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Prove that an expression is divisible by a polynomial Question: Show that the polynomial $f(x)=(x+1)^{2n} +(x+2)^n - 1$
is divisible by $g(x) = x^2+3x+2$, where $n$ is an integer.
I have tried to use mathematical induction. The basis case wasn't that difficult, but when it comes to the inductive step itself, I got a bit confused.
Is it possible to prove this by mathematical induction, and is binominal expansion required at that step?
|
Putting the zeros of $x^2+3x+2=0$ i.e., $-1,-2$ one by one, in $f(x)=(x+1)^{2n}+(x+2)^n-1$
we get, $f(-1)=(-1+1)^{2n}+(-1+2)^n-1=0$
and $f(-2)=(-2+1)^{2n}+(-2+2)^n-1=0$
(i)So, using Remainder Theorem,
$(x+1)\mid f(x)$ and $(x+2)\mid f(x)\implies lcm(x+1,x+2)\mid f(x)$
(ii) Alternatively,
$\frac{(x+1)^{2n}+(x+2)^n-1}{x+1}=(x+1)^{2n-1}+\frac{(x+2)^n-1}{x+1}$
Now $x+1(=x+2-1)\mid \{(x+2)^{2n-1}-1\} $ as $(a-b)\mid (a^n-b^n)--->(1)$
So, $(x+1)\mid f(x).$
$\frac{(x+1)^{2n}+(x+2)^n-1}{x+2}=\frac{\{(x+1)^2\}^n-1}{x+2}+(x+2)^{n-1}$
Using $(1),\{(x+1)^2\}^n-1$ is divisible by $(x+1)^2-1=x^2+2x=x(x+2)$
So, $(x+2)\mid f(x).$
$lcm(x+1,x+2)=(x+1)(x+2)=x^2+3x+2$ (prove this)
|
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"timestamp": "2023-03-29T00:00:00",
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|
Induction: $\sum_{k=0}^n \binom nk k^2 = n(1+n)2^{n-2}$ I found crazy (for me at least) induction example, in fact it just would be nice to prove. (Even have problems with starting) Any hints are highly valued: $$0^2\binom{n}{0}+1^2\binom{n}{1}+2^2\binom{n}{2}+\cdots+n^2\binom{n}{n}=n(1+n)2^{n-2} $$
|
Although a year late, here is a proof using the properties
$$
k\binom{n}{k}=n\binom{n-1}{k-1}\tag{1}
$$
and
$$
\sum_{k=0}^n\binom{n}{k}=2^n\tag{2}
$$
Thus,
$$
\begin{align}
\sum_{k=0}^nk^2\binom{n}{k}
&=\sum_{k=0}^nnk\binom{n-1}{k-1}\tag{3}\\
&=\sum_{k=0}^nn\binom{n-1}{k-1}+\sum_{k=0}^nn(k-1)\binom{n-1}{k-1}\tag{4}\\
&=\sum_{k=0}^nn\binom{n-1}{k-1}+\sum_{k=0}^nn(n-1)\binom{n-2}{k-2}\tag{5}\\
&=n2^{n-1}+n(n-1)2^{n-2}\tag{6}\\[9pt]
&=n(n+1)2^{n-2}\tag{7}
\end{align}
$$
Explanation:
$(3)$: apply $(1)$
$(4)$: $k=1+(k-1)$
$(5)$: apply $(1)$
$(6)$: apply $(2)$ to both sums
$(7)$: $2n+n(n-1)=n(n+1)$
|
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|
Methods to solve a system of many Ax=B equations using least-squares I am working with a force measurement instrument which needs calibration via a calibration matrix. For each of a set of controlled measurements I have a vector $k$ of three known, independent values, and a corresponding vector $m$ containing three measured values, that due to instrument constraints end up being non-orthogonal.
From each measurement, I can assemble an equation in the form
$Cm = k$
where $C$ is a 3x3 calibration matrix for that measurement, found by solving the system.
My goal is to find some calibration matrix that is the least-squares best fit for ALL the measurements in the set of measurements. That would mean I would have a system like:
$Cm_1 = k_1\\
Cm_2 = k_2\\
...\\
Cm_n = k_n$
that I would have to solve for C, and that is my question: how do I solve such a system, where I have a MATRIX as the unknown?
I am not familiar with heavy math notation, and I plan to solve this using Python (Numpy/Scipy), so I'm looking more for a theoretical basis and proper nomenclature of which kind of procedure I should use, and then I could figure out how to implement it numerically.
Any help is much appreciated, thanks for reading!
|
Let's denote $C$ as
\[ C = \begin{pmatrix} c_{11} & c_{12} & c_{13}\\ c_{21} & c_{22} & c_{23} \\ c_{31} & c_{32} & c_{33} \end{pmatrix} \]
We can write the equation $Cm_i = k_i$ as
\begin{align*}
m_{i1} c_{11} + m_{i2}c_{12} + m_{i3} c_{13} &= k_{i1}\\
m_{i1} c_{21} + m_{i2}c_{22} + m_{i3} c_{23} &= k_{i2}\\
m_{i1} c_{31} + m_{i2}c_{32} + m_{i3} c_{33} &= k_{i3}
\end{align*}
or
$$
\begin{pmatrix} m_{i1} & m_{i2} & m_{i3} & 0 & 0 &0 & 0 & 0 &0\\
0 & 0 & 0 & m_{i1} & m_{i2} & m_{i3} & 0 & 0 & 0\\
0 & 0 &0 & 0 & 0 &0 & m_{i1} & m_{i2} & m_{i3}\end{pmatrix}
\begin{pmatrix} c_{11} \\ c_{12}\\c_{13}\\ c_{21} \\ c_{22} \\ c_{23} \\ c_{31} \\ c_{32} \\c_{33} \end{pmatrix} = \begin{pmatrix} k_{i1} \\ k_{i2} \\ k_{i3}\end{pmatrix}
$$
Doing this for all $i$ gives
$$
\begin{pmatrix} m_{11} & m_{12} & m_{13} & 0 & 0 &0 & 0 & 0 &0\\
0 & 0 & 0 & m_{11} & m_{12} & m_{13} & 0 & 0 & 0\\
0 & 0 &0 & 0 & 0 &0 & m_{11} & m_{12} & m_{13}\\
& \vdots & &&&&& \vdots \\
m_{n1} & m_{n2} & m_{n3} & 0 & 0 &0 & 0 & 0 &0\\
0 & 0 & 0 & m_{n1} & m_{n2} & m_{n3} & 0 & 0 & 0\\
0 & 0 &0 & 0 & 0 &0 & m_{n1} & m_{n2} & m_{n3}\end{pmatrix}
\begin{pmatrix} c_{11} \\ c_{12}\\c_{13}\\ c_{21} \\ c_{22} \\ c_{23} \\ c_{31} \\ c_{32} \\c_{33} \end{pmatrix} = \begin{pmatrix} k_{11} \\ k_{12} \\ k_{13} \\ \vdots \\ k_{n1} \\ k_{n2} \\ k_{n3}\end{pmatrix}
$$
Now apply your usual least squares method.
|
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|
absolute value integrated For example I have the function $f(x) = |x^2-1| = \sqrt{(x^2-1)^2}$
$\int \sqrt{(x^2-1)^2}dx = \frac{x(x^2-3)\sqrt{(x^2-1)^2)}}{3 (x^2-1)}+constant$
But plotted it looks like this: Plot 1. There are values < 0.
I have to take the absolute value again to get the correct function.
Plot 2 Why do I have to do it again? I integrated $|x^2-1|$ and not $x^2-1$
Edit: Part 2
|
$x^2-1$ is positive outside of the interval $[-1,1]$.
Therefore,
$$|x^2-1| = \left\{ \begin{array}{rl} -x^2+1, & -1 \le x \le 1, \\ x^2-1, & \textrm{ otherwise}.\end{array}\right.$$
Or more completely
$$|x^2-1| = \left\{ \begin{array}{rl} x^2-1, & \textrm{ if } x < -1, \\
-x^2+1, & \textrm{ if } -1 \le x \le 1, \\ x^2-1, & \textrm{ if } x > 1.\end{array}\right.$$
Then, you integrate. We'll put the constant of integration first, because there's some trickeration involved.
$$\int |x^2-1|\ dx = C + \left\{ \begin{array}{rl} \frac{1}{3}x^3-x+c_1, & \textrm{ if } x < -1 \\ -\frac{1}{3}x^3+x+c_2, & \textrm{ if } -1 \le x \le 1,\\ \frac{1}{3}x^3-x, & \textrm{ if } x > 1. \end{array}\right.$$
Now, what are $c_1$ and $c_2$? Well, if we omitted those, then we would have discontinuities at the points $x=-1$ and $x=1$. So we need to account for a "shift" to make sure the functions are continuous, and indeed, continuous in the first derivative.
Why is there no $c_3$? Because I put the constant of integration first, and we have to choose one branch to be the "anchor". I could have put $c_1$ and $c_2$ on any branch, but I wrote it like this for convenience.
The derivatives are easy to check because the constant terms get blown out. So we must have $\frac{d}{dx}\left(\frac{1}{3}x^3-x\right) = \frac{d}{dx}\left(-\frac{1}{3}x^3+x+c_2\right)$ at $x=1$. This becomes
$$x^2-1 = -x^2+1 \implies 1^2-1=-1^2+1 \implies 0 = 0$$
which is true, so our derivative is continuous.
Then, we must have $\frac{1}{3}x^3-x = -\frac{1}{3}x^3+x+c_2$ at $x=1$. This means that $$\frac{1}{3}-1 = -\frac{1}{3}+1 +c_2 \implies c_2 = \frac{2}{3}-2 = -\frac{4}{3}.$$
Repeating this process, we want
$$\frac{1}{3}x^3-x+c_1 = -\frac{1}{3}x^3+x-\frac{4}{3}$$
at $x = -1$. We can solve again for $c_1$ to obtain
$$\frac{1}{3}-1+c_1 = -\frac{1}{3}+1-\frac{4}{3} \implies c_1 = 0.$$
Testing for continuity of the derivatives is the same as before.
This is exactly what W|A gives, except my result is shifted a bit. But that's OK, because that difference gets thrown into the integration constant, $C$.
In general, the interpretation $|x| = \sqrt{x^2}$, while correct, will sometimes only get you so far. In particular, it is possible to compute different symbolic anti-derivatives of a function such that they are equivalent, and in fact equal, over a subset of the domain of $x$, but not equivalent/not equal over other sub-domains. In such a case, it is erroneous to say that an answer that works on a specific subdomain is the "right answer." In fact, it is only the right answer precisely where it is the right answer, and nowhere else.
In these situations, we need to look at other possible interpretations, if we hope to get an answer that is the "right answer" in more places -- and perhaps everywhere. In this case, the global right answer (that is, the answer that is correct over all of the reals) is the piecewise solution.
|
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|
Decompose a Matrix into a Sum of Tensor Products Given the matrix
$$\phantom{-}
\pmatrix{
0 & a+b& -b+a&0\\
\phantom{-}a+c& a+d&\phantom{-} b+c&b+d\\
-c+a&c+b&-d+a&d+b\\
0&c+d&\phantom{-}d-c&0\\
}.
$$
How could one decompose it into the (smallest) sum of tensor product, like
$M\otimes E_{k}$ and $E_{n}\otimes M$, where $M=\pmatrix{a&b\\c&d}$ and $E_j$ is a matrix with entries $0$ or $\pm 1$, like $\pmatrix{0&1\\0&0}$ or $\pmatrix{1&\phantom{-}1\\1&-1}$.
I tried it with sums of $M\otimes 1_{kn}$ and $1_{kn}\otimes M$ ($1_{kn}$ has a $1$ at element $(k,n)$ and $0$ elsewhere), but haven't found a solution, yet...
In order to minimise the number of summands, feel free to substitue every element by its negative (if a solution exists).
|
You cannot decompose $$A=\phantom{-}
\pmatrix{
0 & a+b& -b+a&0\\
\phantom{-}a+c& a+d&\phantom{-} b+c&b+d\\
-c+a&c+b&-d+a&d+b\\
0&c+d&\phantom{-}d-c&0\\
}
$$
into $X\otimes M+M\otimes Y$. The system $A=X\otimes M+M\otimes Y$ for $X=\pmatrix{x_1&x_2\\x_3&x_4}, \ Y=\pmatrix{y_1&y_2\\y_3&y_4}$ does not have a solution in $x_i,y_i$.
