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Minimize $\frac{2}{1-a}+\frac{75}{10-b}$
Let $a,b>0$ and satisfy $a^2+\dfrac{b^2}{45}=1$. Find the minimum
value of $\dfrac{2}{1-a}+\dfrac{75}{10-b}.$
WA gives the result that $\min\left(\dfrac{2}{1-a}+\dfrac{75}{10-b}\right)=21$ with $a=\dfrac{2}{3},b=5$.
Consider making a transformation like this
$$\dfrac{2}{1-a}+\dfrac{75}{10-b}=\frac{2\cdot (ma)^2}{ma\cdot ma\cdot (1-a)}+\frac{75\cdot(nb)^2}{nb\cdot nb\cdot(10-b)}.$$
If $2m=2n=1$, $ma=1-a$, and $ nb=10-b$ can all hold, we can apply AM-GM inequality as follows
$$\frac{2\cdot (ma)^2}{ma\cdot ma\cdot (1-a)}+\frac{75\cdot(nb)^2}{nb\cdot nb\cdot(10-b)}\geq \frac{2\cdot (ma)^2}{\left(\frac{ma
+ma+(1-a)}{3}\right)^3}+\frac{75\cdot(nb)^2}{\left(\frac{nb+nb+(10-b)}{3}\right)^3}=\cdots$$
But this is invalid, since $n=\frac{1}{2},nb=10-b$ can not satisfy $b=5$.
How to solve it?
|
The Tangent Line method helps.
Let $a=\frac{2}{3}x$ and $b=5y$.
Thus, $$4x^2+5y^2=9$$ and
$$\frac{2}{1-a}+\frac{75}{10-b}-21=3\left(\frac{2}{3-2x}+\frac{5}{2-y}-7\right)=$$
$$=3\left(\frac{2}{3-2x}-2-2(x^2-1)+\frac{5}{2-y}-5-\frac{5}{2}(y^2-1)\right)=$$
$$=\frac{6(x-1)^2(2x+1)}{3-2x}+\frac{15y(y-1)^2}{2(2-y)}\geq0.$$
The equality occurs for $x=y=1,$ which says that $21$ is a minimal value.
There is the following reasoning.
Let $$\frac{2}{1-a}+\frac{75}{10-b}=k.$$
Thus, $$b=10-\frac{75}{k-\frac{2}{1-a}}$$ and find all values of $k$ for which the equation
$$a^2+\frac{\left(10-\frac{75}{k-\frac{2}{1-a}}\right)^2}{45}=1$$ has solutions $0<a<1.$
|
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Find the standard matrix of the linear transformation $T$ A linear transformation $T:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ is defined by $$T((2,11))=(8,5)$$ and $$T((1,5))=(5,3)$$ Find the matrix for this linear transformation.
I did this in a very clunky way - since $T$ is linear, I can write that
$$T((2, 11))=T(2e_1+11e_2)=2T(e_1)+11T(e_2)=(8,5)$$
and
$$T((1,5))=T(e_1+5e_2)=T(e_1)+5T(e_2)=(5,3)$$
I solved this system of two equations for $T(e_1)$ and $T(e_2)$ to get $$T(e_1)=(15,8)$$ $$T(e_2)=(-2,-1)$$
Thus, we get
$$
\begin{pmatrix}
15 & -2 \\
8 & -1 \\
\end{pmatrix}
$$
However, is there a simpler way to get the answer?
|
Here is a bit of an elaboration on top of @G Cab's answer.
Consider the linear transformation $T_1$, which maps components of the identity matrix to vectors specified below.
$$T_1 \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 11 \end{pmatrix}, \, T_1 \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 5 \end{pmatrix}$$
Next, we define a separate linear transformation that maps components of the identity matrix to another set of vectors.
$$T_2 \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 8 \\ 5 \end{pmatrix}, \, T_2 \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 5 \\ 3 \end{pmatrix}$$
Then, there should be some linear transformation $T$ that represents the relationship between these two linear transformations. In other words,
$$T\left(T_1 \begin{pmatrix} 1 \\ 0 \end{pmatrix} \right) = \begin{pmatrix} 8 \\ 5 \end{pmatrix}$$
and
$$T\left(T_1 \begin{pmatrix} 0 \\ 1 \end{pmatrix} \right) = \begin{pmatrix} 5 \\ 3 \end{pmatrix}$$
We can package the basis vectors into a matrix to formulate a more compact expression:
$$T\left(T_1 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right) = \begin{pmatrix} 8 & 5 \\ 5 & 3 \end{pmatrix}$$
Because we intentionally chose very easy basis vectors, namely the components of the identity matrix, we know what $T_1$ and $T_2$ are:
$$T_1 = \begin{pmatrix} 2 & 1 \\ 11 & 5 \end{pmatrix}, \, T_2 = \begin{pmatrix} 8 & 5 \\ 5 & 3 \end{pmatrix}$$
Therefore, the equation above simplifies to the following:
$$T \begin{pmatrix} 2 & 1 \\ 11 & 5 \end{pmatrix} = \begin{pmatrix} 8 & 5 \\ 5 & 3 \end{pmatrix}$$
Can you find the inverse of $T_1$ and go from there?
|
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|
Prove that $\int_1^a \frac{T_n(x) T_n(x/a)}{\sqrt{a^2 - x^2} \sqrt{x^2 - 1^2}} \frac{a}{x} \mathrm{d}x = \frac{\pi}{2}$ In the paper, Representation of a Function by Its Line Integrals, with Some Radiological Applications, A. M. Cormack, Journal of Applied Physics 34, 2722 (1963), an integral identity is expressed which can be reduced to:
$$I_n(a) = \int_1^a \frac{T_n(x) T_n(x/a)}{\sqrt{a^2 - x^2} \sqrt{x^2 - 1^2}} \frac{a}{x} \mathrm{d}x = \frac{\pi}{2}$$
where $T_n(x)$ is the $n^\text{th}$ order Chebyshev polynomial of the first kind.
I'm not quite sure where this result comes from, and was looking to see how it could be derived. The paper notes that it can be shown that
$$I_{n+1} = I_{n-1}, \quad I_0 = I_1 = \frac{\pi}{2}$$
from which then it is apparent that $I_n(a) = \frac{\pi}{2}$
However, I'm not sure where the $I_{n+1} = I_{n-1}$ comes from.
Attempting to substitute the recurrence relationship
$$T_n(x) = \frac{T_{n-1}(x) + T_{n+1}(x)}{2x}$$
seemed like a good start, but ends up converting the integral to a form that's different from $I_{n-1}(a)$ and $I_{n+1}(a)$, and also generates some unwanted cross terms.
Could anyone suggest a push in the right direction?
|
Using the trigonometric definition of the Chebyshev polynomial of the first kind we can write:
$$\small I_{n}(a)=\int_1^a \frac{T_n(x) T_n\left(\frac{x}{a}\right)}{\sqrt{x^2 - 1}\sqrt{a^2 - x^2}}\frac{a}{x}dx=\int_1^a \frac{\cosh(n\operatorname{arccosh} x) \cos\left(n\arccos \left(\frac{x}{a}\right)\right)}{\sqrt{x^2 - 1}\sqrt{1 - \frac{x^2}{a^2}}} \frac{dx}{x}$$
$$I_0(a)=\int_1^a \frac{1}{\sqrt{x^2 - 1}\sqrt{1 - \frac{x^2}{a^2}}} \frac{dx}{x}\overset{x\to \frac{a}{x}}=\int_1^a \frac{\frac{x}{a}\cdot x}{\sqrt{1-\frac{x^2}{a^2}}\sqrt{x^2-1}}\frac{dx}{x}=I_1(a)$$
$$\overset{1-\frac{x^2}{a^2}=t^2}=\int_0^{\sqrt{1-\frac{1}{a^2}}}\frac{1}{\sqrt{1-\frac{1}{a^2}-t^2}}dt=\arcsin\left(\frac{t}{\sqrt{1-\frac{1}{a^2}}}\right)\bigg|_0^{\sqrt{1-\frac{1}{a^2}}}=\frac{\pi}{2}$$
So as the paper mentions, we only need to show that $I_{n-1}(a)=I_{n+1}(a)$ to obtain $I_{n}(a)=\frac{\pi}{2}$.
$$\small T_{n-1}\left(x\right)=\cosh((n-1)\operatorname{arccosh} x)=\color{red}{x\cosh(n\operatorname{arccosh} x)}-\color{blue}{\sqrt{x^2-1}\sinh(n\operatorname{arccosh} x)}$$
$$\small T_{n-1}\left(\frac{x}{a}\right)=\cos\left((n-1)\arccos \left(\frac{x}{a}\right)\right)=\color{blue}{\frac{x}{a}\cos\left(n\arccos \left(\frac{x}{a}\right)\right)}+\color{red}{\sqrt{1-\frac{x^2}{a^2}}\sin\left(n\arccos \left(\frac{x}{a}\right)\right)}$$
$$\small T_{n+1}\left(x\right)=\cosh((n+1)\operatorname{arccosh} x)=\color{red}{x\cosh(n\operatorname{arccosh} x)}+\color{blue}{\sqrt{x^2-1}\sinh(n\operatorname{arccosh} x)}$$
$$\small T_{n+1}\left(\frac{x}{a}\right)=\cos\left((n+1)\arccos \left(\frac{x}{a}\right)\right)=\color{blue}{\frac{x}{a}\cos\left(n\arccos \left(\frac{x}{a}\right)\right)}-\color{red}{\sqrt{1-\frac{x^2}{a^2}}\sin\left(n\arccos \left(\frac{x}{a}\right)\right)}$$
When we rewrite the numerator of $I_{n-1}(a)-I_{n+1}(a)$ (we are trying to show that this vanishes) using the above identities, only the terms of the same color remains (twice), giving:
$$\frac12\left(T_{n-1}(x)T_{n-1}\left(\frac{x}{a}\right)-T_{n+1}(x)T_{n+1}\left(\frac{x}{a}\right)\right)$$
$$\small = \color{red}{x\cosh(n\operatorname{arccosh} x)\sqrt{1-\frac{x^2}{a^2}}\sin\left(n\arccos \left(\frac{x}{a}\right)\right)}-\color{blue}{\sqrt{x^2-1}\sinh(n\operatorname{arccosh} x)\frac{x}{a}\cos\left(n\arccos \left(\frac{x}{a}\right)\right)}$$
So $ I_{n-1}(a)-I_{n+1}(a)$ takes the form of:
$$\small 2\color{chocolate}{\int_1^a \frac{\cosh(n\operatorname{arccosh} x)\sin\left(n\arccos \left(\frac{x}{a}\right)\right)}{\sqrt{x^2-1}}dx}-\frac2a\color{green}{\int_1^a\frac{\sinh(n\operatorname{arccosh} x)\cos\left(n\arccos \left(\frac{x}{a}\right)\right)}{\sqrt{1-\frac{x^2}{a^2}}}dx}$$
$$\small \color{chocolate}{\int_1^a \left(\frac{\sinh(n\operatorname{arccosh} x)}{n}\right)'\sin\left(n\arccos \left(\frac{x}{a}\right)\right)dx}\overset{IBP}=\frac1a\color{green}{\int_1^a\frac{\sinh(n\operatorname{arccosh} x)\cos\left(n\arccos \left(\frac{x}{a}\right)\right)}{\sqrt{1-\frac{x^2}{a^2}}}dx}$$
$$\Rightarrow I_{n-1}(a)-I_{n+1}(a)=0\Rightarrow I_{n-1}(a)=I_{n+1}(a)$$
|
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Proof by induction:$\frac{3}{5}\cdot\frac{7}{9}\cdot\frac{11}{13}\cdots\frac{4n-1}{4n+1}<\sqrt{\frac{3}{4n+3}}$ In the very beginning I'm going to refer to similar posts with provided answers:
Induction Inequality Proof with Product Operator $\prod_{k=1}^{n} \frac{(2k-1)}{2k} \leq \frac{1}{\sqrt{3k+1}}$ (answered by Özgür Cem Birler)
Prove that $\prod\limits_{i=1}^n \frac{2i-1}{2i} \leq \frac{1}{\sqrt{3n+1}}$ for all $n \in \Bbb Z_+$
I examined the solutions and tried to apply the methods used there to make sure whether I understand it or not. I'm concerned about the step of induction.
A task from an earlier exam at my university:
Prove by induction:
$$\frac{3}{5}\cdot\frac{7}{9}\cdot\frac{11}{13}\cdots\frac{4n-1}{4n+1}<\sqrt{\frac{3}{4n+3}}$$
Attempt:
rewritten:
$$\prod_{i=1}^n\frac{4i-1}{4i+1}<\sqrt{\frac{3}{4n+3}}$$
$(1)$ base case: $\tau(1)$
$$\frac{3}{5}=\sqrt{\frac{4}{7}}<\sqrt{\frac{3}{7}}$$
$(2)$ assumption:
Let:$$\frac{3}{5}\cdot\frac{7}{9}\cdot\frac{11}{13}\cdots\frac{4n-1}{4n+1}<\sqrt{\frac{3}{4n+3}}$$
hold for some $n\in\mathbb N$
$(3)$ step: $\tau(n+1)$
$$\frac{4n+3}{4n+5}\cdot\prod_{i=1}^n\frac{4i-1}{4i+1}<\frac{4n+3}{4n+5}\cdot\sqrt{\frac{3}{4n+3}}=\frac{\sqrt{3(4n+3)}}{4n+5}<\sqrt{\frac{3}{4n+7}}$$
$$\frac{12n+9}{16n^2+40n+25}<\frac{3}{4n+7}\iff\frac{48n^2+120n+63-48n^2-120n-75}{\underbrace{(4n+5)^2(4n+7)}_{>0}}<0\iff-\frac{12}{(4n+5)^2(4n+7)}<0$$
Is this combined legitimate?
|
Your calculations are correct.
But I thought it might be helpful also to mention another nice trick to handle such products:
*
*Set $A = \frac 35 \cdot \frac 79 \cdots \frac{4n-1}{4n+1}$
*Let $B = \frac 57 \cdot \frac 9{11} \cdots \frac{4n+1}{4n+3}$
It follows immediately
$$A < B \Rightarrow A^2 < AB = \frac 3{4n+3}$$
Done.
|
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Find all values of $m$ such that the equation $ mx^4 + x^3 + (8m - 1)x^2 + 4x + 16m = 0$ has nonnegative roots.
Find all values of $m$ such that the equation $$\large mx^4 + x^3 + (8m - 1)x^2 + 4x + 16m = 0$$ has nonnegative roots.
For an equation to have nonnegative roots, it mustn't only have negative roots.
Let $y = x^2 - x + 4$ $(y > 0)$, we have that $$mx^4 + x^3 + (8m - 1)x^2 + 4x + 16m = mx^2 + (2m - 1)xy + my^2$$
For the equation $$mx^2 + (2m - 1)xy + my^2 = 0$$ to have only negative roots, it must be satisfied that $\left\{ \begin{align} [(2m - 1)y]^2 - 4m^2y^2 \ge 0\\ (1 - 2m)y < 0\\ my^2 > 0 \end{align} \right.$ $\implies \left\{ \begin{align} (1 - 4m)y^2 \ge 0\\ 1 - 2m < 0\\ m > 0 \end{align} \right.$ $\implies \left\{ \begin{align} 1 - 4m \ge 0\\ m > \dfrac{1}{2}\\ m > 0 \end{align} \right.$ $\iff \left\{ \begin{align} m \le \frac{1}{4}\\ m > \dfrac{1}{2} \end{align} \right. \implies m \in \varnothing$.
Thus for $\forall m \in \mathbb R$, the equation $$mx^4 + x^3 + (8m - 1)x^2 + 4x + 16m = 0$$ has nonnegative roots.
Is the above solution correct? And should my attempt was inaccurate and yours is (hopefully) helpful, please post an answer.
|
Express $m$ as a function of $x\ge 0$,
$$m(x) = - \frac{x(x^2-x+4)}{(x^2+4)^2}\le0\tag 1$$
It is equivalent to find the lower bound of $m(x)$ for $x>0$. Set $m'(x) = 0$ to get
$$x^4-2x^3+8x-16=(x-2)(x+2)(x^2-2x+4) = 0$$
which shows that the minimum is at $x=2$. Plug it into $(1)$ to get $m(2) = -\dfrac3{16}$. Thus, below are all values of $m$ for non-negative roots of $x$,
$$-\frac3{16} \le m \le 0$$
|
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Sum of magnitudes of coefficients of polynomial $(x-1)(x-2)(x-3)\cdots(x-(n-1))$ The title says most of it. I have found that the sum of the coefficients of the polynomial$(x-1)(x-2)(x-3)\cdots(x-(n-1))$ yields $n!$.
For example, the coefficients of the polynomial $(x-1)(x-2)$ sum to $1+3+2=6=3!$ and similarly the coefficients of $(x-1)(x-2)(x-3)$ sum to $1+6+11+6=24=4!$.
My question, then is how might I prove this more generally. I have the feeling that induction might be an optimal way to go about this, but am unsure of the specifics. Any help is appreciated!
|
Since all roots of the polynomial are positive, Descartes' Rule of signs guarantees that the signs of the coefficients alternate, for example with $n=4$:
$(x-1)(x-2)(x-3)=x^3\color{purple}{-}6x^2\color{purple}{+}11x\color{purple}{-}6$
Because of this sign pattern the absolute values of the coefficients are recovered by putting $x=-1$ so that the powers of $x$ correlate with the alternating signs of the coefficients:
$(-1-1)(-1-2)(-1-3)=(-1)^3-6(-1)^2+11(-1)11x-6=-(1+6+11+6)$
and upon taking absolute values
$(1+1)(1+2)(1+3)=1+6+11+6$
where for the general case the product on the left is $n!$ and the sum on the right contains the desired coefficient magnitudes.
|
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Show that $a^2$ cannot be congruent to 2 or 3 mod 5 for any integer a) Show that $a^2$ cannot be congruent to $2$ or $3 \bmod 5$ for any integer.
$a^2≡2,3 \pmod5$
$a≡0,±1,±2\pmod5 )⟹a^2≡0,1,4$.
But $2,3$ are not congruent to $0,1,4\pmod5$.
I am not sure if I did it right, please check it for me
(b) Show that if $5\mid x^2+y^2+z^2$ then either exactly one of, or each of $x, y, z$ are a multiple of $5$.
I really can not figure it out :(
|
For part $(a)$, it suffices to enumerate among all residue classes modulo $5$. So we get:
$$
0^{2} \equiv 0 \pmod{5} \\
1^{2} \equiv 1 \pmod{5} \\
2^{2} \equiv 4 \pmod{5} \\
3^{2} \equiv 4 \pmod{5} \\
4^{2} \equiv 1 \pmod{5} \\
$$
Hence, $a^{2} \not \equiv 2,3 \pmod{5}$, for any integer $a$, since any integer can be written as $5k + t : t \in \{0,1,2,3,4\}$.
For part $(b)$, suppose that exactly two of $x,y,z$ are multiples of $5$. Without loss generality, assume that $x,y$ are multiples of $5$ and $z$ is not. Then $z = 5k + t : t \ne 0$ and $x = 5x_{1}$ and $y = 5y_{1}$ for $x_{1},y_{1} \in \mathbb{Z}$. Then:
$$
x^{2} + y^{2} + z^{2} \equiv t^{2} \not \equiv 0 \pmod{5}
$$
so $5$ cannot divide $x^{2} + y^{2} + z^{2}$. Suppose that none of $x,y,z$ are multiples of $5$. Then:
$$
x^{2} + y^{2} + z^{2} \equiv a^{2} + b^{2} + c^{2} \not \equiv 0 \pmod{5}
$$
Since $a,b,c \in \{1,4\}$.
We proved the claim using the contrapositive.
|
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What is $4{(\frac ab )}^2 + \frac ba - a + 1$
What is $4{(\frac ab )}^2 + \frac ba - a + 1$ when $a, b, c$ are natural number and form geometry sequence with $b/a$ is an integer.
the average of a,b,c is b + 1
$a=a, b = ar, c=ar^2 = b^2 /a $ when $ r = b/a$. Also $c/b=b/a \rightarrow ac = b^2$
$a+b+c = 3b + 3 \rightarrow a+b+ b^2/a = 3b + 3$
|
So $r=b/a\in \mathbb{N}$ and thus $b=ar$ and $c=ar^2$. Now we have $$a-2ar+ar^2=3$$
so $$a(1-r)^2 = 3\implies a= 3\;\;\wedge\;\; 1-r = \pm 1\implies r= 2$$
So $b=6$ and now you can finish.
|
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Find the value of $I=\int_2^3{\frac{dx}{\sqrt{x^3-3x^2+5}}dx}$ Let $I=\int_2^3{\frac{dx}{\sqrt{x^3-3x^2+5}}dx}$ .
Find the value of $[I+\sqrt{3}]$ (where [.] represents the greatest integer function)
I have no idea of calculating such type of problem where integration is complex.
|
$x^3-3x^2+3x-1\ge x^3-3x²+5 \ge 1$. (for x>2)
$\frac{1}{\sqrt{(x-1)^3}} \le \frac{1}{\sqrt{x^3-3x^+5}} \le 1$ (for x>2)
$\int_2^3 \frac{1}{\sqrt{(x-1)^3}}dx \le \int_2^3\frac{1}{\sqrt{x^3-3x^+5}}dx \le \int_2^3 1dx$
$2-\sqrt{2} \le I \le 1$
$2-\sqrt{2} + \sqrt{3} \le I+\sqrt{3} \le 1+\sqrt{3}$
$2 \lt I+\sqrt{3} \lt 3$
Hence answer is 2, i don't think the integral have any nice closed form.
|
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What is $a$ in $\lim_{t \to 0} \left(\frac{a}{t^2} - \frac{\sin 6t}{t^3 \cos^2 3t}\right) = -18$
What is $a$ in $\lim_{t \to 0} \left(\frac{a}{t^2} - \frac{\sin 6t}{t^3 \cos^2 3t}\right) = -18$
We have to make this function defined, by change the cosine into sine or tan. So,
$$\lim_{t \to 0} \left(\frac{a}{t^2} - \frac{2\sin 3t \cos 3t}{t^3 \cos^2 3t}\right) = \lim_{t \to 0} \left(\frac{a}{t^2} - \frac{6}{t^2}\right) = -18$$
But I am confused what should I do next.
|
There's an error in your computation, and you should find first a Laurent expansion:
$$\frac{a}{t^2} - \frac{2\sin 3t \cos 3t}{t^3 \cos^2 3t}=\frac 1{t^2}\biggl(at-2\frac{\tan 3t}t\biggr)$$
Now the Taylor expansion of $\tan x$ at order $3$ is $\;\tan x=x+\dfrac{x^3}{3}+o\bigl(x^3\bigr)$, so
$$
\frac{a}{t^2} - \frac{2\sin 3t \cos 3t}{t^3 \cos^2 3t}=\frac 1{t^2}\biggl(a-\frac 2t\Bigl(3t+9t^3+ o\bigl(t^3\bigr)\Bigr)\biggr) =\frac{a-6}{t^2} -18+o(1).
$$
We conclude there's a limit at $0$ when $a=6$, and in this case the limit is indeed $-18$.
|
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Integer solutions to the product of four consecutive integers $ \bullet \textbf{Question} $
The product of four consecutive integers $ x, x + 1, x + 2, x + 3 $ can be written as the product of two consecutive integers, find all integer solutions for $ x $.
$ \bullet \textbf{Rephrasing} $
I decided to name the other integer as $ y $ so that this equation is a plot on a graph,
$$ y(y + 1) = x(x + 1)(x + 2)(x + 3) \tag{1} $$
$ \bullet \textbf{Attempt} $
To solve for $ y $ I did these simple steps,
$$ y^2 + y = x(x + 1)(x + 2)(x + 3) \tag{2} $$
$$ y^2 + y - x(x + 1)(x + 2)(x + 3) = 0 \tag{3} $$
$$ y = \frac{-1 \pm{\sqrt{1 + 4x(x + 1)(x + 2)(x + 3)}}}{2} \tag{4} $$
I then figured that when,
$$ -1 \pm{\sqrt{1 + 4x(x + 1)(x + 2)(x + 3)}} \equiv 0\pmod{2} \tag{5} $$
then $ y $ will be a integer, so this is just solving for all $ x $, so $ y $ will be a integer but not for only integer $ x $'s.
This then means that
$$ \pm{\sqrt{1 + 4x(x + 1)(x + 2)(x + 3)}} \equiv 1\pmod{2} \tag{6} $$
$$ {\sqrt{1 + 4x(x + 1)(x + 2)(x + 3)}} \equiv 1\pmod{2} \tag{7} $$
This is the end of my knowledge, if you try to square it will include the integer solution for $ y $ but also include non-integer solutions for $ y $.
Am I going in the correct direction, or is this a dead end and I need to apply a new approach?
|
$x(x + 1)(x + 2)(x + 3)=(x^2+3x)(x^2+3x+2)$.
Let $N=x^2+3x+1$, then $N^2-1=m(m+1)$.
If $N=0$
There are no solutions since $m(m+1)$ is even.
If $|N|=1$
Then $m=0$ or $-1$.
If $|N|\ge 2$
Then $N^2-1$ is greater than $|N|(|N|-1)$ but less than $|N|(|N|+1)$. There are no solutions.
The only solutions are when $x(x + 1)(x + 2)(x + 3)=0$ i.e. $x\in\{-3,-2,-1,0\}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve $xy''-y'-x^3y=0$ I want to solve $xy''-y'-x^3y=0$. My solution:
$y = z\cdot \exp(kx^2)$
$y' = z'\cdot \exp(kx^2) + z\cdot 2kx\exp(kx^2)$
$y'' = z''\cdot \exp(kx^2) + z'\cdot 4kx\exp(kx^2) + z\cdot (2k+4k^2x^2)\exp(kx^2)$
Plug in:
$z''\exp(kx^2)+z'(4kx-\frac{1}{x})\exp(kx^2)+z(2k+4k^2x^2-\frac{1}{x}\cdot2kx-x^2)\exp(kx^2)=0
$
$2k+4k^2x^2-\frac{1}{x}\cdot2kx-x^2=0\iff k=\frac{1}{2}$
Plug in:
$z''+z'(2x-x^{-1})=0$
$z'=:u$
$u'+u(2x-x^{-1})=0$
$\frac{\mathrm{d}u}{u}=-(2x-x^{-1})\mathrm{d}x$
$\ln|u|=-(x^2-\ln|x|)+c_1$
$u=\frac{c_1x}{e^{x^2}}$
Substitute back:
$z'=c_1xe^{-x^2}$
$z=-\tfrac{1}{2}c_1e^{-x^2}+c_2$
$y(x)=(-\tfrac{1}{2}c_1e^{-x^2}+c_2)e^{\frac{1}{2}x^2}$
$y(x)=c_2e^{\frac{1}{2}x^2}-\tfrac{1}{2}c_1e^{-\frac{1}{2}x^2}$
Right? Is there another way solving this ode? Thanks!
edit: @Lutz Lehmann: Sorry there was a square missing. Fixed it.
|
Another way to solve it maybe:
$$xy''-y'-x^3y=0$$
Consider $x'$, then the equation becomes:
$$-x''x-x'^2=(x'x)^3y$$
$$-(x'x)'=(x'x)^3y$$
This differential equation is separable:
$$\dfrac {d(x'x)}{(x'x)^3}=-ydy$$
$$\dfrac {1}{(x'x)^2}=y^2+K$$
$$2x'x=\pm \dfrac {2} { \sqrt {y^2+K}}$$
$$(x^2)'= \pm \dfrac {2} { \sqrt {y^2+K}}$$
Integrate :
$$x^2+C=\pm 2 \int \dfrac {dy} { \sqrt {y^2+K}}$$
The last integral depends on the value of the constant K.
|
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|
If $h(x) = \dfrac{f(x)}{g(x)}$, then find the local minimum value of $h(x)$ Let $f(x) = x^2 + \dfrac{1}{x^2}$ and $g(x) = x – \dfrac{1}{x}$
, $x\in R – \{–1, 0, 1\}$. If $h(x) = \dfrac{f(x)}{g(x)}$, then the local minimum value
of $h(x)$ is :
My attempt is as follows:-
$$h(x)=\dfrac{x^2+\dfrac{1}{x^2}}{x-\dfrac{1}{x}}$$
$$h(x)=\dfrac{x^2+x^2-2+2}{x-\dfrac{1}{x}}$$
$$h(x)=x-\dfrac{1}{x}+\dfrac{2}{x-\dfrac{1}{x}}$$
Case $1$: $x-\dfrac{1}{x}>0$
$$\dfrac{x^2-1}{x}>0$$
$$x\in(-1,0) \cup (1,\infty)$$
$$AM\ge GM$$
$$\dfrac{x-\dfrac{1}{x}+\dfrac{2}{x-\dfrac{1}{x}}}{2}>\sqrt{2}$$
$$h(x)\ge 2\sqrt{2}$$
$$x-\dfrac{1}{x}=\dfrac{2}{x-\dfrac{1}{x}}$$
$$x^2+\dfrac{1}{x^2}-2=2$$
$$x^2+\dfrac{1}{x^2}=4$$
$$x^4-4x^2+1=0$$
$$x^2=2\pm\sqrt{3}$$
Only $x=\sqrt{2+\sqrt{3}},-\sqrt{2-\sqrt{3}}$ are the valid solutions.
