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probability that $(a+b\omega+c\omega^2)\cdot (a+b\omega^2+c\omega) = 1$
If $\omega$ is a complex cube root of unity with $0<\arg(\omega)<2\pi. $ A fair die is thrown three times
If $a,b,c$ are the numbers obtained on the dice , Then probability that
$(a+b\omega+c\omega^2)\cdot (a+b\omega^2+c\omega) = 1$
$\bf{Attempt:}$ Given $a,b,c\in \{1,2,3,4,5,6\},$ Then Total number of ways $n(S) = 6^3$
Now for favourable cases $n(A)$
$(a+b\omega+c\omega^2)\cdot (a+b\omega^2+c\omega) = =a^2+b^2+c^2-ab-bc-ca = 1$
$(a-b)^2+(b-c)^2+(c-a)^2 = 2$
Could some help me how to solve it , Thanks
| You are looking for the probability of
$$(a-b)^2+(b-c)^2+(c-a)^2 = 2, \qquad \text{ when }a,b,c \in \{1,2,3,4,5,6\}$$
For the sum of the three squares to be $2$, the only way it can happen is when two of them are $1$ and the third is a $0$. For example, $a=b$, then $c=a \pm 1$. So you are looking for triples of the form $(a,a,a+1)$ or $(a,a,a-1)$ and their cyclic permutations.
For each $a \in \{2,3,4,5\}$, the number of such triplets (including their cyclic permutations) is $24$, then for $a=1$ we can only have $(a,a,a+1)$ and its permutations, likewise for $a=6$ we can only have $(a,a,a-1)$ and its permutations. Therefore
Thus the probability is $\frac{24+3+3}{6^3}=\frac{30}{6^3}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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primality of $\frac{(2^{19}+1)}{3}$ and $\frac{(2^{23}+1)}{3}$ This is from an exercise in Burton's Number theory book. I have a hard time solving this. How could we prove that $\displaystyle\frac{(2^{19}+1)}{3}$ and $\displaystyle\frac{(2^{23}+1)}{3}$ are primes. I know that the primes dividing the numbers are of the form are $38k+1$ and $46k+1$ respectively. How can we proceed without a computer? Any hints. Thanks beforehand.
| In general, if $q>3$ is prime and $p$ a prime divisor of $N=\frac{2^q+1}{3}$, then, as you argued, $p=2qk+1$ for some $k$.
So: $2^{(p-1)/2}\equiv (-1)^k\pmod p$.
If $k$ is even, then $2$ is a square modulo $p$ which means that $p\equiv \pm 1\pmod{8}$ or $2qk\equiv 0,6\pmod{8}$ or $qk\equiv 0,3\pmod{4}$. But since $k$ is even, this requires $k$ to be divisible by $4$, so $p\equiv 1\pmod{8q}$.
If $k$ is odd, then $2$ is not a square modulo $p$ so $p\equiv 3,5\pmod{8}$ so $qk\equiv 1,2\pmod{4}$, which means that $k\equiv q\pmod{4}$. So $p\equiv 1+2q^2\pmod{8q}$.
So you can check only the primes $p\equiv 1,1+2q^2\pmod{8q}$, for $p<\sqrt{N}$.
For $q=19$, $p\equiv 1,115\pmod{152}$. But there are no primes $p$ matching these conditions less than $\sqrt{N}<419.$
For $q=23$, $p\equiv 1,139\pmod{184}$, and $\sqrt{N}<1673$. So the first prime to check is $p=139.$ The next to check is $691$.
In the case of $q=41$, you'd need to check $p\equiv 1,83\pmod{328}$. But now you might have to check up to $\sqrt{N}\approx 857,000$.
| {
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Finding the correct exponential generating function for word counting problem
Find the number of five-letter words that use an odd number of $A$s, at least two $B$s, and at most two $C$s.
The given answer: $65.$
I use the expression $(A + \frac{A^3}{3!})(\frac{B^2}{2!} + \frac{B^3}{3!} + \frac{B^4}{4!})(1 + C + \frac{C^2}{2!}).$ We can't have the term $\frac{A^5}{5!}$ because we must have the term $\frac{B^2}{2!}.$ Everything else looks ok to me, yet I can't derive $65$. Either I am not setting up the g.f. correctly or I am multiplying polynomials incorrectly. Is the g.f. above incorrectly set up? If so, how can I fix it?
|
We are looking for
\begin{align*}
120[z^5]\left(z+\frac{z^3}{3!}+\frac{z^5}{5!}\right)\left(\frac{z^2}{2!}+\frac{z^3}{3!}+\frac{z^4}{4!}+\frac{z^5}{5!}\right)
\left(1+z+\frac{z^2}{2!}\right)
\end{align*}
... can you figure out why?
As you correctly noted, the term $\frac{A^5}{5!}$ does not contribute to the result. So, we can simplify the left-most factor to $\left(z+\frac{z^3}{3!}\right)$.
| {
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Probability of a random selection of $2$ out of $5$ shoes
From a set of five pairs of shoes, two of the shoes are selected at random. Find the probability of each of the following:
a) Both are from the same pair.
My answer: $\dfrac{\binom{10}{1}\binom{1}{1}}{\binom{10}{2}}$
b) One left shoe and one right shoe are selected.
My answer: $\dfrac{\binom{10}{1}\binom{5}{1}}{\binom{10}{2}}$
Note: Solve this problem by counting combinations (i.e $\binom{n}{k}$).
Can someone check if this is correct? If not, can you explain why it is incorrect?
| The numerators in your solutions are incorrect. Considering the two different questions:
*
*Both shoes must be from the same pair. Using probability on the first and second draw, we find: $$\frac{10}{10} \cdot \frac{1}{9} = \frac{1}{9}$$ With your suggested approach, however, we find: $$\frac{10 \cdot 1}{45} = \frac{2}{9}$$ In the numerator, we must only mention the combinations of shoes we pick, not multiply the values for the two different draws. Instead, try writing it as follows: $$\frac{5 \choose 1}{10 \choose 2} = \frac{5}{45} = \frac{1}{9}$$ There are five pairs we can draw, and 45 possible pairs of draws.
*Similarly as for (a), the numerator is incorrect. Using probability on the first and second draw, we find: $$\frac{10}{10} \cdot \frac{5}{9} = \frac{5}{9}$$ Using combinations, you should arrive at the following: $$\frac{{5 \choose 1}{5 \choose 1}}{10 \choose 2} = \frac{25}{45} = \frac{5}{9}$$ In this way, we consider all pairs of draws in which one shoe is drawn from the left shoes and one shoe is drawn from the right shoes.
| {
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Find the Roots of $(x+1)(x+3)(x+5)(x+7) + 15 = 0$ Once I came across the following problem: find the roots of $(x+1)(x+3)(x+5)(x+7) + 15 = 0$.
Here it is how I proceeded:
\begin{align*}
(x+1)(x+3)(x+5)(x+7) + 15 & = [(x+1)(x+7)][(x+3)(x+5)] + 15\\
& = (x^2 + 8x + 7)(x^2 + 8x + 15) + 15\\
& = (x^2 + 8x + 7)[(x^2 + 8x + 7) + 8] + 15\\
& = (x^2 + 8x + 7)^2 + 8(x^2 + 8x + 7) + 15
\end{align*}
If we make the substitution $y = x^2 + 8x + 7$, we get
\begin{align*}
y^2 + 8y + 15 = (y^2 + 3y) + (5y + 15) = y(y+3) + 5(y+3) = (y+5)(y+3) = 0
\end{align*}
From whence we obtain that:
\begin{align*}
y + 5 = 0\Leftrightarrow x^2 + 8x + 12 = 0 \Leftrightarrow (x+4)^2 - 4 = 0\Leftrightarrow x\in\{-6,-2\}\\
\end{align*}
Analogously, we have that
\begin{align*}
y + 3 = 0\Leftrightarrow x^2 + 8x + 10 = 0\Leftrightarrow (x+4)^2 - 6 = 0\Leftrightarrow x\in\{-4-\sqrt{6},-4+\sqrt{6}\}
\end{align*}
Finally, the solution set is given by $S = \{-6,-2,-4-\sqrt{6},-4+\sqrt{6}\}$.
Differently from this approach, could someone provide me an alternative way of solving this problem? Any contribution is appreciated. Thanks in advance.
| One way to solve this problem would be to use the Rational Roots Theorem to find the roots $-2$ and $-6$, then use polynomial division to get a quadratic which is easily solved. However this method will not work in general, as a polynomial does not need to have any rational roots at all
| {
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Algebraic identities Given that $$a+b+c=2$$ and $$ab+bc+ca=1$$
Then the value of
$$(a+b)^2+(b+c)^2+(c+a)^2$$ is how much?
Attempt:
Tried expanding the expression. Thought the expanded version would contain a term from the expression of $a^3+b^3+c^3-3abc$, but its not the case.
| Consider the monic polynomial with roots $a$, $b$, $c$
$$P(x)=(x-a)(x-b)(x-c)$$
Expanding out:
$$P(x)=x^{3}+x^{2}(-a-b-c)+x(ab+bc+ac)-abc$$
(This is essentially Vieta's formula).
Then plugging in your values this polynomial equals:
$$P(x)=x^{3}-2x^{2}+x-abc$$
Further, since $a$, $b$, $c$ are roots
$$P(a)=a(a^{2}-2a+1-bc)=0$$
$$P(b)=b(b^{2}-2b+1-ac)=0$$
$$P(c)=c(c^{2}-2c+1-ab)=0$$
We assume $a,b,c$ are non-zero. Then division gives us:
$$P(a)=a^{2}-2a+1-bc=0$$
$$P(b)=b^{2}-2b+1-ac=0$$
$$P(c)=c^{2}-2c+1-ab=0$$
Adding up the above three, and using $a+b+c=2$ and $ab+bc+ac=1$
$$a^{2}+b^{2}+c^{2}=2(a+b+c)-3+(ab+bc+ca)$$
$$=2$$
Now we can tackle your problem. Just expand:
$$(a+b)^{2}+(b+c)^{2}+(a+c)^{2}$$
$$=2(a^2+b^2+c^2)+2(ab+bc+ca)=2\cdot 2+2=6$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate the area of the triangular region ABC How to calculate the area of the triangle from the figure
| Let $O$ be the centre of the circle and $r$ its radius.
Then $AB=2\sqrt{r^2-(r-1)^2}=2\sqrt{2r-1}$, $BC=2\sqrt{r^2-(r-2)^2}=2\sqrt{4r-4}$, $AC=2\sqrt{r^2-(r-3)^2}=2\sqrt{6r-9}$.
Now depending on what edge we take as base, the area of $ABC$ can be expressed in three forms:
$$\sqrt{2r-1}\cdot (2r-1)=\sqrt{4r-4}\cdot (2r-2)=\sqrt{6r-9}\cdot (2r-3).$$
By squaring, we get
$$8r^3 -12r^2+6r-1 = 16r^3-48r^2+48r-16=24r^3-108r^2+162r-81$$
and from this by subtraction
$$ 8r^3-36r^2+42r-15=0,\quad 8r^3-60r^2+114r-65=0$$
and by another subtraction and division by $2$
$$ 12r^2-36r+25=0.$$
From this,
$$ 0=3\cdot (8r^3-36r^2+42r-15)-(2r+3)\cdot (12r^2-36r+25)=76r-78$$
so that $r=\frac{39}{38}$ -- but unfortunately, this does not fulfil the original equation (so either I made a mistake or there is no solution; but in the former case I assume you can find the correct solution from this)
| {
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Two ways to calculate the primitive of $f(x) =\frac{x^2}{x^4-1}$ but two different results, why? When I search primitives of $f(x)= \dfrac{x^2}{x^4-1}$, using partial fraction decomposition the solutions are
$\boxed {F(x) =\dfrac{1}{4} \ln|x-1| + \dfrac{1}{4} \ln|x+1| + \dfrac{1}{2} \arctan{x} + C}$
But if I use an other way, I've got an another result :
$f(x)=\dfrac{x^2}{x^4-1} =\dfrac{x^2-1+1}{(x^2-1)(x^2+1)} =\dfrac{x^2-1}{(x^2-1)(x^2+1)}+\dfrac{1}{(x^2-1)(x^2+1)} = \\\dfrac{1}{x^2+1} +\dfrac{1}{x^4-1}= \dfrac{1}{x^2+1} -\dfrac{1}{1-(x^2)^2}$
Thus $\boxed {F(x)= \arctan{x} -\operatorname{argth}x^2 +C} $
I think that might be wrong, because that doesn't match with the first result if I plot both on geogebra. I noticed as well their domains are different.
My question is why this two different results? Is one of them wrong?
| $\boxed {F(x) =\dfrac{1}{4} \ln|x-1| - \dfrac{1}{4} \ln|x+1| + \dfrac{1}{2} \arctan{x} + C}$
F is defined over $\mathbb{R}\backslash \{-1,1\}$
Regarding your result, Your function is defined only over $]-1,1[$, why?
Now for the message from Bernard, thank you, I realised I must used the u-substitution. So :
$\displaystyle {F(x)=\int \dfrac{dx}{1-x^4} \implies F(u)=\int \dfrac{du}{2\sqrt{u}(1-u^2)} ,\quad (\text{with} \; u=x^2)}$
This form is not interesting.
Thanks every one
| {
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Finding a Closed Form for $\int_0^\infty A\sin(\frac{2\pi}{T}x) \exp(-bx)dx$ I'm struggling in finding a closed form for,
$$ \int_0^\infty Ae^{-bx}\sin\left(\frac{2\pi}{T}x\right) \,dx $$
My Attempt: Let $H = \int Ae^{-bx}\sin\left(\frac{2\pi}{T}x\right)\,dx$. First I integrate by parts, splitting $f=e^{-bx}$ and $dg=\sin\left(\frac{2\pi}{T}x\right)\,dx$. Then another IBP using $u=e^{-bx}$ and $dv = \cos\left(\frac{2\pi}{T}x\right)\,dx$ and simplify,
\begin{align}
\frac{2\pi}{A\cdot T} H &= -e^{-bx}\cos\left(\frac{2\pi}{T}x\right) - \frac 1 b \int \cos\left(\frac{2\pi}{T}x\right)e^{-bx}\,dx \\
\frac{2\pi}{A\cdot T} H &= -e^{-bx}\cos\left(\frac{2\pi}{T}x\right) - \frac 1 b \left[ e^{-bx}\frac{T}{2\pi}\sin\left(\frac{2\pi}{T}x\right) + \frac{T}{2\pi b}\int e^{-bx}\sin\left(\frac{2\pi}{T}x\right)\,dx \right] \\
\left(\frac{2\pi}{A\cdot T} - \frac{T}{2\pi b^2}\right)H &= e^{-bx}\cos\left(\frac{2\pi}{T}x\right) - \frac{T}{2\pi b} e^{-bx} \sin\left(\frac{2\pi}{T}x\right) \\
H &= -e^{-bx} \left(\cos\left(\frac{2\pi}{T}x\right) + \frac{T}{2\pi b}\sin\left(\frac{2\pi}{T}x\right) \right) \cdot \left( \frac{A\cdot T 2\pi b^2}{(2\pi b)^2 - A\cdot T^2}\right) + c\\
\end{align}
First off, is this horrid looking expression correct? If so, can it be simplified further?
Now I come to evaluate $H|_0^\infty$. $H$ appears to be defined at $x=0$, so I really just need to find $\lim_{x\to \infty^+} H$. Here is where I'm stuck. The denominator grows exponentially for $b\ge 1$, and $T$ and $A$ don't seem to have any effect on the limit. How should I evaluate this limit?
| This is the Laplace Transform of $\sin \left(\dfrac{2 \pi x}{T}\right)$
Its value is $\dfrac{2 \pi A T}{b^2 T^2+4 \pi ^2}$
I just looked on the Laplace transform table
| {
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Find the common roots of $x^3 + 2x^2 + 2x + 1$ and $x^{1990} + x^{200} + 1$ Find the common roots of $x^3 + 2x^2 + 2x + 1$ and $x^{1990} + x^{200} + 1$.
I completed the first part of the question by finding the roots of the first equation. I obtained $3$ roots, one of them being $-1$ and the other two complex. It is evident that $-1$ is not a root of the second equation, but how can I find out whether the other two roots are common or not?
| The solution to the equation $x^2+x+1=0$ is $e^{\frac{2\pi}{3}}$ and $e^{\frac{4\pi}{3}}$ .Having $1990=3\times 663 +1$ and $200=3\times 66 +2 $, we can evaluate the second equation as
if $ x= e^{\frac{2\pi}{3}} $ then $x^{1990}=x^{3\times 663 +1}=x,x^{200}=x^{3\times 66 +2}=x^2$, so $e^{\frac{2\pi}{3}}$ is root of second equation.
by the property of real polynomials, $\textit{conjugate of a root is also a root of that polynomial}$ .
So, the two complex solution of the first equation is also solution for the second one.
| {
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If a and b are two distinct real values such that $F (x)= x^2+ax+b $ And given that $F(a)=F(b)$ ; find the value of $F(2)$
If $a$ and$ b$ are two distinct real values such that
$$F (x)= x^2+ax+b $$
And given that $F(a)=F(b)$ ; find the value of $F(2)$
My try:
$$F(a)=a^2+a^2+b= 2a^2+b,\quad
F(b)=b^2+ab+b $$
$F(a)=F(b)$ implies $2a^2=b^2+ab$, and thus
$F(2)= 4+2a+b $
What Now?
Any help would be appreciated , thank you
| $F(2) = 2^2+2a+b = 4+2a+b$. From $2a^2 = b^2+ab \implies a^2-b^2 = ab - a^2 \implies (a-b)(a+b) = a(b-a) \implies (a-b)(2a+b) = 0\implies a = b$ or $2a+b = 0$. Since $a \neq b, 2a+b = 0 \implies F(2) = 4+0 = 4$.
| {
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Degrees of the irreducible factors of $X^{13}-1$ I need some hints to determine the degrees of irreducible factors of $X^{13}-1$ in $\Bbb F_5[X]$. I know the first factor is $X-1$ and it is a single root. There are not other roots. I also checked from Mathematica that $1+X+ \dots X^{12}$ is not irreducible in $\Bbb F_5[X]$ but I lack proper argument to show it.
I know by Mathematica that $X^{13}-1$ can be factored as
$$(X+4) \left(X^4+X^3+4 X^2+X+1\right) \left(X^4+2 X^3+X^2+2 X+1\right) \left(X^4+3 X^3+3 X+1\right)$$
How do I determine the degrees without Mathematica?
| We know that $X^{5^n} - X$ is the product of all irreducible polynomials over $\mathbb{F}_5$ whose degree divides $n$. You can compute
$$\mathrm{gcd}(X^{13} - 1, X^5 - X) = X - 1,$$
$$\mathrm{gcd}(X^{13} - 1, X^{25} - X) = X - 1,$$
$$\mathrm{gcd}(X^{13} - 1, X^{625} - X) = X^{13} - 1$$ with the Euclidean algorithm and immediately read off that $X^{13} - 1$ has one factors of degree $1$, no factors of degree $2$ and $\frac{12}{4} = 3$ factors of degree $4$.
The fact that $X^{13} - 1$ divides $X^{625} - X$ mod $5$ is because $13 \mid 624$, as suggested in the comments.
| {
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proof that $n^4+22n^3+71n^2+218n+384$ is divisible by $24$ Let $n$ be a positive integer number.
How do we show that the irreducible polynomial $n^4+22n^3+71n^2+218n+384$ is divisible by $24$ for all $n$?
| $$P \equiv n^4-2n^3-n^2+2n \equiv (n-2)\cdot(n-1)\cdot n\cdot(n+1)\equiv 0\mod 24 $$
The last one is because four consecutive numbers must include one number that is divisible by $4$, another number that is divisible by $2$, and one number that is divisible by $3$.
| {
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Evaluating $-\int_0^1\frac{1-x}{(1-x+x^2)\log x}\,dx$ I was trying do variations of an integral representation for $\log\frac{\pi}{2}$ due to Jonathan Sondow, when I am wondering about if it is possible to evaluate
$$\int_0^1-\frac{1-x}{(1-x+x^2)\log x}\,dx,\tag{1}$$
Wolfram Alpha online calculator provide me a closed-form with code
int -(1-x)/((1-x+x^2)log(x)) dx, from x=0 to x=1
Question. Please provide me hints to know how evaluate previous this definite integral as
$$\int_0^1-\frac{1-x}{(1-x+x^2)\log x}\,dx=\log \left(\frac{\Gamma(1/6)}{\Gamma(2/3)}\right)-\frac{\log \pi}{2}$$
as said Wolfram Alpha. Many thanks.
| Let
$$I=\int_0^1\frac{1+x}{1+x^3}\frac{x-1}{\ln(x)}dx$$
and
$$I(a)=\int_0^1\frac{1+x}{1+x^3}\frac{x^a-1}{\ln(x)}dx$$
and notice that $I(1)=I$ and $I(0)=0.$
$$I'(a)=\int_0^1\frac{x^a+x^{a+1}}{1+x^3}dx\overset{x=t^{1/3}}{=}\frac13\int_0^1\frac{t^{\frac{a-2}{3}}+t^{\frac{a-1}{3}}}{1+t}dt$$
By using
$$\int_0^1\frac{x^n}{1+x}dx=\frac12\psi\left(\frac{n+2}{2}\right)-\frac12\psi\left(\frac{n+1}{2}\right),$$
we have
$$I'(a)=\frac16\left[\psi\left(\frac{a+4}{6}\right)-\psi\left(\frac{a+1}{6}\right)+\psi\left(\frac{a+5}{6}\right)-\psi\left(\frac{a+2}{6}\right)\right].$$
Integrate both sides from $a=0$ to $a=1$
$$\int_0^1 I'(a)da=I(a)|_0^1=I(1)-I(0)=I-0$$
$$=\int_0^1 \frac16\left[\psi\left(\frac{a+4}{6}\right)-\psi\left(\frac{a+1}{6}\right)+\psi\left(\frac{a+5}{6}\right)-\psi\left(\frac{a+2}{6}\right)\right]da$$
$$=\left[\ln\Gamma\left(\frac{a+4}{6}\right)-\ln\Gamma\left(\frac{a+1}{6}\right)+\ln\Gamma\left(\frac{a+5}{6}\right)-\ln\Gamma\left(\frac{a+2}{6}\right)\right]_0^1$$
$$=\left[\ln\Gamma\left(\frac{5}{6}\right)-\ln\Gamma\left(\frac{1}{3}\right)+\ln\Gamma\left(1\right)-\ln\Gamma\left(\frac{1}{2}\right)\right]$$
$$-\left[\ln\Gamma\left(\frac{2}{3}\right)-\ln\Gamma\left(\frac{1}{6}\right)+\ln\Gamma\left(\frac{5}{6}\right)-\ln\Gamma\left(\frac{1}{3}\right)\right]$$
$$=-\ln\Gamma\left(\frac{1}{2}\right)-\ln\Gamma\left(\frac{2}{3}\right)+\ln\Gamma\left(\frac{1}{6}\right)$$
$$=\ln\left(\frac{\Gamma\left(\frac16\right)}{\sqrt{\pi}\, \Gamma\left(\frac23\right)}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2360475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Determine if these functions are injective
Determine if the following functions are injective.
$$f(x) = \frac{x}{1+x^2}$$
$$g(x) = \frac{x^2}{1+x^2}$$
My answer:
$f(x) = f(y)$
$$\implies \frac{x}{1+x^2} = \frac{y}{1+y^2}$$
$$\implies x+xy^2 =y+yx^2$$
$$\implies x=y$$
Hence $f(x)$ is injective
$g(x) = g(y)$
$$\implies \frac{x^2}{1+x^2}=\frac{y^2}{1+y^2}$$
$$\implies x^2+x^2y^2=y^2+y^2x^2$$
$$\implies x^2=y^2$$
$$\implies \pm x=\pm y$$
So $g(x)$ is not injective
| Where you had $x+xy^2=y+yx^2,$ you canceled $xy^2$ from one side and $yx^2$ from the other side, but those are not the same.
