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Proof $13 \mid (k\cdot 2^n+1)$ if $n\equiv2 \pmod{12}$ and $k\equiv3 \pmod{13}$?
Proof $13 \mid (k \cdot 2^n + 1)$ if $n\equiv2 \pmod{12}$ and $k\equiv3 \pmod{13}$
Hint: for $k$ odd: $2^n \equiv-k' \pmod p$ and $kk' \equiv1 \pmod p$
My thoughts:
$13\mid(k-3) \Rightarrow k=13a+3$
and
$12|(n-2) \Rightarrow n=12b+2$
so
$\begin{align}k\cdot 2^n+1 &=(13a+3)2^{12b+2}+1 \\
&=4(13a+3)(2^{b})^{12}+1\\
\textrm{ or }
&=(k-3+3)\cdot 2^{n-2+2}+1\\
&=4\cdot 2^{n-2}(k-3+3)+1\\
&=4\cdot 2^{n-2}(k-3)+3\cdot2^n+1\end{align}$
I don't kow how to use the hint :(
|
If $n=12N+2$ and $k=13K+3$ then
$$
k2^n+1=13K2^n+3\cdot 4\cdot \left(\underbrace{2^{12}}_{\equiv 1\bmod{13}}\right)^{N}+1.
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Expand $\frac{1}{z-2}$ on $|z|>2$
Expand $\frac{1}{z-2}$ around $|z|>2$
So there is no singularity points
$$\frac{1}{(z-2)}=-\frac{1}{2-z}=-\frac{1}{2}\frac{1}{1-\frac{z}{2}}=-\frac{1}{2}\sum_{n=0}^{\infty} \frac{z^n}{2^n}=\sum_{n=0}^{\infty}-\frac{z^n}{2^{n+1}}$$
but we can also write:
$$\frac{1}{(z-2)}=\frac{1}{z}\frac{1}{1-\frac{2}{z}}=\frac{1}{z}\sum_{n=0}^{\infty} \frac{2^n}{z^n}=\sum_{n=0}^{\infty} \frac{2^n}{z^{n+1}}$$
So which one is true? the second one as for the first we get $|\frac{z}{2}|<1\Rightarrow |z|<2$ and we are looking at $|z|>2$?
|
You already answered your own question:
To expand
$$\frac{1}{1-q}$$
as you did in both cases, you need $|q|<1$ for the series to converge. Thus, if $|z|<2$, the first one is right, whereas for $|z|>2$, the second one is right.
|
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|
If the roots of an equation are $a,b,c$ then find the equation having roots $\frac{1-a}{a},\frac{1-b}{b},\frac{1-c}{c}$. Actually I have come to know a technique of solving this kind of problem but it's not exactly producing when used in a certain problem.
Say we have an equation
$$2x^3+3x^2-x+1=0$$
the roots of this equation are $a,b,c$. If I were to find the equation having the roots $\frac{1}{2a}, \frac{1}{2b}$ and $\frac{1}{2c}$ then I can use the root co-efficient relations and use the rule of creating equations from roots.
This gives me the result : $$4x^3 + 2x^2 -3x -1 =0$$
A much much simpler way to solve this math is: if we pick that, $f(x)= 2x^3+3x^2-x+1$ and the values of $x$ are $a,b$ and $c$ then $f\left(\frac{1}{2x}\right)$ will denote an equation (if we just write $f\left(\frac{1}{2x}\right)=0$ ) which has the roots $\frac{1}{2a}, \frac{1}{2b}$ and $\frac{1}{2c}$. And this way the result also matches the result of my previous work.
But now I face an equation
$$x^3+3x+1=0$$ and if the roots are $a,b,c$ then I have to find the equation with the roots $\frac{1-a}{a},\frac{1-b}{b},\frac{1-c}{c}$.
If I use the root-coefficient relations then I have the result
$$x^3+6x^2+9x+5=0$$ (which is correct I think).
But here $f\left(\frac{1-x}{x}\right)= 3x^3-6x^2+3x-1$. So writing $f\left(\frac{1-x}{x}\right)=0$ won't give me the actual equation.
Where am I mistaking actually?
|
The equation $x^3+3x+1=0$ has roots $a,b,c$. Find the equation whose roots are $\frac{1-a}{a},\frac{1-b}{b},\frac{1-c}{c}$.
So let $y=\frac{1-x}{x}$ ... invert this equation, we have $x=\frac{1}{1+y}$, now substitute this into the original equation & you get $\color{red}{y^3+6y^2+9y+5=0}$.
Note that in your previous example you should have made the substitution $y=\frac{1}{2x}$ and then inverted this to $x=\frac{1}{2y}$ ... which of course leads like exactly the same thing.
|
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|
A limit involving square roots Transcribed from photo
$$\require{cancel}
\lim_{n\to\infty}\sqrt{n+\tfrac12}\left(\sqrt{2n+1}-\sqrt{2n+3}\right)\\[12pt]
$$
$$
\lim_{n\to\infty}\sqrt{n+\tfrac12}\lim_{n\to\infty}\left(\sqrt{2n+1}-\sqrt{2n+3}\right)\frac{\sqrt{2n+1}+\sqrt{2n+3}}{\sqrt{2n+1}+\sqrt{2n+3}}\\[12pt]
$$
$$
\lim_{n\to\infty}\underbrace{\sqrt{\frac{\cancel{n}1}{\cancel{n}}+\cancelto{0}{\frac1{2n}}}}_1\lim_{n\to\infty}\frac{(2n+1)-(2n+3)}{\sqrt{2n+1}+\sqrt{2n+3}}\\[24pt]
$$
$$
\lim_{n\to\infty}\frac{(\cancel{2n}+1)-(\cancel{2n}+3)}{\sqrt{2n+1}+\sqrt{2n+3}}=\lim_{n\to\infty}\frac{-2}{\sqrt{2n+1}+\sqrt{2n+3}}\cdot\frac1{\sqrt{n}}\\[12pt]
$$
$$
\lim_{n\to\infty}\frac{\cancelto{0}{\frac2{\sqrt{n}}}}{\sqrt{2+\cancelto{0}{\frac1n}}+\sqrt{2+\cancelto{0}{\frac3n}}}=\frac0{2\sqrt2}=0\\[24pt]
$$
$$
0\cdot1=0
$$
I solved it, but i dont know if it is right.
I got $0$
|
$$\lim _{x\to \infty }\left(\sqrt{x+\frac{1}{2}}\right)\left(\sqrt{2x+1}-\sqrt{2x+3}\right)$$
As you said, we can rewrite:
$$\left(\sqrt{2x+1}-\sqrt{2x+3}\right)=\frac{-2}{\sqrt{2x+1}+\sqrt{2x+3}}$$
Try factoring the $x$
$$\frac{-2}{\sqrt{2x+1}+\sqrt{2x+3}}=\frac{-2}{\sqrt{\color{red}{x}(2+1/x)}+\sqrt{\color{red}{x}(2+3/x)}}=\frac{-2}{\color{red}{\sqrt{x}}\sqrt(2+1/x)+\color{red}{\sqrt{x}}\sqrt{2+3/x}}$$
Now factor the $\sqrt{x}$
$$\frac{-2}{\color{red}{\sqrt{x}}\sqrt(2+1/x)+\color{red}{\sqrt{x}}\sqrt{2+3/x}}=\frac{-2}{\color{red}{\sqrt{x}}(\sqrt{(2+1/x)}+\sqrt{2+3/x})}$$
Similarly, factor out the $x$ from $\sqrt{x+1/2}$.
This is pretty trivial, the answer is:
$$\sqrt{x}\sqrt{1+\frac{1}{2x}}$$
Great! Now let's put them together!
$$\lim_{x\to\infty} (\sqrt{x}\sqrt{1+\frac{1}{2x}})\cdot \frac{-2}{\sqrt{x}(\sqrt{(2+1/x)}+\sqrt{2+3/x})}$$
Oh well look at that, the $\sqrt{x}$ cancel! Now we are just left with:
$$\lim_{x\to\infty} (\sqrt{1+\frac{1}{2x}})\cdot \frac{-2}{(\sqrt{(2+1/x)}+\sqrt{2+3/x})}$$
Roughly speaking, since $1/\infty=0$, our limit simplies:
$$\lim_{x\to\infty} \sqrt{1}\cdot \frac{-2}{(\sqrt{2}+\sqrt{2})}=\frac{-2}{2\sqrt{2}}=\frac{-1}{\sqrt{2}}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Eliminating parameter from 2 parametric equations I am given two parametric equations as-
$ x= g\sin^2(t) - f\sin(t)\cos(t) $
$ y = f\cos^2(t) - g\sin(t)\cos(t) $
and I have to eliminate the parameter $t$.
I tried taking $\sin(t)$ common from the first equation and $\cos(t)$ from the second, and divided the two to get $ x/y = -\tan(t) $ but I couldn't proceed from there.
|
\begin{align}
\cos^2 t & = \frac 1 2 + \frac 1 2 \cos(2t) \\[10pt]
\sin^2 t & = \frac 1 2 - \frac 1 2 \cos(2t) \\[10pt]
\sin(t) \cos(t) & = \frac 1 2 \sin(2t) \\[10pt]
x & = g\sin^2 t - f\sin(t) \cos(t) = g\left( \frac 1 2 - \frac 1 2\cos(2t) \right) - f \cdot \frac 1 2 \sin(2t) \\[10pt]
y & = f \cos^2 t - g \sin(t)\cos(t) = f\left( \frac 1 2 + \frac 1 2 \cos(2t) \right) - g\cdot \frac 1 2 \sin(2t) \\[10pt]
fx - gy & = -fg\cos(2t) + \frac{g^2- f^2} 2 \sin(2t) \\[10pt]
gx - fy & = \frac{g^2-f^2} 2 - \frac {g^2+f^2} 2 \cos(2t) \\[10pt]
\sin(2t) & = \cdots \tag A \\[10pt]
\cos(2t) & = \cdots \tag B \\[10pt]
\text{Therefore } & \left( \cdots \right)^2 + \left( \cdots \right)^2 = 1. \text{ (Here insert the expressions in lines (A) and (B).)}
\end{align}
|
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|
Simplify $x^{\frac{2}{3}}+3x^{\frac{1}{3}}-2=0$ to $x^2 +45x-8=0$ Simplify ($x^{\frac{2}{3}}+3x^{\frac{1}{3}}-2=0$) to ($x^2 +45x-8=0$)
These two equations have the same solutions.
If given the first equation how would you go about simplifying such that you end up with the quadratic form ?
I tried cubing the first expression but it got really messy.
|
Let $\alpha$ and $\beta$ be the roots of $y^2+3y-2=0$. So
\begin{eqnarray*}
\alpha+\beta=-3 \\
\alpha \beta =-2.
\end{eqnarray*}
Now calculate the equation whose roots are $\alpha^3$ and $\beta^3$
\begin{eqnarray*}
\alpha^3+\beta^3&=&(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)&=&-45 \\
\alpha^3 \beta^3 &=&(\alpha\beta)^3&=&-8.
\end{eqnarray*}
So $x^2+45x-8=0$ is the equation with roots $\alpha^3$ and $\beta^3$. $x$ and $y$ are linked by $y=x^3$ and $y=x^{\frac{1}{3}}$.
|
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|
Characteristic polynomial of a $7 \times 7$ matrix whose entries are $5$ Avoiding too many steps, what is the characteristic polynomial of the following $7 \times 7$ matrix? And why?
\begin{pmatrix}5&5&5&5&5&5&5\\5&5&5&5&5&5&5\\5&5&5&5&5&5&5\\5&5&5&5&5&5&5\\5&5&5&5&5&5&5\\5&5&5&5&5&5&5\\5&5&5&5&5&5&5\end{pmatrix}
|
Here is a matrix $P,$ the columns are eigenvectors of your matrix. Note that $P$ is not orthogonal, although the columns are pairwise orthogonal.
$$
P =
\left( \begin{array}{rrrrrrr}
1 & -1 & -1 & -1 & -1 & -1 & -1 \\
1 & 1 & -1 & -1 & -1 & -1 & -1 \\
1 & 0 & 2 & -1 & -1 & -1 & -1 \\
1 & 0 & 0 & 3 & -1 & -1 & -1 \\
1 & 0 & 0 & 0 & 4 & -1 & -1 \\
1 & 0 & 0 & 0 & 0 & 5 & -1 \\
1 & 0 & 0 & 0 & 0 & 0 & 6
\end{array}
\right).
$$
The columns of $P$ are of varying lengths; lengths $ \sqrt{7}, \sqrt{2}, \sqrt{6}, \sqrt{12},..$ All that is necessary to make an orthogonal matrix $Q$ out of this is to divide each column by its length.
|
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|
If $f(x)=\frac{1}{x^2+x+1}$, how to find $f^{(36)} (0)$?
If $f(x)=\frac{1}{x^2+x+1}$, find $f^{(36)} (0)$.
So far I have tried letting $a=x^2+x+1$ and then finding the first several derivatives to see if some terms would disappear because the third derivative of $a$ is $0$, but the derivatives keep getting longer and longer. Am I on the right track? Thanks!
|
Let $\omega$ be a complex cube root of $1$. Then Partial Fraction representation of $f(x)$ is given by
$f(x) = \dfrac{1}{x^2+x+1} = \dfrac{1}{(x-\omega)(x-\omega^2)} = \dfrac{1}{\omega - \omega^2}\Big(\dfrac{1}{x-\omega} - \dfrac{1}{x - \omega^2}\Big)$.
Find successive derivatives to show that
$f^{(36)}(x) = \dfrac{1}{\omega - \omega^2}(36! (x-\omega)^{-37} - 36! (x - \omega^2)^{-37})$.
Let $x = 0$ and use $\omega^3 = 1$.
|
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|
In $\triangle ABC$, we have $AB=3$ and $AC=4$. Side $\overline{BC}$ and the median from $A$ to $\overline{BC}$ have the same length. What is $BC$? In $\triangle ABC$, we have $AB=3$ and $AC=4$. Side $\overline{BC}$ and the median from $A$ to $\overline{BC}$ have the same length. What is $BC$?
Alright, so if d is the midpoint then my two triangles are ADB and ADC.
Using the law of cosines I get:
$$3^2=a^2 + 2a^2 - 2\cdot a\cdot 2a\cdot\cos \angle ADB$$
for the first triangle,
$$4^2=a^2 + 2a^2 - 2\cdot a\cdot 2a\cdot\cos\angle ADC$$
for the second one. Then I'm stuck. Any help?
|
You can also practice another method:
Note the areas of $\Delta ABD$ and $\Delta ACD$ are equal.
Labeling $BD=CD=x$ and $AD=2x$, we will use the Heron's formula:
$$S_{\Delta ABD}= \sqrt{\frac{3+3x}{2}\cdot \left(\frac{3+3x}{2}-3\right)\cdot \left(\frac{3+3x}{2}-2x\right)\cdot \left(\frac{3+3x}{2}-x\right)}=$$
$$S_{\Delta ABD}= \sqrt{\frac{4+3x}{2}\cdot \left(\frac{4+3x}{2}-4\right)\cdot \left(\frac{4+3x}{2}-2x\right)\cdot \left(\frac{4+3x}{2}-x\right)} \Rightarrow$$
$$(3x+3)(3x-3)(3+x)(3-x)=(3x+4)(3x-4)(4-x)(4+x)\Rightarrow$$
$$(9x^2-9)(9-x^2)=(9x^2-16)(16-x^2) \Rightarrow$$
$$81x^2-9x^4-81+9x^2=144x^2-9x^4-256+16x^2 \Rightarrow$$
$$70x^2=175 \Rightarrow x=\sqrt{\frac{5}{2}} \Rightarrow BC=AD=2x=\sqrt{10}.$$
|
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|
Algebraic identities Given that $$a+b+c=2$$ and $$ab+bc+ca=1$$
Then the value of
$$(a+b)^2+(b+c)^2+(c+a)^2$$ is how much?
Attempt:
Tried expanding the expression. Thought the expanded version would contain a term from the expression of $a^3+b^3+c^3-3abc$, but its not the case.
|
Hint:
Express
$S_2=a^2+b^2+c^2$ in function of $s=a+b+c$ and $\sigma=ab+bc+ca$ from the algebraic identity for $(a+b+c)^2$.
|
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|
Solving Pinter 7.B.4 with a program Here is exercise 7.B.4 from 'A Book of Abstract Algebra' by Charles C. Pinter.
A solution to this using a C# program is posted.
Is there another good approach using a computer program?
Any language is welcome.
The subgroup of $S_5$ generated by
$
f = \begin{pmatrix}1 & 2 & 3 & 4 & 5\\2 & 1 & 3 & 4 & 5\end{pmatrix}
\qquad
g = \begin{pmatrix}1 & 2 & 3 & 4 & 5\\1 & 2 & 4 & 5 & 3\end{pmatrix}
$
has six elements. List them, then write the table of this group:
$\varepsilon = \begin{pmatrix}1 & 2 & 3 & 4 & 5\\1 & 2 & 3 & 4 & 5\end{pmatrix}$
$f = \begin{pmatrix}1 & 2 & 3 & 4 & 5\\2 & 1 & 3 & 4 & 5\end{pmatrix}$
$g = \begin{pmatrix}1 & 2 & 3 & 4 & 5\\1 & 2 & 4 & 5 & 3\end{pmatrix}$
$h = \begin{pmatrix}1 & 2 & 3 & 4 & 5\\ \, \end{pmatrix}$
$k = \begin{pmatrix}1 & 2 & 3 & 4 & 5\\ \,\end{pmatrix}$
$l = \begin{pmatrix}1 & 2 & 3 & 4 & 5\\ \,\end{pmatrix}$
$
\begin{array}{c|ccc}
\circ & \varepsilon & f & g & h & k & l \\
\hline
\varepsilon \\ f \\ g \\ h \\ k \\ l
\end{array}
$
|
This should be easy to do without a computer, as mentioned.
With a computer, you should be able to do this with any language. Using something designed for algebraic computation is easiest.
In MAGMA, I would type
{x : x in sub<Sym(5)|(1, 2), (3, 4, 5)>};
and it would list out all of the elements (I have them is cycle notation). You can easily form the table from here. You can do something very similar in Sage, GAP, even in Python there is the SymPy package that lets you work with permutation groups.
|
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|
for the cauchy problem , determine unique solution ,or no solution , or infinitely many solution for the cauchy problem , determine unique solution ,or no solution , or infinitely many solution
$u_x-6u_y=y$ with the date $u(x,y)=e^x$ on the line $y=-6x+2$
My attempt:
given $u_x-6u_y=y$ then $\frac{dx}{1}=\frac{dy}{-6}=\frac{du}{u}$
there fore the general solution $\phi(6x+y, y^2+12u)=0$
and hence implicit equation $y^2+12u=F(6x+y)$
but i cant go for further
|
given $u_x-6u_y=y$ then $\frac{dx}{1}=\frac{dy}{-6}=\frac{du}{y}\quad$ not $=\frac{du}{u}\quad$(typo).
$y^2+12u=F(6x+y)\quad$ is OK.
Condition $u(x,y)=e^x$ on the line $y=-6x+2\quad\to\quad y^2+12e^x=F(2)=$constant is impossible. Thus, there is no solution.
Condition $u(x,y)=1$ on the line $y=-x^2\quad\to\quad x^4+12=F(6x-x^2)\quad$ allows to determine a function $F(X)= 12+(3\pm\sqrt{9-X})^4$
$$ u(x,y)=\frac{1}{12}\left(-y^2+ 12+\left(3\pm\sqrt{9-(6x-x^2)}\right)^4\right)$$
Condition $u(x,y)=-4x$ on the line $y=-6x\quad\to\quad (-6x)^2+12(-4x)=F(0)=$constant is impossible. Thus, there is no solution.
|
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|
How to obtain the sum of the following series? $\sum_{n=1}^\infty{\frac{n^2}{2^n}}$ It seems that I'm missing something about this.
First of all, the series is convergent: $\lim_{n\rightarrow\infty}\frac{2^{-n-1} (n+1)^2}{2^{-n} n^2}=\frac{1}{2}$ (ratio test)
What I tried to do is to find a limit of a partial sum $\lim_{n\rightarrow\infty}S_n$ as follows: $S_n=\frac{\frac{1}{6} n (n+1) (2 n+1)}{\frac{1-\left(\frac{1}{2}\right)^n}{2 \left(1-\frac{1}{2}\right)}}$. Still, the limit is $\infty$ and I'm clearly doing something wrong.
|
Suppose $|x|<1$ at the last you put $x=\frac12$
$$s_1=1+x+x^2+x^3+x^4+...=\frac{1}{1-x}$$nw find $s'_1$
$$s'_1=0+1+2x+3x^3+4x^3+5x^4+...+nx^{n-1}+...=(\frac{1}{1-x})'\\1+2x+3x^2+4x^3+...=\frac{1}{(1-x)^2}$$now multiply by $x$
$$S_2=xS'_1=x^1+2x^2+3x^3+4x^4+5x^5+...=\frac{x}{(1-x)^2}$$
now find $S_2'$
$$S'_2=1+2^2x^1+3^2x^2+4^2x^3+5^2x^4+...+(n^2x^{n-1})+...=(\frac{x}{(1-x)^2})'$$finally :multiply by $x$
$$xS'_2=1^2x+2^2x^2+3^2x^3+4^2x^4+...(n^2x^n)+...=x(\frac{x}{(1-x)^2})'\\
\sum_{n=2}^{\infty}n^2x^n=x(\frac{x}{(1-x)^2})'-x$$put $x=\frac12$
$$\sum_{n=2}^{\infty}n^2(\frac{1}{2})^n=\sum_{n=2}^{\infty}\frac{n^2}{2^n}$$
for adjustment $$\sum_{n=2}^{\infty}\frac{n^2}{2^n}=(\sum_{n=1}^{\infty}\frac{n^2}{2^n} )-\frac{1^2}{2^1}$$ you can find $$x(\frac{x}{(1-x)^2})'-x$$ then put $x=\frac12$
|
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|
Find $\lim\limits_{x\to 1}\frac9{1-x^9}-\frac7{1-x^7}$ without using L'Hopital $$\lim_{x\to 1}\frac9{1-x^9}-\frac7{1-x^7}$$
I did this with l'Hôpital's rule, but how can we do this problem other than that? Any hints will be good.
|
The difference of fractions equals $\frac{9(1-x^7) - 7(1-x^9)}{(1-x^7)(1-x^9)}$.
Try to divide out the common $(1-x)^2$ terms in both the bottom and the top.
