Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find number of non-singular matrices
If $w$ is complex cube root of unity $w \ne 1$ then find number of non-singular matrices of the form
$$A=\begin{bmatrix}
1 & a & b\\
w&1 &c \\
w^2 &w & 1
\end{bmatrix}$$ such that $a,b,c$ takes values from the set $S=\left\{w,w^2\right\}$
My Try:
we have $$\det(A)=acw^2-(a+c)w+1$$.
$\det(A)=0$ only if $ac=1$ and $a+c=-1$ since $1+w+w^2=0$.
So $a=w $ ad $c=w^2$ OR $a=w^2$ and $c=w$.
But since $\det(A)$ is independent of $b$ The final possible triplets $(a,b,c)$ are
$1.$ $(w,w^2,w^2)$
$2.$ $(w^2,w,w)$
$3.$ $(w,w,w^2)$
$4.$ $(w^2,w^2,w)$
So total number of singular matrices is $4$.
But total number of matrices $A$ is $8$.
Hence total number of non-singular matrices is $8-4=4$.
But my book answer is $2$. May I know where is the mistake?
| The matrix $A$ is singular iff
$$0=\frac{\det(A)}{w^2}=\frac{1}{w^2}-(a+c)\frac{1}{w}+ac=
\left(\frac{1}{w}-a\right)\left(\frac{1}{w}-c\right)$$
that is iff $a=1/w=w^2$ and $c\in S$ or $c=1/w=w^2$ and $a\in S$. So there are two more cases: $(w^2,w^2,w^2)$ and $(w^2,w,w^2)$.
Hence the total number of non-singular matrices is $2^3-(4+2)=2$.
| {
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Find all primes for which $p^p - 2$ is a perfect square Find all primes for which $p^p - 2$ is a perfect square.
Let $p^p-2=a^2.$
We know $a^2 \equiv 0,1\pmod{3}$
If $a^2 \equiv 1\pmod{3}$, then $p^p \equiv 0\pmod{3}\implies p=3$ which is a solution.
Now, if $a^2 \equiv 0 \pmod{3},$ then $p^p \equiv 2\pmod{3}$.
By Fermat's theorem, $p^p \equiv p \pmod{3}$ as $p$ is coprime to $3$.
So, we have $p$ is of the form $3k+2.$
But what to do next?
| Notice that for $p\neq 3$ must be $p=9k+t$ where $t\in\{1,2,4,5,7,8\}$ we must also have that $p=6n+1$ or $p=6n+5$ since $p\neq 2,3$.
Now by Euler theorem $t^6=1\pmod{9}$ if $p=6n+1$ we have that $$(9k+t)^{6n+1}\equiv t\pmod{9}$$ since the only $t$ such that $t-2$ is a quadratic residue mod $9$ is $t=2$ we must have $t=2$ but $6n+1\neq 9k+2$ hence no solutions.
If $p=6n+5$ we have that $$(9k+t)^{6n+5}\equiv t^5\pmod{9}$$
The only $t$ such that $t-2$ is a quadratic residue mod $9$ is $t=5$ so $p=9k+5$
this implies that the number is of the form $p=18q+5$ but again by euler theorem
$$(18q+5)^{18q+5}$$
Looking mod $4$ we see that $q$ must be even so $(36q+5)^{36q+5}$
Now looking mod $108$ and using Euler we have that
$$(36q+5)^{36q+5}\equiv (36q+5)^5\equiv 65,29,101\pmod{108}$$
But $x^2+2\not \equiv 65,29,101\pmod{108}$ or differently $x^2\not\equiv 63,27,99$
| {
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Finding the inverse of the arc length equation of a parabola I need help in reversing the equation of the arc length of a parabola.
$$y = \frac{1}{2} x \sqrt{1+4\cdot x^2} + \frac{1}{4} \cdot \ln(2\cdot x + \sqrt{1+4\cdot x^2})$$
Thank you.
| This problem seem to somewhat misunderstood. Let me express it this way: Find the function $y(x)$ whose arc length is given by
$$s=\int_0^x\sqrt{1+\left(\frac{dy}{dx} \right)^2}~dx=\frac{1}{2} x \sqrt{1+4\cdot x^2} + \frac{1}{4} \cdot \ln(2\cdot x + \sqrt{1+4\cdot x^2})$$
Differentiating both sides we get
$$\sqrt{1+\left(\frac{dy}{dx} \right)^2}=\sqrt{4x^2+1}\\
1+\left(\frac{dy}{dx} \right)^2=4x^2+1\\
\frac{dy}{dx}=2x\\
\fbox{$y=\pm x^2+C$}
$$
I have verified this solution numerically.
| {
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Coefficient of $x^2$ in a polynomial
Let $p(x)$ be an polynomial given by
$$p(x)=(1+x)(1+3x)(1+5x)\cdots(1+19x)$$ Find the coefficient of $x^2$.
I see that this coefficient is the sum of products of roots of $p(x)$ get eight to eight. But, there is ${10 \choose 8}=45$ terms in this sum. Someone know other way more fast?
| You can write
$$p(x)=(1+10x-9x)(1+10x+9x)\cdots (1+10x-x)(1+10x+x)=((1+10x)^2-(9x)^2)((1+10x)^2-(7x)^2)\cdots ((1+10x)^2-x^2)=(19x^2+20x+1)(51x^2+20x+1)(75x^2+20x+1)(91x^2+20x+1)(99x^2+20x+1)$$
Now you can see that the main coefficients of $x^2$ sum to $19+51+75+91+99=335$ now we can look $\mod x^2$ to get the remaining coefficient of $x^2$ we get $(20x+1)^5$ and the coefficient of the $x^2$ is $4000$ summing those two you get $4335$.
| {
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How to solve in integers the equation, $(x^2-y^2)^2=1+16y$? How to solve in integers the equation,
$$(x^2-y^2)^2=1+16y$$
By observation we can take $y=3,5$
Is there another method to solve the equation?
| We can prove the non-existence of solutions other than those listed in the comments by squeezing $x^2$ between two consecutive squares of integers in the remaining cases. In a sense the key is that unless $y$ is relatively small that $1+16y$ is very small in comparison to the left hand side. It turns out that more often than not it is too small to be the square of the difference between two consecutive squares, and that leads to a solution. Behold.
Clearly $y\ge0$ and if $(x,y)$ is a solution so is $(-x,y)$. So it suffices to find the solutions with $x\ge0, y\ge0$. Also, $x\neq y$.
If $x>y$ then $x^2-y^2>0$. Therefore
$$
x^2=y^2+\sqrt{1+16y}.
$$
For non-negative integers $y$ we have the upper bound $\sqrt{16y}=4\sqrt y\le 4y$ and consequently the inequality $\sqrt{1+16y}\le1+4y$. Therefore $$y^2+\sqrt{1+16y}\le y^2+4y+1<y^2+4y+4=(y+2)^2.$$
This estimate allows us to squeeze:
$$
y^2<x^2=y^2+\sqrt{1+16y}<(y+2)^2
$$
keading to the conclusion $x=y+1$. The equation
$$
(y+1)^2=y^2+\sqrt{1+16y}
$$
has two solutions, namely $y=0$ and $y=3$.
If $0\le x<y$, then we must have
$$
x^2=y^2-\sqrt{1+16y}.
$$
For the right hand side to be non-negative we need $y\ge3$. Assuming that we can estimate
$$
(4y-4)^2=16y^2-32y+16=16y(y-2)+16>16y+1,
$$
so $\sqrt{1+16y}<4y-4$ in the interesting range of values of $y$. Therefore
$$
y^2>x^2=y^2-\sqrt{1+16y}>y^2-(4y-4)=(y-2)^2.
$$
As above, this leaves $x=y-1$ as the only alternative. But the only solution of the equation
$$
(y-1)^2=y^2-\sqrt{1+16y}\Longleftrightarrow 2y-1=\sqrt{1+16y}
$$
is $y=5$.
| {
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Find the range of $(1/2)T^2$ Let $V=M_{2\times 2}(\mathbb{R})$ be the vector space consisting of all $2 \times 2$ real matrices. Define a transformation $T : V → V$ by
$$
T(A)=AP-PA, \mbox{ where }
P = \begin{pmatrix}
0 & 1 \\
1 & 0 \\
\end{pmatrix}.
$$
The transformation $T^2: V → V$ is defined to be $T^2(A) = T(T(A))$, for each $A ∈ V$. Given $c ∈ \mathbb{R}$, the transformation $cT$ is defined to be $(cT)(A) = cT(A)$, for each $A ∈ V$.
Find the Range of $(1/2)T^2$.
This is a question my professor gave me for practice. There were some other questions about proving that $T$ is linear and stuff like that but I have no idea where to even begin with this.
(ps. I apologize if my formatting is not proper but this is the first time I have asked a question on this website).
| Say $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$. Then
$$
T(A) = AP - PA = \begin{pmatrix} b & a \\ d & c \end{pmatrix} - \begin{pmatrix} c & d \\ a & b \end{pmatrix} = \begin{pmatrix} b-c & a-d \\ d-a & c-b \end{pmatrix} .
$$
Continuing,
$$
T^2(A) = T(T(A)) = \begin{pmatrix} (a-d)-(d-a) & (b-c)-(c-b) \\ (c-b)-(b-c) & (d-a)-(a-d) \end{pmatrix} = 2 \begin{pmatrix} a-d & b-c \\ c-b & d-a \end{pmatrix} .
$$
So
$$
\frac{1}{2} T^2(A) = \begin{pmatrix} a-d & b-c \\ c-b & d-a \end{pmatrix} .
$$
Before going on, I'd like to point out a slightly different approach suggested in the comments. Note that $P^2 = I$. Then
$$
T^2(A) = T(A)P - PT(A) = (AP-PA)P - P(AP-PA) = AP^2 - PAP - PAP + P^2A = 2(A-PAP),
$$
so $\frac{1}{2} T^2(A) = A-PAP$ (not $\frac{1}{2}T^2(A)=PAP$). One can compute
$$
A - PAP = \begin{pmatrix} a & b \\ c & d \end{pmatrix} - \begin{pmatrix} d & c \\ b & a \end{pmatrix}
$$
and get the same expression as before.
Now, having an expression for $\frac{1}{2} T^2$ is nice, but the question was to describe the range of this operator. What kind of description would you like? If you want equations for the range, some obvious ones pop out: set
$$
\begin{pmatrix} a-d & b-c \\ c-b & d-a \end{pmatrix} = \begin{pmatrix} x & y \\ z & w \end{pmatrix},
$$
then it's clear that $x+w = 0$ and $y+z = 0$. It's clear that these equations are independent. You already know that $\frac{1}{2} T^2$ is linear (right?) so its range is a linear subspace in $M_{2\times 2}(\mathbb{R})$, which is $4$-dimensional. Since we've written $2$ independent equations, we know the range has dimension at most $2$. A priori, however, there could be additional equations.
To see that there are no other equations we could, for instance, write down $2$ independent elements of the range, such as $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ and $\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ (can you see which matrices $A$ give these as $\frac{1}{2}T^2(A)$?).
So the range is at most $2$-dimensional (because we have $2$ equations and a $4$-dimensional ambient space) and it is also at least $2$-dimensional (because we have these $2$ independent elements). We can conclude that the range is precisely $2$-dimensional, spanned by the two elements given above. And that gives you a second description of the range, if you need a basis for it rather than equations.
For example, having this basis makes it very easy to check that $\frac{1}{2} T^2$ is a projection onto its range.
| {
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Let $n$ natural number. Then prove there exists solutions for $a^2 + 3b^2 = n$ if and only if exists solutions for $x^2 + xy + y^2 = n$. a,b,x,y are integer numbers. First, if i guess there $a^2 + 3b^2 = n$ , then $a^2 + 2b^2 + b^2$, has $x^2 + xy + y^2 form$, then i guess there $a=x$, $b=y$ and $2b^2=xy$, then $2b=a$. I don´t really sure that this solve the first part. With the second part i tried something similar but i don´t arrive to the same result.
| Suppose that $n=x^2+xy+y^2$ for some integers $x$ and $y$. This may be designated as a pair $\{x,y\}$ which is a solution for $n$. Then:
$n=x^2+xy+y^2=(\frac{x-y}{2})^2+3(\frac{x+y}{2})^2$, eq. 1
If $x$ and $y$ are both even or both odd this represents an integer solution for $a$ and $b$. If the parities are different, invoke the identity
$x^2+xy+y^2=(x+y)^2-x(x+y)+x^2$
to identify $\{-x,x+y\}$ as a second solution, and by symmetry $\{-y,x+y\}$ is a third solution. With $x+y$ odd pick the solution out of these two where the other member is odd and apply eq. 1.
Now suppose that $n=a^2+3b^2$. This is also $(b+a)^2+(b+a)(b-a)+(b-a)^2$ proving the converse claim.
| {
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How do I solve this fraction question? If $a = -1/5$, how do I calculate:
$$3 a + 2 a^2$$
I did $3\times(-1/5) + (-1/5) \times (-1/5) \times 2$, but can't figure out what the right way to solve this is.
| That's the correct way.
\begin{align}
\left(3 \times \frac{-1}{5}\right)+\left(\frac{-1}{5} \times 2 \times \frac{-1}{5}\right) &= \frac{-3}{5}+\frac{2}{25} \\
&=\frac{-15}{25}+\frac{2}{25} \\
&= \frac{-13}{25}\\
\end{align}
| {
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If $ax^2+bx+c = 0$ and $bx^2+cx + a = 0$ have a common root and $a\neq 0$, then find $\frac{a^3+b^3+c^3}{abc}$ If $ax^2+bx+c = 0$ and $bx^2+cx + a = 0$ have a common root and $a\neq 0$, then find $$\frac{a^3+b^3+c^3}{abc}$$
I tried that for both equations to have a common root, the expression on left hand sides must be equal, ie $$ax^2+bx+c = bx^2+cx + a$$
for this we must have $x=1$ (i cannot prove this, but it appears to be true). Also both of these must be equal to $0$, so we have:
$$a+b+c=0$$
So using this we say
$$\frac{a^3+b^3+c^3}{abc} = \frac{a^3+b^3+c^3-3abc+3abc}{abc}=\frac{(a+b+c)(...)}{abc}+3$$
So we get the answer as $3$.
How do we say that $x =1$ is the commmon root? Thanks!
| $$0=ax^3+bx^2+cx=ax^3-a$$
which gives $x=1$, $a+b+c=0$ and since $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)=0,$$ we obtain:
$$\frac{a^3+b^3+c^3}{abc}=\frac{3abc}{abc}=3.$$
If you wish the solution for $x\in\mathbb C$ and $\{a,b,c\}\subset\mathbb C$ then since
$$a^3+b^3+c^3-3abc=(a+b+c)(a+\zeta b+\zeta^2c)(a+\zeta^2 b+\zeta c),$$
where $\zeta=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$, we get the same answer.
| {
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Prove that the nth derivative of a function has two real solutions I have the function $f:\mathbb{R} \rightarrow \mathbb{R} f(x)=e^x(x^2-5x+7)$. I need to prove that $\forall n \in \mathbb{N}^*$, the equation $f^{(n)}$ has two real solutions. Where $f^{(n)}$ is the nth derivate of the function.
My idea is that this should be proved by induction but Im not sure.
| Well, let's calculate a few derivatives of function $f$:
*
*$f'(x)=e^x(x^2-5x+7+2x-5)=e^x(x^2-3x+2)$,
*$f''(x)=e^x(x^2-3x+2+2x-3)=e^{x}(x^2-x-1)$,
*$f^{(3)}(x)=e^x(x^2-x-1+2x-1)=e^x(x^2+x-2)$,
*$f^{(4)}(x)=e^x(x^2+x-2+2x+1)=e^x(x^2+3x-1)$
*$f^{(5)}(x)=e^x(x^2+3x-1+2x+3)=e^x(x^2+5x+2)$
So, we can assume that
$$f^{(k)}(x)=e^x\left(x^2+(2k-5)x+7+\sum\limits_{i=0}^{k-1}(2i-5)\right)$$
By induction, we have:
*
*From the above, it is obvious for $k=1$.
*Let us assume it is valid for $k$. Then:
$$\begin{align*}f^{(k+1)}(x)=&e^x\left(x^2+(2k-5)x+7+\sum_{i=0}^{k-1}(2i-5)+2x+2k-5\right)=\\
=&e^x\left(x^2+(2(k+1)-5)x+7+\sum_{i=0}^k(2i-5)\right)
\end{align*}$$
So, we have shown that $f^{(k)}(x)=e^xp_k(x)$, where $p_k(x)=x^2+(2k-5)x+7+\sum\limits_{i=0}^{k-1}(2i-5)$. Since $e^x>0$ for every $x\in\mathbb{R}$, we have to show that $p_k$ has two distinct roots for every $k\in\mathbb{N}$.
For that purpose, we calculate:
$$p_k'(x)=2x+2k-5$$
So:
$$p_k'(x)=0\Leftrightarrow2x+2k-5=0\Leftrightarrow x=-\frac{2k-5}{2}$$
and
$$\begin{align*}p_k\left(\frac{2k-5}{2}\right)=&\frac{(2k-5)^2}{4}-\frac{(2k-5)^2}{2}+7+\sum_{i=0}^{k-1}(2i-5)=\\
=&-\frac{(2k-5)^2}{4}+7+2\sum_{i=0}^{k-1}i-5k=\\
=&-\frac{(2k-5)^2}{4}+7+2\frac{k(k-1)}{2}-5k=\\
=&-\frac{(2k-5)^2}{4}+k^2-6k+7=\\
=&\frac{-4k^2+20k-25}{4}+k^2-6k+4=\\
=&\frac{-4k^2+20k-25+4k^2-24k+16}{4}=\\
=&\frac{-4k-9}{4}<0
\end{align*}$$
Moreover, since $p_k''(x)=2>0$, it comes that $p_k(x)$ has exactly two distinct roots for every $k\in\mathbb{N}$, which was what requested.
Reviewing it, it was a little bit brute-force...
| {
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Evaluating the expression $\cos(\frac{1}{2}\tan^{-1}(-\frac{4}{3}))$ I can prove that $\cos(\tan^{-1}(x)) = +\dfrac{1}{\sqrt{1+x^2}}$, set $y=\tan^{-1}(x)$ and therefore $y=\cos^{-1}(\frac{1}{\sqrt{1+x^2}})$
Dividing both sides by $2$:
$\dfrac{y}{2}=\dfrac{1}{2}\cos^{-1}(\frac{1}{\sqrt{1+x^2}})$
$\cos(\dfrac{1}{2}\tan^{-1}(x))=\cos(\dfrac{1}{2}\cos^{-1}(\frac{1}{\sqrt{1+x^2}}))$
$\cos(\dfrac{1}{2}\tan^{-1}(-\frac{4}{3}))=\cos(\dfrac{1}{2}\cos^{-1}(\frac{3}{5}))$
This is my approach to the problem, are there any better ways to simplify the initial expression and perhaps find an approximation even without using a calculator?
| Take a right-angled triangle with sides $1,x,\sqrt{1+x^2}$. The triangle exists by Pythagoras' Theorem.
The $\tan^{-1}(x)$ is its angle opposite to $x$. The cosine of that angle is $\frac{1}{\sqrt{1+x^2}}$.
| {
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Find the remainder when $7^{7^{7}}$ is divided by 1000 I need help with this problem please
Find the remainder when $7^{7^{7}}$ is divided by $1000$
My try follow
$1000=8×125$ , now
$7 \equiv -1 \;\bmod\; (8)$
$\to$ $7^{7^{7}} \equiv -1 \;\bmod\; (8)$ and
$7^{100}\equiv 1 \;\bmod\; (125)$
Any help to complete this solution?
| Since $7^4=2\,401\equiv1\pmod{400}$ and since $7^3=343$, $7^7\equiv343\pmod{400}$. So, by Fermat-Euler, $7^{7^7}\equiv7^{343}\pmod{1\,000}$.
Since $7^4\equiv401\pmod{1\,000}$, $7^{20}=(7^4)^5\equiv1\pmod{1\,000}$. Therefore$$7^{343}=7^{340+3}\equiv7^3=343\pmod{1\,000}.$$
| {
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For real $a,b,c$ , if $a^2+b^2+c^2=ab+bc+ac$ ,then the value of $\frac{a+b} {c}$ is?
For real $a,b,c$ , if $a^2+b^2+c^2=ab+bc+ac$ ,then the value of
$\frac{a+b} {c}$ is how much?
Ans.
What I could gather:
from the identity,
$$ a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^3-ab+bc+ca)$$
We gather that RHS=0.
$$ =>a^3+b^3+c^3=3abc$$
It would e helpful if someone could tell me how should I proceed.
| For $a=b=c=0$ the needed value does not exist.
If $c\neq0$ then since
$$0=\sum_{cyc}(a^2-ab)=\frac{1}{2}\sum_{cyc}(2a^2-2ab)=\frac{1}{2}\sum_{cyc}(a^2-2ab+b^2)=\frac{1}{2}\sum_{cyc}(a-b)^2,$$
we obtain $a=b=c$ and $\frac{a+b}{c}=2$.
| {
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How many times does $k$ occur in the composition of $n$? How many times does the number $k$ occur in the composition of $n$?
Composition of Integer
In short, the difference between the partition of an integer and composition is the order of numbers. In partition, the order doesn't matter whereas, in composition, it does. That's why Partitions are sometimes called as ordered Compositions.
Example: $k$ = $1$ & $n$ = $5$
The composition of 5 are:
$5$
$4 + 1$
$3 + 2$
$3 + 1 + 1$
$2 + 3$
$2 + 2 + 1$
$2 + 1 + 2$
$2 + 1 + 1 + 1$
$1 + 4$
$1 + 3 + 1$
$1 + 2 + 2$
$1 + 2 + 1 + 1$
$1 + 1 + 3$
$1 + 1 + 2 + 1$
$1 + 1 + 1 + 2$
$1 + 1 + 1 + 1 + 1$
In all $1$ occurs $28$ times in the composition of $5$
Similarly is there any relation between $k$ and $n$ for all $n \geq 0$ & $k \leq n$
| The generating function here is
$$G(z, u) = \sum_{q\ge 1} \left(uz^k - z^k + \frac{z}{1-z}\right)^q.$$
We then have that
$$\left.\frac{\partial}{\partial u} G(z, u) \right|_{u=1}
\\ = \left. \sum_{q\ge 1}
q \left(uz^k - z^k + \frac{z}{1-z}\right)^{q-1} z^k\right|_{u=1}
= z^k \sum_{q\ge 1} q \left(\frac{z}{1-z}\right)^{q-1}
\\ = z^k \frac{1}{(1-z/(1-z))^2}
= z^k \frac{1-2z+z^2}{(1-2z)^2}.$$
Extracting coefficients we find for $n\ge k+2$
$$[z^{n-k}] \frac{1}{(1-2z)^2}
- 2[z^{n-k-1}] \frac{1}{(1-2z)^2}
+ [z^{n-k-2}] \frac{1}{(1-2z)^2}
\\ = 2^{n-k} (n-k+1) - 2^{n-k} (n-k) + 2^{n-k-2} (n-k-1)
\\ = 2^{n-k} + 2^{n-k-2} (n-k-1)
= 2^{n-k-2} (n-k+3).$$
We get for $n=k+1$
$$2^{n-k} (n-k+1) - 2^{n-k} (n-k) = 2$$
which corresponds to $(1,k)$ and $(k,1)$
and for $n=k$
$$2^{n-k} (n-k+1) = 1.$$
These only depend on $n-k$ as pointed out in the comments.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2354225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Telescoping(?) an infinite series Find the value of the sum $\displaystyle \sum\limits_{n=1}^{\infty} \frac{(7n+32) \cdot3^n}{n(n+2) \cdot 4^n}.$
Using partial fraction decomposition, I found the above expression is equivalent to $\displaystyle \sum\limits_{n=1}^{\infty} \frac{25}{n} \cdot \left(\frac{3}{4}\right)^n - \sum\limits_{n=1}^{\infty} \frac{18}{n+2} \cdot \left(\frac{3}{4}\right)^n,$ where I got couldn't find a closed form of either expression because of the $\left(\dfrac{3}{4}\right)^n.$
Similarly, trying to telescope one of the the sums with $\displaystyle \sum\limits_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{n+2}\right) \cdot \left(\frac{3}{4}\right)^n$ fails for the same reason. How can I further simplify the above expression? Thanks.
| Hint: for $|x|<1$ the power series $f(x):=\sum_{n=1}^{\infty}\frac{1}{n}x^n$ is convergent.
