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Are these polynomials irreducible in $\mathbb{Q}[x]$? $$x^3+x^2+x+1$$
$$x^5+x^3+x^2+1$$
$$x^5+x^3+x+1$$
I've tried applying Eisenstein criterion but I can't figure it out.
Thanks in advance.
|
$$x^3+x^2+x+1=(x+1)(x^2+1)$$
$$x^5+x^3+x^2+1=(x^2+1)(x^3+1)$$
For $f(x)=x^5+x^3+x+1$, we have $f(2)=43$, which is prime, so by Cohn's criterion, $f(x)$ is irreducible over $\mathbb{Z}$.
|
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If $p(x)=ax^3 -2x^2 +bx+c$, find $a, b$ and $c$ if $p(0)=12$, $p(-1)=3$ and $p(2)=36$ If $p(x)=ax^3 -2x^2 +bx+c$, find $a, b$ and $c$ if $p(0)=12$, $p(-1)=3$ and $p(2)=36$
Can someone please teach me how to do this question thanks!
|
This is really a problem in basic linear algebra. If we simply plug in the three given values of the independent variable $x$, we obtain the three equations for $a$, $b$, $c$:
$p(0) = 12$ becomes
$c = 12$;
$p(-1) = 3$ becomes
$-a -b +c = 5$;
$p(2) = 36$ becomes
$8a + 2b + c = 44$;
looks like we get $c$ for free! Plugging $c= 12$ into the second and third equations yields
$-a -b = -7$;
$8a + 2b = 32$;
looks like we get
$6a = 18$,
$a = 3$;
and finally,
$b= 4$.
This method actually generalizes to arbitrary polynomials; the powers of the different values of $x$ form a matrix and the coefficients form a vector; the rest is clearly linear algabra. Glad to be of assistance in this matter. Cheers.
|
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smallest $k$ s.t. $(x+y)^2\leq k(x^2-xy+y^2)$ I would appreciate if somebody could help me with the following problem
Q: Find $K$?
$$(x+y)^2\leq k(x^2-xy+y^2)$$.
where $\forall x,y\in \mathbb{R}$
|
$$(x+y)^2\leq k(x^2-xy+y^2)\iff (k-1)x^2-(2+k)xy+(k-1)y^2\geq0$$
using the inequality
$$xy\leq \frac{1}{2}(x^2+y^2)$$
we have
$$(k-1)x^2-(2+k)xy+(k-1)y^2\geq (k-1)x^2-\frac{2+k}{2}(x^2+y^2)+(k-1)y^2\\
=(\frac{k}{2}-2)(x^2+y^2)\geq 0\iff k\geq 4$$
and if we take $x=y=1$ we find that $4\leq k$.
So $k= 4$ answers the question.
|
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If $\alpha +\beta = \dfrac{\pi}{4}$ prove that $(1 + \tan\alpha)(1 + \tan\beta) = 2$
If $\alpha +\beta = \dfrac{\pi}{4}$ prove that $(1 + \tan\alpha)(1 + \tan\beta) = 2$
I have had a few ideas about this:
If $\alpha +\beta = \dfrac{\pi}{4}$ then $\tan(\alpha +\beta) = \tan(\dfrac{\pi}{4}) = 1$
We also know that $\tan(\alpha +\beta) = \dfrac{\tan\alpha + \tan\beta}{1- \tan\alpha\tan\beta}$
Then we can write $1 = \dfrac{\tan\alpha + \tan\beta}{1- \tan\alpha\tan\beta}$
I have tried rearranging $1 = \dfrac{\tan\alpha + \tan\beta}{1- \tan\alpha\tan\beta}$ but it has not been helpful.
I also thought if we let $\alpha = \beta$ then I could write $\tan(\alpha+ \alpha) = 1$
(does this also mean $\tan(2\alpha) = 1$?)
then: $\tan(\alpha + \alpha) = \dfrac{\tan\alpha + \tan\alpha}{1- \tan\alpha\tan\alpha}$
which gives: $1 = \dfrac{2\tan\alpha}{1-\tan^2\alpha}$
Anyway these are my thoughts so far, any hints would be really appreciated.
|
You were almost there:
$$\begin{align}
1 &= \frac{\tan \alpha + \tan\beta}{1 - \tan\alpha \tan\beta}\\
1 - \tan\alpha\tan\beta &= \tan\alpha + \tan\beta\\
2 &= 1 + \tan\alpha + \tan\beta + \tan\alpha\tan\beta\\
2 &= (1+\tan\alpha)(1+\tan\beta)
\end{align}$$
where each equation is equivalent to the preceding/following, and the two can be transformed into each other by a simple step.
*
*First multiply with the denominator (that's $\neq 0$),
*then add $1 + \tan\alpha\tan\beta$ to both sides,
*then write $1 + x + y + xy$ as the product $(1+x)(1+y)$.
|
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Show that $\frac {\sin(3x)}{ \sin x} + \frac {\cos(3x)}{ \cos x} = 4\cos(2x)$ Show that
$$\frac{\sin(3x)}{\sin x} + \frac{\cos(3x)}{\cos x} = 4\cos(2x).$$
|
HINT:
$\sin (3x) = 3\sin x - 4 \sin^3 x$
and
$\cos(3x) = 4\cos^3x - 3\cos x$
|
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Evaluating the limit of a sequence given by recurrence relation $a_1=\sqrt2$, $a_{n+1}=\sqrt{2+a_n}$. Is my solution correct? Problem
The sequence $(a_n)_{n=1}^\infty$ is given by recurrence relation:
*
*$a_1=\sqrt2$,
*$a_{n+1}=\sqrt{2+a_n}$.
Evaluate the limit $\lim_{n\to\infty} a_n$.
Solution
*
*Show that the sequence $(a_n)_{n=1}^\infty$ is monotonic. The statement $$V(n): a_n < a_{n+1}$$
holds for $n = 1$, that is $\sqrt2 < \sqrt{2+\sqrt2}$. Let us assume the statement holds for $n$ and show that $V(n) \implies V(n+1)$. We have that $$a_n < a_{n+1}.$$ Adding 2 to both sides and taking square roots, we have that $$\sqrt{2+a_n} < \sqrt{2+a_{n+1}},$$ that is $a_{n+1} < a_{n+2}$ by definition.
*Find bounds for $a_n$. The statement $$W(n): 0 < a_n < 2$$ holds for $n=1$, that is $0 < \sqrt2 < 2$. Let us assume the statement holds for $n$ and show that $W(n) \implies W(n+1)$. We have that $$0 < a_n < 2.$$ Adding two and taking square roots, we have that $$0 < \sqrt2 < \sqrt{2+a_n} < \sqrt4 = 2.$$
*The limit $\lim_{n\to\infty} a_n$ exists, because $(a_n)_{n=1}^\infty$ is a bounded monotonic sequence. Let $A = \lim_{n\to\infty} a_n$.
*Therefore the limit $\lim_{n \to\infty} a_{n+1}$ exists as well and $\lim_{n \to\infty} a_{n+1} = A$. (For $(n_k)_{k=1}^\infty = (2,3,4, \dots)$, we have that $(a_{n_k})_{k=1}^\infty$ is a subsequence of $(a_n)_{n=1}^\infty$, from which the statement follows.)
*We have that $a_{n+1} = f(a_n)$. That means that $A = \lim_{n\to\infty} a_n = \lim_{n \to\infty} {f(a_n)} = f(\lim_{n \to\infty} a_n) = f(A) = \sqrt{2 + A}$. Solving the equation $A = \sqrt{2 + A}$, we get $A = -1 \lor A = 2$.
*Putting it all together, we get that $A = 2$, because the terms of the sequence are increasing and $a_1 > 0$.
Is my solution correct?
|
Looks great. Here is a fun trick I've seen to answer this question.
Using the half angle formula, notice the following:
$$\cos\left(\frac{\pi}{4}\right)=\frac{1}{2}\sqrt 2\\\cos\left(\frac{\pi}{8}\right)=\sqrt{\frac{1}{2}(1+\frac{1}{2}\sqrt 2)}=\frac{1}{2}\sqrt{2+\sqrt 2}\\\cos\left(\frac{\pi}{16}\right)=\sqrt{\frac{1}{2}(1+\frac{1}{2}\sqrt{ 2+\sqrt2})}=\frac{1}{2}\sqrt{2+\sqrt {2+\sqrt 2}}\\\vdots\\\cos\left(\frac{\pi}{2^{n+1}}\right)=\underbrace{\frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{2+{\ldots}}}}}_\text{n times}=\frac{1}{2}a_{n}$$
Now let $n$ approach infinity.
|
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Show that $\sin2\alpha\cos\alpha+\cos2\alpha\sin\alpha = \sin4\alpha\cos\alpha - \cos4\alpha\sin\alpha$
Show that $\sin2\alpha\cos\alpha+\cos2\alpha\sin\alpha = \sin4\alpha\cos\alpha - \cos4\alpha\sin\alpha$
I know that $\sin2\alpha = 2\sin\alpha\cos\alpha$
so
$$\sin2\alpha\cos\alpha=2\sin\alpha\cos^2\alpha$$
and $\cos2\alpha\sin\alpha$ can be expressed in three ways:
$$(\cos^2\alpha-\sin^2\alpha)\sin\alpha =\sin\alpha\cos^2\alpha-\sin^3\alpha$$
$$(2\cos^2\alpha -1)\sin\alpha = 2\cos^2\alpha\sin\alpha - \sin\alpha$$
$$(1-2\sin^2\alpha)\sin\alpha = \sin\alpha - 2\sin^3\alpha$$
I tried adding these, but nothing came close to the required answer.
So then I tried calculating $\sin4\alpha$ (from the required answer):
$$\sin4\alpha=2\sin(2\alpha)\cdot\cos(2\alpha)$$
$$\sin4\alpha=2\cdot2\sin\alpha\cos\alpha(\cos^2\alpha-\sin^2\alpha)$$
$$\sin4\alpha=4\sin\alpha\cos\alpha(\cos^2\alpha-\sin^2\alpha)$$
so $$\sin4\alpha\cos\alpha= 4\sin\alpha\cos^2\alpha(\cos^2\alpha-\sin^2\alpha)$$
Still looking at the answer, I calculated $\cos4\alpha$
$$\cos4\alpha = 1- \sin^2(2\alpha)$$
If $$\sin2\alpha = 2\sin\alpha\cos\alpha$$
then $$\sin^22\alpha= (2\sin\alpha\cos\alpha)^2$$
and $$\cos4\alpha = 1 - 4\sin^2\alpha\cos^2\alpha$$
and $$\cos4\alpha\sin\alpha=\sin\alpha-4\sin^3\alpha\cos^2\alpha$$
I have tried subtracting my values for $\sin4\alpha\cos\alpha$ and $\cos4\alpha\sin\alpha$, but I have not come close to a solution.
|
You're making this way harder than it needs to be. One is $\sin{(2 \alpha + \alpha)}$. The other is $\sin{(4 \alpha-\alpha)}$.
|
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Maximum and minimum function on circle Find minimum and maximum value of function
$f(x,y) = 3x+4y + |x-y|$
on circle
$$\left\{ (x,y): x^2+y^2 = 1 \right\}$$
I used polar coordinate system. So I have $x = \cos t$ and $y=\sin t$ where $ t \in [0, 2 \pi)$.
Then i exploited definition of absolute function and i got:
$$h(t) = \begin{cases} 4 \cos t + 3 \sin t \quad t \in [0, \frac{\pi}{4}] \cup [\frac{5}{4} \pi,2 \pi) \\ 2 \cos t + 5 \sin t \quad t \in (\frac{\pi}{4}, \frac{5}{4} \pi) \end{cases}$$
Hence i received following critical points (earlier i computed first derivative):
$$\cos t = \pm \frac{4}{5} \vee \cos t = \pm \frac{2}{ \sqrt{29} }$$
Then i computed second derivative and after all i received that in $( \frac{2}{ \sqrt{29} } , \frac{5}{ \sqrt{29} }) $ is maximum equal $\sqrt{29}$ and in $(- \frac{4}{ 5 } ,- \frac{3}{5} )$ is minimum equal $- \frac{23}{5}$.
I examined my resolution in wolfram alpha:
How you can see, this resolutions are very different than my. Even the crital points are other. Could you tell me where i have mistake?
|
We have $\frac{\partial{f}}{\partial{x}}=\frac{x-y}{\sqrt{(x-y)^2}}+3$ and $\frac{\partial{f}}{\partial{y}}=\frac{y-x}{\sqrt{(x-y)^2}}+4$, i.e., $\nabla f=(\frac{x-y}{\sqrt{(x-y)^2}}+3, \frac{y-x}{\sqrt{(x-y)^2}}+4)$. Let $\phi(x,y)=x^2+y^2$ $\Rightarrow$ $\nabla \phi= 2(x,y)$. Now, using Lagrange multipliers we have $\nabla f = \lambda \nabla \phi$, i.e., $(\frac{x-y}{\sqrt{(x-y)^2}}+3, \frac{y-x}{\sqrt{(x-y)^2}}+4) =2 \lambda(x,y)$.
With some calculations, I believe this resolves the problem.
|
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How to get value of $a^2b^2+b^2c^2+c^2a^2$ If $a+b+c=0$ and $a^2+b^2+c^2=36$, what is the value of $a^2b^2+b^2c^2+c^2a^2$?
|
We'll work with what we have until we get what we want:
$$0 = (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2 ac = 36 + 2(ab + bc + ac)$$
And therefore we can conclude that
$$
- 18 = ab + bc + ac
$$
This in turn means that
$$
324 = (ab + bc + ac) ^2 = a^2b^2 + b^2 c^2 + a^2 c^2 + 2(a^2bc + ab^2 c + abc^2) \\\\
=a^2b^2 + b^2 c^2 + a^2 c^2 + 2abc(a + b + c) = a^2b^2 + b^2 c^2 + a^2 c^2 + 0$$
|
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Conversion of sum of series into product form: $\sum_{n=1}^\infty \frac1{n(n+1)} = \frac12 \prod_{n=2}^\infty \left( 1+\frac{1}{n^2-1} \right)$ Show that the following series and product are equivalent:
$$
\sum_{n=1}^\infty \left[ \dfrac{1}{n(n+1)} \right] = \dfrac{1}{2} \prod_{n=2}^\infty \left[ 1+\dfrac{1}{n^2-1} \right]
$$
Thought of solving this through induction as the relation is true for 1....but don't know how to proceed with the n part...Please help!!
|
Hint: First, simplify the fractions to get
$$\prod\limits_{n = 2}^\infty 1 + \frac{1}{n^2 - 1} = \prod\limits_{n = 2}^\infty \frac{n^2}{n^2 - 1}$$
Denote the partial product by $P_n$, and take a logarithm:
$$\log{P_N} = \log{\prod\limits_{n = 2}^N \frac{n^2}{n^2 - 1}} = \sum\limits_{n = 2}^N \log{\frac{n^2}{n^2 - 1}}$$
$$ = \sum\limits_{n = 2}^N \log{n^2} - \log{(n^2 - 1)} = \sum\limits_{n = 2}^N 2\log(n) - \log(n - 1) - \log(n + 1)$$
Now a lot of terms will telescope.
|
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Prove that 360 divides (a-2)(a-1)a.a.(a+1)(a+2) there's a question which asks to prove that
360 | a2(a2-1)(a2-4)
I attempted it in the following manner.
a2(a2-1)(a2-4) = (a-2)(a-1)(a)(a+1)(a+2)(a)
The first 5 terms represent the product of 5 consecutive terms. Hence,
One of them will have a factor of 5
One of them will have a factor of 4
One of them will have a factor of 3
One of them will have a factor of 2
=> 5 x 4 x 3 x 2 = 120 will divide the given expression for sure.
Now, how do i bring a factor of 3. I know that i have missed one power of a. But, i think i am missing something. Help will be appreciated
Thank you
|
We know from here or here, the product $n$ consecutive integers is divisible by $n!$ for integer $n>0$
$$(a-2)(a-1)a\cdot a(a+1)(a+2)$$
$$=(a-2)(a-1)a(a+3-3)(a+1)(a+2)$$
$$=\underbrace{(a-2)(a-1)a(a+1)(a+2)(a+3)}_{\text{ the product of } 6\text{ consecutive integers }}-3\underbrace{(a-2)(a-1)a(a+1)(a+2)}_{\text{ the product of } 5\text{ consecutive integers }}$$
We could divide $a$ as $a-3+3$ as well.
|
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Need help with a number theory question I can't solve this question I got from a Math Olympiad past paper:
Find all integers $a$ such that $\frac{a^2+4}{2a+1}$ is also an integer
I know $a$ can be $0$ and $-1$ but I can't ascertain if there are other values $a$ can take.
|
André's answer is pretty much the best way to do this. A more naïve approach is as follows:
Suppose $\frac{a^2+4}{2a+1} = k\in\mathbb{Z}$. Then:
$a^2 - 2ka + (4-k) = 0$
This is a quadratic equation with solutions:
$a = \frac{2k \pm \sqrt{4k^2 - 4(4-k)}}{2}= k \pm\sqrt{k^2+k-4}$
For $a$ to be an integer we thus need $k^2 + k - 4$ to be a perfect square, i.e. $k^2 + k - 4 = m^2$. Rearrange and get another quadratic equation with solutions:
$k = \frac{-1 \pm \sqrt{1 + 4(4+m^2)}}{2}$
Again we want $k$ to be an integer so need $4m^2 + 17 = n^2$ for some integer $n$.
So we get a difference of two squares equation: $n^2 - (2m)^2 = 17$. Factorising tells us that $(n+2m)(n-2m) = 17$. Solving gives $m=\pm 4$, which tells us that $k=4,-5$ giving $a = -9,-1,0,8$.
You can check all $4$ of these possibilities work and so we are done.
|
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Geometry problem on finding the diameter of a circle $AB$ is the diameter of a circle. Tangents $AD$ and $BC$ are drawn so that $AC$ and $BD$ intersect at a point on the circle. If $|AD|=a$ and $|BC|=b$, $(a \neq b)$ then find the diameter of the circle.
|
Lets say AC and BD intersect at E , so at right angles as diameter subtends right angle at circumference .
Using pythagorous theorem
$$ \displaystyle\boxed{(diameter)^2 = AE^2 + BE^2} $$
So now we should calculate AE and BE to get d
as $$CE \times CA = CB^2 ---> 1$$
and $$DE \times DB = DA^2 ---> 2$$
and as the triangles AED and CEB are similar
$$ \frac{AE}{EC} = \frac{DE}{EB} = \frac{a}{b} $$
$$ \frac{AC}{EC} = \frac{a+b}{b} --------> 3$$
$$ \frac{BD}{DE} = \frac{a+b}{a} --------> 4$$
substitute AC of result 3 in 1 and get $CE^2 $ and from there CE ,
Similaryly get DE
$$ CE = \frac{b*\sqrt{b}}{\sqrt{a+b}} $$
$$ DE = \frac{a*\sqrt{a}}{\sqrt{a+b}} $$
so $$AE = \frac{a}{b} \times CE $$
$$\displaystyle\boxed{AE = \frac{a\sqrt{b}}{\sqrt{a+b}}}$$
similary
$$BE = \frac{b}{a} \times DE $$
$$\displaystyle\boxed{BE = \frac{b\sqrt{a}}{\sqrt{a+b}}}$$
so just substitute AE and BE in $ (diameter)^2 = AE^2 + BE^2 $and find d the diameter
$$ d^2 = \frac{a^2*b}{a+b} + \frac{b^2*a}{a+b} $$
$$ d^2 = \frac{ab*(a+b)}{a+b} $$
this implies
$$ d = \sqrt{ab}$$
|
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Does this hold for three numbers If $a\ge b\ge c\ge0$, does it hold that $\sqrt[3]{\left(a-b+c\right)^{2}}\ge\sqrt[3]{a^{2}}-\sqrt[3]{b^{2}}+\sqrt[3]{c^{2}}$?
Thanks for any help.
|
Consider one case of $a = b \ge c \ge 0$. Both sides are equal, and the inequality holds.
Now, partial differentiate $\sqrt[3]{(a-b+c)^2} - \sqrt[3]{a^2} + \sqrt[3]{b^2} - \sqrt[3]{c^2}$ with respect to a.
$$\frac{2}{3}(a-b+c)^{-\frac{1}{3}} - \frac{2}{3}a^{-\frac{1}{3}}$$
$$=\frac{2}{3\sqrt[3]{a-b+c}} - \frac{2}{3\sqrt[3]{a}}$$
Show this partial derivative is positive (because $-b+c$ is negative), and you are done.
|
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Prove that if $d\cdot e| d(d+1)+e\cdot e$ then $d\cdot (d+1)+e\cdot e=3de$ Prove that if $d\cdot e| d(d+1)+e\cdot e$ then $d\cdot (d+1)+e\cdot e=3de$ where $d$ and $e$ are positive integers.
|
We can simplify the problem a bit. Let $\gamma = \gcd(d,e)$ and write $d = \gamma\delta,\; e = \gamma\varepsilon$. Then the condition becomes
$$\gamma^2\delta\varepsilon \mid \gamma\delta(\gamma\delta+1) + \gamma^2\varepsilon^2\tag{1}$$
and we see that we must have $\gamma \mid \delta$. So write $d = \gamma^2\kappa$ to obtain
$$\begin{align}
\gamma^3\kappa\varepsilon &\mid \gamma^2\kappa(\gamma^2\kappa+1) + \gamma^2\varepsilon^2\\
\iff \gamma\kappa\varepsilon &\mid \kappa(\gamma^2\kappa + 1) + \varepsilon^2.
\end{align}$$
Hence $\kappa \mid \varepsilon^2$. But $\kappa \mid \delta$ and $\gcd(\delta,\varepsilon) = 1$, therefore $\kappa = 1$ and we obtain
$$\gamma\varepsilon \mid \gamma^2 + 1 + \varepsilon^2.\tag{2}$$
That is equivalent to
$$\gamma \mid \varepsilon^2+1 \land \varepsilon \mid \gamma^2+1.\tag{3}$$
Since
$$F_{2k+1}^2 + 1 = F_{2k+3}F_{2k-1},$$
we have the family of consecutive odd-indexed Fibonacci numbers as solutions for $(\gamma,\,\varepsilon)$. For the original $d,\,e$, that means
$$d = F_{2k+1}^2,\quad \text{ and } e \in \{ F_{2k+1}F_{2k-1},\, F_{2k+1}F_{2k+3}\}.$$
It is easily verified that
$$\frac{\gamma^2+1+\varepsilon^2}{\gamma\varepsilon} = 3 = \frac{d(d+1)+e^2}{de}$$
for these pairs.
The above are the only solutions of $(3)$:
Let $1 < a < b$ with $a \mid b^2 + 1$ and $b \mid a^2 +1$. Write $a^2+1 = b\cdot c$ and $b^2+1 = a\cdot k$. Then $ 0 < c < a$ (since $a < b$), we have $c \mid a^2 + 1$, and
$$\begin{align}
c^2 + 1 &= \left(\frac{a^2+1}{b}\right)^2+1\\
&= \frac{(a^2+1)^2 + b^2}{b^2} = \frac{a^4 + 2a^2 + (1 + b^2)}{b^2}\\
&= \frac{a^2(a^2+2) + ak}{b^2}.
\end{align}$$
Since $\gcd(a,b) = 1$, the common factor $a$ in the numerator cannot be cancelled (even partially) by the $b^2$ in the denominator, hence also $a \mid c^2+1$.
So every pair solving $(3)$ gives rise to a smaller pair (whose larger member is the smaller of the original), thus creates a descending chain (and an ascending) necessarily ending in $(1,2)$, therefore it belongs to the Fibonacci family.
|
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|
Some identities with the Riemann zeta function Can someone either help derive or give a reference to the identities in Appendix B, page 27 of this, http://arxiv.org/pdf/1111.6290v2.pdf
Here is a reproduction of Appendix B from Klebanov, Pufu, Sachdev and Safdi's $2012$ preprint (v2) 'Renyi Entropies for Free Field Theories' (from the source at arxiv.org and hoping there is no problem citing it here...(RM)).
B Useful mathematical formulae
In this section we present some useful mathematical formulae. We begin with zeta function identities.
