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Find a and b such that $x^2 + 8x + 7 = (x + a)(x + b)$ for all real values of $x$. Find $a$ and $b$ such that $x^2 + 8x + 7 = (x + a)(x + b)$ for all real values of $x$.
That is what the question states, I think you should factor
Any Ideas on how to begin?
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Use $$x=\frac{-b\pm \sqrt{b^2 - 4ac}}{2a},\quad \textrm{where }ax^2+bx+c=0$$
which gives $x=-1 \textrm{ or } -7$ , so factorize will be $(x+1)(x+7)$.
|
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|
Find all integer solutions to the equation $\frac{xy}{z}+\frac{xz}{y}+\frac{zy}{x}=15$ I was going through one of my Mathematics books and I came to this problem:
Find all integer solutions to the equation: $\frac{xy}{z}+\frac{xz}{y}+\frac{zy}{x}=15$
I tried a few things to begin with but none went in the right direction, any suggestions?
Another thing, if we have the left side of the equation to be equal to 9, there are supposed to be 16 solutions in integers. What way can they be found?
|
*
*First, let us suppose that $x,y,z$ are positive integers satisfying the proposed equality. Then
$$ xyz(15-x-y-z)=\frac{1}{2}\left((xy-yz)^2+(yz-zx)^2+(zx-xy)^2\right)\ge0$$
So we have $x+y+z\le 15$.
*Now, note that squares modulo 5 are $\{0,-1,1\}$. Thus, if the sum of three squares modulo $5 $ is $0$ then either all three of them are multiples of $5$, (the case $0+0+0$) or just exactly one of them is a multiple of $5$ and the other two are equal to $+1$ and $-1$ modulo $5$, ( the case $0+1-1$). In both cases $5$ divides at least one of the three numbers.
The equality
$$(xy)^2+(yz)^2+(zx)^2=15xyz\equiv 0\mod 5\tag1$$
shows then that one of the numbers $xy,yz,zx$ is a multiple of $5$. So, we may and will suppose that $x=5a$. But then $(1)$ proves that $5$ must divide $yz$ and again we may suppose that $ y=5 b$.
*Now, $5+5+1\le 5a+5b+z\le15$ implies that $ a=b=1$, that is $x=y=5$ and the equation implies then that $z=5$.
We conclude that $(5,5,5)$ is the only solution consisting of positive integers.
*The general case. Suppose that $x=sx'$,$y=ty'$ and $z=uz'$ is a solution to the proposed equation where
$x',y',z'$ are positive integers, and $s,t,u$ are $\pm1$. It follows from
$$xyz\left(\frac{1}{x^2}+ \frac{1}{y^2}+\frac{1}{z^2}\right)=15$$
That
$$stu x'y'z'\left(\frac{1}{x'^2}+ \frac{1}{y'^2}+\frac{1}{z'^2}\right)=15$$
This implies that $stu=1$ and
$$ x'y'z'\left(\frac{1}{x'^2}+ \frac{1}{y'^2}+\frac{1}{z'^2}\right)=15$$
So, $x'=y'=z'=5$ and exactly two of $s,t,u$ are equal $-1$, or they are all equal to $+1$. This gives the following set of solutions $$\{(5,5,5),(-5,-5,5),(-5,5,-5),(5,-5,-5)\}.$$
|
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Combinations and Summations" $\sum_{x=0}^{n/k} \binom n{kx}$ How can be calculate this following summation efficiently (is there some shorter formula for it) ?
$$\sum_{x=0}^{\frac{n}{k}}{n\choose k\cdot x}$$
|
$$\sum_{k=0}^{\frac{n}{x}}\binom{n}{x\cdot k} = \binom{n}{x\cdot 1}+\binom{n}{x\cdot 2}+\binom{n}{x\cdot 3}+...+\underbrace{\binom{n}{x\cdot \frac{n}{x}}}_{\binom{n}{n}}$$
$\forall y \in \mathbb{N} : y = \frac{n}{x}$
$$\sum_{k=0}^{n}\binom{n}{x\cdot k} = \binom{n}{x\cdot 1}+\binom{n}{x\cdot 2}+\binom{n}{x\cdot 3}+\underbrace{\binom{n}{x\cdot y}}_{\binom{n}{n}}+\underbrace{\binom{n}{x\cdot (y+1)}+...+\binom{n}{x\cdot n}}_{0}$$
$$\sum_{k=0}^{\frac{n}{x}}\binom{n}{x\cdot k} =\sum_{k=0}^{n}\binom{n}{x\cdot k} $$
|
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Finding a function given a tangent to the curve I have got the following maths problem:
In the curve $y=x^2+ax+b$ where $a$ and $b$ are constant.
The tangent to the curve where $x=1$ is $2x+y=6$. Find the values of $a$ and $b$.
I am just unsure how I would go about answering this.
|
Since the involved function is just a quadratic function, you can try not to use any calculus.
The question can be translated as follows:
The equation $x^2+ax+b= 6-2x$ has a double root at $x=1$,
or even $f(x)=x^2+(a+2)x+(b-6)$ can be factorized as $f(x)=k(x-1)^2$, by factor theorem.
Edit Seems my answer above lead to confusions. Just let me add a few words to elaborate:
The equation $x^2+ax+b= 6-2x$ has a double root at $x=1$.
This line is easy. For a quadratic curves, a tangent line can only intersect the quadratic curve at one and only one point. Thus, after solving $(E)$: $\begin{cases} y=x^2+ax+b \\ 2x+y=6 \end{cases}$, we can obtain one and only one solution. By making $y$ as the subject for the line equation, be obtain $y=6-2x$. Substitute this to another equation, $x^2+ax+b= 6-2x$ has a double (or repeated) real root at $x=1$.
$f(x)=x^2+(a+2)x+(b-6)$ can be factorized as $f(x)=k(x-1)^2$, by factor theorem.
From the first part, we obtain $x^2+(a+2)x+(b-6)=0$ has only one root of multiplicity 2. Write $f(x)=x^2+(a+2)x+(b-6)$. By factor theorem, we can only write $f$ as multiple of $(x-1)^2$. Then $x^2+(a+2)x+(b-6) \equiv kx^2-2kx+k$.
Then we have $k=1, a=-4, b=7$.
No matter what method we used to find the values of $a$ and $b$, we just need to check that the above pair of value $a$ and $b$ is sufficient for $y=x^2+ax+b$ to have a tangent line $2x+y=6$.
Take $f(x)=x^2-4x+7$, $f'(x)=2x-4 \Longrightarrow f(1)=-2$. Then the slope of tangent line $= -2$. $f(1)=1-4+7=4$ The tangent line also passes through $(1,4)$. Then by point slope form, the equation of tangent line is $(-2)(x-1)=y-4 \iff 2x+y=6$.
|
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Find all values of $c$ such that solutions of $x^2 + cx + 6 = 0$ are rational. This is what I've done so far:
Using the formula, we get
$x = \frac{-c \pm \sqrt{c^2 - 24}}{2}$
For solutions to exist, $c \geq 5$.
For solutions to be rational, $c^2 - 24$ has to be a perfect square.
I'm kind of stuck now. I'd appreciate if someone can give me a hint or two on how to solve this question. Thanks!
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Observe that if $x \in \mathbb{Q}$ which is a solution of the equation $x^2 + cx + 6 = 0\text{ } (1)$, then solve for $c$ in terms of $x$ we obtain: $c = -\dfrac{x^2+6}{x}$. Thus if $c = - \dfrac{r^2+6}{r}$ with $r \in \mathbb{Q}\setminus \{0\}$, we prove the solutions of $(1)$ are rational numbers. We have: $x^2+ cx + 6 = 0 \implies x^2 - \left(\dfrac{r^2+6}{r}\right)x + 6 = 0\implies x^2r-r^2x - 6(x-r) = 0\implies (xr-6)(x-r)=0\implies x = r, \dfrac{6}{r}$ which are both rationals.
|
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Prime numbers of the form $(1\times11\times111\times1111\times...)-(1+11+111+1111+...)$ Let
$$R(1) = 1-1,$$
$$R(2) = (1\times11) - (1+11),$$
$$R(3) = (1\times11\times111) - (1+11+111),$$ and so on...
$$R(4)=1355297\quad\text{(a prime number!)}$$
$R(4)$ is the only prime I found of such form up to $R(200)$. Are there anymore primes of such form?
|
I wanted to post this as a comment but it was too long.
I don't know if it will help but consider to rewrite your function in this way:
$$R(n) = (1 \times 11 \times 111 \times \ldots) - (1 + 11 + 111 + 1111 + \ldots)$$
$$R(n) = \left((10^0) \times (10^0 + 10^1) \times (10^0 + 10^1 + 10^2) \times \ldots\right) - \left((10^0) + (10^1 + 10^0) + (10^2 + 10^1 + 10^0) + \ldots \right)$$
Now with a bit of maths you can check that
$$\left((10^0) \times (10^0 + 10^1) \times (10^0 + 10^1 + 10^2) \times \ldots\right) = \prod_{k = 1}^n \frac{10^k - 1}{9}$$
For what concerns the second part, it's more amusing. Indeed we have
$$\left((10^0) + (10^1 + 10^0) + (10^2 + 10^1 + 10^0) + \ldots \right)$$
But as we tend to $n$, we can easily check that there will be $n$ terms of $10^0$, $n-1$ terms of $10^1$, $n-2$ terms of $10^2$ and so on, which means
$$\left((10^0) + (10^1 + 10^0) + (10^2 + 10^1 + 10^0) + \ldots \right) = \sum_{k = 0}^{n-1} (n-k)10^k$$
Now with a bit of help (mathematica rules), the productory gives
$$\prod_{k = 1}^n \frac{10^k - 1}{9} = 9^{-k} {Q}_p (10, 10, k)$$
Where ${Q}_p (10, 10, k)$ is the so called q-Pochammer symbol, whose definition is
$${Q}_p (a, q, k) = \prod_{i = 0}^{k-1} (1 - aq^i)$$
Ref https://en.wikipedia.org/wiki/Q-Pochhammer_symbol
Whereas the sum gives
$$\sum_{k = 0}^{n-1} (n-k)10^k = \frac{1}{81}\left(10^{1+n} - 9n - 10\right)$$
Thus your function is
$$R(n) = 9^{-k} {Q}_p (10, 10, k) - \frac{1}{81}\left(10^{1+n} - 9n - 10\right)$$
As I said, I don't know if this helps, but it's a good way to check $R(n)$ quite fast.
Please If I made some mistake, make a comment. I liked this question and I immediately got involved in trying to find a suitable general form for $R(n)$.
|
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Does not algebraic multiplicity= geometric multiplicity $\Rightarrow$ the matrix is diagonalizable?
let $A=\left(\begin{array}{ccc} -5 & -1 & 6 \\ -2 & -5 & 8 \\ -1 & -1 & 1 \end{array}\right)$ and $B=\left(\begin{array}{ccc} -9 & 3 & -3 \\ -14 & 4 & -7 \\ -2 & 1 & -4 \end{array}\right)$ matrices above $\mathbb{C}$ are they similar?
So I started with finding the eigenvalues and geometric multiplicity.
For $A$ I got $f_{\lambda}(x)=(x+3)^3$ and $m_{\lambda}(x)=(x+3)^3$
For $B$ I got $f_{\lambda}(x)=(x+3)^3$ and $m_{\lambda}(x)=(x+3)^2$
So the jordan normal form of $A$ is $A=J_{1}(-3),J_{1}(-3),J_{1}(-3)$ or $A=\left(\begin{array}{ccc} -3 & 0 & 0 \\ 0 & -3 & 0 \\ 0 & 0 & -3 \end{array}\right)$
And the jordan normal form of $B$ is $B=J_{2}(-3),J_{1}(-3)$ or $B=\left(\begin{array}{ccc} -3 & 1 & 0 \\ 0 & -3 & 0 \\ 0 & 0 & -3 \end{array}\right)$
So $A$ is not similar to $B$ is it correct?
|
A well known theorem state that: two matrices are similar if and only if they have the same Jordan normal form . In your case, the Jordan normal forms are different, so they cannot be similar.
|
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Numbers $p-\sqrt{q}$ having regular egyptian fraction expansions? I remind that the greedy algorithm for egyptian fraction expansion for a positive number $x_0 <1$ goes like this:
$$x_0=\frac{1}{a_0}+\frac{1}{a_1}+\frac{1}{a_2}+\dots$$
$a_n$ are positive integers and are defined:
$$x_n-\frac{1}{a_n}>0$$
$$x_n-\frac{1}{a_n-1}<0$$
And $x_n$ are defined:
$$x_{n+1}=x_n-\frac{1}{a_n}$$
This expansion may rival the simple continued fractions in its importance to the number theory. It's unique for every number and terminating if and only if $x_0$ is rational.
I thought almost no regular GA EF expansions for 'simple' irrationals were known.
The only example I knew from this answer was:
$$\frac{3-\sqrt{5}}{2}=2-\phi=\frac{1}{3}+\frac{1}{21}+\frac{1}{987}+\dots$$
Where the denominators are $2^n$th Fibonacci numbers.
But it turns out that many numbers of the form $p-\sqrt{q}$ I tried have GA EF expansion with a regular pattern, described by $2^n$th terms of a linear second order recurrence.
I summarize the examples below:
$$3-2 \sqrt{2}=\frac{1}{6}+\frac{1}{204}+\frac{1}{235416}+\dots$$
Denominators are $2^n$th terms of the recurrence $A_n=34A_{n-1}-A_{n-2},~A_0=0,~A_1=6$. http://oeis.org/A082405
$$4-2 \sqrt{3}=\frac{1}{2}+\frac{1}{28}+\frac{1}{5432}+\dots$$
Denominators are $2^n$th terms of the recurrence $A_n=14A_{n-1}-A_{n-2},~A_0=0,~A_1=2$. http://oeis.org/A011944
$$3-\sqrt{7}=\frac{1}{3}+\frac{1}{48}+\frac{1}{12192}+\dots$$
Denominators are $2^n$th terms of the recurrence $A_n=16A_{n-1}-A_{n-2},~A_0=0,~A_1=3$. http://oeis.org/A001080
$$4-\frac{1}{3}-\sqrt{11}=\frac{1}{3}+\frac{1}{60}+\frac{1}{23880}+\dots$$
Denominators are $2^n$th terms of the recurrence $A_n=20A_{n-1}-A_{n-2},~A_0=0,~A_1=3$. http://oeis.org/A001084
Is there a general pattern here? How to prove these conjectures?
I know that there is a deep connection between recurrences of this kind and square roots (i.e. Fibonacci numbers and the Golden Ratio), but I don't know what the actual connection is in this case.
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Suppose $u>1$. Then the numbers $c_n := u^n - u^{-n}$
satisfy the linear recurrence
$$
c_{n+1} - (u+u^{-1}) c_n + c_{n-1} = 0.
$$
Moreover,
$$
\frac1{c_n} = \frac{u^n}{u^{2n}-1} = \frac1{u^n-1} - \frac1{u^{2n}-1}.
$$
Hence the sum of the reciprocals of the $2^m$-th terms can be evaluted
as a telescoping sum:
$$
\sum_{m=1}^\infty \frac1{c_{2^m}}
= \sum_{m=1}^\infty \frac1{u^{2^m}-1} - \frac1{u^{2^{m+1}}-1}
= \frac1{u^2-1}.
$$
Now suppose $u+u^{-1} = k > 2$. Then $c_1^2 + 4 = k^2$, so
$c_1 = \sqrt{k^2-4}$, and the $a_n := c_n / c_1$ are polynomials in $k$:
$$
(a_1, a_2, a_3, a_4, \ldots) = (1, k, k^2-1, k^3-2k, \ldots)
$$
and we have
$$
\sum_{m=1}^\infty \frac1{a_{2^m}} =
\sqrt{k^2-4} \sum_{m=1}^\infty \frac1{c_{2^m}} = \frac{\sqrt{k^2-4}}{u^2-1}
= \frac{k-\sqrt{k^2-4}}{2}.
$$
This accounts for all your examples:
$k=3$ gives the Fibonacci sum;
$k=4$ gives the expansion of $2-\sqrt{3}$ multiplied by $2$;
$k=6$ gives the expansion of $3-2\sqrt{2}$;
$k=16$ gives an expansion of $8-3\sqrt{7}$, from which the expansion of
$3-\sqrt{7}$ follows by adding $1$ and dividing by $3$; and
$k=20$ gives an expansion of $10 - 3\sqrt{11}$, from which the
expansion of $4 - \frac13 - \sqrt{11}$ follows by again adding $1$ and
dividing by $3$.
|
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Why are only $n-1$ convergents solution to Pell's Equations Given any equation of form $x^2-ny^2 = 1$, where $n$ is positive, non-square integer, let $[a_0; \overline{a_1, a_2, \dots a_k}]$ denote the continued fraction $\sqrt{n}$.
Let $\frac{p_i}{q_i} = [a_0; a_1, a_2, \dots a_{(k-1)i}]$. Why is it true that the solution of the Pell's equation exists in $\left\{\frac{p_1}{q_1}, \frac{p_2}{q_2}, \dots\right\}$.
More specifically, why doesn't a solution exist outside the set?
By a solution, I mean $p_i^2 - nq_i^2=1$.
|
If $x,y$ are positive integers satisfying $x^2 - ny^2 = 1$, then $\frac{x}{y}$ is a good rational approximation to $\sqrt{n}$. A really good rational approximation:
\begin{align}
&& x^2 - ny^2 &= 1 \\
&\iff& \frac{x^2}{y^2} - n &= \frac{1}{y^2} \\
&\iff& \frac{x}{y} - \sqrt{n} &= \frac{1}{y^2\bigl(\sqrt{n} + \frac{x}{y}\bigr)}
\end{align}
Since $\sqrt{n} + \frac{x}{y} > 2\sqrt{n} > 2$, it is such a good rational approximation to $\sqrt{n}$ that it must be one of the convergents. (Per a theorem of Legendre, if
$$\biggl\lvert \alpha - \frac{p}{q}\biggr\rvert < \frac{1}{2q^2}$$
then $\frac{p}{q}$ is a convergent of $\alpha$. Legendre's more precise theorem easily implies this corollary. Digging a little deeper, if $x^2 - ny^2 = m$ with $\lvert m\rvert < \sqrt{n}$ for positive integers $x,y$, then $\frac{x}{y}$ must be a convergent of $\sqrt{n}$. Note: If $m$ is not squarefree, then $\frac{x}{y}$ need not be in reduced form.)
And if we look at the (simple) continued fraction expansion of $\sqrt{n}$, we can write the $k^{\text{th}}$ complete quotient as
$$\xi_k = \frac{\sqrt{n} + b_k}{c_k}$$
with integers $0 \leqslant b_k < \sqrt{n}$ (and $b_k = 0$ only for $k = 0$) and $0 < c_k < 2\sqrt{n}$, where $b_{k+1}, c_{k+1}$ are given by
\begin{align}
b_{k+1} &= a_kc_k - b_k, \\
c_{k+1} &= \frac{n - b_{k+1}^2}{c_k},
\end{align}
as is seen from
$$\xi_{k+1} = \frac{1}{\xi_k - a_k} = \frac{c_k}{\sqrt{n} + b_k - a_kc_k} = \frac{c_k}{\sqrt{n} - b_{k+1}} = \frac{c_k(\sqrt{n} + b_{k+1})}{n - b_{k+1}^2}\,.$$
One inductively verifies that $c_k$ divides $n - b_{k+1}^2 = n - b_k^2 + 2a_kb_kc_k - a_k^2c_k^2$ and that $b_k,c_k$ satisfy the given inequalities. Further one checks that the period of the continued fraction is complete at the first $k > 0$ with $c_k = 1$.
Now it is an interesting fact that if we write the $r^{\text{th}}$ convergent (in reduced form) as
$$\frac{x_r}{y_r} = [a_0, a_1, \dotsc, a_r]$$
then
$$x_r^2 - n y_r^2 = (-1)^{r+1}\cdot c_{r+1}\,.$$
So the continued fraction of $\sqrt{n}$ has period length $k$, then the positive integer solutions to $x^2 - ny^2 = 1$ are given precisely by the numerator and denominator of the convergents with index $ik - 1$, where $i$ is a positive integer which must be even in case $k$ is odd. If $i$ and $k$ are both odd, we get the solutions to $x^2 - ny^2 = -1$. [So in your question you have a small mistake, it must be $ik - 1$ instead of $(k-1)i$.]
Summary: There are no solutions outside that set because being a solution implies giving such a good rational approximation to $\sqrt{n}$ that it must be a convergent, and the convergents with other indices (than one less than a multiple of the period length) have $\lvert x_r^2 - n y_r^2\rvert > 1$.
|
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Solve $ \begin{cases} z^{8}=|z|^{7} \\ z=\bar z e^{i\frac{\pi }{2}} \end{cases}$ someone can help me with the following system:
$$ \begin{cases}
z^{8}=|z|^{7} \\
z=\bar z e^{i\frac{\pi }{2}}
\end{cases}$$
I found the solution $z=0$
In fact from the second equation:
$$z=i\bar z$$
$$z-i\bar z=0$$
$$z(1-iz\bar z)=0$$
$(1-iz\bar z)=0$ and $z=0$
$$i|z|^{2}=1$$
$$z^8=iz\bar z$$
Now I'm stuck.
Thank you a lot!
|
Say $z=re^{i\theta}$ in the polar form.
Putting this value of $z$ in the 2 given equations, we get that
$$z^{8}=|z|^{7} \implies r^{8}e^{i\cdot 8\theta}=r^{7} \implies e^{i\cdot 8\theta}=\frac{1}{r}\tag1$$ $$z=\bar z e^{i\frac{\pi }{2}} \implies z^2=z\bar z e^{i\frac{\pi }{2}} \implies r^{2}e^{i\cdot 2\theta}=r^2e^{i\frac{\pi }{2}} \tag2$$
So $(2)$ means $\theta=n\pi+\frac{\pi}{4}$ and $(1)$ means $\cos 8\theta=\frac{1}{r}$ and $\sin 8\theta =0 $
Hence $\cos 8\cdot(n\pi+\frac{\pi}{4})=1=\frac{1}{r} \implies r=1$
So $z=1\cdot e^{i(n\pi+\frac{\pi}{4})}=(-1)^n\cdot\frac{1+i}{\sqrt2}$ and apart from this , we have the trivial $z=0$.
|
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|
If the chord $x+y=b$ of the curve ... If the chord $x+y=b$ of the curve $x^2+y^2-2ax-4a^2=0$ subtends a right angle at the origin, prove that: $b(b-a)=4a^2$
My Approach.
Given,
Equation of the chord,
$$x+y=b$$
$$\frac {x+y}{b}=1$$
Now,
Equation of the curve,
$$x^2+y^2-2ax-4a^2=0$$
$$x^2+y^2-2ax=4a^2$$
$$(b-y)^2+(b-x)^2-2ax=4a^2$$
I got stuck at here. Please help me to complete it.
|
First solve the equations simultaneously, and we arrive at the quadratic equation $$2x^2-2x(a+b)+b^2-4a^2=0$$
The roots satisfy $$x_1+x_2=a+b$$ and $$x_1x_2=\frac{b^2-4a^2}{2}$$
The perpendicularity condition can be written as $$x_1x_2+y_1y_2=0$$
$$\implies x_1x_2+(b-x_1)(b-x_2)=0$$
$$\implies 2x_1x_2-b(x_1+x_2)+b^2=0$$
Using the above results for the sum and product of the roots, the result follows immediately.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Solving $\sum_{k=0}^n 2^k \binom{2n-k}{n}=4^n$ using generating functions So I have to show this identity. $$\sum_{k=0}^n 2^k \binom{2n-k}{n}=4^n$$
I tried two techniques to do it. First one would be seeing that it's convolution of two sequences: $\langle 2^n \rangle _{n=0}^\infty$ and $\langle \binom{2n}{n} \rangle _{n=0}^\infty$
That gives me:
$$\lbrack \sum_{n=0}^\infty 2^n x^n \rbrack \times \lbrack \sum_{n=0}^\infty \binom{2n}{n} x^n \rbrack =$$
$$\frac{1}{1-2x} \ \frac{1}{\sqrt{1-4x}} =$$
$$\sum_{n=0}^\infty x^n \sum_{k=0}^n 2^k \binom{2n-k}{n-k}$$
So now we need to show that
$$\lbrack x^n \rbrack \frac{1}{1-2x} \frac{1}{\sqrt{1-4x}} = \frac{1}{1-4x}$$ But i'm unable to do it.
Second method would be snake oil method:
$$\sum_{n=0}^\infty x^n \sum_{k=0}^n 2^k \binom{2n-k}{n} =$$
$$\sum_{k=0}^\infty 2^k \sum_{n=k}^\infty x^n \binom{2n-k}{n} = \text{where l := n - k}$$
$$\sum_{k=0}^\infty 2^k x^k \sum_{l=0}^\infty x^l \binom{2l + k}{l+k} $$
But i'm stuck here too.
I mean i know that
$$\sum_{n=0}^\infty \binom{2n}{n} x^n = \frac{1}{\sqrt{1-4x}}$$
And if i multiply it by $x^k$ i get
$$\sum_{n=0}^\infty \binom{2n+2k}{n+k} x^{n+k} = \frac{x^k}{\sqrt{1-4x}}$$
And it's almost what i need to calculate... but yeah, almost.
I will really appreciate some help on this.
|
Here is a variation based upon the coefficient of method. We use the symbol $[z^n]$ to denote the coefficient of $z^p$ in a series. This way we can write e.g.
\begin{align*}
[z^p](1+z)^q=\binom{q}{p}\tag{1}
\end{align*}
We obtain
\begin{align*}
\color{blue}{\sum_{k=0}^n\binom{2n-k}{n}2^k}&=\sum_{k=0}^{n}[z^n](1+z)^{2n-k}2^k\tag{2}\\
&=[z^n](1+z)^{2n}\sum_{k=0}^n\left(\frac{2}{1+z}\right)^k\tag{3}\\
&=[z^n](1+z)^{2n}\frac{1-\left(\frac{2}{1+z}\right)^{n+1}}{1-\frac{2}{1+z}}\tag{4}\\
&=[z^n]\frac{(1+z)^{2n+1}-2^{n+1}(1+z)^n}{z-1}\\
&=[z^n]\frac{1}{z}\sum_{j=0}^\infty z^{-j}\left((1+z)^{2n+1}-2^{n+1}(1+z)^n\right)\tag{5}\\
&=\sum_{j=0}^n[z^{n+1+j}](1+z)^{2n+1}\tag{6}\\
&=\sum_{j=0}^n\binom{2n+1}{n+1+j}\tag{7}\\
&=\sum_{j=0}^n\binom{2n+1}{j}\tag{8}\\
&=\frac{1}{2}2^{2n+1}\\
&\,\,\color{blue}{=4^n}
\end{align*}
and the claim follows.
Comment:
*
*In (2) we apply the coefficient of operator according to (1).
*In (3) we do some rearrangements and use the linearity of the coefficient of operator.
*In (4) and (5) we use the geometric series expansion.
