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Infinite sum of reciprocal shifted Fibonacci numbers I found on Wikipedia the following infinite sum :
$$\sum_{k=0}^{\infty} \frac{1}{1+F_{2k+1}}=\frac{\sqrt{5}}{2}$$
There is no reference for this sum in the article and I couldn't find it anywhere else .I have no idea how to prove it so I'm asking for help .
Thank you for your time to help me !
|
We can use Binet's formula for $F_{2k+1}$:-
$$F_{2k+1}=\frac{\phi^{2k+1}-(-\phi)^{-(2k+1)}}{\sqrt{5}}$$
where $\phi=\frac{1+\sqrt{5}}{2}$.
The summation (after substitution of $F_{2k+1}$ and partial fraction decomposition) becomes a telescoping sum
$$\begin{align}\sum_{k=0}^{\infty} \frac{1}{1+F_{2k+1}}&=\sum_{k=0}^{\infty} \frac{\sqrt{5}\phi^{2k+1}}{\sqrt{5}\phi^{2k+1}+\phi^{4k+2}+1}\\&=\sum_{k=0}^{\infty} \left(\frac{\sqrt{5}(1+\sqrt{5})}{2\phi^{2k+1}+\sqrt{5}+1}-\frac{\sqrt{5}(\sqrt{5}-1)}{2\phi^{2k+1}+\sqrt{5}-1}\right)\\&=\frac{\sqrt{5}}{2}-x_1+x_1-x_2+x_2+\cdots+\lim_{k\rightarrow\infty} x_k\\&=\frac{\sqrt{5}}{2}\end{align}$$
where $$x_1=\frac{\sqrt{5}(\sqrt{5}-1)}{2\phi^{1}+\sqrt{5}-1}=\frac{\sqrt{5}(1+\sqrt{5})}{2\phi^{3}+\sqrt{5}+1},x_2=\frac{\sqrt{5}(\sqrt{5}-1)}{2\phi^{3}+\sqrt{5}-1}=\frac{\sqrt{5}(1+\sqrt{5})}{2\phi^{5}+\sqrt{5}+1},\cdots$$
and $\lim_{k\rightarrow\infty} x_k=0$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
$25$ colored dice $25$ colored dice are thrown and their results are put at random at a square matrix $5\times 5$. Dice colors are as follows: $5$ Red, $5$ Blue, $5$ Yellow, $5$ Green and $5$ Orange.
I am interested in calculating the following probabilities:
*
*The probability that a bar of length $3$ ($3$ consecutive dice of the same color), vertical or horizontal, appears somewhere in the grid.
*The probability that a bar of length $4$ ($4$ consecutive dice of the same color), vertical or horizontal, appears somewhere in the grid.
*The probability that a bar of length $5$ ($5$ consecutive dice of the same color), vertical or horizontal, appears somewhere in the grid.
|
The third question can be solved without a computer.
We have one condition for each row and one for each column. Multiple conditions can only be satisfied simultaneously if they correspond to the same direction.
The probablity for $k$ particular compatible conditions to be satisfied simultaneously is $\prod_{j=0}^{k-1}\frac{5-j}{\binom{25-5j}5}$.
Thus, by inclusion–exclusion the probability for at least one of the conditions to be satisfied is
$$
10\cdot\frac5{\binom{25}5}-2\cdot\binom52\cdot\frac{5\cdot4}{\binom{25}5\binom{20}5}+2\cdot\binom53\cdot\frac{5\cdot4\cdot3}{\binom{25}5\binom{20}5\binom{15}5}
\\
-2\cdot\binom54\cdot\frac{5\cdot4\cdot3\cdot2}{\binom{25}5\binom{20}5\binom{15}5\binom{10}5}+2\cdot\binom55\cdot\frac{5\cdot4\cdot3\cdot2\cdot1}{\binom{25}5\binom{20}5\binom{15}5\binom{10}5\binom55}
\\
=\frac5{5313}-\frac5{10296594}+\frac5{10306890594}-\frac5{2597336429688}+\frac1{2597336429688}
\\[15pt]
=\frac{32145551}{34175479338}\approx9.406\cdot10^{-4}\;.
$$
Since it's highly unlikely that more than one condition is satisfied, this probability is very well approximated by the expected number of satisfied conditions,
$$
10\cdot\frac5{\binom{25}5}=\frac5{5313}\approx9.411\cdot10^{-4}\;.
$$
|
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|
Evaluation of $\int\frac{\sin 2x}{(3+4\cos x)^3}\,dx$
Evaluation of $$\int\frac{\sin 2x}{(3+4\cos x)^3}\,dx$$
$\bf{My\; Try::}$ Let $$\displaystyle \int\frac{\sin 2x}{(3+4\cos x)^3}dx = 2\int\frac{\sin x\cos x}{(3+4\cos x)^3}dx$$
Now Divide both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $\cos^3 x.$
So we get $$\displaystyle I = 2\int\frac{\sec x\cdot \tan x}{(3\sec x+4)^3}dx$$
Now Put $\displaystyle (3\sec x+4) = t\;,$ Then $3\sec x\cdot \tan xdx = dt$
So we get $$\displaystyle I = \frac{2}{3}\int \frac{1}{t^3}dt = -\frac{1}{3t^2}+\mathcal{C} = -\frac{1}{3(3\sec x+4)^2}+\mathcal{C}$$
But answer is http://www.wolframalpha.com/input/?i=INTEGRATION+OF+%28sin+2x%29%2F%283%2B4cos+x%29%5E3
Where I gave Done Wrong,
Thanks
|
Going by your work so far we have
$$
2\int \frac{\sin \theta \cos \theta}{(3+4\cos \theta)^3}d\theta
$$
Lets set $u = 3+4\cos \theta\to du = -4\sin\theta d\theta$
thus your integral becomes
$$
-\frac{2}{16}\int \frac{u-3}{u^3}du=?
$$
This yields the result you desire! Though the interesting question is how to go from your expression
$$
\begin{align}
-\frac{1}{3}\frac{1}{(3\sec x + 4)^2} + C_1 + C_2 &=& -\frac{1}{3}\left[\frac{\cos^2 x}{(3+4\cos x)^2}-3C_1\right]+ C_2\\
&=&-\frac{1}{3}\left[\frac{\cos^2 x-3C_1(3+4\cos x)^2}{(3+4\cos x)^2}\right]+ C_2\\
&=&-\frac{1}{3}\left[\frac{\cos^2 x-3C_1\left(9+24\cos x + 16\cos^2 x\right)}{(3+4\cos x)^2}\right]+ C_2\\
&=&-\frac{1}{3}\left[\frac{\left(1-3\cdot 16\cdot C_1\right)\cos^2 x-3\cdot 9\cdot C_1-3\cdot 24\cdot C_1\cos x }{(3+4\cos x)^2}\right]+ C_2
\end{align}
$$
To be similar to Wolfram you need to set $\cos^2 x$ coefficient to zero so we have $1-3\cdot 16\cdot C_1 = 0\implies C_1 = \frac{1}{3\cdot 16}$ so we obtain
$$
-\frac{1}{3}\frac{1}{(3\sec x + 4)^2} + C_1 + C_2 = -\frac{1}{3}\left[\frac{- \frac{9}{16}-\frac{24}{16}\cos x }{(3+4\cos x)^2}\right]+ C_2
$$
this becomes
$$
\left[\frac{\frac{3}{16}+\frac{8}{16}\cos x }{(3+4\cos x)^2}\right]+ C_2
$$
which finally becomes
$$
\frac{1}{16}\frac{3+8\cos x }{(3+4\cos x)^2}+ C_2
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Forming a committee from $4$ gentlemen and $4$ ladies with certain conditions
From $4$ gentlemen and $4$ ladies a committee of $5$ is to be formed .
If the committee consists of $1$ president, $1$ vice president and $3$ secretaries.
What will be the number of ways of selecting the committee with at least $3$
women such that at least one women holds the post of president or vice-president?
I tried
Case $1$
$3W2M$
$\dbinom{4}{1}\times \dbinom{4}{1} \times \dbinom{3}{2} \times \dbinom{3}{1}=144 $
$+$
$\dbinom{4}{1}\times \dbinom{3}{1} \times \dbinom{2}{1} \times \dbinom{4}{2}=144 $
Case $2$
$4W1M$
$\dbinom{4}{1}\times \dbinom{3}{1} \times \dbinom{2}{2} \times \dbinom{3}{1}=36 $
$+$
$\dbinom{4}{1}\times \dbinom{4}{1} \times \dbinom{3}{3}=16 $
Total ways=$338$.
But the answer given is $512$.
I look for a short and simple way.
I have studied maths up to $12$th grade.
|
As pointed out in a comment, there are numerous errors in your answer.
Pres-Veep . . . . . Secretaries
W-W . . . . . . . . . 2W,1M or 1W-2M: $\left[{4\choose1 }{3\choose 1}\right]\left[{2\choose2}{4\choose1} + {2\choose 1}{4\choose 2}\right]= 192$
W-M or M-W . . . . 3W or 2W,1M: $\left[2{4\choose1}{4\choose1}\right]\left[{3\choose 3}+{3\choose2}{3\choose1}\right]=320$
|
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|
If $\sin\theta+\cos\theta=1$ prove that $\cos\theta-\sin\theta=\pm1$ So my work,
Squaring both sides $$(\sin\theta+\cos\theta)^2=1$$
$$1+2\sin\theta\cos\theta=1\ \ \ \ \ \text{-------(i)}$$
$$\sin\theta\cos\theta=0 \ \ \ \ \ \text{------(ii)}$$
So reverting back to $(i)$,
$$\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta-4\sin\theta\cos\theta=1-4\sin\theta\cos\theta$$
$$(\cos\theta-\sin\theta)^2=1-4\sin\theta\cos\theta$$
$$\cos\theta-\sin\theta=\pm1$$
But my teacher says that there is a shorter solution than that, so please can someone help me find that?
|
Notice, we have $$\cos \theta+\sin\theta=1$$
$$(\cos \theta+\sin\theta)^2=1$$$$\cos^2\theta+\sin^2\theta+2\sin \theta\cos \theta=1$$ $$1+2\sin \theta\cos \theta=1$$
$$\iff \sin\theta\cos \theta=0\tag 1$$
Now, we have
$$(\cos \theta-\sin\theta)^2=\cos^2\theta+\sin^2\theta-2\sin \theta\cos \theta$$
$$=(\cos^2\theta+\sin^2\theta+2\sin \theta\cos \theta)-4\sin \theta\cos \theta$$
$$=(\cos \theta+\sin\theta)^2-4\sin \theta\cos \theta$$
Substituting the corresponding values
$$=(1)^2-4(0)=1$$ $$\cos\theta-\sin\theta=\pm\sqrt 1=\pm 1$$
|
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|
Coin toss puzzle with 1 biased and 2 unbiased coins Three coins are given: one two-head coin and two fair coins. You randomly choose a coin and the first three tosses give heads. What is the probability that 4-th toss is a head.
I have two solutions that give different results. Please help me to find the error.
1st solution:
$$P(\text{first 3 tosses are heads})=\frac{2}{3}\times\frac{1}{8}+\frac{1}{3}\times 1=\frac{5}{12}$$
$$P(\text{first 4 tosses are heads})=\frac{2}{3}\times\frac{1}{16}+\frac{1}{3}\times 1=\frac{3}{8}$$
$$P(\text{4th is head} \mid \text{first 3 tosses are heads})=\frac{P(\text{first 4 tosses are heads})}{P(\text{first 3 tosses are heads})}=\frac{9}{10}.$$
2nd solution:
$$P(\text{4th is head} \mid \text{first 3 tosses are heads})=$$
$$=\frac{2}{3}\times P(\text{4th is head} \mid \text{first 3 tosses are heads and coin is fair})+$$
$${}+\frac{1}{3}\times P(\text{4th is head}\mid\text{first 3 tosses are heads and coin is two-head})=\frac{2}{3}\times \frac{1}{2}+\frac{1}{3}\times 1=\frac{2}{3}.$$
|
Your second solution is erroneous.
\begin{align}
& \Pr(\text{4th is head} \mid \text{first 3 tosses are heads})\\[10pt]
= {} & \Pr(\text{fair} \mid \text{1st 3 are heads}) \times \Pr(\text{4th is a head} \mid \text{fair & 1st 3 are heads}) \\[4pt]
& {} + \Pr(\text{two-headed}\mid \text{1st 3 are heads} ) \\
& \phantom{{} + \Pr(\text{two-headed})} \times \Pr(\text{4th is a head} \mid \text{two-headed & 1st 3 are heads}) \\[10pt]
= {} & \Pr(\text{fair} \mid \text{1st 3 are heads}) \times \frac 1 2 + \Pr(\text{two-headed}\mid \text{1st 3 are heads} ) \times 1. \tag 1
\end{align}
So we will find $\Pr(\text{fair}\mid\text{1st 3 are heads})$ and $\Pr(\text{two-sided} \mid\text{1st 3 are heads})$.
\begin{align}
& \frac{\Pr(\text{fair}\mid\text{1st 3 are heads})}{\Pr(\text{two-sided} \mid\text{1st 3 are heads})} \\[10pt]
= {} & \frac{\Pr(\text{fair})}{\Pr(\text{two-headed})} \times \frac{\Pr(\text{1st 3 are heads} \mid \text{fair})}{\Pr(\text{1st 3 are heads} \mid \text{two-sided})} \\[10pt]
= {} & \frac{2/3}{1/3} \times \frac{1/8}{1} = \frac 1 4.
\end{align}
If $p/(1-p) = 1/4$ then $p = 1/5$ and $1-p=4/5$.
Hence on line $(1)$ above we get
$$
\left( \frac 1 5 \times \frac 1 2 \right) + \left(\frac 4 5 \times 1\right) = \frac 9 {10}.
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\int \frac{\sqrt{1+x^2}}{1-x^2}dx$ Problem :
$\int \frac{\sqrt{1+x^2}}{1-x^2}dx$
My approach :
Put $x = \tan\theta$
we get $$\int \frac{\sqrt{1+x^2}}{1-x^2}dx = \frac{\frac{\sin^2\theta + \cos^2\theta}{\cos^2\theta \cos\theta}}{\frac{\cos^2\theta -\sin^2\theta}{\cos^2\theta}} d\theta $$
$$= \frac{1}{(\cos^2\theta -\sin^2\theta)\cos\theta}d\theta$$
But is it the right approach please guide will be of great help thanks.
|
Letting $t=\frac{x}{\sqrt{x^{2}+1}}$, then $$
x^{2}=\frac{t^{2}}{1-t^{2}}=\frac{1}{1-t^{2}}-1 \Rightarrow 2 x d x=\frac{2 t d t}{1-t^{2}} \Rightarrow\sqrt{x^{2}+1} d x=\frac{d t}{1-t^2}$$
Plugging them into the integral yields
\begin{aligned}
I &=\int \frac{1}{1-\frac{t^2}{1-t^{2}}} \cdot \frac{d t}{1-t^{2}} \\
&=\int \frac{d t}{1-2 t^{2}} \\
&=\frac{1}{2\sqrt{2}} \ln \left|\frac{\sqrt{2} t+1}{\sqrt{2} t-1}\right|+C\\& =\frac{1}{2\sqrt{2}} \ln \left|\frac{\sqrt{2} x+\sqrt{x^{2}+1}}{\sqrt{2} x-\sqrt{x^{2}+1}}\right|+C
\end{aligned}
|
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|
Finding an angle in a triangle, given the angle bisector and some conditions.
$ABC$ is a triangle in which $\angle B= 2\angle C$. $D$ is a point on side $BC$ such that $AD$ bisects $\angle BAC$ and $AB = CD$.
What is the measure of $\angle BAC? $
I tried using angle bisector theorem, similarity, but the sides, don't include the angles. Also I got a relation, $3x+2\theta=180^{\circ}$, where $\angle ACB=x;\angle BAD=\theta$.
But I cannot derive one more relation, please help.
Please don't use trigonometry, nor any constructions. We have to do without them, simply.
|
Notice, $$\angle ADB=180^\circ -\left(\angle B+\frac{\angle BAC}{2}\right)$$
$$\angle B+\angle C+\angle BAC=180^\circ\iff 2\angle C+\angle C+\angle BAC=180^\circ$$ $$\iff \angle C=\frac{180^\circ-\angle BAC}{3}\tag 1$$$$\iff \angle B=\frac{2(180^\circ-\angle BAC)}{3}\tag 2$$
apply sine rule in $\triangle ABD$ $$\frac{\sin\angle B}{AD}=\frac{\sin \angle ADB}{AB}$$
$$\frac{\sin\angle B}{AD}=\frac{\sin \left(\angle B+\frac{\angle BAC}{2}\right) }{AB}\tag 3$$ Similarly, apply sine rule in $\triangle ADC$
$$\frac{\sin\angle C}{AD}=\frac{\sin \left(\frac{\angle BAC}{2}\right)}{CD}$$
Setting $CD=AB$, we get
$$\frac{\sin\angle C}{AD}=\frac{\sin \left(\frac{\angle BAC}{2}\right)}{AB}\tag 4$$
Dividing (3) by (4), we get
$$\frac{\sin\angle B}{\sin \angle C}=\frac{\sin \left(\angle B+\frac{\angle BAC}{2}\right) }{\sin \left(\frac{\angle BAC}{2}\right) }$$
Setting the values of $\angle B$ & $\angle C$, we get
$$\frac{\sin \left(\frac{2(180^\circ-\angle BAC)}{3}\right)}{\sin \left(\frac{180^\circ-\angle BAC}{3}\right)}=\frac{\sin \left(\frac{2(180^\circ-\angle BAC)}{3}+\frac{\angle BAC}{2}\right) }{\sin \left(\frac{\angle BAC}{2}\right) }$$
$$2\cos \left(60^\circ-\frac{\angle BAC}{3}\right)=\frac{\sin \left(120^\circ-\frac{\angle BAC}{6}\right) }{\sin \left(\frac{\angle BAC}{2}\right) }$$
I hope you can solve for $\angle BAC$
|
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|
Find a vector function represented by the curve of intersection? I'm struggling with the following problem:
Given $\, z = \sqrt{x^2 + y^2}\,$ and $\, z = y+1\,$ find the vector function represented by the curve of intersection of the surfaces using the parametrization $\, x = t$.
I've done problems like this before but this one seems different. The $\,x\,$ is contained in a function with two other variables. My attempt was equating the two functions of $\,z$. Then I solved for $\,y\,$ which got me:
$\,y=\left(t^2-1\right)\big/\,2$.
Not sure how to approach this problem.
|
First, equate $\,z\,$ and square both sides of obtained expression:
$$
z = \sqrt{x^2 + y^2} = y+1
\implies
x^2 + y^2 = y^2 + 2y + 1
\implies
\boxed{y = \dfrac{x^2 - 1}{2}}
$$
Second, apply parametrization $\, x = t$:
\begin{align}
\begin{cases}
y = \dfrac{x^2 - 1}{2}\\
z = y + 1 \\
z = \sqrt{x^2 + y^2}
\end{cases}
\stackrel{x\, =\, t}{\implies}
\begin{cases}
x = t \\
y = \dfrac{t^2 - 1}{2} \\
z = y + 1
\end{cases}
\implies
\begin{cases}
x = t \\
y = \dfrac{t^2 - 1}{2} \\
z = \dfrac{t^2 +1}{2}
\end{cases}
\end{align}
Therefore your vector function will look like
$$
F\left(t\right) =
\begin{pmatrix}
t \\ \dfrac{t^2 - 1}{2} \\ \dfrac{t^2 + 1}{2}
\end{pmatrix}
= \frac 1 2\;
\begin{pmatrix}
2t \\ \ t^2 - 1 \\ t^2 + 1
\end{pmatrix}
$$
|
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|
Number Theoretic Transform (NTT) example not working out I'm reading up on the NTT, which is a generalisation of the DFT. I'm working in $\mathbb{F}_5$ with primitive root $w=2 \mod 5$. Suppose I want to compute the NTT of $x=(1,4)$. So far I have obtained:
$$\hat{x}=\mathcal{N}(x)=\left(\sum_{j=0}^1 2^{jk}x_j\mod 5\right)_{k=0}^1=(1+4,1+2\cdot 4)\equiv(0,4)\mod 5.$$
Applying the inverse NTT should recover $x$. But...
$$\mathcal{N}^{-1}(\hat{x})=\left(\frac{1}{2}\sum_{k=0}^1 2^{-jk}\hat{x}_k\mod 5\right)_{j=0}^1=\left(\frac{1}{2}(0+4),\frac{1}{2}(0+2^{-1}\cdot 4)\right)\equiv(2,1).$$
But this is not the same as $x$. What am I doing/thinking wrong? I'm thinking it could be to do with how I'm working with "inverses", e.g. 1/2, in modular arithmetic?
|
I think this is right. It seems the problem was related to the size of the NNT, $n=2$ in my case, and the order of the primitive root of unity. Since $n=2$ then I needed to choose a $2$-th root of unity, which would mean switching to prime $p=3$, but then this would not work because my vector $(1,4)\equiv(1,1)$ so I lose information given my choices.
To remedy this, I can "pad out" the original vector to obtain $(1,4,0,0)$ and use prime $p=5$. Then $2$ is a $4$-th root of unity in $\mathbb{F}_5$ and $x_i<5$ as required. Note - I need a $4$-th root of unity because my vector is of size $n=4$. Then we have
$$\hat{x}=\mathcal{N}(x)=\left(\begin{array}{llll}2^{0\times 0}&2^{0\times 1}&2^{0\times 2}&2^{0\times 3}\\2^{1\times 0}&2^{1\times 1}&2^{1\times 2}&2^{1\times 3}\\2^{2\times 0}&2^{2\times 1}&2^{2\times 2}&2^{2\times 3}\\2^{3\times 0}&2^{3\times 1}&2^{3\times 2}&2^{3\times 3}\end{array}\right)\left(\begin{array}{l}1\\4\\0\\0\end{array}\right)\stackrel{5}{\equiv}\left(\begin{array}{llll}1&1&1&1\\1&2&4&3\\1&4&1&4\\1&3&4&2\end{array}\right)\left(\begin{array}{l}1\\4\\0\\0\end{array}\right)=\left(\begin{array}{l}5\\9\\17\\13\end{array}\right)\stackrel{5}{\equiv} \left(\begin{array}{l}0\\4\\2\\3\end{array}\right).$$
To check this is correct, we apply the inverse NNT,
$$x=\mathcal{N}^{-1}(\hat{x})=4^{-1}\left(\begin{array}{llll}1&1&1&1\\1&2^{-1}&4^{-1}&3^{-1}\\1&4^{-1}&1^{-1}&4^{-1}\\1&3^{-1}&4^{-1}&2^{-1}\end{array}\right)\left(\begin{array}{l}0\\4\\2\\3\end{array}\right)\stackrel{5}{\equiv}4^{-1}\left(\begin{array}{llll}1&1&1&1\\1&3&4&2\\1&4&1&4\\1&2&4&3\end{array}\right)\left(\begin{array}{l}0\\4\\2\\3\end{array}\right)=4^{-1}\left(\begin{array}{l}4\\1\\0\\0\end{array}\right)\stackrel{5}{\equiv} \left(\begin{array}{l}1\\4\\0\\0\end{array}\right).$$
This article explains further how the order of the root of unity is related to the NTT.
|
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|
Smallest no of balls in the box? A box contains white and black balls. When two balls are drawn without replacement suppose the probability that both are white is 1 /3.
a) Find smallest number of balls in the box ?
b) How small can the total number of balls be if black balls are even in number ?
I considered no of white balls as 'n' and no of black balls as 'm'.
So for first two balls to be white it would be
{n/(m+n)}*{(n-1)/(m+n-1)} = 1/3 . But how to solve further ? we have only one equation
Can anyone please explain in detail.