Hence $f'(x)=\sum_{n=1}^{\infty}x^{n-1}=\frac{1}{1-x}$ for $|x|<1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2355723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
Evaluating $\lim_{x \rightarrow +\infty} \frac{\sqrt x + x^3 -\sqrt{x^6-2x^4} -x}{\log(1+{e}^{3\sqrt x})}$ I'm trying to calculate the following limit:
$$\lim_{x \rightarrow +\infty} \frac{\displaystyle\sqrt x + x^3 -\sqrt{x^6-2x^4} -x}{\displaystyle \log(1+{e}^{3\sqrt x})}$$
For WolframAlpha the result is: $\frac 13$, I've seen its step by step process but I didn't understand the background logic.
Before I got stuck, I did this step:
$\lim_{x \rightarrow +\infty} \frac{\displaystyle\sqrt x + x^3 -\sqrt{x^6-2x^4} -x}{\displaystyle \log[{e}^{3\sqrt x}({e}^{-3\sqrt x}+1)]}$ and then
$\lim_{x \rightarrow +\infty} \frac{\displaystyle\sqrt x + x^3 -\sqrt{x^6-2x^4} -x}{\displaystyle \log({e}^{3\sqrt x})+\log({e}^{-3\sqrt x}+1)}$
so $$\lim_{x \rightarrow +\infty} \frac{\displaystyle\sqrt x + x^3 -\sqrt{x^6-2x^4} -x}{\displaystyle {3\sqrt x}}$$
Someone could give me a hint for solve it?
| First note that $\sqrt{x^6-2x^4} = x^3 \sqrt{1 - 2/x^2} = x^3 \left(1 - 1/x^2 + O(1/x^4)\right) = x^3 - x + O(1/x)$ so
$$\sqrt x + x^3 -\sqrt{x^6-2x^4} - x = \sqrt x + x^3 - \left( x^3 - x + O(1/x) \right) - x \\ = \sqrt x + O(1/x) \sim \sqrt x$$
for large $x$.
Also, for large $x$,
$\log\left(1+e^{3\sqrt x}\right) \sim 3\sqrt x$.
Therefore,
$$\frac{\sqrt x + x^3 -\sqrt{x^6-2x^4} - x}{\log(1+{e}^{3\sqrt x})}
\sim \frac{\sqrt x}{3\sqrt x} = \frac{1}{3}.$$
Thus the limit is $\frac13$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2358805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
How do I show that the arc swept by $(1+\frac{i}{n})^n$ has length 1? I'm working on a tutorial for Euler's identity, and trying to show that the sum of the lengths of the arrows in this picture converges to 1 as $n \rightarrow \infty$
The length of the bottom arrow is $\frac{1}{n}$, and each arrow gets longer by a factor of
$$\left|1 + \frac{i}{n}\right| = \sqrt{1 + \frac{1}{n^2}}$$
so, the total length is the geometric sum
$$
\lim_{n \rightarrow \infty} \sum_{k=0}^{n-1} \frac{\left(\sqrt{1 + \frac{1}{n^2}}\right)^k}{n}
=
\lim_{n \rightarrow \infty} \frac{1 - \left(\sqrt{1 + \frac{1}{n^2}}\right)^n}{\left(1 - \sqrt{1 + \frac{1}{n^2}}\right)n}
$$
I've fiddled with this a lot and I can't seem to get it into a form where I can take the limit.
| As $n\to \infty$ we have $ \frac{1}{ n^2}\to 0$
As you know as $x\to 0$ we have $\sqrt{1+x}\approx 1+\frac{x}{2}$ thus
as $n\to \infty$ we have $\sqrt{1 + \frac{1}{n^2}}\approx 1+\frac{1}{2n^2}$
$\lim_{n \to \infty} \frac{1 - \left(\sqrt{1 + \frac{1}{n^2}}\right)^n}{\left(1 - \sqrt{1 + \frac{1}{n^2}}\right)n}=\lim_{n \to \infty}\frac{1-\left(1+\frac{1}{2n^2}\right)^n}{n\left(1-1-\frac{1}{2n^2}\right)}=\lim_{n \to \infty}\frac{1-\left(1+n\frac{1}{2n^2}+\ldots\right)}{-\frac{1}{2n}}=\lim_{n \to \infty}\frac{-\frac{1}{2n}}{-\frac{1}{2n}}=1$
Terms in parenthesis $(\ldots)$ go to $0$ when $n\to\infty$ because they are $\dfrac{const}{n^k}$
Hope this helps
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2360118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Show that $y = \frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive. Prove, using an algebraic method,that $y=\frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive. Hence, determine the minimum and maximum points $y=\frac{2x}{x^2 +1}$ .
What I tried:
Firstly, I thought of using partial fractions but since $x^2 +1 =(x-i)(x+i)$, I don't think it is possible to show using partial fractions.
Secondly, decided to use differentiation
$y=\frac{2x}{x^2 +1}$
$\frac {dy}{dx} = \frac {-2(x+1)(x-1)}{(x^2 +1)^2 }$
For stationary points:
$\frac {dy}{dx} = 0$
$\frac {-2(x+1)(x-1)}{(x^2 +1)^2 } = 0$
$x=-1$ or $x=1$
When $x=-1,y=-1$
When $x=1,y=1$
Therefore, this implies that $y=\frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive.
^I wonder if this is the correct method or did I leave out something?
The third way was using discriminant
Assume that $y=\frac{2x}{x^2 +1}$ intersects with $y=-1$ and $y=1$
For $\frac{2x}{x^2 +1} = 1$,
$x^2 -2x+1 = 0$
Discriminant = $ (-2)^2 -4(1)(1) = 0 $
For $\frac{2x}{x^2 +1} = -1$,
$x^2 +2x+1 = 0$
Discriminant = $ (2)^2 -4(1)(1) = 0 $
So, since $y=\frac{2x}{x^2 +1}$ touches $y=-1$ and $y=1$, $y=\frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive.
Is the methods listed correct?Is there any other ways to do it?
| use AM-GM:
If $x>0$ then $$\frac{2x}{1+x^2}\le\frac{2x}{2\sqrt{x^2}}=\frac{2x}{2x}=1$$
Similarly $x<0$
$$\frac{2x}{1+x^2}\ge\frac{2x}{2\sqrt{x^2}}=\frac{2x}{2|x|}=-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2361415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 9,
"answer_id": 4
} |
How to find all positive divisors of a given number? For example, I have to find all positive divisors of $372$. The prime factorization of $372$ is $2^2 \cdot 3 \cdot 31$
Now, I wonder if there is a fast method to find all positive divisors of $372$.
| Once you have the prime factorization of a number, say
$$n=p_1^{n_1}\cdot p_2^{n_2}\cdots p_{r}^{n_r},$$
then any positive divisors $d$ of $n$ can be written as
$$d=p_1^{x_1}\cdot p_2^{x_2}\cdots p_{r}^{x_r}$$
where each exponent $x_i$ is a non-negative integer less or equal to $n_i$. The number of such divisors is equal to the product $(n_1+1)\cdot (n_2+1)\cdots (n_r+1)$.
In your example $n=372=2^2 \cdot 3^1 \cdot 31^1$ and therefore the divisors are $3\cdot 2\cdot 2=12$ and they are
$$2^0 \cdot 3^0 \cdot 31^0$$
$$2^0 \cdot 3^0 \cdot 31^1$$
$$2^0 \cdot 3^1 \cdot 31^0$$
$$2^0 \cdot 3^1 \cdot 31^1$$
$$2^1 \cdot 3^0 \cdot 31^0$$
$$2^1 \cdot 3^0 \cdot 31^1$$
$$2^1 \cdot 3^1 \cdot 31^0$$
$$2^1 \cdot 3^1 \cdot 31^1$$
$$2^2 \cdot 3^0 \cdot 31^0$$
$$2^2 \cdot 3^0 \cdot 31^1$$
$$2^2 \cdot 3^1 \cdot 31^0$$
$$2^2 \cdot 3^1 \cdot 31^1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2361609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to solve $x + \sqrt{x + \sqrt{11+x}}=11$ algebraically? How can I solve the following equation algebraically?
$x + \sqrt{x + \sqrt{11+x}}=11$
| Using the same steps as Gio67 in his/her answer, we arrive to the following quartic equation
$$x^4-46 x^3+771 x^2-5567 x+14630=0$$ But, a (tedious) inspection shows that $x_1=14$ is a solution. However, this solution does not satisfy the original equation (it was introduced by the multiple squaring processes).
So, what remains is the cubic $$x^3-32 x^2+323 x-1045=0$$ the discriminant of which being $\Delta=5073$ which implies three distinct real roots. Using the trigonometric method, the three roots are given by
$$x_2=\frac{2\sqrt{55}}{3} \cos (\theta )+\frac{32}{3}\approx 15.5500 $$
$$x_3=\sqrt{\frac{55}{3}} \sin (\theta )-\frac{\sqrt{55}}{3} \cos (\theta
)+\frac{32}{3}\approx 8.89430$$
$$x_4=-\sqrt{\frac{55}{3}} \sin (\theta )-\frac{\sqrt{55}}{3} \cos (\theta
)+\frac{32}{3}\approx 7.55568$$ where $$\theta=\frac{1}{3} \tan ^{-1}\left(\frac{9 \sqrt{1691}}{727}\right)\approx 0.156960$$ But, again, $x_2$ and $x_3$ do not satisfy the original equation.
So, the only root is $x_4\approx 7.55568$ which can be confirmed by a plot of the original function.
For sure, the plot showing a solution close to $x=7$, we could use Newton method using $$f(x)=x+\sqrt{x+\sqrt{x+11}}-11$$
$$f'(x)=\frac{\frac{1}{2 \sqrt{x+11}}+1}{2 \sqrt{x+\sqrt{x+11}}}+1$$ and, starting with $x_0=7$, the successive iterates will be
$$\left(
\begin{array}{cc}
n & x_n \\
0 & 7.000000000 \\
1 & 7.554554415 \\
2 & 7.555683721 \\
3 & 7.555683726
\end{array}
\right)$$ which is the solution for ten significant figures.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2362187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluate: $\lim_{\theta \to \frac {\pi}{4}} \dfrac {\cos \theta - \sin \theta}{\theta - \frac {\pi}{4}}$
Evaluate: $\lim_{\theta \to \frac {\pi}{4}} \dfrac {\cos \theta - \sin \theta}{\theta - \dfrac {\pi}{4}}$.
My Attempt:
\begin{align}
\lim_{\theta \to \frac {\pi}{4}} \dfrac {\cos \theta - \sin \theta }{\theta - \dfrac {\pi}{4}}
&=\lim_{\theta \to \frac {\pi}{4}} \dfrac {\cos \theta - \cos \dfrac {\pi}{4} + \sin \dfrac {\pi}{4} - \sin \theta}{\theta - \dfrac {\pi}{4}}
\\
&=\lim_{\theta \to \frac {\pi}{4}} \dfrac {2\sin \dfrac {\pi-4\theta }{8}\cos \dfrac {\pi+4\theta}{8} - 2\sin \dfrac {4\theta + \pi}{8}\sin \dfrac {4\theta -\pi}{8}}{\theta - \dfrac {\pi}{4}}.
\end{align}
How do I proceed?
| Let $\theta-\frac{\pi}{4}=x$. Hence, $\theta=\frac{\pi}{4}+x$ and
Hence, $$\lim_{\theta \to \dfrac {\pi}{4}} \dfrac {\cos \theta - \sin \theta}{\theta - \dfrac {\pi}{4}}=\lim_{x\rightarrow0}\frac{\cos\left(\frac{\pi}{4}+x\right)-\sin\left(\frac{\pi}{4}+x\right)}{x}=-\sqrt2\lim_{x\rightarrow0}\frac{\sin{x}}{x}=-\sqrt2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2363352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Prove $ \frac{1}{x^{2n}}+x^{2n}$ is an integer for all $n$. Question:
Suppose $ \frac{1}{x^2}+x^2$ is an integer. Prove that $ \frac{1}{x^{2n}}+x^{2n}$ is an integer for all natural $n$.
Hint: Use Strong Induction
My attempt:
Base Case is trivial.
I.H: Assume the result is true for $n = 1,2, ...., k.$
Consider $n = k+1$.
$ \frac{1}{x^{2\left(k+1\right)}}+x^{2\left(k+1\right)}\ =\ \frac{1}{x^{\left(2k+2\right)}}+x^{2k+2}$.
I am not sure what to do from here and how to use the induction hypothesis.
| Observe that
$(\dfrac{1}{x^2} + x^2)(\dfrac{1}{x^{2k}} + x^{2k}) = \dfrac{1}{x^{2(k + 1)}} + x^{2(k - 1)} + \dfrac{1}{x^{2(k - 1)}} + x^{2(k + 1)}$
$= (\dfrac{1}{x^{2(k + 1)}} + x^{2(k + 1)}) + (x^{2(k - 1)} + \dfrac{1}{x^{2(k - 1)}}), \tag{1}$
whence
$\dfrac{1}{x^{2(k + 1)}} + x^{2(k + 1)} = (\dfrac{1}{x^2} + x^2)(\dfrac{1}{x^{2k}} + x^{2k}) - (x^{2(k - 1)} + \dfrac{1}{x^{2(k - 1)}}) \tag{2}$
is an integer if
$(\dfrac{1}{x^2} + x^2)(\dfrac{1}{x^{2k}} + x^{2k}) \tag{3}$
and
$(x^{2(k - 1)} + \dfrac{1}{x^{2(k - 1)}}) \tag{4}$
are integers.
From here, induction will take off and fly on its own.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2365371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Estimate the error that results when $\sqrt{1 + x}$ is replaced by $1 + \frac{1}{2}x$ if $|x| < 0.01$ Question
Estimate the error that results when $\sqrt{1 + x}$ is replaced by $1 + \frac{1}{2}x$ if $|x| < 0.01$
Definition
Taylors formula is $f(x) = P_n(x) + R_n(x)$ where $P_n(x)$ is
\begin{equation}
\begin{aligned}
P_n(x) = f(a)
+ \frac{f'(a)}{1!}(x - a)
+ \frac{f''(a)}{2!}(x - a)^2
& + \ldots
+ \frac{f^{(n)}(a)}{n!}(x - a)^n \\
\end{aligned}
\end{equation}
And $R_n (x) $ is (\emph{where $\xi$ is between $a$ and $x$ })
\begin{equation}
\begin{aligned}
R_n(x) = \frac{f^{(n + 1)}(\xi)}{(n + 1)!}(x - a)^{(n + 1)}
\end{aligned}
\end{equation}
Working
I'm not sure how to go about this, would I say that this is a first order
approximation as
\begin{equation}
\begin{aligned}
P(x) & = 1 - \frac{1}{2}x \\
P'(x) &= - \frac{1}{2} \\
P''(x) &= 0
\end{aligned}
\end{equation}
Then the remainder term would be
\begin{equation}
\begin{aligned}
R_n(x) = \frac{f^{(n + 1)}(\xi)}{(n + 1)!}(x - a)^{(n + 1)}
\end{aligned}
\end{equation}
Where $n + 1 = 2$. For $f(x) = \sqrt{1 + x}$ this would be
\begin{equation}
\begin{aligned}
R_n(x) & = \frac{- \frac{1}{4} (1 + \xi)^{-3/2}}{(3)!}
\end{aligned}
\end{equation}
And $\xi $ is between $-0.01$ and $0.01$
This would give the maximum error as
\begin{equation}
\begin{aligned}
R_n(x) & = \frac{- \frac{1}{4} (1 \pm 0.01 )^{-3/2}}{(3)!} \approx -0.0423
\end{aligned}
\end{equation}
The error is greatest when $\xi = -0.01$.
| expanding about $x=0$
$$ \begin{equation}
\begin{aligned}
P_n(x) = f(0)
+ \frac{f'(0)}{1!}(x - 0)
+ \frac{f''(a)}{2!}(x - 0)^2
& + \ldots
+ \frac{f^{(n)}(0)}{n!}(x - 0)^n \\
\end{aligned}
\end{equation}\\
$$
$$ \begin{equation}
\begin{aligned}
P_n(x) = 1
+ \frac{1}{2}x + \frac{-\frac14}{2!}x^2
& + \ldots
\end{aligned}
\end{equation}\\
$$so w..r.t.$|x| < 0.01$
$$\begin{equation}
\begin{aligned}
R_n(x) & = \frac{- \frac{1}{4(1 + \xi)^{+3/2}} }{2!}
\end{aligned}
\end{equation}\leq \frac{- \frac{1}{4} (1 + (-0.01))^{-3/2}}{2!}=\frac{- \frac{1}{4} (0.99)^{-3/2}}{2!}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2369127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
How to transform $y = 2 \sin(x) - \cos(x)$ into the format $y = z \sin(x - b)$ How can I transform $y = 2 \sin(x) - \cos(x)$ into the format $y = z \sin(x - b)$ for some $z,b\in\Bbb R$? I have been playing with various trig identities and looking through various trigonometry references but cannot so much as find a starting point. Any hints or pushes in the right direction would be much appreciated. Helping a struggling high school study and find myself greatly out of practice.
| \begin{eqnarray}
y&=&a\sin x+b\cos x\\
\frac{y}{\sqrt{a^2+b^2}}&=&\frac{a}{\sqrt{a^2+b^2}}\sin x+\frac{b}{\sqrt{a^2+b^2}}\cos x
\end{eqnarray}
Let $\cos\phi=\frac{a}{\sqrt{a^2+b^2}}$ and $\sin\phi=-\frac{b}{\sqrt{a^2+b^2}}$ then
\begin{eqnarray}
\frac{y}{\sqrt{a^2+b^2}}&=&\sin x\cos\phi-\cos x\sin\phi\\
&=&\sin{(x-\phi)}\\
y&=&(\sqrt{a^2+b^2})\sin{(x-\phi)}
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2369309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Cannot figure out step in solving equation I am reading calculus made easy by Silvanus P. Thompson (1914) and in it the following example was given [p.29 - p.30]:
Find the differential coefficient of $y$ with respect to $x$
$$
ay + bx = by - ax + (x+y)\sqrt{a^2 - b^2}
$$
This is given as an example and is immediately worked out, but I fail to understand a step that the author performed. I quote the solution:
1)
$$
(a-b)y+(a+b)x = (x+y)\sqrt{a^2 - b^2}\\
$$
2)
$$
(a-b)^2y^2 + (a+b)^2x^2+2(a+b)(a-b)xy = (x^2+y^2+2xy)(a^2-b^2)
$$
3)
$$
(a-b)^2y^2+(a+b)^2x^2=x^2(a^2-b^2)+y^2(a^2-b^2)
$$
4)
$$
\lbrack(a-b)^2-(a^2-b^2)\rbrack y^2 = \lbrack(a^2-b^2)-(a+b)^2\rbrack x^2
$$
5)
$$
2b(b-a)y^2 = -2b(b+a)x^2
$$
6)
$$
y = \sqrt{\frac{a+b}{a-b}}x
$$
And finally
$$
\frac{dy}{dx} = \sqrt{\frac{a+b}{a-b}}
$$
The step that I do not understand is the transition from (2) to (3). Namely, where did $2(a+b)(a-b)xy$ (left side) go? And $(x^2+y^2+2xy)$ (right side)?
(the title of this question is probably inappropriate, but I fail to come up with a more appropriate one. I also don't know which tags I should file this question under. Very sorry for the inconvenience)
| $$(a-b)^2y^2 + (a+b)^2x^2+\color{blue}{2(a+b)(a-b)xy} = (x^2+y^2+2xy)(a^2-b^2)$$
$$(a-b)^2y^2+ (a+b)^2x^2+ \color{blue}{2(a^2-b^2)xy} =(x^2+y^2)(a^2-b^2)+\color{blue}{2xy(a^2-b^2)}$$
$$(a-b)^2y^2+ (a+b)^2x^2+ \require{enclose}\enclose{updiagonalstrike}{\color{blue}{2xy(a^2-b^2)}} =(x^2+y^2)(a^2-b^2)+\require{enclose}\enclose{updiagonalstrike}{\color{blue}{2xy(a^2-b^2)}}$$
The rest is group the terms with $x^2$ and $y^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2373594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Cosecants of half angles
The tangent of an angle is $2.4$. Find the cosecant of half the angle.
My tries:
As $\tan A=2.4=\dfrac{12}{5}\implies \sin A=\pm\dfrac{12}{13}=\dfrac{1}{\csc A}$
Also:
$\sin \frac{A}{2}+\cos \frac{A}{2}=\pm\sqrt{1+\sin A}=\pm\dfrac{5}{{\sqrt{13}}}$ , considering $\sin A=\dfrac{12}{13}$
$\sin \frac{A}{2}-\cos \frac{A}{2}=\pm\sqrt{1-\sin A}=\pm\dfrac{1}{{\sqrt{13}}}$, considering $\sin A=\dfrac{13}{13}$
Adding them gives: $\sin \frac{A}{2}=\pm\dfrac{3}{\sqrt{13}}=\dfrac{1}{\csc{\frac A2}}$
I did same by considering $\sin A=-\dfrac{12}{13}$ then also it gave the same result.
But answer provided by the author is $\dfrac{\pm\sqrt{13}}{2}$ and $\dfrac{\pm\sqrt{13}}{3}$.
So what did I miss? please help.
| You found $\sin (A/2)$ but you needed to find $\csc(A/2) = 1/\sin(A/2)$ which is as is in the answer...
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Solving a pair of simultaneous Pell-like equations I'm trying to prove the following.
Conjecture: If $a,b,c$ are positive integers satisfying the system of equations
\begin{align}
a^2+3b^2 &= 4c^2, \\
a^2-3b^2 &= -2,
\end{align}
then $(a,b,c)=(1,1,1)$.
Unfortunately, all my efforts have ended up going around in circles. I’m assuming there’s a relatively easy proof (likely by descent). Any suggestions/hints would be appreciated.
| This is the same problem you once posted 3 years ago. Replacing variable names to avoid confusion, the old system is
\begin{align}
2x^2−1 &= y^2, \\
2x^2+1 &= 3z^2.
\end{align}
Now put $y=a$, $x=c$, and $z=b$, so that the system becomes
\begin{align}
2c^2-1 &= a^2, \\
2c^2+1 &= 3b^2.
\end{align}
Substitute $2c^2=a^2+1$ (as given by the first equation) into the second to get $a^2+2=3b^2$ or $a^2−3b^2=−2$. Now multiply the first equation by $2$ and equate with $-2$ to get $2a^2-4c^2 = a^2-3b^2$, which is $a^2+3b^2=4c^2$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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} |
How many solutions does the equation $a + b + c + d + e = 21$ have in the nonnegative integers if $a \leq 3$, $0 < b < 4$, and $c \geq 15$? How many solutions are there to the equation in the nonnegative integers:
$$a+b+c+d+e = 21$$
Conditions:
a) $ a \le 10$
I understood the solution which is total number of possibilities - the number of possibilities with $a \gt 10$ which is $\binom{5+21-1}{21} - \binom{5+10-1}{10}$.
b) $a \le 3$, $0 \lt b \lt 4$, $c \ge 15$.
How do I solve this part? Thanks in advance.
| We wish to find the number of solutions in the nonnegative integers of the equation
$$a + b + c + d + e = 21 \tag{1}$$
subject to the restrictions $a \leq 3$, $0 < b < 4$, and $c \geq 15$. Since $b$ is an integer, the restriction $0 < b < 4$ is equivalent to the restriction $1 \leq b \leq 3$.
We first address the restrictions $b \geq 1$ and $c \geq 15$. Let $b' = b - 1$ and $c' = c - 15$. Then $b'$ and $c'$ are nonnegative integers. Substituting $b' + 1$ for $b$ and $c' + 15$ for $c$ in equation 1 yields
\begin{align*}
a + b' + 1 + c' + 15 + d + e & = 21\\
a + b' + c' + d + e & = 5 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers with
$$\binom{5 + 5 - 1}{5 - 1} = \binom{9}{4}$$
solutions.
From these, we must subtract those cases in which $a > 3$ or $b > 3$. Observe that the restriction $b > 3 \implies b' = b - 1 > 2$. Notice also that since $4 + 3 = 7 > 5$, these two conditions cannot be violated simultaneously.
Suppose $a > 3$. Let $a' = a - 4$. Then $a'$ is a nonnegative integer. Substituting $a' + 4$ for $a$ in equation 2 yields
\begin{align*}
a' + 4 + b' + c' + d + e & = 5\\
a' + b' + c' + d + e & = 1 \tag{3}
\end{align*}
Equation 3 is an equation in the nonnegative integers with
$$\binom{1 + 5 - 1}{5 - 1} = \binom{5}{4}$$
solutions.
Suppose $b' > 2$. Let $b'' = b' - 3$. Then $b''$ is a nonnegative integer. Substituting $b'' + 3$ for $b'$ in equation 2 yields
\begin{align*}
a + b'' + 3 + c' + d + e & = 5\\
a + b'' + c' + d + e & = 2 \tag{4}
\end{align*}
Equation 4 is an equation in the nonnegative integers with
$$\binom{2 + 5 - 1}{5 - 1} = \binom{6}{4}$$
solutions.