In the bottom we get $(1+x+x^2+\ldots+x^8)(1+x + x^2 + \ldots + x^6)$
In the top $7x^7 + 14x^6 + 12x^5 + 10x^4 + 8x^3 + 6x^2 + 4x + 2$ (multiply by $(1-x)^2$ to check both). Now we can just substitute $x=1$ in both. I get 63 twice OTTOMH.
|
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}
|
If $a,b,c$ are distinct positive numbers , show that $\frac{a^8 + b^8 + c^8}{a^3 b^3 c^3} > \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ If $a,b,c$ are distinct positive numbers , show that $$\frac{a^8 + b^8 + c^8}{a^3 b^3 c^3} > \frac{1}{a} + \frac{1}{b} + \frac{1}{c}.$$ I am thinking of Tchebycheff's inequality for this question, but not able to proceed. How do I solve this?
|
The given inequality can be written as
$$a^8+b^8+c^8 > a^2b^3c^3+b^2a^3c^3+c^2a^3b^3.$$
Since $(8,0,0)$ majorizes $(2,3,3)$, therefore we can apply Muirhead's inequality to get the result.
|
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|
The number of triangles with each side having integral length... The number of triangles with each side having integral length and the longest side is of 11 units is:-? MY ATTEMPT:- I applied following constraints: $12 \leq {a+b} \leq 22$ $a,b \geq 1$ I made different cases that a+b =12, a+b=13 and so on. My answer came was 160 but it is not the correct one.
|
Your conditions are not quite correct. There is no stipulation on what $a + b$ is the only stipulation is that $11$ is the longest sides.
We need $a \le b \le 11$ so we don't count multiple instances the same and so that the maximum side(s) is $11$.
We need $a + b > 11$ or $a + b \ge 12$ to satisfy the (non-trivial) triangle inequality.
So the answer is $\sum\limits_{a=1}^{11}\sum\limits_{b=\max(a,12-a)}^{11}1$
$= \sum\limits_{a=1}^{11}[12- \max(a,12-a)]$
$= \sum\limits_{a=1}^{11}\min(12-a, a)$
$= 1 + 2 + 3 + 4+ 5 + 6 +5 + 4 + 3+2+1 = 21 + 15 = 36$.
I.E. $(a,b) = $
$(1,11)$
$(2,10)..(2, 11)$
$(3,9)....(3,11)$.
.....
$(5,7)....(5,11)$.
$(6,6)....(6,11)$.
$(7,7)....(7,11)$.
......
$(10,11)...(11,11)$
$(11,11)$
|
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|
$\sqrt{9x^2-16}>3x+1$ I'm trying to solve the following inequality:
$$\sqrt{9x^2-16}>3x+1$$
Here's my attempt:
$\sqrt{9x^2-16}>3x+1$
$\longrightarrow 9x^2-16>9x^2+6x+1$
$\longrightarrow -16>6x+1$
$\longrightarrow x<-\frac{17}{6}$
Now, I need to check the constraints:
$9x^2-16 > 0$
$\longrightarrow (3x)^2 > 4^2$
$\longrightarrow \pm3x > 4$
$\longrightarrow 3x > 4$; $-3x > 4$
$\longrightarrow x > \frac{4}{3}$; $x < -\frac{4}{3}$
Making sure the answer meets the constraints:
$\{(-\infty, -\frac{4}{3})\cup(\frac{4}{3}, \infty)\}\cap (-\infty, -\frac{17}{6}) = (-\infty, -\frac{17}{6})$
So, my answer is $x=(-\infty, -\frac{17}{6})$, however verifying on Wolfram|Alpha results in $x=(-\infty, -\frac{4}{3}]$.
Where, what, and why is wrong with my solution?
|
$a > b \implies a^2 > b$ is not true if $b < 0$.
We have two cases:
1) if $3x + 1 \ge 0$ or $x \ge -\frac 13$ you correctly got $x < \frac {-17}6$. But that contradicts $x \ge -\frac 13$
So we must have case 2:
2) $3x + 1 < 0$ and $x < -\frac 13$.
From that we can't square both sides as $a \ge 0 > b$ does/can not tell us anything about $a^2$ in relation to $b^2$ as $|a|$ and $|b|$ can be any relation.
In other words we have no idea if $\sqrt{x^2 - 16}^2 >,<, = (3x + 1)^2$. Any one of those is possible.
But we do have $9x^2 - 16 \ge 0$ and $x < -1/3$. That will get you $x^2 \ge \frac {16}9$ so $|x| \ge \frac 43$ so either $x \ge \frac 43$ which is is a contradiction, or $x \le -\frac 43$. So $x \le -\frac 43$
So we have $x \le -\frac 43$ and $x < -\frac 13$. So solution is $x \le -\frac 43$.
|
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|
Ratio of Height to the Radius of its base A conical tent is of given capacity.For the least amount of canvas required for it,find the ratio of its height to the radius of its base.
(capacity usually means its volume.Only units are different)
|
Let $h$ be an altitude of the tent, $r$ be a radius of the base and $V$ be a volume of the tent.
Also, let $h=rx$.
Thus, $V=\frac{\pi r^2h}{3}$ or
$$V=\frac{\pi r^3x}{3},$$
which gives
$$r=\sqrt[3]{\frac{3V}{\pi{x}}}.$$
Thus, for an area of the canvas by AM-GM we obtain:
$$f(x)=\pi{r}\sqrt{h^2+r^2}=\sqrt[3]{9\pi V^2}\cdot\frac{\sqrt{x^2+1}}{\sqrt[3]{x^2}}=$$
$$=\sqrt[3]{9\pi V^2}\cdot\frac{\sqrt{\frac{x^2}{2}+\frac{x^2}{2}+1}}{\sqrt[3]{x^2}}\geq\sqrt[3]{9\pi V^2}\cdot\frac{\sqrt{3\sqrt[3]{\left(\frac{x^2}{2}\right)^2\cdot1}}}{\sqrt[3]{x^2}}=3\sqrt[3]{\frac{\sqrt3\pi V^2}{2}}.$$
The equality occurs for $\frac{x^2}{2}=1$, which gives $x=\sqrt2$.
Done!
|
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|
Does $\lim_{n\rightarrow\infty} (\frac{1}{n+a} + \frac{1}{n+2a} + \cdots + \frac{1}{n+na}) $ converge? I want to check whether
$\lim_{n\rightarrow\infty} (\frac{1}{n+a} + \frac{1}{n+2a} + \cdots + \frac{1}{n+na}) $
converges or not. (a is a positive constant number.)
If it converges, how to find the value it converges?
And if not, why?
|
$\frac{1}{n+a}+\dots+\frac{1}{n+an} = (\frac{1}{1}+\frac{1}{2}+\dots+\frac{1}{n+na})-(\frac{1}{1}+\frac{1}{2}+\dots+\frac{1}{n+a-1})$
$= \log(n+na)-\log(n+a-1)+o(1) = \log(\frac{n(a+1)}{n+a-1})+o(1)$, so the limit as $n \to \infty$ is $\log(a+1)$, since $\log$ is continuous.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
A sequence defined through an arc tangent and arc cotangent Let $$f:\mathbb{R^*}\to \mathbb{R}, f(x) = \arctan{\dfrac{1}{x}}-\operatorname{arccot}{\dfrac{1}{x}}$$
I found that this function is decreasing on $\mathbb{R^*}$ and a bijection for the same domain and for the range $(-\dfrac{3
\pi}{2};\dfrac{\pi}{2})$.
If we define $x_n$ the sequence of solutions to the equation $$f(x)=\dfrac{1}{n}$$ than prove that $x_n$ is convergent and find its limit.
Here is what I did:
$$\arctan\dfrac{1}{x_n}=\dfrac{1}{n}+\operatorname{arccot}\dfrac{1}{x_n}$$
$$\dfrac{1}{x_n}=\tan \left( \dfrac{1}{n}+\operatorname{arccot}\dfrac{1}{x_n} \right)$$
$$\dfrac{1}{x_n}=\dfrac{\tan\dfrac{1}{n}+x_n}{1-x_n\tan\dfrac{1}{n}}$$
After recombination, we have:
$$(x_n)_{1,2}=\dfrac{-\sin\dfrac{1}{n}+1}{\cos\dfrac{1}{n}}$$
I have already taken the positive solution. For $n$ a natural number both $\sin\dfrac{1}{n}$ and $\cos\dfrac{1}{n}$ are positive.
Obviously the limit of the sequence is $1$, but how do I proove the convergence? Thank you!
|
$f(x) = \arctan{\frac{1}{x}}-arccot{\frac{1}{x}}$
Since
$\arctan(x)+arccot(x)
=\frac{\pi}{2}
$,
this becomes
$f(x)
= 2\arctan{\frac{1}{x}}-\frac{\pi}{2}
$.
If
$f(x_n) = \frac1{n}$.
then
$2\arctan{\frac{1}{x_n}}
=\frac{\pi}{2}+\frac1{n}
$
or
$\begin{array}\\
\frac{1}{x_n}
&=\tan(\frac{\pi}{4}+\frac1{2n})\\
&=\frac{\tan(\frac{\pi}{4})+\tan(\frac1{2n})}
{1-\tan(\frac{\pi}{4})\tan(\frac1{2n})}\\
&=\frac{1+\tan(\frac1{2n})}
{1-\tan(\frac1{2n})}\\
\text{so}\\
x_n
&=\frac{1-\tan(\frac1{2n})}{1+\tan(\frac1{2n})}\\
&=1-\frac{2\tan(\frac1{2n})}{1+\tan(\frac1{2n})}\\
&=1-\frac{2(\frac1{2n}+(\frac1{n^2}))}{1+\frac1{2n}+(\frac1{n^2})}\\
&=1-\frac1{n}(1+\frac1{n})(1-\frac1{2n}+(\frac1{n^2}))\\
&=1-\frac1{n}(1+\frac1{2n}+\frac1{n^2})\\
&\to 1\\
\end{array}
$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that $\frac {1}{\sqrt{5}}[(\frac {1}{x+r_+}) - (\frac {1}{x+r_-}) = \frac {1}{\sqrt{5}x}[(\frac {1}{1-r_{+}x}) - (\frac {1}{1-r_{-}x})] $ I need to manipulate this equation:
$$
\frac {1}{\sqrt{5}}\left(\frac {1}{x+r_+} - \frac {1}{x+r_-}\right)
$$
to show that
$$ \frac {1}{\sqrt{5}}\left(\frac {1}{x+r_+} - \frac {1}{x+r_-}\right) = \frac {1}{\sqrt{5}x}\left(\frac {1}{1-r_{+}x} -\frac {1}{1-r_{-}x}\right)
$$ (*)
where$$r_+=\frac {1+\sqrt{5}}{2} $$and $$r_-=\frac {1-\sqrt{5}}{2}$$
I know that both RHS and LHS of (*) are equal to $\frac{1}{1-x-x^2} $ , but don't know how we get RHS manipulating LHS before someone told us that RHS is also equal to $\frac{1}{1-x-x^2} $
Edit:
Since $$r_+r_-=-1$$, I replace $r_+$ with $r_-$ and $r_-$ with $ r_+$ in the LHS of (*)
So that, I get
$$ \frac {1}{\sqrt{5}}\left(\frac {r_+}{1-r_+x} - \frac {r_-}{1-r_-x}\right)$$ Still, I am not there. Can anybody help me at this step?
After showing the equality above I can continue as
$$
\left(\frac {1}{1-r_{+}x}\right) - \left(\frac {1}{1-r_{-}x}\right)= {\sum_{n\ge0}\ r_+^nx^n}-{\sum_{n\ge0}\ r_-^nx^n}
$$
|
\begin{align*}
\frac {1}{\sqrt{5}}\left(\frac {1}{x+r_+} - \frac {1}{x+r_-}\right)
&= \frac {1}{\sqrt{5}} \frac{x}{x} \left(\frac {1}{x+r_+} - \frac {1}{x+r_-}\right) &&,x \neq 0 \\
&= \frac {1}{\sqrt{5}\, x} \left(\frac {x}{x+r_+} - \frac {x}{x+r_-}\right) &&,x \neq 0 \\
&= \frac {1}{\sqrt{5}\, x} \left(\frac {(x^2 + r_- x) - (x^2 + r_+ x)}{(x+r_+)(x+r_-)}\right) &&,x \neq 0 \\
&= \frac {1}{\sqrt{5}\, x} \left(\frac {r_- x - r_+ x}{(x+r_+)(x+r_-)}\right) &&,x \neq 0 \\
&= \frac {1}{\sqrt{5}\, x} \left(\frac {r_- x - r_+ x}{x^2 + x - 1}\right) &&,x \neq 0 \\
&= \frac {1}{\sqrt{5}\, x} \left(\frac {r_- x - r_+ x}{-(-x^2 - x + 1)}\right) &&,x \neq 0 \\
&= \frac {1}{\sqrt{5}\, x} \left(\frac {-r_- x + r_+ x}{-x^2 - x + 1}\right) &&,x \neq 0 \\
&= \frac {1}{\sqrt{5}\, x} \left(\frac {1-r_- x -1 + r_+ x}{(1-r_+ x)(1-r_- x)}\right) &&,x \neq 0 \\
&= \frac {1}{\sqrt{5}\, x} \left(\frac {(1-r_- x) - (1 - r_+ x)}{(1-r_+ x)(1-r_- x)}\right) &&,x \neq 0 \\
&= \frac {1}{\sqrt{5}\, x} \left(\frac {1}{1-r_+ x} - \frac {1}{1-r_- x}\right) &&,x \neq 0 \\
\end{align*}
For completeness, we verify that the RHS of the equation sought is undefined for $x=0$, so the condition cannot be removed.
|
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|
Compute $\int\limits_{-\infty}^{\infty}\frac{1}{(1+x^2)^{n+1}}dx$ via residue calculus. Let $\Gamma_R$ be the semicircle of radius $R$ in the upper half plane. Then,
\begin{align}
\int\limits_{-\infty}^{\infty}\frac{1}{(1+x^2)^{n+1}}dx
&= \lim_{R\to \infty}\int_{\Gamma_R}\frac{1}{(1+z^2)^{n+1}}dz \\
&= 2\pi i \operatorname{Res}\left(\frac{1}{(1+z^2)^{n+1}},i\right)
\end{align}
The pole of the function at $i$ is of order $n+1$, so the residue is computed by
\begin{align}
\operatorname{Res}\left(\frac{1}{(1+z^2)^{n+1}},i\right) &= \frac{1}{n!}\lim_{z\to i}\frac{d^n}{dz^n}\left(\frac{1}{(z+i)^{n+1}}\right) \\
&= \frac{1}{n!}\lim_{z\to i}(-1)^n\frac{(n+1)(n+2)\cdots(2n+1)}{(z+i)^{2n+1}} \\
&=\frac{(2n+1)!}{i2^{2n+1}(n!)^2}
\end{align}
Hence, $$\int_{-\infty}^{\infty}\frac{1}{(1+x^2)^{n+1}}dx = \pi\frac{(2n+1)!}{2^{2n}(n!)^2}$$ The answer provided is $\frac{1\cdot 3\cdot 5\cdots(2n-1)}{2\cdot 4\cdot 6\cdots (2n)}\pi$. How do I manipulate my answer to obtain this answer?
|
Note that by residuals taking the upper semicircle as trajectory:
\begin{eqnarray*}
\int_{-\infty}^{+\infty} \frac{1}{(1+x^{2})^{n+1}}
& = & 2\pi i\text{Res}(f(z),i)\ i\text{ is a pole of order }n+1\\
% & = & 2\pi i\left(\frac{1}{n!}\frac{d^{n}}{dz^{n}}(z-i)^{n+1}f(z)\right)\\
& = & 2\pi i\left(\lim_{z\to i}\frac{1}{n!}\frac{d^{n}}{dz^{n}}(z-i)^{n+1}\frac{1}{(1+z^{2})^{n+1}}\right)\\
& = & \frac{2\pi i}{n!}\lim_{z\to i}\frac{d^{n}}{dz^{n}}\frac{(z-i)^{n+1}}{(-i^2+z^{2})^{n+1}}\\
& = & \frac{2\pi i}{n!}\lim_{z\to i}\frac{d^{n}}{dz^{n}}\left(\frac{z-i}{z^{2}-i^2}\right)^{n+1}\\
& = & \frac{2\pi i}{n!}\lim_{z\to i}\frac{d^{n}}{dz^{n}}\left(\frac{1}{z+i}\right)^{n+1}\\
& = & \frac{2\pi i}{n!}\lim_{z\to i}\frac{d^{n}}{dz^{n}}(z+i)^{-(n+1)}\\
% & = & \frac{2\pi i}{n!}\lim_{z\to i}\frac{d^{n}}{dz^{n}}(z+i)^{-(n+1)}\\
& = & \frac{2\pi i}{n!}\lim_{z\to i}\frac{d^{n-1}}{dz^{n-1}}(-n-1)(z+i)^{-(n+2)}\\
& = & \frac{2\pi i}{n!}\lim_{z\to i}\frac{d^{n-2}}{dz^{n-2}}(-n-1)(-n-2)(z+i)^{-(n+3)}\\
& = & \dots\dots\\
& = & \frac{2\pi i}{n!}\lim_{z\to i}(-n-1)(-n-2)\cdots(-n-n)(z+i)^{-(n+n+1)}\\
& = & \frac{2\pi i}{n!}\lim_{z\to i}(-1)^{n}(n+1)(n+2)\cdots(2n)(z+i)^{-(2n+1)}\\
& = & \frac{2\pi i}{n!}(-1)^{n}(n+1)(n+2)\cdots(2n)(2i)^{-(2n+1)}\\
& = & \frac{\pi}{n!}(-1)^{n}(n+1)(n+2)\cdots(2n)(2i)^{-2n}\\
& = & \frac{\pi(2n)!}{n!n!}(-1)^{n}(4i^2)^{-n}\\
& = & \frac{\pi(2n)!}{n!n!}\frac{1}{4^n}(-1)^{n}(-1)^{-n}=\frac{\pi(2n)!}{2^{2n}(n!)^2}=\pi\cdot\frac{1}{n!2^{n}}\cdot\frac{(2n)!}{n!2^{n}}\\
\end{eqnarray*}
The important thing is to note that
\begin{equation*}
\frac{(2n)!}{n!2^{n}} = 1\cdot3\cdot5\cdots(2n-1)\text{ and }n!2^{n}=2\cdot4\cdot6\cdots(2n)
\end{equation*}
of what follows the identity sought
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to prove that $2^0 + 2^1 + 2^2 + 2^3 + \cdots + 2^n = 2^{n+1} - 1$ When I'm reading Computer Systems: A Programmer's Perspective, I met the sum of binary numbers and failed to prove it:
$$ 2^0 + 2^1 + 2^2 + 2^3 + \cdots + 2^n = 2^{n+1} - 1 $$
This might be preliminary knowledge, I'm not good at mathematics, any body could give me a hint?
|
You are going to walk a distance of one mile. You walk half the distance and rest. You then walk half the remaining distance and rest. Continue doing this.
So now we can write down everyone's favorite infinite series,
$\frac{1}{2} +\frac{1}{4} +\frac{1}{8} +\frac{1}{16} + \dots = 1$
Multiply both sides by by $2^{n+1}$,
$2^{n} + 2^{n-1} + 2^{n-2} +...+2^2+2^1+2^0+\frac{1}{2} +\frac{1}{4} +\frac{1}{8} +\dots = 2^{n+1}$.
OR
$2^{n} + 2^{n-1} + 2^{n-2} +...+2^2+2^1+2^0+1 = 2^{n+1}$.
Subtracting $1$ from each side,
$2^{n} + 2^{n-1} + 2^{n-2} +...+2^2+2^1+2^0 = 2^{n+1}-1$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate the given limit: $\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$
Evaluate the following limit.
$$\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$$
My Attempt:
$$=\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$$
$$=\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx}) \times \dfrac {\sqrt {x-a} + \sqrt {bx}}{\sqrt {x-a} + \sqrt {bx}}$$
$$=\lim_{x\to \infty} \dfrac {x-a-bx}{\sqrt {x-a} + \sqrt {bx}}.$$
How do I proceed?
|
That is a good idea! Now, intuitively speaking, a square root grows slower than a linear function so you should get "some" infinity. However, it depends on the constants $a$ and $b$ what the result will look like.
E.g. if $a = 0$ and $b=2$, you would get
\begin{align*}
\lim_{x \rightarrow \infty} \frac{x - 2x}{\sqrt{x} + \sqrt{2x}} &= \lim_{x \rightarrow \infty} \frac{-x}{\sqrt{x} + \sqrt{2x}} \\
&= \lim_{x \rightarrow \infty} \frac{- x}{(1 + \sqrt{2}) \sqrt{x}} = - \lim_{x \rightarrow \infty} \frac{\sqrt{x}}{(1 + \sqrt{2}) } \\
&= - \infty.
\end{align*}
But for other choices of constants $a,b$ you might get $+\infty$. So your answer depends on their values.
In general, a way to proceed would be to use L'Hospital's rules to analyse limits of quotients .
|
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"timestamp": "2023-03-29T00:00:00",
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|
Radical problem: $\sqrt{7+2\sqrt{12}}+\sqrt{7-2\sqrt{12}}$ What is the value of
$$\sqrt{7+2\sqrt{12}}+\sqrt{7-2\sqrt{12}}$$
|
Since $(2\pm\sqrt3)^2=7\pm4\sqrt3=7\pm2\sqrt{12}$,$$\sqrt{7+2\sqrt{12}}+\sqrt{7-2\sqrt{12}}=2+\sqrt3+2-\sqrt3=4.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Why does $a^2x^2+(a^2+b^2-c^2)xy+b^2y^2>0,$ imply $(a^2+b^2-c^2)^2-4a^2b^2<0$? Why does $a^2x^2+(a^2+b^2-c^2)xy+b^2y^2>0,$ imply $(a^2+b^2-c^2)^2-4a^2b^2<0$, if $x, y$ are reals greater than $1$, and $a, b, c$ are positive reals?
A proof with all the math to go from one to the other would be nice.
|
Below is old incorrect answer ignoring $x>1$ condition. Explanation:
The discriminant is less than $0$ iff for all $x$ the quadratic has no roots.
However, we only know the LHS has no roots for all $x>1$ so this cannot be used. In fact, the counterexample pointed out by @WillJagy shows this.
$$a^2x^2+(a^2+b^2-c^2)xy+b^2y^2>0$$
$\implies$ (discriminant for LHS quadratic in $x$ is less than $0$ as the LHS has no roots)
$$(a^2+b^2-c^2)^2y^2-4a^2b^2y^2<0$$
$\implies$
$$(a^2+b^2-c^2)^2-4a^2b^2<0$$
|
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|
How to prove $\ln{\frac{n+1}{n}}\le\frac{2}{n+1},\forall n\in\mathbb{N}^+$? I found an inequality: $$\ln\left(\frac{n+1}{n}\right)\le\frac{2}{n+1},\forall n\in\mathbb{N}^+.$$
I tried induction. It is obvious if $k=1$, when $n=k$, $\ln\sqrt{\left(\frac{k+1}{k}\right)^{k+1}}\le 1$, but bogged down for $n=k+1$:
$$\ln\sqrt{\left(\frac{k+2}{k+1}\right)^{k+2}}\le 1$$
|
Your question is equivalent to prove the following form
$$
(2-\frac{n-1}{n})\leq 2^{(1-\frac{n-1}{n+1})} \tag{1}
$$
The relation $(1)$ is true becuse $\frac{n-1}{n} \geq \frac{n-1}{n+1}$.