But the equation can be written as $yx^2 - (y^2+1) x + y = 0,$ and that is $ax^2+bx+c=0,$ where $a=y,$ $b=-(y^2+1),$ and $c=y.$
The solution for $x$ of the equation $ax^2+bx+c=0$ is $x = \dfrac{-b\pm\sqrt{b^2-4ac\ {}}}{2a}.$
So you have $x = \dfrac{y^2+1 \pm \sqrt{(y^2+1)^2 - 4y^2}}{2y} = \dfrac{y^2+1 \pm (y^2-1)}{2y} = \Big( \dfrac 1 y \text{ or } y \Big).$
Since there are two solutions, this is not one-to-one.
| {
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"timestamp": "2023-03-29T00:00:00",
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Given $1 \le |z| \le 7$ Find least and Greatest values of $\left|\frac{z}{4}+\frac{6}{z}\right|$ Given $1 \le |z| \le 7$
Find least and Greatest values of $\left|\frac{z}{4}+\frac{6}{z}\right|$
I have taken $z=r e^{i \theta}$ $\implies$ $1 \le r \le 7$
Now $$\left|\frac{z}{4}+\frac{6}{z}\right|=\left|\frac{r \cos \theta}{4}+\frac{ir \sin \theta}{4}+\frac{6 \cos \theta}{r}-\frac{6 i \sin \theta}{r} \right|$$
So
$$\left|\frac{z}{4}+\frac{6}{z}\right|=\sqrt{\frac{r^2}{16}+\frac{36}{r^2}+3 \cos (2\theta)}$$
any clue from here?
| Let $x = r^2 \implies 1 \le x \le 49 $, and consider $f(x) = \dfrac{x}{16} + \dfrac{36}{x}\implies f'(x) = \dfrac{1}{16}- \dfrac{36}{x^2}\implies f'(x) = 0 \iff x = 24$, and $f(1) = 36+\dfrac{1}{16} = 36.0625, f(49) = \dfrac{49}{16}+\dfrac{36}{49}=3.8, f(24) = 3.$ $3\cos(2\theta)$ has a min of $-3$, and a max of $3$. Put these values together, we have: $\left|\dfrac{z}{4}+\dfrac{6}{z}\right|$ reaches a min of $\sqrt{3.8-3}\approx. 0.89$, and a max of $\sqrt{36.0625+3}=6.25$.
| {
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"timestamp": "2023-03-29T00:00:00",
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On understanding the use of binomial theorem to find asymptotes of a real valued function. I have the function
$$y=\frac{x^3+2x+9}{\sqrt{4x^2+3x+2}}$$
Wolfram Alpha says that it has a non linear asymptote at
$y=\frac{x^2}{2} - \frac{3x}{16}+ \frac{251}{256}$.
I have been told to expand using the binomial theorem to obtain the asymptote at predicting-non-linear-asymptotes-of-a-real-valued-function ; however, I do not understand this. How does expanding binomially give me the asymptote ?
| Consider first the denominator of $$y=\frac{x^3+2x+9}{\sqrt{4x^2+3x+2}}$$ $${\sqrt{4x^2+3x+2}}=2x{\sqrt{1+\frac{3x}{4x^2}+\frac2{4x^2}}}=2x{\sqrt{1+\frac{3}{4x}+\frac1{2x^2}}}$$ Now apply the binomial therorem or Taylor series to get
$${\sqrt{1+\frac{3}{4x}+\frac1{2x^2}}}=1+\frac{3}{8 x}+\frac{23}{128 x^2}+O\left(\frac{1}{x^3}\right)$$ $$y=\frac 1 {2x}\frac{x^3+2x+9}{1+\frac{3}{8 x}+\frac{23}{128 x^2}+O\left(\frac{1}{x^3}\right)}$$ Now, use the long division to get $$y=\frac{x^2}{2}-\frac{3
x}{16}+\frac{251}{256}+O\left(\frac{1}{x}\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\lim_{ x\to \infty} \left( \tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac{\pi}{4}\right)x$
Evaluate
$$\lim_{ x\to \infty} \left( \tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac{\pi}{4}\right)x$$
I assumed $x=\frac{1}{y}$ we get
$$L=\lim_{y \to 0}\frac{\left( \tan^{-1}\left(\frac{1+y}{1+4y}\right)-\frac{\pi}{4}\right)}{y}$$
using L'Hopital's rule we get
$$L=\lim_{y \to 0} \frac{1}{1+\left(\frac{1+y}{1+4y}\right)^2} \times \frac{-3}{(1+4y)^2}$$
$$L=\lim_{y \to 0}\frac{-3}{(1+y)^2+(1+4y)^2}=\frac{-3}{2}$$
is this possible to do without Lhopita's rule
| HINT:
Set $1/x=h$ to get $$F=\lim_{h\to0^+}\dfrac{\tan^{-1}\dfrac{h+1}{4h+1}-\tan^{-1}1}h$$
Now $\tan^{-1}\dfrac{h+1}{4h+1}-\tan^{-1}1=\tan^{-1}\left(\dfrac{\dfrac{h+1}{4h+1}-1}{1+\dfrac{h+1}{4h+1}}\right)=\tan^{-1}\dfrac{-3h}{5h+2}$
$$\implies F=\lim_{h\to0^+}\dfrac{\tan^{-1}\left(\dfrac{-3h}{5h+2}\right)}{\left(\dfrac{-3h}{5h+2}\right)}\cdot\lim_{h\to0^+}\dfrac{-3}{5h+2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $A+B=225^\circ$ then $\frac{\cot A}{1+\cot A}\cdot\frac{\cot B}{1+\cot B}=\ldots$
If $A+B=225^\circ$ then $\frac{\cot A}{1+\cot A}\cdot\frac{\cot B}{1+\cot B}=\ldots$
I tried using $\cot(A+B)$ formula but failed to get a proper answer.
Options are:(A) $1$; (B) $-1$; (C) $0$; (D) $\frac12$.
| $$\frac{\cot{A}}{1+\cot{A}}\cdot\frac{\cot{B}}{1+\cot{B}}=\frac{\cot{A}}{1+\cot{A}}\cdot\frac{\cot\left(45^{\circ}-A\right)}{1+\cot\left(45^{\circ}-A\right)}=$$
$$=\frac{\cos{A}}{\sin{A}+\cos{A}}\cdot\frac{\cos{A}+\sin{A}}{\cos{A}-\sin{A}+\cos{A}+\sin{A}}=\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2368202",
"timestamp": "2023-03-29T00:00:00",
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How do you calculate the expected value of geometric distribution without diffrentiation? Is there any way I can calculate the expected value of geometric distribution without diffrentiation? All other ways I saw here have diffrentiation in them.
Thanks in advance!
| I know at least two ways off hand and there are probably others.
First I'll show you a concrete way to do it. After that I'll show you how to express the same thing exactly. (Together these make up only one of those "two ways". The other one now appears in the answer posted by "Henry".)
$$
\begin{array}{cccccccccccccccccccccccc}
& 0 & & 1 & & 2 & & 3 & & 4 & & 5 & & 6 \\
\hline
& & & p^1 & + & 2p^2 & + & 3p^3 & + & 4p^4 & + & 5p^5 & + & 6p^6 & + & \cdots & {} \\[12pt]
= & & & p^1 & + & p^2 & + & p^3 & + & p^4 & + & p^5 & + & p^6 & + & \cdots \\
& & & & + & p^2 & + & p^3 & + & p^4 & + & p^5 & + & p^6 & + & \cdots \\
& & & & & & + & p^3 & + & p^4 & + & p^5 & + & p^6 & + & \cdots \\
& & & & & & & & + & p^4 & + & p^5 & + & p^6 & + & \cdots \\
& & & & & & & & & & + & p^5 & + & p^6 & + & \cdots \\
& & & & & & & & & & & & + & p^6 & + & \cdots \\
& & & & & & & & & & & & & & + & \cdots \\
& & & & & & & & & & & & & & \vdots
\end{array}
$$
First sum each (horizontal) row. Each is a geometric series. Then sum the remaining series, which is also geometric.
Here is the same method expressed abstractly in the language of algebra:
\begin{align}
\sum_{x=0}^\infty x (1-p) p^x & = \sum_{x=1}^\infty x (1-p) p^x = \sum_{x=1}^\infty \sum_{j=1}^x (1-p)p^x \\[10pt]
& = \sum_{ x,j\, : \, 1 \,\le\, j \, \le \, x} (1-p) p^x = \sum_{j=1}^\infty \sum_{x=j}^\infty (1-p)p^x
\end{align}
Now you're summing a geometric series as $x$ goes from $j-1$ to $\infty,$ and then the outer sum, as $j$ goes from $1$ to $\infty,$ also turns out to be geometric.
(In the very first step above I put $\displaystyle\sum_{x=0}^\infty = \sum_{x=1}^\infty.$ That is justified by the fact that when $x=0,$ the actual term being added is $0$ so it can be dropped.)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Inequality in sum of fractions
Is there a constant $C\ge 0$ such that for any $a,b,c,d>0$ the inequality holds:
$$
\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a} \ge C+\frac{a+b}{b+c}+\frac{b+c}{c+d}+\frac{c+d}{d+a}
$$
My attempt:
$a=b=c=d$ gives $1\ge C$.
Now, take $b=c=d$; this leads to inequality with two variables ($a$ and, let's say, $b$), which for $C=1$ is equivalent to $a^3+2b^3\ge 3ab^2$. This one is true by Cauchy.
The case $a=b$, $c=d$ leads to the same thing.
Is $C=1$ good?
| Using scaling of $a,b,c,d$ you can set $d=1$ without loss of generality.
This gives the function
$$
f(a,b,c) \equiv \left( \frac{a}{b} + \frac{b}{c} + c + \frac{1}{a} \right) - \left( \frac{a+b}{b+c} + \frac{b+c}{c+1} + \frac{c+1}{1+a}\right)
$$
to minimize.
Taking the derivative $\frac{\partial f(a,b,c)}{\partial b}$ gives a result that is linear in $a$, hence equating to zero gives that for the extremum one gets
$$
a = \frac{b^2(b^2 + 2 b c -c^3)}{c^2 (1+c) (2 b +c)}
$$
After that it becomes rather messy.
It is of course cheating but a numerical minimisation of the function $f(a,b,c)$ gives that the minimum is obtained for $a \approx 1.193389$, $b \approx 2.201712$, $c \approx 1.494538$ and results in $C \approx 0.810126$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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is there a formula for this binomial sum? Are there any formulas to calculate the below sum?
$$\sum_{n=1}^{1918}n\binom{2017-n}{99}$$
Or, more generally,
$$\sum_{n=1}^{1918}n\binom{2017-n}{k}$$
| A general solution where $p = 2017$ and $k = 99$ in your case:
First rewrite the sum to look nicer:
\begin{align*}
\sum_{n=0}^{p-k}n\binom{p-n}k &= \sum_{n=k}^p(p-n)\binom{n}k = p\sum_{n=k}^p\binom{n}k - \sum_{n=k}^p n\binom{n}k \\
&= p\sum_{n=0}^{p-k}\binom{n+k}k - \sum_{n=0}^{p-k}(n+k)\binom{n+k}k \\
&= (p-k)\sum_{n=0}^{p-k}\binom{n+k}k - \sum_{n=0}^{p-k}n\binom{n+k}k
\end{align*}
Now the nice thing about this is that this looks like a well known generating function! We have that
$$
\frac1{(1-x)^{k+1}} = \sum_{n\geq0}\binom{n+k}kx^n.
$$
To find the generating function for $n\binom{n+k}k$, we can differentiate both sides and multiply by $x$.
$$
\frac{(k+1)x}{(1-x)^{k+2}} = \sum_{n\geq0}n\binom{n+k}kx^{n-1}\cdot x = \sum_{n\geq0}n\binom{n+k}k x^n
$$
But this is only the generating function for each of the terms, not the sum up to something. The trick for getting the generating function for the first $n$ terms of a sequence from the original generating function $A(x)$ is shifting the terms and adding them up:
$$
A(x) + xA(x) + x^2A(x) + \dots = A(x)\left(1+x+x^2+\dots\right) = \frac{A(x)}{1-x}.
$$
So, the answer you want is the $(p-k)$th coefficient of the generating function
$$
\frac{p-k}{(1-x)^{k+2}} - \frac{(k+1)x}{(1-x)^{k+3}}.
$$
The $n$th coefficient of the first term is $(p-k)\binom{n+k+1}{k+1}$ from before, and the $n$th coefficient of the second term is the $n-1$th coefficient of $\frac{(k+1)}{(1-x)^{k+3}}$, which is $(k+1)\binom{n+k+1}{k+2}$. So, we conclude that the answer is
\begin{align*}
(p-k)\binom{p+1}{k+1}-(k+1)\binom{p+1}{k+2} &= (p-k)\frac{k+2}{(p+1)-(k+1)}\binom{p+1}{k+2}-(k+1)\binom{p+1}{k+2} \\
&= ((k+2)-(k+1))\binom{p+1}{k+2} \\
&= \boxed{\binom{p+1}{k+2}.}
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Intersection of rectangular and polar equations Find the points of intersection for the following relations:
$x^2+y^2-20 = 0$ and $\theta = -3\pi / 4$
So the first one is obviously a circle, and the second one, converting into rectangular form, is y/x = 1, and thus y=x. Substituting y as x in the first equation, I then get $x^2 = 10$, which yields $+- \sqrt {10}$, and thus the points of intersection should then be $(\sqrt{10},\sqrt{10})$ and $(-\sqrt{10},-\sqrt{10})$. However, the answer says that it is $(\sqrt{10},-\sqrt{10})$ and $(-\sqrt{10},\sqrt{10})$. What did I do wrong?
Also, for another problem, I need to find the intersection between these two:
$4x^2+16y^2-64=0$ and $x^2+y^2=9$
Following a similar approach of substitution, I was able to get 4 points: $(20/3, 7/3), (-20/3, 7/3), (-20/3,-7/3), (20/3, -7/3)$. But the answer says $(3,0) and (-3,0)$...
| $x = r\cos\theta, \ y= r\sin\theta$.
$\theta = - 3\pi/4$.
$r = (\sqrt{20})$.
Combining:
$x = (\sqrt{20})\cos(-3\pi/4)$ ;
$y = (\sqrt{20})\sin(-3\pi/4)$ .
$\cos(-3\pi/4) = \cos(3\pi/4) =$
$- (1/2)\sqrt{2}$.
$\sin(-3\pi/4) = - \sin(3\pi/4) =$
$ -(1/2)\sqrt{2}$.
Finally:
$x = -(\sqrt{20})/(\sqrt{2}) = - (\sqrt{10})$;
$y = - (\sqrt{20})/(\sqrt{2})= - (\sqrt{10})$;
For the given $\theta$ value only one point of intersection (in the third quadrant).
| {
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"timestamp": "2023-03-29T00:00:00",
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There exists rational number $a_n$, $(x^2+\frac{1}{2}x+1)\mid(x^{2n}+a_nx^n+1)$ Let $n\in \mathbb{N}$. Prove that there exists a rational number $a_n$ for which $$(x^2+\frac{1}{2}x+1)\mid(x^{2n}+a_nx^n+1)$$
My attempt :
I try $n=2$,
$(x^2+\frac{1}{2}x+1)(x^2-\frac{1}{2}x+1)=x^4+\frac{7}{4}x^2+1$
$a_2 = \frac{7}{4}$
| just an idea
The roots of the left polynom are
$$x_1=\frac{-1+i\sqrt {15} }{4}=e^{it}$$
$$x_2=\frac {-1-i\sqrt {15} }{4}=e^{-it} $$
they are also roots of the right one
$$e^{2int}+a_ne^{int}+1=0$$
$$e^{-2int}+a_ne^{-int}+1=0$$
thus
$$a_n=-\frac {\sin (2nt)}{\sin (nt)} $$
$$=-2\cos (nt) $$
For example,
$$a_2=-2\cos (2t)=2 (1-2\cos^2 (t)) $$
$$=2-4\frac {1}{16}=\frac {7}{4} $$
| {
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"url": "https://math.stackexchange.com/questions/2375822",
"timestamp": "2023-03-29T00:00:00",
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Convert a equation with fractions into whole numbers So I have this equation:
$$\frac{2}{3}a^2-\frac{4}{9}a^2 = 8a$$
So this is a really easy problem, I could just multiply
$$\frac{2}{3}*\frac{3}{3} = \frac{6}{9}$$
Then subtract
$$\frac{6}{9}a^2 - \frac{4}{9}a^2 = \frac{2}{9}a^2=8a$$
$$36a=a^2$$
$$36=a$$
However, I want to solve the equation by getting rid of the fractions right at the beginning:
$$\frac{2}{3}a^2-\frac{4}{9}a^2 = 8a$$
So I thought it'd be much simpler if I could get rid of these fractions by multiplying everything by a single value. Therefore, I thought what value can I multiply 2 and 4 so it gives me a divisible value by 3 and 9? It took me some time but I came up with 9
$$\frac{2}{3}a^2-\frac{4}{9}a^2 = 8a$$
$$9(\frac{2}{3}a^2-\frac{4}{9}a^2) = 9(8a)$$
$$6a^2-9a^2 = 8a$$
My question is if there is an easier way to find this value that when multiplied it eliminates the fractions? It took a few valuable seconds to figure out it was 9, so I was wondering if this process has a name or any rules to find it quicker?
| Method 1: $\frac ab x + \frac cd x = gy$
Multiple by both $b$ and $d$. Multiplying by $bd$ will always eleminate everything. so $bd\frac abx + bd \frac cd x = adx + bc x = bdgy$.
Ex: $\frac 23 a^2 - \frac 49 a^2 = 8a$
$3*9\frac 23 a^2 -3*9\frac 49 a^2 = 3*9*8a$
$18a^2 - 12a^2 = 216 a$
But notice that was over kill. Every term is a multiple of 3.
Method 2: Do it one term at a time:
$\frac ab x + \frac cd x = gy$
$ ax + \frac {bc}dx = by$
$ dax + bcx = bdy$.
Ex:
$\frac 23 a^2 - \frac 49 a^2 = 8a$
$3\frac 23 a^2 - 3\frac 49a^2 = 2a^2 - \frac 43 a^2 = 3*8a = 24a$
$3*2a^2 - 3\frac 43 a^2 = 6a^2 - 4a^2 = 3*24a = 72a$
Notice in this case you avoided the overkill by factoring out the excessive $3$ as it appeared. Had you started with $9$ it wouldn't have come up at all.
$\frac 23 a^2 - \frac 49 a^2 = 8a$
$9*\frac 23 a^2 - 9*\frac 49 a^2 = 9*8a$
$3*2a^2 - 4a^2 = 72a$.
Method 3: The conclusion we can reach. Multiply by the least common multiple.
$\frac ab x + \frac cd x = gy$
$\text{lcm}(bd)\frac ab x + \text{lcm}(bd) \frac cd = \text{lcm}(bd)gy$
$d'a x + b'c x = \text{lcm}(bd) gy$.
Example: $\frac 23 a^2 - \frac 49 a^2 = 8a$. $3=3$ and $9 = 3^2$ so $\text{lcm}(3,9) = 9$.
$9 \frac 23 a^2 + 9\frac 49 a^2 = 9*8a$
$6a^2 - 4a^2 = 9*8a$.
But notice: You not only should get rid of the denominaters you should also get rid of the common factor two.
Final $\frac ab x + \frac cd x = \frac gh y$. Multiply both sides by $\frac {\text{lcm}(b,d,h)}{\gcd(a,c,g)}$
So for $\frac 23 a^2 - \frac 49 a^2 = 8a$. $\text{lcm}(3,9) = 3$ and $\gcd(2,4,8) = 2$.
So multiply everything by $\frac 92$.
$\frac 92\frac 23 a^2 - \frac 92\frac 29 a^2 = \frac 928a$
$3a^2 - 2a^2 = 36a$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the maximum positive integer that divides $n^7+n^6-n^5-n^4$
Find the maximum positive integer that divides all the numbers of the form $$n^7+n^6-n^5-n^4 \ \ \ \mbox{with} \ n\in\mathbb{N}-\left\{0\right\}.$$
My attempt
I can factor the polynomial
$n^7+n^6-n^5-n^4=n^4(n-1)(n+1)^2\ \ \ \forall n\in\mathbb{N}.$
If $n$ is even then exists $\,k\in\mathbb{N}-\left\{0\right\}\,$ such that $n=2k.\;$ so:
$n^4(n-1)(n+1)^2=2^4 k^4(2k-1)(1+2k)^2$
If $n$ is odd then exists $k\in\mathbb{N}$ such that $n=2k+1$ so
$$n^4(n-1)(n+1)^3=2^3(2k+1)^4(k+1)^2$$
Can I conclude that the maximum positive integer that divides all these numbers is $N=2^3?$ (Please, help me to improve my english too, thanks!)
Note: I correct my "solution" after a correction... I made a mistake :\
| We have $3$ consecutive numbers so one of these will give a factor of $3$.
If $n$ is even then $n$ will give a factor of $16$. If $n$ is odd then $(n-1)$ and $(n+1)$ will both be even and if $(n+1)^2$ only gives $2$ factors of $2$ then $(n-1)$ will be divisible by $4$, so $(n-1)(n+1)^2$ will be divisible by $16$.