Thanks in advance!
|
$$\frac{n}{m+n}\frac{n-1}{m+n-1} = \frac{1}{3}$$
$$\frac{n(n-1)}{(m+n)(m+n-1)} = \frac{1}{3}$$
$$\frac{n(n-1)}{m^2 + 2nm + n^2 - m - n} = \frac{1}{3}$$
$$\frac{n(n-1)}{(m+n)^2 - (m+n)} = \frac{1}{3}$$
Let us now set $m+n = B$, the number of total balls
$$3(n^2-n) = B^2 - B$$
$$B^2 - B - 3(n^2-n) = 0$$
$$B = \frac{1 \pm \sqrt{1+12(n^2-n)}}{2}$$
$$m = \frac{1 \pm \sqrt{1+12(n^2-n)}}{2}-n$$
Since $m$ must be an integer, $1+12(n^2-n)$ must be some $k^2$ for some whole number $k$. We throw in some values for $n$ until we get an integer. The first one we get is $n=2$, yielding $k=5$.
$$m = \frac{1 \pm \sqrt{25}}{2}-2$$
$m$ must be positive, throwing out one solution of the quadratic
$$\therefore n=2, \,\,m = 1$$
|
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|
Evaluate the limit of ratio of sums of sines (without L'Hopital): $\lim_{x\to0} \frac{\sin x+\sin3x+\sin5x}{\sin2x+\sin4x+\sin6x}$ Limit to evaluate:
$$\lim_{x \rightarrow 0} \cfrac{\sin{(x)}+\sin{(3x)}+\sin{(5x)}}{\sin{(2x)}+\sin{(4x)}+\sin{(6x)}}$$
Proposed solution:
$$
\cfrac{\sin(x)+\sin(3x)+\sin(5x)}{\sin(2x)+\sin(4x)+\sin(6x)}
\Bigg/ \cdot\ \cfrac{1/x}{1/x}\Bigg/=
\frac{\cfrac{\sin(x)}x + \cfrac{\sin(3x)}{3x} \cdot 3 + \cfrac{\sin(5x)}{5x} \cdot 5}
{\cfrac{\sin(2x)}{2x} \cdot 2 + \cfrac{\sin(4x)}{4x} \cdot 4 + \cfrac{\sin(6x)}{6x} \cdot 6}
$$
Using $\lim_{x \rightarrow 0} \frac{\sin x}x=1$, we get
$$\frac{1+1\cdot 3+1\cdot 5}{1\cdot 2+1\cdot 4+1\cdot 6} = \frac 9{12} = \frac 3 4$$
Please tell me if I am correct.
|
Hint:
$${f(x)\over g(x)}={{f(x)\over x}\over{g(x)\over x}}$$
|
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|
Integral involving cube root and seventh root Find the value of $$\int_{0}^{1} (1-x^7)^{\frac{1}{3}}-(1-x^3)^{\frac{1}{7}}\:dx$$
My Approach:
Let $$I_1=\int_{0}^{1} (1-x^7)^{\frac{1}{3}}dx$$ and
$$I_2=\int_{0}^{1} (1-x^3)^{\frac{1}{7}}dx$$
For $I_1$ substitute $x^7=1-t^3$ so $dx=\frac{-3t^2}{7}(1-t^3)^{\frac{-6}{7}}\:dt$ Hence
$$I_1=\int_{1}^{0}\frac{-3t^3}{7}(1-t^3)^{\frac{-6}{7}}\:dt$$ $\implies$
$$I_1=\frac{-3}{7}\int_{0}^{1}(1-t^3-1)(1-t^3)^{\frac{-6}{7}}\:dt$$ $\implies$
$$I_1=\frac{-3I_2}{7}+\frac{3A}{7}$$
where $A=\int_{0}^{1}(1-t^3)^{\frac{-6}{7}}dt$
Similarly using substitution $x^3=1-t^7$ for $I_2$ and proceeding as above we get
$$I_2=\frac{-7I_1}{3}+\frac{7B}{3}$$ where $B=\int_{0}^{1}(1-t^7)^{\frac{-2}{3}}dt$
But we need to find $I_1-I_2$ so got stuck up here
|
$$I_1=\int_{0}^{1} (1-x^7)^{\frac{1}{3}}dx$$
substitute $t = 1-x^7$, $\,\,dt = -7x^6dx$, and by extension $\,\,x = \sqrt[7]{1-t}$
$$=-\frac{1}{7}\int_{0}^{1} \frac{\sqrt[3]{t}}{(1-t)^\frac{6}{7}}dx$$
Now integrate by parts choosing the factors $u = \sqrt[3]{t}$ $\,\, u' = \frac{1}{3t^{2/3}}$$\,\,dv = \frac{1}{(1-t)^{6/7}}\,\,$ and $v = -7(1-t)^{1/7}$
$$ = -\frac{1}{7}(-\sqrt[3]{t}\sqrt[7]{1-t})\bigg|_0^1 -\frac{1}{7}\int_0^1 \frac{-7\sqrt[7]{1-t}}{3t^{2/3}}dt$$
$$ = -\frac{1}{7}\int_0^1 \frac{-7\sqrt[7]{1-t}}{3t^{2/3}}dt$$
Finally, subsitute $t = v^3$, $\,\, dt = 3v^2dv$
$$ = \int_0^1 \sqrt[7]{1-v^3}dv$$
$$ = I_2$$
So what we have is
$$\int_{0}^{1} (I_2 - I_2)$$
$$=\int_{0}^{1} 0$$
$$=0$$
|
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|
How to calculate $\lim \limits_{x \to 0} \frac{x^2 \sin^2x}{x^2-\sin^2x}$ with $\lim \limits_{x \to 0} \frac{\sin x}{x}=1$? How to calculate
$$\lim \limits_{x \to 0} \frac{x^2 \sin^2x}{x^2-\sin^2x}$$
with
$$\lim \limits_{x \to 0} \frac{\sin x}{x}=1?$$
Yes I know the question has been asked, the answer is $3$, L'Hospital or Taylor series works and they are neat. However the question I meet is that I'm required to do it with the given limit. It may be dull algebra work; however I even don't know how to start with. Any idea will be wonderful.
|
We can proceed as follows
\begin{align}
L &= \lim_{x \to 0}\frac{x^{2}\sin^{2}x}{x^{2} - \sin^{2}x}\notag\\
&= \lim_{x \to 0}\frac{x^{4}}{x^{2} - \sin^{2}x}\cdot\frac{\sin^{2}x}{x^{2}}\notag\\
&= \lim_{x \to 0}\frac{x^{4}}{x^{2} - \sin^{2}x}\cdot 1\notag\\
&= \lim_{x \to 0}\frac{x}{x + \sin x}\cdot\frac{x^{3}}{x - \sin x}\notag\\
&= \lim_{x \to 0}\dfrac{1}{1 + \dfrac{\sin x}{x}}\cdot\frac{x^{3}}{x - \sin x}\notag\\
&= \lim_{x \to 0}\frac{1}{1 + 1}\cdot\frac{x^{3}}{x - \sin x}\notag\\
&= \frac{1}{2}\lim_{x \to 0}\frac{x^{3}}{x - \sin x}\notag
\end{align}
Its difficult to do the last limit without Taylor's series or L'Hospital's Rule. The limit is famous and evaluates to $6$ so that the final answer is $3$.
|
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|
Real part of $\sqrt{ai-1}$ Is there a way to find the real and imaginary parts of
$$
z=\sqrt{ai-1},\qquad a>0
$$
where $i=\sqrt{-1}$. Thanks. I do not know what to to do.
Note
$$
i=e^{i\pi/2}=\cos \frac{\pi}{2}+i\sin \frac{\pi}{2}.
$$
|
for $z=x+iy$ you can right its Polar Representation $z=re^{i\theta}=r(\cos\theta+i\sin\theta)=\color{red}{r\cos\theta}+i\color{blue}{r\sin\theta}$(respectively real part and imaginary part red and blue) where $r=\sqrt{x^2+y^2}$ and $\theta=tan^{-1}\frac{y}{x} $ for $x>0$ and $\theta=tan^{-1}\frac{y}{x}+\pi $ for $x<0$.
For $w = \rho e^{i \phi}$ that $w^n = z$, $\rho = r^{1/n}$, $\phi = \frac{\theta}{n} + \frac{2\pi k}{n}$ for $k=0, 1, ..,n-1$.
here for $z=ai-1$, we have $r=\sqrt{x^2+y^2}=\sqrt{(-1)^2+a^2}$ and $\theta=tan^{-1}\frac{a}{-1}+\pi $
thus answers will be $w=({a^2+1})^\frac{1}{4}e^{i(\frac{\theta}{2} + \frac{2\pi k}{2})}$(use method in used in red and blue part you will have)
real part first answer is $({a^2+1})^\frac{1}{4}cos\frac{\alpha}{2}$ and imaginary part is $({a^2+1})^\frac{1}{4}sin\frac{\alpha}{2}$ where $\alpha=tan^{-1}(-a)+\pi$ and...
|
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|
Solving $\lim_{x\to0}\frac{x}{\sqrt{1-3x}-1}$ without L'Hopital Without L'Hopital:
$$\lim_{x\to0}\frac{x}{\sqrt{1-3x}-1}$$
Rationalize:
$$\frac{x}{\sqrt{1-3x}-1}\cdot \frac{\sqrt{1-3x}+1}{\sqrt{1-3x}+1}$$
$$\frac{x\cdot(\sqrt{1-3x}+1)}{(1-3x)-1}$$
This will still yield $\frac{0}{0}$. Maybe I should now try variable substitution to eliminate the root. Let
$$w^2 = 1-3x$$
So
$$w\to 1$$
Therefore
$$\lim_{w\to1}\frac{x\cdot(\sqrt{w^2}+1)}{(1-3x)-1} = \lim_{w\to1}\frac{x\cdot(w+1)}{(1-3x)-1}$$
We find that
$$x = \frac{1-w^2}{3}$$
So
$$\lim_{w\to1}\frac{\left( \frac{1-w^2}{3}\right)\cdot(w+1)}{\left(1-3\left( \frac{1-w^2}{3}\right)\right)-1}$$
$$\lim_{w\to1}\frac{\left( \frac{1-w^2}{3}\right)\cdot(w+1)}{\left(1-3+3w^2\right)-1}$$
This will still evaluate to $\frac{0}{0}$.
I just usted the two methods that I always use to calculate limits. Rationalize/factorize and variable substitution. I'm not supposed to use L'Hopital.
What did I do wrong, and how should I have done it?
|
Let $\sqrt{1-3x}=1+h$. We know that when $x=0$ the LHS $=1 \Rightarrow h=0$
Then $1-3x=1+2h+h^2$
$-3x=2h+h^2$
$x=\frac{2h+h^2}{-3}$
The expression we want is $\frac x{\sqrt{1-3x}-1}=\frac {\frac{2h+h^2}{-3}}{1+h-1}=\frac{2+h}{-3}$.
As $h$ tends to $0$ this tends to $-\frac23$
|
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|
Solving $\lim_{x\to0}\frac{x}{\sqrt{3+x}-\sqrt{3-x}}$ I'm trying to resolve the $$\lim_{x\to0}\frac{x}{\sqrt{3+x}-\sqrt{3-x}}$$
First answer is $\frac{0}{0}$
Applying formula:
$$\lim_{x\to0}\frac{x}{\sqrt{3+x}-\sqrt{3-x}} = \lim_{x\to0}\frac{x(\sqrt{3+x}+\sqrt{3-x})}{(\sqrt{3+x}-\sqrt{3-x})(\sqrt{3+x}+\sqrt{3-x})}$$
And now:
$$\lim_{x\to0}\frac{x(\sqrt{3+x}+\sqrt{3-x})}{3+x-3+x} = \lim_{x\to0}\frac{x(\sqrt{3+x}+\sqrt{3-x})}{2x} = \lim_{x\to0}\frac{2\sqrt{3}}{x} = \infty$$
What I'm doing wrong? I know that answer is $\sqrt{3}$, but where is my mistake?
|
$$\lim_{x\to0}\frac{x (\sqrt{3+x}+\sqrt{3-x})}{3+x-3+x} = \lim_{x\to0}\frac{x(\sqrt{3+x}+\sqrt{3-x})}{2x} = \lim_{x\to0}\frac{2\sqrt{3}\color{red}x}{\color{red}2x} = \sqrt 3$$
Note the changes in red.
You didn't cancel correctly here.
|
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|
Prove that using induction that $\binom22+\dots+\binom n2 = \binom{n+1}2$ so I have this math problem where I have to prove this using induction. $$\begin{pmatrix}2\\2\end{pmatrix}+\begin{pmatrix}3\\2\end{pmatrix}+\begin{pmatrix}4\\2\end{pmatrix}+\cdots+\begin{pmatrix}n\\2\end{pmatrix}=\begin{pmatrix}n+1\\3\end{pmatrix}\text{ for } n\ge2$$
I start by checking the initial case $P(2)=\begin{pmatrix}2\\2\end{pmatrix}=1$ It's good.
I then assume P(k) is true. $P(k)=\begin{pmatrix}2\\2\end{pmatrix}+\begin{pmatrix}3\\2\end{pmatrix}+\begin{pmatrix}4\\2\end{pmatrix}+\cdots+\begin{pmatrix}k\\2\end{pmatrix}=\begin{pmatrix}k+1\\3\end{pmatrix}$
I then try to prove P(k+1) is also true. $P(k+1)=\begin{pmatrix}2\\2\end{pmatrix}+\begin{pmatrix}3\\2\end{pmatrix}+\begin{pmatrix}4\\2\end{pmatrix}+\cdots+\begin{pmatrix}k+1\\2\end{pmatrix}=\begin{pmatrix}k+2\\3\end{pmatrix}$
I know that $\begin{pmatrix}k+1\\3\end{pmatrix}=\begin{pmatrix}k\\2\end{pmatrix}+\begin{pmatrix}k\\3\end{pmatrix}$
Now I'm stuck here... I'm not sure how to finish this proof. Thanks.
|
To prove $P(k+1)$, start with your left hand side, i.\,e.
$$ \def\P#1{\binom 22 + \binom 32 + \cdots + \binom{#1}2}\P{k+1}$$
Write it as
$$ \P k + \binom{k+1}2 \tag+ $$
Now use, that by the induction hypothesis, we have
$$ P(k): \P k = \binom{k+1}3 $$
Hence, we can write $(+)$ as
$$ \P k + \binom{k+1}2 = \binom{k+1}3 + \binom{k+1}2 $$
But the right hand side here equals (as you write - but for $k+1$ instead of $k$)
$$ \binom{k+1}3 + \binom{k+1}2 = \binom{k+2}3 $$
Collecting everything, you are done.
|
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|
Order of operations in polynomial with exponent I have a simple question about whether or not my approach is correct in simplifying a polynomial, here it is,
$(n(n+1)/2)^2 = ((n^2+n)/2)^2 = 1/4(n^4+2n^3+n^2)$
I apologize if you find that hard to read, I can't write the equations the way I would like.
Many thanks.
|
It is correct, but if desired to write with no parenthesis you may opt to distribute the $\frac{1}{4}$ through the expression to arrive at $$\frac{n^4}{4}+\frac{n^3}{2}+\frac{n^2}{4}$$
A small piece of advice for if you are faced with exponents of higher power than $2$.
There is such a thing known as the Binomial Theorem.
It says:
$(x+y)^n = x^n+\binom{n}{1}x^{n-1}y^1+\binom{n}{2}x^{n-2}y^2+\dots+\binom{n}{r}x^{n-r}y^r+\dots+\binom{n}{n-1}x^1y^{n-1}+y^n$
or more concisely as:
$(x+y)^n = \sum\limits_{r=0}^n \binom{n}{r}x^{n-r}y^r$
If you don't know easily what the symbols $\binom{n}{r}$ mean, that's okay for now. You can figure the small ones out by making Pascals triangle:
To create the next row, add the two numbers that appear above it (except for the far left or far right in which case there is only a one above it and empty space). For example, in the fifth line, the $10$ was created by adding the $4$ and the $6$ above it.
These numbers are the coefficients of having expanded $(x+y)^n$
$$\begin{array}{l&c} (x+y)^0=&1\\
(x+y)^1=&1x+1y\\
(x+y)^2=&1x^2+2xy+1y^2\\
(x+y)^3=&1x^3+3x^2y+3xy^2+1y^3\\
(x+y)^4=&1x^4+4x^3y+6x^2y^2+4xy^3+1y^4\\
(x+y)^5=&1x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+1y^5\\
\vdots\end{array}$$
In your case, since the two pieces are $(\color{red}{n^2}+\color{blue}{n})$, you have $(n^2+n)^2 = (\color{red}{n^2})^2 + 2\color{red}{n^2}\color{blue}{n}+\color{blue}{n}^2$
If you were faced with something like $(n^2+n)^4$ it would be
$$(n^2+n)^4=(\color{red}{n^2})^4+4(\color{red}{n^2})^3\color{blue}{n}+6(\color{red}{n^2})^2\color{blue}{n}^2+4\color{red}{n^2}\color{blue}{n}^3+\color{blue}{n}^4\\=n^8+4n^7+6n^6+4n^5+n^4$$
|
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|
Find the solution of $\lfloor{x^2}\rfloor−\lfloor{3x}\rfloor+2=0$ Is anyone able to help me with the following equation concerned the floor function $\lfloor{x^2}\rfloor−\lfloor{3x}\rfloor+2=0$
I don't know how to deal with the floor terms properly.
|
From the definition of the floor function we have
[correction here]
$$\begin{align}
x^2-1 &< &\lfloor x^2 \rfloor &\le x^2 \\
3x-1 &< &\lfloor 3x \rfloor &\le 3x
\end{align}$$
and so
$$\lfloor x^2 \rfloor - \lfloor 3x \rfloor + 2 > x^2-3x-1+2=x^2-3x+1$$
So all solutions to the floor equation must lie between the two roots of $x^2-3x+1=0$. That is
$$\frac{3-\sqrt5}{2}\le x \le \frac{3+\sqrt5}{2}$$
Now trying various cases:
$$\begin{align}
\lfloor 3x \rfloor=2,&\lfloor x^2 \rfloor=0&\implies x\in\left[\frac{2}{3},1\right)\cap[0,1)=\left[\frac{2}{3},1\right) \\[1em]
\lfloor 3x \rfloor=3,&\lfloor x^2 \rfloor=1&\implies x\in\left[1,\frac{4}{3}\right)\cap[1,\sqrt2)=\left[1,\frac{4}{3}\right) \\[1em]
\lfloor 3x \rfloor=4,&\lfloor x^2 \rfloor=2&\implies x\in\left[\frac{4}{3},\frac{5}{3}\right)\cap[\sqrt2,\sqrt3)=\left[\sqrt2,\frac{5}{3}\right) \\[1em]
\lfloor 3x \rfloor=5,&\lfloor x^2 \rfloor=3&\implies x\in\left[\frac{5}{3},2\right)\cap[\sqrt3,2)=[\sqrt3,2) \\[1em]
\lfloor 3x \rfloor=6,&\lfloor x^2 \rfloor=4&\implies x\in\left[2,\frac{7}{3}\right)\cap[2,\sqrt5)=[2,\sqrt5) \\[1em]
\lfloor 3x \rfloor=7,&\lfloor x^2 \rfloor=5&\implies x\in\left[\frac{7}{3},\frac{8}{3}\right)\cap[\sqrt5,\sqrt6)=\left[\frac{7}{3},\sqrt6\right) \\[1em]
\end{align}$$
So the full solution is:
$$x\in\left[\frac{2}{3},\frac{4}{3}\right)\cup\left[\sqrt2,\frac{5}{3}\right)\cup\left[\sqrt3,\sqrt5\right)\cup\left[\frac{7}{3},\sqrt6\right)$$
|
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|
Inequality involving an exponent I wish to prove the following inequality
$$x^{\frac{3}{x-1}} > 1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3}, \quad x > 1.$$
Graphically the above inequality appears to be true since if one plots
$$g(x) = x^{\frac{3}{x-1}} - \left (1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} \right )$$
it appears as though $g(x) > 0$ for all $x > 1$. I have however been unable to prove analytically this is true.
I know
$$1 < 1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} < 4 \quad \mbox{for all} \,\, x> 1$$
and
$$1 < x^{\frac{3}{x - 1}} < \mathrm{e}^3 \quad \mbox{for all} \,\, x > 1,$$
but neither of these bounds seem to help me very much.
Any pointers in the right direction would be greatly appreciated.
|
Hint: Taking logs, and using $u>\ln (1+u), u> 0,$ we see it suffices to show
$$\ln x \ge (x-1)/3\cdot (1/x + 1/x^2 + 1/x^3), \ \ x> 1.$$
Both sides are $0$ at $x=1,$ so it suffices to show the inequality for the derivatives of each side.
|
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|
Are there $a,b,c,d \in \mathbb N$ such that $\frac{a + b}{a + b + c + d} < \frac{a}{a + c} < \frac{b}{b + d}$? Consider the following $2 \times 2$ contingency table:
\begin{array}{c|cc|c}
& C & \overline C & \Sigma \\ \hline
V & 4000 & 3500 & 7500 \\
\overline V & 2000 & 500 & 2500 \\ \hline
\Sigma & 6000 & 4000 & 10000
\end{array}
In data mining, we informally say that the association rule $C \to V$ is misleading. We could justify this by observing that:
$$
\Pr[V \mid C] = \frac{4000}{6000} = 0.\overline{6} < 0.75 = \frac{7500}{10000} = \Pr[V]
$$
Alternatively, we could justify this by observing that:
$$
\Pr[V \mid C] = \frac{4000}{6000} = 0.\overline{6} < 0.875 = \frac{3500}{4000} = \Pr[V \mid \overline C]
$$
I'm trying to figure out which justification is more compelling. When I change the numbers in the contingency table, both inequalities always seem to be simultaneously satisfied. I can't seem to make one inequality true and the other false.
More generally, consider the contingency table:
\begin{array}{c|cc|c}
& C & \overline C & \Sigma \\ \hline
V & a & b & a+b \\
\overline V & c & d & c + d\\ \hline
\Sigma & a + c & b + d & a + b + c + d
\end{array}
Question: Do there exist $a,b,c,d \in \mathbb N$ such that either one of the following two inequalities are satisfied:
\begin{align*}
\frac{a + b}{a + b + c + d} &< \frac{a}{a + c} < \frac{b}{b + d} \tag{1} \\
\frac{b}{b + d} &< \frac{a}{a + c} < \frac{a + b}{a + b + c + d} \tag{2}
\end{align*}
|
None of the two double-inequalities can hold for any $a,b,c,d\in\Bbb R^+$.
$$\frac{a + b}{a + b + c + d} < \frac{a}{a + c} \iff (a+b)(a+c)<a(a+b+c+d)$$
$$\iff \frac{a}{b}>\frac{c}{d}$$
The same way you get:
$$ \frac{a}{a+c} < \frac{b}{b + d} \iff \frac{a}{b}<\frac{c}{d}$$
The two inequalities contradict each other.
Use the same method for the other double-inequality.
|
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|
how can we show $\frac{\pi^2}{8} = 1 + \frac1{3^2} +\frac1{5^2} + \frac1{7^2} + …$? Let $f(x) = \frac4\pi \cdot (\sin x + \frac13 \sin (3x) + \frac15 \sin (5x) + \dots)$. If for $x=\frac\pi2$, we have
$$f(x) = \frac{4}{\pi} ( 1 - \frac13 +\frac15 - \frac17 + \dots) = 1$$
then obviously :
$$ 1 - \frac13 +\frac15 - \frac17 + \dots=\frac{\pi}{4}$$
Now how can we prove that:
$$\frac{\pi^2}{8} = 1 + \frac1{3^2} +\frac1{5^2} + \frac1{7^2} + \dots$$
|
From the Basel Problem, we have
$$\frac{\pi^2}{6} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2}\dots$$
$$\frac{\pi^2}{24} = \frac{\pi^2}{6\cdot 2^2} = \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2}\dots$$
so that
$$\begin{align}\frac{\pi^2}{8} &= \frac{\pi^2}{6} - \frac{\pi^2}{24}\\&=\frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \dots \end{align}$$
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How many different values can $(x^2 + y^2, x^2 + 2y^2 )$ have mod 4? It is known that if a prime number $p = x^2 + y^2 $ is equivalent to $p \equiv 1 \mod 4$. This is Fermat's theorem on the sum of two squares.
My question is about the value of two simultaneous quadratic forms. Consider the values:
$$ (x,y) \in \mathbb{Z}^2 \longrightarrow (x^2 + y^2 , x^2 + 2 \,y^2) $$
Invidually I can find the value of each coordinate modulo 4:
*
*$x^2 = 0, 1 \mod 4$ e.g. $0,1,4,9,16,25,\dots \equiv 0,1,0,1,0,1,\dots$
*$x^2 + y^2 = \{0 \text{ or } 1\} + \{0 \text{ or } 1\} = \{0,1 \text{ or } 2\}$ modulo $4$.
*$x^2 + y^2 = \{0 \text{ or } 1\} + 2\times\{0 \text{ or } 1\} = \{0,1,2 \text{ or } 3\}$ modulo $4$.
Then $x^2 + y^2$ can take any 3 out of the 4 possible values and $x^2 + 2y^2$ can take all possible values, so that $(x^2 + y^2 , x^2 + 2 \,y^2) $ can be one of $3 \times 4 = 12$ of $4^2 = 16$ possible values, but if we plot:
Only (0,0), (1,1), (1,2) and (2,3)... Why can't we get all $12$ possibilites?
Similarly, we don't get all $11 \times 11 = 121$ possibilities mod $11$.
Question - what fraction of the $m^2$ possible values can we obtain mod $m$?
As a function of $m$ the value do not fit into any OEIS sequence.
1, 4, 4, 4, 9,
16, 16, 9, 16, 36,
36, 16, 49, 64, 36,
16, 81, 64, 100, 36,
64, 144, 144, 36, 121,
196, 121, 64, 225
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$x^2$ and $y^2$ modulo 4 can each only take on 2 values. Thus, there can be at most four distinct combinations of those values, no matter how you arrange them.
Modulo 11, $x^2$ and $y^2$ can take on up to 6 distinct values, so you can have up to 36 possible combinations.
Note that if $(a,b)$ are independent variables, then so are $(a+b,b)$, and likewise $(a+b,a+2b)$. Thus, since $x^2$ and $y^2$ are independent, so must also be $x^2+y^2,x^2+2y^2$.
Thus, all the combinations show up for any $m$.
For generic $m$, the number of possibilities is $f(m)^2$, where $f(m)$ is the number of distinct values that $x^2$ can take modulo $m$. For odd primes, $f(m) = (m+1)/2$.
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|
Area under curves functional analysis question Consider the functions defined inplicitly by te equation $y^{3}-3y+x=0$ on various intervals in the real line . If $x\in(-\infty,2)\cup(2,\infty)$ the equation implicitly defines a unique real valued differentiable function $y=f(x)$ . if $x\in (-2,2)$, the equation implicitly defines a unique real valued differentiable function $y=g(x)$ satisfying $g(0)=0$.
Q1) If $f(-10\sqrt{2}) =2\sqrt{2}$ , then $f''(-10\sqrt{2})$ = ?
