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The $n-th$ term of a sequence is the LCM of the integers from $1$ to $n$
The $n-th$ term of a sequence is the least common multiple (LCM) of the integers from $1$ to $n$. Which term of the sequence is the first one that is divisible by $100$?
How I'll solve this?
Note: This is a problem from BDMO - $2012$
|
For some prime $p$, the greatest $p^n$ that divides the LCM of the integers $1$ through $n$ is the greatest $p^n$ that is less than $n$. For example, for the LCM of the integers $1$ through $50$, the greatest $7^n$ that divides the LCM is $49$ since $7^2=49<50$, but $7^3=343>50$. $7^3$ can not divide the LCM of the integers $1$ through $50$ because there is no integer less than or equal to $50$ that has $7^3$ as a factor.
Thus, for the number $100=2^2*5^2$, we need the LCM to divide $2^2=4$ and $5^2=25$. The maximum of these two prime powers is $25$. Thus, the lowest integer $n$ such that the LCM of $1$ through $n$ is divisible by $100$ is $25$ since it is the smallest LCM that has $5^2$ as a factor.
|
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|
Finding segment in a right triangle. Here is the picture of the question:
*
*$ABC$ is a right triangle.
*$m(CBA)=90^\circ$.
*$m(BAD)=2m(DAC)=2\alpha$.
*$D$ is a midpoint of $[BC]$.
*$E$ is a point on $[AD]$.
*$m(BED)=90^\circ$.
*$|DE|=3$.
*What is $|AB|=x$?
Tried lots of things which gives me some trigonometric identities, but none of them led me to the solution which is $x=6$. And even if it did, i prefer more geometric methods (all solutions are welcome though). Note that this could be an easy question, and i could probably missing something obvious.
|
This is one approach. I don't pretend it's the nicest.
Let $u = \tan\alpha$. Then
$$
\tan2\alpha = \frac{2u}{1-u^2}
$$
$$
\tan3\alpha = \frac{u(3-u^2)}{1-3u^2}
$$
Since $\tan3\alpha = 2\tan2\alpha$, we have
$$
\frac{4u}{1-u^2} = \frac{u(3-u^2)}{1-3u^2}
$$
$$
(1-u^2)(3-u^2) = 4(1-3u^2)
$$
$$
3-4u^2+u^4 = 4-12u^2
$$
$$
1-u^4 = 8u^2
$$
Now let $y = BE$. By similar triangles, we have
$$
\frac{3}{y} = \frac{\sqrt{9+y^2}}{x} = \tan2\alpha = \frac{2u}{1-u^2}
$$
Square all terms to obtain
$$
\frac{9}{y^2} = \frac{9+y^2}{x^2} = \left(\frac{2u}{1-u^2}\right)^2
= \frac{4u^2}{1-2u^2+u^4}
$$
Observe that the outer pair of the equality above gives us
$$
\frac{9+y^2}{y^2} = \frac{9}{y^2}+1 = 1+\frac{4u^2}{1-2u^2+u^4}
= \frac{1+2u^2+u^4}{1-2u^2+u^4}
= \left(\frac{1+u^2}{1-u^2}\right)^2
$$
Multiplying both ends by $\frac{x^2}{9+y^2} \cdot \frac{y^2}{9} = \left(\frac{1-u^2}{2u}\right)^2\left(\frac{1-u^2}{2u}\right)^2$ yields
\begin{align}
\frac{x^2}{9} & = \left(\frac{1+u^2}{2u}\right)^2
\left(\frac{1-u^2}{2u}\right)^2 \\
& = \left(\frac{1-u^4}{4u^2}\right)^2 \\
& = \left(\frac{8u^2}{4u^2}\right)^2
\qquad \longleftarrow 1-u^4 = 8u^2 \\
& = 2^2 = 4
\end{align}
So $x^2 = 36$ and $x = 6$.
Neat question. One almost feels as though the simple and straightforward answer must be obtainable with a correspondingly simple and straightforward approach. Thus far, though, I haven't seen it. Anybody?
|
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|
Calculating determinant of 100x100 matrix I was trying to calculate the determinant of 100x100 matrix:
$$A=\begin{bmatrix}5&5&5&...&5&5&-1\\5&5&5&...&5&-1&5\\5&5&5&...&-1&5&5\\ \\5&5&-1&...&5&5&5\\5&-1&5&...&5&5&5\\-1&5&5&...&5&5&5\end{bmatrix}$$
I thought that using LU decomposition would quickly show a pattern, but the numbers I'm getting after only few transformations quickly get out of hand. What approach might be good for calculating the determinant of such matrix?
|
Let $D_n$ be the determinant of such a matrix that is $n\times n$.
In any one column except the last column, the bottom $n-1$ entries sum to $5n-11$. So add $\frac{-5}{5n-11}$ times each of row 2, row 3, etc. to row 1, and you have
$$\begin{bmatrix}0&0&0&...&0&0&-6\frac{5n-6}{5n-11}\\5&5&5&...&5&-1&5\\5&5&5&...&-1&5&5\\ \\5&5&-1&...&5&5&5\\5&-1&5&...&5&5&5\\-1&5&5&...&5&5&5\end{bmatrix}$$ which doesn't affect the determinant. (Note the upper right entry is $-1-\frac{5}{5n-11}5(n-1)$, simplified.) Now expanding across the top row:
$$\begin{align}D_n&=(-1)^{n+1}(-6)\frac{5n-6}{5n-11}D_{n-1}\\&=(-1)^{n}(6)\frac{5n-6}{5n-11}D_{n-1}\end{align}$$
which expands recursively as a telescoping product down to $$\begin{align}D_n&=(-1)^{n(n+1)/2-1}6^{n-1}\frac{5n-6}{-1}D_{1}\\&=-(-1)^{n(n+1)/2}6^{n-1}(5n-6)\end{align}$$ which perhaps looks nicer as $$\begin{align}D_n&=(-1)^{\binom{n+1}{2}+1}6^{n-1}(5n-6)\\\{D_n\}&=\{-1,24,-324,-3024,24624,\ldots\}\\D_{100}&=-494\cdot6^{99}\end{align}$$
|
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|
A Quadrilateral's area given four sides and a diagonal Assume there exists a quadrilateral called ABCD and AB=5cm,BC=13cm,CD=16cm, DA=20cm and diagonal AC=12cm. The exercise now states that I should calculate the area of a quadrilateral. Thank you for your time!!! :)
|
By the Law of Cosines:
$$AC^2=AB^2+BC^2-2\cdot AB\cdot BC\cdot \cos \angle ABC$$
$$\iff \cos \angle ABC=\frac{5^2+13^2-12^2}{2\cdot 5\cdot 13}=\frac{5}{13}$$
Analogously: $$\cos \angle CDA=\frac{16^2+20^2-12^2}{2\cdot 16\cdot 20}=\frac{4}{5}$$
Now, we know that $\angle ABC$ and $\angle CDA$ are both acute, because $\angle ABC<\angle BAC$ and $\angle CDA<\angle ACD$. Therefore:
$$\sin \angle ABC=\sqrt{1-\cos^2 \angle ABC}=\frac{12}{13}$$
$$\sin\angle CDA=\sqrt{1-\cos^2\angle CDA}=\frac{3}{5}$$
$$A_{ABCD}=A_{\triangle ABC}+A_{\triangle CDA}$$
$$=\frac{\sin \angle ABC\cdot 5\cdot 13}{2}+\frac{\sin\angle CDA\cdot 16\cdot 20}{2}=126$$
|
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|
A sequence to achieve $\frac{1}{a_{2016}}$ It is given that $a_ka_{k-1} + a_{k-1}a_{k-2} = 2a_k a_{k-2}$ , $k\geq3$ and $a_1=1$.
We have $S_q= \sum_{k=1}^{q} \frac{1}{a_k} $ and given that $\frac{S_{2q}}{S_{q}}$ is independent of q then $\frac{1}{a_{2016}}$ is = ?
I think no information is given to find $a_2$. How should we approach this problem?
|
You get some information, I just don't think you get enough (unless I'm misreading the problem).
Given $a_1 = 1, a_2$, we can determine $a_3 = \frac{a_2}{2-a_2}$ and $a_4 = \frac{a_2^2}{3a_2-2a_2^2}$. Based on the conditions imposed on $S_q$, we also have
$$
\frac{S_2}{S_1} = \frac{S_4}{S_2}
$$
$$
1+\frac{1}{a_2} = \frac{1+\frac{1}{a_2}+\frac{2-a_2}{a_2}+\frac{3a_2-2a_2^2}{a_2^2}}{1+\frac{1}{a_2}}
$$
which can be simplified to
$$
\frac{1+a_2}{a_2} = \frac{6-2a_2}{1+a_2}
$$
or
$$
1-4a_2+3a_2^2 = 0
$$
which yields the two solutions $a_2 = 1, a_2 = \frac{1}{3}$. These two solutions yield the two series $1, 1, 1, 1, \ldots$ and $1, \frac{1}{3}, \frac{1}{5}, \frac{1}{7}, \ldots$, and hence two different values of $\frac{1}{a_{2016}}$, so I'm not sure how one should proceed.
|
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|
Laurent series expansion of $f$ Find the Laurent series expansion of $f(z)=\dfrac{1}{2z^2-13z+15}$ about the annulus $\dfrac{3}{2}<|z|<5$.
I did like this :
$f(z)=\dfrac{2}{7}(\dfrac{3}{3-2z}-\dfrac{1}{z-5})$
Then I took $\dfrac{3}{3-2z}=\dfrac{1}{1-\dfrac{2z}{3}}=(1-\dfrac{2z}{3})^{-1}=1+\dfrac{2z}{3}+(\dfrac{2z}{3})^2+...$
I took $\dfrac{1}{z-5}=\dfrac{-1}{5}(1-\dfrac{z}{5})^{-1}=\dfrac{-1}{5}(1+\dfrac{z}{5}+\dfrac{z^2}{25}+...)$
I calculated the co-efficients of $z$ and $z^2$ i.e $a_1,a_2$; I found $a_1=\dfrac{47}{75}$ and $a_2=\dfrac{4}{9}-\dfrac{1}{125}$ but the answer given is $\dfrac{a_1}{a_2}=5$
Why am I wrong?Please give some details
|
There is a small error in your partial fraction decomposition, it should be
$$
f(z) = \frac{1}{7}(\dfrac{2}{3-2z}+\dfrac{1}{z-5})
$$
(However, this does not change the result for $a_1/a_2$, see below.)
The main problem is here:
$$
\frac{3}{3-2z}=\frac{1}{1-\frac{2z}{3}}=(1-\frac{2z}{3})^{-1}=1+\frac{2z}{3}+(\frac{2z}{3})^2+\ldots
$$
converges for $\left|\frac{2z}{3}\right| < 1$, i.e. it is the power (or Laurent) series for $\frac{3}{3-2z}$ in the domain $|z| < \frac 32$.
For $|z| > \frac 32$ you have to develop the expression into
powers of $z^{-1}$:
$$
\frac{3}{3-2z}= -\frac{3}{2z}\frac{1}{1-\frac{3}{2z}}
= -\frac{3}{2z}-(\frac{3}{2z})^2-\ldots
$$
In particular, this Laurent series contains only negative powers
of $z$, which means that the ratio $a_1/a_2$ in the Laurent
series of $f$ can be computed from the Laurent series
of $1/(z-5)$ alone.
|
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|
If $a+b=8$ and $ab+c+d = 23$ and $ad+bc=28$ and $cd=12\;,$ Then $abcd$
If $a+b=8$ and $ab+c+d = 23$ and $ad+bc=28$ and $cd=12\;,$ Then value of
$(1)\;\; a+b+c+d=$
$(2)\;\; ab+bc+cd+da = $
$(3)\;\; abcd=$
My attempt: Let $x=a\;,b$ be the roots of $(x-a)(x-b)=0$ and $x=c\;,d$ be the roots
of $(x-c)(x-d)=0.$
So, $(x-a)(x-b)(x-c)(x-d)=\left[x^2-(a+b)x+ab\right]\cdot \left[x^2-(c+d)x+cd\right]=0.$
Thus, $$\left[x^2-8x+ab\right]\cdot \left[x^2-(c+d)x+12\right]=0.$$
Where do I go from here? Any help would be much appreciated.
|
This problem actually checks your observational skills of algebraic products.
Now take this:
$(x^2+ax+c)(x^2+bx+d) $
$=x^4+(a+b)x^3+(ab+c+d)x^2+(ad+bc)x+cd $
$=x^4+8x^3+23x^2+28x+12=f(x) $
$12$ is the product of roots so by putting $ x=−1 $ we find $f(−1)=0 $
$⟹(x+1) $ is factor of $ f(x) $
Similarly $(x+2) $ and $(x+3)$ are factors of $f(x) $
Hence $f(x)=(x+1)×(x+2)×(x+2)×(x+3) $
This can be written in two ways:
$(x^2+3x+2)×(x^2+5x+6) $
$⟹a=3,b=5,c=2,d=6$
$(x^2+4x+3)×(x^2+4x+4) $
$⟹a=4,b=4,c=3,d=4 $
Hence $2$ sets of solutions are possible
|
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|
$f(x) - f'(x) = x^3 + 3x^2 + 3x +1; f(9) =?$ Given the following $f(x) - f'(x) = x^3 + 3x^2 + 3x +1$
Calculate $f(9) = ?$
I have tried to play with different number of derivatives. Also tried to solve it by equations.
Maybe there is some geometric meaning that could shade the light ?
I feel it is no complex problem at all.
Thanks for your solution
|
It is most convenient to search $f$ among the third degree polynomials with the leading coefficients $1$. Let $f_0 (x) = x^3 + a x^2 + b x + c$. Then the equation becomes $$(f_0 - f'_0) (x) := x^3 + (a - 3) x^2 + (b - 2a) x + (c - b) = x^3 + 3 x^2 + 3 x + 1.$$ Hence, we have $a = 6$, $b = 15$ and $c = 16$. Then $$f_0 (x) = x^3 + 6 x^2 + 15 x + 16.$$ But this is only one from a class of solutions. Now, let $f = f_0 + g$, where $(g - g') (x) \equiv 0$. This gives us $g (x) = c \exp x$ for any constant $c$. Thus, we have $$f (x) = x^3 + 6 x^2 + 15 x + 16 + c \exp x.$$ Hence, $f (9) = 1366 + c e^9$.
|
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|
$ \int\frac{\sin(nx) \sin x}{1-\cos x} \,dx$ by elementary methods What is an elementary way to show that for positive integer $n$
$$
\int\frac{\sin(nx) \sin x}{1-\cos x} \,dx= x + \frac{\sin (nx)}{n} + 2 \sum_{k=1}^{n-1}\frac{\sin(kx)}{k}
$$
This cropped up when trying to answer Proving $\int_0^{\pi } f(x) \, \mathrm{d}x = n\pi$ at a beginning calculus level, that is, without using contour integration or Fourier integrals or Parseval's theorem (but with clever tricks allowed). There, it would suffice to know that the integral from zero to $\pi$ is $\pi$.
I've tried to do this by induction but got nowhere (except that the base for $n=1$ is easy).
I've tried expanding as a series and matching terms but it got pretty messy.
|
To prove
$$\sin(nx) \sin(x)= (1-\cos(x)) \left(1 + \cos (nx) + 2 \sum_{k=1}^{n-1}\cos(kx) \right)(**)$$
by induction, first note that the basis ($N=1)$ is easy: For $n=1$ the sum is vacuously zero, and
$$\sin(1\cdot x) \sin(x)= \sin^2 x = 1-\cos^2 x = (1-\cos(x)) \left(1 + \cos (1\cdot x) \right)$$
Now assume that for some given $n$ it is true that
$$
(1-\cos(x)) \left(1 + \cos (nx) + 2 \sum_{k=1}^{n-1}\cos(kx) \right)
= \sin(nx) \sin(x)
$$
Then The expression on the right of $(**)$, for $n+1$, is
$$
(1-\cos(x)) \left(1 + \cos ((n+1)x) + 2 \sum_{k=1}^{(n+1)-1}\cos(kx) \right) =
(1-\cos(x)) \left(1 + \cos (nx+x) + 2 \sum_{k=1}^{n-1}\cos(kx) + 2 \cos(nx)\right)
= (1-\cos(x)) \left(1 + \cos (nx) \cos x - \sin(nx) \sin x + 2 \sum_{k=1}^{n-1}\cos(kx) + 2 \cos(nx)\right)
\\=1 -\cos x + \cos (nx) \cos x -\cos(nx) \cos^2 x - \sin(nx) \sin x + \sin(nx)\sin x\cos x \\+ 2 (1-\cos x)\sum_{k=1}^{n-1}\cos(kx) + 2 \cos(nx) -2 \cos(nx) \cos x $$
Here we can rearrange terms and us $\cos^2 = 1-\sin^2$ to get
$$
\left(1 + 2 (1-\cos x)\sum_{k=1}^{n-1}\cos(kx) - \sin(nx) \sin x \right) \\- \cos x + \left(\cos (nx) \cos x -2 \cos(nx) \cos x \right)-\cos(nx) (1-\sin^2 x) + 2 \cos(nx) \\+ \sin(nx)\sin x\cos x + 2 \cos(nx) \\
= \left(1 + 2 (1-\cos x)\sum_{k=1}^{n-1}\cos(kx) - \sin(nx) \sin x \right) \\- \cos x -\cos (nx) \cos x +2\cos(nx) -\cos(nx) + \cos(nx)\sin^2 x + \sin(nx)\sin x\cos x \\
=\left(1 + 2 (1-\cos x)\sum_{k=1}^{n-1}\cos(kx) - \sin(nx) \sin x \right) \\-
\cos x \left(1 + \cos(nx) +\sin(nx)\sin x\right)
+\cos(nx) + \cos(nx)\sin^2 x
$$
Now replace $\sin(nx)\sin x$ by the expression in $(**)$, which we can do because by the indcution assumption that holds for $n$:
$$(1-\cos(x)) \left(1 + \cos ((n+1)x) + 2 \sum_{k=1}^{(n+1)-1}\cos(kx) \right) =
(1-\cos(x)) \left(1 + \cos (nx+x) + 2 \sum_{k=1}^{n-1}\cos(kx) + 2 \cos(nx)\right) \\
=\left(1 + 2 (1-\cos x)\sum_{k=1}^{n-1}\cos(kx) - \sin(nx) \sin x \right) \\-
\cos x \left(1 + \cos(nx) +\sin(nx)\sin x\right)
+\cos(nx) + \cos(nx)\sin^2 x \\
= \left(1 + 2 (1-\cos x)\sum_{k=1}^{n-1}\cos(kx) - (1-\cos x) \left(1 + \cos (nx) + 2 \sum_{k=1}^{n-1}\cos(kx) \right) \right) \\-
\cos x \left[1 + \cos(nx) +\sin(nx)\sin x\right]
+\cos(nx) + \cos(nx)\sin^2 x \\
= \left(1 + 2 (1-\cos x)\sum_{k=1}^{n-1}\cos(kx) - 2(1-\cos x) \sum_{k=1}^{n-1}\cos(kx) - (1-\cos x) \left(1 + \cos (nx) \right) \right) \\
-\cos x \left[1 + \cos(nx) +\sin(nx)\sin x\right]
+\cos(nx) - \cos(nx)\sin^2 x \\
= \left(1 - (1-\cos x) \left(1 + \cos (nx) \right) \right) \\
- \cos x \left[1 + \cos(nx) +\sin(nx)\sin x\right]
+\cos(nx) - \cos(nx)\sin^2 x \\
= - \cos nx + \cos x + \cos x \cos(nx) - \cos x - \cos x \cos(nx) \\
-\cos x \sin(nx)\sin x +\cos(nx) - \cos(nx)\sin^2 x \\
= - \cos nx +\cos(nx) + \cos x - \cos x + \cos x \cos(nx) - \cos x \cos(nx) \\ -\cos x \sin(nx) \sin x + \cos(nx)\sin^2 x \\
= \sin x \left[ \sin(nx)\cos x - \cos(nx)\sin x\right]
= \sin x \sin(nx +x) = \sin((n+1)x) \sin x
$$
Thus we have shown that
$$\sin((n+1)x) \sin x = (1-\cos(x)) \left(1 + \cos ((n+1)x) + 2 \sum_{k=1}^{(n+1)-1}\cos(kx) \right)
$$
and induction is established.
IMHO, this was not a "relatively easy" induction question, but it certainly did not use anything sophisticated at a level beyond simple trig.
|
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|
Prove that the equation of the tangent at P is $ \frac {xx_1}{a^2} - \frac {yy_1}{b^2} = 1 $ (Hyperbolas)
Question: Point P ($x_1 , y_1$) is on the hyperbola $\frac {x^2}{a^2}$ - $\frac {y^2}{b^2}$ = 1
Prove that the equation of the tangent at P is
$$ \frac {xx_1}{a^2} - \frac {yy_1}{b^2} = 1 $$
What I have attempted:
Equation of a tangent is:
$$ (y-y_0) = m(x-x_0) $$
In this case we have $x_0 = x_1$ and $y_0 = y_1$ Now we need to find $m$ so I am going to implicitly differentiate the equation of the hyperbola
$$ \frac {x^2}{a^2} - \frac {y^2}{b^2} = 1 $$
$$ \frac {2x}{a^2} - \frac {2y}{b^2} (\frac {dy}{dx})= 1 $$
$$ \frac {dy}{dx}= \frac {xb^2}{ya^2}$$
at $x = x_1$ and $y = y_1$
$$ \frac {dy}{dx}= \frac {x_1b^2}{y_1a^2}$$
With $ (y-y_0) = m(x-x_0) $
$$ (y-y_1) = \frac {x_1b^2}{y_1a^2}(x-x_1) $$
$$ (y-y_1) = \frac {xx_1b^2}{y_1a^2} - \frac {x_1^2b^2}{y_1a^2} $$
$$ (y-y_1) = \frac {xx_1b^2 - x_1^2b^2 }{y_1a^2} $$
$$ yy_1a^2 - y_1^2a^2 = xx_1b^2 - x_1^2b^2 $$
$$ \frac{yy_1}{b^2} -\frac{y_1^2}{b^2} = \frac{xx_1}{a^2} - \frac{x_1^2}{a^2}$$
$$ \frac{x_1^2}{a^2} -\frac{y_1^2}{b^2} = \frac{xx_1}{a^2} - \frac{yy_1}{b^2} $$
Now I am stuck I am trying to get rid of
$ \frac{x_1^2}{b^2} -\frac{y_1^2}{b^2} $ and try to make that equal to $1$ and prove the statement but the equation of the hyperbola is given $\frac {x^2}{a^2}$ - $\frac {y^2}{b^2}$ = 1...
|
Since $(x_1,y_1)$ lie on hyperbola, they satisfy the equation $\frac{x^2}{a^2}-\frac{y^2}{b^2} = 1$
$$ \frac{x_1^2}{a^2}-\frac{y_1^2}{b^2} = 1$$
$$\implies\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$$
|
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|
Find the polynomial $P$ of smallest degree with rational coefficients and leading coefficient $1$ such that $ P(49^{1/3}+7^{1/3})=4 $
Find the polynomial $P$ of smallest degree with rational coefficients
and leading coefficient $1$ such that $$ P(49^{1/3}+7^{1/3})=4 $$
(Source:NYSML)
My attempt
Let $$ 49^{1/3}+7^{1/3}=x$$
Then,
\begin{array}
((49^{1/3})^3 &=(x-7^{1/3})^{3} \\
49 &=x^3-7-3x^2\cdot 7^{1/3}+3x \cdot 7^{2/3} \\
\end{array}
So I have $$x^3+x^2 (-3\cdot 7^{1/3})+x(3\cdot 7^{2/3})-56 =0 $$
Define $$Q(x)=x^3+x^2 (-3\cdot 7^{1/3})+x(3\cdot 7^{2/3})-56 $$
making the substitution $x=7^{1/3} \cdot t $,
I have $$F(t)=\cfrac{Q(7^{1/3} \cdot t)}{7}=t^3+t^2(-21)+21t-56$$
Now I am quite stuck on how I should continue...
|
A fast way to get the polynomial is to observe that
$$(x+y)^3=x^3+3x^2y+3xy^3+y^3=x^3+y^3+3xy(x+y) (**)$$
Now with your $a=49^{1/3}+7^{1/3}$ setting $x=49^{1/3}, y=7^{1/3}$ you have
$$x^3=49, y^3=7, xy=7$$
Replace those in $(**)$ and you are done.
|
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|
constrained stars and bars problem I want to know number of solutions for following equation, where $r_k$'s are non-negative integers, and there is a constraint on $r_k$'s such that $r_1 \geq r_2 \geq \cdots \geq r_K$
\begin{equation}
r_1+r_2+\cdots+r_K=N
\end{equation}
|
EDIT: This response answers a slightly different question, where $K$ (the number of summands) is allowed to vary. I will leave it here for posterity, but it doesn't deserve any upvotes.
