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Let $A$ be a complex $2$ by $2$ matrix having distinct eigenvalues $a, b$. Show that $A^n =\frac{ a^n}{a - b}(A - bI) + \frac{b^n}{b - a}(A - aI)$.
Let $A\in\mathscr{M}_{2\times 2}(\mathbb{C})$ be a matrix having distinct eigenvalues $a\neq b$. Show that, for all $n > 0$,
\begin{equation*}
A^n =\frac{ a^n}{a - b}(A - bI) + \frac{b^n}{b - a}(A - aI).
\end{equation*} (Exercise 707 from Golan, The Linear Algebra a Beginning Graduate Student Ought to Know.)
I proved it by induction, under the assumption that it is true for $A,A^2$,
\begin{equation*}
A^n =\frac{ a^n}{a - b}(A - bI) + \frac{b^n}{b - a}(A - aI)=\frac{(a^n-b^n)A+(ab^n-a^nb)I}{a-b}\\\Rightarrow A^{n+1}=A\frac{(a^n-b^n)A+(ab^n-a^nb)I}{a-b}=\frac{(a^n-b^n)A^2+(ab^n-a^nb)A}{a-b}\cdots
\end{equation*}
but I can't prove it for $A^2$.
I'm also wondering if there is another way to prove it by induction just starting with $A$ or a straight way.
Thanks.
|
i think you can do this with cayley-hamilton theorem which says that a matrix satisfies its characteristic equation. that is $$A^2 -(a+b)A + abI = 0 $$ or $$A^2 = (a+b)A - abI$$ we can use this to write every power of $A$ as a lineay combination of $A, I$. we will the division algorithm to find the remainder. suppose $$x^n = q(x)(x-a)(x-b) + \alpha x + \beta $$ putting $x = a, x = b$ gives $$\pmatrix{a^n\\b^n} = \pmatrix{a &1\\b&1}\pmatrix{\alpha\\\beta} \to \pmatrix{\alpha\\\beta}=\frac{1}{a-b} \pmatrix{1 &-1\\-b&a}\pmatrix{a^n\\b^n}$$
and $$A^n = \frac{1}{a-b}\left(\left(a^n - b^n\right)A + \left(ab^n-a^nb\right)\right)= \frac{1}{a-b}\left(a^n\left(A- bI\right) +b^n \left(aI-A\right)\right).$$
|
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|
Solving a Radical Equation $5(\sqrt{1-x} + \sqrt{1+x}) = 6x + 8\sqrt{1-x^2}$ (squaring doesn't help) How should I approach this problem:
$$ 5(\sqrt{1-x} + \sqrt{1+x}) = 6x + 8\sqrt{1-x^2} $$
I've tried squaring both sides but to get rid of all the radicals requires turning it into a quartic equation, which I don't know how to solve?
Any tips? Thanks!
|
HINT:
Method$\#1:$
Let $2y=\arccos x\implies x=\cos2y,\sqrt{1-x^2}=+|\sin2y|$
As for real $a,\sqrt a\ge0,$
Using the definition of Principal values, $0\le2y\le\pi\implies\sqrt{1-x^2}=+\sin2y$
Again, $0\le2y\le\pi\iff0\le y\le\dfrac\pi2\implies\sin y,\cos y\ge0$
$\implies\sqrt{1-x}=+\sqrt2\sin y,\sqrt{1+x}=+\sqrt2\cos y$
Method$\#2:$
Let $x=\cos2y$
$\sqrt{1-x^2}=|\sin2y|$
$1-x=2\sin^2y\implies\sqrt{1-x}=\sqrt2|\sin y|$ and simialrly $\sqrt{1+x}=\sqrt2|\cos y|$
Now for real $a, |a|=+a$ if $a\ge0,$ else $-a$
Case $\#1:$
$\sin y,\cos y\ge0\implies\sin2y=2\sin y\cos y\ge0$
Case $\#2:$
$\sin y<0,\cos y<0\implies\sin2y=\cdots>0$
Case $\#3:$
$\sin y>0,\cos y<0\implies\sin2y=\cdots$
Case $\#4:$
$\sin y<0,\cos y>0\implies\sin2y=\cdots$
|
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|
What will be the equation of side $BC$.
The equation of two equal sides $AB$ and $AC$ of an isosceles triangle $ABC$ are $x+y=5$ and $7x-y=3$ respectively . What will be the equation of the side $BC$ if the area of the triangle $\triangle ABC$ is $5$ square units.
$a.)x+3y-1=0\\
b.)x-3y+1=0\\
c.)2x-y-5=0\\
\color{green}{d.)x+2y-5=0}\\$
$\quad\\~\\~\\$
Let the slope of the required line be $m$.
I used the slope formula between two lines
$\angle B=\angle C\\~\\\left|{\dfrac{(-1-m)}{(1-m)}}\right|=\left|{\dfrac{(7-m)}{(1+7m)}}\right|\\
\implies m=-3,\dfrac{1}{3}$
But the book is giving right answer as option $d.)$.
By finding $m$ i only found slope not the whole equation.
Also i would like to know if their is clean simple short way , and also using $Area=5$ .
I have studied maths upto $12th$ grade.
Update : by using Geogebra I found that $x-3y+1=0$ fits perfectly
|
the two two lines that bisects the $\angle BAC$ are given by
$$\frac{7x-y-3}{\sqrt{50}} = \pm\frac{x+y-5}{\sqrt2}.$$ they are
$$x-3y+11 = 0, \quad 3x + y - 7=0 \text{ and } A = (1, 4).$$
we will pick a point $$D = (1+2k, 4-6k), k \text{ to be fixed later} $$ on the angle bisector $3x+y-7 = 0.$ the line through $D$ parallel to the other bisector is $$x - 3y = 1+2k -3(4-6k) = 20k-11.\tag 1$$ the point $C$ is the intersection of $(1)$ and $$7x - y = 3 \tag 2 $$ therefore $$C = (1-k, 4-7k). $$ we can now compute $$AD^2 = 4k^2 + 36k^2 = 40k^2\\
CD^2 = 9k^2 + k^2 =10k^2$$ the constraint that the area of the triangle $ABC = 5$ translates to $$AD^2 \times CD^2 = 25\implies 400k^2 = 25, k = \pm \frac14 $$ the line $BC$ are $$x-3y = -6, x-3y = -16. $$
|
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|
Is my algorithm correct? (Polar decomposition) I cant seem to find my mistake. Consider this matrix $T = $\begin{bmatrix}
2 & 1 & 1 \\[0.3em]
-1 & 2 & 0 \\[0.3em]
0 & 1 & -1
\end{bmatrix}
I need the polar decomposition of $T$ ($T=AS$, with $A$ orthogonal and $S$ symmetrical). So, here are my steps:
*
*Calculate $T^TT=$\begin{bmatrix}
5 & 0 & 2 \\[0.3em]
0 & 6 & 0 \\[0.3em]
2 & 0 & 2
\end{bmatrix}
*Calculate eigenvalues of $T^TT$:
$1,6,6$
*Calculate $S=\sqrt{T^TT}=\left[ \begin {array}{ccc} 1&0&2\\ 0&1&0
\\ -2&0&1\end {array} \right]\left[ \begin {array}{ccc} 1&0&0\\ 0&\sqrt{6}&0
\\ 0&0&\sqrt{6}\end {array} \right]\left[ \begin {array}{ccc} \frac{1}{5}&0&-\frac{2}{5}\\ 0&1&0
\\ \frac{2}{5}&0&\frac{1}{5}\end {array} \right]=$\begin{bmatrix}
\frac{1}{5}+\frac{4}{5}\sqrt{6} & 0 & -\frac{2}{5}+\frac{2}{5}\sqrt{6} \\[0.3em]
0 & \sqrt{6} & 0 \\[0.3em]
-\frac{2}{5}+\frac{2}{5}\sqrt{6} & 0 & \frac{4}{5}+\frac{1}{5}\sqrt{6}
\end{bmatrix}
Now I get a really big matrix for $S^{-1}$ and an even bigger one for $A$. When I put my matrices in the equation $T=AS$ an error occurs.
Can you help me find my mistake(s)?
|
What you have for $S$ is exactly correct. To make finding $S^{-1}$ easier, normalize your columns of your matrix of eigenvectors, so that $S = QDQ^{T}$, where $Q$ is an orthogonal matrix. Then $S^{-1} = QD^{-1}Q^{T}$ and all of the matrices in that product are easy to calculate.
In particular,
$$Q = \left[\begin{array}{ccc}
1/\sqrt{5} & 0 & 2/\sqrt{5}\\
0 & 1 & 0\\
-2/\sqrt{5} & 0 & 1/\sqrt{5}\end{array}\right].$$
Then
$$S^{-1} = \left[\begin{array}{ccc}
1/\sqrt{5} & 0 & 2/\sqrt{5}\\
0 & 1 & 0\\
-2/\sqrt{5} & 0 & 1/\sqrt{5}\end{array}\right]
\left[\begin{array}{ccc}
1 & 0 & 0\\
0 & 1/\sqrt{6} & 0\\
0 & 0 & 1/\sqrt{6}\end{array}\right]
\left[\begin{array}{ccc}
1/\sqrt{5} & 0 & -2/\sqrt{5}\\
0 & 1 & 0\\
2/\sqrt{5} & 0 & 1/\sqrt{5}\end{array}\right].$$
Multiplying out gives that
$$S^{-1} = \left[\begin{array}{ccc}
1/5+4/(5\sqrt{6}) & 0 & -2/5+2/(5\sqrt{6})\\
0 & 1/\sqrt{6} & 0\\
-2/5+2/(5\sqrt{6}) & 0 & 4/5+1/(5\sqrt{6})\end{array}\right].$$
Then since $A = TS^{-1}$,
$$A = \left[\begin{array}{ccc}
2 & 1 & 1\\
-1 & 2 & 0\\
0 & 1 & -1\end{array}\right]
\left[\begin{array}{ccc}
1/5+4/(5\sqrt{6}) & 0 & -2/5+2/(5\sqrt{6})\\
0 & 1/\sqrt{6} & 0\\
-2/5+2/(5\sqrt{6}) & 0 & 4/5+1/(5\sqrt{6})\end{array}\right].$$
Multiplying this out gives that
$$ A =
\left[\begin{array}{ccc}
2/\sqrt{6} & 1/\sqrt{6} & 1/\sqrt{6}\\
-1/5-4/(5\sqrt{6}) & 2/\sqrt{6} & 2/5-2/(5\sqrt{6})\\
2/5-2/(5\sqrt{6}) & 1/\sqrt{6} & -4/5-1/(5\sqrt{6})\end{array}\right].$$
|
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|
How to apply the AM-GM inequality? What is the minimum value of $8x^3+36x+54/x+27/x^3 $ for positive real numbers x?
Express your answer in simplest radical form.
I attempted to make an equation between the product of the terms and the original expression, $ 8*36*54*27 = 8x^3+36x+54/x+27/x^3 $
but it seems more complicated than necessary. Is there a better way to apply the inequality?
|
For numbers $a_1,\ldots,a_n > 0$, the AM-GM inequality is $\dfrac{a_1+a_2+\cdots+a_n}{n} \ge \sqrt[n]{a_1a_2\cdots a_n}$.
Equality occurs iff $a_1 = a_2 = \cdots = a_n$.
The trick here is to apply AM-GM to the right number of terms at a time.
Since $x$ is positive, the $4$ terms $8x^3$, $36x$, $\dfrac{54}{x}$, and $\dfrac{27}{x^3}$ are all positive.
Hence, $\dfrac{8x^3+36x+\tfrac{54}{x}+\tfrac{27}{x^3}}{4} \ge \sqrt[4]{8x^3 \cdot 36x \cdot \tfrac{54}{x} \cdot \tfrac{27}{x^3}}$
For equality to occur, you need $8x^3 = 36x = \dfrac{54}{x} = \dfrac{27}{x^3}$. Unfortunately, there is no such $x$ for which all four are equal.
Instead, let's apply AM-GM to two terms at a time:
$\dfrac{8x^3+\tfrac{27}{x^3}}{2} \ge \sqrt{8x^3 \cdot \tfrac{27}{x^3}} = \sqrt{8 \cdot 27} = 6\sqrt{6} \leadsto 8x^3+\tfrac{27}{x^3} \ge 12\sqrt{6}$
$\dfrac{36x+\tfrac{54}{x}}{2} \ge \sqrt{36x \cdot \tfrac{54}{x}} = \sqrt{36 \cdot 54} = 18\sqrt{6} \leadsto 36x+\tfrac{54}{x} \ge 36\sqrt{6}$
Now, add the two together to get $8x^3+36x+\tfrac{54}{x}+\tfrac{27}{x^3} \ge 48\sqrt{6}$.
Finally, equality occurs iff $8x^3 = \tfrac{27}{x^3}$ and $36x = \tfrac{54}{x}$. There is in fact a positive number $x$ for which both are satisfied (specifically $x = \sqrt{\tfrac{3}{2}}$).
Alternatively, if you notice that $8x^3+36x+\tfrac{54}{x}+\tfrac{27}{x^3} = \left(2x+\tfrac{3}{x}\right)^3$ as Alexey Burdin pointed out in the comments, then you can just apply AM-GM to $2x+\tfrac{3}{x}$.
|
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|
Decomposition into partial fractions to compute an integral I'm having problems with:
$$\int_{-\infty}^{\infty}\frac{x^4+1}{x^6+1}dx$$
I was thinking: $\frac{x^4+1}{x^6+1}$ is an even function and the interval $(-\infty,\infty)$ is symmetric about 0, we could write the integral like:
$$2\int_{0}^{\infty}\frac{(x^4+1)}{x^6+1}dx$$
And for $\frac{x^4+1}{x^6+1}$, I will use the partial fractions. I will write $x^6+1$ like a sum of cubes $x^6+1=(x^2)^3+1^3$ and use the formula $a^3+b^3=(a+b)(a^2-ab+b^2)$ and in the end we will have
$$x^6+1=(x^2+1)(x^4-x^2+1)$$ But now I'm stuck and do not know how to decompose $x^4-x^2+1$. I was thinking about $(x^2-1)x^2+1$ or $(x^2-\frac{1}{2})^2+\frac{3}{4}$ but I need a form for that like $-(-1+\sqrt3 x-x^2) (1+\sqrt3 x+x^2)$...
A little help here?
|
HINT:
As $y^3+1=(y+1)(y^2-y+1),$
$$\dfrac{x^4+1}{x^6+1}=\dfrac{x^4-x^2+1}{x^6+1}+\dfrac{x^2}{x^6+1}=\dfrac1{x^2+1}+\dfrac{x^2}{x^6+1}$$
Set $x^3=u$ for the second part
|
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|
why $ \sin \theta = \frac{7}{8} \cos \theta$? I have an example:
$$ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{7}{8} $$
and then this equation is true? why there is cos multiplied?:
$$ \sin \theta = \frac{7}{8} \cos \theta$$
|
You have $\dfrac{\sin \theta}{\cos \theta} = \dfrac{7}{8}$
Multiply both side by $\cos \theta$
You get $\dfrac{\sin \theta}{\cos \theta} \times\cos \theta = \dfrac{7}{8} \cos \theta$
So you get $\sin \theta = \dfrac78 \cos \theta$
|
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|
Difficult inverse tangent identity
Prove that:
$$\arctan\left(\frac{\sqrt{1 + x} - \sqrt{1-x}}{\sqrt{1 + x} + \sqrt{1-x}} \right) = \frac{\pi}{4} - \frac{1}{2}\arccos(x), -\frac{1}{\sqrt{2}} \le x \le 1$$
I'd multiply the inside of $\arctan$ by the conjugate of the denominator.
I get:
$$\arctan\left(\frac{1 - 1\sqrt{1 - x^2}}{x} \right)$$
But that is still very difficult.
Any HINTS, no solutions?
|
Given that $$\tan^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)$$ $$=\tan^{-1}\left(\frac{(\sqrt{1+x}-\sqrt{1-x})(\sqrt{1+x}-\sqrt{1-x})}{(\sqrt{1+x}+\sqrt{1-x})(\sqrt{1+x}-\sqrt{1-x})}\right)$$ $$=\tan^{-1}\left(\frac{1+x+1-x-2(\sqrt{1+x})(\sqrt{1-x})}{(\sqrt{1+x})^2-(\sqrt{1-x})^2}\right)$$ $$=\tan^{-1}\left(\frac{2-2\sqrt{1-x^2}}{1+x-(1-x)}\right)$$$$=\tan^{-1}\left(\frac{1-\sqrt{1-x^2}}{x}\right)$$ Now, let $\color{blue}{x=\sin\theta}$ ($\implies \color{blue}{\theta=\sin^{-1}x=\frac{\pi}{2}-\cos^{-1}x}$ $\forall \color{red}{-1\leq x\leq 1}$) & plug it in the above expression we get $$=\tan^{-1}\left(\frac{1-\sqrt{1-(\sin\theta)^2}}{\sin\theta}\right)$$ $$=\tan^{-1}\left(\frac{1-\cos\theta}{\sin\theta}\right)$$ $$=\tan^{-1}\left(\frac{1-\left(1-2\sin^2\frac{\theta}{2}\right)}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\right)$$ $$=\tan^{-1}\left(\frac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\right)$$ $$=\tan^{-1}\left(\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}\right)=\tan^{-1}\left(\tan\frac{\theta}{2}\right)=\frac{\theta}{2}$$ $$=\frac{1}{2}\sin^{-1}x$$ $$=\frac{1}{2}\left(\frac{\pi}{2}-\cos^{-1}x\right)$$ $$=\color{blue}{\frac{\pi}{4}-\frac{1}{2}\cos^{-1}x}$$
|
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|
How do you solve this quadratic equation? The number of values of a for which
$$ (a^2 - 3a + 2)x^2 + (a^2-5a + 6)x + a^2-4 = 0$$
is an identity in x is?
Here's how much I was able to solve through:-
$$ (a^2 - 3a + 2)x^2 + (a^2-5a + 6)x + a^2-4 = 0$$
$$ ((a-1)(a-2))x^2 + ((a-3)(a-2))x + (a+2)(a-2) = 0$$
$$ (a-2)[(a-1)x^2 + (a-3)x + (a+2)] = 0$$
$$ so \hspace{5pt} (a-2) = 0 \hspace{5pt} or \hspace{5pt} [(a-1)x^2 + (a-3)x + (a+2)] = 0$$
$$ so \hspace{5pt} a = 2$$
$$ Now \hspace{5pt} [(a-1)x^2 + (a-3)x + (a+2)] = 0 $$
I don't know what to do next. Thanks in advance
|
HINT:
If $Ax^2+Bx+C=ax^2+bx+c$ is an identity
$A=a,B=b,C=c$
|
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|
Evaluate $\lim_{n\to\infty}nI_n$ with $I_n=\int_0^1\frac{x^n}{x^2+3x+2}dx$ We have to evaluate: $$\lim_{n\to\infty}nI_n$$ with $$I_n=\int_0^1\frac{x^n}{x^2+3x+2}\:dx.$$
There is an elegant way to solve this problem?
Here is all my steps:
*
*My first ideea was to find a recurrence relation such that:
$$I_{n+2}+3I_{n+1}+2I_n=\frac{1}{n+1},\forall x\in\mathbb{N}$$
*
*Next step I show that $\forall x\in[0,1]\Rightarrow I_{n}\ge I_{n+1}\ge I_{n+2}$
Therefore it involving that: $$6I_n\ge 4I_{n+1}+2I_n\ge\frac{1}{n+1},\forall x\in\mathbb{N}$$
As I said above $$6I_{n+2}\leq 4I_{n+2}+2I_n\leq\frac{1}{n+1}$$
$\Rightarrow \frac{n}{6(n+1)}\leq nI_n\leq\frac{n}{6(n-1)},\forall x\in\mathbb{N}$
Therefore by squeeze thereom:
$$nI_n\to\frac{1}{6}\:as\:n\to\infty$$
|
Substitute $x\mapsto x^{1/(n+1)}$ and use Dominated Convergence:
$$
\begin{align}
n\int_0^1\frac{x^n}{x^2+3x+2}\,\mathrm{d}x
&=\frac{n}{n+1}\int_0^1\frac1{x^2+3x+2}\,\mathrm{d}x^{n+1}\\
&=\frac{n}{n+1}\int_0^1\frac1{x^{2/(n+1)}+3x^{1/(n+1)}+2}\,\mathrm{d}x\\
&\to1\int_0^1\frac1{1+3\cdot1+2}\,\mathrm{d}x\\
&=\frac16
\end{align}
$$
A More Basic Approach
$$
\begin{align}
\frac16-\left(n\int_0^1\frac{x^n}{x^2+3x+2}\,\mathrm{d}x\right)
&=\int_0^1\left(\frac16-\frac{x^{1/n}}{x^{2/n}+3x^{1/n}+2}\right)\,\mathrm{d}x\tag1\\
&=\frac16\int_0^1\left(\frac{x^{2/n}-3x^{1/n}+2}{x^{2/n}+3x^{1/n}+2}\right)\,\mathrm{d}x\tag2\\
&\le\frac1{12}\int_0^1\left(x^{2/n}-3x^{1/n}+2\right)\,\mathrm{d}x\tag3\\[3pt]
&=\frac1{12}\left(\frac{n}{n+2}-3\frac{n}{n+1}+2\right)\tag4\\[6pt]
&=\frac{7n-2}{12(n+1)(n+2)}\tag5
\end{align}
$$
Explanation:
$(1)$: substitute $x\mapsto x^{1/n}$ then bring the $\frac16$ inside the integral
$(2)$: algebra; $x^{2/n}-3x^{1/n}+2=\left(x^{1/n}-1\right)\left(x^{1/n}-2\right)\ge0$ on $[0,1]$
$(3)$: since the integrand is positive, bound it by replacing the denominator with its minimum
$(4)$: integrate
$(5)$: simpllfy
Therefore,
$$
\frac16-\frac{7n-2}{12(n+1)(n+2)}\le\left(n\int_0^1\frac{x^n}{x^2+3x+2}\,\mathrm{d}x\right)\le\frac16\tag6
$$
Apply the Squeeze Theorem to get
$$
\lim_{n\to\infty}n\int_0^1\frac{x^n}{x^2+3x+2}\,\mathrm{d}x=\frac16\tag7
$$
|
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|
Quick Method to Calculate the Maclaurin Series of $\frac{1}{\sqrt{\cos{x}}} $ I am supposed to calculate the maclaurin series for $\frac{1}{\sqrt{\cos{x}}} $ but I can't seem to figure out an efficient way to go about doing this.
|
Hint. You may use, as $x \to 0$,
$$
\cos x =1-\frac{x^2}{2!}+\frac{x^4}{4!}+O(x^6) \tag1
$$ and, as $u \to 0$,
$$
\frac 1{\sqrt{1-u}} =1+\frac{u}{2}+\frac{3}{8}u^2+O(u^3) \tag2
$$ giving $$
\begin{align}
\frac 1{\sqrt{\cos x}}&=\frac 1{\sqrt{1-\frac{x^2}{2!}+\frac{x^4}{4!}+O(x^6)}}\\\\
&=1+\frac12\left(\frac{x^2}{2!}-\frac{x^4}{4!}+O(x^6) \right)+\frac{3}{8}\left(\frac{x^2}{2!}-\frac{x^4}{4!}\right)^2+O(x^6)\\\\
&=1+\frac{x^2}{4}+\frac{7 x^4}{96}+O(x^6)
\end{align}
$$ $(1)$ and $(2)$ have been obtained by Taylor expansions near $0$.
|
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Why cannot $(2x^2 + x)^2$ be simplified to $2x^4 + x^2$? So, I want to simplify an equation : $(2x^2 + x)^2$.
