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What is wrong with this proof of: $2+2 = 5$ I have seen this image and surprised that we can prove $2 + 2 = 5$. can any one tell me what is wrong with this image.
Prove that, $2+2=5$.
We know that, $2+2=4$
$$\begin{align}\Rightarrow2+2&=4-\dfrac92+\dfrac92\\\,\\
&=\sqrt{\left(4-\dfrac92\right)^2}+\dfrac92\\
&=\sqrt{16-2\cdot4\cdot\dfrac92+\left(\dfrac92\right)^2}+\dfrac92\\
&=\sqrt{-20+\left(\dfrac92\right)^2}+\dfrac92\\
&=\sqrt{(5)^2-2\cdot5\cdot\dfrac92+\left(\dfrac92\right)^2}+\dfrac92\qquad\qquad\qquad\\
&=\sqrt{\left(5-\dfrac92\right)^2}+\dfrac92\\
&=5-\dfrac92+\dfrac92\\
&=5\\\,\\&\therefore\,2+2=5\text{ (Proved)}\\
\end{align}$$
|
No,its not possible
your second step is wrong
$4-4.5$ is negative number
So $\sqrt{(4-4.5)^2}$ is not possible
|
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|
List of Three Letters I'm having issues with my logic on this problem:
How many ways are there to form a list of three letters from the letters in the word COMBINATORICS if the letters cannot be used more often than they appear in COMBINATORICS?
I'm trying to think of this as a set with 13 elements (2 Cs, 2 Os, 2 Ms, etc).
Thus, 13*12*11 would give us 1716 possibilities but apparently that isn't correct.
Any help or clarifications would be greatly appreciated.
|
A general method for solving this kind of problem uses exponential generating functions. Let
$$
\begin{aligned}
G(x,C,O,I,M,\ldots)=&\left(1+\frac{Cx}{1!}+\frac{C^2x^2}{2!}\right)\left(1+\frac{Ox}{1!}+\frac{O^2x^2}{2!}\right)\left(1+\frac{Ix}{1!}+\frac{I^2x^2}{2!}\right)\left(1+\frac{Mx}{1!}\right)\\
&\times\left(1+\frac{Bx}{1!}\right)\left(1+\frac{Nx}{1!}\right)\left(1+\frac{Ax}{1!}\right)\left(1+\frac{Tx}{1!}\right)\left(1+\frac{Rx}{1!}\right)\left(1+\frac{Sx}{1!}\right).
\end{aligned}
$$
The variables $C,$ $O,$ $I,$ $M,$ and so on, are there just to indicate which factor counts which letter, and may all be set equal to $1.$ If we do this, we get
$$
G(x):=G(x,1,1,1,1,\ldots)=\left(1+\frac{x}{1!}+\frac{x^2}{2!}\right)^3\left(1+\frac{x}{1!}\right)^7
$$
The coefficient of $x^3$, multiplied by $3!,$ will be the desired number of arrangements.
We get
$$
\begin{aligned}
G(x)&=\left((1+x)^3+\binom{3}{1}\frac{x^2}{2!}(1+x)^2+\binom{3}{2}\left(\frac{x^2}{2!}\right)^2
(1+x)+\binom{3}{3}\left(\frac{x^2}{2!}\right)^3\right)\,\left(1+x\right)^7\\
&=(1+x)^{10}+\binom{3}{1}\frac{x^2}{2!}(1+x)^9+\ldots.
\end{aligned}
$$
The coefficient of $x^3$ in this expression is $\binom{10}{3}+\binom{3}{1}\frac{1}{2!}\binom{9}{1}.$ Multiplying by $3!$ gives $_{10}P_3+3\cdot\binom{3}{1}\binom{9}{1}.$
|
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|
Derivation of the quadratic equation So everyone knows that when $ax^2+bx+c=0$,$$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
But why does it equal this? I learned this in maths not 2 weeks ago and it makes no sense to me
|
For a lot of ways to solve quadratic equations, see Pat Ballew's manuscript Solving Quadratic Equations.
Here's a way to get the quadratic formula if you partially remember it. (I don't know off-hand if this is one of the methods in Pat Ballew's list.)
We remember that the quadratic formula gives us results that can be put into the form $A \pm \sqrt{B}.$ Note that we don't need to incorporate a constant factor with the square root, since we can bring the square of any such factor inside the square root: $7\sqrt{3}$ is equal to $\sqrt{7^2 \cdot 3}.$
Thus, we know the answer will be
$$x \; = \; A \pm \sqrt{B}$$
Isolating the radical and squaring gives:
$$(x - A)^2 \; = \; \left( \pm \sqrt{B} \right)^2$$
$$(x - A)^2 \; = \; B$$
$$(x - A)^2 - B \; = \; 0$$
$$ x^2 - 2Ax + (A^2 - B) \; = \; 0 $$
Now let's solve $x^2 + bx + c = 0.$ [By dividing through by the coefficient of $x^2,$ we may assume the coefficient of $x^2$ is $1.]$
Comparing the coefficients of $x^2 + bx + c = 0$ with the coefficients of the last displayed equation, we see that $x = A \pm \sqrt{B},$ where $-2A = b$ and $(A^2 - B) = c.$ Solving for $A$ and $B,$ we get $A = -\frac{b}{2}$ and $B = \frac{b^2}{4} - c,$ and now we have the solution for $x$ in terms of the equation's coefficients $b$ and $c.$
|
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|
Solve this Diophantine Equation: $a^3-b^3 = ab+61$ "solve this diophantine equation: $a^3-b^3 = ab+61$"
For Diophantine Equations I have always had the question in the form $ax+by = c$ and solved it via Extended Euclidean Algorithm. However, this question is not formatted in such a way.
What I am thinking is that we move $ab$ to the LHS so that we can have a single constant on the RHS. Maybe then we could expand? I'm not sure. Hints/Advice is what I am looking for.
Maybe we can substitute $a$ and $b$ with something else of a form I am more comfortable with? I'm not really sure.
|
Finding all integer solutions to the equation $a^3-b^3=ab+61$ is equal to solving the diophantine equation $a^3-ab-b^3=61$, and placing $-b$ instead of $b$ gives $a^3+ab+b^3=61$.
Since the equation above is symmetric ( that is, the value doesn't changes even if we swap the position of the variables a and b), it can be expressed with the sum and product of a and b. Let c=a+b and d=ab. Then the quadratic equation $t^2-ct+d$ has two zeros, namely a and b. Thus, the discriminant $D=c^2-4d≥0$, which yields $d\le \frac {c^2}{4}$ ( and of course $c, d \in\mathbb Z$). In addition, $a^3+ab+b^3=61 \iff a^3+b^3+ab=61$
$\iff (a+b)^3-3(a+b)ab+ab=61$
$\iff c^3-3cd+d=61$
$\iff d=(c^3-61)/(3c-1)$
Since $d \in\mathbb Z$, $27d = \frac{27c^3-1647}{3c-1} \in\mathbb Z$.
But then, $\frac{1646}{3c-1} \in\mathbb Z$, because $\frac{27c^3-1647}{3c-1} = 9c^2+3c+1\in\mathbb Z$. This means that $3c-1$ must divide $1646=2\cdot 823$ (823 is a prime). So $3c-1$ can be 1, 2, 823, 1646, -1, -2, -823, or -1646. Then $3c$ can be 2, 3, 824, 1647, 0, -1, -822, or -1645. Since c is an integer, the only possible values for c are c=0, 1, 549. This gives us the pair $(c,d) = (0,61), (1, -30), (549, 100528)$, and the only pair that $d\le \frac{c^2}{4}$ holds is $(1, -30)$. Thus, $c=1$ and $d=-30$, and it follows that a and b are the two zeros of the quadratic equation $x^2-x+30=0$. We now can conclude that the only solutions to the Diophantine equation $a^3+ab+b^3=61$ are $(a, b) = (6, -5), (-5, 6)$.
|
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|
Solve for $x: 2/\sqrt{2-x^2} = 4- 2x^2?$ How do you expand and solve for $x$?
It is $2 = 4\sqrt{2-x^2} - 2x^2\sqrt{2-x^2}$
Thank you!
How would I solve for x?
|
Note that $4-2x^2 = 2(2-x^2)$.
Muliplying both sides by $\sqrt{2-x^2}$ shows that any solution of the equation also solves $2 = 2 (2-x^2)^{\frac{3}{2}}$, or equivalently $(2-x^2)^{\frac{3}{2}} = 1$.
This is equivalent to solving $2-x^2 =1$, which yields $x^2 = 1$, from which we get $x = \pm 1$.
|
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|
How to determine whether critical points (of the lagrangian function) are minima or maxima? $f(x,y) = 2x+y$ subject to $g(x,y)=x^2+y^2-1=0$. The Lagrangian function is given by
$$
\mathcal{L}(x,y,\lambda)=2x+y+\lambda(x^2+y^2-1),
$$
with corresponding
$$
\nabla \mathcal{L}(x,y,\lambda)= \begin{bmatrix} 2 + 2\lambda x \\ 1+2\lambda y \\ x^2+y^2-1 \end{bmatrix}.
$$
From the latter we can see that $x=\frac{-1}{\lambda}$ and $y=\frac{-1}{2\lambda}$, which we can substitute into $x^2+y^2=1$ to obtain $\lambda = \pm\sqrt{\frac{5}{4}}$. Meaning that $x = \pm \dfrac{2}{\sqrt{5}}$ and $y= \pm \dfrac{1}{\sqrt{5}}$. We find the critical points $(\frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}})$ and $(\frac{-2}{\sqrt{5}}, \frac{-1}{\sqrt{5}})$.
I am confused on how I should proceed to check wether these points or minima or maxima? I know the Hessian is involved, but which one?
|
Fortunately $\{x^2+y^2-1=0\}$ is compact and $f$ continuous thus $f$ must have a min and max. Just compare the values of $f$ at the critical values. In this case the Hessian is not needed.
|
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|
Math question from the GMATprep If $xy=1$ what is the value of: $2^{(x+y)^2}/2^{(x-y)^2}$
A 1
B 2
C 4
D 16
E 19
$(x+y)^2/(x-y)^2$ because $2$ just cancels out from numerator and denominator, right?
|
Using exponentiation rules, it can be simplified.
$$ \frac{2^{(x+y)^2}}{2^{(x-y)^2}} = 2^{(x+y)^² - (x-y)^2} = 2^{x^2 + 2xy + y^2 - (x^2 - 2xy + y^2)} = 2^{4xy} = \left(2^{4}\right)^{xy} = 16 $$
|
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|
Show that the sum of squares of four consecutive natural numbers may never be a square.
Show that the sum of squares of four consecutive natural numbers may never be a square.
I know (and I have the proof) a theorem that says that every perfect square is congruent to $0, 1$ or $4$ $\pmod8$ and wanted to demonstrate this using it.
|
Let $n$, $n+1$, $n+2$, and $n+3$ be any four consecutive natural numbers. Then we have $$ n^2 + (n+1)^2 + (n+2)^2 + (n+3)^2 = 4n^2 + 12n + 14 = (2n + 3)^2 + 5. $$ Now if this sum were to be the square of a natural number $N$, say, then we must have $$ N^2 - (2n + 3)^2 = 5.$$ However we know that for any two distinct natural numbers $r$ and $k$, whenever $k < r$, we have $k^2 < r^2$ and conversely; in fact, whenever $k < r$, we also have $$ r^2 -k^2 \geq (k+1)^2 - k^2 = 2k+1. $$ That is, the difference between the squres of two natural numbers is at least $1$ more than twice the smaller.
Now $(2n + 3)^2$ is a natural number whenever $n$ is a natural number. And since $$N^2 - (2n + 3)^2 = 5 > 0,$$ we can conclude that $$2n+3 < N.$$ So we must have $$ N^2 - (2n + 3)^2 = 5 \geq 2(2n+3) + 1 = 4n+7.$$ or $$ 5 \geq 4n+7.$$ So $$ -2 \geq 4n,$$ which implies that $$-\frac{1}{2} \geq n,$$ which is clearly impossible because $n$, being a natural number is at least $1$.
|
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|
Spot my error in solving a linear system I almost always get the unit matrix if I try to get to an row reduced echelon form. I probably always make a mistake. Can you spot the error? What illegal operations could a beginner do while trying to solve a linear system?
\begin{array}{}
1 & 1 & 3 \\
-4 & -3 & -8 \\
-2 & -1 & -2 \\
1 & 2 & 7 \end{array}
2.
\begin{array}{}
1 & 1 & 3 \\
0 & 1 & 6 \\
0 & 1 & 4 \\
0 & -1 & -4 \end{array}
3.
\begin{array}{}
1 & 1 & 3 \\
0 & 1 & 6 \\
0 & 1 & 4 \\
0 & 0 & 0 \end{array}
4.
\begin{array}{}
1 & 1 & 3 \\
0 & 1 & 6 \\
0 & 0 & 2 \\
0 & 0 & 0 \end{array}
5.
\begin{array}{}
1 & 1 & 3 \\
0 & 1 & 6 \\
0 & 0 & 1 \\
0 & 0 & 0 \end{array}
6.
\begin{array}{}
1 & 1 & 3 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0 \end{array}
7.
\begin{array}{}
1 & 1 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0 \end{array}
8.
\begin{array}{}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0 \end{array}
|
In the move from $1$ to $2$, you incorrectly obtained $6$ when you should have obtained 4: $4 \times R_1 + R_2 \to R_2 = 0\;\;1\;\; 4\;$
thus giving to identical rows, rows 2 & 3.
\begin{pmatrix}
1 & 1 & 3 \\
-4 & -3 & -8 \\
-2 & -1 & -2 \\
1 & 2 & 7 \end{pmatrix}
2.
\begin{pmatrix}
1 & 1 & 3 \\
0 & 1 & 4 \\
0 & 1 & 4 \\
0 & -1 & -4 \end{pmatrix}
3. \begin{pmatrix}
1 & 1 & 3 \\
0 & 1 & 4 \\
0 & 0 & 0 \\
0 & 0 & 0 \end{pmatrix}
4. \begin{pmatrix}
1 & 0 & -1 \\
0 & 1 & 4 \\
0 & 0 & 0 \\
0 & 0 & 0 \end{pmatrix}
|
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How find this numbers $a,b$ Question
let function $f(x)=ax^2+b$, find all positive real numbers $(a,b)$,such for any real numbers,then we have
$$f(xy)+f(x+y)\ge f(x)f(y)$$
My try:
since $$f(xy)+f(x+y)\ge f(x)f(y)\Longrightarrow a(xy)^2+b+a(x+y)^2+b\ge (ax^2+b)(ay^2+b)$$
|
$f(xy)+f(x+y)≥f(x)f(y)⟹a(xy)^2+b+a(x+y)^2+b≥(ax^2+b)(ay^2+b)$
$ax^2y^2+a(x^2+y^2)+2axy+2b\ge a^2x^2y^2+ab(x^2+y^2)+b^2$
$2b-b^2\ge (a^2-a)x^2y^2+(ab-a)(x^2+y^2)-2axy$....(*)
$\frac{2b-b^2}{a}\ge (a-1)x^2y^2+(b-1)(x^2+y^2)-2xy$
In the last part, we assumed that $a\neq0$. Since the left side is a constant, the right side must have the upper bound which is less than or equal to the left side. If $a-1> 0$, the right side doesn't have the upper bound. If $a-1<0$, $b-1>0$, and $y=0$, the right side still diverges when x-> $\infty$. For similar reason, when $a-1=0$, the right side doesn't have the upper bound. Therefore, we consider the case in which $a-1<0$ and $b-1\le0$. If $x=0$, the maximum value of the right side is $0$; hence, $2b-b^2\ge0$ and $a>0$ or $2b-b^2\le0$ and $a<0$. The first case is equivalent to $2\ge b$ and $a>0$. It implies that the given inequality works if $0< a<1$ and $b\le1$. Needless to say, the second case doesn't work.
|
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|
Find the units digit in the number $7^{9999}$. I have step by step instructions from a previous example to follow, so I figure I know how to get the answer, but I don't understand fully why it works the way it does...
By Euler's theorem, if $(a,m)=1$ then $a^{\varphi(m)}\equiv 1\mod{m}$. Here we have $(7,10)=1$ and $\varphi(10)=4$ Therefore, $7^4\equiv 1\mod{10}$.
$7^4\equiv 1\mod{10}\Longrightarrow 7^{4(2499)}\equiv 1^{2499}\mod{10}$
$7^{4(2499)+3}=(7^{4})^{2499}\cdot 7^3\equiv 1^{2499}\cdot 7^3\mod{10}$
$7^3=343\equiv 3\mod{10}.$
Therefore, the units digit in the number $7^{9999}$ is $3$.
The biggest part I don't understand is why I was able to "cancel" the $7^{4(2499)}$ from both sides of the congruence. The cancellation law states that if $cx\equiv cy\mod{m}$ and $(c,m)=1$ then $x\equiv y\mod{m}$... But I don't see a common $c$ on either side of the congruence.
|
We know that
$$ 7^2 \equiv -1 \pmod{10} $$
and that
$$ 7^{9999} = 7 \cdot 7^{9998} = 7 \cdot (7^2)^{4999}, $$
and that
$$a_1 \equiv b_1 \pmod c \qquad\text{and}\qquad a_2 \equiv b_2 \pmod c \Rightarrow a_1a_2 \equiv b_1b_2 \pmod c.$$
Therefore,
$$ 7^2 \equiv -1 \pmod{10} \Rightarrow 7^{9998} = (7^2)^{4999} \equiv (-1)^{4999} \pmod{10}.$$
Since $(-1)^{4999} = -1$, then $(7^2)^{4999} \equiv -1 \pmod{10}$.
Finally,
$$ 7^{4999} = 7 \cdot 7^{9998} \equiv 7 \cdot (-1) \pmod{10} \equiv 3 \pmod{10},$$
so the units digit is 3.
|
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|
Asymptotics of sequence depending on Tricomi's function I'm dealing with the following sequence
$$ p_n = U(a, a - n, 1)$$
where $a > 0$ and $U$ is Tricomi's function.
I suspect that asymptotically when $n \to \infty$ its behaviour is a power law (evidence)
$$ p_n \sim 1/n^a $$
but I cannot arrive to anything with known properties (I've already checked the NIST handbook).
Any ideas are welcome, thanks in advance.
EDIT: Found stronger evidence: $\beta=a$
EDIT: Specified which asymptote
|
There are some details in the following derivation which will be omitted to keep this answer at a reasonable length. What follows is a set of clotheslines on which a rigorous proof can hopefully be hung.
We will use the integral representation
$$
\begin{align}
U(a,a-n,1) &= \frac{1}{\Gamma(a)} \int_0^\infty e^{-t} t^{a-1} (1+t)^{-n-1}\,dt \\
&= \frac{1}{\Gamma(a)} \int_0^\infty \exp f_{a,n}(t)\,dt.
\end{align}
$$
(see equation 13.2.5 in Abramowitz and Stegun). The function $f_{a,n}(t)$ critical points at $t \approx -n$ and $t \approx \frac{a-1}{n}$, so in the domain of integration the function will be maximized at $t=0$ if $a>1$ for $n$ large enough. One can check explicitly that this is also true when $a=1$. If $a>1$ then $f_{a,n}(t)$ is maximized at the point $t \approx \frac{a-1}{n}$.
Let's consider the case $0 < a \leq 1$. We have
$$
\begin{align}
f_{a,n}(t) &= -t + (a-1)\log t - (n+1)\log(1+t) \\
&= -t + (a-1)\log t - (n+1)t + O(nt^2) \\
&\approx (a-1)\log t - (n+2)t,
\end{align}
$$
near $t=0$, so that as $n \to \infty$ the Laplace method yields
$$
\begin{align}
\int_0^\infty \exp f_{a,n}(t)\,dt &\sim \int_0^\infty \exp\Bigl[(a-1)\log t - (n+2)t\Bigr]\,dt \\
&= \frac{\Gamma(a)}{(n+2)^a} \\
&\sim \frac{\Gamma(a)}{n^a}.
\end{align}
$$
Now let's consider $a>1$. In this case the function $f_{a,n}(t)$ has a maximum inside the range of integration at $t = t^* \approx \frac{a-1}{n}$. Explicitly this is
$$
\begin{align}
t^* &= \frac{-n+a-3}{2}+\frac{1}{2}\sqrt{n^2+(-2a+6)n+a^2-2a+5} \\
&= \frac{-n+a-3}{2}+\frac{n}{2}\sqrt{1+\frac{-2a+6}{n}+\frac{a^2-2a+5}{n^2}} \\
&= \frac{-n+a-3}{2}+\frac{n}{2}\left[1+\frac{1}{2}\left(\frac{-2a+6}{n}+\frac{a^2-2a+5}{n^2}\right) \right. \\
&\qquad\qquad \left.- \frac{1}{8}\left(\frac{-2a+6}{n}+\frac{a^2-2a+5}{n^2}\right)^2 + O\left(\frac{1} {n^3}\right)\right] \\
&= \frac{-n+a-3}{2}+\frac{n}{2}\left[1 + \frac{-a+3}{n} + \frac{2a-2}{n^2} + O\left(\frac{1} {n^3}\right)\right] \\
&= \frac{-n+a-3}{2} + \frac{n}{2} + \frac{-a+3}{2} + \frac{a-1}{n} + O\left(\frac{1} {n^2}\right) \\
&= \frac{a-1}{n} + O\left(\frac{1} {n^2}\right),
\end{align}
$$
where we used the fact that $\sqrt{1+x} = 1 + \frac{x}{2} - \frac{x^2}{8} + O(x^3)$ to get the third equality. We then have
$$
\begin{align}
f_{a,n}(t) &= -t + (a-1)\log t - (n+1)\log(1+t) \\
&= \alpha(a,n) + \beta(a,n)(t-t^*)^2 + \text{higher order terms},
\end{align}
$$
and we calculate
$$
\alpha(a,n) = (a-1)\Bigl[\log(a-1)-\log(n)-1\Bigr] + o(1)
$$
and
$$
\beta(a,n) = -\frac{n^2}{2 (a-1)} + O(n).
$$
By the Laplace method we conclude that
$$
\begin{align}
\int_0^\infty \exp f_{a,n}(t)\,dt &\sim e^{\alpha(a,n)} \int_{-\infty}^{\infty} \exp\Bigl[ \beta(a,n) t^2 \Bigr]\,dt \\
&\sim \exp\Bigl\{(a-1)\Bigl[\log(a-1)-\log(n)-1\Bigr]\Bigr\} \int_{-\infty}^{\infty} \exp\Bigl[ -\frac{n^2}{2 (a-1)} t^2 \Bigr]\,dt \\
&= e^{1-a} \sqrt{\frac{2\pi}{a-1}} \left(\frac{a-1}{n}\right)^a.
\end{align}
$$
In summary,
If $0 < a \leq 1$ then
$$
U(a,a-n,1) \sim \frac{1}{n^a}
$$
and if $a > 1$ then
$$
U(a,a-n,1) \sim \frac{e^{1-a}}{\Gamma(a)} \sqrt{\frac{2\pi}{a-1}} \left(\frac{a-1}{n}\right)^a.
$$
|
{
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$ \lim_{x\to 0}\frac{\tan x-\sin x}{\sin(x^3)}$ $$ \lim_{x\to 0}\frac{\tan x-\sin x}{\sin(x^3)} =[1]\lim_{x\to 0}\frac{\sin x/\cos x-\sin x}{x^3}\\ =[2]\lim_{x\to 0}\frac{1-\cos x}{x^2}\\ =\frac{1}{2} $$
My question is how [1]=[2]?
$\tan x-\sin x=\tan x(1-\cos x)=x(1-\cos x)$?
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We have $\frac{\sin x}{\cos x}-\sin x=\frac{\sin x}{\cos x}(1-\cos x)$.
Thus our function is equal to
$$\frac{1}{\cos x}\frac{\sin x}{x}\frac{1-\cos x}{x^2}.$$
The limit of $\frac{1}{\cos x}$ is $1$. So is the limit of $\frac{\sin x}{x}$.
You did not ask about the limit of $\frac{1-\cos x}{x^2}$. That can be obtained by multiplying top and bottom by $1+\cos x$, getting $\frac{1-\cos x}{x^2}=\frac{1}{1+\cos x}\left(\frac{\sin x}{x}\right)^2$.
Remark: The above answered your specific question. It did not address the question of finding
$$\lim_{x\to 0} \frac{\tan x-\sin x}{\sin(x^3)}.\tag{1}$$
For that, we rewrite (1) as
$$\lim_{x\to 0} \frac{\tan x-\sin x}{x^3}\frac{x^3}{\sin(x^3)}.$$
Since $\lim_{t\to 0}\frac{\sin t}{t}=1$, we have $\lim_{x\to 0}\frac{x^3}{\sin(x^3)}=1$, and we are at the question you asked.
