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Calculate root interval of $x^5+x^4+x^3+x^2+1$ So I have to find an interval (in the real numbers) such that it contains all roots of the following function:
$$f(x)=x^5+x^4+x^3+x^2+1$$
I've tried to work with the derivatives of the function but it doesn't give any information about the interval, only how many possible roots the function might have.
| EDITED for correct function.
$f(x) = 1+x^2+x^3+x^4+x^5$
is positive for x>0.
So consider $x < 0$:
One can write $f(x) = (1+x^2)(1+x^3) + x^4$ which is positive at least for $(1+x^3) >0$, i.e. $x > -1$.
$f'(x) = 2x+3x^2+4x^3+5x^4 = x(2+3x)+x^3(4+5x)$
so this is positive at least for $x < -4/5 $ since then all terms are positive.
This means that $f(x) = 0$ can only happen at $x < -1$ and since $f(x)$ is rising there, we have that there can only be one real root, which is at $x < -1$.
Now this root can be further locked in. Observing e.g. that $f(-1.5) = -85/32 <0$, we can give an interval for the root at (-1.5 , -1). This can obviously be improved, knowing that there is only one real root and that f(x) is monotonously increasing in this interval.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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?
| $$9\left(\frac{2}{3}a^2-\frac{4}{9}a^2\right)=8a\cdot9$$ it's
$$6a^2-4a^2=72a$$ and it's not $6a^2-9a^2=8a$, which you wrote.
The continue is
$$a^2-36a=0$$ or
$$a(a-18)=0,$$
which gives $a=0$ or $a=18$.
| {
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"timestamp": "2023-03-29T00:00:00",
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A problem from the chapter 'theory of equations' Question : Form a biquadratic equation with rational coefficients two of whose roots are √3 + 2 and √3 - 2.
So f(x) = (x- √3-2)(x-√3+2)q(x) where q(x) must be a quadratic equation. This is how far I have done. But I cannot solve it. I am a Bsc student with maths as an ancillary paper. Help would be appreciated.
| Write a quadratic equation which has roots $x_1=\left(\sqrt{3}+2\right)^2=7+4 \sqrt{3};\;x_2=\left(\sqrt{3}-2\right)^2=7-4 \sqrt{3}$
We need the sum and the product of the roots
$x_1+x_2=14;\;x_1x_2=\left(7+4 \sqrt{3}\right)\left(7-4 \sqrt{3}\right)=1$
$$(x-x_1)(x-x_2)=x^2-(x_1+x_2)x+x_1x_2=x^2-14x+1$$
And then substitute $x$ with $x^2$
$$x^4-14x^2+1=0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Maximum value of $a^3 + b^3 + c^3$ If $a+b+c=5$ and $ \frac{1}{a} +\frac{1}{b} +\frac{1}{c} =\frac{1}{5}$ then find the maximum value of $a^3 + b^3 +c^3$ where $a,b,c $ is real numbers.
My Attempt
writing $a+b=5-c$ and $ \frac{1}{a} +\frac{1}{b} =\frac{1}{5} -\frac{1}{c}$ and after algebraic manipulation I obtain $(a+b)(\frac{1}{ab} + \frac{1}{5c})=0$. If $(a+b)=0$ we get $c=5$ and $a^3 + b^3 +c^3 =125$. But I can't analyze the case $(\frac{1}{ab} + \frac{1}{5c})=0$. Can anyone help me from this point. Thanks in advance.
| Here is a lesser known approach to these things, which sometimes helps
Let $xy+yz+zx=p$ so that $x^2+y^2+z^2=(x+y+z)^2-2p=25-2p$.
Set $S_3=a^3+b^3+c^3$
Then clearing fractions in the second equation gives $abc=5p$.
We have also that $a,b,c$ are roots of the polynomial equation $$q(x)=x^3-5x^2+px-5p=0$$
Now observe that $0=q(a)+q(b)+q(c)=S_3-5(a^2+b^2+c^2)+p(a+b+c)-15p$ so that $$S_3-125+10p+5p-15p=0$$ and $S_3=125$
Observe also that we have, with $$S_r=a^r+b^r+c^r$$$$a^{n-3}q(a)+b^{n-3}q(b)+c^{n-3}q(c)=0=S_n-5S_{n-1}+pS_{n-2}-5pS_{n-3}$$which gives us the sums of powers in terms of $p$ using the recurrence relation (note that $S_0=3, S_{-1}=\frac 15$).
| {
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How can one prove $-\arctan\left(\frac{x}{y}\right)+\pi H(y)=2\arctan\left(\frac{y}{\sqrt{x^2+y^2}+x}\right)+\frac{\pi}{2}$? How can one prove the following?
$$
-\arctan\left(\frac{x}{y}\right)+\pi H(y)=2\arctan\left(\frac{y}{\sqrt{x^2+y^2}+x}\right)+\frac{\pi}{2}$$
with $H(x)$ the Heaviside function.
I have the impression that this might follow from the following
$$H(y)=H(-x)+2H(x)H(y)-H(xy)$$
and
$$\arctan\left(\frac{x}{y}\right)+\arctan\left(\frac{y}{x}\right)=\pi H(xy)-\frac{\pi}{2}$$
but I can't quite get it right...
| Using calculus and derivatives is fast but doesn't show the simplicity of this kind of angle relationships.
Start with the case $x>0$. Then consider the following figure
If $y>0$, let the right-angled triangle $\triangle ABC$ have sides $\overline{AB} = y$ and $\overline{BC} = x$, so that
$$\alpha = \arctan \left(\frac{x}{y}\right).$$
Extend $BC$ to a segment $\overline{CD} = \overline{AC} = \sqrt{x^2+y^2}$. Thus
$$\beta = \arctan\left(\frac{y}{\sqrt{x^2+y^2}+x}\right).$$
Using now the fact that $\triangle ACD$ is isosceles you easily get
$$(\alpha + \beta) + \beta = \frac{\pi}{2}$$
which gives
$$2 \arctan\left(\frac{y}{\sqrt{x^2+y^2}+x}\right) + \arctan \left(\frac{x}{y}\right) = \frac{\pi}{2},$$
that is exactly your relationship.
If $y<0$, than just use the same triangle but define $\overline{AB} = -y$ and use the symmetries
$$\alpha = \arctan \left(-\frac{x}{y}\right) = -\arctan \left(\frac{x}{y}\right),$$
and
$$\beta = \arctan\left(-\frac{y}{\sqrt{x^2+y^2}+x}\right)= -\arctan\left(\frac{y}{\sqrt{x^2+y^2}+x}\right).$$
If $x<0$ you can work with the Figure below.
Here $\overline{BC} = -x$, $\overline{AB} = |y|$, and $BC$ is extended to a segment $\overline{BD} = \sqrt{x^2+y^2} + x$. $\triangle ACD$ is isosceles and the relationship between $\alpha$ and $\beta$ is therefore
$$\beta + (\beta - \alpha) = \frac{\pi}{2}.$$
Define $\alpha$ and $\beta$ correctly, depending on the sign of $y$, and you'll get again your relationship.
$\blacksquare$
| {
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Prove that $\sqrt{m-1}$ is an integer
Let $k$ be a fixed odd positive integer. Let $m,n$ be positive integers such that $(2+\sqrt{3})^k=m+n\sqrt{3}$. Prove that $\sqrt{m-1}$ is an integer.
Let $(2+\sqrt{3})^n = a_n+b_n\sqrt{3}$. From $$a_{n+1}+b_{n+1}\sqrt{3} = (a_n+b_n\sqrt{3})(2+\sqrt{3}) = (2a_n+3b_n)+(a_n+2b_n)\sqrt{3},$$ we get $a_{n+1} = 2a_n+3b_n$ and $b_{n+1} = a_n+2b_n$ with $a_1 = 2$ and $b_1 = 1$.
How can we show that $\sqrt{a_n-1}$ is an integer for odd $n$?
| A first step:
Define integer sequences $x_n$ and $y_n$ by
$$ (2+\sqrt{3})^{2n+1} = x_n+y_n\sqrt{3}. $$
Then $x_0=2$ and $y_0=1$, and furthermore you get the recursion
\begin{eqnarray*}
x_{n+1} &=& 7x_n+12y_n\\
y_{n+1} &=& 4x_n+7y_n
\end{eqnarray*}.
The first equation gives $12y_n=x_{n+1}-7x_n$, and plugging this into the second equation gives
$$ y_{n+1} = 4x_n+7y_n = 4x_n+\frac{7}{12}(x_{n+1}-7x_n) =
\frac{1}{12}(7x_{n+1}-x_n).$$
This yields $12y_n=7x_n-x_{n-1}$, and plugging it into the first recursion yields
$$ x_{n+1} = 14x_n-x_{n-1}. $$
Together with $x_0=2$ and $x_1=26$, this nicely describes the sequence of odd-indexed numbers $a_{2n+1}=x_n$. Hence we want to show that every $x_n-1$ is a square.
| {
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How to simplify $\left({{\frac{2207+(2207^2-4)^{1/2}}2}}\right)^{1/8}$ into quadratic form? The 1995 Putnam B4 asks to evaluate some infinite expression and I got this value after some computations:$$\left({{\frac{2207+(2207^2-4)^{1/2}}2}}\right)^{1/8}$$
However I am supposed to write it in form:$$\frac{a+b\sqrt{c}}{d}$$. What do I do?
(The answer is: $\frac{3+\sqrt{5}}{2}$ which is the same as mine if you put in a calculator.)
| Let $\,x \gt 1\,$ be the expression in question, then:
$$\require{cancel}
2x^8=2207 + \sqrt{2207^2-4} \\
(2x^8-2207)^2=2207^2-4 \\
4x^{16}-4\cdot 2207 x^8+\cancel{2207^2}=\cancel{2207^2}-4 \\
x^{16}-2207 x^8 + 1 = 0 \\
x^8 + \frac{1}{x^8}=2207 \\
\left(x^4+\frac{1}{x^4}\right)^2 = 2209 \\
x^4+\frac{1}{x^4} = 47 \\
\left(x^2+\frac{1}{x^2}\right)^2 = 49 \\
x^2+\frac{1}{x^2} = 7 \\
\left(x+\frac{1}{x}\right)^2 = 9 \\
x+\frac{1}{x} = 3 \\
x^2-3x+1=0
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Infinite square root muliplication $(x=3\sqrt{y\sqrt{3\sqrt{y}....}})$ I have this problem which says
$$
x=3\sqrt{y\sqrt{3\sqrt{y\cdots}}}\\
y=3\sqrt{x\sqrt{3\sqrt{x\cdots}}}
$$
What is $x+y$
I tried
$$
x^2=9y\sqrt{x}\\
x^3=81y^2\\
y^3=81x^2\\
x^3+y^3=81(x^2+y^2)\\
(x+y)(x^2-xy+y^2)=81(x^2+y^2)
$$
I don't know what to do or if I am do that right
| Suppose $x \neq 0 \neq y$. From $x^3=81y^2$ and $y^3=81x^2$, it follows that $\frac{x^3}{y^3}=\frac{y^2}{x^2}$, hence $x^5=y^5$, so $x=y$. I think you can take it from there.
| {
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"timestamp": "2023-03-29T00:00:00",
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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?
| So you have a problem to compute the critical points of $f\mid_{S}$ where
$S=g^{-1}(4)$ with $g(x,y)=x^2+y^2$.
You need to solve the equation $\nabla f(x,y)=\lambda \nabla g(x,y)$ for some $\lambda\in\mathbb R$. You get
$$
\nabla f(x,y)=\lambda\nabla g(x,y)\Leftrightarrow \begin{pmatrix}1\\2\end{pmatrix}=\lambda\begin{pmatrix}2x\\2y\end{pmatrix}.
$$
You get $x=\frac1{2\lambda}$ and $y=\frac{1}{\lambda}$. Since $(x,y)\in S$ we conclude
$$
4=x^2+y^2=\frac1{4\lambda^2}+\frac1{\lambda^2}\Rightarrow 4\lambda^2=\frac54\Rightarrow \lambda^2=\frac5{16}\Rightarrow \lambda=\pm\frac{\sqrt{5}}4.
$$
Hence $f\mid_S$ has the critical points
$$
(x,y)\in\left\{\left(\frac{2}{\sqrt{5}},\frac{4}{\sqrt{5}}\right),\left(-\frac{2}{\sqrt{5}},-\frac{4}{\sqrt{5}}\right)\right\}
$$
Since $S$ is compact, one of theses critical points is the global minimum while the other is the global maximum. We compute
$$
f\left(\frac{2}{\sqrt{5}},\frac{4}{\sqrt{5}}\right)=\frac{10}{\sqrt{5}}=2\sqrt{5}
$$
and
$$
f\left(-\frac{2}{\sqrt{5}},-\frac{4}{\sqrt{5}}\right)=-\frac{10}{\sqrt{5}}=-2\sqrt{5}.
$$
These are the maximal and minimal values of $f$ on $x^2+y^2\leq 4$.
| {
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"timestamp": "2023-03-29T00:00:00",
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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..
| hint just use
$$1+\cos (X)=2\cos^2 (\frac {X}{2}) $$
$$1-\cos (X)=2\sin^2 (\frac {X}{2}) $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Induction on Fermat Numbers: $F_n = \prod_{j=0}^{n-1}F_j+2$ Is the Following Proof Correct?
Theorem. Given that $\forall n\in\mathbf{N}(F_n = 2^{2^n}+1)$ show that the following is true
$$\forall n\in{1,2,3...}\left(F_n = \prod_{j=0}^{n-1}F_j+2\right)$$
Proof. We construct the proof by recourse to Mathematical-Induction.
Basis-Step: For $n=1$ we have $F_1=F_0+2=(2^{2^0}+1)+2 = (2+1)+2=5$.
Inductive-Step: Assume now for an arbitrary $k\in\{1,2,3,...\}$ the following
$$F_k = \left(\prod_{j=0}^{k-1}F_j+2\right)\tag{1}$$
consequently using $(1)$ we have
$$\left(\prod_{j=0}^{k}F_j+2\right) = \left(\prod_{j=0}^{k-1}F_j\cdot F_k+2\right)=(F_k-2)\cdot F_k=(F_k)^{2}-2F_k+2\tag{2}$$
thus $(F_k)^{2}-2F_k+2 = (2^{2^k}+1)^2-2(2^k+1)+2$ expanding all terms we have $(2^{2^k}+1)^2-2(2^k+1)+2 = 2^{2^{k+1}}+2\cdot 2^{2^k}+1-2\cdot 2^{2^k}-2+2 = 2^{2^{k+1}}+1 = F_{k+1}$
thus
$$F_{k+1}=\left(\prod_{j=0}^{k}F_j+2\right)\tag{3}$$
$\blacksquare$
| Your proof is fine, but I think it's a little easier if you define the numbers $F_n$ by the recurrence
$$F_n=F_0F_1\cdots F_{n-1}+2$$
and then prove the identity $F_n=2^{2^n}+1$ by induction.
Basis step:
$$F_0=3=2^{2^0}+1$$
Inductive step:
$$F_{n+1}=F_0\cdots F_{n-1}F_n+2$$
$$F_{n+1}-2=(F_0\cdots F_{n-1})F_n=(F_n-2)F_n=\left(2^{2^n}-1\right)\left(2^{2^n}+1\right)=\left(2^{2^n}\right)^2-1=2^{2^{n+1}}-1$$
$$F_{n+1}=2^{2^{n+1}}+1$$
| {
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Inequality : $\sqrt[3]{\frac{x^3+y^3+z^3}{xyz}} + \sqrt{\frac{xy+yz+zx}{x^2+y^2+z^2}} \geq 1+\sqrt[3]{3}$
Let $x>0$, $y>0$ and $z > 0$. Prove that $$\sqrt[3]{\frac{x^3+y^3+z^3}{xyz}} + \sqrt{\frac{xy+yz+zx}{x^2+y^2+z^2}} \geq
1+\sqrt[3]{3}.$$
From Micheal Rozenberg's answer :
$(x+2)\sqrt{x^2+2}\left(\sqrt{x^2+2}+\sqrt{2x+1}\right)\geq\sqrt[3]x\left(\sqrt[3]{(x^3+2)^2}+\sqrt[3]{3x(x^3+2)}+\sqrt[3]{9x^2}\right)$,
Prove that $(x+2)^2(x+2\sqrt{x}+3)\geq9\sqrt[3]{x(x^3+2)^2}$,
LHS :
$(x+2)\sqrt{x^2+2}\left(\sqrt{x^2+2}+\sqrt{2x+1}\right)\geq (x+2) \frac{x+2}
{\sqrt{3}}\left(\frac{x+2}{\sqrt{3}}+\frac{2\sqrt{x}+1}{\sqrt{3}}\right)= \frac{(x+2)^2}{3}(x+2\sqrt{x}+3)$
RHS :
$\sqrt[3]{3x(x^3+2)}\leq \sqrt[3]{(x^3+2)^2}$
$\sqrt[3]{9x^2}\leq \sqrt[3]{(x^3+2)^2}$
so $\sqrt[3]{(x^3+2)^2}+\sqrt[3]{3x(x^3+2)}+\sqrt[3]{9x^2}\leq 3\sqrt[3]{(x^3+2)^2}$
$\sqrt[3]{x}(\sqrt[3]{(x^3+2)^2}+\sqrt[3]{3x(x^3+2)}+\sqrt[3]{9x^2})\leq 3\sqrt[3]{x(x^3+2)^2}$
Thus,
$\frac{(x+2)^2}{3}(x+2\sqrt{x}+3) \geq 3\sqrt[3]{x(x^3+2)^2}$
$(x+2)^2(x+2\sqrt{x}+3) \geq 9\sqrt[3]{x(x^3+2)^2}$
| A proof by using the Sergic Primazon's idea. Dedicated to dear Hans.
Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, we need to prove that
$$\sqrt[3]{\frac{27u^3-27uv^2}{w^3}+3}+\sqrt{\frac{v^2}{3u^2-2v^2}}\geq\sqrt[3]3+1.$$
Now, since $27u^3-27uv^2\geq0$ and $v^4\geq uw^3$, it's enough to prove that
$$\sqrt[3]{\frac{27u^4-27u^2v^2}{v^4}+3}+\sqrt{\frac{v^2}{3u^2-2v^2}}\geq\sqrt[3]3+1.$$
Let $3u^2-2v^2=p^2v^2$, where $p>0$.
Thus, $p\geq1$, $u^2=\frac{p^2+2}{3}v^2$ and we need to prove that
$$\sqrt[3]{3(p^2+2)^2-9(p^2+2)+3}+\frac{1}{p}\geq1+\sqrt[3]3$$ or
$$\sqrt[3]3\left(\sqrt[3]{p^4+p^2-1}-1\right)\geq\frac{p-1}{p}$$ or
$$\frac{\sqrt[3]{3}(p^4+p^2-2)}{\sqrt[3]{(p^4+p^2-1)^2}+\sqrt[3]{p^4+p^2-1}+1}\geq\frac{p-1}{p}$$ or
$$\sqrt[3]3p(p+1)(p^2+2)\geq\sqrt[3]{(p^4+p^2-1)^2}+\sqrt[3]{p^4+p^2-1}+1$$ and since
$$1\leq\sqrt[3]{p^4+p^2-1},$$ it's enough to prove that
$$\sqrt[3]3p(p+1)(p^2+2)\geq3\sqrt[3]{(p^4+p^2-1)^2}$$ or
$$p^3(p+1)^3(p^2+2)^3\geq9(p^4+p^2-1)^2$$ or
$$p^{12}+3p^{11}+9p^{10}+19p^9+21p^8+42p^7+26p^6+36p^5+33p^4+8p^3+18p^2-9\geq0.$$
Done!
| {
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Why is the reciprocal of the power series $1-\frac{z^2}{3!}+\frac{z^4}{5!}+\cdots$ equal to $1+\frac{z^2}{6}+\frac{7z^4}{360}+\cdots$? \begin{align}
\frac1{\sin(z)}
&=\frac1z\frac{z}{\sin(z)}\\
&=\frac1z\left(1-\frac{z^2}{3!}+\frac{z^4}{5!}-\frac{z^6}{7!}+\cdots\right)^{-1}\\
&=\frac1z\left(1+\frac{z^2}{6}+\frac{7z^4}{360}+\frac{31z^6}{15120}+\cdots\right)\\
&=\frac1z+\frac{z}{6}+\frac{7z^3}{360}+\frac{31z^5}{15120}+\cdots
\end{align}
In particular, from lines 2 to 3, what is happening? (eg. why does reciprocating result in a change of signs?)
| The Taylor series of $\frac{\sin z}{z}$ at the origin is straightforward to compute, the Taylor series of the reciprocal function $\frac{z}{\sin z}$ a bit less. However, we may notice that $\frac{z}{\sin z}$ is an even meromorphic function with simple poles at the elements of $\pi\mathbb{Z}\setminus\{0\}$. Additionally
$$ z\prod_{n\geq 1}\frac{1-\frac{z^2}{n^2 \pi^2}}{1-\frac{4z^2}{(2n-1)^2\pi^2}} =\tan(z)\tag{1}$$
leads to:
$$ \frac{1}{\sin z}=\frac{d}{dz}\log\tan\frac{z}{2}\\=\frac{1}{z}+\sum_{n\geq 1}\left(\frac{1}{\pi -2 n \pi -z}+\frac{1}{-2 n \pi +z}-\frac{1}{\pi -2 n \pi +z}+\frac{1}{2 n \pi +z}\right) \tag{2}$$
which simplifies into
$$ \frac{z}{\sin z}=1+2z^2\sum_{n\geq 1}\left(\frac{1}{\left[(2n-1)\pi\right]^2-z^2}-\frac{1}{\left[2n \pi\right]^2-z^2}\right)\tag{3}$$
then into:
$$\begin{eqnarray*} \frac{z}{\sin z}&=&1+2\sum_{n\geq 1}\sum_{h\geq 0}\frac{z^{2h+2}}{\pi^{2h+2}}\left(\frac{1}{(2n-1)^{2h+2}}-\frac{1}{(2n)^{2h+2}}\right)\\&=&1+\sum_{h\geq 0}\frac{z^{2h+2}}{\pi^{2h+2}}\zeta(2h+2)\left(2-\frac{1}{4^{h}}\right)\end{eqnarray*}\tag{4}$$
from which it is clear that the Taylor series of $\frac{z}{\sin z}$ at the origin only has non-negative coefficients.
That can be seen also as a consequence of $\frac{1}{\sin z}=\cot(z/2)-\cot(z)$ and the well-known identity
$$ \sum_{h\geq 1}\zeta(2h)z^{2h}=\frac{1-\pi z\cot(\pi z)}{2}.\tag{5} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2388351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Sum of first $1+3+9+\cdots+3^n$ natural numbers. How to prove that the sum of the first $1+3+9+\cdots+3^n$ natural numbers is equal to $1^2+3^2+9^2+\cdots+(3^n)^2$?
I've tried induction, but I can't get through the induction step. The base is simple, but in the step I can only use the induction hypothesis in a way that would give me the conclusion that the sum of the first $1+3+9+\cdots+3^{n+1}$ is equal to the sum of the first $1+3+9+\cdots+3^n$ numbers, but also the sum of the remaining $3^{n+1}$ numbers, and I don't know how to proceed from here. Any hints would be helpful.
| You don't need induction; you can actually show equality with a relatively well-known (though perhaps at first "tricky") summation technique:
Well first, what is $1+3+9+\cdots+3^n$? Define that quantity to be $N$, and note that $3N = 3+9+\cdots+3^n+3^{n+1} = N + 3^{n+1}-1$, so $2N = 3^{n+1}-1$, and $N = \frac12(3^{n+1}-1)$.
Now that you know how high you are summing, you get:
$$
S = \sum_{i=1}^Ni = \frac12N(N+1) = \frac18(3^{n+1}-1)(3^{n+1}+1) = \frac18(9^{n+1}-1)
$$
Now, consider $S' = 1^2 + 3^2 + 9^2+\cdots + 3^{2n}$. Note that $9S' = 3^2 + 9^2 + \cdots + 3^{2n} + 3^{2n+2} = S' + 3^{2n+2} - 1$, so $8S' = 9^{n+1} - 1$, and $S' = \frac18(9^{n+1}-1)$.
Since $S' = S$, you are done.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $a_{1}>0,a_{2}>0$ and $a_{n}=\frac{2a_{n-1}a_{n-2}}{a_{n-1}+a_{n-2}}, n>2$, then $\{ a_{n}\}$ converges to $\frac{3a_{1}a_{2}}{a_{1}+a_{2}}$. Let $a_{1}>0,a_{2}>0$ and $a_{n}=\dfrac{2a_{n-1}a_{n-2}}{a_{n-1}+a_{n-2}}, n>2$, then $\{ a_{n}\}$ converges to $\dfrac{3a_{1}a_{2}}{a_{1}+a_{2}}$.
