Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
Prove the trigonometric identity $\cos(x) + \sin(x)\tan(\frac{x}{2}) = 1$ While solving an equation i came up with the identity $\cos(x) + \sin(x)\tan(\frac{x}{2}) = 1$.
Prove whether this is really true or not. I can add that $$\tan\left(\frac{x}{2}\right) = \sqrt{\frac{1-\cos x}{1+\cos x}}$$
| Notice, $$LHS=\cos x+\sin x\tan \frac{x}{2}$$
$$=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}+\frac{2\tan \frac{x}{2}}{1+\tan^2\frac{x}{2}}\tan \frac{x}{2}$$
$$=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}+\frac{2\tan^2 \frac{x}{2}}{1+\tan^2\frac{x}{2}}$$
$$=\frac{1-\tan^2\frac{x}{2}+2\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}$$
$$=\frac{1+\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}=1=RHS$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1407794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 9,
"answer_id": 5
} |
Find the sum of all the number formed by 2,4,6, and 8 without repetition.Number may be of any digit like 2, 24, 684, 4862.
Find the sum of all the number formed by 2,4,6, and 8 without repetition.Number may be of any digit like 2, 24, 684, 4862.
My Approach:
single digit no formed = 2,4,6,8
sum= 2+4+6+8= 20
two digit= 24,26,28,42,46,48,62,64,68,82,84,86
sum= 660
three digit no=246+264+426+462+624+642=2664
268+286+628+682+826+862=3552
248+284+428+482+824+842=3108
468+486+648+684+846+864=3996
sum of all 3 digit nos =13310
Similarly for all 4 digit numbers.
| Consider an example by taking digits as $3,4,6,8$.
total combinations possible$=4\times 3\times 2=24$ numbers possible
no of digits$=4$
hence $\dfrac{24}{4}$ each digit comes $6$ times in ones, tens and hundred's place
at ones place=$6\times (3+4+6+8)=126\equiv6$.
at ten's place$=6\times (3+4+6+8)+12(\text{carry})=138\equiv 8$.
At hundred's place$=6\times (3+4+6+8)=139\equiv 9$
so, $3$ will come at thousands place and $1$ at ten thousands place,
Hence, the number is $139864$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1408969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
} |
How many pairs of natural numbers $(x,y)$, satisfy the equation $\space xy=x+y+\gcd(x,y)$. How many pairs of natural numbers $(x,y)$, satisfy the equation
$\space xy=x+y+\gcd(x,y)$. You may assume that $x≤y$.
| If $x=1$ then $\gcd(x,y)=1$ and $x+y+\gcd(x,y)=y+2\ne 1\cdot y$. Hence we may assume $x\ge 2$.
As $\gcd(x,y)\le x$, we have $xy\le y+2x$, so
$$(x-1)(y-2)=xy-y-2x+2\le 2 $$
which leads to $y\le 4$. As wlog. $x\le y$, we need only check the few candidates $(2,2),(2,3),(2,4),(3,3),(3,4),(4,4)$.
Alternatively (again assuming wlog. $x\le y$), we have $(x-y)(y-1)=\gcd(x,y)+1$.
*
*If $\gcd(x,y)=1$ then $(x-1)(y-1)=2$ and from the only factorization of $2$ we get $x=2$,$y=3$.
*If $\gcd(x,y)=2$ then $(x-1)(y-1)=3$ and from the only factorization of $3$ we get $x=2$,$y=4$.
*If $\gcd(x,y)=3$ then $(x-1)(y-1)=4$ and from the only factorizations of $4$ we get $x=2$,$y=5$ (but then $\gcd=1$) or $x=y=3$.
*If $\gcd(x,y)\ge 4$ then $x\ge 4$ and $3y\ge x+y+\gcd(x,y)=xy\ge 4y$, contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1409377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Can this congruence be simplified? $$p(p+1) \equiv -q(q+1) \bmod pq$$
Can this be reduced to an easier format?
| Mod $pq$,
$p(p+1) \equiv -q(q+1)
\iff
p^2+p = -q^2-q
\iff
p^2+q^2 +p+q =0
$
$\begin{array}\\
(p+q+1)^2
&\equiv
p^2+q^2+1+2pq+2p+2q\\
&\equiv
p^2+q^2+1+2p+2q\\
&\equiv
(p^2+q^2+p+q)+p+q+1\\
&\equiv
p+q+1\\
\end{array}
$
Therefore,
if $n = p+q+1$,
$n^2 \equiv n$
or
$n(n-1) \equiv 0
$.
If $p$ and $q$ are distinct primes
(or even just relatively prime),
then
$gcd(n-1, p) = gcd(n-1, q) = 1$,
so that
$n \equiv 0
$.
But $n < pq
$,
so that
this can not hold.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1410565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$z=100^2-x^2$. Then, how many values of $x,z$ are divisible by $6$?
$z=100^2-x^2$. Then, how many values of $x,z$ are divisible by $6$?
My approach:
For $x=1$, $z$ is not divisible by $6$.
For $x=2$, $z$ is divisible by $6$.
For $x=3$, $z$ is not divisible by $6$.
For $x=4$, $z$ is divisible by $6$.
For $x=5$, $z$ is not divisible by $6$.
For $x=6$, $z$ is not divisible by $6$.
For $x=7$, $z$ is divisible by $6$.
For $x=8$, $z$ is divisible by $6$.
I could not identify the pattern in these questions. Also can this problem be solved with a better approach?
| I think a good idea for the problem is using modulos. So, the problem can be written like this
$$
z=4-x^2 \text{ }mod(6),
$$
We know that $100=96+4=4\quad mod(6)$. So $100^2=(100)*(100)=4*4=16=12+4=4\quad mod(6)$. If $z$ is divisible for $6$, then $z=0\quad mod(6)$. For all of this we have to find $x$ such that
$$
x^2=4\quad mod(6).
$$
We have to make a table where we can analize $x^2\quad mod(6)$.
For $x=0$, $x^2=0$ and $x^2=0\quad mod(6)$.
For $x=1$, $x^2=1$ and $x^2=1\quad mod(6)$.
For $x=2$, $x^2=4$ and $x^2=4\quad mod(6)$.
For $x=3$, $x^2=9$ and $x^2=3\quad mod(6)$.
For $x=4$, $x^2=16$ and $x^2=4\quad mod(6)$.
For $x=5$, $x^2=25$ and $x^2=1\quad mod(6)$.
So, the last proposition implies that $x=2 \quad mod(6)$ or $x=4 \quad mod(6)$.
In summary, if $x=2+6m$ or $x=4+6n$ for $m,n\in\mathbb{Z}$. Then, $z$ is divisible by $6$. There are infinity solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1411101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\cos \arctan 1/2 = 2/\sqrt{5}$ How can we prove the following? $$\cos \left( \arctan \left( \frac{1}{2}\right) \right) =\frac{2}{\sqrt{5}}$$
| We know that $\tan (θ)=\frac{y}{x}$.
Thus, for $\tan^{-1}(1/2)$, we have a right triangle whose $x$ and $y$ values are $2$ and $1$, respectively.
Pythagorean theorem in terms of $x$, $y$, and $r$ is written as $x^2+y^2=r^2$. Simply plug in the values $x=1$ and $y=2$ from earlier to get the following:
$$1^2+2^2=r^2$$
$$r^2=5$$
$$r=\sqrt5$$
Now, we know that $\cos(θ)=\frac{x}{r}$. Plugging in $x=2$ and $r=\sqrt5$ from earlier, we obtain the following:
$$\cos(θ)=\frac{2}{\sqrt5}$$
Which matches your answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1411589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
5-Card Poker Two-Pair Probability Calculation Question: What is the probability that 5 cards dealt from a deck of 52 (without replacement) contain exactly two distinct pairs (meaning no full house)?
Solution: $$\frac{\binom{13}{2}\binom{4}{2}\binom{4}{2}\binom{11}{1}\binom{4}{1}}{\binom{52}{5}} = 0.047539$$
Why doesn't ${13\choose 1}{4\choose 2}{12\choose 1}{4\choose 2}{11\choose 1}{4\choose 1}\over{52\choose 5}$ OR ${13\choose 3}{4\choose 2}{4\choose 2}{4\choose 1}\over{52\choose 5}$ work?
| Reduce your problem to counting a simple unique description of the hand. A hand with two pairs of different value can be described as selecting 2 values for the pairs out of 13, 2 suits out of 4 for the lesser pair, 2 suits out of 4 for the higher pair, the value of the fifth card out of the remaining 11 values, and its suit out of 4. Each of these is independent of the others. In all:
$$
\binom{13}{2}
\cdot \binom{4}{2}
\cdot \binom{4}{2}
\cdot \binom{13 - 2}{1}
\cdot \binom{4}{1}
= 123\,552
$$
Breaking down complex problems into such easy-to-count descriptions makes it easier to convince yourself (or others!) that no mistake snuck in
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1414112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Can 720! be written as the difference of two positive integer powers of 3? Does the equation:
$$3^x-3^y=720!$$
have any positive integer solution?
| As alex.jordan writes, $3^x-3^y$ factors as $3^y(3^{x-y}-1)$, so $y$ must be the number of factors of $3$ in $720!$.
I don't actually need to count the number of $3$s in $720!$, so let's just define the notation $720!_3$ for "$720!$ with all of the powers of $3$ divided out". This must yield the other factor $3^{x-y}-1$, so we need to investigate whether $720!_3$ is one less than a power of $3$. To do this we will compute it modulo $3$.
In general we have that
$$ (3k)!_3 \equiv (1\cdot 2)^k \cdot k!_3 \equiv (-1)^k k!_3 \pmod 3 $$
And therefore
$$ \begin{align} 720!_3
\equiv 240!_3
&\equiv 80!_3 \equiv 80\cdot 79 \cdot 78!_3 \equiv - 78!_3 \\
&\equiv -26!_3 \equiv -26\cdot 25 \cdot 24!_3 \equiv 24!_3 \\
&\equiv 8!_3 \equiv 8\cdot 7\cdot 6!_3 \equiv -6!_3 \\
&\equiv -2!_3 \equiv -2 \equiv 1 &\pmod 3
\end{align}$$
which is not one less than a multiple of 3, so certainly not one less than a power of $3$.
So the answer is no.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1414275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
Calculate simple expression: $\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}$ Tell me please, how calculate this expression:
$$
\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}
$$
The result should be a number.
I try this:
$$
\frac{\left(\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}\right)\left(\sqrt[3]{\left(2 + \sqrt{5}\right)^2} - \sqrt[3]{\left(2 + \sqrt{5}\right)\left(2 - \sqrt{5}\right)} + \sqrt[3]{\left(2 - \sqrt{5}\right)^2}\right)}{\left(\sqrt[3]{\left(2 + \sqrt{5}\right)^2} - \sqrt[3]{\left(2 + \sqrt{5}\right)\left(2 - \sqrt{5}\right)} + \sqrt[3]{\left(2 - \sqrt{5}\right)^2}\right)} =
$$
$$
= \frac{2 + \sqrt{5} + 2 - \sqrt{5}}{\sqrt[3]{\left(2 + \sqrt{5}\right)^2} + 1 + \sqrt[3]{\left(2 - \sqrt{5}\right)^2}}
$$
what next?
| Would it help you to know that
${2\pm\sqrt5}=\left(\frac{1\pm\sqrt5}2\right)^3$ ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1416720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Balancing chemical equations using linear algebraic methods I know there are already plenty of questions on this site regarding this topic but I am having difficulty with a particular chemical equation.
I am trying to balance the following:
$$
{ C }_{ 2 }{ H }_{ 2 }{ Cl }_{ 4 }\quad +\quad { C }a{ { (OH }) }_{ 2 }\quad \xrightarrow [ ]{ } \quad { C }_{ 2 }{ H }{ Cl }_{ 3 }\quad +\quad Ca{ Cl }_{ 2 }\quad +\quad { H }_{ 2 }{ O }
$$
The system of linear equations produces the following augmented matrix:
$$
\begin{pmatrix} 2 & 0 & -2 & 0 & 0 & 0 \\ 2 & 2 & -1 & 0 & -2 & 0 \\ 4 & 0 & -3 & -2 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 \\ 0 & 2 & 0 & 0 & -1 & 0 \end{pmatrix}
$$
With the rows in the following order:
Carbon
Hydrogen
Chlorine
Calcium
Oxygen
In row echelon form this reduces to:
$$
\begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \end{pmatrix}
$$
Which would indicate that:
x1 = 0; x2 = 0; x3 = 0; x4 = 0; x5 = 0
which is obviously not correct. What have I done wrong?
| The 4th row of your matrix is missing a minus sign, i.e.
$$
\begin{pmatrix} 2 & 0 & -2 & 0 & 0 & 0 \\ 2 & 2 & -1 & 0 & -2 & 0 \\ 4 & 0 & -3 & -2 & 0 & 0 \\ 0 & 1 & 0 & -1 & 0 & 0 \\ 0 & 2 & 0 & 0 & -1 & 0 \end{pmatrix}
$$
$$ $$
But instead of forming an augmented matrix, you should drop that last column of zeroes, use the next-to-last column as a $b$-vector (equivalent to setting $x_5=1$), and solve for the remaining four coefficients from the linear system
$$
A\,x = b
$$
where
$$ \eqalign{
A &= \begin{pmatrix} 2 & 0 & -2 & 0 \\ 2 & 2 & -1 & 0 \\ 4 & 0 & -3 & -2 \\ 0 & 1 & 0 & -1 \\ 0 & 2 & 0 & 0 \end{pmatrix} \cr\cr
b &= \begin{pmatrix} 0 \\ 2 \\ 0 \\ 0 \\ 1 \end{pmatrix} \cr
}$$
This is an overdetermined system which can be solved as
$$ \eqalign{
x &= (A^TA)^{-1}A^T\,b \cr
&= A^+\,b \cr
}$$
Finally, you can multiply the x-vector by a scalar factor to round any fractional components to whole integers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1418988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to prove $\sum _{k=1}^{\infty} \frac{k-1}{2 k (1+k) (1+2 k)}=\log_e 8-2$? How to prove:
$$\sum _{k=1}^{\infty} \frac{k-1}{2 k (1+k) (1+2 k)}=\log_e 8-2$$
Is it possible to convert it into a finite integral?
| Given how popular these seem to be becoming, it might be worth learning a nice fact to eliminate the need for a lot of cleverness:
$$ \sum_{k=1}^n \frac{1}{k} = \log n + \gamma + O(1/n) $$
where $\gamma$ is Euler's constant.
Partial fractions tells us
$$\frac{k-1}{2k(k+1)(2k+1)} = \frac{3}{2k+1} - \frac{1}{2k} - \frac{1}{k+1} $$
We can use the trick
$$ \begin{align}\sum_{k=1}^n \frac{1}{2k-1}
&= \left(\sum_{k=1}^{2n} \frac{1}{k}\right) - \frac{1}{2} \left( \sum_{k=1}^n \frac{1}{k} \right)
\\&= (\log 2n + \gamma) - \frac{1}{2} (\log n + \gamma) + O(1/n)
\\&= \frac{1}{2}\log n + \frac{1}{2} \gamma + \log 2 + O(1/n)
\end{align}$$
and thus compute the partial sum of the three terms:
$$ \sum_{k=1}^n \frac{3}{2k+1} =
\frac{3}{2}\log(n+1) + \frac{3}{2} \gamma + 3\log 2 - \frac{3}{1} + O(1/n)$$
$$ \sum_{k=1}^n \frac{1}{2k} = \frac{1}{2} \log(n) + \frac{1}{2} \gamma + O(1/n) $$
$$ \sum_{k=1}^n \frac{1}{k+1} = \log(n+1) + \gamma - \frac{1}{1} + O(1/n) $$
combine them, then take the limit. We can even simplify that by noting $\log(n+1) = \log n + O(1/n)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1419166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Proof by induction for "sum-of" Prove that for all $n \ge 1$:
$$\sum_{k=1}^n \frac{1}{k(k+1)} = \frac{n}{n+1}$$
What I have done currently:
Proved that theorem holds for the base case where n=1.
Then:
Assume that $P(n)$ is true. Now to prove that $P(n+1)$ is true:
$$\sum_{k=1}^{n+1} \frac{1}{k(k+1)} = \frac{n}{n+1} + n+1$$
So:
$$\sum_{k=1}^{n+1} \frac{1}{k(k+1)} = \frac{n+1}{n+2}$$
However, how do I proceed from here?
| For n=1:
$$
\sum_{k=1}^{1} 1/(k(k+1))=1/2=1/(1+1)
$$
Let's prove it for $n+1$:
$$
\sum_{k=1}^{n+1} \frac{1}{k(k+1)}= \sum_{k=1}^{n}+\frac{1}{(n+1)(n+2)}\\
$$
But using our assumption for $P(n)$ :
$$
\sum_{k=1}^{n} \frac{1}{k(k+1)}=\frac{n}{n+1}
$$
So the sum will be:
$$
\sum_{k=1}^{n+1} \frac{1}{k(k+1)}=\frac{n}{(n+1)} + \frac{1}{(n+1)(n+2)}\\
=\frac{n(n+2)+1}{(n+1)(n+2)}\\
=\frac{n^2+2n+1}{(n+1)(n+2)}\\
=\frac{{(n+1)}^{2}}{(n+1)(n+2)}\\
=\frac{n+1}{n+2}
$$
Hence proved.
Note that you made a mistake in your last part where you wrote:
$$
\sum_{k=1}^{n+1} \frac{1}{k(k+1)}= \frac{n}{n+1}+n+1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1419749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Find $z$ when $z^4=-i$? Consider $z^4=-i$, find $z$.
I'd recall the fact that $z^n=r^n(\cos(n\theta)+(i\sin(n\theta))$
$\implies z^4=|z^4|(\cos(4\theta)+(i\sin(4\theta))$
$|z^4|=\sqrt{(-1)^2}=1$
$\implies z^4=(\cos(4\theta)+(i\sin(4\theta))$
$\cos(4\theta)=Re(z^4)=0 \iff \arccos(0)=4\theta =\frac{\pi}{2} \iff \theta=\frac{\pi}{8}$
Since $z^n=r^n\cdot e^{in\theta}$, $z^4$ can now be rewritten as $z^4=e^{i\cdot4\cdot\frac{\pi}{8}} \iff z=e^{i\frac{\pi}{8}}$ However, my answer file says this is wrong. Can anyone give me a hint on how to find $z$?
| Remember, that $-i = e^{i\frac{3\pi}{2}}$, then
$$ z^4 = -i = e^{i\frac{3\pi}{2}}$$
The n$^{\text{th}}$ root of a complex number is
$$z^n = re^{i\varphi} \implies z_k = r^{1/n}e^{i\left(\frac{\varphi}{n} + \frac{2\pi k}{n} \right)} \qquad \text{where} \quad k = 0, \dots, n-1$$
In your case:
$$z_0 = e^{i\left(\frac{3\pi}{8} + 0 \right)} = (r, \varphi_0) = \left(1, \frac{3 \pi}{8}\right)$$
$$z_1 = e^{i\left(\frac{3\pi}{8} + \frac{2\pi}{4} \right)} = (r, \varphi_1) = \left(1, \frac{7 \pi}{8}\right)$$
$$z_2 = e^{i\left(\frac{3\pi}{8} + \frac{4\pi}{4} \right)} = (r, \varphi_2) = \left(1, \frac{11 \pi}{8}\right)$$
$$z_3 = e^{i\left(\frac{3\pi}{8} + \frac{6\pi}{4} \right)} = (r, \varphi_3) = \left(1, \frac{15 \pi}{8}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1421839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Evaluation of $5\times 5$ determinant The following $5\times 5$ det. comes from a Russian book. I don't want to expand the det. rather than do some operations on it and extract the result.
Prove:
$$\begin{vmatrix}
-1 &1 &1 &1 &x \\
1& -1 &1 &1 &y \\
1& 1 & -1 & 1 &z \\
1& 1 & 1 & -1 & u\\
x& y & z & u &0
\end{vmatrix}= -4 [x^2+y^2+z^2 +u^2 - 2(xy+zx+zu+yz+yu+\color{red}{x}u)]$$
I have tried many things e.g $R_1 \leftrightarrow R_5+ x R_1$ and of course cyclic ($R_2 \leftrightarrow R_5+ y R_2$ etc) and then adding the last row to the second etc. resulting in:
$$\begin{vmatrix}
0 &y+x &z+x &u+x &x^2 \\
y+x&0 &z+y &u+y &y^2 \\
x+z& z+y & 0 &z+u & z^2\\
u+x& u+y &u+z & 0 &u^2 \\
x &y &z &u &0
\end{vmatrix}$$
but does not look that promising.
On the other hand if I add the the first column to the others I do get a lot of zeros, since the first row for example takes the form $-1, 0, 0, 0, x+1$. This looks more promising. But in either case I get stuck.
May I have some hints or answers?
Edit: The red letters as suggested by Peter.
| Let $Q$ be a real orthogonal matrix whose first column is $(\frac12,\frac12,\frac12,\frac12)^T$. Let also $v=(x,y,z,u)^T$ and $Q^Tv=(a,b,c,d)$. Then the matrix in question is equal to $\pmatrix{Q&0\\ 0&1}A\pmatrix{Q^T&0\\ 0&1}$, where
$$
A=\left[\begin{array}{cccc|c}
2&&&&a\\ &-2&&&b\\ &&-2&&c\\ &&&-2&d\\
\hline
a&b&c&d&0
\end{array}\right].
$$
Hence the required determinant is equal to $\det(A)$. Let $e_1=(1,0,0,0)^T$ and $E\in M_4(\mathbb R)$ be the all-one matrix. By Laplace expansion along the last column of $A$, we see that
$$
\det(A)=8(a^2-b^2-c^2-d^2).
$$
Now we can rewrite this in terms of $x,y,z$ and $u$:
\begin{align}
\det(A)&=8(a^2-b^2-c^2-d^2)\\
&=16a^2-8(a^2+b^2+c^2+d^2)\\
&=16(v^TQe_1)(e_1Q^Tv)-8(v^TQ)(Q^Tv)\\
&=16v^T\left(\frac14E\right)v-8v^Tv\\
&=4(x+y+z+u)^2 - 8(x^2+y^2+z^2+u^2).