Edit:
If the relation $(1)$ be true then we can conclude that:
$$
\left\{
\begin{array}{c}
(2-\frac{n-1}{n})\leq 2^{(1-\frac{n-1}{n+1})} \\
\\
2^{(1-\frac{n-1}{n+1})} \leq e^{(1-\frac{n-1}{n+1})}
\end{array}
\right.
\Rightarrow
(2-\frac{n-1}{n})\leq e^{(1-\frac{n-1}{n+1})} \tag{2}
$$
$$
\ln (2-\frac{n-1}{n})\leq (1-\frac{n-1}{n+1})
\Rightarrow
\ln\left(\frac{n+1}{n}\right)\le\frac{2}{n+1}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Quadratic equations having a common root. I have a cubic polynomial $f(x)=x^3+px^2+qx+72$ which is divisible by both $x^2+ax+b$ and $x^2+bx+a$ (where a,b,p,q are constants and a$ \neq $b).I have to find the sum of the squares of the the roots of the cubic polynomial.
I tried to attempt it like this.
Since the quadratics divide the polynomial its roots must be the same as the roots of the quadratics.
Let the roots of the first quadratic be A and B and that of second quadratic be D and E.
Now for having 3 roots of the cubic polynomial one root of the 2 quadratics must be common.Then suppose B=D.
Now from the first quadratic $A+B=-a$ and $AB=b$.From second quadratic $B+E=-b$ and $BE=a$.
We need the sum of squares of the roots of the cubic polynomial i.e. $A^2+B^2+E^2$.We have $A^2+B^2+E^2=(A+B+E)^2-2(AB+BE+AE)$.We have $A+B+E=-p$ and $AB+BE+AE=q$.So $A^2+B^2+E^2=-p-2q$.
I don't know how to proceed.
|
Suppose $\alpha$ is the common root for the two quadratics $g(x)$ and $h(x)$. Then
\begin{align*}
g(\alpha)=\alpha^2+a\alpha+b &=0\\
h(\alpha)=\alpha^2+b\alpha+a &=0
\end{align*}
Then solving for $\alpha$ gives $\alpha=1$ (assuming $a\neq b$). This also tells us
$$a+b=-1.$$
This means the quadratic equation $g(x)=0$ has roots $1,b$ and the quadratic
equation $h(x)=0$ has roots $1,a$. Thus the three roots of the cubic are $1,a,b$.
So sum of squares is
$$1+a^2+b^2=1+(a+b)^2-2ab=1+1-2(-72)=146.$$
|
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|
In how many ways the sum of 5 thrown dice is 25? What I thought about is looking for the number of solutions to
$$x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=25$$ such that $1\leq x_{i}\leq6$
for every $i$.
Now I know that the number of solutions to this such that
$0\leq x_{i}$ for every $i$ is $${5+25-1 \choose 5-1}={29 \choose 4}$$
How can I continue from here?
Thanks
|
When we multiply polynomials the formula is
$$ \left( \sum_i a_ix^i \right)\left( \sum_j b_jx^j \right) = \sum_k \left( \sum_{i + j = k} a_ib_j \right) x^k. $$
For a product of $5$ polynomials, we have:
$$ \left( \sum_{i_1} a_{1, i_1}x^{i_1} \right)\left( \sum_{i_2} a_{2, i_2}x^{i_2} \right)\left( \sum_{i_3} a_{3, i_3}x^{i_3} \right)\left( \sum_{i_4} a_{4, i_4}x^{i_4} \right)\left( \sum_{i_5} a_{5, i_5}x^{i_5} \right)$$
$$= \sum_k \left( \sum_{i_1 + i_2 + i_3 + i_4 + i_5 = k} a_{1,i_1}a_{2,i_2}a_{3,i_3}a_{4,i_4}a_{5,i_5} \right) x^k. $$
In particular notice the sum is over all solutions to $i_1 + i_2 + i_3 + i_4 + i_5 = k$. Here we want $k = 25$ and we want $a_{1,i_1}a_{2,i_2}a_{3,i_3}a_{4,i_4}a_{5,i_5} = 1$ except when $i_1, i_2, i_3, i_4$ or $i_5$ is $0$ or $> 6$ in which case we want $0$.
Thinking about this for a minute, we know that we want the coefficient on $x^{25}$ in the product
$$ (x + x^2 + x^3 + x^4 + x^5 + x^6)^5. $$
The polynomial $x + x^2 + x^3 + x^4 + x^5 + x^6$ is telling us that each die can be either $1,2,3,4,5$ or $6$ and each number occurs exactly once.
Let $[x^{n}]f(x)$ denote the coefficient of $x^n$ in $f(x)$. Then we have
\begin{align}
[x^{25}](x + x^2 + x^3 + x^4 + x^5 + x^6)^5 &= [x^{25}] \left(\frac{x(1 - x^6)}{1 - x} \right)^5 \\
&= [x^{25}] x^5\left(\frac{1 - x^6}{1 - x} \right)^5 \\
&= [x^{20}] \left(\frac{1 - x^6}{1 - x} \right)^5 \\
&= [x^{20}] (1 - x^6)^5 \frac{1}{(1 - x)^5} \\
&= [x^{20}] \left( \sum_{k = 0}^5 \binom{5}{k} (-1)^kx^{6k} \right) \frac{1}{(1 - x)^5} \\
&= \sum_{k = 0}^5 \binom{5}{k} (-1)^k [x^{20}]x^{6k} \frac{1}{(1 - x)^5} \\
&= \sum_{k = 0}^5 \binom{5}{k} (-1)^k [x^{20 - 6k}] \frac{1}{(1 - x)^5} \\
&= \sum_{k = 0}^3 \binom{5}{k} (-1)^k [x^{20 - 6k}] \frac{1}{(1 - x)^5}
\end{align}
We change the upper limit from $5$ down to $3$ because we need $20 - 6k$ to be $\ge 0$ (equivalently, this says we can only have at most three of the dice equal to $6$). Continuing, we get
$$ \sum_{k = 0}^3 \binom{5}{k} (-1)^k [x^{20 - 6k}] \sum_{j} \binom{5 + j - 1}{j}x^j $$
telling us that $j = 20 - 6k$ and finally this gives us
$$ \sum_{k = 0}^3 \binom{5}{k} (-1)^k \binom{5 + (20 - 6k) - 1}{20 - 6k} = \sum_{k = 0}^3 \binom{5}{k} \binom{24 - 6k}{20 - 6k} (-1)^k = 126. $$
|
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|
Finding solutions for expression to be a perfect square Determine all integers such that $$ n^4- n^2 +64$$ is the square of an integer.
The first two lines of the solution given in the textbook is as below:
Since $n^4-n^2+64>(n^2-1)^2. $ For some non negative integer $k$,
$n^4-n^2+64=(n^2+k)^2$.
I fail to understand what the author tries to say here. Can't this problem be done in another manner?
|
$$(n^2-x)^2=n^4-n^2+64=n^4-2xn^2+x^2$$
where $x$ is an integer.Then,
$$n^2=\dfrac{x^2-64 }{2x-1 }\geq 0$$ hence by multiplying both sides by 4
$$4n^2=\dfrac{4x^2-1+1-256 }{2x-1 }=2x+1-\dfrac{255}{2x-1}$$
Since $255=1\cdot 3\cdot 5\cdot 17 $ then the ratio is an integer only if
$$2x-1=\pm 3^a5^b17^c$$ where $a,b,c \in\{0,1\}$
Which will yield:
$$n\in \{0,\pm 1, \pm 8\}$$
|
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|
Find $ \ \ 8^{504} \equiv \pmod 5$ Find $ \ \ 8^{504} \equiv \pmod 5 $
Answer:
$ 8^{504} \equiv 2^{1512} \pmod 5 $
Now ,
$$ \begin{align}2^4 &\equiv 1 \pmod 5 \\\text{or, } \left(2^{4}\right)^{378} &\equiv 1^{378} \pmod 5 \\ \text{or, } 2^{1512} &\equiv 1 \pmod 5\\\text{or, } 8^{504} &\equiv \ 1 \pmod 5 \end{align}$$
Am I right ?
|
As $gcd(8,5)=1$ so by Euler's theorem $8^4\equiv 1\pmod5$ as $\phi (5)=4$ Now 504 divide by 4 so $8^{504}\equiv 1\pmod 5$
|
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|
Series Solution to the ODE $(x-1)y'' - xy' + y = 0$ with I.C. $y(0) = -3$ and $y'(0)=4$ \begin{align*}
(x-1)y'' - xy' + y = 0 &\iff (x-1)\sum_{n=2}^{\infty} n(n-1)c_n x^{n-2} - x\sum_{n=1}^{\infty} nc_n x^{n-1} + \sum_{n=0}^{\infty} c_n x^n = 0 \\
&\iff \sum_{n=2}^{\infty} n(n-1)c_n x^{n-1} - \sum_{n=2}^{\infty} n(n-1)c_n x^{n-2} - \sum_{n=1}^{\infty} nc_n x^n + \sum_{n=0}^{\infty} c_n x^n = 0 \\
&\iff \sum_{k=1}^{\infty} (k+1)kc_{k+1} x^k - \sum_{k=0}^{\infty} (k+2)(k+1)c_{k+2} x^k - \sum_{k=1}^{\infty} kc_k x^k + \sum_{k=0}^{\infty} c_k x^k = 0 \\
&\iff -2c_2 + c_0 + \sum_{k=1}^{\infty} x^k \left[ k(k+1)c_{k+1} - (k+2)(k+1)c_{k+2} - kc_k + c_k \right] = 0 \\
\end{align*}
Therefore, $-2c_2 + c_0 = 0 \Rightarrow c_2 = \frac{c_0}{2!}$. And,
$$c_{k+2} = \dfrac{c_k (1-k) + k(k+1)c_{k+1}}{(k+2)(k+1)}$$
To find the solution, first we let $c_0 = 1$ and $c_1 = 0$, then $c_2 = \frac{1}{2!}$. Then, for $k=1$ we have:
$$c_3 = \dfrac{2c_2}{3\cdot 2} = \dfrac{1}{3!}$$
For $k=2$ we have:
$$c_4 = \dfrac{-2c_2 + 6c_3}{4 \cdot 3} = \dfrac{1}{4!}$$
And the pattern continues so for this solution say $y_1$ we have:
$$y_1 = c_0 + c_1 x + c_2 x^2 + c_3 x^3 + \dots = 1 + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots = e^x - x$$
Now for $y_2$ we let $c_0 = 0$ and $c_1 = 1$, so that $c_2 = 0$ too. Then, we find that $c_3 = c_4 = \dots = 0$, so that
$$y_2 = c_0 + c_1 x + c_2 x^2 + \dots = x$$
Then, the final solution should be $a_0 y_1 + a_1 y_2 = a_0 e^x + (a_1 - a_0)x$. However, from wolfram it says the solution is $y = a_0 e^x + a_1 e^{-x}$. Where did I go wrong?
|
other approach
The equation can be written as
$$(x-1)(y''-y')-(y'-y)=0$$
or
$$(x-1)z'-z=0$$
with $$z=y'-y $$
the solution is $$z=\lambda (x-1)=-7 (x-1) $$
Now look for series solution of
$$y'-y=-7 (x-1) $$
|
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|
The minimum value of the integral a) Find $a$ and $b$ such that $\displaystyle \int_0^1 |x^2 - ax - b| \,dx$ is minimum.
b) How about $\displaystyle \int_0^1 |x^2 - ax - b|^2 \,dx$?
This is what I did for part a):
If $D=a^2+4b\leq 0$ then $|x^2-ax-b| =x^2-ax-b$ and we have:
$ I= \int_0^1 |x^2-ax-b|\,dx = \int_0^1 (x^2-ax-b)\,dx ={1\over 3}-{a\over 2}-b$
and
$I \geq {1\over 3}-{a\over 2}+{a^2\over 4} = {1\over 12}+{(a-1)^2\over 4} $
so $I_{min}\leq {1\over 12}$ and $I= {1\over 12}$ iff $a= 1$ and $b = -{1\over 4}$.
But, I can not handle easyl case $D>0$.
|
Part (b) is more immediately clear to me:
\begin{align*}
\int_0^1 |x^2+ax+b|^2 \, dx &= \int_0^1 (x^2+ax+b)^2 \, dx\\[0.3cm] &= \int_0^1 (x^4+2ax^3+(a^2+2b)x^2 + 2abx+b^2) \, dx\\[0.3cm] &= 1/5 + a^2/3 + 2 b/3 + b^2 + a/2 + ab
\end{align*}
Let $S=1/5 + a^2/3 + 2 b/3 + b^2 + a/2 + ab.$ This is extremised when $\partial S/\partial a = \partial S/\partial b = 0$ so $2a/3+b+1/2=0$ and $a+2b+2/3 =0$.
Solving we find $a=-1$, $b=1/6$.
As for (a):
Let's assume that both roots of the quadratic are in $[0,1]$, and call them $c$ and $d$ with $c \leq d$. Then
$$ \int_0^1 |(x-c)(x-d)|dx = \int_0^1 (x-c)(x-d) dx - 2\int_c^d |(x-c)(x-d)|dx, $$ which is
$T = 1/3 - c/2 - c^3/3 - d/2 + cd + c^2 d - c d^2 + d^3/3.$
Again we want $\partial T/\partial c = \partial T/\partial d = 0$, so $-1/2 - c^2 + d + 2cd - d^2 = 0$ and $-1/2 + c + c^2 - 2cd + d^2= 0$. This gives two solutions:
$c=1/2$, $d=1/2$ and $c=1/4$, $d=3/4$.
so the two solutions are $a=-1$, $b=1/4$ and $a=-1$, $b=3/16$.
But which is the minimum? Well, $\displaystyle\int_0^1 |x^2-x+1/4| dx = 1/12$ while $$\int_0^1 |x^2-x+3/16| dx=1/16$$
So the correct solution is $a=-1$, $b=3/16$.
|
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|
Let $\ a,b,c$ be integers such that $ \ a^{2} + b^{2} = c^{2}$. If $c$ is divisible by $3$, prove that $a$ and $b$ are both divisible by $3$. Question:
Let $\ a,b,c$ be integers such that $ \ a^{2} + b^{2} = c^{2}$. If $c$ is divisible by $3$, prove that $a$ and $b$ are both divisible by $3$.
My attempt:
Proof by contradiction:
Assume $c$ is divisible by $3$ and $a$ or $b$ is not divisible by $3$.
Since $c$ is divisible by $3$ we can write $c$ as $ \ c = 3m \implies c^{2} = 9m^{2} \implies 9 | c^{2}$.
Since $a$ and $b$ are not divisible by $3$, $\ a = 3k+1$ and $ \ b = 3n+1$ for some integers $\ k,n.$
Then,
$ a^{2} + b^{2} = (3k+1)^{2} + (3n+1)^{2} = 9k^{2} + 6k +9n^{2} + 6n + 2$.
I am stuck here. I can't find a contradiction. How can I show that $ a^{2} + b^{2} $ is not divisible by $9$.
|
Note that $(a,b,c)$ is Pythagorean Triples, then we can choose non-zero integers $k;p,q$ and $\gcd (p,q)=1$ such that $$a=2kpq\\b=k(p^2-q^2)\\c=k(p^2+q^2)$$
Now since $3\mid c,$ we will have different cases:
Case 1:
If $3\mid k$, then $3\mid a$ and $3\mid b$
Case 2:
If $3\nmid k$ and $3\mid (p^2+q^2)$. Note that if $3\mid p$, then also $3\mid p$, which can not be true since $\gcd (p,q)=1$. Now since $3\nmid p$ and $3\nmid q$, $3\mid (p^2-q^2)$ (Why?). Then $3\mid b$.
Now since $3\mid c$ and $3\mid b$, then $3\mid (c^2-b^2)\Rightarrow 3\mid a^2\Rightarrow 3\mid a$ (Why?).
But we can not have $3\mid a$ since we have $3\nmid k$, $3\nmid p$, $3\nmid q$.
Hence Case 2 is not possible.
Hence we only have $3\mid k$ which leads to $3\mid a$ and $3\mid b\space\space\space\space\space\blacksquare$
|
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|
Find $\int e^{x^4}(x+x^3+2x^5)e^{x^2}dx$ Find $\int e^{x^4}(x+x^3+2x^5)e^{x^2}dx$
$$I=\int e^{x^4}(x+x^3+2x^5)e^{x^2}dx$$
$$I=\int e^{x^4}(1+x^2+2x^4)e^{x^2}xdx$$Let $x^2=t$
$I=\int e^{t^2}(1+t+2t^2)e^t\frac{dt}{2}=\frac{1}{2}\int e^{t^2+t}(1+t+2t^2)dt$
I am stuck here.
|
Assuming the answer is given by an elementary function, it has to be a function of the form $p(x) e^{x^2+x^4}$ with $p(x)$ being a quadratic polynomial. We have
$$\frac{d}{dx} x^2 e^{x^2+x^4} = 2(x+x^3+2x^5)e^{x^2+x^4} $$
hence such inspired guess immediately leads to
$$ \int (x+x^3+2x^5) e^{x^2+x^4}\,dx = C+\frac{x^2}{2}e^{x^2+x^4}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find probability that confidence intervals are disjoint Given $X_1, \ldots, X_{100}$ and $Y_1,\ldots,Y_{50}$ which are independent random samples from identical distribution $N(\mu,1)$. Each of two statisticians is trying to build confidence interval for $\mu$ on a confidence level of $0.8$ ($1-\alpha=0.8$). You have to find probability that those intervals are disjoint. I know i have to find $P(|\overline{X}-\overline{Y}|>
\frac{1\cdot1.28}{\sqrt{50}} + \frac{1\cdot1.28}{\sqrt{100}})$ which is $1-P(-0.309 \le \overline{X}-\overline{Y}\le 0.309)$
Question is what should i do next how to calculate this probability?
EDIT: Thus the solution is $1-(\phi(1.28)-(1-\phi(1.28)))\approx0.2$?
|
One interval has endpoints $\bar X \pm 1.28 \dfrac 1 {\sqrt{100}}$ and the other $\bar Y \pm 1.28 \dfrac1 {\sqrt{50}}.$
The intervals are disjoint if $\bar X + 1.28 \dfrac 1 {\sqrt{100}} < \bar Y - 1.28 \dfrac 1 {\sqrt{50}}$ or $\bar Y + 1.28 \dfrac 1 {\sqrt{50}} < \bar X - 1.28 \dfrac 1 {\sqrt{100}}.$
That happens if $\bar Y - \bar X < 1.28 \left( \dfrac 1 {\sqrt{100}} + \dfrac 1 {\sqrt{50}} \right)$ or $\bar X - \bar Y > 1.28 \left( \dfrac 1 {\sqrt{100}} + \dfrac 1 {\sqrt{50}} \right).$ In other words
$$
|\bar X - \bar Y| > 1.28 \left( \dfrac 1 {\sqrt{100}} + \dfrac 1 {\sqrt{50}} \right).
$$
This is equivalent to
$$
\frac{|\bar X - \bar Y|}{\dfrac 1 {\sqrt{100}} + \dfrac 1 {\sqrt{50}}} > 1.28.
$$
You have $\bar X - \bar Y \sim N\left(0, \dfrac 1 {100} + \dfrac 1 {50} \right) = N\left( 0, \dfrac 3 {100} \right),$ so that
$$
\frac{\bar X - \bar Y}{ \sqrt{3/100} } \sim N(0,1).
$$
\begin{align}
& \operatorname{standard deviation}\left( \frac{\bar X - \bar Y}{\dfrac 1 {\sqrt{100}} + \dfrac 1 {\sqrt{50}}} \right) = \frac{\operatorname{standard deviation}\left( \bar X - \bar Y \right)}{\dfrac 1 {\sqrt{100}} + \dfrac 1 {\sqrt{50}}} \\[10pt]
= {} & \dfrac{\sqrt{\dfrac 3 {100}}}{\dfrac 1 {\sqrt{100}} + \dfrac 1 {\sqrt{50}}} = \frac{\sqrt 3}{1 + \sqrt 2} = \sqrt 6 - \sqrt 3.
\end{align}
So
\begin{align}
& \Pr\left( \frac{|\bar X - \bar Y|}{\sqrt{\dfrac 1{100}} + \sqrt{\dfrac 1 {50}}} > 1.28 \right) = \Pr\left( \left. \frac{|\bar X - \bar Y|}{\sqrt{\dfrac 1{100}} + \sqrt{\dfrac 1 {50}}} \right/ (\sqrt6-\sqrt3) > \frac{1.28}{\sqrt6-\sqrt3} \right) \\[10pt]
= {} & \Pr\left( |Z| > \frac{1.28}{\sqrt6-\sqrt 3} \right) \text{ where } Z\sim N(0,1).