My attempt:
\begin{align}
a_{n} &= \frac{2a_{n-1}a_{n-2}}{a_{n-1}+a_{n-2}} \\
&= \frac{2}{\dfrac{1}{a_{n-1}}+\dfrac{1}{a_{n-2}}} \\
& \le \frac{1}{\sqrt{a_{n-1}a_{n-2}}}
\end{align}
I used AM- GM inequality here. I am not able to proceed further. How to solve the question? Please help me.
| Hint: let $b_n=1/a_n$ then $b_{n+1}=(b_n+b_{n-1})/2 \iff (b_{n+1}-b_n)=-(b_n-b_{n-1})/2$. Then:
$$
b_{n+1}-b_n=\frac{-1}{2}(b_n-b_{n-1})=\left(\frac{-1}{2}\right)^2(b_{n-1}-b_{n-2}) = \cdots = \left(\frac{-1}{2}\right)^{n-1}(b_{2}-b_{1})
$$
Next, telescope:
$$
b_{n+1} = b_n + \left(\frac{-1}{2}\right)^{n-1}(b_{2}-b_{1}) = b_{n-1} + \left(\left(\frac{-1}{2}\right)^{n-1}+\left(\frac{-1}{2}\right)^{n-2}\right)(b_2-b_1) = \;\cdots
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2388689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
} |
$\ \log _x\left(\log _{36}\left(2\cdot 9^{2x}-3\cdot 4^{2x}\right)\right)<\:1 $ I have to find x in the inequation below:
$$\ \log _x\left(\log _{36}\left(2\cdot 9^{2x}-3\cdot 4^{2x}\right)\right)<\:1 $$
So I did the following:
$$\ \log _{36}\left(2\cdot \:9^{2x}-3\cdot \:4^{2x}\right)<\:x $$
$$\ 2\cdot \:9^{2x}-3\cdot \:4^{2x}<\:36^x $$
$$\ 2-3\cdot \left(\frac{4}{9}\right)^{2x}<\left(\frac{4}{9}\right)^x $$
$$\ u=\left(\frac{4}{9}\right)^x $$
$$\ 3u^2+u-2>0 $$
Which results in $\ \left(\frac{4}{9}\right)^x=-1 $ and
$\ \left(\frac{4}{9}\right)^x=\frac{2}{3} $ so from the first one x isn't real, and from the second $$\ x=\frac {1}{2} $$
Also verifing that $\ 2\cdot 9^{\left(2x\right)}-3\cdot 4^{\left(2x\right)}>0 $ gives $\ x> \frac{1}{4} $. With the correct answer being $\ \frac{1}{2}<x<1 $. Could you let me know what am I missing and if my approach is good?
| You have to consider two different cases separately: $0<x<1$ and $x>1$. The solution you presented here is only valid for the latter case when $x>1$. But in the case $0<x<1$, logarithm base $x$ is a decreasing function, and thus eliminating it would flip the sign of the inequality — so that your second line will be $\log_{36}(\cdots)\color{red}{>}x$.
By the way, you seem to have made a similar mistake when finishing your solution too. Solving the quadratic inequality gives you $u<-1$ or $u>\frac{2}{3}$. As you pointed out, the equation $\left(\frac{4}{9}\right)^x=-1$ has no solutions; moreover, the inequality $\left(\frac{4}{9}\right)^x<-1$ has no solutions either. But the other root gives you an inequality
$$u>\frac{2}{3} \quad \Longrightarrow \quad \left(\frac{4}{9}\right)^x>\frac{2}{3} \quad \Longrightarrow \quad x\color{red}{<}\frac{1}{2}.$$
Again, the inequality sign flips because the base is less than $1$. Notice that this solution was for the case $x>1$, so in the end we see that this case has no solutions (the outcome $x<\frac{1}{2}$ contradicts the case condition $x>1$). But you still have the other case to solve.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2389246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can't solve this limit using $\frac{\sin x}x$ I've been told to solve the following limit using only $\lim_{x \to 0}\frac{\sin x}{x}=1$:
$$\lim_{x \to 0} \frac{1-\cos(1-\cos x)}{x^4}$$
I don't know how to do it except using L'Hospital's Rule but it's just insane amount of math and not the writer's intention(Symbolab do it but it's a computer so ;)
The answer is: $\frac18$
I'm looking for a way to solve this using $\frac{\sin x}x$.
| $$\begin{array}{rcl}
\displaystyle \lim_{x \to 0} \frac{1-\cos(1-\cos x)}{x^4}
&=&
\displaystyle \lim_{x \to 0} \frac{1-\cos^2(1-\cos x)}{x^4[1+\cos(1-\cos x)]} \\
&=& \displaystyle \lim_{x \to 0} \frac{\sin^2(1-\cos x)}{x^4[1+\cos(1-\cos x)]} \\
&=& \displaystyle \lim_{x \to 0} \frac{(1-\cos x)^2\sin^2(1-\cos x)}{x^4(1-\cos x)^2[1+\cos(1-\cos x)]} \\
&=& \displaystyle \lim_{x \to 0} \left(\frac{\sin(1-\cos x)}{1-\cos x}\right)^2\frac{(1-\cos x)^2}{x^4[1+\cos(1-\cos x)]} \\
&=& \displaystyle \lim_{x \to 0} \left(\frac{\sin(1-\cos x)}{1-\cos x}\right)^2\frac{(1-\cos^2 x)^2}{x^4(1+\cos x)^2[1+\cos(1-\cos x)]} \\
&=& \displaystyle \lim_{x \to 0} \left(\frac{\sin(1-\cos x)}{1-\cos x}\right)^2\frac{(\sin^2 x)^2}{x^4(1+\cos x)^2[1+\cos(1-\cos x)]} \\
&=& \displaystyle \lim_{x \to 0} \left(\frac{\sin(1-\cos x)}{1-\cos x}\right)^2 \left(\frac{\sin x}{x}\right)^4 \frac{1}{(1+\cos x)^2[1+\cos(1-\cos x)]} \\
&=& 1^2 \cdot 1^4 \cdot \dfrac{1}{2^2 \cdot 2} \\
&=& \dfrac18
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2389392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Evaluate $\lim_{x \to 0}\frac{x}{\sqrt{1-\sqrt{1-x^2}}}$ Evaluate $$L=\lim_{x \to 0}\frac{x}{\sqrt{1-\sqrt{1-x^2}}} \tag{1}$$
I have used L Hopital's Rule we get
$$L=\lim_{x \to 0} 2\frac{ \sqrt{1-\sqrt{1-x^2}} \sqrt{1-x^2}}{x}$$
$\implies$
$$L=2 \lim_{x \to 0}\frac{ \sqrt{1-\sqrt{1-x^2}}}{x}$$ and from $(1)$ we get
$$L=\frac{2}{L}$$
$$L=\sqrt{2}$$
is this correct appraoach?
| There is a right limit and a left limit and they are different, which means that the limit doesn't exist because when it exists it is unique.
Let substitute $1-x^2=w^2$
Suppose $x>0$ we have $x=\sqrt{1-w^2}$ and the limit becomes
$$\lim_{w\to 1} \, \dfrac{\sqrt{1-w^2}}{\sqrt{1-w}}=\lim_{w\to 1} \, \frac{\sqrt{1-w} \sqrt{w+1}}{\sqrt{1-w}}=\lim_{w\to 1} \,\sqrt{w+1}= \sqrt{2}$$
If $x<0$ then we have $x=-\sqrt{1-w^2}$
And the previous limit gives $- \sqrt{2}$
Hope this helps
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2390192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Using Lagrange's multiplier method, find the shortest distance between the line y=10-2x and the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$
Using Lagrange's multiplier method, find the shortest distance between the line $y=10-2x$ and the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$.
My work:
Let the point on ellipse be $(2\cos\theta, 3\sin\theta)$
Let $F=(x-2\cos\theta)^2+(y-3\sin\theta)^2+\alpha(2x+y-10)$.
I partially diffentiated $F$ wrt $x$,$y$ and $\theta$ equated to $0$.
I get $\tan\theta=\frac{3}{4}$ and $\alpha=-2$
The point on ellipse I get is $(\frac{8}{5},\frac{9}{5})$ and the point on line I get is $(\frac {18}{5}, \frac{14}{5})$
Is this method correct? Can I take $\theta$ to be an independent variable, and differentiate wrt to it?
| $\frac{x^2}{4} + \frac{y^2}{9}$ can be made into $ 1 = 9x^2 + 4y^2 = 36 $ which can be made into $9cos^2(\theta) + 4sin^2(\theta) = 36$. This should help simplify your calculations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2392883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Elliptic Integrals & Gamma Functions of Rational values. I recently saw this question ... https://math.stackexchange.com/questions/2393668/double-factorial-sum-k-0-infty-left-frac-2k-1-2k-right#2393668 ... & I am unable to show it.
The result
$$1-\left( \frac{1}{2} \right)^{3}+\left( \frac{1\cdot 3}{2\cdot 4} \right)^{3}-\left( \frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6} \right)^{3}+\left( \frac{1\cdot 3\cdot 5\cdot 7}{2\cdot 4\cdot 6\cdot 8} \right)^{3}-...=\frac{\Gamma ^{2}\left( \frac{9}{8} \right)}{\Gamma ^{2}\left( \frac{7}{8} \right)\cdot \Gamma ^{2}\left( \frac{10}{8} \right)}$$
Now I know several results relating to those ratios of double factorials
\begin{eqnarray*}
\int_0^{\frac{\pi}{2}} \sin^{2n} \theta d \theta= \frac{ \pi}{2} \frac{(2n-1)!!}{(2n)!!} \tag{1} \\
\sum_{n=0}^{\infty} \frac{(2n-1)!!}{(2n)!!} y^n = \frac{1}{\sqrt{1-y}} \tag{2}
\end{eqnarray*}
From these two results it is reasonably easy to derive the series for the elliptic integral of the first kind
\begin{eqnarray*}
K(k)=\int_0^{\frac{\pi}{2}} \frac{d \theta}{\sqrt{1-k^2 \sin^{2}(\theta)}} =\frac{\pi}{2} \left(1+ \left(\frac{1}{2}\right)^2+ \left(\frac{1.3}{2.4}\right)^2+ \cdots \right)
\end{eqnarray*}
& it is well known that this can be evaluated for special values of $k$ in terms of Gamma functions whose arguements are rational values. See this question Can $\Gamma(1/5)$ be written in this form? and the reference cited there.
So my first thought is to use $(1)$ three times, sum the geometric plum & we have
\begin{eqnarray*}
\int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}} \frac{d \alpha d \beta d \gamma}{1+ \sin^2 \alpha \sin^2 \beta \sin^2 \gamma}
\end{eqnarray*}
This triple integral looks difficult so ...
Second thoughts ... use $(1)$ twice & then use $(2)$ to get the double integral
\begin{eqnarray*}
\int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}} \frac{d \alpha d \beta }{\sqrt{1+ \sin^2 \alpha \sin^2 \beta }}
\end{eqnarray*}
now substitute $ \sqrt{s} =\sin \alpha$ and $ \sqrt{t} =\sin \beta$ (might have lost a factor of $4$)
\begin{eqnarray*}
\int_{0}^{1} \int_{0}^{1} \frac{d s d t }{\sqrt{s(1-s)t(1-t)(1+st) }}
\end{eqnarray*}
& I am not sure what to do with this.
In both of these attempts I feel I have taken a wrong turn.
Can someone either give me a Big hint or a reference to the original derivation of this result or a reasonably complete solution ?
Bonus question ... why was it stated with $\frac{10}{8}$ instead of $\frac{5}{4}$ ?
The result
$$1-\left( \frac{1}{2} \right)^{3}+\left( \frac{1\cdot 3}{2\cdot 4} \right)^{3}-\left( \frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6} \right)^{3}+\left( \frac{1\cdot 3\cdot 5\cdot 7}{2\cdot 4\cdot 6\cdot 8} \right)^{3}-...=\frac{\Gamma ^{2}\left( \frac{9}{8} \right)}{\Gamma ^{2}\left( \frac{7}{8} \right)\cdot \Gamma ^{2}\left( \color{red}{\frac{5}{4}} \right)}$$
Hint from Jack D'Auirzio : look at equation (6) here http://mathworld.wolfram.com/CompleteEllipticIntegraloftheFirstKind.html and use $2kk'=i$ where $k'$ is the complementary modulus.
| The sum evaluates to $(1-k^{2})(2K(k)/\pi)^{2}$ for $k=\sqrt{2}-1$. The value of $K$ is obtained from here. Using the value of $K$ we can get the desired closed form sum of the given series.
Using hypergeometric transformations we can prove the formula $$\left(\frac{2K(k)}{\pi}\right)^{2}=\sum_{n=0}^{\infty}\left(\frac{1\cdot 3\cdot 5\cdots (2n-1)}{2\cdot 4\cdot 6\cdots (2n)}\right)^{3}(2kk')^{2n}\tag{1}$$ (see complete proof here). Note that the above equation can be thought of as an identity involving the nome $q=e^{-\pi K'/K} $ also and then we can switch from $q$ to $-q$. In terms of Jacobi's theta functions we have $2K/\pi=\vartheta_{3}^{2}(q)$ and replacing $q$ by $-q$ leads us to $$\vartheta_{3}^{2}(-q)=\vartheta_{4}^{2}(q)=\frac{\vartheta_{4}^{2}(q)}{\vartheta_{3}^{2}(q)}\cdot\vartheta_{3}^{2}(q)=k'\cdot\frac{2K}{\pi}$$ Similarly we have $$(2kk')^{2}=4k^{2}k'^{2}=4\cdot\frac{\vartheta_{2}^{4}(q)\vartheta_{4}^{4}(q)}{\vartheta_{3}^{8}(q)}=4\cdot\frac{16q\psi^{4}(q^{2})\vartheta_{4}^{4}(q)}{\vartheta_{3}^{8}(q)}$$ where $\psi$ is one of Ramanujan's theta functions. Now changing $q$ into $-q$ interchanges $\vartheta_{3},\vartheta_{4}$ and hence the above expression becomes $$-4\cdot\frac{16q\psi^{4}(q^{2})\vartheta_{3}^{4}(q)}{\vartheta_{4}^{8}(q)} =-4\cdot\frac{\vartheta_{2}^{4}(q)/\vartheta_{3}^{4}(q)} {\vartheta_{4}^{8}(q)/\vartheta_{3}^{8}(q)}=-\frac{4k^{2}}{k'^{4}} $$ So we have the formula $$k'^{2}\left(\frac{2K}{\pi}\right)^{2}=\sum_{n=0}^{\infty}(-1)^{n}\left(\frac{1\cdot 3\cdots (2n-1)}{2\cdot 4\cdots(2n)}\right)^{3}(2k/k'^{2})^{2n}\tag{2}$$ And putting $k=\sqrt{2}-1$ we get the desired sum.
As indicated in a comment from Jack d'Aurizio the sum in question can also be expressed in terms of arithmetic-geometric-mean as $$\frac{2}{\operatorname {AGM} (\sqrt{2},\sqrt{1+\sqrt{2}})^{2}}\tag{3}$$ Writing in this manner helps to evaluate the sum very accurately with a few iterations of the AGM.
It is also interesting to know that Ramanujan used the series $(1)$ to obtain $$\frac{4}{\pi}=\sum_{n=0}^{\infty}\left(\frac{1\cdot 3\cdots (2n-1)}{2\cdot 4\cdots (2n)}\right)^{3}\cdot\frac{6n+1}{4^{n}}\tag{4}$$ and $$\frac{2}{\pi}=\sum_{n=0}^{\infty}(-1)^{n}(4n+1)\left(\frac{1\cdot 3\cdots (2n-1)}{2\cdot 4\cdots (2n)}\right)^{3}\tag{5}$$ was obtained from series $(2)$. You can get all the details here.
The value of $K(\sqrt{2}-1)$ has been obtained via non-obvious hypergeometric transformations and it would be nice to have an approach based on direct evaluation of integral involved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2393810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
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Matrix similarity is an equivalence relation Recall that for $ A,B \in \mathsf{M}_n$, we say that $A$ is similar to $B$, denoted $A \sim B$, if there is an invertible matrix $S \in \mathsf{M}_n$ such that $A = S B S^{-1}$. Prove that similarity is an equivalence relation on $\mathsf{M}_n$.
I have no idea on what to do here, what i have tried is to say that $A = S A S^{-1} = A I_n = B = BI_n = S B S^{-1} $ which i have a very strong feeling is completely wrong. What should i do? how do i do this proof?
if yall can come up with a better title feel free to replace or change mine.
| I'm not sure how you get any of the equalities in your work, except for $B=BI_n$. Since $S$ is just some invertible matrix, how do you get $A=SAS^{-1}=AI_n$ and $BI_n=SBS^{-1}$? I suspect you are commuting $SAS^{-1}$ into $ASS^{-1}$ and similarly $SBS^{-1}$ into $BSS^{-1}$, but remember that matrix multiplication is not commutative. For example, if
$$
S:=\begin{pmatrix}
2 & 0 \\ 0 & 1
\end{pmatrix}
\quad\text{and}\quad
B:=\begin{pmatrix}
1 & 1 \\ 1 & 1
\end{pmatrix},
$$
then
$$
SBS^{-1} = \begin{pmatrix}
2 & 0 \\ 0 & 1
\end{pmatrix}
\begin{pmatrix}
1 & 1 \\ 1 & 1
\end{pmatrix}
\begin{pmatrix}
1/2 & 0 \\ 0 & 1
\end{pmatrix}
= \begin{pmatrix}
2 & 2 \\ 1 & 1
\end{pmatrix}
\begin{pmatrix}
1/2 & 0 \\ 0 & 1
\end{pmatrix}
=
\begin{pmatrix}
1 & 2 \\ 1/2 & 1
\end{pmatrix}
$$
is not equal to $B$. Also, when you write $AI_n=B$, you seem to be assuming that $A$ is equal to $B$. But this isn't true, even for similar matrices. For instance, if
$$A:=\begin{pmatrix}
0 & 1 \\ 0 & 0
\end{pmatrix}
\quad\text{and}\quad
B:=\begin{pmatrix}
0 & 0 \\ 1 & 0
\end{pmatrix}
\quad\text{and}\quad
S:=\begin{pmatrix}
0 & -1 \\ 1 & 0
\end{pmatrix}
,$$
then $A\ne B$ are similar because $S$ is invertible and
$$
SBS^{-1} =
\begin{pmatrix}
0 & -1 \\ 1 & 0
\end{pmatrix}
\begin{pmatrix}
1 & 0 \\ 0 & 0
\end{pmatrix}
\begin{pmatrix}
0 & -1 \\ 1 & 0
\end{pmatrix}^{-1}
=
\begin{pmatrix}
1 & 0 \\ 0 & 0
\end{pmatrix}
\begin{pmatrix}
0 & 1 \\ -1 & 0
\end{pmatrix}
=\begin{pmatrix}
0 & 1 \\ 0 & 0
\end{pmatrix}
= A.
$$
Now, onto the question.
We need to prove that $\sim$ is reflexive ($A\sim A$ for all $A\in\mathsf{M}_n$), symmetric ($A\sim B$ implies $B\sim A$), and transitive ($A\sim B$ and $B\sim C$ implies $A\sim C$).
I'll let you do reflexivity.
For symmetry, suppose $A\sim B$. Then there exists an invertible matrix $S$ such that $A=SBS^{-1}$. Hence $S^{-1}$ is invertible and $B=S^{-1}A(S^{-1})^{-1}$, so $B\sim A$.
For transitivity, assume $A\sim B$ and $B\sim C$. Then there are invertible matrices $S$ and $T$ such that $A=SBS^{-1}$ and $B=TCT^{-1}$. Then $ST$ is invertible and
$$
A = SBS^{-1} = S(TCT^{-1})S^{-1} = (ST)C(ST)^{-1},
$$
so $A\sim C$.
| {
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"timestamp": "2023-03-29T00:00:00",
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$\sin^3 a\sin(b-c)+\sin^3b\sin(c-a)+\sin^3c\sin(a-b)+\sin(a+b+c)\sin(b-c)\sin(c-a)\sin(a-b)=0$
Verify that$$\sin^3 a\sin(b-c)+\sin^3b\sin(c-a)+\sin^3c\sin(a-b)$$
$$+\sin(a+b+c)\sin(b-c)\sin(c-a)\sin(a-b)=0.$$
I tried to break $\sin^3a$ into $\sin^2\cdot\sin a$ and use it to make sum with $\sin(b-c)$, but eventually messed up everything. Please help.
| $$\sum_{cyc}\sin^3a\sin(b-c)=\frac{1}{4}\sum_{cyc}(3\sin{a}-\sin3a)\sin(b-c)=$$
$$=\frac{1}{8}\sum_{cyc}\left(3\cos(a-b+c)-3\cos(a+b-c)-\cos(3a-b+c)+\cos(3a+b-c)\right).$$
$$\sin(a+b+c)\prod_{cyc}\sin(a-b)=\frac{1}{2}\sin(a+b+c)\sin(b-c)(\cos(2a-b-c)-\cos(b-c))=$$
$$=\frac{1}{4}\sin(a+b+c)(-\sin(2c-2a)-\sin(2a-2b)-\sin(2b-2c))=$$
$$=-\frac{1}{4}\sin(a+b+c)\sum_{cyc}\sin(2b-2c)=-\frac{1}{8}\sum_{cyc}(\cos(3c+a-b)-\cos(3b+a-c))=$$
$$=\frac{1}{8}\sum_{cyc}(\cos(3a+c-b)-\cos(3a+b-c))$$
and since
$$\sum\limits_{cyc}(\cos(a-b+c)-\cos(a+b-c))=\sum\limits_{cyc}(\cos(b-c+a)-\cos(a+b-c))=0,$$
we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
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How to evaluate $\binom{2}{2}\binom{10}{3} + \binom{3}{2}\binom{9}{3} + \binom{4}{2}\binom{8}{3} + \ldots + \binom{9}{2}\binom{3}{3}$ I really can't understand how to approach exercise 41.
I tried by comparing the 5th term of binomial expansion and something like that, but getting 12C5, I think the answer should be 13C6. How to approach?
| We wish to show that
$$\sum_{k = 2}^{9} \binom{k}{2}\binom{10 - k}{3} = \binom{13}{6}$$
The right hand side counts six-element subsets of the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13\}$.
The left-hand side counts six-elements subsets of the set whose third-largest element is $k + 1$. To see this, observe that if the third-largest element is $k + 1$, we must choose two of the $k$ elements smaller than $k + 1$ and three of the $13 - (k + 1) = 12 - k$ elements larger than $k + 1$. Hence, the number of subsets of size six whose third-largest element is $k + 1$ is
$$\binom{k}{2}\binom{10 - k}{3}$$
For $k + 1$ to be the third-largest element, $k$ must be at least $2$ and at most $9$ since there be must be at least two numbers smaller than $k + 1$ and at least three numbers larger than $k + 1$. Since every six-element subset of the set has a third-largest element, the number of all six-element subsets of the set is
$$\sum_{k = 2}^{9} \binom{k}{2}\binom{10 - k}{3}$$
Hence, the identity holds.
| {
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Show that: $\sin ^2 (a+b)+\sin ^2 (a-b)=4(1-\cos ^2 a \cos ^2 b)$ Show that: $$\sin ^2 (a+b)+\sin ^2 (a-b)=4(1-\cos ^2 a \cos ^2 b)$$
I tried two approaches.
Approach 1: $$\sin ^2 (a \pm b)=\sin a\cos b \pm \sin b\cos a$$
This reduced to: $$2(\cos ^2 a +\cos ^2 b -2\cos ^2 a\cos ^2b)$$
I can't see where to go from here.
Approach 2: $$\sin ^2 (a \pm b)=1-\cos ^2 (a \pm b)$$
This reduced to: $$2(\cos ^2 a +\cos ^2 b -2\cos ^2 a\cos ^2b)$$
Both approaches reduced to the same expression which is a good sign, but I still can't figure out how to make the final manipulation.
| hint:$$\sin ^2 (a+b)+\sin ^2 (a-b)=4(1-\cos ^2 a \cos ^2 b)=\\1-\cos(2a+2b)+1-\cos(2a-2b)=\\2-(\cos(2a+2b)+\cos(2a-2b))= $$use sum to product formula
$$=2-(\cos(2a+2b)+\cos(2a-2b))=\\2 -2\cos (\frac{2a+2b+2a-2b}{2}).\cos (\frac{2a+2b-(2a-2b)}{2})=\\2-2\cos (2a).\cos (2b)=\\$$
| {
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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.
| Here's a useful formula for this kind of problems:
$$\sqrt{a \pm \sqrt{b}} = \sqrt{\frac{a + \sqrt{a^2 - b}}{2}} \pm \sqrt{\frac{a - \sqrt{a^2 - b}}{2}}$$
where we have $a,b \ge 0$ and $a^2 > b$
| {
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Show that for any natural number n between $n^2$ and$(n+1)^2$ there exist 3 distinct natural numbers a, b, c, so that $a^2+b^2$ is divisible by c Show that for any natural number n ,one can find 3 distinct natural numbers a, b, c, between $n^2$ and$(n+1)^2$, so that $a^2+b^2$ is divisible by c.
It's easy to prove that such three distinct numbers exist, by supposing the contrary and coming to contradiction(i.e."suppose $(n+1)^2-n^2=0$ -->$n=-1$, $-1$ is not a natural number, and so on.."), but how to show divisibility?
(The task is from 1998 St. Petersburg City Mathematical Olympiad)
| Surprisingly there seems to be another answer to the stricter reading of the problem.
Let $a=(n^2+n)$, $b=(n^2+n+2)$ and $c=(n^2+1)$
and as with @ARoberts Solution if $n\ge 2$ then $n^2 < c < a < b < (n+1)^2$
we have
$a^2+b^2 = (n^2+n)^2 + (n^2 + n + 2)^2 = 2(n^2+1)(n^2+2n+2)$.