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1422846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Another messy integral: $I=\int \frac{\sqrt{2-x-x^2}}{x^2}\ dx$ I found the following question in a practice book of integration:-
$Q.$ Evaluate $$I=\int \frac{\sqrt{2-x-x^2}}{x^2}\ dx$$ For this I substituted $t^2=\frac {2-x-x^2}{x^2}\implies x^2=\frac{2-x}{1+t^2}\implies 2t\ dt=\left(-\frac4{x^3}+\frac 1{x^2}\right)\ dx$. Therefore $$\begin{align}I&=\int\frac {\sqrt{2-x-x^2}}{x^2}\ dx\\&=\int \left(\frac tx\right)\left(\frac{2t\ dt}{-\frac4{x^3}+\frac 1{x^2}}\right)\\&=\int \frac{2t^2\ dt}{\frac{x-4}{x^2}}\\&=\int \frac{2t^2(1+t^2)\ dt}{{x-4}\over{2-x}}\\&=\int \frac{2t^2(1+t^2)(5+4t^2-\sqrt{8t^2+9})\ dt}{\sqrt{8t^2+9}-(8t^2+9)}\end{align}$$ Now I substituted $8t^2+9=z^2 \implies t^2=\frac {z^2-9}8 \implies 2t\ dt=z/4\ dz$. So, after some simplification, you get $$\begin{align}I&=-\frac1{512}\int (z^2-9)(z+1)(z-1)^2\ dz\end{align}$$ I didn't have the patience to solve this integration after all these substitutions knowing that it can be done (I think I have made a mistake somewhere but I can't find it. There has to be an $ln(...)$ term, I believe). Is there an easier way to do this integral, something that would also strike the mind quickly? I have already tried the Euler substitutions but that is also messy.
| Let $$\displaystyle I = \int \frac{\sqrt{2-x-x^2}}{x^2}dx = \int\frac{\sqrt{(x+2)(1-x)}}{x^2}dx$$
Now Let $\displaystyle (x+2) = (1-x)t^2\;,$ Then $\displaystyle x = \frac{t^2-2}{t^2+1} = 1-\frac{3}{t^2+1}$
So $$\displaystyle dx = \frac{6t}{(t^2+1)^2}dt$$ and $$\displaystyle (1-x) = \frac{3}{t^2+1}$$
So Integral $$\displaystyle I = 18\int\frac{t^2}{(t^2+1)\cdot (t^2-2)^2}dt = 6\int\frac{(t^2+1)-(t^2-2)}{(t^2+1)\cdot (t^2-2)^2}dt$$
so we get $$\displaystyle I = 6\int\frac{1}{(t^2-2)^2}dt-6\int\frac{1}{(t^2+1)(t^2-2)}dt$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1423103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Value of $\displaystyle \frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}$
Given
$$
\tag1\frac{1}{\omega+a}+\frac{1}{\omega+b}+\frac{1}{\omega+c} = 2\omega^2
$$
and
$$
\tag2\frac{1}{\omega^2+a}+\frac{1}{\omega^2+b}+\frac{1}{\omega^2+c} = 2\omega
$$
what is the value of $\displaystyle \frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}$, where $\omega$ is a complex cube root of unity?
My Attempt:
Multiply $(1)$ by $\omega$ to get
$$
\frac{\omega}{\omega+a}+\frac{\omega}{\omega+b}+\frac{\omega}{\omega+c} = 2\omega^3 = 2
$$
(because $\omega^3 = 1$).
After simplification , we get
$$
\tag3\frac{a}{\omega+a}+\frac{b}{\omega+b}+\frac{c}{\omega+c} = 1
$$
Now multiply $(2)$ by $\omega^2$ to get
$$
\frac{\omega^2}{\omega^2+a}+\frac{\omega^2}{\omega^2+b}+\frac{\omega^2}{\omega^2+c} = 2\omega^3 = 2
$$
After simplification , we get
$$
\tag4\frac{a}{\omega^2+a}+\frac{b}{\omega^2+b}+\frac{c}{\omega^2+c} = 1
$$
From $(3)$ and $(4)$ we can form a quadratic equation whose roots are $x\in\left\{\omega,\omega^2\right\}$ as
$$
\frac{a}{x+a}+\frac{b}{x+b}+\frac{c}{x+c} = 1\\
\begin{align}
a\left[x^2+(b+c)x+bc\right]+b\left[x^2+(a+c)x+ac\right]&+c\left[x^2+(a+b)x+ab\right]\\
&=(x+a)(x+b)(x+c)\\
(a+b+c)x^2+2(ab+bc+ca)x+3abc&=x^3+(a+b+c)x^2+(ab+bc+ca)x+abc\\
x^3-(ab+bc+ca)x-2abc&=0
\end{align}
$$
How can I solve the problem from this point?
| This turned out to be a cubic. There are three roots. Two of them are $x=\omega$ and $x=\omega^2$.
HINT: Do you know how to find the sum of the roots of a polynomial, without solving the polynomial itself?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1423801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Conjecture $\Re\,\operatorname{Li}_2\left(\frac12+\frac i6\right)=\frac{7\pi^2}{48}-\frac13\arctan^22-\frac16\arctan^23-\frac18\ln^2(\tfrac{18}5)$ I numerically discovered the following conjecture:
$$\Re\,\operatorname{Li}_2\left(\frac12+\frac i6\right)\stackrel{\color{gray}?}=\frac{7\pi^2}{48}-\frac{\arctan^22}3-\frac{\arctan^23}6-\frac18\ln^2\!\left(\frac{18}5\right).$$
It holds numerically with a precision of more than $30000$ decimal digits.
Could you suggest any ideas how to prove it?
Can we find a closed form for $\Im\,\operatorname{Li}_2\left(\frac12+\frac i6\right)$?
Is there a general method to find closed forms of expressions of the form $\Re\,\operatorname{Li}_2(p+iq)$, $\Im\,\operatorname{Li}_2(p+iq)$ for $p,q\in\mathbb Q$?
| First of all we know that:
$$
\operatorname{Li}_2(z) = -\operatorname{Li}_2\left(\frac{z}{z-1}\right)-\frac{1}{2}\ln^2(1-z), \quad z \notin (1,\infty).\tag{$\diamondsuit$}
$$
Furthermore, we have the following relationship between the dilogarithm and the Clausen functions:
$$\operatorname{Li}_2\left(e^{i\theta}\right) = \operatorname{Sl}_2(\theta)+i\operatorname{Cl}_2(\theta), \quad \theta \in [0,2\pi).\tag{$\heartsuit$}$$
where $\operatorname{Cl}_2$ and $\operatorname{Sl}_2$ are the standard Clausen functions, defined as:
$$\begin{align}
\operatorname{Cl}_2(\theta) &= \sum_{k=1}^{\infty}\frac{\sin(k\theta)}{k^2}, \\
\operatorname{Sl}_2(\theta) &= \sum_{k=1}^{\infty}\frac{\cos(k\theta)}{k^2}.
\end{align}$$
Using the relationship between SL-type Clausen functions and Bernoulli polynomials, we have that
$$
\operatorname{Sl}_2(\theta) = \frac{\pi^2}{6}-\frac{\pi\theta}{2}+\frac{\theta^2}{4}, \quad \theta \in [0,2\pi).\tag{$\spadesuit$}
$$
Now let $z:=\tfrac{1}{2}+\tfrac{i}{6}$. Because $\left|\tfrac{z}{z-1}\right| = 1$, the equation
$$
e^{i\theta} = \frac{z}{z-1} = -\frac{4}{5}-\frac{3}{5}i,
$$
has the only solution $\theta = \arctan\left(\tfrac{3}{4}\right) + \pi$ in $[0,2\pi)$.
Because of $(\diamondsuit)$ and $(\heartsuit)$ we have
$$
\operatorname{Li}_2(z) = -\color{red}{\operatorname{Sl}_2(\theta)} - i \color{green}{\operatorname{Cl}_2(\theta)} - \color{blue}{\frac{1}{2}\ln^2(1-z)},
$$
for $z=\tfrac{1}{2}+\tfrac{i}{6}$ and $\theta = \arctan\left(\tfrac{3}{4}\right) + \pi$.
For the logarithm term we get
$$
\Re{\left[\color{blue}{\frac{1}{2}\ln^2(1-z)}\right]} = \frac{1}{8}\left(\ln^2\left(\frac{18}{5}\right)-(\pi-2\arctan 3)^2\right)
$$
and
$$
\Im{\left[\color{blue}{\frac{1}{2}\ln^2(1-z)}\right]} = \frac{1}{4}\ln\left(\frac{18}{5}\right)(\pi-2\arctan 3).
$$
We know that $\color{red}{\operatorname{Sl}_2(\theta)}$ and $\color{green}{\operatorname{Cl}_2(\theta)}$ are real quantities. By using $(\spadesuit)$ for the SL-type Clausen term we get
$$
\color{red}{\operatorname{Sl}_2(\theta)} = \frac{\pi^2}{12}-\frac{1}{4}\arctan^2\left(\frac{3}{4}\right).
$$
Now we could obtain your conjectured closed-form:
$$
\Re\left[\operatorname{Li}_2(z)\right] = -\color{red}{\operatorname{Sl}_2(\theta)} - \Re{\left[\color{blue}{\frac{1}{2}\ln^2(1-z)}\right]} = \frac{7\pi^2}{48}-\frac{\arctan^22}3-\frac{\arctan^23}6-\frac18\ln^2\!\left(\frac{18}5\right).
$$
For the imaginary part we have
$$\begin{align}
\Im\left[\operatorname{Li}_2(z)\right] &= -\color{green}{\operatorname{Cl}_2(\theta)} - \Im{\left[\color{blue}{\frac{1}{2}\ln^2(1-z)}\right]} \\ &= -\operatorname{Cl}_2\left(\arctan\left(\frac{3}{4}\right)+\pi\right)-\frac{1}{4}\ln\left(\frac{18}{5}\right)(\pi-2\arctan 3).
\end{align}$$
By using $(\diamondsuit), (\heartsuit)$ and $(\spadesuit)$, you could generalize this process for all $z \in \mathbb{C}$ such that $\left|\frac{z}{z-1}\right|=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1424600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 3,
"answer_id": 2
} |
Evaluating $\int \frac{dx}{x^3+x+1}$ $$\int \frac{dx}{x^3+x+1}$$
I have no idea how to solve this. How do I evaluate it?
Any advice, hint or well-thought solution will be appreciated.
| Let $\zeta_1,\zeta_2,\zeta_3$ be the roots of $x^3+x+1$. We have:
$$ \text{Res}\left(\frac{1}{z^3+z+1},z=\zeta_i\right) = \frac{1}{3\zeta_i^2+1},\tag{1}$$
hence:
$$ \frac{1}{z^3+z+1}=\sum_{i=1}^{3}\frac{1}{3\zeta_i^2+1}\cdot\frac{1}{z-\zeta_i} \tag{2} $$
and:
$$ \int \frac{dz}{z^3+z+1} = C+\sum_{i=1}^{3}\frac{\log(z-\zeta_i)}{3\zeta_i^2+1}.\tag{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1427163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the matrix A
Find the matrix A that has two rows and two columns and has
$$A\pmatrix{1\\ 1}=\pmatrix{2\\ 1}\text{ and }A\pmatrix{-1\\1}=\pmatrix{1\\-1}.$$
Question:
How do I write out the corresponding system of linear equations with $4$ equations and $4$ unknowns so I can determine $A$?
| If you want to write a system of linear equations, then you can write $$A=\begin{pmatrix}a & b\\ c&d\end{pmatrix}$$
and then see that $$A\begin{pmatrix}1 \\ 1\end{pmatrix}=\begin{pmatrix}a & b\\ c&d\end{pmatrix}\begin{pmatrix}1 \\ 1\end{pmatrix} = \begin{pmatrix}a+b \\ c+d\end{pmatrix}$$
so that gives you the first two equations $a+b=2$ and $c+d=1$.
You do similarly for the other vector and you will have four equations.
If you learned linear algebra before, you may also take an alternative approach. Remember that matrices are really just representations of linear operators given a basis. If you do not know what that means, don't worry. You will probably learn that soon.
You can express $\begin{pmatrix}2 \\ 1\end{pmatrix} = \alpha \begin{pmatrix}1 \\ 1\end{pmatrix} + \beta\begin{pmatrix}-1 \\ 1\end{pmatrix}$
and $\begin{pmatrix}1 \\ -1\end{pmatrix} = \gamma \begin{pmatrix}1 \\ 1\end{pmatrix} + \delta \begin{pmatrix}-1 \\ 1\end{pmatrix}$
and then you can conclude that in the basis $$\left\{\begin{pmatrix}1 \\ 1\end{pmatrix},\begin{pmatrix}-1 \\ 1\end{pmatrix}\right\}$$
the linear operator $\mathcal A$, which in the standard basis is represented by $A$, is represented by a matrix $A' = \begin{pmatrix}\alpha&\beta1 \\ \gamma & \delta\end{pmatrix}$
Then, finding $A$ is just a matter of changing the basis for $\mathcal A$, since $A$ is, by definition, the representation of $\mathcal A$ in the standard basis.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1427938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Inequation: quadratic difference equations Given:
$$\frac{(x - 3)}{(x-4)} > \frac{(x + 4)}{(x + 3)}$$
Step 1:
$$(x + 3)(x - 3) > (x + 4)(x - 4)$$
Step2 : Solving step 1:
$$x^2 - 3^2 > x^2 - 4^2$$
*Step 3:
$ 0 > -16 + 9$ ???
As you see, I can delete the $x^2$, but there is no point in doing that.
What should be the next step?
| One way to do this more carefully is to take $$\frac {x-3}{x-4}-\frac {x+4}{x+3}=\frac 7{(x-4)(x+3)}\gt 0$$ and this is clearly true iff $(x-4)(x+3)\gt 0$
To do this more formally, note that from $a\gt 0$ we have $a^2\gt 0$ and we can deduce $\frac {a}{a^2}=\frac 1a\gt 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1428703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
Where I'm wrong? Why my answer is different from the book? The question is to integrate
$$\int x\cos^{-1}x dx$$
Answer in my book
$$(2x^2-1)\frac{\cos^{-1}x}{4}-\frac{x}{4}\sqrt{1-x^2}+C$$
I'm learning single variable calculus right now and at current about integration with part. I'm confused in a problem from sometime. I don't know where I am wrong. Please have a look at the images.
Solution. $\newcommand{\dd}{\; \mathrm{d}}$
$$
\begin{align}
&\cos^{-1} x \int x \dd x - \int \left[\frac{-1}{\sqrt{1-x^2}}\cdot\frac{x^2}2\right] \dd x=\\
=&\cos^{-1}x \cdot \frac{x^2}2 + \int \frac{x^2}{2\sqrt{1-x^2}} \dd x2=\\
=&\frac{x^2}2 \cdot \cos^{-1}x + \frac12 \int \frac{x}{\sqrt{1-x^2}} \dd x =
\begin{vmatrix}x=\sin t\\ \dd x=\cos t\dd t\end{vmatrix} = \\
=&\frac{x^2\cos^{-1}x}2 + \frac12 \int \frac{\sin^2t \cos t}{\sqrt{1-\sin^2t}} \dd t =\\
=&\frac{x^2\cos^{-1}x}2 + \frac12 \int \frac{\sin^2t \cos t}{\cos t} \dd t =\\
=&\frac{x^2\cos^{-1}x}2 + \frac12 \int \sin^2t \dd t =\\
=&\frac{x^2\cos^{-1}x}2 + \frac12 \int \frac{1-\cos2t}2 \dd t =\\
=&\frac{x^2\cos^{-1}x}2 + \frac12 \int \frac12 \dd t -\frac 14 \int \cos2t \dd t =\\
=&\frac{x^2\cos^{-1}x}2 + \frac14 t - \frac18 \sin 2t + C =\\
=&\frac{x^2\cos^{-1}x}2 + \frac14 \sin^{-1}x - \frac18 \sin2\sin^{-1}x + C =\\
=&\frac{x^2\cos^{-1}x}2 + \frac14 \sin^{-1}x - \frac18 \sin(2\sin^{-1}x) + C
\end{align}
$$
Please help. Thankyou in advance.
| Your answer is not false, but you didn't thoroughly simplify.
This is why both answers are the same:
*
*$\arcsin x+\arccos x=\dfrac\pi2$
*$\sin(2\arcsin x)=2\sin(\arcsin x)\cos(\arcsin x)=2x\sqrt{1-x^2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1430888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Computing $E(XY)$ for finding $Cov(X,Y)$ Consider tossing a cubic die once and let $n$ be the smallest number of dots that appear on top. Define two random variables $X$ and $Y$ such that:
*
*$X=1$ if $n \in \left \{1,2 \right \}$, $X=2$ if $n \in \left \{3,4 \right \}$ and $X=3$ if $n \in \left \{5,6 \right \}$ and
*$Y=0$ if $n \in \left \{3,6 \right \}$, $Y=1$ if $n \in \left \{1,4 \right \}$, and $Y=2$ if $n \in \left \{2,5 \right \}$
Find the $cov(X,Y)$.
The support of $X$ is $\left \{1,2,3 \right \}$ and the support of $Y$ is $\left \{0,1,2 \right \}$.
The marginal probabilities are $f_{x}(X=1)=\frac{1}{3}$, $f_{x}(X=2)=\frac{1}{3}$, and $f_{x}(X=3)=\frac{1}{3}$ and the same for $Y=\left \{ 0,1,2 \right \}$
I have found $E(Y)=\frac{1}{3} (0) + \frac{1}{3} (1) + \frac{1}{3} (2) = 1$ and $E(X)=2$.
The formula for $Cov(X,Y) = E[(X-E(X))(Y-E(Y))]$, so I have $$Cov(X,Y) = E[(X-2)(Y-1)]=E[XY - X - 2Y +2]=E[XY]-E[X]-2E[Y]+2= E[XY] -2-2+2= E[XY] - 2.$$
EDIT: I was stuck on the $E[XY]$ part but have an answer now.
| I think I've figured it out. The support of $E(XY) = \left \{0,1,2,3,4,6 \right \}$.
So $E(XY) = \frac{1}{3} (0) + \frac{1}{3} (1) +\frac{1}{3} (2) + \frac{1}{3} (3) + 0 + 0 = 2$
So $Cov(X,Y) = 2-2=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1433132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Simple limit of a sequence
Need to solve this very simple limit $$ \lim _{x\to \infty
\:}\left(\sqrt[3]{3x^2+4x+1}-\sqrt[3]{3x^2+9x+2}\right) $$
I know how to solve these limits: by using
$a−b= \frac{a^3−b^3}{a^2+ab+b^2}$. The problem is that the standard way (not by using L'Hospital's rule) to solve this limit - very tedious, boring and tiring. I hope there is some artful and elegant solution. Thank you!
| You can use the binomial theorem to expand this.
$$\lim _{x\to \infty
\:}\left(\sqrt[3]{3x^2+4x+1}-\sqrt[3]{3x^2+9x+2}\right)$$
Here 1 and 2 are the smallest terms and they can be ignored.
\begin{align}
&\lim _{x\to \infty \:}\left(\sqrt[3]{3x^2+4x+1}-\sqrt[3]{3x^2+9x+2}\right)
\\=& \lim _{x\to \infty \:}\left(\sqrt[3]{3x^2+4x}-\sqrt[3]{3x^2+9x}\right)
\\=& \lim _{x\to \infty \:}\left(\sqrt[3]{3x^2\left(1+\frac{4}{3x}\right)}-\sqrt[3]{3x^2\left(1+\frac3x\right)}\right)
\\=& \lim _{x\to \infty \:}\sqrt[3]{3x^2}\left(\sqrt[3]{1+\frac{4}{3x}}-\sqrt[3]{1+\frac3x}\right)
\\=& \lim _{x\to \infty \:}\sqrt[3]{3x^2}\left({1+\frac{4}{9x}}-{1-\frac1x}\right)
\\=& \lim _{x\to \infty \:}\sqrt[3]{3x^2}\left({\frac{4-9}{9x}}\right)
\\=& \lim _{x\to \infty \:}\sqrt[3]{3}x^{2/3}\left({\frac{-5}{9x}}\right)
\\=& \lim _{x\to \infty \:}\sqrt[3]{3}x^{-1/3}\left({\frac{-5}{9}}\right)
\\=& \,0
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1433216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
Evaluate $\int \tan^6x\sec^3x \ \mathrm{d}x$ Integrate $$\int \tan^6x\sec^3x \ \mathrm{d}x$$
I tried to split integral to $$\tan^6x\sec^2x\sec x$$ but no luck for me. Help thanks
| You can use the substitution $u = \sec + \tan$.
Note that $\dfrac 1u = \sec - \tan$, because $(\sec + \tan)(\sec - \tan) = \sec^2 - \tan^2 = 1$.