For, $X\otimes M+M\otimes Y=
\pmatrix{
x_1a+ay_1 & x_1b+ay_2& x_2a+by_1&x_2b+by_2\\
x_1c+ay_3& x_1d+ay_4& x_2c+by_3&x_2d+by_4\\
x_3a+cy_1&x_3b+cy_2&x_4a+dy_1&x_4b+dy_2\\
x_3c+cy_3&x_3d+cy_4&x_4c+dy_3&x_4d+dy_4\\
},$ where $x_i,y_i \in \mathbb{Z}.$ The elements $(3,4)$ and $(4,3)$ of this matrix give $x_4=1$ and $x_4=-1↯.$
However, for $E=\pmatrix{1&1\\1&1}$ you have
$$E\otimes M-M\otimes E=\phantom{-}
\pmatrix{
0 & -a+b& -b+a&0\\
-a+c& -a+d& -b+c&-b+d\\
-c+a&-c+b&-d+a&-d+b\\
0&-c+d&-d+c&0\\
}.
$$
|
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|
A couple of formulas for $\pi$ While I was studying sums of polygonal numbers I discovered a couple of formulas for $\pi$. Most of the formulas I found were already known, but I can't seem to find any references to the following four:
$$\pi=\sum_{n=1}^\infty \frac{3}{n(2n-1)(4n-3)},$$
$$\pi=\sum_{n=1}^\infty \frac{3\sqrt{3}}{(3n-1)(3n-2)},$$
$$\pi=4\sqrt{3}\sum_{n=1}^\infty \frac{12n-5}{8n(2n-1)(3n-1)(6n-5)},$$
$$\pi=16\sum_{n=1}^\infty \frac{864n(n-1)+226}{(12n-1)(12n-5)(12n-7)(12n-11)(4n-1)(4n-3)}.$$
Are the above formulas known? I have looked here and here. Because of their simplicity, I find it hard to believe that the first two formulas are unknown.
|
The first formula may be rewritten by changing the lower limit of the summation:
$$\begin{align}
\frac{\pi}{3}
&=\sum_{n=1}^\infty \frac{1}{n(2n-1)(4n-3)} \\
&=\sum_{n=0}^\infty \frac{1}{(n+1)(2(n+1)-1)(4(n+1)-3)} \\
&=\sum_{n=0}^\infty \frac{1}{(n+1)(2n+1)(4n+1)} \\
\end{align}$$
This last equation is given by Lehmer in page 139 here.
A similar one that uses four factors is
$$\pi-3=\sum_{k=0}^\infty \frac{24}{(4k+2)(4k+3)(4k+5)(4k+6)}$$
(see Series and integrals for inequalities and approximations to $\pi$)
A formula which is similar to the third one you give is:
$$\pi=72 \sqrt{3} \sum_{k=0}^\infty \frac{1}{(6 k+1) (6 k+2) (6 k+4) (6 k+5)}$$
which has all terms positive and constant numerator. Taking the first term of the series out of the summation, this proves $\pi>\frac{9}{5}\sqrt{3}\approx 3.12$.
There is also
$$\pi=\frac{23040(17+9 \sqrt{3})}{23} \sum_{k=0}^\infty \frac{1}{(12 k+1) (12 k+3) (12 k+5) (12 k+7) (12 k+9) (12 k+11)}$$
with the denominators of your fourth formula. The first term in this series leads to the approximation
$$\pi \approx \frac{2^9(17+9\sqrt{3})}{5313}=\frac{2^{10}((2^4+1)+(2^3+1)\sqrt{3})}{21·22·23} = \frac{2^{10}}{231(17-9\sqrt{3})}\approx 3.140$$
Using $163$ instead of $\frac{5313}{17+9\sqrt{3}} \approx 163.033$ gets one more correct digit. The approximation
$$\pi \approx \frac{2^9}{163} \approx 3.1411$$
is due to Stoschek (see http://mathworld.wolfram.com/PiApproximations.html)
Also compare to $\pi \approx \frac{2^{13}}{3(-383+560\sqrt{5})}$ given by T.Piezas here.
|
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|
Chebyshev polynomials of first kind I know the chebyshev polynomials of the first kind can be approximated using the cosine function, where $T_n(\cos \theta)=\cos(n \theta)$ and I know that chebyshev polynomials are a family of orthogonal polynomials. How would I prove that $$\int_{-1}^1 \frac{T_h(x)T_i(x)}{\sqrt{1-x^2}}dx=0, h \neq i$$
I can use $x=\cos(\theta)$, but I don't understand how to prove, and what this really means. Isn't it trivial because they're orthogonal?
Can anyone clarify?
Thank you.
|
Using Tschebyscheff differential equation in it's autoadjoint form we have, for $T_n$ and $T_m$
\begin{align}
\frac{d}{dx}\left\{\sqrt{1-x^2} T_n'\right\} + \frac{n^2}{\sqrt{1-x^2}} T_n &= 0 \\
\frac{d}{dx}\left\{\sqrt{1-x^2} T_m'\right\} + \frac{m^2}{\sqrt{1-x^2}} T_m &= 0
\end{align}
Multiplying the first equation by $T_m$ and the second by $T_n$ and substracting them
\begin{align}
\frac{d}{dx}\left\{\sqrt{1-x^2} T_n'\right \}T_m - \frac{d}{dx}\left\{\sqrt{1-x^2} T_m'\right\}T_n + \frac{n^2 - m^2}{\sqrt{1-x^2}} T_m T_n &= 0 \\
\frac{d}{dx}\left\{\sqrt{1-x^2} (T_n'T_m - T_m'T_n)\right\} + \frac{n^2 - m^2}{\sqrt{1-x^2}} T_m T_n &= 0
\end{align}
Now, the wronskian $W(T_n,T_m)$ is
$$
W(T_n,T_m) = T_n T_m' - T_n' T_m = \frac{m \cos(n \theta) \sin(m \theta) - n \sin(n \theta) \cos (m\theta)}{\sqrt{1-x^2}}
$$
and then
$$
\int_{-1}^1 \frac{T_n(x) T_m(x)}{\sqrt{1-x^2}} dx = \int_{-1}^1 \frac{d}{d x} \left\{\frac{m \cos(n \theta) \sin(m \theta) - n \sin(n \theta) \cos (m\theta)}{n^2 - m^2}\right\}dx
$$
If $n \neq m$, it's easy to see that the integral is zero. If $n \to m$, we have that
$$
\lim_{n \to m} \frac{m \cos(n \theta) \sin(m \theta) - n \sin(n \theta) \cos (m\theta)}{n^2 - m^2} = \begin{cases}-\frac{\theta}{2}, & m \neq 0, \\ \\ - \theta, & m = 0 \end{cases}
$$
and then
$$
\int_{-1}^1 \frac{T_n(x) T_m(x)}{\sqrt{1-x^2}} dx = \begin{cases} 0, & m \neq n, \\ \\
\frac{\pi}{2} & m = n \neq 0, \\ \\ \pi & m = n = 0. \end{cases}
$$
|
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|
If $(a, b, c)$ is a Pythagorean triple, then $(a+1, b+1, c+1)$ is not a Pythagorean triple. Pythagoras stated that there exist positive natural numbers, $a$, $b$ and $c$ such that $a^2+b^2=c^2$. These three numbers, $a$, $b$ and $c$ are collectively known as a
Pythagorean triple. For example, $(8, 15, 17)$ is one of these triples as $8^2 + 15^2 = 64 + 225= 289 = 17^2$. Other examples of this triple are $(3, 4, 5)$ and $(5, 12, 13)$.
Using Proof by Contradiction, show that: If $(a, b, c)$ is a Pythagorean triple, then $(a+1, b+1, c+1)$ is not a Pythagorean triple.
|
Suppose $a^2+b^2=c^2$ and $(a+1)^2+(b+1)^2=(c+1)^2$ both hold.
Simplifying the second equation and subtracting the first gives $$2a + 2b = 2c - 1$$ but this is impossible sso they can't both hold!
hint: think about even and odd numbers to complete the proof.
|
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Solve the recurrence $T(n) = 2T(n-1) + n$ Solve the recurrence $T(n) = 2T(n-1) + n$
where $T(1) = 1$ and $n\ge 2$.
The final answer is $2^{n+1}-n-2$
Can anyone arrive at the solution?
|
By subtraction
$$
\begin{array}{rrcl}
& T(n)-2T(n-1) &=& n \\
& T(n-1) - 2T(n-2) &=& n-1 \\
\hline
& T(n)-3T(n-1)+2T(n-2) &=& 1
\end{array}$$
Again, by subtraction
$$
\begin{array}{rrcl}
& T(n)-3T(n-1)+2T(n-2) &=& 1 \\
& T(n-1)-3T(n-2)+ 2T(n-3) &=& 1 \\
\hline
& T(n)-4T(n-1)+5T(n-2)-2T(n-3) &=& 0
\end{array}$$
Characteristic equation of the recursion is
$$x^3-4x^2+5x-2 = 0$$
Roots of the equation are $x_1 = x_2 = 1$ and $x_3 = 2$
So, general solution of the recursion is $$T(n) = C_1\cdot 1^n + C_2\cdot n\cdot 1^n + C_3\cdot 2^n$$ or $$T(n) = C_1 + C_2\cdot n + C_3\cdot 2^n$$
now, from $T(1) = 1$ we get $T(2) = 4$ and $T(3) = 11$.
So, we get system
$$
\begin{array}{ccccccc}
C_1 &+&C_2 &+& 2C_3 &=& 1 \\
C_1 &+&2C_2&+&4C_3 & =& 4 \\
C_1 & +& 3C_2 &+& 8C_3 &=& 11
\end{array}$$
From this system we get that $C_1 = -2$, $C_2 =-1$, and $C_3 = 2$, so particular solution of the recursion is $$T(n) =2\cdot2^n-n-2,$$ or $$T(n) = 2^{n+1}-n-2.$$
|
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|
Proving an equation with invertible matrices I need to prove that if $A$, $B$ and $(A + B)$ are invertible then $(A^{-1} + B^{-1})^{-1} = A(A + B)^{-1}B$
I'm a bit lost with this one,
I can't find a way to make any assumptions about $(A^{-1} + B^{-1})$,
Neither by using $A^{-1}$, $B^{-1}$ and $(A + B)^{-1}$.
If someone could clue me in I'll be grateful.
UPDATE:
Thanks for all of your input - it really helped,
I got a solution, I'd like to know if my way of proving it is valid:
$(A^{-1} + B^{-1})^{-1} = A(A + B)^{-1}B$
multiplying both sides by $A^{-1}$ and $B^{-1}$
$A^{-1}(A^{-1} + B^{-1})^{-1}B^{-1} = (A+B)^{-1}$
now multiplying both sides by $(A + B)$, I get to
$(A+B)(A^{-1}(A^{-1} + B^{-1})^{-1}B^{-1}) = I$
Lets call $(A^{-1}(A^{-1} + B^{-1})^{-1}B^{-1})$ -> $C$
Now I determine that $C$ is $(A+B)^{-1}$,
because $(A+B)C$ equals the Identity Matrix.
So lastly to verify that I place $C$ in the original equation:
$(A^{-1} + B^{-1})^{-1} = A(A^{-1}(A^{-1} + B^{-1})^{-1}B^{-1})B$
and from this I get:
$(A^{-1} + B^{-1})^{-1} = (A^{-1} + B^{-1})^{-1}$
|
How about a purposedly complicated proof, but without matrix inversion?
Let $X = A(A + B)^{-1}B$. By writing the rightmost $B$ as $(A+B)-B$, we get
$$X = A(A + B)^{-1}[(A+B)-A] = \underbrace{A - A(A + B)^{-1}A}_{X_1}.$$
Similarly, if we write the leftmost $A$ as $(A+B)-B$ instead, we get
$$X = [(A+B)-B](A + B)^{-1}B = \underbrace{B - B(A + B)^{-1}B}_{X_2}.$$
Let $Y=A^{-1} + B^{-1}$. Then
\begin{align}
XY &= X(A^{-1} + B^{-1})\\
&= X_1 A^{-1} + X_2B^{-1}\\
&= [I - A(A + B)^{-1}] + [I - B(A + B)^{-1}]\\
&= 2I - (A+B)(A + B)^{-1} = I.
\end{align}
So $XY=I$ and we are done.
|
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|
Solve $x^2 + 10 = 15$ How do I solve the following equation?
$$x^2 + 10 = 15$$
Here's how I think this should be solved.
\begin{align*}
x^2 + 10 - 10 & = 15 - 10 \\
x^2 & = 15 - 10 \\
x^2 & = 5 \\
x & = \sqrt{5}
\end{align*}
I was thinking that the square root of 5 is iregular repeating 2.23606797749979 number. 2.236 multipled by itself equals 5ish.
I've also seen another equation like this:
\begin{align*}
x^2 & = 4 \\
x^2 + 4 & = 0 \\
(x - 2)(x + 2) & = 0 \\
x & = 2 \text{ or } -2
\end{align*}
So I guess I could near the end of my equation do the following:
$$x^2 + 5 = 0$$
and then go from there?
Is my first attempt at solving correct?
|
The idea is there, however be aware of the fact that $x^2 = a$ is equivalent to $x = \sqrt{a}$ OR $x = -\sqrt{a}$.
|
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|
How do you prove a Quadratic Residue? Help Someone please.
Prove or disprove: 28 is a quadratic residue (mod 65).
|
Recall that if $a$ and $m$ are relatively prime, then $a$ is a QR of $m$ if the congruence
$$x^2\equiv a \pmod{m}$$
has a solution.
If the congruence $x^2\equiv 28\pmod{65}$ has a solution, then so do the congruences $x^2\equiv 28\pmod{5}$ and $x^2\equiv 28\pmod{13}$. Indeed the converse also holds, by the Chinese Remainder Theorem.
Let's check whether the congruence $x^2\equiv 28\pmod{5}$ has a solution. So we are asking whether $x^2\equiv 3\pmod{5}$ has a solution. Try everything, there are very few things to try. Modulo $5$, $1^2\equiv 1$, $2^2\equiv 4$, $3^2\equiv 4$, $4^2\equiv 1$. No solution to $x^2\equiv 3\pmod{5}$, and therefore no solution to $x^2\equiv 28\pmod{65}$!