*In (6) we use the rule $[z^{p+q}]A(z)=[z^p]z^{-q}A(z)$ and restrict the upper limit of the sum with $n$ since the exponent $n+1+j\leq2n+1$. The second binomial $(1+z)^n$ does not contribute to $[z^{n+1+j}]$ and can be skipped.
*In (7) we select the coefficient of $z^{n-j}$.
*In (8) we replace $j$ with $n-j$ and apply the binomial identity $\binom{p}{q}=\binom{p}{q-p}$.
Hint: The Cauchy product in your approach is slightly different, since
\begin{align*}
\left(\sum_{n=0}^\infty 2^nx^n\right)\left(\sum_{n=0}^\infty\binom{2n}{n}x^n\right)
=\sum_{n=0}^\infty\left(\sum_{k=0}^n2^k\binom{2n\color{blue}{-2k}}{n-k}\right)x^n
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1916579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
How do we integrate $\dfrac{1}{1+x^4}$? I am still confused with this integral: $\int \dfrac{1}{1+x^4}dx$
There one solution was given by a pupil that we write it as $\dfrac{x^2+1-(x^2-1)}{2(1+x^4)}$
Then split into two terms and proceed by putting $x+{1\over x}=t$ and completing the square.
My question is can we approach this question using other technique, most obviously partial fraction? The roots are complex, so how do we deal with it? Also please tell me a general procedure for such questions! Thank You!
|
For partial fractions factor into quadratics (this is the hard part):
$$x^4+1=(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)$$
Do the standard decomposition method with undetermined cofficients:
$$\frac{1}{x^4+1}=\frac{1}{2\sqrt{2}}\left(\frac{x+\sqrt{2}}{x^2+\sqrt{2}x+1}
-\frac{x-\sqrt{2}}{x^2-\sqrt{2}x+1}
\right)$$
Integrate:
$$\int\frac{dx}{x^4+1}=
\frac{1}{4\sqrt{2}}\ln \left(\frac{x^2+\sqrt{2}x+1}{x^2-\sqrt{2}x+1}\right)
+\frac{1}{2\sqrt{2}}((\arctan(\sqrt{2}x+1)+\arctan(\sqrt{2}x-1))$$
I think its quite beautiful.
|
{
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"url": "https://math.stackexchange.com/questions/1916822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
prove that $a^2(p-q)(p-r)+b^2(q-p)(q-r)+c^2(r-p)(r-q)\ge 0$ If $a,b,c$
are the sides of a triangle and $p,q,r$ are positive real numbers then prove that $a^2(p-q)(p-r)+b^2(q-p)(q-r)+c^2(r-p)(r-q)\ge 0$
After modification. I have to prove $(a^2 p^2+b^2 q^2 + c^2 r^2) \ge pr(a^2+c^2-b^2)+qr(b^2+c^2-a^2)+pq(a^2+b^2-c^2)$
|
Let $x=q-r$, $y=r-p$, and $z=p-q$.
Then $x+y+z=0$, and the problem is equivalent to
$$a^2 yz+b^2 zx+ c^2 xy \leq 0.$$
If we substitute $x$ with $-y-z$, then the above inequality is equivalent to
$$(a^2-b^2-c^2)yz \leq b^2 z^2 + c^2 y^2.$$
Because $b-c<a<b+c$,
$$(a^2-b^2-c^2)yz \leq |a^2-b^2-c^2| |yz| < |2bc||yz| \leq b^2 z^2 + c^2 y^2.$$
The last inequality is AM-GM inequality. Hence the problem is proved.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1918278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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|
What is the maximal value of $2 \sin x - 7 \cos x$? What is the maximal value of $2 \sin x - 7 \cos x$?
How do I calculate this? Do I have to write out the $\sin$ and the $\cos$?
|
Hint: one can use formula
$$\sin x \cos y- \cos x \sin y = \sin(x-y),$$
and use such 'tricky' $y_0$, that
$$
\sin y_0 = \dfrac{7}{?}, \;\cos y_0 = \dfrac{2}{?}.
$$
Let $y_0$ is such that $\sin y_0 = \dfrac{7}{\sqrt{7^2+2^2}} = \dfrac{7}{\sqrt{53}}$, and $\cos y_0 = \dfrac{2}{\sqrt{53}}$, namely $y_0 = \arcsin(7/\sqrt{53})$.
Then $2\sin x - 7\cos x =\sqrt{53}\left(\dfrac{2}{\sqrt{53}}\sin x - \dfrac{7}{\sqrt{53}}\cos x\right)
= \sqrt{53}\sin(x-y_0)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1918362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Find the smallest positive integer that satisfies the system of congruences $N \equiv 2 \pmod{11}, N \equiv 3 \pmod{17}. $ Find the smallest positive integer that satisfies the system of congruences
\begin{align*}
N &\equiv 2 \pmod{11}, \\
N &\equiv 3 \pmod{17}.
\end{align*}
The only way I know to solve this problem is by listing it all out, and so far, it's not working. Is there a faster way? Thanks for posting a solution!
|
By the second constraint $N$ is a number of the form $17k+3$. We may now impose the first constraint:
$$ 17k+3\equiv 6k+3 \equiv 3(2k+1) \equiv 2\pmod{11} $$
leading to $2k+1\equiv 8\pmod{11}$, equivalent to $k\equiv 9\pmod{11}$. It follows that the smallest positive number fulfilling both constraints is given by
$$ \color{red}{N}=17\cdot 9+3 = \color{red}{156}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1918586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Defining the foci of "slanted" ellipse equation How to define the foci ($F_1,F_2$) coordinates of the slanted ellipse $x^2+4xy+9y^2=9$?
|
$x^2 + 4 x y + 9 y^2 = 9$ or
$\begin{pmatrix}x &y\end{pmatrix}\begin{pmatrix}1&2\\2&9\end{pmatrix}\begin{pmatrix}x \\y\end{pmatrix}=9$
Let's rotate by standard theory. The eigenvalues are $\lambda=5\pm 2\sqrt{5}$ and $P=\begin{pmatrix}\frac1{\sqrt{10+4\sqrt{5}}}&-\frac1{\sqrt{10-4\sqrt{5}}}\\\frac{2+\sqrt{5}}{\sqrt{10+4\sqrt{5}}}&\frac{-2+\sqrt{5}}{\sqrt{10-4\sqrt{5}}}\end{pmatrix}$ so that ${\bf x'}^tP^tAP{\bf x'}=\begin{pmatrix}x' &y'\end{pmatrix}\begin{pmatrix}5-2\sqrt{5}&0\\0&5+2\sqrt{5}\end{pmatrix}\begin{pmatrix}x' \\y'\end{pmatrix}=9$, that is $\frac{x'^2}{\frac{9}{5-2\sqrt{5}}}+\frac{y'^2}{\frac{9}{5+2\sqrt{5}}}=1$ which makes the major axis $a=\frac{3}{\sqrt{5-2\sqrt{5}}}$ and the minor axis $b=\frac{3}{\sqrt{5+2\sqrt{5}}}$. The distance from the center $(0,0)$ to the foci is $c=\sqrt{a^2-b^2} = \sqrt{ \frac{9}{5-2\sqrt{5}}-\frac{9}{5+2\sqrt{5}}}=\frac{6}{\sqrt[4]{5}}$. Now we just transform back $(0,\pm \frac{6}{\sqrt[4]{5}})$: $\begin{pmatrix}\frac1{\sqrt{10+4\sqrt{5}}}&-\frac1{\sqrt{10-4\sqrt{5}}}\\\frac{2+\sqrt{5}}{\sqrt{10+4\sqrt{5}}}&\frac{-2+\sqrt{5}}{\sqrt{10-4\sqrt{5}}}\end{pmatrix}\begin{pmatrix}0\\\pm\frac{6}{\sqrt[4]{5}}\end{pmatrix}=(\mp\frac{6}{\sqrt[4]{5}\sqrt{10-4\sqrt{5}}},\pm\frac{6(-2+\sqrt{5})}{\sqrt[4]{5}\sqrt{10-4\sqrt{5}}})\approx(\mp 3.905,\pm 0.922)$, which are the same as Jack's.
|
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"url": "https://math.stackexchange.com/questions/1921399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Expanding $(2^b-1)\cdot(1+2^b+2^{2b}+ \cdots + 2^{(a-1)b})$ Can someone please explain to me how the following works (primarily interested in an explanation of the second step when $2^b$ is expanded?
I understand how each series cancels out to equal $2^n-1$ at the end.
$$\begin{align*}
xy&=(2^b-1) \cdot (1 + 2^b + 2^{2b} + \cdots + 2^{(a-1)b})\\
&=2^b \cdot (1 + 2^b + 2^{2b} + \cdots + 2^{(a-1)b}) - (1 + 2^b + 2^{2b} + \cdots + 2^{(a-1)b})\\
&=(2^b + 2^{2b} + 2^{3b} + \cdots + 2^{ab})-(1+2^b+2^{2b}+ \cdots + 2^{(a-1)b})\\
&=2^{ab}-1\\
&=2^n-1
\end{align*}$$
|
$\begin{align*}
xy&=(2^b-1) \cdot (1 + 2^b + 2^{2b} + \cdots + 2^{(a-1)b})\\
\end{align*}$
Let $A = (1 + 2^b + 2^{2b} + \cdots + 2^{(a-1)b})$ so
$xy = (2^b-1)A = 2^b*A - A$. Let's continue...
$\begin{align*}
&=2^b \cdot (1 + 2^b + 2^{2b} + \cdots + 2^{(a-1)b}) - (1 + 2^b + 2^{2b} + \cdots + 2^{(a-1)b})\\
\end{align*}$
Note: $2^b \cdot (1 + 2^b + 2^{2b} + \cdots + 2^{(a-1)b})=2^b*1 + 2^b2^b + ...+2^b2^{(a-1)b}$
$= 2^b +2^{2b} + .... + 2^{ab}$. And that's pretty much all the detail I can think to provide. Continue...
$\begin{align*}
&=(2^b + 2^{2b} + 2^{3b} + \cdots + 2^{ab})-(1+2^b+2^{2b}+ \cdots + 2^{(a-1)b})\\
&=2^{ab}-1\\
&=2^n-1
\end{align*}$
|
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"url": "https://math.stackexchange.com/questions/1922044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Proving the identity $\frac{\cos^2\theta+\tan^2\theta-1}{\sin^2\theta}=\tan^2\theta$ I am stuck with this trigonometric identity. It appeared in a question paper of mine, and I am wondering whether there is a print error or something, because I absolutely have no idea how to solve this.
$$\frac{\cos^2\theta+\tan^2\theta-1}{\sin^2\theta}=\tan^2\theta$$
I would really appreciate some inputs!
|
Follow the usual process of manipulating only one side to make it look like the other side, and leave the other side completely alone. The LHS is much more complicated, so let's start there and try to make it look like the RHS.
\begin{align*}
\frac{\cos^2\theta + \tan^2\theta - 1}{\sin^2\theta}
&= \frac{\cos^2\theta - 1 + \tan^2\theta}{\sin^2\theta}\\[0.3cm]
&= \frac{\cos^2\theta - 1}{\sin^2\theta} + \frac{\tan^2\theta}{\sin^2\theta}\\[0.3cm]
&= \frac{\cos^2\theta - 1}{1 - \cos^2\theta} + \tan^2\theta \cdot \frac1{\sin^2\theta}\\[0.3cm]
&= \frac{-(1-\cos^2\theta)}{1 - \cos^2\theta} + \frac{\sin^2\theta}{\cos^2\theta} \cdot \frac1{\sin^2\theta}\\[0.3cm]
&= -1 + \frac1{\cos^2\theta}\\[0.3cm]
&= -1 + \sec^2\theta\\[0.3cm]
&= \tan^2\theta
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1922605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
How to calculate $\lim_{x \to 0} \left( \cos(\sin x) + \frac{x^2}{2} \right)^{\frac{1}{(e^{x^2} -1) \left( 1 + 2x - \sqrt{1 + 4x + 2x^2}\right)}}$? I need some help with computing this limit:
$$\lim_{x \to 0} \left( \cos(\sin x) + \frac{x^2}{2} \right)^{\frac{1}{(e^{x^2} -1) \left( 1 + 2x - \sqrt{1 + 4x + 2x^2}\right)}}$$
I'm guessing that I should do a Taylor expansion of $\cos(\sin x)$ in base, but what should I do with the stuff in the exponent?
|
What is the problem?
From first inspection:
*
*$\cos(\sin x) + \frac{x^2}{2} \to 1^-$
*$e^{x^2}-1 \to 0^+$
*$1+2x - \sqrt{ (1+2x)^2 - 2x^2 }\to 0^+$
So we have a case of $1^\infty$, which is indeterminate, and therefore we are going to need series expansions to solve this. Let us develop the following exponent term near $0$:
$$
\frac{ \log\left( \cos(\sin x) + \frac{x^2}{2} \right) }{ (e^{x^2}-1) (1+2x - \sqrt{1+4x+2x^2}) }
$$
Using series expansion
Choosing a precision
From rapid inspection we see that if we choose $\mathcal{O}(x^2)$, we will be left with $\cos(\sin x) + \frac{x^2}{2} \sim 1$, which will lead to $e^{\frac{\log 1}{0^+}}$ and it gets us nowhere. So we need at least $\mathcal{O}(x^3)$. Let us choose $\mathcal{O}(x^4)$ (you will see later why).
Numerator
We have:
\begin{align}
\sin x &\sim x - \frac{x^3}{6} + \mathcal{o}(x^4) \\
\cos(\sin x) &\sim \cos\left( x - \frac{x^3}{6} \right)
\sim 1 - \frac{\left( x - \frac{x^3}{6} \right)^2}{2} + \frac{\left( x - \frac{x^3}{6} \right)^4}{24} + \mathcal{o}(x^4)
\end{align}
Note that the last term is important because it contains $x^4/24$ which is within the precision that we set. Most other terms vanish and we have after simplification:
$$
\cos(\sin x) + \frac{x^2}{2} \sim 1 + \frac{5x^4}{24} + \mathcal{o}(x^4)
$$
Now you can see why $\mathcal{O}(x^3)$ was not sufficient. Finally, we can develop the logarithm further:
$$
\log\left(1 + \frac{5x^4}{24}\right) \sim \frac{5x^4}{24} + \mathcal{o}(x^4)
$$
Denominator
There are two factors in the denominator, one with an exponential, and one with a square-root, the relevant developments are:
\begin{align}
e^{x^2} - 1 &\sim x^2\left(1 + \frac{x^2}{2}\right) + \mathcal{o}(x^4) \\
[1+2x(2+x)]^{1/2} &\sim 1 + x(2+x) - \frac{x^2(2+x)^2}{2} + \mathcal{O}(x^3)\\
&\sim 1 + 2x - x^2 + \mathcal{O}(x^3)
\end{align}
Note that we don't need to develop that last term to full precision because we only need to be able to factor $x^2$ to eliminate the factor $x^4$ in the numerator; once we do that, all other terms $\mathcal{o}(x^2)$ will vanish in the limit. (The actual development is $1+2x-x^2+2x^3-9x^4/2$).
This leads to the following denominator, after simplification:
$$
x^4 \left( 1 + \frac{x^2}{2} \right) \left( 1 +\mathcal{O}(x) \right)
$$
Final limit
Putting all this together, we finally obtain the desired limit:
$$
\exp\left[\frac{5}{24 \left(1+\frac{x^2}{2}\right) \left(1+\mathcal{O}(x)\right)}\right] \to e^\frac{5}{24}
$$
|
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"url": "https://math.stackexchange.com/questions/1925856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
In any triangle $ ABC $ $\frac {sin(B-C)}{sin (B+C)}=\frac {b^2-c^2}{a^2} $. In any triangle $ ABC $ prove that $\frac {sin(B-C)}{sin (B+C)}=\frac {b^2-c^2}{a^2} $.
Please help. Thanks in advance.
|
Note that
\begin{align*}
\frac{\sin(B-C)}{\sin (B+C)}&=\frac{\sin B \cos C-\cos B \sin C}{\sin B \cos C+\cos B \sin C}\\
&=\frac{\frac{\sin B}{\sin C} \cos C-\cos B}{\frac{\sin B}{\sin C} \cos C+\cos B}\\
&=\frac{\frac{b}{c} \frac{c^2-a^2-b^2}{2ab}-\frac{b^2-a^2-c^2}{2ac}}{\frac{b}{c} \frac{c^2-a^2-b^2}{2ab}+\frac{b^2-a^2-c^2}{2ac}}\\
&=\frac{b^2-c^2}{a^2}.
\end{align*}
|
{
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|
How to integrate $ \int^a_0{\cfrac{dx}{x \ + \ \sqrt{a^2 \ - \ x^2}}} $? I am having a little problem with my maths homework. The problem is as follows:
\begin{equation}
\int^a_0{\cfrac{dx}{x \ + \ \sqrt{a^2 \ - \ x^2}}}
\end{equation}
I tried to do the following but got stuck halfway:
Let $\ \ x \ = asin\theta, \ hence, \ dx = acos\theta \ d\theta $
$
\int^a_0{\cfrac{dx}{x \ + \ \sqrt{a^2 \ - \ x^2}}}
$
$
= \int^\frac{\pi}{2}_0{\cfrac{acos\theta}{asin\theta \ + \ \sqrt{a^2 \ - \ a^2sin^2\theta}}}\ d\theta
$
$
= a \cdot \int^\frac{\pi}{2}_0{\cfrac{(cos\theta \ + \ sin\theta) \ + \ (cos\theta \ - \ sin\theta) - \ cos\theta}{asin\theta \ + \ \sqrt{a^2cos^2\theta}}}\ d\theta
$
$
= a \cdot \int^\frac{\pi}{2}_0{\cfrac{(cos\theta \ + \ sin\theta) \ + \ (cos\theta \ - \ sin\theta) - \ cos\theta}{asin\theta \ + \ acos\theta }}\ d\theta
$
$
= \int^\frac{\pi}{2}_0{\cfrac{(cos\theta \ + \ sin\theta) \ + \ (cos\theta \ - \ sin\theta) - \ cos\theta}{sin\theta \ + \ cos\theta }}\ d\theta \\ \\
$
$
= \int^\frac{\pi}{2}_0{\left(1 \ + \ \cfrac{(cos\theta \ - \ sin\theta)}{sin\theta \ + \ cos\theta } - \cfrac{cos\theta}{sin\theta \ + \ cos\theta}\right)}\ d\theta
$
Could someone please advise me how to solve this problem?
|
$$I=\int^a_0{\cfrac{dx}{x \ + \ \sqrt{a^2 \ - \ x^2}}}$$
Let $x^2+y^2=a^2\implies dx=-\frac{y}{\sqrt{a^2-y^2}}dy$
Thus:
$$I=\int^a_0{\cfrac{dy}{y \ + \ \sqrt{a^2 \ - \ y^2}}}\frac{y}{\sqrt{a^2-y^2}}$$
Adding these together,
$$2I=\int^a_0{\cfrac{dx}{x \ + \ \sqrt{a^2 \ - \ x^2}}}\left(1+\frac{x}{\sqrt{a^2-x^2}}\right)=\int_0^a\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\left(\frac{x}{a}\right)\vert_0^a=\frac{\pi}{2}$$
Thus, $I=\frac{\pi}{4}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Closed form of $k$-th term in the recurrence $D_j = 1 + z^2\left(1-\frac{1}{D_{j-1}}\right)$, where $D_0 = 1+z^2$ I have $D_0=1+z^2$ and $D_{j}=1+z^2(1-1/D_{j-1})$ for $j>0$. I want to write an expression for arbitrary $D_k$, and what I have is below.
$D_{k}=1+z^2\Bigg(1-\bigg(1+z^2\big(1-(1+z^2(1-(\cdots(1+z^2)^{-1}\cdots)^{-1})^{-1})^{-1}\big)^{-1}\bigg)^{-1}\Bigg)$
Is there better notation I can use to make this more precise/clear?
|
To expand upon my comment to @BarryCipra's answer ...
Multiplying numerators and denominators by $(1-z^2)$ gives
$$D_0 = \frac{1 - z^4}{1 - z^2} \qquad D_1 = \frac{1 - z^6}{1-z^4} \qquad D_2 = \frac{1-z^8}{1-z^6}$$
We can show that the pattern continues.
Assuming
$$D_j = \frac{1 - z^{2j+4}}{1-z^{2j+2}} \tag{$\star$}$$
we have
$$
\begin{align}
D_{j+1} &= 1 + z^2\left(\; 1 - \frac{1 - z^{2j+2}}{1 - z^{2j+4}} \;\right) \\[6pt]
&= 1 + z^2 \; \frac{z^{2j+2} - z^{2j+4}}{1-z^{2j+4}} \\[6pt]
&=\frac{1 - z^{2j+4} + z^{2j+4} -z^{2j+6}}{1-z^{2j+4}}\\[6pt]
&=\frac{1 -z^{2(j+1)+4}}{1-z^{2(j+1)+2}}
\end{align}$$
so that $(\star)$ holds for all $j$, by induction. $\square$
|
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|
Find all ordered pairs of real numbers $(x, y)$ such that $x^2y^2 + 2xy^2 + 5x^2 + 3y^2 + 10x + 5 = 0$. How can I find all ordered pairs of real numbers $(x, y)$ such that $x^2y^2 + 2xy^2 + 5x^2 + 3y^2 + 10x + 5 = 0$?
I don't see a pattern, I don't know how to proceed. Any answer is greatly appreciated.
|
$$ y^2 \; ( x^2 + 2x + 3) = -5(x^2 + 2 x + 1) $$
$$ y^2 = -5 \; \; \frac{(x+1)^2 }{2 + (x+1)^2} $$
The right hand side is never positive. It is only zero when $x=-1$
|
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|
$\lim_{x\to 0}\frac{x}{x+x^2e^{i\frac{1}{x}}} = 1$ $$\lim_{x\to 0}\frac{x}{x+x^2e^{i\frac{1}{x}}} = 1$$
I've trying seeing that:
$$e^{\frac{1}{x^2}i} = 1+\frac{i}{x^2}-\frac{1}{x^42!}-i\frac{1}{x^63!}+\frac{1}{x^84!}+\cdots\implies$$
$$x^2e^{\frac{1}{x^2}i} = x^2+\frac{i}{1}-\frac{1}{x^22!}-i\frac{1}{x^43!}+\frac{1}{x^64!}+\cdots$$
but I couldn't find a way to prove that the limit goes to $1$
|
We assume that $x \in \mathbb{R}$. One may write, as $x \to 0$,
$$
\left|\frac{x}{x+x^2e^{i\frac{1}{x}}} \right|=\frac{1}{\left|1+xe^{i\frac{1}{x}} \right|}\le \frac{1}{1-\left|xe^{i\frac{1}{x}} \right|}=\frac{1}{1-\left|x\right|} \to 1
$$ since
$$
||a|-|b||\le |a+b|.
$$
|
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|
Prove or disprove $\sum\limits_{cyc}\frac{x^4+y^4}{x+y}\le 3\frac{x^4+y^4+z^4}{x+y+z}$
Let $x,y,z\ge 0$. Prove or disprove
$$\dfrac{x^4+y^4}{x+y}+\dfrac{z^4+y^4}{z+y}+\dfrac{z^4+x^4}{x+z}\le 3\dfrac{x^4+y^4+z^4}{x+y+z}$$
This is what I tried. Without loss of generality, let $x+y+z=1$, then
$$\Longleftrightarrow \sum_{cyc}\dfrac{x^4+y^4}{x+y}\le 3(x^4+y^4+z^4)$$
and
$$(x^4+y^4)=(x+y)(x^3+y^3)-xy(x^2+y^2)=(x+y)(x^3+y^3)-xy(x+y)^2+2x^2y^2$$
which is equivalent to
$$\sum_{cyc}\left((x^3+y^3)-xy(x+y)+\dfrac{2x^2y^2}{x+y}\right)\le 3(x^4+y^4+z^4)$$
or
$$2\sum_{cyc}x^3+2\sum_{cyc}\dfrac{x^2y^2}{x+y}\le 3\sum_{cyc}(x^4+xy(x+y))$$
and now I'm stuck.
|
Let $x\geq y\geq z$. Hence,
$$y^2\left(\frac{3(x^4+y^4+z^4)}{x+y+z}-\sum_{cyc}\frac{x^4+y^4}{x+y}\right)=y^2\sum\limits_{cyc}\left(\frac{3(x^4+y^4)}{2(x+y+z)}-\sum_{cyc}\frac{x^4+y^4}{x+y}\right)=$$
$$=y^2\sum\limits_{cyc}\frac{(x^4+y^4)(x+y-2z)}{2(x+y+z)(x+y)}=y^2\sum\limits_{cyc}\frac{(x^4+y^4)(y-z-(z-x))}{2(x+y+z)(x+y)}=$$
$$=\frac{y^2}{2(x+y+z)}\sum\limits_{cyc}(x-y)\left(\frac{x^4+z^4}{x+z}-\frac{y^4+z^4}{y+z}\right)=$$
$$=\frac{y^2}{2(x+y+z)}\sum\limits_{cyc}\frac{(x-y)^2(xy(x^2+xy+y^2)+z(x+y)(x^2+y^2)-z^4)}{(x+z)(y+z)}\geq$$
$$\geq\frac{y^2}{2(x+y+z)}\sum\limits_{cyc}\frac{(x-y)^2(z(x^3+y^3)-z^4)}{(x+z)(y+z)}=$$
$$=\frac{y^2}{2(x+y+z)\prod\limits_{cyc}(x+y)}\sum\limits_{cyc}(x-y)^2z(x^3+y^3-z^3)(x+y)\geq$$
$$\geq\frac{y^2\left((x-z)^2y(x^3-y^3)(x+z)+(y-z)^2x(y^3-x^3)(y+z)\right)}{2(x+y+z)\prod\limits_{cyc}(x+y)}\geq$$
$$\geq\frac{x^2(y-z)^2y(x^3-y^3)(x+z)+y^2(y-z)^2x(y^3-x^3)(y+z)}{2(x+y+z)\prod\limits_{cyc}(x+y)}=$$
$$=\frac{xy(y-z)^2(x^3-y^3)(x(x+z)-y(y+z))}{2(x+y+z)\prod\limits_{cyc}(x+y)}=$$
$$=\frac{xy(y-z)^2(x-y)^2(x^2+xy+y^2)}{2\prod\limits_{cyc}(x+y)}\geq0.$$
Done!
|
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|
Given $abc=1$ and $0< c \leq b \leq1\leq a$, prove that $8(a+b+c)^2\le9(1+a^2)(1+b^2)(1+c^2)$ I can't make progress with proving this inequality.
I have tried opening the brackets and using $abc=1$ in order to obtain the following:
$$a^2+b^2+ \frac 1{a^2b^2}+18+9a^2b^2+\frac 9{a^2}+\frac9{b^2}\ge 16 \left (ab+\frac 1a+\frac 1b \right)$$
|
We begin by observing that if $a=b=c=1$, then the two expressions are equal to each other.