These are the integer partitions of $N$. There is no closed formula for $p(N)$, the number of such partitions, although there is a nice generating function:
$$
\begin{align}
\sum_{N=1}^\infty p(N) x^N &= \prod_{k=1}^{\infty} \frac{1}{1-x^k} \\
&= (1 + x + x^2 + x^3 + \cdots)
(1 + x^2 + x^4 + x^6 + \cdots)
(1 + x^3 + x^6 + x^9 + \cdots)
\cdots
\end{align}
$$
The beginning of this expansion is
$$
1 + x + 2x^2 + 3x^3 + 5x^4 + 7x^5 + 11x^6 + \cdots
$$
which is another way of saying that for $N = 0, 1, 2, 3, \dots$, the values of $p(N)$ are
$$
1, 1, 2, 3, 5, 7, 11, \dots
$$
|
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|
solution of differential equation $\left(\frac{dy}{dx}\right)^2-x\frac{dy}{dx}+y=0$
The solution of differential equation $\displaystyle \left(\frac{dy}{dx}\right)^2-x\frac{dy}{dx}+y=0$
$\bf{My\; Try::}$ Let $\displaystyle \frac{dy}{dx} = t\;,$ Then Diferential equation convert into $t^2-xt+y=0$
So Its solution is given by $\displaystyle t=\frac{x\pm \sqrt{x^2-4y}}{2}$
So we get $$\frac{dy}{dx} = \frac{x\pm \sqrt{x^2-4y}}{2}$$
Now How can I solve after that, Help me
Thanks
|
Here's another approach. Differentiating gives $y''\left(2y'-x\right)=0$, so $y''=0$ or $y'=\frac{x}{2}$. The former option gives $y=ax+b$ so $a^2-ax+ax+b=0$ and $b=-a^2$. The latter option gives $y=\tfrac{x^2}{4}+c$ so $\frac{x^2}{4}-\frac{x^2}{2}+\tfrac{x^2}{2}+c=0$, which work iff $c=0$. The solution is $y=ax-a^2$ or $y=\frac{x^2}{4}$.
|
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|
Prove that $a^ab^bc^c\geq (\frac{a+b}{2})^{\frac{a+b}{2}} (\frac{b+c}{2})^{\frac{b+c}{2}}(\frac{c+a}{2})^{\frac{c+a}{2}}$
Prove that $$a^ab^bc^c\geq \left(\frac{a+b}{2}\right)^{\frac{a+b}{2}} \left(\frac{b+c}{2}\right)^{\frac{b+c}{2}}\left(\frac{c+a}{2}\right)^{\frac{c+a}{2}}\geq \left(\frac{a+b+c}{3}\right)^{a+b+c}$$ where $a,b,c$ are positive real numbers.
I am able to prove the $\left(\frac{a+b}{2}\right)^{\frac{a+b}{2}} \left(\frac{b+c}{2}\right)^{\frac{b+c}{2}}\left(\frac{c+a}{2}\right)^{\frac{c+a}{2}}\geq \left(\frac{a+b+c}{3}\right)^{a+b+c}$ by using weighted AM$\geq$ GM on the numbers $\frac{a+b}{2}, \frac{b+c}{2}, \frac{c+a}{2}$ with associated weights $\frac{a+b}{2}, \frac{b+c}{2}, \frac{c+a}{2}$.
The problem is to prove $$\bbox[5px,border:2px solid #C0A000]{a^ab^bc^c\geq \left(\frac{a+b}{2}\right)^{\frac{a+b}{2}} \left(\frac{b+c}{2}\right)^{\frac{b+c}{2}}\left(\frac{c+a}{2}\right)^{\frac{c+a}{2}}}$$
Please help me to solve it.
|
For $x > 0 \Rightarrow f(x) = x\ln x \Rightarrow f'(x) = 1 + \ln x \Rightarrow f''(x) = \dfrac{1}{x} > 0\Rightarrow f(a)+ f(b) \geq 2f\left(\dfrac{a+b}{2}\right)\Rightarrow a\ln a + b\ln b \geq 2\left(\dfrac{a+b}{2}\right)\ln\left(\dfrac{a+b}{2}\right)\Rightarrow \ln(a^ab^b) \geq \ln\left(\left(\dfrac{a+b}{2}\right)^{a+b}\right)\Rightarrow a^ab^b \geq \left(\dfrac{a+b}{2}\right)^{a+b}$. Repeat this process for the other pairs $(b,c)$,and $(c,a)$ we have: $a^ab^bb^bc^cc^ca^a \geq \left(\dfrac{a+b}{2}\right)^{a+b}\left(\dfrac{b+c}{2}\right)^{b+c}\left(\dfrac{c+a}{2}\right)^{c+a}\Rightarrow (a^ab^bc^c)^2 \geq \left(\left(\dfrac{a+b}{2}\right)^{\frac{a+b}{2}}\left(\dfrac{b+c}{2}\right)^{\frac{b+c}{2}}\left(\dfrac{c+a}{2}\right)^{\frac{c+a}{2}}\right)^2\Rightarrow \text{QED}$
|
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|
Prove that $\frac{7}{12}<\ln 2<\frac{5}{6}$ using real analysis I studying in Real Analysis 2, but I have no idea how to solve this problem. My guess is to use Mean Value Theorem or a similar theorem? Could any one help me?
Thanks.
|
Consider the following inequalities from Proof that $\frac{2}{3} < \log(2) < \frac{7}{10}$
as suggested by @labbhattacharjee
$$\frac{1}{2}\int_{0}^{1}\frac{x^2(1-x)}{1+x}dx>0$$
$$\frac{1}{2}\int_{0}^{1}\frac{x^5(1-x)}{1+x}dx>0$$
The integrals are positive because the integrands are positive for $0<x<1$.
Let us evaluate the first integral
$$\begin{align}
\frac{1}{2}\int_{0}^{1}\frac{x^2(1-x)}{1+x}dx
&=
\frac{1}{2}\int_{0}^{1}\frac{x^2-x^3}{1+x}dx \\
&=
\frac{1}{2}\int_{0}^{1}\left(-x^2+2x-2+\frac{2}{1+x}\right)dx\\
&=
\left.\frac{1}{2}\left(-\frac{x^3}{3}+x^2-2x+2\log(1+x)\right) \right|_{0}^{1}\\
&=
\frac{1}{2}\left(-\frac{1}{3}+1-2+2\log(2)\right) \\
&=
\log(2)-\frac{2}{3}>0 \\
\end{align}
$$
Similarly,
$$\frac{1}{2}\int_{0}^{1}\frac{x^5(1-x)}{1+x}dx=\frac{7}{10}-\log(2)>0$$
Combining both results,
$$\frac{2}{3}-\log(2)<0<\frac{7}{10}-\log(2)$$
so
$$\frac{2}{3}<\log(2)<\frac{7}{10}$$
To solve your problem we shall compare $\frac{2}{3}$ against $\frac{7}{12}$ and $\frac{7}{10}$ against $\frac{5}{6}$
On the one hand,
$$21<24$$
$$7·3<2·12$$
$$\frac{7}{12}<\frac{2}{3}$$
On the other hand,
$$42<50$$
$$7·6<5·10$$
$$\frac{7}{10}<\frac{5}{6}$$
Finally,
$$\frac{7}{12}<\frac{2}{3}<\log(2)<\frac{7}{10}<\frac{5}{6}$$
so
$$\frac{7}{12}<\log(2)<\frac{5}{6}$$
|
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|
Is there any way to solve integral of $\sqrt{8-x^{2}}$ without using $\sin$ or $\cos$ formulas? I was thinking about the following integral if I could solve it without using trigonometric formulas. If there is no other way to solve it, could you please explain me why do we replace $x$ with $2\sqrt 2 \sin(t)$? I'm really confused about these types of integrals.
$$\int \sqrt{8 - x^2} dx$$
|
Let
$$I=\int\sqrt{8-x^2}\ dx\tag 1$$
using integration by parts,
$$I=\sqrt{8-x^2}\int 1\ dx-\int \left(\frac{-2x}{2\sqrt{8-x^2}}\right)\cdot x\ dx$$
$$I=\sqrt{8-x^2}(x)-\int \frac{(8-x^2)-8}{\sqrt{8-x^2}} \ dx$$
$$I=x\sqrt{8-x^2}-\int \left(\sqrt{8-x^2}-\frac{8}{\sqrt{8-x^2}} \right)\ dx$$
$$I=x\sqrt{8-x^2}-\int\sqrt{8-x^2}\ dx+8\int \frac{1}{\sqrt{8-x^2}}\ dx$$
setting the value from (1),
$$I=x\sqrt{8-x^2}-I+8\int \frac{1}{\sqrt{(2\sqrt 2)^2-x^2}}\ dx$$
$$2I=x\sqrt{8-x^2}+8\sin^{-1}\left(\frac{x}{2\sqrt 2}\right)+c$$
$$I=\color{red}{\frac{1}{2}\left(x\sqrt{8-x^2}+8\sin^{-1}\left(\frac{x}{2\sqrt 2}\right)\right)+C}$$
|
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|
finding real roots by way of complex I was given
$$x^4 + 1$$
and was told to find its real factors. I found the $((x^2 + i)((x^2 - i))$ complex factors but am lost as to how the problem should be approached.
My teacher first found 4 complex roots ( different than mine)
$$( x - \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i)( x - \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i)$$
$$( x + \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i)( x + \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i)$$
He then mltiplied both pairs,resulting in:
$$(x^2 - \sqrt{2}x + 1) and (x^2 + \sqrt{2}x + 1)$$
How do I get the first four imaginary points, or the two distinct because I realize that each has a conjugate.
|
From $$x^4 + 1 = (x^2+i)(x^2-i)$$ we need to find the roots of each factor. Remember that $i = e^{i\pi/2}$, so $x=e^{i\pi/4}$ is a root of $x^2-i$. Hence, so is its conjugate $e^{-i\pi/4}$. By Euler's formula,
$$
e^{i\pi/4} = \cos(\pi/4) + i\sin(\pi/4) = \frac{\sqrt{2}}{2} + i\frac{\sqrt 2}{2}.
$$
I will leave you to do the others.
|
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|
Finding coefficient of polynomial? The coefficient of $x^{12}$ in $(x^3 + x^4 + x^5 + x^6 + …)^3$ is_______?
My Try:
Somewhere it explain as:
The expression can be re-written as: $(x^3 (1+ x + x^2 + x^3 + …))^3=x^9(1+(x+x^2+x^3))^3$
Expanding $(1+(x+x^2+x^3))^3$ using binomial expansion:
$(1+(x+x^2+x^3))^3 $
$= 1+3(x+x^2+x^3)+3*2/2((x+x^2+x^3)^2+3*2*1/6(x+x^2+x^3)^3…..$
The coefficient of $x^3$ will be $10$, it is multiplied by $x^9$ outside, so coefficient of $x^{12}$ is $10$.
Can you please explain?
|
Another way: For $|x|<1$, we have:
$$(x^3+x^4+x^5+...)^3=x^9(1+x+x^2+...)^3=x^9(1-x)^{-1}.$$
Now $(1-x)^{-3}$ is half of the second derivative of $(1-x)^{-1}.$ The second derivative of $(1-x)^{-1}=(1+x+x^2+...)$ is $(1.2+2.3 x+3.4 x^2+4.5 x^3+...). $ Half the co-efficient of $x^3$ in this, which is $(1/2).4.5=10,$ is therefore the co-efficient of $x^{12}$ in $x^9(1+x+x^2+...). $
|
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|
How many numbers between $1$ and $9999$ have sum of their digits equal to $8$? $16$? How many numbers between $1$ and $9999$ have sum of their digits equal to $8$? $16$?
Can someone tell me if I got the right answers? I solved both cases and I've got $148$ for $8$ and $633$ for $16$. I solved this problem using $x_1+x_2+x_3+x_4=8$ then $16$. After that I decided to take the case when $1$ number is bigger than $10$, but I'm not sure if that's the way I should do it for $8$. And then I substract the $4$ cases when $x_1=0$ and $7$ cases when $x_1=9$. Thank you.
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We treat a number with fewer than four digits as a number with leading zeros. For instance, we regard the number $235$ as $0235$ and the number $8$ as $0008$. Then the number of positive integers between $1$ and $9999$ with digit sum $8$ is equal to the number of solutions of the equation
$$x_1 + x_2 + x_3 + x_4 = 8 \tag{1}$$
in the non-negative integers. A particular solution corresponds to the placement of three addition signs in a row of eight ones. For instance,
$$+ + + 1 1 1 1 1 1 1 1$$
corresponds to the choice $x_1 = x_2 = x_3 = 0$ and $x_4 = 8$. Thus, the number of positive integers with up to four digits that have digit sum $8$ is
$$\binom{8 + 3}{3} = \binom{11}{3}$$
since we must choose which three of the eleven symbols (eight ones and three addition signs) will be addition signs.
The number of positive integers between $1$ and $9999$ which have digit sum $16$ is the number of solutions of the equation
$$x_1 + x_2 + x_3 + x_4 = 16 \tag{2}$$
in the non-negative integers subject to the restrictions that $x_k \leq 9$ for $1 \leq k \leq 4$. The number of solutions of equation 2 in the non-negative integers is
$$\binom{16 + 3}{3}$$
From these, we must exclude those solutions in which at least one of the variables exceeds $9$. Notice that since $2 \cdot 10 = 20 > 16$, at most one of the variables may exceed $9$.
Suppose $x_1 > 9$. Let $y_1 = x_1 - 10$. Then $y_1$ is a non-negative integer. Substituting $y_1 + 10$ for $x_1$ in equation 2 yields
\begin{align*}
y_1 + 10 + x_2 + x_3 + x_4 & = 16\\
y_1 + x_2 + x_3 + x_4 & = 6 \tag{3}
\end{align*}
Equation $3$ is an equation in the non-negative integers with
$$\binom{6 + 3}{3} = \binom{9}{3}$$
solutions. By symmetry, there are $\binom{9}{3}$ solutions of equation 2 in which $x_k$ exceeds $9$ for $1 \leq k \leq 4$. Hence, the number of solutions of equation 2 in which one of the variables exceeds $9$ is
$$\binom{4}{1}\binom{9}{3}$$
Hence, the number of positive integers between $1$ and $9999$ which have digit sum $16$ is
$$\binom{19}{3} - \binom{4}{1}\binom{9}{3}$$
|
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|
Multi-variable Limit Question: Does the following limit exist, if so what does it equal?
$$\lim_{(x, y) \to (0,0)} \frac{x^2 y^5}{2x^4 +3y^{10}}$$
-
Solution 1: The limit DOES NOT exist.
Let $ x=y^{5/2}$
$$\lim_{y \to 0} \frac{(y^{5/2})^2 y^5}{2(y^{5/2})^4 +3y^{10}} = \lim_{y \to 0} \frac{y^{10}}{2y^{10} +3y^{10}} = \lim_{y \to 0} \frac{1}{2 +3} = \frac{1}{5} $$
Let $ x=0$
$$\lim_{y \to 0} \frac{(0)^2 y^5}{2(0)^4 +3y^{10}} = \lim_{y \to 0} \frac{0}{0 +3y^{10}} = \lim_{y \to 0} 0 = 0 $$
Thus the limit does not exist
Solution 2: The limit DOES exist.
Change to polar coordinates and find limit as $r \to 0$
$$\lim_{r \to 0} \frac{(r^2cos^2{\theta}) (r^5sin^5\theta)}{2(r^4cos^4{\theta}) +3(r^{10}cos^{10}{\theta})} =\lim_{r \to 0}\frac{r^4}{r^4} \times\frac{r^3cos^2{\theta} sin^5\theta}{2cos^4{\theta} +3r^{6}cos^{10}{\theta}} =\frac{0}{2cos^4{\theta} +0}= 0 $$
Thus the limit does exist and is $0$
-
Problem Can someone please tell me which solution is correct (if any) and why the other is wrong?
|
The first solution is correct, the limit does not exists. The second one is wrong because the "limit"
$$ \frac{0}{0 + 2 \cos^4\theta}$$
is not well-defined if $\theta = \frac\pi 2$.
|
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|
the maximum value of $(a+\frac1b)(b+\frac1c)(c+\frac1d)(d+\frac1a)$
If $a, b, c, d \in [\frac12, 2]$ and $abcd =1$, find the maximum value of
$$(a+\frac1b)(b+\frac1c)(c+\frac1d)(d+\frac1a)$$
Thought: $$\begin{split}(a+\frac1b)(b+\frac1c)(c+\frac1d)(d+\frac1a) &= \frac{(ab+1)(bc+1)(cd+1)(da+1)}{abcd} \\ & \stackrel{\text{C-S}}{\le} \sqrt{(a^2+1)(b^2+1)}\sqrt{(b^2+1)(c^2+1)}\sqrt{(c^2+1)(d^2+1)}\sqrt{(d^2+1)(a^2+1)} \\ & = (a^2+1)(b^2+1)(c^2+1)(d^2+1)\end{split}$$
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I think it can help:
http://www.artofproblemsolving.com/community/c6h621917
See an elim's proof.
|
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|
Prove that $1280000401$ is Composite I tried to prove $N=1280000401$ as composite using complex cube roots of unity:
we can write $$N=1+400+(128*10^{7})$$ which gives
$$N=1+20^2+20^{7}$$
now if $F(x)=1+x^2+x^7$, $w$ and $w^2$ are roots of $F(x)=0$ where $w=\frac{-1+i\sqrt{3}}{2}$ and $w^2$ its conjugate.
Hence $x^2+x+1$ is factor of $1+x^2+x^7$
Hence $1+20+20^2=421$ is a factor of $N$ and hence it is composite.
But how can we prove that without using complex numbers?
|
You can say $1+x^2+x^7=1+x+x^2+(x^7-x)=1+x+x^2+x(x^3-1)(x^3+1)=(1+x+x^2)(1+x(x-1)(x^3+1))$
but your approach is a good one to find this. People have been very clever in finding things to try.
|
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|
Find the factors of $(a+b+c)^3-(b+c-a)^3-(c+a-b)^3-(a+b-c)^3$ Find the factors of
$$(a+b+c)^3-(b+c-a)^3-(c+a-b)^3-(a+b-c)^3$$
the answer is $24abc$
Let $E = (a+b+c)^3-(b+c-a)^3-(c+a-b)^3-(a+b-c)^3$
Since $b+c = a $ makes $E = 0, \therefore (b+c-a)$ is one factor, similarly $(c+a-b)$ and $(a+b-c)$ are factors, but can not proceed further :(
|
@user230452 asks for an enlightened way of doing this.
The expression
$$
E = E(a, b, c) = (a+b+c)^3-(b+c-a)^3-(c+a-b)^3-(a+b-c)^3
$$
is symmetric in $a, b, c$, and homogeneous of total degree $3$, so it must be an integral combination of the elementary symmetric polynomials in $a, b, c$, of the form
$$
E = x \sigma_{1}^{3} + y \sigma_{1} \sigma_{2} + z \sigma_{3},
$$
for suitable integers $x, y, z$. Here $\sigma_{1} = a + b + c$, $\sigma_{2} = ab + ac + bc$, $\sigma_{3} = a b c$ are the elementary symmetric polynomials in $a, b, c$.
Set in order
*
*$a = b = c = 1$,
*$a = b = 1$, $c = 0$,
*$a = 1$, $b = c = 0$
to get
$$
\begin{cases}
3 x + 9 y + z &= E(1, 1, 1) &=24\\
2 x + 2 y &= E(1, 1, 0) &=0\\
x &= E(1, 0, 0) &=0\\
\end{cases}
$$
Hence $x = y = 0$, $z = 24$, and the expression evaluates to $$E = 24 \sigma_{3} = 24 a b c.$$
|
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|
How many distinct combinations of five marbles could be drawn? A bag contains ten marbles of the same size: 3 are identical green marbles, 2 are identical red marbles, and the other 5 are five distinct colors. If 5 marbles are selected at random, how many distinct combinations of five marbles could be drawn?
(A) 41
(B) 51
(C) 62
(D) 72
(E) 82
I am lost in this question, can someone help?
|
Let $k$ be the number of green and red marbles chosen, and let $n$ be the number of other marbles chosen.
Then $k+n=5$, and if $g(k)$ is the number of ways to choose the $k$ marbles,
then $g(0)=g(5)=1,\; \;g(1)=g(4)=2,\;\;g(2)=g(3)=3$.
Since there are $\dbinom{5}{n}$ ways to choose the other $n$ marbles,
there are $\displaystyle1\binom{5}{5}+2\binom{5}{4}+3\binom{5}{3}+3\binom{5}{2}+2\binom{5}{1}+1\binom{5}{0}=82$ selections.
Alternatively, we can find the number of solutions of $x_1+\cdots+x_7=5$ in nonnegative integers
$\hspace .4 in$where $\;x_1\le3, \;x_2\le2,\;$ and $\;x_i\le1\;$ for $3\le i\le 7$.
If we use Inclusion-Exclusion,
letting $E_1$ be the selections with $x_1\ge4$, $E_2$ the selections with $x_2\ge3$, and
$\hspace .4 in E_i$ the selections with $x_i\ge2$ for $3\le i\le7$,
this gives $\binom{11}{6}-\binom{7}{6}-\binom{8}{6}-5\binom{9}{6}+5\binom{6}{6}+\binom{5}{2}\binom{7}{6}=82$ choices.
|
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Proof that the sum of the first $n$ odd numbers is $n^2$. Here is what I have so far:
The $n$th odd number is $2n-1$.
So we prove that $1+3+...+(2n-3)+(2n-1)= n^2$.
Separate the last term and you get: $[1+3+...+(2n-3)]+(2n-1)$
$[1+3+...+(2n-3)]$ is the sum of the first $(n-1)$ odd numbers.