I thought this would simplify to $2x^4 + x^2$
But, if you input a value for $x$, the answers do not equal. For example, if you input $x = 3$, then:
$$(2x^2+x)^2
= 21^2
= 441$$
AND:
$$2x^4 + x^2
= 2(82) + 9
= 173$$
Can anyone explain why this is the case?
|
The commutativity property states that:
*
*For all $\color{red}{a},\color{green}{b}$ we have $\color{red}{a}+\color{green}{b} = \color{green}{b}+\color{red}{a}$
*For all $\color{red}{a},\color{green}{b}$ we have $\color{red}{a}\times\color{green}{b} = \color{green}{b}\times\color{red}{a}$
distributivity property of multiplication over addition states that:
*
*For all $\color{red}{a},\color{green}{b},\color{blue}{c}$ you have: $(\color{red}{a}+\color{green}{b})\times \color{blue}{c} = \color{red}{a}\times \color{blue}{c} + \color{green}{b}\times \color{blue}{c}$
$(2x^2 + x)^2 = (\color{red}{2x^2}+\color{green}{x})\color{blue}{(2x^2+x)}$
For the moment, let us refer to the blue parenthesis as a single piece, and use the distributivity property above:
$=\color{red}{2x^2}\color{blue}{(2x^2+x)}+\color{green}{x}\color{blue}{(2x^2+x)}$
Now, using commutativity and distributivity, and reassigning colors, we see that this is:
$=(\color{red}{2x^2}+\color{green}{x})\color{blue}{2x^2} + (\color{red}{2x^2}+\color{green}{x})\color{blue}{x} = \color{red}{2x^2}\times \color{blue}{2x^2}+\color{green}{x}\times\color{blue}{2x^2}+\color{red}{2x^2}\times\color{blue}{x}+\color{green}{x}\times\color{blue}{x}$
$=4x^4+2x^3+2x^3+x^2=4x^4+4x^3+x^2$
|
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|
Is it possible to have an $n\times n$ real matrix $A$ such that $A^TA$ has an eigenvalue of $-1$?
Question: Is it possible to have an $n\times n$ real matrix $A$ such that $A^TA$ has an eigenvalue of $-1$?
I can prove that it is not possible for $n=1,2$, but I am not sure for the general case.
Case $n=1$: $a^2v=-v$ $\implies$ $(a^2+1)v=0$ $\implies$ $v=0$.
Case $n=2$: Write $A=\begin{pmatrix}a&b \\ c&d\end{pmatrix}$. Then, $A^T A=\begin{pmatrix}a^2+c^2&ab+cd\\ ab+cd&b^2+d^2\end{pmatrix}$, so
$$
\begin{align}
\det(A^TA+1)
&= \begin{vmatrix}a^2+c^2+1&ab+cd\\ ab+cd&b^2+d^2+1\end{vmatrix}\\
&= 1+a^2+b^2+c^2+d^2+(ad-bc)^2\\
&\neq 0.
\end{align}
$$
Now, can we generalize to all $n$?
|
Hint: $A^TA$ is positive semidefinite.
|
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|
simple 2 sides inequality
$$2<\frac{x}{x-1}\leq 3$$
Is the only way is to multiple both sides by $(x-1)^2$?
so we get $2x^2-4x+2<x^2-x $ and $x^2-x<3x^2-6x+3$ which is $-x^2+3x-2$ and $-2x^2+5x-3<0$ so the sloutions are:
$1<x\leq \frac{3}{2}$ and $1<x\leq 2$ so overall it is $1<x\leq\frac{3}{2}$
|
You should just multiply the inequality by (x-1) instead of $(x-1)^2$.
Also, the solution would then be $\frac{3}{2}\le x\lt 2$.
|
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|
What is wrong with this integral reasoning? $$\int\frac{x^{2}+1}{x\sqrt{x^{4}+1}}dx$$
We start by multiplying by $1=\frac{x}{x}$.
$$\int\frac{x^{2}+1}{x^{2}\sqrt{x^{4}+1}}xdx$$
Next, we use the substitution $u=x^{2}$;$\frac{du}{2}=xdx$.
$$\frac{1}{2}\int\frac{u+1}{u\sqrt{u^{2}+1}}du=\frac{1}{2}\int\frac{1}{\sqrt{u^{2}+1}}du+\frac{1}{2}\int\frac{1}{u\sqrt{u^{2}+1}}du=$$
$$=\frac{1}{2}\ln\left(u+\sqrt{u^{2}+1}\right)+\frac{1}{2}\int\frac{1}{u\sqrt{u^{2}+1}}du$$
The problem is now reduced to computing the integral $\frac{1}{2}\int\frac{1}{u\sqrt{u^{2}+1}}du$.
$$\frac{1}{2}\int\frac{1}{u\sqrt{u^{2}+1}}du=\frac{1}{2}\int\frac{1}{u^2\sqrt{1+\frac{1}{u^{2}}}}du$$
We use substitution again $v=\frac{1}{u}$;$-dv=\frac{1}{u^{2}}du$, then we have the next integral.
$$-\frac{1}{2}\int\frac{1}{\sqrt{1+v^{2}}}dv=-\frac{1}{2}\ln\left(v+\sqrt{1+v^{2}}\right)=-\frac{1}{2}\ln\left(\frac{1}{u}+\sqrt{1+\frac{1}{u^{2}}}\right)=$$
$$=-\frac{1}{2}\ln\left(\frac{1+\sqrt{u^{2}+1}}{u}\right)$$
Finally the solution is:
$$\int\frac{x^{2}+1}{x\sqrt{x^{4}+1}}dx=\frac{1}{2}\ln\left(u+\sqrt{1+u^{2}}\right)-\frac{1}{2}\ln\left(\frac{1+\sqrt{u^{2}+1}}{u}\right)=\frac{1}{2}\ln\left(\frac{x^{4}+x^{2}\sqrt{x^{4}+1}}{1+\sqrt{x^{4}+1}}\right).$$
But the problem is, the book has the following solution:
$$\int\frac{x^{2}+1}{x\sqrt{x^{4}+1}}dx=\ln\left(\frac{x^{2}-1+\sqrt{x^{4}+1}}{x}\right)$$
which is obviously different.
|
To explicitly check that they differ by a constant, subtract them:
\begin{align*}
&\frac{1}{2}\log \frac{x^4+x^2\sqrt{x^4+1}}{1+\sqrt{x^4+1}} - \log \frac{x^2-1+\sqrt{x^4+1}}{x}\\
&= \frac{1}{2} \log \frac{1-x^2+x^4-\sqrt{1+x^4} + x^2\sqrt{1+x^4}}{x^2}\\
&\qquad - \frac{1}{2} \log \frac{2-2x^2+2x^4-2\sqrt{1+x^4}+2x^2\sqrt{1+x^4}}{x^2}\\
&=\frac{1}{2}\log \frac{1}{2},
\end{align*}
where I get the first term by rationalizing the denominator and the second by the identity $\log a = \frac{1}{2}\log a^2.$
|
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|
$(1+i)^6$ in polar form $re^{i\theta}$ I used De Moivre's formula and got
$$
\left(\frac{\sqrt{2}}{2}\right)^6 \times
\cos\left(6 \times \frac{1}{4\pi}\right) +
i\sin\left(6 \times \frac{1}{4\pi}\right) =
\frac{1}{8} e^{\frac{3}{2\pi}}.
$$
But the answer is $8e^{\frac{3}{2\pi}}$, can someone explain where the $8$ is from?
Thanks!
|
Hint (you made an arithmetic error):
$$ (1+i)^6 \neq \left(\frac{\sqrt2}{2}\right)^6\left(\frac{\sqrt2}{2} +\frac{\sqrt2}{2}i\right)^6$$
|
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|
Evaluate $\lim\limits_{x\to\ 0}(\frac{1}{\sin(x)\arctan(x)}-\frac{1}{\tan(x)\arcsin(x)})$
Evaluate $$\lim\limits_{x\to\ 0}\left(\frac{1}{\sin(x)\arctan(x)}-\frac{1}{\tan(x)\arcsin(x)}\right)$$
It is easy with L'Hospital's rule, but takes too much time to calculate derivatives. What are some shorter methods?
|
Notice that the limit is
$$
L=\lim_{x \to 0}\frac{1}{\sin x} \left(\frac{1}{\arctan x} - \frac{\cos x}{\arcsin x} \right)
$$
from the series expansion for small values of $x$ one has $\cos x =1- \frac{x^2}{2} + O(x^4)$ and $(\mbox{arc})\sin x = x + O(x^3) = \arctan x$ hence
$$L = \lim_{x \to 0}\frac{1}{(x + O(x^3))} \left(\frac{1}{x + O(x^3) } - \frac{1-x^2/2 + O(x^4)}{x + O(x^3)} \right)=\lim_{x \to 0} \frac{1}{x}\left(\frac{x}{2} + O(x^3) \right) = \frac{1}{2}$$
|
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|
Find a unit tangent vector to a curve that is an intersection of two surfaces. The intersection of the two surfaces given by the Cartesian equations $2x^2+3y^2-z^2=25$ and $x^2+y^2=z^2$ contains a curve $C$ passing through the point $P=(\sqrt{7},3,4)$. These equations may be solved for $x$ and $y$ in terms of $z$ to give a parametric representation of $C$ with $z$ as a parameter.
(a) Find a unit tangent vector $T$ to $C$ at the point $P$ without using an explicit knowledge of the parametric representation.
(b) Check the result in part (a) by determining a parametric representation of $C$ with $z$ as a parameter.
For (b), I solved the two equations for the surfaces given to get $y^2=25-z^2$ and $x^2=2z^2-25$. Since we're looking for the curve that contains $P$, $x$, $y$ should be positive so we get $y=\sqrt{25-z^2}$ and $x=\sqrt{2z^2-25}$. So from this I get the parametric representation $(\sqrt{2z^2-25}, \sqrt{25-z^2},z)$ for the curve $C$.
Is this the correct way of finding the parametrization? Moreover, I do not know how to find the unit targent vector $T$ to $C$ at $P$, without getting a parametrization. How can I find this? The answer to $T$ is $\frac{1}{\sqrt{751}}(24,-4\sqrt{7},3\sqrt{7})$.
I would greatly appreciate any solutions, hints or suggestions.
|
For $(b)$, by
$$
\begin{cases}
2x^2+3y^2-z^2=25 \\
x^2+y^2=z^2 \\
\end{cases}
$$
and working on the first:
$$\underbrace{2x^2+2y^2}_{2 z^2}+y^2-z^2=25\ \Rightarrow \ z^2+y^2=25$$
Then the curve is a circumference in the y-z plane and its parametric representation is:
$$
\begin{cases}
x=5 \cos t \\
y=5 \sin t \\
\end{cases}
$$
for $t\in[0,2\pi]$.
|
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|
Probability of a rectangular card intersecting the lines of a floor I have attempted this problem, but do not understand how the answer was achieved. The question is as follows (coming from Henk Tijms 'Understanding Probability' (3rd edn.) book):
Problem 7.16
Consider the following variant of Buffon's needle problem. A rectangular card with the side lengths $a$ and $b$ is
dropped at random on the floor. It is assumed that the length
$\sqrt{a^2 + b^2}$ of the diagonal of the card is smaller than the
distance $D$ between the parallel lines on the floor. Show that the
probability of the card intersecting one of the lines is given by
$\frac{2(a+b)}{\pi D}$.
I have let $y$ to be the centre of the rectangular card to the closest line of the two, and $x$ be the angle of the diagonal of the card. Thus I have defined the sample space to be $\Omega = \{(x,y): 0 \leq x \leq \pi, 0 \leq y \leq \frac{1}{2}D\}$.
The hypotenuse from the middle of the card to the closest line is then $\frac{y}{\sin(x)}$. If $\frac{y}{\sin(x)} > \frac{1}{2}\sqrt{a^2 + b^2}$ then the card does not touch either of the two lines. So if $\frac{y}{\sin(x)} \leq \frac{1}{2}\sqrt{a^2 + b^2}$ then the card intersects a line of the floor. Thus $y \leq \frac{\sin(x)}{2}\sqrt{a^2 + b^2}$, and so calculating the probability that the card intersects one of the lines is:
$\begin{align} \frac{\int _{0} ^{\pi} {\frac{\sin(x)}{2}\sqrt{a^2 + b^2}}\;dx}{\frac{1}{2}\pi D} &= \frac{\int _{0} ^{\pi} {\sin(x)\sqrt{a^2 + b^2}}\;dx}{\pi D} \\ &= \frac{2\sqrt{a^2 + b^2}}{\pi D} \end{align}$
I'm not sure if I skipped something or made some error here, but thank you in advance for your help.
|
I think the problem is the range of integration for $x$. Assume WLOG that $a\leq b$. There being two diagonals and thus two choices for $x$, we want $x$ to be the angle between a diagonal and the parallel lines that is closest to a right-angle. Then we have $\theta\leq x\leq \theta+\pi/2$ where $\theta$ is the angle between either diagonal and the long side of the rectangle. When the long side of the rectangle is parallel with the parallel lines, $x=\theta$; when the long side is perpendicular to the parallel lines, $x=\theta+\pi/2.$
It's easy to calculate $\theta=\tan^{-1}\left(a/b\right)$.
So the sample space is
$$\Omega = \left\{(x,y)\;:\; \theta\lt x\lt \theta+\frac{\pi}{2},\; 0\lt y\lt \frac{D}{2} \right\}.$$
Now, $x,y$ are independent and uniformly distributed over their range so their density functions are:
$$f_X(x) = \dfrac{2}{\pi} \\ f_Y(y) = \dfrac{2}{D}.$$
And as you found, the region where the card intersects a line satisfies
$$y \lt \dfrac{1}{2}\sqrt{a^2+b^2}\sin{x}.$$
So for the required probability, we integrate as follows
\begin{eqnarray*}
P(\text{intersection}) &=& \int_{x=\theta}^{\theta+\pi/2} \dfrac{2}{\pi} \int_{y=0}^{\frac{1}{2}\sqrt{a^2+b^2}\sin{x}} \dfrac{2}{D}\;dydx \\
&=& \int_{x=\theta}^{\theta+\pi/2} \dfrac{2}{\pi D} \sqrt{a^2+b^2}\sin{x} \;dx \\
&=& \dfrac{2}{\pi D} \sqrt{a^2+b^2}\;\bigg[-\cos{x} \bigg]_{\theta}^{\theta+\pi/2} \\
&=& \dfrac{2}{\pi D} \sqrt{a^2+b^2}\;\left(\cos{\theta} + \sin{\theta} \right) \qquad\qquad\text{(using basic trig identities)} \\
&=& \dfrac{2}{\pi D} \sqrt{a^2+b^2}\;\left(\dfrac{b}{\sqrt{a^2+b^2}} + \dfrac{a}{\sqrt{a^2+b^2}} \right) \\
&=& \dfrac{2(a+b)}{\pi D}.
\end{eqnarray*}
|
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|
Coefficient Problem (polynomial expansion)
Let $C$ be the coefficient of $x^2$ in the expansion of the product $(1 - x)(1 + 2x)(1 - 3x)\cdots(1 + 14x)(1 - 15x).$ Find $|C|.$
Just to begin,
$(1-x)(1+2x) = -2x^2 + x + 1$
$(1-x)(1+2x)(1-3x) = 6x^3 - 5x^2 - 2x + 1$
But expanding on like this take too long.
In the end the terms will be like:
$(1 - 15x)(a_0 + a_1x^1 + a_2x^2 + ....)$
Then just looking at the $x^2$, it will be:
$(a_2x^2 - 15a_1x^2) = x^2(a_2 - 15a_1)$
But it it still a very hard problem, any hints?
|
We get the coefficient of $x^2$ by adding the product of pairs of coefficients of $x$ in the binomials, since the other coefficients would be $1$.
All the coefficients that come from the $1-x$ term multiplied by other $x$ terms add up to
$$-1\cdot 2 + -1\cdot -3 + \ldots + -1\cdot -15$$
$$=-1(2-3+\ldots -15)$$
$$=-1[(-1+2-3+\ldots -15)+1]$$
$$=-1(T+1)$$
$$=-1T-1^2$$
where $T=-1+2-3+\ldots -15$. The coefficients that come from the $1+2x$ term multiplied by other $x$ terms add up to
$$2\cdot -1 + 2\cdot -3 + \ldots + 2\cdot -15$$
$$=2(-1-3+\ldots -15)$$
$$=2[(-1+2-3+\ldots -15)-2]$$
$$=2(T-2)$$
$$=2T-2^2$$
I think you get the idea. The coefficients coming from the $1-15x$ term multiplied by other $x$ terms add up to
$$-15T-15^2$$
Adding all those, to get the twice the sum of all products of pairs of $x$ coefficients (twice since we got each pair twice), we get
$$2C=(-1+2-3+\ldots -15)T-(1^2+2^2+3^2+\ldots +15^2)$$
$$=T^2-(1^2+2^2+3^2+\ldots +15^2)$$
So calculate $T$ and the sum of the first $15$ square numbers, substitute, and you are done. $T$ is easy:
$$T=-1+(2-3)+(4-5)+\ldots +(14-15)$$
$$=-1+-1+-1+\ldots +-1 \quad\text{($8$ times)}$$
$$=-8$$
The sum of the squares is
$$\frac{15\cdot (15+1)(2\cdot 15+1)}6=1240$$
So $C=\frac{(-8)^2-1240}2=-588$.
(Sorry for the late completion of this answer: all internet access in this area was cut off for a while.)
|
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|
Prove that there are no positive integers $a$ and $b$, such that $b^4 + b + 1 = a^4$ So I've been trying to solve this problem for a couple of days now. What I've come up with is this:
By way of contradiction, assume that there are positive integers a and b such that $b^4 + b + 1 = a^4$.
Consider the case when $b \geq a$. Then $a^4 - b^4 = b + 1$, but also $a^4 - b^4 \leq 0$. Thus, $b + 1 \leq 0$. Then, $b \leq -1$, a contradiction.
However, I haven't been able to figure out how to prove it with the case of $b < a$.
Thank you.
|
Indeed, we can show that $b^4+b+1=c^2$ has no solution in positive integers.
This is because if $b>0$ is an integer, then $0<b+1<2b+1\leq 2b^2+1$. Adding $b^4$ to this inequality gives: so $$b^4<c^2=b^4+b+1<b^4+2b^2+1=(b^2+1)^2$$ so $$b^2<c<b^2+1.$$
There are also no solutions with $b\leq -2$, since then:
$$1-2b^2<b+1<0$$ and then add $b^4$ to get: $$(b^2-1)^2<b^4+b+1<b^4$$
There is a $c$ with $b=0,-1$.
|
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Finding the Range of a Trigonometric function The range of $$f(x)=3\cos^2x-8\sqrt3 \cos x\cdot\sin x+5\sin^2x-7$$is given by:(1)$[-7,7]$(2)$[-10,4]$(3)$[-4,4]$(4)$[-10,7]$
ANS: (2)
My Solution
The equation can be written as: $$3\cos^2x-8\sqrt3 \cos x\cdot \sin x+16\sin^2x-11\sin^2x-7 \\\implies (\sqrt3\cos x-4\sin x)^2-11\sin^2x-7$$
So let $y=\sqrt3\cos x+4\sin x$
$$-\sqrt{(\sqrt3)^2+4^2} \le y\le \sqrt{(\sqrt3)^2+4^2} \implies 19 \le y^2 \le 19 \implies y^2\in[0,19]$$
CASE 1: When $y^2=0$
$$f(x)=0-11\sin^2x-7 \text{ ,taking } \sin^2 x=0\text{, minimum value of }f(x)= -7$$
CASE 2: When $y^2=19$
$$f(x)=19-11\sin^2x-7 \text{ ,taking } \sin^2 x=1\text{, minimum value of }f(x)= 1$$
So my range is $y\in[-7, 1]$
where is the problem?
|
First use the double angle formulas to lower the degree
$$3\frac{\cos(2x)+1}2-\frac82\sqrt3 \sin(2x)+5\frac{1-\cos(2x)}2-7
=-\cos(2x)-4\sqrt3 \sin(2x)-3.$$
The dot product $$(\cos(2x),\sin(2x))\cdot(-1,-4\sqrt3)$$ equals $$1\cdot\sqrt{(-1)^2+(-4\sqrt3)^2}\cdot\cos(\phi)$$ where $\phi$ is the angle between the vectors, hence the range is $$[-3-7,-3+7].$$
|
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|
How do you solve the summation of $2-4+8-16+32- \dots 2^{48}$? This is a summation problem but I can't seem to figure out how to solve this with the mix of subtraction and addition.
|
Split it into a positive series and a negative series:
$\color\red{2}-\color\green{4}+\color\red{8}-\color\green{16}+\color\red{32}-\ldots-\color\green{2^{48}}=$
$\color\red{2^1}-\color\green{2^2}+\color\red{2^3}-\color\green{2^4}+\color\red{2^5}-\ldots-\color\green{2^{48}}=$
$\color\red{\sum\limits_{n=0}^{23}2^{2n+1}}-\color\green{\sum\limits_{n=0}^{23}2^{2n+2}}=$
$\color\red{2\sum\limits_{n=0}^{23}2^{2n}}-\color\green{4\sum\limits_{n=0}^{23}2^{2n}}=$
$\color\red{2\sum\limits_{n=0}^{23}4^n}-\color\green{4\sum\limits_{n=0}^{23}4^n}=$
$-2\sum\limits_{n=0}^{23}4^n=$
$-2\cdot\dfrac{4^{24}-1}{4-1}$
|
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|
inequalities with fraction problem x $\frac{1}{x} > \frac{2x} {x^2 +2}$ solving this inequalities:
My long solution (wrong) :
multiplying $(x^2 + 2)^2 (x)^2 \dots$ (multiplying square of each denominator, getting rid of the > or < 0)
$x (x^2 +2)^2 > 2x(x^2) ( x^2 +2)$
$x(x^4+4x^2 +4)>2x^2(x^2+2)$
$x^5+4x^3+4x>2x^4+4x^2$
$x^5+4x^3+4x-2x^4-4x^2>0$
$x ( x^4 + 4x^2 + 4 +2x^3 -4x) >0 $
$x>0$ or ^ ...
This solving method doesn't look right
|
Break this up into cases.
Case 1: $x > 0$.
Multiplying both sides by $x$, we have $1 > \dfrac{2x^2}{x^2 + 2}$ (since $x$ is positive, the sign doesn't change), and so $x^2 + 2 > 2x^2$ (since $x^2 + 2$ is positive, the sign doesn't change).
Then $2 > x^2$. Now, just find the positive $x$ for which this is true.
Case 2: $x < 0$.
Multiplying both sides by $x$, we have $1 < \dfrac{2x^2}{x^2 + 2}$ (since $x$ is negative, the sign changes), and so $x^2 + 2 < 2x^2$.
Then $2 < x^2$. Now, just find the negative $x$ for which this is true.
Edit: Also, to gain some intuition to see if whatever approach you choose works, you could plot the two functions, say by putting the following in the search bar in Chrome: "y = (2x/(x^2 + 2)) and y = 1/x". This won't give you an exact answer, but it'll let you know if you're barking up the right tree.
|
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|
Calculate $\lim_{x \to \infty} (\sqrt{x^2 + 3x} - \sqrt{x^2 + 1})$ without using L'Hospital's rule Question:
Calculate
$$\lim_{x \to \infty} (\sqrt{x^2 + 3x} - \sqrt{x^2 + 1})$$
without using L'Hospital's rule.
Attempted solution:
First we multiply with the conjugate expression:
$$\lim_{x \to \infty} (\sqrt{x^2 + 3x} - \sqrt{x^2 + 1}) = \lim_{x \to \infty} \frac{x^2 + 3x - (x^2 + 1)}{\sqrt{x^2 + 3x} + \sqrt{x^2 + 1}}$$
Simplifying gives:
$$\lim_{x \to \infty} \frac{3x - 1}{\sqrt{x^2 + 3x} + \sqrt{x^2 + 1}}$$
Breaking out $\sqrt{x^2}$ from the denominator and $x$ from the numerator gives:
$$\lim_{x \to \infty} \frac{x(3 - \frac{1}{x})}{x(\sqrt{1 + 3x} + \sqrt{1 + 1})} = \lim_{x \to \infty} \frac{3 - \frac{1}{x}}{\sqrt{1 + 3x} + \sqrt{2}}$$
The result turns out to be $\frac{3}{2}$, but unsure how to proceed from here.
|
$$(\sqrt{x^2 + 3x} - \sqrt{x^2 + 1}) =(x \sqrt{1 + 3/x} - x\sqrt{1+ x^{-2} })$$
$$
=x\left( \sqrt{1 + 3/x} - \sqrt{1+ x^{-2} }\right)
$$
$$
\left( \sqrt{1 + 3/x} - \sqrt{1+ x^{-2} }\right) =1 +\frac{3}{2}\frac{1}{x} + O(x^{-2}) - (1+ O(x^{-2}))
$$
using Taylor's theorem.