We have:
$$u^2 + 1 = [(\sec + \tan)^2] + [1] \\
= [\sec^2 + 2 \sec \tan + \tan^2] + [\sec^2 - \tan^2]\\
= 2\sec^2 + 2 \sec \tan \\
= \sec[2(\sec + \tan)] \\
= \sec(2u) \\
\implies \dfrac {(u^2 + 1)}{2u} = \sec
$$
Also:
$$
\begin{align}
u^2 - 1 &= [(\sec + \tan)^2] - [1]\\
&= [\sec^2 + 2 \sec \tan + \tan^2] - [\sec^2 - \tan^2]\\
&= 2\tan^2 + 2 \sec \tan\\
&= \tan[2(\sec + \tan)]\\
&= \tan(2u)\\
\end{align}
$$
$$\implies \dfrac {u^2 - 1}{2u} = \tan$$
Finally:
$$
\begin{align}
du &= (\sec + \tan)' dx \\
&= (\sec\tan + \sec^2) dx\\
&= \sec(\sec + \tan) dx\\
&= \dfrac {u^2 + 1}{2u}u dx\\
&= (u^2 + 1)\dfrac {dx}2\\
\end{align}
$$
$$\implies dx =\dfrac { 2du }{(u^2 + 1)}$$
So in our integral:
$$\sec \to \dfrac {u^2 + 1}{2u}\\
\tan \to \dfrac {u^2 - 1}{2u}\\
dx \to \dfrac { 2}{(u^2 + 1)} du
$$
This technique gives us a series of powers of $u$
for any integrand $\tan^m\sec^{n+1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1435755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Use the inclusion-exclusion principle to determine Use the inclusion-exclusion principle to determine :
(a) the number of ways there are to choose nineteen balls (identical apart from their colour) from a pile of red, blue, yellow and green balls if there have to be at most seven balls of each colour?
(b) the number of arrangements of length 9 of the numbers 1, 2, 3 and 4 (repetitions allowed) in which each number occurs at least once.
I have no idea how to approach part (a). For part (b), I think that the sets, say Ai, should be set up to be arrangements that do not have 1 or 2 or 3 or 4. Example : A1 be the set of length 9 of the numbers 2,3,4 and A2 be the set of length 9 of the numbers 1,3,4 and so on. Would this be correct and, if it is, how would I proceed with it?
| $a)$
How many ways are there to select them so that two colors have more than $7$ balls? There are $\binom{4}{2}=6$ ways to choose the two colors. After that we can go ahead and take out $8$ balls of each of those colors. We must now select $19-16=3$ balls out of $4$ colors. By stars and bars there are $\binom{6}{3}=20$ ways to do this. So $6\cdot20=120$ ways in total.
How many ways are there to select them so that exactly one color has more than $7$ balls? First select the color in $4$ ways,let us assume it is red. After this take out $8$ balls of that color. We must now select $19-8=11$ balls out of $4$ colors, by stars and bars there are $\binom{14}{3}=364$ ways to do so. However out of those $364$ ways there are $20$ ways in which red and blue appear more than seven times, $20$ ways in which red and green appear more than $7$ times and $20$ ways in which red and yellow appear more than $7$ times. Therefore there are $364-3\cdot20=304$ ways in which only red is repeated. Therefore there are $4\cdot304=1216$ colorings in which exactly one color appears more than seven times.
Once we have this we can count the number of colorings in which no color apears more than $7$ times. This is because if we are allowed to color freely then by stars and bars there are $\binom{22}{3}=1540$ colorings. Therefore there are $1540-120-1216=204$ coloring in which no color appears more than $7$ times.
$b)$
This one is easier, we solve for an arbitrary length $n$ of words, where $n\geq 4$:
There are $1+1+1+1$ sequences with exactly one number.
How many sequences have exactly $2$ numbers? there are $6$ ways to choose the numbers and then $2^{n}-2$ sequences, so $6(2^n-2)$ total.
How many sequences have exactly $3$ numbers? There are $4$ ways to choose them. After this there are $3^n$ sequences with those $3$ numbers, out of these $3$ use exactly $1$ number and $3(2^n-2)$ use exactly $2$ numbers. Hence $3^n-3\cdot2^n-3$ use all of those three numbers. So $4(3^n-3\cdot2^n+3)$ total.
Therefore there are $4^n-4-6(2^n-2)-4(3^n-3\cdot2^n+3)$ words that use all four numbers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1436598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
When is $f(y)$ prime? Find all prime numbers that can be expressed as :
$f(y)=y^{2015}+y+1$, where $y$ is a natural number. $y=1$ gives us 3, but how do we find others or prove that there can be no other??
| Hint
Use the fact that
$$x^{2015}+x+1 = x^{2015}-x^2+(x^2+x+1)=x^2(x^{3\times 671}-1)+(x^2+x+1) = x^2(x^{3}-1)(x^{3\times 670}+x^{3\times 669}+\dots+x^3+1)+(x^2+x+1) = (x^2+x+1)(x^2(x-1)A+1)$$
where $A = x^{3\times 670}+x^{3\times 669}+\dots+x^3+1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1437873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Rearange an expression I am learning induction. At one step I have to show that:
$$ 1 - \frac{1}{(1+n)} + \frac{1}{(n+1)(n+2)} $$
can be transformed to
$$ 1 - \frac{1}{n+2} $$
Theese are the steps for the transformation, but I cant understand them:
$$ 1 - \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} = 1 - \frac{(n+2) -1}{(n+1)(n+2)} = 1 - \frac{n+1}{(n+1)(n+2)} = 1 - \frac{1}{n+2} $$
Could you explain to me what happens there?
| $$\begin{align}1-\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}&=1-\frac{n+2}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)}\\&=1+\frac{-(n+2)}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)}\\&=1+\frac{-(n+2)+1}{(n+1)(n+2)}\\&=1+\frac{-(n+1)}{(n+1)(n+2)}\\&=1+\frac{-1}{n+2}\\&=1-\frac{1}{n+2}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1438393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Verify Mean Value Theorem for $\,f\left(x\right)=x^3\,$ I am trying to verify the Mean-Value Theorem for $\,f\left(x\right)=x^3.\,$ So far I have:
$$f'\left(c\right)=\frac{b^3-a^3}{b-a}=b^2+ab+a^2=3c^2$$
Question: since we know that $b>a$, how can we show that at least one solutions of $c$ lies between $a$ and $b$?
| I could not find a simpler solution than the following one.
Mean Value Theorem: There exists $c\in (a,b)$ such that $3c^2=a^2+ab+b^2$.
Denote $c_1 = \sqrt{\frac{a^2+ab+b^2}{3}} > 0$ and $c_2 = -\sqrt{\frac{a^2+ab+b^2}{3}} < 0$. We need to prove that: at least one of the following two inequalities holds:
$$(1) \quad a<c_1<b$$
$$(2)\quad a<c_2<b$$
Consider 3 cases:
*
*$0\le a <b$: In this case $(1)$ is true because $c_1^2-a^2 = \frac{1}{3}(b-a)(b+2a) >0$ and $b^2-c_1^2 = \frac{1}{3}(b-a)(2b+a) >0$.
*$a < b \le 0$: Similarly, in this case $(2)$ is true because $a^2-c_2^2 >0$ and $c_2^2-b^2 >0$.
*$a\le 0 <b$: in this case $a<c_1$ and $c_2<b$ are already true, we only need to show that at least one of the following two are true: $c_1<b$ or $c_2>a$, or equivalently: $c_1^2<b^2$ or $c_2^2 < a^2$. This is easy by contradiction: if $c_1^2 \ge b^2$ AND $c_2^2 \ge a^2$ then $c_1^2+c_2^2 \ge a^2+b^2 \Leftrightarrow 2(a^2+ab+b^2)\ge 3(a^2+b^2) \Leftrightarrow 0 \ge (a-b)^2$, obviously false.
We are done :)
P/s: a remark for those who are trying to apply the intermediate value theorem:
If we denote $g(x) = 3x^2 - (a^2+ab+b^2)$ then we have $g(a)g(b) = -(a-b)^2(2a+b)(2b+a)$, not always negative.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1439213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Condition for quartic polynomial coefficients given at least one real root
Find the minimum possible value of $a^2+b^2$ where $a$, $b$ are two real numbers such that the polynomial
$$x^4+ax^3+bx^2+ax+1,$$
has at least one real root.
My attempt: Let p be a real root.
Therefore $p^4 + a(p^3) + b(p^2) + ap + 1 = 0$.
Divide both sides by $p^2$, noting that $p$ cannot be zero since $P(0) = 1$, we get
$$p^2 + ap + b + a/p + 1/(p^2) = 0.$$
I rearranged this into a quadratic; i.e.
$$(p+1/p)^2 + a(p+1/p) + (b-2) = 0.$$
If $p$ is real, $p+1/p$ is real, so discriminant is non-negative.
setting discriminant $\geq0$, I get:
$$a^2 - 4(b-2)\geq0,$$
$$a^2 \geq 4b-8,$$
$$a^2 + b^2 \geq b^2 + 4b - 8 = (b+2)^2 - 12 \geq -12$$
But this is useless because it is obvious $a^2 + b^2 \geq 0$.
I believe this is because I also need to use the fact that $|p+1/p| \geq 2$, however I am unsure how to use this inequality with the discriminant.
Thanks in advance
| Replacing $a$ with $-a$ only changes the signs of the zeros, so w.l.o.g. we can assume that $a\ge0$. The quadratic with the unknown $p+1/p$ (nice trick, BTW!) that you derived gives
$$
p+\frac1p=\frac{-a\pm\sqrt{a^2-4(b-2)}}2.
$$
Given the assumption $a\ge0$ we see that of these two alternatives the solution with a minus sign gives the larger value to $|p+1/p|$. Therefore the condition $|p+1/p|\ge2$, now rewritten as $p+1/p\le-2$, gives us another constraint
$$
-a-\sqrt{a^2-4(b-2)}\le-4.
$$
This is equivalent to
$$
4-a\le\sqrt{a^2-4(b-2)}.\qquad(*)
$$
We shall see that there are solutions with $a^2+b^2<16$, so w.l.o.g. we can further assume that $a<4$, so $4-a>0$ and we can square both sides of $(*)$ arriving at
$$
16-8a+a^2\le a^2-4b+8\implies a\ge (b+2)/2.
$$
Again, later developments will reveal that $b+2$ must be positive at the sought minimium, so to minimize $a^2$ we must have equality here, i.e. $a=(b+2)/2$. Thus we really want to minimize
$$
a^2+b^2=\frac14[(b+2)^2+4b^2]=\frac14[5b^2+4b+4].
$$
It is trivial to show that this has a minimum at $b=-2/5$. The corresponding
$a=(b+2)/2=4/5$, and at this point we have
$$
a^2+b^2=\frac{4^2+2^2}{25}=\frac45.
$$
If either the assumption $4-a\ge0$ or the assumption $b+2\ge0$ (that I made while looking for this candidate point) were invalid, then clearly $a^2+b^2$ would have a larger value, so we can dismiss those possibilities.
As a final check we see that when $a=4/5, b=-2/5$ your polynomial has a double root at $x=-1$, not unexpectedly matching with $p+1/p=-2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1439501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
If $8$ does not divide $x^2-1$, then $x$ is even; prove by contrapositive If $8$ does not divide $x^2-1$, then $x$ is even
proof by contrapositive
the contrapositive of this is :
if $x$ is odd, then $8$ divides $x^2-1$
proof by contrapositive:
Assume $x$ is odd
by definition of odd $∃k∈ℤ$ such that $x=2k+1$
Well, $x^2-1$ = $(2k+1)^2-1$
= $4k^2+4k$
= $4k(k+1)$
Therefore, $k(k+1)$ is an even integer
( i also know the definition of division : $∃m∈ℤ$ such that $x^2-1$ = $8m$)
$x^2-1$ = $4 · 2 · k(k+1)/2$
Therefore,
$8(k^2+1)/2$ is divisible by 8
| Rewrite $$(x^2-1)=(x-1)(x+1)$$ If 8 does not divide $(x^2-1)=(x-1)(x+1)$, then 2 does not devide $x^2-1=(x-1)(x+1)$. But that is the same as saying that neither $x-1$ nor $x+1$ are even. Hence $x$ must be even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1439872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If $\frac{x^2+y^2}{x+y}=4$, then what are the possible values of $x-y$?
If $\frac{x^2+y^2}{x+y}=4$,then all possible values of $(x-y)$ are given by
$(A)\left[-2\sqrt2,2\sqrt2\right]\hspace{1cm}(B)\left\{-4,4\right\}\hspace{1cm}(C)\left[-4,4\right]\hspace{1cm}(D)\left[-2,2\right]$
I tried this question.
$\frac{x^2+y^2}{x+y}=4\Rightarrow x+y-\frac{2xy}{x+y}=4\Rightarrow x+y=\frac{2xy}{x+y}+4$
$x-y=\sqrt{(\frac{2xy}{x+y}+4)^2-4xy}$, but I am not able to proceed. I am stuck here. Is my method wrong?
| The condition $\frac{x^2+y^2}{x+y}=4$ is equivalent to $(x+y)^2+(x-y)^2=8(x+y)$. Let $x+y=s$ and $x-y=d$. Note that this induces a bijection from $\Bbb R^2$ to itself, meaning that for every pair $(s,d)$ there exist corresponding $x,y$. We have $0\leq d^2=8s-s^2$. The nonnegative values that $8s-s^2$ can take are $[0,16]$, so $d$ takes values in $[-4,4]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1443441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Solving $\lim _{x\to 1}\left(\frac{1-\sqrt[3]{4-3x}}{x-1}\right)$ $$\lim _{x\to 1}\left(\frac{1-\sqrt[3]{4-3x}}{x-1}\right)$$
So
$$\frac{1-\sqrt[3]{4-3x}}{x-1} \cdot \frac{1+\sqrt[3]{4-3x}}{1+\sqrt[3]{4-3x}}$$
Then
$$\frac{1-(4-3x)}{(x-1)(1+\sqrt[3]{4-3x})}$$
That's
$$\frac{3\cdot \color{red}{(x-1)}}{\color{red}{(x-1)}(1+\sqrt[3]{4-3x})}$$
Finally
$$\frac{3}{(1+\sqrt[3]{4-3x})}$$
But this evaluates to
$$\frac{3}{2}$$
When the answer should be
$$1$$
Where did I fail?
| the problem is that you used
(a-b)(a+b) to simplify a cubic root instead of a simple root
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1447217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
How do I solve $\lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x^2}}{\sqrt{1+x}-1}$ indeterminate limit without the L'hospital rule? I've been trying to solve this limit without L'Hospital's rule because I don't know how to use derivates yet. So I tried rationalizing the denominator and numerator but it didn't work.
$\lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x^2}}{\sqrt{1+x}-1}$
What is wrong with $\lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x^2}-1+1}{\sqrt{1+x}-1} = \lim_{x \to 0} \frac{\sqrt{1+x}-1}{\sqrt{1+x}-1} + \lim_{x \to 0} \frac{1-\sqrt{1-x^2}}{\sqrt{1+x}-1}$ = 1 + DIV?
| HINT 1:
$$\begin{align}
\sqrt{1+x}-\sqrt{1-x^2}&=\sqrt{1+x}\left(1-\sqrt{1-x}\right)\\\\
&=\sqrt{1+x}\,\,\left(\frac{x}{1+\sqrt{1-x}}\right)
\end{align}$$
HINT 2:
$$\begin{align}
\frac{1}{\sqrt{1+x}-1}&=\frac{\sqrt{1+x}+1}{x}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1449739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Permutation: Distribute 10 distinct items in 3 boxes: one box contain odd number, one box even number object , and all boxes at least one item.
I wish to distribute $10$ distinct toys to my children $A, B$ and $C$. Each child must get at least one toy, but $A$ must receive an even number of toys, while $C$ must receive an odd number.
How many ways can I go about my distribution?
It's from an exam question. I had stumbled upon it doing my revision, but I still couldn't figure out how to go about solving it.
| Let's do it for $n$ toys.
Each term in $(A+B+C)^n$, expanded without making use of the commutativity of $A$, $B$, and $C$, corresponds to a way of assigning the toys. We want to eliminate terms in which the total degree of $A$ is odd or the total degree of $C$ is even. This can be done by computing
$$
\frac{1}{4}\left[(A+B+C)^n+(-A+B+C)^n-(A+B-C)^n-(-A+B-C)^n\right].
$$
Now we want to eliminate terms in which the total degree of any of $A$, $B$, or $C$ is $0$. Since the total degree of $C$ is now guaranteed odd, we only have to worry about $A$ and $B$. We can eliminate terms that are do not contain any factor of $A$ or any factor of $B$ by subtracting from the expression above the expression with $A$ set to $0$, then subtracting the expression with $B$ set to $0$, and finally by adding the expression with both $A$ and $B$ set to $0$ (to compensate for double-subtraction). We get
$$
\begin{aligned}
&\frac{1}{4}\left[(A+B+C)^n+(-A+B+C)^n-(A+B-C)^n-(-A+B-C)^n\right]\\
&\quad-\frac{1}{2}\left[(B+C)^n-(B-C)^n\right]-\frac{1}{4}\left[(A+C)^n+(-A+C)^n-(A-C)^n-(-A-C)^n\right]\\
&\quad+\frac{1}{2}\left[C^n-(-C)^n\right].
\end{aligned}
$$
The number of terms in the resulting expression is found by setting $A=B=C=1$. This gives
$$
\frac{1}{4}\left[3^n-(-1)^n\right]-\frac{1}{2}\left[2^n-0^n\right]-\frac{1}{4}\left[2^n+0^n-0^n-(-2)^n\right]+\frac{1}{2}\left[1^n-(-1)^n\right].
$$
This formula works for any $n$, including $n=0$ with the convention $0^0=1$. If you assume $n>0$, you can make the formula slightly more concise; likewise, if you restrict to odd $n$ or to even $n>0$ you can obtain somewhat shorter formulas:
$$
\begin{cases}
0 & n=0,\\
\frac{1}{4}\left[3^n+1\right]-2^n+1 & \text{$n$ odd,}\\
\frac{1}{4}\left[3^n-1\right]-2^{n-1} & \text{$n>0$, even.}
\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1449946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
coefficient on $s^{14}$ in generating function I have $(s+s^2+s^3+s^4+s^5+s^6)^7$, and I'm trying to find the coefficient on $s^{14}$. I've tried using the multinomial theorem, but that leads to the problem of finding all $k_1, k_2, \ldots , k_6$ such that $\sum_{n=1}^6 k_n = 7$ and $\sum_{n=1}^6nk_n = 14$, and that doesn't seem to put me any closer to an answer.
I tried rewriting it as $s^7(1+s+s^2+s^3+s^4+s^5)^7$ and looking for the coefficient of $s^7$ in the right half, $(1+s+s^2+s^3+s^4+s^5)^7$, but that doesn't make things much easier.
| Binomial Series Approach
Using the Binomial Theorem and negative binomial coefficients,
$$
\begin{align}
\left(s+s^2+s^3+s^4+s^5+s^6\right)^7
&=s^7\left(\frac{1-s^6}{1-s}\right)^7\\
&=s^7\sum_{k=0}^7\binom{7}{k}\left(-s^6\right)^k\sum_{j=0}^\infty\binom{-7}{j}(-s)^j\\
&=s^7\sum_{k=0}^7(-1)^k\binom{7}{k}s^{6k}\sum_{j=0}^\infty\binom{j+6}{j}s^j\tag1
\end{align}
$$
Using Cauchy Products, the coefficient of $s^n$ in $(1)$ is
$$
\sum_{k=0}^7(-1)^k\binom{7}{k}\binom{n-6k-1}{n-6k-7}\tag2
$$
Plug $n=14$ into $(2)$:
$$
\bbox[5px,border:2px solid #C0A000]{\binom{7}{0}\binom{13}{7}-\binom{7}{1}\binom{7}{1}=1667}\tag3
$$
The Sum in $\bf{(2)}$ represents a Polynomial
When $n\ge43$, $n-6k-1\ge0$ for all $k\le 7$. This means that
$$
\sum_{k=0}^7(-1)^k\binom{7}{k}\binom{n-6k-1}{n-6k-7} = \sum_{k=0}^7(-1)^k\binom{7}{k}\binom{n-6k-1}{6}
$$
which is an order $7$ repeated difference of a degree $6$ polynomial in $k$. Therefore, for $n\ge43$,
$$
\sum_{k=0}^7(-1)^k\binom{7}{k}\binom{n-6k-1}{n-6k-7}=0
$$
That is, for $n\ge43$, the coefficient of $x^n$ vanishes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1451693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
What is the sum of the cube of the roots of $ x^3 + x^2 - 2x + 1=0$? I know there are roots, because if we assume the equation as a function and give -3 and 1 as $x$:
$$
(-3)^3 + (-3)^2 - 2(-3) + 1 <0
$$
$$
1^3 + 1^2 - 2(1) + 1 > 0
$$
It must have a root between $[-3,1]$. However, the root is very hard and it appeared on a high school test. How can I solve it simply?
The given options were -10, -5, 0 , 5 and 10. Note: we didn't even learn the cube root formula. The test had just logic problems and I didn't use any calculus or complicated stuff. So there must be an easier way without using cube root concepts or formulas.
| You know,
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2+ab+bc+ca)$
$a^3+b^3+c^3=(a+b+c)[(a+b+c)^2-3(ab+bc+ca)]+3abc$
Now I hope you know the relation between roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1453824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
In how many ways can we add $1$'s and $2$'s to get $11$ (when the order matters)? Examples:
$1+1+1+1+1+1+1+1+1+1+1$,
$2+2+2+2+2+1$,
$1+2+2+2+2+2$ (order matters)
I tried solving it with permutations but I realized it won't work
| Our first rather clumsy solution takes advantage of the fact that $11$ is a quite small number.
We could have $0$ $2$'s, $1$ way.
We could have $1$ $2$ and $9$ $1$'s, total $10$ numbers. The location of the $2$ can be chosen in $\binom{10}{1}$ ways.
We could have $2$ $2$'a and $7$ $1$'s, total of $9$ entries. The location of the $2$'s can be chosen in $\binom{9}{2}$ ways.
We could have $3$ $2$'s and $5$ $1$'s. There are $\binom{8}{3}$ choices.
For $4$ $2$'s there are $\binom{7}{4}$ choices, and for $5$ there are $\binom{6}{5}$ choices.
Add up.