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What is the smallest integer $n$ greater than $1$ such that the root mean square of the first $n$ integers is an integer? What is the smallest integer $n$ greater than $1$ such that the root mean square of the first $n$ integers is an integer?
The root mean square is defined as:
$$\sqrt{\left(\frac{a_1^2 + a_2^2 + a_3^2+\dots+a_n^2}{n}\right)}$$
I started off by the relation:
$$\sum^n_{i = 1}i^2 = \frac{n(n+1)(2n+1)}{6}$$
We need an $n$ such that:
$$\sqrt{\frac{n(n+1)(2n+1)}{6n}} = k$$
$$\implies (n+1)(2n+1) = 6k^2$$
$$\implies 2n^2 + 3n + (1-6k^2) = 0$$
$$\implies n = \frac{-3 \pm \sqrt{9 - 8(1-6k^2)}}{4} = \frac{-3 \pm \sqrt{1 + 48k^2}}{4}$$
For the equation to have rational solutions $1 + 48k^2$ must be a perfect square. So, the problem boils down to solving the smallest $k$ such that $1 + 48k^2$ is a perfect square.
I have no idea how to proceed.
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Starting with $1+48k^2= m^2$ or $m^2-48k^2 = 1$, it makes sense to try convergents to $\sqrt{48}$, which are $\left\{6,7,\frac{90}{13},\frac{97}{14},\frac{1254}{181},\frac{1351}{195},\frac{17466}{2521}, ...\right\}$.
We spot solutions $(m, k) = (7, 1), (97, 14), (1351, 195), \;\dots $ corresponding to $n = 1, \frac{47}2, 337, ...$
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Did I do this limit right? $\lim_{x \to 0} \frac{(\sqrt{x^2+2}-\sqrt{2})}{\sqrt{x^2+1}-1}$ This is how I did this limit:
$$\lim_{x \to 0} \frac{(\sqrt{x^2+2}-\sqrt{2})}{\sqrt{x^2+1}-1} =\lim_{x \to 0}\frac{(\sqrt{x^2+2}-\sqrt{2})(\sqrt{x^2+1}+1)}{(\sqrt{x^2+1}-1)(\sqrt{x^2+1}+1)} = \lim_{x \to 0} \frac{(\sqrt{x^2+2}-\sqrt{2})(\sqrt{x^2+1}+1)(\sqrt{x^2+2}+\sqrt{2})}{(x^2+1-1)(\sqrt{x^2+2}+\sqrt{2})} = \lim_{x \to 0}\frac{({x^2+2}-2)(\sqrt{x^2+1}+1)}{(x^2+1-1)(\sqrt{x^2+2}+\sqrt{2})} =\frac{(\sqrt{0^2+1}+1)}{(\sqrt{0^2+2}+\sqrt{2})}=\frac{1}{2\sqrt{2}}$$
But Wolframalpha gave me another answer!
So did I do it right?
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As a backup check, you could use $\sqrt{x+a} = \sqrt{a}+\frac{1}{2\sqrt{a}} x +o(x)$, or $\sqrt{x+a} - \sqrt{a} =\frac{1}{2\sqrt{a}} x +o(x)$.
Then $\frac{\sqrt{x^2+2}-\sqrt{2}}{ \sqrt{x^2+1}-\sqrt{1} } = \frac{ \frac{1}{2\sqrt{2}} x^2 +o(x^2) }{ \frac{1}{2\sqrt{1}} x^2 +o(x^2) } =
\frac{ \frac{1}{2\sqrt{2}} +\frac{ o(x^2) }{x^2} }{ \frac{1}{2\sqrt{1}} +\frac{ o(x^2) }{x^2} }$. Taking limits gives the desired result.
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Integral of quartic function in denominator I'm sorry, I've really tried to use MathJaX but I can't get integrals to work properly.
indefinite integral
$$\int {x\over x^4 +x^2 +1}$$
I set it up to equal
$$x\int {x\over x^4 +x^2 +1} - \int {x\over x^4 +x^2 +1}$$
$$\text{so } (x-1)\int {1\over x^4 +x^2 +1}$$
OKAY, now I set the denominator to $(X^2 +.5)^2 + \frac 34$
So I multiplied the top and bottom by $\frac 43$
then I absorbed it into the squared quantity by dividing it (within the parenthesis) by $\sqrt 3\over 2$
so
$${1\over \left({X^2 +.5\over {\sqrt 3\over 2}}\right)^2 + 1}$$
so $\arctan\left({X^2 +.5\over{\sqrt 3\over 2}}\right)$
FINAL ANSWER: $(x-1)\arctan\left({X^2 +.5\over{\sqrt 3\over 2}}\right) + C$
Thanks for reading, I have no way of checking this work... tutors are always have too many people wanting help.
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$$
\int\frac{x}{x^4+x^2+1}dx=\int\frac{x}{x^4+2x^2+1-x^2}dx=\int\frac{x}{\left(x^2+1\right)-x^2}dx=
$$
$$
=\int\frac{x}{\left(x^2+x+1\right)\left(x^2-x+1\right)}dx=\frac{1}{2}\int\frac{\left(x^2+x+1\right)-\left(x^2-x+1\right)}{\left(x^2+x+1\right)\left(x^2-x+1\right)}dx=
$$
$$
=\frac{1}{2}\left(\int\frac{dx}{x^2-x+1}-\int\frac{dx}{x^2+x+1}\right)=\frac{1}{2}\left(\int\frac{\left(\frac{2x-1}{2}+\frac{1}{2}\right)dx}{x^2-x+1}-\int\frac{\left(\frac{2x+1}{2}-\frac{1}{2}\right)dx}{x^2+x+1}\right)
$$
$$
=\frac{1}{4}\int\frac{(2x-1)dx}{x^2-x+1}+\frac{1}{4}\int\frac{dx}{(x-\frac{1}{2})^2+\frac{3}{4}}-\frac{1}{4}\int\frac{(2x+1)dx}{x^2+x+1}-\frac{1}{4}\int\frac{dx}{(x+\frac{1}{2})^2+\frac{3}{4}}
$$
$$
=\frac{1}{4}\ln|x^2-x+1|-\frac{1}{4}\ln|x^2+x+1|+\frac{1}{4}\cdot\frac{1}{\frac{\sqrt 3}{2}}\arctan \frac{\frac{2x-1}{2}}{\frac{\sqrt 3}{2}}-\frac{1}{4}\cdot\frac{1}{\frac{\sqrt 3}{2}}\arctan \frac{\frac{2x+1}{2}}{\frac{\sqrt 3}{2}}
$$
$$
=\frac{1}{4}\ln\left\vert\frac{x^2-x+1}{x^2+x+1}\right\vert+\frac{1}{2\sqrt 3}\left(\arctan\frac{2x-1}{\sqrt 3}-\arctan\frac{2x+1}{\sqrt 3}\right)
$$
$$
=\frac{1}{4}\ln\left\vert\frac{x^2-x+1}{x^2+x+1}\right\vert+\frac{1}{2\sqrt 3}\arctan\frac{\frac{2x-1}{\sqrt 3}-\frac{2x+1}{\sqrt 3}}{1+\frac{2x-1}{\sqrt 3}\cdot\frac{2x+1}{\sqrt 3}}
$$
$$
=\frac{1}{4}\ln\left\vert\frac{x^2-x+1}{x^2+x+1}\right\vert+\frac{1}{2\sqrt 3}\arctan\left(\frac{-\sqrt 3}{2x^2+1}\right)+C
$$
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|
Solving this logarithm equation? How do I solve this equation using common logarithms?
$\log x = 1-\log(x-3)$
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Let $1 = \log 10$ then, $$\log x = \log 10 - \log (x - 3).$$ For all $x$ and $y$ and a constant $a$, one always has $$\log_a xy = \log_a x + \log_a y.\tag1$$ This is one of the most important logarithm rules. From this rule, we can express $\log\dfrac xy$. $$\log \frac xy = \log x\bigg(\frac 1y\bigg),$$ and from $(1)$, we see that $$\log \frac xy = \log x + \log \frac 1y = \log x + \log y^{-1}.$$ Let $\log y^{-1} = d$ then $10^d = y^{-1}\Leftrightarrow \dfrac{1}{10^d} = \dfrac{1}{y^{-1}} \Leftrightarrow 10^{-d} = y$. $$\begin{align}\therefore \log y &= -d \\ \Leftrightarrow -\log y &= d.\end{align}$$ Since we asserted that $d = \log y^{-1}$ then, $$\log\frac xy = \log x - \log y.$$ Do you notice how useful the rule $(1)$ is? This implies that $$\begin{align} \log x &= \log 10 - \log (x - 3) \\ &\ \vdots \\ \Leftrightarrow x^2 - 3x - 10 &= 0. \end{align}$$ We can solve for $x$ in this trinomial using the quadratic formula, such that $$\begin{align} x &= \frac{3\pm 7}{2} \\ \Leftrightarrow x = -2 \ &\text{ or } \ x = 5.\end{align}$$ If you are not familiar with the quadratic formula, comment below.
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Prove geometric sequence question Prove that $x+2x^2+3x^3+4x^4+...+nx^n = \frac{nx^{n+2}-(n+1)x^{n+1}+x}{(x-1)^2}$
I see that this can be written as $$\sum_{n=1}^n nx^n = n\sum_{n=1}^n x^n$$
$$\sum_{n=1}^n x^n = \frac{x(x^n-1)}{(x-1)}$$
$$\therefore n\sum_{n=1}^n x^n = n(\frac{x(x^n-1)}{(x-1)})$$
Following this I squared top and bottom so the denominator would be the same as that n the question. But I cannot seem to show the top as that in the question.
Can I have a hint on how to proceed? Also in the 1 line, is there a proof of the fact that a constant can be taken out of the summation?
Thanks
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Multiply $x + 2x^2 + 3x^3 + 4x^4+...+nx^n$ with $x-1$. You'll get:
$$(x + 2x^2 + 3x^3 + 4x^4+...+nx^n)(x-1) = x(x + 2x^2 + 3x^3 + 4x^4+...+nx^n) - (x + 2x^2 + 3x^3 + 4x^4+...+nx^n)$$
Expand it and you'll obviously end up with sort of telescopic series. Every number in the first term, except the last one will be of the form $nx^{n+1}$ and will have a "companion" of the second term of the form $(n+1)x^{n+1}$. So their difference will be $-x^{n-1}$. As we said the last of the first term and the first term of the second sequence will be "alone" and you'll end up with:
$$nx^{n+1} - x^{n-1} -...-x^2 - x$$
Now multiply again with $x-1$, so we'll have:
$$(nx^{n+1} - x^{n-1} -...-x^2 - x)(x-1) = nx^{n+1}(n-1) - (x^{n-1} +...-x^2 + x)(x-1)$$
Expand the first term. The second term is well know identity and is equal to $x^n - x$, anyway you can prove that using the same telescoping method we used previously. So we have:
$$nx^{n+2} - nx^{n-1} - x^n + x = nx^{n+2} - (n+1)x^{n+1} + x$$
Q.E.D.
Note that if the initial LHS was infinite the we could have written as:
$$\frac{1}{(x-1)^2} - 1$$
Using the formula:
$$\frac{1}{(1-x)^{k+1}} = \sum_{n=0}^{\infty} \binom{n+k}{n}x^n$$
But because it's finite series we have to add something instead of the 1 to make that boundary. Simular as we make $1 + x + x^2 + x^3 + x^4+... = \frac{1}{x-1}$ bounded with:
$1 + x + x^2 + x^3+...+x^{n-1} = \frac{x^n - 1}{x-1}$
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Inequality. $\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{a}}\geq3$ Let $a,b,c \in (0, \infty)$, with $a+b+c=3$. How can I prove that:
$$\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{a}}\geq3 ?$$.
I try to use Cauchy-Schwarz rewriting the inequality like :
$$\sum_{cyc}\frac{a\sqrt{b}}{b} \geq \frac{(\sum_{cyc}{\sqrt[4]{a^2b}})^2}{a+b+c}$$ but I don't obtain anything.
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Rewrite it as
$$\sum_{cyc} \frac{a}{\sqrt{b}} + (\sqrt{b} - 2\sqrt{a}) = \sum_{cyc} \frac{a}{\sqrt{b}} -\sum_{cyc} \sqrt{a} \ge 3 - \sum_{cyc} \sqrt{a} = \sqrt{3(a+b+c)} - \sum_{cyc} \sqrt{a}$$
LHS becomes
$$\sum_{cyc} \frac{(\sqrt{a} - \sqrt{b})^2}{\sqrt{b}}$$
RHS becomes
$$\frac{3(a+b+c) - (\sum_{cyc} \sqrt{a})^2}{\sqrt{3(a+b+c)} + \sum_{cyc} \sqrt{a}} = \frac{\sum_{cyc} (\sqrt{a} - \sqrt{b})^2}{\sqrt{3(a+b+c)} + \sum_{cyc} \sqrt{a}}$$
The inequality then follows since
$\sqrt{3(a+b+c)} + \sum_{cyc} \sqrt{a} \ge \sqrt{a}, \sqrt{b}$ and $\sqrt{c}$.
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Can't solve this problem $lim_{x \to 0}\frac{3^{5x}-2^{7x}}{\arcsin\left(2x\right) - x}$(Without using L'Hospital's rule) I don't see the way to solve this limit.
$$\lim_{x \to 0}\frac{3^{5x}-2^{7x}}{\arcsin\left(2x\right) - x} $$
My attempt is
1) Divide the numerator by $3^{5x}$
$$\lim_{x \to 0}\frac{3^{5x}-2^{7x}}{\arcsin\left(2x\right) - x} = \lim_{x \to 0} \frac {1- \left(2/3 \right)^{5x}\,\,2^{2x}}{\arcsin\left(2x\right)-x} $$
And I am unable to do next step. So how this limit can be solved?
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Using L'Hospital's rule, we find
\begin{align*}
\lim_{x \to 0} \frac{3^{5x} - 2^{7x}}{\arcsin{2x} - x} &= \lim_{x \to 0} \frac{3^{5x} (5 \ln{3}) - 2^{7x}(7 \ln{2})}{\frac{2}{\sqrt{1 - (2x)^2}} - 1} \\
&= \frac{3^0 (5 \ln{3}) - 2^0 (7 \ln{2})}{2 - 1} \\
&= 5 \ln{3} - 7 \ln{2}
\end{align*}
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If a line makes angles $\alpha, \beta, \gamma$ with the $x, y, z$ axes, then $\sin^2{\alpha} + \sin^2{\beta} + \sin^2{\gamma} = 2 $ The following is the question in my textbook:-
If a straight line makes angle $\alpha$, $\beta$, $\gamma$ with the $x, y, z$ axes respectively, then show that $\sin^2{\alpha} + \sin^2{\beta} + \sin^2{\gamma} = 2 $?
Here is what I have done:-
Since $\alpha$, $\beta$, $\gamma$ are the angles made by the line with the axes so $\cos\alpha$, $\cos\beta$, $\cos\gamma$ are the direction cosines of the line
now
$$\sin^2\theta + \cos^2\theta = 1$$
$$\sin^2\theta = 1 - \cos^2\theta$$
so
$$\sin^2\alpha + \sin^2\beta + \sin^2\gamma = 1 - \cos^2\alpha + 1 - \cos^2\beta + 1 - \cos^2\gamma$$$$ =3 - (\cos^2\alpha + \cos^2\beta + \cos^2\gamma)$$
now since $\cos\alpha$, $\cos\beta$, $\cos\gamma$ are the direction cosines of the line so $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$
so
$$3 - \cos^2\alpha + \cos^2\beta + \cos^2\gamma= 3 - 1$$
$$=2- answer$$
My question was that have I done it correctly and if not what is the correct way of doing it.
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Yes, the solution is correct. Turning $\sin^2$ into $1-\cos^2 $ reduces the problem to the sum of squared cosines being $1$, and the latter fact is essentially the distance formula. Indeed, move the line so that it passes through the origin and pick a point $(u,v,w)$ on the line at distance $1$ from the origin. Then $u^2+v^2+w^2=1$. On the other hand, $u,v,w$ are exactly the direction cosines.
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Fermat's 2 Square-Like Results from Minkowski Lattice Proofs
Minkowski's Convex Body Theorem for lattices in the plane: Suppose $\mathfrak{L}$ is a lattice in $\mathbf{R}^2$ defined as $\mathfrak{L}=\{m\vec{v_1}+n\vec{v_2}:m,n\in\mathbf{Z}\}$, where $\vec{v_1}$ and $\vec{v_2}$ are linearly independent. Suppose $d$ is the area of a fundamental parallelogram of $\mathfrak{L}$. Then, if $\mathcal{S}$ is a convex and origin-symmetric region with $Area(\mathcal{S})>4d$, then $\mathcal{S}$ contains some point $q\neq 0$ such that $q\in\mathfrak{L}$.
There is a particularly nice proof of Fermat's two square theorem - an odd prime $p$ is expressible as $x^2+y^2$ for $x,y\in\mathbf{Z}$ $\iff$ $p\equiv 1\pmod{4}$ - using this fact. Here is my proof below:
Choose $a\in\mathbf{Z}$ such that $p|a^2+1$. Now, let $\mathfrak{L}$ be a lattice in $\mathbf{R}^2$ defined as $\mathfrak{L}=\{m\vec{v_1}+n\vec{v_2}:m,n\in\mathbf{Z}\}$ where $\vec{v_1}=(a,1)$ and $\vec{v_2}=(p,0)$. It is easy to see that a fundamental parallelogram $\mathfrak{F}$ of $\mathfrak{L}$ has area $p$. Now, suppose $(x,y)\in\mathfrak{L}$. Then we have $x=ma+np, y=m$, so \begin{align*}x^2+y^2 &= (ma+np)^2 + m^2 \\ &= m^2a^2+2manp+n^2p^2+m^2\\ &= m^2(a^2+1) + 2manp + n^2p^2 \\ &\equiv m^2(a^2+1)\pmod{p} \\ &\equiv 0\pmod{p} \end{align*} due to our choice of $a$. Now, let $\mathfrak{C}$ be an $O$-symmetric circle with radius $\sqrt{2p}$, so $\mathfrak{C}=\{(x,y)\in\mathbf{R}^2 : x^2+y^2<2p\}$ Then $$Area(\mathfrak{C})=2p\pi > 4p = 4\cdot Area(\mathfrak{F}).$$ Thus, by Minkowski, $\exists$ a lattice point $(j,k)\in\mathfrak{C}\setminus (0,0)$. By the above work, $p|j^2+k^2$, and by definition of $\mathfrak{C}$, $j^2+k^2<2p$. Moreover, $j^2+k^2$ is positive, so $0<j^2+k^2<2p$, but in order for $j^2+k^2$ to be divisible by $p$, we must have $p=j^2+k^2$, so we are done.
Are there any other results that can be derived in a similar manner? Specifically, what primes can be expressed as $x^2+2y^2$? $x^2+3y^2$? Any ideas are greatly appreciated.
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Unfortunately I'm answering my own question, but I was able to prove that if $p\equiv 1\pmod{8}$ or $p\equiv 3\pmod{8}$, then $p$ is expressible as $x^2+2y^2$ for $x,y\in\mathbf{Z}$, and if $p\equiv 1\pmod{3}$, then $p$ is expressible as $x^2+3y^2$ for $x,y\in\mathbf{Z}$. For posterity, here it is.
Lemma: $p\equiv 1\pmod{8}$ or $p\equiv 3\pmod{8}\iff p|a^2+2$ for some $a\in\mathbf{Z}$.
Proof: The latter statement is equivalent to $\left(\frac{-2}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{2}{p}\right)=1$. It is not hard to see that this holds if and only if $p$ is of the shape $8k+1$ or $8k+3$ due to CRT. $\Box$
By our lemma, we can choose $a\in\mathbf{Z}$ such that $p|a^2+2$. Define a lattice $\mathfrak{L}\in\mathbf{R}^2$ as $\mathfrak{L}=\{m\vec{v_1}+n\vec{v_2}:m,n\in\mathbf{Z}\}$ where $\vec{v_1}=(a,1)$ and $\vec{v_2}=(p,0)$. The fundamental parallelogram $\mathfrak{F}$ of $\mathfrak{L}$ has area $p$. Let $(x,y)\in\mathfrak{L}$. Then $x=ma+np$ and $y=m$, so \begin{align*}x^2+2y^2 &= (ma+np)^2+2m^2 \\ &= m^2a^2+2manp+n^2p^2+2m^2 \\ &= m^2(a^2+2)+2manp+n^2p^2 \\ &\equiv m^2(a^2+2)\pmod{p}\\ &\equiv 0\pmod{p}\end{align*} due to our choice of $a$. Now, let $\mathfrak{E}$ be a convex, origin symmetric ellipse with semi-major axis length $\sqrt{2p}$ and semi-minor axis length $\sqrt{p}$, so $\mathfrak{E}=\{(x,y)\in\mathbf{R}^2:x^2+2y^2<2p\}$. Then, we have $$Area(\mathfrak{E})=p\pi\sqrt{2}>4p=4\cdot Area(\mathfrak{F}).$$ Thus, by Minkowski there exists a lattice point $(j,k)\in\mathfrak{E}\setminus(0,0).$ By the above work, $p|j^2+2k^2$ and by definition of $\mathfrak{E}$, $j^2+2k^2<2p$. $j^2+2k^2$ is obviously positive, so $0<j^2+2k^2<2p$, so in order for $p|j^2+2k^2$ we must have $p=j^2+2k^2$, which completes the proof.
UPDATE: I have included my proof of the proposition for primes expressible as $x^2+3y^2$ for $x,y\in\mathbf{Z}$.
Lemma: $p\equiv 1\pmod{3} \iff p|a^2+3$ for some $a\in\mathbf{Z}$.
Proof: The latter statement is equivalent to $\left(\frac{-3}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)=1$. It is well known that this occurs if and only if $p\equiv 1\pmod{6}$. The proof is a basic application of the Quadratic Reciprocity law and the Chinese Remainder Theorem. Since all primes congruent to $1$ modulo $6$ are exactly those congruent to $1$ modulo $3$, we are done. $\Box$
By our lemma, we can choose $a\in\mathbf{Z}$ such that $p|a^2+3$. Define a lattice $\mathfrak{L}\in\mathbf{R}^2$ as $\mathfrak{L}=\{m\vec{v_1}+n\vec{v_2}:m,n\in\mathbf{Z}\}$ where $\vec{v_1}=(a,1)$ and $\vec{v_2}=(p,0)$. The fundamental parallelogram $\mathfrak{F}$ of $\mathfrak{L}$ has area $p$. Let $(x,y)\in\mathfrak{L}$. Then $x=ma+np$ and $y=m$, so \begin{align*}x^2+3y^2 &= (ma+np)^2+3m^2 \\ &= m^2a^2+2manp+n^2p^2+3m^2 \\ &= m^2(a^2+3)+2manp+n^2p^2 \\ &\equiv m^2(a^2+3)\pmod{p}\\ &\equiv 0\pmod{p}\end{align*} due to our choice of $a$. Now, let $\mathfrak{E}$ be a convex, origin symmetric ellipse with semi-major axis length $\sqrt{3p}$ and semi-minor axis length $\sqrt{p}$, so $\mathfrak{E}=\{(x,y)\in\mathbf{R}^2:x^2+3y^2<3p\}$. Then, we have $$Area(\mathfrak{E})=p\pi\sqrt{3}>4p=4\cdot Area(\mathfrak{F}).$$ Thus, by Minkowski's Convex Body Theorem, there exists a lattice point $(j,k)\in\mathfrak{E}\setminus(0,0).$ By the above work, $p|j^2+3k^2$ and by definition of $\mathfrak{E}$, $j^2+3k^2<3p$. $j^2+3k^2$ is obviously positive, so $0<j^2+3k^2<3p$, so in order for $p|j^2+3k^2$ we must have $p=j^2+3k^2$ or $2p=j^2+3k^2$. In order to show that the latter possibility is impossible, consider the equation modulo $3$. We get, since $p\equiv 1\pmod{3}$, $j^2+3k^2\equiv j^2\equiv 2p\equiv 2\pmod{3}$, but this is impossible since the only quadratic residues modulo $3$ are $0$ and $1$. Hence proved. $\Box$
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Intersection of two tangents on a parabola proof There are two tangent lines on a parabola $x^2$. The $x$ values of where the tangent lines intersect with the parabola are $a$ and $b$ respectively. The point where the two tangent lines intersect has an $x$ value of $c$.