Case $2$: $x-\dfrac{1}{x}<0$
$$x\in(-\infty,-1) \cup (0,1)$$
$$h(x)=-\left(\dfrac{1}{x}-x+\dfrac{2}{\dfrac{1}{x}-x}\right)$$
By $AM\ge GM$, $h(x)\ge-2\sqrt{2}$
We will get this minimum value at $-\sqrt{2+\sqrt{3}},\sqrt{2-\sqrt{3}}$
So answer should have been $-2\sqrt{2}$ but actual answer is $2\sqrt{2}$. What am I missing here.
|
If you are dealing with $h(x)=f(x)/g(x)$ then:
$$h'(x_0)=0=\frac{f'(x_0)g(x_0)-f(x_0)g'(x_0)}{[g(x_0)]^2}\to \frac{f'(x_0)}{g'(x_0)}=\frac{f(x_0)}{g(x_0)} \quad \quad (1)$$
but, $f(x)=[g(x)]^2+2$, so
$$f'(x)=2g'(x)g(x)\to \frac{f'(x)}{g'(x)}=2g(x)\quad \quad (2)$$
From $(1)$ and $(2)$,
$$f(x_0)=2[g(x_0)]^2=[g(x_0)]^2+2$$
then,
$$g(x_0)=\pm \sqrt{2}=x_0-\frac{1}{x_0}$$
Can you finish?
|
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|
Limits and infinity minus infinity I understand that we need to rationalize when we have infinity minus infinity like here
$\lim_{x\to \infty}\left(\sqrt{x^2 + 1} - \sqrt{x^2 + 2}\right)$
My question is why can I not just split the limits like this
$\lim_{x\to \infty}\left(\sqrt{x^2 + 1}\right) - \lim_{x\to \infty}\left(\sqrt{x^2 + 2}\right)$
and then
$\lim_{x\to \infty}\sqrt{x^2} * \lim_{x\to \infty}\sqrt{1 + \frac 1x} - \lim_{x\to \infty}\sqrt{x^2} * \lim_{x\to \infty}\sqrt{1 + \frac 2x}$
which gives
$\lim_{x\to \infty}\sqrt{x^2} - \lim_{x\to \infty}\sqrt{x^2} = 0$
because $\frac 1x$ and $\frac 2x$ tend to $0$
Where am i wrong ?
|
You can not subtract
$$
\lim_{x\to\infty}(\sqrt{x^2+1}-\sqrt{x^2+2})=\lim_{x\to\infty}\sqrt{x^2+1}-\lim_{x\to\infty}\sqrt{x^2+2},
$$
because the left hand side is $\infty-\infty$ which is not definable.
Instead, you can obtain that
$$
\sqrt{x^2+1}-\sqrt{x^2+2}=\frac{(\sqrt{x^2+1}-\sqrt{x^2+2})(\sqrt{x^2+1}+\sqrt{x^2+2})}{\sqrt{x^2+1}+\sqrt{x^2+2}}\\ =\frac{(x^2+1)-(x^2+2)}{\sqrt{x^2+1}+\sqrt{x^2+2}}=-\frac{1}{\sqrt{x^2+1}+\sqrt{x^2+2}}\to 0,
$$
where the properties of limits apply.
|
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|
Trouble with trig substitution I have a few questions and a request for an explanation.
I worked this problem for a quite a while last night. I posted it here.
Here is the original question: $$\int\frac{-7 x^2}{\sqrt{4x-x^2}} dx$$
And here is the work that I did on it:
Help with trig sub integral
Sorry that the negative sometimes gets cut off in the photo, and yes I know it's not fully simplified there.
\begin{align}
-7 \int\frac{x^2}{\sqrt{4x-x^2}} dx
&= -7 \int\frac{x^{3/2}}{\sqrt{4-x}} dx \\
&= -7 \int\frac{8\sin^3\theta\ 2\cos\theta}
{\sqrt{4-4\sin^2\theta}} d\theta
&& \begin{array}{c}
2\sin\theta = \sqrt x \\
2\cos\theta = dx \\
(2\sin\theta)^3 = x^{3/2}
\end{array} \\
&= -7 \cdot 8 \int\frac{\sin^3\theta\ 2\cos\theta}
{2\sqrt{1 - \sin^2\theta}} d\theta \\
&= -56 \int \sin^3\theta\, d\theta \\
&= -56 \int (1 - \cos^2\theta) \sin\theta\, d\theta \\
&= -56 \int(\sin\theta - \cos^2\theta\sin\theta)\,d\theta \\
&= 56 \cos\theta - 56 \int u^2\, du &&
\begin{array}{c}
u = \cos\theta \\
du = \sin\theta
\end{array}\\
&= 56 \cos\theta - 56 \frac{\cos^3\theta}{3} + C \\
&= 56 \left(\frac{\sqrt{4-x}}{2}
- \frac13 \left( \frac{\sqrt{4-x}}{2}\right)^3 \right)+ C
&& \cos\theta = \frac{\sqrt{4-x}}{2}\\
&= 56 \frac{\sqrt{4-x}}{2}
\left(1 - \frac{4-x}{12}\right) + C\\
\end{align}
My first question is the more involved one: Is the algebra in my original work sound? If it is, why doesn't it work in this instance?
My second question is: is this a correct solution?
$$14\left(\frac{\sqrt{4x-x^2}(x-2)}{2}-2\sqrt{4x-x^2}-3\arcsin\left(\frac{x-2}{2}\right)\right)+C$$
It is for webwork, and I used two out of three chances. I'd prefer to keep my perfect webwork mark, obviously ;p
Finally, I was kind of impressed with Ans4's square completion and had to run it through to see that it was correct. That's such a useful skill. Do you have some specific advice about how I could improve my math tricks to that point?
|
You do not have a "$\theta$" on your diagram, so it is not possible to verify that $\theta$ and $x$ have a correct relationship. You do not write "$x = \dots$", so I have to guess that you mean
$$ x = 4 \sin^2 \theta \text{.} $$
The original integrand is only defined on $x \in [0,4]$, so happily this choice of substitution for $x$ is capable of representing the entire valid range of $x$s in the original integral.
From this,
$$ \mathrm{d}x = 8 \cos \theta \sin \theta \,\mathrm{d}\theta \text{,} $$
which is not equivalent to your "$2 \cos \theta = \mathrm{d}x$".
The derivative of your proposed solution with respect to $x$ is
$$ \frac{-7x^2 + 84x - 126}{\sqrt{4x-x^4}} \text{,} $$
which is not (up to a constant of integration) equivalent to the given integrand.
Using only algebra, substitution, and trig substitution, I would attack your integral as \begin{align*}
\int & \; \frac{-7 x^2}{\sqrt{4x-x^2}} \,\mathrm{d}x \\
&= \int \; \frac{-7 x^2}{\sqrt{-(x^2 - 4x + 4 - 4)}} \,\mathrm{d}x \\
&= \int \; \frac{-7 x^2}{\sqrt{-((x-2)^2 - 4)}} \,\mathrm{d}x &\begin{bmatrix} u = x-2 \\ \mathrm{d}u = \mathrm{d}x \end{bmatrix} \\
&= \int \; \frac{-7 (u+2)^2}{\sqrt{4 - u^2}} \,\mathrm{d}u &\begin{bmatrix} u = 2\cos\theta \\ \mathrm{d}u = -2\sin\theta\,\mathrm{d}\theta\end{bmatrix} \\
&= \int \; \frac{-7 (2\cos\theta+2)^2}{\sqrt{4 - 4\cos^2 \theta}} (-2 \sin\theta) \,\mathrm{d}\theta \\
&= \int \; \frac{-7 (2\cos\theta+2)^2}{2 \sin \theta} (-2 \sin\theta) \,\mathrm{d}\theta \\
&= 7 \int \; (2\cos\theta+2)^2 \,\mathrm{d}\theta \text{.}
\end{align*}
Then multiply out the binomial and split the sum into three integrals. The constant integral is easy. The cosine integral is easy. The squared cosine integral is easy if you use reduction formulas (the fifth and sixth entries in the table here) or the identity $\cos 2x = 2 \cos^2 x - 1$ to double the angle and lower the power.
|
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|
Solve the recurrence relation $a_n - 5a_{n-1} +8a_{n-2} -4a_{n-3} = 3^n$ Solve the recurrence
$a_n - 5a_{n-1} +8a_{n-2} -4a_{n-3} = 3^n$
with initial conditions $a_1 = 1, a_2 =5$ and $a_3 = 17$ How this was calculated? any hint or idea highly appreciated
|
Hint: I'd suggest using characteristic polynomials. This particular recurrence can be made homogeneous linear from:
$$a_n - 5a_{n-1} +8a_{n-2} -4a_{n-3} = 3^n$$
$$a_{n-1} - 5a_{n-2} +8a_{n-3} -4a_{n-4} = 3^{n-1}$$
and
$$a_n - 5a_{n-1} +8a_{n-2} -4a_{n-3} = 3^n$$
$$3a_{n-1} - 15a_{n-2} +24a_{n-3} -12a_{n-4} = 3^n$$
and subtracting
$$a_n-8a_{n-1}+23a_{n-2}-28a_{n-3}+12a_{n-4}=0$$
which leads to the following characteristic polynomial:
$$x^4-8x^3+23x^2-28x+12=0$$
Integers solutions (if any, rational roots theorem) should be divisors of $12$ and indeed $1,2,3$ are the solutions. Or
$$x^4-8x^3+23x^2-28x+12=(x-1)(x-2)^2(x-3)$$
Everything else is mechanics, here and here are a few examples to help you finish the exercise.
|
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|
Find a closed-form solution for the following recurrence $$T(n) =\begin{cases}5, & \text{if $n=1$} \\2T(n-1)+3n+1, & \text{if $n\geq 2$}\end{cases}$$
Now the answer is as such:
*
*$T(n)=5\cdot 2^{n-1}+\sum_{i=2}^n2^{n-i}(3i+1)$, which I certainly understand.
*$=(3n+6)2^{n-1}-3n-1-3\sum_{i=1}^{n-2}i\cdot 2^i=15\cdot 2^{n-1}-3n-7$, which throws me a big curve ball right immediately without any sense.
Could anyone explain to me the drastic transition algebraically?
|
A change of variable $j=n-i$ gives:
\begin{align}
T(n)
&=5\cdot 2^{n-1}+\sum_{i=2}^n2^{n-i}(3i+1)\\
&=5\cdot 2^{n-1}+\sum_{j=0}^{n-2}2^j(3(n-j)+1)\\
&=5\cdot 2^{n-1}+(3n+1)\sum_{j=0}^{n-2}2^j-3\sum_{j=0}^{n-2}j2^j\\
&=5\cdot 2^{n-1}+(3n+1)(2^{n-1}-1)-3\sum_{j=0}^{n-2}j2^j\\
\end{align}
Now you need:
\begin{align}
\sum_{i=0}^{n-1}ix^{i-1}
&=\frac d{dx}\sum_{i=0}^{n-1}x^i\\
&=\frac d{dx}\frac{x^n-1}{x-1}\\
&=\frac{nx^{n-1}(x-1)-(x^n-1)}{(x-1)^2}\\
&=\frac{(n-1)x^n-nx^{n-1}+1}{(x-1)^2}
\end{align}
which for $x=2$ become:
$$\sum_{i=0}^{n-1}i2^{i-1}=(n-1)2^n-n2^{n-1}+1$$
|
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|
For what values of $a\in \mathbb{R}$ is the integral $\int\int_{\mathbb{R}^2} \frac{1}{(1+x^4+y^4)^a}dxdy$ convergent? Well, first notice that the above integral equals to $2\int_{0}^\infty \int_0^\infty \frac{1}{(1+x^4+y^4)^a}dxdy$ since the integrand is an even function of both $x$ and $y$.
There are two approaches which I thought to do here.
*
*Change of variables: $x^2 = r\cos \theta , y^2 = r\sin \theta$; which I am not sure it's valid since there are value of theta which give the rhs to be negative, and we are in real calculus.
*The legitimate approach is to break the integral into:
$$\int_0^1 \int_0^1+\int_0^1\int_1^\infty + \int_1^\infty\int_0^1+\int_1^\infty \int_1^\infty$$
I thought to compare the integrand with $x^2+y^2$, i.e when $x,y \in [0,1]$ we know that $x^2 \ge x^4$ and when $x>1$ then the opposite follows.
Seems a bit long calculation.
Can anyone help me with this?
Thanks!
|
Assume $a > 0$, since divergence is obvious for $a \leqslant 0$. Changing to polar coordinates, $x = r \cos \theta, \,y = r \sin \theta$, we have
$$\int_0^\infty\int_0^\infty \frac{1}{(1+ x^4 + y^4)^a}\, dx\, dy =\int_0^\infty\int_0^{\pi/2} \frac{r\,dr\, d\theta }{(1+ r^4\cos^4 \theta + r^4 \sin^4 \theta)^a}\, $$
Since $\cos^4 \theta + \sin^4 \theta$ attains a minimum value of $1/2$ at $\theta = \pi/4$ and a maximum value of $1$ at $\theta = 0, \pi/2$, it follows that
$$\int_0^\infty\int_0^\infty \frac{1}{(1+ x^4 + y^4)^a}\, dx\, dy \leqslant\int_0^\infty\int_0^{\pi/2} \frac{1}{(1+ r^4/2 )^a}\, r\,dr\, d\theta \\= \frac{\pi}{2}\int_0^\infty \frac{r}{(1+ r^4/2 )^a}\,dr, \\ \int_0^\infty\int_0^\infty \frac{1}{(1+ x^4 + y^4)^a}\, dx\, dy \geqslant\int_0^\infty\int_0^{\pi/2} \frac{1}{(1+ r^4 )^a}\, r\,dr\, d\theta \\= \frac{\pi}{2}\int_0^\infty \frac{r}{(1+ r^4 )^a}\,dr$$
Note that $\dfrac{r}{(1+cr^4)^a} = \mathcal{O}(r^{1-4a})$ as $r \to \infty$. You should easily determine the values of $a$ for which the integral converges / diverges using the above comparisons.
|
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|
Let $G = \{1,x,y\}$ be a group with identity element $1$. Prove that $x^2 = y, y^2 = x,$ and $xy = 1.$
Let $G = \{1,x,y\}$ be a group with identity element $1$. Prove that $x^2 = y$, $y^2 = x$, and $xy = 1$.
Every time I try this I end up with pages of nonsense. Any help would be appreciated.
|
Let's fill in the Cayley table! Recall that Cayley tables are latin squares meaning that every element of the group appears exactly once in every row and every column (not counting the row and column headers). Let's start with a blank canvas:
$$\begin{array}{|c|c|c|}\hline & \color{blue}1 & \color{blue}x & \color{blue}y \\ \hline \color{blue}1 & & & \\ \hline \color{blue}x & & & \\ \hline \color{blue}y & & & \\ \hline\end{array}$$
First, we have that $1$ is the identity. Thus $1 \cdot 1 = 1$, $1 \cdot x = x \cdot 1 = x$, and $1 \cdot y = y \cdot 1 = y$, so
$$\begin{array}{|c|c|c|}\hline & \color{blue}1 & \color{blue}x & \color{blue}y \\ \hline \color{blue}1 & 1 & x & y \\ \hline \color{blue}x & x & & \\ \hline \color{blue}y & y & \color{red}{?} & \\ \hline\end{array}$$
What can we fill in for the red question mark$\color{red}?$ Note that $y$ already appears in the given row, so it's not $y$. Similarly, $x$ already appears in the given column, so it's not $x$. Thus, it must be $1$.
In more concrete terms, $y \cdot x$ cannot equal $y$, as
$$y \cdot x = y \implies y^{-1} \cdot y \cdot x = y^{-1} \cdot y \implies x = 1,$$
and similarly,
$$y \cdot x = x \implies y = 1.$$
Updating our Cayley table,
$$\begin{array}{|c|c|c|}\hline & \color{blue}1 & \color{blue}x & \color{blue}y \\ \hline \color{blue}1 & 1 & x & y \\ \hline \color{blue}x & x & & \color{red}{?} \\ \hline \color{blue}y & y & 1 & \\ \hline\end{array}$$
Similar logic works again here
$$\begin{array}{|c|c|c|}\hline & \color{blue}1 & \color{blue}x & \color{blue}y \\ \hline \color{blue}1 & 1 & x & y \\ \hline \color{blue}x & x & & 1 \\ \hline \color{blue}y & y & 1 & \\ \hline\end{array}$$
We can now finish the whole Cayley table by filling in the blanks. The second row/column is missing a $y$, and the third row/column is missing an $x$. Thus,
$$\begin{array}{|c|c|c|}\hline & \color{blue}1 & \color{blue}x & \color{blue}y \\ \hline \color{blue}1 & 1 & x & y \\ \hline \color{blue}x & x & y & 1 \\ \hline \color{blue}y & y & 1 & x \\ \hline\end{array}$$
We can now read off the fact that $x^2 = y$ and $y^2 = x$.
|
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|
A and B commute. Delete the same row and column from each. Do they still commute? Suppose that two matrices $A$ and $B$ commute. If I delete the $i$th row and $i$th column from each, do they necessarily commute? Does the answer change if we guarantee that $A$ and $B$ are Hermitian?
For example, consider the matrices
$A=
\begin{bmatrix}
\frac{4}{3} & \frac{1}{3\sqrt{2}} & -\frac{1}{\sqrt{6}} \\
\frac{1}{3\sqrt{2}} & \frac{7}{6} & -\frac{1}{2\sqrt{3}}\\
-\frac{1}{\sqrt{6}} & -\frac{1}{2\sqrt{3}} & \frac{3}{2}
\end{bmatrix}
$
and
$B=
\begin{bmatrix}
\frac{5}{3} & \frac{\sqrt{2}}{3} & -\frac{\sqrt{2}}{\sqrt{3}} \\
\frac{\sqrt{2}}{3} & \frac{4}{3} & -\frac{1}{\sqrt{3}}\\
-\frac{\sqrt{2}}{\sqrt{3}} & -\frac{1}{\sqrt{3}} & 2
\end{bmatrix}
$.
A patient calculation shows that $AB - BA = 0$. Let's construct the truncated matrices $A_r$ and $B_r$ formed by deleting the third column and third row.
Then
$A_r=
\begin{bmatrix}
\frac{4}{3} & \frac{1}{3\sqrt{2}} \\
\frac{1}{3\sqrt{2}} & \frac{7}{6}
\end{bmatrix}
$
and
$B_r=
\begin{bmatrix}
\frac{5}{3} & \frac{\sqrt{2}}{3} \\
\frac{\sqrt{2}}{3} & \frac{4}{3}
\end{bmatrix}
$.
Another patient, but faster calculation shows that $A_r B_r - B_r A_r=0$. Then for these particular matrices, the answer to the question is yes.
A similar calculation can be done for the non-Hermitian $A=
\begin{bmatrix}
\frac{11}{7} & 0 & -\frac{4}{7} \\
-\frac{1}{7} & 1 & \frac{1}{7}\\
-\frac{3}{7} & 0 & \frac{10}{7}
\end{bmatrix}$ and $B=
\begin{bmatrix}
\frac{15}{7} & 0 & -\frac{8}{7} \\
-\frac{2}{7} & 1 & \frac{2}{7}\\
-\frac{6}{7} & 0 & \frac{13}{7}
\end{bmatrix}$, again affirming yes for this particular example.
This question has a physical motivation from quantum mechanics, where whether or not two Hermitian matrices commute determines whether or not they are simultaneously observable. I'm imagining the case where we are unable to access a certain state, which would correspond to deleting the row and column corresponding to that state.
|
Without loss of generality, you may assume that the first columns and the first rows of the two matrices are deleted. Let
$$
A=\pmatrix{a&u^\ast\\ v&W},\ B=\pmatrix{b&x^\ast\\ y&Z}.
$$
Then
$$
AB=\pmatrix{\ast&\ast\\ \ast&vx^\ast+WZ},
\ BA=\pmatrix{\ast&\ast\\ \ast&yu^\ast+ZW}.
$$
The equality $AB=BA$ therefore implies that $vx^\ast+WZ=yu^\ast+ZW$. It follows that $WZ=ZW$ if and only if $vx^\ast=yu^\ast$.
In case $A$ and $B$ are Hermitian, we have $u=v$ and $x=y$. Therefore, under the assumption that $AB=BA$, the equality $WZ=ZW$ holds if and only if $ux^\ast=xu^\ast$, i.e. if and only if $x$ and $u$ are linearly dependent.
|
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|
What is the locus of $z^2+\bar{z}^2=2$? I have to prove that it's the equation of an equilateral hyperbola
$$z^2+z^{-2}=2$$
I try this
$z^2+z^{-2} +2 = 2+2$
$(z^2+1/z^2 +2 ) = 4$
$(z+1/z)^2=4$
$ z+1/z= 2 $
$ x+yi + 1/(x+yi) = 2 $
$ ((x+yi)^2+1)/(x+yi)=2$
$ x^2+2xyi-y^2+1=2x+2yi$
$ x^2-2x+1 +2xyi -y^2=2yi $
$(x-1)^2 +2xyi-y^2= 2yi$
I don't know how to end it.
|
Hint One can rewrite this quickly using polar coordinates: For $z = r e^{i \theta}$, we have
$$2 = z^2 + \bar z^2 = (r e^{i\theta})^2 + (r e^{-i\theta})^2 = r^2 (e^{2 i \theta} + e^{-2 i \theta}) = r^2 \cdot 2 \cos 2 \theta .$$
Now, divide both sides by $2$, apply the double angle identity for $\cos$, and rewrite in terms of the components $x, y$ of $z = x + i y$.
Doing so gives $$1 = r^2 (\cos^2 \theta - \sin^2 \theta) = (r \cos \theta)^2 - (r \sin \theta)^2 = x^2 - y^2 .$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\lim_{n\to\infty}\sum_{k=1}^n\frac{k^3}{\sqrt{n^8+k}}$ & $\lim_{n\to\infty}n\bigg(\sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}-\frac{1}{4}\bigg)$. This is a homework question. I have to find two limits:
i. $$\lim_{n\to \infty} \sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}$$
ii. $$\lim_{n\to \infty} n\bigg(\sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}-\frac{1}{4}\bigg)$$
I was able to solve the first one:
$\sqrt{n^8+1}\le \sqrt{n^8+k}\le \sqrt{n^8+n}$
$\implies \displaystyle\dfrac{1}{\sqrt{n^8+1}}\ge \dfrac{1}{\sqrt{n^8+k}}\ge \dfrac{1}{\sqrt{n^8+n}}$
$\implies \displaystyle\dfrac{k^3}{\sqrt{n^8+1}}\ge \dfrac{k^3}{\sqrt{n^8+k}}\ge \dfrac{k^3}{\sqrt{n^8+n}}$
$\implies \displaystyle\sum_{k=1}^n\dfrac{k^3}{\sqrt{n^8+1}}\ge \sum_{k=1}^n\dfrac{k^3}{\sqrt{n^8+k}}\ge \sum_{k=1}^n\dfrac{k^3}{\sqrt{n^8+n}}$
$\implies \displaystyle\dfrac{n^2(n+1)^2}{4\sqrt{n^8+1}}\ge \sum_{k=1}^n\dfrac{k^3}{\sqrt{n^8+k}}\ge \dfrac{n^2(n+1)^2}{4\sqrt{n^8+n}}$
The upper fraction goes to $\frac{1}{4}$ and the lower as well. From the Squeeze theorem:
$$\lim\limits_{n\to \infty} \sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}} = \frac{1}{4}$$
The same method did not work for the second limit. Can I get a clue or a hint, please?
|
Note first that $\sum_{k=1}^n k^3=\frac14n^2(n+1)^2$. Now note that $\sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}=\frac1n\sum_{k=1}^n \frac{(k/n)^3}{\sqrt{1+k/n^8}}$ and hence $$\frac1n\sum_{k=1}^n \frac{(k/n)^3}{\sqrt{1+n/n^8}}\leq\sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}\leq\frac1n\sum_{k=1}^n (k/n)^3$$ implying that $\lim_{n\to\infty}\sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}=\frac14$.
For part 2 again see that
$$n\left(\frac1n\sum_{k=1}^n \frac{(k/n)^3}{\sqrt{1+n/n^8}}-\frac14\right)\leq n\left(\sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}-\frac14\right)\leq n\left(\frac1n\sum_{k=1}^n (k/n)^3-\frac14\right)$$ which directly implies the second limit your are after is $\frac12$.
|
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|
Integral $\int \arcsin \left(\sqrt{\frac{x}{1-x}}\right)dx$ $$\int \arcsin\left(\sqrt{\frac{x}{1-x}}\right)dx$$
I'm trying to solve this integral from GN Berman's Problems on a course of mathematical analysis (Question number 1845)
I tried substituting $x$ for $t^2$:
$$2\int t\arcsin\left(\frac{t}{\sqrt{1-t^2}}\right) dt$$
And then evaluating it by parts, which made it very complicated.
So how do I go about solving this?
|
$$I=\int \sin^{-1} \sqrt{\frac{x}{1-x}} dx$$
Let $$\frac{x}{1-x}=t^2 \implies x=\frac{t^2}{1+t^2} \implies dx=\frac{2t}{(1+t^2)^2}$$
Then $$I=\int\frac{2t \sin^{-1}t}{(1+t^2)^2} ~dt $$
Integrate it by parts taking $\sin^{-1} t$ as the first function.
Then $$I=\sin^{-1} t \int \frac{ 2t dt}{(1+t^2)^2}=\frac{-\sin^{-1}t}{1+t^2}+\int \frac{dt}{(1+t^2)\sqrt{1-t^2}}=f(t)+I_1$$
Let $t=1/u \implies du=- dt/t^2$ and then let $v=u^2$, then
$$I_1=-\int \frac{u du}{(1+u^2)\sqrt{u^2-1}}=-\int \frac{v dv}{2(1+v)\sqrt{v-1}}+C$$
Next take $v=w^2 \implies dv=2wdw$, then
$$I_1=-\int \frac{dw}{2+w^2}=-\frac{1}{\sqrt{2}} \tan^{-1} \frac{w}{\sqrt{2}}=-\frac{1}{\sqrt{2}}\tan^{-1} \frac{1-2x}{2x}$$
Finally, $$I=(x-1)\sin^{-1} \sqrt{\frac{x}{1-x}}-\frac{1}{\sqrt{2}}\tan^{-1} \frac{1-2x}{2x}$$
|
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|
Given a right angled triangle ABC, where AP is the median and Q is a point on AB then find AQ in terms of given.
Problem: In a triangle $ABC$, right angled at $B$, $AP$ is the median. $Q$ is a point taken on side $AB$ such that $\angle ACQ =\alpha $ and $\angle APQ =\beta $ and let $CP = BP = x$, Find $AQ$ in terms of the given parameters.
What I have tried is draw perpendiculars from $Q$ on $AC$ and on $AP$ to get some trigonometry going but I am unable to proceed so I graphed it on the coordinate plane but I'm unable to proceed there too. So I looked at the options (its a multiple choice) and all of them seem to include $\cot \alpha$ and $\cot \beta$.
Rough Figure:
Edit: Thanks for the help everyone. Also, I am sorry to learn that I forgot to mention that I had assumed the distance $|AB|$ and it isn't given in the question itself.
|
Given $\angle QCA=\phi$, $\angle QPA=\psi$ and $x=|BP|=|CP|$, find
$c=|AB|$ and $q=|AQ|$.
Let $|AC|=b,\ |AP|=u,\ |CQ|=v,\ |PQ|=w$.
\begin{align}
b^2&=4\,x^2+c^2
,\quad
v^2=4\,x^2+(c-q)^2
,\\
u^2&=\phantom{4\,}x^2+c^2
,\quad
w^2=\phantom{4\,}x^2+(c-q)^2
,\\
b^2-u^2&=
v^2-w^2=3\,x^2
,\\
b^2-v^2&=u^2-w^2
=2\,c\,q-q^2
.
\end{align}
Considering squared areas of $\triangle CAQ$
$\triangle PAQ$, which share the same base $|AQ|=q$
and $h_{CAQ}=2h_{PAQ}$, we have
\begin{align}
v^2\,b^2\,\sin^2\phi
&=
4\,u^2\,w^2\,\sin^2\psi
\tag{1}\label{1}
.
\end{align}
\begin{align}
\triangle CAQ:\quad
2\,x\,q&=
b\,v\,\sin\phi
\tag{2}\label{2}
,\\
\triangle PAQ:\quad
\phantom{2\,}x\,q&=
u\,w\,\sin\psi
\tag{3}\label{3}
,\\
u\,w&=
b\,v\,\frac{\sin\phi}{2\,\sin\psi}
\tag{4}\label{4}
.
\end{align}
By cosine rule
\begin{align}
\triangle CAQ:\quad
q^2&=
b^2+v^2-2\,b\,v\,\cos\phi
\tag{5}\label{5}
,\\
\triangle CAQ:\quad
q^2&=
u^2+w^2-2\,u\,w\,\cos\psi
\tag{6a}\label{6a}
\\
&=u^2+w^2-b\,v\,\sin\phi\,\cot\psi
\tag{6b}\label{6b}
,\\
b\,v&=
\frac{b^2+v^2-u^2-w^2}{2\,\cos\phi-\sin\phi\cot\psi}
\tag{7}\label{7}
,\\
b\,v&= \frac{6\,x^2}{2\,\cos\phi-\sin\phi\cot\psi}
\tag{8}\label{8}
,\\
u\,w&=
\frac{3\,x^2}{2\,\sin\psi\,\cot\phi-\cos\psi}
\tag{9}\label{9}
.