For $0 < a \leq 1$ we have the identity
$$\tag{B.1} \zeta(z, a) = \frac{2 \Gamma(1 - z)}{(2 \pi)^{1-z}} \left[\sin \frac{z \pi}{2} \sum_{n=1}^\infty \frac{\cos 2 \pi a n}{n^{1-z}}
+ \cos \frac{z \pi}{2} \sum_{n=1}^\infty \frac{\sin 2 \pi a n}{n^{1-z}} \right] \,$$
Taking derivatives at $z=0, -1, -2$ gives
\begin{align}
\zeta'(-2, a) &= - \frac{1}{4 \pi^2} \sum_{n=1}^\infty \frac{\cos 2 \pi a n}{n^3}
- \frac{1}{4 \pi^3} \sum_{n=1}^\infty \frac{(2 \log (2 \pi n) + 2 \gamma - 3) \sin 2 \pi a n}{n^3} \,, \\
\tag{B.2}\zeta'(-1, a) &= \frac{1}{4 \pi} \sum_{n=1}^\infty \frac{\sin 2 \pi q n}{n^2}
- \frac{1}{2 \pi^2} \sum_{n=1}^\infty \frac{(\log (2 \pi n) + \gamma - 1) \cos 2 \pi a n}{n^2} \,, \\
\zeta'(0, a) &= \frac{1}{2} \sum_{n=1}^\infty \frac{\cos 2 \pi a n}{n}
+ \frac{1}{\pi} \sum_{n=1}^\infty \frac{(\log (2 \pi n) + \gamma) \sin 2 \pi a n}{n} \,.
\end{align}
Two other useful identities are the regularized sums
\begin{align}
\tag{B.3}\sum_{n \in \mathbb{Z}} \log \left( \frac{n^2}{q^2} + a^2 \right)
&= 2 \log \left[2 \sinh (\pi q |a|) \right] \,, \\
\sum_{n \in \mathbb{Z} + \frac 12} \log \left(\frac{n^2}{q^2} + a^2 \right)
&= 2 \log \left[2 \cosh (\pi q |a|) \right] \,.
\end{align}
These sums follow from the more general formula
$$\tag{B.4}\sum_{n \in \mathbb{Z}} \log \left( \frac{(n + \alpha)^2}{q^2} + a^2 \right)
= \log \left[2 \cosh (2 \pi q |a|) - 2 \cos (2 \pi \alpha) \right] \,.$$
This relation in turn follows from the Poisson summation formula
$$\tag{B.5} \frac{1}{ 2 \pi q} \sum_{n \in \mathbb{Z}} \hat f \left( \frac{n + \alpha}{q} \right) =\sum_{k \in \mathbb{Z}} e^{-i 2 \pi k \alpha} f(2 \pi q k) \,$$
applied to
$$\tag{B.6}\hat f(\omega) = \log \left( \omega^2 + a^2 \right) \,.$$
For $t \neq 0$ one can simply calculate the inverse Fourier transform of $\hat f$:
$$\tag{B.7}f(t) = \int_{-\infty}^\infty \frac{d\omega}{2 \pi} e^{-i \omega t} \log \left(\omega^2 +a^2 \right) = - \frac{e^{-|a|\;|t|}}{|t|} \,.$$
The case $t=0$ requires special care because the expression for $f(0)$ is divergent and requires regularization:
$$\tag{B.8}f(0) = \int_{-\infty}^\infty \frac{d\omega}{2 \pi} \log \left(\omega^2 +a^2 \right)
= -\frac{d}{ds} \int \frac{d\omega}{2 \pi} \frac{1}{\left(\omega^2 +a^2 \right)^s} \Biggr\rvert_{s=0}
= |a| \,.$$
Using $(B.6)-(B.8)$ one can show that $(B.5)$ reduces to $(B.4)$.
|
The first identity is essentially the functional equation for the Hurwitz Zeta Function. You can find a proof and associated results here, where the identity appears on page 2 of the pdf (pg 1916) as equation (2) where the $F$ series is defined on the previous page.
|
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|
$a+b+c+d=4n$; where $a,b,c$ lie between $0$ and $n $ and $d$ lies between $0$ and $2n.$ In the examination the max marks for each of three papers is $n$ and the fourth paper is $2n$. find the number of ways in which a candidate can gets $3n$ marks ?
|
Rather than applying inclusion-exclusion, you can let generating functions do the work for you. The possible (integral) marks that can be obtained for a single paper with maximum mark$~m$ is given by the polynomial $1+X+\cdots+X^m=\frac{1-X^{m+1}}{1-X}$. You want to get the coefficient of $X^{4n}$ (or maybe of $X^{3n}$) in the product of those polynomials for $m=n,n,n,2n$. Multiplying the numerators gives $(1-X^{n+1})^3(1-X^{2n+1})$, which gives $1-3X^{n+1}-X^{2n+1}+3X^{2n+2}+3X^{3n+2}-X^{3n+3}$ after dropping term of degree${}>4n$ that do not interest us. The denominator is $(1-X)^4$, so we must multiply the numerator by $(1-X)^{-4}=\sum_{k\in\Bbb N}\binom{k+3}3X^k$. Then the coefficient of $X^{4n}$ in the product is
$$
\binom{4n+3}3-3\binom{3n+2}3-\binom{2n+2}3+3\binom{2n+1}3+3\binom{n+1}3-\binom n3=\frac{n^3+6n^2+11n+6}6.
$$
One can check this result as being the unique cubic polynomial in $n$ that gives $1,4,10,20$ (found by explicit counting) for $n=0,1,2,3$, respectively.
|
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|
How to expand $\left(1+px+qx^2\right)^{-2}$ The coefficients of $x$ and $x^2$ in the expansion of $\left(1+px+qx^2\right)^{-2}$ in ascending powers of $x$ are 4 and 14 respectively. Find the values of $p$ and $q$.
How should you go about solving this?
|
In general
$$(1+y)^{-2} = \sum_{k=0}^{\infty} (-1)^k (k+1) \, y^k$$
so that
$$(1+p x+q x^2)^{-2} = 1 - 2 (p x+q x^2) + 3 (p x + q x^2)^2 - \cdots$$
Out to $O(x^2)$:
$$(1+p x+q x^2)^{-2} = 1 -2 p x + (3 p^2 - 2 q) x^2 + \cdots$$
$$-2 p = 4 \implies p=-2$$
$$3 p^2-2 q=14 \implies -2 q=2 \implies q = -1$$
You can also consider the derivatives of $(1+y)^{-2}$ at $y=0$:
$$\frac{d}{dy} (1+y)^{-2} = -2 (1+y)^{-3}$$
$$\frac12 \frac{d^2}{dy^2} (1+y)^{-2} = 3 (1+y)^{-4}$$
|
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|
Determine the minimum of $a^2 + b^2$ if $a,b\in\mathbb{R}$ are such that $x^4 + ax^3 + bx^2 + ax + 1 = 0$ has at least one real solution I just wanted the solution, a hint or a start to the following question.
Determine the minimum of $a^2 + b^2$ if $a$ and $b$ are real numbers for which
the equation
$$x^4 + ax^3 + bx^2 + ax + 1 = 0$$
has at least one real solution.
Thanks in advance.
|
As has already been observed, if $c$ is a root of your equation, so is $c^{-1}$. So suppose your equation has roots $c,d,c^{-1},d^{-1}$.
We have
$$a=-(c+c^{-1} + d + d^{-1})$$
and
$$b=cc^{-1}+ cd+cd^{-1} +c^{-1}d +c^{-1}d^{-1} + dd^{-1}$$
$$=2+(c+c^{-1})(d+d^{-1})$$
Now if $c\in \mathbb{R}$, we must also have $d+d^{-1}\in \mathbb{R}$. This means $d=i$ or $d\in \mathbb{R}$.
Thus if we write $z=c+c^{-1}$ and $w=d + d^{-1}$, we see that your problem is a question of minimising
$$f=(w+z)^2 + (2+wz)^2$$
subject to the constraint $|w|\geq 2$ and $|z|\geq 2$ or $z=0$.
If $z=0$ the $f=w^2+4\geq8$. Consider instead $|z|\geq 2$. In this case $|zw|\geq 4$ so $f\geq (2-zw)^2\geq 4$. Thus we see that the minimal value of $f$ is 4. This is obtained when $z=2,w=-2$. And the equation is
$$x^4-2x^2+1$$
|
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|
Integral using Euler's substitution Question:
Solve this:
$$\displaystyle \int \frac {2x-\sqrt {4x^2-x+1} } {x-1} \,dx$$
Our solution:
$$16 \ln|t+4|-\ln|t+0.25|+C$$ when $t=\sqrt {4x^2-x+1}-2x$ (using Euler's formula)
But wolfram's solution was totally different. We were wondering if these two are equal.
We found this question very helpful
Integral( using Euler substitution)
but we were wondering how could one check that hyperbolic solution and Euler solution are equal?
|
Tools that might help for a direct check if the two solutions are equal up to a constant ... and perhaps useful in the future.
*
*We have that the expression $\sqrt{4x^2-x+1}-2x$ can be re-written as $\frac{(\sqrt{4x^2-x+1}-2x)(\sqrt{4x^2-x+1}+2x)}{\sqrt{4x^2-x+1}-2x}=\frac{(4x^2-x+1)-4x^2}{\sqrt{4x^2-x+1}+2x}$.
*Moreover, if $\frac{(4x^2-x+1)-4x^2}{\sqrt{4x^2-x+1}+2x}$ is inside a logarithm it can be re-written as $\log(1-x)-\log(\sqrt{4x^2-x+1}+2x)$.
*The inverse of $\sinh$ can be computed in terms of logarithms: $y:=\sinh(x)=\frac{e^x-e^{-x}}{2}$, from where $e^{2x}-2ye^x-1=0$, and then $e^x=\frac{2y+\sqrt{4y^2+4}}{2}$. Then we get $\sinh^{-1}(x)=\log(y+\sqrt{y^2+1})$.
|
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|
Forget about the sin and cos functions, show that $(x-x^3/3!+x^5/5!-x^7/7!+...)^2+ (1-x^2/2!+x^4/4!-x^6/6!+...)^2=1$. Forget about the $\sin$ and $\cos$ functions, are there possibly some brilliant way to show that
$$\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots\right)^2+
\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots\right)^2=1$$
?
I've thought for a long time, without making much progress. Can someone help me? Thanks.
|
If we denote
$$f(x)=\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots\right)^2+
\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots\right)^2$$
then by differentiation term by term we see that $f'(x)=0$ so $f(x)=f(0)=1$.
|
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|
Finding minima and maxima of $\sin^2x \cos^2x$ I am trying to find the turning points of $\sin^2x \cos^2x$ in the range $0<x<\pi$.
So far I have:
$\begin{align} f'(x) & =\cos^2x \cdot 2\sin x \cdot \cos x + \sin^2x \cdot 2\cos x \cdot -\sin x \\
& = 2 \cos^3x \sin x - 2\sin^3x \cos x \\
& = 2 \sin x \cos x (\cos^2x - \sin^2x) \\
& = 2 \sin x \cos x (\cos x + \sin x) (\cos x - \sin x)
\end{align}$
And, after lots of simplification, I think I've found that:
$f''(x) = 2 \left[\left( \cos^2x-\sin^2x \right)^2 - 4\sin^2x \cos^2 x\right]$
My questions are:
*
*How can I evaluate $0 = \cos x + \sin x$ and $0 = \cos x - \sin x$ without resorting to graph plotting?
*Are there trigonometric identities that I could have used to simplify either derivative?
|
One might broadly interpret "algebraically" to mean "without calculus." This is actually quite doable, and in my opinion less complicated than dealing with messy trigonometric equations.
Note the following identities:
$f(x) = \sin^2 x \cos^2 x = \sin^2 x - \sin^4 x = \sin^2 x (1-\sin^2 x) \ge 0$
Equality holds when $\sin x = 0$ or $\pm 1,$ which occurs only for $x = \frac{\pi }{2}$ on $(0,\pi ).$
I claim that $f$ is maximized at $\frac{\pi }{6}.$ Since $0\le \sin^2 x \le 1$ for all real $x,$ it suffices to show that the function $g(x) = x (1-x)$ is maximized at $x=\frac{1}{2}$ on $[0,1].$ We can see this by noting that
$\frac{1}{4} - x (1-x) = x^2 - x + \frac{1}{4} = (x-\frac{1}{2})^2 \ge 0$
|
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|
Sum of the series $1+\frac{1\cdot 3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+\cdots$ Decide if the sum of the series
$$1+\frac{1\cdot3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+ \cdots$$
is: (i) $\infty$, (ii) $1$, (iii) $2$, (iv) $4$.
|
Note that
$$\binom{n}{r} = \frac{n(n-1)(n-2)\dots(n-r+1)}{1\cdot2\cdot3\cdots r}$$
and so
$$\binom{1/2}{r} = \frac{\frac12 \cdot \frac{-1}2 \cdot \frac{-3}{2} \cdot \frac{-5}{2} \cdots \frac{-(2r-3)}{2}}{1\cdot2\cdot3\cdots r} = \frac{(-1)^{r-1}1 \cdot 3 \cdot 5 \cdots (2r-3)}{2^r r!}$$
Now, in this problem, each term after the "$1$" term follows a pattern. Let the last factor in the denominator of a term be $2r$, then the general term is:
$$
\frac{1 \cdot 3 \cdot 5 \cdots (2r - 3)}{6 \cdot 8 \cdot 10 \cdots (2r)}
= \frac{1 \cdot 3 \cdot 5 \cdots (2r - 3)}{2^{r-2} 3 \cdot 4 \cdot 5 \cdots r}
= \frac{1 \cdot 3 \cdot 5 \cdots (2r - 3)}{2^{r-3} r!}
= 8 (-1)^{r-1} \binom{1/2}{r} \\
= -8\binom{1/2}{r}(-1)^r
$$
We know from the binomial theorem that $\sum_{r=0}^{\infty} \binom{n}{r} x^r = (1 + x)^n$ for $|x| < 1$, and with $x = -1$, we also know that for positive integer $n$, at least, we have $\sum_{r=0}^{\infty} \binom{n}{r} (-1)^r = (1-1)^n = 0$. In light of these, it is not hard to believe that $\sum_{r=0}^{\infty} \binom{1/2}{r} (-1)^r = 0$ as well: I'm not exactly sure how to prove this, but it would follow from a Tauberian theorem considering we can prove that the sum converges.
So our sum in this problem is
$$
1 + \sum_{r=3}^{\infty} -8 \binom{1/2}{r} (-1)^r
= 1 - 8\sum_{r=3}^{\infty} \binom{1/2}{r} (-1)^r
$$
The sum inside is almost the sum we said above is $0$, except that the $r = 0, 1, 2$ terms are missing. In other words, our sum in this problem is:
$$
\begin{align}
&1 - 8\Bigg(0 - \Big(1 + \frac12(-1) + \frac{(1/2)(-1/2)}{2}\Big)\Bigg) \\
&= 1 + 8\left(\frac38\right) \\
&= 4.
\end{align}
$$
|
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|
A problem on limit It seems
$\lim_{n \rightarrow \infty} \sum_{r=1}^{n-1}\binom{n}{r-1}
\sum_{j=0}^{n-r+1} (-1)^j \binom{n-r+1}{j} \frac{1}{r+j+1}$
is $\frac{1}{2}$. Here's a plot of the function for $n \leq 300$:
But I can not prove this. Any hints?
|
$\begin{align}
S(n)
&=\sum_{r=1}^{n-1}\binom{n}{r-1}
\sum_{j=0}^{n-r+1} (-1)^j \binom{n-r+1}{j} \frac{1}{r+j+1}\\
&= \sum_{r=0}^{n-2}\binom{n}{r}
\sum_{j=0}^{n-r} (-1)^j \binom{n-r}{j} \frac{1}{r+j+2}\\
\text{(replace }r \text{ by } n-r)\quad&= \sum_{r=2}^{n}\binom{n}{r}
\sum_{j=0}^{r} (-1)^j \binom{r}{j} \frac{1}{n-r+j+2}\\
(D \text{ described below) }\quad&= D+\sum_{r=0}^{n}\binom{n}{r}
\sum_{j=0}^{r} (-1)^j \binom{r}{j} \frac{1}{n-r+j+2}\\
\text{(replace } j \text{ by } r-j)\quad&= D+\sum_{r=0}^{n}\binom{n}{r}
\sum_{j=0}^{r} (-1)^{r-j} \binom{r}{j} \frac{1}{n-j+2}\\
\text{(change order of summation) }\quad&= D+\sum_{j=0}^{n}\sum_{r=j}^{n}\binom{n}{r}
(-1)^{r-j} \binom{r}{j} \frac{1}{n-j+2}\\
\quad&=D+ \sum_{j=0}^{n}\frac{(-1)^{j}}{n-j+2}\sum_{r=j}^{n}
(-1)^{r}\binom{n}{r}\binom{r}{j} \\
\text{ (put } n-r \text{ for } r)\quad&=D+ \sum_{j=0}^{n}\frac{(-1)^{j}}{n-j+2}
\sum_{r=0}^{n-j}(-1)^{n-r}\binom{n}{n-r}\binom{n-r}{j} \\
\text{ (put } n-j \text{ for } j)\quad&= D+\sum_{j=0}^{n}\frac{(-1)^{n-j}}{j+2}
\sum_{r=0}^{j}(-1)^{n-r}\binom{n}{n-r}\binom{n-r}{n-j} \\
&= D+\sum_{j=0}^{n}\frac{(-1)^{j}}{j+2}
\sum_{r=0}^{j}(-1)^{r}\frac{n!(n-r)!}{(n-r)!r!(n-j)!(j-r)!} \\
&= D+n!\sum_{j=0}^{n}\frac{(-1)^{j}}{(n-j)!(j+2)}
\sum_{r=0}^{j}(-1)^{r}\frac{1}{r!(j-r)!} \\
&= D+n!\sum_{j=0}^{n}\frac{(-1)^{j}}{j!(n-j)!(j+2)}
\sum_{r=0}^{j}(-1)^{r}\frac{j!}{r!(j-r)!} \\
&= D+\sum_{j=0}^{n}\binom{n}{j}\frac{(-1)^{j}}{j+2}
\sum_{r=0}^{j}(-1)^{r}\binom{j}{r} \\
\text{(since }\sum_{r=0}^{j}(-1)^{r}\binom{j}{r}=0 \text{ for }j > 0)\quad
&= D+\frac{1}{2}\\
\end{align}
$
$D=
-\sum_{r=0}^1\binom{n}{r}\sum_{j=0}^{r} (-1)^{r-j} \binom{r}{j} \frac{1}{n-j+2}
$.
If $r=0$,
$\binom{n}{r}\sum_{j=0}^{r}(-1)^{r-j} \binom{r}{j} \frac{1}{n-j+2}
=\binom{n}{0}\binom{0}{0} \frac{1}{n+2}
=\frac{1}{n+2}
$.
If $r=1$,
$\binom{n}{r}\sum_{j=0}^{r} (-1)^{r-j} \binom{r}{j} \frac{1}{n-j+2}
=\binom{n}{1}\left(-\binom{1}{0} \frac{1}{n+2}+\binom{1}{1} \frac{1}{n+1}\right)
=n\left(-\frac{1}{n+2}+\frac{1}{n+1}\right)
=n\left(\frac{1}{(n+1)(n+2)}\right)
=\frac{n}{(n+1)(n+2)}
$.
So $D
=-\frac{1}{n+2}
-\frac{n}{(n+1)(n+2)}
=\frac{-(n+1)-n}{(n+1)(n+2)}
=\frac{-2n-1}{(n+1)(n+2)}
=-\frac{2n+1}{(n+1)(n+2)}
$.
Therefore
$S(n) = \frac12-\frac{2n+1}{(n+1)(n+2)}
$.
OMG!!!!!
I can't believe that
all this algebra
came up with something that
looks correct!
Note:
to compare my remainder
with user71352's,
$\frac{2n+1}{(n+1)(n+2)}-\frac{1}{n+1}
=\frac{2n+1-(n+2)}{(n+1)(n+2)}
=\frac{n-1}{(n+1)(n+2)}
$.
A computation could decide which is correct,
but, considering the relative complexity of the solutions,
mine has a higher chance of being wrong.
(Added later)
But it wasn't.
Surprisingly (to me),
mine was correct.
Feels good.
I knew it would.
|
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|
Showing Whether a Sequence is Bounded Above or Not I am trying to solve the following problem about a sequence:
Consider the sequence ${a_n}$ where $a_n = 1 + \frac{1}{1 \cdot 3} + \frac {1}{1 \cdot 3 \cdot 5} + \frac {1}{1 \cdot 3 \cdot 5 \cdot 7} + ... + \frac{1}{1 \cdot 3 \cdot ... \cdot (2n-1)}.$ Decide whether ${a_n}$ is bounded above or not, and prove your answer is correct.
I started to solve it in the following manner:
Let $a_n \geq a_{n+1} \Rightarrow \frac{1}{1 \cdot 3 \cdot ... \cdot (2n-1)} \geq \frac{1}{1 \cdot 3 \cdot ... \cdot (2n-1) \cdot 2(n+1)-1} \Rightarrow \frac{1}{1 \cdot 3 \cdot ... \cdot (2n-1)} \geq \frac{1}{1 \cdot 3 \cdot ... \cdot (2n-1) \cdot 2n+1}.$ Multiplying by the reciprocal of $a_n$, we have $1 \geq \frac{1}{2n+1}$ which shows that $a_n$ is increasing.
This is where I am stuck since I do not know how to proceed in showing it is bounded from what I have so far. Any assistance or criticism is welcome.
I am using the following textbook: Introduction to Analysis by Arthur Mattuck.
|
If you want a direct proof that doesn't use geometric series outright, note that if you consider the terms in $a_n$ shifted forward one position and then compare to the terms in $a_{n+1}$, you get $a_{n+1} \leq 1 + a_n/3$. Also note $a_1 = 1 < 3/2$. If $a_n < 3/2$ then $a_{n+1} \leq 1 + a_n/3 < 1 + 1/2 = 3/2$ as well. So the entire sequence is bounded above by $3/2$ by induction.
|
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Roots of biquadratic equation This question also was a part of my today's maths olympiad paper:
If squares of the roots of $x^4 + bx^2 + cx + d = 0$ are $\alpha, \beta, \gamma, \delta$ then prove that:
$64\alpha\beta\gamma\delta - [4\Sigma \alpha\beta - (\Sigma \alpha)^2]^2 = 0$
I found value of $\alpha\beta\gamma\delta$ = $d^2$ and $\Sigma \alpha$ = -2b.
How to move on ?
|
Complete solution of the problem:
Given that α,β,γ & δ are squares of the roots of $x^4+bx^2+cx+d=0$ hence $\sqrt{α},\sqrt{β},\sqrt{γ}$ & $\sqrt{δ}$ are the roots of the given equation.