Q2) $\int^{1}_{-1}g'(x)dx$ =
Q3) The area of the region bounded by the curve $y=f(x)$, the x-axis and the lines $x=a$ and $x=b$ where $-\infty<a<b<-2$, ?
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$(1)\;\;$ Given $$y^3-3y+x=0\;,...........(1)$$ Now Differentiate both side w. r to $x\;,$ We get
$$\displaystyle 3y^2\cdot y'-3y'+1=0....................(2)\Rightarrow y'=\frac{1}{3-3y^2}\;,$$ Where $$\displaystyle y=f(x)\;\;, y'=\frac{dy}{dx}=f'(x)\;\;,y''=f''(x)=\frac{d^2y}{dx^2}$$
Now Given $\displaystyle y=f\left(-10\sqrt{2}\right)=2\sqrt{2}.$
So we get $$\displaystyle y' = \frac{1}{3-3y^2}=\frac{1}{3-3(2\sqrt{2})^2} = -\frac{1}{21}$$
Now Again Differentiate $(2)\;,$ We get
$$\displaystyle 3y^2\cdot y''+y'\cdot 6yy'-3y''=0\Rightarrow 6y(y')^2+3y^2y''-3y''=0$$
Now put $\displaystyle y = 2\sqrt{2}$ and $\displaystyle y'=-\frac{1}{21}\;,$ We get $\displaystyle y''= \frac{2y(y')^2}{1-3y^2}=-\frac{6\cdot 2\sqrt{2}}{(21)^3} = -\frac{4\sqrt{2}}{243\times9}$
So we get $\displaystyle f''\left(-10\sqrt{2}\right) = -\frac{4\sqrt{2}}{243\times 9}$
$(2)\;\;$ We have $$\displaystyle g'(x) = \frac{dy}{dx} = \frac{1}{3[1-(f(x))^2]}$$ (Even function)
So $$\displaystyle \int_{-1}^{1} g'(x) = 2\int_{0}^{1}g'(x)dx = 2\left[g(x)\right]_{0}^{1} = 2[g(1)-g(0)] = 2g(0)\;\;,$$ bcz given $g(0)=0$
$(3)\;\;$ The Required Area $$\displaystyle = \int_{a}^{b}f(x)\cdot 1 dx = \left[x\cdot f(x)\right]_{a}^{b}-\int_{a}^{b}xf'(x)dx$$
$$\displaystyle =bf(b)-af(a)+\int_{a}^{b}\frac{x}{3\left[(f(x))^2-1\right]}dx$$
Bcz from equation no. $(2)\;,$ We have $\displaystyle y'=\frac{dy}{dx}=\frac{1}{3-3y^2}$
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|
Integration of elementary function $\int \frac {\log x}{(1+x)^3}\,{\rm d}x $
The question is to find the integral
$$\int \frac {\log x}{(1+x)^3}\,{\rm d}x $$
It can be easily be solved by integration by parts, but I want to solve it without using integration by parts.
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It can be done by a sort of "undetermined coefficients". Suppose we can guess that the solution is of the form
$$ F(x) = a(x) + b(x) \ln(1+x) + c(x) \ln(x) $$
where $a, b, c$ are rational functions. Taking the derivative and comparing to $\ln(x)/(1+x)^3$, we see
$$ \eqalign{c' &= \dfrac{1}{(1+x)^3}\cr
b' &= 0\cr
a' &= -\dfrac{b}{1+x} - \dfrac{c}{x}\cr} $$
From the first equation,
$$c = c_0 - \dfrac{1}{2(1+x)^2}$$
with $c_0$ constant. From the second, $b$ is constant.
And then, using partial fractions
$$ a' = -\dfrac{b}{1+x} - \dfrac{c}{x} = \dfrac{1-2c_0}{2x} - \dfrac{1}{2(1+x)^2} - \dfrac{1+2b}{2(1+x)}$$
In order for $a$ to be a rational function, the terms in $x^{-1}$ and
$(1+x)^{-1}$ must vanish, so $b=-1/2$ and $c_0 = 1/2$. Then we get
$$ \eqalign{a' &= - \dfrac{1}{2(1+x)^2}\cr a &= \dfrac{1}{2(1+x)} + C\cr}$$
so that
$$ \int \dfrac{\ln x}{(1+x)^3} = \dfrac{1}{2(1+x)} - \dfrac{\ln(1+x)}{2} + \left(\dfrac{1}{2} - \dfrac{1}{2(1+x)^2}\right) \ln(x) + C $$
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Doing $\lim_{x\to0}\frac{\sin x - \tan x}{x^2\cdot\sin 2x}$ without L'Hopital Without L'Hopital,
$$\lim_{x\to0}\frac{\sin x - \tan x}{x^2\cdot\sin 2x}$$
This is
$$\frac{\sin x -\frac{\sin x}{\cos x}}{x^2\cdot\sin 2x} = \frac{\frac{\sin x \cdot \cos x - \sin x}{\cos x}}{x^2\cdot\sin 2x} = \frac{\sin x\cdot \cos x - \sin x}{x^2\cdot\sin 2x\cdot \cos x}$$
Split that:
$$\frac{\sin x\cdot \cos x}{x^2\cdot\sin 2x\cdot \cos x} - \frac{\sin x}{x^2\cdot\sin 2x\cdot \cos x}$$
In the left side, we can cancel the $\cos x$ and also apply $\frac{\sin x}{x} = 1$ once:
$$\frac{1}{x\cdot\sin 2x} - \frac{\sin x}{x^2\cdot\sin 2x\cdot \cos x}$$
That was probably a bad idea, since $x \cdot \sin2x$ will definitely be $0$... But anyway, let's keep going with the right side. There, we can apply the identity $\frac{\sin x}{x} = 1$ again:
$$\frac{1}{x\cdot\sin 2x} - \frac{1}{x\cdot\sin 2x\cdot \cos x}$$
Hey, I could get rid of the $\sin 2x$ on the left side if I multiply and divide by $2x$... the same on the right side:
$$\frac{1}{2x^2} - \frac{1}{2x^2\cdot \cos x}$$
Looking pretty, but sadly that's not going anywhere. What can I do?
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$$\lim_{x\to0}\frac{\sin x - \tan x}{x^2\cdot\sin 2x} =-\lim_{x\to0}\frac{\sin x(1-\cos x)}{2x^2\sin x\cos^2x} =-\lim_{x\to0}\frac{(1-\cos x)}{2x^2}\cdot\frac1{\lim_{x\to0}\cos^2x}$$
Now $$\lim_{x\to0}\frac{(1-\cos x)}{2x^2}=\lim_{x\to0}\frac{(1-\cos x)(1+\cos x)}{2x^2}\cdot\dfrac1{\lim_{x\to0}(1+\cos x)}=?$$
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Evaluate Integral $\int\frac{1+x}{1+x^2}\ dx$ $\int\frac{1+x}{1+x^2}\ dx$
Let $u=1+x^2$
Then $du = 2x\ dx$
Here is my work.
Split integral $\int\frac{1}{1+x^2}\ dx$ + $\int\frac{x}{1+x^2}\ dx$
Integrate first integral term:
$\int\frac{1}{1+x^2}\ dx=tan^{-1}x$
$\int\frac{x}{1+x^2}\cdot\frac{du}{2x} $
I am stuck when it comes to the second integral term (I hope that is the right term). I have $\frac{du}{2x}$ It does get rid of the x. Should I carry the 2 over by placing outside the second integral.
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HINT:
$$\int \frac{1}{u}\,du=\log|u|+C$$
So, let $u=1+x^2$, $du=2x\,dx$ and ...
SPOILER ALERT Scroll over the highlighted area to reveal the answer
$$\int\frac{1+x}{1+x^2}\,dx=\int\frac{1}{1+x^2}\,dx+\int\frac{x}{1+x^2}\,dx=\arctan(x)+\frac12 \log(1+x^2)+C$$where for the second integral, we used the HINT and substituted $u=1+x^2$, $du=2xdx$, so that $$\int\frac{x}{1+x^2}\,dx=\frac12\int\frac{1}{u}\,du=\frac12\log |u|+C=\frac12\log (1+x^2)+C$$
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Expansion of $\frac{1}{\sqrt{1-4x}}$ Expand $\frac{1}{\sqrt{1-4x}}$ in ascending of power of $x$, up to and including the term in $x^2$, simplifying the coefficient. Hence find the coefficient of $x^2$ in the expansion of $\frac{1+2x}{\sqrt{4-16x}}$
My attempt, $(1-4x)^{-\frac{1}{2}}$
$=1+(-\frac{1}{2})(-4x)+\frac{(-\frac{1}{2})(-\frac{3}{2})}{2!}(-4x)^2$
$=1+2x+6x^2$
How to proceed to the next part by using the result of first part?
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$$ [x^2]\frac{1+2x}{\sqrt{4-16 x}} = \frac{1}{2}[x^2]\frac{1}{\sqrt{1-4x}}+[x]\frac{1}{\sqrt{1-4x}}\tag{1} $$
and since:
$$ \frac{1}{\sqrt{1-4x}} = 1+2x+6x^2+O(x^3)\tag{2} $$
we have:
$$ [x^2]\frac{1+2x}{\sqrt{4-16 x}} = \frac{1}{2}\cdot 6+2=\color{red}{5}.\tag{3}$$
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The product of $2×653×733×977$ has each digit exactly once except for one, which one is it? The product of $2×653×733×977$ is a number with nine digits and it contains every digit once except for one, which digit is that?
I noticed that it is a product of primes, but so far I cannot solve it without multiplying it out. I suspect I can solve this using modular arithmetic but I have no idea how. Any help?
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In mod 9 we have
2*653*733*977 = 2*5*4*5 = 2 mod 9
Now we make the following table :
0+1+2+3+4+5+6+7+8+() = 36 = 0 mod 9
0+1+2+3+4+5+6+7+()+9 = 37 = 1 mod 9
0+1+2+3+4+5+6+()+8+9 = 38 = ...
0+1+2+3+4+5+()+7+8+9 = 39 = ...
0+1+2+3+4+()+6+7+8+9 = 40 = ...
0+1+2+3+()+5+6+7+8+9 = 41 = ...
0+1+2+()+4+5+6+7+8+9 = 42 = ...
0+1+()+3+4+5+6+7+8+9 = 43 = ...
0+()+2+3+4+5+6+7+8+9 = 44 = ...
()+1+2+3+4+5+6+7+8+9 = 45 = ...
After that, the missing number is .....
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Evaluation of $ I = \int_{0}^{1}\frac{2x-(1+x^2)^2\cdot \cot^{-1}(x)}{(1+x^2)[1-(1+x^2)\cot^{-1}(x)]}dx$
If $\displaystyle I = \int_{0}^{1}\frac{2x-(1+x^2)^2\cdot \cot^{-1}(x)}{(1+x^2)[1-(1+x^2)\cot^{-1}(x)]}dx\;,$ Then value of $100(I-\ln 2) =$
$\bf{My\; Try::}$ Let $\cot^{-1}(x)=t\;,$ Then $\displaystyle \frac{1}{1+x^2}dx = -dt$ and changing limit, We get
$$\displaystyle I = -\int_{\frac{\pi}{2}}^{\frac{\pi}{4}}\frac{2\cot t-\csc^4 t\cdot t}{1-\csc^2 t\cdot t}dt = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\left(\frac{2\cot t-\csc^4 t\cdot t}{1-\csc^2 t\cdot t}\right)dt$$
Now How can I solve after that, Help Required
Thanks
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HINT:
$$I=\int\frac{2x-(1+x^2)^2\cdot \cot^{-1}(x)}{(1+x^2)[1-(1+x^2)\cot^{-1}(x)]}dx =\int\dfrac{\dfrac{2x}{(1+x^2)^2}-\cot^{-1}x}{\dfrac1{1+x^2}-\cot^{-1}(x)}dx$$
As $\dfrac{d\left(\dfrac1{1+x^2}-\cot^{-1}(x)\right)}{dx}=-\dfrac{2x}{(1+x^2)^2}+\dfrac1{1+x^2},$ write
$$I=\int\dfrac{\dfrac{2x}{(1+x^2)^2}-\dfrac1{1+x^2}}{\dfrac1{1+x^2}-\cot^{-1}(x)}dx+\int dx$$
Set $\dfrac1{1+x^2}-\cot^{-1}(x)=u$
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$\binom{n}{0}-\frac{1}{(2+\sqrt3)^2}\binom{n}{2}+\frac{1}{(2+\sqrt3)^4}\binom{n}{4}-\frac{1}{(2+\sqrt3)^6}\binom{n}{6}+.....=$ If $n=12m$, where $m\in\Bbb{N}$, prove that
$$\binom{n}{0}-\frac{1}{(2+\sqrt3)^2}\binom{n}{2}+\frac{1}{(2+\sqrt3)^4}\binom{n}{4}-\frac{1}{(2+\sqrt3)^6}\binom{n}{6}+.....=(-1)^m\left(\frac{2\sqrt2}{1+\sqrt3}\right)^n$$
I tried to solve it but stuck after few steps:
$\binom{n}{0}-\frac{1}{(2+\sqrt3)^2}\binom{n}{2}+\frac{1}{(2+\sqrt3)^4}\binom{n}{4}-\frac{1}{(2+\sqrt3)^6}\binom{n}{6}+.....$
$=1-\frac{1}{(2+\sqrt3)^2}\frac{n(n-1)}{2}+\frac{1}{(2+\sqrt3)^4}\frac{n(n-1)(n-2)(n-3)(n-4)(n-5)}{6!}+....$
How should I prove this question?
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Using $\displaystyle (1+x)^n = \binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+\binom{n}{3}x^3+............$
Now Put $\displaystyle x=\frac{i}{2+\sqrt{3}}$ and $\displaystyle x=-\frac{i}{2+\sqrt{3}}$ Respectively, We get
$\displaystyle \left(1+\frac{i}{2+\sqrt{3}}\right)^{n} = \binom{n}{0}+\binom{n}{1}\cdot \frac{i}{2+\sqrt{3}}+\binom{n}{2}\cdot \frac{i^2}{(2+\sqrt{3})^2}+............(1)$
$\displaystyle \left(1-\frac{i}{2+\sqrt{3}}\right)^{n} = \binom{n}{0}-\binom{n}{1}\cdot \frac{i}{2+\sqrt{3}}+\binom{n}{2}\cdot \frac{i^2}{(2+\sqrt{3})^2}+...........(2)$
Now Add these two equations, We get
$\displaystyle \frac{[1+i(2-\sqrt{3})]^n+[1-i(2-\sqrt{3})]^n}{2} =\binom{n}{0}-\binom{n}{2}\cdot \frac{1}{(2+\sqrt{3})^2}+\binom{n}{4}\cdot \frac{1}{(2+\sqrt{3})^4}$
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|
binomials product alternating sum calculation I need to somehow prove that $\sum\limits_{k = 0}^{n - 1} {n \choose k} {3 n - k - 1 \choose 2 n - k}(-1)^k = (-1)^{n + 1} {2 n - 1 \choose n}$.
I didn't manage to do it using induction or any combinatorial ideas. Could someone help me?
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Suppose we seek to verify that
$$\sum_{k=0}^{n-1}
{n\choose k} {3n-k-1\choose 2n-k} (-1)^k
= (-1)^{n+1} {2n-1\choose n}$$
which is the same as
$$\sum_{k=0}^{n}
{n\choose k} {3n-k-1\choose 2n-k} (-1)^k
= \left((-1)^n+(-1)^{n+1}\right) {2n-1\choose n} = 0.$$
Introduce
$${3n-k-1\choose 2n-k} = {3n-k-1\choose n-1} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n}}
(1+z)^{3n-k-1} \; dz.$$
This yields for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n}}
(1+z)^{3n-1}
\sum_{k=0}^n {n\choose k} (-1)^k \frac{1}{(1+z)^k} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n}}
(1+z)^{3n-1} \left(1-\frac{1}{1+z}\right)^n \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n}}
(1+z)^{3n-1} \frac{z^n}{(1+z)^{n}} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
(1+z)^{2n-1} \; dz = 0.$$
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If there exist real numbers $a,b,c,d$ for which $f(a),f(b),f(c),f(d)$ form a square on the complex plane.Find the area of the square. For all real numbers $x$,let the mapping $f(x)=\frac{1}{x-i},\text{where} i=\sqrt{-1}$.If there exist real numbers $a,b,c,d$ for which $f(a),f(b),f(c),f(d)$ form a square on the complex plane.Find the area of the square.
$f(a)=\frac{1}{a-i}=\frac{a+i}{a^2+1}=(\frac{a}{a^2+1},\frac{1}{a^2+1})$
Similarly,$f(b)=\frac{1}{b-i}=\frac{b+i}{b^2+1}=(\frac{b}{b^2+1},\frac{1}{b^2+1})$
Similarly,$f(c)=\frac{1}{c-i}=\frac{c+i}{c^2+1}=(\frac{c}{c^2+1},\frac{1}{c^2+1})$
Similarly,$f(d)=\frac{1}{d-i}=\frac{d+i}{d^2+1}=(\frac{d}{d^2+1},\frac{1}{d^2+1})$
Now the area of the square$=(\frac{a}{a^2+1}-\frac{b}{b^2+1})^2+(\frac{1}{a^2+1}-\frac{1}{b^2+1})^2$
But this expression simplifies to $\frac{(a-b)^2}{(1+a^2)(1+b^2)}$.
How should i prove the area of the square to be $\frac{1}{2}$.Is my approach not correct?Please help me.Thanks.
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$$\displaystyle f(x)=\frac { 1 }{ x-i }=\frac { x+i }{ x^{ 2 }+1 } $$
Now in Argand plane $f(a),f(b),f(c),f(d)$ all lie on a curve whose parametric coordinates
is given by $$\displaystyle x=\frac { t }{ t^{ 2 }+1 } ,y=\frac { 1 }{ t^{ 2 }+1 }$$
Now eliminating variable $t\;,$ We get $$x^{ 2 }+y^{ 2 }=y\Rightarrow x^2+y^2-y=0$$
Or we can write it as $$(x-0)^2+\left(y-\frac{1}{2}\right)^2=\frac{1}{4}$$
So, they lie on a circle of radius $\displaystyle \frac { 1 }{ 2 }.$ So the area of the square is $\displaystyle \frac { 1 }{ 2 } $
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How to compute the coefficients of this generating function Working on some combinatorial problem, I arrived at the following generating function
$$K_m(x) = \sum_{n\geq 0}K_{mn}x^n =\frac{x}{1-\sqrt{1+x^2}\cdot\frac{\displaystyle{y_+(x)^{m+1}+y_-(x)^{m+1}}}{\displaystyle{y_+(x)^{m+1}-y_-(x)^{m+1}}}}$$
with
$$y_\pm(x) =x\pm\sqrt{1+x^2}.$$
I aim to compute the coefficients $K_{mn}$ in a closed form. I solved many problems with generating functions, but this one I tried for days, and I'm not sure if it is impossible at all, or if I lack an important skill. For comparison I give the first coefficients, which I computed by hand
$$K_1(x) = -2x^2-4x^3-4x^4+8x^6+16x^7+O(x^8) \\
K_2(x) = x+3x^2+9x^3+19x^4+33x^5+59x^6+121x^7+O(x^8)\\
K_3(x) = -4x^2-16x^3-40x^4-64x^5-32x^6+192x^7+O(x^8)\\
K_4(x) = x+5x^2+25x^3+85x^4+225x^5+541x^6+1385x^7+O(x^8)$$
also the dependence of the first coefficients on $m$ for the first orders $n$ is given here for odd $m$
$$K_m(x) = (-m-1)x^2-(m+1)^2x^3-\frac{2}{3}m(m+2)(m+1)x^4-\frac{1}{3}(m+1)^2(m-1)(m+3)x^5-\frac{1}{15}(2m^4+8m^3-13m^2-42m-15)(m+1)x^6-\frac{2}{45}m(m+2)(m^2+2m-33)(m+1)^2x^7+O(x^8)$$
and here for even $m$
$$K_m(x) = x+(m+1)x^2+(m+1)^2x^3+\frac{1}{3}(m+1)(2m^2+4m+3)x^4+\frac{1}{3}(3+m^2+2m)(m+1)^2x^5+\frac{1}{15}(m+1)(2m^4+8m^3+27m^2+38m+15)x^6+\frac{1}{45}(2m^4+8m^3+62m^2+108m+45)(m+1)^2x^7+O(x^8)$$
I decided to not post the underlying combinatorial problem, as the point of my question is really to see, if a generating function approach is possible here.
One idea, which I did not finish, however, would be to use the substitution $$x=i\sin(u)$$ which transforms the generator into
$$K_m(u) = \frac{\sin(u)}{\cos(u)\tan((m+1)u)-i}$$
which looks much simpler. Possibly, one could compute the coefficients $$K_m(u)=\sum_{n\geq0}R_{mn}u^n$$ and then transform the $R_{mn}$ into the $K_{mn}$ somehow, but I'm not sure if any of these two steps is possible, and even simpler than directly computing $K_{mn}$.
Any suggestions are highly appreciated.
|
In terms of Lucas sequences with the parameters $(P,Q)=(2x,-1)$, we have
$$y_+(x)^{m+1} + y_-(x)^{m+1} = V_{m+1},$$
$$\frac{y_+(x)^{m+1} - y_-(x)^{m+1}}{\sqrt{1+x^2}} = 2U_{m+1}.$$
It follows that
$$K_m(x) = \frac{x}{1-\frac{V_{m+1}}{2U_{m+1}}} = \frac{xU_{m+1}}{U_{m+1}-\frac12V_{m+1}}.$$
Since the denominator value at $x=0$ is (-1)^m, we have the following expansion:
$$K_m(x) = (-1)^m xU_{m+1} \sum_{k=0}^\infty (1+(-1)^{m+1}U_{m+1}+(-1)^m\frac12V_{m+1})^k.$$
Correspondingly, the coefficient of $x^n$ in $K_m(x)$ (i.e. $K_{m,n}$) equals the coefficient of $x^{n-1}$ in
$$(-1)^m\sum_{i+j\leq n-1} \binom{i+j}{i} \binom{n}{i+j+1} \frac{(-1)^{(m+1)i+mj}}{2^j} U_{m+1}^{i+1} V_{m+1}^j.$$
To get an explicit formula for $K_{m,n}$, we will need the power expansions:
$$V_{m+1}^{k} = \frac12 \sum_{t=0}^{k} \binom{k}t (-1)^{(m+1)t} V_{(m+1)(k-2t)}$$
and
$$U_{m+1}^{k} = \frac1{2(4+4x^2)^{\lfloor k/2\rfloor}} \sum_{t=0}^{k} \binom{k}t (-1)^{mt}
\begin{cases}
V_{(m+1)(k-2t)} & \text{if $k$ is even},\\
U_{(m+1)(k-2t)} & \text{if $k$ is odd},
\end{cases}
$$
and the product formulae: $V_kV_\ell = V_{k+\ell} + (-1)^\ell V_{k-\ell}$ and $V_kU_\ell = U_{k+\ell} + (-1)^\ell U_{k-\ell}$.
Putting all together and letting $u_{k,n}$ and $v_{k,n}$ denote the coefficient of $x^n$ in $U_k$ and $V_k$ respectively, we obtain the following explicit formula:
$$K_{m,n} = (-1)^m\sum_{i+j\leq n-1} \binom{i+j}{i} \binom{n}{i+j+1} \frac{(-1)^{(m+1)i+mj}}{2^{j+2}} \sum_{t=0}^j \binom{j}t (-1)^{(m+1)t}$$
$$\times\frac{1}{4^{\lfloor (i+1)/2\rfloor}} \sum_{\ell=0}^{\lfloor (n-1)/2\rfloor}\binom{-\lfloor (i+1)/2\rfloor}{\ell} \sum_{z=0}^{i+1}\binom{i+1}z (-1)^{mz}$$
$$\times\begin{cases}
v_{(m+1)(i+j+1-2t-2z,n-1-2\ell} + (-1)^{(m+1)j} v_{(m+1)(i+1-j+2t-2z),n-1-2\ell} & \text{if $i$ is odd},\\
u_{(m+1)(i+j+1-2t-2z,n-1-2\ell} + (-1)^{(m+1)j} u_{(m+1)(i+1-j+2t-2z),n-1-2\ell} & \text{if $i$ is even}.
\end{cases}
$$
P.S. It is worth to notice that our Lucas sequences can be also expressed in terms of Chebyshev polynomials as
$U_k = (-I)^{k-1} {\mathcal U}_{k-1}(Ix)$ and $V_k = 2(-I)^k {\mathcal T}_k(Ix)$, where $I$ is the imaginary unit, which give handy formulas for $u_{k,n}$ and $v_{k,n}$.
Here is a Sage code that implements the above explicit formula for $K_{m,n}$.
|
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|
$2=1$ Paradoxes repository I really like to use paradoxes in my math classes, in order to awaken the interest of my students. Concretely, these last weeks I am proposing paradoxes that achieve the conclusion that 2=1. After one week, I explain the solution in the blackboard and I propose a new one. For example, I posted the following one some months ago: What is wrong with the sum of these two series?