Prove that $c=(a+b)/2$
I have tried taking the derivative of $x^2$ and using the point slope form to find each tangent line's equation, and then making each tangent equal to each other to find that $c$ is equal to $a+b$, but each time I try it out I can't get anything out of the algebra
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Consider the curve $C$ with the equation $y=x^2$. When $x=a$, we have $y=a^2$. The derivative is given by $y'=2x$ and so the gradient at the is exactly $2a$. The equation of the tangent line is
$$y-a^2=2a(x-a)$$
Expanding this gives $y-2ax=-a^2$. Similarily, the equation of the tangent line to $C$ at the point with $x$-coordinate $x=b$ is given by $y-2bx=-b^2$. We are left needing to solve
\begin{eqnarray*}
2ax-y&=&a^2 \\ 2bx-y&=&b^2
\end{eqnarray*}
Subtracting the second equation from the first gives $2(a-b)x=a^2-b^2$. Notice that $a^2-b^2$ factorises to give $(a-b)(a+b)$ and hence: $2(a-b)x=(a-b)(a+b)$. Assuming $a \neq b$ we can divide both sides by $a-b$ to give $2x=a+b$ or, better still $x=\tfrac{1}{2}(a+b).$
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|
Evaluating Artin symbol Consider the field $K=\mathbb Q(\sqrt{2})$. Let $mp=\frac{(2+\sqrt{2})^{p}-1}{1+\sqrt{2}}$.
For the field extension, $K(-1+2\sqrt{2})/K$, and $p\equiv 5\bmod{6}$ how can one show
$ \frac{(2+\sqrt{2})^{p}-1}{1+\sqrt{2}}\equiv 1\pmod{-1+2\sqrt{2} }$
and hence, Artin symbol corresponding to $ K(-1+2\sqrt{2})/K$, is trivial,
where as over $ K(-1-2\sqrt{2})/K$
$ \frac{(2+\sqrt{2})^{p}-1}{1+\sqrt{2}}\not\equiv 1\pmod{-1-2\sqrt{2} }$. I workedout as to take, $-7=(-1+2\sqrt{2})(-1-2\sqrt{2})$ and evaluate $mp$ mod at 7. Any Hint is also welcomed.
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The Galois group of the extension $K(-1+2\sqrt2)/K$ is trivial, so all the Frobenius automorphisms are equal to the identity for lack of alternatives.
In the specific example from comments we can also check it as follows. You already showed that $(-1+2\sqrt2)\mid 7$. Therefore
$$2+\sqrt2\equiv2-6\sqrt2=-1-3(-1+2\sqrt2)\equiv-1\pmod{(-1+2\sqrt2)}.$$
Hence
$$
(2+\sqrt2)^7\equiv(-1)^7\equiv(-1)\equiv2+\sqrt2\pmod{(-1+2\sqrt2)},
$$
that is compatible with the Frobenius being the identity.
|
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|
Evaluating $\int_0^1 \frac{x \arctan x \log \left( 1-x^2\right)}{1+x^2}dx$ I am trying to prove that
$$\int_0^1 \frac{x \arctan x \log \left( 1-x^2\right)}{1+x^2}dx = -\frac{\pi^3}{48}-\frac{\pi}{8}\log^2 2 +G\log 2$$
where $G$ is the Catalan's Constant. Numerically, it's value is $-0.199739$.
Using the substitution $x=\tan \theta$, it can be written as
$$
\begin{align*}
I &= \int_0^{\frac{\pi}{4}}\theta \tan \theta \log(\cos 2\theta) d\theta-2\int_0^{\frac{\pi}{4}}\theta \tan \theta \log(\cos \theta)d\theta \end{align*}
$$
Can anyone suggest a good approach to evaluate it?
|
A solution proposed by Cornel Ioan Valean
As also shown in this post, an immediate connection can be made with the integral in the current OP. So, integrating by parts, rearranging, and employing the needed results, we arrive at once at the desired closed form
$$\int_0^1 \frac{x \arctan(x) \log(1-x^2)}{1+x^2}\textrm{d}x=\frac{1}{2}\int_0^1 \left(\log\left(\frac{1+x^2}{2}\right)\right)' \arctan(x) \log(1-x^2)\textrm{d}x$$
$$=\frac{1}{2}\log(2)\underbrace{\int_0^1 \frac{\log(1-x^2)}{1+x^2}\textrm{d}x}_{\displaystyle \log(2)\pi/4-G}-\underbrace{\int_0^1 \frac{x\arctan(x) \log(2/(1+x^2))}{1-x^2}\textrm{d}x}_{\displaystyle \pi^3/192}-\frac{1}{2}\underbrace{\int_0^1\frac{\log(1-x^2)\log(1+x^2)}{1+x^2}\textrm{d}x}_{\displaystyle(\pi ^3+16 \log ^2(2)\pi -96 \log (2) G)/32}=\log(2)G-\log^2(2)\frac{\pi}{8}-\frac{\pi^3}{48}.\tag1$$
The first remaining integral in $(1)$ may be viewed as a sums of two well-known classical (and trivial) integrals, one of them being a Putnam question, that is $\displaystyle\int_0^1 \frac{\log(1+x)}{1+x^2}\textrm{d}x=\frac{\pi}{8}\log(2)$.
The second resulting integral in $(1)$ is calculated magically here.
As regards the last integral in $(1)$, we have the simple (and magical) fact that
$$\int_0^1 \frac{\log(1-x^2)\log(1+x^2)}{1+x^2}\textrm{d}x$$
$$
=\lim_{b\to 0} \, \left(\frac{1}{16} \frac{\partial ^2}{\partial b^2}B\left(\frac{1}{4},b\right)-\frac{1}{16} \frac{\partial ^2}{\partial b^2}B\left(\frac{3}{4},b\right)\right)-\frac{1}{64} \left(\lim_{a\to \frac{1}{2}} \, \frac{\partial ^2}{\partial a^2}B\left(a,\frac{1}{2}\right)\right)$$
$$=\frac{1}{32} \left(\pi ^3+16 \pi \log ^2(2)-96 \log (2) G\right).$$
A note: Observe that for the last integral it is enough to use the algebraic identity $ab=1/4 ((a + b)^2 - (a - b)^2)$, where $a=\log(1-x^2)$ and $b=\log(1+x^2)$. Thus, we have
$$\int_0^1 \frac{\log(1-x^2)\log(1+x^2)}{1+x^2}\textrm{d}x$$
$$=\frac{1}{4} \int_0^1 \left(\frac{\log ^2\left(1-x^4\right)}{1-x^4}-\frac{x^2 \log ^2\left(1-x^4\right)}{1-x^4}\right) \textrm{d}x-\frac{1}{4} \int_0^1 \frac{\displaystyle \log^2((1-x^2)/(1+x^2))}{1+x^2} \textrm{d}x,$$
where in the first integral we may let the variable change $x^4=y$, and in the second integral we may act in various ways, like letting $(1-x^2)/(1+x^2)=t^{1/2}$, which together lead us to the Beta function form above. So, a straightforward story essentially.
End of story
More details will be given in the sequel of (Almost) Impossible Integrals, Sums, and Series.
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How to prove this inequality $\sin{\left(\frac{\pi}{2}ab\right)}\le\sin{\left(\frac{\pi}{2}a\right )}\sin{\left(\frac{\pi}{2}b\right)}$? Let $$0\le a\le 1,0\le b\le 1$$
Prove or disprove
$$\sin{\left(\dfrac{\pi}{2}ab\right)}\le\sin{\left(\dfrac{\pi}{2}a\right )}\sin{\left(\dfrac{\pi}{2}b\right)}$$
My try:
Since
$$\sin{x}\sin{y}=\dfrac{1}{2}[\cos{(x-y)}-\cos{(x+y)}]$$
So
$$\sin{\left(\dfrac{\pi}{2}a\right)}\sin{\left(\dfrac{\pi}{2}b\right)}=\dfrac{1}{2}[\cos{\dfrac{\pi}{2}(a-b)}-\cos{\dfrac{\pi}{2}(a+b)}]$$
Then I can't,Thank you very much!
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The proof is related to the concavity of the $\sin\left(\frac\pi2 x\right)$ curve.
Consider the following blue curve, which shows $y = \sin\left(\frac\pi2 x\right)$. The red dot $P$ is at $\left(A, \sin\left(\frac\pi2 A\right)\right)$. Now I fit (the red curve) a smaller version of $y = \sin\left(\frac\pi2 x\right)$ ending at $P$, so that a point $Q$ on the red curve split the width of the red rectangle to $B:(1-B)$ and the height to $\sin\left(\frac\pi2 B\right):\left[1-\sin\left(\frac\pi2 B\right)\right]$. The red curve is never below the blue curve. When we evaluate at $x=AB$, the green point $Q$ gives $\left(AB,\sin\left(\frac\pi2 A\right)\sin\left(\frac\pi2 B\right)\right)$ and the blue point $R$ gives $\left(AB,\sin\left(\frac\pi2 AB\right)\right)$.
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Factorize the polynomial $x^3+y^3+z^3-3xyz$ I want to factorize the polynomial $x^3+y^3+z^3-3xyz$. Using Mathematica I find that it equals $(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$. But how can I factorize it by hand?
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\begin{align}
x^3+y^3+z^3-3xyz\\
&= x^3+y^3+3x^2y+3xy^2+z^3-3xyz-3x^2y-3xy^2\\
&= (x+y)^3+z^3-3xy(x+y+z)\\
&= (x+y+z)((x+y)^2+z^2-(x+y)z)-3xy(x+y+z)\\
&= (x+y+z)(x^2+2xy+y^2+z^2-yz-xz-3xy)\\
&= (x+y+z)(x^2+y^2+z^2-xy-yz-zx)
\end{align}
|
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|
True or false: $a^2+b^2+c^2 +2abc+1\geq 2(ab+bc+ca)$ Is this inequality true?
$a^2+b^2+c^2 +2abc+1\ge2(ab+bc+ca)$, where $a,b,c\gt0$.
Can you find a counterexample for this or not?
|
EDIT: I found a much better solution:
WLOG assume that $b,c$ are on the same side of $1$. Bringing the RHS over to the LHS, it suffices to show that
$$(a-1)^2+2a(b-1)(c-1)+(b-c)^2\ge 0$$
But since $b,c$ are on the same side of $1$, this is true.
Old solution:
We prove the inequality for $a,b,c\ge 0$.
The inequality is equivalently
$$
a^2+2(bc-b-c)a+b^2+c^2+1-2bc\ge 0
$$
If $bc>b+c$, then the expression is minimized when $a=0$. But in this case it is only left to prove $b^2+c^2-2bc+1=(b-c)^2+1\ge 0$, which is clear.
If $bc\le b+c$, then then expression is minimized when $a = b+c-bc$. We must show
$$
-(bc-b-c)^2+b^2+c^2+1\ge 2bc\Leftrightarrow 2(b+c)bc+1\ge 4bc+(bc)^2
$$
But letting $bc=k$, we have
$$
2(b+c)bc+1\ge 4k^{3/2}+1
$$
And so we want to prove
$$
k^2-4k^{3/2}+4k-1=(\sqrt{k}-1)^2(k-2\sqrt{k}-1)\le 0
$$
given $bc\le b+c\le 2\sqrt{bc}\Rightarrow 0\le k\le 4$. But then $k-2\sqrt{k}-1=\sqrt{k}(\sqrt{k}-2)-1\le -1<0$, so the inequality holds.
Equality holds when $a=b=c=1$.
|
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How to compute $\frac{t}{t+1}$ to the form $1-\frac{1}{t+1}$? How to compute $\frac{t}{t+1}$ to the form $1-\frac{1}{t+1}$?
What else? Well.
Well can you use long division?
|
Use method add something and take something.
$$\frac{t}{t-1}= \frac{t-1+1}{t-1}$$
Then splits equation on 2 parts.
$$\frac{t-1}{t-1}+ \frac{1}{t-1} $$
Then reduce fraction and you get: $$1+\frac{1}{t-1}$$
If this is true
$$\frac{t}{t-1}= 1-\frac{1}{t-1}$$
Than you get:
$$\frac{t}{t-1}= \frac{t-1-1}{t-1} $$
Than this equation multiply with $(t-1)$
$$t=t-1-1 \Rightarrow t=t-2 \Rightarrow 0t=-2$$
So we take $t=t-2$
$$\frac{t-2}{t-2-1}= \frac{t-2-1+1}{t-2-1}$$
$$\frac{t-2-1}{t-2-1}+ \frac{1}{t-2-1}$$
We get solution which is not the same as we previously calculated
$$1 + \frac{1}{t-2-1}=1 + \frac{1}{t-3}$$
So
$$\frac{t}{t-1} \neq 1 -\frac{1}{t-1}$$
$$\frac{t}{t+1} = \frac{t+1-1}{t+1} = \frac{t+1}{t+1} -\frac{1}{t+1}= 1 -\frac{1}{t+1} $$
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How to efficiently compute $17^{23} (\mod 31)$ by hand? I could use that $17^{2} \equiv 10 (\mod 31)$ and express $17^{23}$ as $17^{16}.17^{4}.17^{3} = (((17^2)^2)^2)^2.(17^2)^2.17^2.17$ and take advantage of the fact that I can more easily work with powers of ten ($17^2 \equiv 10 (\mod 31), (17^2)^2 \equiv 100 (\mod 31) \equiv 7 (\mod 31)$, etc.). While this makes the computation easier, I'm thinking there might be a better way to do this.
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We can do away with manipulating the powers of $17$ as follows
Observe that $17\equiv-14\pmod{31}$
$\displaystyle\implies 17^{23}\equiv(-14)^{23}\equiv-14^{23}\pmod{31}$
Now, $7^2=49\equiv18\pmod{31}, 7^3\equiv18\cdot7\equiv2\pmod{31}$
$\displaystyle\implies14^{23}=2^{23}\cdot7^{23}=2^{23}(7^3)^7\cdot7^2\equiv2^{23}\cdot2^7\cdot7^2\equiv2^{30}\cdot49\pmod{31}$
Using Fermat's Little Theorem, $2^{30}\equiv1\pmod{31}$
Alternatively, $2^5=32\equiv1\pmod{31}\implies 2^{30}=(2^5)^6\equiv1^6\pmod{31}\equiv1$
So, $\displaystyle-14^{23}\equiv-49\pmod{31}\equiv13$
|
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Quadratic equation and trig If the quadratic equation $ax^2+bx+c=0$ has equal roots where $a, b$ and $c$ denote the lengths of the sides opposite to vertices A, B and C of a triangle ABC respectively, then find the sum of integers in the range of
$$\left(\frac{\sin A}{\sin C}+\frac{\sin C}{\sin A}\right)$$
Just some hints will do.
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Let $s=\frac ab,t=\frac cb$. Then, we have
$$b^2-4ac=0\Rightarrow 4st=1.$$
So, we have
$$\begin{align}&a+b\gt c,\quad b+c\gt a,\quad c+a\gt b\\&\iff s+1\gt t,\quad 1+t\gt s,\quad t+s\gt 1\\&\iff s+1\gt \frac{1}{4s},\quad 1+\frac{1}{4s}\gt s,\quad \frac{1}{4s}+s\gt 1\\&\iff \frac{\sqrt 2-1}{2}\lt s\lt \frac 12\quad\text{or}\quad \frac 12\lt s\lt \frac{1+\sqrt 2}{2}\end{align}$$
Now note that we can write
$$\frac{\sin A}{\sin C}+\frac{\sin C}{\sin A}=\frac ac+\frac ca=\frac{a^2+c^2}{ac}=\frac{a^2+c^2}{\frac{b^2}{4}}=4s^2+4t^2=4s^2+4\left(\frac{1}{4s}\right)^2.$$
Let $f(s)=4s^2+\frac{1}{4s^2}$. Then,
$$f'(s)=\frac{16s^4-1}{2s^3}.$$
Hence, $f(s)$ is increasing only when $s\ge \frac 12$. Thus, we have
$$2= f\left(\frac 12\right)\lt f(s)\lt f\left(\frac{\pm 1+\sqrt 2}{2}\right)=6.$$
Thus, the answer is $3+4+5=12$. (note that $f(s)\not=2$.)
|
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How to prove this inequality $\frac{x^y}{y^x}+\frac{y^z}{z^y}+\frac{z^x}{x^z}\ge 3$
let $x,y,z$ be positive numbers, and such $x+y+z=1$
show that
$$\dfrac{x^y}{y^x}+\dfrac{y^z}{z^y}+\dfrac{z^x}{x^z}\ge 3$$
My try:
let
$$a=\ln{\dfrac{x^y}{y^x}},b=\ln{\dfrac{y^z}{z^y}},c=\ln{\dfrac{z^x}{x^z}}$$
so
$$a=y\ln{x}-x\ln{y},b=z\ln{y}-y\ln{z},c=x\ln{z}-z\ln{x}$$
and we note
$$az+bx+yc=(y\ln{x}-x\ln{y})z+(z\ln{y}-y\ln{z})x+(x\ln{z}-z\ln{x})y=0$$
so
$$\dfrac{x^y}{y^x}+\dfrac{y^z}{z^y}+\dfrac{z^x}{x^z}=e^a+e^b+e^c$$
so
$$\Longleftrightarrow e^a+e^b+e^c\ge 3$$
But then I can't prove it.
If this problem is to prove
$$ze^a+xe^b+ye^c\ge 3,$$
I can prove it,because
$$ze^a+xe^b+ye^c\ge=\dfrac{z}{x+y+z}e^a+\dfrac{x}{x+y+z}e^b+\dfrac{y}{x+y+z}e^c$$
so
use Jensen's inequality,we have
$$ze^a+xe^b+ye^c\ge e^{\dfrac{az+bx+yc}{x+y+z}}=3$$
This problem comes from How prove this $\dfrac{x^y}{y^x}\ge (1+\ln{3})x-(1+\ln{3})y+1$?
Thank you very much!
|
Another solution.
Without loss of generality, we can assume that $0<x\le y\le z$ and so $x=az$, $y=bz$, $0<a\le b\le1$. By the AM-GM inequality, we find that
\begin{align}
\frac{x^y}{y^x}+\frac{y^z}{z^y}+\frac{z^x}{x^z}&\ge 3\sqrt[3]{x^{y-z}y^{z-x}z^{x-y}}=3\sqrt[3]{(az)^{(b-1)z}(bz)^{(1-a)z}z^{(a-b)z}}=3\sqrt[3]{a^{(b-1)z}b^{(1-a)z}}=\\
&=3(a^{b-1}b^{1-a})^{z/3}
\end{align}
so we only need to prove that $a^{b-1}b^{1-a}\ge1$, which Hansen did in the answer below. We also see that the condition $x+y+z=1$ is not needed.
|
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|
limit $\lim_{x\to-2}\frac{\sqrt[3]{x - 6}+2}{x + 2}$ Can you help me with
$$\lim_{x\to-2}\frac{\sqrt[3]{x - 6}+2}{x + 2}$$
I've already tried to multiply at conjugate expression but I failed.
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$$ \lim_{x\to-2}\frac{\sqrt[3]{x - 6}+2}{x + 2} \\$$
Let $ a= a=\sqrt[3]{x-6} $ and $ b= {2} $ . Then $ a^3+b^3 = (a+b)(a^2-ab+b^2)$ . So $ a^3+b^3 = x+2 $ and $ a+b = \sqrt[3]{x - 6}+2 $ . So $ \frac{\sqrt[3]{x - 6}+2}{x + 2} $ = $ \frac{a+b}{a^3+b^3}$ = $\frac{1}{ a^2 - ab +b^2} $ . Hope so this helps a lot .
Here at $ x =-2 $ , the value of a is -2 . So at $ x= -2 $ the value of $ a^2-ab+b^2 = 4-(-2)2+4 = 12 $ .
So the ultimate value of $$ \lim_{x\to-2}\frac{\sqrt[3]{x - 6}+2}{x + 2} $$ is $ \frac{1}{12} $
|
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evaluation of $\int_{0}^{1}\frac{1-x^2}{\left(1+x^2\right)\sqrt{1+x^4}}dx$ $\displaystyle \int_{0}^{1}\frac{1-x^2}{\left(1+x^2\right)\sqrt{1+x^4}}dx$
By Using Substution $\sqrt{1+x^4} = (1+x^2)\cdot \cos \theta$
I have Tried it without using the given substution.
$\bf{My\; Try}::$ $\displaystyle \int_{0}^{1}\frac{1-x^2}{\left(1+x^2\right)\sqrt{1+x^4}}dx = \int_{0}^{1}\frac{1-x^2}{x\cdot \left(x+\frac{1}{x}\right)\cdot x \cdot \sqrt{\left(x^2+\frac{1}{x^2}\right)}}dx$
$\displaystyle = -\displaystyle \int_{0}^{1}\frac{\left(1-\frac{1}{x^2}\right)}{\left(x+\frac{1}{x}\right)\cdot \sqrt{\left(x+\frac{1}{x}\right)^2-\left(\sqrt{2}\right)^2}}dx$
Now Let $\displaystyle \left(x+\frac{1}{x}\right) = t$ and $\displaystyle \left(1-\frac{1}{x^2}\right)dx = dt$
$\displaystyle = -\int_{0}^{1}\frac{1}{t\sqrt{t^2-\left(\sqrt{2}\right)^2}}dt = -\frac{1}{\sqrt{2}}\left[\sec^{-1}\left(\frac{x+\frac{1}{x}}{\sqrt{2}}\right)\right]_{0}^{1} = -\frac{1}{\sqrt{2}}\left(\sec^{-1}\left(\sqrt{2}\right)-\sec^{-1}\left(\infty \right)\right)$
$\displaystyle = \frac{1}{\sqrt{2}}\left(\frac{\pi}{2}-\frac{\pi}{4}\right) = \frac{\pi}{4\sqrt{2}}$
Would anyone explain me how can i solve using the substution $\sqrt{1+x^4} = (1+x^2)\cdot \cos \theta$
Help Required
Thanks
|
Nicely done. So they suggest setting
$$
\cos \theta =\frac{\sqrt{1+x^4}}{1+x^2} \quad\Rightarrow\quad \frac{2x(1-x^2)}{\sqrt{1+x^4}(1+x^2)^2}dx=\sin \theta \,d\theta
$$
and, since $\sin\theta\geq 0$ here:
$$
\sin\theta =\sqrt{1-\cos^2\theta}=\frac{\sqrt{2} \;x}{1+x^2} \quad\Rightarrow \quad \frac{1-x^2}{(1+x^2)\sqrt{1+x^4}} dx=\frac{d\theta}{\sqrt{2}}
$$
This is indeed a valid change of variable $\theta =\arccos\left( \frac{\sqrt{1+x^4}}{1+x^2}\right)$. The bounds become $\arccos(1)=0$ and $\arccos(\sqrt{2}/2)=\pi/4$. This yields
$$
\int_0^1 \frac{1-x^2}{(1+x^2)\sqrt{1+x^4}} dx=\int_0^{\pi/4}\frac{d\theta}{\sqrt{2}}=\frac{\pi}{4\sqrt{2}}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$ \frac1{bc-a^2} + \frac1{ca-b^2}+\frac1{ab-c^2}=0$ implies that $ \frac a{(bc-a^2)^2} + \frac b{(ca-b^2)^2}+\frac c{(ab-c^2)^2}=0$ $a ,b , c$ are real numbers such that $ \dfrac1{bc-a^2} + \dfrac1{ca-b^2}+\dfrac1{ab-c^2}=0$ , then how do we prove
(without routine laborious manipulation) that $\dfrac a{(bc-a^2)^2} + \dfrac b{(ca-b^2)^2}+\dfrac c{(ab-c^2)^2}=0$
|
$$\frac 1 {b c - a^2 } + \frac 1 {c a - b^2} + \frac 1 {a b - c^2} =
\frac {
(b c + a c + a b) (bc+ac+ab-a^2-b^2-c^2)
}
{
(bc-a^2)(ac-b^2)(ab-c^2)
}
$$
$$ \frac a {(bc - a^2)^2} + \frac b {(ca - b^2)^2} + \frac c {(ab - c^2)^2} =
\frac{
(bc+ac+ab)(bc+ac+ab-a^2-b^2-c^2)
}
{
(bc-a^2)(ac-b^2)(ab-c^2)
}
\times
\frac{ (a+b+c)(a^2bc+ab^2c+abc^2-b^2c^2-a^2c^2-a^2b^2)
}{
(bc-a^2)(ac-b^2)(ab-c^2)
}
$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Non-Homogeneous System [Problem] "Find a general solution of the system and use that solution to find a general solution of the associated homogeneous system and a particular solution of the given system."