\end{align}
From $\triangle PAQ$
\begin{align}
2\,x\,q&=
b\,v\sin\phi
\tag{10}\label{10}
,\\
q&=\frac{b\,v\sin\phi}{2\,x}
=
\frac{3\,x\,\sin\phi}{2\,\cos\phi-\sin\phi\cot\psi}
=
\frac{3\,x}{2\,\cot\phi-\cot\psi}
\tag{11}\label{11}
.
\end{align}
\begin{align}
b^2+v^2&=
q^2+2\,b\,v\,\cos\phi
\tag{12}\label{12}
,\\
b^2+v^2&=
q^2
+\frac{12\,x^2}{2\cos\phi-\sin\phi\cot\psi}
\tag{13}\label{13}
\end{align}
Substitution of \eqref{1} gives a quadratic equation in $c$:
\begin{align}
c^2-q\,c+4\,x^2-\frac{6\,x^2\,\cos\phi}{2\,\cos\phi-\sin\phi\cot\psi}
&=0
\tag{14}\label{14}
,\\
c^2-q\,c+\tfrac23\,q\,x\,(\cot\phi-2\,\cot\psi)&=0
\tag{15}\label{15}
\end{align}
with a suitable root
\begin{align}
c&=\frac{q}{2}\,\left(1
+\sqrt{1-\frac{8\,x}{3\,q}\,(\cot\phi-2\,\cot\psi)}
\right)
\tag{16}\label{16}
,\\
c&=\frac x2\cdot\frac{3+\sqrt{9+8\,\cot\phi\cot\psi-16\,(\cot\phi-\cot\psi)^2}}
{2\,\cot\phi-\cot\psi}
\tag{17}\label{17}
.
\end{align}
|
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|
Find the value of $k$ in this question Suppose $x_1$ and $x_2$ are solutions to $x^2+x+k=0$, if $x_1^2+x_1x_2+x_2^2=2k^2$, find the value of $k$.
$x_1+x_2 = -1$
$x_1x_2 = k$
$(x_1+x_2)^2 = x_1^2+2(x_1x_2)+x_2^2=2k^2$
$=\frac {x_1^2}{2}+x_1x_2+\frac {x_2^2}{2}=2k^2=\frac{1}{2}$
$2k^2=\frac{1}{2}$
$k^2=\frac{1}{4}$
$k=\sqrt{\frac{1}{4}}$
Could someone please confirm if this is correct and perhaps identify any errors I may have made?
|
You are wrong in this line $~(x_1+x_2)^2 = x_1^2+2(x_1x_2)+x_2^2\color{red}{=2k^2}~$. The value in the left hand side is not equal to $~{2k^2}~$ due to the present of $~\bf 2~$ in the middle term. Here is the complete solution (as per the process you follow).
$$(x_1+x_2)^2 = x_1^2+2x_1x_2 +x_2^2$$
$$\implies (-1)^2=(x_1^2+x_1x_2 +x_2^2)+x_1x_2$$
$$\implies 1=2k^2+k$$
$$\implies 2k^2+k-1=0$$
$$\implies k=\dfrac{-1\pm \sqrt{1^2-4\cdot 2\cdot(-1)}}{4}$$
$$\implies k=\dfrac{-1\pm 3}{4}$$
$$\implies k=-1,~~\dfrac 12$$
|
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|
Rabin Cryptosystem Proof The system is explained here: https://cryptography.fandom.com/wiki/Rabin_cryptosystem
I am trying to prove that the roots
$r \equiv y_p \cdot p \cdot m_q + y_q \cdot q \cdot m_p \pmod{N}\\
-r \equiv N - r \pmod{N}\\
s \equiv y_p \cdot p \cdot m_q - y_q \cdot q \cdot m_p \pmod{N}\\
-s \equiv N - s \pmod{N}$
are in fact the roots you get in decryption. Where $N=pq; p, q$ are two primes.
So far I have
let
\begin{align*}
r &\equiv y_p \cdot p \cdot m_q + y_q \cdot q \cdot m_p \pmod{N}\\
r^2 &\equiv (y_p \cdot p \cdot m_q + y_p \cdot p \cdot m_q)(y_p \cdot p \cdot m_q + y_q \cdot q \cdot m_p) \pmod{N} \\
&\equiv y_p^2 \cdot p^2 \cdot m_q^2 + 2(y_p \cdot p \cdot m_q \cdot y_q \cdot q \cdot m_p) + y_q^2 \cdot q^2 \cdot m_p^2 \pmod{N} \\
&\equiv y_p^2 \cdot p^2 \cdot m_q^2 + y_q^2 \cdot q^2 \cdot m_p^2 \pmod{N}
\end{align*}
I don't know that the next steps are to get to
$r^2 \equiv m \pmod{N}$
|
Note that $m_p^2 \equiv c \pmod{p}$ by construction and also $m_q^2 \equiv c \pmod{q}$.
Also by EEA, we chose $x_p,x_q$ so that $x_p p + y_q q=1 $ so it follows that
$$x_p \cdot p \equiv 1 \pmod{q} \text{ and } x_q \cdot q \equiv 1 \pmod{p}$$
So using the CRT, look at what $r^2 \pmod{p}$, $r^2 \pmod{q}$ equal:
$$r^2 \equiv m_p^2 \equiv c \pmod{p}\tag{1}$$ and
$$r^2 \equiv m_q^2 \equiv c \pmod{q}\tag{2}$$
so CRT's uniqueness clause says that $r^2 \equiv c \pmod{pq}$
When $r$ has the property so as $-r \equiv n-r \pmod{n}$ as well, of course, in any ring.
The case for $s$ is similar, the cross term cancels out mod $N$ anyway.
Note that the construction of $r$ and $s$ closely follows the construction in the proof of the CRT for two moduli.
|
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|
Does $\int_{0}^{\infty} \frac{2x +3}{\sqrt {x^3 + 2x + 5}} \,dx $ converge? Does the following integral converge? I will post my solution, but I am unsure if it is true.
$$\int_{0}^{\infty} \frac{2x +3}{\sqrt {x^3 + 2x + 5}} \,dx $$
My solution
Let
$$ g(x) = \frac{2x}{\sqrt {x^3}}$$
$$ f(x) = \frac{2x +3}{\sqrt {x^3 + 2x + 5}} $$
Then $$\lim_{k \to \infty} \frac{f(x)}{g(x)} = 1$$
Therefore whatever one does, so does the other.
$$ \int_{0}^{\infty} g(x) = \int_{0}^{\infty} \frac{2}{\sqrt {x}} = +\infty $$
Therefore g(x) diverges, thus
$$\int_{0}^{\infty} f(x) = \int_{0}^{\infty} \frac{2x +3}{\sqrt {x^3 + 2x + 5}} \,dx = + \infty$$
diverges too
|
Notice that:
$$\frac{2x+3}{\sqrt{x^3+2x+5}}\approx x^{-0.5}$$ for large $x$ and we know that:
$$\int_0^\infty\frac{1}{x^n+c}dx$$
only converges for $c>0,n>1$
|
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|
Find an equation for the tangent line of $y = e^{4x}$ that is parallel to the linear equation $y = 5 x + 17$. I don't know how to solve this problem:
Find an equation for a tangent line of $y = e^{4x}$ that is parallel to the line given by $y = 5x + 17$.
I have tried with multiple approaches and have gotten
$$y-e^4(4e^{20})=5(x- 4e^{20}).$$
And my answer remains incorrect any help would be greatly appreciated
|
The tangent line will have gradient $5$, so you have to find an $x$ such that
$\frac{d (e^{4x})}{dx} = 5$. Since $\frac{d (e^{4x})}{dx} = 4e^{4x}$, setting
$x = \frac{\log\left(\frac{5}{4}\right)}{4}$ works. Since $\frac{dy}{dx} = 4y$ when $y = e^{4x}$, then $y$ will be $\frac{5}{4}$ when the gradient is $5$.
The line is $y = 5x + c$, where $y$ is $\frac{5}{4}$ when $x$ is $\frac{\log\left(\frac{5}{4}\right)}{4}$. Plugging these values in
gives $\frac{5}{4} = 5\left(\frac{\log\left(\frac{5}{4}\right)}{4}\right) + c$. This means that $c = \frac{5}{4}\left(1 - \log\left(\frac{5}{4}\right)\right)$. So the line is $y = 5x + \frac{5}{4}\left(1 - \log\left(\frac{5}{4}\right)\right)$
|
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|
solve for $(\cos x)(\cos2x)(\cos3x)=\frac{1}{4}$ solve for $(\cos x)(\cos2x)(\cos3x)=\frac{1}{4}$. what is the general solution for $x$.
I wrote $\cos2x$ as $2\cos^2x-1$ and $\cos3x$ as $4\cos^3x-3\cos x$. Then the expression gets reduced to $\cos x(2\cos^2x-1)(4\cos^3x-3\cos x)=\frac{1}{4}$. substituted $\cos x=t$ and got
$t(2t^2-1)(4t^3-3t)=\frac{1}{4}$. how do i proceed further?
|
Expanding the equation in $t$ gives
$$32 t^6-40 t^4+12 t^2-1=0$$ Let $y=t^2$ to make
$$32 y^3-40 y^2+12y-1=0$$ By inspection $y=\frac 14$ is a solution; so,
$$\left(y-\frac 14 \right)(8y^2-8y+1)=0$$
I am sure that you will easily finish.
|
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|
Closed formula for $\sum_{k=0}^{n}\binom{n+k}{k}x^k$ What would be the following sum:
$$\sum_{k=0}^{n}\binom{n+k}{k}x^k$$
I want this closed formula to prove
:
$$\sum_{k=0}^{n}\binom{n+k}{k}\frac{1}{2^k}=2^n$$
My try:
I could not find any closed form of this expression, but a recurrence relation would be like that:
$$\color{blue}{S_n}:=\sum_{k=0}^{n}\binom{n+k}{k}x^k=\frac{1}{x^n}\sum_{k=n}^{2n}\binom{k}{n}x^k$$$$=\frac{1}{x^n}\left[\sum_{k=n}^{2n}\binom{k+1}{n+1}x^k-\sum_{k=n}^{2n}\binom{k}{n+1}x^k \right]$$$$=\frac{1}{x^n}\left[\frac{1}{x}\sum_{k=n+1}^{2n+1}\binom{k}{n+1}x^k-\sum_{k=n}^{2n}\binom{k}{n+1}x^k \right]$$$$=\frac{1}{x^n}\left[\left(\frac{1}{x}-1\right)\sum_{k=n+1}^{2n}\binom{k}{n+1}x^k+\binom{2n+1}{n+1}x^{2n}\right]$$$$=\frac{1}{x^n}\left[\left(\frac{1}{x}-1\right)\sum_{k=n+1}^{2(n+1)}\binom{k}{n+1}x^k+\binom{2n+1}{n+1}x^{2n}-\left(\frac{1}{x}-1\right)\left(\binom{2n+1}{n+1}x^{2n}+\binom{2n+2}{n+1}x^{2n+1}\right)\right]$$$$=\frac{1}{x^n}\left[\left( 1-x \right)x^{n}S_{n+1}+x^{\left(2n-1\right)}\left(2x^{2}-1\right)\binom{2n+1}{n}\right]$$$$=\color{blue}{\left(1-x\right)S_{n+1}+x^{\left(n-1\right)}\left(2x^{2}-1\right)\binom{2n+1}{n}}$$
Using this recurrence relation we have:
$$\bbox[5px,border:2px solid #00A000]{S_{n}=\frac{S_{n-1}-x^{\left(n-2\right)}\left(2x^{2}-1\right)\binom{2(n-1)+1}{n}}{\left(1-x\right)}}\; \;\;\;\;\;\;\;\;\;\;\; \left(n \in \mathbb N^{*} \;\;,\;\; x \neq 1\right)$$
But I'm not sure if that's right.
|
$$\sum_{k=0}^{n}\binom{n+k}{k}x^k = \sum_{k=0}^{n}x^k\cdot[t^n](1+t)^{n+k}=[t^n](1+t)^n\sum_{k=0}^{n}(x(1+t))^k=[t^n](1+t)^n\frac{1-x^n(1+t)^n}{1-x(1+t)}$$
if $x=\frac{1}{2}$ equals
$$ [t^n](1+t)^n \frac{1-\left(\frac{1+t}{2}\right)^n}{\frac{1-t}{2}}=\frac{2}{2^n}[t^n]\frac{2^n(1+t)^n-(1+t)^{2n}}{1-t}=\frac{2}{2^n}\sum_{k=0}^{n}[t^k]\left(2^n(1+t)^n-(1+t)^{2n}\right)$$
or
$$ \frac{2}{2^n}\sum_{k=0}^{n}\left(2^n\binom{n}{k}-\binom{2n}{k}\right)=\frac{2}{2^n}\left(4^n-\frac{4^n}{2}\right)=\color{red}{2^n}. $$
|
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Integral $\int_{0}^{\frac{\pi}{4}}x\ln(\cos x)dx$ I would like to learn more about this integral:
$$\int_{0}^{\frac{\pi}{4}}x\ln(\cos x)dx=\frac{1}{128}\Big(16\pi \operatorname{C}-21\zeta(3)-4\pi^2\ln(2)\Big)$$
where $\operatorname{C}$ is the Catalan's constant.
I have tried integration by parts , substitution $y=\cos x$, etc. But, it reveals nothing. WA gives the anti-derivative in polylogarithmic functions that I don't know about.
I think it is a hard integral. If you have an elementary method to guide me, it would be nice.
Thanks a lot!
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Not an answer but some transformations which might be useful. Didn't go beyond double series.
$$\int_{0}^{\frac{\pi}{4}}x\ln(\cos(x))dx= \color{blue}{(\cos x=t)} $$
$$ =\int_{1/\sqrt{2}}^1 \arccos t \ln t \frac{dt}{\sqrt{1-t^2}}=\color{blue}{(1-t^2=u^2)} $$
$$ = \frac12 \int_0^{1/\sqrt{2}} \frac{du}{\sqrt{1-u^2}} \arcsin u \ln (1-u^2)=$$
$$=\frac12 \int_0^{1/\sqrt{2}} \int_0^1 \frac{udv du}{\sqrt{1-u^2}\sqrt{1-u^2 v^2}} \ln (1-u^2)= \color{blue}{(u=\sqrt{w})}$$
$$=\frac14 \int_0^{1/2} \int_0^1 \frac{dv dw}{\sqrt{1-w}\sqrt{1-w v^2}} \ln (1-w)= $$
$$ =-\frac14 \sum_{k=1}^\infty \frac{1}{k} \int_0^1\int_0^{1/2} \frac{w^k dw dv}{\sqrt{1-w}\sqrt{1-w v^2}}=\color{blue}{(w=\frac{s}{2})} $$
$$=-\frac18 \sum_{k=1}^\infty \frac{1}{k2^k} \int_0^1\int_0^1 \frac{s^k ds dv}{\sqrt{1-s/2}\sqrt{1-s v^2/2}}=$$
$$=-\frac18 \sum_{k=1}^\infty \frac{1}{k2^k} \sum_{n=0}^\infty \binom{2n}{n} \frac{1}{8^n}
\int_0^1\int_0^1 \frac{s^{k+n} v^{2n} ds dv}{\sqrt{1-s/2}}=$$
$$=-\frac18 \sum_{k=1}^\infty \frac{1}{k2^k} \sum_{n=0}^\infty \binom{2n}{n} \frac{1}{(2n+1)8^n} \int_0^1 \frac{s^{k+n} ds }{\sqrt{1-s/2}}=$$
$$=-\frac14 \sum_{k=1}^\infty \frac{1}{k} \sum_{n=0}^\infty \binom{2n}{n} \frac{B_{1/2} \left(\frac12,k+n+1 \right)}{(2n+1)4^n}=$$
$$=- \sum_{k=1}^\infty \frac{4^k}{k} \sum_{n=0}^\infty \binom{2n}{n} \frac{B_{1/2-1/\sqrt{8}} \left(k+n+1,k+n+1 \right)}{2n+1}$$
Where we have incomplete Beta function. Note that in this case incomplete Beta is a polynomial with order $n+k+2$.
|
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|
Given unit roots $ω$, find $\frac ω{1+ω^2} + \frac{ω^2}{1+ω^4} + \frac{ω^3}{1+ω} + \frac{ω^4}{1+ω^3}$ Let ω be a complex number such that $ω^5=1$ and $ω≠1$. Find
$$\frac ω{1+ω^2} + \frac{ω^2}{1+ω^4} + \frac{ω^3}{1+ω} + \frac{ω^4}{1+ω^3}$$
I've been having trouble with this unit, need help with solving this problem. (Sorry, I don't know how to format it correctly.)
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Note that,
$$\frac ω{1+ω^2}\cdot\frac{ω^3}{ω^3} = \frac{ω^4}{1+ω^3}, \>\>\>\>\>\>\>
\frac{ω^3}{1+ω}\cdot\frac{ω^4}{ω^4} = \frac{ω^2}{1+ω^4}$$
Thus,
$$\begin{aligned}
& \frac ω{1+ω^2} + \frac{ω^2}{1+ω^4} + \frac{ω^3}{1+ω} + \frac{ω^4}{1+ω^3} \\
= &2\left(\frac ω{1+ω^2} + \frac{ω^3}{1+ω} \right)
=2\cdot \frac{ω(1+ω)+ω^3(1+ω^2)}{(1+ω^2)(1+ω)}
=2\cdot \frac{ω+ω^2+ω^3+1}{1+ω^2+ω+ω^3}=2
\end{aligned}$$
|
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How to prove that $a^{2^n} - 1$ is divisible by $4\cdot2^n$ by induction Given that $a$ is any odd number and $n$ is any integer
I got to $(2k + 3)^{2^k}\cdot(2k + 3)^{2^k} - 1$ at the $(k+1)$-th step.
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For $n=1$ this is obvious, because $a^2-1=(a-1)(a+1)$, and both factors are even, so $2^2 = 4\cdot 2^0$ divides $a^2-1$.
Assume this is true for $n$, so $4\cdot 2^n$ divides $a^{2^n}-1$. Then
\begin{align*}
a^{2^{n+1}}-1 & = a^{2^n\cdot2}-1 \\
& = (a^{2^n}-1)(a^{2^n}+1) \\
& = 4\cdot 2^n \cdot k\cdot(a^{2^n}+1),
\end{align*}
for $k$ an integer. But now $a^{2^n}$ is odd, so $a^{2^n}+1$ is even, hence $a^{2^n}+1 = 2h$ for $h$ an integer. Therefore
$$a^{2^{n+1}}-1 = 4\cdot 2^n \cdot k \cdot 2h = 4\cdot 2^{n+1} \cdot hk$$
and the claim follows.
|
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How to solve this quartic? I have the following equation, which I've put into the form of a normal quartic:
$$x^4+4ax^3+(4a^2+1)x^2-1=0$$
I'm trying to find the solutions for $x$ in terms of $a$. Is there an easier way then using the quartic formula?
|
Solving by Resultant.
$Res_x(x^4 + 4 a x^3 + (4 a^2 + 1) x^2 - 1, x^2 + (y + a) x + A)=\\
1 - a^2 - a^4 + 2 A - A^2 + 5 a^2 A^2 + 4 a^4 A^2 - 2 A^3 + 4 a^2 A^3 + A^4\\
-2 a (1 + 2 A + A^2 + 4 a^2 A^2 + 2 A^3)y + (-1 + 2 a^2 + 4 A + A^2 + 4 a^2 A^2) y^2 - y^4$
Let linear term equal zero, then:
1) find $A$ from cubic $1 + 2 A + A^2 + 4 a^2 A^2 + 2 A^3=0$
2) find $y$ from biquadratic
$1 - a^2 - a^4 + 2 A - A^2 + 5 a^2 A^2+ 4 a^4 A^2 - 2 A^3 + 4 a^2 A^3 + A^4\\
+ (-1 + 2 a^2 + 4 A + A^2 + 4 a^2 A^2) y^2 - y^4=0$
3) find $x$ from quadratic $x^2 + (y + a) x + A=0$
|
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Integer Solution to $x^3+y^2=z^2$ What is the non-zero integer general solution to $x^3+y^2=z^2$ ? I guess it is already solved in some book or paper, in that case plz help me to find that.
Edit:
*
*One solution is -
$n^3 = [(n)(n+1)/2]^2 - [(n)(n-1)/2]^2$
Found in here.
*$(x,y,z)=\left(abuv,\frac{ab(bu^3-av^3)}{2},\frac{ab(bu^3+av^3)}{2}\right)$ where $bu^3\equiv av^3\pmod{2}$
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There are a lot of solutions. For example, parameterize a set of such solutions.
We have modulo $7$ the only cubes are $0$ and $\pm1$ and besides $t^6=1$ for all $t$ non divisible by $7$. Take for example the cube $1$ so we have
$$1\equiv(z-y)(z+y)\pmod7$$ We can choose for example $z+y=t^5$ and $z-y=t$ so we get
$$z=\frac{t^5+t}{2}\hspace{15mm}y=\frac{t^5-t}{2}$$ This allow us to get the parameterization in integers
$$\begin{cases}x=t^2\\y=\dfrac{t^5-t}{2}\\z=\dfrac{t^5+t}{2}\end{cases}$$in relation with the identity $$(t^2)^3+\left(\frac{t^5-t}{2}\right)^2=\left(\frac{t^5+t}{2}\right)^2$$
|
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Manual square root method explained I was investigating some methods of computing square roots by hand and I found a certain script which presents the following method of computing:
NOTE: the following notation represents positional notation of a number, e.g. $$\overline{ab} = 10a + b$$
The algorithm is presented with an example of a square of a 4-digit number:
$$\overline{abcd}^2 = (10^3a + 10^2b + 10c + d)$$
$$= a^210^6 + b^210^4 +c^210^2 + d^2 + 2ab10^5 + 2ac10^4 +2ad10^3 + 2bc10^3 + 2bd10^2 +2cd10$$
$$= a^2 \cdot 10^6 + b \cdot 10^4 \cdot (2a \cdot 10 + b) + c \cdot 10^2(2 \cdot (10a + b) \cdot 10 + c) + d(2 \cdot (a \cdot 10^2 + b \cdot 10 + c) \cdot 10 + d)$$
$$= a^2 \cdot 10^6 + \overline{(2a)b} \cdot b \cdot 10^4 + \overline{(2\overline{ab})c} \cdot c \cdot 10^2 + \overline{(2\overline{abc})d} \cdot d$$
The final equality is used in the algorithm. Now it is shown on the example of $\sqrt{14175225}$:
$$14175225 = 9 \cdot 10^6 + 5175225$$
$$= 9 \cdot 10^6 + 67 \cdot 7 \cdot 10^4 + 485225$$
$$= 9 \cdot 10^6 + 67 \cdot 7 \cdot 10^4 + 746 \cdot 6 \cdot 10^2 + 37625$$
$$= 9 \cdot 10^6 + 67 \cdot 7 \cdot 10^4 + 746 \cdot 6 \cdot 10^2 + 7525 \cdot 5$$
$$= 3765^2$$
So the process basically finds the numbers $a,b,c,d$ so they satisfy the final equality. Now there is no other explanation for this, for example the author just says that $a = 3$ with no further logic behind it. Why couldn't it be $a=2$ for example? How can one know at the start that $a = 3$ and then continue the process by simply knowing it will satisfy the equation? Also how can we know that a number such as $14175225$ will have a 4-digit square root?
None of these things are explained in the script so I would like an explanation if someone has any insight into this.
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the author just says that a=3 with no further logic behind it.
Why couldn't it be a=2 for example?
Pairing from the implied decimal point, the number is "14 17 52 25".
The value of "a" is chosen as the largest digit whose square isn't greater than 14.
4×4 is 16, which is too big, but 3×3 is 9, which fits, so a = 3.
If someone simply asked you what the square root of 14 million is, you'd likely think that 4000 is slightly too big and 3000 is a lot too small, so you'd reply something like "in the high 3000s".
|
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|
Comparing $2$ infinite continued fractions
$A = 1 +\dfrac{1}{1 + \frac{1}{1 + \frac{1}{\ddots}}} \\
B = 2 +\dfrac{1}{2 + \frac{1}{2 + \frac{1}{\ddots}}}$
Given the two infinite continued fractions $A$ and $B$ above, which is larger, $2A$ or $B?$
I used the golden ratio on the $2$ and came up with:
$A = 1 + \dfrac{1}{A} \\
B = 2 + \dfrac{1}{B}$
Converting to quadratic equations:
$A^2 - A - 1 = 0 \\
B^2 -2B - 1 = 0$
Resulting to:
$2A = 1 + \sqrt{5} > 1 + \sqrt{2} = B$
My Question is:
Are there any more ways to solve this type of problem?
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Clearly $A,B,2B-A>0$.
Note that $B=2+\dfrac{1}{B}$ and $A=1+\dfrac{1}{A}\iff 2A=2+\dfrac{2}{A}$.
It follows, $2A-B=\dfrac{2}{A}-\dfrac{1}{B}=\dfrac{2B-A}{AB}>0$.
|
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Calculate limit of this sequence $I_n = \int_{0}^{\pi/2} (\tan x)^{1/n} dx$ I need to find the limit of the sequence :
$$I_n = \int_{0}^{\pi/2} (\tan x)^{1/n} dx$$
The only thing I have done is : take substitution $\tan x = z$
this gives
$$dx = \dfrac{1}{1 + z^2}\,dz$$
So, integral becomes :
$$I_n = \int_{0}^{\infty} \frac{z^{1/n}}{1 + z^2}\,dz$$
After this I am stuck how should I proceed ?
Can someone help me here ?
Thank you.
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While the estimation done above is sufficient, it might be interesting to see how to obtain the analytical expression by elementary principles from which the result is immediately seen. Rewrite the integral using the symmetry
\begin{align}
I_n &= \int_0^1 \frac{x^{1/n} + x^{-1/n}}{x^2+1} \, {\rm d}x \\
&=\sum_{k=0}^\infty (-1)^k \int_0^1 \left(x^{1/n} + x^{-1/n}\right) x^{2k} \, {\rm d}x \\
&=\sum_{k=0}^\infty (-1)^k \left\{ \frac{n}{2nk+n-1} + \frac{n}{2nk+n+1} \right\} \\
&=\sum_{k=0}^\infty \left\{ \frac{n}{4nk+n-1} + \frac{n}{4nk+n+1} - \frac{n}{4nk+3n-1} - \frac{n}{4nk+3n+1} \right\} \\
&=\sum_{a\in \{-1,1\}} \sum_{k=0}^\infty \left\{ \frac{n}{4nk+n-a} - \frac{n}{4nk+3n+a} \right\} \\
&=\sum_{a\in \{-1,1\}} \sum_{k=0}^\infty \frac{\frac{n+a}{2n}}{(2k+1)^2-\left(\frac{n+a}{2n}\right)^2} \, . \tag{0}
\end{align}
Now let's analyse the function
$$\sum_{k=0}^\infty \frac{x}{(2k+1)^2-x^2}$$
in an analytic sense. It has simple poles at the odd integers $x=\pm (2k+1)$ and converges for all other $x\in \mathbb{C}$. Therefore we should write it as
$$\sum_{k=0}^\infty \frac{x}{(2k+1)^2-x^2} = \frac{f(x)}{\cos\left(\frac{\pi x}{2}\right)} \tag{1}$$
with some analytic function $f(x)$ not vanishing at the odd integers. Calculating the residues on both sides at $x=2n+1$ it follows
$$-\frac{1}{2} = (-1)^{n+1} \frac{2}{\pi} \, f(2n+1) \quad \Rightarrow \quad f(2n+1)=(-1)^n \frac{\pi}{4} \, .$$
We can thus write $$f(x)=\frac{\pi}{4} \, \sin\left(\frac{\pi x}{2}\right) \, g(x)$$
for some analytic function $g(x)$ which obeys $g(2n+1)=1$.
Now besides $g(x)$ being analytic, $g(x)$ is also bounded by (1) and the fact that $|\tan(\pi x/2)|$ is bounded $\forall x\in \mathbb{C} \setminus \mathbb{Z}$ by some constant $C>0$, and from Liouvilles theorem it follows it must be a constant $c$. To obtain that constant we go back to (1) and calculate
$$\lim_{x\rightarrow 0} \sum_{k=0}^\infty \frac{1}{(2k+1)^2-x^2} = \lim_{x\rightarrow 0} c \, \frac{\pi}{4} \, \frac{\tan\left(\frac{\pi x}{2}\right)}{x} \\
\frac{\pi^2}{8} = c \, \frac{\pi^2}{8} \, ,$$
hence $c=1$.
Back into (0) it follows
$$\sum_{a \in \{-1,1\}} \frac{\pi}{4} \, \tan\left(\frac{\pi(n+a)}{4n}\right) = \frac{\pi}{2\cos\left(\frac{\pi}{2n}\right)}$$
by the addition theorem of the tangens.
|
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The maximum for $xy \sin \alpha + yz \sin \beta +zx \sin \gamma$. Question:
Deduce the maximum of $xy \sin \alpha + yz \sin \beta +zx \sin \gamma$ if $x,y,z$ are real numbers that satisfy $x^2+3y^2+4z^2=6$ with $0<\alpha,\beta,\gamma<\pi$ such that $\alpha+\beta+\gamma=2\pi$.
Currently, I am not very sure how to approach the problem. I had an idea to consider the area of a triangle made of 3 smaller triangles with areas $\frac{1}{2}xy \sin \alpha$, $\frac{1}{2}yz \sin \beta$ and $\frac{1}{2}zx \sin \gamma$ respectively. However, that kinda got me no where as I did not have any good ideas on how to use the condition $x^2+3y^2+4z^2=6$. Moreover, that would have assumed $x,y,z\geq0$ which might not be the case.