Hence, the number of solutions of equation 1 subject to the restrictions $a \leq 3$, $0 < b < 4$, and $c \geq 15$ is
$$\binom{9}{4} - \binom{5}{4} - \binom{6}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2379635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Splitting over some extension of $\mathbb Z_3$ I want to write $x^3 + 2x + 1$ as a product of linear polynomials over some extension field of $\mathbb Z_3$.
I know that the roots will lie in a field that will be isomorphic to the field $\mathbb Z_3[x]/<x^3 + 2x + 1>$ but not sure how to proceed further.
| As your polynomial is irreducible, the Frobenius automorphism $a \mapsto a^3$ gives $$T^3+2T+1 = (T-\alpha)(T-\alpha^3)(T-\alpha^{3^2})$$ where $\alpha = x + (x^3+2x+1)$ is one of its roots in $\mathbb{Z}_3[x]/(x^3+2x+1)$.
Then you compute $\alpha^3 = x^3+(x^3+2x+1)= x+2 +(x^3+2x+1), \alpha^{3^2} = (x+2)^3 + (x^3+2x+1) = x^3+2^3 +(x^3+2x+1) = x+1+(x^3+2x+1) $ and hence
$\qquad\qquad T^3+2T+1 = (T-x)(T-x+1)(T-x+2)$ in $\mathbb{Z}_3[x]/(x^3+2x+1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2384781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Evaluating limits Evaluate the limit without L’Hôpital rule:
$$
\lim_{x \to 0}\frac{\sin^2{x}+2\ln\left(\cos{x}\right)}{x^4}
$$
My work is:
\begin{align}
L&=\lim_{x \to 0}\frac{\sin^2{x}-x^2}{x^4}+\lim_{x \to 0} \frac{x^2+2\ln\left(\cos{x}\right)}{x^4}\\
&=
\lim_{x \to 0}\frac{\sin{x}-x}{x^3}
\lim_{x \to 0}\frac{\sin{x}+x}{x}+
\lim_{x \to 0}\frac{x^2+2\ln\left(\cos{x}\right)}{x^4}\\
&=
\frac{-1}{6}\left[\lim_{x \to 0}\frac{\sin x}{x}+1\right]
+\lim_{x \to 0}\frac{x^2+2\ln\left(\cos{x}\right)}{x^4}\\
&=\frac{-1}{6}\left(2\right)+\lim \limits_{x \to 0}\frac{x^2+2\ln\left(\cos{x}\right)}{x^4}\\
&=
\frac{-1}{3}+\lim_{x \to 0}\frac{x^2+2\ln\left(\cos{x}\right)}{x^4}
\end{align}
I could not evaluate the second limit
| We evaluate the limit in question as follows
\begin{align}
L&=\lim_{x\to 0}\frac{\sin^{2}x+2\log\cos x}{x^{4}}\notag\\
&=\lim_{x\to 0} \frac{\log(1-\sin^{2}) +\sin^{2}x}{\sin^{4}x}\cdot\frac{\sin^{4}x}{x^{4}}\notag\\
&=\lim_{t\to 0^{+}}\frac{\log(1-t)+t}{t^{2}}\text{ (putting } t=\sin^{2}x)\notag\\
&=-\frac{1}{2}\text{ (via Taylor series or L'Hospital's Rule)} \notag
\end{align}
I have left the last step (which is almost routine).
| {
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"question_score": "3",
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"answer_id": 4
} |
Roots of $(x-\alpha)(x-\beta)(x-\gamma)+d=0$?
Let $\alpha,\beta,\gamma$ be the roots of $$(x-a)(x-b)(x-c)=d,d\neq 0$$ then the roots of $$(x-\alpha)(x-\beta)(x-\gamma)+d=0?$$
My tries:
Clearly $$(x-a)(x-b)(x-c)=-(x-\alpha)(x-\beta)(x-\gamma)=d\tag{1}$$
As $\alpha,\beta,\gamma$ are roots of equation on left, also if we'll put $x=a,b,c$ gives $0$ on $LHS\implies$ $a,b,c$ are roots of: $$(x-\alpha)(x-\beta)(x-\gamma)=0\\\underbrace{(x-a)(x-b)(x-c)}_{0\ at\ x=a,b,c}+(x-\alpha)(x-\beta)(x-\gamma)=0\rightarrow a,b,c\ \text{satisfy this.}\\(x-\alpha)(x-\beta)(x-\gamma)+d=0\ \rightarrow a,b,c\ \text{satisfy this.}$$.
Hence the result, $a,b,c$ is the answer.
But when we put in $x=\alpha,\beta,\gamma$ in $(1)$, gives $$(x-a)(x-b)(x-c)=0=d$$ but $d\neq 0$.
What I'm mistaking, please help.
| From $(x-a)(x-b)(x-c)=d$ we have $$a+b+c=\alpha+\beta+\gamma\\ab+bc+ca=\alpha\beta+\beta\gamma+\gamma\alpha\\abc+d=\alpha\beta\gamma$$
Now $(x-\alpha)(x-\beta)(x-\gamma)+d=x^3-(\alpha+\beta+\gamma)x^2+(\alpha\beta+\beta\gamma+\gamma\alpha)x-\alpha\beta\gamma+d\\=x^3-(a+b+c)x^2+(ab+bc+ca)x-abc-d+d\\=x^3-(a+b+c)x^2+(ab+bc+ca)x-abc=(x-a)(x-b)(x-c)$
Hence roots are $a,b,c\space\space\space\space\space\space\blacksquare$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $n! +5$ is not a perfect square for $n\in\mathbb{N}$ I have a proof of this simple problem, but I feel that the last step is rather clunky:
For $n=1,2,3,4$ we have $n!+5=6,7,11,29$ respectively, none of which are square. Now assume that $n\geq 5$, then:
$$\begin{aligned}
n! +5 & \;=\; n(n-1)\cdots 6\cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 + 5 \\[0.2cm]
& \;=\; 5\left[ \frac{}{} n(n-1)\cdots 6\cdot 4 \cdot 3 \cdot 2 \cdot 1 + 1 \right] \\[0.2cm]
&\;=\; 5(3k+1)
\end{aligned}$$
for some $k\in\mathbb{N}$. Since $5(3k+1)=15k+5\equiv 5\,\text{mod} \, 15$ and all perfect squares are congruent either $0,1,4,6, 9$ or $10\,\text{mod}\,15$, the result follows. $\;\blacksquare$
It took a pretty tedious exhaustive search of squares modulo 15 in the last step; is there a theorem I am missing that means the last step follows immediately? Your comments are much appreciated!
| If $n! + p = k^2$ for some prime it's not too difficult to conclude $n < p$ and $k$ has no prime factors less than $p$, or $p \le n < 2p$, $\frac {n!}p + 1$ is divisible by $p$ and $\frac {\frac {n!}p + 1}p$ is a perfect square to no prime factors less than $p$[$*$].
We can go a bit further. $(p-1)! \equiv -1 \mod p$ by Wilson's Theorem so for $p \le n < 2p$ with $\frac {n!}p + 1\equiv 0 \mod p$ we can conclude $\frac {n!}p = (p-1)!\prod (p + k) \equiv -1 \mod p$ so $(n-p)! \equiv 1 \mod p$. And $n = p+1$ or $n = p + (p-2)= 2p-2$ so only two cases to check, if $n \ge p$.
So if $n < 5$ and $n! + 5 = k^2$ then $n! + 5 \ge 7^2$ which is simply not possible for $n\le 4; n! \le 24$
If $5\le n < 10$. then $\frac {n!}5 \equiv -1 \mod 5$. If $n = 5+k$ then $\frac {n!}5 = 4!*(6....n) \equiv 4!*k! \equiv -k! \equiv -1$ so $k! \equiv 1\mod 5$. So $k =1, 3$ and $n=6,8$. Which can be checked.
Which is probably much more clunky than your proof. But it could be useful for generalizing.
====
[$*$] $k \le n$ implies $k|n!$ so $k\not \mid n! +p$ unless $k = p$. If so $p|n! + p$ and $n! + p = p(\frac {n!}p + 1)$. If $n \ge 2p$ then $p|\frac {n!}p$ and $p \not \mid \frac{n!}p + 1$. If $n! + p$ is a perfect square and $p|n! + p$ then $p^2|n!+p$ and so $p \le n < 2p$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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On largest box not containing integer vector solutions I am trying to understand the largest cube in $\Bbb Z^4$ around origin not containing integer vector $(a,b,c,d)$ solutions to $$aBD+bBC+cAD+dAC=0$$ where $gcd(AB,CD)=gcd(AC,BD)=1$ when $n<A,B,C,D<2n$ and $n/8<|A-B|,|C-D|,|A-C|,|A-D|,|B-C|,|B-D|<n/4$ holds?
Is the box $c'n^{2/3}$ in length around origin for some constant $0<c'$? That is a vector $(a,b,c,d)\in\Bbb Z^4$ with $|a|,|b|,|c|,|d|<c'n^{2/3}$ exists?
This is where I get the $2/3$ value.
$$aBD+bBC+cAD+dAC=0$$
$$B(aD+bC)=-A(cD+dC)$$
$$cD+dC=\ell B$$
$$aD+bC=-\ell A$$
If size of $\ell$ is $n^\eta$ then size of $\ell B$ is size of $cD+dD$ is $n^{1+\eta}$. We have $n^{2\eta}$ values in $cd$ and $n^{2\eta}$ values in $ab$ and we want to hit $\ell B$ and $\ell A$ (a total of $2n^{\eta}$ possibilities) of a total of $n^{2(1+\eta)}$ possibilities. The expected intersection is
$$\frac{n^{2\eta}n^{2\eta}2n^{\eta}}{n^{2(1+\eta)}}=\frac{2n^{3\eta}}{n^2}$$
If $3\eta>2$ or $2/3<\eta$ we have expected intersection $\gg1$.
| October 4, 2017
You had a question for a few hours about doubling the dimension; maybe that was yesterday. We are looking for a basis for the lattice inside $\mathbb Z^8$ of vectors orthogonal to
$$ (EBD, EBC, EAD, EAC,FBD, FBC, FAD, FAC ) $$
where we define integers such that
$$ MB - NA =1, $$
$$ JD-KC=1. $$
A 7 by 8 matrix with rows giving a basis is
$$
R =
\left(
\begin{array}{rrrrrrrr}
A & 0 & -B & 0 & 0 & 0 & 0 & 0 \\
0& -A &0 & B &0 & 0 & 0 &0 \\
0& 0 &0 &0 & A &0 & -B & 0 \\
0& 0 & 0 &0 & 0 & -A &0 & B \\
MC& -MD & -NC & ND & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & MC & -MD & -NC & ND \\
FMJ & -FMK & -FNJ & FNK & -EMJ & EMK & ENJ & -ENK
\end{array}
\right)
$$
The resulting Gram matrix is
$$
G = R R^T =
\scriptsize
\left(
\begin{array}{ccccccc}
A^2 + B^2 & 0 & 0 & 0 & (AM+BN)C & 0 & (AM+BN)FJ \\
0& A^2 + B^2 &0 & 0 & (AM+BN)D & 0 & (AM+BN)FK \\
0& 0 & A^2 + B^2 &0 & 0 & (AM+BN)C & -(AM+BN)EJ \\
0& 0 & 0 & A^2 + B^2 & 0 & (AM+BN)D &-(AM+BN)EK \\
(AM+BN)C& (AM+BN)D & -NC & ND & (M^2 + N^2)(C^2 + D^2) & 0 & (M^2 + N^2)(JC+KD)F\\
0 & 0 & (AM+BN)C & (AM+BN)C & 0 &(M^2 + N^2)(C^2 + D^2) & -(M^2 + N^2)(JC+KD)E \\
(AM+BN)FJ & (AM+BN)FK & -(AM+BN)EJ &-(AM+BN)EK & (M^2 + N^2)(JC+KD)F &-(M^2 + N^2)(JC+KD)E & (M^2 + N^2)(J^2 + K^2)(E^2 + F^2)
\end{array}
\right)
$$
The amazing thing is all the cancellation as
$$ \det G = (A^2 + B^2)(C^2 + D^2)(E^2 + F^2) $$
As before, the smallest element of the indicated quadratic form is no more than an explicit constant times
$$ \left( \det G \right)^{1/7}. $$
Each element of the vector in $\mathbb Z^8$ that achieves that quadratic form value is no larger than a constant times
$$ \left( \det G \right)^{1/14}. $$
It is pretty clear that this process continues; repeat to get, in dimension 16,
$$ \det G = (A^2 + B^2)(C^2 + D^2)(E^2 + F^2)(H^2 + I^2) $$
with exponents given as $1/15$ then $1/30.$
Here are pages 136-137 of Cassels, Rational Quadratic Forms, which you should purchase.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Under what conditions $z=x+iy$ in $3^{rd}$ quadrant implies $\frac{\bar{z}}{z}$ in $3^{rd}$ quadrant?
If $z=x+iy$ is in the $3^{rd}$ quadrant, then $\frac{\bar{z}}{z}$ also lies in the $3^{rd}$ quadrant if ?
How do I obtain the solution which is given as $x<y<0$.
My Attempt:
$x=|z|\cos\theta<0$ & $y=|z|\sin\theta<0$
Since $\frac{\bar{z}}{z}=\frac{|\bar{z}|}{|z|}[\cos(-2\theta)+i\sin(-2\theta)]$ is in the $3^{rd}$ quadrant,
$$
\cos(-2\theta)<0 \text{ & } \sin(-2\theta)<0\implies \cos^2\theta-\sin^2\theta<0\text{ & }\sin\theta\cos\theta<0
$$
From $\sin\theta\cos\theta<0$ we only take $x<0$ & $y<0$ since it is already stated. Now,
$$
\cos^2\theta-\sin^2\theta<0\implies\cos^2\theta<\sin^2\theta\implies|\cos\theta|<|\sin\theta|\implies|x|<|y|
$$
Since both $x,y<0$, I think $|x|=-x$ & $|y|=-y$
$$
|x|<|y|\implies-x<-y\implies x>y
$$
So the solution I obtain is $y<x<0$.
Is this the right way to approach the inequalities of this kind ?. What am I doing wrong ?
| Note that a complex number $a+ib$ is in the third quadrant (not including axes) iff $a,b<0$. Therefore we can assume that $x,y<0$ in order to have $z$ in the third quadrant, and we want to find conditions so that
$\frac{\bar{z}}{z}$ is in the third quadrant. Now
$$
\frac{\bar{z}}{z} = \frac{x-iy}{x+iy} = \frac{x^2-y^2-ixy}{x^2+y^2}
= \frac{x^2-y^2}{x^2+y^2} +i\frac{-xy}{x^2+y^2},
$$
so we need to have $x^2-y^2<0$ and $-xy<0$. The latter is true because $x,y<0$, and the former is equivalent to $y<x$ since we know $x$ and $y$ are negative.
We conclude that $y<x<0$ are the desired conditions.
The conditions $x<y<0$, which appear in the question, don't work. Indeed, consider $z=-2-i$. Then $z$ is in the third quadrant, but
$$
\frac{\bar{z}}{z} = \frac{-2+i}{-2-i} = \frac{3-4i}{4+1} = \frac{3}{5}-\frac{4}{5}i
$$
is in the fourth quadrant.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integrate $\int{\frac{1}{x^2-x+1}dx}$ Original question was to integrate $$\int{\frac{(x+1)}{x^2-x+1}dx}$$
But I was able to break it into 2 parts: $$\frac{1}{2}\int{\frac{2x-1}{x^2-x+1}dx}+\frac{3}{2}\int{\frac{1}{x^2-x+1}dx} $$
The first part $\frac{1}{2}\int{\frac{2x-1}{x^2-x+1}dx}$ could be easily integrated by substituting $u={x^2-x+1}$ and thus, getting $\frac{\ln(x^2-x+1)}{2}$ as the answer. But I have no idea on how to integrate part 2 of the equation.
Any help will be appreciated.
| Here is a shorter path, using your own idea to start:
$$\int \frac{x+1}{x^2-x+1}dx = \dfrac{1}{2}\int \frac{(2x - 1) + 3}{x^2-x+1}dx$$
$$= \dfrac{1}{2}\int \frac{2x - 1}{x^2-x+1}dx + \dfrac{3}{2}\int \frac{1}{(x-\frac 1 2)^2 +(\frac{\sqrt{3}}{2})^2 }dx$$
$$= \frac{1}{2}\ln|x^2-x+1| + \frac{3}{2}\cdot \dfrac{2}{\sqrt{3}}\tan^{-1}\left(\dfrac{x-\frac{1}{2}}{\sqrt{3}/2}\right) +C$$
| {
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Why does $(a + b)^3$ expand to $a^3 + 3ab^2 + 3a^2b + b^3$
Why does $(a + b)^3$ expand to $a^3 + 3ab^2 + 3a^2b + b^3$
Why does this work? I am confused as to why this happens.
| In overly pedantic detail:
\begin{align}
(a+b)^3
&= (a+b)[(a+b)(a+b)] \tag{by definition} \\
&= (a+b)[a(a+b) + b(a+b)] \tag{distribution} \\
&= (a+b)[a^2 + ab + ba + b^2] \tag{distribution} \\
&= (a+b)[a^2 + 2ab + b^2] \tag{commutativity} \\
&= a[a^2 + 2ab + b^2] + b[a^2 + 2ab + b^2] \tag{distribution} \\
&= a^3 + a2ab + ab^2 + ba^2 + b2ab + b^3 \tag{distribution} \\
&= a^3 + 2a^2b + ab^2 + a^2b + 2ab^2 + b^3 \tag{commutativity} \\
&= a^3 + 3a^2b + 3ab^2 + b^3. \tag{collect like terms}
\end{align}
| {
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For $g(X) = \frac{(x^2+y^4)^3}{1+x^6y^4}$ , show $\lim\limits_{|x| \rightarrow \infty} g(x,ax) = \infty? $ for any real number $a$
Let $$g(X) = \frac{(x^2+y^4)^3}{1+x^6y^4}.$$ (1) Show that $\lim\limits_{|x| \rightarrow \infty} g(x,ax) =\infty $ for any real number $a$.
(2) Does $\lim\limits_{|X| \rightarrow \infty } g(X) = \infty$?
(1) Considering $g(x,ax)$:
$$g(x,ax) = \frac{(x^2+(ax)^4)^3}{1+x^6(ax)^4}= \frac {x^6+3a^4x^8+3x^{10}a^8+a^{12}x^{12}} {1+x^{10}a^4} = \frac {\frac{1}{x^4}+\frac{3a^4}{x^2}+3a^8+a^{12}x^{2}} {\frac{1}{x^{10}}+a^4}$$
It follows that as $|x| \rightarrow \infty$, we have $g(x,ax) \rightarrow \infty$
(2) Does $\lim\limits_{|X| \rightarrow \infty } g(X) = \infty$?
$$g(X) = \frac{(x^2+y^4)^3}{1+x^6y^4}.$$ I believe I am supposed to factor out $|X|^2=|x^2+y^2|$ but I do see how.
I d like to know if I am in the right direction. What would be a $\epsilon,\delta$ approach is this case?
Much appreciated for your help/explanation.
| For (2),
let
$x=t^u, y=t^v$
and I will try to choose
$u$ and $v$
to make things interesting.
$\begin{array}\\
g(X)
&= \dfrac{(x^2+y^4)^3}{1+x^6y^4}\\
&= \dfrac{(t^{2u}+t^{4v})^3}{1+t^{6u}t^{4v}}\\
&\sim \dfrac{(t^{\max(2u, 4v)})^3}{t^{6u+4v}}\\
&= \dfrac{t^{\max(6u, 12v)}}{t^{6u+4v}}\\
&= t^{\max(6u, 12v)-6u-4v}\\
\end{array}
$
If $12 v < 6u+4v$,
then $g(X) \to 0$.
This is
$8v < 6u$.
This works if,
for example,
$u=2, v=1$.
Try
$x = t^2, y=t$.
Then
$g(X)
=\dfrac{(t^4+t^4)^3}{1+t^{12}t^4}
=\dfrac{8t^{12}}{1+t^{16}}
\to 0
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2398798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Sum of the recursive series Let $\langle a_n \rangle$ be a recursive sequence given by $a_1>2$ and,
$a_{n+1}=a_n^2-2$ for $n \in \mathbb N$
Show that
$\sum_{n=1}^\infty \frac{1}{a_1a_2\cdots a_n} = \frac{a_1-\sqrt{a_1^2-4}}{2}$
I have reached this step:
$\frac{1}{a_1a_2a_3\cdots a_n}=\frac{1}{2} (\frac{a_n}{a_1a_2\cdots a_n-1} -\frac{a_n-1}{a_1a_2\cdots a_n})$
But I am not able to obtain the final expression .Please help me to obtain it. Thanks for help in advance.
| Let $P_n = a_1 a_2 \cdots a_n$, and let $b_n = a_{n+1}/P_n$.
From the recursion, note that $$a_{n+1} - 2 = {a_n}^2 - 4 \implies a_n + 2 = \frac{a_{n+1} - 2}{a_n - 2},$$ hence we have the telescoping product $${P_n}^2 = \prod_{k=2}^{n+1} (a_k+2) = \prod_{k=2}^{n+1} \frac{a_{k+1} - 2}{a_k - 2} = \frac{a_{n+2} - 2}{a_2 - 2}$$
It follows that $${b_n}^2 = (a_2 - 2) \frac{{a_{n+1}}^2}{a_{n+2} - 2} = (a_2 - 2) \frac{a_{n+2} + 2}{a_{n+2} - 2} \implies \lim_{n\to\infty} b_n = \sqrt{a_2 - 2} = \sqrt{{a_1}^2 - 4}$$
Next, note that $$b_{k-1} - b_{k} = \frac{{a_k}^2}{P_k} - \frac{a_{k+1}}{P_k} = \frac{2}{P_n} \implies 2\sum_{k=2}^{n} \frac{1}{P_k} = b_1 - b_n $$ $$\implies 2\sum_{k=1}^{n} \frac{1}{P_k} = \frac{a_2+2}{a_1} - b_n = a_1 - b_n$$
We conclude that $$\sum_{k=1}^{\infty} \frac{1}{P_k} = \lim_{n\to\infty} \sum_{k=1}^{n} \frac{1}{P_k} = \lim_{n\to\infty} \frac{a_1 - b_n}{2} = \frac{a_1 - \sqrt{{a_1}^2 - 4}}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Points of Intersection between a function and its inverse Say we have this function: $g(x) = 1+ \frac{1}{x+1}$ for $x>-1$.
The question asks to solve for $g(x) = g^{-1}(x)$. I intuitively thought to first calculate the inverse function's equation which turned out to be $y=\frac{-x+2}{x-1}$ and then to simply equate it with the original function (which I thought was the normal way to find POI's of functions). However, while this did produce an answer ($x=0,-2$) they were not correct (the answer is $x=\sqrt{2}$.Could I have some direction on this please.
| Your method is correct, but you made a mistake when finding the inverse of $g$. To see why, note that $g(0)=2$ but
$$
-\frac{(2)+2}{(2)-1} = -4 \ne 0.
$$
It should be
$$
g^{-1}(x) = \frac{2-x}{x-1}.
$$
Let's verify that this is correct. Note that
$$
(g\circ g^{-1})(x) = 1+\frac{1}{\frac{2-x}{x-1}+1}
= 1+\frac{1}{\frac{1}{x-1}} = 1+(x-1)=x
$$
and
$$
(g^{-1}\circ g(x) = \frac{2-(1+\frac{1}{x+1})}{(1+\frac{1}{x+1})-1}
= \frac{1-\frac{1}{x+1}}{\frac{1}{x+1}}
= \frac{1}{\frac{1}{x+1}} -1=(x+1)-1=x.
$$
Continuing with the solution, we need to solve $g(x)=g^{-1}(x)$. That is:
\begin{align*}
1+\frac{1}{x+1} &= \frac{2-x}{x-1} \\
(x+1)(x-1) + (x-1) &= (x+1)(2-x) \\
(x-1)(x+2) &= (x+1)(2-x) \\
x^2+x-2 &= -x^2+x+2 \\
2x^2-4&=0 \\
x^2 &= 2 \\
x &= \pm\sqrt{2}
\end{align*}
Since the domain excludes $x=-\sqrt{2}$, we are left with only $\sqrt{2}$.
| {
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"url": "https://math.stackexchange.com/questions/2399268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Remainder Theorem Problem
When the polynomial $p(x)$ is divided by $(x^2+3x+2)$ the remainder is $5x+1$. Find the remainder when $p(x)$ is divided by $(x+2)$.
