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Solving the definite integral: $\int_0^1 x\sqrt{px + 1}\,dx$ with $p>0$?
How to integrate $$I = \int_0^1 x\sqrt{px + 1}\,dx$$ , $p>0, p\in\mathbb{R}$ ?
I tried this:
$$ t = \sqrt{px + 1} \implies x = \frac{t^2 - 1}{p}$$
$$ x = 0 \implies t = 1$$
$$ x = 1 \implies t = \sqrt{p+1}$$
so I have:
$$ I = \int_1^\sqrt{p+1} \frac{t^2 - 1}{p}t \,dt = \frac{1}{p}\int_1^\sqrt{p+1} t^3 - t \,dt = \frac{1}{p}\bigg[\frac{t^4}{4}\bigg\vert_1^\sqrt{p+1} - \frac{t^2}{2}\bigg\vert_1^\sqrt{p+1}\bigg]$$
$$ I = \frac{1}{p} \bigg[\frac{1}{4}(p^2 + 2p + 1 - 1) - \frac{1}{2}(p + 1 - 1) \bigg] = \frac{1}{p}\frac{1}{4}p^2 = \frac{1}{4}p $$
So for $p=4$, the integral should have the value $1$, but WA doesn't agree: http://wolfr.am/5p-i9uOA
I looked at it for quite a while but I can't locate the mistake. Since $p>0$, everything seems fine to me but apparently it isn't.
| You may write
$$
\begin{align}
I &= \int_0^1 x\sqrt{px + 1}\,dx\\\\
&= \frac1p\int_0^1 ((px+1)-1)\sqrt{px + 1}\,dx\\\\
&= \frac1p\int_0^1 (px+1)^{3/2}\,dx-\frac1p\int_0^1 (px+1)^{1/2}\,dx\,dx\\\\
&= \frac1{p^2}\int_0^1 (px+1)^{3/2}\,d(px+1)-\frac1{p^2}\int_0^1 (px+1)^{1/2}\,d(px+1)\\\\
&= \frac1{p^2}\left[\frac{(px+1)^{3/2+1}}{3/2+1}\right]_0^1 -\frac1{p^2}\left[\frac{(px+1)^{1/2+1}}{1/2+1}\right]_0^1\\\\
&= \frac{1}{15p^2}\left(4-4 \sqrt{1+p}+2 p \sqrt{1+p}+6 p^2 \sqrt{1+p}\right).
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1330334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Parameter Matrix Determinant $A=\begin{bmatrix} 7x+42&x-21&x-21&x-21&x-21\\x-21&7x+42&x-21&x-21&x-21\\x-21&x-21&7x+42&x-21&x-21\\x-21&x-21&x-21&7x+42&x-21\\x-21&x-21&x-21&x-21&7x+42\end{bmatrix}$
I am trying to find the determinant of $A$, so far without success. I tried reducing it to a triangular matrix but wasn't successful. I would greatly appreciate any help.
| Let me write $a=7x+42$ and $b=x-21$. Then by subtracting the first columns from all other columns we obtain
$$
\det(A) = \det\pmatrix{ a & b & b & b & b \\ b & a & b & b & b \\ b & b & a & b & b \\ b & b & b & a & b \\ b & b & b & b & a }
= \det\pmatrix{ a & b-a & b-a & b-a & b-a \\ b & a-b & 0 & 0 & 0 \\ b & 0 & a-b & 0 & 0 \\ b & 0 & 0 & a-b & 0 \\ b & 0 & 0 & 0 & a-b }
\\
= (a-b)^4 \det\pmatrix{ a & -1 & -1 & -1 & -1 \\ b & 1 & 0 & 0 & 0 \\ b & 0 & 1 & 0 & 0 \\ b & 0 & 0 & 1 & 0 \\ b & 0 & 0 & 0 & 1 }
$$
Now developing with respect to the first column gives
$$
\det(A) = (a-b)^4(a+b+b+b+b) = (a-b)^4 (a+4b) = (6x+63)^4(11x-42).
$$
If you instead develop with respect to the last row (or column), you wil find a recursive way to compute the determinant, which can be generalized to matrices of any size.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Sum of real solutions on equation $\sqrt{\sin^2{x} + {1 \over 2}} + \sqrt{\cos^2{x} + {1 \over 2}} = 2$ in interval $[0,2\pi]$ is? I know that solution is $4\pi$ but I do not know how do they get to this solution. I always get that $x \in R$ and that $-1 < \cos 2x < 1$ when converting it to double angle.
EDIT : So ok, I tried to do this and at the end I get that $\sin^2{2x} = 1$ which when I solve I get that solutions are $2x = {\pi\over2}$ and $2x = {3\pi\over2}$ but when I sum up these two I get that result is $\pi$, which is not the solution by my booklet $(4\pi)$.
| Using the Cauchy-Schwarz inequality. which states that for any 2 vectors $( a, b )$ and $( c, d )$ with $a,b,c,d \ge 0$:
$$(ac + bd)^2 \le (a^2 + b^2)(c^2 + d^2) $$
Let $a = \sqrt{\sin^2 x + \frac{1}{2}}$, $b = \sqrt{\cos^2 x + \frac{1}{2}}$ and $c = d = 1$ we have
$$ \left(\sqrt{\sin^2 x + \frac{1}{2}} + \sqrt{\cos^2 x + \frac{1}{2}}\right)^2 \le \left( \sin^2 x + \frac{1}{2} + \cos^2 x + \frac{1}{2}\right)(1^2+1^2) = 4 $$
or
$$ \sqrt{\sin^2 x + \frac{1}{2}} + \sqrt{\cos^2 x + \frac{1}{2}} \le 2 $$
The equality happens only when $\frac{a}{c} = \frac{b}{d}$, or $\sin^2 x = \cos^2 x = \frac{1}{2}$. That's 4 solutions on the interval $[0, 2\pi]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
If real numbers $x$ and $y$ satisfy the equation $\frac {2x+i}{y+i}= \frac {1+i\sin{\alpha}}{1-i\sin{3\alpha}}$ then quotient $\frac xy$ is equal to?
If real numbers $x$ and $y$ satisfy the equation $\frac {2x+i}{y+i}= \frac {1+i\sin{\alpha}}{1-i\sin{3\alpha}}$, then quotient $\frac xy$ is equal to?
Other conditions are ($\alpha \neq k\pi,\ \alpha \neq \frac \pi2+ k\pi,\ k\in\mathbb Z,\ i^2 = -1$). I tried this by making nominator the difference of squares but it does not lead me anywhere. Solution for this task is $-2-4\cos{2\alpha}$
| $$2x+\sin3a+i(1-2x\sin3a)=y-\sin a+i(1+y\sin a)$$
Eauting the imaginary parts, $1-2x\sin3a=1+y\sin a\iff \dfrac xy=\dfrac{\sin a}{-2\sin3a}$
Now $\sin3a=\sin a(3-4\sin^2a)$ and $\cos2a=1-2\sin^2a\iff2\sin^2a=1-\cos2a$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1333611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to integrate $\frac{x^2}{x^2+9}$? I know you can use division to get $1-\frac{p}{x^2+9}$ and then use the result for arctan, but I was trying to do this using substitution and nothing seems to work? Is there a substitution that could be used for this?
| If one chooses not to write $\frac{x^2}{x^2+9}=1-\frac{9}{x^2+9}$, then one can make still directly substitute $x=3 \tan y$ with $dx=3 \sec^2y \,dy$. Thus, we have
$$\begin{align}
\int \frac{x^2}{x^2+9}dx&=\int\frac{9\tan^2y}{9\tan^2y+9}3\sec^2y\,dy\\\\
&=3\int \tan^2y\,dy\\\\
&=3(\tan y-y)+C\\\\
&=x-3\arctan (x/3)+C
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1333780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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} |
Solve the follwing system of equations for $x, y$ and $z$ $$\frac{y+z}{5}=\frac{z+x}{8}=\frac{x+y}{9}$$ and $$6(x+y+z)=11$$
My teacher told me that I would have to get $3$ different equations to get $x, y$ and $z$. I've tried many methods and I'm confused as to how to do this problem.
| Hint
$$\frac{y+z}{5} =\frac{z+x}{8} = \frac{x+y}{9}$$ is the same as
$\color{red}{\large{\frac{y+z}{5} =\frac{z+x}{8}}}$ and $\color{red}{\large{\frac{z+x}{8} = \frac{x+y}{9}}}$
Which is also the same as $\color{blue}{\large{\frac{y+z}{5} = \frac{x+y}{9}}}$ and $\color{blue}{\large{\frac{z+x}{8} = \frac{y+z}{5}}}$
And so you can break it into two equations.
$\color{blue}{\large{\frac{z+x}{8} = \frac{y+z}{5}} \implies 5(z+x)=8(y+z)} \implies 5z + 5x =8y + 8z \implies 5x -8y -3z = 0$
and $\color{blue}{\large{\frac{y+z}{5} = \frac{x+y}{9}}} \implies 9(y+z) = 5(x+y) \implies 9y + 9z = 5x + 5y \implies 5x + 9z + 4y = 0$
and you have the 3rd equation which is $6(x+y+z) = 11 \implies x + y + z = \frac{11}{6}$.
Now can you solve this system of equations
$$5x -8y -3z = 0, 5x + 9z + 4y= 0, x + y + z = \frac{11}{6}$$
Now you can use this link here linear algebra tool kit to check your work
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1335582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find coefficient of $x^{12}$ in the expansion of $(1+x+x^2+x^3+...+x^n)^4$ How to find coefficient of $x^{12}$ in the expansion of $(1+x+x^2+x^3+...+x^n)^4$
I tried this : Since $(1+x+x^2+x^3+...+x^n)$ is in GP its sum will be $(x^{n+1}+1)(x-1)^{-1}$ now ACQ we have to expand $((x^{n+1}+1)(x-1)^{-1})^4$ next i am not able to do this problem please hep !
| Continuing from where you left off but with a corrected formula for the geometric series, use the negative binomial theorem:
$$
(1-x)^{-n}=\sum_{k=0}^\infty \binom{n+k-1}{k} x^k
$$
In your case, we have
$$
(1-x)^{-4}=\sum_{k=0}^\infty \binom{k+3}{3} x^k = 1+\frac{2\cdot 3 \cdot 4}{1 \cdot 2\cdot 3}x + \frac{3\cdot 4 \cdot 5}{1 \cdot 2\cdot 3}x^2+ \dots
$$
and you can then multiply this by $(1-x^{n+1})^4=1-4x^{n+1}+6x^{2n+2}-4x^{3n+3}+x^{4n+4}$ to get your original expression in a more agreeable form. Upon multiplying, we'll get a degree-$12$ contribution for each term this expression, when it's paired off with the corresponding term in the infinite sum such that the total degree of the paired terms is $12$. Thus the coefficient of $x^{12}$ in the product will be
$$
\binom{15}{3}-4 \binom{14 - n}{3}+6 \binom{13 -2n}{3}-4\binom{12-3n}{3}+\binom{11-4n}{3}
$$
where we adopt the (somewhat nonstandard) convention that $\binom{n}{k}=0$ for all $n<k$, even $n$ negative.
In particular, whenever $n \geq 12$, the coefficient is always just $\binom{15}{3}=455$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1336661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
} |
find coordinate on line at given distance from given coordinate I got two coordinates of a straight line $(-2,-4)$ and $(3,4)$. How can i find a coordinate that lies on this line and is $5$ units away from the $(-2,-4)$ coordinate?
| The equation of the line passing through the points $(-2, -4)$ & $(3, 4)$ is given as
$$y-(-4)=\frac{4-(-4)}{3-(-2)}(x-(-2))$$ $$\implies y+4=\frac{8}{5}(x
+2)$$ $$\implies 5y+20=8x+16$$ $$\implies 8x-5y-4=0$$
Let the coordinates be $(p, q)$ on the line: $8x-5y-4=0$ Then this point will satisfy the equation of the line as follows $$8p-5q-4=0 \implies q=\frac{8p-4}{5}$$ Now, the distance of the point $(p, q)$ from $(-2, -4)$ is $5$ hence we have
$$\sqrt{(p-(-2))^2+(q-(-4))^2}=5$$ $$(p+2)^2+\left(\frac{8p-4}{5}+4\right)^2=25$$
$$(p+2)^2+\frac{64}{25}(p+2)^2=25 $$$$ 89(p+2)^2=25\times 25$$ $$ p+2=\pm\frac{25}{\sqrt{89}}$$ $$\implies p=\pm\frac{25}{\sqrt{89}}-2$$
$$\text{if}\quad \color{blue}{p=\frac{25}{\sqrt{89}}-2} \implies \color{blue}{q=\frac{40}{\sqrt{89}}-4}$$
$$\text{if}\quad \color{blue}{p=-\frac{25}{\sqrt{89}}-2} \implies \color{blue}{q=-\frac{40}{\sqrt{89}}-4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1339965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
} |
Alternative Quadratic Formula Well the formula for solving a Quadratic equation is :
$$\text{If }\space ax^2+bx+c=0$$
then
$$x=\dfrac{-b \pm \sqrt{b^2 -4ac} }{2a}$$
But looking at this : [Wolfram Mathworld] (And also in other places)
They give An Alternate Formula:
$$x=\dfrac{2c}{-b \pm \sqrt{b^2 -4ac}}$$
How does one get this?
Also in the first formula (the one we know) , $a \neq 0$ ... but here is it still the case?
Please help, Thanks!
| Consider the quadratic $ax^2 + bx + c$. If $a = 0$, then finding the root is easy and boring, so we only look at the case when $a \neq 0$.
One way to get at the original quadratic formula is to complete the square. That is, note that
$$\begin{align}
ax^2 + bx + c &= a\left(x^2 + \frac{b}{a} x + \frac{c}{a}\right) \\
&= a \left(x^2 + \frac{b}{a} x + \frac{b^2}{4a^2} - \frac{b^2}{4a^2} + \frac{c}{a} \right) \\
&= a \left( \left( x + \frac{b}{2a} \right)^2 + \frac{4ac - b^2}{4a^2} \right).
\end{align}$$
So when looking for solutions to $ax^2 + bx + c = 0$, we are looking at solutions to
$$ 0 = a \left( \left( x + \frac{b}{2a} \right)^2 + \frac{4ac - b^2}{4a^2} \right) = a\left( x + \frac{b}{2a}\right)^2 + a\frac{4ac - b^2}{4a^2},$$
or equivalently
$$ \left(x + \frac{b}{2a} \right)^2 = \frac{b^2 - 4ac}{4a^2}.$$
Taking square roots and solving for $x$ gives the original quadratic formula.
Now let us derive the alternate form. Notice that if $c = 0$ then $x = 0$ is a root and $x = -b/a$ is the other root. So let us now consider $c \neq 0$. We now know that $x = 0$ is not a root. This means that if $r$ is a root of $ax^2 + bx + c = 0$, then $\frac{1}{r}$ will be a root of $a \frac{1}{x^2} + b \frac{1}{x} + c = 0$. Multiplying through by $x^2$ (which makes sense because we are not interested in $x = 0$), we are interested in reciprocals of the roots of
$$ a + bx + cx^2 = 0.$$
The original quadratic formula applies, and we know the roots are
$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2c}.$$
Since we are interested in the reciprocals, we find the roots to be
$$ x = \frac{2c}{-b \pm \sqrt{b^2 - 4ac}}.$$
This isn't necessarily the best derivation, but I like it. $\diamondsuit$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding an isomorphism between polyomial quotient rings Let $F_1 = \mathbb{Z}_5[x]/(x^2+x+1)$ and $F_2 = \mathbb{Z}_5[x]/(x^2+3)$. Note neither $x^2+x+1$ nor $x^2+3$ has a root in $\mathbb{Z}_5$, so that each of the above are fields of order 25, and hence they are isomorphic from elementary vector space theory, however I'm tasked with constructing an explicit isomorphism between the two fields, so I applied the technique I've learned here and elsewhere as follows.
First, let $\phi: F_1 \rightarrow F_2$ by $x \mapsto ax+b$ (as the base field is fixed it suffices to find the image of $x$). Then we should have that $\phi(x^2+x+1) = x^2 + 3$, and the left hand side reduces to $\phi(x)^2 + \phi(x) + 1$ as $\phi$ is a homomorphism. Simplifying, we have $(ax+b)^2 + (ax+b) + 1 = a^2x^2 + 2abx + b^2 + ax+b+1 = a^2x^2 + (2ab+a)x + (b^2+b+1) = x^2+3$.
However, this appears to be a problem. Notably, since $a^2 = 1$ mod 5, we have $a = 1,4$. Picking the former, we have $2b+1 = 0$ and thus $b = 2$, a problem as $2^2+2+1 = 7 \neq 1$ mod 5. Trying $a=4$, we have $3b+4 = 0$ so $b = 2$, and again the same problem.
Are there flaws in my technique or arithmetic that I don't see? If not, what other options do I have to attempt to construct the isomorphism?
| As per Jyrki Lahtonen's hint, I will instead try $\phi(x^2+x+1) = c(x^2+3)$. So then since $x \mapsto ax+b$, we have $a^2x^2 + (2ab+a)x + b^2+b+1 = cx^2 + 3c$, so $a^2 = c$, $2ab+a = 0$, and $b^2+b+1 = 3c$. Since $a^2 = c$ and the only squares mod 5 are 1 and 4, we have that $c = 1,4$. Having already gleaned that 1 will not suffice, let us try $c = 4$.
Then $a^2 = 4$ so $a = 2,3$. The former did not yield nice results, so let $a = 3$. Then we have that $b + 3 = 0$ so that $b = 2$, and from the last relation $7 = 12$ which is true mod 5.
Therefore $a = 3$ and $b = 2$, so $x \mapsto 3x+2$ is an isomorphism.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the last two digits of $33^{100}$
Find the last two digits of $33^{100}$
By Euler's theorem, since $\gcd(33, 100)=1$, then $33^{\phi(100)}\equiv 1 \pmod{100}$. But $\phi(100)=\phi(5^2\times2^2)=40.$
So $33^{40}\equiv 1 \pmod{100}$
Then how to proceed?
With the suggestion of @Lucian:
$33^2\equiv-11 \pmod{100}$ then $33^{100}\equiv(-11)^{50}\pmod{100}\equiv (10+1)^{50}\pmod{100}$
By using the binomial expansion, we have:
$33^{100}\equiv (10^{50}+50\cdot 10^{49}+ \cdots + 50\cdot 10+1)\pmod{100}$
$\implies 33^{100}\equiv (50\cdot 10+1)\pmod{100}\equiv 01 \pmod{100}$
| Hint: $33^2\equiv-11\bmod100$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the value of this series what is the value of this series $$\sum_{n=1}^\infty \frac{n^2}{2^n} = \frac{1}{2}+\frac{4}{4}+\frac{9}{8}+\frac{16}{16}+\frac{25}{32}+\cdots$$
I really tried, but I couldn't, help guys?
| Since $n^2 = 2\binom{n}{2}+\binom{n}{1}$ we have:
$$ S=\sum_{n\geq 1}\frac{n^2}{2^n}=\left.\left(2\frac{x^2}{(1-x)^3}+\frac{x}{(1-x)^2}\right)\right|_{x=\frac{1}{2}}=\color{red}{6}.$$
As an alternative to the negative binomial series, we may also use:
$$ S = 2S-S = \sum_{n\geq 1}\frac{n^2}{2^{n-1}}-\sum_{n\geq 1}\frac{n^2}{2^n}=1+\sum_{n\geq 1}\frac{(n+1)^2-n^2}{2^n}=2+2\sum_{n\geq 1}\frac{n}{2^n}$$
so that, in the same way:
$$ \sum_{n\geq 1}\frac{n}{2^n} = T = 2T-T = \sum_{n\geq 1}\frac{n}{2^{n-1}}-\sum_{n\geq 1}\frac{n}{2^n} = 1+\sum_{n\geq 1}\frac{1}{2^n}=2 $$
and $S=2+2\cdot 2=\color{red}{6}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1343721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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What is the maximum value of the LCM of three numbers $\leq n$, as a function of $n$? Given $n \geq 3$, what maximum LCM of any three numbers $\leq n$ can we obtain?
Now, if $n$ is odd, the answer would be
$$n(n - 1)(n - 2)$$
because $\newcommand{\lcm}{\operatorname{lcm}}$
$$\begin{align*}
\lcm(a, b, c) &= \lcm(a, \lcm(b, c))\\\\
&= \lcm\left(a, \frac{bc}{\gcd(b, c)}\right)\\\\
&=\frac{abc}{\gcd(b, c) \gcd(a, \frac{b c}{\gcd(b, c)})}
\end{align*}$$
Now, if $b = a + 1$ and $c = a + 2$ where $a$ is odd, then
$$\gcd(a, c) = \gcd(a, a + 2) = 1$$
Also, we know that $\gcd(a, a + 1) = 1$ for any $a$. We can simplify the
formula as
$$\begin{align*}
\frac{abc}{\gcd(b, c) \gcd(a, \frac{b c}{\gcd(b, c)})} &= \frac{a (a + 1) (a + 2)}{1 \cdot \gcd(a, \frac{b c}{1)})}\\\\
&= \frac{a (a + 1) (a + 2)}{\gcd(a, bc))}
\end{align*}$$
Also since, $a$ and $b$ are coprime and $a$ and $c$ are coprime,
we must have $a$ and $bc$ coprime too, i.e., $\gcd(a, bc) = 1$.
So, the maximum value of LCM for odd $n$ would be
$$n (n - 1)(n - 2)$$
Now, how should I proceed for even $n$?
| If $n$ is even and $\ge 4$, and not divisible by $3$, use $n(n-1)(n-3)$.
If $n$ is even and divisible by $3$, use $(n-1)(n-2)(n-3)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Simultaneous equation with fractional solutions. How do you get to find $x$ when $y$ is a fraction ? Anyone mind to explain it step by step for the clearest explanation.=)
$$-7x +2y = 2$$
$$14x + 3y = -5$$
Answer: $x=?, y=-1/7$
| Inserting $y=-\frac{1}{7}$ in the first equation
$-7x-2\cdot \frac{1}{7}=2$
Transforming 2 into $\frac{14}{7}$ to have a common denominator with $2\cdot \frac{1}{7}$
$-7x-\frac{2}{7}=\frac{14}{7} \quad \quad |+\frac{2}{7}$
Adding $\frac{2}{7}$ on both sides.
$-7x\underbrace{-\frac{2}{7}+\color{blue}{\frac{2}{7}}}_{=0}=\frac{14}{7} +\color{blue}{\frac{2}{7}}$
$-7x+0=\frac{14}{7} +\color{blue}{\frac{2}{7}}$
$-7x=\frac{14}{7}+\frac{2}{7}$
$-7x=\frac{16}{7}$
$-x=\frac{16}{7\cdot 7}$
$x=-\frac{16}{49}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1350126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Finding $\sum\limits_{k=0}^n k^2$ using summation by parts Sorry to bother you guys again, but I still have some doubts. I do think I'm making some progress, though.
So, again, the formula that I'm using for summation by parts is
$\sum\limits_{k=o}^n f(k)g(k) = g(n)(\sum\limits_{k=0}^n f(k)) - \sum\limits_{k=0}^{n-1} [\Delta g(k) \sum\limits_{i=0}^k f(i)]$
Using this, I'm trying to find $\sum\limits_{k=0}^n k^2$. I think I almost got it, but there are some problems that I still can't get around.
Plugging in $g(k) = k$ and $f(k)=k$, we end up with the following equations:
\begin{align}
\sum\limits_{k=0}^n k^2 &= \sum\limits_{k=0}^n kk\\
\sum\limits_{k=0}^n k^2 &= n (\sum\limits_{k=0}^n k) - \sum\limits_{k=0}^{n-1}[\sum\limits_{i=0}^k i]\\
\sum\limits_{k=0}^n k^2 &= n(\frac{n(n+1)}{2}) - \sum\limits_{k=0}^{n-1}[\frac{k(k+1)}{2}]\\
\sum\limits_{k=0}^n k^2 &= \frac{n^3 + n^2}{2} - \frac{\sum\limits_{k=0}^{n-1} k^2+k}{2}\\
2\sum\limits_{k=0}^n k^2 &= n^3 + n^2 - \sum\limits_{k=0}^{n-1} k^2+k\\
2\sum\limits_{k=0}^n k^2 &= n^3 + n^2 - \sum\limits_{k=0}^{n-1} k^2+ \sum\limits_{k=0}^{n-1}k
\end{align}
Anyway, my problem is, I need the upper indexes of the sums in the right hand side to be $n$, not $n-1$ (I've already checked that, if the indexes are $n$, then I can find the desired result). Is there any trick I'm missing to make those indexes go up by $1$? Or am I missing an obvious step?
| I would solve it like this: $p_3(n+1)^3 + p_2(n+1)^2 + p_1(n+1) + p_0 - (p_3n^3 + p_2n^2 + p_1n + p_0) = 1n^2$
Then treat the polynomial as a vector space, applying binomial theorem and writing on matrix form solving a linear equation system, we get:
$${\bf p} = \left[\begin{array}{r}p_3\\p_2\\p_1\\p_0\end{array}\right] = \left(\left[\begin{array}{cccc}1&0&0&0\\3&1&0&0\\3&2&1&0\\1&1&1&1\end{array}\right] - \left[\begin{array}{cccc}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{array}\right]\right)^{-1}
\left[\begin{array}{c}0\\1\\0\\0\end{array}\right] =
\left[\begin{array}{r}1/3\\-1/2\\1/6\\0\end{array}\right]$$
Now I "cheated" ( let octave solve this equation system for me ) so I'm sure it should be -1/2 on the square factor and not 1/2.