\end{align}
If the variance were unknown but the variances of the two populations were equal then we would be working with a t-distribution.
|
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|
Prove the inequality $\sum_{cyc}\frac{a}{1+\left(b+c\right)^2}\le \frac{3\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2+12abc}$
Let $a>0$, $b>0$ and $c>0$ such that $a+b+c=3$. Prove the inequality
$$\frac{a}{1+\left(b+c\right)^2}+\frac{b}{1+\left(c+a\right)^2}+\frac{c}{1+\left(a+b\right)^2}\le \frac{3\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2+12abc}.$$
$$LHS=\frac{a}{1+\left(3-a\right)^2}+\frac{b}{1+\left(3-b\right)^2}+\frac{c}{1+\left(3-c\right)^2}$$
We have inequality: $$\frac{a}{1+\left(3-a\right)^2}\le \frac{9}{25}a-\frac{4}{25}\Leftrightarrow -\frac{\left(a-1\right)^2\left(9a-40\right)}{25\left(\left(a-3\right)^2+1\right)}\le 0\forall 0<a\le 1$$
$$\Rightarrow LHS\le \frac{9}{25}\left(a+b+c\right)-\frac{4}{25}\cdot 3=\frac{3}{5}$$
Need prove: $$\frac{3}{5}\le \frac{3\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2+12abc}$$
$$\Leftrightarrow a^2+b^2+c^2+12abc\le 5\left(a^2+b^2+c^2\right)$$
$$\Leftrightarrow 12abc\le 4\left(a^2+b^2+c^2\right)\Leftrightarrow 3abc\le a^2+b^2+c^2$$
By AM-GM: $$3abc\le \frac{\left(a+b+c\right)^3}{9}=3=\frac{\left(a+b+c\right)^2}{3}\le a^2+b^2+c^2$$
Right or Wrong. I think it's wrong. Help me
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We need to prove that
$$\sum_{cyc}\left(\frac{a}{1+(b+c)^2}-a\right)\leq\frac{3(a^2+b^2+c^2)}{a^2+b^2+c^2+12abc}-3$$ or
$$\sum_{cyc}\frac{a(b+c)^2}{1+(b+c)^2}\geq\frac{36abc}{a^2+b^2+c^2+12abc}.$$
Now, by C-S
$$\sum_{cyc}\frac{a(b+c)^2}{1+(b+c)^2}=\sum_{cyc}\frac{a}{1+\frac{1}{(b+c)^2}}=$$
$$=\sum_{cyc}\frac{a^2}{a+\frac{a}{(b+c)^2}}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}\left(a+\frac{a}{(b+c)^2}\right)}=\frac{9}{3+\sum\limits_{cyc}\frac{a}{(b+c)^2}}.$$
Thus, it remains to prove that
$$a^2+b^2+c^2+12abc\geq4abc\left(3+\sum\limits_{cyc}\frac{a}{(b+c)^2}\right)$$ or
$$a^2+b^2+c^2\geq4abc\sum\limits_{cyc}\frac{a}{(b+c)^2}$$ or
$$\sum_{cyc}\left(\frac{a}{bc}-\frac{4a}{(b+c)^2}\right)\geq0$$ or
$$\sum_{cyc}\frac{a(b-c)^2}{bc(b+c)^2}\geq0.$$
Done!
|
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|
Prove $\frac{1}{\log_2\pi }+\frac{1}{\log_5\pi}>2$
Knowing that $\pi^2 < 10$. Prove that:
$$\frac{1}{\log_2\pi}+\frac{1}{\log_5\pi}>2.$$
I have tried to do this the following way:
$\log_2\pi+\log_5\pi>\frac{1}{2} \Leftrightarrow \log_2\pi+\frac{\log_2\pi}{\log_25}>\frac{1}{2}\Leftrightarrow 2\log_2\pi-\log_25>\frac{1}{2}$
Is this the right way to proceed?
|
Notice that:
\begin{align*}
\frac{1}{\log_2 \pi} + \frac{1}{\log_5 \pi}
&= \frac{1}{\left(\frac{\log_\pi \pi}{\log_\pi 2} \right)} + \frac{1}{\left(\frac{\log_\pi \pi}{\log_\pi 5} \right)} & \text{change of base} \\
&= \frac{\log_\pi 2}{\log_\pi \pi} + \frac{\log_\pi 5}{\log_\pi \pi} \\
&= \log_\pi 2 + \log_\pi 5 \\
&= \log_\pi 10 \\
&> \log_\pi (\pi^2) &\text{since $10 > \pi^2$ and $\log_\pi(x)$ is strictly increasing} \\
&= 2
\end{align*}
|
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|
Equation solving involving terms inside roots How do I solve
$$\left(x-\frac{1}{x}\right)^{1/2}+\left(1-\frac{1}{x}\right)^{1/2} = x$$
My Try I tried to take the first term as $t$ but then I had to square both sides twice and that led to a complex bi quadratic. I'm not sure even that'll solve the problem.
guys please let me know what are different possible ways to tackle problems like these, and how would you go about on solving this particular one.
|
$$\left(x-\frac{1}{x}\right)^{1/2}+\left(1-\frac{1}{x}\right)^{1/2} = x\iff \left(x-\frac{1}{x}\right)^{1/2} = x-\left(1-\frac{1}{x}\right)^{1/2}.$$ Squaring
$$x-\frac{1}{x} = x^2+1-\frac{1}{x}-2x\left(1-\frac{1}{x}\right)^{1/2}.$$ Simplifying
$$x = x^2+1-2x\left(1-\frac{1}{x}\right)^{1/2}.$$ That is
$$2x\left(1-\frac{1}{x}\right)^{1/2}=x^2-x+1.$$ Squaring again
$$4x^2\left(1-\frac{1}{x}\right)=x^4-2x^3+3x^2-2x+1.$$ Thus, we get
$$x^4-2x^3-x^2+2x+1=0.$$ That is
$$(x^2-x-1)^2=0.$$ This must be easy to solve.
|
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|
Find the $[\Bbb{Q}(\sqrt2+\sqrt3)(\sqrt5):\Bbb{Q}(\sqrt2+\sqrt3)]$ We want to find the $[\Bbb{Q}(\sqrt2+\sqrt3)(\sqrt5):\Bbb{Q}(\sqrt2+\sqrt3)]$.
My first thought is to find the minimal polynomial of $\sqrt5$ over $\Bbb{Q}(\sqrt2+\sqrt3)$. And from this, to say that $[\Bbb{Q}(\sqrt2+\sqrt3)(\sqrt5):\Bbb{Q}(\sqrt2+\sqrt3)]=\deg m_{\sqrt5,\Bbb{Q}(\sqrt2+\sqrt3)}(x).$
We take the polynomial $m(x)=x^2-5\in\Bbb{Q}(\sqrt2+\sqrt3)[x].$ This is a monic polynomial which has $\sqrt5\in \Bbb{R}$ as a root. We have to show that this is irreducible over $\Bbb{Q}(\sqrt2+\sqrt3)$, in order to say that $m(x)=m_{\sqrt5,\Bbb{Q}(\sqrt2+\sqrt3)}(x)$. The roots of $m(x)$ are $\pm \sqrt5\in \Bbb{R}.$ So,
$$m(x) \text{ is irreducible over } \Bbb{Q}(\sqrt2+\sqrt3) \iff \pm \sqrt5 \notin \Bbb{Q}(\sqrt2+\sqrt3)=\Bbb{Q}(\sqrt2,\sqrt3)$$
because $\deg m(x) =2.$
And this is the point I stack. I tried with the use of the basis of the $\Bbb{Q}$-vector space $\Bbb{Q}(\sqrt2+\sqrt3)$:
$$A:= \{1,\sqrt2,\sqrt3,\sqrt6 \}$$
in order to claim that $\nexists a,b,c,d\in \Bbb{Q}:\sqrt5=a+b\sqrt2+c\sqrt3+d\sqrt6$ but this doesn't help.
Any ideas please?
Thank you in advance.
|
Here is an idea that avoids most of the theory.
The correspondence:
$$a+ b \sqrt{2} + c\sqrt{3} + d \sqrt{6} \mapsto a+ b \sqrt{2} - c\sqrt{3} - d \sqrt{6}$$ preserves sums and products. So if
$$\sqrt{5} = a+ b \sqrt{2} + c\sqrt{3} + d \sqrt{6}$$ then
$$(a+ b \sqrt{2} + c\sqrt{3} + d \sqrt{6})^2= 5$$ so
$$(a+ b \sqrt{2} - c\sqrt{3} - d \sqrt{6})^2 = 5$$ so
$$a+ b \sqrt{2} - c\sqrt{3} - d \sqrt{6} = \pm (a+ b \sqrt{2} + c\sqrt{3} + d \sqrt{6})$$ so either $a = b = 0$ or $c = d = 0$. Now it should be easy, but really, the same trick applies: If $\sqrt{5} = c\sqrt{3} + d\sqrt{6}$ then $c\sqrt{3} - d\sqrt{6}= \pm( c\sqrt{3} + d \sqrt{6})$, so again, $c= 0$ or $d=0$. In the end we get a contradiction.
Note that this procedure can be generalized, for more radicals, provided we have the uniqueness of the writing. We haven't really used $5$, it really shows which square roots are in such a field.
|
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|
Is there an analytic solution for the equation $\log_{2}{x}+\log_{3}{x}+\log_{4}{x}=1$?
I am looking for a close form solution for below equation.
$$\log_{2}{x}+\log_{3}{x}+\log_{4}{x}=1.$$
I solve it by graphing, but I don't know is there a way to find $x$ analytically ?
|
$$\tag1
\log_{2}{x}+\log_{3}{x}+\log_{4}{x}=1$$
$$\tag2
\frac{\log{x}}{\log{2}}+\frac{\log{x}}{\log{3}}+\frac{\log{x}}{\log{4}} = 1$$
$$\tag3
\left(\frac{1}{\log{2}}+\frac{1}{\log{3}}+\frac{1}{\log{4}}\right)\log{x} = 1$$
$$\tag4
\left(\frac{1}{\log{2}}+\frac{1}{\log{3}}+\frac{1}{\log{4}}\right) = \frac{1}{\log{x}}$$
$$\tag5
\left(\frac{1}{\log{3}} + \frac{3}{\log{4}}\right) = (\log{x})^{-1}$$
$$\tag6
\left(\frac{1}{\log{3}} + \frac{3}{\log{4}}\right)^{-1} = \log{x}$$
$$\tag7
10^{\left(\frac{1}{\log{3}} + \frac{3}{\log{4}}\right)^{-1}}= x$$
$$10^{\frac{1}{\frac{1}{\log{3}} + \frac{3}{\log{4}}}}= x$$
|
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|
Show that the integral is divergent $\int_0^\infty\frac{\ dx}{1+x^2\sin^2x}$
Show that the integral is divergent
$$\int_0^\infty\frac{\ dx}{1+x^2\sin^2x}$$
It has no point of discontinuity in range of integration. Also, I have found $\int_0^\infty\frac{\ dx}{1+x^2\sin^2x}>\frac{\pi}{2}$, but that seems to be of no consequence. I don't how to move ahead with this.
|
We have that
\begin{align*}\int_0^\infty\frac{dx}{1+x^2\sin^2x}
&=\sum_{n=0}^{\infty}
\int_0^{\pi}\frac{dx}{1+(x+n\pi)^2\sin^2(x+n\pi)}
\\&\geq \sum_{n=0}^{\infty}
\int_0^{\pi/2}\frac{2dx}{1+\pi^2(1+n)^2\sin^2(x)}\\
&=\sum_{n=0}^{\infty}\frac{\pi}{\sqrt{1+\pi^2(n+1)^2}}\\
&\geq \frac{\pi}{\sqrt{1+\pi^2}}\sum_{n=0}^{\infty}\frac{1}{n+1}=+\infty
\end{align*}
where we used the fact that
$$\int\frac{\ dx}{1+a^2\sin^2(x)}=\frac{\arctan(\sqrt{1+a^2}\tan(x))}{\sqrt{1+a^2}}+C.$$
P.S. We can also use the inequality $\sin^2(x)\leq x^2$ and evaluate
$$\int_0^{\pi/2}\frac{2dx}{1+\pi^2(1+n)^2x^2}=\frac{2\arctan\left((n+1)\pi^2/2\right)}{\pi(n+1)}.$$
|
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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.
|
$$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=?$$
Now $$(a^2)^2+(b^2)^2+(c^2)^2=(a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+c^2a^2)$$
$$a^2b^2+b^2c^2+c^2a^2=(ab+bc+ca)^2-abc(a+b+c)=?$$
|
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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?
|
Consider $f(x)=e^{1+x}$. By your reasoning, since $1+x=1+x+O(x^2)$, you get that:
$$f(x)=1+(1+x)+O(x^2)$$
But that isn't correct. If it was correct, then $f(0)=2$, but we know $f(0)=e$.
Now, if you did the full substitution you'd get:
$$f(x)=1+(1+x)+\frac{(1+x)^2}{2!}+\frac{(1+x)^3}{3!}+\cdots$$
The problem is that the terms $\frac{(1+x)^k}{k!}$ keep adding to the constant and linear terms.
What you really should get is $$e^{1+x+O(x^2)}=e^{1}\left(1+\left(x+O(x^2)\right)+\frac{\left(x+O(x^2)\right)^2}{2!}+\cdots\right)$$
Now, $\left(x+O(x^2)\right)^2=O(x^2)$. So you get:
$$e^{1+x}=e+ex+O(x^2)$$
In your case, you get:
$$\begin{align}e^{1+\frac{x^2}{2}-\frac{x^4}{8}+\mathcal{O}(x^6)
}&=e\left(1+\left(\frac{x^2}{2}-\frac{x^4}{8}\right)+\frac{\left(\frac{x^2}{2}-\frac{x^4}{8}\right)^2}{2!}+O(x^6)\right)\\
&=e\left(1+\frac{x^2}{2}-\frac{x^4}{8}+\frac{x^4}{8}+O(x^6)\right)\\
&=e+\frac{e}{2}x^2+O(x^6)
\end{align}$$
And you get $a=e,b=\frac{e}{2},$ and $\lim = 0$.
|
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|
Find polynomial : $P(x)=x^3+ax^2+bx+c$ Find all polynomials in $\mathbb{Q}[x]$ is of the form $P(x)=x^3+ax^2+bx+c$ which has $a, b, c$ as its roots.
Is my answer below correct?
|
$x^3+ax^2+bx+c = (x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+bc+ca)x-abc$
so $a=-(a+b+c)$, $b=ab+bc+ca$, $c=-abc$
Case 1 : $c=0$ so $b=ab$ then $a=1$ or $b=0$
if $a=1$, we get $(a,b,c)=(1,-2,0)$
if $b=0$, we get $(a,b,c)=(0,0,0)$
Case 2 : $c\not=0$ so $ab=-1$ then $-1+c(a+b)=b$ ---[1]
Since $a=-(a+b+c)$, so $c=-2a+\frac{1}{a}$
substitute $c=-2a+\frac{1}{a}$ and $b=-\frac{1}{a}$ in [1]
we have $1+(a-\frac{1}{a})(2a-\frac{1}{a})=\frac{1}{a}$, then $(a-1)(2a^2(a+1)-1)=0$
If $2a^2(a+1)-1=0$, then $2a^3+2a^2-1=0$, by Rational root theorem, a $\not \in \mathbb{Q}$
so $a=1$, we get $(a,b,c)=(1,-1,-1)$
Answer : $P(x)=x^3$, $x^3+x^2-2x$, $x^3+x^2-x-1$
|
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|
Proving that for $\{ a,b\} \subset \Bbb{R^{+}}$; $a+b=1 \implies a^2+b^2 \ge \frac{1}{2}$
For $\{ a,b\} \subset \Bbb{R^{+}}$; $a+b=1 \implies a^2+b^2 \ge
\frac{1}{2}$
I'm trying to prove this in the following way, but I'm not sure if it's correct. Could anyone please check it and see if it's okay?
$a+b=1 \implies (a+b)^2 = 1^2 = 1 \implies (a+b)-(a+b)^2 = 1-1 =0$ (1)
$(a-b)^2 \ge 0$
So by (1) we have:
$(a-b)^2 \ge (a+b)-(a+b)^2$
$(a^2-2ab+b^2) \ge (a+b) - (a^2+2ab+b^2)$
$(a^2-2ab+b^2) + (a^2+2ab+b^2) \ge (a+b) $
$ a^2+a^2+b^2+b^2+2ab-2ab \ge (a+b)$
$2(a^2+b^2) \ge (a+b)$
$2(a^2+b^2) \ge 1$
$(a^2+b^2) \ge \frac{1}{2} $
$\blacksquare$
|
For fun:
$a + b = 1$;
$a := \cos^2t;$ $b := \sin^2t$ , $ 0 \le t \lt 2π$.
$(a+b)^2 = a^2 +2ab +b^2 = $
$\cos^4t +2\cos^2t \sin^2t + \sin^4t =1$.
$\cos^4t + \sin^4t = 1 - (1/2)(\sin2t)^2$.
$\Rightarrow \cos^4t + \sin^4t \ge 1 - 1/2$ ,
since $ (\sin2t)^2 \le 1$.
Finally substituting back:
$a^2 + b^2 \ge 1/2$.
Recall: $\sin(2t) = 2 \sin t \cos t$.
|
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|
Let $h(4x-1)=2x+7$. For what value of $x$ is $h(x)=x$? Let $h(4x-1)=2x+7$. For what value of $x$ is $h(x)=x$?
If $h(a)=a$, then $4x-1=2x+7$ which implies $x=4$. So $a=15$ when I substitute $x=4$ into both linear equations. Is the value of $x$ $15$?
|
The posted proof looks good. For an alternative one, let $4x-1=y \iff x = \frac{y+1}{4}\,$, then:
$$\,h(y) = h(4x-1)=2x+7=2\,\frac{y+1}{4}+7=\frac{y}{2}+\frac{15}{2}\,$$
Therefore $\,h(x)=\frac{x}{2}+\frac{15}{2}\,$, and $h(x)=x \iff x = \frac{x}{2}+\frac{15}{2} \iff x=15\,$.
|
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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 $x\in(\sqrt{2},2)$
|
Hint :$$\frac{1}{x}<\frac{x}{2} \to x\in (-\sqrt 2,0) \cup(\sqrt 2,+\infty)\tag{1}$$
$$\frac{x}{2}<\frac{2}{x}\to x\in(-\infty,-2)\cup (0,2) \tag{2}$$ then find $(1) \cap (2)$
|
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|
Simple logarithmic differentiation Given the equation $\frac{dy}{dx} = \frac{-x}{y}$
How can you solve for $\frac{d^2y}{d^2x}$ using logarithmic differentiation?
Here is my work:
$ln(\frac{dy}{dx}) = ln(\frac{-x}{y})$
$ln(\frac{dy}{dx}) = ln(-x) - ln(y)$
(Take derivative of both sides)
$(\frac{d^2y}{d^2x})/(\frac{dy}{dx}) = \frac{1}{x} - \frac{1}{y}$
(Then solve for $\frac{d^2y}{d^2x}$)
$\frac{d^2y}{d^2x} = \frac{\frac{dy}{dx}}{x} - \frac{\frac{dy}{dx}}{y}$
(Then plug in $\frac{-x}{y}$ for $\frac{dy}{dx}$)
$\frac{d^2y}{d^2x} = \frac{-x}{yx} + \frac{x}{y^2}$
And simplify to get:
$\frac{d^2y}{d^2x} = \frac{-1}{y} + \frac{x}{y^2}$
$ = \frac{(x-y)}{y^2}$
However, when I solve the derivative using the quotient rule, I get:
$\frac{x\frac{-x}{y} - y}{y^2}$
$= \frac{\frac{-x^2}{y}-y}{y^2}$
What am I doing wrong?
|
You made a mistake. In particular, differentiating both sides of $$\log(\frac{dy}{dx}) = \log(-x)-\log(y)$$ gives
$$\frac{\frac{d^2y}{dx^2}}{\frac{dy}{dx}} = \frac{1}{x} - \frac{1}{y} \color{red}{\frac{dy}{dx}}.$$
Here $\frac{d}{dx} \log(y) = \frac{1}{y} \color{red}{\frac{dy}{dx}}$ comes from the chain rule.
|
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|
Show that $11^{10} \equiv 1 (\mod 100)$ $11^{10} \equiv 1 \pmod{100}$
I tried to solve by using euler's theorem, But I got stuck.
$\gcd(11, 100) = 1$
$11^{φ(100)} \equiv 1 \pmod{100}$
$11^{40} \equiv 1 \pmod{100}$
I don't know how to go on as $11^{40}$ is bigger than $11^{10}$
|
$$\qquad{11^2\equiv 121\equiv 21\mod 100\\\\
(11^2)\times 11\equiv 231\equiv 31\mod 100\\
(11^4)\equiv 31\times 11\equiv 341\equiv 41\mod 100\\
(11^5)\equiv 41\times 11\equiv 451\equiv 51\mod 100\\
(11^6)\equiv 51\times 11\equiv 61\mod 100\\
(11^7)\equiv 61\times 11\equiv 71\mod 100\\
(11^8)\equiv 71\times 11\equiv 81\mod 100\\
(11^9)\equiv 81\times 11\equiv 91\mod 100\\
(11^{10})\equiv 91\times 11\equiv 1001\equiv 1\mod 100\\}$$
|
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|
Finding the $nth$ term of a sequence Okay so I'm asking this quesion knowing a thing or two about sequences and general terms
What is the sum of the series :
$$1+\frac{1\cdot3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+\cdots$$
My Try: I tried calculating the general term $T_{n}$ for the sequence but I'm not able to understand how to include every single term of the sequence in a single general term. Writing the series again below the original series with the first term one ahead of the first term of the original series and subtraction didn't help too.How can I get this done? I get a feeling it could be a telescopic sum but unless I have a general term, I can't be sure of it.
Please suggest me any way of doing this. Thanks for giving this your time.
|
$$1+\frac{1\cdot3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+\cdots$$
Numerator is the product of odd positive integers which can be written as
$1 \cdot 3\cdot 5=\dfrac{1 \cdot 2\cdot 3\cdot 4 \cdot 5 }{2 \cdot 4}=\dfrac{5! }{2^2\left( 1\cdot 2\right)}=\dfrac{(2\cdot 2 +1)!}{2^2\cdot 2!}$
$n-$th numerator will be $\dfrac{(2n+1)!}{2^n \,n!}$
denominator is $6\cdot 8$ and can be written as $(2\cdot 3)(2\cdot 4)=2^2\cdot (3\cdot 4)=\dfrac{2^2\cdot 4!}{2}$
next denominator will be $6\cdot 8 \cdot 10=(2\cdot 3)(2\cdot 4)(2\cdot 5)=2^3(3\cdot 4\cdot 5)=\dfrac{2^3\cdot 5!}{2}$
$n-$th denominator will be $2^{n-1} (n+2)!$
Then the $n-$th term for $n\ge 2$ will be
$a_n=\dfrac{\dfrac{(2n+1)!}{2^n \,n!}}{2^{n-1} (n+2)!}=\dfrac{(2n+1)!}{2^n \,n!\,2^{n-1} (n+2)!}=\dfrac{(2 n+1)!}{2^{2 n-1} n! (n+2)!}$
And the series
$$1+\sum _{n=1}^{\infty } \dfrac{(2 n+1)!}{2^{2 n-1} n! (n+2)!}=4$$
|
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|
$|x^2-5x+2|\leq 4$
Solve the following inequality.
$$|x^2-5x+2|\leq 4.$$
I know how to interpret $|x-a|$ as the distance between $x$ and $a$ along the $x$-axis, but how does one interpret an absolute value of $|ax^2+bx+c|$?