Remark: For substantially bigger moduli, we need theory, lots of it. There are good algorithms for determining whether $a$ is a QR modulo a prime.
|
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|
What am I doing wrong in calculating this determinant? I have matrix:
$$
A = \begin{bmatrix}
1 & 2 & 3 & 4 \\
2 & 3 & 3 & 3 \\
0 & 1 & 2 & 3 \\
0 & 0 & 1 & 2
\end{bmatrix}
$$
And I want to calculate $\det{A}$, so I have written:
$$
\begin{array}{|cccc|ccc}
1 & 2 & 3 & 4 & 1 & 2 & 3 \\
2 & 3 & 3 & 3 & 2 & 3 & 3 \\
0 & 1 & 2 & 3 & 0 & 1 & 2 \\
0 & 0 & 1 & 2 & 0 & 0 & 1
\end{array}
$$
From this I get that:
$$
\det{A} = (1 \cdot 3 \cdot 2 \cdot 2 + 2 \cdot 3 \cdot 3 \cdot 0 + 3 \cdot 3 \cdot 0 \cdot 0 + 4 \cdot 2 \cdot 1 \cdot 1) - (3 \cdot 3 \cdot 0 \cdot 2 + 2 \cdot 2 \cdot 3 \cdot 1 + 1 \cdot 3 \cdot 2 \cdot 0 + 4 \cdot 3 \cdot 1 \cdot 0) = (12 + 0 + 0 + 8) - (0 + 12 + 0 + 0) = 8
$$
But WolframAlpha is saying that it is equal 0. So my question is where am I wrong?
|
Sarrus's rule works only for $3\times 3$-determinants. So you have to find another way to compute $\det A$, for example you can apply elementary transformations not changing the determinant, that is e. g. adding the multiple of one row to another:
\begin{align*}
\det \begin{bmatrix}
1 & 2 & 3 & 4 \\
2 & 3 & 3 & 3 \\
0 & 1 & 2 & 3 \\
0 & 0 & 1 & 2
\end{bmatrix} &=
\det \begin{bmatrix}
1 & 2 & 3 & 4 \\
0 & -1 & -3 & -5 \\
0 & 1 & 2 & 3 \\
0 & 0 & 1 & 2
\end{bmatrix}\\
&=
\det \begin{bmatrix}
1 & 2 & 3 & 4 \\
0 & -1 & -3 & -5 \\
0 & 0 & -1 & -2 \\
0 & 0 & 1 & 2
\end{bmatrix}\\
&= \det \begin{bmatrix}
1 & 2 & 3 & 4 \\
0 & -1 & -3 & -5 \\
0 & 0 & -1 & -2 \\
0 & 0 & 0 & 0
\end{bmatrix}
\end{align*}
To compute the determinant of a triagonal matrix, we just have to multiply the diagonal elements, so
$$ \det A = \det \begin{bmatrix}
1 & 2 & 3 & 4 \\
0 & -1 & -3 & -5 \\
0 & 0 & -1 & -2 \\
0 & 0 & 0 & 0
\end{bmatrix} = 1 \cdot (-1)^2 \cdot 0 = 0.
$$
|
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|
p(n) is count of all n-digit numbers... Let $n$ be an arbitrary positive integer, and $p(n)$ be the number of $n$-digit numbers which consist only of the digits $1,2,3,4,5$, and in which each two neighbour digits differ by $2$ or more.
My task is to prove that this inequality is true for any positive integer $n$:
$$\ 5 \cdot (2,4)^{n-1} ≤ p(n) ≤ 5 \cdot (2,5)^{n-1} $$
Any ideas?
|
Just an idea (not a complete answer). Let $p_k(n)$ be the number of $n$-digit numbers of the requested form that end in the digit $k$. Then $p_k(1) = 1$ for all $k \in \{1, \dotsc, 5\}$ and $p_k(n+1)$ can be computed recursively by
$$
\begin{pmatrix}
p_1(n+1) \\
p_2(n+1) \\
p_3(n+1) \\
p_4(n+1) \\
p_5(n+1)
\end{pmatrix} = \begin{pmatrix}
0 & 0 & 1 & 1 & 1 \\
0 & 0 & 0 & 1 & 1 \\
1 & 0 & 0 & 0 & 1 \\
1 & 1 & 0 & 0 & 0 \\
1 & 1 & 1 & 0 & 0
\end{pmatrix} \begin{pmatrix}
p_1(n) \\
p_2(n) \\
p_3(n) \\
p_4(n) \\
p_5(n)
\end{pmatrix}
$$
The largest eigenvalue of the matrix on the left is $\approx 2.48$ for a strictly positive eigenvector. Not all other eigenvalues have norm less than $1$ though, so a more refined argument is still needed.
|
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|
Why does $\frac{a}{\frac{b}{x}} = x \times \frac{a}{b}$? As much as it embarasses me to say it, but I always had a hard time understanding the following equality:
$$
\frac{a}{\frac{b}{x}} = x \times \frac{a}{b}
$$
I always thought that the left-hand side of the above equation was equivalent to
$$
\frac{a}{\frac{b}{x}} = \frac{a}{b} \div \frac{x}{1} = \frac{a}{b} \times \frac{1}{x}
$$
What am I doing wrong, here?
|
$\frac{a}{\frac{b}{x}}$ means a divided by $\frac{b}{x}$.
Note that
$$\frac{b}{x}\frac{x}{b} =1$$
This means that
$$a \frac{b}{x}\frac{x}{b} = a$$
Now divide both sides by $\frac{b}{x}$ and you get
$$a\frac{x}{b} = \frac{a}{ \frac{b}{x}}$$
The mistake you make is confusing $\frac{a}{\frac{b}{x}}$ with $\frac{\frac{a}{b}}{x}$. In general
$$\frac{a}{\frac{b}{x}}\neq \frac{\frac{a}{b}}{x}$$
as they have different meanings.
|
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|
Show $2(x+y+z)-xyz\leq 10$ if $x^2+y^2+z^2=9$ If $x,y,z$ are real and $x^2+y^2+z^2=9$, how can we prove that $2(x+y+z)-xyz\leq 10$?
Please provide a solution without the use of calculus. I know the solution in that way.
|
without loss of generality we can assume that $|x|\le |y|\le |z|$,and Note that $3z^2\ge x^2+y^2+z^2$, which implies that $z^2\ge 3$
we have
\begin{align}
[2(x+y+z)-xyz]^2&=[2(x+y)+z(2-xy)]^2\\
&\le [(x+y)^2+z^2][4+(2-xy)^2]\\
&=(9+2xy)(8-4xy+(xy)^2))\\
&=2(xy)^3+(xy)^2-20(xy)+72\\
&=(xy+2)^2(2xy-7)+100
\end{align}
from $z^2\ge 3$ it follows that $2xy\le x^2+y^2=9-z^2\le 6$,then
$$[2(x+y+z)-xyz]^2\le 100$$
or equivalently
$$2(x+y+z)-xyz\le 10$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Property of Fejer kernel Let
$$
F_n(x) = \frac{1}{n} \left( \frac{ \sin(\frac{1}{2} n x ) } { \sin(\frac{1}{2} x ) } \right)^2
$$
be the n-th Fejer-Kernel. Then
$$
\forall \epsilon > 0, r < \pi : \exists N \in \mathbb{N} : \forall n \ge N : \int_{[-\pi,\pi] \setminus [-r, r]} F_n(t) \, \mathrm{d} t < \epsilon
$$
The proof of this goes like this
\begin{align*}
\int_{[-\pi,\pi] \setminus [-r, r]} F_n(t) \, \mathrm{d} t
& = \int_{[-\pi,\pi] \setminus [-r, r]} \frac{1}{n} \left( \frac{ \sin(\frac{1}{2} n x ) } { \sin(\frac{1}{2} x ) } \right)^2 \, \mathrm{d} t \\
& \le \frac{1}{n} \int_{[-\pi,\pi] \setminus [-r, r]} \frac{1}{\sin^2(\frac{1}{2}t)} \, \mathrm{d} t \\
& \le \frac{1}{n} \int_{[-\pi,\pi]} \frac{1}{\sin^2(\frac{1}{2}r)} \, \mathrm{d} t \qquad (*) \\
& = \frac{2\pi}{n} \frac{1}{\sin^2(\frac{1}{2}r)} \\
& < \epsilon \quad \textrm{for certain } ~ n \ge N.
\end{align*}
The step i marked with (*) is totally unclear to me, why could there replaced $t$ by $r$, because it need not to be the case that $\sin^2(0.5*r) < \sin^2(0.5*t)$ as i think?
|
Note that you're integrating over the set $r < |x| \le \pi$.
Let $f(x) = \sin^2(\frac{1}{2}x)$. Differentiating, we get that $f'(x) = \frac{\sin(x)}{2}$. Now if $x \in (0,\pi]$, then $\sin(x) > 0$, and so $f$ is increasing on $(0,\pi]$. Thus $\sin^2(\frac{1}{2} r) < \sin^2(\frac{1}{2}x)$ for $0 < r < x \le \pi$. Similarly, we see that since $\sin(x) < 0$ for $x \in [-\pi, 0)$, so $f$ is decreasing on this interval. So if you pick some $-\pi \ge x > -r > 0$, it follows that $\sin^2(\frac{1}{2}r) < \sin^2(\frac{1}{2}x)$, which gives the desired inequality.
It may be somewhat instructive to look at a graph
You can easily see that the inequality should hold over the desired interval. The idea of checking this using calculus is a standard tool for these sorts of estimates.
|
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|
Can't argue with success? Looking for "bad math" that "gets away with it" I'm looking for cases of invalid math operations producing (in spite of it all) correct results (aka "every math teacher's nightmare").
One example would be "cancelling" the 6's in
$$\frac{64}{16}.$$
Another one would be something like
$$\frac{9}{2} - \frac{25}{10} = \frac{9 - 25}{2 - 10} = \frac{-16}{-8} = 2 \;\;.$$
Yet another one would be
$$x^1 - 1^0 = (x - 1)^{(1 - 0)} = x - 1\;\;.$$
Note that I am specifically not interested in mathematical fallacies (aka spurious proofs). Such fallacies produce shockingly wrong ends by (seemingly) valid means, whereas what I am looking for all cases where one arrives at valid ends by (shockingly) wrong means.
Edit: fixed typo in last example.
|
The problem : solve the equation $$e^{(x-2)}+e^{(x+8)}=e^{(4-x)}+e^{(3x+2)}$$
Incorrect method
As everybody knows $$e^a+e^b=e^{ab}$$
So let's apply it to solve the equation :
$\begin{array}{lll}
e^{(x-2)}+e^{(x+8)}=e^{(4-x)}+e^{(3x+2)} & \iff & e^{(x-2)(x+8)}=e^{(4-x)(3x+2)} \\
& \iff & (x-2)(x+8)=(4-x)(3x+2) \\
& \iff & x^2+6x-16=10x-3x^2+8 \\
& \iff & 4x^2-4x-24=0 \\
& \iff & x^2-x-6=0 \\
& \iff & (x+2)(x-3)=0 \\
\end{array}$
Thus $x=-2$ or $x=3$.
Correct method:
Let's set $X=e^x$ and $a=e^2$ and substitute
$\frac Xa+a^4X=\frac {a^2}X+aX^3\iff X^2+a^5X^2=a^3+a^2X^4\iff(a^2X^2-1)(X^2-a^3)=0$
Leading also to $x=-2$ and $x=3$.
The problem : simplify the expression $\sin(x+y)\sin(x-y)$
Incorrect method
As everybody knows $$f(a+b)f(a-b)=f(a)^2-f(b)^2$$
Applying the formula gives immediately
$\sin(x+y)\sin(x-y)=\sin^2(x)-\sin^2(y)$
Correct method
$\begin{array}{lll}
\sin(x+y)\sin(x−y) &=& \big[\sin(x)\cos(y)+\cos(x)\sin(y)\big]\ \big[\sin(x)\cos(y)−\cos(x)\sin(y)\big]\\
&=&\sin^2(x)\cos^2(y)−\cos^2(x)\sin^2(y)\\
&=&\sin^2(x)(1−\sin^2(y))−(1−\sin^2(x))\sin^2(y)\\
\require{cancel}&=&\sin^2(x)−\cancel{\sin^2(x)\sin^2(y)}−\sin^2(y)+\cancel{\sin^2(x)\sin^2(y)}\\
&=&\sin^2(x)−\sin^2(y)\\
\end{array}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
What kind of software can be used to solves systems of equations? For example, I have to solve the following equations:
$$\left\{\begin{align*}
&x^2 + y^2 + z^2 = 1\\
&Ax + By + Cz = 0
\end{align*}\right.$$
for $y$ and $z$, where $x$, $A$, $B$ and $C$ are known values. I need this for production. I guess I could solve this by hand, but there are many other of those and, if did all manually, my production would stop. Is there a software that solves this kind of situation?
|
Hot off the Mathematica presses:
> Solve[{x^2 + y^2 + z^2 - 1 == 0, A x + B y + C z == 0}, {y, z}] // Simplify
$$\begin{align}
\Bigg\{\Bigg\{y & \to -\frac{A B x+\sqrt{-C^2 \left(A^2 x^2+B^2 \left(-1+x^2\right)+C^2 \left(-1+x^2\right)\right)}}{B^2+C^2}, \\
z & \to \frac{-A C^2 x+B \sqrt{-C^2 \left(A^2 x^2+B^2 \left(-1+x^2\right)+C^2 \left(-1+x^2\right)\right)}}{C \left(B^2+C^2\right)}\Bigg\}, \\
\Bigg\{y & \to \frac{-A B x+\sqrt{-C^2 \left(A^2 x^2+B^2 \left(-1+x^2\right)+C^2 \left(-1+x^2\right)\right)}}{B^2+C^2}, \\
z & \to -\frac{A C^2 x+B \sqrt{-C^2 \left(A^2 x^2+B^2 \left(-1+x^2\right)+C^2 \left(-1+x^2\right)\right)}}{C \left(B^2+C^2\right)}\Bigg\}\Bigg\} \\
\end{align}$$
I have intentionally not simplified these equations, so as to give you a feel what Mathematica solutions are like.