Thus, we have
sum of the roots
$$=\sqrt{α}+ \sqrt{β}+\sqrt{γ}+ \sqrt{δ}=\frac{-B}{A}=\frac{0}{1}=0$$ &
sum of the products of two roots
$$= \sqrt{αβ}+\sqrt{αγ}+\sqrt{αδ}+\sqrt{βγ}+\sqrt{βδ}+\sqrt{γδ}=\frac{C}{A}=\frac{b}{1}=b$$
now, taking the square of the first expression, we get
$$(\sqrt{α}+ \sqrt{β}+\sqrt{γ}+ \sqrt{δ})^2=0 $$
$$α+β+γ+δ+2(\sqrt{αβ}+\sqrt{αγ}+\sqrt{αδ}+\sqrt{βγ}+\sqrt{βδ}+\sqrt{γδ})=0$$
$$∑α+2b=0 $$ $∑α=-2b\tag1$
Now, the product of the roots
$$=\sqrt{α} \sqrt{β} \sqrt{γ} \sqrt{δ}=\frac{E}{A}=\frac{d}{1}=d$$
$$ \sqrt{αβγδ}=d $$ $$ αβγδ=d^2 \tag2$$
Now, satisfying the given equation $x^4+bx^2+cx+d=0$ by its roots $\sqrt{α},\sqrt{β},\sqrt{γ}$ & $\sqrt{δ}$ one by one, we get the following
$$\begin{align}
α^2+bα+c\sqrt{α}+d=0\\
β^2+bβ+c\sqrt{β}+d=0\\
γ^2+bγ+c\sqrt{γ}+d=0\\
δ^2+bδ+c\sqrt{δ}+d=0\\
\end{align}$$
Now by adding above four expressions column-wise, we get
$$α^2+β^2+γ^2+δ^2+b(α+β+γ+δ)+c(\sqrt{α}+ \sqrt{β}+\sqrt{γ}+ \sqrt{δ})+4d=0$$
$$α^2+β^2+γ^2+δ^2+b(-2b)+c(0)+4d=0$$
$$α^2+β^2+γ^2+δ^2=2b^2-4d \tag3$$
Now by squaring both the sides of eq(1),we have
$$(α+β+γ+δ)^2=(-2b)^2$$
$$α^2+β^2+γ^2+δ^2+2(αβ+αγ+αδ+βγ+βδ+γδ)=4b^2$$
By substituting the value from eq(3), we get
$$2b^2-4d+2∑αβ=4b^2 $$
$$2∑αβ= 2b^2+4d $$
$$ 4∑αβ= 4b^2+8d \tag4$$
Now, by substituting all the corresponding values from eq(1), (2) & (4) in the LHS of the expression to be proved, we get
$$LHS=64αβγδ-[4∑αβ-(∑α)^2 ]^2=64d^2-[4b^2+8d-(-2b)^2 ]^2$$
$$ =64d^2-[8d]^2=64d^2-64d^2=0=RHS $$
|
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|
Compute $ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx$ Compute $$ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx$$
Of course we have $$ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx = 2 \int_{0}^{\infty} \frac{x^2}{(1+x^2)^2} dx = 2 \int_{0}^{\infty} \left( \frac{x}{1+x^2} \right) ^2 dx = \lim_{ A \to \infty } \int_{0}^{A} \left( \frac{x}{1+x^2} \right) ^2 dx $$
I think that firstly I should compute $ \int \left( \frac{x}{1+x^2} \right) ^2 dx $ but I don't have idea.
|
HINT:
$$I=\int\frac{x^2dx}{(x^2+1)^2}=\int x\cdot \frac x{(x^2+1)^2}dx$$
Integrating by parts,
$$I=x\int \frac x{(x^2+1)^2}dx-\int\left(\frac{d(x)}{dx}\cdot \frac x{(x^2+1)^2}dx\right)dx$$
$$\text{As }\int \frac x{(x^2+1)^2}dx=\frac12\int\frac{d(x^2+1)}{(x^2+1)^2}=-\frac1{2(x^2+1)}$$
$$I=-x\cdot\frac1{2(x^2+1)}+\int \frac1{2(x^2+1)}dx=-\frac x{2(x^2+1)}+\frac{\arctan x}2+C $$
Can you take it from here?
|
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|
How do I write this sum in summation notation? $$\sum_{n=?}^\infty \left(\frac{x^n}{?}\right) = \frac{x^0}{1} + \frac{x^1}{x^2 -1}+\frac{x^2}{x^4 - x^2 +1}+\frac{x^3}{x^6 -x^4 + x^2 -1}+\frac{x^4}{x^8-x^6 +x^4 - x^2 +1}+\cdots$$
I am pretty sure I have the numerator of the summation, $x^n$ correct, but don't know how to write the denominator because of the alternating signs in each term. As for the starting point, $n=?$, I think its going to be either $0$ or $1$, depending on the denominator - I don't want to be kicked from this site by attempting to divide by $0$ .
|
Once you realise that the denominators are sums of powers of $-x^2$, this becomes easy. You have
$$
\sum_{n=0}^\infty\frac{x^n}{\sum_{i=0}^n(-x^2)^i}.
$$
Actually that denominator is a geometric series, that can be written explicitly without summation (provided that $X=-x^2\neq1$) using
$$
\sum_{i=0}^nX^i=\frac{1-X^{n+1}}{1-X},
$$
which gives a fraction in the denominator that can be removed by simple algebra
$$
\sum_{n=0}^\infty\frac{x^n}{\sum_{i=0}^n(-x^2)^i}
=\sum_{n=0}^\infty\frac{x^n}{\frac{1-(-x^2)^{n+1}}{1-(-x^2)}}
=\sum_{n=0}^\infty\frac{x^n(1+x^2)}{1-(-x^2)^{n+1}}
=(1+x^2)\sum_{n=0}^\infty\frac{x^n}{1-(-x^2)^{n+1}}.
$$
After this I'm not sure whether it can be further simplified.
|
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|
How to show $\sqrt{4+2\sqrt{3}}-\sqrt{3} = 1$ I start with $x=\sqrt{4+2\sqrt{3}}-\sqrt{3}$, then
$\begin{align*}
x +\sqrt{3} &= \sqrt{4+2\sqrt{3}}\\
(x +\sqrt{3})^2 &= (\sqrt{4+2\sqrt{3}})^2\\
x^2 + (2\sqrt{3})x + 3 &= 4+ 2\sqrt{3}\\
x^2 + (2\sqrt{3})x - 1 - 2\sqrt{3} &= 0
\end{align*}$
So I have shown that there is some polynomial whose solution is $x=\sqrt{4+2\sqrt{3}}-\sqrt{3}$, but I have not shown it to be 1.
|
Hint: $$4+2\sqrt{3} = 1+2\sqrt{3}+\sqrt{3}^2 = (1+\sqrt{3})^2.$$
|
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IMO 1988, problem 6 In 1988, IMO presented a problem, to prove that $k$ must be a square if $a^2+b^2=k(1+ab)$, for positive integers $a$, $b$ and $k$.
I am wondering about the solutions, not obvious from the proof. Beside the trivial solutions a or $b=0$ or 1 with $k=0$ or $1$, an obvious solution is $a=b^3$ so that the equation becomes $b^6+b^2=b^2(1+b^4)$ . Are there any other solutions?
|
Supposing that
$$ \frac{a^2 + b^2}{ab + 1} = k$$ then
$$ a^2 - a(kb) + (b^2 - k) = 0 $$
So using quadratic formula gives: $$ a = \frac{kb \pm \sqrt{k^2b^2 -4(b^2 -k)}}{2}$$
The solutions are when $k = b^2 $ and thus $a= kb = b^3$
So $$ b = 1 , a = 1 , k= 1$$
$$b = 2 , a = 8, k = 4$$
$$ b= 3, a= 27, k = 9 $$ and so on....
There are other ways of getting more solutions than these such as b = 8, a = 30 and k = 4 where k is not $b^2$ or $a = b^3$ and of course the zero one. I have not yet found a way to find more solutions than this.
|
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|
Left and Right Ideal Generated by Two Matrices. Let $R= {\rm Mat}_2(\Bbb R)$ be the ring (with $1$) of $2\times2$-matrices with entries in $\Bbb R$. Let $$M = \left\{\begin{pmatrix}1&0 \\0&0\end{pmatrix},\begin{pmatrix}0&1\\0&0\end{pmatrix}\right\},$$ i.e. a set of two matrices. What are the left and right ideal generated by $M$ in $R$?
I tried to write my answer by it won't accept it. I don't know how to write Matrices (and their product) in the right format.
Is this correct:
For left ideal: $$\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}1&0\\0&0\end{pmatrix}=\begin{pmatrix}a&0\\c&0\end{pmatrix} \quad \text{ and } \quad \begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}0&1\\0&0\end{pmatrix}=\begin{pmatrix}0&a\\0&c\end{pmatrix}$$
hence $L = \begin{pmatrix}a&a\\c&c\end{pmatrix}$.
For right ideal:
$$\begin{pmatrix}1&0\\0&0\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}a&b\\0&0\end{pmatrix} \quad \text{ and } \quad \begin{pmatrix}0&1\\0&0\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}c&d\\0&0\end{pmatrix}$$
hence $R = \begin{pmatrix}a+c&b+d\\0&0\end{pmatrix}$.
|
Sorry, there are some issues here caused by your choice of notation. By multiplying the same $a,b,c,d$ matrix with both generators, you've overlooked that there's nothing wrong with multiplying the generators with two different matrices and adding, which will produce many more possiblities.
In the left ideal generated by those two things, a general element will look like this:
$\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}1&0\\0&0\end{bmatrix}+\begin{bmatrix}e&f\\g&h\end{bmatrix}\begin{bmatrix}0&1\\0&0\end{bmatrix}=\begin{bmatrix}a&e\\c&g\end{bmatrix}$.
Notice how we didn't recycle the $a,b,c,d$ matrix for both generators. Since you can pick $a,e,c,g$ to be whatever you want, you can see that the left ideal generated by these two things is the entire matrix ring.
Try to apply similar reasoning to the right ideal generated by these two things. You will get a different answer than the left ideal, and the answer you gave is not incorrect. It's just that you can express what the right ideal looks like more simply.
|
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|
Evaluate $\lim_{x\to0} \frac{6}{x}-\frac{42}{x^2+7x}$ Evaluate $\lim_{x\to0} \frac{6}{x}-\frac{42}{x^2+7x}$
I'm I want to say that you cross multiply to get the same denominator, but I could be wrong.
Please Help!!
|
First, we find the common denominator:
$\require{cancel}$
$$\frac ab - \frac c{bd} = \frac{ad - c}{bd}$$
Applying the "general rule" in your case:
$$\begin{align}
\frac 6x - \frac{42}{x^2 + 7x} & = \frac 6x - \frac{42}{x(x + 7)} \\ \\
& = \frac{6(x + 7) - 42}{x(x + 7)} \\ \\ & = \frac{6x}{x(x+7)} \\ \\ & = \frac{6\color{blue}{\cancel{x}}}{\color{blue}{\cancel{ x}}(x + 7)} \\ \\ & = \frac{6}{x + 7}
\end{align}$$
And so we have that $$\lim_{x\to 0} \frac 6x - \frac{42}{x^2 + 7x} \quad = \quad \lim_{x\to 0} \frac 6{x+7} \quad = \quad\frac 67$$
|
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|
Proving Muirhead-like inequalities Let $T_{m,n,p}(x,y,z)=\sum_{Sym} x^m y^n z^p$.
For $x,y,z>0$, prove
$2T_{6,3,0}(x,y,z)+T_{3,3,3}(x,y,z)+3T_{4,4,1}(x,y,z)\geq 6T_{5,2,2}(x,y,z)$.
I tried to prove that by using AM-GM inequality, without success.
Is there a general way to prove these "Muirhead-like" inequalities?
|
This is a very old question, but I'll answer it anyway.
We should prove that:
$$ \sum_{sym}2x^6 y^3 + x^3 y^3 z^3 + 3 x^4 y^4 z^1 - 6 x^5 y^2 z^2 \ge 0$$
If we divide the inequality by $x^3 y^3 z^3$, we get:
$$ \sum_{sym} 2 \frac{x^3}{z^3} + 1 + 3 \frac{x y}{z^2} - 6 \frac{x^2}{y z} \ge 0$$
Apply the following subsitution:
$$\frac{x}{z}=a, \frac{z}{y}=b, \frac{y}{x}=c$$
The inequality becomes:
$$ \sum_{sym} 2 a^3 + 1 + 3 \frac{a}{b} - 6 \frac{a}{c} \ge 0$$
Note that $abc=1$ so we can homogenize the last inequality in the following way:
$$ \sum_{sym} 2 a^3 + abc + 3 \frac{a}{b} abc - 6 \frac{a}{c} abc \ge 0$$
$$ \sum_{sym} 2 a^3 + abc + 3 a^2 c - 6 a^2b \ge 0$$
If we note that:
$$ \sum_{sym}a^2 b \equiv \sum_{sym}a^2 c $$
...the inequality reduces to:
$$ \sum_{sym} 2 a^3 + abc - 3 a^2b \ge 0$$
...or, by using "Muirhead-like" notation:
$$2T_{3, 0, 0} + T_{1, 1, 1} - 3T_{2, 1, 0} \ge 0$$
...which is obvious, because:
$$T_{3, 0, 0} \ge T_{2, 1, 0} \quad (Muirhead)$$
$$T_{3, 0, 0} + T_{1, 1, 1} \ge T_{2, 1, 0} \quad (Schur)$$
|
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Expression of an Integer as a Power of 2 and an Odd Number (Chartrand Ex 5.4.2[a])
Let $n$ be a positive integer. Show that every integer $m$ with $ 1 \leq m \leq 2n $ can be expressed as $2^pk$, where $p$ is a nonnegative integer and $k$ is an odd integer with $1 \leq k < 2n$.
I wrote out some $m$ to try to conceive the proof. I observed:
$\bbox[5px,border:1px solid grey]{\text{Case 1: $m$ odd}}$ Odd numbers $\neq 2p$, thus the only choice is to put $p = 0$ and $k = m$.
$\bbox[5px,border:1px solid grey]{\text{Case 2: $m$ even and power of 2}}$ Then $p$ can be determined by division or inspection to "square with" the power of $2$. This requires $k = 1$. Is an explicit formula for $p$ necessary?
$\bbox[5px,border:1px solid grey]{\text{Case 3: $m$ even and NOT A power of 2}}$
$\begin{array}{cc}
\\
\boxed{m = 6}: 6 = 2^1 \cdot 3 \qquad \qquad & \boxed{m = 10}: 10 = 2^1 \cdot 5 \qquad \qquad & \boxed{m = 12}: 12 = 2^2 \cdot 3 \\
\boxed{m = 14}: 14 = 2 \cdot 7 \qquad \qquad & \boxed{m = 18}: 18 = 2^1 \cdot 9 \qquad \qquad & \boxed{m = 20}: 20 = 2^2 \cdot 5\\
\boxed{m = 22}: 22 = 2 \cdot 11 \qquad \qquad & \boxed{m = 24}: 24 = 2^3 \cdot 3 \qquad \qquad & \boxed{m = 26}: 26 = 2^1 \cdot 13
\end{array}$
Solution's Proof by Contradiction: $\color{#0073CF}{\Large{\mathbb{[}}}$Let $p$ be the greatest nonnegative integer
such that $2^p | m. \color{#0073CF}{\Large{\text{]}}}$ $\color{red}{\Large{\text{[}}}$ Then $m= 2^pk$ for some positive integer $k$. Necessarily $k$ is odd,
for otherwise this would contradicts the definition of $p$. $\color{red}{\Large{\text{]}}}$
$\Large{\text{1.}}$ Could someone please expound on the answer? It differs from my work above?
$\Large{\text{2.}}$ Is there a general formula/pattern for Case $3$?
I referenced 1. Source: Exercise 5.42(a), P125 of Mathematical Proofs, 2nd ed. by Chartrand et al
|
I personally found this problem easier to do if you consolidate case 2 and 3 into one case and then prove by contradiction:
Assume m is even and that m cannot be written in the form $2^p \bullet k$ where $p$ is a nonnegative integer and k is an odd integer with $1 \le k \le 2n$. Thus, $m \neq 2^p \bullet k$. We can rewrite $2^p \bullet k = 2(2^{p-1} \bullet k)$. (Note that because m is even $p \ge 1$ and so $p-1 \ge 0$). This implies that m is not even, which directly contradicts our assumption that m is even. Therefore, we must conclude that m can be written in the form $2^p \bullet k$ where $p$ is a nonnegative integer and k is an odd integer with $1 \le k \le 2n$.
|
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Series of modified Bessel functions There is a known identity to evaluate a sum of the form
$$\sum_{n\geq1} \rho^n I_n(\omega) $$
Where $\rho>0$, $\omega >0$ and $I_n$ is the modified Bessel function of the first kind.
???
Thanks.
|
$$ S = \sum_{n=1}^{\infty} x^nI_n(y) = \sum_{n=1}^{\infty} x^n \sum_{k = 0}^{\infty} \frac{\left(\frac{y}{2} \right)^{2k + n}}{k!\Gamma(n+k+1)}$$
$$ = \sum_{n=1}^{\infty} \left( \frac{xy}{2} \right)^n \sum_{k=0}^{\infty} \frac{\left( \frac{y^2}{4} \right)^k }{k!(n+k)!} = \sum_{n=1}^{\infty} \frac{\left( \frac{xy}{2} \right)^n}{n!} + \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{\left( \frac{xy}{2} \right)^n \left( \frac{y^2}{4} \right)^k}{k!(n+k)!} $$
we can interchange
$$ = e^{\frac{xy}{2}} - 1 + \sum_{k=1}^{\infty} \frac{\left( \frac{y^2}{4} \right)^k}{k!} \sum_{n=1}^{\infty} \frac{\left(\frac{xy}{2} \right)^n}{(n+k)!} $$
$$ \sum_{n=1}^{\infty} \frac{\left(\frac{xy}{2} \right)^n}{(n+k)!} = \frac{\gamma \left(k,\frac{xy}{2} \right)}{\left(\frac{xy}{2} \right)^k (k-1)!e^{-\frac{xy}{2}}} - \frac{1}{k!}$$
$$ \Rightarrow S = e^{\frac{xy}{2}} - 1 + e^{\frac{xy}{2}}\sum_{k=1}^{\infty} \frac{k\gamma \left(k,\frac{xy}{2} \right)\left(\frac{y}{2x} \right)^k}{k!^2} - \sum_{k=1}^{\infty} \frac{\left(\frac{y^2}{4} \right)^k}{k!^2} $$
if this is right i think it became kinda easier
please edit if you saw anything wrong
|
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find this limit $\lim_{x\to0^{+}}\frac{\tan{(\tan{x})}-\tan{(\sin{x})}}{\tan{x}-\sin{x}}$ find the limit.
$$\lim_{x\to0^{+}}\dfrac{\tan{(\tan{x})}-\tan{(\sin{x})}}{\tan{x}-\sin{x}}$$
my try:
$$\tan{x}=x+\dfrac{1}{3}x^3+o(x^3),\sin{x}=x-\dfrac{1}{6}x^3+o(x^3)$$
so
$$\tan{(\tan{x})}=\tan{x}+\dfrac{1}{3}(\tan{x})^3+o(\tan^3{x})=x+\dfrac{1}{3}x^3+\dfrac{1}{3}(x+\dfrac{1}{3}x^3)^3+o(x^3)$$
$$\tan{(\sin{x})}=\sin{x}+\dfrac{1}{3}(\sin{x})^3+o(\sin^3{x})=x-\dfrac{1}{6}x^3+\dfrac{1}{3}(x-\dfrac{1}{6}x^3)^3+o(x^3)$$
so
$$\tan{(\tan{x})}-\tan{(\sin{x})}=\dfrac{1}{2}x^3+o(x^3)$$
$$\tan{x}-\sin{x}=\dfrac{1}{2}x^3+o(x^3)$$
Have other metods? Thank you
|
I think this identity is useful here:
$$\tan(a-b)=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}$$ and the fact that $\tan(a)\sim a$ while $a\sim 0$.
|
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|
Prove $3^n \ge n^3$ by induction Yep, prove $3^n \ge n^3$, $n \in \mathbb{N}$.
I can do this myself, but can't figure out any kind of "beautiful" way to do it.
The way I do it is:
Assume $3^n \ge n^3$
Now,
$(n+1)^3 = n^3 + 3n^2 + 3n + 1$,
and $\forall{} n \ge 3$,
$3n^2 \le n^3, \,\, 3n + 1 \le n^3$
Which finally gives $(n+1)^3 \le 3n^3 \le 3^{n+1}$ by our assumption.
Now just test by hand for n=1,2,3 and the rest follows by induction.
Anyone got anything simpler?
|
Basis:
$n = 1$
$$3^1 \ge 1^3 \implies 3 \ge 1\text{, which is true}$$
Inductive hypothesis
Let $n=k$ and also let this inequality hold:
$$3^k \ge k^3$$
Inductive step
We'll prove that it also holds when $n=k+1$
$$3^{k+1} \ge (k+1)^3$$
$$3^k \cdot 3 \ge k^3 + 3k^2 + 3k + 1$$
$$3^k + 3^k + 3^k \ge k^3 + 3k^2 + 3k + 1$$
Now obviously as $3^k \ge k^3$ by the inductive hypothesis it's enough to show that
$$2 \cdot 3^k \ge 3k^2 + 3k + 1$$
Note that for every $k \ge 2$, this inequality holds $3k^2 \ge 3k + 1$
$$2 \cdot 3^k \ge 6k^2$$
$$3^k \ge 3k^2$$
This is true for every $k \ge 3$
Now you can prove the other 2 cases $n = 1,2$ to give a complete proof.
|
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How to evaluate the limit $\lim\limits_{x \to 1} \left(\frac{2}{1-x^2} - \frac{3}{1-x^3}\right) $, and others? $$ \lim_{x \to 1} \left( \frac{2}{1-x^2} - \frac{3}{1-x^3} \right)$$
In my opinion the function is not defined at $ x = 1 $ but somehow when I look at the graph, it's continuous and there is no break. I learned to look for points where my function is not defined due to division by zero.
So my example here is: I should look for the limit as $x \rightarrow 1$ but I don't know how to do this. I'm not allowed to use L'Hospital. I know how to look for limits if my variable ($n$ or $x$) "runs" to infinity. I know that $\frac{1}{n}$ when $n \to \infty$ is $0$. I just "know" that. And if $n$ would "go" or "run" or "tend" (what is the right way to call it) to $1$, the limit would be $1$ just by "imagining the $n$ as $1$". Or is that the wrong way? I'm still learning, so please let me know the right way to do it.
Back to the example. Is there an approach of calculating the limit? In my book they suggest to replace $x$ with a sequence $ x_n$ and $ x_n \rightarrow 1$ and then let $ n \rightarrow \infty $
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Consider the expression
$$\lim_{x \to 1}
\left(
\frac1{1-x}
-
\frac1{1-x}
\right)
$$
This is "obviously" zero,
even though both terms blow up
as $x \to 1$,
because the singularities cancel.
The same thing happens here.
The basic identity needed here is
$1-x^n
=(1-x)(1+x+x^2+...+x^{n-1})
$.
Using this,
$\begin{align}
\frac{2}{1-x^2} - \frac{3}{1-x^3}
&=\frac{2}{(1-x)(1+x)} - \frac{3}{(1-x)(1+x+x^2)}\\
&=\frac1{1-x}\left(\frac{2}{1+x} - \frac{3}{1+x+x^2}\right)\\
&=\frac1{1-x}\left(\frac{2(1+x+x^2)-3(1+x)}{(1+x)(1+x+x^2)}\right)\\
&=\frac1{1-x}\left(\frac{-1-x+2x^2}{(1+x)(1+x+x^2)}\right)\\
&=\frac1{1-x}\left(\frac{(2x+1)(x-1)}{(1+x)(1+x+x^2)}\right)\\
&=\frac{(2x+1)(x-1)}{(1-x)(1+x)(1+x+x^2)}\\
&=\frac{-(2x+1)}{(1+x)(1+x+x^2)} \quad \text{ (except where } x=1)\\
\end{align}
$
This final expression has no problem
being evaluated as $x \to 1$.
Its value is
$\frac{-3}{2\cdot 3}
=-\frac12
$.
Another way to do this
is to let
$x = 1+y$,
so $y \to 0$
is the same as $x \to 1$.
I find it easier to see a limit
with a variable going to $0$,
so I make this kind of transformation
whenever possible.
Since
$1-x^2
= 1-(1-y)^2
=1-(1-2y+y^2)
=2y-y^2
$
and
$1-x^3
= 1-(1-y)^3
=1-(1-3y+3y^2-y^3)
=3y-3y^2+y^3
$,
$\begin{align}
\frac{2}{1-x^2} - \frac{3}{1-x^3}
&=\frac{2}{2y-y^2}- \frac{3}{3y-3y^2+y^3}\\
&=\frac1{y}\left(\frac{2}{2-y}- \frac{3}{3-3y+y^2}\right)
\quad \text{ (factoring }y\text{ from each denominator)}\\
&=\frac1{y}\frac{2(3-3y+y^2)-3(2-y)}{(2-y)(3-3y+y^2)}\\
&=\frac1{y}\frac{-3y+2y^2}{(2-y)(3-3y+y^2)}\\
&=\frac{-3+2y}{(2-y)(3-3y+y^2)}\\
\end{align}
$
Again,
this has no problem as $y \to 0$,
and gives $-\frac12$ as before
(as it should).
