\frac12\int_0^\infty\frac{(1+y^2)(\ln2+\ln y)}{1+y^4}\ dy\\
=\frac{\ln2}{2}\int_0^\infty\frac{1+y^2}{1+y^4}\ dy+\frac12\int_0^\infty\frac{(1+y^2)\ln y}{1+y^4}\ dy$$
where
$$\int_0^\infty\frac{1+y^2}{1+y^4}\ dy=\int_0^\infty\frac{1/y^2+1}{1/y^2+y^2}\ dy=\int_0^\infty\frac{1+1/y^2}{(y-1/y)^2+2}\ dy\\=\left.\frac1{\sqrt{2}}\tan^{-1}\frac{(y-1/y)}{\sqrt{2}}\right|_0^\infty=\frac{\frac{\pi}{2}-(-\frac{\pi}{2})}{\sqrt{2}}=\frac{\pi}{\sqrt{2}}$$
and $$\int_0^\infty\frac{(1+y^2)\ln y}{1+y^4}\ dy\overset{y\mapsto 1/y}{=}-\int_0^\infty\frac{(1+y^2)\ln y}{1+y^4}\ dy\Longrightarrow 2\int_0^\infty\frac{(1+y^2)\ln y}{1+y^4}\ dy=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3414066",
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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How to express $f(x)=\frac{1}{x-1} + \frac{1}{x-3}$ as a power series? I am working through my Calculus 2 course problems, and I have reached..
Express the function $f(x)=\frac{2x-4}{x^2-4x+3}$ as the sum of a power series by first using partial fractions.
I can do the partial fractions..
$$\frac{2x-4}{x^2-4x+3}=\frac{2(x-2)}{(x-1)(x-3)} =\frac{A}{x-1}+\frac{B}{x-3}$$
$$2(x-2)=A(x-3)+B(x-1)$$
$$\to -2=-2A \to1=A$$
$$\to2=2B\to 1 = B$$
$$\frac{1}{x-1} + \frac{1}{x-3}$$
I am not sure how to approach creating a power-series with two fractions like this.
| Separately,
$$\frac{1}{1-x} = 1 + x + x^2 + x^3 +\> ...$$
$$\frac{1}{3-x} =\frac13 \left(1 + \frac x3 + \frac {x^2}{9} + \frac {x^3}{27} +\> ... \right) $$
Then, together,
$$\frac{1}{x-1} +\frac{1}{x-3} = -(1+\frac13) - (1+\frac19)x - (1+\frac{1}{27})x^2
- (1+\frac{1}{81})x^3 -\> …$$
or, in the compact form,
$$\frac{1}{x-1} +\frac{1}{x-3} =- \sum\limits_{n=0}^\infty
\left(1+\frac{1}{3^{n+1}}\right)x^n$$
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the smallest multiple of $3^{1000}$, which has only 1’s in the digits?(Like 1, 11, 111 etc) I think the answer is $3^{1000}$ ones,
and I want to prove it with induction.
$3^0 | 1$ and $1$ is the smallest.
$ 3^1 | 111 $; it’s the smallest too.
Let’s say it's true for $ 3^k | 111...111 $ ($3^k$ ones). How can I prove to $3^{k+1} $ ?
| Note that
$$
\underbrace{111\cdots11}_{3^{k}} = \underbrace{111\cdots11}_{3^{k-1}}\cdot(1+10^{3^{k-1}} + 10^{2\cdot 3^{k-1}})
$$
and $1+10^{3^{k-1}} + 10^{2\cdot 3^{k-1}}$ is divisible by $3$, so the left-hand side is indeed divisible by $3^{k}$ by induction.
But is it the smallest such number? For simplicity, let $a_n$ for $n\geq 0$ be the sequence defined by
$$
a_n = \underbrace{111\cdots11}_{3^n}
$$
and assume for induction that $a_{k-1}$ is the smallest number consisting only of $1$s that is divisible by $3^{k-1}$ (you have already checked that this is the case for $k = 0$ and $k = 1$, so the base cases are done). Let $x$ be the smallest number consisting only of $1$s that is divisible by $3^k$. We know $x$ exists, as $a_k$ is one such number.
By induction hypothesis, and the first paragraph, we have $a_{k-1}\leq x\leq a_k$. And $x$ can't be $a_{k-1}$ itself, as
$$
a_{k-1} = a_{k-2}(1 + 10^{3^{k-2}} + 10^{2\cdot 3^{k-2}})
$$
and $a_{k-2}$ is not divisible by $3^{k-1}$, and $1 + 10^{3^{k-2}} + 10^{2\cdot 3^{k-2}}$ is divisible by $3$ but not $9$. Thus $a_{k-1}$ is not divisible by $3^k$.
Therefore we have $a_{k-1}<x\leq a_k$. Now consider $$y = \frac{x - a_{k-1}}{10^{3^{k-1}}}$$ It is a strictly positive integer consisting only of $1$s which is divisible by $3^{k-1}$. So we have $a_{k-1}\leq y$. Can we have equality? If that were the case, then we would have
$$
x = a_{k-1}(1+10^{3^k})
$$
which isn't divisible by $3^k$. So we must have $a_{k-1}<y$. So we may repeat the process and consider $$z = \frac{y - a_{k-1}}{10^{3^{k-1}}}$$ Once again $z$ is a strictly positive integer consisting only of $1$s which is divisible by $3^{k-1}$. That means that $a_{k-1}\leq z$. Unnesting this final inequality gives us $a_k\leq x$, and we are done.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating Convergence (Uniform) Hi Guys was attempting this question and was wondering if I was doing the question correctly?