So, is there a way to deduce the maximum without a calculus approach?
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Okay I finally arrived at an answer. First, lets reduce the cases we need to look at. Clearly $\sin \alpha, \sin \beta, \sin \gamma >0$. So, if one/two of the variables $x,y$ or $z$ is negative, the expression $xy \sin \alpha +yz \sin \beta + zx \sin \gamma$ will not be maximum. Clearly, if all three $x,y,z$ are negative, it is the same when they are positive for the expression $xy \sin \alpha +yz \sin \beta + zx \sin \gamma$. This means we can reduce to the problem to only when $x,y,z>0$.
Consider the following triangle;
Let $CD=x, BD=y, AD=z$ with $\angle CDB=\alpha,\angle BDA=\beta$ and $\angle ADC=\gamma$. This implies that the area of $ABC$, I'll denote it as $[ABC]$, will be $\frac{1}{2}(xy \sin \alpha +yz \sin \beta + zx \sin \gamma)$.
Let $DE,CF$ be altitudes with respect to $AB$, where $E,F$ are on $AB$. Let $AE=m, BE=n, DE=h$ with $CF=H$. This implies that $z^2=m^2+h^2$ and $y^2=n^2+h^2$.
Okay now we are gonna bash this with inequalities,
$$\begin{align}
6 &=x^2+3y^2+4z^2\\
&=x^2+3h^2+3n^2+4h^2+4m^2\\
&=x^2+7h^2+3n^2+4m^2\\
& \geq \frac{(x+h)^2}{1+\frac{1}{7}}+\frac{(n+m)^2}{\frac{1}{3}+\frac{1}{4}}\tag{1}\label{eq1} \\
& \geq 2\sqrt{\frac{(x+h)^2(n+m)^2}{\frac{8}{7}\times\frac{7}{12}}} \tag{2}\label{eq2} \\
& =2\sqrt{\frac{3}{2}}(x+h)(m+n)\\
&\geq \sqrt{6} (H)(m+n)\tag{3}\label{eq3} \\
&=2\sqrt{6} [ABC].
\end{align}$$
Now $\eqref{eq1}$ is due to Cauchy Schwarz Inequality, $\eqref{eq2}$ is due to AM-GM Inequality and $\eqref{eq3}$ is due to Triangle Inequality. I guess equality does occur for each case.
So this implies that;
$$\begin{align}
& 6\geq 2\sqrt{6} [ABC]\\
\Longleftrightarrow & \sqrt{6} \geq 2[ABC]=xy \sin \alpha +yz \sin \beta + zx \sin \gamma.
\end{align}$$
Therefore, the maximum of $xy \sin \alpha +yz \sin \beta + zx \sin \gamma=\sqrt{6}$.
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Find the sum of all positive integers $k$ for which $5x^2-2kx+1<0$ has exactly one integral solution.
Find the sum of all positive integers $k$ for which $5x^2-2kx+1<0$ has exactly one integral solution.
My attempt is as follows:
$$\left(x-\dfrac{2k-\sqrt{4k^2-20}}{10}\right)\left(x-\dfrac{2k+\sqrt{4k^2-20}}{10}\right)<0$$
$$\left(x-\dfrac{k-\sqrt{k^2-5}}{5}\right)\left(x-\dfrac{k+\sqrt{k^2-5}}{5}\right)<0$$
$$x\in\left(\dfrac{k-\sqrt{k^2-5}}{5},\dfrac{k+\sqrt{k^2-5}}{5}\right)$$
As it is given that it has got only one integral solution, so there must be exactly one integer between $\dfrac{k-\sqrt{k^2-5}}{5}$ and $\dfrac{k+\sqrt{k^2-5}}{5}$
Let $x_1=\dfrac{k-\sqrt{k^2-5}}{5}$ and $x_2=\dfrac{k+\sqrt{k^2-5}}{5}$ , then $[x_2]-[x_1]=1$ where [] is a greater integer function.
But from here, how to proceed? Please help me in this.
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hint
Observe that
$$x_2=\frac{k+\sqrt{k^2-5}}{5}=\frac{1}{k-\sqrt{k^2-5}}$$
$ x_2 $ is integral if $ (k -\sqrt{k^2-5})=\pm 1$ ,
By the same, $ x_1 $ is integral if
$ (k+\sqrt{k^2-5}) =\pm 1$.
This gives the possible values for $ k$ :
$-4, -3, 3, 4$.
the sum of positives values is $7$.
|
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prove that: $\sqrt{2}=\frac{F_n^2+F_{n+1}^2+F_{n+2}^2}{\sqrt{F_n^4+F_{n+1}^4+F_{n+2}^4}}$ Pythagoras's constant in Fibonacci number!
How do I show that?
$$\sqrt{2}=\frac{F_n^2+F_{n+1}^2+F_{n+2}^2}{\sqrt{F_n^4+F_{n+1}^4+F_{n+2}^4}}\tag1$$
Where $F_n$ is Fibonacci sequence.
$$2(F_n^4+F_{n+1}^4+F_{n+2}^4)^2=F_n^2+F_{n+1}^2+F_{n+2}^2$$
$$2(F_n^8+F_{n+1}^8+F_{n+2}^8+2F_n^2F_{n+1}^2+2F_n^2F_{n+2}^2+F_{n+1}^2F_{n+2}^2)=F_n^2+F_{n+1}^2+F_{n+2}^2$$
This is getting too messy. I hope there is an easy way of proving this formula.
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Let $x=F_{n+2}$,$y=F_{n+1}$ and $z=F_{n}$. Squaring both sides, we have:
$$2(x^4+y^4+z^4)-(x^2+y^2+z^2)^2=0$$
Factoring out, we have:
$$(x-y-z)(x+y-z)(x-y+z)(x+y+z)=0$$
Now, by the product cancellation law, we can say that all the product is equal to zero when: $$x-y-z=0\leftrightarrow x=y+z$$
Remebring the substituitons done, we are in the context of Fibonacci's sequence, namely:
$$x=y+z\rightarrow F_{n+2}=F_{n+1}+F_n=F_k=F_{k-1}+F_{k-2}$$for $k=n-2, n\geq 2$.
Note that this proof holds for all different sequences such that $F_n=F_{n-1}+F_{n-2}$. For ecample, consider this case:$$\sqrt2=\frac{1^2+2^2+3^2}{\sqrt{1^4+2^4+3^4}}$$
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Isosceles triangle $ABC$ with an inside point $M$, find $\angle BMC$ We have an isosceles $\triangle ABC, AC=BC, \measuredangle ACB=40^\circ$ and a point $M$ such that $\measuredangle MAB=30^\circ$, $\measuredangle MBA=50^\circ$.
Find $\measuredangle BMC$.
Starting with $\angle ABC=\angle BAC=70^\circ \Rightarrow \angle CBM=20 ^\circ$. Let us construct the equilateral $\triangle ABH$. If we look at $\triangle ACH, \angle ACH=20^\circ$ and $\angle CAH=10^\circ$. Can we show $\triangle AHC \cong CHB$? Any other ideas?
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Construct the equilateral triangle $AHB$. Given that $AC = BC, AH = BH$ and the shared $CH$, the triangles $AHC$ and $BHC$ are congruent. Then, $\angle BCH = \dfrac12\angle ACB = 20^\circ$.
Since $AH = BH$ and $\angle BAM = \angle HAM = 30^\circ$, the triangles $BAM$ and $HAM$ are congruent, which yields $\angle HBM = \angle BHM = \angle HBC = 10^\circ$ and $HM || CB$.
Then, the triangles $CHB$ and $BHC$ have the same altitudes $h$ with respect to the base $BC$. Since $\angle BCH = \angle CBM = 20^\circ$, we have $CH = BM = h\cot 20^\circ$.
As a result, the triangles $CHB$ and $BMC$ are congruent, which leads to,
$$\angle BMC = \angle CHB = 180^\circ - \angle CBH - \angle BCH = 180^\circ - 10^\circ - 20^\circ = 150^\circ$$
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|
How to show that : $4(-1)^nL_n^2+L_{4n}-L_n^4=2$ How can we prove that:
$$4(-1)^nL_n^2+L_{4n}-L_n^4=2$$
Where $L_n$ is Lucas number
We got $L_n=\phi^n+(-\phi)^{-n}$
$4(-1)^nL_n^2=8(-1)^n\phi^{2n}+8$
$L_{4n}=\phi^{4n}+(-\phi)^{-4n}$
$L_n^4=4\phi^{4n}+4(-1)^n\phi^{2n}+4$
$$8(-1)^n\phi^{2n}+8+\phi^{4n}+(\phi)^{-4n}-4\phi^{4n}-4(-1)^n\phi^{2n}-4=2$$
$$3\phi^{4n}-4(-1)^n\phi^{2n}-\phi^{-4n}=2$$
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$4(-1)^nL_n^2=8(-1)^n\phi^{2n}+8$
$L_n^4=4\phi^{4n}+4(-1)^n\phi^{2n}+4$
These are incorrect.
We have
$$\begin{align}4(-1)^nL_n^2&=4(-1)^n(\phi^n+(-\phi)^{-n})^2
\\\\&=4(-1)^n(\phi^{2n}+2\phi^n(-\phi)^{-n}+(-\phi)^{-2n})
\\\\&=4(-1)^n(\phi^{2n}+2(-1)^n+\phi^{-2n})
\\\\&=4(-1)^n\phi^{2n}+8+4(-1)^n\phi^{-2n}
\\\\&=4(-1)^n\phi^{2n}+4(-1)^n\phi^{-2n}+8\end{align}$$
and
$$\begin{align}L_n^4&=(L_n^2)^2
\\\\&=(\phi^{2n}+2(-1)^n+\phi^{-2n})^2
\\\\&=\phi^{4n}+4+\phi^{-4n}+4(-1)^n\phi^{2n}+2+4(-1)^n\phi^{-2n}
\\\\&=\phi^{4n}+\phi^{-4n}+4(-1)^n\phi^{2n}+4(-1)^n\phi^{-2n}+6\end{align}$$
It follows from these that
$$\begin{align}&4(-1)^nL_n^2+L_{4n}-L_n^4
\\\\&=4(-1)^n\phi^{2n}+4(-1)^n\phi^{-2n}+8+(\phi^{4n}+\phi^{-4n})
\\&\qquad\qquad -(\phi^{4n}+\phi^{-4n}+4(-1)^n\phi^{2n}+4(-1)^n\phi^{-2n}+6)
\\\\&=8-6
\\\\&=2\end{align}$$
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On the integral $\int_0^{\sqrt{2}/2} \frac{\arctan \sqrt{1-2t^2}}{1+t^2} \, \mathrm{d}t$ I'm having a difficult time evaluating the integral
$$\mathcal{J} = \int_0^{\sqrt{2}/2} \frac{\arctan \sqrt{1-2t^2}}{1+t^2} \, \mathrm{d}t$$
This is integral arose after simplifying the integral $\displaystyle \int_{0}^{\pi/4 } \arctan \sqrt{\frac{1-\tan^2 x}{2}} \, \mathrm{d}x$;
\begin{align*}
\require{cancel.js}
\int_{0}^{\pi/4} \arctan \sqrt{\frac{1-\tan^2 t}{2}}\, \mathrm{d}t &\overset{1-\tan^2 t \mapsto 2t^2}{=\! =\! =\! =\! =\! =\!=\!=\!} \int_{0}^{\sqrt{2}/2} \frac{t \arctan t}{\sqrt{1-2t^2} \left ( 1-t^2 \right )} \, \mathrm{d}t \\
&=\cancelto{0}{\left [ - \arctan \sqrt{1-2t^2} \arctan t \right ]_0^{\sqrt{2}/2}} + \int_{0}^{\sqrt{2}/2} \frac{\arctan \sqrt{1-2t^2}}{1+t^2} \, \mathrm{d}t
\end{align*}
My main guess is that differentiation under the integral sign is the way to go here. Any ideas?
|
\begin{align}J&=\int_0^{\frac{1}{\sqrt{2}}} \frac{\arctan\left(\sqrt{1-2x^2}\right)}{1+x^2}\,dx\\
&\overset{x=\frac{1}{\sqrt{2}}\sin u}=\frac{1}{\sqrt{2}}\int_0^{\frac{\pi}{2}}\frac{\cos u\arctan(\cos u)}{1+\frac{1}{2}\sin^2 u}\,du\\
&=\sqrt{2}\int_0^{\frac{\pi}{2}}\frac{\cos u\arctan(\cos u)}{2+\sin^2 u}\,du\\
&=\left[\arctan\left(\frac{1}{\sqrt{2}}\sin u\right)\arctan(\cos u)\right]_0^{\frac{\pi}{2}}+\int_0^{\frac{\pi}{2}}\frac{\arctan\left(\frac{1}{\sqrt{2}}\sin u\right)\sin u}{1+\cos^2 u}\,du\\
&=\int_0^{\frac{\pi}{2}}\frac{\arctan\left(\frac{1}{\sqrt{2}}\sin u\right)\sin u}{1+\cos^2 u}\,du\\
&=\int_0^{\frac{\pi}{2}}\int_0^{\frac{1}{\sqrt{2}}}\left(\frac{\sin^2 u}{(1+\cos^2 u)(1+a^2\sin^2 u)}\,da\right)\,du\\
&=\int_0^{\frac{1}{\sqrt{2}}}\left[\frac{\sqrt{2}\arctan\left(\frac{1}{\sqrt{2}}\tan u\right)}{2a^2+1}-\frac{\arctan\left(\sqrt{1+a^2}\tan u\right)}{(2a^2+1)\sqrt{1+a^2}}\right]_{u=0}^{u=\frac{\pi}{2}}\,da\\
&=\frac{\pi}{2}\int_0^{\frac{1}{\sqrt{2}}}\frac{\sqrt{2}}{2a^2+1}\,da-\frac{\pi}{2}\int_0^{\frac{1}{\sqrt{2}}}\frac{1}{(2a^2+1)\sqrt{1+a^2}}\,da\\
&=\frac{\pi}{2}\Big[\arctan\left(\sqrt{2}a\right)\Big]_0^{\frac{1}{\sqrt{2}}}-\frac{\pi}{2}\left[\arctan\left(\frac{a}{\sqrt{1+a^2}}\right)\right]_0^{\frac{1}{\sqrt{2}}}\\
&=\frac{\pi}{2}\times \frac{\pi}{4}-\frac{\pi}{2}\times \frac{\pi}{6}\\
&=\boxed{\frac{\pi^2}{24}}
\end{align}
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|
Find the locus of the complex number from the given data If $|\sqrt 2 z-3+2i|=|z||\sin (\pi/4+\arg z_1) + \cos (3\pi/4-\arg z_1)|$, where $z_1=1+\frac{1}{\sqrt 3} i$, then the locus of $z$ is ...
$$\arg~z_1 =\frac{\pi}{6}$$
So RHS of the equation is $\;|z|\dfrac{1}{\sqrt 2}$.
Then
$\;|\sqrt 2 z-3+2i|=|z|\dfrac{1}{\sqrt 2}$.
How should I proceed?
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$Arg[1+i/\sqrt{3}]= \pi/6$
$$|z\sqrt{2}-3-2i|=|z||\sin(\pi/4+\pi/6)+cos(3\pi/4-\pi/6)|$$
$$\implies |z\sqrt{2}-3-2i|=|z|\sqrt{\frac{1}{2}} \implies
\left|\frac{2z-(3+2i)\sqrt{2}}{z}\right|=1$$
The locus of $z$ is a circle. Let $z=x+iy$,
we have $$(2x-3\sqrt{2})^2+(2y-2\sqrt{2})^2=(x^2+y^2)$$
tHe simplified eq. of circle is
$$(x-2 \sqrt{2})^2+(y-4\sqrt{2}/3)^2=10$$
|
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|
Show that $(n+1)^{2/3} -n^{2/3} <\frac{2}{3} n^{-1/3}$ for all positive integers
I am trying to prove
$$(n+1)^{2/3} -n^{2/3} <\frac{2}{3} n^{-1/3}$$
for all positive integers.
My attempts so far have been to Taylor expand the left hand side:
$$(n+1)^{2/3} -n^{2/3}\\
=n^{2/3}\big((1+1/n)^{2/3} -1\big)\\
=n^{2/3}\left(\sum_{\alpha=0}^\infty\frac{\frac{2}{3}(\frac{2}{3}-1)\cdots(\frac{2}{3}-\alpha+1)}{\alpha!} n^\alpha-1\right)\\
=n^{2/3}\sum_{\alpha=1}^\infty\frac{\frac{2}{3}(\frac{2}{3}-1)\cdots(\frac{2}{3}-\alpha+1)}{\alpha!} n^\alpha$$
I also tried proof by induction. Assume that it's true for n=k, so that
$$(n+1)^{2/3} -n^{2/3} < \frac{2}{3}n^{-1/3}\\
n^{2/3}\big((1+1/n)^{2/3} -1\big)<\frac{2}{3} n^{-1/3}\\
(1+1/n)^{2/3} -1<\frac{2}{3} n^{-1}$$
Then I want to prove that $(n+2)^{2/3} -(n+1)^{2/3} < \frac{2}{3}(n+1)^{-1/3}$. The left hand side is:
$$(n+2)^{2/3} -(n+1)^{2/3}\\
=(n+1)^{2/3}\left[\left(1+\frac{1}{n+1}\right)^{2/3}-1^{2/3}\right]\\
<(n+1)^{2/3}\cdot \frac{2}{3} n^{-1}\\
=\frac{2}{3}\frac{(n+1)^{2/3}}{n^{-1}}$$
But this is bigger than $\frac{2}{3}(n+1)^{-1/3}$, so I am stumped!
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It seems nobody can resist this one :)
$$\begin{align}(n+1)^{2/3}-n^{2/3}&=\left((n+1)^{1/3}-n^{1/3}\right)\left((n+1)^{1/3}+n^{1/3}\right)\\
&=\frac{\left(n+1-n\right)\left((n+1)^{1/3}+n^{1/3}\right)}{(n+1)^{2/3}+n^{1/3}(n+1)^{1/3}+n^{2/3}}\\
&=\frac1{n^{1/3}\left(\frac{(n+1)^{2/3}}{n^{1/3}(n+1)^{1/3}+n^{2/3}}+1\right)}\\
&=\frac1{n^{1/3}\left(\frac{\left(\frac{n+1}{n}\right)^{1/3}}{1+\left(\frac{n}{n+1}\right)^{1/3}}+1\right)}\end{align}$$
And there you have it:
$\frac n{n+1}<1$, so $1+\left(\frac n{n+1}\right)^{1/3}<2$ and
$$\frac{\left(\frac{n+1}{n}\right)^{1/3}}{1+\left(\frac{n}{n+1}\right)^{1/3}}+1>\frac{\left(\frac{n+1}n\right)^{1/3}}2+1>\frac12+1=\frac32$$
So
$$(n+1)^{2/3}-n^{2/3}<\frac1{\frac32n^{1/3}}$$
|
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|
How to show that $\sum_{n=0}^{N}\frac{(-1)^n}{{(N+n)!}(N-n)!}=\frac{1}{2N!^2}$?
How to show that $\displaystyle\sum_{n=0}^{N}\frac{(-1)^n}{{(N+n)!}(N-n)!}=\frac{1}{2N!^2}$?
How show that $(1)$ is:
$$\sum_{n=0}^{N}\frac{(-1)^n{2n \choose n}{N+n \choose N-n}}{[(n+1)(n+1)(n+2)(n+3)\cdots(n+N)]^2}=\frac{1}{2N!^2}\tag1$$
using : $\frac{\Gamma(n+N+1)}{\Gamma(n+1)}=(n+1)(n+2)\cdots(n+N)$
simplify to:
$$\frac{{2n \choose n}{N+n \choose N-n}}{[(n+1)(n+1)(n+2)(n+3)\cdots(n+N)]^2}=\frac{1}{(N-n)!(N+n)!}$$
$$\sum_{n=0}^{N}\frac{(-1)^n}{{(N+n)!}(N-n)!}\tag2$$
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Multiplying both sides by $2(2N)!$, we see the equality is equivalent to
\begin{align}
&2\sum_{n=0}^{N}(-1)^n\binom{2N}{{N+n}}=\binom{2N}{N}.
\end{align}
Now write the left side
\begin{align}
2\sum_{n=0}^{N}(-1)^n\binom{2N}{{N+n}} &=\left(\binom{2N}{N}-\binom{2N}{N+1}+\dots\right) +\left(\binom{2N}{N}-\binom{2N}{N+1}+\dots\right)\\
&=\left(\binom{2N}{N}-\binom{2N}{N-1}+\dots\right)+\left(\binom{2N}{N}-\binom{2N}{N+1}+\dots\right)\\
&=(-1)^{N}\cdot\left(\binom{2N}{0}-\binom{2N}{1}+\dots+\binom{2N}{2N}\right) + \binom{2N}{N}\\
&=(-1)^{N}\cdot(1-1)^{2N} + \binom{2N}{N}\\
&=\binom{2N}{N}\\
\end{align}
|
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|
The maximum an complex expression If Z is a complex number such that :
$|Z+4|\leq3$ , find the maximum value of $|Z+1|$
My turn:
$|(x+4) + yi|\leq 3$
represents the surface of acircle whose center is (-4 ,0) and its radius is 3 , then
$-7\leq x \leq -1$ , then
$-6 \leq (x+1)\leq 0$
Then
$0\leq(x+1)^2 \leq 36 $
And
$-3\leq y \leq 3$
Then
$0\leq y^2 \leq 9$
Then
$0\leq (x+1)^2 + y^2 \leq 45$
Then
$|Z +1|\leq 3\sqrt{5}$
Then the required maximum value is $3\sqrt{5}$
Is the solution correct ?
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As seen from the graph, the maximum $|z+1|$ is at $z=-7$, i.e.
$$|z+1| \le |-7+1| = 6$$
which can also be shown analytically as follows,
$$|z+1|=|z+4+(-3)| \le |z+4| + |-3| \le 3 +3 = 6$$
|
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|
Integrating $\int _0^1\frac{\ln \left(1-x\right)}{x^2+1}\:dx$ When i came across this one i thought of using the substitution $x=\frac{1-t}{1+t}$ that lead to an easy expression.
$$\int _0^1\frac{\ln \left(1-x\right)}{x^2+1}\:dx\overset{x=\frac{1-t}{1+t}}=\ln \left(2\right)\int _0^1\frac{1}{t^2+1}\:dt+\int _0^1\frac{\ln \left(t\right)}{t^2+1}\:dt-\int _0^1\frac{\ln \left(t+1\right)}{t^2+1}\:dt$$
$$=\frac{\pi }{4}\ln \left(2\right)-G-\frac{\pi }{8}\ln \left(2\right)=\boxed{\frac{\pi }{8}\ln \left(2\right)-G}$$
But then i tried to solve it using feynman's trick with the following parameter:
$$I=\int _0^1\frac{\ln \left(1-x\right)}{x^2+1}\:dx$$
$$I\left(a\right)=\int _0^1\frac{\ln \left(1-ax\right)}{x^2+1}\:dx$$
$$I'\left(a\right)=-\int _0^1\frac{x}{\left(x^2+1\right)\left(1-ax\right)}\:dx=\frac{1}{a^2+1}\int _0^1\frac{a-x}{\left(x^2+1\right)}-\frac{a}{1-ax}\:dx$$
$$=\frac{1}{a^2+1}\left(\frac{\pi a}{4}-\frac{\ln \left(2\right)}{2}+\ln \left(1-a\right)\right)\:$$
$$\int _0^1I'\left(a\right)\:da=\frac{\pi }{4}\int _0^1\frac{a}{a^2+1}\:da-\frac{\ln \left(2\right)}{2}\int _0^1\frac{1}{a^2+1}\:da+\int _0^1\frac{\ln \left(1-a\right)}{a^2+1}\:da\:$$
$$I=\frac{\pi }{8}\ln \left(2\right)-\frac{\pi }{8}\ln \left(2\right)+I$$
And everything cancels.
¿Am i doing something wrong?, ¿Should i be using another parameter or other upper and lower bounds?
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You are not allowed to switch the integral and derivative. The conditions of the theorem about switching those are not met. Denote $\frac{\ln(1-ax)}{x^2+1}$ by $f(x,a)$, one of the conditions is that there must exist a function $g:[0,1] \rightarrow \mathbb{C} $ such that $\int_0^1 |g(x)|dx < \infty$ and $| \frac{\partial f}{\partial a}(x,a) |\leq g(x)$ for all $a$.
|
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Given that $a+b+c=0$, show that $2(a^4+b^4+c^4)$ is a perfect square Given that $a+b+c=0$. Show that: $2(a^4+b^4+c^4)$ is a perfect square
MY ATTEMPTS:
I found that when $a+b+c=0$, $a^3+b^3+c^3=3abc$
So I did:
$(a^3+b^3+c^3)(a+b+c)$ -- $a^4+b^4+c^4=-(a^3c+a^3b+ab^3+b^3c+c^3a+c^3b)$
And then I tried to substitute $a^4+b^4+c^4$, but I found nothing that I thought relevant to the question
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From $a+b+c=0$ it follows $c=-a-b$, and then
\begin{align*}
c^4&=a^4+4a^3b+6a^2b^2+4ab^3+b^4\\
\implies \qquad 2(a^4+b^4+c^4)&=4a^4+8a^3b+12a^2b^2+8ab^3+4b^4\\
&=(2a^2+2ab+2b^2)^2
\end{align*}
|
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|
Prove by mathematical induction that $3^n>2n^3$ I'm having trouble with this question:
"Prove by mathematical induction that for all integers $n\ge 6$, $3^n>2n^3$".
I got to $P(k)=2k^3<3^k$ and $P(k+1)=2(k+1)^3<3^{k+1}=2k^3+6k^2+6k+2<3^k*3$,
but I dont know how I can get $P(k+1)$ from $P(K)$...
Thanks
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Note that $f(k) = \frac{k}{k+1}$ is a positive, strictly increasing function for positive $k$ (since $\frac{k}{k+1} = 1 - \frac{1}{k+1}$ and $\frac{1}{k+1}$ is strictly decreasing). Thus, for $k \ge 6$, you have
$$\begin{equation}\begin{aligned}
f^3(k) & = \left(\frac{k}{k+1}\right)^3 \\
& \ge \left(\frac{6}{7}\right)^3 \\
& = \frac{216}{343} \\
& \gt \frac{1}{3}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Using this, you get with the induction part that
$$\begin{equation}\begin{aligned}
3^{k+1} & = 3(3^{k}) \\
& > 3(2k^3) \\
& = 2(3)\left(\frac{k}{k+1}\right)^3(k+1)^3 \\
& \gt 2(3)\left(\frac{1}{3}\right)(k+1)^3 \\
& = 2(k + 1)^3
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Thus, this shows that if $P(k)$ is true, then so is $P(k+1)$.
|
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|
Finding all Lines Passing Through a Point Given the Product of Intercepts How does one find the equation of all lines passing through a point (Ex. $(6, -1)$), satisfying the condition that the product of their $x$ and $y$ intercepts must equal some number $c$ (Ex. $3$)?
As far as I understand this can be conceptualized as finding the equation of the line containing the points $(6,-1), (a,0), (0,b)$ where $a$ is the $x$-intercept, $b$ is the $y$-intercept and $ab=3$.
I've tried finding the slope with $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$ and have gotten $m=\frac{-1}{6-a}$ and $m=\frac{-1-b}{6}$. I've also solved for $a$ and $b$ in terms of $m$ and tried susbstituting these values into the equation of a line ($y=mx+b$) but I just cant eliminate enough variables to solve for anything useful. Feel like I'm missing something obvious, I'm not even sure if there is more than one equation that satisfies the conditions.
|
So I came across a similar issue and I couldn't find any answers which were satisfactory. So I dug up as much information as I could and I found a solution. Here goes...
We have a known point, $(6,-1)$. For any line that passes through this point, it must have an x-intercept of $(a,0)$ and a y-intercept of $(0,b)$. We don't know what $a$ and $b$ are as of yet, but we do know that their product is 3. By this definition, we also know that any line cannot be horizontal or vertical, since those lines would not have an x-intercept and y-intercept, respectively.
Let's take a look at our line equation: $y=mx+b$. We know that $a⋅b=3$ so with some rearranging we have $b=\frac{3}{a}$. We can sub this in for $b$, such that:
$y=mx+\frac{3}{a}$
We now need to find the slope. We can't explicitly define what the slope is yet, since that's part of what the question is asking. But we can substitute in what we do know. The slope is defined as:
$m=\frac{y_1-y_0}{x_1-x_0}$
This is just the distance between our two points along the graph, in each axis. We only know one point $(6, -1)$ but we can substitute in our x-intercept placeholder $(a,0)$ so our slope becomes:
$m=\frac{0-(-1)}{a-6}=\frac{1}{a-6}$
We can now sub this into our line equation as well, such that:
$y=\frac{1}{a-6}x+\frac{3}{a}$
Then we myltiply by $x$ which is really just $\frac{x}{1}$ such that:
$y=\frac{x}{a-6}+\frac{3}{a}$
We now have a fraction addition, and we do this just like any other: making the denominators the same. Multiply each fraction by the equivalent of $\frac{1}{1}$, using the denominator of the other fraction:
$y=\frac{x(a)}{(a-6)(a)}+\frac{3(a-6)}{a(a-6)}$
$y=\frac{a(x) + 3(a-6)}{a(a-6)}$
After some distribution we come to:
$y=\frac{ax+3a-18}{a^2-6a}$
Multiplying both sides by $(a^2-6a)$ and a bit more distribution we get:
$y(a^2-6a)=ax+3a-18$
$a^2y-6ay=ax+3a-18$
For this next step, we can sub in our known point $(6,-1)$ for $x$ and $y$. Remember that in our original formula $y=mx+b$, we already knew what $(x,y)$ was, as it was given to us as part of the question. The information we're missing are the slope $(m)$ and the y-intercept $(b)$. In fact, we could've substituted in our known point for $(x,y)$ right from the very start.