$\begin{bmatrix}3 & 4 & 1 & 2 \\ 6 & 8 & 2 & 5\\9 & 12 & 3 & 10 \end{bmatrix}\begin{bmatrix}x_1 \\ x_2 \\ x_3 \\ x_4\end{bmatrix} = \begin{bmatrix}3 \\ 7 \\ 13\end{bmatrix}$
So we solve it like its a homogenous system. I end up getting to $\begin{bmatrix} 3 & 4& 1 & 2\\ 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\end{bmatrix}$.So would $x_2$ and $x_3$ be free variables? I can set them to $x_2 = \alpha$ and $x_3 = t$, $x_4 = 0$ then we set up:
$\begin{cases}3x_1 + 4x_2 + x_3 + 2x_4 = 0\\0+0+0+x_4 = 0 \\0+0+0+0 = 0\end{cases}$
So, $\underline{v} = \begin{pmatrix}3 \\ 7 \\ 13\end{pmatrix} + \begin{pmatrix}\frac{1}{3}(-4\alpha -t - 2_4) \end{pmatrix}$
I don't think this is right. What would $x_4$ be? I'm confused
|
The method we were taught was to set up an augmented matrix like so:
$$
\left[\begin{array}{cccc|c}
3 & 4 & 1 & 2 & 3\\
6 & 8 & 2 & 5 & 7\\
9 & 12 & 3 & 10 & 13\\
\end{array}\right]
$$
And then solve the matrix to the left of the vertical bar like a homogeneous system.
$$
\left[\begin{array}{cccc|c}
3 & 4 & 1 & 2 & 3\\
0 & 0 & 0 & 1 & 1\\
0 & 0 & 0 & 0 & 0\\
\end{array}\right] \rightarrow \left[\begin{array}{cccc|c}
3 & 4 & 1 & 0 & 1\\
0 & 0 & 0 & 1 & 1\\
0 & 0 & 0 & 0 & 0\\
\end{array}\right]
$$
So $3x_1+4x_2+x_3 = 1$, and $x_4 = 1$
|
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|
Find a matrix $X$ given $X^4$ Find the matrix $X$ such that
$$X^4=\begin{bmatrix}
3&0&0\\
0&3&1\\
0&0&0
\end{bmatrix}$$
This problem I can't work,and I think let the matrix the eigenvalue is $\lambda$,then
$\lambda^4$ is $$\begin{bmatrix}
3&0&0\\
0&3&1\\
0&0&0
\end{bmatrix}$$
eigenvalue?Thank you for your help.
|
Hint: Write the Jordan Normal Form., then take:
$$X = S.J^{1/4}.S^{-1}$$
So, we have:
$$\begin{bmatrix} 3&0&0\\ 0&3&1\\ 0&0&0 \end{bmatrix}$$
We can write this in Jordan Normal Form as:
$$S.J.S^{-1} = \begin{bmatrix} 0&0&1\\ -1&1&0\\ 3&0&0 \end{bmatrix}.\begin{bmatrix} 0&0&0\\ 0&3&0\\ 0&0&3 \end{bmatrix}.\begin{bmatrix} 0&0&\dfrac{1}{3}\\ 0&1&\dfrac{1}{3}\\ 1&0&0 \end{bmatrix}$$
Now, just take the fourth roots of the diagonal entries of $J$ and then multiply out with the other two matrices.
Spoiler - Do Not Peek (do the work and then look)
$X = S.J^{1/4}.S^{-1} = \begin{bmatrix} 0&0&1\\ -1&1&0\\ 3&0&0 \end{bmatrix}.\begin{bmatrix} 0&0&0\\ 0&3^{1/4}&0\\ 0&0&3^{1/4} \end{bmatrix}.\begin{bmatrix} 0&0&\dfrac{1}{3}\\ 0&1&\dfrac{1}{3}\\ 1&0&0 \end{bmatrix}=\begin{bmatrix} 3^{1/4}&0&0\\ 0&3^{1/4}&\dfrac{1}{3^{3/4}}\\ 0&0&0 \end{bmatrix}$
|
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|
Compute $\int_0^\infty \frac{dx}{1+x^3}$ Problem
Compute $$\displaystyle \int_0^\infty \frac{dx}{1+x^3}.$$
Solution
I do partial fractions
$$\frac{1}{x^3+1}= \frac{2-x}{3 \left( x^{2}-x+1 \right)}+\frac{1}{3 \left( x+1 \right)}.$$
But we could simplify the left one $$\frac{2-x}{3\left( x^{2}-x+1 \right)} = \frac{2}{3}\cdot \frac{1}{x^{2}-x+1}-\frac{x}{x^{2}-x+1}$$
From here, I do this see images.
But I find the wrong primitive functions. Why am I wrong, and how do I find the correct one?
|
Since
$$
\frac{1}{x^3+1}=\frac{1}{(x+1)(x^2-x+1)},
$$
we can find some $a,b,c \in \mathbb{R}$ such that
$$
\frac{1}{x^3+1}=\frac{a}{x+1}+\frac{bx+c}{x^2-x+1}.
$$
A simple computation shows that
$$
a=-b=\frac{c}{2}=\frac13,
$$
i.e.
\begin{eqnarray}
\frac{1}{x^3+1}&=&\frac13\cdot\frac{1}{x+1}-\frac13\cdot\frac{x-2}{x^2-x+1}\\
&=&\frac13\cdot\frac{1}{x+1}-\frac16\cdot\frac{2x-1}{x^2-x+1}+\frac12\cdot\frac{1}{x^2-x+1}\\
&=&\frac13\cdot\frac{1}{x+1}-\frac16\cdot\frac{2x-1}{x^2-x+1}+\frac12\cdot\frac{1}{\left(x-\frac12\right)^2+\frac34}.
\end{eqnarray}
It follows that
\begin{eqnarray}
F(r)&=&\int_0^r\frac{1}{x^3+1}\,dx=\frac13\ln(1+r)-\frac16\ln(r^2-r+1)+\frac{1}{\sqrt{3}}\arctan\frac{2r-1}{\sqrt{3}}+\frac{1}{\sqrt{3}}\arctan\frac{1}{\sqrt{3}}\\
&=&\frac16\ln\frac{(r+1)^2}{r^2-r+1}+\frac{1}{\sqrt{3}}\arctan\frac{2r-1}{\sqrt{3}}+\frac{\pi}{6\sqrt{3}}\\
&=&\frac16\ln\frac{r^2+2r+1}{r^2-r+1}+\frac{1}{\sqrt{3}}\arctan\frac{2r-1}{\sqrt{3}}+\frac{\pi}{6\sqrt{3}}.
\end{eqnarray}
Thus
$$
\int_0^\infty\frac{1}{x^3+1}\,dx=\lim_{r\to\infty}F(r)=\frac{\pi}{2\sqrt{3}}+\frac{\pi}{6\sqrt{3}}=\frac{2\pi}{3\sqrt{3}}.
$$
|
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|
How to evaluate the limit $\lim_{x\to0}\frac{\sqrt{x+1}-\sqrt{2x+1}}{\sqrt{3x+4}-\sqrt{2x+4}}$ First I tried direct substitution, which resulted in the indeterminate form. Then because of the square roots, I tried rationalizing (both numerator and denominator), but still get the indeterminate form. I can't use L'Hopital's rule because we haven't learned that yet (not sure if it would work anyway). Don't really know what else to try algebraically... Any hints?
|
$\lim_{x\to0}\frac{\sqrt{x+1}-\sqrt{2x+1}}{\sqrt{3x+4}-\sqrt{2x+4}}\cdot\frac{\sqrt{3x+4}+\sqrt{2x+4}}{\sqrt{3x+4}+\sqrt{2x+4}}\cdot\frac{\sqrt{x+1}+\sqrt{2x+1}}{\sqrt{x+1}+\sqrt{2x+1}}=\lim_{x\to0}\frac{-x(\sqrt{3x+4}+\sqrt{2x+4})}{x(\sqrt{x+1}+\sqrt{2x+1})}=-\frac{4}{2}=-2$.
|
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|
Ordinary generating function for $\binom{3n}{n}$ The ordinary generating function for the central binomial coefficients, that is, $$\displaystyle \sum_{n=0}^{\infty} \binom{2n}{n} x^{n} = \frac{1}{\sqrt{1-4x}} \, , \quad |x| < \frac{1}{4},$$
can be derived by using the duplication formula for the gamma function and the generalized binomial theorem.
But what about the ordinary generating function for $ \displaystyle \binom{3n}{n}$?
According to Wolfram Alpha, $$ \sum_{n=0}^{\infty} \binom{3n}{n} x^{n} = \frac{2\cos \left(\frac{1}{3} \arcsin \left(\frac{3 \sqrt{3x}}{2} \right)\right)}{\sqrt{4-27x}} \, , \quad |x| < \frac{4}{27}. $$
Any suggestions on how to prove this?
EDIT:
Approaching this problem using the fact that $$ \text{Res} \Big[ \frac{(1+z)^{3n}}{z^{n+1}},0 \Big] = \binom{3n}{n},$$
I get $$ \sum_{n=0}^{\infty} \binom{3n}{n} x^{n} = -\frac{1}{2 \pi i x} \int_{C} \frac{dz}{z^{3}+3z^{2}+3z - \frac{z}{x}+1},$$
where $C$ is a circle centered at $z=0$ such that every point on the circle satisfies $ \displaystyle\Big|\frac{x(1+z)^{3}}{z} \Big| < 1$.
Evaluating that contour integral would appear to be quite difficult.
|
The following is a proof of the hypergeometric identity $$_2F_1\left(a,1-a;\frac{1}{2};\sin^2(x)\right)=\frac{\cos[(2a-1)x]}{\cos x} \, , \quad - \frac{\pi}{2} < x < \frac{\pi}{2} \tag{1}.$$
This identity is used in R. J. Mathar's answer.
Similar to my answer here, I will first use the generalized binomial theorem to show that $$_2F_1\left(a,1-a;\frac{1}{2};-z^{2}\right) = \frac{1}{2 \sqrt{1+z^{2}}} \left[\left(\sqrt{1+z^{2}}+z \right)^{2a-1} + \left(\sqrt{1+z^{2}}-z \right)^{2a-1} \right] \, , \quad |z| <1. $$
Identity $(1)$ then follows if $z$ is replaced with $i \sin (x)$.
\begin{align} &\frac{1}{2\sqrt{1+z^{2}}} \left[\left(\sqrt{1+z^{2}}+z \right)^{2a-1} + \left(\sqrt{1+z^{2}}-z \right)^{2a-1}\right] \\ &= \frac{1}{2\sqrt{1+z^{2}}} \left[\sum_{k=0}^{\infty} \binom{2a-1}{k} z^{k}\left(\sqrt{1+z^{2}} \right)^{2a-1-k} + \sum_{j=0}^{\infty} \binom{2a-1}{j} (-z)^{j} \left(\sqrt{1+z^{2}} \right)^{2a-1-j}\right ] \\ &= \sum_{k=0}^{\infty} \binom{2a-1}{2k} z^{2k} (1+z^{2})^{a-1-k}\\ &= \sum_{k=0}^{\infty} \binom{2a-1}{2k} z^{2k} \sum_{l=0}^{\infty} \binom{a-1-k}{l}z^{2l}\\ & \stackrel{(2)}= \sum_{n=0}^{\infty} \sum_{m=0}^{n}\binom{2a-1}{2m} \binom{a-1-m}{n-m} z^{2n} \\ &=\sum_{n=0}^{\infty} \frac{\Gamma(2a)}{\Gamma(a-n)}\sum_{m=0}^{n} \frac{\Gamma(a-m)}{\Gamma(2m+1)\Gamma(2a-2m) \Gamma(n-m+1)} \, z^{2n} \\ & \stackrel{(3)}= \sum_{n=0}^{\infty} \frac{\Gamma(2a)}{\Gamma(a-n)} \sum_{m=0}^{n} \frac{\pi}{2^{2a-1}} \frac{1}{\Gamma \left(m+\frac{1}{2}\right) \Gamma(m+1) \Gamma(a-m+\frac{1}{2}) \Gamma(n-m+1)} \, z^{2n} \\ &= \sum_{n=0}^{\infty} \frac{\Gamma(2a)}{\Gamma(a-n)} \frac{\pi}{2^{2a-1}} \frac{1}{\Gamma(n+ \frac{1}{2}) \Gamma(a+ \frac{1}{2})} \sum_{m=0}^{n} \binom{n-\frac{1}{2}}{n-m} \binom{a-\frac{1}{2}}{m} \, z^{2n} \\ & \stackrel{(4)}= \sum_{n=0}^{\infty} \frac{\Gamma(2a)}{\Gamma(a-n)} \frac{\pi}{2^{2a-1}}\frac{1}{\Gamma(n+ \frac{1}{2}) \Gamma(a+ \frac{1}{2})} \binom{a+n-1}{n} \, z^{2n} \\ & \stackrel{(5)} = \sum_{n=0}^{\infty} \frac{\Gamma(a+n)}{\Gamma(a-n)} \frac{\sqrt{\pi}}{\Gamma (n+\frac{1}{2})} \, \frac{z^{2n}}{n!} \frac{\Gamma(a)}{\Gamma(a)} \\ & \stackrel{(6)} = \sum_{n=0}^{\infty} \frac{\Gamma(a+n)}{\Gamma(a)} \frac{\Gamma(1-a+n)}{\Gamma(1-a)} \frac{\Gamma(\frac{1}{2})}{\Gamma(n+ \frac{1}{2})} \frac{(-z^{2})^{n}}{n!} \\ &= \, _2F_{1} \left(a, 1-a; \frac{1}{2}; -z^{2} \right) \end{align}
$(2)$: Cauchy product
$(3)$: Duplication formula for the gamma function
$(4)$: Chu-Vandermonde identity
$(5)$: Duplication formula again
$(6)$: $\frac{\Gamma(a)}{\Gamma(a-n)}= (-1)^{n} \frac{\Gamma(1-a+n)}{\Gamma(1-a)}$
|
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|
Proving combinatorial identity $\sum_s (-1)^s\binom{p+s-1}{s}\binom{2m+2p+s}{2m+1-s}2^s=0$ I need to prove following combinatorial identities:
$$
\sum\limits_s(-1)^s\binom{p+s-1}{s}\binom{2m+2p+s}{2m+1-s}2^s=0
$$
$$
\sum\limits_s(-1)^s\binom{p+s-1}{s}\binom{2m+2p+s-1}{2m-s}2^s=(-1)^m\binom{p+m-1}{m}
$$
given the fact that
$$
(1-x)^{2k}\left(1+\frac{2x}{(1-x)^2}\right)^k = (1+x^2)^k
$$
for either $k=p$ or $k=-p$.
And for the first one I cannot understand where $\binom{2m+2p+s}{2m+1-s}$ is emerging from (different signs for $s$ on top and bottom seem strange to me). I'm trying to prove first identity as following: if we transform given equation and let $k=p$ we get something like this:
$$
(1-x)^{2p}(1+2x+4x^2+\dots)^p=(1+x^2)^p
$$
Let's find coefficient for $x^{2m+1}$ for both sides. For the right side it is always equal to $0$, as only even powers are present there. For the left side let's take $x^s$ from second bracket and $x^{2m+1-s}$ from first. Getting $x^s$ from second bracket is equal to splitting $s$ into $p$ addends with zeroes allowed, so the coefficient is equal to $2^s\binom{p+s-1}{p-1} = 2^s\binom{p+s-1}{s}$. So we get some of needed multipliers for our identity. But now if we take $2m+1-s$ from first bracket we get coefficient like $(-1)^{2m+1-s}\binom{2p}{2m+1-s} = (-1)^s\binom{2p}{2m+1-s}$. And the final result is:
$$
\sum\limits_s(-1)^s\binom{p+s-1}{s}\binom{2p}{2m+1-s}2^s=0
$$
And I see no way to transform it to required identity.
For the second equality I do not even understand where right side is taken from.
Thanks in advance for any help.
|
Suppose we seek to verify that
$$\sum_{q=0}^{2m} (-1)^q {p-1+q\choose q}
{2m+2p+q-1\choose 2m-q} 2^q =
(-1)^m {p-1+m\choose m}.$$
Introduce
$${2m+2p+q-1\choose 2m-q} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2m-q+1}} (1+z)^{2m+2p+q-1} \; dz.$$
Observe that this controls the range being zero when $q\gt 2m$ so we
may extend $q$ to infinity to obtain for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2m+1}} (1+z)^{2m+2p-1}
\sum_{q\ge 0} {p-1+q\choose q} (-1)^q 2^q z^q (1+z)^q \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2m+1}} (1+z)^{2m+2p-1}
\frac{1}{(1+2z(z+1))^p} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2m+1}} (1+z)^{2m+2p-1}
\frac{1}{((1+z)^2+z^2)^p} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2m+1}} (1+z)^{2m-1}
\frac{1}{(1+z^2/(1+z)^2)^p} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2m}} (1+z)^{2m} \frac{1}{z(1+z)}
\frac{1}{(1+z^2/(1+z)^2)^p} \; dz.$$
Now put $$\frac{z}{1+z} = u
\quad\text{so that}\quad z=\frac{u}{1-u}
\quad\text{and}\quad dz=\frac{1}{(1-u)^2} du$$
to obtain for the integral
$$\frac{1}{2\pi i}
\int_{|u|=\gamma}
\frac{1}{u^{2m}} \frac{1}{u/(1-u)\times 1/(1-u)}
\frac{1}{(1+u^2)^p} \frac{1}{(1-u)^2} \; du
\\ = \frac{1}{2\pi i}
\int_{|u|=\gamma}
\frac{1}{u^{2m+1}} \frac{1}{(1+u^2)^p} \; du.$$
This is
$$[u^{2m}] \frac{1}{(1+u^2)^p}
= [v^{m}] \frac{1}{(1+v)^p}
= (-1)^m {m+p-1\choose m},$$
as claimed.
For the conditions on $\epsilon$ and $\gamma$ we require convergence of the geometric series with $|2z(1+z)|\lt 1$ which holds for $\epsilon \lt (-1+\sqrt{3})/2.$ Note that with $u=z+\cdots$ the image of $|z|=\epsilon$ makes one turn around zero. The closest it comes to the origin is at $\epsilon/(1+\epsilon)$ so we must choose $\gamma \lt \epsilon/(1+\epsilon)$ e.g. $\gamma=\epsilon^2/(1+\epsilon)$ for $|w|=\gamma$ to be entirely contained in the image of $|z|=\epsilon.$ Taking $\epsilon = 1/5$ will work.
|
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|
Residue/Contour integration problem Supposedly,
$\displaystyle\int_{-\infty}^\infty \frac{\cos ax}{x^4+1}dx=\frac{\pi}{\sqrt{2}}e^{-a/\sqrt{2}}\left(\cos\frac{a}{\sqrt{2}}+\sin\frac{a}{\sqrt{2}}\right)$, $a>0$.
Using Residues/Contour integrals, I have
$\displaystyle\int_{-\infty}^\infty \frac{e^{iax}}{x^4+1}dx=\int_{-\infty}^\infty \frac{\cos ax}{x^4+1}dx+i\int_{-\infty}^\infty \frac{\sin ax}{x^4+1}dx=\frac{\pi}{\sqrt{2}}e^{-a/\sqrt{2}}\left(\cos\frac{a}{\sqrt{2}}+i\sin\frac{a}{\sqrt{2}}\right)$
I don't see how to draw the final conclusion.
|
Here, because $a >0$, it is easiest to use a contour $C$ that is a semicircle of radius $R$ in the upper half-plane in the complex plane. Thus, consider
$$\oint_C dz \frac{e^{i a z}}{z^4+1} $$
This contour integral is equal to
$$\int_{-R}^{R} dx \frac{e^{i a x}}{x^4+1} + i R \int_0^{\pi} d\theta \, e^{i \theta} \frac{e^{-a R \sin{\theta}} e^{i a R \cos{\theta}}}{R^4 e^{i 4 \theta}+1} $$
As $R \to \infty$, the magnitude of the second integral vanishes as
$$\frac{2}{R^3} \int_0^{\pi/2} d\theta\, e^{-a R \sin{\theta}} \le \frac{2}{R^3} \int_0^{\pi/2} d\theta e^{-2 a R/\pi} \le \frac{\pi}{R^4}$$
The contour integral is also equal to $i 2 \pi$ times the sum of the residues of the poles within $C$. These poles are at $z=e^{i \pi/4}$ and $e^{i 3 \pi/4}$. The integral we seek is therefore found through the following equality:
$$\begin{align}\int_{-\infty}^{\infty} dx \frac{e^{i a x}}{x^4+1} &= i 2 \pi \left [\frac{e^{-a/\sqrt{2}} e^{i a/\sqrt{2}}}{4 e^{i 3 \pi/4}} + \frac{e^{-a/\sqrt{2}} e^{-i a/\sqrt{2}}}{4 e^{i 9 \pi/4}} \right ]\\ &= i\frac{\pi}{2} e^{-a/\sqrt{2}} e^{-i \pi/2} \left [e^{i(a/\sqrt{2}-\pi/4)} + e^{-i(a/\sqrt{2}-\pi/4)}\right ]\\ &= \pi \, e^{-a/\sqrt{2}} \cos{\left (\frac{a}{\sqrt{2}}-\frac{\pi}{4} \right )} \\ &= \pi \, e^{-a/\sqrt{2}} \frac{1}{\sqrt{2}} \left (\cos{\frac{a}{\sqrt{2}}} + \sin{\frac{a}{\sqrt{2}}} \right )\end{align}$$
Because cosine is even and sine is odd, the desired integral is equal to the above result, as was to be shown.
|
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Why is $(2+\sqrt{3})^{50}$ so close to an integer? I just worked out $(2+\sqrt{3})^{50}$ on my computer and got the answer
$39571031999226139563162735373.999999999999999999999999999974728\cdots$
Why is this so close to an integer?
|
Let $$x=2+\sqrt { 3 }, \\$$then$$ { x }^{ 2 }=7+4\sqrt { 3 }, \\ 4x=8+4\sqrt { 3 } ,$$ therefore $${ x }^{ 2 }+1=4x,\\ x+\frac { 1 }{ x } =4.$$ Take the square of both side and repeat the process $${ x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } ={ 4 }^{ 2 }-2\\ { x }^{ 4 }+\frac { 1 }{ { x }^{ 4 } } ={ \left( { 4 }^{ 2 }-2 \right) }^{ 2 }-2\\ { x }^{ 8 }+\frac { 1 }{ { x }^{ 8 } } ={ \left( { \left( { 4 }^{ 2 }-2 \right) }^{ 2 }-2 \right) }^{ 2 }-2$$ You can see that $$\lim _{ n\rightarrow \infty }{ \frac { 1 }{ { x }^{ { 2 }^{ n } } } } =0,$$ because $x=2+\sqrt { 3 }>1.$ Although you can intuitively say $ \frac { 1 }{ { x }^{ 50 } } \approx0,$ you can verify it by saying $x=y^{50/2^n}$, where $y>1$.
|
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|
What is the summation of the following expression? What's the summation of the following expression;
$$\sum_{k=1}^{n+3}\left(\frac{1}{2}\right)^{k}\left(\frac{1}{4}\right)^{n-k}$$
The solution is said to $$2\left(\frac{1}{4} \right)^{n}\left(2^{n+3}-1\right)$$
But I'm getting $$\left(\frac{1}{4} \right)^{n}\left(2^{n+3}-1\right).$$
How is this possible?
$$\sum_{k=1}^{n+3}\left(2 \times\frac{1}{4}\right)^{k}\left(\frac{1}{4}\right)^n\left(\frac{1}{4}\right)^{-k} \rightarrow \left(\frac{1}{4}\right)^n \sum_{k=1}^{n+3} 2^k\left(\frac{1}{4}\right)^k \left(\frac{1}{4}\right)^{-k}\rightarrow \left(\frac{1}{4}\right)^n\left(2^{n+3}-1\right)$$
|
The formula $$\sum_{k=1}^{n}ar^{k}=a\left(\frac{1-r^{n}}{1-r}\right)$$ is incorrect. The correct formula is $$\sum_{{\Large k=}{\Huge0}}^{\Huge n-1}ar^{k}=a\left(\frac{1-r^{n}}{1-r}\right)$$ or $$\sum_{k=1}^{n}ar^{k}=a{\Large{r}}\left(\frac{1-r^{n}}{1-r}\right)$$. With $a=1$ and $r=2$, you should have $\sum_{k=1}^{n+3}1\cdot2^{k}=2\left(\frac{1-2^{n+3}}{1-2}\right)=1\cdot2\left(2^{n+3}-1\right)$
|
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|
Use partial fractions to find the integral. Find the integral using partial factions.
$$\int\frac{(2x^2+5x+3)}{(x-1)(x^2+4)}\,dx$$
So do I do...$$\frac{2x^2+5x+3}{(x-1)(x^2+4)}=\frac{A}{x-1} + \frac{Bx+C}{x^2+4},$$
then get
\begin{align*}
2x^2+5x+3 &= A(x^2+4)+(Bx+C)(x-1) \\
2x^2+5x+3 &= Ax^2+4A+Bx^2-Bx+Cx-C?