So again $c | a^2 + b^2$
| {
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Show that if $a+b+c=0$, $2(a^4 + b^4+ c^4)$ is a perfect square
Show that for $\{a,b,c\}\subset\Bbb Z$ if $a+b+c=0$ then $2(a^4 + b^4+ c^4)$ is a perfect square.
This question is from a math olympiad contest.
I started developing the expression $(a^2+b^2+c^2)^2=a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2$ but was not able to find any useful direction after that.
Note: After getting 6 answers here, another user pointed out other question in the site with similar but not identical content (see above), but the 7 answers presented include more comprehensive approaches to similar problems (e.g. newton identities and other methods) that I found more useful, as compared with the 3 answers provided to the other question.
| Since $$2(a^2b^2+a^2c^2+b^2c^2)-a^4-b^4-c^4=(a+b+c)(a+b-c)(a+c-b)(b+c-a)=0,$$
we obtain
$$2(a^4+b^4+c^4)=a^4+b^4+c^4+2(a^2b^2+a^2c^2+b^2c^2)=(a^2+b^2+c^2)^2.$$
Done!
| {
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find a and b such that the limit will exist and find the limit I am given this question $$\lim_{x\rightarrow 0} \frac{e^{\sqrt{1+x^2}}-a-bx^2}{x^4}$$
And asked to find the $a$ and $b$ such that the limit exists and also to compute the limit
My solution:
Let $t$=$\sqrt{1+x}$.
Then the Maclaurin polynomial is : $$\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+\mathcal{O}(x^3)$$
Now plugging in $x^2$ for x we get $$\sqrt{1+x^2}=1+\frac{x^2}{2}-\frac{x^4}{8}+\mathcal{O}(x^6)$$
and from the common Maclaurin polynomial we have that $e^t=1+t+\frac{t^2}{2}+ \mathcal{O}(t^3)$. Plugging in $\sqrt{x^2+1}$ for $t$ we get:
$$e^{1+\frac{x^2}{2}-\frac{x^4}{8}+\mathcal{O}(x^6)
}$$
which in turn is:
$$e^{1+\frac{x^2}{2}-\frac{x^4}{8}+\mathcal{O}(x^6)
}=1+\left(1+\frac{x^2}{2}-\frac{x^4}{8}+\mathcal{O}(x^6)\right)+\mathcal{O}(x^6)$$
hence we have:
$$\lim_{x\rightarrow 0} \frac{1+\left(1+\frac{x^2}{2}-\frac{x^4}{8}+\mathcal{O}(x^6)\right)+\mathcal{}O(x^6)-a-bx^2}{x^4}$$
And i argue that $a=2, b=1/2$ and the $\lim=-1/8 $
However the book disagrees with me and argues that the they should be $a=e, b=e/2$ and limit $=0$
i mean i can see how theyd done it, by not expanding $e^t$ but by only expanding $\sqrt{x^2+1}$ however what i dont understand how come we didnt get the same values or at least the same value for the limit?
| Ignoring the squares, we have
$$\lim_{x\to0}\frac{e^{\sqrt{1+x}}-a-bx}{x^2}.$$
By adjusting $a$ and $b$, we can certainly get rid of the linear terms and the limit will exist.
So let's compute the Taylor development of $f(x):=e^{\sqrt{1+x}}$.
$$f(0)=e=a,$$
$$f'(x)=\frac{e^{\sqrt{1+x}}}{2\sqrt{1+x}},f'(0)=\frac e2=b,$$
$$f''(x)=\frac{e^{\sqrt{1+x}}}{4(1+x)}-\frac{e^{\sqrt{1+x}}}{4(1+x)^{3/2}},f''(0)=0.$$
Hence the limit is $0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Wrong result for a $1^\infty$ limit I wanna know the $$\lim_{x\rightarrow\infty}(\sqrt{x^2+2x+4}-x)^x .$$
I applied $1^\infty$ algorithm and I have in last sentence $$\lim_{x\rightarrow\infty} \left(\frac{2x+4-\sqrt{x^2+2x+4}}{\sqrt{x^2+2x+4}+x}\right)^x.$$ After that I used $x=e^{\ln x}$ and got this: $$\lim_{x\rightarrow\infty} \left(\frac{x^2+4x-x\sqrt{x^2+2x+4}}{\sqrt{x^2+2x+4}+x}\right).$$ The result for this, I think is $2$, but I'm not really sure of this, Wolfram-Alpha says that it's $\frac{3}{2}$. Can explain somebody why, but without L'Hospital?
Thks a lot!
| Sometimes WA gives absolutely wrong results, but in our case he gave a right answer.
Since $f(x)=e^x$ is a continuous function, we obtain:
$$\lim_{x\rightarrow+\infty}(\sqrt{x^2+2x+4}-x)^x =\lim_{x\rightarrow+\infty}\left(\frac{2x+4}{\sqrt{x^2+2x+4}+x}\right)^x =$$
$$=\lim_{x\rightarrow+\infty}\left(1+\frac{2x+4}{\sqrt{x^2+2x+4}+x}-1\right)^{\frac{1}{\frac{2x+4}{\sqrt{x^2+2x+4}+x}-1}\cdot x\left(\frac{2x+4}{\sqrt{x^2+2x+4}+x}-1\right)} =$$
$$=e^{\lim\limits_{x\rightarrow+\infty}x\left(\frac{2x+4}{\sqrt{x^2+2x+4}+x}-1\right)}=e^{\lim\limits_{x\rightarrow+\infty}\frac{x( x+4-\sqrt{x^2+2x+4})}{\sqrt{x^2+2x+4}+x}}=$$
$$=e^{\lim\limits_{x\rightarrow+\infty}\frac{x(6x+12)}{\left(\sqrt{x^2+2x+4}+x\right)\left( x+4+\sqrt{x^2+2x+4}\right)}}=e^{\frac{6}{2\cdot2}}=e^{\frac{3}{2}}.$$
| {
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Solve $\lfloor \frac{2x-1}{3} \rfloor + \lfloor \frac{4x+1}{6} \rfloor=5x-4$ I came across this problem and it doesn't seem so tricky, but I didn't do it.
Solve the equation
$$\lfloor {\frac{2x-1}{3}}\rfloor + \lfloor {\frac{4x+1}{6}}\rfloor=5x-4.$$
My thoughts so far:
Trying to use the inequality $k-1 < \lfloor k\rfloor \leq k$
$5x-4$ is integer, so $x=\frac{a}{5},$ where $a$ is from $\mathbb Z,$ then replace in the original equation, to get
$$\lfloor \frac{4a-10}{30}\rfloor + \lfloor \frac{4a+5}{30}\rfloor =a-4$$
Let $k= \frac{4a-10}{30}$
so $\lfloor k\rfloor +\lfloor k+\frac{1}{2}\rfloor =a-4$
but $\lfloor k\rfloor =\lfloor k+\frac{1}{2}\rfloor $ or $\lfloor k\rfloor +1=\lfloor k+\frac{1}{2}\rfloor $, so we have 2 cases
I strongly think that a shorter solution exists, I would appreciate a hint!
| Let $\lfloor{\frac{2x-1}{3}}\rfloor + \lfloor{\frac{4x+1}{6}}\rfloor=n$, where $n\in\mathbb Z$.
Hence, $5x-4=n$, which gives $x=\frac{n+4}{5}$.
Thus,
$$\lfloor{\frac{2n+13}{15}}\rfloor + \lfloor{\frac{4x+21}{30}}\rfloor=n,\tag{1}$$
which gives
$$n\leq\frac{2n+13}{15}+ \frac{4x+21}{30}<n+2,\tag{2}$$ which gives
$0\leq n\leq2$.
Now, for $n=0$ we get $x=\frac{4}{5}$ and a substitution in the original equation gives $1=1$,
which says that $1$ is the root.
$n=1$ gives $0=1$, $n=2$ gives $0=2$,
which says that the answer is
$$\left\{\frac{4}{5}\right\}.$$
| {
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Solve $\frac{1}{x}<\frac{x}{2}<\frac{2}{x}$. First I consider the case $\frac{1}{x}<\frac{x}{2}$, manipulating to $\frac{(x-\sqrt{2})(x+\sqrt{2})}{2x}<0\Leftrightarrow x < -\sqrt{2}.$ Did this by subtracting $\frac{x}{2}$ from both sides and writing everything with a common denominator.
Second case to consider is $\frac{x}{2}<\frac{2}{x}\Leftrightarrow\frac{(x-2)(x+2)}{2x}<0\Leftrightarrow x<-2.$
Since both of these inequalities have to be satisfied simultaneously, one can combine them to get $x<-2.$ Correct answer is $x\in(\sqrt{2},2)$
| There are some good answers up already but let me address specifically what's wrong with your attempted solution.
1) $1/x < x/2$ is not equivalent to $\frac{(x-\sqrt{2})(x+\sqrt{2})}{2x} < 0$. The inequality is going the wrong way. Subtracting $x/2$ on the left side gives
$$\frac{1}{x} - \frac{x}{2} = \frac{2 - x^2}{2x} = \frac{(\sqrt{2}-x)(\sqrt{2}+x)}{2x},$$
so this is what is $<0$.
2) Meanwhile this inequality is not equivalent to $x<-\sqrt{2}$. The numerator is positive when $x$ is between $-\sqrt{2}$ and $\sqrt{2}$ and negative otherwise. The denominator changes sign with $x$. Thus the fraction is pos/neg = neg when $-\sqrt{2}<x<0$, and neg/pos = neg when $x>\sqrt{2}$, so overall it is negative when either $-\sqrt{2}<x<0$ or $x>\sqrt{2}$.
3) For $x/2< 2/x$, this time it is equivalent to $(x-2)(x+2)/2x<0$, as you wrote. (It's different from the last one because the $x$ is in the numerator on the left side.) But again, this is not equivalent to $x<-2$. Now the numerator is positive when $x<-2$ or $x>2$ and negative between $-2$ and $2$, while again the denominator changes sign with $x$. Thus the whole fraction is pos/neg = neg when $x<-2$, and it is neg/pos = neg when $0<x<2$. So this inequality yields $x<-2$ or $0<x<2$.
| {
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If $a$ is a root of $x^2+ x + 1$, simplify $1 + a + a^2 +\dots+ a^{2017}.$
If $a$ is a root of $x^2 + x + 1$ simplify $$1 + a + a^2 + a^3 + \cdots + a^{2017}.$$
my solution initially starts with the idea that $1 + a + a^{2} = 0$ since $a$ is one of the root then using the idea i grouped $$1 + a + a^2 + a^3 + \cdots + a^{2017}$$ just like this $$(1 + a + a^{2}) + a^3(1 + a + a^2) + a^6(1 + a + a^2) + \cdots + a^{2013} + (1 + a + a^2) + a^{2016} + a^{2017}$$
which is equivalent to $$(0) + a^3(0) + a^6(0) + \cdots + a^{2013}(0) + a^{2016} + a^{2017}$$ and finally simplified into $$a^{2016} + a^{2017}$$. That's my first solution, but eventually I noticed that since $1 + a + a^2 = 0$ then it can be $a + a^2 = -1 $ using this idea I further simplified $$a^{2016} + a^{2017}$$ into $$a^{2015}(a + a^{2})$$ ==> $$a^{2015}(-1)$$ ==> $$-a^{2015}$$ But answers vary if I start the grouping at the end of the expression resulting to $1 + a$ another solution yields to $-a^{2017}$ if I set $1 + a = -a^2$ several solutions rise when the grouping is being made anywhere at the body of the expression... my question is, Is $$a^{2016} + a^{2017} = 1 + a = -a^{2015} = -a^{2017} \text{ etc.?}$$ How does it happen? I already forgot how to manipulate complex solutions maybe that's why I'm boggled with this question...any help?
| You are so close to figuring this out on your own!
Here's a "meta"-question and some ideas for you.
Meta-question: What do think the purpose of this exercise is? Other than busy work, of course. What do you think will be learned?
ideas: 1) If $a$ is a root to $1 + x + x^2$ then as $1+x+x^2$ is a quadratic there are two possible values for $a$ which we can figure out via the quadratic equation. Because $b^2 - 4ac = (1)^2 - 4(1)(1) = - 3< 0$ neither of these values are real.
It could be illuminating to actually figure out the values of $a$, to graph them on the complex plane, and to figure out what $a^k$ would be. But that's not actually the point of the problem.
2) If $a$ is a root to $1 + x + x^2$, then $a$ is also a root to $(1-x)(1 + x + x^2) = x^3 -1$.
$x^3 -1$ is a third degree polynomial and has $3$ roots. One of them is the real $x = 1$. The other non-real two roots are the two possible values of $a$.
And that is the answer to the meta-question. $a$ is what is called a third root of unity. The polynomial $x^k - 1=0$ will have $k$ complex roots. $x=1$ will be one of them. If $k$ is even $x = -1$ will be another. The rest will be complex and they will be roots of the polyomial $1 + x + x^2 + ..... + x^{k-1}$.
There will be $n-1$ such roots and they will all but so that $x^k = 1$.
3) Another way of thinking about 2) above:
$1 + a + a^2 = 0$
$a(1 + a + a^2) = 0*a = 0$
$a + a^2 + a^3 = 0$
$1 + a + a^2 + a^3 = 1 + 0 = 1$
$(1 + a + a^2) + a^3 = 1$
$a^3 = 1$.
So $1 + a + a^2 + a^3 + ....... + a^{2017} = $
$(1+a + a^2)(1 + a^3 + a^6 + ....... + a^{2013}) + a^{2016} + a^{2017}=$
$0*(1+1+........ + 1) + a^{2016}(1 + a) = $
$1(1+a)= 1 + a$
=== post script ====
Okay. Back to 1)...
$a^2 + a + 1 = 0$
$a = \frac {-1 \pm \sqrt{-3}}2 = -\frac 12 \pm i\frac {\sqrt{3}}2$.
(This should look familiar. If $x = -\frac 12$ and $y=\pm \frac {\sqrt{3}}2$ what kind of angle is being formed by $(x,y)$ and $(0,0)$ and the $x$-axis?)
(Food for thought: What kind of shape is made be the three points $a, -a$ and $1$. i.e. what shape is made by the three third roots of unity?)
(What shape do you think will be made by the $k$ $k$-roots of unity?)
An $1 + a + .... +a^{2017} = 1 + a = \frac 12 \pm i\frac {\sqrt{3}}2$.
One thing very much worth noting:
$a^0 = 1$
$a^1 = a = -\frac 12 \pm i\frac {\sqrt{3}}2$
$a^2 = (-\frac 12 \pm i\frac {\sqrt{3}}2)^2 =$
$\frac 14 - \frac 34 \pm i*2(-\frac 12)(\frac{\sqrt{3}}2) =$
$= -\frac 12 \mp i\frac {\sqrt{3}}2 = -a$.
$a^3 = (-\frac 12 \pm i\frac {\sqrt{3}}2)(-\frac 12 \mp \frac {\sqrt{3}}2) =$
$= \frac 14 - (-\frac 34) = \frac 14 + \frac 34 = 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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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 $481$?????
| $481_{[9]} = 4\,\cdot\,81 +8\,\cdot9+1\,\cdot\,1 = 324 + 72 + 1 = 397_{[10]}$
The conversion you presented is incorrect, you converted $397_{[9]}$ into $331_{[10]}$, which is not what you want.
To convert it properly:
$397/9 = 44$ (remainder = $1$)
$44/9 = 4$ (remainder = $8$)
$4/9 = 0$ (remainder = $4$)
The remainders give the result: $\boxed{397_{[10]}=481_{[9]}}$
| {
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Interesting way to evaluate $ \int \cos^3 x\ dx$ I have read these days a nice way for integrating $\cos^3x$:
First differentiate:
$$f=\cos^3(x)$$$$f'=-3\cos^2(x)\sin(x)$$$$f''=6\cos(x)\sin^2(x)-3\cos^3(x)$$$$f''=6\cos(x)(1-\cos^2(x))-3f$$$$f''=6\cos(x)-6\cos^3(x)-3f$$$$f''=6\cos(x)-9f$$
Then integrate:
$$f'= 6\sin(x) -9\int f(x)dx$$
Therefore,
$$\int f(x)dx=\frac 23 \sin(x) - \frac {f'} 9$$
Do you know any other way to calculate easily this integral?
| \begin{eqnarray*}
\int \cos^3(x) dx &=& \int (1-\sin^2(x)) \cos(x) dx = \int \cos(x) dx - \int \sin^2(x) \cos(x) dx \\
\int \cos(x) dx &=& \sin(x) + C_1
\end{eqnarray*}
Now, we need to integrate $\sin^2(x) \cos(x)$. We do this with the substitution $u = \sin(x)$.
\begin{eqnarray*}
\int \sin^2(x) \cos(x) dx &=& \int u^2 du = \frac{u^3}{3} + C_1 \\
\int \sin^2(x) \cos(x) dx &=& \frac{\sin^3(x)}{3} + C_1 \\
\int \cos^3(x) dx &=& \sin(x) - \frac{\sin^3(x)}{3} + C \\
\end{eqnarray*}
| {
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Quadratic Diophantine Equation$(x^2+x)(y^2-1)=240$ For whole numbers $x$ and $y$, $$x,y | (x^2+x)(y^2-1)= 240$$
Find the biggest and smallest value for $x-y$.
How do you proceed with such a question?
Are their formulas or something for that type of equation?
I'd appreciate any help.
| Notice that $y^2-1$ is adivisor of $240$;
also we know that $-1 \leq y^2-1$,
by checking throuh all divisors of $240$,
we get the folowing possibilities for $y^2-1$:
*
*$y^2-1=-1$ but there is no pair in this case.
*$y^2-1=0$ but there is no pair in this case.
*$y^2-1=3$ and $x^2+x=80$;
but notice that there is no solution to the equation $x^2+x=80$; so there is no pair $(x,y)$ for the main equation, in this case.
*$y^2-1=8$ and $x^2+x=30$;
which gives $(x,y)=(5,\pm3)$ and $(x,y)=(-6,\pm3)$.
*$y^2-1=15$ and $x^2+x=16$;
but notice that there is no solution to the equation $x^2+x=16$; so there is no pair $(x,y)$ for the main equation, in this case.
*$y^2-1=24$ and $x^2+x=10$;
but notice that there is no solution to the equation $x^2+x=10$; so there is no pair $(x,y)$ for the main equation, in this case.
*$y^2-1=48$; and $x^2+x=5$;
but notice that there is no solution to the equation $x^2+x=5$; so there is no pair $(x,y)$ for the main equation, in this case.
*$y^2-1=80$; and $x^2+x=3$;
but notice that there is no solution to the equation $x^2+x=3$; so there is no pair $(x,y)$ for the main equation, in this case.
*$y^2-1=120$ and $x^2+x=2$;
which gives $(x,y)=(1,\pm11)$ and $(x,y)=(-2,\pm11)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Prove that inequality $\sum _{cyc}\frac{a^2+b^2}{a+b}\le \frac{3\left(a^2+b^2+c^2\right)}{a+b+c}$
Let $a>0$,$b>0$ and $c>0$. Prove that:
$$\dfrac{a^2+b^2}{a+b}+\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}\le \dfrac{3\left(a^2+b^2+c^2\right)}{a+b+c}.$$
$$\Leftrightarrow \frac{(a^2+b^2)(a+b+c)}{a+b}+\frac{(b^2+c^2)(a+b+c)}{b+c}+\frac{(c^2+a^2)(a+b+c)}{c+a}\leq 3(a^2+b^2+c^2)$$
$\Leftrightarrow 2(a^2+b^2+c^2)+\frac{c(a^2+b^2)}{a+b}+\frac{a(b^2+c^2)}{b+c}+\frac{b(a^2+c^2)}{a+c}\leq 3(a^2+b^2+c^2)$
$\Leftrightarrow \frac{c(a^2+b^2)}{a+b}+\frac{a(b^2+c^2)}{b+c}+\frac{b(a^2+c^2)}{a+c}\leq a^2+b^2+c^2$
$\Leftrightarrow \frac{c(a+b)^2-2abc}{a+b}+\frac{a(b+c)^2-2abc}{b+c}+\frac{b(a+c)^2-2abc}{a+c}\leq a^2+b^2+c^2$
$\Leftrightarrow 2(ab+bc+ac)\leq a^2+b^2+c^2+2abc\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)$
I can't continue. Help me
| The following quantity is clearly positive
\begin{eqnarray*}
\sum_{perms} a^2 b (a-b)^2 \geq 0
\end{eqnarray*}
This can be rearranged to
\begin{eqnarray*}
3(a^2+b^2+c^2)(a+b)(b+c)(c+a) \geq (a+b+c) \left( \sum_{cyc} (a^2 +b^2)(c+a)(c+b) \right)
\end{eqnarray*}
Now divide this by $(a+b+c)(a+b)(b+c)(c+a)$ and we have
\begin{eqnarray*}
\frac{3(a^2+b^2+c^2)}{(a+b+c)} \geq \frac{a^2 +b^2}{a+b}+ \frac{b^2 +c^2}{b+c}+ \frac{c^2 +a^2}{c+a}.
\end{eqnarray*}
| {
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For what value of $k$ is one root of the equation
For which value of $k$ is one root of the equation $x^2+3x-6=k(x-1)^2$ double the other?
My Attempt:
$$x^2+3x-6=k(x-1)^2$$
$$x^2+3x-6=k(x^2-2x+1)$$
$$x^2+3x-6=kx^2-2kx+k$$
$$(1-k)x^2+(3+2k)x-(6+k)=0$$
| Hint: Let $a$ and $2a$ be the two roots.
Then
$$
-\dfrac{3+2k}{1-k}=3a,
\qquad
-\dfrac{6+k}{1-k}=2a^2
$$
| {
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"answer_id": 2
} |
Max and min of $f(x)=\sin^2{x}+\cos{x}+2$. My first try was to look for when $\cos{x}$ and $\sin{x}$ attain their min and max respectively, which is easy using the unit circle. So the minimum of $\sin{x}+\cos{x}$ has maximum at $\sqrt{2}$ and minimum at $-\sqrt{2}.$ But if the sine-term is squared, how do I find the minimum of $\sin^2{x}+\cos{x}$?
I then tried the differentiation method:
$$f'(x)=2\sin{x}\cos{x}-\sin{x}=(2\cos{x}-1)\sin{x}=0,$$
which in the interval $(0,2\pi)$, gave me the the roots $\left(\frac{\pi}{2},\frac{3\pi}{2},\frac{\pi}{3},\frac{5\pi}{3}\right)$. So
$$f(\pi/2)=3, \quad f(3\pi/2)=3, \quad f(\pi/3)=\frac{13}{4}, \quad f(5\pi/3)=\frac{13}{4}.$$
So $f_{\text{max}}=\frac{13}{4}$ and $f_{\text{min}}=3.$ But the minimum should be $1$.
| Credit to @iamwhoiam for suggesting this solution.
There's nothing wrong with your method - you just missed a root. When you find the roots of the first derivative, either $2 \cos x - 1$ or $\sin x$ should equal $0$, so we have:
First case:
$2 \cos x - 1 = 0$, $\cos x = \frac{1}{2}$.
Since $\cos(x)$ is an even function, $\cos(x) = \cos(-x)$. Since we know that one solution is $x = \frac{\pi}{3}$, $x$ can also be $\frac{5\pi}{3}$.
Second case:
$\sin x = 0, \pi$. Using the properties of $\sin x$ we know these are the only solutions in the range $[0,2\pi)$.
Therefore the solutions in the range $[0, 2\pi)$ are: $x = 0, \frac{\pi}{3}, \frac{5\pi}{3}, \pi$. Substituting each in the original equation gives $3, \frac{13}{4}$, $\frac{13}{4}$ and $1$ respectively. The minimum of these critical points is $1$, which is the minimum value of the original function.
| {
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Find $\lim\limits_{n \to \infty} \frac{\sqrt 1 + \sqrt 2 + \dots + \sqrt{n}}{n\sqrt{n}}$.
$$\lim_{n \to \infty} \dfrac{\sqrt 1 + \sqrt 2 + \dots + \sqrt{n}}{n\sqrt{n}}$$
$$\lim_{n \to \infty} \dfrac{\sqrt 1 + \sqrt 2 + \dots + \sqrt{n}}{n\sqrt{n}} =\lim_{n\to \infty} \dfrac1{n}\sum^{n}_{k = 1} \sqrt{\dfrac k n} $$
While searching this question I found,
Turning infinite sum into integral.
Like in the accepted answer I first compared my series to LRAM,
$$\int_a^b f(x)\,dx=\lim_{n\to\infty}\frac 1n\sum_{i=1}^n f\left(a+\frac{b-a}n i \right)$$
I got $a = 0$, $b = 1$ and $f(x) =\sqrt{x}$ so,
$$\int_0^1 \sqrt{x}\ dx = \dfrac2 3$$ should be the answer.