I would like to increase my repertoire of fake-proofs. I would be glad to read your proposals and discuss them! My students are 18 years old, so don't be too cruel :) Here is my own contribution:
\begin{equation}
y(x) = \tan x
\end{equation}
\begin{equation}
y^{\prime} = \frac{1}{\cos^{2} x}
\end{equation}
\begin{equation}
y^{\prime \prime} = \frac{2 \sin x}{\cos^{3} x}
\end{equation}
This can be rewritten as:
\begin{equation}
y^{\prime \prime} = \frac{2 \sin x}{\cos^{3} x} = \frac{2 \sin x}{\cos x \cdot \cos^{2} x} = 2 \tan x \cdot \frac{1}{\cos^{2} x} = 2yy^{\prime} = \left( y^{2} \right)^{\prime}
\end{equation}
Integrating both sides of the equation $y^{\prime \prime} = \left( y^{2} \right)^{\prime}$:
\begin{equation}
y^{\prime} = y^{2}
\end{equation}
And therefore
\begin{equation}
\frac{1}{\cos^{2} x} = \tan^{2} x
\end{equation}
Now, evalueting this equation at $x = \pi / 4$
\begin{equation}
\frac{1}{(\sqrt{2}/2)^{2}} = 1^{2}
\end{equation}
\begin{equation}
2 = 1
\end{equation}
|
Here is one of my favorites.
Proof that $1=0$
Let's consider for real $x$ the function $f(x)=xe^{-x^2}$. Note, the following integral representation of $f$ is valid (substitute: $u=x^2/y$).
\begin{align*}
\int_{0}^{1}\frac{x^3}{y^2}e^{-x^2/y}\,dy
=\left[xe^{-x^2/y}\right]_0^1
=xe^{-x^2}
\end{align*}
We obtain for all $x$ the following relationship
\begin{align*}
e^{-x^2}(1-2x^2)&=\frac{d}{dx}\left(xe^{-x^2}\right)\\
&=\frac{d}{dx}\int_0^1\frac{x^3}{y^2}e^{-x^2/y}\,dy\\
&=\int_0^1\frac{\partial}{\partial x}\left(\frac{x^3}{y^2}e^{-x^2/y}\right)\,dy\\
&=\int_0^1e^{-x^2/y}\left(\frac{3x^2}{y^2}-\frac{2x^4}{y^3}\right)\,dy
\end{align*}
and observe by setting $x=0$ the left-hand side is one while the right-hand side is zero.
\begin{align*}
\text{LHS: }\qquad e^0(1-0)&=1\\
\text{RHS: }\qquad \int_0^1 0\,dy&=0
\end{align*}
Note: This example can be found in Counterexamples in Analysis by B.R. Gelbaum and J.H.M. Holmsted.
|
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|
Prove inequality $\sqrt{\frac{2a}{b+a}} + \sqrt{\frac{2b}{c+b}} + \sqrt{\frac{2c}{a+c}} \leq 3$ How to prove the following inequality :
$$
\sqrt{\frac{2a}{b+a}} + \sqrt{\frac{2b}{c+b}} + \sqrt{\frac{2c}{a+c}} \leq 3
$$
with $a>0,\ b>0$ and $c>0$.
|
since $$9(a+b)(b+c)(a+c)\ge 8(ab+bc+ac)(a+b+c)$$
By Cauchy-Schwarz inequality we have
$$\sum_{cyc}\sqrt{\dfrac{2a}{a+b}}\le\sqrt{\left[\sum(c+a)\right]\left[\sum_{cyc}\dfrac{2a}{(a+b)(c+a)}\right]}
=\sqrt{\dfrac{8(a+b+c)(ab+bc+ac)}{(a+b)(b+c)(a+c)}}\le 3$$
|
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|
Minimum number of fractions to be summed up to $\frac45$ What is the minimum number of fractions having numerator 1 and a natural number as denominator to be summed up to $\frac 45$?
I have tested with 2 fractions: $\frac1a + \frac1b = \frac45$ and get into the diophantine equation: $5(a+b)=4ab$ and it seems this should have some solutions but can't find one!!
|
Considering prime factorization gives that $5 \mid ab$ and, by relabeling if necessary, we can assume that $5 \mid a$, and in particular $\frac{1}{a} \leq \frac{1}{5}$. Substituting in the two-fraction equation gives $b \leq \frac{5}{3}$, so we must have $b = 1$, but this gives $\frac{1}{a} + 1 = \frac{4}{5}$, and the solution $a$ to this equation is negative, so there is no decomposition of $\frac{4}{5}$ as a sum of two such fractions.
On the other hand, if we permit ourselves three such fractions, we can see immediately that one of the fractions must be $\frac{1}{2}$ or $\frac{1}{3}$. (If not, each of the three fractions would be $\leq \frac{1}{4}$ and so would have sum $\leq 3 \cdot \frac{1}{4} < \frac{4}{5}$.) So, this reduces the problem to finding decompositions of either $\frac{3}{10}$ or $\frac{7}{15}$ into two such fractions. There turn out to be two solutions. (Incidentally, fractions of the form $\frac{1}{n}$, where $n$ is a positive integer, are often called Egyptian fractions.)
|
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|
Does $c\cdot\sum\limits_{n=k}^{\infty}a_{n}= \sum\limits_{n=k}^{\infty}c\cdot a_{n}$ provided that the series converge? I am struggling to find what is wrong about this reasoning when calculating a series that does not start at $n=0$.
For instance, let $S = \sum\limits_{n=2}^{\infty}\left(\frac{1}{2}\right)^n$.
Then $\left(\frac{1}{2}\right)^2S=\left(\frac{1}{2}\right)^2\sum\limits_{n=2}^{\infty}\left(\frac{1}{2}\right)^n$ (I)
And:
$$\left(\frac{1}{2}\right)^2S=\left(\frac{1}{2}\right)^2\sum\limits_{n=2}^{\infty}\left(\frac{1}{2}\right)^n=\sum\limits_{n=2}^{\infty}\left(\left(\frac{1}{2}\right)^n\cdot\left(\frac{1}{2}\right)^2\right)=\sum\limits_{k=0}^{\infty}\left(\frac{1}{2}\right)^k=2$$
But that means (from I):
$$\left(\frac{1}{2}\right)^2S = 2 \Rightarrow \dfrac{\frac{1}{4}}{\frac{1}{4}}S=\dfrac{2}{\frac{1}{4}}\Rightarrow S = 8$$
Which is clearly wrong since:
$$\sum\limits_{n=2}^{\infty}\left(\frac{1}{2}\right)^n=\sum\limits_{n=0}^{\infty}\left(\frac{1}{2}\right)^n-\sum\limits_{n=0}^{1}\left(\frac{1}{2}\right)^n=\sum\limits_{n=0}^{\infty}\left(\frac{1}{2}\right)^n - 1 -\frac{1}{2}=\frac{1}{2}$$
I suspect that there is something behind this step $\left(\frac{1}{2}\right)^2\sum\limits_{n=2}^{\infty}\left(\frac{1}{2}\right)^n=\sum\limits_{n=2}^{\infty}\left(\left(\frac{1}{2}\right)^n\cdot\left(\frac{1}{2}\right)^2\right)$ that I am missing.
So my question is:
Considering that $\sum\limits_{n=k}^{\infty}a_{n}$ is well defined (converges?), does:
$$c\cdot\sum\limits_{n=k}^{\infty}a_{n}= \sum\limits_{n=k}^{\infty}c\cdot a_{n}$$
And if it does, why does the reasoning presented above is wrong?
|
You reindexed the sum incorrectly when you changed from $n$ to $k$ as dummy variable.
Just look at the first terms: when $n=2$, the term is $(\frac12)^2\cdot(\frac12)^2=\frac1{16}$; but when $k=0$, your first term is $(\frac12)^0=1$. Your first index for $k$ should be $4$, not $0$.
|
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|
Prove by induction: $\sum\limits_{k=1}^{n}\frac{1}{2^k}\tan\frac{x}{2^k}=\frac{1}{2^n}\cot\frac{x}{2^n}-\cot x,x\neq k\pi,k\in \mathbb{Z}$ $\sum\limits_{k=1}^{n}\frac{1}{2^k}\tan\frac{x}{2^k}=\frac{1}{2^n}\cot\frac{x}{2^n}-\cot x,x\neq k\pi,k\in \mathbb{Z}$
Base Case: For $n=1$, $\frac{1}{2}\tan\frac{x}{2}=\frac{1}{2}\cot\frac{x}{2}-\cot x$
Proof:
$$\frac{1}{2}\cot\frac{x}{2}-\cot x=\frac{1}{2\tan\frac{x}{2}}-\frac{1}{\frac{2\tan\frac{x}{2}}{1-\tan^2\frac{x}{2}}}=\frac{1-1+\tan^2\frac{x}{2}}{2\tan\frac{x}{2}}=\frac{1}{2}\tan\frac{x}{2}$$
Induction Hypothesis:
Assume true that for $n = m$,
$$\sum\limits_{k=1}^{m}\frac{1}{2^k}\tan\frac{x}{2^k}=\frac{1}{2^m}\cot\frac{x}{2^m}-\cot x$$
Inductive Step: For $n=m+1$
$\sum\limits_{k=1}^{m+1}\frac{1}{2^{k}}\tan\frac{x}{2^{k}}=\frac{1}{2^{m+1}}\cot\frac{x}{2^{m+1}}-\cot x$
Proof:
$$\frac{1}{2^m}\cot\frac{x}{2^m}-\cot x+\frac{1}{2^{m+1}}\tan\frac{x}{2^{m+1}}=\frac{1}{2^m}\cot\frac{x}{2^m}-\cot x+\frac{1}{2^{m+1}\cot\frac{x}{2^{m+1}}}$$
I don't know how to transform $\frac{1}{2^m}\cot\frac{x}{2^m}-\cot x+\frac{1}{2^{m+1}\cot\frac{x}{2^{m+1}}}$ into $\frac{1}{2^{m+1}}\cot\frac{x}{2^{m+1}}-\cot x$
|
Proving that $$\frac{1}{2^k}\cot\frac{x}{2^k}-\cot x+\frac{1}{2^{k+1}}\tan\frac{x}{2^{k+1}}=\frac{1}{2^{k+1}}\cot\frac{x}{2^{k+1}}-\cot x$$ is equivalent to prove that $$\frac{1}{2^k}\cot\frac{x}{2^k}+\frac{1}{2^{k+1}}\tan\frac{x}{2^{k+1}}=\frac{1}{2^{k+1}}\cot\frac{x}{2^{k+1}}$$ by omitting $-\cot x$. Then if you multiply both sides by $2^{k+1}$ we get $$2\cot\frac{x}{2^k}+\tan\frac{x}{2^{k+1}}=\cot\frac{x}{2^{k+1}}$$
You proved earlier (case n=1) that $$\tan\frac{y}{2}=\cot\frac{y}{2}-2\cot y$$ Thus if you take $y=\frac{x}{2^k}$ you get $$\tan\frac{x}{2^{k+1}}=\cot\frac{x}{2^{k+1}}-2\cot\frac{x}{2^k}$$
Remark: $k$ is already used for indexation in $\sum$ so you should take a different variable like $i$ or even simply work with $n$ and then you take $n+1$ in the inductive step.
|
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|
Find the last three digits of $17^{102}$.
Find the last three digits of $17^{102}$.
That is we have to find out $17^{102}\equiv ?\pmod{1000}$.
Now if we solve it through general congruence relation , then the process becomes very difficult , computational and laborious. Again from Euler's theorem , $$17^{\phi(1000)}\equiv 1 \pmod{1000}.$$ But $\phi(1000)=400(>102)$. So it can't help me.
Does there any simplest way to solve this congruence relation easily ?
|
The units digit of $17^1$ is $7$.
The units digit of $17^2$ is $9$ ($7 \times 7 = 49$).
The units digit of $17^3$ is $3$ ($7 \times 9 = 63$).
The units digit of $17^4$ is $1$ ($7 \times 3 = 21$).
Calculating,
$\; 17^4 = 289 \cdot 289 = (280+9)^2 = 280^2 + 18 \cdot 280 + 81 \equiv$
$\quad\quad\quad 400 + 800 + 600 + 640 + 81 \equiv 521 \pmod{1000}$
Calculating,
$\; 17^{102} \equiv 17^2 \cdot 521^{25} \equiv 289 \cdot (1 + 520)^{25} \equiv 289 \cdot (1 + 2^3 \cdot 5^1 \cdot 13^1)^{25} \equiv$
$\quad\quad\quad\; \, \; 289 \; \bigr(1 + \binom{25}{1}\,2^3 \cdot 5^1 \cdot 13^1 + \binom{25}{2}\,2^6 \cdot 5^2 \cdot 13^2 + 0\bigr) \equiv$
$\quad\quad\quad\; \, \; 289 \; \bigr(1 + 0 + 0 + 0 \bigr) \equiv 289 \pmod{1000}$
|
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|
Closed form for ${\large\int}_0^1x\,\operatorname{li}\!\left(\frac1x\right)\ln^{1/4}\!\left(\frac1x\right)dx$ Let $\operatorname{li}(x)$ denote the logarithmic integral:
$$\operatorname{li}(x)=\int_0^x\frac{dt}{\ln t}.$$
How can we prove the following conjectured closed form?
$${\large\int}_0^1x\,\operatorname{li}\!\left(\frac1x\right)\ln^{1/4}\!\left(\frac1x\right)dx\stackrel{\color{gray}?}=\left(\frac12-\frac{\operatorname{arctan}\left(\sqrt[4]2\right)+\operatorname{arcoth}\left(\sqrt[4]2\right)}{4\sqrt[4]2}\right)\cdot\Gamma\!\left(\frac14\right)$$
Related questions: [1][2].
|
Let $x = e^{-y}$ and denote the value of the integral by $I$. Then we have
\begin{equation}
I = \int_0^\infty y^{1/4} e^{-2y} \text{ li}(e^y) \, dy.
\end{equation}
Consider the parameter
\begin{equation}
I(b) = \int_0^\infty y^{1/4} e^{-2y} \text{ li}(e^{by}) \, dy.
\end{equation}
Differentiating we obtain
\begin{equation}
I'(b) = \frac{1}{b} \int_0^\infty y^{1/4} e^{-(2-b)y} \, dy.
\end{equation}
Letting $u = (2-b)y$, we obtain
\begin{equation}
I'(b) = \frac{\Gamma\left(\frac{1}{4} \right)}{4 b(2-b)^{5/4}},
\end{equation}
where we require $b < 2$ for convergence of the integral. Using mathematica to evaluate the integral we obtain
\begin{equation}
I(b) = \frac{\Gamma\left(\frac{1}{4} \right)}{4} \left \{\frac{2}{(2-b)^{1/4}} + \frac{\arctan \left[ \left(1 - \frac{b}{2} \right) \right]^{1/4} - \text{arctanh} \left[ \left(1 - \frac{b}{2} \right) \right]^{1/4}}{2^{1/4}} \right \} + C.
\end{equation}
The constant of integration $C$ can be determined by letting $b \to -\infty$ which gives
\begin{equation}
C = - \frac{\pi \Gamma \left(\frac{1}{4}\right) e^{i \pi/4}}{4 \cdot 2^{3/4}}.
\end{equation}
Finally, if we note $I = I(1)$ we obtain
\begin{equation}
I(1) = \Gamma\left(\frac{1}{4} \right) \left [ \frac{1}{2} + \frac{\arctan \left( 2^{-1/4} \right) - \text{arctanh} \left( 2^{-1/4} \right)}{4 \cdot 2^{1/4}} - \frac{\pi e^{i \pi/4}}{4 \cdot 2^{3/4}} \right ].
\end{equation}
Taking the real part results in the correct numerical value when plugged into mathematica (as does your result).
|
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|
(-8)^(4/3) is equals with 16 or (-16)*(-1)^(1/3)? 1. $(-8)^{4/3}=\bigl((-8){^4\bigr)^{1/3}}=4096^{1/3}=16$.
2.
$$
\begin{align*}
(-8)^{4/3} &= (-8)^{1+1/3} \\
&= -8\times(-8)^{1/3} \\
&= -8\times (-1)^{1/3}\times 8^{1/3} \\
&= -2\times 8\times (-1)^{1/3} \\
&= -16\times (-1)^{1/3}.
\end{align*}
$$
So, which is the correct?
|
Using the following rules, with $a,b \in \mathbb{R}$:
*
*$\left|a^b\right|=\left|a\right|^b$;
*$\arg\left(a^b\right)=\tan^{-1}\left(\cos(b\cdot\arg(a)),\sin(b\cdot\arg(a))\right)$
$$(-8)^{\frac{4}{3}}=$$
$$\left|(-8)^{\frac{4}{3}}\right|e^{\arg\left((-8)^{\frac{4}{3}}\right)i}=$$
$$\left|-8\right|^{\frac{4}{3}}e^{\arg\left((-8)^{\frac{4}{3}}\right)i}=$$
$$8^{\frac{4}{3}}e^{\arg\left((-8)^{\frac{4}{3}}\right)i}=$$
$$16e^{\arg\left((-8)^{\frac{4}{3}}\right)i}=$$
$$16e^{\tan^{-1}\left(\cos\left(\frac{4\pi}{3}\right),\sin\left(\frac{4\pi}{3}\right)\right)i}=$$
$$16e^{\tan^{-1}\left(-\frac{1}{2},-\frac{\sqrt{3}}{2}\right)i}=$$
$$16e^{-\frac{2\pi}{3}i}=$$
$$16\cos\left(-\frac{2\pi}{3}\right)+16\sin\left(-\frac{2\pi}{3}\right)i=$$
$$16\cdot \left(-\frac{1}{2}\right)+\left(16\cdot -\frac{\sqrt{3}}{2}\right)i=$$
$$-8+(-8\sqrt{3})i=$$
$$-8-8\sqrt{3}i$$
So:
$$(-8)^{\frac{4}{3}}=-8-8\sqrt{3}i$$
|
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|
If $p\equiv 3\pmod{4}$ is a prime, then $\frac{p-1}{2}! \equiv \pm1 \pmod p$
If $p\equiv 3\pmod{4}$ is a prime, then $\frac{p-1}{2}! \equiv \pm1 \pmod p$.
I don't know how to prove this statement.
$p=4m+3$, so $(2m+1)! \equiv \pm1\pmod p$
This is all I did.
|
$\left( \frac{p-1}{2}! \right)^2 = \prod \limits_{n=1}^{\frac{p-1}{2}}n·\prod \limits_{n=1}^{\frac{p-1}{2}}n=\prod \limits_{n=1}^{\frac{p-1}{2}}n·\prod \limits_{n=1}^{\frac{p-1}{2}}-n \equiv\prod \limits_{n=1}^{\frac{p-1}{2}}n·\prod \limits_{n=\frac{p+1}{2}}^{p-1}n=(p-1)! \equiv 1$
You can change $n$ for $-n$ because there is an even number of factors from $1$ to $\frac{n-1}{2}$
Therefore, $ \frac{p-1}{2}!$ is congruent to either $1$ or $-1$
|
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"url": "https://math.stackexchange.com/questions/1485760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Find all five solutions of the equation $z^5+z^4+z^3+z^2+z+1 = 0$ $z^5+z^4+z^3+z^2+z+1 = 0$
I can't figure this out can someone offer any suggestions?
Factoring it into $(z+1)(z^4+z^2+1)$ didn't do anything but show -1 is one solution.
I solved for all roots of $z^4 = -4$ but the structure for this example was more simple.
|
Here is another way to pursue this (just for fun!).
Recall that for $|x| < 1$ we have:
$$\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots$$
Multiplying both sides by $x^6$ yields:
$$\frac{x^6}{1-x} = x^6 + x^7 + x^8 + x^9 + \cdots$$
Now we subtract the former line from the latter:
$$\frac{x^6 - 1}{1-x} = -(1 + x + x^2 + x^3 + x^4 + x^5)$$
We now multiply both sides by $-1$ to obtain your original expression:
$$\frac{x^6 - 1}{x-1} = 1 + x + x^2 + x^3 + x^4 + x^5$$
In one way, this is no different than the suggestion in your accepted answer. But I think (hope) seeing the connection arise in a different manner is of interest. Anyway, at this point you seek to find the roots of $x^6 - 1 = (x^3 + 1)(x^3 - 1)$, and you can now factor the latter two parenthetical expressions as the sum and difference of cubes, respectively, to find the roots.
(Noting that/why $x\neq 1$.)
|
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"timestamp": "2023-03-29T00:00:00",
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|
Formula for how many combinations of powers of 2 sum to $2^n$ Given a number $2^n, n\in\mathbb{Z}\gt 0$, I would like to find a formula for how many unique sets of powers of $2$ sum to that number. This is related to the triangular numbers but excludes non-power-of-$2$ terms.
For example,
$$\begin{align}2^3
&= 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1\\
&= 2 + 1 + 1 + 1 + 1 + 1 + 1\\
&= 2 + 2 + 1 + 1 + 1 + 1\\
&= 2 + 2 + 2 + 1 + 1\\
&= 2 + 2 + 2 + 2\\
&= 4 + 1 + 1 + 1 + 1\\
&= 4 + 2 + 1 + 1\\
&= 4 + 2 + 2\\
&= 4 + 4\\
&= 8\end{align}
$$
so $f(3) = 10$
|
Define $g(m)$ to be the number of ways of writing $m$ as sums of powers. Then $f(n)=g(2^n)$. The sequence $g$ is OEIS sequence A018819.
There's a little more at the OEIS page, but it doesn't list a closed form.
The generating function for $g$ is:
$$G(x)=\sum_{m=0}^\infty g(m)x^m = \prod_{k=0}^\infty \frac{1}{1-x^{2^k}}$$ We see that $G(x^2)=G(x)(1-x)$.
|
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|
Let $p$ be prime not equal to 2 or 5. Show $p^2+1$ or $p^2-1$ is divisible by 10. I can do half the proof but can not think of a way to finish. If $p$ is prime then both $p-1$ and $p+1$ are even. In cases where either $p-1$ or $p+1$ is divisible by 5 that implies $(p-1)(p+1)=p^2-1=10n$ for some positive integer $n$. I am not sure if I am on the right track with this.
|
Since $p \neq 5$ and $p$ is prime, $p$ cannot be divisible by $5$, so $p \neq 0 \mod 5$. Now,
$1^2 = 1 \mod 5$
$2^2 = 4 \mod 5$
$3^2 = 4 \mod 5$
$4^2 = 1 \mod 5$
That is, $p^2 \mod 5$ can be either $1$ or $4$. If $p^2=1\mod 5$, then $p^2-1=0\mod 5$. If $p^2=4\mod 5$,then $p^2+1=0\mod 5$. That is, either $p^2-1$ or $p^2+1$ is divisible by $5$. Since both are even, it implies that one of them that is divisible by $5$ is actually divisible by $10$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
If $|ax^2+bx+c|\le 1\ \forall |x|\le 1$, then what is the maximum possible value of $\frac 83a^2+2b^2$? Let $f(x) = ax^2 + bx + c$ ; $a,b,c\in\mathbb R$
It is given that $|f(x)| \le 1$ $\forall |x| \le 1$
Q1) The possible value of $|a+c|$, if $\displaystyle \frac{8}{3} a^2 + 2b^2$ is maximum, is given by:
a) $0$
b) $1$
c) $2$
d) $3$
Q2) The possible value of $|a+b|$, if $\displaystyle\frac{8}{3} a^2 + 2b^2$ is maximum, is given by:
a) $0$
b) $1$
c) $2$
d) $3$
Q3) The maximum possible value of $\displaystyle\frac{8}{3} a^2 + 2b^2$ is given by:
a) $32$
b) $\displaystyle\frac{32}{3}$
c) $\displaystyle\frac{2}{3}$
d) $\displaystyle\frac{16}{3}$
I have no idea how to go about this question and any help will be appreciated.
|
I'm going to write an answer because the given answer seems to have some errors.
First of all, noting that $a,b,c$ can be written as
$$a=\frac 12\left(f(-1)+f(1)-2f(0)\right),\quad b=\frac 12\left(f(1)-f(-1)\right),\quad c=f(0)$$
might make things easy.
Let us consider first Q3).
$$\begin{align}\frac 83a^2+2b^2&=\frac 43\left(2a^2+\frac 32b^2\right)\\&=\frac 43\left(a^2+2ab+b^2+a^2-2ab+b^2-\frac 12b^2\right)\\&=\frac 43\left((a+b)^2+(a-b)^2-\frac 12b^2\right)\\&=\frac 43\left(\left(f(1)-f(0)\right)^2+\left(f(-1)-f(0)\right)^2-\frac 12b^2\right)\\&=\frac 43\left(|f(1)-f(0)|^2+|f(-1)-f(0)|^2-\frac 12b^2\right)\\&\le \frac 43\left(\left(|f(1)|+|f(0)|\right)^2+\left(|f(-1)|+|f(0)|\right)^2-\frac 12 b^2\right)\\&\le \frac 43\left(\left(1+1\right)^2+\left(1+1\right)^2-\frac 12\cdot 0^2\right)\\&=\frac{32}{3}\end{align}$$
This is attained if and only if
$$(f(-1),f(0),f(1))=(1,-1,1),(-1,1,-1),$$
i.e.
$$(a,b,c)=(2,0,-1),(-2,0,1).$$
So, the maximum possible value of $\frac 83a^2+2b^2$ is $\color{red}{\frac{32}{3}}$.
Q1) $|a+c|=|\pm 2\mp 1|=\color{red}{1}$.
Q2) $|a+b|=|\pm 2+0|=\color{red}{2}$.
|
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|
How do I integrate $x^5(4+x^2)^{-1/2}$? I just started learning substitution and I can't seem to solve this exercise.