Expand everything. Use the equation $abc=1$ to get rid of compound terms. Gather terms with each variable separately. Use the preceding observation to factor each nominator.
$$\begin{align} & 9(1+a^2)(1+b^2)(1+c^2)-8(a+b+c)^2 \\ &=9\bigl(1+a^2+b^2+c^2+a^2b^2+b^2c^2+c^2a^2+a^2b^2c^2\bigr)-8\bigl(a^2+b^2+b^2+2ab+2bc+2ca\bigr) \\ &= 9\Bigl(1+a^2+b^2+c^2+{1\over c^2}+{1\over a^2}+{1\over b^2}+1\Bigr)-8\Bigl(a^2+b^2+c^2-{2\over c}-{2\over a}-{2\over b}\Bigr) \\ & ={a^4+6a^2-16a+9 \over a^2}+{b^4+6b^2-16b+9 \over b^2}+{c^4+6c^2-16c+9 \over c^2} \\ & ={(a-1)^2(a^2+2a+9) \over a^2}+{(b-1)^2(b^2+2b+9) \over b^2}+{(c-1)^2(c^2+2c+9) \over c^2}\end{align}$$
It is now obvious that this last expression is always $\geq 0$, and that it is $=0$ only in the case when $a=b=c=1$.
|
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|
How to find the square roots of $z = 5-12i$ I am asked the following question:
Find the square roots of $z = 5-12i$
I know that this problem can be easily solved by doing the following:
$$
z_k^2 = 5-12i\\
(a+bi)^2 = 5-12i\\
(a^2-b^2) + i(2ab) = 5-12i\\
\\
\begin{cases}
a^2 - b^2 = 5\\
2ab = -12
\end{cases} \quad \Rightarrow \quad z_1 = -3+2i \quad z_2 = 3-2i
$$
My question is: can the following method (below) be used to solve the problem above? Motivation for this question: if I were to find the cubic roots of the number given, I couldn't use the first method.
I will use this "other method" it in a different problem.
Find the square roots of $ z = 2i $
The method:
Since $ \rho = 2 $ and $ \theta = \frac{\pi}{2} $ we have to find a complex number such that
\begin{align*}
z_k^2 &= 2i\\
z_k^2 &= 2 \left( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right)\\
\rho_k^2 \left( \cos 2\theta_k + i \sin 2 \theta_k \right) &= 2 \left( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right)\\
\end{align*}
\begin{cases}
\rho^2 &= 2\\
2\theta_k &= \frac{\pi}{2} + 2k\pi
\end{cases}
\begin{cases}
\rho &= \sqrt{2}\\
\theta_k &= \frac{\pi}{4} + 2k\pi
\end{cases}
\begin{align*}
z_0 &= \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right) = 1+i\\
z_1 &= \sqrt{2} \left( \cos \frac{5\pi}{4} + i \sin \frac{5\pi}{4} \right) = -1-i
\end{align*}
Using this method for the first example seems like a dead end, specially because of the fact that the angle of $z$ is not as straightforward as the angle of the second example.
Thank you.
|
You can apply de Moivre's theorem (second example) on the first example without doubt. Though you need a calculator or a trigonometrical table handy to check the sin and cos of a certain angle which feels kinda painful, at least for me.
|
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|
Easier way to discover the area of a right triangle In the following right triangle: $y-x=5$ and the altitude to the hypotenuse is 12. Calculate its area.
I've managed to discover its area using the following method, but it ends up with a 4th degree equation to solve. Is there an easier way to solve the problem?
$ha=xy \Rightarrow 12 \sqrt{x^2+y^2} = xy$
Substitute $y=5+x$ and square both sides:
$144 (x^2 + (5+x)^2)=x^2 (5+x)^2 \Rightarrow x^4+10x^3-263x^2-1440x-3600=0$
Which only positive solution is $x=15$ and therefore $y=20$ and the area is $\frac{15 \cdot 20 }{2}=150$
Thanks in advance.
|
Square both sides of relationship $(x-y)=5$; then apply Pythagoras, giving
$\tag{1}x^2+y^2-2xy=25 \Leftrightarrow a^2-2xy=25$
Besides, the area $S$ of the triangle can be computed in two ways :
$\tag{2}S=\frac{xy}{2}=\frac{12a}{6}=6a$
Plugging the value of $a$ taken from (2) in (1), one gets a quadratic equation in variable $S$ which yields the looked for value for $S$. This equation is
$$\left(\frac{S}{6}\right)^2-4S=25 \ \ \Leftrightarrow \ \ S^2-144S-900=0$$
whose roots are $S=150$ (the unique answer) and $S=-6$, this one having no geometrical meaning.
|
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|
Solve 6th degree polynomial: $(x^2 - 3x - 4)(x^2 - 5x + 6)(x^2 + 2x) + 30$ I came across what seems to be a very difficult "solve for $x$" type of
problem, primarily because there should be $6$ real roots of this problem:
$$(x^2 - 3x - 4)(x^2 - 5x + 6)(x^2 + 2x) + 30 = 0.$$
My first step was to (tediously) expand this into the degree $6$ polynomial
that it is:
$$x^6-6 x^5+x^4+36 x^3-20 x^2-48 x+30 = 0.$$
I would imagine that the first step would be to simplify this into a set of
smaller roots, i.e.
$$(ax + b)(cx^5 + dx^4 + ex^3 + fx^2 + gx + h) = 0,$$
and then perform a similar factorization on the polynomial of degree $5$.
However, after trying hard to find the constant terms for this factorization,
I can never quite get the correct factorization. Any recommendations on
problems like these? I would really like to solve this without using Wolfram.
|
Let $P(x)$ denote the LHS. Note that the first summand factors nicely:
$$
P(x)=(x-2)(x-3)(x-4)x(x+1)(x+2)+30.
$$
From this it is clear that $P(x)$ is symmetric around $x=1$. Explicitly, setting $y=x-1$ we have
$$\begin{eqnarray*}
P(x)&=&(y-1)(y-2)(y-3)(y+1)(y+2)(y+3)+30\\
&=&(y^2-1)(y^2-4)(y^2-9)+30.
\end{eqnarray*}$$
Note that this is a cubic in $y^2$, so it is possible to solve for $y^2$ using the formula for roots of a cubic, and hence find the roots of $P$ in terms of radicals.
|
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|
Expressing $\sqrt[3]{7+5\sqrt{2}}$ in the form $x+y\sqrt{2}$ Express $\sqrt[3]{(7+5\sqrt{2})}$ in the form $x+y\sqrt{2}$ with $x$ and $y$ rational numbers.
I.e. Show that it is $1+\sqrt{2}$.
|
$$(1+\sqrt{2})^3=(1+2\sqrt{2}+2)(1+\sqrt{2})=(3+2\sqrt{2})(1+\sqrt{2})=3+3\sqrt{2}+2\sqrt{2}+4$$
$$\therefore (1+\sqrt{2})^3=7+5\sqrt{2}$$
$$\Rightarrow 1+\sqrt{2}=\sqrt[3]{7+5\sqrt{2}}$$
|
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|
Closed form solution for infinite summation Thomas, Bruckner & Bruckner, Elementary Real Analysis.
Prove that for all r > 1,
$$\frac{1}{r - 1} = \frac{1}{r+1} + \frac{2}{r^2 + 1} + \frac{4}{r^4 + 1} + \frac{8}{r^8 + 1} + \cdots$$
So far I have $$ \frac{1}{r-1} -\frac{1}{r+1} = \frac{2}{r^2 -1} $$
$$\sum_{n=1}^\infty \frac{2^n}{r^{2^n} + 1} = \sum_{n=1}^\infty \left(\frac{2^n}{r^{2^n}} - \frac{2^n}{r^{4^n} + r^{2^n}}\right)$$
|
Notice that
$$\frac 1 {r-1} - \left(\frac 1 {r+1} + \dots + \frac {2^n} {r^{2^n} + 1} \right) = \left( \frac 1 {r-1} - \frac 1 {r+1} \right) - \left(\frac 2 {r^2 + 1} + \dots + \frac {2^n} {r^{2^n} + 1} \right) = \\
\frac 2 {r^2 - 1} - \left(\frac 2 {r^2 + 1} + \dots + \frac {2^n} {r^{2^n} + 1} \right) = \frac 4 {r^4 - 1} - \left(\frac 4 {r^4 + 1} + \dots + \frac {2^n} {r^{2^n} + 1} \right) \dots = \\
\frac {2^n} {r^{2^n} - 1} - \frac {2^n} {r^{2^n} + 1} = \frac {2^{n+1}} {r^{2^{n+1}} - 1} .$$
Since $r>1$ we now have that
$$\lim _{n \to \infty} \frac {2^{n+1}} {r^{2^{n+1}} - 1} = \lim _{t \to \infty} \frac t {r^t - 1} = 0 ,$$
which means that
$$\lim _{n \to \infty} \left( \frac 1 {r-1} - \left(\frac 1 {r+1} + \dots + \frac {2^n} {r^{2^n} + 1} \right) \right) ,$$
i.e. that
$$\frac 1 {r-1} = \sum _{n=0} ^\infty \frac {2^n} {r^{2^n} + 1} .$$
|
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|
Solving the equation $(x^2-3x+1)^2=4x^2-12x+9$ I need to solve the following equation:
$$(x^2-3x+1)^2=4x^2-12x+9.$$
I think I need to bring everything to one side but I don't know anything else.
|
(x$^2$ - 3x + 1)$^2$ = (2x - 3)$^2$
(x$^2$ - 3x + 1)$^2$ - (2x - 3)$^2$ = 0
(x$^2$-3x+1-2x+3)(x$^2$-3x+1+2x-3) = 0
(x$^2$-5x+4)(x$^2$-x-2) = 0
(x-1)(x-4)(x+1)(x-2) = 0
Therefore, solution is x = 1, x = 4, x = -1 or x = 2
|
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|
Find the base numeric system Find the numeric base we are using if $ x = 4 $ and $ x = 7 $ are zeros of $ 5x^2 - 50x + 118. $
When I plug in $ x = 4 $ and $ x = 7, $ I receive $ -2 $ and $ 13, $ respectively, but how do I proceed from that?
|
Here's a different way to do it.
$f (x)=5x^2 - 50_bx +118_b =$
$5x (x-b)+118_b $
$f (4)=20 (4-b) +118_b$
$f (7)=35 (7-b)+118_b $
So $118_b= 20 (b-4)=35 (b-7) $
So $15b =7*35-80$
$b=11$.
Or we could note $35|118_b $ and $20|118_b $ so $140|118_b $ and $7|b-4$ and $4|b-7$. If $140 =118_b $ then $b=11$ is a good guess.
And here's a third way:
$5*4^2 - 4*(5b+0) + (b^2+b+8)=0$
$b^2 -19b +88=0$
$b= \frac {19 \pm \sqrt {19^2- 4*88}}{2}= 8;11$
$5*7^2 -7*5b+(b^2+b+8)=0$
$b^2 -34b+253=0$
$b= \frac {34 \pm \sqrt {34^2- 4*253}}{2}= 11;23$
So $b=11$
Still my favorite way is $5 (x-4)(x-7) =5x^2-55x+140= 5x^2 -50_bx + 118_b$ So $50_b=5b=55$ and $118_b =b^2+b+8=140$ so $b=11$.
|
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|
Supremum of rounded function Can you please help me to find and prove $\sup\{\sqrt{n}- \left\lfloor \sqrt{n}\right\rfloor : n\in N\}$?
I assume that it is something like $1$ or some other constant, but not quiet sure.
|
To prove that $$\sup \left\{ \sqrt{n} - \left\lfloor \sqrt{n} \right \rfloor : n \in \Bbb{N} \right\} = 1$$
(note that I use the more modern symbol $\lfloor x \rfloor$ in place of $[x]$ to mean the highest integer not exceeding $x$)
Step 1: For $n \in \Bbb{N}$ or indeed any real value of $n$, $\sqrt{n} - \left\lfloor \sqrt{n} \right \rfloor <1$. Because if $\sqrt{n} - \left\lfloor \sqrt{n} \right \rfloor = 1+\alpha$ with $\alpha > 0$, then $\left\lfloor \sqrt{n} \right \rfloor +1 = \sqrt{n}-\alpha < \sqrt{n}$ so $\left\lfloor \sqrt{n} \right \rfloor$ is not the greatest integer not exceeding $\sqrt{n}$ and this contradicts the definition of $\lfloor x \rfloor$.
Step 2: For any $\epsilon >0$, there exists some $n \in \Bbb{N}$ such that
$$
\sqrt{n} - \left\lfloor \sqrt{n} \right \rfloor > 1-\epsilon
$$
Proof: Choose any $k > \frac1\epsilon + 1$ and consider $n = k^2 -1$. Then
$\left\lfloor \sqrt{n} \right \rfloor = k-1$ and $\sqrt{n} = (k-1) + x$ for some $0<x<1$. Now of any $x : |x|\leq 1$,
$$
1+\frac{x}2-\frac{x^2}8 = \sqrt{1+x-\frac{x^3}{8}+\frac{x^4}{64}} < \sqrt{1+x}
$$
In particular, choose $x = \frac{2}{k-1}$ so that
$$
\sqrt{1+\frac2{k-1}}> 1+\frac1{k-1}-\frac1{2(k-1)^2} \\
(k-1)\sqrt{1+\frac2{k-1}}> (k-1)\left(1+\frac1{k-1}-\frac1{2(k-1)^2} \right)\\
\sqrt{k^2-2k+1+2(k-1)}> k-1+1-\frac1{2(k-1)} \\
\sqrt{k^2-1)}> k-\frac1{2(k-1)} > k-\frac1{2/\epsilon} = k-\frac{\epsilon}{2} > k-\epsilon
$$
So for our arbitrary $\epsilon$ we have demonstrated an $n$ such that
$$\left\{ \sqrt{n} - \left\lfloor \sqrt{n} \right \rfloor \right\}> 1-\epsilon$$
The combination of steps 1 and 2 establishes that $1$ is the desired suprenum.
|
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|
If $f(x+1) = x^2 + 3x +5$, then find $f(x)$ A challenge problem from precalc class. I don't know what to do with the one from $f(x+1)$, it doesn't factor well, I could do completing the square but then how do I find $f(x)$? Just stuck on this.
|
Just to be different.
$$f(x+1) = x^2 + 3x + 5$$
\begin{align}
f(x) - f(x+1)
&= f((x-1)+1) - f(x) \\
&= (x-1)^2 + 3(x-1) + 5) - (x^2 + 3x + 5) \\
&= ((x-1)^2 - x^2) + (3(x-1) - 3x) + (5-5) \\
&= (x-1-x)(x-1+x) - 3\\
&= (-1)(2x-1) - 3 \\
&= -2x - 2
\end{align}
So $f(x) = f(x+1) - 2x - 2 = x^2 + x + 3$
proof by induction
It is not unreasonable to suppose that $f(x) = x^2 + ax + b$ for some real numbers $a$ and $b$. We can use inductive reasoning to prove that this is the case and to find the values of $a$ and $b$ at the same time.
Our hypothesis will be $f(x) = x^2 + ax + b$ for some real numbers $a$ and $b$.
$f(0) = f(-1+1) = (-1)^2 + 3(-1) + 5 = 3$
By our hypothesis, $f(0) = b$. Hence $b = 3$
So our hypothesis is now $f(x) = x^2 + ax + 3$
Again, by our hypothesis, $f(x+1) = (x+1)^2 + a(x+1) + 3$
Hence
\begin{align}
x^2 + 3x + 5 &= (x+1)^2 + a(x+1) + 3 \\
x^2 + 3x + 5 &= (x^2+2x+1) + (ax + a) + 3 \\
x^2 + 3x + 5 &= x^2 + (2 + a)x + (a + 4) \\
\end{align}
And we see that this is true when $a = 1$.
Hence, by mathematical induction, $f(x) = x^2 + x + 3$.
Now that I look at this, I realize that I have only shown that $f(n) = n^2 + n + 3$ for $n = 0, 1, 2, \dots$. It is true however that three points uniquely determine a parabola. And we have agreement at an infinite number of points. So this is still a proof.
|
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"timestamp": "2023-03-29T00:00:00",
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|
System of equations involving square roots Hallo :) I am hopeless with this exercise:
Solve the system of equations over the positive real numbers
$$\sqrt{xy}+\sqrt{xz}-x=a$$
$$\sqrt{zy}+\sqrt{xy}-y=b$$
$$\sqrt{xz}+\sqrt{yz}-z=c$$
where $a, b, c$ are positive real numbers.
I tried to +,- and / the equations with one another, but I din`t see any reasonable result.
I also rised them to the power of two and then count,,, I had this solution
$z+\sqrt{yz}-\sqrt{xy}-x=\frac{a^{2}}{2x}-\frac{c^{2}}{2z}$
a firend of mine tried to count all three equations together and he got this
$2(\sqrt{xy}+...)-(x+...)=a +b+c$
$2(\sqrt{xy}+...)-((\sqrt x+\sqrt y+\sqrt z)^2-2(\sqrt {xy}+...))=a +b+c$
$4(\sqrt {xy}+...)-(\sqrt x+\sqrt y+\sqrt z)^2=a+b+c$
But we both don`t know what to do with that.
Do you know some reasonable method how to solve this system?
Thank you wery much!
|
We can solve the system in the following way (though I'm not sure if it is "reasonable") :
We have
$$\sqrt y+\sqrt z-\sqrt x=\frac{a}{\sqrt x}\tag1$$
$$\sqrt z+\sqrt x-\sqrt y=\frac{b}{\sqrt y}\tag2$$
$$\sqrt x+\sqrt y-\sqrt z=\frac{c}{\sqrt z}\tag3$$
From $(1)$,
$$\sqrt z=\sqrt x-\sqrt y+\frac{a}{\sqrt x}\tag4$$
From $(2)(4)$,
$$\sqrt x-\sqrt y+\frac{a}{\sqrt x}+\sqrt x-\sqrt y=\frac{b}{\sqrt y},$$
i.e.
$$2\sqrt x-2\sqrt y+\frac{a}{\sqrt x}-\frac{b}{\sqrt y}=0$$
Multiplying the both sides by $\sqrt{xy}$ gives
$$2x\sqrt y-2y\sqrt x+a\sqrt y-b\sqrt x=0,$$
i.e.
$$y=\frac{2x\sqrt y+a\sqrt y-b\sqrt x}{2\sqrt x}\tag5$$
From $(3)(4)$,
$$\sqrt x+\sqrt y-\left(\sqrt x-\sqrt y+\frac{a}{\sqrt x}\right)=\frac{c}{\sqrt x-\sqrt y+\frac{a}{\sqrt x}},$$
i.e.
$$\frac{2\sqrt{xy}-a}{\sqrt x}=\frac{c\sqrt x}{x-\sqrt{xy}+a}$$
Multiplying the both sides by $\sqrt x\ (x-\sqrt{xy}+a)$ gives
$$2x\sqrt{xy}-2xy+3a\sqrt{xy}-ax-a^2=cx,$$
i.e.
$$y=\frac{2x\sqrt{xy}+3a\sqrt{xy}-ax-a^2-cx}{2x}\tag6$$
From $(5)(6)$,
$$\frac{2x\sqrt y+a\sqrt y-b\sqrt x}{2\sqrt x}=\frac{2x\sqrt{xy}+3a\sqrt{xy}-ax-a^2-cx}{2x},$$
i.e.
$$\sqrt y=\frac{a^2+(a-b+c)x}{2a\sqrt x}\tag7$$
From $(5)(7)$,
$$\left(\frac{a^2+(a-b+c)x}{2a\sqrt x}\right)^2=\frac{(2x+a)\frac{a^2+(a-b+c)x}{2a\sqrt x}-b\sqrt x}{2\sqrt x},$$
i.e.
$$\frac{(a^2+(a-b+c)x)^2}{4a^2x}=\frac{(2x+a)(a^2+(a-b+c)x)-2abx}{4ax}$$
Multiplying the both sides by $4a^2x$ gives
$$(a^2+(a-b+c)x)^2=a((2x+a)(a^2+(a-b+c)x)-2abx),$$
i.e.
$$x((a^2-b^2+2bc-c^2)x+a^3-a^2b-a^2c)=0$$
Finally, from $(7)(4)$,
$$\color{red}{x=\frac{a^2(b+c-a)}{(a+b-c)(c+a-b)},\quad y=\frac{b^2(c+a-b)}{(b+c-a)(a+b-c)},\quad z=\frac{c^2(a+b-c)}{(b+c-a)(c+a-b)}}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Necessary conditions for a Sudoku puzzle to have no repetitions Is it true that if a Sudoku puzzle has the following features there will be no repetitions in rows, columns and $3 \times 3$ subsquares?
*
*The sum of each row must be $45$
*The sum of each column must be $45$
*The sum of each $3 \times 3$ subsquare must be $45$
If so, why? Is there a mathematical proof? If not, why? Is there a case where these conditions are satisfied, but is there at least one repetition?
Thanks!
|
No. For instance, this "sudoku" fulfills your conditions, but has some repetitions:
$$
\begin{array}{|ccc|ccc|ccc|}
\hline 5&5&5&5&5&5&5&5&5\\
5&5&5&5&5&5&5&5&5\\
5&5&5&5&5&5&5&5&5\\
\hline 5&5&5&5&5&5&5&5&5\\
5&5&5&5&5&5&5&5&5\\
5&5&5&5&5&5&5&5&5\\
\hline 5&5&5&5&5&5&5&5&5\\
5&5&5&5&5&5&5&5&5\\
5&5&5&5&5&5&5&5&5\\
\hline
\end{array}
$$
|
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|
Lemma on switching between mod $p$ and mod $p^2$ or mod $p^3$ Can someone help me prove the following lemma? Also can it be strengthened?
Let $p\geq 5$ be a prime number. Prove that if $p|a^2+ab+b^2$, then $p^3|(a+b)^p-a^p-b^p$
Here is what I tried:
We want to show $$p^3|a^{p-1}\binom{p}{1}b+a^{p-2}\binom{p}{2}b^2+\dots+a\binom{p}{p-1}b^{p-1}$$
It is easy to see that all the terms in the expression divide $p$, so we want to show:
$$\begin{align*} a^{p-1}b\frac{(p-1)!}{(p-1)!}+a^{p-2}b^2\frac{(p-1)!}{2!(p-2)!}+a^{p-3}b^3\frac{(p-1)!}{3!(p-3)!}+\dots &\equiv a^{p-1}b-a^{p-2}b^2\frac{1}{2!(p-2)!}\dots \\ &\equiv 0\pmod{p} \end{align*}$$
from Wilson's Theorem. But I do not know how to do that, as the expression is quite ugly. Also, we ultimately want to show it is divisble by $p^3$ and not just $p$. Finally, I could not find a way to use the given condition. Any ideas are appreciated.
I found the lemma in a solution to the problem here:
http://artofproblemsolving.com/community/c6h514444p2890151
|
Here is the proof anticipated by Stefan4024, based on this question linked to in his answer.
We first show that if $p \equiv 1 \pmod 6$, then
$$
p(a^2 + ab + b^2)^2 \,\mid\, (a+b)^p - a^p - b^p .
$$
Consider some fixed $b$, and let $f(x)$ be the polynomial
$$
f(x) = (x + b)^p - x^p - b^p.
$$
Let $\omega = e^{2\pi i / 3}$ be a primitive third-root of unity, and note that since $1 + \omega + \omega^2 = 0$, that we have that $1 + \omega = -\omega^2$ is a sixth-root of unity.
We will show that $\omega b$ is a root of both $f(x)$, and its derivative $f^\prime (x)$.
We have that $f(\omega b)$ is equal to
$$
(\omega b + b)^p - (\omega b)^p - b^p = b^p \left( (1 + \omega)^p - \omega^p - 1 \right)
$$
Since $\omega^3 = (1 + \omega)^6 = 1$, and $p-1$ is divisible by $6$, we have that $\omega^p = \omega$, and $(1 + \omega)^p = (1 + \omega)$. Thus we have that
$$
f(\omega b) = b^p \left( (1 + \omega) - \omega - 1 \right) = 0.
$$
Thus $\omega b$ is a root of $f$. Similarly,
$$
f^\prime (\omega b) = p (\omega b + b)^{p-1} - p(\omega b)^{p-1} = p b^{p-1} \left( (1 + \omega)^{p-1} - \omega^{p-1} \right) = p b^{p-1} (1 - 1) = 0.
$$
Thus $\omega b$ is a root of both $f$ and $f^\prime$, from which it follows that $(x - \omega b)^2$ is a factor of $f$. Since $f$ has real coefficients, $(x - \bar{\omega})^2$ is also a factor of $f$, and we see that $(x^2 + xb + b^2)^2$ is a factor of $f$.
Now it is a well-known fact that for $1 \leq k \leq (p-1)$, that the binomial coefficient $\binom{p}{k}$ is divisible by $p$, and so we see by the binomial theorem that all of the coefficients of $f$ are divisible by $p$. Thus $\frac{1}{p} f$ is a polynomial with integer coefficients, and is divisible by $(x^2 + xb + b^2)^2$, which is a monic polynomial with integer coefficients. It follows that we can write
$$
\frac{1}{p} f(x) = (x^2 + xb + b^2)^2 \cdot g(x)
$$
where $g(x)$ is some polynomial with integer coefficients. From this it follows easily that $f(a) = (a + b)^p - a^p - b^p$ is divisible by $p(a^2 + ab + b^2)^2$ if $p \equiv 1 \pmod 6$.
Now suppose that $p \,\mid\, a^2 + ab + b^2$. We will show that $p \equiv 1 \pmod 6$, so that
$$
p^3 \,\mid\, p(a^2 + ab + b^2)^2 \,\mid\, (a + b)^p - a^p - b^p .
$$
We note that
$$
p \,\mid\, 4a^2 + 4ab + 4b^2 = (2a + b)^2 + 3b^2.
$$
If $p \,\mid\, b$, then we see that we must also have that $p \,\mid\, a$, and so $(a + b)^p - a^p - b^p$ is divisible by $p^p$, and so is certainly divisible by $p^3$. Suppose now that $b$ is not divisible by $p$. Then we have that
$$
\left((2a + b) \cdot b^{-1} \right)^2 \equiv -3 \pmod p
$$
where $b^{-1}$ is the multiplicative inverse of $b$ modulo $p$, and so we see that $-3$ is a quadratic residue modulo $p$.
Thus
$$
\left( \frac{-1}{p} \right)\left( \frac{3}{p} \right) = \left( \frac{-3}{p} \right) = 1
$$
where $\left( \frac{\;}{} \right)$ is the Jacobi symbol. If $p \equiv 1 \pmod 4$, then
$$
\left( \frac{-1}{p} \right) = 1
$$
and by quadratic reciprocity,
$$
\left( \frac{3}{p} \right) = \left( \frac{p}{3} \right) =
\begin{cases}
1 & \text{ if } p \equiv 1 \pmod 3 \\
-1 & \text{ if } p \equiv 2 \pmod 3
\end{cases}.
$$
We see that in this case, we must have that $p \equiv 1 \pmod 3$.
On the other hand, if $p \equiv 3 \pmod 4$, then we know that
$$
\left( \frac{-1}{p} \right) = -1
$$
and so we must have
$$
\left( \frac{3}{p} \right) = -1.
$$
Since $3$ and $p$ are both $3$ mod $4$, quadratic reciprocity in this case gives us that
$$
-1 = \left( \frac{3}{p} \right) = -\left( \frac{p}{3} \right),
$$
and so we again have that $p \equiv 1 \pmod 3$.