EDIT: It turns out that I calculated $n+1$ and not $n$. If I had started with $P(n) - P(n-1) = n^2$ instead of $P(n+1)-P(n) = n^2$ we would have gotten 1/2 for the second degree. To check that this is the case we have the perfect excuse to make another matrix exercise, this time using the binomial expansion of $(n-1)^k$:
$$\left[\begin{array}{rrrr}
1&0&0&0\\
-3&1&0&0\\
3&-2&1&0\\
-1&1&-1&1
\end{array}\right]^{-1}\left[\begin{array}{r}1/3\\-1/2\\1/6\\0\end{array}\right] = \left[\begin{array}{r}1/3\\1/2\\1/6\\0\end{array}\right]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1351843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Determine whether $\sum \frac{2^n + n^2 3^n}{6^n}$ converges For the series $$\sum_{n=1}^{\infty}\dfrac{2^n+n^23^n}{6^n},$$ I was thinking of using the root test? so then I would get $(2+n^2/n+3)/6$ but how do I find the limit of this?
| $$\sum\limits_{n=1}^{\infty}\frac{2^n+n^2 3^n}{6^n}$$
I wouldn't recommend the root test for this series. However, here are the steps
$$r=\limsup\limits_{n\to\infty}\sqrt[n]{\left|\frac{2^n+n^2 3^n}{6^n}\right|}$$
$$=\limsup\limits_{n\to\infty}\frac{\sqrt[n]{2^n+n^2 3^n}}{\sqrt[n]{6^n}}$$
$$=\frac16\limsup\limits_{n\to\infty}\left(2^n+n^2 3^n\right)^{\frac{1}{n}}$$
$$=\frac16\limsup\limits_{n\to\infty}\exp\left(\log\left(2^n+n^2 3^n\right)^{\frac{1}{n}}\right)$$
$$=\frac16\exp\left(\limsup\limits_{n\to\infty}\frac{1}{n}\log\left(2^n+n^2 3^n\right)\right)$$
$$=\frac16\exp\left(\limsup\limits_{n\to\infty}\frac{2^n\log 2+2n3^n+n^2 3^n\log 3}{2^n+n^2 3^n}\right)$$
$$=\frac16\exp\left(\limsup\limits_{n\to\infty}\frac{\frac{2^n\log 2}{n^2 3^n}+\frac{2}{n}+\log 3}{\frac{2^n}{n^2 3^n}+1}\right)$$
$$=\frac16\exp\left(\log 3\right)=\frac36=\frac12$$
Note that $r\lt 1$, therefore
$$\sum\limits_{n=1}^{\infty}\frac{2^n+n^2 3^n}{6^n}=\mbox{convergent} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1352763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Let $a$, $b$, and $c$ be positive real numbers. Let $a$, $b$, and $c$ be positive real numbers. Prove that
$$\sqrt{a^2 - ab + b^2} + \sqrt{a^2 - ac + c^2} \ge \sqrt{b^2 + bc + c^2}$$
Under what conditions does equality occur? That is, for what values of $a$, $b$, and $c$ are the two sides equal?
| The expressions $a^2 - ab + b^2$, $a^2 - ac + c^2$, and $b^2 + bc + c^2$ should remind you of the law of cosines. Let $\vec{a}$, $\vec{b}$, $\vec{c}$ be vectors such that $\vec{a}$ and $\vec{b}$ are separated by 60 degrees, and $\vec{a}$ and $\vec{c}$ are separated by 60 degrees. Consider the triangle formed by the three vectors. This triangle has side lengths
$$\sqrt{a^2 - ab + b^2}, \sqrt{a^2 - ac + c^2}, \sqrt{b^2 + bc + c^2}$$
according to the law of cosines. Then the inequality follows by the triangle inequality.
Equality occurs when the three vectors are collinear. The conditions for this shouldn't be too hard to find using basic geometric methods.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1353184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Parabolic range conditions proof This problem is getting the better of me, since I have no idea where to start:
The equation of a curve is $y=ax^2-2bx+c$, where a, b and c are constants with $a>0$.
Given that the vertex of the curve lies on the line $y=x$, find an expression for $c$ in terms of $a$ and $b$. Show that in this case, whatever the value of $b$, $ c\ge -\frac{1}{4a} $.
My working so far:
Vertex:$\left(\frac ba,c-\frac {b^2}a\right)$ (Completion of square)
$$
y=x
$$$$
c- \frac {b^2}a = \frac ba
$$Thus,$$
c=\frac{b^2+b}a
$$When $a>0$
The discriminant of the original equation comes to $-4b$
Thanks in advance :-)
| $$c=\frac{b^2+b}{a}=\frac{4b^2+4b}{4a}=\frac{(4b^2+4b+1)-1}{4a}=\frac{(2b+1)^2-1}{4a}\ge -\frac{1}{4a}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1354306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How can I find the minimum value for $F(x,y,z,w)=x^2+y^2+z^2+w^2+xy+zw-xz-yw-yz$ Let $x,y,z,w$ be integer numbers,and $xw=yz+1$
Find this minimum of the value
$$x^2+y^2+z^2+w^2+xy+zw-xz-yw-yz$$
This is how did it and I would like to know if I made a mistake
Let $$F(x,y,z,w)=x^2+y^2+z^2+w^2+xy+zw-xz-yw-yz$$
we have that
$$2F(x,y,z,w)=(x-z)^2+(y-w)^2+(x+y)^2+(w+z)^2-2yz$$
| let $z=-v ,xw=1-yv,f=x^2+y^2+v^2+w^2+xy-vw+xv-yw+yv=(x-w)^2+2xw+(x-w)(y+v)+(y+v)^2-yv=(x-w)^2+(x-w)(y+v)+(y+v)^2+2-3yv, \\ yv \le \dfrac{(y+v)^2}{4} \implies -3yv \ge -\dfrac{3(y+v)^2}{4} \implies f \ge (x-w)^2+(x-w)(y+v)+\dfrac{(y+v)^2}{4}+2=(x-w+\dfrac{y+v}{2})^2+2 \ge 2 $
when $y=v ,x-w+y=0 \cap xw=1-y^2 ,f $ get min
now we prove the condition is reachable.
$x-w\pm \sqrt{1-xw}=0 \cap xw \le 1 \implies (x-w)^2=1-xw \iff $
$\\x^2-xw+w^2-1=0 \\ \tag 1\\x=\dfrac{w \pm \sqrt{4-3w^2}{}}{2}$
so there is infinity solution because from (1), we always have $0= x^2+w^2-xw-1 \ge xw-1 \iff xw \le 1$. details of $x-w$ curve see picture below:
if you want integer ,there are 4 pairs for $x,w$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1354997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Probability that no two consecutive heads occur?
A fair coin is tossed $10$ times. What is the probability that no two
consecutive tosses are heads?
Possibilities are (dont mind the number of terms):
$H TTTTTTH$, $HTHTHTHTHTHTHT$.
But except for those,
let $y(n)$ be the number of sequences that start with $T$
$T _$, there are two options, $T$ and $H$ so,
$y(n) = y(n - 1) + x(n-1) = y(n - 1) + y(n - 2)$
Let $x(n)$ be the number of sequences that start with $H$,
$H _$ the next option is only $T$ hence,
$x(n) = y(n - 1)$
The total number of sequences $F(n)$ is:
$$F(n) = y(n) + x(n) = 2y(n - 1) + y(n - 2)$$
We are after $F(10)$,
$$F(10) = 2y(9) + y(8)$$
$$F(3) = 2y(2) + y(1) = 1 + 4 = 5$$
$$F(4) = 2y(3) + y(2) = 6 + 2 = 8$$
But I'm not quite sure where this will take me though.
| $F (n)$ is in fact a fibonacci sequence:
\begin{align}
F (n+1) &= 2 y (n) + y (n-1)\\
&= 2 [y (n-1) +y (n-2)] + y (n-2) + y (n-1)\\
&= 2 y (n-1) + y (n-2) + 2 y(n-2) +y (n-3)\\
&= F (n) +F (n-1)\\
\end{align}
Now $F (1) = |\{H,T\}|=2$
And $F (2) = |\{HT,TH,TT\}|=3$
So $ F (10) = 144$
There are $ 2^{10}=1024$ possible events.
Thus the required probability is $\frac {144}{1024} $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1355511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Differentiate the Function: $y=e^{k\ tan\sqrt{x}}$ $y=e^{k\tan\sqrt{x}}$
$=e^{k\tan\sqrt{x}}\cdot [{k\tan\sqrt{x}}]'$
$=e^{k\tan\sqrt{x}}\cdot\ (k)\cdot[\tan x^{\frac{1}{2}}]'+(\tan x^{\frac{1}{2}})\cdot[k]'$
$=e^{k\tan\sqrt{x}}\cdot\ (k)\cdot (\frac{1}{2}tanx)\cdot \sec^2x + (\tan x^{\frac{1}{2}})\cdot (1)$
Is my process correct?
| Your work is not quite correct.
$$\begin{align}
(e^{k\tan\sqrt x})' &= e^{k\tan\sqrt x}\cdot(k\tan\sqrt x)' \\
&= e^{k\tan\sqrt x}\cdot k\cdot (\tan\sqrt x)' \\
&= e^{k\tan\sqrt x}\cdot k\cdot\sec^2\sqrt x\cdot (\sqrt x)' \\
&= e^{k\tan\sqrt x}\cdot k\cdot\sec^2\sqrt x\cdot \frac 1{2\sqrt x} \\
&= \frac{k\sec^2\sqrt x\cdot e^{k\tan\sqrt x}}{2\sqrt x}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1356986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the roots of the summed polynomial
Find the roots of: $$x^7 + x^5 + x^4 + x^3 + x^2 + 1 = 0$$
I got that:
$$\frac{1 - x^8}{1-x} - x^6 - x = 0$$
But that doesnt make it any easier.
| Clever factoring: $x^7+x^5+x^4+x^3+x^2+1 =(x+1)(x^2 - x + 1)(x^4+x^2+1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1358551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Help with this Trigonometry integral I've to find this integral:
$$\int_{\frac{\pi}{365}}^{\frac{365} {\pi}}\left(\frac{4^\pi}{\tan(x)}+\tanh^{-1}(x)-4\sec^2(x)\right)dx$$
So far as I know to go:
$$\int_{\frac{\pi}{365}}^{\frac{365} {\pi}}\left(\frac{4^\pi}{\tan(x)}+\tanh^{-1}(x)-4\sec^2(x)\right)dx=$$
$$\int_{\frac{\pi}{365}}^{\frac{365} {\pi}}\left(4^\pi\cdot \frac{1}{\tan(x)}+\tanh^{-1}(x)-4\sec^2(x)\right)dx=$$
$$\int_{\frac{\pi}{365}}^{\frac{365} {\pi}}\left(4^\pi \cot(x)+\tanh^{-1}(x)-4\sec^2( x)\right)dx$$
Thanks a lot!
| $$\int_{\frac{\pi}{365}}^{\frac{365}{\pi}}\left(\frac{4^\pi}{\tan(x)}+\tanh^{-1}(x)-4\sec^2(x)\right)dx$$
$$\int_{\frac{\pi}{365}}^{\frac{365}{\pi}}\left(\frac{4^\pi}{\tan(x)}+\tanh^{-1}(x)-4\sec^2(x)\right)dx=$$
$$\int_{\frac{\pi}{365}}^{\frac{365}{\pi}}\left(4^\pi\cdot \frac{1}{\tan(x)}+\tanh^{-1}(x)-4\sec^2(x)\right)dx=$$
$$\int_{\frac{\pi}{365}}^{\frac{365}{\pi}}\left(4^\pi \cot(x)+\tanh^{-1}(x)-4\sec^2(x)\right)dx=$$
$$\int_{\frac{\pi}{365}}^{\frac{365}{\pi}}\left(4^\pi \cot(x)\right)dx+\int_{\frac{\pi}{365}}^{\frac{365}{\pi}}\left(\tanh^{-1}(x)\right)dx+\int_{\frac{\pi}{365}}^{\frac{365}{\pi}}\left(4\sec^2(x)\right)dx=$$
$$4^\pi\cdot\int_{\frac{\pi}{365}}^{\frac{365}{\pi}}\left( \cot(x)\right)dx+\int_{\frac{\pi}{365}}^{\frac{365}{\pi}}\left(\tanh^{-1}(x)\right)dx+4\cdot\int_{\frac{\pi}{365}}^{\frac{365}{\pi}}\left(\sec^2(x)\right)dx=$$
$$4^\pi\cdot\left[\ln(\sin(x))\right]_{\frac{\pi}{365}}^{\frac{365}{\pi}}+\left[\frac{1}{2}\ln(1-x^2)+x\tanh^{-1}(x)\right]_{\frac{\pi}{365}}^{\frac{365}{\pi}}+4\cdot\left[\tan(x)\right]_{\frac{\pi}{365}}^{\frac{365}{\pi}}=$$
$$4^\pi\left(a\right)+\left(b\right)+4\left(c\right)\Longrightarrow$$
$$a=\ln\left(\sin\left(\frac{365}{\pi}\right)\right)-\ln\left(\sin\left(\frac{\pi}{365}\right)\right)$$
$$b=\left(\frac{1}{2}\ln\left(1-\left(\frac{365}{\pi}\right)^2\right)+\frac{365}{\pi}\cdot\tanh^{-1}\left(\frac{365}{\pi}\right)\right)-\left(\frac{1}{2}\ln\left(1-\left(\frac{\pi}{365}\right)^2\right)+\frac{\pi}{365}\cdot\tanh^{-1}\left(\frac{\pi}{365}\right)\right)$$
$$c=\tan\left(\frac{365}{\pi}\right)-\tan\left(\frac{\pi}{365}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1359590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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This inequality $a+b^2+c^3\ge \frac{1}{a}+\frac{1}{b^2}+\frac{1}{c^3}$
Let $0\le a\le b\le c,abc=1$, then show that
$$a+b^2+c^3\ge \dfrac{1}{a}+\dfrac{1}{b^2}+\dfrac{1}{c^3}$$
Things I have tried so far:
$$\dfrac{1}{a}+\dfrac{1}{b^2}+\dfrac{1}{c^3}=\dfrac{b^2c^3+ac^3+ab^2}{bc^2}$$
Since $abc=1$, it suffices to prove that
$$c+b^3c^2+bc^5\ge b^2c^3+ac^3+ab^2$$
then the problem is solved. I stuck in here.
| hint: $c\ge 1, a \le 1$
when $b\le 1 $. replace $a$,get $f(c)=$LHS-RHS,prove $f'(c)\ge0$ with
$ bc=\dfrac{1}{a} \ge 1,$ then use $ c\ge \dfrac{1}{b^2} $ to prove $f(c=\dfrac{1}{b^2} )\ge 0$
Edit: here is full solution:
case 1: $b \le 1, c \ge \dfrac{1}{b^2},bc=\dfrac{1}{a} \ge 1$
LHS-RHS$=b^2c^6-b^3c^4+b^4c^3-c^3+bc^2-b^2=f(c),\\f'(c)=6b^2c^5-4b^3c^3+3b^4c^2-3c^2+2bc=4b^2c^3(c^2-b)+2b^2c^5-3c^2+2bc+ 3b^4c^2 >0 $
becasue $c^2 \ge1 \ge b, 2b^2c^5+2bc-3c^2 > 2c^3-3c^2+1=(c-1)(2c^2-c-1)\ge 0 $
$f(c)_{min}=f(\dfrac{1}{b^2})=\dfrac{(1-b^3)(b^9+b^6-b^5-b^4+b^3+1 )}{b^{10}} \ge 0$
when $a=b=1$ get min $0$
case 2: $b \ge 1, \implies b^2 -\dfrac{1}{b^2} \ge 0,a=\dfrac{1}{bc} \ge \dfrac{1}{c^2}$
it is trivial $g(a)=a-\dfrac{1}{a}$ is mono increasing function,$\implies g(a) \ge \dfrac{1}{c^2} -c^2$
LHS-RHS$=a-\dfrac{1}{a}+b^2-\dfrac{1}{b^2}+c^3-\dfrac{1}{c^3} \ge a-\dfrac{1}{a}+c^3-\dfrac{1}{c^3} \ge c^3-c^2+\dfrac{1}{c^2}-\dfrac{1}{c^3}=\dfrac{c^2(c-1)(c^5+1)}{c^5} \ge 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1360349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that if $a^2+b^2$ is a multiple of three, then a and b are multiples of three I have attempted to prove the above. I am uncertain about the correctness of my proof:
Both numbers have to be multiples of three, i.e. $3a+3b=3n$, $\ 3(a+b)=3n$
It is not possible to arrive at an integer that is a multiple of three without adding two integers that are multiples of three.
Assumption: Suppose that $b$ is not a multiple of three, then it can be expressed as $(3v \pm 1)$, therefore we have:
\begin{align*} a^2+(3v \pm 1)^2=3n\\ a^2= 3n-9v^2 \mp 6v -1\\ a=\sqrt{3(n-3v^2\mp2v-\frac{1}{3})}
\end{align*}
which is not a multiple of three? (or is it).
As mentioned before, $(a+b) \ne 3m \ $ if either $a$ or $b$ is not a multiple of 3, in which case assumption that $b$ is not a multiple of 3 is false. And hence it is a multiple of three, so is $a$.
| Working in $\pmod{3}$ we can just break this into cases.
If $a\equiv b \equiv 0\pmod 3$, then $a^2+b^2\equiv 0 \pmod 3$
If $a=0$, $b=1$,then $a^2+b^2 \equiv 1 \pmod 3$, and same if $a$ and $b$ are switched.
If $a=0$, $b=2$, then $a^2+b^2 \equiv 1 \pmod 3$
Just fill in the rest and note that the only time $a^2+b^2 \equiv 0 \pmod 3$ is when $a\equiv b\equiv 0 \pmod 3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1362554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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The maximum and minimum values of the expression Here is the question:find the difference between maximum and minimum values of $u^2$ where $$u=\sqrt{a^2\cos^2x+b^2\sin^2x} + \sqrt{a^2\sin^2x+b^2\cos^2x}$$
My try:I have just normally squared the expression and got
$u^2=a^2\cos^2x+b^2\sin^2x + a^2\sin^2x+b^2\cos^2x +2\sqrt{a^2\cos^2x+b^2\sin^2x} \sqrt{a^2\sin^2x+b^2\cos^2x}$
$u^2=a^2+b^2 +2\sqrt{a^2\cos^2x+b^2\sin^2x} .\sqrt{a^2\sin^2x+b^2\cos^2x}$
I am not getting how to solve the irrational part,so how should we do it.Is there some general way to solve such questions?
| For $\min$
Using $\triangle$ Inequality::
Let $z_{1} = a\cos x+i b\sin x$ and $z_{2} = b\cos x+i a\sin x$
So $$|z_{1}|+|z_{2}|\geq |z_{1}+z_{2}|$$
So $$\sqrt{a^2\cos^2 x+b^2 \sin^2 x}+\sqrt{a^2 \sin^2 x+b^2 \cos^2 x}\geq \sqrt{(a+b)\cos^2 x+(a+b)^2\sin^2 x}=|a+b|$$
For $\max$ Same as Deepsea
$$\left[\left(\sqrt{a^2\cos^2 x+b^2 \sin^2 x}\right)^2+\left(\sqrt{a^2\sin^2 x+b^2 \cos^2 x}\right)^2\right]\cdot \left[1^2+1^2\right]\geq \bigg(\sqrt{a^2\cos^2 x+b^2 \sin^2 x}+\sqrt{a^2 \sin^2 x+b^2 \cos^2 x}\bigg)^2$$
So $$\sqrt{a^2\cos^2 x+b^2 \sin^2 x}+\sqrt{a^2 \sin^2 x+b^2 \cos^2 x}\leq \sqrt{2(a^2+b^2)}$$
So $$|a+b|\leq \sqrt{a^2\cos^2 x+b^2 \sin^2 x}+\sqrt{a^2 \sin^2 x+b^2 \cos^2 x}\leq \sqrt{2(a^2+b^2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1364170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How can I prove the following equality I have the following equality :
$$I_1=-\frac{ab}{2\pi}\int_0^\pi \frac{\cos(2t)}{a^2\sin^2(t)+b^2\cos^2(t)}dt=\frac{a}{a+b}-\frac{1}{2}$$
$$I_2=-\frac{ab}{2\pi}\int_0^\pi \frac{\sin(2t)}{a^2\sin^2(t)+b^2\cos^2(t)}dt=0$$
where $0 <b \leq a$.
I used the residues but I could not prove this equality
| Let $$I=I_1+iI_2.$$
Then
\begin{eqnarray}
I&=&-\frac{ab}{2\pi}\int_0^\pi \frac{e^{2it}}{a^2\sin^2(t)+b^2\cos^2(t)}dt\\
&=&-\frac{ab}{4\pi}\int_0^{2\pi} \frac{e^{it}}{a^2\frac{1-\cos t}{2}+b^2\frac{1+\cos t}{2}}dt\\
&=&-\frac{ab}{2\pi}\int_0^{2\pi} \frac{e^{it}}{(a^2+b^2)+(b^2-a^2)\cos t}dt\\
&=&-\frac{ab}{2\pi i}\int_{|z|=1} \frac{1}{(a^2+b^2)+(b^2-a^2)\frac12(z+\frac{1}{z})}dz\\
&=&-\frac{ab}{\pi i}\int_{|z|=1} \frac{z}{2(a^2+b^2)z+(b^2-a^2)(z^2+1)}dz\\
&=&-\frac{ab}{\pi i}(2\pi i)\text{Res}\left(\frac{z}{2(a^2+b^2)z+(b^2-a^2)(z^2+1)},\frac{a-b}{a+b}\right)\\
&=&-\frac{ab}{\pi i}2\pi i\frac{a-b}{4ab(a+b)}\\
&=&\frac{b-a}{2(a+b)}\end{eqnarray}
and hence
$$ I_1=\frac{b-a}{2(a+b)}, I_2=0. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1364751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Value of an expression with cube root radical What is the value of the following expression?
$$\sqrt[3]{\ 17\sqrt{5}+38} - \sqrt[3]{17\sqrt{5}-38}$$
| \begin{align}
x&=\sqrt[3]{17\sqrt{5}+38} - \sqrt[3]{17\sqrt{5}-38}
\end{align}
Note that
\begin{align}
17\sqrt{5}-38&=\frac{1}{17\sqrt{5}+38}.
\end{align}
Let $a=\sqrt[3]{17\sqrt{5}+38}$.
Then we have
\begin{align}
x^3&=\left(a-\frac1a\right)^3
\\
x^3&= a^3-3a+\frac3a-\frac{1}{a^3}
\\
x^3+3\left(a-\frac1a\right)&= a^3-\frac{1}{a^3}
\\
x^3+3x-76&=0
\\
(x-4)(x^2+4x+19)=0,
\end{align}
The only real solution $x=4$ is the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1365489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 6
} |
Nature of the roots of quadratic equation Here is the problem that I need to prove:
If $x$ is real and $\displaystyle{\ p = \frac{3(x^2+1)}{(2x-1)}}$, prove that $\ p^2-3(p+3) \geq 0$
Here is what I did:
\begin{align*}
p(2x-1)=3(x^2+1) \\
3x^2 - 2px + (p+3)=0 \\
b^2 - 4ac = 4(p^2-3(p+3))
\end{align*}
By inspection I can see that $p^2 > -3(p+3)$ for almost all values of $p \ $, therefore
$p^2-3(p+3) > 0 $. However, the question asks to show that $\ p^2-3(p+3) \geq 0$
If I make $p^2 = 3(p+3)$ I can find roots and so $\ p^2-3(p+3) = 0$, when $\displaystyle{p = \frac{3 \pm \sqrt{45}}{2}}$. Therefore $\ p^2-3(p+3) \geq 0$
Having done this, how can I mathematically show that $p^2$ is never $<$ than $3(p+3)$? Because I am not satisfied with just saying that by inspection $p^2$ is greater than $3(p+3)$.
Thank you
| It must be $b^2-4ac\geq 0$ because it is given $x$ is real so equation must have real roots which may be either equal or unequal
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Vector equation with no solution There exists a real number $k$ such that the equation
$\begin{pmatrix} -1 \\ -2 \end{pmatrix} + t\begin{pmatrix} 3 \\ -2 \end{pmatrix} = \begin{pmatrix} 5 \\ 0 \end{pmatrix} + s\begin{pmatrix} -4 \\ k \end{pmatrix}$
does not have any solutions in $t$ and $s$. Find $k$.
I am confused as to how I will find k so this equation has NO solutions. Any help is appreciated.
Thanks
| Given $$\begin{bmatrix}-1 \\ -2\end{bmatrix} + t\begin{bmatrix}3\\-2\end{bmatrix} = \begin{bmatrix}5 \\ 0\end{bmatrix} + s\begin{bmatrix}-4\\k\end{bmatrix}$$ implies
$$-s\begin{bmatrix}-4\\k\end{bmatrix} + t\begin{bmatrix}3 \\- 2\end{bmatrix} = \begin{bmatrix}5\\0\end{bmatrix} - \begin{bmatrix}-1\\-2\end{bmatrix}$$ which is equivalent to $$s\begin{bmatrix}4\\-k\end{bmatrix} + t\begin{bmatrix}3 \\- 2\end{bmatrix} = \begin{bmatrix}5\\0\end{bmatrix} + \begin{bmatrix}1\\2\end{bmatrix}$$ or rather $$\begin{bmatrix}4s + 3t\\-ks - 2t\end{bmatrix} = \begin{bmatrix}6\\2\end{bmatrix}.$$ In matrix form, this system is equivalent to $$\begin{bmatrix}4 & 3 & 6\\-k & -2 & 2\end{bmatrix}$$ Applying row reduction to this system allows us to get the reduced row echelon form $$\text{rref}\begin{bmatrix}4 & 3 & 6\\-k & -2 & 2\end{bmatrix} = \begin{bmatrix}1 & 0 & {3 \over 2} - {3(3k+4) \over 2(3k-8)}\\0 & 1 & {2(3k+4) \over 3k-8}\end{bmatrix}.$$
The system should have a solution for all $k \in \mathbb{R}\backslash \{{8 \over 3}\}$ because $k = {8 \over 3}$ makes the system undefined. So $k = {8 \over 3}$ makes the system have no solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Prime factorization number theory Let $n$ be a positive integer, and let
$ 1=d_1<d_2<\dots <d_6=n $
be all of its divisors. Find all $n$ that satisfy
$ \frac 1{d_1} +\frac 1{d_2} + \dots + \frac 1{d_6 } = 2. $
I started by noting $n=a^2b$ where $a,b$ are primes, and tried to manipulate the second equation. Not getting anywhere
solution would help.
| As you wrote, $n$ can be expressed as $n=a^2b$ where $a,b$ are distinct primes. Now we have
$$\frac{1}{1}+\frac{1}{a}+\frac{1}{b}+\frac{1}{a^2}+\frac{1}{ab}+\frac{1}{a^2b}=2,$$
i.e.
$$a^2b+ab+a^2+b+a+1=2a^2b,$$
i.e.
$$b=1+\frac{2a+2}{a^2-a-1}$$
Here, since $(2a+2)/(a^2-a-1)\lt 1$ for $a\ge 4$, we have $a=2,3$.
