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Proving $\frac{1}{\cos^2\frac{\pi}{7}}+ \frac {1}{\cos^2\frac {2\pi}{7}}+\frac {1}{\cos^2\frac {3\pi}{7}} = 24$ Someone gave me the following problem, and using a calculator I managed to find the answer to be $24$.
Calculate $$\frac {1}{\cos^2\frac{\pi}{7}}+ \frac{1}{\cos^2\frac{2\pi}{7}}+\frac {1}{\cos^2\frac{3\pi}{7}}\,.$$
The only question left is, Why? I've tried using Euler's Identity, using a heptagon with Law of Cosine w/ Ptolemy's, etc. but the fact that the cosine values are all squared and in the denominator keeps getting me stuck.
If $\zeta=e^{\frac{2\pi i}{7}}$, then the required expression is
$$4\left(\frac{\zeta^2}{(\zeta+1)^2}+\frac{\zeta^4}{(\zeta^2+1)^2}+\frac{\zeta^6}{(\zeta^3+1)^2}\right).$$
How do we simplify this result further?
| First notice that since $\cos(x)=\cos(\pi-x)$, we have
$$1+2\left(\frac{1}{\cos(\frac{\pi}{7})^2}+\frac{1}{\cos(\frac{2\pi}{7})^2}+\frac{1}{\cos(\frac{3\pi}{7})^2}\right)=\sum_{k=0}^6 \frac{1}{\cos(\frac{k\pi}{7})^2}$$
Now, $x\mapsto 2x$ is a bijection of the integers mod $7$, so we may make the summands $\cos(\frac{2\pi k}{7})^{-2}$.
Using $\cos(\frac{2\pi k}{n})=(\zeta^k+\zeta^{-k})/2$ where $\zeta=e^{2\pi i/n}$ combined with the geometric sum formula
$$\frac{a^n+b^n}{a+b}=\sum_{r=0}^{n-1} a^{(n-1)-r}b^r,$$
and the fact that for $n$th roots of unity $\xi$,
$$\sum_{k=0}^{n-1} \xi^k =\begin{cases} n & \xi=1 \\ 0 & \xi\ne1 \end{cases} $$
we may derive
$$\sum_{k=0}^{n-1}\frac{1}{\cos(\frac{2\pi k}{n})^m}=\sum_k \left(\frac{2}{\zeta^k+\zeta^{-k}}\right)^m=\sum_k \left(\frac{\zeta^{nk}+\zeta^{-nk}}{\zeta^k+\zeta^{-k}}\right)^m $$
$$=\sum_k\left(\sum_{r=0}^{n-1}(-1)^r \zeta^{-(2r+1)k}\right)^m=\sum_k \sum_{\substack{r_1,\cdots,r_m \\ \sum r_i=r}}(-1)^r\zeta^{-(2r+m)k}$$
$$ =\sum_{\substack{r_1,\cdots,r_m \\ \sum r_i=r}} (-1)^r \sum_k (\zeta^{-2r-m})^k=n(A-B).$$
Therefore in conclusion we have
Theorem.
$$\sum_{k=0}^{n-1} \frac{1}{\cos(\frac{2\pi k}{n})^m}=n(A-B)$$
where $A$ and $B$ count the solutions to $r_1+\cdots+r_m\equiv -m/2$ mod $n$ with $\sum_i r_i$ even and odd respectively (and $0\le r_1,\cdots,r_m<n$).
As a special case, if $m=2$ we see that $A=n$ and $B=0$, yielding the corollary
$$\sum_{k=0}^{n-1}\frac{1}{\cos(\frac{2\pi k}{n})^2}=n^2.$$
| {
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The sum of more than two consecutive natural numbers cannot be prime.
The sum of more than two consecutive natural numbers cannot be prime.
Is the statement true and is there any way to prove it?
I was able to prove that the sum of an odd amount of consecutive numbers cannot be prime:
So, since the sum of consecutive integers is $x+(x+1)+(x+2)+(x+3)$ etc...
we can also write this as
$$nx + n(n-1)/2 = n(x + (n-1)/2)$$
with $n$ as the amount of numbers and $x$ the first number in the row.
So, with an odd number as $n\neq 1$, we will get a product which will never result in a prime.
Any way to prove this for all $n \ge 2$? Thanks for all the help.
| For a sum of three or more consecutive positive integers
$S = x + (x+1) + (x + 2) + ..... + (x + n -1)$ $x > 0; n > 2$
$S = (x + n-1) + (x+n - 1) + (x + n - 2) + ..... + (x + 1)+x$
Add 'em together.
$2S = (2x + n -1) + (2x + n-1) + .... (2x + n-1) = n(2x + n-1)$
Case one: $n$ is even. Then $S=\frac n2(2x + n -1)$ is not prime as $n/2 > 1$ and $2x + n - 1 > 2$
Case 2: $n$ is odd. Then $2x + n - 1$ is even and $S = n\frac{2x + n - 1}2$ which is not prime as $n > 1$ and $(2x + n - 1)/2 > 1$.
You were 90% of the way there. You just needed to hit it with your paddle a few more times.
| {
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Fully simplify $\sqrt {14 + 8\sqrt {3}}$. Please help me with this problem: Fully simplify $\sqrt{14 + 8\sqrt {3}}$.
I've tried to assume that the radical is in the form of $a+b\sqrt{3}$, so $a+b\sqrt{3}=\sqrt {14 + 8\sqrt {3}}$, then I squared both sides and got $$a^2+3b^2+2ab\sqrt{3}=14+8\sqrt{3}.$$
So I assumed that $a^2+3b^2$ must be equalto $14$, and $2ab$ must be equal to $-8$. So $ab$ must equal to $-4$. So now I know that $a$ and $b$ must be factors of $-4$. But now I am stuck because the cases $a=-1$, $b=4$, $a=1$, $b=-4$, $a=4$, $b=-1$, and $a=-4$, $b=1$ do not work because of the fact that $a^2+3b^2$ must equal to $14$. (We have that $(-1)^2+3\cdot4^2=1+3\cdot16$ which is clearly not $14$, same goes with $a=1$, $b=-4$; $4^2+3(-1)^2=16+3=19$ which is also not $14$.)
So now I am out of ideas to simplify this expression, or maybe this is already fully simplified? Please help!
| Note
$$\sqrt {14 + 8\sqrt {3}}=\sqrt{2(7+2\sqrt{12})}
=\sqrt{2(\sqrt4+\sqrt3)^2}=\sqrt2(2+\sqrt3)
$$
| {
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How to simplify this expression I'm trying to find a way to simplify $\sqrt{5+\sqrt{24}}$. I know that this expression is equivalent to $\sqrt{2}+\sqrt{3}$ because they are both roots of the equation: $x^4-10x^2+1$ (and the decimal equivalents are the same using a calculator). My question is , how does one get from $\sqrt{5+\sqrt{24}}$ to $\sqrt{2}+\sqrt{3}$ algebraically?
| If you don't want guess-and-check or hope-you-see-it approaches, suppose that $a+\sqrt{b} = \sqrt{5+\sqrt{24}}$ for some $a,b$. (Note that $a$ could be a square root.. it often works out where one of the two terms is an integer and the other is a root - which is why I wrote it in this way.) Squaring both expressions gives us
$$a^2+b + 2a\sqrt{b} = 5+\sqrt{24}.$$
Thus $a^2 + b = 5$ and $2a\sqrt{b} = \sqrt{24}$ or $4a^2b = 24.$ Thus $a^2 = 5-b$ and substituting into the other expression gives
$$4(5-b)b = 24 \Longrightarrow (5-b)b = 6 \Longrightarrow b^2-5b+6 = 0.$$
You will get two solutions to this. I will let you take it from here.
| {
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Indefinite integral of a rational function with linear denominator: $ \int \frac{ x^7}{(x+1)}{dx} $ $$ \int \frac{ x^7}{(x+1)}{dx} $$
$$ \int \frac{ \left(x^7 + x^6 - x^6 - x^5 + x^5 + x^4 -x^4 - x^3 + x^3 + x^2 - x^2 -x^1 + x^1 +1 -1\right ) }{\left(x+1\right)}{dx}$$
$$ \int { \left(x^6 - x^5 + x^4 - x^3 + x^2 -x +1 - \frac{1}{x+1}\right)}{dx}$$
$$\frac{x^7}{7} - \frac{x^6}{6} + \frac{x^5}{5} -\frac{x^4}{4}+ \frac{x^3}{3} - \frac{x^2}{2} + \frac{x^1}{1} -\frac{\log(x+1)}{1} +C $$
Is there any other good way to do this ?
| Let
$$I=\int\frac{x^7}{x+1}dx$$
Apply $x+1\to y$ to get
$$I=\int\frac{(y-1)^7}{y}dy$$
Using the binomial theorem
$$I=\int\frac{1}{y}\sum_{k=0}^{7}\binom{7}{k}y^k(-1)^{7-k}dy$$
$$=\int\sum_{k=0}^{7}\binom{7}{k}y^{k-1}(-1)^{7-k}dy$$
$$=\sum_{k=0}^{7}\int\binom{7}{k}y^{k-1}(-1)^{7-k}dy$$
$$=C-\ln|y|+\sum_{k=1}^{7}\frac{1}{k}\binom{7}{k}y^k(-1)^{7-k}$$
$$=C-\ln|x+1|+\sum_{k=1}^{7}\frac{1}{k}\binom{7}{k}(x+1)^k(-1)^{7-k}$$
At which point you have by most standards a satisfactory answer.
EDIT
If you teacher did some geometric progression stuff this might have been what he did:
$$I=\int\frac{x^7}{x+1}dx$$
$$=\int\frac{y^7}{1-y}dy\qquad(x\to -y)$$
$$=-\int\frac{-1+1-y^7}{1-y}dy$$
$$=\int\frac{1}{1-y}dy-\int\frac{1-y^7}{1-y}dy$$
$$=\ln|1-y|-\int\left(\sum_{n=0}^6y^n\right)dy$$
$$=\ln|1-y|-\sum_{n=0}^6\int y^ndy$$
$$=\ln|1-y|-\sum_{n=1}^7\frac{y^n}{n}$$
$$=\ln|1+x|-\sum_{n=1}^7\frac{(-x)^n}{n}$$
That method there makes use of the partial sum formula for a geometric series.
| {
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Decompose $5^{1985}-1$ into factors
Decompose the number $5^{1985}-1$ into a product of three integers, each of which is larger than $5^{100}$.
We first notice the factorization $x^5-1 = (x-1)(x^4+x^3+x^2+x+1)$. Now to factorize $x^4+x^3+x^2+x+1$ we get $$(x^2+ax+1)(x^2+bx+1) = x^4+(a+b)x^3+(ab+2)x^2+(a+b)x+1 = x^4+x^3+x^2+x+1$$ implies $a+b = 1,ab+2 = 1$. Thus, $$x^4+x^3+x^2+x+1 = (x^2+\left(\frac{1+\sqrt{5}}{2}\right)x+1)(x^2+\left(\frac{1-\sqrt{5}}{2}\right)x+1).$$ Is it possible to continue from this approach because now the factors I have aren't integers or is there a better way?
| I would try Aurifeuillian factorization here.
Let $f(x)=x^4+x^3+x^2+x+1$ be the fifth cyclotomic polynomial. The Aurifeuillian stuff says that $f(5x^2)$ factors into a product as follows
$$
f(5x^2)=\left(25 x^4-25 x^3+15 x^2-5 x+1\right) \left(25 x^4+25 x^3+15 x^2+5
x+1\right).
$$
Call those two factors on the RHS $g_1(x)$ and $g_2(x)$.
Your number is
$$
\begin{aligned}
5^{1985}-1&=(5^{397}-1)f(5^{397})\\
&=(5^{397}-1)g_1(5^{198})g_2(5^{198}),
\end{aligned}
$$
and this fits the bill.
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If $x$ is positive, then why does $\frac{1}{\sqrt{x+1} + \sqrt{x}} = \sqrt{x+1} - \sqrt{x}$? Given that $x$ is positive, $\frac{1}{\sqrt{x+1} + \sqrt{x}} = \sqrt{x+1} - \sqrt{x}$
I've been trying to convert the left side of the equation to the right side:
$$ \frac{1}{\sqrt{x+1} + \sqrt{x}}$$
But then how can I flip this round to be what I have on the right side?
I know that $\frac{1}{\sqrt{x}} = x^{-\frac{1}{2}}$, so I would think that this would give me
$$(x+1)^{-1/2} + x^{-1/2}$$
Which I thought would then convert to
$$-\sqrt{x+1} + (-\sqrt{x}) $$
so I'm not sure how the first part $\sqrt{x+1}$ got to be positive
| You need to use a little trick with fractions. Recall that for any number $n$ that is not zero, $\frac{n}n=1$, and 1 times any number is itself.
So start with the left side: $\sqrt{1+x}-\sqrt{x}$. If we choose our $n$ to be $\sqrt{1+x}+\sqrt{x}$, then:
$$\sqrt{1+x}-\sqrt{x}=\frac{\sqrt{1+x}-\sqrt{x}}1\cdot\frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}+\sqrt{x}}$$
Then "FOIL" the top:
$$\frac{\sqrt{1+x}^2+\sqrt{x}\sqrt{1+x}-\sqrt{x}\sqrt{1+x}-\sqrt{x}^2}{\sqrt{1+x}+\sqrt{x}}=\frac{1+x-x}{\sqrt{1+x}+\sqrt{X}}=\frac1{\sqrt{1+x}+\sqrt{x}}$$
| {
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Solution of Inequality $ \frac{1}{x-6}\le 3$
Solve the inequality: $\displaystyle \frac{1}{x-6}\le 3$
solution:
\begin{align*}\frac{1}{x-6}& \le 3 \\ x-6& \le \frac{1}{3}
\\x& \le 6+\frac{1}{3}\\
x&\le19/3\end{align*}
but, for values of $x\le 6$ also, inequality holds true as left hand side provides some negative value. but, i don't find this from the solution. what am i missing?
| If $x < 6$, then $x - 6$ is negative, so multiplying both sides by $x-6$ reverses the inequality. Thus $1 \ge 3(x - 6)$, or $19 \ge 3x$, which is true for all $x < 6$.
| {
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How to factorise $(x-1)^2 - (x-5)^2$? My attempt:
$a = (x-1)$
$c = (x-5)$
$a^2 - c^2$
which is equal to:
$$((x-1) - (x-5))((x-1)+(x-5))$$
But the correct answer is :
$8(x-3)$
Can you explain, please?
| Your doing is correct, you just need to expand the expressions inside the parenthesis:
\begin{align}((x-1) - (x-5))((x-1)+(x-5))&=(x-1-x+5)(x-1+x-5)\\ & =4(2x-6)\\ & =8(x-3)\end{align}
| {
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How to solve $\int \frac{1}{x^2+4x+7} dx$? How to solve $\int \frac{1}{x^2+4x+7} dx$?
I think the first step is to write it in the following form: $$\int \frac{1}{(x+2)^2+3} dx$$
| With the substitution $u=\frac{x+2}{\sqrt{3}}$:
\begin{align*}\int \frac{1}{(x+2)^2+3} \mathrm{d}x &= \int \frac{1}{3(\frac{x+2}{\sqrt{3}})^2+3} \mathrm{d}x \\ &= \frac13\int \frac{1}{(\frac{x+2}{\sqrt{3}})^2+1} \mathrm{d}x \\ &= \frac{1}{3}\sqrt{3}\int \frac{1}{u^2+1} \mathrm{d}u \\ &= \frac{1}{3}\sqrt{3} \arctan(u)+C \\ &= \frac{1}{3}\sqrt{3} \arctan\left(\frac{x+2}{\sqrt{3}}\right)+C \end{align*}
| {
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Proving that $\lim_{(x,y) \to (0,0)} (x^2 +y^2 -x^3 y^3)/(x^2 +y^2) =1$ How can I go about proving that $$\lim_{(x,y) \to (0,0)} \frac{x^2 +y^2 -x^3 y^3}{x^2 +y^2} = 1 ?$$
I checked some lines along $x, y$ and $x=y$ and it all gave $1$
| An alternative solution without polar coordinate:
$$\lim_{(x,y) \to (0,0)} \frac{x^2 +y^2 -x^3 y^3}{x^2 +y^2} = \lim_{(x,y) \to (0,0)} 1- \frac{x^3 y^3}{x^2 +y^2}$$
To show $\lim_{(x,y) \to (0,0)} \frac{x^3 y^3}{x^2 +y^2} = 0$ is equivalent to show $\lim_{(x,y) \to (0,0)} |\frac{x^3 y^3}{x^2 +y^2}| = 0 $:
Since $$0 \leq |\frac{x^3 y^3}{x^2 +y^2}| = \frac{|x|^3 |y|^3}{x^2 +y^2} \leq^{[1]} \frac{|x|^3 |y|^3}{2|x||y|} = \frac{|x|^2 |y|^2}{2} $$
Where the inequality $[1]$ is with AM-GM inequality.
By Squeeze Theorem, $ 0 \leq \lim_{(x,y) \to (0,0)} |\frac{x^3 y^3}{x^2 +y^2}| \leq \lim_{(x,y) \to (0,0)}\frac{|x|^2 |y|^2}{2} = 0$.
So $\lim_{(x,y) \to (0,0)} |\frac{x^3 y^3}{x^2 +y^2}| = 0 $ which means $\lim_{(x,y) \to (0,0)} \frac{x^3 y^3}{x^2 +y^2} = 0 $
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Proof Verification - Spivak's Calculus - Chapter 1 Question 1.3 In this post, I asked for help getting started on a proof in Spivak's Calculus Fourth Edition, Chapter 1 Question 1.3
Prove that if $x^2 = y^2$, then $x = y$ or $x = −y$ using only the following properties of numbers:
A lot of awesome people responded with hints to get me started:
P7 implies that if $ab = 0$ then $a = 0$ or $b = 0$. Various of the others show that if $x^2 = y^2$ then $\left( x + y \right) \left( x − y \right) = 0$.
Based on that hint, I came up with the following proof:
Note 1
$$
\begin{array} { c c }
x \cdot 0 + x \cdot 0 = x \left( 0 + 0 \right) & By\ P9 \\
x \cdot 0 + x \cdot 0 = x \cdot 0 & By\ P2 \\
x \cdot 0 + x \cdot 0 + \left( -x \cdot 0 \right) = x \cdot 0 + \left( -x \cdot 0 \right) & By\ Addition \\
x \cdot 0 + 0 = 0 & By\ P3 \\
\mathbf{ x \cdot 0 = 0} & By\ P2 \\
\end{array}
$$
Note 2
$$
\begin{array} { c c }
\left( -a \right) \cdot b + a \cdot b = \left[ \left( -a \right) + a \right] \cdot b & By\ P9 \\
\left( -a \right) \cdot b + a \cdot b = 0 \cdot b & By\ P3 \\
\left( -a \right) \cdot b + a \cdot b = 0 & By\ Note\ 1 \\
\left( -a \right) \cdot b + a \cdot b + - \left( a \cdot b \right) = 0 + - \left( a \cdot b \right) & By\ Addition \\
\left( -a \right) \cdot b + 0 = 0 + - \left( a \cdot b \right) & By\ P3 \\
\mathbf{ \left( -a \right) \cdot b = - \left( a \cdot b \right)} & By\ P2 \\
\end{array}
$$
Note 3
$$
\begin{array} { c c }
\left[ x + y \right] \left[ x + (-y) \right] = 0 & \text{By assumption} \\
x \left( x + y \right) + \left( -y \right) \left( x + y \right) = 0 & \text{By P9} \\
x^2 + xy + x \left( -y \right) + y \left( -y \right) = 0 & \text{By P9} \\
x^2 + x \left[ y + \left( -y \right) \right] + y \left( -y \right) = 0 & \text{By P9} \\
x^2 + x \cdot 0 + y \left( -y \right) = 0 & \text{By P3} \\
x^2 + 0 + y \left( -y \right) = 0 & \text{By Note 1} \\
x^2 + y \left( -y \right) = 0 & \text{By P2} \\
x^2 + \left( -y^2 \right) = 0 & \text{By Note 2 for all a = y and b = y} \\
x^2 + \left( -y^2 \right) + y^2 = 0 + y^2 & \text{By Addition} \\
x^2 + 0 = 0 + y^2 & \text{By P3} \\
x^2 = y^2 & \text{By P2} \\
\end{array}
$$
Finally, note that the last line of Note 3 implies the first line of Note 3, and the first line of Note 3 implies $x + y = 0$ or $x - y = 0$. Additionally, P2 and P3 can then be used to show that $x = y$ or $x = -y$, QED.
I feel like I've jumped to a conclusion without proof--namely that the first line of Note 3 is true if and only if the last line of Note 3 is true.
Should I have started at the last line of Note 3 and worked backward to the first Line of Note 3?
| In logic, the statement $p \to q$ does not necessarily mean $q \to p$. ( That is, if $p$ implies $q$, $q$ does not necessarily imply $p$. In other words, if $p$ is true, $q$ is also true--but there are some cases in the universe of discourse when $q$ is true but $p$ is not. )
In the case of this particular question from Spivak's book, the proof above does say that if $\left[ x + y \right]\left[ x + \left( -y \right) \right] = 0$, then $x^2 + y^2 = 0$ is true. And you're right to suspect that--until we prove otherwise--it may be possible there are sets of numbers for which $x^2 + y^2 = 0$ while $\left[ x + y \right]\left[ x + \left( -y \right) \right] \neq 0$.
So, it is important to show that one can get from $x^2 + y^2 = 0$ to $\left[ x + y \right]\left[ x + \left( -y \right) \right] = 0$ using only the number properties given.
$$
\begin{array} { c c }
x^2 = y^2 & Given \\
x^2 + (-y^2) = y^2 + (-y^2) & \text{By Addition} \\
x^2 + (-y^2) = 0 & \text{By P3} \\
x^2 + (-y^2) + 0 = 0 + 0 & \text{By Addition} \\
x^2 + (-y^2) + x \cdot 0 = 0 + 0 & \text{By Note 1} \\
x^2 + (-y^2) + x \cdot 0 = 0 & \text{By P2} \\
x^2 + (-y^2) + x \left[ y + (-y) \right] = 0 & \text{By P3 Substitution} \\
x^2 + (-y^2) + y \cdot x + (-y) \cdot x = 0 & \text{By P9} \\
x \cdot x + y \cdot (-y) + x \cdot y + x \cdot (-y) = 0 & \text{By Note 2} \\
x \cdot x + x \cdot y + x \cdot (-y) + y \cdot (-y) = 0 & \text{By P4} \\
x ( x + y ) + (-y)( x + y ) = 0 & \text{By P9} \\
\left[ x + (-y) \right] ( x + y ) = 0 & \text{By P9} \\
\left[ x + (-y) = 0 \right] \lor \left[ x + y = 0 \right] & \text{By Note 1} \\
\left[ x + (-y) + y = 0 + y \right] \lor \left[ x + y + (-y) = 0 + (-y) \right] & \text{By Addition} \\
\left[ x + 0 = 0 + y \right] \lor \left[ x + 0 = 0 + (-y) \right] & \text{By P3} \\
\left[ x = y \right] \lor \left[ x = (-y) \right] & \text{By P2} \\
QED
\end{array}
$$
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Proving that $\frac{\sqrt{1-x^2_{1}}+\sqrt{1-x^2_{2}}+\ldots+\sqrt{1-x^2_{17}}}{x_{1}+x_{2}+\ldots+x_{17}}\geq 4$
If $x_{1},x_{2},x_{3},x_{4},\ldots,x_{17}\in \left[0,1\right]$ and $x^2_{1}+x^2_{2}+x^2_{3}+\ldots+x^2_{17} = 1$ then prove that $$\frac{\sqrt{1-x^2_{1}}+\sqrt{1-x^2_{2}}+\ldots+\sqrt{1-x^2_{17}}}{x_{1}+x_{2}+\ldots+x_{17}}\geq 4$$
$\bf{My\; Try:}$ Let $y^2_{i}=1-x_{i}^2,\forall i = 1,2,3,\ldots,17$
So $y^2_{1}+y^2_{2}+\ldots+y_{17}^2 = 16$
So here $y^2_{i}\leq y_{i},\forall i = 1,2 ,3,...,17$
So $$y^2_{1}+y^2_{2}+\ldots+y_{17}^2\leq y_{1}+y_{2}+y_{3}+\ldots+y_{17}$$
So $$y_{1}+y_{2}+y_{3}+\ldots+y_{17}\geq 16$$
Now how can I solve it after that? Help required.
| Let $x_i=\frac{a_i}{\sqrt{17}}$.
Hence, we need to prove that $\sum\limits_{i=1}^{17}\sqrt{17-a_i^2}\geq4\sum\limits_{i=1}^{17}a_i$, where $\sum\limits_{i=1}^{17}a_i^2=17$.
Indeed, we need to prove that $\sum\limits_{i=1}^{17}\left(\sqrt{17-a_i^2}-4a_i\right)\geq0$ or
$\sum\limits_{i=1}^{17}\frac{1-a_i^2}{4a_i+\sqrt{17-a_i^2}}\geq0$ or $\sum\limits_{i=1}^{17}\left(\frac{1-a_i^2}{4a_i+\sqrt{17-a_i^2}}+\frac{1}{8}\left(a_i^2-1\right)\right)\geq0$ or
$\sum\limits_{i=1}^{17}\frac{\left(a_i^2-1\right)\left(\sqrt{17-a_i^2}+4a_i-8\right)}{4a_i+\sqrt{17-a_i^2}}\geq0$ or $\sum\limits_{i=1}^{17}\frac{\left(a_i^2-1\right)\left(\sqrt{17-a_i^2}-4a_i+8a_i-8\right)}{4a_i+\sqrt{17-a_i^2}}\geq0$ or
$\sum\limits_{i=1}^{17}\frac{\left(a_i-1\right)^2\left(a_i+1\right)\left(8-\frac{17\left(a_i+1\right)}{4a_i+\sqrt{17-a_i^2}}\right)}{4a_i+\sqrt{17-a_i^2}}\geq0$ or $\sum\limits_{i=1}^{17}\frac{\left(a_i-1\right)^2\left(a_i+1\right)\left(15a_i+8\sqrt{17-a_i^2}-17\right)}{\left(4a_i+\sqrt{17-a_i^2}\right)^2}\geq0$,
for which it remains to prove that $8\sqrt{17-a_i^2}\geq17-15a_i$.
If $17-15a_i<0$ so the previous inequality is obvious.
But for $0\leq a_i\leq\frac{17}{15}$ it remains to prove that $64\left(17-a_i^2\right)\geq\left(17-15a_i\right)^2$ or
$47+30a_i-17a_i^2\geq0$, which is obvious for $0\leq a_i\leq\frac{17}{15}$. Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find the remainder of $2^n+n^2$ modulus 6
Find the remainder of $2^n+n^2$ modulus 6 given that $2^n+n^2$ is a prime and $n\geq2$($n$ positive integer)
I tried to solve this but failed!I just know that $n$ must be odd. No progress at all!!
| $n$ can't be even because then $n^2+2^n$ is even and larger than $2$.
Therefore $n$ is odd and $2^n\equiv 2\bmod 3$. Notice that $n^2$ can only be $1$ or $0\bmod 3$. In the first case $3$ divides $2^n+n^2$ and $n>3$, implying $2^n+n^2$ is not prime.
We conclude $n^2\equiv 0\bmod 3$. Therefore $2^n+n^2$ is odd and $2\bmod 3$. So $2^n+n^2\equiv 5\bmod 6$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1879053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Which number is more near to $S$ Which number is more near to $S$?
$S=\frac{(2^3-1)(3^3-1)\dots (100^3-1)}{(2^3+1)(3^3+1)\dots (100^3+1)}$
1.$0.6$
2.$0.67$
3.$0.667$
4.$0.6667$
I personaly can not do anything special I have tried to calculate some easy case but no resaults any hints?
| \begin{align*}
\prod_{k=2}^{n} \frac{k^3-1}{k^3+1} &=
\prod_{k=2}^{n} \frac{(k-1)(k^2+k+1)}{(k+1)(k^2-k+1)} \\
&= \prod_{k=2}^{n} \frac{k-1}{k+1} \times \frac{k^2+k+1}{(k-1)^2+(k-1)+1} \\
&= \left(
\frac{1}{3} \cdot \frac{2}{4} \cdot \frac{3}{5} \cdot \frac{4}{6} \cdot
\ldots \cdot \frac{n-3}{n-1} \cdot \frac{n-2}{n} \cdot \frac{n-1}{n+1}
\right) \times \frac{n^2+n+1}{3} \\
&= \frac{2(n^2+n+1)}{3n(n+1)} \\
S &= \frac{2(100^2+100+1)}{3(100\times 101)} \\
&= \frac{2}{3}\left(1+\frac{1}{10100} \right) \\
&\approx 0.6667
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluation of $I=\int \frac{1}{\sin (x-a) \sin(x-b) \sin(x-c)}.dx$ Evaluate the given integral:
$$I=\int \frac{1}{\sin (x-a) \sin(x-b) \sin(x-c)}.dx$$
When it was $I=\int \frac{1}{\sin (x-a) \sin(x-b)}.dx$, I solved it by multiplying and dividing by $\sin (b-a)$. But I am not getting how to proceed here..