In either case, we see that $p \equiv 1 \pmod 3$, and so $p \equiv 1 \pmod 6$, and the result follows.
|
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|
Prove there exists a point O such that $ OB \leq \frac {2}{\sqrt {3}} BY, OD \leq \frac {2}{\sqrt {3}} DZ, OF \leq \frac {2}{\sqrt {3}} FX$
Let the incircle of BDF touch DF, FB, BD at X, Y, Z respectively. Prove there exists a point O such that $$ OB \leq \frac {2}{\sqrt {3}} BY, OD \leq \frac {2}{\sqrt {3}} DZ, OF \leq \frac {2}{\sqrt {3}} FX$$.
It seem interesting,How prove it?
|
I'm a big fan of the tangent half-angle formula. Without loss of generality, the incircle is the unit circle, rotated such that none of $X,Y,Z$ lies at $(-1,0)$. Then one may choose coordinates as
\begin{align*}
X&=\frac1{1+x^2}\begin{pmatrix}1-x^2\\2x\end{pmatrix}&
Y&=\frac1{1+y^2}\begin{pmatrix}1-y^2\\2y\end{pmatrix}&
Z&=\frac1{1+z^2}\begin{pmatrix}1-z^2\\2z\end{pmatrix}\\
B&=\frac1{1+yz}\begin{pmatrix}1-yz\\y+z\end{pmatrix}&
D&=\frac1{1+xz}\begin{pmatrix}1-xz\\x+z\end{pmatrix}&
F&=\frac1{1+xy}\begin{pmatrix}1-xy\\x+y\end{pmatrix}
\end{align*}
I'll assume that the point $O$ you're after is the Gergonne point, where $BX$, $DY$ and $FZ$ intersect. Its coordinates are
$$O=\frac1{x^2y^2-x^2yz-xy^2z+x^2z^2-xyz^2+y^2z^2+x^2-xy+y^2-xz-yz+z^2}
\\[4ex]
\cdot\begin{pmatrix}
-x^2y^2+x^2yz+xy^2z-x^2z^2+xyz^2-y^2z^2+x^2-xy+y^2-xz-yz+z^2\\
x^2y+xy^2+x^2z-6xyz+y^2z+xz^2+yz^2
\end{pmatrix}$$
From this you can compute
\begin{align*}
\lVert OB\rVert^2 &= \scriptsize\frac{(z-y)^{4}\cdot(x^2+1)\cdot(x^2y^2+2x^2yz-4xy^2z+x^2z^2-4xyz^2+4y^2z^2+4x^2-4xy+y^2-4xz+2yz+z^2)}{(yz+1)^2\cdot(x^2y^2-x^2yz-xy^2z+x^2z^2-xyz^2+y^2z^2+x^2-xy+y^2-xz-yz+z^2)^2} \\
\lVert BY\rVert^2 &= \frac{(y-z)^2}{(yz+1)^2}
\end{align*}
Now you can translate your first inequality into
\begin{align*}
0 &\le 4\lVert BY\rVert^2 - 3\,\lVert OB\rVert^2 \\&=
\scriptsize\frac{(-y+z)^2\cdot(-x^2y^2+4x^2yz-2xy^2z-x^2z^2-2xyz^2+2y^2z^2+2x^2-2xy-y^2-2xz+4yz-z^2)^2}{(yz+1)^2\cdot(x^2y^2-x^2yz-xy^2z+x^2z^2-xyz^2+y^2z^2+x^2-xy+y^2-xz-yz+z^2)^2}\\&=
\scriptsize\left(\frac{(-y+z)\cdot(-x^2y^2+4x^2yz-2xy^2z-x^2z^2-2xyz^2+2y^2z^2+2x^2-2xy-y^2-2xz+4yz-z^2)}{(yz+1)\cdot(x^2y^2-x^2yz-xy^2z+x^2z^2-xyz^2+y^2z^2+x^2-xy+y^2-xz-yz+z^2)}\right)^2
\end{align*}
So that difference is a square of a real number, and therefore non-negative, proving your inequality. The same argument holds for the other two inequalities, simply by a change in variable names. Obviously this proof relies a lot on polynomial computations, and can benefit massively from the aid of computer algebra systems.
One could assume $y=0$, still without loss of generality, but the fact that generality is preserved is somewhat harder to see in this case, and the symmetry of the setup is lost. So use that if you have to do the computations by hand, but the above if you can trust a computer to do the tedious work for you.
|
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|
How to prove the limit of a sequence by definition: $\lim((n^2 + 1)^{1/8} - n^{1/4}) = 0$? I have to prove the limit of a sequence using $\epsilon - N$ definition: $$\lim_{n\to\infty}((n^2 + 1)^{1/8} - n^{1/4}) = 0$$.
Attempt:
We want to show: $\forall \epsilon>0$ $\exists N \in \mathbb{N}$, s.t. if $n \ge N$, then $|(n^2 + 1)^{1/8} - n^{1/4} - 0|<\epsilon$. So we need to find N as a function of $\epsilon$, s.t. N > $f(\epsilon)$.
$(n^2 + 1)^{1/8} - n^{1/4}$ = $((n^2 + 1)^{1/8} - n^{1/4}) * \frac{(n^2 + 1)^{1/8} + n^{1/4}}{(n^2 + 1)^{1/8} + n^{1/4}}$ = $\frac{(n^2 + 1)^{1/4} - n^{1/2}}{(n^2 + 1)^{1/8} + n^{1/4}}$.
$\frac{(n^2 + 1)^{1/4} - n^{1/2}}{(n^2 + 1)^{1/8} + n^{1/4}} < \frac{(n^2 + 1)^{1/4} - n^{1/2}}{n^{1/4}}$, so any n, solving $\frac{(n^2 + 1)^{1/4} - n^{1/2}}{n^{1/4}} < \epsilon$, will suffice.
Then $\frac{(n^2 + 1)^{1/4} - n^{1/2}}{n^{1/4}}$ = $(n + \frac{1}{n})^{1/4} - n^{1/4}$.
I do not know how to proceed further. I tried the same trick with multiplying the last equation by $\frac{(n + \frac{1}{n})^{1/4} + n^{1/4}}{(n + \frac{1}{n})^{1/4} + n^{1/4}}$, but I did not get anything meaningful.
|
Our expression equals $(n^2+1)^{1/8} - (n^2)^{1/8}.$ By the mean value theorem,
$$\tag 1 (n^2+1)^{1/8} - (n^2)^{1/8} =(1/8)c_n^{-7/8}\cdot 1,$$
where $c_n\in (n^2, n^2+1).$ It follows that $(1) \le (1/8)(n^2)^{-7/8} = (1/8)n^{-7/4}.$ You're now set up for an $\epsilon$-$N$ proof.
|
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|
Change of polynomial basis exercise Can someone please check if my solution is correct? Thanks.
Considering the following basis:
$$
B= \left\{ x^2,1+x^2,x-1 \right\}\\
C= \left\{ x,1-x,x^2 \right\}
$$
a) Find the change of basis matrices from $B$ to $C$ and $C$ to $B$.
b) Find the coordinates of the vector $x+1$ in $C$.
c) Find a basis $D$ of $R^2[x]$ such that the change of basis matrix from $D$ to $C$ is the same as the change from $C$ to $B$.
My Solution:
a)To reach what I've done, I decomposed the vectors in one basis in terms of a linear combination of the spam of the other basis...
From $B$ to $C$:
$$
\begin{pmatrix}
0 & 1 & 0\\
0 & 1 & -1\\
1 & 1 & 0\\
\end{pmatrix}
$$
From $C$ to $B$:
$$
\begin{pmatrix}
-1 & 0 & 1\\
1 & 0 & 0\\
1 & -1 & 0\\
\end{pmatrix}
$$
b) x+1 in $C$:
$$
\alpha x + \beta (1-x) + \gamma x^2 = x+1\\
\gamma = 0\\
\alpha - \beta = 1\\
\beta = 1\\
\left[\begin{matrix}
2\\
1\\
0
\end{matrix}\right]
$$
c)Basis $D$:
First vector is going to be the first column of this $C$ to $B$ matrix, times the vectors in basis $C$:
$$
\left[\begin{matrix}
-1\\
1\\
1\\
\end{matrix}\right] \cdot \left[\ x,1-x,x^2\right]\ =
-x+1-x+x^2 = \\
x^2-2x+1
$$
Same for second and third vectors:
$$
\left[\begin{matrix}
0\\
0\\
-1\\
\end{matrix}\right] \cdot \left[\ x,1-x,x^2\right]\ = -x^2\\
\left[\begin{matrix}
1\\
0\\
0\\
\end{matrix}\right] \cdot \left[\ x,1-x,x^2\right]\ = x\\
$$
So the basis $D$ is going to be:
$$
D= \left\{ x^2-2x+1,-x^2,x \right\}
$$
Is everything correct?!
Thanks.
|
Everything looks correct. The only thing I would say is that you should make sure D is in fact a basis by checking the vectors are linearly independent and checking whether they in fact span $R^2[x]$. This should be fairly easy by looking at the dimension of the subspace or the linear independence of the coordinates.
|
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|
Find the range of $a$ such $|f(x)|\le\frac{1}{4},$ $|f(x+2)|\le\frac{1}{4}$ for some $x$.
Let $$f(x)=x^2-ax+1.$$
Find the range of all possible $a$ so that there exist $x$ with
$$|f(x)|\le\dfrac{1}{4},\quad |f(x+2)|\le\dfrac{1}{4}.$$
A sketch of my thoughts: I write
$$f(x)=\left(x-\dfrac{a}{2}\right)^2+1-\dfrac{a^2}{4}\ge 1-\dfrac{a^2}{4}$$
so if $1-\dfrac{a^2}{4}>\dfrac{1}{4}$ or $-\sqrt{3}<a<\sqrt{3}$
this case impossible
But I don't know how to prove the other case, or if this there are better ideas.
|
Let $$f(x)=x^2-ax+1.$$
Find the range of all possible $a$ so that there exist $x$ with
$$|f(x)|\le\dfrac{1}{4},\quad |f(x+2)|\le\dfrac{1}{4}.$$
There is a solution if and only if one of the two equations $x^2-ax+1=\pm\dfrac14$ has two roots $r,s$ and $|r-s|\ge 2$ and the solution of $f(x+2)=f(x)$ satisfy $\left|f\left(\frac a2-1\right)\right|\le \frac 14.$
Part 1
$$\left|f\left(\frac a2-1\right)\right|\le \frac 14\iff 7\le a^2 \le 9.$$
Part 2
Let's work with $x^2-ax+1=\dfrac14.$ We have that $$x^2-ax+1=\dfrac 14\iff x=\dfrac{-a\pm\sqrt{a^2-3}}{2}.$$ So, there are two roots if and only if $a^2>3.$ Now
$$\left|\dfrac{a+\sqrt{a^2-3}}{2}-\dfrac{a-\sqrt{a^2-3}}{2}\right|\ge 2\iff \sqrt{a^2-3}\ge 2.$$ That is, $a^2\ge 7.$ So $a$ must satisfy three conditions: $a^2> 3,$ $a^2\ge 7$ and $7\le a^2\le 9.$
Part 3
If we work with $x^2-ax+1=-\dfrac14$ we get that $$x^2-ax+1=-\dfrac 14\iff x=\dfrac{-a\pm\sqrt{a^2-5}}{2}.$$ So, there are two roots if and only if $a^2>5.$ Now
$$\left|\dfrac{a+\sqrt{a^2-5}}{2}-\dfrac{a-\sqrt{a^2-5}}{2}\right|\ge 2\iff \sqrt{a^2-5}\ge 2.$$ That is, $a^2\ge 9.$ So $a$ must satisfy three conditions: $a^2> 5,$ $a^2\ge 9$ and $7\le a^2\le 9.$
Part 4
Thus the solution is
$$a\in [-3,-\sqrt 7]\cup [\sqrt 7,3].$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Minimal value of $\sqrt{(x-a)^2 + b^2} +\sqrt{(x-c)^2 + d^2}$ without derivatives and without distance formula Let $f(x) = \sqrt{(x-a)^2 + b^2} +\sqrt{(x-c)^2 + d^2}$, where all coefficients are real.
It can be shown using a distance formula that the minimal value of $f(x)$ is
$D = \sqrt{(a-c)^2+(|b|+|d|)^2}$.
Show that result without derivatives and without a distance formula. At what value of $x$ does the minimum of $f(x)$ occur?
Hint: this is a generalization of this question.
|
Applying Fermat's Principle on reflection:
Assume $a<c$,
\begin{align*}
T(x) &= \sqrt{(x-a)^2+b^2}+\sqrt{(x-c)^2+d^2} \\
T'(x) &=\frac{x-a}{\sqrt{(x-a)^2+b^2}}-\frac{c-x}{\sqrt{(x-c)^2+d^2}} \\
0 &= \sin i-\sin r \\
i &= r \\
\end{align*}
If you've accepted the law of reflection, then
\begin{align*}
\frac{x-a}{b} &= \frac{c-x}{d} \\
x &= \frac{ad+bc}{b+d}
\end{align*}
where $b,d>0$
See another answer for the case of refraction.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
Trying to find shortest distance from point $(2,0-3)$ to $x+y+z=1$ I'm trying to find shortest distance from the point $(2,0-3)$ to $x+y+z=1$.
I found $d^2=(x-2)^2+y^2+(z+3)^2$
Substituting for $z$: $d^2=(x-2)^2+y^2+(-x-y+1+3)^2$
$f_x=2x-4-2(-x-y+4)$ and $f_y=2y-2(-x-y+4)$ Next would be finding zeros as the critical points and using them as the $xyz$ values in the formula for $d$ but I'm having issues finding those points.
|
We have $4x+2y=12$ and $2x+4y=8$. First equation reduces to $y=6-2x$. Plugging that into the second gives $2x+4(6-2x)=8$ which solves to $2x+24-8x=8$ or $x=8/3$. Plugging this answer into $y=6-2x$ yields $y=2/3$. This is one of the many ways to solve such a system. Anyway, with this info you can find $z$ and from here you can find the shortest distance.
|
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|
Prove that the chord of the ellipse passes through a fixed point Variable pairs of chords at right angles are drawn through a point $P$ (forming an angle of $\pi/4$ with the major axis) on the ellipse $\frac {x^2}{4}+y^2=1$, to meet the ellipse at two points $A $ and $B $. Prove that the line joining these two points passes through a fixed point.
I am writing the equations and they are just as filthy as they could be. It's just tedious. I am sure there is something I am missing or some better way to approach it. Thanks.
|
This is a property of general ellipses, so let's consider the ellipse with major and minor radii $a$ and $b$. For a given point $P = (a \cos 2\theta, b \sin 2\theta)$, we'll identify the point $Q$ common to all chords $\overline{AB}$ such that $\overline{AP}\perp\overline{BP}$.
It's straightforward to find the coordinates of $Q$ at the intersection of two convenient chords, namely: $\overline{A_0 B_0}$, the "other" diagonal of the inscribed rectangle with vertex $P$; and $\overline{PN}$, along the normal to the ellipse at $P$ (which serves as the degenerate case).
$$\begin{align}
\overline{A_0 B_0}:&\quad b x \sin 2\theta + a y \cos 2\theta = 0 \\
\overline{PN}:&\quad a x \sin 2\theta - b y \cos 2\theta = ( a^2 - b^2 ) \cos 2\theta \sin 2\theta
\end{align}$$
$$Q := \overline{A_0 B_0} \cap \overline{PN} = \frac{a^2-b^2}{a^2+b^2} \;\left( a \cos 2\theta, - b \sin 2\theta \right)$$
Now, consider a generic chord $\overline{AB}$ where $A = (a \cos2\alpha, b \sin 2\alpha)$ and $B = (a \cos2\beta, b \sin 2\beta)$. The condition that $\overline{AP}\perp\overline{BP}$ gives rise to this equation
$$\begin{align}
0 &= (A-P)\cdot(B-P) \\
&= 4 \sin(\alpha-\theta) \sin(\beta-\theta) \left(\;
a^2 \sin(\alpha + \theta) \sin( \beta + \theta )
+ b^2 \cos(\alpha + \theta) \cos(\beta + \theta)\;\right)
\end{align}$$
where we may safely ignore the initial factors, so that
$$a^2 \sin(\alpha + \theta) \sin( \beta + \theta )
+ b^2 \cos(\alpha + \theta) \cos(\beta + \theta) = 0 \tag{1}$$
On the other hand, the equation for the line containing $\overline{AB}$ is
$$b x \cos(\alpha + \beta) + a y \sin(\alpha + \beta) = a b \cos(\alpha - \beta) $$
so that the condition that $Q$ lies on the chord becomes
$$\frac{a^2-b^2}{a^2+b^2}\;\left(\;a b \cos(\alpha + \beta) \cos 2\theta - a b \sin(\alpha + \beta) \sin 2\theta \; \right) = a b \cos(\alpha - \beta) $$
whereupon
$$\left(a^2-b^2\right)\cos(\alpha + \beta + 2\theta) = \left(a^2+b^2\right) \cos(\alpha - \beta) \tag{2}$$
Verification that (1) and (2) are equivalent is left as an easy exercise for the reader. $\square$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "7",
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|
How to find results for $-z + \frac {3} {\overline {z}}=2$ Can you just please help me solve this problem, because i don't know how to solve it: $$-z + \frac {3} {\overline z}=2$$
|
$$-z + \frac {3} {\overline z}=2$$
$$-z \overline z+3=2\overline z$$
As $z \overline{z}=\left|{z^2}\right|$,
$$
\begin{align}
-\left|{z^2}\right|+3&=2 \overline{z}\\
&= 2a-2bi \\
\end{align}
$$
Equating imaginary parts,
$$2\Im{(z)}=0\implies 2b=0\implies b=0$$
Equating real parts,
$$
\begin{align}
2 \Re{(z)} &= -(a^2+b^2)+3 \\
2a &= -a^2+3 \\
a^2+2a-3 &= 0 \\
(a+3)(a-1) &= 0 \\
a &= -3,\,1 \\
\end{align}
$$
So $z=-3$ or $z=1$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
If two consecutive numbers are removed from the series $1+2+3+\ldots+n$ the average becomes $99/4$. Find the two numbers. The initial average will be $\frac{n+1}{2}$. If the two numbers are $k$ and $k+1$ then the new average will be $\frac{n(n+1)/2-(2k+1)}{n-2}$. I couldn't figure further even though I got the relation between $n$ and $k$ in many different ways.
If the question is not clear, here is an example to explain it.
If $n=10$, the initial average will be $5\cdot 5$ {$(1+2+\cdots + 10)/10$}
Now if two consecutive numbers like $2,3$ or $8,9$ are removed from this series, the new average changes, and this new average has been given to be $99/4$, however we also don't know the value of $n$, so the question seems to be pretty difficult.
|
We have, that the sum of $n+1$ terms, excluding the $m$-th and $m+1$-th, is:
$$
\begin{gathered}
S(n + 1,m) = \sum\limits_{1\, \leqslant \,k\, \leqslant \,m - 1} k + \sum\limits_{m + 2\, \leqslant \,k\, \leqslant \,n + 1} k = \sum\limits_{1\, \leqslant \,k\, \leqslant \,m - 1} k + \sum\limits_{1\, \leqslant \,k\, \leqslant \,n - m} {\left( {k + m + 1} \right)} = \hfill \\
= \left( \begin{gathered}
m \\
2 \\
\end{gathered} \right) + \left( {m + 1} \right)\left( {n - m} \right) + \left( \begin{gathered}
n + 1 - m \\
2 \\
\end{gathered} \right) = \hfill \\
= \frac{1}
{2}m\left( {m - 1} \right) + \left( {m + 1} \right)\left( {n - m} \right) + \frac{1}
{2}\left( {n + 1 - m} \right)\left( {n - m} \right) = \hfill \\
= \frac{1}
{2}\left( {m\left( {n - 1} \right) + \left( {n + 3} \right)\left( {n - m} \right)} \right) = \hfill \\
= \frac{1}
{2}\left( {n\left( {n - 1} \right) + 4\left( {n - m} \right)} \right) = \frac{{n\left( {n + 3} \right)}}
{2} - 2m \hfill \\
\end{gathered}
$$
So we shall have:
$$
\begin{gathered}
\frac{{S(n + 1,m)}}
{{n - 1}} = \frac{{99}}
{4}\quad \Rightarrow \quad \left\{ \begin{gathered}
n - 1 = 4\,q \hfill \\
S(n + 1,m) = \frac{{n\left( {n + 3} \right)}}
{2} - 2m = 99\;q \hfill \\
\end{gathered} \right.\quad \Rightarrow \hfill \\
\Rightarrow \quad \left\{ \begin{gathered}
1 \leqslant m \leqslant n = 4\,q + 1 \hfill \\
n\left( {n + 3} \right) = 198\;q + 4m \hfill \\
\end{gathered} \right. \hfill \\
\end{gathered}
$$
The last gives:
$$
\begin{gathered}
4 \leqslant 4\left( {4q + 1} \right)\left( {q + 1} \right) - 198\;q = 4m \leqslant 4\left( {4\,q + 1} \right) \hfill \\
0 \leqslant \left( {4q + 1} \right)\left( {q + 1} \right) - \frac{{198}}
{4}\;q - 1 \leqslant 4\,q \hfill \\
0 \leqslant q^{\,2} - \frac{{178}}
{{16}}\;q \leqslant \,q \hfill \\
0 \leqslant q - \frac{{178}}
{{16}} \leqslant \,1 \hfill \\
\end{gathered}
$$
i.e.
$$
\frac{{178}}
{{16}} \leqslant q \leqslant \,\frac{{194}}
{{16}}\quad \Rightarrow \quad \left\lceil {\frac{{178}}
{{16}}} \right\rceil \leqslant q \leqslant \,\left\lfloor {10 + \frac{{34}}
{{16}}} \right\rfloor \quad \Rightarrow \quad 12 \leqslant q \leqslant 12
$$
In conclusion, so we have:
$$
\left\{ \begin{gathered}
q = 12 \hfill \\
n = 4\,q + 1 = 49 \hfill \\
m = \frac{1}
{4}\left( {n\left( {n + 3} \right) - 198\;q} \right) = 43 \hfill \\
\end{gathered} \right.
$$
which in fact gives:
$$
\frac{{S(n + 1,m)}}
{{n - 1}} = \frac{{\frac{{n\left( {n + 3} \right)}}
{2} - 2m}}
{{n - 1}} = \frac{{\frac{{49 \cdot 52}}
{2} - 86}}
{{48}} = \frac{{1188}}
{{48}} = \frac{{99}}
{4}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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|
Proof of $\sum_\limits{n = 1}^{\infty} {k\choose n}{b-1 \choose n-1}x^n \sim \frac{x^b k^b}{\Gamma(1+b)} \quad \text{as}~k \to \infty$ I want to prove the following aymptotic result:
$$\sum_{n = 1}^{\infty} {k\choose n}{b-1 \choose n-1}x^n \sim \frac{x^b k^b}{\Gamma(1+b)} \quad \text{as}~k \to \infty,$$
where $ k \in \mathbb{C}$, $b \in [0,1]$, and $x \in [0,1].$
I tried using some results of infinite series for combinatorics, but could not prove it. One of the results I tried is
$$\sum_{n = 0}^{\infty} {k \choose n}x^n = (1+x)^k.$$
Another way I tried is to use
$$ \sum_{n = 1}^{\infty} {k\choose n}{b-1 \choose n-1}x^n = xk ~_2F_1(1-k, 1-b;2;x).$$
|
Define
$$
f_k(x)=\sum_{n=1}^\infty\binom{k}{k-n}\binom{b-1}{n-1}x^{n-1}\tag{1}
$$
then we are looking for the asymptotic expansion of $x\,f_k(x)$.
Vandermonde's Identity and Gautschi's Inequality say
$$
\begin{align}
f_k(1)
&=\sum_{n=1}^\infty\binom{k}{k-n}\binom{b-1}{n-1}\\
&=\binom{k+b-1}{k-1}\\
&=\frac{\Gamma(k+b)}{\Gamma(k)\Gamma(b+1)}\\[4pt]
&=\frac{k^b}{\Gamma(b+1)}\left(1+O\!\left(\frac1k\right)\right)\tag{2}
\end{align}
$$
Let $(n)_m$ be the Falling Factorial, then the $m^{\text{th}}$ derivative of $f_k$ at $1$ is
$$
\begin{align}
f_k^{(m)}(1)
&=\sum_{n=1}^\infty\binom{k}{k-n}\binom{b-1}{n-1}(n-1)_m\\
&=\sum_{n=1}^\infty\binom{k}{k-n}\binom{b-m-1}{n-m-1}(b-1)_m\\
&=\binom{k+b-m-1}{k-m-1}(b-1)_m\\
&=\frac{k^b(b-1)_m}{\Gamma(1+b)}\left(1+O\!\left(\frac1k\right)\right)\tag{3}
\end{align}
$$
For $|x-1|\lt1$, Taylor's Theorem and the Generalized Binomial Theorem say
$$
\begin{align}
x\,f_k(x)
&=x\sum_{m=0}^\infty\frac{f_k^{(m)}(1)}{m!}(x-1)^m\\
&=\frac{k^bx}{\Gamma(1+b)}\left(1+O\!\left(\frac1k\right)\right)\sum_{m=0}^\infty\frac{(b-1)_m}{m!}(x-1)^m\\
&=\bbox[5px,border:2px solid #C0A000]{\frac{k^bx^b}{\Gamma(1+b)}\left(1+O\!\left(\frac1k\right)\right)}\tag{4}
\end{align}
$$
Asymptotic Expansion of Fractional Binomial Coefficients
Using the Euler-Maclaurin Sum Formula, we get
$$
\begin{align}
\log\binom{n+\alpha}{n}
&=\sum_{k=1}^n\log\left(1+\frac\alpha{k}\right)\\
&=\alpha\sum_{k=1}^n\frac1k-\frac{\alpha^2}2\sum_{k=1}^n\frac1{k^2}+\frac{\alpha^3}3\sum_{k=1}^n\frac1{k^3}-\frac{\alpha^4}4\sum_{k=1}^n\frac1{k^4}+\cdots\\
&=\alpha\left(\log(n)+\gamma+\frac1{2n}-\frac1{12n^2}+O\!\left(\frac1{n^4}\right)\right)\\
&-\frac{\alpha^2}2\left(\zeta(2)-\frac1n+\frac1{2n^2}+O\!\left(\frac1{n^3}\right)\right)\\
&+\frac{\alpha^3}3\left(\zeta(3)-\frac1{2n^2}+O\!\left(\frac1{n^3}\right)\right)\\
&-\frac{\alpha^4}4\left(\zeta(4)+O\!\left(\frac1{n^3}\right)\right)\\
&=\alpha\log(n)-\log(\Gamma(1+\alpha))+\frac{\alpha+\alpha^2}{2n}-\frac{\alpha+3\alpha^2+2\alpha^3}{12n^2}+O\!\left(\frac1{n^3}\right)\tag{5}
\end{align}
$$
Therefore, as $n\to\infty$,
$$
\binom{n+\alpha}{n}=\frac{n^\alpha}{\Gamma(1+\alpha)}\left(1+\frac{\alpha+\alpha^2}{2n}-\frac{2\alpha+3\alpha^2-2\alpha^3-3\alpha^4}{24n^2}+O\!\left(\frac1{n^3}\right)\right)\tag{6}
$$
and so, for fixed $m$,
$$
\begin{align}
\binom{n-m+\alpha}{n-m}
&=\frac{(n-m)^\alpha}{\Gamma(1+\alpha)}\left(1+\frac{\alpha+\alpha^2}{2n}+O\!\left(\frac1{n^2}\right)\right)\\
&=\frac{n^\alpha}{\Gamma(1+\alpha)}\left(1-\frac mn\right)^\alpha\left(1+\frac{\alpha+\alpha^2}{2n}+O\!\left(\frac1{n^2}\right)\right)\\
&=\frac{n^\alpha}{\Gamma(1+\alpha)}\left(1+\frac{(1-2m)\alpha+\alpha^2}{2n}+O\!\left(\frac1{n^2}\right)\right)\tag{7}
\end{align}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find $24 \cot^2 x$
Suppose $x$ is in the interval $[0,\pi/2]$ and $\log_{24 \sin x} (24 \cos x) = \frac{3}{2}$. Find $24 \cot^2 x$.