I get that $(x+2)$ is a factor of $(x^2+3x+2)$ but I'm not sure how to use that fact to get the question's answer.
| Think of it this way:
$p(x)$ divided $x^2 + 3x + 2$ has remainder $5x+1$ means:
$$p(x) = q(x)(x^2 + 3x + 2) + 5x+1$$ where $q(x)$ is a polynomial.
But we have $x^2 + 3x + 2$ so
$$p(x) = q(x)(x+1)(x+2) + 5x + 1$$
$$= s(x)(x+2) + 5x+1$$
Where $s(x) = q(x)(x+1)$
So we just need the divide $5x + 1$ by $x+2$ to get $5x + 1 = t(x)(x+2) + r(x)$.
As $5x + 1 = 5(x+2) - 9$ we have
$$p(x) = q(x)(x+1)(x+2) + 5(x+2) - 9$$
$$= [q(x)(x+1) + 5](x+2) - 9$$.
And the remainder is $9$.
===
This is actually the exact same logic as it would be with numbers.
If $M$ divided by $36$ has remainder $17$, then what is the remainder of $M$ divided by $12$.
As $12$ divides $36$ we know that $12$ will divide "just as evenly" as $36$ and the remainder will be the same $17$ but further divided by $12$ to have remainder: $5$.
Formally: $$M = k*36 + 17$$
$$= (3k)*12 + (12 + 5)$$
$$= (3k+1)*12 + 5$$.
The remainder is $5$ because the remainder of $17$ is $5$.
It's no different with polynomials.
There remainder is $-9$ because the remainder of $5x + 1$ is $-9$.
| {
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"url": "https://math.stackexchange.com/questions/2399371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Expressing the roots of $y^5 + 11y^4 - 77y^3 + 132y^2 - 77y + 11 = 0$ in terms of of $\zeta_{11}$? Is there a concise way to express the roots of,
$$x^3 + 7x^2 + 7x - 7 = 0$$
using the root of unity $\zeta_7$?
Similarly, is there an analogous expression for the solvable quintic,
$$y^5 + 11y^4 - 77y^3 + 132y^2 - 77y + 11 = 0$$
in terms of of $\zeta_{11}$?
P.S. Note the similar-looking,
$$z^5 + 11z^4 + 44z^3 + 77z^2 + 55z + 11 = 0$$
has $z = \big(\zeta_{11}-\zeta_{11}^{-1}\big)^2$.
| If $\;t^7=1\;$ and $\;t\ne1,\;$ then $\;x=-1+2/(t+t^{-1})\;$ is a root of $x^3 + 7x^2 + 7x - 7 = 0\;$ but the six primitive $7$th roots come in conjugate pairs giving the three roots of the cubic in $x$.
If $\;t^{11}=1\;$ and $\;t\ne1,\;$ then $\;z=t\!+\!t^{-1}\!-\!2$ is a root of $z^5 + 11z^4 + 44z^3 + 77z^2 + 55z + 11 = 0$ but the ten primitive $11$th roots come in conjugate pairs giving the five roots of the quintic in $z$ and the five roots of $\;y^5 + 11y^4 - 77y^3 + 132y^2 - 77y + 11 = 0\;$ are from $\;y=(2-u)(1-u+u^3)\;$ where $u:=t+t^{-1}.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Range of the integral of a trigonometric function Given $$g(x) = \int_0^{x} \sqrt{1+\sin t} dt -\sqrt{1+\sin x}$$
I would like to show that $g(x) \geq 2\sqrt{2} -10,\,\forall x\in \left(0,\infty\right)$.
Any help is appreciated.
| Simply
$\int \sqrt{1+\sin t} \, dt=\int \dfrac{\sqrt{1+\sin t}\sqrt{1-\sin t}}{\sqrt{1-\sin t}}\,dt=\int \dfrac{\cos t}{\sqrt{1-\sin t}}\,dt$
Substitute $\sin t=u \to \cos t\, dt = du$
$\int \dfrac{du}{\sqrt{1-u}}=-2 \sqrt{1-u}+C=-2\sqrt{1-\sin t}+C$
thus $\int_0^x \sqrt{1+\sin t} \, dt=-2 \sqrt{1-\sin x}+2$
therefore $g(x)=-2 \sqrt{1-\sin x}+2-\sqrt{\sin x+1}$
$g'(x)=-\frac{1}{2} \left(\sqrt{1-\sin x}-2 \sqrt{\sin x+1}\right)$
and we have $g'(x)=0$ when
$\sqrt{1-\sin x}-2 \sqrt{\sin x+1}=0$
that is $\sqrt{1-\sin x}=2 \sqrt{\sin x+1}=0$
and squaring both sides
$1-\sin x=4(\sin x +1)\to \sin x =-\dfrac{3}{5}$
$g''(x)=\frac{1}{4} \left(2 \sqrt{1-\sin (x)}+\sqrt{1+\sin (x)}\right)$
and $g''\left(-\arcsin \dfrac{3}{5}\right)=\sqrt{\dfrac{2}{5}}+\dfrac{1}{2 \sqrt{10}}>0$ so at $\sin x =-\dfrac{3}{5}$ the function $g(x)$ has a minimum
$g\left(-\arcsin \dfrac{3}{5}\right)=2-\sqrt{10}$
and in conclusion we can say that
$g(x)\geq 2-\sqrt{10}$ for any $x>0$
as $ 2\sqrt{2} -10<2-\sqrt{10}$ we can also say that
$g(x) \geq 2\sqrt{2} -10$
even if I suspect that there is a typo somewhere
Hope this helps
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving this inequality $\left(1+\frac{1}{n}\right)^n < 3$ How would I prove this, particularly using a method using a geometric series and binomial coefficient.
I'm having trouble giving a reason for $(\star)$ especially.
This is what I did
For $n=1$, LHS = $(1+1)^1 = 2 < 3$ = RHS.
Thus, for $n=2$,
$$\left(1+\frac{1}{n}\right)^n \equiv {\large{\sum_{k=0}^n}} \begin{pmatrix}n \\ k \end{pmatrix}\frac{1}{n^k} \\ = {\large{\sum_{k=0}^n}}\frac{n!}{k!(n-k)!n^k} \\ < \sum_{k=0}^n \frac{1}{k!} \\ < \sum_{k=0}^\infty \frac{1}{k!} \\ < \sum_{k=0}^\infty \frac{1}{(\frac{3}{2})^k} \qquad (\star)\\ = 3. $$
$(\star)$ since $k>2 > \frac{3}{2} \implies k! > 2 > \frac{3}{2}$ and $k! > (\frac{3}{2})^k$ for $k>2$.
I just know this to be true (the last step of raising 1.5 to the power $k$) by checking my calculator, but I don't see a reason why the factorial grows faster or becomes larger than the exponential.
| I am not sure if this will work for $\frac{2}{3}$ but something similar does for $\frac{1}{2}$ ...
For $k \geq 2$ we have $ k! \geq 2^{k-1}$
\begin{eqnarray*}
\sum_{k=0}^{\infty} \frac{1}{k!} = 1+1+ \sum_{k=2}^{\infty} \frac{1}{k!} < 2+\sum_{k=2}^{\infty} \left(\frac{1}{2}\right)^{k-1} =2+ \frac{\frac{1}{2}} {1-\frac{1}{2}} =3.
\end{eqnarray*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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What is the value of this expression: $\log(\tan 6^\circ) + \log (\tan 12^\circ) + \log (\tan 18^\circ)+\cdots+\log(\tan 84^\circ)$ I am not able to solve this problem. Please try it to solve. The log base is 10 and angles are in degree.
Which of the following option is true for the given expression?
(A) Whole number
(B) An irrational number
(C) An negative number
| We know $\tan(x^\circ) \tan(90^\circ-x^\circ)=1 \\\tan(x^\circ)=\cot(90^\circ-x^\circ)$
$$6,12,18,\ldots,84\\\text{number of terms}=\frac{\text{last}-\text{first}}{\text{step}}+1=\frac{84-6}{6}+1=13+1=14$$ those are coupled $$6+84=90\\12+78=90\\18+72=90\\\vdots$$so $$\log(\tan 6^\circ) + \log (\tan 12^\circ) + \log (\tan 18^\circ)+\cdots+\log(\tan 84^\circ)=\\
\log(\tan 6^\circ\cdot\tan 12^\circ\cdot\tan 18^\circ\cdots\tan 84^\circ)
=\\\log(\underbrace{\tan 6^\circ\cdot\tan 84^\circ\cdot\tan 12^\circ\cdot\tan 78^\circ\tan 18^\circ\cdot\tan 72^\circ\cdots}_{14 \text{terms}})=\\
\log(\underbrace{1\cdot1\cdot1\cdots}_{7 \text{ terms}})=\\\log(1)=0$$
| {
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"url": "https://math.stackexchange.com/questions/2404131",
"timestamp": "2023-03-29T00:00:00",
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How to simplify $\frac{1}{3^{-2}}$ properly? Saw how it was simplified and I was wondering what allows you to rewrite this as $3^2 = 9$?
| $\frac{1}{3^{-2}}=\frac{1}{3^{-2}}\cdot 1 = \frac{1}{3^{-2}}\cdot\frac{3^2}{3^2}=\frac{1\cdot 3^2}{3^{-2}\cdot 3^2}=\frac{3^2}{3^{-2+2}}=\frac{3^2}{3^0}=\frac{3^2}{1}=3^2$
In general, you can skip many of these steps and use the result $x^n=\frac{1}{x^{-n}}$ for any nonzero $x$.
| {
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Prove $\frac{a(a^2+2)}{3}$ is an integer for all integer $a\geqslant 1$ If $\frac{a(a^2+2)}{3}$ then $3\mid{a(a^2+2)}$.
By induction:
Lets define the set, $S=\left\{a\in N:a\geqslant1, 3\mid a(a^2+2) \right\}$
If $a=1$ then, $1\in S$
So we have to prove that if $k(k^2+2)=3m$ then $(k+1)((k+1)^2+2)=3n$ with $m,n\in Z$
If $k(k^2+2)=3m$ then,
$\begin{align*}k(k^2+2)+3(k^2+k+1)=&3m+3(k^2+k+1)\\=&3(m+k^2+k+1)\end{align*}$
Also,
$\begin{align*}k(k^2+2)+3(k^2+k+1)=&k^3+2k+3k^2+3k+3\\=&k^3+2k^2+k^2+2k+3k+3\\=&k^2(k+2)+k(k+2)+3(k+1)\\=&(k+2)(k^2+k)+3(k+1)\\ =&k(k+2)(k+1)+3(k+1)\\=&(k+1)(k(k+2)+3)\\ =&(k+1)(k^2+2k+1+2)\\ =&(k+1)((k+1)^2+2)\\ =&3(m+k^2+k+1) \end{align*}$
where $n=m+k^2+k+1$
Therefore,
$3\mid(k+1)((k+1)^2+2)$
Can I do it simplier using induction?
| Work backwards. It's far more natural.
$(k+1)((k+1)^2+2)=$
$(k+1)(k^2+2k+1+2)=$
$(k+1)(k^2+2)+(k+1)(2k+1)=$
$k (k^2+2)+(k^2+2)+(k+1)(2k+1)=$
$3m+(k^2+2)+(2k^2+3k+1)=$
$3m+3k^2+3k+3$.
It is sometimes easier to subtract results.
$(k+1)((k+1)^2+2)-k (k^2+2)=$
$k (k+1)^2+(k+1)^2+2k+2-k^3-2k=$
$k^3+2k^2+k +k^2+2k+1+2-k^3=$
$3k^2+3k+3$
is a multiple of 3.
But it's probably easiest not to use induction at all.
Let $a \equiv i \mod 3;i=0,\pm 1$
If $i=0$ then $3|a $ so $3|a (a^2+2) $
Otherwise $i=\pm 1$
So $a^2+2\equiv i^2+2\equiv 1+2\equiv 0\mod 3$
So $3|a^2+2$ so $3|a (a^2+2) $.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding values of $x,y,z$ in three rows of quadratic equations $x^2+xy+y^2=25$
$x^2+xz+z^2=64$
$y^2+yz+z^2=49$
$x,y,z>0$ $x+y+z=?$
First i have named the rows in order $R_1,R_2,R_3$ as i have tried
$R_3-R_1$ $R_2-R_3$ and $R_2-R_1$ give reasonable results. Then i had the following
$(z-x).(x+y+z)=24$
$(x-y).(x+y+z)=15$
$(z-y).(x+y+z)=39$
It is obvious that if I add the first two rows it equals to the value of the third row. So i got back to where i started. Maybe the solution can move further from here, i don't know but there is another idea i had.
We know that $\dfrac{a_1+a_2+\cdots+a_n}{n}\ge\sqrt[n]{a_1a_2\cdots a_n}$ In other words Arithmetic mean is greater or equal to Geometric mean.
Proceeding $x^2+xy+y^2\ge\sqrt[3]{x^3y^3}\rightarrow\dfrac{25}{3}\ge xy$
Using the same method $\dfrac{64}{3}\ge xz$ and $\dfrac{49}{3}\ge yz$
I multiplied these expressions and came to the conclusion that $\dfrac{280}{9}\ge xyz$
$x+y+z\ge 3.\sqrt[3]{xyz}$ Approximately
$x+y+z\ge 11,3313$ That is the approximate value of $\sqrt{129}$
In order to fully solve this problem i need to prove that $x+y+z$ is certainly equal. And my question is this;
How can I do that?
| Let $\Delta ABC$ such that $AB=c=7$, $AC=b=8$ and $BC=a=5$.
Easy to see that this triangle is an acute-angled triangle,
which says that there is a point $F$ inside the triangle for which
$\measuredangle AFB=\measuredangle AFB=\measuredangle AFB=120^{\circ}$.
This point names Fermat point of the triangle.
Now, let $FA=z$, $FB=y$ and $FC=x$.
Thus, by law of cosines we obtain your system and we need to find a value of $FA+FB+FC$.
Indeed, by law of cosines again we have $\cos\alpha=\frac{8^2+7^2-5^2}{2\cdot8\cdot7}=\frac{11}{14}$.
We'll use the rotation $R^{60^{\circ}}_{A}$ of $\Delta ABC$ (around $A$ on $60^{\circ}$).
Let $R^{60^{\circ}}_{A}(F)=F'$ and $R^{60^{\circ}}_{A}(C)=C'$.
Thus, since $\Delta AFF'$ is an equilateral triangle,
$$x+y+z=CF+BF+AF=C'F'+BF+FF'=C'F'+F'F+FB=C'B,$$
$AF'=AC=b=8$ and $\measuredangle F'AB=60^{\circ}+\alpha$.
Thus, by law of cosines for $\Delta C'AB$ we obtain:
$$x+y+z=\sqrt{b^2+c^2-2bc\cos(60^{\circ}+\alpha)}=$$
$$=\sqrt{8^2+7^2-2\cdot8\cdot7\left(\frac{1}{2}\cdot\frac{11}{14}-\frac{\sqrt3}{2}\cdot\frac{5\sqrt3}{14}\right)}=\sqrt{129}.$$
Done!
| {
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Proof of $e^x (\ln x+\frac{1}{x})>\ln 8$ Prove that
$$e^x \left(\ln x+\frac{1}{x}\right)>\ln 8$$
I found that the minimum of $e^x \left(\ln x+\frac{1}{x}\right)$ is close to $\ln 8$, then how do we prove that it's greater than $\ln 8$?
| $f(x)=e^x \left(\dfrac{1}{x}+\log x\right)$
$f'(x)=\dfrac{e^x}{x^2} \left(x^2 \log x+2 x-1\right)$
$f'(x)=0\to x^2 \log x+2 x-1=0\to x\approx 0.59$
Taylor polynomial at $x=\dfrac{1}{2}$
$f(x)=\sqrt e \left[x^2 \left(5 -\dfrac{1}{8} \log 16\right)+x \left(-5 -\dfrac{1}{8} \log 16\right)+\dfrac{13 }{4}-\dfrac{1}{8} \log 32\right]+O(x^3)$
$f(x)\approx 7.6722 x^2-8.81501 x+4.64409$ in a neighbourhood of $x=\frac12$
To estimate the error we need the third derivative
$f^{(3)}(x)=\dfrac{e^x}{x^4} \left(x^4 \log (x)+4 x^3-6 x^2+8 x-6\right)$
On the interval $[0.4,0.6]$ we have $|f^{(3)}(x)|\le 36.0234$
thus the error is $R_3(x)\le \dfrac{ |f^{(3)}(x)| \cdot \left|\,x-\dfrac{1}{2}\right|}{3!}\approx 0.006$
Now as
$7.6722 x^2-8.81501 x+4.64409>\log 8;\;\forall x\in\mathbb{R}$
we can conclude that $f(x)>\log 8$ for any $x\in\mathbb{R}$
Hope this helps
Edit
A graph can explain better. Remember that the minimum is at $x\approx 0.59$
| {
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Inequality with $ab + bc + ca = 1$
As in the title, let $a>0$, $b>0$ and $c > 0$ be such that $ab + bc + ca = 1.$ Define $S = \frac{a+b+c}{\sqrt{bc} + \sqrt{ca} + \sqrt{ab}}.$ Prove that
$$S^2 \ge \sum_{\text{cyc}} \frac{1}{(2a+b)(2a+c)} \ge 1.$$
Edit: this is from an old mock usamo, which no longer exists on the internet, and to which I don't have solutions. Inequalities are not my strong point, so I'm hoping this will be entertaining to someone here :)
| The right inequality.
$$\sum_{cyc}\frac{1}{(2a+b)(2a+c)}-1=\sum_{cyc}\frac{1}{(2a+b)(2a+c)}-\frac{1}{ab+ac+bc}=$$
$$=\sum_{cyc}\left(\frac{1}{(2a+b)(2a+c)}-\frac{1}{3(ab+ac+bc)}\right)=$$
$$=\frac{1}{3}\sum_{cyc}\frac{ab+ac+2bc-4a^2}{(2a+b)(2a+c)}=$$
$$=\frac{1}{3}\sum_{cyc}\frac{(c-a)(2a+b)-(a-b)(2a+c)}{(2a+b)(2a+c)}=$$
$$=\frac{1}{3}\sum_{cyc}(a-b)\left(\frac{1}{2b+a}-\frac{1}{2a+b}\right)=$$
$$=\sum_{cyc}\frac{(a-b)^2}{3(2a+b)(2b+a)}\geq0.$$
The left inequality.
We'll use the identity
$$\sum_{cyc}\frac{1}{(2a+b)(2a+c)}-1=\frac{1}{3}\sum_{cyc}\frac{(a-b)^2}{(2a+b)(2b+a)},$$
which we got during the proof of the right inequality.
Indeed, let $a\geq b\geq c$.
Thus, by C-S $$b^2\left(\frac{(a+b+c)^2}{(\sqrt{ab}+\sqrt{ac}+\sqrt{bc})^2}-\sum_{cyc}\frac{1}{(2a+b)(2a+c)}\right)\geq$$
$$\geq b^2\left(\frac{(a+b+c)^2}{3(ab+ac+bc)}-\sum_{cyc}\frac{1}{(2a+b)(2a+c)}\right)=$$
$$= b^2\left(\frac{(a+b+c)^2}{3(ab+ac+bc)}-1-\left(\sum_{cyc}\frac{1}{(2a+b)(2a+c)}-1\right)\right)=$$
$$= b^2\left(\frac{\sum\limits_{cyc}(a^2-ab)}{3(ab+ac+bc)}-\frac{1}{3}\sum_{cyc}\frac{(a-b)^2}{(2a+b)(2b+a)}\right)=$$
$$= b^2\left(\frac{\sum\limits_{cyc}(a-b)^2}{6(ab+ac+bc)}-\frac{1}{3}\sum_{cyc}\frac{(a-b)^2}{(2a+b)(2b+a)}\right)=$$
$$=\frac{b^2}{6}\sum_{cyc}\frac{(a-b)^2((2a+b)(2b+a)-2(ab+ac+bc))}{(2a+b)(2b+a)}=$$
$$=\frac{b^2}{6}\sum_{cyc}\frac{(a-b)^2(2a^2+2b^2+3ab-2ac-2bc)}{(2a+b)(2b+a)}\geq$$
$$\geq\frac{b^2}{3}\sum_{cyc}\frac{(a-b)^2(a^2+b^2+ab-ac-bc)}{(2a+b)(2b+a)}\geq$$
$$\geq\frac{b^2(a-c)^2(a^2+c^2+ac-ab-bc)}{3(2a+c)(2c+a)}+\frac{b^2(b-c)^2(b^2+c^2+bc-ab-ac)}{3(2b+c)(2c+b)}\geq$$
$$\geq\frac{a^2(b-c)^2(a^2+ac-ab-bc)}{3(2a+c)(2c+a)}+\frac{b^2(b-c)^2(b^2+bc-ab-ac)}{3(2b+c)(2c+b)}=$$
$$=\frac{(b-c)^2(a-b)}{3}\left(\frac{a^2(a+c)}{(2a+c)(2c+a)}-\frac{b^2(b+c)}{(2b+c)(2c+b)}\right)\geq$$
$$\geq\frac{(b-c)^2(a-b)}{3}\left(\frac{a^2(b+c)}{(2a+c)(2c+a)}-\frac{b^2(b+c)}{(2b+c)(2c+b)}\right)=$$
$$=\frac{c(b+c)(a-b)^2(b-c)^2(5ab+2ac+2bc)}{3(2a+c)(2c+a)(2b+c)(2c+b)}\geq0.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Trignometric Equality Proof
Prove that $$\frac{\cos 3A + Cos 3B}{2\cos(A-B) -1} =(\cos A + \cos B)\cos(A+B)-(\sin A+\sin B)\sin(A+B)$$
I used $\cos 3A=4{\cos ^3A}-3\cos A$, but it is getting more and more complicated.
| Using Prosthaphaeresis Formulas,
$$(\cos A+\cos B)\cos(A+B)-(\sin A+\sin B)\sin(A+B)$$
$$=2\cos\dfrac{A+B}2\cos\dfrac{A-B}2\cos(A+B)-2\sin\dfrac{A+B}2\cos\dfrac{A-B}2\sin(A+B)$$
$$=2\cos\dfrac{A-B}2\cos\left(\dfrac{A+B}2+A+B\right)$$
Now apply Prosthaphaeresis Formulas on $\cos3A+\cos3B$
Finally use $\cos3x=\cos x(4\cos^2x-3)=\cos x\{2(1+\cos2x)-3\}=\cos x(2\cos2x-1)$ for $2x=A-B$
| {
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"url": "https://math.stackexchange.com/questions/2406459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove that $1+2=3, 4+5+6=7+8,... $ ad infinitum? Given this set of equations:
$$
1+2=3\\
4+5+6=7+8\\
9+10+11+12=13+14+15\\
\ldots
$$
How can I prove that this is true for all continuations of this sequence?
I would put it in the form of:
$$
(k,m)\in \{n^2,n|\in\Bbb N\}\\
\sum_{i=k}^{k+m} i=\sum_{i=k+m+1}^{k+2m}i
$$
However, I have problems in formulating and solving the inductions step, which I think should be to go from $n$ to $n+1$
| The first row is true:
$$1+2=3.$$
Swap the sides of the first row and add to the second row to get:
$$3+4+5+6=1+2+7+8.$$
This is true, because in the arithmetic progression $1,2,3,4,5,6,7,8$, the sums of terms equidistant from the center are equal.
Similarly, swap the sides of row $2$ and add to row $3$:
$$7+8+9+10+11+12=4+5+6+13+14+15.$$
Again the sum of the central terms is equal to the sum of the external terms in the AP: $4,5,6,\cdots,13,14,15$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2408850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
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find limit of a sequence I have a sequence:
$$ X_n\:=\:\cos\left(\left(\frac{3^n+\pi^n}{3^n+\left(\pi-\frac{1}{4} \right)^n} \right)^{1/n}\right) $$
I have to find the limit when $n \to \infty $ where $n \in \mathbb{N}$.