Here is where I get stuck. The textbook says that the sum of the first $(n-1)$ odd numbers is $(n-1)^2$, but why is that the case? It seems like a recursive explanation because we are trying to prove that the sum of the first $n$ odd numbers is $n^2$. Since we have not yet proved that, how can one say with certainty that the sum of the first $(n-1)$ odd numbers is $(n-1)^2$ ?
|
Another way of looking at is ...
$$\begin{align}&1 + 3 + 5 + {\dots} + 2n - 1 \\
&= (n - (n - 1)) + {\dots} + (n - 4) + (n - 2) + n + (n + 2) + (n + 4) + {\dots} + (n + n - 1)\\
&= n + n +n + {\dots} +n (n times)\\
&= n^{2}\end{align}$$
It can also be proved with the sum of arithmetic progression. The sum of an $A.P.$ with $n$ terms, difference $d$, and initial term $a$.
$$\begin{align}T_1 &=a\\
T_2 &=a + 2d\\
.\\
.\\
.\\
T_n &= a + (n-1)d\end{align}$$
Adding all these terms,
$$\begin{align}S &= na + d(1+2+{\dots}+n-1)\\
S &= na + \frac{n(n-1)d}{2}\\
S &= \frac{n}{2}(a + a + (n-1)d)\\
S &= \frac{n}{2}(a + T_n)\end{align}$$
Applying this to our series,
$$\begin{align}S &= \frac{n}{2}(1+2n - 1)\\
S &= n^{2}\end{align}$$
This property of odd numbers was used to generate Pythagorean triplets by Fibonacci.
$1$. Pick an odd number and square it. The number chosen in $m$.
$2$. $m^2$ also lies in the series. Take the sum of all odd numbers before $m^2$. It is $(\frac{(m^2-1)}{2})^2$
$3$. Adding these two numbers will also be a perfect square. $m^2 +(\frac{m^2-1}{2})^2 = \frac{m^4 +2m^2+ 1}{4} = (\frac{m^2 + 1}{2})^2$
The numbers, $m, \frac{m^2-1}{2}, \frac{m^2+1}{2}$ are a Pythagorean triplet, wherever m is odd.
|
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|
How to find Ratio And proportion If $\frac{4x+3y}{4x-3y}=\frac{7}{5}$.Find the Value of the $\frac{2x^2-11y^2}{2x^2+11y^2}$.
Okay I just want hint how to solve this problem.
|
Just notice that
$$\frac{4x+3y}{4x-3y}=\frac{7}{5}\qquad\implies\qquad 5(4x+3y)=7(4x-3y)\qquad\implies\qquad 36y=8x$$
So $y=\frac{2}{9}x$, then
$$\frac{2x^2-11y^2}{2x^2+11y^2}=\frac{2x^2-\frac{44}{81}x^2}{2x^2+\frac{44}{81}x^2}=\frac{2\cdot 81-44}{2\cdot 81+44}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
limit of sum defined sequence Let $x_n=\displaystyle \sum_{k=1}^n \sqrt{1+\frac{k}{n^2}}, n\ge1$. Prove that $\displaystyle \lim_{n \rightarrow \infty} n (x_n-n-\frac{1}{4})=\frac{5}{24}$.
What I've done:it's easy to show that $\displaystyle \lim_{n \rightarrow \infty} \frac{x_n}{n}=1$ and $\displaystyle \lim_{n \rightarrow \infty} (x_n-n)=\frac{1}{4}$
|
Expand the square root in a taylor series to three terms, to get $1+k/2n^2 -k^2/8 n^4+ k^3/16 n^6,$ then sum.
The sum of the $1$s gives you $n$. The sum of $k/n^2$ gives $n(n+1)/4n^2,$ the sum of $k^2/n^4$ gives you $(1+n)(2+n)/48 n^3,$ and the next term is $O(1/n^2).$
|
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|
Factoring $({16+4t^2+\frac{t^4}{4}})^{1/4}$ How do you go about factoring $({16+4t^2+\frac{t^4}{4}})^{1/4}$
Which factoring method would be the best?
|
Notice that $t^4/4 = (t^2/2)^2$ so let $u = t^2/2$
Then $({16+4t^2+\frac{t^4}{4}})^{1/4} = (16 + 8u + u^2)^{1/4}$
$16 + 8u + u^2 = (4+u)^2$ and $(a^2)^{1/4} = a^{1/2}$
So $({16+4t^2+\frac{t^4}{4}})^{1/4} = (16 + 8u + u^2)^{1/4} = (4 +u)^{1/2} = (4+t^2/2)^{1/2}$
Is that simplified enough?
|
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|
Primes of the form $x^3+y^3+z^3 - 3xyz$ Do quadruplets $(x,y,z,p)$ of positive integers exist for which $p$ is a prime number and $$x^3+y^3+z^3 = 3xyz + p?$$
I've tried looking for solutions in mathematica for $x,y,z<1000$, without finding any. Unfortunately, looking at the equation $\mod 3$ or $\mod 9$ has yielded no result. Is there anyone who can help? Thanks.
EDIT: I just discovered a fatal flaw in my mathematica code. vrugtehagel rightfully pointed out that $8^3+8^3+7^3 - 3\cdot8\cdot8\cdot7 = 23$ is a valid solution to this problem.
|
We use the identity
$$x^3 + y^3 + z^3 - 3xyz = (x+y+z)\left( \frac{ (x-y)^2 + (y-z)^2 + (z-x)^2 }{2}\right)$$
to try to find some solutions using linear and quadratic equations with the following system:
$$\begin{cases}x+y+z=1\\(x-y)^2+(y-z)^2+(z-x)^2=2p\end{cases}$$
Putting $z=1-x-y$ one gets
$$(x-y)^2 + (y-z)^2 + (z-x)^2=2(x^2+y^2+z^2)-2(xy+xz+yz)=2p\Rightarrow 3(x^2+y^2+xy-x-y)+1=p$$
We look for primes in the arithmetic progression $\{3n+1\}_{n\in \mathbb N}$ so we get the equation $$x^2+y^2+xy-x-y=n$$ and we look for integer points in ellipses.
We look at the six first primes $p=7,13,19,31,37,43$ of the form $3n+1$ corresponding to $n=2,4,6,10,12,14$ respectively.
$$x^2+y^2+xy-x-y=2\qquad (*)$$
Solutions $(x,y)=(0,-1),(2,-1),(2,0)$ yield the solution $(x,y,z)=(2,-1,0)$ giving an example for
$$x^3+y^3+z^3-xyz=7$$
$$x^2+y^2+xy-x-y=4\qquad (**)$$
Solutions $(x,y)= (1,\pm2),(2,-2)$ yield the solutions $(x,y,z)=(1,\pm 2,\mp 2),(2,-2,1)$ giving examples for $$x^3+y^3+z^3-xyz=13$$
$$x^2+y^2+xy-x-y=6\qquad (***)$$ Solutions $(x,y)=(0,-2),(3,-2),(3,0)$ leads to examples for $p=19$.
Similarly $$x^2+y^2+xy-x-y=10, 12, 14\qquad (****)$$ admit respectively the solutions
$\{(-1,-2),(4,-2),(4,-1)\},\{(0,-3),(4,-3),(4,0)\}$ and $\{(2,-4),(3,-4),(3,2)$ which give solutions for
$$x^3+y^3+z^3-xyz=31,37\space \text{and} \space 43$$
FINAL NOTE.- I wanted just to exhibit some examples of solutions like the case $p=23$ given by the O.P.(note that prime $23=3\cdot 7+2$ it is not of the form $3n+1$). However I add a question and a pertinent quote to finish.
►“It is very difficult to prove the non-existence of integer solutions of the general equation
$$ax^3+by^3+cz^3-dxyz=0”\space \text{(L.J.Mordell)}$$
►Is there always solution for $x^3+y^3+z^3-xyz=p$ when $p=3n+1$? (for the first 6 examples of such primes, I have found without difficulty solutions).
|
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|
Prove that for positive integer $ n$ we have $169| 3^{3n+3}-26n-27$.
Prove that for positive integer $n$ we have $169| 3^{3n+3}-26n-27$.
It is easy to show that if we take the expression modulo $13$ we have $3^{n+3}-26n-27 \equiv 1^{n+1}-1 \equiv 1-1 \mod 13 = 0 \mod 13$. How do I prove it is divisible by $13^2$?
|
$A(n) = 3^{3n+3}-26n-27$
$A(n+1) = 3^{3n+6}-26(n+1)-27$
$A(n+1) - A(n) = 3^{3n+3}(3^3-1) - 26(n+1)-27 + 26n + 27 = 3^{3n+3} \cdot 26 - 26 = (3^{3n+3} - 1) \cdot 26 = (27^{n+1} - 1) \cdot 26$
From here, it's quite obvious now, since 13 divides both multipliers.
$13 / (27^{n+1} - 1) $ and
$13 / 26 $
Hence $13^2 / (A(n+1) - A(n))$
Finally all you have to check is that $13^2 / A(0)$
|
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Proof for $\forall x\in \mathbb{Z}^+, \exists t \in \mathbb{Z}, 5 \nmid x \to ((x^2= 5t + 1) \vee (x^2 = 5t – 1))$ I am trying to write a proof for the following: If $x$ is a positive integer that is not divisible by $5$, then $x^2$ can be written either as $x^2 = 5t+1$ or $x^2 = 5t−1$ for some integer $t$. So far, I have managed to get that if $x$ is not divisible by $5$, then it is one of the following cases, where $n$ represents a positive integer:
\begin{align*}
x & = 5n + 1\\
x & = 5n + 2\\
x & = 5n + 3\\
x & = 5n + 4
\end{align*}
I have tried substituting for $x$, but with two different variables $n$ and $t$ and everything, I'm not sure how I would prove they can be equal. Any help would be much appreciated. Thank you!
|
If $5 \not \mid x$, then
*
*$x \equiv 1\mod 5$,
*$x \equiv 2 \mod 5$,
*$x \equiv 3 \mod 5$ or
*$x \equiv 4 \mod 5$.
Now $1^2 \equiv 4^2 \equiv 1\mod 5$ and $2^2 \equiv 3^2 \equiv -1\mod 5$ and therefore $x^2 \equiv 1 \mod 5$ or $x^2 \equiv -1 \mod 5$.
|
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Strengthen inequality It is known that for positive integers $ a, b, c, d $ and $n $, the inequalities $ a + c <n $ and $ \frac ab + \frac cd <1 $ hold. Prove that
$$ \frac ab + \frac cd <1 - \frac {1} {n ^ 3} $$
My work so far:
$ a, b, c, d, n -$ positive integers $\Rightarrow a+c \le n-1$
and $ad+bc \le bd-1.$
|
For convenience, let us to think that $a\le c$ and $t=d-c\in\mathbb{N}$.
1. Not hard to see that if $\frac{a}{b}+\frac{c}{c+t}<1$ then
\begin{equation}
\frac{a}{b}+\frac{c}{c+t}\le\frac{a}{\left\lfloor \frac{ac}{t}\right\rfloor+a+1} + \frac{c}{c+t} < 1.
\end{equation}
2. Now, let us fix only $a\le c$ and let $t\ge1$ is variable. We can show that the sum
$$\frac{a}{\left\lfloor \frac{ac}{t}\right\rfloor+a+1}+\frac{c}{c+t}$$
has the maximal value for $t=1$, or by other words
$$\frac{t}{c+t}-\frac{a}{\left\lfloor \frac{ac}{t}\right\rfloor+a+1}\geq\frac{1}{c+1}-\frac{a}{a(c+1)+1}=\frac{1}{(c+1)(a(c+1)+1)}.$$
2.a Let $t\le c^2$. Note, that $\left\lfloor \frac{ac}{t}\right\rfloor = \frac{ac}{t} - \frac{ac(\mathrm{mod\ }t)}{t}\ge \frac{ac}{t} - \frac{t-1}{t}$, so
$$\frac{t}{c+t}-\frac{a}{\left\lfloor \frac{ac}{t}\right\rfloor+a+1}\ge\frac{t}{c+t}-\frac{at}{a(t+c)+1} = \frac{t}{(c+t)(a(t+c)+1)},$$
and we have to prove that
$$\frac{t}{(c+t)(a(t+c)+1)}\ge\frac{1}{(c+1)(a(c+1)+1)}$$
or
$$(t-1)\left(t-\frac{ac^2+c}{a}\right)\le0$$
or
$$t < c^2+\frac{c}{a}.$$
Case 2.a is done.
2.b Now, let $t \ge c^2+1$. Since $a\le c$, $\left\lfloor\frac{ac}{t}\right\rfloor=0$, therefore we have to prove that
$$\frac{t}{c+t}-\frac{a}{a+1}\geq\frac{1}{c+1}-\frac{a}{a(c+1)+1 }.$$
In view of $t\ge c^2+1$, it suffices to prove
$$\frac{1}{1+a}-\frac{1}{1+c}\ge\frac{1}{1+c+\frac{1}{c}}-\frac{1}{1+c+\frac{1}{a}}.$$
It is easy to see that for the function $f(x)=\frac1x$ the following property is true:
*
*If $y\ge x>0$ then for all $\delta > 0$ : $f(x)-f(x+\delta)\ge f(y)-f(y+\delta)$.
From the last statement and from $\frac{1}{c}+c-a\geq \frac{1}{a}$ we can conclude
$$\frac{1}{1+a}-\frac{1}{1+c}\ge\frac{1}{1+c+\frac{1}{c}}-\frac{1}{1+c+\frac{1}{c}+(c-a)}\ge \frac{1}{1+c+\frac{1}{c}}-\frac{1}{1+c+\frac{1}{a}}$$
Now 2. is done.
3. From 1. and 2. :
$$\frac{a}{b}+\frac{c}{c+t}\le \frac{a}{\left\lfloor\frac{ac}{t}\right\rfloor+a+1}+\frac{c}{c+t}\le \frac{a}{a(c+1)+1}+\frac{c}{c+1}=1-\frac{1}{(c+1)(a(c+1)+1)}\le 1-\frac{1}{(a+c)^3} < 1-\frac{1}{n^3}$$
I hope that more laconic solvation exists:)
|
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What is the maximum of $\ 2(a+b)-ab$ if we have: $\ a^2+b^2=8\ (a,b\ real)$ What is the maximum of $\quad 2(a+b)-ab\ $ if we have: $a^2+b^2=8 \ (a,b\ real) $
My work is as follows:
According to AM-GM inequality:
$$\frac{a^2+b^2}2\ge\sqrt{a^2b^2} \Rightarrow $$
$$ab\le4 $$
On the other hand , from $\ a^2+b^2=8\ ,\ $we have: $\ 2(a+b)-ab=2\sqrt{8+2ab}-ab$
We can get this: $$2\sqrt{8+2ab}\le8$$
Is it possible to somehow combine these two inequalities to find the maximum value of $\ 2(a+b)-ab$ ?
I think in this solution technique we must find the minimum value of $ab\ $ , but how???
|
Using differentiation: Let $f(t)=2\sqrt{8+2t}-t$, then $f'(t)=\frac{2}{\sqrt{8+2t}}-1$. Then $f(t)$ increases when $-4<t<-2$ and decreases when $-2<t<4$. Thus $f(t)$ has a maximum at $t=-2$ and $f(-2)=2\sqrt{8-4}+2=6$. Since $|ab|\le 4$, $2(a+b)-ab$ has a maximum $6$.
Alternative method: Let $a=2\sqrt{2}\cos\theta$ and $b=2\sqrt{2}\sin\theta$, then
\begin{align}
2(a+b)-ab&=-4(\cos\theta+\sin\theta)^2+4\sqrt{2}(\cos\theta+\sin\theta)+4\\
&=-4((\cos\theta+\sin\theta)^2-\sqrt{2}(\cos\theta+\sin\theta)-1)\\
&=-4\left(\left(\cos\theta+\sin\theta-\frac{1}{\sqrt{2}}\right)^2-\frac{3}{2}\right)
\end{align}
Therefore, $2(a+b)-ab$ has maximum $6$ when $\sin\theta+\cos\theta=\frac{1}{\sqrt{2}}$.
|
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|
Using the usual notation for a triangle write $\sin^2A$ in terms of the sides a, b and c. This is an A-level trigonometric problem.
Using the usual notation for a triangle write $\sin^2A$ in terms of the sides $a$, $b$ and $c$.
Answer: $$\frac{(a+b-c)(a-b+c)(a+b+c)(-a+b+c)}{4b^2c^2}$$
The last step of the solution requires going from:
Expression 1: $$-a^4-b^4-c^4+2a^2b^2+2b^2c^2+2a^2c^2$$
to:
Expression 2: $$(a+b-c)(a-b+c)(a+b+c)(-a+b+c)$$
I can see that these two expressions are equal by working in reverse and multiplying out the second expression to get the first but how does one go from Exp.1 to Exp.2?
|
$$\begin{align}&-a^4-b^4-c^4+2a^2b^2+2b^2c^2+2a^2c^2\\&=-a^4+(2b^2+2c^2)a^2-b^4+2b^2c^2-c^4\\&=-a^4+2(b^2+c^2)a^2-(b^2-c^2)^2\\&=\color{red}{-a^4+2(b^2+c^2)a^2-(b^2+c^2)^2}+(b^2+c^2)^2-(b^2-c^2)^2\\&=\color{red}{-(a^2-(b^2+c^2))^2}+4b^2c^2\\&=(2bc)^2-(a^2-b^2-c^2)^2\end{align}$$
Now use $A^2-B^2=(A+B)(A-B)$.
|
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Find seventh term of the evaluation of $(2a+b^2)^8$ I need to find the seventh term of
$(2a+b^2)^8$
But, at this moment, the only way to solve this I can make up is to distribute the brackets..
Is there any other way to do this?
|
See here.
Using the binomial theorem, we can expand $(a+b)^8$ to be
${8 \choose 0}a^8 + {8 \choose 1}a^7b + {8 \choose 2}a^6b^2 + {8 \choose 3}a^5b^3 + {8 \choose 4}a^4b^4 + {8 \choose 5}a^3b^5 + {8 \choose 6}a^2b^6+{8 \choose 7}ab^7 + {8 \choose 8}b^8$
From this we get the seventh term to be ${8 \choose 6}a^2b^6$ or $28a^2b^6$.
Note that in the above expansion, we used a and b. To fit your formula of $(2a + b^2)^8$ we simply replace a and b for the answer of ${8 \choose 6}(2a)^2(b^2)^6$ or $7168a^2b^{12}$.
|
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|
Inverse Fourier Transform I've got a problem where I need to find the IFT of
$$F(\omega) = \frac{1 + i\omega}{6-\omega^2+5i\omega}$$
I've been trying to solve it through partial fractions, but that gives us
$$\frac{1 + i\omega}{6-\omega^2+5i\omega} = \frac{1 + i\omega}{(-\omega + 3i)(\omega-2i)} = \frac{i(\omega - i)}{(-\omega+3i)(\omega-2i)}= \frac{A(\omega-2i) + B(-\omega+3i)}{(-\omega + 3i)(\omega-2i)} $$
and I'm not entirely certain how to solve for A and B here.
Substituting it into the IFT formula give us
$$\frac 1{2\pi} \int_{-\infty}^\infty \frac{1 + i\omega}{6-\omega^2+5i\omega} \cdot e^{i\omega t} \ d\omega$$
or any combination of the above factorizations, but I have no idea how to even begin integrating something like this.
Anything, even a hint at what I should do in such a situation like this would help.
Thanks a lot for taking a look!
|
By using partial fractions we have
\begin{align}
\frac{i\omega +1}{6-\omega^2+5i\omega}&=\frac{i\omega+1}{(i\omega+2)(i\omega+3)}\\
&=\frac{A}{i\omega+3}+\frac{B}{i\omega+2}
\end{align}
Where $A$ and $B$ are constants such that
\begin{align}
A(i\omega+2)+B(i\omega+3)&=i\omega+1\\
(A+B)i\omega+2A+3B&=i\omega+1
\end{align}
Then, solving the linear system of equations
\begin{align}
A+B&=1\\
2A+3B&=1
\end{align}
we get $\color{red}{A=2}$ and $\color{red}{B=-1}$.
Hence
\begin{align}
\frac{i\omega +1}{6-\omega^2+5i\omega}&=\frac{2}{i\omega+3}-\frac{1}{i\omega+2}
\end{align}
From here you can apply the inverse FT:
\begin{align}
\mathscr{F}^{-1}\left\{\frac{i\omega +1}{6-\omega^2+5i\omega}\right\}&=2\mathscr{F}^{-1}\left\{\frac{1}{i\omega+3}\right\}-\mathscr{F}^{-1}\left\{\frac{1}{i\omega+2}\right\}\\[5pt]
&=\color{blue}{\boxed{\left(2e^{-3t}-e^{-2t}\right)u(t)}}
\end{align}
|
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|
Find the value of $\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7}\sin\frac{\pi}{14}\sin\frac{3\pi}{14}\sin\frac{5\pi}{14}$
Find the value of $$\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7}\sin\frac{\pi}{14}\sin\frac{3\pi}{14}\sin\frac{5\pi}{14}$$
$$\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7}\sin\frac{\pi}{14}\sin\frac{3\pi}{14}\sin\frac{5\pi}{14}$$
$$=\sin\frac{2\pi}{14}\sin\frac{4\pi}{14}\sin\frac{6\pi}{14}\sin\frac{\pi}{14}\sin\frac{3\pi}{14}\sin\frac{5\pi}{14}$$
$$=\sin\frac{\pi}{14}\sin\frac{2\pi}{14}\sin\frac{3\pi}{14}\sin\frac{4\pi}{14}\sin\frac{5\pi}{14}\sin\frac{6\pi}{14}$$
How can I compute $$\prod\limits_{r=1}^{n}\sin\frac{r\pi}{c}$$
|
It can be seen as an instance of the well-known identity:
$$ \prod_{k=1}^{n-1}\sin\left(\frac{k\pi}{n}\right)=\frac{2n}{2^n} \tag{1} $$
that for $n=14$ gives:
$$ \left[\prod_{k=1}^{6}\sin\left(\frac{k\pi}{14}\right)\right]^2 = \frac{14}{2^{13}} \tag{2}$$
from which:
$$ \prod_{k=1}^{6}\sin\left(\frac{k\pi}{14}\right) = \color{red}{\frac{\sqrt{7}}{64}}.\tag{3} $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1680159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Extracting the divergent part of an integral I want to evaluate the integral
$$ \int_0^1 \frac{2x(x-2)(1-x)}{(1-x)^2 + ax} \, \mathrm{d}x$$
in the limit of small $a$. For $a = 0$ this integral is divergent due to the $1/(1-x)$ pole. The exact expression for this integral is, according to Mathematica, a horrible combination of logs and inverse hyperbolic functions. However, I don't really care about all of the terms that go to zero as $a$ goes to zero. In, other words, I want to deduce that this integral is
$$1 + \ln a + \mathcal{O}(a)$$
(I believe this is the correct answer) by making some approximation at an early stage that saves me from having to compute the entire integral and then take the small $a$ limit.
|
The integral can be rewritten explicitely as
$$
-2\int_0^1 \frac{x(x^2-3x+2)}{x^2-(2-a)x+1}\mathrm{d}x=-2 \int_0^1 x\left(1-\frac{(a+1)x-1}{x^2-(2-a)x+1}\right)\mathrm{d}x,
$$
hence it is equal to
$$
-1+2\int_0^1 \frac{(a+1)x^2-x}{x^2-(2-a)x+1}\mathrm{d}x=-1+2(a+1)+2\int_0^1 \frac{(-a^2+a+1)x-(a+1)}{x^2-(2-a)x+1}\mathrm{d}x
$$
$$
=1+2a+(-a^2+a+1)\int_0^1 \frac{2x-\frac{2(a+1)}{-a^2+a+1}}{x^2-(2-a)x+1}\mathrm{d}x
$$
$$
=1+2a+(-a^2+a+1)\int_0^1 \frac{2x-(2-a)}{x^2-(2-a)x+1}\mathrm{d}x+(-a^2+a+1)\left(2-a-\frac{2(a+1)}{-a^2+a+1}\right)\int_0^1\frac{1}{x^2-(2-a)x+1}\mathrm{d}x .
$$
We conclude that it is equal to
$$
1+2a+(-a^2+a+1)\ln|a|+(-a^2+a+1)\left(2-a-\frac{2(a+1)}{-a^2+a+1}\right)\int_0^1\frac{1}{x^2-(2-a)x+1}\mathrm{d}x.
$$
You can calculate it explicitely, anyway it already proves your approximation
$$
1+\ln a+\mathcal{O}(a)
$$
holds true when $a \to 0^+$.
|
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|
How to solve the non-homogeneous linear recurrence $ a_{n+1} - a_n = 2n+3 $, given that $a_0=1$? The problem: $ a_{n+1} - a_n = 2n+3 $, given that $a_0=1$
First I solved the associated homogeneous recurrence and got $a_n = A(1)^n = A$, where A is a constant, but I got stuck solving the rest. My final answer was $a_n=2n^2+n+1$ while my textbook has $a_n=n^2+2n+1$
Here is my work:
$ a_{n+1} - a_n = 2n+3 $
$pn=nd_1+d_2$ for all n
plug in and solve for $d_1$:
$(n+1)d_1+d_2-nd_1-d_2=2n+3$
$d_1=2n+3$
then solve for $d_2$:
$a_0=1=a_{n+1}-2n-3=(n+1)d_1+d_2-2n-3$
$d_2=1-n(2n+3)$
therefore
$p_n=nd_1+d_2 = 2n^2+n+4$ and $a_n=A+2n^2+n+4$
then solve for A:
$a_0=1=A +0+0+4$ so $A=-3$
therefore
$a_n=2n^2+n+1$
|
You got in trouble when you set up $p_n$. Since the forcing term is linear, we expect $p_n$ to be quadratic in $n$; $p_n=d_0n^2+d_1n+d_2$. Now we have
$$\begin{align*}
2n+3&=p_{n+1}-p_n\\
&=d_0(n+1)^2+d_1(n+1)+d_2-d_0n^2-d_1n-d_2\\
&=d_0(2n+1)+d_1\\
&=2d_0n+(d_0+d_1)\;.