Could someone help me with this?
| I looked at the partial feaction decomposition method. It is not particularly elegant, but it reduces the calculus problem to a linear algebra one. It is likely a more efficient method for this integral exists, but I am not sure what it would be.
$$\int\frac{dx}{\sin(x-a)\sin(x-b)\sin(x-c)}$$
$$=\int\left(\frac{A}{\sin(x-a)}+\frac{B}{\sin(x-b)}+\frac{C}{\sin(x-c)}\right)dx$$
If $A$, $B$, and $C$ are constants, we can use the identity $\int\csc x\ dx=\ln\left(\tan\frac x2\right)+k.$
$$=A\ln\left(\tan\frac{x-a}2\right)+B\ln\left(\tan\frac{x-b}2\right)+C\ln\left(\tan\frac{x-c}2\right)+k$$
In order to find these constants, we can rewrite $\sin(x-a)=\sin x\cos a-\sin a\cos x.$
$$=\int\frac{\begin{array}A(\sin x\cos b-\sin b\cos x)(\sin x\cos c-\sin c\cos x)\\+B(\sin x\cos a-\sin a\cos x)(\sin x\cos c-\sin c\cos x)\\+C(\sin x\cos b-\sin b\cos x)(\sin x\cos c-\sin c\cos x)\end{array}}{(\sin x\cos a-\sin a\cos x)(\sin x\cos b-\sin b\cos x)(\sin x\cos c-\sin c\cos x)}dx$$
Rearranging, and rewiring some sines using the sum identity:
$$=\int\frac{\begin{array}(A\cos b\cos c+B\cos a\cos c+C\cos a\cos\ b)\sin^2x\\-(A\sin(b+c)+B\sin(a+c)+C\sin(a+b))\sin x\cos x\\+(A\sin b\sin c+B\sin a\sin c+C\sin a\sin b)\cos^2x\end{array}}{(\sin x\cos a-\sin a\cos x)(\sin x\cos b-\sin b\cos x)(\sin x\cos c-\sin c\cos x)}dx$$
The double angle identity can be used to show that $$\alpha\sin^2x+\beta\sin x\cos x+\gamma\cos^2x=\frac12(\alpha+\gamma)+\frac12(-\alpha+\gamma)\cos 2x+\beta\sin 2x,$$ which is equal to $1$ for all $x$ if and only if $\alpha=1,\beta=0,\gamma=1$. This gives a linear system of equations.
$$A\cos b\cos c+B\cos a\cos c+C\cos a\cos b=1$$
$$A\sin b\sin c+B\sin a\sin c+C\sin a\sin b=1$$
$$A\sin(b+c)+B\sin(a+c)+C\sin(a+b)=0$$
Rewriting as a matrix:
$$\begin{bmatrix}\cos b\cos c & \cos a\cos c & \cos a\cos b\\
\sin b\sin c & \sin a\sin c & \sin a\sin b\\
\sin(b+c) & \sin(a+c) & \sin(a+b)\\\end{bmatrix}\begin{bmatrix}A\\B\\C\\\end{bmatrix}=\begin{bmatrix}1\\1\\0\\\end{bmatrix}$$
For specific values of $a$, $b$, and $c$, this equation above can be solved by row reduction. Technically, the inverse of the $3\times3$ matrix above can be computed to find the general solution, but it would be a rather cumbersome process.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
What is the square root of $3 + 2\sqrt{10}i$? I need to compute the square root of $3 + 2\sqrt{10}i$.
I know how to solve it, but for some reason I'm not getting the correct answer. I attempted to solve it like this:
$$
\sqrt{3 + 2\sqrt{10}i} = x + iy
\quad \longrightarrow \quad
3 + 2\sqrt{10}i = x^2 - y^2 +2xyi
$$
and so forth, but my answer isn't correct.
| You want $z=a+bi$ where $a,b\in\mathbb{R}$ such that
$$z^2=a^2-b^2+2abi=3+2\sqrt{10}i.$$
Comparing coefficients, you need $a^2-b^2=3$ and $ab=\sqrt{10}$. So $a=\sqrt{5}$ and $b=\sqrt{2}$, or $a=-\sqrt{5}$ and $b=-\sqrt{2}$. Thus the square root of $3+2\sqrt{10}$:
$$\pm(\sqrt{5}+\sqrt{2}i).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Evaluating the indefinite integral $\int\frac{x\arctan x}{(1+x^2)^2}dx$ The integral again is $$\int\frac{x\arctan x}{(1+x^2)^2}dx$$
I'm actually pretty much just stumped on this one. It looks like it should be amenable to a substitution kinda like $\arctan x$
but that obviously does not work.
I also thought maybe by parts, since
$$
\int\frac{\arctan x}{1+x^2}dx
$$
Isn't too bad to anti differentiate but this approach seems to complicate things on the differentiation side.
| Here is a way by direct integration by parts:
Let $u = \arctan x, dv = \frac{x}{(1+x^2)^2}dx$, then $du = \frac{1}{x^2+1}dx, v = -\frac{1}{2(x^2+1)}$.
Then $$\int\frac{x\arctan x}{(1+x^2)^2}dx =-\frac{1}{2(x^2+1)}\arctan x+\int \frac{1}{2(x^2+1)}\frac{1}{x^2+1}dx=\int \frac{1}{2(x^2+1)^2}dx$$
Try some trigonometric subsititution for $\int \frac{1}{2(x^2+1)^2}dx$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1886601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How many $5$-digit numbers can be made from digits in the number $75226522$? I'm having troubles with this one:
How many $5$-digit numbers can be made from digits in the number $75226522$ ?
So there is: one seven, one six, two fives, and four twos in that number.
So I guess I need to use combination with repetition but I'm having trouble with understanding this.
I know how many $8$-digit numbers there are: $$ \frac{8!}{4!2!}$$ But I'm clueless in $5$-digit case
| Add up the following:
*
*The number of ways to rearrange $\color\red{2222}\color\green{ 5}\color\orange{ }\color\lightblue{ }$, which is $\frac{5!}{\color\red{4!}\color\green{1!}\color\orange{ }\color\lightblue{ }}= 5$
*The number of ways to rearrange $\color\red{2222}\color\green{ }\color\orange{6}\color\lightblue{ }$, which is $\frac{5!}{\color\red{4!}\color\green{ }\color\orange{1!}\color\lightblue{ }}= 5$
*The number of ways to rearrange $\color\red{2222}\color\green{ }\color\orange{ }\color\lightblue{7}$, which is $\frac{5!}{\color\red{4!}\color\green{ }\color\orange{ }\color\lightblue{1!}}= 5$
*The number of ways to rearrange $\color\red{ 222}\color\green{55}\color\orange{ }\color\lightblue{ }$, which is $\frac{5!}{\color\red{3!}\color\green{2!}\color\orange{ }\color\lightblue{ }}=10$
*The number of ways to rearrange $\color\red{ 222}\color\green{ 5}\color\orange{6}\color\lightblue{ }$, which is $\frac{5!}{\color\red{3!}\color\green{1!}\color\orange{1!}\color\lightblue{ }}=20$
*The number of ways to rearrange $\color\red{ 222}\color\green{ 5}\color\orange{ }\color\lightblue{7}$, which is $\frac{5!}{\color\red{3!}\color\green{1!}\color\orange{ }\color\lightblue{1!}}=20$
*The number of ways to rearrange $\color\red{ 222}\color\green{ }\color\orange{6}\color\lightblue{7}$, which is $\frac{5!}{\color\red{3!}\color\green{ }\color\orange{1!}\color\lightblue{1!}}=20$
*The number of ways to rearrange $\color\red{ 22}\color\green{55}\color\orange{6}\color\lightblue{ }$, which is $\frac{5!}{\color\red{2!}\color\green{2!}\color\orange{1!}\color\lightblue{ }}=30$
*The number of ways to rearrange $\color\red{ 22}\color\green{55}\color\orange{ }\color\lightblue{7}$, which is $\frac{5!}{\color\red{2!}\color\green{2!}\color\orange{ }\color\lightblue{1!}}=30$
*The number of ways to rearrange $\color\red{ 22}\color\green{ 5}\color\orange{6}\color\lightblue{7}$, which is $\frac{5!}{\color\red{2!}\color\green{1!}\color\orange{1!}\color\lightblue{1!}}=60$
*The number of ways to rearrange $\color\red{ 2}\color\green{55}\color\orange{6}\color\lightblue{7}$, which is $\frac{5!}{\color\red{1!}\color\green{2!}\color\orange{1!}\color\lightblue{1!}}=60$
Hence the total number of ways is $5+5+5+10+20+20+20+30+30+60+60=265$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1886901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Functional Equation $f(f(x))=\frac{x}{2}+3$ Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ of class $C^1$ where $f\circ f(x)=\dfrac{x}{2}+3$.
I need help with this problem.
| Let $f^{\circ n}$ be the shorthand for composing $f$ with itself for $n$ times.
Let $g(x) = f(x+6)-6$. Since $f \in C^1$, so does $g$. Notice
$$g\left(\frac{x}{2}\right) = f\left(\frac{x+6}{2}+3\right)-6 = f(f^{\circ 2}(x+6)))-6
= f^{\circ 2}(f(x+6)) - 6\\ = \frac{f(x+6)}{2} + 3 - 6 = \frac{f(x+6)-6}{2} = \frac{g(x)}{2}$$
When $x = 0$, this implies $g(0) = 0$.
When $x \ne 0$, divide both sides by $\frac{x}{2}$ gives us
$\displaystyle\;\frac{g(x)}{x} = \frac{g(x/2)}{x/2}$.
Repeat applying this, we find
$\displaystyle\;\frac{g(x)}{x} = \frac{g(x/2^k)}{x/2^k}$ for any integer $k > 0$.
As a result,
$$\frac{g(x)}{x} = \lim_{k\to\infty}\frac{g(x/2^k)}{x/2^k}
\stackrel{\color{blue}{\because\text{ RHS exists}}}{=} \lim_{h\to 0} \frac{g(h)}{h} = \lim_{h\to 0}\frac{g(h)-g(0)}{h} \stackrel{\color{blue}{g \in C^1}}{=} g'(0)
$$
This means for some constant $a$,
$$g(x) = ax \quad\implies\quad f(x) = a(x-6)+6
\quad\implies\quad f(f(x)) = a^2(x-6) + 6
$$
Compare rightmost expression with the functional equation $f(f(x)) = \frac{x}{2} + 3$,
we find
$$a = \pm \frac{1}{\sqrt{2}}
\quad\implies\quad f(x) = \pm\frac{1}{\sqrt{2}}(x-6) + 6$$
This means the original functional equation has two and only two solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1887028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Prove that $\lnot (p \implies q)$ is equivalent to $p \land \lnot q$? By equivalent I mean the biconditional, as in
$$\lnot (p \implies q) \iff p \land \lnot q$$
Given the definition of implication, I understand why this is true, but I need a bit of help showing this with a formal proof using rules like $\lor-\text{Elim}$ and $\bot-\text{Intro}$.
| $$\newcommand{\DeductionBox} [1]{\begin{array} {l|} #1 \\ \hline \end{array}}$$
The reverse direction is straightforward:
$$\begin{array} {l}
\begin{array} {rlr}
(1) & p \land \lnot q & \text{Given} \\
(2) & p & \land\text{-Elimination of }(1) \\
(3) & \lnot q & \land\text{-Elimination of }(1) \\
\end{array}\\
\\
\DeductionBox{
\begin{array} {rlr}
\quad (4) & p \implies q & \text{New Assumption} \\
\quad (5) & q & \implies\text{-Elimination of }(4) \text{ and } (2) \\
\quad (6) & \bot & \bot\text{-Introduction of }(5) \text{ and }(3)
\end{array} \\
} \\
\\
\begin{array} {rlr}
(7) & (p \implies q) \implies \bot & \quad\implies-\text{Introduction of }(4) \text{ to } (6) \\
(8) & \lnot (p \implies q) & \lnot-\text{Introduction of }(7) \\
\end{array}
\end{array}$$
$$~$$
$$~$$
The foward direction can be done using the law of the excluded middle:
$$\begin{array} {l}
\begin{array} {rlr}
(1) & \lnot (p \implies q) & \text{Given} \\
(2) & \bot \implies \lnot q & \text{Vaccous Implication } \\
\end{array}\\
\\
\DeductionBox{
\begin{array} {rlr}
\quad (3) & p & \text{New Assumption} \\
\end{array} \\
\\
\DeductionBox{
\begin{array} {rlr}
\quad \quad (4) & q & \text{New Assumption} \\
\end{array} \\
\\
\DeductionBox{
\begin{array} {rlr}
\quad \quad \quad (5) & p & \text{New Assumption} \\
\quad \quad \quad (6) & q & \text{Copy of }(4) \\
\end{array} \\
}\\
\\
\begin{array} {rlr}
\quad \quad (7) & p \implies q & \implies\text{-Introduction of }(5) \text{ to }(6) \\
\quad \quad (8) & \bot & \bot\text{-Introduction of }(7) \text{ and }(1) \\
\quad \quad (9) & \lnot q & \implies\text{-Elimination of }(2) \text{ and }(8) \\
\quad \quad (10) & p \land \lnot q & \land\text{-Introduction of }(3) \text{ and }(9) \\
\end{array} \\
}\\
\\
\DeductionBox{
\begin{array} {rlr}
\quad \quad (11) & \lnot q & \text{New Assumption} \\
\quad \quad (12) & p \land \lnot q & \land\text{-Introduction of }(3)\text{ and }(11) \\
\end{array} \\
}\\
\\
\begin{array} {rlr}
\quad (13) & q \implies (p \land \lnot q) & \implies\text{-Elimination of }(4)\text{ to }(10) \\
\quad (14) & \lnot q \implies (p \land \lnot q) & \implies\text{-Elimination of }(11)\text{ to }(12) \\
\quad (15) & q \lor \lnot q & \text{Law of the Excluded Middle} \\
\quad (16) & p \land \lnot q & \lor\text{-Elimination of }(15),(13),(14) \\
\end{array} \\
} \\
\\
\DeductionBox{
\begin{array} {rlr}
\quad (17) & \lnot p & \text{New Assumption } \\
\end{array}\\
\\
\DeductionBox{
\begin{array} {rlr}
\quad \quad (18) & p & \text{New Assumption} \\
\quad \quad (19) & \bot & \bot\text{-Introduction of }(18)\text{ and }(17) \\
\quad \quad (20) & \bot \implies q & \text{Vaccuous Implication} \\
\quad \quad (21) & q & \implies\text{-Elimination of }(20)\text{ and }(19) \\
\end{array} \\
}\\
\\
\begin{array} {rlr}
\quad (22) & p \implies q & \implies\text{-Introduction of }(18)\text{ to }(21) \\
\quad (23) & \bot & \implies\text{-Elimination of }(1)\text{ and }(22) \\
\quad (24) & \lnot q & \implies\text{-Elimination of }(2)\text{ and }(23) \\
\quad (25) & p \land \lnot q & \land\text{-Introduction of }(3)\text{ and }(24) \\
\end{array}\\
}\\
\begin{array} {rlr}
(26) & p \implies (p \land \lnot q) & \implies\text{-Elimination of }(3)\text{ to }(16) \\
(27) & \lnot p \implies (p \land \lnot q) & \implies\text{-Elimination of }(17)\text{ to }(25) \\
(28) & p \lor \lnot p & \text{Law of the Excluded Middle} \\
(29) & p \land \lnot q & \lor\text{-Elimination of }(28),(26),(27) \\
\end{array}\\
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1887802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Which of the following sets are orthonormal? I have three sets to determine whether they're orthonormal or not. These are;
(a) $\{1,\frac{x}{\sqrt2}+\frac{x^3}{\sqrt2},\frac{x^2}{\sqrt2}-\frac{x^4}{\sqrt2}$}
(b) $\{\frac{x}{2}-\frac{x^2}{2}-\frac{x^3}{2}, -\frac{x}{2}-\frac{x^2}{2}+\frac{x^4}{2}, \frac{x^2}{2}-\frac{x^3}{2}$}
(c) $\{\frac{x^3}{3}-\frac{2x^4}{3}+\frac{2x^6}{3}, \frac{x^2}{3}+\frac{2x^5}{3}-\frac{2x^7}{3}$}
The inner product for two arbitrary polynomials is
$$a.b= \sum_{k=0}^n a_kb_k$$
I know that to prove that they are I need to do the dot product of each pair and for them to all equal 0 and that the magnitude of each should equal 1 for the sets to be determined as orthonormal but I'm having trouble starting off with this exercise, the x's are throwing me off and I'm wondering if there's something else I have to do before working out the magnitude and scalar product. Thanks in advance.
| In polynomial space, the standard orthonormal basis (with respect to the inner-product you gave above) is $\{1, x, x^2, x^3, ...\}$ and so, for example, a polynomial such as $2 + 3x^2 - 4x^4$ can be represented by the vector $(2, 0, 3, 0, -4, 0, ...)$ since
$$2\cdot 1 + 0 \cdot x + 3 \cdot x^2 + 0 \cdot x^3 + (-4) \cdot x^4 = 2 + 3x^2 - 4x^4.$$
In your case, for (a), you then have the identifications:
$$
\left\{1,\frac{x}{\sqrt2}+\frac{x^3}{\sqrt2},\frac{x^2}{\sqrt2}-\frac{x^4}{\sqrt2}\right\} \longrightarrow \left\{ \begin{pmatrix}1 \\0\\0\\0\\0\\ 0\\\vdots\end{pmatrix}, \begin{pmatrix}0\\1/\sqrt{2}\\0\\1/\sqrt{2}\\0\\0\\\vdots\end{pmatrix}, \begin{pmatrix} 0 \\0\\1/\sqrt{2}\\0\\-1/\sqrt{2}\\ 0\\\vdots\end{pmatrix} \right\}
$$
Now that you have these understood componentwise, you can use the inner-product you have given.
| {
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} |
Find $x$ and $y$ where $20!=\overline{24329020081766xy\dots}$ Find $x$ and $y$ where $20!=\overline{24329020081766xy\dots}$(without using calculator.)
My attempt:I first find how many zeroes does it have:
$$\left\lfloor {\frac{20}{5}} \right\rfloor=4.$$
It can be solved easily if we know that after $y$ there are only three digits then we can know:
$$y=0.$$
Then $\overline {6x}$ is divisible by $4$ which gives us:
$$x=4\ \ \ \text{ or }x=8.$$
Then if we check divisiblity role of $8$ we will get that $\overline {66x}$ is divisible by $8$ that tells to us $x$ can only be $4$. Thus
$$x=4.$$
But know the biggest problem is that we don't know how many digits are there after $y$.
Or in a bigger amount how many digits are there in $20!$.
Thanks.
| Some judicious pairing of the numbers from $1$ to $20$ leads to the rough estimate
$$\begin{align}
20!&=(20\cdot1)(17\cdot3)(19\cdot2)(12\cdot10)(15\cdot4)(18\cdot5)(16\cdot6)(14\cdot7)(13\cdot8)(11\cdot9)\\
&\approx20\cdot50\cdot40\cdot120\cdot60\cdot100\cdot100\cdot100\cdot100\cdot100\\
&\approx1000\cdot5000\cdot60\cdot10^{10}\\
&=10^3\cdot300000\cdot10^{10}\\
&=3\cdot10^{18}
\end{align}$$
Some extra work could probably establish rigorous upper and lower bounds
$$2\cdot10^{18}\lt20!\lt3\cdot10^{18}$$
Knowing that $20!$ ends with $4$ $0$'s and counting that there are $14$ digits before the $xy$, we see that $y$ is the first of the trailing $0$'s. Finally, knowing that $9$ divides $20!$, we have
$$(2+4+3)+2+(9+0)+2+(0+0+8+1)+7+6+6+x\equiv5+x\equiv0\mod9$$
implies $x=4$.
Added later: Here is a second approach, using that fact that $20!$ is divisible by both $9$ and $11$.
We have
$$20!\lt20^{10}\cdot10!\lt2^{10}\cdot10^{10}\cdot(4\cdot10^6)=4096\cdot10^{16}\lt5\cdot10^{19}$$
so $20!$ has at most $20$ digits. Since it ends in $4$ $0$'s, everything to the right of the $xy$ is a $0$. Using the digit sum test for divisibility by $9$, we get
$$x\equiv4-y\mod9$$
Using the alternating digit sum test for divisibility by $11$, we get
$$x\equiv4+y\mod 11$$
The restriction $0\le x,y\le9$ makes it easy to check that $x=4, y=0$ is the only solution.
| {
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Counting multiple of $3$ in a triangular arrangement of the integers. I write
:$$\mathbb{N}_\triangle = \begin{matrix}
&&&&&21&\ldots \\
&&&&15&20&\ldots \\
&&&10&14&19&\ldots \\
&&6&9&13&18&\ldots \\
&3&5&8&12&17&\ldots \\
1&2&4&7&11&16&\ldots
\end{matrix}$$
Can we determine how many numbers in any given column of $\mathbb{N}_\triangle$ are divisible by $3$. I need help proving the following claim:
If $f_3(n)$ counts the numbers in the $n^{th}$ column of
$\mathbb{N}_\triangle$ that are divisible by $3$ then $$f_3(n)
=\Bigg\lfloor{n+2 \above 1.5pt 3}\Bigg\rfloor$$
For example $$f_3(6) = \Bigg\lfloor{6+2 \above 1.5pt 3}\Bigg\rfloor=\Bigg\lfloor{8 \above 1.5pt 3}\Bigg\rfloor=\Bigg\lfloor 2.66666\Bigg\rfloor =2$$ and surely there are $2$ numbers in column six that are divisible by $3$. I suspect there are no multiples of $3$ on the base of the triangle since possibly $$\Bigg(3, 1+ {n(n-1) \above 1.5pt 2}\Bigg)=1$$ In fact if $(3,T_n)=1$ then I suspect that there are no numbers in the $n^{th}$ row that are divisible by $3$. Surely there are consecutive triangular numbers divisible by $3$.
| The number of multiples of $3$ in the interval $[1,N]$ is $\lfloor N/3\rfloor$.
Moreover the $n^{th}$ column of $\mathbb{N}_\triangle$ is the set of consecutive integers from $T_{n-1}+1$ to $T_n$ with $T_n=n(n+1)/2$.
Hence, the number of multiples of $3$ in the $n^{th}$ column of
$\mathbb{N}_\triangle$ is
$$f_3(n)=\lfloor T_{n}/3\rfloor-\lfloor T_{n-1}/3\rfloor=
\left\lfloor \frac{n(n+1)}{6}\right\rfloor-
\left\lfloor \frac{n(n-1)}{6}\right\rfloor
=\left\lfloor \frac{n+1}{3}\right\rfloor.$$
The last equality can be proven easily by letting $n=6k+r$. In fact, after the substitution, we obtain
$$\left\lfloor \frac{r(r+1)}{6}\right\rfloor-
\left\lfloor \frac{r(r-1)}{6}\right\rfloor
=\left\lfloor \frac{r+1}{3}\right\rfloor$$
which can be verified by hand for $r=0,1,2,3,4,5$.
| {
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} |
coefficient of $x^9$ in binominal expansion Find the co efficient of $x^9$ in the expansion of $$\left[\frac{(1+x)^3}{(1-x)^3}\right]^\frac{1}{2}$$
I get answer as $\frac{3}{128}$
But in my book the answer is given as $\frac{1}{128}$
I was very much confused .someone help me please.
| You could notice that $$\left(\frac{(1+x)^3}{(1-x)^3}\right)^\frac{1}{2}=\left(\frac{1+x}{1-x}\right)^\frac{3}{2}=\left(1+2\sum_{i=1}^\infty x^i\right)^\frac{3}{2}$$ Now, being patient, you could use the generalized binomial theorem.
If you are not, Taylor series (tedious too) would lead to
$$\left(\frac{(1+x)^3}{(1-x)^3}\right)^\frac{1}{2}=1+3 x+\frac{9 x^2}{2}+\frac{11 x^3}{2}+\frac{51 x^4}{8}+\frac{57 x^5}{8}+\frac{125
x^6}{16}+\frac{135 x^7}{16}+\frac{1155 x^8}{128}+\frac{1225 x^9}{128}+\frac{2583
x^{10}}{256}+O\left(x^{11}\right)$$ and the result given by Wolfram Alpha, as commented by Arthur.
This coefficient is in fact $\frac {3472875}{9!}=\frac{1225 }{128}$.
| {
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"answer_count": 1,
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A simpler way to solve this determinant equation-$2$
Question Statement:-
Solve the following determinant equation
$$\begin{vmatrix}
x & 2 & 3 \\
6 & x+4 & 4 \\
7 & 8 & x+8 \\
\end{vmatrix}=0$$
My Solution:-
$$\begin{vmatrix}
x & 2 & 3 \\
6 & x+4 & 4 \\
7 & 8 & x+8 \\
\end{vmatrix}=
\begin{vmatrix}
x+13 & x+14 & x+15 \\
6 & x+4 & 4 \\
7 & 8 & x+8 \\
\end{vmatrix}\tag{$R_1\rightarrow R_1+R_2+R_3$}$$
$$(x+13)\begin{vmatrix}
1 & 1 & 1 \\
6 & x+4 & 4 \\
7 & 8 & x+8 \\
\end{vmatrix}+
\begin{vmatrix}
0 & 1 & 2 \\
6 & x+4 & 4 \\
7 & 8 & x+8 \\
\end{vmatrix}\tag{1}$$
Now lets solve the two matrices obtained in the last step above seperately.
$$(x+13)\begin{vmatrix}
1 & 1 & 1 \\
6 & x+4 & 4 \\
7 & 8 & x+8 \\
\end{vmatrix}$$
$$=(x+13)\begin{vmatrix}
0 & 0 & 1 \\
2 & x & 4 \\
-(x+1) & -x & x+8 \\
\end{vmatrix}\hspace{3cm}(C_1\rightarrow C_1-C_3, C_2\rightarrow C_2-C_3)$$
$$=x(x+13)\begin{vmatrix}
0 & 0 & 1 \\
2 & 1 & 4 \\
-(x+1) & -1 & x+8 \\
\end{vmatrix}=x(x+13)(x-1)$$
And the second matrix can be simplified as
$$\begin{vmatrix}
0 & 1 & 0 \\
6 & x+4 & -4-2x \\
7 & 8 & x-8 \\
\end{vmatrix}\tag{$C_3\rightarrow C_3-C_2$}$$
$$=-1\times(6x-48+28+14x)=-20(x-1)$$
So, we get
$$\begin{vmatrix}
x & 2 & 3 \\
6 & x+4 & 4 \\
7 & 8 & x+8 \\
\end{vmatrix}=0\implies x(x+13)(x-1)-20(x-1)=0\\
\implies (x-1)(x^2+13x-20)=0\implies x=1,\dfrac{-13\pm\sqrt{249}}{2}$$
As can be seen, my solution is way long and it is no better than opening the determinant. So if anyone can provide me with a better, simpler and a more intuitive solution I will be very thankful.
| Your solution is fine but here is a more simple way to solve this problem:
$$\begin{vmatrix}
x & 2 & 3 \\
6 & x+4 & 4 \\
7 & 8 & x+8 \\
\end{vmatrix} $$
= $$-\begin{vmatrix}-\begin{pmatrix}
0 & 2 & 3 \\
6 & 4 & 4 \\
7 & 8 & 8 \\
\end{pmatrix} - x I)\end{vmatrix} = 0$$ where $I$ is the identity matrix.
It lookes similar to the equation for eigenvalue.