After some more rearranging and collecting like terms, we can get to an expression in standard polynomial form:
$a^2(-1)-6a(-1)=a(6)+3a-18$
$-a^2+6a=9a-18$
$-a^2-3a+18=0$
It's at this point we can use the quadratic formula to solve for a. But for clarity, let's substitute x for a:
$-x^2-3x+18=0$
For a polynomial in standard form $(ax^2+bx+c=0)$ the quadratic formula is defined as:
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
Looking at our equation so far, we can sub in our known values of $(a,b,c)$ as $(-1,-3,18)$ into the quadratic formula, such that:
$x=\frac{-(-3)\pm\sqrt{(-3)^2-4(-1)(18)}}{2(-1)} \Rightarrow x=\frac{3\pm\sqrt{9+72}}{-2} \Rightarrow x=\frac{3\pm\sqrt{81}}{-2} \Rightarrow x=\frac{3\pm9}{-2}=-6,3$
At last, we have two points at which any line satisfying our conditions intersects the x-axis: $(-6,0)$ and $(3,0)$. We now have enough information to draw our lines and find the equations. For $y=mx+b$ the two pieces of information we need to find are $m$ and $b$.
For $m$, we just use the slope formula from before:
$m=\frac{y_1-y_0}{x_1-x_0}$
But now we can substitute all values for both our slopes. $(x_0,y_0)$ is our original known point, and $(x_1,y_1)$ are the points we just calculated. Since both points are the x-intercept, their y-coordinate will be 0:
$m_1=\frac{0-(-1)}{-6-6} \Rightarrow m_1=-\frac{1}{12}$
$m_2=\frac{0-(-1)}{3-6} \Rightarrow m_2=-\frac{1}{3}$
Finally, we just need to find $b$. To do so, we just need to sub in a known $(x,y)$ and solve for $b$. Since we know that both lines pass through $(6,-1)$ we can use it for both:
$y=mx+b$
$-1=-\frac{1}{12}(6)+b_1 \Rightarrow b_1=\frac{6}{12}-1 \Rightarrow b_1=-\frac{1}{2}$
$-1=-\frac{1}{3}(6)+b_2 \Rightarrow b_2=\frac{6}{3}-1 \Rightarrow b_2=2-1 \Rightarrow b_2=1$
We now have all the information we need to write our equations, for all lines which pass through $(6,-1)$ and for which the products of their $x/y$ intercepts is 3:
$y=-\frac{1}{12}x-\frac{1}{2}$, and $y=-\frac{1}{3}x+1$
You can see this Desmos graph to get a visualisation.
|
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|
What are the integer solutions to $a^{b^2} = b^a$ with $a, b \ge 2$ I saw this in quora.
What are all the
integer solutions to
$a^{b^2} = b^a$
with $a, b \ge 2$?
Solutions I have found so far:
$a = 2^4 = 16, b = 2,
a^{b^2}
= 2^{4\cdot 4}
=2^{16},
b^a = 2^{16}
$.
$a = 3^3, b = 3,
a^{b^2} = 3^{3\cdot 9}
=3^{27},
b^a = 3^{3^3}
=3^{27}
$.
In the general case,
$a$ and $b$ have the
same set of prime divisors,
so let
$a
=\prod_P p_i^{a_i}$,
$b
=\prod_P p_i^{b_i}
$
with each
$a_i \ge 1, b_i \ge 1$.
$b^a
=b^{\prod p_i^{a_i}}
=(\prod p_j^{b_j})^{\prod p_i^{a_i}}
=\prod p_j^{b_j\prod p_i^{a_i}}
$
$a^{b^2}
=a^{\prod p_i^{2b_i}}
=(\prod p_j^{a_j})^{\prod p_i^{2b_i}}
=\prod p_j^{a_j\prod p_i^{2b_i}}
$
Therefore,
for each $p_j$,
$b_j\prod p_i^{a_i}
=a_j\prod p_i^{2b_i}
$.
I haven't gotten any further than this.
I conjecture that
there are no other solutions.
|
IMO 1997, Problem B2
Find all pairs $(a, b)$ of positive integers that satisfy $a^{b^2} = b^a$.
Answer
$(1,1)$, $(16,2)$, $(27,3)$.
Solution
Notice first that if we have $a^m = b^n$, then we must have $a = c^e$, $b = c^f$, for some $c$, where $m=fd$, $n=ed$ and $d$ is the greatest common divisor of $m$ and $n$.
[Proof: express $a$ and $b$ as products of primes in the usual way.]
In this case let $d$ be the greatest common divisor of $a$ and $b^2$, and put $a = de$, $b^2 = df$. Then for some $c$, $a = ce$, $b = cf$. Hence $f c^e = e c^{2f}$. We cannot have $e = 2f$, for then the $c$'s cancel to give $e = f$. Contradiction.
Suppose $2f > e$, then $f = e c^{2f-e}$. Hence $e = 1$ and $f = c^{2f-1}$. If $c = 1$, then $f = 1$ and we have the solution $a = b = 1$. If $c ≥ 2$, then $c^{2f-1} ≥ 2^f > f$, so there are no solutions.
Finally, suppose $2f < e$. Then $e = f c^{e-2f}$. Hence $f = 1$ and $e = c^{e-2}$. $c^{e-2} ≥ 2^{e-2} ≥ e$ for $e ≥ 5$, so we must have $e = 3$ or $4$ ($e > 2f = 2$). $e = 3$ gives the solution $a = 27$, $b = 3$. $e = 4$ gives the solution $a = 16$, $b = 2$.
P.S. Since the website I referred to in the comment above has been moved in the past, I don't know how permanent this link will be. So I quoted the solution here.
|
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|
How can I solve this geometry problem without trigonometry?
Let $ABC$ be a triangle with $D$ on the side $AC$ such that $\angle DBC=42^\circ$ and $\angle DCB=84^\circ$. If $AD = BC$, find $x = \angle DAB$.
It's supposed to be solved with constructions, but I couldn't figure out. See the trigonometric version here.
|
Let $E$ be such that $BCDE$ is an isosceles trapezoid with bases $BC$, $DE$. Then
$$\angle EBD = \angle EBC - \angle DBC = 84^\circ - 42^\circ = 42^\circ = \angle DBC = \angle BDE$$
thus triangle $EBD$ is isosceles with $BE=DE$.
Note that $ED=BE=DC$, $AD=BC$, and $\angle EDA = \angle BCD$. Hence triangle $EDA$ is congruent to $DCB$. Therefore, if we let $F$ be such that $DAFE$ is an isosceles trapezoid with bases $AD$, $EF$ then $ADEF$ is congruent to $BCDE$ and in particular $DF=DB$.
Let $G$ be a point such that triangle $BDG$ is equilateral. So, $B, G, F$ lie on a circle with center $D$. Hence $$\angle BFG = \frac 12 \angle BDG = 30^\circ.$$ Also, since $EF=EB$, we have $$\angle EFB = 90^\circ - \frac 12 \angle BEF = 90^\circ - 84^\circ = 6^\circ.$$
Thus $$\angle EFG=\angle EFB+\angle BFG=6^\circ+30^\circ=36^\circ.$$
On the other hand, clearly $\angle GED = 180^\circ -\frac 12\angle DEB = 180^\circ - \frac 12 96^\circ = 132^\circ$, hence
$$\angle GEF = \angle GED - \angle FED = 132^\circ - 96^\circ = 36^\circ.$$
So $\angle EFG = \angle GEF$. It follows that $FG=EG$. Therefore $G$ lies on the perpendicular bisector of $EF$ which coincides with the perpendicular bisector of $AD$. Thus $GA=GD$.
Finally, since $GA=GD=GB$, points $A,D,B$ lie on a circle with center $G$. Hence
$$\angle DAB = \frac 12 \angle DGB =\frac 12 60^\circ = 30^\circ.$$
|
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|
Evaluate $\sum _{n=1}^{\infty } \frac{1}{n^5 2^n \binom{3 n}{n}}$ in terms of elementary constants How can we evaluate
$$\sum _{n=1}^{\infty } \frac{1}{n^5 2^n \binom{3 n}{n}}$$
Related: $\sum _{n=1}^{\infty } \frac{1}{n^4 2^n \binom{3 n}{n}}$ is solved recently (see here for the solution) by elementary method, so I wonder if a generalization to weight $5$ can be made.
|
This is by no means a solution but an extended comment which shows that this sum - generalized to a generating function - belongs to the well-known class of special functions, the hypergeometric functions. Maybe this can be of some use.
Defining the generating function
$$g(q,z) = \sum _{n=1}^{\infty } \frac{z^n}{n^q\binom{3 n}{n} }\tag{1}$$
the left hand side of the identity in question is the generating function for $q=5$ at the point $z=\frac{1}{2}$.
$$\sum _{n=1}^{\infty } \frac{1}{n^5 2^n \binom{3 n}{n}} = g(5,\frac{1}{2})\tag{2}$$
We have
$$g(0,z)= \sum _{n=1}^{\infty } \frac{z^n}{\binom{3 n}{n}}=\frac{z}{3} \; _3F_2\left(1,\frac{3}{2},2;\frac{4}{3},\frac{5}{3};\frac{4 z}{27}\right)\tag{3}$$
and, recursively, with the indefinite integral
$$g(q+1,z)=\int \frac{1}{z} g(q,z)\,dz\tag{4}$$
Explicitly,
$\begin{align}
& g(1,z) = \frac{z}{3} \; _3F_2\left(1,1,\frac{3}{2};\frac{4}{3},\frac{5}{3};\frac{4 z}{27}\right)\\\\
& g(2,z) =\frac{z}{3} \; _4F_3\left(1,1,1,\frac{3}{2};\frac{4}{3},\frac{5}{3},2;\frac{4 z}{27}\right)\\\\
&g(3,z)= \frac{z}{3} \; _5F_4\left(1,1,1,1,\frac{3}{2};\frac{4}{3},\frac{5}{3},2,2;\frac{4 z}{27}\right)\\\\
\ldots\\\\
& g(q,z) =\frac{z}{3} \; _P F_Q\left(1_{1},1_{2},1_{3},\ldots,1_{q+1},\frac{3}{2};\frac{4}{3},\frac{5}{3},2_{1},2_{2},\ldots,2_{q-1};\frac{4 z}{27}\right), \\\\
& P=q+2, Q=q+1\\
\end{align}\tag{5}$
In the end we are interested in these functions at $z=\frac{1}{2}$.
Hopefully there are some means of expanding these hypergeometric functions in terms of the more common function types like zeta, polylog, polygamma (specifically harmonic number).
Extension
Similarly we can study the family of sums
$$g_{k,q}(z)=\sum _{n=1}^{\infty }\frac{1}{n^q} \frac{z^n}{\binom{k\; n}{n}}\tag{6}$$
These generating functions are also expressible in terms of hypergeometric functions.
|
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If $\log_8 3 = P$ and $\log_3 5 = Q$, express $\log_{10} 5$ in terms of $P$ and $Q$. If $\log_8 3 = P$ and $\log_3 5 = Q$, express $\log_{10} 5$ in terms of $P$ and $Q$. Your answer should no longer include any logarithms.
I noted that $\log_5 10=\frac{1}{\log_{10} 5}.$
I also noted that $\log_{5} 10=\log_5 2+\log_5 5=\log_5 2+1.$ I don't know how to continue, how do I finish this problem using my strategy?
|
An attempt. But I'm not sure whether the assumption I make at the beginning is correct, or whether it is arbitrary.
We have :
$P = \log_8 3 \iff 8^P = 3$
and
$Q = log_3 5 \iff 3^Q = 5$.
Suppose that : $log_{10} 5 = kPQ$
$log_{10} 5 = kPQ$
$\rightarrow 10^{log_{10} 5} = 10^{kPQ}$
$\rightarrow 5 = 10^{kPQ}$
$\rightarrow 3^Q = 10^{kPQ}$
$\rightarrow {(8^P)}^Q = 10^{kPQ}$
$\rightarrow 8^{PQ} = 10^{kPQ}$
$\rightarrow 8^{PQ} = {(8^{\log_{8}{10}})}^{kPQ}$
$\rightarrow 8^{PQ} = {8^{(\log_{8}{10\times kPQ)}}}$
$\rightarrow PQ = \log_{8}{10\times kPQ}$
$\rightarrow 1 = \log_{8}{10\times k}$ ( Dividing by $PQ$ on both sides).
$\rightarrow k = \frac {1} { \log_{8}{10}}$
Therefore, $\log_{10}5 = k\times PQ = \frac {1} { \log_{8}{10}} \times PQ$
Symbolab gives $\frac {1} { \log_{8}{10}} \times PQ = \frac {\ln 5} {\ln 10}$
But $\frac {\ln 5} {\ln 10} = \log_{10} 5$ using the change of base formula , with base $e$.
So, number $k$ seems to do the job.
|
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|
If a,b,c be roots of $2x^3+x^2+x-1=0$ show that some expression is equal to 16. If $a,b,c$ are roots of $2x^3+x^2+x-1=0,$ show that
$$\bigg(\frac{1}{b^3}+\frac{1}{c^3}-\frac{1}{a^3}\bigg)\bigg(\frac{1}{c^3}+\frac{1}{a^3}-\frac{1}{b^3}\bigg)\bigg(\frac{1}{a^3}+\frac{1}{b^3}-\frac{1}{c^3}\bigg)=16$$
My attempt:
Let $\frac{1}{a}=p,\frac{1}{b}=q,\frac{1}{c}=r$
$p+q+r=1$
$pqr=2$
$$pq+qr+rp=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}$$
$$=\frac{ab+bc+b^2}{(abc)^2}$$
$$=4\bigg(\frac{1}{pq}+\frac{1}{qr}+\frac{1}{q^2}\bigg)$$
$$pq+qr+rp=4\bigg(\frac{pq+qr+rp}{pq^2r}\bigg)$$
$$pq^2r=4$$
$$\implies q=2 \implies b=\frac{1}{2}$$
So p, r are roots of $x^2+x+1=0$
$\implies p^3=q^3=1$
But this condition gives a different value of the required expression, so what am I doing wrong? Please tell me the right solution.
|
Hint:
Set $x^3=y$
$$(2x^3-1)^3=-(x^2+x)^3$$
$$(2y-1)^3=-(y)^2-y-2y\iff?\ \ \ \ (2)$$ whose roots are $a^3,b^3,c^3$
Now $$z=\dfrac1{a^3}+\dfrac1{b^3}-\dfrac1{c^3}=\dfrac1{a^3}+\dfrac1{b^3}+\dfrac1{c^3}-\dfrac2{c^3}=\dfrac{a^3b^3+b^3c^3+c^3a^3}{(abc)^3}-\dfrac2{c^3}$$
$$\implies\dfrac2{c^3}=?\iff c^3=?$$
Now as $c^3$ is a root of $(2),$ replace the value of $c^3$ in terms of $z$ to form a cubic equation $z$
|
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|
Integration of Exponential with Polynomial in its Exponent I seek to solve the following integral:
$$
\int_0^T dt \frac{1}{\sqrt{4 \pi s^2 t}}\cdot \exp\left({-\frac{(x-vt)^2}{4 s^2t}}\right)
$$
My first idea was to substitute $1/\sqrt{t}$ using $u=\frac{1}{2} \sqrt{t}$ and $du = dt \frac{1}{\sqrt{t}}$, yielding
$$
\int_0^{\sqrt{T}/2} du \exp \left(- \frac{(x - 4u^2 v)^2}{16 s^2 u^2}\right)
$$
From there, I thought about expanding the term according to the binomial formula and obtain three exponential terms, one that has $u^2$ in its power, one with $u^0$ that I can pull out of the integral and one with $u^{-2}$.
However, partially integrating the $u^2$ term using further substitutions did not work out, and in the end everything cancels and yields 0.
I am aware that the error function is involved with the following term: $\int_0^b du \exp(-u^2) = \sqrt{\pi}/2 \cdot \operatorname{erf}(b)$
I appreciate any help! Thank you!
|
For $a,b>0$ we have
\begin{align}
\int \limits_0^a \mathrm{e}^{-b \left(x-\frac{1}{x}\right)^2} \, \mathrm{d} x &= \frac{1}{2} \int \limits_0^a \mathrm{e}^{-b \left(x-\frac{1}{x}\right)^2} \left(1 + \frac{1}{x^2} + 1 - \frac{1}{x^2}\right) \mathrm{d} x \\
&= \frac{1}{2} \int \limits_0^a \mathrm{e}^{-b \left(x-\frac{1}{x}\right)^2} \left(1 + \frac{1}{x^2}\right) \mathrm{d} x + \frac{1}{2} \mathrm{e}^{4 b}\int \limits_0^a \mathrm{e}^{-b \left(x+\frac{1}{x}\right)^2} \left(1 - \frac{1}{x^2}\right) \mathrm{d} x \\
&= \frac{1}{2} \frac{\sqrt{\pi}}{2 \sqrt{b}} \left(\operatorname{erf}\left[\sqrt{b} \left(a - \frac{1}{a}\right)\right] +1\right) + \frac{1}{2} \mathrm{e}^{4 b} \frac{\sqrt{\pi}}{2 \sqrt{b}} \left(\operatorname{erf}\left[\sqrt{b} \left(a + \frac{1}{a}\right)\right] - 1\right) \\
&= \frac{\sqrt{\pi}}{4 \sqrt{b}} \left(\operatorname{erfc}\left[\sqrt{b} \left(\frac{1}{a} - a\right)\right] - \mathrm{e}^{4b} \operatorname{erfc}\left[\sqrt{b} \left(\frac{1}{a} + a\right)\right]\right) \, .
\end{align}
Now let $T,s>0$ and (for now) $x,v>0$. Then your integral (divided by $T$ to make the result a density) is
$$ \frac{1}{T} \int \limits_0^T \frac{\mathrm{e}^{-\frac{(x-vt)^2}{4s^2 t}}}{\sqrt{4 \pi s^2 t}} \, \mathrm{d} t \stackrel{t = \frac{x}{v} \xi^2}{=} \sqrt{\frac{x}{\pi v}} \frac{1}{sT} \int \limits_0^{\sqrt{\frac{v T}{x}}} \mathrm{e}^{- \frac{v x}{4 s^2} \left(\xi - \frac{1}{\xi}\right)^2}\, \mathrm{d} \xi = \frac{\operatorname{erfc} \left(\frac{x - vT}{2 s \sqrt{T}}\right) - \mathrm{e}^{\frac{v x}{s^2}} \operatorname{erfc} \left(\frac{x + vT}{2 s \sqrt{T}}\right)}{2 v T} \, . $$
For $x < 0$ the exponent becomes $- \frac{(\lvert x \rvert + v t)^2}{4s^2t}$ and a few signs change, but the calculation is similar. The final result for $x \in \mathbb{R} \setminus \{0\}$ is
\begin{align}
\frac{1}{T} \int \limits_0^T \frac{\mathrm{e}^{-\frac{(x-vt)^2}{4s^2 t}}}{\sqrt{4 \pi s^2 t}} \, \mathrm{d} t &= \operatorname{sgn}(x) \frac{\operatorname{erfc} \left(\operatorname{sgn}(x) \frac{x - vT}{2 s \sqrt{T}}\right) - \mathrm{e}^{\frac{v x}{s^2}} \operatorname{erfc} \left(\operatorname{sgn}(x) \frac{x + vT}{2 s \sqrt{T}}\right)}{2 v T} \\
&= \frac{\operatorname{sgn}(x)-\operatorname{erf} \left( \frac{x - vT}{2 s \sqrt{T}}\right) - \mathrm{e}^{\frac{v x}{s^2}} \left[\operatorname{sgn}(x) -\operatorname{erf} \left(\frac{x + vT}{2 s \sqrt{T}}\right)\right]}{2 v T} \, .
\end{align}
The second expression is correct for $x = 0$ as well.
For $v < 0$ the result remains unchanged and for $v=0$ we can let $\frac{x^2}{4 s^2 t} = \frac{1}{u^2}$ and integrate by parts to find
$$ \frac{1}{T} \int \limits_0^T \frac{\mathrm{e}^{-\frac{x^2}{4s^2 t}}}{\sqrt{4 \pi s^2 t}} \, \mathrm{d} t = \frac{\mathrm{e}^{-\frac{x^2}{4s^2 T}}}{\sqrt{\pi s^2 T}} - \frac{\rvert x\lvert}{2 s^2 T} \operatorname{erfc} \left(\frac{\rvert x\lvert}{2 s \sqrt{T}}\right) \, .$$
|
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|
Proving $\sum\limits_{\rm cyc} 1/(a^2 -b+4) \geq3/4$ Suppose $a,b,c\in\mathbb R^+$ with $a+b+c=3.$ Prove that $$\frac{1}{a^2- b+4}+\frac{1}{b^2-c+4}+\frac{1}{c^2- a+4}\geqslant\frac{3}{4}.$$
I tried various approaches, but nothing seems to work. Whenever I give a lower bound using $C-S$ or $AM-GM$ it becomes very weak. And I am seeking solution that uses not so much computation.
|
By C-S
$$\sum_{cyc}\frac{1}{a^2-b+4}=\sum_{cyc}\frac{(a+5b+3c)^2}{(a+5b+3c)^2(a^2-b+4)}\geq\frac{81(a+b+c)^2}{\sum\limits_{cyc}(a+5b+3c)^2(a^2-b+4)}.$$
Thus, it's enough to prove that:
$$\frac{81(a+b+c)^2}{\sum\limits_{cyc}(a+5b+3c)^2(a^2-b+4)}\geq\frac{3}{4},$$ which is true and nice!
After homogenization we need to prove that
$$\sum_{cyc}(17a^4+23a^3b+23a^3c-78a^2b^2+15a^2bc)\geq0$$ or
$$17\sum_{cyc}(a^4-a^3b-a^3c+a^2bc)+\sum_{cyc}(40a^3b+40a^3c-78a^2b^2-2a^2bc)\geq0,$$ which is true by Schur and Muirhead.
|
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|
Proving the limit of $\frac{n+2}{n^3 -4n-2}$ using first principles. I'm trying to prove the limit of the sequence of $x_n=\frac{n+2}{n^3 -4n-2}$ using first principles. Here's what I've managed.
First I find the proposed limit using the algebra of limits.
$$x_n=\frac{n+2}{n^3 -4n-2} = \frac{\frac{1}{n^2}+\frac{2}{n^3}}{1-\frac{4}{n^2}-\frac{2}{n^3}}$$
Therefore as $n \rightarrow \infty$ the limit is proposed to be $0$.
Fix $\epsilon >0 $ and $N \in \mathbb{N}$,
$$ n\geq N \implies \bigg| \frac{n+2}{n^3 -4n-2} -0\bigg| < \epsilon$$
For all $n \geq 3$.
$$\bigg| \frac{n+2}{n^3 -4n-2} -0\bigg| = \frac{n+2}{n^3 -4n-2}$$
$$\frac{n+2}{n^3 -4n-2} \leq \frac{2}{n} < \epsilon \iff \frac{2}{\epsilon} < n $$
Therefore choose $N$ greater than $Max\{3,\frac{2}{\epsilon}\}$, has the required property.
Any feedback would be much appreciated
|
Your proof is fine, except you have not provided details of how $\frac{n+2}{n^3-4n-2} \leq \frac 2n$ for $n \geq 3$.
How would I prove this? Cross multiply first, to see that it is enough to prove $n^2+2n \leq 2n^3-8n -4$ for $n \geq 3$. At $n=3$, this inequality is true by checking, and if it is true for $n=k$, then when I move to $k+1$ the LHS increases by $2k+3$ while the RHS increases by $6k^2+6k+2 - 16$, which by usual theory for quadratics is seen to be bigger than $2k-2$ for $k \geq 3$.
For an alternate proof, using the heuristic, you would notice :
$$
\frac{\frac 1{n^2} + \frac 2{n^3}}{1 - \frac 4{n^2} - \frac 2{n^3}} = \frac {a_n}{b_n}
$$
where $a_n$ goes to $0$ monotonically and $b_n$ goes to $1$ monotonically as $n \to \infty$. Therefore, you "mock" the proof of quotient of limits as follows : for $\epsilon > 0$, choose $N$ large enough so that $a_n < \frac \epsilon 2$ for all $n > N$, and $b_n > \frac 12$ for all $n > N$ (by taking a maximum). The quotient is then at most $\epsilon$ (it has to be positive for such large $N$, so must lie in $(0,\epsilon)$) . Conclude.
|
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Find the minimum value of $a^2+b^2+c^2+2abc$ when $a+b+c=3$ and $a,b,c\geq0$.
Given $a,b,c\geq0$ such that $a+b+c=3$, find the minimum value of $$P=a^2+b^2+c^2+2abc.$$
It seems like the minimum value of $P$ is $5$ when $a=b=c=1$, but I can find at least one example where $P<5$.
My attempt:
Without loss of generality, I can suppose that $a\geq b\geq c$ and so $a\geq 1$. Therefore I have:
$$P\geq a^2+b^2+c^2+2bc=a^2+(b+c)^2\geq \frac{(a+b+c)^2}{2}=\frac{9}{2}.$$
The issue with this is that equality doesn't occur with this method.
What's your take on the problem?
|
It occurs for $c=0$ and $a=b=\frac{3}{2}$ and we obtain a value $\frac{9}{2}$.
Also, you proved that it's a minimal value.
I like the following way.
We need to prove that
$$(a+b+c)(a^2+b^2+c^2)+6abc\geq\frac{1}{2}(a+b+c)^3$$ or
$$\sum_{cyc}(a^3-a^2b-a^2c+2abc)\geq0,$$ which is true by Schur.
|
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|
Fastest way to solve $x^3\equiv x \pmod{105}$ $$x^3\equiv x \pmod{105}$$
I'm trying to solve this equation. Here's what I tried so far:
$$x^3\equiv x \pmod{105} \iff x^2\equiv 1 \pmod{105}$$
Then, applying the Chinese remainder theorem, I got the system:
$$\cases{x^2 \equiv 1 \pmod{5}\\x^2 \equiv 1 \pmod{7}\\x^2 \equiv 1 \pmod{3}}$$
With the following solutions:
$$\cases{x \equiv \pm1 \pmod{5}\\x \equiv \pm1 \pmod{7}\\x \equiv \pm1 \pmod{3}}$$
At this point, I think I need to pretty much solve these eight systems:
$$\cases{x \equiv 1 \pmod{5}\\x \equiv 1 \pmod{7}\\x \equiv 1 \pmod{3}}
\cases{x \equiv 1 \pmod{5}\\x \equiv 1 \pmod{7}\\x \equiv -1 \pmod{3}}
\cases{x \equiv 1 \pmod{5}\\x \equiv -1 \pmod{7}\\x \equiv 1 \pmod{3}}
\cases{x \equiv -1 \pmod{5}\\x \equiv 1 \pmod{7}\\x \equiv 1 \pmod{3}}$$$$
\cases{x \equiv -1 \pmod{5}\\x \equiv 1 \pmod{7}\\x \equiv -1 \pmod{3}}
\cases{x \equiv -1 \pmod{5}\\x \equiv -1 \pmod{7}\\x \equiv 1 \pmod{3}}
\cases{x \equiv 1 \pmod{5}\\x \equiv -1 \pmod{7}\\x \equiv -1 \pmod{3}}
\cases{x \equiv -1 \pmod{5}\\x \equiv -1 \pmod{7}\\x \equiv -1 \pmod{3}}$$
Here's how I solved the first one:
Considering the first two equations, we get:
$$x=5k+1=7h+1$$ from which $k = 7+7y, h = 5+5y$, with $y \in \mathbb{Z}$. Therefore, $$x=36+35y\iff x\equiv1\pmod{35}$$
Adding in the third equation, we have that $36+35y = 1+3 w$, from which $x = 1281 + 35w \iff x \equiv1\pmod{105}$.
However, this one seems like a really tedious method as I'd have to do the same calculations for seven more systems. Is there anything I'm missing? Is there a faster way to do this?
|
Hint $ $ It's always true mod $3,\,$ so by CRT we need only combine all roots $\{0,\pm1\}$ mod $5$ and $7,\,$ and $\,x\equiv a\pmod{\!5},\,x\equiv b\pmod{\!7}\!\iff\! x\equiv b+14(b-a)\pmod{\!35}.\,$ For $\,a,b\in \{0,\pm1\}$ this yields $\,x\equiv \pm \{0,1,6,14,15\}\pmod{\!35}$
|
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|
Find all real solutions for $x$ in $ 2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2 . $ Find all real solutions for $x$ in
$$ 2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2 . $$
I noticed that $$2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2=2(2^x−1)(x^2-1)+(2^{x^2}−2)x=0.$$
I know that $2^{n+1}-1$ will always be greater than $2^n-1$ for all positive integers. $2^{n-1}-1$ will always be less than $2^n-1$.
How do I solve this problem?
|
$1$, $-1$ and $0$ are roots.
Let $x\notin\{0,1,-1\}$.