So we get
$$
\frac{3}{2} + O(x^{-1})
$$
which converges to $3/2.$
Note that this approach is more easily generalizable to cases of different powers, eg cube roots.
|
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|
How does one calculate: $\left(\frac{z}{2!}-\frac{z^3}{4!}+\frac{z^5}{6!}-\cdots\right)^2$ How does one calculate:
$$\left(\frac{z}{2!}-\frac{z^3}{4!}+\frac{z^5}{6!}-\cdots\right)^2$$
Is the best way to just take the first term times the following two, and the second two times the next two to see the pattern?
First term will contribute:
$$\left(\frac{z^2}{2!2!}-\frac{2z^4}{2!4!}+\frac{2z^6}{2!6!}-\cdots\right)$$
$$\left(\frac{z^2}{4}-\frac{z^4}{4!}+\frac{z^6}{6!}-\cdots\right)$$
Second term will contribute:
$$\left(-\frac{z^4}{4!}+\cdots\right)$$
This already seems wrong, so expanding via the first term seems good, and I hazard a guess that
$$\left(\frac{z}{2!}-\frac{z^3}{4!}+\frac{z^5}{6!}-\cdots\right)^2=\left(\frac{z^2}{4}-\frac{z^4}{4!}+\frac{z^6}{6!}-\cdots\right)$$
|
Multiply each term inside the parenthesis by $z$. You'll find it is simply:
$$\frac{(1-\cos z)^2}{z^2}.$$
|
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|
Calculating determinant of a Vandermonde type matrix of order n Det$
\begin{bmatrix}
1 & 2 & 3 &\ldots &n\\
1& 2^3& 3^3& \ldots & n^3\\
1 &2^5& 3^5& \ldots & n^5\\
\vdots & \vdots& & \vdots \\
1&2^{2n-1}& 3^{2n-1}& \ldots &n^{2n-1}
\end{bmatrix}
$
If the powers were consecutively increasing down the rows than we could use the Vandermonde’s identity. However I am quite uncertain about how to tackle this. I am convinced that elementary row operations will be to no avail. Any hints?
|
Call $\Delta$ your determinant. It seems to me that $$\Delta = n! \det \left( \begin{array}{ccccc}
1 & 1 & 1 & \dots & 1 \\
1 & 4 & 9 & \cdots & n^2 \\
1 & 4^2 & 9^2 & \cdots & (n^2)^2 \\
\vdots & \vdots & \vdots & \cdots & \vdots \\
1 & 4^{n-1} & 9^{n-1} & \cdots & {(n^2)}^{(n-1)}
\end{array} \right) = n! \left( \begin{array}{ccccc}
1 & 1 & 1 & \dots & 1 \\
x_1 & x_2 & x_3 & \cdots & x_n \\
x_1^2 & x_2^2 & x_3^2 & \cdots & x_n^2 \\
\vdots & \vdots & \vdots & \cdots & \vdots \\
x_1^{n-1} & x_2^{n-1} & x_3^{n-1} & \cdots & x_n^{n-1}
\end{array} \right)$$
were $x_1 = 1, x_2=4, \cdots, x_i = i^2 , \cdots, x_n = n^2$. From which you can use the usual Vandermonde.
|
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|
Prove that $ a^2-4b \neq2$ if $ a,b \in \mathbb{ Z}$ My solution :
We suppose that is true. Then by contradiction:
$a^2-4b-2=0$
$a^2=4b+2$
$a=2(b+1/2) ^{0.5}$
then $(b+1/2)$ is fraction and rooted by $0.5$ so the square root of any fraction $+$ any-Integer will give fraction so then $a$ must be fraction, not an integer. Contradiction.
Then $a^2 -4b \neq 2$?
Is that a good proof?
|
Suppose
$a^2=4b+2
$.
$a$ must be of the form
$2c$ or $2c+1$
(i.e., even or odd).
If $a=2c$,
then
$2 = (2c)^2-4b
= 4c^2-4b
= 4(c^2-b)
$.
But 4 divides the right side but not the left,
so this is impossible.
If $a=2c+1$,
then
$2 = (2c+1)^2-4b
= 4c^2+4c+1-4b
= 4(c^2+c-b)+1
$.
But the left side is even
and the right side is odd,
so this is also impossible.
Note that this works,
with minor changes,
for
$a^2 = 4b+3
$.
|
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|
Finding the limit of $(1-\cos x)/x^2$ $$\lim _{x \to 0}{1-\cos x\over x^2}={2\sin^2\left(\frac{x}{2}\right)\over x^2}={\frac{2}{x^2}\cdot {\sin^2\left(\frac{x}{2}\right)\over \left(\frac{x}{2}\right)^2}}\cdot\left(\frac{x}{2}\right)^2$$
now $$\lim_{x \to 0}{\sin^2\left(\frac{x}{2}\right)\over \left(\frac{x}{2}\right)^2}=\left({\sin\left(\frac{x}{2}\right)\over \left(\frac{x}{2}\right)}\right)^2=1^2$$
So we have $$\frac{2}{x^2}\cdot \left(\frac{x}{2}\right)^2=\frac{2}{x^2}\cdot \left(\frac{x^2}{4}\right)=\frac{1}{2}$$
Are the moves right?
|
Correct, but too complicated (and missing several $\lim_{x\to0}$).
$$
\lim_{x\to0}\frac{1-\cos x}{x^2}=
\lim_{x\to0}\frac{2\sin^2(x/2)}{4(x/2)^2}=
\lim_{x\to0}\frac{1}{2}\left(\frac{\sin(x/2)}{(x/2)}\right)^{\!2}=
\frac{1}{2}
$$
Alternative way:
$$
\lim_{x\to0}\frac{1-\cos x}{x^2}=
\lim_{x\to0}\frac{1-\cos^2 x}{x^2(1+\cos x)}=
\lim_{x\to0}\frac{1}{1+\cos x}\left(\frac{\sin x}{x}\right)^{\!2}
$$
|
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|
$a^2+b$ and $a+b^2$ prime implies $\gcd(ab+1,a+b)=1$ Let $a,b>1$ be integers such that $a^2+b$ and $a+b^2$ are prime. Prove that $\gcd(ab+1,a+b)=1$.
Clearly $a$ and $b$ are of different parities; suppose $a$ is odd and $b$ even. If a prime $p\neq 2$ divides $ab+1$ and $a+b$, then it also divides $(ab+1)+(a+b)=(a+1)(b+1)$ and $(ab+1)-(a+b)=(a-1)(b-1)$. So either $p$ divides both $a+1,b-1$ or $p$ divides both $a-1,b+1$.
|
Suppose a prime $p$ divides both $ab+1$ and $a+b$. Then, writing $b\equiv -a \pmod p$, and plugging this in $ab+1$, we get $p| 1-a^2$, which (since $p$ is prime) implies $a\equiv 1$ or $-1$ modulo $p$.
i) $a\equiv 1 \pmod p$: Then, $b\equiv -1 \pmod p$, which implies $a^2+b\equiv 0 \pmod p$, so $a^2+b=p$, but since they're positive integers, $b$ is at least $p-1$, so $a$ can be at most $1$, which contradicts to the given assumption.
ii) $a\equiv -1 \pmod p$: Now, $b\equiv 1 \pmod p$. Similarly $a+b^2=p$ in this case, and again similarly since $a$ is at least $p-1$, $b$ can be at most $1$, which gives a contradiction.
So, there is no such a prime $p$, and $\gcd(ab+1,a+b)=1$.
|
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|
A trig ratio integral Consider the two integrals, where $a$ and $n$ are integers,
\begin{align}
I_{1} &= \int_{-n \pi}^{n \pi} \frac{\tan^{2a}x \, dx}{\tan^{2a}x + \cot^{2a}x} \\
I_{2} &= \int_{-(2n+1)\pi/2}^{(2n+1)\pi/2} \frac{\tan^{2a}x \, dx}{\tan^{2a}x + \cot^{2a}x}.
\end{align}
It would seem that $I_{2}$ would have a finite value.
Questions:
*
*What is the finite value of $I_{2}$ ?
*What is the best way to evaluate the integral $I_{1}$ ?
|
inside $I_1$ it is even function so integral will be
\begin{align}
I_{1} &=2 \int_{0}^{n \pi} \frac{\tan^{2a}x \, dx}{\tan^{2a}x + \cot^{2a}x} \\
\end{align}
then what function written inside the integral holds $f(x)=f(n\pi-x)$
then integral $I_1$ will be
\begin{align}
I_{1} &=4 \int_{0}^{\frac{n \pi}{2}} \frac{\tan^{2a}x \, dx}{\tan^{2a}x + \cot^{2a}x} \\
\end{align}
then use property \begin{align}
I &= \int_{a}^{{b}} f(x) \\
\end{align}\begin{align}
I &= \int_{a}^{{b}} f(a+b-x) \\
\end{align}and add both, you will get\begin{align}
2I_{1} &=4 \int_{0}^{\frac{n \pi}{2}} {\, dx}{ } \\
\end{align}
and this is real easy to do.in similar you can do $I_2$
|
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|
$xy + yz + zx + 2xyz = 1$ implies $4x+y+z\geq 2$ Let $x,y,z>0$ satisfy $$xy + yz + zx + 2xyz = 1.$$ Prove that $4x+y+z\geq 2$.
The condition invites the factoring $(1+x)(1+y)(1+z)+xyz-2=x+y+z$, but having the factor $4$ in the desired inequality makes things more difficult.
|
$x(y+z)+yz(1+2x)=1,yz\le (\dfrac{y+z}{2})^2 ,p=y+z \implies xp+(1+2x)\dfrac{p^2}{4} \ge 1 \\ \implies x \ge \dfrac{1-\frac{p^2}{4}}{p+\frac{p^2}{2}}=\dfrac{1}{p}-\dfrac{1}{2}$
$4x+y+z \ge \dfrac{4}{p}-2+p \ge 2\sqrt{\dfrac{4}{p}*p}-2=2$
|
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|
When is $a \space \sin^2(x) + b \space \cos^2(x) \le 1$? When is the above expression less than or equal to $1$, meaning for what values of $a$ and $b$ will the above expression be less than or equal to $1$?
|
Well, clearly it equals 1 when $a=b=1$. Rewrite $b=a+c$. Then you are asking when
$$a\sin^2 x+b\cos^2x=a\sin^2x+(a+c)\cos^2x=a(\sin^2x+\cos^2x)+c\cos^2x=a+c\cos^2x>1$$
or when
$$\cos^2x>\frac{1-a}{c}=\frac{1-a}{b-a}$$
|
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|
How to proceed with evaluating $\int\frac{dx}{\sqrt{9+4x^2}}$ and $\int\tan^2(3x)dx$
*
*$\displaystyle\int\frac{dx}{\sqrt{9+4x^2}}$
*$\displaystyle\int\tan^2(3x)dx$
For the first one i'm not sure if I did it correctly, here is what I did:
Let $2x=3\tan(t)$, so $x=\frac{3}{2}\tan(t)$ and $dx=\frac{3}{2}\sec^2(t)dt$. So by substitution,
$$\begin{align}\displaystyle\int\frac{dx}{\sqrt{9+4x^2}}&=\int\frac{\frac{3}{2}\sec^2(t)}{3\sec(t)}dt\\
&=\frac{1}{2}\int\sec(t)dt\\
&=\frac{1}{2}\left(\ln|\sec(t)+\tan(t)| + C\right)\\
&=\frac{1}{2}\left(\ln\bigg|\frac{\sqrt{9+4x^2}}{3}+\frac{2x}{3}\bigg|+C\right)\\
\end{align}$$
For the second one, i'm unable to proceed, what I did was
$\displaystyle\int\tan^2(3x)dx=\int\frac{\sin^2(3x)}{\cos^2(3x)}dx=\int\frac{\frac{1}{2}(1-\cos(6x)}{\frac{1}{2}(1+\cos(6x)}dx=\int\frac{(1-\cos(6x)}{(1+\cos(6x)}dx$
is this the right way to proceed?
Thanks
|
For the first one, rewrite the integral as follows:
$$ \int \frac{1}{\sqrt{9+4x^2}} \, \mathrm{d} x = \frac{1}{3} \int \frac{1}{\sqrt{1+4x^2/9}} \, \mathrm{d} x = \frac{1}{3} \int \frac{1}{\sqrt{1+(2x/3)^2}} \, \mathrm{d} x, $$ now let $v = 2x/3$ so we have:
$$ \int \frac{1}{\sqrt{9+4x^2}} \, \mathrm{d} x = \frac{1}{2} \int \frac{1}{\sqrt{1 + v^2} } \, \mathrm{d}v = \frac{1}{2} \textrm{asinh}{v},$$
where for the last integral we recall that [link].
|
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|
calculate the $1/6+1/12+1/24+1/48 \ldots $. Wolfram is wrong? I am trying to calculate the following sum
$$
S = \frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48} + \cdots
$$
so
$$
S+\frac{1}{3} = \frac{1}{3} + \frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48} + \cdots = \frac{1}{3} \cdot \sum_{k=0}^{\infty} \frac{1}{2^k}
$$
This is equal to
$$
\frac{1}{3} \cdot \frac{1}{1-0.5} = \frac{2}{3}
$$
so
$$
S + \frac{1}{3} = \frac{2}{3}
$$
and thus $S = \frac{1}{3}$.
So why in Wolfram Alpha when I do that [EDIT: new link ]
I get the following answer:
$$
0.3450320298895586027335724702689612099836962897927387
$$
|
You seem to have found a bug in Wolfram Alpha. It appears to first interpret the sum $$\dfrac{1}{6} + \dfrac{1}{12} + \dfrac{1}{24} + \dfrac{1}{48} + \ldots $$ as $$\sum_{n=1}^\infty \dfrac{1}{n((n-3)n+8)} \approx 0.3450320299$$
(which does start $ \dfrac{1}{6} + \dfrac{1}{12} + \dfrac{1}{24} + \dfrac{1}{48}$, but the next term is $\dfrac{1}{90}$)
then after a few seconds "changes its mind" and writes the sum as
$$\sum_{n=1}^\infty \dfrac{1}{3 \times 2^n} $$
but leaves the rest of the answer unchanged.
|
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|
Product of $A$ with the adjoint of $A$: why are all nondiagonal elements zero? Let \begin{align*} A = \begin{pmatrix} 1 & 2 & 4 \\ 3 & 2 & 1 \\ 6 & 8 & 2 \end{pmatrix}. \end{align*} We have $\det(A) = 44$. The cofactor matrix corresponding with $A$ is \begin{align*} C = \begin{pmatrix} -4 & 0 & 12 \\ 28 & -22 & 4 \\ -6 & 11 & -4 \end{pmatrix}. \end{align*} Hence the adjugate matrix of $A$ is \begin{align*} C^T = \begin{pmatrix} -4 & 28 & -6 \\ 0 & -22 & 11 \\ 12 & 4 & -4 \end{pmatrix}. \end{align*} Then \begin{align*} A C^T = \begin{pmatrix} 1 & 2 & 4 \\ 3 & 2 & 1 \\ 6 & 8 & 2 \end{pmatrix} \begin{pmatrix} -4 & 28 & -6 \\ 0 & -22 & 11 \\ 12 & 4 & -4 \end{pmatrix} = \begin{pmatrix} 44 & 0 & 0 \\ 0 & 44 & 0 \\ 0 & 0 & 44 \end{pmatrix} = \begin{pmatrix} \det(A) & 0 & 0 \\ 0 & \det(A) & 0 \\ 0 & 0 & \det(A) \end{pmatrix}. \end{align*}
Now I'm trying to understand in an intuitive manner why all the elements $a_{ij}$ for $i \neq j$ are zero. I understand why the diagonal elements are all $44$. For $a_{11}$ this is just Laplace expansion along the first row, i.e. \begin{align*} \sum_{j=1}^3 (-1)^{1+j} a_{1j} \det(M_{1j}) = \sum_{k=1}^3 a_{1k} C_{1k}, \end{align*} with $C_{1k} = (-1)^{1+j} \det(M_{1j})$.
Now, let's say we compute $a_{12}$. Then we multiply row $1$ of $A$ with column $2$ of $C^T$, or equivalently with row $2$ of the cofactor matrix $C$, since it is the transpose. I don't see why this should always equal zero? Why is $ \begin{pmatrix} 1 & 2 & 4 \end{pmatrix} \cdot \begin{pmatrix} 28 & -22 & 4 \end{pmatrix} = 0 $?
|
Because this is the Laplace expansion for the determinant of a matrix with two identical rows.
|
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|
Substituting a value of sine function in a trigonometric equation I am trying to really understand trigonometric equations and I've stumbled upon a rather confusing example. Solve the following equation:
$\sin x= 2|\sin x|+ {\sqrt{3}}\cos x$
First step is to define the absolute $\sin x$:
$$|\sin x| =
\begin{cases}
\sin x & \sin x \in [0, 1]\\
-\sin x & \sin x \in [-1, 0)
\end{cases}
$$
Let's see what happens in the first case:
$$\sin x \in [0,1] \Rightarrow\, x \in {[2k{\pi}, (2k+1){\pi}]}$$
which means x is in the first or second quadrant.
Bringing everything to the right side:
$$\sin x+{\sqrt{3}}\cos x=0$$
Now we notice terms ${\sqrt{3}}$ with $\cos x$ and $1$ with $\sin x$ and do the following:
\begin{align*}
\sin x+{\sqrt{3}}\cos x& =0 \quad /:2\\
\frac{1}{2}\sin x+{\frac{\sqrt{3}}{2}}\cos x& = 0
\end{align*}
Now, the goal is to try and substitute integer terms with trigonometric functions they are solutions to, in order to apply one of the trigonometric identities for solving the problem. Let's aim at the following formula: $\sin(x+y)=\sin x\cos y + \sin y\cos x$ .We have several options:
*
*$\frac {1}{2} $is value of $\cos$ for either $x=\frac {\pi}{3} + 2k\pi$ OR $x=\frac {5\pi}{3} + 2k\pi$. Of all of these angles, only $x=\frac {\pi}{3} $ falls under our domain.
*${\frac {\sqrt{3}}{2}} $is value of $\sin$ for either $x=\frac {\pi}{3} + 2k\pi$ OR $x=\frac {2\pi}{3} + 2k\pi$. Out of all of these angles, there are two of them that fall under our domain. $x=\frac {\pi}{3}$ and $x=\frac {2\pi}{3}$.
This is where the problem is. I feel I should consider both of the $\sin$ values and solve two different versions of the equation. The first one is by substituting $\frac {1}{2} = \cos\frac {\pi}{3}$ and $\frac {\sqrt{3}}{2} = \sin\frac {\pi}{3}$, which is easily solved using the identity mentioned above. The second case is when substituting $\frac {\sqrt{3}}{2} = \sin\frac {2\pi}{3}$. This does not conform to any trig. identity. I tried treating $\sin\frac {2\pi}{3}$ as $\sin$ of double angle, but ended up at the beginning. I typed this equation into Symbolab and examined their step-by-step solution. It turns out that at some point they divided the equation with $\cos x$. Even when $\cos x$ is definitely not zero, I was taught dividing a trigonometric equation with anyting other than integers was a risky move very likely to result in a loss of solutions. So, my questions are:
Should you just ignore possible angles that won't make you a simple trigonometric identity?
Can you divide trigonometric equation with some trigonometric function and when?
I hope I explained myself clearly enough. Thank you in advance.
|
the function $y = 2|\sin x| + \sqrt 3 \cos x$ is an even $2\pi$-periocic function and is also symmetric about $x = \pi.$ the equation $$\sin x = 2|\sin x| + \sqrt 3 \cos x = \begin{cases} 2\sin x + \sqrt 3 \cos x & \text{ if } 0 \le x \le \pi \\
-2\sin x + \sqrt 3 \cos x & \text{ if } \pi \le x \le 2\pi \\\end{cases}$$ which is equivalent to
$$\tan x = \begin{cases} 1/\sqrt 3 & \text{ if } 0 \le x \le \pi \\
\sqrt 3 & \text{ if } \pi \le x \le 2\pi \\\end{cases}$$
there are two roots $x = \pi/6, 5\pi/3$ in the interval $0 \le x \le 2\pi.$
|
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|
Find all complex numbers so that $Im(\frac{z+2}{2-i})=1$ and $Re(z^2+1)=1$ and for $z$ which is in the first quadrant find $\sqrt{z}$
Find all complex numbers so that $Im(\frac{z+2}{2-i})=1$ and $Re(z^2+1)=1$ and for $z$ which is in the first quadrant find $\sqrt{z}$.
If $z=x+iy$ then $$\frac{z+2}{2-i}=\frac{x+2+iy}{2-i}\times \frac{2+i}{2+i}=\frac{2x-y+4+i(x+2+2y)}{5}=\frac{2x-y+4}{5}+i\frac{x+2+2y}{5}$$
$Im(\frac{z+2}{2-i})=\frac{x+2+2y}{5}=1$
This gives the equation $$x+2y=3$$
$$z^2+1=(x+iy)^2+1=x^2+2xyi-y^2=(x^2-y^2)+2xyi$$
$Re(z^2+1)=x^2-y^2=1$
Solving linear system of equations
$$
\left\{
\begin{array}{c}
x+2y=3 \\
x^2-y^2=1
\end{array}
\right.
$$
I get $$x_1=\frac{4\sqrt3-3}{3}$$ $$y_1=\frac{6-2\sqrt3}{3}$$ $$x_2=\frac{-3-4\sqrt3}{3}$$ $$y_2=\frac{6+2\sqrt3}{3}$$
Now,
$$z_1=\frac{4\sqrt3-3}{3}+i\frac{6-2\sqrt3}{3}$$
$$z_2=\frac{-3-4\sqrt3}{3}+i\frac{6+2\sqrt3}{3}$$
Is this correct, because I don't get good values for $arg(z)$ and $module(z)$?
Complex number $z_1$ is in the first quadrant, so $\sqrt{z_1}=?$
I found the module of $z_1$ as $$\rho=\sqrt{\frac{35-16\sqrt3}{3}}$$ and argument of $z_1$ as $$cos\phi=0.84$$ $$sin\phi=0.54$$ which I think is wrong. Approximately, this would be $$\phi=\frac{\pi}{6}$$
Now, using Moivre's formula,
$$\sqrt[n]{z_k}=\sqrt[n]{\rho}(cos\frac{\phi+2k\pi}{n}+isin\frac{\phi+2k\pi}{n})$$ there are two roots for $\sqrt{z_1}$ for $k=0$ and $k=1$
$$\sqrt{z_0}=\sqrt{\rho}(cos\frac{\pi}{12}+isin\frac{\pi}{12})$$ and
$$\sqrt{z_1}=\sqrt{\rho}(cos\frac{13\pi}{12}+isin\frac{13\pi}{12})$$
Could someone check this?
Thanks for replies.
|
Note that
$$z^2+1=(x+iy)^2+1=x^2+2xyi-y^2+1=x^2-y^2\color{red}{+1}+2xyi.$$
|
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|
Solving $x^2=17\pmod{128}$ I'm attempring to solve a congruence $x^2 \equiv 17\pmod{128}$ but not quite sure how to go about it. I see that $128 = 2^7$, but the Chinese Remainder Theorem doesn't apply to $\gcd > 1$. I found one solution quite easily by finding solutiong to $x^2 \equiv\pmod{32}$ which was $x \equiv 23\pmod{32}$ and it turned out that $23 + 128x_0 \equiv 17\pmod{128}$. How do I find the other solution ans what's the proper way of doing it?
|
The series for the square root is
\begin{eqnarray}
\sqrt{1+16 t} =1 + 8 t - 32 t^2 + 256 t^3 - 2560 t^4 + 28672 t^5 - 344064 t^6+\\ +
4325376 t^7 - 56229888 t^8 + 749731840 t^9 - 10196353024 t^{10}+ \cdots
\end{eqnarray}
The first three terms truncation $1 + 8 t - 32t^2$ has square $1 + 16 t - 512 t^3 + 1024 t^4$. Therefore, for $t=1$ we obtain
$$(1+8-32)^2 = 17 -512 + 1024= 17 + 512 \equiv 17 \!\!\!\!\mod 512$$
therefore, $23^2 \equiv 17 \!\!\!\!\mod 128$.