Thus, the answer is $n=2^2\cdot 7=28$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
} |
Fixed points of $\phi_a(z)$
Prove that $\phi_a(z)=\frac{a-z}{1-\bar az}$ , $0<|a|<1$ has exactly two fixed points ; one inside the unit disc and the other outside the unit disc.
Putting $\phi_a(z)=z$ I find that there are exactly two fixed points which are $$\frac{1\pm \sqrt{1-|a|^2}}{\bar a}.$$
I am unable to find out the second part.
Let $z_1=\frac{1- \sqrt{1-|a|^2}}{\bar a}$. From $0<|a|<1$ , we get $z_1<\frac{1}{\bar a}=\frac{a}{|a|^2}$. Then , $|z_1|<\frac{1}{|a|}$. But how I show that $|z_1|<1$ so that I can say that $z_1$ lies inside the circle ?
| $z_1\overline{a}=1-\sqrt{1-x^2}, x = |a|\to (|z_1||\overline{a}|)^2 =1-2\sqrt{1-x^2}+1-x^2\to bx^2= 2-x^2-2\sqrt{1-x^2}$. Thus we can prove: $2-x^2-2\sqrt{1-x^2} < x^2 \iff 1-x^2 < \sqrt{1-x^2}$ which is true since $0 < 1-x^2 < 1$. Thus $b = |z_1| < 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1367153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is this a correct solution to the linear congruence? I want to solve this linear congruence:
$$2x \equiv 5 \pmod{9}$$
Backward substitution:
$$9 = 4 \cdot 2 + 1$$
$$4(-2) + 9 = 1$$
Therefore, the inverse is: $-2$
Now multiply the linear congruence with $-2$
$$(2)(-2)x \equiv (-2)5 \pmod{9}$$
$$x \equiv -10 \pmod{9}$$
So:
$$x = 8 + 9k$$ for an integer $k$
EDIT:
With the answers given below, the solution is:
Therefore, the inverse is: $-4$
Now multiply the linear congruence with $-4$
$$(2)(-4)x \equiv (-4)5 \pmod{9}$$
$$x \equiv -20 \pmod{9}$$
$$x \equiv 7 \pmod{9}$$
So:
$$x = 7 + 9k$$ for an integer $k$
| $$
2x\equiv 5\pmod 9 \Longrightarrow 2x=9n + 5, n\in\mathbb Z
$$
We can solve linear diophantine equation now, but LHS divides by $2$; hence,
$$
0\equiv 9n + 5 \equiv 1 \cdot n + 1\pmod 2\Longrightarrow n = 2m + 1, m\in\mathbb Z
$$
So,
$$
2x=9(2m+1)+5\Longrightarrow x = 9m + 7
$$
If you know that $2^{-1}\equiv 5 \pmod 9$, then it's much more simpler:
$$
5\cdot 2x\equiv 25\equiv 7\pmod 9\Longrightarrow x\equiv 7\pmod 9
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1367763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solve triangle point given base, point height and difference of sides I have the intuition that one should be able to calculate the position of the circle in the image below (or the equivalent, solve a and b).
We have the following information:
h and d is known as well as the difference between the lengths a and b.
Looking at the sine law:
$$ \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$$
and cosine law:
$$c^2 = a^2 + b^2 - 2ab\cos(\gamma)\,$$
I don't see how we have information to use those formulas.
How do I solve the position of the dot?
| Let $a=x,b-a=t$. Here, note that $b=x+t$ and that $b\lt a+d\Rightarrow t\lt d$.
Then, by the Pythagorean theorem for the biggest triangle, we have
$$\begin{align}&(x+t)^2=h^2+\left(\sqrt{x^2-h^2}+d\right)^2\\&\Rightarrow x^2+2tx+t^2=h^2+x^2-h^2+2d\sqrt{x^2-h^2}+d^2\\&\Rightarrow 2tx+t^2-d^2=2d\sqrt{x^2-h^2}\\&\Rightarrow (2tx+t^2-d^2)^2=4d^2(x^2-h^2)\\&\Rightarrow (4d^2-4t^2)x^2+(-4t^3+4td^2)x-4d^2h^2-t^4-d^4+2t^2d^2=0\\&\Rightarrow a=x=\frac{t^3-d^2t+d\sqrt{(d^2-t^2)(d^2+4h^2-t^2)}}{2(d^2-t^2)}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1367832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving a system of five polynomials I am trying to solve the following system of equations for tuple $\left(a,b,c,d,t\right) \in \mathbb{R}^{4} \times [0,1]$, with parameter $\ell\in\mathbb{R}$.
$$
\begin{eqnarray}
a\frac{t^{2}}{2} - bt + 1 = 0 \qquad (1)\\
a \frac{t^{3}}{6} - b \frac{t^{2}}{2} + t - \ell = 0 \qquad (2)\\
c \frac{t^{2}}{2} - dt + \left(d - \frac{c}{2} - 1\right) = 0 \qquad (3)\\
c \frac{t^{3}}{6} - d \frac{t^{2}}{2} + \left(d - \frac{c}{2} - 1\right)t + \left(\frac{c}{3} - \frac{d}{2} + 1 - \ell\right) = 0 \qquad (4)\\
at - b - ct + d = 0 \qquad (5)
\end{eqnarray}
$$
This arose while solving an optimal control problem, whose solution is known. That known solution tells that if $\frac{1}{6} \leq \ell \leq \frac{1}{4}$, then the above system of equations has unique solution given by $a = 24(1-4\ell)$, $b = 8(1-3\ell)$, $c = -24(1-4\ell)$, $d = 8(1-3\ell) - 24(1-4\ell)$, $t = \frac{1}{2}$. Otherwise, (when $\ell$ does not satisfy the stated inequality) the above system has no solution. While this seems like an easy exercise, I am not able to recover these results. I feel that these facts are not as difficult to establish as I am making it ... so any help or insight in proving them would be appreciated.
So far I have tried these (other than verifying the claimed solution):
*
*Brute force: from (5), substitute $t = \frac{b-d}{a-c}$ in the first four equations and try to solve the resulting four polynomials for $(a,b,c,d)$. As you can see, this soon leads to complications, specially in establishing the bound for $\ell$, and in using the fact that $0 \leq \frac{b-d}{a-c} \leq 1$.
*Let us call the LHS of (2), a cubic polynomial in $t$, as $p(t)$, and the LHS of (4), another cubic, as $q(t)$. Then (1) and (2) say that $p(t) = p^{\prime}(t) = 0$, meaning $p(t)$ has (at least) a double real root $t$. This, using (1), tells that the quadratic discriminant $= b^{2} - 2 a \geq 0$, and using (2), tells that the cubic discriminant $ = \frac{3a^{2}}{4}\ell^{2} + \frac{1}{2}\left(3ab + b^{3}\right)\ell + \left(\frac{2a}{3} - \frac{b^{2}}{4}\right) = 0$. Similarly, (3) and (4) tells that $q(t) = q^{\prime}(t) = 0$, meaning $q(t)$ has (at least) a double real root $t$, which in turn, yields two similar conditions on the quadratic discriminant of (3) and on the cubic discriminant of (4). It is not clear to me how to use the resulting relations together with (5).
*Consider the quadratic (in $t$) equation obtained by subtracting (1) from (3), the cubic (in $t$) equation obtained by subtracting (2) from (4), and the original equation (5), which is linear in $t$. If we call the cubic (in $t$) LHS obtained by subtracting (2) from (4), as $r(t)$, then the resulting three equations are $r(t) = r^{\prime}(t) = r^{\prime\prime}(t) = 0$, which tells that the cubic $r(t)$ has a real root with multiplicity 3, with the root being $\frac{b-d}{a-c}$. Again, not sure how to go from here to recover the desired results.
| Seems like a perfect opportunity to use computer algebra (such as Mathematica), where one imposes the range of values of $l$ explicitly:
Assuming[1/6 <= l <= 1/4,
Solve[{a t^2/2 - b t + 1 == 0, a t^3/6 - b t^2/2 + t - l == 0,
c t^2/2 - d t + (d - c/2 - 1) == 0,
c t^3/6 - d t^2/2 + (d - c/2 - 1) t + (c/3 - d/2 + 1 - l) == 0,
a t - b - c t + d == 0}, {a, b, c, d, t}]
]
$\left\{\left\{b\to \frac{1}{3} \left(-\frac{\sqrt{1-12
l}}{l}+\frac{1}{l}+6\right),a\to \frac{-6 \sqrt{1-12 l} l-\sqrt{1-12
l}+1}{9 l^2},d\to \frac{1}{9} \left(-\frac{\sqrt{1-12
l}}{l^2}-\frac{1}{l^2}-\frac{3 \sqrt{1-12
l}}{l}+\frac{3}{l}+18\right),c\to \frac{-6 \sqrt{1-12 l} l-\sqrt{1-12
l}-1}{9 l^2},t\to \frac{1}{2} \left(\sqrt{1-12
l}+1\right)\right\},\left\{b\to \frac{\sqrt{1-12 l}}{3 l}+\frac{1}{3
l}+2,a\to \frac{6 \sqrt{1-12 l} l+\sqrt{1-12 l}+1}{9 l^2},d\to
\frac{\sqrt{1-12 l}}{9 l^2}-\frac{1}{9 l^2}+\frac{\sqrt{1-12 l}}{3
l}+\frac{1}{3 l}+2,c\to \frac{6 \sqrt{1-12 l} l+\sqrt{1-12 l}-1}{9
l^2},t\to \frac{1}{2} \left(1-\sqrt{1-12 l}\right)\right\},\left\{b\to
-8 (3 l-1),a\to -24 (4 l-1),d\to 8 (9 l-2),c\to 24 (4 l-1),t\to
\frac{1}{2}\right\}\right\}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding $\frac {a}{b} + \frac {b}{c} + \frac {c}{a}$ where $a, b, c$ are the roots of a cubic equation, without solving the cubic equation itself Suppose that we have a equation of third degree as follows:
$$
x^3-3x+1=0
$$
Let $a, b, c$ be the roots of the above equation, such that $a < b < c$ holds. How can we find the answer of the following expression, without solving the original equation?
$$
\frac {a}{b} + \frac {b}{c} + \frac {c}{a}
$$
| Required,
$= a/b + b/c + c/a$
By Cross multiplication,
$= (a^2bc+b^2ac+c^2ab)/(abc)$
$= abc (a+b+c) /(abc)$
$ = (a+b+c) ..........1$
Well known
Property Relation:-
{If α1, α2,α3 ... αn are the roots of the equation
$f(x)= a_0x_n +a_1x_{n-1} +a_2x_{n-2} +...+a_{n-1}x + a_n =0$, then
$f(x)= a_0 (x-α_1)(x-α_2)(x-α_3)... (x-α_n)$
Equating both the RHS terms we get,
$a_0x_n +a_1x_{n-1} +a_2x_{n-2} +...+a_{n-1}x + a_n = a_0(x-α_1)(x-α_2)(x-α_3)... (x-α_n)$
Comparing coefficients of $x_{n-1}$ on both sides, we get
$S1 = α_1 + α_2+α_3 +... + α_n = ∑α_i = -a_1/ a_0$
or, S1= - coeff. of $x_n-1$/coeff. of $x_n$
Comparing coefficients of xn-2 on both sides, we get
$S2 = α_1 α_2+ α_1α_3 +... = ∑α_i α_j = (-1)^2a_2/ a_0$
$i≠ j$
or, S2= (-1)2 coeff. of $x_{n-2}$/coeff. of $x_n$
Comparing coefficients of xn-3 on both sides, we get
$S3 = α_1 α_2α_3+ α_2α_3α_4 +... = ∑α_i α_j α_k = (-1)^3a_3/ a_0$
If $a$ $b$ $c$ are the roots of
$x^3-3x+1=0$
then
$abc = 1$
$a+b+c = 0$
$ab + bc + cd = -1$
Substituting in eqn 1, we get
Ans: 0
| {
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The expression $(1+q)(1+q^2)(1+q^4)(1+q^8)(1+q^{16})(1+q^{32})(1+q^{64})$ where $q\ne 1$, equals The expression
$(1+q)(1+q^2)(1+q^4)(1+q^8)(1+q^{16})(1+q^{32})(1+q^{64})$
where $q\ne 1$, equals
(A) $\frac{1-q^{128}}{1-q}$
(B) $\frac{1-q^{64}}{1-q}$
(C) $\frac{1-q^{2^{1+2+\dots +6}}}{1-q}$
(D) none of the foregoing expressions
What I have done
$(1+q)(1+q^2)(1+q^4)(1+q^8)(1+q^{16})(1+q^{32})(1+q^{64})$ is a polynomial of degree $127$. Now the highest degree of the polynomial in option (A), (B) and (C) is $127$, $63$ and $41$ respectively. And therefore (A) is the correct answer.
I get to the correct answer but I don't think that my way of doing is correct. I mean what if, if option (B) was $\frac{1+q^{128}}{1-q}$.
Please show how should I approach to the problem to get to the correct answer without any confusion.
| Let $$P = (1+q)(1+q^2)(1+q^4)(1+q^8)(1+q^{16})(1+q^{32})(1+q^{64}).$$
One has $$(1-q)P = (1-q)(1+q)(1+q^2)(1+q^4)(1+q^8)(1+q^{16})(1+q^{32})(1+q^{64})
= 1 - q^{128}.
$$
So, one gets $$P = \frac{1-q^{128}}{1-q}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solution of a quadratic diophantine equation I try to solve the Diophantine quadratic equation:
$$X^2+Y^2+Z^2=3W^2.$$ Obviously, there is a non-trivial solution: $(1,1,1,1)$. So I tried to apply Jagy's method: Solutions to $ax^2 + by^2 = cz^2$ .
I consider integers $t,p,q,r$ and the point $P=(1,1,1)$ of the sphere $X^2+Y^2+Z^2=3$. I look for a second point on the sphere in the form $(1+pt,1+qt,1+rt)$. That gives $p^2t+q^2t+r^2t+2p+2q+2r=0$. This implies that
$$\left(1-\frac{2p(p+q+r)}{p^2+q^2+r^2},1-\frac{2q(p+q+r)}{p^2+q^2+r^2},1-\frac{2r(p+q+r)}{p^2+q^2+r^2}\right)$$
is a rational point of the sphere $X^2+Y^2+Z^2=3$. But what to do with that? Thanks in advance.
| From above, the below mentioned equation has parametric solution,
$x^2+y^2+z^2=3w^2$
$x=k^2-6k+1$
$y=k^2-2k+5$
$z=5k^2-2k+1$
$w=3k^2-2k+3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integral $\int_0^1\frac{\log(x)\log(1+x)}{\sqrt{1-x}}\,dx$ I'm trying to evaluate this definite integral:
$$\int_0^1\frac{\log(x) \log(1+x)}{\sqrt{1-x}} dx$$
It's clear that the result can be expressed in terms of derivatives of a hypergeometric function with respect to its parameters. I obtained the following form:
$$4 \left(1 - \log 2\right){_2F_1}^{(0,1,0,0)}\left(1, 0; \tfrac{3}{2}; -1\right) - 2 {_2F_1}^{(1,1,0,0)}\left(1, 0; \tfrac{3}{2}; -1\right) - 2{_2F_1}^{(0,1,1,0)}\left(1, 0; \tfrac{3}{2}; -1\right)$$
Is it possible to expand these derivatives to some explicit form and further simplify this result? Or maybe you could suggest a different way to evaluate this integral that gives a simpler result without going through hypergeometric functions?
| The integral may readily be decomposed into a sum of integrals of products of log-linear terms:
$$\begin{align}
\mathcal{I}
&=\int_{0}^{1}\frac{\ln{\left(x\right)}\ln{\left(1+x\right)}}{\sqrt{1-x}}\,\mathrm{d}x\\
&=2\int_{0}^{1}\ln{\left(1-y^2\right)}\ln{\left(2-y^2\right)}\,\mathrm{d}y;~~~\small{\left[\sqrt{1-x}=y\right]}\\
&=2\int_{0}^{1}\ln{\left(1-y\right)}\ln{\left(\sqrt{2}-y\right)}\,\mathrm{d}y\\
&~~~~~+2\int_{0}^{1}\ln{\left(1-y\right)}\ln{\left(\sqrt{2}+y\right)}\,\mathrm{d}y\\
&~~~~~+2\int_{0}^{1}\ln{\left(1+y\right)}\ln{\left(\sqrt{2}-y\right)}\,\mathrm{d}y\\
&~~~~~+2\int_{0}^{1}\ln{\left(1+y\right)}\ln{\left(\sqrt{2}+y\right)}\,\mathrm{d}y.\\
\end{align}$$
Each of these four integrals can be resolved in terms of dilogarithms in a systematic manner, for instance by using the general closed forms for two integrals I derive below. Since a final result has already been provided in another response, I leave the plugging-and-chugging step as an exercise to the fearless reader.
Suppose $0<a\land0<a+b$. Then we find:
$$\begin{align}
J{(a,b)}
&=\int_{0}^{1}\ln{\left(1-y\right)}\ln{\left(a+by\right)}\,\mathrm{d}y\\
&=\int_{0}^{1}\ln{\left(w\right)}\ln{\left(a+b-bw\right)}\,\mathrm{d}w;~~~\small{\left[1-y=w\right]}\\
&=\small{-\int_{0}^{1}\frac{1}{w}\left[\frac{\left(a+b\right)\ln{\left(a+b\right)}}{b}-\frac{bw+\left(a+b-bw\right)\ln{\left(a+b-bw\right)}}{b}\right]\,\mathrm{d}w}\\
&=\int_{0}^{1}\frac{bw-bw\ln{\left(a+b\right)}+\left(a+b-bw\right)\ln{\left(1-\frac{b}{a+b}w\right)}}{bw}\,\mathrm{d}w\\
&=1-\ln{\left(a+b\right)}+\int_{0}^{1}\frac{\left(a+b-bw\right)\ln{\left(1-\frac{b}{a+b}w\right)}}{bw}\,\mathrm{d}w\\
&=1-\ln{\left(a+b\right)}+\int_{0}^{1}\frac{\left(1-cw\right)\ln{\left(1-cw\right)}}{cw}\,\mathrm{d}w;~~~\small{\left[c:=\frac{b}{a+b}\right]}\\
&=\small{1-\ln{\left(a+b\right)}-\int_{0}^{1}\ln{\left(1-cw\right)}\,\mathrm{d}w+\frac{1}{c}\int_{0}^{1}\frac{\ln{\left(1-cw\right)}}{w}\,\mathrm{d}w}\\
&=1-\ln{\left(a+b\right)}+1+\frac{1-c}{c}\ln{\left(1-c\right)}-\frac{1}{c}\,\operatorname{Li}_{2}{\left(c\right)}\\
&=2-\ln{\left(a+b\right)}+\frac{1-c}{c}\ln{\left(1-c\right)}-\frac{1}{c}\,\operatorname{Li}_{2}{\left(c\right)}\\
&=2+\frac{a\ln{\left(a\right)}-\left(a+b\right)\ln{\left(a+b\right)}-\left(a+b\right)\operatorname{Li}_{2}{\left(\frac{b}{a+b}\right)}}{b}\\
\end{align}$$
Suppose $0<a\land0<a+b\land0<a-b$. Then we find:
$$\begin{align}
K{(a,b)}
&=\int_{0}^{1}\ln{\left(1+y\right)}\ln{\left(a+by\right)}\,\mathrm{d}y\\
&=\small{\ln{(2)}\ln{\left(a+b\right)}-\int_{0}^{1}y\left[\frac{b\ln{\left(1+y\right)}}{a+by}+\frac{\ln{\left(a+by\right)}}{1+y}\right]\,\mathrm{d}y}\\
&=\ln{(2)}\ln{\left(a+b\right)}\\
&~~~~~\small{-\int_{0}^{1}\left[\ln{\left(1+y\right)}-\frac{a\ln{\left(1+y\right)}}{a+by}+\ln{\left(a+by\right)}-\frac{\ln{\left(a+by\right)}}{1+y}\right]\,\mathrm{d}y}\\
&=\small{\ln{(2)}\ln{\left(a+b\right)}-\left(2\ln{(2)}-1\right)+\frac{b+a\ln{\left(a\right)}-\left(a+b\right)\ln{\left(a+b\right)}}{b}}\\
&~~~~~\small{+\int_{0}^{1}\frac{a\ln{\left(1+y\right)}}{a+by}\,\mathrm{d}y+\int_{0}^{1}\frac{\ln{\left(a+by\right)}}{1+y}\,\mathrm{d}y}\\
&=\small{2-2\ln{(2)}+\ln{(2)}\ln{\left(a+b\right)}+\frac{a\ln{\left(a\right)}-\left(a+b\right)\ln{\left(a+b\right)}}{b}}\\
&~~~~~\small{+\int_{0}^{1}\frac{a\ln{\left(1+y\right)}}{a+by}\,\mathrm{d}y+\ln{(2)}\ln{\left(a+b\right)}-\int_{0}^{1}\frac{b\ln{\left(1+y\right)}}{a+by}\,\mathrm{d}y}\\
&=\small{2-2\ln{(2)}+2\ln{(2)}\ln{\left(a+b\right)}+\frac{a\ln{\left(a\right)}-\left(a+b\right)\ln{\left(a+b\right)}}{b}}\\
&~~~~~\small{+\left(a-b\right)\int_{0}^{1}\frac{\ln{\left(1+y\right)}}{a+by}\,\mathrm{d}y}\\
&=\small{2-2\ln{(2)}+2\ln{(2)}\ln{\left(a+b\right)}+\frac{a\ln{\left(a\right)}-\left(a+b\right)\ln{\left(a+b\right)}}{b}}\\
&~~~~~\small{+\left(a-b\right)\int_{1}^{2}\frac{\ln{\left(w\right)}}{a-b+bw}\,\mathrm{d}w};~~~\small{\left[1+y=w\right]}\\
&=\small{2-2\ln{(2)}+2\ln{(2)}\ln{\left(a+b\right)}+\frac{a\ln{\left(a\right)}-\left(a+b\right)\ln{\left(a+b\right)}}{b}}\\
&~~~~~\small{+\int_{0}^{2}\frac{\ln{\left(w\right)}}{1+\frac{b}{a-b}w}\,\mathrm{d}w-\int_{0}^{1}\frac{\ln{\left(w\right)}}{1+\frac{b}{a-b}w}\,\mathrm{d}w}\\
&=\small{2-2\ln{(2)}+2\ln{(2)}\ln{\left(a+b\right)}+\frac{a\ln{\left(a\right)}-\left(a+b\right)\ln{\left(a+b\right)}}{b}}\\
&~~~~~\small{+2\int_{0}^{1}\frac{\ln{\left(2u\right)}}{1+\frac{2b}{a-b}u}\,\mathrm{d}u-\frac{a-b}{b}\operatorname{Li}_{2}{\left(-\frac{b}{a-b}\right)}};~~~\small{\left[w=2u\right]}\\
&=\small{2-2\ln{(2)}+2\ln{(2)}\ln{\left(a+b\right)}+\frac{a\ln{\left(a\right)}-\left(a+b\right)\ln{\left(a+b\right)}}{b}}\\
&~~~~~+2\ln{(2)}\int_{0}^{1}\frac{\mathrm{d}u}{1+\frac{2b}{a-b}u}+2\int_{0}^{1}\frac{\ln{\left(u\right)}}{1+\frac{2b}{a-b}u}\,\mathrm{d}u\\
&~~~~~-\frac{a-b}{b}\operatorname{Li}_{2}{\left(-\frac{b}{a-b}\right)}\\
&=2-2\ln{(2)}+2\ln{(2)}\ln{\left(a+b\right)}\\
&~~~~~+\frac{a\ln{\left(a\right)}-\left(a+b\right)\ln{\left(a+b\right)}+\left(a-b\right)\ln{(2)}\ln{\left(\frac{a+b}{a-b}\right)}}{b}\\
&~~~~~+\frac{a-b}{b}\left[\operatorname{Li}_{2}{\left(-\frac{2b}{a-b}\right)}-\operatorname{Li}_{2}{\left(-\frac{b}{a-b}\right)}\right].\\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 3,
"answer_id": 2
} |
$f(x) =ax^6 +bx^5+cx^4+dx^3+ex^2+gx+h $, find $f(7)$. Problem :
Given a polynomial $f(x) =ax^6 +bx^5+cx^4+dx^3+ex^2+gx+h$
such that $f(1)= 1, f(2) =2 , f(3) = 3, f(4) =4, f(5)=5, f(6) =6$.
Find $f(7)$ in terms of $h$.
My approach:
We can put the values of $f(1) = 1$ in the given equation and $f(2) = 2$, etc. But this is quite time consuming by making six different equations and then solve them to get the values of $a,b,c,d,e,g,h$. Please suggest some alternate solution for this.
| For a method which is intermediate in cleverness between the one given by the other answers and just solving the system of simultaneous equations, you could use successive differences. We write the following table:
$$
\begin{array}{ccccccccccccccc}
h & & 1 & & 2 & & 3 & & 4 & & 5 & & 6 & & x_6 \\
& 1-h & & 1 & & 1 & & 1 & & 1 & & 1 & & x_5 \\
&& h && 0 && 0 && 0 && 0 && x_4 \\
&&& -h && 0 && 0 && 0 && x_3 \\
&&&& h && 0 && 0 && x_2 \\
&&&&& -h && 0 && x_1 \\
&&&&&& h && x_0
\end{array}$$
where each number is the difference of the two directly above it. Then each row of this table represents a polynomial whose degree is one lower than the row above it; since we started with a degree-6 polynomial, the bottom row is constant, and so $x_0=h$. Working our way back up the table, we then have
$$
x_1=0+x_0=h\\
x_2=0+x_1=h\\
x_3=0+x_2=h\\
x_4=0+x_3=h\\
x_5=1+x_4=h+1\\
x_6=6+x_5=h+7
$$
and so $f(7)=h+7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 0
} |
Complex number identity by trigonometry Show that $\lvert e^{i\theta} - 1\rvert = 2\lvert\sin(\theta/2)\rvert$ by using the geometry of the triangle with vertices 0, 1, and the midpoint of the line joining 0 and $e^{i\theta}$.