Thus, $$2(2^x-1)x^2+2x(2^{x^2-1}-1)=2(2^x-1)$$ or
$$(2^x-1)(x^2-1)+(2^{x^2-1}-1)x=0$$ or
$$\frac{2^x-1}{x}+\frac{2^{x^2-1}-1}{x^2-1}=0,$$ which has no real roots because for any $a\neq0$ we have $$\frac{2^a-1}{a}>0.$$
|
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|
Question about $\lim\limits_{x \to2} \frac{\sqrt[3]{x} - \sqrt[3]{2}}{\sqrt{x} - \sqrt{2}}$ $\lim\limits_{x \to2} \frac{\sqrt[3]{x} - \sqrt[3]{2}}{\sqrt{x} - \sqrt{2}}$
Let $x=t^6$, then it becomes $\lim\frac{t^2 - (\sqrt[6]{2})^2}{t^3 - (\sqrt[6]{2})^3}$=$\lim\frac{(t+\sqrt[6]{2})(t-\sqrt[6]{2})}{(t-\sqrt[6]{2})(t^2+t\sqrt[6]{2}+(\sqrt[6]{2})^2}$=$\lim\frac{(t+\sqrt[6]{2})}{(t^2+t\sqrt[6]{2}+(\sqrt[6]{2})^2}$
But when $x\to 2$, t can go to $\sqrt[6]{2}$ or -$\sqrt[6]{2}$, which gives two different limits, where I was wrong? Thanks!
|
Let $x=y+2$
$$\frac{\sqrt[3]{x} - \sqrt[3]{2}}{\sqrt{x} - \sqrt{2}}=\frac{\sqrt[3]{y+2} - \sqrt[3]{2}}{\sqrt{y+2} - \sqrt{2}}$$ and use the binomial theorem or Taylor series around $y=0$. Using Taylor, you would have
$$\frac{\frac{y}{3\ 2^{2/3}}-\frac{y^2}{18\ 2^{2/3}}+O\left(y^3\right) } {\frac{y}{2 \sqrt{2}}-\frac{y^2}{16 \sqrt{2}}+O\left(y^3\right) }$$ Divide top and bottom by $y$ and long division to get
$$\frac{2^{5/6}}{3}-\frac{y}{36 \sqrt[6]{2}}+O\left(y^2\right)$$ which shows the limit and also how it is approached.
|
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Show that $\int \frac{dx}{(a+b\cos x)^n}=\frac{A\sin x}{(a+b\cos x)^{n-1}}+B\int {dx \over (a+b\cos x)^{n-1}}+C\int {dx \over (a+b\cos x)^{n-2}}$ I've been trying to develop this integral in parts twice, but I can't come up with anything meaningful,
Could you give me some hints on how to develop this exercise?
Prove that
$$\int \frac{dx}{(a+b\cos x)^n}=\frac{A\sin x}{(a+b\cos x)^{n-1}}+B\int {dx \over (a+b\cos x)^{n-1}}+C\int {dx \over (a+b\cos x)^{n-2}}, \hspace{1cm} (|a|\neq |b|). $$
and determine the coefficients $A,B$ and $C$, if $n$ is a natural number greater than the unit
|
Note,
$$\frac b{\sin x} \left(\frac{\sin^2x}{(a+b\cos x)^{n-1}}\right)'
=\frac{(n-1)(b^2-a^2)}{(a+b\cos x)^n}+\frac{2a(n-2)}{(a+b\cos x)^{n-1}}
-\frac{n-3}{(a+b\cos x)^{n-2}}$$
Integrate both sides and denote $I_n= \int \frac{dx}{(a+b\cos x)^n}$,
$$(n-1)(b^2-a^2)I_n + 2a(n-2)I_{n-1} -(n-3)I_{n-2} \\
= \int \frac b{\sin x} d\left(\frac{\sin^2x}{(a+b\cos x)^{n-1}}\right)
= \frac{b\sin x}{(a+b\cos x)^{n-1}}-aI_{n-1} +I_{n-2} $$
which leads to
$$I_n = \frac{A\sin x}{(a+b\cos x)^{n-1}}+BI_{n-1}+CI_{n-2} $$
where,
$$A = \frac b{(n-1)(b^2-a^2)},\>\>\>
B = -\frac {(2n-3)a}{(n-1)(b^2-a^2)},\>\>\>
C = \frac {n-2}{(n-1)(b^2-a^2)}$$
|
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|
Is there a simple proof for the behaviour of this solution? Let $0 <s \le 1$, and suppose that $0 <b \le a$ satisfy
$$ ab=s,a+b=1+\sqrt{s}.$$
Then $a \ge 1$.
I have a proof for this claim (see below), but I wonder if there are easier or alternative proofs.
In particular, my proof is based on explicit computation of $a,b$ in terms of $s$ (solving the quadratic). Can we avoid that?
The proof:
We have $a+\frac{s}{a}=1+\sqrt s$, or
$$ a^2-(1+\sqrt s)a+s=0,$$
which implies (since we assumed $a \ge b$) that
$$ a=\frac{1}{2}(1+\sqrt s+\sqrt{1+2\sqrt s-3s}).$$
Thus, $a \ge 1$ iff $$\sqrt s+\sqrt{1+2\sqrt s-3s} \ge 1 \iff \\
1+2\sqrt s-2s+2\sqrt s \sqrt{1+2\sqrt s-3s} \ge 1 \iff \\ \sqrt s-s+\sqrt s \sqrt{1+2\sqrt s-3s} \ge 0 \iff \\
1-\sqrt s+ \sqrt{1+2\sqrt s-3s} \ge 0 \iff \\ \sqrt{1+2\sqrt s-3s} \ge \sqrt s-1.$$
(We passed from the first line to the second line by squaring).
The last inequality clearly holds, since the LHS $\ge 0$, and the RHS is $\le 0$.
(since we assumed $s \le 1$).
|
Hint: $(a-1)(b-1) = ab - a - b + 1 = s - \sqrt{s} \leq 0 $.
Hence, conclude that $ a \geq 1 \geq b $, with equality when $ s = 1$.
|
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|
Evaluating $\cos\frac{17\pi}{18} \cos\frac{7\pi}{9}+ \sin\frac{17\pi}{18} \sin\frac{7\pi}{9}$ with a sum or difference formula
Use a sum or difference formula to find the exact value of the following:
$$\cos\frac{17\pi}{18} \cos\frac{7\pi}{9}+ \sin\frac{17\pi}{18} \sin\frac{7\pi}{9}$$
The answer I got for this was $0.8660254$. Please let me know if my answer is correct.
Any help is greatly appreciated.
|
Yes; using the difference formula, it's $\cos\left(\dfrac{17\pi}{18}-\dfrac{14\pi}{18}\right)=\cos
\left(\dfrac\pi6\right)=\dfrac{\sqrt3}2.$
|
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Finding $\lim_{n\to\infty}\prod_{n=1}^{\infty}\left(1-\frac{1}{n(n+1)}\right)$ I have a trouble with this limit of the infinite product:
$$\lim _{n \to\infty}\left(1-\frac{1}{1 \cdot 2}\right)\left(1-\frac{1}{2 \cdot 3}\right) \cdots\left(1-\frac{1}{n(n+1)}\right)$$
My attempt:
We have
$$\prod_{n=1}^{\infty}\left(1-\frac{1}{n(n+1)}\right)=\prod_{n=1}^{\infty}\left(\frac{n^{2}+n-1}{n^{2}+n}\right)=\prod_{n=1}^{\infty} \frac{\left(n-a_{1}\right)\left(n-a_{2}\right)}{n \left(n+1\right)},$$
where $a_{1}=\dfrac{-1+\sqrt{5}}{2}$, $a_{2}=\dfrac{-1-\sqrt{5}}{2}$.
So I would just like a hint as to how to proceed. Any help would be appreciated.
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By means of Weierstrass product of the gamma function，observe that
$$\frac{1}{\Gamma(z)}=e^{\gamma z} z \prod_{n=1}^{\infty}\left(1+\frac{z}{n}\right) e^{-z / n}$$
where $\gamma $ is the Euler–Mascheroni constant.
Then
$$\prod_{n=1}^{\infty} \frac{\left(n-a_{1}\right)\left(n-a_{2}\right)}{n(n+1)}=\frac{\Gamma\left(1\right)\Gamma\left(2\right)}{\Gamma\left(1-a_{1}\right)\Gamma\left(1-a_{2}\right)}=\frac{\Gamma(1) \Gamma(2)}{\Gamma\left(\frac{3-\sqrt{5}}{2}\right) \Gamma\left(\frac{3+\sqrt{5}}{2}\right)}.$$
Since $\Gamma(x+1)=x \Gamma(x)$ and $\Gamma(n)=(n-1) !$ for any positive integer $n$,
$$\frac{\Gamma(1) \Gamma(2)}{\Gamma\left(\frac{3-\sqrt{5}}{ 2}\right) \Gamma\left(\frac{3+\sqrt{5}}{2}\right)}=\frac{1}{\Gamma\left(\frac{3-\sqrt{5}}{2}\right) \Gamma\left(\frac{3+\sqrt{5}}{2}\right)}.$$
Using the relation
$\Gamma(x) \Gamma(1-x)=\frac{\pi}{\sin \pi x}$, thus we obtain
$$\frac{1}{\Gamma\left(\frac{3-\sqrt{5}}{2}\right) \Gamma\left(\frac{3+\sqrt{5}}{2}\right)}=\frac{1}{(\frac{1-\sqrt{5}}{2})(\frac{1+\sqrt{5}}{2})}\frac{1}{\Gamma\left(\frac{1-\sqrt{5}}{2}\right) \Gamma\left(\frac{1+\sqrt{5}}{2}\right)}=-\frac{\sin \left(\frac{(1+\sqrt{5})\pi}{2}\right)}{\pi}$$
|
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How to prove the number of three-order $3\times 3$ matrices with row and column sums both equal to $r$ is $H_3(r) = \binom{r+5}{5} - \binom{r+2}{5}$? The combinatorial problem is as follows:
Let $H_3(r)$ denote the number of $3\times 3$ matrices with nonnegative integer
entries such that each row and each column sum to $r$.
Show that $$H_3(r) = \binom{r+5}{5} - \binom{r+2}{5}$$
Theorem. (Birkhoff-von Neumann). Every $n \times n$ magic square with row and column sum $r$ is a sum of $r$ permutation matrices (of size $n \times n)$.
Using this theorem and the fact that the number of $3\times 3$ permutation matrices is $3! = 6$. I find that if there are no "repeated" cases, the number is $\binom{r+5}{5}$. But there are, for example, $r = 3$:
$$\begin{aligned}
&\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]+\left[\begin{array}{lll}
0 & 1 & 0 \\
0 & 0 & 1 \\
1 & 0 & 0
\end{array}\right]+\left[\begin{array}{lll}
0 & 0 & 1 \\
1 & 0 & 0 \\
0 & 1 & 0
\end{array}\right]=\\
&\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0
\end{array}\right]+\left[\begin{array}{lll}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0
\end{array}\right]+\left[\begin{array}{lll}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{array}\right]
\end{aligned}$$
Question: How to prove that the number of superfluous matrices is exactly $\binom{r+2}{5}$ for a general $r$ so that we can subtract it safely from $\binom{r+5}{5}$?
I tried to think of it but failed. Any help is appreciated.
|
Let
$$A=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\qquad
B=\begin{bmatrix}0&1&0\\0&0&1\\1&0&0\end{bmatrix}\qquad
C=\begin{bmatrix}0&0&1\\1&0&0\\0&1&0\end{bmatrix}$$
$$D=\begin{bmatrix}0&0&1\\0&1&0\\1&0&0\end{bmatrix}\qquad
E=\begin{bmatrix}0&1&0\\1&0&0\\0&0&1\end{bmatrix}\qquad
F=\begin{bmatrix}1&0&0\\0&0&1\\0&1&0\end{bmatrix}$$
By the theorem, every magic square with nonnegative entries and row/column sum $r$ can be expressed as
$$aA+bB+cC+dD+eE+fF$$
where $a+b+c+d+e+f=r$. But we have
$$A+B+C=D+E+F$$
so to canonicalise the representation of a magic square, we will require that $d,e,f$ cannot all be positive (at least one has to be zero). This is the only deduplication we can make.
The number of ordered tuples of nonnegative integers $(a,b,c,d,e,f)$ that sum to $r$ and have $d,e,f\ge1$ is the same as the number of tuples $(a,b,c,d',e',f')$ that sum to $r-3$ where $d',e',f'\ge0$ like $a,b,c$. This count can be determined by stars and bars as $\binom{r+2}5$, and is the number of superfluous representations that we need to subtract from $\binom{r+5}5$ to get the correct final answer.
|
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What is $\frac{a b \sin x}{2\sqrt{(a^2 + b^2 + 2 ab \cos x ) \cdot (a+b - \sqrt{a^2+b^2+ 2 ab \cos x})}}$ for $x \rightarrow 0$? I am trying to find the limit
$\lim_{x \rightarrow0} \frac{a b \sin x}{2\sqrt{(a^2 + b^2 + 2 ab \cos x ) \cdot (a+b - \sqrt{a^2+b^2+ 2 ab \cos x})}}$
for $a,b \in \rm{I\!R}_{+}$. Applying L'Hospital's rule leads to
$\lim_{x \rightarrow0}\frac{\cos x \cdot \sqrt{(a^2 + b^2 + 2 ab \cos x ) \cdot (a+b - \sqrt{a^2+b^2+ 2 ab \cos x})}}{-2 \sin x \cdot (a+b - \sqrt{a^2+b^2+ 2 ab \cos x}) + \frac{\sin x \cdot (a^2 + b^2 + 2 ab \cos x )}{\sqrt{a^2+b^2+ 2 ab \cos x}} }$.
However, this remains with both, cosine and sine. Maybe one could use a trigonometric identity, which I cannot find.
|
Note
$$\begin{align}
L = & \lim_{x \rightarrow0} \frac{a b \sin x}{2\sqrt{(a^2 + b^2 + 2 ab \cos x ) \cdot (a+b - \sqrt{a^2+b^2+ 2 ab \cos x})}}\\
= & \lim_{x \rightarrow0} \frac{1}{2\sqrt{a^2 + b^2 + 2 ab \cos x }}
\cdot \lim_{x \rightarrow0} \frac{a b \sin x}{\sqrt{ a+b - \sqrt{a^2+b^2+ 2 ab \cos x}}}\\
=&\frac{1}{2(a+b)}\cdot L_1 \tag1\\
\end{align}$$
where
$$\begin{align}
L_1= & \lim_{x \rightarrow0} \frac{a b \sin x}{\sqrt{a+b - \sqrt{a^2+b^2+ 2 ab \cos x}}} \\
= & \lim_{x \rightarrow0} \frac{a b \sin x\sqrt{a+b + \sqrt{a^2+b^2+ 2 ab \cos x}}}{\sqrt{(a+b)^2 - (a^2+b^2+ 2 ab \cos x)}} \\
=& \lim_{x \rightarrow0} \frac{a b \sin x\sqrt{a+b + \sqrt{a^2+b^2+ 2 ab \cos x}}}{\sqrt{ 2ab(1-\cos x)}} \\
=& \lim_{x \rightarrow0} \frac{2a b\sin \frac x2\cos\frac x2\sqrt{a+b + \sqrt{a^2+b^2+ 2 ab \cos x}}}{\sqrt{ 2ab\cdot 2\sin^2\frac x2}} \\
=& \lim_{x \rightarrow0} \sqrt{a b}\cos\frac x2 \sqrt{a+b + \sqrt{a^2+b^2+ 2 ab \cos x}} \\
=& \sqrt{a b}\sqrt{a+b + \sqrt{a^2+b^2+ 2 ab}} \\
=& \sqrt{2ab(a+b)} \\
\end{align}$$
Plug $L_1$ into (1)
$$\begin{align}
L=\frac{1}{2(a+b)}\cdot \sqrt{2ab(a+b)} = \sqrt{\frac{ab}{2(a+b)}} \\
\end{align}$$
|
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|
Does this limit converge on e? Playing around with some math in python today I came across what appears to be an interesting pattern:
Starting at n=1 as n approaches positive infinity, take (n+1)^(n+2)/n^(n+1) and get a list of ratios of exponential expressions. At first glance the ratios between the numbers appeared to be converging to something so...
Next, I took the difference between consecutive ratios, e.g. (n+2)^(n+3)/(n+1)^(n+2)-(n+1)^(n+2)/n^(n+1).
The differences appear to be approaching e (2.718...) as n gets larger.
The first few ratios rounded to the third decimal place are...
2^3/1^2 = 8
3^4/2^3 = 10.125
4^5/3^4 = 12.642
5^6/4^5 = 15.259
6^7/5^6 = 17.916
...
With their differences being...
10.125 - 8 = 2.125
12.642 - 10.125 = 2.517
15.259 - 12.642 = 2.617
17.916 - 15.259 = 2.657
...
After the 13th iteration you get 2.711, and it looks like the series will converge on e as it gets arbitrarily large. That or likely positive infinity and my hunch is off!
Can anyone with better knowledge of limits tell me if I've stumbled across a novel way of calculating e or not?
Here's the python code for those curious (first loop stops at 15 because that's all my cheap phone could handle):
import numpy as np
ratios = []
i = 1
while i < 15:
a = np.power(i,i+1)
b = np.power(i+1,i+2)
print(a)
ratios.append(b/a)
i+=1
print(ratios)
x=0
diffs = []
while x < len(ratios) - 1:
temp = ratios[x+1] - ratios[x]
diffs.append(temp)
x+=1
print(diffs)
This reminds of the time I thought I discovered a novel formula for exponents about the golden ratio. I didn't, it was already known and I think this may be the case too but my brief search hasn't turned up anything yet.
Thanks!
|
We have
\begin{align*}
&\left( {1 + \frac{1}{{n + 1}}} \right)^{n + 2} (n + 2) - \left( {1 + \frac{1}{n}} \right)^{n + 1} (n + 1) \\ & = \left( {1 + \frac{1}{n}} \right)^{n + 1} \left[ {\left( {\frac{{(n + 2)n}}{{(n + 1)^2 }}} \right)^{n + 1} \frac{{(n + 2)^2 }}{{(n + 1)^2 }} - 1} \right](n + 1)
\\ &=
\left( {1 + \frac{1}{n}} \right)^{n + 1} \left[ {\left( {1 - \frac{1}{n} + \frac{3}{{2n^2 }} + \mathcal{O}\!\left( {\frac{1}{{n^3 }}} \right)} \right)\frac{{(n + 2)^2 }}{{(n + 1)^2 }} - 1} \right](n + 1)
\\ &= \left( {1 + \frac{1}{n}} \right)^{n + 1} \left( {\frac{1}{n} - \frac{3}{{2n^2 }} + \mathcal{O}\!\left( {\frac{1}{{n^3 }}} \right)} \right)(n + 1) \\ &= \left( {1 + \frac{1}{n}} \right)^{n + 1} \left( {1 - \frac{1}{{2n}} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right)} \right) = e + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right).
\end{align*}
|
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|
Find the first four non zero terms of the Taylor series for $f(x) = \int_0^x \frac{1-\cos(t)}{t}dt$
Find the first four non zero terms of the Taylor series for $f(x) =
\int_0^x \frac{1-\cos(t)}{t}dt$
$$f(x) = \int_0^x \frac{1-\cos(t)}{t}dt = \int_0^x \frac{1- (1 + \sum_{n=1}^{\infty}\frac{(-1)^n t^{2n}}{(2n)!})}{t} dt = \int_0^x\sum_{n=1}^{\infty}\frac{(-1)^{n+1}t^{2n -1}}{(2n)!} dt=\sum_{n=1}^{\infty}\int_0^x\frac{(-1)^{n+1}t^{2n -1}}{(2n)!} dt = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{(2n)!}\int_0^x t^{2n-1}dt =\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^{2n}}{(2n)! \cdot (2n)} = \frac{x^2}{4} - \frac{x^4}{4\cdot4!} + \frac{x^6}{6\cdot 6!} - \frac{x^8}{8\cdot 8!} +\dots$$
So the first four terms are:
$\frac{x^2}{4}$; $\;- \frac{x^4}{4\cdot4!}$;$\; \frac{x^6}{6\cdot 6!}$;$\;- \frac{x^8}{8\cdot 8!}$
Is this correct?
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It is correct, and a quite fast way of solving it. Another (not as optimal) way of doing it is by differentiating the function and using the fundamental theorem of calculus.
$$f(x) = \int_0^x \frac{1-\cos(t)}{t}dt \Rightarrow f'(x) = \frac{1-\cos(x)}{x}.$$
Evaluating the term above at $x=0$ yields zero, and therefore the first order term is zero.
$$f''(x) = \frac{\sin(x)}{x} - \frac{1-\cos(x)}{x^2}.$$
At $x=0$ the first summand yields $1$ and the second yields $1/2$. This gives $\frac{f''(0)}{2} = \frac{1}{4}.$
I proceeded with the differentiations in Mathematica and obtained exactly what you have,
$$\frac{x^2}{4};-\frac{x^4}{4\cdot 4!};\frac{x^6}{6\cdot 6!};\frac{x^8}{8\cdot 8!}.$$
Congratulations for the quick solution!
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Evaluate $\lim_{x \to 0} \frac{\sin\left(a+2x\right)-2\sin\left(a+x\right)+\sin\left(a\right)}{x^{2}}$ without L'Hôpital This limit is one of the "Problems Plus" from Stewart Calculus:
$$\lim_{x \to 0} \frac{\sin\left(a+2x\right)-2\sin\left(a+x\right)+\sin\left(a\right)}{x^{2}}$$
Note that the limit is of the indeterminate form $\frac{0}{0}$. The problem appears several chapters before L'Hôpital's rule is discussed, so I assume there is a solution without using L'Hopital.
Looking at a graph, the local behavior of the function near $0$ appears to be $-\sin(a+x)$, which, of course, suggests a limit of $-\sin(a)$. Using L'Hôpital's rule twice confirms this guess:
$$\begin{align}
& \lim_{x \to 0} \frac{\frac{d}{dx} [\sin\left(a+2x\right)-2\sin\left(a+x\right)+\sin\left(a\right)]}{ \frac{d}{dx} x^{2}} \\
& = \lim_{x \to 0} \frac{2\cos(a+2x)-2\cos(a+x)}{2x} \\
& = \lim_{x \to 0} \frac{-4\sin(a+2x)+2\sin(a+x)}{2} \\
& = -\sin(a).
\end{align}$$
Can anyone give a hint or solution for evaluating this limit without L'Hôpital?
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Let's first expand the $\sin$ functions using the addition formulae,
$$\sin(a+2x) = \sin(a) \cos(2x) + \cos(a) \sin(2x)$$
and similarly
$$\sin(a+x) = \sin a \cos x + \cos a \sin x$$
Expand $x$-dependent $\sin$ and $\cos$ terms to second order,
$$\sin x = x + O(x^3), \quad \cos x = 1 - \frac{x^2}{2} + O(x^4)$$
so that we have (writing $S = \sin a$ and $C = \cos a$ for readability)
$$\sin(a+2x) = S \cdot \left(1-2x^2\right) + 2Cx + o(x^2)$$
and
$$\sin(a+x) = S \cdot \left(1-\frac{x^2}{2}\right) + Cx + o(x^2)$$
Now substitute these into the numerator of interest,
\begin{align*}
\sin(a+2x)-2\sin(a+x)+\sin(a) &= S \cdot\left(1-2x^2\right) + 2Cx - 2\left[S\left(1-\frac{x^2}{2}\right)+Cx\right]+S+o(x^2) \\
&= \left(S - 2S + S\right) + x\left[ 2C-2C \right] + x^2 \left[ -2S + S \right] + o(x^2) \\
&= -Sx^2 + o(x^2)
\end{align*}
where we've collected coefficients by powers of $x$ (which is a somewhat messy experience, but it collects nicely) and hence we have
$$\lim_{x \to 0} \frac{\sin(a+2x)-2\sin(a+x)+\sin(a)}{x^2} = \lim_{x \to 0} \left[ -S + \frac{o(x^2)}{x^2} \right] = -S$$
which gives the desired result.
As a side note, L'Hôpital is essentially doing all this expansion and cancelling for us by the machinery of differentiation. In particular, if $f(x) = Ax^2 + o(x^2)$, then
$$\frac{f(x)}{x^2} \to A \text{ as } x \to 0$$
which we can "extract" by noting that
$$f''(0) = 2A$$ so that
$$\lim_{x \to 0} \frac{\frac{d^2}{dx^2} f(x)}{\frac{d^2}{dx^2} x^2} = \lim_{x \to 0} \frac{2A + o(1)}{2} = A$$
Of course, the true story is somewhat more involved (since denominators need not be $x^k$), but I find this a useful intuition to have. If you know about complex residues, I also like to think of this as essentially analogous to residues at higher order poles (since we're extracting a coefficient of an expansion).
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Prove that for all positives $a, b$ and $c$, $(\sum_{cyc}\frac{c + a}{b})^2 \ge 4(\sum_{cyc}ca)(\sum_{cyc}\frac{1}{b^2})$.
Prove that for all positives $a, b$ and $c$, $$\left(\frac{b + c}{a} + \frac{c + a}{b} + \frac{a + b}{c}\right)^2 \ge 4(bc + ca + ab) \cdot \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right)$$
Let $ca + ab = m$, $ab + bc = n$ and $bc + ca = p$, we have that $$\left(\frac{m}{a^2} + \frac{n}{b^2} + \frac{p}{c^2}\right)^2 \ge 2(m + n + p) \cdot \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right)$$
$$\iff \left(\frac{m}{a^2} + \frac{n}{b^2} + \frac{p}{c^2} - 1\right)^2 \ge 2 \cdot \sum_{cyc((m, n, p), (a, b, c))}\left[n \cdot \left(\frac{1}{c^2} + \frac{1}{a^2}\right)\right] + 1$$
Expanding $\displaystyle \sum_{cyc((m, n, p), (a, b, c))}\left[n \cdot \left(\frac{1}{c^2} + \frac{1}{a^2}\right)\right]$ gives $$2 \cdot \sum_{cyc}\frac{ca}{b^2} + \left(\frac{b + c}{a} + \frac{c + a}{b} + \frac{a + b}{c}\right)$$
Let $\dfrac{b + c}{a} = m'$, $\dfrac{c + a}{b} = n'$ and $\dfrac{a + b}{c} = p'$, we have that $$(m' + n' + p' - 1)^2 \ge 2 \cdot \left[2 \cdot \sum_{cyc}\frac{ca}{b^2} + (m' + n' + p')\right] + 1$$
Moreover, $$(m')^2 + (n')^2 + (p')^2 = \sum_{cyc}\left[\left(\frac{c + a}{b}\right)^2\right] \ge 2 \cdot \sum_{cyc}\frac{ca}{b^2}$$
$$\implies (m' + n' + p' - 1)^2 \ge 2 \cdot \left[(m')^2 + (n')^2 + (p')^2 + m' + n' + p'\right] + 1$$
$$\iff -[(m')^2 + (n')^2 + (p')^2] + 2(m'n' + n'p' + p'm') - 4(m' + n' + p') \ge 0$$, which is definitely not correct.
Another attempt, let $(0 <) \ a \le b \le c \implies ab \le ca \le bc \iff ca + ab \le ab + bc \le bc + ca$
$\iff m \le n \le p$ and $a^2 \le b^2 \le c^2 \iff \dfrac{1}{a^2} \ge \dfrac{1}{b^2} \ge \dfrac{1}{c^2}$.
By the Chebyshev inequality, we have that $$3 \cdot \left(\frac{m}{a^2} + \frac{n}{b^2} + \frac{p}{c^2}\right) \le (m + n + p) \cdot \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right)$$
Any help would be appreciated.
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The inequality follows from the famous inequality Ukraine MO 2001.
If $ a,\,b,\,c$ and $x,\,y,\,z$ non-negative real numbers, then
$$ [x(b+c) + y(c+a) + z(a+b)]^2 \geqslant 4(ab+bc+ca)(xy+yz+zx). \quad (1)$$
Now, using $(1)$ with $x=a^2,\,y=b^2,\,z=c^2$ we get
$$ [a^2(b+c) + b^2(c+a) + c^2(a+b)]^2 \geqslant 4(ab+bc+ca)(a^2b^2+b^2c^2+c^2a^2),$$
or
$$\left(\frac{b + c}{a} + \frac{c + a}{b} + \frac{a + b}{c}\right)^2 \geqslant 4(ab+bc+ca)\left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right).$$
Note. The sum of squares by Darij Grinberg
$$[ab(a+b)+bc(c+a)+ca(c+a)]^2-4(ab+bc+ca)(a^2b^2+b^2c^2+c^2a^2)$$
$$=\sum a^2b^2(a-b)^2 + \sum 2abc \sum a(a-b)(a-c).$$
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|
How to find the expected value of infinite discrete random variable? Question: A fair six-sided die is rolled repeatedly. Let $X$ be the number of times it is rolled until two different numbers are seen (so, for example, if the first two rolls were each $4$ and the third roll was $5$ then $X=3$).
How to I find the expected value? I realize that this is a geometric distrition and this is the expression I have for $\mathsf{E}[X]$:
$\mathsf{E}[X]=2\left(\frac 56\right) + 3\left(\frac 56\right)\left(\frac 16\right) + 4\left(\frac 56\right)\left(\frac 16\right)^2 + 5\left(\frac 56\right)\left(\frac 16\right)^3 + ...$
I wrote down an expression in terms of summation and this seems to be a airthmetico-geometric series but there has got be an easier way to compute the expected value. What am I missing here?