From the given equation we have $24 \cos{x} = (24 \sin{x})^{\frac{3}{2}}$ and so $24\cot^2{x} = 24^2\sin{x}$. How do we continue?
|
You have $(24\cos x)^2 =(24\sin x)^3,\;$ or $\cos^2 x =24\sin^3 x.\;$
With $y=\sin x$ this gives the cubic $24y^3=1-y^2$ with the real solution
$y_1 = \frac{1}{3} \;$ or
$$x=\arcsin\left(\frac{1}{3}\right) \approx 0.3398369$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove $10^{3n+1}+10^{6n+2}+1=111k$ I'm stuck to prove this one can anyone help me please?
Prove that $10^{3n+1}+10^{6n+2}+1=111k$.
I'm not sure what exactly I should do.
n is a natural number and k is an integer
|
The OP has corrected the problem. The corrected answer is below.
Assuming that $k$ must be an integer (not stated by OP), this is the same as showing that
$$
10^{n+1}+10^{6n+2}+1\equiv 0\pmod{111}.
$$
Observe that $10^3=1000=999+1=1+9(111)$. Therefore, there are only three values for the first and second terms, $10$, $100$, and $1$. In particular, when
$$
10^{n+1}=\begin{cases}10&n+1\equiv 1\pmod 3\\100&n+1\equiv 2\pmod 3\\1&n+1\equiv 0\pmod 3.\end{cases}=\begin{cases}10&n\equiv 0\pmod 3\\100&n\equiv 1\pmod 3\\1&n\equiv 2\pmod 3.\end{cases}
$$
On the other hand, there is only one value for $10^{6n+2}\pmod{111}$ since $10^{6n}=(10^3)^{2n}\equiv 1^{2n}=1\pmod{111}$. Therefore, the second term is equivalent to $100$ modulo $111$.
This appears to be a problem, since (modulo $111$) there are exactly three sums $10+11+1$, $100+100+1$, and $1+100+1$, only the first one is equivalent to $0$ modulo $111$. In other words, if we let $n=1$, then the sum is $100,000,101$, which is not a multiple of $111$.
To the OP, please check the original statement. The statement would be correct if the first term were $10^{3n+1}$, however.
When the first term is $10^{3n+1}$ instead of $10^{n+1}$, observe that $10^{3n}=(10^3)^n\equiv 1^n=1\pmod{111}$, therefore, the sum can be reduced to
$$
10^{3n+1}+10^{6n+2}+1\equiv 10+10^2+1\equiv 0\pmod{111}.
$$
|
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|
Modular arithmetic three variables Show that if the integers $x, y,$ and $z$ satisfy $x^3 + 3y^3 = 9z^3$
then $x = y = z = 0.$
How should I interpret this question and how to proceed?
I am thinking about the Euclidean algorithm but it becomes confusing when $x,y,z$ comes like variables?
|
First notice that if $d=\mbox{gcd}(x,y,z)$ then $d^3$ can be factored out of the equation. So we can assume that $d=1$. Then $x^3 = 9z^3-3y^3$, so $3$ divides $x$, say $ x=3k$. So we have $3^3k^3 = 9z^3-3y^3$ and we can divide everything by $3$ to get $9k^3= 3 z^3-y^3$. A similar argument shows $3$ divides $y$. Repeat to show $3$ divides $z$. This contradicts that $d=1$.
|
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|
Matrix with unknown coefficient $a$ - determine number of solutions for each $a$ I have a matrix that looks like this:
\begin{array}{cccc|c}
a & 1 & 0 & 0 & 1 \\
1 & a & 0 & 0 & 2 \\
0 & 0 & a & 2 & 1 \\
0 & 0 & 2 & a & 1
\end{array}
I then proceeded to row reduce the matrix so it looks like this:
\begin{array}{cccc|c}
1 & 0 & 0 & 0 & a-2/a^2-1 \\
0 & 1 & 0 & 0 & 2a-1/a^2-1 \\
0 & 0 & 1 & 0 & 1/a+2 \\
0 & 0 & 0 & 1 & 1/a+2
\end{array}
I need to determine the number of solutions for each $a$ (one, infinite or none).
I can clearly see that there are no solutions when a is either $1$, $-1$ or $2$ because than you would have a division by zero in atleast one of the expressions on the right hand side. But how do I know which $a$ will have an infinite number of solutions. I have never worked with matrices that have unknown coefficients so I have no idea if I am doing anything correctly here.
The original equation system looked like this:
$$ax+y+0+0=1$$
$$x+ay+0+0=2$$
$$0+0+az+2w=1$$
$$0+0+2z+aw=1$$
Maybe I am missing something really obvious. This looks like a very simple task...
|
Swap rows 1 and 2, and rows 2 and 4. You obtain an augmented matrix for which it's easier to apply pivot's method:
\begin{align}
\begin{bmatrix}
1&a&0&0&2\\a&1&0&0&1\\0&0&2&a&1\\0&0&a&2&1
\end{bmatrix}\rightsquigarrow
\begin{bmatrix}
1&a&0&0&2\\0&1-a^2&0&0&1-2a\\0&0&2&a&1\\0&0&0&4-a^2&2-a
\end{bmatrix}
\end{align}
Thus, if $a\ne \pm1,\pm2$, the matrix of the system and the augmented matrix have the same, maximal rank ($4$). This means there is exactly one solution.
If $a=\pm1$, the augmented matrix is
$$\begin{bmatrix}
1&\pm1&0&0&2\\0& 0&0&0& \scriptstyle\begin{cases}-1\\3\end{cases}\\0&0&2&\pm1&1\\0&0&0&3& \scriptstyle\begin{cases}1\\3\end{cases}
\end{bmatrix}$$
The second row shows there's no solution.
If $a=\pm2$, the augmented matrix is
$$\begin{bmatrix}
1&\pm2&0&0&2\\0&-3&0&0& \scriptstyle\begin{cases}-3\\5\end{cases}\\0&0&2&\pm2&1\\
0&0&0&0& \scriptstyle\begin{cases}0\\4\end{cases}
\end{bmatrix}$$
The last row shows there is no solution if $a=2$ and is an affine subspace of dimension $1$ if $a=-2$, because both matrices have the same rank ($3$).
|
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|
Is it inequality true Please help me prove the inequality:
$$\sqrt[3]{3+\sqrt[3] 3}+\sqrt[3]{3-\sqrt[3] 3}<2\sqrt[3] 3$$
Thanky for your help and your attention.
|
We can use $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$.
Since $a^2+b^2+c^2-ab-ac-bc=0\Leftrightarrow a=b=c$
and $\sqrt[3]{3+\sqrt[3]3}\neq\sqrt[3]{3-\sqrt[3]3}$ , we need to prove that
$$3+\sqrt[3]3+3-\sqrt[3]3-8\cdot3+3\cdot2\sqrt[3]3\sqrt[3]{9-\sqrt[3]9}<0$$ or
$$\sqrt[3]{27-3\sqrt[3]9}<3$$
which is obvious.
|
{
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Vieta Jumping: Related to IMO problem 6, 1988: If $ab + 1$ divides $a^2 + b^2$ then $ab + 1$ cannot be a perfect square. The famous IMO problem 6 states that if $a,b$ are positive integers, such that $ab + 1$ divides $a^2 + b^2$, then $\frac{ a^2 + b^2}{ab + 1 }$ is a perfect square, namely, $gcd(a,b)^2$. How about a modification of this problem:
*
*If $a,b$ are (strictly) positive integers, such that $ab + 1$ divides $a^2 + b^2$, then $ab + 1$ cannot be a perfect square.
I am looking for a proof of the claim above, or a counter-example.
One possible approach towards a proof is the following:
Suppose there is such a pair $(a,b)$ as above, then by the famous IMO problem 6 from 1988, $\frac{ a^2 + b^2}{ab + 1 } = g^2$ where $g = gcd(a,b)$. Since $ab + 1$ is a perfect square, $a^2 + b^2 = c^2$ for some integer $c$. So that $(a,b,c)$ is a Pythagorean tripple, therefore there exists positive integers $n,m,l$ with $n$ coprime to $m$, such that $a = l(n^2 - m^2)$ and $b = 2lnm$ - by Euclids formula.
Then by plugging in $a$ and $b$ in terms of $n,m,l$ into the original equation, and solving for $l$, it is possible to obtain the following:
There exists positive coprime integers $n,m$ such that
$\frac{(n^2 + m^2 + 1)(n^2 + m^2 - 1)}{2mn(n+m)(n-m)}$ is a perfect square.
If we put the quotient above into a program, as in this python code snipet:
N = 1000
for n in range(1,N):
for m in range(n+1, N):
A = (n*n + m*m + 1)*(n*n + m*m - 1)
B = 2*m*n*(n+m)*(m-n)
if A % B == 0:
print(A/B)
The quotient always is 2, regardless of whether or not $n$ and $m$ are co prime! So if the very strong implication
*
*$2mn(n+m)(n-m) \mid (n^2 + m^2 + 1)(n^2 + m^2 - 1) \implies \frac{(n^2 + m^2 + 1)(n^2 + m^2 - 1)}{2mn(n+m)(n-m)} = 2$
holds, then the original problem will be solved.
|
Here is a bit of a hacky answer to your question in the affirmative
You observe that $k = \mathrm{gcd}(a,b)^2$. After plodding through various resources, it is because one can Viete Jump:
$$ (a,b) \mapsto \big(a_1, b_1\big) \mapsto \dots \mapsto (k,0) $$
and the gcd is conserved. Once I agree with you, let $a = \text{gcd}(a,b) \, c$ and $b = \text{gcd}(a,b) \, d$ so that
\begin{eqnarray} ab+1 &=& \frac{a^2+b^2}{\mathrm{gcd}(a,b)^2}= c^2 + d^2 \\
&=& \,\text{gcd}(a,b)^2 \, cd + 1
\end{eqnarray}
In this way there are two conditions $c, d$ might solve (where the two $\square$ are different) and $\text{gcd}(c,d)=1$:
\begin{eqnarray*}
c^2 - \square \, cd + d^2 &=& 1 \\
c^2 + d^2 &=& \square
\end{eqnarray*}
Hopefully these two equations leads to a contradiction.
Added 11/15 The answer is definitely no. Let $k = \mathrm{gcd}(a,b)$ We are trying to solve in integers:
\begin{eqnarray} c^2 - k\; cd + d^2 &=& 1 \\
c^2 + d^2 = \square
\end{eqnarray}
As I learned, the first can be solve with $(c,d) = (1,0)$ or $(k,1)$ and there are an infinite family of solutions using consecutive terms of a recursive sequence [2, 3]
$$ x_{n+1} = k \, x_n + x_{n-1} $$
There are sometimes ways to link Pythagorean triples to Pell equations [1] (Modular Tree of Pythagoras)
$$ x_{n+1}^2 + \frac{1}{2} x_n^2 < \sqrt{x_{n+1}^2 + x_n^2 } = x_{n+1}^2 \sqrt{1 + (x_n/x_{n+1})^2} < x_{n+1}^2 + \frac{1}{2} x_n^2 + 1$$
This cannot be an integer. So any time we solve the Pell equation, we cannot also solve Pythagoras. $\quad\quad\square$
Old Answer
This is discussed on Wikipedia's article on Vieta Jumping:
Nobody of the six members of the Australian problem committee could solve it. Two of the members were George Szekeres and his wife, both famous problem solvers and problem creators. [...] The problem committee submitted it to the jury of the XXIX IMO marked with a double asterisk, which meant a superhard problem, possibly too hard to pose. After a long discussion, the jury finally had the courage to choose it as the last problem of the competition. Eleven students gave perfect solutions.
Among the eleven was Bau Chau Ngô (Fields Medal 2010). His work on the Fundamental Lemma also has a jumpy flavor [1, 2, 3] but is quite advanced.
The discussion on YouTube is helpful as well. These videos give a thorough discussion of different ways to solve
*
*The Legend of Question Six (Numberphile)
*The Return of the Legend of Question Six (Numberphile2)
These may not directly solve your problem but provide historical context and indicate possible strategies.
In the Wikipedia article, the example of Viete Jumping is IMO 1988/6 -- the same as asked in the question:
Let $a,b$ be positive integers such that $ab+1$ divides $a^2 + b^2$ show that $ \frac{a^2 + b^2}{ab+1}$ is a perfect square.
and the solution goes in three steps
#1 Let $a, b \geq 0$ be solutions to $\frac{a^2 + b^2}{ab+1} = k$ such that $k$ is not a perfect square: $k \neq \square$
#2 Starting from $(a,b)$ we can try to generate another solution $(x,b)$ which solves the quadratic equation:
$$ x^2 - kb\, x +(b^2 - k) = 0 $$
The map $(a,b) \mapsto (a_1,b)$ is our Vieta jumping Since both $a, a_1$ are acceptable solutions we have:
$$ (x-a)(x-a_1) = x^2 - (a + a_1) x + aa_1 = 0$$
By the Viete equations (comparing the coefficients. We find out two things:
*
*$ a + a_1 = kb $ so that $a_1 = kb - a \in \mathbb{Z}$ (it is an integer)
*$ aa_1 = b^2 - k $ so that $a_1 = \frac{b^2 - k}{a} \neq 0$
#3 If $a \geq b$ we can deduce that $a_1 \geq 0$ (is positive) and additionally $ a > a_1 \ge 0 $
*
*From #2 $ a_1 = \frac{b^2 - k }{a} < \frac{b^2}{a} < \frac{a^2}{a}=a $
*$\frac{a_1^2 + b^2}{ab+1}= k > 0 $ implies that $ a_1b+1 > 0$ or $a_1 > - \frac{1}{b}$ but $a \in \mathbb{Z}$ so $a_1 \geq 0$.
Summary We've show that given two positive numbers $a,b$ solving $\frac{a^2+b^2}{ab+1}=k$ with $k \neq \square$ we can always find another solution $(a_1,b)$ solving the same equation with $a > a_1$.
Then the Viete jump consists of a map:
$$ (a,b) \mapsto \left\{\begin{array}{rc}
(b,a) & \text{ if } a \leq b \\
(\frac{b^2-k}{a},b) & \text{ if } a \geq b
\end{array}\right.$$
While this does not solve your problem -- to show that $ab+1 \neq \square$ -- it does indicate possibly where to start and some possible resources.
A quick use of Bezout formula shows that $ab+1$ should also divide
$$ \big[(a^2 + b^2) + 2(ab+1)\big] +
\big[(a^2 + b^2) - 2(ab+1)\big] = (a+b)^2 +(a-b)^2 $$
and this could lead to your contradiction.
|
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Solve Recurrence Relation for Maximum and Minimum in an Array. I know that recursive equation of this algo is
$T\left ( n \right )=2T\left ( \frac{n}{2} \right )+2 $
Given that $T\left ( 1 \right )=0,T\left ( 2 \right )=1 $
and its solution is also given here, ijust want to clear my doubt where i am stuck at..
I solved this eqtn as--:
$\Rightarrow T\left ( n \right )=2\left ( 2*T\left ( \frac{n}{4} \right )+2 \right )+2 $
$\Rightarrow T\left ( n \right )=2^{2}+T\left ( \frac{n}{2^{2}} \right )+2^{2}+2^{1}$
$\vdots$
$\Rightarrow T\left ( n \right )=2^{k}*T\left ( \frac{n}{2^{k}} \right )+2^{k}+\cdots 2^{2}+2^{1}$
Taking $T\left ( 1 \right )=0 $
$\frac{n}{2^{k}}=1 \Rightarrow k=\log_{2}n $
And our equation becomes
$2^{k}+2^{k-1}+\cdots 2^{2}+2^{1} = \frac{2*\left ( 1-2^{k} \right )}{1-2}=2n-2 ,\left \{ 2^{k}=2^{log_{2}n}=n\right \} $
But answer is $\frac{3*n}{2}-2$..please correct me in this equation.
|
You can check as followed that your solution fits the recurrence like this:
$T(1)=2\cdot 1-2=0$
$T(n) = 2n-2 = 2(2\frac{n}{2}-2) +2 = 2\cdot T(\frac{n}{2}) + 2$
Similarly you can check that the answer you have been given fits:
$T(2)=1$
$T(n) = 2\cdot T(\frac{n}{2}) + 2$
How to reconcile these 2 facts? Well the 2 base-cases aren't compatible. Try:
$T(2) = 2\cdot T(\frac{2}{2}) + 2 = 2\cdot T(1) + 2$
If we assume $T(1)=0$, we get:
$T(2) = 2\cdot 0 + 2 = 2$
If we assume $T(2)=1$, we get:
$1=T(2) = 2\cdot T(1) + 2 \implies T(1)=-\frac{1}{2}$
So the 2 different base cases give 2 different solutions for the recurrence. If you want the other solution, use the other base-case in your derivation.
edit: to see what happens, we use:
$\frac{n}{2^k}=2\implies k = (\log_2{n})-1$
And the equation becomes:
$2^{k}\cdot 1+2^{k}+2^{k-1}+\cdots 2^{2}+2^{1} =2^{k}+ \frac{2*\left ( 1-2^{k} \right )}{1-2}=\frac{n}{2}+2\frac{n}{2}-2=\frac{3\cdot n}{2}-2 ,\left \{ 2^{k}=2^{(log_{2}n)-1}=\frac{n}{2}\right \} $
|
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How prove this $BC$ always passes through a fixed point with $\frac{x^2}{4}+y^2=1$
if the point $A(0,1)$ on the ellipse $\Gamma:$ $\dfrac{x^2}{4}+y^2=1$ and the circle $\tau:$ $(x+1)^2+y^2=r^2(0<r<1)$,if $AB,AC$ tangent the circle $\tau$ ,$B,C\in \Gamma$,show that the line $BC$ always passes through a fixed point
I try Let $AB:y=kx+1$ then
$$\begin{cases}
\dfrac{x^2}{4}+y^2=1\\
y=kx+1
\end{cases}
$$
so we have
$$x^2+4(kx+1)^2=4\Longrightarrow (4k^2+1)x^2+8kx=0$$
so we have
$$B(-\dfrac{8k}{4k^2+1},\dfrac{1-4k^2}{1+4k^2})$$
But for $C$ it hard to find it.
|
The equation of the line passing through $A(0,1)$ is given by $mx-y+1=0$.
Since we want this line to be tangent to the circle, we have
$$r=\frac{|m(-1)-0+1|}{\sqrt{m^2+(-1)^2}},$$
i.e.
$$(r^2-1)m^2+2m+r^2-1=0\implies m_1+m_2=\frac{2}{1-r^2},\quad m_1m_2=1\tag1$$
Eliminating $y$ from $mx-y+1=0$ and $x^2/4+y^2=1$ gives
$$\frac{x^2}{4}+(mx+1)^2=1\implies x(x+4m^2x+8m)=0\implies x=0,\frac{-8m}{4m^2+1}$$
and
$$y=mx+1=m\cdot \frac{-8m}{4m^2+1}+1=\frac{-4m^2+1}{4m^2+1}$$
So we can write $B\left(\frac{-8m_1}{4m_1^2+1},\frac{-4m_1^2+1}{4m_1^2+1}\right),C\left(\frac{-8m_2}{4m_2^2+1},\frac{-4m_2^2+1}{4m_2^2+1}\right)$.
Using $(1)$, we have
$$\frac{b_y-c_y}{b_x-c_x}=\frac{\frac{-4m_1^2+1}{4m_1^2+1}-\frac{-4m_2^2+1}{4m_2^2+1}}{\frac{-8m_1}{4m_1^2+1}-\frac{-8m_2}{4m_2^2+1}}=-\frac{m_1+m_2}{3}=\frac{2}{3(r^2-1)}$$and
$$\frac{-c_xb_y+b_xc_y}{b_x-c_x}=\frac{-\frac{-8m_2}{4m_2^2+1}\cdot\frac{-4m_1^2+1}{4m_1^2+1}+\frac{-8m_1}{4m_1^2+1}\cdot\frac{-4m_2^2+1}{4m_2^2+1}}{\frac{-8m_1}{4m_1^2+1}-\frac{-8m_2}{4m_2^2+1}}=-\frac{5}{3}$$
So, the equation of the line $BC$ is given by
$$y=\frac{b_y-c_y}{b_x-c_x}x+\frac{-c_xb_y+b_xc_y}{b_x-c_x}=\frac{2}{3(r^2-1)}x-\frac 53$$
which passes through $(0,-5/3)$.
|
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Show that $5\cdot10^n+10^{n-1}+3$ is divisible by 9 Prove by induction the following:
$5*10^n+10^{n-1}+3$ is divisible by 9
Base case: $n=1$ $5*10+10^{1-1}+3=5*10+10^0+3=50+1+3=54$
$9|54=6$
Inductive Hypothesis: If $k$ is a natural number such that $9|5*10^k+10^{k-1} +3$
Inductive step:
Show that $S_k$ is true $\Rightarrow$ $S_{k+1}$ is true
$S_{k+1}$: $9|5*10^{k+1}+10^{k} +3$
$9|10(5*10^{k+1}+10^{k} +3)$
$9|5*10^{k+2}+10^{k+1} +10*3$
$9|5*(10^{k+1}*10^1)+(10^{k}*10^1) +(9+1)*3$
$9|5*(10^{k+1}*(9+1))+(10^{k}*10*(9+1)) +(9+1)*3$
$9|5(9*10^{k+1}+10^{k+1})+9*10^k+10^k+((9*3)+(1*3))$
This is were I am stuck for the last day try to figure out what move next would speed up the inductive proof as I have a feeling it can be finished up. Anyone help me see what I am unable to find.
|
The best way to show this kind of an equality, is to show that $S_{k-1} - S_k$ is divisible by $9$.
\begin{split} S_{k+1} - S_k & = (5 \times 10^{k+1} + 10^{k} + 3) - (5 \times 10^k + 10^{k-1} + 3)\\ & = 5(10^{k+1}-10^k) + (10^k-10^{k-1}) \\ & = 5\times 9 \times 10^k + 9 \times 10^{k-1} \\ & = 9 \times (5 \times 10^k + 10^{k-1}) \end{split}
Hence, $S_{k+1} = S_k + 9 \times (5 \times 10^k + 10^{k-1})$. Since $9|S_k$ and $9|9 \times (5 \times 10^k + 10^{k-1})$, it is clear that $9|S_{k+1}$. Hence, the induction is complete.
It is also nice to note that the pattern of numbers produced by $S_k$ go like $54$,$513$,$5103$,$51003$ etc. all of which have sum of digits the same i.e. $9$, so all are divisible by $9$ (this needn't be proved inductively at all)
|
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Right Triangle: Hypotenuse and Side differ by 1 So I have search to the best of my abilities but cannot find mathematically why this is true and if it is called something specific, the closest thing would be Pythagorean Triples but this is not the case although a (3,4,5) is a triple. I will do my best to explain and apologize in advance as math is not my strength.
So in any right triangle I could find the following holds true. If (a,b,c) and b differs -1 from c the other angle is the sum of a = sqrt{b+c}.
For example:
(3,4,5) a = sqrt{4+5} = sqrt{9} = 3 = sqrt{9} = sqrt{25-16} = sqrt{5^2 - 4^2}
(a,17,18) a = sqrt{17+18} = sqrt{35} = sqrt{324-289} = sqrt{18^2 - 17^2}
The only way I understand it is the difference between two squared numbers who base differs by 1 will always be the sum of both numbers.
So assuming b is always a+1, b^2 - a^2 = a + b. Any explanation or pointing to where I can read about more would be much appreciated.
|
For all $a,b$
$$b^2 - a^2 = (b+a)(b-a)$$
Assuming as you do that $b=a+1$
$$b^2 - a^2 = (b+a)(b-a) = (2a+1)(1)=2a+1$$
and
$$a+b=a+a+1=2a+1$$
so yes, when $b=a+1$, $b^2 - a^2 = a + b$
|
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Determinant with variables What is the following determinant?
$$\begin{vmatrix}1+a & b & c & d \\a & 1+b & c & d \\a & b & 1+c & d \\a & b & c & 1+d \end{vmatrix}$$
I calculated it as $0$ but I do not think it is right. Thanks in advance.
|
$$\begin{vmatrix}1+a& b & c & d \\a & 1+b & c & d \\a & b & 1+c & d \\a & b & c & 1+d \end{vmatrix}=\begin{vmatrix}1 & b & c & d \\0 & 1+b & c & d \\0 & b & 1+c & d \\0 & b & c & 1+d \end{vmatrix}+\begin{vmatrix}a & b & c & d \\a & 1+b & c & d \\a & b & 1+c & d \\a & b & c & 1+d \end{vmatrix}\\
=\begin{vmatrix}1+b & c & d \\b & 1+c & d \\ b & c & 1+d \end{vmatrix}+\begin{vmatrix}a & b & c & d \\0 & 1 & 0 & 0 \\0 &0 & 1 & 0 \\0&0 & 0 & 1\end{vmatrix}\\
=\begin{vmatrix}1 & c & d \\0 & 1+c & d \\ 0 & c & 1+d \end{vmatrix}+\begin{vmatrix}b & c & d \\b & 1+c & d \\ b & c & 1+d \end{vmatrix}+a\\
=\begin{vmatrix}1+c & d \\ c & 1+d \end{vmatrix}+\begin{vmatrix}b & c & d \\0& 1 & d \\ 0 & 0& 1 \end{vmatrix}+a\\
=\begin{vmatrix}1 & d \\ 0 & 1+d \end{vmatrix}+\begin{vmatrix}c & d \\ c & 1+d \end{vmatrix}+b+a\\
=\begin{vmatrix}1+d \end{vmatrix}+\begin{vmatrix}c & d \\ 0& 1 \end{vmatrix}+b+a\\
=1+d+c+b+a.$$
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If $\sin \alpha +\cos \alpha =1.2$, then what is $\sin^3\alpha + \cos^3\alpha$? If $\sin \alpha +\cos \alpha =1.2$, then what is $\sin^3\alpha + \cos^3\alpha$?