So $x$ is eigenvalue for matrix:$$A :=\begin{pmatrix}
0 & -2 & -3 \\
-6 & -4 & -4 \\
-7 & -8 & -8 \\
\end{pmatrix}$$
Because $trace(A) = -12$ and $det(A) = -20$ so the characteristic polynomial of form: $-x^3 -12 x^2 +B x -20$. Where $B$ is an real number.
How to find $B$?
Let $x = 2$ into $det(-A-xI)$,then $det(-A-xI) = -10 = -76+2B \implies B=33$.
So it becomes solve this polynomial:$-x^3 -12 x^2 +33 x -20=0$
EDIT: Another way to find B is to use $x = 1$ as subsitution, as Francesco suggested in comment. Since the row become dependent then the determinant of matrix becomes zero so $-1-12-20+B =0 \implies B = 33$
| {
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Prove that $b^2-4ac \equiv 0 \pmod{p}$
Let $a,b,c,r$ be integers and $p \geq 5$ be a prime such that $ax^2+bx+c \equiv a(x-r)^2 \pmod{p}$. Prove that $b^2-4ac \equiv 0 \pmod{p}$. Does the same hold if $p = 3$?
I wasn't sure how to go about proving this. We can rearrange the congruence to get $$x(b+2ar)+c-ar^2 \equiv 0 \pmod{p}.$$ How do we continue from here?
| Since $ax^2+bx+c \equiv a(x-r)^2 \pmod{p}$ is a "forall" statement for all the x in least residue system with respect to p,
$x(b+2ar)+c-ar^2 \equiv 0 \pmod{p}$ also satisfies this condition.
Since $0$ and $1$ are in the least residue system with respect to p, Therefore
$$0 \times (b+2ar)+c-ar^2 \equiv c-ar^2 \equiv 0 \pmod{p} \tag{1}$$
$$1 \times (b+2ar)+c-ar^2 \equiv (b+2ar) \equiv 0 \pmod{p} \tag{2}$$
Then from $(1)$ and $(2)$, $b^2 - 4ac \equiv (-2ar)^2 - 4a(ar^2) \equiv 4a^2r^2- 4a^2r^2 \equiv 0 \pmod{p}$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
find $f^{(15)}(0)$ if $f = \frac{10x^2+12x+4}{(x+2) (x^2 +1)}$ $$f = \frac{10x^2+12x+4}{(x+2)(x^2 +1)}=\frac{4}{x+2}+\frac{6x}{x^2+1}=\frac{2}{1-\left(-\frac x 2 \right)}+\frac{6x}{1-(-x^2)}$$
So we end up with $$2 \sum_{n=0}^{15} \left(-\frac x 2 \right)^n + 6x \sum_{n=0}^{15} (-x^2)^n$$ but because we are looking for $f^{(15)}(0)$ we get the result of $2$?
| We have
\begin{align*}
f(x) = \frac{4}{x+2} + \frac{3}{x+i} + \frac{3}{x-i}
\end{align*}
and hence
\begin{align*}
f^{15}(x) = -(15!)(4(x+2)^{-16} + 3(x+i)^{-16} + 3(x-i)^{-16}
\end{align*}
and
\begin{align*}
f^{15}(0) = -(2^{-14}+6)\cdot 15!
\end{align*}
| {
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"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
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Show that, for all $n > 1: \log \frac{2n + 1}{n} < \frac1n + \frac{1}{n + 1} + \cdots + \frac{1}{2n} < \log \frac{2n}{n - 1}$ I'm learning calculus, specifically limit of sequences and derivatives, and need help with the following exercise:
Show that for every $n > 1$,
$$\log \frac{2n + 1}{n} < \frac1n + \frac{1}{n + 1} + \cdots + \frac{1}{2n} < \log \frac{2n}{n - 1} \quad \quad (1)$$
Important: this exercise is the continuation of a previous problem showing that, for every $n > 1$,
$$\frac{1}{n + 1} < \log(1 + \frac1n) < \frac1n \quad \quad (2)$$
A detailed solution of the latter inequality using the MVT can be found here.
Now back to inequality $(1)$. My first guess was to use mathematical induction in order to prove it but I didn't get far. I think I should make use of inequality $(2)$ from the previous exercise but I'm stuck here.
| For the LHS it is $$\frac { 1 }{ n } +\frac { 1 }{ n+1 } +\cdots +\frac { 1 }{ 2n } =\int _{ n }^{ n+1 }{ \frac { 1 }{ n } dx } +\int _{ n+1 }^{ n+2 }{ \frac { 1 }{ n+2 } dx } +...+\int _{ 2n }^{ 2n+1 }{ \frac { 1 }{ 2n } dx } >\\ >\int _{ n }^{ n+1 }{ \frac { 1 }{ x } dx } +\int _{ n+1 }^{ n+2 }{ \frac { 1 }{ x } dx } +...+\int _{ 2n }^{ 2n+1 }{ \frac { 1 }{ x } dx } =\int _{ n }^{ 2n+1 }{ \frac { 1 }{ x } } dx=\log \frac { 2n+1 }{ n } $$
and for the RHS it is $$\frac { 1 }{ n } +\frac { 1 }{ n+1 } +\cdots +\frac { 1 }{ 2n } =\int _{ n }^{ n+1 }{ \frac { 1 }{ n } dx } +\int _{ n+1 }^{ n+2 }{ \frac { 1 }{ n+2 } dx } +...+\int _{ 2n }^{ 2n+1 }{ \frac { 1 }{ 2n } dx } <\\ <\int _{ n-1 }^{ n }{ \frac { 1 }{ x } dx } +\int _{ n }^{ n+1 }{ \frac { 1 }{ x } dx } +...+\int _{ 2n-1 }^{ 2n }{ \frac { 1 }{ x } dx } =\int _{ 2n-1 }^{ 2n }{ \frac { 1 }{ x } } dx=\log \frac { 2n }{ n-1 } \\ $$
| {
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If $A+B+C=\frac {\pi}{2}$ then prove that.... If $A+B+C=\frac {\pi}{2}$ thrn prove that:$$\tan (A+B) [\tan A-\tan B]=\sqrt {1+\cot^2 C} (\sec A\cos B-\cos A\sec B)$$
My attempt
Here, $$A+B+C=\frac {\pi}{2}$$
$$A+B=\frac {\pi}{2} - C$$
Now,
$$L.H.S.=\tan(A+B) [\tan A-\tan B]
= \tan\left(\frac{\pi}{2} - C\right) \left(\frac{\sin A}{\cos A} - \frac{\sin B }{\cos B} \right)$$
$$=\cot C \frac{\sin (A-B)}{\cos A \cos B}$$
now, plz help me to continue from here.
| \begin{align*}
\cot C \frac{\sin (A-B)}{\cos A \cos B} &= \frac{\cos C}{\sin C}\frac{\sin (A-B)}{\cos A \cos B} \\
&= \frac{\sin(A+B)}{\cos(A+B)}\frac{\sin (A-B)}{\cos A \cos B} \\
&=\frac{\frac{\cos 2B - \cos 2A}{2}}{\sin C(\cos A \cos B)}\\
&= \frac{\cos^2 B - \cos^2 A}{\sin C \cos A \cos B}\\
&= \frac{1}{\sin C}(\sec A\cos B-\cos A\sec B)\\
&= \text{cosec }C(\sec A\cos B-\cos A\sec B)\\
&= \sqrt{1+\cot^2 C}(\sec A\cos B-\cos A\sec B)
\end{align*}
| {
"language": "en",
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"source": "stackexchange",
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How do I find the tangent of an equation, when the $x$-coordinate is $1$? Find the equation of the tangent to the graph $y = x^2 + 4x + 3$, whose $x$ coordinate is $1$.
Write your answer in form $y = mx + c$
I got the equation $y = 2x^2 + 2x + 2$.
Is this correct?
| $$y=x^2+4x+3$$
$$\frac{dy}{dx}=2x+4$$
At $x=1$, y is 8 and the derivative is 6, so write
$y=mx+c\rightarrow8=6(1)+c$; we get $c=2$. Hence the tangent at $x=1$ is $y=6x+2$.
| {
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What is the probability of selecting three different integers, without replacement, from the numbers $1$ to $15$ so their sum is divided by $3$?
What is the probability of selecting three different integers, without replacement, from the numbers $1$ to $15$, inclusive, so that their sum can be exactly divided by $3$?
My solution:
Number of ways of selecting $3$ numbers (so that their sum can be exactly divided by $3$) either,
*
*$3$ numbers that are $0$ modulo $3$
*$3$ numbers that are $1$ modulo $3$
*$3$ numbers that are $2$ modulo $3$
*$3$ numbers such that one is $0$ modulo $3$, one is $1$ modulo $3$ and one is $2$ modulo $3$
so, there are $3\cdot {_5\mathbf C}_3 + 5\cdot5\cdot5 = 155$ favorable outcomes, ${_{15}\mathbf C }_3 = 455$
so, the probability is $\frac{155}{455}$
isn't the procedure correct?
| Correct. Here it is a high-tech alternative. Let we consider:
$$q(x,y)=(1+yx)(1+yx^2)(1+yx^3)\cdot\ldots(1+yx^{15}).$$
If we take the coefficient of $y^3$ in the expansion of $q(x,y)$ we get a polynomial of the $x$-variable, that evaluated at $x=1$ gives the number of subsets of $\{1,2,\ldots,15\}$ with three elements:
$$ [y^3](1+y)^{15} = \binom{15}{3}.$$
If we consider the coefficient of $y^3$ in $q(x,y)$ and we evaluate it at $x=1, x=\omega=\exp\left(\frac{2\pi i}{3}\right)$ and $x=\omega^2$, we get that the number of subsets of $\{1,2,\ldots,15\}$ with three elements adding to a multiple of three is given by:
$$ \frac{1}{3}\left[\binom{15}{3}+2[y^3](1+y)^5(1+\omega y)^5(1+\omega^2 y)^5\right]$$
that is:
$$ \frac{1}{3}\left[\binom{15}{3}+2[y^3](1+y^3)^5\right]=\frac{1}{3}\left[\binom{15}{3}+10\right]=\color{red}{155}.$$
| {
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"source": "stackexchange",
"question_score": "2",
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Find the co-efficient of $x^3y^3zw^2$ in the expression of $(x-y+2z-2w)^9$.
Find the co-efficient of $x^3y^3zw^2$ in the expression of $(x-y+2z-2w)^9$.
I have clearly understood how to use the bin theorem for $(a+b)^n$, but am unable to extend the theorem to $4$ terms .
| It is also possible to apply the binomial theorem iteratively. We focus on one term and keep the other terms together. This way we reduce the number of terms step by step by one, till we have only two left at the end.
To keep the calculations manageable it is convenient to use the coefficient of operator $[x^j]$ to select the coefficient of $x^j$ of a polynomial or series. This way we can write e.g.
\begin{align*}
[x^j](x+y+z)^n&=[x^j]\sum_{k=0}^n\binom{n}{k}x^k(y+z)^{n-k}\\
&=\binom{n}{j}(y+z)^{n-j}\\
\end{align*}
Using this technique we obtain
\begin{align*}
[x^3y^3zw^2]&(x-y+2z-2w)^9\\
&=[x^3y^3zw^2]\sum_{k=0}^9\binom{9}{k}x^k(-y+2z-2w)^{9-k}\tag{1}\\
&=\binom{9}{3}[y^3zw^2](-y+2z-2w)^6\tag{2}\\
&=\binom{9}{3}[y^3zw^2]\sum_{k=0}^6\binom{6}{k}(-y)^k(2z-2w)^{6-k}\tag{3}\\
&=\binom{9}{3}\binom{6}{3}(-1)^3[zw^2](2z-2w)^3\tag{4}\\
&=-\binom{9}{3}\binom{6}{3}[zw^2]\sum_{k=0}^3\binom{3}{k}(2z)^k(-2w)^{3-k}\tag{5}\\
&=-\binom{9}{3}\binom{6}{3}\binom{3}{1}2^1(-2)^2\tag{6}\\
&=-84\cdot20\cdot3\cdot8\\
&=-40320
\end{align*}
Comment:
*
*In (1), (3) and (5) we apply the binomial theorem
*In (2), (4) and (6) we select the coefficient of $x^3,y^3,z$ and $w^2$
| {
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"question_score": "1",
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Sum from $n=1$ to $N$ of $\sin(nx)/2^n$ The question states:
I'm not sure what to do here. I used the geometric sum to do $2^{-n}$, but I can't think of a way to sum $\sin(n\theta)$ to get the required RHS.
Further working
Apologies for maybe being slow or missing something obvious here... I don't know how to get it into required form.
| The solution below is just to give a basic proof without using Euler's identity.
\begin{align*}
&\quad \sum_{n=1}^N \frac{\sin(n\theta)}{2^n} \cdot 2^N(5-4\cos\theta) \\
&= \left( \sum_{n=1}^{N} 2^{n-1}\sin(N-n+1)\theta \right)(1+4-4\cos \theta) \\
&= \left( \sum_{n=1}^{N} 2^{n-1}\sin(N-n+1)\theta \right) + \left( \sum_{n=1}^{N} 2^{n+1}\sin(N-n+1)\theta \right) \\
&\quad - \left( \sum_{n=1}^{N} 2^{n+1}\sin(N-n+1)\theta\cos\theta \right) \\
&= \left( \sum_{n=1}^{N} 2^{n-1}\sin(N-n+1)\theta \right) + \left( \sum_{n=1}^{N} 2^{n+1}\sin(N-n+1)\theta \right) \\
&\quad - \left( \sum_{n=1}^N 2^n [\sin(N-n+2)\theta + \sin(N-n)\theta] \right) \\
&= \left[ 2^{N-1}\sin\theta + \sum_{n=2}^{N-1}2^{n-1}\sin(N-n+1)\theta + \sin(N\theta) \right] \\
&\quad + \left[ 2^{N+1}\sin\theta + \sum_{n=2}^{N-1}2^{n+1}\sin(N-n+1)\theta + 2^2\sin(N\theta) \right] \\
&\quad - \left[ 2\sin(N+1)\theta + 2^2\sin(N\theta) + \sum_{n=2}^{N-1}2^{n+1}\sin(N-n+1)\theta + \sum_{n=2}^{N-1}2^{n-1}\sin(N-n+1)\theta + 2^{N-1}\sin\theta \right] \\
&= 2^{N+1}\sin\theta + \sin(N\theta) - 2\sin((N+1)\theta).
\end{align*}
But I do spend almost half an hour to write it, hah.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1900142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Finding the determinant of a skew-symmetric matrix $K$
Find the determinant of the skew-symmetric matrix $K$
$$K = \begin{bmatrix}
0 & 1 & 3\\
-1 & 0 & 4 \\
-3 & -4 & 0 \\
\end{bmatrix}$$
My Attempted Solution:
I performed the following row operations to reduce $K$ into upper-triangular form $U$
$R_2 \leftrightarrow R_1$
$R_3 - (l_{31} = 3)R_1$
$R_3 \leftrightarrow R_2$
$R_3 - (l_{32} = -4) R_2$
$$U = \begin{bmatrix}
-1 & 0 & 4 \\
0 & -4 & -12 \\
0 & 0 & -45
\end{bmatrix}$$
From this I got
$$\begin{align}
\det(K) &= \pm \ \det(U) \\
&= + \det(U) & \text{(Even no. of row exchanges)} \\
& = (-1)(-4)(-45) \\
&= 180
\end{align}$$
However the correct answer is $\det(K) = 0$. What could I have done wrong, I wouldn't think it would've been the row operations as the row operations apart from the row exchanges don't affect the $\det(K)$? Any hints or suggestiong are greatly appreciated
| $$\begin{array}{rl} \det \left[\begin{array}{cc|c} 0 & 1 & 3\\ -1 & 0 & 4 \\ \hline -3 & -4 & 0\end{array}\right] &= \det \begin{bmatrix} 0 & 1\\ -1 & 0\end{bmatrix} - \det \left(\begin{bmatrix} 0 & 1\\ -1 & 0\end{bmatrix} - \begin{bmatrix} 3\\ 4\end{bmatrix} \begin{bmatrix} 3\\ 4\end{bmatrix}^T\right)\\ &= 1 - \det \begin{bmatrix} -9 & -11\\ -13 & -16\end{bmatrix}\\ &= 1 - \det \begin{bmatrix} 9 & 11\\ 13 & 16\end{bmatrix}\\ &= 1 - (\underbrace{144 - 143}_{=1}) = 0\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1904093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Prove that $\cos^6 \beta - \sin^6 \beta=\frac {1}{16} (15 \cos 2\beta + \cos 6\beta)$.
Prove the given trigonometric identity:
$$\cos^6 \beta - \sin^6 \beta=\frac {1}{16} (15 \cos 2\beta + \cos 6\beta)$$
My Approach:
\begin{align*}
\text{L.H.S.} &=\cos^6 \beta - \sin^6 \beta\\
&=(\cos^2 \beta)^3 - (\sin^2 \beta)^3\\
&=\cos^32\beta+3\cos^2\beta\cdot\sin^2\beta\cdot\cos2\beta
\end{align*}
Please help me to continue further.
| Use $$\cos2A=2\cos^2A-1=1-2\sin^2A$$ to find
$$8(\cos^6 \beta - \sin^6 \beta)=(2\cos^2\beta)^3-(2\sin^2\beta)^3=(1+\cos2\beta)^3-(1-\cos2\beta)^3=?$$
Finally use $$\cos3B=4\cos^3B-3\cos B\iff4\cos^3B=?$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If the sum of the digits of $a+b, a+c, a+b$ are bounded by $k$, what is the maximum sum of digits of $a+b+c$?
Question: For $n \in \mathbb{N}$, let $S(n)$ denote the sum of the base 10 digits of $n$. Assuming that $a,b,c \ge 0$ and
$$
S(a+b), S(b+c),S(a+c) \le k,
$$
What is the maximum possible value of $S(a+b+c)$ (as a function of $k$)?
This is motivated by this question.
We know that the answer is between $15k-9$ and $15k$, inclusive. Both of these bounds are found by adapting the answers to the motivating question. Specifically, we can achieve $15k-9$ by taking
\begin{align*}
a &= \underbrace{444\ldots4}_{k-1} \;\underbrace{555\ldots 5}_{k-1}\;\underbrace{555\ldots 5}_{k-1} \;5 \\\
b &= \underbrace{555\ldots 5}_{k-1}\;\underbrace{444\ldots 4}_{k-1}\;\underbrace{555\ldots 5}_{k-1}\;5\\
c &= \underbrace{555\ldots 5}_{k-1}\;\underbrace{555\ldots 5}_{k-1}\;\underbrace{444\ldots 4}_{k-1}\;5.
\end{align*}
And we can also prove that $S(x+y) \le S(x) +S(y)$ and $S(x) \le 5S(2x)$ (see my answer to the motivating question), so that
$$
S(a+b+c) \le 5S(2a+2b+2c) \le 5[S(a+b) + S(a+c) + S(b+c)] \le 15k.
$$
In the comments, it was also determined that we may assume either $S(a+b) = S(a+c) = S(b+c) = k$ and $S(a+b+c) = 15k$, or $S(a+b) = S(a+c) = k$, $S(b+c) = k-1$, and $S(a+b+c) = 15k - 5$.
I expect we can get an exact answer, and that that answer is $15k - 9$ (for $k \ge 1$).
| We will show that the maximum is indeed $15k - 9$.
First, let's investigate the inequalities which you used, and in particular, what the effect is if either is not an equality.
In your previous answer, it was shown that $S(x) \leq 5S(2x)$ essentially by noting that this is true for each digit, and then noting that $S(2x)$ is equal to the sum over the digits $d$ of $x$ of $S(2d)$. Now we note that for individual digits, equality occurs in $S(d) \leq 5S(2d)$ precisely when $d$ is either $0$ or $5$. It follows that if any digit of $x$ (say the digit is $d$) is not $5$ or $0$, then we in fact have that $S(x) \leq 5S(2x) - 5S(2d) + S(d)$. But if $d$ is not $5$ or $0$, then we can check that $5S(2d) - S(d)$ is at least $9$, and so $S(x) \leq 5S(2x) - 9$.
The consequence of this is that if the digits of $a + b + c$ are not each equal to either $5$ or $0$, then we get that $S(a + b + c) \leq 5S(2a + 2b + 2c) - 9 \leq 15k - 9$.
We see that if $S(a + b + c) > 15k - 9$ then the digits of $a + b + c$ is each either $0$ or $5$.
If we look at your proof that $S(x + y) \leq S(x) + S(y)$, we note that equality occurs if and only if there are no carries when adding $x$ and $y$, and that in fact if there are any carries, then we can see that $S(x + y) \leq S(x) + S(y) - 9$. (A carry decreases the sum of the digits by at least $9$.)
We see that if there are any carries involved in adding $a + b$, $b + c$ and $c + a$ to get $2(a + b + c)$, then in fact $$S(a + b + c) \leq 5S(2a + 2b + 2c) \leq 5(S(a + b) + S(b + c) + S(c + a) - 9) \leq 15k - 45.$$
This then shows that if $S(a + b + c) > 15k - 9$, then there are no carries involved in adding together $a + b$, $b + c$ and $c + a$.
Now suppose that $a$, $b$, and $c$ are as in the conditions of the problem, and that $S(a + b + c) > 15k - 9$. We can assume that $a$, $b$ and $c$ do not all end in a zero, as otherwise we could remove the trailing zero from each one. Let $M = a + b + c$, and let
$$ a = \sum_{i = 0}^\infty a_i 10^i $$
be the decimal expansion of $a$, and analogously for $b$ and $c$.
Since the last digit of $M$ is a $0$ or a $5$, the last digit of $2M$ must be a $0$. Since there are no carries involved in the addition, each of $a + b$, $b + c$, and $c + a$ must end in a $0$. We thus have that each of $a_0 + b_0$, $b_0 + c_0$, and $c_0 + a_0$ must be $0$ or $10$, and they are not all $0$. If, say, $a_0 + b_0$ were $0$, then this would imply $a_0 = b_0 = 0$. But then since at least one of $b_0 + c_0$ and $c_0 + a_0$ is not $0$, we would have that $c_0 = 10$, a contradiction. It follows that $a_0 + b_0 = b_0 + c_0 = c_0 + a_0 = 10$, and so $a_0 = b_0 = c_0 = 5$.
For will now show by induction that for each $n$, we have that $9 \leq a_n + b_n + c_n \leq 19$, and that when adding $a + b + c$ to get $M$, there is a carry of $1$ for each digit. This is true for $n = 0$ as shown above. Now suppose that for some $n$ that the claim is true for $n-1$. We will show that it is also true for $n$.
Since the digits of $M$ are each either $0$ or $5$, we have that the $n^\text{th}$ digit of $2M$ is either a $0$ or a $1$. Since there are no carries involved in the addition of $a + b$, $b + c$ and $c + a$, we have that the $n^\text{th}$ digit of each of $a + b$, $b + c$ and $c + a$ is either a $0$ or a $1$.
Now the $n^\text{th}$ digit of $a + b$ is either the units digit of $a_{n} + b_{n}$, or the units digit of $a_n + b_n + 1$. (The carry for each digit when adding $a$ and $b$ is at most $1$)
We thus have that $a_n + b_n$ must be one of $0, 1, 9, 10$ or $11$. Similarly, each of $b_n + c_n$ and $c_n + a_n$ is one of $0, 1, 9, 10$ or $11$.
Now by the inductive hypothesis, the carry involved when calculating the $n^\text{th}$ digit of $M$ is exactly $1$, and so the $n^\text{th}$ digit of $M$ is equal to the units digit of $a_n + b_n + c_n + 1$. The units digit of this must be either $0$ or $5$, and so we get that $a_n + b_n + c_n$ is one of $4, 9, 14, 19$ or $24$. Thus $2(a_n + b_n + c_n)$ is one of $8, 18, 28, 38$ or $48$. Since $a_n + b_n$, $b_n + c_n$ and $c_n + a_n$ are each at most $11$, we can eliminate $38$ and $48$ as possibilities because they are too large. It is also easy to see that $8$ is not possible, since this would require each of $a_n + b_n$, $b_n + c_n$ and $c_n + a_n$ to be less than $9$, and hence each would be at most $1$.
The remaining possibilities for $2(a_n + b_n + c_n)$ are $18$ and $28$, and so $a_n + b_n + c_n$ is one of $9$ or $14$. Each lies in the range $9$ to $19$, and together with the carry of $1$ from the previous column, we see that there will also be a carry of $1$ into the next column, so our claim is proven.
But the claim above implies that for each $n$, at least one of $a_n$, $b_n$ or $c_n$ is not zero, which is impossible since $a$, $b$, and $c$ are finite. We thus have a contradiction, and hence we must have that $S(a + b + c) \leq 15k - 9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1905610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find $\frac{x^{3333}+x^{333}+x^{33}+x^{3}+1996}{100(x^2+x)}$ if $x^2+x+1=0$
Find $$\frac{x^{3333}+x^{333}+x^{33}+x^{3}+1996}{100(x^2+x)}$$ if $x^2+x+1=0$
My work so far:
1)$x^2+x=-1$, then $100(x^2+x)=-100$
2)$x^2+x+1=0$
$x=\frac{-1\pm\sqrt{3}i}{2}$
| $x^2+x+1=0\implies$
$x=-(-1)^{1/3},(-1)^{2/3}\implies$
$x^3=1\implies$
$\cfrac{x^{3333}+x^{333}+x^{33}+x^{3}+1996}{-100}=\cfrac{1^{1111}+1^{111}+1^{11}+1^{1}+1996}{-100}=-20$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1906012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to understand the following result? I am beginning to study probability and I found the following problem:
What is the probability of to get a number twice after two throwings?
The answer was that since we have six distinct results in a fair die every possible result has $1/6$ of probability of happen and since that the result is:
$$ P(\text{rolling the same number twice in a row on a fair die} ) = (1/6)(1/6).$$
I understand till the part of the probabilities, I know that since every event is independent of the other we are applying the multiplication rule, However I would like to go more deep in the analysis and not to take this as a strict rule, I would like to understand why we had to multiply the numbers, I mean why not to sum them or divide?, I need to understand why the rule of multiplication works, I would like to receive a empirical or formal explanation of this result in order to understand more, thanks for the support.
I would like to state that my question is different since I know that if I have the scenario of only one die, thrown out in two different times my result is correct but I don't understand why we have to multiply the probabilities I know that we are using the multiplication rule but why it works? what is the reason to multiply the numbers I mean why not to divide it or to sum it, I would like to go more deep in the analysis.
| The probability of drawing a $1$ is $1/6$, because there is a single favorable outcome and six possible ones.
The probability of drawing two $1$s in a row is $1/36=1/6^2=(1/6)^2$ because there are now thirty-six possible outcomes. The product appears for combinatorial reasons. Specifically, the set of possible cases is the Cartesian product of the possible outcomes at the first drawing and those at the second drawing.
So just like $6^2=36$, the probabilites are multiplied, $1/6^2=(1/6)^2$. For three $1$s in a row, the probability would be $1/6^3=(1/6)^3$.
Now assume that you throw a die and a coin. The outcomes are from the sets $\{1,2,3,4,5,6\}$ and $\{H,T\}$. By the Cartesian product, there are $6\cdot2$ combinations and the probability of drawing a one and a head is $1/(6\cdot2)=(1/6)(1/2)$.
In the example with a die, probabilites aren't just multiplied, because there are several favorable cases ($6$ ways to draw the same value).
The probabilities for $1,2$ and $3$ identical drawings would be
$$\frac66=1,\\\frac6{6^2}=\frac16,\\\frac6{6^3}=\frac1{36}.$$
More "difficult": we throw the die twice and the coin three times and we look for equal numbers and sides.
As regards the dies, there are $6^2$ possible outcomes, and $2^3$ for the coin. Among these, $6$ combinations are favorable for the die and $2$ for the coin, forming a total of $6\cdot2$ distinct configurations (again by the Cartesian product).