\end{align*}
|
Note that we can simply add zero: $2x^2 = 2(x^2 + 4) - 8 = 2x^2 + 8 - 8$. Then
$$\frac{2x^2 + 5x + 3}{(x - 1)(x^2 + 4)} = \frac{2(x^2 + 4) - 8 + 5x + 3}{(x - 1)(x^2 + 4)} = \frac{2}{x - 1} + \frac{5x - 5}{(x - 1)(x^2 + 4)}.$$
Do you think you can integrate
$$\frac{2x^2 + 5x + 3}{(x - 1)(x^2 + 4)}$$
now?
Hint:
$$\frac{5x - 5}{(x - 1)(x^2 + 4)} = \frac{5(x - 1)}{(x - 1)(x^2 + 4)} = \frac{5}{x^2 + 4}$$
and
$$\int \frac{1}{x^2 + a^2} = \frac{1}{a}\tan^{-1}\frac{x}{a} + C, \quad a \ne 0.$$
|
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|
How to find the limit of $\frac{1−\cos 5x}{x^2}$ as $x\to 0$? What is the right approach to calculate the Limit of $(1-\cos(5x))/x^2$ as $x \rightarrow 0$?
From Wolfram Alpha, I found that:
$$\lim_{x \to 0} \frac{1- \cos 5x}{x^2} = \frac{25}{2}.$$
How do I get that answer?
|
$\cos(2\alpha) = 1-2\sin ^2 \alpha$
Therefore
$\frac{1-\cos (5x)}{x^2}=\frac{2 \sin ^2 \left( \frac{5x}{2}\right)}{x^2} = \frac{25}{2} \left( \frac{\sin \frac{5}{2}x}{\frac{5}{2}x} \right)^2 \rightarrow \frac{25}{2}$
|
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|
Show that $x^3+y^3 = 3\mod 9$ has no solutions Not even sure how to go about this. I tried $x^3 = 3-y^3 \mod 9$, but not sure what that does.
|
Modulo 9, we have:
$$0^3=0, 1^3=1, 2^3=-1, 3^3=0, 4^3=1, 5^3=-1, 6^3=0, 7^3=1, 8^3=-1$$
No pair of these sums to 3.
In fact, we can use this table to conclude the stronger statement that $$x^3+y^3+z^3=4\pmod{9}$$ has no solutions.
|
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Limit: $\lim_{n\to \infty} \frac{n^5}{3^n}$ I need help on a homework assignment. How to show that $\lim_{n\to\infty} \left(\dfrac{n^5}{3^n}\right) = 0$? We've been trying some things but we can't seem to find the answer.
|
Here's an "elementary" proof that doesn't use any theorems such as l'Hôpital. Start off by writing $\frac{n^5}{3^n} = \frac{n^5}{2^n} \left(\frac{2}{3}\right)^n$. The term $\left(\frac{2}{3}\right)^n$ tends to zero as $n \rightarrow \infty$, so we would be done if we showed that the term $\frac{n^5}{2^n}$ is bounded as $n \rightarrow \infty$. It is clear that this term is always positive. The derivative of the function given by $f(n) = \frac{n^5}{2^n}$ is equal to $f'(n) = -2^{-n} n^4 (n \log(2) - 5)$. For $n > \frac{5}{\log 2}$, we have $f'(n) < 0$, so $f(n)$ is decreasing there. In particular, for all $n > \frac{5}{\log 2}$ we get $0 < f(n) \leq f\left(\frac{5}{\log 2}\right)$ and hence $0 < \frac{n^5}{3^n} \leq f\left(\frac{5}{\log 2}\right) \left(\frac{2}{3}\right)^n$. The right hand side tends to zero as $n \rightarrow \infty$ and we can conclude that $\lim_{n \rightarrow \infty} \frac{n^5}{3^n} = 0$.
|
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|
Finding polynomal function with given zeros and one zero is a square root I've been having trouble with this problem: Find a polynomial function of minimum degree with $-1$ and $1-\sqrt{3}$ as zeros. Function must have integer coefficients.
When I tried it, I got this:
\begin{align}
(x+1)(x-(1-\sqrt{3}))=& x^2 - x(1-\sqrt{3}) + x - 1 + \sqrt{3}\\
=& x^2 - x + x(\sqrt{3}) + x - 1 + \sqrt{3}\\
=& x^2 + x(\sqrt{3}) - 1 + \sqrt{3}
\end{align}
This looks completely wrong to me and it probably is as my teacher said the answer should not be a quadratic. I'm not looking for the answer to be given to me; but if I could get some guidance, it would be greatly appreciated.
|
If you're looking for a polynomial that has those two roots and integer coefficients, you'll need to add another root.
Substitute $w = x-1$ so that $(x+1)(x-(1-\sqrt{3})) = (w+2)(w+\sqrt{3})$.
Now multiply by $(w-\sqrt{3})$ to get
$(w+2)(w+\sqrt{3})(w-\sqrt{3}) = (w+2)(w^2-3)$.
Then substitute back in $x$ and you have your polynomial, which is a cubic with integer coefficients:
$(x+1)(x^2-2x-2) = x^3 -2x^2 - 2x + x^2 - 2x - 2 = x^3 - x^2 - 4x - 2$.
|
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|
Double Integral $\int\limits_0^b\int\limits_0^x\sqrt{a^2-x^2-y^2}\,dy\,dx$ What is the best method for evaluating the following double integral?
$$
\int_{0}^{b}\int_{0}^{x}\,\sqrt{\,a^{2} - x^{2} - y^{2}\,}\,\,{\rm d}y\,{\rm d}x\,,
\qquad a > \sqrt{\,2\,}\,\,b
$$
Is there exist an easy method?
My try:
$$\int_0^b\int_0^x\sqrt{a^2-x^2-y^2}dy\,dx=\int_0^{\frac{\pi}{4}}\int_0^{b\sec(\theta)}r\sqrt{a^2-r^2}dr\,d\theta$$
$$=\int_0^{\frac{\pi}{4}}\frac{-1}{3}\left[(a^2-r^2)\sqrt{a^2-r^2}\right]_0^{b\sec(\theta)}d\theta$$
$$=\frac{1}{3}\int_0^{\frac{\pi}{4}}\left[a^3-(a^2-b^2\sec^2(\theta))\sqrt{a^2-b^2\sec^2(\theta)}\right]d\theta$$
but evaluating above integral is very difficult and antiderivative is very complexity! see here.
|
(The following is basically obsolete because already answered in
https://math.stackexchange.com/questions/571967/integral-inta2-b2-sec2x-sqrta2-b2-sec2xdx)
The first integral of Harry Peter is easy with the substitution $x^2=t$.
The second, attacked with partial integration, requires to integrate
$(a^2-x^2)/2$ which is $x(3a^2-x^2)/6$ and to differentiate the $\arcsin$
which gives $a^2/\sqrt{a^2-2x^2}/(a^2-x^2)$. So we need besides the
preintegrated term essentially
$$\int dx \frac{x(3a^2-x^2)}{6}\frac{a^2}{\sqrt{a^2-2x^2}(a^2-x^2)}$$
which turns with $x^2=t$ in something proportional to
$$\int dx \frac{(3a^2-t)}{6}\frac{a^2}{\sqrt{a^2-2t}(a^2-t)}$$.
By expansion of $(3a^2-6)/(a^2-t)$ into partial fractions only
the format $1/[\sqrt{a^2-2t}(a^2-t)]$ is required which seems with $y=a^2-2t$
to be covered by $$\int \frac{dy}{(a+by)\surd y} =\frac{2}{\sqrt{ab}}\arctan\sqrt{\frac{by}{a}}$$
|
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How to evaluate a limit with subtractions $\lim_{x \rightarrow -1}(\frac{3}{x^3+1}-\frac{1}{x+1})$? I'm having trouble thinking of a way to solve this.
$$\lim_{x \rightarrow -1}\left(\frac{3}{x^3+1}-\frac{1}{x+1}\right)$$
|
Because$$x^3+1=(x+1)(x^2-x+1)$$we get
$$\frac{3}{x^3+1}-\frac{1}{x+1}=\frac{3}{(x+1)(x^2-x+1)}-\frac{1}{x+1}$$
$$=\frac{3-(x^2-x+1)}{(x+1)(x^2-x+1)}=\frac{x-x^2+2}{(x+1)(x^2-x+1)}=$$
$$=\frac{x+1-(x^2-1)}{(x+1)(x^2-x+1)}=\frac{(x+1)(2-x)}{(x+1)(x^2-x+1)}=\frac{2-x}{x^2-x+1}$$
|
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|
Finding exactly one real solution to the system
I dont know how to go about finding the one real solution to the system where k is a real number
Thank you
|
You can write $n=\frac{6}{m}$. Substituting in the first equation, we get $m+\frac{6}{m}=k$, and then $m^2-km+6=0$.
This is a quadratic equation, which we can solve using the Quadratic Formula. We get
$$m=\frac{k\pm\sqrt{k^2-24}}{2}.$$
There is a single solution if $k^2-24=0$. There is no real solution if $k^2-24\lt 0$.
Another way: Note that $(m-n)^2=(m+n)^2-4mn=k^2-24$. There is no real solution if $k^2-24\lt 0$, since the square of a real number cannot be negative.
If $k^2-24\gt 0$, we can take the square roots, and we get $m-n=\pm\sqrt{k^2-24}$. If $k^2-24=0$, we get $m=n$, and a single solution. If $k^2-24\gt 0$, there are two possible values of $m-n$, and therefore two possible values of $(m,n)$.
Remark: We can do the exactly one part with less computation. Note that the system of equations is symmetric in $m$ and $n$. Therefore if $(x,y)$ is a solution, so is $(y,x)$. That gives two solutions unless $x=y$. So our only chance at one solution is if $x=y$. That gives $x=y=\pm\sqrt{6}$, and therefore $k=x+y=\pm2\sqrt{6}$.
|
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How find this equation solution $2\sqrt[3]{2y-1}=y^3+1$ find this equation roots:
$$2\sqrt[3]{2y-1}=y^3+1$$
My try: since
$$8(2y-1)=(y^3+1)^3=y^9+1+3y^3(y^3+1)$$
then
$$y^9+3y^6+3y^3-16y+9=0$$
Then I can't.Thank you someone can take hand find the equation roots.
|
Another approach:
Put $x=\sqrt[3]{2y-1}$ then $x^3=2y-1$. So, $x^3+1=2y$
Then we have the system of equation:
$\left\{\begin{matrix} x^3+1=2y\\y^3+1=2x \end{matrix}\right.$
Subtracting these two equations from each other, we have
$(x-y)(x^2+xy+y^2+2)=0$
which gives $x=y$ as ($x^2+xy+y^2+2>0 $ $\forall$ $x,y$)
Hence we have $y=\sqrt[3]{2y-1}$
$\Rightarrow y^3-2y+1=0$
$\Leftrightarrow(y-1)(y^2+y-1)=0$
Therefore, $y=1$ or $y=\frac{-1\pm \sqrt{5} }{2}$
|
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|
I need to find the second derivative! I need to find the second derivative of
$$f(x)=\frac{\sqrt{x}}{x+4}.$$
I think the first derivative is
$$f'(x)=\frac{4-x} {2 \sqrt{x} (x+4)^2}.$$
EDIT: I got the second derivative to be:
$$f''(x)=-\frac{1}{4 x^{\frac{3}{2}} (x+4) }-\frac{1}{\sqrt{x}(x+4)^2 }+\frac{2 \sqrt{x}}{(x+4)^3}.$$
I'm asking because I need to find the concavity and am very confused.
((just a remark: I have no idea how to comment on other people's posts other than edit my main questions or create more questions, I sincerely apologize for any illicit posts on here or if it goes against guidelines, I am just a college freshman struggling in Calculus, and genuinely appreciate all answers and tips! Thanks again to ANYONE who's helped!!))
Does this mean the only real root is 8.62 because -0.62 would be eliminated due to restrictions of domain?
|
Hint.
$$\begin{align}
\frac{d}{dx} \frac{4 - x}{2\sqrt{x}(x + 4)^2} ~ = ~ & \frac{2\sqrt{x}(x + 4)^2\frac{d}{dx}(4 - x) - (4 - x)\frac{d}{dx}(2\sqrt{x}(x + 4)^2)}{(2\sqrt{x}(x + 4)^2)^2} \\
~ = ~ & \frac{2\sqrt{x}(x + 4)^2\frac{d}{dx}(4 - x) - 2(4 - x)((x + 4)^2\frac{d}{dx}\sqrt{x} + \sqrt{x}\frac{d}{dx}(x + 4)^2)}{(2\sqrt{x}(x + 4)^2)^2}
\end{align}$$
|
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For any real numbers $a,b,c$ show that $\displaystyle \min\{(a-b)^2,(b-c)^2,(c-a)^2\} \leq \frac{a^2+b^2+c^2}{2}$ For any real numbers $a,b,c$ show that:
$$ \min\{(a-b)^2,(b-c)^2,(c-a)^2\} \leq \frac{a^2+b^2+c^2}{2}$$
OK. So, here is my attempt to solve the problem:
We can assume, Without Loss Of Generality, that $a \leq b \leq c$ because of the symmetry.
$a \leq b \leq c$ implies that $0 \leq b-a \leq c-a$. Since both sides are positive and $y=x^2$ is an increasing function for positive numbers we conclude that $(b-a)^2 \leq (c-a)^2$ and since $(a-b)^2=(b-a)^2$ we obtain $(a-b)^2 \leq (c-a)^2$.
Therefore:
$\min\{(a-b)^2,(b-c)^2,(c-a)^2\}=$ $$\min\{(b-c)^2,\min\{(a-b)^2,(c-a)^2\}\}=
\min\{(a-b)^2,(b-c)^2\}.$$
So, we have to prove the following inequality instead:
$$\min\{(a-b)^2,(b-c)^2\} \leq \frac{a^2+b^2+c^2}{2}$$
Now two cases can happen:
*
*$$(a-b)^2 \leq (b-c)^2 \implies |a-b| \leq |b-c|=c-b \implies b-c \leq a-b \leq c-b \implies 2b-c \leq a \implies b \leq \frac{a+c}{2}.$$
*$$(b-c)^2 \leq (a-b)^2 \implies |c-b| \leq |a-b|=b-a \implies a-b \leq c-b \leq b-a \implies c \leq 2b-a \implies b \geq \frac{a+c}{2}.$$
So, this all boils down to whether $b$ is greater than the arithmetic mean of $a$ and $b$ are not.
Now I'm stuck. If $a,b,c$ had been assumed to be positive real numbers it would've been a lot easier to go forward from this step. But since we have made no assumptions on the signs of $a,b$ and $c$ I have no idea what I should do next.
Maybe I shouldn't care about what $\displaystyle \min\{(a-b)^2,(b-c)^2\}$ is equal to and I should continue my argument by dealing with $(a-b)^2$ and $(b-c)^2$ instead. I doubt that using the formula $\displaystyle \min\{x,y\}=\frac{x+y - |x-y|}{2}$ would simplify this any further. Any ideas on how to go further are appreciated.
|
Note that, since you can change $a,b,c$ to $-a,-b,-c$, you can assume that at least two of the $a,b,c$ are nonnegative. So, first start by supposing that two of your numbers are nonnegative, and after that order them. Therefore, without loss of generality, assume that $a\leq b\leq c$, with $0\leq b\leq c$. Then, for the first case, equivalently you want $$2a^2+2b^2-4ab\leq a^2+b^2+c^2\Rightarrow (a-2b)^2\leq c^2+3b^2.$$ But, $a-2b=a-b-b\leq -b\leq 0$, so $$|a-2b|=2b-a\leq c\Rightarrow (a-2b)^2\leq c^2,$$ so the first case is done. For the second case, $$2b^2+2c^2-4bc\leq a^2+b^2+c^2\Rightarrow(2b-c)^2\leq a^2+3b^2.$$ Now, if $2b-c\leq 0$, since $2b-c\geq a$, you have that $|2b-c|\leq |a|\Rightarrow (2b-c)^2\leq a^2$, and, if $2b-c\geq 0$, you have that $|2b-c|=2b-c\leq 2b-b=b\Rightarrow (2b-c)^2\leq b^2$. So, the second case is also done.
Edit: There is actually no need to consider that two of them are nonnegative. In your first case, equivalently you want $(a-2b)^2\leq c^2+3b^2$. So, if $2b-a\geq 0$, then, since $2b-a\leq c$, you have that $(a-2b)^2\leq c^2$, and, if $2b-a\leq 0$, then $|a-2b|=a-2b\leq-b$, so $|a-2b|^2\leq b^2$. You treat the other case similarly.
|
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Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that $a^2 + b^2 + c^2 \ge a + b + c$.
Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that
$a^2 + b^2 + c^2 \ge a + b + c$.
I'm supposed to prove this use AM-GM, but can't figure it out. Any hints?
|
since
$$a^2+1\ge 2a$$
so
$$a^2+b^2+c^2+3\ge a+b+c+(a+b+c)\ge a+b+c+3\sqrt[3]{abc}=a+b+c+3$$
so
$$a^2+b^2+c^2\ge a+b+c$$
|
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Finding value of equation without solving for a quadratic equation How do I go about solving this problem:
If $α$ and $β$ are the roots of $x^2+2x-3=0$, without solving the equation, find the values of $α^6 +β^6$.
In my thoughts: I commenced by expanding $(α +β)^6$, such that:
$$(α +β)^6 =α^6+6α^5β+15α^4β^2+20α^3β^3+15α^2β^4+6αβ^5+β^6$$ which when I reorganise:
$$(α +β)^6 =(α^6+β^6)+6α^5β+15α^4β^2+20α^3β^3+15α^2β^4+6αβ^5$$
when I isolate $(α^6+β^6)$ on one side:
$$(α^6+β^6) = (α +β)^6-6α^5β-15α^4β^2-20α^3β^3-15α^2β^4-6αβ^5$$
where does all this end for me to get a solution?
|
HINT:
$$a^6+b^6=(a^3)^2+(b^3)^2=(a^3+b^3)^2-2(ab)^3\text{ and } a^3+b^3=(a+b)^3-3ab(a+b)$$
or
$$a^6+b^6=(a^2)^3+(b^2)^3=(a^2+b^2)^3-3(ab)^2(a^2+b^2) \text{ and } a^2+b^2=(a+b)^2-2ab$$
|
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|
Integration with substitution I want to integrate (2x+1)/((x^2 - 6x + 14)^3)
I'm guessing you use substitution but im unsure what to substitute, is it best to make u = x^2 or X^2-6x or even x^2 - 6x + 14
I find that it makes turning the top of the fraction into terms of u very difficult
|
Using Trigonometric substitution
put $x-3=\sqrt5\tan\theta$ as $\displaystyle x^2-6x+14=(x-3)^2+(\sqrt5)^2$
$$\int \frac{2x+1}{(x^2-6x+14)^3}dx=\int\frac{2(\sqrt5\tan\theta+3)+1}{(5\sec^2\theta)^3}\sqrt5\sec^2\theta d\theta$$
$$=\frac1{25\sqrt5}\int(2\sqrt5\tan\theta+7)\cos^4\theta d\theta=\cdots$$
$$=\frac2{25}\int\sin\theta\cos^3\theta d\theta+\frac7{25\sqrt5}\int\cos^4\theta d\theta$$
The first integral can be managed by substituting $\cos\theta =u$
For the second one use Double-Angle Formulas , $\displaystyle \cos^4\theta=\frac{(2\cos^2\theta)^2}4=\frac{(1+\cos2\theta)^2}4=\frac14+\frac{\cos2\theta}2+\frac{\cos^22\theta}4$
Again, $\displaystyle2\cos^22\theta=1+\cos4\theta$
As the principal value $\arctan$ lies $\left[-\frac\pi2,\frac\pi2\right]$
$\displaystyle\cos\theta=+\frac1{\sqrt{1+\tan^2\theta}}=\frac{\sqrt5}{\sqrt{x^2-6x+14}}$
and $\displaystyle\sin\theta=\tan\theta\cdot\cos\theta$
|
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|
Triangle - coordinate geometry problem Let ABC be a triangle. Let BE and CF be internal angle bisectors of B and C
respectively with E on AC and F on AB. Suppose X is a point on the segment CF
such that AX is perpendicular to CF; and Y is a point on the segment BE such that AY perpendicular BE. Prove
that XY = $\frac{b+c-a}{2}$, where BC = a , CA = b, AB = c.
Please give a proof using coordinate geometry.
My solution:
I took a general triangle with two vertices lying on x-axis and one vertex lying on y-axis.
Found the equation of angle bisectors of B and C.
Found the foot of perpendicular from A on CF and BE as X and Y respectively.
Found XY but the answer was not matching . Please help.
Again I request, please give a proof using coordinate geometry only.
|
The following is a diagram. And $B$ is the origin and assume each point $P$ has coordinate $(p_1,p_2)$.
*
*Point $A,B,C$
From the length three edges. We can immediately write $B(0,0)$ and $C(a,0)$. Then solve $\displaystyle A(\frac{a^2-b^2+c^2}{2a},\frac{\sqrt{(b+c-a)(c+a-b)(a+b-c)(a+b+c)}}{2a})$ from $a_1^2+a_2^2=c^2,~(a_1-a)^2+a_2^2=b^2$.
*
*Line $BE,CF$
Noting that
$$\displaystyle k_{BE}=\tan\frac B2=\frac{\tan B}{1+\sqrt{1+\tan^2B}}=\frac{k_{AB}}{1+\sqrt{1+k_{AB}^2}}=\frac{a_2}{a_1+\sqrt{a_1^2+a_2^2}}=\frac{\sqrt{(b+c-a)(c+a-b)(a+b-c)(a+b+c)}}{(a+c)^2-b^2}$$
$$\displaystyle k_{CF}=\frac{-a_2}{(a-a_1)+\sqrt{(a-a_1)^2+(-a_2)^2}}=\frac{\sqrt{(b+c-a)(c+a-b)(a+b-c)(a+b+c)}}{(a+b)^2-c^2}$$
The equation of $BE$ is $k_{BE}x-y=0$ and $CF$, considering the line passes point $C$, $k_{CF}(x-a)-y=0$
*
*Point $X,Y$
Let $X$ be $\displaystyle (x_1,k_{CF}(x_1-a))$, $Y$ be $\displaystyle (y_1,k_{BE}y_1)$ since they are on lines $CF$, $BE$ respectively. According to orthogonal relation, we have
$\displaystyle AX\perp CF\Longleftrightarrow k_{AX}k_{CF}=\frac{a_2-k_{CF}(x_1-a)}{a_1-x_1}k_{CF}=-1\\
\displaystyle AY\perp BE\Longleftrightarrow k_{AY}k_{BE}=\frac{a_2-k_{BE}y_1}{a_1-y_1}k_{BE}=-1$
We obtain $\displaystyle X(\frac{3a^2-2ab-b^2+c^2}{4a},\frac{\sqrt{(b+c-a)(c+a-b)(a+b-c)(a+b+c)}}{4a}),Y(\frac{a^2-b^2+2ac+c^2}{4a},\frac{\sqrt{(b+c-a)(c+a-b)(a+b-c)(a+b+c)}}{4a})$ and hence the distance
$$d(X,Y)=y_1-x_1=\frac {b+c-a}2$$
|
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|
How we got $z\cdot(x+y)=x\cdot y$ This is from "Test of math at 10+2 level":
A vessel contains $x$ gallons of wine and another contains $y$ gallons of water. From each vessel $z$ gallons are taken out and transferred to the other. From the resulting mixture in each vessel, $z$ gallons are again taken out and transferred to the other. If after a second transfer, the quantity of wine in each vessel remains the same as it was after the first transfer, then show that $z\cdot(x+y)=x\cdot y.$
*My Try:*$$\text{Originally:}\begin{array}{|c|c|c|}
\text{Vessel:} & \text{1} & \text{2} \\ \hline
\text{Wine} & x & 0 \\ \hline
\text{Water} & 0 & y \\ \hline
\end{array}$$
$$\text{After first transfer:}\begin{array}{|c|c|c|}
\text{Vessel:} & \text{1} & \text{2} \\ \hline
\text{Wine} & x-z & z \\ \hline
\text{Water} & z & y-z \\ \hline
\end{array}$$
$$\text{After second transfer:}\begin{array}{|c|c|c|}
\text{Vessel:} & \text{1} & \text{2} \\ \hline
\text{Wine} & x-z-\left(\frac{x-z}{x}\right)\cdot z & z+\left(\frac{x-z}{x}\right)\cdot z \\ \hline
\text{Water} & z-\left(\frac{z}{x}\right)\cdot z & y-z+\left(\frac{z}{x}\right)\cdot z \\ \hline
\end{array}$$
And because, the quantity of wine in each vessel remains the same as it was after the first transfer, $$x-z=x-z-\left(\frac{x-z}{x}\right)\cdot z\tag1,$$$$z=z+\left(\frac{x-z}{x}\right)\cdot z\tag2,$$ which implies $x=z$. Where am I wrong?
|
The problem is symmetric in $x$ and $y$. That means that at any step, you should be able to reverse:
1) $x$ and $y$
2) Wine and Water
3) vessel 1 and vessel 2
And the result is the exact same relation.