Let $x$ any other solution of $x^2 \equiv 17 \!\!\!\!\mod 128$. Then
$y\colon =\frac{x}{23} \!\!\!\mod 128$ satisfies $y^2\equiv 1 \!\!\!\mod 128$
and for this we have $4$ solutions $\pm 1$, $\pm 65 \!\!\!\mod 128$.
Therefore, the solutions are $\pm 23$, $\pm 23 \cdot 65 \!\!\!\mod 128$, that is
$$23, 41, 87, 105$$
|
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|
Calculus 2 - $\int(\sqrt{72+36x^2}dx$ I have done this problem several times and this is the only answer i ever come to. My schools webwork gives me incorrect for my answer (answer is not simplified but it should be accepted in this format). Did i do this correctly?
Here is my work:
\begin{align}
\int \sqrt{72+36x^2}\, dx&=\sqrt {36}\int \sqrt{2+x^2}\,dx\\
&=6\int \sqrt{2+x^2}\,dx\\
&=6\int \sqrt 2 \sec \theta \sqrt 2 \sec^2 \theta \, d\theta\\
&=12 \int sec^3 \theta=12\left[\frac{\tan \theta \sec \theta}2 +\frac 12 \int \sec\theta \, d\theta\right]\\
&=12\left[\frac{\tan\theta\sec\theta}2+\frac 12 \ln|\sec\theta+\tan\theta|\right]+C\\
&=6\tan \theta\sec\theta+6\ln|\sec\theta+\tan\theta|+C\\
&=6\tan\left(\tan^{-1}\frac{x}{\sqrt 2}\right)\sec\left(\tan^{-1}\frac{x}{\sqrt 2}\right)\\
&+6\ln \left|\sec\left(\tan^{-1}\frac{x}2\right)+\tan\left(\tan^{-1}\frac{x}{\sqrt 2}\right)\right|+C
\end{align}
Any help is appreciated. Thanks
|
I believe that your work is correct. However, at the end you are most likely asked to put this in a nicer form. The way to simplify $trig_1(trig_2^{-1}(\frac{a}{b}))$ is form a right triangle that fits your $trig^{-1}$ conditions and then compute $trig_1(angle)$.
As an example, simplifying $\tan{(\arctan{\frac{x}{\sqrt{2}}})}$ means forming a right triangle where the opposite side from an angle (we'll call $\theta$) is $x$ and the adjacent side from $\theta$ is $\sqrt{2}$. The hypotenuse of this triangle is obviously then $\sqrt{2+x^2}$. Therefore,
$$\tan{(\arctan{\frac{x}{\sqrt{2}}})}=\tan{\theta} = \frac{x}{\sqrt{2}}$$
The above could also be seen by recognizing that tangent and arctangent are inverse functions. Using this same triangle, we also have:
$$\sec{(\arctan{\frac{x}{\sqrt{2}}})} = \sec{\theta} = \frac{\sqrt{2+x^2}}{\sqrt{2}}$$
I'll leave you to simplify the rest.
|
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|
What is the logic/theorem/derivation behind finding the exponent of p in n! By [n/p] + [n/p^2] + [n/p^3] + ....? The exponent of prime number of 3 in 100! is 48.
It means 100! is divisible by $3^48$
$$E_3(100!) = \left\lfloor\frac{100}3\right\rfloor + \left\lfloor\frac{100}{3^2}\right\rfloor + \left\lfloor\frac{100}{3^3}\right\rfloor + \left\lfloor\frac{100}{3^4}\right\rfloor
= 33+11+3+1 = 48$$
What is the derivation/math behind the above logic?
|
suppose $p$ is prime number
$$1,2,3,4,5,...,p,p+1,p+2,....,2p,....3p,3p+1,...,p^2,p^2+1,....2p^2,...p^3,p^3+1,...$$ obviously : every p number has multiple of p
every $p^2 $number has multiple of $p^2$
every $p^3 $number has multiple of $p^3$
and so on
sum of the p multiple is :sum of $$\left \lfloor \frac{n}{p^1} \right \rfloor+\left \lfloor \frac{n}{p^2} \right \rfloor +\left \lfloor \frac{n}{p^3} \right \rfloor +...$$
|
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|
Complex numbers - roots of unity
Let $\omega$ be a complex number such that $\omega^5 = 1$ and $\omega \neq 1$. Find
$$\frac{\omega}{1 - \omega^2} + \frac{\omega^2}{1 - \omega^4} + \frac{\omega^3}{1 - \omega} + \frac{\omega^4}{1 - \omega^3}.$$
I have tried adding the first two and the second two separately, then adding those sums but how do I get a numerical value as the answer?
Thanks
|
The sum of the first and fourth terms is
\begin{align*}
\frac{\omega}{1 - \omega^2} + \frac{\omega^4}{1 - \omega^3} &= \frac{\omega (1 - \omega^3) + \omega^4 (1 - \omega^2)}{(1 - \omega^2)(1 - \omega^3)} \\
&= \frac{\omega - \omega^4 + \omega^4 - \omega^6}{(1 - \omega^2)(1 - \omega^3)} \\
&= \frac{\omega - \omega^4 + \omega^4 - \omega}{(1 - \omega^2)(1 - \omega^3)} \\
&= 0,
\end{align*}and the sum of the second and third terms is
\begin{align*}
\frac{\omega^2}{1 - \omega^4} + \frac{\omega^3}{1 - \omega} &= \frac{\omega^2 (1 - \omega) + \omega^3 (1 - \omega^4)}{(1 - \omega^4)(1 - \omega)} \\
&= \frac{\omega^2 - \omega^3 + \omega^3 - \omega^7}{(1 - \omega^4)(1 - \omega)} \\
&= \frac{\omega^2 - \omega^3 + \omega^3 - \omega^2}{(1 - \omega^4)(1 - \omega)} \\
&= 0.
\end{align*}Therefore, the sum of all four terms is $\boxed{0}$.
|
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|
System of equations in a,b,c,d $a,b,c,d$ are complex numbers satisfying
\begin{cases}
a+b+c+d=3 \\
a^2+ b^2+ c^2+ d^2=5 \\
a^3+ b^3+ c^3+ d^3=3 \\
a^4+ b^4+ c^4+ d^4=9
\end{cases}
Find the value of the following: $$a^{2015} + b^{2015} + c^{2015} + d^{2015}$$.
|
Hint
Let $a_{n}=a^n+b^n+c^n+d^n$,then
$$a_{n+3}=(a+b+c+d)a_{n+2}-(ab+ac+ad+bc+bd+cd)a_{n+1}+(abc+abd+acd+bcd)a_{n}-abcd\cdot a_{n-1}$$
and you have only find this $ab+ac+ad+bc+bd+cd,abc+abd+acd+bcd,abcd$ this is not hard to find it
|
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|
What is the area of the part of the surface $z=yx$ bounded by $x^2+y^2=1$? A parametrization of the part of the surface $z=yx$ bounded by $x^2+y^2=1$ is
\begin{align}
x &= u \cos v \\
y &= u \sin v \\
z &= \frac12 u^2 \sin 2v,
\end{align}
or
$$r(u,v)=u \cos v \, {\bf i} + u \sin v \, {\bf j} + \frac12 u^2 \sin 2v \, {\bf k}, \quad 0<v<2\pi, 0<u<1.$$
The norm of the cross product of $r_u$ and $r_v$ is $$\sqrt {u^4+u^2} = u\sqrt {u^2+1},$$
so
$$\int^{2\pi}_0\int_0^1u\sqrt {u^2+1}dudv=\int^{2\pi}u\sqrt {u^2+1}=\frac13((4\pi^2+1)^\frac23-1)$$
However, the answer showed in my text book is $\frac{2\pi}{3}(2\sqrt 2-1)$, so I don't know whether I did it wrong or there is a mistake in my textbook.
|
$I=\int \int_Vxy\ dx\ dy = \int_{x=-1}^1\int_{y=-\sqrt{1-x^2}}^\sqrt{1-x^2} xy\ dy\ dx$
Change of coordinates -
$x=r\cos \theta$
$y=r\sin \theta$
Jacobian -
$I=\int \int r^2\cos\theta \sin\theta\ \left\|d\frac{x,y}{r,\theta}\right\|\ d\theta\ dr$
$\left\|d\frac{x,y}{r,\theta}\right\|=\left|\begin{array}{c}\cos \theta&\sin\theta\\ -r\sin \theta&r\cos \theta\end{array}\right|=r$
$I=\int_0^1 \int_0^{2\pi} r^3\cos\theta \sin\theta\ d\theta\ dr=\int r^3\cdot0\ dr=0$
In retrospect, it is easy to see this is zero geometrically, since $xy+(-x)y=0$.
|
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|
Finding the roots of a different Quadratic equation from the roots of a Given Quadratic equation The Question:
If $\alpha$ and $\beta$ are the roots of the equation $ax^2+bx+c=0$...
Then find the roots of the equation $ax^2-bx(x-1)+c(x-1)^2=0$
My Attempt:
The new equation can be made into a quadratic as:
$$(a-b+c)x^2+(b-2c)x+c=0$$
Now $$\text{Sum of roots}=\dfrac{-b}{a} = \dfrac{2c-b}{a-b+c}$$
And $$\text{Product of roots}=\dfrac{c}{a}=\dfrac{c}{a-b+c}$$
But I don't seem to be going anywhere with the way I'm proceeding
Please Help! Thanks!
|
Let $\alpha = \dfrac{-b+\sqrt{b^2-4ac}}{2a}$ and $\beta = \dfrac{-b-\sqrt{b^2-4ac}}{2a}$.
Note that $\alpha + \beta = -\dfrac{b}{a}$ and $\alpha\beta = \dfrac{c}{a}$.
We want to calculate:
$$\dfrac{2c-b\pm \sqrt{(b-2c)^2 - 4(a-b+c)c}}{2(a-b+c)} $$
In terms of $\alpha$ and $\beta$.
$$\dfrac{2c-b\pm \sqrt{(b-2c)^2 - 4(a-b+c)c}}{2(a-b+c)} = \dfrac{2c-b\pm \sqrt{b^2-4bc+4c^2 -4ac + 4bc -4c^2}}{2(a-b+c)} = $$
$$= \dfrac{a}{a-b+c}\cdot\dfrac{2c-b\pm \sqrt{b^2-4ac}}{2a} =\dfrac{\frac{c}{a}}{1-\frac{b}{a}+\frac{c}{a}} +\dfrac{1}{1-\frac{b}{a}+\frac{c}{a}}\cdot\dfrac{-b\pm \sqrt{b^2-4ac}}{2a} = $$
$$ = \dfrac{\alpha\beta}{1+\alpha+\beta + \alpha\beta} +\dfrac{1}{1+\alpha+\beta + \alpha\beta}\cdot\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$$
So the roots are:
$$\dfrac{\alpha\beta + \alpha}{1+\alpha+\beta + \alpha\beta}\qquad \dfrac{\alpha\beta + \beta}{1+\alpha+\beta + \alpha\beta}$$
Simplifying:
$$\dfrac{\alpha}{\alpha+1} \qquad \dfrac{\beta}{\beta+1}$$
|
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|
How do I prove that $\lim_{(x,y)→(0,0)}\frac{1-\cos(x^2+y^2)}{\sqrt{x^2+y^2}} = 0$ $$\lim_{(x,y)→(0,0)}\frac{1-\cos(x^2+y^2)}{\sqrt{x^2+y^2}} = 0$$
I believe this is correct since I couldn't find a directional limit that won't validate this.
From what I know, I have to prove that
$$\forall\epsilon\gt 0, \exists\delta\gt 0$$
$$ \mbox{such that}$$
$$0 \lt \|(x,y)\| \lt \delta\Rightarrow \frac{1-\cos(x^2+y^2)}{\sqrt{x^2+y^2}} \lt \epsilon$$
I know that $$\sqrt{x^2+y^2} = \|(x,y)\|$$
So I take $$\frac{1-\cos(x^2+y^2)}{\sqrt{x^2+y^2}} \le \frac{2}{\|v\|}$$ and I kind of get stuck there. I greatly appreciate some help.
|
Here's yet another way
$$\lim\limits_{(x,y)→(0,0)}\frac{1-\cos\left(x^2+y^2\right)}{\sqrt{x^2+y^2}} $$
Using polar coordinates, we have
$$\lim\limits_{r\to 0^+} \frac{1-\cos\left(r^2\cos^2\phi+r^2\sin^2\phi\right)}{\sqrt{r^2\cos^2\phi+r^2\sin^2\phi}} $$
$$=\lim\limits_{r\to 0^+} \frac{1-\cos\left(r^2\left(\cos^2\phi+\sin^2\phi\right)\right)}{\sqrt{r^2}\sqrt{\cos^2\phi+\sin^2\phi}} $$
$$=\lim\limits_{r\to 0^+} \frac{1-\cos\left(r^2\right)}{r} $$
$$=\left(\lim\limits_{r\to 0^+} r\right)\left(\lim\limits_{r\to 0^+} \frac{1-\cos\left(r^2\right)}{r^2} \right)=0$$
|
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|
Side limits of the derivative of this function $f:\mathbb{R}\to \mathbb{R}$ with $f\left(x\right)=\left(x^3+3x^2-4\right)^{\frac{1}{3}}$
Calculate side limits of this function's derivative, $f'_s\:and\:f'_d$, in $x_o=-2$
The answer key says I should get $\infty $ and $-\infty$ but I'm not getting that. The derivative I get is $\frac{x\left(x+2\right)}{\left(\left(x-2\right)^2\left(x+2\right)^4\right)^{\frac{1}{3}}}$ and by doing the multiplication from the denominator I would get something with $x^2$.
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It might help to observe that $f(x) \; = \; (x+2)^{2/3}(x-1)^{1/3}.$ To obtain this factorization, note that $x^3 + 3x^2 - 2$ equals zero when $x = -2,$ so we know $x+2$ is a factor of $x^3 + 3x^2 - 2.$ Use long division, or use synthetic division, or note that $x^3 + 3x^2 - 2 = x^3 + 2x^2 + x^2 - 2$ (and factor by grouping), and you'll find that $x^3 + 3x^2 - 2 = (x+2)(x^2 + x - 2).$ Now factor the quadratic. Since this "looks like" the graph of $y = -3^{1/3}(x+2)^{2/3}$ when $x$ is close to $-2,$ the graph will look like that of $y = x^{2/3}$ translated left $2$ units and reflected about the $x$-axis and slightly stretched, and thus we'd expect the left and right limits of the derivative to be what you said the answer is. However, two graphs can be very close to each other and still have vastly different derivative behaviors. Consider, for example, the graph of $y = x^2$ and the graph of $y = x^2 + 10^{-100}W(x),$ where $W(x)$ is the Weierstrass nowhere differentiable function. So let's carry out these limits explicitly.
Assuming $x \neq -2$ and using the product rule, we get
$$f'(x) \;\; = \;\; \frac{2}{3}(x+2)^{-\frac{1}{3}}(x-1)^{\frac{1}{3}} \;\; + \;\; \frac{1}{3}(x+2)^{\frac{2}{3}}(x-1)^{-\frac{2}{3}}$$
The two limits are now straightforward.
For $x \rightarrow -2$ from the left we get
$$f'(x) \;\; \longrightarrow \;\; \frac{2}{3}(\rightarrow -\infty)(\rightarrow -3^{\frac{1}{3}}) \;\; + \;\; \frac{1}{3}(\rightarrow 0)(\rightarrow 3^{\frac{2}{3}}) \;\; = \;\; +\infty $$
and for $x \rightarrow -2$ from the right we get
$$f'(x) \;\; \longrightarrow \;\; \frac{2}{3}(\rightarrow +\infty)(\rightarrow -3^{\frac{1}{3}}) \;\; + \;\; \frac{1}{3}(\rightarrow 0)(\rightarrow 3^{\frac{2}{3}}) \;\; = \;\; -\infty $$
|
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How to show that that the following three consecutive numbers $3^{2^{10}} − 1$, $3^{2^{10}}$,$3^{2^{10}}+ 1$ are the sum of two squares? Show that the following three consecutive numbers:
$$
3^{2^{10}} − 1, 3^{2^{10}} , 3^{2^{10}} + 1
$$
can be represented as sums of two integer squares.
|
the second and third can be expressed as $(3^{2^{9}})^2+0^2$ and $(3^{2^9})^2+1^2$
For the first:
$3^{2^{10}}-1=$
$(3^{2^{9}}+1)(3^{2^8}+1)(3^{2^7}+1)(3^{2^6}+1)(3^{2^5}+1)(3^{2^4}+1)(3^{2^3}+1)(3^{2^2}+1)(3^{2}+1)8$
Each of the factors is trivially a sum of two squares, So all we need to conclude is the fact that the product of two numbers that are the sum of two squares is also the sum of two squares, this is immediate by the Brahmagupta–Fibonacci identity.
|
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Find eigenvalues and eigenvectors of this matrix Problem: Let \begin{align*} A = \begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{pmatrix}. \end{align*} Compute all the eigenvalues and eigenvectors of $A$.
Attempt at solution: I found the eigenvalues by computing the characteristic polynomial. This gives me \begin{align*} \det(A - x \mathbb{I}_4) = \det \begin{pmatrix} 1-x & 1 & 1 & 1 \\ 1 & 1-x & 1 & 1 \\ 1 & 1 & 1-x & 1 \\ 1 & 1 & 1 & 1-x \end{pmatrix} = -x^3 (x-4) = 0 \end{align*} after many steps. So the eigenvalues are $\lambda_1 = 0$ with multiplicity $3$ and $ \lambda_2 = 4$ with multiplicity $1$.
Now I was trying to figure out what the eigenvectors are corresponding to these eigenvalues. Per definition we have $Av = \lambda v$, where $v$ is an eigenvector with the corresponding eigenvalue. So I did for $\lambda_2 = 4$: \begin{align*} \begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} = 4 \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} \end{align*} I think this is only possible when $x_1 = x_2 = x_3 = x_4 = 1$. So am I right in stating that all the eigenvectors corresponding to $\lambda_2$ are of the form $t \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix}$ with $t$ some number $\neq 0$? For $\lambda_1 = 0$, I'm not sure how to find the eigenvectors. The zero vector is never an eigenvector. This means $x_1, x_2, x_3$ and $x_4$ can be anything aslong as they add to zero?
|
the matrix you have has rank one and can be written as $$A = uu^\top \text{ where } u = \pmatrix{1,1,1,1}^\top.$$ now, $$Av = uu^\top v=(u^\top v)u.$$ note that $u^\top v$ is a scalar, therefore $A$ has eigenvalue $0$ of multiplicity $3$ corresponding eigenvector any vector orthogonal to $u$ which has dimension $3.$ $A$ has also the nonzero eigenvalue $u^\top u = 4$ and the corresponding eigenvector $u.$
the same ideas can be used to find the eigenvalues and eigenvectors of any rank one matrix $ab^\top.$
|
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|
How to find this limit : $x\sin{f(x)}$ How to find the limit
$$\lim_{x\to\infty}x\sin f(x)$$
where $$f(x)=\left(\sqrt[3]{x^3+4x^2}-\sqrt[3]{x^3+x^2}\right)\pi\ ?$$
Is it possible to solve without L'Hospital's rule ?
|
Write
$\begin{array}\\
(x^3+ax^2)^{1/3}
&=x(1+a/x)^{1/3}\\
&=x(1+(a/3x)+(1/3)(-2/3)(a/x)^2/2 + O(1/x^3))\\
&=x+(a/3)-(a^2/9x) + O(1/x^2))\\
\end{array}
$
so
$\begin{array}\\
(x^3+ax^2)^{1/3}-(x^3+bx^2)^{1/3}
&=(x+(a/3)-(a^2/9x) + O(1/x^2)))-(x+(b/3)-(b^2/9x) + O(1/x^2)))\\
&=(a-b)/3-(a^2-b^2)/(9x) + O(1/x^2)\\
\end{array}
$
Therefore,
setting $a=4, b=1$,
$\begin{array}\\
\sin f(x)
&= \sin(\pi(1-15/(9x)+O(1/x^2)))\\
&=\sin(\pi(5/(3x)+O(1/x^2)))\\
&=\pi(5/(3x)+O(1/x^2)\\
\end{array}
$
so
$x \sin f(x)
=5\pi/3+O(1/x)
.$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to find variance of k+1 elements if variance of k elements is known? I need to find the variance of k+1 elements given the variance of k elements. I can also store other features for k elements like mean ($\mu_n$) etc. So, given the below function's value,
$$
\frac{1}{n}\sum\limits_{i=1}^n(a_i-\mu_{n})^2
$$
I need to find
$$
\frac{1}{n+1}\sum\limits_{i=1}^{n+1}(a_i-\mu_{n+1})^2
$$
where $\mu_{n}$ is the mean of k elements and $\mu_{n+1}$ is the new mean of k+1 elements.
|
$$\sigma_{n+1}^2 = \frac{1}{n+1}\sum_{k=1}^{n+1}((a_{k} - \mu_{n}) + (\mu_{n} - \mu_{n+1}))^2 $$
$$ = \frac{1}{n+1}\sum_{k=1}^{n+1}[(a_k-\mu_n)^2 - 2(a_k-\mu_n)(\mu_{n+1} - \mu_n) + (\mu_{n+1} - \mu_n)^2]$$
$$ = \frac{1}{n+1}[\sum_{k=1}^{n}[(a_k-\mu_n)^2] + (a_{n+1} - \mu_n)^2 - 2 (\mu_{n+1} - \mu_n)\sum_{k=1}^{n+1}[(a_k-\mu_n)] + (n+1)(\mu_{n+1}-\mu_n)^2]$$
$$ = \frac{n\sigma_n^2 + (a_{n+1} - \mu_n)^2}{n+1} + \frac{- 2 (\mu_{n+1} - \mu_n)}{n+1}((n+1)(\mu_{n+1}-\mu_n)) + (\mu_{n+1}-\mu_n)^2 $$
$$ = \frac{n\sigma_{n}^2 + (a_{n+1} - \mu_{n})^2}{n+1} - 2(\mu_{n+1}-\mu_n)^2 + (\mu_{n+1}-\mu_n)^2 $$
$$= \frac{n\sigma_{n}^2 + (a_{n+1} - \mu_{n})^2}{n+1} - (\mu_{n+1}-\mu_{n})^2$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Using equation to find value of $1/x - 1/y$ $$\left(\frac{48}{10}\right)^x=\left(\frac{8}{10}\right)^y=1000$$
What is the value of $\frac{1}{x}-\frac{1}{y}$?
I have already used that when $48$ divided by $10$ then it becomes $4.8$ and when $8$ divided by $10$ then it becomes $0.8$ by getting $10^x$ and $10^y$ to make it simpler but I cannot proceed further.
|
$$\left(\frac{48}{10}\right)^x=1000 \Longleftrightarrow $$
$$\left(\frac{24}{5}\right)^x=1000 \Longleftrightarrow x=\frac{\log_{10}(1000)}{\log_{10}\left(\frac{24}{5}\right)}$$
$$\left(\frac{8}{10}\right)^x=1000 \Longleftrightarrow $$
$$\left(\frac{4}{5}\right)^x=1000 \Longleftrightarrow y=\frac{\log_{10}(1000)}{\log_{10}\left(\frac{4}{5}\right)}$$
So:
$$\frac{1}{x}-\frac{1}{y}\Longrightarrow$$
$$\frac{1}{\frac{\log_{10}(1000)}{\log_{10}\left(\frac{24}{5}\right)}}-\frac{1}{\frac{\log_{10}(1000)}{\log_{10}\left(\frac{4}{5}\right)}}=$$
$$\frac{\log_{10}\left(\frac{24}{5}\right)}{\log_{10}(1000)}-\frac{\log_{10}\left(\frac{4}{5}\right)}{\log_{10}(1000)}=$$
$$\frac{\log_{10}\left(\frac{24}{5}\right)}{3}-\frac{\log_{10}\left(\frac{4}{5}\right)}{3}=$$
$$\frac{\log_{10}\left(\frac{24}{5}\right)-\log_{10}\left(\frac{4}{5}\right)}{3}=$$
$$\frac{\log_{10}(6)}{3}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Trigonometric integration 3. $\int \frac{\sqrt{3cos2x-1}}{cosx}$
ATTEMPT:
I did the following substitution
Let $sinx=t$
$cosxdx=dt$
$I=\frac{\sqrt{3(1-2t^2)-1}}{1-t^2}=\frac{\sqrt{2-6t^2}}{1-t^2}$
Now let $t=\frac{1}{z}$
$dt=-\frac{1}{z^2}dz$
Now $I=\color{red}{-} \frac{\sqrt{2z^2-6}}{z(z^2-1)}$
Lastly substitute $2z^2-6=p^2$
Finally $I=\color{red}{-}2\int \frac{p^2}{(p^2+6)(p^2+4)}dp$ which can be easily be done by using Partial fractions.