Is there any simpler method to do this sum ? I have not learnt this method to do infinite sums so I can't use it.
| picture......................................
$$ \frac{2}{3} n \sqrt n < \mbox{SUM} < \frac{2}{3} \left( \; (n+1) \sqrt {n+1} \; - \; 1 \right) \; < \; \frac{2}{3} \left( \; (n+1) ( 1 +\sqrt n) \; - \; 1 \right) = \frac{2}{3} \left( \; n \sqrt n + n + \sqrt n \right) $$
$$ \frac{2}{3} n \sqrt n < \mbox{SUM} < \frac{2}{3} \left( \; n \sqrt n + n + \sqrt n \right) $$
For anyone worried about the little estimate above,
$$ n + 1 < n + 2 \sqrt n + 1, $$
$$ \sqrt {n+1} \; \; < \; \; 1 + \sqrt n. $$
| {
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A.P.: Sum of numbers in a particular group Question: A series of odd positive integers are divided into the following groups $(1), (3,5), (7,9,11),\dotsm$ . Prove that the sum of numbers in the $n^{\text{th}}$ group is $n^3$.
My attempt:
It is observed that the number of terms in the $n^{\text{th}}$ group is equal to $n$. The common difference is same in all groups and is equal to 2. If we do not consider the grouping, then the resulting sequence $1,3,5,\dotsm$ is in A.P. with first term $t_1$ equal to 1 and common difference equal to 2. Now the first term of the first group is equal to $t_1=t_{1+0}$, that of the second group is equal to $t_2=t_{1+1}$, that of the third group is equal to $t_4=t_{1+1+2}$, that of the fourth group is equal to $t_7=t_{1+1+2+3}$ and so on. Thus the first term of the $n^{\text{th}}$ group is equal to $t_{1+S_{n-1}}=t_{1+\frac{(n-1)n}{2}}=t_{\frac{n^2-n+2}{2}}=t_1+(\frac{n^2-n+2}{2}-1)2=t_1+\frac{n^2-n}{2}\times2=t_1+n^2-n=1+n^2-n$
Here $S_{n-1}$ denotes the sum of first $(n-1)$ natural numbers.
Therefore the first term of the $n^{\text{th}}$ group is equal to $n^2-n+1$.
Sum of all numbers in the $n^{\text{th}}$ group $=S^{'}_n=\frac{n}{2}[2(n^2-n+1)+(n-1)2]=\frac{n}{2}[2n^2-2n+2+2n-2]=n^3$.
My problem: I am looking for other methods to prove this result.
| There are $\color{red}n$ terms in the $n$-th group of numbers.
The "middle" term (for groups with even terms, this refers to the average of the two centre-most terms) of each group is $\color{red}{n^2}$*, i.e.
$$\overset{\overset{\;1^2}{\downarrow}}{\underbrace{(1)}_{1\text{ term}}}, \overset{\overset{\;2^2}{\downarrow}}{\underbrace{(3,5)}_{2\text{ terms}}}, \overset{\overset{\;3^2}{\downarrow}}{\;\underbrace{(7,9,11)}_{3\text{ terms}}}, \overset{\overset{\;4^2}{\downarrow}}{\underbrace{(13,15,17,19)}_{4\text{ terms}}},\cdots$$
As each group is an AP, the sum of all numbers in the group is the number of terms multiplied by the "middle" term, i.e. $n\cdot n^2=\color{red}{n^3} $.
*It can be easily deduced that the numbers of each group are:
$$\begin{aligned}
&1^2\\
&2^2\pm 1\\
&3^2\pm 2\\
&4^2\pm 1,\pm 3\\
&5^2\pm2, \pm 4\\
&6^2\pm 1,\pm 3,\pm 5\\
&7^2\pm2, \pm4,\pm6\\
&\qquad \vdots\\
&n^2\pm \cdots, \pm (n-3),\pm (n-1)
\end{aligned}$$
| {
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"question_score": "3",
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complex number relations I can structure a solution using geometry but I am struggling to find on using algebra.
In an Argand diagram, the complex numbers $z, w$ and $z+w$ are represented by the points $P, Q$ and $S$ respectively.
Show that:
$$| z + w | ≥ | z | − | w |$$
Thanks for any help
| Let $z=x+yi$ and $w=a+bi$, where $\{x,y,a,b\}\subset\mathbb R$.
Thus, we need to prove that
$$\sqrt{(x+a)^2+(y+b)^2}+\sqrt{a^2+b^2}\geq\sqrt{x^2+y^2}$$ or
$$\sqrt{(a^2+b^2)((x+a)^2+(y+b)^2)}\geq-a^2-b^2-ax-by.$$
If $-a^2-b^2-ax-by<0$ then the inequality is obviously true, but for
$-a^2-b^2-ax-by\geq0$ it's enough to prove that
$$(a^2+b^2)((x+a)^2+(y+b)^2)\geq(a^2+b^2+ax+by)^2,$$ which is C-S:
$$(a^2+b^2)((x+a)^2+(y+b)^2)\geq(a(x+a)+b(y+b))^2.$$
Done!
If you don't like the C-S then for the proof of the last inequality we need to prove that
$$(a^2+b^2)(a^2+b^2+2(ax+by)+x^2+y^2)\geq(a^2+b^2)^2+2(a^2+b^2)(ax+by)+(ax+by)^2$$ or
$$(a^2+b^2)(x^2+y^2)\geq(ax+by)^2$$ or
$$(ay-bx)^2\geq0.$$
| {
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Find $\int \frac{\sqrt{2-3x}\,dx}{x^2+1}$ Find $$\int \frac{\sqrt{2-3x}\,dx}{x^2+1}$$
I used substitution $x=\frac{2}{3} \sin^2 y$ So we get $dx=\frac{4}{3} \sin y \cos y dy$
hence
$$I= \frac{36\sqrt{2}}{3} \int \frac{\sin y \cos^2 y\, dy}{4 \sin^4 y+9}$$
Now if again i use substitution $\cos y=t$ we get
$$I=-12\sqrt{2} \int \frac{t^2 \,dt}{4t^4-8t^2+13}$$
$$I=-12 \sqrt{2} \int \frac{dt}{4t^2+\frac{13}{t^2}-8}$$
Any other approach?
| hint: let $u = \sqrt{2-3x}\implies 2-3x=u^2\implies x = \dfrac{2-u^2}{3}, dx = -\dfrac{2udu}{3}$. Then do a fraction decomposition. It can be done but its not short !
| {
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Finding all $a,b,c$ such that $(ax)^2+(b+cy)^2 \leq 1$ whenever $x^2+y^2 \leq 1$. Suppose $x,y,a,b,c \in \mathbb R$ and suppose that $$(ax)^2+(b+cy)^2 \leq 1$$ for all $x,y$ such that $x^2+y^2 \leq 1$. What are the allowed values of $a,b,c$?
Plugging in $x=1$, $y=0$ we get $a^2+b^2 \leq 1$. Similarly for $x=0,y=1$ we get $(b+y)^2 \leq 1$. Are there any additional constraints imposed on $a,b,c$?
| I played around for a bit
and came to a point
where I didn't want
to go further.
I'll show what I did
in the hope that
someone else
might find this useful.
Assume that
$x$ and $y$
are on the boundary,
so
$x^2+y^2 = 1
$.
This becomes
$\begin{array}\\
1
&\ge a^2x^2+b^2+2bcy+c^2y^2\\
&=a^2(1-y^2)+b^2+2bcy+c^2y^2\\
&=a^2+b^2+2bcy+(c^2-a^2)y^2\\
\end{array}
$
or
$1-a^2-b^2
\ge 2bcy+(c^2-a^2)y^2
$.
The equality case for this is
$\begin{array}\\
y
&=\dfrac{-2bc\pm\sqrt{4b^2c^2+4(c^2-a^2)(1-a^2-b^2)}}{2(c^2-a^2)}\\
&=\dfrac{-bc\pm\sqrt{b^2c^2+(c^2-a^2)(1-a^2-b^2)}}{(c^2-a^2)}\\
&=\dfrac{-bc\pm\sqrt{a^4 + a^2 b^2 - a^2 c^2 - a^2 + c^2}}{(c^2-a^2)}\\
\end{array}
$
And that's where
I stop.
| {
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Find maximum of $f(x)=3\cos{2x}+4\sin{2x}.$ Find maximum of $f(x)=3\cos{2x}+4\sin{2x}.$ The shortest way to do this is to only consider $2x\in[0,2\pi)$, set $t=2x$ and note that
*
*Max$(\cos{t})=1$ for $t=0.$
*Max$(\sin{t})=1$ for $t=\frac{\pi}{2}.$
Max of these two functions added is when $t$ equals the angle exactly in the middle of $[0,\pi/2]$, which is $t=\pi/4.$ For this $t$ we have that $\cos{\pi/4}=\sin{\pi/4}=\sqrt{2}/2.$ So; $$f\left(\frac{\pi}{4}\right)=3\frac{\sqrt{2}}{2}+4\frac{\sqrt{2}}{2}=\frac{7\sqrt{2}}{2}.$$
Correct answer: $5$. My answer is slightly less than $5$. Why?
| With $\tan t=\dfrac43$ then $\cos t=\dfrac{1}{\sqrt{1+(\dfrac43)^2}}=\dfrac35$ so
\begin{align}
3\cos{2x}+4\sin{2x}
&=3(\cos{2x}+\tan t\sin{2x})\\
&=\dfrac{3}{\cos t}(\cos t\cos{2x}+\sin t\sin{2x})\\
&=\dfrac{3}{\cos t}\cos(t-2x)\\
&=5\cos(t-2x)\\
&\leqslant5
\end{align}
| {
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Finding $\int\frac{x}{\sqrt{3-2x-x^2}} dx$. I was looking for the integral of
$$\int\frac{x}{\sqrt{3-2x-x^2}} dx$$
My work:
$$\int \frac{x}{\sqrt{3-2x-x^2}} dx = \int \frac{x}{\sqrt{(3)+(-2x-x^2)}} dx $$
$$ = \int \frac{x}{\sqrt{(3)-(2x+x^2)}} dx $$
$$ = \int \frac{x}{\sqrt{(3)-(1+2x+x^2) +1}} dx $$
$$\int \frac{x}{\sqrt{3-2x-x^2}} dx = \int \frac{x}{\sqrt{(4)-(x+1)^2}} dx $$
Then I often remember this integral $\frac{u}{\sqrt{a^2 - x^2}} du$. So I modified the above integral to
look like the integral $\frac{u}{\sqrt{a^2 - x^2}}$.
$$\int \frac{x}{\sqrt{3-2x-x^2}} dx = \int \frac{x+1-1}{\sqrt{(4)-(x+1)^2}} dx $$
$$\int \frac{x}{\sqrt{3-2x-x^2}} dx = \int \frac{x+1}{\sqrt{(4)-(x+1)^2}} dx + \int \frac{-1}{\sqrt{(4)-(x+1)^2}} dx$$
I recognized the the last integral $\int \frac{-1}{\sqrt{(4)-(x+1)^2}} dx$ has the form $\int \frac{1}{\sqrt{a^2-u^2}} du$, where $a =2$ and $u = x+1$.
It's corresponding integral would be $\arcsin \left( \frac{u}{a}\right) + c$.
Evaluating $\int \frac{-1}{\sqrt{(4)-(x+1)^2}} dx$, it would be $-\arcsin \left( \frac{x+1}{2}\right)$
Here's the problem: I couldn't find the integral of $\int \frac{x+1}{\sqrt{(4)-(x+1)^2}} dx, $ because my Table of Integral doesn't show what is the
integral of $\frac{u}{\sqrt{a^2 - x^2}} du$.
How to evaluate the integral of $\frac{x}{\sqrt{3-2x-x^2}} dx$ properly?
| Integration strategy:
By adding/subtracting a constant to the numerator, you can let the derivative of the polynomial under the radical appear and you can integrate the ratio.
Then remains the inverse of the square root of this quadratic polynomial. By completing the square, you see that by translating by $1$ ($\to4-u^2$) and scaling the variable by $2$, you can reduce to $1-v^2$, giving an arc sine.
Hence the answer will be of the form
$$a\sqrt{3-2x-x^2}+b\arcsin\left(\frac{x+1}2\right).$$
By differentiating this expression (mentally) you can even guess that $a=-1$ to yield $x$ at the numerator. As this numerator will be $x+1$, the $+1$ is compensated with $b=-1$ (as $\arcsin'(u/2)=1/\sqrt{4-u^2}$ there is no extra factor).
| {
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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.
| Your proof is correct (fleablood's comment). Let me rephrase a bit:
$A(n):= n^2 +4$; $B(r,q) = 7 q^2 + 2rq;$
$A(n) = 7B(r,q) + (r^2 +4)$.
Fairly simple to show that:
$A(n)$ is divisible by $7 \iff $
$(r^2 +4)$ is divisible $ n.$
By inspection
$(r^2 +4) , r = 0,1,2,3,4,5,6,$
is not divisible by $7$.
$\Rightarrow$: $A(n)$ is not divisible by $7$.
| {
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How to split $\frac{x^3}{(x^2+5)^2}$ into partial fractions?
Split $\frac{x^3}{(x^2+5)^2}$ into partial fractions.
I have tried letting $$\frac{x^3}{(x^2+5)^2}=\frac{ax+b}{x^2+5}+\frac{cx^3+dx^2+ex+f}{(x^2+5)^2},$$ but this yields a system of only four simultaneous equations.
I have also tried looking at $$\frac{x^3}{(x^2+5)^2} = x \cdot \left( \frac{x}{x^2+5} \right) ^2,$$ but sadly in vain.
| If we format it into the partial fractions 'template':
$$\frac{x^3}{(x^2+5)^2}=\frac{Ax+B}{(x^2+5)}+\frac{Cx+D}{(x^2+5)^2}$$
Multiplying both sides by the denominator,
$$\frac{x^3\left(x^2+5\right)^2}{\left(x^2+5\right)^2}=\frac{\left(B+Ax\right)\left(x^2+5\right)^2}{x^2+5}+\frac{\left(D+Cx\right)\left(x^2+5\right)^2}{\left(x^2+5\right)^2}$$
Simplifying, we find that:
$$x^3=\left(B+Ax\right)\left(x^2+5\right)+Cx+D$$
so
$$x^3=Ax^3+Bx^2+5Ax+Cx+5B+D$$
$$x^3=Ax^3+Bx^2+x\left(5A+C\right)+\left(5B+D\right)$$
We have the system of equations
$$\begin{bmatrix}5B+D=0\\ 5AC=0\\ B=0\\ A=1\end{bmatrix}$$
Solving this gives $D=0,\:C=-5,\:B=0,\:A=1$.
Putting this in our original 'template' gives us:
$$\frac{x}{x^2+5}-\frac{5x}{\left(x^2+5\right)^2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to do this Orthogonality Integral? I am solving a question, where
$$ p_1(x) = x , Q_0(x) = \frac {\ln \frac {1+x} {1-x}}{2}$$
are solutions of legedre's differential equation corresponding to different eigen values.
I have to evaluate their orthogonality integral
$$ \int \limits_{-1}^{1} x \frac {\ln \frac {1+x} {1-x}}{2}$$
While doing this, I came to a point where I have to do
$$ \int \limits_{-1}^{1} \ln(1+x)dx $$
which leads to
$$ [(1+x)\ln(1+x) - (1+x)]_{-1}^{1} $$ which gives a term
$$ (1-1)\ln(1-1) -(1-1) $$
What is it's value ? Is it zero? The space here is Hilbert space.
| Ignoring limits,
$$
\int x\ln\left(\frac{1+x}{1-x}\right) dx \\
= \frac{1}{2}\int \frac{d}{dx}(x^2-1)\ln\left(\frac{1+x}{1-x}\right)dx \\
= \frac{1}{2}(x^2-1)\ln\left(\frac{1+x}{1-x}\right)-\frac{1}{2}\int(x^2-1)\frac{d}{dx}\ln\left(\frac{1+x}{1-x}\right)dx \\
= \frac{1}{2}(x^2-1)\ln\left(\frac{1+x}{1-x}\right)-\frac{1}{2}\int(x^2-1)\left[\frac{1}{1+x}+\frac{1}{1-x}\right]dx \\
= \frac{1}{2}(x^2-1)\ln\left(\frac{1+x}{1-x}\right)-\int(x^2-1)\frac{1}{1-x^2}dx \\
= \frac{1}{2}(x^2-1)\ln\left(\frac{1+x}{1-x}\right)+x+C.
$$
Evaluating the right side between $1^{-}$ and $(-1)^{+}$ gives 2. Therefore,
$$
\langle p_1,Q_0\rangle = \lim_{\epsilon\downarrow 0}\int_{-1+\epsilon}^{1-\epsilon}p_1(x)Q_0(x)dx = 1.
$$
| {
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Prove $\lim_{x\to 1} \frac{x+2}{x^2+1}=\frac{3}{2}$ The title is quite clear. I am required to prove $\lim_{x\to 1} \frac{x+2}{x^2+1}=\frac{3}{2}$ using the epsilon-delta definition of the limit.
Given any $\varepsilon \gt 0$, there exists a $\delta =$
Such that $0 \lt \lvert x-1 \rvert \lt \delta \Rightarrow \lvert\frac{x+2}{x^2+1} -\frac{3}{2} \rvert \lt\varepsilon$
$\frac{x+2}{x^2+1} -\frac{3}{2}= \frac{-3x^2+2x+1}{2(x^2+1)}$
I have managed to express both the numerator and the denominator as such
$-3x^2+2x+1 = -(x-1)(3x+1)\\
2(x^2+1) = 2x^2+2= 2(x-1)^2 + 4(x-1) +4$
Returning back to the fraction $\frac{-(x-1)(3x+1)}{2(x-1)^2 + 4(x-1) +4}$
I am unsure on how to continue and would really appreciate some guidance.
| Your fraction, in absolute value, equals
$$\left|\frac{(x-1)(3x+1)}{2(x^2+1)} \right|<\left|\frac{(x-1)(3x+1)}{x^2} \right|.$$
Now, if $|x-1|<\delta$, then $|x|<\delta+1$. So if $\delta\leq1$, then $|x|<2$ and $|x^2|<4$. So if we choose $\delta=\min\{1,\frac{7\epsilon}{4}\}$, then the above inequality is less than $\epsilon$, which is what we wanted to show.
| {
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Prove the sequence $\frac{n^2}{n^2-n-5}$ converges I know the basic procedure for proving convergence of sequences, but I'm having difficulty reducing this sequence. I believe the sequence converges to 1, so I've set up the following two approaches:
1:
$\left|\frac{n^2}{n^2 - n - 5} - 1\right| = \left|\frac{n^2}{n^2 - n - 5} -
\frac{n^2 - n - 5}{n^2 - n - 5}\right| = \frac{n + 5}{n^2 - n - 5}$
2:
$\left|\frac{n^2}{n^2-n-5} - 1\right| = \left|\frac{1}{1-\frac{1}{n}-\frac{5}{n^2}} - \frac{1-\frac{1}{n}-\frac{5}{n^2}}{1-\frac{1}{n}-\frac{5}{n^2}}\right| = \frac{\frac{1}{n}+\frac{5}{n^2}}{1-\frac{1}{n}-\frac{5}{n^2}}$
Both assume $n \geq 3$. Neither of these approaches appear to be leading me in the right direction. I can technically solve these for $f(n) < f(\epsilon)$, but both result in something very nasty, and I feel like I'm missing something simple.
| Your first approach should ring a bell since the nominator is first degree while the denominator is second degree.
You can do the formals by applying an estimate for the nominator and denominator. For $n>5$ we have that $0 < n+5 < 2n$ and $n^2 - n - 5 > n^2 - n^2/5 - n^2/5 = 3n^2/5$. This means that for $n>5$ we have
$$0<{n+5\over n^2-n-5}< {2n\over 3n^2/5} = {10\over 3}{1\over n}\to 0$$
Your second approach is also workable as the nominator $\to 0$ and the denominator $\to 1$ os $n\to \infty$. If you want/need to do $\delta$-$\epsilon$ you'd have to be a bit more elaborate: you estimate the denominator to be larger than a positive constant and then estimate the nominator with an factor of $1/n$.
| {
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Is it possible to have a 3rd number system based on division by zero? Is it possible in mathematics to use a third number line based on division by zero; in addition to the real and imaginary number lines?
This is because some solutions blow up when there is a division by zero. Would it be possible to solve them with this new number line?
$\therefore$ on the z axis we would have $\frac{1}{0}$ , $\frac{2}{0}$ , $\frac{3}{0}$ , etc. where, $p = \frac{1}{0}$ and $p \cdot 0 = 1$ .
division by zero graph
Is this a viable number system?
A similar question to this one: Is there a third dimension of numbers?
| One way to approach this problem could be to imagine zero as a very small number.
So to start with, let us define,
$$small \to 0$$
$$\therefore$$
instead of $$ (2 \cdot 0) \cdot p = 1 $$
as above @Jim H. Use:
$$2 \cdot small \cdot \frac{1}{small} = 2$$
As
$$(2 \cdot small) \cdot \frac{1}{small} \neq small \cdot \frac{1}{small} = 1$$
$$= 2 \cdot \frac{small}{small} \cdot 1 = 2$$
Do not use $(2 \cdot small) \to 0$ during multiplication and division, otherwise it will produce nonsense such as 2 = 1.
$\underline {The\, p\, number\, system}$
If a very small number, $$s \to 0$$ $$p = \frac{1}{s} \to \infty$$
then, $$\frac{p}{p} = \frac{1}{s} \cdot \frac{s}{1} = 1$$
$$\therefore p \cdot p = \frac{1}{s} \cdot \frac{1}{s} = \frac{1}{s^2} = p^2$$
Also, we could invent a p number,
$$4 + 3p$$
multiply it by s,
$$(4 + 3p) \cdot s = 4 \cdot s + 3 \cdot \frac{1}{s} \cdot s$$
$$4 \cdot s \to 0$$
$$\therefore (4 + 3p) \cdot s = 3$$
$\underline {Adding\, and\, multiplying\, fractions\, in\, p\, number\,system}$
$$p + p = \frac{1}{s} + \frac{1}{s} = \frac{1 + 1}{s} = 2p$$
$$p - p = \frac{1}{s} - \frac{1}{s} = \frac{1 - 1}{s} = \frac{0}{s} = 0$$
$$\frac{2}{3} + \frac{p}{3} = \frac{2 + p}{3}$$
Instead of $$\frac{0}{0} = 0$$ use $$\frac{s}{s} = 1$$
$$\frac{2}{3} + \frac{s}{s} = \frac{2}{3} + 1 = \frac{5}{3}$$
$\underline {Adding\, two\, p\, numbers}$
$$(3 + 2p) + (5 + p) = 8 + 3p$$
| {
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If $a,b$ are positive integers and $x^2+y^2\leq 1$ then find the maximum of $ax+by$ without differentiation. If $x^2+y^2\leq 1$ then maximum of $ax+by$
Here what I have done so far.
Let $ax+by=k$ . Thus $by=k-ax$.
So we can have that $$b^2x^2+(k-ax)^2 \leq b^2$$
$$b^2x^2+k^2-2akx +a^2x^2-b^2\leq 0 $$
By re-writing as a quadratic of $x$ ,
$$(a^2+b^2)x^2-2akx +k^2-b^2\leq 0 $$
Since $a^2+b^2$ is positive , above quadratic has a minimum. Thus it to be negeative it must have roots.
So $$(-2ak)^2-4(a^2+b^2)(k^2-b^2) \geq 0$$
$$a^2k^2-(a^2+b^2)k^2+(a^2+b^2)b^2 \geq 0$$
$$(a^2+b^2) \geq k^2$$
So maximum of $k$ is $\sqrt{a^2+b^2}$
Is this correct ?
If it is correct any shorter method ? Thanks in advance.
| We can solve this by geometry,
$$x^2+y^2\le 1$$ is area bounded by circle with centre at origin and radius 1
&
$$ax+by=c$$
Is a line
We want to find max value of c,
Note: a and b are positive so slope is negative.
We want to find (x,y) such that it lies in circle and on line but also that line gives us maximum intersect at y-axis (which is $=\frac{c}{b}$)
If i draw the circle, and couple of lines
In above picture we see that y-intersect is max when line is tangent to circle
Using basic coordinate knowledge, we can deduce the web of line such that it is tangent to circle
and then get c.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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} |
Eavaluating the roots of quadratic equation If $b>a , c>0$
Determine the intervals that the roots of the equation
$(x-a)(x-b) -c =0$ belong to
My work is to get the values of the roots in terms of a , b and c using the general form but i couldn't determine those intervals
| The given equation is $x^2-(a+b)x+ab-c=0$. But $$\Delta=(a+b)^2-4(ab-c)=(a-b)^2+4c>0.$$ So the equation has two distinct real roots. The two roots are $\frac{(a+b)+\sqrt{(a-b)^2+4c}}{2}$ and $\frac{(a+b)-\sqrt{(a-b)^2+4c}}{2}$. Hence the roots are in the interval $[\frac{(a+b)-\sqrt{(a-b)^2+4c}}{2},\frac{(a+b)+\sqrt{(a-b)^2+4c}}{2}]$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integration changing limits - does the question have an error? I am asking about changing the limits of integration.
I have the following integral to evaluate -
$$\int_2^{3}\frac{1}{(x^2-1)^{\frac{3}{2}}}dx$$ using the substitution $x = sec \theta$.
The problem states
Use the substitution to change the limits into the form $\int_a^b$ where $a$ and $b$ are multiples of $\pi$.