Note that,
by doing it as $y \to 0$,
we do not need to know
the factorizations
of $1-x^2$ and $1-x^3$.
The fact that $1-x$ divides these
becomes the more obvious fact
that $y$ divides the expressions
for $1-(1-y)^2$
and $1-(1-y)^3$.
|
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|
Compute $\int_0^n \left[\frac x {x+1} + \frac x {2x+4} + \frac x {3x+9} + \cdots\right] \, dx$ for $x>0$ I want to compute $\displaystyle \int_0^n \left[\frac x {x+1} + \frac x {2x+4} + \frac x {3x+9} + \cdots \right] \, dx$ for $x>0$
My attempt: The integrand can be written as a sum of $\displaystyle f_k(x)=\frac x{kx+k^2}$ which is positive for $x>0$, can we interchange sum and integral here? If so, then $\displaystyle \int_0^n \frac x{kx+k^2}=1-\log(2)$, so the given integral diverges?
|
We have
\begin{align}
f(x) & = \sum_{k=1}^{\infty} \dfrac{x}{kx+k^2} = \sum_{k=1}^{\infty} \left(\dfrac1k - \dfrac1{k+x} \right) = \sum_{k=1}^{\infty} \int_0^1 \left(t^{k-1} - t^{k+x-1}\right)dt\\
& = \int_0^1 (1-t^x) \left(\sum_{k=1}^{\infty} t^{k-1} \right) dt = \int_0^1 \dfrac{1-t^x}{1-t} dt
\end{align}
We now have
$$I_n = \int_0^n f(x) dx = \int_0^n \int_0^1 \dfrac{1-t^x}{1-t} dt dx = \int_0^1 \dfrac{n + \dfrac{1-t^n}{\log(t)}}{1-t}dt$$
From WolframAlpha for $k=1,2,3,4,5$, we get the value of
$$J_k = \int_0^1 \left(\dfrac1{1-t} + \dfrac{t^{k-1}}{\log(t)} \right) dt = \gamma + \log(k)$$
Assuming this is true in general and noting that $I_n = \displaystyle \sum_{k=1}^n J_k$, we get that,
$$I_n = \sum_{k=1}^n J_k = n \gamma + \log(n!)$$
|
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|
How prove two segments are congruent If $\angle ACB=2\angle ABC=4\angle BAD$,and $D$ is on $BC$, such $CG\parallel AB$, $CG=CD$, $HG \parallel AC$.
show that
$$BD=HG$$
My idea: let $\angle BAD=x$, then $\angle ABC=2x$, $\angle ACB=4x$.
Lemma 1：in $\Delta ABC,AB=c,AC=b,BC=a$ if $\angle A=2\angle B$,then $a^2=b(b+c)$
So use this lemma:
in $\Delta ABC$, we have
$$AB^2=AC(AC+BC)$$
and in $\Delta ABD$, we have
$$AD^2=BD(BD+AB)$$
Then I can't prove $$BD=HG.$$
Can someone help me?
|
The statement can be proved by applying the sine theorem four times.
If $\widehat{BAD}=\theta,\widehat{ABC}=2\theta,\widehat{ACB}=4\theta$, then $$\widehat{DAC}=\pi-7\theta,\widehat{DCG}=2\theta,\widehat{ADC}=3\theta,\widehat{DHG}=7\theta,\widehat{CDG}=\widehat{CGD}=\frac{\pi}{2}-\theta,\\ \widehat{HDG}=\frac{\pi}{2}-2\theta.$$
Now we consider the triangles $ABD,ADC,CDG,GDH$, in order to have:
$$ BD = \frac{\sin\theta}{\sin 2\theta}AD = \frac{\sin\theta \sin 4\theta}{\sin 2\theta \sin 7\theta}CD = \frac{\sin\theta \sin 4\theta \cos \theta}{\sin 2\theta \sin 7\theta \sin 2\theta}DG = \frac{\sin\theta \sin 4\theta \cos \theta \sin 7\theta}{\sin 2\theta \sin 7\theta \sin 2\theta \cos2\theta}HG,$$
that implies $BD=HG$ since $\sin 4\theta = 2\sin 2\theta\cos 2\theta$ and $\sin 2\theta = 2 \sin\theta \cos\theta$.
|
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|
Different Ways of Integrating $3\sin x\cos x$ I am asking this question for my son who is in (equivalent) twelfth
grade and I failed to answer his query.
When he tries to integrate $3\sin x\cos x$, he finds that
this can be done in at least following three ways.
And these three ways do not produce equivalent results.
ONE
Let us assume, $\sin x = z$.
This gives,
\begin{align*}
\cos x &= \frac{dz}{dx}\\
\cos x dx &= dz
\end{align*}
So, we can write,
\begin{align*}
\int 3\sin x\cos x dx &=3 \int zdz\\
&=3 \frac{z^2}{2}\\
&=\frac{3}{2} \sin^2 x\\
&=\frac{3}{4}\times 2\sin^2 x\\
&=\frac{3}{4} (1 -\cos 2x)\\
\end{align*}
TWO
Let us assume, $\cos x = z$.
This gives,
\begin{align*}
-\sin x &= \frac{dz}{dx}\\
\sin x dx &= -dz
\end{align*}
So, we can write,
\begin{align*}
\int 3\sin x\cos x dx &=-3 \int zdz\\
&=-3 \frac{z^2}{2}\\
&=-\frac{3}{2} \cos^2 x\\
&=-\frac{3}{4}\times 2\cos^2 x\\
&=-\frac{3}{4} (1 +\cos 2x)\\
\end{align*}
THREE
\begin{align*}
\int 3\sin x\cos x dx &=\frac{3}{2}\int 2\sin x\cos x dx\\
&=\frac{3}{2}\int \sin 2x dx\\
&=-\frac{3}{2}\times\frac{1}{2} \cos 2x\\
&=-\frac{3}{4} \cos 2x\\
\end{align*}
The results found in above three methods are not the same.
If we try a simple approach of evaluating the integration results at, $x = \frac{\pi}{6}$, we get as follows.
From the first one,
$\frac{3}{4} (1 -\cos 2x) = \frac{3}{4} (1 -\cos \frac{2\pi}{6})
= \frac{3}{4} (1 -\cos \frac{\pi}{3})
= \frac{3}{4} (1 - \frac{1}{2})
= \frac{3}{4}\times\frac{1}{2}
= \frac{3}{8}$
From the second one,
$-\frac{3}{4} (1 +\cos 2x) = -\frac{3}{4} (1 +\cos \frac{2\pi}{6})
= -\frac{3}{4} (1 +\cos \frac{\pi}{3})
= -\frac{3}{4} (1 + \frac{1}{2})
= -\frac{3}{4}\times\frac{3}{2}
= -\frac{9}{8}$
From the third one,
$-\frac{3}{4} \cos 2x=-\frac{3}{4} \cos \frac{2\pi}{6}
= -\frac{3}{4} \cos \frac{\pi}{3}
= -\frac{3}{4} \times \frac{1}{2}
= -\frac{3}{8} $
Clearly, we are getting some nonequivalent results. We have failed to find
the mistakes or explanations behind this. Your help will be appreciated.
|
All three answers are correct provided you add a constant to each one of those.
Because from the very definition of integration, it is the area under the curve, so it requires bounds to give a unique value.
You can't evaluate the value of an indefinite integral without including constant.
And I am sure that in the examination, your son won't be asked to evaluate the value of an integral without providing limits of integration or providing its value at some other point.
For instance, in question it may be mentioned that evaluate the value of expression at x=π/6 , given its value at x=0 is 1.
So in this case, all three answers will give the correct value i.e. 11/8
|
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|
Prove limit of $\frac1{n+1}+\frac1{n+2}+\ldots+\frac1{n+n}$ exists and lies between $0$ and $1$. Prove limit of $\displaystyle \frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{n+n}$ exists and lies between $0$ and $1$.
So far I have $\displaystyle \lim_{n\rightarrow\infty}\sum_{k=1}^n\frac{1}{k+n}=L$ for some $L>0$.
Then given any $\epsilon >0,\exists N>0$ such that if $n>N$, then $\displaystyle\left|\sum_{k=1}^n\frac{1}{k+n}-L\right|<\epsilon$
A hint would be appreciated!
|
It's a monotone rising sequence: if $$s_n = \frac{1}{n+1} + ... + \frac{1}{2n},$$ then we have $$s_{n+1} - s_n = \frac{1}{2n+2} + \frac{1}{2n+1} - \frac{1}{n+1} = \frac{1}{2n+1} - \frac{1}{2n+2} > 0.$$
It is bounded below by $0$; and it's bounded above by $1$, because $$s_n = \frac{1}{n+1} + ... + \frac{1}{2n} < \frac{1}{n} + ... + \frac{1}{n} = 1.$$ So it converges to a limit between $0$ and $1$.
|
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|
Mathematics max and min $x$ and $y$ values of an ellipse I am trying to find the max and min x and y values of an ellipse. The ellipse is given by $$x^2-xy+y^2=3$$
So I need to find the max points, so I need the derivative of the function.
I got $$\dfrac{dy}{dx} = \dfrac{(y-2x)}{(2y-x)}$$
So I then thought that I should get the x value and the y value so I got $x=-(y/2)$ and $y=-2x$
But now I'm not sure what to do?
|
(originally posted on gamedev)
The general equation of an ellipse centered at the origin is:
$$(ax+by)^2+(cx+dy)^2=r^2$$
Expand w.r.t $x$ and $y$:
$$(a^2+c^2)x^2+(b^2+d^2)y^2+2(ab+cd)xy=r^2$$
An horizontal line has equation $y=k$. Either it intersects with the ellipse in two points, or it intersects in only one point (it's tangent), or it doesn't intersect at all. We find the intersection by solving the trinomial (in x):
$$(a^2+c^2)x^2+2(ab+cd)kx+(b^2+d^2)k^2-r^2=0$$
There is only one intersection when the discriminant is zero, hence you want to find $k$ such that
$$4(ab+cd)^2k^2-4(a^2+c^2)(b^2+d^2)k^2+4(a^2+c^2)r^2=0$$
Expand and simplify:
$$\left(a^2b^2+c^2d^2+2abcd-a^2b^2-a^2d^2-b^2c^2-c^2d^2\right)k^2+(a^2+c^2)r^2=0$$
$$(2abcd-a^2d^2-b^2c^2)k^2+(a^2+c^2)r^2=0$$
$$(ad-bc)^2k^2=(a^2+c^2)r^2$$
$$k=\pm\frac{r\sqrt{a^2+c^2}}{|ad-bc|}$$
The positive $k$ is the maximum value of $y$, the negative the minimum value. By symmetry of the ellipse, they are opposite.
The extreme values of $x$ can be found the same way, using a vertical line $x=l$. The same computation will lead to
$$l=\pm\frac{r\sqrt{b^2+d^2}}{|ad-bc|}$$
The positive $l$ is the maximum value of $x$, the negative the minimum value. Again, by symmetry of the ellipse, they are opposite.
The expressions found are not defined when $ad=bc$, but then the two squares in the original equation are proportional and the conic is degenerate (the equation leads to two parallel lines).
|
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|
Improving bound on $\sqrt{2 \sqrt{3 \sqrt{4 \ldots}}}$ An old challenge problem I saw asked to prove that $\sqrt{2 \sqrt{3 \sqrt{4 \ldots}}} < 3$. A simple calculation shows the actual value seems to be around $2.8$, which is pretty close to $3$ but leaves a gap. Can someone find $C < 3$ and prove $\sqrt{2 \sqrt{3 \sqrt{4 \ldots}}} < C$?
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Put $K=\log C$. The problem is then to found a tight upper bound for:
$$ K = \sum_{n=1}^{+\infty}\frac{\log(1+n)}{2^n}. $$
Note that:
$$ K/2=K-K/2 = \sum_{n=1}^{+\infty}\frac{\log\left(1+\frac{1}{n}\right)}{2^n}. $$
By writing $K/4$ as $K/2-K/4$ just like above, we get:
$$\frac{K}{4}=\frac{\log 2}{2}-\sum_{n=1}^{+\infty}\frac{\log\left(1+\frac{1}{n(n+2)}\right)}{2^{n+1}},$$
equivalent to:
$$(\heartsuit)\quad K = \log 3 -\sum_{n=1}^{+\infty}\frac{\log\left(1+\frac{1}{(n+1)(n+3)}\right)}{2^n} = \log 3 +\sum_{n=1}^{+\infty}\frac{\log\left(1-\frac{1}{(n+2)^2}\right)}{2^n}.$$
Now using the Bernoulli inequality we have:
$$\log 3-\sum_{n=1}^{+\infty}\frac{1}{(n+1)(n+3)2^n}\leq K \leq \log 3-\sum_{n=1}^{+\infty}\frac{1}{(n+2)^2 2^n}, $$
that gives:
$$ 1 < 3\log 2 + \log 3 -\frac{13}{6} \leq K \leq \log 3+\frac{1}{12}\left(27-4\pi^2+24\log^2 2\right), $$
so $K$ is between $1.0113\ldots$ and $1.0196$, and $C$ is between $2.7494\ldots$ and $2.7722\ldots$.
The next step of the acceleration process leads us to:
$$ K = 3\log 2-\log 3+\sum_{n=1}^{+\infty}\frac{\log\left(1+\frac{5+2n}{(n+1)(n+3)^3}\right)}{2^n},$$
that converges faster but is way less appealing than $(\heartsuit)$.
|
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|
When is $2^n \pm 1$ a perfect power Is there an easy way of showing that $2^n \pm 1$ is never a perfect power, except for $2^3 + 1 = 3^2 $?
I know that Catalan's conjecture (or Mihăilescu's theorem) gives the result directly, but I'm hopefully for a more elementary method.
I can show that it is never a square, except for $2^3 + 1$.
Proof: Cases $n=1, 2, 3$ are easily dealt with. Henceforth, let $n\geq4$.
$2^n -1 \equiv 3 \pmod{4}$ hence is never a square.
If $2^n +1 =x^2$, then $2^n = (x-1)(x+1)$ and both of these are powers of 2. Thus we must have $(x-1) = 2, (x+1) = 4$. This gives the solution of $2^3 + 1 = 3^2$.
Let's now do an odd prime.
Say $2^n +1 = x^p$. Then $2^n = x^p - 1 = (x-1)(x^{p-1}+x^{p-2} + \ldots +1)$, so both terms are powers of 2. We have $ x = 2^m+1$ is odd. But the other term is the sum of $p$ odd numbers, hence is odd. Since this is clearly greater than 1, we get a contradiction.
Say $2^n -1 = x^p$. Then $2^n = x^p + 1 = (x+1)(x^{p-1} - x^{p-2} + \ldots -x +1 )$, so both terms are powers of 2. We have $x = 2^m -1$ is odd. But the other term is the sum of $p$ odd numbers, hence is odd. Since $x^p + 1 \neq x+1$ except for $p=1$, this means that the term is greater than 1. Hence we get a contradiction.
|
(This proof was completed with an insightful comment from Gottfried. I'm placing it as an answer so that it is readily seen that an answer exists, as opposed to just leaving it in the question.)
Suppose we have $ 2^n \pm 1 = x^k$ for some positive integers $x, k$. Cases $n=1, 2, 3$ are easily dealt with. $n=3$ yields the solution $2^3 + 1 = 3^2$. Henceforth, let $n\geq4$. Furthermore, a simple check shows that $x\neq 1, 2$.
First, we will deal with the power $k=2l$ being even. Rewriting $x^{2l}$ as $(x^l)^2$, we may assume that $k=2$.
Proof: $2^n -1 \equiv 3 \pmod{4}$ hence is never a square.
If $2^n +1 =x^2$, then $2^n = (x-1)(x+1)$ and both of these are powers of 2 that differ by 2. Thus we must have $(x-1) = 2, (x+1) = 4$, which gives $2^n = 8$ so $n=3$ (reject). $_\square$
Now, we deal with $k$ odd.
Proof: Say $2^n +1 = x^k$. Then $2^n = x^k - 1 = (x-1)(x^{k-1}+x^{k-2} + \ldots +1)$, so both terms are powers of 2. We have $ x -1$ is a power of 2 greater than 1, hence is even. Thus $x$ is odd. But the other term is the sum of $k$ odd numbers, hence is odd. Since this is clearly greater than 1, we get a contradiction.
Say $2^n -1 = x^k$. Then $2^n = x^k + 1 = (x+1)(x^{k-1} - x^{k-2} + \ldots -x +1 )$, so both terms are powers of 2. We have $x+1$ is a power of 2 that is greater than 1, hence is even. Thus $x$ is odd. But the other term is the sum of $p$ odd numbers, hence is odd. Since $x^p + 1 \neq x+1$ except for $p=1$, this means that the term is greater than 1. Hence we get a contradiction. $ _\square$
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Calculus limit homework problem $$ \lim_{n→\infty} \frac1 n \left(\left(a + \frac 1 n\right)^2 + \left(a + \frac 2 n\right)^2 + ... + \left(a + \frac{n-1}{n}\right)^2\right)$$
$$ \text{hint: }\ 1^2 + 2^2 + ... + n^2 = \frac{n(n+1)(2n+1)}{6} $$
I can't figure out how to find this problem's limit. Does anybody have ideas?
|
Ignore the limit for now. We will find the expression first, and then evaluate the limit.
$$\frac{1}{n}\left((a^2+2a/n+1/n^2)+(a^2+4a/n+4/n^2) + \ldots + (a^2 + (k) 2a/n + k^2/n^2) + \ldots + (a^2+(n-1)2a/n) + (n-1)^2/n^2)\right).$$
This simplifies to $$\frac{1}{n}((n-1)a^2 + 2a/n (1 + 2 + \ldots + n - 1) + 1/n^2 (1^2 + 2^2 + \ldots + (n-1)^2).$$
Use $$1 + 2 + \ldots + n - 1 = n(n - 1)/2$$ and the other identity given to you, and you should be set.
|
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Prove $\sqrt{s_n+1} = \frac{1}{2}(1+\sqrt{5})$ This is to prove how the limit of $s_n$ converges to $\frac{1}{2}(1+\sqrt{5})$.
Assume: $s_1 = 1$; for $n \geq 1$, $s_{n+1} = \sqrt{s_n + 1}$.
How to prove this converges to $\frac{1}{2}(1+\sqrt{5})$?
|
Argue by induction:
*
*The terms of the sequence are all positive.
*They are all bounded above (by $3$): The base case is clear, and if $s_n<3$, then $s_{n+1}=\sqrt{s_n+1}<\sqrt{3+1}=2<3$.
*The sequence is increasing: $s_2=\sqrt2>1=s_1$, and if $s_n<s_{n+1}$, then $s_n+1<s_{n+1}+1$, so $s_{n+1}=\sqrt{s_n+1}<\sqrt{s_{n+1}+1}=s_{n+2}$.
It follows that the sequence converges (to its supremum). Call its limit $L$, so $L=\lim_n s_{n+1}=\lim_n\sqrt{s_n+1}=\sqrt{L+1}$ (by continuity of $f(x)=\sqrt{x+1}$).
Solving the equation $L=\sqrt{L+1}$ gives us that $L^2-L-1=0$, so $L=\frac12(1\pm\sqrt5)$, and the sign must be $+$ rather than $-$ since the terms of the sequence are all positive, so also their limit $L$ is non-negative, $L\ge0$.
|
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|
Taylor Series Expansion for $\sin^2(\omega t)$ What are the first few terms for the Taylor Series Expansion for $\sin^2(\omega t)$? $(\omega$=$2\pi f$)
If you could show some working, that would be helpful
|
Sometimes a little sledgehammering is good practice. I prefer Andrés solution, which is elegant, but this can also be a good exam question to test various Taylor expansion skills by insisting that no short cuts be used.
Applying basic Taylor expansion
Consider $f(x) = \sin^2 x$:
\begin{align*}
f'(x) &= 2\sin x\cos x = \sin2x\\
f''(x) &= 2\cos^2 x - 2\sin^2 x = 2\cos2x\\
f'''(x) &= -4\cos x\sin x - 4\sin x\cos x = -8 \sin x\cos x = -4\sin2x\\
f^{iv}(x) &= -8\cos^2x + 8\sin^2x = -8\cos2x\\
\cdots &= \cdots
\end{align*}
There's a definite pattern emerging here, but in fact after $f'(x) = \sin2x$, the long way isn't really necessary. A shortened version is:
\begin{align*}
f(x) &= \sin^2x \\
f'(x) &= 2\sin x\cos x \\
f'(x) &= \sin2x \\
f''(x) &= 2\cos2x \\
f'''(x) &= -4\sin2x \\
f^{iv}(x) &= -8\cos2x \\
\cdots &= \cdots
\end{align*}
Now expand around $0$ for the M$^c$Laurin Series
\begin{align*}
f(0) &= 0 \\
f'(0) &= 0 \\
f''(0) &= 2 \\
f'''(0) &= 0 \\
f^{iv}(0) &= -8\\
f^{v}(0) &= 0 \\
f^{vi}(0) &= 32 \\
\cdots &= \cdots
\end{align*}
$$\sin^2 x = 2x^2/2! - 8x^4/4! + 32x^6/6! - \cdots = \sum_{n=1}^\infty \frac {(-1)^{n+1}2^{2n-1}x^{2n}}{(2n)!}$$
|
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|
How to find the type of triangle when given the ratio of it's sides?
Q.The sides of a triangle are in ratio 4 : 6 : 7, then the triangle is:
(A) acute angled
(B) obtuse angled
(C) right angled
(D) impossible
It's definitely not (C) right-angled since $7^2 ≠ 6^2+4^2$
Is it possible to use trigonometry here even though the triangle is not right-angled?
|
Let $ABC$ be a triangle with sides $a,b,c$.
If $c^2=a^2+b^2$, then the angle at $C$ is a right angle.
If $c^2\lt a^2+b^2$, then the angle at $C$ is acute.
If $c^2\gt a^2+b^2$, then the angle at $C$ is obtuse.
We can think of these facts as coming from the Cosine Law
$$c^2=a^2+b^2-2ab\cos(\angle C).$$
Remark: But we don't need trigonometry to see the answer. Take two sticks of length $a$ and $b$, and join them by a hinge. When the hinge puts the two sticks at right angles, the square of the distance between their two ends is $a^2+b^2$.
If we spread the ends further apart, then the angle at the hinge becomes greater than $90^\circ$, and in that case if $c$ is the distance between the two ends, we have $c^2\gt a^2+b^2$.
If on the other hand we bring the two ends closer together, the angle at the hinge becomes less than $90^\circ$, and $c^2\lt a^2+b^2$.
|
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|
minimal polynomial of a matrix with some unknown entries Question is to prove that :
characteristic and minimal polynomial of $ \left( \begin{array}{cccc}
0 & 0 & c \\
1 & 0 & b \\
0 & 1 & a \end{array} \right) $ is $x^3-ax^2-bx-c$.
what i have done so far is :
characteristic polynomial of a matrix $A$ is given by $\det(A-xI)$
in case of $A= \left( \begin{array}{cccc}
0 & 0 & c \\
1 & 0 & b \\
0 & 1 & a \end{array} \right)$ we have $\det(A-xI)=\det\left( \begin{array}{cccc}
-x & 0 & c \\
1 & -x & b \\
0 & 1 & a-x \end{array} \right)=-(x^3-ax^2-bx-c)$
So, i have got the characteristic polynomial as $x^3-ax^2-bx-c$.
Now, the problem is how do i find minimal polynomial.
As $a,b,c$ are arbitrary, I can not factorize $x^3-ax^2-bx-c$ so as to see which factor gives me minimal polynomial.
I am confused.
please suggest me some hint.
EDIT : This is just after Mr.Will Jagyy's hint :
I have $A= \left( \begin{array}{cccc}
0 & 0 & c \\
1 & 0 & b \\
0 & 1 & a \end{array} \right)$ then, $A^2= \left( \begin{array}{cccc}
0 & 0 & c \\
1 & 0 & b \\
0 & 1 & a \end{array} \right)\left( \begin{array}{cccc}
0 & 0 & c \\
1 & 0 & b \\
0 & 1 & a \end{array} \right)=\left( \begin{array}{cccc}
0 & c & ac \\
0 & b & c+ab \\
1 & a & b+a^2 \end{array} \right)$
Now,
$A^2+rA+sI=\left( \begin{array}{cccc}
0 & c & ac \\
0 & b & c+ab \\
1 & a & b+a^2 \end{array} \right)+r\left( \begin{array}{cccc}
0 & 0 & c \\
1 & 0 & b \\
0 & 1 & a \end{array} \right)+s\left( \begin{array}{cccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \end{array} \right)=\left( \begin{array}{cccc}
s & c & * \\
r & b+s & * \\
1& * & * \end{array} \right)$
As element of $3^{rd}$ row $1^{st}$ column is $1$ in above matrix, this can never be $0$
i.e., $A^2+rA+sI$ can never be $0$.