|
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|
Fourier transform of $f(x)=\frac{1}{x^2+6x+13}$ How to find the Fourier transform of the following function:
$$f(x)=\frac{1}{x^2+6x+13}$$
|
Breaking down to partial fractions yields
$$
\begin{align}
\frac1{x^2+6x+13}
&=\frac1{(x+3)^2+4}\\
&=\frac1{(x+3-2i)(x+3+2i)}\\
&=\frac1{4i}\left(\frac1{x+3-2i}-\frac1{x+3+2i}\right)
\end{align}
$$
Therefore, for $\xi\ge0$,
$$
\begin{align}
\int_{-\infty}^\infty\frac1{x^2+6x+13}e^{-i2\pi x\xi}\,\mathrm{d}x
&=\frac1{4i}\int_{-\infty}^\infty\left(\frac1{x+3-2i}-\frac1{x+3+2i}\right)e^{-i2\pi x\xi}\,\mathrm{d}x\\
&=\frac1{4i}\int_{\gamma}\left(\frac1{z+3-2i}-\frac1{z+3+2i}\right)e^{-i2\pi z\xi}\,\mathrm{d}z\\
&=\frac{2\pi i}{4i}e^{-i2\pi(-3-2i)\xi}\\
&=\frac\pi2e^{(-4+6i)\pi\xi}\\
&=\frac\pi2e^{(-4|\xi|+6i\xi)\pi}\quad\text{for all }\xi
\end{align}
$$
where $\gamma$ is the curve passing left to right along the real axis then circles back clockwise around the lower half-plane.
Thus, the Fourer Transform is
$$
\frac\pi2e^{-4\pi|\xi|}(\cos(6\pi\xi)+i\sin(6\pi\xi))
$$
|
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|
Prove every odd integer is the difference of two squares I know that I should use the definition of an odd integer ($2k+1$), but that's about it.
Thanks in advance!
|
Proof for odd numbers: Divide any odd number, $x$, in half, then let $C$ be the half rounded up and $B$ be the half rounded down. Then, $C^2 - B^2 = x$. Let's examine this algebraically:
$$C = \frac{x+1}{2}$$
$$B = \frac{x-1}{2}$$
$$C^2 - B^2 = \biggl(\frac{x+1}{2}\biggr)^2 - \biggl(\frac{x-1}{2}\biggr)^2 = \frac{x^2+2x+1}{4} - \frac{x^2-2x+1}{4} = \frac{x^2-x^2+2x+2x+1-1}{4} = \frac{4x}{4} = x$$
Then, because even/odd integers alternate, $x \pm 1$ will always result in an even integer when $x$ is odd and an odd integer when $x$ is even.
Because only even integers are divisible by two (without incurring a decimal), $x \pm 1$ must be even to be divisible by 2, so $x$ must be odd. Otherwise, if $x$ were even, then $C$ and $B$ would not be integers.
Example for odd numbers: Take the number 31. Let $C=\frac{31+1}{2}=16$ and $B=\frac{31-1}{2}=15$. Then, $16^2 - 15^2 = 256 - 225 = 31$.
Proof for all composite numbers: every pair of $x$'s factors, $p$ and $q$, including composite factor pairs, where the oddity of $p$ matches the oddity of $q$, corresponds to a separate representation as the difference of two squares:
$$p \times q = x$$
$$C = \frac{p + q}{2}$$
$$B = \frac{|p - q|}{2}$$
$$C^2 - B^2 = \biggl( \frac{p + q}{2} \biggl)^2 + \biggl( \frac{|p - q|}{2} \biggl)^2 = \frac{p^2 + 2pq + q^2}{4} - \frac{p^2 - 2pq + q^2}{4} = \frac{p^2 - p^2 + q^2 - q^2 + 4pq}{4} = pq = x$$
Example for every-other even numbers: Because half of every-other even number (that is, 4, 8, 12, etc.) has an even oddity, it can be shown that at least every-other even number can be expressed as the difference of two squares. For $x=48$, we get $p=2$ and $q=\frac{x}{p}=24$. Then, $C=\frac{2+24}{2}=13$ and $B=13-2=11$. We verify $C^2 - B^2 = 169 - 121 = 48$.
Conclusion: The only numbers that cannot be expressed as the difference of two squares are the sequence $4n+2$: 2, 6, 10, 14, 18, 22, etc.
|
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|
What is the proper convergence proof for this sequence? Limit of $n^{1/2}(n^{1/n}-1)=0$, as $n$ approaches infinity.
I need a strict math proof that this sequence converges to zero, but without using L'Hôpital's rule, because I am not allowed to use it yet. Thanks in advance.
|
Denote $a_n=n^\tfrac{1}{n}-1, \;\; b_n={\sqrt{n}\left(n^\tfrac{1}{n}-1\right)}.$
Obviously, $a_n \geqslant {0}.$ Rewrite $n^{\tfrac{1}{n}}=1+a_n $ and apply the binomial formula
\begin{gather}n=(1+a_n)^n= \\
=1+na_n+\dfrac{n(n-1)}{2!}a_n^2+\dfrac{n(n-1)(n-2)}{3!}a_n^3+\ldots+a_n^n \geqslant \\
\geqslant \dfrac{n(n-1)(n-2)}{3!}a_n^3.
\end{gather}
For $n\geqslant{4}$
$$n-1 \geqslant \dfrac{n}{2}, \\
n-2 \geqslant \dfrac{n}{2}.
$$
Therefore, $$n \geqslant \dfrac{n^3}{4 \cdot 3!}a_n^3=\dfrac{n^3a_n^3}{24} {\;\;}\Rightarrow {\;\;}a_n^3 \leqslant \dfrac{24}{n^2}.$$
Then ${a_n}\leqslant \dfrac{{\sqrt[3]{24}}}{n^{\tfrac{2}{3}}}$ and $${0} \leqslant{b_n} =\sqrt{n}a_n \leqslant \sqrt{n}\dfrac{\sqrt[3]{24}}{n^{\tfrac{2}{3}}}={\sqrt[3]{24}}n^{-\tfrac{1}{6}} \underset{n\to\infty} {\to} {0}.$$
|
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|
Prove that $\cot^2{(\pi/7)} + \cot^2{(2\pi/7)} + \cot^2{(3\pi/7)} = 5$ Prove that $\cot^2{(\pi/7)} + \cot^2{(2\pi/7)} + \cot^2{(3\pi/7)} = 5$ .
I am sure this is derived from using roots of unity and Euler's complex number function, but I am very uncomfortable in these areas so some help would be great. It is evident that $(a + b + c)^2 - 2(ab + ac + bc) = a^2 + b^2 + c^2$ .
So, using a polynomial of degree 3 and the coefficients on the $x^2$ and $x$ terms will get where we need to be.
|
The following approach is intimately related to Ofir's; you can think of this as the linear algebraic variation of his route.
Consider the perturbed Toeplitz tridiagonal matrix
$$\begin{pmatrix}2&-1&&&\\-1&2&-1&&\\&-1&\ddots&\ddots&\\&&\ddots&2&-1\\&&&-1&1\end{pmatrix}$$
which has the characteristic polynomial
$$(-1)^n\left((x-1)U_{n-1}\left(\frac{x-2}{2}\right)-U_{n-2}\left(\frac{x-2}{2}\right)\right)$$
(where $U_n(x)$ is the Chebyshev polynomial of the second kind) and the eigenvalues
$$\mu_k=4\cos^2\left(\frac{k\pi}{2n+1}\right),\qquad k=1\dots n$$
We find then that the matrix $\mathbf H=4\mathbf I-\mathbf T$ has the eigenvalues
$$\eta_k=4\sin^2\left(\frac{k\pi}{2n+1}\right),\qquad k=1\dots n$$
Now, take the case $n=3$:
$$\mathbf H=\begin{pmatrix}2&1&\\1&2&1\\&1&3\end{pmatrix}$$
We have
$$\mathbf U=\mathbf H^{-1}=\frac17\begin{pmatrix}5&-3&1\\-3&6&-2\\1&-2&3\end{pmatrix}$$
The eigenvalues of $\mathbf U$ are
$$\xi_k=\frac14\csc^2\left(\frac{k\pi}{7}\right),\qquad k=1\dots 3$$
which means the eigenvalues of $\mathbf W=4\mathbf U-\mathbf I$ are
$$\lambda_k=\cot^2\left(\frac{k\pi}{7}\right),\qquad k=1\dots 3$$
Now, with
$$\mathbf W=\frac17\begin{pmatrix}13&-12&4\\-12&17&-8\\4&-8&5\end{pmatrix}$$
you can see that the trace of $\mathbf W$ (which is also the sum of the eigenvalues of $\mathbf W$) is $5$.
|
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|
where is wrong in the sum of series $\frac{1}{3}+\frac{1}{4}\cdot\frac{1}{2!}+\frac{1}{5}\cdot\frac{1}{3!}+\cdots$ I came through two types of solutions of the series $\frac{1}{3}+\frac{1}{4}\cdot\frac{1}{2!}+\frac{1}{5}\cdot\frac{1}{3!}+\cdots$
$$\begin{align*}
\frac{1}{3}+\frac{1}{4}\cdot\frac{1}{2!}+\frac{1}{5}\cdot\frac{1}{3!}+\cdots
&=\sum_{n=1}^{\infty}\frac{1}{n+2}\cdot\frac{1}{n!}\\
&=\sum_{n=1}^{\infty}\frac{n+1}{(n+2)!}\\
&=\sum_{n=1}^{\infty}[\frac{1}{(n+1)!}-\frac{1}{(n+2)!}]\\
&=\frac{1}{2}
\end{align*}$$
$$\begin{align*}
\frac{1}{3}+\frac{1}{4}\cdot\frac{1}{2!}+\frac{1}{5}\cdot\frac{1}{3!}+\cdots
&=\sum_{n=1}^{\infty}\frac{1}{n+2}\cdot\frac{1}{n!}\\
&=\sum_{n=1}^{\infty}\int_{0}^{1}\frac{x^{n+1}}{n!}dx\\
&=\int_{0}^{1}\sum_{n=1}^{\infty}\frac{x^{n+1}}{n!}dx\\
&=\int_{0}^{1}x\sum_{n=1}^{\infty}\frac{x^{n}}{n!}dx\\
&=\int_{0}^{1}x(e^x-1)dx\\
&=\frac{-1}{2}
\end{align*}$$
where am I getting wrong please help!
|
You have $$\int^{1}_{0}x(e^{x}-1)dx=[e^{x}(x-1)-\frac{x^{2}}{2}]|^{1}_{0}=\frac{1}{2}$$
|
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|
How to compute the determinant of a tridiagonal matrix with constant diagonals? How to show that the determinant of the following $(n\times n)$ matrix
$$\begin{pmatrix}
5 & 2 & 0 & 0 & 0 & \cdots & 0 \\
2 & 5 & 2 & 0 & 0 & \cdots & 0 \\
0 & 2 & 5 & 2 & 0 & \cdots & 0 \\
\vdots & \vdots& \vdots& \vdots & \vdots & \vdots & \vdots \\
0 & \cdots & \cdots & 0 & 2 & 5 & 2 \\
0 & \cdots & \cdots & \cdots & \cdots & 2 & 5
\end{pmatrix}$$
is equal to $\frac13(4^{n+1}-1)$?
More generally:
How does one compute the determinant of the following tridiagonal matrix (where the three diagonals are constant)?
$$M_n(a,b,c) = \begin{pmatrix}
a & b & 0 & 0 & 0 & \cdots & 0 \\
c & a & b & 0 & 0 & \cdots & 0 \\
0 & c & a & b & 0 & \cdots & 0 \\
\vdots & \vdots& \vdots& \vdots & \vdots& \vdots & \vdots \\
0 & \cdots & \cdots & 0 & c & a & b \\
0 & \cdots & \cdots & \cdots & \cdots & c & a
\end{pmatrix}$$
Here $a,b,c$ can be taken to be real numbers, or complex numbers.
In other words, the matrix $M_n(a,b,c) = (m_{ij})_{1 \le i,j \le n}$ is such that
$$m_{ij} = \begin{cases}
a & i = j, \\
b & i = j - 1, \\
c & i = j + 1, \\
0 & \text{otherwise.}
\end{cases}$$
There does not seem to be an easy pattern to use induction: the matrix is not a diagonal block matrix of the type $M = \bigl(\begin{smallmatrix} A & C \\ 0 & B \end{smallmatrix}\bigr)$ (where we could use $\det(M) = \det(A) \det(B)$ for the induction step), and there are no lines or columns with only one nonzero entry, so Laplace expansion gets complicated quickly.