Another way: Let $a_n$ be the number of ways to represent $n$ as an ordered sum of $1$'s and/or $2$'s. We have $a_1=1$ and $a_2=2$.
There are two types of sequences of length $n+1$, (i) the ones that end in $1$ and (ii) the ones that end in $2$. There are $a_n$ of Type (i), and $a_{n-1}$ of Type (ii). It follows that
$$a_{n+1}=a_n+a_{n-1}.$$
We can now use the above recurrence to find $a_3$ then $a_4$ and so on. We get the familiar Fibonacci sequence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1456583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
trigonometry equation $3\cos(x)^2 = \sin(x)^2$ I tried to solve this equation, but my solution is wrong and I don't understand why. the answer in the book is: $x = \pm60+180k$. my answer is: $x= \pm60+360k$.
please help :)
3cos(x)^2 = sin(x)^2
3cos(x)^2 = 1 - cos(x)^2
t = cos(x)^2
3t=1-t
4t=1
t=1/4
cos(x)^2 = 1/4
cos(x) = 1/2
cos(x) = cos(60)
x = +-60+ 360k
edited:
cos(x) = -1/5
cos(x) = cos(120)
x = +-120 + 360k
I still don't get the answer in the book
| Since $\sin^2 x=1-\cos^2 x$ the given equation is equivalent to
\begin{align}
3\cos^2 x&=1-\cos^2 x\\
\iff \quad \cos^2 x&=\frac{1}{4}\\
\iff \cos x&\in\left\{-\frac{1}{2},\frac{1}{2}\right\}
\end{align}
Hence $$x=180^{\circ}\cdot k \pm 60^{\circ}$$
where $k$ is an integer number.
Silas2033: Notice that in this problem $\cos x$ can be either $\frac{1}{2}$ or $-\frac{1}{2}$, then you must regard two cases:
1) $\cos x = \cos y$ giving us $x=\pm y +360^{\circ}k=\pm y +180^{\circ}(2n)$ and
2) $\cos x = -\cos y$ giving us $x=180^{\circ}-(\pm y +360^{\circ}k)=\mp y - 180^{\circ}(2k-1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1456648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Calculate $\int {\frac{{\sqrt {x + 1} - \sqrt {x - 1} }}{{\sqrt {x + 1} + \sqrt {x - 1} }}} dx $ Calculate
$$\int {\frac{{\sqrt {x + 1} - \sqrt {x - 1} }}{{\sqrt {x + 1} + \sqrt {x - 1} }}} dx
$$
My try:
$$\int {\frac{{\sqrt {x + 1} - \sqrt {x - 1} }}{{\sqrt {x + 1} + \sqrt {x - 1} }}} dx = \left| {x + 1 = {u^2}} \right| = 2\int {\frac{{(u - \sqrt { - 2 + {u^2}} )}}{{u + \sqrt { - 2 + {u^2}} }}du}
$$
I tried to do first Euler substitute:
$$\sqrt { - 2 + {u^2}} = {u_1} - u
$$
But it did not lead me to the goal. Any thoughts will be appriciated.
| Try
$$\left (\sqrt{x+1}+\sqrt{x-1} \right )\left (\sqrt{x+1}-\sqrt{x-1} \right ) = 2$$
Then the integral is
$$\frac12 \int dx \, \left (\sqrt{x+1}-\sqrt{x-1} \right )^2$$
which is
$$\int dx \left (x - \sqrt{x^2-1}\right ) $$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1456834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Proof that $3^c + 7^c - 2$ by induction I'm trying to prove the for every $c \in \mathbb{N}$, $3^c + 7^c - 2$ is a multiple of $8$. $\mathbb{N} = \{1,2,3,\ldots\}$
Base case: $c = 1$
$(3^1 + 7^1 - 2) = 8$ Base case is true.
Now assume this is true for $c=k$.
Now I prove this holds for $c=k+1$ $(3^{k+1}+7^{k+1}-2)$.
$(3^{k+1}+7^{k+1}-2)$
$(3^k*3+7^k*7-2)$
But now I'm stuck...
| By the induction hypothesis,
$$
3^k+7^k-2=8m
$$
for some integer $m$. Then $3^k=8m+2-7^k$ and so
\begin{align}
3^{k+1}+7^{k+1}-2
&=3\cdot 3^k+7^{k+1}-2\\
&=3(8m+2-7^k)+7\cdot7^{k}-2\\
&=24m+6-3\cdot7^k+7\cdot7^{k}-2\\
&=4(6m+1+7^k)
\end{align}
Can you finish up?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1457515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Estimate the value $ \int_{1}^{2}\int_{3}^{4}{f(x,y)dydx} $ Using the Gauss-Legendre form. Estimate the value of$$\displaystyle\int_{1}^{2}\displaystyle\int_{3}^{4}{f(x,y)dydx}$$
where, $f(x,y)=x^3y$.
My approach: We can approximate the integral $\int_{-1}^{1}{f(x)dx}=\sum_{i=1}^{n}{A_{i}f(x_{i})}$ with the form. Gauss-Legendre, by orthogonals polynomials $$p_{n}(x)=\dfrac{(-1)^{n}}{2^{n}n!}\dfrac{d^{n}}{dx^{n}}[(1-x^2)^n]$$ $$A_{i}=\dfrac{2}{(1-x_{i}^2)[p'_{n}(x_{i})]^{2}}$$ If n=2, then $x_{1}=0.577350=-x_{2}$, and $A_{i}=1$. And ,for instance $[1,2]\to [-1,1]$, this implies that $y=2x-3$ as I continue?
| Some pretty good progress. Since the integrand is at most cubic in $x$ and $y$ the result should be exact for the Gauss $2$-point formula. In the $x$-dimension, we want $u=-1$ when $x=1$ and $u=1$ when $x=2$. So
$$\frac{x-1}{u+1}=\frac{2-1}{1+1}=\frac12$$
So $x=\frac12u+\frac32$. In the $y$-dimension we want $v=-1$ when $y=3$ and $v=1$ when $y=4$:
$$\frac{y-3}{v+1}=\frac{4-3}{1+1}=\frac12$$
And we have $y=\frac12v+\frac72$. Then
$$\begin{align}\int_1^2\int_3^4x^3y\,dy\,dx&=\int_{-1}^1\int_{-1}^1\left(\frac{u+3}2\right)^3\left(\frac{v+2}{2}\right)\frac{dv}2\frac{du}2\\
&=\sum_{i=1}^2\sum_{j=1}^2A_iA_j\frac1{64}(x_i+3)^3(x_j+7)\\
&=\frac1{64}(0.577350+3)^3(0.577350+7)+\frac1{64}(0.577350+3)^3(-0.577350+7)\\
&\quad+\frac1{64}(-0.577350+3)^3(0.577350+7)+\frac1{64}(-0.577350+3)^3(-0.577350+7)\\
&=5.420279376+4.594291848+1.683485587+1.426941966\\
&=13.12499878\end{align}$$
This compares well with the exact value of $\frac{105}8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1459002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is there an easier way to solve this logarithmic equation? $$2\log _{ 8 }{ x } =\log _{ 2 }{ x-1 } $$
Steps I took:
$$\frac { \log _{ 2 }{ x^{ 2 } } }{ \log _{ 2 }{ 8 } } =\log _{ 2 }{ x-1 } $$
$$\frac { \log _{ 2 }{ x^{ 2 } } }{ 3 } =\log _{ 2 }{ x-1 } $$
$$\log _{ 2 }{ x^{ 2 } } =3\log _{ 2 }{ x-1 } $$
$$2\log _{ 2 }{ x } =3\log _{ 2 }{ x-1 } $$
$$\log _{ 2 }{ x } =\frac { 3 }{ 2 } \log _{ 2 }{ x-1 } $$
$$\log _{ 2 }{ x } =\log _{ 2 }{ (x-1)^{ \frac { 3 }{ 2 } } } $$
This method seems to be very inefficient and I don't know how I would go from here. Can someone please point me in the right direction. Hints only please. No actual solution.
| Using a change of base: $$\log_b a = \frac{\log_n a}{\log_n b}$$
Change $\log_8 x^2$ base to 2: $$\log_8 x^2 = \frac{\log_2 x^2}{\log_2 8} = \frac{2\log_2 x}{3} $$
Let $\log_2 x = a$
$$ \frac{2a}{3}= a-1 \ \implies \ a=3$$
$$\log_2 x = 3\ \implies \ x=2^3 = 8$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1462075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Where am I going wrong while trying to solve this logarithmic equation? $$\log _{ 0.2 }{ x } +\log _{ \sqrt { 5 } }{ x } =\log _{ 25 }{ x } +1$$
Steps I took:
$$\log _{ \frac { 1 }{ 5 } }{ x } +\frac { \log _{ \frac { 1 }{ 5 } }{ x } }{ \log _{ \frac { 1 }{ 5 } }{ \sqrt { 5 } } } =\frac { \log _{ \frac { 1 }{ 5 } }{ x } }{ \log _{ \frac { 1 }{ 5 } }{ 25 } } +1$$
$$\log _{ \frac { 1 }{ 5 } }{ x } +-\frac { \log _{ \frac { 1 }{ 5 } }{ x } }{ 2 } =-\frac { \log _{ \frac { 1 }{ 5 } }{ x } }{ 2 } +1$$
I can keep going, but this doesn't seem to lead me to the correct answer. Where did I go wrong?
| Finding a common base might be the best approach. Re-write your equation as
$$\frac{\log x}{\log 1/5} +\frac{\log x}{\log 5^{1/2}} = \frac{\log x}{\log 5^2} +1. $$
$$\log x \left(\frac{1}{\log 1/5} +\frac{1}{\log 5^{1/2}}-\frac{1}{\log 5^2}\right) = 1 $$
Then the solution is $x = e^{1/a}$ in which $a$ is the numerical factor on the left above. As $a$ simplifies to $a = 1/\log 25$ the answer is $x = 25.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1463452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Fundamental Identities I was given a task to prove:
$$\frac{\sec\theta}{\sec\theta\tan\theta} = \sec\theta(\sec\theta + \tan\theta)$$
Then I replaced them with their Ratio and Reciprocal Identities
\begin{align*}
\sec\theta & = \frac{1}{\cos\theta}\\
\tan\theta & = \frac{\sin\theta}{\cos\theta}
\end{align*}
so I came up with this:
$$\frac{\frac{1}{\cos\theta}}{\frac{1}{\cos\theta} \cdot \frac{\sin\theta}{\cos\theta}} = \frac{1}{\cos\theta}\left(\frac{1}{\cos\theta} + \frac{\sin
\theta}{\cos\theta}\right)$$
then,
$$\frac{\frac{1}{\cos\theta}}{\frac{\sin\theta}{\cos\theta}} = \frac{1}{\cos\theta}\left(1+\frac{\sin\theta}{\cos\theta}\right)$$
and I had the reciprocal,
$$\frac{1}{\cos\theta} \cdot \frac{\cos\theta}{\sin\theta} = \frac{1}{\cos\theta}\left(1+\frac{\sin\theta}{\cos\theta}\right)$$
and I don't know what to do next, can someone explain to me how? T_T
| It's wrong.
Here's why:
$$\text{LHS}=\frac{\sec\theta}{\sec\theta\tan\theta}=\frac1{\tan\theta}=\frac{\text{adjacent}}{\text{opposite}}\phantom{....}$$
\begin{align}
\text{RHS}&=\sec\theta(\sec\theta+\tan\theta) \\[0.5ex]
&=\frac1{\cos\theta}\left(\frac 1{\cos\theta}+\frac{\sin\theta}{\cos\theta}\right) \\[0.6ex]
&=\frac1{\cos\theta}\left(\frac{1+\sin\theta}{\cos\theta}\right) \\[0.6ex]
&=\frac{1+\sin\theta}{\cos^2\theta}=\frac{\text{hypotenuse}+\text{opposite}}{\text{adjacent}}
\end{align}
Now, for $\frac{\text{adjacent}}{\text{opposite}}$ to be equal to $\frac{\text{hypotenuse}+\text{opposite}}{\text{adjacent}}$, $``\text{opposite"}$ must be equal to $``\text{adjacent"}$ and $``\text{hypotenuse"}$ must be $0$. Since this cannot be true, I suppose it is wrong.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1463740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to sketch the subset of a complex plane? The question asks to sketch the subset of $\{z\ \epsilon\ C : |Z-1|+|Z+1|=4\}$
Here is my working:
$z=x+yi$
$|x+yi-1| + |x+yi+1|=4$
$\sqrt{ {(x-1)}^2 + y^2} + \sqrt{{(x+1)}^2+y^2}=4$
${ {(x-1)}^2 + y^2} + {{(x+1)}^2+y^2}=16$
$x^2 - 2x+1+y^2+x^2+2x+1+y^2=16$
$2x^2+2y^2+2=16$
$x^2+y^2=7$
$(x-0)^2+(y-0)^2=\sqrt7$
=This is a circle with center $0$ and radius $\sqrt7$
My answer is different from the correct answer given: "This is an ellipse with foci at $-1$ and $1$ passing through $2$"
I have no idea how to get to this answer. Could someone please help me here?
| Hint:
The equations
$$\sqrt{{(x-1)}^2 + y^2} + \sqrt{{(x+1)}^2+y^2}=4$$
$${{(x-1)}^2 + y^2} + {{(x+1)}^2+y^2}=16$$
are not equivalent.
Squaring the first one yields
\begin{align}
\left[\sqrt{{(x-1)}^2 + y^2} + \sqrt{{(x+1)}^2+y^2}\right]^2&=4^2\\
(x-1)^2+y^2+2\sqrt{{(x-1)}^2 + y^2}\sqrt{{(x+1)}^2+y^2}+(x+1)^2+y^2&=16
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1464152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
How to solve irrational inequality? Irrational inequality wolfram alpha
I want to find $x$ such that
$\sqrt{x-3}+(9-x)^{1/4}>\sqrt{3}$.
Yeah, I know the answer but I don't know how to find this answer
| The existence of roots implies that $3\le x\le 9$. Put $y=(9-x)^{1/2}$. Then $0\le y\le \sqrt{6}$ and
$$\sqrt{6-y^2}+\sqrt{y}>\sqrt{3}$$
Squaring both sides we obtain
$$6-y^2+y+2\sqrt{6y-y^3}>3$$
$$2\sqrt{6y-y^3}>y^2-y-3$$
Then $y^2-y-3\le 0$ or $4(6y-y^3)>(y^2-y-3)^2$.
*
*$y^2-y-3\le 0$ implies $\frac{1-\sqrt{13}}2<0\le y\le\frac{1+\sqrt{13}}2<\sqrt{6}$.
*$4(6y-y^3)>(y^2-y-3)^2$ implies $f(y)<0$, where $f(y)=y^4+2y^3-5y^2-18y+9$.
The graph of the function $f(y)$ at the segment $\left[\frac{1+\sqrt{13}}2;\sqrt{6}\right]$
suggests that an equation $f(y)=0$ has a unique root $y_0\simeq 2.44366$ at this segment.
So the answer is $0\le y<y_0$, which yields an answer of the initial inequality $$3.0285\simeq 9-y_0^2 <x\le 9.$$
The equation $f(x)=0$ has fourth degree, so, theoretically, it can be solved in radicals (see, for instance, “Курс высшей алгебры” Куроша), but practically it is too complicated (for instance, in my practice which holds for more than quarter of a century, I never used Ferrari algorithm by hand). A form of Wolfram Alpha’s answer also suggests that there is no simple solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1466046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Min of $\frac{a^{10}+ b^{10}}{a^{7}+ b^{7}} +\frac{b^{10}+ c^{10}}{b^{7}+ c^{7}} +\frac{c^{10}+ a^{10}}{c^{7}+ a^{7}} $ I got this problem I tried several time to solve it by many inequalities but I got stuk. My question is how I get the minimum value of $$ \frac{a^{10}+ b^{10}}{a^{7}+ b^{7}} +\frac{b^{10}+ c^{10}}{b^{7}+ c^{7}} +\frac{c^{10}+ a^{10}}{c^{7}+ a^{7}} $$ if you know that $a, b, c \in (0, \infty) $ and $ a+b+c=1$? Any hint?
| $2(a^{10}+b^{10})\ge (a^7+b^7)(a^3+b^3)$ with chebishev inequality, then $$\frac{a^{10}+b^{10}}{a^7+b^7}+\frac{b^{10}+c^{10}}{b^7+c^7}+\frac{a^{10}+c^{10}}{a^7+c^7} \ge a^3+b^3+c^3 \ge 3 (\frac{a+b+c}{3})^3=\frac{1}{9}$$ because with power mean we have $ (\frac{a^3+b^3+c^3}{3})^\frac{1}{3} \ge (\frac{a+b+c}{3})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1467379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Find A of the matrix Find $$(2A)^T = \left[\begin{array}{cc}1&-1\\2&3\end{array}\right]^{-1}$$
My solution
$$2A^T = \left[\begin{array}{cc}3&-2\\1&1\end{array}\right] = \frac12\left[\begin{array}{cc}3&-2\\1&1\end{array}\right], A = \left[\begin{array}{cc}3&-2\\1&1\end{array}\right]$$
Apparently this is completely wrong and the answer is
$$\frac{1}{10}\left[\begin{array}{cc}3&-2\\1&1\end{array}\right]$$ I do not understand where they got the $1/10$.
| Determinant = ad - bc = (3 $\times$ 1) - (-2 $\times$ 1) = 5
The inverse of the matrix = $\frac{1}{determinant}$adj(matrix) = $\frac {1}{5}$adj(matrix)
Bring the 2 over, it becomes $\frac{1}{10}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1468813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
The number of solutions of the equation $4\sin^2x+\tan^2x+\cot^2x+\csc^2x=6$ in $[0,2\pi]$ The number of solutions of the equation $4\sin^2x+\tan^2x+\cot^2x+\csc^2x=6$ in $[0,2\pi]$
$(A)1\hspace{1cm}(B)2\hspace{1cm}(C)3\hspace{1cm}(D)4$
I simplified the expression to $4\sin^6x-12\sin^4x+9\sin^2x-2=0$
But i could not solve it further.Please help me.Thanks.
| Let $t=\sin^2 x$. Then, by your simplified expression,
$$p(t)=4t^3 - 12t^2 + 9t - 2 =0 -----(*)$$
The polynomial $p$ factors as
$$p(t)=(2t-1)^2(t-2).$$
So, the solution to $(*)$ is
$$t=\frac{1}{2} \mbox{ or } t=2.$$
Now, we try to solve
$$\sin ^2x = \frac{1}{2} \mbox{ or } \sin^2 x = 2.$$
Well, the rightmost equation does not have a solution. So, we only have to solve the leftmost equation. Taking square roots,
$$\sin x = \pm\frac{1}{\sqrt{2}}.$$
Restricting $x$ to be in $[0,2\pi]$, we see that
$$x=\frac{(2k+1)\pi}{4} \mbox{ , for } k=0,1,2,3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1470150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Proving the limit of a sequence using definitions This is a review that my professor posted and I want to make sure I'm on the right path as I study
*
*$\cdot \lim \limits_{n \to \infty} n - \sqrt{2n^2+1} = n -\sqrt{2n^2+1}*\frac{n+\sqrt{2n^2+1}}{n+\sqrt{2n^2+1}} = \frac{-n^2-1}{n+\sqrt{2n^2+1}}$
So can I say that $M<\frac{-n^2-1}{n+\sqrt{2n^2+1}} \le \frac{n^2}{2n}\le \frac{n}{2} $ and let my function be $N(M) = \frac{2}{M}$ when $n>N$
*$\cdot \lim \limits_{n \to \infty} \frac{10^n}{n!} $ so can I say for a large n, maybe $n>20$ $\frac{10^n}{n!}\le \frac{1}{n!}$ therefore the function $N(\epsilon) = \max${$20, \epsilon!$} I feel like I'm very off track with this one
*This is the reverse of 2, but wouldn't it be the same approach but to find $N(M) $ such that $M>n!$
*Finally,
$\cdot \lim \limits_{n \to \infty} \frac{n}{3n-2000} = \frac{\frac{n}{n}}{\frac{3n}{n}-\frac{2000}{n}} = \frac{1}{3-\frac{2000}{n}}$ so then find $N(M)$ such that $M> \frac{1}{3} - \frac{n}{2000}$ therefore $N(M) = 2000(M-\frac{1}{3})$ or would $N(M) = Max${$\frac{2000}{3}, 2000M$}
| For the first sequence: If $n \geq 1$, then
$$
n - \sqrt{2n^{2}+1} = \frac{n^{2}-2n^{2}-1}{n+\sqrt{2n^{2}+1}} = \frac{-n^{2}-1}{n + \sqrt{2n^{2}+1}} < \frac{-n^{2}}{n + \sqrt{2n^{2}+1}} < \frac{-n^{2}}{n+4n} = \frac{-n}{5};
$$
given any $M < 0$, we have $\frac{-n}{5} < M$ if $n > 5|M|$; so for all $n \geq \lceil 5|M| \rceil +1$ we have $n - \sqrt{2n^{2}+1} < M$.
For the second sequence: If $n \geq 11$, then
$$
\frac{10^{n}}{n!} = \frac{10^{n}}{1\cdot 2 \cdot \cdots n} \leq \frac{10^{10}}{10!}\cdot \frac{10}{n} = \frac{10^{10}}{9!n};
$$
taking any $\varepsilon > 0$, we have $10^{10}/9!n < \varepsilon$ if $n \geq \lceil \frac{10^{10}}{9!\varepsilon} \rceil + 1$; so if $n \geq \max \{11, \lceil \frac{10^{10}}{9!\varepsilon} \rceil + 1 \}$, then $\frac{10^{n}}{n!} < \varepsilon$.