So $\color{red}{48}$ will always divide $n^7+n^6-n^5-n^4$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Definite integral for a 4 degree function
The integral is:
$$\int_0^a \frac{x^4}{(x^2+a^2)^4}dx$$
I used an approach that involved substitution of x by $a\tan\theta$. No luck :\ . Help?
| One way to solve this problem is to do integration by parts.
$$\int{x^3 \frac{x}{(x^2+a^2)^4}dx}=-\frac{x^3}{6(x^2+a^2)^3}+\frac{1}{2}\int{x\frac{x}{(x^2+a^2)^3} dx}.$$
Continuing further
$$\int{x\frac{x}{(x^2+a^2)^3} dx} = -\frac{1}{4} \frac{x}{(x^2+a^2)^2}+\frac{1}{4}\int{\frac{1}{(x^2+a^2)^2}dx}.$$
Now you can substitute $x=a\tan{\theta}$ to get the final result.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Inequality $\frac{1}{a+b}+\frac{1}{a+2b}+...+\frac{1}{a+nb}<\frac{n}{\sqrt{a\left( a+nb \right)}}$
Let $a,b\in \mathbb{R+}$ and $n\in \mathbb{N}$. Prove that:
$$\frac{1}{a+b}+\frac{1}{a+2b}+...+\frac{1}{a+nb}<\frac{n}{\sqrt{a\left( a+nb \right)}}$$
I have a solution using induction, but without induction I can not think of any way to prove this.
| For positives $a$ and $b$ By C-S we obtain:
$$\left(\frac{1}{a+b}+\frac{1}{a+2b}+\cdots+\frac{1}{a+nb}\right)^2$$
$$\leq n\left(\frac{1}{(a+b)^2}+\frac{1}{(a+2b)^2}+\cdots+\frac{1}{(a+nb)^2}\right)$$
$$<n\left(\frac{1}{(a+\frac{b}{2})(a+\frac{3b}{2})}+\frac{1}{(a+\frac{3b}{2})(a+\frac{5b}{2})}+\cdots+\frac{1}{(a+\frac{(2n-1)b}{2})(a+\frac{(2n+1)b}{2})}\right)$$
$$=\frac{n}{b}\left(\frac{1}{a+\frac{b}{2}}-\frac{1}{a+\frac{3b}{2}}+\frac{1}{a+\frac{3b}{2}}-\frac{1}{a+\frac{5b}{2}}+\cdots+\frac{1}{a+\frac{(2n-1)b}{2}}-\frac{1}{a+\frac{(2n+1)b}{2}}\right)$$
$$=\frac{n}{b}\left(\frac{1}{a+\frac{b}{2}}-\frac{1}{a+\frac{(2n+1)b}{2}}\right)=\frac{n^2}{(a+\frac{b}{2})(a+\frac{(2n+1)b}{2})}<\frac{n^2}{a(a+nb)}.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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If $ x \in \left(0,\frac{\pi}{2}\right)$. Then value of $x$ in $ \frac{3}{\sqrt{2}}\sec x-\sqrt{2}\csc x = 1$
If $\displaystyle x \in \left(0,\frac{\pi}{2}\right)$ then find a value of $x$ in $\displaystyle \frac{3}{\sqrt{2}}\sec x-\sqrt{2}\csc x = 1$
$\bf{Attempt:}$ From $$\frac{3}{\sqrt{2}\cos x}-\frac{\sqrt{2}}{\sin x} = 1.$$
$$3\sin x-2\cos x = \sqrt{2}\sin x\cos x$$
$$(3\sin x-2\cos x)^2 = 2\sin^2 x\cos^2 x$$
$$9\sin^2 x+4\cos^2 x-12 \sin x\cos x = 2\sin^2 x\cos^2 x$$
Could some help me to solve it, thanks
| The function is increasing in $\left(0, \dfrac{\pi}{2}\right)$ which can be seen easily by differentiating and seeing that the derivative is positive (in the given interval).
It is easy to see that $x = \dfrac{\pi}{4}$ is a root and so is the only root in $\left(0, \dfrac{\pi}{2}\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2379567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
If $\frac{\sin x}{\sin y}= {3}$ and $\frac{\cos x}{\cos y}= \frac{1}{2}$, then find $\frac{\sin2x}{\sin2y}+\frac{\cos2x}{\cos2y}$
Let $x$ and $y$ be real numbers such that $\dfrac{\sin x}{\sin y}= {3}$ and $\dfrac{\cos x}{\cos y}= \dfrac{1}{2}$. Find a value of $$\dfrac{\sin2x}{\sin2y}+\dfrac{\cos2x}{\cos2y}.$$
My attempts:
Using the given condition and double angle formula,
$\dfrac{\sin2x}{\sin2y} = \dfrac{3}{2}$
Now I am struggling to find the value of $\dfrac{\cos2x}{\cos2y}$
The simplest form I could reach was:
$\dfrac{2\cos^2x-1}{8\cos^2x-1}$
How do I continue from here? Simpler methods to solve the problem are welcome.
PS: The answer is $\dfrac{49}{58}$
| hint
$$\sin (y)=\frac {1}{3}\sin(x) $$
$$\cos (y )=2\cos (x) $$
$$\implies \sin^2(x )+36\cos^2 (x)=9$$
$$\implies 35\cos^2 (x)=8$$
$$\implies \cos^2 (x)=\frac {8}{35} $$
the second expression is
$$\frac {16-35}{64-35}=-19/29$$
the final result is
$$3/2-19/29=49/58$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Maximize linear function over disk of radius $2$
Maximize $f(x, y) = x + 2y$ with constraint $x^2 + y^2 \le 4$.
$f(x, y)$ has no CP's so thats something gone.
I considered the boundary/edge of $g(x, y) = x^2 + y^2$. And I got the point $(0, 2), (2, 0)$, which gave $f(0, 2) = f(2, 0) = 4$. However, the point $(2/\sqrt(5), 4/\sqrt(5))$ gives a larger max.
How do I approach this to get this point?
|
Max $z=x+2y$ subject to $x^2+y^2\le4$.
The contour lines: $y=-\frac{1}{2}x+\frac{1}{2}z$, the largest value of $z$ occurs when $y$ is tangent to the constraint circle $x^2+y^2=4$ or $y=\sqrt{4-x^2}$.
Thus the slope of tangent line must be equal to the slope of $y=-\frac{1}{2}x+\frac12z:$
$$y'=\frac{-2x}{2\sqrt{4-x^2}}=-\frac12 \Rightarrow x=\frac{2}{\sqrt{5}} \Rightarrow y=\frac{4}{\sqrt{5}}.$$
Graphical illustration:
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Rationalize $\frac{1}{2 + 2\sqrt[6]{2} + 2\sqrt[3]{2}}$ I am having trouble rationalizing the denominator of $$\frac{1}{2 + 2\sqrt[6]{2} + 2\sqrt[3]{2}}$$
I tried grouping the denominator as $(2 + 2\sqrt[6]{2}) + 2\sqrt[3]{2}$ and multiplying top and bottom by $(2 + 2\sqrt[6]{2})^2-(2 + 2\sqrt[6]{2})(2\sqrt[3]{2})+(2\sqrt[3]{2})^2$ to obtain $$\frac{(2+2\sqrt[6]{2})^2-(2 + 2\sqrt[6]{2})(2\sqrt[3]{2})+(2\sqrt[3]{2})^2}{(2+2\sqrt[6]{2})^3+16}$$
However, expanding the denominator produces $24 + 24\sqrt[6]{2}+24\sqrt[3]{2}+8\sqrt{2}$ which doesn't look any better than the original.
So what should I do differently or is there another way to approach this problem? Any advice would be appreciated. Thanks.
| Let $a=\sqrt[6]2$. Then your expression is$$\frac1{2+2a+2a^2}=\frac{1-a}{2(1+a+a^2)(1-a)}=\frac{1-a}{2(1-a^3)}=\frac{1-\sqrt[6]2}{2\bigl(1-\sqrt2\bigr)}=\frac{\bigl(1-\sqrt[6]2\bigr)\bigl(1+\sqrt2\bigr)}{-2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2383577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Please prove $\frac{1 + \sin\theta - \cos\theta}{1 + \sin\theta + \cos\theta} = \tan \left(\frac \theta 2\right)$
Prove that $\dfrac{1 + \sin\theta - \cos\theta}{1 + \sin\theta + \cos\theta} = \tan\left(\dfrac{\theta}{2}\right)$
Also it is a question of S.L. Loney's Plane Trignonometry
What I've tried by now:
\begin{align}
& =\frac{1+\sin\theta-\sin(90-\theta)}{1+\cos(90-\theta)+\cos\theta} \\[10pt]
& =\frac{1+2\cos45^\circ \sin(\theta-45^\circ)}{1+2\cos45^\circ \cos(45-\theta)} \end{align}
Cause I do know
\begin{align} & \sin c + \sin d = 2\sin\left(\frac{c+d}{2}\right)\cos\left(\frac{c-d}{2}\right) \\[10pt] \text{and } & \cos c + \cos d = 2\cos\left(\frac{c+d}{2}\right)\sin\left(\frac{c-d}{2}\right)
\end{align}
I can't think of what to do next..
| $$
\begin{aligned}
& \frac{1+\sin \theta-\cos \theta}{1+\sin \theta+\cos \theta}\\=& \frac{(1-\cos \theta)+\sin \theta}{(1+\cos \theta)+\sin \theta}\\
=& \frac{2 \sin ^{2}\left(\frac{\theta}{2}\right)+2 \sin \left(\frac{\theta}{2}\right) \cos \left(\frac{\theta}{2}\right)}{2 \cos ^{2}\left(\frac{\theta}{2}\right)+2 \sin \left(\frac{\theta}{2}\right) \operatorname{cor}\left(\frac{\theta}{2}\right)} \\
=& \frac{\sin \left(\frac{\theta}{2}\right)}{\cos \left(\frac{\theta}{2}\right)} \cdot \frac{\sin \left(\frac{\theta}{2}\right)+\cos \left(\frac{\theta}{2}\right)}{\cos \left(\frac{\theta}{2}\right)+\sin \left(\frac{\theta}{2}\right)} \\
=& \tan \left(\frac{\theta}{2}\right)
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2384719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Calculation of a partial derivative I have to find the $\frac{\partial}{\partial x}\left( f(x,y)\right)=\frac{\partial}{\partial x}\left( \frac{1}{\sqrt{x^2+y^2}}\right)$.
1) Thinking of that as $\frac{\partial}{\partial x}\left( (x^2+y^2)^{-\frac{1}{2}}\right)$ and $\frac{\partial}{\partial a}x^a = ax^{a-1}$ I get $\frac{\partial}{\partial x}\left( \frac{1}{\sqrt{x^2+y^2}}\right) = -\frac{x}{\Vert (x,y) \Vert^3}$.
2) Thinking of that as $\frac{\partial}{\partial x}\left( (\sqrt{x^2+y^2})^{-1}\right)$ and $\frac{\partial}{\partial a}\sqrt{a}=\frac{1}{2\sqrt{a}}$ I get $x \cdot (x^2+y^2)$.
3) Thinking of that as $\frac{\partial}{\partial x}\left( \frac{1}{\sqrt{x^2+y^2}}\right)$ and $\frac{\partial}{\partial a}\frac{1}{a}=-\frac{1}{a^2}$ I get $-\frac{2x}{x^2+y^2}$.
Why 2) and 3) are not correct?
| Note:
$$\frac{\partial}{\partial x}\left( \left(\sqrt{x^2+y^2}\right)^{-1}\right)=(-1)\left(\sqrt{x^2+y^2}\right)^{-2}\frac{\partial}{\partial x}\sqrt{x^2+y^2}$$
$$\frac{\partial}{\partial x}\left( \frac{1}{\sqrt{x^2+y^2}}\right)=\frac{0-1\cdot \frac{\partial}{\partial x}\sqrt{x^2+y^2}}{\left(\sqrt{x^2+y^2}\right)^2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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A question related to the existence of a prime Let $I_m=[m!+2,m!+m]\cap\mathbb N$ an "interval" of $\mathbb N$; it can obviously be as long as we want and it is easy to prove $I_m$ does not contain any prime. Prove the following:
$$\text { if }n^2+(n+1)^2\in I_m\text{ then }4n^2+1\notin I_M$$
Note that if in a large interval $I_m$ could exist $n$ denying what is proposed here, then we would have found a counterexample to the conjecture in here
| Remark(I):
At first notice that:
$\dfrac{3+\sqrt{9+4}}{2} \leq \dfrac{4+4}{2}=4$ ,
so for $4 \leq n$ we have:
$$0 \leq n^2-3n+1 \ \ \Longrightarrow \ \
3n^2+3n+2 \leq 4n^2+1 \ \ \Longrightarrow \ \
\\
\dfrac{3}{2}\Big(2n^2+2n+1 \Big) < 4n^2+1 \ \ \Longrightarrow \ \
\dfrac{3}{2}\Big(n^2+(n+1)^2 \Big) < 4n^2+1 .
$$
Remark(II):
On the other hand let $4 \leq m$, then we have:
$2 < (m-1)!$, i.e. $1 < \dfrac{(m-1)!}{2}$
multiplying both sides by $m$ we get:
$$m < \dfrac{m!}{2} \ \ \Longrightarrow \ \
m!+m < m!+ \dfrac{m!}{2} = \dfrac{3}{2} m! \ \ \ \ .
$$
Suppose on contrary that $4n^2+1 \in I_m $.
Now notice that sicce both of $n^2+(n+1)^2$ and $4n^2+1$
belongs to the interval $[m!+2,m!+m]$,
therefor we have:
$$ \ \ \ \ \ \ \ \ \ \ \ \ \
m! < m!+2 \leq n^2+(n+1)^2
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{(III)} \ ,
\ \ \ \ \ \ \ \ \ \ \ \text {and} \ \ \ \
\\
4n^2+1 \leq m!+m
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{(IV)} .$$
$\color{Red}{\text{First case}}$:
Let $m$ and $n$ are both greater or equal than $4$,
i.e. $4 \leq m$ and $4 \leq n$.
In this case we have:
$$\dfrac{3}{2} m!
\overset{ \tiny{ \text{III} } }{<}
\dfrac{3}{2}\Big(n^2+(n+1)^2 \Big)
\overset{ \tiny{ \text{Rmk(I)} } }{<}
4n^2+1
\overset{ \tiny{ \text{IV} } }{\leq}
m!+m
\overset{ \tiny{ \text{Rmk(II)} } }{<}
\dfrac{3}{2} m! \ \ \ \ ,
$$
so we have: $\dfrac{3}{2} m! < \dfrac{3}{2} m!$ ,which is an obvious contradiction. So this case is immpossible!
$\color{Red}{\text{Second case}}$:
Let $4 \leq m$ and $n \leq 3$.
In this case by the (III) inequality we have:
$$26=
4!+2
\leq
m! + 2
\overset{ \tiny{ \text{III} } }{\leq}
\Big(n^2+(n+1)^2 \Big)
\leq
9+16=25 \ ,
$$
which is again an obvious contradiction!
$\color{Red}{\text{Third case}}$:
Let $m \leq 3$ and $4 \leq n$.
In this case by the (IV) inequality we have:
$$265=
4.(4)^2+1
\leq
4n^2+1
\overset{ \tiny{ \text{IV} } }{\leq}
m!+m
\leq
3!+3=9 \ ,$$
which is again an obvious contradiction!
$\color{Red}{\text{Fourth case}}$:
Let $m \leq 3$ and $n \leq 3$.
In this case we have the following sub-cases:
*
*$m=3$, then we have: $I_3=[8,9]=\{ 8, 9 \}$.
So $n^2+(n+1)^2=8$ or $n^2+(n+1)^2=9$, but none of them have a solution.
*$m=2$, then we have: $I_2=[6,6]=\{ 6 \}$.
So $n^2+(n+1)^2=6$ , but it does'nt have a solution.
*$m=1$, then we have: $I_1=[3,2]=\phi$.
At the end, it looks, that it was better; if I have been organized the cases as follows:
$\color{Green}{\text{First case}}$:
$\color{Yellow}{4 \leq m}$ and $4 \leq n$.
$\color{Green}{\text{Second case}}$:
$\color{Yellow}{4 \leq m}$ and $n \leq 3$.
$\color{Purple}{\text{Third case}}$:
$\color{Yellow}{m} \color{Orange}{\leq} \color{Yellow}{3}$
| {
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"url": "https://math.stackexchange.com/questions/2391825",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Find the determinant of the following $4 \times 4$ matrix
Use a cofactor expansion across a row or column to find the determinant of the following matrix
$$B=\begin{pmatrix}1 &c&0&0\\0&1&c&0\\0&0&1&c\\c&0&0&1\end{pmatrix}$$
Clearly indicate the steps you take.
I have tried
$$\det B = 1 \det \begin{pmatrix}1 &c&0\\0&1&c\\0&0&1\end{pmatrix}+(-c) \det \begin{pmatrix}c &0&0\\1&c&0\\0&1&c\end{pmatrix}$$
$$ = \det\begin{pmatrix}1 &c\\0&1\end{pmatrix}+c(-c)det\begin{pmatrix}c &0\\1&c\end{pmatrix}$$
$$=1-c^4$$
| Expanding $\det B$ on first column we have that
$$\det B=1\det\begin{pmatrix}1 &c&0\\0&1&c\\0&0&1\end{pmatrix}-c\begin{pmatrix}c &0&0\\1&c&0\\0&1&c\end{pmatrix}=1-c^4$$
The determinant of triangular matrix is the product of the diagonal entries
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2392090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Inequality : $\sum_{cyc}\frac{\sqrt{a^3c}}{2\sqrt{b^3a}+3bc}\geq \frac{3}{5}$
Let $a$, $b$ and $c$ be positive real numbers. Prove that:
$$\frac{\sqrt{a^3c}}{2\sqrt{b^3a}+3bc}+\frac{\sqrt{b^3a}}{2\sqrt{c^3b}+3ca}+\frac{\sqrt{c^3b}}{2\sqrt{a^3c}+3ab}\geq \frac{3}{5}$$
My attempt :
Substitute $a=x^2, b=y^2, c=z^2$, we have to prove that
$\displaystyle\sum_{cyc}\frac{x^3z}{2y^3x+3y^2z^2} \geq \frac{3}{5}$
By C-S,
$\left(\displaystyle\sum_{cyc}\frac{x^3z}{2y^3x+3y^2z^2}\right)\left(\displaystyle\sum_{cyc}(2y^2x+3y^2z^2)\frac{x}{z}\right) \geq \left(\displaystyle\sum_{cyc}x^2\right)^2$
$\left(\displaystyle\sum_{cyc}\frac{x^3z}{2y^3x+3y^2z^2}\right)\left(\displaystyle\sum_{cyc}(x^2y^2+3y^2xz\right) \geq \left(\displaystyle\sum_{cyc}x^2\right)^2$
| I think your start is good.
Let $a=x^2$, $b=y^2$ and $c=z^2$, where $x$, $y$ and $z$ are positives.
Hence, by C-S
$$\sum_{cyc}\frac{\sqrt{a^3c}}{2\sqrt{b^3a}+3bc}=\sum_{cyc}\frac{x^3z}{y^2(3z^2+2xy)}=\frac{1}{x^2y^2z^2}\sum_{cyc}\frac{x^5z^3}{3z^2+2xy}=$$
$$=\frac{1}{x^2y^2z^2}\sum_{cyc}\frac{x^6z^4}{xz(3z^2+2xy)}\geq\frac{(x^3z^2+y^3x^2+z^3y^2)^2}{x^2y^2z^2\sum\limits_{cyc}(3x^3y+2x^2yz)}.$$
Thus, it's enough to prove that
$$5(x^3z^2+y^3x^2+z^3y^2)^2\geq3\sum_{cyc}3x^5y^3z^2+2x^4y^3z^3)$$ or
$$\sum_{cyc}(5x^6z^4+x^5y^3z^2-6x^4y^3z^3)\geq0,$$
which is true by Rearrangement:
$$\sum_{cyc}x^6z^4=x^4y^4z^4\sum_{cyc}\frac{x^2}{y^4}\geq x^4y^4z^4\sum_{cyc}\frac{x^2}{x^4}=$$
$$=x^4y^4z^4\sum_{cyc}\frac{1}{x^2}\geq x^4y^4z^4\sum_{cyc}\frac{1}{xy}=\sum_{cyc}x^4y^3z^3$$
and
$$\sum_{cyc}x^5y^3z^2=x^3y^3z^3\sum_{cyc}\frac{x^2}{z}\geq x^3y^3z^3\sum_{cyc}\frac{x^2}{x}=\sum_{cyc}x^4y^3z^3.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove the the general equation of the family of circles through the intersection points of two circles? Let $ \mathscr{C_{1}} $ and $ \mathscr C_{2} $ be two intersecting circles determined by the equations $$ x^2 + y^2 + A_{1}x + B_{1}y + C_{1} = 0 $$ and $$ x^2 + y^2 + A_{2}x + B_{2}y + C_{2} = 0. $$ For any number $ k≠−1 $, show that $$ x^2 + y^2 + A_{1}x + B_{1}y + C_{1} + k ( x^2 + y^2 + A_{2}x + B_{2}y + C_{2} ) = 0 $$ is the equation of a circle through the intersection points of $\mathscr{C_{1}}$ and $\mathscr C_{2}$. Show, conversely, that every such circle may be represented by such an equation for a suitable $ k $.