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There are faster ways, but one relatively simple approach is
$\mathsf{E}[X]=2\left(\frac 56\right) + 3\left(\frac 56\right)\left(\frac 16\right) + 4\left(\frac 56\right)\left(\frac 16\right)^2 + 5\left(\frac 56\right)\left(\frac 16\right)^3 + \cdots$
so $\frac16 \mathsf{E}[X]=2\left(\frac 56\right)\left(\frac 16\right) + 3\left(\frac 56\right)\left(\frac 16\right)^2 + 4\left(\frac 56\right)\left(\frac 16\right)^3 + 5\left(\frac 56\right)\left(\frac 16\right)^4 + \cdots$
and subtracting $\frac56 \mathsf{E}[X]=2\left(\frac 56\right) + \left(\frac 56\right)\left(\frac 16\right) + \left(\frac 56\right)\left(\frac 16\right)^2 + \left(\frac 56\right)\left(\frac 16\right)^3 + \cdots$
i.e. $\mathsf{E}[X]=2 + \left(\frac 16\right) + \left(\frac 16\right)^2 + \left(\frac 16\right)^3 + \cdots$
This is essentially $1$ more than a geometric series but you can in fact do the same thing again
so $\frac16 \mathsf{E}[X]=2\left(\frac 16\right) + \left(\frac 16\right)^2 + \left(\frac 16\right)^3 + \left(\frac 16\right)^4 + \cdots$
and subtracting $\frac56 \mathsf{E}[X]=2 -\frac 16 = \frac{11}{6}$
and thus $\mathsf{E}[X] = \frac{11}{5}=2.2$
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|
Repeated roots: $f^{1992}(x) = x$ for $f(x) = 4x(1-x)$ Given a function $f : [ 0 , 1 ] \to \ [ 0 , 1 ]$ defined by $f(x) = 4x(1-x)$ for all $x \in [ 0 , 1 ]$. How many distinct roots does the equation $f^{1992}(x) = x$ has. ($f^n(x) = f\left(f^{n-1}(x)\right)$)
I've already know that $\deg\left(f^{1992}(x)\right) = 2^{1992}$. So, it has $1992$ roots. But, I don't know how to deal with repeated roots. Can anyone give me some hint or a solution to this problem? Thanks in advance!
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It is possible to solve $f_n(x) = x$, $ \ \ f_n(x)=f(f_{n-1}(x))$ in a general way, for any integer $n$:
$ $
$$f_{n+1}(x)=4 f_{n}(x)(1-f_{n}(x))$$
Solving recursively for $f_n(x)$:
$$f_{n}(x)=\frac{1}{2}-\frac{1}{2}\cos(2^n c_x)$$
$$f_1(x)=\frac{1}{2}-\frac{1}{2}\cos(2 c_x)=4x(1-x).$$
solving for $c_x$ and $x\in [0,1]$:
$$c_x=\pm\frac{1}{2} \arccos \big(1+8x(1-x)\big)+2k\pi $$
$$f_{n}(x)=\frac{1}{2}\left(1-\cos\left(2^{n-1}\arccos\left(1+8x\left(x-1\right)\right)\right)\right)$$
$$f_{n}(x)=\sin^2\left(2^{n-2}\arccos\left(1+8x\left(x-1\right)\right)\right).$$
For $\theta \in[-\frac{\pi}{2},\frac{\pi}{2}]$, $x=\sin(\theta)^2$
$$\arccos\left(1+8\sin(\theta)^2\left(\sin(\theta)^2-1\right)\right)=\arccos\left(\cos(4 \theta)\right)=4\theta$$
therefore we have that $f_n(x)=x$ is equivalent to $\sin(\theta)^2=\sin(2^n \theta)^2$
$$\theta=\left\{\frac{\pm \pi+4k }{1+2^n},\frac{2 \pi(\pm 1+2 k)}{1+2^n},\frac{\pm 4 \pi k}{\pm 1+2^n},\frac{\pi(1 \pm 4 k)}{\pm (1-2^n) }\right\}$$
$k$ such that
$$-\frac{\pi}{2}\leq\left\{\frac{\pm \pi+4k }{1+2^n},\frac{2 \pi(\pm 1+2 k)}{1+2^n},\frac{\pm 4 \pi k}{\pm 1+2^n},\frac{\pi(1 \pm 4 k)}{\pm (1-2^n) }\right\}\leq \frac{\pi}{2}$$
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Let $x, y \in \mathbb R$ such that $x^2+y^2=2x-2y+2$. Find the largest possible value of $x^2+y^2$
Let $x, y \in \mathbb R$ s. t . $x^2+y^2=2x-2y+2$. Find the largest possible value of $x^2+y^2-\sqrt{32}$
I know this is a duplicate of another question, but that question has solutions involving calculus and geometry, while I want a solution relying on algebra and basic inequalities only to solve this problem.
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We have: $$6+ \underbrace{\left( 1+\frac{\,\sqrt {2}}{2} \right) \left( {x}^{2}+{y}^{2}-2\,x+2\,y-2
\right)}_{=\, 0} -({x}^{2}+{y}^{2}-\sqrt {32})$$
$$=\frac{\,\sqrt {2}}{4} \left( \sqrt {2}\,x-\sqrt {2}-2 \right) ^{2}+\frac{\sqrt{2}}{4}\,
\left( \sqrt {2}y+\sqrt {2}+2 \right) ^{2}\geqq 0$$
Therefore: $$x^2 +y^2 -\sqrt{32} \leqq 6$$
Use this useful identity: $$ax^2 +bx+c =a(x+\frac{b}{2a})^2 +\frac{4ac-b^2}{4a}$$
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Decomposing an inverse Laplace transform The given inverse Laplace transform is:
$$\mathscr{L}^{-1}\left[\frac{5s^2+12s-4}{s^3-2s^2+4s-8} \right]$$
First split it up into three separate fractions and factorize the denominator
$$\mathscr{L}^{-1} \left[\frac{5s^2}{(s-2)(s^2+4)} \right]+\mathscr{L}^{-1} \left[\frac{12s}{(s-2)(s^2+4)} \right]-\mathscr{L}^{-1} \left[\frac{4}{(s-2)(s^2+4)} \right]$$
Now to compile them into partial fraction decomp. form:
$$\mathscr{L}^{-1} \left[\frac{5s^2}{(s-2)(s^2+4)} \right]=\frac{A}{(s-2)}+\frac{Bs+c}{(s^2+4)}$$
$$\mathscr{L}^{-1} \left[\frac{12s}{(s-2)(s^2+4)} \right]=\frac{A}{s-2}+\frac{Bs+c}{s^2+4}$$
$$\mathscr{L}^{-1} \left[\frac{4}{(s-2)(s^2+4)} \right]=\frac{A}{s-2}+\frac{Bs+c}{s^2+4}$$
Then following through:
$$5s^2=A(s^2+4)+(Bs+c)(s-2)$$
$$12s=A(s^2+4)+(Bs+c)(s-2)$$
$$4=A(s^2+4)+(Bs+c)(s-2)$$
Now to solve for the variables, first with $s=2$:
$$5*4=8A \to A=\frac{20}{8}$$
$$12(2)=8A \to A=3$$
$$4=8A \to A=\frac{1}{2}$$
Now subbing in for $s=0$:
$$0=\frac{20}{8}(0^2+4)+(B(0)+C)(0-2) \to 0=10-2C \to C=5$$
$$12(0)=3(4)+(B(0)+C)(-2) \to 0=12-2C \to C=6$$
$$4=\frac{1}{2}(4)-2C \to 4=2-2C \to C=-1$$
Now to solve for $B$; $s=1$
$$5=\frac{100}{8}-B-5 \to B=\frac{5}{2}$$
$$12=3(1+4)+(B+6)(-1) \to B=-3$$
$$4=\frac{1}{2}(1+4)+(B-1)(-1) \to B=-\frac{1}{2}$$
So now our Laplace transforms are:
$$\mathscr{L}^{-1} \left[\frac{5s^2}{(s-2)(s^2+4)} \right]=\frac{20}{8(s-2)}+\frac{5s+5}{2(s^2+4)}$$
$$\mathscr{L}^{-1} \left[\frac{12s}{(s-2)(s^2+4)} \right]=\frac{3}{s-2}-\frac{3s+6}{s^2+4}$$
$$\mathscr{L}^{-1} \left[\frac{4}{(s-2)(s^2+4)} \right]=\frac{1}{2(s-2)}-\frac{s-1}{2(s^2+4)}$$
Are my calculations correct or have I over complicated the problem?
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Note that the denominator is:
$$(s-2)(s^2+4)$$
The numerator can be rewritten as :
$$f(s)=5s^2+12s-4$$
$$f(s)=5(s^2+4)-24+12s$$
$$f(s)=5(s^2+4)+12(s-2)$$
Back to our function :
$$h(t)=\mathscr{L}^{-1}\left[\frac{5s^2+12s-4}{s^3-2s^2+4s-8} \right]
$$
$$h(t)=\mathscr{L}^{-1}\left[\dfrac 5 {s-2}+\dfrac{12}{s^2+4} \right]$$
Finally:
$$\boxed {h(t)=5e^{2t}+6 \sin (2t)}$$
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Using triangle Inequality to solve inequality. Question from a textbook.
Show that for non negative $x,y,z$ that
$$ (x+y+z) \sqrt{2} \leq \sqrt{x^2 + y^2} + \sqrt{ y^2 + z^2} + \sqrt{x^2 +z^2} $$
and that for $ 0< x \leq y \leq z$,
$$ \sqrt{y^2 +z^2} \leq x\sqrt{2} + \sqrt{(y-x)^2 +(z-x)^2} $$
,which the hint was to utilize the triangle inequality to "an appropriate sum". How does these approaches work? Is there a way to "know" beforehand what these sums are?
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*
*$(\sqrt{x^2 + y^2} + \sqrt{ y^2 + z^2} + \sqrt{x^2 +z^2} )^2= 2(x^2 + y^2+ z^2) + \;\;non-negative\;\; terms\geq 2(x^2 + y^2+ z^2)$
which yields the first inequality.
*$(y-x)^2 + (z-x)^2=y^2+z^2+2x^2-2x(y+z)\geq y^2+z^2+2x^2-2\sqrt{2}x(\sqrt{y^2+z^2})=(\sqrt{y^2+z^2}-\sqrt{2}x)^2$
yields the second inequality.
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Lower bound $\frac{\det(A+B+C)}{\det(A+C)}$ in terms of $\frac{\det(A+B)}{\det(A)}$ Assume matrix $A$ is symmetric and positive definite, and matrices B and C are symmetric and positive semi-definite. Originally I have ratio between determinants:
$$\frac{\det(A+B)}{\det(A)}$$
By adding another matrix C inside the determinant on both numerator and determinator, we obtain a new ratio between determinants:
$$\frac{\det(A+B+C)}{\det(A+C)}$$
From this question, the new ratio is proved to be upper bounded by the original one:
$$\frac{\det(A+B+C)}{\det(A+C)} \leq \frac{\det(A+B)}{\det(A)}$$
Now the question is, can we prove a lower bound in terms of the original ratio as well. For example:
$$ s(A,B,C) \frac{\det(A+B)}{\det(A)} \leq \frac{\det(A+B+C)}{\det(A+C)} \leq \frac{\det(A+B)}{\det(A)}$$
where $s(A,B,C) \in [0,1]$ is some scalar value that may depend on matrices $A, B, C$. Intuitively, if $C$ is a zero matrix, then $s(A,B,C)$ should be equal to 1, making the lower bound equal to upper bound.
My initial attemp is shown below:
\begin{aligned}
\frac{\det(A+B)}{\det(A)} &=\det(I+A^{-1}B)\\
&= \det(I+(A+C)^{-1}(A+C)A^{-1}B)\\
&= \det(I+(A+C)A^{-1}B(A+C)^{-1}) \quad (\text{[Weinstein–Aronszajn identity][2]})
\end{aligned}
I am wondering if the following inequality holds:
$$\det(I+(A+C)A^{-1}B(A+C)^{-1}) \leq \det((A+C)A^{-1})\det(I+B(A+C)^{-1})$$
If it holds, then we can show that:
$$\frac{\det(A+B)}{\det(A)}=\det(I+(A+C)A^{-1}B(A+C)^{-1}) \leq \frac{\det(A+C)}{\det(A)}\frac{\det(A+B+C)}{\det(A+C)}$$
Therefore,
$$\frac{\det(A)}{\det(A+C)}\frac{\det(A+B)}{\det(A)} \leq \frac{\det(A+B+C)}{\det(A+C)}$$
So $s(A,B,C)=\frac{\det(A)}{\det(A+C)}$
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Turns out it is very easy to show the inequality holds.
\begin{aligned}
\frac{\det(A+B)}{\det(A)} &= \frac{det(A+C)}{det(A)}\frac{det(A+B)}{det(A+C)}\\
& \leq \frac{det(A+C)}{det(A)}\frac{det(A+B+C)}{det(A+C)}\\
\end{aligned}
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Finding $\frac{c\sin(A-B)}{a^2-b^2}-\frac{b\sin(C-A)}{c^2-a^2}$
If $a, b$ and $c$ (all distinct) are the sides of a triangle ABC opposite to the angles $A, B$ and $C$, respectively, then $\frac{c\sin(A-B)}{a^2-b^2}-\frac{b\sin(C-A)}{c^2-a^2}$ is equal to $?$
By opening, $\sin(A-B)$ as $\sin A\cos B-\cos A\sin B$ and then $\sin A$ as $\frac{a}{2R}$ and $\cos A$ as $\frac{b^2+c^2-a^2}{2bc}$, I am able to get the answer as zero. But I am looking for a shorter solution, maybe by putting values of angles and sides.
My first instinct was to assume the triangle to be equilateral. But the question invalidates that case. Then I thought of a right angled triangle with pythogorean triplet as $3,4,5$. But here, I don't know other two angles.
I wonder if I could just keep a dummy triangle handy whose sides and angles I know, which I could quickly use to solve such questions. Any help please? Thanks.
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One can easily show that
$$
\frac{c\sin(A-B)}{a^2-b^2}=\frac1{2R},\tag1
$$
where $R$ is the radius of the circumscribed circle.
Indeed substituting in LHS of (1)
$$a=2R\sin A,\quad b=2R\sin B,\quad c=2R\sin C $$
one obtains:
$$\begin{align}
\frac{c\sin(A-B)}{a^2-b^2}&=\frac1{2R}\frac{\sin C\sin(A-B)}{\sin^2A-\sin^2B}\\
&=\frac1{2R}\frac{\sin(A+B)\sin(A-B)}{\sin^2A-\sin^2B}\\
&=\frac1{2R}\frac{\frac12(\cos 2B-\cos 2A)}{\frac12(\cos 2B-\cos 2A)}.
\end{align}
$$
The same result one obtains of course for $\frac{b\sin(C-A)}{c^2-a^2}$ as well.
|
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|
Find the minimum value of $\frac ab+\frac {b}{a+b+1}+\frac {b+1}{a}$ when $a,b>0$ I was trying to solve this question and I observed that if I use partial derivative then it would be calculating. Moreover, I observed that $\frac ab\times\frac {b}{a+b+1}\times\frac {b+1}{a}=\frac {b+1}{a+b+1}$. So I thought to use AM-GM inequality and got
$3(\frac {b+1}{a+b+1})^\frac13 \leq \frac ab+\frac {b}{a+b+1}+\frac {b+1}{a}$.
Now the equality holds if $\frac ab=\frac {b}{a+b+1}=\frac {b+1}{a}$.
From here I also got these relations $\frac {b+1}{a+b+1}=\frac a{a+b}$ and $\frac ab=\frac {b}{a+b+1}=\frac {b+1}{a}=\frac {a+b+1}{a+b}$. I am hoping that I will get a finite expression of $(\frac {b+1}{a+b+1})^\frac13 $ and then we are done as the particular values of $a,b$ are not asked.
Help me from here. Am I in the correct way? Do you have any other suggestions?
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For $a>0$ and $b>0$ let $1=t$ and $f(t)=\frac{a}{b}+\frac{b}{a+b+t}+\frac{b+t}{a},$ where $t\geq0.$
Thus, $f'(t)=\frac{1}{a}-\frac{b}{(a+b+t)^2}>0,$ which says $$f(t)\geq f(0)=\frac{a}{b}+\frac{b}{a+b}+\frac{b}{a}.$$
Now, let $a=bx$, where $x>0$.
Thus, we need to find $\min\limits_{x>0}g,$ where $$g(x)=x+\frac{1}{x+1}+\frac{1}{x}.$$
But $$g'(x)=1-\frac{1}{(x+1)^2}-\frac{1}{x^2}=\frac{x^4+2x^3-x^2-2x-1}{x^2(x+1)^2}=$$
$$=\frac{(x^2+x-1)^2-2}{x^2(x+1)^2}=\frac{(x^2+x-1-\sqrt2)(x^2+x+\sqrt2-1)}{x^2(x+1)^2},$$ which gives $$x_{min}=\frac{-1+\sqrt{5+4\sqrt2}}{2}.$$
Can you end it now?
I got that the infimum is equal to $$\frac{2\sqrt{10+8\sqrt2}-\sqrt{5+4\sqrt2}-1}{2}=\frac{(2\sqrt2-1)\sqrt{5+4\sqrt2}-1}{2}=$$
$$=\frac{\sqrt{(9-4\sqrt2)(5+4\sqrt2)}-1}{2}=\frac{\sqrt{13+16\sqrt2}-1}{2}\approx2.4844...$$
We can use also the following reasoning.
Let $a=bx$.
Thus, $$\frac{a}{b}+\frac{b}{a+b+1}+\frac{b+1}{a}=x+\frac{b}{bx+b+1}+\frac{b+1}{bx}=$$
$$=x+\frac{1}{x+1+\frac{1}{b}}+\frac{1}{x}+\frac{1}{bx}>x+\frac{1}{x+1}+\frac{1}{x}$$ because
$$x+\frac{1}{x+1+\frac{1}{b}}+\frac{1}{x}+\frac{1}{bx}-\left(x+\frac{1}{x+1}+\frac{1}{x}\right)=$$
$$=\frac{1}{bx}-\left(\frac{1}{x+1}-\frac{1}{x+1+\frac{1}{b}}\right)=\frac{1}{bx}-\frac{\frac{1}{b}}{(x+1)\left(x+1+\frac{1}{b}\right)}>0.$$
Id est, it's enough to find an infimum of $x+\frac{1}{x+1}+\frac{1}{x}$ and the rest is the same.
We see the for $b\rightarrow+\infty$ the expression $\frac{a}{b}+\frac{b}{a+b+1}+\frac{b+1}{a}$ is closed to $x+\frac{1}{x+1}+\frac{1}{x}$, which says that $$\inf_{a>0,b>0}\left(\frac{a}{b}+\frac{b}{a+b+1}+\frac{b+1}{a}\right)=\inf_{x>0}\left(x+\frac{1}{x+1}+\frac{1}{x}\right)=\frac{\sqrt{13+16\sqrt2}-1}{2}.$$
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On Radical Mersenne Prime Containing Series, and why the number 5 doesn't appear here? As part of a project on operators I was studying the function:
$$ L =\left( \sqrt{x^3} - x \right) + \left( \sqrt{\sqrt{x^3}} - \sqrt{x} \right) + \left( \sqrt{\sqrt{\sqrt{x^3}}} - \sqrt{\sqrt{x}}\right) + ... $$
It can be written explicitly as $$L(x) = \sum_{k=1}^{\infty} \left( x^{\frac{3}{2^n}} - x^{\frac{1}{2^{n-1}} }\right) $$
In order to get a better understanding of the function I decided to taylor expand it around the point $x=1$. This leads to the series:
$$ L(x) = (x-1) - \frac{2}{7}(x-1)^2 + \frac{8}{21}(x-1)^3 - \frac{80}{93}(x-1)^4 + \frac{5360}{1953} (x-1)^5 -\frac{133760}{11811}(x-1)^6... $$
The coefficients in the series were bizarre looking and the numerators didn't appear to be part of any known sequences in OEIS so I decided to study the denominators in depth, specifically I prime factorized each denominator (and numerator too for those curious).
$$ L(x) = (x-1) - \frac{2}{7}(x-1)^2 + \frac{2^3}{3\times 7}(x-1)^3 - \frac{2^5 \times 5}{3\times 31}(x-1)^4 + \frac{2^5 \times 5 \times 67}{3^2 \times 7 \times 31} (x-1)^5 -\frac{2^7 × 5 × 11 × 19}{3 \times 31 \times 127}(x-1)^6... $$
This was continued for about 50 terms surprisingly the terms had a LOT of common structure. The mersenne primes appear in all the terms (and after skipping ahead to the 100th term, it appears the first 10 mersenne primes definitely are present although the 9th mersenne prime shows as early as term 60).
The Question:
I wanted to start with something very simple before going after crazier conjectures involving this sequence of denominators: how can I quickly prove that the number 5 will never be a divisor of any of the denominators?
An approach:
If we let the first coefficent be 1. Then the generally we have
$$ a(n) = \sum_{i=1}^{\infty} \left[ \prod_{k=0}^{n-1} \left[ \frac{3}{2^i} - k \right] \right] - \sum_{i=1}^{\infty} \left[ \prod_{k=0}^{n-1} \left[ \frac{2}{2^i} - k \right] \right] $$
Example:
$$ a(2) = \frac{3}{2} \frac{1}{2} + \frac{3}{4}(-\frac{1}{4}) + \frac{3}{8}(-\frac{5}{8}) + \frac{3}{16} (-\frac{13}{16}) ... - \frac{1}{2}(-\frac{1}{2}) - \frac{1}{4} (-\frac{3}{4}) - \frac{1}{8}(-\frac{7}{8}) ... $$
[Intentionally left unsimplified so the pattern is easier to deduce]
Somehow we need to get a closed from to these product series and then show that after simplification the denominator term will never contain a 5. Getting a closed form feels out of reach for me at the moment.
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$$\prod_{k=0}^{n-1} \left(\frac{3}{2^i}-k\right) = \sum_{j=1}^n \frac{3^j}{2^{ij}} \left[{n \atop j}\right]$$
where $\left[{n \atop j}\right]$ is a Stirling number of the first kind.
Then
$$\sum_{i=1}^\infty \prod_{k=0}^{n-1} \left(\frac{3}{2^i}-k\right) = \sum_{j=1}^n \frac{3^j}{2^j-1} \left[{n \atop j}\right] $$
Similarly
$$ \prod_{k=0}^{n-1} \left(\frac{2}{2^i}-k\right) = \sum_{j=1}^n \frac{2^j}{2^j-1} \left[{n \atop j}\right] $$
so your
$$a(n) = \sum_{j=1}^n \frac{3^j-2^j}{2^j-1} \left[{n \atop j}\right] $$
Now the $5$-adic order of $2^j - 1$ is
$i > 0$ iff $j$ is divisible by $4 \cdot 5^{i-1}$ but not by $4 \cdot 5^i$, and the same is true for $3^j-2^j$. Therefore the
$5$-adic order of $a(n)$ is nonnegative, i.e. $5$ does not appear in the denominator of $a(n)$.
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Probability problem on umbrellas
Two absent-minded roommates, mathematicians, forget their umbrellas in some way or another. $A$ always takes his umbrella when he goes out, while $B$ forgets to take his umbrella with probability $1/2$. Each of them forgets his umbrella at a shop with probability $1/4$. After visiting three shops, they return home. Find the probability that they have only one umbrella.
My Attempt
$A_i/\bar{A}_i$ : $A$ remembers or forgets umbrella at shop $i$
$B_i/\bar{B}_i$ : $B$ remembers or forgets umbrella at shop $i$
$B_o/\bar{B}_o$ : $B$ remembers or forgets umbrella at shop home
$A_i,,B_i=3/4;\bar{A}_i,\bar{B}_i=1/4$ and $B_o,\bar{B}_o=1/2$
$$
\text{Req. Prob.}=A_1A_2A_3\bar{B}_o+A_1A_2A_3B_o\bar{B}_1+A_1A_2A_3B_oB_1\bar{B}_2+A_1A_2A_3B_oB_1B_2\bar{B}_3+\bar{A}_1B_oB_1B_2B_3+A_1\bar{A}_2B_oB_1B_2B_3+A_1A_2\bar{A}_3B_oB_1B_2B_3\\
=\Big(\frac{3}{4}\Big)^3\Big(\frac{1}{2}\Big)+\Big(\frac{3}{4}\Big)^3\Big(\frac{1}{2}\Big)\Big(\frac{1}{4}\Big)+\Big(\frac{3}{4}\Big)^3\Big(\frac{1}{2}\Big)\Big(\frac{3}{4}\Big)\Big(\frac{1}{4}\Big)+\Big(\frac{3}{4}\Big)^3\Big(\frac{1}{2}\Big)\Big(\frac{3}{4}\Big)^2\Big(\frac{1}{4}\Big)+\Big(\frac{1}{4}\Big)\Big(\frac{1}{2}\Big)\Big(\frac{3}{4}\Big)^3\\
+\Big(\frac{3}{4}\Big)\Big(\frac{1}{4}\Big)\Big(\frac{1}{2}\Big)\Big(\frac{3}{4}\Big)^3+\Big(\frac{3}{4}\Big)^2\Big(\frac{1}{4}\Big)\Big(\frac{1}{2}\Big)\Big(\frac{3}{4}\Big)^3\\
=\frac{27}{128}+\frac{27}{128*4}+\frac{81}{128*16}+\frac{243}{128*64}+\frac{27}{128*4}+\frac{81}{128*16}+\frac{243}{128*64}\\
=\frac{3726}{128*64}=\frac{3726}{8192}
$$
But my reference gives the solution $\frac{7278}{8192}$, so what am I missing in my attempt ?
Note: Please check $b$ part of link for a similar attempt to solve the problem to obtain the solution as in y reference.
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A forgets his umbrella along the way with probability $p=1-\left(\frac{3}{4}\right)^3=\frac{37}{64}$. B forgets his umbrella along the way with probability $q=\frac{1}{2}\left(1-\left(\frac{3}{4}\right)^3\right)=\frac{37}{128}$. Therefore, the probability that exactly one umbrella remains when both A and B return home is $$\begin{split}p(1-q)+q(1-p)&=p+q-2pq=\\&=\frac{37}{64}+\frac{37}{128}-\frac{37\cdot37}{4096}=\frac{37\cdot(64+32-37)}{4096}=\frac{37\cdot59}{4096}=\frac{2183}{4096}.\end{split}$$
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|
$a+b+c+d=1, a,b,c,d ≠ 0$, then prove that $(a + \frac{1}{a})^2 + (b + \frac{1}{b})^2 + (c + \frac{1}{c})^2 + (d + \frac{1}{d})^2 \ge \frac{289}{4} $ If $a+b+c+d=1$, $a,b,c,d ≠ 0$, prove that $$\left(a + \frac{1}{a}\right)^2 + \left(b + \frac{1}{b}\right)^2 + \left(c + \frac{1}{c}\right)^2 + \left(d + \frac{1}{d}\right)^2 \ge \frac{289}{4} $$
I tried expanding the entire LHS and I got
$a^2 + b^2 + c^2 + d^2 + \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} + \frac{1}{d^2} + 8$ on the LHS. Then I tried applying AM-GM to the first 4 terms and the last 4 terms and I got
$4 \sqrt{abcd} + \frac{4}{\sqrt{abcd}}$ but that didnt seem useful.
Any hints?
Thank you in advance.
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If you consider $a,b,c,d$ are all positive
According to cauchy-schwarz inequality
$(1+1+1+1)*((a+1/a)^2+(b+1/b)^2+(c+1/c)^2+(d+1/d)^2)>=$
$(a+b+c+d+1/a+1/b+1/c+1/d)^2=$
$(a+b+c+d+4+b/a+a/b+c/a+a/c+d/a+a/d+c/d+d/c+c/b+b/c+d/b+b/d)^2>=(5+12)^2$[because $a/b+b/a>=2 and a+b+c+d=1$]
So we get.... $(a+1/a)^2+(b+1/b)^2+(c+1/c)^2+(d+1/d)^2>= 289/4$.
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If $z^2+z+2=0$, then $z^2 + \frac{4}{z^2} = -3$
If $z^2+z+2=0$, for $z\in\mathbb C$, demonstrate:
$$z^2 + \frac{4}{z^2} = -3$$
$z^2 = - 2 - z$, but it didn't help me.
Is there any other elegant solution?
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$z^2+z+2=0$ means $z^2=-z-2$ and $1+z+\dfrac 2z=0$ so $\dfrac2z=-1-z$ so $\dfrac4{z^2}=z^2+2z+1$.
Therefore, $z^2+\dfrac4{z^2}=z^2+z-1=z^2+z+2-3=-3$.
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The polynomial $x^{2n} - 2x^{2n-1} + 3x^{2n-2}- \cdots -2nx + 2n + 1$ has no real roots.
Prove that the polynomial $$x^{2n} - 2x^{2n-1} + 3x^{2n-2}- \cdots -2nx + 2n + 1$$ has no real roots.
Here is the solution
Transcribed from this image
*For $x \leq 0$ we have obviously $p(x)>0$. Let $x>0$. We transform the polynomial in the same way as a geometric series:
$$
\begin{aligned}
p(x) &=x^{2 n}-2 x^{2 n-1}+3 x^{2 n-2}-\cdots-2 n x+2 n+1 \\
x p(x) &=x^{2 n+1}-2 x^{2 n}+3 x^{2 n-1}-4 x^{2 n-2}+\cdots+(2 n+1) x.
\end{aligned}
$$
Adding, we get
$$
\begin{aligned}
x p(x)+p(x) &=x^{2 n+1}-x^{2 n}+x^{2 n-1}-x^{2 n-2}+\cdots,+x+2 n+1 \\
(1+x) p(x) &=x \cdot \frac{1+x^{2 n+1}}{1+x}+2 n+1.