For the third sequence, by initial inspection we may guess that the sequence converges to $\frac{1}{3}$. To prove this, note that, if $n \geq 1$, then
$$
\bigg| \frac{n}{3n-2000} - \frac{1}{3} \bigg| = \frac{2000}{9n-6000};
$$
taking any $\varepsilon > 0$, if $n \geq \lceil \frac{2000(1+3\varepsilon)}{9\varepsilon} \rceil + 1$, then $\frac{2000}{9n-6000} < \varepsilon$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1472678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How do you factorize quadratics when the coefficient of $x^2 \gt 1$? So I've figured out how to factor quadratics with just $x^2$, but now I'm kind of stuck again at this problem:
$2x^2-x-3$
Can anyone help me?
| You can either use the quadratic formula for a general quadratic $ax^2+bx+c$ which is
$$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$ which I will say no more of as JMoravitz has rigorously explained it.
Or you can complete the square, or write
$$2x^2 \color{blue}{-x}-3$$
$$=2x^2+\color{blue}{2x-3x}-3$$
$$=2x(x+1)-3(x+1)$$
$$=\color{red}{(2x-3)(x+1)}$$
Notice, in the part marked $\color{blue}{\mathrm{blue}}$ I have simply rewritten $-x$ as $2x-3x$.
From then on you simply take out common factors.
Note that this method is only valid when you have two numbers whose product is $-6$ and sum is $-1$.
Or, put in another way for the general quadratic equation $ax^2 +bx +c$, this inspection method is only valid iff you can find two numbers whose product is $ac$ and sum is $b$. Otherwise using the quadratic formula is the best method as it suits all scenarios.
For completing the square:
$$2x^2 -x-3$$
$$=2\left(x^2 -\frac{x}{2}-\frac{3}{2}\right)$$
$$=2\left(\left(x -\frac{1}{4}\right)^2-\frac{1}{16}-\frac{3}{2}\right)$$
$$=2\left(\left(x -\frac{1}{4}\right)^2-\left(\frac{5}{4}\right)^2\right)\tag{1}$$
$$=2\left(\left(x -\frac{1}{4}+\frac{5}{4}\right)-\left(x -\frac{1}{4}-\frac{5}{4}\right)\right)\tag{2}$$
$$=2\left((x+1)\left(x -\frac{3}{2}\right)\right)$$
$$=\color{red}{(2x-3)(x+1)}$$
as before.
From $(1)$ to $(2)$ I used the difference of two squares formula such that $a^2-b^2=(a+b)(a-b) \space \space\forall a,b \in \mathbb{R}$
Hope this helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1477307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
How to solve this limit: $\lim_{n \to \infty} \frac{(2n+2) (2n+1) }{ (n+1)^2}$ $$
\lim_{n\to\infty}\frac{(2n+2)(2n+1)}{(n+1)^{2}}
$$
When I expand it gives:
$$
\lim_{n\to\infty} \dfrac{4n^{2} + 6n + 2}{n^{2} + 2n + 1}
$$
How can this equal $4$? Because if I replace $n$ with infinity it goes $\dfrac{\infty}{\infty}$ only.
| To handle limits involving fractions you factor out the dominant term from top and bottom such that they cancel, by dominant term; I mean the term of highest degree, in this case it's $n^2$: $$\lim_{n\to\infty} \frac{4n^2 + 6n + 2}{n^2 + 2n + 1}$$
$$=\require\cancel\lim_{n\to\infty} \frac{2\cancel{n^2}\left(2 + \frac{3}{n} + \frac{1}{n^2}\right)}{\cancel{n^2}\left(1 + \frac{2}{n} + \frac{1}{n^2}\right)}$$
$$=\lim_{n\to\infty} \frac{2\left(2 + \frac{3}{n} + \frac{1}{n^2}\right)}{1 + \frac{2}{n} + \frac{1}{n^2}}\tag{1}$$
$$=\frac{2\left(2 + 0 + 0\right)}{1 + 0 + 0}\tag{2}$$
$$=\color{blue}{4}$$
You get from $(1)$ to $(2)$ by making the observation that each of the fractions with $n$ or $n^2$ in the denominator will equal zero in the limit as $n$ tends to infinity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1477946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Using arithmetic mean>geometric mean Prove that if a,b.c are distinct positive integers that
$$a^4+b^4+c^4>abc(a+b+c)$$
My attempt:
I used the inequality A.M>G.M to get two inequalities
First inequality
$$\frac{a^4+b^4+c^4}{3} > \sqrt[3]{a^4b^4c^4}$$
or
$$\frac{a^4+b^4+c^4}{3} > abc \sqrt[3]{abc}$$ or new --
first inequality
$$\frac{a^4+b^4+c^4}{3abc} > \sqrt[3]{abc}$$
second inequality:
$$\frac{a+b+c}{3} > \sqrt[3]{abc}$$
I am seeing the numbers in the required equation variables here but am not able to manipulate these to get the inequality I want?? Please direct me on which step should I take after this??
| an other way $$2a^4+b^4+c^4\geq 4\sqrt[4]{a^8b^4c^4}=4a^2bc$$ $$2b^4+a^4+c^4\geq 4\sqrt[4]{b^8a^4c^4}=4b^2ac$$ $$2c^4+a^4+b^4\geq 4\sqrt[4]{c^8a^4b^4}=4c^2ab$$ adding this ineqalitis we get desired one
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1480000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Express the following product as a single fraction: $(1+\frac{1}{3})(1+\frac{1}{9})(1+\frac{1}{81})\cdots$
I'm having difficulty with this problem:
What i did was:
I rewrote the $1$ as $\frac{3}{3}$
here is what i rewrote the whole product as:
$$\left(\frac{3}{3}+\frac{1}{3}\right)\left(\frac{3}{3}+\frac{1}{3^2}\right)\left(\frac{3}{3}+\frac{1}{3^4}\right)\cdots\left(\frac{3}{3}+\frac{1}{3^{2^n}}\right)$$
but how would i proceed after this?
| Recall the difference of squares formula $$a^2 - b^2 = (a+b)(a-b).$$ With the special case $a = 1$, $b = x^n$, we would get $$1 - x^{2n} = (1-x^n)(1+x^n).$$ What does this suggest?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1480214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
How can I evaluate $ \int \frac{dx}{x(1+x^3)(1+3x^3)}$? I want to solve this problem but I have no clue on break the denominator:
$$ \int \frac{dx}{x(1+x^3)(1+3x^3)}$$
I have tried breaking the denominator into partial fractions but failed to do so.
| $$\int \frac{1}{x(1+x^3)(1+3x^3)} \text{d}x =$$
Substitue $u=x^3$ and $\text{d}u=3x^2\text{d}x$:
$$\frac{1}{3} \int \frac{1}{u(1+u)(1+3u)} \text{d}u =$$
$$\frac{1}{3} \int \left(\frac{1}{2(u+1)}-\frac{9}{2(3u+1)}+\frac{1}{u}\right) \text{d}u =$$
$$\frac{1}{3} \left(\int\frac{1}{2(u+1)}\text{d}u-\int\frac{9}{2(3u+1)}\text{d}u+\int\frac{1}{u}\text{d}u\right) =$$
$$\frac{1}{3} \left(\frac{1}{2}\int\frac{1}{u+1}\text{d}u-\frac{9}{2}\int\frac{1}{3u+1}\text{d}u+\int\frac{1}{u}\text{d}u\right) =$$
Substitue $s=3u+1$ and $\text{d}s=3\text{d}u$:
$$\frac{1}{3} \left(\frac{1}{2}\int\frac{1}{u+1}\text{d}u-\frac{3}{2}\int\frac{1}{s}\text{d}s+\int\frac{1}{u}\text{d}u\right) =$$
$$\frac{1}{3} \left(\frac{1}{2}\int\frac{1}{u+1}\text{d}u-\frac{3}{2}\ln(s)+\ln(u)\right) =$$
Substitue $p=u+1$ and $\text{d}p=\text{d}u$:
$$\frac{1}{3} \left(\frac{1}{2}\int\frac{1}{p}\text{d}p-\frac{3}{2}\ln(s)+\ln(u)\right) =$$
$$\frac{1}{3} \left(\frac{1}{2}\ln(p)-\frac{3}{2}\ln(s)+\ln(u)\right)+C =$$
$$\frac{1}{3} \left(\frac{1}{2}\ln(u+1)-\frac{3}{2}\ln(3u+1)+\ln(u)\right)+C =$$
$$\frac{1}{3} \left(\frac{1}{2}\ln(1+x^3)-\frac{3}{2}\ln(1+3x^3)+\ln(x^3)\right)+C $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1481625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the inverse of the cubic function What is the resulting equation when $y=x^3 + 2x^2$ is reflected in the line $y=x$ ?
I have tried and tried and am unable to come up with the answer.
The furthest I was able to get without making any mistakes or getting confused was $x= y^3 + 2y^2$. What am I supposed to do after that step?
| The general solution to the cubic equation
$$a x^3 + b x^2 + c x + d = 0$$
can be written
$$x = -\frac{1}{3 a} \left( b + \sigma C - \sigma \frac{\Delta_0}{C} \right)$$
where
$$
\Delta_0 = b^2 - 3 a c \\
\Delta_1 = 2 b^3 - 9 a b c + 27 a^2 d \\
C = \sqrt[3]{\frac{\Delta_1 \pm \sqrt{\Delta_1^2 - 4 \Delta_0^3}}{2}} \\
\sigma = 1, \frac{1}{2} \pm i \frac{\sqrt{3}}{2}
$$
Note that if you don't care about complex solutions, then you can just ignore $\sigma$.
Comparing this to our equation
$$x = y^3 + 2 y^2 \\
y^3 + 2 y^2 - x = 0$$
we see that all we need to to turn the general form at the top into our problem is substitute $y$ for $x$, $1$ for $a$, $2$ for $b$, $0$ for $c$, and $-x$ for $d$. Plugging these values into the formula and simplifying yields our solution:
$$
\Delta_0 = (2)^2 - 3 (1) (0) = 4 \\
\Delta_1 = 2 (2)^3 - 9 (1) (2) (0) + 27 (1)^2 (-x) = -27 x + 16 \\
C = \sqrt[3]{\frac{(-27 x + 16) \pm \sqrt{(-27 x + 16)^2 - 4 (4)^3}}{2}} = \sqrt[3]{-\frac{27}{2} x + 16 \pm \sqrt{\frac{729}{4} x^2 - 108 x}} \\
y = -\frac{1}{3 (1)} \left((2) + \sigma C - \sigma \frac{(4)}{C}\right) \\
y = -\frac{2}{3} - \frac{\sigma}{3} \left(\textstyle \sqrt[3]{-\frac{27}{2} x + 16 \pm \sqrt{\frac{729}{4} x^2 - 108 x}} \displaystyle - \frac{4}{\sqrt[3]{-\frac{27}{2} x + 16 \pm \sqrt{\frac{729}{4} x^2 - 108 x}}}\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1483839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Points on ellipsoid with maximum Gaussian curvature/mean curvature.
Find the points on the ellipsoid $$x^2/a^2+y^2/b^2+z^2/c^2=1$$ with maximum Gaussian curvature and mean curvature respectively.
I parametrized it as $(a\sin u\cos v,b\sin u\sin v, c\cos v)$ and managed to compute its Guassian and mean curvature, which are both very messy:$$
K(u,v)=\frac{a^2b^2c^2}{[a^2b^2\cos^2v+c^2(b^2\cos^2u+a^2\sin^2u)\sin^2v]^2 }
$$
and
$$H(u,v) =\frac{abc[3(a^2+b^2)+2c^2+(a^2+b^2-2c^2)\cos(2v)-2(a^2-b^2)\cos(2u)\sin^2v]}{8[a^2b^2\cos^2v+c^2(b^2\cos^2u+a^2\sin^2u)\sin^2v]^{3/2}}$$
How can I cleverly find points which maximize the complicated $K$ or $H$? (Perhaps in less than 2 pages of computation, by some geometric reasoning?)
| Using the implicit definition of this surface may be a better approach. Let $$F(x,y,z) = \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} - 1.$$ The ellipsoid is then defined by $F(x,y,z) = 0$. Then it is known that the Gaussian curvature at a point $(x,y,z)$ is given by
$$\kappa = \frac{g H^{ad}g^T}{|g|^4}.$$
and the mean curvature is given by
$$h = \frac{gHg^T - \operatorname{trace}(H)|g|^2}{2|g|^3}.$$
where $g = \nabla F$ is the gradient, $H = \nabla g$ is the Hessian matrix, and $H^{ad}$ is its adjoint. Here
$$
g = (2x/a^2, 2y/b^2, 2z/c^2), \, H = \begin{pmatrix} 2/a^2 & 0 & 0 \\ 0 & 2/b^2 & 0 \\ 0 & 0 & 2/c^2 \end{pmatrix} , \quad
H^{ad} = \begin{pmatrix}4/(b^2c^2) & 0 & 0 \\
0 & 4/(a^2c^2) & 0 \\
0 & 0 & 4/(a^2b^2) \end{pmatrix}
$$
and therefore
$$
|g|^4 = 16\left( (x^2/a^4 + y^2/b^4 + z^2/c^4) \right)^2, \quad
gH^{ad} g^T = 16\frac{x^2/a^2 + y^2/b^2 + z^2/c^2}{a^2b^2c^2} = \frac{16}{a^2b^2c^2} \, .
$$
The Guassian curvature can now be maximized as follows. We obtain
$$
\kappa = \frac{1}{a^2b^2c^2\left( (x^2/a^4 + y^2/b^4 + z^2/c^4) \right)^2} \, .
$$
Now it is enough to find the extrema of the denominator, subject to the constraint $F(x,y,z) = 0$ which leads to the Lagrange multiplier equations
$$
2x/a^4 = 2\lambda x/a^2, \quad 2y/b^4 = 2\lambda y/b^2, \quad 2z/c^4 = 2\lambda z/c^2\, .
$$
In the generic case where $a,b,c$ are all different, this means $x(\lambda - a^2) = y(\lambda - b^2) = z(\lambda - c^2) = 0$. This is only possible if any two of the unknowns are zero and the third one isn't. Therefore the critical points of $\kappa$ on the ellipsoid are $(\pm a,0,0), (0,\pm b, 0,0), (0,0,\pm c)$ with Gaussian curvatures $ \frac{a^2}{b^2c^2}, \frac{b^2}{a^2c^2},\frac{c^2}{a^2b^2}$. Pick the largest one and that's the maximum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1484420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Infinite series of alternating reciprocals $\frac 1{1\cdot3}-\frac 1{3\cdot 5}+\frac 1{5\cdot 7}-\cdots $ Having misread the recent question here as
$$\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)(2n+3)}$$
and having solved it, I thought that I would post it as a question instead. It has a rather interesting answer!
Edit
Now that we have nice solutions from Jack D'Aurizio and Michael Biro, I'd like to point out that what struck me was the fact that
$$\begin{align}
\sum_{n=0}^\infty=\frac \pi 4-\frac 12&=\sum_{n=0}^\infty (-1)^n\frac 1{2n+1}-\frac 1{2^{n+2}}\\
\color{red}{\frac 1{1\color{black}{\cdot 3}}-\frac 1{3\color{black}{\cdot 5}}+\frac 1{5\color{black}{\cdot 7}}-\frac 1{7\color{black}{\cdot 9}}+\cdots }&=
\left(\color{red}{\frac 11-\frac 13+\frac15-\frac17+\cdots}\right)-\frac 12\\
&=\left(\color{red}{\frac 11-\frac 13+\frac15-\frac17+\cdots}\right)-\left(\frac 14+\frac 18+\frac 1{16}+\frac 1{32}+\cdots \right)
\end{align}$$
Is it possible to reduce the LHS expansion to the RHS expansion directly without first knowing the answer? If so then this would be another solution method.
| Combine terms to get
$\sum \frac{(-1)^n}{(2n+1)(2n+3)} = \frac{1}{1 \cdot 3} - \frac{1}{3 \cdot 5} + \dots = \frac{4}{1 \cdot 3 \cdot 5} + \frac{4}{5 \cdot 7 \cdot 9} + \dots = \sum \frac{4}{(4n+1)(4n+3)(4n+5)}$
Partial fractions (and some questionable rearrangement) gives:
$\sum \frac{4}{(4n+1)(4n+3)(4n+5)} = \sum \frac{1}{2(4n+1)} + \frac{1}{2(4n+5)} - \frac{1}{4n+3} $
$= \frac{1}{2}(1 + \frac{1}{5} + \frac{1}{9} + \dots) + \frac{1}{2}(\frac{1}{5} + \frac{1}{9} + \dots) - (\frac{1}{3} + \frac{1}{7} + \dots)$
$=\frac{1}{2} + (-\frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots) = \frac{1}{2} + (\arctan(1) - 1 )=\frac{1}{2} + (\frac{\pi}{4} - 1) = \frac{\pi}{4} - \frac{1}{2}$
Rigor is a little suspect, but hey, it works! :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1484738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Laplace for a function I have problems to understand Laplace for:
$$\frac{s-1}{(s^2+4)^2}$$
I found that fraction can be wrote with Laplace like this:
$$\frac{s-1}{(s^2+4)^2}=A\cdot\frac{s}{s^2+4}+B\cdot\frac{2}{s^2+4}+C\cdot\frac{4s}{(s^2+4)^2}+D\cdot\frac{s^2-4}{(s^2+4)^2}$$
Can someone please explain me from what are those elements at numerator?
| Sal, Alistar .
Just break your initial expression in two pieces and make use of a Laplace Transform table. The identities that encapsulate our particular case are :
$$ \bullet \mathcal{L}[t\sin(at)] = \frac{2as}{(s^2+a^2)^2} $$
$$ \bullet \mathcal{L}[\sin(at) - at\cos(at)] = \frac{2a^3}{(s^2+a^2)^2} $$
So, in order to get the desired original function, we use a little algebraic manipulation :
$$\frac{s}{(s^2+4)^2}-\frac{1}{(s^2+4)^2} = \frac{1}{4} \cdot\frac{4s}{(s^2+4)^2} + \frac{1}{16}\cdot\frac{16}{(s^2+4)^2} $$
This finally yields :
$$\mathcal{L}^{-1}\bigg[\frac{1}{4} \cdot\frac{4s}{(s^2+4)^2} + \frac{1}{16}\cdot\frac{16}{(s^2+4)^2}\bigg] = \frac{t\sin(2t)}{4} - \frac{\sin(2t)-2t\cos(2t)}{16}$$
For a more hands-on approach, one might use a Bromwich contour as a support of integration. Sure that would require a little complex analysis, but it'd be a bit more fun than just scrolling through the entries of a table.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1485947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Continuity of $\frac{x^5-4x^3y^2-xy^4}{(x^2+y^2)^2}$ at (0, 0) I am having trouble proving that $\dfrac{x^5-4x^3y^2-xy^4}{(x^2+y^2)^2}$ is continuous at $(0, 0)$ if we set the value at $(0, 0)$ to be $0$.
I don't see a way to prove this as I cannot factor this into partial fractions.
| It is easy to estimate for instance:
$$
\begin{aligned}
|x^5-4x^3y^2-xy^4|
&\le |x^5|+|-4x^3y^2|+|-xy^4| \\
&= |x|\cdot( x^4+4x^2y^2+y^4) \\
&\le |x|\cdot( \ (x^2+y^2)^2 + 2(x^2+y^2)^2 + (x^2+y^2)^2\ )\\
&= 4|x|\; (x^2+y^2)^2\ .
\end{aligned}
$$
So is $f$ denotes the given function we have $|f(x,y)|\le 4|x|$, which holds also in zero (which is $(0,0)$). This estimation leads to continuity in zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1491859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Number of points $C=(x,y)$ on the circle $x^2+y^2=16$ such that the area of the triangle whose vertices are $A,B$ and $C$ is a positive integer Let $A(-4,0)$ and $B(4,0)$.Number of points $C=(x,y)$ on the circle $x^2+y^2=16$ such that the area of the triangle whose vertices are $A,B$ and $C$ is a positive integer,is...
I found the area of triangle $ABC=4|y|$ but i cannot count in how many cases the area will be a positive integer.The answer given is $62$.Is there a method by which i can solve this.Please help me.Thanks.
| Hint: $-4 \leq y \leq 4$, so the possible $y$ values are $$\frac{1}{4}, -\frac{1}{4}, \frac{2}{4}, -\frac{2}{4}, \frac{3}{4}, - \frac{3}{4}, \dots, \frac{16}{4}, -\frac{16}{4}$$
Then, how many times is each $y$ value realized on the circle?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1492619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to use implicit differentiation to prove $\sinh^{-1}(3x)$ equal to something?
I understand that you let $y = \sinh^{-1}(3x)$ and thus $\sinh(y) = 3x$, but not sure where to go from there.
| Start from basic principles. The hyperbolic sine function can be expressed as
$$\sinh y=\frac{e^y-e^{-y}}{2}\tag 1$$
We can easily invert $(1)$ to find that
$$y=\log\left(\sinh y+\sqrt{\sinh^2 y+1}\right)$$
Letting $y=\text{arsinh}(x)$ reveals
$$\text{arsinh}(x)=\log\left(x+\sqrt{x^2+1}\right)$$
Note that
$$\begin{align}
\cosh (\text{arsinh}(x))&=\frac12\left(x+\sqrt{x^2+1}+\frac{1}{x+\sqrt{x^2+1}}\right)\\\\
&=\frac12\left(x+\sqrt{x^2+1}+\frac{1}{x+\sqrt{x^2+1}}\frac{x-\sqrt{x^2+1}}{x-\sqrt{x^2+1}}\right)\\\\
&=\sqrt{x^2+1} \tag 1
\end{align}$$
Therefore, using the chain rule we have for $3x=\sinh(y(x))$
$$\begin{align}
\frac{d(3x)}{dx}&=3\\\\
&=\frac{d\sinh(y(x))}{dx}\\\\
&=\cosh (y(x))y'(x) \tag 2
\end{align}$$
whereupon solving $(2)$ for $y'$ and using $(1)$ yields
$$\bbox[5px,border:2px solid #C0A000]{y'(x)=\frac{3}{\sqrt{(3x)^2+1}}=\frac{1}{\sqrt{x^2+\frac19}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1493317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
A polynomial problem related to lx^2 + nx + n If the roots of $lx^2 + nx + n = 0$ are in the ratio $p:q$, find the value of $\sqrt{\frac{p}{q}}$ + $\sqrt{\frac{q}{p}}$ + $\sqrt{\frac{n}{l}}$.