Substituting back from $p$ to $z$ to $t$ and finally to $sinx.$
I got $I=2arctan\sqrt{\frac{2-6sin^2x}{4sin^2x}}\color{red}{-}\sqrt{6}arctan\sqrt{\frac{2-6sin^2x}{6sin^2x}}$
But the answer given in the text is $\sqrt{6}arcsin(\sqrt{3}sinx)-2arctan(\frac{2sinx}{\sqrt{3Cos2x-1}})$
Where did i go wrong?
|
Not sure I can follow your last substitution, but I think the mistake was in dropping a sign when you changed variables from $t$ to $z$. It appears your answer is off by a sign.
Draw a right triangle and label one degree $\alpha$.Make the side opposite $\alpha$ a length of $\sqrt3\sin x$ and the hypoteneuse a length of $1$. Then $\sin\alpha=\sqrt3\sin x$ and $\alpha=\sin^{-1}(\sqrt3\sin x)$. By the Pythagorean theorem, the third side has length $\sqrt{1-3\sin^2x}.$ Note that $\tan\alpha=\frac{\sqrt3\sin x}{\sqrt{1-3\sin^2x}}=\sqrt\frac{3\sin^2x}{1-3\sin^2x}$. The other acute angle, however has a tangent equal to $\sqrt\frac{1-3\sin^2x}{3\sin^2x}.$ Therefore, $\tan^{-1}\sqrt\frac{1-3\sin^2x}{3\sin^2x}=\frac\pi2-\alpha$. Similarly, you can find that
$$\tan^{-1}\sqrt\frac{2-6\sin^2x}{4\sin^2x}=\tan^{-1}\frac{\sqrt{3\cos2x-1}}{2\sin x}=\frac\pi2-\tan^{-1}\frac{2\sin x}{\sqrt{3\cos2x-1}}$$
Therefore,
$$\sqrt6\tan^{-1}\sqrt\frac{1-3\sin^2x}{3\sin^2 x}-2\tan^{-1}\sqrt\frac{1-3\sin^2x}{2\sin^2x}=$$
$$\sqrt6(\frac\pi2-\sin^{-1}(\sqrt3\sin x))$$
Still editing... Why won't this refresh?
|
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|
Faulhaber Formula Identity I have to show the following identity:
$$ S_n^p := 1^p+2^p+...+n^p $$
$$ (p+1)S_n^p+\binom{p+1}{2}S_n^{p-1}+\binom{p+1}{3}S_n^{p-2}+...+S_n^0=(n+1)^{p+1}-1 $$
What I did first is to use the binomial theorem on the term the right side of the equation, which results in:
$$ (n+1)^{p+1}-1=\binom{p+1}{1}n+\binom{p+1}{2}n^{2}+...+\binom{p+1}{p}n^{p}+n^{p+1} $$
From here I am not quite sure how both of these terms are equal. Any hint on how I could proceed?
|
Suppose we have
$$S_n^p = \sum_{q=1}^n q^p$$
and we seek to evaluate
$$\sum_{q=1}^{p+1} {p+1\choose q} S_n^{p+1-q}
= - S_n^{p+1}
+ \sum_{q=0}^{p+1} {p+1\choose q} S_n^{p+1-q}.$$
Observe that
$$q^p = \frac{p!}{2\pi i}
\int_{|z|=\epsilon} z^{p-1} \exp(q/z) \; dz.$$
Introduce the generating function
$$f(w) = \sum_{p\ge 0} S_n^p \frac{w^p}{p!}.$$
We thus have for $f(w)$
$$f(w) =
\sum_{p\ge 0} \frac{w^p}{p!}
\frac{p!}{2\pi i}
\int_{|z|=\epsilon} z^{p-1}
\sum_{q=1}^n \exp(q/z) \; dz
\\ = \sum_{q=1}^n
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \exp(q/z) \frac{1}{z}
\sum_{p\ge 0} z^p w^p \; dz
\\ = \sum_{q=1}^n
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \exp(q/z) \frac{1}{z}
\frac{1}{1-zw} \; dz.$$
Note that the quantity we seek to evaluate is
$$-S_n^{p+1} + (p+1)! [w^{p+1}] f(w) \exp(w)$$
where the product term is given by the integral
$$\frac{(p+1)!}{2\pi i}
\int_{|w|=\epsilon} \frac{1}{w^{p+2}} \exp(w)
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \sum_{q=1}^n \exp(q/z)
\frac{1}{z}
\frac{1}{1-zw}
\; dz \; dw.$$
We evaluate the integral in $z$ using the negative of the sum of the
residues at $z=1/w$ and $z=\infty.$
For the residue at $z=1/w$ re-write the integral as follows:
$$-\frac{(p+1)!}{2\pi i}
\int_{|w|=\epsilon} \frac{1}{w^{p+3}} \exp(w)
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \sum_{q=1}^n \exp(q/z)
\frac{1}{z}
\frac{1}{z-1/w}
\; dz \; dw.$$
This yields
$$-\frac{(p+1)!}{2\pi i}
\int_{|w|=\epsilon} \frac{1}{w^{p+3}} \exp(w)
\sum_{q=1}^n \exp(qw) \times w \; dw
\\ = -\frac{(p+1)!}{2\pi i}
\int_{|w|=\epsilon} \frac{1}{w^{p+2}}
\sum_{q=1}^n \exp((q+1)w) \; dw.$$
The negative of this is
$$\sum_{q=1}^n (q+1)^{p+1}.$$
For the residue at infinity we get
$$-\mathrm{Res}_{z=0}
\frac{1}{z^2}
\sum_{q=1}^n \exp(qz) \times z \times \frac{1}{1-w/z}
\\ = -\mathrm{Res}_{z=0}
\sum_{q=1}^n \exp(qz) \times \frac{1}{z-w} = 0.$$
We have shown that
$$- S_n^{p+1}
+ \sum_{q=0}^{p+1} {p+1\choose q} S_n^{p+1-q}
= - S_n^{p+1} + \sum_{q=1}^n (q+1)^{p+1}
= (n+1)^{p+1} - 1,$$
as claimed.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solving $\cos(2x+\frac{\pi}{4})= -1/2 $ My suggestion:
$$\cos\left(2x+\frac{\pi}{4}\right)= -\frac{1}{2}$$
$$ 2x+\frac{\pi}{4} = \frac{2\pi}{3} \pm 2\pi n, n\in\mathbb{Z}$$
$$ x= \frac{\left( \frac{2\pi}{3} - \frac{\pi}{4} \right)}{2} \pm 2\pi n, n\in\mathbb{Z}$$
My answer:
$$ x = \frac{5\pi}{24} \pm 2\pi n, n\in\mathbb{Z} $$
But correct answer is:
$$-\frac{\pi}{8} \pm \frac {\pi}{3} + \pi n $$
What's my mistake?
|
HINT:
$\dfrac12=\cos\dfrac\pi3\implies-\dfrac12=\cos\left(\pi-\dfrac\pi3\right)$ as $\cos(\pi-u)=-\cos u$
$\implies2x+\dfrac\pi4=2m\pi\pm\dfrac{2\pi}3$ where $m$ is any integer
|
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|
Why does this sum converge? I know that the following sum converges to 2 via WolframAlpha, but I am not sure why.
$$\sum_{k=1}^\infty k \left[\frac{2}{k} - \frac{4}{k+1} + \frac{2}{k+2}\right] = 2$$
WolframAlpha gives the following partial sum formula:
$$\sum_{k=1}^n k \left[\frac{2}{k} - \frac{4}{k+1} + \frac{2}{k+2}\right] = \frac{2n}{n+2}$$
I would intuitively guess that the result of the partial sum formula is 2 for $n = \infty$. Where did that partial sum formula come from? Can someone help me build some intuition here?
|
Let the sum of series be denoted by $S$, the term corresponding to each k be $s_k$. Thus each $s_k$ is made of three terms.
$S=\sum s_k=\sum k(\frac{2}{k}-\frac{4}{k+1}+\frac{2}{k+2})$
$=\sum\limits_{k=1}^{\infty} (2-\frac{4k}{k+1}+\frac{2k}{k+2})$
Notice that last term of each $s_k$ and first 2 terms of $s_{k+1}$ have a sum $0$ as seen below: $\frac{2k}{k+2}+2-\frac{4(k+1)}{k+2}=2 +\frac{-2k-4}{k+2}=0$
Taking first two terms of $s_0$ and last term of $s_n$ as $n$ goes to infinity:
$S=\lim_{n \to \infty} (2-\frac{4*1}{1+1}+\frac{2n}{n+2})=\lim_{n \to \infty} \frac{2n}{n+2}=2$
|
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|
Show that $4$ does not divide $x^3-2$ Show that $4$ does not divide $x^3-2$ is what I need to prove.
I think I should put $4k$ is $x^3-2$ and then contradict it somehow.
Alternatively is to factor it out as $x^3$ is $x(x+2)(x-2)$ but I am not sure of that.
Do you know how to show this?
|
*
*Using modular arithmetic:
If $x \equiv 0 \mod 4$, then $x^3 \equiv 0 \mod 4$ and $x^3-2 \equiv 2 \mod 4$.
If $x \equiv 1 \mod 4$, then $x^3 \equiv 1 \mod 4$ and $x^3-2 \equiv 3 \mod 4$.
If $x \equiv 2 \mod 4$, then $x^3 \equiv 8 \equiv 0 \mod 4$ and $x^3-2 \equiv 2 \mod 4$.
If $x \equiv 3 \mod 4$, then $x^3 \equiv 27 \equiv 3 \mod 4$ and $x^3-2 \equiv 1 \mod 4$.
*Using a divisibility argument:
If $x$ is odd, then $x^3$ is odd and $x^3-2$ is odd, so it is not divisible by 4.
If $x$ is even, then $x^3$ is a multiple of 8 and thus of 4. So $x^3-2$ is not a multiple of 4.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Derive Cartesian cubic Möbius strip from parametric The following link:
http://mathworld.wolfram.com/MoebiusStrip.html
shows the Möbius strip parametrized as
\begin{eqnarray}
x = [ R + s \cos \left ( \frac{1}{2} t \right ) ] \cos t \\
y = [ R + s \cos \left ( \frac{1}{2} t \right ) ] \sin t \\
z = s \sin \left ( \frac12 t \right )
\end{eqnarray}
The symbols for $R$ and $s$ and angle $t$ are explained there.
Then they say that from this parametrization we can derive the cubic.
\begin{equation}
-R^2 y + x^2 y + y^3 - 2 R x z - 2 x^2 z + y z^2 = 0.
\end{equation}
Any ideas about how to do this? I have tried with no success.
Thanks.
This is what I have done so far:
Square the first two equations above and add to find
\begin{equation}
x^2 + y^2 = \left ( R + s \left ( \cos \frac{t}{2} \right ) \right )^2
\end{equation}
Take the square root of this
\begin{equation}
\sqrt{x^2 + y^2 }= R + s \left ( \cos \frac{t}{2} \right ) \Longrightarrow
\sqrt{x^2 + y^2 } - R = s \left ( \cos \frac{t}{2} \right )
\end{equation}
Square this and the third equation (for $z$) and add to find
\begin{equation}
s^2 = \left ( \sqrt{x^2+y^2}- R \right )^2 + z^2
\end{equation}
Now, let us divide the second by the first equation
That is
\begin{equation}
\frac{y}{x} = \tan t = \frac{2 \tan (t/2)}{1 - \tan^2 (t/2)}
\end{equation}
multiply numerator and denominator by $\cos^2 (t/2)$
\begin{equation}
\frac{y}{x} = \frac{2 \sin(t/2) (\sqrt{1-\sin^2(t/2)} }{\cos^2 (t/2) - \sin^2(t/2)}
= \frac{2 \sin(t/2) \sqrt{1 - \sin^2(t/2)}}{1 - 2 \sin^2(t/2)}.
\end{equation}
Multiply numerator and denominator by $s^2$, then
\begin{equation}
\frac{y}{x} = \frac{2 z \sqrt{s^2 - z^2}}{s^2 - 2 z^2}
\end{equation}
That is
\begin{equation}
\frac{y}{x} = \frac{2 z (\sqrt{x^2 + y^2} - R)}{ ( \sqrt{x^2 + y^2} - R)^2 - z^2}
\end{equation}
or
\begin{equation}
\frac{y}{x} = \frac{2 z (\sqrt{x^2 + y^2} - R)}{ (x^2 + y^2) + R^2 - 2 R \sqrt{x^2+y^2}
- z^2}
\end{equation}
Or
\begin{equation}
y (x^2 + y^2) + y R^2 - 2 R y \sqrt{x^2+y^2} - z^2 y =
2 z x \sqrt{x^2 + y^2} -2 R x z
\end{equation}
|
I'd just substitute the given coordinates into the equation and show that it's satisfied. The terms can be grouped according to the powers of $R$ and $s$ they contain, with the two exponents always summing to $3$, and the equation has to be satisfied for all four groups separately; that makes the calculation more manageable. For example, for $R^3s^0$ I get
$$-\sin t+\cos^2t\sin t+\sin^3t=0\;,$$
which is indeed satisfied.
|
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|
Solving $6 \cos x - 5 \sin x = 8$ My attempt:
Using the formula for linear combinations of sine and cosine:
$$A \cos x+B \sin x=C \sin (x+\phi)$$
$$
\sqrt{51} \left(\frac{6}{\sqrt{51}} \cos x - \frac{5}{\sqrt{51}}\sin x\right) = 8
$$
$$
\frac{6}{\sqrt{51}} \cos x - \frac{5}{\sqrt{51}}\sin x = \frac{8}{\sqrt{51}}
$$
And then assume:
$$
\frac{6}{\sqrt{51}}= \cos \psi ; \frac{5}{\sqrt{51}}= \sin\psi ;
$$
$$
\cos \psi \cos x - \sin \psi \sin x = \cos (x+ \psi) = \cos(x + \arccos ( \frac{6}{\sqrt{51}}))
$$
$$
x + \arccos\left(\frac{6}{\sqrt{51}}\right) = \arcsin\left( \frac{8}{\sqrt{51}}\right)
$$
$$
x \approx 12^\circ
$$
But answer is:
$$
-\frac{\pi}{4} + (-1)^n \frac{\pi}{4} + \pi n , n\in\Bbb Z
$$
|
$$6\cos { x } -5\sin { x } =8\\ 6\cos ^{ 2 }{ \frac { x }{ 2 } -6\sin ^{ 2 }{ \frac { x }{ 2 } -10\sin { \frac { x }{ 2 } \cos { \frac { x }{ 2 } =8 } \cos ^{ 2 }{ \frac { x }{ 2 } } +8\sin ^{ 2 }{ \frac { x }{ 2 } } } } } \\ 14\sin ^{ 2 }{ \frac { x }{ 2 } } +10\sin { \frac { x }{ 2 } \cos { \frac { x }{ 2 } } } +2\cos ^{ 2 }{ \frac { x }{ 2 } } =0\\ \cos ^{ 2 }{ \frac { x }{ 2 } } \neq 0$$
$$14\tan ^{ 2 }{ \frac { x }{ 2 } +10\tan { \frac { x }{ 2 } +2=0 } } \\ \tan { \frac { x }{ 2 } =a } \\ 7{ a }^{ 2 }+5a+1=0\\ $$
can you take from here?
|
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|
Simplifying $ 2\cos^2(x)\sin^2(x) + \cos^4(x) + \sin^4(x)$ who can simplify the following term in a simplest way?
$$ 2\cos^2(x)\sin^2(x) + \cos^4(x) + \sin^4(x)$$
(The answer is 1). Thanks for any suggestions.
|
Although the answer is straight forward but here are some steps if you find helpful
Notice, $$2\cos^2 x\sin^2 x+\cos^4 x+\sin^4 x$$ $$=(\cos^2 x)^2+(\sin^2 x)^2+2(\cos^2 x)(\sin^2 x)$$ Now, assume $\alpha =\cos^2 x$ & $\beta=\sin^2 x$ & apply $\alpha^2+\beta^2+2\alpha \beta=(\alpha+\beta)^2$$$=(\cos^2x+\sin^2 x)^2$$$$=(1)^2=1 \quad (\text{since,}\ \cos^2 x+\sin^2 x=1)$$
|
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|
$\lim_{n \to \infty} (\sqrt{9n^2 + 2n + 1} - 3n) = \frac{1}{3}$ Prove
$$\lim_{n \to \infty} (\sqrt{9n^2 + 2n + 1} - 3n) = \frac{1}{3}$$
I got this problem in Harro Heuser's "Lehrbuch der Analysis Teil 1". It is surely smaller than 1 because $\sqrt{9n^2 + 2n + 1} < \sqrt{9n^2 + 6n + 1} = 3n + 1$, but I cannot get closer than that, although it looks very simple...
|
Notice, $$\lim_{n\to \infty}(\sqrt{9n^2+2n+1}-3n)$$ Let, $n=\frac{1}{t} \implies t\to 0\ as\ n\to \infty$ $$=\lim_{t\to 0}\left(\sqrt{9\left(\frac{1}{t}\right)^2+2\left(\frac{1}{t}\right)+1}-3\left(\frac{1}{t}\right)\right)$$ $$=\lim_{t\to 0}\frac{\sqrt{t^2+2t+9}-3}{t}$$ Now, applying L-hospital's rule for $\frac{0}{0}$ form $$=\lim_{t\to 0}\frac{\frac{d}{dt}(\sqrt{t^2+2t+9}-3)}{\frac{d(t)}{dt}}$$ $$=\lim_{t\to 0}\frac{\frac{(2t+2)}{2\sqrt{t^2+2t+9}}-0}{1}$$
$$=\lim_{t\to 0}\frac{2(t+1)}{2\sqrt{t^2+2t+9}}$$$$=\lim_{t\to 0}\frac{t+1}{\sqrt{t^2+2t+9}}$$ $$=\frac{0+1}{\sqrt{0+2(0)+9}}$$ $$=\frac{0+1}{3}=\color{blue}{\frac{1}{3}}$$
|
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}
|
Show that $\frac{1}{b-a_{1}} + \frac{1}{b-a_2}+ ...+ \frac{1}{b-a_n} \geq \frac{n}{b-\frac{1}{n}(a_1+a_2+...+a_n)}$ Let, $b> \max\{a_1,a_2,...,a_n\}.$ Show that $\frac{1}{b-a_{1}} + \frac{1}{b-a_2}+ ...+ \frac{1}{b-a_n} \geq \frac{n}{b-\frac{1}{n}(a_1+a_2+...+a_n)}$
f is convex if $f(t_1x_1+t_2x_2+...+t_nx_n)\leq t_1f(x_1)+t_2f(x_2)+...+t_nf(x_n)$
Let, $t_i = 1/n \Rightarrow \sum\limits_{i=1}^n t_i = 1$
Then, $f(\frac{a_1+a_2+...+a_n}{n}) \leq \frac{1}{n}[f(a_1)+f(a_2)+...+f(a_n)]$
NTS: $\frac{1}{b-\frac{1}{n}(a_1+a_2+...+a_n)} \leq \frac{1}{n}[\frac{1}{b-a_1}+...+\frac{1}{b-a_n}]$
Let$f(x) = \frac{1}{b-x}$ In order to prove the inequality, we have to show that $f''(x) \geq 0.$ So, $f''(x) = \frac{2}{(b-x)^3}$
Now, how can we show that $f''(x) \geq 0$? $b$ can be a negative number, can it?
|
I would use the function $g(x)=\frac{1}{x}$ defined on $(0,\infty)$; you can readily verify that $g''(x)>0$ on its domain. (The issue with your $f$ is that $f''(x)>0$ only for $x<b$.) The rest of the argument proceeds similarly: convexity gives
$$
\frac{1}{n}\times\text{LHS}=\frac{1}{n}\sum_{i=1}^ng(b-a_i)\geq g\left(b-\sum_ia_i/n\right)=\frac{1}{n}\times\text{RHS}\implies\text{LHS}\geq \text{RHS}.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1370668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Wanted : for more formulas to find the area of a triangle? I know some formulas to find a triangle's area, like the ones below.
*
*Is there any reference containing most triangle area formulas?
*If you know more, please add them as an answer
$$s=\sqrt{p(p-a)(p-b)(p-c)} ,p=\frac{a+b+c}{2}\\s=\frac{h_a*a}{2}\\s=\frac{1}{2}bc\sin(A)\\s=2R^2\sin A \sin B \sin C$$
Another symmetrical form is given by :$$(4s)^2=\begin{bmatrix}
a^2 & b^2 & c^2
\end{bmatrix}\begin{bmatrix}
-1 & 1 & 1\\
1 & -1 & 1\\
1 & 1 & -1
\end{bmatrix} \begin{bmatrix}
a^2\\
b^2\\
c^2
\end{bmatrix}$$
Expressing the side lengths $a$, $b$ & $c$ in terms of the radii $a'$, $b'$ & $c'$ of the mutually tangent circles centered on the triangle's vertices (which define the Soddy circles)
$$a=b'+c'\\b=a'+c'\\c=a'+b'$$gives the paticularly pretty form $$s=\sqrt{a'b'c'(a'+b'+c')}$$
If the triangle is embedded in three dimensional space with the coordinates of the vertices given by $(x_i,y_i,z_i)$ then $$s=\frac{1}{2}\sqrt{\begin{vmatrix}
y_1 &z_1 &1 \\
y_2&z_2 &1 \\
y_3 &z_3 &1
\end{vmatrix}^2+\begin{vmatrix}
z_1 &x_1 &1 \\
z_2&x_2 &1 \\
z_3 &x_3 &1
\end{vmatrix}^2+\begin{vmatrix}
x_1 &y_1 &1 \\
x_2&y_2 &1 \\
x_3 &y_3 &1
\end{vmatrix}^2}$$
When we have 2-d coordinate $$ s=\frac{1}{2}\begin{vmatrix}
x_a &y_a &1 \\
x_b &y_b &1 \\
x_c &y_c & 1
\end{vmatrix}$$
In the above figure, let the circumcircle passing through a triangle's vertices have radius $R$, and denote the central angles from the first point to the second $q$, and to the third point by $p$ then the area of the triangle is given by:
$$ s=2R^2|\sin(\frac{p}{2})\sin(\frac{q}{2})\sin(\frac{p-q}{2})|$$
|
Using the law of sines, one can get
$$s=\frac{1}{2}bc \sin(\alpha)=\frac{1}{2}bc \frac{a}{2R}=\frac{abc}{4R}$$
where $R$ is the radius of the circumscribed circle.
|
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"timestamp": "2023-03-29T00:00:00",
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|
$ x^2 + \frac {x^2}{(x-1)^2} = 2010 $ I found this question from last year's maths competition in my country. I've tried any possible way to find it, but it is just way too hard.
Given $$ x^2 + \frac {x^2}{(x-1)^2} = 2010,$$
find $\dfrac {x^2} {x-1}.$
(A) $1+\sqrt {2011}$ (B) $ 1-\sqrt {2011}$ (C) $1\pm \sqrt{2011} $ (D) $\sqrt {2011}$
So I multiply them with $(x-1)$
$$x^2(x-1) + \frac {x^2} {x-1} = 2010(x-1)$$
$$\frac {x^2} {x-1} = (x-1)(2010-x^2)$$
and I stuck in here, dont know how to remove $x$ in there
|
Hint:
$$a^2 + b^2 = (a+b)^2-2ab$$
Going forward,
$$ \left(x + \frac{x}{x-1}\right)^2 - \frac{2x^2}{x-1} = 2010$$
$$ \left(\frac{x^2}{x-1}\right)^2 - \frac{2x^2}{x-1} = 2010$$
Let $\frac{x^2}{x-1} = n$
$$ n^2 -2n -2010 =0 $$
You should be able to solve this using the quadratic formula or otherwise to get a value for $ \frac{x^2}{x-1}$
Indeed, $\frac{x^2}{x-1} = 1\pm \sqrt{2011}$
|
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|
Find $x$ if $\frac {1} {x} + \frac {1} {y+z} = \frac {1} {2}$ I found this question from past year's maths competition in my country. I've tried any possible way to find it, but it is just way too hard.