I'm stuck in this question, developing the given equation I could show that it assumes the form of
$$ x^2 + y^2 + \frac{A_{1}+kA_{2}}{1+k}x + \frac{B_{1}+kB_{2}}{1+k}y + \frac{C_{1}+kC_{2}}{1+k} = 0 $$
which proves that for $ k≠−1 $ the given equation is the equation of a circle. But I'm not entirely sure if it really proves that it is the equation of the family of circles through the intersection points of $ \mathscr{C_{1}} $ and $ \mathscr C_{2} $. I have no clue about how to prove the converse statement.
| Let the two points of the intersection be A and B and consider the straight line AB then, since the equation $x^2 + y^2 + \frac{A_{1}+kA_{2}}{1+k}x + \frac{B_{1}+kB_{2}}{1+k}y + \frac{C_{1}+kC_{2}}{1+k} = 0$ is the equation of a circle of centre $(\frac{A_{1}+kA_{2}}{2(1+k)}, \frac{B_{1}+kB_{2}}{2(1+k)})$ pasing through the intersection points, you have to prove that the set of all those points for $k \in \mathbb{R} - \left\{1\right\}$ is a straight line except for a point (because since all the circles pass through the points, then it should be the bisection). Well, actually, we are forgetting the case $k = -1$ but check where is the centre.
| {
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$\ f(x)=\log _{x-2}\left(\frac{2x+3}{7-x}\right)<1 $ $$\ f(x)=\log _{x-2}\left(\frac{2x+3}{7-x}\right) $$
Which is the solution of$\ f(x)<1 $ ? I did:
$$\ \frac{2x+3}{7-x}>0 $$
$\ x<7 $ results in $\ x> \frac{-3}{2} $ whereas $\ x>7 $ results in $\ x< \frac{-3}{2} $ so x=($\ \frac{-3}{2} $,7) and also
$\ x-2>0 $ so $\ x>2 $
$$\ x=(2,7)-\{3\} $$
So after finding out where $\ f $ is defined, I did this:
$$\log _{x-2}\left(\frac{2x+3}{7-x}\right) <\log \:_{x-2}\left(x-2\right) $$
$$\ \frac{2x+3}{7-x} <x-2 $$
$$\ x^2-7x+17<0 $$
And I figured I must have done a mistake somewhere as $\ d<0 $ and $\ x^2 $'s coefficient is $\ >0 $ so $\ f $ would be $\ >0$. Could you let me know where is my mistake? The correct answer is $\ x=(2,3) $
| $$\log _{(x-2)}\left(\frac{2x+3}{7-x}\right)<1$$
Hint
$1)$ $x-2>0$ and $x-2\ne 1$
$2)$ $\frac{2x+3}{7-x}>0$
$2)$ If $x-2>1$ then
$$\frac{2x+3}{7-x}<x-2$$
$3)$ If $0<x-2<1$ then
$$\frac{2x+3}{7-x}>x-2$$
Now you can solve it.
| {
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Solve $1^n+2^n+ \ldots +n^n=k!$ over positive integers If $k,n \in \mathbb{N}^*$, solve the following equation
$1^n+2^n+ \ldots +n^n=k!$, where $k! $ denotes $1 \cdot 2 \cdot 3 \cdots k$.
| I will answer for $n$ odd, because barto above answered the question for $n$ even.
We have $1^n+2^n+ \ldots+n^n=1+(2^n+n^n)+(3^n+(n-1)^n) + \ldots \equiv 1 \pmod {n+2}$.
So, $k! \equiv 1 \pmod {n+2}$. If $k \geqslant n+2$, we have $k! \equiv 0 \pmod {n+2}$.
So, $n<k<n+2$, which yields $k=n+1$.
We have now $(n+1)! \equiv 1 \pmod {n+2}$.
If $n+2$ is a prime, let $n+2=p$ we have $(p-1)! \equiv 1 \pmod p$ and by Wilson $(p-1)! \equiv -1 \pmod p$, so there are no solutions.
If now $n+2$ is not a prime, let $n+2=ab, (a,b)=1, a,b \leqslant n+1$ we have that $a,b \mid (n+1)! \Rightarrow ab \mid (n+1)! \Rightarrow n+2 \mid (n+1)!$, which is a contradiction.
So, we do not have solutions if $n \geqslant 3$ odd .
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the inverse of the function $f$ defined by $f(x) = x^2 - 4x$ for $x \geq 2$ Let $f$ be the function defined by $$f(x)=x^2-4x$$ for $x\ge 2$. Find the inverse function indicating its domain and the range.
When I try to make $x$ the subject I get $+$and $-$ two answers for the inverse function. Is it the proper answer?
The answer I got is $$ x=2
\pm \sqrt{y+4}.$$
Is it correct and how do I determine the domain and range of the function? Thanks
| The function $f$ defined by $f(x) = x^2 - 4x$ for $x \geq 2$ has domain $\text{Dom}_f = [2, \infty)$. We can find its range by completing the square.
\begin{align*}
f(x) & = x^2 - 4x\\
& = (x^2 - 4x + 4) - 4\\
& = (x - 2)^2 - 4
\end{align*}
The graph of $y = (x - 2)^2 - 4$ is a parabola with vertex $(2, -4)$ that opens upwards. The restriction that $x \geq 2$ means that the graph consists only of the right half of the parabola. Since the graph is continuous and increases without bound as $x$ approaches $\infty$, the range of $f$ is $\text{Ran}_f = [-4, \infty)$.
Solving for the inverse yields
\begin{align*}
y & = (x - 2)^2 - 4\\
y + 4 & = (x - 2)^2\\
\sqrt{y + 4} & = |x - 2|
\end{align*}
Since $x \geq 2$, $|x - 2| = x - 2$. Thus,
\begin{align*}
\sqrt{y + 4} & = x - 2\\
2 + \sqrt{y + 4} & = x
\end{align*}
Therefore, the inverse function is
$$f^{-1}(x) = 2 + \sqrt{x + 4}$$
The domain of the inverse is $\text{Dom}_{f^{-1}} = [-4, \infty)$ and the range of the inverse is $\text{Ran}_{f^{-1}} = [2, \infty)$.
Notice that the domain of the function is the range of its inverse and the range of the function is the domain of its inverse. Consequently, the graphs of $f$ and $f^{-1}$ are symmetric with respect to the line $y = x$, as shown in the figure below.
| {
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Simplify $\sqrt{6-\sqrt{20}}$ My first try was to set the whole expression equal to $a$ and square both sides. $$\sqrt{6-\sqrt{20}}=a \Longleftrightarrow a^2=6-\sqrt{20}=6-\sqrt{4\cdot5}=6-2\sqrt{5}.$$
Multiplying by conjugate I get $$a^2=\frac{(6-2\sqrt{5})(6+2\sqrt{5})}{6+2\sqrt{5}}=\frac{16}{2+\sqrt{5}}.$$
But I still end up with an ugly radical expression.
| Let $\sqrt{6 - \sqrt{20}} = \sqrt{a} - \sqrt{b}$, where $a$ and $b$ are rational numbers. Squaring both sides of the equation
$$\sqrt{a} - \sqrt{b} = \sqrt{6 - \sqrt{20}}$$
yields
\begin{align*}
a - 2\sqrt{ab} + b & = 6 - \sqrt{20}\\
a - 2\sqrt{ab} + b & = 6 - 2\sqrt{5}
\end{align*}
Matching rational and irrational parts yields the system of equations
\begin{align*}
a + b & = 6 \tag{1}\\
-2\sqrt{ab} & = -2\sqrt{5} \tag{2}
\end{align*}
Solving equation 2 for $b$ yields
\begin{align*}
-2\sqrt{ab} & = -2\sqrt{5}\\
\sqrt{ab} & = \sqrt{5}\\
ab & = 5\\
b & = \frac{5}{a}
\end{align*}
Substituting $5/a$ for $b$ in equation 1 yields
\begin{align*}
a + \frac{5}{a} & = 6\\
a^2 + 5 & = 6a\\
a^2 - 6a + 5 & = 0\\
(a - 1)(a - 5) & = 0
\end{align*}
Hence, $a = 1$ or $a = 5$.
If $a = 1$, then $b = 6 - a = 5$, in which case
$$\sqrt{6 - \sqrt{20}} = \sqrt{1} - \sqrt{5} < 0$$
which is impossible since the principal square root of a positive number must be positive.
Thus, $a = 5$ and $b = 6 - a = 1$, so
$$\sqrt{6 - \sqrt{20}} = \sqrt{5} - \sqrt{1} = \sqrt{5} - 1$$
Check: Observe that $\sqrt{5} - 1 > 0$. Moreover,
$$(\sqrt{5} - 1)^2 = 5 - 2\sqrt{5} + 1 = 6 - 2\sqrt{5} = 6 - \sqrt{20}$$
Hence,
$$\sqrt{6 - \sqrt{20}} = \sqrt{5} - 1$$
as claimed.
| {
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A quicker approach to the integral $\int{dx\over{(x^3+1)^3}}$ Source: A question bank on challenging integral problems for high school students.
Problem:
Evaluate the indefinite integral
$$\int{dx\over{(x^3+1)^3}}$$
Seems pretty compact but upon closer look, no suitable substitution comes to mind. I can use partial fractions but it would be VERY time consuming and altogether boring. I am unable to find an alternate solution other than partials. Can anyone lead me towards a quicker approach, because the exam I'm preparing for is time bound and I can't afford to spend much time on a single problem. Thanks!
Edit 1:
I looked upon reduction formulas. I guess we can generalise this by using:
$$I_n = \int {dx\over{(x^3+1)^n}}$$
$$I_n = \int{(x^3+1)^{-n}dx}$$
Will try and solve it. Maybe we can reduce it to a simpler integral!
Edit 2:
Ok so I got the reduction formula. Can anybody verify if its right, like if you've solved it?
$$I_{n+1} = \frac {x}{3n(x^3+1)^n} + \frac{(3n-1)}{3n}I_n$$
Edit 3: Now I have reached the solution nearly
take $n=2$
$$I_3 = \frac{x}{6(x^3+1)^2}+\frac{5}{6}I_2$$
now take $n=1$
$$I_2 = \frac{x}{3(x^3+1)}+\frac{2}{3}I_1 $$
now we see that $I_1$ is nothing but
$$I_1= \int{\frac{dx}{x^3+1}}$$
which simplifies to
$$I_1 = \int{\frac{dx}{(x+1)(x^2+1-x)}}$$
Now this integral is cake, solve using Partial Fractions
$$\frac{1}{(x+1)(x^2+1-x)} \equiv \frac{A}{x+1} + \frac{Bx+C}{(x^2+1-x)}$$
And eventually $I_1$ is:
http://www.wolframalpha.com/input/?i=integrate+1%2F((x%2B1)(x%5E2%2B1-x))
This way we get $I_1$. Putting in equation above we get $I_2$
Substitute $I_2$ in the equation above and obtain an expression for $I_3$ !
I'll verify ASAP :)
Final edit: Yeas! Reached the answer. Matches term to term!
Final answer is :
http://www.wolframalpha.com/input/?i=integrate+1%2F((x%2B1)(x%5E2%2B1-x))
Use the above approach or any alternatives that are more quick are welcome!
| In order to show the reduction formula, we use differentiation.
$$
\begin{aligned}
\frac{d}{d x}\left(\frac{x}{3 n\left(x^{3}+1\right)^{n}}\right) &=\frac{\left(x^{3}+1\right)^{n}-x n\left(x^{3}+1\right)^{n-1}\left(3 x^{2}\right)}{3 n\left(x^{3}+1\right)^{2 n}} \\
&=\frac{\left(x^{3}+1\right)^{n}-3 nx^{3}\left(x^{3}+1\right)^{n-1}}{3 n\left(x^{3}+1\right)^{2 n}} \\
&=\frac{x^{3}+1-3 n x^{3}}{3 n\left(x^{3}+1\right)^{n+1}} \\
&=\frac{1}{3 n\left(x^{3}+1\right)^{n}}-\frac{x^{3}+1-1}{\left(x^{3}+1\right)^{n+1}} \\
&=\frac{1}{3 n\left(x^{3}+1\right)^{n}} - \frac{1}{\left(x^{3}+1\right)^{n}}+\frac{1}{\left(x^{3}+1\right)^{n-1}}\\&= \frac{1-3 n}{3 n\left(x^{3}+1\right)^{n}}+\frac{1}{\left(x^{3}+1\right)^{n+1}}
\end{aligned}
$$
Integrating both sides yields $$
\begin{aligned}
\frac{x}{3 n\left(x^{3}+1\right)^{n}}&=\int \frac{1-3 n}{3 n\left(x^{3}+1\right)^{n}} d x+\int \frac{1}{\left(x^{3}+1\right)^{n+1}} d x \\
I_{n+1}&=\frac{x}{3 n\left(x^{3}+1\right)^{n}}+\frac{3 n-1}{3 n} I_{n}
\end{aligned}
$$
Wish it helps.
| {
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The quadratic equation with two unknowns: $ x^2 - 2axy + by^2 = 0 $ Please, could you help me with the next problem:
The problem:
Determine real parameters $a, b \in \Bbb{R}$, such that with:
$$ \langle x, y \rangle = x_1y_1 - ax_2y_1 - ax_1y_2 + bx_2y_2 $$
the inner product is defined in (or is it 'on') the vector space $\Bbb{R}^2$.
1) In order to $\langle x, x \rangle \ge 0$, i got that it should be $a \in \left[ -1, 1\right]$ and $b \ge 1$.
2) In order to $\langle x, x \rangle = 0 \,\Leftrightarrow\, x = 0$, i came to the equation $x^2 - 2axy + by^2 = 0$ (x = x_1, y= x_2).
My solution is next:
$$x^2 - 2axy + by^2 = 0 \,\Leftrightarrow\, x^2 + by^2 = 2axy \,\Leftrightarrow\, \frac{x^2}{y^2} + b = \frac{2ax}{y} \,\Leftrightarrow\, $$
$$\,\Leftrightarrow\, b = \frac{2ax}{y} - \frac{x^2}{y^2} \,\Leftrightarrow\, \frac{2ax}{y} - \frac{x^2}{y^2} \ge 1 \,\Leftrightarrow\, \frac{2ax}{y} \ge \frac{x^2}{y^2} \setminus :\frac{y^2}{x^2} \,\Leftrightarrow\,$$
$$\,\Leftrightarrow\, \frac{2ay}{x} \ge 1.$$
Now, because $x$ and $y$ are arbitrary, it's easy to find such $x$ and $y$, that for those numbers final inequality isn't correct.
So, my conclusion is that there isn't solution for this equation. Am i right?
Thank you, in advance, for your time!
P.S. I'm sorry for not writing the whole problem in the first place. I just wasted my and yours time.
| Here's an answer building on the comments.
\begin{align}
\langle x,x \rangle &= x_1^2 - 2ax_1x_2 + bx_2^2 \\
&= x_1^2 - 2ax_1x_2 + a^2x_2^2 + (b-a^2)x_2^2 \\
&= \underbrace{(x_1 - ax_2)^2}_{\geq 0} + \underbrace{(b-a^2)x_2^2}_{\geq 0 \text{ if }a^2\leq b}
\end{align}
So we are guaranteed that $\langle x,x \rangle\geq 0$ for all $x\in\mathbb{R}^2$ if $a^2 \leq b$.
Next, we can write:
\begin{align}
\langle x,x \rangle &= 0\\
x^2_1 - 2ax_1x_2 + bx_2^2 &= 0 \\
\frac{x_1^2}{x_2^2} - 2a\frac{x_1}{x_2} + b &= 0 \\
z^2 - 2az + b &= 0 \\
z = \frac{1}{2}\left( 2a\pm \sqrt{4a^2 - 4b} \right) &= \frac{x_1}{x_2}\\
x_2\left(a\pm\sqrt{a^2-b}\right) &= x_1
\end{align}
since $x=(x_1,x_2)\in\mathbb{R}^2$, we have $a^2\geq b$.
But since $a^2 \geq b$ and $a^2 \leq b$, we get $a^2 = b$.
| {
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Three positive integers $a$, $b$, and $c$ satisfy $abc=8!$ where $aThree positive integers $a$, $b$, and $c$ satisfy $abc=8!$ where $a<b<c$. What is the smallest possible value of $c-a$?
I know that $40,320=8!=8*7*6*5*4*3*2*1=7*5*3^2*2^7*1$.
I'm trying to see how I can choose $a$, $b$, and $c$ such that $abc=8!$ and $a<b<c$. I don't see it yet though.
| One has $8!=2^7\cdot 3^2\cdot 5\cdot 7$ and ${\root3\of 8!}\approx34.3$. We now should factor $8!$ into three factors as equal as possible, which means: as near to $34.3$ as possible.. It seems that $a=2^5=32$, $b=5\cdot 7=35$, $c=2^2\cdot3^2=36$ is the best we can do. Note that neither $33$ nor $34$ can be attained with the primes at disposal. The minimal possible value of $c-a$ therefore is $4$.
| {
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Find all positive integers $n > 1$ such that the polynomial $P(x)$ belongs to the ideal generated by the polynomial $x^2 +x +1$ in $\Bbb Z_n[x]$
Find all positive integers $n > 1$ such that the polynomial $x^4 + 3x^3 + x^2 + 6x + 10$ belong to the ideal generated by the polynomial $x^2 + x + 1$ in $\Bbb Z_n[x]$.
My attempt: I was using the Division Algorithm
$$P(X)= X^4 + 3x^3 + X^2 +6X + 10 = (X^2 + x + 1)( x^2 + 2x -2) + (6x + 12).$$
Here I got the remainder $6x + 12$ not equal to $0$, so $P(x)$ is irreducible over $\Bbb Z_n[x]$ because it cannot be factored into the product of two non-constant polynomials.
My thinking is that $\Bbb Z_3[x]$ is only the satisfied $x^2 + x + 1$
$$(1)^2 + 1 + 1 =3$$ and we if we divide $3/3 =1$ and remainder $= 0$.
Therefore $3$ is the only positive integer $n > 1$ such that the polynomial $P(x)$ belong to the ideal generated by the polynomial $x^2 + x + 1$ in $\Bbb Z_n[x]$.
Is my answer is correct or not? I would be more thankful to rectifying my mistake.
| Hint $\ 6x\!+\!12 = 0\,$ in $\,\Bbb Z_n[x]\iff 6=0=12\in \Bbb Z_n\!\iff n\mid 6,12\iff n\mid (6,12) = 6.$
| {
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Rudin Chapter 3 exercise 14 For exercise 14c of Baby Rudin:
If $\{s_n\}$ is a complex sentence, define its arithmetic means $\sigma_n$ by
$$\sigma_n = \frac{s_0 + s_1 + \cdots + s_n}{n + 1},$$
where $n = 0, 1, 2, ...$
14c: Can it happen that $s_n>0$ for all $n$ and that $\limsup s_n = \infty$, although $\lim \sigma_n = 0$?
Does my example of $s_n = \sqrt{n}$ work?
My attempt:
It is clear that $\limsup s_n = \infty$ since $\sqrt{n} \rightarrow \infty$ as $n \rightarrow \infty$.
And $\sigma_n = \frac{1 + \sqrt{2} + \cdots + \sqrt{n}}{n + 1}$, so $$\begin{split}
\lim_{n\to\infty} \sigma_n &= \lim_{n\rightarrow\infty} \frac{1 + \sqrt{2} + \cdots + \sqrt{n}}{n + 1}\\
&= \lim_{n\rightarrow\infty} \frac{1/\sqrt{n} + 2/\sqrt{n}+ \cdots + 1}{\sqrt{n} + 1/\sqrt{n}} = 0
\end{split}$$
| Your example does not work since
$$\frac{1}{n+1} \sum_{k=1}^n \sqrt{k} \geqslant \frac{1}{n+1}\int_0^n \sqrt{x} \, dx = \frac{2n^{3/2}}{3(n+1)}= \frac{2 \sqrt{n}}{3 + 3/n} \to \infty$$
| {
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Conversion to base $9$ Where am I going wrong in converting $397$ into a number with base $9$?
$397$ with base $10$ $$= 3 \cdot 10^2 + 9 \cdot 10^1 + 7 \cdot 10^0$$ To convert it into a number with base $9$
$$3 \cdot 9^2 + 9 \cdot 9^1 + 7 \cdot 9^0 = 243 + 81 + 7 = 331$$
But answer is $481$?????
| As I explained in 2011, one easy way is to write the number in Horner form in the original base, then do the base conversion from the inside-out, e.g. below where $\rm\color{#c00}{red}$ means radix $9$ notation
$$\begin{align} 397
\,&=\, (3\cdot 10\, +\, 9)\,10 +7\\
&=\, (\color{#c00}{3\cdot 11+10})10+7\\
&=\qquad\quad\ \ \color{#c00}{43\cdot 11}+7\\
&=\qquad\qquad\ \ \ \color{#c00}{473}+7\\
&=\qquad\qquad\ \ \ \color{#c00}{481}\end{align}$$
| {
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How can I prove the following sequence converges Given the following n-th term sequence:
$$a_{n} = \sqrt[n]{1^2+2^2+...+n^2}$$
You're asked to evaluate the limit of the given sequence, justifying your operations.
What strategy should I take on this? I have considered taking some inequality in order to, eventually, be able to use the Squeeze Theorem.
I've tried exploring some initial terms, viz:
$$
a_{1} = \sqrt[n]{1^2} = 1^\frac{1}{n}\\
a_{2} = \sqrt[n]{1^2 + 2^2} = 5^\frac{1}{n}\\
a_{3} = \sqrt[n]{1^2 + 2^2 + 3^2} = 14^\frac{1}{n}\\
\vdots\\
a_{n} = \sqrt[n]{1^2+2^2+...+n^2} = \sqrt[n]{k + n^2} = (k + n^2)^\frac{1}{n}
$$
Supposing $k$ is the sum of all the $n-1$ terms of the sequence, rightly before $n²$. We can see that $a_{n}$ is always smaller than $a_{n+1}$ for any $n$ strictly positive.
I'm not sure, though, what else I can try. I would appreciate a hand here.
Answer is:
\begin{align} 1 \end{align}
| Hint
Show by induction that
$$\sum_{k=1}^n k^2=\dfrac{n^3}{3}+\dfrac{n^2}{2}+\dfrac n6. $$ Thus
$$\sqrt[n]{1^2+2^2+...+n^2}\le \sqrt[n]{\dfrac{5n^3}6 }.$$
Can you conclude from here?
| {
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Find $\iiint (x^2+y^2+z^2)~dV$, above the cone $z=\sqrt{3(x^2+y^2)}$, inside the sphere $x^2+y^2+z^2\leq a^2$.
Find
$$\iiint (x^2+y^2+z^2)~dV$$
above the cone $z=\sqrt{3(x^2+y^2)}$, and inside the sphere $x^2+y^2+z^2\leq a^2$.
I think that the solution for this problem is following:
Use cylindrical coordinates:
$x = r\cdot \cos(\theta)$
$y = r\cdot \sin(\theta)$
$z = z$
and we will get:
$$\int_0^a \int_0^{\pi} \int_0^{a/2} (r^2+z^2)r dr d\theta dz$$
Is this approach correct?
| $\mathbf{HINT}$: First we have $\sqrt{3(x^2+y^2)}\leq z \leq \sqrt{a^2-x^2-y^2} $.
Second solve system: $z=\sqrt{3(x^2+y^2)}$, $x^2 + y^2 + z^2 = a^2 $. Solution: $D:x^2+y^2 = \frac{a^2}{4}$.