Now, $A+rI=\left( \begin{array}{cccc}
0 & 0 & c \\
1 & 0 & b \\
0 & 1 & a \end{array} \right)+r\left( \begin{array}{cccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \end{array} \right)=\left( \begin{array}{cccc}
r & *& * \\
\color{magenta}{1} & * & * \\
* & \color{magenta}{1} & * \end{array} \right)\neq 0$ if $r\neq 0$
Thus, $A^2+rA+sI\neq 0$ and $A+rI\neq 0$ for any $r,s$.
Thus, minimal polynomial for $A$ can not be of order less than $3$.
Thus, minimal polynomial for $A$ has to be $x^3-ax^2-bx-c$.
I have written this just to make sure i have tried in correct way as i can not write this in a comment.
I would be thankful if there is any other way to proceed further.. Thank you :)
|
It is very easy to see what the minimal polynomial is for this matrix, even without knowing let alone factoring the characteristic polynomial. The argument even works for matrices of this type (called companion matrices) of arbitrary size $n$. Since $A\cdot e_i=e_{i+1}$ for $0<i<n$, where $e_i$ is the standard basis vector, it is clear that $(c_0I+c_1A+c_2A^2+\cdots+c_{n-1}A^{n-1})\cdot e_1=c_0e_1+c_1e_2+\cdots+c_{n-1}e_n$, which is obviously nonzero unless all the coefficients $c_i$ are zero. Therefore no nonzero polynomial of degree${}<n$ in$~A$ has $e_1$ in its kernel, and a fortiori no such polynomial can be the zero matrix. Therefore the minimal polynomial of$~A$ must have degree at least$~n$, so that it must coincide with the characteristic polynomial by Cayley-Hamilton.
You computed the characteristic polynomial in your example, but it would be a bit harder (though doable) for larger companion matrices. For the minimal polynomial it is easy by reasoning similarly to above. If the entries down the last column are $a_1,a_2,\ldots,a_n$, then $A\cdot e_n=a_1e_1+a_2e_2+\cdots+a_ne_n$. Now the minimal polynomial of$~A$ must be of the form $\mu=c_0+c_1X+\cdots+c_{n-1}X^{n-1}+X^n$, and computing as above one has $\mu[A]\cdot e_1=c_0e_1+c_1e_2+\cdots+c_{n-1}e_n+A^n\cdot e_1$, which must be zero; given that $A^n\cdot e_1=A\cdot e_n$ has the value just computed, it must be that $c_i+a_i=0$ for all $i$, so $c_i=-a_i$ fo all $i$ and hence $\mu=-a_0-a_1X+\cdots-a_{n-1}X^{n-1}+X^n$. You can conclude (using Cayley-Hamilton) that this is also the characteristic polynomial of$~A$. You can also prove this by direct calculation, either by induction on$~n$ and development of the determinant by the first row, or (without induction) by development of the determinant by the final column.
|
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|
If $ k = \sqrt{n\cdot (n+1)\cdot(n+2)\cdot (n+3)}$. Then $\lfloor k \rfloor =$ If $ k = \sqrt{n\cdot (n+1)\cdot(n+2)\cdot (n+3)}$, where $n\in \mathbb{N}$. Then $\lfloor k \rfloor = $
$\underline{\bf{My\; Try}}::$ We can write the expression $n\cdot (n+1)\cdot(n+2)\cdot (n+3) = (n^2+3n).(n^2+3n+2)$
$ = (n^2+3n)^2+2\cdot (n^2+3n)+1-1 = (n^2+3n+1)^2-1<(n^2+3n)^2$
Now I Did not Understand How can i solve further
Help Required.
Thanks
|
We want to determine
$\lfloor v \rfloor$,
where
$v = \sqrt{n\cdot (n+1)\cdot(n+2)\cdot (n+3)}$.
Let $x = n+3/2$,
so
$x^2 = n^2+3n+9/4$.
Then
$\begin{align}
v^2
&=(x-3/2)(x-1/2)(x+1/2)(x+3/2)\\
&=(x^2-(3/2)^2)(x^2-(1/2)^2)\\
&=x^4-5x^2/2+9/16\\
&=(x^2-5/4)^2-25/16+9/16\\
&=((n^2+3n+9/4)-5/4)^2-1\\
&=(n^2+3n+1)^2-1\\
\end{align}
$
Therefore
$v < n^2+3n+1$
and
$v > n^2+3n$
(since
$u^2-1 > (u-1)^2$
for $u > 1$)
so
$\lfloor v \rfloor = n^2+3n$.
|
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|
How find the maximum of $\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{z}{1+z^2}$ let $x,y,z$ are postive numbers,and such
$$xy+zx+yz=1$$
find the maxum of
$$\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}+\dfrac{z}{1+z^2}$$
my try
note
$$x^2+1=x^2+xy+yz+xz=(x+y)(x+z)$$
$$y^2+1=(y+x)(y+z)$$
$$z^2+1=(z+x)(z+y)$$
$$\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}+\dfrac{z}{1+z^2}=\dfrac{y+z+x+z+z(x+y)}{(x+y)(y+z)(x+z)}$$
I think we must use
$$8(a+b+c)(ab+bc+ac)\le 9(a+b)(b+c)(a+c)$$
$$LHS\le \dfrac{9}{8}\dfrac{x+y+z+z(x+y+1)}{x+y+z}=\dfrac{9}{8}(1+\dfrac{z(x+y+1)}{x+y+z})$$
then follow I can't works,Thank you
|
Trigonometric substitution seems easier here. We may set $x = \tan\frac{A}{2}, y = \tan\frac{B}{2}, z = \tan\frac{C}{2}$, where $A, B, C$ are angles of a triangle, so that the constraint is satisfied. Then we have to find the maximum of
$$f = \dfrac{1}{1+\tan^2\frac{A}{2}}+\dfrac{1}{1+\tan^2\frac{B}{2}}+\dfrac{\tan\frac{C}{2}}{1+\tan^2\frac{C}{2}} = \cos^2\dfrac{A}{2} + \cos^2\dfrac{B}{2}+ \dfrac{\sin C}{2}$$
Using $\cos \theta = 2 \cos^2 \frac{\theta}{2} - 1$,
$$f = \dfrac{\cos A + 1}{2} + \dfrac{\cos B + 1}{2}+ \dfrac{\sin C }{2} = 1+\dfrac{\cos A + \cos B + \sin C}{2}$$
So we need to find the maximimum of $\cos A + \cos B + \sin C$, where $A, B, C$ are angles of a triangle. Now,
$$\cos A + \cos B + \sin C = \cos A + \cos B + \sin (\pi - (A+B)) \\
= \frac{2}{\sqrt3}\left(\frac{\sqrt3}{2} \cos A + \frac{\sqrt3}{2} \cos B\right) + \frac{1}{\sqrt3}\left(\sqrt3 \sin A \cos B + \sqrt 3 \cos A \sin B\right)\\
\le \frac{1}{\sqrt3}\left(\frac{3}{4} + \cos^2 A + \frac{3}{4}+ \cos^2 B\right) + \frac{1}{2\sqrt3}\left(3 \sin^2 A +\cos^2 B + \cos^2 A +3\sin^2 B\right)\\
= \frac{\sqrt3}{2}+\frac{\sqrt3}{2}(\sin^2 A + \cos^2A) + \frac{\sqrt3}{2}(\sin^2 B + \cos^2 B) = \frac{3\sqrt 3}{2}$$
Hence $f \le 1 + \frac{3\sqrt3}{4}$, and equality is obtained when the AM-GMs we used have equality, i.e. $A = B = \frac{\pi}{6}, C = \frac{2\pi}{3}$ or $x = y = 2 - \sqrt{3}, z = \sqrt{3}$
|
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|
How find this equation $\prod\left(x+\frac{1}{2x}-1\right)=\prod\left(1-\frac{zx}{y}\right)$ let $x,y,z\in(0,1)$, find the pairs of $(x,y,z)$ such
$$\left(x+\dfrac{1}{2x}-1\right)\left(y+\dfrac{1}{2y}-1\right)\left(z+\dfrac{1}{2z}-1\right)=\left(1-\dfrac{xy}{z}\right)\left(1-\dfrac{yz}{x}\right)\left(1-\dfrac{zx}{y}\right)$$
my try:use
$$x+\dfrac{1}{2x}-1\ge 2\sqrt{\dfrac{1}{2}}-1=\sqrt{2}-1$$
so
$$LHS\ge (\sqrt{2}-1)^3$$
so I guess
$$RHS\le (\sqrt{2}-1)^3$$
But I can't prove it.Thank you
|
The only $(x,y,z)$ is $(\frac12,\frac12,\frac12)$.
For all other $x,y,z \in (0,1)$ the left-hand side is larger.
We prove this by showing that if we fix the average $a=\frac13(x+y+z)$ then
(i) the left-hand side is minimized when $x=y=z=a$, while
(ii) the right-hand side is maximized when $x=y=z=a$.
Hence it is enough to show the claimed inequality when $x=y=z=a$, which is
$$
\left(a + \frac1{2a} - 1\right)^3 \geq (1-a)^3
$$
with equality iff $a=1/2$. Equivalently, we claim
$a + \frac1{2a} - 1 \geq 1-a$ with the same equality condition;
and this easy because the difference between the sides is $(2a-1)^2/(2a)$.
It remains to prove assertions (i) and (ii). For (i),
the OP already noted that each factor is bounded below by $\sqrt 2 - 1$,
so in particular the factors are all positive.
We compute that $\log(x + \frac1{2x} - 1)$ is convex upwards by calculating
$$
\frac{d^2}{dx^2} \log\left(x + \frac1{2x} - 1\right)
= \frac{1-4x+8x^2-4x^4}{x^2(1-2x+2x^2)^2}
$$
and showing that the numerator is positive for $0<x<1$:
$$
1 - 4x + 8x^2 - 4x^4 > 1 - 4x + 8x^2 - 4x^3
= 1 - 4x(1-x)^2 > 1-4x(1-x) = (2x-1)^2.
$$
For (ii), we may assume that each of the factors is positive:
at most one of $xy/z$, $yz/x$, and $xz/y$ can be $\geq 1$
(if two of them are, then so is their product, which is
$x^2$, $y^2$, or $z^2$, contradicting $x,y,z \in (0,1)$);
and if exactly one factor is not positive, then
the right-hand side is $\leq 0$ and we're done.
Once all three factors are positive, we have
(because $\log(1-x)$ is concave downwards)
$$
\left( 1 - \frac{xy}{z} \right)
\left( 1 - \frac{yz}{x} \right)
\left( 1 - \frac{xz}{y} \right)
\leq (1-c)^3
$$
where $c$ is the average of $xy/z$, $yz/x$, and $xz/y$.
But $c \geq a$ with equality iff $x=y=z=a$: we have
$\frac12(\frac{xy}{z} + \frac{yz}{x}) \geq y$
by the inequality on arithmetic and geometric means, and likewise
$\frac12(\frac{yz}{x} + \frac{zx}{y}) \geq z$ and
$\frac12(\frac{zx}{y} + \frac{xy}{z}) \geq x$;
and summing these three inequalities yields
$$
\frac{xy}{z} + \frac{yz}{x} + \frac{xz}{y} \geq x + y + z = 3a
$$
as claimed. Hence $(1-c)^3 \leq (1-a)^3$ and we're done.
|
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|
Show $\lim_{x\to3} \frac{2x+3}{4x-9}=3$ by the definition of the limit. Show $\lim_{x\to3} \frac{2x+3}{4x-9}=3$ by the definition of the limit.
The definition of the limit: if for any given $\epsilon>0$ $\exists\delta>0$ such that if $x\in D(f(x))$ and $0<|x-c|<\delta$, then $|f(x)-L|<\epsilon$ where $L$ is the limit.
Following an example from the textbook, I have
$|f(x)-3|=\left|\frac{2x+3}{4x-9}-3\right|=\left|\frac{-6x+36}{3(4x-9)}\right|=\frac{6|-x+6|}{3|4x-9|}\cdot\frac{|x-3|}{|x-3|}=\frac{2|-x+6|}{|4x^2-21x+27|}\cdot|x-3|$
Restrict $|x-3|$ to the bounds $1<x<4$. Then $|f(x)-3|\leq\frac{2|-(4)+6|}{|4(1)^2-21(1)+27|}\cdot|x-3|=\frac{2|2|}{|4-21+27|}\cdot|x-3|=\frac{4}{7}|x-3|$
Choose $\delta(\epsilon)=\inf\{1,\frac{7\epsilon}{4}\}$. Then, is $0<|x-3|<\delta(\epsilon)$, we have $|f(x)-3|\leq\frac{4}{7}|x-3|<\epsilon$.
Thus, by the definition of the limit, $\lim_{x\to3} \frac{2x+3}{4x-9}=L=3$.
I don't understand what the point of all this was. You can get the limit by essentially plugging in $x=3$. What exactly is being done? Why do we restrict $|x-3|$ to $1<x<4$? Why choose $\delta(\epsilon)=\inf\{1,\frac{7\epsilon}{4}\}$?
I get that if we end up with a form of $|f(x)-L|<\epsilon$, we then know $L$, but that's about all I grasped.
Any help would be greatly appreciated.
|
We have that
$$|f(x)-3|=\left|\frac{2x+3}{4x-9}-3\right|=\left|\frac{10(x-3)}{4x-9}\right|=\frac{10}{|4x-9|}|x-3|.$$
We want to simplify the denominator, ideally getting something like
$$|f(x)-3|<(...)|x-3|.$$
Restrict $x$ to a range around $3$ where $|4x-9|$ is bounded below, like $x\in(\frac{5}{2},\frac{7}{2})$. When $x$ is in that range, $|4x-9|>1$ so that
$$|f(x)-3|<10|x-3|.$$
Therefore if we are given $\epsilon>0$, and we want to provide a $\delta$ such that $|x-3|<\delta$ implies $|f(x)-3|$, we can take
$$\delta=\frac{\epsilon}{10}$$
so that $|f(x)-3|<10|x-3|<\epsilon.$
What about the restriction we placed on $x$? We've already done the hard work. When $\epsilon$ gets small, we have to shrink our interval $(3-\delta,3+\delta)$ so as to not let the function get too big there. We succeeded in quantifying that shrinking: $\delta=\epsilon/10$. When $\epsilon$ gets really big, we don't want to use that formula anymore: it gives us a big $\delta$ which may eventually include the asymptote at $x=9/4$. Fortunately, big $\epsilon$ is satisfied by the $\delta$ from any smaller $\epsilon$. Then we just cut off: if the formula would predict a $\delta>1/2$, don't even bother. Take
$$\delta=\textrm{min}\left(\frac{1}{2},\frac{\epsilon}{10}\right).$$
|
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|
Inequality....(RMO $1994$...question $8$) If $a$, $b$, $c$ are positive real numbers such that $a+b+c = 1$, prove that
$$(1+a)(1+b)(1+c) ≥ 8(1-a)(1-b)(1-c)\text{.}$$
|
We can easily prove the inequality: $(a+b)(b+c)(c+a)\ge 8abc$.
Replacing $a$ by $(1-a)$, $b$ by $(1-b)$ and $c$ by $(1-c)$ we get:
$$(2-(a+b))(2-(c+b))(2-(c+a))\geq 8(1-a)(1-b)(1-c)$$
but $a + b + c= 1$ therefore:
$$a+b=1-c,~ b+c=1-a, ~\text{and}~ c+a=1-b$$
substituting these values in LHS we get :
$$(1+a)(1+b)(1+c)\ge8(1−a)(1−b)(1−c)$$
which proves the inequality.
|
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|
Use mathematical induction to prove that 9 divides $n^3 + (n + 1)^3 + (n + 2)^3$; Looking for explanation, I already have the solution. I have the solution for this but I get lost at the end, here's what I have so far.
basis $n = 0$; $9 \mid 0^3 + (0 + 1)^3 + (0 + 2)^2 ?$
$9 \mid 1 + 8$ = true
Induction: Assume $n^3 + (n + 1)^3 + (n + 2)^3 = k * 9$ // Why set it equal to $k * 9$? I know it works but why not just make the assumption => $n^3 + (n + 1)^3 + (n + 2)^3$ for some $n = k \ge 0$
Then, //and here's where I get lost
$(n + 1)^3 + (n + 2) + (n + 3)^3 = k * 9 + [(n + 3)^3 - n^3] = 9 (k + n^2 + 3n + 3)$
.
I've done similar examples but none like this. What am I not seeing?
|
We will show that $(n+1)^3+(n+2)^3+(n+3)^3$ is divisible by $9$. The following compuation shows that it is indeed so:
$(n+1)^3+(n+2)^3+(n+3)^3= n^3+(n+1)^3+(n+2)^3+9(n^2+3n+3)$
since the sum of first 3 terms on the RHS is divisible by $9$ by assumption.
|
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|
Check whether $36^{36}+41^{41}$ a multiple of $77$ Let $$a=\frac{36^{36}+41^{41}}{77}.$$
Is $a$ an integer?
I know that:
*
*The last digit ofr $41^{41}$ is $1$.
*The last digit of $36^{36}=6^{72}$ is $6$.
How can I use this information to answer my question, or do I need another thing to help me?
|
Let $n=36^{36}+41^{41}$. Three elementary steps:
*
*What happens modulo $7$: Note that $36\equiv 1\pmod{7}$ and that $41\equiv -1\pmod{7}$ hence $n \equiv 1^{36}+(-1)^{41}=1+(-1)=0\pmod{7}$.
*What happens modulo $11$: Note that $36\equiv 3\pmod{11}$ and that $41\equiv -3\pmod{11}$ hence $n\equiv 3^{36}+(-3)^{41}=3^{36}-3^{41}=3^{36}\cdot(1-3^5)\pmod{11}$. Since $3^5=9^2\cdot 3\equiv (-2)^2\cdot 3=12\equiv1\pmod{11}$, $n\equiv 0\pmod{11}$.
*And the Grand Finale: Since $\gcd(7,11)=1$, $n\equiv 0\pmod{7}$ and $n\equiv 0\pmod{11}$ together imply (actually, they are equivalent to the fact) that $n\equiv 0\pmod{7\cdot11}$, QED.
|
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How prove this inequality $a^2+b^2+c^2+d^2\ge abcd$ let $a,b,c,d$ are positive numbers,and such
$$2(a+b+c+d)\ge abcd$$
show that
$$a^2+b^2+c^2+d^2\ge abcd$$
My try:if $a,b,c,d\le 16$,then we have
$$a^2+b^2+c^2+d^2\ge 4\sqrt{abcd}$$
|
You are on the right track. So if, $abcd\le 16,$ then $4\sqrt{abcd}\ge abcd.$ What if $abcd\ge 16?$ Well, then you need another estimate for $a^2+b^2+c^2+d^2.$ The most natural one is $$a^2+b^2+c^2+d^2\ge \frac{(a+b+c+d)^2}{4}\ge \frac{(abcd)^2}{16}\ge abcd. $$
|
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How to prove $\sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-\cdots}}}}}}} = \frac{2+\sqrt 5 +\sqrt{15-6\sqrt 5}}{2}$ Ramanujan stated this radical in his lost notebook:
$$\sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-\cdots}}}}}}} = \frac{2+\sqrt 5 +\sqrt{15-6\sqrt 5}}{2}$$
I don't have any idea on how to prove this.
Any help appreciated.
Thanks.
|
The correct period has length 4, namely (+,+,+,-)
$$x_1=\small+\sqrt{5+\sqrt{5+\sqrt{5\color{red}-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5\color{red}-\cdots}}}}}}} = \frac{2+\sqrt 5 +\sqrt{15-6\sqrt 5}}{2}=2.7472\dots$$
The other roots of the quartic in $x$ are given by the patterns $\small(+,+,-,+),\; (+,-,+,+),\; (-,+,+,+)$, respectively
$$x_2=\small+\sqrt{5+\sqrt{5\color{red}-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5\color{red}-\sqrt{5+\cdots}}}}}}} = \frac{2-\sqrt 5 +\sqrt{15+6\sqrt 5}}{2}=2.5473\dots$$
$$x_3=\small+\sqrt{5\color{red}-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5\color{red}-\sqrt{5+\sqrt{5+\cdots}}}}}}} = \frac{2+\sqrt 5 -\sqrt{15-6\sqrt 5}}{2}=1.4888\dots$$
$$x_4=\small\color{red}-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5\color{red}-\sqrt{5+\sqrt{5+\sqrt{5+\cdots}}}}}}} = \frac{2-\sqrt 5 -\sqrt{15+6\sqrt 5}}{2}=-2.7833\dots$$
This immediately implies that the four roots obey the system,
$$\begin{aligned}
x_1^2 &= x_2+5\\
x_2^2 &= x_3+5\\
x_3^2 &= x_4+5\\
x_4^2 &= x_1+5\\
\end{aligned}$$
also studied by Ramanujan. (See this related post.) More generally, using any of the $2^4=16$ possible periods,
$$x = \pm\sqrt{a\pm \sqrt{a\pm \sqrt{a\pm \sqrt{a\pm\dots}}}}$$
will be the absolute value of a root of the 16th deg eqn,
$$x = (((x^2 - a)^2 - a)^2 - a)^2 - a\tag{1}$$
In his Notebooks IV (p.42-43), Ramanujan stated that (1) was a product of 4 quartic polynomials, one of which is the reducible,
$$(x^2-x-a)(x^2+x-a+1)=0\tag{2}$$
and the other three had coefficients in the cubic,
$$y^3+3y = 4(1+ay)\tag{3}$$
Using Mathematica to factor (1), we find that it is indeed a product of (2) and a 12th deg eqn with coefficients in a. After some manipulation, the 12 roots are,
$$x_n = -\frac{y-z}{4}\pm\frac{1}{2}\sqrt{\frac{(y-2)(y+z)z}{2y}}\tag{4}$$
where,
$$z =\pm\sqrt{y^2+4}\tag{5}$$
Since there are 4 sign changes and (3) gives 3 choices for $y$, this yields the 12 roots.
Note: For $a=5$ (as well as $a=2$), the cubic factors over $\mathbb{Q}$, hence no cubic irrationalities are involved, and one of the $x_n$ will give the value of the appropriate infinite nested radical.