Find $x$ if \begin{align}\frac {1} {x} + \frac {1} {y+z} &= \frac {1} {2}\\ \frac {1} {y} + \frac {1} {x+z} &= \frac{1}{3}\\ \frac {1} {z} + \frac {1} {x+y} &= \frac {1} {4}\end{align}
$(A)\;\frac 32$
$(B)\;\frac {17}{10}$
$(C)\;\frac {19}{10}$
$(D)\;\frac {21}{10}$
$(E)\;\frac {23}{10}$
EDIT: I'm very sorry guys, it should be $\frac {1} {x+y}$ not $xy$, I'm sorry for the typos (idk what is wrong with me)
|
multiplying (2) by $x$ $(x \ne 0)$ we obtain
$$\frac{1}{z}=\frac{1}{4}-\frac{1}{xy}$$
with (3) we obtain $$y=\frac{12(x^2-1)}{4x^2-3x}$$ (I)
from (1) and (3) we get
$$\frac{x^2y}{3}-x^2+1=\frac{xy}{4}$$
plugging (I) in this equation and simplifying we get
$$-60 x^5+119 x^4+156 x^3-288 x^2-72 x+144=0$$
with five real solutions, no of them from (A) to (E)
the system after the correction has the solution $$x=\frac{23}{10}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
$\int\dfrac{dx}{x^2(x^4+1)^{3/4}}$
Evaluate $$\large{\int\dfrac{dx}{x^2(x^4+1)^{3/4}}}$$
I thought of rewriting this as $$\large{\int\dfrac{dx}{x^5(1+\frac{1}{x^4})^{3/4}}}$$ and substituting $$u^4=\left(1+\frac{1}{x^4}\right)\Rightarrow u=\left(1+\frac{1}{x^4}\right)^{1/4}$$ and subsequently I got $$du=\dfrac{1}{4}\left(1+\frac{1}{x^4}\right)^{-3/4}\times (-4x^{-5})dx$$
However, I can not think of how to proceed further. Any help would be truly appreciated. Many thanks in advance!
|
We have $$\int\frac{1}{x^{2}\left(x^{4}+1\right)^{3/4}}dx\overset{u=1/x^{4}}{=}-\frac{1}{4}\int\frac{1}{\left(u+1\right)^{3/4}}du\overset{u+1=v}{=}-\frac{1}{4}\int\frac{1}{v^{3/4}}d=
$$ $$=-\sqrt[4]{v}+C=-\frac{\sqrt[4]{x^{4}+1}}{x}+C.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1373612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Evaluate the Integral: $\int(x^5+5^x)\ dx$ $\int(x^5+5^x)\ dx$
I made the the terms within the parenthesis u
$u=x^5+5^x$
$du=5x^4+5^xln\ 5$
$du=5x^4+5^x\ ln\ 5 dx$
$\frac{u}{5x^4+5^xln\ 5}\ du$
I am stuck at this point. Is there a better way to attack this problem? I know I have to use this formula
$\int\ a^x dx=\frac{a^x}{ln\ a}+C$
But how?
|
If you want to evaluate this integral, you should simplify like this:
$\int{(x^5 + 5^x) dx} = \int{x^5 dx} + \int{5^x dx}$. You can always split up an expression of the same variable like this to simplify your computation. Now, generally, $\int{x^ndx} = \frac{x^{n+1}}{n+1} +C$, and $\int{a^xdx} = \frac{a^x}{ln{(a)}} +C$.
So, our result is: $\int{x^5 dx} + \int{5^x dx} = \frac{x^6}{6} + \frac{5^x}{ln{(5)}} +C$. $C$ is the constant of integration, due to the integral being indefinite.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate the Integral: $\int \frac{\log_{10}\ x}{x}\ dx$ $\int \frac{\log_{10}\ x}{x}\ dx$
$du=x\ln10\ dx$
$\log_{10}x\ \ln10+ C$
Is this answer correct? If not what step should I take to convert the log into a term I can manipulate?
|
$$\int { \frac { \log _{ 10 }{ x } }{ x } dx } =\int { \log _{ 10 }{ x } d\ln { x } =\log _{ 10 }{ x } \ln { x } -\int { \frac { \ln { x } }{ x\ln { 10 } } +C= } } \\ =\log _{ 10 }{ x } \ln { x } -\frac { 1 }{ \ln { 10 } } \int { \ln { x } d\ln { x } +C= } \log _{ 10 }{ x } \ln { x } -\frac { 1 }{ \ln { 10 } } \ln { x } +C=\\ =\ln { x\left( \log _{ 10 }{ x } -\frac { 1 }{ \ln { 10 } } \right) +C } $$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Simultaneous equations, $\frac{1}{x}+\frac{1}{y}=1$,$x+y=a$,$\frac{y}{x}=m$ By eliminating $x$ and $y$ from the following equations, I need to find the relation between $m$ and $a$.
\begin{align*}
\frac{1}{x}+\frac{1}{y}=1 \\
x+y=a \\
\frac{y}{x}=m
\end{align*}
I tried different ways, but cannot arrive at the answer. I end up with a quadratic.
This is what I have.
\begin{align*}
\frac{1}{x}=1-\frac{1}{y} \\
x(y-1)=y \\
x=\frac{y}{y-1}
\end{align*}
Now using second equation:
\begin{align*}
\frac{y}{y-1}+\frac{y(y-1)}{y-1}=a \\
\frac{y^2}{y-1}=a \\
y^2-ay+a=0
\end{align*}
Now roots are: $y_{1,2}=\displaystyle{\frac{a\pm\sqrt{a^2-4a}}{2}}$
Then it follows that: $x_{1,2}=\displaystyle{\frac{a\pm\sqrt{a^2-4a}}{2}}$
Hence there are three scenarios for the relation between $a$ and $m$.
First: $\frac{4a}{4}=m$, $a=m$
Second: $\displaystyle{\frac{(a+\sqrt{a^2-4a})^2}{4}=m}$
Third: $\displaystyle{\frac{(a-\sqrt{a^2-4a})^2}{4}=m}$
Is this correct? Thank you
|
Since you need to find the relation between $a$ and $m$, you can consider that you face a problem of three equations for three unknowns $x,y,a$ for which the solution is unique $$\left\{x= \frac{m+1}{m},y= m+1,a= \frac{(m+1)^2}{m}\right\}$$
|
{
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|
$f(x) =ax^6 +bx^5+cx^4+dx^3+ex^2+gx+h $, find $f(7)$. Problem :
Given a polynomial $f(x) =ax^6 +bx^5+cx^4+dx^3+ex^2+gx+h$
such that $f(1)= 1, f(2) =2 , f(3) = 3, f(4) =4, f(5)=5, f(6) =6$.
Find $f(7)$ in terms of $h$.
My approach:
We can put the values of $f(1) = 1$ in the given equation and $f(2) = 2$, etc. But this is quite time consuming by making six different equations and then solve them to get the values of $a,b,c,d,e,g,h$. Please suggest some alternate solution for this.
|
Hint:
$$f(x)=K(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)+x$$
For some constant K. Plug in $x=0$ to get K and then plug $x=7$ to get answer.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proving $\sum_{i=1}^n\frac{1}{i(i+1)(i+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$ for $n\geq 1$ by mathematical induction Prove using mathematical induction that
$$\frac{1}{1\cdot 2\cdot 3} + \frac{1}{2\cdot 3\cdot 4} + \cdots + \frac{1}{n(n+1)(n+2)} = \frac{n(n+3)}{4(n+1)(n+2)}.$$
I tried taking $n=k$,
so it makes
$$\frac{1}{1\cdot 2\cdot 3} + \frac{1}{2\cdot 3\cdot 4} + \cdots + \frac{1}{n(n+1)(n+2)} = \frac{k(k+3)}{4(k+1)(k+2)}.$$
Then proving the statement for $n=k+1$:
$$\frac{k(k+3)}{4(k+1)(k+2)}+ \frac{1}{(k+1)(k+2)(k+3)} = \frac{(k+1)(k+1+3)}{4(k+1+1)(k+1+2) }.$$
|
Actually, mathematical induction is not necessary for this formula. A direct calculation with a telescoping decomposition into partial fractions will show where the formula comes from.
Indeed one checks that
$$\frac1{k(k+1)(k+2)}=\frac12\frac1k-\frac1{k+1}+\frac12\frac1{k+2}.$$
Thus the sum is:
\begin{align*}&\Bigl(\frac12-\frac12+\frac12\frac13\bigr)+\Bigl(\frac12\frac12-\frac13+\frac12\frac14\Bigr)+\Bigl(\frac12\frac13-\frac14+\frac12\frac15\bigr)+\dotsm\dotsm\\&\color{red}+\frac1{2(n-2)}-\frac1{n-1}+\frac1{2n}\color{red}+\frac1{2(n-1)}-\frac1n+\frac1{2(n+1)}\color{red}+\frac1{2n}- \frac1{n+1} +\frac1{2(n+2)}\\
&=\frac14+\frac12\Bigl(\frac1{n+2}-\frac1{}\Bigr)=\frac{(n+1)(n+2)-2}{4(n+1(n+2)} = \frac{n(n+3)}{4(n+1(n+2)}.
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to rotate a whole rectangle by an arbitrary angle around the origin using a transformation matrix? Suppose, I have a 2D rectangle ABCD like the following:
$A(0,0)$, $B(140,0)$, $C(140,100)$, $D(0,100)$.
I want to rotate the whole rectangle by $\theta = 50°$.
I want to rotate it around the Z-axis by an arbitrary angle using a rotation transformation matrix.
How to do that?
I know that, $$ A =
\begin{bmatrix}
\
0 & 0 & 1 \\
\end{bmatrix};
B =
\begin{bmatrix}
\
140 & 0 & 1 \\
\end{bmatrix};
C =
\begin{bmatrix}
\
140 & 100 & 1 \\
\end{bmatrix};
D =
\begin{bmatrix}
\
0 & 100 & 1 \\
\end{bmatrix}.
$$
And, I know that the rotation matrix is, $$R =
\begin{bmatrix}
\
cos \theta & -sin \theta & 0 \\
sin \theta & cos \theta & 0 \\
0 & 0 & 1 \\
\end{bmatrix}$$
Now, what is the calculation?
I have tried the following ways,
$$Rotation =
\begin{bmatrix}
\
0 & 0 & 1 \\
140 & 0 & 1 \\
140 & 100 & 1 \\
0 & 100 & 1 \\
\end{bmatrix}.\begin{bmatrix}
\
cos \theta & -sin \theta & 0 \\
sin \theta & cos \theta & 0 \\
0 & 0 & 1 \\
\end{bmatrix}$$
And,
$$
A' = \begin{bmatrix}
\
0 & 0 & 1 \\
\end{bmatrix} . \begin{bmatrix}
\
cos \theta & -sin \theta & 0 \\
sin \theta & cos \theta & 0 \\
0 & 0 & 1 \\
\end{bmatrix}\\
B' = \begin{bmatrix}
\
140 & 0 & 1 \\
\end{bmatrix} . \begin{bmatrix}
\
cos \theta & -sin \theta & 0 \\
sin \theta & cos \theta & 0 \\
0 & 0 & 1 \\
\end{bmatrix}\\
C' = \begin{bmatrix}
\
140 & 100 & 1 \\
\end{bmatrix} . \begin{bmatrix}
\
cos \theta & -sin \theta & 0 \\
sin \theta & cos \theta & 0 \\
0 & 0 & 1 \\
\end{bmatrix}\\
D' = \begin{bmatrix}
\
0 & 100 & 1 \\
\end{bmatrix} . \begin{bmatrix}
\
cos \theta & -sin \theta & 0 \\
sin \theta & cos \theta & 0 \\
0 & 0 & 1 \\
\end{bmatrix}
$$
For example,
$$B' = \begin{bmatrix}
\
140 & 0 & 1 \\
\end{bmatrix} . \begin{bmatrix}
\
cos 50^\circ & -sin 50^\circ & 0 \\
sin 50^\circ & cos 50^\circ & 0 \\
0 & 0 & 1 \\
\end{bmatrix} = \begin{bmatrix}
\
89.99 & -107.24 & 1 \\
\end{bmatrix}\\
C' = \begin{bmatrix}
\
140 & 100 & 1 \\
\end{bmatrix} . \begin{bmatrix}
\
cos 50^\circ & -sin 50^\circ & 0 \\
sin 50^\circ & cos 50^\circ & 0 \\
0 & 0 & 1 \\
\end{bmatrix} = \begin{bmatrix}
\
166.59 & -42.96 & 1 \\
\end{bmatrix}$$
What is the right way to work with?
|
When you show $B'$ as $(5,7)$ it is rotated by about 54.5^\circ and the length of the side is now $\sqrt{74} \approx 8.6$, not $7$ This is a problem with your expectation, not the code. You also have the sign of the sine backwards if you are using row vectors-rotate $(1,0,1)$ by $45^\circ$ and it should be $(\frac 12\sqrt 2, \frac 12\sqrt 2,1)$ but you put a minus sign on the second coordinate. A proper rotation of $B$ by $50^\circ$ is $(7 \cos 50^\circ, 7 \sin 50^\circ)\approx (4.5,5.36)$. $C'=(7 \cos 50^\circ-5 \sin 50^\circ, 7 \sin 50^\circ + 5 \cos 50^\circ)\approx(0.6693,8.576)$
|
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|
Prove that a trigonometric equation has six distinct roots Show that,in general,the equation $A \sin^3x+B\cos^3x+c=0 $has six distinct roots,no two of which differ by $2\pi$,and that the tangent of their semi-sum is $-\frac{A}{B}$.
My attempt:
I tried to express it as sixth degree equation.
$A \sin^3x+B\cos^3x+C=0 $
$A \sin^3x+B\cos^3x=-C $
Squaring both sides,
$A^2 \sin^6x+B^2\cos^6x+2AB\sin^3x\cos^3x=C^2 $
Does this prove that this equation has six distinct roots,no two of which differ by $2\pi$ and second part i could not prove.Can someone enlighten me in this problem?Thanks in advance..
|
By setting $u=\tan\frac{x}{2}$, we have to solve:
$$ A\left(\frac{2u}{1+u^2}\right)^3+B\left(\frac{1-u^2}{1+u^2}\right)^3 = -c $$
that is equivalent to finding the roots of the sixth-degree polynomial:
$$ p(u)=(c+B)+3(c-B)u^2+8A u^3+3(c+B)u^4+(c-B)u^6. $$
I do not agree that this polynomial has in general six real roots: by looking at the coefficients of $u^0,u^1,u^2$ we have that Newton's inequality does not hold, so $p(u)$ may have at most $\color{red}{4}$ real roots, since complex roots come in conjugated pairs.
|
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|
Simplifying transfer functions in Z domain I have difficulties to check whether the below transfer function is recursive or non-recursive:
$$H(z)=\frac{1-z^{-1}+z^{-2}-3z^{-3}}{z^{-2}(1-z^{-1})}$$
I know that I have to multiply the num and denum by ${z^3}$ but the problem here is the denum. I think it must be first simplified even more, or maybe it should be flatted like:
$$z^{-2}(1-z^{-1}) = z^{-2} - 2z^{-3}$$
I cant go further from here...can you please tell me what is the correct procedure?
|
Starting from
$$H(z)=\frac{1-z^{-1}+z^{-2}-3z^{-3}}{z^{-2}(1-z^{-1})}$$
then
\begin{align}
H(z) &= \frac{z^{3}}{z^{3}} \cdot \frac{1-z^{-1}+z^{-2}-3z^{-3}}{z^{-2}(1-z^{-1})} \\
&= \frac{z^{3} - z^{2} + z - 3}{z-1} = \frac{z^{2} (z-1) + (z-1) -2}{ z-1} \\
&= 1 + z^{2} + \frac{2}{1-z} = 1 + z^{2} + 2 \, \sum_{n=0}^{\infty} z^{n} \\
&= 3 + 2 z + 3 z^{2} + 2 z^{3} + 2 z^{4} + \cdots.
\end{align}
|
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|
Solve the differential equation $\{xy\log \frac{x}{y}\}dx+\{y^2-x^2 \log \frac{x}{y}\}dy=0$ given that $y(1)=0$. Solve the differential equation $\{xy\log \frac{x}{y}\}dx+\{y^2-x^2 \log \frac{x}{y}\}dy=0$ given that $y(1)=0$.
I was able to find the solution to it by the method of solving homogeneous differential equations, which is as follows:
$\frac {x^2}{2y^2} \log \frac{x}{y}-\frac {x^2}{4y^2}+ \log y = C$.
But I am not able to find the value of $C$, I am not getting how to substitute for $y$ because it seems to give not-defined values. Please help.
According to the answer hint, $C=1- \frac{3}{4e^2}$.
Thanks in advance...
EDIT:
Since most of you are asking for the complete steps I followed to get the final solution, I am posting it as follows:
$\{xy\log \frac{x}{y}\}dx+\{y^2-x^2 \log \frac{x}{y}\}dy=0$
Let $x=vy \implies \frac{dx}{dy}=v+y \frac{dv}{dy}$
Therefore, $\frac{dx}{dy}=-(\frac{y^2-x^2 \log \frac{x}{y}}{xy\log \frac{x}{y}})$
$\implies v+y \frac{dv}{dy}= \frac{v^2 \log v-1}{v \log v}$
$\implies y \frac{dv}{dy}= \frac{v^2 \log v-1}{v \log v}-v=\frac{-1}{v \log v}$
$\implies v \log v dv + \frac{1}{y} dy = 0$
Integrating the equation we get,
$\frac{v^2}{2} \log v-\frac{v^2}{4}+\log y=C$
On replacing value of $v=\frac{x}{y}$
$\frac {x^2}{2y^2} \log \frac{x}{y}-\frac {x^2}{4y^2}+ \log y = C$
Now, $y(1)=0$
Therefore, $\frac {1^2}{2\times 0^2} \log \frac{1}{0}-\frac {1^2}{4\times 0^2}+ \log 0 = C$
How do I solve for $C$? Please help...
|
$$\left(xy\log\frac{x}{y}\right)dx+\left(y^2-x^2\log\frac{x}{y}\right)=0$$
$$\left(xy\log\frac{y}{x}\right)dx=\left(y^2+x^2\log\frac{y}{x}\right)dy$$ $$\left(\frac{y}{x}\log\frac{y}{x}\right)dx=\left(\left(\frac{y}{x}\right)^2+\log\frac{y}{x}\right)dy$$ Let $\frac{y}{x}=u \implies y=ux \implies \frac{dy}{dx}=u+x\frac{du}{dx}$
$$\left(u\log u\right)=\left(u^2+\log u\right)\left(u+x\frac{du}{dx}\right)$$
$$x\frac{du}{dx}=\frac{u\log u}{u^2+\log u}-u$$
$$x\frac{du}{dx}=\frac{u\log u-u^3-u\log u}{u^2+\log u}=\frac{-u^3}{u^2+\log u}$$
$$\frac{(u^2+\log u)du}{u^3}=-\frac{dx}{x}$$
Can you take it from here?
|
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"timestamp": "2023-03-29T00:00:00",
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|
Behaviour of $\zeta(s)$ near $s=1$ I would appreciate if somebody could run this over and see if it works out? any suggestions or pointers would be appreciated. I denote the standard eta function $\eta$ by $\zeta^{*}$. I have not used big O notation and just used general well behaved functions. I do not wish to express the full error term, but instead ,just the principal part.
Behaviour of $\zeta(s)$ near $1$
From Abel's Theorem we can see that when $s=1$, $ \zeta^{*}(1) = \log(2)$. Now looking at $(1-2^{1-s})$ we can write it in terms of an exponential like so,
\begin{equation}
1-2^{1-s} = 1 - e^{(1-s)\log(2)}
\end{equation}
The power series expansion of $e^{z}$ is,
\begin{equation}
e^{z}= \sum_{n=0}^{\infty} \frac{z^{n}}{n!}\\
\Rightarrow 1-2^{1-s} = - e^{\log(2)(s-1)}= - \sum_{n=0}^{\infty} \frac{((1-s)\log(2))^{n}}{n!}
\end{equation}
We can ignore the term when $n=0$ due to it being zero and sum from $n=1$ instead,
\begin{equation}
1-2^{1-s} = 0 - \sum_{n=1}^{\infty} \frac{(1-s)^{n}\log(2)^{n}}{n!}
\end{equation}
Expanding this sum and multiplying in the negative sign we have,
\begin{equation}
1-2^{1-s}= (s-1) \Bigg( \log(2) - \frac{\log(2)^{2}}{2!}(s-1) + \cdot \cdot \cdot \Bigg )
\end{equation}
Factorizing the $\log(2)$ term out,
\begin{equation*}
(s-1)\log(2)\Bigg [ 1 - \bigg( \frac{\log(2)}{2!}(s-1) + \frac{\log(2)}{3!}(s-1)^{2} - \cdot \cdot \cdot \bigg ) \Bigg ]
\end{equation*}
By the geometric series formula, for $|s| < 1$,
\begin{equation}
\frac{1}{\bigg[1 - \bigg( \frac{\log(2)}{2!}(s-1) + \cdot \cdot \cdot \bigg ) \bigg ] }= 1 + \Bigg( \frac{\log(2)}{2!}(s-1) + \cdot \cdot \cdot \Bigg ) + \Bigg( \frac{\log(2)}{2!}(s-1) + \cdot \cdot \cdot \Bigg )^{2} + \cdot \cdot \cdot
\end{equation}
The terms of this geometric series decrease rapidly, so we are only interested in keeping the first terms while letting a well-behaved and analytic function $g$ represent the remaining terms as a function in $s$.
\begin{equation}
\frac{1}{\bigg[1 - \bigg( \frac{\log(2)}{2!}(s-1) + \cdot \cdot \cdot \bigg ) \bigg ] } = 1 + \frac{\log(2)(s-1)}{2} + (s-1)^{2}\cdot g(s).
\end{equation}
We can now return to $\frac{1}{1-2^{1-s}}$, and express it in terms of what we have learned.
\begin{equation}
\frac{1}{1-2^{1-s}} = \frac{1}{\log(2)(s-1)} \Bigg( 1 + \frac{\log(2)(s-1)}{2} + (s-1)^{2}\cdot g(s) \Bigg )
= \frac{1}{\log(2)} \cdot \Bigg [ \frac{1}{s-1} + \frac{\log(2)}{2} + (s-1)g(s)\Bigg ]
\end{equation}
We can now study $\zeta(s)$ when $s$ is near to $1$.
\begin{equation}
\zeta(s) = \frac{\zeta^{*}(s)}{1-2^{1-s}} = \frac{\zeta^{*}(s)}{\log(2)} \cdot \Bigg [ \frac{1}{s-1} + \frac{\log(2)}{2} + (s-1)g(s)\Bigg ]
= \frac{\zeta^{*}(s)}{\log(2)} \cdot \frac{1}{s-1} + \frac{\zeta^{*}(s)}{2 \log(2)} \log(2) + \frac{\zeta^{*}(s)(s-1)g(s)}{\log(2)}
\end{equation}
As we know already, $\zeta^{*}(1) = \log(2)$ is analytic, so $\zeta^{*}(s)$ can be expanded as a series around $1$,
\begin{equation}
\zeta^{*}(s) = \log(2) + (s-1)a_1 + (s-1)^{2}a_2 + \cdot \cdot \cdot
= \log(2) + (s-1) h(s)
\end{equation}
for a well behaved and analytic $h$.
Near $s=1$ and by just looking at the principal terms,
\begin{equation}
\zeta(s) = \frac{\zeta^{*}(s)}{1-2^{1-s}} = \frac{ \log(2) +(s-1)h(s) }{\log(2)(s-1)}
= \frac{1}{s-1} + \frac{h(s)}{\log(2)}
\end{equation}
|
It looks fine to me. Maybe it's interesting to note that from $$\zeta\left(s\right)=s\int_{1}^{\infty}\frac{\frac{1}{2}-\left\{ x\right\} }{x^{s+1}}dx+\frac{1}{s-1}+\frac{1}{2},\,\,\textrm{Re}\left(s\right)>0$$ where $\left\{ x\right\}$ is the fractional part of $x$, we can easily get $$\zeta\left(s\right)=\frac{1}{s-1}+\gamma+O\left(\left|s-1\right|\right)$$ when $s$ is near to $1$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find min & max of $f(x,y) = x + y + x^2 + y^2$ when $x^2 + y^2 = 1$ Problem: Find the maximum and minimal value of $f(x,y) = x + y + x^2 + y^2$ when $x^2 + y^2 = 1$.