I have been able to show this identity through other means, however I am stuck on how to utilize this particular triangle.
| $|e^{i\theta}-1|=|\cos\theta+i\sin\theta-(1+0i)|=|(\cos\theta-1)+i\sin\theta|=\sqrt{(\cos\theta-1)^2+(\sin\theta)^2}=\sqrt{\cos^2\theta+1-2\cos\theta+\sin^2\theta}=\sqrt{2-2\cos\theta}=\sqrt{2(1-\cos\theta)}=\sqrt{2(\sin^2\frac{\theta}{2}+\cos^2\frac{\theta}{2}-\cos^2\frac{\theta}{2}+\sin^2\frac{\theta}{2})}=\sqrt{4sin^2\frac{\theta}{2}}=2|\sin\frac{\theta}{2}|$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
evaluate $\frac 1{1+\sqrt2+\sqrt3} + \frac 1{1-\sqrt2+\sqrt3} + \frac 1{1+\sqrt2-\sqrt3} + \frac 1{1-\sqrt2-\sqrt3}$
Evaluate $\frac 1{1+\sqrt2+\sqrt3} + \frac 1{1-\sqrt2+\sqrt3} + \frac 1{1+\sqrt2-\sqrt3} + \frac 1{1-\sqrt2-\sqrt3}$
How to evalute this equation without using calculator?
| We have, $$\underbrace{\frac {1}{1+\sqrt2+\sqrt3} + \frac {1}{1-\sqrt2+\sqrt3}} + \underbrace{\frac {1}{1+\sqrt2-\sqrt3} + \frac {1}{1-\sqrt2-\sqrt3}}$$ $$=\left(\frac {1}{1+\sqrt2+\sqrt3} + \frac {1}{1-\sqrt2+\sqrt3}\right) +\left( \frac {1}{1+\sqrt2-\sqrt3} + \frac {1}{1-\sqrt2-\sqrt3}\right)$$ $$=\left(\frac {1-\sqrt2+\sqrt3+1+\sqrt2+\sqrt3}{(1+\sqrt2+\sqrt3)(1-\sqrt2+\sqrt3)} \right) +\left( \frac {1-\sqrt2-\sqrt3+1+\sqrt2-\sqrt3}{(1+\sqrt2-\sqrt3)(1-\sqrt2-\sqrt3)} \right)$$ $$=\left(\frac {2+2\sqrt3}{(1+\sqrt3)^2-(\sqrt{2})^2} \right) +\left( \frac {2-2\sqrt{3}}{(1-\sqrt3)^2-(\sqrt{2})^2} \right)$$ $$=\frac{2+2\sqrt{3}}{1+3+2\sqrt{3}-2}+\frac{2-2\sqrt{3}}{1+3-2\sqrt{3}-2}$$ $$=\frac{2+2\sqrt{3}}{2+2\sqrt{3}}+\frac{2-2\sqrt{3}}{2-2\sqrt{3}}$$ $$=1+1=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
} |
Modular Quadratic Equation I'm trying to solve that equation:
$x^2-3x-5\equiv0\pmod{343}$
I've completed the square as follows:
$x^2-3x-5 \equiv x^2+340x-5\equiv(x+170)^2-170^2-5\pmod{343}\\
(x+170)^2 \equiv 93\pmod{343}\\
y^2 \equiv 93 \pmod{343}$
But I have no idea how to move on. How can I use the fact that $343=7^3$?
| $$y^2\equiv93\pmod{343}\implies y^2\equiv93\pmod{49}\implies y^2\equiv93\pmod{7}$$
From the last congruence we get $y\equiv\pm3\pmod 7$ or $y=7z\pm 3$. Now, plug this into the second congruence to get
$$\pm21z+9\equiv-5\pmod{49}$$
or
$$\pm3z\equiv-2\pmod 7\iff z\equiv\mp3\pmod 7$$
so $z=7u\mp 3$ and $y=7(7u\mp 3)\pm 3=49u\mp21\pm 3$. So $y=49u\pm 18$.
Finally, put this into the first congruence:
$$\pm49u+324\equiv 93\pmod{343}$$
The rest should be quite straightforward.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Sum of cosines of complementary/suplementary angles Why are $(\cos(2^{\circ})+\cos(178^{\circ})), (\cos(4^{\circ})+\cos(176^{\circ})),.., (\cos(44^{\circ})+\cos(46^{\circ}))$ all equal zero?
Could you prove it by some identity?
| Notice, the following trigonometric equation $$\color{blue}{\cos A+\cos B=2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)}$$ Hence, if angles $A$ & $B$ are supplementary angles i.e. $A+B=180^\circ$ $$\implies \cos\left(\frac{A+B}{2}\right)=\cos\left(\frac{180^\circ}{2}\right)=\cos\left(90^\circ\right)=0$$ Hence, we get
$$\cos(2^\circ)+\cos(178^\circ)=2\cos\left(\frac{2^\circ+178^\circ}{2}\right)\cos\left(\frac{2^\circ-178^\circ}{2}\right)=2\cos(90^\circ)\cos(88^\circ)=0$$
$$\cos(4^\circ)+\cos(176^\circ)=2\cos\left(\frac{4^\circ+176^\circ}{2}\right)\cos\left(\frac{4^\circ-176^\circ}{2}\right)=2\cos(90^\circ)\cos(86^\circ)=0$$
$$\cos(44^\circ)+\cos(136^\circ)=2\cos\left(\frac{44^\circ+136^\circ}{2}\right)\cos\left(\frac{44^\circ-136^\circ}{2}\right)=2\cos(90^\circ)\cos(46^\circ)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
Find the area of a triangle given the radius of its incircle and a tangential point A friend gave me recently the following interesting problem and I would like to share a couple of solutions. Any additional contributions are welcome.
A triangle $\vartriangle ABC$ is given and we know the radius $r$ of its incircle $(O, r)$. Let $D$ be the tangential point of the incircle on $AC$ which partitions $AC$ into $AD=\alpha$ and $DC=\beta$. Determine the area of $\vartriangle ABC$ as a function of $\alpha$, $\beta$ and $r$.
| Solution 1: Draw $AO$, $AO$ and $CO$. We know that these are the bisectors of $\angle BAC$, $\angle ABC$ and $\angle BCA$ respectively.
Let $\angle OAC = \frac{1}{2} \angle BAC =: \phi_A$ and $\angle OBE = \frac{1}{2} \angle ABC =: \phi_B$ and $\angle OCA = \frac{1}{2}\angle BCA =: \phi_C$.
We have that $\angle ODA$ is a right angle (and so is $\angle ODC$). Thus $\tan \phi_A = \frac{r}{\alpha}$ and $\tan \phi_C = \frac{r}{\beta}$. Therefore $\angle AOD = \frac{\pi}{2} - \phi_A$.
The two right-angled triangles $\vartriangle ADO$ and $\vartriangle AEO$ are equal (indeed, they share a common side $AO$ and $OE=OD=r$). Thus, $\angle EOA = \angle DOA = \frac{\pi}{2} - \phi_A$.
Similarly, $\angle COD = \frac{\pi}{2}-\phi_C$ and since $\vartriangle COD = \vartriangle COF$ it is $\angle COF = \angle COD = \frac{\pi}{2}-\phi_C$.
Using the fact that
$$
\angle AOD + \angle DOC + \angle COF + \angle FOE + \angle EOA = 2\pi,\\
2(\frac{\pi}{2}-\phi_A) + 2(\frac{\pi}{2}-\phi_C) + \angle FOE = 2\pi,\\
\angle FOE = 2(\phi_A + \phi_C).
$$
Because of the equality of $\vartriangle BEO$ and $\vartriangle BFO$ we have $\angle EOB = \frac{\angle FOE}{2} = \phi_A + \phi_C$, so
$$
\tan \angle EOB = \frac{BE}{r},\\
BE = r \tan(\phi_A + \phi_C) = r\frac{\tan\phi_A + \tan\phi_C}{1-\tan\phi_A\tan\phi_C}=\\
=r\frac{\frac{r}{\alpha}+\frac{r}{\beta}}{1-\frac{r}{\alpha}\frac{r}{\beta}}=r\frac{\frac{r(\alpha+\beta)}{\alpha\beta}}{\frac{\alpha\beta-r^2}{\alpha\beta}}=r^2\frac{\alpha+\beta}{\alpha\beta-r^2}
$$
From the aforementioned triangle equalities we have $AE = \alpha$, $CF = \beta$ and $BF = BE$, therefore, the semiperimeter of $\vartriangle ABC$ is
$$
\tau = \alpha + \beta + BE = \alpha + \beta + r^2\frac{\alpha+\beta}{\alpha\beta-r^2}=\frac{\alpha\beta(\alpha+\beta)}{\alpha\beta-r^2},
$$
and the area of the triangle is
$$S = r\tau = r\frac{\alpha\beta(\alpha+\beta)}{\alpha\beta-r^2}$$
Solution 2: This is a solution without trigonometry and is less elegant. Since $\angle BOE = \phi_A + \phi_C$ there is a point $M$ on $BE$ so that $\angle MOE = \phi_A$ and $\angle MOB = \phi_C$ as in the following figure:
It is easy to see that the two triangles $\vartriangle MEO$ and $\vartriangle OEA$ are similar (they have all their angles equal to one another), so,
$$
\frac{EM}{EO}=\frac{EO}{EA},\\
EM = \frac{r^2}{\alpha}.
$$
We now need to determine $BM$. What is the same, $\vartriangle BMO$ and $\vartriangle BOC$ are similar (all angles equal to one another). Note that (Pythagorean theorem on $\vartriangle BEO$)
$$BO=\sqrt{BE^2 + r^2} = \sqrt{(BM+ME)^2 + r^2},$$
and (Pythagorean theorem on $\vartriangle OCD$)
$$OC = \sqrt{r^2 + \beta^2}.$$
We also have
$$
MO = \sqrt{r^2 + ME^2} = \sqrt{r^2 + \frac{r^4}{\alpha^2}} = r\sqrt{1+(r/a)^2}
$$
Using the fact that $\vartriangle BMO$ and $\vartriangle BOC$ are similar (all angles equal to one another), we arrive at
$$
\frac{BM}{BO}=\frac{MO}{OC},\\
\frac{BM}{\sqrt{(BM+\frac{r^2}{\alpha^2})^2 + r^2}}=\frac{r\sqrt{1+\frac{r^2}{\alpha^2}}}{\sqrt{r^2 + \beta^2}},\\
\frac{BM^2}{(BM+\frac{r^2}{\alpha^2})^2 + r^2}=\frac{r^2 (1+\frac{r^2}{\alpha^2})}{r^2 + \beta^2},
$$
which is a quadratic equation one can solve to derive a formula for $BM$ (the non-positive solution should be discarded) in terms of $r$, $\alpha$ and $\beta$. Having determined $BM$ and $ME$ we can easily determine the semiperimeter of $\vartriangle ABC$, $\tau = \alpha + \beta + BM + ME$ and the area of $\vartriangle ABC$ is again $S = r\tau$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Does $\frac{x+y}{2}>\frac{a+b}{2}$ hold? $a$ and $b$ are two real positive numbers. Given that $x=\sqrt{ab}$ and $y=\sqrt{\frac{a^2+b^2}{2}}$, which one has a higher value, $\frac{x+y}{2}$ or $\frac{a+b}{2}$?
We know that $y=\sqrt{\frac{a^2+b^2}{2}}>\frac{a+b}{2}>x=\sqrt{ab}$ by inequality, and at this point I'm stuck.
| we know $$\sqrt{\frac{x^2+y^2}{2}} \geq \frac{x+y}{2} $$ so plug x,y $$ \sqrt{\frac{x^2+y^2}{2}} \geq \frac{x+y}{2}\\\sqrt{\frac{(\sqrt{ab})^2+(\sqrt{\frac{a^2+b^2}{2}})^2}{2}} \geq \frac{x+y}{2}\\\sqrt{\frac{ab+\frac{a^2+b^2}{2}}{2}} \geq \frac{x+y}{2}\\\sqrt{\frac{\frac{a^2+b^2+2ab}{2}}{2}} \geq \frac{x+y}{2}\\\sqrt{\frac{(a+b)^2}{4}} \geq \frac{x+y}{2}\\ \frac{a+b}{2} \geq \frac{x+y}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to integrate $\int_{l1}^{l2}\frac{e^{\pm i a x}}{\sqrt{bx^2+cx+d}}dx$ I have the above mentioned integral
$$
\int_{l_1}^{l_2}\frac{e^{\pm i a x}}{\sqrt{bx^2+cx+d}}dx
$$
which I want to solve. I expect some special functions in its solution, but so far I am out of ideas now.
I have tried the following:
I substitute $t=\sqrt{bx^2+cx+d}$ and get an expression for $x$ that I can insert in the exponential:
$$
t=\sqrt{bx^2+cx+d}
$$
$$
\frac{t^2}{b}=x^2+\frac{c}{b}x+\frac{d}{b}=\left (x+\frac{c}{2b}\right )^2 + \frac{d}{b} - \left (\frac{c}{2b}\right )^2
$$
$$
\sqrt{\frac{t^2}{b}-\frac{d}{b} + \left (\frac{c}{2b}\right )^2}-\frac{c}{2b}=x
$$
The derivative of the substitution t is:
$$
\frac{dt}{dx}=\frac{2bx+c}{2\sqrt{bx^2+cx+d}}=\frac{2bx+c}{2t}
$$
inserting everything in the integral gives
$$\int_{t(l_1)}^{t(l_2)}\frac{e^{\pm i a \left( \sqrt{\frac{t^2}{b}-\frac{d}{b} + \left (\frac{c}{2b}\right )^2}-\frac{c}{2b} \right )}}{t}\frac{2t}{2bx+c}dt=2
\int_{t(l_1)}^{t(l_2)}\frac{e^{\pm i a \left( \sqrt{\frac{t^2}{b}-\frac{d}{b} + \left (\frac{c}{2b}\right )^2}-\frac{c}{2b} \right )}}{2b\left ( \sqrt{\frac{t^2}{b}-\frac{d}{b} + \left (\frac{c}{2b}\right )^2}-\frac{c}{2b}\right ) + c}dt
$$
This looks nearly like an exponential integral. I would expect a solution involving this kind of integral, but this is where I am stuck right now. Is it possible to give a closed form solution involving special functions?
| Too long for a comment (a work in progress)
$$
\int \frac{\mathrm{e}^{\pm iax}}{\sqrt{bx^2+cx+d}}dx
$$
Focusing on the denominator
$$
b\left(x^2+\frac{c}{b}x+\frac{d}{b}\right) =b\left[\left(x+\frac{c}{2b}\right)^2+\frac{d}{b}-\left(\frac{c}{2b}\right)^2\right]
$$
lets change variables
$$
t = \frac{x+\frac{c}{2b}}{\sqrt{\left(\frac{c}{2b}\right)^2-\frac{d}{b}}}\implies x = -\frac{c}{2b} + \left(\sqrt{\left(\frac{c}{2b}\right)^2-\frac{d}{b}}\right) t
$$
leads to
$$
b\left(x^2+\frac{c}{b}x+\frac{d}{b}\right) = b\left[\left(\frac{c}{2b}\right)^2-\frac{d}{b}\right]\left(t^2-1\right)
$$
therefore we have
$$
\left(\sqrt{\left(\frac{c}{2b}\right)^2-\frac{d}{b}}\right)\int \frac{\mathrm{e}^{\pm ia\left(-\frac{c}{2b} + \left(\sqrt{\left(\frac{c}{2b}\right)^2-\frac{d}{b}}\right) t\right)}}{\sqrt{b\left[\left(\frac{c}{2b}\right)^2-\frac{d}{b}\right]\left(t^2-1\right)}}dt
$$
let $t=\cosh \theta$ we have
$$
t^2-1 = \sinh^2 \theta\\
dt = \sinh \theta d\theta
$$
thus we get
$$
\frac{\mathrm{e}^{\mp i\frac{ac}{2b}}}{\sqrt{b}}\int_{\bar{l_1}}^{\bar{l_2}} \mathrm{e}^{\pm i\lambda_1\cosh \theta}d\theta
$$
where
$$
\lambda_1 = a\left[-\frac{c}{2b} + \sqrt{\left(\frac{c}{2b}\right)^2-\frac{d}{b}}\right]
$$
I am still working on the last integral.
| {
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"url": "https://math.stackexchange.com/questions/1381879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding $P$ such that $P^TAP$ is a diagonal matrix
Let $$A = \left(\begin{array}{cc} 2&3 \\ 3&4 \end{array}\right) \in
M_2(\mathbb{C})$$
Find $P$ such that $P^TAP = D$ where $D$ is a diagonal matrix.
So here's the solution:
$$A = \left(\begin{array}{cc|cc} 2&3&1&0\\ 3&4&0&1 \end{array}\right) \sim \left(\begin{array}{cc|cc} 2&0&1&-3/2\\ 0&-1/2&0&1 \end{array}\right)$$
Therefore, $$P = \left(\begin{array}{cc} 1&-3/2\\ 0&1 \end{array}\right) \\ P^TAP = \left(\begin{array}{cc} 2&0\\ 0&-1/2 \end{array}\right) $$
What was done here exactly? I'd be glad elaborate about the process.
Thanks.
| Hermite Reduction.
SEE ALSO Orthogonal basis for this indefinite symmetric bilinear form
Transforming quadratic forms, how is this theorem called?
What is the difference between using $PAP^{-1}$ and $PAP^{T}$ to diagonalize a matrix?
When you have a symmetric matrix of integers, you may use Hermite's method for diagonalizing, the order they want is $P^t A P = D.$ Alright, I will need to do an inverse at the end.
Make a column vector
$$
V =
\left(
\begin{array}{c}
x \\
y
\end{array}
\right)
$$
and write out
$$ V^T A V = 2 x^2 + 6 xy + 4 y^2 $$
Next, we cancel out all $x$ terms using
$$ \left( x + \frac{3}{2} y \right)^2 = x^2 + 3 xy + \frac{9}{4} y^2, $$ and
$$ 2 \left( x + \frac{3}{2} y \right)^2 = 2x^2 + 6 xy + \frac{9}{2} y^2. $$
As a result,
$$ 2 \left( x + \frac{3}{2} y \right)^2 - \frac{1}{2} y^2 = 2 x^2 + 6 xy + 4 y^2 . $$
MORE TYPING TO COME !!!!
In matrices, the direction I did is
$$
\left(
\begin{array}{cc}
1 & 0 \\
\frac{3}{2} & 1
\end{array}
\right)
\left(
\begin{array}{cc}
2 & 0 \\
0 & -\frac{1}{2}
\end{array}
\right)
\left(
\begin{array}{cc}
1 & \frac{3}{2} \\
0 & 1
\end{array}
\right) =
\left(
\begin{array}{cc}
2 & 3 \\
3 & 4
\end{array}
\right)
$$
With
$$ Q =
\left(
\begin{array}{cc}
1 & \frac{3}{2} \\
0 & 1
\end{array}
\right)
$$
notice that the rows correspond exactly to the linear substitutions, the first row means $x + \frac{3}{2} y$ and the second row means $y.$
EVEN MORE EXCITING TYPING ANY MINUTE !!!!!!!!!!
What I did so far is in the order $Q^T D Q = A.$ All we need to do is take $p= Q^{-1},$ which is easier than usual because $\det Q = 1.$ The result is
$$
\left(
\begin{array}{cc}
1 & 0 \\
-\frac{3}{2} & 1
\end{array}
\right)
\left(
\begin{array}{cc}
2 & 3 \\
3 & 4
\end{array}
\right)
\left(
\begin{array}{cc}
1 & -\frac{3}{2} \\
0 & 1
\end{array}
\right) =
\left(
\begin{array}{cc}
2 & 0 \\
0 & -\frac{1}{2}
\end{array}
\right)
$$
The second example in the question, with 3 by 3 matrix, is
$$ x^2 + 4 y^2 + 4 z^2 + 16 yz + 4 zx + 4 xy. $$ This is an example where an extra trick must be used:
$$ (x+2y+2z)^2 = x^2 + 4 y^2 + 4 z^2 + 8 yz + 4 zx + 4 xy. $$
All that remains to construct is $8yz$ because we used up the $y^2$ and $z^2.$ The trick is that $(y+z)^2 - (y-z)^2 = 4yz,$ so
$$ (x+2y+2z)^2 + 2 (y+z)^2 -2 (y-z)^2= x^2 + 4 y^2 + 4 z^2 + 16 yz + 4 zx + 4 xy. $$
Thus the diagonal matrix gets entries $1,2,-2$ and, in this direction,
$$
Q =
\left(
\begin{array}{ccc}
1 & 2 & 2 \\
0 & 1 & 1 \\
0 & 1 & -1
\end{array}
\right)
$$
and then $P = Q^{-1}$
$$
P =
\left(
\begin{array}{ccc}
1 & -2 & 0 \\
0 & \frac{1}{2} & \frac{1}{2} \\
0 & \frac{1}{2} & - \frac{1}{2}
\end{array}
\right)
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
simplifying complex fractions I have this expression
${2 \over x^2 - 4}$ + ${1 \over x + 2}$
So I would take the LCM to ${x^2 - 4}$ or using the difference of 2 squares to be ${(x + 2)(x - 2)}$
So I would simplify this to:
${2 + (x - 2) \over (x - 2)(x + 2)}$ which would simplify further to
${2\over x + 2}$
But the answer to the question is ${x \over x^2 - 4}$
I'm not sure how the real answer is achieved and how I am wrong
| Notice, we have $$\frac{2}{x^2-4}+\frac{1}{x+2}$$ $\color{red}{\text{apply} \ a^2-b^2=(a-b)(a+b)}$ $$=\frac{2}{(x-2)(x+2)}+\frac{1}{x+2}$$ $\color{red}{\text{take L.C.M of denominators }}$ $$=\frac{2+x-2}{(x-2)(x+2)}$$$$ =\frac{x}{(x-2)(x+2)}$$ Thus we have following simplified form $$\bbox[5px, border: 2px solid #C0A000]{\color{red}{\frac{2}{x^2-4}+\frac{1}{x+2}=\frac{x}{(x-2)(x+2)}}}$$
| {
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How does this algebraic trick regarding partial fraction works? Suppose I have to evaluate the integral $$\int \frac{x}{(x-1)(2x+1)(x+3)} \, dx $$
I write it as $$\frac{a_1}{x-1} +\frac{a_2}{2x+1} +\frac{a_3}{x+3}$$
where $a_1$, $a_2$, $a_3$ are constants. once I have foud relationaships between them I solve for the $a_{i}$ and subsequently substitute them into the integral.
It can be observed that for finding this constants we may use following trick.
Suppose we have to find $a_1$ then for time being remove the term beneath $a_1$ from the relation $$\frac{a_1}{x-1} +\frac{a_2}{2x+1} +\frac{a_3}{x+3}$$ now put the value of $x$ that would appear if the removed term is equated to $0$ in $\frac{x}{(x-1)(2x+1)(x+3)}$. This the value of constant $a_1$.
How does this works ?
| As you wrote, $$\frac x {(x-1)(2x+1)(x+3)}=\frac {a_1}{x-1}+\frac {a_2}{2x+1}+\frac {a_3}{x+3}$$ As Peter commented, multiply everything by $(x-1)(2x+1)(x+3)$; this gives $$x=a_1(2x+1)(x+3)+a_2(x-1)(x+3)+a_3(x-1)(2x+1)$$ First make $x=1$ so $$1=a_1\times3\times 4=12a_1$$ Do the same with $x=-3$ and $x=-\frac 12$. You immediately get the value of each of the $a_i$'s without needing to solve any equation.
Remember the trick : it is extremely useful.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\lim\limits_{x\to 0}\frac{\sqrt{2(2-x)}(1-\sqrt{1-x^2})}{\sqrt{1-x}(2-\sqrt{4-x^2})}$ I use L'Hospitals rule, but can't get the correct limit.
Derivative of numerator in function is
$$\frac{-3x^2+4x-\sqrt{1-x^2}+1}{\sqrt{(4-2x)(1-x^2)}}$$
and derivative of denominator is
$$\frac{-3x^2+2x-2\sqrt{4-x^2}+4}{2\sqrt{(1-x)(4-x^2)}}$$
Now, L'Hospitals rule must be applied again. Is there some easier way to compute the limit?
Limit should be $L=4$
| For $a>0,$
$$\lim_{x\to0}\dfrac{a-\sqrt{a^2-x^2}}{x^2}=\lim_{x\to0}\dfrac{a^2-(a^2-x^2)}{x^2(a+\sqrt{a^2-x^2})}=\dfrac1{2a}$$ as $x\ne0$ as $x\to0$
Or set $x=a\sin2y\implies y\to0\implies\sin y\to0$
$$\lim_{x\to0}\dfrac{a-\sqrt{a^2-x^2}}{x^2}=a\lim_{y\to0}\dfrac{1-\cos2y}{(a\sin2y)^2}=\dfrac1a\lim_{y\to0}\dfrac{1-\cos2y}{(2\sin y\cos y)^2}$$
$$=\dfrac1a\lim_{y\to0}\dfrac{2\sin^2y}{(2\sin y\cos y)^2}=\dfrac1{2a}$$ as $\sin y\ne0$ as $\sin y\to0$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find all solutions of the equation $\cos 2θ + 7 \cos θ = 8$ Find all solutions of the equation in the interval [0, 2π) (express your answer in terms of $k$, where $k$ is any integer).
$$\cos 2\theta + 7 \cos\theta = 8$$
Did I make a mistake somewhere? My final answer was $0+2\pi k$
Here is my work. I went from step 1 to step 2 by using the trig identity $\cos 2\theta = 2\cos^2\theta -1$ and then solved it like a quadratic. I also set $\cos \theta = x$ because it's easier for me to visualize the equation that way.
| Notice, we have $$\cos 2\theta+7\cos \theta=8$$ $$2\cos^2\theta-1+7\cos \theta=8$$ $$2\cos^2\theta+7\cos \theta-9=0$$ $$\underbrace{2\cos^2\theta+9\cos \theta}\underbrace{-2\cos \theta-9}=0$$
$$\cos\theta(2\cos \theta+9)-(2\cos \theta+9)=0$$ $$(\cos \theta-1)(2\cos \theta+9)=0$$ $$\implies \cos\theta-1=0\iff \cos\theta=1=\cos 0$$ Writing the general solution, we get $$\theta=2k\pi\pm 0=2k\pi$$
$$\implies 2\cos\theta+9=0\implies \cos\theta \neq \frac{-9}{2}$$ this case does not give any solution.