Which is the best way to find the answer ? Can I reduce or use the Squeeze theorem in that case ?
| Almost as a reflex, I'd go for Taylor expansions. (Because it works. Not always the most elegant, bu at least it gets us somewhere.) When $n\to\infty$, letting $x\stackrel{\rm def}{=} \frac{\pi -\frac{1}{4}}{3}$ and $y\stackrel{\rm def}{=} \frac{3}{\pi}$, with $x>y$ (so that $y^n=o(x^n)$)
\begin{align}
\cos\left(\left(\frac{3^n+\pi ^n}{3^n+\left(\pi -\frac{1}{4}\right)^n}\right)^{\frac{1}{n}}\right)
&= \cos\left(\frac{\pi}{3}\left(\frac{1+\left(\frac{3}{\pi}\right)^n}{1+\left(\frac{\pi -\frac{1}{4}}{3}\right)^n}\right)^{\frac{1}{n}}\right)\\
&= \cos\left(\frac{\pi}{3}\left(\left(1+y^n\right)\left(1+x^n + o(x^{n})\right)\right)^{\frac{1}{n}}\right)\\
&= \cos\left(\frac{\pi}{3}\left(1+y^n + x^n +o(x^n)\right)^{\frac{1}{n}}\right) \\
&= \cos\left(\frac{\pi}{3}\left(1+x^n +o(x^n)\right)^{\frac{1}{n}}\right) \tag{as $x>y$}\\
&= \cos\left(\frac{\pi}{3}\exp\left(\frac{1}{n}\ln\left(1+x^n +o(x^n)\right)\right)\right) \\
&= \cos\left(\frac{\pi}{3}\exp\left(\frac{x^n}{n}+o\left(\frac{x^n}{n}\right)\right)\right) \\
&= \cos\left(\frac{\pi}{3}\left(1+\frac{x^n}{n}+o\left(\frac{x^n}{n}\right)\right)\right) \\
&\xrightarrow[n\to\infty]{} \cos\frac{\pi}{3} = \frac{1}{2}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Quartic equation with integer solutions I am trying to find the solutions of the following quartic equation given that all the solutions are integers.
$$x^4+22x^3+172x^2+552x+576=0$$
Below is the original phrasing of the problem with hints building up to this equation. I can prove all of the results it asks for before the final part of the question, but I am struggling to actually find the solutions to the equation and require help with explaining the process as well.
I then know that:
(1) $k_1k_2k_3k_4 = 576 = (1)(2^6)(3^2)$
(2) $(k_1+1)(k_2+1)(k_3+1)(k_4+1) = 1323=(1)(3^3)(7^2)$
(3) $(k_1-1)(k_2-1)(k_3-1)(k_4-1) = 175 = (1)(5^2)(7)$
However, I do not know how to proceed from here and the solution to this problem didn't explain in enough detail for me to either understand the solution, or the approach.
| Observe that the given equations imply that $k_1,k_2,k_3$ and $k_4$ should be non-zero even integers. Now consider $$(k_1-1)(k_2-1)(k_3-1)(k_4-1)=1\times 5^2 \times 7.$$
Since, $k_i$'s are integers, one of the $k_i-1$ must be $1$ ($k_i-1 \neq -1$ as $k_i \neq 0$). Let $k_1-1=1 \implies \color{red}{k_1 =2}$.
We are now left with $$(k_2-1)(k_3-1)(k_4-1)= 5^2 \times 7.$$
Since we have $5^2$ on the RHS, at least one of the $k_i-1$ (say $k_2-1$) cannot be a multiple of $5$. Consequently, $k_2-1 = 1$ or $k_2-1=\pm 7$ (i.e., $k_2 =2,8$ or $-6$). But $k_2 + 1$ must be a multiple of $3$ or $7$ (due to the second constraint). So $k_2 \neq -6$. Therefore, $k_2=2$ or $k_2=8$.
Now suppose $k_2=2$. Then the constraints simplify to
$$k_3 k_4 = 2^4 \times 3^2$$
$$(k_3+1)(k_4+1) = 3 \times 7^2$$
$$(k_3-1)(k_4-1) = 5^2 \times 7$$
It can be verified that these equations do not have any solution.
Therefore, we are left with $\color{red}{k_2 = 8}$ only. In that case, the equations simplify to
$$k_3 k_4 = 2^2 \times 3^2$$
$$(k_3+1)(k_4+1) = 7^2$$
$$(k_3-1)(k_4-1) = 5^2$$
These equations can easily be solved to result in $\color{red}{k_3 =k_4 = 6}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Review. Prove: if $a$ and $b$ are odd integers then $8\mid (a^2-b^2)$ Please notice that this post is not a duplicate from the ones posted by @Lil or @LizW since I want to use another result.
Proof:
Lets prove the next statement first:
If $n$ is an odd integer, then $n$ is equal to $4k+1$ or $4k+3$
Any integer $n$ divided by $4$ has remainder 0, 1, 2 or 3 using the quotient remainder theorem. Therefore any integer can be written as $4k, 4k+1, 4k+2$ and $4k+3$ for some integer $k$. Clearly, $4k$ and $4k+2$ are even numbers. Therefore, all odd integer numbers are $4k+1$ or $4k+3$.
Using this result, if $a$ and $b$ are odd integers then, $a=4m+1$ and $b=4n+1$ for some integers $m$ and $n$.
By algebra, $a^2=(4m+1)^2$ and $b^2=(4n+1)^2$
Then,
$\begin{align*}a^2-b^2=&(4m+1)^2-(4n+1)^2\\ =&16m^2+8m+1-16n^2-8n-1\\ =&8(2m^2+m-2n^2-n)\end{align*}$
This shows that $8\mid (a^2-b^2)$
Should I prove it for $4k+3$?
| If $a$ and $b$ are odd integers, then for some integers $m$ and $n$,
$a^2 - b^2 = (2m+1)^2 - (2n+1)^2 = 4(m^2 + m - n^2 - n) = 4(m(m+1) - n(n+1))$
Since $2 \mid m(m+1)$ and $2 \mid n(n+1)$, then $2 \mid (m(m+1) - n(n+1))$
It follows that $8 \mid a^2-b^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show $5n^3+7n^5 \equiv 0$ mod 12 for all integers $n$. I've been stuck on this question for quite a while now. I could obvious show this by setting $n$ to all numbers from 0-11, however this is not very efficient.
Note $5n^3+7n^5 = n^3(5+7n^2)$. We could use the Chinese remainder theorem, we show $n^3(5+7n^2) \equiv 0 $ mod 3 and mod 4, but this doesn't seem to be any easier. What would be the best way to solve this problem?
| Oh, geez..... $7+5 = 12$ so $5n^3 + 7n^5 \equiv 5n^3 - 5n^5 \equiv -7n^3 + 7n^5$.
Suffices to show $12|n^5 - n^3 = n^3(n^2 - 1)= n^3(n+1)(n-1)$.
.....
$n-1, n, n+1$ are three consecutive integers so $3$ divides one of them.
$n-1,n,n+1$ are two consecutive integers so either $2|n$ in which case $8|n^3$, or $2|n-1$ and $2|n+1$ in which case $4|(n-1)(n+1)$. In either case $4|n^3(n+1)(n-1)$.
So $12|n^3(n+1)(n-1)$ so
$n^3(n+1)(n-1) \equiv 0 \mod 12$
$n^5 - n^3 \equiv 0 \mod 12$
$7n^5 - 7n^3 \equiv 0 \mod 12$
$7n^5 + 5n^3 \equiv 0 \mod 12$.
| {
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"timestamp": "2023-03-29T00:00:00",
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proof of $\arctan(x)=\frac{i}{2}(\ln(x+i)-\ln(x-i)-i\pi)$ without solving $\int\frac{1}{x^2+1}dx$ So I know $\int\frac{1}{x^2+1}dx=\arctan(x)+c$
Now when i tried to integral $\frac{1}{x^2+1}$ in a different way i got this:
$$\frac{1}{x^2+1}=\frac{1}{(x-i)(x+i)}=\frac{A}{x-i}+\frac{B}{x+i}\\\implies1=Ax+Ai+Bx-Bi=x(A+B)+i(A-B)\\\implies
\begin{cases}
A+B=0
\\[2ex]
A-B=\frac{1}{i}=-i
\end{cases}\\\implies2A=-i\implies A=\frac{-i}{2}\implies B=\frac{i}{2}$$
So in the end i get:$$\int\frac{1}{x^2+1}dx=\int\frac{i}{2(x+i)}-\frac{i}{2(x-i)}dx=\frac{i}{2}\int\frac{1}{(x+i)}-\frac{1}{(x-i)}dx$$
After integrating this i get $\frac{i}{2}(\ln(x+i)-\ln(x-i)+C)$ after comparing this to $\arctan(x)$ at $x=0$ i find $C=-i\pi$
$\therefore~\arctan(x)=\frac{i}{2}(\ln(x+i)-\ln(x-i)-i\pi)$
now my question is how can i prove that without using the integral
| Perhaps using Euler?
$$ z= \tan w = \frac{\sin w}{\cos w} = \frac{1}{i}\frac{e^{iw}-e^{-iw}}{e^{iw}+e^{iw}} = \frac{1}{i} \frac{e^{2iw}-1}{e^{2iw}+1}$$
so for any $k\in {\Bbb Z}$ $$e^{2iw} = e^{2i(w -k)} = \frac{1+iz}{1-iz}$$
whence
$$ w = \frac{1}{2i} \log \left( \frac{1+iz}{1-iz} \right) + k \pi$$
where the choice of $k\in {\Bbb Z}$ depends upon the cut you want.
$k=0$ corresponds to standard choices of $\log(1)=0$ and $\arctan 0=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove: $\frac{\tan{x}}{1-\cot{x}} + \frac{\cot{x}}{1-\tan{x}}=\sec{x}\csc{x}+1$
Prove that:$$\frac{\tan{x}}{1-\cot{x}} + \frac{\cot{x}}{1-\tan{x}}=\sec{x}\csc{x}+1.$$
My trying:$$L.H.S.=\frac{\tan{x}+\cot{x}-\tan^2{x}-\cot^2{x}}{(1-\cot{x})(1-\tan{x})}$$
Changing to $\sin{x}$ and $\cos{x}$ and simplifying gives (not typing the entire simplifying, its very lengthy)
$$\frac{\sin{x}\cos{x}+2\sin{x}\cos{x}-1}{(\sin{x}\cos{x})(2\sin{x}\cos{x}-1)}$$
Tried many times but no luck getting anywhere close to the expression on R.H.S.
| Since $a^3-b^3=(a-b)(a^2+ab+b^2)$ and $\sin^2x+\cos^2x=1$, we obtain:
$$\frac{\tan{x}}{1-\cot{x}} + \frac{\cot{x}}{1-\tan{x}}=\frac{\sin^2x}{\cos{x}(\sin{x}-\cos{x})}-\frac{\cos^2x}{\sin{x}(\sin{x}-\cos{x})}=$$
$$=\frac{\sin^3x-\cos^3x}{\sin{x}\cos{x}(\sin{x}-\cos{x})}=\frac{1+\sin{x}\cos{x}}{\sin{x}\cos{x}}=\sec{x}\csc{x}+1.$$
| {
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"url": "https://math.stackexchange.com/questions/2415438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\displaystyle{\lim_{n\to \infty}}(1-\frac 12 +\frac 13 - \frac 14 + \cdots + \frac{1}{2n-1}-\frac{1}{2n}) $ I want to calculate this limit.
$$\lim_{n\to \infty}(1-\frac 12 +\frac 13 - \frac 14 + \cdots + \frac{1}{2n-1}-\frac{1}{2n}) $$
I tried to pair the terms of the sum in order to reduce each other but without any succes. By writing the sum:
$$\frac{1}{1*2} + \frac{1}{3*4}+\frac{1}{5*6}+\cdots+\frac{1}{2n(2n-1)}$$
I have not reached anything useful. Could you help me?
| Hint.
$$
x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots=\int_0^x\left(1-t+t^2-t^3-t^4+\cdots\right)\,dt=\int_0^x\frac{dt}{1+t}=\log(1+x)
$$
From this, one can obtain that
$$
1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots=\ln 2
$$
| {
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"url": "https://math.stackexchange.com/questions/2416623",
"timestamp": "2023-03-29T00:00:00",
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Calculate $\int_0^1\ln{(\sqrt{1+x}+\sqrt{1-x})}dx$ I am looking for the fastest possible way to calculate:
$$\int_0^1\ln{(\sqrt{1+x}+\sqrt{1-x})}dx$$
The integral appeared on an integration bee where only a few minutes are given to calculate it, so please do not hesitate to use shortcuts in your solution, even if they involve advanced mathematics. Thank you in advance.
| $\begin{align}J=\int_0^1\ln{(\sqrt{1+x}+\sqrt{1-x})}dx\end{align}$
Perform the change of variable $\displaystyle y=\frac{1-x}{1+x}$,
$\begin{align}J&=\ln 2\int_0^1 \frac{1}{(1+x)^2}\,dx-\int_0^1 \frac{\ln(1+x)}{(1+x)^2}\,dx+2\int_0^1 \frac{\ln(1+\sqrt{x})}{(1+x)^2}\,dx\\
&=\ln 2\left[\frac{-1}{1+x}\right]_0^1+\left(\left[\frac{\ln(1+x)}{1+x}\right]_0^1-\int_0^1\frac{1}{(1+x)^2}\,dx\right)+2\int_0^1 \frac{\ln(1+\sqrt{x})}{(1+x)^2}\,dx\\
&=\left(-\frac{\ln 2}{2}+\ln 2\right)+\frac{\ln 2}{2}+\left[\frac{1}{1+x}\right]_0^1+2\int_0^1 \frac{\ln(1+\sqrt{x})}{(1+x)^2}\,dx\\
&=\ln 2-\frac{1}{2}+2\int_0^1 \frac{\ln(1+\sqrt{x})}{(1+x)^2}\,dx\\
&=\ln 2-\frac{1}{2}+2\left(\left[-\frac{\ln(1+\sqrt{x})}{1+x}\right]_0^1+\int_0^1\frac{1}{2(1+x)(1+\sqrt{x})\sqrt{x}}\,dx\right)\\
&=-\frac{1}{2}+\int_0^1\frac{1}{(1+x)(1+\sqrt{x})\sqrt{x}}\,dx\\
\end{align}$
Perform the change of variable $\displaystyle y=\sqrt{x}$,
$\begin{align}J&=-\frac{1}{2}+\int_0^1\frac{2}{(1+x^2)(1+x)}\,dx\\
&=-\frac{1}{2}+\int_0^1\left(\frac{1}{1+x}+\frac{1}{1+x^2}-\frac{x}{1+x^2}\right)\,dx\\
&=-\frac{1}{2}+\Big[\ln(1+x)\Big]_0^1+\Big[\arctan x\Big]_0^1-\frac{1}{2}\Big[\ln(1+x^2)\Big]_0^1\\
&=-\frac{1}{2}+\ln 2+\dfrac{\pi}{4}-\frac{1}{2}\ln 2\\
&=\boxed{\dfrac{\pi}{4}+\frac{1}{2}\ln 2-\frac{1}{2}}
\end{align}$
| {
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"timestamp": "2023-03-29T00:00:00",
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$n$-th derivative of $y$ in $x^y=e^{x-y}$ Question:
If $x^y=e^{x-y}$, find a general formula for $\displaystyle\frac{d^ny}{dx^n}$.
My Attempt:
$$x^y=e^{x-y}$$
$$\Rightarrow\ \ \ \ \ y\ln x=x-y$$
$$\Rightarrow\ \ \ \ \ y=\frac{x}{\ln x+1}$$
So
$$\begin{align}\frac{dy}{dx}&=\frac{\ln x+1-1}{(\ln x+1)^2}\\
&=\frac{\ln x}{(\ln x+1)^2}
\end{align}$$
Then
$$\begin{align}\frac{d^2y}{dx^2}&=\frac{\frac{1}{x}(\ln x+1)^2+\frac{2\ln x}{x}(\ln x+1)}{(\ln x+1)^4}
\end{align}$$
Now, I can see that the denominator satisfies a straight-forward pattern $(\ln x+1)^{2^n}$. So the general formula for $\displaystyle\frac{d^ny}{dx^n}$ is
$$\frac{d^ny}{dx^n}=\frac{g(n)}{(\ln x+1)^{2^n}}$$But I can't derive any formula for the numerator $g(n)$, nor can I see any regular pattern in the structure. Should I differentiate the initial expression $x^y=e^{x-y}$ implicity? Can anyone help me in that?
| Here is a formula of the $n$-th derivative of $\frac{f}{g}$ in terms of derivatives of $f$ and $g$ which might be helpful.
Let $D_x$ represent differentiation with respect to $x$. Hence $D^n_x f(x)$ is the $n$-th derivative of $f$ with respect to $x$. The following holds true for $n$ times differentiable functions $f$ and $g$
\begin{align*}
D_x^n\left(\frac{f}{g}\right)=\sum_{k=0}^n\sum_{j=0}^{k} (-1)^j\binom{n}{k}\binom{k+1}{j+1}\frac{1}{g^{j+1}}
D_x^{n-k}\left(f\right) D_x^{k}\left( g^j\right)\tag{1}
\end{align*}
This formula is based upon the Hoppe Form of Generalized Chain Rule and a proof for it is given in this MSE post.
In the current situation we have $f(x)=x$ and $g(x)=\ln(x)+1$ and we obtain from (1) if we change the order of summation of the outer sum by replacing $k$ with $n-k$:
\begin{align*}
\color{blue}{D_x^n\left(\frac{x}{\ln x+1}\right)}
&=\sum_{k=0}^n\sum_{j=0}^{n-k}(-1)^j\binom{n}{k}\binom{n-k+1}{j+1}
\frac{1}{(\ln x + 1)^{j+1}}D_x^k(x)D_x^{n-k}\left((\ln x + 1)^j\right)\\
&=\color{blue}{\sum_{j=0}^{n}(-1)^j\binom{n+1}{j+1}
\frac{x}{(\ln x+1)^{j+1}}D_x^{n}\left((\ln x+1)^j\right)}\\
&\qquad\color{blue}{+\sum_{j=0}^{n-1}(-1)^jn\binom{n+1}{j+1}
\frac{1}{(\ln x+1)^{j+1}}D_x^{n-1}\left((\ln x+1)^j\right)}\\
\end{align*}
Since $D_x^k(x)=0$ if $k>1$ it is sufficient to consider the first two terms $k=0,1$ of the outer sum.
Let's look at a small example in order to see the formula in action
Example: $D^1_x\left(\frac{x}{\ln x + 1}\right)$
\begin{align*}
\color{blue}{D_x^1\left(\frac{x}{\ln x +1}\right)}
&=\sum_{j=0}^1(-1)^j\binom{2}{j+1}\frac{x}{(\ln x+1)^{j+1}}D_x^1\left((\ln x+1)^j\right)\\
&\qquad +\sum_{j=0}^0(-1)^j\binom{1}{j+1}\frac{1}{(\ln x+1)^{j+1}}D_x^0\left((\ln x+1)^j\right)\\
&=(-1)^0\binom{2}{1}\frac{x}{\ln x+1}D_x(1)+(-1)^1\binom{2}{2}\frac{x}{(\ln x+1)^2}D_x(\ln x + 1)\\
&\qquad +(-1)^0\binom{1}{1}\frac{1}{\ln x+1}D_x^0(1)\\
&=0-\frac{x}{(\ln x+1)^2}\cdot \frac{1}{x}+\frac{1}{\ln x+1}\\
&\color{blue}{=\frac{\ln x}{(\ln x+1)^2}}
\end{align*}
in accordance with OPs result.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
Show that $\lfloor(1+\sqrt{3})^{2n-1}\rfloor$ is divisible by $2^n$
Show that $\lfloor(1+\sqrt{3})^{2n-1}\rfloor$ is divisible by $2^n$ where $n$ is a positive integer.
We have \begin{align*}\lfloor(1+\sqrt{3})^{2n-1}\rfloor &= (\sqrt{3}+1)^{2n-1}-(\sqrt{3}-1)^{2n-1}\\&=\dfrac{(4+2\sqrt{3})^n}{1+\sqrt{3}}+\dfrac{(4-2\sqrt{3})^n}{1-\sqrt{3}}\\&=2^n\left(\dfrac{(2+\sqrt{3})^n}{1+\sqrt{3}}+\dfrac{(2-\sqrt{3})^n}{1-\sqrt{3}}\right).\end{align*} Thus we must show that the second term is an integer. How can we continue?
| $$(\sqrt{3}+1)^{2n-1}-(\sqrt{3}-1)^{2n-1}$$
Take n = 2
$$(\sqrt{3}+1)^{3}-(\sqrt{3}-1)^{3}$$
Look at, binomial expansion of the powers
First term,
$$ 3\sqrt(3) + 3*3 + 3\sqrt(3) + 1 $$
Minus second term,
$$ 3\sqrt(3) - 3*3 + 3\sqrt(3) - 1 $$
Gives
$$ 2*3*3 + 2 $$
So you get only every second term. Generalize to higher powers and get a general formula. See that it divides by $2^n$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Is it possible find the function $f(n)$? Is it possible find the function $f(n)$ ?
$$\lim_{n\to\infty}\frac{2^{8 n - 3} - 2^{2n-1}- 3}{2^{4 f(n) - 1} - 2^{f(n)} -1} = 9$$
| $\lim \limits_{n \to \infty} \frac{2^{8n-3}}{2^{4f(n)-1)}} =9 $ since both are leading terms in numerator and denominator,
So $2^{8n-3-4f(n)+1} = 9$ or $2^{8n-3-4f(n)+1} = 2^{\log_2(9)}$ which means that
$8n-2-4f(n) = \log_2(9) $ so $4f(n) = 8n-2-\log_2(9)$ so $f(n) = 2n-\frac{1}{2} -\frac{1}{4} \log_2(9) \approx 2n -0.5788$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Limit of a sequence : $x_n = \frac{1}{1\cdot 2\cdot 3} + \frac{1}{2\cdot 3\cdot 4} + \ldots + \frac{1}{n(n+1)(n+2)}$ Can someone help me with this problem?
Finding the limit $\lim_{n \to \infty}\ x_n$ where
$$x_n = \frac{1}{1\cdot 2\cdot 3} + \frac{1}{2\cdot 3\cdot 4} + \ldots + \frac{1}{n(n+1)(n+2)},\quad n\in\mathbb{N}.$$
I don't have a clue how to do this.
| HINT: use that $$\frac{1}{i(i+1)(i+2)}=1/2\, \left( i+2 \right) ^{-1}- \left( i+1 \right) ^{-1}+1/2\,{i}^{-1}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
$\sum_{n=1}^{2017}\left(\left((n+2)^4\bmod{(n+1)^4}\right)\bmod{4}\right)$
What's
$$\sum_{n=1}^{2017}\left(\left((n+2)^4\bmod{(n+1)^4}\right)\bmod{4}\right)$$
What have I tried?
$$(n+2)^4=n^4+8n^3+24n^2+32n+16$$
$$(n+1)^4=n^4+4n^3+6n^2+4n+1$$
Remainder:
$$4n^3+18n^2+28n+15$$
mod:
$$2n^2-1\pmod{4}$$
I can compute $\sum x^2$ but I don't know what to do with $$\sum_{n=1}^{2017}\left(2n^2-1\mod{4}\right)$$
|
Here
$$\{\}\text{ is the fractional part function}$$
So we can write, (if you have problem with this part please comment)
$$\text{a mod b}= \{\frac{a}{b}\} b$$
$$\sum_{n=1}^{2017} \{\frac{2 n^2-1}{4}\} 4$$
$$\sum_{n=1}^{2017} \{\frac{ n^2}{2}-\frac{1}{4}\} 4$$
$$\{\frac{ n^2}{2}-\frac{1}{4}\}= \frac {1}{4} \text { for } n\in odd$$
$$ =\frac{3}{4} \text{ for }n \in even $$
$$\sum_{n=1}^{1009} \frac{1}{4}4 +\sum_{n=1}^{1008} \frac{3}{4} 4$$
$$\sum_{n=1}^{1009} 1 +\sum_{n=1}^{1008} 3 $$
$$1009 + 3 × 1008$$
$$4033$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $\cos(z-x) + \cos(y-z) + \cos(x-y) = -\frac{3}{2}$, then $\sin x + \sin y + \sin z = 0 = \cos x + \cos y + \cos z $.
If $$\cos(z-x) + \cos(y-z) + \cos(x-y) = -\frac{3}{2}$$ then how can I show that the sum of cosines of each angle ($x$, $y$, $z$) and sines of each angle sum up to zero? i.e. $$\sin x + \sin y + \sin z = 0 = \cos x + \cos y + \cos z $$
I tried:
• Expanded using $\cos(A-B) = \cos A\cos B+\sin A\sin B $, but it did nothing.