I'm using $x = 2sinh(t)$
Full Solution:
$32\int(sinh(t)*(cosh^2(t)-1)^2=$
$32\int(sinh(t)*(cosh^4(t)-2cosh^2(t)+1)=$
$32\int(cosh^4(t)sinh(t)-2cosh^2(t)sinh(t)+sinh(t))=$
$32(\int(cosh^4(arcsinx)sinh(arcsinx)-2\int(cosh^2(arcsinx)sinh(arcsinhx)+\int(sinh(arcsinx)))=$
$32(\int((1-x^2)^2*x) - 2\int((1-x^2)*x) + \int(x))=$
|
Notice,
let, $x=2\tan \theta\ \implies dx=2\sec^2\theta \ d\theta $
$$\int \frac{x^5}{\sqrt{4+x^2}}\ dx$$$$=\int \frac{(2\tan\theta)^5}{\sqrt{4+4\tan^2\theta}}(2\sec^2\theta \ d\theta)$$
$$=32\int \frac{\tan^5\theta \sec^2\theta}{|\sec\theta|}\ d\theta$$
Assuming $0<\theta<\pi/2$
$$=32\int \frac{\tan^5\theta \sec^2\theta}{\sec\theta}\ d\theta$$
$$=32\int (\tan^2\theta)^2 \sec\theta\tan\theta\ d\theta$$
$$=32\int (\sec^2\theta-1)^2 \sec\theta\tan\theta\ d\theta$$
$$=32\int (\sec^4\theta-2\sec^2\theta+1) \ d(\sec\theta)$$
$$=32\left(\frac{\sec^5\theta}{5}-\frac{2\sec^3\theta}{3}+\sec\theta\right)+C$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Probability of hitting a target number on x number of dice with y number of sides I have a complicated probability question. I want to know how to calculate the probability of reaching a target number (or greater) on a variable number of dice with each die having a variable number of sides. For example, the chance of rolling a 10 or higher on 2d6 and 1d4. I can calculate the answer by just writing down all the possibilities for any given set of dice but I need a formula that can work this out on its own if given the inputs. I want to use this formula as part of a google spreadsheet, (similar to Microsoft Excel). If it helps, I wont ever need more than 10 dice at a time and the dice can only have 4, 6, 8, 10, or 12 sides.
Thanks.
|
One way is to use generating functions.
If you have $n$ dice, with sides $a_1,\dots,a_n$, then the coefficient on $x^k$ of the polynomial
$$
\prod_{i=1}^n \left( \frac{1}{a_i} \sum_{j=1}^{a_i} x^i \right)
$$
yields the probability of a sum of $k$ with these dice.
Summing the appropriate values will give you the probability of a sum above a given value.
For example, if you have a three-sided, a four-sided, and a six-sided die, then we want to look at
$$
\left(\frac{1}{3}\sum_{i=1}^3 x^i\right)\left(\frac{1}{4}\sum_{i=1}^4 x^i\right)\left(\frac{1}{6}\sum_{i=1}^6 x^i\right)=\frac{1}{72} x^{13}
+ \frac{1}{24} x^{12}
+ \frac{1}{12} x^{11}
+ \frac{1}{8} x^{10}
+ \frac{11}{72} x^9
+ \frac{1}{6} x^8
+ \frac{11}{72} x^7
+ \frac{1}{8} x^6
+ \frac{1}{12} x^5
+ \frac{1}{24} x^4
+ \frac{1}{72} x^3.
$$
Then, for example, the probability that the sum is greater than $10$ is $\frac{1}{12}+\frac{1}{24}+\frac{1}{72}=\frac{5}{36}$.
Another example. For the probability of throwing 9 or greater with two six-sided dice, we look at
$$
\left(\frac{1}{6}\sum_{i=1}^6 x^i\right)\left(\frac{1}{6}\sum_{i=1}^6 x^i\right)=\frac{1}{36} x^{12}
+ \frac{1}{18} x^{11}
+ \frac{1}{12} x^{10}
+ \frac{1}{9} x^9
+ \frac{5}{36} x^8
+ \frac{1}{6} x^7
+ \frac{5}{36} x^6
+ \frac{1}{9} x^5
+ \frac{1}{12} x^4
+ \frac{1}{18} x^3
+ \frac{1}{36} x^2
$$
Summing the coefficients on $x^9$, $x^{10}$, $x^{11}$, and $x^{12}$, we find the probability of throwing $9$ or higher is $\frac{5}{18}$.
|
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|
Show that $f'(x) = \lim\limits_{h \rightarrow 0} \dfrac{f(x+h)-f(x-h)}{2h}$ I have to prove that if a function $f$ is differentiable on $(a,b)$, then
\begin{align*}
f'(x) = \lim\limits_{h \rightarrow 0} \dfrac{f(x+h)-f(x-h)}{2h}
\end{align*}
Using the fact that $f'(x) = \lim\limits_{h \rightarrow 0}\dfrac{f(x+h)-f(x)}{h}$, I wrote my proof in the following manner:
\begin{align*}
\lim\limits_{h \rightarrow 0}f(x+h) - 2f(x) = \lim\limits_{h \rightarrow 0} - f(x-h)
\\
\lim\limits_{h \rightarrow 0} \dfrac{f(x+h)}{2h} - \dfrac{f(x)}{h} = \lim\limits_{h \rightarrow 0}\dfrac{-f(x-h)}{2h}
\\
\lim\limits_{h \rightarrow 0}\dfrac{f(x+h)}{h} - \dfrac{f(x)}{h} = \lim\limits_{h \rightarrow 0} \dfrac{f(x+h)}{2h} - \dfrac{f(x-h)}{2h}
\\
\lim\limits_{h \rightarrow 0} \dfrac{f(x+h) - f(x)}{h} = \lim\limits_{h \rightarrow 0} \dfrac{f(x+h)-f(x-h)}{2h}
\end{align*}
However, I believe that it is actually incorrect, because when I divide by $2h$ I am potentially making the limit undefined. How would I go about correcting my proof?
|
\begin{align*}
f'(x) = \lim\limits_{h \rightarrow 0} \dfrac{f(x+h)-f(x)}{h}
\end{align*}
but also
\begin{align*}
f'(x) = \lim\limits_{h \rightarrow 0} \dfrac{f(x)-f(x-h)}{h}
\end{align*}
sum them up and divide by 2 to get
\begin{align*}
f'(x) = \lim\limits_{h \rightarrow 0}\frac{ \dfrac{f(x+h)-f(x)}{h} + \dfrac{f(x)-f(x-h)}{h}}2
=\lim\limits_{h \rightarrow 0} \dfrac{f(x+h)-f(x-h)}{2h}
\end{align*}
|
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|
Show difference between values of turning points of $ f(x) = (c-\frac{1}{c}-x)(4-3x^2) $ How would you show that the difference between the values of the turning points of
$$
f(x) = (c-\frac{1}{c}-x)(4-3x^2)
$$
is
$$
\frac{4}{9}(c+\frac{1}{c})^3
$$
$c>0$
I have attempted to compute the derivative and set it to $0$ (which is the standard approach), but things get messy (in my opinion).
|
Starting with
$$f(x) = (c-\frac{1}{c}-x)(4-3x^2)$$
to make computation a bit more manageable, set $\boxed{\gamma=c-\frac{1}{c}}$, so
$$f(x)=4\gamma-4x-3\gamma x^2+3x^3$$
You had the correct idea to set $f'(x)=0$ at the turning points. So
$$f'(x)=-4-6\gamma x+9x^2=0\quad(\text{at turning points})$$
By the quadratic formula
$$x=\frac{6\gamma\pm\sqrt{36\gamma^2+144}}{18}=\frac{\gamma\pm\sqrt{\gamma^2+4}}{3}$$
At this stage, see if the quantity under the square root can be simplified:
$$\gamma^2+4=\left(c-\frac{1}{c}\right)^2+4=\left(c^2-2+\frac{1}{c^2}\right)+4=c^2+2+\frac{1}{c^2}=\left(c+\frac{1}{c}\right)^2$$
Often this sort of simplification will work as the goal of such a question is not typically to test whether you can perform messy number crunching, e.g. computing $f(1+\frac{1}{\sqrt5})$.
So the two roots are given by
$$x_{1,2}=\frac{\left(c-\frac{1}{c}\right)\pm\left(c+\frac{1}{c}\right)}{3} \implies x_1=-\frac{2}{3c},\:x_2=\frac{2c}{3}$$
I will omit the computation of $f(x_1)$ and $f(x_2)$.
|
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|
Find : $\sqrt[6]{\frac{\sqrt{2}+(-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)^7}{(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)^{11}}}$ in its algebraic form. Find : $$\sqrt[6]{\frac{\sqrt{2}+(-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)^7}{(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)^{11}}}$$ in its algebraic form.
Now, I kinda think it would not be wise to try to expand this, but rather apply de Moivre formula on the complex number in the numerator and denominator, then simplify that complex number within the root, and once again apply moivres formula.
I have tried but then I get the expression:$\sqrt[6]{\frac{\sqrt{2}+\cos{\frac{21 \pi}{4}+i\sin{\frac{21 \pi }{4}}}}{\cos \frac{11\pi}{4}+i\sin\frac{11 \pi}{4}}}$ and don't know what to do with it.
|
Here's my attempt at a solution using complex exponentials...
$$\begin{align*}
\sqrt[6]{\frac{\sqrt{2}+(-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)^7}{(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)^{11}}} &=
\sqrt[6]{\frac{\sqrt{2}+\left(e^{i3\pi/4}\right)^7}{\left(e^{i\pi/4}\right)^{11}}}
\\
& = \sqrt[6]{\frac{\sqrt{2}+e^{i21\pi/4}}{e^{i11\pi/4}}}
\\
& = \sqrt[6]{\sqrt{2}e^{-i11\pi/4}+e^{i10\pi/4}}
\\
& = \sqrt[6]{\sqrt{2}e^{-i3\pi/4}+e^{i\pi/2}}
\\
& = \sqrt[6]{\sqrt{2}\left(-\frac{\sqrt{2}}{2}\color{red}{-}\frac{\sqrt{2}}{2}i\right)+i}
\\
& = \color{green}{\sqrt[6]{-1}}
\end{align*} $$
Now take the root, again using complex exponentials. Take care, though, as the root is multivalued. You ought to find six distinct points, as this is a sixth root of unity. Let me know if anything looks iffy.
EDIT: I checked the answer with Wolfram. I made a sign error above (red). The green result works out.
|
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|
Prove that $(1 + \sqrt2)^{2n} + (1 - \sqrt{2})^{2n}$ is an even integer. Prove that $(1+\sqrt2)^{2n} + (1-\sqrt2)^{2n}$ is an even integer.
I'm not sure how to prove that it is an even integer. What would I do for the Inductive Step? And for the basic step, can I plug in zero and prove something from that?
|
Note that
$(1+\sqrt2)(1-\sqrt2)
=-1
$
and
$(1+\sqrt2)^2
=3+2\sqrt{2}
$.
Therefore,
if $a
=3+2\sqrt{2}
$,
then
$1/a = 3-2\sqrt{2}
$,
so that
$a+1/a
=6
$
and
$(1+\sqrt2)^{2n} + (1-\sqrt2)^{2n}
=a^n+1/a^n
$.
We now use the identity
true for any $a$
that
$a^{n+1}+1/a^{n+1}
=(a+1/a)(a^n+1/a^n)-(a^{n-1}+a^{n-1})
$.
Therefore,
for this particular $a$,
$a^{n+1}+1/a^{n+1}
=6(a^n+1/a^n)-(a^{n-1}+a^{n-1})
$.
Since
$a^n+1/a^n$
is an integer
for $n=0$
and
$n=1$,
it is an integer
for all $n$.
Explicitly,
if
$u_n = a^n+1/a^n
=(1+\sqrt2)^{2n} + (1-\sqrt2)^{2n}
$,
$u_{n+1}
=6u_n-u_{n-1}
$
with
$u_0 = 1$
and
$u_1 = 6$.
|
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|
Write out the Form of the Partial Fraction Decomposition: $\int\frac{x^6}{x^2-4}$ Write out the Form of the Partial Fraction Decomposition: $\int\frac{x^6}{x^2-4}$
The book says use long division my answer was $x^3+\frac{4x^3}{x^2-4}$
The answer manual is $\frac{x^6}{x^2-4}=x^4+4x^2+16+\frac{64}{(x+2)(x-2)}$
|
The first step is dividing $x^6$ by $x^2$ to get $x^4$.
Then you subtract $x^6-x^4(x^2-4)=4x^4$ (as you also got). But you need to continue.
Second step is to divide $4x^4$ by $x^2$ to get $4x^2$.
Then you subtract $4x^4-4x^2(x^2-4)=16x^2$. Again you need to continue and dividing $16x^2$ by $x^2$ to get $16$.
Then you subtract $16x^2-16(x^2-4)=64$.
And your final answer is
$$\frac{x^6}{x^2-4}=x^4+4x^2+16+\frac{64}{x^2-4}.$$
|
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|
If $x^2-3x+9=0$, can we say $(x+3)(x^2-3x+9)$ is also $0$ hence $x^3=-27$? I was solving this question
If $\dfrac x3 + \dfrac 3x = 1$ then find the value of $x^3$.
I solved it as. Cube both sides and substitute $x^3$ with $t$,
$$ \dfrac{t}{27} + \dfrac{27}{t}=-2$$
$$\implies t^2 + {27}^2 + 2\cdot 27 \cdot t = 0$$
$$\implies (t+27)^2=0$$
$$\implies t=x^3=-27$$
Someone over facebook put another solution as,
$$\dfrac x3 + \dfrac 3x = 1$$
$$\implies x^2-3x+9=0$$
Multiplying both the sides with $(x+3)$,
$$(x+3)(x^2-3x+9)=0$$
$$\implies x^3+27=0$$
Hence $x^3=-27$.
My question is, is it correct to multiply both sides with $0$? ( $x+3$ is $0$ here because $x^3=-27 \implies x=-3$ is one possible solution and other two being complex numbers.)
|
Yes there is nothing as such wrong with this process since it really helps in identifying the roots.
BUT one thing to keep in mind is that when you multiply the expression with $(x+3)$, you are introducing extraneous roots. That is, you had a quadratic equation with 2 roots and you have now made it cubic with 3 roots. So after solving, you must carefully eradicate this extraneous root you had introduced.
In the given problem, $x=-3$ is the extraneous root.
However you could have approached the problem using the discriminant.Then you get the roots easily.
$$x^2-3x+9=0 \Rightarrow x=\frac{3 \pm \sqrt{9-36}}{2}=\frac{3 \pm 3\sqrt{3}i}{2}$$
|
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|
Am I correctly finding the standard matrix?
Question:
Let $F: \Bbb R^3 \to \Bbb R^3$ be the linear transformation satisfying
$F(1,0,1)=(-3,-3,1)$,$F(0,1,0)=(0,1,1)$, and $F(0,1,1)=(2,-2,1)$. Find
the standard matrix $A$ of $F$.
My Approach:
I used a method that I haven't been taught, so I'm not sure if I am correct and allowed to do this. Can you guys see if this is acceptable.
Since:
$$F\begin{pmatrix}1\\0\\1\end{pmatrix}=\begin{pmatrix}3\\-3\\1\end{pmatrix}$$
$$F\begin{pmatrix}0\\1\\0\end{pmatrix}=\begin{pmatrix}0\\1\\1\end{pmatrix}$$
$$F\begin{pmatrix}0\\1\\1\end{pmatrix}=\begin{pmatrix}2\\-2\\1\end{pmatrix}$$
We want to find:
$$\text{standard matrix:}F\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}=?$$
So what I did was:
$$\therefore \text{standard matrix:}F\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}=\begin{pmatrix}3&0&2\\-1&0&-1\\-2&1&-1\end{pmatrix}$$
Is this method correct?
|
Your matrix does not map $(0,1,0)^t$ to $(0,1,1)^t$, so I believe it is wrong.
Instead you could use
$$
A X = Y \iff \\
A = Y X^{-1}
$$
where $X$ has the given argument vectors and $Y$ the corresponding image vectors, assuming the argument vectors are linear independent.
Applying the above I get
$$
X =
\left(
\begin{array}{rrr}
1 &0 & 0 \\
0 &1 & 1 \\
1 &0 & 1
\end{array}
\right)
\quad\quad
Y=
\left(
\begin{array}{rrr}
3 & 0 & 2 \\
-3 & 1 & -2 \\
1 & 1 & 1
\end{array}
\right)
$$
which leads to
$$
A=
\left(
\begin{array}{rrr}
1&0&2\\
0&1&-3\\
1&1&0
\end{array}
\right)
$$
The matrix you calculated is $B = X^{-1} Y$. It solves $X B = Y$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1515520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Solving an equation by telling the value of $x^2+y^2$. I have a problem solving an equation. The equation is:
$xy+x+y=44$ and $x^2y+xy^2=448$
and we have to tell the value of $x^2+y^2$
First I tried solving this by doing the following:
$xy+x+y=44~\to~x+y=44-xy~\to~x^2+2xy+y^2=44^2-88xy+x^2y^2~\Rightarrow$
$\Rightarrow~x^2+y^2=44^2-90xy+x^2y^2$
But from here I didn't know what to do. Could you help me in solving this equation?
|
Here is one suggestion. Let $a=x+y, b=xy$ then $a+b=44, ab=448$
This gives $a=16, b=28, x^2+y^2=a^2-2b=200$ or $a=28, b=16, x^2+y^2= 752$
The intermediate step is to formulate and solve the quadratic satisfied by $a$ and $b$
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Which natural numbers satisfy $2^n > n^2$?
Which natural numbers satisfy $2^n > n^2$ ?
My work. Step 1: $n = 1 $, $2^1 > 1^2$. True.
For $n = k$, $2^k > k^2$. For $n = k+1$,
$$ 2^{(k+1)} > (k+1)^2 \\
2\cdot 2^k>k^2+2k+1 \\
2^k+2^k > k^2+2k+1$$
*
*$2^k > k^2 \text{ - from step 1}$
*$2^k > k^2+2k+1$
How I can find the numbers now?
|
For $n = 1$ is true, but for $n = 2,3,4$ is not. but for $n \geq 5$ is true.
Hence, the solution set is $\{1\}\cup[5,\infty)$.
Now we assume is true for $n = k$, that is
$2^k > k^2$, we will have to show that it is true for $n = k+1$.
$2^{k+1} = 2(2^k)$, but $2^k > k^2$ by induction hypothesis then $2^{k+1} > 2k^2 > k^2$.
Hence by the induction principle it is true for all $n \geq 5$ (also for $n = 1$).
|
{
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"url": "https://math.stackexchange.com/questions/1518774",
"timestamp": "2023-03-29T00:00:00",
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|
Mathematical induction $\sum_{i=0}^{n}{n \choose i} = 2^n $ Prove by mathematical induction:
$\sum_{i=0}^{n}{n \choose i} = 2^n ; n \ge 0$
Step 1: n = 0
${0 \choose 0}=2^0$
Step 2:
for n = k
$\sum_{i=0}^{k}{k \choose i} = 2^k$
assumption: for n = k+1
$\sum_{i=0}^{k+1}{k+1 \choose i} = 2^{k+1}$
Step 3:
$\sum_{i=0}^{k+1}{k+1 \choose i} = \sum_{i=0}^{k}{k \choose i} + {k+1 \choose k+1} = 2^k+1$
Assumption:
$2^{k+1}$
Result of mathematical induction:
$2^k+1$
Is that correct or is somewhere the mistake ?
|
Your base case is essentially correct, although I would prefer to see the justification that both sides are equal to $1$. In step 2, your assumption should be $$\sum_{i = 0}^{k} \binom{k}{i} = 2^k$$ since you must show that $P(k + 1)$ holds whenever $P(k)$ holds. In step 3, the statement
$$\sum_{i = 0}^{k + 1} \binom{k + 1}{i} = \sum_{i = 0}^{k} \binom{k}{i} + \binom{k + 1}{k + 1}$$ is false. Furthermore, if $k$ is a positive integer, $2^{k + 1} = 2 \cdot 2^k = 2^k + 2^k > 2^k + 1$.
An induction proof is provided below:
Proof. Let $P(n)$ be the statement that
$$\sum_{k = 0}^{n} \binom{n}{k} = 2^n$$
Let $n = 0$. Then
$$\sum_{k = 0}^{0} \binom{0}{k} = \binom{0}{0} = 1 = 2^0$$
Hence, $P(0)$ holds.
Since $P(0)$ holds, we may assume there exists a nonnegative integer $m$ such that $P(m)$ holds. Then
$$\sum_{k = 0}^{m} \binom{n}{k} = 2^m$$
Let $n = m + 1$. We must show that $P(m) \Rightarrow P(m + 1)$.
\begin{align*}
\sum_{k = 0}^{m + 1} \binom{m + 1}{k} & = \binom{m + 1}{0} + \sum_{k = 1}^{m} \binom{m + 1}{k} + \binom{m + 1}{m + 1}\\
& = 1 + \sum_{k = 1}^{m} \binom{m + 1}{k} + 1\\
& = 1 + \sum_{k = 1}^{m} \left[\binom{m}{k} + \binom{m}{k - 1}\right] + 1 & \text{by Pascal's Identity}\\
& = 1 + \sum_{k = 1}^{m} \binom{m}{k} + \sum_{k = 1}^{m} \binom{m}{k - 1} + 1\\
& = 1 + \sum_{k = 1}^{m} \binom{m}{k} + \sum_{j = 0}^{m - 1} \binom{m}{j} + 1 & \text{where $j = k - 1$}\\
& = \binom{m}{0} + \sum_{k = 1}^{m} \binom{m}{k} + \sum_{j = 0}^{m - 1} \binom{m}{j} + \binom{m}{m}\\
& = \sum_{k = 0}^{m} \binom{m}{k} + \sum_{j = 0}^{m} \binom{m}{j}\\
& = 2^m + 2^m & \text{induction hypothesis}\\
& = 2 \cdot 2^{m}\\
& = 2^{m + 1}
\end{align*}
Since $P(0)$ holds and $P(m) \Rightarrow P(m + 1)$ for each nonnegative integer $m$, $P(n)$ holds for all nonnegative integers.$\blacksquare$
Note. The binomial coefficient $$\binom{n}{k}$$ represents the number of subsets of size $k$ that can be selected from a set of size $n$. Thus, the summation $$\sum_{k = 0}^{n} \binom{n}{k}$$ counts the number of subsets of a set with $n$ elements. There are $2^n$ such subsets since a subset is determined by the decision of whether or not to include each of the $n$ elements.
|
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|
How to solve this limit without using L'Hospital's Rule? $$
\lim\limits_{x\to{\infty}}\frac{\arctan \frac{3}{x}}{|\arctan \frac{2}{x}|}
$$
Can anybody help me to solve this one ?
I ve done somethig like this but im not sure if it is the correct aproach.
$$
\lim\limits_{x\to{\infty}}\frac{\arctan\frac{3}{x}}{|\arctan \frac{2}{x}|} =
\lim\limits_{x\to{\infty}}\frac{\arctan \frac{3}{x}}{\arctan \frac{2}{x}} =
\lim\limits_{x\to{\infty}} \frac{\arctan \frac{3}{x}}{\arctan \frac{2}{x}} * \frac{\frac{\frac{3}{x}}{\frac{3}{x}}}{\frac{\frac{2}{x}}{\frac{2}{x}}} = \frac{\lim\limits_{x\to{\infty}}\frac{\arctan \frac{3}{x}}{\frac{3}{x}}}{\lim\limits_{x\to{\infty}}\frac{\arctan \frac{2}{x}}{\frac{2}{x}}}*\lim\limits_{x\to{\infty}}\frac{\frac{3}{x}}{\frac{2}{x}}
$$
and then for each limit with arctan i ve substitued 3/x and 2/x by tan y and tan z
$$
\frac{\lim\limits_{y\to{0^+}}\frac{\arctan \tan y}{\tan y}}{\lim\limits_{z\to{0^+}}\frac{\arctan \tan z}{\tan z}}*\lim\limits_{x\to{\infty}}\frac{\frac{3}{x}}{\frac{2}{x}}
$$
and then for each limit which goes to 0^+ i ve done this
$$
\lim\limits_{y\to{0^+}}\frac{\arctan \tan y}{\tan y} = \lim\limits_{y\to{0^+}}\frac{y}{\tan y} = \lim\limits_{y\to{0^+}}\frac{y}{\frac{\sin y}{\cos y}} = 1
$$
so in the end
$$
\frac{1}{1}*\frac{3}{2} = \frac{3}{2}
$$
|
Your approach is correct assuming you already know $\lim \frac{\sin x}{x}=1$. Make sure you justify your steps: you can only separate the big limit in a product/quotient of separate limits assuming those exist (which you prove at the end), and you can say that there exists a $y\in(0,\frac \pi2)$ such that $\tan y = \frac 3x$ by bijectivity of $\tan:(0,\frac \pi 2)\to (0,\infty)$.
|
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|
A field with an element of order $12$: $(a+a^{-1})^2=3.$
Let $K$ be a field and $a\in K^*$ of order $12.$ I need to prove that $(a+a^{-1})^2=3.$
'Progress': I expand $(a+a^{-1})^2=a^2+a^{-2}+2$ so equivalently I need to prove that $a^2+a^{-2}=1$, or the inverse of $a^2$ is $1+a^{-4}.$ I tried also Binomial theorem and working with the inverse but no success.