All I know is that $\sin^{3}a+\cos^{3}a$ is equal to
$$(\sin a + \cos a)(\sin^{2} a - \sin a \cos a + \cos^{2} a)= \dfrac{6}{5}(\sin^{2} a - \sin a \cos a + \cos^{2} a)$$ But now, I'm stuck. Solutions are greatly appreciated.
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${x^3+y^3 \over x+y} = x^2+y^2-xy$.
$(x+y)^2 = x^2+y^2+ 2 xy$, and so
${x^3+y^3 \over x+y} = x^2+y^2 -{1 \over 2} ((x+y)^2-(x^2+y^2))$, from which we get
$x^3+y^3 = (x+y) (x^2+y^2 -{1 \over 2} ((x+y)^2-(x^2+y^2)))$.
Since $x+y = 1.2, x^2+y^2 = 1$ we get $x^3+y^3 = (1.2)(1-{1 \over 2}((1.2^2-1)) = {117 \over 125}$.
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Find the inverse of 17 mod 41 Questions
(1) Find the inverse of $17 \mod 41$.
(2) Find the smallest positive number n so that $17n \equiv 14 \pmod{41}$
For the first question, my attempt is as follows:
$$41-17\cdot2=7$$
$$17-7\cdot2=3$$
$$7-3\cdot2=1$$
$$7-2(17-7\cdot2)=1$$
$$7-2\cdot17=1$$
$$41-17\cdot2-2\cdot17=1$$
$$41-4\cdot17=1$$
So the inverse of $17$ is $-4$.
That is, the inverse of $17$ is $37$
Am I right?
|
Using Fermat's little theorem, $$17^{40}\equiv 1\pmod{41}\Rightarrow 17^{39}\cdot17\equiv 1\pmod{41}$$ Hence $17^{39}$ is the searched inverse. We have $$17^2\equiv 2\pmod{41}\Rightarrow
17^{30}\equiv 2^{15}\equiv 9\pmod{41}$$
$$17^9\equiv(17^2)^4\cdot 17\equiv 16\cdot17\equiv26\pmod{41}$$Hence $$17^{39}\equiv9\cdot26\equiv 29\pmod{41}$$
Thus the inverse is $29$.
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How to calculate width and height of a 45° rotated ellipse bounded by a square? I'm coming from a programming background so I apologies if this is blindingly simple or I misuse terms. I have an ellipse bounded by a square. For simplicity the centre of the square and ellipse is the origin (0,0) while the square is 2 width and 2 height.
The ellipse is rotated -45° or +45° (angle in image) and I can easily work this out. The ellipse touches all sides of the square, and I also know the intersection points. In the image A is the distance between the corner and the intersection while B is the length of the long section (the other points are symmetric).
What is the width and height, as described in the image, of the ellipse?
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Consider an square formed by the sides of $\pm x\pm y=m$ (i.e. side length $m\sqrt2$).
Now find an ellipse which touches the square. Consider the top right quadrant. Edge of square is $$L: \quad x+y=m$$Assume point of tangency is $P(h,k)$.
Ellipse, $E$:
$$\begin{align}
\frac {x^2}{a^2}+\frac {y^2}{b^2}&=1\\
\frac {dy}{dx}&=-\frac {b^2x}{a^2y}=-\frac {b^2h}{a^2k}=-1\quad\text{at }P\\
\frac h{a^2}&=\frac k{b^2}
\end{align}$$
$P$ lies on L, hence $h+k=m$, which gives
$$\begin{align}
h&=\left(\frac {m}{a^2+b^2}\right)a^2\\
k&=\left(\frac {m}{a^2+b^2}\right)b^2
\end{align}$$
$P$ lies on $E$, hence
$$\begin{align}
\left(\frac {m}{a^2+b^2}\right)^2\left[\frac {(a^2)^2}{a^2}+\frac {(b^2)^2}{b^2}\right]&=1\\
m&=\sqrt{a^2+b^2}
\end{align}$$
This gives
$$\begin{align}
h&=\frac {a^2}m\\
k&=\frac {b^2}m\end{align}$$
$P$ divides edge of square into side lengths $$B=h\sqrt{2}=\frac {a^2\sqrt{2}}m\\ A=k\sqrt{2}=\frac {b^2\sqrt{2}}m$$
In the question posted, side length of square is $2$, hence $m=\sqrt{2}$, which gives
$$A=b^2\\
B=a^2$$
"Height" (major axis) and "Width" (minor axis) of $E$ are given by
$$\text{Height (major axis)=}\color{red}{2a=2\sqrt{B}}\\
\text{Width (minor axis)=}\color{red}{2b=2\sqrt{A}}$$
Note
$\hspace{3cm}$
From this solution to another recent MSE question, note that the ellipse
$$\frac{x^2}{u}+\frac {y^2}{1-u}=1$$
has a nice property in that its point of tangency with the line $L: x+y=1$ is $(u,1-u)$ which divides the line segment $V(0,1), H(1,0)$ on $L$ in the ratio $u:1-u$. Note that $0<u<1$. The line segment $VH$ has length $\sqrt2$. The semi-major and semi-minor axes of the ellipse are $\sqrt{u}, \sqrt{1-u}$ respectively.
|
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|
How can I solve $T(n) = \lfloor \frac {n} {2}\rfloor + T(\lceil \frac{n}{2} \rceil)$? I came u with the following equation:
$T(2) = 1$
$T(n) = \lfloor \frac {n} {2}\rfloor + T(\lceil \frac{n}{2} \rceil) | n ∈ \mathbb N$
But I can't find a way to solve it. Is there one?
I need to proof that the solution is right, so I can't just look at the numerical sequence of the equation.
|
You can solve this as follows. First calculate the first few values
\begin{align}
T(3) = 1 + T(2) = 2 \\
T(4) = 2 + T(2) = 3 \\
T(5) = 2 + T(3) = 4 \\
T(6) = 3 + T(3) = 5
\end{align}
Ok, it might be that T(n) = n - 1. Let's try to prove this by induction. Firstly, we have $T(2) = 1$ so this is obviously true for $n = 2$. Now assume it is true for $n = 2, 3 , \dots k$ and we would now like to prove it for $n= k+1$.
\begin{equation}
T(k+1) = \left\lfloor \frac{k+1}{2} \right\rfloor + T \left( \left\lceil \frac{k+1}{2}\right\rceil\right).
\end{equation}
Now, we must consider two different cases. Either $k+1$ is even or odd. If $k$ is even the above equation becomes
\begin{equation}
T(k+1) = \frac{k+1}{2} + T\left( \frac{k+1}{2} \right) = \frac{k+1}{2} + \left( \frac{k+1}{2} - 1 \right) = (k+1) - 1 .
\end{equation}
Above I used that we have assumed that $T(n) = n-1$ for $n\leq k$ in the second step. Now assume that $k+1$ is odd. We then get
\begin{equation}
T(k+1) = \frac{k+1}{2} - 0.5 + T\left(\frac{k+1}{2} + 0.5 \right) =
\frac{k+1}{2} - 0.5 + \left(\frac{k+1}{2} + 0.5 -1 \right) = (k+1)-1.
\end{equation}
Hence, by the induction principle, this is true for all $n$.
|
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|
Unique pair of positive integers $(p,n)$ satisfying $p^3-p=n^7-n^3$ where $p$ is prime
Q. Find all pairs $(p,n)$ of positive integers where $p$ is prime and $p^3-p=n^7-n^3$.
Rewriting the given equation as $p(p+1)(p-1)=n^3(n^2+1)(n+1)(n-1)$, we see that $p$ must divide one of the factors $n,n+1,n-1,n^2+1$ on the $\text{r.h.s}$.
Now, the $\text{l.h.s}$ is an increasing function of $p$ for $p\ge1$. This implies that for any given $n\ge1$, there is exactly one real $p$ for which $\text{l.h.s}=\text{r.h.s}$. For $p=n^2$, we get $\text{l.h.s}=n^6-n^2<n^7-n^3=\text{r.h.s}.$ This means that either $p>n^2$ or $p<n^2$ must hold.
Assuming $p>n^2$, it follows that the prime $p$ cannot divide any of $n,n+1,n-1$. So $p$ must divide $n^2+1$ and hence $p=n^2+1\quad (\because p>n^2)$.
Substituting the value of $p$ in the given equation we get, $n^2+2=n^3-n\implies n^3-n^2-n=2$. As the factor $n$ on the $\text{l.h.s}$ must divide $2$, the above equation has a unique integer solution $n=2$.
Finally, we get $(5,2)$ as the solution to the given equation.
But how do I conclude this is the only solution possible? Also, why does'nt $p<n^2$ (the case which I ignored) hold? As a bonus question, I would like to ask for any alternative/elegant solution (possibly using congruence relations) to the problem.
|
Given that $p,n\in\mathbb{Z}^+$, where $p$ is a prime, such that $p^3-p=n^7-n^3$.
This implies that $p(p^2-1)=n^3(n^2-1)(n^2+1)\implies p|n^3$ or $p|n^2-1$ or $p|n^2+1$.
Case 1: $p|n^3$. This implies that $p|n\implies n=pk$ for some positive integer $k$. Therefore we have $n^7=p^7k^7$ and $n^3=p^3k^3$. Therefore $$p^3-p=p^7k^7-p^3k^3\\ \implies p^2-1=p^6k^7-p^2k^3\\\implies p^2-1=p^2(p^4k^7-k^3)\\\implies p^2|p^2-1, \text{ which is a contradiction.}$$ Hence $p\not|n^3.$
Case 2: $p|n^2-1.$ Observe that for any prime $p_1, p_1^3-p_1>0\implies p^3-p>0\implies n^7-n^3=n^3(n^2-1)(n^2+1)>0.$ Now for any $n_1\in\mathbb{N}, n_1^3>0$ and $n_1^2+1>0$. This implies that $n^3>0$ and $n^2+1>0$, which in turn implies that we must have $n^2-1>0$.
Therefore, $p|n^2-1$ implies that $pk=n^2-1$, for some positive integer $k$. Therefore, $n^2+1=pk+2$ and $n^3=pkn+n$.
Therefore we have $$p^3-p=n^7-n^3=n^3(n^2-1)(n^2+1)=(pkn+n)(pk)(pk+2)\\\implies p^2-1=(pk^2+2k)(pkn+n).$$
Now $pk^2+2k>p$ and $pkn+n>p$, which in turn implies that $(pk^2+n)(pkn+n)>p^2$. Therefore we have $$p^2-1>p^2\\\implies -1>0, \text{ which is a contradiction.}$$ Hence $p\not |n^2-1.$
Case 3: $p|n^2+1$. This implies that $p\le n^2+1$. Now since we have $$p^3-p=n^7-n^3\implies p^3-p=n^3(n^4-1)\implies n^3|p^3-p\implies n^3\le p^3-p\implies n^3<p^3\implies n<p.$$
Therefore we have $n<p\le n^2+1$. Thus $n^2<p^2\le (n^2+1)^2\implies n^2-1< p^2-1\le n^4+2n^2.$
Hence we have $$n^3-n<p^3-p\le (n^2+1)(n^4+2n^2)\\ \implies n^3-n<n^7-n^3\le n^6+3n^4+2n^2\\ \implies n^2-1<n^6-n^2\le n^5+3n^3+2n\hspace{0.5 cm}...(*)$$
Observe that $(*)$ is satisfied only by $n=1$ and $n=2$.
Now of course $n=1$ yields no solution, since $p^3-p>0$.
Now when $n=2$, we have $p^3-p=(p-1)p(p+1)=120.$ Observe that only $p=5$ satisfies the equation.
Therefore, the only pair $(p,n)$ that satisfies the equation $p^3-p=n^7-n^3$ is $(p,n)=(5,2)$.
|
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|
convergence of $\sum_{n=1}^{\infty}(n^2+1)\log^\alpha \left ( \frac{ \sinh \frac{1}{n}}{\sin\frac{1}{n}} \right )$ $ \sum_{n=1}^{\infty}(n^2+1)\log^\alpha \left ( \frac{\sinh\frac{1}{n}}{\sin\frac{1}{n}} \right ) $
Find the value of $\alpha$ for which the series converges
please, I have no idea to approach the solution
thanks.
|
Hint: use Taylor series, again, again, and again.
In detail:
From a Taylor series around $0$, since $\frac{1}{n}\xrightarrow[n\to\infty]{}0$:
$$\sinh \frac{1}{n} = \frac{1}{n} + \frac{1}{6n^3} + o\left(\frac{1}{n^3}\right) \tag{1}$$
and
$$\sin \frac{1}{n} = \frac{1}{n} - \frac{1}{6n^3} + o\left(\frac{1}{n^3}\right)\tag{2}$$
from which
$$
\frac{\sinh \frac{1}{n}}{\sin \frac{1}{n}}
= \frac{1+\frac{1}{6n^2}+ o\left(\frac{1}{n^2}\right)}{1-\frac{1}{6n^2}+ o\left(\frac{1}{n^2}\right)}
= 1+\frac{1}{3n^2}+ o\left(\frac{1}{n^2}\right).
$$
(Using $\frac{1}{1+x}= 1-x+o(x)$ around $0$). From the Taylor series $\log(1+x)=x+o(x)$ at $0$, we get
$$
\log^\alpha \frac{\sinh \frac{1}{n}}{\sin \frac{1}{n}}
= \log^\alpha\left( 1+\frac{1}{3n^2}+ o\left(\frac{1}{n^2}\right)\right)
= \left( \frac{1}{3n^2}+ o\left(\frac{1}{n^2}\right)\right)^\alpha
= \frac{1}{3^\alpha n^{2\alpha}}+ o\left(\frac{1}{n^2\alpha}\right)
$$
and finally
$$
(n^2+1)\log^\alpha \frac{\sinh \frac{1}{n}}{\sin \frac{1}{n}}
= \frac{1}{3^\alpha n^{2(\alpha-1)}}+ o\left(\frac{1}{n^{2(\alpha-1)}}\right).
$$
Theorems of comparison for positive series should then lead you to the conclusion: you need $2(\alpha-1)> 1$.
|
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|
If $a_{ij}=\max(i,j)$, calculate the determinant of $A$
If $A$ is an $n \times n$ real matrix and
$$a_{ij}=\max(i,j)$$
for $i,j = 1,2,\dots,n$, calculate the determinant of $A$.
So, we know that
$$A=\begin{pmatrix}
1 & 2 & 3 & \dots & n\\
2 & 2 & 3 & \dots & n\\
3 & 3 & 3 & \dots & n\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
n& n & n & \dots & n
\end{pmatrix}$$
but what do I do after?
|
Apply row operations $R_{j} \leftarrow R_{j} - R_{1}, \, \forall j \in \left \{2, ...n\right \}$ on $A$ to get the following matrix:
$A_1 = \begin{pmatrix}
1& 2 & 3 & ... & n-1 & n\\
1& 0& 0 & ... & 0& 0\\
2& 1& 0 & ... & 0& 0\\
...& ...& ... & ... & ...& ...\\
n-2& n-3 & n-4 & ...& 0& 0\\
n-1& n-2& n-3& ...& 1& 0
\end{pmatrix}$
Now, apply the following row operations in the order of $j = 2,3,...,n-1$,
$R_{k} \leftarrow R_{k} - (k-1)R_{j}, \, \forall k \in \left \{j+1, j+2, ..., n\right \}$
to get the following matrix:
$A_2 = \begin{pmatrix}
1& 2 & 3 & ... & n-1 & n\\
1& 0& 0 & ... & 0& 0\\
0& 1& 0 & ... & 0& 0\\
...& ...& ... & ... & ...& ...\\
0& 0 & 0 & ...& 0& 0\\
0& 0& 0& ...& 1& 0
\end{pmatrix}$
Now apply the following row operations $R_1 \leftarrow (j-1)R_j, \, \forall j \in {2,3,...,n}$ to get the following matrix:
$A_3 = \begin{pmatrix}
0& 0 & 0 & ... & 0 & n\\
1& 0& 0 & ... & 0& 0\\
0& 1& 0 & ... & 0& 0\\
...& ...& ... & ... & ...& ...\\
0& 0 & 0 & ...& 0& 0\\
0& 0& 0& ...& 1& 0
\end{pmatrix}$
Note that $A_3$ has a value of $n$ at $(1,n)$th cell, all $1$s on the first lower off-diagonal and all other elements are $0$s. Now, compute the determinant along the first row to get:
$|A_3| = (-1)^{n+1}n$
Since, all the above row operations do not change the value of the determinant, we obtain,
$|A| = |A_1| = |A_2| = |A_3| = (-1)^{n+1}$
|
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|
Probability of no complete losses and no complete wins In the NFL, a division consists of 4 teams, each of which plays each other team twice.
Assume that in any game, either team is equally likely to win (and there are no ties). What is the probability that, at the end of the season, the division has neither a perfect team with 6 wins nor a futile team with 6 losses?
Enter your answer as a fraction in simplest form.
I have approached this question using a combination of casework and complementary counting.
I set up a table for a team winning all games and another table for a team winning absolutely no games.
|
You need to consider the important outcomes and their probabilities
*
*The probability that Team A wins all $6$ matches is $\dfrac{1}{2^6}$
*The probability that Team D loses all $6$ matches is $\dfrac{1}{2^6}$
*The probability that Team A wins all $6$ matches and Team D loses all $6$ is $\dfrac{1}{2^{10}}$
*The probability that a team wins all $6$ matches is $\displaystyle 4 \times \dfrac{1}{2^6}$
*The probability that a team loses all $6$ matches is $\displaystyle 4 \times \dfrac{1}{2^6}$
*The probability that a team wins all $6$ matches and another team loses all $6$ is $\displaystyle 12 \times \dfrac{1}{2^{10}}$
*The probability that a team wins all $6$ matches but no team loses all $6$ is $\displaystyle 4 \times \dfrac{1}{2^6} - \displaystyle 12 \times \dfrac{1}{2^{10}}$
*The probability that no team wins all $6$ matches but a team loses all $6$ is $\displaystyle 4 \times \dfrac{1}{2^6} - \displaystyle 12 \times \dfrac{1}{2^{10}}$
*The probability that no team wins all $6$ matches and no team loses all $6$ is $$1- \left(4 \times \dfrac{1}{2^6} - 12 \times \dfrac{1}{2^{10}}\right)- \left(4 \times \dfrac{1}{2^6} - 12 \times \dfrac{1}{2^{10}}\right) - 12 \times \dfrac{1}{2^{10}} =\dfrac{227}{256}$$
|
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|
Knowing a quartic has a double root, how to find it? I have a depressed quartic polynomial with three free parameters in the real numbers:
$x^{4}+qx^{2}+rx+s$
Furthermore, the discriminant is constrained to be zero and there are four real roots, exactly two of which are equal. I am only interested in the double root.
Wikipedia has the general solution. However, is there a simpler formula for this special case?
|
Yes, there is. You know that your roots (including multiplicities) can be expressed as $\{a, a, -a+δ, -a-δ\}$ (because the sum is zero after the normalization). Expand
$$(x-a)^2(x+a-δ)(x+a+δ) = x^4 - (2a^2 + δ^2)x^2 + 2aδ^2 x + a^4-δ^2a^2$$
and compare with your form. Next, note that the equations
$$\begin{aligned}
- (2a^2 + δ^2) &= q \\
a^4-δ^2a^2 &= s
\end{aligned}$$
can be both expressed using
$$A = a^2, D = δ^2$$
and they only lead to a quadratic equation for $A$. It's easy from there. You'll get $a^2$ but the sign of $a$ is correlated with the sign of the linear term (which otherwise only contains factors of a 2 and a square).
Generally:
$$\bbox[7px,border:2px solid]{a = (\mathop{\rm sgn} r)\cdot \sqrt{\frac{-q \pm \sqrt{q^2 + 12s}}6}}\ .$$
The sign inside the square root needs to be chosen so that $2aD = -2a(q+2a^2) = r$. For $s > 0$ only + guarantees a real $a$ but if $s < 0$ then both signs lead to a real solution and one of them corresponds to a quartic with a different $r$.
Example:
$$\begin{aligned}
p(x) &= x^4 - 19x^2 + 6x + 72 \\
⇒ 2A + D &= 19 \\
A^2 - AD &= 72 \\
⇒ A^2 - A(19-2A) = 3A^2 - 19A &= 72 \\
⇒ A &= \frac{19 \pm \sqrt{1225}}{6} = \frac{19 \pm 35}6, \quad D = 19-2A \\
A ≥ 0 ⇒ A &= 9, \quad D = 1 \\
⇒ a &= 3, d = 1
⇒ p(x) = (x-3)^2 (x+2) (x+4)
\end{aligned}$$
|
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|
Evaluating $\sum _{n=1}^{\infty } \sinh ^{-1}\left(\frac{1}{\sqrt{2^{n+1}+2}+\sqrt{2^{n+2}+2}}\right)$ How can we evaluate:
$$\sum _{n=1}^{\infty } \sinh ^{-1}\left(\frac{1}{\sqrt{2^{n+1}+2}+\sqrt{2^{n+2}+2}}\right)$$
I tried to transform it into $$\sinh ^{-1}\left(\frac{1}{\sqrt{2^{n+1}+2}+\sqrt{2^{n+2}+2}}\right)$$$$=\ln \left(\sqrt{\frac{1}{\left(\sqrt{2^{n+1}+2}+\sqrt{2^{n+2}+2}\right)^2}+1}+\frac{1}{\sqrt{2^{n+1}+2}+\sqrt{2^{n+2}+2}}\right)$$
So that the original expression becomes:
$$\sum _{n=1}^{\infty } \sinh ^{-1}\left(\frac{1}{\sqrt{2^{n+1}+2}+\sqrt{2^{n+2}+2}}\right)$$
$$=\ln\left(\prod_{n=1}^\infty\left(\sqrt{\frac{1}{\left(\sqrt{2^{n+1}+2}+\sqrt{2^{n+2}+2}\right)^2}+1}+\frac{1}{\sqrt{2^{n+1}+2}+\sqrt{2^{n+2}+2}}\right)\right)$$
But it doesn't look good at all.
The final answer is close to 0.658479, which is $\ln\sqrt{2+\sqrt{3}}$
|
Note that $$\sinh^{-1} x-\sinh^{-1} y=
\sinh^{-1} \left( x\sqrt{1+y^2}-y\sqrt{1+x^2} \right)$$
\begin{align*}
\frac{1}{\sqrt{2^{n+2}+2}+\sqrt{2^{n+1}+2}} &=
\frac{\sqrt{2^{n+2}+2}-\sqrt{2^{n+1}+2}}
{(2^{n+2}+2)-(2^{n+1}+2)} \\
&= \frac{\sqrt{2^{n+2}+2}-\sqrt{2^{n+1}+2}}
{2^{n+1}} \\
&= \sqrt{\frac{1}{2^{n}}
\left( 1+\frac{1}{2^{n+1}} \right)}-
\sqrt{\frac{1}{2^{n+1}}
\left( 1+\frac{1}{2^{n}} \right)} \\
\sinh^{-1}
\left(
\frac{1}{\sqrt{2^{n+1}+2}+\sqrt{2^{n+2}+2}}
\right) &=
\sinh^{-1} \frac{1}{\sqrt{2^{n}}}-\sinh^{-1} \frac{1}{\sqrt{2^{n+1}}} \\
\sum_{n=1}^{\infty} \sinh^{-1}
\left(
\frac{1}{\sqrt{2^{n+1}+2}+\sqrt{2^{n+2}+2}}
\right) &=
\sinh^{-1} \frac{1}{\sqrt{2}} \\
&= \ln \left( \frac{1+\sqrt{3}}{\sqrt{2}} \right) \\
&= \ln \sqrt{2+\sqrt{3}}
\end{align*}
|
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|
Solve the equation $\sqrt{3x^2-7x-30}-\sqrt{2x^2-7x-5}=x-5$
Problem Statement:-
Solve the equation $\sqrt{3x^2-7x-30}-\sqrt{2x^2-7x-5}=x-5$
Attempt at solution:-
Let $\alpha=\sqrt{3x^2-7x-30}\;\;\; \text{&} \;\;\;\beta=\sqrt{2x^2-7x-5}$.
Then, we have $$\alpha-\beta=x-5\tag{1}$$
And, we have $$\alpha^2-\beta^2=x^2-25\tag{2}$$
From $(1)$ and $(2)$, we have
$$\alpha+\beta=\dfrac{(x-5)(x+5)}{x-5}=x+5\qquad(\therefore x\neq5)\tag{3}$$
From $(1)$ and $(3)$, we get
$$\alpha=\sqrt{3x^2-7x-30}=x\qquad\qquad\beta=\sqrt{2x^2-7x-5}=5$$
On solving any one of $\alpha=x$ or $\beta=5$, we get $x=6,-\dfrac{5}{2}$.
On putting $x=-\dfrac{5}{2}$, $\alpha=x$ is not satisfied. Hence, $x=6$ is the only solution.
But as in $(3)$, we have ruled out $x=5$, as a solution then we can't put $x=5$ in the original equation. But, if we do put $x=5$ in the original equation we see that it is indeed satisfied. So, I tried a solution which also gives $x=5$ as a solution. So I picked the last solution from the $(2)$ without cancelling $(x-5)$.
$$(x-5)(\alpha+\beta)=x^2-25\implies (x-5)(\alpha+\beta-(x+5))=0$$.
Now, I am pretty much stuck here.
Edit-1:- I just saw now and feel pretty stupid about it.
If we proceed from the last step i.e. $(x-5)(\alpha+\beta-(x+5))=0$, we get
$$(x-5)(\alpha+\beta-(x+5))=0\implies (x-5)=0\;\;\;\text{ or }\;\;\;\alpha+\beta=x+5$$
So we have the following equations to be solved:-
$$\alpha-\beta=x-5\\
\alpha+\beta=x+5\\
x-5=0$$
Which results in $x=5,6$.
Now, since I have put so much effort in posting this question, I might as well ask for better solutions if you can come up with one. And, please don't post a solution which includes a lot of squaring to result into a quartic equation.
|
As $(\sqrt{x^2-7x-30})^2-(\sqrt{2x^2-7x-5})^2=(x+5)(x-5),$
$$\sqrt{3x^2-7x-30}-\sqrt{2x^2-7x-5}=x-5$$
$$\implies \sqrt{3x^2-7x-30}+\sqrt{2x^2-7x-5}=x+5$$
Adding we get $$x=\sqrt{3x^2-7x-30}\ \ \ \ (1)$$
Squaring $(1)$ we get $$0=2x^2-7x-30=(2x+5)(x-6)$$
Now for real $x,x=\sqrt{3x^2-7x-30}\ge0$
|
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|
How can I derive what is $1\cdot 2\cdot 3\cdot 4 + 2\cdot 3\cdot 4\cdot 5+ 3\cdot 4\cdot 5\cdot 6+\cdots + (n-3)(n-2)(n-1)(n)$ ?? I'd like to evaluate the series $$1\cdot 2\cdot 3\cdot 4 + 2\cdot 3\cdot 4\cdot 5+ 3\cdot 4\cdot 5\cdot 6+\cdots + (n-3)(n-2)(n-1)(n)$$
Since I am a high school student, I only know how to prove such formula's (By principal of mathematical induction). I don't know how to find result of such series.