So, $$\iiint (x^2+y^2+z^2) dV = \iint_{D} \int_{\sqrt{3(x^2+y^2)}}^{\sqrt{a^2-x^2-y^2}} (x^2+y^2+z^2) dz dxdy,$$
then we use cylindrical coordinates: $x=r\cos(\phi), y=r\sin(\phi),z=z$ and we have
$$\int_{0}^{2 \pi} \int_{0}^{a/2} \int_{\sqrt{3r^2}}^{\sqrt{a^2-r^2}} (r^2+z^2)r dz dr d\phi=...=\frac{\pi a^5}{5}$$
| {
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How to find the closed form of $\sum_{k=2}^{\infty}{\lambda(k)-1\over k}?$ Given that:
$$\sum_{k=2}^{\infty}{\lambda(k)-1\over k}\tag1$$
Where $\lambda(k)$ is Dirichlet Lambda Function
We are seeking to determine the closed form $(1)$ and came close to estimates it to $1-{\frac12}\left(\gamma+\ln{\pi}\right)$.
where $\gamma=0.5772156...$ is Euler-Mascheroni Constant
How can we evalauate the exact closed form for $(1)$?
| You have found it. Interchange the order of summation, and a few rewrites plus Wallis's product yield the result:
\begin{align}
\sum_{k = 2}^{\infty} \frac{1}{k}\bigl(\lambda(k) - 1\bigr)
&= \sum_{k = 2}^{\infty} \frac{1}{k} \sum_{n = 1}^{\infty} \frac{1}{(2n+1)^k} \\
&= \sum_{n = 1}^{\infty} \sum_{k = 2}^{\infty} \frac{1}{k(2n+1)^k} \\
&= \sum_{n = 1}^{\infty}\Biggl(-\frac{1}{2n+1} - \log \biggl(1 - \frac{1}{2n+1}\biggr)\Biggr) \\
&= \sum_{n = 1}^{\infty} \Biggl(\log \biggl(1 + \frac{1}{2n}\biggr) - \frac{1}{2n+1}\Biggr) \\
&= \sum_{n = 1}^{\infty} \Biggl(\frac{1}{2}\log \biggl(1 + \frac{1}{n}\biggr) - \frac{1}{2n} + \frac{1}{2n} - \frac{1}{2n+1} \\
&\qquad\quad - \frac{1}{2}\biggl(\log\biggl(1 + \frac{1}{n}\biggr) - 2\log \biggl(1 + \frac{1}{2n}\biggr)\biggr)\Biggr) \\
&= -\frac{1}{2}\gamma + 1 - \log 2 - \frac{1}{2} \log \prod_{n = 1}^{\infty} \frac{(2n+1)^2n}{(2n)^2(n+1)} \\
&= - \frac{1}{2}\gamma + 1 - \log 2 - \frac{1}{2} \log \Biggl(\prod_{k = 2}^{\infty}\biggl(1 - \frac{1}{k^2}\biggr)\prod_{m = 1}^{\infty}\biggl(1 - \frac{1}{(2m)^2}\biggr)^{-1}\Biggr) \\
&= - \frac{1}{2}\gamma + 1 - \log 2 - \frac{1}{2} \log \biggl( \frac{1}{2}\cdot \frac{\pi}{2}\biggr) \\
&= 1 - \frac{1}{2}\bigl(\gamma + \log \pi\bigr).
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2420683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove identity holds Just wondering if there is any other way to show that for each positive integer $n$ holds
$$2\left(\sqrt{n} - 1\right) < 1 + \frac{1}{\sqrt{2}} +\cdots+\frac{1}{\sqrt{n}} < 2\sqrt{n}$$
other than by mathematical induction~
| For the right inequality:
$$\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{n}}\right) = \frac{1}{n}\left(\sqrt{n}+\sqrt{\frac{n}{2}}+\cdots+\sqrt{\frac{n-1}{n}} + \sqrt{\frac{n}{n}}\right)$$
This is a Riemann sum of a function $f(x) = \frac{1}{\sqrt{x}}$ on the segment $[0,1]$ with the equidistant subdivision with endpoints $\left\{0, \frac{1}{n}, \frac{2}{n}, \ldots, 1\right\}$.
So:
$$\lim_{n\to\infty} \frac{1}{n}\left(\sqrt{n}+\sqrt{\frac{n}{2}}+\cdots+ \sqrt{\frac{n}{n}}\right) = \int_{0}^1 \frac{dx}{\sqrt{x}} = 2$$
Since the Riemann sums increase towards the integral, by multiplying with $\sqrt{n}$ we obtain the desired inequality.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $a+b\sqrt{2}=0$ iff $a=b=0$ I want to prove for $a,b \in \mathbb{Q}$
\begin{align}
a+b\sqrt{2}=0 \quad \Leftrightarrow \quad a=b=0
\end{align}
For $\Leftarrow$ is trivial, now I am in trouble of showing $\Rightarrow$
Simply I can say $\sqrt{2}$ is irrational and go proceed, but I want some formal mathematical proof.
| Assume:($a\neq b\neq 0$)
$$a+b\sqrt{2}=0$$
$$(a+b\sqrt{2})(\frac{a-b\sqrt{2}}{a-b\sqrt{2}})=0$$
$$\frac{a^2-2b^2}{a-b\sqrt{2}}=0$$
$$\left\{\begin{matrix}
a^2-2b^2=0 & & \\
a-b\sqrt{2}\neq 0& &
\end{matrix}\right.$$
$$\left\{\begin{matrix}
\frac{a}{b}=\sqrt{2} & & \\
\frac{a}{b}\neq \sqrt{2}& &
\end{matrix}\right.$$
Contradiction
So:($a=b=0$) for $a+b\sqrt{2}=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2421046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Evaluating $\int^\infty _1 \frac{1}{x^4 \sqrt{x^2 + 3} } dx$ I was evaluating
$$\int^\infty _1 \frac{1}{x^4 \sqrt{x^2 + 3} } dx$$
My work:
I see that in the denominator, there is the radical $\sqrt{x^2 + 3}$. This reminds me of the trigonometric substitution $\sqrt{u^2 + a^2}$ and letting
$ u = a \tan \theta$. With this in mind: I let $x = \sqrt{3} \tan \theta$ .Moving on, $dx = \sqrt{3} (\sec \theta )^2 $. And.....if $\theta$ is an
acute angle, we see in the illustration below that...
$\tan \theta = \frac{x}{\sqrt{3}}$
Then...getting the equivalent form of $\sqrt{x^2 + 3}$ in terms of $\theta$:
$$\sqrt{x^2 + 3}$$
$$ =\sqrt{(\sqrt{3} \tan \theta)^2 + 3}$$
$$= \sqrt{3(\tan \theta)^2 + 3}$$
$$= \sqrt{3((\tan \theta)^2 + 1)}$$
$$= \sqrt{3} \sqrt{(\sec \theta)^2} $$
$$\sqrt{x^2 + 3} = \sqrt{3} \sec \theta $$
Getting the equivalent form of $x^4$ in terms of $\theta$:
$$x^4 = (\sqrt{3} \tan \theta)^4$$
$$ = 9 (\tan \theta)^4 $$
Getting the equivalent form of $dx$ in terms of $\theta$:
$$dx = \sqrt{3} (\sec \theta )^2 d\theta$$
Substituting these equivalent expressions to the integral above, we get:
$$\int \frac{1}{x^4 \sqrt{x^2 + 3} } dx = \int \frac{1}{9 (\tan \theta)^4 \sqrt{3} \sec \theta } \sqrt{3} (\sec \theta )^2 d\theta$$
$$ = \int \frac{1}{9 \sqrt{3} (\tan \theta)^4 \sec \theta } \sqrt{3} (\sec \theta )^2 d\theta$$
$$ = \int \frac{\sqrt{3} (\sec \theta )^2}{9 \sqrt{3} (\tan \theta)^4 \sec \theta } d\theta$$
$$ = \int \frac{\sec \theta}{9 (\tan \theta)^4 } d\theta$$
$$ = \int \frac{\sec \theta}{9 ((\tan \theta)^2)^2 } d\theta$$
$$ = \int \frac{\sec \theta}{9 ((\sec \theta)^2 - 1 )^2 } d\theta$$
$$ = \int \frac{\sec \theta}{9 ((\sec \theta)^2 - 1 )^2 } d\theta$$
$$ = \frac{1}{9}\int \frac{\sec \theta}{((\sec \theta)^2 - 1 )^2 } d\theta$$
$$\int \frac{1}{x^4 \sqrt{x^2 + 3} } dx = \frac{1}{9}\int \frac{\sec \theta}{ (\sec \theta)^4 -2(\sec \theta)^2 +1 } d\theta$$
My problem is....I don't know how to evaluate $\frac{1}{9}\int \frac{\sec \theta}{ (\sec \theta)^4 -2(\sec \theta)^2 +1 } d\theta$.
I can't go forward. I'm stuck. How to evaluate $\int^\infty _1 \frac{1}{x^4 \sqrt{x^2 + 3} } dx$ properly?
| By letting $x=1/t$ and by integration by parts, we get
\begin{align*}
\int_1^{\infty} \frac{1}{x^4 \sqrt{x^2 + 3} } dx
&=
\int_0^1\frac{t^3}{\sqrt{3t^2 + 1}}\,dt=
\frac{1}{3}\int_0^1t^2\frac{d(\sqrt{3t^2 + 1})}{dt}\,dt\\
&=\frac{1}{3}\left[t^2\sqrt{3t^2 + 1}\right]_0^1-
\frac{2}{3}\int_0^1 t\sqrt{3t^2 + 1} dt\\
&=\frac{2}{3}-\frac{2}{27}\left[(3t^2 + 1)^{3/2}\right]_0^1=\frac{4}{27}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2422416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that for any integer $n, n^2+4$ is not divisible by $7$. The question tells you to use the Division Theorem, here is my attempt:
Every integer can be expressed in the form $7q+r$ where $r$ is one of $0,1,2,3,4,5$ or $6$ and $q$ is an integer $\geq0$.
$n=7q+r$
$n^2=(7q+r)^2=49q^2+14rq+r^2$
$n^2=7(7q^2+2rq)+r^2$
$n^2+4=7(7q^2+2rq)+r^2+4$
$7(7q^2+2rq)$ is either divisible by $7$, or it is $0$ (when $q=0$), so it is $r^2+4$ we are concerned with.
Assume that $r^2+4$ is divisible by 7. Then $r^2+4=7k$ for some integer $k$.
This is the original problem we were faced with, except whereas $n$ could be any integer, $r$ is constrained to be one of $0,1,2,3,4,5$ or $6$.
Through trial and error, we see that no valid value of $r$ satisfies $r^2+4=7k$ so we have proved our theorem by contradiction.
I'm pretty sure this is either wrong somewhere or at the very least not the proof that the question intended. Any help would be appreciated.
| The question boils down to 3 being a quadratic residue modulo 7 or non-residue
$$n^2+4 \equiv 0 \; (mod\;7)$$
$$n^2\equiv 3 \;(mod\;7)$$
For this we have Legendre symbol
$$\left ( \frac{a}{p} \right )=\begin{cases}
& 1\text{ if }a\text{ is a quadratic residue modulo }p\text{ and }p\text{ does not divide }a \\
& -1\text{ if }a\text{ is a quadratic non-residue modulo }p \\
& 0\text{ if }p\text{ divides }a
\end{cases}$$
calculated as
$$\left ( \frac{a}{p} \right )=a^{\frac{p-1}{2}}$$
So
$$3^{\frac{7-1}{2}} \equiv 3^{3} \equiv 27 \equiv -1 \;(mod\;7)$$
making $3$ a non-residue modulo $7$.
Thus there is no $n$ that solves the congruence so $n^2+4$ is not divisible by $7$ for any $n$.
A little bit more advanced step that applies the Law of quadratic reciprocity would give that $3$ is a residue of $p$ if and only if
$$p \equiv \; 1, \; 11 \; (mod \; 12)$$
and $7$ does not belong to this group obviously.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2422879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
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Distance between two lines in parametric form
a) Parametrize the line $L$ through $P = (2, 1, 2)$ that intersects the
line
$x = 1 + t$, $y = 1 − t$, $z = 2t$
perpendicularly.
b) Parametrize the $z$ axis.
c) What is the distance from this line $L$ to the $z$-axis?
My work For the part a), I got the equation was $x = 2 + \frac{t}{6}$ ; $y = 1 + \frac{5t}{6}$ ; $z = 2 + \frac{t}{3}$.
Stuck with the last 2 parts. Anyone want to give this math novice a hand? :D
| Let $B(1+t,1-t,2t)$ and $A(2,1,2)$.
Thus, $$\vec{AB}\perp\vec{(1,-1,2)}$$ or
$$\vec{(t-1,-t,2t-2)}\vec{(1,-1,2)}=0$$ or
$$t-1+t+4t-4=0$$ or
$$t=\frac{5}{6}$$ and since
$$\vec{\left(\frac{5}{6}-1,-\frac{5}{6},\frac{5}{3}-2\right)}=\vec{\left(-\frac{1}{6},-\frac{5}{6},-\frac{1}{3}\right)},$$
we obtain:
$$L:(2,1,2)+s(1,5,2),$$
which you got by yourself.
Now, let $\pi$ be a plain in which placed $z$-axis such that $\pi||L.$
Thus, the parametric equation of $\pi$ it's
$$(x,y,z)=(0,0,0)+t(0,0,1)+s(1,5,2).$$
Let $\vec{n}(a,b,c)$ be a normal of $\pi$.
Thus,
$$0\cdot a+0\cdot b+1\cdot c=0$$ and
$$a+5b+2c=0,$$
which gives that we can assume that $\vec{n}(5,-1,0)$ and we got an equation of the plain $\pi$:
$$5(x-0)-1(y-0)+0(z-0)=0$$ or
$$5x-y=0.$$
Now, we can calculate our distance:
$$\frac{|5\cdot2-1\cdot1+0\cdot2|}{\sqrt{5^2+1^2+0^2}}=\frac{9}{\sqrt{26}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2423846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding generating function for a given sequence The task is to find the generating function for the sequence $$d_n = (\sum_{k=0}^n \binom{n}{k})*(\sum_{k=0}^n \frac{(k-1)^2}{2^k})$$
let's call this function $D(x)$.
I marked $a_n = \frac{1}{2^n}$; $b_n = \frac{-2n}{2^n}$; $c_n = \frac{n^2}{2^n}$ with respective generating functions $A(x)$, $B(x)$, $C(x)$.
So I found $$A(x) = \frac{1}{1-\frac{x}{2}}$$ and
$$B(x) = \frac{-x}{(1-\frac{x}{2})^2}$$
$$C(x) = \frac{2x+3x^2}{4(1-\frac{x}{2})^3}$$
using derivatives.
now the generating function for $a_n + b_n +c_n$ should be $A(x)+B(x)+C(x)$ and it's $$\frac{x^2}{(1-\frac{x}{2})^3}$$.
Next by using convolution (with let's say $f_n = 1$) we'll get that the function for the series $$u_n = \sum_{k=0}^n \frac{(k-1)^2}{2^k})$$ is $$\frac{x^2}{(1-\frac{x}{2})^3(1-x)}$$
Now since $h_n = \sum_{k=0}^n \binom{n}{k} = 2^n$ the appropriate function is $$H(x) = \frac{1}{1-2x}$$.
Again with using convolution with $h_n$ and $u_n$ we'll get that $$D(x) = H(x)U(x) = \frac{x^2}{(1-\frac{x}{2})^3(1-x)(1-2x)}$$
Is my answer good? or shall i keep expanding it to partial fractions.
| Hint: The situation is somewhat different.
Write
\begin{align*}
d_n &= \sum_{k=0}^n \binom{n}{k}\sum_{k=0}^n \frac{(k-1)^2}{2^k}\\
&=\sum_{k=0}^n(k-1)^22^{n-k}
\end{align*}
and consider the Cauchy-product $\sum_{k=0}^na_kb_{n-k}$ when multiplying generating functions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2424357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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$ 3^{2^n }- 1 $ is divisible by $ 2^{n+2} $ Prove that if n is a positive integer, then $ \ \large 3^{2^n }- 1 $ is divisible by $ \ \large 2^{n+2} $ .
Answer:
For $ n=1 \ $ we have
$ \large 3^{2^1}-1=9-1=8 \ \ an d \ \ 2^{1+2}=8 $
So the statement hold for n=1.
For $ n=2 $ we have
$ \large 3^{2^2}-1=81-1=80 \ \ and \ \ 2^{2+2}=16 \ $
$Also \ \ \ 16 /80 $ .
Thus the statement hold for $ n=2 $ also.
Let the statement hold for $ n=m \ $
Then $ a_m=3^{2^m}-1 \ \ is \ \ divisible \ \ by \ \ b_m=2^{m+2} \ $
We have to show that $ b_{m+1}=2^{m+3} \ $ divide $ \ \large a_{m+1}=3^{2^{m+1}}-1 \ $
But right here I am unable to solve . If there any help doing this ?
Else any other method is applicable also.
| We can write,
$$(3)^{2^n}-1$$
$$(4-1)^{2^n}-1$$
Using binomial,
$$=\color{red}{\binom{2^n}{0}}-\color{blue}{\binom{2^n}{1}4}+\color{green}{\binom{2^n}{2}4^2}-\cdots-\color{orange}{\binom{2^n}{2^n-1} 4^{2^n-1}}+\color{purple}{\binom{2^n}{2^n}4^{2^n}}-\color{red}{1}$$
$$$$
$$\binom{n}{r}=n (r+1) \binom{n-1}{r+1}$$
$$-\color{blue}{ 2^n 2 \binom{2^n-1}{2}4}+\color{green}{2^n 3\binom{2^n-1}{3}4^2}-\cdots+\color{magenta}{2^n (2^n-1) \binom{2^n-1}{2^n-1}4^{2^n-2}} -\color{orange}{2^n 4^{2^n-1}} +\color{purple}{4^{2^n}}$$
$$$$
$$- \color{red}{2^{n+2}}(-\color{blue}{2\binom{2^n-1}{2}}+\color{green}{3\binom{2^n-1}{3}4}-\cdots+\color{magenta}{(2^n-1)\binom{2^n-1}{2^n-1}4})-\color{orange}{2^n4^{2^n-1}}+\color{purple}{4^{2^n}}$$
$$$$
$$\color{red}{2^{n+2}}(-\color{grey}{\binom{2^n-1}{1}}+\color{blue}{2\binom{2^n-1}{2}}-\color{green}{3\binom{2^n-1}{3}4}-\cdots-\color\magenta{(2^n-1)\binom{2^n-1}{2^n-1}4^{2^n-2}}-\color{grey}{2^n +1})-\color{orange}{2^n4^{2^n-1}}+\color{purple}{4^{2^n}}$$
$$$$
$$\color{red}{2^{n+2}}(f'(4)-\color{grey}{2^n+1})-\color{orange}{2^n4^{2^n-1}}+\color{purple}{4^{2^n}}$$
(f(x) being $=(x-1)^{2^n-1}$)$$$$
Hence$$f'(4)=(2^n-1)3^{2^n-2}$$
$$\color{red}{2^{n+2}}((2^n-1)3^{2^n-2}+\color{grey}{2^n+1}-\color{orange}{4^{2^n-3}}+\color{purple}{2^{2^n}2^{2^n-n-2}})$$
$$$$
$$\color{red}{2^{n+2}}C$$
$$$$
Hence.......
OR
$$3^{2^{m+1}}-1$$
$$3^{2^m×2}-1$$
$$(3^{2^m}-1)(3^{2m}+1)$$
$$k×2^{m+2} ×(3^{2^m}-1+2)$$
$$2^{m+2} ×k× (k2^{m+2}+2)$$$$2^{m+2} ×k× 2×(k2^{m+1}+1)$$$$2^{m+3} ×\gamma$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solution of $\sqrt{5-2\sin x}\geq 6\sin x-1$
Solve the following inequality. $$\sqrt{5-2\sin x}\geq 6\sin x-1.$$
My tries:
As $5-2\sin x>0$ hence we do not need to worry about the domain.
Case-1: $6\sin x-1\leq0\implies \sin x\leq\dfrac{1}{6}\implies -1\leq\sin x\leq\dfrac{1}{6}\tag*{}$
Case-2:$6\sin x-1>0\implies \dfrac{1}{6}<\sin x<1\tag*{}$
$\implies 5-2\sin x\geq36\sin^2x+1-12\sin x\implies 18\sin^2x-5\sin x-2\leq0$
$\implies(2\sin x-1)(9\sin x+2)\leq0$
$\implies\sin x\ \epsilon\ \bigg(\dfrac{1}{6},\dfrac{1}{2}\bigg]$
All of above implies $\sin x\ \epsilon\ \bigg[-1,\dfrac{1}{2}\bigg]$.
Answer is given in the form: $\bigg[\dfrac{\pi(12n-7)}{6},\dfrac{\pi(12n+1)}{6}\bigg]\ (n\epsilon Z)$
How do I reach the form given in options? I even don't know what I've is correct or not.
Please help.
| solving the inequality $$18\sin(x)^2-5\sin(x)-2\le 0$$ we have
$$2 \pi c_1-\sin ^{-1}\left(\frac{2}{9}\right)\leq x\leq \frac{1}{6} \left(12 \pi c_1+\pi
\right)\lor \frac{1}{6} \left(12 \pi c_1+5 \pi \right)\leq x\leq 2 \pi c_1+\pi +\sin
^{-1}\left(\frac{2}{9}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2429305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$\frac{3}{abc} \ge a + b + c$, prove that $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \ge a + b + c$ Let $a$, $b$ and $c$ be positive numbers such that $\frac{3}{abc} \ge a + b + c$. Prove that: $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \ge a + b + c.$$
Any way I use - I get stuck after 2 or 3 steps...
| $$\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\ge 3\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\ge abc(a+b+c)\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=(a+b+c)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2429512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the coordinates of the points on $y=-(x+1)^2+4$ that have a distance of $\sqrt {14}$ to $(-1,2)$ Create a function that gives the distance between the point $(-1,2)$ the graph of $$y=-(x+1)^2+4.$$ Find the coordinates of the points on the curve that have a distance of $\sqrt {14}$ units from the point $(-1,2)$.
I know that the $x$-intercepts are $x=1$ and $x=-3$, and that the vertex is $(-1,4)$. I'm trying to use the distance formula by equating $\sqrt{14}$ to the function, but that is not getting me anywhere.
$$
d(x) = \sqrt{(x+1)^2 + \big(-(x+1)^2 + 2\big)^2}
$$
| You need to solve $$\sqrt{(x+1)^2 + \big(-(x+1)^2 + 2\big)^2}=\sqrt{14}.$$ Squaring both parts and substituting $u=(x+1)^2$ we get a quadratic equation $u^2-3u-10=0$ which has two roots, $u=5$ and $u=-2$. Only $(x+1)^2=5$ is possible for real numbers so we get $x=\sqrt5-1$ and $x=-\sqrt5-1$
| {
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"timestamp": "2023-03-29T00:00:00",
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Compute: $\arctan{\frac{1}{7}}+\arctan{\frac{3}{4}}.$
Compute: $\arctan{\frac{1}{7}}+\arctan{\frac{3}{4}}.$
I want to compute this sum by computing one term at a time. It's clear that
$$\arctan{\frac{1}{7}}=A \Longleftrightarrow\tan{A}=\frac{1}{7}\Longrightarrow A\in\left(0,\frac{\pi}{2}\right).$$
Drawing a right triangle with sides $1$, $7$ and $5\sqrt{2},$ I get that $$\begin{array}{lcl}
\sin{A} & = & \frac{1}{5\sqrt{2}}\Longleftrightarrow A= \arcsin{\frac{1}{5\sqrt{2}}} \\
\cos{A} & = & \frac{7}{5\sqrt{2}}\Longleftrightarrow A= \arccos{\frac{7}{5\sqrt{2}}} \\
\end{array}$$
But this will not get me standard angles for $A.$
| I think of these sorts of things in terms of complex numbers. Let $\theta = \arctan 1/7$. Then $e^{i\theta} = \cos \theta + i \sin \theta$. We need $\sin \theta / \cos \theta = 1/7$, so this will be a multiple of $7 + i$, but it must have length 1, so $e^{i \theta} = (7+i)/\sqrt{50}$. Similarly let $\phi = \arctan 3/4$; by the same argument $e^{i \phi} = (4 + 3i)/5$. Then we have
$$e^{i(\theta + \phi)} = {(7+i) (4+3i) \over 5 \sqrt{50}} = {28 + 4i + 21i - 3 \over 25 \sqrt{2}} = {25 + 25i \over 25 \sqrt{2}} = {1+i \over \sqrt{2}}$$
from which we can see that $\theta + \phi = \arctan 1 = \pi/4$.