For the inequality $|x^2-5x+2|\leq 4$ I factored the LHS and got $$\left|\left(x+\frac{\sqrt{7}-5}{2}\right)\left(x+\frac{5-\sqrt{7}}{2}\right)\right|\leq 4.$$
I don't know how to case-divide the LHS.
|
It's just $$-4\leq x^2-5x+2\leq4,$$
which is $$x^2-5x-2\leq0$$ (which gives
$\frac{5-\sqrt{33}}{2}\leq x\leq\frac{5+\sqrt{33}}{2}$)
and $$x^2-5x+6\geq0$$ (which is
$x\geq3$ or $x\leq2$).
Finally, we obtain:
$$\left[\frac{5-\sqrt{33}}{2},2\right]\cup\left[3,\frac{5+\sqrt{33}}{2}\right]$$
|
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|
If the quadratic equation $x^2+px+q=0$ and $x^2+qx+p=0$ have a common root
If quadratic equations $x^2+px+q=0$ and $x^2+qx+p=0$ have a common root, prove that: either $p=q$ or $p+q+1=0$.
My attempt:
Let $\alpha $ be the common root of these equations. Since one root is common, we know:
$$(q-p)(p^2-q^2)=(q-p)^2.$$
How do I get to the proof from here?
|
Let $\alpha$ be the common root.
\begin{align}
\alpha^2 + p \alpha + q &= 0 \\
\alpha^2 + q \alpha + p &= 0 \\
\hline
(p-q)\alpha + (q-p) &= 0 \\
(p-q)(\alpha-1) &= 0
\end{align}
$\alpha = 1$ or $p=q$.
If $\alpha = 1$, then $\alpha^2 + p \alpha + q = 0$ becomes
$1+p+q = 0$
|
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|
Finding the minimum value of $a^2+b^2+c^2$
Let $a$, $b$ and $c$ be $3$ real numbers satisfying $2 \leq ab+bc+ca$. Find the minimum value of $a^2+b^2+c^2$.
I've been trying to solve this, but I don't really know how to approach this. I thought of $(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca)$, but that gives me $a+b+c$, which is unknown. How can I solve this?
|
Again from Cauchy-Schwarz from a different perspective
$\sqrt{a^2c^2}+\sqrt{b^2c^2}+\sqrt{a^2c^2} \leq \sqrt{a^2+b^2+c^2}\sqrt{a^2+b^2+c^2}$
$a^2+b^2+c^2 \ge ab+bc+ac \ge 2$
$a^2+b^2+c^2 \ge 2$
İnequality holds for $a=b=c=\sqrt{\dfrac{2}{3}}$
|
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|
Determinant of a symmetric matrix of order $2018$
Find out the determinant of the following matrix :
$$\begin{bmatrix}1^{2016} & 2^{2016} & ... & 2018^{2016}\\2^{2016} & 3^{2016} & ... & 2019^{2016}\\.. & .. & ... & ..\\2018^{2016} & 2019^{2016} & ... & 4035^{2016}\end{bmatrix}$$
Through examples of order $2\times 2$ as $\displaystyle \begin{bmatrix}1^0 & 2^0\\2^0 & 3^0\end{bmatrix}$ and $4\times 4$ as $\displaystyle \begin{bmatrix}1^2 & 2^2 & 3^2 & 4^2\\2^2 & 3^2 & 4^2 & 5^2\\3^2 & 4^2 & 5^2 & 6^2\\4^2 & 5^2 & 6^2 & 7^2\end{bmatrix}$, I found that in both case the answer is $0$. But I want to know the procedure to find such determinant.
|
Suppose $p_1, p_2...,p_n $ are polynomials of degree $m-1 $ with atleast one has degree exactly $m-1$ . Let us consider a matrix $A$ whose $(i,j)-$th matrix has entry $p_i(u_j)$ , where $u_1,u_2,.......,u_n$ are $ n$ distinct number
let $p_i(x) = a_{i,1} + a_{i,2} x +.....+a_{i,m}x^{m-1} $
Now we can easily show that $A=\displaystyle \begin{bmatrix}a_{1,1} & a_{1,2} .......&a_{1,m}\\ . & ....... & .\\. & ....... & .\\a_{n,1} & a_{1,2} .......&a_{n,m}\end{bmatrix}\displaystyle \begin{bmatrix}1 & 1 & ..... & 1\\u_1 & u_2 & ..... & u_n\\. & .& ..... & .\\. & .& ..... & .\\u_1^{m-1} & u_2^{m-1} & .... & u_n^{m-1}\end{bmatrix}$,
This show that in particular if max $deg(p_i) = m_i < n$ , (this mean , the first matrix is of size $n \times m$ and$ m < n$ ) , then $rank (A) < n$ and so determinant is $0$
For the given matrix $p_i(x) = (i-1 +x)^{2016}$ and points are $ 1,....., 2018$. Since the matrix is of size $2018 \times 2018$, the determinant is $0$
|
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|
Equation of one variable
Solve the following equation.
$$(8-x)\cdot(x^2-2x+16)^2+2x^4\cdot(x^2-2x+16)=16x^7,$$
where $x\in \mathbb{R}$
I already know that we need to prove $x=2$, but don't know how to show it...
I'd be grateful for some hints or solutions ;)
|
It's $$(x-2)(16x^6+30x^5+65x^4+86x^3+240x^2+128x+1024)=0$$
and since
$$16x^6+30x^5+65x^4+86x^3+240x^2+128x+1024=$$
$$=(16x^6+30x^5+15x^4)+(50x^4+86x^3+37x^2)+(203x^2+128x+1024)>0,$$
we get the answer: $\{2\}$.
|
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|
Rules of Double Sums What are the (most important) rules of double sums? Below are some rules I encountered - are they all correct and complete? Offerings of clear intuition or proofs (or other additions) are most welcome to remove confusion.
*
*General case: $$\sum_{i=1}^m[x_i] \cdot \sum_{j=1}^n[y_j] = \sum_{i=1}^m\sum_{j=1}^n[x_iy_j]$$
*Less general case ($m=n$): $$\sum_{i=1}^n[x_i] \cdot \sum_{j=1}^n[y_j] = \sum_{i=1}^n\sum_{j=1}^n[x_iy_j] = \sum_{i=1}^n[x_iy_i] + \sum_{i=1}^n\sum_{j=1 \\ j \neq i}^n[x_iy_j]$$
*Special case ($m=n,x_i=y_i$): $$\left(\sum_{i=1}^n[x_i]\right)^2 = \sum_{i=1}^n[x_i] \cdot \sum_{j=1}^n[x_j] = \sum_{i=1}^n\sum_{j=1}^n[x_ix_j] = \sum_{i=1}^n[x_i^2] + \sum_{i=1}^n\sum_{j=1 \\ j \neq i}^n[x_ix_j]$$
Question related to (3): why suddenly an index $j$ appears (initially, we had only $i$)?
Furthermore, in relation to (3) there is a theorem given in my textbook without intuition/proof:
When we work with double sums, the following theorem is of special interest; it is an immediate consequence of the multinominal expansion of $(x_1 + x_2 + \ldots + x_n)^2$:
Theorem: $$\sum_{i<j}\sum[x_ix_j] = \frac{1}{2}\left[\left(\sum_{i=1}^n[x_i]\right)^2 - \sum_{i=1}^n[x_i^2]\right], \text{ where } \sum_{i<j}\sum[x_ix_j] = \sum_{i=1}^{n-1}\sum_{j=i+1}^n[x_ix_j].$$
What is the special interest/purpose of this theorem (when is it useful?) and what is the relation with (3) above?
|
Explanation to (3): Given relation is
\begin{align}
\left(\sum_{i = 1}^{n} x_{i} \right)^{2} & = \sum_{i = 1}^{n} x_{i} \sum_{j = 1}^{j} x_{j} \tag{1} \\
& = \sum_{i = 1}^{n} \sum_{j = 1}^{n}x_{i}x_{j} \tag{2} \\
& = \sum_{i = 1}^{n}x_{i}^{2} + \sum_{i = 1}^{n}\sum_{\substack{j = 1 \\ j \neq i}}^{n} x_{i}x_{j} \tag{3}
\end{align}.
To understand this consider a simple (still general) example of $n = 3$. So you have
\begin{align}
\left(\sum_{i = 1}^{n = 3} x_{i} \right)^{2} & = \left( x_{1} + x_{2} + x_{3} \right)^{2} = x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + 2 x_{1}x_{2} + 2x_{2}x_{3} + 2x_{3}x_{1} \\
& = \underbrace{x_{1}^{2} + x_{2}^{2} + x_{3}^{2}}_{\text{squared terms}} + \underbrace{x_{1}x_{2} + x_{1}x_{2} + x_{2}x_{3} + x_{2}x_{3} + x_{3}x_{1} + x_{3}x_{1}}_{\text{cross-product terms}} \tag{4}
\end{align}
Now consider the following
\begin{align}
\sum_{i = 1}^{n = 3} x_{i} \sum_{j = 1}^{j = 3} x_{j} & = (x_{1} + x_{2} + x_{3}) (x_{1} + x_{2} + x_{3}) \\
& = x_{1}^{2} + x_{1}x_{2} + x_{1}x_{3} + x_{2}x_{1} + x_{2}^{2} + x_{2}x_{3} + x_{3}x_{1} + x_{3}x_{2} + x_{3}^{2} \\
& = \underbrace{x_{1}^{2} + x_{2}^{2} + x_{3}^{2}}_{\text{squared terms}} + \underbrace{x_{1}x_{2} + x_{1}x_{2} + x_{2}x_{3} + x_{2}x_{3} + x_{3}x_{1} + x_{3}x_{1}}_{\text{cross-product terms}} \tag{5}
\end{align}
Looking at (4) and (5) you can see that equality in (1) is true and the proof can be extended to any arbitrary number $n$.
Similarly,
\begin{align}
& \sum_{i = 1}^{n = 3} \sum_{j = 1}^{n = 3}x_{i}x_{j} \\
= & \sum_{i = 1}^{n = 3} \left[ x_{i} x_{1} + x_{i} x_{2} + x_{i}x_{3} \right] \\
= & \left[ x_{1} x_{1} + x_{1} x_{2} + x_{1}x_{3} \right] + \left[ x_{2} x_{1} + x_{2} x_{2} + x_{2}x_{3} \right] + \left[ x_{3} x_{1} + x_{3} x_{2} + x_{3}x_{3} \right] \\
= & \underbrace{x_{1}^{2} + x_{2}^{2} + x_{3}^{2}}_{\text{squared terms}} + \underbrace{x_{1}x_{2} + x_{1}x_{2} + x_{2}x_{3} + x_{2}x_{3} + x_{3}x_{1} + x_{3}x_{1}}_{\text{croos-product terms}} \tag{6}
\end{align}
Hence (6) and (4) explains the equality in (2). Now equality in (3) is merely the separation of the squared terms and the cross product terms.
\begin{align}
& \sum_{i = 1}^{n = 3}x_{i}^{2} + \sum_{i = 1}^{n = 3}\sum_{\substack{j = 1 \\ j \neq i}}^{n} x_{i}x_{j} \\
= & \underbrace{(x_{1}^{2} + x_{2}^{2} + x_{3}^{2})}_{\text{squared terms}} + \underbrace{\left[ \underbrace{x_{1} (x_{2} + x_{3})}_{i = 1, j \neq 1} + \underbrace{x_{2}(x_{1} + x_{3})}_{i = 2, j \neq 2} + \underbrace{x_{3}(x_{1} + x_{2})}_{{i = 3, j \neq 3}}\right]}_{\text{cross-product terms}} \tag{7}
\end{align}
Hence (7) and (4) explains the equality in (3) and this completes the explanation.
|
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|
Induction: Bounds on sum of inverses of first $n$ square roots I am trying to show by induction that $2(\sqrt{n}-1)<1+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}<2\sqrt{n}$.
For the upper bound, I have that if $1+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}<2\sqrt{n}$, then $1+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n+1}}<2\sqrt{n}+\dfrac{1}{\sqrt{n+1}}=$
$2\sqrt{n}+\dfrac{2}{2\sqrt{n+1}}<2\sqrt{n}+\dfrac{2}{\sqrt{n}+\sqrt{n+1}}=2\sqrt{n}+2(\sqrt{n+1}-\sqrt{n})=2\sqrt{n+1}$.
However, despite hours of trying, I cannot figure out how to show that $2(\sqrt{n+1}-1)<1+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n+1}}$.
|
Try to use the induction method for $2(\sqrt{n}-1+\frac{1}{\sqrt{n}}) < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}}$
Check it for base cases like $n=1,2,3,4,5,6,7,8,9,10$ and it will fail for all $n=1,2,3,4,5,6,7$ and will hold for $n=8,9,10$.
Induction step : assume that $2(\sqrt{n}-1+\frac{1}{\sqrt{n}}) < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}}$ is true for the number $n$.
Proving step : we need to prove that $2(\sqrt{n+1}-1+\frac{1}{\sqrt{n+1}}) < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}$
From the induction step we know that $2(\sqrt{n}-1+\frac{1}{\sqrt{n}}) < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}}$ thus replacing $ \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}}$ with $2(\sqrt{n}-1+\frac{1}{\sqrt{n}})$ will strengthen the inequality.
We arrive at $2(\sqrt{n+1}-1+\frac{1}{\sqrt{n+1}}) < 2(\sqrt{n}-1+\frac{1}{\sqrt{n}})+\frac{1}{\sqrt{n+1}}$, (i will leave this for you to prove).
Now since its obvious that $2(\sqrt{n}-1+\frac{1}{\sqrt{n}}) > 2(\sqrt{n}-1)$ then the inequality $2(\sqrt{n}-1) < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}}$ will be a corollary to this induction proof.
you just have to check basic cases for $2(\sqrt{n}-1) < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}}$ which are $n=1,2,3,4,5,6,7$ and thus completing the proof.
|
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|
$\lim\limits_ {n \to \infty} \left( \cos ^{2n} \left(\frac{k \pi}{3}\right)-\cos ^{2n}\left(\frac{k \pi}{5}\right)\right )=0$ For how many values of $k=1,2,3,....200$ $$\lim\limits_ {n \to \infty} \left( \cos ^{2n} \left(\frac{k \pi}{3}\right)-\cos ^{2n}\left(\frac{k \pi}{5}\right)\right )=0$$
My Try:
Let $A=\cos\left(\frac{k \pi}{3}\right)$ and $B=\cos\left(\frac{k \pi}{5}\right)$
The limit approaches zero only if
$$A=B=1$$ OR
$$A=B=-1$$ OR
$$A=B=0$$
if $A=B=1$ or $A=B=-1$ $\implies$
$$\frac{k \pi}{3}=2m \pi$$ and
$$\frac{k \pi}{5}=2n \pi$$
OR
$$\frac{k \pi}{3}=(2m+1) \pi$$ and
$$\frac{k \pi}{5}=(2n+1) \pi$$
$\implies$ $k$ is a multiple of $15$ which are $13$ in number.
Finally if $A=B=0$ we have
$$\frac{k \pi}{3}=(2m+1) \frac{\pi}{2}$$ and
$$\frac{k \pi}{5}=(2n+1)\frac{\pi}{2}$$ $\implies$
Any idea of how to find number of $k's$ in this case?
|
A different approach. It is worth mentioning that
$$0\leq \cos^2\left(\frac{k\pi}{3}\right)\leq1 \text{ and } 0\leq \cos^2\left(\frac{k\pi}{5}\right)\leq1$$
for $\forall k \in \mathbb{Z}$. As a result ($0\leq a \leq 1 \Rightarrow 0\leq a^n \leq 1, n\in \mathbb{N}$)
$$0\leq \cos^{2n}\left(\frac{k\pi}{3}\right)\leq1 \text{ and } 0\leq \cos^{2n}\left(\frac{k\pi}{5}\right)\leq1$$
It is also known (or can easily be shown) that
$$\lim\limits_{n \rightarrow \infty}a^n=0, 0\leq a<1 \tag{1}$$
As a result
$$\lim\limits_{n\rightarrow \infty}\left(\cos^{2n}\left(\frac{k\pi}{3}\right)−\cos^{2n}\left(\frac{k\pi}{5}\right)\right)=0 \tag{2}$$
is true when
*
*$\color{red}{3 \nmid k \text{ and }5 \nmid k}$, because in this case $0\leq \cos^2\left(\frac{k\pi}{3}\right)< 1 \text{ and } 0\leq \cos^2\left(\frac{k\pi}{5}\right)< 1$. It results from $(1)$ and $$\lim\limits_{n\rightarrow \infty}\left(\cos^{2n}\left(\frac{k\pi}{3}\right)−\cos^{2n}\left(\frac{k\pi}{5}\right)\right)=\\
\lim\limits_{n\rightarrow \infty}\left(\cos^{2n}\left(\frac{k\pi}{3}\right)\right)−\lim\limits_{n\rightarrow \infty}\left(\cos^{2n}\left(\frac{k\pi}{5}\right)\right)=\\
0-0=0$$
*$\color{red}{3 \mid k \text{ and } 5 \mid k \text{ or simply } 15 \mid k}$, because in this case $\cos^2\left(\frac{k\pi}{3}\right)=1 \text{ and } \cos^2\left(\frac{k\pi}{5}\right)=1$ and
$$\lim\limits_{n\rightarrow \infty}\left(\cos^{2n}\left(\frac{k\pi}{3}\right)−\cos^{2n}\left(\frac{k\pi}{5}\right)\right)=\\
\lim\limits_{n\rightarrow \infty}\left(\cos^{2n}\left(\frac{k\pi}{3}\right)\right)−\lim\limits_{n\rightarrow \infty}\left(\cos^{2n}\left(\frac{k\pi}{5}\right)\right)=\\
1-1=0$$
In all the other case, the limit is either $1$ or $-1$. Those "other cases" are $\color{red}{3 \mid k \text{ or } 5 \mid k \text{ but } 15 \nmid k}$ or playing with inclusion–exclusion principle $\left \lfloor \frac{200}{3} \right \rfloor + \left \lfloor \frac{200}{5} \right \rfloor - 2\left \lfloor \frac{200}{15} \right \rfloor=80$ of cases when $(2)$ is not true. Or $200-80=120$ cases when $(2)$ is true.
|
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Discrete math $A\,\triangle\, B = C$ implies that $A\,\triangle\, C = B$ $A\,\triangle\, B = C$ implies that $A\,\triangle\, C = B$
I understand that the delta is the symmetric difference and that the symmetric difference of $A$ and $B$ is the set of elements that belong to exactly one of $A$ and $B$. How do I prove the above statement?
|
We need to prove that given $C = A \ \Delta \ B$, then $B = A \ \Delta \ C$
A simple and elementary proof is to just use the definition of the symmetric difference, or $XOR$ as I like to call it, and setup some truth tables:
$\begin{array}{|c|c|c|} \hline
A & B & \mathbf{C} \\ \hline
0 & 0 & \mathbf{0} \\ \hline
0 & 1 & \mathbf{1} \\ \hline
1 & 0 & \mathbf{1} \\ \hline
1 & 1 & \mathbf{0} \\ \hline
\end{array}$
$\begin{array}{|c|c|c|c|} \hline
A & B & \mathbf{C} & \mathbf{A \ \Delta \ C} \\ \hline
0 & 0 & \mathbf{0} & \mathbf{0}\\ \hline
0 & 1 & \mathbf{1} & \mathbf{1}\\ \hline
1 & 0 & \mathbf{1} & \mathbf{0}\\ \hline
1 & 1 & \mathbf{0} & \mathbf{1}\\ \hline
\end{array}$
In the last truth table we can clearly see that $B = A \ \Delta \ C$, completing our proof.
|
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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?
|
This is a well-known problem: the maximum of $a\cos x+b\sin x$ is
$\sqrt{a^2+b^2}$. One way to see this is to use Cauchy-Schwarz:
$$a\cos x+b\sin x\le\sqrt{a^2+b^2}\sqrt{\cos^2x+\sin^2x}=\sqrt{a^2+b^2}$$
and then get equality by finding an $x$ with $\cos x=a/\sqrt{a^2+b^2}$
and $\sin x=b/\sqrt{a^2+b^2}$.
|
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|
Integration: $\int_{2}^{\infty} \frac{\sqrt x}{x^2-1} dx$ I am having a problem integrating this term, I am not able to solve it by substitution either.
|
$$\mathscr{\text{Let } t= \sqrt x}$$
$$\mathscr{\text{.:} dt=\frac {1}{2\sqrt x} dx}$$
$$\mathscr{\text{.:} dx= 2t dt}$$
$$\int_0^\infty \frac{t}{t^4-1} 2t dt$$
$$=\int_0^\infty \frac {2t^2}{t^4-1} dt$$
$$= \int_0^\infty \frac{2t^2}{(t^2-1)(t^2+1)} dt$$
$$= \int_0^\infty \frac{(t^2-1)+(t^2+1)}{(t^2-1)(t^2+1)} dt$$
$$= \int_0^\infty (\frac{1}{t^2-1} +\frac{1}{t^2+1})dt$$
$$= \int_0^\infty \frac{(t+1)-(t-1)}{2(t-1)(t+1)} dt +\int_0^\infty \frac{1}{t^2+1} dt$$
$$= \frac{1}{2} \int_0^\infty (\frac{1}{t-1} -\frac{1}{t+1})dt +\tan^{-1} x$$
$$= \frac{1}{2} \int_0^\infty \frac{1}{t-1} dt - \frac{1}{2} \int_0^\infty \frac{1}{t+1} dt +\tan^{-1} x$$
$$=[\frac{1}{2} (\ln (|t-1|) -\ln (|t+1|)) +\tan^{-1} t]_0^\infty$$
$$=[\frac{1}{2} \ln (|\frac{\sqrt x -1}{\sqrt x+ 1}|) +\tan^{-1} \sqrt x]_0^\infty$$
$$=(lim_{x\to \infty} \frac{1}{2} \ln( \frac{\sqrt x - 1}{\sqrt x+1}) -\tan^{-1} (\infty))-(\frac{1}{2} \ln (1) - \tan^{-1} (0))$$
$$= (lim_{x\to \infty} \frac{1}{2} \ln (\frac{1-\frac{1}{\sqrt x}}{1+\frac{1}{\sqrt x}}) -\frac {π}{2})- 0) $$
$$= 0-\frac{π}{2} $$
$$=- \frac{π}{2} $$
$$ \mathrm{\int_0^\infty \frac {\sqrt x}{x^2-1} dx = - \frac {π}{2} }$$
$$\frac{}{}$$
Edit : The question was changed from
$\int_0^\infty \frac {\sqrt x}{x^2-1} dx \to \int_2^\infty \frac {\sqrt x}{x^2-1} dx $
$$\text{You can similarly solve for this limit}$$
|
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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?
|
use the Eulersubstution and set $$\sqrt{3-2x-x^2}=xt\pm \sqrt{3}$$
|
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|
Why $\sqrt{x^2}$ is not equal to $\big(\sqrt{x}\big)^2$? I'm reading Precalculus from James Stewart. In the book, the author says that $\sqrt{x^2}$ is not equal to $\big(\sqrt{x}\big)^2$. I was performing a couple of proofs and I ended up here:
If you have, for example:
$\sqrt{5^2} ≟ \big(\sqrt{5}\big)^2 \implies (\sqrt{5})² = \sqrt{5}\cdot\sqrt{5} = \sqrt{5}\cdot5 = \sqrt{25} = \big(\sqrt{5}\big)^2$
So, for me, $\sqrt{5^2}= \big(\sqrt{5}\big)^2$.