How to go about this problem?
| Let $x_1$ and $x_2$ be the roots. we know that $x_1x_2=\frac n l$ or that $x_1=\frac{1}{x_2} \frac nl$. Since we have to calculate $\sqrt{\frac n l}$ then $n$ and $l$ should have the same sign. Let them both be positive.
Now if $\frac{x_1}{x_2}=\frac pq$ then we have that $\frac p q=\frac 1 {x^2} \frac n l$ therefore
\begin{align}
\sqrt{\frac{p}{q}} + \sqrt{\frac{q}{p}} + \sqrt{\frac{n}{l}}&=\sqrt{\frac 1 {x_2^2} \frac n l} + \sqrt{x_2^2 \frac l n} + \sqrt{\frac{n}{l}}\\
&=\sqrt{\frac{n}{l}}\Big(\sqrt{\frac 1 {x_2^2} } + \frac l n\sqrt{x_2^2 } +1\Big)\\
\end{align}
now we should argue about the sign of $x_2$. If the roots are real and $n$ and $l$ are of the same sign, then both roots should either be negative of positive. Since we assumed that $n>0$, both roots will be negative. Hence
\begin{align}
\sqrt{\frac 1 {x_2^2} } + \frac l n\sqrt{x_2^2 }&=-\frac 1 {x_2} -\frac 1 {x_1}\\
&=1
\end{align}
you could check for yourself what happens if both $n$ and $l$ are negative :-)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1493495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Chance that last digit in the product is $1,3,7 $ or $9$
If $4$ whole numbers are taken at random and multiplied
together ,what is the chance that last digit in the
product is $1,3,7 $ or $9$ ?
$a.)\ \dfrac{15}{653} \\
b.)\ \dfrac{12}{542} \\
c.)\ \color{green}{\dfrac{16}{625}} \\
d.)\ \dfrac{17}{625} $
I did $\dfrac{\dbinom{4}{4}}{\dbinom{10}{4}}=\dfrac{1}{210}$, but the answer given in book is $c.)$
I look for a short and simple way.
I have studied maths upto $12$th grade.
| The last digits won't be evenly distributed. If any number ends in $5$ or an even number then the product ends in a $5$ or an even number. To end in a $1,3,7,$ or $9$ all numbers must end with $1,3,7,$ or $9$. The probability of that is $(4/10)^4 = (2/5)^4 = 16/625$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1494520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find all the solutions of diophantine eq: $x^3-2xy^2+y^3-s^2=0$ Given $x,y,s$ are natural numbers:
$$x^3-2xy^2+y^3-s^2=0$$
I found the solutions using wolfram alpha
$$(x,y,s) = (1,2,1), (6,10,4), (4,8,8)$$
But how do I prove these are the only solutions? Any tips or reference to papers that study this diophantine equation would be much appreciated.
| An infinite set of solutions is found when $y=2x$. In this case the equation $x^3 - 2xy^2 + y^3 = s^2$ reduces to:
$$x^3 = s^2$$
So we can choose $x=a^2$ for any positive integer $a$, so then $a^6 = s^2$, so $s=a^3$. So the following combination works for any positive integer $a$:
$$(x,y,s) = (a^2, 2a^2, a^3)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1494816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solve using AM GM Inequality if possible Let x, y, z be non-zero real numbers such that $\dfrac{x}{y} + \dfrac{y}{z} + \dfrac{z}{x} = 7$ and $\dfrac{y}{x} + \dfrac{z}{y} + \dfrac{x}{z} = 9$, then $\dfrac{x^3}{y^3} + \dfrac{y^3}{z^3} + \dfrac{z^3}{x^3}$ is equal to?
I don't really know how to solve this. any methods would be welcome. I was solving a couple of AM GM inequality questions and I'm assuming this should be solved in a similar way. Correct me.
| Let $$\frac{x}{y} = a,\,\,\frac{y}{z} = b,\,\,\frac{z}{x} = c$$ where $$abc = \frac{x}{y} \cdot \frac{y}{z} \cdot \frac{z}{x} = 1$$
So we have $$\left\{ \begin{gathered}
a + b + c = 7 \hfill \\
ab + ac + bc = 9 \hfill \\
\end{gathered} \right.$$
Using the identity $${a^3} + {b^3} + {c^3} - 3abc = (a + b + c)({a^2} + {b^2} + {c^2} - (ab + ac + bc))$$ which is equivalent to $${a^3} + {b^3} + {c^3} = {(a + b + c)^3} - 3(ab + ac + bc)(a + b + c) + 3abc$$ Now substituting the values we get $${a^3} + {b^3} + {c^3} = {7^3} - 3 \cdot 9 \cdot 7 + 3 = 343 - 189 + 3 = 157$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1495357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Get matrix $A^{n}$ problem with eigenvalues and eigenvectors i have problem with some matrix
$A = \begin{bmatrix}
\frac{1}{2} & 0 \\
2 & \frac{1}{2}
\end{bmatrix}$
To get $A^{n} = P J^{n} P^{-1}$
$\begin{bmatrix}
\frac{1}{2} - \lambda & 0 \\
2 & \frac{1}{2} - \lambda
\end{bmatrix}$
$J^{n} =\begin{bmatrix}
(\frac{1}{2})^{n} & n(\frac{1}{2})^{n-1} \\
0 & (\frac{1}{2})^{n}
\end{bmatrix}$
i'm not sure how to create this matrix
So the eigenvaues are $\lambda_{1} = \lambda_{2} = \frac{1}{2}$
then
$\begin{bmatrix}
0 & 0 \\
2 & 0\\
\end{bmatrix}$
$\begin{bmatrix}
x\\
y\\
\end{bmatrix}$ =
$\begin{bmatrix}
0\\
0\\
\end{bmatrix}$
The eigenvector are : $\begin{bmatrix}
x\\
0\\
\end{bmatrix}$
The question is: How look like $P$ matrix?
edit: So the answer will be:
$A^{n} =\begin{bmatrix}
0 & \frac{1}{2}\\
1 & 1\\
\end{bmatrix}$
$\begin{bmatrix}
(\frac{1}{2})^{n} & n(\frac{1}{2})^{n-1} \\
0 & (\frac{1}{2})^{n}
\end{bmatrix}$
$\begin{bmatrix}
-2 & 1\\
2 & 0\\
\end{bmatrix}$
Is that right?
| The eigenvalues are indeed $\lambda_1 = \lambda_2 = \lambda =\frac 12$. The matrix $P$ contains as its columns the right hand (generalized) eigenvectors of $A$, which are linearly independent. Solving the system $$(A- \lambda I) \mathbf{v} =\mathbf 0\iff \begin{bmatrix} 0 &0 \\ 2 & 0 \end{bmatrix}\cdot \begin{bmatrix} x \\ y\end{bmatrix} = \begin{bmatrix} 0 \\ 0\end{bmatrix}$$ yields that $x = 0$ and $y \in \mathbb R^*$ (since the zero vector cannot be an eigenvector).Thus, we can consider the eigenvector $\mathbf{v} = \begin{bmatrix} 0\\1\end{bmatrix}.$
Due to $\dim V_{\lambda} = 1,$ in order to define matrix $P$ we need to find a generalized eigenvector $\mathbf w$ of rank $2$, such that $$\left(A-\frac 12 I\right)^2\mathbf w = \mathbf 0\\[2ex]
\text {and} \\[2ex]
\left(A- \frac 12 I\right)\mathbf w \neq \mathbf 0.$$ For that reason we define the vector $\mathbf w $ as the solution of the system $$\left(A- \frac 12 I\right) \mathbf w = \mathbf v.$$
That means we have to solve the linear system $$\begin{bmatrix} 0 & 0 \\2 & 0 \end{bmatrix} \cdot\begin{bmatrix} w_1 \\ w_2 \end{bmatrix} =\begin{bmatrix}0 \\ 1\end{bmatrix}.$$
The matrix $P$ is equal to $P = \begin{bmatrix} 0 & w_1 \\ 1 & w_2 \end{bmatrix}$. Can you fill in the details?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1495945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find all non-negative integers $n$ satisfying $2^{n}\equiv n^{2} mod\, 5$ I'm trying to find all non-negative integers $n$ satisfying $2^{n}\equiv n^{2}\pmod{5}$.
So far, all the progress I've made is figuring out that $n^{2} mod \, 5$ for $n=1$ to $5$ has the pattern "$1,\, 4,\, 4,\, 1,\, 0$" which repeats $mod \, 5$. Then, I started looking at powers of $2$ to see if there was a pattern to when they are $0,\,1,\,\text{and}\, -1\, mod\, 5$. I saw that when $n = 4k$, where $k$ is an integer $\geq 0$, $2^{n}\,mod\,5 = 1$; for $n=5k$, where $k$ is an integer $\geq 0$, $2^{n} \, mod\, 5 = -1$; and there exists no $n$ such that $2^{n} \,mod\, 5 = 0$.
But, I found no pattern as to when $2^{n}\mod 5 = n^{2}\,\mod\,5$, so then I can find all the $n$'s that give me the divisibility I want.
No hints, please, I'm not a number theorist and am really out of my comfort zone. Just complete answers, so I can figure out how problems that look like this are supposed to be done.
Thank you in advance
| The powers of $2$ mod $5$ repeat with period $4$, and the squares repeat with period $5$. Thus $(2^n - n^2) \mod 5$ repeats with period $20$.
For $0 \le n \le 19$ you get
$$ \matrix{n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 \cr
n^2 \mod 5 & 0 & 1 & 4 & 4 & 1 & 0 & 1 & 4 & 4 & 1 & 0 & 1 & 4 & 4 & 1 & 0 & 1 & 4 & 4 & 1 \cr
2^n \mod 5 & 1 & 2 & 4 & 3 & 1 & 2 & 4 & 3 & 1 & 2 & 4 & 3 & 1 & 2 & 4 & 3 & 1 & 2 & 4 & 3 \cr}$$
Thus $2^n \equiv n^2 \mod 5$ for $n \equiv 2, 4, 16$, or $18 \mod 20$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1497183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Expanding $(x-2)^3$ I was trying to expand $(x-2)^3$. This is what I did
*
*Expanded the term so $(x-2)(x-2)(x-2)$
*Multiplied the first term and second term systematically through each case
The answer I got did not match the one at the back of the book, can someone show me how to do this please?
| If you are not familiar with the Binomial Theorem, you can expand $(x - 2)^3$ as follows:
\begin{align*}
(x - 2)^3 & = (x - 2)[(x - 2)(x - 2)]\\
& = (x - 2)[x(x - 2) - 2(x - 2)]\\
& = (x - 2)(x^2 - 2x - 2x + 4)\\
& = (x - 2)(x^2 - 4x + 4)\\
& = x(x^2 - 4x + 4) - 2(x^2 - 4x + 4)\\
& = x^3 - 4x^2 + 4x - 2x^2 + 8x - 8\\
& = x^3 - 6x^2 + 12x - 8
\end{align*}
The key idea here is that we can only multiply two factors at once, so we first find the product of two of the factors, then multiply the result by the third factor.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1501496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to find the determinant of this $5 \times 5$ matrix? How can I find the determinant of this matrix?
I know in matrix $3 \times 3$
$$A= 1(5\cdot 9-8\cdot 6)-2 (4\cdot 9-7\cdot 6)+3(4\cdot 8-7\cdot 5) $$
but how to work with a $5\times 5$ matrix?
| The Laplace expansion of the determinant can be done using any row or column of a square matrix. In each row, we multiply the $a_{ij}$ component of the matrix with the determinant of the matrix formed by deleting the $i$th row and $j$th column of our original matrix. This new matrix is called the $ij$-minor of $A$ and is commonly given the symbol $M_{ij}$. This term is then given a sign that depends on the sum of the two indices of the associated component: $(-1)^{i+j}$. So, in total, each term is
$$(-1)^{i+j}a_{ij}M_{ij}$$
We add all of these terms for an entire row, or an entire column. No matter which we choose, we will get the same number, called the determinant of the matrix.
To decrease the amount of terms involved, we usually try to expand along a row or column with the most zeroes. In your matrix, the third row has the most zeroes, with only one non-zero component: $a_{33} = 4$.
So the only non-zero term in the Laplace expansion of the determinant of your matrix is
$$(-1)^{3+3}a_{33}\left|\begin{array}{cccc}1&2&4&1\\0&-1&4&2\\-3&-6&-12&4\\0&0&1&1\end{array}\right| = 4\left|\begin{array}{cccc}1&2&4&1\\0&-1&4&2\\-3&-6&-12&4\\0&0&1&1\end{array}\right|$$
Now we do the same exercise for the 4x4 matrix. Let's call it $B$. The 4th row has the most zeroes, so we choose that for our expansion. The only non-zero components are $b_{43} = 1$ and $b_{44} = 1$. So our expansion is:
$$4\left((-1)^{4+3}b_{43}\left|\begin{array}{ccc}1&2&1\\0&-1&2\\-3&-6&4\end{array}\right| + (-1)^{4+4}b_{44}\left|\begin{array}{ccc}1&2&4\\0&-1&4\\-3&-6&-12\end{array}\right|\right) = 4\left(-\left|\begin{array}{ccc}1&2&1\\0&-1&2\\-3&-6&4\end{array}\right| + \left|\begin{array}{ccc}1&2&4\\0&-1&4\\-3&-6&-12\end{array}\right|\right)$$
The first column is fine for both of these 3x3 determinant expansions. I'll omit the cofactor symbolism since you already know how to expand 3x3 determinants:
$$4\left(-\left(\left|\begin{array}{cc}-1&2\\-6&4\end{array}\right| - 3\left|\begin{array}{cc}2&1\\-1&2\end{array}\right|\right) + \left(\left|\begin{array}{cc}-1&4\\-6&-12\end{array}\right| - 3\left|\begin{array}{cc}2&4\\-1&4\end{array}\right|\right)\right)$$
And finally, we expand the 2x2 determinants:
$$4(-((-4-(-12)) - 3(4 - (-1))) + ((12-(-24)) - 3(8-(-4)))) = 28$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1502099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
In the real number system,the equation $\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$ has how many solutions? In the real number system,the equation $\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$ has how many solutions?
I tried shifting the second term to the rhs and squaring.Even after that i'm left with square roots.No idea how to proceed.Help!
| Set $x=z^2+1$. Then:
$$ \sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+3-8\sqrt{x-1}} = \sqrt{(z-2)^2}+\sqrt{(z-3)^2} = |z-2|+|z-3| $$
equals one for every $z\in[2,3]$, hence for every $x\in[5,10]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1503958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
To show that group G is abelian if $(ab)^3 = a^3 b^3$ and the order of $G$ is not divisible by 3
Let $G$ be a finite group whose order is not divisible by $3$.
suppose $(ab)^3 = a^3 b^3$ for all $a,b \in G$. Prove that $G$ must be abelian.
Let$ $G be a finite group of order $n$. As $n$ is not divisible by $3$ ,$3$ does not divide $n$ thus $n$ should be relatively prime to $n$. that is gcd of an $n$ should be $1$.
$n = 1 ,2 ,4 ,5 ,7 ,8 ,10 ,11, 13 ,14 ,17,...$
further I know that all groups upto order $5$ are abelian and every group of prime order is cyclic. when it remains to prove the numbers which are greater than $5$ and not prime are abelian.
Am I going the right way?
please suggest me the proper way to prove this.
| I have followed the similar method as previous answer with a few changes (without defining the $f$, thought it would be easier:) )
As $(ab)^3=a^3b^3$ for all $a,b\in G$
we have \begin{align*} ababab&=aaabbb\\
\Rightarrow baba&=aabb\\
\Rightarrow (ba)^2&=a^2b^2 \end{align*}
Consider,
\begin{align*}(ab)^4&=((ab)^2)^2\\
&=(b^2a^2)^2\\
&=a^4b^4
\\
&=aaaabbbb\end{align*}
Also
\begin{align*}
(ab)^4&=abababab\\
&=a(ba)^3b
\end{align*}
Therefore, we get
\begin{align}aaaabbbb&=a(ba)^3b\\
\Rightarrow (ba)^3&=(ab)^3\\
\Rightarrow (ab)^{-3}(ba)^3&=e
\end{align}
Where $e$ is identity in $G$
$$\Rightarrow [(ab)^{-1}(ba)]^3=e$$
Now for $x=(ab)^{-1}(ba)$ ,$|x|$ divides $3$ by which $|x|$ can be $3$ or $1$.
Now $|x|$ can not be $3$ (as by Lagrange's theorem if $|x|$=3 then $3$ divides $|G|$ which is not true).
Thus,
\begin{align*}
|x|&=1\\
\Rightarrow x&=e\\
\Rightarrow (ab)^{-1}(ba)&=e
\end{align*}
Multiplying by $ab$ from left,
$$\Rightarrow ba=ab$$ for all $a,b\in G$.
Thus, $G$ is abelian.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1505189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Can a pre-calculus student prove this?
a and b are rational numbers satisfying the equation $a^3 + 4a^2b = 4a^2 + b^4$
Prove $\sqrt a - 1$ is a rational square
So I saw this posted online somewhere, and I kind of understand what the question is saying. I'm interesting in doing higher order mathematics but don't quite have the mathematical skills yet to grasp them (hence, precalculus) but is this problem way more complex than it seems or are the tools within the reach of a precalculus student?
| Here's another solution that proceeds mostly by the use of divisibility properties. All steps can be justified if one assumes the fundamental theorem of arithmetic, whose statement (but not proof) will be familiar to many precalculus students. I personally, however, would have found it tough-going to follow this proof as a precalculus student.
Write $a=\frac{w}{x}$, $b=\frac{y}{z}$, where $w$ and $x$ are relatively prime integers, $y$ and $z$ are relatively prime integers, and $x$ and $z$ are positive. We then have
$$
a^2(a+4b-4)=b^4\Rightarrow w^2(wz+4xy-4xz)=\frac{y^4x^3}{z^3}.
$$
Since the left side is an integer and $y$ and $z$ are relatively prime, $\frac{x^3}{z^3}=\left(\frac{x}{z}\right)^3$ is an integer and therefore $\frac{x}{z}$ is an integer. Write $x=nz$, where $n$ is a positive integer. Since $w$ is relatively prime to $x$, it is also relatively prime to $z$ and $n$. Substituting $x=nz$ gives
$$
w^2z(w+4ny-4nz)=y^4n^3.
$$
Since $n$ is relatively prime to $w$, it is relatively prime to $w+4ny-4nz$, and therefore $z$ is divisible by $n^3$. Write $z=n^3m$, where $m$ is a positive integer. Since $y$ is relatively prime to $z$, it is relatively prime to $n$ and $m$. Substituting $z=n^3m$ gives
$$
w^2m(w+4ny-4n^4m)=y^4.
$$
Since $m$ is a positive integer, relatively prime to $y$, but $m$ divides $y^4$, the only value $m$ can take is $1$. Hence $x=n^4$ and $z=n^3$. This leads to
$$
w^2(w+4ny-4n^4)=y^4.
$$
Assuming now that $y\ne0$, $\frac{y^4}{w^2}$ is an integer, and therefore $\frac{y^2}{w}$ is an integer. Setting $y^2=kw$, with $k$ a nonzero integer, and eliminating $w$ yields
$$
y^2+4nky-4n^4k=k^3.
$$
Solving the quadratic for $y$ gives
$$
y=-2nk\pm\sqrt{4n^2k^2+4n^4k+k^3}=-2nk\pm\sqrt{k(4n^4+4n^2k+k^2)}=-2nk\pm\sqrt{k(2n^2+k)^2}.
$$
So either $2n^2+k=0$, which implies $k=-2n^2$ and $y=4n^3$, or $k$ is a perfect integer square, say $k=r^2$, and
$$
y=-2nr^2+r(2n^2+r^2)=r(2n^2-2nr+r^2)=r(n^2+(n-r)^2).
$$
If the former, then
$$
a=\frac{w}{x}=\frac{y^2/(-2n^2)}{n^4}=\frac{-8n^4}{n^4}=-8,\qquad b=\frac{y}{z}=\frac{4n^3}{n^3}=4.
$$
If the latter, then
$$
a=\frac{w}{x}=\frac{y^2/r^2}{n^4}=\frac{(n^2+(n-r)^2)^2}{n^4},\qquad b=\frac{y}{z}=\frac{r(n^2+(n-r)^2)}{n^3},\qquad n,r\in\mathbf{Z}, n\ge1,
$$
and so $\sqrt{a}-1=\frac{(n-r)^2}{n^2}$.
In conclusion, either $y=0$ (and hence $b=0$ and $a\in\{0,4\}$), or $a=-8$ and $b=4$, or $\sqrt{a}-1$ is a rational square.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1505661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 8,
"answer_id": 4
} |
Determine the indefinite integral for $x\tan^{-1}x^{2}$ I'm not sure how to move on after using integration by parts to arrive at
$$\frac{x^{2}\tan^{-1}x^{2}}{2} - \frac{1}{2}\int \frac{x^{2}}{x^{4} + 1}\,dx$$
| $$\int x\arctan x^2\ dx=\frac{1}{2}x^2\arctan x^2 -\int \frac{x^3}{1+x^4}\ dx$$
by integration by parts, where u=arctan(x^2) and dv=xdx.
next use a simple u-substition where $u=1+x^4$.
$$\int \arctan x^2\ dx=\frac{1}{2} x^2 \arctan x^2-\int \frac{x^3}{1+x^4}\ dx
=\frac{1}{2}x^2\arctan x^2 -\frac{1}{4}\ln(1+x^4)+C $$
qed
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1507186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
To determine if a polynomial has real solution I have the following polynomial : $x^{7}+x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1$
I must determine if this polynomial has at least 1 real solution and justify why. We have a theorem which says that all polynomials with real coefficients can be decomposed in a product of polynomials of real coefficients with degree 1 or 2.