This illustrates the general point that lots of trig identities are in some sense just identities about complex numbers.
| {
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"url": "https://math.stackexchange.com/questions/2431754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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distance between 2 lines in 3d
Calculate the distance between the lines
$$L_1:x=4+2t,y=3+2t,z=3+6t$$
$$L_2: x=-2+3s ,y=3+4s ,z=5+9s$$
I tried subtraction $L_1$ from $L_2$ then multiplying the resting vector by the $t$'s and $s$'s original values and trying to find value for $t$ to or $s,$ but I found $t=\frac{19}{12} s$ and I don't know how to keep solving this.
| The distance between the lines
$L_1:x=4+2t,y=3+2t,z=3+6t$ and $L_2: x=-2+3s ,y=3+4s ,z=5+9s$ is
$d=2 \sqrt{10}$
Indeed consider the function which gives the distance between a generic point of the first line and a generic point of the second line
$f(t,s)=\sqrt{(2 t-4 s)^2+(-3 s+2 t+6)^2+(-9 s+6 t-2)^2}$
and set to zero the partial derivatives $\partial f_t=0,\partial f_s=0$
$\left\{
\begin{array}{l}
4 (2 t-4 s)+4 (-3 s+2 t+6)+12 (-9 s+6 t-2)=0 \\
-8 (2 t-4 s)-6 (-3 s+2 t+6)+18 (-9 s+6 t-2)=0 \\
\end{array}
\right.
$
$\left\{
\begin{array}{l}
17 s-11 t=0 \\
53 s-34 t=0 \\
\end{array}
\right.$
which has one solution
$t= 0,\;s=0$
we have
$f(0,0)=2 \sqrt{10}$
Hope it helps
| {
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"question_score": "2",
"answer_count": 3,
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Is sum of digits of $3^{1000}$ divisible by $7$? I'm working on a little exercise I found in my high school book (printed in 2007) which is pretty complicated.
Is the sum of digits of $3^{1000}$ a multiple of $7$?
Do you have any advice to solve this type of problem (without programming of course!)
PS :
We are a group of 3 french people working on it since 2007.
The sum calculated with Python is 2142, this number is a multiple of 7 BUT we want a mathematical answer.
All the results below are mathematically proved !!
$3^{1000}$ has 478 digits and the sum of digits of $3^{1000}$ can't be superior to 4302 (9*478).
This sum is a multiple of 3 and 9.
The last digits of $3^{1000}$ are 0001 (math proof not a result of a computer calculation).
The one who created this exercise doesn't know the answer.
Please help us with any clue!
Cross-posted at https://mathoverflow.net/q/282035/22954 on MathOverflow.
| I am not answering the question but the post asks for clues so here it is a couple of ideas.
If $a_0,a_1,a_2, \cdots , a_{477} $ are the decimal digits of $3^{1000}$ then the numbers
$$b_i=a_{6i}+a_{6i+1}\cdot 10+a_{6i+2}\cdot 10^2+a_{6i+3}\cdot 10^3+a_{6i+4}\cdot 10^4+a_{6i+5}\cdot 10^5$$
for $i=0,1,2,\cdots, 79$ are the digits of $3^{1000}$ in base $10^6$ ($80$ is the nearest integer above $477 \div 6$ so there are $80$ digits numbered $0,1,2,\cdots,79$)
In other words:
$$ 3^{1000} = b_0 + b_1 \cdot 10^6+ \cdots + b_{79} \cdot (10^6)^{79}$$
Now, if we resort to modular arithmetic we see that
$$ 3^0=1, 3^1=3 , 3^2=2, 3^3=6, 3^4=4, 3^5=5, 3^6=1 $$ (all the equalities taken modulo $7$).
Also $$3^{1000}=(3^6)^{166}\cdot 3^4= 1 \cdot 4 = 4$$(all the equalities taken modulo $7$).
Now if we note that $10^6=1$ (modulo 7) the expression of $3^{1000}$ in base $10^6$ reads (modulo 7)
$$ 4=b_0+b_1+ \cdots +b_{79}$$
So we can assert that the sum of digits of $3^{1000}$ in base one milion gives a remainder of $4$ when divided by $7$.
Another partial result comes from the decimal expansion read modulo 7:
$$3^{1000}= a_0+ a_1 \cdot 10 + \cdots +a_{477} \cdot 10^{477} =1 = a_0+3 a_1+ 2a_2 + 6 a_3 + 4 a_4 + 5 a_ 5 + \cdots $$
So, given that $a_0=1$ we can say that this particular linear combination of the remaining digits is divisible by $7$.
| {
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Given $a^3+b^3+c^3=(a+b+c)^3$, show that $a^{2n+1}+b^{2n+1}+c^{2n+1}=(a+b+c)^{2n+1}$
Assume that $\{a,b,c\} \subset \Bbb R$, $n\in\mathbb N$ and $a^3+b^3+c^3=(a+b+c)^3$.
Show that $a^{2n+1}+b^{2n+1}+c^{2n+1}=(a+b+c)^{2n+1}.$
This is from a list of problems used for training a team for a math olympics. I tried to use known Newton identities and other symmetric polynomial results but without success (perhaps wrong approach). Sorry if it is a duplicate. Hints and answers welcomed.
| $$(a+b+c)^3-a^3-b^3-c^3=\sum_{cyc}(3a^2b+3a^2c+2abc)=3(a+b)(a+c)(b+c)$$ and the rest is smooth.
| {
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Is there any sequence $\{a_n:n\in \mathbb N\}$ with each non-zero terms s.t. $\displaystyle \sum_{n=1}^{\infty}a_n=0$
Is there any sequence $\{a_n:n\in \mathbb N\}$ with each non-zero terms s.t. $\displaystyle \sum_{n=1}^{\infty}a_n=0$?
I'm finding an infinite linearly dependent set which has no finite linearly dependet subset. If I could find the above then I could say so.
Edit: sorry I wrote first 'strictly positive' terms instead of 'non-zero'.
| Let convergence be understood in the $\|\cdot\|_\infty$ norm.
Consider the familiy of sequences $$\mathcal{F} = \left\{-\left(\underbrace{0, 0, \ldots, 0}_{n-1}, \frac{1}{2^n}, 0, 0, \ldots\right) : n \in \mathbb{N}\right\} \cup \left\{\left(1, \frac{1}{2}, \frac{1}{4}, \ldots, \frac{1}{2^n}, \frac{1}{2^{n+1}}, \ldots\right)\right\} \subseteq c_0$$
Notice that every finite subset of $\mathcal{F}$ is linearly independent, since the last sequence has infinite support, and the others have finite disjoint supports.
Also,
$$\sum_{x \in \mathcal{F}}x = \left(1, \frac{1}{2}, \frac{1}{4}, \ldots, \frac{1}{2^n}, \frac{1}{2^{n+1}}, \ldots\right) - \sum_{n=1}^\infty\left(\underbrace{0, 0, \ldots, 0}_{n-1}, \frac{1}{2^n}, 0, 0, \ldots\right) = 0$$
since:
$$\left\|\left(1, \frac{1}{2}, \frac{1}{4}, \ldots, \frac{1}{2^n}, \frac{1}{2^{n+1}}, \ldots\right) - \sum_{n=1}^N\left(\underbrace{0, 0, \ldots, 0}_{n-1}, \frac{1}{2^n}, 0, 0, \ldots\right)\right\|_\infty$$
$$=\left\|\left(1, \frac{1}{2}, \frac{1}{4}, \ldots, \frac{1}{2^n}, \frac{1}{2^{n+1}}, \ldots\right) - \left(1, \frac{1}{2}, \frac{1}{4}, \ldots, \frac{1}{2^{N}}, 0, 0, \ldots\right)\right\|_\infty$$
$$=\left\|\left(0, 0, \ldots, 0, \frac{1}{2^{N+1}}, \frac{1}{2^{N+2}}, \ldots\right)\right\|_\infty = \frac{1}{2^{N+1}}\xrightarrow{N\to\infty} 0$$
This property is sometimes called $\omega$-independence:
$$\sum_{n=1}^\infty\alpha_n a_n = 0 \implies \alpha_n = 0, \,\forall n\in\mathbb{N}$$
Thus, $\mathcal{F}$ is (finitely) linearly independent, but it is not $\omega$-independent.
| {
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If four dice are thrown, what is the probability that the sum of numbers thrown up will be 15? How about 16?
If four dice are thrown, what is the probability that the sum of numbers thrown up will be 15? How about 16?
What's a good method of finding the answer? I know that overall there are 1296 possibilities.
| Quoting from this answer:
It is the coefficient of $x^{15}$ in the expansion of the generating function $$(x^6+x^5+x^4+x^3+x^2+x^1)^4=\left(\frac{x(1-x^6)}{1-x}\right)^4$$ which is $$1\cdot{{x}^{24}}+4\cdot {{x}^{23}}+10\cdot {{x}^{22}}+20\cdot {{x}^{21}}+35\cdot {{x}^{20}}+56\cdot {{x}^{19}}+80\cdot {{x}^{18}}\\+104\cdot {{x}^{17}}+125\cdot {{x}^{16}}+140\cdot {{x}^{15}}+146\cdot {{x}^{14}}+140\cdot {{x}^{13}}+125\cdot {{x}^{12}}+104\cdot {{x}^{11}}\\+80\cdot {{x}^{10}}+56\cdot {{x}^{9}}+35\cdot {{x}^{8}}+20\cdot {{x}^{7}}+10\cdot {{x}^{6}}+4\cdot {{x}^{5}}+1\cdot{{x}^{4}}$$
so the answer is $140$.
| {
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"source": "stackexchange",
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A formula for $1^4+2^4+...+n^4$ I know that
$$\sum^n_{i=1}i^2=\frac{1}{6}n(n+1)(2n+1)$$
and
$$\sum^n_{i=1}i^3=\left(\sum^n_{i=1}i\right)^2.$$
Here is the question: is there a formula for
$$\sum^n_{i=1}i^4.$$
| We can get the formula by the following way.
$$(n+1)^5-1=\sum_{k=1}^n((k+1)^5-k^5)=\sum_{k=1}^n(5k^4+10k^3+10k^2+5k+1).$$
Thus,
$$\sum_{k=1}^nk^4=$$
$$=\frac{1}{5}\left((n+1)^5-1-10\cdot\frac{n^2(n+1)^2}{4}-10\cdot\frac{n(n+1)(2n+1)}{6}-5\cdot\frac{n(n+1)}{2}-n\right)=$$
$$=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}.$$
| {
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"source": "stackexchange",
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Find the value of $P(X < KY)$ The random variables X and Y are independent and have the pdf's as follows:
$f_X(x) = \begin{cases}
xe^{-\frac{1}{2}x^2}& \text{if $x\geq0$} \\
0 & \text{otherwise}
\end{cases}$
and
$f_Y(y) = \begin{cases}
ye^{-\frac{1}{2}y^2}& \text{if $y\geq0$} \\
0 & \text{otherwise}
\end{cases}$
Find the probability that X is less than or equal to KY where K is a constant.
My Approach:
\begin{align*}
P(X \leq KY) &= \int_0^{KY}\int_0^{Y}xe^{-\frac{1}{2}x^2}ye^{-\frac{1}{2}y^2}\,dx\,dy
\\ & = \frac{Y^2}{2}\int_{0}^{KY} xe^{-x^2}\,dx
\\ & = \frac{-1}{2}e^{-K^2Y^2} + \frac{1}{2}
\end{align*}
I just want to know whether my approach is correct or not. Thanks in advance.
| if $K \leq 0$, $P(X \leq KY ) = 0$.
If $K>0$, the integral should be
\begin{align*}
P(X \leq KY) &= \color{blue}{\int_0^\infty}\color{red}{\int_0^{Ky}}xe^{-\frac{1}{2}x^2}ye^{-\frac{1}{2}y^2}\, \color{red}{dx}\,\color{blue}{dy}
\end{align*}
Your final solution should be an expression that is depending only on $K$, it shouldn't involve $Y$.
| {
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Evaluating limits without L'Hopital's I need to evaluate without using L'Hopital's Rule. They say you don't appreciate something until you lose it. Turns out to be true.
*
*$\lim_{x \to 0} \frac{\sqrt[3]{1+x}-1}{x}$
*$\lim_{x \to 0} \frac{\cos 3x - \cos x}{x^2}$
I have tried to "rationalise" Q1 by using the identity $a^3-b^3=(a-b)(a^2+ab+b)$ but I still end up with a case of $\frac{0}{0}$.
For Q2, I tried to apply the formula and express $\cos 3x-\cos x =2\cos 2x \cos x$ which didn't help.
I would really be grateful for some advice on how I can proceed with the evaluating.
| As regards 2), note that
$$\frac{\cos(3x) - \cos(x)}{x^2}=\frac{1 - \cos x}{x^2}-9\cdot\frac{1 - \cos (3x)}{(3x)^2}.$$
Hence
$$\lim_{x \to 0} \frac{\cos(3x) - \cos(x)}{x^2}=
\lim_{x \to 0}\frac{1 - \cos x}{x^2}-9\cdot\lim_{x \to 0}\frac{1 - \cos (3x)}{(3x)^2}\\=(1-9)\lim_{x \to 0}\frac{1 - \cos x}{x^2}
=-8\lim_{x \to 0}\frac{1 - \cos^2 x}{x^2(1+\cos x)}=-\frac{8}{2}\lim_{x \to 0}\left(\frac{\sin x}{x}\right)^2=-4.$$
| {
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"source": "stackexchange",
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Multivariable $\epsilon-\delta$ proof verification Prove that $\lim\limits_{(x,y) \to (1,1)} xy=1$
Of course, I am aware that this is "obvious", but I want to add some rigor to it. When I searched around for multivariable limits using $\epsilon-\delta$, most of the examples had $(x,y) \rightarrow (0,0)$, but in this case I have $x$ and $y$ approaching something else.
$(x,y) \rightarrow (1,1) \Leftrightarrow \lvert\lvert (x,y)-(1,1)\lvert\lvert \rightarrow 0$ which can be written as $0 < \sqrt {(x-1)^2+(y-1)^2} < \delta$ for some arbitrarily small $\delta >0$.
Goal: show that $\forall$ $\epsilon>0$ $\exists$ $\delta>0$ such that
$0 < \sqrt {(x-1)^2+(y-1)^2}<\delta\Rightarrow0<|xy-1|<\epsilon$
Proof:
If $0 < \sqrt {(x-1)^2+(y-1)^2}<\delta$, then
$|xy-1|=|xy-x-y+1+x+y-2|=|(x-1)(y-1)+(x-1)+(y-1)|$
$\le|(x-1)(y-1)|+|x-1| +|y-1|=|x-1||y-1|+|x-1|+|y-1|$
$=(\sqrt{(x-1)^2})(\sqrt{(y-1)^2})+\sqrt{(x-1)^2}+\sqrt{(y-1)^2}$
$\le(\sqrt{(x-1)^2+(y-1)^2})^2+2\sqrt{(x-1)^2+(y-1)^2}<\delta^2+2\delta$
If $\delta=$$\epsilon \over4$, then $\delta^2+2\delta=\frac{\epsilon^2}{16}+\frac{8\epsilon}{16}=\frac{\epsilon^2+8\epsilon}{16}$
Now, $\frac{\epsilon^2+8\epsilon}{16}<\epsilon \Leftrightarrow \epsilon^2+8\epsilon<16\epsilon\Leftrightarrow\epsilon(\epsilon-8)<0$
Since $\epsilon>0$, we have $\epsilon(\epsilon-8)<0$ if and only if $\epsilon<8$
So if we have $\epsilon<8$, then we can pick $\delta=\frac{\epsilon}{4}$ which gives us $\delta^2+2\delta<\epsilon$. If $\epsilon \ge8$, then we can pick $\delta=2$ which gives us $\delta^2+2\delta\le\epsilon$.
Therefore, if we pick $\delta =$ min {${\frac{\epsilon}{4}, 2}$}, then $0 < \sqrt {(x-1)^2+(y-1)^2}<\delta$ implies $0<|xy-1|<\epsilon$
Thus, $\lim\limits_{(x,y) \to (1,1)} xy=1$
| Your proof is correct and nice.
Take $\delta<1$
By doing this a priori you will find the delta you need more easily and save some time from the calculations
You proved that $$|xy-1| \leq |x-1||y-1|+|x-1|+|y-1|$$
Thus $$|x-1|=\sqrt{(x-1)^2} \leq \sqrt{(x-1)^2+(y-1)^2} < \delta$$
$$|y-1|=\sqrt{(y-1)^2} \leq \sqrt{(x-1)^2+(y-1)^2} < \delta$$
Thus $|f(x,y)-1|=|xy-1| <\delta^2 +2\delta=\delta(\delta+2) <3 \delta$
Choose $0<\delta <\min\{ \frac{\epsilon}{3},1\}$ and you have the result
| {
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"question_score": "6",
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Using the Chinese Remainder theorem to solve the systems of congruence's. We got introduced to the Chinese Reminder Theorem, but I still haven't quite grasped it and I have a problem that asks:
Find $x \bmod 77$ if $x \equiv 2 \pmod 7$ and $x \equiv 4 \pmod {11}$.
My attempt was like so:
$$B = 7 * 77 * 11 = 5929$$
$$B_{1} = \frac{5929}{77} = 77$$
Using $B_{1}*x_{1} \equiv 1 \pmod {77}$:
$$77x_{1} \equiv 1 \pmod {77} $$
$$x_{1} \equiv 78 \pmod {77}$$
So then would the $x$ be $78$?
| Let us use the Euclidean algorithm to express $\gcd(7,11)$ as a linear combination:
$$4 = 11 - 7\\
3 = 7 - 4 = 2 \cdot 7 - 11 \\
1 = 4 - 3 = 2 \cdot 11 - 3 \cdot 7.$$
Now, $x = (2 \cdot 11 - 3 \cdot 7) x = 2 (11x) - 3 (7x)$. By the given congruences, $11x \equiv 11 \cdot 2 = 22 \pmod{77}$ and $7x \equiv 7 \cdot 4 = 28 \pmod{77}$. Therefore, $x \equiv 2 \cdot 22 - 3 \cdot 28 = -40 \equiv 37 \pmod{77}$.
(This is hopefully closer to what you were hinting at in your comments as the method of solving CRT problems that your source is using.)
| {
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Solving for equation of a circle when tangents to it are given. Given: $\mathrm L(x) = x - 2$, $\mathrm M(y) = y - 5$ and $\mathrm N(x, y)= 3x - 4y - 10$.
To find: All the circles that are tangent to these three lines.
Outline of the method :
If we parameterised any line $\mathrm Z$ in terms of $x(t) = pt + q$ and $y(t) = rt + s$, then for a general circle $\mathrm C(x, y) = (x - \alpha)^2 + (y - \beta)^2 - \gamma^2$ we can substitute the the parameters of the line and get
$$\mathrm C(x(t), y(t)) = a t^2 + bt + c = 0,$$
where $a,b,c \in \Bbb R$ and they depend on $p$,$q$.
Now if this quadratic has $2$ roots then the circle intersects the line at $2$ points and same for $1$ and $0$ roots.
If we are given that $\mathrm Z$ is tangent then we can using completing the square and get $(t - d(p, q) ) ^2 + e(p, q) = 0$.
So our condition for the line tangent to the circle would be $e(p, q) = 0$.
Parameterising the three equations,
$\mathbf L^\prime (l) = (2,0) + l (0,1)$, $\mathbf M^\prime(m) = (0, 5) + m(-1, 0)$ and $\mathbf N^\prime(n) = (2 , -1) + n(4,3)$.
Substituting these in the general equation of a circle, $\mathrm C(x, y) = (x - h)^2 + (y - k)^2 - r^2$.
$$\left\{ \begin{align} \mathrm{C(L_x, L_y)} &= (2-h)^2 +(l-k)^2 -r^2 \\ \mathrm{C(M_x, M_y)} &= (m+h)^2 +(5-k)^2 -r^2 \\ \mathrm{C(N_x, N_y)} &= (2+4n-h)^2 +(3n-1-k)^2 -r^2 \end{align} \right.$$
As per the above method,
From first equation we get $r^2 = (2 - h) \iff r = |2 - h|$ and from second equation, $r = |5 - k|$.
Completing the square on the third equation we get,
$$\mathrm{C(N_x, N_y)} = \left(5n + \dfrac{5 - 4h - 3k}{5}\right) + 5 - 4h + 2k - r^2 + h^2 + k^2 - \left[1 - \dfrac{4h - 3k}{5}\right]^2.$$
Therefore,
$$ 2 + 2k + h^2 + k^2 = r^2 + 4h + \left[1 - \dfrac{4h - 3k}{5}\right]^2.$$
So here we got three equations:
\begin{align} 2 +2k +h^2 +k^2 &= r^2 +4h +\left[1 - \dfrac{4h-3k}{5}\right]^2 \\ r &= |5-k| \\ r &= |2-h|. \end{align}
I don't mind solving the third equation after substituting for $r$ from first in it, it is just a quadratic. What I don't understand is whether I should take $r = + (5 - k)$ or $r = - (5 - k)$ and $r = +(2 - h)$ or $r = - (2 - h)$?
Which two out of those should I take? And why?
| Consider $\triangle ABC$,
$A=(2,5),B=(2,-1),C=(10,5)$
and its incircle and three excircles,
\begin{align}
a&=10,\quad
b=8,\quad
c=6,\quad
\\
r&=2,\quad
r_a=12,\quad
r_b=6,\quad
r_c=4
,\\
O_i&=(4,3),\quad
O_a=(14,-7),\quad
O_b=(8,11),\quad
O_c=(-2,1)
.