\end{align*}$$
This has to hold for all $n$, so we must have $2=2d_0$ and $3=d_0+d_1$. Solving this system is easy: $d_0=1$, so $d_1=2$, and $p_n=n^2+2n+d_2$. Now we want $p_0=a_0=1$, so $d_2=1$, and
$$p_n=n^2+2n+1\;.$$
Finally, $a_n=p_n+A=n^2+2n+1+A$, and setting $n=0$ shows that we must have $A=0$.
Alternatively, we could have left $d_2$ temporarily undetermined, writing $p_n=n^2+2n+d_2$, and observed that then
$$a_n=n^2+2n+d_2+A\;;$$
setting $n=0$ yields the conclusion that $d_2+A=1$ and hence $a_n=n^2+2n+1$. That is, it wasn’t actually necessary to solve separately for $d_2$ and $A$, since in the end we care only about their sum anyway.
|
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|
Suppose $n$ divides $3^n + 4^n$. Show that $7$ divides $n$. Suppose $n \geq 2$ and $n$ is a divisor of $3^n + 4^n$. Prove that $7$ is a divisor of $n$.
My work so far:
I had a hypothesis that if $n| 3^n + 4^n$, then $n = 7^k$ for some $k\in\mathbb{N}$. But this is not necessarily so. Take $n = 7⋅379$, where $3^7 + 4^7 = 7^2⋅379$. Then, $3^7+4^7$ divides $3^n + 4^n$, and since $n$ divides $3^7+4^7$, we must have $n|3^n+4^n$.
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Assume $\enspace n\mid 3^n+4^n$, for some $n\in\mathbb{N}$, $n\geq 2$.
Hence, $$3^n+4^n=nm, \quad for \enspace m\in\mathbb{N}.$$
Since $3^n+4^n$ is odd $\enspace (odd\cdot odd=odd, \enspace even\cdot even=even, \enspace odd+even=odd)$, and since $\enspace n \mid 3^n+4^n$, we know that $n$ is also odd. Thus, we can express $3^n+4^n$ as
$$3^n+4^n=(3+4)\bigg(3^{n-1}-3^{n-2}4+3^{n-3}4^2-\ldots +3^24^{n-3}-34^{n-2}+4^{n-1}\bigg)$$
$$3^n+4^n=7k \qquad$$
whereby $\enspace k=\big(3^{n-1}-3^{n-2}4+3^{n-3}4^2-\ldots +3^24^{n-3}-34^{n-2}+4^{n-1}\big).$
Since $\enspace 7 \mid 7k$, it follows that $\enspace 7 \mid 3^n+4^n$.
Thus, so far we have,
$$3^n+4^n=nm=7k$$
and hence,
$$n=7\bigg(\frac{k}{m} \bigg)$$
If $\enspace m \mid k$, then great, we are done!
So, let's assume the contrary that $\enspace m \nmid k$. Since $\enspace m \nmid k$, there exists $q\in\mathbb{N}, q>0$ such that $m=7q$ and $n=\frac{k}{q}$, whereby $q\mid k$. $\enspace 7\nmid n$, since $\frac{n}{7}=\frac{k}{m}$. Hence, $gcd(n,7)=1$. We have already shown that $7\mid 7k$ and $n\mid 7k$. Thus,
$$gcd(n,7)=1, \enspace 7\mid 7k, \enspace n\mid 7k \enspace \Longrightarrow \enspace 7n\mid 7k \enspace \Longrightarrow \enspace n\mid k \enspace \Longrightarrow \enspace \frac{k}{q}\mid k$$
Since $m\nmid k$, $k=mb+r$ for some $b,r\in \mathbb{Z}, \enspace 0\leq r < m$. For some $s\in\mathbb{N}$,
$$\frac{k}{q}\mid k \enspace \Longrightarrow \enspace \frac{mb+r}{q}\mid k \enspace \Longrightarrow \enspace \big(7b+\frac{r}{q}\big)\mid k \enspace \Longrightarrow \enspace k=s\big(7b+\frac{r}{q}\big)$$
$$k=7sb+\frac{rs}{q}=mb+r \enspace \Longrightarrow \enspace b(7s-m)=r\big(1-\frac{s}{q}\big) \enspace \Longrightarrow \enspace \frac{7bq}{r}=\frac{q-s}{s-q} \enspace \Longrightarrow \enspace \frac{7bq}{r}=-1 \enspace \Longrightarrow \enspace 7bq=-r \enspace \Longrightarrow \enspace r=-mb \enspace \Longrightarrow \enspace k=mb-mb=0$$
But we already know that $k>0$, hence a contradiction. Thus, our assumption is false, and $m\mid k$.
Hence, since $n=7\big(\frac{k}{m}\big)$ and $m \mid k$, therefore $7\mid n$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to prove $\small 2\sin{(2m+1)x} -\sin{(2m-1)x}=\sin{x}\left(1+2\cos{(2mx)}+2\sum\limits_{k=1}^{m}\cos{(2kx)}\right)$? By chance I found:
$$
2\sin(2m+1)x - \sin(2m-1)x=\sin(x)\left(1+2\cos(2mx)+2\sum_{k=1}^{m}\cos(2kx)\right)
$$
Any idea how to prove it?
|
it's simple if u know this identity:
$$\sum_{i=0}^n cos(nb) = \frac{sin(\frac{(n+1)b}{2}) \times cos(\frac{nb}{2})}{sin(\frac{b}{2})}$$
I'll proof this identity if u want but first we use this to solve the above problem:
substituting $b = 2x$ and $n=m$ we get:
$$1+2\sum_{k=1}^{m}\cos(2kx) =-1+ 2 \sum_{k=0}^{m}\cos(2kx) = -1+\frac{sin((m+1)x) \times cos(mx)}{sin(x)}$$
then we have:
$$\sin(x)\left(1+2\cos(2mx)+2\sum_{k=1}^{m}\cos(2kx)\right) = -sin(x)+2sin((m+1)x)cos(mx)+2sin(x)cos(2mx)$$
which is the left side and proof is done.
now we change product to summation using $sinAcosB = 1/2(sin(A+B)+sin(A-B))$
then we get:
$$-sin(x) + sin((2m+1)x) +six(x) + sin((1+2m)x)+sin((1-2m)x)=
2sin((2m+1)x)-sin((2m-1)x)$$
and now proving the identity:
$$S = \sum_{i=0}^{n}\cos(a+nb)$$
multiply and divide S by $2sin(\frac{b}{2})$ then use product to sum identity $sinAcosB = \frac{1}{2}(sin(A+B)+sin(A-B))$ and u see this:
$$2sin(\frac{b}{2}) \times S = [sin(\frac{b}{2}+a)+sin(\frac{b}{2}-a)]+[sin(\frac{3b}{2}+a)+sin(-\frac{b}{2}-a)] + [sin(\frac{5b}{2}+a)+sin(-\frac{3b}{2}-a)]+ ... +[sin(\frac{(2n+1)b}{2}+a)+sin(-\frac{(2n-1)b}{2}-a)]$$
after simplifying negative and positive terms we get:
$$2sin(\frac{b}{2}) \times S = sin(\frac{b}{2}+a)+sin(\frac{(2n+1)b}{2}+a)$$
and using sum to product Identity $sinA + sinB = 2sin(\frac{A+B}{2})cos(\frac{A-B}{2})$ yields the result.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Multiple Nested Radicals $\sqrt{9-2\sqrt{23-6\sqrt{10+4\sqrt{3-2\sqrt{2}}}}}$
I have no idea how to unnest radicals, can anyone help?
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Observe that
$$(\sqrt{2}-1)^2=2-2\sqrt{2}+1=3-2\sqrt{2}\qquad\implies\qquad\color{blue}{\sqrt{3-2\sqrt{2}}=\sqrt{2}-1}$$
Then,
$$\color{blue}{\sqrt{10+4\sqrt{3-2\sqrt{2}}}}=\sqrt{10+4(\sqrt{2}-1)}=\sqrt{6+4\sqrt{2}}=\sqrt{4+4\sqrt{2}+2}=\color{blue}{\sqrt{(2+\sqrt{2})^2}}$$
So
\begin{align}
\sqrt{23-6\sqrt{10+4\sqrt{3-2\sqrt{2}}}}&=\sqrt{23-6(2+\sqrt{2})}\\
&=\sqrt{11-6\sqrt{2}}\\
&=\sqrt{9-6\sqrt{2}+2}\\
&=\sqrt{(3-\sqrt{2})^2}\\
&=3-\sqrt{2}
\end{align}
Finally
\begin{align}
\sqrt{9-2\sqrt{23-6\sqrt{10+4\sqrt{3-2\sqrt{2}}}}}&=\sqrt{9-2(3-\sqrt{2})}\\
&=\sqrt{3+2\sqrt{2}}\\
&=\sqrt{2}+1
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to prove that $\binom{n}{1}\binom{n}{2}^2\binom{n}{3}^3\cdots \binom{n}{n}^n \leq \left(\frac{2^n}{n+1}\right)^{\binom{n+1}{2}}$?
How can we prove that $$\binom{n}{1}\binom{n}{2}^2\binom{n}{3}^3\cdots \binom{n}{n}^n \leq \left(\frac{2^n}{n+1}\right)^{\binom{n+1}{2}}$$
$\bf{My\; Try::}$ Using $\bf{A.M\geq G.M\;,}$ we get
$$\binom{n}{0}+\binom{n}{1}+\binom{n}{2} + \cdots+\binom{n}{n}\geq (n+1)\cdot \left[\binom{n}{0}\cdot \binom{n}{1}\cdot \binom{n}{2} \cdots \binom{n}{n}\right]^{\frac{1}{n+1}}$$
So $$2^n\geq (n+1)\left[\binom{n}{1}\cdot \binom{n}{2}\cdots \binom{n}{n}\right]^{\frac{1}{n+1}}$$
How can I solve it after that? Help me.
Thanks.
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Hint: You are close to the answer. Here is the right way: Apply the AM-GM:
$(a_1+2a_2+\cdots + na_n)^{\binom{n+1}{2}} \geq \binom{n+1}{2}^{\binom{n+1}{2}}a_1a_2^2\cdots a_n^n$, with $a_k = \binom{n}{k}$, and the left side is a popular sum that can be calculated by several methods one of which is using derivative of $(1+x)^n$
|
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|
Finding the order of $2 \times 2 $ matrix. Finding the order of $2 \times 2 $ matrix.
$$\begin{pmatrix}1&-1\\1&0\end{pmatrix}$$
Is there an easier way to find the order of this matrix?
I have been multiplying the matrix for five times or so, and I still haven't found the order yet.
|
Not that long:
$$A^2=\begin{pmatrix}1&\!-1\\1&0\end{pmatrix}\begin{pmatrix}1&\!-1\\1&0\end{pmatrix}=\begin{pmatrix}0&\!-1\\1&\!-1\end{pmatrix}$$
$$A^4=\begin{pmatrix}0&\!-1\\1&\!-1\end{pmatrix}\begin{pmatrix}0&\!-1\\1&\!-1\end{pmatrix}=\begin{pmatrix}-1&1\\\!-1&0\end{pmatrix}$$
$$A^6=\begin{pmatrix}0&\!-1\\1&\!-1\end{pmatrix}\begin{pmatrix}-1&1\\\!-1&0\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}$$
|
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|
Evaluating $\lim_{x \to 3} \frac{x - 3}{\sqrt {x - 2} - \sqrt {4 - x}}$ The question is to evaluate the given limit :
$$\lim_{x \to 3} \frac{x - 3}{\sqrt {x - 2} - \sqrt {4 - x}}$$
When I'm trying to evaluate this I'm getting $1$ as my answer but the answer given in the text is $0$. I'm wondering how can this limit be equal to $0$.
I'm using $x = 3 + h$ (assuming $h$ is approaching zero) and now converting this limit in the form of $h$. Lim (h tends to zero) and substituting $x = 3 + h$.
In this way numerator equals $h$ and now after rationalizing the denominator I will get $2h$ and the numerator will also become $2h$ both of them will cancel each other and the limit will be equal to $1$.
I don't know where I'm going wrong.
Please edit if you can for proper reading because I don't know how to edit this in mathematical format.
|
We also have
\begin{align*}
\lim_{x \to 3} \frac{x - 3}{\sqrt {x - 2} - \sqrt {4 - x}}&=\lim_{x \to 3} \frac{x - 3}{\sqrt {x - 2} - \sqrt {4 - x}}\cdot\frac{\sqrt{x-2}+\sqrt{4-x}}{\sqrt{x-2}+\sqrt{4-x}}\\
&=\lim_{x \to 3}\frac{(x-3)(\sqrt{x-2}+\sqrt{4-x})}{(x-2)-(4-x)}\\
&=\lim_{x \to 3}\frac{(x-3)(\sqrt{x-2}+\sqrt{4-x})}{2x-6}\\
&=\lim_{x \to 3}\frac{(x-3)(\sqrt{x-2}+\sqrt{4-x})}{2(x-3)}\\
&=\lim_{x \to 3}\frac{\sqrt{x-2}+\sqrt{4-x}}{2}\\
&=\frac{\sqrt{3-2}+\sqrt{4-3}}{2}\\
&=\frac{\sqrt{1}+\sqrt{1}}{2}\\
&=\frac{2}{2}\\
&=1.
\end{align*}
|
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|
Induction Proof $k^2 \times 2^k$ I need help on this proof. I am not able to do it after setting m=m+1.
Prove by induction on n that sum of $k^2 \times 2^k$ from $k=1$ to $n$ is equal to $(n^2-2n+3) \times 2^{n+1}-6$
Base case:
Let $k=1$ so L.H.S side is $2$
Let $n=1$ so R.H.S side is $2$
Inductive hypothesis:
Let $n=m$ so $(m^2-2m+3) \times 2^{m+1}-6$
Proof:
Let $n=m+1$ so prove that $((m+1)^2-2(m+1)+3) \times 2^{m+2}-6=(m^2-2m+3) \times 2^{m+1}-6$
But I am not able to prove that they are equal.
|
For the inductive step the computations would be as follows:
$$
\begin{array}{rcl}
\sum_{k = 1}^{m+1} k^{2} 2^{k} & = & (m+1)^{2}2^{m+1} + \sum_{k = 1}^{m} k^{2}2^{k}\\
& = & (m+1)^{2}2^{m+1} + (m^2 - 2m + 3)2^{m+1} - 6\\
& = & 2^{m+1} \left((m+1)^2+m^2 - 2m +3\right) - 6\\
& = & 2^{m+1} (2m^2 + 4) - 6\\
& = & 2^{m+2} (m^2 + 2) - 6 \\
& = & 2^{m+2} (m^2 + 2m + 1 -2m - 2 + 3) - 6\\
& = & \left( (m+1)^{2} - 2(m+1) + 3 \right) - 6.
\end{array}
$$
|
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|
Find $\lim_{(x,y) \to (0,0)} \frac{\sin(x^3y^2)}{(x^2+y^2)^2}$
Find $$\lim_{(x,y) \to (0,0)} \frac{\sin(x^3y^2)}{(x^2+y^2)^2}$$
I tried multiplying $x^3y^3$ the nominator and denominator, didn't work.
I tried the polar way too, $x = r \cos\phi$, $y=r \sin\phi$,$x^2 + y^2 = r^2$ too. which gives me $\frac{"0"}{"0"}$
I tried proving that this limit doesn't exist but wolfram shows that its limit is zero.
Can anyone give me a hint on how to approach this?
|
Switching to polar is a good idea and gives
$$\lim_{r \to 0} \frac{\sin\left(r^5 \cos^3t\sin^2t \right)}{r^4}$$
You probably know that $\sin(a)/a$ tends to 1 for $a \to 0$; so:
$$\begin{array}{cc}
\displaystyle \lim_{r \to 0} \frac{\sin\left(r^5 \cos^3t\sin^2t \right)}{r^4} & \displaystyle
= \lim_{r \to 0} \left( \frac{\sin\left(r^5 \cos^3t\sin^2t \right)}{r^5 \cos^3t\sin^2t}(r \cos^3t\sin^2t) \right) \\ \\
& \displaystyle
= \underbrace{\lim_{r \to 0} \frac{\sin\left(r^5 \cos^3t\sin^2t \right)}{r^5 \cos^3t\sin^2t}}_{\to 1} \underbrace{\lim_{r \to 0} \left( r \cos^3t\sin^2t \right)}_{\to \ldots}
\end{array}$$
Can you take it from here?
Alternatively:
$$\lim_{(x,y)\to(0,0)} \frac{\sin(x^3y^2)}{(x^2+y^2)^2}
=\lim_{(x,y)\to(0,0)} \frac{\sin(x^3y^2)}{x^3y^2}\frac{x^3y^2}{(x^2+y^2)^2}
=\lim_{(x,y)\to(0,0)} \frac{x^3y^2}{(x^2+y^2)^2} = \cdots$$
And use $x^2+y^2 \ge 2xy$.
|
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|
Derive a polynomials formula by setting $n = 1$, $n = 2$, $n = 3$ to determine the coefficients. Derive a formula for the sum of squares $1^2 + 2^2 + 3^2 + … + n^2$.
Hint: assume the formula is a polynomial of degree 3, i.e. $an^3 + bn^2 + cn + d$, and use the cases of $n=0, n=1, n=2$, and $n=3$ to determine its coefficients.
|
.I'll answer your question, but I sincerely encourage you to have an attempt at your question before you ask a question in this website.
So we assume that the sum of squares up till $n$ is given by a cubic polynomial $an^3+bn^2+cn+d$. What we do is put n=0,1,2,3 on both sides of the equation:
$$
\sum_{i=1}^n i^2 = an^3+bn^2+cn+d
$$
We get four equations, one each for $0,1,2,$ and $3$, as follows:
$$
n=0 \implies d=0
$$
$$
n = 1 \implies a+b+c+d=1^2=1
$$
$$
n=2 \implies 8a+4b+2c+d=1^2+2^2=5
$$
$$
n=3 \implies 27a+9b+3c+d=1^2+2^2+3^2=14
$$
So we have to solve the system:
$$
a+b+c=1 ; 8a+4b+2c=5;27a+9b+3c=14.
$$
Eliminte $c $ from the second and third equations using the first equation directly:
$$
6a+2b=3;24a+6b=11
$$
Solving, $a=\frac{1}{3}$,$b=\frac{1}{2}$,and then by substituting back,$c=\frac{1}{6}$. So our final formula is $\frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6} = \sum_{i=1}^{n} i^2$
|
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"timestamp": "2023-03-29T00:00:00",
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|
differential equation of family of circles passing through origin How do I find the DE of all circles passing through origin? I tried something like this
The family of circles passing through the origin is given by
$$
(x- r \cos \theta )^2 + (y- r\sin \theta)^2 = r^2
$$Differentiating once, we get
$$
2 ( (x- r \cos \theta) + (y - r \sin \theta) y') = 0
$$
Differentiating again, we get
$$
1 + y'^2 + (y - r \sin \theta) y'' = 0
$$
How to get rid of $r$ and $\theta$ algebraically? Is there any other approach?
|
Let the the circle be of radius $r$. Then,
$$\tag1(x-r\cos\theta)^2+(y-r\sin\theta)^2=r^2$$
$$\tag22(x-r\cos\theta)+2(y-r\sin\theta)y'=0$$
$$2x-2r\cos\theta+2yy'-2r\sin\theta y'=0$$
$$1+yy''+(y')^2-r\sin\theta y''=0$$
$$\tag3r\sin\theta=\frac{1+yy''+(y')^2}{y''}$$
Substitute $(3)$ in $(1)$,
$$(x-r\cos\theta)^2+(y-r\sin\theta)^2=r^2$$
$$x^2-2rx\cos\theta+y^2-2ry\sin\theta+r^2\sin^2\theta=r^2\sin^2\theta$$
$$x^2-2rx\cos\theta+y^2-2ry\sin\theta=0$$
$$\tag4r\cos\theta=\frac{x^2+y^2-2ry\sin\theta}{2x}$$
Now substitute $(3)$ in $(4)$ and then $(3)$ and $(4)$ in $(2)$.
It is likely that there may be an easier way than this.
|
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|
What is a mathematical expression for the sequence $\{1,1,-1,-1,1,1,-1,-1,\dots\}$? What is a mathematical expression for the sequence $\{1,1,-1,-1,1,1,-1,-1,\dots\}$, that is $1$ and $-1$, two at a time alternating?
|
The sequence $a_0=1$, $a_1=1$, $a_2=-1$, $a_3=-1$, $a_4=1$ and so on satisfies the recursion
$$
a_0=1,\quad a_1=1,\qquad a_{n+2}=-a_n
$$
so its characteristic equation is $t^2+1=0$. Thus the general solution is of the form
$$
xi^n+y(-i)^n
$$
The initial conditions tell that $x+y=1$ and $xi-yi=1$, thus
$$
\begin{cases}
x+y=1\\[4px]
x-y=-i
\end{cases}
$$
that gives
$$
x=\frac{1-i}{2},\quad y=\frac{1+i}{2}
$$
Since $i=\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}$, we can write
\begin{align}
a_n
&=\frac{1}{2}\bigl(
(1-i)(\cos\tfrac{n\pi}{2}+i\sin\tfrac{n\pi}{2})+
(1+i)(\cos\tfrac{n\pi}{2}-i\sin\tfrac{n\pi}{2})
\bigr)\\[6px]
&=\cos\frac{n\pi}{2}+\sin\frac{n\pi}{2}
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Let $A$ be the $n\times n$ matrix with a $1$ in every entry. What are the eigenvalues of $A$? Is $A$ diagonalizable? Let $A$ be the $n\times n$ matrix with a $1$ in every entry. What are the eigenvalues of $A$ and a basis for each eigenspace? Is $A$ diagonalizable?