Since $x^2 > x$ (edit $x^2 \geq x$) for all $x \in \mathbb{R}$, $f$ is bowl-ish with a minimal value in the bottom.
This is a critical point which means that we can set partial derivatives of $f$ equal to $0$ and try to solve for $x$ and $y$
$\nabla f = (1+2x, 1+2y) = (0,0) \implies (x,y) = (\frac{-1}{2}, \frac{-1}{2})$
So we get the minimal value $f(\frac{-1}{2}, \frac{-1}{2}) = \frac{-1}{2} + \frac{-1}{2} + (\frac{-1}{2})^2 \frac{-1}{2})^2 = -\frac{1}{2}$
But how about the maximal value? How does $x^2 + y^2 = 1$ restrict $f$?
|
If $x^2 + y^2 = 1$ then
$$
f(x,y) = 1 + x + y.
$$
Now, let $x=\cos t$, $y=\sin t$. So,
$$
f(t) = 1 + \cos t + \sin t = 1 + \sqrt2 \sin(t + \pi/4).
$$
Could you proceed?
|
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|
10th derivative of a function I want to find $f^{(10)}(0)$ where $f(x)=\ln(2+x^2)$.
I know that it can be done "by hand", but I believe there is a smarter way.
I think I should use Taylor series and the fact that $f^{(n)}(0)=a_n*n!$ , but I'm not sure how.
|
We have
$$
\frac{2x}{x^2+2} = \frac{x}{1+\frac{x^2}{2}} = x\bigg(1-\frac{x^2}{2} + \frac{x^4}{4} - \frac{x^6}{8} + \frac{x^8}{16}+\cdots \bigg).
$$
Then integrating we get
$$
\log(x^2+2)-\log2 = \int_0^x \frac{2t}{t^2+2}\; dt = \frac{x^2}{2}-\frac{x^4}{2\cdot 4} + \frac{x^6}{4\cdot 6} - \frac{x^8}{8\cdot 8} + \frac{x^{10}}{10\cdot 16}+\ldots
$$
Answer: $22680$ :)
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove $ \ \frac{a}{x^3 + 2x^2 - 1} + \frac{b}{x^3 + x - 2} \ = \ 0 \ $ has a solution in $ \ (-1,1) $ If $a$ and $b$ are positive numbers, prove that the equation
$$\frac{a}{x^3 + 2x^2 - 1} + \frac{b}{x^3 + x - 2} = 0$$
has at least one solution in the interval $ \ (-1,1) \ $ .
The question is from the exercises section of a textbook chapter on limits/continuity.
I've been stumped on this one for a couple of days. I've been trying to calculate $\lim _{x \to -1}$ and $\lim _{x \to 1}$ and then show the function is continuous to show a root must lie in the interval. Factorising the denominators gives...
$$\frac{a}{(x+1)(x^2+x-1)} + \frac{b}{(x-1)(x^2+x+2)} = 0$$
So of course $x = 1$ and $x = -1$ are undefined and so the limits will be one-sided. Playing around with equation I haven't been able to find an equivalent function across $x \neq -1, x \neq 1$.
The only thing I have been able to show is
$$\frac{a}{b} = - \frac{(x+1)(x^2+x-1)}{(x-1)(x^2+x+2)}$$
and so
$$\lim _{x \to -1} \frac{a}{b} = 0, \lim _{x \to 1} \frac{b}{a} = 0$$
but I'm not sure if this is significant or I'm overthinking things. Could anyone point me in the right direction?
|
To solve this problem, first of all think about the x values for which the following function is undefined.
We have,
$$f(x) = \frac{a}{x^3 + 2x^2 - 1} + \frac{b}{x^3 + x - 2}$$
which can be written as,
$$f(x) = \frac{a}{(x+1)(x^2+x-1)} + \frac{b}{(x-1)(x^2+x+2)}$$
Note that the above function is undefined for
$$x = \pm 1\ , \ \frac{-1\pm \sqrt{5}}{2} \ ...(1)$$
Our interval is $(-1,1)$, so forget about $\frac{-1 - \sqrt{5}}{2}$ which is approximately $-1.62$.
Now, let's calculate the following limits,
$$\lim_{x \to 1^{-}}f(x) =\lim_{x \to 1^{-}}\left( \frac{a}{x^3 + 2x^2 - 1} + \frac{b}{x^3 + x - 2}\right) = -\infty $$ as the first term is simply $\frac{a}{2}$ and the function $x^3 + x - 2$ goes to $0$ from behind as x approaches $1$ from the left. So, the second term in the function is a huge negative number.
And,
$$\lim_{x \to {\left( \frac{-1 + \sqrt{5}}{2}\right)}^{+}}f(x) =\lim_{x \to {\left( \frac{-1 + \sqrt{5}}{2}\right)}^{+}}\left( \frac{a}{x^3 + 2x^2 - 1} + \frac{b}{x^3 + x - 2}\right) = \infty $$ as the function $x^3 + 2x^2 - 1$ approaches $0$ from above from x approaches $\frac{-1 + \sqrt{5}}{2}$ from the right. So, the first term is actually a big positive number and the second term which is actually defined is negligible.
Now, the reason for calculating these limits was to get to the answer by saying that between x = $\frac{-1 + \sqrt{5}}{2}$ and x = $1$, $f(x)$ is continuous everywhere (see $(1)$). The limits for these values of x show that the function has to cross the x-axis between these values of x, that is, there should be at least one root!
The given graph for $ \frac{2}{x^3 + 2x^2 - 1} + \frac{3}{x^3 + x - 2}$
illustrates the above proof:
|
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|
If $\alpha, \beta, \gamma$ solutions for the equation If $\alpha, \beta, \gamma$ solutions for the equation
$$x^{3}+x^{2}+1=0$$
Then
$$\frac{1+\alpha }{2-\alpha }+\frac{1+\beta }{2-\beta }+\frac{1+\gamma }{2-\gamma }=??$$
I know that the answer is $\frac{9}{13}$, It's just for sharing a new ideas, thanks :)
|
Using the root-coefficient relationship, we have the following:
$$\alpha + \beta + \gamma = -1,$$
$$\alpha\beta\gamma = -1,$$
$$\alpha\beta + \beta\gamma + \gamma\alpha = 0.$$
The sum in question can be evaluated easily now.
\begin{align*}
\frac{1+\alpha }{2-\alpha }+\frac{1+\beta }{2-\beta }+\frac{1+\gamma }{2-\gamma } &= \frac{3(\alpha\beta + \beta\gamma + \gamma\alpha - \alpha\beta\gamma) -12}{\alpha\beta\gamma -2 (\alpha\beta +\beta\gamma + \gamma\alpha) + 4(\alpha + \beta + \gamma) -8} \\
&= \frac{3(0 + 1) -12}{-1 -2 (0) + 4(-1) -8} \\
&= \frac{9}{13}
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
how to solve $3 - \cfrac{2}{3 - \cfrac {2}{3 - \cfrac {2}{3 - \cfrac {2}{...}}}}$ $$A = 3 - \cfrac{2}{3 - \cfrac {2}{3 - \cfrac {2}{3 - \cfrac {2}{...}}}}$$
My answer is:
$$\begin{align}
&A = 3 - \frac {2}{A}\\
\implies &\frac {A^2-3A+2}{A}=0\\
\implies &A^2-3A+2=0\\
\implies &(A-1)\cdot(A-2)=0\\
\implies &A=1\;\text{ or }\; A=2
\end{align}$$
I should note that I'm not sure if the above answer is true. Because I expected just one answer for A (A is a numeric expression), but I found two, $1$ and $2$. This seems to be a paradox.
|
Let us define two series. The first is
\begin{align}
a_1 &= 3 \\
a_2 &= 3 - \frac{2}{3} \\
a_3 &= 3 - \frac{2}{3 - \frac{2}{3}} \\
a_4 &= 3- \frac{2}{3 - \frac{2}{3 - \frac{2}{3}}} \\
&\vdots \\
a_{n+1} &= 3 - \frac{2}{a_n} \quad (*)
\end{align}
and
\begin{align}
b_1 &= 3 - 2 \\
b_2 &= 3 - \frac{2}{3-2} \\
b_3 &= 3 - \frac{2}{3 - \frac{2}{3-2}} \\
b_4 &= 3 - \frac{2}{3 - \frac{2}{3 - \frac{2}{3-2}}} \\
&\vdots \\
b_{n+1} &= 3 - \frac{2}{b_n} \quad (**) \\
\end{align}
Note: This is the same recurrence relation $(*)$ or $(**)$ but with different start value $a_1 = 3$ and $b_1 = 1$.
For convergence we need $a_{n+1} - a_{n} \to 0$ or $a_n \to a$.
This way (in case of convergence) equations $(*)$ and $(**)$ have a limit
$$
a = 3 - \frac{2}{a} \quad (\#)
$$
which has indeed the solutions $a = 1$ and $a = 2$.
However that means we could also try
$$
c_n = 3 - \frac{2}{c_{n+1}}
$$
or
$$
c_{n+1} = \frac{2}{3 - c_n} \quad (\#\#)
$$
because it has the limit form $(\#)$.
Note that equation $(\#\#)$ is quite different from equation $(*)$ (see image below).
And indeed this recurrence relation $(\#\#)$ works too. Using $c_1 = 1$ will give $c_n \to 1$, Using $c_1 = 2$ will give $c_n \to 2$. Using $c_1 = 1000$ will give $c_n \to 1$.
So why is this? Still two solutions and the start value decides the limit.
Here is an image:
The green graph is related to $(*)$:
$$
f(x) = 3-\frac{2}{x}
$$
the blue graph is related to $(\#\#)$:
$$
g(x) = \frac{2}{3-x}
$$
and the red graph is the identity function:
$$
\mbox{id}(x) = x
$$
We see that both $f$ and $g$ hit the identity at $x=1$ and $x=2$. Those points are fixed points of $f$ and $g$:
\begin{align}
x^* &= f(x^*) \\
x^* &= g(x^*)
\end{align}
And one could now try to apply the theory of fixed points, esp. properties of fixed point iterations.
\begin{align}
x_{n+1} &= f(x_n) \quad (\$) \\
x_{n+1} &= g(x_n)
\end{align}
The fixed point iteration of $f$ is like the the iteration of original continued fractions (compare $(\$)$ with $(*)$ or $(**)$).
The theory behind can now help with statements about convergence and the dependency of start values.
|
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|
Quadratic formula does not work If I put the equation: $5x^2-x-4 =0$ in the quadratic formula, than I get
$x = 1$ or $x = \frac{-4}{5}$
but the real zeros are: $x = -1$ or $x = \frac{4}{5}$
Can somebody explain me if the quadratic formula fails or me?
|
Given $5x^2 −x−4=0 $ you must use, $a=5, b=-1, c=-4$ in the formula$$\begin{align}x & =\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\[2ex] & = \frac{-(-1)\pm\sqrt{(-1)^2-4(5)(-4)}}{2(5)} \\[2ex] & = \frac{1\pm\sqrt{1+80}}{10}\\[2ex] & = \frac{1\pm 9}{10}\\ \therefore x\in \{-\tfrac {4}{5}, 1\}\end{align}$$
|
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|
Is there another way to compute this Laurent series? $\frac{1}{(1+z)^2}$ Consider $g(z) = \frac{1}{(1+z)^2}$. To compute a Laurent series of $g$ at $0$ on the region $|z| > 1$, I let $w = \frac{1}{z}$, so $$g(z) = \frac{w^2}{(1+z)^2w^2} = w^2 \frac{1}{(1+w)^2} $$ Since $|w| < 1$, we have $\frac{1}{1+w} = 1 - w + w^2 - \cdots$, so $$\frac{1}{(1+w)^2} = (1 - w + w^2 - w^3 + \cdots)(1 - w + w^2 - w^3 + \cdots)$$ $$ = 1 - 2w + 3w^2 - 4w^3 + \cdots$$ and therefore $$g(z) = w^2 \sum\limits_{n=0}^{\infty} (-1)^n (n+1)w^n = \sum\limits_{n=0}^{\infty} (-1)^n (n+1)z^{-n-2}$$ My question is, is there a way to do this problem without having to resort to formal power series multiplication?
|
HINT:
Note that $\frac{d}{dz}\left(\frac{1}{1+z}\right)=-\frac{1}{(1+z)^2}$. Find the Laurent series for $\frac{1}{1+z}$ (this is pretty easy), differentiate term by term, and multiply by $-1$ should do it.
|
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"timestamp": "2023-03-29T00:00:00",
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|
evaluate the integral Evaluate the integral:
$$\int_0^{\pi} \frac{\cos 2\theta}{1 -2a\cos \theta +a^2}d\theta$$
The way I approach this problem is:
since, $\cos \theta = \frac{e^{it} + e^{-it}}{2}$; and $cos 2\theta = Re(z^2)$. Then, the integral will be written as follow:
$$\frac{1}{2}\int_0^{2 \pi} \frac{Re(z^2)}{1 - 2a \left( \frac{z + z^{-1}}{2}\right) +a^2}dz$$
$$ = \frac{1}{2}\int_0^{2\pi} \frac{-Re(z^2)}{(z-z_1)(z-z_2)}dz$$
where:
$$z_1 = \frac{(2a^2 - 2) + \sqrt{4(a^4+1)}}{4a}; \; z_2 = \frac{(2a^2 - 2) - \sqrt{4(a^4+1)}}{4a}$$
until here, I don't know how to use the Residue theorem to evaluate the integral. Can someone show me ?
|
Notice, the following expression
$$\frac{\cos 2\theta}{1-2a\cos \theta+a^2}=\frac{2\cos^2 \theta-1}{1-2a\cos \theta+a^2}=\frac{A\cos \theta(1-2a\cos \theta+a^2)+B(1-2a\cos \theta+a^2)+C}{1-2a\cos \theta+a^2}$$$$=A\cos \theta+B+\frac{C}{1-2a\cos \theta+a^2}$$
solving for $A, B, C$, we get $$A=-\frac{1}{a}, \ B=-\frac{a^2+1}{2a^2}, \ C=\frac{a^4+1}{2a^2}$$
Now we have
$$\int_{0}^{\pi}\frac{\cos 2\theta}{1-2a\cos \theta+a^2}d\theta$$ $$=\int_{0}^{\pi}\frac{2\cos^2 \theta-1}{1-2a\cos \theta+a^2}d\theta$$
$$=\int_{0}^{\pi}\left(-\frac{1}{a}\cos\theta-\frac{a^2+1}{2a^2}+\frac{a^4+1}{2a^2}\frac{1}{1-2a\cos \theta+a^2}\right)d\theta$$
$$=-\frac{1}{a}\int_{0}^{\pi}\cos\theta d\theta-\frac{a^2+1}{2a^2}\int_{0}^{\pi} d\theta+\frac{a^4+1}{2a^2}\int_{0}^{\pi}\frac{1}{1-2a\cos \theta+a^2}d\theta$$
$$=0-\frac{(a^2+1)\pi}{2a^2}+\frac{a^4+1}{2a^2}\int_{0}^{\pi}\frac{1}{(a^2+1)-2a\cos \theta}d\theta$$
$$=-\frac{(a^2+1)\pi}{2a^2}+\frac{a^4+1}{4a^3}\int_{0}^{\pi}\frac{1}{\frac{(a^2+1)}{2a}-\cos \theta}d\theta$$
Now, set $\cos\theta=\frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}$.
I hope you can take it from here.
|
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|
The range of the function $f(x)=\frac{\sin x}{\sqrt{1+\tan^2x}}-\frac{\cos x}{\sqrt{1+\cot^2x}}$ Find the range of the function $f(x)=\frac{\sin x}{\sqrt{1+\tan^2x}}-\frac{\cos x}{\sqrt{1+\cot^2x}}$.
By simply looking at the problem and simplifying trigonometrically,it looks as if range is zero but not.I think there is some trick to solve this.I am perplexed.I tried to find the domain first but not successful.Please help....
|
Using $\displaystyle |\sin x|=\left\{\begin{matrix}
\displaystyle \sin x \;,& \displaystyle 0 \leq x\leq \frac{\pi}{2} \\\\
\displaystyle \sin x \;,& \displaystyle \frac{\pi}{2} \leq x\leq \pi \\\\
\displaystyle -\sin x\;, & \displaystyle\pi \leq x\leq \frac{3\pi}{2} \\\\
\displaystyle -\sin x \;,& \displaystyle\frac{3\pi}{2} \leq x\leq 2\pi
\end{matrix}\right.$ and $\displaystyle |\cos x|=\left\{\begin{matrix}
\displaystyle \cos x \;,& \displaystyle 0 \leq x\leq \frac{\pi}{2} \\\\
\displaystyle -\cos x \;,& \displaystyle \frac{\pi}{2} \leq x\leq \pi \\\\
\displaystyle -\cos x \;,& \displaystyle\pi \leq x\leq \frac{3\pi}{2} \\\\
\displaystyle \cos x \;,& \displaystyle\frac{3\pi}{2} \leq x\leq 2\pi
\end{matrix}\right.$
So $$\displaystyle f(x)=\frac{\sin x}{\sqrt{1+\tan^2x}}-\frac{\cos x}{\sqrt{1+\cot^2x}} = \sin x\cdot \left|\cos x\right|-\cos x\cdot \left|\sin x\right|$$
Here Function $f(x)$ is Periodic With Time Period $= 2\pi.$
So we will Calculate for Only one Time Period.
In $\bullet \displaystyle \; 0 \leq x\leq \frac{\pi}{2}\;,$ We get $f(x) = \sin x\cdot \cos x-\cos x\cdot \sin x = 0$
In $\bullet \displaystyle \; \frac{\pi}{2} \leq x\leq \pi\;,$ We get $f(x)=-\sin 2x .$ So $0 \leq f(x)\leq 1$
In $\bullet \displaystyle \; \pi \leq x\leq \frac{3\pi}{2}\;,$ We get $f(x)=0 .$
In $\bullet \displaystyle \; \frac{3\pi}{2} \leq x\leq 2\pi\;,$ We get $f(x)=\sin 2x .$ So $-1 \leq f(x)\leq 0$
So Here We get $\displaystyle -1 \leq f(x)\leq 1$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proving $\frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \cdots + \frac{1}{n\cdot(n+2)} = \frac{3}{4} - \frac{(2n+3)}{2(n+1)(n+2)}$ by induction for $n\geq 1$ I'm having an issue solving this problem using induction. If possible, could someone add in a very brief explanation of how they did it so it's easier for me to understand?
$$\frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \cdots + \frac{1}{n\cdot(n+2)} = \frac{3}{4} - \frac{(2n+3)}{2(n+1)(n+2)}$$
How do I prove the above equation for all integers where $n\geq1$?
|
Let $$p(n):\displaystyle\frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \cdots + \frac{1}{n\cdot(n+2)} = \frac{3}{4} - \frac{(2n+3)}{2(n+1)(n+2)}$$
Put $n=1\;,$ We get $$\displaystyle \frac{1}{1\cdot 3} = \frac{3}{4}-\frac{5}{2\cdot 2\cdot 3} = \frac{4}{12}$$
So it is true for $n=1$
Now Put $n=k\;,$ We get $$\displaystyle \frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \cdots + \frac{1}{k\cdot(k+2)}=\frac{3}{4} - \frac{(2k+3)}{2(k+1)(k+2)}$$
Now Using $p(k)\;,$ We will prove for $p(k+1)$
So $$\displaystyle p(k+1):\frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \cdots + \frac{1}{(k+1)\cdot(k+3)}=\frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \cdots + \frac{1}{k\cdot(k+2)}+\frac{1}{(k+1)\cdot (k+3)}$$
So $$\displaystyle p(k+1) = \frac{3}{4} - \frac{(2k+3)}{2(k+1)(k+2)}+\frac{1}{(k+1)(k+3)}= \frac{3}{4}-\frac{1}{(k+1)}\left\{\frac{2k+3}{2(k+2)}-\frac{1}{(k+3)}\right\}$$
$$\displaystyle = \frac{3}{4}-\frac{1}{(k+1)}\left\{\frac{2k^2+9k+9-2k-4}{2(k+2)(k+3)}\right\} = \frac{3}{4}-\frac{1}{2(k+1)(k+2)(k+3)}\cdot (2k+5)\cdot (k+1)=\frac{3}{4}-\frac{(2k+5)}{2(k+2)(k+3)}$$
So $p(k)$ We have prove for $p(k+1).$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\frac{\pi}2 < \sum_0^\infty \frac{1}{1+n^2} < \frac{3\pi}4 $ Prove that:
$\frac{\pi}2 < \sum_0^\infty \frac{1}{1+n^2} < \frac{3\pi}4 $
What I've tried:
I solved the improper integral: $\int_0^\infty \frac{1}{1+x^2} = \lim_{b\to \infty} \arctan b -\arctan 0 = \frac{\pi}2 $.
Now, everything in the sum (and in the integral) is positive so the sum must be lower than the integral (it has more members), so the claim is wrong (originally it was a "Prove/Disprove" problem...). However, if we only add the first 3 members we get 1.7 which is bigger than $\frac{\pi}2 $.
Also, I checked with wolfram and the sum is somewhere around 2 so the claim is true. I have no idea how to prove it.
Someone suggested I would read Basel's problem ,in which the integral is also smaller than the sum (starting from 1) but it led me nowhere and I don't think that this kind of complex solution is needed here.
|
An advanced approach using zeta function values $\zeta(2)$ and $\zeta(4)$ to get even better bounds.
For $n>1$, $$\frac{1}{1+n^2}=\frac{1}{n^2}\frac{1}{1+\frac{1}{n^2}} = \frac1{n^2}-\frac{1}{n^4}+\frac{1}{n^6}-\frac{1}{n^8}+...$$
So $$\frac{1}{n^2}-\frac{1}{n^4}<\frac{1}{n^2+1}<\frac{1}{n^2}$$
So:
$$\frac{3}{2}+\left(\zeta(2)-\zeta(4)-\left(\frac{1}{2^2}-\frac{1}{2^4}\right)\right)=1+\frac{1}{2}+\sum_{n=2}^\infty \left(\frac{1}{n^2} -\frac{1}{n^4}\right)<\sum_{n=0}^{\infty}\frac{1}{n^2+1} < \frac{1}{2} + \zeta(2)$$
Now, $$\frac{1}{2}+\zeta(2)=\frac 12+\frac{\pi^2}{6} <\frac{3\pi}{4}$$
And:
$$\frac{21}{16}+\zeta(2)-\zeta(4) = \frac{21}{16} + \frac{\pi^2}{6} - \frac{\pi^4}{90}>\frac{\pi}{2}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
solving integral How can I solve this integral to get the result as follow:
$${\sqrt{\alpha} \over 2\pi} \int_{0}^{t} {1\over \sqrt{r^{3}(t-r)}}[\sin({\alpha\over2r})+\cos({\alpha\over2r})] \mathrm{d}r= {1\over{\sqrt{\pi t}}} \cos({\alpha\over2t}) $$
Thanks!
|
Let
$$I(t) = {\sqrt{\alpha} \over 2\pi} \int_{0}^{t} {1\over \sqrt{r^{3}(t-r)}}[\sin({\alpha\over2r})+\cos({\alpha\over2r})] \mathrm{d}r $$
then let $x = (\alpha/2 r)$ to obtain
\begin{align}
I(t) &= \frac{\alpha \, \sqrt{\alpha}}{4 \pi} \left(\frac{\alpha}{2}\right)^{3/2} \, \int_{\alpha/2t}^{\infty} \left(\sin(x) + \cos(x) \right) \, \frac{dx}{\sqrt{xt - \alpha/2}} \\
&= \frac{1}{\sqrt{2 t} \, \pi} \, \int_{0}^{\infty} \frac{\sin(x) + \cos(x)}{\sqrt{x - \frac{\alpha}{2t}}} \, dx
\end{align}
Let $x = u^{2} + \frac{\alpha}{2t}$ to obtain
\begin{align}
I(t) &= \sqrt{\frac{2}{\pi}} \, \frac{1}{\sqrt{\pi \, t}} \, \int_{0}^{\infty} \left[ \sin\left(u^{2} + \frac{\alpha}{2 t} \right) + \cos\left(u^{2} + \frac{\alpha}{2t} \right) \right] \, du
\end{align}
Now using
\begin{align}
\int_{0}^{\infty} \sin(u^{2} + a) \, du &= \frac{1}{2} \, \sqrt{\frac{\pi}{2}} \, \left( \sin a + \cos a \right) \\
\int_{0}^{\infty} \cos(u^{2} + a) \, du &= \frac{1}{2} \, \sqrt{\frac{\pi}{2}} \, \left(\cos a - \sin a \right)
\end{align}
then the desired result becomes
\begin{align}
I(t) = \frac{1}{\sqrt{\pi \, t}} \, \cos\left(\frac{\alpha}{2 \, t}\right)
\end{align}
Thus,
\begin{align}
{\sqrt{\alpha} \over 2\pi} \int_{0}^{t} {1\over \sqrt{r^{3}(t-r)}}[\sin({\alpha\over2r})+\cos({\alpha\over2r})] \mathrm{d}r
= \frac{1}{\sqrt{\pi \, t}} \, \cos\left(\frac{\alpha}{2 \, t}\right)
\end{align}
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
If $x^3+y^3=72$ and $xy=8$ then find the value of $x-y$. I recently came across a question,
If $x^3+y^3=72$ and $xy=8$ then find the value of $(x-y)$.