P.S. Interestingly, for period length $n> 4$, not all the roots of the deg $2^n$ equation will be expressible as finite radical expressions for general $a$. The exception is $a=2$ where the solution involves roots of unity as discussed in this post.
|
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|
A conjectural closed form for $\sum\limits_{n=0}^\infty\frac{n!\,(2n)!}{(3n+2)!}$ Let
$$S=\sum\limits_{n=0}^\infty\frac{n!\,(2n)!}{(3n+2)!},\tag1$$
its numeric value is approximately $S \approx 0.517977853388534047...$${}^{[more\ digits]}$
$S$ can be represented in terms of the generalized hypergeometric function:
$$S={_3F_2}\left(\frac12,1,1;\ \frac43,\frac53;\ \frac4{27}\right)\cdot\frac12.\tag2$$
Let $\sigma$ be the closed-form expression constructed from integers and elementary functions as follows:
$$\sigma=3\,\alpha\,\ln(2\,\alpha+1)-\sqrt{\beta\,}\arccos\gamma,\tag3$$
where
$$\alpha=\frac{\sqrt[3]{3\,}}{6\,\sqrt[3]{2\,}}\left(\sqrt[3]{9-\sqrt{69}}+\sqrt[3]{9+\sqrt{69}}\right),\tag4$$
$$\beta=\frac1{4\,\sqrt[3]{2\,}}\left(\sqrt[3]{25+3\,\sqrt{69}}+\sqrt[3]{25-3\,\sqrt{69}}\right)-\frac12,\tag5$$
$$\gamma=\frac1{6\,\sqrt[3]{2\,}}\left(\sqrt[3]{57\,\sqrt{69}-459}-\sqrt[3]{57\,\sqrt{69}+459}\right)+\frac12\tag6$$
are the unique real roots of the following cubic equations:
$$8\,\alpha^3-2\,\alpha-1=0,\tag7$$
$$64\,\beta^3+96\,\beta^2+36\,\beta-23=0,\tag8$$
$$8\,\gamma^3-12\,\gamma^2+16\,\gamma+11=0.\tag9$$
Equivalently,
$$\sigma = \frac{3\,p}{2}\,\ln\big(p+1\big)-\frac{1}{2}\sqrt{\frac{3-p}{p}}\arccos\Big( \frac{p-6}{6p+2}\Big)\tag{10}$$
where $p$ is the plastic constant or the real root of
$$p^3-p-1=0\tag{11}$$
It can be numerically checked that the following inequality holds:
$$\Big|S-\sigma\Big|<10^{-10^5},\tag{12}$$
I conjecture that the actual difference is the exact zero, and thus $S$ has an elementary closed form:
$$\sum\limits_{n=0}^\infty\frac{n!\,(2n)!}{(3n+2)!}\stackrel?=3\,\alpha\,\ln(2\,\alpha+1)-\sqrt{\beta\,}\arccos\gamma,\tag{13}$$
I am asking for you help in proving this conjecture.
|
partial answer
Use the method of LINK
Since
$$
\int_0^1 t^n (1-t)^{2n}\;dt = \frac{n!(2n)!}{(3n+1)!} ,
$$
If we write
$$
S(x) = \sum_{n=0}^\infty \frac{n!(2n)!}{(3n+2)!}\;x^{3n+2}
$$
then
$$
S(x) = \int_0^1 \left(\sum_{n=0}^\infty t^n(1-t)^{2n} \frac{x^{3n+2}}{3n+2}\right)dt
$$
and $S = S(1)$.
But the derivative with respect to $x$ of $\sum_{n=0}^\infty t^n(1-t)^{2n} \frac{x^{3n+2}}{3n+2}$ is a geometric series. Its sum is a rational function, which may be integrated (with some work or a CAS). Plug in $x=1$. Result (if I copied right):
$$
S = \int_0^1\frac{F(t)}{18(1-t)^{4/3}t^{2/3}}\;dt
$$
where
$$
F(t) = \pi \,\sqrt {3}-6\,\sqrt {3}\arctan
\left( 2/\sqrt {3}\cdot \left( 1-t \right) ^{2/3}{t}^{1/3}+1/
\sqrt {3} \right) \\
-6\,\ln \left( ({1-t})^{1/3}
{t}^{2/3} -\left( 1-t \right) t \right) +3\,\ln \left( t \left( 1-t \right) ^{2}+ \left( 1-t \right) ^{4/3}{t}^{2/3}+ \left( 1-t \right) ^{2/3}{t}^{1/3}
\right) \\
+6\,\ln \left( ({1-t})^{1/3}{t}^{2/3} \right) -3\,\ln
\left( \left( 1-t \right) ^{2/3}{t}^{1/3} \right)
$$
|
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|
If both the sum and sum of squares of two rationals are integers, the two rationals are integers too There are two rational numbers $\alpha, \beta$ such that $\alpha + \beta,\ \alpha^2 + \beta^2$ are both integers. Prove that $\alpha, \beta$ are integers.
I started off by assuming that $\alpha = \frac{a}{b}, \beta = \frac{c}{d}$ such that $\gcd(a, b) = \gcd(c, d) = 1$ and $b, d \neq 1$.
$\frac{a}{b} + \frac{c}{d}$ is an integer. Let $\frac{a}{b} - \lfloor\frac{a}{b}\rfloor = x , \frac{c}{d} - \lfloor\frac{c}{d}\rfloor = y$
$\implies x + y = 1$ and $\left(\lfloor\frac{a}{b}\rfloor+x\right)\left(\lfloor\frac{c}{d}\rfloor+y\right)$ is an integer
$\implies \lfloor\frac{a}{b}\rfloor \lfloor\frac{c}{d}\rfloor + \lfloor\frac{a}{b}\rfloor y + \lfloor\frac{c}{d}\rfloor x + xy$ is also an integer
Since the leading term is an integer, the rest must be an integer too. Substituting $y = 1-x$ we get
$\lfloor\frac{a}{b}\rfloor - \lfloor\frac{a}{b}\rfloor x + \lfloor\frac{c}{d}\rfloor x + x - x^2$
Again, the leading term is an integer, so we can remove it. Simplifying the rest, we get...
$x\left(\lfloor\frac{c}{d}\rfloor - \lfloor\frac{a}{b}\rfloor + 1 - x\right)$ is an integer.
I couldn't think of any other suitable transformation and couldn't hit a contradiction.
I would prefer a solution that elaborates on what I have done above.
|
Hint: If $\alpha+\beta=k$ and $\alpha=\dfrac{a}{b}$, then $\alpha^2+(k-\alpha)^2-k^2=2\alpha^2-2\alpha k=\dfrac{2a}{b}\left(\dfrac{a}{b}-k\right)$ must be integer.
|
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|
Find the equations of the tangent line(s) that are parallel to the line $y= -4x +3$ Find the equations of the tangent line(s) to $\;f(x) = x^3 -4x^2 + 2\;$ that are parallel to the line $y= -4x +3$.
I know the slope is -4 and $f'(x)= 3x^2-8x.$
|
if $f'(x)=-4$ then the tangent line to that point will be parallel to your line.
Set $f'(x)=-4$ then solved for $x$.
$$-4=3x^2-8x$$
$$3x^2-8x+4=0$$
$$x=\dfrac{8\pm\sqrt{(-8)^2-4(3)(4)}}{2(3)}$$
$$x=\dfrac{8\pm\sqrt{64-48}}{6}$$
$$x=\dfrac{8\pm4}{6}$$
$$x=\dfrac{8+4}{6}\text{ or }\dfrac{8-4}{6}$$
$$x=2\text{ or }\dfrac23$$
Now for both of these points we find the equation of the line that is tangent to them.
For $x=2$:
$$f(x)=2^3-4(2)^2+2$$
$$=8-16+2$$
$$=-6$$
Now using the point slope form we can find the equation of the line tangent to that point:
$$y-y_1=m(x-x_1)$$
$$y-(-6)=-4(x-2)$$
$$y+6=-4x+8$$
$$y=-4x+2$$
That is the equation of the line that is tangent to $x=2$
For $x=\dfrac23$:
$$f(x)=\left(\dfrac23\right)^3-4\left(\dfrac23\right)^2+2$$
$$=\dfrac8{27}-\dfrac{16}{9}+2$$
$$=\dfrac{-40}{27}+2$$
$$=\dfrac{14}{27}$$
Now using the point slope form we can find the equation of the line tangent to that point:
$$y-y_1=m(x-x_1)$$
$$y-\left(\dfrac{14}{27}\right)=-4\left(x-\left(\dfrac23\right)\right)$$
$$y-\left(\dfrac{14}{27}\right)=-4x+\dfrac83$$
$$y=-4x+\dfrac{86}{27}$$
That is the equation of the line that is tangent to $x=\dfrac23$
So the equations of the two lines are:$$y=-4x+\dfrac{86}{27}$$ and $$y=-4x+2$$
|
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|
Given $f(x+1)=x^2-3x+2$, how can I find $f(x)$? Given $f(x+1)=x^2-3x+2$, how can I find $f(x)$?
|
Let $y=x+1$. We then have $x=y-1$. Hence,
\begin{align}
f(y) & = f(x+1) = x^2 -3x+2 = (y-1)^2 - 3(y-1)+2\\
& = y^2-2y+1-3y+3+2 = y^2-5y+6
\end{align}
|
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|
The sum of three square roots bounded below by $\sqrt{82}$
Let $a,b,c >0$ and $a+b+c \le 1$. Prove: $$\sqrt{a^2+1/a^2}+\sqrt{b^2+1/b^2}+\sqrt{c^2+1/c^2} \ge \sqrt{82}$$
Progress: I tried to have 3 vectors $(a,1/a)$, $(b,1/b)$ and $(c,1/c)$, played around with the vector inequalities but no success.
|
Assume first $a+b+c=1$.
Let $x=(a,1/a)$, $y=(b,1/b)$, $z=(1/c)$. Then the triangle inequality for the Euclidean norm tells us that
$$
\|x+y+z\|\leq\|x\|+\|y\|+\|z\|.
$$
This looks like
$$
\sqrt{1+\left(\frac1a+\frac1b+\frac1c\right)^2}\leq\sqrt{a^2+\frac1{a^2}}+\sqrt{b^2+\frac1{b^2}}+\sqrt{c^2+\frac1{c^2}}.
$$
So to prove the inequality in the question we need to show that
$$
\frac1a+\frac1b+\frac1c\geq9.
$$
Using the Cauchy-Schwarz inequality:
$$9=\left\|( \sqrt{a}, \sqrt{b}, \sqrt{c})\cdot(\frac{1}{\sqrt{a}}, \frac{1}{\sqrt{b}}, \frac{1}{\sqrt{c}})\right\|^2\leq (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} . $$
Finally, when $a+b+c <1$, let $d=1-(a+b) $. Then $c <d <1$; as $\sqrt {x^2+1/x^2} $ is decreasing on $(0,1) $, we have $\sqrt {c^2+1/c^2}>\sqrt {d^2+1/d^2} $. Also, $a+b+d=1$ and so
$$
\sqrt{a^2+\frac1{a^2}}+\sqrt{b^2+\frac1{b^2}}+\sqrt{c^2+\frac1{c^2}}\geq
\sqrt{a^2+\frac1{a^2}}+\sqrt{b^2+\frac1{b^2}}+\sqrt{d^2+\frac1{d^2}}\geq\sqrt {82}.
$$
|
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|
How do you factor $x^3 - 8 = 0$? I factored $x^3 - 8 = 0$ and I only got $x = 2$, but the answer said it's $x=2$ and $x=-1 \pm \sqrt{3}i$? How do you get $x=-1 \pm\sqrt{3}i$?
|
Let $x = y-1$.
Then $x^3 - 8 = y^3 - 3y^2 + 3y - 9 = (y-3)(y^2 + 3)$.
Setting this last expression equal to zero, we find all three roots:
One is $y = 3$ and the other two are $y = \pm i\sqrt{3}$.
Since $x = y-1$, we conclude the roots are $x = 2$ and $x = \pm i\sqrt{3} - 1$.
"QED"
|
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How prove this $a+b\le 1+\sqrt{2}$ let $0<c\le b\le 1\le a$, and such $a^2+b^2+c^2=3$, show that
$a+b\le 1+\sqrt{2}$
My try: let $ c^2=3-(a^2+b^2)\le b$
|
As Sun stated, given a valid tuple of $(a,b,c)$, replace it with $(A, b, 0)$ where $ A^2 = a^2 + c^2$. Observe that $a + b \leq A + b$, hence it remains to show that $ A + b \leq 1 + \sqrt{2}$.
Squaring this, we need to show that $ A^2 + 2Ab + b^2 \leq 3 + 2 \sqrt{2}$ or that $AB \leq \sqrt{2}$.
But since $A^2 + 2b^2 \leq 4$, hence $ 2 \sqrt{2} Ab \leq A^2 + 2b^2 \leq 4$, hence we do have $Ab \leq \sqrt{2}$.
|
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|
Proof by Math Induction I have 3 math induction proofs I have been struggling with for a while. I understand how to do summation proofs but these ones, I can't find a general pattern to solve. Please help.
1) $D(n) = {n(n-3) \over2}$ for all $n \ge 3$ This is for the n diagonals of a polygon.
2)$\binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \ldots + \binom{n}{n} = 2^n$ for all $n\ge 0$
3)$n^2 + 1 \ge 3n$ for all $n \ge 3$; For this one, I got stuck when I did this:
$$\begin{align}
k^2 + 1 &\ge 3k \\
(k+1) ^ 2 +1 &\ge 3(k+1) \\
(k+1) ^ 2 &\ge k^2 \ge 3k \\
(k+1) ^ 2 &\ge k^2 + 3 \ge 3k + 3 \,\, \text{(? I don't know how to proceed from here ?)}
\end{align}$$
Please give me possible solutions and techniques to solve these kind of problems.
|
I assume that $D(n)$ is the number of diagonals in an $n$-sided convex polygon. Descriptively: each vertex of the polygon can be connected to $n-3$ other (connecting it with itself or its neighbors will not produce diagonals). That way, you make each diagonal twice, hence the formula.
If you want to use induction, then check for $n=3$, and then assume it works for all $k$-sided convex polygons where $k < n$. Look at the $n$-sided one.
We can pick any $n-1$ vertices, and the polygon they span has $(n-1)((n-1)-3)/2$ diagonals (by the assumption). But, one of its sides will also become a diagonal of the bigger polygon. Now, we're just missing $n-3$ diagonals from the $n$-th vertex, so in total we have:
\begin{align*}
\frac{(n-1)((n-1)-3)}{2} + 1 + (n-3) &= \frac{(n-1)(n-4) + 2 + 2(n-3)}{2} \\
&= \frac{n^2 - 5n + 4 + 2 + 2n - 6}{2} = \frac{n^2 - 3n}{2} \\
&= \frac{n(n-3)}{2}.
\end{align*}
If you want the second one using induction, use the recursive formula for binomial coefficients:
$$\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}.$$
Edit: Since you asked how to do this (in comment to nsanger), I'm expanding that part of the answer. The idea is to note that the two terms on the right hand side are the same, only with a shift in $k$. In other words:
$$\sum_{k=1}^n \binom{n-1}{k-1} = \sum_{k=\color{red}{0}}^{\color{red}{n-1}} \binom{n-1}{\color{red}{k}}.$$
Now, we just need to play with the edge cases to get the sums to have the same limits:
\begin{align*}
\sum_{k=0}^n \binom{n}{k} &= \binom{n}{0} + \binom{n}{n} + \sum_{k=\color{red}{1}}^{\color{red}{n-1}} \binom{n}{k} = 1 + 1 + \sum_{k=1}^{n-1} \binom{n-1}{k} + \sum_{k=1}^{n-1} \binom{n-1}{k-1} \\
&= \binom{n-1}{0} + \sum_{k=1}^{n-1} \binom{n-1}{k} + \binom{n-1}{n-1} + \sum_{k=1}^{n-1} \binom{n-1}{k-1} \\
&= \sum_{k=\color{red}{0}}^{n-1} \binom{n-1}{k} + \sum_{k=1}^{\color{red}{n}} \binom{n-1}{k-1} \\
&= \Big\{ \text{Use the above formula for the second sum} \Big\} \\
&= \sum_{k=0}^{n-1} \binom{n-1}{k} + \sum_{k=0}^{n-1} \binom{n-1}{k} = 2\sum_{k=0}^{n-1} \binom{n-1}{k} \\
&= \Big\{ \text{Use the induction hypothesis} \Big\} \\
&= 2 \cdot 2^{n-1} = 2^n.
\end{align*}
The third one was answered by nsanger, so I won't be repeating that.
|
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Finding the remainder when a polynomial is divided by a product of numbers whose remainders are known We have a polynomial $f(x)$ with rational roots that leaves remainders $15, 2x + 1$ when divided by the polynomials $x - 3, (x-1)^2$ respectively. What is the remainder when $f(x)$ is divided by $(x - 3)(x-1)^2$?
From the remainder theorem we know that,
$f(3) = 15$.
I thought quite hard about this and tried various methods, but couldn't come up with anything useful.
|
Write $$f(x)=A(x-3)(x-1)^2+B(x-1)^2+C(x-3)+D$$ where $A$ is an integer polynomial or constant
$\implies 15=f(3)=4B+D\implies D=15-4B$
$\implies f(x)=A(x-3)(x-1)^2+B(x-1)^2+C(x-3)+15-4B$
Now, $f(x)\equiv C(x-3)+15-4B\pmod{(x-1)^2}\equiv Cx+15-4B-3C $
But $f(x)\equiv 2x+1\pmod{(x-1)^2}$
$\implies C=2$ and $15-4B-3C=1$ and so on
HINT:
Write $$f(x)=A(x-3)(x-1)^2+Bx^2+Cx+D$$
$\implies 15=f(3)=9B+3C+D\ \ \ \ (1)$
Now, $f(x)\equiv Bx^2+Cx+D\pmod{(x-1)^2}\equiv(2B+C)x+D-B\pmod{(x-1)^2} $
But $f(x)\equiv2x+1\pmod{(x-1)^2}$
$\implies 2B+C=2\ \ \ \ (2)$ and $D-B=1\ \ \ \ (3)$
Can you solve for $B,C,D$ from $(1),(2),(3)$?
|
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|
Characteristic polynomial of a matrix is monic? Given a $n \times n$ matrix A, I need to show that its characteristic polynomial, defined as $P_A (x) = det (xI-A)$ is monic. I am trying induction. But no clue after induction hypothesis.
|
For $n=1$,
\begin{equation}
P_A(x)=det(xI_1-A)=|s-a_{11}|=s-a_{11}.
\end{equation}
So, $P_A(x)$ is monic for $n=1$.
Assume $P_A(x)$ is monic for $n=k$, i.e.
\begin{equation}
P_A(x)=det(xI_k-A)=\begin{vmatrix}
(x-a_{11})&-a_{12}&...&-a_{1k}\\
-a_{21}&(x-a_{22})&...&-a_{2k}\\
\vdots&\vdots&\ddots&\vdots\\
-a_{k1}&-a_{k2}&...&(x-a_{kk})
\end{vmatrix}=x^k+...
\end{equation}
We are now left to show that; $P_A(x)$ is monic for $n=k+1$, so
\begin{equation}
\begin{split}
P_A(x)=&det(xI_{k+1}-A)\\
=&\begin{vmatrix}
(x-a_{11})&-a_{12}&...&-a_{1k}&-a_{1,k+1}\\
-a_{21}&(x-a_{22})&...&-a_{2k}&-a_{2,k+1}\\
\vdots&\vdots&\ddots&\vdots\\
-a_{k1}&-a_{k2}&...&(x-a_{kk})&-a_{1,k+1}\\
-a_{k+1,1}&-a_{k+1,2}&...&-a_{k+1,k}&(x-a_{k+1,k+1})
\end{vmatrix}\\
=&\underbrace{(-a_{k+1,1})(-1)^{(k+1)+1}\begin{vmatrix}
-a_{12}&...&-a_{1k}&-a_{1,k+1}\\
(x-a_{22})&...&-a_{2k}&-a_{2,k+1}\\
\vdots&\ddots&\vdots\\
-a_{k2}&...&(x-a_{kk})&-a_{1,k+1}
\end{vmatrix}}_{order\ of\ x\ is\ k-1}\\
+&\underbrace{(-a_{k+1,2})(-1)^{(k+1)+2}\begin{vmatrix}
(x-a_{11})&-a_{13}&...&-a_{1k}&-a_{1,k+1}\\
-a_{21}&-a_{23}&...&-a_{2k}&-a_{2,k+1}\\
-a_{31}&(x-a_{33})&...&-a_{3k}&-a_{3,k+1}\\
\vdots&\ddots&\vdots\\
-a_{k1}&-a_{k3}&...&(x-a_{kk})&-a_{1,k+1}
\end{vmatrix}}_{order\ of\ x\ is\ k-1}\\
+&...\\
+&(x-a_{k+1,k+1})(-1)^{\overbrace{(k+1)+(k+1)}^{2(k+1)}}det(xI_k-A)\\
=&(x-a_{k+1,k+1})(x^{k}+...)+...\\
=&x^{k+1}+...
\end{split}
\end{equation}
So, $P_A(x)$ is also found monic for $n=k+1$.
$\therefore$ $P_A(x)$ is monic.$\square$
|
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|
How to solve this first-order ODE $\frac{dy}{dx}=\frac{y^6-2x^2}{2x^2y+2y^3-y}$? solve this ODE equation
$$\dfrac{dy}{dx}=\dfrac{y^6-2x^2}{2x^2y+2y^3-y}$$
My try:
$$\dfrac{y~dy}{dx}=\dfrac{y^6-2x^2}{2x^2+2y^2-1}$$
let
$$u=y^2$$
then
$$\dfrac{du}{dx}=\dfrac{2u^3-4x^2}{2x^2+2u-1}$$
then I can't work. Thank you, this problem is from ODE equation excise book,
and this book only take this answer:
$$(y^3-3x)^7(y^3+2x)^3=cx^{15}$$
|
You did it correctly.
Now for $\dfrac{du}{dx}=\dfrac{2u^3-4x^2}{2x^2+2u-1}$ ,
$(2u+2x^2-1)\dfrac{du}{dx}=2u^3-4x^2$
Let $v=u+x^2-\dfrac{1}{2}$ ,
Then $u=v-x^2+\dfrac{1}{2}$
$\dfrac{du}{dx}=\dfrac{dv}{dx}-2x$
$\therefore2v\left(\dfrac{dv}{dx}-2x\right)=2\left(v-x^2+\dfrac{1}{2}\right)^3-4x^2$
$2v\dfrac{dv}{dx}-4xv=2v^3-6\left(x^2-\dfrac{1}{2}\right)v^2+6\left(x^2-\dfrac{1}{2}\right)^2v-2\left(x^2-\dfrac{1}{2}\right)^3-4x^2$
$2v\dfrac{dv}{dx}=2v^3-6\left(x^2-\dfrac{1}{2}\right)v^2+\left(6\left(x^2-\dfrac{1}{2}\right)^2+4x\right)v-2\left(x^2-\dfrac{1}{2}\right)^3-4x^2$
$v\dfrac{dv}{dx}=v^3-3\left(x^2-\dfrac{1}{2}\right)v^2+\left(3\left(x^2-\dfrac{1}{2}\right)^2+2x\right)v-\left(x^2-\dfrac{1}{2}\right)^3-2x^2$
This belongs to an Abel equation of the second kind.
In fact all Abel equation of the second kind can be transformed into Abel equation of the first kind.
Let $v=\dfrac{1}{w}$ ,
Then $\dfrac{dv}{dx}=-\dfrac{1}{w^2}\dfrac{dw}{dx}$
$\therefore-\dfrac{1}{w^3}\dfrac{dw}{dx}=\dfrac{1}{w^3}-3\left(x^2-\dfrac{1}{2}\right)\dfrac{1}{w^2}+\left(3\left(x^2-\dfrac{1}{2}\right)^2+2x\right)\dfrac{1}{w}-\left(x^2-\dfrac{1}{2}\right)^3-2x^2$
$\dfrac{dw}{dx}=\left(\left(x^2-\dfrac{1}{2}\right)^3+2x^2\right)w^3-\left(3\left(x^2-\dfrac{1}{2}\right)^2+2x\right)w^2+3\left(x^2-\dfrac{1}{2}\right)w-1$
Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2
|
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|
How to find the $f^{-1}(x)$ of $f(x)=x^{3}-12x+\frac{48}{x}-\frac{64}{x^{3}}$ It is a question from a quiz.
The following is the whole question.
Let
\begin{eqnarray}
\\f(x)=x^{3}-12x+\frac{48}{x}-\frac{64}{x^{3}} , \space x\in (-\infty ,0),
\end{eqnarray}
find $f^{-1}(x)$. Hint : $f(x)$ can be written in the form, like$(A+B)^{3}$.
The first thing I think is $(A+B)^{3}=A^3+3A^2B+3AB^2+B^3$, then try to make it become the the form of $A^3+3A^3B+3AB^3+B^3$. However, it it so difficult to obtain this form.
I need help.
Update :
Now I have $\left(x - \frac 4x\right)^3$ but how to find the $f^{-1}(x)$ of $f(x)=\left(x - \frac 4x\right)^3$?
Thank you for your attention
|
for $x \in (-\infty,0)$ the inverse function has the form
$$
f^{-1}(x)=\frac{1}{2}\left( \sqrt[3]{x}-\sqrt{\sqrt[3]{x^2}+16} \right)
$$
As mentioned above we have
$y = \left(x - \dfrac4x\right)^3 \Longrightarrow x^2 - x\sqrt[3]{y} - 4 = 0$.