You can also check at each step that wine in vessel1 plus water in vessel1 is x, and that wine in vessel2 plus water in vessel2 is y.
Notice that after the second transfer, the triple reverse doesn't give you back the same equation. The sums do not work out either. That tells you where a mistake is.
$$
\text{After second transfer:}\begin{array}{|c|c|c|}
\text{Vessel:} & \text{1} & \text{2} \\ \hline
\text{Wine} &
x-z-\underbrace{\left(\frac{x-z}{x}\right)\cdot z}_{\text{loss of wine}}
+ \underbrace{\frac z y z}_{\text{gain of wine}} &
z+\underbrace{\left(\frac{x-z}{x}\right)\cdot z}_{\text{gain of wine}} - \underbrace{\frac z y z}_{\text{loss of wine}} \\ \hline
\text{Water} &
z-\underbrace{\left(\frac{z}{x}\right)\cdot z}_{\text{Loss of water}} + \underbrace{\frac z y (y - z)}_{\text{Gain of water}} &
y-z+\underbrace{\left(\frac{z}{x}\right)\cdot z}_{\text{Gain of water}} - \underbrace{\frac z y (y - z)}_{\text{Loss of water}} \\ \hline
\end{array}
$$
From the wine amount being the same in the first vessel:
$$x - y = x-z-\left(\frac{x-z}{x}\right)\cdot z + \frac z y z \tag{Wine is same in vessel 1}$$
$$x - y = x-z-z+\frac{z^2} x + \frac {z^2} y$$
$$2z = y +\frac{z^2} x + \frac {z^2} y$$
$$2 = \frac y z +\frac{z} x + \frac {z} y \tag{R1}$$
As it turns out, $R1$ isn't necessary to solve the problem. Sometimes that happens, you get extra information. Let's look at the second vessel.
Edit: there is a mistake in the initial equation, it should be $x - z = $ rather than $x - y =$ so a corrected form of R1 is useful.
$$ z = z+\left(\frac{x-z}{x}\right)\cdot z - \frac z y z \tag{Wine is same in vessel 2}$$
$$ z = z+z-\frac{z^2}{x} - \frac {z^2} y$$
$$ z = \frac{z^2}{x} + \frac {z^2} y$$
$$ 1 = \frac{z}{x} + \frac {z} y$$
$$x \cdot y = z \cdot(x + y)$$
|
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|
Exponential map of 2x2 matrix I saw the following formula (in (7.25) of Lie Groups, Physics, and Geometry)
$$
\exp\left(
\begin{array}{c}
w & z\\
0 & 1\\
\end{array}
\right)
=
\left(
\begin{array}{ccc}
e^w & (e^w-1)z/w\\
0 & 1 \\
\end{array}
\right).
$$
Is this correct or some kind of typo? At least for $z=0$, it does not seem right?
|
The formula is indeed incorrect, including the case when $z=0$.
In the simple case that $z=0$, you have a diagonal matrix, and the exponential of a diagonal matrix $\exp\begin{pmatrix}a&0\\0&b\end{pmatrix}$ is the diagonal of the exponentials $\begin{pmatrix}e^a&0\\0&e^b\end{pmatrix}$. In your case it would give
$$\exp\begin{pmatrix}w&0\\0&1\end{pmatrix}=\begin{pmatrix}e^w&0\\0&e\end{pmatrix}.\tag1$$
In the general case that $z\neq0$, you need a general expression for the $n$th power of you matrix $M=\begin{pmatrix}w&z\\0&1\end{pmatrix}$, which is easy to see is of the form $M^n=\begin{pmatrix}w^n&u_n\\0&1\end{pmatrix}$, where $u_1=z$ and $u_{n+1}=wu_n+z$, so therefore $u_{n+1}=z\sum_{k=0}^nw^k=z\frac{w^{n+1}-1^{n+1}}{w-1}$.
Since the matrix exponential is defined as $\exp M=\sum_{n=0}^\infty\frac1{n!} M^n$, everything stays just as in (1) except the right-hand corner, which is given by
$$
\left(\exp M\right)_{12}=\sum_{n=0}^\infty\frac{u_n}{n!}
=\frac{z}{w-1}\left[\sum_{n=0}^\infty\frac{w^n}{n!}-\sum_{n=0}^\infty\frac{1^n}{n!}\right]
=z\frac{e^w-e}{w-1}.
$$
The final result is then $$\boxed{\exp\begin{pmatrix}w&z\\0&1\end{pmatrix}=\begin{pmatrix}e^w&z\frac{e^w-e}{w-1}\\0&e\end{pmatrix}.}\tag2$$
|
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|
Finding Length Of Segment
$AB=12$ , $BC=24$ , $CA=20$
$\dfrac{BF}{CG}=\dfrac{3}{5}$ , $\angle FAG=\angle CAG$
Find the length of $FG$.
It seems simple but is not easy for me.
|
let $FG=x,CG=y,AF=z,\dfrac{z}{20}=\dfrac{x}{y}, z^2=20^2+(x+y)^2-2*20*(x+y)cosC <1>$
$z^2-20^2=20^2\dfrac{x^2-y^2}{y^2},cosC=\dfrac{20^2+24^2-12^2}{2*20*24}=\dfrac{13}{15}$
so $<1>$ can be simplify as:
$\dfrac{20^2(x-y)}{y^2}=(x+y)-\dfrac{2*20*13}{15} <2>$
note $BF=\dfrac{3CG}{5}=\dfrac{3y}{5}, BF+FG+CG=24 \implies \dfrac{3y}{5}+x+y=24 \implies y=\dfrac{120-5x}{8}$
$<2>$ becomes: $\dfrac{75*x^3}{512}-\dfrac{2825*x^2}{192}-\dfrac{1575*x}{8}+1575=0$
there are $3$ real roots,but only $x=5.7054$ is available.
|
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|
$\text{Let }y=\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5-...}}}} $, what is the nearest value of $y^2 - y$? I found this question somewhere and have been unable to solve it. It is a modification of a very common algebra question.
$\text{Let }y=\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5-...}}}} $, what is the
nearest value of $y^2 - y$?
$\textbf {(1) } 1 \qquad \textbf {(2) } \sqrt5 \qquad \textbf {(3) } 4 \qquad \textbf {(4) } 2\sqrt5$
I worked the problem and got $$y^2 - y = 5+ \sqrt{5-\sqrt{5+\sqrt{5-...}}} -\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5-...}}}} \\ = 5+ \sqrt{5-y} - y = 5-y + \sqrt{5-y}$$
What do I do next?
|
You were on the right track. Put
$$x=\sqrt{5+\sqrt{5-\sqrt{5+\ldots}}}\implies x^2=5+\sqrt{5-\sqrt{5+\ldots}}\implies$$
$$x^4-10x^2+25=5-\sqrt{5+\sqrt{5-\ldots}}\implies$$
$$x^4-10x^2+20=-\sqrt{5+\sqrt{5-\ldots}}=-x\implies\;\ldots\ldots$$
|
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|
$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}$ approximation Is there any trick to evaluate this or this is an approximation, I mean I am not allowed to use calculator.
$$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}$$
|
$$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7...}}}}}=7^\frac{1}{2}\cdot7^\frac{1}{4}\cdot 7^\frac{1}{8}\cdots=7^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots}=7^{\frac{\frac{1}{2}}{1-\frac{1}{2}}}=7$$
|
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|
why is $ \sqrt x - \sqrt2 = \sqrt{(x-2)} $ when x tends to 2 $ \sqrt x - \sqrt2 = \sqrt{(x-2)} $ when x tends to $ 2^+ $
i have this problem
lim when x tends to $ 2^+ $
$ \frac{\sqrt x -\sqrt2 +\sqrt{x-2}}{\sqrt{x^2-4}} $
i know i must group $ \sqrt x - \sqrt 2 $ into $ \sqrt{x-2} $ only because that is true when x tends to 2 but then i separate the sums and simplify but the calculator gives me another result here's what i did
$ \frac{\sqrt{x-2} +\sqrt{x-2}}{\sqrt{x^2 -4}} $
$\frac{\sqrt{x-2}}{\sqrt{x^2 -4}} + \frac{\sqrt{x-2}}{\sqrt{x^2 -4}} $
$ \frac{\sqrt{x-2}}{\sqrt{x-2}\sqrt{x+2}} + \frac{\sqrt{x-2}}{\sqrt{x-2}\sqrt{x+2}} $
simplifying
$ \frac{1}{\sqrt{x+2}} + \frac{1}{\sqrt{x+2}} $
this is when x tends to $ 2^+ $
$ \frac{2}{\sqrt{4}} = 1 $
but the calculator gets me $ \frac{\sqrt{2}}{2} $
where did i go wrong?
|
The only way you can do is to make $\sqrt{x}-\sqrt{2}$ into $x-2$ by $$\frac{(\sqrt{x}-\sqrt{2})(\sqrt{x}+\sqrt{2})}{\sqrt{x}+\sqrt{2}}=\frac{x-2}{\sqrt{x}+\sqrt{2}}$$ So the main fraction can be changed to $$\sqrt{\frac{x-2}{x+2}}\times\frac{1}{\sqrt{x}+\sqrt{2}}+\frac{1}{\sqrt{x+2}},~~x\neq 2$$
|
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|
Let $a;b;c>0$. Prove : $\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}\geq \frac{b^{2}}{b+c}+\frac{c^{2}}{c+a}+\frac{a^{2}}{a+b}$ Let $a;b;c>0$. Prove :
$\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}\geq \frac{b^{2}}{b+c}+\frac{c^{2}}{c+a}+\frac{a^{2}}{a+b}$
P/s : Only use AM-GM and Cauchy-Schwarz
In the cases only use AM-GM and Cauchy-Schwarz; i don't have any thoughts about this problem !!
|
I don't see how to do it with AM/GM and CS, but here's an alternative method: by homogeneity, we may assume wlog that $a+b+c=1$, so we are to prove that
$$ \frac{a^2}{1-a} + \frac{b^2}{1-b} + \frac{c^2}{1-c}
\ge \frac{b^2}{1-a} + \frac{c^2}{1-b} + \frac{a^2}{1-c} $$
Since $x\mapsto x^2$ and $x\mapsto\frac1{1-x}$ are increasing functions for $x\in(0,1)$, the tuples $(a^2,b^2,c^2)$ and $(\frac1{1-a},\frac1{1-b},\frac1{1-c})$ are in the same order (e.g., if $a\ge b\ge c$ then $a^2\ge b^2\ge c^2$ and $\frac1{1-a}\ge\frac1{1-b}\ge\frac1{1-c}$), so the desired inequality follows from the rearrangement inequality.
|
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|
Solve the equation $\lfloor x^2\rfloor-3\lfloor x \rfloor +2=0$
Solve the equation
$$
\lfloor x^2\rfloor-3\lfloor x \rfloor +2=0
$$
where $\lfloor x\rfloor $ denotes floor function.
My Attempt:
Let $x = n+f$, where $n= \lfloor x \rfloor \in \mathbb{Z} $, $f=x-\lfloor x \rfloor = \{x\} $, and $0\leq f<1$. Then using $\lfloor n+f \rfloor = n$ gives us
$$
\begin{align}
\lfloor (n+f)^2 \rfloor -3 \lfloor n+f \rfloor +2 &= 0\\
n^2+\lfloor f^2+2nf \rfloor -3n+2 &= 0
\end{align}
$$
How can I solve the equation from here?
|
First note that $\lfloor x^2 \rfloor \ge \lfloor x \rfloor ^2$. Now, if we let $y = \lfloor x \rfloor$, then $y^2 -3y+2 \le 0$ and $y $ is an integer, so $y=1$, or $2$.
Then, let $x = y+f$, with $f \in (0,1)$. If $y=1$, then
$$\lfloor2f+f^2\rfloor = 0 \iff 2f+f^2 <1 \iff f < \sqrt{2}-1$$
If $y=2$, then
$$\lfloor4f+f^2\rfloor = 0 \iff 4f+f^2 <1 \iff f < \sqrt{5}-2$$
So the solutions for $x$ comprise exactly the set $[1,\sqrt2) \cup [2,\sqrt 5)$.
|
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|
How to show that $2730\mid n^{13}-n\;\;\forall n\in\mathbb{N}$
Show that $2730\mid n^{13}-n,\;\;\forall n\in\mathbb{N}$
I tried, $2730=13\cdot5\cdot7\cdot3\cdot2$
We have $13\mid n^{13}-n$, by Fermat's Little Theorem.
We have $2\mid n^{13}-n$, by if $n$ even then $n^{13}-n$ too is even; if $n$ is odd $n^{13}-n$ is odd.
And the numbers $5$ and $7$, how to proceed?
|
One Approach
If $k\mid n$, then $x^{k+1}-x\mid x^{n+1}-x$. Therefore,
$$
\begin{array}{}
13&\mid&n^{13}-n\\
7&\mid&n^7-n&\mid&n^{13}-n&\text{since }6\mid12\\
5&\mid&n^5-n&\mid&n^{13}-n&\text{since }4\mid12\\
3&\mid&n^3-n&\mid&n^{13}-n&\text{since }2\mid12\\
2&\mid&n^2-n&\mid&n^{13}-n&\text{since }1\mid12\\
\end{array}
$$
Since the factors are all relatively prime, we have that
$$
2730=2\cdot3\cdot5\cdot7\cdot 13\mid n^{13}-n
$$
Another Approach
Decomposing into a sum of combinatorial polynomials
$$
\begin{align}
n^{13}-n
&=2730\left[\vphantom{\binom{n}{2}}\right.3\binom{n}{2}+575\binom{n}{3}+22264\binom{n}{4}+330044\binom{n}{5}\\
&+2458368\binom{n}{6}+10551552\binom{n}{7}+28055808\binom{n}{8}+47786112\binom{n}{9}\\
&+52272000\binom{n}{10}+35544960\binom{n}{11}+13685760\binom{n}{12}+2280960\binom{n}{13}\left.\vphantom{\binom{n}{2}}\right]\\
\end{align}
$$
|
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|
Divisibility induction proof: $8\mid 7^n+3^n-2$ I'm stuck on the following proof by induction: $$8\mid3^n +7^n -2$$
And this is how far I've gotten:
$$\begin{aligned}3&\cdot3^n+7\cdot7^n-2\\3&(3^n+7^n-2)+7^n(7-3)-2\end{aligned}$$
Any help on where to go after this would be great!
|
You did a mistake:
$$3\cdot3^n+7\cdot7^n-2=3(3^n+7^n-2)+7^n(7-3)+4$$
Bow by $P(n)$ you know that $8|3(3^n+7^n-2)$. Thus, to complete the problem, you have to prove that
$$8|7^n\cdot 4+4$$
you can do this either by induction or by simply observing that $7^n+1$ is even, thus $4(7^n+1)$ is a multiple of $8$.
Added $P(n+1)$ contains $3\cdot3^n+7\cdot7^n-2$. Our goal is to create $3(3^n+7^n-2)$.
So, we start by adding and subtracting the missing terms:
$$3\cdot3^n+7\cdot7^n-2 =3 \cdot3^n +\left( 3 \cdot 7^n +3 \cdot (-2)- 3 \cdot 7^n -3 \cdot (-2) \right)+7\cdot7^n-2\\
=3 \left(\cdot3^n + 7^n -2\right)- 3 \cdot 7^n +6 +7\cdot7^n-2\\
=3 \left(\cdot3^n + 7^n -2\right)+7\cdot7^n- 3 \cdot 7^n +6 -2$$
An alternate way to figure this is the following:
You have
$$A=3\cdot3^n+7\cdot7^n-2$$
and need
$$B=3 \cdot \left(3^n + 7^n -2\right)$$
Then you can write
$$A=B+(A-B) \,.$$
Thus, what you need to add to $3 \cdot \left(3^n + 7^n -2\right)$ is
$$3\cdot3^n+7\cdot7^n-2-3 \cdot \left(3^n + 7^n -2\right)=3\cdot3^n+7\cdot7^n-2-3\cdot3^n-3\cdot7^n-6\\
=7^n(7-3)+4$$
|
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|
How do I find the Jordan normal form of a matrix with complex eigenvalues? I'm trying to obtain the Jordan normal form and the transformation matrix for the following matrix:
$A = \begin{pmatrix} 1 & 0 & 0 & 0 \\\ 1 & 0 & 0 & 1 \\\ 0 & 1 & 0 & 0 \\\ 0 & 0 & 1 & 0 \end{pmatrix}$
I've calculated its characteristic and minimum polynomials as $(λ - 1)^2(λ^2 + λ + 1)$, and thus the eigenvalues are $λ = 1$ (with an algebraic multiplicity of $2$) and $λ = \frac{-1 \pm i\sqrt{3}}{2}$.
An eigenvector for $λ = 1$ is $\begin{pmatrix} 0 \\\ 1 \\\ 1 \\\ 1 \end{pmatrix}$.
Since the minimum polynomial contains two identical factors, there must be at least a $2 x 2$ Jordan block associated with the eigenvalue $λ = 1$, and so the Jordan normal form must look something like the following:
$A = \begin{pmatrix} 1 & 1 & 0 & 0 \\\ 0 & 1 & 0 & 0 \\\ 0 & 0 & \frac{-1 + i\sqrt{3}}{2} & 0 \\\ 0 & 0 & 0 & \frac{-1 - i\sqrt{3}}{2} \end{pmatrix}$
However, I don't know how to derive a transformation matrix $P$ such that $PJ = AP$. How would I go about solving for $P$?
|
Well, the dimension of the eigenspace corresponding to $\lambda=1$ is two and therefore there are two Jordan blocks for eigenvalue 1. Further calculation is unnecessary as we know that distinct eigenvalues gives rise to distinct Jordan blocks. Therefore in this case the Jordan normal form is,
$$A = \begin{pmatrix} 1 & 0 & 0 & 0 \\\ 0 & 1 & 0 & 0 \\\ 0 & 0 & \frac{-1 + i\sqrt{3}}{2} & 0 \\\ 0 & 0 & 0 & \frac{-1 - i\sqrt{3}}{2} \end{pmatrix}$$
|
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|
proving $ \frac{a^2}{c-1}+\frac{b^2}{d-1}+\frac{c^2}{e-1}+\frac{d^2}{a-1}+\frac{e^2}{b-1}\geq 20$ If $a,b,c,d,e>1$, Then prove that $\displaystyle \frac{a^2}{c-1}+\frac{b^2}{d-1}+\frac{c^2}{e-1}+\frac{d^2}{a-1}+\frac{e^2}{b-1}\geq 20$
$\bf{My\; Try}::$ Using Cauchy- Schtwartz Inequality
$\displaystyle \frac{a^2}{c-1}+\frac{b^2}{d-1}+\frac{c^2}{e-1}+\frac{d^2}{a-1}+\frac{e^2}{b-1}\geq \frac{(a+b+c+d+e)^2}{(c+d+e+a+b)-5}$
Now I did not understand how can i solve it
Help Required
Thanks
|
Apply CS inequality to get
$$LHS > \frac{(a+b+c+d+e)^2}{a+b+c+d+e - 5} = M$$
Then let $y = a + b + c + d + e - 5$, then $y > 0$, and so
$$M = \frac{(y + 5)^2}{y}= 10 + y + \frac{25}{y} \ge 10 + 2\sqrt{25} \overset{\strut\text{AM-GM}}= 10 + 10 = 20$$
|
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|
Prove the following trigonometric identity. $$\frac{\sin x - \cos x +1}{\sin x + \cos x -1}=\frac{\sin x +1}{\cos x}$$
I tried substituting $\sin^2x+\cos^2x = 1$ but I cannot solve it.
|
$(\sin x- \cos x+1)\cos x = \sin x \cos x -\cos^2 x +\cos x$
$(\sin x+ \cos x-1)(\sin x +1) = \sin^2 x + \sin x \cos x +\cos x -1 = \sin x \cos x +\cos x +(\sin^2 x -1)= \sin x \cos x -\cos^2 x +\cos x$
|
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|
Get the last two digits of $16^{100}$ and $17^{100}$ I try to get the last two digits of $16^{100}$ and $17^{100}$
I started with:
$6^{1} = 6$
$6^{2} = 36$
$6^{3} = 216$ means last digit for $6$ is always $6$
That is why the last digit for $16^{100}$ is $6$
And for $17^{100}$ i calculated:
$7^{1} = 7$
$7^{2} = 49$
$7^{3} = 343$
$7^{4} = 2401$
$7^{5} = 16807$
And because $4*25 = 100$ or better said $100mod4 = 0$
The last digit for $17^{100}$ is $1$
But how do i get the penultimate digits? Thanks
|
As $(17,100)=1,$ and using Carmichael function $\displaystyle\lambda(100)=20$
$\displaystyle\implies17^{20}\equiv1\pmod{100}$
$16^{100}=(2^4)^{100}=2^{400}$
As $(2^{400},100)=4=2^2\ne1$ let us find $2^{400-2}\pmod{25}$
As $\displaystyle\lambda(25)=\phi(25)=20, 2^{20}\equiv1\pmod{25}$
and $\displaystyle400-2\equiv18\pmod{20}\implies 2^{400-2}\equiv2^{18}\pmod{25}$
Now, $\displaystyle2^9=512\equiv12\pmod{25}\implies 2^{18}=(2^9)^2\equiv12^2=144\equiv19\pmod{25}$
$\displaystyle\implies 2^{400-2}\equiv19\pmod{25}\ \ \ \ (1)$
Now as $\displaystyle a\equiv b\pmod m\implies a\cdot n\equiv b\cdot n\pmod{m\cdot n}$ for any integer $n,$
multiply either sides of $(1)$ by $4$
|
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|
Prove $1 + \cot^2\theta = \csc^2\theta$ Prove the following identity:
$$1 + \cot^2\theta = \csc^2\theta$$
This question is asked because I am curious to know the different ways of proving this identity depending on different characterizations of cotangent and cosecant.
|
Assuming the First Pythagorean Trigonometric Identity,
$$\sin^2\theta + \cos^2\theta = 1$$
Dividing by $\sin^2\theta$,
$$\Rightarrow \frac{\sin^2\theta}{\sin^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \frac{1}{\sin^2\theta}$$
$$\Rightarrow \left(\frac{\sin\theta}{\sin\theta}\right)^2 + \left(\frac{\cos\theta}{\sin\theta}\right)^2 = \left(\frac{1}{\sin\theta}\right)^2$$
Since $\cot\theta = \large \frac{1}{\tan\theta} = \large\frac{\cos\theta}{\sin\theta}$ and $\csc\theta = \large\frac{1}{\sin\theta}$,
$$\Rightarrow 1 + \cot^2\theta = \csc^2\theta .$$
|
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|
Where did I make a mistake
In a class, 2/3 of the students are girls. 1/4 of the girls are asleep, and 2/5 of the boys are asleep. What fraction of the class is awake.
This is what I did:
Let $x%$ be the total number of students.
Thus the number of girls is $\frac{2}{3}x$ and the number of boys is $\frac{1}{3}x$.
Since $1/4$ of the girls are asleep, the number of sleeping girls is $\frac{2}{3}\cdot \frac{1}{4}x\; =\; \frac{1}{6}x$
Similarly, the number of sleeping boys is $\frac{1}{3}\cdot \frac{2}{5}x\; =\; \frac{2}{15}x$
Since $\frac{1}{6}x$ is the number of sleeping girls, $\frac{5}{6}x$ is the number of girls that are awake. Similarly, since the number of sleeping boys is $\frac{2}{15}x$, the number of boys that are awake is $\frac{13}{15}x$.
This gives a total of $\frac{13}{15}x + \frac{5}{6}x$ sleeping students, but this simplifies to $\frac{17}{10}x$, and that is obviously wrong.