Please help. I shall be thankful if you guys can provide me general solution (Since I have been told that there exist a general solution by my friend who gave me this question).
|
Can you prove that $(n-3)(n-2)(n-1)(n) = \frac{(n-3)(n-2)(n-1)(n)(n+1)}{5} - \frac{(n-4)(n-3)(n-2)(n-1)(n)}{5}?$
From there you can substitute different values of $n$ and derive a formula for your expression, and the final summation will be given by
$$ \frac{(n-3)(n-2)(n-1)(n)(n+1)}{5}$$
|
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|
How can we show $\cos^6x+\sin^6x=1-3\sin^2x \cos^2x$? How can we simplify $\cos^6x+\sin^6x$ to $1−3\sin^2x\cos^2x$?
One reasonable approach seems to be using $\left(\cos^2x+\sin^2x\right)^3=1$, since it contains the terms $\cos^6x$ and $\sin^6x$. Another possibility would be replacing all occurrences of $\sin^2x$ by $1-\cos^2x$ on both sides and then comparing the results.
Are there other solutions, simpler approaches?
I found some other questions about the same expression, but they simplify this to another form: Finding $\sin^6 x+\cos^6 x$, what am I doing wrong here?, Alternative proof of $\cos^6{\theta}+\sin^6{\theta}=\frac{1}{8}(5+3\cos{4\theta})$ and Simplify $2(\sin^6x + \cos^6x) - 3(\sin^4 x + \cos^4 x) + 1$. Perhaps also Prove that $\sin^{6}{\frac{\theta}{2}}+\cos^{6}{\frac{\theta}{2}}=\frac{1}{4}(1+3\cos^2\theta)$ can be considered similar. The expression $\cos^6x+\sin^6x$ also appears in this integral Find $ \int \frac {\tan 2x} {\sqrt {\cos^6x +\sin^6x}} dx $ but again it is transformed to a different from then required here.
Note: The main reason for posting this is that this question was deleted, but I still think that the answers there might be useful for people learning trigonometry. Hopefully this new question would not be closed for lack of context. And the answers from the deleted question could be moved here.
|
$$\cos^6x+\sin^6x=(\cos^2x)^3+(\sin^2x)^3=$$
$$=(\cos^2x+\sin^2x)(\cos^4x-\cos^2x\sin^2x+\sin^4x)=$$
$$=\cos^4x-\cos^2x\sin^2x+\sin^4x=$$
$$=((\cos^2x)^2+(\sin^2x)^2+2\cos^2x\sin^2x)-3\cos^2x\sin^2x=$$
$$=(\cos^2x+\sin^2x)^2-3\cos^2x\sin^2x=$$
$$=1-3\cos^2x\sin^2x$$
|
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|
Prove that if $a$ is odd, then $a^2\equiv 1\pmod 8$ Prove that if $a$ is odd, then $a^2\equiv 1\pmod 8$I got this question in my discrete mathematics class, can anyone help me?Thanks.
|
The clever way to do it is:
$a$ odd $\implies a -1$ and $a+1$ are two consecutive even numbers. As they are consecutive one of them is divisible by $4$ and the other is divisible by $2$. We have no way of knowing which is which but it doesn't matter. One is divisible by $2$ and the other is divisible by $4$ and multiplied together $(a-1)(a+1)$ is divisible by $8$.
So $(a-1)(a+1) \equiv 0 \mod 8$
So $a^2 -1 \equiv 0 \mod 8$
So $a^2 \equiv 1 \mod 8$.
=====
The brute force way is easier.
$a$ is odd means $a \equiv 1, 3, 5, \text{ or } 7 \mod 8$ so
$a^2 \equiv 1,9, 25, \text{ or } 49 \mod 8 \equiv 1 \mod 8$.
===
An intermediary method might by to realize $a = 2k + 1$ so $a^2 = 4k^2 + 4k + 1 = 4(k^2 + k) + 1$.
If $k$ is odd then $k^2$ is odd and $k^2 + k$ is even.
If $k$ is even then $k^2$ is even and $k^2 + k$ is even.
So $k^2 + k$ is divisible by $2$.
So $4(k^2 + k)$ is divisible by $8$.
So $4(k^2 + k) \equiv 0 \mod 8$
so $a^2 = 4(k^2 + k) + 1 \equiv 1 \mod 8$.
|
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Find all $n\in\mathbb N$ such that $\sqrt{n^2+8n-5}$ is an integer. $n^2+8n-5$ has to be a perfect square.
How to find all $n$?
|
$y^2=n^2+8n-5 = (n+4)^2 - 21$ implies $21 = x^2-y^2$, for $x=n+4$.
Write $21 = x^2-y^2= (x-y)(x+y)$. Since $x=n+4 \ge 4$, we must have $x+y=7$ or $x+y=21$.
*
*$x+y=7$ implies $x-y=3$ and so $x=5$ and $n=1$
*$x+y=21$ implies $x-y=1$ and so $x=11$ and $n=7$
|
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Prove that $N = \frac{b^n-1}{b-1}$ is a pseudoprime number to the base b. Let $n$ a pseudoprime-number to the base $b$ and $\gcd(b - 1, n) = 1$
Prove that $N = \frac{b^n-1}{b-1}$ is a pseudoprime number to the base b.
My Atachment:
Proof. Note that $Φ_p(b) = M_p (b)$ and $\gcd(p, Φ_p(b)) = 1$.
So $P_p(b) = M_p(b)$
Let $N > 2$ and $P_N (b) = \frac{Φ_N (b)}{ gcd (N, ΦN (b))}$.
If $P_N (b)$ is
composite, then $P_N (b)$ is an overpseudoprime to base $b$.
Source: https://arxiv.org/pdf/1206.0606.pdf
is that right?
|
I have an alternative approach.
We want to see that:
\begin{equation*}
b^{N-1}\equiv 1 ( \text{ mod } N)
\end{equation*}
lets begin by proving that $n|N-1$.
$n$ is pseudoprime, i.e
$$b^{n-1} \equiv 1 ( \text{ mod } n)$$
Therefore $n|( b^{n-1} -1 )$ which is equivalent to say that $n| b^n-b$.
On the other hand, we know that
$N-1=\frac{b^n-1}{b-1}-1=\frac{b^n-b}{b-1}$
Now we use the fact that $gcd(b-1,n)=1$ to confirm that $(b-1)$ and $n$ don't have any common divisor, and since $n$ divides the numerator, he obtain $n|N-1$.
We can express $N-1=n*k$.
Now we have:
\begin{equation*}
b^{N-1}\equiv (b^{n})^k ( \text{ mod } N)
\end{equation*}
but $b^n \equiv 1 ($ mod $N)$, since
\begin{equation*}
b^n-1\equiv N \cdot (b-1) \equiv 0 \cdot (b-1) \equiv 0 (\mod N)
\end{equation*}
Thus,
\begin{equation*}
b^{N-1}\equiv (b^{n})^k\equiv 1^k\equiv 1 ( \text{ mod } N)
\end{equation*}
|
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|
If $S=\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\cdots+\frac{n}{1+n^2+n^4}$, then calculate $14S$.
If $$S=\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\cdots+\frac{n}{1+n^2+n^4}\,$$ find the value of $14S$.
The question can be simplified to:
Find $S=\sum\limits_{k=1}^n\,t_k$ if $t_n=\dfrac{n}{1+n^2+n^4}$.
As $1+n^2+n^4$ forms a GP,
$$t_n=\frac{n(n^2-1)}{n^6-1}\,.$$
But I can't figure out how to solve further. It would be great if someone could help.
|
From the first four terms, you can easily guess the pattern: The numerators are $1,3,6,10$ which look like $n(n+1)/2$ and the denominators are $3,7,13,21$ which look to have a constant second difference of $2$ and are thus probably $n^2+n+1$.
This motivates trying to prove, by induction, that
$$
S_n = \frac{n(n+1)}{2(n^2+n+1)}
$$
The basis is trivial to establish for $n=1$ and the induction is fairly simple algebra.
|
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Conjecture for the value of $\int_0^1 \frac{1}{1+x^{p}}dx$ While browsing the post Is there any integral for the golden ratio $\phi$?, I came across this nice answer,
$$ \int_0^\infty \frac{1}{1+x^{10}}dx=\frac{\pi\,\phi}5$$
it seems the general form is just
$$p \int_0^\infty \frac{1}{1+x^{p}}dx=\color{blue}{\frac{\pi}{\sin\big(\tfrac{\pi}{p}\big)}}$$
I wondered about
$$\int_0^\color{red}1 \frac{1}{1+x^p}dx=\,?$$
Mathematica could find messy closed-forms for $p=5,7$. After some laborious simplification,
$$5\int_0^1 \frac{1}{1+x^5}dx=\frac{\pi\sqrt{\phi}}{5^{1/4}}+\ln2+\sqrt{5}\ln\phi$$
Question 1: In general, is it true that for any $p$ ,
$$2p\,\int_0^1 \frac{1}{1+x^p}dx=\color{blue}{\frac{\pi}{\sin\big(\tfrac{\pi}{p}\big)}}+2\ln2-\psi\big(\tfrac{1}{p}\big)+\psi\big(\tfrac{p-1}{2p}\big)+\psi\big(\tfrac{p+1}{2p}\big)-\psi\big(\tfrac{p-1}{p}\big)$$
where $\psi(z)$ is the digamma function?
Note: The four digammas, implemented in Mathematica as PolyGamma[z], can be expressed as a sum of cosines x logarithms for odd $p=2m+1$. Let $k=\frac{2n-1}{p}\pi$, then,
$$-\psi\big(\tfrac{1}{p}\big)+\psi\big(\tfrac{p-1}{2p}\big)+\psi\big(\tfrac{p+1}{2p}\big)-\psi\big(\tfrac{p-1}{p}\big)=-4\sum_{n=1}^m \cos (k)\ln\big(\sin\tfrac{k}{2}\big)$$
Question 2: For even $p$, can we can also avoid the digamma by using cosines and logarithms?
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I'm only going to address Question 1.
The expression proposed in Question 1 is true. However, it is a little bit too complicated than necessary. A simpler version of the expression is
$$2p\int_0^1 \frac{dx}{1+x^p} = \psi\left(\frac{1}{2p} + \frac12\right) - \psi\left(\frac{1}{2p}\right)$$
From reflection formula, take logarithm and differentiate, we get
$$\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin\pi z} \implies
\psi(z) - \psi(1-z) = \pi\cot\pi z$$
This leads to
$$\frac{\pi}{\sin z} = \pi\cot\frac{\pi z}{2} - \pi\cot\pi z = \psi\left(\frac{z}{2}\right) - \psi\left(1-\frac{z}{2}\right) - \psi(z) + \psi(1-z)
$$
From duplication formula, take logarithm and differentiate, we get
$$\Gamma(z)\Gamma\left(z+\frac12\right) = 2^{1-2z}\sqrt{\pi}\Gamma(2z)
\implies \psi(z) + \psi\left(z + \frac12\right) = -2\log 2 + 2\psi(2z)
$$
Apply these to RHS of Question 1, we can expose RHS to following mess
$$
\left[ \color{red}{\psi\left(\frac{1}{2p}\right)} - \psi\left(1 - \frac{1}{2p}\right) - \color{green}{\psi\left(\frac{1}{p}\right)} + \color{blue}{\psi\left(1 - \frac{1}{p}\right)} \right]
+ \left[ \color{green}{2\psi\left(\frac1p\right)} - \color{red}{\psi\left(\frac{1}{2p}\right)} - \color{magenta}{\psi\left(\frac{p+1}{2p}\right)}\right]\\
- \color{green}{\psi\left(\frac{1}{p}\right)} + \psi\left(\frac{p-1}{2p}\right) + \color{magenta}{\psi\left(\frac{p+1}{2p}\right)} - \color{blue}{\psi\left(\frac{p-1}{p}\right)}
$$
After massive cancellation, we can simplify RHS to
$$
\psi\left(\frac{p-1}{2p}\right) - \psi\left(1 - \frac{1}{2p}\right)
= \psi\left( 1 - \left(\frac{1}{2p} + \frac12\right)\right) - \psi\left(1 - \frac{1}{2p}\right)
= \psi\left(\frac{1}{2p} + \frac12\right) - \psi\left(\frac{1}{2p}\right)
$$
Recall following expansion of digamma function
$$\psi(z) = \frac{1}{z} + \sum_{n=1}^\infty \left(\frac{1}{z+n} - \frac{1}{n}\right)$$
We find
$$\begin{align}
\text{RHS}
&= \frac{1}{\frac{1}{2p}} - \frac{1}{\frac{1}{2p} + \frac12} + \sum_{n=1}^\infty\left(\frac{1}{\frac{1}{2p}+n} - \frac{1}{\frac{1}{2p} + n + \frac12}\right)\\
&= 2\sum_{n=0}^\infty\frac{(-1)^n}{\frac{1}{p}+n}
= 2\sum_{n=0}^\infty\int_0^1 (-1)^n t^{\frac{1}{p}+n-1} dt
= 2\int_0^1 \sum_{n=0}^\infty (-1)^n t^{\frac{1}{p}+n-1} dt\\
&= 2 \int_0^1 \frac{t^{\frac{1}{p}-1}}{1+t} dt
= 2p\int_0^1 \frac{dx}{1+x^p} = \text{LHS}
\end{align}
$$
|
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Prove:$x^2+2xy+3y^2-6x-2y\ge-11\;\;\forall(x,y)\in\Bbb{R}$
Problem Statement:-
Prove that for all real values of $x$ and $y$
$$x^2+2xy+3y^2-6x-2y\ge-11$$
I have no idea how to approach this question all I could think on seeing it was tryin to find the linear factors, turns out that the determinant
$$\begin{vmatrix}
a & h & g\\
h & b & f\\
g & f & c\\
\end{vmatrix}\neq0$$
So finding linear factor just flew out of the window, so I plotted the equation $x^2+2xy+3y^2-6x-2y=0$ turns out it is an ellipse. I could conclude no further as to how to approach this problem.
|
We have
$$\begin{align}x^2+2xy+3y^2-6x-2y+11&=x^2+(2y-6)x+3y^2-2y+11\\\\&=(x+y-3)^2-(y-3)^2+3y^2-2y+11\\\\&=(x+y-3)^2+2y^2+4y+2\\\\&=(x+y-3)^2+2(y+1)^2\end{align}$$
|
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|
(CHMMC #8 Individual, 2016) Find the smallest $n$ such that $n^2 \,\% \, 5 < n^2 \,\%\, 7 < n^2 \,\%\, 11 < n^2 \,\%\, 13?$
Define $n\,\%\,d$ as the remainder when n is divided by d. What is the smallest
positive integer n, not divisible by 5, 7, 11, or 13, for which $n^2 \,\%
\, 5 < n^2 \,\%\, 7 < n^2 \,\%\, 11 < n^2 \,\%\, 13?$
So basically I tried using quadratic residues, but as each mod generates 3,4,6, or 7 quadratic residues, I would have to bash out all $3\cdot 4\cdot6\cdot7$ possibilities to find an answer. Also, through a computer program, I found that $n=19$. Is there an easier method or a method to solve this without a computer?
|
I found an elementary solution only using congruences and trial-and-error. Given $x\in\Bbb{N}$ such that $5\not\mid x$ and $7\not\mid x$, then $x^2\equiv 1,4\pmod 5$ and $x^2\equiv 1,2,4\pmod 7$. So if $n^2\equiv 4\pmod 5$ it's impossible to have $n^2 \,\%
\, 5 < n^2 \,\%\, 7$. Now, $n^2\equiv 4\pmod 5$ iff $n\equiv2, 3\pmod 5$, therefore if $n$ holds the required condition, $n$ must satisfy $n\equiv \pm1\pmod 5$.
It's obvious that if $n^2\le 13$ it's impossible to have the required condition, so $n^2>13$, which give us $n\ge 4$. We now check case by case:
If $n=4$, then $16\,\% \, 11=5$, but $16\,\% \, 13=3$.
If $n=5$, then $n$ is divisible by $5$.
If $n=6$, then $36\,\% \, 5=36\,\% \, 7=1$.
If $n=7$, then $n$ is divisible by $7$.
If $n=8$, then $8\not\equiv \pm1\pmod 5$.
If $n=9$, then $81\,\% \, 7=81\,\% \, 11=4$.
If $n=10$, then $n$ is divisible by $5$.
If $n=11$, then $n$ is divisible by $11$.
If $n=12$, then $12\not\equiv \pm1\pmod 5$.
If $n=13$, then $n$ is divisible by $13$.
If $n=14$, then $n$ is divisible by $7$.
If $n=15$, then $n$ is divisible by $5$.
If $n=16$, then $256\,\% \, 7=4$, but $256\,\% \, 11=3$.
If $n=17$, then $17\not\equiv \pm1\pmod 5$.
If $n=18$, then $18\not\equiv \pm1\pmod 5$.
If $n=19$, then $361\,\% \, 5=1$, $361\,\% \, 7=4$, $361\,\% \, 11=9$ and $361\,\% \, 13=10$.
Hence, the smallest number which satisfies the required condition is $n=19$.
|
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Find all solutions to $x^2\equiv 1\pmod {91},\ 91 = 7\cdot 13$ I split this into $x^2\equiv 1\pmod {7}$ and $x^2\equiv 1\pmod {13}$.
For $x^2\equiv 1\pmod {7}$, i did:
$$ (\pm1 )^2\equiv 1\pmod{7}$$ $$(\pm2 )^2\equiv 4\pmod{7}$$ $$(\pm3 )^2\equiv 2\pmod{7}$$ Which shows that the solutions to $x^2\equiv 1\pmod {7}$ are $\pm1$.
For $x^2\equiv 1\pmod {13}$, i did:
$$ (\pm1 )^2\equiv 1\pmod{13}$$ $$(\pm2 )^2\equiv 4\pmod{13}$$ $$(\pm3 )^2\equiv 9\pmod{13}$$ $$ (\pm4 )^2\equiv 3\pmod{13}$$ $$(\pm5 )^2\equiv {-1}\pmod{13}$$ $$(\pm6 )^2\equiv 10\pmod{13}$$Which shows that the solutions to $x^2\equiv 1\pmod {13}$ are $\pm1$.
Thus, I concluded that the solutions to $x^2\equiv 1\pmod {91}$ must be $\pm1$. I thought that $\pm1$ were the only solutions, but apparently I am incorrect! How do I go about finding the other solutions to this congruence?
|
Hint: Consider the possibility that $x \equiv 1 \pmod 7$ but $x \equiv -1 \pmod {13}$, and so on. (In other words, your mistake was to assume that the $\pm1$ modulo $7$ was the same sign as $\pm1$ modulo $13$.)
Also note that for any prime $p$, if $x^2 \equiv 1 \pmod p$, then we can rewrite this as $$x^2 - 1 \equiv (x+1)(x-1) \equiv 0 \pmod p.$$
Thus we get $x \equiv \pm 1 \pmod p$, showing that it isn't necessary to run through all the values of $x^2$ to find the solution.
|
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|
Find $\sin \theta$ if $\tan \theta +\sec \theta =1.5 $ Find $\sin \theta$ if $\tan \theta +\sec \theta =1.5 $
$$\tan \theta +\sec \theta =1.5 $$
$$2\tan \theta +2\sec \theta =3 $$
$$2\sec \theta =3-2\tan \theta$$
$$4\sec^2 \theta =(3-2\tan \theta)^2$$
$$4+4\tan^2 \theta =9-12\tan \theta+4\tan^2 \theta$$
So I get $$\tan \theta = \frac{5}{12}$$
Thus $$\sin\theta=\frac{5}{13}$$
But If I do like this , $$\frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta} =\frac{3}{2}$$
$$ 2\sin\theta +2=3\cos \theta$$
$$ (2\sin\theta +2)^2=9\cos^2 \theta$$
$$ 4\sin^2\theta+8\sin\theta +4=9-9\sin^2\theta$$
$$13\sin^2\theta+8\sin\theta-5=0$$
Therefore I get two answers $$\sin\theta=\frac{5}{13} , \sin\theta =-1$$
What is the reason behind this ? Why am I getting two answers in one method and one in another ?
|
No one would have any doubt in first solution. In the second solution we have no problem with $\sin\theta=5/13$, but $\sin\theta=-1$ gives $cos\theta=0$ and when you apply $cos\theta=0$ in the expression $\frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta} =\frac{3}{2}$ you will find that you have done a mistake because the function $\frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta} =\frac{3}{2}$ is defined only when $cos\theta\ne0$.
|
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|
Trouble with a substitution I'm struggling to show that
$$ \int_{-1}^{1} \frac{(1-x^2)^{1/2}}{1+x^2} dx $$
to
$$ -\pi + \int_{-\pi}^{\pi} (1+\cos^2(\theta))^{-1}d\theta$$
with $x=\cos(\theta)$
I'm aware I'm missing something obvious but I end up with a stray $\sin(\theta)$
|
After the usual substitution, you get
$$
\int \frac{\sqrt{1-x^2}}{1+x^2}dx = \int_0^\pi \frac{\sin^2\theta}{1+\cos^2 \theta}d\theta = \int_0^\pi \frac{1-\cos^2\theta}{1+\cos^2 \theta}d\theta = \int_0^\pi -1 + \frac{2}{1+\cos^2\theta} = \\-\pi + \int_{-\pi}^\pi \frac{1}{1+\cos^2\theta}d\theta.
$$
In the last steps I was using that
$$
\frac{1-x^2}{1+x^2} = -1 + \frac{2}{1+x^2}
$$
and that $\frac{1}{1+\cos^2\theta}$ is an even function.
|
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|
Solve $\frac{5}{x+3}+\frac{2}{2x+6}=4$ I want to solve $\frac{5}{x+3}+\frac{2}{2x+6}=4$ for $x$
I try this;
$5(2x+6)+2(x+3)=4(x+3)(2x+6)$
$12x+36 = 4(2x^2+12x+18)$
$8x^2+36x+36=0$
Where I would then go to the formula, however the answer says -1.5, what am I doing wrong.
|
We have $5\over{x+3}$ + $2\over{2x+6}$ = 4.
This can be rewritten as $10\over{2x+6}$ + $2\over{2x+6}$ = 4.
Multiplying by 2x + 6 on each side, we get: 10 + 2 = 4(2x + 6)
Dividing by 4 on each side: 3 = 2x + 6
$\Rightarrow$ -3 = 2x $\Rightarrow$ x = $-3\over{2}$
|
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|
the sum of consecutive odd numbers If the sum of consecutive odd numbers starting with $-3$ until $2k+1$ equals $21$
What is the value of $k$ ?
I can solve this by trying the numbers $-3-1+1+3+5+7+9=21$ , so the last term is $7th$ so the $k$ value is $3$
But I could not solve this with formula, I know the odd numbers come in the form of $2k+1$ but could not get much further.
|
You can reduce this sequence: $-3-1+1+3+5+7+9=21$.
$-3-1+1+3=0$. So you are being left with:
$5+7+9=21$. Since the last term is $9$ and the last term in your question is $2k+1$ we have:
$2k+1=9$
$2k=9-1$
$k=8:2$
$k=4$
Edit:
I would like to add, that your confusion might come from how you are understand "until $2k+1$". One might say: until some term or until some number. In the first instance you are right: $k=3$. In the second: $k=4$. Two following steps can help:
We can write an Arithmetic Sequence as a rule:
$X_n=a_1 + d(n-1)$
$a_1$ - first term
$d$ - common difference
$n$ - number of terms
Since you have $7$ terms, last term is equal:
$X_7=-3+(7-1)2=-3+12=9$
Before you can calculate the above first we need to find this $7$, therefore to sum up the terms of this arithmetic sequence you can use formula:
$\sum_{k=1}^{n}(a+kd)=\frac{n}{2}(2a+(n-1)d)$
$21=\frac{n}{2}(-6+2n-2)$
$21=-3n+n^2-n$
$n^2-4n-21=0$
Δ $=16-4*1*(-21)=16+84=100$
$n_1=7$ and $n_2=-3$
|
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|
Prove that $\frac1{a+b+1}+\frac1{b+c+1}+\frac1{c+a+1}\le1$ If $abc=1$ then
$$\frac1{a+b+1}+\frac1{b+c+1}+\frac1{c+a+1}\le1$$
I have tried AM-GM and C-S and can't seem to find a solution. What is the best way to prove it?
|
Also we can use the following reasoning.
For positives $a$, $b$ and $c$ let $a=x^3$, $b=y^3$ and $c=z^3$.
Hence, $\sum\limits_{cyc}\frac{1}{a+b+1}=\sum\limits_{cyc}\frac{1}{x^3+y^3+xyz}\leq\sum\limits_{cyc}\frac{1}{x^2y+y^2x+xyz}=\sum\limits_{cyc}\frac{z}{x+y+z}=1$,
but I think the first way is better.
|
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|
Calculating $\int_0^{\pi/3}\cos^2x+\dfrac{1}{\cos^2x}\mathrm{d}\,x$ I've been given the following exercise:
Show that the exact value of $$\int_0^{\pi/3}\cos^2x+\frac{1}{\cos^2x}\,\mathrm{d}x = \frac{\pi}{6}+\frac{9}{8}\sqrt{3}$$
Can someone help me with this?
|
This is an answer that requires the knowledge of simple functions such as $\tan$, you can do it without knowing $\sec$ and e.c. Other answers are great, but I am unsure you know all the functions they use.
You can also notice that $\frac{d}{dx}\tan(x)=\frac{1}{\cos(x)^2}$ and that $\cos(x)^2= \frac{1+cos(2x)}{2}$ Thus giving us:
$$\int^{\frac{\pi}{3}}_{0}\cos(x)^2+\frac{1}{\cos(x)^2}dx=\int^{\frac{\pi}{3}}_{0}\frac{1+\cos(2x)}{2}dx+ \left [ \tan(x) \right ]_0^{\frac{\pi}{3}}=\frac{1}{2}\int^{\frac{\pi}{3}}_{0}\cos(2x)dx+\frac{1}{2}\int^{\frac{\pi}{3}}_{0}dx+ \sqrt{3}$$
Now make a variable change:
$u=2x \rightarrow du=2dt$ so $\frac{du}{2}=dt$. We have now:
$$\frac{1}{4}\int^{u(\frac{\pi}{3})}_{0}\cos(u)dx+\frac{1}{2}\int^{\frac{\pi}{3}}_{0}dx+ \sqrt{3}=\frac{1}{4}\left [ sin(x) \right ]_{0}^{\frac{2\pi}{3}}+\frac{1}{2}\left [ x \right ]_0^{\frac{\pi}{3}}+ \sqrt{3}=\frac{\sqrt{3}}{8}+\frac{\pi}{6} + \sqrt{3}=\frac{9\sqrt{3}}{8}+\frac{\pi}{6}$$
|
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|
Solving $(1+x^2)y' - 2xy = (1+x^2)\arctan(x)$ I'm asked to solve the differential equation: $$(1+x^2)y' - 2xy = (1+x^2)\arctan(x).$$
I rewrite it:
$$y' - \frac{2x}{1+x^2}y = \arctan(x).$$
The integrating factor is:
$$e^{-\int{\frac{2x}{1+x^2}}dx} = e^{-\ln{1+x^2}} = (e^{ln{1+x^2}})^{-1} = \frac{1}{1+x^2}.$$
$$\frac{1}{1+x^2}y' - \frac{1}{1+x^2}\frac{2x}{1+x^2}y = \arctan(x)\frac{1}{1+x^2}$$
$$\frac{d}{dx}(\frac{1}{1+x^2}y) = \arctan(x)\frac{1}{1+x^2}$$
So to solve this equation I need to find the primitive of
$$\arctan(x)\frac{1}{1+x^2}$$
But I don't think I can do that yet, is there any other way to solve this differential equation?
|
The derivative of $\arctan$ is $x\mapsto \frac{1}{1+x^{2}}$.