Can you proof that I'm wrong?
|
You are not wrong about $\sqrt{(5)^2} = \left( \sqrt{5}\right) ^2$.
But if you think that this proves that $\sqrt{(-5)^2}$ is equal to $\left( \sqrt{-5}\right) ^2$, then you are mistaken.
$\sqrt{(-5)^2} = \sqrt{25} = 5$
While $\left( \sqrt{-5}\right) ^2 = \left( \sqrt{-5}\right)\left( \sqrt{-5}\right) = -5$ (or it does not exist if our domain of discourse if the real numbers.)
|
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|
If the partial sums $s_n $of $\sum a_n $ are bounded, show that the series $\sum \frac{a_n}{n} $ converges to $\sum \frac{s_n}{n (n+1)} $ My try: $\left(\frac{1}{n}\right)_{n=1}^\infty $is decreasing and $\lim_{n\to\infty} \frac{1}{n}=0$ with $\sum a_n $is bounded so by Dirichlet's Test $\sum \frac{a_n}{n}$ is convergent.
To prove the given statement I am trying to show $|\sum \frac{a_n}{n}-\sum \frac{s_n}{n(n+1)}|$ tends to $0$.
\begin{align}
|\sum \frac{a_n}{n}-\sum \frac{s_n}{n(n+1)}|
&=\left|a_1 +\frac{a_2}{2} +\frac{a_3}{3} +\cdots-\frac{a_1}{1\cdot2} -\frac{a_1+a_2}{2\cdot3}-\frac{a_1+a_2+a_3}{3\cdot4} -\cdots\right|\\
&=|a_1-\frac{a_1}{1\cdot2} + \frac{a_2}{2} -\frac{a_1+a_2}{2\cdot3} +\frac{a_3}{3} -\frac{a_1+a_2+a_3}{3\cdot4} +\cdots|\\
&\leq\left|a_1-\frac{a_1}{1\cdot2}\right| + \left|\frac{a_2}{2} -\frac{a_1+a_2}{2\cdot3}\right| +\left|\frac{a_3}{3} -\frac{a_1+a_2+a_3}{3\cdot4}\right| + \cdots & (1)
\end{align}
Now From the $n$th term
$$\left|\frac{na_n-a_1-a_2-...-a_{n-1}}{n (n-1)}\right|\geq\left|\frac{na_n-M}{n(n-1)}\right|$$
($M$ is an upper bound of $\sum a_n$)
Also $\left|\frac{na_n+M}{n(n-1)}\right|$ is greater than the nth term. Taking $n$ tends to infinity we see that the $n$th term goes to $0$. Using this in $(1)$ we note that tail of both the series coincide and then $1$st series converges to $2$nd series. Am I correct?
|
Try the usual summation by parts as suggested in the comments.
Another approach:
$$\begin{align}\sum_{n=1}^N \frac{s_n}{n(n+1)} &= \sum_{n=1}^N \sum_{k=1}^n\frac{a_k}{n(n+1)} \\&= \sum_{n=1}^N \sum_{k=1}^N\frac{a_k\, 1_{k \leqslant n}}{n(n+1)} \\&= \sum_{k=1}^N \sum_{n=1}^N\frac{a_k\, 1_{k \leqslant n}}{n(n+1)} \\ &= \sum_{k=1}^N a_k\sum_{n=k}^N\frac{1 }{n(n+1)} \\ &= \sum_{k=1}^N a_k \left(\frac{1}{k} - \frac{1}{N+1} \right)\\ &= \sum_{k=1}^N \frac{a_k}{k} - \frac{s_N}{N+1} \end{align}$$
Now take limits as $N \to \infty$ using the fact that $s_N$ is bounded.
|
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|
How to calculate the determinant of this n by n matrix? Find the determinant of this n by n matrix.
$$
\begin{pmatrix}
0 & x_1 & x_2 & \cdots& x_k \\
x_1 & 1 & 0 & \cdots & 0 \\
x_2 & 0 & 1& \cdots & 0 \\
\vdots& \vdots& \vdots& \ddots & \vdots\\
x_k & 0 & 0 & \cdots& 1 \\
\end{pmatrix}
$$
where, $$ k=n-1 $$.
I am new to matrices and determinants, but this is what I did:
I developed the determinant using the second column:
$$
(-1)^2*x_1
\begin{pmatrix}
x_1 & 0 & \cdots & 0 \\
x_2 & 1& \cdots & 0 \\
\vdots& \vdots& \ddots& \vdots \\
x_k & 0 & \cdots& 1 \\
\end{pmatrix} + (-1)^3 *1
\begin{pmatrix}
0 & x_2 & \cdots & x_k \\
x_2 & 1& \cdots & 0 \\
\vdots& \vdots& \ddots& \vdots \\
x_k & 0 & \cdots& 1 \\\end{pmatrix}
$$
the first determinant is triangular, so its equal to $ x_1 $ but this is where I got stuck. I don't know what to do with the second determinant. Any help is appriciated. Thanks
|
You can do one thing
$$
\begin{pmatrix}
0 & x_1 & x_2 & \cdots& x_k \\
x_1 & 1 & 0 & \cdots & 0 \\
x_2 & 0 & 1& \cdots & 0 \\
\vdots& \vdots& \vdots& \ddots & \vdots\\
x_k & 0 & 0 & \cdots& 1 \\
\end{pmatrix}
$$
Taking $x_r$ common from $(r+1)^{th}$ row for r=1 to k
$$
(\prod_{i=1}^k x_k)
\begin{pmatrix}
0 & x_1 & x_2 & \cdots& x_k \\
1 & \frac{1}{x_1} & 0 & \cdots & 0 \\
1 & 0 & \frac{1}{x_2}& \cdots & 0 \\
\vdots& \vdots& \vdots& \ddots & \vdots\\
1 & 0 & 0 & \cdots& \frac{1}{x_k} \\
\end{pmatrix}
$$
And multiply $x_r$ to $(r+1)^{th}$ column for r=1 to k
$$
\begin{pmatrix}
0 & x_1^2 & x_2 ^2 & \cdots& x_k ^2\\
1 & 1 & 0 & \cdots & 0 \\
1 & 0 & 1& \cdots & 0 \\
\vdots& \vdots& \vdots& \ddots & \vdots\\
1 & 0 & 0 & \cdots& 1 \\
\end{pmatrix}
$$
Now substracting each column from column no. 1
$$
\begin{pmatrix}
-\sum_{i=1}^k x_i^2 & x_1^2 & x_2 ^2 & \cdots& x_k ^2\\
0 & 1 & 0 & \cdots & 0 \\
0 & 0 & 1& \cdots & 0 \\
\vdots& \vdots& \vdots& \ddots & \vdots\\
0 & 0 & 0 & \cdots& 1 \\
\end{pmatrix}
$$
You know what to do next
|
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|
Greatest Integer function limit problem I need help solving this problem without redefining the greatest interger function, is that possible?
$$\lim_{x \, \to \,\frac {1}{2}^-}[3x-\frac {1}{2}]$$
What about this rule
|
Since we are interested in the behavior of $\left\lfloor 3x-\frac{1}{2}\right\rfloor$ as $x$ tends to $\frac{1}{2}$ from the left, we may assume that $\frac{1}{6}<x<\frac{1}{2}$. Then
$$\frac{3}{6}<3x<\frac{3}{2}$$
so
$$\frac{3}{6}-\frac{1}{2}<3x-\frac{1}{2}<\frac{3}{2}-\frac{1}{2}$$
which reduces to $0<3x-\frac{1}{2}<1$. It follows that $\left\lfloor 3x-\frac{1}{2}\right\rfloor=0$, so
$$\lim_{x\to\frac{1}{2}^-}\left\lfloor 3x-\frac{1}{2}\right\rfloor=0$$
|
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|
Find maximize and minimize of $P=x+y$
For $\{x,y\}\subset\mathbb R$ such that $\sqrt{x+1}+\sqrt{y+1}=\sqrt{2}\left(x+y\right)$ find maximize and minimize of $P=x+y$
I found the maximize but minimize I have no idea. Help me.
|
By AM-GM
$$2(x+y)^2=x+y+2+2\sqrt{(x+1)(y+1)}\leq x+y+2+(x+1+y+1)=2(x+y+2).$$
Thus, $$(x+y)^2-(x+y)-2\leq0$$ or
$$-1\leq x+y\leq2.$$
In the right inequality the equality occurs for $x=y=1$,
which says that $2$ is a maximal value of $P$.
Now, $\sqrt{.}$ is a concave function.
Thus, since $(x+1+y+1,0)\succ(x+1,y+1)$, by Karamata we obtain:
$$\sqrt2(x+y)=\sqrt{x+1}+\sqrt{y+1}\geq\sqrt{0}+\sqrt{x+1+y+1}$$ or
$$2(x+y)^2-(x+y)-2\geq0$$ or
$$x+y\geq\frac{1+\sqrt{17}}{4}.$$
The equality occurs for $y=-1$ and $x=\frac{5+\sqrt{17}}{4}$,
which says that $\frac{1+\sqrt{17}}{4}$ is a minimal value.
Done!
|
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|
Partial Differentiation Ex 2.2 Eng Mathematics by H.K. Das Test for continuity when
$f(x,y) = \frac{x^3\,y^3}{x^3+y^3}$ when $x\neq0, y\neq0$ and f(x,y)=0 when $x=0,y=0$.
|
The domain of $f$ is $D=\mathbb{R^2}\setminus \{(x,-x):x\in\mathbb{R}\}$.
In $D$, let $y=x$ then, as $x\to 0$, also $y\to 0$ and
$$f(x,y)=\frac{x^3y^3}{x^3+y^3}=\frac{x^3}{2}\to 0.$$
In $D$, let $y=-x+x^4$ then, as $x\to 0$, also $y\to 0$ and
$$f(x,y)=\frac{x^3y^3}{x^3+y^3}=\frac{x^3(-1+x^3)^3}{1+(-1+x^3)^3}=\frac{-x^3+o(x^3)}{1-1+3x^3+o(x^3)}\to -\frac{1}{3}.$$
Since the two limits are different it follows that $f$ cannot be extended to a continuous function in $D\cup\{(0,0)\}$.
|
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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.
|
Use Cauchy Буняко́вський Schwarz, $|\langle (a,b), (x,y) \rangle| \le \|(a,b)\| \|x,y\|$. Choosing $(x,y)= {1 \over \|(a,b)\|} (a,b)$ shows equality.
|
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|
How to integrate $\int \frac {e^y}{y} dy$? The question is to evaluate $$\iint_R \frac {x}{y} e^y dx dy$$ where R is the region bounded by $0 \leq x \leq 1$ and $x^2 \leq y \leq x$.
So i write it as $$\int_0^1 \int_{x^2}^{x} \frac{x}{y} e^y dy dx$$.
The thing is, how do i evaluate $I=\int_{x^2}^{x} \frac{1}{y} e^y dy$? I tried integration by parts but failed. Then i tried to use infinite series.
Use $$e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!}$$
we get \begin{align}
I & =\int_{x^2}^{x} \frac{1}{y}+1+\frac{y}{2!}+\frac{y^2}{3!}+\cdots dy \\
&=ln(\frac{1}{x})+(x-x^2)+\frac{1}{4}(x^2-x^4)+\frac{1}{18}(x^3-x^6)+ \cdots
\end{align}
Now get back to $\int_0^1 I dx$. We need to find $P=\int_0^1 ln(\frac{1}{x})$ and $Q=\int_0^1 (x-x^2)+\frac{1}{4}(x^2-x^4)+\frac{1}{18}(x^3-x^6)+ \cdots$
integrate by parts, we find $P=0$.
But for Q,we have $\left((\frac{1}{2}x^2-\frac{1}{3}x^3)+\frac{1}{4}(\frac{1}{3}x^3-\frac{1}{5}x^5)+ \cdots \right)_0^1$
Can we simplify the infinite sum in the bracket as a simple result?
|
This is the Exponential integral, and it can't be expressed in terms of standard functions. However, one can first integrate over $x$ (with slightly more complicated bounds), and then the integration in $y$ becomes much more straightforward.
|
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|
A floor function equation I want to solve below equation analytically $$\lfloor \frac{2x+1}{3}\rfloor +\lfloor \frac{\lfloor4x\rfloor+2}{3}\rfloor=1$$ but I have no idea to start .
Implicit: I solve it by graphing and the solution was $x\in[\frac {1}{4},1)$
I am thankful for any idea in advance .
I try to use $x=n+p \\0\leq p<1 $ but can't go further $$\lfloor \frac{2n+2p+1}{3}\rfloor +\lfloor \frac{\lfloor4n+4p\rfloor+2}{3}\rfloor=1\\0\leq 2p<2 \\0\leq 4p<4 $$
|
If $x \geqslant 1,$ then $\lfloor{\frac{2 x+1}{3}} \rfloor + \lfloor{\frac{\lfloor{4 x \rfloor}+2}{3}} \rfloor \geqslant 1+2>1.$ If $x<.25,$ then $\lfloor{\frac{2 x+1}{3}} \rfloor + \lfloor{\frac{\lfloor{4 x \rfloor}+2}{3}} \rfloor < 0+1,$ so solutions will not be found in these intervals. Conversely, if $.25 \leqslant x < 1,$ then $\lfloor{\frac{2 x+1}{3}} \rfloor + \lfloor{\frac{\lfloor{4 x \rfloor}+2}{3}} \rfloor = 1.$
For a more direct solution, consider the possible (integer) values of $\lfloor{4 x \rfloor}.$
Case $\lfloor{4 x \rfloor} \leqslant -1:$ This means $x <0.$ In this case,
$\lfloor{\frac{2 x+1}{3}} \rfloor + \lfloor{\frac{\lfloor{4 x \rfloor}+2}{3}} \rfloor \leqslant 0+0=0$
Case $\lfloor{4 x \rfloor}=0:$ This means $0 \leqslant x <\frac{1}{4}.$ In this case,
$\lfloor{\frac{2 x+1}{3}} \rfloor + \lfloor{\frac{\lfloor{4 x \rfloor}+2}{3}} \rfloor = 0+0=0$
Case $\lfloor{4 x \rfloor}=1:$ This means $\frac{1}{4} \leqslant x <\frac{1}{2}.$ In this case,
$\lfloor{\frac{2 x+1}{3}} \rfloor + \lfloor{\frac{\lfloor{4 x \rfloor}+2}{3}} \rfloor = 0+1=1$
Case $\lfloor{4 x \rfloor}=2:$ This means $\frac{1}{2} \leqslant x <\frac{3}{4}.$ In this case,
$\lfloor{\frac{2 x+1}{3}} \rfloor + \lfloor{\frac{\lfloor{4 x \rfloor}+2}{3}} \rfloor =0+1=1 $
Case $\lfloor{4 x \rfloor}=3:$ This means $\frac{3}{4} \leqslant x <1.$ In this case,
$\lfloor{\frac{2 x+1}{3}} \rfloor + \lfloor{\frac{\lfloor{4 x \rfloor}+2}{3}} \rfloor =0+1=1$
Case $\lfloor{4 x \rfloor} \geqslant 4:$ This means $1 \leqslant x.$ In this case,
$\lfloor{\frac{2 x+1}{3}} \rfloor + \lfloor{\frac{\lfloor{4 x \rfloor}+2}{3}} \rfloor \geqslant 1+2=3.$
Thus the solution is $\frac{1}{4} \leqslant x < 1.$
|
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|
Prove that $\mathbb{Z}[\omega]$ is an integral domain
Prove that $\mathbb{Z}[\omega]=\{ a+b \omega | a,b \in \mathbb{Z} \}$ is an integral domain, where $\omega^3=1$.
My solution:
Let $(a+b\omega)(c+d\omega)= (ac-bd)+(ad+bc-bd)\omega=0$
$\Rightarrow ac=bd \; \& \; ad+bc=bd. $
If $a\neq 0$, then put $c= bd/a$ to solve the equation and get
\begin{equation}
\begin{split}
&\quad d(a^2+b^2-ab)=0 \\
&\quad \Rightarrow d= 0 \; \text{or} \; a^2+b^2=ab.
\end{split}
\end{equation}
The first case $d=0$ solves the problem, but I am not able to figure out, how to reject case 2, when $a^2+b^2=ab$.
|
Since $ab = a^2 + b^2 \geq a^2 > 0$, we must have $b\neq 0$ and $a$ and $b$ must have the same sign. This gives
$$
a^2 + b^2 = ab\\
\frac{a}{b} + \frac{b}{a} = 1
$$
$a/b$ is positive, and therefore has a (possibly irrational) square root. We get
$$
0\leq\left(\sqrt\frac ab - \sqrt\frac ba\right)^2 = \frac{a}{b} + \frac{b}{a}-2 = 1-2
$$
which is a contradiciton
|
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|
Let $A$ be a $4\times2$ matrix and $B$ be a $2\times4$ matrix. Find $BA$ when $AB$ is given.
Let $A$ be a $4\times2$ matrix and $B$ be a $2\times4$ matrix so that
$$AB=
\begin{pmatrix}
1 & 0 & -1 & 0 \\
0 & 1 & 0 & -1 \\
-1 & 0 & 1 & 0 \\
0 & -1 & 0 & 1
\end{pmatrix}.
$$
Find $BA$.
I have the answer, which should be
$BA=
\begin{pmatrix}
2 & 0 \\
0 & 2
\end{pmatrix}
$, but how do I show that this is the only possible solution or is it sufficient (according to some property which I'm not aware of) to have one match for $A$ and $B$ and therefore no other outcome for $BA$ is possible.
|
Let $C = (c_{ij}) = AB$, so $c_{ij} = a_{i1}b_{1j}+a_{i2}b_{2j}$, that means each entry $c_{ij}$ can be interpreted as the scalar product of row $i$ of $A$ and column $j$ of $B$. Let $a_{i:}$ be the $i$th row of $A$ and $b_{:j}$ the $j$-th column of $B$. Now we see that
$$a_{1:} \perp b_{:2} \text{ and } a_{1:} \perp b_{:4}$$
$$a_{3:} \perp b_{:2} \text{ and } a_{3:} \perp b_{:4}$$
$$a_{2:} \perp b_{:1} \text{ and } a_{4:} \perp b_{:1}$$
$$a_{2:} \perp b_{:3} \text{ and } a_{4:} \perp b_{:3}$$
and therefore $a_{1:} \perp b_{:2} \perp a_{3:}$ which implies $a_{1:} \Vert a_{3:}$ and similarly $a_{2:} \Vert a_{4:}$ and $b_{:1} \Vert b_{:3}$ and $b_{:2} \Vert b_{:4}$.
So we can conlcude $a_{1:} = c a_{3:}$ for some factor $c$. But we know that $a_{1:} \cdot b_{:1} = 1$ and $ca_{1:} \cdot b_{:1} = a_{3:} \cdot b_{:1} = -1$ so $c=-1$.
In the very same way we can conclude that $a_{2:} = -a_{4:}$ and $b_{:1} = - b_{:3}$ and $b_{:2} = - b_{:4}$. To summarize what we've got so far:
$$A = \begin{bmatrix}
a_{11} & a_{12} \\
a_{21} & a_{22} \\
-a_{11} & -a_{12} \\
-a_{21} & -a_{22}
\end{bmatrix} \quad B = \begin{bmatrix}
b_{11} & b_{12} & -b_{11} & -b_{12} \\
b_{21} & b_{22} & -b_{21} & -b_{22}
\end{bmatrix}$$
Now consider the blocks $$\tilde A = \begin{bmatrix}
a_{11} & a_{12} \\
a_{21} & a_{22} \end{bmatrix} \text{ and }\tilde B = \begin{bmatrix}
b_{11} & b_{12} \\
b_{21} & b_{22} \end{bmatrix} .$$
Note that $A = \begin{bmatrix} \tilde A \\ - \tilde A \end{bmatrix}$ and $B = \begin{bmatrix} \tilde B & -\tilde B\end{bmatrix}$.
You immediately see from the given equation that $\tilde A \tilde B = I$ so $\tilde A = \tilde B^{-1}$.
And therefore $BA = \tilde B \tilde A + (-\tilde B)(-\tilde A) = 2I$ which is your desired result.
|
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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 $3n^2 + 3n +2$.
|
Because for $n\geq4$ we have:
$$3^n=(1+2)^n\geq1+2n+\frac{n(n-1)}{2}\cdot2^2+\frac{n(n-1)(n-2)}{6}\cdot2^3+$$
$$+\frac{n(n-1)(n-2)(n-3)}{24}\cdot2^4>n^3+n.$$
The last inequality is true because it's
$$3+6n+6n^2-6n+4n^3-12n^2+8n+2n^4-12n^3+22n^2-12n>3n^3+3n$$ or
$$2n^4-11n^3+16n^2-7n+3>0$$ or
$$2n^4-8n^3-3n^3+12n^2+4n^2-16n+9n+3>0$$ or
$$(n-4)(2n^3-3n^2+4n)+9n+3>0,$$
which is obvious for $n\geq4$.
|
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|
If $ (x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1$, show that $x+y=0$
For $\{x,y\}\subset \Bbb R$, $(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1.$
Prove that $x+y=0.$
Problem presented in a book, as being from Norway Math Olympiad 1985. No answer was presented. My developments are not leading to a productive direction. Sorry if this is a duplicate. Hints and answers are welcomed.
|
Multiplying by $(x-\sqrt{x^2 +1})(y-\sqrt{y^2 +1})$ we get
$$(x-\sqrt{x^2 +1})(y-\sqrt{y^2 +1})=1$$
and thus $$(x+\sqrt{x^2 +1})(y+\sqrt{y^2 +1})=(x-\sqrt{x^2 +1})(y-\sqrt{y^2 +1})$$
hence
$$x(\sqrt{x^2 +1} +\sqrt{y^2 +1} )=-y(\sqrt{x^2 +1} +\sqrt{y^2 +1} )$$
therefore $$x=-y$$
|
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|
$\arctan{x}=\arccos{2x}.$ First, let's determine some domains. By definition, it follows that $\arccos:[-1,1]\rightarrow[0,\pi]$ and $\arctan:\mathbb{R}\rightarrow\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. Since the domain of $\arctan$ is the entire reals, the restricting factor here should be the domain of arccos. This means that $2x\in[-1,1]\Rightarrow x\in\left[-\frac{1}{2},\frac{1}{2}\right]$.