Determine whether or not the sequence of functions is uniformly convergent:-
$$g_n:(0,1)\to \mathbb{R}$$
$$g_n(x) = \frac{n^3+1}{n^3x^2+1}, x\in(0,1)$$
Checking point wise convergence first
$$\lim_{n\to \infty}g_n(x) = \lim_{n\to \infty}\frac{n^3+1}{n^3x^2+1}$$
Dividing by $n^3$gives the following :-
$$\lim_{n\to \infty}g_n(x) = \lim_{n\to \infty}\frac{1+\frac{1}{n^3}}{x^2+\frac{1}{n^3}}$$
Taking the Limit as n $\to \infty$ gives the following
$$\lim_{n\to \infty}g_n(x) = \frac{1+\frac{1}{\infty^3}}{x^2+\frac{1}{\infty^3}}$$
$$\lim_{n\to \infty}g_n(x) = \frac{1+0}{x^2+1} = \frac{1}{x^2}$$
Therefore by point wise convergence the sequence of functions converges to the previous function.
In order to determine the uniform convergence we must analyze the follwing
$$M_n = sup|f_n(x)-f(x)|,x\in \mathbb{R}$$
$$|f_n(x)-f(x)|$$
$$|\frac{n^3+1}{n^3x^2+1} - \frac{1}{x^2}|$$
$$\frac{(n^3x^2+x^2)-(n^3x^2+1)}{(n^3x^2+1)(x^2)}$$
$$|\frac{x^2-1}{(n^3x^2+1)(x^2)}|$$
The mod gives
$$\frac{x^2+1}{(n^3x^2+1)(x^2)}$$
is it accurate to say the following when checking to see uniform convergence
$$\frac{x^2+1}{(n^3x^2+1)(x^2)} < \frac{1}{n^3}$$
$$\lim{n \to \infty} $$
Therefore I can conlclude that $$sup|f_n(x)-f(x)|\to 0$$
Therefore the function is uniformly convergent? Oh am i wrong in my evaluation?
| For $ x\in (0,1)$ and $ n$ large enough,
$$g_n(x)=|f_n(x)-\frac{1}{x^2}|=\frac{|x^2-1|}{(n^3x^2+1)x^2}$$
Now take the sequence $ (x_n) $ such that
$$n^3x_n^2=1$$ or
$$x_n=n^{-\frac 32}=\frac{1}{n^{\frac 32}}$$
Then, $ x_n\in (0,1)$ and
$$|f_n(x_n)-f(x_n)|=\frac{|n^{-3}-1|}{2n^{-3}}$$
$$=\frac 12|1-n^{3}| \to +\infty$$
But
$$|f_n(x_n)-f(x_n)|\le \sup_{(0,1)}|f_n-f|$$
thus
$$\lim_{n\to+\infty}\sup_{(0,1)}|f_n-f|=+\infty$$
The convergence is not uniform at $(0,1)$.
It is uniform at $(a,1) $ with $0<a<1$.
| {
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Which numbers can be realized in the form $a^x - b^x$ for $a, b$ of opposite parity and $x$ even?
How can we check whether a number can be represented in the form of $a^x - b^x$, where $a$ and $b$ are integers of opposite parity and $x$ is positive and even integer.
I thougth of using $$a^n - b^n = (a - b) (a^{n-1} + a^{n-2}b + \cdots + b^{n-2}a +b^{n-1}) .$$
| If $a,b$ have different parity then $a^x - b^x$ must be odd.
And if $x$ is even then $a^x - b^x = (a^{\frac x2} - b^{\frac x2})(a^{\frac x2} + b^{\frac x2})$
So for any odd $2k-1$, the $a = k; b=k-1; x= 2$ and $2k-1= (k-(k-1))(k+(k-1)) =k^2 - (k-1)^2$
So all odd numbers and only odd numbers can be so represented.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve in $\mathbb R$ : $x^4-2x^{3}-3x^{2}+4x+\frac{15}{16}=0$ Problem :
Solve in $\mathbb R$ :
$x^4-2x^{3}-3x^{2}+4x+\frac{15}{16}=0$
The roots are : (by Wolfram Alpha )
$x_{i}=\frac{5}{2},\frac{3}{2},-\frac{1+\sqrt 2}{2},-\frac{1-\sqrt 2}{2}$
I don't know how to start because in the first place it is not simple to try roots like $\frac{3}{2}$
So I need some ideas or hints to factorization this cubic equation
| Suppose that instead of a polynomial in $x$ we had a polynomial in $2x$.
$$\begin{align}
x^4 - 2x^3-3x^2+4x+\frac{15}{16}&=0\\
16x^4 - 16\cdot2x^3-16\cdot3x^2+16\cdot4x+15&=16\cdot0\\
(2x)^4 - 4(2x)^3-12(2x)^2+32(2x)+15&=0\\
y^4 - 4y^3-12y^2+32y+15&=0\\
\end{align}$$
Where $y=2x$.
Can you take it from here? :)
| {
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"timestamp": "2023-03-29T00:00:00",
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Subsets and Splits
Fractions in Questions and Answers
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