So this means we have four scenarios :
Factors : 2+2+2+1 , 2+2+1+1+1, 2+1+1+1+1+1, 1+1+1+1+1+1+1
In all these cases, we have atleast one factor of degree 1, so there is atleast one real solution in each case. What do you think ?
| $\begin{align*}
x^7+x^6+x^5+x^4+x^3+x^2+x+1 &= (x^4+1)(x^3+x^2+x+1) \\
&= (x^4+1)(x^2+1)(x+1)
\end{align*}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1507262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
A quotient of trigonometric expressions in complex analysis How is $$\frac{\cos\frac{20\pi}{3}+i\sin\frac{20\pi}{3}}{\cos{\frac{15\pi}{4}}{+i\sin\frac{15 \pi}{4}}}=\cos\frac{35\pi}{12}+i\sin\frac{35 \pi }{12}\ \ \ ?$$
I think that I am having trouble understanding a fundamental concept in complex analysis, but cannot pinpoint which.
| $$\frac{\cos\frac{20\pi}{3}+i\sin\frac{20\pi}{3}}{\cos{\frac{15\pi}{4}}{+i\sin\frac{15 \pi}{4}}}=\frac{e^{i\frac{20\pi}{3}}}{e^{i\frac{15\pi}{4}}}$$
$$=e^{i\frac{20\pi}{3}-i\frac{15\pi}{4}}$$
$$=e^{i\frac{35}{12}}$$
$$=\cos\frac{35\pi}{12}+i\sin\frac{35 \pi }{12}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1507839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
$\epsilon-N$ for $\lim_{n\to\infty} \frac{3n+1}{2n+5}=\frac32$. I want to prove using $\epsilon-N$ that $\lim_{n\to\infty} \frac{3n+1}{2n+5}=\frac32$.
Firstly we will find a sufficently large $n\in \Bbb N$:
\begin{align*}
\epsilon\quad\gt&\quad\left|\frac{3n+1}{2n+5} - \frac32 \right|\\
=&\quad \left|\frac{3n+1 -3n-7.5}{2n+5}\right|\\
=&\quad \left|\frac{-13}{4n+10}\right|\\
\implies&\quad \epsilon \gt \frac{13}{4n+10}\\
\implies&\quad \frac1\epsilon \lt \frac{4n+10}{13}\\
\implies&\quad \frac14\left(\frac{13}\epsilon-10\right) \lt n\\
\end{align*}
So $\forall \epsilon\gt 0, \exists N(\epsilon)>\frac14\left(\frac{13}\epsilon-10\right)$ such that $\forall n\in N, n>N$, $a_n=|\frac{3n+1}{2n+5}-\frac32|\lt \epsilon$
Proof: $\left|\frac{3n+1}{2n+5}-\frac{3}{2}\right| \color{red}{\lt} \left|\frac{3N+1}{2N+5}-\frac{3}{2}\right|\lt\epsilon$
Assuming this is even right. Everything feels good except the $\color{red}{\lt}$ in red, and then when I check, since we have
\begin{align*}
&\left|\frac{3n+1}{2n+5}-\frac{3}{2}\right| \color{red}{\lt}\left|\frac{3N+1}{2N+5}-\frac{3}{2}\right|,\quad n\gt N\\
&\implies \left|\frac{3n+1}{2n+5} - \frac32\right| - \left|\frac{3(n+1)+1}{2(n+1)+5} - \frac{3}{2}\right|\gt 0
\end{align*}
But this fails for all $n$? What did I do wrong?
Also, since $N\in\Bbb N$, we have to choose our $N$ so it is a natural number simultaneous to being greater than $\frac14\left(\frac{13}\epsilon-10\right)$, how do I ensure that subbing in $\frac14\left(\frac{13}\epsilon-10\right)$ is greater than subbing in the natural number that is the ceiling of this?
| $\lim_{n\to\infty} \frac{3n+1}{2n+5}=\frac32$.
Let $\epsilon >0$ be arbitrary.
Now by the archemedian property there exists $N \in \mathbb N$ such that $N>\frac{13}{\epsilon}$.
Observe that for each $n>N $
$|\frac{3n+1}{2n+5}-\frac{3}{2}|=\frac{13}{4n+10}< \frac{13}{n}<\frac{13}{N}<\epsilon$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1509081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Let $a,b,c$ be non-negative real numbers , then $\frac{(a+b+c)^2}{3}\ge a\sqrt{bc}+b\sqrt{ca}+c\sqrt{ab}$? Let $a,b,c$ be non-negative real numbers , then is it true that
$\dfrac{(a+b+c)^2}{3}\ge a\sqrt{bc}+b\sqrt{ca}+c\sqrt{ab}$ ?
| Yes, note that
$$ab+bc\ge 2 b\sqrt{ac},...$$
so
$$ab+bc+ca\ge a\sqrt{bc}+b\sqrt{ca}+c\sqrt{ab}.$$
The inequality then follows from
$$\frac{(a+b+c)^2}3\ge ab+bc+ca.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1514592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to integrate this fraction: $\int\frac{1}{1-2x^2}dx$? I'm not sure how to integrate this:
$$\int\frac{1}{1-2x^2}dx$$
I think it has to be this:
$$ -2\cdot \arctan(x)$$
Or this:
$$\arctan(\sqrt{-2x^2})$$
| $$ \int \frac{1}{1-2z^2} = \frac{1}{2} \int \frac{1}{1-\sqrt{2}z} + \frac{1}{1+\sqrt{2}z} = \frac{1}{2} (-\frac{1}{\sqrt{2}} \log|1-\sqrt{2}z| + \frac{1}{\sqrt{2}} \log|1+\sqrt{2}z|) = \frac{1}{2\sqrt{2}} \log|\frac{1+\sqrt{2}z}{1-\sqrt{2}z}| $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1517816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
how to solve the following mordell equation:$ y^2 = x^3 - 3$ i just started solving mordell's equations and get a little bit stuck. For example:
$ y^2 = x^3 - 3$.
I know that $x$ must be odd, for if $x$ is even $y^2 \equiv 5 \pmod{8}$. So $x \equiv \{1,3\} \pmod{4}$. Now note the following if we add 4 at both sides:
$y^2 + 4 = (x+1)(x^2 -x +1)$. The right hand side is the same as $(x-1)^2 + x$
I now know $(x^2 -x +1)$ is divisible by a prime divisor $3 \pmod{4}$ if $x \equiv 3 \pmod {4}$ and divisible by $1 \pmod{4}$ if $x \equiv 1 \pmod{4}$.
So we get $y^2 + 4 \equiv 0\pmod{p}$, which is solvable if $\left(\frac{-2}{p}\right)$ = 1. (I don't think this step is valid). this is only true if the following occurs:
$p \equiv 1 \pmod{4}$ and $p \equiv +- 1\pmod{8}$.
or
$p \equiv 3 \pmod{4}$ and $p \equiv +- 3\pmod{8}$.
so $p \equiv \{1,3\}\pmod{8}$
but $y^2 + 4$ is either $4 \pmod{8}$ or $5 \pmod {8}$, so there are no solutions?
Am i solving this the correct way or is there a shorter path?
Kees
| Your solution is flawed, so here's a complete solution: As you said, $x$ is odd, because if $x$ were even, then $y^2\equiv -3\pmod{8}$, which is not a quadratic residue. Then $y=2k$ is even.
$4(k^2+1)=(x+1)\left(x^2-x+1\right)$.
$x^2-x+1$ is always odd. Let $p$ be a prime divisor of $x^2-x+1$. Then $p\equiv 1\pmod{4}$, because if $p\equiv 3\pmod{4}$, then $k^2\equiv -1\pmod{p}$, contradiction (by Quadratic Reciprocity). Therefore $x^2-x+1\equiv 1\pmod{4}$, so $4\mid x(x-1)$, so $4\mid x-1$, i.e. $x\equiv 1\pmod{4}$. But then $x+1\equiv 2\pmod{4}$, contradiction, because this implies $4\nmid (x+1)\left(x^2-x+1\right)$ (while $4$ divides the LHS). No solutions exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1520668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate the Integral: $\int^\pi_0\cos^6\theta\ d\theta$ $\int^\pi_0\cos^6\theta\ d\theta$
So I split the trig value into:
$\int^\pi_o\cos^5\theta\ cos\theta\ d\theta$
Then I utilized the Pythagorean theorem for $cos^5\theta$
$\int^\pi_o(1-sin^5\theta)\ cos\theta$
I utilized u-substitution:
$u=sin\ \theta$
$du=cos\ \theta$
Thus:
$\int^{x=\pi}_{x=0}\ (1-u^5)\ d\theta$
I intergated
$(\frac{1}{6}u^6)+(\frac{1}{6}u^6)$
$-(\frac{\pi^6}{6})+(0)$
$-(\frac{\pi^6}{6})$
Is my answer right?
| $$\int^\pi_0\cos^6\theta\ d\theta=\int^\pi_0(\frac{1+\cos2\theta}{2})^3 d\theta$$
$$\frac{1}{8}\int^\pi_0(1+\cos2 \theta)^3d\theta=\frac{1}{8}\int^\pi_0(1+3\cos2\theta+3\cos^2 2\theta+\cos^32\theta)d\theta $$
$$\frac{1}{8}\int^\pi_0(1+3\cos2\theta+\frac{3}{2}(1+cos 4\theta)+\cos2\theta(1-\sin^22\theta)))d\theta$$
see, all terms can be integrated directly
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1521105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Prove $ \forall x >0, \quad \sqrt{x +2} - \sqrt{x +1} \neq \sqrt{x +1}-\sqrt{x}$
I would like to prove
$$ \forall x >0, \quad \sqrt{x +2} - \sqrt{x +1} \neq \sqrt{x +1}-\sqrt{x}$$
*
*I'm interested in more ways of proving it
My thoughts:
\begin{align}
\sqrt{x+2}-\sqrt{x+1}\neq \sqrt{x+1}-\sqrt{x}\\
\frac{x+2-x-1}{\sqrt{x+2}+\sqrt{x+1}}&\neq \frac{x+1-x}{\sqrt{x +1}+\sqrt{x}}\\
\frac{1}{\sqrt{x+2}+\sqrt{x+1}}&\neq \frac{1}{\sqrt{x +1}+\sqrt{x}}\\
\sqrt{x +1}+\sqrt{x} &\neq \sqrt{x+2}+\sqrt{x+1}\\
\sqrt{x} &\neq \sqrt{x+2}\\
\end{align}
*
*Is my proof correct?
*I'm interested in more ways of proving it.
| By the MVT:
$$\sqrt {x+2} - \sqrt {x+1} = \frac{1}{2\sqrt {c_x}}\cdot 1, \ \ \ \ \sqrt {x+1} - \sqrt {x} = \frac{1}{2\sqrt {d_x}}\cdot 1.$$
Here $c_x\in (x+1,x+2), d_x\in (x,x+1).$ Because $1/\sqrt x$ strictly decreases, the left term minus the right term is negative.
Concavity: Slopes of successive chords on a strictly concave graph are strictly decreasing, and $\sqrt x$ is strictly concave. Therefore $\sqrt{x +2} - \sqrt{x +1} < \sqrt{x +1}-\sqrt{x}$ for all $x\ge 0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1523179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 3
} |
Convergence of the series $\sum\limits_{n = 4}^\infty {\frac{{n + 1}}{{(n + 5)(n + 4)(n - 3)}}}$? To analyze the convergence of the
$$\sum\limits_{n = 4}^\infty {\frac{{n + 1}}{{(n + 5)(n + 4)(n - 3)}}}$$
series I used the criterion of integral $$\displaystyle\int_4^\infty {\frac{{x + 1}}{{(x + 5)(x + 4)(x - 3)}}dx},$$ but calculate this improper integral is a very laborious task.
Is there a shorter way? What criteria of convergence would be most effective or simple?
| $$\dfrac{n+1}{(n+5)(n+4)(n-3)} = -\dfrac12\cdot\dfrac1{n+5} + \dfrac37\cdot\dfrac1{n+4} + \dfrac1{14} \cdot\dfrac1{n-3}$$
Hence,
\begin{align}
\sum_{n=4}^{m} \dfrac{n+1}{(n+5)(n+4)(n-3)} & = - \dfrac12 \cdot \sum_{n=4}^{m} \dfrac1{n+5} + \dfrac37 \cdot \sum_{n=4}^{m} \dfrac1{n+4} + \dfrac1{14} \cdot \sum_{n=4}^{m} \dfrac1{n-3}\\
& = - \dfrac12 \sum_{n=9}^{m+5} \dfrac1n + \dfrac37 \sum_{n=8}^{m+4} \dfrac1n + \dfrac1{14} \sum_{n=1}^{m-3}\dfrac1n\\
& = - \dfrac12 \sum_{n=9}^{m+5} \dfrac1n + \dfrac37 \sum_{n=9}^{m+4} \dfrac1n + \dfrac1{14} \sum_{n=9}^{m-3}\dfrac1n + \dfrac37 \sum_{n=8}^{8} \dfrac1n + \dfrac1{14} \sum_{n=1}^{8}\dfrac1n\\
& = \dfrac{971}{3920} + \dfrac37 \sum_{m-2}^{m+4} \dfrac1n - \dfrac12 \sum_{m-2}^{m+5}\dfrac1n
\end{align}
Taking $m \to \infty$, we have $$\lim_{m \to \infty}\left(\dfrac37 \sum_{m-2}^{m+4} \dfrac1n - \dfrac12 \sum_{m-2}^{m+5}\dfrac1n \right) = 0$$
Hence,
$$\sum_{n=4}^{\infty} \dfrac{n+1}{(n+5)(n+4)(n-3)} = \dfrac{971}{3920}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1526806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Probability of roots of equation are real Let $A$, $B$, and $C$ be independent random variables, uniformly distributed over $[0,9], [0,3]$ and $[0,5]$ respectively. What is the probability that both roots of the equation $Ax^2+Bx+C=0$ are real?
I know I need $B^2>4AC$ but not sure how to get probability.
| This is more of a rough sketch:
With a slight abuse of notation:
$|A| = 9-0 = 9$
$|B| = 3-0 = 3$
$|C| = 5-0 = 5$
First the probability distribution of $B^2$:
$P(B^2 < x) = P(-\sqrt{x} < B < \sqrt{x}) = P(0 < B < \sqrt{x}) = \frac{\sqrt{x}}{|B|}$
$\implies f_{B^2}(x) = \frac{1}{2\sqrt{x}\ |B|}$
Now consider:
\begin{align*}
P(B^2-4AC > 0) &= P(C<\frac{B^2}{4A})\\
&=P(C < \frac{B^2}{4A}, \frac{B^2}{4A} > |C|) + P(C < \frac{B^2}{4A}, \frac{B^2}{4A} < |C|)\\
&=P(C < |C|, \frac{B^2}{4A} > |C|) + P(C < \frac{B^2}{4A}, \frac{B^2}{4A} < |C|)\\
&= P( C < |C| | \frac{B^2}{4A} > |C|) \times P(\frac{B^2}{4A} > |C|) + P (C < \frac{B^2}{4A}) \times P(\frac{B^2}{4A} < |C|) \\
&= 1 \times P({B^2} > 4A|C|) + \int \int \frac{B^2}{4A|C|} \times P(B^2 < 4A|C|) dA dB
\end{align*}
Let $B'=B^2$
\begin{align*}
P(B^2 < 4A|C|) &= \int_0^{|B'|} \int_{\frac{B'}{4|C|}}^{|A|} \frac{\sqrt{4|C|}}{|B'|} \sqrt{A}\ dA\ dB'\\
&= \frac{\sqrt{4|C|}}{|B'|} \int_0^{|B'|} \frac{2}{3} A^{3/2}\big|_{\frac{B'}{4|C|}}^{|A|} dB'\\
&= \frac{2}{3}\frac{\sqrt{4|C|}}{|B'|} \big(|A|^{3/2}B' - \frac{2}{5}\frac{B'^{5/2}}{(4|C|)^{3/2}} \big)\big|_{0}^{|B'|}\\
&= \frac{2}{3}\frac{\sqrt{4|C|}}{|B'|} \big(|A|^{3/2}|B'| - \frac{2}{5}\frac{|B'|^{5/2}}{8|C|^{3/2}} \big)\\
\end{align*}
\begin{align*}
\int \int \frac{B^2}{4A|C|} \times P(B^2 < 4A|C|) &= \frac{\sqrt{4|C|}}{4|C||B'|} \int_0^{|B'|} \int_{\frac{B'}{4|C|}}^{|A|} \frac{B'}{A} \sqrt{A} \ dA \ dB'\\
&= \frac{\sqrt{4|C|}}{4|C||B'|} \int_0^{|B'|} {2B'}\sqrt{A}\big|_{\frac{B'}{4|C|}}^{|A|} \ dB'\\
&= \frac{\sqrt{4|C|}}{2|C||B'|} \int_0^{|B'|} B'(\sqrt{|A|}-\sqrt{\frac{B'}{4|C|}}) \ dB'\\
&= \frac{\sqrt{4|C|}}{2|C||B'|} \big( \sqrt{|A|} \frac{|B'|^2}{2} - \frac{2}{5} \frac{|B'|^{5/2}}{\sqrt{4|C|}} \big)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1528677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Given Two Fibonacci numbers, predicting the median Fibonacci number Wolfram Alpha gives the $100$th fibonacci number to be $354224848179261915075$ and the $104$th fibonacci number to be $2427893228399975082453$. Just from this, can we deduce what the $102$th fibonacci will be? Is it at all possible, and if it is possible, can we use it to predict the $n$th fibonacci number?
|
Generalization:
$$F_{n+4} = F_{n+3} + F_{n+2} = (F_{n+2} + F_{n+1}) + F_{n+2} = 2\cdot F_{n+2} + F_{n+1}$$
Since
$$F_{n+2} = F_{n+1} + F_{n} \implies F_{n+1} = F_{n+2} - F_{n}$$
$$\therefore F_{n+4} = 2\cdot F_{n+2} + (F_{n+2} - F_{n}) = 3\cdot F_{n+2} - F_{n}$$
$$\implies F_{n+2} = \dfrac{F_{n+4} + F_{n}}{3}$$
Now substituting $n = 100$ we have:
$$F_{102} = \dfrac{F_{104} + F_{100}}{3} = \dfrac{2427893228399975082453 + 354224848179261915075}{3} = 927372692193078999176 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1529311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Finding $\int _0^a\sqrt{1+\frac{1}{4x}}dx$ to calculate arclength So I'm trying to find the arclength of $x^{0.5}$ and its tougher than I thought. Tried substitutions like $\dfrac{\cot^2x}{4}$ and some other trig subs but they got me nowhere. Any tips?
$$\int _0^a\sqrt{1+\frac{1}{4x}}dx$$
Edit:
This is what I got so far: $\int_{0}^{a}\sqrt{1+\left(\left(\sqrt{x}\right)'\right)^{2}}dx=\int_{0}^{a}\sqrt{1+\frac{1}{4x}}dx=\left[\begin{array}{cc}
t^{2}=1+\frac{1}{4x} & \sqrt{1+\frac{1}{4N}},\sqrt{1+\frac{1}{4a}}\\
2tdt=-\frac{1}{8x^{2}}dx & x=\frac{1}{4\left(t^{2}-1\right)}
\end{array}\right]=\\\lim_{N\to0}-\int_{\sqrt{1+\frac{1}{4N}}}^{\sqrt{1+\frac{1}{4a}}}\frac{t^{2}dt}{\left(t^{2}-1\right)^{2}}=\lim_{N\to0}\int_{\sqrt{1+\frac{1}{4N}}}^{\sqrt{1+\frac{1}{4a}}}\left(\frac{1}{4\left(t+1\right)}-\frac{1}{4\left(t+1\right)^{2}}-\frac{1}{4\left(t-1\right)}-\frac{1}{4\left(t-1\right)^{2}}\right)dt=\\=\lim_{N\to0}\frac{1}{4}\left[\ln\left(t+1\right)+\frac{1}{t+1}-\ln\left(t-1\right)+\frac{1}{t-1}\right]_{\sqrt{1+\frac{1}{4N}}}^{\sqrt{1+\frac{1}{4a}}}=\\=\left[\ln\left(\frac{\sqrt{1+\frac{1}{4a}}+1}{\sqrt{1+\frac{1}{4a}}-1}\right)-\frac{2\sqrt{1+\frac{1}{4a}}}{1+\frac{1}{4a}-1}\right]-\lim_{N\to0}\left[\ln\left(\frac{\sqrt{1+\frac{1}{4N}}+1}{\sqrt{1+\frac{1}{4N}}-1}\right)-\frac{2\sqrt{1+\frac{1}{4N}}}{1+\frac{1}{4N}-1}\right]=\\=\left[\ln\left(\frac{\sqrt{1+\frac{1}{4a}}+1}{\sqrt{1+\frac{1}{4a}}-1}\right)-8a\sqrt{1+\frac{1}{4a}}\right]-\lim_{N\to0}\left[\ln\left(\frac{\sqrt{1+\frac{1}{4N}}+1}{\sqrt{1+\frac{1}{4N}}-1}\right)-8N\sqrt{1+\frac{1}{4N}}\right]=\\=\ln\left(\frac{\sqrt{1+\frac{1}{4a}}+1}{\sqrt{1+\frac{1}{4a}}-1}\right)-8a\sqrt{1+\frac{1}{4a}}-0+0=\ln\left(4a\left(\sqrt{\frac{1}{a}+4}+2\right)+1\right)-8a\sqrt{1+\frac{1}{4a}}$
But it doesn't seem right... any Ideas what went wrong?
| We have $$\int_{0}^{a}\sqrt{1+\frac{1}{4x}}dx=\frac{1}{4}\int_{0}^{a}\left(\frac{1}{\sqrt{x}\sqrt{1+4x}}+\frac{1+8x}{\sqrt{x}\sqrt{1+4x}}\right)dx
$$ for the first integral use the substitution $4x=t^{2}
$ (the second is trivial) to get $$\int_{0}^{a}\sqrt{1+\frac{1}{4x}}dx=\frac{1}{4}\left(\textrm{arcsinh}\left(2\sqrt{a}\right)+2\sqrt{a}\sqrt{1+4a}\right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1531672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Solving the recurrence $a_{n+2} = 3a_{n+1} - 2a_n, a_0 = 1, a_1 = 3$ using generating functions
Solve the following recurrence using generating functions: $a_{n+2} = 3a_{n+1} - 2a_n, a_0 = 1, a_1 = 3$.