Now, this is what I did.
$$ x= \sec \theta$$
$$\frac{dx}{d\theta} = \sec\theta \tan\theta$$
$$dx = sec\theta tan\theta \ d\theta$$
$$\begin{align}\int_2^{3}\frac{1}{(x^2-1)^{\frac{3}{2}}}\,dx \\
&= \int\frac{1}{(\sec^2\theta-1)^{\frac{3}{2}}}\sec\theta \tan\theta \,d\theta \\
&= \int\frac{\sec\theta \tan\theta}{(\tan^2\theta)^{\frac{3}{2}}} \,d\theta \\
&= \int\frac{\sec\theta \tan\theta}{\tan^3\theta} \, d\theta \\
&= \int\frac{\sec\theta}{\tan^2\theta} \, d\theta \\
&= \int\frac{\cos\theta}{\sin^2\theta} \, d\theta \\
&= \int \csc\theta \cot\theta \, d\theta
\end{align}$$
But here is my problem.
I know that when $x = 2$,
$$2 = \sec \theta$$
$$\frac{1}{2} = \cos \theta$$
$$\frac{\pi}{3} = \theta$$
but when $x = 3$
$$3 = \sec \theta$$
$$\frac{1}{3} = \cos \theta$$
$$\arccos\left(\frac{1}{3}\right) = \theta = $$
but this does not give me a definite result in $\pi$.
The book says the following -
where $\arccos\left(\frac{1}{3}\right) = \frac{\pi}{3}$
Am I the only one or is the book wrong in this instance?
| For simplicity, write $a=\arccos\frac{1}{2}$ and $b=\arccos\frac{1}{3}$. One can take care of them at the end.
It is true that $a=\pi/3$, but it's definitely wrong that $b=\pi/3$. Actually, $b\approx1.230959$, but you won't need that.
The integral becomes
$$
\int_a^b-\frac{1}{(\sec^2\theta-1)^{3/2}}\sec^2\theta\sin\theta\,d\theta
$$
Now it's better to write the integrand in terms of sine and cosine, recalling that the interval of integration is contained in $(0,\pi/2)$. Thus
$$
\sec^2\theta-1=
\frac{1-\cos^2\theta}{\cos^2\theta}=
\frac{\sin^2\theta}{\cos^2\theta}
$$
so the integrand is
$$
-\frac{\cos^3\theta}{\sin^3\theta}\frac{1}{\cos^2\theta}\sin\theta=
-\frac{\cos\theta}{\sin^2\theta}
$$
Thus we get
$$
\int_2^{3}\frac{1}{(x^2-1)^{3/2}}\,dx=
\int_a^b-\frac{\cos\theta}{\sin^2\theta}\,d\theta=
\int_a^b-\frac{1}{\sin^2\theta}\,d(\sin\theta)=
\left[\frac{1}{\sin\theta}\right]_a^b
$$
Since $a=\arccos(1/2)$ and $b=\arccos(1/3)$, we get
$$
\sin a=\sqrt{1-\cos^2a}=\frac{\sqrt{3}}{2}
\qquad
\sin b=\sqrt{1-\cos^2b}=\frac{2\sqrt{2}}{3}
$$
with no “sign uncertainty”, because $a$ and $b$ lie in $(0,\pi/2)$.
Thus the integral is
$$
\frac{2\sqrt{2}}{3}-\frac{\sqrt{3}}{2}
$$
| {
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Probability, without replacement An urn contains 11 balls, of which 4 are white. Three players $A$, $B$, and $C$, successively draw from the urn. $A$ first, then $B$, then $C$, then $A$ again, and so on. The winner is the first one to draw a white ball. Find the probability of winning for each player, assuming the balls that are withdrawn are not replaced.
Find $\mathbf{P}(A \text{ wins})$
Find $\mathbf{P}(B \text{ wins})$
Find $\mathbf{P}(C \text{ wins})$
I have completed the same problem with replacement of the marbles. I do not know how to start without replacement.
| Assume $4$ white and $7$ red balls in the box. Then:
$$P(A)=P(W_A)+P(R_AR_BR_CW_A)+P(R_AR_BR_CR_AR_BR_CW_A)=\frac{4}{11}+\frac{7\cdot 6\cdot 5\cdot 4}{11\cdot 10\cdot 9\cdot 8}+\frac{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 4}{11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5}\approx 0.4818.$$
Similarly for $P(B), P(C)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $ \forall n \ge 4$, $n^{3} + n < 3^{n}$
Prove that $\forall n \ge 4$, $n^{3} + n < 3^{n}$
My attempt: Base case is trivial.
Suppose $ \ n \ge 4$, $n^{3} + n < 3^{n}$. Then,
$$ (n+1)^{3} + (n+1) = n^3 + n + 3n^2 + 3n + 1 +1 < 3^{n} + 3n^2 + 3n + 2 \\< 3\cdot 3^{n} + 3n^2 + 3n +2.$$
Not sure how to get rid of $3n^2 + 3n +2$.
| By the binomial theorem, $3^n = (1+2)^n \ge \binom{n}{1}+ \binom{n}{4}2^4$ for $n \ge 4$.
So, it's enough to prove that $\binom{n}{1}+ \binom{n}{4}2^4 > n^3+n$ (*).
This simplifies to $2 n^3 - 15 n^2 + 22 n - 12 > 0$.
The largest critical point of $2 x^3 - 15 x^2 + 22 x - 12$ is at $x \approx 4.1$. Also (*) is true for $n=6$. Therefore, (*) is true for all $n \ge 6$.
The cases $n=4$ and $n=5$ are easily handled separately.
| {
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$1/\cos x$ integration Question: The aim of this question is to integrate $\frac{1}{\cos x}$
a. Write an integral as $\int \frac{\cos x}{\cos^2(x)}dx$ [DONE]
b. Use $u= \sin x$ [DONE]
c. Use partial fraction to integrate [DONE]
d. Multiply a fraction top and bottom by $1+\sin x$ to simplify it.
My answer to part c is $\frac{\ln(1+\sin x)}{2} - \frac{\ln(1-\sin x)}{2} +c$ but I could not simplify it by multiplyting $1 + \sin x$. Is there a way I could do it?
| From your answer in part (c)
$ A =\frac{1}{2} \ln(1+\sin x ) - \frac{1}{2}\ln (1-\sin x) $
$ = \ln (\sqrt{1+\sin x}) - \ln (\sqrt{1-\sin x}) $
$ = \ln\left(\frac{\sqrt{1+\sin x }}{\sqrt{1-\sin x}}\right) $
$= \ln\left(\sqrt{\frac{1+\sin x }{1-\sin x} \cdot \frac{1+\sin x}{1+\sin x}}\right) $
$= \ln\left(\frac{1+\sin x }{\sqrt{1-\sin^2 x}}\right)$
$= \ln\left(\frac{1+\sin x }{\cos x}\right)$
$= \ln\left(\sec x + \tan x \right) .$
| {
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Cantor's "original" proof of $\mathbb{R}$ uncountable Found this problem in a set of notes I'm self studying from. I believe there might be an error.
"Consider $\{x_n\}_{n∈N}$ a sequence in $[0, 1]$. Construct new sequences $a_n$ and $b_n$ as follows: $a_0$ and $b_0$ are the first two elements in the sequence ${x_n}$ such that $a_0 < b_0$. For $n ≥ 0$, take the first two unique elements of $\{x_n\}$ that lie in $(a_n, b_n)$, and call then $a_{n+1}$ and $b_{n+1}$, where $a_{n+1} < b_{n+1}$ (if there are none, simply let the sequence
stagnate at $(a_n, b_n)$). Prove that $\bigcap_{n≥0}(a_n, b_n)$ is nonempty and does not appear in ${x_n}$ (Hint: any element $x_n$ has the property that either there exists m such that $x_n ≤ a_m$ or $x_n ≥ b_m$). Conclude that it is not possible to enumerate $[0, 1]$."
Shouldn't it be : Prove that $\bigcap_{n≥0}[a_n, b_n]$ is nonempty? Because otherwise, we can consider the following sequence:
$(\frac{1}{2} - \frac{1}{3}, \frac{1}{2} + \frac{1}{3} , \frac{1}{2} - \frac{1}{4}, \frac{1}{2} + \frac{1}{4} , \frac{1}{2} - \frac{1}{5}, \frac{1}{2} + \frac{1}{5}, \dots ,\frac{1}{2} - \frac{1}{n}, \frac{1}{2} + \frac{1}{n}, \dots)$
This would give us the nested open intervals $\bigcap_{n \geq 3} \left( \frac{1}{2} - \frac{1}{n}, \frac{1}{2} + \frac{1}{n} \right) = \emptyset$
| Nope, it is correct as stated. Your intersection $\bigcap_{n \geq 3} \left( \frac{1}{2} - \frac{1}{n}, \frac{1}{2} + \frac{1}{n} \right)$ is actually not empty, since it contains $\frac{1}{2}$. Indeed, for any $n\geq 3$, $\frac{1}{2}-\frac{1}{n}<\frac{1}{2}<\frac{1}{2}+\frac{1}{n}$.
(That said, the proof would work just as well if you used $[a_n,b_n]$ instead of $(a_n,b_n)$, though you might have to say things a bit differently in the case that your sequence stagnates.)
| {
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Prove that for $n \ge 2$ the follow inequality holds $\frac {4^n}{n+1} \lt \frac{(2n)!}{(n!)^2}$. As already shown above I need to prove that $$\frac {4^n}{n+1} \lt \frac{(2n)!}{(n!)^2}\qquad \forall n \ge 2.$$
What I've come up with is the following:
$\underline{n=2:}\qquad$ $$\frac {16}{3} \lt 6$$
$\underline{n=k:}\qquad $$$\frac {4^k}{k+1} \lt \frac{(2k)!}{(k!)^2} $$
$\underline{n+1}:$ $$\frac {4^{k+1}}{k+2} \lt \frac{(2(k+1))!}{((k+1)!)^2} $$
$$\frac{4^k \cdot 4}{k+2} \lt \frac {(2k+2)!}{(k!)^2\cdot(k+1)\cdot(k+1)}$$
$$\frac{4^k \cdot 4}{k+2} \lt \frac{(k+1)\cdot \ldots \cdot (k + (k-1))\cdot{2k} \cdot (2k+1)\cdot(2k+2)}{(k!)^\require{enclose}\enclose{updiagonalstrike}2\cdot(k+1)\cdot(k+1)}$$
$$\frac{4^k \cdot 4}{k+2} \lt 2 \cdot\frac{\require{enclose}\enclose{updiagonalstrike}{(k+1)}\cdot \ldots \cdot (k + (k-1))\cdot{2k} \cdot (2k+1)\cdot\require{enclose}\enclose{updiagonalstrike}{(k+1)}}{(k!)\cdot\require{enclose}\enclose{updiagonalstrike}{(k+1)}\cdot\require{enclose}\enclose{updiagonalstrike}{(k+1)}}$$
$$\frac{4^k \cdot 4}{k+2} \lt 2 \cdot\frac{{(k+2)}\cdot \ldots \cdot (k + (k-1))\cdot{2k} \cdot (2k+1)}{(k!)}$$
From here on I'm stuck. Can somebody help me please?
| As your first steps of mathematical induction are right, I'll just perform a proof of inductive thesis for $n=k+1$.
First notice, that for $k>1$
$$\frac{4(k+1)}{(k+2)} < \frac{(2k+2)(2k+1)}{(k+1)^2}$$
proof:
$$(k-1)k(k+1)^2(k+2)>0\\
\frac{k(k-1)}{(k+1)^2(k+2)}>0\\
\frac{4k^2-4k}{(k+1)^2(k+2)}>0\\
\frac{-4k^2+4k}{(k+1)^2(k+2)}<0\\
\frac{4(k^3+3k^2+3k+1)-(4k^3+16k^2+8k+4)}{(k+1)^2(k+2)}<0\\
\frac{4(k+1)^3}{(k+1)^2(k+2)}<\frac{4k^3+16k^2+8k+4}{(k+1)^2(k+2)}\\
\frac{4(k+1)^2}{(k+1)^2(k+2)}<\frac{(2k+2)(2k+1)(k+2)}{(k+1)^2(k+2)}\\
\frac{4(k+1)}{(k+2)}<\frac{(2k+2)(2k+1)}{(k+1)^2}$$
Thus, since $\frac{4(k+1)}{(k+2)}$ and $\frac{4^k}{k+1}$ are both positive, we have:
$$\frac{4^{k+1}}{k+2}=\frac{4(k+1)}{(k+2)}\frac{4^k}{k+1} < \frac{4(k+1)}{(k+2)}\frac{(2k)!}{(k!)^2}<\frac{(2k+2)(2k+1)}{(k+1)^2}\frac{(2k)!}{(k!)^2} = \frac{(2k+2)!}{(((k+1)!)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2437448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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} |
Find all the points that lie $ 3$ units from each of the points $ (2,0,0), (0,2,0), \text{ and }(0,0,2)$
I calculated the result $$ \{(x,y,z) \in\mathbb{R^3}: x=y=z\}.$$
I'm wondering whether I did this problem correctly and if I did how to draw the set of solutions.
I used the euclidean distance formula with the square root (the euclidean norm) and set up three equations for each point $(2,0,0)$, $(0,2,0)$ and $(0,0,2)$ that made sure the distance from arbitrary $(x,y,z)$ to these three points equal to $3$ and arrived at the solution set. –
| To avoid forgetting solutions it is best to keep the whole system together.
$\begin{cases}
(x-2)^2+y^2+z^2=9\\
x^2+(y-2)^2+z^2=9\\
x^2+y^2+(z-2)^2=9\\
\end{cases}\iff\begin{cases}
(-4x+4)+x^2+y^2+z^2=9\\
(-4y+4)+x^2+y^2+z^2=9\\
(-4z+4)+x^2+y^2+z^2=9\\
\end{cases}$
Now replacing row $(2)$ by $(2)-(1)$ you get $4x=4y$ and replacing row $(3)$ by $(3)-(1)$ you get $4x=4z$.
But you still have to carry equation $(1)$ for the system to stay complete. This is probably what you forget to do in your solution.
Or maybe you also replaced $(1)$ by a combination of rows, and lost information. In Gauss elimination at least one row should stay untouched at each step.
$\begin{cases} (-4x+4)+x^2+y^2+z^2=9\\ x=y\\ y=z\\ \end{cases} \iff
\begin{cases} (-4x+4)+3x^2=9\\ x=y\\ y=z\\ \end{cases}\iff
\begin{cases} 3x^2-4x-5=0\\ x=y\\ y=z\\ \end{cases}$
Which solves to $\displaystyle x=y=z=\frac{2\pm\sqrt{19}}3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2439824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Prove the limit $\lim_{x\to 1+}\frac{1}{\sqrt{x}}=1$, using epsilon-delta definition. $$\lim_{x\to 1+}\frac{1}{\sqrt{x}}=1$$
The proof that I have:
Let $\varepsilon > 0$, we must show that
$$\exists \delta >0: 0<x-1<\delta \Rightarrow \left | \frac{1}{\sqrt{x}}-1\right|<\epsilon$$
So usually, when doing these $\varepsilon,\delta$ proofs, I would write $\ldots 0<|x-1|<\delta \ldots$ , but as $x\to 1+$, $x-1$ should always be greater than $0$. Is that correct? Can I just take $|x-1|=x-1$?
We see that
$$ \left | \frac{1}{\sqrt{x}}-1\right|= \left | \frac{1-\sqrt{x}}{\sqrt{x}}\right|=\left | \frac{-(\sqrt{x}-1)}{\sqrt{x}}\right|=\left | \frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}+1)}\right|=\left | \frac{x-1}{\sqrt{x}(\sqrt{x}+1)}\right|\left(<\varepsilon\right)$$
We know that $0<x-1<\delta$. But we don't know the estimation for $\frac{1}{\sqrt{x}(\sqrt{x}+1)}$. Let $\delta \leq 1$, so
$$
\begin{align}
0<&x-1<\delta\leq 1\\
0<&x-1<1\\
1<&x<2\\
1<&\sqrt{x}<\sqrt{2}\\
\frac{1}{\sqrt{2}}<&\frac{1}{\sqrt{x}}<1
\end{align}
$$
If I had $|x-1|<\delta$, I would take $\delta\leq\frac{1}{2}$ and get
$$
\begin{align}
&|x-1|<\frac{1}{2}\\
-\frac{1}{2}<&x-1<\frac{1}{2}\\
\frac{1}{2}<&x<\frac{3}{2}\\
\frac{1}{\sqrt{2}}<&\sqrt{x}<\frac{\sqrt{3}}{\sqrt{2}}\\
&\vdots
\end{align}
$$
The estimation for $\frac{1}{\sqrt{x}}$ is known, let's do the same for $\frac{1}{\sqrt{x}+1}$:
$$
\begin{align}
1<&\sqrt{x}<\sqrt{2}\\
2<&\sqrt{x}+1<\sqrt{2}+1\\
\frac{1}{\sqrt{2}+1}<&\frac{1}{\sqrt{x}+1}<\frac{1}{2}
\end{align}
$$
So the estimation for $\frac{1}{\sqrt{x}(\sqrt{x}+1)}$ is
$$
\frac{1}{\sqrt{x}(\sqrt{x}+1)}<\frac{1}{2}
$$
Finally we get
$$
\left | \frac{x-1}{\sqrt{x}(\sqrt{x}+1)}\right|<\frac{\delta}{2}\leq \varepsilon $$
$$\delta:=\min\{1,2\varepsilon\}$$
How to prove it? Is my approach correct? Any pointers, when doing one-sided limit proofs?
| Coming from the right, you just have to show that $1-\frac{1}{\sqrt x} \longrightarrow$ as $x \longrightarrow 1$. To show the limit, you could define $y=x-1$.
Then $$1-\frac{1}{\sqrt x} = \frac{\sqrt{1+y} - 1}{\sqrt{1+y}}$$.
Note that when $y < \delta$
$$\frac{\sqrt{1+\delta} - 1}{\sqrt{1+\delta}} \leq \sqrt{1+\delta} - 1 \leq 1+\frac{\delta}{2} - 1 = \frac{\delta}{2}$$ where the last inequality is by noting that the concave square root function lies below its linear approximation at $x=1$. So, set $ \delta = 2\epsilon $ to meet the epsilon-delta definition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2441520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Probability of a target being hit I don't have the answer of the following question so I wanted to cross check my solution.
$A$ can hit a target 3 times in 5 shots, $B$ 2 times in 5 shots and $C$ 3 times in 4 shots. Find the probability of the target being hit at all when all of them try.
my method
P(target being hit) = $$ \frac{3}{5}*\frac{3}{5}*\frac{1}{4} + \frac{3}{5}*\frac{3}{5}*\frac{3}{4} + \frac{3}{5}*\frac{2}{5}*\frac{1}{4} + \frac{3}{5}*\frac{2}{5}*\frac{3}{4} + \frac{2}{5}*\frac{3}{5}*\frac{3}{4} + \frac{2}{5}*\frac{2}{5}*\frac{1}{4} = 0.82$$
$$ = HMM + HMH + HHM + HHH + MMH + MHM$$
where, H = hit & M = miss
| There is also a way we can use we take all the probable situations and subtract the ones that we don't succeed hitting;
For $A$ it is $\dfrac{2}{5}$
For $B$ it is $\dfrac{3}{5}$
And for $C$ it is $\dfrac{1}{4}$
$$1-\frac{2}{5}\frac{3}{5}\frac{1}{4}=\frac{47}{50}$$
I accidentally wrote this answer without seeing the first one, but I am not going to delete my effort. (Sorry:()
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
$\frac{a^3}{b^2} + \frac{b^3}{c^2} + \frac{c^3}{a^2} \geq 3 \, \frac{a^2 + b^2 + c^2}{a + b + c}$ Proposition
For any positive numbers $a$, $b$, and $c$,
\begin{equation*}
\frac{a^3}{b^2} + \frac{b^3}{c^2} + \frac{c^3}{a^2}
\geq 3 \, \frac{a^2 + b^2 + c^2}{a + b + c} .
\end{equation*}
I am requesting an elementary, algebraic explanation to this inequality. (I suppose the condition for equality is that $a = b = c$.) I am not familiar with symmetric inequalities in three variables. I would appreciate any references.
| Actually this is not a symmetric inequality, since the LHS is cyclic but not symmetric.
Anyway, here's the proof I've come up with. From Cauchy-Schwarz on $\displaystyle \left(\frac{a^2}{b\sqrt{a}}, \: \frac{b^2}{c\sqrt{b}}, \: \frac{c^2}{a\sqrt{c}}\right)$ and $\displaystyle \left(b\sqrt{a}, \: c\sqrt{b}, \: a\sqrt{c}\right)$ we get $$\left(\sum_{cyc} \frac{a^3}{b^2}\right)\left(\sum_{cyc} ab^2\right) \ge (a^2 + b^2 + c^2)^2 \implies \frac{a^3}{b^2} + \frac{b^3}{c^2} + \frac{c^3}{a^2} \ge \frac{(a^2 + b^2 + c^2)^2}{ab^2 + bc^2 + ca^2}$$
Hence it suffices to show that the inequality $\displaystyle (a + b + c)(a^2 + b^2 + c^2) \ge 3(ab^2 + bc^2 + ca^2)$ holds. Thi can be proved rearranging the terms on the left and applying AM-GM: $$(a + b + c)(a^2 + b^2 + c^2) = \sum_{cyc} (b^3 + a^2b + ab^2) \ge \sum_{cyc} (2\sqrt{b^3 \cdot a^2b} + ab^2) = \\ = 3(ab^2 + bc^2 + ca^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2446312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find all $4 \times 4$ real matrices such that $A^3=I$ Find all $4 \times 4$ real matrices such that $A^3=I$.
The minimal polynomial must divide $x^3-1$. Since the matrix is real, the minimal polynomial must be either $x-1$ or $x^3-1$ (i.e., if it contains one of the complex roots of unity, it must contain the conjugate). Of course, we already know the identity matrix satisfies this equation. If $A$ is any other non-identity real matrix satisfying the above properties, then its characteristic polynomial must be $(x-1)(x^3-1)$, by similar reasoning above. This is as far as I could get.
edit: I missed a few cases, as pointed out in the comments. The minimal polynomial can also be $x^2+x+1$ which yields additional possible characteristic polynomials.
| Since $p(A) = 0$ where $p(x) = x^3 - 1 = (x-1) (x^2 + x + 1)$, and $x-1$ and $x^2 + x + 1$ are irreducible over $\mathbb{R}[x]$, then $\mathbb{R}^4$ can be decomposed as a direct sum of cyclic subspaces for $A$ corresponding to the polynomials $x-1$ and $x^2 + x + 1$. Now, we know that a cyclic subspace with annihilator $x-1$ has matrix representation of the restricted operator equal to $\begin{bmatrix} 1 \end{bmatrix}$; and similarly, a cyclic subspace with annihilator $x^2 + x + 1$ has matrix representation of the restricted operator equal to $\begin{bmatrix} 0 & -1 \\ 1 & -1 \end{bmatrix}$. Therefore, any $4 \times 4$ matrices $A$ with $A^3 = I$ is similar to one of:
$$ \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix},
\begin{bmatrix} 0 & -1 & 0 & 0 \\ 1 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix},
\begin{bmatrix} 0 & -1 & 0 & 0 \\ 1 & -1 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & -1 \end{bmatrix}. $$
(And conversely, it's clear that any matrix similar to one of these matrices satisfies $A^3 = I$.)
| {
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"url": "https://math.stackexchange.com/questions/2446574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Prove that $\frac{ab}{a+b} + \frac{cd}{c+d} \leq \frac{(a+c)(b+d)}{a+b+c+d}$ $$\frac{ab}{a+b} + \frac{cd}{c+d} \leq \frac{(a+c)(b+d)}{a+b+c+d}$$
I tried applying a.m. g.m inequality to l.h.s and tried to find upper bound for l.h.s and lower bound for r.h.s but i am not getting answer .
| I suppose that $a,b,c,d$ are $>0$. An idea is to put
$$F(x)=\frac{(x+c)(b+d)}{x+b+c+d}-\frac{xb}{x+b}-\frac{cd}{c+d}$$
and to compute the derivative:
$$F^{\prime}(x)=\frac{(b+d)^2}{(x+b+c+d)^2}-\frac{b^2}{(x+b)^2}$$
This show that the minimum of $F$ on $]0,+\infty[$ is obtained for $x=bc/d$, and to finish you have to compute this minimum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2448601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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} |
$a^2b^2 + b^2c^2 + c^2d^2 + d^2a^2$ can be expressed as the sum of two squares in at least two different ways The question: Every expression of the form $a^2b^2 + b^2c^2 + c^2d^2 + d^2a^2$ can be expressed as the sum of two squares in at least two different ways. Find any one of the three possible ordered pairs of
positive integers $(x, y)$, with $x > y$, that satisfies $x^2 + y^2 = 44^2 \cdot 10^2 + 10^2 \cdot 33^2 + 33^2 \cdot 5^2 + 5^2 \cdot 44^2$.
What I found: From factorizing, we see that the first expression is just $(a^2 + c^2)(b^2 + d^2)$. $1 = (0 + 1)(0 + 1)$ and $2 = (0 + 1)(1 + 1)$ are of this form, but they cannot be expressed two different ways. Also, if $2$ is the sum of two squares, then it's also implied that $4 = 2 \cdot 2$, $8 = 4 \cdot 2$, and generally all powers of two should be expressible as a sum of two squares two different ways, but $8$ is another counterexample.