Hence the probability
$$\frac{6\cdot2}{6^2\cdot2^3}=\left(\frac6{6^2}\right)\left(\frac2{2^3}\right).$$
$$\begin{matrix}
&& HHH & THH & HTH & TTH & HHT & THT & HTT & TTT \\
& 11 &\times & & & & & & &\times \\
& 12 & & & & & & & & \\
& 13 & & & & & & & & \\
& 14 & & & & & & & & \\
& 15 & & & & & & & & \\
& 16 & & & & & & & & \\
& 21 & & & & & & & & \\
& 22 &\times & & & & & & &\times \\
& 23 & & & & & & & & \\
& 24 & & & & & & & & \\
& 25 & & & & & & & & \\
& 26 & & & & & & & & \\
& 31 & & & & & & & & \\
& 32 & & & & & & & & \\
& 33 &\times & & & & & & &\times \\
& 34 & & & & & & & & \\
& 35 & & & & & & & & \\
& 36 & & & & & & & & \\
& 41 & & & & & & & & \\
& 42 & & & & & & & & \\
& 43 & & & & & & & & \\
& 44 &\times & & & & & & &\times \\
& 45 & & & & & & & & \\
& 46 & & & & & & & & \\
& 51 & & & & & & & & \\
& 52 & & & & & & & & \\
& 53 & & & & & & & & \\
& 54 & & & & & & & & \\
& 55 &\times & & & & & & &\times \\
& 56 & & & & & & & & \\
& 61 & & & & & & & & \\
& 62 & & & & & & & & \\
& 63 & & & & & & & & \\
& 64 & & & & & & & & \\
& 65 & & & & & & & & \\
& 66 &\times & & & & & & &\times \\
\end{matrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1907412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can we find incentre of a triangle by using equation of lines? My question is, in the manner which we find the orthocentre of a triangle merely by using it's side's line equations, can we also find the incentre?
For example, consider,
$$3x+4y-7=0$$
$$4x-3y+19=0$$
$$18x-6y+7=0$$
We can find the orthocentre of the triangle made by these lines using the fact that in a right angled triangle, the vertex containing the right angle is the orthocentre.
So, for any triangle made by three straight lines, can we find the incentre as well?
I will appreciate both short and long answers. Any thoughts are welcome. Thanks!
EDIT :I will appreciate if someone gives me a method where I don't have to solve for the vertices.
| The intersection point (Incenter) of the internal bisectors can be obtained through a formula with the cofactors, coefficients and constants of the equations.
$A·x+B·y+C=0$
$D·x+E·y+F=0$
$G·x+H·y+I=0$
$M=\left( \begin{array}{} A & B & C \\ D & E & F \\ G & H & I \\ \end{array} \right) $
$p=\sqrt{A^2+B^2}$
$q=\sqrt{D^2+E^2}$
$r=\sqrt{G^2+H^2}$
$xI=\frac{(\frac{p·cA }{ sgn(cC)} +\frac{q· cD}{sgn(cF)}+ \frac{r ·cG}{sgn(cI)})}{p·|cC| + q· |cF| + r ·|cI|}$
$yI=\frac{(\frac{p·cB }{ sgn(cC)} +\frac{q· cE}{sgn(cH)}+ \frac{r ·cG}{sgn(cI)})}{p·|cC| + q· |cF| + r ·|cI|}$
where cA, cB, cC ... cI are the cofactors of the matrix M
calculating the Incenter with its equations
$M=\left( \begin{array}{} -3 & 8 & -6 \\ 4 & -4 & 14 \\ 18 & -6 & 7 \\ \end{array} \right)$
$CofactorsM=\left( \begin{array}{} 56 & 224 & 48 \\ -20 & 87 & 126 \\ 88 & 18 & -20 \\ \end{array} \right) $
$p=\sqrt{(-3)^2+8^2}=\sqrt{73}$
$q=\sqrt{4^2+(-4)^2}=\sqrt{32}$
$r=\sqrt{18^2+(-6)^2}=\sqrt{360}$
$xI=\frac{\frac{\sqrt{73}·56}{1}+\frac{\sqrt{32}·(-20)}{1}+\frac{\sqrt{360}·88}{-1}}{\sqrt{73}·|48|+\sqrt{32}·|126|+\sqrt{360}·|-20|}=-0.868$
$ yI=\frac{\frac{\sqrt{73}·(224)}{1}+\frac{\sqrt{32}·(87)}{1}+\frac{\sqrt{360}·(18)}{-1}}{\sqrt{73}·|48|+\sqrt{32}·|126|+\sqrt{360}·|-20|}=1.374$
alternative formula:
$xI2=\frac{\left| \left ( \begin{array}{} cA \\ cD \\ cG \end{array} \right)^T \left( \begin{array}{} \frac{p}{sgn(cC)} \\ \frac{q}{sgn(cF)} \\ \frac{r}{sgn(cI)}) \end{array} \right ) \right| }{{\left| \left ( \begin{array}{} cC \\ cF \\ cI \end{array} \right )^T \left ( \begin{array}{} \frac{p}{sgn(cC)} \\ \frac{q}{sgn(cF)} \\ \frac{r}{sgn(cI)} \end{array} \right ) \right| }}$
$yI2=\frac{\left| \left ( \begin{array}{} cB \\ cE \\ cH \end{array} \right)^T \left( \begin{array}{} \frac{p}{sgn(cC)} \\ \frac{q}{sgn(cF)} \\ \frac{r}{sgn(cI)}) \end{array} \right ) \right| }{{\left| \left ( \begin{array}{} cC \\ cF \\ cI \end{array} \right )^T \left ( \begin{array}{} \frac{p}{sgn(cC)} \\ \frac{q}{sgn(cF)} \\ \frac{r}{sgn(cI)} \end{array} \right ) \right| }}$
IncenterThreeLinesEquations
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove the limit of a function with two variables is zero I have to prove as homework that
$$\lim_{(x,y)\to (0,0)} \frac{\sin\left(x^2y\right)}{x^2+y^2}=0$$
and that
$$\lim_{(x,y)\to (0,0)} \frac{\tan\left(x^3\right)}{x^2+y^2}=0.$$
For the second limit here's what I did:
let $\epsilon >0$ we want $\delta >0$ such that if $\|(x,y)\|<\delta$ then $|tan(x^3)/(x^2+y^2)|<\epsilon$
So suppose $\|(x,y)\|<\delta$, $|tan(x^3)/(x^2+y^2)|= |tan(x^3)/(x^3)||(x^3)/(x^2+y^2)|< |tan(x^3)/(x^3)||x|=|sin(x^3)/(x^3)||1/cos(x^3)||x|<|x|/|cos(x^3)|$
For the first limit I don't know how to begin.
| Hint: Use the fact that $\sin(x)$ and $\tan(x)$ are odd to reduce to the case that $x,y\ge0$. Then
$$
x^2\le x^2+y^2\quad\text{and}\quad y\le\left(x^2+y^2\right)^{1/2}
$$
therefore,
$$
\frac{x^2y}{x^2+y^2}\le\left(x^2+y^2\right)^{1/2}
$$
Then use the fact that for $x\ge0$, $\sin(x)\le x$.
For the second inequality, you can use the fact that $\tan(x)$ is convex on $\left[0,\frac\pi2\right)$and $\tan\left(\frac\pi4\right)=1$ to see that for $0\le x\le\frac\pi4$
$$
\tan(x)\le\frac4\pi x
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to come up with ways of simplifying equations? Consider the cubic equation
$$ x^3 + bx^2 + cx + d = 0$$
If one replaces $x$ with $x - {b\over 3}$ then this reduces to
$$ x^3 + px + q = 0$$
If one knows what to substitute with then checking the claim is easy. But my question is:
How does one see which substitution to make? How do I know that it
simplifies if I replace $x$ with $x - {b\over 3}$?
| Recall that $-b$ is the sum of the roots of $ x^3 + bx^2 + cx + d = 0.$ That is, if the roots are $r_{1},\,r_{2},\,r_{3},$ then $(r_1 + r_2 + r_3) = -b.$ See Vieta's formulas if this is new to you.
Now note that if the graph is shifted right by $\frac{b}{3},$ which can be done by replacing $x$ with $x - \frac{b}{3},$ then each of the roots of the new equation is additively increased by $\frac{b}{3}.$ Thus, the roots of the equation you get by replacing $x$ with $x - \frac{b}{3}$ are $r_{1} + \frac{b}{3} ,\,r_{2} + \frac{b}{3},\,r_{3} + \frac{b}{3},$ and the sum of these roots is
$$\left(r_{1} + \frac{b}{3}\right) \; + \; \left(r_{2} + \frac{b}{3}\right) \; + \; \left(r_{3} + \frac{b}{3}\right) \;\; = \;\; (r_1 + r_2 + r_3) \; + \; \left(\frac{b}{3} + \frac{b}{3} + \frac{b}{3}\right) = \; -b + b = 0 $$
Since the sum of the roots of the new equation is $0,$ it follows (again, by Vieta's formulas) that the coefficient of $x^2$ in the new cubic polynomial is $0.$
In general, if you replace $x$ with $x - \frac{b}{n}$ in the equation $x^n + bx^{n-1} + \cdots = 0,$ then the coefficient of the $x^{n-1}$ term is $0$ for the polynomial you get after the replacement. For what it's worth, note also that what you're doing is shifting right by the arithmetic mean of the roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1910915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Determine all real $x$ that satisfies $\sqrt{3-x} - \sqrt{x+1} > \frac{1}{2}$ I have this equation that I have to solve:
Determine all real $x$ that satisfies $\sqrt{3-x} - \sqrt{x+1} > \frac{1}{2}$
Maybe it involves means? The square roots are leaning towards it... Could someone help me?
| Note that
*
*$\sqrt{3-x}$ is defined only for $x\le3$
*$\sqrt{x+1}$ is defined only for $x\ge-1$
*The first square root is strictly decreasing and the second strictly increasing, so the whole LHS must be strictly decreasing (to see why, graph the functions)
Thus it remains to solve for $x$ in the case of equality, and take the interval as $[-1,x)$.
$$\sqrt{3-x}-\sqrt{x+1}=\frac12$$
$$\sqrt{3-x}=\frac12+\sqrt{x+1}$$
$$3-x=\frac14+\sqrt{x+1}+x+1$$
$$\frac74-2x=\sqrt{x+1}$$
$$\frac{49}{16}-7x+4x^2=x+1$$
$$4x^2-8x+\frac{33}{16}=0$$
$$x=1\pm\frac{\sqrt{31}}8$$
Checking, we see that the negative sign must be used for $x$, as only that produces an equality. Hence the range of $x$ is $[-1,1-\frac{\sqrt{31}}8)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1911239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Positive integers $a$ and $b$ are such that $a + b =\frac{ a}{b} +\frac{ b}{a}$. What is the value of $a^2 + b^2$? I tried to take $a = 1$ and $b = 2$ and arrive at the conclusion that $a^2 + b^2 = 2$ which indeed is the answer.
However, I wanted to prove this equation algebraically without using the replacement technique.
| If $a>1$ or $b>1$, then it follows that $a>\frac{a}{b}$ or $b>\frac{b}{a}$ so $$a+b=\frac{a}{b}+\frac{b}{a}<a+b$$
Which is a contradiction.
Thus we conclude that $a,b \le 1$, and thus $a=b=1$. Therefore, the answer is $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1912782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
homework - probability problem A dice is thrown 3 times and the sum of the 3 numbers thrown is 15. Then, what is the probability that the first throw was a four?
My Attempt:
The sample space for the above problem, $S = 6^3$.
Combinations of thrown numbers such that their sum is $15$ and the first digit is $4 = [\{4, 5, 6\}, \{4, 6, 5\}]$
Thus, the total number of possible sets are 2. Thus, the probability for the required event (say $A$),
$$P(A) = \frac{2}{216}$$
$$\therefore P(A) = \frac{1}{108}$$
In my book, the answer given is $\frac{1}{5}$. How's is my answer?
| Since the sum of the 3 numbers thrown is 15 it follows that the sample space is given by 10 elements: 15=6+6+3 ($3!/2!=3$ ways), 15=6+5+4 (3!=6 ways), 15=5+5+5 ($3!/3!=1$ way).
There are two cases when $4$ is the first value 4,5,6 and 4,6,5.
Therefore the probability is $p=2/10=1/5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1913428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Symmetric and antisymmetric matrices as subspaces of $M_{3 \times 3}(\Bbb R)$ Show that the set $S$ compound of $3 \times 3$ symmetric matrices and the set $A$ of the antisymmetric matrices, are subspaces of $M_{3\times 3}(\Bbb R)$. Determine basis for $S$ and $A$. Show, using a $\dim$ relation, that $S \oplus A = M_{3\times 3}(\Bbb R)$.
My solution:
Set $S$ can be described as any matrix in the follow pattern:
$$
\begin{pmatrix}
a & d & e \\
d & b & f \\
e & f & c
\end{pmatrix}
$$
Set $A$ follows this other pattern:
$$
\begin{pmatrix}
0 & d & e \\
-d & 0 & f \\
-e & -f & 0
\end{pmatrix}
$$
Proof for both sets being subspaces of $M_{3\times 3}(\Bbb R)$:
To summarize this post I'll affirm that set $S$ and $A$ are closed under multiplication, addition, and contains the $ \left\{ 0 \right\} $, which make them subspaces. It's easy to see that.
Vector space $M_{3 \times 3}(\Bbb R)$ follows this pattern:
$$
\begin{pmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{pmatrix}
$$
To see that the set $S$ is included there, take $b=d, c=g, h=f$.
To see that the set $A$ is included there, take $d=-b, h=-c, f=-h$ and $a=e=i=0$
Hence, because of that and the affirmation made previously, we can assume that $S$ and $A$ are subspaces of $M_{3 \times 3}(\Bbb R)$
Basis:
For $S$:
$$
B_{s}= \left\{
\begin{pmatrix}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}
, \begin{pmatrix}
0 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 0
\end{pmatrix}
, \begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 1
\end{pmatrix}
, \begin{pmatrix}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}
, \begin{pmatrix}
0 & 0 & 1 \\
0 & 0 & 0 \\
1 & 0 & 0
\end{pmatrix} ,
\begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0
\end{pmatrix}
\right\} \\
\dim(S) = 6
$$
For $A$:
$$
B_{a}= \left\{
\begin{pmatrix}
0 & 1 & 0 \\
-1 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}
, \begin{pmatrix}
0 & 0 & 1 \\
0 & 0 & 0 \\
-1 & 0 & 0
\end{pmatrix}
, \begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & 1 \\
0 & -1 & 0
\end{pmatrix}
\right\}\\
\dim(A) = 3
$$
Proof for $S \oplus A = M_{3x3}(R)$ using a dimension relation:
Using that $\dim$ relation:
$$
\dim(a+b) = \dim(a) + \dim(b) - \dim(a \cap b)
$$ We can assume that:
$$
\dim(S) + \dim(A) = 9
$$ Since $\dim(M_{3 \times 3}(\Bbb R)) = 9$ we get:
$$
\dim(M_{3 \times 3}(\Bbb R)) = \dim(S) + \dim(A)
$$ Hence:
$$
S \cap A = \left\{ 0 \right\}\\
S \oplus A = M_{3 \times 3}(\Bbb R)
$$
Is my approach to the answer correct? Do I need to be more rigorous?
Thanks
| Any $n \times n$ matrix $B$ can be written uniquely as a sum of a symmetric matrix and a skew symmetric matrix:
\begin{align*}
B = \frac{1}{2}(B+B^t) + \frac{1}{2}(B-B^t)
\end{align*}
where $B^t$ denotes the transpose of $B$. Thus $M_{n\times n}(\mathbb{R}) = S + A$. If $B \in S \cap A$, then $B^t = B$ and $B^t = -B$ and hence $B$ is the zero matrix. It follows that $S \oplus A = M_{n \times n}(\Bbb R)$.
We haven't used dimension argument. However, it is not difficult to see that
\begin{align*}
\dim(S) = \frac{n(n+1)}{2}, \qquad \dim(A) = \frac{n(n-1)}{2}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1913820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
RMO inequality problem If $a,b,c,d,e>1$, then prove that:
$$\frac{a^2}{b-1} + \frac{b^2}{c-1} + \frac{d^2}{e-1} + \frac{c^2}{a-1} + \frac{e^2}{d-1} \ge 20. $$
I don't know how to begin. What should be the approach?
| Note that $x^2-4(x-1) = (x-2)^2\ge 0$ for all $x\in\mathbb{R}$, and hence $x^2\ge 4(x-1)$ for all $x\in\mathbb{R}$. As such, we have
$$ \frac{a^2}{b-1} + \frac{b^2}{c-1} + \frac{d^2}{e-1} + \frac{c^2}{a-1} + \frac{e^2}{d-1} \ge 4\left(\frac{a-1}{b-1} + \frac{b-1}{c-1} + \frac{c-1}{a-1} + \frac{d-1}{e-1} + \frac{e-1}{d-1}\right). $$
So it suffices to show
$$\frac{a-1}{b-1} + \frac{b-1}{c-1} + \frac{c-1}{a-1} + \frac{d-1}{e-1} + \frac{e-1}{d-1}\ge 5.$$
Can you see how to finish?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1913873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
} |
Find the ratio in which the perpendicular Find the ratio in which the perpendicular from $(4,1)$ to the line segment joining the points $(2,-1)$ and $(6,1)$ divides the segment.
My Approach:
Equation of the line joining the points $(2,-1)$ and $(6,1)$ is given by:
$$y-y_1=\frac {y_2-y_1}{x_2-x_1} (x-x_1)$$
$$y+1=\frac {2}{4} (x-2)$$
$$y+1=\frac {1}{2} (x-2)$$
$$2y+2=x-2$$
$$x-2y-4=0$$.
I got stuck at here. Please help me to complete.
| You have a line $x-2y-4=0$, and you can fix it to the form $y=\frac{1}{2}x-2$.
The perpendicular line will have a slope of $-\frac{1}{a}$ or in this case $-2$ and will pass via $(4,1)$.
Hence:
$$1=4\times (-2)+b \implies b=9$$
So the perpendicular line:
$$y=-2x+9$$
Now let us find the point these two line intersect:
$$-2x+9=\frac{1}{2}x-2$$
$$11=\frac{5}{2}x$$
$$x=\frac{22}{5}$$
$$y=(-2)\times \frac{22}{5}+9=\frac{1}{5}$$
Th line $y=\frac{1}{2}x-2$ will be divided by the perpendicular at the point $(x,y)=(\frac{22}{5},\frac{1}{5})$
Do you know how to carry on?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1914268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve $3^x \equiv 43 \pmod {97}$ using Pohlig-Hellman Problem: Solve $3^x \equiv 43 \pmod {97}$ using the Pohlig-Hellman algorithm.
Here is my solution, but the problem is that I don't get it right at the end.
First of all, I calculate $\phi(97) = 96 = 2^5 \cdot 3$.
Now, let $x=a_0 + 2^5a_1=a_0+32a_1$. Then
$$ {3^{(a_0+32a_1)}}^3 \equiv 43^3 \pmod{97}$$
$$27^{a_0} \equiv 64 \pmod {97}.$$
Trial and error yields $a_0 = 6$. So now we have $x = 6 + 32a_1$, meaning $x \equiv 6 \pmod {32}$.
Now, let $x=b_0 + 3b_1$. Then
$$ {3^{(b_0 + 3b_1)}}^{32} \equiv 43^{32} \pmod{97}$$
$$35^{b_0} \equiv 35\pmod {97}.$$
Obviously, $b_0=1$. So now we get $x = 1 + 3b_1$, meaning $x \equiv 1 \pmod {3}$.
Using CRT, we can now solve:
$$x \equiv 6 \pmod {32}$$
$$x \equiv 1 \pmod {3}.$$
Solving the first equivalence:
$$3x_1 \equiv 6 \pmod {32}.$$
The multiplicative inverse of $3$ modulo $32$ is $11$.
$$11 \cdot 3x_1 \equiv 11 \cdot 6 \pmod {32}$$
$$x_1 \equiv 11 \cdot 6 \equiv 2 \pmod {32}.$$
Now for the second one:
$$32x_2 \equiv 1 \pmod {3}$$
$$2x_2 \equiv 1 \pmod {3}$$
$$x_2 \equiv 2 \pmod 3.$$
And at last:
$$x_0 = 3 \cdot 2 + 32 \cdot 2 = 70 \equiv 70 \pmod {96}.$$
Now $3^{70} \pmod {97}$ is not $43$, and I cannot find where is my error.
But, if I got $a_0=22$, then I would get $x_1 \equiv 18 \pmod {32}$ and $x_0 \equiv 22 \pmod {96}$, and that would be correct answer.
| $3^{70} \mod 97$ IS 43, and your solution is fine. The only problem is that it's not unique - because the order of $3 \mod 97$ is 48 instead of 96.
So morally speaking, you should look at $x \mod 48$ instead. Using Chinese remainder theorem, you would split that up into $x \mod 16$ and $x \mod 3$. Your calculations were correct - the answer would be $x \equiv 6 \mod 16$ and $x \equiv 1 \mod 3$, and from there you see that the solution is $x \equiv 22 \mod 48$. In particular, 70 IS also a correct solution - it just doesn't exhaust everything.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1915508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $A+B+C+D=2\pi$, prove that: If $A+B+C+D=2\pi$, prove that: $$\cos A+\cos B+\cos C+\cos D=4\cos\frac {A+B}{2}\cdot\cos\frac {A+C}{2}\cdot\cos\frac {B+C}{2}$$.
My Approach:
Here,
$$A+B+C+D=2\pi$$
$$A+B=2\pi - (C+D)$$
$$ \sin(A+B)=\sin(2\pi-(C+D))$$
$$\sin(A+B)=-\sin(C+D)$$
Again,
$$\cos(A+B)=\cos(2\pi-(C+D))$$
$$\cos(A+B)=\cos(C+D)$$
Now,
$$L.H.S=\cos A+\cos B+\cos C+\cos D$$
$$=2 \cos\frac {A+B}{2}\cdot\cos\frac {A-B}{2} + 2 \cos\frac {C+D}{2}\cdot \cos\frac {C-D}{2}$$.
I got stuck at here. Please help me to complete the proof.
| $$cos\frac{C+D}{2}=cos\frac{2\pi-(A+B)}{2}=-cos\frac{A+B}{2}$$
$$cos\frac{C-D}{2}=cos\frac{C+A+B+C-2\pi}{2}=-cos\frac{A+C+B+C}{2}$$
Now go to your last line
$$2cos\frac{A+B}{2}(cos\frac{A-B}{2}+cos\frac{A+C+B+C}{2})$$
Use the rule to turn an addition of cosinuses to a multiplication and you will get your proof
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1916560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Solve algebraically and graphically: $\arcsin(x) + \arccos\left(\frac{x}{2}\right) = \frac{5\pi}{6}$ Solve:
$$\arcsin(x) + \arccos\left(\frac{x}{2}\right) = \frac{5\pi}{6}$$
I think the algebraic solution should start like:
$$\arcsin(x) = \frac{5\pi}{6} - \arccos\left(\frac{x}{2}\right)$$
$$x = \sin\left(\frac{5\pi}{6} - \arccos\left(\frac{x}{2}\right)\right)$$
at that stage probably I should use some trigonometric relation or property of $\arccos(x)$ to convert it to $\arcsin(x)$, however I can't figure it out.
As for the solution by graph I can't even think what steps should I follow to build it.
| $$\arcsin x + \arccos\frac{x}{2} = \frac{5\pi}{6}$$
$$\arcsin x = \frac{5\pi}{6} - \arccos\frac{x}{2}$$
$$x = \sin{\left(\frac{5\pi}{6} - \arccos\frac{x}{2}\right)}$$
$$x=\sin\frac{5\pi}6\cos \left(\arccos\frac{x}{2}\right)-\cos\frac{5\pi}6 \sin\left(\arccos\frac{x}{2}\right)$$
$$x=\frac12\cdot\frac x2+\frac{\sqrt3}{2} \cdot\sqrt{1-\left(\frac x2\right)^2}$$
Answer: $x=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1916903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove $\int_0^{\pi/2}x^2\sqrt{\cot x} \ dx=\frac{\sqrt2}{8}\left( \frac{5\pi^{3}}{12}-\pi^2\ln2-\pi \ln^22 \right)$ I came across the follwing integral:
$$\int_0^{\pi/2}x^2\sqrt{\cot x} \ dx=\frac{\sqrt2}{8}\left( \frac{5\pi^{3}}{12}-\pi^2\ln2-\pi \ln^22 \right)$$
I try to do it like the following:
Consider
$$I(a,b)=\int_0^{\pi} \frac{\cos ax}{\sin^b x}\ dx=2\int_0^{\pi/2} \frac{\cos 2ax}{\sin^b 2x}\ dx$$
Then
$$I''(a,b)=-8\int_0^{\pi/2}x^2 \frac{\cos 2ax}{\sin^b 2x}\ dx$$
Let $a=1/2,b=1/2$
$$I''(1/2,1/2)=-8\int_0^{\pi/2}x^2 \frac{\cos x}{\sqrt{\sin 2x}}\ dx=-\frac{8}{\sqrt2}\int_0^{\pi/2}x^2\sqrt{\cot x} \ dx $$
Back to $I(a,b),\quad I(a,b) $can be expressed by beta fuction by Wolfram Mathematica.
$$I(a,b)=\int_0^{\pi} \frac{\cos ax}{\sin^b x}\ dx=\frac{\pi \cdot 2^b\cdot\cos (\pi a/2)\cdot \Gamma(1-b) }{\Gamma(a/2-b/2+1) \cdot \Gamma(-a/2-b/2+1)} dx$$
Then we can get $I''(a,b)$ approach from other way.
Finally,we can get $\int_0^{\pi/2}x^2\sqrt{\cot x} \ dx$,but it seems a little complex.
For a similar integral:
$$\int_0^{\pi/2}x\cdot\tan^p x \ dx=\frac{\pi}{4\sin (p\pi/2)}\left(\Psi\left(\frac{1}{2}\right)-\Psi\left(\frac{1-p}{2}\right) \right)$$
The above integral can be solved by method of parametric development.Let $p=-\frac{1}{2}$,We can get
$$\int_0^{\pi/2}x\sqrt{\cot x} \ dx=\frac{\pi\left(\pi-2\ln 2\right)}{4\sqrt2}$$
But to this one, it seems to be difficult with method of parametric development. Could you suggest any ideas how to prove this?
| A slightly easier form for $I(a, \frac{1}{2})$ is
$$I(a, \frac{1}{2}) = 2^{a-\frac{3}{2}} \beta\left(\frac{1+2a}{4}\right) (1 + \sin \pi a + \cos \pi a), $$
where $\beta(p) = \Gamma(p)^2/\Gamma(2p)$ is the central beta function. (This can be easily obtained by applying the Euler reflection formula and the Legendre duplication formula to OP's representation.) Now from the relation
$$ \frac{d}{dp} \log \beta(p) \bigg|_{p=\frac{1}{2}} = 2\psi(1/2) - 2\psi(1) = 2 \sum_{n=0}^{\infty} \left( \frac{1}{n+1} - \frac{1}{n+\frac{1}{2}} \right) = -4 \log 2 $$
and
$$ \frac{d^2}{dp^2} \log \beta(p) \bigg|_{p=\frac{1}{2}} = 2\psi^{(1)}(1/2) - 4\psi^{(1)}(1) = \frac{\pi^2}{3}, $$
we can compute $\frac{\partial^2 I}{\partial a^2}(\frac{1}{2}, \frac{1}{2})$ as follows:
\begin{align*}
&\frac{\partial^2 I}{\partial a^2} (1/2, 1/2) \\
&= \frac{1}{4} \beta''(1/2) - \left(\frac{\pi}{2}-\log 2\right) \beta'(1/2) - \left(\frac{\pi^2}{2} + \pi\log 2-\log^2 2\right) \beta(1/2) \\
&= \frac{1}{4} \pi \left( 16\log^2 2 + \frac{\pi^2}{3} \right) \\
&\qquad - \left(\frac{\pi}{2}-\log 2\right) (-4\pi \log 2)
- \left(\frac{\pi^2}{2} + \pi\log 2-\log^2 2\right) \pi \\
&= -\left(\frac{5\pi^3}{12} - \pi^2\log2 - \pi \log^2 2 \right).
\end{align*}
I am also trying a more direct approach. Using contour integral, we can check that
$$ \int_{0}^{\frac{\pi}{2}} x^2 \sqrt{\cot x} \, \mathrm{d}x
= \frac{1}{3\sqrt{2}} \left( \frac{\pi^3}{4} - \frac{3\pi}{2} \int_{0}^{1} \frac{\operatorname{artanh}^2 t}{\sqrt{t}} \, \mathrm{d}t \right). $$
So I am tackling the last integral, but have no good news at this point.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1921384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 2,
"answer_id": 0
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Why is it intuitively "obvious" that: $2^0+2^1+2^2+...+2^n=2^{n+1}-1$ Why is it "obvious" that: $$2^0+2^1+2^2+...+2^n=2^{n+1}-1$$
And how about: $$2^{-1}+2^{-2}+2^{-3}+...+2^{-n}=1-2^{-n}$$
I am looking for some intuition about these facts.