By solving the equation we get
$$
x=\frac{1}{2}\left( \sqrt[3]{y}-\sqrt{\sqrt[3]{y^2}+16} \right),
$$
and
$$
x=\frac{1}{2}\left( \sqrt[3]{y}+\sqrt{\sqrt[3]{y^2}+16} \right).
$$
The first expression is always negative (note $x \in (-\infty,0)$ but second is positive and not suitable for us. Thus, changing the variables, we get the inverse function.
|
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|
Prove that there are no positive integers $a,b,c,d$ such that $a^2 + b^2 = c^2$ and $a^2 - b^2 = d^2$ Prove that there are no positive integers $a,b,c,d$ such that $a^2 + b^2 = c^2$ and $a^2 - b^2 = d^2$?
Could you show me the proof?
|
Dividing by $(a,b)$, we can assume that $(a,b)=1$.
Multiplying the two equations yields
$$
a^4-b^4=c^2d^2\tag{1}
$$
whereby
$$
b^4+(cd)^2=a^4\tag{2}
$$
This answer characterizes all primitive Pythagorean Triples; that is, there must be a pair of positive integers $(m,n)$ so that $(m,n)=1$ and $m+n$ is odd so that
$$
x=m^2-n^2,\quad y=2mn,\quad z=m^2+n^2\tag{3}
$$
Suppose that $(x,y,z)$ is the Pythagorean triple with the smallest hypotenuse so that at least two sides are a square or twice a square.
Suppose that $z$ and $x$ are squares (since they are odd, they can't be twice a square). Then $(\sqrt{xz},n^2,m^2)$ has a smaller hypotenuse.
Suppose that $z$ is a square and $y$ is a square or twice a square. Then $m$ and $n$ must both be either a square or twice a square. Then $(m,n,\sqrt{z})$ has a smaller hypotenuse (since $x,y\ge1$, $z\gt1$).
Suppose that $x$ is a square and $y$ is a square or twice a square. Then $m$ and $n$ must both be either a square or twice a square. Then $(\sqrt{x},n,m)$ has a smaller hypotenuse.
Thus, there is no primitive Pythagorean Triple with at least two sides a square or twice a square.
If $(2)$ were true, there would be a primitive Pythagorean Triple with two sides a square.
|
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|
Find the limit of $\sum\limits_{k=1}^n\left(\sqrt{1+\frac{k}{n^2}}-1\right)$
$$\lim_{n\rightarrow\infty}\sum_{k=1}^n\left(\sqrt{1+\frac{k}{n^2}}-1\right)$$
Note that $\forall x\ge 0, \sqrt{x}-1\le\sqrt{1+x}-1\le x$
Then
$$\sum_{k=1}^n\left(\sqrt{\frac{k}{n^2}}-1\right)\le S_n\le \sum_{k=1}^n\frac{k}{n^2}=\frac{1}{2}-\frac{1}{2n}$$
How do I evaluate the sum on the left?
|
For every $x$ in $(0,1)$, $\frac12x-x^2\leqslant\sqrt{1+x}-1\leqslant\frac12x$ hence
$$
\frac1{2n^2}T_n-\frac1{n^4}R_n\leqslant S_n\leqslant\frac1{2n^2}T_n,\qquad T_n=\sum_{k=1}^nk,\quad R_n=\sum_{k=1}^nk^2.
$$
One knows that $T_n=\frac12n(n+1)$ and $R_n\leqslant\sum\limits_{k=1}^nn^2=n^3$ hence
$$
\frac{n+1}{4n}-\frac1n\leqslant S_n\leqslant\frac{n+1}{4n},
$$
from which the limit of $S_n$ is easy to guess.
|
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|
Proving any product of four consecutive integers is one less than a perfect square Prove or disprove that : Any product of four consecutive integers is one less than a
perfect square.
OK so I start with $n(n+1)(n+2)(n+3)$ which can be rewritten $n(n+3)(n+1)(n+2)$
After multiplying we get $(n^2 + 3n)(n^2 + 3n + 2)$
How do I proceed from here to end up with something squared $- 1$?
|
Observe that the product $p$ of four consecutive integers can be written as $p=(x-\frac{3}{2})(x-\frac{1}{2})(x+\frac{1}{2})(x+\frac{3}{2})$ where $x=n+\frac{1}{2}$ for some integer $n$. Then $p=(x^2-\frac{9}{4})(x^2-\frac{1}{4}) = (x^2-\frac{5}{4}+1)(x^2-\frac{5}{4}-1) = (x^2-\frac{5}{4})^2-1$. It remains to show that $x^2-\frac{5}{4}$ is an integer, which is an easy bit of arithmetic.
|
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|
What is the tens digit of $3^{100}$? Is there a general formula to calculate the n-th digit of any big number?
|
We would like to calculate $3^{100}\pmod{100}$. This will tell us the last two digits of $3^{100}$, which includes the tens digit.
We proceed as follows: $3^4 = 81$, so:
$$\begin{align}
3^8 & \equiv (81)^2 \equiv 61 \pmod{100} \\
3^{16} & \equiv (61)^2 \equiv 21\pmod{100} \\
3^{24} & \equiv 61\cdot 21 \equiv 81\pmod{100}\\
3^{48} & \equiv (81)^2 \equiv 61 \pmod{100} \\
3^{50} & \equiv 9\cdot 61 \equiv 49\pmod{100} \\
3^{100} & \equiv (49)^2\equiv 01 \pmod {100}
\end{align}$$
So the last two digits are 01.
In general it's very quick to calculate $a^b\pmod n$. You calculate $a^{\lfloor b/2 \rfloor}\pmod n$ (using this method recursively if necessary) and square it, again mod $n$; if $b$ is odd you multiply the result by $a$ and you are done.
|
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|
Is i true that $x+y\le(2x^2+2y^2)^{1/2}\le...\le(2^{p-1}x^p+2^{p-1}y^p)^{1/p}$ Is it true that,
$$
x+y\le(2x^2+2y^2)^{1/2}\le...\le(2^{p-1}x^p+2^{p-1}y^p)^{1/p}
$$
How to prove or disprove it?
|
The first link is OK. $x+y \le |x+y|$, and then note we have
$$|x+y| \le (2x^2+2y^2)^{1/2} \tag{1}$$
iff $(x-y)^2 \ge 0$, after squaring both (nonnegative) sides of $(1),$ rearranging and factoring. So this gives the first inequality of the list
$x+y \le (2x^2+2y^2)^{1/2}.$
The second inequality (if I see the pattern) would be
$$(2x^2+2y^2)^{1/2} \le (4x^3+4y^3)^{1/3},$$
which fails if $x=y=-1$, so maybe you should restrict to $x,y>0$ for a possible string of inequalities.
|
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|
Matrix Equation $A^3-3A=\begin{pmatrix}-7 & -9\\ 3 & 2\end{pmatrix}$ How can I solve in $\mathcal{M}_{2}(\mathbb{Z})$ the equation $$A^3-3A=\begin{pmatrix}-7 & -9\\ 3 & 2\end{pmatrix}?$$
I try to use $$A^2-Tr(A)A+detA\cdot I_2=O_2$$ but I don't still obtain anything.
thanks.
|
$$
A=\pmatrix{{2}&{3}\\{-1}&{-1}}.$$
Note that $a^3+b^3=(a+b)^3-3ab(a+b)$.
Let $a,b$ be eigenvalues of $A$. Then $trA=a+b$, $detA=ab$, and they are integers.
Let $X=trA$.
We see from $A(A^2-3I)$ has determinant $13$, we have $detA | 13$.
From the identity in the beginning, we have $X^3-3(ab+1)X+5=0$ , which we can obtain from taking traces of original equation.
Just plug in the four possible values for $ab$, namely $\pm 1, \pm 13$, we find that only possible value is $ab=1$, and consequently
$X^3-6X+5=0$.
Solving for integers, we obtain $X=1$.
Thus, $A$ should satisfy $trA=1$, $detA=1$.
By Cayley-Hamilton, we have $A^2-A+I=0$, this forces $A^3=-I$.
Now, plug in to original equation to get
$$
-3A=\pmatrix{{-6}&{-9}\\{3}&{3}}.$$
Then we arrive at the answer.
|
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|
Prove : $\frac{1}{a}+\frac{1}{b}-\frac{1}{c}< \frac{1}{abc}$ $a;b;c>0$ such that $a^2+b^2+c^2=\frac{5}{3}$. Prove :
$\frac{1}{a}+\frac{1}{b}-\frac{1}{c}< \frac{1}{abc}$
|
$$
c(a+b)\leq \frac{c^2+(a+b)^2}{2}=\frac{a^2+b^2+c^2}{2}+ab=\frac 56+ab
$$
which is the rearrangement of your inequality and even the stronger one.
$$
c(a+b)\leq \frac 56+ab \implies \\
\frac 1a+\frac 1b\leq \frac 5{6abc}+\frac 1c\implies
\frac 1a+\frac 1b-\frac 1c\leq \frac 5{6abc}
$$
|
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|
Solving the recurrence $T(n) = 2T\left(\frac{n}{2}\right) + \frac{n}{2}\log(n)$ Please help me solve the recurrence
$$ T(n) = 2T\left(\frac{n}{2}\right) + \frac{n}{2}\log(n) $$
|
Here is an exact solution of a related recurrence that has the same complexity as yours. We take $T(0)=0$ and the recurrence is
$$T(n) = 2 T(\lfloor n/2 \rfloor) + \frac{1}{2} n \; (\lfloor\log_2 n\rfloor+1).$$
Let the binary digits of $n$ be given by
$$n = \sum_{k=0}^{\lfloor\log_2 n\rfloor} d_k 2^k.$$
Then the exact solution to the recurrence is
$$T(n) = \frac{1}{2}\sum_{k=0}^{\lfloor\log_2 n\rfloor} 2^k
(\lfloor\log_2 n\rfloor +1 - k)\sum_{j=k}^{\lfloor\log_2 n\rfloor} d_j 2^{j-k}.$$
Now an upper bound on $T(n)$ is given by the case where we have a string of ones, producing
$$T(n) \le \frac{1}{2}\sum_{k=0}^{\lfloor\log_2 n\rfloor} 2^k
(\lfloor\log_2 n\rfloor +1 - k)\sum_{j=k}^{\lfloor\log_2 n\rfloor} 2^{j-k}
\\= \frac{1}{2}\times
\left(\lfloor\log_2 n\rfloor^2 2^{\lfloor\log_2 n\rfloor}
+ 3 \lfloor\log_2 n\rfloor 2^{\lfloor\log_2 n\rfloor}
- 2 \times 2^{\lfloor\log_2 n\rfloor} + \lfloor\log_2 n\rfloor +3\right).$$
We get a lower bound when $n$ is a one digit followed by zeros, giving
$$T(n)\ge \frac{1}{2}\sum_{k=0}^{\lfloor\log_2 n\rfloor} 2^k
(\lfloor\log_2 n\rfloor +1 - k) 2^{\lfloor\log_2 n\rfloor-k}
= \frac{1}{2}\times 2^{\lfloor\log_2 n\rfloor} \sum_{k=0}^{\lfloor\log_2 n\rfloor}
(\lfloor\log_2 n\rfloor +1 - k)
\\ = \frac{1}{2}\times 2^{\lfloor\log_2 n\rfloor}
\left(\frac{1}{2} \lfloor\log_2 n\rfloor^2
+\frac{3}{2} \lfloor\log_2 n\rfloor +1\right).$$
Taking the two dominant terms from the upper and lower bound we get the asymptotic complexity as
$$\Theta\left( \lfloor\log_2 n\rfloor^2 2^{\lfloor\log_2 n\rfloor}\right)
= \Theta(\log^2 n\times n).$$
This link points to a series of similar calculations.
|
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|
How find this integral $I=\int_{0}^{\infty}\frac{\log{\cos^2{x}}}{1+e^{2x}}dx$ find the value
$$I=\int_{0}^{\infty}\dfrac{\log{\cos^2{x}}}{1+e^{2x}}dx$$
My try: let $$e^{2x}=u\Longrightarrow x=\dfrac{1}{2}\log{u}$$
then
$$I=\int_{1}^{\infty}\dfrac{\log{(\cos^2{(\dfrac{1}{2}\log{u})})}}{2u(1+u)}du$$
then I can't.Thank you
|
Referring to the identity
$$ \log|\cos x| = - \log 2 + \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \cos (2nx) $$
(which can be easily obtained by considering the real part of $\log (1 + e^{2ix})$), we find that
\begin{align*}
I&= \int_{0}^{\infty} \frac{2}{1+e^{2x}} \left\{ - \log 2 + \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \cos (2nx) \right\} \, dx \\
&= \int_{0}^{\infty} \frac{1}{1+e^{x}} \left\{ - \log 2 + \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \cos (nx) \right\} \, dx \quad (2x \mapsto x) \\
&= - \int_{0}^{\infty} \frac{\log 2}{1+e^{x}} \, dx + \int_{0}^{\infty} \left\{ \sum_{m=1}^{\infty} (-1)^{m-1}e^{-mx} \right\} \left\{ \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \cos (nx) \right\} \, dx \\
&= -\log^{2} 2 + \sum_{n=1}^{\infty}\sum_{m=1}^{\infty} \frac{(-1)^{m+n}}{n} \int_{0}^{\infty} \cos (nx) e^{-mx} \, dx \\
&= -\log^{2} 2 + \sum_{n=1}^{\infty}\sum_{m=1}^{\infty} \frac{(-1)^{m+n}}{n} \frac{m}{m^{2} + n^{2}} \\
&=: -\log^{2} 2 + S.
\end{align*}
Now let us consider the sum $S$. Interchanging the role of $m$ and $n$, we have
$$ S = \sum_{n=1}^{\infty}\sum_{m=1}^{\infty} \frac{(-1)^{m+n}}{m} \frac{n}{m^{2} + n^{2}}. $$
Averaging,
$$ S
= \frac{1}{2} \sum_{n=1}^{\infty}\sum_{m=1}^{\infty} \left(\frac{m}{n} + \frac{n}{m} \right) \frac{(-1)^{m+n}}{m^{2} + n^{2}}
= \frac{1}{2} \left( \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \right)^{2}
= \frac{1}{2}\log^{2} 2. $$
Therefore the answer is
$$ I = -\frac{1}{2}\log^{2} 2. $$
|
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|
How to derive this second derivative using the quotient rule? If a given first derivative is: $\ {dy \over dx} = {-48x \over (x^2+12)^2} $
What are the steps using the quotient rule to derive the second derivative: $\ {d^2y \over dx^2} = {-144(4-x^2) \over (x^2+12)^3} $
My Steps:
\begin{align*}
{d^2y \over dx^2}
&= {-48(x^2 +12)^2 - 2(x^2+12)(2x)(-48x)\over (x^2+12)^4} \\
&=-48{(x^2 +12)^2 - 4x^2(x^2+12)\over (x^2+12)^4} \\
&= -48{(x^2 +12)( - 4x^2 +(x^2+12))\over (x^2+12)^4} \\
&= -48{(x^2 +12)( - 4x^2)\over (x^2+12)^3} \\
&= {???}
\end{align*}
|
We can factor out the constant term to make life easier (and just multiply by that $-48$ at the end of our calculation) as:
$$-48 \dfrac{x}{(x^2+12)^2}$$
This makes it easier to use the quotient and chain rule (I will assume you know these).
The derivative of $\ {dy \over dx} = {-48x \over (x^2+12)^2} $, using the quotient and chain rule is:
$\dfrac{d^2y}{dx^2} = -48 \dfrac{(1)(x^2+12)^2 - 2(x^2+12)(2x)x}{(x^2+12)^4} = -48 \dfrac{(x^2+12) - 2(2x)x}{(x^2+12)^3} = \dfrac{-144(4-x^2)}{(x^2+12)^3}$
|
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|
Calculate $a^3+b^3+c^3+d^3$ for the real roots of $x^4+2x^3-3x^2-3x+2$ $a,b,c,d \in \mathbb{R}$ are the real roots of $x^4+2x^3-3x^2-3x+2$.
Calculate $a^3+b^3+c^3+d^3$.
With approximation i found out, that $a^3+b^3+c^3+d^3 = -17$, but how can I proof that without calculating the roots exactly?
Cheers
|
Express $a^3 + b^3 + c^3 + d^3$ in terms of elementary symmetric polynomials and use Vieta's formulas.
Also, here is an alternative approach for the fans of linear algebra. Consider a matrix
$$
A = \left(\matrix{0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -2 & 3 & 3 & -2}\right).
$$
A trivial check shows that its characteristic polynomial is equal to
$$
\det (A - \lambda I) = \lambda^4 + 2\lambda^3 - 3\lambda^2 - 3\lambda + 2.
$$
So $a,b,c,d$ are the characteristic roots of $A$. Then their cubes are characteristic roots of $A^3$ (as easily follows from, say, the existence of a Jordan normal form). Then
$$
a^3 + b^3 + c^3 + d^3 = \operatorname{Tr} A^3 = -17.
$$
|
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|
Show that $\sum_{n=0}^\infty r^n e^{i n \theta} = \frac{1- r\cos(\theta)+i r \sin(\theta)}{1+r^2-2r\cos(\theta)}$ Show that $$\sum_{n=0}^\infty r^n e^{i n \theta} = \frac{1- r\cos(\theta)+i r \sin(\theta)}{1+r^2-2r\cos(\theta)},$$ where $0\leq r <1$.
Using this, prove that $\sum_{n=0}^\infty r^n \cos(n\theta)$ and $\sum_{n=0}^\infty r^n \sin(n\theta)$ are convergent.
|
The first sum is a geometric series
$$\sum_{n=0}^{\infty}{r^ne^{in\theta}} = \sum_{n=0}^{\infty}(re^{i\theta})^n$$
$$=\frac{1}{1-re^{i\theta}} = \frac{1}{1-r\cos\theta -ir\sin{\theta}}\cdot\frac{1-r\cos\theta + ir\sin\theta}{1-r\cos\theta + ir\sin\theta} =\frac{1-r\cos\theta + ir\sin\theta}{1+r^2-2r\cos\theta}$$
For the next two sums notice that
$$\sum_{n=0}^{\infty}{r^ne^{in\theta}}=\sum_{n=0}^{\infty}{r^n\cos(n\theta) + i\sum_{n=0}^{\infty}r^n\sin(n\theta)}$$
So $$\displaystyle{\sum_{n=0}^{\infty}{r^n\cos(n\theta)}} = \frac{1-r\cos\theta}{1+r^2-2r\cos\theta}$$ $$\displaystyle{\sum_{n=0}^{\infty}{r^n\sin(n\theta)}} = \frac{r\sin{\theta}}{1+r^2-2r\cos\theta}$$
|
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|
Calculate $\sum\limits_{k=0}^{\infty}\frac{1}{{2k \choose k}}$ Calculate $$\sum \limits_{k=0}^{\infty}\frac{1}{{2k \choose k}}$$
I use software to complete the series is $\frac{2}{27} \left(18+\sqrt{3} \pi \right)$
I have no idea about it. :|
|
Consider the function
$$f(x) = \frac{\arcsin{x}}{\sqrt{1-x^2}}$$
$f(x)$ has a Maclurin expansion as follows:
$$f(x) = \sum_{n=0}^{\infty} \frac{2^{2 n}}{\displaystyle (2 n+1) \binom{2 n}{n}} x^{2 n+1}$$
Differentiating, we get
$$f'(x) = \frac{x \, \arcsin{x}}{(1-x^2)^{3/2}} + \frac{1}{1-x^2} = \sum_{n=0}^{\infty} \frac{2^{2 n}}{\displaystyle \binom{2 n}{n}} x^{2 n}$$
Evaluate at $x=1/2$
$$\sum_{n=0}^{\infty} \frac{1}{\displaystyle \binom{2 n}{n}} = \frac{\frac12 \arcsin{\frac12}}{3 \sqrt{3}/8} + \frac{4}{3} = \frac{2\sqrt{3} \pi+36}{27}$$
ADDENDUM
There are many derivations here of the above result for the Maclurin series for $f(x)$; I refer you to this one.
|
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|
Find the Area of the ellipse Given $$\frac{x^2}{a^2} + \frac{y^2}{b^2}=1$$
where $a>0$, $b>0$
I tried to make $y$ the subject from the equation of the ellipse and integrate from $0$ to $a$. Then multiply by $4$ since there are $4$ quadrants.
$$Area=4\int^a_0\left(b^2-x^2\left(\frac{b^2}{a^2}\right)\right)^\frac{1}{2}dx$$
I can't get the answer $\pi ab$
|
In order to find the the area inside the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, we can use the transformation $(x,y)\rightarrow(\frac{bx}{a},y)$ to change the ellipse into a circle. Since the lengths in the $x$-direction are changed by a factor $b/a$, and the lengths in the $y$-direction remain the same, the area is changed by a factor $b/a$. Thus $$\text{Area of circle} = \frac{b}{a}\times \text{Area of
ellipse},$$
which gives the area of the ellipse as $(a/b\times\pi b^2)$, that is $\pi ab$.
|
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|
Let $a,b,c \in \mathbb{R^+}$, does this inequality holds $\frac{a}{na + kb} + \frac{b}{nb+kc} + \frac{c}{nc + ka} \ge \frac{3}{k+n}$? Does the following statement/inequality holds for $a,b,c \in \mathbb{R^+}$?
$$\frac{a}{na + kb} + \frac{b}{nb+kc} + \frac{c}{nc + ka} \ge \frac{3}{k+n}$$
I've been thinking for hours and I couldn't find a solution to this inequality. I tried to use Cauchy-Schwarz Inequality for fraction:
$$\frac{x^2}{a} + \frac{y^2}{b} + \frac{z^2}{c} \ge \frac{(z+x+y)^2}{a+b+c}$$
But I only get:
$$LHS \ge \frac{(\sqrt{a} + \sqrt{b} + \sqrt{c})^2}{(n+k)(a+b+c)}$$
And then I can't prove that:
$$(\sqrt{a} + \sqrt{b} + \sqrt{c})^2 \ge 3(a+b+c)$$
$$a + b + c + 2\sqrt{ab} + 2\sqrt{ac} + 2\sqrt{bc} \ge 3(a+b+c)$$
$$\sqrt{ab} + \sqrt{ac} + \sqrt{bc} \ge a+b+c$$
Which simplifyies to:
$$\sqrt{a}(\sqrt{b} - \sqrt{a}) + \sqrt{b}(\sqrt{c} - \sqrt{b}) + \sqrt{c}(\sqrt{a} - \sqrt{c}) \ge 0$$
But if $a\neq b \neq c$, one of the terms must be negative, so nothing here.
And at last I plug it into Wolfram Alpha, and for some random numbers $n$ and $k$ it's true. I wonder maybe the fact that the LHS is cyclic we make the LHS to have minimal values when $a=b=c$?
|
If $k\ge 2n$ then by CS, the inequality is true as proved by @medicu. If $k<2n$, strictly, then letting $a\to 0$, we have
$$\frac{b}{nb+kc} \ge \frac{3}{k+n}-\frac{1}{n}$$
But this is not true because the right hand side is positive, while we can make the left hand side goes to zero when $c\to \infty$.
So the necessary and sufficient condition is $k\ge 2n$.
|
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|
How to find $f$ if $f(f(x))=\frac{x+1}{x+2}$ let $f:\mathbb R\to \mathbb R$,and such
$$f(f(x))=\dfrac{x+1}{x+2}$$
Find the $f(x)$
My try
I found $f(x)=\dfrac{1}{x+1}$
because when $f(x)=\dfrac{1}{x+1}$,then
$$f(f(x))=f\left(\dfrac{1}{x+1}\right)=\dfrac{1}{\dfrac{1}{x+1}+1}=\dfrac{x+1}{x+2}$$
so $f(x)=\dfrac{1}{x+1}$ such this condition,But $f(x)$ Have other form? Thank you
|
You can easily check that if we define
$$f_{A} = \frac{ax+b}{cx+d} \quad \text{for} \quad A = \begin{pmatrix}a & b \\ c & d \end{pmatrix}, $$
then $f_{A} \circ f_{B} = f_{AB}$. Thus any matrix $A$ satisfying
$$ A^{2} = k \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}, \quad \text{for some} \ k \neq 0 $$
gives rise to a solution. Mathematica yields two different solutions
$$ f(x) = \frac{1}{x+1} \quad \text{and} \quad f(x) = \frac{2x+1}{x+3}, $$
but I'm not sure if other solutions exist.
|
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|
Prove that $\frac{n^2+(-1)^nn+2}{7n^2+3}$ converges to $\frac{1}{7}$ I want to show that $\frac{n^2+(-1)^nn+2}{7n^2+3}$ converges to $\frac{1}{7}$ using the definition of convergence.