Am I misinterpreting the question, or is it wrong? Also, how do I actually tackle finding $x$ and $y$ for the last sentence of the question?
| $$\text{First way} $$
$$a^2b^2 + b^2c^2 + c^2d^2 + d^2a^2=\color{red}{a^2b^2 + c^2d^2 + b^2c^2+ d^2a^2+2(ab)(cd) - 2(bc)(da)}=\color{blue}{(ab+cd)^2+(bc-da)^2} $$
$$\text{Second way} $$
$$a^2b^2 + b^2c^2 + c^2d^2 + d^2a^2= \color{red}{a^2b^2+c^2d^2 -2(ab)(cd)+b^2c^2+ d^2a^2+2(bc)(da)}=\color{blue}{(ab-cd)^2+(bc+da)^2}$$
$$\text{Our desired minimum two ways}$$
${\text{Now,}}$
$$x^2 + y^2 = 44^2 \cdot 10^2 + 10^2 \cdot 33^2 + 33^2 \cdot 5^2 + 5^2 \cdot 44^2$$ $$\text{Use the reasoning which has been used till now:}$$
$$x^2 + y^2 = 44^2 \cdot 10^2 + 10^2 \cdot 33^2 + 33^2 \cdot 5^2 + 5^2 \cdot 44^2=\color{red}{(44^2 \cdot 10^2) + ( 33^2 \cdot 5^2 ) +2\cdot (44 \cdot 10) \cdot (33 \cdot 5)+ (10^2 \cdot 33^2)+(5^2 \cdot 44^2 )-2( 10 \cdot 33)( 5 \cdot 44)}$$
$$\color{blue}{(44\cdot10+33\cdot5)^2+(33\cdot10-44\cdot5)^2=?+?}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2449930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do you solve quadratic inequalities/functions?
Find the least value of $n$ such that $(n-2)x^2+8x+n+4 > 0$, where $x \in \mathbb R$ and $n \in \mathbb N$.
What I did -
As $x$ is real, $D>0$
$64-4(n+4)(n-2)>0$
$(n-4)(n-6)<0$
$n \epsilon (-6,4)$
But as $n \epsilon N$
The least value is 3
According to the book's method,
$n-2>0$
$n>2$
And,
$64-4(n-2)(n+4)<0$
$(n-4)(n+6)>0$
$n>4$
And least value of n comes out to be 5.
I don't understand why D<0 when x is real?
Thanks.
| the case $n=0$ gives $$x^2-4x-2<0$$ which is not fulfilled for all real $x$
$$n=1$$ gives $$x^2-8x-5<0$$
$$n=2$$ gives $$8x+6>0$$
for $$n>2$$ we get
$$x^2+2\left(\frac{4}{n-2}\right)x+\left(\frac{4}{n-2}\right)^2+\frac{n+4}{n-2}-\left(\frac{4}{n-2}\right)^2>0$$ and we get
$$\left(x+\frac{4}{n-2}\right)^2+\frac{n^2+2n-24}{(n-2)^2}>0$$
finally we get the answer $$n=5$$ then we have
$$\left(x+\frac{4}{3}\right)^2+\frac{11}{9}>0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2451981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $\sin x - \cos x = \frac{1}{2}$ then determine: $\sin^4 x + \cos^4 x$ If $$\sin x - \cos x = \frac{1}{2}$$ then determine: $$\sin^4 x + \cos^4 x$$
I tried making it $(\sin^2 x)^2+(\cos^2 x)^2$ but then I get nothing that can help. What is the trick to this?
| You have
$$
\frac{1}{4}=(\sin x-\cos x)^2=\sin^2x-2\sin x\cos x+\cos^2x
$$
so
$$
\sin x\cos x=\frac{1}{2}\left(1-\frac{1}{4}\right)=\frac{3}{8}
$$
Then
$$
\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-\dotsb
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2452142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\int \frac{dx}{x\sqrt{4x^2+1}}$ I was evaluating $\int \frac{dx}{x\sqrt{4x^2+1}}$ using Table of Integrals.
My work
I was evaluating $\int \frac{dx}{x\sqrt{4x^2+1}}$ using Table of Integrals. I found in the Table of Integrals an integral that is akin to
$\int \frac{dx}{x\sqrt{4x^2+1}}$. The integral I was talking about is $$\int \frac{du}{u\sqrt{u^2+a^2}} = -\frac{1}{a} ln \left( \frac{a+\sqrt{u^2+a^2}}{u} \right)$$
With that in mind, I need to modify $\int \frac{dx}{x\sqrt{4x^2+1}}$ to look like $\int \frac{du}{u\sqrt{u^2+a^2}}$.
Modifying now:
$$\int \frac{dx}{x\sqrt{4x^2+1}} = \int \frac{2dx}{2x\sqrt{4x^2+1}} = \int \frac{2dx}{2x\sqrt{(2x)^2+(1)^2}}$$
The modified integral to be evaluated would be $\int \frac{2dx}{2x\sqrt{(2x)^2+(1)^2}}$.
Now getting the integral:
$$\int \frac{dx}{x\sqrt{4x^2+1}} = \int \frac{2dx}{2x\sqrt{(2x)^2+(1)^2}} = -2\left(\left(\frac{1}{(1)}\right) ln \left( \frac{(1)+\sqrt{(2x)^2+(1)^2}}{(2x)} \right) \right)$$
$$ \int \frac{2dx}{2x\sqrt{(2x)^2+(1)^2}} = -2 ln \left( \frac{1+\sqrt{4x^2+1}}{2x} \right) $$
$$ \int \frac{2dx}{2x\sqrt{(2x)^2+(1)^2}} = ln \left(\left( \frac{1+\sqrt{4x^2+1}}{2x} \right)^{-2} \right)$$
$$ \int \frac{2dx}{2x\sqrt{(2x)^2+(1)^2}} = \frac{1}{ln \left(\left( \frac{1+\sqrt{4x^2+1}}{2x} \right)^2 \right)}$$
$$ \int \frac{2dx}{2x\sqrt{(2x)^2+(1)^2}} = ln \left( \frac{4x^2}{1+\sqrt{4x^2+1}}\right)$$
So...the integral of $\frac{dx}{x\sqrt{4x^2+1}}$ would be $ln \left( \frac{4x^2}{1+\sqrt{4x^2+1}}\right)$. But in the book I used, it is
$ln \left( \frac{x}{1+\sqrt{4x^2+1}}\right)$. It was so close to my answer, yet I don't know where I messed up.
Where did I messed up?
| After "Now getting the integral", when you use the formula, you introduce a $2$ that should not be there. You would get
$$
-\ln\left(\frac{1+\sqrt{4x^2+1}}{x}\right)
=\ln\left(\frac{x}{1+\sqrt{4x^2+1}}\right).
$$
You also need to review the rules for logarithms. It is not true that $\ln 1/x=1/\ln x$ (you used it twice, so it actually "cancelled itself", but it is stil wrong).
| {
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Solving the equation $c=\dfrac{x^2+y^2-1}{x^2+(y+1)^2}$ Solving the equation $c=\dfrac{x^2+y^2-1}{x^2+(y+1)^2}$
$$c=\dfrac{x^2+y^2-1}{x^2+(y+1)^2}$$
$$c{x^2+c(y+1)^2}={x^2+y^2-1}$$
$$c{x^2+cy^2+2cy+c}={x^2+y^2-1}\text{ [expanded]}$$
$$1+c=x^2-cx^2+y^2-cy^2-2cy\text{ [moved to other side]}$$
$$1+c=(1-c)x^2+\color{red}{(1-c)y^2-2cy}\text{ [factor, will complete square of red]}$$
$$1+c=(1-c)x^2+\color{red}{(1-c)(y^2-\dfrac{2cy}{c-1}+(\dfrac{2c}{c-1})^2)-(\dfrac{2c}{c-1})^2}\text{ [completed square]}$$
$$1+c+\color{red}{(\dfrac{2c}{c-1})^2}=(1-c)x^2+\color{red}{(1-c)(y-\dfrac{2c}{c-1})^2}\text{ [done]}$$
So we see that this is an ellipse, stretched in the $y$ direction by a factor of $1-c$, and translated both in $x$ and $y$.
$c\neq 1$.
Consider $c=2$
$19=-x^2-(y+16)^2$
$-19=x^2+(y+16)^2$
But this not a circle nor an ellipse. It's supposed to be a circle? What is the issue here?
| You do not need all those calculations!
Starting from $c=\dfrac{x^2+y^2-1}{x^2+(y+1)^2}$
you get
$\left(x^2+y^2-1\right)-c \left(x^2+(y+1)^2\right)=0$
expand
$-c x^2-c y^2-2 c y-c+x^2+y^2-1=0$
collect
$(1-c) x^2+(1-c) y^2-2 c y-c-1=0$
and divide each term by $c-1$ provided that $c\ne 1$
$x^2+y^2-\dfrac{2 c }{1-c}\,y-\dfrac{1+c}{1-c}=0$
which is the equation of a circle with centre $\left(0,\dfrac{2 c }{1-c}\right)$
and radius $r=\dfrac{1}{|c-1|}$
| {
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The intersection of $5$ planes If the intersection of planes $a^{k}x+2^{k}y+3^{k}z+d^{k}=0(0\leq k\leq 4)$ is a line, then $a=?$ and $d=?$.
| The matrix of coefficients is
$$A=\left(
\begin{array}{cccc}
1 & 1 & 1 & 1 \\
a & 2 & 3 & d \\
a^2 & 4 & 9 & d^2 \\
a^3 & 8 & 27 & d^3 \\
a^4 & 16 & 81 & d^4 \\
\end{array}
\right)$$
The system has One and only one solution, that is one point, if $\text{rank }A=4$
We know that the planes intersect along a line, and this means that $\text{rank }A=3$ which happens if determinants of order $4$ are all zero.
$$\det \left|
\begin{array}{cccc}
1 & 1 & 1 & 1 \\
a & 2 & 3 & d \\
a^2 & 4 & 9 & d^2 \\
a^3 & 8 & 27 & d^3 \\
\end{array}
\right|=\\=-a^3d^2+5 a^3 d-6 a^3+a^2 d^3-19 a^2 d+30 a^2-5 a d^3+19 a d^2-36 a+6 d^3-30 d^2+36 d$$
can be easily factored as $$(a-3) (a-2) (d-3) (d-2) (d-a)=0$$
$(a=2,\;d=2);\;(a=2,\;d=3);\;(a=3,\;d=2);\;(a=3,\;d=3)$
The solution $a=d$ leads to planes intersecting in one point so it's been eliminated
Hope this helps
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2454173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find numbers $a\leq b$ that maximize the value of the integral $\int_a^ba-x-x^2dx$ I'm trying to find numbers $a\leq b$ that maximize the value of the integral $\int_a^ba-x-x^2dx$. I tried computing the integral and i got the function in two variables
$$f(a,b)=\int_a^ba-x-x^2dx=ab-\frac{b^2}2-\frac{b^3}3-\frac{a^2}2+\frac{a^3}3$$
Then I applied multivariate calculus and I obtained that the only critical point satisfying $a\leq b$ is $(0,0)$. But I can't classify that point since the determinant of the second partial derivatives is zero, so If it is going to be a maximum I have to show that if $a\leq b$ are numbers not both zero then the integral is negative. Is this argument correct? If it is, how do I show that that is a negative number?
| We have that $$ab - \frac{b^2}{2} - \frac{b^3}{3} - \frac{a^2}{2} + \frac{a^3}{3} \leq 0 \iff ab + \frac{a^3}{3} \leq \frac{a^2}{2} + \frac{b^2}{2} + \frac{b^3}{3} \ \ \ \ \ \ \ (I)$$
To prove this inequality holds, note that
$$0 \leq \left( \frac{a}{\sqrt{2}} - \frac{b}{\sqrt{2}} \right)^2 = \frac{a^2}{2} - ab + \frac{b^2}{2},$$
which is equivalent to
$$ab \leq \frac{a^2}{2} + \frac{b^2}{2}.$$
From this we have that
$$ab + \frac{a^3}{3} \leq \frac{a^2}{2} + \frac{b^2}{2} + \frac{a^3}{3} \leq \frac{a^2}{2} + \frac{b^2}{2} + \frac{b^3}{3}.$$
Using the fact that $\frac{a^3}{3} \leq \frac{b^3}{3}$. This proves the inequality $(I)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2458197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding minimal polynomial of $x=a+b\sqrt[3]{2}+c\sqrt[3]{4}$ I want to find the minimal polynomial of $x=a+b\sqrt[3]{2}+c\sqrt[3]{4}$
For simple case like $x=a+b\sqrt[3]{2}$, I can find
\begin{align}
(x-a)^3 = 2b^3 \qquad \Rightarrow \qquad (x-a)^3-2b^3 =0
\end{align}
I tried to do similar tings such as
\begin{align}
(x-a)^3 = (\sqrt[3]{2}(b+c\sqrt[3]{2}))^3 = 2 (b+c\sqrt[3]{2})^3
\end{align}
but this does not seems good.
From the argument of
\begin{align}
x^3-2 = (x-\sqrt[3]{2})(x-\sqrt[3]{2}w)(x-\sqrt[3]{2}w^2)
\end{align}
where $w^2+w+1=0$,
I came up with some idea, and do the computation via mathematica. And i figure out the minimal polynomial for that is
\begin{align}
&(x-(a+b\sqrt[3]{2}+c\sqrt[3]{4}))(x-(a+bw\sqrt[3]{2}+cw^2\sqrt[3]{4}))(x-(a+bw^2\sqrt[3]{2}+cw\sqrt[3]{4})) \\
&=x^3 +3 a^2 x-3 a x^2-6 b c x-2 b^3-4 c^3 +6 a b c-a^3
\end{align}
The question is how to obtain this polynomial from the starting point.
| An alternative approach.
Unless $b=c=0$, $\alpha=a+b\sqrt[3]2+c\sqrt[3]4$ has degree $3$ and so
a cubic minimum polynomial. Let's assume this. Observe that
$$\pmatrix{a&b&c\\2c&a&b\\2b&2c&a}\pmatrix{1\\\sqrt[3]2\\\sqrt[3]4}=\alpha\pmatrix{1\\\sqrt[3]2\\\sqrt[3]4}.$$
Therefore $\alpha$ is an eigenvalue of
$\pmatrix{a&b&c\\2c&a&b\\2b&2c&a}$. Its characteristic polynomial is
therefore the minimum polynomial of $\alpha$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2461150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Derivation of fourth-order accurate formula for the second derivative I am trying to derive / prove the fourth order accurate formula for the second derivative:
$f''(x) = \frac{-f(x + 2h) + 16f(x + h) - 30f(x) + 16f(x - h) - f(x -2h)}{12h^2}$.
I know that in order to do this I need to take some linear combination for the Taylor expansions of $f(x + 2h)$, $f(x + h)$, $f(x - h)$, $f(x -2h)$. For example, when deriving the the centered-difference formula for the first derivative, the Taylor expansion of $f(x + h)$ minus $f(x-h)$ can be computed to give the desired result of $f'(x)$, in that case.
In what way would I have to combine these Taylor expansions above to obtain the required result?
| $$
f(x+h) =
f(x) + h f'(x) + \frac{h^2}{2} f''(x)
+ \frac{h^3}{6} f'''(x)
+ O(h^4)
$$
$$
f(x-h) =
f(x) - h f'(x) + \frac{h^2}{2} f''(x)
- \frac{h^3}{6} f'''(x)
+ O(h^4)
$$
$$
f(x+2h) =
f(x) + 2h f'(x) + 2 h^2 f''(x)
+ \frac{4 h^3}{3} f'''(x) + O(h^4)
$$
$$
f(x-2h) =
f(x) - 2h f'(x) + 2 h^2 f''(x)
- \frac{4 h^3}{3} f'''(x)
+ O(h^4)
$$
Calculate:
$$
-f(x + 2h) + 16f(x + h) - 30f(x) + 16f(x - h) - f(x -2h)
$$
Which is
$$
\begin{aligned}
&
- \left[
f(x) + 2h f'(x) + 2 h^2 f''(x)
+ \frac{4 h^3}{3} f'''(x)
\right]
\\
&
+16
\left[
f(x) + h f'(x) + \frac{h^2}{2} f''(x)
+ \frac{h^3}{6} f'''(x)
\right]
\\
&
-30
f(x)
\\
&
+16
\left[
f(x) - h f'(x) + \frac{h^2}{2} f''(x)
- \frac{h^3}{6} f'''(x)
\right]
\\
&
-
\left[
f(x) - 2h f'(x) + 2 h^2 f''(x)
- \frac{4 h^3}{3} f'''(x)
\right]
\\
&
+ O(h^4)
\end{aligned}
$$
Which evaluates to
$
12 h^2
$
to give the required result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2461506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\int_{0}^{\infty} \frac{\ln x}{x^2+6x+10}dx$ Evaluate $$\int_{0}^{\infty} \frac{\ln x}{x^2+6x+10}dx$$
The given answer is $0.370429$. Is there any method to solve this? Thanks in advance.
| Note
$$ \int_0^\infty\frac{x^a}{x+1}dx=-\frac{\pi}{\sin(a\pi)}$$
and hence
$$ \int_0^\infty\frac{x^a}{x+b}dx=-\frac{b^a\pi}{\sin(a\pi)}. $$
So
\begin{eqnarray}
&&\int_{0}^{\infty} \frac{\ln x}{x^2+6x+10}dx\\
&=&\lim_{a\to0}\frac{d}{da}\int_{0}^{\infty} \frac{x^a}{x^2+6x+10}dx\\
&=&\lim_{a\to0}\frac{d}{da}\int_{0}^{\infty} \frac{x^a}{(x+3)^2+1}dx\\
&=&\lim_{a\to0}\frac{i}{2}\frac{d}{da}\int_{0}^\infty\left(\frac{x^a}{x+3+i}-\frac{x^a}{x+3-i}\right)dx\\
&=&\lim_{a\to0}\frac{i}{2}\frac{d}{da}\bigg[-\frac{(3+i)^a}{\sin(a\pi)}+\frac{(3-i)^a}{\sin(a\pi)}\bigg]\\
&=&\lim_{a\to0}\frac{i}{2}\frac{d}{da}\frac{(3-i)^a-(3+i)^a}{\sin(a\pi)}\bigg]\\
&=&\lim_{a\to0}\frac{i}{2}\frac{[(3-i)^a \ln (3-i)-(3+i)^a \ln (3+i)]\sin(a\pi)-[(3-i)^a-(3+i)^a]\cos(a\pi)}{\sin^2(a\pi)}\\
&=&\frac{i}{4}[\ln^2(3-i)-\ln^2(3+i)]\\
&=&\frac{i}{4}\ln[(3-i)(3+i)]\ln[(3-i)/(3+i)]\\
&=&\frac{1}{2}\ln10\arctan(\frac13).
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2463122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
How to prove that $21\mid a^2+b^2\Rightarrow 441 \mid a^2 + b^2$ How to prove that $441 \mid a^2 + b^2$ if it is known that $21 \mid a^2 + b^2$.
I've tried to present $441$ as $21 \cdot 21$, but it is not sufficient.
|
Lemma. If a prime $p \equiv 3\pmod{4}$ and $p$ divides $a^2+b^2$ then $p$ divides $a,b$.
Proof. Assume the contrary that $p \nmid a,p \nmid b$.
By Fermat's little theorem, we have $a^{p-1} \equiv 1 \pmod{p}, b^{p-1} \equiv 1 \pmod{p}$ so $a^{p-1}+b^{p-1} \equiv 2 \pmod{p}$.
On the other hand, note that $x+y$ divides $x^{2k+1}+y^{2k+1}$ for all integer $x,y,k$ and $k \ge 0$. Hence,since $p \equiv 3 \pmod{4}$ so $(p-1)/2$ is odd, we obtain $a^2+b^2$ divides $(a^2)^{(p-1)/2}+(b^2)^{(p-1)/2}$. Since $p \mid a^2+b^2$ so $p$ divides $(a^2)^{(p-1)/2}+(b^2)^{(p-1)/2}=a^{p-1}+b^{p-1}$, a contradiction since $a^{p-1}+b^{p-1}\equiv 2 \pmod{p}$ and $p \ge 3$. $\square$
Back to the problem, since $3,7 \equiv 3 \pmod{4}$ are primes so if $7 \mid a^2+b^2$ then $7 \mid a, 7 \mid b$ which means $7^2 \mid a^2+b^2$. Similarly, $3^2 \mid a^2+b^2$. Thus, $441 \mid a^2+b^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2467327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find the $n^{th}$ partial sum of a telescoping series $$
a_k = \frac{6^k}{(3^{k+1}+2^{k+1})(3^k + 2^k)}
$$
Been stuck on this for a while, I started by looking at values of $a_k$ for varying values of $k$ and looking for a pattern, but haven't come to anything useful yet. The next step, I think, is to try and find a pattern in $\sum a_k$ for varying values of k, but I'm having trouble simplifying some of the crazy expressions that result from that. The second part of the question asks what is the sum to infinity, but I think that once I find the $k^{th}$ partial sum, I can find the limit as $k\rightarrow\infty$.
But yeah, and help with this problem would be appreciated.
| This is just a supplement on how to obtain hint given by Qing Zhang.
Consider the following
$$\frac{6^k}{(3^{k+1}+2^{k+1})(3^k+2^k)} = \frac{A \cdot 3^k + B \cdot 2^k}{3^{k+1} +2^{k+1}} + \frac{C\cdot 3^k + D\cdot 2^k}{3^k+2^k}$$
where $A,B,C,D$ are constants to be found.
By making the same denominator on right side fractions and comparing numerators, we have
$$6^k = (A\cdot 3^k + B\cdot 2^k)(3^k+2^k) + (C\cdot 3^k+D\cdot2^k)(3^{k+1}+2^{k+1})$$
$$=(A \cdot 3^{2k} + A \cdot 6^k + B\cdot 6^k + B\cdot 2^{2k}) + (3C\cdot 3^{2k} + 2C\cdot 6^{k} + 3D\cdot 6^k +2D\cdot 2^{2k}).$$
Therefore, by comparing coefficient of $6^k,2^{2k}$ and $3^{2k},$ we have
$$1 = A + B + 2C+3D,$$
$$0 = B + 3C+2D,$$
$$0=A+3C.$$
By solving simultaneous equations, we have
$$A=0,B=-2,C=0,D=1.$$
Therefore,
$$\frac{6^k}{(3^{k+1}+2^{k+1})(3^k+2^k)}=\frac{2^k}{3^k+2^k}-\frac{2^{k+1}}{3^{k+1}+2^{k+1}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2469969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solving for $k$: $\sqrt{k-\sqrt{k+x}}-x = 0$
$$\sqrt{k-\sqrt{k+x}}-x = 0$$
Solve for $k$ in terms of $x$
I got all the way to
$$x^{4}-2kx^{2}-x+k^{2}-x^{2}$$
but could not factor afterwards. My teacher mentioned that there was grouping involved
Thanks Guys!
Edit 1 : The exact problem was solve for $x$ given that $$\sqrt{4-\sqrt{4+x}}-x = 0$$ with a hint of substitute 4 with k
| As discovered $\sqrt{k - \sqrt{k+x}} = x$ leads to the equation $k^2 - (2 \, x^2 + 1) \, k + (x^4 - x) = 0$. Now,
\begin{align}
k &= \frac{1}{2} \, \left[(2 \, x^2 + 1) \pm \sqrt{ (2 \, x^2 + 1)^2 - 4 \, (x^4 - x) } \right] \\
&= \frac{1}{2} \, [(2 \, x^2 + 1) \pm (2 x + 1)] \\
&= \begin{cases}{ x^2 + x + 1 = \frac{1 - x^3}{1-x} \\ x(x-1) } \end{cases}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2470062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
If $z = \tan(x/2)$, what is $\sin(x)$ and $\cos(x)$? While reading mathematical gazette, I noticed an interesting "theorem". If $z = \tan(x/2)$, then $\sin(x) = \frac{2z}{1+z^2}$ and $\cos(x) = \frac{1-z^2}{1+z^2}$.
How can I derive these so I don't have to remember them?
| \begin{align}
z & = \tan \frac x 2 & & \frac z 1 = \tan = \frac{\text{opposite}}{\text{adjacent}} \\[10pt]
x & = 2\arctan z \\[10pt]
\sin x & = \sin(2\arctan z) \\
& = \sin(2\theta) = 2\sin\theta\cos\theta \\[10pt]
& = 2\sin(\arctan z)\cos(\arctan z) \\[10pt]
& = 2 \frac{\text{opposite}}{\text{hypotenuse}} \cdot \frac{\text{adjacent}}{\text{hypotenuse}} \\[10pt]
& = 2\frac{z}{\sqrt{1+z^2}} \cdot\frac{1}{\sqrt{1+z^2}} \\[12pt]
& = \frac{2z}{1+z^2}.
\end{align}
The fact that $\text{hypotenuse} = \sqrt{1+z^2}$ comes from the Pythagorean theorem. And $\cos x$ is handled similarly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
In how many ways can $m$ employees be assigned to $n$ projects if every project is assigned to at least one employee? In how many ways can we assign $m$ employees to $n$ projects so that every employee is assigned to exactly one project and every project is assigned to at least one employee? Also, $m>n$.
My answer:
Let's choose $n$ employees from $m$ and assign them projects such that each project has exactly one employee working under it. Thus, $mPn$. For the remaining $m-n$ employees, each employee can work under $n$ project options. Thus, $n^{m-n}$. Thus, total answer $= mPn \cdot n^{m-n}$.
For $m = 5$ and $n = 4$, my answer $= 480$ and expected answer $= 240$.