Is there a similar pattern when the base is $3$?
| Adding another answer to the very good proofs below : you are already familiar with this concept each time you use numbers.
$999_{10} + 1 = 1000_{10}$ is the same than $(9 \times 10^2)+(9 \times 10^1)+(9 \times 10^0) +1 = 10^3$.
If you set $b=9_{10}$ you wrote $(b-1) \times b^2+ (b-1) \times b^1+(b-1) \times b^0 +1 = b^3 = (b-1) ( b^2+b^1+b^0 )$
In base 2, it is the same :
$111_2 = (1 \times 2^2)+(1 \times 2^1)+(1 \times 2^0) +1 = 2^3$.
If you set $b=2_{10}$ you wrote $(b-1) \times b^2+ (b-1) \times b^1+(b-1) \times b^0 +1 = b^3 = b^2+b^1+b^0+1$
Not a proof, just a recall of a knowledge already acquired ...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1921892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
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Existence of $A$ and $B$ such that $A^2-4A+4I=0$, $A+B=\begin{pmatrix} 4 & 1 \\ -3 & 4\end{pmatrix}$ and $AB=$...
Question: Prove or disprove that $A$ and $B$ exist such that
\begin{align}
&A^2-4A+4I=0\\
&A+B=\begin{pmatrix} 4 & 1 \\ -3 & 4\end{pmatrix}\\
&AB=\begin{pmatrix} 1 & 1 \\ -9 & 3\end{pmatrix}\\
\end{align}
(*) I've found what I've done wrongly below. (see the red and blue).
If I follow abnry's method instead then I get the right answer because $C-4I$ has its inverse matrix in this case.
What I did is as follows:
\begin{align}
&\text{Assume that such }A\text{ and }B\text{ exist.}\\
&\text{Then }A\text{ satisfies both of below equations:}\\
&X^2-4IX+4I=0\tag1\\
&\color{red}{(X-A)(X-B)=X^2-(A+B)X+AB=0\leftarrow Wrong}\\
&\color{blue}{(X-A)(X-B)=X^2-(AX+XB)+AB=0}\\
&\\
&(1)-(2):\\
&(A+B-4I)X=AB-4I\\
&\begin{pmatrix} 0 & 1 \\ -3 & 0\end{pmatrix}X=\begin{pmatrix} -3 & 1 \\ -9 & -1\end{pmatrix}\\
&X=\begin{pmatrix} 0 & 1 \\ -3 & 0\end{pmatrix}^{-1}\begin{pmatrix} -3 & 1 \\ -9 & -1\end{pmatrix}=\begin{pmatrix} 0 & -\frac13 \\ 1 & 0\end{pmatrix}\begin{pmatrix} -3 & 1 \\ -9 & -1\end{pmatrix}=\begin{pmatrix} 3 & \frac13 \\ -3 & 1\end{pmatrix}\\
&\\
&\text{As }X\text{ is uniquely found, so }A=X\\
&B=\begin{pmatrix} 4 & 1 \\ -3 & 4\end{pmatrix}-A=\begin{pmatrix} 1 & \frac23 \\ 0 & 3\end{pmatrix}\\
&\text{Now calculating }AB,\\
&\\
&AB= \begin{pmatrix} 3 & \frac13 \\ -3 & 1\end{pmatrix}\begin{pmatrix} 1 & \frac23 \\ 0 & 3\end{pmatrix}=\begin{pmatrix} 3 & 3 \\ -3 & 1\end{pmatrix}\ne\begin{pmatrix} 1 & 1 \\ -9 & 3\end{pmatrix}\\
&\\
&\text{Therefore, such }A\text{ and }B\text{ do not exist.}\\
\end{align}
But as I don't have much knowledge on linear algebra (or at least I forgot all of them), I don't know a good explanation on why this happened, and on what condition $A$ and $B$ exist or not. I think there must be much better way to know it only by looking at those two equations without actually trying to find $A$ and $B$. Could someone help here?
| The second two equations are of the form $A+B=C$ and $AB=D$.
Then $A^2+AB=AC$ and so $A^2=AC-AB=AC-D$.
Substitute for $A^2$ in the original equation to get
$$AC-D-4A+4I=0$$
or
$$A(C-4I)=D-4I.$$
So if the inverse of $C-4I$ exists, this gives you a unique $A$. From this you can compute $B=C-A$ (and verify that indeed $AB=D$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1923301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
limit of $\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4}$ I have question. I want to solve this limit. it's $\frac{0}{0}$ so we have to change it. there is two way with two different value.
$\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4}$
First way:
before that we know that $\lim_{x\to 0} \frac{\sin x}{x}$ or $\lim_{x\to 0} \frac{(\sin x) ^ 2}{x ^ 2}$ is equal to 1 so:
$\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4} = \lim_{x\to 0} \frac{x^2 - x ^ 2 (\cos x) ^ 2}{x^4} = \lim_{x\to 0} \frac{x^2 (1 - (\cos x) ^ 2)}{x^4} = \lim_{x\to 0} \frac{(\sin x) ^ 2}{x^2} = 1$
second way:
before that we know that $\sin x \sim x - \frac{x^3}{6}$ and $\cos x \sim 1 - \frac{x^2}{2}$ so:
$\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4} = \lim_{x\to 0} \frac{(\sin x) - (x \cos x)}{x^3} * \frac{(\sin x) + (x \cos x)}{x} = \lim_{x\to 0} \frac{x - \frac{x^3}{6} - x + \frac{x^3}{2}}{x^3} * \frac{x - \frac{x^3}{6} + x - \frac{x^3}{2}}{x} = (\frac{1}{2} - \frac{1}{6}) * 2 = \frac{2}{3}$
Update:
Is it possible to explain more? We have limit and we solve like this and that's work but in this limit we can't use $\lim_{x\to0}\left(\frac{\sin x}x\right)^2=1$.
this in another limit:
$\lim_{x\to0} \frac{1 - \cos 2x}{x^2}= \lim_{x\to0} \frac{2 (\sin x)^2}{x^2} = 2 * \lim_{x\to0} (\frac{\sin x}{x})^2 = 2 * 1 = 2$
Which way is true? Is it possible to help me?
I'm sorry for bad English.
Thanks.
| Note that we have
$$\begin{align}
\frac{\sin^2(x)-x^2\cos^2(x)}{x^4}&=\frac{\left(\frac{\sin(x)}{x}\right)^2-1}{x^2}+\frac{\sin^2(x)}{x^2}\\\\
&=\color{blue}{\underbrace{\left(\frac{\sin(x)}{x}+1\right)}_{\to 2\,\,\text{as}\,\,x\to 0}}\left(\frac{\sin(x)-x}{x^3}\right)+\color{red}{\underbrace{\frac{\sin^2(x)}{x^2}}_{\to 1\,\,\text{as}\,\,x\to 0}}\\\\
\end{align}$$
Evaluating the limit of interest boils down to evaluating the limit of $\frac{\sin(x)-x}{x^3}$.
Given that $\sin(x)-x=-\frac16 x^3+O(x^5)$, the limit is easily seen to be $-\frac16$.
Putting it all together, we find the limit of interest is
$$\lim_{x\to 0}\left(\frac{\sin^2(x)-x^2\cos^2(x)}{x^4}\right)=\frac23$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1924213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 3
} |
find value of determinant $f(n)=β^n+α^n$
then $$
\begin{vmatrix}
3 & 1+f(1) & 1+f(2) \\
1+f(1) & 1+f(2) & 1+f(3) \\
1+f(2) & 1+f(3) & 1+f(4) \\
\end{vmatrix}
$$
I don't quite understand which property of determinant can we use here? i tried using linear property but i did not get the required answer.
| There is a trick, indeed. The values defined by $f(n)=\alpha^n+\beta^n$ obey the recurrence relation
$$ f(n+2) - (\alpha+\beta)\,f(n+1) + (\alpha\beta)\,f(n) = 0$$
hence the original determinant equals
$$ (1-\alpha)(1-\beta)\det\begin{pmatrix}1+f(0) & 1+f(1) & 1+f(2) \\ 1+f(1) & 1 + f(2) & 1+f(3) \\ 1 & 1 & 1\end{pmatrix}$$
or
$$ (1-\alpha)(1-\beta)\det\begin{pmatrix}f(0) & f(1) & f(2) \\ f(1) & f(2) & f(3) \\ 1 & 1 & 1\end{pmatrix}$$
or
$$ (1-\alpha)^2(1-\beta)^2\det\begin{pmatrix}f(0) & f(1) \\ f(1) & f(2)\end{pmatrix}$$
or
$$ \boxed{\det\begin{pmatrix}1+f(0) & 1+f(1) & 1+f(2) \\ 1+f(1) & 1 + f(2) & 1+f(3) \\ 1+f(2) & 1+f(3) & 1+f(4)\end{pmatrix}=\color{red}{(1-\alpha)^2(1-\beta)^2 (\alpha-\beta)^2}.}\tag{1}$$
The properties exploited are
*
*$\det\begin{pmatrix} r_1 \\ r_2 \\ r_3\end{pmatrix}=\det\begin{pmatrix} r_1 \\ r_2 \\ r_3-j r_2+k r_1\end{pmatrix}=\det\begin{pmatrix}r_1-r_3 \\ r_2-r_3 \\ r_3\end{pmatrix}$
*$\det\begin{pmatrix} r_1 \\ r_2 \\ \alpha r_3\end{pmatrix}=\alpha\det\begin{pmatrix} r_1 \\ r_2 \\ r_3\end{pmatrix}$
*$\det M^T = \det M$
*$\det\begin{pmatrix}a & b & 0 \\ c & d & 0 \\ e & f & 1\end{pmatrix}=\det\begin{pmatrix}a & b \\ c & d\end{pmatrix}.$
Another tricky approach may be the following one. One may notice that the determinant is a polynomial in $\mathbb{Z}[\alpha,\beta]$ with degree $4+2=6$, and that the given matrix has rank one if $\alpha=1,\beta=1$ or $\alpha-\beta$. It follows that the determinant is an integer multiple of $(1-\alpha)^2(1-\beta)^2(\alpha-\beta)^2$ and by performing an explicit computation at $\alpha=2,\beta=3$ the claim $(1)$ follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1924279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solution of simultaneous equations using properties of symmetric polynomials
Solve the following system of equations.
$$ x + y + z = 5 $$
$$ \frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx} = 5 $$
$$ x^3 + y^3 + z^3 = 53 $$
How would I use properties of symmetric polynomials (e) to solve this?
| .From $\frac{1}{yz} + \frac{1}{xz} + \frac{1}{xy} = 5$, we get:
$$
\frac{x+y+z}{zyx} = 5 \implies zyx = 1
$$
Now, $$
x^3+y^3+z^3 - 3xyz= (x+y+z)(x^2+y^2+z^2 - xy-yz-zx) \\= (x+y+z)((x+y+z)^2-3(xy+yz+zx))
$$
Solving using values, this gives $xy+yz+zx = 5$.
Now, note that:
$$
(t-x)(t-y)(t-z) = t^3 - (xy+yz+zx)t^2 + (x+y+z)t - xyz \\= t^3 - 5t^2 + 5t-1
$$
Hence, if we are able to find the roots of this equation, we can find the values of $x,y,z$.
By inspection, $t=1$ works, hence $t-1$ ia a factor.
After the division ,we are left with $t^2-4t+1$, which gives $t = 2 \pm \sqrt{3}$.
Thus, $x=1, y,z = 2 \pm \sqrt{3}$, and since the equation is symmetric, any two of these can be switched.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1924707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
expanding and factoring in algebra Expanding very simple process, just open round brackets and multiply each with each for example
$(x-6)(x+3) = x^2 + 3x - 6x -18 = x^2 -3x -18$
Factoring is confused me a little for example I have $x^2 -3x -18$ I try to brute force all combinations in my mind and get the result like this
$(x-6)(x+3)$
Cold you please recommend some algorithm, how to find factoring in the easy way.
Thanks.
| Here is another alternative
$$
\begin{align}
x^2-3x-18 &= \frac{1}{4}\bigg(4x^2-4\cdot 3x-4\cdot 18\bigg )\\
&= \frac{1}{4}\bigg((2x)^2-2\cdot 3(2x)-72\bigg )\\
\end{align}$$
Now suppose we define $t=2x$
$$\begin{align}
\frac{1}{4}\bigg(t^2-2\cdot 3t-72\bigg )&= \frac{1}{4}\bigg(t^2-2\cdot 3t+9-9-72\bigg )\\
&= \frac{1}{4}\bigg((t-3)^2-81\bigg )\\
&= \frac{1}{4}\bigg((t-3)^2-9^2\bigg )\\
&= \frac{1}{4}\bigg((t-3)-9\bigg )\bigg((t-3)+9\bigg )\\
&= \frac{1}{4}\bigg(t-12\bigg )\bigg(t+6\bigg )\\
&= \frac{1}{4}\bigg(2x-12\bigg )\bigg(2x+6\bigg )\\
&= \bigg(x-6\bigg )\bigg(x+3\bigg )\\
\end{align}$$
This method is probably overkill, but it shines when used to derive the quadratic formula.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1925037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove $2^{n+3}|(p^2-1)$ Olympiad problem:
let odd prime nunmber $p$ such $$p|(2+\sqrt{3})^{2^n}+(2-\sqrt{3})^{2^n}$$
show that:
$$2^{n+3}|(p^2-1)$$
I want to take advantage of this conclusion：
if prime factor $ p $ of a Fermat number $ 2^{2^n} + 1 $ ,then $ p \equiv 1 \pmod{2^{n + 2}} $
But I failed
| If $p\equiv 1\pmod{2^{n+2}}$, then either $p\equiv 1\pmod{2^{n+3}}$ or $p\equiv 2^{n+2}+1\pmod{2^{n+3}}$. In either case, squaring yields $p^2\equiv 1\pmod{2^{n+3}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1925109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluation of $\int^{\frac{\pi}{2}}_{0}\frac{\cos^6 x}{\sin x+\cos x}dx$
Evaluation of $\displaystyle \int^{\frac{\pi}{2}}_{0}\frac{\cos^6 x}{\sin x+\cos x}dx$
$\bf{My\; Try::}$ Let $\displaystyle I = \int^{\frac{\pi}{2}}_{0}\frac{\cos^6 x}{\sin x+\cos x}dx = \int^{\frac{\pi}{2}}_{0}\frac{\sin^6 x}{\sin x+\cos x}dx$
Above we have used $\displaystyle \int^{a}_{0}f(x)dx = \int^{a}_{0}f(a-x)dx$
So $$2I = \int^{\frac{\pi}{2}}_{0}\frac{\sin^6 x+\cos^6 x}{\sin x+\cos x}dx = \int^{\frac{\pi}{2}}_{0}\frac{(\sin^2 x+\cos^2 x)(\sin^4 x+\cos^4 x-\sin^2 x\cos^2 x)}{\sin x+\cos x}dx$$
So $$2I = \int^{\frac{\pi}{2}}_{0}\frac{\sin^4 x+\cos^4 x-\sin^2 x\cos^2 x}{\sin x+\cos x}dx = \int^{\frac{\pi}{2}}_{0}\frac{1-3\sin^2 x\cos^2 x}{\sin x+\cos x}dx$$
Now How can i solve it after that, Help Required, Thanks
| Let
$$ A=\int^{\frac{\pi}{2}}_{0}\frac{\cos^6 x}{\sin x+\cos x}dx, B=\int^{\frac{\pi}{2}}_{0}\frac{\sin^6 x}{\sin x+\cos x}dx. $$
Clearly $A=B$ by changing variable from $x$ to $\frac{\pi}{2}-x$. Now
\begin{eqnarray}
A+B&=&\int^{\frac{\pi}{2}}_{0}\frac{\cos^6 x+\sin^6x}{\sin x+\cos x}dx\\
&=&\int^{\frac{\pi}{2}}_{0}\frac{(\cos^2 x+\sin^2x)(\cos^4-\cos^2x\sin ^2x+\sin^4x)}{\sin x+\cos x}dx\\
&=&\int^{\frac{\pi}{2}}_{0}\frac{1-3\cos^2x\sin ^2x}{\sin x+\cos x}dx\\
&=&\int^{\frac{\pi}{2}}_{0}\frac{1-\frac34\sin ^2(2x)}{\sqrt{2}\sin(x+\frac{\pi}{4})}dx\\
&=&\int^{\frac{3\pi}{24}}_{\frac{\pi}{4}}\frac{1-\frac34\sin ^2[2(u-\frac{\pi}{4})]}{\sqrt{2}\sin u}du\\
&=&\int^{\frac{3\pi}{24}}_{\frac{\pi}{4}}\frac{1-\frac34\cos ^2(2u)}{\sqrt{2}\sin u}du\\
&=&\frac1{4\sqrt2}\int^{\frac{3\pi}{4}}_{\frac{\pi}{4}}\frac{4-3\cos ^2(2u)}{\sin u}du\\
&=&\frac1{4\sqrt2}\int^{\frac{3\pi}{4}}_{\frac{\pi}{4}}\frac{4-3(1-2\sin ^2u)^2}{\sin u}du\\
&=&\frac1{4\sqrt2}\int^{\frac{3\pi}{4}}_{\frac{\pi}{4}}(\frac{1}{\sin u}+12\sin ^2u-12\sin^4 u)du\\
&=&\frac1{4\sqrt2}\left[\ln\tan(\frac{u}{2})-3\cos u-3\cos (3u)\right]\bigg|^{\frac{3\pi}{4}}_{\frac{\pi}{4}}\\
&=&\frac12+\frac1{2\sqrt{2}}\ln(\cot(\frac{\pi}{8}))\\
&=&\frac12+\frac1{2\sqrt{2}}\coth^{-1}(\frac{1}{\sqrt2})
\end{eqnarray}
and hence
$$ A=B=\frac14+\frac1{4\sqrt{2}}\coth^{-1}(\frac{1}{\sqrt2}). $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1929308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Polar to cartesian form of r=sin(3θ) $$\sin(2θ) = \sinθ\cosθ + \cosθ\sinθ = 2\sinθ\cosθ $$
$$\cos(2θ) = \cosθ\cosθ - \sinθ\sinθ = \cos²θ - \sin²θ $$
$$\sin(3θ) = \sin(2θ+θ) $$
$$= \sin(2θ)\cosθ + \cos(2θ)\sinθ $$
$$= 2\sinθ\cos²θ + \cos²θ\sinθ - \sin³θ
$$ $$= 2\sinθ(1-\sin²θ) + (1-\sin²θ)\sinθ - \sin³θ
$$ $$= 2\sinθ - 2\sin³θ + \sinθ - \sin³θ - \sin³θ
$$ $$r=\sin(3θ)= 3\sinθ - 4\sin³θ$$
How to deal with the differing exponents? I could add an $r$ to both sides or $r^3$ but not both.
$$r^4= 3r^3\sinθ - 4r^3\sin^{3}θ $$
$$r^4= 3r^2y - 4y^3$$
$$(x^2+y^2)^{4/2}= 3(x^2+y^2)^{2/2}y - 4y^3$$
$$(x^2+y^2)^{2}= 3(x^2+y^2)y - 4y^3$$
$$(x^2+y^2)^{2}= 3yx^2-y^3$$
| $$\sin(3\theta) = 3\sin\theta - 4\sin^3\theta$$
$$\sin^3\theta = \sin\theta\cdot \sin^2\theta$$
$$r = 3\sin\theta - 4\sin\theta\sin^2\theta$$
Hence
$$r^2 = 3r\sin\theta\left(1 - \frac{4}{3}\sin^2\theta\right)$$
$$r^2 = 3y\left(1 - \frac{4}{3}\frac{r^2}{r^2}\sin^2\theta\right) = 3y\left(1 - \frac{4}{3r^2}y^2\right)$$
So
$$r^2 = 3y - \frac{4y^3}{r^2}$$
And again
$$r^4 = 3yr^2 - 4y^3 \to r^4 = 3y(x^2 + y^2) - 4y^3$$
Arranging the terms you get
$$r^4 = 3x^2y - y^3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1930388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Show that $arg(z_1) \equiv \arg(z_2) \pmod{2\pi}$ assuming $|a|^2 - 4|b| \leq 0$ and $\arg b \equiv 2\arg a \pmod{2\pi}$ Let $a, b \in \mathbb{C}^{*}$ and $(E)$ the following equation : $z^2 - az + b = 0$.
Assuming :
$\left\{
\begin{equation}
\begin{aligned}
& \arg b \equiv 2\arg a \pmod{2\pi} \\
& |a|^2 - 4|b| \geq 0
\end{aligned}
\end{equation}
\right.
$
How should I start in order to show that $z_1$ and $z_2$ (roots of $(E)$) would have the same arguments ($\arg z_1 = \arg z_2 \pmod{2\pi}$).
I have proved the first implication by assuming $z_1$ and $z_2$ have the same arguments, hence the set of conditions above is true.
For the reciprocal, I tried to start again from $b = z_1 z_2$ and $a = z_1 + z_2$ and manipulating these equations, but I have difficulty in wrapping my mind around the simultaneous usage of these two conditions.
I wonder also if that's provable through counter example?
| I think the condition you want is actually $|a|^2-4|b|\ge 0$ (otherwise, take $a = b = 2$, and the roots of $z^2-2z+2$ are actually $1\pm i$, which have different arguments)
If the roots have the same argument, then they should have the same argument as their sum, i.e. $a$. As such, let $y = z/a$. We then have
$$ z^2-az+b = (ay)^2-a(ay)+b = a^2y^2 - a^2y + b = a^2\left(y^2 - y + \frac{b}{a^2}\right).$$
Note that if $y_1$ and $y_2$ are the roots of $y^2-y+\frac{b}{a^2}$, then $y_1 = z_1/a$ and $y_2 = z_2/a$. It would suffice to show that $y_1$ and $y_2$ are both positive reals to conclude that $z_1$ and $z_2$ have the same argument.
Since $\arg b\equiv 2\arg a \equiv \arg(a^2)\mod 2\pi$, it follows that $b$ and $a^2$ have the same argument, and hence $\frac{b}{a^2}$ is real. Furthermore, by the quadratic formula, the roots $y_1$ and $y_2$ are given by
$$ y_{1,2} = \frac{1\pm\sqrt{1-4\left(\frac{b}{a^2}\right)}}{2}. $$
Now use the condition $|a|^2-4|b|\ge 0$ to conclude that both roots are real and positive.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1930601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the relation between $\lfloor x+y \rfloor$ and $\lfloor x \rfloor + \lfloor y\rfloor$ for any reals $x,y$ What is the relation between $\lfloor x+y \rfloor$ and $\lfloor x \rfloor + \lfloor y\rfloor$ for any reals $x,y$?
My effort:
We have $x+y-1\lt \lfloor x+y\rfloor \leq x+y$. What can I do after this? Please help
| For some real number $0 \le \alpha < 1,\quad x = \lfloor x \rfloor + \alpha$
That is really all you need to show that $$x-1 < \lfloor x \rfloor \le x < \lfloor x \rfloor + 1$$
and
$$\lfloor \lfloor x \rfloor + y \rfloor = \lfloor x \rfloor + \lfloor y \rfloor$$
One thing you can use it for is, for some $0 \le \alpha, \beta < 1$
\begin{align}
\lfloor x + y \rfloor
&= \lfloor
(\lfloor x \rfloor + \alpha) +
(\lfloor y \rfloor + \beta)
\rfloor \\
&= \lfloor x \rfloor + \lfloor y \rfloor +
\lfloor \alpha + \beta \rfloor \\
\end{align}
Since $0 \le \alpha + \beta < 2$, it follows that $0 \le \lfloor \alpha + \beta \rfloor \le 1$. So
$$\lfloor x \rfloor + \lfloor y \rfloor
\le \lfloor x + y \rfloor
\le \lfloor x \rfloor + \lfloor y \rfloor + 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1931463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
A similar Somos sequence problem prove $A_{2n}B_{n+3}$ is integer sequence I have read some interesting with Somos sequence,Recently I met a similar question：
For a fixed positive integer $k$, there are two sequences $A_n$ and $B_n$.
They are defined inductively, by the following recurrences.
$$A_1 = k,A_2 = k,A_{n+2} = A_{n}A_{n+1}$$
$$B_1 = 1,B_2 = k, B_{n+2} = \frac{B^3_{n+1}+1}{B_{n}}$$
Prove that for all positive integers $n$, $A_{2n}B_{n+3}$ is an integer.
I try to Solve this problem,following is my some works.
since
$$A_{n+2}=A_{n}A_{n+1}\Longrightarrow \ln{A_{n+2}}=\ln{A_{n}}+\ln{A_{n+1}}$$
so we have
$$A_{n}=k^{f_{n}}$$
where $f_{n}$ be the Fibonacci sequece,which satisfies $f_{1}=f_{2}=1,f_{n}=f_{n-1}+f_{n-2}$
for $B_{n}$ It is difficult to handle, even for a general term
If we choose to use mathematical induction
(1):if $n=1$ since
$$B_{3}=k^3+1,\Longrightarrow B_{4}=\dfrac{(k^3+1)^3+1}{k}$$
so we have
$$A_{2}B_{4}=(k^3+1)^3+1$$ is integer
if $n=2$,we have
$$B_{5}=\dfrac{B^3_{4}+1}{B_{4}}=\dfrac{\left(\dfrac{(k^3+1)^3+1}{k}\right)^3+1}{k^3+1}=\dfrac{((k^3+1)^3+1)^3+k^3}{(k^3+1)k^3}$$
and $A_{3}=k^2,A_{4}=k^3$,so we have
$$A_{4}B_{5}=\dfrac{((k^3+1)^3+1)^3+k^3}{k^3+1}=\dfrac{(k^3+1)^9+3(k^3+1)^6+3(k^3+1)^3+(k^3+1)}{k^3+1}
=(k^3+1)^8+3(k^3+1)^5+3(k^3+1)^2+1\in N^{+}$$
Now Assmue that $A_{2n}B_{n+3}(n\le k)$ is integer,or $n^{f_{n}}B_{n+3}(n\le k)$ is integrthen consider $$A_{2k+2}B_{k+4}=(k+1)^{f_{k+1}}\cdot\dfrac{B^3_{k+3}+1}{B_{k+2}}$$
Now the problem is equlivant to prove
$$\dfrac{(B^3_{k+3}+1)(k+1)^{f_{k+1}}}{B_{k+2}}\in N^{+}~~~?$$
| Considering the sequence (for $n \ge 2$):
$$
A_{2(n-1)} B_{(n-1)+3}, \space A_{2n} B_{n+3}, \space A_{2(n+1)} B_{(n+1)+3} \space \rightarrow \space A_{2n-2} B_{n+2}, \space A_{2n} B_{n+3}, \space A_{2n+2} B_{n+4}
$$
According to definitions, we have:
$$
{ A_{2n+3} = A_{2n+2} A_{2n+1} = (A_{2n+1} A_{2n}) A_{2n+1} = A^2_{2n+1} A_{2n} = A^3_{2n} A^2_{2n-1} \brace A_{2n+3} = A_{2n+2} A_{2n+1} = A_{2n+2} (A_{2n} A_{2n-1}) = A_{2n+2} A^2_{2n-1} A_{2n-2} }
$$
$$
\Rightarrow A^3_{2n} = A_{2n+2} A_{2n-2}
$$
And,
$$
B_{n+4} = \frac{B^3_{n+3}+1}{B_{n+2}} \Rightarrow B^3_{n+3}+1 = B_{n+4} B_{n+2}
$$
Thus,
$$
A^3_{2n} (B^3_{n+3}+1) = (A_{2n+2} A_{2n-2}) (B_{n+4} B_{n+2})
$$
$$
\Rightarrow (A_{2n} B_{n+3})^3 + A^3_{2n} = (A_{2n+2} B_{n+4}) (A_{2n-2} B_{n+2})
$$
By checking the first two terms: $(A_{2n-2} B_{n+2} = A_2 B_4 \in N^{+}), \space (A_{2n} B_{n+3} = A_4 B_5 \in N^{+})$ and because all $A_k \in N^{+}$
$ \Rightarrow $ Every next term in the sequence is an integer.
$$
(A_{2n} B_{n+3})^3 + A^3_{2n} = (A_{2n+2} B_{n+4}) (A_{2n-2} B_{n+2}) \Rightarrow (Int)^3 + (Int)^3 = (A_{2n+2} B_{n+4}) \cdot (Int)
$$
$$
\Rightarrow A_{2n+2} B_{n+4} \in N^{+}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1932492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
Evaluating $\lim_{x\to 0} \frac{1-\cos x\sqrt{\cos 2x}\sqrt[3]{\cos 3x}}{x^2}$ I want to evaluate:
$$\lim_{x\to 0} \frac{1-\cos x\sqrt{\cos 2x}\sqrt[3]{\cos 3x}}{x^2}.$$
Here's what I did. We know that as soon as ${x\to 0}$
$$1 - \cos x = \frac{x^2}{2} + O(x^4)$$
Therefore
$$\cos x = 1 - \frac{x^2}{2} + O(x^4)$$
Now we apply this to cosines in limit
$$\cos 2x = 1 - 2x^2 + O(x^4)$$
$$\cos 3x = 1 - \frac{9x^2}{2} + O(x^4)$$
Then use equality
$$ (1 + x)^n = 1 + xn + o(x)$$
It yields us
$$\lim_{x\to 0} \frac{1 - (1 - \frac{x^2}{2})(1 - x^2)(1 - \frac{3x^2}{2})}{x^2}$$
After simplification 1s cancel and we get $-(-1/2 - 1 - 3/2) = 3$.
| Since in a neighbourhood of the origin $\cos(x)= 1-\frac{x^2}{2}+o(x^2)$, it follows that $\cos(2x)= 1-2x^2+o(x^2)$ and $\cos(3x)= 1-\frac{9x^2}{2}+o(x^2)$. Since $\sqrt{1+z}=1+\frac{z}{2}+o(z)$ and $\sqrt[3]{1+z}=1+\frac{z}{3}+o(z)$ we get that:
$$ \cos(x)= 1-\frac{x^2}{2}+o(x^2),\quad \sqrt{\cos(2x)}= 1-x^2+o(x^2),\quad \sqrt[3]{\cos(3x)}= 1-\frac{3x^2}{2}+o(x^2) $$
hence their product behaves like $1-\left(\frac{1}{2}+1+\frac{3}{2}\right)x^2=1-3x^2$ and the wanted limit is $\color{red}{\large 3}$.