\end{align}
| {
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Why does the augmented matrix method for finding an inverse give different results for different orders of elementary row operations? Why does the augmented matrix method for finding an inverse give different results for different orders of elementary row operations?
Consider the example of elements from the Heisenberg group:
$\begin{bmatrix}1 & a & b\\ 0 & 1 & c\\ 0 & 0 &1 \end{bmatrix}$.
Now, after augmenting this matrix,
$
\left[\begin{array}{rrr|rrr}
1 & a & b & 1 & 0 & 0\\
0 & 1 & c & 0 & 1 & 0\\
0 & 0 & 1 & 0 & 0 & 1
\end{array}\right]
$
if you subtract c times the third row from the second,
$
\left[\begin{array}{rrr|rrr}
1 & a & b & 1 & 0 & 0\\
0 & 1 & 0 & 0 & 1 & -c\\
0 & 0 & 1 & 0 & 0 & 1
\end{array}\right]
$
then b times the third row from the first,
$
\left[\begin{array}{rrr|rrr}
1 & a & 0 & 1 & 0 & -b\\
0 & 1 & 0 & 0 & 1 & -c\\
0 & 0 & 1 & 0 & 0 & 1
\end{array}\right]
$
and then a times the new second row from the first,
$
\left[\begin{array}{rrr|rrr}
1 & 0 & 0 & 1 & -a & -b\\
0 & 1 & 0 & 0 & 1 & -c\\
0 & 0 & 1 & 0 & 0 & 1
\end{array}\right]
$
you then get the following matrix for the inverse:
$$\begin{bmatrix}1 & -a & -b\\ 0 & 1 & -c\\ 0 & 0 &1 \end{bmatrix}.$$
But, this is not the inverse.
I thought order of elementary row operations doesn't matter.
| $$\left[ \begin{array}{ccc|ccc} 1 & a & b & 1 & 0 & 0 \\ 0 & 1 & c & 0 & 1 & 0\\ 0 & 0 & 1 & 0 & 0 & 1\end{array} \right]$$
I believe there is a typo in your working, I believe you mean subtract $c$ times the third row and add it to the second row
$$\left[ \begin{array}{ccc|ccc} 1 & a & b & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & -c\\ 0 & 0 & 1 & 0 & 0 & 1\end{array} \right]$$
After which,
$$\left[ \begin{array}{ccc|ccc} 1 & a & 0 & 1 & 0 & -b \\ 0 & 1 & 0 & 0 & 1 & -c\\ 0 & 0 & 1 & 0 & 0 & 1\end{array} \right]$$
and for the last step:
$$\left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & 1 & -a & ac-b \\ 0 & 1 & 0 & 0 & 1 & -c\\ 0 & 0 & 1 & 0 & 0 & 1\end{array} \right]$$
Edit:
Your mistake is at the last step
subtract $a$ times the new second row from the first,
It should involve $(-a)(-c)+(-b)$
| {
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} |
Solve $\arcsin{(2x^2-1)}+2\arcsin{x} = -\frac{\pi}{2}$ First of, we have to restrict the domain of the equation by looking at the argument of the first term. The domain for $\arcsin$ is $[-1,1]$, so the inequality $ -1\leq 2x^2-1 \leq1$ has to hold. So $x\geq0$ and $-1\leq x\leq 1$. It follow that the domain is $[-1,1]\cap[0,\infty)=[0,1].$ So the real domain of the equation remains $[0,1]$ and if there exists $x\in\mathbb{R}$ that solves the equation it should belong to that domain.
Lets solve it by rewriting $$\arcsin{(2x^2-1)}=-\frac{\pi}{2}-2\arcsin{x}=-\left(\frac{\pi}{2}+2\arcsin{x}\right).$$
Sine of both sides gives $$2x^2-1=\sin\left(\frac{\pi}{2}+2\arcsin{x}\right)=\cos{(2\arcsin{x})}=\cos^2{(\arcsin{x})}-\sin^2{(\arcsin{x})}.$$
Using $\cos^2{x}=1-\sin^2{x}$ we get
$2x^2-1=1-2\sin^2(\arcsin{x})=1-2x^2 \Longleftrightarrow2x^2-1\Longleftrightarrow x_{1,2}=\pm\frac{1}{\sqrt{2}}.$
Only $x=\frac{1}{\sqrt{2}}\in{[0,1]},$ which is the only root.
According to wolfram alpha, the only real solutions are $x=0$ and $x=-1.$ Where did I go wrong?
| Let $f(x)=\arcsin{(2x^2-1)}+2\arcsin{x}+\frac{\pi}{2}$.
Thus, $$f'(x)=\frac{2x}{|x|\sqrt{1-x^2}}+\frac{2}{\sqrt{1-x^2}}=0$$
for all $-1<x<0$.
Thus, $f(x)=conctant$ for $-1\leq x\leq0$ and since $f(0)=0$ we obtain $[-1,0]$.
In another hand $f'(x)>0$ for all $0<x<1$.
Thus, $f(x)>f(0)=0$ and on $[-1,0]$ we got all solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2443040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
For positive numbers $a,b,c$ that $a+b+c=3$ prove that: For positive numbers $a,b,c$ that $a+b+c=3$ prove that:
$\frac{a(a+b-2c)}{ab+1}+\frac{b(b+c-2a)}{bc+1}+\frac{c(c+a-2b)}{ac+1} \ge 0$
This problem wad in the first part of the inequality book(AM-GM inequality)But it doesn't seem to have a solution by AM_GM inequality since we have negative parantheses.Any hints?
| C0nsider $(1.2,1.2,0.6)$ one has $LHS = \frac{1.2^2}{1.2^2 + 1} + \frac{-0.6\times 1.2}{1.2\times 0.6 + 1} + \frac{- 0.6\times 0.6}{1.2\times 0.6 + 1} < 0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2448518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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About the property of a sequence of polynomials
There is a sequence of polynomials $P_n(x)$ such that $P_0(x) = x^3 - 4x$ and, for $n>0$,
$$P_n(x) = P_{n-1}(1+x)P_{n-1}(1-x) - 1.$$
Prove that $x^{2016}$ divides $P_{2016}(x)$.
I tried induction, but then I had to prove that $x^n \vert P_n(x)P_n(2 - x) + P_n(-x)P_n(2+x)$
|
We show by induction that $x^n(x-2)(x+2)$ divides $P_n(x)$ for any even integer $n$.
For $n=0$, $P_0=x(x+2)(x-2)$ and the property holds. Let $n>0$. Note that
$$P_{n+2}(x)=(P_{n}(2+x)P_{n}(-x)-1)(P_{n}(2-x)P_{n}(x)-1)-1.$$
If $x^n(x-2)(x+2)$ divides $P_n(x)$ then, for some polynomial $Q_n(x)$,
$$P_{n}(2+x)P_{n}(-x)=(x+2)^{n+1}x^{n+1}(x-2)Q_n(x)$$
and
$$P_{n}(2-x)P_{n}(x)=-(x+2)x^{n+1}(x-2)^{n+1}Q_n(-x).$$
Therefore, for some polynomial $R_n(x)$,
\begin{align*}
P_{n+2}(x)&=((x+2)^{n+1}x^{n+1}(x-2)Q_n(x)-1)(-(x+2)x^{n+1}(x-2)^{n+1}Q_n(-x)-1)-1\\
&=x^{n+1}(x+2)(x-2)\left(x^{n+1}R_n(x)-(x+2)^nQ_n(x)+(x-2)^nQ_n(-x)\right).
\end{align*}
Since $n$ is even,
$$\left.\left(x^{n+1}R_n(x)-(x+2)^nQ_n(x)+(x-2)^nQ_n(-x)\right)\right|_{x=0}=0-2^nQ_n(0)+(-2)^nQ_n(0))=0$$
which implies that we have an extra factor $x$ (or, as remarked by Nick Pavlov, since $P_{n+2}$ is even, if the odd polynomial $x^{n+1}$ divides $P_{n+2}$ then $x^{n+2}$ divides it as well).
Hence $x^{n+2}(x+2)(x-2)$ divides $P_{n+2}$ and the proof of the inductive step is complete.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2452714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find value of $\frac{\sum_{k=1}^{399} \sqrt{20+\sqrt{k}}}{\sum_{k=1}^{399} \sqrt{20-\sqrt{k}}}$ Find value of
$$S=\frac{\sum_{k=1}^{399} \sqrt{20+\sqrt{k}}}{\sum_{k=1}^{399} \sqrt{20-\sqrt{k}}}$$
I started with $$S+\frac{1}{S}=\frac{\sum_{k=1}^{399} \sqrt{20+\sqrt{k}}}{\sum_{k=1}^{399} \sqrt{20-\sqrt{k}}}+\frac{\sum_{k=1}^{399} \sqrt{20-\sqrt{k}}}{\sum_{k=1}^{399} \sqrt{20+\sqrt{k}}}$$
$\implies$
$$S+\frac{1}{S}=\frac{\sum_{i=1}^{399}(20+\sqrt{i})+2S_1+\sum_{i=1}^{399}(20-\sqrt{i})+2S_2}{\left(\sum_{k=1}^{399}\sqrt{20+\sqrt{k}}\right) \times \left(\sum_{k=1}^{399}\sqrt{20-\sqrt{k}}\right)}$$ $\implies$
$$S+\frac{1}{S}=\frac{15960+2S_1+2S_2}{\left(\sum_{k=1}^{399}\sqrt{20+\sqrt{k}}\right) \times \left(\sum_{k=1}^{399}\sqrt{20-\sqrt{k}}\right)}$$
where $$S_1=\sum_{i \ne j=1}^{399}\left(\sqrt{20+\sqrt{i}}\right)\left(\sqrt{20+\sqrt{j}}\right)$$ and like wise
$$S_2=\sum_{i \ne j=1}^{399}\left(\sqrt{20-\sqrt{i}}\right)\left(\sqrt{20-\sqrt{j}}\right)$$
Any way to proceed from here?
| Amazingly, it appears that
$$ \dfrac{\sum_{k=1}^{m^2-1} \sqrt{m + \sqrt{k}}}{\sum_{k=1}^{m^2-1} \sqrt{m - \sqrt{k}}} = 1 + \sqrt{2} $$
for all integers $m \ge 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2456831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find all points on which a function is discontinuous. $ f(x,y) = \begin{cases} \dfrac{x^3+y^3}{x^2+y^2} &\quad\text{if} [x,y] \neq [0,0]\\[2ex] 0 &\quad\text{if}[x,y] = [0,0]\\ \end{cases} $
The only point it could be discontinuous in is [0,0]. How do I find the limit of the function for $(x,y) \rightarrow (0,0)$? $ \lim_{(x,y) \rightarrow (0,0)} \frac{x^3+y^3}{x^2+y^2} $ seems pretty hard to analyse.
| For $x,y \in \mathbb R \setminus \{0\}$ we have
$$\vert f(x,y)\vert = \left\vert\frac{x^3 + y^3}{x^2+ y^2}\right\vert \leq \left\vert \frac{x^3}{x^2+ y^2} \right\vert + \left\vert \frac{y^3}{x^2+ y^2}\right\vert \leq \left\vert \frac{x^3}{x^2} \right\vert + \left\vert \frac{y^3}{y^2}\right\vert = \vert x \vert + \vert y \vert \to 0$$
for $(x, y) \to (0, 0)$. If $x = y = 0$ we have $f(x,y) = 0$. Thus it follows that $\lim_{(x,y) \to (0,0)} f(x,y) = 0$. Therefore we can deduce that $f$ is continuous at $(0,0)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2456976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
SAT II Geometry Find the missing side length
I'm thinking the answer is choice A but I want someone to back up my reasoning/check. So since DE and DF are perpendicular to sides AB and AC respectively that must make EDFA a rectangle. Therefore AF must be 4.5 and AE must be 7.5. Since they state AB = AC that must mean EB is 4.5 and CF is 7.5. Therefore CF rounded to the nearest whole number is 8
| Since $\triangle ABC$ is isosceles, $\angle ABC \cong \angle ACB$. Since $\angle BED$ and $\angle CFD$ are right angles, they are congruent. Hence, $\triangle BED \sim \triangle CFD$ by the Angle-Angle Similarity Theorem. Hence,
$$\frac{|CD|}{|BD|} = \frac{|CF|}{|BE|} = \frac{|FD|}{|ED|} = \frac{7.5}{4.5} = \frac{5}{3}$$
Since $|CD| + |BD| = 24$, we obtain
\begin{align*}
|CD| + |BD| & = 24\\
\frac{5}{3}|BD| + |BD| & = 24\\
\frac{8}{3}|BD| & = 24\\
|BD| & = 9
\end{align*}
Thus,
$$|CD| = \frac{5}{3}|BD| = 15$$
Since
$$\frac{|DF|}{|CD|} = \frac{7.5}{15} = \frac{1}{2}$$
we may conclude that $\angle FCD = 30^\circ$ and $\angle FDC = 60^\circ$. Thus, $\triangle CDF$ is a $30^\circ, 60^\circ, 90^\circ$ right triangle. Since the ratio of the side lengths in a $30^\circ, 60^\circ, 90^\circ$ right triangle is $1: \sqrt{3}: 2$, $\overline{DF}$ is opposite the $30^\circ$ angle, and $\overline{CF}$ is opposite the $60^\circ$ angle,
$$|CF| = \sqrt{3}|DF| = 7.5\sqrt{3} \approx 12.99$$
so the best answer is $13$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2458313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\lim_{(x,y) \to (0,0)} \frac{x^3+y^3}{x^2+y^2}$ assuming it exists.
Find $\lim_{(x,y) \to (0,0)} \frac{x^3+y^3}{x^2+y^2}$ assuming it
exists.
Since limit exists, we can approach from any curve to get the limit...
if we approach (0,0) from y=x
$\lim_{(x,y) \to (0,0)} \frac{x^3+y^3}{x^2+y^2} \Rightarrow \lim_{x \to 0} \frac{x^3+x^3}{x^2+x^2} = x = 0$
is this method correct?
| You want to prove that for every $\epsilon>0$ there exists $\delta>0$ such that
$$0<\sqrt{x^2+y^2}<\delta\implies\left|\frac{x^3+y^3}{x^2+y^2}\right|<\epsilon.$$
Let $\delta=\frac{\epsilon}{2}$. Then
$$\left|\frac{x^3+y^3}{x^2+y^2}\right|\leq\left|\frac{x^3}{x^2+y^2}\right|+\left|\frac{y^3}{x^2+y^2}\right|$$
$$=|x|\underbrace{\left|\frac{x^2}{x^2+y^2}\right|}_{\leq 1}+|y|\underbrace{\left|\frac{y^2}{x^2+y^2}\right|}_{\leq 1}$$
$$\leq|x|+|y|\leq\sqrt{x^2+y^2}+\sqrt{x^2+y^2}$$
$$=2\sqrt{x^2+y^2}<2\cdot\frac{\epsilon}{2}=\epsilon$$
This proves that the limit exists and is equal to $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2460071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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How to show that $\sum_{n=1}^{\infty}{2^n\over (2^n-1)(2^{n+1}-1)(2^{n+2}-1)}={1\over 9}?$ How to show that
$${2\over (2-1)(2^2-1)(2^3-1)}+{2^2\over (2^2-1)(2^3-1)(2^4-1)}+{2^3\over (2^3-1)(2^4-1)(2^5-1)}+\cdots={1\over 9}?\tag1$$
We may rewrite $(1)$ as
$$\sum_{n=1}^{\infty}{2^n\over (2^n-1)(2^{n+1}-1)(2^{n+2}-1)}={1\over 9}\tag2$$
$${2^n\over (2^n-1)(2^{n+1}-1)(2^{n+2}-1)}={A\over 2^n-1}+{B\over 2^{n+1}-1}+
{C\over 2^{n+2}-1}\tag3$$
$${2^n\over (2^n-1)(2^{n+1}-1)(2^{n+2}-1)}={1\over 3(2^n-1)}-{1\over 2^{n+1}-1}+
{2\over 3(2^{n+2}-1)}\tag4$$
$${1\over 3}\sum_{n=1}^{\infty}{1\over 2^n-1}-\sum_{n=1}^{\infty}{1\over 2^{n+1}-1}+{2\over 3}\sum_{n=1}^{\infty}{1\over 2^{n+2}-1}={1\over 9}\tag5$$
| Well done so far. Now shift the indices on the last two sums:
$${1\over 3}\sum_{n=1}^{\infty}{1\over 2^n-1}-\sum_{n=1}^{\infty}{1\over 2^{n+1}-1}+{2\over 3}\sum_{n=1}^{\infty}{1\over 2^{n+2}-1}=\\
{1\over 3}\sum_{n=1}^{\infty}{1\over 2^n-1}-\sum_{n=2}^{\infty}{1\over 2^{n}-1}+{2\over 3}\sum_{n=3}^{\infty}{1\over 2^{n}-1}$$
and note that all the terms with $n \ge 3$ add to zero, so we just keep the first few terms:
$$=\frac 13 \cdot \frac 1{2-1}+\frac 13\cdot \frac 1{4-1}-\frac 1{4-1}=\frac 19$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2460845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
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Prove $\frac{x+1}{y+1}+\frac{y+1}{z+1}+\frac{z+1}{x+1}\leq\frac{x-1}{y-1}+\frac{y-1}{z-1}+\frac{z-1}{x-1}$ Let $x,y,z$ be real numbers all greater than $1$, then prove that
$$\frac{x+1}{y+1}+\frac{y+1}{z+1}+\frac{z+1}{x+1}\leq\frac{x-1}{y-1}+\frac{y-1}{z-1}+\frac{z-1}{x-1}$$
My Attempt:
I am trying to use $A.M>GM$ but am not able to do it
| To prove it, let's subtract LHS from RHS and call it $\Delta$:
\begin{align}
\Delta&=\frac{x-1}{y-1}-\frac{x+1}{y+1}+\frac{y-1}{z-1}-\frac{y+1}{z+1}+\frac{z-1}{x-1}-\frac{z+1}{x+1}\\
&= 2\left(\frac{x-y}{y^2-1}+\frac{y-z}{z^2-1}+\frac{z-x}{x^2-1}\right)\\
\end{align}
Assuming $1<x\leq y\leq z$, we can see:
$$\frac{z-x}{x^2-1}\geq \frac{z-x}{y^2-1}$$
and also since $y-z\leq 0$:
$$\frac{y-z}{z^2-1}\geq \frac{y-z}{y^2-1}$$
Therefore:
$$\Delta\geq \frac{2}{y^2-1} \left(x-y+y-z+z-x\right)=0$$
For the other case where $1<x\leq z\leq y$, you can see:
$$\frac{z-x}{x^2-1}\geq \frac{z-x}{z^2-1}$$
and since $x-y\leq0$:
$$\frac{x-y}{y^2-1}\geq \frac{x-y}{z^2-1}$$
and again results into:
$$\Delta\geq 0$$
which completes the proof.
| {
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"url": "https://math.stackexchange.com/questions/2462971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Determine $\textrm{sign}[x + \ln y]$ I need to determine the sign of
\begin{align}
f(a,b) = \textrm{sign}\left[\frac{(a-1)(b-1)^2}{ba^2(b(a-1)+1)} + \ln\left(\frac{b(a-1)+1}{ab}\right)\right] = ?
\end{align}
with $a \geq 1$, and $b > 0$.
We have $f(a,b=1) = 0$. I conjecture also that $f(a,b<1) > 0$ and $f(a,b>1) < 0$. Cannot show it, however, because of the $\ln$. Any hints are appreciated.
Idea: There is only one root for $f$, namely $f(a,b) = 0$ iff $b=1$ for all $a$. Let $a=1$ such that
\begin{align}
f(1,b) = \ln\left(\frac{1}{b}\right)
\begin{cases}
> 0 \quad &\textrm{if } b < 1\\
< 0 \quad &\textrm{if } b > 1.
\end{cases}
\end{align}
| for $0<b<1$:
$$
[\ldots]=\left[\underbrace{\ldots}_{> 0}+\ln\Big(\underbrace{1+\frac{1-b}{ab}}_{> 1}\Big)\right]>0.
$$
for $b>1$: use the fact that $\ln(1+x)\le x$, for $x>-1$, i.e.
$$
[\ldots]\le\frac{a-1}{a-1+1/b}\cdot\Big(\frac{b-1}{ab}\Big)^2
-\frac{b-1}{ab}=\epsilon\cdot q^2-q<0
$$
since $0<\epsilon<1$ and $0<q<1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2464226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate the sum ${1 - \frac{1}{2} {n \choose 1} + \frac{1}{3} {n \choose 2} + \ldots + (-1)^n \frac{1}{n+1} {n \choose n}}$ Evaluate the sum $${1 - \frac{1}{2} {n \choose 1} + \frac{1}{3} {n \choose 2} + \ldots + (-1)^n \frac{1}{n+1} {n \choose n}}.$$
I have tried comparing this to the similar problem here.
I believe I need to differentiate or integrate? But I'm not sure how that might work.
Any ideas? Thanks.
| $$\begin{align}
S&=1-\frac 12\binom n1+\frac 13 \binom n2-\cdots+(-1)^n\frac 1{n+1}\binom nn\\
\times (n+1):\hspace{1cm}\\
(n+1)S&=(n+1)-\frac {n+1}2\binom n1+\frac {n+1}3\binom n2-\cdots+(-1)^n\frac {n+1}{n+1}\binom nn\\
&=\binom {n+1}1-\binom {n+1}2+\binom {n+1}3-\cdots +(-1)^n\binom {n+1}{n+1}\\
&=\color{blue}{\binom {n+1}0}\underbrace{\color{blue}{-\binom {n+1}0}+\binom {n+1}1-\binom {n+1}2+\binom {n+1}3-\cdots +(-1)^n\binom {n+1}{n+1}}_{=-\sum_{r=0}^{n+1}\binom {n+1}r(-1)^r=-(1-1)^{n+1}=0}\\
&=1\\
S&=\color{red}{\frac 1{n+1}}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2465407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Calculating residue of $f(z) = \frac{(z^2 - 1)^4}{z^5}$ at $z=0$
Calculate the residue of $$f(z) = \frac{(z^2 - 1)^4}{z^5}$$ in $z = 0$
I let $g(z) = (z^2 - 1)^4.$
I'm using a theorem which states:
Suppose $g$ is holomorphic around $z = \alpha$ and that $N$ is a
positive integer, then $$RES_{z=\alpha}\frac{g(z)}{(z-\alpha)^N} =
\frac{g^{(N-1)}(\alpha)}{(N-1)!}$$
and I get the correct answer (since $g(z) / (z-0)^5 = f(z))$, however, it is really annoying to have to differentiate that many times. Is there a smarter method?
| $$
\frac{(z^2 - 1)^4}{z^5}
= \frac{z^8 - 4 z^6 + 6 z^4 - 4 z^2 + 1}{z^5}
= \cdots + \frac{6}{z} + \cdots
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2465909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Epsilon- Delta Proof $$
\lim_{x\to 1} \frac{1}{x^2+2} = \frac{1}{3}
$$
I'm having a problem proving this using $\epsilon-\delta$ proofs. For some reason When I solve for $\epsilon$, I get a negative number. Since this value is supposed to equal $\delta$ and $\delta$ can't be negative I'm not sure how to move forward.