I now take cosine of both sides and get $$\cos{(\arctan{x})}=\cos{(\arccos{2x})}=2x.$$
I use the formular $\cos{a}=\frac{1}{\sqrt{1+\tan^2{a}}},$ and rewrite as
$$\frac{1}{\sqrt{1+\tan^2{(\arctan{x})}}}=\frac{1}{\sqrt{1+x^2}}=2x.$$
Squaring both sides, simplifying and setting $t=x^2$ gives the quadratic equation
$$t^2+t-\frac{1}{4}=0\Longleftrightarrow \left\{
\begin{array}{rcr}
t_1 & = & -\frac{1}{2}-\frac{\sqrt{2}}{2} <0 \\
t_2 & = & -\frac{1}{2}+\frac{\sqrt{2}}{2} >0\\
\end{array}
\right.$$
Since $t_1<0$, it is a false root since a square of a real number $x$ can not become a negative real number $t$. So the final two roots in terms of $x$ are
$$\left\{\begin{array}{rcr}
x_1 & = & +\sqrt{-\frac{1}{2}+\frac{\sqrt{2}}{2}} = \sqrt{\frac{1}{2}(\sqrt{2}-1)} \\
x_2 & = & -\sqrt{-\frac{1}{2}+\frac{\sqrt{2}}{2}} = -\sqrt{\frac{1}{2}(\sqrt{2}-1)} \\
\end{array}
\right.$$
Clearly, both $x_1,x_2\in[-\frac{1}{2},\frac{1}{2}]$. Only $x_1$ is correct however. I did something wrong setting up the domain didin't I? I have a feeling that the correct domain for the solutions is $[0,\frac{1}{2}],$ I don't know how to see it.
|
Let $\arctan x=\arccos(2x)=u$
Using the definition of Principal values of Inverse trigonometric functions,
$-\dfrac\pi2\le u\le\dfrac\pi2,0\le u\le\pi\implies0\le u\le\dfrac\pi2\implies x\ge0$
We have $x=\tan u,2x=\cos u$
$$\dfrac1{(2x)^2}-x^2=1\implies4x^4+4x^2-1=0$$
$$\implies x^2=\dfrac{-1\pm\sqrt2}2$$
For real $x,$
$$x^2=\dfrac{\sqrt2-1}2$$
and we know $x\ge0$
|
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|
Calculating the limit $\lim\limits_{x \to 0^+} \frac{\sqrt{\sin x}-\sin\sqrt{ x}}{x\sqrt{x}}$
Calculate $$\lim\limits_{x \to 0^+} \dfrac{\sqrt{\sin x}-\sin \sqrt{x}}{x\sqrt{x}}$$
without use Taylor serie and L'Hôpital.
$$\lim\limits_{x \to 0^+} \dfrac{\sqrt{\sin x}-\sin \sqrt{x}}{x\sqrt{x}}\cdot\dfrac{\sqrt{\sin x}+\sin \sqrt{x}}{\sqrt{\sin x}+\sin \sqrt{x}}=\lim\limits_{x \to 0^+} \dfrac{\sin x-\sin^2\sqrt{x}}{x\sqrt{x}(\sqrt{\sin x}+\sin \sqrt{x})}$$
now what ?
|
The hint.
Prove that for all $x>0$ we have:
$$-\frac{1}{6}<\frac{\sin{x}-x}{x^3}<-\frac{1}{6}+\frac{x^2}{120}$$ and from this we obtain:
$$\lim_{x\rightarrow0^+}\frac{\sin{x}-x}{x^3}=-\frac{1}{6}.$$
Let $f(x)=\sin{x}-x+\frac{1}{6}x^3$, where $x>0$.
Thus, $f'''(x)=-\cos{x}+1\geq0$, which gives
$f''(x)>f''(0)=0$, $f'(x)>f'(0)=0$ and $f(x)>f(0)=0$,
which gives a proof of the left inequality.
By the same way we can prove the right inequality.
|
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|
Number of $6$ digit natural numbers with $3$ odd and $3$ even digits I know this has been asked before but I am stuck and need help.
For arranging $6$ digits there are $6! = 720$ ways.
One way is $3$ odd first then $3$ even, which is
$$5 \times 4 \times 3 \times 5 \times 4 \times 3.$$
This is one way, so we multiply this by $720$ (no. of ways) then deduct the numbers where $0$ comes first, but the answer is coming huge, more than total no. of $6$ digit numbers. Where am I wrong?
|
For arranging $6$ digits, there are $6! = 720$ ways.
This claim is true only if all six digits are distinct. However, $434152$ is a six-digit number with three even and three odd digits.
Find the number of six-digit natural numbers with three even and three odd digits.
Method 1: We add the number of six-digit numbers whose leading digit is odd to those whose leading digit is even.
If the leading digit is odd, two of the remaining five digits are odd. We choose their positions in $\binom{5}{2}$ ways. Each of the three odd digits can be filled in $5$ ways. Each of the three even digits can be filled in $5$ ways. Hence, there are
$$\binom{5}{2} \cdot 5^3 \cdot 5^3 = \binom{5}{2}5^6$$
six-digit natural numbers whose leading digit is odd.
If the leading digit is even, it cannot be zero, so it can be filled in $4$ ways. Two of the remaining five digits are even. Their positions can be selected in $\binom{5}{2}$ ways. Each of those positions can be filled in $5$ ways, as can each of the three positions that are filled with odd digits. Hence, there are
$$\binom{5}{2} 4 \cdot 5^2 \cdot 5^3 = \binom{5}{2} \cdot 4 \cdot 5^5$$
Hence, the total number of six-digit natural numbers with three even and three odd digits is
$$\binom{5}{2}(5^6 + 4 \cdot 5^5) = 5^5\binom{5}{2}(4 + 4) = 9 \cdot 5^5\binom{5}{2} = 90 \cdot 5^5$$
Method 2: We count six-digit sequences with three even and three odd digits, then multiply by $9/10$ to account for the fact that the leading digit cannot be zero.
There are $\binom{6}{3}$ ways to choose the positions of the even digits. The three even digits can each be filled in $5$ ways. The three odd digits can each be filled in $5$ ways. Hence, the number of permissible sequences is
$$\frac{9}{10} \binom{6}{3} \cdot 5^3 \cdot 5^3 = \frac{9}{10} \cdot 20 \cdot 5^6 = 9 \cdot 2 \cdot 5 \cdot 5^5 = 90 \cdot 5^5$$
|
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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?
|
For $k+1$ it is equivalent to
$$
\frac{4^k}{k+1}<\frac{(2k+1)(k+2)(2k)!}{2(k+1)^2(k!)^2}$$
It is enough to prove that
$$\frac{(2k+1)(k+2)}{2(k+1)^2}>1$$
Notice that the above fraction is monotonically decreasing and its limit at $k\to\infty$ is $1$, therefore it is always greater than $1$.
And also, you have a mistake, as Harry noticed, $(2k)!\neq2k!$.
Edit
I have no idea what is Jaroslaw talking about. There is step-by-step solution. Suppose it is true for $k$. Then it is true for $k+1$ if and only if
$$\frac{4^{k+1}}{k+2}<\frac{(2k+2)!}{\left((k+1)!\right)^2}\\\frac{4^k}{k+1}\cdot\frac{4(k+1)}{k+2}<\frac{(2k+2)(2k+1)(2k)!}{(k+1)^2(k!)^2}\\\frac{4^k}{k+1}\cdot\frac{4(k+1)}{k+2}<\frac{(2k)!}{(k!)^2}\cdot\frac{(2k+2)(2k+1)}{(k+1)^2}$$
It is enough to prove that:
$$\frac{4(k+1)}{k+2}<\frac{(2k+2)(2k+1)}{(k+1)^2}$$
It is equivalent to
$$\frac{(2k+2)(2k+1)}{(k+1)^2}>\frac{4(k+1)}{k+2}\\\frac{2(k+1)(2k+1)(k+2)}{4(k+1)(k+1)^2}>1\\\frac{(2k+1)(k+2)}{2(k+1)^2}>1$$
|
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|
help in solving $z^2+|z|^2=1+2i$ I can't solve this:
$$z^2+|z|^2=1+2i$$
I've found $\sqrt{ \frac{2}{5}}$ for the modulus of z but I can't explicit about the angle of z .
The result of the book is, $ \pm (\frac{1}{\sqrt{2}}+i*\sqrt{2})$ but there are so many mistakes that I'm afraid is not very reliable
|
$$z^2+|z|^2=(a+bi)^2+a^2+b^2=a^2+2abi-b^2+a^2+b^2=2a^2+2abi$$
So $$2a^2+2abi=1+2i$$
Equating real and imaginary parts:
$$\Re:2a^2=1\Rightarrow a=\pm\frac{1}{\sqrt{2}}$$
$$\Im:2(\pm\frac{1}{\sqrt{2}})bi=2i\Rightarrow\pm\frac{1}{\sqrt{2}}b=1\Rightarrow b=\pm\sqrt{2}$$
Thus,$z=\pm\frac{1}{\sqrt{2}}\pm\sqrt{2}i$
|
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|
Finding all values such that column vector is a linear combination Question:
For which values(s) of $ \ a$ is the column $ \ c = \begin{bmatrix} a\\a^2\\0\\a+1\end{bmatrix}$ a linear combination of the columns of,
$ \ x = \begin{bmatrix} 1\\1\\1\\1\end{bmatrix}$, $\ y = \begin{bmatrix} 1\\0\\-1\\0\end{bmatrix}$, $ \ z= \begin{bmatrix} 0\\1\\0\\1\end{bmatrix}$.
My attempt:
$ \ c = \begin{bmatrix} a\\a^2\\0\\a+1\end{bmatrix} = \ r\begin{bmatrix} 1\\1\\1\\1\end{bmatrix} + s \begin{bmatrix} 1\\0\\-1\\0\end{bmatrix} + t \begin{bmatrix} 0\\1\\0\\1\end{bmatrix} $
We know its a linear combination $\iff $ $ \ r+s = a, r+t = a^2, r-s = 0, r+t = a+1$ . I am not sure how to find all values of $ \ a$ now.
|
A necessary condition is:
$$a^2 = r+t = a+1 \implies a = \frac{1\pm\sqrt{5}}{2}$$
Then you have $2r = a \implies r = s = \frac{1\pm\sqrt{5}}{4}$ so in that case the rows are indeed linearly dependent.
Thus, the rows are linearly dependent iff $a = \frac{1\pm\sqrt{5}}{2}$
|
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How would I solve the following? $$M = \begin{pmatrix}1&-4\\ -6&-6\end{pmatrix}$$
Find $c_1$ and $c_2$ such that $M^2 + c_1M +c_2I_2 = 0$, where $I_2$ is the identity $2 \times 2$ matrix.
The matrix above is $M$.
I got stuck after I merged the $c_1$ and $c_2$ into their respective matrices and calculated the square of $M$.
|
$M^2=\left(
\begin{array}{cc}
25 & 20 \\
30 & 60 \\
\end{array}
\right)$
$M^2 + c_1M +c_2I_2 = 0$ becomes
$\left(
\begin{array}{cc}
25 & 20 \\
30 & 60 \\
\end{array}
\right)+\left(
\begin{array}{cc}
c_1 & -4 c_1 \\
-6 c_1 & -6 c_1 \\
\end{array}
\right)+\left(
\begin{array}{cc}
c_2 & 0 \\
0 & c_2 \\
\end{array}
\right)=\left(
\begin{array}{cc}
0 & 0 \\
0 & 0 \\
\end{array}
\right)$
that is
$\left(
\begin{array}{rr}
c_1+c_2+25=0 & 20-4 c_1=0 \\
30-6 c_1=0 & -6 c_1+c_2+60=0 \\
\end{array}
\right)$
from $20-4c_1=0$ we get $c_1=5$ and substituting both in $c_1+c_2+25=0$ and in $-6c_1+c_2+60=0$ we get $c_2=-30$
Solution is $c_1=5;\;c_2=-30$
|
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|
Determine the derivative of $\arctan$ function Find $f'(x)$ for the function:
$f(x)= \arctan(\frac{a+x}{1-ax}))$ , $a\in R$
So this is what I've done:
$f(x) = \arctan x$
$f'(x) = \frac{1}{1+x^2}$
$x= \frac{a+x}{1-ax}$
$f'(x) = \frac{1}{1+\frac{(a+x)^2}{(1-ax)^2}}$
$f'(x) = \frac{(1-ax)^2}{(1-ax)^2+(a+x)^2}$
Is this correct?
|
you must use the chain rule: $${1 \left( \left( -ax+1 \right) ^{-1}+{\frac { \left( a+x \right) a}{
\left( -ax+1 \right) ^{2}}} \right) \left( {\frac { \left( a+x
\right) ^{2}}{ \left( -ax+1 \right) ^{2}}}+1 \right) ^{-1}}
$$
simplified to
$$\frac{1}{1+x^2}$$
|
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|
No. of isosceles triangles possible of integer sides with sides $\leq n$ Prove that the no. of isosceles triangles with integer sides, no sides exceeding $n$ is $\frac{1}{4}(3n^2+1)$ or $\frac{3}{4}(n^2)$ according as n is odd or even, n is any integer.
How to do it? I found that under these conditions no. of triangles possible may be
${n\choose 2}$
|
Fix the length of the two equal sides, say $k$. In how many ways can you choose the length of the base $b(k)$? Obviously $b(k) \ge 1$ and, for the triangle inequality, $b(k) < 2k$. But, since no side can exceed $n$, $b(k) \le n$. Putting these things together, we conclude that:
*
*when $2k - 1 \le n$, that is, when $\displaystyle k \le \left\lfloor\frac{n+ 1}{2}\right\rfloor$, $b(k)$ can range from $1$ to $2k - 1$ inclusive;
*when $2k - 1 > n$, which means $k > \left\lfloor\frac{n + 1}{2}\right\rfloor$, $b(k)$ can range from $1$ to $n$ inclusive.
Hence the total number of isosceles triangles is $$\sum_{k = 1}^n b(k) = \sum_{k = 1}^{\left\lfloor\frac{n + 1}{2}\right\rfloor} (2k - 1) + \sum_{k = \left\lfloor\frac{n + 1}{2}\right\rfloor + 1}^n n = \left\lfloor\frac{n + 1}{2}\right\rfloor^2 + n\left(n - \left\lfloor\frac{n + 1}{2}\right\rfloor\right)$$
Now when $n$ is odd $\displaystyle \left\lfloor\frac{n + 1}{2}\right\rfloor = \frac{n + 1}{2}$, so the expression above reduces to $$\left(\frac{n + 1}{2}\right)^2 + n\left(n - \frac{n + 1}{2}\right) = \frac{n^2 + 2n + 1 + 4n^2 - 2n^2 - 2n}{4} = \boxed{\frac{3n^2 + 1}{4}}$$ while, when $n$ is even, $\displaystyle \left\lfloor\frac{n + 1}{2}\right\rfloor = \frac{n}{2}$ and the formula is $$\left(\frac{n}{2}\right)^2 + n\left(n - \frac{n}{2}\right) = \frac{n^2 + 4n^2 - 2n^2}{4} = \boxed{\frac{3n^2}{4}}$$
|
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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.
|
Also, SOS helps!
$$\text{LHS-RHS} = \sum\limits_{cyc}\left\{\frac{a^3}{b^2}-a-2(a-b)\right\}-\frac{3(a^2+b^2+c^2)-(a+b+c)^2}{a+b+c}$$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\sum\limits_{cyc} {\frac { \left( a+2\,b \right) \left( a-b \right) ^{2}}{{b}^{2}}} - \sum\limits_{cyc} \frac{(a-b)^2}{a+b+c}=\frac{1}{a+b+c} \sum\limits_{cyc} {\frac { \left( a-b \right) ^{2} \left( {a}^{2}+3\,ab+ac+{b}^{2}+2\,bc
\right) }{{b}^{2} }}\geqq 0
$
Done!
|
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|
In a circle, parallel chords of length $2$, $3$, and $4$ I was helping my comrade answering some questions taken from review classes when I stumbled upon this question. It looks like this:
In a circle, parallel chords of length $2$, $3$, and $4$ determine central angles of $A$, $B$, and $A+B$ radians, respectively, where
$A+B < \pi$. Express the $\cos A$ into a fraction in lowest term and find the sum of the numerator and denominator of the fraction.
My work
The problem sounds difficult for us, so we came here to seek help. How do you answer the above problem?
|
Note: this question is basically identical to the 1985 AIME Problem 9, so I'll summarise the solution below.
Because all chords of a given length in a given circle subtend the same arc and therefore the same central angle, we can rearrange our chords into a triangle with the circle as its circumcircle.
This triangle has a semiperimeter of $\frac{9}{2}$, and so by Heron's formula, this triangle has area $\sqrt{\frac{9}{2}\times\frac{5}{2}\times\frac{3}{2}\times\frac{1}{2}}=\frac{3}{4}\sqrt{15}$
Now, the area of a triangle with circumradius $R$ and sides $a,b,c$ is given by $\frac{abc}{4R}$, so $\frac{3}{4}\sqrt{15}=\frac{6}{R}$, which means $R=\frac{8}{\sqrt{15}}$.
Now, consider the triangle formed by two radii and the chord of length $2$ (see Fig. 2). Because this isosceles triangle has vertex angle $\alpha$, by the Law of Cosines, $$2^2=R^2+R^2-2R^2\cos\alpha\implies\cos\alpha=\frac{2R^2-4}{2R^2}=\frac{17}{32}$$
So $\alpha$ in lowest terms is $\frac{17}{32}$, and the sum of the numerator and denominator is equal to $17+32=49$
|
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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 .
|
Not a beautiful proof I have to admit, but it is a proof:
Multiply 2 sides of equation with $(a+b)(c+d)(a+b+c+d)$, we get to the equivalent form:
$$
ab(c+d)(a+b+c+d) + cd(a+b)(a+b+c+d) \leq (a+c)(b+d)(a+b)(c+d).
$$
Work out both sides of the equation above to reduce it to the following equivalent form:
$$
2abcd \leq a^2d^2 + b^2c^2,
$$
which is obviously true due to the AM-GM inequality.
|
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|
$f(x)=?$ if we have $f\left(\frac{x}{x^2+x+1}\right)=\frac{x}{x^2-x+1}$ $f(x)=?$
If we have $$f\left(\frac{x}{x^2+x+1}\right)=\frac{x}{x^2-x+1}$$ to fractions are very similar. I don't have an idea to find $f(x)$. Can someone show me a clue ?
|
$$f\left( \dfrac{x}{x^{2}+x+1}\right)=\dfrac{x}{x^{2}-x+1}\rightarrow f\left( \dfrac{1}{x+\frac{1}{x}+1}\right)=\dfrac{1}{x+\frac{1}{x}-1}\rightarrow$$
$$f\left( \dfrac{1}{z+1}\right)=\dfrac{1}{z-1}$$
$$\dfrac{1}{z+1}=t\rightarrow \dfrac{1}{t}=z+1\rightarrow z=\dfrac{1}{t}-1\rightarrow z=\dfrac{1-t}{t}\rightarrow$$
$$f(t)=\dfrac{1}{\frac{1-t}{t}-1}=\dfrac{1}{\frac{1-2t}{t}}=\dfrac{t}{1-2t}\rightarrow f(x)=\dfrac{x}{1-2x}$$
Testing:
$$f\left( \frac{x}{x^{2}+x+1}\right)=\dfrac{\frac{x}{x^{2}+x+1}}{1-2\left( \frac{x}{x^{2}+x+1}\right)}=\dfrac{\frac{x}{x^{2}+x+1}}{\frac{x^{2}+x+1-2x}{x^{2}+x+1}}=\dfrac{x}{x^{2}-x+1}$$
|
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|
Distributing 10 different folders between 7 people such that John gets at least 1 folder I did this by choosing 1 folder to give John ${10 \choose 1}=10$ and then saying that each of the remaining 9 folders can have 7 "states": They have to go to one of the 7 people. So that's $7^9$
So my answer is $10 \times 7^9$
My friend, however, says that the answer should be counting all the possibilites and substracting the ones in which John doesn't get any folders. That'd be $7^{10} - 6^{10}$
So who is wrong here and why? Thanks in advance
|
Your friend is correct. You are counting cases in which John receives more than one folder multiple times.
If John receives exactly $k$ folders, then there are $6^{10 - k}$ ways to distribute the remaining $10 - k$ folders. Since John can receive exactly $k$ folders in $\binom{10}{k}$ ways, the number of ways to distribute the folders so that John receives at least one is
$$\sum_{k = 1}^{10} \binom{10}{k}6^{10 - k} = \binom{10}{1}6^9 + \binom{10}{2}6^8 + \binom{10}{3}6^7 + \binom{10}{4}6^6 + \binom{10}{5}6^5 + \binom{10}{6}6^4 + \binom{10}{7}6^3 + \binom{10}{8}6^2 + \binom{10}{9}6^1 + \binom{10}{10}6^0 = 7^{10} - 6^{10}$$
Say each folder is a different color and John receives blue, red, and green folders. You count this distribution three times, once for each way of designating one of the colors as the one John receives and the other two as additional folders. In general, if John receives $k$ folders, you count this case $k$ times, once for each way of designating a particular folder as the one reserved for John. Notice that
$$\sum_{k = 1}^{10} k\binom{10}{k}6^{10 - k} = 1\binom{10}{1}6^9 + 2\binom{10}{2}6^8 + 3\binom{10}{3}6^7 + 4\binom{10}{4}6^6 + 5\binom{10}{5}6^5 + 6\binom{10}{6}6^4 + 7\binom{10}{7}6^3 + 8\binom{10}{8}6^2 + 9\binom{10}{9}6^1 + 10\binom{10}{10}6^0 = 10 \cdot 7^9$$
|
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If $\alpha,a,b$ are integers and $b\neq-1$, then prove that, if $\alpha$ satisfies the equation $x^2+ax+b+1=0$, $a^2+b^2$ must be composite.
Let $\alpha,a,b$ be integers such that $b\neq-1$. Assume that $\alpha$ satisfies the equation $x^2+ax+b+1=0$. Prove that the integer $a^2+b^2$ must be composite.
$\alpha=\frac{-a\pm\sqrt{a^2-4(b+1)}}{2}$is an integer. But manipulating this expression is leading me to nowhere. Please help.
|
We have
\begin{eqnarray*}
\alpha^2 + a \alpha +b+1=0.
\end{eqnarray*}
Multiply by $4$ and add $a^2$
\begin{eqnarray*}
(2 \alpha + a)^2 =a^2 -4(b+1).