Is there a general pattern that one could use? Or is the answer only known on a case-by-case basis? It's possible to compute the determinant by hand for small $n$:
$$\begin{align}
\det(M_1(a,b,c)) & = \begin{vmatrix} a \end{vmatrix} = a \\
\det(M_2(a,b,c)) & = \begin{vmatrix} a & b \\ c & a \end{vmatrix} = a^2 - bc \\
\det(M_3(a,b,c)) & = \begin{vmatrix} a & b & 0 \\ c & a & b \\ 0 & c & a \end{vmatrix} = a^3 - 2abc
\end{align}$$
But there is no readily apparent pattern and the computation becomes very difficult when $n$ gets large.
|
We will generalize Calvin Lin's answer a bit. Let
$$A_n = \begin{bmatrix} a & b & 0 & 0 & \cdots & 0\\ c & a & b & 0 & \cdots & 0\\ 0 & c & a & b & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \cdots & 0\\ \vdots & \vdots & \vdots & \vdots & a & b\\ 0 & 0 & 0 & \cdots & c & a \end{bmatrix}.$$
We then have, by using Laplace expansion twice,
$$\det(A_n) = a \det(A_{n-1}) - bc \det(A_{n-2}).$$
Calling $\det(A_n) = d_n$ we have the following linear homogeneous recurrence relation:
$$d_n = a d_{n-1} - bc d_{n-2}.$$
The characteristic equation is
$$\begin{align}
x^2 - ax + bc = 0 & \implies \left(x - \frac{a}2 \right)^2 - \left(\frac{a}2 \right)^2 + bc = 0 \\
& \implies x = \frac{a \pm \sqrt{a^2-4bc}}2.
\end{align}$$
(This assumes a square roots exist. It's always the case in $\mathbb{C}$.)
Case 1: $a^2 - 4bc \neq 0$
In this case the characteristic polynomial has two distinct roots, so we have (for some constants $k_1$, $k_2$): $$d_n = k_1 \left( \dfrac{a + \sqrt{a^2-4bc}}2\right)^n + k_2 \left( \dfrac{a - \sqrt{a^2-4bc}}2\right)^n.$$
We have $d_1 = a$ and $d_2 = a^2 - bc$. We then get that $d_0 = 1$. Hence,
$$k_1 + k_2 = 1.$$
$$a (k_1 + k_2) + (k_1 - k_2)\sqrt{a^2-4bc} = 2a \implies k_1 - k_2 = \dfrac{a}{\sqrt{a^2-4bc}}.$$
Hence,
$$\begin{align}
k_1 & = \dfrac{a + \sqrt{a^2-4bc}}{2\sqrt{a^2-4bc}}, & k_2 & = -\dfrac{a-\sqrt{a^2-4bc}}{2\sqrt{a^2-4bc}}
\end{align}$$
And finally:
$$\color{red}{\det(A_n) = \dfrac1{\sqrt{a^2-4bc}} \left( \left( \dfrac{a + \sqrt{a^2-4bc}}2\right)^{n+1} - \left( \dfrac{a - \sqrt{a^2-4bc}}2\right)^{n+1}\right)}.$$
Plug in $a = 5$ and $b=c=2$ ($a^2 - 4 bc \neq 0$), to get
$$\det(A_n) = \frac{1}{3} ( 4^{n+1} - 1)$$
Case 2: $a^2 - 4bc = 0$
If the characteristic polynomial has a double root $x = a/2$, there exist constants $k_1$, $k_2$ such that:
$$d_n = (k_1 + k_2 n) \bigl(\frac{a}{2}\bigr)^n.$$
The initial conditions are $d_0 = 1$ and $d_1 = a$, thus:
$$\begin{align}
k_1 & = 1 & (k_1 + k_2) a = 2a
\end{align}$$
If $a = 0$, then $4bc = a^2$ implies either $b$ or $c$ is zero, and $d_n = 0$ for $n \ge 1$. Otherwise $$(k_1 + k_2) a = 2a \implies k_1 + k_2 = 2 \implies k_2 = 1.$$
And finally:
$$\color{red}{\det(A_n) = (n+1) \bigl(\frac{a}{2}\bigr)^n}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\frac{n}{2} < 1+\frac{1}{2} + \frac{1}{3}+ \ldots +\frac{1}{2^n-1}How to show that for $n\geqslant 2$
$$\frac{n}{2} < 1+\frac{1}{2} + \frac{1}{3}+ \ldots +\frac{1}{2^n-1}<n$$
|
HINT: $$\frac12<2^n\left(\frac1{2^{n+1}-1}\right)<\sum_{k=2^n}^{2^{n+1}-1}\frac1k<2^n\left(\frac1{2^n}\right)=1$$
|
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|
Smart demonstration to the formula $ \sum_{n=1} ^{N} \frac{n}{2^n} = \frac{-N + 2^{N+1}-2}{2^n}$ Someone could give me a smart and simple solution to show the folowing identity?
$$ \sum_{n=1} ^{N} \frac{n}{2^n} = \frac{-N + 2^{N+1}-2}{2^n}$$
|
Proceed by induction.
Base case is true with $N=1$.
Inductive (I replaced the actual sum with $\sum$):
$$ \frac{N+1}{2^{N+1}} + \sum^N = \sum^{N+1} $$
$$ \frac{N+1}{2^{N+1}} + \frac{-N+2^{N+1}-2}{2^N} = \frac{-(N+1) + 2^{N+2}-2}{2^{N+1}} $$
$$ \frac{N + 1 - 2N + 2^{N+2} - 4}{2^{N+1}} = \frac{-N-3+2^{N+2}}{2^{N+1}} $$
$$ \frac{- N - 3 + 2^{N+2}}{2^{N+1}} = \frac{-N-3+2^{N+2}}{2^{N+1}} $$
Which is true, so the summation is true. This is a good general way for proving summation formulas.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Alternate proof that for every natural number $n,\ \left\lfloor\left(\frac{7+\sqrt{37}}{2}\right)^n\right\rfloor$ is divisible by $3$ Original Problem:
Prove that for every natural number $n$,$$\left\lfloor\left(\frac{7+\sqrt{37}}{2}\right)^n\right\rfloor$$ is divisible by $3$.
I found the problem in the book Winning Solutions. I do have a solution at hand which I was able to construct duly from the given hints: At first, take $a_n=\left(\frac{7+\sqrt{37}}{2}\right)^n+\left(\frac{7-\sqrt{37}}{2}\right)^n$ and then prove that $a_{n+2}=7a_{n+1}-3a_n$ which in turn, can be used to prove that $3\ |\ a_n-1$.The proof is complete once we can prove that $a_n-1=\left\lfloor\left(\frac{7+\sqrt{37}}{2}\right)^n\right\rfloor$.
Questions:
*
*Why would anyone think of coming up with such an ingenious recurrence? I am simply at a loss to understand the motivation behind the whole series of hints. They led me to a solution but I doubt I would have come up with it by myself.
*Can anyone please show me a more natural proof?
Thanks!
|
Considering the number $\frac{7+\sqrt{37}}{2}$ and its quadratic conjugate $\frac{7-\sqrt{37}}{2}$:
$$
\left(\frac{7+\sqrt{37}}{2}\right)\left(\frac{7-\sqrt{37}}{2}\right)=\frac{49-37}{4}=3\tag{1}
$$
$$
\left(\frac{7+\sqrt{37}}{2}\right)+\left(\frac{7-\sqrt{37}}{2}\right)=7\tag{2}
$$
Thus, we get that both of these conjugates satisfy
$$
x^2-7x+3=0\tag{3}
$$
That is, the recurrence
$$
a_n=7a_{n-1}-3a_{n-2}\tag{4}
$$
is satisfied by
$$
a_n=\left(\frac{7+\sqrt{37}}{2}\right)^n+\left(\frac{7-\sqrt{37}}{2}\right)^n\tag{5}
$$
Using $(5)$, we get $a_0=2$ and $a_1=7$.
Furthermore, note that $(4)$ says that $a_n\equiv a_{n-1}\pmod{3}$ for $n\ge2$. Since $a_1\equiv1\pmod{3}$, we must have $a_n\equiv1\pmod{3}$ for $n\ge1$. That is, for $n\ge1$,
$$
\left(\frac{7+\sqrt{37}}{2}\right)^n+\left(\frac{7-\sqrt{37}}{2}\right)^n\equiv1\pmod{3}\tag{6}
$$
Since $0<\left(\frac{7-\sqrt{37}}{2}\right)^n<1$ for $n\ge1$, we must have that
$$
\left\lfloor\left(\frac{7+\sqrt{37}}{2}\right)^n\right\rfloor\equiv0\pmod{3}\tag{7}
$$
The method above is similar to many problems that deal with recursive sequences. This sort of process is often first encountered when studying the Fibonacci sequence, which satisfies the recurrence
$$
F_n=F_{n-1}+F_{n-2}\tag{8}
$$
The recurrence $(8)$ is related to the equation
$$
x^2-x-1=0\tag{9}
$$
since both roots of $(9)$ satify $x_i^n=x_i^{n-1}+x_i^{n-2}$, which is $(8)$ after substituting $F_n=x_i^n$. Since the recursion $(8)$ is linear, the general solution of $(8)$ is
$$
F_n=c_1x_1^n+c_2x_2^n
$$
With enough experience with these types of equations, the answer above is pretty simple.
|
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Evaluating $\int^{\infty}_{0}{\frac{\ln x}{(1+x^2)^2}dx}$ $$\int^{\infty}_{0}{\frac{\ln x}{(1+x^2)^2}dx}$$
I've done a similar one: $\int^{\infty}_{1}{\frac{\ln^n x}{x^2}dx}$ using IBP, but in this case I've tried IBP multiple times, differentiating all of the possible choices, and it's only getting more convoluted. Also, no substitution I've tried like $x=e^u$ or $u=1+x^2$made the integral simpler. I'd appreciate a hint at this point, since I don't think it can be that hard.
|
Let $x=1/y$. We then get
$$I = \int_0^{\infty} \dfrac{\ln(x)}{(1+x^2)^2} dx = \int_0^{1} \dfrac{\ln(x)}{(1+x^2)^2} dx + \int_1^{\infty} \dfrac{\ln(x)}{(1+x^2)^2} dx\\
= \int_0^{1} \dfrac{\ln(x)}{(1+x^2)^2} dx - \underbrace{\int_0^{1} \dfrac{x^2\ln(x)}{(1+x^2)^2} dx}_{x \to 1/x}\\
= \int_0^{1} (1-x^2)\left(\sum_{k=0}^{\infty} (-1)^{k}(k+1)x^{2k}\right) \log(x) dx\\
= \sum_{k=0}^{\infty}(-1)^k(k+1)\left(\int_0^1 x^{2k} \log(x) dx - \int_0^1 x^{2k+2} \log(x) dx \right)\\
= \sum_{k=0}^{\infty}(-1)^k(k+1)\left(- \dfrac1{(2k+1)^2} + \dfrac1{(2k+3)^2} \right) = -\dfrac{\pi}4$$
The last equality can be seen from below.
$$\int_0^1 x^m \log(x) dx = -\dfrac1{(m+1)^2}$$
\begin{align}
\sum_{k=0}^{\infty}(-1)^k(k+1)\left(- \dfrac1{(2k+1)^2} + \dfrac1{(2k+3)^2} \right) & = 1\left(-\dfrac1{1^2} + \dfrac1{3^2} \right)\\
& -2 \left(-\dfrac1{3^2} + \dfrac1{5^2} \right)\\
& +3 \left(-\dfrac1{5^2} + \dfrac1{7^2} \right)\\
& -4 \left(-\dfrac1{7^2} + \dfrac1{9^2} \right) + \cdots
\end{align}
$$\sum_{k=0}^{\infty}(-1)^k(k+1)\left(- \dfrac1{(2k+1)^2} + \dfrac1{(2k+3)^2} \right) = -\dfrac1{1^2} + \dfrac{3}{3^2} - \dfrac{5}{5^2} + \dfrac7{7^2} \mp \cdots\\
= -1 + \dfrac13 - \dfrac15 + \dfrac17 \mp \cdots = -\dfrac{\pi}4$$
|
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Algebra question involving fractions How would I perform the indicated operation.
$$\frac{t+2}{t^2+5t+6}+\frac{t-1}{t^2+7t+12}-\frac{2}{t+4}.$$
I simplified it to
$$
\frac{t+2}{(t+3)(t+2)} + \frac{t-1}{(t+4)(t+3)}-\frac{2}{t+4}.
$$
Then I did the lowest common denominator but I still have problems.
According to my book the final answer $\;\dfrac{-3}{(t+3)(t+4)},\;$ but I cannot seem to get it.
|
Note that $\dfrac{t+2}{t^2+5t+6} = \dfrac{t+2}{(t+2)(t+3)} = \dfrac1{t+3}$ for $t \neq -2$.
Hence, $\dfrac{t+2}{t^2+5t+6} - \dfrac2{t+4} = \dfrac1{t+3} - \dfrac2{t+4} = \dfrac{t+4-2t-6}{(t+3)(t+4)} = - \dfrac{t+2}{(t+3)(t+4)}$.
Hence, $$\dfrac{t+2}{t^2+5t+6} - \dfrac2{t+4} + \dfrac{t-1}{t^2+7t+12} = \dfrac{t-1}{t^2+7t+12} - \dfrac{t+2}{(t+3)(t+4)}\\ = \dfrac{t-1-t-2}{(t+3)(t+4)} = -\dfrac3{(t+3)(t+4)}$$
|
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|
Understanding the solution of $\int\left(1-x^{p}\right)^{\frac{n-1}{p}}\log\left(1-x^{p}\right)dx$ I would like to solve the integral
$$\int\left(1-x^{p}\right)^{\frac{n-1}{p}}\log\left(1-x^{p}\right)dx.$$
My problem is that I arrived at a solution via wolfram alpha, but I would like to understand how one would arrive there by hand.