\end{aligned}
$$
From here we see that $p(x)>0$ for $x>0$.
Can anyone explain how the last line of the solution implies that the polynomial has no real roots?
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Solution shows that $p(x)>0, \text{ for all } x$, which means that there is no any value of $x$ such that $p(x)=0$
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Probability: 3 Urns, pulling out 3 balls Question: There are 3 urns: A, B and C.
Urn A has 2 white balls and 4 red balls
Urn B has 8 white balls and 4 red balls
Urn C has 1 white ball and 3 red balls
We pull one ball from each urn, and we get that 2 of them are white.
What is the probability that the ball we pulled from urn A is white?
This is my approach:
Given 2/3 of the balls are white, what is the probability we pulled a white ball from A:
$$\frac{P( \text{white from A})}{P(\text{white from A,B})+P(\text{white from A,C})+ P(\text{white from B,C})}$$
$=$
$$\frac{\frac{1}{3}}{(\frac{1}{3} \cdot \frac{8}{12} \cdot \frac{3}{4}) + (\frac{1}{3} \cdot \frac{4}{12} \cdot \frac{1}{4}) + (\frac{4}{6} \cdot \frac{8}{12} \cdot \frac{1}{4})} ~~~~~ (\geq 1)$$
However what I get is a number bigger than $1$ .. where am I wrong with this logic? If it is given $2/3$ balls are white then we need to calculate all the different ways we can get 2 white balls out of 3 urns.. that is: white from A and B OR white from A and C OR white from B and C..
I would appreciate your help!
Thank you!
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Let $P(E)$ be the probability that the ball we pulled from urn A is white. There are $6 \cdot 12 \cdot 4 = 288$ possible combinations of balls drawn in our sample space.
If $P(F)$ is the probability that exactly two balls are chosen from the sample space then: $$P(F) = \frac{\text{Number of combinations of exactly 2 white balls}}{288}$$
$$= \frac{(2 \cdot 8 \cdot 3) + (2 \cdot 4 \cdot 1) + (4 \cdot 8 \cdot 1)}{288} = \frac{88}{288} = \frac{11}{36}$$
If $P(E \cap F)$ is the probability that the ball chosen from Urn A is white then: $$P(E \cap F) = \frac{\text{Number of combinations of F where the ball from Urn A is white}}{288}$$
$$= \frac{(2 \cdot 8 \cdot 3) + (2 \cdot 4 \cdot 1)}{288} = \frac{56}{288} = \frac{7}{36}$$
Then $P(E \mid F)$ probability that one of the 2 white balls we pulled is from urn A: $$P(E \mid F) = \frac{P(E \cap F)}{P(F)} = \frac{7/36}{11/36} = \frac{7}{11}$$
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Prove this inequality with $xyz=1$ let $x,y,z>0$ and such $xyz=1$,show that
$$f(x)+f(y)+f(z)\le\dfrac{1}{8}$$
where $f(x)=\dfrac{x}{2x^{x+1}+11x^2+10x+1}$
I try use this $2x^x\ge x^2+1$,so we have
$$2x^{x+1}+11x^2+10x+1\ge x^3+11x^2+11x+1=(x+1)(x^2+10+1)$$
It need to prove
$$\sum_{cyc}\dfrac{x}{(x+1)(x^2+10x+1)}\le\dfrac{1}{8},$$where $xyz=1$
then I can't
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Now, let $x=\frac{a}{b},$ $y=\frac{b}{c}$, where $a$, $b$ and $c$ are positives.
Thus, $z=\frac{c}{a}$ and since $$x^x\geq x,$$ it's enough to prove that:
$$\sum_{cyc}\frac{ab}{13a^2+10ab+b^2}\leq\frac{1}{8},$$
which is true by BW.
Indeed, let $a=\min\{a,b,c\}$, $b=a+u$ and $c=a+v$.
Thus, we need to prove that:
$$384(u^2-uv+v^2)a^4+192(2u^3+7u^2v-5uv^2+2v^3)a^3+$$
$$+16(13u^4+82u^3v+39u^2v^2-62uv^3+13v^4)a^2+$$
$$+4(91u^3+258u^2v-162uv^2+13v^3)uva+13(13u^2+2uv+v^2)u^2v^2\geq0.$$
Now, we can show that:
$$384(u^2-uv+v^2)\geq384uv,$$
$$192(2u^3+7u^2v-5uv^2+2v^3)\geq768\sqrt{u^3v^3},$$
$$16(13u^4+82u^3v+39u^2v^2-62uv^3+13v^4)\geq-32u^2v^2,$$
$$4(91u^3+258u^2v-162uv^2+13v^3)\geq-384\sqrt{u^3v^3}$$ and
$$13(13u^2+2uv+v^2)\geq112uv.$$
Now, let $a=t\sqrt{uv}.$
Thus, it's enough to prove that:
$$384t^4+768t^3-32t^2-384t+112\geq0,$$ which is smooth.
Can you end it now?
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Find solutions of recursive equations Stuck... Find solutions of recursive equations using generating functions.
$$
x_{n+2} = 14x_{n+1} - 49x_n + n7^n, n\ge 0\\
x_0 = 1\\
x_1=14
$$
What I tried:
$$
x_{n} = 14x_{n-1} - 49x_{n-2} + (n-2)7^{n-2}, n\ge 2\\
a_{n} = 14a_{n-1} - 49a_{n-2} + (n-2)7^{n-2}\\
F(x) = \sum_{n=0}a_nx^n = 1 + \sum_{n=1}a_nx^n\\
F(x) - 1 = \sum_{n=1}a_nx^n$$
\begin{align}
F(x) &= 1 + 14x + \sum_{n=2}^\infty (14a_{n-1} - 49a_{n-2} + (n-2)7^{n-2})x^n\\
&= 1 + 14x + 14\sum_{n=2}^\infty a_{n-1}x^n -49\sum_{n=2}^\infty a_{n-2}x^n + \sum_{n=2}^\infty (n-2)7^{n-2}x^n \\
&= 1 + 14x + 14x\sum_{n=2}^\infty a_{n-1}x^{n-1} -49x^2\sum_{n=2}^\infty a_{n-2}x^{n-2} + x^2\sum_{n=2}^\infty (n-2)7^{n-2}x^{n-2} \\
&= 1 + 14x + 14x\sum_{n=1}^\infty a_{n}x^{n} -49x^2\sum_{n=0}^\infty a_{n}x^{n} + x^2\sum_{n=0}^\infty n7^{n}x^{n} \\
&= 1 + 14x + 14x(F(x) - 1) - 49x^2F(x) + x^2\sum_{n=0}^\infty n7^{n}x^{n}
\end{align}
And there is a problem. Is this correct? If yes, how to transform
$
x^2\sum_{n=0}^\infty n7^{n}x^{n}
$
into something like $x^2F(x)$?
It's a method which our teacher showed us to get generating function, but he didn't show us how make it with inhomogeneous.
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This looks correct so far. To proceed, note that $$\sum_{n \ge 0} n z^n = \sum_{n \ge 1} n z^n = z \sum_{n \ge 1} n z^{n-1} = z \frac{d}{dz}\sum_{n \ge 1} z^n = z \frac{d}{dz}\frac{z}{1-z} = \frac{z}{(1-z)^2}$$
|
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|
Suppose $f \circ g(x) = \frac{x^2-6x+2}{x+1}$ and $g(x) = 1 - x$. Then $f(-1)$ is equal to... I tried to substitute the value of $g(x)$ to every $x$ in $f \circ g(x) = \frac{x^2-6x+2}{x+1}$ and ended up with $\frac{x^2+4x-3}{-x+2}$.
Although honestly I do not know if that helps, or what I could do next.
I thought maybe this question just requires me to substitute $-1$ to every $x$ in $f\circ g(x) = \frac{x^2-6x+2}{x+1}$?
|
Given that $f(1-x)=\frac{x^2-6x+2}{x+1}$
Put $x=1-u$, we get:
$f(u)=\frac{(1-u)^2-6(1-u)+2}{(1-u)+1}$
Set $u=-1$, we get:
$f(-1)=\frac{2^2-2\cdot 6+2}{2+1}=\frac{-6}{3}=-2$
|
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|
How to integrate $\int\frac{dx}{\sqrt{x+1}+\sqrt{x-1}}$? Firstly, I tried to multiply the denominator and numerator by $\sqrt{x+1}$ but to no avail. Then I tried to take $\sqrt{x+1} = u$; now $\sqrt{x-1}=u-2$ but I couldn't find any success here either.
How can I integrate this expression?
|
Setting $\small\left\lbrace\begin{aligned}x&=\cosh{\left(2y\right)}\\ \mathrm{d}x&=2\sinh{\left(2y\right)}\,\mathrm{d}y\end{aligned}\right. $, we get : \begin{aligned}\int{\frac{\mathrm{d}x}{\sqrt{1+x}+\sqrt{x-1}}}&=\frac{1}{\sqrt{2}}\int{\frac{2\sinh{\left(2y\right)}}{\sinh{y}+\cosh{y}}\,\mathrm{d}y}\\ &=\frac{1}{\sqrt{2}}\int{\mathrm{e}^{-y}\left(\mathrm{e}^{2y}-\mathrm{e}^{-2y}\right)\mathrm{d}y}\\ &=\frac{1}{\sqrt{2}}\int{\mathrm{e}^{y}\,\mathrm{d}y}-\frac{1}{\sqrt{2}}\int{\mathrm{e}^{-3y}\,\mathrm{d}y}\\ &=\frac{\mathrm{e}^{y}}{\sqrt{2}}+\frac{\mathrm{e}^{-3y}}{3\sqrt{2}}+C\\ &=\frac{\mathrm{e}^{\frac{\cosh^{-1}{\left(x\right)}}{2}}}{\sqrt{2}}+\frac{\mathrm{e}^{-\frac{3\cosh^{-1}{\left(x\right)}}{2}}}{3\sqrt{2}}+C\\ &=\frac{1}{\sqrt{2}}\sqrt{x+\sqrt{x^{2}-1}}+\frac{1}{3\sqrt{2}\left(x+\sqrt{x^{2}-1}\right)\sqrt{x+\sqrt{x^{2}-1}}}+C\\ &=\frac{\sqrt{x+1}+\sqrt{x-1}}{2}+\frac{1}{3\left(x+\sqrt{x^{2}-1}\right)\left(\sqrt{x+1}+\sqrt{x-1}\right)}+C\end{aligned}
|
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|
Technique for simplifying, e.g. $\sqrt{ 8 - 4\sqrt{3}}$ to $\sqrt{6} - \sqrt{2}$ How to find the square root of an irrational expression, to simplify that root. e.g.:
$$
\sqrt{ 8 - 4\sqrt{3} } = \sqrt{6} - \sqrt{2}
$$
Easy to verify:
\begin{align}
(\sqrt{6} - \sqrt{2})^2 = 6 - 2\sqrt{12} +2
= 8 - 4 \sqrt{3}
\end{align}
But how to work it out in the first place? I feel there's a standard technique (Completing-the-square? Quadratic formula?), but don't recall it or what it's called...
BTW: this came up in verifying equivalence of different calculations of $\cos{75°}$ (the above divided by $4$), as $\cos{\frac{90°+60°}{2}}$ vs $\cos{(45°+30°)}$, from 3Blue1Brown's lockdown video on complex numbers and trigonometry.
|
I don't think there's a name for this procedure but let's apply it to $\sqrt{8-4\sqrt3}$.
If you suspect this equals $\sqrt a\pm\sqrt b$ with rationals $a$ and $b$, then
$$8-4\sqrt3=(\sqrt a\pm\sqrt b)^2=(a+b)\pm2\sqrt{ab}$$
so you want to solve simultaneously $a+b=8$ and $-4\sqrt{3}=\pm2\sqrt{ab}$.
So you need the minus sign, and $ab=12$. Then $a$ and $b$ are roots of the
quadratic equation
$$(X-a)(X-b)=X^2-(a+b)X+ab=X^2-8X+12.$$
This does has rational roots: $2$ and $6$.
|
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|
The binomial coefficient $\left(\begin{array}{l}99 \\ 19\end{array}\right)$ is $ 107,196,674,080,761,936, x y z $ , Find $x y z$
The binomial coefficient $\left(\begin{array}{l}99 \\ 19\end{array}\right)$ is a 21 -digit number:
$
107,196,674,080,761,936, x y z
$
Find the three-digit number $x y z$
I showed that $\left(\begin{array}{l}99 \\ 19\end{array}\right) \equiv 2(\bmod 4)$
and $\left(\begin{array}{l}99 \\ 19\end{array}\right) \equiv 19(\bmod 25)$
Now how to combine them to find last two digit (y and z)??
because we can only combine when $a \equiv b(\bmod n)$
$a \equiv b(\bmod m)$ then if (n,m)=1 then
$a \equiv b(\bmod mn)$ but here we have different b's ...
and also can someone tell some easier method to find $\left(\begin{array}{l}99 \\ 19\end{array}\right) \equiv 2(\bmod 4)$
and $\left(\begin{array}{l}99 \\ 19\end{array}\right) \equiv 19(\bmod 25)$ my approach takes me too long, so i want to see some easier method...
|
Since $99 \equiv -1 \pmod {25}$, we have $99 \cdot 98 \cdots 81 \equiv (-1)^{19}19! \pmod {25}$. What we would like to do is to simply divide by $19!$ and be done, but you'll notice that $19! \equiv 0 \pmod{25}$ because of the multiples of $5$. So instead, we treat the multiples of $5$ separately and this gives
$$ \binom{99}{19} \equiv (-1)^{19 - 3} \frac{95 \cdot 90 \cdot 85}{15 \cdot 10 \cdot 5} \pmod{25}.$$
Now we simplify:
$$ (-1)^{19 - 3} \frac{95 \cdot 90 \cdot 85}{15 \cdot 10 \cdot 5} = 3 \cdot 17 \cdot 19 = 51 \cdot 19 \equiv 19 \pmod{25}.$$
|
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|
How would you find the number of solutions to $m_1+2m_2+3m_3=n$? Assuming $m_i$ are nonnegative integers
I understand that we need to use $C(n+k-1,n-1)$ here but I am not sure how the coefficients of $m_i$ affect the equation?
For example to find the number of solutions to $m_1+m_2+m_2=n$, we could use the equation above to find that it is $C(n+2,2)=\frac{1}{2}(n+2)(n+1)$
|
We can write a statement that is not easy to calculate and try to simplify:
We know $0\le m_3 \le \left\lfloor \dfrac{n}{3}\right\rfloor$ and once we have chosen $m_3$, $0\le m_2 \le \left\lfloor\dfrac{n-3m_3}{2}\right\rfloor$. And for every choice of $m_3$ and $m_2$, there is exactly one choice of $m_1$, which yields the summation:
$$\sum_{m_3=0}^{\left\lfloor \tfrac{n}{3}\right\rfloor}\sum_{m_2=0}^{\left\lfloor \tfrac{n-3m_3}{2}\right\rfloor} 1 = \sum_{m_3=0}^{\left\lfloor \tfrac{n}{3}\right\rfloor}\left(\left\lfloor\dfrac{n-3m_3}{2}\right\rfloor+1\right)$$
Plugging this into Wolframalpha gives:
$$ = \left\lfloor \dfrac{n}{3} \right\rfloor + \left\lfloor \dfrac{n - 3}{2} \right\rfloor \left(\left\lfloor \dfrac{ \left\lfloor \dfrac{n}{3} \right\rfloor-1 }{2} \right\rfloor+1\right) + \dfrac{1}{2} \left( -3 \left\lfloor \dfrac{ \left\lfloor \dfrac{n}{3} \right\rfloor -1 }{2} \right\rfloor^2 - 3 \left\lfloor \dfrac{ \left\lfloor \dfrac{n}{3} \right\rfloor -1 }{2} \right\rfloor - 3 \left\lfloor \dfrac{ \left\lfloor \dfrac{n}{3} \right\rfloor }{2}\right\rfloor^2 - 3 \left\lfloor\dfrac{ \left\lfloor \dfrac{n}{3} \right\rfloor }{2} \right\rfloor + 2 \left\lfloor\dfrac{n}{2}\right\rfloor \left( \left\lfloor \dfrac{ \left\lfloor \dfrac{n}{3}\right\rfloor }{2}\right\rfloor + 1\right) + 2\right)$$
|
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|
Integrability of $\frac{1}{(x^2+y^2+z^2)^a}$ on $E=\{(x,y,z)\in \mathbb{R}^3: z>1, \ z^2(x^2+y^2)<1 \}$ Let $E=\{(x,y,z)\in \mathbb{R}^3: z>1, \ z^2(x^2+y^2)<1 \}$ and $$f_{a}(x)=\frac{1}{(x^2+y^2+z^2)^a}$$
I need to find all $a\in \mathbb{R}$ such that $f_a\in L^1(E).$
I already know a solution to this problem, and I want to understand why mine is wrong.
My attempt
1) [Measurability check]
First of all $E$ is open in $\mathbb{R}^3$ being a union over all $z>1$ of open disks of radius $\frac{1}{z}$: $E=\bigcup_{z>1} E_z=\bigcup_{z>1}\{(x,y): x^2+y^2<\frac{1}{z^2}\}.$ Next, since $z>1,$ $f_a$ looks continous on every point of $E,$ thus is measurable.
2) [Positivity]
The function $f_a$ is strictly positive for every $a\in \mathbb{R},$ thus to check integrability we can reduce ourselves to compute
$$\int_E f_a \ d(x,y,z)$$
and by positivity of the function we can apply Tonelli's theorem, rewriting the integral as
$$\int_1^{+\infty}\int_{x^2+y^2<\frac{1}{z^2}} f_a \ \ d(x,y) \ d(z)$$
3)[Polar coordinates]
We now compute the inner integral using polar coordinates to obtain
$$\int_{x^2+y^2<\frac{1}{z^2}} f_a \ \ d(x,y)= \int_0^{2\pi} \int_0^{1/z} \frac{r}{(r^2+z^2)^a} \ dr d\theta= 2\pi \int_0^{1/z} \frac{r}{(r^2+z^2)^a} \ dr =...$$
changing variable $r\mapsto \ t= \phi(r)=r^2+z^2 $
$$...=\pi \int_{z^2}^{1/z^2+z^2} \frac{1}{t^a} \ dt= F(z,a) $$
4) [Computation by cases]
4.i)
For $a=1,$ we have $F(1,z)= \log(1/z+z^2)-\log(z^2)$ and $$\int_1^{+\infty}\log(1/z^2+z^2)-\log(z^2) \ dz= \int_1^{+\infty} \log(z^2(1/z^4+1))-\log(z^2) \ dz= \int_1^{+\infty}\log(1+1/z^4) \ dz < +\infty$$
and so $f_{a=1}\in L^1(E).$
4.ii)
Now we let $a \neq 0,$ and we have
$$\int_{z^2}^{1/z^2+z^2}t^{-a} \ dt=\frac{(1/z^2+z^2)^{1-a}}{1-a}-\frac{(z^2)^{1-a}}{1-a}$$
and we have to evaluate
$$\int_1^{+\infty}\frac{(1/z^2+z^2)^{1-a}}{1-a}-\frac{(z^2)^{1-a}}{1-a} \ dz $$
But
$$\int_1^{+\infty}\frac{(1/z^2+z^2)^{1-a}}{1-a}-\frac{(z^2)^{1-a}}{1-a}
\ dz \geq \int_1^{+\infty}\frac{(1/z^2)^{1-a}}{1-a}+
\frac{(z^2)^{1-a}}{1-a} -\frac{(z^2)^{1-a}}{1-a} \ dz=
\int_1^{+\infty}\frac{(1/z^2)^{1-a}}{1-a} \ dz > +\infty$$
for $0<2-2a\leq 1 \iff 0<1-a \leq \frac{1}{2} \iff a \geq \frac{1}{2}$ so that $f_a \notin L^1(E)$ for $a \geq \frac{1}{2};$
on the other hand, we have
$$\int_1^{+\infty}\frac{(1/z^2+z^2)^{1-a}}{1-a}-\frac{(z^2)^{1-a}}{1-a}
\ dz \leq \int_1^{+\infty}\frac{(1/z^2+z^2)^{1-a}}{1-a} \ dz$$
and since in a ngbh of $+\infty$ $$1/z^2+z^2 \sim z^2 $$ the latter is finite if and only if $$\int_1^{+\infty}\frac{(1/z^2)^{1-a}}{1-a}<+\infty$$
if and only if
$$2-2a >1 \iff a<\frac{1}{2}$$
Can someone proof-read and tell me if there are any mistakes?
The other solution I have, using spherical coordinates, has $f_a $ integrable $\iff$ $a=3/2$ or $a>-\frac{1}{2}, $ while my solution gives $f_a$ integrable iff $a=1$ or $a<\frac{1}{2}$. I want to understand what went wrong with my solution.
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Estimating $\int_{z^2}^{z^2+z^{-2}} \frac{dt}{t^a}$ correctly is how you would get the correct range for convergence. For $a\ge 0$, the integrand is decreasing, so we can estimate
$$ \int_{z^2}^{z^2+z^{-2}} \frac{dt}{t^a} \le \frac1{(z^{2})^a}(z^2 + z^{-2} - z^2) = \frac1{z^{2a+2}}\in L^1_{\text{loc}}(1,\infty),$$
and this is integrable at infinity if $2a+ 2>1$, which gives $a>-1/2$, so $a\ge 0$. (Note that this estimate works better than just trying to use that $z^2+z^{-2}\sim z^2$ as a black box.)
For $a<0$, the integrand is increasing, we instead go
$$\int_{z^2}^{z^2+z^{-2}} \frac{dt}{t^a} \le \frac1{z^2+z^{-2}}(z^2 + z^{-2} - z^2) = \frac1{1+z^{2\alpha + 2}}$$
and the same analysis applies, giving the remaining cases $-1/2 < a < 0$.
As for the divergence part, I don't believe its true that
$$ (a+b)^s \ge a^s + b^s$$
where you used this for $s\in[0,1/2]$. Actually in this regime, you can use the concavity of $x\mapsto x^s$ to deduce that the opposite inequality holds. Also, the original 3D integral is integrable on the larger set $\{z>1\}\supset E$ when $2a>3$, so that's already a sign that you can't prove it diverges for the range you claimed.
For a correct proof, actually the above proof gives lower bounds as well, by using the other integration bound. So $\int_{z^2}^{z^2+z^{-2}} \frac{dt}{t^a} \sim \frac1{z^{2a+2}}$ at infinity, and this diverges if $a\le 1/2$.
|
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|
I don’t understand how to reduce this fraction to the stated solution: The fraction is as follows:
$$
\frac{9 \cdot 11 + 18 \cdot 22 + 27 \cdot 33 + 36 \cdot 44 }{
22 \cdot 27 + 44 \cdot 54 + 66 \cdot 81 + 88 \cdot 108}
$$
That’s all fine. What I don’t get is that my textbook says this reduces to the following:
$$\frac{9\times 11 + (1^2 + 2^2 + 3^2 + 4^2)}{22 \times 27 \times (1^2 + 2^2 + 3^2 +4^2)}$$
I don’t understand how the sum of consecutive squares can be deduced from that fraction, or why the denominator contains $22\times 27 \times\dots $ as opposed to the numerator which is $9 \times 11 + \dots$”
Any insight would be really appreciated!
|
There seems to be a mistake:
\begin{align*}
\frac{9\cdot 11 + 18 \cdot 22 + 27 \cdot 33 + 36 \cdot 44}{22\cdot 27 + 44\cdot 54 + 66 \cdot 81 + 88 \cdot 108} & = \frac{9\cdot 11(1\cdot 1+2\cdot 2+3\cdot 3 + 4\cdot 4)}{22\cdot 27(1\cdot 1+2\cdot 2+3\cdot 3+4\cdot 4)} \\
& = \frac{9\cdot 11(1^2+2^2+3^2 + 4^2)}{22\cdot 27(1^2+2^2+3^2+4^2)} = \frac{99}{594} < 1,
\end{align*}
and this is different from $99+\frac{1}{594} > 99$, which is the result you have.
|
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|
Using the Maclaurin series for $\frac{1}{1-x}$ to find $\frac{x}{1+x^2}$ Suppose I know the Maclaurin series for $$\frac{1}{1-x}=1+x+x^2+x^3+...= \sum_{n=0}^{\infty}x^n \tag{1}$$
then I can find the Maclaurin series for $\frac{1}{(1-x)^2}$ by the substitution $x\to x(2-x)$, which is obtained by solving the following equation for $x$:
$$1-x=(1-u)^2$$ $$x=u(2-u)$$
and replacing $u$ by and $x$, and $=$ by $\to$ , which leads to
$$\frac{1}{(1-x)^2}=1+x(2-x)+x^2(2-x)^2+x^3(2-x)^3+... \tag{2}$$
Succumbing to the same 'substitution'approach with $\frac{x}{1+x^2}$, I put
$$\frac{1}{1-x}=\frac{u}{1+u^2}$$ $$x=1-\frac{1}{u}-u$$
so with $x \to 1-\frac{1}{x}-x$
I should have:
$$\frac{x}{1+x^2}=1+\left(1-\frac{1}{x}-x\right)+\left(1-\frac{1}{x}-x\right)^2+\left(1-\frac{1}{x}-x\right)^3+...\tag{3}$$
but this does not seem correct (at least according to my Desmos graph for $|x|<1$).
Can someone please explain what my conceptual errors are?
|
Your series $(1)$ is valid for $x\in(-1,1)$.
So series $(2)$ is valid for $x(2-x)\in(-1,1)$, i.e. for $x\in(1-\sqrt 2,1)\cup(1,1+\sqrt 2)$.
But series $(3)$ is only valid if $1-\frac{1}{x}-x\in(-1,1)$. And it turns out that this is not true for any $x\in\Bbb R$.
|
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|
Prove $(a^2+b^2+c^2)^3 \geqq 9(a^3+b^3+c^3)$ For $a,b,c>0; abc=1.$ Prove$:$ $$(a^2+b^2+c^2)^3 \geqq 9(a^3+b^3+c^3)$$
My proof by SOS is ugly and hard if without computer$:$
$$\left( {a}^{2}+{b}^{2}+{c}^{2} \right) ^{3}-9\,abc \left( {a}^{3}+{b} ^{3}+{c}^{3} \right)$$
$$=\frac{1}{8}\, \left( b-c \right) ^{6}+{\frac {117\, \left( b+c \right) ^{4} \left( b+c-2\,a \right) ^{2}}{1024}}+{\frac {3\,{a}^{2} \left( 40\,{a }^{2}+7\,{b}^{2}+14\,bc+7\,{c}^{2} \right) \left( b-c \right) ^{2}}{ 32}}$$
$$+{\frac {3\, \left( b+c \right) ^{2} \left( 3\,a-2\,b-2\,c \right) ^{2} \left( b-c \right) ^{2}}{32}}+\frac{3}{16}\, \left( a+2\,b+2\,c \right) \left( 4\,a+b+c \right) \left( b-c \right) ^{4}$$
$$+{\frac { \left( 16\,{a}^{2}+24\,ab+24\,ac+11\,{b}^{2}+22\,bc+11\,{c}^{ 2} \right) \left( 4\,a-b-c \right) ^{2} \left( b+c-2\,a \right) ^{2} }{1024}} \geqq 0$$
I think$,$ $uvw$ is the best way here but it's not concordant for student in The Secondary School.
Also$,$ BW helps here, but not is nice, I think.
So I wanna nice solution for it! Thanks for a real lot!
|
Suppose $a = \max\{a,b,c\}.$ By the AM-GM inequality we have
$$9abc(a^3+b^3+c^3) \leqslant \left(ab+ca+\frac{a^3+b^3+c^3}{3a}\right)^3.$$
Therefore, we need to prove
$$a^2+b^2+c^2 \geqslant ab+ca+\frac{a^3+b^3+c^3}{3a},$$
equivalent to
$$\frac{(2a-b-c)(a^2+b^2+c^2-ab-bc-ca)}{3a} \geqslant 0.$$
which is true.
|
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|
Find three vectors in ${R^3}$ such that the angle between all of them is pi/3? Is there a simple way to do this? I have found that ${(a.b)/|a||b|}$ must be equal to ${1/2}$ but from there I am stuck how to proceed. Any help?
P.s. This is from the MIT 2016 Linear Algebra course and is not homework.
|
Take an orthonormal basis $(i,j,k)$ of $\mathbb R^3$, $v_1 = \cos \frac{\pi}{6} i + \sin \frac{\pi}{6} j$ and $v_2 = \cos \frac{\pi}{6} i - \sin \frac{\pi}{6} j$. We have by construction $\angle(v_1, v_2) = \frac{\pi}{3}$.
Now let's find $\alpha$ such that $v_3 = \cos \alpha i + \sin \alpha k$ solve the problem.
That will be the case providing that $\angle(v_1,v_3) = \frac{\pi}{3}$, i.e. if $\cos \frac{\pi}{6} \cos \alpha = \cos \frac{\pi}{3}$, i.e. $\cos \alpha = \frac{1}{\sqrt 3}$ and $\sin \alpha = \sqrt{1 - \cos^2 \alpha} = \frac{\sqrt2}{\sqrt{3}}$.
Finally $$\begin{cases}
v_1 &= \frac{\sqrt 3}{2} i + \frac{1}{2}j\\
v_2 &= \frac{\sqrt 3}{2} i - \frac{1}{2}j\\
v_3 &= \frac{1}{\sqrt 3}(i + \sqrt 2 k)
\end{cases}$$
is a solution.
|
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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.