Not sure how can I use the fact that the order of $a$ is $12$. Any ideas?
|
We have $(a^6-1)(a^6+1)=0$. But $a^6\ne 1$, so $a^6+1=0$. Thus $(a^2+1)(a^4-a^2+1)=0$. Because the order of $a$ does not divide $4$, we have $a^4-a^2+1=0$. This is equivalent to what you want to show.
|
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|
Prove $3^n > n^2$ by induction Prove that $$3^n > n^2$$
I am using induction and I understand that when $n=1$ it is true. The induction hypothesis is when $n=k$ so $3^k>k^2$. So for the induction step we have $n=k+1$ so
$3^{k+1} > (k+1)^2$ which is equal to $3\cdot3^k > k^2+2k+1$. I know you multiple both sides of the induction hypothesis by $3$ but I'm not sure what to do next.
|
You start with knowing that $3^k > k^2$ and you prove that if you know that then you prove that $3^{k+1} > (k+1)^2$.
One straightforward and Doy! way is to note:
$3^k > k^2$
$3^k + 3^k + 3^k> k^2 + k^2+k^2$.
Since $k \ge 2$ then $k^2 \ge 2k$ and as $k \ge 2$ then $k^2 > 1$
So
$3^k + 3^k + 3^k > k^2 + 2k + 1$
$3^{k+1} = 3*3^k > (k+1)^2$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Conditions for symmetry for pair of commuting matrices Let A = $\left(\begin{array}{ccc}
2 & -1 & 0 \\
-1 & 2 & -1 \\
0 & -1 & 2
\end{array} \right)$
Show that every real matrix $B$ such that $AB = BA$ has the form $B = aI + bA + cA^2$
My attempt: If we assume that $AB = BA$ are simultaneously diagonalizable by $P$, then since $A = PDP^{-1}$we have the RHS expression to be $aPIP^{-1} + bPDP^{-1} + cPD^2P^{-1}$. Computing $D$, we observe that $I$ and $D$ and $D^2$ are linearly independent, and thus they span all 3X3 diagonal matrices, diagonalized B included.
But $A$ and $B$ are simultaneously diagonalizable iff the commute and they are both diagonalizable. Now $A$ is real symmetric so it is diagonalizable, but what can we say about $B$?
It looks like $B$ wants to be symmetric. But A symmetric and $AB=BA$ does not imply B is symmetric, take $A = I$ for instance. So I have two questions:
1) Under what conditions for A will the statement ($A$ symmetric and $AB= BA) \Rightarrow B$ symmetric hold?
2) If my approach does not work, does anyone have any other way of proceeding?
|
We can prove the statement by brute force.
Search a matrix $B$ the commute with $A$:
$$
\begin{bmatrix}
2&-1&0\\
-1&2&-1\\
0&-1&2
\end{bmatrix}
\begin{bmatrix}
x_1&y_1&z_1\\
x_2&y_2&z_2\\
x_3&y_3&z_3
\end{bmatrix}=
\begin{bmatrix}
x_1&y_1&z_1\\
x_2&y_2&z_2\\
x_3&y_3&z_3
\end{bmatrix}
\begin{bmatrix}
2&-1&0\\
-1&2&-1\\
0&-1&2
\end{bmatrix}
$$
wit a bit of work, equating the corresponding entries of the two products, we find that $B$ has the form:
$$
B=\begin{bmatrix}
x&y&z\\
y&x+z&y\\
z&y&x
\end{bmatrix}
$$
Now:
$$aI+bA+cA^2=
a\begin{bmatrix}
1&0&0\\
0&1&0\\
0&0&1
\end{bmatrix}+b
\begin{bmatrix}
2&-1&0\\
-1&2&-1\\
0&-1&2
\end{bmatrix}+c
\begin{bmatrix}
5&-4&1\\
-4&6&-4\\
1&-4&5
\end{bmatrix}=
\begin{bmatrix}
2a+2b+5c&-b-4c&c\\
-b-4c&a+2b+6c&-b-4c\\
c&-b-4c&a+2b+5c
\end{bmatrix}
$$
tha has the form of $B$ for $x=2a+2b+5c$, $y=-b-4c$ and $z=c$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find polynomial $f(x)$ based on divisibility properties of $f(x)+1$ and $f(x) - 1$ $f(x)$ is a fifth degree polynomial. It is given that $f(x)+1$ is divisible by $(x-1)^3$ and $f(x)-1$ is divisible by $(x+1)^3$. Find $f(x)$.
|
Another ad hoc solution is the following:
$$
32 = \Big((x+1)-(x-1)\Big)^5 \\
= (x+1)^5 -5(x+1)^4(x-1) +10(x+1)^3(x-1)^2 \qquad \\ \qquad -10(x+1)^2(x-1)^3 +5(x+1)(x-1)^4
-(x-1)^5 \\
= (x+1)^3\Big((x+1)^2-5(x+1)(x-1)+10(x-1)^2\Big) \qquad \\ \qquad
-(x-1)^3\Big(10(x+1)^2-5(x+1)(x-1)+(x-1)^2\Big)
$$
so
$$
f(x) =
-\frac1{16} (x+1)^3\Big((x+1)^2-5(x+1)(x-1)+10(x-1)^2\Big) +1 =
-\frac1{16} (x-1)^3\Big(10(x+1)^2-5(x+1)(x-1)+(x-1)^2\Big) -1
$$
is $a$ solution.
For solving the general problem, we can repeat the usual proof of the Chinese Remainder Theorem. To solve the system
$$
f(x) \equiv r_1(x) \pmod{m_1(x)} \\
f(x) \equiv r_2(x) \pmod{m_2(x)}
$$
where $m_1$ and $m_2$ are co-prime, using Euclid's algorithm we construct some polynomials $p(x)$ and $q(x)$ such that
$$ p(x)\cdot m_1(x) + q(x)\cdot m_2(x) = 1. $$
Then the solution will be
$$
f = (q \cdot m_2 \cdot r_1 + p \cdot m_1 \cdot r_2) \mod m_1m_2.
$$
|
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|
Is the numerator of $\sum_{k=0}^n \frac{(-1)^k}{2k+1}\binom{n}{k}$ a power of $2$? I stumbled on something numerically, and was just starting to work on it, but it seemed fun enough to share.
Let $$f(n)=\sum_{k=0}^{n} \frac{(-1)^{k}}{2k+1}\binom{n}{k}$$
It appears, from the first few values, that $f(n)$ always has numerator equal to a power of $2$. Is this true? If so, why?
The first values:
$$\frac{1}{1},
\frac{2}{3},
\frac{8}{15},
\frac{16}{35},
\frac{128}{315},
\frac{256}{693},
\frac{1024}{3003},
\frac{2048}{6435},
\frac{32768}{109395},\\
\frac{65536}{230945},
\frac{262144}{969969},
\frac{524288}{2028117},
\frac{4194304}{16900975}
$$
Alternative ways you can see this value: $$f(n)=\int_0^{1}(1-x^2)^n\,dx=\int_{0}^{\pi/2}\cos^{2n+1} t\,dt=\frac{1}{2^{2n}}\sum_{k=0}^{n}\frac{(-1)^k}{2k+1}\binom{2n+1}{n-k}$$
To summarize some of the comments, the sequence above appears to match:
$$\begin{align}
f(n)&=\frac{(2n)!!}{(2n+1)!!} \\
&= \frac{2}{3}\cdot \frac{4}{5}\cdot \frac{6}{7}\cdots \frac{2n}{2n+1}
\end{align}$$
Thus, if this is correct (and the answer below proves it is,) we have $$f(n)=f(n-1)\cdot \frac{2n}{2n+1}=f(n-1)\left(1-\frac{1}{2n+1}\right).$$
There might be a proof of this recursion using integration by parts for one of the integral forms above.
|
$$\begin{eqnarray*} f(n) = \sum_{k=0}^{n}\frac{(-1)^k}{2k+1}\binom{n}{k}&=&\int_{0}^{1}\sum_{k=0}^{n}(-1)^k x^{2k}\binom{n}{k}\,dx\\&=&\int_{0}^{1}(1-x^2)^n\,dx\\&=&\frac{1}{2}\int_{0}^{1}z^{-1/2}(1-z)^{n}\,dz\\&=&\frac{\Gamma\left(\frac{1}{2}\right)\Gamma(n+1)}{2\,\Gamma\left(n+\frac{3}{2}\right)}\\&=&\frac{4^n\,n!^2}{(2n+1)!}\\&=&\frac{2^{2n}}{(2n+1)\binom{2n}{n}}\end{eqnarray*}$$
hence it is enough to check that $\nu_2\left(\binom{2n}{n}\right)<2n$ to prove your conjecture.
As a matter of fact,
$$\nu_2\left(\binom{2n}{n}\right) = \sum_{k\geq 1}\left(\left\lfloor\frac{2n}{2^k}\right\rfloor-2\left\lfloor\frac{n}{2^k}\right\rfloor\right)\leq 2+\log_2(n),$$
since the terms of the last sum can be only $0$ or $1$, and they are zero as soon as $k$ is big enough.
As stated by the OP in the comments, a careful analysis of the previous formula reveals that $\nu_2\left(\binom{2n}{n}\right)$ is just the number of non-zero bits in the binary representation of $n$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
convergence of series $ {n}^{p} \cdot \left\{ \frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\right\}$ $$ {n}^{p} \cdot \left\{ \frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\right\}$$
determine the convergence of the series
$$ {n}^{p} \cdot \left\{ \frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\right\} < {n}^{p} \cdot \frac{1}{\sqrt{n-1}\sqrt{n}} < \frac{{n}^{p}}{n-1} < \frac{1}{{n}^{1-p}-{n}^{-p}} < \frac{1}{{n}^{1-p}} $$
so converges when p<0 how about other cases?
|
$$\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}=\frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n}\sqrt{n-1}}=\frac{1}{\sqrt{n^2-n}(\sqrt{n}+\sqrt{n-1})}=\frac{1}{n^{3/2}}\frac{1}{\sqrt{1-\frac{1}{n}}\left(1+\sqrt{1-\frac{1}{n}}\right)}\approx \frac{1}{2n^{3/2}}$$
So use the limit comparison test between $n^p\left(\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\right)$ and $n^{p-3/2}$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Question on inequality Let $a,b,c$ be real numbers such that $a+b+c = 0$ and $a^2 + b^2 + c^2 = 1$. Prove that $$a^2b^2c^2\le \frac1{54}$$
|
By calculation $$ab+bc+ca=-1/2$$ and
$$1/4=(ab+bc+ca)^2=2abc(a+b+c)+a^2 b^2+b^2 c^2+c^2 a^2=a^2 b^2+b^2 c^2+c^2 a^2$$.
Let $K=a^2b^2c^2$, then $a^2,b^2,c^2$ are the solutions of
$$t^3-t^2+\frac{1}{4}t-K=0$$
Thus
$$K=t^3-t^2+\frac{1}{4}t$$
By taking derivative, we know the maximal value $K=1/54$ is achieved at $t=1/6$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Spectral Decomposition for matrices with eigenvalue(s) 0 When expressing the spectral decomposition of a matrix that has eigenvalues of 0, do you express the corresponding matrices with factor '0', or leave them out altogether?
EDIT: let's consider some matrix M, where
$$
M =
\begin{bmatrix}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1 \\
\end{bmatrix}
$$
Then, would the spectral decomposition of this matrix M be
(i)
$$
\begin{bmatrix}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1 \\
\end{bmatrix}
=
3
\begin{bmatrix}
\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\
\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\
\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\
\end{bmatrix}
+
0
\begin{bmatrix}
\frac{1}{2} & -\frac{1}{2} & 0 \\
-\frac{1}{2} & \frac{1}{2} & 0 \\
0 & 0 & 0 \\
\end{bmatrix}
+
0
\begin{bmatrix}
\frac{1}{6} & \frac{1}{6} & -\frac{1}{3} \\
\frac{1}{6} & \frac{1}{6} & -\frac{1}{3} \\
-\frac{1}{3} & -\frac{1}{3} & \frac{2}{3} \\
\end{bmatrix}
$$
or (ii)
$$
\begin{bmatrix}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1 \\
\end{bmatrix}
=
3
\begin{bmatrix}
\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\
\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\
\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\
\end{bmatrix}
$$
I hope this clarifies the problem.
EDIT 2: Maybe the terminology in my textbook is incorrect, but the form I am talking about looks like this: if the matrix M was orthogonally diagonalized by
$$
P =
\begin{bmatrix}
\bar{u_1} & \bar{u_2} & \bar{u_3} \\
\end{bmatrix}
$$
and
$$
D =
\begin{bmatrix}
\lambda_1 & 0 & 0 \\
0 & \lambda_2 & 0 \\
0 & 0 & \lambda_3 \\
\end{bmatrix}
$$
where $\bar{u_1}$, $\bar{u_2}$, and $\bar{u_3}$ are unit eigenvectors of M, then M could be rewritten as
$$M = \lambda_1\bar{u_1}\bar{u_1}^T + \lambda_2\bar{u_2}\bar{u_2}^T + \lambda_3\bar{u_3}\bar{u_3}^T$$
If the answerer happens to know the name of this form, I'd be much obliged.
|
What your book calls spectral decomposition is just the equation you obtain after carrying out the matrix product in what is usually called a spectral decomposition. This seems very odd to me, because usually the term decomposition is used to refer as a way to write a matrix as a product of other matrices, not as a sum.
Anyway, a spectral decomposition of a matrix $M$ is an equation of the form
$$
M = Q \Delta Q^{-1} \label{eq:1} \tag{1}
$$
where $\Delta$ is the diagonal matrix with diagonal elements $\lambda_1,\dotsc,\lambda_n$, the eigenvalues of $M$ counted with their respective multiplicity, and $Q$ is a matrix whose columns $q_1,\dotsc,q_n$ form a basis of eigenvectors of $M$ such that $q_i$ corresponds to $\lambda_i$.
Now, if you choose an orthonormal basis of eigenvectors, then $Q$ is an orthogonal matrix and $Q^{-1} = Q^T$. Thus $\eqref{eq:1}$ becomes
$$
\begin{align}
M &=
\left(
\begin{array}{c | c | c}
q_1 & \dotsc & q_n
\end{array}
\right)
\begin{pmatrix}
\lambda_1 & & \\
& \ddots & \\
& & \lambda_n
\end{pmatrix}
\left(
\begin{array}{c}
q_1^T \\
\hline
\vdots \\
\hline
q_n^T
\end{array}
\right) \\
&= \lambda_1 q_1 q_1^T + \dotsc + \lambda_n q_n q_n^T
\end{align}
$$
TL;DR: No, you don't need to include the summands with eigenvalue $0$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluating $\lim_\limits{x\to 1 }\bigl( (2^x x + 1)/(3^x x)\bigr)^{\tan(\pi x/2)}$ I have to calculate limit
$$\lim_{x\to 1 } \left(\frac{2^x x + 1}{3^x x}\right)^{\tan(\frac{\pi x}{2})}.$$
I know $\tan(\frac{\pi x}{2})$ is undefined in $x = 1$, but can I just put $x = 1$ into $\frac{x\cdot 2^x + 1}{x\cdot3^x}$ and get
$$\lim_{x\to 1 } (1)^{\tan(\frac{\pi x}{2})} = 1.$$
Is the answer $1$ correct?
It's forbidden to use L'Hôpital's rule.
|
Let $$y = \left( \frac{x 2^x + 1}{x 3^x}\right)^{\tan\left(\frac{\pi x}{2}\right)}$$
Then $$\ln(y) = \tan\left(\frac{\pi x}{2}\right)\ln\left( \frac{x 2^x + 1}{x 3^x}\right) = \frac{\ln\left( \frac{x 2^x + 1}{x 3^x}\right)}{\frac{1}{\tan\left(\frac{\pi x}{2}\right)}}$$
Apply L'Hôpital's rule to obtain your answer, it is still a lot of work. Maybe there is a faster way to go about this problem.
|
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|
Least squares and QR factorization I have a full-column-rank matrix $A \in \mathbb{R}^{N \times n} $ ($N >> n$):
$Q^{T} A = \begin{bmatrix}
R & w \\
0 & v \\
\end{bmatrix} , Q^{T} = \begin{bmatrix}
c \\
d \\
\end{bmatrix} $,
with $R \in \mathbb{R}^{(n-1)\times(n-1)}, w \in \mathbb{R}^{n-1}, v \in \mathbb{R}^{N-n+1}, c \in \mathbb{R}^{n-1}$ and $d \in \mathbb{R}^{N-n+1}$
Now I have to show that
$ \min_x ||Ax - b||^²_2 = ||d||^2_2 - (\frac{v^T d}{||v||_2})^2$
Anyone a idea how to approach this? I don't know how to start with this problem.
What I know is that with a standard least squares problem $\min_x ||y - Fx||_2^2$ with a QR factorization of $F$ and the application of $Q^T$ to $y$
you have the following solution
$\hat{x} = R^{-1} d_1$ and $||y-Fx||_2^2 = ||d_2||_2^2$
with
$
\begin{bmatrix}
Q_1^T \\
Q_2^T \\
\end{bmatrix} y = \begin{bmatrix}
d1 \\
d2 \\
\end{bmatrix}$
How to rewrite things to have a format more like this?
|
$x \in R^n$ must be true, otherwise the matrices cannot be subtracted form each other.
\begin{equation}
\begin{split}
\displaystyle \min_{x} ||Ax - b||_2^2 = \displaystyle \min_{x} ||Q^T(Ax - b)||_2^2 \\
= \displaystyle \min_{x} ||Q^TAx - Q^Tb||_2^2 \\
= \displaystyle \min_{x} || \begin{pmatrix} Rx & wx \\ 0 & vx \end{pmatrix} - \begin{pmatrix} c \\ d \end{pmatrix} ||_2^2 \\
\end{split}
\end{equation}
This can be written as:
\begin{multline}
\displaystyle \min_{x} || \begin{pmatrix} Rx & wx \\ 0 & vx \end{pmatrix} - \begin{pmatrix} 0 & 0 \\ 0 & vx \end{pmatrix} - \begin{pmatrix} c \\ d \end{pmatrix} ||_2^2 - \displaystyle \min_{x} || \begin{pmatrix} 0 & 0 \\ 0 & vx \end{pmatrix} - \begin{pmatrix} c \\ d \end{pmatrix} ||_2^2 \\
\end{multline}
Solving the first part of equation 2 following LSQR-Theorom :
\begin{equation}
\begin{split}
\displaystyle \min_{x} || \begin{pmatrix} Rx & wx \\ 0 & vx \end{pmatrix} - \begin{pmatrix} 0 & 0 \\ 0 & vx \end{pmatrix} - \begin{pmatrix} c \\ d \end{pmatrix} ||_2^2 \\
= \displaystyle \min_{x} || \begin{pmatrix} Rx & wx \\ 0 & 0 \end{pmatrix} - \begin{pmatrix} c \\ d \end{pmatrix} ||_2^2 \\
= ||d||_2^2
\end{split}
\end{equation}
Than solving the second part of equation 2 following LS-Theorom :
\begin{equation}
\begin{split}
\displaystyle \min_{x} || \begin{pmatrix} 0 & 0 \\ 0 & vx \end{pmatrix} - \begin{pmatrix} c \\ d \end{pmatrix} ||_2^2 \\
= \displaystyle \min_{x} || vx - d||_2^2 \\
= ((v^Tv)^{-1}v^td)^2 = \left(\frac{v^Td}{||v||_2}\right)^2
\end{split}
\end{equation}
Substituting equation 3 and 4 into equation 2 results in the asked answer.
\begin{equation}
\displaystyle \min_{x} ||Ax - b||_2^2 = ||d||_2^2 - \left(\frac{v^Td}{||v||_2}\right)^2
\end{equation}
Any chance you are a stundent at TU delft following filtering and identification? Since I had the same exact exercise as homework :).
This is what i think works, not the best explanation. It's still fuzzy in my own head. Plus I have doubts about it being $\left(\frac{v^Td}{||v||_2}\right)^2$ because I would expect it being $\frac{v^Td}{||v||_2^2}$. Maybe you have any ideas on that! goodluck and let me know if you think it can be written out better.
|
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|
A new approach to find value of $x^2+\frac{1}{x^2}$ When I was teaching in a college class, I write this question on board.
If we now $\bf{x+\dfrac{1}{x}=4}$ show the value of $\bf{x^2+\dfrac{1}{x^2}=14}$
Some student asks me for a multi idea to show or prove that.
I tried these : $$ $$1 :$$\left(x+\frac{1}{x}\right)^2=4^2\\x^2+\frac{1}{x^2}+2x\times \frac{1}{x}=16\\x^2+\frac{1}{x^2}=16-2 $$
2:solving quadratic equation ,and putting one of roots$$x+\frac{1}{x}=4\\\frac{x^2+1}{x}=4\\x^2+1=4x\\x^2-4x+1=0\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-(-4)\pm \sqrt{(-4)^2-4(1)(+1)}}{2}=\\x=\frac{4\pm \sqrt{12}}{2}=2\pm \sqrt{3}\\x=2+\sqrt3 \to x^2=4+4\sqrt3+3=7+4\sqrt3\\x^2+\frac{1}{x^2}=7+4\sqrt3+\frac{1}{7+4\sqrt3}=\\7+4\sqrt3+\frac{1}{7+4\sqrt3}\cdot\frac{7-4\sqrt3}{7-4\sqrt3}=\\7+4\sqrt3+\frac{7-4\sqrt3}{49-16\cdot 3}=\\7+4\sqrt3+\frac{7-4\sqrt3}{1}=14$$
3: visual approach .assume side length of a square is $x+\frac{1}{x}=4$
now I am looking for new idea to proof.Any hint will be appreciated.(more visual proof - geometrical - trigonometrical - using complex numbers ...)
|
You can simplify this by noting $$x^2+\frac1{x^2}=x^2+x^{-2}=e^{2\ln(x)}+e^{-2\ln(x)}=\cos(2i\ln(x))$$
Now you can solve for $x$ very easily.
$$x^2+\frac1{x^2}=y$$$$\cos(2i\ln(x))=y$$$$x=e^{\frac{\arccos(y)}{2i}}$$
|
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|
Mathematics Olympiad Question $a+b+c=7$, ... Given $a+b+c=7$ and $\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} = 0.7$, need to find $\frac{c}{a+b} + \frac{a}{b+c} + \frac{b}{a+c}$.
I have noted that these two differ by a factor of $10$. So I divided the first equation by $10$ and equated the two. But that did not lead me anywhere. I have also tried to multiply one by the other, but the result obtained is identical to just multiplying the second expression by $7$.
|
$$\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{a+c}$$
$$=\frac{7-(a+b)}{a+b}+\frac{7-(b+c)}{b+c}+\frac{7-(a+c)}{a+c}$$
$$=\frac{7}{a+b}+\frac{7}{b+c}+\frac{7}{a+c}-3$$
$$=7\cdot 0.7-3=1.9$$
|
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|
put the following in standard form how do I compute the following by putting it in standad form?
\begin{equation}
(2-2\sqrt{3i})^{20}
\end{equation}
what I have tried to do is use de Moivre theorem
but that requires me to put it in polar form, as in z=...
but I don't know how to do that either.
|
Let $2 - 2\sqrt 3 i = 4(\frac 1 2 - \frac {\sqrt 3} 2 i)$.
Now a neat thing happens when we raise $(\frac 1 2 - \frac {\sqrt 3} 2 i)$ to powers.
$(\frac 1 2 - \frac {\sqrt 3} 2 i)^2 = \frac 1 4 - \frac 3 4 - \frac {\sqrt 3} 2 i = - \frac 1 2 - \frac {\sqrt 3} 2 i$
So $(\frac 1 2 - \frac {\sqrt 3} 2 i)^3 = (- \frac 1 2 - \frac {\sqrt 3} 2 i)(\frac 1 2 - \frac {\sqrt 3} 2 i) = - (\frac 1 2)^2 + (\frac {\sqrt 3} 2 i)^2 = - \frac 1 4 - \frac 3 4 = -1$
$(\frac 1 2 - \frac {\sqrt 3} 2 i)^{20} = ((\frac 1 2 - \frac {\sqrt 3} 2 i)^3)^6*(\frac 1 2 - \frac {\sqrt 3} 2 i)^2 = 1*(- \frac 1 2 - \frac {\sqrt 3} 2 i) = - \frac 1 2 - \frac {\sqrt 3} 2 i$
======
So $(2 - 2\sqrt 3 i)^{20} = 4^{20}(\frac 1 2 - \frac {\sqrt 3} 2 i)^{20} = 2^{40}(- \frac 1 2 - \frac {\sqrt 3} 2 i) = -2^{39}(1 + \sqrt 3 i)$
+++++++++++++++
And we all could have saved ourselves a lot of time is we recognize $(1/2, -\sqrt3 /2)$ as the height and base of a 30-60-90 triangle and thus the number is a 6th root of 1 and a cube root of -1.
|
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|
quadratic equation problem - proving a statement I was given that $ax^2+2bx+c=0$
Using $y=x+\frac{1}{x}$ I need to prove that $acy^2+2b(c+a)y+(a-c)^2+4b^2=0$
Tried to make the pattern $x+\frac{1}{x}$ and to substitute $y$, but couldn't prove it.
|
Given that $y = x + \frac1x,$ if you make this substitution for $y$ in
$acy^2+2b(c+a)y+(a-c)^2+4b^2,$ you get
$$ac\left(x + \frac1x\right)^2
+ 2b(c+a)\left(x + \frac1x\right) + (a-c)^2 + 4b^2.$$
Use the binomial theorem, multiplication of polynomials, the distributive law,
or any other methods you know in order to convert this to an expression
with no parentheses. Then look for collections of terms within that expression
that have factors of $ax^2 + 2bx + c.$
Here's a really big hint: since $ax^2 + 2bx + c = 0,$
it follows that
$$\left(c + \frac{a}{x^2} + \frac{2b}{x}\right)(ax^2 + 2bx + c) = 0.$$
|
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|
Summation $1-4x+9x^2-16x^3 + \cdots ?$ Taylor series gives $$\frac 1 {(1+x)^2}=1-2x+3x^2-4x^3+\cdots$$
is there a nice expression for $1-4x+9x^2-16x^3 + \cdots ?$
It would be helpful for a problem I am trying to solve.
|
For $|x| < 1$, we have
$$\begin{align}
\sum_{n=0}^\infty (n+1)^2 (-x)^n
&= \sum_{n=0}^\infty \left(x\frac{d}{dx} + 1 \right)^2 (-x)^n
= \left(x\frac{d}{dx} + 1 \right)^2 \sum_{n=0}^\infty (-x)^n\\
&= \left(x\frac{d}{dx} + 1 \right)^2 \frac{1}{1+x}
= \frac{d}{dx}\left[ x \frac{d}{dx}\left(\frac{x}{1+x}\right)\right]
= \frac{d}{dx}\left[\frac{x}{(1+x)^2}\right]\\
&= \frac{1-x}{(1+x)^3}
\end{align}
$$
|
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|
Write $1, 2, \dots, n^2$ into a $n \times n$ square grid such that sum of each row and column is a power of 2 Let $n$ be a positive integer. Show that one cannot fill a $n \times n$ square grid with numbers $1, 2, \dots, n^2$ such that sum of numbers on each row and column is a power of 2.