Where did I make a mistake?
|
Number of girls, who are awake is not $x- \dfrac{x}6$. It is $\color{red}{\dfrac{2x}{3}} - \dfrac{x}6$.
Similarly, number of boys, who are awake is not $x- \dfrac{2x}{15}$. It is $\color{blue}{\dfrac{x}{3}} - \dfrac{2x}{15}$.
|
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|
Improper Integral $\int\limits_0^{\frac12}(2x - 1)^6\log^2(2\sin\pi x)\,dx$ How can I find a closed form for the following integral
$$\int_0^{\frac12}(2x - 1)^6\log^2(2\sin\pi x)\,dx?$$
|
Here is a proof of Cleo's claim that
\begin{align}
\int^\frac{1}{2}_0(2x-1)^6\ln^2(2\sin(\pi x))\ {\rm d}x=\boxed{\displaystyle\frac{11\pi^2}{360}+\frac{60}{\pi^4}\zeta^2(3)-\frac{720}{\pi^6}\zeta(3)\zeta(5)}
\end{align}
It doesn't take much to show that the integral is equivalent to
$$\mathscr{J}=\frac{64}{\pi^7}\int^\frac{\pi}{2}_0x^6\ln^2(2\cos{x})\ {\rm d}x$$
Use the identity $\displaystyle\ln^2(2\cos{x})={\rm Re}\ln^2(1+e^{i2x})+x^2$ to get
\begin{align}
\color{red}{\Large{\mathscr{J}}}
&=\frac{\pi^2}{72}+\frac{64}{\pi^7}{\rm Re}\int^\frac{\pi}{2}_0x^6\ln^2(1+e^{i2x})\ {\rm d}x\\
&=\frac{\pi^2}{72}+\frac{1}{2\pi^7}{\rm Im}\int^{-1}_1\frac{\ln^{6}{z}\ln^2(1+z)}{z}{\rm d}z\\
&=\frac{\pi^2}{72}-\frac{3}{\pi^6}\int^1_0\frac{\ln^5{z}\ln^2(1-z)}{z}{\rm d}z+\frac{10}{\pi^4}\int^1_0\frac{\ln^3{z}\ln^2(1-z)}{z}{\rm d}z-\frac{3}{\pi^2}\int^1_0\frac{\ln{z}\ln^2(1-z)}{z}{\rm d}z\\
&=\frac{\pi^2}{72}+\frac{720}{\pi^6}\sum^\infty_{n=1}\frac{H_n}{(n+1)^7}-\frac{120}{\pi^4}\sum^\infty_{n=1}\frac{H_n}{(n+1)^5}+\frac{6}{\pi^2}\sum^\infty_{n=1}\frac{H_n}{(n+1)^3}\\
&=\frac{\pi^2}{72}+\frac{720}{\pi^6}\left(\frac{\pi^8}{7560}-\zeta(3)\zeta(5)\right)-\frac{120}{\pi^4}\left(\frac{\pi^6}{1260}-\frac{1}{2}\zeta^2(3)\right)+\frac{6}{\pi^2}\left(\frac{\pi^4}{360}\right)\\
&=\boxed{\displaystyle\color{red}{\Large{\frac{11\pi^2}{360}+\frac{60}{\pi^4}\zeta^2(3)-\frac{720}{\pi^6}\zeta(3)\zeta(5)}}}
\end{align}
Generalized Euler sum $\sum^\infty_{n=1}\frac{H_n}{n^q}$
|
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|
Calculate the inverse of a complex matrix I am trying to calculate the inverse of a given matrix but somewhere I have an error in my calculation that I cannot find
$$\begin{array}{ccc}
&&
\left(
\begin{array}{ccc|ccc}
1-i & 2 & 1+i & 1 & 0 & 0 \\
0 & 1+3i & 1+4i & 0 & 1 & 0 \\
4+i&6+2i&2+i & 0 & 0 & 1
\end{array}
\right)
\begin{array}{cl}
| & \cdot\, (1+i)\\
&\\
|& \cdot\, (8-2i)
\end{array}\\
&\leadsto&
\left(
\begin{array}{ccc|ccc}
2 & 2+2i & 2i & 1+i & 0 & 0 \\
0 & 1+3i & 1+4i & 0 & 1 & 0 \\
34&52+4i&18+4i & 0 & 0 & 8-2i
\end{array}
\right)
\begin{array}{cl}
&\\
&\\
|&-17\cdot I
\end{array}
\\
&\leadsto&
\left(
\begin{array}{ccc|ccc}
2 & 2+2i & 2i & 1+i & 0 & 0 \\
0 & 1+3i & 1+4i & 0 & 1 & 0 \\
0&18-30i&18-30i & -17-17i & 0 & 8-2i
\end{array}
\right)
\begin{array}{cl}
&\\
|&\cdot\,(1-3i)\\
|&\cdot\,(18+30i)
\end{array}\\
&\leadsto&
\left(
\begin{array}{ccc|ccc}
2 & 2+2i & 2i & 1+i & 0 & 0 \\
0 & 4 & 13+i & 0 & 1-3i & 0 \\
0&1224&1224& 204-816i & 0 & 204+204i
\end{array}
\right)
\begin{array}{cl}
&\\
&\\
|& :\,102
\end{array}
\\
&\leadsto&
\left(
\begin{array}{ccc|ccc}
2 & 2+2i & 2i & 1+i & 0 & 0 \\
0 & 4 & 13+i & 0 & 1-3i & 0 \\
0&12&12 & 2-8i & 0 & 2+2i
\end{array}
\right)
\begin{array}{cl}
&\\
&\\
|&-3\cdot II
\end{array}\\
&\leadsto&
\left(
\begin{array}{ccc|ccc}
2 & 2+2i & 2i & 1+i & 0 & 0 \\
0 & 4 & 13+i & 0 & 1-3i & 0 \\
0&0&-27-3i & 2-8i & -3+9i & 2+2i
\end{array}
\right)
\end{array}
$$
Now i would "devide" by $\left(-27-3i\right)$ and then for example the second entry of the third line on the right sight should be $-i$ (see Wolfram Alpha) but obviously I get a different result.
Can you spot any error in my calculations?
|
Here is an error that I found. In the step where you multiply row two by $1-3i$, the second row should become $(0,\ 10,\ 13+i)$. You have it as $(0,\ 4,\ 13+i)$.
|
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|
two short doubts about the inverse function in a point the function is $F(x,y,z)=(y^2+z^2, z^2+x^2, x^2+y^2)$ the point is (-1,1,-1)
task: find the local inverse of F in that point.
I have already proved that F is actually invertible there. then i solved the system:
$$
\begin{cases}
y^2+z^2=a\\
z^2+x^2=b\\
x^2+y^2=c
\end{cases}
$$
which gave me
$$
x=\pm \sqrt{\frac{c-a+b}{2}}\\
y=\pm \sqrt{\frac{a-b+c}{2}}\\
z=\pm \sqrt{\frac{b-c+a}{2}}\\
$$
The doubt I have is: how should I choose the signs? I thought of looking at the values of x,y,z in the point, which means x,z=negative y=positive
so i picked
$$
x= -\sqrt{\frac{c-a+b}{2}}\\
y= \sqrt{\frac{a-b+c}{2}}\\
z= -\sqrt{\frac{b-c+a}{2}}\\
$$
finding the inverse G:
$$ G(a,b,c)=(-\sqrt{\frac{c-a+b}{2}},\sqrt{\frac{a-b+c}{2}},-\sqrt{\frac{b-c+a}{2}})$$
the second doubt is:
should i leave it written that way or is it better:
$G(y^2+z^2, z^2+x^2, x^2+y^2)=(-x,y,-z)$ ? (with the same doubt about the signs?)
thank you in advance
|
First doubt: you are right
Secon doubt: the first way you have written is good, the other does not make sense.
|
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|
Inequality in triangle involving medians Let $ABC$ be a triangle and $M$ a point on $(BC)$, $N$ a point on $(CA)$ and $P$ a point on $(AB)$ such that the triangles $ABC$ and $MNP$ have the same centroid.
Does the following inequality hold: $$AM^2+BN^2+CP^2\ge AA'^2+BB'^2+CC'^2 ?$$ where $A',B'$ and $C'$ are the midpoints of $(BC), (AC)$ and $(AB)$.
EDIT: $AA'^2+BB'^2+CC'^2=3/4(AB^2+BC^2+CA^2)$ and from MB/MC=NC/NA=PA/PB follows that the triangles ABC and MNP have the same centroid, by Pappus Theorem.
|
If $MNP$ has the same centroid than $ABC$ then $$\frac{M+N+P}3=\frac{A+B+C}3$$ (if seen each point as a vector in $\mathbb R$.)
Also, as $M$ is in $BC$ then $$M=mB+(1-m)C$$ for some $0\le m\le1$. Analogous:
$$N=nC+(1-n)A,$$
$$P=pA+(1-p)B.$$
This means that:
$$N+M+P=(1+p-n)A+(1+m-p)B+(1+n-m)C$$
Where $A+B+C=N+M+P$, therefor
$$0=(p-n)A+(m-p)B+(n-m)C$$
Having $\lambda=p-n$ and $\mu=m-p$, then $n-m=-(\lambda+\mu)$, and we get
$$0=\lambda(A-C)+\mu(B-C).$$
Unless $ABC$ are colinear, this only happens when $\lambda=\mu=0$ which means $m=n=p$.
So:
\begin{align}
M &= mB+(1-m)C \\
M-A &= mB+C-mC-A \\
M-A &= m(B-C)+(C-A) \\
AM^2=(M-A)(M-A) &= m^2(B-C)(B-C)+m(B-C)(C-A)\\&\qquad+m(C-A)(B-C)+(C-A)(C-A)^2 \\
&= m^2CB^2-2mCB\,AC\cos C+AC^2
\end{align}
By law of cosins: $AB^2=AC^2+CB^2-2CB\,AC\cos C$, therefor $2CB\,AC\cos C = AC^2+CB^2-AB^2$, replacing we have:
\begin{align}
AM^2&=m^2CB^2-m(AC^2+CB^2-AB^2)+AC^2
\\&= mAB^2 + (m^2-m)CB^2 + (1-m)AC^2
\\\text{analogously}\\
BN^2&= mBC^2 + (m^2-m)AC^2 + (1-m)BA^2\\
CP^2&= mCA^2 + (m^2-m)BA^2 + (1-m)CB^2\\
AM^2+BN^2+CP^2 &= (1-m+m^2)(AB^2 + BC^2 + CA^2)
\end{align}
Where the minimum value of $1-m+m^2$ is where $m=\frac12$ as can be prove with a little calculus. And $m=\frac12$ is the value for $M=A',N=B',P=C'$.
|
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|
Evaluate the limit $\lim\limits_{n \to \infty} \frac{1}{1+n^2} +\frac{2}{2+n^2}+ \ldots +\frac{n}{n+n^2}$ Evaluate the limit
$$\lim_{n \to \infty} \dfrac{1}{1+n^2} +\dfrac{2}{2+n^2}+ \ldots+\dfrac{n}{n+n^2}$$
My approach :
If I divide numerator and denominator by $n^2$ I get :
$$\lim_{ n \to \infty} \dfrac{\frac{1}{n^2}}{\frac{1}{n^2} +1} +\dfrac{\frac{2}{n^2}}{\frac{2}{n^2} +1} + \ldots+ \dfrac{\frac{1}{n}}{\frac{1}{n} + 1}=0$$
but the answer is $\dfrac{1}{2}$ please suggest how to solve this.. thanks.
|
Answer
$$\lim_{n \to \infty} \dfrac{1}{1+n^2} +\dfrac{2}{2+n^2}+ \ldots+\dfrac{n}{n+n^2}$$
$$\lim_{n \to \infty} \dfrac{1}{n^2}(\dfrac{1}{1/n^2 + 1} +\dfrac{2}{2/n^2+1}+ \ldots+\dfrac{n}{n/n^2+1})$$
$$\lim_{n \to \infty} \dfrac{1}{n^2}\dfrac{n*(n+1)}{2}$$
$$\lim_{n \to \infty}\dfrac{1}{2}*(1+\dfrac{1}{n})$$
$$=\dfrac {1}{2}$$
|
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|
Need help simplifiying a rational expression There's a math question on an online test which asks the following
Multiply the following expression, and simplify:
$\frac{x^2-16y^2}{x} * \frac{x^2+4xy}{x-4y}$
But no matter how I try I keep getting the answer incorrect with a message telling me to simplify my answer. I can't seem to figure out how to simplify it enough to get it right.
$\frac{x^2-16y^2}{x} * \frac{x^2+4xy}{x-4y}$
equals, $\frac{x^4 + 4x^3y - 16x^2y^2 - 64xy^3}{x^2-4xy}$. I then factored x out of the numerator and denominator to get $\frac{x(x^3 + 4x^2y - 16xy^2 - 64y^3)}{x(x-4y)}$ and cancelled out the factored x's to get $\frac{x^3 + 4x^2y - 16xy^2 - 64y^3}{x-4y}$. I don't know what to do from here though.
I've managed to get enough marks to be able to pass it but since it's a readiness test I want to understand all of the material going in.
|
The cancellation make sense if instead of $x^2+16y^2$ we use $x^2-16y^2$
$$\frac{x^2-16y^2}{x}\frac{x^2+4xy}{x-4y}=\frac{x^2-(4y)^2}{x}\frac{x(x+4y)}{x-4y}=$$
$$=\frac{(x-4y)(x+4y)}{x}\frac{x(x+4y)}{x-4y}=(x+4y)^2$$
|
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|
Finding the maximum value of a function on an ellipse Let $x$ and $y$ be real numbers such that $x^2 + 9 y^2-4 x+6 y+4=0$. Find the maximum value of $\displaystyle \frac{4x-9y}{2}$.
My solution: the given function represents an ellipse. Rewriting it, we get $\displaystyle (x-2)^2 + 9(y+\frac{1}{3})^2=1$. To find the maximum of $\displaystyle \frac{4x-9y}{2}$, $x$ should be at its maximum and $y$ at its minimum. Solving the equations, I get that $x = 3$ and $\displaystyle y = -\frac{2}{3}$, but the answer i get is wrong. What am I doing wrong?
|
HINT : Letting $k=\frac{4x-9y}{2}$, this represents a line $y=\frac{4}{9}x-\frac{2}{9}k.$
Then, find the max $k$ such that the line is tangent to the ellipse.
You can set $y=\frac{4}{9}x-\frac{2}{9}k$ in the equation of the ellipse, then you'll get an equation of $x$. Then, find the condition that the equation has only one real solution. This gives you two $ks$. Then, the bigger $k$ is what you want.
|
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|
Question on Sets Given a finite nonempty set $S$ of integers, let $p(S)$ denote the
product of all integers in $S$ (for example, if $S = \{3, 11, 61\}$,
then $p(S) = 2013$). Determine the least positive integer $n$ such
that in any sequence $A_1, A_2, \cdots A_n$ of $n$ nonempty subsets of
$\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$, there exist sets $A_i
,A_j$ with $i\not=j$ and $p(A_i) = p(A_j)$.
I have made an attempt to answer the problem. Could someone let me know if this is correct?
The way I have the arranged the solution is to find the equivalence of factors that might satisfy $p(A_i) = p(A_j)$
If $A_i$ = {1, any combination of from 2 to 9} is equivalent to $A_j$ = { similar combination of from 2 to 9}
= 2*(${8\choose1} +{8\choose2} + \cdots + {8\choose8})$ = $2*(2^{8}-1)$ = 510
Next would be the equivalence of 2*3*Any combination of 4,5,7,8,9 = 6*Any combination of (4,5,7,8,9)
= 2*(${5\choose1} +{5\choose2} + \cdots + {5\choose5})$ = $2*(2^{5}-1)$ = 62
Next would be the equivalence of 2*4*Any combination of 3,5,6,7,9 = 8*Any combination of 3,5,6,7,9
= 2* (${5\choose1} +{5\choose2} + \cdots + {5\choose5})$ = $2*(2^{5}-1)$ = 62
Next would be the equivalence of 2*9*Any combination of 5,6,7,8 = 3*6*Any combination of (5,7,8,9)
= 2*(${4\choose1} +{4\choose2} + \cdots + {4\choose4})$ = $2*(2^{4}-1)$ = 30
Some double counting among equivalence would 2*3*4*(any combination of 5,7,9)
= (${3\choose0} +{3\choose1} +\cdots + {3\choose3})$ = $(2^{3})$ = 8
The least n would then be
= 510-62-62-30+8 = 364
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This is not an answer. The following examples show that $n$ has to be larger than $48$.
$$\{1\},\{2\},\cdots,\{9\}$$
$$\{2,5\},\{2,6\},\{2,7\},\{2,8\},\{2,9\},\{3,7\},\{3,8\},\{3,9\},\{4,7\},\{4,8\},\{4,9\},\{5,8\},\{5,9\},\{6,8\},\{6,9\},\{7,8\},\{7,9\},\{8,9\}$$
$$\{2,8,9\},\{3,8,9\},\cdots,\{7,8,9\}$$
$$\{2,7,8,9\},\{3,7,8,9\},\cdots,\{6,7,8,9\}$$
$$\{2,6,7,8,9\},\{3,6,7,8,9\},\cdots, \{5,6,7,8,9\}$$
$$\{2,5,6,7,8,9\},\{3,5,6,7,8,9\},\{4,5,6,7,8,9\}$$
$$\{2,4,5,6,7,8,9\},\{3,4,5,6,7,8,9\}$$
$$\{2,3,4,5,6,7,8,9\}$$
|
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|
Show the identity $\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}=-\frac{a-b}{a+b}\cdot\frac{b-c}{b+c}\cdot\frac{c-a}{c+a}$ I was solving an exercise, so I realized that the one easiest way to do it is using a "weird", but nice identity below. I've tried to found out it on internet but I've founded nothingness, and I wondering how to show easily this identity ? As a matter of the fact, it's a beautiful identity.
$$-\frac{a-b}{a+b}\cdot\frac{b-c}{b+c}\cdot\frac{c-a}{c+a}=\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}$$
The one way I think about this I'll let here :
$$\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}=\frac{(a-b)(b+c)(c+a)+(b-c)(c+a)(a+b)+(c-a)(c+b)(b+a)}{(a+b)(b+c)(c+a)}$$
Now, I'll take one of the term of RHS
$(a −b)(b +c)(c + a)\\ =(a −b)(b−c + 2c)(a −c + 2c)\\ =(a −b)(b −c)(a −c)+ 2c(a −b)(b−c + a −c + 2c) \\=(a −b)(b −c)(a −c)+ 2c(a −b)(a +b)$
Similarly, We'll do that with the remainder
$(b−c)(a+b)(c + a)+(c − a)(a +b)(b+ c)\\=(a +b)[(b −c)(c + a)+(c −a)(b +c)]\\=(a +b)(bc +ba −c^{2} −ca +cb+c^{2} −ab −ac)\\= (a +b)(2bc − 2ac)\\= −2c(a −b)(a +b)$
and we get :
$$\boxed{\frac{a-b}{a+b}\cdot\frac{b-c}{b+c}\cdot\frac{c-a}{c+a}}$$
So it's too many work to show this identity. I just want to know if there's a simple way to show that or I don't know. I'm questing that because I didn't find anything on internet.
|
Bringing all to one side and clearing denominators, we wish to prove zero
$$f(c) = (a\!-\!b)(b\!-\!c)(c\!-\!a)+(a\!-\!b)(b\!+\!c)(c\!+\!a)+(b\!-\!c)(a\!+\!b)(c\!+\!a) + (c\!-\!a)(a\!+\!b)(b\!+\!c)$$
Note $\ f(a) = (a\!-\!b)(b\!+\!a)2a\!+\!(b\!-\!a)(a\!+\!b)2a = 0.\ $ By symmetry $f(b) = 0.$
The coefficent of $c^2$ is $\ {-}(a\!-\!b) \!+\! (a\!-\!b) -(a\!+\!b)\!+\!(a\!+\!b) = 0.$
Thus $f(c)$ is a polynomial in $\,c\,$ of degree $\le 1$ with $\,2\,$ roots, so $f = 0$.
|
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|
Prove that $(\frac{bc+ac+ab}{a+b+c})^{a+b+c} \ge \sqrt{(bc)^a(ac)^b(ab)^c}$ Prove that $(\frac{bc+ac+ab}{a+b+c})^{a+b+c} \ge \sqrt{(bc)^a(ac)^b(ab)^c}$
I tried it to do using $AM \ge GM$ but don't know how to proceed.
Please help.
|
There is a nice solution of this using convexity. Jensen's inequality states that if $f$ is a concave function and $p_1, ... ,p_n$ is a distribution, then for any numbers $x_1, ... ,x_n$ we have:
$$
f\left(\sum_{i=1}^n{p_i x_i}\right) \geq \sum_{i=1}^n{p_i f(x_i)} .
$$
We shall use this and the fact that $\log$ is a concave function. Let us take the log of the left hand side and see where that takes us. We get:
$$
(a+b+c) \log\left(\frac{ab+bc+ac}{a+b+c}\right) =
$$
$$
(a+b+c) \log\left(\left(\frac{a}{a+b+c}\right) b +
\left(\frac{b}{a+b+c}\right) c + \left(\frac{c}{a+b+c}\right) a \right) \geq
$$
Now we use Jensen.
$$
(a+b+c) \left[ \frac{a}{a+b+c} \log(b) + \frac{c}{a+b+c} \log(a) +
\frac{b}{a+b+c} \log(c)\right] =
$$
$$
a\log(b)+b\log(c) + c\log(a) = \log\left(b^a c^b a^c\right).
$$
This gives us the inequality
$$
\left(\frac{ab+bc+ac}{a+b+c}\right)^{a+b+c} \geq b^a c^b a^c .
$$
In exactly the same way we can show that
$$
\left(\frac{ab+bc+ac}{a+b+c}\right)^{a+b+c} \geq a^b b^c c^a ,
$$
And so the inequality holds for the geometric mean of these two quantities.
$$
\left(\frac{ab+bc+ac}{a+b+c}\right)^{a+b+c} \geq \sqrt{b^a c^b a^c a^b b^c c^a} =
\sqrt{(ab)^c(ac)^b(bc)^a} .
$$
|
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|
Find $\sum_{n=0}^\infty\frac{2^n}{3^{2^{n-1}}+1}$ Find $\sum_{n=0}^\infty\frac{2^n}{3^{2^{n-1}}+1}$.
I already proved that it converges but I can't find the sum, Although I think it should be $1+\frac{1}{\sqrt3 +1}$, from calculations.
|
There's a very simple way to evaluate this sum. Let
$$a_n = \frac{2^n}{3^{2^{n-1}}-1}$$
Then
$$\begin{align}a_n-a_{n+1} &= \frac{2^n}{3^{2^{n-1}}-1} - \frac{2^{n+1}}{3^{2^{n}}-1} \\ &= \frac{2^n}{3^{2^{n-1}}-1} \left [1-\frac{2}{3^{2^{n-1}}+1} \right ]\\ &= \frac{2^n}{3^{2^{n-1}}-1} \frac{3^{2^{n-1}}-1}{3^{2^{n-1}}+1}\\ &= \frac{2^n}{3^{2^{n-1}}+1}\end{align}$$
Thus,
$$\sum_{n=0}^{\infty}\frac{2^n}{3^{2^{n-1}}+1} = a_0-a_{\infty}$$
where
$$a_0 = \frac{1}{\sqrt{3}-1}$$
and
$$a_{\infty} = \lim_{n\to\infty} \frac{2^n}{3^{2^{n-1}}-1} = 0$$
Therefore the sum is $\frac{1}{\sqrt{3}-1}$. Note that
$$1+\frac1{\sqrt{3}+1} = 1+\frac{\sqrt{3}-1}{2} = \frac{\sqrt{3}+1}{2} = \frac1{\sqrt{3}-1}$$
|
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|
Prove that $ 1^2 + 3^2 + ... + (2n-1)^2 = \displaystyle \frac{4n^3 -n}{3} $ Prove that $ 1^2 + 3^2 + ... + (2n-1)^2 = \displaystyle \frac{4n^3 -n}{3} $. I provide the answer below.
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We prove that $ 1^2 + 3^2 + ... + (2n-1)^2 = \displaystyle \frac{4n^3 -n}{3} $.
Base Case: Let $ n=1 $. Then
$ 1^2 + 3^2 + ... + (2n-1)^2 = (2*1-1)^2 = 1 = \displaystyle \frac{4*1^3 - 1}{3} = \frac{4n^3 -n}{3} $.