Having some trouble with this one. I tried using the fact that $\lambda$ is an eigenvalue of $A$ iff there exist non-zero solutions to $Ax=\lambda x$. Well, clearly the $1\times n$ non-zero vector $x$ with all entries equal is an eigenvector of $A$, with corresponding eigenvalue $\lambda=n$, since $Ax=nx$. But I can't seem to get much further than that in terms of finding eigenvalues for $A$, any hints/suggestions?
|
Here is a matrix $P$ that I made up some time ago. Note that $P$ is not orthogonal, although the columns are pairwise orthogonal and therefore independent. The columns are eigenvectors for your matrix, in the 10 by 10 case. Multiply your matrix of all ones (10 by 10) on the left, this matrix on the right. What happens?
$$
\left( \begin{array}{rrrrrrrrrr}
1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\
1 & 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\
1 & 0 & 2 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\
1 & 0 & 0 & 3 & -1 & -1 & -1 & -1 & -1 & -1 \\
1 & 0 & 0 & 0 & 4 & -1 & -1 & -1 & -1 & -1 \\
1 & 0 & 0 & 0 & 0 & 5 & -1 & -1 & -1 & -1 \\
1 & 0 & 0 & 0 & 0 & 0 & 6 & -1 & -1 & -1 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 7 & -1 & -1 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 8 & -1 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 9
\end{array}
\right).
$$
|
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|
How to calculate $\lim_{(x,y) \rightarrow (1,-1)} \frac{x^2 - y^2}{x^3 + y^3}$ $$\lim_{(x,y) \rightarrow (1,-1)} \frac{x^2 - y^2}{x^3 + y^3}$$
By approaching along x = 1 or y = -1 and using L'Hopitals rule, I can establish that if the limit does exist, then it is $\frac{2}{3}$.
This matches the result given by Wolfram Alpha and is consistent with a 3D graph of the function.
However I am struggling to prove that this limit exists along all approach paths.
I have tried $x = -y$ and the Sandwich Theorem but still seem no closer to a general proof.
Can you give me a hint for to the correct method for proving that this limit exists?
|
\begin{align}
\lim_{(x,y)\to (1,-1)}\frac{x^2-y^2}{x^3+y^3}&=\lim_{(x,y)\to(1,-1)}\frac{(x+y)(x-y)}{(x+y)(x^2-xy+y^2)}\\
&=\lim_{(x,y)\to(1,-1)}\frac{x-y}{x^2-xy+y^2}\\
&=\frac{1-(-1)}{1-1\cdot(-1)+1}\\
&=\frac{2}{3}
\end{align}
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Find a formula for $f^{(n)}(x)$ where $f(x) = \dfrac{1}{x^2-a^2}$
Let $a$ be a constant. Find a formula for $f^{(n)}(x)$ where $f(x) = \dfrac{1}{x^2-a^2}$.
After computing a few derivatives, the derivatives seem to have factorials in them sometimes and other times not. For example, $\dfrac{d^5}{dx^5} \left (\dfrac{1}{x^2-a^2} \right) = -\dfrac{240(3a^4x+10a^2x^3+3x^5)}{(x^2-a^2)^6}$ while $\dfrac{d^6}{dx^6} \left (\dfrac{1}{x^2-a^2} \right) = -\dfrac{720(a^6+21a^4x^2+35a^2x^4+7x^6)}{(x^2-a^2)^6}$, so I am not immediately seeing the pattern.
|
$$
{1 \over {x^{\,2} - a^{\,2} }} = {1 \over {\left( {x + a} \right)\left( {x - a} \right)}} = {1 \over {2\,a}}\left( {{1 \over {\left( {x - a} \right)}} - {1 \over {\left( {x + a} \right)}}} \right)
$$
... then it goes easily ..
|
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|
Solve $ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $
Question: Solve $$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $$
$$ 0 \le x \le 360^{\circ} $$
My attempt:
$$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $$
$$ 6(\frac{1}{2} - \frac{\cos(2x)}{2}) + \sin(x)\cos(x) -(\frac{1}{2} + \frac{\cos(2x)}{2}) = 5 $$
$$ 3 - 3\cos(2x)+ \sin(x)\cos(x) - \frac{1}{2} - \frac{\cos(2x)}{2} = 5$$
$$ \frac{7\cos(2x)}{2} - \sin(x)\cos(x) + \frac{5}{2} = 0 $$
$$ 7\cos(2x) - 2\sin(x)\cos(x) + 5 = 0 $$
$$ 7\cos(2x) - \sin(2x) + 5 = 0 $$
So at this point I am stuck what to do, I have attempted a Weierstrass sub of $\tan(\frac{x}{2}) = y$ and $\cos(x) = \frac{1-y^2}{1+y^2}$ and $\sin(x)=\frac{2y}{1+y^2} $ but I got a quartic and I was not able to solve it.
|
$7cos(2x)-sin(2x)+5=0$
$7cos(2x)/\sqrt{50}-sin(2x)/\sqrt{50}=-5/\sqrt{50}$
Let $\alpha=sin^{-1}(7/\sqrt{50})$
Then
$sin(\alpha)cos(2x)+cos(\alpha)sin(2x)=-5/\sqrt{50}$
$sin(2x+\alpha)=-5/\sqrt{50}$
$2x+\alpha=sin^{-1}(-1/\sqrt{2})$
So
$x=(sin^{-1}(-1/\sqrt{2})-\alpha)/2$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Geometric series $ar^n$ where $n \ne 1,2,3,4 \cdots$ The probability of rolling a seven on two dice is as follows
$$p=1/6+1/6(5/6)^2+1/6(5/6)^4 + \cdots$$
what is the probability of rolling a $7$?
Is there an advantage to rolling first?
My attempt at a solution; $n=1,2,3,4,5 \cdots$
$$\sum_{n=1}^\infty \frac 1 6 \left ( \frac 5 6 \right)^{2n2} = \frac 1 6 + \frac 1 6 \left( \frac 5 6 \right)^2 + \frac 1 6 \left( \frac5 6 \right )^4 + \cdots$$
this is an example of a geometric series where $a=(1/6)$, $r=(5/6)$
we can say
$$1/6\sum_{n=1}^{\infty}(5/6)^{2n-2}=1/6+1/6(5/6)^2+1/6(5/6)^4...$$
let $$n_1=2n-2$$
therefore
$$S_n=1/6\sum_{n_1=0}^{\infty}(5/6)^{n_1}$$
$$6S_n=\sum_{n_1=0}^{\infty}(5/6)^{n_1}=\frac{5/6}{1-5/6}$$
$$S_n=(1/6)\frac{5/6}{1-(5/6)}=\frac{5}{1-(5/6)}=\frac{5}{1/6}=30$$
The probability of rolling a seven is 30 percent.
I think rolling first does give an advantage.
Am I understanding the rules of series properly??
Thanks
|
You are complicating things, note that :
$$p=\frac{1}{6} [1+(\frac{5}{6})^2+(\frac{5}{6})^4+\cdots ]$$
This is a geometric progression, where $a=1, r=\frac{25}{36}$ The infinite sum of which is:
$$\frac{a}{6(1-r)}=\frac{1}{6}\cdot\frac{36}{36-25}=\color {blue}{\frac{6}{11}}$$
|
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|
Prove $ \frac{\sin^3(x)-\cos^3(x)}{\sin(x)+\cos(x)} = \frac{\csc^2(x) -\cot(x) -2\cos^2(x)}{1-\cot^2(x)} $
Question: Prove $$ \frac{\sin^3(x)-\cos^3(x)}{\sin(x)+\cos(x)} = \frac{\csc^2(x) -\cot(x) -2\cos^2(x)}{1-\cot^2(x)} $$
RHS: $$ \frac{\csc^2(x) -\cot(x) -2\cos^2(x)}{1-\cot^2(x)} $$
$$ ⇔ \frac{\frac{1}{\sin^2(x)} +\frac{\cos(x)}{\sin(x)}-2\cos^2(x)}{1-\frac{\cos^2(x)}{\sin^2(x)}} $$
$$ ⇔ \frac{\frac{1}{\sin^2(x)} +\frac{\cos(x)\sin(x)}{\sin^2(x)}-\frac{2\cos^2(x)\sin^2(x)}{\sin^2(x)}}{\frac{\sin^2(x)-\cos^2(x)}{\sin^2(x)}} $$
$$ ⇔ \frac{\frac{1+\cos(x)\sin(x)-2\cos^2(x)\sin^2(x)}{\sin^2(x)}}{\frac{\sin^2(x)-\cos^2(x)}{\sin^2(x)}} $$
$$ ⇔ \frac{1+\cos(x)\sin(x)-2\cos^2(x)\sin^2(x)}{\sin^2(x)} \times {\frac{\sin^2(x)}{\sin^2(x)-\cos^2(x)}} $$
$$ ⇔ \frac{1+\cos(x)\sin(x)-2\cos^2(x)\sin^2(x)}{\sin^2(x)-\cos^2(x)} $$
$$ ⇔ \frac{1+\cos(x)\sin(x)-2\cos^2(x)\sin^2(x)}{(\sin(x)-\cos(x))(\sin(x)+\cos(x))} $$
Now I am stuck
|
HINT:
Writing $\cos x\sin x=u$
$$1-u-2u^2=(1-2u)(1+u)$$
$1-2u=(\sin x-\cos x)^2$
|
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|
Let $a$ and $b$ be the coefficient of $x^3$ in $(1+x+2x^2+3x^3)^4$ and $(1+x+2x^2+3x^3+4x^4)^4$ respectively. Let $a$ and $b$ be the coefficient of $x^3$ in $(1+x+2x^2+3x^3)^4$ and $(1+x+2x^2+3x^3+4x^4)^4$ respectively.Find $(a-b).$
I tried to factorize $(1+x+2x^2+3x^3)$ and $(1+x+2x^2+3x^3+4x^4)$ into product of two binomials,but i could not.And i do not have any other method to solve it.
|
Here is a slightly different variation of the theme. If we introduce the polynomial
\begin{align*}
p(x)=1+x+2x^2+3x^3
\end{align*}
and write $[x^n]$ to denote the coefficient of $x^n$ of a polynomial,
we observe
\begin{align*}
&[x^3]\left(p(x)\right)^4-[x^3]\left(p(x)+4x^4\right)^4\\
&\quad=[x^3]\left(p(x)\right)^4-[x^3]\left(p(x)\right)^4\tag{1}\\
&\quad=0
\end{align*}
In (1) we use the fact that multiplication of $4x^4$ with $p(x)$ gives always terms with power of $x$ greater or equal to $4$ and so there is no contribution to $[x^3]$.
|
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|
Relation between two sequences or summations Let us define two sequences
\begin{equation}
G_{n}=\sum_{i=1}^n a^{-i}t_{i-1}
\end{equation}
and
\begin{equation}
g_{n}=\sum_{i=1}^n t_{i-1}
\end{equation}
where $a$ is an integer and $t_n$ is an another sequence.
Can we find a relation between $G_n$ and $g_n$ ?
|
Here are relationships of $G_n$ and $g_n$ in terms of the delta operator $\Delta$ and of generating functions which could be helpful.
We define the generating functions
\begin{align*}
G(z)=\sum_{n\geq 1}G_{n}z^n\qquad\text{and}\qquad
g(z)=\sum_{n\geq 1}g_{n}z^n
\end{align*}
and show
The following is valid
\begin{align*}
\Delta G_n&=\frac{1}{a^{n+1}}\Delta g_n\qquad\qquad n\geq 1\tag{1}\\
&\\
g(z)&=\frac{1-az}{1-z}G(az)\tag{2}
\end{align*}
$$ $$
Ad (1) We obtain
\begin{align*}
\Delta G_n&=G_{n+1}-G_n\\
&=\sum_{i=1}^{n+1}a^{-i}t_{i-1}-\sum_{i=1}^{n}a^{-i}t_{i-1}\\
&=a^{-(n+1)}t_{n}\\
&=\frac{1}{a^{n+1}}\left(\sum_{i=1}^{n+1}t_{i-1}-\sum_{i=1}^{n}t_{i-1}\right)\\
&=\frac{1}{a^{n+1}}\left(g_{n+1}-g_n\right)\\
&=\frac{1}{a^{n+1}}\Delta g_n\\
&\qquad\qquad\qquad\qquad\qquad\qquad\Box
\end{align*}
Ad (2) We obtain
\begin{align*}
g(z)&=\sum_{n=1}^\infty g_nz^n=\sum_{n=1}^\infty\left(\sum_{i=1}^{n}t_{i-1}\right)z^n\\
&=\sum_{n=1}^\infty\left(\sum_{i=0}^{n-1}t_{i}\right)z^n=\sum_{n=0}^\infty\left(\sum_{i=0}^{n}t_{i}\right)z^{n+1}\tag{3}\\
&=\frac{z}{1-z}\sum_{n=0}^\infty t_nz^{n}=\frac{z}{1-z}\sum_{n=0}^\infty \frac{t_n}{a^n}\left(az\right)^n\tag{4}\\
&=\frac{z}{1-z}(1-az)\sum_{n=0}^\infty\left(\sum_{i=0}^n\frac{t_i}{a^i}\right)(az)^n\tag{5}\\
&=\frac{z}{az}\frac{1-az}{1-z}\sum_{n=0}^\infty\left(\sum_{i=0}^n\frac{t_i}{a^i}\right)(az)^{n+1}\\
&=\frac{z}{az}\frac{1-az}{1-z}\sum_{n=1}^\infty\left(\sum_{i=0}^{n-1}\frac{t_i}{a^i}\right)(az)^{n}\\
&=\frac{1-az}{1-z}\sum_{n=1}^\infty\left(\sum_{i=1}^{n}\frac{t_{i-1}}{a^i}\right)(az)^{n}\\
&=\frac{1-az}{1-z}G(az)\\
&\qquad\qquad\qquad\qquad\qquad\qquad\Box
\end{align*}
Comment:
*
*In (3) we shift the index $i$ by $1$ and then $n$ by $1$.
*In (4) we use
\begin{align*}
\frac{1}{1-z}\sum_{n=0}^{\infty}a_nz^n=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n}a_k\right)z^n
\end{align*}
*In (5) we do it similarly as we did it in (4) but the other way round.
|
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|
Problem with showing convergence or divergence of $\sum_{n=2}^\infty\frac{(-1)^{n+1}}{3n+4}$ Determine if the following series is absolutely convergent, conditionally convergent or divergent.
$$\sum_{n=2}^\infty\frac{(-1)^{n+1}}{3n+4}$$
So I showed that$\sum_{n=2}^\infty\bigl|\frac{(-1)^{n+1}}{3n+4}\bigr|$ is divergent using limit comparison test.
Now for $\sum_{n=2}^\infty\frac{(-1)^{n+1}}{3n+4}$, I considered 2 ways:
*
*Alternating Series Test
$$a_n:=\frac{1}{3n+4}$$
$$3n+4>0\implies \frac{1}{3n+4}>0\implies a_n\ge0$$
$$\lim_{n\to \infty} a_n=\lim_{n\to \infty}\frac{1}{3n+4}=\lim_{n\to \infty}\frac{\frac{1}{n}}{3+\frac{4}{n}}=\frac{0}{3+0}=0$$
$$3n+4\le 3n+7\implies a_n=\frac{1}{3n+4}\ge \frac{1}{3n+7}=a_{n+1}$$
Therefore $\sum_{n=2}^\infty\frac{(-1)^{n+1}}{3n+4}$ is convergent.
*Comparison Test (Note: $3n+4\ge0$)
$$-1\le(-1)^{n+1}\le 1 \implies \frac{-1}{3n+4} \le \frac{(-1)^{n+1}}{3n+4} \le \frac{1}{3n+4}$$
But I have already showed that $\sum_{n=2}^\infty\frac{1}{3n+4}$ diverges.
and consequently, $-\sum_{n=2}^\infty\frac{1}{3n+4}$ diverges as well. So that would imply that $\sum_{n=2}^\infty\frac{(-1)^{n+1}}{3n+4}$ is divergent!
Did i make a computational error somewhere? I can't seem to find it.
|
The alternating test is OK.
Now, there is no contradition with your second point, because
$$
\frac{-1}{3n+4} \le \frac{(-1)^{n+1}}{3n+4} \le \frac{1}{3n+4}
$$ yields
$$
-\infty \le \sum_{n\geq1}\frac{(-1)^{n+1}}{3n+4} \le +\infty
$$ which does not exclude convergence of the initial series.
|
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|
The number of possible options to get colored balls I have a bag with 5 white balls and 3 blue balls. We getting out 3 balls (without returning).
How many possible ways is to get at least 1 white ball?
My question is why this is not the answer:
$${{5}\choose{1}}*{7 \choose 2}$$
|
The number of ways of selecting $k$ of the five white balls and $3 - k$ of the three blue balls is
$$\binom{5}{k}\binom{3}{3 - k}$$
Thus, the number of ways of selecting at least one white ball is
$$\binom{5}{1}\binom{3}{2} + \binom{5}{2}\binom{3}{1} + \binom{5}{3}\binom{3}{0}$$
Alternatively, we could find the same total by subtracting the number of ways of selecting zero white balls from the total number of ways of selecting three of the eight available balls. That yields
$$\binom{8}{3} - \binom{5}{0}\binom{3}{3}$$
Why did your calculation lead to an incorrect result?
You selected one of the five white balls and two of the seven other balls. However, this counts those selections in which more than one white ball is selected more than once.
Suppose you selected two particular white balls. You counted each selection twice, once for each way you could have selected one of those two white balls as the white ball you selected from the five white balls while choosing the other white ball from the remaining seven balls.
You counted each selection of three particular white balls three times, once for each of the three ways you could have selected one of those three white balls as the white ball you selected from the five white balls while selecting the other two selected white balls from the remaining seven balls.
Notice that
$$\binom{5}{1}\binom{3}{2} + 2\binom{5}{2}\binom{3}{1} + 3\binom{5}{3}\binom{3}{0} = \binom{5}{1}\binom{7}{2}$$
|
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|
determinant of 3x3 matrix algebra not matching CAS Find
$$\begin{vmatrix}
i & 2 & -1
\\ 3 & 1+i & 2
\\ -2i & 1 & 4-i \end{vmatrix}$$
we will expand along first row
$$ \begin{aligned}
\begin{vmatrix}
i & 2 & -1
\\ 3 & 1+i & 2
\\ -2i & 1 & 4-i \end{vmatrix}
&=i \begin{vmatrix} 1+i & 2 \\ 1 & 4-i \end{vmatrix}
-2 \begin{vmatrix} 3 & 2 \\ -2i& 4-i \end{vmatrix}
-1 \begin{vmatrix} 3 & 1+i \\ -2i & 1 \end{vmatrix}
\\&=i[(1+i)(4-i)-2]
-2[3(4-i)-2(-2i)]
-1[3-(1+i)(-2i)]
\\&=i[1(4-i)+i(4-i)-2]
-2[12-3i+4i]
-1[3+(2i)(1+i)]
\\ &= i[4-i+4i-i^2-2]
-2[12+i]
-1[3+2i+2i^2]
\\ &=i[4-3i+1-2]-2[12+i]-1[3+2i-2]
\\ &= i[3-3i]-2[12+i]-1[1+2i]
\\ &=i(3-3i)-2(12+i)-1(1+2i)
\\ &=3i-3i^2-24-2i-1-2i
\\ &=3i+3-24-2i-1-2i
\\ &=3-24-1+3i-2i-2i
\\& =22-i
\end{aligned}
$$
Did something wrong. Some silly algebra mistake, did not use the cofactor expansion formula right?? Cannot catch and is time to ask for help.
This is what maxima spits out it say the det is -28-i
|
$$ \begin{aligned}
\begin{vmatrix}
i & 2 & -1
\\ 3 & 1+i & 2
\\ -2i & 1 & 4-i \end{vmatrix}
&=i \begin{vmatrix} 1+i & 2 \\ 1 & 4-i \end{vmatrix}
-2 \begin{vmatrix} 3 & 2 \\ -2i& 4-i \end{vmatrix}
-1 \begin{vmatrix} 3 & 1+i \\ -2i & 1 \end{vmatrix}
\\&=i[(1+i)(4-i)-2]
-2[3(4-i)-2(-2i)]
-1[3-(1+i)(-2i)]
\\&=i[1(4-i)+i(4-i)-2]
-2[12-3i+4i]
-1[3+(2i)(1+i)]
\\ &= i[4-i+4i-i^2-2]
-2[12+i]
-1[3+2i+2i^2]
\\ &=i[4 \color{red}{+3i}+1-2]-2[12+i]-1[3+2i-2]
\\ &= i[3\color{red}{+3i}]-2[12+i]-1[1+2i]
\\ &=i(3\color{red}{+3i})-2(12+i)-1(1+2i)
\\ &=3i\color{red}{+3i^2}-24-2i-1-2i
\\ &=3i\color{red}{-3}-24-2i-1-2i
\\ &=\color{red}{-3}-24-1+3i-2i-2i
\\& =\color{red}{-28}-i
\end{aligned}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1707810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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|
Simplify $2(\sin^6x + \cos^6x) - 3(\sin^4 x + \cos^4 x) + 1$ Here is the expression:
$$2(\sin^6x + \cos^6x) - 3(\sin^4 x + \cos^4 x) + 1$$
The exercise is to evaluate it.
In my text book the answer is $0$
I tried to factor the expression, but it got me nowhere.
|
Consider $a^3+b^3=(a+b)^3-3ab(a+b)$ and $a^2+b^2=(a+b)^2-2ab$. For $a=\sin^2x$ and $b=\cos^2x$ we have
$$
2(\sin^6x+\cos^6x)-3(\sin^4x+\cos^4x)+1=
2(a+b)^3-6ab(a+b)-3(a+b)^2+6ab+1
$$
However, $a+b=\sin^2x+\cos^2x=1$, so the expression simplifies to
$$
2-6ab-3+6ab+1=0
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1710182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
}
|
Find the ratio of $\frac{\int_{0}^{1} \left(1-x^{50}\right)^{100} dx}{\int_{0}^{1} \left(1-x^{50}\right)^{101} dx}$ $$I_1=\int_{0}^{1} \left(1-x^{50}\right)^{100} dx$$ and $$I_2=\int_{0}^{1} \left(1-x^{50}\right)^{101} dx$$ Then find $\frac{I_1}{I_2}$
I tried by subtracting $I_1$ and $I_2$
$$I_1-I_2=\int_{0}^{1}\left(1-x^{50}\right)^{100}\left(1-(1-x^{50}\right))dx$$ so
$$I_1-I_2=\int_{0}^{1} \left(1-x^{50}\right)^{100} x^{50} dx$$ Now using Integration by Parts we get
$$I_1-I_2= \left(1-x^{50}\right)^{100} \times \frac{x^{51}}{51}\bigg|_{0}^{1} -\int_{0}^{1} 100 \left(1-x^{50}\right)^{99} \times -50 x^{49} \times \frac{x^{51}}{51} dx$$
So
$$I_1-I_2=\frac{5050}{51} \times \int_{0}^{1}\left(1-x^{50}\right)^{99} x^{100} dx$$
Now $x^{100}=\left(1-x^{50}\right)^2-(1-x^{50}-x^{50})$ so
$$\frac{51}{5050}(I_1-I_2)=\int_{0}^{1} \left(1-x^{50}\right)^{99} \times \left(\left(1-x^{50}\right)^2-(1-x^{50})+x^{50}\right)dx=I_2-I_1+\int_{0}^{1} \left(1-x^{50}\right)^{99} x^{50} dx$$
Need a hint to proceed further.
|
Factoring the integrand gives
$$
\int_0^1\left(1-x^{50}\right)^{101}\,\mathrm{d}x
=\int_0^1\left(1-x^{50}\right)^{100}\,\mathrm{d}x
-\int_0^1\left(1-x^{50}\right)^{100}x^{50}\,\mathrm{d}x
$$
Integration by parts gives
$$
\int_0^1\left(1-x^{50}\right)^{101}\,\mathrm{d}x
=101\cdot50\int_0^1\left(1-x^{50}\right)^{100}x^{50}\,\mathrm{d}x
$$
Combining yields
$$
\frac{\int_0^1\left(1-x^{50}\right)^{100}\,\mathrm{d}x}{\int_0^1\left(1-x^{50}\right)^{101}\,\mathrm{d}x}=\frac{5051}{5050}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1710265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
}
|
Definition of modulo division Could anyone tell me why
$3/7\equiv 9 \mod12$? How is this operation defined and why sometimes modulo division is impossible, like in the case of $3/4 \mod 12$?