\end{eqnarray*}
Now add & subtract $b^2$ & rearrange to
\begin{eqnarray*}
a^2+b^2=(2 \alpha + a+b+2)(2 \alpha + a-b-2).
\end{eqnarray*}
|
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|
Basic arithmetic with matrices We have matrices.
$$A=\begin{bmatrix}
-2 & 0 \\
-5 & 6 \\
\end{bmatrix}
B^{-1}=\begin{bmatrix}
-7 & 8 \\
2 & -8 \\
\end{bmatrix}
C=\begin{bmatrix}
-15 & -2 \\
-8 & -14 \\
\end{bmatrix}
$$
We need to solve matrix $X$ from equation:
$$A^{-1}XB-C=0$$
$$X=AB^{-1}+C$$
$$X=\begin{bmatrix}
-2 & 0 \\
-5 & 6 \\
\end{bmatrix}
\begin{bmatrix}
-7 & 8 \\
2 & -8 \\
\end{bmatrix}
+
\begin{bmatrix}
-15 & -2 \\
-8 & -14 \\
\end{bmatrix}
$$
$$
X=\begin{bmatrix}
-1 & -18 \\
39 & -102 \\
\end{bmatrix}
$$
This doesn't seem to be correct solutions to this equations ?
|
As point out by you that the order matters:
Pre-multiply and post-multiply are different.
Starting from
$${A^{ - 1}}XB - C = 0$$
Add $C$ on both sides
$${A^{ - 1}}XB = C$$
Now post-multiply by $B^{-1}$ on both sides
$${A^{ - 1}}X{B}{B^{ - 1}} = C{B^{ - 1}}$$
$${A^{ - 1}}X{I} = C{B^{ - 1}}$$
$${A^{ - 1}}X = C{B^{ - 1}}$$
Now pre-multiply by $A$ on both sides to get the final form
$${A}{A^{ - 1}}X = AC{B^{ - 1}}$$
$$X = AC{B^{ - 1}}$$
|
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|
How can find domain and range of $f(x)=\sqrt{\sqrt 2 \lfloor x \rfloor - \lfloor\sqrt 2x \rfloor}$? How can find $D_{f(x)},R_{f(x)}$ when $$f(x)=\sqrt{\sqrt 2 \lfloor x \rfloor - \lfloor\sqrt 2x \rfloor} \ ?$$
I help some problem like this for a question here ... (Domain and range of a floor function)
But I get stuck on this to find Domain and Range of above function .
If You can help me to find out , I will be very thankful .
|
So, with the standard notation for the floor function, we are to study
$$ \bbox[lightyellow] {
y = \sqrt {\sqrt 2 \left\lfloor x \right\rfloor - \left\lfloor {\sqrt 2 x} \right\rfloor }
} \tag {1}$$
Let's first introduce some additional notation concerning the fractional part $\{x\}$
$$
x = \left\lfloor x \right\rfloor + \left\{ x \right\}\quad 0 \le \left\{ x \right\} < 1
$$
and the Iverson bracket
$$
\left[ P \right] = \left\{ {\begin{array}{*{20}c}
1 & {P = TRUE} \\
0 & {P = FALSE} \\
\end{array} } \right.
$$
Then the radicand can be written as
$$ \bbox[lightyellow] {
\eqalign{
& r = \sqrt 2 \left\lfloor x \right\rfloor - \left\lfloor {\sqrt 2 x} \right\rfloor = \cr
& = \sqrt 2 \left( {x - \left\{ x \right\}} \right) - \left( {\sqrt 2 x - \left\{ {\sqrt 2 x} \right\}} \right) = \cr
& = \left\{ {\sqrt 2 x} \right\} - \sqrt 2 \left\{ x \right\} \cr}
} \tag {2}$$
and, since the fractional part ranges within $[0,1)$, the radicand is bound to
$$ \bbox[lightyellow] {
- \,\sqrt 2 < r = \left\{ {\sqrt 2 x} \right\} - \sqrt 2 \left\{ x \right\} < 1
} \tag {3}$$
If the function is defined in the real field, then its Domain of definition will be given by the values of $x$ such that
$$ \bbox[lightyellow] {
x:\;\;0 \le \left\{ {\sqrt 2 x} \right\} - \sqrt 2 \left\{ x \right\}\quad \Leftrightarrow \quad \sqrt 2 \left\{ x \right\} \le \left\{ {\sqrt 2 x} \right\}
} \tag {4}$$
It can be easily seen that the fractional part of a sum can be written as
$$ \bbox[lightyellow] {
\eqalign{
& \left\{ {z + y} \right\} = \left\{ {\left\{ z \right\} + \left\{ y \right\}} \right\} = \cr
& = \left\{ z \right\} + \left\{ y \right\} - \left\lfloor {\left\{ z \right\} + \left\{ y \right\}} \right\rfloor = \cr
& = \left\{ z \right\} + \left\{ y \right\} - \left[ {1 \le \left\{ z \right\} + \left\{ y \right\}} \right] \cr}
} \tag {5}$$
so that the RHS of the inequality (4) above becomes
$$ \bbox[lightyellow] {
\eqalign{
& \left\{ {\sqrt 2 x} \right\} = \left\{ {\sqrt 2 \left\lfloor x \right\rfloor + \sqrt 2 \left\{ x \right\}} \right\} = \cr
& = \left\{ {\sqrt 2 \left\lfloor x \right\rfloor } \right\} + \left\{ {\sqrt 2 \left\{ x \right\}} \right\} - \left[ {1 \le \left\{ {\sqrt 2 \left\lfloor x \right\rfloor } \right\} + \left\{ {\sqrt 2 \left\{ x \right\}} \right\}} \right] \cr}
} \tag {6}$$
and the inequality transforms into
$$ \bbox[lightyellow] {
\eqalign{
& \sqrt 2 \left\{ x \right\}\; \le \;\left\{ {\sqrt 2 \left\lfloor x \right\rfloor } \right\} + \left\{ {\sqrt 2 \left\{ x \right\}} \right\} - \left[ {1 \le \left\{ {\sqrt 2 \left\lfloor x \right\rfloor } \right\} + \left\{ {\sqrt 2 \left\{ x \right\}} \right\}} \right] \cr
& \quad \Downarrow \cr
& \left\lfloor {\sqrt 2 \left\{ x \right\}} \right\rfloor + \left\{ {\sqrt 2 \left\{ x \right\}} \right\}\; \le \;\left\{ {\sqrt 2 \left\lfloor x \right\rfloor } \right\} + \left\{ {\sqrt 2 \left\{ x \right\}} \right\} - \left[ {1 \le \left\{ {\sqrt 2 \left\lfloor x \right\rfloor } \right\} + \left\{ {\sqrt 2 \left\{ x \right\}} \right\}} \right] \cr
& \quad \Downarrow \cr
& 0 \le \left\lfloor {\sqrt 2 \left\{ x \right\}} \right\rfloor + \left[ {1 \le \left\{ {\sqrt 2 \left\lfloor x \right\rfloor } \right\} + \left\{ {\sqrt 2 \left\{ x \right\}} \right\}} \right]\; \le \;\left\{ {\sqrt 2 \left\lfloor x \right\rfloor } \right\} < 1 \cr}
} \tag {7}$$
Now, the Poisson bracket cannot be one (because of the <1), so it shall be null, i.e. its inequality false.
Also, the floor before it (which is either $0$ or $1$) cannot be but $0$.
Thus we can conclude that the Domain of definition (y real) is given by
$$ \bbox[lightyellow] {
\left\{ \matrix{
\left\{ {\sqrt 2 \left\lfloor x \right\rfloor } \right\} + \left\{ {\sqrt 2 \left\{ x \right\}} \right\} < 1 \hfill \cr
\left\lfloor {\sqrt 2 \left\{ x \right\}} \right\rfloor = 0 \hfill \cr} \right.\quad \Rightarrow \quad \left\{ \matrix{
\left( {0 \le } \right)\left\{ x \right\} < 1/\sqrt 2 \hfill \cr
\left( {0 \le } \right)\left\{ {\sqrt 2 \left\lfloor x \right\rfloor } \right\} + \sqrt 2 \left\{ x \right\} < 1 \hfill \cr} \right.
} \tag {7.a}$$
and the Codomain is
$$ \bbox[lightyellow] {
0 \le y < 1
} \tag {7.b}$$
Because of the irrationality of $\sqrt{2}$ inequality (7.a) does not add much to the original (4), and cannot be further simplified.
|
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|
how to $\int_0^{ +\infty} \frac{\sin(x)}{x+1}\, dx \leq \frac e5 \ln(\pi)$? I need to prove irreproachably that $$\int_0^{ +\infty} \frac{\sin(x)}{x+1}\, dx \leq \frac e5 \ln(\pi)$$ .
With an approximate calculation $\int_0^{ +\infty} \frac{\sin(x)}{x+1}\, dx\approx 0.62145$ and $\frac e5 \ln(\pi)\approx 0.62233$
We can see by Laplace transform that $$\int_{0}^{+\infty}\frac{\sin x}{1+x}\,dx = \int_{0}^{+\infty}\frac{e^{-s}}{1+s^2}\,ds $$ and deduce $$\int_{0}^{+\infty}\frac{\sin x}{1+x}\,dx = \int_{0}^{+\infty}\frac{e^{-s}}{1+s^2}\,ds\leq \sqrt{\int_{0}^{+\infty}\frac{ds}{(1+s^2)^2 } } \sqrt{\int_{0}^{+\infty} e^{-2s} ds} = \sqrt{\frac{\pi}{8}}.$$
But $\frac e5 \ln(\pi)< \sqrt{\frac{\pi}{8}}$
|
The given integral equals
$$\begin{eqnarray*} \int_{0}^{\pi}\sin(x)\sum_{n\geq 0}\frac{(-1)^n}{x+n\pi+1}\,dx &=& \frac{1}{2\pi}\int_{0}^{\pi}\sin(x)\left[\,\psi\left(\tfrac{x+1}{2\pi}+\tfrac{1}{2}\right)-\psi\left(\tfrac{x+1}{2\pi}\right)\right]\,dx\\&=&\int_{\frac{1}{2\pi}}^{\frac{\pi+1}{2\pi}}\sin(2\pi z-1)\left[\,\psi\left(z+\tfrac{1}{2}\right)-\psi(z)\right]\,dz\end{eqnarray*}$$
which can be efficiently approximated by using integration by parts and the Kummer-Malmsten Fourier series:
$$ \log\Gamma(z) = \left(\tfrac12 - z\right)(\gamma + \log 2) + (1 - z)\ln\pi
- \tfrac12\log\sin(\pi z) + \frac{1}{\pi}\sum_{n=1}^\infty \frac{\sin(2\pi n z)\log n} n$$
holding for any $z\in(0,1)$. Indeed the original integral equals
$$ 2\pi \int_{0}^{1/4}\left[\log\,\frac{\Gamma\left(\tfrac{3}{4}+\tfrac{1}{2\pi}+z\right)\,\Gamma\left(\tfrac{1}{4}+\tfrac{1}{2\pi}-z\right)}{\Gamma\left(\tfrac{3}{4}+\tfrac{1}{2\pi}-z\right)\,\Gamma\left(\tfrac{1}{4}+\tfrac{1}{2\pi}+z\right)}\right]\sin(2\pi z)\,dz $$
where the $\log$ term is a very regular function on the interval $\left(0,\frac{1}{4}\right)$, convex and $$\leq \left(8\log\,\frac{\Gamma\left(\tfrac{1}{2\pi}\right)}{\Gamma\left(\tfrac{1}{2\pi}+\tfrac{1}{2}\right)}-4\log(2\pi)\right)z.$$
|
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|
The sum of series $\frac14+\frac{1\cdot3}{4\cdot6}+\cdots$ The problem I have here today is the following;
$$\frac{1}{4}+\frac{1\cdot3}{4\cdot6}+\frac{1\cdot3\cdot5}{4\cdot6\cdot8}+\cdots$$ the problem is exactly phrased like this (I can't say that the $\infty$ sign is a bit unnecessary at the end),
My Attempts
We can generalize this sum by noticing that each time indices get greater the denominator and numerators are multiplied by $n+2$ for each $n$ either in numerator or denominator, we take $\dfrac{1}{4}$ out of the sum first, so the sum is equal to $$\dfrac{1}{4}+\sum_{k=4} \frac{(k-3)(k-1)}{k(k+3)}$$ then we open up the brackets and we get; ....
Then I was a little stuck here because when I opened up these brackets and try to get the partitions of the sum one of them was logical $\displaystyle\sum\frac{3}{k(k+3)}=\sum\frac{1}{k}-\frac{1}{k+3}$. I couldn't carry it out longer. What do you suggest?
Is this a problem way above elementary solutions?
|
The solution below is inspired by this other solution here.
Note that
$$\begin{align}
f(r)&=\frac {1\cdot 3\cdot 5\cdot\cdots \cdot(2r+1)}{4\cdot 6\cdot 8\cdot \cdots \cdot(2r+4)}\\
&=2\cdot \underbrace{\boxed{\frac {1\cdot 3\cdot 5\cdot\cdots \cdot(2r+1)}{2\cdot 4\cdot 6\cdot\cdots \cdot(2r+2)}}}_{A_r}\cdot \frac 1{2r+4}\\
&=2\ A_r\ \left(1-\frac {2r+3}{2r+4}\right)\\
&=2\left(A_r-A_{r+1}\right)\\
\frac 14+\frac {1\cdot 3}{4\cdot 6}+\frac {1\cdot 3\cdot 5}{4\cdot 6\cdot 8}+\cdots&=\sum_{r=0}^\infty f(r)\\
&=2\sum_{r=0}^\infty A_r-A_{r+1}\\
&=2\left(A_0-\lim_{r\to\infty}A_{r+1}\right)\\
&=2\left(\frac 12-0\right)\\
&=\color{red}1\end{align}$$
See also this for the limit of $A_{r+1}$.
|
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|
Find the volume between $x^2+y^2=9 , z=9-x^2-y^2, x^2+y^2+(z-16)^2=9$ So we have the surfaces$$x^2+y^2=9 $$$$ z=9-x^2-y^2$$$$ x^2+y^2+(z-16)^2=9$$
I was thinking this integral may be easier with cylindrical coordinates with the limits $0\le r\le 3$; $0\le\theta\le 2\pi$; and $9-r^2\le z\le S$.
Not sure how to define the superior function of $z$ ($S$) and then set the integral
|
Your approach by polar coordinates is correct.
Let $x=r\cos \theta, y=r\sin \theta$, then the first equation gives integrating area, while the second and third equation give the lower and upper bound:
$$
\begin{align}
0&\leq r\leq3 \\
9-r^2&\leq z\\
z&\leq 16-\sqrt {9-r^2}
\end{align}
$$
so we can write the double integral as
$$
\int^{2\pi}_0 \int^3_0 [ 16-\sqrt {9-r^2}-(9-r^2)]rdrd\theta
$$
which gives us $171\pi/2$
|
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|
Equation of the tangent to a graph where the point is not on the graph. If $f(x)=(x+1)^{3/2}$, provided $x\geq -1$, I am asked to find the equation of all tangent lines to $f(x)$ at the point $(\frac{4}{3},3)$.
Simple enough. I first took the derivative which is:
$$f'(x)= \frac {3\sqrt{x+1}}{2}$$
Since $(\frac{4}{3},3)$ is not on the graph of $f(x)$, I need to find the point of tangency.
So I assumed the point of tangency is at some point $(k,(k+1)^{\frac{3}{2}})$ It also has to pass the point $(\frac{4}{3},3)$ so thus:
$y-y_{1} = m(x-x_{1})$
$3-(k+1)^{\frac{3}{2}} = \frac{3}{2}\sqrt{k+1}(\frac{4}{3}-k)$
$3-\sqrt{(k+1)^2(k+1)} = 2\sqrt{k+1}-\frac{3}{2}k\sqrt{k+1}$
$3-(k+1)\sqrt{k+1} = 2\sqrt{k+1}-\frac{3}{2}k\sqrt{k+1}$
$3-k\sqrt{k+1}-\sqrt{k+1}=2\sqrt{k+1}-\frac{3}{2}k\sqrt{k+1}$
$3+\frac{1}{2}k\sqrt{k+1}-3\sqrt{k+1}=0$
Here is where I am stuck. How do I solve for $k$ (And hence $x$)?
I tried letting $\sqrt{k+1}$ equal $a$ and then try and solve the corresponding cubic but that's not getting me anywhere... Any help on this? Thanks!
|
HINT
Use the fact that your line passes through $(k,f(k))$ and $(4/3,3)$, so the slope must also be
$$
m = \frac{\Delta y}{\Delta x} = \frac{3 - f(k)}{4/3 - k}
$$
and an alternate equation would be
$$
y - 3 = m(x - 4/3)
$$
but the equations have to match exactly...
|
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Calculate $\iint_D x dxdy$ using polar coordinates Using polar coordinates, I want to calculate $\iint_D x dxdy$, where $D$ is the disk with center $(2,3)$ and radius $2$.
$$$$
I have done the following:
We have $D=\{(x,y)\mid (x-2)^2+(y-3)^2\leq 4\}$.
We use $(x,y)=(r\cos \theta, r\sin \theta)$.
From the inequality $$(x-2)^2+(y-3)^2\leq 4\Rightarrow x^2-4x+4+y^2-6y+9\leq 4 \Rightarrow x^2+y^2-4x-6y\leq -9$$ we get $$r^2\cos^2\theta+r^2\sin^2\theta-4r\cos\theta-6r\sin\theta\leq -9 \Rightarrow r^2-r(4\cos\theta-6\sin\theta)+9\leq 0$$
To find for which values of $r$ that inequality is true, we have to find first the roots of $r^2-r(4\cos\theta-6\sin\theta)+9=0$.
The roots are $$2\cos \theta+3\sin\theta\pm \sqrt{12\cos\theta\sin\theta-5\cos^2\theta}$$ Therefore, we get the inequality $r^2-r(4\cos\theta-6\sin\theta)+9\leq 0$ for $$2\cos \theta+3\sin\theta-\sqrt{12\cos\theta\sin\theta-5\cos^2\theta}\leq r\\ \leq 2\cos \theta+3\sin\theta+\sqrt{12\cos\theta\sin\theta-5\cos^2\theta}$$ or not?
So, at the integral do we use these limits for $r$ ? And what about $\theta$ ? Does it hold that $0\leq \theta\leq 2\pi$ ?
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OFC your area of $\theta$ has to be restricted to the area where it "hits" the circle and as you already figured out your ansatz leads you to a very complicated calculation.
The reason is that your circle is not centered so first shift your integration area that you have a centered circle by substituting $\overline{x} = x-2, \overline{y} = y-3$ so you get $$\iint_D x dxdy = \iint_\overline{D} (\overline{x} + 2)\;d\overline{x}d\overline{y}$$ with $$\overline{D} =\{(\overline{x},\overline{y})\mid \overline{x}^2+\overline{y}^2\leq 4\}$$
What can easily be solved by using polar coordinates.
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"url": "https://math.stackexchange.com/questions/2465787",
"timestamp": "2023-03-29T00:00:00",
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How to find $\lim_{x\to1}\left(\tan\frac{\pi x}{4}\right)^{\tan\frac{\pi x}{2}}$?
Find
$$\lim_{x\to1}\left(\tan\frac{\pi x}{4}\right)^{\tan\frac{\pi x}{2}}$$
My attempt: ON THE basis of This post
$$\lim_{x\to1}\tan\frac{\pi x}{4} =1,\quad \lim_{x\to1}\tan\frac{\pi x}{2}=\infty$$
$$\implies\lim_{x\to1}\left(\tan\frac{\pi x}{4}\right)^{\tan\frac{\pi x}{2}}= e^{\lim_{x\to1}\left[\tan\frac{\pi x}{4}-1\right]\tan\frac{\pi x}{2}}$$
Now I need to solve $\lim_{x\to1}\left[\tan\frac{\pi x}{4}-1\right]\tan\frac{\pi x}{2}$, but I don't know how to go on.
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$$\lim_{x\rightarrow1}\left(\tan\frac{\pi x}{4}\right)^{\tan\frac{\pi x}{2}}=\lim_{x\rightarrow1}\left(1+\tan\frac{\pi x}{4}-1\right)^{\frac{1}{\tan\frac{\pi x}{4}-1}\cdot\tan\frac{\pi x}{2}\left(\tan\frac{\pi x}{4}-1\right)}=$$
$$=e^{-\lim\limits_{x\rightarrow}\frac{2\tan\frac{\pi x}{4}}{1+\tan\frac{\pi x}{4}}}=e^{-1}=\frac{1}{e}.$$
I used $\tan2\alpha=\frac{2\tan\alpha}{1-\tan^2\alpha}=-\frac{2\tan\alpha}{(\tan\alpha-1)(1+\tan\alpha)}.$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2469040",
"timestamp": "2023-03-29T00:00:00",
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Length of the chord of an ellipse whose midpoint is given Find the length of the chord of the ellipse $$\frac{x^2}{25}+\frac{y^2}{16}=1\tag1$$where mid point is $(\frac{1}{2},\frac{2}{3})$
My attempt: I know the equation of the chord of an ellipse whose mid-point is given is $T=S_1$,where T is $\left(\frac{x.x_1}{25}+\frac{y.y_1}{16} - 1\right)$ and $S_1$ is $\left(\frac{x_1^2}{25}+\frac{y_1^2}{16} - 1\right)$. Using this I get the equation of the chord as $$\frac{x}{50}+\frac{y}{24}=\frac{17}{450}\tag2$$ Now by solving (1) & (2), I should get the point of intersections and eventually the length of the chord. I want to know if there is a better way to solve this question.
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Line passing through midpoint $(\tfrac{1}{2}, \tfrac{2}{3})$ is
$$p(r)=\langle\tfrac{1}{2} + r \cos t, \tfrac{2}{3} + r \sin t\rangle \tag 1$$
where $r$ is the distance along the line and $t$ is the angle of slope of line.
Using your equation of chord, we can calculate $\cos t$ and $\sin t$:
$$\frac{x}{50}+\frac{y}{24}=\frac{17}{450} \tag 2$$
$\tan t =\frac{-24}{50} = \frac{-12}{25}$, so $\sin t= \frac{12}{\sqrt{25^2+12^2}}$ and $\cos t = \frac{-25}{\sqrt{25^2 + 12^2}}$
Substitute $\cos t$ and $\sin t$ in equation $1$, substitute $p(r)$ in the ellipse equation and solve for $r$. This gives a quadratic in $r$:
$$r^2 = \dfrac{332977}{15300}$$
So the length of chord is $2r$, ie
$$2r = \frac{1}{15}\sqrt{\dfrac{332977}{17}} \approx 9.3302$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2469193",
"timestamp": "2023-03-29T00:00:00",
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