After spending one hour to this problem, I thought that there must be a shorter and ideal way.
| By the angle-difference identity you mention, the given equation is equivalent to
$$\cos x \cos y + \sin x \sin y + \cos y \cos z + \sin y \sin z + \cos z \cos x + \sin z \sin x = -\frac32 \tag{1}$$
Thus,
$$3 + 2 (\cos x \cos y + \cdots ) + 2(\sin x \sin y + \cdots ) = 0 \tag{2}$$
But,
$$3 = 1 + 1 + 1 = \left(\cos^2 x + \sin^2 x \right) + \left( \cos^2 y + \sin^2 y \right) + \left( \cos^2 z + \sin^2 z\right) \tag{3}$$
So, (2) becomes
$$\begin{align}
0 &= \cos^2 x + \cos^2 y + \cos^2 z + 2 \cos x \cos y + 2 \cos y \cos z + 2\cos z \cos x \\
&+ \sin^2 x + \sin^2 y + \sin^2 z + 2 \sin x \sin y + 2 \sin y \sin z + 2 \sin z \sin x \\
&= \left( \cos x + \cos y + \cos z \right)^2 + \left( \sin x + \sin y + \sin z \right)^2
\end{align} \tag{4}$$
Now, the sum of two squares can be zero only if each square is itself zero, and we are done. $\square$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Number of Unordered Partition of $[2n]$ into $n$ pairs First the notation $[2n] = \{1,2,....,2n\}$
Since partition is a set of disjoint and non-empty subset of [2n],
I assign each element in partition one element in [2n] : $2nCn$
Now we have left n elements to assign, and each n elements has n possibilities to be assigned: $n^n$
Therefore, the answer is $2nCn * n^n$
However, the solution says $(2n-1)!!$.
Is the number $2nCn*n^n = (2n-1)!!$ ?
| Your answer is incorrect.
Method 1: Choose two of the $2n$ elements for the first subset, two of the $2n - 2$ remaining elements for the second subset, two of the $2n - 4$ remaining elements for the third subset, and so forth. That yields
$$\binom{2n}{2}\binom{2n - 2}{2}\binom{2n - 4}{2} \ldots \binom{2}{2}$$
ways to divide $2n$ elements into $n$ subsets of size $2$. However, the order in which we select the subsets does not matter. Hence, we must divide the above result by $n!$ to obtain
$$\frac{1}{n!}\binom{2n}{2}\binom{2n - 2}{2}\binom{2n - 4}{2} \ldots \binom{2}{2}$$
Method 2: We list the elements of $[2n]$ from smallest to largest. We have $2n - 1$ ways of selecting the element that will be matched with $1$. That leaves $2n - 2$ elements. We have $2n - 3$ ways of matching an element with the smallest element remaining. Continuing in this way, we can divide the set $[2n]$ into $n$ subsets of size $2$ in
$$(2n - 1)!! = (2n - 1)(2n - 3) \ldots (3)(1)$$
ways.
Observe that our answers are equivalent since
\begin{align*}
& \frac{1}{n!}\binom{2n}{2}\binom{2n - 2}{2}\binom{2n - 4}{2} \ldots \binom{2}{2}\\
\qquad & = \frac{1}{n!} \cdot \frac{(2n)!}{2!(2n - 2)!} \cdot \frac{(2n - 2)!}{2!(2n - 4)!} \cdot \frac{(2n - 4)!}{2!(2n - 6)!} \cdots \frac{2!}{0!2!}\\
\qquad & = \frac{1}{n!} \cdot \frac{(2n)!}{2!^n}\\
\qquad & = \frac{1}{n!} \cdot \frac{(2n)(2n - 1)(2n - 2)(2n - 3)(2n - 4) \ldots (4)(3)(2)(1)}{2^n}\\
\qquad & = \frac{1}{n!} \cdot n(2n - 1)(n - 1)(2n - 3)(n - 2) \cdots (2)(3)(1)(1)\\
\qquad & = (2n - 1)(2n - 3) \cdots (3)(1)
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove inequality involving square roots of trigonometric functions For $N$=even and $x\geq0$, prove that
\begin{eqnarray}
&&\sum^{\frac{N}{2}}_{j=1}\sqrt{x^2-2x\cos\frac{(2j-1)\pi}{N}+1}\geq1+\sum^{\frac{N}{2}-1}_{j=1}\sqrt{x^2-2x\cos\frac{2j\pi}{N}+1}
\end{eqnarray}
where the equality holds for $x=0$.
Update 1: As shown by Y.Guo (see the answer below), the original inequality is equivalent to
\begin{eqnarray}
f(x)\equiv\sum^{\frac{N}{2}}_{j=1}\frac{2\cos\frac{(2j-1)\pi}{N}-2\cos\frac{2j\pi}{N}}{\sqrt{x^2-2x\cos\frac{(2j-1)\pi}{N}+1}+\sqrt{x^2-2x\cos\frac{2j\pi}{N}+1}}\leq 1.
\end{eqnarray}
It is easy to show that $f(0)=1$ and $f(1)<1$. When $x\geq \cos\frac{\pi}{N}$, each term in the sum is a monotonically decreasing function, which proves the inequality for $x\in[\cos\frac{\pi}{N},\infty)$.
However, numerical tests indicate that $f(x)$ also decreases monotonically on $x\in[0,\cos\frac{\pi}{N}]$. I have no idea how to prove this since some terms in the sum are not monotonical functions on this interval.
Update 2: The inequality for $0<x<1$ can be proved by combining the Perron-Frobenius theorem in matrix theory with the fermion representation of the one-dimensional quantum Ising model.
This can be viewed as a kind of “physical mathematics” in some sense. See the following paper for details:
https://arxiv.org/abs/2001.00511
| A Proof when $x\ge1$
$$
\begin{aligned}
\sum_{j=1}^{\frac{N}{2}}\sqrt{x^{2}-2x\cos\frac{(2j-1)\pi}{N}+1}&\ge1+\sum_{j=1}^{\frac{N}{2}-1}\sqrt{x^{2}-2x\cos\frac{2j\pi}{N}+1}\\
\iff\sum_{j=1}^{\frac{N}{2}}\sqrt{x^{2}-2x\cos\frac{(2j-1)\pi}{N}+1}+x&\ge{x}+1+\sum_{j=1}^{\frac{N}{2}-1}\sqrt{x^{2}-2x\cos\frac{2j\pi}{N}+1}\\
\iff\sum_{j=1}^{\frac{N}{2}}\sqrt{x^{2}-2x\cos\frac{(2j-1)\pi}{N}+1}+x&\ge\sum_{j=1}^{\frac{N}{2}}\sqrt{x^{2}-2x\cos\frac{2j\pi}{N}+1}\\
\iff\sum_{j=1}^{\frac{N}{2}}(\sqrt{x^{2}-2x\cos\frac{(2j-1)\pi}{N}+1}&-\sqrt{x^{2}-2x\cos\frac{2j\pi}{N}+1})+x\ge0\\
\end{aligned}
$$
$$
\begin{aligned}
\iff\sum_{j=1}^{\frac{N}{2}}&\frac{2x\cos\frac{2j\pi}{N}-2x\cos\frac{(2j-1)\pi}{N}}{\sqrt{x^{2}-2x\cos\frac{(2j-1)\pi}{N}+1}+\sqrt{x^{2}-2x\cos\frac{2j\pi}{N}+1}}+x\ge0
\end{aligned}
$$
$$
\begin{aligned}
\iff\sum_{j=1}^{\frac{N}{2}}&\frac{2\cos\frac{2j\pi}{N}-2\cos\frac{(2j-1)\pi}{N}}{\sqrt{x^{2}-2x\cos\frac{(2j-1)\pi}{N}+1}+\sqrt{x^{2}-2x\cos\frac{2j\pi}{N}+1}}+1\ge0\\
\iff\sum_{j=1}^{\frac{N}{2}}&\frac{2\cos\frac{(2j-1)\pi}{N}-2\cos\frac{2j\pi}{N}}{\sqrt{x^{2}-2x\cos\frac{(2j-1)\pi}{N}+1}+\sqrt{x^{2}-2x\cos\frac{2j\pi}{N}+1}}\le1
\end{aligned}
$$
Notice that the left hand side decreases as $x$ increases when $x\ge1$
($\forall 1\le{j}\le\frac{N}{2}:2\cos\frac{(2j-1)\pi}{N}-2\cos\frac{2j\pi}{N}>0$ and $x^{2}-2x\cos\frac{k\pi}{N}+1$ increases for all $1\le{k}\le{N}$ $x\ge1$)
So if $x=1$ the inequality holds, so does $x\ge1$
When $x=1$
$$
\begin{aligned}
LHS&=\sum_{j=1}^{\frac{N}{2}}\frac{2\cos\frac{(2j-1)\pi}{N}-2\cos\frac{2j\pi}{N}}{\sqrt{2-2\cos\frac{(2j-1)\pi}{N}}+\sqrt{2-2\cos\frac{2j\pi}{N}}}\\
&=\sum_{j=1}^{\frac{N}{2}}\sqrt{2-2\cos\frac{(2j-1)\pi}{N}}-\sqrt{2-2\cos\frac{2j\pi}{N}}\\
&=\sum_{j=1}^{\frac{N}{2}}2(\sin\frac{2j\pi}{2N}-\sin\frac{(2j-1)\pi}{2N})\\
&=\frac{\cos(\frac{\pi}{2N})-\cos(\frac{\pi}{2}+\frac{\pi}{2N})}{\sin(\frac{\pi}{2N})}-\frac{1}{\sin(\frac{\pi}{2N})}\\
&=\frac{\cos(\frac{\pi}{2N})+\sin(\frac{\pi}{2N})-1}{\sin(\frac{\pi}{2N})}\\
&\le1
\end{aligned}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Using Square Numbers... To Make Squares I came across a question in an exam today (don't worry, I have already completed the test), and was wondering whether there was a better solution to mine.
The question, as I remember, is as follows:
The teacher gives 12 students squares of side lengths 1, 2, 3, 4 ... 11, 12. All students receive a distinct square. Then, he asks the students to cut up the squares into unit squares (of side length 1). He challenges the students to arrange their squares adjacently to create a larger square, with no gaps. Of course, they find that it is impossible.
Alice has a square of side length $a$. She exclaims that if she doesn't use any of her unit squares, then the class is able to create a square.
Similarly, Bill has a square of side length $b$, and says the same thing.
Given neither Alice nor Bill is lying, what is the value of $ab$?
That was a mouthful of a question, and I have attempted it, with very messy results. This is how I started:
\begin{align}
1^2 + 2^2 + 3^2 ... 11^2 + 12^2 -a^2 & = x^2 \\
1^2 + 2^2 + 3^2 ... 11^2 + 12^2 -b^2 & = y^2 \\
\end{align}
Then, by subtracting the latter equation from the first,
\begin{align}
b^2 - a^2 & = x^2 - y^2 \\
(b+a)(b-a) & = (x+y)(x-y) \\
\end{align}
From there, my brain switched off and I just turned to trial & error (which is one of my favourite problem solving techniques).
We know: $1^2 + 2^2 + 3^2 ... 11^2 + 12^2 = 650$, therefore the square created by the class must be $650 - 12^2 < x^2 < 650$.
I got the answers 5 and 11 which multiply for 55, which I believe is correct, but can someone please show me a 'proper' way of completing this question.
| Hint. Note that $650=2\cdot 5^2\cdot 13$ (it is not a perfect square) can be written as a sum of $2$ squares in three ways:
$$650=5^2+25^2=11^2+23^2=17^2+19^2.$$
(actually, in your case, it suffices to check for representations $x^2+y^2$ such that $1\leq x\leq 12\leq y$).
Since $5$ and $11$ are $\leq 12$, it follows that $a\cdot b=5\cdot 11=55$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Multivariable Chain Rule. Where is my mistake? Say $x=\sqrt{z^2+y^2}$.
For a function $f(y,z)$, we have
$$
\tag{1}
\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y}\frac{\partial y}{\partial x} + \frac{\partial f}{\partial z}\frac{\partial z}{\partial x} \ .
$$
By definition of $x$, we have
\begin{align}
\frac{\partial y}{\partial x} = \frac{x}{\sqrt{x^2 - z^2}} = \frac{x}{y} \ , \tag{2}\\
\frac{\partial z}{\partial x} = \frac{x}{\sqrt{x^2 - y^2}} = \frac{x}{z} \ .\tag{3}\\
\end{align}
Therefore
$$
\tag{4}
\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y}\frac{x}{y} + \frac{\partial f}{\partial z}\frac{x}{z} \ .
$$
Now assume $f(y,z) = e^{\sqrt{y^2 + z^2}}$. We have
\begin{align}
\frac{\partial f}{\partial y} = e^{\sqrt{y^2 + z^2}} \frac{y}{\sqrt{y^2 + z^2}} = e^x\frac{y}{x}\ , \tag{5}\\
\frac{\partial f}{\partial z} = e^{\sqrt{y^2 + z^2}} \frac{z}{\sqrt{y^2 + z^2}} = e^x\frac{z}{x}\ . \tag{6}\\
\end{align}
By $(4)$, we have
$$
\tag{7}
\frac{\partial f}{\partial x} = e^x\frac{y}{x}\frac{x}{y} + e^x\frac{x}{z}\frac{z}{x} = 2e^x\ .
$$
But that can't be right, because $f = e^x$, so it must be
$$
\tag{8}
\frac{\partial f}{\partial x} = e^x \neq 2e^x \ .
$$
So where is my mistake?
| What you've written really doesn't make sense. You start with a function $f(u)=e^u$ — a function of a single variable. Then you have $u=g(y,z)$. So, letting $F(y,z) = f(g(y,z))$, we have
$$\frac{\partial F}{\partial y} = f'(g(y,z))\frac{\partial g}{\partial y},$$
and similarly for the $z$ partial.
With $f(u) = e^{\sqrt u}$, this gives
$$\frac{\partial F}{\partial y} = \frac1{2\sqrt{y^2+z^2}}e^{\sqrt{y^2+z^2}}\cdot
2y = \frac y{\sqrt{y^2+z^2}}e^{\sqrt{y^2+z^2}}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Calculate $\sum_\limits{k=1}^{100}\left \lfloor \sqrt{k}+\frac{1}{k} \right \rfloor$ Determine: $\sum_\limits{k=1}^{100}\left \lfloor \sqrt{k}+\frac{1}{k} \right \rfloor$.
It's well-known that $x-1<\lfloor x \rfloor \leq x, \forall x \in \mathbb{R}$, but this didn't help me at all.
I computed it and the sum equals 627, but I found no useful properties.
| $f(k)=\left \lfloor \sqrt{k}+\frac{1}{k} \right \rfloor$. We have $f(1)=2,\space f(2)=1,\space f(3)=2$ so $$\sum_\limits{k=1}^{100}\left \lfloor \sqrt{k}+\frac{1}{k} \right \rfloor=5+\sum_\limits{k=4}^{100}\left \lfloor \sqrt{k}+\frac{1}{k} \right \rfloor$$
Now, between $n^2$ and $(n+1)^2-1$ one has $\sqrt{k}+\dfrac{1}{k}=n$ from which
$$\sum_\limits{k=4}^{100}\left \lfloor \sqrt{k}+\frac{1}{k} \right \rfloor=\sum_\limits{k=2}^{9}n(2n+1)+10=622$$
Thus $$\sum_\limits{k=1}^{100}\left \lfloor \sqrt{k}+\frac{1}{k} \right \rfloor=622+5=\color{red}{627}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove or disprove convergence for the series: $\sum_{n=2}^{\infty}\left((1+\frac{1}{n})^n-e\right)^{\sqrt{\log(n)}}$ Because $(1+(1/n))^n$ converges to $e$, I was thinking of comparing the sum to the series $p^{\sqrt{\log(n)}}$, but this series apparently diverges according to wolfram so I'm at a loss. Can someone give me a clue as to where to go from here?
| Note that
\begin{align}
\left(
1+\frac{1}{n}
\right)^{n}
=&
\frac{n!}{0!(n-0)!}\frac{1}{n^0}
+
\frac{n!}{1!(n-1)!}\frac{1}{n}
+
\cdots
+
\frac{n!}{n!(n-n)!}\frac{1}{n^n}
\\
=&
\frac{1}{0!}
+
\frac{1}{1!}\left( 1-\frac{1}{n}\right)
+
\frac{1}{2!}\left( 1-\frac{1}{n}\right)\left( 1-\frac{2}{n}\right)
+
\cdots
+
\frac{1}{n!}\left( 1-\frac{1}{n}\right)\cdot \cdots\cdot \left( 1-\frac{n-1}{n}\right)
\\
=&
\sum_{k=0}^{n}
\frac{1}{k!}
\prod_{i=0}^{k}
\left(
1-\frac{i}{n}
\right)
\end{align}
implies
$$
\left| \;
e-\left(1+\frac{1}{n} \right)^n
\;\right|
=
\sum_{k>n}
\frac{1}{k!}
\prod_{i=1}^{k}
\left(
1-\frac{i}{n}
\right)
\leq
\sum_{k>n}
\frac{1}{k!}
=\frac{1}{n^2}
\sum_{k>n}
\frac{n^2}{k!}
$$
Here $\prod_{i=0}^{0}
\left(
1-\frac{k}{n}
\right)=1$.
Since $n$ is constant with respect to $k$,
$
\sum_{k>n}
\frac{n^2}{k!}
$ is limited say by a constant $L$. Then we have
$$
\left| \;
e-\left(1+\frac{1}{n} \right)^n
\;\right|
<\frac{1}{n^2}\cdot L.
$$
Recall that $1 - \frac1x \leq \log x \leq x-1$ for all $ x > 0$. Then
\begin{align}
\sum_{n=2}^{\infty}
\left|e -
\left(
1+\frac{1}{n}
\right)^n
\right|^{\sqrt{\log(n)}}
&\leq
\sum_{n=2}^{\infty}
\left(\frac{L}{n^2}\right)^{\sqrt{\log(n)}}
\\
&\leq
\sum_{n=2}^{\infty}
\left(\frac{L}{n^2}\right)^{\sqrt{1-1/n}}
\\
&\leq
\sum_{n=2}^{\infty}
\left(\frac{L}{n^2}\right)^{1/4}
\\
&=
\sum_{n=2}^{\infty}
\left(\frac{L}{n^{3/4}}\right)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2431091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
How can I prove this inequality: $\frac{a^2}{3}+b^2+c^2>ab+bc+ca$? If $abc=1$ and $a^3>36$ prove that,
$\frac{a^2}{3}+b^2+c^2>ab+bc+ca$
I tried to use the general proof method.
$\frac{a^2}{3}+b^2+c^2-ab-bc-ca>0$
Symbolically notation:
$\frac{a^2}{3}+b^2+c^2-ab-bc-ca>0 \Rightarrow(x+y+z+...)^2>0$
But, after trying too much, I accepted the defeat.
| I remember I solved this inequality a few years ago. Here is my proof. It may look complicated but the idea is actually very simple - we want to make $b,c$ closer to each other keeping the inequality $a^2\ge 36bc$ valid and the sum $b^2+c^2$ fixed so the lefthandside remains fixed and the righthandside increases. The process of making $b,c$ closer to each other will terminate either in the moment when $36bc$ becomes equal to $a^2$ (this corresponds to the second case below) or when $b=c$ (this is the first case).
Let $d=\sqrt{\frac{b^2+c^2}2}$ so that $2d^2=b^2+c^2$. Observe that $d^2 = \frac{b^2+c^2}2 \ge bc$ and $2d = \sqrt{2(b^2+c^2)}\ge b+c$. Thus $$d^2+2ad \ge bc+a(b+c)=ab+bc+ca.$$
It is enough to prove that $$\frac{a^2}3+2d^2 > d^2+2ad.$$
This is true if $a\ge 6d$ because $$\frac{a^2}3+2d^2 - (d^2+2ad) = \frac{a(a-6d)}{3}+d^2>0.$$
We are left with the case $a<6d$.
Due to $a<6d$ there are $m,n>0$ such that $m^2+n^2=2d^2$ and $36mn=a^2$. Then $36bc<a^2=36mn$ so $bc<mn$ and also $$b+c=\sqrt{b^2+2bc+c^2}=\sqrt{2d^2+2bc}<\sqrt{2d^2+2mn}=\sqrt{m^2+2mn+n^2}=m+n.$$
It follows that $$ab+bc+ca=a(b+c)+bc<a(m+n)+mn=6(m+n)\sqrt{mn}+mn$$
so we only have to prove that $$\frac{a^2}3+b^2+c^2=12mn + n^2+m^2 \ge 6(m+n)\sqrt{mn} + mn,$$
which is true because $$12mn + n^2+m^2 - (6(m+n)\sqrt{mn} + mn) = (m+n-3\sqrt{mn})^2\ge 0.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Complete factorization of $64y^6 - 1$ If we use the difference of two cubes identity, then we have
$$64y^6 - 1 = (4y^2-1)(16y^4+4y^2+1) = (2y+1)(2y-1)(16y^4+4y^2+1)$$ which is the suggested solution of the textbook.
But if we use the difference of two squares identity, we then have
$$64y^6 - 1 = (8y^3+1)(8y^3-1) = (2y+1)(2y-1)(4y^2-2y+1)(4y^2+2y+1).$$
I then came up with the question: what is complete factorization? Of course we can do this trick:
$$16y^4+4y^2+1 = 16y^4+8y^2+1 - 4y^2 = (4y^2+1 - 2y)(4y^2+1+2y).$$
How do we know when to stop factorizing?
| Are you looking for the factorization in $\mathbb{R}[x]$?
Note that for all real values $y$,
$$4y^2+1 - 2y=3y^2+(y-1)^2>0\quad 4y^2+1+2y=3y^2+(y+1)^2>0$$
so they cannot be factored as $a(y-y_1)(y-y_2)$ with $a,y_1,y_2\in \mathbb{R}$, otherwise they should be zero for $y_1$ and $y_2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Show that the roots of the quadratic equation
Show that the roots of the quadratic equation $$(b-c)x^2+2(c-a)x+(a-b)=0$$
are always reals.
My Attempt:
$$(b-c)x^2+2(c-a)x+(a-b)=0$$
Comparing above equation with $Ax^2+Bx+C=0$
$$A=b-c$$
$$B=2(c-a)$$
$$C=(a-b)$$
Now,
$$B^2-4AC=[2(c-a)]^2-4(b-c)(a-b)$$
$$=4(c-a)^2 - 4(ab-b^2-ac+bc)$$
$$=4c^2-8ac+4a^2-4ab+4b^2+4ac-4bc$$
$$=4(a^2+b^2+c^2-ab-bc-ca)$$
How do I proceed further?
| You need to prove that
$$a^2+b^2+c^2\ge ab+bc+ca.$$
This is a well-known inequality, and can be deduced from the
two-variable AM/GM inequality, which states
$$a^2+b^2\ge2ab.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2437091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Arithmetic and Geometric progression in 3 numbers Suppose
$$a,b,c \textrm{ is an arithmetic progression}$$ and
$$a^2,b^2,c^2\textrm{ is a geometric progression}$$
$$a+b+c = \frac{3}{2}.$$
From these equations I get
$$2b=a+c \textrm{, from A.P.}$$
$$b^4=a^2c^2 \textrm{, from G.P.}$$
and finally $$a=b=c=\frac{1}{2}.$$
Can there be any such triplet such that $a<b<c$ ?
| If $a^2,b^2,c^2$are in geometric progression,
Then
$$\Big(\frac{a}{b}\Big)^2=\Big(\frac{b}{c}\Big)^2$$
$$\color{red}{1. }$$
$$\frac{a}{b}=\frac{b}{c}$$
Hence $a,b,c$ are also in GP.
Let $a,b,c$ be $f,fg,fg^2$
For $a,b,c$ to be in AP,
$$f+fg^2=2fg$$
$$g^2+1=2g$$
$$(g-1)^2=0$$
$$g=1$$
If g=1 then $$a=b=c$$ hence there is no possibility for$$a\neq b\neq c$$
$$\color{red}{2. }$$
$$\frac{a}{b}=-\frac{b}{c}$$
$$b^2=a(-c)$$
Hence a,b,-c are in GP,
Let them be $f,fg,fg^2$
So now a,b,c are in AP,
$$a+c=2b$$
$$f-fg^2=2fg$$
$$g^2+2g-1=0$$
$$(g+1)^2=2$$
$$g=\sqrt{2} - 1 \ , or\, -(\sqrt{2} + 1)$$
Hence $a,b,-c$ are,
$$f,(\sqrt{2}-1)f,(3-2\sqrt{2})f$$
Or
$$f,-(\sqrt{2}+1)f,(3+2\sqrt{2})f$$
For $a,b,c$ to be in AP
$$f-3f+2\sqrt{2}f=2(\sqrt{2}-1)f$$
$$2(\sqrt{2}-1)f=2(\sqrt{2}-1)f$$
Or
$$f-3f-\sqrt{2}f=-2(\sqrt{2}+1)f$$
$$-2(\sqrt{2}f+1)f=-2(\sqrt{2}+1)f$$
Which is true, hence such pairs to exist
In you case, alternate solutions are,
$$f+(\sqrt{2}-1)f-3f+2\sqrt{2}f=\frac{3}{2}$$
$$f(3\sqrt{2}-3)=\frac{3}{2}$$
$$f=\frac{\sqrt{2}+1}{2}$$
Hence one of the possible outcomes studying your equations is,
$$\frac{\sqrt{2}+1}{2},\frac{1}{2},\frac{1-\sqrt{2}}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2437923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find k if the two polynomials have one common root The equations $x^2−4x+k=0$ and $x^2+kx−4=0$, where $k$ is a real number, have exactly one common root. What is the value of $k$?