By trial and error I found that $x=4$ and $y=2$ satisfies both the conditions. But in general how can I solve it analytically? I tried using $a^3+b^3=(a+b)(a^2+b^2-ab)$ and also $a^3+b^3=(a+b)^3-3ab(a+b)$. But both ways aren't working.
Please explain how do I solve these types of questions analytically.
|
It's two equations, with two variables. We can do what we always do. You can write $y= 8/x$, and substitute into the first equation, obtaining $x^3 + 8^3/x^3 =72.$
From here, we want to write this as a polynomial, so we multiply through $x^3$ so there are no negative powers, and then move everything to the left. This gives $x^6-72x^3+512.$ This is a pretty large polynomial, but since all the exponents are multiples of 3, all is not lost. We'll set $u=x^3$ and obtain $u^2-72u + 512$. The quadratic formula gives the factorization $(u-8)(u-64)=0,$ which of course implies that $u=8$ or that $u=64$.
Now we just go through the substitution in $u$. We write $x^3 = 8$ or $x^3 = 64$. There is only one cube root when dealing with real numbers, so we have $x=2$ or $x=4$. If $x=2$, then we have $y=8/2=4$. If $x=4$, then we have $y=8/4=2$.
So there are two solutions then, depending on the case. Either $x-y = 2$ or $x-y=-2$ depending on which pair of solutions you work with.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Triangle Area problem I've been trying to solve the following:
Let $ABC$ be a triangle with sides $a, b $ and $ c$, inradius $r$ and exradii $r_a, r_b$ and $r_c$. If $A'B'C'$ is another triangle with sides $\sqrt{a}, \sqrt{b}$ and $\sqrt{c}$ show that $Area(A'B'C')=\frac {\sqrt{r(r_a+r_b+r_c)}} {2}$.
I tried to combine various formulas and theorems involving exradii and inradii but it got me nowhere so far.. maybe there's a better way to approach the problem! Any advice or help will be appreciated!!
|
Using
$$\text{area($\triangle{ABC}$)}=\frac{a+b+c}{2}r=\frac{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}{4}$$
$$\Rightarrow \frac{a+b+c}{2}r\cdot\frac{1}{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}=\frac 14$$
and
$$\text{area($\triangle{ABC}$)}=\frac{a+b+c}{2}r=\frac{-a+b+c}{2}r_a=\frac{a-b+c}{2}r_b=\frac{a+b-c}{2}r_c$$$$\Rightarrow rr_a=\frac{a+b+c}{-a+b+c}r^2,\quad rr_b=\frac{a+b+c}{a-b+c}r^2,\quad rr_c=\frac{a+b+c}{a+b-c}r^2$$we have
$$\begin{align}\text{area($\triangle{A'B'C'}$)}&=\frac{\sqrt{(\sqrt a+\sqrt b+\sqrt c)(-\sqrt a+\sqrt b+\sqrt c)(\sqrt a-\sqrt b+\sqrt c)(\sqrt a+\sqrt b-\sqrt c)}}{4}\\&=\frac{\sqrt{2ab+2bc+2ca-a^2-b^2-c^2}}{4}\\&=\frac{a+b+c}{2}r\cdot \frac{\sqrt{2ab+2bc+2ca-a^2-b^2-c^2}}{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}\\&=\frac r2\sqrt{\frac{(a+b+c)(2ab+2bc+2ca-a^2-b^2-c^2)}{(-a+b+c)(a-b+c)(a+b-c)}}\\&=\frac r2\sqrt{(a+b+c)\left(\frac{1}{-a+b+c}+\frac{1}{a-b+c}+\frac{1}{a+b-c}\right)}\\&=\frac 12\sqrt{\frac{a+b+c}{-a+b+c}r^2+\frac{a+b+c}{a-b+c}r^2+\frac{a+b+c}{a+b-c}r^2}\\&=\frac{\sqrt{rr_a+rr_b+rr_c}}{2}\end{align}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve trigonometric equation $ \cot x + \cos x = 1 + \cot x \cos x $ Solve trigonometric equation: $$ \cot (x) + \cos (x) = 1 + \cot (x) \cos (x) $$
I tried to multiply both sides with $\sin x$ (which I'm not sure if I can multiply with sin).
|
Notice, we have $$\cot x+\cos x=1+\cot x\cos x$$
$$\frac{\cos x}{\sin x}+\cos x=1+\frac{\cos x}{\sin x}\cos x $$
$$\cos x+\sin x\cos x=\sin x+\cos^2 x $$
$$\cos x+\sin x\cos x-\sin x-\cos^2 x=0 $$
$$\underbrace{\cos x-\sin x}+\underbrace{\sin x\cos x-\cos^2 x}=0 $$
$$(\cos x-\sin x)-\cos x(\cos x-\sin x)=0 $$
$$(1-\cos x)(\cos x-\sin x)=0 $$
$$1-\cos x=0\vee \cos x-\sin x=0$$
$$\cos x=1\vee \sin x=\cos x\iff \tan x=1$$
$$x=2k\pi\vee x=k\pi+\frac{\pi}{4}$$
For all, $k$ is any integer
|
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"timestamp": "2023-03-29T00:00:00",
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|
Why is this sum zero? I have been looking at the following sum (for any positive integer $n$)
$$\left(1-\frac{1^2}{n}\right) + \left(1-\frac{1}{n}\right)\left(1-\frac{2^2}{n}\right) + \left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\left(1-\frac{3^2}{n}\right) + \ldots $$
Note that the $i$th term in the sum has $i$ factors and is
$$\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\left(1-\frac{3}{n}\right)\dots \left(1-\frac{i-1}{n}\right)\left(1-\frac{i^2}{n}\right).$$
It seems, amazingly, that the answer is 0. How can one show this?
|
The partial sums are
$$\begin{eqnarray}
1/2&0\\
2/3&4/9&0\\
3/4&12/16&18/64&0\\
4/5&24/25&72/125&96/625&0\\
5/6&40/36&180/216&480/1296&600/6^5&0\end{eqnarray}$$
The ratios from one partial sum to the next are
$$\begin{eqnarray}0\\
2/3&0\\
4/4&(3/2)/4&0\\
6/5&(6/2)/5&(4/3)/5&0\\
8/6&(9/2)/6&(8/3)/6&(5/4)/6&0\end{eqnarray}$$
So is the sum of the first $k$ terms, with $n$ as the key variable,
$$\frac{\frac{(n-1)!}{(n-1-k)!}k}{n^k}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Intersection between two three-dimensional planes The intersection of the planes defined by
$x \bullet \begin{pmatrix} 8 \\ 1 \\ -12 \end{pmatrix} = 35$
and
$x \bullet \begin{pmatrix} 6 \\ 7 \\ -9 \end{pmatrix} = 70$
is a line. Find an equation of this line.
I've been attempting this problem for hours and cannot solve it. I would love a few comprehensive hints so I can solve this problem.
|
Let $x=(a,b,c)$. Then we can rewrite your equations as
$$\begin{align}
8a+1b-12c&=35 \\[2 ex]
6a+7b-9c&=70
\end{align}$$
Now we want to solve those simultaneous equations. Subtracting $3/4$ of the first equation from the second we get
$$\begin{align}
8a+1b-12c&=35 \\[2 ex]
0a+\frac{25}4b-0c&=\frac{175}4
\end{align}$$
Dividing the first equation by $8$ and multiplying the second equation by $\frac 4{25}$ we get
$$\begin{align}
1a+\frac 18b-\frac 32c&=\frac{35}8 \\[2 ex]
0a+1b-0c&=7
\end{align}$$
Subracting $\frac 18$ of the second equation from the first we get
$$\begin{align}
1a+0b-\frac 32c&=\frac{7}2 \\[2 ex]
0a+1b-0c&=7
\end{align}$$
Our final solution is then
$$a=\frac 72+\frac 32c,\quad b=7$$
with no restriction on $c$. If we let our parameter be $t$ and let that be $c$, we see that our line has the points
$$\begin{align}
x&=\left(\frac 72+\frac 32t, 7, t\right) \\[2 ex]
&=\left(\frac 72,7,0\right)+t\cdot\left(\frac 32,0,1\right)
\end{align}$$
If you understand matrices, you could have gotten here faster by putting the coefficients of the two equations into a $2\times 4$ matrix and reducing it to row-reduced echelon form.
If you dislike fractions, you could choose another base point on the line where $t=1$ and multiply the direction vector by $2$, getting
$$\begin{align}
x&=(5+3t, 7, 1+2t) \\[2 ex]
&=(5,7,1)+t\cdot (3,0,2)
\end{align}$$
|
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|
What techniques would be use to prove $6x^2 +12x +8$ cannot be perfect cube for integer x > 0 I'm wondering if there are any basic techniques.
|
Let $(x,y)$ be any integer solution to $y^3 = 6x^2 + 12x + 8$.
Since RHS is even, so does LHS and $y$. This implies $\text{RHS} \equiv 0 \pmod 8$.
It is easy to check
$$\text{LHS} \equiv
\begin{cases} 0,& x \text{ even}\\2, & x \text{ odd}\end{cases}
\pmod 8$$
So $x$ is also even. This implies existence of integers $X,Y$ such that $x = 2X$ and $y = 2Y$.
In terms of $X, Y$, we have
$$Y^3 = 3X^2 + 3X + 1 = (X+1)^3 - X^3
\quad\iff\quad X^3 + Y^3 = (X+1)^3$$
By Fermat's Last theorem, at least one of $X, Y$ or $X+1$ is $0$.
This leads to three and only three set of possibilities:
*
*$X = 0 \rightarrow (x,y) = (0,2)$
*$Y = 0 \rightarrow \text{ no solution}$.
*$X = -1 \rightarrow (x,y) = (-2,2)$
Since all of these have $x \le 0$, $6x^2 + 12x + 8$ is never a cube for $x > 0$.
|
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|
To compute improper integral $\int_3^{5}\frac{x^{2}\, dx}{\sqrt{x-3}{\sqrt{5-x}}}$ I am given improper integral as
$$\int_3^{5}\frac{x^{2}}{\sqrt{x-3}{\sqrt{5-x}}}dx$$
DOUBT
I see that problem is at both the end points, so i need to split up the integral. But problem seems to me that when i split up integral one of terms in denominator which is ${5-x}$ becomes negative. So how do i split up the integral ? Kindly help
Thanks
|
Beta Function Approach
Substituting $x\mapsto2x+3$,
$$
\begin{align}
&\int_3^5\frac{x^2\,\mathrm{d}x}{\sqrt{(x-3)(5-x)}}\\
&=\int_0^1\frac{(2x+3)^2\,\mathrm{d}x}{\sqrt{x(1-x)}}\\
&=\int_0^1\frac{(3(1-x)+5x)^2\,\mathrm{d}x}{\sqrt{x(1-x)}}\\
&=9\int_0^1\frac{(1-x)^2\,\mathrm{d}x}{\sqrt{x(1-x)}}
+30\int_0^1\frac{(1-x)x\,\mathrm{d}x}{\sqrt{x(1-x)}}
+25\int_0^1\frac{x^2\,\mathrm{d}x}{\sqrt{x(1-x)}}\\
&=9\operatorname{B}\left(\frac52,\frac12\right)
+30\operatorname{B}\left(\frac32,\frac32\right)
+25\operatorname{B}\left(\frac12,\frac52\right)\\
&=9\cdot\frac38\pi
+30\cdot\frac18\pi
+25\cdot\frac38\pi\\
&=\frac{33}2\pi
\end{align}
$$
using the Beta Function.
Trigonometric Subsitution
After the substitution $x\mapsto2x+3$, we can use $x\mapsto\sin^2(x)$
$$
\begin{align}
&\int_3^5\frac{x^2\,\mathrm{d}x}{\sqrt{(x-3)(5-x)}}\\
&=9\int_0^1\frac{(1-x)^2\,\mathrm{d}x}{\sqrt{x(1-x)}}
+30\int_0^1\frac{(1-x)x\,\mathrm{d}x}{\sqrt{x(1-x)}}
+25\int_0^1\frac{x^2\,\mathrm{d}x}{\sqrt{x(1-x)}}\\
&=9\int_0^{\pi/2}2\cos^4(x)\,\mathrm{d}x
+30\int_0^{\pi/2}2\sin^2(x)\cos^2(x)\,\mathrm{d}x
+25\int_0^{\pi/2}2\sin^4(x)\,\mathrm{d}x\\
&=9\cdot\frac38\pi
+30\cdot\frac18\pi
+25\cdot\frac38\pi\\
&=\frac{33}2\pi
\end{align}
$$
|
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|
Modular maths: How do I find the remainder? How do I find the remainder of $5^{22} \pmod{25}$?
And also how do I find the remainder of $3^{16} + 7 \pmod{5}$?
|
Try to see which power of $5$ can give $1$ or $-1$ modulo $7$. This will reduce your computations quite significantly. For example, $5^2 \equiv 4 \pmod{7}$ so $5^3 \equiv 20 \equiv 6 \equiv -1 \pmod{7}$. Thus
$$5^{22} \equiv 5^{21} \cdot 5 \equiv (5^{3})^7 \cdot 5 \equiv -5 \equiv 2 \pmod{7}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
How to express $\phi$ in terms of $R\text{, }x\text{ and }\theta$ Let $S$ be a circle with radius $R$ and center at $O$.
Let $P$ be any arbitrary point inside circle such that its distance from $O$ is $x$ and the ray $\overrightarrow{OP}$ cuts the circle $S$ at $M$.
Let $N$ be any other point on the circle such that $\angle NPM = \theta$ and $\angle NOM = \phi$.
How can we express $\phi$ in terms of $R\text{, }x \text{ and }\theta$.
|
First, it's important to realise that we can choose a coordinate system such that $P$ and $M$ are on the horizontal axis passing through the centre $O$ (see figure below). This can be done without any loss of generality.
With respect to the centre $O$, then, the coordinates of the various points of interest are:
$$
P = (x, 0)
\,,\qquad
M = (r, 0)
\,,\qquad
N = (r\cos\phi, r\sin\phi)
$$
But the coordinates of $N$ can be expressed in another way as well, namely,
$$
N = (x + \overline{PN}\cos\theta, \overline{PN}\sin\theta)
$$
Therefore,
$$
\overline{PN}\cos\theta = r\cos\phi - x
\qquad\mbox{and}\qquad
\overline{PN}\sin\theta = r\sin\phi
$$
Eliminating $\overline{PN}$ by dividing the second equation by the first results in
$$
\tan\theta = \frac{r\sin\phi}{r\cos\phi - x}
$$
which, after dividing by $\cos\phi$, can be rewritten as
$$
r\tan\theta - x\tan\theta\sec\phi = r\tan\phi
\qquad(1)
$$
Now we need an equation relating $\sec\phi$ with $\tan\phi$:
$$
\tan^2\phi = \sec^2\phi - 1
\qquad(2)
$$
So, squaring $(1)$,
$$
r^2\tan^2\theta - 2rx\tan^2\theta\sec\phi + x^2\tan^2\theta\sec^2\phi = r^2\tan^2\phi
$$
Using $(2)$, we have
$$
r^2\tan^2\theta - 2rx\tan^2\theta\sec\phi + x^2\tan^2\theta\sec^2\phi = r^2\,(\sec^2\phi - 1) = r^2\sec^2\phi - r^2
$$
which, finally, reduces to a quadratic equation for $\sec\phi$:
$$
(x^2\tan^2\theta - r^2)\,\sec^2\phi - 2rx\tan^2\theta\,\sec\phi + r^2\,(1 + \tan^2\theta) = 0
$$
whose solution is:
$$
\sec\phi = \frac{rx\,\tan^2\theta \pm r\sqrt{(x\tan^2\theta)^2 - (x^2\tan^2\theta - r^2)\,(1 + \tan^2\theta)}}{(x^2\tan^2\theta - r^2)}
$$
The quantity inside the radical simplifies to:
$$
(x\tan^2\theta)^2 - (x^2\tan^2\theta - r^2)\,(1 + \tan^2\theta) =
r^2 + (r^2 - x^2)\tan^2\theta
$$
and we find
$$
\sec\phi = \frac{rx\,\tan^2\theta \pm r\sqrt{r^2 + (r^2 - x^2)\tan^2\theta}}{(x^2\tan^2\theta - r^2)}
$$
Since $\cos\phi = 1/\sec\phi$,
$$
\cos\phi = \frac{x^2\tan^2\theta - r^2}{rx\,\tan^2\theta \pm r\sqrt{r^2 + (r^2 - x^2)\tan^2\theta}}
$$
To fix the sign, consider that $\theta = 0 \Rightarrow \phi = 0$, so
$$
\cos\phi = \frac{x^2\tan^2\theta - r^2}{rx\,\tan^2\theta - r\sqrt{r^2 + (r^2 - x^2)\tan^2\theta}}
$$
Since $x \le r$, the value inside the radical is non-negative and shouldn't cause any trouble. The only possible source of trouble is $\theta = \pi/2$, for which $\tan\theta$ is infinitely large. However, we can see that the solution above can also be written as
$$
\cos\phi = \frac{x^2 - r^2/\tan^2\theta}{rx - r\sqrt{r^2/\tan^4\theta + (r^2 - x^2)/\tan^2\theta}}
$$
Thus, in the limit $\theta \to \pi/2$,
$$
\cos\phi = \frac{x^2}{rx} = \frac{x}{r}
$$
which is the result we'd expect by looking at the figure.
Therefore the final solution is:
$$
\cos\phi = \frac{x}{r} \qquad\mbox{if } \theta = \frac{\pi}{2}
$$
$$
\cos\phi = \frac{x^2\tan^2\theta - r^2}{rx\,\tan^2\theta - r\sqrt{r^2 + (r^2 - x^2)\tan^2\theta}} \qquad\mbox{if } \theta \ne \frac{\pi}{2}
$$
The solution for the region below the horizontal axis can be obtained from the above, by symmetry.
Edit: Before I start getting down-votes, let me show that my answer agrees with Rory's answer. His answer is:
$$
\phi=\theta-\sin^{-1}\left(\frac xR\,\sin\theta\right)
$$
or
$$
\frac{x}{R}\,\sin\theta = \sin(\theta-\phi) = \sin\theta\cos\phi - \cos\theta\sin\phi
$$
But this, upon division by $\cos\theta\cos\phi$ and multiplication by $R$, is exactly my equation $(1)$ above so the two solutions are equivalent. Mine just happens not to be as elegant as Rory's.
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"language": "en",
"url": "https://math.stackexchange.com/questions/1399371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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|
Find inverse of 15 modulo 88. Here the question: Find an inverse $a$ for $15$ modulo $88$ so that $0 \le a \le 87$; that is, find an integer $a \in \{0, 1, ..., 87\}$ so that $15a \equiv1$ (mod 88).
Here is my attempt to answer:
Find using the Euclidean Algorithm, we need to find $\gcd(88, 15)$, that must equal to $1$ to be possible to find an inverse of $15 \pmod{88}$.
\begin{align*}
88 & = 5 \times 15 + 13\\
15 & = 1 \times 13 + 2\\
13 & = 6 \times 2 + 1\\
2 & = 2 \times 1 + 0
\end{align*}
So,
$$\gcd(88, 15) = 1$$
Now, we need to write this into the form:
$$\gcd(88, 15) = 88x + 15y.$$
And find $x$ and $y$.
\begin{align*}
1 & = 13(1) + 2(-6)\\
& = 13(7) + 15(-6)\\
& = 88(7) + 15(-41)
\end{align*}
So, $x = 7$ and $y = -41$.
So, an inverse of $15 \pmod{88} = -41$. Now, I need to find an inverse that is between $0$ and $87$. What is a good easy approach to find other inverses? Any ideas please?
|
work Translation(unnecessary once you know how to do this.)
| 88 1 0 88 = 1(88) + 0(15)
-6 | 15 0 1 15 = 0(88) + 1(15), add -6 x this row to the above row
7 | -2 1 -6 -2 = 1(88) - 6(15), add 7 x this row to the above row
| 1 7 -41 1 = 7(88) - 41(15)
Hence -41(15) ≡ 1 (mod 88)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1399743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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|
Find $\lim\limits_{x\to0}\frac{\sqrt{1+\tan x}-\sqrt{1+x}}{\sin^2x}$ Find:
$$\lim\limits_{x\to0}\frac{\sqrt{1+\tan x}-\sqrt{1+x}}{\sin^2x}$$
I used L'Hospital's rule, but after second application it is still not possible to determine the limit. When applying Taylor series, I get wrong result ($\frac{-1}{6}$). What method to use?
Result should be $\frac{1}{4}$
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If you know that $$\lim_{x \to 0}\frac{\tan x - x}{x^{2}} = 0\tag{1}$$ then it is easy to give a simple evaluation for the limit in question
\begin{align}
L &= \lim_{x \to 0}\frac{\sqrt{1 + \tan x} - \sqrt{1 + x}}{\sin^{2}x}\notag\\
&= \lim_{x \to 0}\frac{\sqrt{1 + \tan x} - \sqrt{1 + x}}{x^{2}}\cdot\frac{x^{2}}{\sin^{2}x}\notag\\
&= \lim_{x \to 0}\frac{\sqrt{1 + \tan x} - \sqrt{1 + x}}{x^{2}}\notag\\
&= \lim_{x \to 0}\frac{\tan x - x}{x^{2}\{\sqrt{1 + \tan x} + \sqrt{1 + x}\}}\notag\\
&= \frac{1}{2}\lim_{x \to 0}\frac{\tan x - x}{x^{2}}\notag\\
&= 0\notag
\end{align}
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"language": "en",
"url": "https://math.stackexchange.com/questions/1400657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
}
|
Solve the sequences inequality If $a_1=1$ and $a_n=a_{n-1}+\dfrac{1}{a_{n-1}}$ for $n≥2$ , then prove that $12 < a_{75} < 15$ ?
I have tried solving this by:
$$a_{75} - a_1 = \frac{1}{a_1}+\frac{1}{a_2}....+\frac{1}{a_{74}}$$
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Use the hint given by Did:
We have that $a_n^2=a_{n-1}^2+\frac{1}{a_{n-1}^2}+2$. Thus $a_n^2\geq a_{n-1}^2+2$ for all $n\geq 2$. Hence $a_n^2\geq 2\cdot (n-1) +1$. Hence $a_{75}\geq \sqrt{2\cdot 74+1}\geq \sqrt{2\cdot 72}=\sqrt{144}=12$.
For the other inequality, use the following reasoning:
$$a_n^2=a_{n-1}^2+\frac{1}{a_{n-1}^2}+2\leq a_{n-1}^2+\frac{1}{2(n-2)+1}+2.$$
Here we used the inequality $a_{n-1}^2\geq 2\cdot (n-2) +1$ in the denominator. Now replace $a_{n-1}^2$ by $a_{n-2}^2+\frac{1}{a_{n-2}^2}+2$ and again use the inequality in the denominator. Continue in this fashion to obtain the result. (Do this yourself, it requires a bit of work).
All credits go to @Did for his brilliant hint!
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"language": "en",
"url": "https://math.stackexchange.com/questions/1401132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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|
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