The neglected parts in the above manipulations are $o(x^2)$, hence they do not affect the final outcome.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1933292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Question about Diophantine equations? Solve the Diophantine equation.
$6x+ 9y = 15$
We can see that $x = 1$ and $y=1$ solves the equation, so all solutions are given by $x = 1 + 3n$ and $y = 1 - 2n, n \in \mathbb{Z}$. By using:
$$x=x_0 + n \times \frac{b}{gcd(a,b)}$$ and $$y=y_0 - n \times \frac{a}{gcd(a,b)}$$
However, solving it as my textbook says I get:
I get that $gcd(9,6)=3$
$9 = 1 \times 6 + 3$
$6 = 2 \times 3 + 0 $
Then,
$3 = 9 - (1\times 6)$
So,
$3=6(-1) + 9(1) \implies 15 = 6(-5) + 9(5)$
All solutions are given by (as it is stated in my textbook) $$x=x_0 - n \times \frac{b}{gcd(a,b)}$$ and $$y=y_0 + n \times \frac{a}{gcd(a,b)}$$
Thus,
$x = - 5 - 3n$ and $y = 5 + 2n, n \in \mathbb{Z}$
My textbook give the answer as $ x=1+3n, y= 1-2n$, how did they arrive at that by using $$x=x_0 - n \times \frac{b}{gcd(a,b)}$$ and $$y=y_0 + n \times \frac{a}{gcd(a,b)}$$
and $$3=6(-1) + 9(1)?$$
All I get is $x=-1 -3n, y= 1 + 2n,n \in \mathbb{Z}$ which gives $3$ and not $15$, or $x = - 5 - 3n, y = 5 + 2n, n \in \mathbb{Z}$ which gives $15$ but looks ugly. What am I missing?
| For any $x_0,y_0$ that are solutions, you can use the formula:
$$x=x_0+n\cdot\frac{b}{\gcd(a,b)}\ ,\ y=y_0-n\cdot\frac{a}{\gcd(a,b)}$$
But you can also change the sign, and use the formula like this:
$$x=x_0-n\cdot\frac{b}{\gcd(a,b)}\ ,\ y=y_0+n\cdot\frac{a}{\gcd(a,b)}$$
It doesn't matter.
Now, the textbook used $(1,1)=(x_0,y_0)$ and he used the first formula, so he got $x=1+3n\ ,\ y=1-2n$. You used $(-5,5)=(x_0,y_0)$, and the second formula, so you have got $x=-5-3n\ ,\ y=5+2n$, but it's all the same.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1934048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Are there different integers so that $ab=c^2$ and $a^2+b^2=2d^2$? I wish to find 2 integers $a,b\ \ (a\neq b)$ so that $GM(a,b)$ and $RMS(a,b)$ are integers.
So I write down:
$$\sqrt{ab}=c\ \ \rightarrow ab=c^2\\\sqrt{\frac{a^2+b^2}{2}}=d\ \ \rightarrow a^2+b^2=2d^2$$
By lots of trial and error I think there aren't any solutions, but I didn't find a way to prove.
Is that true that there aren't solutions? And how do I prove it?
| If $a$ and $b$ are equal, then of course the above are satisfied.
$a^2+b^2 + 2ab = (a+b)^2 = 2(d^2+c^2)$ and $(a-b)^2 = 2(d^2-c^2)$. Hence, both $2(d^2-c^2)$ and $2(d^2+c^2)$ are perfect squares.
Let $e^2 = 2(d^2-c^2)$ and $f^2=2(d^2+c^2)$. Then, $ e^4 - f^4 = (e^2-f^2)(e^2+f^2) = 16d^2c^2$. In other words,
$$
e^4 = f^4 + (4dc)^2
$$
This is a solution to the equation $x^4-y^4 = z^2$, with $x=e,y=f,z=4dc$.
I claim that $x^4-y^4=z^2$ has no non-trivial solutions.
Suppose that $x,y,z$ is the smallest (the smallest means that $x^2+y^2$ is the minimized quantity) non-trivial solution of $x^4-y^4=z^2$ (which exists if there is a non-trivial solution).
Then $z,y^2,x^2$ are Pythagorean triplets. Furthermore, they are primitive (which means that there is no common factor they share), otherwise we can get a smaller solution. Then $z,y,x$ can be written in the form (assuming $z$ is even)$ z = 2pq, y^2 = p^2-q^2, x^2 = p^2+q^2$ ($p$ and $q$ of different parity and coprime, otherwise $z,y,x$ would have common factor $2$). Plug this back in, and you get $p^4 - q^4 = (xy)^2$. Hence, we can remove all the factors of $2$ from $z$, and assume that it is odd, in which case $y^2 = 2pq$ and $z = p^2-q^2$.
If $z$ is odd, then $y$ is even (as $y^2$ must then be of them form $2pq$). Now, since $y^2=2pq$, we can write $q=2u^2$ and $p=v^2$, because $p$ and $q$ were assumed coprime.
Now, since $p^2+q^2=x^2$, we write $p=r^2-s^2,q=2rs,x=r^2+s^2$ for $r,s$ coprime . Hence, we get that $2u^2=2rs$, so we can write $r=g^2$ and $s=h^2$.
Finally,
$$
p=v^2=r^2-s^2 = g^4-h^4 \implies v^2 = g^4-h^4
$$
Note that:
$$
g^2+h^2=r+s<(r+s)(rs)(r−s)=\frac{1}{2}(r^2-s^2)(2rs)=\frac{pq}{2}=\frac{y^2}4 \leq y^2 < x^2+y^2
$$
This gives a contradiction,since we were assuming that the solution minimized $x^2+y^2$. Hence, the above problem (and your problem) have no non-trivial solutions.
Hence the solutions to your problem are only when $a=b$, so that the values of $c$ and $d$ are also trivial.
Finally, I would like to point out that $x^4+y^4 = z^4 \implies z^4-y^4=(x^2)^2$, which we showed has no solutions. Hence, we just proved Fermat's last theorem for the case $n=4$ as well.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1935328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\sin^4\theta+\cos^4\theta=\frac{3+\cos4\theta}{4}$ I have a question that goes exactly like this:
By considering $(\sin^2\theta+\cos^2\theta)^2$ and $(\sin^2\theta+\cos^2\theta)^3$ prove that
$$
\text{a) }\sin^4\theta+\cos^4\theta=\frac{3+\cos4\theta}{4},\qquad\text{b) }\sin^6\theta+\cos^6\theta=\frac{5+3\cos4\theta}{8}
$$
I have not idea how to do this. Please help.
| There is a general systematic method to prove this kind of linearisation identities. It's not as swift as the previous answers, bit it's purely automatic.
You use the Euler formulae
$$\cos \theta = \frac{e^{i\theta} + e^{-i\theta}}2 \qquad \text{et}\qquad \sin \theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}$$
and you expand the powers to get an expression with terms in $\cos(p\theta)$ and $\sin(q\theta)$.
\begin{align*}
\cos^4(\theta) &= \left(\frac{e^{i\theta}+e^{-i\theta}}2\right)^4\\
&= \frac{e^{i4\theta} + 4 e^{i2\theta} + 6 + 4 e^{-2i\theta} + e^{-i4\theta}}{16}\\
&= \frac{\cos 4\theta}8 + \frac{\cos 2\theta}2 + \frac 38.\\
\sin^4(\theta) &= \left(\frac{e^{i\theta}-e^{-i\theta}}{2i}\right)^4\\
&= \frac{e^{i4\theta} - 4 e^{i2\theta} + 6 - 4 e^{-2i\theta} + e^{-i4\theta}}{16}\\
&= \frac{\cos 4\theta}8 - \frac{\cos 2\theta}2 + \frac 38.\\
\text{therefore}\qquad \cos^4\theta+\sin^4\theta &= \frac{\cos 4\theta}4 + \frac 34.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1936121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 4
} |
Proving nth difference of squares I'm a little stuck on proving $$(b^n-a^n)=(b-a)(a^0b^{n-1}+a^1b^{n-1}+ \cdots + a^{n-1}b^{0}).$$A solution I came across gave this as an answer:
$$(b-a)(b^n + b^{n-1} a+...+ba^{n-1} +a^{n} )\\
=(b-a)b^n + (b-a)b^{n-1} a+...+(b-a)ba^{n-1} +(b-a)a^n\\
=b^n+1 -b^n a+b^n a-b^{n-1} a^2 +...+b^2 a^{n-1} -ba^n +ba^n -a^{n+1}\\
=b^{n+1} -a^{n+1}$$
Since for each $i=0,\ldots,n$, notice that for each term $b^{n-i} a^{i}$ there is a
$-b^{n-i} a^{i}$, so everything cancels except for $b^{n+1} -a^{n+1}$ , so
$$(b-a)(b^n + b^{n-1} a+\cdots +ba^{n-1} +a^n )= b^{n+1} - a^{n+1}$$
But how does proving that $b^{n+1} - a^{n+1}$ is true prove the formula?
| To prove that
$$\begin{align}
b^n-a^n &= (b-a)(a^0b^{n-1}+a^1b^{n-1} + \cdots + a^{n-1}b^{0})\\
&=(b-a)\sum_{k=0}^{n-1}b^ka^{n-1-k}
\end{align}$$
It suffices to show that $$b^n-1 = (b-1)\sum_{k=0}^{n-1}b^k,$$
as follows:
$$\begin{align}
(b-1)\sum_{k=0}^{n-1}b^k &=b\sum_{k=0}^{n-1}b^k-\sum_{k=0}^{n-1}b^k\\
&=\sum_{k=0}^{n-1}b^{k+1}-\sum_{k=0}^{n-1}b^k\\
&=\sum_{k=1}^{n}b^{k}-\sum_{k=0}^{n-1}b^k\\
&=b^{n}-1\\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1938027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to show this conjecture：$\frac{1}{\sqrt{a_{1}}}+\frac{1}{\sqrt{a_{2}}}+\cdots+\frac{1}{\sqrt{a_{n}}}\ge\frac{2n}{\sqrt{a_{1}}+\sqrt{a_{n}}}$ I have prove this inequality
$$\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}>2\sqrt{n+1}-2$$
because
$$\sum_{k=1}^{n}\dfrac{1}{\sqrt{k}}>\sum_{k=1}^{n}\dfrac{2}{\sqrt{k}+\sqrt{k+1}}=2\sum_{k=1}^{n}(\sqrt{k+1}-\sqrt{k})=2\sqrt{n+1}-2$$
so
$$\sum_{k=1}^{n}\dfrac{1}{\sqrt{k}}>2\sqrt{n+1}-2$$
Now I conjecture：
if postive arithmetic progression $\{a_{n}\}$ ,and the common difference of successive members is $d>0$.have
$$\dfrac{1}{\sqrt{a_{1}}}+\dfrac{1}{\sqrt{a_{2}}}+\cdots+\dfrac{1}{\sqrt{a_{n}}}\ge\dfrac{2n}{\sqrt{a_{1}}+\sqrt{a_{n}}}$$
| Using the same method, I can show the following slightly weaker result.
Let $a_i = a_0 + id$, for $1 \le i \le n+1$. Since $d > 0$, we have $a_{i+1} > a_i$. Hence
$$ \frac{1}{\sqrt{a_i}} > \frac{2}{\sqrt{a_{i+1}} + \sqrt{a_i}} = \frac{2(\sqrt{a_{i+1}} - \sqrt{a_i})}{a_{i+1} - a_i} = \frac{2(\sqrt{a_{i+1}} - \sqrt{a_i})}{d}. $$
This gives
$$ \sum_{i=1}^n \frac{1}{\sqrt{a_i}} > \frac{2}{d} \sum_{i=1}^n \left( \sqrt{a_{i+1}} - \sqrt{a_i} \right) = \frac{2(\sqrt{a_{n+1}} - \sqrt{a_1})}{d}.$$
Finally, we simplify, getting
$$ \sum_{i=1}^n \frac{1}{\sqrt{a_i}} > \frac{2(a_{n+1} - a_1)}{d(\sqrt{a_{n+1}} + \sqrt{a_1})} = \frac{2n}{\sqrt{a_1} + \sqrt{a_{n+1}}}.$$
| {
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"url": "https://math.stackexchange.com/questions/1939123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How can I evaluate: $\lim_{x \to \infty} \sqrt{x^2 + 4x} - \sqrt{x^2 - 5x}$ I need to find this limit:
$$\lim_{x \to \infty} \sqrt{x^2 + 4x} - \sqrt{x^2 - 5x}$$
The answer I got from using the limit laws is $\sqrt{\infty} - \sqrt{\infty}$.
How do I proceed now?
Added I took the conjugate of the function and I got a new and probably better function to work with: $$\lim_{x\to \infty}\frac{9x}{\sqrt{x^2 + 4x} + \sqrt{x^2-5x}}$$
| Multiply and divide by $$\sqrt{x^2+4x} +\sqrt{x^2-5x}$$
$$\lim_{x\to \infty} \frac{9x}{\sqrt{x^2+4x} +\sqrt{x^2-5x}}$$
$$\lim_{x\to \infty} \frac{9}{\frac{\sqrt{x^2+4x}}{x} +\frac{\sqrt{x^2-5x}}{x}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1941003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Estimate $ \int_\sqrt{n}^{\sqrt{n+1}} \sin (2\pi x^2) \; dx$ using the Trapezoid rule As a follow-up to my previous question. What is the integral of $\sin (2\pi x^2)$ as $x \in [\sqrt{n}, \sqrt{n+1}]$ we get this:
$$ \int_\sqrt{n}^{\sqrt{n+1}} \sin (2\pi x^2) \; dx \tag{$\ast$}$$
Since $\sin 2\pi n = 0$ for all $n \in \mathbb{Z}$ this integral should be close to zero. I could approximate by adding over the four corners:
$$
\int_\sqrt{n}^{\sqrt{n+1}} \sin (2\pi x^2) \; dx \approx \frac{1}{4} \left[ \sin (2\pi n) + \sin 2\pi (\sqrt{n}+ \frac{1}{4})^2
+ \sin 2\pi (\sqrt{n}+ \frac{1}{2})^2 + \sin 2\pi (\sqrt{n}+ \frac{3}{4})^2\right]
$$
This might be hard to estimate since we get unpredictible terms like. And I may have misaplied the trapezoid rule
$$ \sin 2\pi (n + \frac{1}{2} \sqrt{n} + \frac{1}{4})$$
if we place the mesh and $\sqrt{n} < \sqrt{n + \frac{1}{4}} < \sqrt{n + \frac{1}{2}}< \sqrt{n + \frac{3}{4}}< \sqrt{n + 1}$ the trapeoid rule is:
$$\tiny \sin(2\pi n ) \frac{ \sqrt{n+\tfrac{1}{4}} - \sqrt{n}}{2}
+ \sin (2\pi \sqrt{n + \frac{1}{4}})\left[ \sqrt{n+ \frac{1}{2} }- \sqrt{n}\right]
+ \sin (2\pi \sqrt{n + \frac{1}{2}})\left[ \sqrt{n+ \frac{3}{4} }- \sqrt{n+ \frac{1}{4}} \right]
+ \sin (2\pi \sqrt{n + \frac{3}{4}})\left[ \sqrt{n+ \frac{1}{2} }- \sqrt{n+ 1}\right]
+ \sin (2\pi \sqrt{n + 1})\left[ \frac{\sqrt{n+ \frac{3}{4} }- \sqrt{n+ 1}}{2} \right] $$
I think there is a mistake what should the trapezoid rule be? Maybe if I said:
$$ \sqrt{n+ \frac{1}{4}} - \sqrt{n} = \sqrt{n} \left( \sqrt{1 + \frac{1}{4} } - 1 \right)= \frac{\sqrt{n}}{8} $$
Then after fixing all my mistakes I get something of order $\frac{1}{\sqrt{n}}$. How does this look?
\begin{eqnarray}\small \int \dots dx &\approx& \frac{1}{8} \left[ \sqrt{n }\cdot 0
+ 2\sqrt{n + \tfrac{1}{4}}\cdot 1
+ 2\sqrt{n + \tfrac{1}{2}}\cdot 0
+ 2\sqrt{n + \tfrac{3}{4}}\cdot (-1)
+ \sqrt{n+1}\cdot 0\right] \\ \\
&=& \frac{1}{4} \left[ \sqrt{n+ \frac{1}{2}} - \sqrt{n + \frac{3}{4}}\right] = O(\frac{1}{\sqrt{n}})\end{eqnarray}
| One approach, following @Leg's answer:
$$I = \int_0^{1} \sin(2\pi t) \dfrac{dt}{2\sqrt{n+t}}=\int_0^{1/2} \sin(2\pi t) \dfrac{dt}{2\sqrt{n+t}}+\int_{1/2}^{1} \sin(2\pi t) \dfrac{dt}{2\sqrt{n+t}}$$
$$=\int_0^{1/2} \sin(2\pi t) \dfrac{dt}{2\sqrt{n+t}}-\int_{0}^{1/2} \sin(2\pi t) \dfrac{dt}{2\sqrt{n+t+1/2}}$$
$$=\int_0^{1/2}\sin{(2\pi t)}\frac{1}{2}\left(\dfrac{1}{\sqrt{n+t}}-\dfrac{1}{\sqrt{n+t+1/2}}\right)\,dt$$
Now, because:
$$\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x+h}}=\frac{\sqrt{x+h}-\sqrt{x}}{\sqrt{x}\sqrt{x+h}}=\frac{h}{\sqrt{x}\sqrt{x+h}(\sqrt{x+h}+\sqrt{x})}$$
Thus
$$I=\int_0^{1/2}\sin{(2\pi t)}\frac{1}{2}\frac{1}{2\sqrt{n+t}\sqrt{n+t+1/2}(\sqrt{n+t}+\sqrt{n+t+1/2})}$$
and taking simple bounds on the non-$\sin$ part of the integrand, we have:
$$\int_0^{1/2}\frac{\sin{(2\pi t)}}{8(n+1)^{3/2}}<I<\int_0^{1/2}\frac{\sin{(2\pi t)}}{8n^{3/2}}$$
Observing that $\int_0^{1/2}\sin(2\pi t)\,dt=\frac{1}{\pi}$, we see that:
$$\frac{1}{8\pi (n+1)^{3/2}}<I<\frac{1}{8\pi n^{3/2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1943837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $1+5^n + 5^{2n} + 5^{3n} + 5^{4n}$ is never prime for n > 0 integer Show that $1+5^n + 5^{2n} + 5^{3n} + 5^{4n}$ is never prime for n > 0 integer
Trying cases on a computer I can see that the equation is divisible by 11 when $n \neq 5^k$. That's easy to prove if you write the sum as a fraction using the fact that the series is geometric.
Is there a more general way to solve the problem, one which includes a solution for $n=5k$ and doesn't require ansatz?
edit: $ n = 5k$ not $ n = 5^k $
| Case $1:n=2k$
Then our sum is $\frac{5^{10k}-1}{5^{2k}-1}=\frac{(5^{5k}-1)(5^{5k}+1)}{5^{2k}-1}$. Since the denominator is smaller than each of the factors we conclude the number is not prime.
Case $2: n=2k+1$
Let $a=5^{2k+1}$ and notice $1+a+a^2+a^3+a^4=(a^2+3a+1)^2-5a(a+1)^2$.
Since $a=5^{2k+1}$ this is equal to: $(a^2+3a+1)^2-5^{2k+2}(a+1)^2=(a^2+3a+1)^2-(5^{k+1}(a+1))^2$
We recognize this as a difference of squares and obtain:
$((a^2+3a+1)-5^{k+1}(a+1))(a^2+3a+1)+5^{k+1}(a+1))$.
It is very easy to see the first factor is greater than $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1944434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Nth power of the following matrix The matrix is
$ \begin{pmatrix}0&1\\-1&0\end{pmatrix}^n $
for $n=2 \implies \left(\begin{matrix}-1 & 0\\ 0 &-1\end{matrix}\right)$ for $n=3 \implies \begin{pmatrix}1&0\\0&1\end{pmatrix}$ for $n=4 \implies \begin{pmatrix}-1&0\\0&-1\end{pmatrix}$
so I assume that for every $n=2k $, where k is a natural number and bigger than $0$ the matrix will be
$\begin{pmatrix}-1&0\\0&-1\end{pmatrix}$ and for every $n=2k+1$ where k is a natural number and bigger than $0 $the matrix will be $\begin{pmatrix}1&0\\0&1\end{pmatrix}$
How can I prove it? probably with induction and how can I get easily the inverses of the matrices?
| $${{\left( \begin{matrix}
\cos \,\phi & -\sin \,\phi \\
\sin \,\phi & \cos \,\phi \\
\end{matrix} \right)}^{n}}=\left( \begin{matrix}
\cos \,n\phi & -\sin \ n\phi \\
\sin \,n\phi & \cos \,n\phi \\
\end{matrix} \right)$$
We know $A=\begin{pmatrix}0&\!-1\\1&0\end{pmatrix}$
then set $\phi=\frac{\pi}{2}$.
| {
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"url": "https://math.stackexchange.com/questions/1944689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Arrangements of a,a,a,b,b,b,c,c,c in which no three consecutive letters are the same Q: How many arrangements of a,a,a,b,b,b,c,c,c are there such that
$\hspace{5mm}$ (i). no three consecutive letters are the same?
$\hspace{5mm}$ (ii). no two consecutive letters are the same?
A:(i). 1314. ${\hspace{5mm}}$ (ii). 174.
I thought of using the General Principle of Inclusion and Exclusion along with letting $p_i$ denote a property that.. , and doing so, I will evaluate $E(0)$, which gives us the number of arrangements without any of the properties.
How do I go about doing that? I am trying to use the method stated above in solving this question, but I am unable to generalize the properties $p_i$.
| (i)
For the first question, your idea of using the Inclusion-Exclusion Principle is the correct approach.
The number of distinguishable arrangements of the letters $a, a, a, b, b, b, c, c, c$ is
$$\frac{9!}{3!3!3!}$$
where the factors of $3!$ in the denominator are attributable to the fact that interchanging identical letters in a given arrangement does not produce an arrangement that is distinguishable from the original arrangement. Alternatively, we can choose to fill three of the nine positions with $a$'s, three of the remaining six positions with $b$'s, and the remaining three positions with $c$'s, which can be done in
$$\binom{9}{3}\binom{6}{3}\binom{3}{3} = \frac{9!}{3!6!} \cdot \frac{6!}{3!3!} \cdot \frac{3!}{3!0!} = \frac{9!}{3!3!3!}$$
ways.
From these, we must exclude those arrangements in which there are three consecutive identical letters.
Next, we count the number of arrangements in which a block of three consecutive identical letters appears. There are three ways to choose which letter appears in the block of three consecutive letters. Treating the block of three identical letters as one object, we have seven objects to arrange, the block and the other six letters. Since those six letters contain two sets of three identical letters, the number of arrangements containing a block of three consecutive letters is
$$\binom{3}{1}\frac{7!}{3!3!}$$
However, we have counted arrangements in which two blocks of consecutive identical letters appear twice.
Next, we count arrangements in which two blocks of three consecutive identical letters appear. There are three ways to choose which two letters appear in blocks of three consecutive letters. If we treat each block as an object, we have five objects to arrange, the two blocks and the other three letters. Since the other three letters are identical, the number of arrangements containing two blocks of three consecutive letters is
$$\binom{3}{2}\frac{5!}{3!}$$
If we subtract the number of arrangements in which two blocks of three identical letters appear from our initial count of the number of arrangements in which a block of three identical letters appears, we will not have counted arrangements in which three blocks of three identical letters appear since we first counted them three times, then subtracted them three times.
The number of arrangements with three blocks of three consecutive identical objects is $3!$.
By the Inclusion-Exclusion Principle, the number of arrangements in which at least one block of three consecutive letters appears is
$$\binom{3}{1}\frac{7!}{3!3!} - \binom{3}{2}\frac{5!}{3!} + \binom{3}{3}3!$$
so the number of arrangements in which no three consecutive letters are identical is
$$\frac{9!}{3!3!3!} - \binom{3}{1}\frac{7!}{3!3!} + \binom{3}{2}\frac{5!}{3!} - \binom{3}{3}3!$$
(ii)
The second problem can also be handled by Inclusion-Exclusion Principle.
We wish to exclude pairs of identical letters.
One pair of consecutive identical letters:
There are three ways to choose the repeated letter. Treating that pair of identical letters as a block leaves us with eight objects, including two sets of three identical letters. The number of ways these objects can be arranged is
$$\binom{3}{1}\frac{8!}{3!3!}$$
Two pairs of consecutive identical letters:
There are two cases. We can have three consecutive identical letters (two overlapping pairs) or two distinct pairs of consecutive identical letters (such as $aa$ and $bb$). We found above that there are
$$\binom{3}{1}\frac{7!}{3!3!}$$
arrangements with three consecutive identical letters.
For the other case, there are three ways of choosing the letters of the alphabet from which the pairs of consecutive identical letters are chosen. This leaves us with seven objects, the two blocks of consecutive identical letters, the third letter from each chosen type, and the three letters of the third type. Since the three letters of the third type are indistinguishable, the number of such arrangements is
$$\binom{3}{2}\frac{7!}{3!}$$
In total, the number of arrangements with two pairs of consecutive identical letters is
$$\binom{3}{1}\frac{7!}{3!3!} + \binom{3}{2}\frac{7!}{3!}$$
Three pairs of consecutive identical letters:
Again, there are two cases. We could choose three pairs of consecutive identical letters from different letters of the alphabet or three consecutive identical letters (two overlapping pairs) and a pair of consecutive identical letters from a different letter of the alphabet.
If each pair of consecutive identical letters is from a distinct letter of the alphabet, we have six objects to arrange, $aa$, $bb$, $cc$, $a$, $b$, and $c$. This can be done in
$$6!$$
ways.
For the other case, there are three ways of selecting the letter of the alphabet from which three consecutive identical letters are selected, and two ways of selecting the letter of the alphabet from which two consecutive identical letters are selected. Again, we have six objects, the block of three consecutive identical letters, the block of two consecutive identical letters, and the other four letters. Three of those other four letters are indistinguishable, so the number of such arrangements is
$$\binom{3}{1}\binom{2}{1}\frac{6!}{3!}$$
Therefore, the total number of arrangement with three consecutive identical letters is
$$6! + \binom{3}{1}\binom{2}{1}\frac{6!}{3!}$$
Four pairs of consecutive identical letters:
Again, we have two cases. We have two blocks of three consecutive identical letters (two sets of two overlapping pairs), or we have one block of three consecutive identical letters (two overlapping pairs) with a pair of consecutive identical letters from each of the other letters of the alphabet.