What I did is to use the substitution $x=\left(1-\exp(z)\right)^{\frac{1}{p}}$. This yields
$$-\frac{1}{p}\int\left(1-\exp(z)\right)^{\frac{1}{p}-1}\exp\left(\frac{n-1+p}{p}z\right)zdz$$ for which wolfram alpha yields the integral
\begin{align}
&\frac{\exp\left(z\frac{(n+p-1)}{p}\right)}{\left(n+p-1\right)^{2}}p\,_{3}F_{2}\left(1-\frac{1}{p},\frac{n}{p}-\frac{1}{p}+1,\frac{n}{p}-\frac{1}{p}+1;\frac{n}{p}-\frac{1}{p}+2,\frac{n}{p}-\frac{1}{p}+2;\exp(z)\right)\\&-z\frac{\exp\left(z\frac{(n+p-1)}{p}\right)}{\left(n+p-1\right)}\,_{2}F_{1}\left(\frac{p-1}{p},\frac{n+p-1}{p};\frac{n+2p-1}{p};\exp z\right).
\end{align}
I would really like to understand how it arrived at the generalized hypergeometric function in the solution because I think it would be useful to spot it in an integral (like spotting possible solutions in terms of a gamma function or beta function in integrals).
Does anybody have a hint how it arrived at the solution?
|
You made a substitution $u=1-x^p$, transforming the integral as
$$
\int \left(1-x^p\right)^{(n-1)/p} \log\left(1-x^p\right) \mathrm{d} x = - \frac{1}{p} \int \left(1-u\right)^a u^b \log\left(u\right) \mathrm{d} u
$$
where $a = \frac{1}{p}-1$ and $b=\frac{n-1}{p}$. In order to integrate this we shall take advantage of the following differentiation property of the Gauss hypergeometric function:
$$
\left( \frac{1}{\alpha_1} z \frac{\mathrm{d}}{\mathrm{d} z} + 1 \right) {}_2 F_1\left( \left. \begin{array}{cc} \alpha_1 & \alpha_2 \cr \beta&- \end{array} \right| z \right) = {}_2 F_1\left( \left. \begin{array}{cc} \alpha_1 +1 & \alpha_2 \cr \beta& - \end{array} \right| z \right)
$$
Choosing $\beta=\alpha_1 + 1$ we get
$$
\left( \frac{1}{\alpha_1} z \frac{\mathrm{d}}{\mathrm{d} z} + 1 \right) {}_2 F_1\left( \left. \begin{array}{cc} \alpha_1 & \alpha_2 \cr &\alpha_1+1& \end{array} \right| z \right) = {}_1 F_0\left( \left. \begin{array}{c} \alpha_2 \\ - \end{array} \right| z \right) = \left(1-z\right)^{-\alpha_2} \tag{1}
$$
Now, using
$$\left( \frac{1}{\alpha_1} z \frac{\mathrm{d}}{\mathrm{d} z} + 1 \right) f(z) = \frac{1}{\alpha_1} z^{1-\alpha_1} \frac{\mathrm{d}}{\mathrm{d} z} \left( z^{\alpha_1} f(z) \right) \tag{2}
$$
Combining eqs. $(1)$ and $(2)$ and choosing $\alpha_2 = -a$ and $\alpha_1 = b +1$ we arrive at
$$
\int \left(1-u\right)^a u^b \mathrm{d}u = \frac{u^{b+1}}{b+1} {}_2 F_1\left( \left. \begin{array}{cc} -a & b+1 \cr &b+2& \end{array} \right| u \right) \tag{3}
$$
We can now integrate by parts:
$$ \begin{eqnarray}
\int \left(1-u\right)^a u^b \log(u) \mathrm{d}u &=& \frac{u^{b+1}}{b+1} \log\left(u \right) \cdot {}_2 F_1\left( \left. \begin{array}{cc} -a & b+1 \cr &b+2& \end{array} \right| u \right) \\ && - \frac{1}{b+1} \int u^b \cdot {}_2 F_1\left( \left. \begin{array}{cc} -a & b+1 \cr &b+2& \end{array} \right| u \right) \mathrm{d}u
\end{eqnarray}
$$
The latter integral is integrated in a similar way by using differentiation properties of generalized hypergeometric function:
$$
\frac{\mathrm{d}}{\mathrm{d} z} \left( z^{\alpha_1} \cdot {}_3 F_2\left( \left. \begin{array}{ccc} \alpha_1 & \alpha_2 & \alpha_3 \cr \beta_1& \beta_2 & \end{array} \right| z \right) \right) = \alpha_1 z^{\alpha_1-1} \cdot {}_3 F_2\left( \left. \begin{array}{ccc} \alpha_1+1 & \alpha_2 & \alpha_3 \cr \beta_1& \beta_2 & \end{array} \right| z \right)
$$
we get
$$\begin{eqnarray}
\int \left(1-u\right)^a u^b \log(u) \, \mathrm{d} u &=& \frac{u^{b+1}}{b+1} \log\left(u \right) \cdot {}_2 F_1\left( \left. \begin{array}{cc} -a & b+1 \cr &b+2& \end{array} \right| u \right) \\ && - \frac{u^{b+1}}{(b+1)^2} \cdot {}_3 F_2\left( \left. \begin{array}{ccc} -a & b+1 & b+1 \cr b+2& b+2 & \end{array} \right| u \right)
\end{eqnarray} \tag{4}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/275158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
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|
a question on $O$- notation I'm trying to understand $O$-notationbetter. I've found a really helpful answer : Big O notation, $1/(1-x)$ series
But I have trouble with quotient. Let me discuss about this example function:
$$f(x)=a+bx+\frac{c+dx}{e+fx^2}e^{ax}$$
If I calculate for the zero order error $O(x^0)$, I'll get
$$a+\frac{c}{e}=0$$
...for the first order error $O(x^1)$, I'll get
$$b+\frac{d}{2fx}ae^{ax}|_{x=0}=0$$
which seems a bit or more then a bit silly:(
Could you please explain me how to calculate for the first order error $O(x^1)$?
|
Taylor series get you a long way in practice. Using the Taylor series for the exponential function and $\frac{1}{1 + x}$ you can get as many order terms exactly as you need
$$\begin{align} f(x) &=a+bx+\frac{c+dx}{e+fx^2}e^{ax} \\
&= a + bx + \frac{1}{e} \cdot \frac{c+dx}{1 + \frac{f}{e}x^2} \cdot \left(1 + ax + \frac{a^2}{2} x^2 + O(x^3)\right) \\
&= a + bx + \frac{1}{e} \cdot (c+dx) \cdot \left(1 - \frac{f}{e}x^2 + O(x^4)\right) \cdot \left(1 + ax + \frac{a^2}{2} x^2 + O(x^3)\right) \\
&= a + bx + \frac{1}{e} \cdot \left(c + (ac + d)x + \left(\frac{a^2 c}{2} + ad - \frac{cf}{e}\right)x^2 + O(x^3)\right) \\
&= \left(a + \frac{c}{e}\right) + \left(b + \frac{ac + d}{e}\right)x + \left(\frac{a^2 c}{2e} + \frac{ad}{e} - \frac{cf}{e^2}\right)x^2 + O(x^3). \end{align}$$
|
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"url": "https://math.stackexchange.com/questions/277264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
What will be the one's digit of the remainder in: $\left|5555^{2222} + 2222^{5555}\right|\div 7=?$ What will be the ones digit of the remainder in: $$\frac{\left|5555^{2222} + 2222^{5555}\right|} {7}$$
|
Hint: $ 5555 \equiv 4 \pmod{7}$, $2222 \equiv 3 \pmod{7}$.
Edit: Calculate that $4^3 \equiv 1, 3^6 \equiv 1 \pmod{7}$. Hence, this implies that $4^{3k} \equiv (4^3)^k \equiv 1^k \equiv 1 \pmod{7}, 3^{6j} \equiv (3^6)^j \equiv 1^j \equiv \pmod{7}$.
Now, $2222 \equiv 2 \pmod{3}$, and $5555 \equiv 5 \pmod{6}$. Hence,
$$ 5555^{2222} + 2222^{5555} \equiv 4^{2222} + 3^{5555} \equiv 4^ 2 + 3^ 5 \pmod{7}.$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
find the sum of the alternating series How to find the sum of the infinite series
$$\frac{1}{12}-\frac{1\cdot 4}{12 \cdot 18 } + \frac{1\cdot 4\cdot 7}{12\cdot 18\cdot 24} - \frac{1 \cdot 4 \cdot 7\cdot 10}{12 \cdot 18 \cdot 24 \cdot 30}+...$$
I understood the answer posted in Yahoo Answer till the last but one step:
That is how did he get: $ \lim_{n \to \infty} S_n = 0 $
Other steps I understood.
Thanks in advance
|
We can rewrite the series as
$$
\begin{align}
-3\sum_{n=2}^\infty\binom{2/3}{n}(1/2)^n
&=-3\left((1+1/2)^{2/3}-1-1/3\right)\\
&=4-3\ (3/2)^{2/3}
\end{align}
$$
A Bit of Explanation
By the Generalized Binomial Theorem, we have
$$
\sum_{n=0}^\infty\binom{2/3}{n}(1/2)^n=(1+1/2)^{2/3}
$$
The first two terms are $\binom{2/3}{0}(1/2)^0=1$ and $\binom{2/3}{1}(1/2)^1=1/3$. Subtracting the first two terms yields
$$
\sum_{n=2}^\infty\binom{2/3}{n}(1/2)^n=(1+1/2)^{2/3}-1-1/3
$$
The General Term
$$
\begin{align}
\binom{2/3}{n}(1/2)^n
&=\frac{2/3(-1/3)(-4/3)\dots(5/3-n)}{1\cdot2\cdot3\cdots n}\frac1{2^n}\\
&=\frac{2(-1)(-4)(-7)\dots(5-3n)}{6\cdot12\cdot18\cdot24\cdots(6n)}
\end{align}
$$
which is $-1/3$ of the general term of the series.
|
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|
Green's function for $\frac{d^2}{dx^2} + \frac{1}{4}$ I am trying to solve the following differential equation $$y''+\frac{1}{4}y=\sin(2x)~~~y(0)=y(\pi)=0,$$ using Green's function. I have found the Green's function for the operator $y''+\frac{1}{4}$ to be $$G(x,\xi)= \sin\left(\frac{1}{2}(x-\xi)\right).$$ However, when integrating
$$y(x)=\int_0^\pi \sin\left(\frac{1}{2}(x-\xi)\right) \cdot \sin(2 \xi)~d\xi =\frac{8}{15}\left(\sin(x/2)+\cos(x/2)\right)$$ which is very different from the actual answer of $$y(x)=-\frac{4}{15} \sin(2x).$$
Obviously, I am not doing something right. Where am I going wrong?
|
Let's start from the beginning. You want to express the problem as solving for a function $G(x,y)$ that satisfies, for $x \in [0,\pi]$
$$\frac{d}{dx^2} G(x,x') + \frac{1}{4} G(x,x') = \delta(x-x') $$
where $\delta(x-x') = 0 \, \forall \, x \ne x'$ and $\int_0^{\pi} dx' \: \delta(x-x') = 1 $. The solution you seek, $y(x)$, satisfies the boundary conditions $y(0) = y(\pi) = 0$, and may be expressed in terms of $g$ as
$$ y(x) = \int_0^{\pi} dx' \: G(x,x') \sin{2 x'} $$
To find $G$, we assume $x \ne x'$ and write down the general solution to the homogeneous equation:
$$G(x,x') = A \cos{\frac{x}{2}} + B \sin{\frac{x}{2}} $$
The boundary conditions on the solution may be expressed in terms of the relation of $x$ to $x'$. For example, the condition on $y(0)$ translates into a condition for $G(x,x') \, \forall \, x < x'$; that is, $G(0,x') = A = 0$ when $x < x'$. Similarly, the condition $y(\pi) = 0 \implies G(\pi,x') = B = 0$ when $x > x'$. We may then write
$$G(x,x') = \displaystyle \begin{cases} B \sin{\frac{x}{2}} & x < x' \\ A \cos{\frac{x}{2}} & x > x' \end{cases}$$
(This is where I think you went off the rails.) To find $A$ and $B$, we impose 2 conditions. The first is that $G(x,x')$ be continuous when $x=x'$. This leads to the relationship $A = B \tan{\frac{x}{2}}$. The second is that the derivative of $G(x,x')$ with respect to $x$ is discontinuous at $x=x'$ and satisfies
$$\lim_{\epsilon \rightarrow 0} \left [ \frac{\partial}{\partial x} G(x'+\epsilon,x') - \frac{\partial}{\partial x} G(x'-\epsilon,x') \right ] = 1$$
This relationship may be seen from integration of the differential equation defining $G$ above. Plugging in the above expression for $G(x,x')$, we get a second relation for $A$ and $B$: $A \sin{\frac{x'}{2}} + B \cos{\frac{x'}{2}} = -2$. We may solve for $A$ and $B$ and find that $A = -2 \sin{\frac{x'}{2}}$ and $B = -2 \cos{\frac{x'}{2}}$. (A little manipulation of trig identiies is needed to get this.) We may now write
$$G(x,x') = \displaystyle \begin{cases} -2 \cos{\frac{x'}{2}} \sin{\frac{x}{2}} & x < x' \\ -2 \sin{\frac{x'}{2}} \cos{\frac{x}{2}} & x > x' \end{cases}$$
We are now ready to compute the solution $y(x)$ as written above. Because of the different functional forms for $G$ about $x=x'$, we need to split the integral defining $y$ into two pieces:
$$y(x) = -2 \cos{\frac{x}{2}} \int_0^x dx' \: \sin{\frac{x'}{2}} \sin{2 x'} -2 \sin{\frac{x}{2}} \int_0^x dx' \: \cos{\frac{x'}{2}} \sin{2 x'}$$
The evaluation of these integrals is made possible through the trigonometric addition formulas $\cos{(a-b)} - \cos{(a+b)} = 2 \sin{a} \sin{b}$ and $\sin{(a+b)} + \sin{(a-b)} = 2 \sin{a} \cos{b}$. It does get a little messy, but you may verify that the solution you seek,
$$y(x) = - \frac{4}{15} \sin{2 x} $$
is the result of the evaluation of the above integrals.
|
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|
limit of the sum $\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n} $ Prove that : $\displaystyle \lim_{n\to \infty} \frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2n}=\ln 2$
the only thing I could think of is that it can be written like this :
$$ \lim_{n\to \infty} \sum_{k=1}^n \frac{1}{k+n} =\lim_{n\to \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{\frac{k}{n}+1}=\int_0^1 \frac{1}{x+1} \ \mathrm{d}x=\ln 2$$
is my answer right ? and are there any other method ?(I'm sure there are)
|
Set $a_n
= 1/(n+1) + 1/(n+2)+···+ 1/(2n)
=\sum_{k=1}^n \dfrac1{n+k}
$.