See here for more details, I'm refering to page 95.
I can't grasp why $3/7\equiv 9 \mod12$. I divide $3$ by $7$, what I get is $3 = 0 \times 7 + 3$, the remainder is $3$. Now how is it equal to $9$ mod $12$?
|
Think of it like this:
$$\frac{3}{7} = 3 \cdot \frac{1}{7} = 3 \cdot 7^{-1}.$$
So then
\begin{align}
\frac{3}{7} \mod 12 &= (3 \cdot 7^{-1}) \mod 12\\
&= (3 \mod 12) \cdot (7^{-1} \mod 12).
\end{align}
$3 \mod 12$ is just 3. What's $7^{-1} \mod 12$? In other words, find $x$ so that $7x \equiv 1 \mod 12$. Then multiply this $x$ by 3 and evaluate modulo 12 again.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1711724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
$\lim_{x\to 0} \frac{1-\cos(x)}{x\cos(x)} $ without L'hopital Please help me to solve this limit without using L'Hôpital's rule. I don't know what other method can't be used to solve this limit.
$$\lim_{x\to 0} \frac{1-\cos(x)}{x\cos(x)} $$
|
Use Taylor series
$$\cos(x) \approx 1 - \frac{x^2}{2}$$ thence
$$\frac{1 - \cos(x)}{x\cos(x)} = \frac{1 - \left(1 - \frac{x^2}{2}\right)}{x\cdot\left(1 - \frac{x^2}{2}\right)} = \frac{x^2}{2x - x^3} = \frac{x^2}{x^2\left(\frac{2}{x} - x\right)} = \frac{1}{\frac{2}{x}} = \frac{x}{2}$$
And since $x\to 0$ the limit is
$$\boxed{0}$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
How to prove that: if $q= b+d$, then $p = a+c$? Let $a,b,c,d,p$, and $q$ be natural numbers such that $ad-bc = 1$ and $\frac{a}{b} > \frac{p}{q} > \frac{c}{d}$.
How to prove that: if $q= b+d$, then $p = a+c$? Is there a simple way?
|
Quite an easy problem. while it is informed that all the variables in the problem is natural numbers so, $p = a+c$ is equivalent to $a+c-1 < p <a+c+1$, that is to say $\frac{a+c-1}{b+d}<\frac{c}{d}$ and $\frac{a+c+1}{b+d}>\frac{a}{b}$.in fact:
$$
ad - d<bc
$$
$$
ad+cd -d <bc+cd
$$
$$
\frac{a+c-1}{b+d}<\frac{c}{d}
$$
what's more,
$$
bc+b>ad
$$
$$
ab+bc+b>ab+ad
$$
$$
\frac{a+c+1}{b+d}>\frac{a}{b}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1714050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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|
Inequality $(x-1)(y-1)(z-1)\geq 8$ provided that $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1$ How can I prove that if $$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1,$$ then $(x-1)(y-1)(z-1) \geq 8$?
Edit:
$x,y,z \in \mathbb R_{>0} $
Thanks
|
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$$
$$xyz=xy+yz+zx$$
$$AM \ge GM \Rightarrow (x+y+z)\left (\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right)\ge 3\sqrt[3]{xyz}\cdot 3\sqrt[3]{\frac 1{xyz}}=9 \Rightarrow$$
$$\Rightarrow (x+y+z)\cdot 1 \ge 9$$
Now $$(x-1)(y-1)(z-1)=(xy-x-y+1)(z-1)=$$
$$=(xyz-xz-yz-xy)+(x+y+z)-1=0+(x+y+z)-1\ge 9-1=8$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1714212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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|
Can you find a series whose sum is $.... = (a) *\sqrt{\frac{1}{1-n}}$ I discovered here that $ \frac{1}{n-1}= (n-1)^-1$ is the sum om a geometric series:
$$\frac{1}{n}+\frac{1}{n^2}+\frac{1}{n^3}.... = \frac{1}{n-1}$$
can you tell me if $(\sqrt{1-n})^-1$ $$\sqrt{\frac{1}{1-n}}$$
can be considered the result/sum of any kind of series, sequence, progression
Edit:
consider $|n| <1$, specify what happens if we have a factor $a$ before the square root and, if you please, case consider the particular case when the sum of the series is $$0.7071 * \sqrt{\frac{1}{1-.99}}$$
|
$\sum_{n=m}^{\infty} (a_{n}-a_{n+1})
=a_m
$,
so
$\begin{array}\\
\frac1{\sqrt{m}}
&=\sum_{n=m}^{\infty} (\frac1{\sqrt{n}}-\frac1{\sqrt{n+1}})\\
&=\sum_{n=m}^{\infty} (\frac1{\sqrt{n}}-\frac1{\sqrt{n+1}})\frac{\frac1{\sqrt{n}}+\frac1{\sqrt{n+1}}}{\frac1{\sqrt{n}}+\frac1{\sqrt{n+1}}}\\
&=\sum_{n=m}^{\infty} \frac{\frac1{n}-\frac1{n+1}}{\frac1{\sqrt{n}}+\frac1{\sqrt{n+1}}}\\
&=\sum_{n=m}^{\infty} \frac{1}{n(n+1)(\frac1{\sqrt{n}}+\frac1{\sqrt{n+1}})}\\
&=\sum_{n=m}^{\infty} \frac{1}{(n+1)\sqrt{n}+n\sqrt{n+1}}\\
\end{array}
$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1715297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Find $\lim_{x\to0}\frac{(\tan(\frac{\pi}{4}+x))^{\frac{1}{x}}-e^2}{x^2}$ Find $\lim_{x\to0}\frac{(\tan(\frac{\pi}{4}+x))^{\frac{1}{x}}-e^2}{x^2}$
My attempt:
$\lim_{x\to0}(\tan(\frac{\pi}{4}+x))^{\frac{1}{x}}=e^2$
So the required limit is in $\frac{0}{0}$ form.
Then i used L hospital form.
$\lim_{x\to0}\frac{e^2(\frac{2}{x\cos 2x}-\frac{1}{x^2}\log(\frac{1+\tan x}{1-\tan x}))}{2x}$
I am stuck here.
|
In the same spirit as Olivier Oloa's answer.
By Taylor $$\tan(\frac{\pi}{4}+x))=1+2 x+2 x^2+\frac{8 x^3}{3}+\frac{10 x^4}{3}+\frac{64 x^5}{15}+O\left(x^6\right)$$ $$\log\Big(\tan(\frac{\pi}{4}+x)) \Big)=2 x+\frac{4 x^3}{3}+\frac{4 x^5}{3}+O\left(x^6\right)$$ $$\frac 1x\log\Big(\tan(\frac{\pi}{4}+x)) \Big)=2+\frac{4 x^2}{3}+\frac{4 x^4}{3}+O\left(x^5\right)$$ $$\exp\Big(\frac 1x\log\Big(\tan(\frac{\pi}{4}+x)) \Big)\Big)=\Big(\tan(\frac{\pi}{4}+x)) \Big)^{\frac 1x}=e^2+\frac{4 e^2 x^2}{3}+\frac{20 e^2 x^4}{9}+O\left(x^5\right)$$ $$\frac{(\tan(\frac{\pi}{4}+x))^{\frac{1}{x}}-e^2}{x^2}=\frac{4 e^2}{3}+\frac{20 e^2 x^2}{9}+O\left(x^3\right)$$ which shows the limit and how it is approached.
To show how good is the approximation : using $x=\frac 1{10}$, the expression is $\approx 10.0191$ while the approximation gives $\frac{61 e^2}{45}\approx 10.0163$ that is to say a relative error of less than $0.03$%.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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|
Solution of the first order differential equation $y'=\frac{x(x^2+y^2)^2}{4y}$. I m stuck in finding the solution of the following differential equation
$$y'=\frac{x(x^2+y^2)^2}{4y}.$$
Please give me some hint.
|
$\frac{4yy'}{(x^2+y^2)^2}=x$
then define $x^2+y^2=h^2 \Rightarrow 2hh'=2yy'+2x \Rightarrow y'=\frac{hh'-x}{y}$
then by substituting we get: $\frac{4(hh'-x)}{h^4}=x \Rightarrow4(hh'-x)=xh^4 \Rightarrow \frac{4hh'}{4+h^4}=x$
now define $g=h^2$ and solve the problem: $\Rightarrow g'=2hh'$
$$\frac{2g}{4+g^2}=x \Rightarrow \tan^{-1}(\frac{g}{2})=\frac{x^2}{2}+C \Rightarrow g=2\tan(\frac{x^2}{2}+C)$$
then we get: $$h=\pm \sqrt{2\tan(\frac{x^2}{2}+C)} \Rightarrow y=\pm \sqrt{ 2\tan(\frac{x^2}{2}+C)-x^2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1717313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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|
Find the interval of convergence for $\sum_{n=1}^∞ sin(\frac{1}{n}) tan(\frac{1}{n})x^n$ Find the interval of convergence for:
$$\sum_{n=1}^{\infty} \sin(\frac{1}{n})\tan(\frac{1}{n})x^n$$
I have no idea how to do this question. Any help would be greatly appreciated
|
Here $a_n=\sin(\frac{1}{n})\tan(\frac{1}{n})x^n$
So
\begin{align*}
\Big|\frac{a_{n+1}}{a_n}\Big|&=\Big|\frac{\sin\Big(\frac{1}{n+1}\Big)\tan\Big(\frac{1}{n+1}\Big)x^{n+1}}{\sin\Big(\frac{1}{n}\Big)\tan\Big(\frac{1}{n}\Big)x^n}\Big|\\
&=\Big|\frac{\sin\Big(\frac{1}{n+1}\Big)}{\frac{1}{n+1}}\frac{\tan\Big(\frac{1}{n+1}\Big)}{\frac{1}{n+1}}\frac{\frac{1}{n}}{\sin\Big(\frac{1}{n}\Big)}\frac{\frac{1}{n}}{\tan\Big(\frac{1}{n}\Big)}x|\stackrel{n\to\infty}{\longrightarrow}|x|
\end{align*}
Therefore rate of convergence = $-1<x<1$
You also need to check at the endpoints.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1718200",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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|
find two different generalized inverse of the given matrix Definition:
For a given matrix $A_{m\times n}$, a matrix $G_{n\times m}$ is said to be a generalized inverse of $A$, if it satisfies
$$AGA=A.$$
Question:
Find two different generalized inverse of the given matrix
$$\begin{pmatrix}
1 & 0 &-1 & 2\\2 & 0 &-2 & 4 \\-1 & 1 & 1 & 3\\
-2 & 2 & 2 & 6
\end{pmatrix}$$
Work done:
Since the echelon form of the matrix is,
$$
\left(\begin{array}{rrrr}
1 & 0 & -1 & 2 \\
0 & 1 & 0 & 5 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{array}\right)$$ rank is 2.
since there are two distinct $2\times 2$ minors,
one of the generalized inverse is,
$$\left(\begin{array}{rrrr}
0 & 0 & 0 & 0 \\
\frac 1 2 &0 & 0 & 0 \\
\frac 1 2 & 1 & 0 & 0 \\
0 & 0 & 0 & 0
\end{array}\right)$$
and the other one is,
$$\left(\begin{array}{rrrr}
0 & 0 & 0 & 0 \\
0 &0 & \frac 3 {10} & -\frac 4{10} \\
0& 0 & \frac 1 {10} & \frac 2 {10} \\
0 & 0 & 0 & 0
\end{array}\right)$$
Luckily we get two different solutions,
But if the question is to find 5 different generalized inverses, How to do that?
As we know there are plenty of generalized inverses are there for this given matrix, different possible ways are welcome.
Thanks in advance.
|
The matrix and its Moore-Penrose pseudoinverse are
$$
\mathbf{A} =
\left[
\begin{array}{rrrr}
1 & 0 & -1 & 2 \\
2 & 0 & -2 & 4 \\
-1 & 1 & 1 & 3 \\
-2 & 2 & 2 & 6 \\
\end{array}
\right],
\qquad
\mathbf{A}^{\dagger} = \frac{1}{40}
\left[
\begin{array}{rrrr}
8 & 16 & -5 & -10 \\
-2 & -4 & 3 & 6 \\
-8 & -16 & 5 & 10 \\
6 & 12 & 5 & 10 \\
\end{array}
\right].
$$
The pseudoinverse satisfies all four requirements:
*
*$\mathbf{A} \, \mathbf{A}^{\dagger} \mathbf{A} = \mathbf{A}$
*$\mathbf{A}^{\dagger} \mathbf{A} \, \mathbf{A} = \mathbf{A}^{\dagger}$
*$\left( \mathbf{A} \, \mathbf{A}^{\dagger} \right)^{*} = \mathbf{A} \, \mathbf{A}^{\dagger}$
*$\left( \mathbf{A}^{\dagger} \mathbf{A} \right)^{*} = \mathbf{A}^{\dagger} \mathbf{A}$
The inverses presented in the question are problematic:
$$
\mathbf{G}_{1} = \left(
\begin{array}{cccc}
0 & 0 & 0 & 0 \\
\frac{1}{2} & 0 & 0 & 0 \\
\frac{1}{2} & 1 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{array}
\right), \qquad
%
\mathbf{A} \, \mathbf{G}_{1} \mathbf{A} =
\left(
\begin{array}{rrrr}
-\frac{5}{2} & 0 & \frac{5}{2} & -5 \\
-5 & 0 & 5 & -10 \\
3 & 0 & -3 & 6 \\
6 & 0 & -6 & 12 \\
\end{array}
\right)
\ne \mathbf{A}
$$
$$
\mathbf{G}_{2} = \left(
\begin{array}{cccr}
0 & 0 & 0 & 0 \\
0 & 0 & \frac{3}{10} & -\frac{2}{5} \\
0 & 0 & \frac{1}{10} & \frac{1}{10} \\
0 & 0 & 0 & 0 \\
\end{array}
\right), \qquad
%
\mathbf{A} \, \mathbf{G}_{2} \mathbf{A} =
\left(
\begin{array}{rrrr}
\frac{3}{10} & -\frac{3}{10} & -\frac{3}{10} &
-\frac{9}{10} \\
\frac{3}{5} & -\frac{3}{5} & -\frac{3}{5} & -\frac{9}{5}
\\
\frac{1}{5} & -\frac{1}{5} & -\frac{1}{5} & -\frac{3}{5}
\\
\frac{2}{5} & -\frac{2}{5} & -\frac{2}{5} & -\frac{6}{5}
\end{array}
\right)
\ne \mathbf{A}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
"answer_count": 2,
"answer_id": 1
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|
Does the sequence $\frac{n!}{1\cdot 3\cdot 5\cdot ... \cdot (2n-1)}$ converge? I'm trying to determine if this sequence converges as part of answering whether it's monotonic:
$$
\left\{\frac{n!}{1\cdot 3\cdot 5\cdot ... \cdot (2n-1)}\right\}
$$
First, I tried expanding it a bit to see if I could remove common factors in the numerator and denominator:
$$
\left\{\frac{1\cdot 2\cdot 3\cdot 4\cdot 5\cdot ...\cdot n}{1\cdot 3\cdot 5\cdot 7\cdot 9 \cdot ...\cdot (2n-1)}\right\}
$$
Second, I tried looking at elements of the sequence with common factors removed:
$$
1, \frac{2}{3}, \frac{2}{5}, \frac{2\cdot 4}{5\cdot 7}, \frac{2\cdot 4}{7\cdot 9}, ...
$$
Third, I tried looking at the elements again as fractions without simplifications:
$$
\frac{1}{1}, \frac{2}{3}, \frac{6}{15}, \frac{24}{105}, \frac{120}{945}, ...
$$
Last, I tried searching for similar questions on Stack Exchange and I found one for $\frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)}$ but I didn't understand how that might apply to my question. So, any hints would be much appreciated.
|
Just manipulate the Wallis product for $\pi$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Maximum value of $f(x) = \cos x \left( \sin x + \sqrt {\sin^2x +\sin^2a}\right)$ Can we find maximum value of $$f(x) = \cos x \left( \sin x + \sqrt
{\sin^2x +\sin^2a}\right)$$
where '$a$' is a given constant.
Using derivatives makes calculation too complicated.
|
the first derivative is given by $$f'(x)=-\sin \left( x \right) \left( \sin \left( x \right) +\sqrt { \left(
\sin \left( x \right) \right) ^{2}+ \left( \sin \left( a \right)
\right) ^{2}} \right) +\cos \left( x \right) \left( \cos \left( x
\right) +{\frac {\sin \left( x \right) \cos \left( x \right) }{\sqrt
{ \left( \sin \left( x \right) \right) ^{2}+ \left( \sin \left( a
\right) \right) ^{2}}}} \right)
$$ simplifying this you have to solve the equation for $x$:
$$- \left( \sin \left( x \right) +\sqrt { \left( \sin \left( x \right)
\right) ^{2}+ \left( \sin \left( a \right) \right) ^{2}} \right)
\left( \sin \left( x \right) \sqrt { \left( \sin \left( x \right)
\right) ^{2}+ \left( \sin \left( a \right) \right) ^{2}}- \left(
\cos \left( x \right) \right) ^{2} \right)
=0$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Show that $||X||$ is a norm in $\mathbb{R}^2$ Let $a,b \in \mathbb{R}$ fixed numbers and the following norm in $\mathbb{R}^2$:
$$ ||x|| = \sqrt{\frac{x_1^2}{a^2}+\frac{x_2^2}{b^2}} \qquad x=(x_1,x_2)\in\mathbb{R}^2$$
Prove that it is, in fact, a norm.
Taking the rules of a norm, I get the following:
*
*$||x||\geq0$: since we take $x_1^2$ and $x_2^2$, it will never be negative.
*$||x||=0 \iff x=0$: same applies here. It is only $0$ if $x_1$ and $x_2$ are $0$.
*$||\lambda x||=|\lambda|||x||$: $||\lambda x||=\lambda \sqrt{\frac{x_1^2}{a^2}+\frac{x_2^2}{b^2}}=\sqrt{ \frac{\lambda^2x_1^2}{a^2}+ \frac{\lambda^2x_2^2}{b^2}}=\sqrt{\lambda^2 (\frac{x_1^2}{a^2}+\frac{x_2^2}{b^2})}=|\lambda|\sqrt{\frac{x_1^2}{a^2}+\frac{x_2^2}{b^2}}=|\lambda| ||x||$
*$||x+y||\leq||x||+||y||:$The triangle inequality is the point where I got stuck. Not sure if $||x+y||= \sqrt{\frac{x_1^2}{a^2}+\frac{x_2^2}{b^2}+\frac{y_1^2}{a^2}+\frac{y_2^2}{b^2}}$ or $\sqrt{\frac{x_1^2}{a^2}+\frac{x_2^2}{b^2}}+\sqrt{\frac{y_1^2}{a^2}+\frac{y_2^2}{b^2}}$.
Any help or suggestion is appreciated!
|
Notice that $x+y=(x_1+y_1,x_2+y_2)$ so we have that $$||x+y||=\sqrt{\frac{(x_1+y_1)^2}{a^2}+\frac{(x_2+y_2)^2}{b^2}}$$ is the correct first step. You then need to manipulate it to show that it is greater than $$||x||+||y||=\sqrt{\frac{x_1^2}{a^2}+\frac{x_2^2}{b^2}}+\sqrt{\frac{y_1^2}{a^2}+\frac{y_2^2}{b^2}}$$
|
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|
What is the smallest value of $n$ such that the final digit of $13^n$ is one more than the digit adjacent to it?
What is the smallest value of $n$ such that the final digit of $13^n$ is one more than the digit adjacent to it?
If you have a computer, it is easy to check that the answer to this question is $14$, but I was wondering if there is a way to do it without brute-forcing or using a computer.
|
I suppose you want to find $n$ such that $13^n$ is of the form $*(x)(x+1)$ for some $x\in\{0,\cdots,9\}$.
This is equivalent to finding $n$ such that $13^n\equiv 11x+1\pmod{100}.$
Then notice that $\phi(100)=\phi(4)\times\phi(25)=2\times20=40,$ so, by Fermat's little theorem, $13^{40}\equiv 1\pmod{100}.$ In fact, $$\begin{cases}13\equiv1\pmod4\\13^{20}=13^{\phi(25)}\equiv1\pmod{25}\end{cases},$$ hence $13^{20}\equiv1\pmod{100}.$ (Furthermore, $13^{10}\equiv13^{2}\times((13^{2})^2)^2\equiv-6\times-4\equiv-1\pmod{25},$ so the order of $13$ modulo $25,$ hence modulo $100,$ is $20.$)
So we are trying to solve $$\begin{cases}1^n\equiv-x+1\pmod4\\13^n\equiv11x+1\pmod{25}\end{cases}.$$
Thus $x$ is divisible by $4,$ say $x=4\times y,$ and we are to solve $13^n\equiv44y+1\equiv-6y+1\pmod{25}, y\in\{0, 1, 2\}.$ So $13^n\equiv1, -5, -11\pmod{25}.$
The first case is answered by $n=20.$
Now notice that $\begin{cases}13^2\equiv-6\\13^3\equiv-3\\13^4\equiv11\\13^{10}\equiv-1\end{cases}\pmod{25}.$ So $13^{14}
\equiv-11\pmod{25}.$ And this solves the problem.
Why is this the minimal one? As the order of $13$ modulo $100$ is $20,$ the first $20$ powers of $13$ cannot repeat itself, and it is easy to see that any power of $13$ cannot be divisible by $5,$ thus $14$ is the minimal one.
Hope this helps.
|
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|
All 3 digit numbers the sum of whose digits is not greater than 16 My working: Let the digits of the numbers be $x,y,z$ where
\begin{align}1\leq x&\leq 9\\0\leq y,z&\leq 9\\x+y+z &\leq 16 \end{align}
I tried to solve it by making different cases here and using counting but it got really complicated and i couldnt get very far.
So how do i solve such questions?
|
Let $k\in\{1,2,...,16\}$ a fixed number. If you count the number of triplets $(x,y,z)$ such that $x+y+z=k$, say $S_k$, then you are searching for $S_0+S_1+...+S_{16}$,
Now, by generating functions method, we use the polinomyal $p(x)=(x+x^2+...+x^9)(1+x+x^2+...+x^9)^2$ and we want to find the coefficient of $x^k$.
But
$\begin{eqnarray}
p(x)&=&(x+x^2+...+x^9)(1+x+x^2+...+x^9)^2\\
&=&x\left(\frac{1-x^{9}}{1-x}\right)\left(\frac{1-x^{10}}{1-x}\right)^2\\
&=&x(1-x^9)(1-x^{10})^2(1-x)^{-3}\\
&=&x(1-x^{9})(1-2x^{10}+x^{20})\sum_{r=0}^\infty\binom{2+r}{2}x^r
\end{eqnarray}$
Now, it is easy to see that for $k=1,...,9$, the coefficient of $x^k$ is $\binom{2+k-1}{2}=\binom{k+1}{2}$.