We found above that the number of arrangements with two blocks of three consecutive identical letters is
$$\binom{3}{2}\frac{5!}{3!}$$
For the remaining case, there are three ways of choosing the letter of the alphabet with three consecutive identical letters. We have five objects, the block of three consecutive identical letters, the two blocks of two consecutive identical letters, and the two remaining single letters. They can be arranged in
$$\binom{3}{1}5!$$
ways. The total number of arrangements with four pairs of consecutive identical letters is then
$$\binom{3}{2}\frac{5!}{3!} + \binom{3}{1}5!$$
Five pairs of consecutive identical letters:
For this to occur, there must be two blocks of three consecutive identical letters (two sets of two overlapping pairs) and a block of two consecutive identical letters drawn from the other available letter. There are three ways of selecting the two letters from which the blocks of three identical consecutive letters are drawn. We now have four objects to arrange, the two blocks of three consecutive identical letters, the block of two consecutive identical letters, and the remaining letter. The number of such arrangements is
$$\binom{3}{2}4!$$
Six pairs of consecutive identical letters:
The only ways for this to occur is if we have three blocks of three consecutive identical letters (three sets of two overlapping pairs). We saw above that there are
$$3!$$
such arrangements.
By the Inclusion-Exclusion Principle, the number of arrangements of the letters $a, a, a, b, b, b, c, c, c$ in which no two identical letters are consecutive is
$$\frac{9!}{3!3!3!} - \binom{3}{1}\frac{8!}{3!3!} + \left[\binom{3}{1}\frac{7!}{3!3!} + \binom{3}{2}\frac{7!}{3!}\right] - \left[6! + \binom{3}{1}\binom{2}{1}\frac{6!}{3!}\right] + \left[\binom{3}{2}\frac{5!}{3!} + \binom{3}{1}5!\right] - \binom{3}{2}4! + \binom{3}{3}3! = 174$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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Submonoid of $SL(2,\mathbb{Z})$ generated by two matrices. I'm trying to figure out which kind of matrices can be written as $A^{k_{1}}B^{k_{2}}...A^{k_{n}}B^{k_{n}}$ where
A=$\begin{pmatrix} 1&1\\0&1\end{pmatrix}$ and B=$\begin{pmatrix} 2&1\\1&1\end{pmatrix}$.
I did it for $C=\begin{pmatrix} 1&2\\0&1\end{pmatrix}$ and $D=\begin{pmatrix} 1&0\\2&1\end{pmatrix}$ and got that any word on $\{C,D\}$ appears as $\begin{pmatrix} a&2b\\2c&d\end{pmatrix}$with $a$ and $d$ odd and obviously $2b$ and $2c$ even. I can show this for $\{C,D\}$, but can't find an expression for $B^{k}$ so I can't go on and look for something similar.
| Regarding $B^k$:
$B$ is Hermitian and thus diagonalizable. See this for details. So, we have that:
$$B = \overbrace{\begin{pmatrix}
\frac{1+\sqrt{5}}{2} & \frac{1-\sqrt{5}}{2} \\ 1 & 1
\end{pmatrix}}^{P}
\overbrace{\begin{pmatrix}
\frac{3+\sqrt{5}}{2} & 0 \\ 0 & \frac{3-\sqrt{5}}{2}
\end{pmatrix}}^{D}
\overbrace{\begin{pmatrix}
\frac{1}{\sqrt{5}} & \frac{-1+\sqrt{5}}{2\sqrt{5}} \\ -\frac{1}{\sqrt{5}} & \frac{1+\sqrt{5}}{2\sqrt{5}}
\end{pmatrix}}^{P^{-1}}
$$
So, we can write $B = PDP^{-1}$ where $D$ is diagonal. It follows that $B^k = PD^kP^{-1}$, which you should be able to calculate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1945144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $\sum_{k=0}^{n}\binom{n}{k}^2=\frac{n+1}{n}\sum_{k=1}^{n}\binom{n}{k}\binom{n}{k-1}$ Show that $\sum_{k=0}^{n}\binom{n}{k}^2=\frac{n+1}{n}\sum_{k=1}^{n}\binom{n}{k}\binom{n}{k-1}$
I came across this result
while trying to solve this:
inductive proof for $\binom{2n}{n}$
My proof is cumbersome,
so I hope that
someone can come up
with a more elegant proof.
Note:
I know that
$\sum_{k=0}^{n}\binom{n}{k}^2
=\binom{2n}{n}
$.
| Apply four equation:
$$\sum_{k=0}^\infty\binom{r}{k}\binom{s}{n-k}=\binom{r+s}{n} [1]$$
and
$$\binom{n}{k}=\frac{n}{n-k}\binom{n-1}{k} [2]$$
and
$$\binom{n}{k}=\binom{n}{n-k} [3]$$
and (for k!= 0)
$$\binom{n}{k}=\frac{n}{k}\binom{n-1}{k-1} [4]$$
Apply [1] and [2] we have:
$$\sum_{k=0}^{n}\binom{n}{k}^2 = \binom{n+n}{n}=\binom{2n}{n}$$
$$=\frac{2n}{2n-n}\binom{2n-1}{n}=2\binom{2n-1}{n}$$
Note that $\binom{n}{n+1}=0$, Apply [3] and [1] and [4] we have
$$\frac{n+1}{n}\sum_{k=1}^{n}\binom{n}{k}\binom{n}{k-1} = \frac{n+1}{n}\sum_{k=0}^{n+1}\binom{n}{k}\binom{n}{n+1-k}$$
$$ = \frac{n+1}{n}\binom{n+n}{n+1}=\frac{n+1}{n}\binom{2n}{n+1} $$
$$ = \frac{n+1}{n}\frac{2n}{n+1}\binom{2n-1}{n} = 2\binom{2n-1}{n}$$
Conclusion:
$$\sum_{k=0}^{n}\binom{n}{k}^2=\frac{n+1}{n}\sum_{k=1}^{n}\binom{n}{k}\binom{n}{k-1} = 2\binom{2n-1}{n}$$
| {
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Find the equation of the circle passing ... Find the equation of the circle passing through the points $P(5,7)$, $Q(6,6)$ and $R(2,-2)$.
My Attempt:
Let the equation of the circle be:
$$x^2+y^2+2gx+2fy+c=0$$
The point $P(5,7)$ lies on the circle then,
$$5^2+7^2+10g+14f+c=0$$
$$10g+14f+c=-74$$-----(1)
The point $Q(6,6)$ lies on the circle then,
$$6^2+6^2+12g+12f+c=0$$
$$12g+12f+c=-72$$-------(2)
The point $R(2,-2)$ lies on the circle the,
$$2^2+(-2)^2+4g-4f+c=0$$
$$4g-4f+c=0$$-----(3).
Now, please help me from here.
| Subtract $2^{nd}$ equation from $3$ times the $3^{rd}$ to get
$$
c-12f=36
$$
Then subtract $2$ times the first from $5$ times the third to get:
$$
3c-48f=148
$$
From the last two, subtract $3$ times the first from the second to get $f=-10/3$. Plugging back in one them gives $c=-4$. Now pluging in one of the first three, gives $g=-7/3$.
| {
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How many natural solutions to $x_1+x_2+x_3+x_4=100$ with $x_1 \neq x_2 \neq x_3 \neq x_4$? I have invented a little exercise of combinatorics that I can't solve (without brute forcing): in how many ways can I choose 4 non-negative integers, the sum of which is 100 and they are all different? So:
$x_1+x_2+x_3+x_4=100$
$x_1, x_2, x_3, x_4 \in \mathbb N$
$x_1 \neq x_2 \neq x_3 \neq x_4$
I have thought that the result have to be the number of total combination $\binom {100 + 4 - 1} {4 - 1}$ minus the ways I can solve $2x_1 + x_2 + x_3 = 100$. Anyway, I am not able to solve it either.
The solution should be:
161664 or $\binom {100 + 4 - 1} {4 - 1} - 15187 = 176851 - 15187 = 161664$
Does anyone know how to calculate the combinations of this problem?
| Here is a solution with generating functions. Let $a_n$ be the number of objects you wish to count, namely the ordered quadruples of distinct non-negative integers with a sum of $n$, and let $f(x)=\sum_{n\ge0}a_nx^n$ be the ordinary generating function of this sequence. Since the integers in the sum are distinct, this generating function must be
$$f(x)=\left[\frac{y^4}{4!}\right]\prod_{n\ge0}(1+yx^n).$$
Then the problem of computing $a_n$ is reduced to finding the coefficient $[x^n]f(x)$. You can do this with standard techniques, but it does take some persistence! The result is an explicit formula for $a_n$, and in particular for $a_{100}$, as requested.
First, since sums are more familiar to me than products, I'll change the product to a sum by way of exponentials and logarithms:
$$f(x)=\left[\frac{y^4}{4!}\right]\exp\bigl(\sum_{n\ge0}\log(1+yx^n)\bigr).$$
Since we'll eventually extract the coefficient of $y^4$, we can expand the logarithm in its Taylor series up through the 4-th order powers of $y$, using $\log(1+t)\approx t-\frac12t^2+\frac13t^3-\frac14t^4$, and then sum the resulting geometric series:
\begin{eqnarray*}
% \nonumber to remove numbering (before each equation)
f(x) &=& \left[\frac{y^4}{4!}\right]\exp\bigl(\sum_{n\ge0}yx^i-\frac12y^2x^{2i}+\frac13y^3x^{3i}-\frac14y^4x^{4i} \bigr)\\
&=& \left[\frac{y^4}{4!}\right]\exp\bigl(\frac y{1-x}-\frac12\,\frac{y^2}{1-x^2}+\frac13\,\frac{y^3}{1-x^3}-\frac14\,\frac{y^4}{1-x^4}\bigr).
\end{eqnarray*}
At this stage, it's slightly easier to replace the exponential of the sum with the product of the exponentials. Then we'll expand each exponential in its Taylor series (again through the 4-th order powers of $y$), using $\exp(t)\approx1+t+\frac12t^2+\frac1{3!}t^3+\frac1{4!}t^4$:
\begin{eqnarray*}
% \nonumber to remove numbering (before each equation)
f(x) &=& \left[\frac{y^4}{4!}\right]\exp\bigl(\frac y{1-x}\bigr) \exp\bigl(-\frac12\frac{y^2}{1-x^2}\bigr)
\exp\bigl(\frac13\frac{y^3}{1-x^3}\bigr) \exp\bigl(-\frac14\frac{y^4}{1-x^4}\bigr) \\
&=& \left[\frac{y^4}{4!}\right]\left( 1+\frac y{1-x}+ \frac12\frac{y^2}{(1-x)^2}+ \frac16\frac{y^3}{(1-x)^3}+\frac1{24}\frac{y^4}{(1-x)^4}\right)\times\\
&&\quad \left( 1-\frac12\,\frac{y^2}{1-x^2}+\frac18\frac{y^4}{(1-x^2)^2}\right) \times\left( 1+\frac13\,\frac{y^3}{1-x^3}\right) \times\left( 1-\frac14\,\frac{y^4}{1-x^4}\right) .
\end{eqnarray*}
In the expansion of this product, there are just five terms with a coefficient of $y^4$, so
$$ f(x)=\frac 8{(1-x)(1-x^3)}-\frac6{(1-x)^2(1-x^2)}
+\frac1{(1-x)^4}+\frac3{(1-x^2)^2}-\frac6{1-x^4}.$$
Using basic facts about series expansions of rational functions, it is straightforward to compute the coefficient of $x^n$ in each of the five terms, resulting in
\begin{eqnarray*}
% \nonumber to remove numbering (before each equation)
a_n &=&[x^n]f(x) \\
&=& 8\left(\lfloor n/3\rfloor+1\right) -\frac32\bigl(\{1\text{ if }2|n\}+ (n+3)(n+1)\bigr)+\\
&&\frac{(n+3)(n+2)(n+1)}{6}+3\left\{n/2+1\text{ if } 2|n\right\}-6\left\{1\text{ if }4|n\right\}.
\end{eqnarray*}
As in other calculations, this gives $a_{100}=161\,664$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Bulgarian Mathematical Olympiad Round 4 problem 1. Problem: Prove that there exists a unique triple of positive integers greater than 1 such that product of any two increased by one is divisible by the third number.
I found $(2,3,7)$ to satisfy this but cant prove this. Please help.
| Clearly all numbers must be pairwise coprime, let them be $a<b<c$
Notice that $abc| ab+ac+bc+1$ and so $abc\leq ab+ac+bc+1$.
Notice that if $a\geq 3$ then $abc\geq 3bc> ab+ac+bc+1$ and so $a=2$.
It follows that $b|2c+1$ and $c|2b+1$ and so $bc|2b+2c+1$ In particular $bc\leq 2b+2c+1$.
It $b\geq 4$ then $bc\geq 4c>2b+2c+1$ and so $b=3$.
Hence $a=2,b=3$. And finally we must have $c|2\times 3+1$. So $c=7$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What will be the value of the following determinant without expanding it? $$\begin{vmatrix}a^2 & (a+1)^2 & (a+2)^2 & (a+3)^2 \\ b^2 & (b+1)^2 & (b+2)^2 & (b+3)^2 \\ c^2 & (c+1)^2 & (c+2)^2 & (c+3)^2 \\ d^2 & (d+1)^2 & (d+2)^2 & (d+3)^2\end{vmatrix} $$
I tried many column operations, mainly subtractions without any success.
| I wanted to present a different view of the problem that takes advantage of the sameness of the rows in a less computational way.
We can view the entries of each row as the evaluations of 4 polynomials of degree 2: $x^2$, $(x+1)^2$, $(x+2)^2$ and $(x+3)^2$ at a point.
Since the set of polynomials of degree at most 2, $\mathbb{P}_2$, forms a 3 dimensional vector space over $\mathbb{R}$, any collection of 4 polynomials will be linearly dependent. Hence there is some dependence relation among the columns. This dependence relation, $\alpha_0x^2+\alpha_1(x+1)^2+\alpha_2(x+2)^2+\alpha_3(x+3)^2=0$ means some combination of elementary column operations will create an all 0's column. Hence the matrix has linearly dependent columns.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof $\sum_{n=0}^\infty \frac{n^2+3n+2}{4^n} = \frac{128}{27}$
$\sum_{n=0}^\infty \frac{n^2+3n+2}{4^n} = \frac{128}{27}$ Given hint: $(n^2+3n+2) = (n+2)(n+1)$
I've tried converting the series to a geometric one but failed with that approach and don't know other methods for normal series that help determine the actual convergence value. Help and hints are both appreciated
| One may start with the standard finite evaluation:
$$
1+x+x^2+...+x^n=\frac{1-x^{n+1}}{1-x}, \quad |x|<1. \tag1
$$ Then by differentiating $(1)$ we have
$$
1+2x+3x^2+...+nx^{n-1}=\frac{1-x^{n+1}}{(1-x)^2}+\frac{-(n+1)x^{n}}{1-x}, \quad |x|<1, \tag2
$$ by differentiating once more one gets
$$
2\times 1+3\times 2 x^2+...+n\times (n-1)x^{n-2}=\frac{2-x^{n-1}\left(n+n^2 (1-x)^2+2x-nx^2\right)}{(1-x)^3},\tag3
$$ then by making $n \to +\infty$ in $(3)$, using $|x|<1$, one obtains
$$
\sum_{n=0}^\infty(n+1)\cdot n \cdot x^n=\frac{2 x}{(1-x)^3},\quad |x|<1, \tag4
$$
from which one gets the desired series by putting $x:=\dfrac14.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Conditional probability selecting a number and flipping a fair coin. One of the numbers 1, 2 or 3 is selected at random. Then a fair coin is flipped that number of times. What is the probability that the number 3 was selected given:
*
*no heads on the coin flip(s)
*1 head
*2 heads
*3 heads
I know for three heads, the probability is of course $1$ or $100\%$ but I can't wrap my head around finding it for the other 3 conditions.
| I think I solved my own problem here:
For the first case of no heads:
*
*There is $\frac{1}{2}$ chance to get no heads if the number picked is 1.
*There is $\frac{1}{4}$ chance to get no heads if the number picked is 2.
*There is $\frac{1}{8}$ chance to get no heads if the number picked is 3.
So we can set this up: $\frac{\frac{1}{8}}{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}}=\frac{1}{7}$
For the second case of one head:
*
*There is $\frac{1}{2}$ chance to get 1 head if the number picked is 1.
*There is $\frac{1}{2}$ chance to get 1 head if the number picked is 2.
*There is $\frac{3}{8}$ chance to get 1 head if the number picked is 3.
So we can set this up: $\frac{\frac{3}{8}}{\frac{1}{2}+\frac{1}{2}+\frac{3}{8}}=\frac{3}{11}$
For the third case of two heads:
*
*There is $\frac{0}{2}$ chance to get 2 heads if the number picked is 1.
*There is $\frac{1}{4}$ chance to get 2 heads if the number picked is 2.
*There is $\frac{3}{8}$ chance to get 2 heads if the number picked is 3.
So we can set this up: $\frac{\frac{3}{8}}{\frac{0}{2}+\frac{1}{4}+\frac{3}{8}}=\frac{3}{5}$
For the last case, its clear that the probability is 1.
| {
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Find all $0^\circ\leq A\leq360^\circ$ with $\tan A + \tan 2A + \tan 3A = 0$ solve:
$\tan A+ \tan 2A+ \tan 3A=0$
My Attempt:
$$\tan A+\tan 2A+\tan 3A=0$$
$$\tan A+\frac {2\tan A}{1-\tan^2A}+\frac {3\tan A-\tan^3A}{1-3\tan^2A}=0$$
$$\frac {\tan A-\tan^3A+2\tan A}{1-\tan^2A}+\frac {3\tan A-\tan^3A}{1-3\tan^2A}=0$$
What should I do further. Please suggest.
| $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\tan\pars{A} + \tan\pars{2A} + \tan\pars{3A} = 0\,,\qquad
A \in \bracks{0,2\pi}}$.
Lets $\ds{\tan\pars{A} \equiv x}$ such that
$\ds{\tan\pars{2A} = {2\tan\pars{A} \over 1 - \tan^{2}\pars{A}} =
{2x \over 1 - x^{2}}}$.
\begin{align}
0 & = \tan\pars{A} + \tan\pars{2A} + \tan\pars{3A} =
x + {2x \over 1 - x^{2}} + {\tan\pars{2A} +
\tan\pars{A} \over 1 - \tan\pars{2A} \tan\pars{A}}
\\[5mm] & =
x + {2x \over 1 - x^{2}} + {2x/\pars{1 - x^{2}} + x \over
1 - \bracks{2x/\pars{1 - x^{2}}}x} =
{2x\pars{2x^{4} - 7x^{2} + 3} \over 3x^{4} - 4x^{2} + 1}
\end{align}
Then,
\begin{equation}
x = \tan\pars{A}\ \in\ \Omega \equiv\braces{\pm\root{3},\pm{\root{2} \over 2},0}
\label{1}\tag{1}
\end{equation}
From \eqref{1}, you can deduce the values of $\ds{A \in \bracks{0,2\pi}}$. Note that
$\ds{\pars{3x^{4} - 4x^{2} + 1}_{\ x\ \in\ \Omega} \not= 0}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1958135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Equation for Distance of the Straight line from the Origin. By reduction of the equation $ax + by + c = 0 $ of a straight line to the normal form , we get
$$\left(\frac{-a}{\sqrt{a^2 + b^2}}\right)x + \left(\frac{-b}{\sqrt{a^2 + b^2}}\right)y = \frac{c}{\sqrt{a^2 + b^2}}$$
And,
$$p= \frac{∣c∣}{\sqrt{a^2 + b^2}}$$
And my textbook says that $p$ is the distance of the straight line from the origin. I don't know why we are getting it as a distance from origin?
I know $p= x\cos\theta + y\sin\theta$ ( where $p$ is distance of line from origin).
Also I want to know how can we relate both equations?
| By simple identification, the two equations are identical when
$$-\frac a{\sqrt{a^2+b^2}}=\cos\theta,\\
-\frac b{\sqrt{a^2+b^2}}=\sin\theta,\\
\frac c{\sqrt{a^2+b^2}}=p.$$
This is coherent as you verify $\cos^2\theta+\sin^2\theta=1$. If $c$ is negative, change all signs.
As one can verify by substitution,
$$x=p\cos\theta-t\sin\theta,\\y=p\sin\theta+t\cos\theta.$$ describes any point along the line, by varying $t$.
The distance from the origin to this point is given by, after simplification,
$$d=\sqrt{x^2+y^2}=\sqrt{p^2+t^2}.$$
The minimum value is indeed $p$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A function with domain that is asking for sum of a+b+c+d Let $f$ be a function such that $$ \sqrt {x - \sqrt { x + f(x) } } = f(x) , $$for $x > 1$. In that domain, $f(x)$ has the form $\frac{a+\sqrt{cx+d}}{b},$ where $a,b,c,d$ are integers and $a,b$ are relatively prime. Find $a+b+c+d.$
| Let $y = f(x)$. Our first goal is to find all pairs of real numbers $(x,y)$ that satisfy
[\sqrt{x - \sqrt{x + y}} = y.]We observe that $y$ must be nonnegative. Squaring both sides, we get
[x - \sqrt{x + y} = y^2,]so
[\sqrt{x + y} = x - y^2.]Squaring both sides again, we get
[x + y = x^2 - 2xy^2 + y^4,]so
[y^4 - 2xy^2 - y + x^2 - x = 0.]This is a quartic equation in $y$, which has no obvious solutions. However, we can re-write it as a quadratic equation in $x$, which we can solve:
[x^2 - (2y^2 + 1)x + y^4 - y = 0.]By the quadratic formula,
\begin{align*}
x &= \frac{(2y^2 + 1) \pm \sqrt{(2y^2 + 1)^2 - 4(y^4 - y)}}{2} \\
&= \frac{(2y^2 + 1) \pm \sqrt{4y^4 + 4y^2 + 1 - 4y^4 + 4y}}{2} \\
&= \frac{(2y^2 + 1) \pm \sqrt{4y^2 + 4y + 1}}{2} \\
&= \frac{(2y^2 + 1) \pm \sqrt{(2y + 1)^2}}{2} \\
&= \frac{2y^2 + 1 \pm (2y + 1)}{2},
\end{align*}so
[x = \frac{2y^2 + 1 + 2y + 1}{2} = y^2 + y + 1,]or
[x = \frac{2y^2 + 1 - (2y + 1)}{2} = y^2 - y.]We consider these solutions separately.
Case 1: $x = y^2 + y + 1$.
In this case,
[x + y = y^2 + 2y + 1 = (y + 1)^2,]so
[\sqrt{x + y} = |y + 1|.]We know that $y \ge 0$, so $y + 1 \ge 1$. Hence, $\sqrt{x + y} = y + 1$. Then
[x - \sqrt{x + y} = y^2 + y + 1 - (y + 1) = y^2,]so
[\sqrt{x - \sqrt{x + y}} = \sqrt{y^2} = |y| = y.]Thus, if $x = y^2 + y + 1$, then $(x,y)$ satisfies the given equation. We also know that $y \ge 0$, so $x \ge 1$. From the equation $x = y^2 + y + 1$, we have that $y^2 + y + 1 - x = 0$, so by the quadratic equation,
[y = \frac{-1 \pm \sqrt{1 - 4(1 - x)}}{2} = \frac{-1 \pm \sqrt{4x - 3}}{2}.]Since $y \ge 0$, we must take the root with the plus sign. Hence,
[y = \frac{-1 + \sqrt{4x - 3}}{2}.]
Case 2: $x = y^2 - y$.
Recall above that we also found $\sqrt{x+y}=x-y^2.$ Substituting $x=y^2-y$ gives
[\sqrt{y^2}=-y,]which is only satisfied when $y=0$, in which case $x = 0$, but we are given that $x > 0$.
We conclude that the function $f(x)$ is given by
$$f(x) = \frac{-1 + \sqrt{4x - 3}}{2}.$$This gives $a = -1, b = 2, c = 4, d = -3,$ or $a+b+c+d = \boxed{ 2 }.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1960328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integrating $\frac{x}{1+x^2}$ I need to integrate $\frac{x}{1+x^2}$ and I tried it like this:
$$f(x)=\frac{x}{1+x^2}=\frac{x}{(x+1)(x-1)}=\frac{x+1-1}{(x+1)(x-1)}= \frac{x+1}{(x+1)(x-1)}-\frac{1}{(x+1)(x-1)}=\frac{1}{x-1}-\frac{1}{1+x^2}$$
After I simplified it I integrate it!
$$\int{f(x)}=\int{\frac{1}{x-1}}-\int{\frac{1}{1+x^2}}$$
$\int{\frac{dx}{1+x^2}}$ becomes arctan(x) and I substitute $\int{\frac{dx}{x-1}}$ to $\int{\frac{du}{u}}$ now it becomes $ln(x-1)$ after this been done I get:
$$\int{f(x)}=ln(x-1)-arctan(x)$$
but when I type integrate x/(1+x^2) into wolframAlpha I it tells me the answer is: $$\int{f(x)}=\frac{1}{2}log(x^2+1)$$
where did I go wrong or what should I have done instead ?
Thank you for your help in advance,
Raavgo
| Easiest way is to use $$\int \frac{f'(x)}{f(x)} dx = \log(f(x)) + c$$
| {
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"source": "stackexchange",
"question_score": "2",
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Prove $(3+\sqrt{11})^{1/3}$ is irrational. I can't say I've gotten very far. You can show $3 + \sqrt{11}$ is irrational, call it $a$. Then I tried supposing it's rational, i.e.:
$a^{1/3}$ = $\frac{m}{n}$ for $m$ and $n$ integers.
You can write $m$ and $n$ in their canonical factorizations, then cube both sides of the equation...but I can't seem to derive a contradiction.
| Rational Root Theorem
If $P(x) = a_nx^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + \cdots + a_1x + a_0$, where all of the $a_i$ are integers, and $P\left( \dfrac uv \right)=0$ where $u$ and $v$ are integers, then $v \mid a_n$ and $u \mid a_0$.
Let $x = (3 + \sqrt{11})^{\frac 13}$. Then
\begin{align}
x^3 &= 3 + \sqrt{11} \\
x^3-3 &= \sqrt{11}\\
x^6 - 6x^3 + 9 &= 11 \\
x^6 - 6x^3 - 2 &= 0 \\
\end{align}
According to the rational root theorem, the only possible rational zeros of the polynomial $P(x) = x^6 - 6x^3 - 2$ are $1, -1, 2,$ and $-2$.
We find $P(1) = -7$, $P(-1)=-5$, $P(2)=14$ and $P(-2)=110$
Hence $P(x) = x^6 - 6x^3 - 2$ has no rational roots. It follows that
$x = (3 + \sqrt{11})^{\frac 13}$ is irrational.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Existence of integer solutions to $a^2 + b^2 + c^2 = d^2$ for specified $d$ Previous questions contained cubes instead of squares, that was my mistake, sorry.
How to tell if $a^2 + b^2 + c^2 = d^2$ has any integer, non-zero solutions for specified positive integer $d$? Is there any criterium or algorithm for this?
EDIT: To clarify, $a, b, c$ are non-zero integers and $d$ is specified and it is a positive integer.
| 1) As regards the original post: existence of integer solutions to
$a^3 + b^3 + c^3 = d^3$ for a given $d$.
Note that the following identity holds:
$$(9t^4)^3+(1-9t^3)^3 +(3t-9t^4)^3=1.$$
Therefore, given a positive integer $d$ there is always an integer solution,
$$(9dt^4)^3+(d(1-9t^3))^3+(d(3t-9t^4))^3 = d^3.$$
For more details, take a look at this page (with references): https://ckrao.wordpress.com/2012/04/10/integers-equal-to-the-sum-of-three-cubes/
2) As regards the new post: existence of integer solutions to
$a^2 + b^2 + c^2 = d^2$ for a given $d$.
Note that by a theorem of Legendre, if we allow $a,b,c$ to be zero, then
$$a^2 + b^2 + c^2=N$$
iff $N$ is NOT of the form $4^n(8m+7)$.
For numbers that are the sum of three non zero squares see the OEIS sequence A000408.
In the list we find the squares 9, 36,49, 81 (but not 4,25,64,100).