$\begin{array}\\
a_{n+1}-a_n
&=\sum_{k=1}^{n+1} \dfrac1{n+1+k}-\sum_{k=1}^n \dfrac1{n+k}\\
&=\sum_{k=2}^{n+2} \dfrac1{n+k}-\sum_{k=1}^n \dfrac1{n+k}\\
&=\dfrac1{n+n+2}+\dfrac1{n+n+1}-\dfrac1{n+1}\\
&=\dfrac1{2n+2}+\dfrac1{2n+1}-\dfrac1{n+1}\\
&=-\dfrac1{2 (n + 1) (2 n + 1)}\\
&< 0\\
\text{and}\\
a_{n+1}-a_n
&=-\dfrac1{2 (n + 1) (2 n + 1)}\\
&>-\dfrac1{2n (n + 1)}\\
\text{so}
&\text{if } m > 0\\
a_{m+n}-a_n
&< 0\\
\text{and}\\
a_{m+n}-a_n
&=\sum_{k=0}^{m-1}(a_{n+k+1}-a_{n+k})\\
&>-\sum_{k=0}^{m-1}(\dfrac1{2(n+k+1) (n + k+2)})\\
&=-\dfrac12\sum_{k=0}^{m-1}(\dfrac1{(n+k+1)}-\dfrac1{(n + k+2)})\\
&=-\dfrac12(\dfrac1{(n+1)}-\dfrac1{(n +m+1)})\\
&>-\dfrac1{2(n+1)}\\
\end{array}
$
so,
for $m > 0$,
$|a_{m+n}-a_n|
\lt \dfrac1{2(n+1)}
\to 0
$
as $n \to \infty$
so,
by Cauchy,
$a_n$
converges.
|
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|
How to transform $2^{n-2}\frac{(2n-5)(2n-7)...(3)(1)}{(n-1)(n-2)...(3)(2)(1)}$ into $\frac{1}{n-1}\binom{2n-4}{n-2}$? Just an algebraic step within the well known solution for the number of triangulations of a convex polygon!
|
First, rewrite the denominator as $(n-1)\cdot(n-2)!$, then multiply numerator and denominator by $(n-2)!$ :
$$2^{n-2}\frac{(2n-5)(2n-7)\ldots(3)(1)(n-2)(n-3)\ldots(2)(1)}{(n-1)(n-2)!(n-2)!}$$
Next, distribute the $n-2$ factors of $2$ over the $n-2$ 'elements' of $(n-2)!$ in the numerator:
$$\frac{(2n-5)(2n-7)\ldots(3)(1)(2n-4)(2n-6)\ldots(4)(2)}{(n-1)(n-2)!(n-2)!}$$
Now, just interleave the even and odd values in the numerator:
$$\frac{(2n-4)(2n-5)(2n-6)(2n-7)\ldots(3)(2)(1)}{(n-1)(n-2)!(n-2)!}$$
But the numerator is just $(2n-4)!$, so the overall expression is $\dfrac{(2n-4)!}{(n-1)(n-2)!(n-2)!}$, which is exactly your second expression.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integration by parts $ \int \sqrt{x-x^2} dx $ I have to integrate $$ \int \sqrt{x-x^2} dx $$
The answer on my textbook is $ \frac 14 \left( \arcsin(\sqrt x)-(1-2x)\sqrt{x-x^2} \right) $ but I want to solve this by myself so can you please give me a clue ? :) Thank you.
|
for calculation of Integral $\displaystyle \int \sqrt{a^2-x^2}dx$
We will Use Integration by parts method
Let $\displaystyle \mathbb{I} = \int \sqrt{a^2-x^2}.xdx = \sqrt{a^2-x^2}.x-\int\frac{1}{2\sqrt{a^2-x^2}}.-2x.xdx$
$\displaystyle \mathbb{I}= x.\sqrt{a^2-x^2}-\int\frac{(a^2-x^2)-a^2}{\sqrt{a^2-x^2}}dx$
$\displaystyle \mathbb{I} = x.\sqrt{a^2-x^2}-\mathbb{I}+a^2.\sin^{-1}\left(\frac{x}{a}\right)$
So $\displaystyle \mathbb{I}=\sqrt{a^2-x^2}dx = \frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}.\sin^{-1}\left(\frac{x}{a}\right)+\mathbb{C}$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Calculating the limit $\lim \limits_{n \to \infty} \sqrt[n]{4n + \sin \sqrt{n} + \cos (\frac{1}{n^2}) + 17}$ I am trying to calculate this limit:
$$
\lim \limits_{n \to \infty} \sqrt[n]{4n + \sin \sqrt{n} + \cos (\tfrac{1}{n^2}) + 17}$$
I understand I should use squeeze theorem but I am having some trouble applying it to this particular formula.
|
$\lim \limits_{n \to +\infty} \sqrt[n]{4n + \sin \sqrt{n} + \cos (\tfrac{1}{n^2}) + 17}$
$4n+15 \le 4n + \sin \sqrt{n} + \cos (\tfrac{1}{n^2}) + 17 \le 4n+19$
$\sqrt[n]{4n+15} \le \sqrt[n]{4n + \sin \sqrt{n} + \cos (\tfrac{1}{n^2}) + 17} \le \sqrt[n]{4n+19}$
$\lim \limits_{n \to +\infty} \sqrt[n]{4n+15} \le \lim \limits_{n \to +\infty} \sqrt[n]{4n + \sin \sqrt{n} + \cos (\tfrac{1}{n^2}) + 17} \le \lim \limits_{n \to +\infty} \sqrt[n]{4n+19}$
|
{
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"url": "https://math.stackexchange.com/questions/287358",
"timestamp": "2023-03-29T00:00:00",
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|
Is the sum always bigger than $n^2$? Let $s(n)$ an arithmetical function defined as
$$s(n)=(p_1+1)^{e_1} (p_2+1)^{e_2} \cdots (p_m+1)^{e_m}$$
where prime factorization of $n$ is $n=p_1^ {e_1} p_2 ^{e_2} \cdots p_m^{e_m}$.
(For example, $s(49)=s(7^2)=8^2=64$, $s(60)=s(2^2 \times 3^1 \times 5^1)=3^2 \times 4^1 \times 6^1=216$)
Prove or disprove the following proposition;
$\phantom{a}\bullet$ There exists a positive integer $M$ such that
$$\forall N>M(N\in\mathbb{N})\phantom{;}; \phantom{;}\sum_{n= 1}^{N}s(n)>N^2$$
|
This is true. If $\nu_p(n)$ is maximal such that $p^{\nu_p(n)}\mid n$, then
$$
s(n)\ge (\frac32)^{\nu_2(n)} (\frac43)^{\nu_3(n)}n.
$$
Therefore, for all $i=1$, $\dots$, $2^5 3^3$,
$$
s(2^5 3^3n+i)\ge (\frac32)^{\min(\nu_2(i),5)} (\frac43)^{\min(\nu_3(i),3)} (2^5 3^3 n+i).
$$
Summing over $i=1$, $\dots$, $2^5 3^3$ gives
\begin{eqnarray*}
\sum_{1\le i\le 2^5 3^3} s(2^5 3^3n+i)&\ge&1555910n+ 785732\\
&\ge& 2.08 \sum_{1\le i\le 2^5 3^3}( 2^5 3^3n + i).
\end{eqnarray*}
This proves that
$$
\sum_{1\le n\le N} s(n)\ge 2.08 \frac{N(N+1)}{2}>N^2
$$
whenever $N$ is a multiple of $2^5 3^3=864$. If $N$ is not a multiple of $864$, write $N=N'+N''$, where $N'$ is a multiple of $864$ and $1\le N''\le 863$. Then, since $s(n)\ge n$ for all $n$,
$$
\sum_{1\le n\le N} s(n)\ge 2.08 \frac{N'(N'+1)}{2} + N' N'' + \frac{N''(N''+1)}{2}
$$
which will be bigger than
$$
N^2 = N'^2 + 2 N' N'' + N''^2
$$
whenever $N'\ge 22464$. Also, computation shows that
$$
\sum_{1\le n\le N} s(n)> N^2
$$
whenever $24\le N\le 22463$. So, we can take $M=23$.
Asymptotically, $s(n)/n$ should behave similarly to the random variable
$$
W:=\prod_p (1+\frac{1}{p})^{Z_p-1},
$$
where the $Z_p$'s are independent geometrically distributed random variables with success probability $1-\frac{1}{p}$, and
$$
{\Bbb E}(W) = \prod_p \frac{p(p-1)}{p^2-p-1} \approx 2.67411.
$$
|
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|
Bell Numbers: How to put EGF $e^{e^x-1}$ into a series? I'm working on exponential generating functions, especially on the EGF for the Bell numbers $B_n$.
I found on the internet the EGF $f(x)=e^{e^x-1}$ for Bell numbers. Now I tried to use this EGF to compute $B_3$ (should be 15). I know that I have to put the EGF into a series and have a look at the coefficients. Using $e^{f(x)}=1+f(x)+\frac{f(x)^2}{2!}+\frac{f(x)^3}{3!}+\ldots$ I get
\begin{eqnarray*}
e^{e^x-1}&=&1+(e^x-1)+\frac{(e^x-1)^2}{2!}+\frac{(e^x-1)^3}{3!}\\
&=&1+e^x-1+\frac{e^{x^2}-2e^x+1}{2!}+\frac{e^{x^3}-3e^{x^2}+3e^x-1}{3!}\\
&=&e^x+\frac{e^{x^2}}{2!}-e^x+\frac{1}{2!}+\frac{e^{x^3}}{3!}-\frac{e^{x^2}}{2!}+\frac{e^{x}}{2!}-\frac{1}{3!}\\
&=&\frac{e^{x}}{2!}+\frac{e^{x^3}}{3!}-\frac{1}{2!}+\frac{1}{3!}\\
&=&\frac{1}{2!}e^x+\frac{1}{3!}e^{x^3}-\frac{1}{3}\\
&=&\frac{1}{2!}\left( 1+x^2+\frac{x^4}{2!}+\frac{x^8}{3!} \right)+\frac{1}{3!}\left( 1+x^3+\frac{x^6}{2!}+\frac{x^9}{3!}\right)\\
&=&\frac{1}{2}+\frac{x^2}{2}+\frac{x^4}{4}+\frac{x^8}{24}+\frac{1}{6}+\frac{x^3}{3!}+\frac{x^6}{12}+\frac{x^9}{36}\\
&=&\frac{1}{2}+\frac{x^2}{2}+\frac{x^3}{6}+\ldots
\end{eqnarray*}
I think I can stop here, because the coefficient in front of $\frac{x^3}{3!}$ is not $15$.
Perhaps someone can help me out and give a hint to find my mistake?
|
You probably don't want to use the series expansion for the exponential as you've done, because you will receive contributions to each coefficient from all orders of this expansion, so you'll still have infinite sums left to evaluate. Instead, you should use the fact that term $n$ in a sequence is given by $F^{(n)}(0)$, where $F(x)$ is the sequence's exponential generating function (and $F^{(n)}$ is the $n$-th derivative of $F$). If $F(x)=e^{f(x)}$, then its derivatives are given by
$$
F(0)=e^{f(0)} \\
F'(0)=f'(0)e^{f(0)} \\
F''(0)=(f''(0)+f'(0)^2)e^{f(0)} \\
F'''(0)=(f'''(0)+3f''(0)^2 f'(0) + f'(0)^3)e^{f(0)} \\
F^{(4)}(0)=(f^{(4)}(0)+f'''(0)f'(0)+6f'''(0)f''(0)f'(0)+3f''(0)^2+3f''(0)f'(0)^2+f'(0)^4)e^{f(0)}
$$
and so on. If $f(x)=e^x-1$, then $e^{f(0)}=1$ and $f^{(n)}(0)=1$ for any $n \ge 1$. Adding up the terms, then, you have $B_0=B_1=1$, $B_2=2$, $B_3=5$, $B_4=15$, and so on.
|
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|
primes of the form $a^2+b^2=x^2-xy+y^2$? Let $a,b,x,y$ be strict positive integers.