Thanks!
| The problem is that “solving for $\epsilon$” is too restrictive in general. Let's unpack @mfl's hint.
$$
\left| \frac{1}{x^2+2} - \frac{1}{3} \right|
= \left|\frac{1-x^2}{x^2+2}\right|
= \left|\frac{(x-1)(x+1)}{x^2+2}\right|
= |x-1|\cdot |x+1| \cdot \frac{1}{x^2+2}
$$
We want to make this product smaller than $\epsilon$. The only factor we can control directly is $|x-1|$; we can choose any $\delta$ we want to make that small. The trick is to control the other factors based on the assumption that $|x-1|<\delta$.
One factor is easy: $x^2 \geq 0$ for all $x$, so $x^2 + 2 \geq 2$ for all $x$, so $\frac{1}{x^2+2} \leq \frac{1}{2}$ for all $x$.
To control $|x+1|$, notice that
\begin{gather*}
|x-1| < \delta \implies -\delta < x-1 < \delta \\
\implies 1-\delta < x < 1+\delta \\
\implies 2-\delta < x+1 < 2+\delta \\
\end{gather*}
If we know that $\delta \leq 1$, we can say $1 < x+1 < 3$, so $|x+1|<3$.
[We could use any positive number in place of $1$, but $1$ is the traditional choice.]
This would give us
$$
\left| \frac{1}{x^2+2} - \frac{1}{3} \right|
< \delta \cdot 3 \cdot \frac{1}{2} = \frac{3}{2}\delta
$$
If we know also that $\delta \leq \frac{2}{3}\epsilon$, we can conclude that the right-hand side above is $<\epsilon$, and that's exactly what we want.
We can choose $\delta$ as small as we need; how can we ensure that both $\delta \leq 1$ and $\delta \leq \frac{2}{3}\epsilon$? By setting
$$
\delta = \min\left\{1,\frac{2}{3}\epsilon\right\}
$$
This is not the end; you still need to write it up “forwards.” But hopefully it does explain how to work around quotients and other factors.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving that: $ (a^3+b^3)^2\le (a^2+b^2)(a^4+b^4)$ This problem is from Challenge and Thrill of Pre-College Mathematics:
Prove that $$ (a^3+b^3)^2\le (a^2+b^2)(a^4+b^4)$$
It would be really great if somebody could come up with a solution to this problem.
| \begin{eqnarray*}
\color{blue}{a^2b^2(a-b)^2} \geq 0 \\
\color{red}{a^6} +\color{blue}{a^4b^2+a^2 b^4} +\color{red}{b^6} \geq \color{red}{a^6} +\color{blue}{2 a^3b^3} +\color{red}{b^6} \\
(a^2+b^2)(a^4+b^4) \geq (a^3+b^3)^2.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2468155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Help with homework of 3rd order equation I can't solve this equation :
$$t^3-4t^2-5=0$$
I think it should be solved by cover it to $(a+b)^3$, but I can't figure it out
| Let $u=t-\frac{4}{3}$. One then has $$(u+\frac{4}{3})^3-4(u+\frac{4}{3})^2-5 = 0$$
$$u^3-\frac{16}{3}u - \frac{263}{27}=0.$$
Now, let $x+y = u$, then $x^3+y^3 = (x+y)^3-3xy(x+y) = u^3 - 3xyu$.
One can set $xy = \frac{16}{9}$, $x^3+y^3= \frac{263}{27}$. One can solve to find $x$, $y$ then $u= x+y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2468471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}=\sin\theta+i\cos\theta$
Prove that $$\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}=\sin\theta+i\cos\theta$$
I tried to rationalize the denominator but I always end up with a large fraction that won't cancel out. Is there something I'm missing?
Thanks in advance
| $$\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}=\frac{2\cos^2\left(\frac{\pi}{4}-\frac{\theta}{2}\right)+2i\sin\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\cos\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}{2\cos^2\left(\frac{\pi}{4}-\frac{\theta}{2}\right)-2i\sin\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\cos\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}=$$
$$=\frac{\cos\left(\frac{\pi}{4}-\frac{\theta}{2}\right)+i\sin\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}{\cos\left(\frac{\pi}{4}-\frac{\theta}{2}\right)-i\sin\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}=\left(\cos\left(\frac{\pi}{4}-\frac{\theta}{2}\right)+i\sin\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\right)^2=\sin\theta+i\cos\theta.$$
I used $$1+\cos\alpha=2\cos^2\frac{\alpha}{2};$$
$$\sin\alpha=2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}$$ and
$$\sin\alpha=\cos\left(\frac{\pi}{2}-\alpha\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2470541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 2
} |
How to differentiate $(x+1)^{3/2}$ using the limit definition of the derivative? I want to use the definition of the derivative to find $f'$ of $f(x)=(x+1)^{3/2}$.
I solved it using the chain rule. Would like to try to solve it using the definition of a derivative:
$$
\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}.
$$
Don't really know how I should start it. Should I make it $\sqrt{(x+1)^3}$? Or keep the exponent as $3/2$? Other guidance would be appreciated.
| Let $f(x)=(x+1)^{3/2}$, then $$f'(x)= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}=\lim_{h \to 0} \frac{(x+1+h)^{3/2}-(x+1)^{3/2}}{h}=\lim_{h \to 0} \frac{(x+1+h)^{3/2}-(x+1)^{3/2}}{h} \cdot \frac {(x+1+h)^{3/2}+(x+1)^{3/2}}{(x+1+h)^{3/2}+(x+1)^{3/2}}= \lim_{h \to 0} \frac{(x+1+h)^3 -(x+1)^3}{h[(x+1+h)^{3/2}+(x+1)^{3/2}]} = \lim_{h \to 0} \frac{h[h^2 + 3hx + 3h + 3x^2 + 6x + 3]}{h[(x+1+h)^{3/2}+(x+1)^{3/2}]} = \lim_{h \to 0} \frac{h^2 + 3hx + 3h + 3x^2 + 6x + 3}{(x+1+h)^{3/2}+(x+1)^{3/2}} = \frac{3x^2+6x+3}{2(x+1)^{3/2}} = \frac{3}{2} \frac{(x+1)^2}{(x+1)^{3/2}}=\frac{3}{2}(x+1)^{1/2}$$
which is exactly what we get from the chain rule:
$$\frac d{dx}(x+1)^{3/2}=\frac{3}{2}(x+1)^{1/2} \cdot\frac{d}{dx}(x)=\frac 32 \sqrt{x+1}.$$ The key step is in the third equality above, where you multiply the numerator and denominator by the conjugate to get rid of the square root.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2471529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Foot of perpendicular proof. I know there is few answer on Foot of perpendicular, but my doubt is different, so please don't mark this as the duplicate.
My book says:
Foot of the perpendicular from a point $(x_1,y_1)$ on the line $ax+by+c=0$ is $$\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=-\dfrac{ax_1+by_1+c}{a^2+b^2}$$
My proof:
Slope of line $\perp$ to $ax+by+c=0$ is $\frac{b}{a}\implies\tan\theta=\frac{b}{a}\implies\begin{cases}\cos\theta &=\pm\frac{a}{\sqrt{a^2+b^2}}\\ \sin\theta &=\frac{b}{\sqrt{a^2+b^2}}\end{cases}$
Let $L$ passes through point $(x_ 1,y_1)$ perpendicular to $ax+by+c=0$.
Let $r$ be the algebraic distance of $(x_ 1,y_1)$ from $ax+by+c=0$ $\implies r=\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}$.
Now co-ordinate of any point on $L$ distance $r$ from point $(x_1,y_1)$ can be given as:$$\bigg(x_1+r\cos \theta,\ y_1+r\sin\theta\bigg)$$, where $\theta$ is angle $L$ makes with positive direction of $x$-axis.
Substituting the values:$$\bigg(x_1+\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\cdot\frac{a}{\sqrt{a^2+b^2}},\ y_1+\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\cdot\frac{b}{\sqrt{a^2+b^2}}\bigg)$$ Now I didn't wrote $\pm$ with $\cos$, whose sign can be calculated as of $\tan$.
Now camparing gives :
$x=x_1+\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\cdot\frac{a}{\sqrt{a^2+b^2}}$, $\ \ \ y=y_1+\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\cdot\frac{b}{\sqrt{a^2+b^2}}$
Note, I didn't mixed the denominators $\bigg(\sqrt{a^2+b^2}\bigg)$ to $\bigg(a^2+b^2\bigg)$, as I've calculated $\frac{a}{\sqrt{a^2+b^2}}$ from $\tan\theta$ with sign.
When I applied this on some problems gives me the correct answers but is not in the bookish form please help me to do this.
| First the algebraic distance $r$ should be $\dfrac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}$ depending on the position of $(x_1;y_1)$ for a given system of coordinates.
The correct formula is $$\dfrac{x-x_1}{a}=-\dfrac{ax_1+by_1+c}{a^2+b^2}.$$
To see this take the scalar product to be zero for $(x,-\dfrac{ax}{b}-\frac{c}{b})$ is on the line of direction $\overrightarrow{v}(1;-\frac{a}{b})$:
$$(x_1-x)\times 1+(y_1-(-\dfrac{ax}{b}-\frac{c}{b}))(\dfrac{-a}{b})=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find a + b + c + bc. My math trainer friend ask my help to look for the shortest possible solution for this problem:
Let $a$, $b$ and $c$ be positive integers such that
$$\left \{\begin{matrix}a + b + ab = 15 \\
b + c + bc = 99 \\
c + a + ca = 399\end{matrix}\right. $$
Find $a + b + c + bc.$
I tried elimination but it took us for about 3 min to do it. This question was intended for 15 seconds only.
| $(a+1)(b+1) = 16 $
$(b+1)(c+1) = 100$
$(c+1)(a+1) = 400$
Multiply all $3$ equations to get value of $(a+1)(b+1)(c+1) = 800$ then
$c+1 = 50,\quad a+1 = 8,\quad b+1 = 2$ , required is $a+(99) = 106$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to find : $\lim_{x \to \pi/2}\frac{1-\sin x}{\sin x+\sin 3x}$ How to find :
$$\lim_{x \to \pi/2}\frac{1-\sin x}{\sin x+\sin 3x}$$
My Try :
$$x-\pi/2=t \to x=t+\frac{\pi}{2}$$
And:
$$ \sin (t+\frac{\pi}{2})=\cos t \\ \sin 3(t+\frac{\pi}{2})=-\cos 3t $$
So :
$$\lim_{t \to 0}\frac{1-\cos t}{\cos t-\cos 3t}=\\\lim_{t \to 0}\frac{1-\cos t}{t^2} \cdot \frac{t^2}{\cos t-\cos 3t}$$
Now what ?
| $$\\ \lim _{ t\to 0 } \frac { 1-\cos t }{ t^{ 2 } } \cdot \frac { t^{ 2 } }{ \cos t-\cos 3t } =\lim _{ t\to 0 } \frac { \sin ^{ 2 }{ \frac { t }{ 2 } } }{ 2{ \left( \frac { t }{ 2 } \right) }^{ 2 } } \cdot \frac { t\cdot t }{ 2\sin { t } \sin { 2t } } =\\ =\lim _{ t\to 0 } \frac { \sin ^{ 2 }{ \frac { t }{ 2 } } }{ 2{ \left( \frac { t }{ 2 } \right) }^{ 2 } } \cdot \frac { 1 }{ 4\frac { \sin { t } }{ t } \frac { \sin { 2t } }{ 2t } } =\frac { 1 }{ 8 } \\ $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
How to show that $\int_{0}^{\pi}\sin^3(x){\mathrm dx\over (1+\cos^2(x))^2}=1?$ How can we show that $(1)$
$$\int_{0}^{\pi}\sin^3(x){\mathrm dx\over (1+\cos^2(x))^2}=1?\tag1$$
$\sin^3(x)={3\over 4}\sin(x)-{1\over 4}\sin(3x)$
$1+\cos^2(x)=2-\sin^2(x)$
$$\int_{0}^{\pi}[{3\over 4}\sin(x)-{1\over 4}\sin(3x)]{\mathrm dx\over (2-\sin^2(x))^2}=1\tag2$$
$2-\sin^2(x)={3\over 2}+{1\over 2}\cos(2x)$
$$\int_{0}^{\pi}{3\sin(x)-\sin(3x)\over 3+\cos(2x)}\mathrm dx\tag3$$
$$\int_{0}^{\pi}{3\sin(x)\over 3+\cos(2x)}\mathrm dx-\int_{0}^{\pi}{\sin(3x)\over 3+\cos(2x)}\mathrm dx\tag4$$
$\cos(2x)={1-\tan^2(x)\over 1+\tan^2(x)}$
$\sin(x)={2\tan(x/2)\over 1+\tan^2(x/2)}$
| For the integrand, use the trigonometric identitity $\sin^2(x)=1-\cos^2(x)$ and consider the two changes of variables \begin{align*}
u=\cos(x), &\,\,du=-\sin(x)\,dx\\
u=\tan(s), &\,\,du=\sec^2(s)\,ds
\end{align*}
to get\begin{align*}
\int_{0}^{\pi}\sin^3(x){\mathrm dx\over (1+\cos^2(x))^2}&=\int_0^\pi\frac{\sin(x)(1-\cos^2(x))}{(\cos^2(x)+1)^2}\,dx=-\int_{-1}^1\frac{1-u^2}{(u^2+1)^2} \,du\\
&=-\int_{-1}^1\frac{1}{u^2+1}\,du+2\int_{-1}^1\frac{1}{(u^2+1)^2}\,du\\
&=\left[-\tan^{-1}(u) \right]_{-1}^1+2\int_{-1}^1\frac{1}{(u^2+1)^2}\,du\\
&=-\frac{\pi}{2}+2\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac12\cos(2s)+\frac12\,ds\\
&=-\frac{\pi}{2}+1+\frac{\pi}{4}+\frac{\pi}{4}=1.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2476520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Need Clarification on Trig Substitution $$\int\frac{x}{(1+x^2)^\frac{3}{2}}\ \mathrm dx$$
becomes
$$\sin\theta = \frac{x}{\sqrt{x^2+1}}$$
I am aware that we would use $x = a^2\tan^2\theta$. and that our answer before we plug in from the triangle becomes $\sin\theta$.
I just took a test and got the answer of
$$\sin\theta = \frac{x}{(x^2+1)^\frac{3}{2}}$$
Why does the $\dfrac{3}{2}$ become $\dfrac{1}{2}$ when we are plugging in for $\sin\theta$?
| It doesn't need any trigonometric substitution $$\int \frac { x }{ \left( 1+x^{ 2 } \right) ^{ \frac { 3 }{ 2 } } } dx=\frac { 1 }{ 2 } \int \frac { d\left( 1+{ x }^{ 2 } \right) }{ \left( 1+x^{ 2 } \right) ^{ \frac { 3 }{ 2 } } } =\frac { 1 }{ 2 } \int { { \left( 1+{ x }^{ 2 } \right) }^{ -\frac { 3 }{ 2 } } } d\left( 1+{ x }^{ 2 } \right) =\\ =\frac { 1 }{ 2 } \frac { { \left( 1+{ x }^{ 2 } \right) }^{ -\frac { 3 }{ 2 } +1 } }{ -\frac { 3 }{ 2 } +1 } =-\frac { 1 }{ \sqrt { 1+{ x }^{ 2 } } } +C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2479030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If and when a linear system has exactly three solutions
Does the following system have exactly three solutions?
$$\left\{ \begin{array}{l}
2x - y +3z = 1 \\
x + 4y - 2z = -7 \\
3x + y -z = 4 \\
\end{array} \right.$$
I marked this answer as True.
I proceeded to row reduce it and obtained the resultant matrix as -
$$\left[ \begin{array}{ccc|c}
1 & -1/2 & 3/2 & 4 \\
0 & 1 & 9 & 13/9 \\
0 & 0 & 1 & 10/28 \\
\end{array} \right]$$
I'm pretty sure I made some mistake while row-reducing it. I answered this question in an exam setting. Then I gave the explanation as follows -
Form row reduction, we know that the system is consistent and the rank of the matrix is 3 which is equal to the number of variables. So, the system of equations has 3 distinct solutions.
Can someone point out where I went wrong?
| Start with the augmented matrix:
\begin{eqnarray}
\left(\begin{array}{cccc} 2&-1&3 &: 1\\ 1&4&2&:-7\\ 3&1&-1&:4\end{array}\right)
\end{eqnarray}
First: Interchange rows 1 and 2 to get:
$$\left(\begin{array}{cccc} 1&4&-2&:-7\\ 2&-1&3 &: 1\\ 3&1&-1&:4\end{array}\right)$$
Second: Use the first element of the first row as a pivot to eliminate the 2 and 3 of the first column to get:
$$ \left(\begin{array}{cccc} 1&4&-2&:-7\\ 0&-9&7 &: 15\\ 0&-11&5&:25\end{array}\right) $$
Then you can replace Row2 by Row2-Row3 to get:
$$\left(\begin{array}{cccc} 1&4&-2&:-7\\ 0&2&2&: -10\\ 0&-11&5&:25\end{array}\right)$$
then you can divide by 2 the Row2 to get:
$$\left(\begin{array}{cccc} 1&4&-2&:-7\\ 0&1&1&: -5\\ 0&-11&5&:25\end{array}\right)$$
Then you can eliminate the -11 in the third row:
$$\left(\begin{array}{cccc} 1&4&-2&:-7\\ 0&1&1&: -5\\ 0&0&16&:-30\end{array}\right)$$
So $z=\frac{-30}{16}=\frac{-15}{8}$. Because you have the equivalent augmentedmatrix
$$\left(\begin{array}{cccc} 1&4&-2&:-7\\ 0&1&1&: -5\\ 0&0&1&:-\frac{15}{8}\end{array}\right)$$
To get $y$ you simply replace Row2 with Row2-Row3 and you will get:
$$\left(\begin{array}{cccc} 1&4&-2&:-7\\ 0&1&0&: -5+\frac{15}{8}\\ 0&0&1&:-\frac{15}{8}\end{array}\right)$$
and $y=-5+\frac{15}{8}=\frac{-40+15}{8}=\frac{-25}{8}$. I leave $x$ for you to compute.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2479400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Maximum value of diagonals angle in parallelogram Let $ABCD$ be a parallelogram with sides a and b $(a>b)$ and let $r=a/b$ the ratio of the two sides.
The two diagonals form 4 angles which are by two congruent. Let's call the acute angle φ and the obtuse θ.
Find the maximum value that φ can take, in relation to r.
I have managed to find a relationship for the angles φ and θ, but they also contain the diagonals:
$\cosφ = (D_1^2+D_2^2-4b^2)/2D_1D_2$ and $\cos θ = (D_1^2+D_2^2-4a^2)/2D_1D_2$ but I don't know how to get rid of the diagonals and somehow take into account the ratio a/b.
| Let $\alpha$ be a measure of the angle of our parallelogram.
Thus, $S_{ABCD}=ab\sin\alpha=rb^2\sin\alpha$.
In another hand, by law of cosines we obtain:
$$a^2=\frac{AC^2}{4}+\frac{BC^2}{4}-\frac{AC\cdot BC\cos(180^{\circ}-\varphi)}{2}$$ and
$$b^2=\frac{AC^2}{4}+\frac{BC^2}{4}-\frac{AC\cdot BC\cos\varphi}{2},$$
which gives
$$a^2-b^2=AC\cdot BC\cos\varphi$$ or
$$AC\cdot BC=\frac{a^2-b^2}{\cos\varphi}$$ and
$$S_{ABCD}=\frac{(a^2-b^2)\tan\varphi}{2}=\frac{b^2(r^2-1)\tan\varphi}{2}.$$
Id est, $$\frac{b^2(r^2-1)\tan\varphi}{2}=rb^2\sin\alpha$$
or $$\tan\varphi=\frac{2r\sin\alpha}{r^2-1}.$$
We see that $0^{\circ}<\varphi<90^{\circ}$ and $\phi$ can get all these values,
which says that the maximum does not exist and the supremum is $90^{\circ}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2480178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to prove combinatoric identity How would you prove this identity using combinatorics? Any hints or advice?
For all positive integers $n>1$,
$\sum_{k=0}^{n} \frac{1}{k+1} {n\choose k} (-1)^{k+1}=\frac{-1}{n+1} $
|
We obtain
\begin{align*}
\color{blue}{\sum_{k=0}^n\frac{1}{k+1}\binom{n}{k}(-1)^{k+1}}
&=\frac{1}{n+1}\sum_{k=0}^n\binom{n+1}{k+1}(-1)^{k+1}\tag{1}\\
&=\frac{1}{n+1}\sum_{k=1}^{n+1}\binom{n+1}{k}(-1)^{k}\tag{2}\\
&=\frac{1}{n+1}(1-1)^{n+1}-\frac{1}{n+1}\tag{3}\\
&\color{blue}{=-\frac{1}{n+1}}
\end{align*}
Comment:
*
*In (1) we use the binomial identity $\binom{n+1}{k+1}=\frac{n+1}{k+1}\binom{n}{k}$.
*In (2) we shift the index to start from $k=1$.
*In (3) we apply the binomial summation formula and subtract $\frac{1}{n+1}$ as compensation for the lower limit starting with $1$ instead of $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2486705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Factored $z^4 + 4$ into $(z^2 - 2i)(z^2 + 2i)$. How to factor complex polynomials such as $z^2 - 2i$ and $z^2 + 2i$? I was wondering how to factor complex polynomials such as $z^2 - 2i$ and $z^2 + 2i$?
I originally had $z^4 + 4$, which I factored to $(z^2 - 2i)(z^2 + 2i)$ by substituting $X = z^2$ and using the quadratic formula. However, I'm not sure if there exists any method to factor $z^2 - 2i$ and $z^2 + 2i$?
I would greatly appreciate it if people could please explain how one would go about this.
| For example:
$$z^2-2i=z^2-(1+i)^2=(z-1-i)(z+1+i).$$
I think the following a bit of better.
$$z^4+4=z^4+4z^2+4-4z^2=(z^2+2)^2-(2z)^2=$$
$$=(z^2-2z+2)(z^2+2z+2)=((z-1)^2-i^2)((z+1)^2-i^2)=$$
$$=(z-1-i)(z-1+i)(z+1-i)(z+1+i).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2488626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
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