For $k=10$, the coefficient is $\binom{2+9}{2}-\binom{2+0}{2}=\binom{11}{2}-1$
For $k=11,...,16$, the coefficient is $\binom{2+k-1}{2}-2\binom{2+k-11}{2}-\binom{2+k-10}{2}=\binom{k+1}{2}-2\binom{k-9}{2}-\binom{k-8}{2}$.
So, the answer is
$\begin{eqnarray}
S_0+...+S_{16}&=&\sum_{k=1}^{16}\binom{k+1}{2}-1-2\sum_{k=11}^{16}\binom{k-9}{2}-\sum_{k=11}^{16}\binom{k-8}{2}\\
&=&816-1-2*56-83\\
&=&620
\end{eqnarray}$
|
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|
Find all integer solutions to $\frac{1}{x} + \frac{1}{y} = \frac{2}{3}$
Find all integer solutions $(x, y)$ of the equation
$$\frac{1}{x} + \frac{1}{y} = \frac{2}{3}$$
What have done is that:
$$\frac{1}{x}= \frac{2y-3}{3y}$$
so,
$$x=\frac{3y}{2y-3}$$
If $2y-3 = +1 \text{ or } {-1}$, $x$ will be an integer, so we choose integer $y$
to make $2y-3=1 \text{ or } {-1}$. $y = 2$ or $y = 1$ is such a solution.
Also, $2y - 3$ can be deleted by numerator $3$, so $2y - 3$ can be $3$ or $-3$ too. This gives $y = 3$ or $y = 0$, but $y$ can not be $0$. So far, we have $y=1,2,3$. Finally, $2y-3$ can be deleted by numerator $y$, but how can we find such a $y$?
|
Another answer that does not follow your approach is like this:
Given integers $x,y$ such that $\frac{1}{x} + \frac{1}{y} = \frac{2}{3}$:
$3x+3y = 2xy$.
Thus $4xy - 6x - 6y = 0$ and hence $(2x-3)(2y-3) = 9$.
Now it remains to find all factorizations of $9$ as a product of two integers.
|
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|
Find the sum of $-1^2-2^2+3^2+4^2-5^2-6^2+\cdots$
Find the sum of $$\sum_{k=1}^{4n}(-1)^{\frac{k(k+1)}{2}}k^2$$
By expanding the given summation,
$$\sum_{k=1}^{4n}(-1)^{\frac{k(k+1)}{2}}k^2=-1^2-2^2+3^2+4^2-5^2-6^2+\cdots+(4n-1)^2+(4n)^2$$
$$=(3^2-1^2)+(4^2-2^2)+(7^2-5^2)+(8^2-6^2)+\cdots+[(4n-1)^2-(4n-3)^2]+((4n)^2-(4n-2)^2)$$
$$=2(4)+2(6)+2(12)+2(14)+2(20)+2(22)+\cdots+2(8n-4)+2(8n-2)$$
$$=2[4+6+12+14+20+22+\cdots+(8n-4)+(8n-2)]$$
How should I proceed further?
|
\begin{equation*}
\begin{aligned}
\sum_{k=1}^{4n}(-1)^{\frac{k(k+1)}{2}}k^2 &= \sum_{k=1}^{n}[((4k - 1)^2 - (4k - 3)^2) + ((4k)^2 - (4k - 2)^2)]\\
&= \sum_{k=1}^{n}[2(8k - 4) + 2(8k - 2)] \\
&= 4\sum_{k=1}^{n}[(8k - 3)] \\
&= -12n + 32\sum_{k=1}^{n}k\\
&= -12n + \frac{32n(n+1)}{2} \\
&= 16n^2 + 4n. \\
\end{aligned}
\end{equation*}
|
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|
Calculating exponential limit I've been breaking my mind over this one.
Find the limit.
$\lim\limits_{n \to \infty} (\frac{n^2+3}{n^2+5 n-4})^{2n} $
I know it equals $\frac{1}{e^{10}} $ but can't figure out how to find it.
Help?
|
Divide your fraction by $n^2$ and look at the asymptotics for $n\rightarrow \infty$
$$\frac{n^2+3}{n^2+5n-4}=\frac{1+\frac{3}{n^2}}{1+\frac{5}{n}-\frac{4}{n^2}}
= 1 - \frac{5}{n} + O(n^{-2}) \sim 1 - \frac{5}{n}$$
And now for the power:
$$\left(\frac{n^2+3}{n^2+5n-4}\right)^{2n}
\sim \left(1 - \frac{5}{n}\right)^{2n}
\sim \left(1 - \frac{10}{2n}\right)^{2n} \rightarrow e^{-10} $$
Another way is to use logarithms
$$2n \ln \frac{n^2+3}{n^2+5n-4}
=2n \ln \left(1 + \frac{7-5n}{n^2+5n-4}\right)$$
$$\sim 2n \frac{7-5n}{n^2+5n-4} = \frac{14n-10n^2}{n^2+5n-4} \rightarrow -10
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Applying root test to sequence $\frac{1}{2} + \frac{1}{3} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots$ The following is an example from Principles of Mathematics, by Rudin. I've been trying to understand the example but haven't quite grasped it because it seems I can solve it differently.
Given the following sequence: $\displaystyle \frac{1}{2} + \frac{1}{3} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{2^3} + \frac{1}{3^3} + \cdots$
Using the Ratio Test:
$$\lim \inf_{n \to \infty} \frac{a_{n+1}}{a_{n}} = \lim_{n \to \infty}
(\frac{2}{3})^n = 0$$
$$\lim \sup_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty}
\frac{1}{2}(\frac{3}{2})^n = +\infty$$
Using the Root Test:
$$\lim \inf_{n \to \infty} (a_n)^{\frac{1}{n}} = \lim_{n \to \infty} (\frac{1}{3^n})^{\frac{1}{2n}} = \frac{1}{\sqrt{3}}$$
$$\lim \sup_{n \to \infty} (a_{n})^{\frac{1}{n}} = \lim_{n \to \infty} (\frac{1}{2^n})^{\frac{1}{2n}} = \frac{1}{\sqrt{2}}$$
What I don't understand is how to find the $\lim \sup$ and $\lim \inf$ for the ratio test. I also don't understand why for the root test, we are looking at the $2n^\text{th}$ root. Where does this 2 come from? Furthermore, are we looking at $a_n$ as alternating between $\frac{1}{2^m}$ and $\frac{1}{3^m}$ or is $a_{n}$ actually $\frac{1}{2^m} + \frac{1}{3^m}$?
As a side note, I do know how to solve this question if asked whether or not this series converges. I simply don't understand the book went around solving it.
|
As Abstraction says, the book uses $n$ for two different things.
$a_n$ alternates between $\frac1{2^m}$ and $\frac1{3^m}$. So the first, third, fifth... terms are powers of $1/2$, and the second,fourth,sixth... are powers of $1/3$.
For the root test, for example, $\frac1{3^6}$ is the twelfth term, so you look at $\left(\frac1{3^6}\right)^{1/12}=\frac1{\sqrt{3}}$. For the eleventh term, you should look at $\left(\frac1{2^6}\right)^{1/11}$, but the book looks at $\left(\frac1{2^6}\right)^{1/12}=\frac1{\sqrt2}$. The true root is $\left(\frac1{\sqrt2}\right)^{12/11}=\left(\frac1{\sqrt2}\right)^{1+\frac1{
11}}$. This root approaches $\frac1{\sqrt2}$ as you go further in the sequence.
For the ratio test, $\frac{a_{2n}}{a_{2n-1}}$ approaches zero, and $\frac{a_{2n+1}}{a_{2n}}$ grows without limit.
|
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|
How do I determine the transformation matrix T of the coordinate transformation from the base E to the base B? In $\mathbb{R}^3$ the canonical basis $E=\left (\mathbf{e_1},\mathbf{e_2},\mathbf{e_3} \right )$ and $B=\left (\mathbf{b_1},\mathbf{b_2},\mathbf{b_3} \right )$ with
$\mathbf{b_1}=(1,2,4)^T$, $\mathbf{b_2}=(0,-1,1)^T$ and $\mathbf{b_3}=(2,3,8)^T$.
How do I determine for vector $\mathbf{v}=2\mathbf{e_1}+\mathbf{e_2}+2\mathbf{e_3}$ coordinates $\left \lfloor \mathbf{v} \right \rfloor_E$ and $\left \lfloor \mathbf{v} \right \rfloor_b$ ?
a) $\left \lfloor \mathbf{v} \right \rfloor_E$ =$\begin{pmatrix} 2\\ 1\\
2\end{pmatrix}$
$\left \lfloor \mathbf{v} \right \rfloor_b$ =$\begin{pmatrix} -16\\ 6\\
9\end{pmatrix}$
How do I determine for vector $\mathbf{w}=\mathbf{b_1}+2\mathbf{b_2}+3\mathbf{b_3}$ coordinates $\left \lfloor \mathbf{w} \right \rfloor_E$ and $\left \lfloor \mathbf{w} \right \rfloor_b$ ?
b)
$\left \lfloor \mathbf{w} \right \rfloor_E$ =$\begin{pmatrix} 7\\ 9\\
30\end{pmatrix}$
$\left \lfloor \mathbf{v} \right \rfloor_b$ =$\begin{pmatrix} 1\\ 2\\
3\end{pmatrix}$
Where I need help?
Determine the transformation matrix T of the coordinate transformation from the base E to the base B, where the old coordinates on E and the new coordinates refer to B. Note: Determine the matrix T such that applies
$\left [ \vec{x} \right ]_E=T\left [ \vec{x} \right ]_B $
I really don't understand how to do it?
d) How can I calculate $\left [ \vec{x} \right ]_b$ from $\left [ \vec{x} \right ]_E$. Take care that $\left [ \vec{x} \right ]_E=\vec{x}$.
|
Hint:
The matrix
$$ M=
\begin{bmatrix}
1 & 0 & 2\\
2 & -1 & 3\\
4 & 1 & 8
\end{bmatrix}
$$
represents the transformation:
$$
\mathbf{e_1}\to \mathbf{b_1} \qquad \mathbf{e_2}\to \mathbf{b_2} \qquad \mathbf{e_3}\to \mathbf{b_3}
$$
and its inverse:
$$M^{-1}
\begin{bmatrix}
-11 & 2 & 2\\
-4 & 0 & 1\\
6 & -1 & -1
\end{bmatrix}
$$
represents the transformation:
$$
\mathbf{b_1}\to \mathbf{e_1} \qquad \mathbf{b_2}\to \mathbf{e_2} \qquad \mathbf{b_3}\to \mathbf{e_3}
$$
Use $M^{-1}$ to substitute $\mathbf{e_i}$ in the vector $\mathbf{v}$ and $M$ to substitute $\mathbf{b_i}$ in the vector $\mathbf{w}$
Note that the columns of $M$ are the vectors $\mathbf{b_i}$ in the standard basis, so $M\mathbf{e_i}=\mathbf{b_i}$. In the same way the columns of $M^{-1}$ are the vectors of the standard basis expressed in the basis $\mathbf{b_i}$. So, by linearity, your vector $\mathbf{v}$ that in the standard basis is $\mathbf{v}=2\mathbf{e_1}+\mathbf{e_2}+2\mathbf{e_3}$, in the basis $B$ is:
$$
M^{-1}\mathbf{v}=
\begin{bmatrix}
-11 & 2 & 2\\
-4 & 0 & 1\\
6 & -1 & -1
\end{bmatrix}
\begin{bmatrix}
2\\
1\\
2
\end{bmatrix}=
\begin{bmatrix}
-16\\
-6\\
9
\end{bmatrix}
$$
and the vector $\mathbf{w}$ that in the basis $B$ is $[1,2,3]^T$ , in the standard basis is:
$$
M\mathbf{w}=
\begin{bmatrix}
1& 0 & 2\\
2 & -1 & 3\\
4 & 1 & 8
\end{bmatrix}
\begin{bmatrix}
1\\
2\\
3
\end{bmatrix}=
\begin{bmatrix}
7\\
9\\
30
\end{bmatrix}
$$
|
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|
How many ordered pairs $(x,y)$ are there such that ${1\over\sqrt x}+{1\over\sqrt y}={1\over\sqrt {20}}$, where both $x$ and $y$ are positive integers $${1\over \sqrt x}+{1\over \sqrt y}={1\over \sqrt {20}}$$
I could find one ordered pair that satisfies above equation, that is $(80,80)$. But the answer says that there are $3$ ordered pairs satisfying above equation.
|
If you manipulate the expression, we get
$$
\frac{1}{\sqrt{y}}=\frac{1}{\sqrt{20}}-\frac{1}{\sqrt{x}}=\frac{\sqrt{x}-\sqrt{20}}{\sqrt{20x}}.
$$
Therefore, $y=\frac{20x}{x-2\sqrt{20x}+20}.$ In order for there to be any hope of $y$ being an integer, $\sqrt{20x}=2\sqrt{5x}$ must be an integer. Therefore, the only options that need to be considered are when $x$ is five times a square. Moreover, since $(80,80)$ is a solution, you only need to check $x$ values up to $80$.
In other words, the only values for $x$ to check are $x=5,20,45$. Of these, $x=5$ and $x=20$ are too large, so we are only left with considering $x=45$.
When $x=45$, $y=\frac{20\cdot 45}{45-2\sqrt{20\cdot 45}+20}=180.$
Therefore, $(45,180)$, $(80,80)$, and $(180,45)$ are the pairs.
|
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|
Write $4\cos x - 3 \sin x$ in the form $k\sin(x - \alpha)$ where $0 \le \alpha \le 360$
Write $4\cos x - 3 \sin x$ in the form $k\sin(x - \alpha)$ where $0 \le \alpha \le 360$.
$4\cos x - 3 \sin x = k\sin(x - \alpha)$
=> $k(\sin x \cos \alpha - \cos x \sin \alpha)$
=> $k\cos \alpha \sin x - \sin \alpha \cos x$
Equating coefficients:
$k\cos \alpha = - 3$
$k\sin \alpha = 4$
$k = \sqrt{(-3)^2 + 4^2} = 5$
$\alpha$ is in the fourth quarter because $ \cos \alpha$ is positive and $\sin \alpha$ is negative.
$\alpha = \arctan\frac{4}{-3} = -53.1$
$\alpha = 360 - 53.1 = 306.9$
$4\cos x - 3 \sin x = 5\sin(x - 306.9)$
But the answer is $4\cos x - 3 \sin x = 5\sin(x - 233.1)$
|
$$4\cos x - 3 \sin x = k\sin(x - \alpha)$$
$$= k(\sin x \cos \alpha - \cos x \sin \alpha)$$
$$=k\cos \alpha \sin x - k\sin \alpha \cos x$$
Hence:
$$k\cos \alpha \sin x = - 3 \sin x$$
$$- k\sin \alpha \cos x = 4\cos x$$
so:
$$k\cos \alpha = -3$$
$$\color{red}{k\sin \alpha = - 4}$$
|
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|
Evaluation of limit at infinity: $\lim_{x\to\infty} x^2 \sin(\ln(\cos(\frac{\pi}{x})^{1/2}))$ $$\lim_{x\to\infty} x^2 \sin(\ln(\cos(\frac{\pi}{x})^{1/2}))$$
What I tried was writing $1/x=t$ and making the limit tend to zero and writing the cos term in the form of sin
|
Notice that
\begin{align*}
\ln\big(\cos(\pi/x)^{1/2}\big) &= \frac 1 2 \ln \cos \frac{\pi}{x} \\
&= \frac 1 2 \ln \left(1 - \frac 1 2\frac{\pi^2}{x^2} + O\left(\frac 1 {x^4}\right)\right) \\
&= -\frac 1 4 \frac{\pi^2}{x^2} + O \left( \frac 1 {x^4}\right)
\end{align*}
Now do you see how to show that the desired limit is $-\pi^2/4$?
|
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|
Considering the power series $\sum_{n=1}^\infty(-1)^{n-1}\frac{x^{2n+1}}{(2n+1)(2n-1)}$. Consider the power series $\sum_{n=1}^\infty(-1)^{n-1}\frac{x^{2n+1}}{(2n+1)(2n-1)}$. Find a closed form expression for all x which converge and hence evaluate $\sum_{n=1}^\infty\frac{(-1)^{n-1}}{(2n+1)(2n-1)}$.
Attempt at the solution: The radius of convergence is 1. We can rewrite the summands by:
\begin{eqnarray}
\sum_{n=1}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)(2n-1)} &=& \frac{1}{2}\Big[(x^3-\frac{x^3}{3}) - (\frac{x^5}{3} - \frac{x^5}{5}) + (\frac{x^7}{5} - \frac{x^7}{7}) + \dots \Big]\\
&=& \frac{1}{2}\Big[(x^3 -\frac{x^5}{3}+ \frac{x^7}{5} + \dots ) + (\frac{-x^3}{3} + \frac{x^5}{5}-\frac{x^7}{7} +...)\Big]\\
&=& \frac{1}{2}\Big[x^2\int\frac{1}{1+x^2}dx + \int\frac{1}{1+x^2}dx-x\Big]\\
&=& \frac{x^2}{2}\arctan(x)+\frac{1}{2}\arctan(x) -\frac{x}{2}
\end{eqnarray}
Substituting $x=1$ then gives $\frac{\pi}{4}-\frac{1}{2}$ .
The issue I have is two fold. Firstly, when dealing with evaluations at the boundary, term by term differentiation may not be valid. In particular, we used the fact that $\arctan(x) = x-\frac{x^3}{3} + ...$ by integrating power series for $\frac{1}{1+x^2}$, valid for |x|<1. This means that the arctan formula can only be guaranteed to hold within the interior (-1,1). What are the conditions needed to talk about power series validity at boundary points?
(Abelian/Tauberian theorems came to mind at first, but the conditions in this problem weren't strong enough. Alternatively, I noted that uniform convergence of the terms meant that the limit function of $x-\frac{x^3}{3} + ...$ had to be continuous. So $\arctan(1) = \frac{\pi}{4}$ by continuous extension. Do correct me if I'm wrong.
The other issue that I have not been able to justify is that of conditional convergence. Clearly, the arctan series is conditionally convergent at $x=1$. How do we justify the rearrangments carried out above then?
|
First, by using the standard power series,
$$
\sum_{n=1}^\infty(-1)^{n-1}\frac{t^{2n-1}}{2n-1}=\arctan t, \quad |t|<1,\tag1
$$ we multiply $(1)$ by $t$ then we are alowed to integrate termwise:
$$
\sum_{n=1}^\infty(-1)^{n-1}\frac{x^{2n+1}}{(2n-1)(2n+1)}=\int_0^x t\arctan t\: dt, \quad |x|<1,\tag2
$$ then integrating by parts on the right hand side, gives
$$
\int_0^x t\arctan t\: dt=\frac12\left(1+x^2\right)\arctan x-\frac{x}2, \tag3
$$ thus
$$
\sum_{n=1}^\infty(-1)^{n-1}\frac{x^{2n+1}}{(2n-1)(2n+1)}=\frac12\left(1+x^2\right)\arctan x-\frac{x}2, \quad |x|<1.\tag4
$$
Second, noticing the following absolute convergence,
$$
\sum_{n=1}^\infty\left|(-1)^{n-1}\frac{1}{(2n-1)(2n+1)}\right|<\infty,
$$ one may use Abel's theorem in $(4)$ with $x \to 1^-$ to obtain
$$
\sum_{n=1}^\infty\frac{(-1)^{n-1}}{(2n-1)(2n+1)}=\frac{\pi}4-\frac12 \tag6
$$
as announced.
|
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|
Real solution of the equation $\sqrt{a+\sqrt{a-x}} = x\;,$ If $a>0$
For a real number $a>0\;,$ How many real solution of the equation $\sqrt{a+\sqrt{a-x}} = x$
$\bf{My\; Try::}$ We can Write $\sqrt{a+\sqrt{a-x}} = x$ as $a+\sqrt{a-x}=x^2$
So we get $(x^2-a)=\sqrt{a-x}\Rightarrow (x^2-a)^2 = a-x\;,$ Where $x<a$
So we get $x^4+a^2-2ax^2=a-x\Rightarrow x^4-2ax^2+x+a^2-a=0$
Now How can i solve it after that, Help me
Thanks
|
You get a quadratic equation in terms of "$a$". Using the quadratic formula for $a$,
$$
a=\frac{2x^2+1\pm|2x-1|}{2}
$$
From this, we get $a=x^2+x$ or $a=x^2-x+1$, so
$$
x=\frac{-1\pm \sqrt{1+4a}}{2}\text{ or } x=\frac{1\pm\sqrt{-3+4a}}{2}
$$
Now we must get rid of possiblities:
*
*If $x=\frac{-1-\sqrt{1+4a}}{2}$, it violates $x\ge 0$.
*If $x=\frac{-1+\sqrt{1+4a}}{2}$, it must satisfy $\frac{-1+\sqrt{1+4a}}{2}\ge \sqrt{a}$, which has no roots.
*If $x=\frac{1-\sqrt{-3+4a}}{2}$, it must satisfy $\frac{1-\sqrt{-3+4a}}{2}\ge\sqrt{a}$, which has no roots.
The only remaining possibility is $x=\frac{1+\sqrt{-3+4a}}{2}$. Solve below inequality:
$$
\sqrt{a}\le \frac{1+\sqrt{-3+4a}}{2}\le a
$$
Solving left inequality, we get $a\ge 1$, and solving right inequality, we get $4(a-1)^2\ge 0$. Thus the root of given equality when $a\ge 1$ is
$$
x=\frac{1+\sqrt{-3+4a}}{2}\;(a\ge 1)
$$
If $0<a<1$, then there are no roots.
|
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|
Why is there a difference of $\frac{-1}{6}$ when integrating $(1-2x)^2$ in two different ways? I integrated $(1-2x)^2$ by expanding the expression first, that is $1 - 4 x + 4 x^2$ and I get $$x - 2 x^2 + \frac{4 x^3}{3} +C$$
When I integrate by substitution I get $$\frac{-(1-2x)^3}{6} + C$$
Expanding $(-(1-2x)^3)/6 +C$ get $$-\frac{1}{6} + x - 2 x^2 + \frac{4 x^3}{3} +C$$
Why the difference in the two answers?
|
in $x - 2 x^2 + \frac{4 x^3}{3} +C$ you did right and got a constant C.
in $-\frac{1}{6} + x - 2 x^2 + \frac{4 x^3}{3} +C$ what happened is that you got different C, you can call it D. and it is a constant so $-\frac{1}{6}$ is also a part of that same constant. your expression should be $x - 2 x^2 + \frac{4 x^3}{3} +E$ where E=D$-\frac{1}{6}$. This is a property of indefinite integrals, if you integrated on some interval then your substitution would change your boundaries of integration and you would get the same result in the end.
|
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|
How do I show that $\frac xy + \frac yz + \frac zx \ge 1 + \frac {z + x}{x + y} + \frac {x + y}{z + x}$? Show that
$$\frac xy + \frac yz + \frac zx \ge 1 + \frac {z + x}{x + y} + \frac {x + y}{z + x}$$
for $x, y, z \gt 0$.
I observed that this is a homogeneous inequality so normalization might work. I tried to set $x = 1$ or $xyz = 1$ or $x + y + z = 1$, but none of these yields a solution. How do I tackle this problem? In particular, what can I do with the term $1$? Any hints will be appreciated.
|
Using CS inequality in slightly different way,
$$\frac{x}y + \frac{y}z + \frac{z}x + 1 \geqslant \frac{(x+y+z+x)^2}{xy+yz+zx+x^2} = \frac{(\color{red}{x+y}\:+\:\color{blue}{z+x})^2}{(\color{red}{x+y})\cdot(\color{blue}{z+x})} = \frac{x+y}{z+x}+2 + \frac{z+x}{x+y}$$
The cyclic counterparts of the inequality can be shown similarly.