My attempt: Assume the contrary. Let the sum of the columns be $2^{a_i}, 1\leq i \leq n$ and the sum of the rows be $2^{b_i}, 1 \leq i \leq n$. Observe that $n$ must be even (otherwise we have sum of all the numbers on the grid is $\dfrac{n^2(n^2+1)}{2}$ not divisible by 2). Then from the identity $\displaystyle\sum_{i=1}^n2^{a_i} = \sum_{i=1}^n2^{b_i}$ I think of the binary representation of this number, but it doesn't seem to work since the $a_i$'s are not necessarily distinct. Then I got stucked from here.
|
The sum is $\frac{n^2(n^2+1)}{2}$. For $n\gt 1$, $n$ odd is impossible.
Let $2^k$ be the highest power of $2$ that divides $n$. Then the highest power of $2$ that divides our sum is $2^{2k-1}$, which is $\le \frac{n^2}{2}$.
Consider the smallest row or column sum. Say it is a row sum, and is equal to $2^l$. Then all row sums are $\ge 2^l$, and powers of $2$. So the sum of the row sums, that is, $1+2+\cdots +n^2$, is congruent to $0$ modulo $2^l$.
The smallest possible row sum is $\ge 1+2+\cdots +n$, which is $\frac{n(n+1)}{2}$. This is greater than $\frac{n^2}{2}$, contradicting the fact that the highest power of $2$ that divides $1+2+\cdots +n^2$ is $\le \frac{n^2}{2}$.
|
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|
Exponential generating function and number of balls
Use exponential generating functions to determine the number $a_n$ of ordered choices of $n$ balls such that there are $2$ or $4$ red balls, an even number of green balls, and an arbitrary number of blue balls.
If we denote red with $r$, green with $g$, and blue with $b$. I guess we will have $r^2 + r^4$ for red balls, $1 + b + b^2 + \ldots = \frac{b}{1-b}$ for blue balls. But how can I express the green balls, and how can I then solve the problem using exponential generating functions?
|
Exponential generating functions are useful for this sort of problem because we can find the answer by multiplying the EGF for each of the types of balls:
$$\color{red}{\left(\frac{x^2}{2!}+\frac{x^4}{4!}\right)}\color{green}{\frac12\left(e^x+e^{-x}\right)}\color{blue}{e^x}.$$
This expands to
$$\frac12\cdot\frac{x^2}{2!} + \frac12\cdot\frac{x^4}{4!} + \sum_{n=2}^\infty \frac1{16}2^nn(n-1)\frac{x^n}{n!} + \sum_{n=4}^\infty\frac1{768}2^nn(n-1)(n-2)(n-3)\frac{x^n}{n!}. $$
Simplification yields
$$\frac{x^2}{2!}+3\frac{x^3}{3!}+13\frac{x^4}{4!} + \sum_{n=5}^\infty \frac1{768} 2^nn(n-1)(n^2-5n+54)\frac{x^n}{n!}. $$
Hence
$$
a_n =\begin{cases}
1,& n=2\\
3,& n=2\\
13,& n=4\\
\frac1{768} 2^nn(n-1)(n^2-5n+54),& n\geqslant 5.
\end{cases}
$$
|
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|
Easy way of Compute this limit Easy way to compute: $$\lim_{x\to \:0\:}\left(\left(\frac{a^x-x\cdot \ln\left(a\right)}{b^x-x\cdot \ln\left(b\right)}\right)^{\frac{1}{x^2}}\right)$$
|
For any $a,b > 0$ and as $x \to 0$:
$\def\l{\!\left}$
$\def\r{\right}$
$a^x = \exp(x\ln(a)) \in 1 + x\ln(a) + \frac{1}{2}(x\ln(a))^2 + o(x^2)$.
$b^x = \exp(x\ln(b)) \in 1 + x\ln(b) + \frac{1}{2}(x\ln(b))^2 + o(x^2)$.
$\l( \dfrac{a^x - x\ln(a)}{b^x - x\ln(b)} \r)^\dfrac{1}{x^2} = \exp\l( \dfrac{1}{x^2}\ln\l( \dfrac{a^x - x\ln(a)}{b^x - x\ln(b)} \r) \r)$
$ \in \exp\l( \dfrac{1}{x^2}\ln\l( \dfrac{ 1 + \frac{1}{2}\ln(a)^2 x^2 + o(x^2) }{ 1 + \frac{1}{2}\ln(b)^2 x^2 + o(x^2) } \r) \r)$
$ \subseteq \exp\l( \dfrac{1}{x^2}\ln\l( 1 + \frac{1}{2}(\ln(a)^2-\ln(b)^2) x^2 + o(x^2) \r) \r)$
$ \subseteq \exp\l( \dfrac{1}{x^2} \l( \frac{1}{2}(\ln(a)^2-\ln(b)^2) x^2 + o(x^2) \r) \r)$
$ \subseteq \exp\l( \frac{1}{2}(\ln(a)^2-\ln(b)^2) + o(1) \r) = \exp\l( \frac{1}{2}(\ln(a)^2-\ln(b)^2) \r) ( 1 + o(1) )$
$ \to \exp\l( \frac{1}{2}(\ln(a)^2-\ln(b)^2) \r)$.
|
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|
How to solve the following pde? How to solve the following PDE?
For an arbitrary continuously differentiable function $f$ , which of the following is a general solution of $\;$
$z(px-qy)=y^2-x^2$?
1)$\;$ $x^2+y^2+z^2=f(xy)$
2)$\;$ $(x+y)^2+z^2=f(xy)$
3)$\;$ $x^2+y^2+z^2=f(y-x)$
4)$\;$ $x^2+y^2+z^2=f((x+y)^2+z^2)$
Here options $1)$$\;$ $2)$ and $ 4)$ are correct
but I am getting only first option as answer.
Is there any general method to solve such pdes?
Please help because I don't have any teacher who can help me and I am learing pde without any teacher.
Thank you very much for giving me your precious time.
|
$z(xz_x-yz_y)=y^2-x^2$
$xz_x-yz_y=\dfrac{y^2-x^2}{z}$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dx}{dt}=x$ , letting $x(0)=1$ , we have $x=e^t$
$\dfrac{dy}{dt}=-y$ , letting $y(0)=y_0$ , we have $y=y_0e^{-t}=\dfrac{y_0}{x}$
$\dfrac{dz}{dt}=\dfrac{y^2-x^2}{z}=\dfrac{y_0^2e^{-2t}-e^{2t}}{z}$ , we have $z^2=f(y_0)-y_0^2e^{-2t}-e^{2t}=f(xy)-y^2-x^2$ , i.e. $x^2+y^2+z^2=f(xy)$
$\therefore$ 1) is obviously correct.
But 2) is in fact also correct since $(x+y)^2=x^2+2xy+y^2$ .
From 2), $g((x+y)^2+z^2)=xy$
$\therefore x^2+y^2+z^2=f(g((x+y)^2+z^2))$ , equivalent to $x^2+y^2+z^2=f((x+y)^2+z^2)$
Hence 4) is in fact also correct.
See rule 6 for further details.
|
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|
Prove congruence rule Say you have integers $a$ and $b$
If $a \equiv b \pmod 5$, we know that $a \pmod 5 = b \pmod 5$.
Let $a \pmod 5 = c$ and $x,y$ be some whole numbers satisfying
$$a = 5x + c \quad \text{and} \quad b = 5y + c$$
Then, $a-b = 5x + c -(5y + c) = 5(x-y)$, so $5$ is a factor in $a-b$.
Now my question is, if we start with only knowing that $5$ is a factor in $a-b$, how can we show that $a \pmod 5 = b \pmod 5$?
|
You correctly showed that $a\bmod 5=b\bmod 5\implies 5\mid a-b$ and now want to prove $5\mid a-b\implies a\bmod 5=b\bmod 5$.
Let $a=5x+(a\bmod 5)$, $b=5y+(b\bmod 5)$. If $5\mid a-b$, then $$5\mid 5(x-y)+(a\bmod 5-b\bmod 5)\iff 5\mid a\bmod 5-b\bmod 5$$
But $a\bmod 5,b\bmod 5\in\{0,1,2,3,4\}$, so $a\bmod 5=b\bmod 5$.
|
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"url": "https://math.stackexchange.com/questions/1555340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Integral of $\frac{1}{x^2+4}$ Different approach underneath is a brief method of partial fractions integration on the problem given in the title
Using a standard trigonometric result it is known that:
$$ \int \frac{1}{x^2+4}dx=\frac{1}{2}\tan^{-1}(\frac{x}{2})+C$$
But also:
$$\frac{1}{x^2+4}=\frac{A}{x+2i}+\frac{B}{x-2i}$$
Hence using partial fractions,
$$1=A(x-2i)+B(x+2i)$$
Let $x=2i$
$$\therefore 1=4Bi$$
$$\text{Hence } B=\frac{-i}{4}$$
Let $x=-2i$
$$\therefore 1=-4Ai$$
$$\text{Hence } A=\frac{i}{4}$$
Hence,
$$\frac{1}{x^2+4}=\frac{i}{4(x+2i)}-\frac{i}{4(x-2i)}$$
Or,
$$\frac{1}{x^2+4}=\frac{i}{4}(\frac{1}{x+2i}-\frac{i}{x-2i})$$
Hence, it is quite easy to see that:
\begin{align*}
\int \frac{1}{x^2+4}dx &=\int \frac{i}{4}(\frac{1}{x+2i}-\frac{1}{x-2i}) dx\\
&=\frac{i}{4}\int \frac{1}{x+2i}-\frac{1}{x-2i} dx\\
&=\frac{i}{4}(\log(x+2i)-\log(x-2i))+C\\
\end{align*}
We know that the principal value of log of a complex number can be calculated by the following formula:
$$\log(x+yi)=\log(x^2+y^2)+arg(x+yi)$$
Hence,
\begin{align*}
\int \frac{1}{x^2+4}dx &=\frac{i}{4}(\log(x+2i)-\log(x-2i))+C\\
&=\frac{i}{4}(\log(x^2+4)+\arg(x+2i)-\log(x^2+4)-\arg(x-2i)+C\\
&=\frac{i}{4}(\arg(x+2i)-\arg(x-2i))+C\\
\end{align*}
Now, consider cases. If x is bigger than 0, the argument of a complex number is always defined as $\arg(x+yi)=\tan^{-1}(\frac{y}{x})$
So our integral becomes
$$\int \frac{1}{x^2+4}dx=\frac{i}{4}(\tan^{-1}(\frac{2}{x})-\tan^{-1}(\frac{-2}{x}))+C$$
And since arctan is an odd function
\begin{align*}
\int \frac{1}{x^2+4}dx&=\frac{i}{4}(2\tan^{-1}(\frac{2}{x}))+C\\&=\frac{i}{2}\tan^{-1}(\frac{2}{x})+C\\
\end{align*}
Now, if $x$ is less than $0$ and $y$ is less than $0$ then argument of a complex number becomes $\tan^{-1}(\frac{y}{x})-\pi$ and if $x$ is less than $0$ and $y$ is more than $0$ the argument becomes $\tan^{-1}(\frac{y}{x})+\pi$ Also as in a previous case becasue arctan is an odd function, the integral becomes
\begin{align*}
\int \frac{1}{x^2+4}dx&=\frac{i}{4}(\tan^{-1}(\frac{2}{x})-\tan^{-1}(\frac{-2}{x})+2\pi)+C\\
&=\frac{i}{4}(2\tan^{-1}(\frac{2}{x})+2\pi)+C\\
&=\frac{i}{2}\tan^{-1}(\frac{2}{x})+D
\end{align*}
Hence
$$\int \frac{1}{x^2+4}dx=\frac{i}{2}(\tan^{-1}\frac{2}{x}), x\in\mathbb R\ \land x\neq0$$
Now, obviously, it is not the same, as the trig identity, it has a pole at x=0 while the original integral doesn't and most importantly it is not a real number. This was done by me purely for recreational purposes but now it frustrates me.
Can it be done this way? Did I basically did a mistake or maybe I missed something crucial?. thanks in advance.
EDIT
After using the correct definition of principal value of log and nice property about arctan (both given to me by you guys)
we know that
$$\log(x+yi)=\frac{1}{2}\log(x^2+y^2)+i\arg(x+yi)$$
and hence
*** becomes
$$\frac{i}{4}(i(\arg(x+2i)-\arg(x-2i)))$$
=$$\frac{-1}{4}((\arg(x+2i)-\arg(x-2i)))$$
So the answe becomes
$$\int \frac{1}{x^2+4}dx=\frac{-1}{2}\tan^{-1}(\frac{2}{x})+D$$
And since $$\tan^{-1}(\frac{2}{x})=\frac{\pi}{2}-\tan^{-1}(\frac{x}{2})$$
the result is $$\int \frac{1}{x^2+4}dx=\frac{-1}{2}(\frac{\pi}{2}-\tan^{-1}(\frac{x}{2}))+D$$
OR$$\int \frac{1}{x^2+4}dx=\frac{1}{2}\tan^{-1}(\frac{x}{2})+E$$ as required
Thanks :)
|
For those kinds of integrals, for how many ideas one might have, the simplest way to proceed is to collect first the constant you have in the denominator:
$$\frac{1}{x^2 + 4} = \frac{1}{4\cdot \left(\frac{x^2}{4} + 1\right)}$$
Remember about the $\frac{1}{4}$ factor, that you bring out of the integral.
Now use the substitution $y = \frac{x}{2}$ so $\text{d}y = \frac{1}{2}\text{d}x$
Result integration: $\int \frac{1}{y^2 + 1}\text{d}y = \arctan(y)$
The final result will be then
$$\frac{1}{2}\arctan\left(\frac{x}{2}\right)$$
Moreover your expression about the complex logarithm is wrong. The right one is:
$$\boxed{\log(a + ib) = \frac{1}{2}\log(a^2 + b^2) + i\ \text{Arg}(x + ib)}$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
How do I choose a free variable? I have a question regarding the Gaussian method for solving linear equations. I had to solve 2 equations with 3 unknowns and naturally with the elimination process I had 2 variables left. I thought that it didn't matter which variable I chose as a free variable, but apparently after inputting random values after choosing the last variable as the free one, the values didn't match. So is there any method of how to choose a free variable?
\begin{cases}
& 4x_{1} + 2x_{2} + 3x_{3} = -2 \\
& 2x_{1} + 8x_{2} - x_{3} = -10
\end{cases}
\begin{pmatrix}
4 & 2 & 3 & | & -2 \\
2 & 8 & -1 & | & -10
\end{pmatrix}
\begin{pmatrix}
2 & 8 & -1 & | & -10 \\
4 & 2 & 3 & | & -2
\end{pmatrix}
\begin{pmatrix}
2 & 8 & -1 & | & -10 \\
0 & -14 & 5 & | & 18
\end{pmatrix}
$$
-14x_{2} + 5x_{3} = 18
$$
$$
x_{3} = C
$$
$$
x_2 = \frac{18-5C}{-14}
$$
$$
4x_1 = -2 -3C - 2*\frac{18-5C}{-14}
$$
$$
4x_1 = -2 -3C - \frac{18-5C}{-7}
$$
$$
x_1 = \frac{-14-21C-18+5C}{28}
$$
$$
x_1 = \frac{-16C-32}{28} = \frac{-8C-16}{14}
$$
this is what I did, but using x3 as a free variable was not a correct choice, because after inputting a random value, the first equation didn't equal to -2. so I want to find out why do I necessarily have to choose x2 instead of x3.
|
As noted by Bye_World in the comments, in Reduced Row Echelon Form, the matrix specified becomes:
\begin{pmatrix}
1 & 0 & \frac{13}{14} & | & \frac{1}{7} \\
0 & 1 & -\frac{5}{14} & | & -\frac{9}{7}
\end{pmatrix}
You then note that you can use $x_3$ as the free variable. So you get:
$$x_1 = -\frac{13}{14}x_3 + \frac{1}{7}$$
$$x_2 = \frac{5}{14}x_3 + -\frac{9}{7}$$
Hope this helps!
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Finite Element Method for the 1d wave equation I'm solving the 1D wave equation
\begin{equation}
\frac{\partial^2 \eta}{\partial t ^2} - \frac{\partial^2 \eta}{\partial x ^2} = 0
\end{equation}
with boundary conditions
\begin{equation}
\frac{\partial \eta}{\partial x} = 0 \qquad \qquad \text{on} \qquad \qquad x = 0, 1
\end{equation}
using Finite Element Method.
Weak formulation, where $w(x)$ is the weight function, yields:
\begin{equation}
\int_{0}^{1} w(x) \frac{\partial^2 \eta}{\partial t ^2} \mathrm{d}x - \int_{0}^{1} w(x) \frac{\partial^2 \eta}{\partial x ^2} \mathrm{d}x = 0
\end{equation}
\begin{equation}
\int_{0}^{1} w \frac{\partial^2 \eta}{\partial t ^2} \mathrm{d}x
- \int_{0}^{1} \frac{\partial }{\partial x }\left( w \frac{\partial \eta}{\partial x }\right) \mathrm{d}x
+ \int_{0}^{1} \frac{\partial w}{\partial x } \frac{\partial \eta}{\partial x } \mathrm{d}x = 0
\end{equation}
the middle integral vanishes through integration and use of B.C.'s
Expressing $\eta$ and $w$ in terms of basis functions:
\begin{align}
\eta = \eta_j (t) \phi_j(x) \\
w = \phi_i(x)
\end{align}
($\phi_i(x)$ are hat functions in my case) gives
\begin{equation}
\left(\int_{0}^{1} \phi_i \phi_j \mathrm{d}x \right) \frac{d^2 \eta_j}{d t ^2}
+ \left(\int_{0}^{1} \frac{d \phi_i}{d x } \frac{d \phi_j}{d x } \mathrm{d}x \right)\eta_j= 0
\end{equation}
which can be written in matrix form as follows
\begin{equation}
M_{ij} \frac{d^2 \eta_j}{d t ^2} = - S_{ij} \eta_j
\end{equation}
TASK: Introduce an auxiliary variable $p_j = \frac{d \eta_j}{d t }$, using Crank-Nicolson time discretization formulate the final system $A \mathbf{x} = \mathbf{b}$.
Introducing the auxiliary variable gives rise to a system of first order equations in time:
\begin{align*}
\frac{d \eta_j}{d t } &= p_j \\
M_{ij} \frac{d p_j}{d t} &= - S_{ij} \eta_j
\end{align*}
Using C-N time discretization:
\begin{align*}
\frac{\eta_j^{n+1} - \eta_j^{n}}{\Delta t } &= \frac{1}{2} (p_j^{n+1}+p_j^{n} )\\
M_{ij} \frac{p_j^{n+1} - p_j^{n}}{\Delta t } &= -\frac{1}{2} S_{ij} \left(\eta_j^{n+1} + \eta_j^{n} \right)
\end{align*}
Here's where I got stuck. I tried eliminating $\eta_j^{n+1}$ and $p_j^{n+1}$ from right hand sides of above equations and obtained:
\begin{align}
\left[M_{ij} + \left(\frac{\Delta t}{2} \right)^2 S_{ij} \right] p_j^{n+1} &= \left[M_{ij} - \left(\frac{\Delta t}{2} \right)^2 S_{ij} \right] p_j^{n} - \Delta t S_{ij} \eta_j^{n} \\
\left[M_{ij} + \left(\frac{\Delta t}{2} \right)^2 S_{ij} \right] \eta_j^{n+1} &= \left[M_{ij} - \left(\frac{\Delta t}{2} \right)^2 S_{ij} \right] \eta_j^{n} + \Delta t M_{ij} p_j^{n}
\end{align}
Out of these I can form a single matrix equation, or just implement as they are. However, neither approach has worked - the numerical solution does not make much sense.
Questions: Have missed something? Is my method valid? Does anybody have experience/advice on solving these kind of problems numerically?
Thanks.
|
I think I found a way...
Rearranging the Crank-Nicolson formulation as follows:
\begin{align*}
\eta_j^{n+1} - \frac{\Delta t }{2} p_j^{n+1} &= \eta_j^{n} + \frac{\Delta t }{2} p_j^{n} \\
M_{ij} p_j^{n+1} +\frac{\Delta t}{2} S_{ij} \eta_j^{n+1} &= M_{ij} p_j^{n} -\frac{\Delta t}{2} S_{ij} \eta_j^{n}
\end{align*}
and solving the matrix form:
\begin{equation}
\begin{bmatrix}
I_{(N+1)\times(N+1)} & - \frac{\Delta t }{2} I_{(N+1)\times(N+1)} \\
\frac{\Delta t }{2} S & M
\end{bmatrix}
\begin{bmatrix}
\mathbf{\eta}^{n+1} \\
\mathbf{p}^{n+1}
\end{bmatrix} =
\begin{bmatrix}
I_{(N+1)\times(N+1)} & \frac{\Delta t }{2} I_{(N+1)\times(N+1)} \\
-\frac{\Delta t }{2} S & M
\end{bmatrix}
\begin{bmatrix}
\mathbf{\eta}^{n} \\
\mathbf{p}^{n}
\end{bmatrix}
\end{equation}
where $I_{(N+1)\times(N+1)}$ is an $(N+1)\times(N+1)$ identity matrix, produces sensible results!
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
If three distinct integers are chosen at random, show that there will exist two among them, say $a$ and $b$, such that $30 | (a^3b-ab^3)$ Problem:
If three distinct integers are chosen at random, show that there will exist two among them, say $a$ and $b$, such that $30 | (a^3b-ab^3)$
My work:
$a^3b-ab^3=ab(a+b) (a-b)$ and if $30 | (a^3b-ab^3)$, then each of $2,3,5$ divides $(a^3b-ab^3)$ since $2\cdot3\cdot5=30$ and $(2,3,5)=1$
Case for 2:
If 2 divides either $a$ or $b$, then all good.
If not, then $a$ and $b$ are both odd and so $a+b$=sum of odds = even is divisible by 2. Done.
Case for 3:
If 3 divides a or b, good.
Else, let:
*
*$a=3k+1$ and $b=3k+1$. Then, $3|(a-b)$
*$a=3k+1$ and $b=3k+2$. Then, $3|(a+b)$
*$a=3k+2$ and $b=3k+2$. Then, $3|(a-b)$
*$a=3k+2$ and $b=3k+1$. Then, $3|(a+b)$
So far, so good.
Case for 5:
Completely stuck here. Five possible remainders ($0,1,2,3,4$) seem to be impossible to do. Any hints?
Questions:
*
*Need hints for last case.
*Why do we need to pick three integers when we only need to check with two?
|
Let $p,q$ be remainders when $5$ divides $a,b$ respectively. If any of the following holds:
*
*$p=0$
*$q=0$
*$p=q$
*$p+q = 5$
Then, $5|a$, $5|b$, $5|(a-b)$ or $5|(a+b)$ respectively.
So, now check for $(p,q)=$
*
*$(1,2)$ Now, if $c = 5k+1$, then $5|(a-c)$, so repeat the question taking the two numbers $a,c$ noting that these two already answer the case for $5$. If, $c=5k+2$, then $5|(b-c)$, proceed similarly as above. If $c=5k+3$, then $5|(b+c)$, proceed similarly as above. If $c=5k+4$, then $5|(a+c)$, proceed similarly as above. If $c=5k$, then $5|c$ so take the numbers $a,c$ or $b,c$ and proceed similar to above.
*For the remaining tuples, repeat similar to what we did in point 1 just above.
Q.E.D.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Intersection of three period functions Let $f(x)=\frac{1}{2}-|\frac{\sqrt{3}}{2}x-1/2|$ for $x\in [0,\frac{2}{\sqrt{3}}]$ and $f(x+\frac{2}{\sqrt{3}})=f(x)$ for all $x\in\mathbb{R}$.
$g(x)=\frac{1}{2}-|\frac{1}{2}x-1/2|$ for $x\in [0,2]$ and $g(x+2)=g(x)$ for all $x\in\mathbb{R}$.