For $m = 6$ and $n = 3$, my answer $= 3240$ and expected answer $= 540$.
| For $m=5$, $n=4$ . . .
Exactly one project gets two people, the others get one.
*
Choose the project with $2$ people: $\binom{4}{1}=4$ choices.
Choose the $2$ people for that project: $\binom{5}{2}=10$ choices.
Assign the remaining $3$ people to the remaining $3$ projects: $3!=6$ choices.
Hence the number of valid assignments is $(4)(10)(6)=240$.
For $m=6$, $n=3$ . . .
Up to reordering, the number of people for the $3$ projects is one of
\begin{align*}
1,1,4\\
1,2,3\\
2,2,2\\
\end{align*}
Consider each case separately . . .
Case: $1,1,4$.
*
Choose the project with $4$ people: $\binom{3}{1}=3$ choices.
Choose the $4$ people for that project: $\binom{6}{4}=15$ choices.
Assign the remaining $2$ people to the remaining $2$ projects: $2!=2$ choices.
Hence the number of valid assignments is $(3)(15)(2)=90$.
Case: $1,2,3$.
*
Choose which project gets how many people: $3!=6$ choices.
Choose the $3$ people for the $3$-person project: $\binom{6}{3}=20$ choices.
Choose the $2$ people for the $2$-person project: $\binom{3}{2}=3$ choices.
Hence the number of valid assignments is $(6)(20)(3)=360$.
Case: $2,2,2$.
*
Choose the $2$ people for the first project: $\binom{6}{2}=15$ choices.
Choose the $2$ people for the second project: $\binom{4}{2}=6$ choices.
Hence the number of valid assignments is $(15)(6)=90$.
Thus, the total number of valid assignments is $90+360+90=540$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2477051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove $\lim_{x\rightarrow 0}\frac{\sqrt{9-x}-3}{x}=-\frac{1}{6}$ with epsilon-delta I am looking to show that $$f:(0,1)\rightarrow \mathbb{R} \lim_{x\rightarrow 0}\frac{\sqrt{9-x}-3}{x}=-\frac{1}{6}$$ Normally for these types of problems I have been doing epsilon-delta proofs but I cannot figure out how to define my $\delta$.
Essentially I need with finding an appropriate $\delta$. I am not looking for an application of L'Hopital's Rule.
So far with determining a $\delta$ from $\varepsilon$ I have $$\left| \frac{\sqrt{9-x}-3}{x}+\frac{1}{6} \right|<\varepsilon $$
$$\left| \frac{-1}{\sqrt{9-x}-3}+\frac{1}{6} \right|<\varepsilon $$
$$\left| \frac{-1}{\sqrt{9-x}-3}\right|+\frac{1}{6} \leq\varepsilon $$
by the triangle inequality
$$\left| \frac{-1}{\sqrt{9-x}-3}\right| \leq\varepsilon -\frac{1}{6}$$
$$ \frac{1}{\left|\sqrt{9-x}-3\right|} \leq\varepsilon -\frac{1}{6}$$
$$ \frac{1}{\varepsilon -\frac{1}{6}} \leq\left|\sqrt{9-x}-3\right|$$
$$ \frac{1}{\varepsilon -\frac{1}{6}} \leq\left|\sqrt{9-x}\right|+3$$
$$ \left(\frac{1}{\varepsilon -\frac{1}{6}}\right)-3 \leq\left|\sqrt{9-x}\right|$$ because the square root is always positive
$$ \left(\frac{1}{\varepsilon -\frac{1}{6}}\right)-3 \leq\sqrt{9-x}$$
$$ \left(\left(\frac{1}{\varepsilon -\frac{1}{6}}\right)-3\right)^2 \leq9-x$$
$$ 9-\left(\left(\frac{1}{\varepsilon -\frac{1}{6}}\right)-3\right)^2 \geq x$$
| With
$$\frac{\sqrt{9-x}-3}{x}=\frac{-1}{\sqrt{9-x}+3}$$
then
\begin{align}
\left|\frac{\sqrt{9-x}-3}{x}+\frac{1}{6}\right|
&= \left|\frac{-1}{\sqrt{9-x}+3}+\frac{1}{6}\right|\\
&= \left|\frac{\sqrt{9-x}-3}{6(\sqrt{9-x}+3)}\right|\\
&= \left|\frac{-x}{6(\sqrt{9-x}+3)^2}\right|\\
&\leq\dfrac{1}{54}|x|
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2477969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Inequality question: If $a + b + c =1$, what is the minimum value of $\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}$. If $a + b + c =1$, what is the minimum value of $\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}$. I've tried AM-HM but it gave $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \geq 9$ which gives $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} + 2 (\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} )\geq 81$
| Assuming $a,b,c$ are positive reals,
$$\sum\limits_{\textrm{cyc}}\frac 1{ab}=\frac 1{abc}\sum\limits_{\textrm{cyc}}a=\frac 1{abc}\geq \frac 1{\left(\frac{a+b+c}3\right)^3}=27$$
with equality when $a=b=c=1/3$ giving $3\cdot\dfrac 1{(1/3)^3}=27$
where the penultimate step follows from the AM-GM inequality
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2478189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Sum of trigonometric secants I don't know what formula to use to solve the next sum
$\textrm{sec}\left ( \frac{2\pi }{7} \right )+\textrm{sec}\left ( \frac{4\pi }{7} \right )+\textrm{sec}\left ( \frac{6\pi }{7} \right )$
Please, give me some advice. Thanks for your help.
| $$\textrm{sec}\left ( \frac{2\pi }{7} \right )+\textrm{sec}\left ( \frac{4\pi }{7} \right )+\textrm{sec}\left ( \frac{6\pi }{7} \right )=\frac{\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}+\cos\frac{2\pi}{7}\cos\frac{6\pi}{7}+\cos\frac{4\pi}{7}\cos\frac{6\pi}{7}}{\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{6\pi}{7}}=$$
$$=\frac{\cos\frac{6\pi}{7}+\cos\frac{2\pi}{7}+\cos\frac{6\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{2\pi}{7}}{2\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{8\pi}{7}}=$$
$$=\frac{\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}}{\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{8\pi}{7}}=$$
$$=\frac{\frac{2\sin\frac{\pi}{7}\cos\frac{2\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{4\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{6\pi}{7}}{2\sin\frac{\pi}{7}}}{\frac{8\sin\frac{2\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{8\pi}{7}}{8\sin\frac{2\pi}{7}}}=$$
$$=\frac{\frac{\sin\frac{3\pi}{7}-\sin\frac{\pi}{7}+\sin\frac{5\pi}{7}-\sin\frac{3\pi}{7}+\sin\frac{7\pi}{7}-\sin\frac{5\pi}{7}}{2\sin\frac{\pi}{7}}}{\frac{\sin\frac{16\pi}{7}}{8\sin\frac{2\pi}{7}}}=\frac{-\frac{1}{2}}{\frac{1}{8}}=-4.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2478907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Use induction to prove that $3|(4^n − 1) $ for any integer $n \geq 0$ Use induction to prove that :
$ 3 |(4^n − 1)$ for any integer $n ≥ 0$
Hint: If $k \geq 0$ is an integer then
$4^{(k+1)} = 4\cdot4^k = 3 \cdot4^k + 4^k$.
Honestly have no idea how to even start this one
| The case $k = 0$ is self-evident: $3 \mid 0$.
Consider that
$3 \mid 4 - 1, \tag 1$
which is the case $k = 1$;. then since we have
$4(4^k - 1) + 3 = 4^{k + 1} - 4 + 3 = 4^{k + 1} - 1, \tag 2$
if we assume
$3 \mid 4^k - 1, \tag 3$
and recall as in (1) that
$3 \mid 3, \tag 4$
we see that
$3 \mid 4(4^k - 1) + 3 \tag 5$
directly follows. Thus, via (2), we obtain
$3 \mid 4^{k + 1} - 1; \tag 6$
So much for an inductive proof of (6). As simple as this is, an even simpler (non-inductive) proof may be had by observing that
$4^{k + 1} - 1 = (2^2)^{k + 1} - 1 = (2^{k + 1})^2 - 1 = (2^{k + 1} - 1)(2^{k + 1} + 1), \tag 7$
and that $3$ must divide at least one of the three consecutive integers
$2^{k + 1} - 1, 2^{k + 1}, 2^{k + 1} + 1; \tag 8$
since $3 \not \mid 2^{k + 1}, \tag 9$
we must have
$3 \mid 2^{k + 1} - 1 \vee 3 \mid 2^{k + 1} + 1, \tag{10}$
and in either case we see that
$3 \mid (2^{k + 1} - 1)(2^{k + 1} + 1), \tag{11}$
from which, by (7), (6) becomes evident as well.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2479652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
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Showing that $\frac{\sin(a_{n-1}) + 1}{2}$ is a Cauchy sequence In my homework set, I have the following question:
Show that $$ a_n = \frac{\sin(a_{n-1}) + 1}{2}, \quad a_1=0 $$
satisfies the definition of Cauchy sequence.
As we went over the concept of Cauchy sequences a bit too quickly in class, I'm puzzled about how I should go about showing this. Conceptually, I understand what a Cauchy sequence is, and I know that in $\mathbb{R}$ it is the same as convergence, but I find it hard to apply to the above sequence.
I'd appreciate some hints about how to approach this problem.
Edit: I've been working on this with help from responses below so I thought I'd update my work. It's verbose but I thought that might show my thinking process better.
My work so far:
$$
\begin{align}
|a_{n+1} - a_n| &= \left | \frac{\sin(a_n) + 1}{2} - \frac{\sin(a_{n-1}) + 1}{2} \right | \\
\\
&= \frac{1}{2} \left | \sin(a_n) + 1 - (\sin(a_{n-1}) + 1)\right | \\
\\
&= \frac{1}{2} \left | \sin(a_n) - \sin(a_{n-1}) \right | \\
\\
&= \frac{1}{2} \left | 2 \sin(\frac{a_n - a_{n-1}}{2}) \times \cos(\frac{a_n + a_{n-1}}{2}) \right | \\
\\
&\leq \frac{1}{2} \left | \sin(\frac{a_n - a_{n-1}}{2}) \times \cos(\frac{a_n + a_{n-1}}{2}) \right | \\
\\
&\leq \frac{1}{2} \left | \sin(\frac{a_n - a_{n-1}}{2}) \right | \\
\\
&\leq \frac{\left | a_n - a_{n-1} \right |}{2}
\end{align}
$$
This set of inequalities work because $|sin(x)| \leq |x|, \ \forall x \in \mathbb{R}$.
Hence, we have:
$$
\begin{align}
|a_{n+1} - a_n| &\leq \frac{1}{2} \left |a_n - a_{n-1} \right | \leq \frac{1}{2^2} \left | a_{n-1} - a_{n-2} \right | \leq \dots \leq \frac{1}{2^{n-1}} \left |a_{2} - a_{1} \right |
\end{align}
$$
So we learn that the distance between consecutive terms is becoming smaller, and that:
$$
\begin{align}
\left | a_{n+1} - a_n \right | &\leq \frac{1}{2^{n-1}} \times \left | \frac{1}{2} - 0 \right | = \frac{1}{2} \times \frac{1}{2^{n-1}} = \frac{1}{2^n}
\end{align}
$$
Now, for $n>m$ and $k=m$ we have:
$$
\begin{align}
\left | a_n - a_m \right | &= \sum_{k=m}^{n-1} \left | a_{k+1} - a_k \right | \\
\\
&= \sum_{k=m}^{n-1} \frac{1}{2^k} \\
\\
&= \left ( \frac{1}{2^m} + \frac{1}{2^{m+1}} + \dots + \frac{1}{2^{n-2}} + \frac{1}{2^{n-1}} \right ) \\
\\
&= \frac{1}{2^{m-1}} \left ( \frac{1}{2} + \frac{1}{2^2} + \dots + \frac{1}{2^{n-m-2}} + \frac{1}{2^{n-m-1}} \right ) \\
\\
&\leq \frac{1}{2^{m-1}}
\end{align}
$$
So, if we have $|a_n - a_{n-1}| < \epsilon$, then $\frac{1}{2^{m-1}} < \epsilon$. If we solve for $m$, we get that $m > \frac{\log(\frac{2}{\epsilon})}{\log(2)}$ and therefore $N > \frac{\log(\frac{2}{\epsilon})}{\log(2)}$.
So we have found $|a_n - a_m| < \epsilon$ for $n, m > N$. Hence, $(a_n)$ is Cauchy.
Resources used:
*
*Answers given below.
*This video was very helpful
*Understanding the definition of Cauchy sequence
*Proving that a sequence such that $|a_{n+1} - a_n| \le 2^{-n}$ is Cauchy
*Showing a recursive sequence is Cauchy
*How do I find the limit of the sequence $a_n=\frac{n\cos(n)}{n^2+1}$ and prove it is a Cauchy sequence?
| I think it is not necessary to define such a function defined by Robert Z.
$\displaystyle |a_{n+1}-a_n|\le \frac{1}{2}|a_n-a_{n-1}|$ for all $n=1,2,\cdots$
So , $\displaystyle |a_{n+1}-a_n|\le \frac{1}{2}.|a_n-a_{n-1}|\le \frac{1}{2^2}.|a_{n-1}-a_{n-2}|\le \cdots \le \frac{1}{2^{n-1}}.|a_2-a_1| \text{ for all } n=1,2,\cdots$
Now , for any positive integer $p$ , $$|a_{n+p}-a_n|\\\le |a_{n+p}-a_{n+p-1}|+\cdots +|a_{n+2}-a_{n+1}|+|a_{n+1}-a_n|\\\le \left( \frac{1}{2^{n+p-2}}+\cdots+\frac{1}{2^{n}}+\frac{1}{2^{n-1}}\right).|a_2-a_1|\\=\frac{1}{2^{n-2}}.[1-(1/2)^p].|a_2-a_1|\\ \to 0 \text { , as } n\to \infty \text{ for all } p=1,2,\cdots $$
So , $\{a_n\}$ is Cauchy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2481450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
The maximum possible value of $x^2+y^2-4x-6y$ subject to the condition $|x+y|+|x-y|$ The maximum possible value of
$x^2+y^2-4x-6y$
subject to the condition $|x+y|+|x-y|$=4
My workout...
Now if we add 13 to the equation we get
$x^2+y^2-4x-6y+13-13$
or,$x^2+y^2-4x-6y+4+9-13$
or,$(x-2)^2+(y-3)^2-13$
Are there any methods other than function.
|
Solution is x=-2, y=-2.
Let us call $S(x,y)=x^2+y^2-4x-6y$.
Let us analyze our condition $|x+y|+|x-y|=4$. If we now rise this to quadrat :
$$
(|x+y|+|x-y|)^2=4^2
$$
$$
x^2+y^2+2xy+x^2+y^2-2xy+2x^2-2y^2=4x^2=4^2
$$
$$
x=+2,-2
$$
2 Solutions, or
$$
x^2+y^2+2xy+x^2+y^2-2xy-2x^2+2y^2=4y^2=4^2
$$
$$
y=+2,-2
$$
For x=2 or x=-2, we have for y a range between -2 and 2. But because function is convex, a maximal value can be only at y=-2, or y=2 but not between. Same if happen fixing y : For y=2 or y=-2, we have for x a range between -2 and 2, but it need to be -2 or 2. So we only need to check all 4 possibilities :
S(-2,2)=4
S(2,2)=-12
S(-2,-2)=28
S(2,-2)=12
Clearly a maximal value appear for S(x=-2,y=-2)=28.
greatings, Daniel
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2484326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Parametric equation of the intersection between $x^2+y^2+z^2=6$ and $x+y+z=0$ I'm trying to find the parametric equation for the curve of intersection between $x^2+y^2+z^2=6$ and $x+y+z=0$. By substitution of $z=-x-y$, I see that $x^2+y^2+z^2=6$ becomes $\frac{(x+y)^2}{3}=1$, but where should I go from here?
| When you substitute $z=-(x+y)$ into $x^2+y^2+z^2=6$ we get
\begin{eqnarray*}
x^2+y^2+(x+y)^2=6 \\
x^2+xy+y^2 =3.
\end{eqnarray*}
Now multiply this by $4$ & complete the square
\begin{eqnarray*}
(2x+y)^2+3y^2=12.
\end{eqnarray*}
This is easily parameterised by $2x+y = \sqrt{12} \cos ( \theta)$ and $y = 2 \sin ( \theta)$.
\begin{eqnarray*}
x &=& \frac{\sqrt{12} \cos ( \theta)-2 \sin ( \theta)}{2} \\
y &=& 2 \sin ( \theta) \\
z &=& \frac{-\sqrt{12} \cos ( \theta)- 2\sin ( \theta)}{2}
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2484914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Addition of two piecewise function Suppose I have function
$$y(x)=\begin{cases}
x+1\qquad & 0\leq x\leq1 \\
2-x\qquad & 1<x \leq2 \\
0\qquad & \mathrm{elsewhere}
\end{cases}$$
I need to find the function $g(x)=y(x+2)+2y(x+1)$
$$y(x+2)=\begin{cases}
x+3\qquad &-2\leq x\leq-1 \\
-x\qquad & -1<x \leq0 \\
0\qquad &\mathrm{elsewhere}
\end{cases}$$
$$2y(x+1)=\begin{cases}
2x+4\qquad & -1\leq x\leq0\\
2-2x\qquad & 0<x \leq1 \\
0\qquad &\mathrm{elsewhere}
\end{cases}$$
Now,
$$g(x)=y(x+2)+2y(x+1)=\begin{cases}
x+3\qquad & -2\leq x\leq -1 \\
x+4\qquad & -1<x \leq0 \\
2-2x\qquad &0<x\leq1\\
0\qquad &\mathrm{elsewhere}
\end{cases}$$
Now the result looks Okay,
But the problem is the value of $y(x+2)$ at $x=-1$ is $2$ and $2y(x+2)$ at $x=-1$ is also $2$ so if I add both then $y(x+2)+2y(x+1)$ at $x=-1$ will be $2+2=4$,but the value of $g(x)$ at $x=-1$ is $-1+3=2$
| In your calculation of $g(x)$, the first line, $x+3$, was obtained by adding $y(x+2)=x+3$ and $2y(x+1)=0$. The first of these two equations, $y(x+2)=x+3$, is valid for $x$ in the range $-2\leq x\leq-1$. The second, $2y(x+1)=0$, is valid when $x<-1$ and also when $x>1$. In particular, it is not valid when $x=-1$. So the $x+3$ line in your formula for $g(x)$ is not valid for $x=-1$ but only for $-2\leq x<-1$.
More generally, when you combine several formulas, each f which is valid for a certain range of $x$ values, then the combination will in general only be valid for those $x$ values that are in all the relevant ranges simultaneously.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2485305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Simplest way to get the lower bound $\pi > 3.14$ Inspired from this answer and my comment to it, I seek alternative ways to establish $\pi>3.14$. The goal is to achieve simpler/easy to understand approaches as well as to minimize the calculations involved. The method in my comment is based on Ramanujan's series $$\frac{4}{\pi}=\frac{1123}{882}-\frac{22583}{882^{3}}\cdot\frac{1}{2}\cdot\frac{1\cdot 3}{4^{2}}+\frac{44043}{882^{5}}\cdot\frac{1\cdot 3}{2\cdot 4}\cdot\frac{1\cdot 3\cdot 5\cdot 7}{4^{2}\cdot 8^{2}}-\dots\tag{1}$$ This is quite hard to understand (at least in my opinion, see the blog posts to convince yourself) but achieves the goal of minimal calculations with evaluation of just the first term being necessary.
On the other end of spectrum is the reasonably easy to understand series
$$\frac\pi4=1-\frac13+\frac15-\cdots\tag2$$
But this requires a large number of terms to get any reasonable accuracy. I would like a happy compromise between the two and approaches based on other ideas apart from series are also welcome.
A previous question of mine gives an approach to estimate the error in truncating the Leibniz series $(2)$ and it gives bounds for $\pi$ with very little amount of calculation. However it requires the use of continued fractions and proving the desired continued fraction does require some effort.
Another set of approximations to $\pi$ from below are obtained using Ramanujan's class invariant $g_n$ namely $$\pi\approx\frac{24}{\sqrt{n}}\log(2^{1/4}g_n)\tag{3}$$ and $n=10$ gives the approximation $\pi\approx 3.14122$ but this approach has a story similar to that of equation $(1)$.
| If we consider the Beuker-like integral
$$ 0<\int_{0}^{1}\frac{x^8(1-x)^8}{1+x^2}\,dx = 4\pi-\frac{188684}{15015} $$
we get, through partial fraction decomposition and few operations in $\mathbb{Q}$,
$$ \pi > \frac{47171}{15015} > 3.14159.$$
Inspired by Professor Vector's brilliant approach, I am adding a further approach.
By the Shafer-Fink inequality we have $\arctan(x)>\frac{3x}{1+2\sqrt{1+x^2}}$ for any $x>0$, hence by evaluating both sides at $x=\frac{1}{\sqrt{3}}$ we get $\pi>\frac{18}{13}(4-\sqrt{3})=3.140237\ldots$ A refinement of the previous inequality is
$$\forall x>0,\qquad \arctan(x)>\frac{6x}{1+\sqrt{1+x^2}+2\sqrt{2}\sqrt{1+x^2+\sqrt{1+x^2}}} $$
and the evaluation at $x=\frac{1}{\sqrt{3}}$ produces the sharper bound
$$ \pi > \frac{36}{2+\sqrt{3}+4 \sqrt{2+\sqrt{3}}} > 3.1415.$$
Yet another approach. The inequality $\arctan(x)>\frac{5x(21+11x^2)}{105+90x^2+9x^4}$ for any $x\in(0,1)$ comes from the Gauss continued fraction / the Padé approximants for the arctangent function. By replacing $x$ with $\frac{x}{1+\sqrt{1+x^2}}$, then evaluating at $x=\frac{1}{\sqrt{3}}$, we get the nice and tight approximation:
$$ \pi > \color{blue}{\frac{5}{601}\left(944-327\sqrt{3}\right)}>3.141592.$$
There is also a nice geometric argument leading to an acceleration of Vieta's formula. Let $PQ$ be a side of a regular $n$-agon inscribed in a unit circle centered at $O$. Let $M$ be the midpoint of the minor arc $PQ$. We may consider the unique parabola through $P,M,Q$ and approximate the area of the circle sector delimited by $P,O,Q$ through $[POQ]$ plus the area of a parabolic segment, $\frac{4}{3}[PMQ]:$
It follows that if $A_n$ is the area of the inscribed $2^n$-agon, we have
$$ \pi \geq A_n+\frac{4}{3}\left(A_{n+1}-A_n\right) = \frac{4}{3}A_{n+1}-\frac{1}{3}A_n $$
with
$$ A_n = 2^{n-1} \sin\frac{\pi}{2^{n-1}} $$
and $\{A_n\}_{n\geq 2}$ being computable through a simple recursion,
$$ A_{n+1} = 2^n \sqrt{\frac{1-\sqrt{1-\frac{A_n^2}{4^{2n-2}}}}{2}}$$
where
$$ \frac{4}{3}A_5-\frac{1}{3}A_4 = \frac{4}{3}\left[8\sqrt{2-\sqrt{2+\sqrt{2}}}-\sqrt{2-\sqrt{2}}\right]=\color{green}{3.141}44\ldots $$
By considering a $12$-agon and a $24$-agon we get the simpler
$$ \pi > 4\sqrt{2}(\sqrt{3}-1)-1 = \color{green}{3.141}10\ldots $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2485558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "55",
"answer_count": 11,
"answer_id": 4
} |
Find all posible $x$ given the congruence $(x+y)^p-y^p \equiv 1 \pmod{q}$ for both $pA few examples.
$$\begin{array}{c|c|l}
\text{prime } p & \text{prime } q & \text{possible values } x \\ \hline
3 & 5 & [1, 2, 4] \\
5 & 13 & [1, 3, 5, 8, 9, 10] \\
13 & 17 & [1, 4, 7, 10, 14, 15, 16] \\
23 & 37 & [1, 2, 3, 6, 11, 12, 15, 18, 19, 22, 29, 30, 35] \\
\end{array}$$
We notice that the number of solutions is limited. $x=1$ is always a solution if we take $y=q-1$.
Is there a general way of finding possible solutions for $x$ ?
| If $z = x+y$, you are solving $z^p \equiv 1 + y^p \mod q$.
If $\gcd(p, q-1) = 1$, the map $t \mapsto t^p$ is one-to-one on $\mathbb Z/q \mathbb Z$, so there is always one $z$. If $\gcd(p, q-1) = g > 1$, the map is neither one-to-one nor onto, so for some $y$ there is no solution and for others there are several. The nonzero $p$'th powers mod $q$ form a subgroup of
the multiplicative group mod $q$.