Skratch work:
I need $\mid\frac{n^2+(-1)^nn+2}{7n^2+3}-\frac{1}{7}\mid<\epsilon$.
So I take $\frac{n^2+(-1)^nn+2}{7n^2+3}-\frac{1}{7}=\frac{7(-1)^nn+11}{7(7n^2+3)}$.
For $n=1$, $\frac{7(-1)^nn+11}{7(7n^2+3)}=\frac{-4}{70}$
For $n=2$, $\frac{7(-1)^nn+11}{7(7n^2+3)}=\frac{24}{196}$
For $n=4$, $\frac{7(-1)^nn+11}{7(7n^2+3)}=\frac{39}{805}$.
So I think I can safely say that $\frac{7(-1)^nn+11}{7(7n^2+3)}<\frac{1}{n}$.
Now I can start the proof:
Let $\epsilon>0$. And choose $N>\frac{1}{\epsilon}$. Hence, $\frac{1}{N}<\epsilon$. So $n>N \implies \mid\frac{n^2+(-1)^nn+2}{7n^2+3}-\frac{1}{7}\mid < \frac{1}{n}< \frac{1}{N} < \epsilon$.
Does this proof look correct? I think I may have made a mistake in choosing $N$. Please let me know. Thanks
|
you can choose n in better way
you can use sup of a(n) or a bit more to find n>N ,which satisfy that relation
|
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|
How prove $\sum\frac{1}{2(x+1)^2+1}\ge\frac{1}{3}$ let $x,y,z>0$ and such $xyz=1$ show that
$$\dfrac{1}{2(x+1)^2+1}+\dfrac{1}{2(y+1)^2+1}+\dfrac{1}{2(z+1)^2+1}\ge\dfrac{1}{3}$$
My try: I will find a value of the $k$ such
$$\dfrac{1}{2(x+1)^2+1}\ge\dfrac{1}{9}+k\ln{x}$$
note $\ln{x}+\ln{y}+\ln{z}=0$,so
$$\sum_{cyc}\dfrac{1}{2(x+1)^2+1}\ge\dfrac{1}{3}+k(\ln{x}+\ln{y}+\ln{z})=\dfrac{1}{3}$$
so let
$$f(x)=\dfrac{1}{2(x+1)^2+1}-k\ln{x}-\dfrac{1}{9}$$
$$\Longrightarrow f'(x)=\dfrac{-4x-4}{(2x^2+4x+3)^2}-\dfrac{k}{x}$$
let $f'(1)=0\Longrightarrow k=-\dfrac{8}{81}$
so
$$f'(x)=\dfrac{-4x-4}{(2x^2+4x+3)^2}+\dfrac{8}{81x}=\dfrac{4(x-1)(8x^3+40x^2+15x-18)}{81x(2x^2+4x+3)^2}$$
so note when $1>x>\dfrac{1}{2}$ then
$$f'(x)=\dfrac{4(x-1)(8x^3+40x^2+15x-18)}{81x(2x^2+4x+3)^2}<0$$
$x>1,f'(x)>0$
so
$$f(x)\ge f(1)=0$$
so if $x,y,z>\dfrac{1}{2}$ we have prove done.
But for other case,How prove it? Thank you
|
Hint:
Use Tringnometric transformation such that
x=sinu, y=secu and z = cotu
For any value of u you can find x,y,z >0 such that xyz = 1 by definition.
For a value of u = pi/6, x=.5, y = 1.154 and z = 1.732
Further, I computed LHS, for u = pi/4 and its value is .3365
Just to double check, I computed LHS for u = pi/3 and its value is .34552
In both sample cases, LHS is >= (1/3).
Thanks
Satish
|
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|
Function range problem $(sin^2\theta +\sin\theta -1)/(sin^2\theta -\sin\theta+ 2)$ .
It is asked to find the range of this function.
Here i was assumed a variable
$z=sin\theta$ so that i get
$-1<=z<=1 $
and then i obtained a quadratic equation as :-
$(y-1)z^2 -(y+1)z + 2y +1 =0$ from here i need some help or if there's some other way for this problem that might also help. Thanks!
|
Let $f(x)=\frac{\sin^2 x+\sin x-1}{\sin^2 x-\sin x+2}$
Then $g(z):=f(\sin^{-1} z)=\frac{z^2+z-1}{z^2-z+2}$ $$g'(x)=\frac{(2x+1)(x^2-x+2)-(2x-1)(x^2+x-1)}{(x^2-x+2)^2}=\frac{-2x^2+6x+1}{(x^2-x+2)^2}$$
In order to find maxima and minima, set $g'(x)=0$: $-2x^2+6x+1=0$. Thus $x=\frac{3\pm\sqrt{11}}{2}$. Note that $\frac{3+\sqrt{11}}{2}>1$ and is not an element of the domain.
You should be able to finish the thought process.
|
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problem on solving equations with three variables Find all positive real numbers $x,y,z$ which satisfy the following equations simultaneously.
$x^3+y^3+z^3=x+y+z$
$x^2+y^2+z^2=xyz$
|
This is an answer that I am writing to give an alternative, brainless, path for these kind of problems, as an answer using means will soon appear (has already appeared).
Check the details as I have to go now.
First we use Newton's identities to write everything in terms of the elementary symmetric polynomials.
Denote
$e_1=x+y+z$, $p_1=x+y+z$
$e_2=xy+xz+yz$, $p_2=x^2+y^2+z^2$
$e_3=xyz$, $x^3+y^3+z^3$.
Then $e_2=(e_1^2-p_2)/2$
The given equations are $p_3=e_1$ and $p_2=e_3$. So, $e_2=(e_1^2-e_3)/2$
So $Q(t):=(t-x)(t-y)(t-z)=t^3-e_1t^2+\frac{(e_1^2-e_3)}{2}t-e_3$
We also have $p_3=e_1p_2-e_2p_1-3e_3=e_1e_3-\frac{(e_1^2-e_3)}{2}e_1-3e_3$. So, from the given equation, we get $e_1e_3-\frac{(e_1^2-e_3)}{2}e_1-3e_3=e_1$, from where we can solve for $e_3$ to get $e_3=\frac{e_1^3/2}{e_1-e_1/2-3}$.
Then $$Q(t)=t^3-e_1t^2+\frac{(e_1^2-\frac{e_1^3/2}{e_1-e_1/2-3})}{2}t-\frac{e_1^3/2}{e_1-e_1/2-3}$$
Or $$Q(t)=t^3-e_1t^2+\frac{(\frac{-3e_1^2}{e_1/2-3})}{2}t-\frac{e_1^3/2}{e_1/2-3}$$
Now, we try to impose that this polynomial has three positive real roots.
For example: From Descartes' rule of signs, if $e_1/2-3\geq0$ then $Q$ has only one positive root, but it must have 3 or them. So $e_1<3/2$.
We can now compute the discriminant and impose the condition that the roots are reals.
|
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Prove that $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1} $ Prove that $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1} $
Thanks in advance, my professor asked us to this a couple weeks ago, but I was enable to get to the right answer.
Good luck!
Here is what I got up to;
$\frac{(n+1)!}{(n-r)!(r+1)!} = \frac{(n)!}{(r)!(n-r)!} + \frac{(n)!}{(r+1)!(n-r-1)!} $
|
Even though it seems a little far-fetched I will use the Binomial Theorem. The definition of number $\binom{n}{r}$ has a reason to come to exist in the development of $(x + y)^n$. So I think the natural and instructive. For all $x,y\in\mathbb{R}$ we have,
\begin{array}{rrl}
\hspace{2cm}&( x+y)^{n+1}= & (x+y)\cdot ( x+y)^n,
\\
\Longleftrightarrow & \sum_{k=0}^{n+1}\binom{n+1}{k}x^{(n+1)-k}y^{k}= & (x+y)\cdot \sum_{k=0}^{n}\binom{n}{k}x^{n-k}y^{k},
\\
\Longleftrightarrow & \sum_{k=0}^{n+1}\binom{n+1}{k}x^{(n+1)-k}y^{k}= & \sum_{k=0}^{n}\binom{n}{k}x^{n-k}y^{k+1}+ \sum_{k=0}^{n}\binom{n}{k}x^{n-k+1}y^{k}.
\end{array}
Then for all $k=1,2,\ldots n$ we have
\begin{array}{rl}
\binom{n+1}{k}x^{(n+1)-k}y^{k}= & \binom{n}{(k-1)}x^{n-(k-1)}y^{(k-1)+1}+ \binom{n}{k}x^{n-k+1}y^{k}.
\end{array}
For $x=1$ and $y=1$,
\begin{array}{rl}
\binom{n+1}{k}= & \binom{n}{(k-1)}+ \binom{n}{k},\qquad k=1,2,\ldots n.
\end{array}
Setting $k = r +1$ then
\begin{array}{rl}
\binom{n+1}{r+1}= & \binom{n}{r}+ \binom{n}{r+1},\qquad r=0,1,2,\ldots n-1.
\end{array}
|
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|
Elementary proof that $\pi < \sqrt{5} + 1$ I wanted to show that
$$ \frac{\pi}{4\phi} < \frac{1}{2} $$
Where $\phi$ is the golden ratio. I have confirmed the results numerically, and by
simple algebra the inequality simplifies down to
$$
\pi < \sqrt{5} + 1
$$
This is a weaker relation than what was shown here. Prove that $\dfrac{\pi}{\phi^2}<\dfrac{6}5 $.
By squaring my inequality (valid since both sides are positive), and dividing by $6$ I obtain.
$$ \frac{\pi^2}{6} < 1 + \frac{\sqrt{5}}{3} $$
Where the left handside has a very neat series representation, alas the same does not hold for the right handside. However this is far from an elementary solution. Does someone have a relative simple proof for the equality? To be precise something that is not using advanced knowledge of series. =)
|
Just use that $\pi ^2 \leq \frac{61}{6}$ which comes from the series representation of $\frac{\pi^2}{6}$.
Then
$
\begin{align}
\pi^2 <5+1+2\cdot \sqrt5&\Leftarrow\\
\frac{61}{6} <6+2\cdot \sqrt5&\Leftrightarrow\\
\frac{5^2}{6}<2\sqrt 5&\Leftrightarrow\\
\frac{5^3}{36}<4\Leftrightarrow\\
125<144
\end{align}
$
$$$$
Indeed you have that $\frac{\pi^2}{6}= \sum _{k\geq1}\frac{1}{k^2}$
then
$
\begin{align}
\frac{\pi^2}{6}-1-\frac{1}{4}-\frac{1}{9}= \sum _{k\geq 4}\frac{1}{k^2} &\leq \\
\sum _{k\geq 4}\frac{1}{(k-1)k} &=\\= \sum _{k\geq4}\left ( \frac{1}{k-1} - \frac{1}{k} \right)&=\frac{1}{3}\\
\end{align}
$
|
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|
Find a polynomial $p$ of degree $3$ if its value in $4$ points is given Find a polynomial $p$ of degree $3$ such that
\begin{align*}
p(−4) &= −142, \\
p(1) &= −2, \\
p(−5) &= −242, \\
p(4) &= 10.
\end{align*}
Then use your polynomial to approximate $p(2)$.
\begin{align*}
p(x) &= ?, \\
p(2) &= ?.
\end{align*}
I can't find this sort of question in the textbook so I'm having trouble.
Please teach me how to solve this question and, perhaps, fill in point 2?
Thank you.
|
Hint: Find constants $a$, $b$, $c$ and $d$ such that the polynomial
$$a(x-1)(x+5)(x-4)+b(x+4)(x+5)(x-4)+c(x+4)(x-1)(x-4)+d(x+4)(x-1)(x-5)$$
takes on the desired values.
For example, to compute $a$, we want $a(-4-1)(-4+5)(-4-4)=-142$.
|
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|
Differentiation of $2\arccos \left(\sqrt{\frac{a-x}{a-b}}\right)$ Okay so the question is:
Show that the function
$$2\arccos \left(\sqrt{\dfrac{a-x}{a-b}}\right)$$
is equal to
$$\frac{1}{\sqrt{(a-x)(x-b)}} .$$
I started by changing the arccosine into inverse cosine, then attempted to apply chain rule but I didn't get very far.
Then I tried substituting the derivative for arccosine in and then applying chain rule. Is there another method besides chain rule I should use? Any help is appreciated.
|
$$\dfrac{d}{du} 2\arccos u = - 2\dfrac{1}{\sqrt{1 - u^2}} ~du$$
See the Proof Wiki for a proof of this.
In this problem, we have, $u = \sqrt{\dfrac{a-x}{a-b}}$, and we need to find $dx$, so we have:
$$ \dfrac{d}{dx} \left(\sqrt{\dfrac{a-x}{a-b}} \right) = -\dfrac{\sqrt{\dfrac{a-x}{a-b}}}{2 (a-x)} = -\dfrac{1}{2 \sqrt{(a - b)(a - x)}}$$
So, lets put these two together.
$\dfrac{d}{du}\left(2 \arccos u \right) =-2 \dfrac{1}{\sqrt{1 - u^2}} ~du = -\dfrac{2}{\sqrt{1 - \left(\sqrt{\dfrac{a-x}{a-b}}\right)^2}} \left(-\dfrac{1}{2 \sqrt{(a - b)(a - x)}} \right)$
We can reduce this to:
$$\dfrac{d}{dx} \left(2 \arccos \left(\sqrt{\dfrac{a-x}{a-b}}\right)\right)=\dfrac{1}{\sqrt{(a-x)(x-b)}}$$
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"timestamp": "2023-03-29T00:00:00",
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|
Arithmetic sequence of tangent values I have two angles $A_1, A_2 > 0$, and $A_1+A_2 < \pi$, is it possible to find an $A_0$ such that
$$\tan(A_0),\ \tan(A_0+A_1),\ \tan(A_0+A_1+A_2)$$
forms an arithmetic sequence on the same continuous range of tangent?
I have been looking for a formula for sum of tangents (not tangent of sum), but I have not been successful so far.
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(Resolving my own question)
Let the first term $\tan(A_0) = a$, term difference be $d$, then
$$\begin{align*}
\cot A_1 &= \cot(A_0+A_1-A_0)\\
&= \frac{1+\tan(A_0+A_1)\tan A_0}{\tan(A_0+A_1)-\tan A_0}\\
&= \frac{1+(a+d)a}{d}\\
\cot A_2 &= \cot(A_0+A_1+A_2-A_0-A_1)\\
&= \frac{1+\tan(A_0+A_1+A_2)\tan(A_0+A_1)}{\tan(A_0+A_1+A_2)-\tan(A_0+A_1)}\\
&= \frac{1+(a+2d)(a+d)}{d}\\
\end{align*}$$
then
$$\begin{align*}
\cot A_2 - \cot A_1 &= \frac{1+(a+2d)(a+d)}d - \frac{1+(a+d)a}d\\
&= \frac{2d(a+d)}d\\
&= 2(a+d)\\
&= 2\tan(A_0+A_1)
\end{align*}$$
A value of $A_0$ can be taken as
$$A_0 = \arctan \left(\frac{\cot A_2 - \cot A_1}2\right)-A_1$$
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"language": "en",
"url": "https://math.stackexchange.com/questions/562028",
"timestamp": "2023-03-29T00:00:00",
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|
How to prove this inequality $(a^2+bc^4)(b^2+ca^4)(c^2+ab^4) \leq 64$? Question:
If $a,b,c$ are nonnegative real numbers such that $a+b+c=3,$ then
$$(a^2+bc^4)(b^2+ca^4)(c^2+ab^4) \leq 64$$
My try: I found the equality holds only if $(a,b,c)=(2,0,1)$ or all of its permutations.
But I can't prove this inequality it. I would appreciate very much a proof.
This problem comes from:http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=562119
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→ → The three dimensional problem can be simplified to a two dimensional problem by introducing
(again
and again)
suitable triangle coordinates:
$$
\left[ \begin{array}{c} a \\ b \\ c \end{array} \right] =
\left[ \begin{array}{c} 3 \\ 0 \\ 0 \end{array} \right] +
\left[ \begin{array}{c} -3 \\ 3 \\ 0 \end{array} \right] x +
\left[ \begin{array}{c} -3 \\ 0 \\ 3 \end{array} \right] y
$$
Then the equation $(a^2+bc^4)(b^2+ca^4)(c^2+ab^4) \leq 64$ does not so much "simplify", but anyway becomes an equation in two
variables (2-D). And the equation $\;a + b + c = 3\;$ corresponds with a normed 2-D triangle, with vertices $(0,0),(1,0),(0,1)$ .
The insides of both can easily be visualized, as has been done in the above picture in the middle: $\color{red}{red}$ for $\;a + b + c = 3\;$
and $\color{green}{green}$ for $\;(a^2+bc^4)(b^2+ca^4)(c^2+ab^4) \leq 64\;$ .The transformed inequality is:
$$\left[ 3^2\left( 1-x-y \right)^{2} + 3^5\,x{y}^{4} \right]
\left[ 3^2\,{x}^{2} + 3^5\,y \left( 1-x-y \right)^{4} \right]
\left[ 3^2\,{y}^{2} + 3^5\, \left( 1-x-y \right){x}^{4} \right] \le 64
$$
It is seen in the same picture that the edge $\;y=1-x\;$ of the triangle maybe is tangent
to the curve $(a^2+bc^4)(b^2+ca^4)(c^2+ab^4) = 64$ . Indeed, if we substitute $y=1-x$ into (the transformation of) that equation and simplify, then we get:
$$
3^9\, x^3 (1-x)^6 - 64 = 0
$$
The same sort of equation is found with the substitutions $\,x=0\,$ or $\,y=0\,$, for the other two edges.
And, as has been found by others, there is only one solution of that equation, within the specified range, namely $x=1/3$,
corresponding with $y=2/3$ and hence $(a,b,c) = (0,1,2)$ . And of course any cyclic permutation of this, due to symmetry.
The rest of the (red) triangle $\;a + b + c = 3\;$ is well within the (green) area of $\;(a^2+bc^4)(b^2+ca^4)(c^2+ab^4) \leq 64$ . Which can be shown by plotting the triangle first: then it becomes absorbed by the green area (see picture on the right).
Analytically, the proof is completed by considering the function $\;f(x) = 3^9\, x^3 (1-x)^6$ .Its extreme values are found for $\;f'(x) = x^2(1-x)^5(3-9x)=0$ , giving $\;x = \{0,1,1/3\}$ , with the maximum $\;f(1/3)=64$ .The picture on the right shows the the inequality as observed in the plane $\;a + b + c = 3\;$ of the triangle in 3-D (picture on the left). Mind the symmetries.
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|
Inequality: $ab^2+bc^2+ca^2 \le 4$, when $a+b+c=3$. Let $a,b,c $ are non-negative real numbers, and $a+b+c=3$.
How to prove inequality
$$
ab^2+bc^2+ca^2\le 4.\tag{*}
$$
In other words, if $a,b,c$ are non-negative real numbers, then how to prove inequality
$$
27(ab^2+bc^2+ca^2)\le 4(a+b+c)^3.\tag {**}
$$
$\color{gray}{\mbox{(Without using "universal" Lagrange multipliers method).}}$
Thanks!
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I prove this stronger inequality,
With loss of out,let $a=\min{(a,b,c)}$
\begin{align*}&4(a+b+c)^3-27(a^2b+b^2c+c^2a+abc)\\
&=9a(a^2+b^2+c^2-ab-bc-ac)+(4b+c-5a)(a+b-2c)^2\ge 0
\end{align*}
other nice methods:
with out loss of let $b=mid{(a,b,c)}$,then $(b-a)(b-c)\le 0$,so
$$a^2b+b^2c+c^2a+abc\le b(a^2+c^2+ac)+abc=b(a+c)^2=2b(a+c)(a+c)/2\le\dfrac{4}{27}(a+b+c)^3$$
|
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"url": "https://math.stackexchange.com/questions/566768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Someone who is calculating $43434343^2$. The answer is $18865ab151841649$, where the two digits $a$ and $b$ were lost. So I used congruence $\bmod 9$ and $\bmod 11$. First, $43434343^2 \pmod{11}$ is $5$ and I did the answer $18865ab15184169 \pmod{11}$ and got $-b+a+3 \pmod{11}$. I also did $\bmod 9$ and got $a+b+67 \pmod 9$. I am confused as to what to do next.
|
$\let\cong\equiv$
So you know that
$$43434343^2 \cong 5 \pmod{11}$$
and
$$43434343^2 \cong 1 \pmod{9}$$
and you found that
$$\begin{aligned}
18865ab15184169 &\cong a-b+3 \pmod{11}, \\
18865ab15184169 &\cong a+b+67 \pmod{9}.
\end{aligned}$$
You want $18865ab15184169$ to be equal to $43434343^2$, so the congruence classes of these two numbers in both $(\bmod 9)$ and $(\bmod 11)$ also need to match:
$$\begin{aligned}
a-b+3 &\cong 5 \pmod{11}, \\
a+b+67 &\cong 1 \pmod{9}.
\end{aligned}$$
This can be simplified further as
$$\begin{aligned}
a-b &\cong 2 \pmod{11}, \\
a+b &\cong 6 \pmod{9}.
\end{aligned}$$
At the same time, both $a$ and $b$ are just decimal digits and thus constrained to $\{0,1,2,3,4,5,6,7,8,9\}$. This leaves not many possibilities how to satisfy the first congruence: either $a-b=2$ (note that this is an actual equality, not a congruence) or $a-b=-9$, which could only mean $a=0$, $b=9$.
We discard the latter quickly by finding that $0+9 \not\cong 6 \pmod{9}$. The other possibility would mean $a = b+2$, whatever $b$ is, so plug that into the remaining congruence:
$$\begin{aligned}
(b+2)+b &\cong 6 \pmod{9} \\
2b + 2 &\cong 6 \pmod{9} \\
2b &\cong 4 \pmod{9}
\end{aligned}$$
Out of all digits, only $b=2$ can satisfy this. Then use $a = b+2$ again to find $a$ and you're done.
Another way:
When at the point where you have the congruences
$$\begin{aligned}
a-b &\cong 2 \pmod{11}, \\
a+b &\cong 6 \pmod{9},
\end{aligned}$$
see what we can find for the sum of the two LHS's and for their difference:
$$\begin{aligned}
2a &\cong 2+6 \pmod{11}, \\
2b &\cong 6-2 \pmod{9}.
\end{aligned}$$
The solution $a=4$, $b=2$ follows immediately.
|
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|
Find $\lim _{n\rightarrow \infty }\frac {na^{2}_{n}+1}{\sum ^{n}_{k=1}\left( 1+2+3+\ldots +k\right) }$ Define $\left\{ a_{n}\right\} $ is an arithmetic sequence that all terms are positive integers. If $a_{10}-a_{1}=225$, find $$\lim _{n\rightarrow \infty }\dfrac {na^{2}_{n}+1}{\sum\limits^{n}_{k=1}\left( 1+2+3+\ldots +k\right) }$$
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HINT:
If $d$ is the common difference, $a_n=a_1+(n-1)d\implies a_n-a_1=(n-1)d,$
putting $n=10,225=9d\implies d=25\implies a_n=a_1+(n-1)25=25n+a_1-25$
$$na_n^2+1=n(25n+a_1-25)^2+1=625n^3+50n^2(a_1-25)+n(a_1-25)^2+1=625n^3+O(n^2)$$
As $1+2+\cdots+k-1+k=\frac{k^2+k}2$
$$\sum_{1\le k\le n}(1+2+\cdots+k-1+k)=\frac{\sum_{1\le k\le n}(k^2+k)}2=\frac{\sum_{1\le k\le n}k^2+\sum_{1\le k\le n}k}2$$
$$=\frac{\frac{n(n+1)(2n+1)}6+\frac{n(n+1)}2}2=\frac{n^3}6+O(n^2)$$
|
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|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.