Addendum: the inequality above is perhaps seen simplest by
$$\frac{x}y + \frac{y}z + \frac{z}x + 1 = \frac{x^2}{xy}+\frac{y^2}{yz}+\frac{z^2}{zx}+\frac{x^2}{x^2} \geqslant \frac{(x+y+z+x)^2}{xy+yz+zx+x^2} $$
the CS form is called "Titu's lemma" and is a useful tool to have.
|
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|
Find all solutions of the equation $3 \cdot 2^{x+2}+5^x=8\cdot 3^x+5$
Find all solutions of the equation
$$3 \cdot 2^{x+2}+5^x=8\cdot 3^x+5$$
My work so far:
Let $f(x)=3 \cdot 2^{x+2}+5^x-8\cdot 3^x-5$
$f'(x)=12\cdot 2^x\ln2+5^x\ln5-8\cdot3^x\ln3$
|
Your equation can be changed to $$5(5^{x-1}-1)=24(3^{x-1}-2^{x-1})$$
At least $x=0$, $x=1$ and $x=3$ are solutions.
If your question is a Diophantine problem:
With $n_1,n_2\in\mathbb{N}$ you have the conditions $$3^{x-1}-2^{x-1}=5n_1$$ $$5^{x-1}-1=24n_2$$
EDIT:
$x\in\mathbb{R}$ o.k.
But with the values 0,1,3 you can check
$$5(5^{x-1}-1)<,=,>24(3^{x-1}-2^{x-1})$$
with $x<0$, $0<x<1$, $1<x<3$ and $x>3$
|
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|
How do I go about solving this? I have tried substitution, but it is not working for me.
$$
\int_0^\pi \frac{dx}{\sqrt{(n^2+1)}+\sin(x)+n\cos(x)}=\int_0^\pi \frac{n dx}{\sqrt{(n^2+1)}+n\sin(x)+\cos(x)}=2
$$
General form of this integral is
$$
\int_0^\pi \frac{dx}{\sqrt{(n^2+m^2)}+m\sin(x)+n\cos(x)}=\frac{2}{m}
$$
|
This seems so easy in polar form. Let $1=r\cos\phi$, $n=r\sin\phi$, then $r=\sqrt{1+n^2}$, $\phi=\tan^{-1}n$ and then let $x+\phi=y+\frac{\pi}2$
$$\begin{align}\int_0^{\pi}\frac{dx}{\sqrt{1+n^2}+\sin x+n\cos x}&=\int_0^{\pi}\frac{dx}{\sqrt{1+n^2}(1+\sin(x+\phi))}\\
&=\int_{-\frac{\pi}2+\phi}^{\frac{\pi}2+\phi}\frac{dy}{\sqrt{1+n^2}(1+\cos y)}\\
&=\left.\frac{\sin y}{\sqrt{1+n^2}(1+\cos y)}\right|_{-\frac{\pi}2+\phi}^{\frac{\pi}2+\phi}\\
&=\frac1{\sqrt{1+n^2}}\left(\frac{\cos\phi}{1-\sin\phi}+\frac{\cos\phi}{1+\sin\phi}\right)\\
&=\frac1{\sqrt{1+n^2}}\left(\frac1{\sqrt{1+n^2}-n}+\frac1{\sqrt{1+n^2}+n}\right)\\
&=2\end{align}$$
The second integral, after the substitution $y=x-\phi$ becomes
$$\begin{align}\int_0^{\pi}\frac{n\,dx}{\sqrt{1+n^2}+n\sin x+\cos x}&=\int_0^{\pi}\frac{n\,dx}{\sqrt{1+n^2}(1+\cos(x-\phi))}\\
&=\int_{-\phi}^{\pi-\phi}\frac{n\,dy}{\sqrt{1+n^2}(1+\cos y)}\\
&=2\end{align}$$
|
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|
For which primes $p\not=2$ is $5$ a square mod $p$? For which primes $p\not=2$ is $5$ a square mod $p$?
Using the Legendre symbol, $5$ is a square modulo $p$ if
$$\left(\frac{5}{p}\right)=5^{\dfrac{p-1}{2}} \equiv 1 \pmod{p}$$
Now we have
$$5^{\dfrac{p-1}{2}} =
(5^{p-1})^{1/2} \equiv 1\pmod{p}$$
So $5$ is a square for any $p$. But this doesn't seem correct to me. Can anyone check this and if it's wrong, explain why.
|
You can't take the square root like this. By Fermat's little theorem, we know that for $p$ prime we have that $5^{p-1} \equiv 1 \mod p$. This means that $p \mid 5^{p-1} -1$, and hence $$p \mid \left(5^{\tfrac{p-1}{2}} -1\right)\left(5^{\tfrac{p-1}{2}} +1\right)$$
Therefore $5^{\tfrac{p-1}{2}} \equiv 1 \mod p$ or $5^{\tfrac{p-1}{2}} \equiv -1 \mod p$.
Note that if $p$ is not prime, there can be even more possibilities.
Note that for every odd prime $p$ we have $2 \mid p-1$. Also $4 \mid 5-1$. So $2 \mid \left(\frac{p-1}{2} \cdot \frac{5-1}{2} \right)$. Hence by quadratic reciprocity we have $$\left( \frac{p}{5} \right)\left( \frac{5}{p} \right) = (-1)^{\frac{p-1}{2} \cdot \frac{5-1}{2}} = 1$$
So $\left( \frac{p}{5} \right) =1$ if $\left( \frac{5}{p} \right)=1$, and the latter happens when $p\equiv 1 \mod 5$ or $p \equiv 4 \mod 5$.
Note that one can not use the law if $p=5$, since the law of quadratic reciprocity requires $p$ and $q$ to be different primes. Whether $p=5$ is a quadratic residue depends on the definition of quadratic residue.
|
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|
If $a,A,b,B,c,C$ are non negative reals such that $a+A=b+B=c+C=k$ Prove that $aB+bC+cA \le k^2$ If $a,A,b,B,c,C$ are non negative reals such that $a+A=b+B=c+C=k$ Prove that $aB+bC+cA \le k^2$
I substituted $B=k-b,C=k-c,A=k-a$ and plugged them to get a quadratic of $k$ which I had to show positive .SO, the discriminant should be $<0$ but actually its not coming
|
Think of a equilateral triangle $\triangle ABC$ where each side is $k$, then let $$\overline {AD}=A, \overline {DB}=a, \overline {BE}=B, \overline {EC}=b, \overline {CF}=C, \overline {FA}=c$$Where $D,E,F$ are points on $\overline {AB}, \overline {BC}, \overline {CA}$ respectively. Now, notice $$\frac{\sqrt 3}{4}Ac+\frac{\sqrt{3}}{4}Cb+\frac{\sqrt{3}}{4}Ba=\triangle {ADF}+\triangle {DBE}+\triangle {ECF} \le \triangle {ABC}=\frac{\sqrt{3}}{4}k^2$$We have our desired result.
|
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|
Solving $\arcsin\left(2x\sqrt{1-x^2}\right) = 2 \arcsin x$ If we have
$$\arcsin\left(2x\sqrt{1-x^2}\right) = 2 \arcsin x$$
Then, what will be the set of $x$ for which this equation is true?
I tried to solve it by putting $x = \sin a$ or $\cos a$ but got no result.
I am totally stuck on how to do it.
|
Consider the function
$$
f(x)=\arcsin(2x\sqrt{1-x^2})-2\arcsin x
$$
which is defined in the domain where
$$
\begin{cases}
|2x\sqrt{1-x^2}|\le 1\\[4px]
|x|\le 1
\end{cases}
$$
The first inequality becomes, writing $t=|x^2|$,
$$
4t-4t^2\le 1
$$
that's satisfied for every $t$. So our function $f$ is defined over $[-1,1]$.
The derivative of $f$ is
$$
f'(x)=\frac{1}{\sqrt{1-4x^2(1-x^2)}}\left(2\sqrt{1-x^2}-\frac{2x^2}{\sqrt{1-x^2}}\right)-\frac{2}{\sqrt{1-x^2}}
$$
that can be simplified to
$$
f'(x)=\frac{2}{\sqrt{1-x^2}}
\left(\frac{1-2x^2}{|1-2x^2|}-1\right)
$$
The derivative exists in the set
$$
(-1,-1/\sqrt{2})\cup(-1/\sqrt{2},1/\sqrt{2})\cup(1/\sqrt{2},1)
$$
and it can be simplified further as
$$
f'(x)=\begin{cases}
-\dfrac{4}{\sqrt{1-x^2}} & \text{for $1/\sqrt{2}<|x|<1$}\\[8px]
0 & \text{for $|x|<1\sqrt{2}$}
\end{cases}
$$
Thus the function is constant over $[-1/\sqrt{2},1/\sqrt{2}]$ (also at the extremes, by continuity). Since $f(0)=0$, we have that the given identity is valid on this interval and nowhere else, because over $[-1,-1/\sqrt{2}]$ and $[1/\sqrt{2},1]$ the function is strictly decreasing.
|
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|
Definite integral of a positive continuous function equals zero? Let's calculate $$\int_0^{\frac\pi 2} \frac {dx}{\sin^6x + \cos^6x}$$
We have
$$\int \frac {dx}{\sin^6x + \cos^6x} = \int \frac {dx}{1 - \frac 34 \sin^2{2x}}$$
now we substitute $u = \tan 2x$, and get
$$\int \frac {dx}{1 - \frac 34 \sin^2{2x}} = \frac{1}{2} \int \frac {du}{1 + \frac 14 u^2} = \tan^{-1}\frac u2 + C= \tan^{-1}\left(\frac12 \tan 2x\right) + C = F(x)$$
Now, evaluating the primitive function at $x = 0, x = \frac{\pi}2$, we get
$$\int_0^{\frac\pi 2} \frac {dx}{\sin^6x + \cos^6x} = F\left(\frac {\pi}2\right) - F(0) = 0 - 0 = 0$$
But the integrand is positive and continuous, so the integral should be positive!! What have I done wrong?
|
Hint
$$\sin^6(x)+\cos^6(x)=(\sin^2(x)+\cos^2(x))((\sin^2(x)+\cos^2(x))^2-\sin^2(x) \cos^2(x))= \\ = 1-\frac{\sin^2(2x)}{4}$$
note I have reduced $a^3+b^3$ and then it's easy
|
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|
How many values does the expression $1 \pm 2 \pm 3 \pm \cdots \pm n$ take?
How many different values does the expression $1 \pm 2 \pm 3 \pm \cdots \pm n$ take?
I was wondering about this problem and didn't think it was immediately obvious. The answer can't be $2^{n-1}$ since the combinations all might not be unique. I was thinking of looking at partitions of $\{1,2,\ldots,n, k\}$ where $1 \pm 2 \pm 3 \pm \cdots \pm n = k$ and thus the sum must be even. Therefore, $\dfrac{n(n+1)}{2}+k = \dfrac{n(n+1)+2k}{2}$ must be even which means $n(n+1) + 2k$ must be a multiple of $4$ so $n(n+1)+2k = 4m$. Thus if $n \equiv 1,2 \bmod 4$, then $k \equiv 1,3 \bmod 4$ and if $n \equiv 3,0 \bmod 4$ then $k \equiv 0,2 \bmod 4$.
|
Let us use generating functions
$$G_n(X)=X\prod_{k=2}^n(X^{-k}+X^k)=\dfrac{1}{X^p}\prod_{k=2}^n(1+X^{2k}) \ \ \ \text{with} \ \ \ p=\frac{n(n+1)}{2}-2$$
(very close to the proposal of @Jack D'Aurizio). In this way, we transfer the issue into counting how many combinations as positive or negative exponents of indeterminate $X$ are present in the expansion of $G_n(X)$.
Let us manipulate the first values of $n$ which is enough to get a good perception of the method.
We begin by the case $n=3$:
$$G_3(X)=X(X^{-2} + X^2)(X^{-3} + X^3)=$$
$$X^{1-2-3} + X^{1+2-3} + X^{1-2+3} +X^{1+2+3}$$
$$=X^{-4} + X^0 + X^2 + X^6$$
Note the central symmetry (with respect to position $1$) of the exponents $-4,0,2,6$. This central symmetry will exist for all values of $n$.
Let us now consider the cases $n=4, 5, 6$:
$n=4$:
$$G_4(X)=X(X^{-2} + X^2)(X^{-3} + X^3)(X^{-4} + X^4) =$$
$$X^{-8} + X^{-4} + X^{-2} + X^0+ X^2 + X^4 + X^6 + X^{10}$$
Till now, there are no repetitions. Here they come for $n \geq 5$:
$n=5$:
$$G_5(X)=X(X^{-2} + X^2)(X^{-3} + X^3)(X^{-4} + X^4)(X^{-5} + X^5)$$
$$=X^{-13} + X^{-9} + X^{-7} + X^{-5} + 2X^{-3} +
X^{-1} + 2\,X + X^3 + 2\,X^5 + X^7 + X^9 + X^{11} +
X^{15}$$
This time, not all coefficients are one; the reason is clear: for example, monomial $^2X^5$ means that $5$ is obtainable in two ways, i.e., $1+2+3+4-5$ and $1-2-3+4+5$. This is an "added value" of the approach by generating functions.
A last case, $n=6$, in order to see the differences between the cases $n$ odd and $n$ even:
$$X(X^{-2} + X^2) \cdots (X^{-5} + X^5)(X^{-6} + X^6)=$$
$$=X^{-19} + X^{-15} + X^{-13} + X^{-11} + 2X^{-9} +
2X^{-7} + 2X^{-5} + 2X^{-3} +
3X^{-1} + 2\,X + 3\,X^3 + 2\,X^5 + 2\,X^7 + 2\,X^9 +
2\,X^{11} + X^{13} + X^{15} + X^{17} + X^{21}$$
I will not go beyond this point because there is a clear recurrence relationship, but developing it would look to the reader like paraphrasing the (very detailed) redaction of @David K .
|
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|
Which irrational number represents the infinite simple continued fraction [0;7]? Which irrational number represents the infinite simple continued fraction [0;7]?
-So from my current understanding [o;7] can be represented as the following:
$ = \frac{1}{7 + \frac{1}{7 + \frac{1}{7 +...}}}$
So, x would be $x = \frac{1}{7 + \frac{1}{7 + x}}$
$\frac{7(7+x)}{7+x} + \frac{1}{7+x}$
$\frac{49 + 7x}{7+x} + \frac{1}{7+x}$
$ \frac{50 + 7x}{7+x}$
From here we get:
$x(50 + 7x) = 7 + x$
$50x = 7x^2 = 7 + x$
$49x + 7x^2 = 7$
$ 49x = 7x^2 - 7 = 0$
$x = \frac{-49\pm \sqrt{2205}}{14}$
*
*UPDATED, the square root looked a bit weird to me wanted to confirm my answer or suggestions, any help is appreciated.
|
Let $x$ be the value of the continued fraction $[0;7]$. We have that
$$ x = \frac{1}{7 + \frac{1}{7 + \frac{1}{7 +...}}}$$
Now we notice that $x$ appears in this continued fraction, so we can write it as:
$$x = \frac{1}{7 + x}$$
This simplifies to $x^2+7x-1=0$, and solving for $x$ gives $x=\frac{-7}{2}\pm \frac{\sqrt{53}}{2}$
|
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|
How to properly find supremum of a function $f(x,y,z)$ on a cube $[0,1]^3$? Solving an applied problem I was faced with the need to find supremum of the following function
$$f(x,y,z)=\frac{(x-xyz)(y-xyz)(z-xyz)}{(1-xyz)^3}$$
where $f\colon\ [0,1]^3\backslash\{(1,1,1)\} \to\mathbb{R}$.
I used Wolfram Mathematica and it resulted with "there is no maximum in the region in which the objective function is defined" and that the supremum is $\frac{8}{27}$ when $x,y,z\to 1$. I was able to verify some of it:
$$\lim\limits_{x,y,z\to 1}f(x,y,z)
=
\lim\limits_{x\to 1}f(x,x,x)
=
\lim\limits_{x\to 1}\frac{(x-x^3)^3}{(1-x^3)^3}
=
\lim\limits_{x\to 1}\frac{x^3(x+1)^3}{(x^2+x+1)^3}
=
\frac{8}{27}
$$
though I'm not sure how to justify the first equality. Can I neglect direction of derivative in this case?
And secondly, I don't know how to show that $f(x,y,z)< \frac{8}{27}$ for all $x,y,z\in[0,1]^3\backslash\{(1,1,1)\}$. When I tried to do that straightforwardly I drowned in computations.
|
tl; dr: The supremum of $f$ can be found by restricting to the diagonal of the cube and approaching $(1, 1, 1)$.
Partition the cube $C$ into level sets of $xyz$. If $0 < c < 1$, the level set
$$
L_{c} = \{(x, y, z) \in C : f(x, y, z) = c\}
$$
is compact, so $f$ achieves absolute extrema in $L_{c}$. (In fact, $L_{c}$ is homeomorphic to a closed disk, with three hyperbola arcs as boundary.)
To establish the bound
$$
f(x, y, z) \leq f(\sqrt[3]{xyz}, \sqrt[3]{xyz}, \sqrt[3]{xyz})
= f(\sqrt[3]{c}, \sqrt[3]{c}, \sqrt[3]{c})\quad\text{on $L_{c}$,}
\tag{1}
$$
it suffices to show that the maximum of $f$ subject to $xyz = c$ occurs where $x = y = z$.
Assume from now on that $0 < c < 1$. Introduce the function $g(u) = u - \frac{1}{u}$ with domain $0 < u < 1$, and note that $ug'(u) = u + \frac{1}{u}$ is monotone for $0 < u < 1$.
On the surface $xyz = c$, we have $xy = \frac{c}{z}$, $xz = \frac{c}{y}$, and $yz = \frac{c}{x}$, so
\begin{align*}
f(x, y, z)
&= \frac{(x - c)(y - c)(z - c)}{(1- c)^{3}} \\
&= \frac{c - c(yz + xz + xy) + c^{2}(x + y + z) - c^{3}}{(1- c)^{3}} \\
&= \frac{c - c^{2}(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}) + c^{2}(x + y + z) - c^{3}}{(1- c)^{3}} \\
&= \frac{c + c^{2}\bigl(g(x) + g(y) + g(z)\bigr) - c^{3}}{(1- c)^{3}}. \\
\end{align*}
Since the denominator is positive, this is maximized when $g(x) + g(y) + g(z)$ is maximized subject to $xyz = c$.
Lagrange multiplers gives $g'(x) = \lambda yz$, or $xg'(x) = c\lambda$, and similarly for $y$ and $z$. That is, the only critical points of $f(x, y, z)$ on the surface $xyz = c$ occur where
$$
xg'(x) = yg'(y) = zg'(z).
$$
Since $0 < x, y, z < 1$ on the cube and $u \mapsto ug'(u)$ is monotone (hence injective) in $(0, 1)$, the preceding implies $x = y = z$. That is, the only critical point of $f$ in the interior of $L_{c}$ (and hence the only possible interior extremum) is $(\sqrt[3]{xyz}, \sqrt[3]{xyz}, \sqrt[3]{xyz}) = (\sqrt[3]{c}, \sqrt[3]{c}, \sqrt[3]{c})$.
To conclude (1) holds, it remains to show that the boundary values of $f$ do not exceed the right-hand side. Because $f$ is symmetric, we may as well assume $x = 1$ and $yz = c$, so that
$$
f(x, y, z)
= \frac{(y - c)(z - c)}{(1- c)^{2}}
= \frac{c - c(y + z) + c^{2}}{(1 - c)^{2}}.
$$
This is maximized when $y + z$ is minimized, which is easily seen to occur where $y = z = \sqrt{c}$. Since
$$
f(1, \sqrt{c}, \sqrt{c}) < f(\sqrt[3]{c}, \sqrt[3]{c}, \sqrt[3]{c}),
$$
(a straightforward, if slightly tedious, one-variable calculation), (1) holds.
|
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"url": "https://math.stackexchange.com/questions/1766888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Solve by factoring, then perform synthetic division. $${x^4 - 3x^3 + 3x^2 - x = 0}$$
$${x(x^3 - 3x^2 + 3x - 1)}$$
Now, we have to perform synthetic division, if I'm correct we get the 1 to divide by from the X outside of the parentheses from the second equation, which is above this paragraph, and is also the factored form of the first equation.
$$\frac{x - 3 + 3 - 1}{1}$$
The answer is ${x = 0, 1}$. Does the 0 come from the remainder, and do the 1 come from the divisor?
|
Synthetic division is not the greatest tool, in my opinion, because it makes the division process very mysterious. It's a shorthand for long division, and I would recommend writing out the steps of long division instead.
Anyway, to use synthetic division here, you can notice by plugging in a few values of $x$ that $x=1$ makes the cubic factor disappear: $1^3 - 3(1)^2 + 3(1) - 1 = 0$. On to synthetic division, with $1$ as the magic number:
$$\begin{array}{c|cccc}
& 1 & -3 & 3 & -1\\
1 & & 1 & -2 & 1\\
\hline
& 1 & -2 & 1 & 0
\end{array}$$
This means that $x^3 - 3x^2 + 3x - 1 = (x-1)(x^2 - 2x +1)$. (There is no remainder term since the last number in the bottom right is $0$. For that matter, we knew in advance that there would be no remainder since $x=1$ is a zero.)
Now you could repeat the process with $x^2-2x+1$, but hopefully you recognize that $x^2-2x+1 = (x-1)^2$. Therefore, we have $x^3 - 3x^2 + 3x - 1 = (x-1)^3$. Coming back to the original polynomial of degree $4$, we can see that:
$$x^4 - 3x^3 + 3x^2 - x = x(x^3 - 3x^2 + 3x - 1) = x(x-1)^3.$$
The solutions $x=0,x=1$ come from the factors in this last expression ($x$ and $x-1$, respectively).
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1768630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
How can I make sense of this basic algebraic manipulation backwards? $$\frac{1}{n^2-1} = \frac12\left(\frac1{n-1}-\frac1{n+1}\right)$$
It is easy for me to understand this algebraic from right to left ←, but I struggle to find a reasonable way to work from left to right.
Regarding expected difficulty, the level of algebraic manipulated is here meant for a Calculus II course.
Thanks all.
|
Write $$R = \frac{1}{n^2 - 1} = \frac{1}{(n - 1)(n + 1)}$$
Now, we wish to decompose $R$ into something with denominator $(n - 1)$ and denominator $(n + 1)$
The standard way to do this is by a method knows as "partial fractions". We assign $R$ to
$$R = \frac{A}{n - 1} + \frac{B}{n + 1} = \frac{1}{(n - 1)(n + 1)}$$
Cross multiplying,
$$R = \frac{A (n + 1) + B (n - 1)}{(n - 1)(n + 1)} = \frac{1}{(n - 1)(n + 1)}
$$
Since the denominators are equal, we can simply equate the numerators and see what values of A, B we need.
$$
A(n + 1) + B(n - 1) = 1 + 0n \\
(A - B) + n(A + B) = 1 + 0n
$$
Clearly, we have two inequalities (by assigining the coefficients of the constant term and $n$ to each other.
$$
A - B = 1 \\
A + B = 0
$$
Substituting $B = -A$ in the first equation,
$$
A - (-A) = 1 \\
2A = 1 \\
A = \frac{1}{2}
$$
Substituting $A = \frac{1}{2}$ in $A + B = 0$ gives $B = \frac{-1}{2}$
Hence, $$R = \frac{\frac{1}{2}}{n - 1} + \frac{\frac{-1}{2}}{n + 1} \\
= \frac{1}{2}\left(\frac{1}{n - 1} - \frac{1}{n + 1} \right)$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1769262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.