For example, if $p = 5$ and $q = 11$, the $5$'th powers mod $11$ are
$0, 1, 10$. $z^5 \equiv y^5 + 1$ has solutions $y^5 = 0, z^5 = 1$ corresponding to $y=0, z = 1,3,4,5,9$, and $y^5=10, z^5 = 0$ corresponding to $z=0, y=2,6,7,8,10$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2486569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Prove the following inequality by induction for all $n\in \Bbb N$ I know how to prove the base case for this. But how would I continue from there?
| Set $$S_n =\sum_{i=1}^n \frac{1}{\sqrt{i}} \geq \sqrt{n}$$
then $$S_{n+1} =\sum_{i=1}^{n+1} \frac{1}{\sqrt{i}} \geq \sqrt{n} =S_n +\frac{1}{\sqrt{n+1}}$$
Obviously $S_1\ge 1 =\sqrt1.$
If we assume that $$S_n\ge \sqrt{n}$$
then
we have that
$$S_{n+1} =S_n +\frac{1}{\sqrt{n+1}} \ge \sqrt{n}+\frac{1}{\sqrt{n+1}} =\frac{\sqrt{n^2+n} +1}{\sqrt{n+1}} $$
But we know that,
$$n^2\le 4n^2 +4n \Longleftrightarrow n\le 2 \sqrt{n^2+n}\Longleftrightarrow n+\color{blue}{n^2+n+1} \le 2 \sqrt{n^2+n}+\color{blue}{n^2+n+1} $$
that $$(n+1)^2 = n^2 +2n +1 \le 2 \sqrt{n^2+n}+\color{blue}{n^2+n+1} = (\sqrt{n^2+n}+1)^2$$
we just prove that, $$\sqrt{n^2+n}+1\ge n+1 \Longleftrightarrow \frac{\sqrt{n^2+n} +1}{\sqrt{n+1}}\ge \sqrt{n+1}$$
Hence,
$$S_{n+1}\ge \sqrt{n}+\frac{1}{\sqrt{n+1}} =\frac{\sqrt{n^2+n} +1}{\sqrt{n+1}} \ge \sqrt{n+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2487195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to show that $\int_{0}^{\infty}e^{-x}(x^2+3x+3)\cos(x){\mathrm dx\over (1+x)^3}=1?$ How to show that $(1)$ is
$$\int_{0}^{\infty}e^{-x}(x^2+3x+3)\cos(x){\mathrm dx\over (1+x)^3}=1?\tag1$$
$$x^2+3x+3=\left(x+{3\over 2}\right)^2+{3\over 4}$$
$$\int_{0}^{\infty}e^{-x}\cos(x){\mathrm dx\over (1+x)^3}+\int_{0}^{\infty}e^{-x}\left(x+{3\over 2}\right)^2\cos(x){\mathrm dx\over (1+x)^3}=I_1+I_2\tag2$$
$$(1+x)^{-3}=\sum_{n=0}^{\infty}(-1)^n{n+2\choose n}x^n$$
$$\sum_{n=0}^{\infty}(-1)^n{n+2\choose n}\int_{0}^{\infty}x^{n}e^{-x}\cos(x)\mathrm dx=I_1\tag3$$
$$\sum_{n=0}^{\infty}(-1)^n{n+2\choose n}\int_{0}^{\infty}x^{n}e^{-x}\left(x+{3\over 2}\right)^2\cos(x)\mathrm dx=I_2\tag4$$
| By partial integration, we have
\begin{align}
&\int_0^{+\infty}\frac{e^{-x}\cos x}{1+x}dx
=\int_0^{+\infty}\frac{e^{-x}}{1+x}d\sin x\\
&=\left.\frac{e^{-x}\sin x}{1+x}\right|_0^{+\infty}
+\int_0^{+\infty}e^{-x}\sin x\left[\frac{1}{1+x}+\frac1{(1+x)^2}\right]dx\\
&=0-\int_0^{+\infty}e^{-x}\left[\frac{1}{1+x}+\frac1{(1+x)^2}\right]d\cos x\\
&=\left.e^{-x}\cos x\left[\frac{1}{1+x}+\frac{1}{(1+x)^2}\right]\right|_0^{+\infty}
-\int_0^{+\infty}e^{-x}\cos x\left[\frac{1}{1+x}+\frac2{(1+x)^2}+\frac2{(1+x)^3}\right]dx\\
&=2-\int_0^{+\infty}e^{-x}\cos x\left[\frac{1}{1+x}+\frac2{(1+x)^2}+\frac2{(1+x)^3}\right]dx,
\end{align}
then
$$
\int_0^{+\infty}e^{-x}\cos x\frac{x^2+3x+3}{(1+x)^3}dx
=\int_0^{+\infty}e^{-x}\cos x\left[\frac{1}{1+x}+\frac1{(1+x)^2}+\frac1{(1+x)^3}\right]dx
=1.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2488730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
For how many positive integers $n$ is $4^n − 1$ a prime number? This was the initial question and one of the methods to work it out include this:
The second method uses the fact when $n$ is a positive integer $x − 1$ is a factor of $x^n − 1$,
since, for $n ≥ 2$,
$$x^n − 1 = (x − 1)(x^{n−1} + x^{n−2} + \ldots + x + 1).$$
Hence, putting $x = 4$, we deduce that, for each positive integer $n, 3$ is a factor of $4^n − 1$. Therefore $4^n −1$ is not prime except when $n = 1$ and $4^n −1 = 3$. So there is just one positive integer $n$ for which $4^n − 1$ is a prime number.
However, I don't understand the initial method and how they deduced that $x-1$ is a factor of $x^n - 1$
| For the question at hand, if $P(x)$ is a polynomial and $P(r)=0$, then $P(x)=(x-r)Q(x)$ for some polynomial $Q$. If the coefficients of $P$ and $r$ are all integers, then the coefficients of $Q$ are also all integers.
The specific case of $x^n-1$ might be best understood by example:
$$\begin{align}
(x-1)(x^3+x^2+x+1)
&=(x-1)x^3+(x-1)x^2+(x-1)x+(x-1)\\
&=(x^4-x^3)+(x^3-x^2)+(x^2-x)+(x-1)\\
&=x^4+(x^3-x^3)+(x^2-x^2)+(x-x)-1\\
&=x^4-1
\end{align}$$
The key step is the "telescopic" rearrangement of the parentheses. This is a standard technique (aka trick) when working with sums and series; you'll run into it again and again. (The term "telescopic" refers to old-style spyglasses, like you see in pirate movies, which collapse down to compact form.)
Another, possibly easier, way of seeing that $n=1$ is the only exponent for which $4^n-1$ is prime:
$$4^n-1=2^{2n}-1=(2^n-1)(2^n+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2490762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Conditional Joint Distribution Concern Random variables $X$ and $Y$ are jointly distributed with density
$$f(x,y)=
\begin{cases}
3x & 0 <y<2x<2 \\
0 & \text{otherwise}
\end{cases}
$$
(i) Find $P\left(X<\frac{1}{2} \Big| Y < \frac{1}{2}\right)$
$$
P\left(X<\frac{1}{2} \Big| Y < \frac{1}{2}\right) = \frac{P\left(X < \frac{1}{2} \cap Y < \frac{1}{2} \right)}{P\left( Y < \frac{1}{2}\right)} \\ = \frac{ \int_0^{\frac{1}{2}} \int_{\frac{y}{2}}^{\frac{1}{2}} 3x \, dx\,dy}{\int_0^{\frac{1}{2}} \int_{\frac{y}{2}}^{2} 3x \, dx\,dy}$$
Would someone help me out in verifying my work and if it has anything wrong to direct me to the right way.
| In the denominator you should have
$\int_{0}^{\frac{1}{2}} \int_{\frac{y}{2}}^{1} 3x dx dy$
and in the numerator you should have
$\int_{0}^{\frac{1}{2}} \int_{\frac{y}{2}}^{\frac{1}{2}} 3x dy dx$
In other words, you should have $$\dfrac{\int_{0}^{\frac{1}{2}} \int_{\frac{y}{2}}^{\frac{1}{2}}3x dx dy}{\int_{0}^{\frac{1}{2}} \int_{\frac{y}{2}}^{1} 3x dx dy}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2493844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$x+y+z=3$, prove the inequality For $x,y,z>0$ and $x+y+z=3$ prove that $\frac{x^3}{(y+2z)^2}+\frac{y^3}{(z+2x)^2}+\frac{z^3}{(x+2y)^2}\ge \frac{1}{3}$.
QM, AM, GM, HM suggested ;)
| $$\sum_{\text{cyc}}(y+2z)=3(x+y+z)=9$$
By Holder inequality:
$$\sum_{\text{cyc}}\frac{x^3}{(y+2z)^2}=$$
$$=\frac{\sum_{\text{cyc}}(y+2z)\sum_{\text{cyc}}(y+2z)\sum_{\text{cyc}}\frac{x^3}{(y+2z)^2}}{81}\ge$$
$$\ge \frac{(x+y+z)^3}{81}=\frac{1}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2497192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Binomial summation problem
Show that
$$ \frac{\binom{n}{0}}{1} - \frac{\binom{n}{1}}{4} +\dots + (-1)^n \frac{\binom{n}{n}}{3n+1} = \frac{3^n \cdot n!}{ 1\cdot 4\cdot 7\cdots(3n+1)}.$$
I don't know how to proceed in such type of problems. Any help or hint will be much appreciated.
| Note $$(1-x^{3})^{n} = \binom{n}{0} x^{0} - \binom{n}{1}x^{3} + \binom{n}{2}x^{6} + \cdots +(-1)^{n}\binom{n}{n}x^{3n}$$ Then what you need is $$\int_{0}^{1}(1-x^{3})^{n} \ dx = \frac{\Gamma\frac{4}{3} \cdot n!}{\Gamma\left(n+\frac{4}{3}\right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2498077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the number of terms of a product to make it 1/2 How to find $k$ such that:
$$\prod_{j=1}^{k-1} \Big(1 - \frac jn \Big) = \frac 12$$
?
According to a book I'm reading, this should hold approximatively when $k \approx \sqrt{n}$ (no proof is given).
Indeed, it seems to work. The equation is equivalent to
$$\begin{align}\sum_{j=1}^{k-1} \log\Big(1 - \frac jn \Big) = - \log(2)\end{align}$$
i.e.
$$\begin{align}\sum_{j=1}^{k-1} \Big( \frac {-j}n + O\Big(\frac {k^2}{n^2}\Big) \Big) = - \log(2) \end{align}$$
or
$$\frac {1}{n}\begin{align}\sum_{j=1}^{k-1} j + O\Big(\frac {k^3}{n^2}\Big) = \log(2) \end{align}$$
Let's suppose $k^3 = o(n^2)$. Then
$$\frac 1 n \frac{k(k-1)}{2} + o(1) = \log(2)$$
and this proves that:
$$\frac{k^2}{n} + O\Big(\frac k n\Big) + o(1) = 2 \log(2)$$
This means that $k$ should be taken around $c \sqrt{n}$, with $c = \sqrt{2\log2}$.
This seems to work, but is there a cleaner proof that leads to a more precise result?
Application: If $k$ people choose a random number in $\{1, 2, \cdots, n \}$, then the probability of at least two people having chosen the same number (i.e. probability of a "collision") is $1/2$, when $k = c \sqrt{n}$.
See birthday problem.
| Starting with
$\sum_{j=1}^{k-1} -\log\Big(1 - \frac jn \Big)
= \log(2)
$
we can use,
for $0< x < 1$,
$-\log(1-x)
=\sum_{m=1}^{\infty} \dfrac{x^m}{m}
$.
Then,
using the first $h$ terms,
$-\log(1-x)
\gt\sum_{m=1}^{h} \dfrac{x^m}{m}
$.
Looking at the remaining terms,
$\sum_{m=h+1}^{\infty} \dfrac{x^m}{m}
\lt \sum_{m=h+1}^{\infty} \dfrac{x^m}{h+1}
=\dfrac1{h+1} \sum_{m=h+1}^{\infty}x^m
=\dfrac{x^{h+1}}{(h+1)(1-x)}
$
so that
$-\log(1-x)
\lt\sum_{m=1}^{h} \dfrac{x^m}{m}+\dfrac{x^{h+1}}{(h+1)(1-x)}
$.
Setting $h=1$,
$x < -\log(1-x)
\lt x+\dfrac{x^{2}}{2(1-x)}
$.
Using the lower bound,
if $n > k$,
$\log(2)
\gt \sum_{j=1}^{k-1} j/n
=\dfrac{k(k-1)}{2n}
$.
Using the upper bound,
if $n > k$,
$\begin{array}\\
\log(2)
&\lt \sum_{j=1}^{k-1} (j/n+\dfrac{j^2/n^2}{2(1-j/n)})\\
&<\dfrac{k(k-1)}{2n}+\sum_{j=1}^{k-1}\dfrac{j^2/n^2}{2(1-k/n)}\\
&=\dfrac{k(k-1)}{2n}+\dfrac{1}{2n(n-k)}\sum_{j=1}^{k-1}j^2\\
&=\dfrac{k(k-1)}{2n}+\dfrac{1}{2n(n-k)}\dfrac{(k-1)k(2k-1)}{6}\\
&<\dfrac{k(k-1)}{2n}+\dfrac{(k-1)k^2}{3n(n-k)}\\
&=\dfrac{k(k-1)}{2n}(1+\dfrac{2k}{3(n-k)})\\
\end{array}
$
so that
$1
\lt \dfrac{2n\log(2)}{k(k-1)}
\lt 1+\dfrac{2k}{3(n-k)}
$.
From the first,
$n
> \dfrac{k(k-1)}{2\log(2)}
> \dfrac{k(k-1)}{2}
$.
From the second,
$\dfrac{k}{n-k}
\lt \dfrac{k}{k(k-1)/2-k}
= \dfrac{1}{(k-1)/2-1}
= \dfrac{1}{(k-3)/2}
= \dfrac{2}{k-3}
$
so
$ \dfrac{2n\log(2)}{k(k-1)}
\lt 1+\dfrac23\dfrac{2}{k-3}
= 1+\dfrac{4}{3(k-3)}
$.
Therefore
$\dfrac{k(k-1)}{2\log(2)}
\lt n
\lt \dfrac{k(k-1)}{2\log(2)}+\dfrac{2k(k-1)}{3\log(2)(k-3)}
$.
In particular,
$n
=\dfrac{k^2}{2\log(2)}+O(k)
$
or
$k
=\sqrt{2n\log(2)}+O(1)
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2500187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solving $a \sin\theta + b \cos\theta = c$ Could someone help me with the steps for solving the below equation $$a \sin\theta + b \cos\theta = c$$
I know that the solution is $$\theta = \tan^{-1} \frac{c}{^+_-\sqrt{a^2 + b^2 - c^2}} - \tan^{-1} \frac{a}{b} $$
I just can't figure out the right steps to arrive at this solution.
| here is a trick:
write
$$\frac{a}{\sqrt{a^2+b^2}}\sin(\theta)+\frac{b}{\sqrt{a^2+b^2}}\cos(\theta)=\frac{c}{\sqrt{a^2+b^2}}$$
Setting $$\cos(\phi)=\frac{a}{\sqrt{a^2+b^2}}$$ and $$\sin(\phi)=\frac{b}{\sqrt{a^2+b^2}}$$ then you will get
$$\sin(\phi+\theta)=\frac{c}{\sqrt{a^2+b^2}}$$
so $$\theta=\arcsin\left(\frac{c}{\sqrt{a^2+b^2}}\right)-\phi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2502976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Star operator in matrices and new math to find inverse? Example Calculation
I realized I could make some progress on an open problem if I introduced a new operation: $f(*) x = f(x)$, where $f$ is a function and $x$ is a variable, then I could make progress. It seems initially just like some new notation but its not as trivial. Consider the following transformation:
$$
\left( \begin{array}{cc}
x \\
y
\end{array} \right)=
%
\left( \begin{array}{cc}
e^{*} & 0 \\
0 & e^{*}
\end{array} \right) \cdot(
\left( \begin{array}{cc}
\ln(*) &\ln(\cos(*)) \\
\ln(*) & \ln(\sin(*))
\end{array} \right)
\left( \begin{array}{cc}
r \\
\theta
\end{array} \right))
$$
Note:
$$
\left( \begin{array}{cc}
e^{*} & 0 \\
0 & e^{*}
\end{array} \right) \cdot
\left( \begin{array}{cc}
\ln(*) &\ln(\cos(*)) \\
\ln(*) & \ln(\sin(*))
\end{array} \right)
\neq
\left( \begin{array}{cc}
* &\cos(*) \\
* & \sin(*)
\end{array} \right)$$
As one gets the wrong answer then as it seems associative property of matrices breaks down. Why? Because if one assumes it doesn't then one gets the answer:
$ x = r + \cos \theta$ and $ y = r + \sin \theta$ whereas if one solves the bracketed matrices first one gets:
$ x = r \cos \theta$ and $ y = r \sin \theta$. But it seems we can multiply by inverses that takes precedence over other operations (inverse precedence conjecture):
Multiplying both sides by an inverse of the leftmost matrix on the R.H.S:
$$
\left( \begin{array}{cc}
\ln(*) & 0 \\
0 & \ln(*)
\end{array} \right) \cdot
\left( \begin{array}{cc}
x \\
y
\end{array} \right)=
%
\left( \begin{array}{cc}
\ln(*) &\ln(\cos(*)) \\
\ln(*) & \ln(\sin(*))
\end{array} \right)
\left( \begin{array}{cc}
r \\
\theta
\end{array} \right)
$$
Once again, multiplying with an inverse (the inverse was using a clever guess in this case):
$$
(\left( \begin{array}{cc}
1 & 0 \\
0 & \tan^{-1} e^*
\end{array} \right) \cdot(
\left( \begin{array}{cc}
\exp(*)^2 & \exp(*)^2 \\
-1 & 1
\end{array} \right) \cdot
(\left( \begin{array}{cc}
\ln(*) & 0 \\
0 & \ln(*)
\end{array} \right) \cdot
\left( \begin{array}{cc}
x \\
y
\end{array} \right))))=
\left( \begin{array}{cc}
r \\
\theta
\end{array} \right)
$$
Questions
Is there any case where the inverse precedence conjecture fails? Is there a general procedure find an inverse of a matrix with $*$ operators in it?
| Lets take a 'general case' of an mxn matrix 'multiplying' a n element column vector, that is
$$
\left(
\begin{array}{cccc}
f_{11}(\star) & f_{12}(\star) & \ldots & f_{1n}(\star) \\
\vdots \\
f_{m1}(\star) & f_{m2}(\star) & \ldots & f_{mn}(\star)
\end{array}
\right)
\left(
\begin{array}{c}
x_1\\
\vdots \\
x_n
\end{array}
\right)
=
\left(
\begin{array}{cccc}
f_{11}(x_1) + f_{12}(x_2) +\ldots + f_{1n}(x_n) \\
\vdots \\
f_{m1}(x_1) + f_{m2}(x_2) + \ldots + f_{mn}(x_n)
\end{array}
\right)
$$
Thus we see that this notation is equivalent to a function
$$
f
\left(\left(
\begin{array}{c}
x_1\\
\vdots \\
x_n
\end{array}
\right) \right)
=
\left(
\begin{array}{cccc}
f_{11}(x_1) + f_{12}(x_2) +\ldots + f_{1n}(x_n) \\
\vdots \\
f_{m1}(x_1) + f_{m2}(x_2) + \ldots + f_{mn}(x_n)
\end{array}
\right)
$$
and we are considering functions of column vector expressable as a sum of functions of each element, the matrix notation being useful as book-keeping.
To obtain an inverse function of $f$ is not possible in general. However, if $m=n$ and the function-matrix has one non-zero element per row and collumn, so only $f_{1\sigma(1)},\ldots,f_{n\sigma(n)}$ are non zero (and invertable), where $\sigma$ is a permutation function on $1,\ldots,n$, then
$$
f
\left(\left(
\begin{array}{c}
x_1\\
\vdots \\
x_n
\end{array}
\right) \right)
=
\left(
\begin{array}{cccc}
f_{1\sigma(1)}(x_{\sigma(1)})\\
\vdots \\
f_{n\sigma(n)}(x_{\sigma(n)} )
\end{array}
\right)
$$
$$
f^{-1}
\left(\left(
\begin{array}{c}
y_1\\
\vdots \\
y_n
\end{array}
\right) \right)
=
\left(
\begin{array}{c}
f_{\sigma^{-1} (1) 1 }^{-1}(y_{\sigma^{-1} (1)})\\
\vdots \\
f_{\sigma^{-1} (n) n }^{-1}(y_{\sigma^{-1} (n)})
\end{array}
\right)
$$
One of the cases considered in the opening question corrosponds to the case $n=2$, $\sigma(i)=i$.
Explicit example:
Consider the case $n=m=2$ where we have a 'matrix'
$$
\left(
\begin{array}{cc}
f_{11}(\star) & f_{12}(\star) \\
f_{21}(\star) & f_{22}(\star)
\end{array}
\right)
=
\left(
\begin{array}{cc}
0 & \sin(\star) \\
\cos(\star) & 0
\end{array}
\right)
$$
so that there is exactly one non-zero element in each row and column. The action of this matrix can be summed up in a function
$$
f
\left(\left(
\begin{array}{c}
x_1\\
x_2
\end{array}
\right) \right)
=
\left(
\begin{array}{cc}
f_{11}(\star) & f_{12}(\star) \\
f_{21}(\star) & f_{22}(\star)
\end{array}
\right)
\left(
\begin{array}{c}
x_1\\
x_2
\end{array}
\right)
=
\left(
\begin{array}{cc}
0 & \sin(\star) \\
\cos(\star) & 0
\end{array}
\right)
\left(
\begin{array}{c}
x_1\\
x_2
\end{array}
\right)
=
\left(
\begin{array}{c}
\sin(x_2)\\
\cos(x_1)
\end{array}
\right)
$$
To find an inverse function we write
$$
\left(
\begin{array}{c}
y_1\\
y_2
\end{array}
\right)
=
\left(
\begin{array}{c}
\sin(x_2)\\
\cos(x_1)
\end{array}
\right)
$$
$$
\Rightarrow
\left(
\begin{array}{c}
x_1\\
x_2
\end{array}
\right)
=
\left(
\begin{array}{c}
\arccos(y_2)\\
\arcsin(y_1)
\end{array}
\right)
$$
$$
\Rightarrow
f^{-1}
\left( \left(
\begin{array}{c}
y_1\\
y_2
\end{array}
\right) \right)
=
\left(
\begin{array}{c}
\arccos(y_2)\\
\arcsin(y_1)
\end{array}
\right)
=
\left(
\begin{array}{cc}
0 & \arccos(\star)\\
\arcsin(\star) & 0
\end{array}
\right)
\left(
\begin{array}{c}
y_1\\
y_2
\end{array}
\right)
$$
$$
\Rightarrow
\left(
\begin{array}{cc}
0 & \sin(\star) \\
\cos(\star) & 0
\end{array}
\right)^{-1}
=
\left(
\begin{array}{cc}
0 & \arccos(\star)\\
\arcsin(\star) & 0
\end{array}
\right)
$$
The important things to note:
*
*the inverse of the 'matrix' contains the inverse functions
*the functions are permuted (I expect this corresponds to the matrix being 'transposed') to preserve function action
Tackling manipulation
Let us suppose now that we have some 'matrices' $F(\star)$ and $G(\star)$ with corresponding functions $f$ and $g$. Moreover, we have an expression
$$
y = F(\star) G(\star) x
$$
where $x$ and $y$ are column vectors. If we can find inverse functions for $f$ and $g$ then we can perform manipulations
$$
y = F(\star) G(\star) x = f(g(x))
$$
$$
\Rightarrow
f^{-1}(y) = g(x)
$$
$$
\Rightarrow
g^{-1}(f^{-1}(y)) = x
$$
If these inverse functions can be written in matrix form then
$$
F^{-1}(\star) y = G(\star) x
$$
and
$$
G^{-1}(\star)F^{-1}(\star)y = x
$$
Hopefully this sheds some light on your 'inverse precedence conjecture'
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2504894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$a_1 = a_2 = 1$ and $a_n = \frac{1}{2} \cdot (a_{n-1} + \frac{2}{a_{n-2}})$. Prove that $1 \le a_n \le 2: \forall n \in \mathbb{N} $
Let $a_n$ be a sequence satisfying $a_1 = a_2 = 1$ and $a_n = \frac{1}{2} \cdot (a_{n-1} + \frac{2}{a_{n-2}})$. Prove that $1 \le a_n \le 2: \forall n \in \mathbb{N} $
Attempt at solution using strong induction:
Base cases: $n = 1$ and $n = 2 \implies a_1 = a_2 = 1 \implies 1\le 1 \le 2$
Inductive assumption (strong induction): Assume that for all $m\in \mathbb{N}$ such that $1\le m \le k$, where $k\in \mathbb{N}$, the condition $1\le a_m \le 2$ holds True.
Show that $m = k+1$ holds true
$a_{k+1} = \frac{1}{2} \cdot (a_k + \frac{2}{a_{k-1}})$
I know that $a_k$ and $a_{k-1}$ are satisfying $ 1\le a_m\le2$ but I am not sure how to use that to prove that $a_{k+1}$ holds true for the condition.
| $$a_{n+1} = \frac{1}{2}a_{n} + \frac1{a_{n-1}}$$
Assume that for all $k \leq n$ we have $1 \leq a_k \leq 2$. Under those constraints it's easy to see that the above formula for $a_{n+1}$ takes on maximal value of $\frac{1}{2}\cdot 2 + \frac{1}{1} = 2$ and minimal value of $\frac{1}{2} \cdot 1 + \frac{1}{2} = 1$. But both of these extremes still maintain $1\leq a_{n+1} \leq 2$.
Thus when for all $k \leq n$ we have $1 \leq a_k \leq 2$ we also have $1 \leq a_{n+1} \leq 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2506658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
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