$h(x)=\frac{1}{2}-|\sqrt{2}x-1/2|$ for $x\in [0,\frac{1}{\sqrt{2}}]$ and $f(x+\frac{1}{\sqrt{2}})=f(x)$ for all $x\in\mathbb{R}$.
Are there real $t>0$ such that $f(t)=g(t)=h(t)$ ?
If it true, find the minimum value of $t$.
Thanks in advance.
|
Let's assume there exists such $t>0$. Then for some $p,q,r\in\mathbb N$,
$$t-\frac2{\sqrt3} p\in\left[0,\frac2{\sqrt3}\right],\quad t-2q\in\left[0,2\right],\quad t-\frac1{\sqrt2}q\in\left[0,\frac1{\sqrt2}\right]$$
then
$$f\left(t-\frac2{\sqrt3} p\right)=g\left(t-2q\right)=h\left(t-\frac1{\sqrt2}r\right)$$
And
\begin{align}
f(x)=\frac{\sqrt3}2x\quad \text{or} \quad 1-\frac{\sqrt3}2x&\rightarrow f\left(t-\frac2{\sqrt3} p\right)=\frac{\sqrt3}2t-p\quad \text{or} \quad1-\frac{\sqrt3}2t+p\\
g(x)=\frac12x\quad \text{or} \quad 1-\frac12x&\rightarrow g\left(t-2q\right)=\frac12t-q\quad \text{or} \quad1-\frac12t+q\\
h(x)=\sqrt2x\quad \text{or} \quad 1-\sqrt2x&\rightarrow h\left(t-\frac1{\sqrt2}r\right)=\sqrt2t-r\quad \text{or} \quad1-\sqrt2t+r
\end{align}
From $f\left(t-\frac2{\sqrt3} p\right)=g\left(t-2q\right)$,
$$t=\frac{2(p-q)}{\sqrt3-1}\quad \text{or}\quad t=\frac{2(p+q+1)}{\sqrt3+1}$$
From $g\left(t-2q\right)=h\left(t-\frac1{\sqrt2}r\right)$,
$$t=\frac{2(r-q)}{2\sqrt2-1}\quad \text{or}\quad t=\frac{2(r+q+1)}{2\sqrt2+1}$$
i) If $t=\frac{2(p-q)}{\sqrt3-1}=\frac{2(r-q)}{2\sqrt2-1}$, and $t\ne0$, then
$\frac{p-q}{r-q}=\frac{\sqrt3-1}{2\sqrt2-1}$.
ii) If $t=\frac{2(p-q)}{\sqrt3-1}=\frac{2(r+q+1)}{2\sqrt2+1}$, then
$\frac{p-q}{r+q+1}=\frac{\sqrt3-1}{2\sqrt2+1}$.
iii) If $t=\frac{2(p+q+1)}{\sqrt3+1}=\frac{2(r-q)}{2\sqrt2-1}$, and $t\ne0$, then
$\frac{p+q+1}{r-q}=\frac{\sqrt3+1}{2\sqrt2-1}$.
iv) If $t=\frac{2(p+q+1)}{\sqrt3+1}=\frac{2(r+q+1)}{2\sqrt2+1}$, then
$\frac{p+q+1}{r+q+1}=\frac{\sqrt3+1}{2\sqrt2+1}$.
Any of these four cases is not possible because LHS's are rational and RHS's are irrational.
Therefore, such $t$ does not exist.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How can we evaluate the following limit?
How can this problem be solved?
$$
\lim_{(n,r) \rightarrow (\infty, \infty)} \frac{\prod\limits_{k=1}^{r} \left( \sum\limits_{i=1}^{n} i^{2k-1} \right)}{n^{r+1} \prod\limits_{k=1}^{r-1} \left( \sum\limits_{i=1}^{n} i^{2k} \right)}
$$
|
The sum of the $k$ powers of the first $n$ natural numbers is a polynomial of degree $k+1$ with leading coefficient $\frac{1}{k+1}$, that is,
$$
\sum_{i=1}^n i^k
=
\frac{1}{k+1}n^{k+1} + p_k(n)
$$
where $p_k(n) \in O(n^k)$ is a polynomial of degree (at most) $k$.
It follows that
\begin{align}
\frac{
\prod\limits_{k=1}^{r}
\left(\sum\limits_{i=1}^{n} i^{2k-1} \right)
}{
n^{r+1}
\prod\limits_{k=1}^{r-1}
\left(\sum\limits_{i=1}^{n} i^{2k} \right)
}
= {} &
\frac{
\prod\limits_{k=1}^{r}
\left(\frac{1}{2k}n^{2k} + O(n^{2k-1}) \right)
}{
n^{r+1}
\prod\limits_{k=1}^{r-1}
\left(\frac{1}{2k+1}n^{2k+1} + O(n^{2k}) \right)
}
\end{align}
When passing to the limit for $n\to\infty$ only the highest order powers are relevant, so that this simplifies to
\begin{align}
\lim_{(n,r)\to(\infty,\infty)}
\frac{
\prod\limits_{k=1}^{r}
\left(\sum\limits_{i=1}^{n} i^{2k-1} \right)
}{
n^{r+1}
\prod\limits_{k=1}^{r-1}
\left(\sum\limits_{i=1}^{n} i^{2k} \right)
}
= {} &
\lim_{(n,r)\to(\infty,\infty)}
\frac{
\prod\limits_{k=1}^{r}
\frac{1}{2k}n^{2k}
}{
n^{r+1}
\prod\limits_{k=1}^{r-1}
\frac{1}{2k+1}n^{2k+1}
}
\\
= {} &
\lim_{(n,r)\to(\infty,\infty)}
\frac{
\frac{1}{2^r r!}n^{r(r+1)}
}{
n^{r+1}
\frac{2^{r-1}(r-1)!}{(2r-1)!}
n^{r^2-1}
}
\\
= {} &
\lim_{(n,r)\to(\infty,\infty)}
\frac{1}{2^{2r-1}}
\binom{2r-1}{r}
\end{align}
Above, I used the following equalities:
*
*$\quad\displaystyle
\prod_{k=1}^r n^{2k}
=
n^{\sum_{k=1}^{r} 2k}
=
n^{r(r+1)}
$
*similarly, $\quad\displaystyle
\prod_{k=1}^{r-1} n^{2k+1}
=
n^{\sum_{k=1}^{r-1} (2k+1)}
=
n^{r^2-1}
$
*and $\quad\displaystyle
\prod_{k=1}^{r-1}(2k+1)
=
\frac{(2r-1)!}{2^{r-1}(r-1)!}
$
I'll leave the evaluation of the limit to you now... :)
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding a diagonalizing matrix associated with Jordan normal form Find the Jordan normal form $J$ of the upper triangular matrix
$$A = \begin{pmatrix}2 & 0 & 1 & 2 \\ 0 & 2 & 2 & 1 \\ 0 & 0 & 2 & 1\\ 0 &0 & 0& 3 \end{pmatrix}$$
and find a matrix $M$ such that $M^{-1}AM = J$.
Note that the characteristic polynomial is $\det(A- \lambda I) = (2- \lambda)^{3}(3-\lambda)$. We have
$$(A - 2I)(A-3I) = \begin{pmatrix}0 & 0 & -1 & 1 \\ 0 & 0 & 0 & 2 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix} \neq 0_{4 \times 4}$$
and $(A - 2I) \neq 0_{4 \times 4}$ and $(A - 3I) \neq 0_{4 \times 4}$, but $(A-2I)^{2}(A-3I) = 0_{4 \times 4}$. Hence, the minimal polynomial is: $m_{A}(x) = (x-2)^{2}(x -3)$.
It follows that the other invariant factor must be $(x-2)$. This gives us the Jordan form as
$$J = \begin{pmatrix}2 & 0 & 0 & 0 \\ 0 & 2 & 1 & 0 \\ 0 & 0 & 2 & 0\\ 0 &0 & 0& 3 \end{pmatrix}.$$
Now, $v_{4} = \begin{pmatrix} 3 \\ 3 \\ 1 \\ 1 \end{pmatrix}$ is an eigenvector associated with $\lambda = 3$.
We note that for $k \geq 2$, we have
$$B_{k} = (A - 2I)^{k} = \begin{pmatrix}0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 1 \end{pmatrix}.$$
The null space of $B_{k}$ is three-dimensional with the standard basis vectors as a basis for instance. My difficulty stems from not being able to find a seed vector $v_{i}$ in this basis such that I am able to generate the three generalized eigenvectors in the usual manner: $v_{i-1} = (A - 2I)^{i}v_{i}$. I will appreciate a fix to this of course in a generalizable way though.
|
You have found that there are two Jordan blocks for the eigenvalue $2$, so $\dim\ker(A-2I)=2$. There is a $1\times1$-block and a $2\times2$-block, which means that $\dim\ker(A-2I)^2=3$. Now pick a vector $v_3\in\ker(A-2I)^2$ that is not containd in $\ker(A-2I)$. Then $v_2=(A-2I)v_3$ is an eigenvector with eigenvalue $2$, and you can pick another $v_1\in\ker(A-2I)$ to complete your basis.
|
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|
Dividing the squares $1^2,2^2,\ldots,54^2$ into three equal groups with the same total sum Is it possible to divide the squares $1^2,2^2,\ldots,54^2$ into three groups, each of which contains $18$ squares, such that the sum of squares within each group is the same for all three groups?
|
We can partition $9$ consecutive squares $n^2, (n+1)^2, \ldots,(n+8)^2$ in an almost equal fashion:
$$\begin{align}
(n+0)^2+(n+4)^2+(n+8)^2&=3n^2+24n+80\\
(n+1)^2+(n+5)^2+(n+6)^2&=3n^2+24n+62\\
(n+2)^2+(n+3)^2+(n+7)^2&=3n^2+24n+62\end{align} $$
Each group has three members and only one group has a sum too large by $18$.
By rotating the role of the too large partition, we can partition any $27$ consecutive squares into three groups of 9 members each and of equal sum.
As we can do this with $1^2,\ldots 27^2$ as well as $28^2,\ldots, 54^2$, the answer to the problem statement is: yes.
|
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|
exponential difference inequality Im asked to prove the inequality when $0\leq a<b$ and $x>0$:
$$
a^x(b-a)<{b^{x+1}-a^{x+1}\over{x+1}}<b^x(b-a)
$$
So far I have seen that obviously:
$$a^x(b-a)<b^x(b-a)$$
and that
$$b^{x+1}-a^{x+1} = (b-a)(b^x+b^{x-1}a+...+ba^{x-1}+a^x) > a^x(b-a)$$
This has done no good for me yet but it seems related to it.
$$$$
I was thinking it may have to do with $a<{a+b\over{2}}<b$ but I cant see how. If you think a hint may help me out I would prefer that over a straight solution. But any help is appreciated thank you.
|
Take user math110's idea:
$$b^x+b^{x-1}a+\cdots+ba^{x-1}+a^x>a^x+a^x+...+a^x=(x+1)a^x$$
and
$$b^x+b^{x-1}a+\cdots+ba^{x-1}+a^x<b^x+b^x+\cdots+b^x=(x+1)b^x$$
From this and since $0\leq a<b$
$$
(x+1)a^x<(b^x+b^{x-1}a+...+ba^{x-1}+a^x)<(x+1)b^x
$$
Divide by $x+1$ and we get,
$$
a^x<{b^x+b^{x-1}a+...+ba^{x-1}+a^x\over{x+1}}<b^x
$$
Multiply by $b-a$, the middle term becomes,
$$
{(b-a)(b^x+b^{x-1}a+...+ba^{x-1}+a^x)\over{x+1}}={b^{x+1}-a^{x+1}\over{x+1}}
$$
and we have,
$$
a^x(b-a)<{b^{x+1}-a^{x+1}\over{x+1}}<b^x(b-a)
$$
|
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|
Is this a sufficient proof of a math contest problem? Problem:
If a,b,c,d are real, prove that
$$a^2+b^2=2$$
$$c^2+d^2=2$$
$$ac=bd$$
Is true if and only if
$$a^2+c^2=2$$
$$b^2+d^2=2$$
$$ab=cd$$
My proof is as follows:
Note that each of the second set of equations nearly corresponds to the lengths of the sides of a right triangle. The equations may be equivalently written as:
$$\sqrt{a^2+c^2}=\sqrt{2}$$
$$\sqrt{b^2+d^2}=\sqrt{2}$$
$$\frac{c}{a}=\frac{b}{d}$$
It should also be noted that $Arctan(\frac{c}{a})=Arctan(\frac{b}{d})=\theta$. Represented geometrically, the two triangles represented by these equations would look like this:
Note that the two triangles are similar, and that they share the hypotenuse length $\sqrt{2}$. This means that the triangles are congruent, which means that the sides opposite the angle are as well, and the sides adjacent the angle are also congruent.
So one can surmise that $b=c$ and $a=d$. When substituted into the given equations, you get the resulting equations.
This covers the cases of $|a,b,c,d|>0$. If $a=0$, then we know that $b=\sqrt{2}$. If $b=\sqrt{2}$ and $a=0$, then $d\sqrt{2}=0$, so $d=0$. Because $d=0$, we know that $c=\sqrt{2}$.
By symmetry, the same argument for a=0 can apply to b=0 if a and b switch places. Sorry if that's a bit lazy. The remaining arguments are easier, though tedious to prove, and I'm primarily concerned with the geometric argument.
Please let me know if this is sufficient, and if not, where I went wrong. Thank you!
|
The difficulty with the geometric argument is that you cannot actually conclude that $b=c$ and $a=d$. For example, if the second set of equations is true, then you can have $$a = \sqrt{1/2}, c = \sqrt{3/2}, \quad b = -\sqrt{3/2}, d = -\sqrt{1/2},$$ which satisfies the conditions but it is clear that $b \ne c$. This is because your triangles do not distinguish lengths of opposite sign.
A fully rigorous geometric argument would probably be inelegant due to the need to handle such cases where the variables do not have the same sign, or one or more variables is zero. So an algebraic approach may be more appropriate. We would note $(ab)^2 = (cd)^2$ as a result of squaring the third equation; then substituting $a^2 = 2-c^2$ and $d^2 = 2-b^2$, we get $$(2-c^2)b^2 = c^2(2-b^2),$$ or $2b^2 = 2c^2$, or $b^2 = c^2$. Similarly, we easily conclude $a^2 = d^2$. Then using these relationships in the first two equations gives $$a^2 + b^2 = 2, \quad c^2 + d^2 = 2.$$ The last desired condition readily follows since given $ab = cd$ and $b^2 = c^2$, we have $$b(cd) = (ab)b = ab^2 = ac^2,$$ and if $c \ne 0$, cancellation gives $bd = ac$ as desired. If $c = 0$, then $b = 0$ as well, thus $ac = 0 = bd$. This completes the proof.
|
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|
Convergence of an integral $\int_1^{+\infty} \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx$ Convergence of an integral $\int_1^{+\infty} \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx$
$\int_1^{+\infty} \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx=\lim\limits_{t\to\infty}\int_1^{t} \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx$
Partial integration can't solve the integral $\int \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx$.
What substitution (or other methods) would you suggest?
|
The integral $$\int_1^{+\infty} \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx$$ converges.
Indeed $$\sum_{n=1}^\infty {\frac{1}{n\sqrt[3]{n^2+1}}} $$ converges. Indeed $$ {\frac{1}{n\sqrt[3]{n^2+1}}}\sim_{\infty} {\frac{1}{n\sqrt[3]{n^2}}} $$ that converges
|
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|
$\int 1/(x^3-25x) \mathrm{d}x$ Integrate $\displaystyle \frac{1}{x^3-25x}$
Between limits of $4,3$.
Think it involves the 'Difference in squares' substitution.
I ended up with an answer of $0.3242$ which is wrong as it is $0.023$.
Any help and pointers would be greatly appreciated.
Matt
|
John's approach is correct.
$\frac{1}{x^3-25x} = \frac{1}{x(x+5)(x-5)} = \frac{A}{x} + \frac{B}{x+5} + \frac{C}{x-5}$.
Continuing, we generate the following equation: $$A(x+5)(x-5)+Bx(x-5)+Cx(x+5)=1$$
By strategically choosing our $x$ values (I recommend $x=-5,0,$ and $5$), we can derive that $A=\frac {-1}{25}, B= \frac{1}{50},$ and $C= \frac {1}{50}$. Thus, we rewrite:
$\int \frac {1}{x^3-25x}= \int \frac{-1/25}{x} + \frac{1/50}{x+5} + \frac{1/50}{x-5} \mathrm{d} x$.
I suggest breaking the integral into three separate integrals and proceed by setting the denominator of each equal to some dummy variable. Then, all three very nicely fit the form $\int \frac {\mathrm{d} u} {u}$. Can you proceed from here?
|
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|
Find the Area bounded by $y=-x^2+4$, $y=x+2$ I set the two polynomials equal to each other and after multiplying everything by $-1$ I got $x^2+x-2=0$
My points of intersection are $x=1$ and $x=-2$
However, the graph of the polynomial is concave up. Why?
|
Update: Setting your new equations equal yields $x^2+x-2=0$, which factors as $(x+2)(x-1)$. Now, your intersection points are correct. Also, $-x^2+4>x+2$ from $-2$ to $1$. Now, take
$$\int_{-2}^{1} -x^2-4-(x+2) \mathrm{d}x= \frac{-39}{2}$$
|
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|
Limit of $\frac{1-2\cos(x)+\cos^2(2x)}{x^2}$ I tried to find the value of this limit without L'Hopital , but no luck
$$\lim_{x\to 0}\frac{1-2\cos(x)+\cos^2(2x)}{x^2}$$
|
$$\frac{1-2\cos(x)+\cos^2(2x)}{x^2}=\frac{4\sin^2\frac{x}{2}-\sin^2(2x)}{x^2}$$
and since $\lim_{x\to 0}\frac{\sin x}{x}=1$, your limit equals $1-4=\color{red}{-3}$.
|
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|
Computing the Integral $\int \frac{\sqrt{x}}{x^2+x} dx$
Find $\displaystyle \int \dfrac{\sqrt{x}}{x^2+x} dx$.
What would be the best way to integrate this? I saw the answer to this and it looked simple so that might mean the steps would be too?
|
Method 1. Algebraic substitution:
let $\sqrt x=u\implies \frac{dx}{2\sqrt x}=du$ or $dx=2u\ du$, hence $$\int \frac{\sqrt x}{x^2+x}\ dx=\int \frac{u}{u^4+u^2}(2u\ du)=2\int \frac{1}{u^2+1}\ du=2\tan^{-1}(u)+C=2\tan^{-1}(\sqrt x)+C$$
Method 2. Trigonometric substitution:
let $\sqrt x=\tan\alpha\implies \frac{dx}{2\sqrt x}=\sec^2\alpha\ d\alpha$ or $dx=2\tan\alpha\sec^2\alpha \ d\alpha$, hence $$\int \frac{\sqrt x}{x^2+x}\ dx=\int \frac{\tan\alpha}{\tan^4\alpha+tan^2\alpha}(2\tan\alpha\sec^2\alpha \ d\alpha)=2\int \frac{\tan^2\alpha\sec^2\alpha \ d\alpha}{\tan^2\alpha(\tan^2\alpha+1)}$$
$$=2\int \frac{\tan^2\alpha\sec^2\alpha \ d\alpha}{\tan^2\alpha\sec^2\alpha}=\int d\alpha=\alpha+C=2\tan^{-1}(\sqrt x)+C$$
|
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|
Find $ \int \frac{1}{2\sin(x)-3\cos(x)}dx$.
Find $\displaystyle \int \dfrac{1}{2\sin(x)-3\cos(x)}dx$.
My book said to solve this by saying $u = \tan \left(\dfrac{x}{2} \right)$ since $\cos(x) = \dfrac{1-u^2}{1+u^2}$ and $\sin(x) = \dfrac{2u}{1+u^2}$. I don't see how this will help since $du = \dfrac{1}{\cos(x)+1}dx $. How will we get that in the integrand?
|
Notice, $$\int \frac{1}{2\sin x-3\cos x}\ dx$$
$$=\int \frac{1}{2\frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}-3\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}}\ dx$$
$$=\int \frac{1+\tan^2\frac{x}{2}}{3\tan^2\frac{x}{2}+4\tan\frac{x}{2}-3}\ dx$$
$$=\frac13\int \frac{\sec^2\frac{x}{2}}{\tan^2\frac{x}{2}+\frac{4}{3}\tan\frac{x}{2}-1}\ dx$$
$$=\frac23\int \frac{d\left(\tan\frac{x}{2}\right)}{\left(\tan\frac{x}{2}+\frac23\right)^2-\left(\frac{\sqrt{13}}{3}\right)^2}\ dx$$
$$=\frac{2}{3\times 2\frac{\sqrt {13}}{3}}\ln\left|\frac{\tan\frac{x}{2}+\frac23-\frac{\sqrt{13}}{3}}{\tan\frac{x}{2}+\frac23+\frac{\sqrt{13}}{3}}\right|+C$$
$$=\frac{1}{\sqrt {13}}\ln\left|\frac{3\tan\frac{x}{2}+2-\sqrt{13}}{3\tan\frac{x}{2}+2+\sqrt{13}}\right|+C$$
|
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|
I want to calculate the limit of $\lim_{x \to +\infty} (\sqrt{(x+a)(x+b)}-x)$ I want to calculate the limit of: $$\lim_{x \to +\infty} (\sqrt{(x+a)(x+b)}-x) $$ $$(a,b ∈ R)$$
Now I know the result is $\frac{1}{2}(a+b)$, but I am having trouble getting to it.
|
Notice, $$\lim_{x\to \infty}\left(\sqrt{(x+a)(x+b)}-x\right)$$
$$=\lim_{x\to +\infty}\frac{\left(\sqrt{(x+a)(x+b)}-x\right)\left(\sqrt{(x+a)(x+b)}+x\right)}{\left(\sqrt{(x+a)(x+b)}+x\right)}$$
$$=\lim_{x\to +\infty}\frac{x^2+(a+b)x+ab-x^2}{\left(\sqrt{(x+a)(x+b)}+x\right)}$$
$$=\lim_{x\to +\infty}\frac{(a+b)+\frac{ab}{x}}{\left(\sqrt{\left(1+\frac{a}{x}\right)\left(1+\frac{a}{x}\right)}+1\right)}$$
$$=\frac{a+b+0}{\left(\sqrt{\left(1+0\right)\left(1+0\right)}+1\right)}$$
$$=\frac{a+b}{1+1}=\frac{a+b}{2}$$
|
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|
Evaluate the Integral: $\int\frac{dx}{\sqrt{x^2+16}}$ I want to evaluate $\int\frac{dx}{\sqrt{x^2+16}}$.
My answer is: $\ln \left| \frac{4+x}{4}+\frac{x}{4} \right|+C$
My work is attached:
|
I prefer to use this substitution
$$\begin{align}
x &= 4 \sinh u \\
dx &= 4 \cosh u \, du
\end{align}$$
and hence the integral becomes
$$\begin{align}
I &= \int \frac{1}{\sqrt{16(1+\sinh^2u)}} 4 \cosh u du\\
&= \int \frac{4 \cosh u}{4\sqrt{\cosh^2u}} du \\
&= \int 1 \, du \\
&= u+C \\
&= \sinh^{-1} \frac{x}{4} + C\\
&= \ln \left( \sqrt{(1+(\frac{x}{4})^2}+\frac{x}{4} \right) + C
\end{align}$$
|
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|
Is there a name for a binomial expansion without coefficients? I am investigating a problem from George E. Andrews Number Theory (Dover, 1971), discussed previously here:
Induction Proof that $x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\ldots+xy^{n-2}+y^{n-1})$
I was led astray for a bit, because I misunderstood the rhs to be the difference of $x$ and $y$ times the binomial expansion $(x+y)^{n-1}$.
But this is wrong... the rhs is actually the difference of $x$ and $y$ times the binomial expansion $(x+y)^{n-1}$ "without the coeffients", i.e.,
$$\sum_{i=0}^{n-1}x^{n-1-i}y^i$$
(also briefly discussed here: Binomial summation without coefficient)
Is there a name for this summation?
|
There is no specific name for the bivariate polynomial
\begin{align*}
x^{n-1}+x^{n-2}y+x^{n-3}y^2+\cdots+x^2y^{n-3}+xy^{n-2}+y^{n-1}
\end{align*}
but the nice expression
\begin{align*}
(x-&y)(x^{n-1}+x^{n-2}y+x^{n-3}y^2+\cdots+x^2y^{n-3}+xy^{n-2}+y^{n-1})\\
\end{align*}
is called a telescoping sum since all terms besides the first and the last cancel out.
\begin{align*}
(x-&y)(x^{n-1}+x^{n-2}y+x^{n-3}y^2+\cdots+x^2y^{n-3}+xy^{n-2}+y^{n-1})\\
&=x(x^{n-1}+x^{n-2}y+x^{n-3}y^2+\cdots+x^2y^{n-3}+xy^{n-2}+y^{n-1})\\
&\qquad-y(x^{n-1}+x^{n-2}y+x^{n-3}y^2+\cdots+x^2y^{n-3}+xy^{n-2}+y^{n-1})\\
&=\color{blue}{x^{n}}+x^{n-1}y+x^{n-2}y^2+\cdots+x^3y^{n-3}+x^2y^{n-2}+xy^{n-1}\\
&\qquad\ \;-x^{n-1}y-x^{n-2}y^2-\cdots-x^3y^{n-3}-x^2y^{n-2}-xy^{n-1}\color{blue}{-y^{n}}\\
&=\color{blue}{x^n-y^n}\\
\end{align*}
|
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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.