Now, observe that for a differentiable function $f$, we have:
$$\int f(x)f'(x)\text{d}x=\frac{1}{2}(f(x))^{2}+K$$
where $K\in\mathbb{R}$. Taking $f=\arctan$, we deduce what you asked.
|
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|
What are roots of higher power like $n^9=512$? We know a value of $n$ in $n^2 = 4$ is $\pm2$ (two roots), then what about $n^9 = 512$, generic answer is $+2$, but it should have $9$ roots. What are those $8$ other roots and how can we find them ?
|
$$n^9-a^9=(n^3)^3-(a^3)^3=(n^3-a^3)(n^6+a^3n^3+a^6)=(n-a)(n^2+an+a^2)(n^6+a^3n^3+a^6)$$
|
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|
Algebra question grade 9 I am a a student and I am having difficulty with answering this question. I keep getting the answer wrong. Please may I have a step by step solution to this question so that I won't have difficulties with answering these type of questions in the future.
Question: Rearrange the formula below to make n the subject.
$$x = \sqrt{\frac{1+n}{1-n}}.$$
I got rid of the square roots first:
$$x^2(1-n) = 1+n.$$
However I don't know what to do next.
Kind Regards. Help would be appreciated
|
Your first step is correct. So, we have
$$x^2 = \frac{1+n}{1-n}$$
Now, the question is: how do we get rid of this fraction and simplify two $n$'s to one? The natural step is to cross multiply like this:
\begin{align}
{x^2} \cdot \color{#f00}{1} &= \frac{1+n}{1-n} \cdot \color{#f00}1\\
{x^2} \cdot \color{red}{\left(\frac{1-n}{1-n} \right)} &= \frac{1+n}{1-n} \cdot \color{red}{\left(\frac{1-n}{1-n}\right)}
\end{align}
The above is how cross multiplication works. Both sides are always being multiplied by $1$. We can rewrite the 1 as anything as long as the 1 simplifies to 1.
Which gives us what you came up with:
$$x^2 \cdot (1-n) = (1+n)$$
Now, it's a matter of just expanding and bringing all of our $n$'s to one side.
\begin{align}
x^2 \cdot (1-n) &= (1+n) \\
x^2 - nx^2 &= 1 + n \\
x^2 - 1 &= n + nx^2 \\
...
\end{align}
There are two more steps left which I have left for your to complete.
Some times it pays to just expand even if it looks ugly to you and then try to factor things based on whatever you are trying to rearrange for.
Let me know if you have any further questions :)
EDIT:
So, you have $x^2 - 1$ on one side and $n + nx^2$ on another side. Since you have all of your $n$'s in one place you can try separate the $n$'s from everything else because remember, you want to isolate for $n$. You can do this by factoring $n$ from $n + nx^2$. So you will have $n (1 + x^2)$.
Since you want $n$ by itself, you can divide both sides by $(1 + x^2)$ and come to your answer.
What you did is:
\begin{align}
\require{cancel}
x^2 - 1 &= n + nx^2 \\
\frac{\cancel{x^2} - 1}{\cancel{x^2}} &= \frac{n + n\cancel{x^2}} {\cancel{x^2}} \color{red}{Wrong.} \\
\end{align}
Remember, when you have more than one term in the numerator like you do above and divide by $1$ thing i.e. $x^2$ then what you are really saying is:
\begin{align}
x^2 - 1 &= n + nx^2 \\
\frac{x^2 - 1}{x^2} &= \frac{n + nx^2}{x^2} \\
\frac{x^2}{x^2} - \frac{1}{x^2} &= \frac{n}{x^2} + \frac{n^2x^2}{x^2}
\end{align}
I hope that clarifies your problem.
|
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|
If $f(x)\mod (x-1)=5$, $f(x)\mod (x+1)=3$, $f(x)\mod (x+2)=5$ then find the remainder when $f(x)$ is divided by $x^3+2x^2-x-2$
If $f(x)\mod (x-1)=5$, $f(x)\mod (x+1)=3$, $f(x)\mod (x+2)=5$ then find the remainder when $f(x)$ is divided by $x^3+2x^2-x-2$
I don't have any idea as to how to proceed with this question. I have figured out that $x^3+2x^2-x-2=(x-1)(x+1)(x+2)$.
But I don't know how to use this information to find the remainder $x^3+2x^2-x-2$ would give. A small hint as to how to proceed would be enough.
|
From the information provided, we can write
$$ f(x) = p(x)(x-1) + 5,$$
$$ f(x) = q(x)(x+1) + 3,$$
and
$$ f(x) = r(x) (x+2) + 5.$$
where $p(x)$, $q(x)$, and $r(x)$ are polynomials such that
$$\mathrm{gcd} (p(x), x-1 ) = \mathrm{gcd} (q(x), x+1 ) = \mathrm{gcd} (r(x), x+2 ) = 1.$$
Now as
$$ x^3 + 2x^2 -x-2 = (x+1)(x-1)(x+2),$$
and as
$$\mathrm{gcd}(x-1, x+1) = \mathrm{gcd}(x+1, x+2) = \mathrm{gcd}(x+2, x-1) = 1, $$
so we can conclude that ...
Can you proceed from here?
|
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|
Linear least squares in $\mathbb{R}^{3}$ with three data points. Given data points in the form $(x,y,f(x,y)) = (1,1,7),(1,3,0),(1,-1,8)$, find the least squares solution $\hat x$ to the system of equations $Ax=b$.
Is there enough information to use least squares? The solution created $A$ using the $x$ and $y$ coordinates and $b$ using $f(x,y)$. I do not understand this problem. How can you solve for $\hat x$ when you do not know anything about the function? Isn't it presumptuous to assume a linear relationship?
|
We are given a sequence of measurements $\left\{ x_{k}, y_{k}, f_{k} \right\}_{k=1}^{3}$. Generally, we could hope to find three fit parameters, a trial function like
$$
f(x,y) = c_{0}g_{0}(x,y) + c_{1}g_{1}(x,y) + c_{2}g_{2}(x,y)
$$
where the functions $g$ are linearly independent.
You asked about the linear case where
$$
f(x,y) = c_{0} + c_{1}x + c_{2}y.
$$
That problem looks like this
$$
\begin{align}
\mathbf{A} c &= F \\[5pt]
%
\left[
\begin{array}{ccc}
1 & x_{1} & y_{1} \\
1 & x_{2} & y_{2} \\
1 & x_{3} & y_{3}
\end{array}
\right]
\left[
\begin{array}{c}
c_{0} \\
c_{1} \\
c_{2}
\end{array}
\right]
&=
\left[
\begin{array}{c}
f_{1} \\
f_{2} \\
f_{3}
\end{array}
\right] \\[5pt]
%
\left[
\begin{array}{ccr}
1 & 1 & 1 \\
1 & 1 & 3 \\
1 & 1 & -1
\end{array}
\right]
\left[
\begin{array}{c}
c_{0} \\
c_{1} \\
c_{2}
\end{array}
\right]
&=
\left[
\begin{array}{c}
7 \\
0 \\
7
\end{array}
\right]
\end{align}
$$
The is no solution vector $c$ which satisfies this equation, so instead of asking that $\mathbf{A}c-F = 0$, we ask instead that $\mathbf{A}c-F$ be as small as possible.To measure length, we must select a norm, and the choice here is the $2-$norm of least squares.
The least squares solution is
$$
c_{LS} = \mathbf{A}^{\dagger}F + \left(\mathbf{I}_{3} - \mathbf{A}^{\dagger}\mathbf{A} \right) y, \quad y\in\mathbb{C}^{3}
$$
There is a problem with this model in that the first two column vectors of $\mathbf{A}$ are identical. This matrix has rank $\rho = 2.$ We can't use the normal equations.
Using the SVD,
$$
\mathbf{A}^{\dagger} = \frac{1}{24}
\left[
\begin{array}{ccr}
4 & 1 & 7 \\
4 & 1 & 7 \\
0 & 6 & -6 \\
\end{array}
\right]
$$
and the full solution is
$$
c_{LS} = \frac{1}{24}
\left[
\begin{array}{r}
77 \\
77 \\
-42
\end{array}
\right]
+ \alpha
\left[
\begin{array}{r}
-1 \\
1 \\
0
\end{array}
\right].
$$
The least squares fit is the best fit, but it may not be a good fit. That is, given the trial function, the method will find the best parameters. But the trial function may be bad.
The residual error vector
$$
r(c_{LS}) = \frac{1}{6}
\left[
\begin{array}{r}
-14 \\
7 \\
7
\end{array}
\right].
$$
The total error is $r\cdot r = \frac{49}{6} \approx 8.2.$
|
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|
Double Integrals - Finding the volume of a unit disk under a function For a Unit Disk
$x^2 + y^2 \ge 1$
And for a function of $x$ and $y$
$f(x, y) = 3 + y - x^2$
I want to find the volume underneath the function bound by the unit disk.
At first I integrated the function with respect to $y$, and made the upper bound $\sqrt{1 - x^2}$ and made the lower bound $-\sqrt{1 - x^2}$. I did this using Pythagoras' theorem to express the upper and lower bounds of the circle in terms of $x$. The resulting expression is $6\sqrt{1-x^2}$
Then I have to integrate the expression I just found with respect to $x$, and the upper and lower bounds are $1$ and $-1$ respectively.
However, the expression given is as follows:
I do not understand why the expression has changed.
The expression that I worked out $6\sqrt{1-x^2}$ represents the area of a slice with a constant $x$ value, and surely I just integrate this expression with respect to x between the bounds of $1$ and $-1$ to find the total volume.
|
Note that
$$\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}(3+y-x^2)dy=\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}ydy+\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}(3-x^2)dy.$$ Now
$$\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}ydy=0$$ and $$\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}(3-x^2)dy=2(3-x^2)\sqrt{1-x^2}.$$
Thus the volume is given by
$$\int_{-1}^1 2(3-x^2)\sqrt{1-x^2}dx,$$ which is the expression you have.
|
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|
How can the equation be positive in the interval $ -1 < x < \frac{7-\sqrt{73}}{2}$? I've got the inequality:
$$\frac{x^2-8x-7}{x+1}\geq -1$$
For the inequality to be larger than zero we need either both numerator and denominator to be positive OR both numerator and denominator to be negative.
I draw the number line with all the zeros in question, i.e, -1 (root of the bottom equation) and $\frac{7\pm\sqrt{73}}{2}$ (roots of the quadratic). This way I have the intervals of interest. I "move" the 1 to the left hand side and then simplifying I obtain $x^2-7x-6$ in the numerator and $x-1$ in the denominator. Let's call the quadratic 'A' and the denominator equation 'B'.
For $x \leq-1$, I get that A is positive and B is negative. For $-1\leq x \leq \frac{7-\sqrt{73}}{2}$ I get A is negative and B is positive. For $\frac{7-\sqrt{73}}{2} \leq x \leq \frac{7+\sqrt{73}}{2}$ I get that A is negative and B is positive. And finally for $x\geq\frac{7+\sqrt{73}}{2}$ I get that both A and B are positive.
Thus, the only interval that satisfies my inequality is the last one, because both numerator and denominator are positive. But in the official result, the interval between $-1$ and $\frac{7-\sqrt{73}}{2}$ is also included in the result. But if you replace X with any value from that interval you'll get a negative result, which would render the whole equation negative. What am I missing here?
Thank you
|
It should be as follows :$$\frac { x^{ 2 }-8x-7 }{ x+1 } \geq x+1\\ \frac { x^{ 2 }-8x-7 }{ x+1 } -x-1\ge 0\\ \frac { x^{ 2 }-8x-7-{ x }^{ 2 }-2x-1 }{ x+1 } \ge 0\\ \frac { -10x-8 }{ x+1 } \ge 0\\ \frac { 5x+4 }{ x+1 } \le 0\\ \frac { \left( x+1 \right) \left( 5x+4 \right) }{ { \left( x+1 \right) }^{ 2 } } \le 0\\ \left( x+1 \right) \left( 5x+4 \right) \le 0\\ -1<x\le -\frac { 4 }{ 5 } \\ \\ $$
EDIT version.
$$\frac { x^{ 2 }-8x-7 }{ x+1 } \geq -1\\ \frac { x^{ 2 }-8x-7 }{ x+1 } +1\ge 0\\ \frac { x^{ 2 }-8x-7+x+1 }{ x+1 } \ge 0\\ \frac { x^{ 2 }-7x-6 }{ x+1 } \ge 0\\ \frac { \left( x+1 \right) \left( x^{ 2 }-7x-6 \right) }{ { \left( x+1 \right) }^{ 2 } } \ge 0\\ \left( x+1 \right) \left( x^{ 2 }-7x-6 \right) \ge 0\\ \left( x+1 \right) \left( x-\frac { 7-\sqrt { 73 } }{ 2 } \right) \left( x-\frac { 7+\sqrt { 73 } }{ 2 } \right) \ge 0\\ \\ \frac { 7-\sqrt { 73 } }{ 2 } \approx -0.772001872659\\ \frac { 7+\sqrt { 73 } }{ 2 } \approx 7.77200187266\\ -1<x\le \frac { 7-\sqrt { 73 } }{ 2 } \quad and\quad \frac { 7+\sqrt { 73 } }{ 2 } \le x<+\infty \\ \\ \\ $$
|
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|
If $3x^2+2\alpha xy+2y^2+2ax-4y+1$ can be resolved into two linear factors, prove that $\alpha$ is the root of the equation $x^2+4ax+2a^2+6=0$.
If $3x^2+2\alpha xy+2y^2+2ax-4y+1$ can be resolved into two linear factors, prove that $\alpha$ is the root of the equation $x^2+4ax+2a^2+6=0$.
I know this question has already been asked on Math.SE here, but is there a way to solve this by only using the concepts of quadratic equations? I am not very clear with the solution given there.
|
The discriminant of the equation $p(x,y) = 3x^2+(2\alpha y+2a)x +2y^2-4y+1=0$ is $\delta = (2\alpha y+2a)^2-12(2y^2+4y+1)$.
Thus, $$p(x,y) = 3\left(x-\frac{-(2\alpha y+a) +\sqrt{\delta}}{6}\right)\left(x-\frac{-(2\alpha y+a)-\sqrt{\delta}}{6}\right)$$
Now $p(x,y)$ is a perfect square iff the discriminant $\delta_1$ of $\delta =0$ is $0$. We can verify $\delta_1 =0 \Leftrightarrow \cdots \Leftrightarrow \alpha^{2}+4\alpha a+ 2a^2+6=0$. Equivalently, $\alpha$ is a root of $x^2+4ax+2a^2+6=0$.
|
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|
Show that ${(F_n^2+F_{n+1}^2+F_{n+2}^2)^2\over F_{n}^4+F_{n+1}^4+F_{n+2}^4}=2$
If $F_n$ is the $n$-th Fibonacci number ($1,1,2,3,5,8,\dots$), show that
$${(F_n^2+F_{n+1}^2+F_{n+2}^2)^2\over F_{n}^4+F_{n+1}^4+F_{n+2}^4}=2$$
I have tested with a lot of Fibonacci numbers and it seem to obey the ruse, but I don't know how simplify it to 2.
I try:
Let $a=F_n$, $b=F_{n+1}$ and $c=F_{n+2}$
$a^4+b^4+c^4+2(ab)^2+2(ac)^2+2(bc)^2=2a^4+2b^4+2c^4$
$2(ab)^2+2(ac)^2+2(bc)^2=a^4+b^4+c^4$
I am not sure, what to do next. Can anyone help by completing the prove?
|
Setting $x=a$, $y=b$, and $z=-(a+b)$ we have to evaluate $\frac{(x^2+y^2+z^2)^2}{x^4+y^4+z^4}$
Since $x+y+z= 0$, $x,y,z$ are roots of $t^3 + t(\sum xy)-xyz = 0$
Hence $x,y,z$ satisfy $t^4 + t^2(\sum xy)-xyzt = 0$
Setting $t=x,y,z$ successively and adding the resulting equations we obtain
$\sum x^4 +\sum x^2 \sum xy -xyz \sum x = 0 \implies \sum x^4 +\sum x^2 \sum xy = 0$
Since $\sum x = 0$ we get that $2\sum xy = - \sum x^2$
Now its easy to see that $\frac{(x^2+y^2+z^2)^2}{x^4+y^4+z^4} = 2$
|
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|
Trying to understand trickier textbook question from less trickier textbook question - number theory Question (difficult):
Show that if $p$ is odd prime, $\binom{p^2}{p} = p \bmod p^3$
Similar question (easier):
Show that if $p$ is odd prime, $\binom{p^2}{p} = p \bmod p^2$
My attempt to answer "easier" question:
$x = (p^2-1)(p^2-2)\dots(p^2-p+1) \to (-1)^{p-1}\times1\times2\times\dots\times(p-1) \bmod p^2$ and since $p-1$ is even, then $x \equiv (p-1)! \bmod p^2$
$\binom{p^2}{p} \equiv \frac{p^2!}{p!\times(p^2-p)!} \equiv \frac{p^2\times x}{p\times(p-1)!} \equiv p \bmod p^2$
How to solve the $\bmod p^3$
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Notice that the calculation which you did actually shows that $$\frac{\binom{p^2}{p}}{p}=\frac{p^2\cdot x}{p^2(p-1)!}=\frac{x}{(p-1)!}\equiv 1\pmod{p^2}.$$ That is, $\frac{\binom{p^2}{p}}{p}-1$ is divisible by $p^2$. Thus $\binom{p^2}{p}-p$ is divisible by $p^3$, so $\binom{p^2}{p}\equiv p\pmod{p^3}$.
In fact, by a more refined calculation, you can show that $\binom{p^2}{p}\equiv p\pmod{p^4}$ as well. To make it easier to understand what's going on, let's write $y=p^2$ and expand $$x=(y-1)(y-2)\cdots(y-(p-1)).$$ Grouping terms according to their power of $y$, we get $$x=(-1)^{p-1}(p-1)!+(-1)^{p-2}\left(\frac{(p-1)!}{1}+\frac{(p-1)!}{2}+\dots+\frac{(p-1)!}{p-1}\right)y+\text{terms with $y^2$ and higher powers of $y$}.$$ Let's now think about the integer $$s=\frac{(p-1)!}{1}+\frac{(p-1)!}{2}+\dots+\frac{(p-1)!}{p-1}$$ mod $p$. Mod $p$, we can think of the $k$th term of $s$ as $(p-1)!$ multiplied by the multiplicative inverse of $k$ mod $p$. The multiplicative inverses of $1,\dots,p-1$ mod $p$ are just all the nonzero residues mod $p$ in some different order, so $$s\equiv (p-1)!(1+2+\dots+(p-1))\pmod{p}.$$ Since $p$ is odd, $1+2+\dots+(p-1)=\frac{p(p-1)}{2}$ is divisible by $p$, so $s$ is divisible by $p$.
Now consider $x$ mod $p^3$. Since $y=p^2$, all the terms of $x$ with $y^2$ or higher powers are $0$ mod $p^3$. Since $s$ is divisible by $p$, the term $(-1)^{p-2}sy$ is also $0$ mod $p^3$. Thus $$x\equiv (-1)^{p-1}(p-1)!=(p-1)!\pmod{p^3}$$ and now as in the argument above we find that $$\binom{p^2}{p}\equiv p\pmod{p^4}.$$
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"language": "en",
"url": "https://math.stackexchange.com/questions/2065340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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$\dfrac{2}{\pi} = \dfrac{\sqrt 2}{2} \cdot \dfrac{\sqrt {2+\sqrt 2}}{2} \cdot\dfrac{\sqrt {2+\sqrt {2+\sqrt 2}}}{2} \cdots $ One can show inductively that
$$
\cos \frac{\pi}{2^{n+1}}\ = \frac{\sqrt {2+\sqrt {2+\sqrt {2+\sqrt {\cdots+\sqrt {2 }}}}}}{2},
$$
with $n$ square roots in the right side of the equation.
The second part of the question was to deduct the following from the first part:
$$\frac{2}{\pi} = \frac{\sqrt 2}{2} \cdot \frac{\sqrt {2+\sqrt 2}}{2} \cdot\frac{\sqrt {2+\sqrt {2+\sqrt 2}}}{2} \cdot \cdots $$
with the hint to use the following limit:
$$\lim_{n\to \infty}\cos\Big(\frac{t}{2}\Big)\cos\Big(\frac{t}{2^2}\Big)\cdots\cos\Big(\frac{t}{2^n}\Big) = \frac{\sin t}{t}.$$
A hint or some general intuition will be appreciated.
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Using the trogonometric identity
$$
\sin (2a)=2\sin a\,\cos a\qquad\text{or}\qquad \cos a=\frac{\sin 2a}{2\sin a},
$$
provided that $\,\sin a\ne 0,\,$ we obtain that
$$
\cos(x/2)\cos(x/4)\cdots\cos(x/2^n)=\frac{\sin x}{2\sin(x/2)}\frac{\sin (x/2)}{2\sin(x/4)}\cdots\frac{\sin (x/2^{n-1})}{2\sin(x/2^n)}=\frac{\sin x}{2^n\sin(x/2^n)}.
$$
Hence
$$
\lim_{n\to\infty}\cos(x/2)\cos(x/4)\cdots\cos(x/2^n)=\frac{\sin x}{x},
$$
since
$$
\lim_{t\to 0}\frac{\sin(tx)}{t}=x.
$$
In particular,
$$
\prod_{n=1}^\infty \cos\left(\frac{\pi}{2^{n+1}}\right)=\frac{2}{\pi}.
$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2067042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
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How to find the projection of $x^2+y^2+z^2+2xyz=1$ on $S^2$? The primary problem is to find the singular points of $g$ when $g$ defines on $S^2$ that $g:S^2 \rightarrow R^3,(x,y,z)\rightarrow(yz-x,zx-y,xy-z)$.
$Note:$The singular point in the problem is the point when the rank of $dg_{p}：T_p S^2 \rightarrow T_{g(p)}R^3$ is less than 2.
I tried to find the when the 3 determinant of 2-order minor of the Jacobian of $(y\sqrt{1-x^2-y^2}-x,\sqrt{1-x^2-y^2}-y,xy-\sqrt{1-x^2-y^2})$ are all $0$.
The equation is hard to solve.So I turned to find the point in $R^3 \setminus (0,0,0)$ whose rank is less than 3.
It is equivalent to find when the determinat of matrix
$$
\left[
\begin{matrix}
-1 & z & y \\
z & -1 & x \\
y & x & -1
\end{matrix}
\right]
$$
is $0$.
And I get the equation $x^2+y^2+z^2+2xyz=1$
The next,as I think, is to find the projection of $x^2+y^2+z^2+2xyz=1$ on$S^2$ through the map $(x,y,z)\rightarrow(\dfrac{x}{\sqrt{x^2+y^2+z^2}},\dfrac{y}{\sqrt{x^2+y^2+z^2}} ,\dfrac{z}{\sqrt{x^2+y^2+z^2}})$.
The intersection of $x^2+y^2+z^2+2xyz=1$ and $x^2+y^2+z^2=1$ is obvious a solution.
$x^2+y^2=1,z=0;y^2+z^2=1,x=0;x^2+z^2=1,y=0$
But I have no idea how to find the other projections.What should I do next?
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Let $f \colon \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be defined by the same equation as $g$ so $f(x,y,z) = (yz - x, zx - y, xy - z)$ and $g = f|_{S^2}$. Let $p = (x_0,y_0,z_0) \in S^2$ (so that $x_0^2 + y_0^2 + z_0^2 = 1$). The tangent plane $T_{p}S^2$ can be naturally identified with the plane
$$ T_p S^2 = \{ (x,y,z)^T \in \mathbb{R}^3 \, | \, x \cdot x_0 + y \cdot y_0 + z \cdot z_0 = 0 \} \subseteq \mathbb{R}^3. $$
By calculation, we have
$$ df|_p = \begin{pmatrix} -1 & z_0 & y_0 \\
z_0 & -1 & x_0 \\
y_0 & x_0 & -1 \end{pmatrix}$$
and so
$$ \det(df|_p) = -1(1 - x_0^2) - z_0(-z_0 - x_0 y_0) + y_0(z_0 x_0 + y_0) \\
= -1 + x_0^2 + z_0^2 + x_0 y_0 z_0 + y_0 z_0 x_0 + y_0^2 = 2 x_0 y_0 z_0. $$
Thus, unless $x_0 y_0 z_0 = 0$, we have $\operatorname{rank} (df|_p) = 3$ and so $\operatorname{rank} (dg|_p) = \operatorname{rank} \left( df|_p \right)|_{T_p S^2} = 2$. We conclude that the only suspicious critical points are the points on $S^2$ satisfying $x_0 y_0 z_0 = 0$.
Note that if we permute the coordinates of $\mathbb{R}^3$ then the coordinates of $f$ gets permuted in the same way and so it is enough to analyze the case $z_0 = 0$. For such points,
$$ T_pS = \operatorname{span} \{ (0,0,1)^T, (-y_0, x_0, 0)^T \} $$
and
$$ df|_p \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 & 0 & y_0 \\
0 & -1 & x_0 \\
y_0 & x_0 & -1 \end{pmatrix} \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} y_0 \\ x_0 \\ -1 \end{pmatrix}, \\
df|_p \begin{pmatrix} -y_0 \\ x_0 \\ 0 \end{pmatrix} = \begin{pmatrix} -1 & 0 & y_0 \\
0 & -1 & x_0 \\
y_0 & x_0 & -1 \end{pmatrix} \begin{pmatrix} -y_0 \\ x_0 \\ 0 \end{pmatrix} = \begin{pmatrix} y_0 \\ x_0 \\ x_0^2 - y_0^2 \end{pmatrix}. $$
The vectors above will be linearly dependent if and only if $x_0^2 - y_0^2 = -1$. Together with $x_0^2 + y_0^2 = 1$, we see that the only solutions are $(0,1,0)^T,(0,-1,0)^T$. The other cases can be deduced by symmetry.
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"language": "en",
"url": "https://math.stackexchange.com/questions/2067157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Finding the coefficient of $x^{10}$ in $(1+x^2-x^3)^8$ The coefficient of $x^{10}$ in $(1+x^2-x^3)^8$. I tried to factor it into two binomials but it became way to long to solve by hand.
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$10=2\cdot 5$ or $2\cdot 3+2\cdot 2$. Hence
$$
\binom{8}{5}+\binom{8}{2}\binom{6}{2}=476.
$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2067334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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