My partial solution:
We can rewrite $a_{n+2} = 3a_{n+1} - 2a_n$, as $a_{n+2} - 3a_{n+1} + 2a_n = 0$, and we let $A(z) = \sum a_n z^n$. The goal is to compute $A(z)$ as this can be done as follows: $$A(z) - a_0 - a_1z - 3z(A(z) - a_0) + 2z^2A(z) = 0$$
$$(1-3z+2z^2)A(z) = a_0 + a_1z -3a_0z$$
$$A(z) = \frac{a_0 + (a_1 - 3a_0)z}{1-3z+2z^2}$$
$$\quad \quad = \frac{a_0 + (a_1 - 3a_0)z}{(1-z)(1-2z)}$$
$$\quad \quad = \frac{C}{(1-z)}+\frac{D}{(1-2z)}$$
And, I don't know how to continue, I cannot figure out the remaining. I'm pretty sure it is obvious, but I just cannot see it. If someone can help me I would be glad.
| Simple trick for partial fractions: Say yu have:
$$
\frac{a_0 + (a_1 - 3 a_0) z}{(1 - z) (1 - 2 z)} = \frac{A}{1 - z} + \frac{B}{1 - 2 z}
$$
This is supposed to be an identity, so for instance:
$\begin{align*}
\lim_{z \to 1/2} \left(
(1 - 2 z)
\cdot \frac{a_0 + (a_1 - 3 a_0) z}{(1 - z) (1 - 2 z)}
\right)
&= \lim_{z \to 1/2} \left(
(1 - 2 z)
\cdot \frac{A}{1 - z}
\right)
+ \lim_{z \to 1/2} \left(
(1 - 2 z)
\cdot \frac{B}{1 - 2 z}
\right) \\
\left. \frac{a_0 + (a_1 - 3 a_0) z}{(1 - z)} \right|_{z = 1/2}
&= 0 + B \\
\frac{a_0 + (a_1 - 3 a_0) \cdot 1/2}{1/2}
&= B
\end{align*}$
This simplifies to $B = a_1 - a_0$.
The same works for $A$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1532413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Combinatorics Identity about Catalan numbers: $\sum_{k=0}^n \frac{1}{k+1}\binom{2k}k \binom{2n-2k}{n-k}=\binom{2n+1}n$ I need to prove this identity:
$\sum_{k=0}^n \frac{1}{k+1}{2k \choose k}{2n-2k \choose n-k}={2n+1 \choose n}$
without using the identity:
$C_{n+1}=\sum_{k=0}^n C_kC_{n-k}$.
Can't figure out how to.
| Suppose we seek to prove that
$$\sum_{k=0}^n \frac{1}{k+1} {2k\choose k}
{2n-2k\choose n-k} = {2n+1\choose n}.$$
We get for the LHS
$$[z^n] (1+z)^{2n} \sum_{k=0}^n \frac{1}{k+1} {2k\choose k}
\frac{z^k}{(1+z)^{2k}}.$$
Here the coefficient extractor $[z^n]$ combined with the factor $z^k$
enforces the upper limit of the sum which we may thus extend to
infinity:
$$[z^n] (1+z)^{2n} \sum_{k\ge 0} \frac{1}{k+1} {2k\choose k}
\frac{z^k}{(1+z)^{2k}}.$$
Now the generating function of the Catalan numbers is
$$\frac{1-\sqrt{1-4z}}{2z}$$ so this simplifies to
$$\begin{align*}
& [z^n] (1+z)^{2n} \frac{1-\sqrt{1-4z/(1+z)^2}}{2z/(1+z)^2}
\\ = & [z^{n}] (1+z)^{2n+1}
\frac{1+z-\sqrt{(1+z)^2-4z}}{2z}
\\ = & [z^{n}] (1+z)^{2n+1}
\frac{1+z-\sqrt{(1-z)^2}}{2z}
\\ = & [z^n] (1+z)^{2n+1}
= {2n+1\choose n}
\end{align*}$$
as claimed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1533362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How to show that $a_{1}=a_{3}=a_{5}=\cdots=a_{199}=0$ Let $$(1-x+x^2-x^3+\cdots-x^{99}+x^{100})(1+x+x^2+\cdots+x^{100})=a_{0}+a_{1}x+\cdots+a_{200}x^{200}$$
show that
$$a_{1}=a_{3}=a_{5}=\cdots=a_{199}=0$$
I have one methods to solve this problem:
Let$$g(x)=(1-x+x^2-x^3+\cdots-x^{99}+x^{100})(1+x+x^2+\cdots+x^{100})$$
Note $$g(x)=g(-x)$$
so $$a_{1}=a_{3}=\cdots=a_{199}=0$$
there exist other methods?
| HInt 1:
$$g(x)=(1+x^2+\cdots+x^{100})^2-x^2(1+x^2+x^4+\cdots+x^{98})^2$$
Hint 2: since
$$g(x)=\dfrac{1+x^{101}}{1+x}\cdot\dfrac{1-x^{101}}{1-x}=\dfrac{1-x^{202}}{1-x^2}=1+x^2+\cdots+x^{198}+x^{200}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1533555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Express $\frac{9x}{(2x+1)^2(1-x)}$ as a sum of partial fractions with constant numerators. Answer doesn't match with solution provided in book. Express $\frac{9x}{(2x+1)^2(1-x)}$ as a sum of partial fractions with constant numerators. Answer doesn't match with solution provided in book.
My method:
$\frac{9x}{(2x+1)^2(1-x)}\equiv\frac{A}{2x+1}+\frac{B}{(2x+1)^2}+\frac{C}{1-x}$
Comparing L.H.S. numerators with R.H.S. numerators:
$9x=A(2x+1)+B(1-x)+C(2x+1)^2$
$x=-\frac12, B=-3$
$x=0, A+B+C=0 \Rightarrow A=3+C$
$x=1, 3A+9C=9 \Rightarrow 3(3+C)+9C=9\Rightarrow C=0$
This is the point where my answer differs from the one given in the book. And I don't understand why.
Answer in book: $\frac{1}{1-x}+\frac{2}{2x+1}-\frac{3}{(2x+1)^2}$
| You have to have
$$9x=A(2x+1)\color{red}{(1-x)}+B(1-x)+C(2x+1)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1535277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Subtracting square roots I'm studying trigonometry but I don't understand one example I have here:
$\cos(\frac{5}{12}\pi) = \cos(\frac{1}{6}\pi + \frac{1}{4}\pi)$
$= \cos(\frac{1}{6}\pi)\cos(\frac{1}{4}\pi) - \sin(\frac{1}{6}\pi)\sin(\frac{1}{4}\pi)$
$=(\frac{1}{2}\sqrt{3})(\frac{1}{2}\sqrt{2})-(\frac{1}{2})(\frac{1}{2}\sqrt{2})$
$=\frac{1}{4}\sqrt{2}(\sqrt{3}-1)$
I got the result $\frac{1}{4}\sqrt{6}-\frac{1}{4}\sqrt{2}$, which is the same as the result above. However, I don't know how this calculation became $=\frac{1}{4}\sqrt{2}(\sqrt{3}-1)$.
Please, anyone can explain to me?
| As fast as light:
$$\frac{1}{4}\sqrt{6} - \frac{1}{4}\sqrt{2} = \frac{1}{4}(\sqrt{6} - \sqrt{2})$$
Now you will use the property $\sqrt{6} = \sqrt{3\cdot 2} = \sqrt{3}\cdot \sqrt{2}$ to get the result, i.e.
$$\frac{1}{4}\sqrt{2}(\sqrt{3} - 1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1539713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluating the limit $\mathop {\lim }\limits_{x \to 0} \frac{{x(1 - 0.5\cos x) - 0.5\sin x}}{{{x^3}}}$ For evaluating the limit $\lim\limits_{x \to 0} \frac{x(1 - 0.5\cos x) - 0.5\sin x}{x^3}$, I proceeded as follows:
$$\lim_{x \to 0} \left(\frac{x(1 - 0.5\cos x)}{x^3} - \frac{0.5}{x^2}\left(\frac{\sin x}{x}\right)\right)$$
Using the fact that $\lim\limits_{x \to 0} \frac{\sin x}{x}=1$ and $\lim\limits_{x \to a} \{ f(x) - g(x)\} = \lim\limits_{x \to a} f(x) - \lim\limits_{x \to a} g(x)$, I got
$$\lim\limits_{x \to 0} \frac{1 - 0.5\cos x}{x^2} - \lim\limits_{x \to 0} \frac{0.5}{x^2}$$
$$ = \lim_{x \to 0} \left(\frac{1 - 0.5\cos x - 0.5}{x^2} \right)$$
Now, after applying L'Hopital's Rule I got the final answer as $0.25$. However, on evaluating the original limit using Mathematica, I got the answer as $\frac{1}{3}$.
Can someone please tell me where am I going wrong.
Somehow, I believe that you cannot use the fact that $\lim\limits_{x \to a} \{ f(x) - g(x)\} = \lim \limits_{x \to a} f(x) - \lim\limits_{x \to a} g(x)$ in case of indeterminate forms.
Thanks in advance!
| Apply iteratedly l'Hôpital Rule to $$\frac{x(2-\cos x)-\sin x}{2x^3}$$ We get successively $$\frac{2-2\cos x+x\sin x}{6x^2}$$ $$\frac{3\sin x+x\cos x}{12x}$$ $$\frac{3\cos x+\cos x-x\sin x}{12}$$
Thus the searched limit is the limit of $$\frac{4\cos x-x\sin x}{12}$$ when $x$ tends to $0$ which is clearly $\frac{4}{12}=\frac 13$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1541165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
What is $\frac{x^{10} + x^8 + x^2 + 1}{x^{10} + x^6 + x^4 + 1}$ given $x^2 + x - 1 = 0$?
Given that $x^2 + x - 1 = 0$, what is $$V \equiv \frac{x^{10} + x^8 + x^2 + 1}{x^{10} + x^6 + x^4 + 1} = \; ?$$
I have reduced $V$ to $\dfrac{x^8 + 1}{(x^4 + 1) (x^4 - x^2 + 1)}$, if you would like to know.
| Applying polynomial long division, we have
$\begin{array}{rlllllllllll}
&~~1x^8-1x^7+3x^6+\dots\\
\hline
x^2+x-1&|x^{10}+0x^9+x^8+0x^7+0x^6+0x^5+0x^4+0x^3+1x^2+0x+1\\
&~x^{10}+x^9-x^8\\
\hline
&~~~~~-x^9+2x^8+0x^7\\
&~~~~~-x^9-x^8+x^7\\
\hline
&~~~~~~~~~~~~~~~~~3x^8+\dots\\
&~~~~~~~~~~~~~~~~~~\vdots
\end{array}$
eventually arriving at $$x^{10}+x^8+x^2+1\\=(x^8-x^7+3x^6-4x^5+7x^4-11x^3+18x^2-29x+48)(x^2+x-1)+49-77x\\=49-77x$$
Similarly, applying polynomial long division to the denominator, it simplifies to
$42-66x$.
So the fraction in general simplifies to
$\frac{49-77x}{42-66x}=\frac{7}{6}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1542378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
Solving $a_n=3a_{n-1}-2a_{n-2}+3$ for $a_0=a_1=1$ I'm trying to solve the recurrence $a_n=3a_{n-1}-2a_{n-2}+3$ for $a_0=a_1=1$. First I solved for the homogeneous equation $a_n=3a_{n-1}-2a_{n-2}$ and got $\alpha 1^n+\beta 2^n=a_n^h$. Solving this gives $a_n^h =1$. The particular solution, as I understand, will be $a_n^*=B$ since $f(n)=3\times 1^n$. But then I get $B=a^*=3a^*_{n-1}-2a^*_{n-2}+3=B+3$. This has to be a mistake, but I don't see what I did wrong.
| $$a_n - a_{n-1}+3=2(a_{n-1}-a_{n-2}+3)$$
$$a_n - a_{n-1}+3=2^{n-1}(a_1-a_0+3)=3\cdot 2^{n-1}$$
$$\\$$
$$a_n = a_{n-1} + 3\cdot (2^{n-1}-1)$$
$$a_{n-1} = a_{n-2} + 3\cdot (2^{n-2}-1)$$
$$...$$
$$a_1 = a_0 + 3\cdot (2^0-1)$$
Adding side by side,
$$a_n=a_0+3\cdot (2^0+2^1+...+2^{n-1}-n)$$
$$=1+3\cdot (2^n-1-n)=3\cdot2^n-3n-2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1543804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find GCD$(A_0,A_1,...,A_{2015})$ where $A_n=2^{3n}+3^{6n+2}+5^{6n+2}$ ; $n=0,...,2015$ Find GCD$(A_0,A_1,...,A_{2015})$ where $A_n=2^{3n}+3^{6n+2}+5^{6n+2}$ ; $n=0,...,2015$
Is there some intuitive method or formula for finding GCD of $n$ integers?
| $n=0$ gives $1+9+25 = 35$, so it's either $1, 5, 7$ or $35$.
$n=1$ gives $8+ 3^8 + 5^8 = 397194 = 2 \cdot 3 \cdot 7^3 \cdot 193$ so it could be $1$ or $7$.
Let's see what happens mod $7$ (this means we look at all numbers ignoring powers of $7$, so basically any number becomes $0,1,2,3,4,5$ or $6$ according to the rest of its division by $7$)
$2^{3n} = 8^n = 1^n = 1$
$3^{6n+2} = (9)(3^{6n}) = 2(2^{3n}) = 2(1)^n = 2$
$5^{6n+2} = (25)(5^{6n}) = 4(4^{3n}) = 4(64^n) = 4(1)^n = 4$
$A_n = 1 + 2 + 4 = 7 = 0$
So any of them is divisible by $7$, so it's $7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1544893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
partial fraction for complex roots While solving Laplace transform using Partial fraction expansion. I have confusion in solving partial fraction for complex roots.
I have this equation $$ \frac {2s^2+5s+12} {(s^2+2s+10)(s+2)}$$
Please anyone help to tell me to understand the steps for solving partial fraction for complex roots
| It is the same story as with real roots.
Consider $$A=\frac {2s^2+5s+12} {(s^2+2s+10)(s+2)}$$ the first partial fraction decomposition gives $$A=\frac{s+1}{s^2+2 s+10}+\frac{1}{s+2}$$ So, now, just focus on $$B=\frac{s+1}{s^2+2 s+10}=\frac{s+1}{(s+1+3i)(s+1-3i)}=\frac \alpha {(s+1+3i)}+\frac \beta {(s+1-3i)} $$ Reducing to same denominator $$s+1=\alpha(s+1+3i)+\beta (s+1-3i)=(\alpha+\beta)(s+1)+3(\alpha-\beta)i$$ Idenitfyng the real and imaginary parts than gives $$\alpha+\beta=1 \, \, \, \quad \alpha-\beta=0$$ that is to say $\alpha=\beta=\frac 12$.
Edit
To clarify the first partial fraction decomposition $$\frac {2s^2+5s+12} {(s^2+2s+10)(s+2)}=\frac a{s+2}+\frac {b+cs}{s^2+2s+10}$$ Reducing to same denominator and removing the common denominator $$2s^2+5s+12=a(s^2+2s+10)+(b+cs)(s+2)$$ Expanding the rhs and grouping terms $$2s^2+5s+12=(a+c)s^2+ (2 a+b+2 c) s+(10 a+2 b)$$ Comparing the coefficients $$a+c=2 \quad, 2a+b+2c=5 \quad, 10a+2b=12$$ Solving the three equations for the three unknowns leads to $a=b=c=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1546961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How to Evaluate $\lim_{x\to0}\frac{(1-x)^{1/3}-1}{4^x-3^x}$? How to find this limit without using L'Hospital rule
$$\lim_{x\to0}\frac{(1-x)^{1/3}-1}{4^x-3^x}$$
| $Step\,1.$ $(1+x)^a = 1+ax + O(x^2)$ for all a and all $x$ with $|x|<1$
$Step\,2.$ $ 4^x - 3^x= 4^x\left(1-\left(\frac34\right)^x\right)$
$Step\,3.$ $(3/4)^x = \exp(x \ln(3/4)) = 1+ x \ln(3/4)+ \frac{x^2}2 \ln^2(3/4)+ O(x^3)$.
$Step\,4.$ The limit $L= \lim_{x \to 0} \frac{(1-\frac13x + O(x^2)-1)}{4^x ( -x \ln(3/4)+O(x^2))}= \lim_{x \to 0} \frac{-\frac13 x}{ -x\ln3/4 }=-\frac1{3\ln4/3}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1547720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the sum of the series $1+\frac{1}{3}\cdot\frac{1}{4}+\frac{1}{5}\cdot\frac{1}{4^2}+\frac{1}{7}\cdot\frac{1}{4^3}+\cdots$ Find the sum of the series : $$1+\frac{1}{3}\cdot\frac{1}{4}+\frac{1}{5}\cdot\frac{1}{4^2}+\frac{1}{7}\cdot\frac{1}{4^3}+\cdots$$
| This can be transformed to $$\sum_{n=1}^{\infty} \frac{2}{(2n-1)2^{2n-1}}$$
Let $$f(x)=\sum_{n=1}^{\infty} \frac{x^{2n-1}}{(2n-1)}$$
Then, we have $f'(x)=\sum_{n=1}^{\infty} x^{2n-2} = \frac{1}{1-x^2}$.
Therefore, we have $$f(x)=\int \frac{1}{1-x^2} = \frac{1}{2} \ln \frac{x+1}{1-x}+C$$
It is clear that $C=0$.
Now plugging $x=\frac{1}{2}$ in this equation, we have $\sum_{n=1}^{\infty} \frac{1}{(2n-1)2^{2n-1}} = \frac{1}{2} \ln 3$, so the desired answer is double that number, or $\ln 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1548665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Find generating function of sequences: $a_n=(n^2+n+1)_{n\ge0},b_n=(2^{1+[\frac{n}{3}]})_{n\ge 0}$ Find generating function of sequences: $a_n=(n^2+n+1)_{n\ge0},b_n=(2^{1+[\frac{n}{3}]})_{n\ge 0}$
For the first function, generating function is trivial:
$$f(x)=\sum\limits_{n=0}^{\infty}a_nx^n=\sum\limits_{n=0}^{\infty}(n^2+n+1)x^n=\frac{n^2+n+1}{1-x}$$
Second sequence: $2,2,2,4,4,4,8,8,8,...$
$$2+2x+2x^2+4x^3+...=(2+2x^2+4x^4+...)+x(2+4x^2+4x^4+...)$$
I don't know how to find close form of these partial sums.
Could someone give a hint?
| $a_n$:
$$
f(x)=\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}\\
\sum_{n=0}^{\infty}\left(1+n+n^2\right)x^n=f(x)+x\cdot f'(x)+x\left(x\cdot f'(x)\right)'$$
$b_n$:
$$\sum_{n=0}^{\infty}2^{n+1}\left(1+x+x^2\right)x^{3n}\\
=2\left(1+x+x^2\right)\sum_{n=0}^{\infty}\left(2x^3\right)^n\\
=2\left(1+x+x^2\right)\frac{1}{1-2x^3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1550241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integrate $\int \frac{\arctan\sqrt{\frac{x}{2}}dx}{\sqrt{x+2}}$ $$\int \frac{\arctan\sqrt{\frac{x}{2}} \, dx}{\sqrt{x+2}}$$
I've tried substituting $x=2\tan^2y$, and I've got:
$$\frac{1}{\sqrt2}\int\frac{y\sin y}{\cos^4 y} \, dy$$
But I'm not entirely sure this is a good thing as I've been unable to proceed any further from there.
| Let $\alpha=\arctan\sqrt{\frac{x}{2}}$
$$I=\int \frac{\alpha dx}{\sqrt{x+2}}=\int\alpha d(2\sqrt{x+2})=2\alpha\sqrt{x+2}-2\int \sqrt{x+2}\space d\alpha $$
The calculation gives $$d\alpha=\frac{\sqrt {2} dx}{\sqrt x(x+2)}$$
Hence $I=2\alpha\sqrt{x+2}-2\sqrt 2\int \frac{\sqrt{x+2}\space dx}{(x+2)\sqrt x}$
$I=2\alpha\sqrt{x+2}-2\sqrt 2\int\frac{dx}{\sqrt{x(x+2)}}$
Now $\int \frac{dx}{\sqrt{x(x+2)}}=2$ $\sin h^{-1}{\sqrt\frac{x}{2}}=2\ln(\sqrt {\frac{x}{2}}+ \sqrt{\frac{x}{2}+1})$
Thus $$I=2\alpha\sqrt{x+2}-4\sqrt2\ln \left(\sqrt {\frac{x}{2}}+ \sqrt{\frac{x}{2}+1}\right)+ Constant$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1550405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.