Hence, the general solution of given equation: $\cos 2\theta+7\cos \theta=8$ is $$\bbox[5px, border:2px solid #C0A000]{\color{red}{\theta=2k\pi}}$$
Where, $\color{red}{k}$ is any integer.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Distance formula for points in the Poincare half plane model on a "vertical geodesic". In comment at https://math.stackexchange.com/a/1381829/88985 at
Distance of two hyperbolic lines
is says (as i interpreted it) that the distance between two points $(a,r)$ and $(a, R)$ in the Poincaré half-plane model ( https://en.wikipedia.org/wiki/Poincar%C3%A9_half-plane_model ) is $ \ln(R) - \ln(r) $.
I tried to deduct this formula from the formulas:
$$ \operatorname{dist} (\langle a, r \rangle, \langle a, R \rangle) = \operatorname{arcosh} \left( 1 + \frac{ {(R - r)}^2 }{ 2 r R } \right) $$
and
$$\operatorname{arcosh} {x} =\ln \left(x + \sqrt{x^2 - 1} \right) $$
but failed miserably, (especially the bit under the square root didn't want to simpify)
can somebody show me the deduction?
ADDED LATER:
following https://math.stackexchange.com/users/208255/user24142
's suggestion (below) and other comments I come to
$$ \operatorname{dist} (\langle a, 1 \rangle, \langle a, R \rangle) = \operatorname{arcosh} \left(\frac{1}{2} \left( R + \frac{1}{R} \right) \right)$$
$$ = \ln \left(\frac{1}{2} \left( R + \frac{1}{R} \right) +\sqrt{\frac{1}{4}\left( R + \frac{1}{R} \right)^2 - 1 } \right) = $$
$$ = \ln \left(\frac{1}{2} \left( R + \frac{1}{R} \right) +\frac{1}{2}\sqrt{ R^2 - 2 +\frac{1}{R^2} \ } \right) = $$
$$ = \ln \left(\frac{1}{2} \left( R + \frac{1}{R} \right) +\frac{1}{2}\sqrt{ (R - \frac{1}{R} )^2 } \right) = $$
$$ = \ln \left(\frac{1}{2} \left( R + \frac{1}{R} \right) +\frac{1}{2} \left( R - \frac{1}{R} \right) \right) = \ln (R)$$
THANKS
| Compute first the inverse of
$$y=\operatorname {arcosh}(x)=\frac{e^x+e^{-x}}2$$
for positive $x$'s.
Whit the substitution $u=e^x$ we have $2y=u+\frac1u$. Multiplying both sides by $u$ we get the following quadratic equation:
$$u^2-2yu+1=0,$$
the solutions of which are
$$u^+=y+\sqrt{y^2-1}\ \text{ and } \ u^-=y-\sqrt{y^2-1}.$$
We need only the positive part
$$e^x=y+\sqrt{y^2-1}$$
and if $y\ge 1$ then we'll have real results.
Taking the natural logarithm of both sides we get
$$x=\ln\left(y+\sqrt{y^2-1}\right), \ y\ge 1.$$
In our case $y= 1 + \frac{ {(R - 1)}^2 }{ 2R }
\ge 1$, so we may substitute it in $y+\sqrt{y^2-1}.$ The result will be $R$.
This is why we may say that for $R\ge0$
$$\operatorname{arcosh}\left(1 + \frac{ {(R - 1)}^2 }{ 2R }\right)=\ln(R).$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given functional equation $f(x,y)=f(x,y+1)+f(x+1,y)+f(x-1,y)+f(x,y-1)$, show that $f(0,0)=0$ Let $f:\mathbb{Z}\times \mathbb{Z}\to \mathbb{R}$, such that
for all $x,y$
$$f(x,y)=f(x,y+1)+f(x+1,y)+f(x-1,y)+f(x,y-1),$$
and if $m,n\in \mathbb{Z},mn\neq 0$,we have $f(2m,2n)=0$.
Show that
$$f(0,0)=0.$$
| Sketching a solution since there is a lot of tedium required. This method finds all solutions $f$; perhaps there is a shortcut to show only $f(0, 0) = 0$?
*
*Show that if $f$, $g$ are solutions, then $f + g$ is a solution as well. Easy from the condition.
*Show that if$$f(3, 4) = f(4, 3) = f(3, 6) = f(6, 3) = 0,$$then $f(x, y) = 0$ for all $x$, $y$. To show this, notice that$$f(3, 3) + f(5, 3) = 0$$from $(x, y) = (4, 3)$ in the condition, and$$f(3, 3) + f(3, 5) = 0$$from $(x, y) = (3, 4)$ in the condition. So $f(3, 5) = f(5, 3)$. Then since$$f(5, 3) + f(5, 5) = f(5, 4)$$and $$f(3, 5) + f(5, 5) = f(4, 5),$$we have $f(5, 4) = f(4, 5)$. Finally,$$f(5, 4) + f(4, 5) = 0$$from $(x, y) = (4, 4)$ in the condition, so$$f(5, 4) = f(4, 5)=0.$$This logic is the hardest step needed; showing all the other values have to be $0$ is long, but each step is either repeating what we just did or applying the original condition where four of the terms are $0$.
*Show that if we treat $f$ as a function on the plane, then the following $4 \times 6$ array is a solution when tessellated infinitely, aligned so the top left corner is $(0, 0)$. Here, $a$ is an arbitrary real.$$\begin{matrix} 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & a & a & 0 & -a & -a \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & -a & -a & 0 & a & a\end{matrix}$$
*Show that if $f(3, 4)$, $f(4, 3)$, $f(3, 6)$, $f(6, 3)$ are arbitrarily set to real numbers, then there exists a solution. This is easy using step 1 and various transposes/translations of step 3.
*Using steps 1 and 2, we have that if $f(3, 4)$, $f(4, 3)$, $f(3, 6)$, $f(6, 3)$ are specified, then there is at most one solution $f$. This means that step 4 describes all possible solutions. These solutions all satisfy $f(2m, 2n) = 0$ for all $m$, $n$, so we are done.
| {
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"url": "https://math.stackexchange.com/questions/1386770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 1
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Solve this trigonometric equation $\cos3x\cos x=\sin 3x$ Solve this equation $$ \cos 3x\cos x=\sin 3x$$
I tried converting product into sum but with no results. I think they forgot to add $\sin x$ after $\sin 3x$.
| Hint:
Notice, $$\cos 3x\cos x=\sin 3x$$ $$(4\cos^3x-3\cos x)\cos x=3\sin x-4\sin^3x$$
$$4\cos^4x-3\cos^2 x=3\sin x-4\sin^3x$$ $$3\sin x-4\sin^3x=4(1-\sin^2 x)^2-3(1-\sin^2x)$$ $$3\sin x-4\sin^3x=4+4\sin^4 x-8\sin^2x-3+3\sin^2x$$
$$4\sin^4 x+4\sin^3 x-5\sin^2x-3\sin x+1=0$$ Let, $\sin x=t$, we get $$4t^4+4t^3-5t^2-3t+1=0$$
I hope you can take it from here.
| {
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"url": "https://math.stackexchange.com/questions/1390616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate the limit $\lim_{x \to 0} \left(\frac{1}{x^{2}}-\frac{1}{\tan^{2}x}\right)$
Evaluate the limit $$\lim_{x \to 0}\left( \frac{1}{x^{2}}-\frac{1}{\tan^{2}x}\right)$$
My attempt
So we have $$\frac{1}{x^{2}}-\frac{\cos^{2}x}{\sin^{2}x}$$
$$=\frac{\sin^2 x-x^2\cos^2 x}{x^2\sin^2 x}$$
$$=\frac{x^2}{\sin^2 x}\cdot\frac{\sin x+x\cos x}{x}\cdot\frac{\sin x-x\cos x}{x^3}$$
Then I have $3$ limits to evaluate
$$\lim_{x \to 0}\frac{x^2}{\sin^2 x}=\left(\lim_{x \to o}\frac{x}{\sin x}\right)^2=1^2=1$$
$$\lim_{x \to 0}\frac{\sin x+x\cos x}{x}=\lim_{x \to 0}\left(\frac{\sin x}{x} + \cos x\right)=1+1=2$$
Now I'm having trouble with the last one which is
$$\lim_{x \to 0}\frac{\sin x-x\cos x}{x^3}=?$$
Thanks for any help.
| I think this limit would be considerably easier using Taylor Series instead of LHR.
$ \frac{\sin^2(x)-x^2\cos^2(x)}{x^2\sin^2(x)} \approx \frac{(x-\frac{x^3}{6})^2-x^2(1-\frac{x^2}{2})^2}{x^2(x)^2}=\frac{\frac{2x^4}{3}+O(x^5)}{x^4} \rightarrow \frac{2}{3} $
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Finding the equations of the tangents where a quadratic equation cuts the $x$-axis and the angle between the tangents (differentiation involved) Calculate the equations of the tangent where $y=x^2-5x-24$ cuts the $x$-axis.
$(x-8)(x+3)$ factorising
$x=8, x=-3 $
$y'(x)=2x-5$
$y'(8)=11$
$y'(-3)=-11$
$y=11x+c$
$0=11(8)+c$
And then I find, $c$, and repeat for the other tangent equation which gives:
$y=11x-88$ and $y=-11x-33$
The second question is what is the angle between the tangents and I don't know how to find it. I know it has something to do with tan-theta.
Could someone also check if my arithmetic is correct for the first part.
| Notice, the slope of tangent at general point of the curve $y=x^2-5x-24$ $$\frac{dy}{dx}=\frac{d}{dx}(x^2-5x-24)=2x-5$$ Now, the point where curve $y=x^2-5x-24$ cuts the x-axis has $y=0$ thus we have $$0=x^2-5x-24$$ Solving the quadratic equation for the values of $x$ as follows $$x=\frac{-(-5)\pm\sqrt{(-5)^2-4(1)(-24)}}{2(1)}$$ $$=\frac{5\pm 11}{2} \iff x=8, -3$$ Hence, we get two points $(8, 0)$ & $(-3, 0)$ where curve intersects x-axis
Now, the slope of the tangent at $(8, 0)$ is $$=\frac{dy}{dx}|_{x=8}=2\times 8-5=11$$ Hence its equation $$y-0=11(x-8)$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{y=11x-88}}$$
Similarly, the slope of the tangent at $(-3, 0)$ is $$=\frac{dy}{dx}|_{x=-3}=2\times (-3)-5=-11$$ Hence its equation $$y-0=-11(x-(-3))$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{y=-11x-33}}$$
Hence, the angle between the tangents is given as $$\tan \theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right|$$ now, setting $m_1=11$ & $m_2=-11$, we get
$$\tan \theta=\left|\frac{11-(-11)}{1+11(-11)}\right|$$
$$=\left|\frac{11}{-60}\right|=\frac{11}{60}$$ Hence, $$\theta=\tan^{-1}\left(\frac{11}{60}\right)$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{acute angle:}\ \theta=\tan^{-1}\left(\frac{11}{60}\right)\approx10.39^\circ}}$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{obtuse angle:}\ \theta=\pi-\tan^{-1}\left(\frac{11}{60}\right)\approx 169.61^\circ}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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maximum value of $a+b+c$ given $a^2+b^2+c^2=48$? How can i get maximum value of this $a+b+c$ given $a^2+b^2+c^2=48$ by not using AM,GM and lagrange multipliers .
| We prove if $a+b+c=r$ then $a^2+b^2+c^2>\frac{r^2}{3}$ and equality is reached only when $a=b=c$.
Suppose the minimum is achieved with values $a,b,c$ and we do not have $a=b=c$. then without loss of generality we have $a\neq b$ let $a+b=m$ and write $a$ as $\frac{m+n}{2}$ and $b$ as $\frac{m-n}{2}$. Then the sum of the squares is $\frac{m^2+n^2}{2}+c^2$ which is greater than what we would get with $\frac{m}{2},\frac{m}{2},c$. Hence the minimum is reached when $a=b=c$.
When $r=12$ we get the minimum value of $a^2+b^2+c^2$ is $\frac{144}{3}=48$. When $r$ is larger $\frac{r^2}{3}$ is larger than $48$. Hence the maximum value of $a+b+c$ is $12$ and is reached only when $a=b=c=4$
| {
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"answer_id": 3
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$\lim _{ x\to 1 }{ \frac { \sqrt { x } +\sqrt [ 3 ]{ x } -2 }{ x-1 } } $ I have been trying to algebraically solve this limit problem without using L'Hospital's rule but was whatsoever unsuccessful:
$$\lim _{ x\to 1 }{ \frac { \sqrt { x } +\sqrt [ 3 ]{ x } -2 }{ x-1 } } $$
Thanks in advance!
| A scheme sometimes useful is to make the limit at zero...
Let $x=1+t$ so that $x \to 1$ means $t \to 0$. Then
$$
\frac{\sqrt{x}+\sqrt[3]{x}-2}{x-1} = \frac{\sqrt{1+t}+\sqrt[3]{1+t}-2}{t}
$$
so compute some asymptotics:
$$
\sqrt{1+t} = 1+\frac{1}{2}t+o(t)
\\
\sqrt[3]{1+t} = 1+\frac{1}{3}t+o(t)
\\
\sqrt{1+t}+\sqrt[3]{1+t}-2 = \frac{5}{6}t + o(t)
\\
\frac{\sqrt{1+t}+\sqrt[3]{1+t}-2}{t} = \frac{5}{6}+o(1)
$$
and your limit is $5/6$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Prove $\sum\limits^m_{k=0} \frac{2n-k\choose k}{2n-k\choose n}\frac{2n-4k+1}{2n-2k+1}2^{n-2k}=\frac{n\choose m}{2n-2m\choose n-m}2^{n-2m}$ for- Let $n$ be a positve integer. Prove that$$\sum\limits^m_{k=0} \frac{2n-k\choose k}{2n-k\choose n}\frac{2n-4k+1}{2n-2k+1}2^{n-2k}=\frac{n\choose m}{2n-2m\choose n-m}2^{n-2m}$$ for each non-negative integers $m\leq n$.
I know this is to be done by induction. Base case is easy. But I can't simplify expression for $(m+1)$. Please guide.
Edit: $\large P(m+1)=P(m)+\frac{2n-(m+1)\choose m+1}{2n-(m+1)\choose n}\cdot\frac{2n-4(m+1)+1}{2n-2(m+1)+1}\cdot2^{n-2(m+1)}$
$\Large=\frac{n\choose m}{2n-2m \choose n-m}.2^{n-2m}+\frac{2n-m-1\choose m+1}{2n-m-1\choose n}\cdot\frac{2n-4m-3}{2n-2m-1}.2^{n-2m-2}$
It must be equal to $\Large\frac{n\choose m+1}{2n-2m-2\choose n-m-1}.2^{n-2m-2}$
So lets try to simplify it-
$\Large=\frac{n!(n-m)!}{m!(2n-2m)!}.2^{n-2m}+\frac{n!(n-m-1)!}{(m+1)!(2n-2m-2)!}\cdot\frac{2n-4m-3}{2n-2m-1}\cdot2^{n-2m-2}$
$\Large=\frac{2^{n-2m-2}\cdot n!(n-m-1)!}{m!(2n-2m-2)!(2n-2m-1)}[\frac{2}{1}+\frac{1}{(m+1)}\cdot\frac{2n-4m-3}1]$
$\Large=\frac{2^{n-2m-2}\cdot n!(n-m-1)!}{m!(2n-2m-2)!(2n-2m-1)}[\frac{2n-2m-1}{m+1}]$
$\Large=\frac{2^{n-2m-2}\cdot n!(n-m-1)!}{(m+1)!(2n-2m-2)!}$
$\Large=\frac{\frac{n!}{(m+1)!(n-m-1)!}}{\frac{(2n-2m-2)!}{(n-m-1)!(n-m-1)!}}\cdot2^{n-2m-2}$
$\Large=\Large\frac{n\choose m+1}{2n-2m-2\choose n-m-1}.2^{n-2m-2}$
I think it is right. But is there any shorter way also?
| This reduces to a telescoping sum
$$
\begin{align}
&\sum_{k=0}^m\frac{\binom{2n-k}{k}}{\binom{2n-k}{n}}\frac{2n-4k+1}{2n-2k+1}2^{n-2k}\\
&=\sum_{k=0}^m\frac{(2n-k)!}{k!\,(2n-2k)!}\frac{n!\,(n-k)!}{(2n-k)!}\left(1-\frac{2k}{2n-2k+1}\right)2^{n-2k}\\
&=\sum_{k=0}^m\left[\frac{n!\,(n-k)!}{k!\,(2n-2k)!}2^{n-2k}
-\frac{n!\,(n-k)!}{(k-1)!\,(2n-2k+1)!}2^{n-2k+1}\right]\\
&=\sum_{k=0}^m\left[\frac{n!\,(n-k)!}{k!\,(2n-2k)!}2^{n-2k}
-\frac{n!\,(n-k+1)!}{(k-1)!\,(2n-2k+2)!}2^{n-2k+2}\right]\\
&=\frac{n!\,(n-m)!}{m!\,(2n-2m)!}2^{n-2m}\\
&=\frac{\binom{n}{m}}{\binom{2n-2m}{n-m}}2^{n-2m}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
} |
Convergence of improper integral $\int_{0}^{\frac{\pi}{6}}\dfrac{x}{\sqrt{1-2\sin x}}dx$ I'm trying to determine whether the following improper integral is convergent or divergent.
$$
\int_{0}^{\pi/6}\frac{x}{\sqrt{1-2\sin x}}dx
$$
At first, I substituted $t=\dfrac{\pi}{2} - x $
and then I used $1-\dfrac{1}{2}x^2 \le \cos x$.
But I couldn't determine. :-(
$$$$
Second attempt, I used $\sin x\le x $ on $[ 0, \frac{\pi}{6} ]$.
But I couldn't determine. :-[
Could you give me some advice?
Thanks in advance.
| Using the right hand side of the well known Jordan inequality that states $\sin x < x$ for all $x \in \left[{0,\frac{\pi}{2}}\right]$ we have $1-2\sin x > 1-2x$ so that $\frac{1}{1-2\sin x} < \frac{1}{1-2x}$ so that for all $0\le x\le \frac{\pi}{6}$ we have $\frac{x}{\sqrt{1-2\sin x}} < \frac{x}{\sqrt{1-2x}}$.
Thus,
\begin{align}
\int_{0}^{\frac{\pi}{6}}{\frac{x}{\sqrt{1-2\sin x}}dx} < \int_{0}^{\frac{\pi}{6}}{\frac{x}{\sqrt{1-2x}}dx}
\end{align}
since $0\le x\le \frac{\pi}{6}$ we may consider the one-to-one substitution $u^2=1-2x$, so that $x= \frac{1-u^2}{2}$ and $dx=-u du$. Also, when $x=0$ say $u=u_0$ and $x=\frac{\pi}{6}$ say $u=u_1$, therefore,
\begin{align}
\int_{0}^{\frac{\pi}{6}}{\frac{x}{\sqrt{1-2\sin x}}dx} < \int_{0}^{\frac{\pi}{6}}{\frac{x}{\sqrt{1-2x}}dx}&= \int_{u_0}^{u_1}{\frac{\frac{1-u^2}{2}}{u}(-u)du}
\\
&= \int_{u_0}^{u_1}{\frac{u^2-1}{2}du} = \frac{u^3_1-u^3_0}{6}-u_1 +u_0<\infty
\end{align}
which means that $\int_{0}^{\frac{\pi}{6}}{\frac{x}{\sqrt{1-2\sin x}}dx}$ converges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1397735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
} |
Evaluation of Gaussian integrals. In Bender and Orszag's section on applying Laplace's method they state the equality
$$\int_{0}^{\infty}e^{-xt^2}\frac{1}{3}xt^4 dt = \frac{1}{2}\sqrt{\frac{\pi}{x}}\left(\frac{1}{4x}\right)$$
How does one calculate this integral ?
| Notice, the Laplace transform $$L[t^n]=\int_{0}^{\infty}e^{-st}t^ndt=\frac{\Gamma(n+1)}{s^{n+1}}$$ & $\Gamma\left(\frac{1}{2}\right)=\sqrt {\pi}$
Now, we have $$\int_{0}^{\infty}e^{-xt^2}\frac{1}{3}xt^4dt$$
$$=\frac{1}{3}x\int_{0}^{\infty}e^{-xt^2}t^4dt$$
Let, $t^2=u\implies 2tdt=du\iff dt=\frac{du}{2\sqrt u}$, we get
$$\frac{1}{3}x\int_{0}^{\infty}e^{-xu}u^2\frac{du}{2\sqrt u}$$
$$=\frac{1}{6}x\int_{0}^{\infty}e^{-xu}u^{3/2}du$$
$$=\frac{1}{6}xL[u^{3/2}]_{s=x}$$
$$=\frac{1}{6}x\left[\frac{\Gamma\left(\frac{3}{2}+1\right)}{s^{\frac{3}{2}+1}}\right]_{s=x}$$
$$=\frac{1}{6}x\left[\frac{\frac{3}{2}\frac{1}{2}\Gamma\left(\frac{1}{2}\right)}{s^{5/2}}\right]_{s=x}$$
$$=\frac{x}{8}\left[\frac{\sqrt{\pi}}{x^{5/2}}\right]$$
$$=\frac{1}{2}\sqrt{\frac{\pi}{x}}\frac{1}{(4x)}$$ $$=\frac{1}{8x}\sqrt{\frac{\pi}{x}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1399088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Coeff. of $x^{97}$ in $f(x) = (x-1)\cdot (x-2)\cdot (x-3)\cdot (x-4)\cdot ........(x-100)$
If $f(x) = (x-1)\cdot (x-2)\cdot (x-3)\cdot (x-4)\cdot ........(x-100)\;,$ Then Coefficient of $x^{99}$ and
Coefficient of $x^{98}$ and Coefficient of $x^{97}$ in $f(x).$
$\bf{My\; try::}$ We can write $f(x)$ as
$$\displaystyle f(x) = x^{100}-\left(\sum_{i=1}^{100}i\right)x^{99}+\left(\mathop{\sum^{100}\sum^{100}}_{i=1\ \ j=1\ i<j}i\cdot j\right)x^{98}-\left(\mathop{\sum^{100}\sum^{100}\sum^{100}}_{i=1\ j=1\ k=1\ i<j<k}i\cdot j\cdot k\right)x^{97}+..$$
So $\bf{Coefficients}$ of $$\displaystyle x^{99} = -\sum_{i=1}^{100}\left(1+2+3+......+100\right) = -\frac{100\cdot 101}{2} = -5050$$
Similarly Coeff. of $$\displaystyle x^{98} = \mathop{\sum^{100}\sum^{100}}_{i=1\ \ j=1\ i<j}i\cdot j = \frac{1}{2}\left[\left(\sum_{i=1}^{100}i\right)^2-\sum_{i=1}^{100}i^2\right] = \frac{1}{2}\left[\left(\frac{100\cdot 101}{2}\right)^2-\frac{100\cdot 101\cdot 201}{6}\right]$$
But I did not understand How can i calculate Coeff. of $\displaystyle x^{97} =-\mathop{\sum^{100}\sum^{100}\sum^{100}}_{i=1\ j=1\ k=1\ i<j<k}i\cdot j\cdot k$
Help me, Thanks
| For the coefficient of $x^{97}$ you can apply a method similar to the one you used to find the coefficient of $x^{98}$. You already noted that the coefficient $C$ of $x^{97}$ is
$$C:=-\mathop{\sum^{100}\sum^{100}\sum^{100}}_{i=1\ j=1\ k=1\ i<j<k}i\cdot j\cdot k=-\sum_{i=1}^{98}\sum_{j=i+1}^{99}\sum_{k=j+1}^{100}i\cdot j\cdot k.$$
We can compute $C$ from the identity
$$\left(\sum_{i=1}^{100}i\right)^3=-2\sum_{i=1}^{100}i^3+3\left(\sum_{i=1}^{100}i^2\right)\left(\sum_{j=1}^{100}j\right)-6C.$$
Rewriting the above yields
\begin{eqnarray*}
C&=&-\frac{1}{6}\left[\left(\sum_{i=1}^{100}i\right)^3+2\sum_{i=1}^{100}i^3-3\left(\sum_{i=1}^{100}i^2\right)\left(\sum_{j=1}^{100}j\right)\right],\\
&=&-\frac{1}{6}\left[\left(\tfrac{1}{2}\cdot100\cdot101\right)^3+2\cdot\left(\tfrac{1}{4}\cdot100^2\cdot101^2\right)-3\cdot\left(\tfrac{1}{6}\cdot100\cdot101\cdot201\right)\cdot\left(\tfrac{1}{2}\cdot100\cdot101\right)\right]\\
&=&-20618771250,
\end{eqnarray*}
modulo miscalculations.
| {
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"url": "https://math.stackexchange.com/questions/1399182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx$ $$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx$$
I tried to solve it.
$$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx=\int\frac{1+x^2}{(x^2+1)^2+3x^3+x^2-3x+1}dx=\int\frac{1+x^2}{(x^2+1)^2+3x(x^2-1)+x^2+1}dx$$
But this does not seem to be solving.Please help.