I know the answer but can it be done with the relation of roots.
$a$ and $b$ are roots of equation 1 and $a$ and $c$ are roots of equation 2. So the relations are :-
*
*$a + b = 4$
*$ab = k$
*$a + c = (-k)$
*$ac= (-4)$
I tried doing all the stuff but couldn't get it. Can we find it using these 4 equations?
| Using your equations . . .
*
*$\;a + b = 4$
*$\;ab = k$
*$\;a + c = (-k)$
*$\;ac= (-4)$
Adding equations $(1)$ and $(4)$, we get
$$a + b + ac=0\tag{5}$$
Adding equations $(2)$ and $(3)$, we get
$$a + c + ab=0\tag{6}$$
Subtracting equation $(6)$ from equation $(5)$, we get
\begin{align*}
&b-c + ac - ab = 0\\[4pt]
\implies\;&b-c+a(c-b) = 0\\[4pt]
\implies\;&(b-c)(1-a)= 0\\[4pt]
\implies\;&b=c\;\;\text{or}\;\;a=1\\[4pt]
\end{align*}
If $b=c$, then the left-hand-sides of equations $(2)$ and $(4)$ are equal, hence the right-hand-sides must also be equal, which yields $k=-4$. But for $k=-4$, the two given equations are identical, hence have two common roots, not one.
Hence we must have $a=1$.
Using $a=1$ in equation $(1)$ yields $b=3$, and then equation $(2)$ yields $k=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2438585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Prove this inequality $3(abc+4)\ge 5(ab+bc+ca)$
Let $\{a,b,c\}\subset\mathbb R$ such that $a+b+c=3$ and $abc\ge -4$. Prove that: $$3(abc+4)\ge 5(ab+bc+ca).$$
*) $ab+bc+ca<0$ This ineq is right
*) $ab+bc+ca\ge 0$ then in $ab,bc, ca$ at least a non-negative number exists assume is $ab$
$\Rightarrow \displaystyle f(ab)=(3c-5)ab+5c^2-15c+12$
+)$\displaystyle 3c-5 > 0\Rightarrow \displaystyle f \geq 5c^2-15c+12=5(c-\frac{3}{2})^2+\frac{3}{4} > 0$
+) $\displaystyle 3c-5 \leq 0$. And we have:
$\displaystyle \Rightarrow \frac{(3-c)^2}{4}+c(3-c) \geq ab+bc+ca \geq 0\displaystyle \Leftrightarrow -1 \leq c \leq \frac{5}{3}$
$\Rightarrow \displaystyle f \geq (3c-5)\frac{(3-c)^2}{4}+5c^2-15c+12 \geq 0$
$\displaystyle \Leftrightarrow (c-1)^2(c+1) \geq 0$
Help me to check up and post your solution. Thanks very much
| We have $a+ b+ c= 3$, in $a,\,b,\,c$ always at least a positive number exists. Assume $a> 0$, hypothesis
$$abc\geqq -\,4\,\therefore\,bc\geqq -\,\dfrac{4}{a}$$
On the other hand, we also have $t= bc\leqq \dfrac{(\!b+ c\!)^{\!2}}{4}= \dfrac{(\!3- a\!)^{\!2}}{4}$ and ${\rm A}\!=\!(\!3\,abc+ 4\!)- 5(\!ab+ bc+ ca\!)$
$$f(t)= 3at+ 12- 5t- 5a(\!3- a\!)= (\!3a- 5\!)t- 15a+ 5a^{2}+ 12$$ with $-\,\dfrac{4}{a}\leqq t\leqq \dfrac{(\!3- a\!)^{\,2}}{4}$. Consider two following cases
*
*$3\,a- 5= 0$ or $a= \dfrac{5}{3}$ then now
$$f(t)= -\,15\,.\,\frac{5}{3}+ 5\left ( \frac{5}{3} \right )^{\,2}+ 12- \frac{8}{9}> 0$$
*$3\,a- 5\neq 0$ then now $f(t)$ is a linear function for $t$ with $-\,\dfrac{4}{a}\leqq t\leqq \dfrac{(\!3- a\!)^{\,2}}{4}$, so to prove $f(t)\geqq 0$ is to prove $f\left ( -\,\dfrac{4}{a} \right )\geqq 0$ and $f\left ( \dfrac{(\!3- a\!)^{\,2}}{4} \right )\geqq 0$. Indeed, we have
$$f\left ( -\,\frac{4}{a} \right )= -\,\frac{4}{a}(\!3a- 5\!)- 15a+ 5a^{2}+ 12= \frac{1}{a}(\!5a^{3}- 15a^{2}+ 20\!)= \frac{5}{a}(\!a+ 1\!)(\!a- 2\!)^{2}\geqq 0$$
$$f(\!\dfrac{(\!3- a\!)^{\!2}}{4}\!)\!=\!\frac{(\!3a- 5\!)(\!3- a\!)^{\!2}}{4}\!-\!15a\!+\!5a^{2}\!+\!12\!=\!\frac{1}{4}(\!3a^{3}- 3a^{2}- 3a+ 3\!)\!=\!\frac{3}{4}(\!a+ 1\!)(\!a- 1\!)^{\!2}\!\geqq\!0$$
Therefore $f(t)\geqq 0$ with $-\dfrac{4}{a}\leqq t\leqq \dfrac{(\!3- a\!)^{\,2}}{4}$. Due to the same role of $a, b, c$ in ${\rm A}\geqq 0$, therefore
$$\therefore\,3(\!abc+ 4\!)\geqq 5(\!ab+ bc+ ca\!)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2439140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int_0^{π/2} \frac{x\sin x\cos x}{\sin^4 x+\cos^4 x}\,dx$ Find the value of
$$\int_0^{\pi/2} \frac{x\sin x\cos x}{\sin^4 x+\cos^4 x}\,dx$$
I'm having a very tough time solving this question, can any body give me some hints?
| Let $f(x) = \dfrac{\sin x\cos x}{\sin^4 x + \cos^4 x}$ and let (capital) $F$ by some antiderivative of (lower-case) $f.$ Then
$$
\int_0^{\pi/2} x f(x)\,dx = \overbrace{\int x\,dv = xv-\int v\,dx}^{\Large\text{integration by parts}} = \left. xF(x)\vphantom{\frac 11} \, \right|_0^{\pi/2} - \int_0^{\pi/2} F(x)\, dx.
$$
To find this, we need to find $F(x).$
\begin{align}
F(x) & = \int \frac{\sin x\cos x}{\sin^4 x+ \cos^4 x}\, dx = \int \frac u {u^4 + (1-u^2)^2} \,du & & \text{where } u = \sin x, \\[10pt]
& = \int \frac{dw/2}{w^2 + (1-w)^2} & & \text{where } w=u^2, \\[10pt]
& = \frac 1 2 \int \frac{dw}{2w^2 - 2w+1}.
\end{align}
And then:
$$
2w^2-2w+1 = \frac 1 2 \Big( (2w-1)^2+ 1 \Big) = \frac 1 2 (v^2 + 1), \qquad dw = dv/2
$$
So you get an arctangent. And $\arctan\left( 2\sin^2 x - 1\right) = -\arctan( \cos(2x)).$
And then you've got some more work to do.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2439244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Binomial Theorem and $n(n+1)2^{n-2} = \sum_{i=1}^ni^2\binom{n}{i}$ $n(n+1)2^{n-2} = \sum_{i=1}^ni^2\binom{n}{i}$
I had proved this combinatorially but also trying to derive this identity using binomial theorem.
From the bionomial theorem, one could easily get $2^n = \sum_{k=0}^n\binom{n}{k}$
Using this LHS could be equated with$\binom{n+1}{2}{1\over2} \sum_{k=0}^n\binom{n}{k}$
I'd like to go further to make this form closer to $\sum_{i=1}^ni^2$ but it's hard to imagine where to go.
Any advice?
| Let,
$$(1+x)^n=\binom{n}{0}+\binom{n}{1}x+\binom{n}{2} x^2\cdots +\binom{n}{n} x^n$$$$$$
Differentiating wrt x,
$$n(1+x)^{n-1}=\binom{n}{1}+2\binom{n}{2}x+3\binom{n}{3} x^2+\cdots +n\binom{n}{n} x^{n-1}$$$$$$
Multiplying both sides by x,
$$n(1+x)^{n-1} x=\binom{n}{1}x+2\binom{n}{2}x^2+3\binom{n}{3}x^3+\cdots +n\binom{n}{n}x^n$$$$$$
Differential wrt x,
$$\frac{d}{dx}(n(1+x)^{n-1}x)=\binom{n}{1}+2^2\binom{n}{2}x+3^2\binom{n}{3}x^2+\cdots +n^2\binom{n}{n} x^{n-1}$$$$$$
$$n((n-1)x(1+x)^{n-2}+(1+x)^{n-1})=\binom{n}{1}+2^2\binom{n}{2}x+3^2\binom{n}{3}x^2+\cdots +n^2\binom{n}{n} x^{n-1}$$$$$$
$$n(1+x)^{n-2}(nx+1)=\binom{n}{1}+2^2\binom{n}{2}x+3^2\binom{n}{3}x^2+\cdots +n^2\binom{n}{n} x^{n-1}$$$$$$
Put $x=1$,
$$n(1+1)^{n-2}(n+1)=\binom{n}{1}+2^2\binom{n}{2}+3^2\binom{n}{3}+\cdots +n^2\binom{n}{n}$$$$$$
$$n(n+1)2^{n-2}=\binom{n}{1}+2^2\binom{n}{2}+3^2\binom{n}{3}+\cdots +n^2\binom{n}{n}$$
Hence...
$$n(n+1)2^{n-2} = \sum_{i=1}^ni^2\binom{n}{i}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2439878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Calculate eigenvalues and eigenvectors I have this matrix below, and I need to find the eigenvalues.\begin{bmatrix}
2 & -1 & 0 \\
5 & 2 & 4 \\
0 & 1 & 2 \\
\end{bmatrix}
This is what I have done so far:
I used $\det(A-λI)=0$ and reached this form \begin{bmatrix}
2-λ & -1 & 0 \\
5 & 2-λ & 4 \\
0 & 1 & 2-λ \\
\end{bmatrix}
I have done some simplifications:
$(2-λ)[(2-λ)(2-λ)-4]-5(-(2-λ))=0$
$(2-λ)[(2-λ)(2-λ)-4]-5(-1)=0$
$(2-λ)[4-4λ+λ^2-4+5]=0$
$(2-λ)[λ^2-4λ+5]=0$
$λ^2(2-λ)-4λ(2-λ)+5(2-λ)=0$
$2λ^2-λ^3-8λ+4λ^2+10-5λ=0$
$-λ^3+6λ^2-13λ+10=0$
or $-λ(λ^2-6λ+13)+10=0$
$-λ (λ-(3+{\sqrt 22})) (λ+(3-{\sqrt 22}))+10=0$
Am I doing it right and if so:
I checked the answer on Symbolab and it was $2,2-i,2+i$, How so? and what
is $i$?
And is the matrix will be like this when I want to calculate the eigenvector for $2-i$ ??\begin{matrix}
i & -1 & 0 \\
5 & i & 4 \\
0 & 1 & i \\
\end{matrix}
| It happens that, if $\lambda=2$, then $-\lambda^3+6\lambda^2-13\lambda+10=0$. In fact, $-\lambda^3+6\lambda^2-13\lambda+10=(2-\lambda)(\lambda^2-4\lambda+5)$. The roots of the polynomial $\lambda^2-4\lambda+5$ are complex non-real numbers: $2+i$ and $2-i$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2440541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Find the coefficient of $x^{17}$ in $(x^2+x^3+x^4+x^5+x^6+x^7+...)^3.$ This is #5 from Section 6.2 of Applied Combinatorics by Alan Tucker:
Find the coefficient of $x^{17}$ in $(x^2+x^3+x^4+x^5+x^6+x^7+...)^3.$
The books answer: $C(11+3−1,11)−C(3,1)×C(5+3−1,5)$.
I'm not sure where the second term comes from? Here is my work:
$$\begin{array}{l}
(x^2+x^3+x^4+x^5+x^6+x^7+\ldots)^3\\
=x^6(1 + x + x^2 + \ldots)^3\\
=x^6\dfrac{1}{(1-x)^3}\\
=x^6\left(1 + C(1+3-1,1)x + C(2 +3-1,2)x^2+C(3+3-1,3)x^3+ \ldots \right)
\end{array}$$
So, the coefficient of $x^{17}$ is really going to be the coefficient of $x^{11}$. So our answer is: $$C(11+3-1,11).$$
But that's only half the book answer stated above. Why?
| Your answer is right, your book is wrong. Here are two alternative ways to prove it:
*
*The coefficient of $x^{11}$ is the number of ways to put 11 $x$'s in 3 baskets. This is the "stars and bars" problem, and the answer is $\binom{11+3-1}{3-1}$.
*Expand and look at the $x^{11}$ term.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2442613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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How to evaluate $\lim_{x\to0}\frac{\sin^2\left(\frac x2\right)-\frac{x^2}4}{e^{x^2}+e^{-x^2}-2}$? $$\begin{align*}
\lim_{x \to 0} \frac{\sin^2 \left(\frac{x}{2}\right) - \frac{x^2}{4}}{e^{x^{2}} + e^{-x^{2}} - 2} &\overset{L}{=} \lim_{x \to 0} \frac{\sin \frac{x}{2} \cos \frac{x}{2} - \frac{1}{2}x}{2xe^{x^{2}} + (-2x)e^{-x^{2}}} \\
&= \lim_{x \to 0} \frac{\sin \frac{x}{2} \cos \frac{x}{2} - \frac{1}{2}x}{2xe^{x^{2}} -2xe^{-x^{2}}} \\
&\overset{L}{=} \lim_{x \to 0} \frac{\frac{1}{2}\cos^2 \frac{x}{2} - \frac{1}{2}\sin^2 \frac{x}{2} - \frac{1}{2}}{(2x)(2x)e^{x^{2}} - (2x)(-2x)(e^{-x^{2}})} \\
&= \lim_{x \to 0} \frac{\frac{1}{2}\cos^2 \frac{x}{2} - \frac{1}{2}\sin^2 \frac{x}{2} - \frac{1}{2}}{4x^2 e^{x^{2}} + 4x^2 e^{-x^{2}}} \\
&\overset{L}{=} \lim_{x \to 0} \frac{\frac{1}{2}\left( -\sin \frac{x}{2} \cos \frac{x}{2} \right) - \frac{1}{2} \left( \sin \frac{x}{2} \cos \frac{x}{2} \right)}{(4x^2)(2x)e^{x^{2}} + (4x^2)(-2x)(e^{-x^{2}})} \\
&= \lim_{x \to 0} \frac{-\sin \frac{x}{2} \cos \frac{x}{2}}{8x^3e^{x^{2}} - 8x^3 e^{-x^{2}}} \\
\end{align*}$$
After evaluating the limit as $x \to 0$, I noticed that the problem comes up to be in an indeterminate form of $0/0$. I immediately utilized the L'Hospital Rule by differentiating both the numerator and denominator.
However, after using L'Hospital rule for 5-6 times, I noticed that the question will go through a loop of $0/0$ indeterminants.
In my second attempt,
I have tried multiplying $\exp(x^2)$ in both the numerator and denominator with hopes to balance out the $\exp(x^{-2})$. However, an indeterminant is $0/0$ still resulting.
Any help would be appreciated, thank you all!
| You only have to find equivalents for the numerator and the denominator. We'll begin with rewriting them, and use classical Taylor's expansions:
Numerator:
$$\sin^2\dfrac x2-\dfrac{x^2}4=\dfrac{1-\cos x}{2}-\dfrac{x^2}4=\biggl[\frac12-\Bigl(\frac12-\frac{x^2}{4}+\frac{x^4}{48}+o\bigl(x^4\bigr)\Bigr)\biggr]-\frac{x^2}4=-\frac{x^4}{48}+o\bigl(x^4\bigr)$$
so the numerator is equivalent near $\,0\;$ to $-\dfrac{x^4}{48}$.
Denominator:
We know $\sinh u\sim_0 u$, so
$$\mathrm e^{x^2}+\mathrm e^{-x^2}-2=1+x^2+\frac{x^4}2+o(x^4)+1-x^2+\frac{x^4}2+o(x^4)-2=x^4+o(x^4)$$
so the denominator is equivalent to $x^4$. There results that
$$\frac{\sin^2\dfrac x2-\dfrac{x^2}4}{\mathrm e^{x^2}+\mathrm e^{-x^2}-2}\sim_0\frac{-\cfrac{x^4}{48}}{x^4}=-\frac1{48}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find the remainder when $f(x)$ divided by $(x^2 + x + 1)(x+1)$. When a polynomial $f(x)$ is divided by $x^2 + x + 1$ and $(x+1)^2$, the remainder are $x+5$ and $x-1$ respectively. Find the remainder when $f(x)$ is divided by $(x^2 + x + 1)(x+1)$.
First, I let the remainder be $Ax^2 + Bx +C$, then I try to find the values of $A$, $B$ and $C$. There is $3$ unknowns so we need three equations but I can only get two equations.
| The remainder when dividing by $(x+1)^2$ was $x-1$, so the remainder when dividing by just $x+1$ must be $-2$.
We know that $f(x)$ is of the form
$$
f(x) = q(x)(x^2 + x + 1)(x+1) + Ax^2 + Bx + C
$$
where $q(x)$ is a polynomial and $Ax^2 + Bx + C$ has the same remainders as $f$ does when divided by $x+1$ and by $x^2 + x + 1$. Now we just calculate:
$$
\frac{Ax^2 + Bx + C}{x+1} = Ax + B-A+\frac{C-B+A}{x+1}
$$
so we must have $C-B+A = -2$. That's one equation down. Now we deal with $x^2 + x + 1$:
$$
\frac{Ax^2 + Bx + C}{x^2 + x + 1} = Ax + \frac{(B-A)x + C-A}{x^2 + x + 1}
$$
so we know that $(B-A)x + C-A = x+5$. As I alluded to in the comments to the question above, this is actually two equations in one, as this is an equality of functions. The slopes must be equal, and the constant terms must be equal. This gives you the remaining two equations $B-A = 1$ and $C-A = 5$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrate $\int^\frac{T}{2}_0\frac{dt}{\pi ^2 t^2 (\csc^2 \frac{\pi t}{T}) -T^2}$.
Evaluate
$$\int^\frac{T}{2}_0 \frac{dt}{\pi ^2 t^2 (\csc^2 \frac{\pi t}{T})-T^2}.$$
My work. Let $u=\frac{\pi t}{T}$, then $du=\frac{\pi}{T}dt$ and
$$\frac{T}{\pi} \int^\frac{\pi}{2}_0\frac{du}{u^2 T^2 (\csc^2 u)-T^2}=\frac{1}{\pi T} \int^\frac{\pi}{2}_0\frac{du}{u^2 (\csc^2 u) -1}.$$
Please help! I do not know what to do next.
| Yes, you are on the right track. For $T>0$,
$$\int^\frac{T}{2}_0 \frac{dt}{\pi ^2 t^2 (\csc^2 \frac{\pi t}{T})-T^2}=\frac{1}{\pi T} \int^\frac{\pi}{2}_0\frac{du}{u^2 \csc^2 u -1}.$$
Now, note that in a right neighbourhood of $0$:
$$0<\frac{1}{u^2 (\csc^2 u) -1}=\frac{\sin^2 u}{u^2-\sin^2 u}=\frac{u^2+o(u^2)}{u^2-(u^2-2\frac{u\cdot u^3}{3!}+o(u^4))}=\frac{u^2+o(u^2)}{\frac{u^4}{3}+o(u^4)}\sim \frac{3}{u^2}$$
so the integral diverges to $+\infty$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Summation of a series-combination of a GP and another series Find the sum of the series:
$$\frac{1}{2}\biggl(\frac{1}{2}\biggl)+\frac{2}{3}\biggl({\frac{1}{2}\biggl)}^2+\frac{3}{4}\biggl({\frac{1}{2}\biggl)}^3+\cdots+
n^{th}terms$$
I calculated the $n^{th}$ term to be $\frac{n}{n+1}\frac{1}{2^n}$
$$\sum_{k=1}^n \frac{k}{k+1}\frac{1}{2^k}=\sum_{k=1}^n \frac{1}{2^k}\biggl(1-\frac{1}{k+1}\biggl)=\sum_{k=1}^n \frac{1}{2^k}-\sum_{k=1}^n \frac{1}{2^k(k+1)} $$
The first summation is a GP. But then I could not solve the second summation as it was not telescoping. Please help.
| If we consider the more general case of $$S_n(x)=\sum_{k=1}^n \frac{k}{k+1}{x^k}=\sum_{k=1}^n x^k\biggl(1-\frac{1}{k+1}\biggl)=\sum_{k=1}^n x^k-\sum_{k=1}^n \frac{x^k}{(k+1)}$$ the problem looks quite simple when the summation is done up to $\infty$.
For finite $n$, at least to me, the problem seems to be difficult since we have $$\sum_{k=1}^n x^k=\frac{x \left(1-x^n\right)}{1-x}$$
$$\sum_{k=1}^n \frac{x^k}{(k+1)}=-1-\frac{\log (1-x)}{x}-x^{n+1} \Phi (x,1,n+2)$$ where appears the Lerch transcendent function. This makes
$$S_n(x)=1+\frac{x \left(1-x^n\right)}{1-x}+\frac{\log (1-x)}{x}+x^{n+1} \Phi (x,1,n+2)$$
$$S_n\left(\frac 12\right)=2-2\log(2)+\frac{\Phi \left(\frac 12,1,n+2\right)}{2^{n+1}}$$ As shown below, the last term decreases very fast
$$\left(
\begin{array}{cc}
n & \frac{\Phi \left(\frac 12,1,n+2\right)}{2^{n+1}}\\
1 & 0.136294 \\
2 & 0.052961 \\
3 & 0.021711 \\
4 & 0.00921103 \\
5 & 0.00400269 \\
6 & 0.00177055 \\
7 & 0.000793989 \\
8 & 0.000359961 \\
9 & 0.000164649 \\
10 & 0.0000758704
\end{array}
\right)$$ and a quick and dirty nonlinear regression (done for $1\leq n \leq 20$) seems to show a quite good approximation by $$\frac{\Phi \left(\frac 12,1,n+2\right)}{2^{n+1}}\approx 0.414102\, e^{-1.11129\, n^{0.888026}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2447219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proof verification: $\lim_{x\to2} \frac{1}{x} = \frac{1}{2}$ Is either of the following methods correct?
Prove
$$\lim_{x\to2} \frac{1}{x} = \frac{1}{2}$$
First Method
Preliminary Analysis:
We know that $a = 2$, $L= \frac{1}{2}$, and $f(x)= \frac{1}{x}$. By the precise definition of limit we have the following:
$$
0<|x-a|<\delta \implies |f(x)-L|<\varepsilon\\
0<|x-2|<\delta \implies \left|\frac{1}{x}-\frac{1}{2}\right|<\varepsilon \implies \left|\frac{2}{2x}-\frac{x}{2x}\right| <\varepsilon \implies \left|\frac{2-x}{2x}\right| <\varepsilon \\
\implies \left|\frac{-(-2+x)}{2x}\right| <\varepsilon \implies \left|\frac{x-2}{2x}\right| <\varepsilon \implies \frac{\left|x-2\right|}{\left|2x\right|} <\varepsilon \implies \left|x-2\right|<\varepsilon \left|2x\right|
$$
let $\delta = \varepsilon\left|2x\right|$ but we need to simplify it further because delta should be in terms of $\varepsilon$ only.
Assume $|x-a| < 1$
$$ |x-2| < 1 \implies -1 <x -2<1 \implies -1+2<x<1+2 \implies 1<x<3$$
Then we have to simplify $|2x|$ as well, which ends up being
$$ 2<2x<6 \implies -6<2x<6 \implies |2x| <6$$
Now consider the inequality we discovered, specifically, $\left|x-2\right|<\varepsilon \left|2x\right|$ this inequality is also valid for $\left|x-2\right|<\varepsilon\cdot6$.