Since
$$t^2+(t+1)^2+(t(t+1))^2=(t^2+t+1)^2$$
it follows that if $d$ has a divisor of the form $t^2+t+1$ with $t\geq 1$ (like 3, 7, 13, 31, 43) then there is a solution.
According to this link, there is a finite set $T$ such that any positive integer is a sum of three non-zero squares unless $n$ is of the form $4^n(8m+7)$ or of the form $4^am$ where $m\in T$.
It seems that
$$T=\{1,2,5,10,13,25,37,58,85,130\}$$
therefore $a^2 + b^2 + c^2=d^2$ has a solution iff $d\not= 2^n$ or $d\not= 2^n\cdot 5$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Need help in indefinite integral elimination I'm trying to get a easy and short answer for that indefinite integral:
$$\int\frac{\sqrt{x^2+x+1}}{(x+1)^2}\,dx$$
But everything I find is very big and difficult to understand (too much so it seems like a wrong result). Is there any way to solve it in a more beautiful way?
| Let's calculate by parts first
$\int{\frac{\sqrt{x^2+x+1}}{(1+x)^2}dx} = -\frac{\sqrt{x^2+x+1}}{1+x} + \int{\frac{1}{x+1}\cdot\frac{2x+1}{2\sqrt{x^2+x+1}}dx}\\
\int{\frac{\sqrt{x^2+x+1}}{(1+x)^2}dx} = -\frac{\sqrt{x^2+x+1}}{1+x} +\int{\frac{(2x+2)-1}{2(x+1)\sqrt{x^2+x+1}}dx}\\
\int{\frac{\sqrt{x^2+x+1}}{(1+x)^2}dx} = -\frac{\sqrt{x^2+x+1}}{1+x} +\int{\frac{1}{\sqrt{x^2+x+1}}dx} - \frac{1}{2}\int{\frac{1}{(x+1)\sqrt{x^2+x+1}}dx}
$
Integral $\int{\frac{1}{\sqrt{x^2+x+1}}dx}$ can be calculated with substitution $\sqrt{x^2+x+1} = u - x$
and integral $\int{\frac{1}{(x+1)\sqrt{x^2+x+1}}dx}$
can be calculated with substitution $\sqrt{x^2+x+1} = u\cdot x - 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1967094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the range of $a$ such $|f(x)|\le\frac{1}{4},$ $|f(x+2)|\le\frac{1}{4}$ for some $x$.
Let $$f(x)=x^2-ax+1.$$
Find the range of all possible $a$ so that there exist $x$ with
$$|f(x)|\le\dfrac{1}{4},\quad |f(x+2)|\le\dfrac{1}{4}.$$
A sketch of my thoughts: I write
$$f(x)=\left(x-\dfrac{a}{2}\right)^2+1-\dfrac{a^2}{4}\ge 1-\dfrac{a^2}{4}$$
so if $1-\dfrac{a^2}{4}>\dfrac{1}{4}$ or $-\sqrt{3}<a<\sqrt{3}$
this case impossible
But I don't know how to prove the other case, or if this there are better ideas.
| To start off a lot of time can be saved by only handling $a\geq0$ since both $|x^2-ax+1|\leq \frac{1}{4}$ and $|(x+2)^2-a(x+2)+1|\leq\frac{1}{4}$ are odd inequalities all of solutions that we find for $a\geq0$ will be the same except with opposite signs.
As a first step we should try to isolate $a$ from both of the inequalities. Starting with $|x^2-ax+1|\leq \frac{1}{4}$ we remove the absolute value and rewrite the single inequality as a double inequality, $-\frac{1}{4}\leq x^2-ax+1\leq \frac{1}{4}$ from there working step by step we can solve for $a$.
$$-\frac{1}{4}\leq x^2-ax+1\leq \frac{1}{4}$$
$$-\frac{5}{4}\leq x^2-ax\leq-\frac{3}{4}$$
$$-\frac{5}{4x}\leq x-a\leq-\frac{3}{4x}$$
$$-\frac{5}{4x}-x\leq -a\leq-\frac{3}{4x}-x$$
$$\frac{5}{4x}+x\geq a\geq\frac{3}{4x}+x$$
For the second inequality a similar path can be taken to obtain $$\frac{5}{4(x+2)}+x+2\geq a \geq \frac{3}{4(x+2)}+x+2$$
From here we can find the overlap of these two double inequalities and the range of $a$ for which they overlap which will be our solution. The upper bound of the over lap can be found by finding the intersection of the corresponding upper bonds of each double inequality.
$$\frac{5}{4x}+x=\frac{5}{4(x+2)}+x+2$$
$$\frac{5}{4x}=\frac{5}{4(x+2)}+2$$
Assuming that $x\neq0$,$x\neq-2$
$$\frac{5(x+2)}{4x}=\frac{5}{4}+2(x+2)$$
$$\frac{5(x+2)}{4}=\frac{5x}{4}+2x(x+2)$$
$$\frac{5x+10}{4}=\frac{5x}{4}+2x^2+4x$$
$$\frac{5x+10}{4}=\frac{5x}{4}+2x^2+4x$$
$$5x+10=5x+8x^2+16x$$
$$0=8x^2+16x-10$$
$$0=4x^2+8x-5$$
Using the quadratic formula
$$x=\frac{-8\pm\sqrt{64-4(4)(-5)}}{8}$$
$$x=\frac{-8\pm\sqrt{144}}{8}$$
$$x=\frac{-8\pm12}{8}$$
$$x=-1\pm1.5$$
$$x=\{.5,-2.5\}$$
We can discard the negative solution for now since for now we are only considering $a\geq0$ which also means that $x\geq0$.
Using the upper bound from the first inequality to calculate the corresponding value of $a$ gives us $a=3$ as our maximum for $a$'s range.
Working through the same problem for our lower bound gives the values $x=\{\frac{1}{2}\sqrt{7}-1,-\frac{1}{2}\sqrt{7}-1\}$, discarding the negative one for the same reason as before and plugging in $\frac{1}{2}\sqrt{7}-1$ into the lower bound equations of either the first or second double inequality gives us $a=\sqrt{7}$ as our lower bound for the range of $a$. That gives us the range for $a\geq0$ as $[\sqrt{7},3]$. Since both inequalities are odd we can extrapolate that for $a\leq0$ the range is $[-3,-\sqrt{7}]$, combining the two gives use that the range for $a$ is $[\sqrt{7},3]\cup[-3,-\sqrt{7}]$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Internal angle bisectors and right-angled triangle? Imagine a right angled triangle. Two of its internal bisectors are of length $7cm$ and $4cm$, respectively. The first internal angle bisector belongs to the angle of the side $a$ and the second one belongs to the angle of the side $b$ Calculate the length of the hypotenuse of the right angled triangle.
| Case 1: Suppose that $a,b$ are the legs of the right triangle. Let then $c$ be the length of the hypotenuse, and $C = \frac{\pi}{2}, A, B$ the respective angles.
It follows by inspection that:
$$ 7 = \frac{c \; \cos A}{\cos A/2} \quad \text{and} \quad 4 = \frac{c \; \cos B\; }{\cos B/2}$$
Since $\cos \frac{x}{2}=\sqrt\frac{1 + \cos x}{2}$ and $B = \frac{\pi}{2} - A$, the above gives:
$$7^2 = \frac{2 c^2 \cos^2 A}{1+ \cos A} \quad \text{and} \quad 4^2 = \frac{2 c^2 \sin^2 A}{1+ \sin A}$$
After dividing the two equalities:
$$\left(\frac{4}{7}\right)^2 = \tan^2 A \;\frac{1+ \cos A}{1+ \sin A}$$
In terms of $t=\tan A/2$ (with $\cos A = \frac{1 - t^2}{1+t^2}, \sin A = \frac{2t}{1+t^2}$) the equation becomes:
$$\require{cancel}\frac{2 \cdot \bcancel{8}}{49}=\frac{\bcancel{4} t^2}{(1-t^2)^2} \cdot \frac{\bcancel{2}}{\cancel{1+t^2}} \cdot \frac{\cancel{1+t^2}}{1+2 t+t^2}$$
$$2(1-t^2)^2(1+t)^2 - 49 t^2 = 0$$
$$\Big(\sqrt{2}(1-t^2)(1+t) - 7 t\Big)\Big(\sqrt{2}(1-t^2)(1+t) + 7 t\Big) = 0$$
$$\Big(\sqrt{2} + (\sqrt{2}-7) t- \sqrt{2} t^2-\sqrt{2} t^3\Big)\Big(\sqrt{2} + (\sqrt{2}+7) t- \sqrt{2} t^2-\sqrt{2} t^3\Big) = 0$$
Since $A/2 \in (0, \pi / 4)$ the roots of interest are $t = \tan A/2 \in (0,1)$ and it can be shown that the only such root belongs to the first cubic. The value can be calculated exactly by solving the cubic using Cardano's formula, for example, though the resulting expression is not pretty.
Numerically, the eligible root is $t \approx 0.23579$ corresponding to $A \approx 26.53447^{\circ}$ and the hypotenuse $c \approx 7.61533$ (which agrees with the Maple solver result posted by Han de Brujin in his answer).
Case 2: Suppose that one of $a,b$ is the hypotenuse. Since the shortest bisector corresponds to the largest angle, the right angle bisector must be $4$ and side $b$ must be the hypotenuse. Then (by the bisector length formula for the former):
$$ 4^2 = \frac{2 a^2 c^2}{(a+c)^2} = \frac{2 \;b^2 \;\sin^2 A \;\cos^2 A}{(\sin A + \cos A)^2} \quad \text{and} \quad 7 = \frac{b \;\cos A}{\cos A/2}$$
$$ 4^2 = \frac{2 \;b^2 \;\sin^2 A \;\cos^2 A}{1 + 2 \;\sin A \;\cos A} \quad \text{and} \quad 7^2 = \frac{2 b^2 \;\cos^2 A}{1 + \cos A}$$
Similar to the previous case, dividing and rationalizing in $t = \tan A/2$ gives:
$$\left(\frac{7}{4}\right)^2 = \frac{1 + 2 \;\sin A \;\cos A}{(1 + \cos A) \sin^2 A} = \frac{(1+t^2)^2 + 4 t (1-t^2)}{(1+t^2)^2} \cdot \frac{(1+t^2)^3}{8 t^2}$$
$$\cdots$$
$$2 t^6-8 t^5+6 t^4-43 t^2+8 t+2 = 0$$
The sextic can be shown to have a unique real root in $(0,1)$ which is numerically $t \approx 0.32999$ and corresponds to $A \approx 36.52438^{\circ}$ and the hypotenuse $b \approx 8.27202$.
| {
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Find a line which is tangent to the curve $y=x^4-4x^3$ at 2 points How can I solve this? Should I set the two points as $a^4-4a^3$ and $b^4-4b^3$?
| Suppose a line $y=mx+c$ is tangent to the curve $y=x^4-4x^3$ at two points. Then the difference between these two equations must be a quartic polynomial with two double roots:
$$x^4-4x^3-mx-c=(x-a)^2(x-b)^2$$
$$=x^4-2(a+b)x^3+(a^2+4ab+b^2)x^2-2ab(a+b)x+(ab)^2$$
Comparing coefficients, we have
$$2(a+b)=4$$
$$a^2+4ab+b^2=(a+b)^2+2ab=0$$
$$m=2ab(a+b)$$
$$c=-(ab)^2$$
The first two relations yield values for $a+b$ and $ab$:
$$a+b=\frac42=2$$
$$ab=-\frac{(a+b)^2}2=-\frac{2^2}2=-2$$
The last two relations allow $m$ and $c$ to be evaluated:
$$m=2\cdot-2\cdot2=-8$$
$$c=-(-2)^2=-4$$
Hence the line tangent to $y=x^4-4x^3$ at two points is $y=-8x-4$.
| {
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Fourier series of $\cos\mu x$ Find the Fourier series for the $2\pi$-periodic function $F(x)=\cos\mu x$, $x\in[-\pi,\pi]$ and where $\mu \notin \mathbb Z$. Simplify the result for $\mu=\frac{1}{4}$. Hence show that
$$\pi\sqrt{2}=A-\sum^\infty_{n=1}\frac{(-1)^n}{Bn^2-\frac{1}{8}},$$
where $A$ and $B$are integers to be determined.
$a_n,a_0=0$
\begin{align}
b_n & =\frac{2}{\pi}\int^\pi_0 \cos\mu x\sin nx\, dx \\[10pt]
& =\frac{1}{\pi} \left[-\frac{\cos{(n-\mu)x}}{n-\mu}-\frac{\cos{(n+\mu)x}}{n+\mu}\right] \\[10pt]
& =\frac{1}{\pi} \left[\frac{(-1)^n(-2n)}{n^2-\mu^2}+\frac{2n}{n^2-\mu^2}\right] \\[10pt]
& =\frac{2n}{\pi(n^2-\mu^2)}[-(-1)^n+1]
\end{align}
$$-(-1)^n+1=2 \text{ if } n=2k+1$$
$$b_n=\frac{4(2k+1)}{\pi[(2k+1)^2-\mu^2]}$$
Putting $\mu=\frac{1}{4}$,
$$\cos{\frac{1}{4}x}=\frac{4}{\pi}\sum^{\infty}_{k=1} \frac{2k+1}{(2k+1)^2-\frac{1}{16}} \cdot \sin{nx}$$
This does not look like what I have to prove.
| Hint: Once you get the correct Fourier cosine series
$$\cos(\mu x)=\frac{\sin(\mu \pi)}{\mu \pi}+\frac{2\mu\sin(\mu\pi)}{\pi}\sum_{n\geq 1}\frac{(-1)^n}{\mu^2-n^2}\cos(nx)\tag{1}$$
by setting $\mu=\frac{1}{4}$ you get
$$\cos\frac{x}{4}=\frac{2\sqrt{2}}{\pi}+\frac{\sqrt{2}}{4\pi}\sum_{n\geq 1}\frac{(-1)^n}{\frac{1}{16}-n^2}\cos(nx)\tag{2}$$
and by evaluating $(2)$ at $x=0$
$$ 1 = \frac{2\sqrt{2}}{\pi}+\frac{\sqrt{2}}{4\pi}\sum_{n\geq 1}\frac{(-1)^n}{\frac{1}{16}-n^2}\tag{3} $$
so:
$$\boxed{ \pi\sqrt{2}=\color{red}{4}-\sum_{n\geq 1}\frac{(-1)^n}{\color{red}{2}n^2-\frac{1}{8}}}\tag{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1971226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Discrete mathematics, divisibility How can I prove that for all $n\in\mathbf{N}$ that $6 | n^5 + 5n$?
I tested for $n = 2$ and got $6 | 32 + 10 = 42$.
| Use modular arithmetic and solve for the finite number of cases $n \equiv 0, 1, 2, 3, 4, 5 \pmod{6}$.
For example, $4 \cdot 4 = 16 \equiv 4 \pmod{6}$, so that $4^5 \equiv 4 \pmod{6}$, and $n^5 + 5n \equiv 4 + 20 \pmod{6} \equiv 24 \pmod{6} \equiv 0 \pmod{6}$.
The five other cases are similar.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1974310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Compute $ \sqrt{n^2 +3} - \sqrt{n^2 +2} $. I have a question concerning determining a basic limit of $ \sqrt{n^2 +3} - \sqrt{n^2 +2} $ as $n \rightarrow \infty $. Upon using the difference of squares, this yields: $ \frac{1}{\sqrt{n^2 +3} + \sqrt{n^2 +2}} $Now the limit seems intuitive, but I cant find a formal way of determining it. Any hints about what the next step should be would be appreciated.
| Let $f(x) = \sqrt{1+3x} -\sqrt{1+2x}$.
Then $f(0) = 0$ and $f'(0) = {1 \over 2}$, hence for $x$ sufficiently
small, we have $|f(x)| \le |x|$.
Hence for $n$ sufficiently large, we have
$|\sqrt{n^2+3} -\sqrt{n^2+2} | = n |f({1 \over n^2})| \le {1 \over n}$,
from which the limit follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1976042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
General form of a matrix that is both centrosymmetric and orthogonal A centrosymmetric matrix is of the form $JAJ=A$ where $J$ is the counter identity matrix, i.e. $$J=
\begin{bmatrix}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0 \\
\end{bmatrix}
$$And an orthogonal matrix is $A$ such that $A^{-1}=A^T$
The only such matrix I could think of having these property is the identity matrix $$I=
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{bmatrix}
$$Is there a general form that A takes that has these properties?
| Note that $J^2 = I$ and so the condition $JAJ = A$ is equivalent to the condition $JA = AJ$. That is, $A$ and $J$ should commute. The matrix $J$ is orthogonally diagonalizable with eigenvalues $1,-1$ and eigenspaces
$$ V_{1} = \operatorname{span} \{ e_2, \frac{1}{\sqrt{2}}(e_1 + e_3) \}, V_{-1} = \operatorname{span} \{ \frac{1}{\sqrt{2}}(e_1 - e_3) \} $$
where $e_i$ are the standard basis vectors.
Any matrix that commutes with $J$ must keep the eigenspaces of $J$ invariant and, since $J$ is diagonalizable, any matrix that satisfies $AV_{1} \subseteq V_{1}$ and $AV_{-1} \subseteq V_{-1}$ will indeed commute with $J$. Set
$$ v_1 = e_2, v_2 = \frac{1}{\sqrt{2}}(e_1 + e_3), v_3 = -\frac{1}{\sqrt{2}}(e_1 - e_3), \\
P = \begin{pmatrix} v_1 & | & v_2 & | & v_3 \end{pmatrix} $$
so $P$ is the orthogonal $3 \times 3$ matrix whose columns are the eigenvectors of $J$. A matrix $A$ will commute with $J$ if and only if we have
$$ P^T A P = \begin{pmatrix} a & b & 0 \\ c & d & 0 \\ 0 & 0 & e \end{pmatrix} $$
for some $a,b,c,d,e \in \mathbb{R}$. Finally, $A$ will be orthogonal if and only if $P^T A P$ will be orthogonal and so $P^T A P$ must be in one of the forms
$$ P^T A P = \begin{pmatrix} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{pmatrix},
\begin{pmatrix} \sin \theta & \cos \theta & 0 \\ \cos \theta & -\sin \theta & 0 \\ 0 & 0 & -1 \end{pmatrix}, \\
\begin{pmatrix} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & -1 \end{pmatrix}, \begin{pmatrix} \sin \theta & \cos \theta & 0 \\ \cos \theta & -\sin \theta & 0 \\ 0 & 0 & +1 \end{pmatrix} $$
where $\theta \in \mathbb{R}$. The first case corresponds to rotation matrices around the $v_3$ axis and $A$ can be written explicitly as
$$ A = P(P^TAP)P^T = \begin{pmatrix} \frac{\cos \theta + 1}{2} & \frac{\sin \theta}{\sqrt{2}} & \frac{\cos \theta - 1}{2} \\
-\frac{\sin \theta}{\sqrt{2}} & \cos \theta & -\frac{\sin \theta}{\sqrt{2}} \\
\frac{\cos \theta - 1}{2} & \frac{\sin \theta}{\sqrt{2}} & \frac{\cos \theta + 1}{2}.
\end{pmatrix} $$
One can obtain similar expression for the other three cases.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1976654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove that $ 2/1!+4/3!+6/5!+8/7!+\dots = e$
Prove that $ 2/1!+4/3!+6/5!+8/7!+\dots = e$
My attempt:
I googled the problem, and I found that $\sum_{(n=1)}^∞ \cfrac{2 n}{(2 n-1)!} = e$
I also found that $\sum _{n=0}^{\infty \:}\cfrac{2n+2}{\left(2n+1\right)!}$ is equal to $e$.
How can I prove this?
| Let $f(x)= x \sin(x)$. Then
$$
f(x) = \frac{x^2}{1!} - \frac{x^4}{3!} + \frac{x^6}{5!} -\frac{x^8}{7!} + \cdots
$$
and so
$$
f'(x) = \frac{2x}{1!} - \frac{4x^3}{3!} + \frac{6x^5}{5!} -\frac{8x^7}{7!} + \cdots
$$
Therefore
$$
f'(i) = \frac{2i}{1!} + \frac{4i}{3!} + \frac{6i}{5!} +\frac{8i}{7!} + \cdots = iS
$$
On the other hand,
$$
f'(x) = \sin(x) + x \cos(x)
$$
and so
$$
f'(i) = \sin(i) + i \cos(i) = i (\cos(-i) + i \sin(-i)) = ie^{i(-i)} = ie
$$
Therefore, $S=e$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1979848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Find matrix $A$. How many solutions are there? The matrices below is matrix multiplcation of $AB=C$ but someone has changed the order of the coloumns in $C$ and erased what was in $A$. What is matrix $A$? Is it the only one? I have no idea how to start. I appreciate all the help I can get.
$$\begin{bmatrix} & & & \\ & & & \\ & & &\\& & & & \end{bmatrix}
\cdot\begin{bmatrix}2 & 0 &-1 & -2&-4&-4\\-1 & 1 & 0 & -1&2&-1\\ -4& 0 & 1 & 3&8&5\\ -1&0&0&1&2&1\end{bmatrix}=\begin{bmatrix}1 & 1 &3 & -1&0&2\\0 & 1 & 1 & 0&0&0\\0 & 0 & 0&-1&0& 2\\ 0&0&0&0&1&0\end{bmatrix}$$
| There is a unique solution to this problem. The problem states
that $AB=C$ holds, but that someone has changed the order of the
columns in $C$ and erased what was in $A$. We are told what the matrix
$B$ is, but not told what $C$ is. We are only told what $C$ looks like
after its columns have been permuted. That is, we are told what
$CP$ is for some permutation matrix $P$ which has not been specified.
The reduced row echelon matrix for $B$ (and $C$)
plays a role, and I will label it $R$. I will write
$L$ and $M$ for the $4\times 4$-matrices for which
$LB=R=MC$. (These are uniquely determined, since $B$ and $C$ have rank 4.)
It can be shown that the matrices are:
$$
L = \left[\begin{array}{rrrr}
-1&0&-1&1\\
-2&1&-2&3\\
-1&0&0&-2\\
-1&0&-1&2
\end{array}\right],\quad
B = \left[\begin{array}{rrrrrr}
2&0&-1&-2&-4&-4\\
-1&1&0&-1&2&-1\\
-4&0&1&3&8&5\\
-1&0&0&1&2&1
\end{array}\right],
$$
$$
M = \left[\begin{array}{rrrr}
0&0&-1&0\\
0&0&0&1\\
1&-1&-1&0\\
0&1&0&0
\end{array}\right],\quad
C = \left[\begin{array}{rrrrrr}
-1&0&1&1&2&3\\
0&0&0&1&0&1\\
-1&0&0&0&2&0\\
0&1&0&0&0&0
\end{array}\right],
$$
$$
R = LB = MC
= \left[\begin{array}{rrrrrr}
1&0&0&0&-2&0\\
0&1&0&0&0&0\\
0&0&1&0&0&2\\
0&0&0&1&0&1
\end{array}\right],
$$
and
$$
A = \left[\begin{array}{rrrr}
-1&0&0&-1\\
-1&0&-1&2\\
1&0&1&-1\\
-2&1&-2&3
\end{array}\right],\quad
P = \left[\begin{array}{rrrrrr}
0&0&0&1&0&0\\
0&0&0&0&1&0\\
1&0&0&0&0&0\\
0&1&0&0&0&0\\
0&0&0&0&0&1\\
0&0&1&0&0&0
\end{array}\right].
$$
Justification:
We are told that $AB=C$. We are given $B$, but not $A$. We
are given a matrix obtained by $C$ by permuting the columns of $C$,
that is we are given $CP$ for some $6\times 6$ permutation matrix,
but we are not told what $P$ is.
Write $\textrm{Null}(X)$ for the nullspace of matrix $X$.
Since $AB=C$, it follows that $ABP=CP$, so
$x\in\textrm{Null}(CP)=\textrm{Null}(ABP)$
iff $P^{-1}x\in\textrm{Null}(AB)$.
But $B$ and $C$ have rank $4$, so $A$ must be invertible, so
$\textrm{Null}(AB)=\textrm{Null}(B)$.
To summarize, $x\in\textrm{Null}(CP)$ iff
$P^{-1}x\in \textrm{Null}(B)$. Equivalently,
$P\cdot \textrm{Null}(B)=\textrm{Null}(CP)$.
Now we are given the matrices
$B$ and $CP$, so we can compute these nullspaces and
try to determine which permutation matrices satisfy
$P\cdot \textrm{Null}(B) =
\textrm{Null}(CP)$.
There is only one such matrix, and it is the one
I have written above. Here I only sketch the kind of reasoning I used
to determine $P$.
Vectors $x\in \textrm{Null}(B)$ must satisfy
the following conditions.
*
*The second coordinate of $x$ must be $0$.
*The third coordinate of $x$ must be $+2$ times
the fourth coordinate of $x$ and $-2$ times the sixth coordinate of $x$.
*The first coordinate of $x$ must be $+2$ times the fifth
coordinate, but does not have to be $-2$ times another coordinate.
Now vectors in $\textrm{Null}(CP)$ must satisfy
the same conditions with different coordinate orders,
and comparing the coordinate orders we can determine $P$.
Once $P$ is known, we are basically done. We were given
$CP$, so with $P$ we can compute $C$.
From $B$ and $C$ we can solve for $A$.
I solved for $A$
by row reducing $[B|I]$ and $[C|I]$ to $[R|L]$ and $[R|M]$ respectively,
thereby finding matrices
$L$ and $M$ such that $LB=R=MC$. Then $(M^{-1}L)B=C$, forcing
$A=M^{-1}L$. This I computed by row reducing $[M|L]$ to $[I|M^{-1}L]$.
I then used Maple to check that $AB$ is indeed a matrix that differs from $CP$ by a permutation of columns.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1980645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Quadratic equation with absolute values Solve $|x^{2}+x-2|+|x^{2}-x-2|=2$
My attempt:
$|x^{2}+x-2|=
x^{2}+x-2, x\in (-\infty,-2]\cup[1,\infty)$
$|x^2+x-2|=-x^{2}-x+2, x\in(-2,1)\\$
$|x^{2}-x-2|=x^{2}-x-2, x\in (-\infty,-1]\cup[2,\infty)$
$|x^{2}-x-2|=-x^{2}+x+2, x\in (-1,2)\\$
$1)$ $x\in(-\infty,-2]\cup[1,\infty)$ and $x\in(-\infty,-1]\cup[2,\infty)$
$x^{2}+x-2+x^{2}-x-2=2$
$\Rightarrow 2x^{2}-6=0$
$\Rightarrow x^{2}=3$
$\Rightarrow x_1=\sqrt{3}, x_2=-\sqrt{3}$
$2)$ $x\in(-\infty,-2]\cup[1,\infty)$ and $x\in(-1,2) \Rightarrow x\in[1,2)$
$x^{2}+x-2-x^{2}+x+2=2$
$\Rightarrow 2x=2$
$\Rightarrow x=1$
$3)$ $x\in(-2,1)$ and $x\in(-\infty,-1]\cup[2,\infty)$ $\Rightarrow x\in(-2,-1]$
$-x^{2}-x+2+x^{2}-x-2=2$
$\Rightarrow x=-1$
$4)$ $x\in(-2,1)$ and $x\in(-1,2)$ $\Rightarrow x\in(-1,1)$
$-x^{2}-x+2-x^{2}+x+2=2$
$\Rightarrow -2x^{2}+2=0$
$\Rightarrow -2x^{2}=-2$
$\Rightarrow x^{2}=1$
$\Rightarrow x=\pm{1}$ but that's not in the interval $(-1,1)$ so I can throw away that solution.
My question is how do I find the intersection in $1)$ $x\in(-\infty,-2]\cup[1,\infty)$ and $x\in(-\infty,-1]\cup[2,\infty)$ since $-\sqrt{3}$ and $\sqrt{3}$ should not be the solutions.
| The derivative of the function is
$$(2x+1)\text{sign}|x^2+x-2|+(2x-1)\text{sign}|x^2-x-2|.$$
Depending on the signs, this reduces to $\pm4x$ or $\pm2$ so that the only possible root occurs at $x=0$. The changes of sign can occur at $-2,-1,1,2$ so that the function is certainly monotonous in the intervals
$$]-\infty,-2,-1,0,1,2,\infty[.$$
The endpoints correspond to the function values
$$\infty,4,\color{green}2,4,\color{green}2,4,\infty$$ and you clearly see the two solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1981701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.