| Writing the polynomial $x^4+3x^3+3x^2-3x+1$ as a product of two polynomials of degree two, i.e,. $$x^4+3x^3+3x^2-3x+1= (x^2+Ax+B)(x^2+Cx+D)$$, then we will get 4 equations $A+C=3,B+D+AC=3,AD+BC=-3,BD=1$, using wolframalpha.com to solve this eqs. we get
Thus, $$\int{\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx}=\int{\frac{1+x^2}{(x^2+Ax+B)(x^2+Cx+D)}dx}$$
Using partial fraction, we get $$\frac{1+x^2}{(x^2+Ax+B)(x^2+Cx+D)}=\frac{Px+Q}{x^2+Ax+B}+\frac{Tx+S}{x^2+Cx+D}$$
Solving for $P,Q,T$ and $S$ ...etc
Therefore,
$$\int{\frac{1+x^2}{(x^2+Ax+B)(x^2+Cx+D)}dx}=\int{\frac{Px+Q}{x^2+Ax+B}dx}+\int{\frac{Tx+S}{x^2+Cx+D}dx}=\text{constant}_1 \ln|x^2+Ax+B|+\text{constant}_2 \ln|x^2+Cx+D|+ CONSTANT$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403985",
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"source": "stackexchange",
"question_score": "6",
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"answer_id": 5
} |
$\int_{0}^{\infty}f\left(\frac{a}{x}+\frac{x}{a}\right)\frac{\ln x}{x}dx=\ln a\int_{0}^{\infty}f\left(\frac{a}{x}+\frac{x}{a}\right)\frac{dx}{x}$ Show that $\displaystyle \int_{0}^{\infty}f\left(\frac{a}{x}+\frac{x}{a}\right)\frac{\ln x}{x}dx=\ln a\int_{0}^{\infty}f\left(\frac{a}{x}+\frac{x}{a}\right)\frac{dx}{x}$
What substitution should i make for this.Both sides are looking alike,how to transform one
into another.Putting $\displaystyle \frac{a}{x}+\frac{x}{a}=t$ will not work,i think.
| Let $$\displaystyle I = \int_{0}^{\infty} f\left(\frac{a}{x}+\frac{x}{a}\right)\cdot \frac{\ln x}{x}dx\;,$$ Now Put $\displaystyle \frac{a^2}{x} = t\;,$ Then $\displaystyle x= \frac{a^2}{t}$ and $\displaystyle dx = -\frac{a^2}{t^2}dt$
and Changing Limit, We get
$$\displaystyle I = \int_{\infty}^{0}f\left(\frac{t}{a}+\frac{a}{t}\right)\cdot \frac{t}{a^2}\cdot \ln\left(\frac{a^2}{t}\right)\cdot -\frac{a^2}{t^2}dt = \int_{0}^{\infty}f\left(\frac{t}{a}+\frac{a}{t}\right)\cdot \frac{1}{t}[2\ln a-\ln t]dt$$
So $$\displaystyle I = 2\ln a\int_{0}^{\infty}f\left(\frac{a}{t}+\frac{t}{a}\right)\cdot\frac{1}{t}-I\Rightarrow I = \ln a\int_{0}^{\infty}f\left(\frac{a}{t}+\frac{t}{a}\right)\cdot\frac{1}{t}dt$$
So $$\displaystyle I = \ln a\int_{0}^{\infty}f\left(\frac{a}{x}+\frac{x}{a}\right)\cdot\frac{1}{x}dx$$
Using the formula $\displaystyle \int_{a}^{b}f(t)dt = \int_{a}^{b}f(x)dx$
| {
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"source": "stackexchange",
"question_score": "4",
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Upper and lower bounds for $S(n) = \sum_{i=1}^{2^{n}-1} \frac{1}{i} = 1+\frac{1}{2}+ \cdots +\frac{1}{2^n-1}.$
For a positive integer $n$ let $S(n) = \sum_{i=1}^{2^{n}-1} \frac 1i = 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+ \cdots +\frac{1}{2^n-1}.$
Then which of the following are true.
*
*(a) $S(100)\leq 100$.
*(b) $S(100)>100$.
*(c) $S(200)\leq 100$.
*(d) $S(200)>100$.
My attempt
*
*For the upper bound
$$\begin{align}
S(n) &= 1 + \left( \frac 12 + \frac 13 \right) + \left( \frac 14 + \frac 15 + \frac 16 + \frac 17 \right) + \cdots + \left( \frac 1{2^{n-1}} + \frac 1{2^{n-1}+1} + \cdots + \frac 1{2^n-1} \right)
\\ &< 1 + \left( \frac 12 + \frac 12 \right) + \left( \frac 14 + \frac 14 + \frac 14 + \frac 14 \right) + \cdots + \left( \frac 1{2^{n-1}} + \frac 1{2^{n-1}} + \cdots + \frac 1{2^{n-1}} \right) \\
&= \underbrace{1 + 1 + \cdots + 1}_{n\text{ times}} \\
&= n.
\end{align}$$
So we get $S(n) < n$ (for $n > 1$), and in particular $S(100) < 100$.
Now I did not understand how to calculate a lower bound, or if there is any other method by which we can solve this.
| This is not an answer but it is too long for a comment.
Since $$S(n)=\sum_{i=1}^{2^n-1}\frac 1i=H_{2^n-1}$$ you could be interested by this paper in which the author, Mark B. Villarino, developed extremely sharp bounds for the harmonic numbers.
The first and simplest set of bounds (Theorem $1$) is given by $$\frac{1}{2 m+\frac{1}{1-\gamma }-2}+\log (m)+\gamma\leq H_m \lt \frac{1}{2 m+\frac{1}{3}}+\log (m)+\gamma$$
Applied to the cases given by Dr. MV, the bounds coincide to more than $50$ significant figures.
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum of powers of a matrix with primitive polynomial modulo $2^{r}$ I need to prove an statement in the matrix form, which leads to the following equality modulo $2^{r}$. Which I couldn't prove but with computer simulation for lots of primitive polynomial, it seems to be true.
Question.Prove
$$ (I+A+A^2+\ldots+A^{2^r(2^m-1)-1}) \equiv 0 \pmod{2^{r}} $$
Where $A^{m\times m}$ is a matrix with elements in $\mathbb{N}$ defined as:
$$
A=\left(\begin{array}{cccc}
a_1 & \ldots & a_{m-1}& a_m\\
1 & \ldots & 0 & 0\\
\vdots & \ddots & \vdots &\vdots\\
0 & \ldots &1 & 0
\end{array}\right)
$$
$a_i's$ are the coefficients of a primitive polynomial mod 2,
$$P(x)=x^{m}-a_1x^m-\ldots-a_{m-1}x-a_m$$
Examples:
$$m=1\,,r\rightarrow A=[1]\rightarrow 1+A+\ldots+A^{2^r-1}\equiv 2^r\equiv 0 \pmod{2^r}$$
$$m=2\,,r=1\rightarrow A=\left(\begin{array}{cc}
1 & 1\\
1 & 0
\end{array}\right)\rightarrow I+A+A^2+A^3+A^4+A^5\equiv\\
\left(\begin{array}{cc}
1 & 0\\
0 & 1
\end{array}\right)+\left(\begin{array}{cc}
1 & 1\\
1 & 0
\end{array}\right)+\left(\begin{array}{cc}
2 & 1\\
1 & 1
\end{array}\right)+\left(\begin{array}{cc}
3 & 2\\
2 & 1
\end{array}\right)+\left(\begin{array}{cc}
5 & 3\\
3 & 2
\end{array}\right)+\left(\begin{array}{cc}
8 & 5\\
5 & 3
\end{array}\right)\equiv \left(\begin{array}{cc}
20 & 12\\
12 & 8
\end{array}\right)\equiv \left(\begin{array}{cc}
0 & 0\\
0 & 0
\end{array}\right) \pmod{2}$$
$$m=3\,,r=1\rightarrow A=\left(\begin{array}{cc}
1 & 0 & 1\\
1 & 0 & 0\\
0 & 1 & 0
\end{array}\right)\rightarrow I+A+A^2+\ldots+A^{12}+A^{13}\equiv \left(\begin{array}{cc}
276 & 128 & 188\\
188 & 88 & 128\\
128 & 60 & 88
\end{array}\right)\equiv \left(\begin{array}{cc}
0& 0& 0\\
0& 0& 0\\
0& 0& 0
\end{array}\right)\pmod{2}$$
| $A$ is the companion matrix of $P(x)$. As $P(x)$ was assumed to be primitive
$$
B:=A^{2^m-1}\equiv I\pmod 2.
$$
Furthermore, $2^m-1$ is the lowest positive exponent for which this congruence holds. So $B=I+2C$ for some integer matrix $C$. We prove by induction that
$$
B^{2^\ell}\equiv I\pmod{2^{\ell+1}}.
$$
The base case $\ell=0$ was already settled. If we already know that for some natural number $k$ we have
$$
B^{2^k}=I+2^{k+1}D
$$
for some integer matrix $D$, then squaring this equation gives
$$
B^{2^{k+1}}=I+2^{k+2}D+2^{2(k+1)}D^2\equiv I\pmod{2^{k+2}}.
$$
So we know that $B^{2^{r}}=A^{2^r(2^m-1)}\equiv I\pmod{2^{r+1}}$.
Again by primitivity $I-A$ is invertible modulo two. By Hensel lifting it is also invertible modulo $2^{r+1}$. Your identity can thus be multiplied by $I-A$. By geometric sum formula it is thus equivalent to
$$I-A^{2^r(2^m-1)}\equiv0\pmod{2^r}.$$
We have just shown that this congruence actually holds modulo $2^{r+1}$.
| {
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Solve the following If tangent at $(a,b)$ passes through $(a_1,b_1)$ for the curve $x^3+y^3=c^3$ and $\frac{a_1}{a} + \frac{b_1}{b} = -k$ then find the value of k.(a_1,b_1) lies on curve as well?
solution :$ \frac{b_1-b}{a_1-a} = -\frac{a^2}{b^2} , a^3+b^3=c^3$ and $a_1^3+b_1^3 = c^3$
I got stuck here how to solve it?
| We have $$ 0= (a^3+b^3)-(a_1^3+b_1^3)$$ $$=(a^3-a_1^3)+(b^3-b_1^3)$$ $$=(a-a_1)(a^2+aa_1+a_1^2)+(b-b_1)(b^2 +bb_1+b_1^2).$$Now ,as $ (a-a_1)/(b-b_1)=b^2/(-a^2)$, this gives$$ 0=b^2(a^2+aa_1+a_1^2)+(-a^2)(b^2+bb_1-b_1^2)$$ $$=(ba_1-ab_1)(ba+ba_1+ab_1)$$ $$=(ba_1-ab_1)(ba)(1-k).$$ I believe you can do the rest.
| {
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Maximize $xy^2$ on the ellipse $x^2+4y^2=4$ I was using Lagrange multiplier, any steps gone wrong?
$$f(x,y)=xy^2$$
$$c(x,y)=x^2+4y^2$$
Partial Derivatives
$$\frac {\partial f}{\partial x} = y^2 $$
$$\frac {\partial f}{\partial y} = 2xy $$
$$\frac {\partial c}{\partial x} = 2x $$
$$\frac {\partial c}{\partial y} = 8y $$
Find Lambda
$$y^2=2x\lambda$$
$$\lambda=\frac {y^2}{2x} $$
$$ 2xy=8y\lambda $$
$$\lambda= \frac{x}{4} $$
Let
$$ \frac {y^2}{2x}=\frac {x}{4} $$ yields $$y=\frac {x}{\sqrt 2}$$ and $$y=-\frac {x}{\sqrt 2}$$
Bringing $$y=\frac {x}{\sqrt 2}$$ into C
$$ x^2 + 2x^2 = 4 $$
$$ 3x^2=4 $$ yields $$ x=\frac {2}{\sqrt 3} and -\frac {2}{\sqrt 3} $$
Since $$ y=\frac {x}{\pm\sqrt 2} $$
plugging $x=\frac {2}{\pm\sqrt 3}$ into y I got 4 points:
$(\frac {2}{\sqrt 3},\frac {2}{\sqrt 6}),(-\frac {2}{\sqrt 3},\frac {2}{\sqrt 6}),(-\frac {2}{\sqrt 3},-\frac {2}{\sqrt 6}),(\frac {2}{\sqrt 3},-\frac {2}{\sqrt 6})$
is every step ok up to this point?
| $$f = xy^2$$
with
$$4=x^2+4y^2$$
becomes
$$f = xy^2 = x\left(1-\frac{1}{4}x^2\right) = x - 0.25x^3$$
maximum via derivative
$$\frac{d}{dx}f = 0 = 1 - 0.75x^2$$
$$x_{1,2}=\pm\sqrt{\frac{4}{3}}$$
$$\frac{d^2}{dx^2}f(x_1) < 0 $$
$$\frac{d^2}{dx^2}f(x_2) > 0 $$
and with $4=x^2+4y^2$ again:
$$y_{1,2} = \pm\sqrt{\frac{2}{3}}$$
I think that's somewhat plausible, as the $y^2$ creates a symmetry around the x axis. The two points are $(x_1,y_1)$ and $(x_1,y_2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1406260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove $\lim \frac{3^n + 2^n}{5\cdot3^n + 7\cdot2^n} = \frac{1}{5}$ Prove using limit definition.
$$\lim \frac{3^n + 2^n}{5\cdot 3^n + 7\cdot 2^n} = \frac{1}{5}
$$
My try:
$$\left| {\frac{3^n + 2^n}{5\cdot 3^n + 7\cdot 2^n} - \frac{1}{5}} \right| < \varepsilon
$$
$$ \Leftrightarrow \left| \frac{3^n + 2^n - 5\cdot 3^n - 7\cdot 2^n}{5(5\cdot3^n + 7\cdot2^n)} \right| < \varepsilon \
$$
$$ \Leftrightarrow \frac{4\cdot 3^n + 5\cdot 2^n}{5(5\cdot 3^n + 7\cdot 2^n)} < \varepsilon
$$
I can increase numerator next and simplify like this:
$$\frac{4\cdot 3^n + 5\cdot 2^n}{5(5\cdot 3^n + 7\cdot 2^n)} < \,\frac{5\cdot 3^n + 7\cdot 2^n}{5\cdot (5\cdot 3^n + 7\cdot 2^n)} = \frac{1}{5}
$$
But in this case I'm loosing the $n$ in the dominator.
| If you know equivalents in Asymptotic analysis, that is easy:
$\;3^n+2^n\sim3^n$, $\sim5\times3^n$, hence
$$\frac{3^n+2^n}{5\times3^n+7\times2^n}\sim\frac{3^n}{5\times3^n}=\frac15.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1408198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Thue equation $ x^4 - 6 x^3 y - x^2 y^2 + 6 x y^3 - y^4=-1 $ I need help solving the Thue equation
$ x^4 - 6 x^3 y - x^2 y^2 + 6 x y^3 - y^4=-1 $.
It can be written as
$ x(x-y)(x+y)(x-6y) = (y-1)(y+1)( y^2 +1) $.
From this I found 8 solutions (0,1),(0,-1),(1,1),(-1,-1),(-1,1),(1,-1),(6,1) and (-6,-1). But there are two more solutions (15,17) and (-15,-17), and I don't know how to get them.
I would be grateful for any kind of suggestions.
| OP seems to be looking for a simple method that would ensure an easy discovery of all solutions listed. One such simple method is to let $y$ run through $\pm 1, \pm 2, \pm 3, ...$ and calculate $(y−1)(y+1)(y^2+1) = y^4-1$. From the left hand side, we can see that $x$ must be a divisor of $y^4-1$ and we simply check all divisors of $y^4-1$. For $ y = \pm 17$ we get that $y^4-1 = 83520$ which is divisible by $15$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Prove that $1^2 + 2^2 + ..... + (n-1)^2 < \frac {n^3} { 3} < 1^2 + 2^2 + ...... + n^2$ I'm having trouble on starting this induction problem.
The question simply reads : prove the following using induction:
$$1^{2} + 2^{2} + ...... + (n-1)^{2} < \frac{n^3}{3} < 1^{2} + 2^{2} + ...... + n^{2}$$
| Perhaps simpler, the induction step is done if you show
$$3n^2<(n+1)^3-n^3<3(n+1)^2$$
But $(n+1)^3-n^3=3n^2+3n+1$ which is clearly in between $3n^2$ and $3(n+1)^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1416472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Why doesn't quadratic formula lead to a the correct factored form of the original equation? Applying the quadratic formula to $2x^2-3x+1$ we have
\begin{eqnarray*}
a&=&2 \\
b&=&-3 \\
c&=&1
\end{eqnarray*}
which gives me two roots:
\begin{eqnarray*}
x_1&=&1 \\
x_2&=&\tfrac{1}{2}
\end{eqnarray*}
Therefore you can re-write the original quadratic as:
$$(x-1)\left(x-\tfrac{1}{2}\right)$$
However, if you actually multiply this out then you get:
$x^2-\frac{3}{2}x + \frac{1}{2}$ which is not the same as $2x^2-3x+1$.
So why the discrepancy? Aren't you supposed to get the original equation?
| The quadratic formula gives the solution to the equation
$$ax^2+bx+c=0.$$
But the formula is derived by completing the square, and to do so the coefficient of $x^2$ must be 1. Therefore, the first step is to divide by $a$ (it is assumed without loss of generality that $a>0$). This gives
$$x^2+\tfrac{b}{a}x+\tfrac{c}{a}=0.$$
Now notice that
$$x^2+\tfrac{b}{a}x+\tfrac{c}{a}=\left(x-\left(\frac{-b+\sqrt{b^2-4ac}}{2a}\right)\right)\left(x-\left(\frac{-b-\sqrt{b^2-4ac}}{2a}\right)\right).$$
Long story short, the quadratic formula gives the solutions of $ax^2+bx+c=0$ by finding the solutions to the $equivalent$ equation $x^2+\tfrac{b}{a}x+\tfrac{c}{a}=0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1417467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
What is $\lim_{n\to\infty}2^n\sqrt{2-\sqrt{2+\sqrt{2+\dots+\sqrt{p}}}}$ for $negative$ and other $p$? This was inspired by similar posts like this one. Define the function,
$$F(p) = \lim_{n\to\infty}2^n\sqrt{2-\underbrace{\sqrt{2+\sqrt{2+\dots+\sqrt{p}}}}_{n \textrm{ square roots}}}$$
We know that,
$$F(2) = \frac{\pi}{2},\quad F(3) = \frac{\pi}{3}$$
I was wondering what it evaluates to if we use other integers. Some numerical computation and the Inverse Symbolic Calculator suggests that,
$$\begin{aligned}
F(5) &= 2\ln\big(\tfrac{1+\sqrt{5}}{2}\big)\,i\\
F(6) &= \ln\big(2+\sqrt{3}\big)\,i\\
F(7) &= \ln\big(\tfrac{5+\sqrt{3\times7}}{2}\big)\,i\\
\vdots\\
F(11) &= \ln\big(\tfrac{9+\sqrt{7\times11}}{2}\big)\,i\\
\vdots\\
F(17) &= \ln\big(\tfrac{15+\sqrt{13\times17}}{2}\big)\,i
\end{aligned}$$
Note that the radical arguments are fundamental units. If we use negative $p$,
$$\begin{aligned}
F(-1) &= \pi-2\ln\big(\tfrac{1+\sqrt{5}}{2}\big)\,i\\
F(-2) &= \pi-\ln\big(2+\sqrt{3}\big)\,i\\
F(-3) &= \pi-\ln\big(\tfrac{5+\sqrt{3\times7}}{2}\big)\,i\\
\vdots\\
F(-7) &= \pi-\ln\big(\tfrac{9+\sqrt{7\times11}}{2}\big)\,i\\
\vdots\\
F(-13) &= \pi-\ln\big(\tfrac{15+\sqrt{13\times17}}{2}\big)\,i
\end{aligned}$$
and so on. It seems $F(2+m)+F(2-m) = \pi$. I also observed that if $m\pm2$ are primes, then,
$$F(2+m) = \pi-F(2-m) = \ln\Big(\tfrac{m+\sqrt{(m-2)(m+2)}}{2}\Big)\,i\tag1$$
though the form of $(1)$ is only conjectural.
Question: What is then the formula for $F(p)$ using general $p$?
| The (hyperbolic) cosine bisection formula gives:
$$2\cos\frac{x}{2}=\sqrt{2+2\cos x},\qquad 2\cosh\frac{x}{2}=\sqrt{2+2\cos x}$$
hence assuming $a_0=2\cosh(u_0)=\sqrt{p}$ and $a_{n+1}=\sqrt{2+a_n}$ we have:
$$ a_n = 2 \cosh\left(\frac{u_0}{2^n}\right),\quad \sqrt{2-a_n}= 2\sinh\left(\frac{u_0}{2^{n+1}}\right)$$
so:
$$ \lim_{n\to +\infty} 2^n\sqrt{2-a_n} = u_0 = \text{arccosh}\left(\frac{\sqrt{p}}{2}\right). $$
If $p$ is a positive real number less than $4$ it is enough to replace $\cosh$ with $\cos$ and $\text{arccosh}$ with $\arccos$. If $p$ is negative, we have to be careful in defining what the square root of a complex number is, but the trick is just the same.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1418513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
$a,b,c\in \Bbb Z$ and $a\cdot b\cdot c$ is a root of $ax^2+bx+c$. I was curious if there are quadratic equations where $a,b,c\in \Bbb Z$ and $a\cdot b\cdot c$ is a root of $ax^2+bx+c$.
So trivially if $c=0$, $a$ and $b$ can be arbitrary, and if either $a$ or $b$ is zero, this implies that $c=0$, and the other arbitrary.
Is there a way to find other solutions, if any exist?
So the equation to solve in integers is $$a^3b^2c^2+ab^2c+c=0.$$
Dividing out by $c$ gives $$a^3b^2c+ab^2+1=0.$$
We could factor to give $$ab^2(a^2c+1)=-1$$
So either
*
*$ab^2=1$ and $a^2c+1=-1$ or
*$ab^2=-1$ and $a^2c+1=1$
For 1. $a=1$ and $b=\pm1$ and $c=-2$
For 2. $a=-1$ and $b=\pm1$ and $c=-2$ and $c=0$ is forced.
Is this analysis correct?
Are the non-trivial solutions:
$x^2+x-2$ and $x^2-x-2$, with roots $-2$ and $2$ respectively?
| The proof looks correct to me.
However, it might be good to say "However, we assumed $c \neq 0$" after point 2.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
} |
When is y increasing? Arctan is the inverse of tan.
I tried to normally differentiate and solve for y' but the LCM is complicated.
Is there any easy substitution simplifies the problem?
$$y=(\frac{1}{3})\textrm{log}(\frac{x+1}{\sqrt{x^2-x+1}}) + \frac{1}{\sqrt{3}}\textrm{arctan}(\frac{2x+1}{\sqrt3})$$
| $$y=\frac{1}{3}\ln{\frac{x+1}{\sqrt{x^2-x+1}}}+\frac{1}{\sqrt{3}}\tan^{-1}{\frac{2x+1}{\sqrt{3}}}$$
Before differentiating, we will manipulate the function to make it easier to differentiate.
$$y=\frac{\ln(x+1)}{3}-\frac{\ln(x^2-x+1)}{6}+\frac{1}{a}\tan^{-1}\frac{u}{a}$$where $a=\sqrt3$
$$\frac{\delta y}{\delta x}=\frac{1}{3(x+1)}-\frac{2x-1}{6(x^2-x+1)}+\frac{\delta u}{a^2+u^2}$$
The derivative of the inverse tangent part of the function follows by reversing the standard integration tables.
$$\frac{\delta y}{\delta x}=\frac{2x^2-2x+2-2x^2-2x+x+1}{6(1+x^3)}+\frac{2}{4x^2+4x+1+3}=\frac{-3x+3}{6(1+x^3)}+\frac{1}{2(x^2+x+1)}=\frac{1-x}{2(1+x^3)}+\frac{1}{2(x^2+x+1)}=\frac{1-x^3+1+x^3}{2(1+x^3)(1+x+x^2)}$$
$$\frac{\delta y}{\delta x}=\frac{1}{(1+x)(1-x+x^2)(1+x+x^2)}$$
The only term which has a change in sign is $(1+x)$, the other two quadratic factors are positive for all real $x$.
Hence, your function has a positive gradient for all $x>-1$ and therefore, it is strictly increasing in its domain for all $x>-1$. We do not take the equality, since that is a singularity of the function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1423819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to find a formula for the expression $2^0+2^1+\cdots+2^n$? Is there a way to simply this equation and express it in terms of $n$?
$$2^0+2^1+2^2+2^3+\cdots+2^n$$
| $$\sum\limits_{i=0}^{n}2^i = 1+2+2^2+2^3+\cdots + 2^n = 1 + 2\left(1+2+2^2+\cdots + 2^{n-1}\right)=1+2\sum\limits_{i=0}^{n-1}2^i\text{.}$$
Now subtract $\sum\limits_{i=0}^{n-1}2^i$ from both sides:
$$\sum\limits_{i=0}^{n}2^i-\sum\limits_{i=0}^{n-1}2^i=2^n$$
and
$$1+2\sum\limits_{i=0}^{n-1}2^i-\sum\limits_{i=0}^{n-1}2^i=1+\sum\limits_{i=0}^{n-1}2^i\text{.}$$
This gives $$2^n-1=\sum\limits_{i=0}^{n-1}2^i$$
or
$$2^{n+1}-1=\sum\limits_{i=0}^{n}2^i\text{.}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1426144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
A simple double inequality with roots Prove or disprove for integer $n$ greater than $1$:
$$\sqrt{1} + \sqrt{2} + ...+ \sqrt{n-1} < \frac{2n\sqrt{n}}{3} < \sqrt{1} + \sqrt{2} + ...+ \sqrt{n} $$
I think I know a solution, but I am looking for different approaches.