Let $\delta = \min{\{1, \varepsilon\cdot6}\}$
Proof:
Given $\varepsilon > 0$ let $\delta = \min{\{1, \varepsilon\cdot6}\}$ if $ 0<|x-2|<\delta \implies |x-2|<1 \implies 1 <x < 3 \implies 2 < 2x < 6 \implies - 6 < 2x < 6 \\ \implies |2x| < 6$
We also have $|x - 2| < \varepsilon \cdot6$.
$$
\left|\frac{1}{x}-\frac{1}{2}\right|\implies \left|\frac{2}{2x}-\frac{x}{2x}\right| \implies \left|\frac{2-x}{2x}\right| \\
\implies \left|\frac{-(-2+x)}{2x}\right|\implies \left|\frac{x-2}{2x}\right| \implies \frac{\left|x-2\right|}{\left|2x\right|} < \frac{\varepsilon \cdot 6}{6} = \varepsilon
$$
By the precise definition of limit $$\lim_{x\to2} \frac{1}{x} = \frac{1}{2}$$
Second Method
Proof by Definition/Property:
By the direct substitution property if $f$ is a polynomial or a rational function and $a$ is in the domain of $f$, then
$$\lim_{x \to a} f(x) = f(a)$$
Then by the direct substituon property of limit: $$\lim_{x\to2} \frac{1}{x} = \frac{1}{2}$$
Are either of the above methods correct? (I am putting the question here as well in case someone misses it)
| For the second method, it is ok since the rational function is continuous on it definition area.
| {
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"timestamp": "2023-03-29T00:00:00",
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Understanding the proof for the divergence of the harmonic series I'm trying to understand the following proof for the divergence of the harmonic series:
The parts which I can't seem to understand are:
$$1. \sum_{i=0}^{n-1} \sum_{r=2^i+1}^{2^{i+1}} \frac{1}{r} > \sum_{i=0}^{n-1}
\sum_{r=2^i+1}^{2^{i+1}} \frac{1}{2^{i+1}}$$
$$2. \sum_{i=0}^{n-1} \sum_{r=2^i+1}^{2^{i+1}} \frac{1}{2^{i+1}} = \sum_{i=0}^{n-1} (2^{i+1} - 2^i)\frac{1}{2^{i+1}}$$
I think understanding 2. follows from understanding 1. but I'm not entirely sure. Anyway for 1. I couldn't understand if the series on the RHS of the equality was $(\frac{1}{2^3})+(\frac{1}{2^3} + \frac{1}{2^4})+(\frac{1}{2^5} + \frac{1}{2^6}+\frac{1}{2^7}+\frac{1}{2^8})+...$ or $(\frac{1}{2^1})+(\frac{1}{2^2} + \frac{1}{2^3})+(\frac{1}{2^4} + \frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7})+...$ . And for part I simply could not understand 2. (probably because I didn't understand 1.). So I'm wondering what the series on the RHS of 1. actually is and how it equals $\sum_{i=0}^{n-1} (2^{i+1} - 2^i)\frac{1}{2^{i+1}}$ .
| It's useful to look just at the difference between each term in the sequence $\{S_{2^n}\}$, i.e.,
$$S_{2^{n+1}} - S_{2^n} = \sum_{k = 2^n+1}^{2^{n+1}} \frac{1}{k} \geq \sum_{k = 2^n+1}^{2^{n+1}} \frac{1}{2^{n+1}} = \frac{1}{2^{n+1}} \sum_{k = 2^n+1}^{2^{n+1}} 1 = \frac{2^n}{2^{n+1}} = \frac{1}{2}.$$
So $S_{2^{n+1}}\geq S_{2^n} + \frac{1}{2} $ for all $n$. If the partial sums increase by at least $\frac{1}{2}$ each time, the series must diverge to infinity.
| {
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"timestamp": "2023-03-29T00:00:00",
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Hypotenuse on a right triangle in a right triangle, the side $a = 3$ and the subtraction the sides $b - c =\sqrt{3}$, What is the value of the hypotenuse?
Can someone help me? I need understand how to arrive at the answer.
| The sides are $3, c, c+\sqrt{3}$. Now consider possible pythagorean pairs.
Either $3$ can be hypotenuse, or $c+\sqrt{3}$
*
*Using 3 as hypotenuse, we get:
$$9 = 2c^2 + 2c\sqrt{3} +3\\
c^2 + c\sqrt{3} -3=0\\
c = \frac{\sqrt 3}{2}$$
*Using $c+\sqrt{3}$ as hypotenuse, you have:
$$c^2 + 2c\sqrt 3+3=c^2+9 \\
c = \frac{1}{\sqrt{3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2450560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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An integral from 0 to infinity I am trying to show that:
$$\int_0^\infty \frac{\sin^{2n-1}x}{x}dx = \int_0^\infty \frac{\sin^{2n}x}{x^2}dx$$
for any positive integer $n$.
This eqn struck me when I was evaluating $$\int_0^\infty \frac{\sin^{2n-1}x}{x}dx $$ and $$ \int_0^\infty \frac{\sin^{2n}x}{x^2}dx$$ for different values of $n$ on Wolfram Alpha, and I noticed that they always came out to be equal.
I verified this eqn to hold from $n=1$ to $n= 26$ on Wolfram Alpha so I generalised it for all positive integer $n$. I tried to prove this using Integration by parts as follows:
\begin{align*}
& \int_0^\infty \frac{\sin^{2n}x}{x^2}dx = \int_0^\infty 1.\frac{\sin^{2n}x}{x^2}dx \\
&= \left[ \frac{sin^{2n}x}{x^2}.x\right]_{0}^{\infty}-\int_0^\infty \frac{2n\sin^{2n-1}x.cosx.x^2-2x.sin^{2n}x}{x^4}.xdx\\
&= 0-2n\int_0^\infty \frac{\sin^{2n-1}x.cosx}{x}dx + 2\int_0^\infty \frac{\sin^{2n}x}{x^2}dx \\
&\Rightarrow \int_0^\infty \frac{\sin^{2n}x}{x^2}dx = 2n \int_0^\infty \frac{\sin^{2n-1}x.cosx}{x}dx\\
\end{align*}
Now I am stuck at the process of proving:
$$\int_0^\infty \frac{\sin^{2n-1}x}{x}dx = 2n\int_0^\infty \frac{\sin^{2n-1}x.cosx}{x} dx$$
How do I proceed? Please help.
| For the evaluation of
$$\int_0^\infty \frac{\sin^{2n-1}x}{x}dx\quad\mbox{and}\quad
\int_0^\infty \frac{\sin^{2n-1}x\cos x}{x}d x,$$
apply the Power-reduction formulae,
$$\sin^{2n-1}x=\frac{1}{4^{n-1}}\sum_{k=0}^{n-1}(-1)^{k}\binom{2n-1}{n+k}\sin((2k+1)x),$$
note that
$$\sin((2k+1)x)\cos x=\frac{1}{2}\left(\sin((2k+2)x)+\sin(2kx)\right),$$
and use the fact that for $a>0$,
$$\int_{0}^{+\infty}\frac{\sin(ax)}{x}\,dx=\frac{\pi}{2}.$$
Hence
$$\int_0^\infty \frac{\sin^{2n-1}x}{x}dx=
\frac{1}{4^{n-1}}\sum_{k=0}^{n-1}(-1)^{k}\binom{2n-1}{n+k}\int_0^\infty\frac{\sin((2k+1)x)}{x}dx=\frac{n\pi}{4^n(2n-1)}\binom{2n}{n}$$
and
$$\int_0^\infty \frac{\sin^{2n-1}x\cos x}{x}dx=\frac{\pi}{2\cdot4^n(2n-1)}\binom{2n}{n}.$$
| {
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Prove that $\frac{a}{b+c} +\frac{b}{c+a}+\frac{c}{a+b} \ge \frac{3}{2}$ Prove that $\frac{a}{b+c} +\frac{b}{c+a}+\frac{c}{a+b} \ge \frac{3}{2}$
My steps
A= $\frac{a}{b+c}$,B= $\frac{b}{c+a}$ &C=$\frac{c}{a+b}$
A.M$\ge$H.M
$\frac{A+B+C}{3} \ge \frac{3ABC}{AB+BC+AC}$
I am struck after this step
${A+B+C} \ge \frac{9}{\frac{a}{c}+\frac{c}{a}+\frac{b}{a}+\frac{a}{b}+\frac{c}{b}+\frac{b}{c}}$
| This is Nesbitt's inequality. You find many proofs of that classic one. See here.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve system of equation, $\sqrt{x-1}+\sqrt{y-1}=4\sqrt 3$, $\sqrt{y-4}+\sqrt{z-4}=4\sqrt3$ and $ \sqrt{x-9}+\sqrt{z-9}=4\sqrt3$ . $$\begin{cases}\sqrt{x-1}+\sqrt{y-1}=4\sqrt 3\\\sqrt{y-4}+\sqrt{z-4}=4\sqrt3\\\sqrt{x-9}+\sqrt{z-9}=4\sqrt3\end{cases}$$ I tried somthing,like go to the power of two , and change of variables... but it became more complicated . Is there an idea to solve this system of equation ?
Thanks in advance
| Solving $1$:-
$\sqrt{x-1}+\sqrt{y-1}=4\sqrt3\\$
$\text{Squaring}$
$\begin{align}x-1+y-1+2\sqrt{(x-1)(y-1)}&=48\\to\\50-x-y&=2\sqrt{(x-1)(y-1)}\\\end{align}$
$\text{Squaring}$
$\begin{align}x^2 + y^2+ 2 x y- 100 x - 100 y + 2500&=4 x y - 4 x - 4 y + 4\\to\\x^2 + y^2 + 2496 &= 96 x+ 96 y +2 x y \end{align}$
Continue for the other equations, then solve.
| {
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Ten books are made into two piles. In how many ways can this be done if books as well as piles may or may not be distinguishable? Require that neither pile be empty.
Both not distinguishable: $10=1+9=8+2=7+3=6+4=5+5$. Then there are $5$ ways.
Piles distinguishable but books not: $5 \cdot 2-1 = 9$ (we subtract $1$ since $5+5=5+5$).
Books distinguishable but piles not: I say that it must be
$$\binom{10}{1}+\binom{10}{2}+\binom{10}{3}+\binom{10}{4}+\binom{10}{5}=637$$ where, for example, $\binom{10}{3}$ represents the different number of piles of three books which can occupy the first pile. I don't sum up to $\binom{10}{10}$ because, for example, when I am selecting piles of 4 for the first pile, I'm automatically making piles of 6 in the second pile. However, the result in the book is said to be $$\dfrac{2^{10}-2}{2}=511$$ What's the problem?
| Method 1: Suppose one of the books is Shakespeare's King Lear. The piles are distinguished by which of the books are in the same pile as King Lear. There are $2^9$ subsets of the other books, each of which can be placed in the same pile as King Lear except the entire pile. Hence, the number of permissible ways of distributing the books is $2^9 - 1 = 512 - 1 = 511$.
Method 2: Each of the ten books is placed in the first pile or second pile, which can be done in $2^{10}$ ways. However, the two empty piles are prohibited, so there are $2^{10} - 2$ ways to distribute the books into two non-empty labeled piles. Since the piles are not labeled, we divide by $2$ to obtain
$$\frac{2^{10} - 2}{2} = 511$$
unlabeled piles.
Books are distinguishable but piles not: I say that it must be
$$\binom{10}{1} + \binom{10}{2} + \binom{10}{3} + \binom{10}{4} + \binom{10}{5} = 637$$
where, for example, $\binom{10}{3}$ represents the number of piles of three books which can occupy the first pile. I don't sum up to $\binom{10}{10}$ because, for example, when I am selecting piles of $4$ for the first pile, I'm automatically making piles of $6$ in the second pile.
Your reasoning is correct when you select one, two, three, or four books for the first pile since the other pile has a different number of books than the first pile. If the piles were labeled, there would be $\binom{10}{5}$ ways of selecting five books for the first pile. However, when we select its complement, the same five books are in the second pile. Since the piles are not labeled, selecting a particular set of five books for the first pile counts each such selection twice, once when we select a particular set of five books and once when we select its complement. Hence, you should have
$$\binom{10}{1} + \binom{10}{2} + \binom{10}{3} + \binom{10}{4} + \frac{1}{2}\binom{10}{5} = 10 + 45 + 120 + 210 + 126 = 511$$
| {
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Determine sinus and cosinus : $2\sin x + 3\cos x = 3$ $$2\sin x + 3\cos x = 3$$
Seemed easy at first but I have no idea how to determine them. I tried replacing the 3 with $3(\sin^2 x + \cos x^2x)$ But it doesnt work. I also tried switching sides from almost everything and no luck. Please help!
| \begin{align}
2\sin x + 3\cos x & = \sqrt{2^2+3^2} \left( \frac 2 {\sqrt{2^2+3^2}} \sin x + \frac 3 {\sqrt{2^2+3^2}} \cos x \right) \\[10pt]
& = \sqrt{13} (\cos\alpha\sin x + \sin\alpha \cos x) \\[10pt]
& = \sqrt{13} \,\sin(\alpha+x) = 3. \\[12pt]
\frac 3 {\sqrt{13}} & = \sin(\alpha+x).
\end{align}
Since $3/\sqrt{13} < 1,$ this can be solved.
Since $\sin\alpha/\cos\alpha = 3/2,$ the number $\alpha$ must be a number whose tangent is $3/2$ and whose sine and cosine are both positive, i.e. in the first quadrant.
| {
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"url": "https://math.stackexchange.com/questions/2452111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Prove that $\frac{1}{a^{3}(b+c)} +\frac{1}{b^{3}(a+c)} +\frac{1}{c^{3}(a+b)} \ge \frac{3}{2}$ Prove that $\frac{1}{a^{3}(b+c)} +\frac{1}{b^{3}(a+c)} +\frac{1}{c^{3}(a+b)} \ge \frac{3}{2}$ if abc=1
My approach
A= $\frac{1}{a^{3}(b+c)}$,B= $\frac{1}{b^{3}(c+a)}$ &C=$\frac{1}{c^{3}(a+b)}$
A.M$\ge$G.M
$\frac{A+B+C}{3} \ge ({ABC})^{\frac{1}{3}}$
As abc=1
$\frac{A+B+C}{3} \ge ({\frac{1}{(b+c)(a+c)(a+b)}})^\frac{1}{3}$
I am struck after this step, I even tried to compare G.M$\ge$H.M but not able to eliminate the terms
| By C-S and AM-GM we obtain:
$$\sum_{cyc}\frac{1}{a^3(b+c)}=\sum_{cyc}\frac{b^2c^2}{a(b+c)}\geq\frac{(ab+ac+bc)^2}{\sum\limits_{cyc}(ab+ac)}=\frac{ab+ac+bc}{2}\geq\frac{3}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2452275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving that $\lim\limits_{(x,y) \to (0,0)} \frac{x^2y^3}{x^2+y^2}=0$ I need to show that $\lim\limits_{(x,y) \to (0,0)} \frac{x^2y^3}{x^2+y^2}=0$. My attempted method was showing that as the distance between $(x,y)$ and $(0,0)$ approaches $0$, so does the function in the limit. I tried to show $\lvert{\frac{x^2y^3}{x^2+y^2}}\rvert \leq \sqrt{x^2+y^2}$ for $(x,y) \neq 0$. I expanded the terms to get $\frac{x^2\lvert y^3\rvert}{x^2+y^2} \leq \frac{(x^2+y^2)\sqrt{x^2+y^2}}{x^2+y^2}$, or $(x^2+y^2)^\frac{3}{2}\geq x^2\lvert y^3\rvert$. I am unsure of how to prove this and whether this method is valid or would work in showing the existence of the above limit at all.
| Note : $x^2+y^2 \ge |2xy|$.
Let $ |x^2+y^2|^{1/2} \lt \delta, $
Choose $\epsilon = \delta^3 /4$.
$(x,y) \ne (0,0)$.
$|\dfrac{x^2y^3}{x^2+y^2}| \le$
$ (1/4)|\dfrac{y(x^2+y^2)^2}{x^2+y^2}| =$
$ (1/4)|y||(x^2+y^2)| \lt $
$ (1/4) (\delta)(\delta)^2 = \epsilon$.
Used: $ |y| \le (x^2+y^2)^{1/2} \lt \delta $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2452387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $(a^2 + 1)(b^2 + 1)(c^2 + 1) \ge (a + b)(b + c)(c + a)$ for $a, b, c \in \mathbb{R}$ How to prove that $(a^2 + 1)(b^2 + 1)(c^2 + 1) \ge (a + b)(b + c)(c + a)$ for $a, b, c \in \mathbb{R}$ ? I have tried AM-GM but with no effect.
| By C-S we obtain:
$$\prod_{cyc}(a^2+1)=\sqrt{\prod_{cyc}(a^2+1)(1+b^2)}\geq\sqrt{\prod_{cyc}(a+b)^2}=$$
$$\geq\left|\prod_{cyc}(a+b)\right|\geq\prod_{cyc}(a+b)$$
and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2452783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How many $4$-digit even numbers can be formed from a certain set? Let's imagine a set with $6$ numbers, of which $3$ are even (including $0$) and $3$ are odd. How many $4$-digit even numbers can be formed? Digits cannot be repeated.
Zero is present: $5 \cdot 4 \cdot 3 \cdot 1$
Zero is not present: $4 \cdot 3 \cdot 2 \cdot 2$
Summing both outputs gives $108$ cases, but it is not correct.
| We have $4$ positions, $p_{1},p_{2},p_{3},p_{4}$.
*
*We can put $0$ in $p_{4}$. In this case, we have $P(5,3)$ for remaining digits.
*We can put $0$ in $p_{2}$ or $p_{3}$. In this case, $2$ even digits remains and we must put one of them in $p_{4}$.Thus, $C(2,1) \times C(2,1) \times P(4,2)$.
*We can have a number without $0$. In this case, $C(2,1)\times P(4,3)$.
Therefore, the result is $$P(5,3) + C(2,1) \times C(2,1) \times P(4,2) + C(2,1)\times P(4,3)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2452971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Complete solution for a system of polynomial equations A research question led to the following system of polynomial equations.
\begin{align*}
& x^3 - x^2y + x^2z + x^2 - xy + xz + y^2 - yz - y = 0 \\
& x^3z - 3x^2y + 3xy^2 + xz^2 - y^3 + y^2 - 2yz = 0 \\
& x^4 - x^3y - 2x^3z + x^3 + x^2yz - x^2y + x^2z + xy^2 + xyz - xy - xz^2 - yz^2 = 0 \\
& x^4z + x^4 - x^3y + x^2yz - x^2y + 2x^2z^2 + x^2z - x^2 - xy^2z + xy^2 + xy - y^2z + z^3 - z = 0
\end{align*}
I used the poly_system function in SymPy (a Python library for symbolic mathematics) to solve this system, and it returned the following solutions:
$$[(-1, 0, 0), (0, 0, -1), (0, 0, 0), (0, 0, 1), (0, 1, 0), (1, 1, 0), (1, 2, 1), (2, 3, 0), (3, 6, 1)]$$
This was better than I had hoped, but I am concerned that the function may fail to find all solutions. How can I be certain that no other solutions exist?
| It looks like I can answer my own question. SymPy reports that the ideal generated by these polynomials has the following Gröbner basis.
\begin{align*}
& 2xyz^2 + 2xyz - 2y^2z + 2xz^2 - yz^2 \\
& y^2z^2 + 2xyz - 2y^2z + 2xz^2 - yz^2 \\
& 2xz^3 - yz^2 \\
& yz^3 - yz^2 \\
& z^4 + 2xyz - y^2z + 2xz^2 - yz - z^2 \\
& 4x^3 + 2xyz - 2y^2z + 2xz^2 + 5yz^2 - 4z^3 + 4x^2 - 8xy + 4z \\
& 2x^2y + 2xyz - 2y^2z + 4xz^2 + 3yz^2 - 2z^3 - 2xy - 2y^2 - 2xz + 2y + 2z \\
& 2xy^2 + 6xyz - 6y^2z + 12xz^2 + 7yz^2 - 4z^3 - 2xy - 4y^2 - 2xz - 4yz + 4y + 4z \\
& y^3 + 6xyz - 7y^2z + 12xz^2 + 7yz^2 - 3z^3 - 4y^2 - 4yz + 3y + 3z \\
& 4x^2z + 2xyz - 2y^2z + 6xz^2 + yz^2 - 4yz
\end{align*}
The Wikipedia article on the Gröbner basis has the following statement.
Given the Gröbner basis G of an ideal I, it has only a finite number
of zeros, if and only if, for each variable x, G contains a polynomial
with a leading monomial that is a power of x (without any other
variable appearing in the leading term). If it is the case the number
of zeros, counted with multiplicity, is equal to the number of
monomials that are not multiple of any leading monomial of G. This
number is called the degree of the ideal.
Since the above condition is satisfied, we know that the system has exactly nine solutions.
| {
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"url": "https://math.stackexchange.com/questions/2453702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Simplest way to solve radical equation $\sqrt{3x+1}-\sqrt{x+4}=1$ I have the following equation:
$$\sqrt{3x+1}-\sqrt{x+4}=1$$
I can get the answer $x=5$ through tedious and long algebraic manipulation with quite a few extraneous solutions. It's not elegant. Is there a simple, straightforward way to solve this equation?
| What do you consider long and tedious.
I usually isolate one radical to one side by itself and square, repeat, keep track of signage and it usually works well
$\sqrt{3x+1}-\sqrt{x+4}=1$
$\sqrt{3x + 1} = 1 + \sqrt{x+4}$ [make note $3x+1 \ge 1; x+4 \ge 0$]
$3x + 1 = (1 + \sqrt{x+4})^2 = 1 + 2\sqrt{x+4} + x+4$
$2x -4 = 2\sqrt{x+4}$
$x -2 = \sqrt{x+4}$ [Make note $x-2 \ge 0$
$(x-2)^2 = x + 4$
$x^2 - 4x +4 = x + 4$
$x^2 -5x = 0$.
Quadratic equation or in this case simple factoring yields:
$x(x-5)= 0$
So $x = 0$ or $x = 5$. As $0 - 2 < 0$ so that's out. So $x = 5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2453811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Determine whether $\sum_n(2n!)/(n!(2n)^n)$ converges or diverges The question asks to determine whether the following series converges or diverges:
$$\sum_{n=1}^\infty \frac{(2n)!}{n!(2n)^n}$$
applying the ratio test:
$$\begin{align} \lim_{n \to \infty} | \frac{a_{n+1}}{a_n}| &= \lim_{n \to \infty} |\frac{(2(n+1))!}{(n+1)!(2(n+1))^{n+1}} \frac{n!(2n)^n}{(2n)!}| \\ &= \frac{1}{2}\lim_{n \to \infty} |\frac{2(n+1)n!}{(n+1)n!2^n(n+1)^{n+1}} \frac{n!2^nn^n}{2(n+1)n!}| \\ &= \frac{1}{2} \lim_{n \to \infty} | \frac{n^n}{(n+1)^{n+2}} | \end{align}$$
from here i am not sure what to do. Following the limit is a term of the form $\frac{\infty^\infty}{\infty^\infty}$ and i am finding this problematic.
Thanks in advance,
edit:
as cronos2 kindly points out $(2(n+1)!) \text{ is not } 2(n+1)n! \text{ but rather } (2n+2)(2n+1)(2n)!$ so making this correction we then have
$$\begin{align} &= \lim_{n \to \infty} |\frac{(2n+2)(2n+1)(2n)!}{(n+1)(n)!(2n+1)^{n+1}} \frac{(n)!(2n)^n}{(2n)!}| \\ &= \lim_{n \to \infty} \frac{(2n+1)(2n)^n}{(2n+2)^n} \\ &=\lim_{n \to \infty} |(2n+1)(\frac{2n}{2n+2})^n| \\ &=\lim_{n \to \infty} |(2n+1)(\frac{1}{1+\frac{1}{n}})^n| \\ &= \lim_{n \to \infty} |\frac{(2n+1)}{(1+\frac{1}{n})^n}| \\ &= \infty ? \end{align}$$
My answer book states that this series is convergent and not divergent?
| Using the ratio test, we have
\begin{align}\frac{a_{n+1}}{a_n}&=\frac{(2n)!(2n+2)(2n+1)}{n!(n+1)(2n+2)^{n+1}}\cdot\frac{n!(2n)^n}{(2n)!}\\
&=\cdots=\frac{(2n+1)n^n}{(n+1)^{n+1}}.\end{align}
Taking the limit yields:
\begin{align}
\underset{n\to\infty}\lim\frac{(2n+1)n^n}{(n+1)^{n+1}}&=\underset{n\to\infty}\lim\frac{(2n)n^n}{(n+1)^{n+1}}\\
&=2\underset{n\to\infty}\lim\frac{n^{n+1}}{(n+1)^{n+1}}\\
&=2\underset{n\to\infty}\lim\left(\frac1{1+1/n}\right)^{n+1}\\
&=2\underset{n\to\infty}\lim\left(1+\frac 1{n-1}\right)^{-n}=\frac 2e<1.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2454131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Subsets and Splits
Fractions in Questions and Answers
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