Inductive Step: Assume that $ 1^2 + 3^2 + ... + (2k-1)^2 = \displaystyle \frac{4k^3 -k}{3} $ for some $ k \in \textbf{N} $.
We show that $ 1^2 + 3^2 + ... + (2k-1)^2 + (2(k+1)-1)^2 = \displaystyle \frac{4(k+1)^3-(k+1)}{3} $.
Due to the assumption, it is sufficient to show that
$ \displaystyle \frac{4k^3-k}{3} + (2(k+1)-1)^2 = \displaystyle \frac{4(k+1)^3-(k+1)}{3} $.
Note the following expansions
$ 4(k+1)^3 - (k+1) = 4k^3 + 12k^2 + 12k + 4 - k - 1 = 4k^3 + 12k^2 + 11k + 3 $
and
$ (2(k+1)-1)^2 = (2k+1)^2 = 4k^2 +4k + 1 $.
So $ \displaystyle \frac{4k^3-k}{3} + (2(k+1)-1)^2 = \frac{4k^3 - k + 12k^2 + 12k + 3}{3} = \frac{4k^3 + 12k^2 + 11k + 3}{3} $
$ = \displaystyle \frac{4(k+1)^3-(k+1)}{3} $.
|
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|
Prove $ \cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right) + \cos\left(\frac{6\pi}{5}\right)+ \cos\left(\frac{8\pi}{5}\right) = -1 $ $$
\cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right) + \cos\left(\frac{6\pi}{5}\right)+ \cos\left(\frac{8\pi}{5}\right) = -1
$$
Wolframalpha shows that it is a correct identity, although I can't prove it.
I've tried to use the formula
$$ \cos(z) = \frac{e^ {iz} - e^ {iz}}{2} $$ but without any satisfying result.
This exercise is from chapter on series.
EDIT: I corrected a mistake in the formula I wanted to use.
|
Let $\omega = \exp\left(\frac{2\pi i}{5}\right) = \cos\left(\frac{2\pi}{5}\right) + i\sin\left(\frac{2\pi}{5}\right)$. Then $\omega$ is a fifth root of unity ($\omega^5 = 1$). Then $$1 + \omega + \omega^2 + \omega^3 + \omega^4 = 0.$$ By taking the real part of both sides (after applying De Moivre's Theorem), we obtain $$1 + \cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right) + \cos\left(\frac{6\pi}{5}\right) + \cos\left(\frac{8\pi}{5}\right) = 0.$$ Subtracting one from both sides gives the identity you're after.
|
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|
Smallest Prime-Factor of $4^{52} + 52^{2013} + 2013^{52}$ I would like to find out the smallest prime factor of $4^{52} + 52^{2013} + 2013^{52}$ by hand.
Thanks in advance
|
The given number is odd, so it cannot be divided by $2$.
In mod $3$, we have
$$\begin{align}4^{52}+52^{2013}+2013^{52}\equiv 1+1+0\equiv 2\end{align}$$
In mod $5$, we have
$$\begin{align}4^{52}+52^{2013}+2013^{52}&\equiv (-1)^{52}+2^{2013}+3^{52}\\&\equiv 1+2\cdot 2^{2012}+(-1)^{26}\\&\equiv 1+2\cdot {16}^{503}+1\\&\equiv 1+2+1\\&\equiv 4\end{align}$$
In mod $7$, we have
$$\begin{align}4^{52}+52^{2013}+2013^{52}&\equiv (-3)^{52}+3^{2013}+4^{52}\\&\equiv \color{red}{2^{26}}+3\cdot 2^{1006}+\color{red}{2^{26}}\\&\equiv \color{red}{2^{27}}+3\cdot 4^{503}\\&\equiv 2\cdot 4^{13}+3\cdot 4\cdot {2}^{251}\\&\equiv 2\cdot 4\cdot {2}^6+5\cdot 2^{251}\\& \equiv 1\cdot 2^6+5\cdot 2\cdot 4^{125}\\&\equiv 64+3\cdot 4\cdot 2^{62}\\&\equiv 1+5\cdot 4^{31}\\&\equiv 1+5\cdot 4\cdot 2^{15}\\&\equiv 1+6\cdot 2\cdot {4}^{7}\\&\equiv 1+5\cdot 4\cdot 2^{3}\\&\equiv 1+6\cdot 1\\&\equiv 0 \end{align}$$
Hence, the answer is $7$.
|
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|
How many integers less than $1000$ can be expressed in the form $\frac{(x + y + z)^2}{xyz}$? How many integers less than $1000$ can be expressed in the form $$\frac{(x + y + z)^2}{xyz}$$ where $x, y, z$ are positive integers?
|
You need to check if the question restricted to only positive integers $x, y, z$, otherwise it is quite trivial to get any desired number.
For zero, take $x = 2, y =-1, z = -1$.
For any positive number $N$, take $x = -N, y = 1, z = -1$.
For any negative number $-N$, take $x = N, y = 1, z = -1$. Clearly all negative numbers are less than $1000$, so you have an infinite number of them.
Added for the revised question with only positive $x, y, z$.
WLOG, we can assume $x \ge y \ge z$. From all such triples which satisfy $\displaystyle \frac{(x+y+z)^2}{xyz} = n$, where $n$ is a fixed positive integer, choose the triple $(a, b, c)$ with the minimum first component. Now we have the quadratic
$$x^2 +(2(b+c)- nbc)x + (b+c)^2 = 0 \tag{1}$$
for which $a$ is a root, and let the other root be $\alpha$. As $a \alpha = (b+c)^2$, $\alpha > 0$, and from our choice of triple, $\alpha \ge a \ge b \ge c$.
We have $(b+c)^2 = \alpha \,a \ge \alpha (b+c)/2 \implies \alpha \le 2(b+c)$.
Also we have $a^2 \le a\alpha = (b+c)^2$, so $a \le b+c$. Using these in the formula for sum of roots, $nbc - 2(b+c) = \alpha+a \le 3(b+c)$ gives $n bc \le 5(b+c)$. As $b, c$ are positive integers,
$$\implies n \le 5\left(\frac1c + \frac1b\right) \le 10 \tag{2}$$
This gives us a bound to restrict our search. It is easily observed that for $n = 10$, the bound $(2) \implies b = c = 1$, which leads to the quadratic $(1): x^2 -6x+4=0$ which does not have integer roots.
A quick search (well it was quick in Mathematica) yields solutions for $n = 1, 2, 3, 4, 5, 6, 8, 9$ - for e.g. in order, $(9, 9, 9), (9, 6, 3), (3, 3, 3), (4, 2, 2), (9, 5, 1), (3, 2, 1), (2, 1, 1), (1, 1, 1)$.
For $n=7$, we have $(2) \implies \frac75 \le \frac1b + \frac1c$, for which the only ordered positive integer solutions possible are $b=c=1$ and $b=2, c=1$. In the first case we have the quadratic $x^2 - 3x + 4=0$ which has no real roots, and the second case gives the quadratic $x^2 - 8x + 9 = 0$, which has no integer roots.
Hence there are only $8$ possible integers that can be expressed as desired.
|
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|
compute limit (no l'Hospital rule) I need to compute
$$\lim_{x\to 0} \frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^2x}.$$
I can not use the l'Hospital rule.
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As lcm $(2,3)=6, $ let $(\cos x)^\frac16=y$
$\displaystyle\implies (\cos x)^\frac12=y^3,(\cos x)^\frac13=y^2$ and $\displaystyle \cos x=y^6\implies\sin^2x=1-y^{12}$
$$\lim_{x\to0}\frac{(\cos x)^\frac12-(\cos x)^\frac13}{\sin^2x}=\lim_{y\to1}\frac{y^3-y^2}{1-y^{12}}=\lim_{y\to1}\frac{-y^2(1-y)}{(1-y)(1+y+\cdots+y^{10}+y^{11})}$$
We can cancel $1-y$ as $1-y\ne0$ as $y\to1$
Alternatively, $$\lim_{y\to1}\frac{y^3-y^2}{1-y^{12}}=\lim_{y\to1}(-y^2)\cdot\frac1{\lim_{y\to1}\frac{y^{12}-1}{y-1}}$$
Again, $$\lim_{y\to1}\frac{y^{12}-1}{y-1}=\lim_{y\to1}\frac{y^{12}-1^{12}}{y-1}=\frac{d(y^{12})}{dy}_{(\text{ at }y=1)}$$
|
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|
Determinant of a special $n\times n$ matrix Compute the determinant of the nun matrix:
$$
\begin{pmatrix}
2 & 1 & \ldots & 1 \\
1 & 2 & \ldots & 1\\
\vdots & \vdots & \ddots & \vdots\\
1 & 1 &\ldots & 2
\end{pmatrix}
$$
For $n=2$, I have$$
\begin{pmatrix}
2 & 1 \\
1 & 2
\end{pmatrix}
$$
Then $det = 3$.
For $n=3$, we have
$$
\begin{pmatrix}
2 & 1 & 1\\
1 & 2 & 1\\
1 & 1 & 2 \\
\end{pmatrix}
$$
Then $det = 4$.
For $n=4$ again we have
$$
\begin{pmatrix}
2 & 1 & 1 & 1 \\
1 & 2 & 1 & 1\\
1 & 1 & 2 & 1\\
1 & 1 & 1 & 2
\end{pmatrix}
$$
Then $det = 5$
How can I prove that the determinant of nun matrix is $n+1$.
|
Let $$v=(1,1,1,1...1)^T$$
Your matrix is $$
I + v v^T$$
This has $n-1$ eigenvalues equal to $1$ and one with value $n+1$. Hence the result.
|
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|
If $n$ is a positive integer greater than 1 such that $3n+1$ is perfect square, then show that $n+1$ is the sum of three perfect squares. If $n$ is a positive integer greater than 1 such that $3n+1$ is perfect square, then show that $n+1$ is the sum of three perfect squares.
My work:
$3n+1=x^2$
$3n+3=x^2+2$
$3(n+1)=x^2+2$
$(n+1)=\dfrac{x^2+2}{3}$
I have no clue what to do next. Please help!
|
We are given that $3n+1 = a^2 $.
We want to show that $n+1$ is the sum of 3 perfect squares.
Note that $a$ is not a multiple of 3.
If $ a \equiv 1 \pmod{3}$, then observe that $9n+9 = 3a^2 + 6 = (a-1)^2 + (a-1)^2 + (a+2)^2$, and hence
$$ n+1 = \left( \frac{a-1}{3} \right)^2 + \left( \frac{a-1}{3} \right)^2 + \left( \frac{a+2}{3} \right)^2. $$
If $ a \equiv 2 \pmod{3}$, then observe that $9n+9 = 3a^2 + 6 = (a+1)^2 + (a+1)^2 + (a-2)^2$, and hence
$$ n+1 = \left( \frac{a+1}{3} \right)^2 + \left( \frac{a+1}{3} \right)^2 + \left( \frac{a-2}{3} \right)^2 $$
Thus the result is true.
The motivation behind the solution is: We want to show that $n+1$ is the sum of 3 squares, and the only thing that we have to work with is $a^2$, and possibly things around it. Since $3a^2$ (the naive sum of 3 squares) is so close to $9n+9$, this suggests that we have some sort of wriggle room. Remembering that we have to account for the factor of 3 then greatly restricts our options.
As an extension, show that $n+3$ can also be written as the sum of 3 perfect squares, using a similar approach.
|
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|
How to prove this formula I want to know how to prove this:
$$
\sum_{k=1}^\infty \arctan\left( \frac{x}{k^2+a} \right)
= \pi \left\lfloor \frac{b}{\sqrt{2}} \right\rfloor
+ \arctan\left( \frac{x}{b^2} \right)
- \arctan\left[ \tanh\left( \frac{\pi}{\sqrt{2}} \frac{x}{b} \right)
\cot\left( \frac{\pi}{\sqrt{2}} b \right) \right]
$$
for $0 \le a, b = \sqrt{\sqrt{x^2 + a^2} - a}$.
At least in the special case $x=1$,$a=0$
Thank you ..
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Let $\;\displaystyle u + iv = \pi\sqrt{a+ix} = \frac{\pi}{\sqrt{2}}\left(\frac{x}{b} + ib\right)$ and follow nearly exactly the same step as in this answer, you get
$$\begin{align}
\sum_{k=1}\tan^{-1}\left(\frac{x}{k^2+a}\right)
&= \tan^{-1}\left(\frac{\tan v}{\tanh u}\right) - \tan^{-1}\left(\frac{v}{u}\right) + \pi N\\
&= \tan^{-1}\left(\frac{x}{b^2}\right) - \tan^{-1}\left(\tanh\left(\frac{\pi x}{\sqrt{2}{b}}\right)\cot\left(\frac{\pi b}{\sqrt{2}}\right)\right) + \pi N
\end{align}$$
for some integer $N$. The only thing left is to figure out what $N$ is...
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|
Convergence of $ 1+1+\frac{1}{3} + \frac{1}{2} + \frac{1}{3^2} + \frac{1}{2^2} + ...$ How does one use the comparison test to prove that $ 1+1+\frac{1}{3} + \frac{1}{2} + \frac{1}{3^2} + \frac{1}{2^2} + \frac{1}{3^3} + \frac{1}{2^3} + \; ...$ converges?
Is the following argument valid?
$\quad 1+1+\frac{1}{3} + \frac{1}{2} + \frac{1}{3^2} + \frac{1}{2^2} + \frac{1}{3^3} + \frac{1}{2^3} + \; ...$
$< 1 + 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^3} + \;... $
$= 2 + 1 + \frac{1}{2} + \frac{1}{2^2} + \;... $
$= \frac{2}{1-1/2} = 4 $
So the series is bounded above and since every term is strictly positive, the sequence of partial sums is monotone increasing and hence the series converges.
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$$S=1+1+\frac{1}{3} + \frac{1}{2} + \frac{1}{3^2} + \frac{1}{2^2} + \frac{1}{3^3} + \frac{1}{2^3} + \; ...$$
$$=\sum_{0\le r<\infty} \left(\frac12\right)^r+\sum_{0\le r<\infty} \left(\frac13\right)^r$$
Using summation formula of Infinite Geometric Series
$$S=\frac1{1-\frac12}+\frac1{1-\frac13}$$
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|
pre algebraic factoring with polynomials I really need help solving this particular problem.
$$\frac14x^2y(x-1)^3-\frac54xy(x-1)^2$$
I need help factoring this. It seems like I need to get rid of the fraction but I really just need a little boost.
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The given equation is
$$\dfrac{1}{4}x^2y(x - 1)^3 - \dfrac{5}{4}xy(x - 1)^2$$
Since the GCF (greatest common factor) is $\frac{1}{4}xy(x - 1)^2$, write the expression as
$$\begin{aligned}
\underbrace{\dfrac{1}{4}xy(x - 1)^2 \cdot x(x - 1)}_{\frac{1}{4}x^2y(x - 1)^3} - \underbrace{\dfrac{1}{4}xy(x - 1)^2 \cdot 5}_{\frac{5}{4}xy(x - 1)^2}&= \underbrace{\dfrac{1}{4}xy(x - 1)^2}_{\text{GCF}}\left(x(x - 1) - 5 \right)\\
&= \dfrac{1}{4}xy(x - 1)^2(x^2 - x - 5)
\end{aligned}$$
So the answer is
$$\dfrac{1}{4}xy(x - 1)^2(x^2 - x - 5)$$
Explanations for each step I have shown for your problem
*
*Extract the common factor out from each term you see, which is $\frac{1}{4}xy(x - 1)^2$ (Noticed that both terms have common terms $\frac{1}{4}$, $x$, $y$ and $(x - 1)^2$). This is called the GCF. Underneath each factored term, you see the original terms in the original equation.
*Factor out that term, leaving the uncommon factors (namely $x(x - 1)$ and $-5$) in the parentheses. (Multiple each term in the parentheses by $\frac{1}{4}xy(x - 1)^2$. You should get the same answer.)
*Simplify the expression in the parentheses.
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|
Show that $\sum_{i=1}^n \frac{1}{i^2} \le 2 - \frac{1}{n}$ So I am able to calculate the given problem and prove $P(K) \implies P(k + 1)$; it's been sometime since I did proofs
and I perform my steps I get what Wolfram Alpha shows as an alternate solution.
Any help is greatly appreciated
The problem is the following:
Show that
$$\sum_{i=1}^n \frac{1}{i^2} \le 2 - \frac{1}{n}$$
What I have:
$P(1)$:
$$\sum_{i=1}^n \frac{1}{i^2} \le 2 - \frac{1}{n}$$
Replace n with 1
$$\frac{1}{1} \le 2 - \frac{1}{1}$$
Conclusion
$$1 \le 1$$
Prove:
$$\sum_{i=1}^{k+1} \frac{1}{i^2} \le 2 - \frac{1}{k+1}$$
P(K) Assume
$$\sum_{i=1}^k \frac{1}{i^2} \le 2 - \frac{1}{k}$$
$P(K) \implies P(k + 1)$
Performed Steps:
Working the LHS to match RHS
$$2 - \frac{1}{k} + \frac{1}{(k+1)^2}$$
Edit: Fixed error on regrouping
$$2 - \left[\frac{1}{k} - \frac{1}{(k+1)^2}\right]$$
Work the fractions
$$2 - \left[\frac{1}{k} \frac{(k+1)^2}{(k+1)^2} - \frac{1}{(k+1)^2} \frac{k}{k} \right]$$
$$2 - \left[\frac{(k+1)^2 - k}{k(k+1)^2} \right]$$
$$2 - \left[\frac{k^2 + 2k + 1 - k}{k(k+1)^2} \right]$$
$$2 - \left[\frac{k^2 + k + 1}{k(k+1)^2} \right]$$
$$2 - \left[\frac{k(k+1) + 1}{k(k+1)^2} \right]$$
$$2 - \left[\frac{k(k+1)}{k(k+1)^2} + \frac{1}{k(k+1)^2} \right]$$
$$2 - \frac{1}{(k+1)} - \frac{1}{k(k+1)^2}$$
EDIT: I fixed my mistake of my regrouping and signs; had completely missed the regrouping.
This is the final step I got to. I am hung on where to go from here. The answers given have been really helpful and I'm happy with them. I'd just like to know the mistake I made or next step I should take.
$$\sum_{i=1}^{k+1} \frac{1}{i^2} \le 2 - \frac{1}{k+1}$$
Thanks for the help
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You can approximate the sum by an integral to obtain the inequality
$$
\sum_{i=1}^n\frac1{i^2}=1+\sum_{i=2}^n\frac1{i^2}\le1+\int_1^n\frac1{x^2}\mathrm dx=2-\frac1n.
$$
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|
Limit with geometric sequence I computed
$$\lim_{n \rightarrow \infty } \sqrt[n]{n^n+n^{n+1}+\cdots+n^{2n}} \cdot\left(1-\cos{\frac{3}{n}}\right)$$
$$=\lim_{n \rightarrow \infty } n^2 \sqrt[n]{n^{-n}+n^{-n+1}+\cdots+1} \cdot \left(1-\cos \frac{3}{n} \right) $$
$$=\lim_{n \rightarrow \infty } \sqrt[n]{n^{-n}+n^{-n+1}+ \cdots +1} \cdot \frac{1-\cos \frac{3}{n} }{\frac{9}{n^2}}\cdot 9=\frac{9}{2} $$
1) Is it correct to extract $n^{2n}$ from $\sqrt[n]{n^n+n^{n+1}+\cdots+n^{2n}}$ the way I did?
2)How do I show that $\lim_{n \rightarrow \infty } \sqrt[n]{n^{-n}+n^{-n+1}+\cdots+1}=1$?
I came up with this:
for geometric sequence the sum can be computed as $s_n=a_1\frac{q^n-1}{q-1}$, but I learned this in high school and if I recall correctly we used it for finite number of terms. So I don't know if I can use it here as $n\rightarrow \infty$.
$$\lim_{n \rightarrow \infty } \sqrt[n]{n^{-n}+n^{-n+1}+\cdots+1} = \lim_{n \rightarrow \infty } \sqrt[n]{n^{-n}\cdot \frac{n^n-1}{n-1}+1}$$
So I can estimate upper/lower bound:
$$\sqrt[n]{1} \le \sqrt[n]{n^{-n}\cdot \frac{n^n-1}{n-1}+1} \le \sqrt[n]n $$
Then:
$$\lim_{n \rightarrow \infty } \sqrt[n]{n^{-n}+n^{-n+1}+\cdots+1}=1$$
Thank you.
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We have these inequalities
$$n^2\le(n^n+\cdots+n^{2n})^{1/n}\le((n+1)n^{2n})^{1/n}=n^2(n+1)^{1/n}\sim_\infty n^2$$
and by Taylor series
$$1-\cos\left(\frac 3 n\right)\sim_\infty\frac 1 2\left(\frac{3}{ n}\right)^2=\frac{9}{2n^2}$$
hence
$$\lim_{n \rightarrow \infty } {\sqrt[n]{n^n+n^{n+1}+...+n^{2n}}\cdot(1-\cos{\frac{3}{n}})}=\lim_{n \rightarrow \infty }n^2\frac{9}{2n^2}=\frac{9}{2}$$
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|
Why does $2^{-n} = 5^n \times 10^{-n}$? If we look at the decimal equivilents of $2^{-n}$, we see they resemble $5^n$ with a decimal point in front of them:
$\begin{align}
2^{-1} &= 0.5 \\
2^{-2} &= 0.25 \\
2^{-3} &= 0.125 \\
2^{-4} &= 0.0625 \\
2^{-5} &= 0.03125 \\
...
\end{align}$
It looks like it's as simple as saying $2^{-n} = 5^n \times 10^{-n}$, and when we calculate that out, it's correct:
$\begin{align}
5^1 \times 10^{-1} &= 5 \times 0.1 = 0.5 \\
5^2 \times 10^{-2} &= 25 \times 0.01 = 0.25 \\
5^3 \times 10^{-3} &= 125 \times 0.001 = 0.125 \\
5^4 \times 10^{-4} &= 625 \times 0.0001 = 0.0625 \\
5^5 \times 10^{-5} &= 3125 \times 0.00001 = 0.03125 \\
...
\end{align}$
I calculated this out for $n = [0, 10]$ and it works out, but I have no idea how to prove it fully.
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You want to prove that
$$2^{-n} = 5^n \times 10^{-n}.$$
And
$$
5^n10^{-n} = \frac{5^n}{10^n} = \left(\frac{5}{10}\right)^n = \left(\frac{1}{2}\right)^n = \frac{1}{2^{n}} = 2^{-n}.
$$
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|
The beginning of a series in a limit using integrals Question:
evaluate:
$\lim \limits_{n \to \infty} n\left(\dfrac 1{(n+2)^2}+\dfrac 1{(n+3)^2}+1\ldots+\dfrac 1{(2n+1)^2}\right)$
What we did
We found that the limit (using the integral of $\displaystyle \int \limits_1^2\dfrac 1{x}\mathrm dx$) is $0.5$.
But that is true only if the sum also includes $\dfrac 1 {(n+1)^2} $ as well. The question is - can you ignore the fact that element is missing since it's "the beginning" of an endless series?
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If you plot the graph of the function : $x \mapsto \frac{1}{x^2}$ you can easily be convinced by $$\int_{k}^{k+1} \frac{\text{d}x}{x^2} \leqslant \frac{1}{k^2} \leqslant \int_{k-1}^{k} \frac{\text{d}x}{x^2} $$ for all $k \geqslant 2$. Therefore, if you sum this from $k=n+2$ to $k=2n+1$ we got
$$ \int_{n+2}^{2n+2} \frac{\text{d}x}{x^2} \leqslant \sum_{k=n+2}^{2n+1} \frac{1}{k^2} \leqslant \int_{n+1}^{2n+1} \frac{\text{d}x}{x^2}$$
which is also
$$ \frac{1}{n+2} - \frac{1}{2n+2}\leqslant \sum_{k=n+2}^{2n+1} \frac{1}{k^2} \leqslant \frac{1}{n+1} - \frac{1}{2n+1}$$
Multiplying by $n \geqslant 0$,
$$ \frac{n}{n+2} - \frac{n}{2n+2}\leqslant n\sum_{k=n+2}^{2n+1} \frac{1}{k^2} \leqslant \frac{n}{n+1} - \frac{n}{2n+1}$$
And since $$ \lim_{n\to+\infty}\frac{n}{n+2} - \frac{n}{2n+2} = \frac12$$
and $$ \lim_{n\to+\infty}\frac{n}{n+1} - \frac{n}{2n+1} = \frac12$$
The squeeze theorem tells us that
$$\lim_{n\to+\infty} n\sum_{k=n+2}^{2n+1} \frac{1}{k^2} = \frac 12$$
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.