EDIT: Inspired by Prove that $1^2 + 2^2 + ..... + (n-1)^2 < \frac {n^3} { 3} < 1^2 + 2^2 + ...... + n^2$.
| by induction :
for LHS: if $n=k ,\sum _{i=1}^k \sqrt{i-1} \le \dfrac{2k\sqrt{k}}{3}$
when $n=k+1$, we have to prove $\dfrac{2k\sqrt{k}}{3}+\sqrt{k} \le \dfrac{2(k+1)\sqrt{k+1}}{3} \iff 3k\ge 4$ which is true when $k\ge2$
similar method to RHS.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1426367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Number Theory - Remainders A number is of the form $13k_1+12$ and of the form $11k_2+7$
That is $N = 13k_1 + 12 = 11k_2 + 7$
Now why must N also equal $(13 \times 11)k_3 + 51$ ?
Thanks
| Finding a Solution
To find an $N$ so that
$$
\begin{align}
N&\equiv12&\pmod{13}\\
N&\equiv7&\pmod{11}
\end{align}\tag{1}
$$
we can start by solving
$$
13x+11y=1\tag{2}
$$
using the Extended Euclidean Algorithm. The implementation described in this answer gives
$$
\begin{array}{r}
&&1&5&2\\\hline
1&0&1&-5&11\\
0&1&-1&6&-13\\
13&11&2&1&0\\
\end{array}\tag{3}
$$
which implies
$$
13(-5)+11(6)=1\tag{4}
$$
Using $(4)$, we can show
$$
\begin{align}
66&\equiv1&\pmod{13}\\
66&\equiv0&\pmod{11}
\end{align}\tag{5}
$$
and
$$
\begin{align}
-65&\equiv0&\pmod{13}\\
-65&\equiv1&\pmod{11}
\end{align}\tag{6}
$$
$12$ times $(5)$ plus $7$ times $(6)$ gives
$$
\begin{align}
337&\equiv12&\pmod{13}\\
337&\equiv7&\pmod{11}
\end{align}\tag{7}
$$
Subtracting $2\cdot11\cdot13=286$ from the left sides of $(7)$ gives
$$
\begin{align}
51&\equiv12&\pmod{13}\\
51&\equiv7&\pmod{11}
\end{align}\tag{8}
$$
Finding a General Solution
If $N_1$ and $N_2$ are any two solutions to $(1)$, then
$$
\begin{align}
N_1-N_2&\equiv0&\pmod{13}\\
N_1-N_2&\equiv0&\pmod{11}
\end{align}\tag{9}
$$
Therefore,
$$
N_1-N_2\equiv0\pmod{11\cdot13}\tag{10}
$$
Putting together $(8)$ and $(10)$, we get that all the solutions of $(1)$ are given by
$$
N\equiv51\pmod{143}\tag{11}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1427579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Finding a perpendicular plane Question
Plane $B$ contains the points $(4, 2, 1)$ and $(4, 1, -6)$, and it is perpendicular to the plane $7x+9y+4z=18$ . What is the equation of the plane $B$?
What I tried:
I know that to find the equation of plane $B$ I need a point on the plane and a perpendicular vector. The point part is already covered $(4, 1, -6)$, and the cross product of $\langle4-4, 2-1, 1-(-6)\rangle$ and $\langle7, 9, 4\rangle$ should give me the perpendicular vector; then it's just a matter of plugging the numbers in the formula. However, this approach did not work and I don't know what I am doing wrong.
| Your approach is both correct and simple.
Your plane has to be parallel to both the vectors
$$\begin{pmatrix}4\\2\\1\end{pmatrix}-\begin{pmatrix}4\\1\\-6\end{pmatrix}=\begin{pmatrix}0\\1\\7\end{pmatrix}
\qquad\text{and}\qquad
\begin{pmatrix}7\\9\\4\end{pmatrix}$$
so its normal vector can be computed as
$$\begin{pmatrix}0\\1\\7\end{pmatrix}\times
\begin{pmatrix}7\\9\\4\end{pmatrix}=
\begin{pmatrix}-59\\49\\-7\end{pmatrix}$$
Plug in one of your two points
$$-59\cdot4+49\cdot2-7\cdot1=-145$$
so if you also get rid of some minus signs, the equation of the plane is
$$59\,x-49\,y+7\,z=145$$
To verify the required properties, you can plug in both your points and see that they satisfy the equation, and you can also compute the dot product
$$\left<\begin{pmatrix}59\\-49\\7\end{pmatrix},
\begin{pmatrix}7\\9\\4\end{pmatrix}\right>=
59\cdot7-49\cdot9+7\cdot4=0$$
which proves that the normals are perpendicular, therefore the planes are perpendicular.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1427656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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A sequence related to squares of Fibonacci nubers Let $f(n)$ be defined by
$f(n)=f(n-1)+f(n-3)+f(n-4)$, for $n \ge 5$,
$f(1)=1, f(2)=1, f(3)=2, f(4)=4$.
First few terms of the sequence $(f(1), f(2), f(3), \ldots$) look like
$(1, 1, 2, 4, 6, 9, 15, 25, 40, 64, 104, 169, \ldots)$
Here we recognize that the subsequence consisting of even numbered terms looks like
$(f(2), f(4), f(6), f(8), f(10), f(12), \ldots)=(1^2, 2^2, 3^2, 5^2, 8^2, 13^2, \ldots)$
which tells that
$f(2k)=F_{k+1}^2$ for $k=1, 2, 3, 4, 5, 6$, where $F_k$ stands for the $n$th Fibonacci number.
Does this hold for all $k=1, 2, \ldots$?
| This is simple to solve, surprisingly. Define $g(z) = \sum_{n \ge 0} f(n) z^n$,
shift indices to get:
$\begin{align}
f(n + 4)
= f(n + 3) + f(n + 1) + f(n)
\end{align}$
Multiply by $z^n$, sum over $n \ge 0$ and recognize the resulting sums:
$\begin{align}
\frac{g(z) - f(0) - f(1) z - f(2) z^2 - f(3) z^3}{z^4}
= \frac{g(z) - f(0) - f(1) z - f(2) z^2}{z^3}
+ \frac{g(z) - f(0)}{z}
+ g(z)
\end{align}$
Running the recurrence backwards gives $f(0) = 1$. With the initial values we can solve for $g(z)$:
$\begin{align}
g(z)
&= \frac{1}{1 - z - z^3 - z^4} \\
&= \frac{3 + z}{5 (1 - z - z^2)}
- \frac{2 + z}{5 (1 + z^2)}
\end{align}$
The Fibonacci numbers are defined by:
$\begin{align}
F_{n + 2} = F_{n + 1} + F_n \qquad F_0 = 0, F_1 = 1
\end{align}$
Their generating function is:
$\begin{align}
F(z) = \frac{z}{1 - z - z^2}
\end{align}$
The generating function for the sequence $F_{n + 1}$ is just:
$\begin{align}
\frac{F(z) - F_0}{z}
= \frac{1}{1 - z - z^2}
\end{align}$
This gives the first term. For the second one:
$\begin{align}
\frac{2 + z}{1 + z^2}
&= \frac{2 - \mathrm{i}}{2} \cdot \frac{1}{1 - \mathrm{i} z}
+ \frac{2 + \mathrm{i}}{2} \cdot \frac{1}{1 + \mathrm{i} z}
\end{align}$
Note that the coefficients are complex conjugates, so that:
$\begin{align}
[z^n] \frac{2 + z}{5 (1 + z^2)}
&= 2 \Re\left( \frac{2 - \mathrm{i}}{10} \cdot \mathrm{i}^n \right) \\
&= \frac{\sqrt{5}}{5}
\Re\left(\exp\left(\arctan(1/2) \mathrm{i}\right)
\cdot \exp\left(\frac{\pi n \mathrm{i}}{2}\right)
\right) \\
&= \frac{\sqrt{5}}{5}
\cos\left(\frac{\pi n}{2} + \arctan\left( \frac{1}{2}\right)\right)
\end{align}$
Pulling all together:
$\begin{align}
f(n) = \frac{3}{5} F_{n + 1} + \frac{1}{5} F_n
+ \frac{\sqrt{5}}{5}
\cos\left(
\frac{\pi n}{2} + \arctan\left( \frac{1}{2}\right)
\right)
\end{align}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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In how many ways can a selection be done of $5$ letters?
In how many ways can a selection be done of $5$ letters out of $5 A's, 4B's, 3C's, 2D's $ and $1 E$.
$ a) 60 \\
b) 75 \\
\color{green}{c) 71} \\
d.) \text{none of these} $
Number of ways of selecting $5$ different letters = $\dbinom{5}{5} = 1$ way
Number of ways to select $2$ similar and $3$ different letter = $\dbinom{4}{1}\times \dbinom{4}{3}=16$.
Number of ways of selecting $2$ similar + $2$ more similar letter and $1$ different letter = $\dbinom{4}{2}\times \dbinom{3}{1}=18$.
Number of ways to select $3$ similar and $2$ different letter = $\dbinom{4}{2}\times \dbinom{3}{1}=18$.
Number of ways to select $3$ similar and another $2$ other similar = $\dbinom{3}{1}\times \dbinom{3}{1}=9$
Number of ways to select $4$ similar and $1$ different letter = $\dbinom{2}{1}\times \dbinom{4}{1}=8$
Ways of selecting
$5$ similar letters = $1$
Total ways = $1+16+18+18+9+8+1= 71$
Well I have the solution But I am not able to fully understand it.
Or if their could be an $\color{red}{\text{alternate way}}$ than it would be great.
I have studied maths up to $12$th grade.
| Let $x_i$ be the number of letter $i$ chosen for $1\le i\le 5$, where we number the letters in alphabetical order.
We want to find the number of solutions in nonnegative integers to the equation $x_1+\cdots+x_5=5$
$\hspace{.5 in}$with the restrictions $x_1\le5,\; x_2\le4, \;x_3\le3,\; x_4\le2, \;x_5\le1$.
Let $S$ be the set of all solutions, and let $A_i$ be the set of solutions with $x_i\ge7-i$ for $i=2,\cdots,5$.
Using Inclusion-Exclusion, we have that
$\hspace{.15 in}\displaystyle\big|\overline{A_2}\cap\overline{A_3}\cap\overline{A_4}\cap\overline{A_5}\big|=|S|-\sum_{i}|A_i|+\sum_{i<j}|A_i\cap A_j|-\sum_{i<j<k}|A_i\cap A_j\cap A_k|+\cdots$
$\hspace{1.53 in}=|S|-|A_2|-|A_3|-|A_4|-|A_5|+|A_4\cap A_5|$
$\displaystyle\hspace{1.53 in}=\binom{9}{4}-\binom{4}{4}-\binom{5}{4}-\binom{6}{4}-\binom{7}{4}+\binom{4}{4}=\color{red}{71}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1431903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 1
} |
Consider the following system and find the values of b for which the system has a solution So I have this system:
$$\left\{\begin{array}{c}
x_1 &− x_2 &+ 2x_3 &= 2 \\
x_1 &+ 2x_2 &− x_3 &= 2 \\
x_1 &+ x_2 & &= 2 \\
x_1 & & +x_3 &= α
\end{array}\right.$$
And we are asked to find the values of $\alpha$ for which the system has a solution. When I do the coefficient matrix and a Gauss-Jordan elimination i get two entirely zero rows with one of the augmented matrix solutions being $\alpha-2$ so am I correct in saying that the system will have an infinite number of solutions if $\alpha=2$?
The thing that concerns me is that we are then asked to find the value of $\alpha$ if the degree of freedom is $2$?
Any help would be greatly appreciated!
PS. Sorry for any formatting errors!
| Gaussian elimination.
Downwards:
$$\tag 1 \begin{array} {l}
x_1 & -x_2 & + 2 x_3 & 0 & = 2 \\
x_1 & +2x_2 & - x_3 & 0 & = 2 \\
x_1 & +1x_2 & 0 & 0 & = 2 \\
x_1 & 0 &+1 x_3 & -\alpha & = 0 \\
\end{array} $$
Subtract row 1 from rows 2 to 4:
$$\tag 2 \begin{array} {l}
x_1 & -x_2 & + 2 x_3 & 0 & = 2 \\ \hline
0 & 3x_2 & - 3 x_3 & 0 & = 0 \\
0 & 2x_2 & -2x_3 & 0 & = 0 \\
0 & 1x_2 & -1x_3 & -1\alpha & = -2 \\
\end{array} $$
Cancel the 3 and the 2 because on the rhs are zeros in row 2 and 3:
$$\tag 3 \begin{array} {l}
x_1 & -x_2 & + 2 x_3 & 0 & = 2 \\ \hline
0 & x_2 & - x_3 & 0 & = 0 \\
0 & x_2 & -x_3 & 0 & = 0 \\
0 & x_2 & -x_3 & -1\alpha & = -2 \\
\end{array} $$
Subtract row 2 from rows 3 and 4:
$$ \begin{array} {l}
x_1 & -x_2 & + 2 x_3 & 0 & = 2 \\
0 & x_2 & - x_3 & 0 & = 0 \\ \hline
0 & 0 & 0 & 0 & = 0 \\
0 & 0 & 0 & -1\alpha & = -2 \\
\end{array} $$
It follows $\alpha = 2$. Also, from row 2 that $x_2=x_3$
From the remaining gaussian-eiminate upwards: add row 2 to row 1
$$ \begin{array} {l}
x_1 & 0 & + x_3 & 0 & = 2 \\ \hline
0 & x_2 & - x_3 & 0 & = 0 \\ \hline
0 & 0 & 0 & 0 & = 0 \\
0 & 0 & 0 & -1\alpha & = -2 \\
\end{array} $$
We get $x_1 =2-x_3$ with arbitrary many solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1432632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
Assumption about the form of solutions to a recurrence relation Basically, when solving such recurrence relations, we try to find solutions of the form $a_n = r_n$, where $r$ is a constant.
$a_n = r^n$ is a solution of the recurrence relation
$a_n = c_1a_{n-1} + c_2a_{n-2} + … + c_ka_{n-k}$ if and only if
$r^n = c_1r^{n-1} + c_2r^{n-2} + … + c_kr^{n-k}$.
Divide this equation by $r^{n-k}$ and subtract the right-hand side from the left:
$r^k - c_1r^{k-1} - c_2r^{k-2} - … - c_{k-1}r - c_k = 0$.
This is called the characteristic equation of the recurrence relation.
Why do we think that the solutions are of the form(we try to find solutions of the form)
$a_n = r^n$ and not some other form?
Is there an intuitive explanation?
| One clean explanation (and a uniform way to solve such recurrences) is to use generating functions. Say you have:
$\begin{align}
a_{n + k} = c_{k - 1} a_{n + k - 1}
+ \dotsb + c_0 a_n
\end{align}$
Define the generating function $A(z) = \sum_{n \ge 0} a_n z^n$, multiply the recurrence by $z^n$ and sum over $n \ge 0$, noting that e.g.:
$\begin{align}
\sum_{n \ge 0} a_{n + s} z^n
= \frac{A(z) - a_0 - a_1 z - \dotsb - a_{s - 1} z^{s - 1}}{z^s}
\end{align}$
to get:
$\begin{align}
\frac{A(z) - a_0 - \dotsb - a_{k - 1} z^{k - 1}}{z^k}
= c_{k - 1}
\frac{A(z) - a_0 - \dotsb - a_{k - 2} z^{k - 2}}{z^{k - 1}}
+ c_{k - 2}
\frac{A(z) - a_0 - \dotsb - a_{k - 3} z^{k - 3}}{z^{k - 2}}
+ \dotsb
+ c_0 A(z)
\end{align}$
Multiply through by $z^k$ and collect terms to get:
$\begin{align}
A(z) (1 - c_{k - 1} z - \dotsb - c_0 z^k)
= b_{k - 1} z^{k - 1} + \dotsb + b_0
\end{align}$
Here the $b_i$ are messy combinations of the initial values $a_0$ through $a_{k - 1}$. The critical point is that:
$\begin{align}
A(z)
= \frac{b_{k - 1} z^{k - 1} + \dotsb + b_0}
{1 - c_{k - 1} z - \dotsb - c_0 z^k}
\end{align}$
This can be split into partial fractions. By that technique you know that a zero $1/r$ of multiplicity $m$ of the denominator gives rise to terms:
$\begin{align}
\frac{A_m}{(1 - r z)^m} + \dotsb + \frac{A_1}{1 - r z}
\end{align}$
Now, by the generalized binomial theorem, for $s \in \mathbb{N}$:
$\begin{align}
(1 - r z)^{-s}
&= \sum_{n \ge 0} (-1)^n \binom{-s}{n} r^n z^n \\
&= \sum_{n \ge 0} \binom{n + s - 1}{s - 1} r^n z^n
\end{align}$
Noting that $\binom{n + s - 1}{s - 1}$ is a polynomial of degree $s - 1$ in $n$, you see that a zero $1/r$ of multiplicity $m$ gives rise to a set of terms that add up to $p(n) r^n$, with $p(n)$ a polynomial of degree (up to) $m - 1$ in $n$ ("up to" as $1 - r z$ might be a factor of the numerator).
In case you have complex zeros, they come in conjugate pairs $r$, $\overline{r}$, and the coefficients of the terms are also conjugates (otherwise the result wouldn't be real). Thus you get a bunch of terms like:
$\begin{align}
\alpha n^s r^n + \overline{\alpha} n^s \overline{r}^n
= 2 \Re\left(\alpha n^s r^n\right)
\end{align}$
These terms can be expressed in trigonometric terms by expressing the values as complex exponentials.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1433651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simplifying $\tan100^{\circ}+4\sin100^{\circ}$ The answer is $-\sqrt3$.
I was wondering if this is just a coincidence?
Also, is there a relation between $$\tan(100^{\circ}+20^{\circ})=\frac{\tan100^{\circ}+\tan20^{\circ}}{1-\tan100^{\circ}.\tan20^{\circ}}=-\sqrt3$$ and the given expression? Or is there a more elegant method of solving the question?
| One has $$\tan 100^\circ + 4\sin 100^\circ = \frac{\sin 100^\circ + 2\sin 200^\circ}{\cos 100^\circ} = \frac{\sin 100^\circ - 2\sin 20^\circ}{\cos 100^\circ} = \frac{2\cos 60^\circ\sin 40^\circ - \sin 20^\circ}{\cos 100^\circ} = \frac{\sin 40^\circ - \sin 20^\circ}{\cos 100^\circ} = \frac{2\cos 30^\circ\sin 10^\circ }{\cos 100^\circ} = -\sqrt{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1434561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Can I show $\frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n} \leq 1$ by induction? It is known that $a_{n+1} > a_n$. I tried to prove
$$\frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n} \leq 1$$
via induction.
Base case: $a_0 = \frac{1}{0+1} = 1 \leq 1$
Inductive step: assume $a_n \leq 1$, prove $a_{n+1} \leq 1$.
$a_n = \frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n} \leq 1$
$\frac{1}{n+2} + \frac{1}{n+3}+ \cdots + \frac{1}{2n} + \frac{1}{2n+1} + \frac{1}{2n+2} \leq 1-\frac{1}{n+1} + \frac{1}{2n+1} + \frac{1}{2n+2}$
But $-\frac{1}{n+1} + \frac{1}{2n+1} + \frac{1}{2n+2}$ is positive, so I'm not sure where to go from here.
Normally, the inductive step would involve a change on the right hand side, but in this case, it remains the same. Makes me wonder if induction is even the correct approach?
| According to Wolfram Alpha, this series is equivalent to $ \psi^{(0)}(2n+1)-\psi^{(0)}(n+1)$ where $ \psi^{(0)}(x)$ is the $n^{th}$ derivative of the digamma function. Using this, you could show the continuity for the function for values of $n > 0$, show that the limit as n approaches infinity is $log(2)$ (easily proven), and show that the function is continuously increases (by showing derivative is positive for $n>0$?)
Note that this actually shows that the series is always $< log(2)$, a stronger statement than that given by the induction argument.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1435944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the domain of the function $f(x) = \frac{x}{\sqrt{x^2+9}}$ For some reason I can't find domain to this function, but it is pretty clear that it's (ALL X)
$$f(x) = \frac{x}{\sqrt{x^2+9}}$$
I can say that: $$\space x^2+9 > 0\space$$
$$x^2 > -9$$
$$x = \pm \sqrt{-9}$$
I don't get it..
| $$f(x) = \frac{x}{\sqrt{x^2+9}}$$
Only $\sqrt{x^2+9}=0$ could be a problem for your domain. But as $x^2+9>0$ (sum of two squares), this cannot happen. Hence, the domain of $f$ is $R$.
If you also need to find the map of your function:
First of all we evaluate the derivative of $f(x)$
$$f'(x)=\frac{1\cdot\sqrt{x^2+9}-x\frac{2x}{2\sqrt{x^2+9}}}{(x^2+9)}$$
$$=\frac{\sqrt{x^2+9}-\frac{x^2}{\sqrt{x^2+9}}}{(x^2+9)}=\frac{\frac{(x^2+9)-x^2}{\sqrt{x^2+9}}}{(x^2+9)}=\frac{9}{\sqrt{x^2+9}(x^2+9)}$$
As $9>0$, $(x^2+9)>0$ and $\sqrt{x^2+9}>0$, we can conclude that the function is strictly monotonic for all $x \in R$.
Now we need to evaluate the limits of $f(x)$ for $x \to \pm \infty$
$$\lim_{x\to \infty}\frac{x}{\sqrt{x^2+9}}=\lim_{x\to \infty}\frac{x}{|x|\sqrt{1+\frac{9}{x^2}}}=\lim_{x\to \infty}\frac{x}{x\sqrt{1+\frac{9}{x^2}}}=1$$
$$\lim_{x\to -\infty}\frac{x}{\sqrt{x^2+9}}=\lim_{x\to \infty}\frac{x}{|x|\sqrt{1+\frac{9}{x^2}}}=\lim_{x\to \infty}\frac{x}{-x\sqrt{1+\frac{9}{x^2}}}=-1$$
Hence, we conclude the function maps from $R \to (-1,1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1438638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Legendre functions $Q_n(x)$ of the second kind Legendre functions $Q_n(x)$ of the second kind
\begin{equation*}
Q_n(x)=P_n(x) \int \frac{1}{(1-x^2)\cdot P_n^2(x)}\, \mathrm{d}x
\end{equation*}
what to do after this step?
how can I complete ?
I need to reach this formula
\begin{equation*}
Q_n(x)=\frac{1}{2} P_n(x)\ln\left( \frac{ 1+x}{1-x}\right)
\end{equation*}
| So you want to prove that
$$Q_n(x)= P_n(x) \int^x \frac{1}{(1-x^2) \, \left[P_n(x)\right]^2} \, dx = \frac{1}{2} P_n(x) \ln\left(\frac{1+x}{1-x}\right)$$.
Let's first compute for the case $n=0$, knowing that $P_0(x) = 1$ we would have
$$Q_0(x) = \int^x \frac{1}{(1-x^2)} \, dx = \frac{1}{2} \int \left[\frac{1}{1+x} + \frac{1}{1-x} \right] = \frac{1}{2} \, \ln \left(\frac{1+x}{1-x} \right)$$
where we decomposed the fraction to compute the integral.
Let us now compute for $n=1$, where $P_1(x) = x$.
$$Q_1(x) = x \int^x \frac{1}{(1-x^2)\,x^2} \, dx = \frac{x}{2} \, \ln \left(\frac{1+x}{1-x} \right) - 1$$
where we decomposed the fraction again to compute for the integral.
For $n=2$ we use the same process again, leading us to the result
$$Q_2(x) = \frac{1}{2} P_2(x) \ln \left( \frac{1+x}{1-x} \right) - \frac{3x}{2}$$
So it seems that it only works for $n=0$ and that the true formula is
$$Q_n(x) = \frac{1}{2} P_n(x) \ln \left( \frac{1+x}{1-x} \right) - C_n(x)$$
for some value of $C_n(x)$.
As an alternative you could just use the recurrence formulas for $Q_n(x)$ which are
$$(n+1)Q_{n+1}(x) - (2n+1)x \, Q_n(x) + n \, Q_{n-1}(x) = 0$$
$$(2n+1)Q_n(x) = Q'_{n+1}(x) -Q'_{n-1}(x)$$
To show that
$$Q_n(x) = \frac{1}{2} P_n(x) \ln \left(\frac{1+x}{1-x}\right) - \frac{2n-1}{n}P_{n-1}(x) - \frac{2n-5}{3(n-1)} P_{n-3}(x) - \cdots$$
If you want to read more about the subject then I suggest you take a look at some mathematical physics books like Mathematical Methods for Physicists by Arfken (Some of my friends say that they dumbed down the seventh edition, but I still use the seventh anyways) or Mathematical Methods in the Physical Sciences by Boas.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1439890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Problem with multivariable calculus: $\lim_{(x,y)\to (0,0)} \frac{x^3 + y^3}{x^2 + y}$ Anyone can help me with this limit?
$$\lim_{(x,y)\to (0,0)} \frac{x^3 + y^3}{x^2 + y}$$
I'm having trouble with proving that this limit really goes to $0$
thank you
| Let $y=(x^3-x^6)^2-x^2$. $\,$Then
\begin{align*}
\frac{x^3+y^3}{x^2+y} &= \frac{x^3+(x^3-x^6)^6 - x^6 + 3x^4(x^3-x^6)^2 - 3 x^2 (x^3-x^6)^4}{(x^3-x^6)^2}\\
&= \frac{1}{x^3-x^6} + (x^3-x^6)^4+3x^4-3x^2(x^3-x^6)^2\\
&\rightarrow \infty.
\end{align*}
That is,
\begin{align*}
\lim_{(x,y)\rightarrow (0, 0)} \frac{x^3+y^3}{x^2+y}
\end{align*}
does not exist.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
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Implicit differentiation of $3x^2 + 4y^2 = 12$
The equation $3x^2 + 4y^2 = 12$ defines $y$ implicitly as two functions of $x$ if $|x| < 2$ and $|x|=2$. Assuming
the second derivative $y''$ exists, show that it satisfies the equation $4y^3 y'' = -9$.
Progress. I derived $6x+8yy'=0$ then $y'= \frac{-3x}{4y}$ then derived again $$y'' = \frac{-3(y-xy')}{4y^2}= \frac{-3(y-xy')y}{4y^2y}= \frac{-3(y^2-xyy')}{4y^3} $$ Then $4y^3y''= -3(y^2-xyy')$, so I want $-9 = -3(y^2-xyy')$... and don't know the rest.
| It turns out that
if
$ax^2+by^2 = c$,
then
$b^2y^3y''
=-ac
$.
In this case,
$16y^2y''
= -36$
or
$4y^2y''
= -9$.
I use implicit differentiation
twice.
From
$ax^2 + by^2 = c
$,
as you did
$0
=2ax+2byy'
=ax+byy'
$,
so
$y'
=-\frac{ax}{by}
$.
Doing it again,
$0
= a+b((y')^2+yy'')
$
or
$-\frac{a}{b}
=(y')^2+yy''
=(\frac{ax}{by})^2+yy''
$
or
$-aby^2
=a^2x^2+b^2y^3y''
$
or
$b^2y^3y''
=-aby^2-a^2x^2
=-a(by^2+ax^2)
=-ac
$.
In your case,
$c = ab$,
so
$b^2y^3y''
=-a^2b
$
or
$by^3y''
=-a^2
$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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How many pairs of positive integers $(x,y)$ satisfy the equation $x^2 - 10! = y^2$? I have a question about number theory:
How many pairs of positive integers $(x,y)$ satisfy the following equation? $$x^2 - 10! = y^2$$
My attempt:
Move the $y^2$ from right to the left and 10! From left to the right such that:
$$x^2-y^2= 10!$$
$$(x-y)(x+y)=10!$$
$$(x-y)(x+y)= 10.9.8.7.6.5.4.3.2$$
Until this step, I think I will get many possibility to get the answer, such that:
$$(x-y)= 10.9$$ and $$(x+y)= 8!$$
Also
$$(x-y)= 10.9.8$$ and $$(x+y)= 7!$$
And another possibilities, like $$(x-y)=10.9.8.7$$ and $$(x+y)=6!$$
And so on until $$(x-y)=10.9.8.7.6.5.4.3.2$$ and $$(x+y)=1$$
And I think solving all the possibility is a bit tedious work. Can somebody help me to find a better solution to solve this kind of problem?
Thanks
| $105$ pairs.
As marty cohen said: "$(x+y)(x-y)=10!=ab$
with
$x+y=a$ and $x-y=b$.
Then
$x = (a+b)/2$ and
$y = (a-b)/2$.
So $a$ and $b$
have to have the same parity."
$10!=2^{\color{red}{8 }}\cdot 3^{\color{red}{4}}\cdot 5^{\color{red}{2}} \cdot 7^{\color{red}{1}} \implies (\color{red}{8}+1)(\color{red}{4}+1)(\color{red}{2}+1)(\color{red}{1}+1)=270 \ $divisors in $10! \ $. Since $10!$ is an even number, each divisor must be even so that $x$ and $y$ will come out as integers when the divisors are added or subtracted and subsequently divided by $2$.
So we need
*
*all the ways $p=p_1\cdot p_2=3^4\cdot 5^2 \cdot 7^1$ can be split into two divisors, ... crossed with
*all the ways we can distribute the $2^8$ into two factors so that each factor has at least one $2$
For,
*
*$p=3^4\cdot 5^2 \cdot 7^1$ can be split into $\frac{270}{(\color{red}{8}+1)}=30$ different factors, that's $\frac{30}{2}=15$ unique choices of factor-pairs $p_1 \cdot p_2\ $ since the order of $p_1$ and $p_2$ doesn't change the outcome.
And for
*There's $(2^7p_1 \cdot 2^1p_2) \cdots (2^1p_1 \cdot 2^7p_2)$ for each choice of $p_1 \cdot p_2$, and I see $7$ of them.
So I see $7\cdot 15 =105$ different ways we can write $(x,y)$ down so that $x^2-y^2=10!$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Evaluation of $ \int\frac{x^2-1}{\sqrt{x^4+4x^3+7x^2+4x+1}}dx$
Evaluation of $\displaystyle \int\frac{x^2-1}{\sqrt{x^4+4x^3+7x^2+4x+1}}dx$
$\bf{My\; Try::}$ Let $\displaystyle I = \int\frac{x^2-1}{\sqrt{x^4+4x^3+7x^2+4x+1}}dx = \int\frac{x^2-1}{\sqrt{(x+1)^4+x^2}}dx$
So $\displaystyle I = \int\frac{x^2-1}{x\cdot \sqrt{\left(\frac{(x+1)^2}{x}\right)^2+1}}dx\;,$ ow Put $\displaystyle \frac{(x+1)^2}{x}=t\;,$
Then $\displaystyle \frac{2x(x+1)-(x+1)^2}{x^2}dx = dt\Rightarrow \left(\frac{x^2-1}{x^2}\right)dx = dt$
Can we solve the Integral in terms of elementry function.
If yes then plz explain here,
Thanks
| It looks to me that an elliptic integral is unavoidable.
By replacing $\sqrt{x}+\frac{1}{\sqrt{x}}$ with $y$ the problem boils down to computing:
$$ \int \left(y^3-2y+y^2\sqrt{y^2-4}\right)\,\frac{dy}{\sqrt{1+y^4}}=\frac{\sqrt{1+y^4}}{2}+\text{arcsinh}(y^2)+\frac{1}{2}\int\sqrt{\frac{z^2-4z}{z^2+1}}\,dz $$
where in the last step I set $z=y^2$.
| {
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"url": "https://math.stackexchange.com/questions/1446658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $\sqrt{\frac{n+15}{n+1}}\in\mathbb Q$, then $n=17$ How can one show that :
If $\sqrt{\frac{n+15}{n+1}}\in\mathbb Q$ so $n=17$
I tried using the fact that any number $a\in\mathbb Q$ so $a=\frac{x}{y}$ such that $\gcd(x,y)=1$
So $\frac{n+15}{n+1}=\frac{x^2}{y^2}$
But here I'm stuck.
| Essentially, you know $x^2(n+1)=y^2(n+15)$, let $m=n+1$ for convenience, then you know that since $x$ and $y$ are relatively prime, $y^2 \mid m$
and $x^2\mid m+14$. Since $\frac{m+14}{m}=\frac{x^2}{y^2}$, we then have
that $m=y^2a$ for some $a$ and $m+14=x^2a$ for that same $a$. Then $y^2a+14=x^2a$. Then $14=(x^2-y^2)a$. Now there are only a few possibilities. $x^2-y^2=1,2,7,14$. But we factor, so that we have $(x-y)(x+y) =1,2,7,14$. Then if $(x-y)(x+y)=2k$ for an odd number $k$, we have that since $2$ divides exactly one of the factors, $(x-y)+(x+y)$ is odd, but $x-y+x+y=2x$, which is even, contradiction. Thus $x^2-y^2=7$ or $x^2-y^2=1$. In the first case no matter how we assign $x+y$ and $x-y$ to $1$ and $7$ or $-1$ and $-7$, their difference is always $\pm 6$, so that $y=\pm 3$. But then $m=3^2\cdot 2=18$, which is fine, so now we just need to check the other case, $x^2-y^2=1$. The difference of the factors will always be $0$, since if $(x+y)(x-y)=1$, $x+y=x-y$. Thus $y=0$. But then $m=0$, which is impossible.
Hence $m$ is always 18, or $n$ is always 17.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove by induction $\sum \frac {1}{2^n} < 1$ Prove by induction $\sum \frac {1}{2^n} < 1$
Well supposing the base case has been shown to be true, I start with the induction step:
Suppose true for n = k:
$$ \frac{1}{2} + \frac{1}{4} + ....\frac{1}{2^k} < 1$$
Want to show this is true for:
$$ \frac{1}{2} + \frac{1}{4} + ....\frac{1}{2^k} + \frac{1}{2^{k+1}} < 1$$
Now this is where i am getting stuck, should i be tryig to show that
$$ \frac{1}{2} + \frac{1}{4} + ....\frac{1}{2^k} + \frac{1}{2^{k+1}} < 1 +\frac{1}{2^{k+1}}$$ or should i be attempting to show that
$1 + \frac{1}{2^{k+1}} < 1$ which is utter nonsense. So i am stuck on what exactly to prove here.
Note i also thought of maybe using the geometric series of $\frac {1}{2} $ as some upper bound but then that would provide me a value that was greater than 1 so it didn't work out
| Use induction to prove that $$\frac 12 + \frac 14 + \cdots + \frac 1{2^n} = 1 - \frac{1}{2^n}$$ for all $n$. This will give you the result immediately since $1 - \frac{1}{2^n} < 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate limit as x approaches infinity of $\lim_{x\to\infty}\frac{\sqrt{x^3 +7x}}{\sqrt{4x^3+5}}$ I am having trouble figuring out how to answer this question by determining the degree of the numerator and/or denominator:
$$\lim_{x\to\infty}\frac{\sqrt{x^3 +7x}}{\sqrt{4x^3+5}}$$
I have tried deriving the first coefficient of the numerator and denominator, but not sure how to proceed to find the limit as $x \to \infty$.
| You factor out the dominant term from the numerator such that it cancels with the dominant term in the denominator. In this case we take out a factor of $x^3$ since $x^3$ is the highest degree or dominant term in both the numerator and denominator.
Hence, showing all intermediate steps:
$\color{blue}{\lim_{x\rightarrow \infty}\cfrac{\sqrt{x^3 +7x}}{\sqrt{4x^3+5}}=\lim_{x\rightarrow \infty}\cfrac{x^\frac{3}{2}\sqrt{1 +7x^{-2}}}{x^\frac{3}{2}\sqrt{4+5x^{-3}}}}=\lim_{x\rightarrow \infty}\color{red}{\cfrac{\sqrt{1 +7x^{-2}}}{\sqrt{4+5x^{-3}}}}=\cfrac{\sqrt{1 +7(0)}}{\sqrt{4+5(0)}}=\cfrac{1}{2}$
$\color{blue}{\mathrm{This}}$ step is easier to see if you write $$\cfrac{\sqrt{x^3 +7x}}{\sqrt{4x^3+5}}= \cfrac{(x^3 +7x)^\frac{1}{2}}{({4x^3+5})^\frac{1}{2}}= \cfrac{\left(x^3(1 +\frac{7}{x^2})\right)^\frac{1}{2}}{\left({x^3(4+\frac{5}{x^3}})\right)^\frac{1}{2}}=\cfrac{x^\frac{3}{2}(1 +\frac{7}{x^2})^\frac{1}{2}}{{x^\frac{3}{2}(4+\frac{5}{x^3}})^\frac{1}{2}}=\color{red}{\cfrac{\sqrt{1 +7x^{-2}}}{\sqrt{4+5x^{-3}}}}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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Solving $\lim_{x\to0}\frac{1-\sqrt{\cos x}}{x^2}$ I've been trying to solve this over and over without L'Hopital but keep on failing:
$$\lim_{x\to0}\frac{1-\sqrt{\cos x}}{x^2}$$
My first attempt involved rationalizing:
$$\frac{1-\sqrt{\cos x}}{x^2} \cdot \frac{1+\sqrt{\cos x}}{1+\sqrt{\cos x}} = \frac{1-\cos x}{x\cdot x \cdot (1+\sqrt{\cos x})}$$
Using the rule $\frac{1-\cos x}{x} = 0$ for $x\to0$ is useless because we would end up with
$$\frac{0}{0\cdot x \cdot \sqrt{\cos x}} = \frac{0}{0}$$
But hey, perhaps we can rationalize again?
$$\frac{1-\cos x}{(x^2+x^2\sqrt{\cos x})}\cdot \frac{(x^2-x^2\sqrt{\cos x})}{(x^2-x^2\sqrt{\cos x})}$$
Resulting in
$$\frac{(1-\cos x)(x^2-x^2\sqrt{\cos x})}{x^4 - x^4\cdot\cos x} = \frac{(1-\cos x)(x^2-x^2\sqrt{\cos x})}{x^4 \cdot (1-\cos x)}$$
Cancelling
$$\frac{(x^2-x^2\sqrt{\cos x})}{x^4} = \frac{1-\sqrt{\cos x}}{x^2}$$
Well that was hilarious. I ended up at the beginning! Dammit.
My second attempt was to use the definition $\cos x = 1 - 2\sin \frac{x}{2}$:
$$\lim_{x\to0}\frac{1-\sqrt{\cos x}}{x^2} = \frac{1-\sqrt{1 - 2\sin \frac{x}{2}}}{x^2}$$
And then rationalize
$$\frac{1-\sqrt{1 - 2\sin \frac{x}{2}}}{x^2} \cdot \frac{1+\sqrt{1 - 2\sin \frac{x}{2}}}{1+\sqrt{1 - 2\sin \frac{x}{2}}}$$
$$\frac{1-(1 - 2\sin \frac{x}{2})}{x^2+x^2\sqrt{1 - \sin \frac{x}{2}}} = \frac{- 2\sin \frac{x}{2}}{x^2\left(1+\sqrt{1 - \sin \frac{x}{2}}\right)}$$
I want to make use of the fact that $\frac{\sin x}{x} = 1$ for $x\to0$, so I will multiply both the numerator and denominator with $\frac{1}{2}$:
$$\frac{-\sin \frac{x}{2}}{\frac{x}{2}\cdot x\left(1+\sqrt{1 - \sin \frac{x}{2}}\right)}$$
Then
$$\frac{-1}{x\left(1+\sqrt{1 - \sin \frac{x}{2}}\right)}$$
Well clearly that's not gonna work either. I will still get $0$ in the denominator.
The correct answer is $\frac{1}{4}$. I can kind of see why is the numerator $1$, but no idea where is that $4$ going to come out of.
I don't know how am I supposed to solve this without L'Hopital.
| After your first rationalization, you showed that the original expression is equivalent to
$$\lim_{x\to 0} \frac{1-\cos x}{x^2(1+\sqrt{\cos x})}=\lim_{x \to 0}\color{green}{\frac{1-\cos x}{x^2}} \cdot \color{blue}{\frac {1}{1+\sqrt{\cos x}}}$$
Since the $\color{green}{\text{green}}$ part of the limit converges to $\frac 12$ (it's a notable limit) and the $\color{blue}{\text{blue}}$ coloured function of $x$ is continuous in a neighborhood of $0$, we can conclude that the limit exists and its value is
$$ \lim_{x \to 0} \frac12 \cdot \frac{1}{1+\sqrt{\cos x}}= \frac 12 \cdot \frac{1}{1+\sqrt{\cos(0)}}=\frac 12 \cdot \frac12= \color{red}{\frac14}$$
Note: in case you did not know the aforementioned notable limit, here is a quick proof of it using the fact that $\lim_{x \to 0} \frac {\sin x}{x}=1$ :
$$ \lim_{x\to 0} \frac{1- \cos x}{x^2}$$
$$ = \lim_{x\to 0} \frac{1- \cos x}{x^2} \cdot \color{red}{\frac{1+ \cos x}{1+\cos x}}$$
$$ = \lim_{x \to 0} \frac{\sin^2 x}{x^2}\cdot \frac{1}{1+\cos x}= \frac 12$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Binomial transform of Catalan numbers formula How to prove that OEIS A007317 Binomial transform of Catalan numbers $a_{n}: 1, 2, 5, 15, 51, 188, 731, 2950, 12235, 51822, .. (n = 1, 2, ..)$ has a recurrence formula: $(n+2)a_{n+2} = (6n+4)a_{n+1} - (5n)a_{n}$ ?
The sequence is defined as: $a_{n+1} = \sum_{k=0}^{n} \binom{n}{k}c_{k}$ where $c_{n} = \frac{1}{n+1}\binom{2n}{n}$ is OIES A000108 Catalan numbers: $1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, .. (n = 0, 1, ..)$
| We prove the recurrence formula with the help of generating functions. Let $A(z)$ be the generating functions of the $a_n$
\begin{align*}
A(z)=\sum_{n=1}^\infty a_nz^n=\sum_{n=0}^\infty a_{n+1}z^{n+1}
=\sum_{n=0}^\infty \left(\sum_{k=0}^n\binom{n}{k}C_k \right)z^{n+1}
\end{align*}
and let $C(z)$ be the (well-known) generating function of the Catalan numbers $C_n$
\begin{align*}
C(z)&=\sum_{n=0}^{\infty} C_nz^n=\sum_{n=0}^{\infty} \frac{1}{n+1}\binom{2n}{n}=\frac{1}{2z}\left(1-\sqrt{1-4z}\right)
\end{align*}
We prove the claim in three steps. Since the $a_n$ are a binomial transform of $C_n$, we derive at first a corresponding functional equation between $A(z)$ and $C(z)$. From this functional equation we obtain an explicit expression of $A(z)$. In the last step we use $A(z)$ to show that the corresponding generating function of the left hand and the right hand side of the recurrence relation coincide.
Step 1: Functional equation between $A(z)$ and $C(z)$
The following is valid
\begin{align*}
A(z)=\frac{z}{1-z}C\left(\frac{z}{1-z}\right)\tag{1}
\end{align*}
We obtain
\begin{align*}
\frac{z}{1-z}C\left(\frac{z}{1-z}\right)&=\frac{z}{1-z}\sum_{k=0}^{\infty}C_k\left(\frac{z}{1-z}\right)^k\\
&=\sum_{k=0}^{\infty}C_k\left(\frac{z}{1-z}\right)^{k+1}\\
&=\sum_{k=0}^{\infty}C_kz^{k+1}(1-z)^{-(k+1)}\\
&=\sum_{k=0}^{\infty}C_kz^{k+1}\sum_{l=0}^{\infty}\binom{-(k+1)}{l}(-z)^l\tag{2}\\
&=z\sum_{k=0}^{\infty}C_kz^{k}\sum_{l=0}^{\infty}\binom{k+l}{l}z^l\tag{3}\\
&=z\sum_{n=0}^{\infty}\left(\sum_{{k+l=n}\atop{k,l\geq 0}}\binom{k+l}{l}C_k\right)z^n\tag{4}\\
&=z\sum_{n=0}^{\infty}\left(\sum_{k=0}^n\binom{n}{n-k}C_k\right)z^n\tag{5}\\
&=\sum_{n=0}^{\infty}\left(\sum_{k=0}^n\binom{n}{k}C_k\right)z^{n+1}\\
&=A(z)\\
\end{align*}
Comment:
*
*In (2) we use the binomial series
*In (3) we use the relation $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$
*In (4) we multiply the series (Cauchy product)
*In (5) we replace $l$ with $n-k$
Step 2: Explicit expression of $A(z)$
The following is valid
\begin{align*}
A(z)=\frac{1}{2}\left(1-\sqrt{\frac{1-5z}{1-z}}\right)
\end{align*}
Based upon the generating function $C(z)=\frac{1}{2z}\left(1-\sqrt{1-4z}\right)$ of the Catalan numbers we obtain from (1)
\begin{align*}
A(z)&=\frac{z}{1-z}C\left(\frac{z}{1-z}\right)\\
&=\frac{z}{1-z}\cdot\frac{1}{2\left(\frac{z}{1-z}\right)}\left(1-\sqrt{1-\frac{4z}{1-z}}\right)\\
&=\frac{1}{2}\left(1-\sqrt{\frac{1-5z}{1-z}}\right)\\
\end{align*}
$$ $$
Step 3: Recurrence relation
The following recurrence relation is valid
\begin{align*}
&(n+2)a_{n+2}=(6n+4)a_{n+1}-5na_n\qquad\qquad n\geq 0\\
&a_0=0, a_1=1
\end{align*}
We take the left hand and right hand side of the recurrence relation, use an Ansatz with the generating function
\begin{align*}
A(z)=\sum_{n=1}^{\infty}a_n z^n=\frac{1}{2}\left(1-\sqrt{\frac{1-5z}{1-z}}\right)
\end{align*}
and show that both sides coincide. We will also use the derivative of $A(z)$
\begin{align*}
\frac{d}{dz}A(z)=\sqrt{\frac{1-z}{1-5z}}\frac{1}{(1-z)^2}
\end{align*}
We start with the left hand side
\begin{align*}
\sum_{n=0}^\infty(n+2)a_{n+2}z^n&=\sum_{n=2}^\infty na_{n}z^{n-2}\\
&=\frac{1}{z}\frac{d}{dz}\left(\sum_{n=2}^{\infty}a_nz^n\right)\\
&=\frac{1}{z}\frac{d}{dz}\left(A(z)-z\right)\\
&=\frac{1}{z}\sqrt{\frac{1-z}{1-5z}}\frac{1}{(1-z)^2}-\frac{1}{z}\tag{6}
\end{align*}
and we obtain from the right hand side
\begin{align*}
\sum_{n=0}^\infty&(6n+4)a_{n+1}z^n-\sum_{n=0}^{\infty}5na_nz^n\\
&=\sum_{n=1}^\infty(6n-2)a_{n}z^{n-1}-5z\sum_{n=0}^{\infty}na_nz^{n-1}\\
&=6\sum_{n=1}^\infty na_{n}z^{n-1}-2\sum_{n=1}^\infty a_{n}z^{n-1}-5z\sum_{n=0}^{\infty}na_nz^{n-1}\\
&=6\frac{d}{dz}\sum_{n=1}^{\infty}a_nz^n-\frac{2}{z}A(z)-5z\frac{d}{dz}\sum_{n=1}^{\infty}a_nz^n\\
&=(6-5z)\frac{d}{dz}A(z)-\frac{2}{z}A(z)\\
&=\frac{6-5z}{(1-z)^2}\sqrt{\frac{1-z}{1-5z}}-\frac{1}{z}\left(1-\sqrt{\frac{1-5z}{1-z}}\right)\\
&=\sqrt{\frac{1-z}{1-5z}}\left(\frac{6-5z}{(1-z)^2}+\frac{1}{z}\cdot\frac{1-5z}{1-z}\right)-\frac{1}{z}\\
&=\frac{1}{z}\sqrt{\frac{1-z}{1-5z}}\frac{1}{(1-z)^2}-\frac{1}{z}\tag{7}
\end{align*}
We see the generating functions (6) and (7) of the left-hand and right-hand side coincide showing the validity of the recurrence relation.
| {
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"url": "https://math.stackexchange.com/questions/1456848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to simplify $\sin^4 x+\cos^4 x$ using trigonometrical identities? $\sin^{4}x+\cos^{4}x$
I should rewrite this expression into a new form to plot the function.
\begin{align}
& = (\sin^2x)(\sin^2x) - (\cos^2x)(\cos^2x) \\
& = (\sin^2x)^2 - (\cos^2x)^2 \\
& = (\sin^2x - \cos^2x)(\sin^2x + \cos^2x) \\
& = (\sin^2x - \cos^2x)(1) \longrightarrow\,
= \sin^2x - \cos^2x
\end{align}
Is that true?
| Let $$\displaystyle y=\sin^4 x+\cos^4 x = \left(\sin^2 x+\cos^2 x\right)^2-2\sin^2 x\cdot \cos^2 x = 1-\frac{1}{2}\left(2\sin x\cdot \cos x\right)^2$$
Now using $$ \sin 2A = 2\sin A\cos A$$
So, we get $$\displaystyle y=1-\frac{1}{2}\sin^2 2x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1458305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Convert $r^2= 9 \cos 2 \theta$ into a Cartesian equation This is how I tried so far...
$r^2= 9 \cos 2 ( \theta)$
$\cos (2 \theta) = \cos ^2 (\theta) - \sin^2 (\theta)$ and
$r^2= x^2 + y^2$
so, it will become
$x^2 + y^2 = 9 [\cos^2 (\theta) - \sin^2 (\theta) ]$
| multiply both side by $r^2$ and replace $r.cos\theta=x$ and $r.sin\theta=y$
and $r^2=x^2+y^2$ as $cos^2\theta-sin^2\theta=(cos\theta+sin\theta)(cos\theta-sin\theta)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1459152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Mathematical Induction Question About A Basis Step Use mathematical induction to show that:
$1^3 + 3^3 + 5^3 + ... + (2n + 1)^3 = (n+1)^2(2n^2+4n+1)$ whenever $n$ is a positive integer.
I am able to solve this problem, but the thing that confuses me on this is the Basis Step.
P(1): $(2(1) + 1)^3$ $\neq$ $4(7)$, does this mean that this problem can not be proved using mathematical induction, or am I missing something? Thanks.
| Method -1: Mathematical Induction
Notice, the following steps by Mathematical Induction
*
*Setting $n=1$, we get
$$1^3+(2\cdot 1+1)^3=(1+1)^2(2(1)^3+4(1)+1)$$
$$28=28$$
The identity holds good for $n=1$.
*Assume that it holds for $n=k$ then we have
$$1^3+3^3+5^3+\ldots +(2k+1)^3=(k+1)^2(2k^2+4k+1)$$
*Setting $n=k+1$, we get
$$1^3+3^3+5^3+\ldots +(2k+1)^3+(2k+3)^3=(k+1+1)^2(2(k+1)^2+4(k+1)+1)$$
$$\implies 1^3+3^3+5^3+\ldots +(2k+1)^3=(k+2)^2(2k^2+8k+7)-(2k+3)^3$$
$$=2k^4+8k^3+11k^2+6k+1$$ $$=(k+1)^2(2k^2+4k+1)$$
Which is true from (2)
Hence, the given identity holds true for all positive integers $n\geq 1$
Method-2
Notice, we have
$$1^3+3^3+5^3+\dots +(2n+1)^3$$
$$=(1^3+2^3+3^3+4^3+5^3+\dots +(2n)^3+(2n+1)^3)-(2^3+4^3+6^3+\ldots +(2n)^3)$$
$$=(1^3+2^3+3^3+4^3+5^3+\dots +(2n)^3+(2n+1)^3)-8(1^3+2^3+3^3+\ldots +n^3)$$
$$=\left(\frac{(2n+1)((2n+1)+1)}{2}\right)^2-8\left(\frac{n(n+1)}{2}\right)^2$$
$$=(n+1)^2(2n+1)^2-2n^2(n+1)^2$$
$$=(n+1)^2[(2n+1)^2-2n^2]$$
$$=(n+1)^2[4n^2+1+4n-2n^2]$$
$$=(n+1)^2(2n^2+4n+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1460608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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