Im intrested in primes $p$ such that $p=a^2+b^2=x^2-xy+y^2$.
What is the analogue PNT for these type of primes ? I think these primes are all the primes $p \equiv 1 \pmod{12}$.
|
By Fermat's theorem on sums of two squares, $p$ can be written as $a^2+b^2$ iff $p \equiv 1 \pmod 4$ or $p = 2$. By this question, $p$ can be written as $x^2-xy+y^2$ iff $p \equiv 1 \pmod 3$ or $p = 3$. A prime $p$ satisfies both if and only if $p \equiv 1 \pmod {12}$, by the Chinese Remainder Theorem.
|
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|
Proving $ (1)(2) + (2)(3) + (3)(4) + \cdots+ (n) (n+1) = \frac{(n)(n+1)(n+2)}{3} $ by induction I have the Following Proof By Induction Question:
$$ (1)(2) + (2)(3) + (3)(4) + \cdots+ (n) (n+1) = \frac{(n)(n+1)(n+2)}{3} $$
Can Anybody Tell Me What I'm Missing.
This is where I've Gone So Far.
Show Truth for N = 1
LHS = (1) (2) = 2
RHS = $$ \frac{(1)(1+1)(1+2)}{3} $$
Which is Equal to 2
Assume N = K
$$ (1)(2) + (2)(3) + (3)(4) + \cdots+ (k) (k+1) = \frac{(k)(k+1)(k+2)}{3} $$
Proof that the equation is true for N = K + 1
$$ (1)(2) + (2)(3) + (3)(4) + \cdots+ (k) (k+1) + (k+1) (k + 2)$$
Which is Equal To:
$$ \frac{(k)(k+1)(k+2)}{3} + (k+1) (k + 2)$$
This is where I've went so far
If I did the calculation right the Answer should be
$$\frac{(k+1)(k+2)(k+3)}{3}$$
|
Your proof is fine, but you should show clearly how you got to the last expression.
$\dfrac{k(k+1)(k+2)}{3}+(k+1)(k+2)$
$=\dfrac{k}{3}(k+1)(k+2)+(k+1)(k+2)$
$=(\dfrac{k}{3}+1)(k+1)(k+2)$
$=\dfrac{k+3}{3}(k+1)(k+2)$
$=\dfrac{(k+1)(k+2)(k+3)}{3}$.
You should also word your proof clearly. For example, you can say "Let $P(n)$ be the statement ... $P(1)$ is true ... Assume $P(k)$ is true for some positive integer $k$ ... then $P(k+1)$ is true ... hence $P(n)$ is true for all positive integers $n$".
|
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|
Find the spanning set of the range of the linear transformation $T(x)=Ax$. Let
$$
A=
\begin{bmatrix}
-4 & -4 & 12 & 0 \\
-4 & -4 & 12 & 0 \\
4 & -2 & 0 &-6 \\
1 &-4 &7 &-5 \\
\end{bmatrix}
$$
Find the spanning set of the range of the linear transformation $T(x)=Ax$.
I have found the row reduced echelon form of A.
$$
RREF(A)=
\begin{bmatrix}
1 & 0 & -1 & -1 \\
0 & 1 & -2 & 1 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{bmatrix}
$$
I don't know what to do with it after.
|
It $T(v)=Av,~~~v\in\mathbb K^4$, then $$
T(v)=
\begin{pmatrix}
1 & 0 & -1 & -1 \\
0 & 1 & -2 & 1 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{pmatrix}\begin{pmatrix}
x \\
y \\
z \\
t \\
\end{pmatrix}=\begin{pmatrix}
x-z-t \\
y-2z+t \\
0 \\
0 \\
\end{pmatrix}
$$ but $$\begin{pmatrix}
1 & 0 & -1 & -1 \\
0 & 1 & -2 & 1 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{pmatrix}\approx\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{pmatrix} $$ so $$T(v)=
\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{pmatrix}\begin{pmatrix}
x \\
y \\
z \\
t \\
\end{pmatrix}=\begin{pmatrix}
x \\
y \\
0 \\
0 \\
\end{pmatrix}=x\begin{pmatrix}
1 \\
0 \\
0 \\
0 \\
\end{pmatrix}+y\begin{pmatrix}
0 \\
1 \\
0 \\
0 \\
\end{pmatrix}$$
|
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|
Find the maximum and minimum of $\cos x\sin y\cos z$.
Given $x\geq y\geq z\geq\pi/12$, $x+y+z=\pi/2$, find the maximum and minimum of $\cos x\sin y\cos z$.
I tried using turn $\sin y$ to $\cos(x+z)$, and Jensen Inequality, but filed. Please help. Thank you.
*p.s. I'm seeking for a solution without calculus.
|
Let $$P=\cos x\sin y\cos z=\frac{\cos z}{2}\bigg[2\cos x \sin y\bigg] = \frac{\cos z}{2}\bigg[\sin(x+y)-\sin(x-y)\bigg]\leq \frac{\cos z}{2}\cdot \sin (x+y)$$
So $$P\leq \frac{\cos z \cdot \cos z}{2}=\frac{1}{4}(1+\cos 2z)\leq \frac{1}{4}\bigg[1+\cos 2\cdot \frac{\pi}{12}\bigg] = \frac{2+\sqrt{3}}{8}$$
Above equality hold when $\sin (x-y) = 0\Rightarrow x=y$ and given $\displaystyle x+y = \frac{\pi}{2}-z$ and $\displaystyle x\geq y \geq z\geq \frac{\pi}{12}$
And for $\max(P),$ We must have $\displaystyle z = \frac{\pi}{12}$ and $\displaystyle x=y = \frac{5\pi}{24}$
|
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|
Deriving $e^{ix} = 1 + ix + \frac{(ix)^2}{2!}+...$ I know how to prove equation:
$$e = \lim \limits_{x \rightarrow \infty} \left( 1 + \frac{1}{x} \right)^x$$
How can I now derive the series:
$$e^{ix} = 1 + ix + \frac{(ix)^2}{2!}+\frac{(ix)^3}{3!}+...$$
Those two seem very similar to me...
|
Use the binomial theorem on the following limit:
$$e^z = \lim_{x \rightarrow \infty} \left( 1 + \frac{z}{x} \right)^x$$
i.e.,
$$\begin{align} \left( 1 + \frac{z}{x} \right)^x &= 1 + \frac{x}{1!} \frac{z}{x} + \frac{x (x-1)}{2!} \left (\frac{z}{x} \right )^2 + \frac{x (x-1) (x-2)}{3!} \left (\frac{z}{x} \right )^3 + \ldots \\ &= 1 + \frac{z}{1!} + \frac{z^2}{2!} \left ( 1-\frac{1}{x} \right ) + \frac{z^3}{3!} \left ( 1-\frac{1}{x} \right ) \left ( 1-\frac{2}{x} \right ) + \ldots \\ \end{align} $$
As $x \rightarrow \infty$, the product terms approach $1$. Therefore
$$\lim_{x \rightarrow \infty} \left( 1 + \frac{z}{x} \right)^x = \sum_{k=0}^{\infty} \frac{z^k}{k!} = e^z$$
Now, for $z=i x$,
$$e^{i x} = \sum_{k=0}^{\infty} \frac{(i x)^k}{k!} = 1 + i x + \frac{(i x)^2}{2!} + \ldots$$
|
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|
Show $\mathbb{Q}[\sqrt[3]{2}]$ is a field by rationalizing I need to rationalize $\displaystyle\frac{1}{a+b\sqrt[3]2 + c(\sqrt[3]2)^2}$
I'm given what I need to rationalize it, namely $\displaystyle\frac{(a^2-2bc)+(-ab+2c^2)\sqrt[3]2+(b^2-ac)\sqrt[3]2^2}{(a^2-2bc)+(-ab+2c^2)\sqrt[3]2+(b^2-ac)\sqrt[3]2^2}$
But I fail to see the motivation behind this. Can someone give a more general explanation of how to come to this expression?
|
You can multiply it by $1+d\sqrt[3]2+e\sqrt[3]4$ and insist the product not have any factors of $\sqrt[3]2$ or $\sqrt[3]4$. So
$$(a+b\sqrt[3]2+c\sqrt[3]4)(1+d\sqrt[3]2+e\sqrt[3]4)=\\a+ebe+2cd+\sqrt[3]2(ad+b+2ce)+\sqrt[3]4(ae+bd+c)$$
So we need $ad+b+2ce=0, ae+bd+c=0$, giving $$e=-\frac {ad+b}{2c}\\-a^2d-ab+2cbd+2c^2=0\\(a^2-2bc)d=2c^2-ab\\(a^2-2bc)e=b^2-ac$$ Then we make the leading term $a^2-bc$ to clear the fractions
|
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|
How do I integrate $\int \frac{x+1}{2\sqrt{x+1}} \mathrm dx$? I know how to do the bottom part, but I can't figure it out how to get $x+1$ on top
$$\int \frac{x+1}{2\sqrt{x+1}} \mathrm dx$$
|
$$\int \frac {x+1}{2(x+1)^{1/2}} = \int\frac{x+1}{2\sqrt{x+1}}\,dx.$$ Let $u=(x+1)^{1/2}$, so $du=\dfrac{1}{2(x+1)^{1/2}}\,dx.\;$ Substitution gives us $$\int u^2\,du\;\;=\;\;\frac{u^3}{3}+C.$$
Back substitute, and our answer is $$\frac {((x+1)^{1/2})^3}{3} \;=\;\frac{(x+1)^{3/2}}{3}+C.$$
This can be approached in a much more straightforward manner by noting that:
$$\frac{x+1}{2(x+1)^{1/2}}\;= =\;\frac12(x+1)^{ 1 - (1/2)}\;=\frac12(x+1)^{1/2}$$
Then since $\dfrac{d}{dx}(x+ 1) = 1$, we simply integrate: $$\int \frac12(x+1)^{1/2} \,dx \;\;=\;\; \frac12\frac{(x+1)^{3/2}}{\frac32} + C \;\;= \;\;\frac{(x+1)^{3/2}}{3} + C$$
|
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|
probability distribution $X, Y$ and $X+Y$ A box contains $5$ ticket, $\{ 0 , 0 , 0 , 4 , 4\}$.
Drawing two tickets at random w/o replacement.
$X$ be the sum of the first two draws and Y be the outcome of the first draw.
Question: Find distribution of $X + Y$?
*
*What I did --
I found the distribution of $X$. There are only three possible sums: $0 , 4, 8$. And there are $10$ ways to get these sums: $(0,0), (0,0), (0,4),(0,4),(0,0),(0,4),(0,4),(0,4),(0,4) (4,4)$. Therefore $P(0) = 3/10 , P(4) = 6/10, P(8) = 1/10$.
My other question: Is there another way (different from what I did) to get the distribution of $X$? And I have no idea how to get distribution of $Y$.
|
For the distribution of $X$, I do not think there is a substantially better way than yours.
For the distribution of $X+Y$, we do something similar. If it helps to keep track of things, draw a tree diagram. Let $W=X+Y$.
Case $1$: Maybe the first pick was $0$. This has probability $\frac{3}{5}$. Given that this happened, the probability of picking a $0$ next is $\frac{2}{4}$. Then $W=0$. The probability of piking a $4$ is $\frac{2}{4}$. Then $W=4$.
Case $2$: Maybe the first pick was a $4$. This has probability $\frac{2}{5}$. In that case, the probability of next getting $0$ is $\frac{3}{4}$, and we get $W=8$. The probability of getting a $4$ is $\frac{1}{4}$. In that case $W=12$.
So $W$ takes on values $0$, $4$, $8$, $12$.
For the probability that $W=0$, locate all the ways $W$ can be $0$. This only happens in one way, $0$ then $0$. We can see that the probability is $\frac{3}{5}\frac{2}{4}$.
Similarly, the probability that $W=4$ is $\frac{3}{5}\frac{2}{4}$.
Similarly, $\Pr(W=8)=\frac{2}{5}\frac{3}{4}$ and the probability that $W=12$ is $\frac{2}{5}\frac{1}{4}$.
As a partial correctness check, add up our four numbers. We should get $1$, and we do.
|
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|
Simplifying $\frac{1}{x} + \frac{5+x}{x+1} - \frac{7x^2 + 3}{(x+2)^2}$ I'm having trouble simplifying this expression:
$$\frac{1}{x} + \frac{5+x}{(x+1)} - \frac{7x^2 + 3}{(x+2)^2}$$
Would you first do the addition or subtraction?
What's the steps to solve this?
The final answer is
$$\frac{-6x^4 + 3x^3 + 26x^2 + 25x + 4}{x^4 + 5x^3 + 8x^2 + 4x}.$$
Thanks.
|
HINT
The guiding idea is the same as when you're evaluating $\frac{1}{3} + \frac{3}{4}$, which is to say that you find a common denominator. In my example, it would be $3\cdot 4$. In yours, it would be...
|
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|
How many solutions to prime = $a^3+b^3+c^3 - 3abc$ Let $a,b,c$ be integers.
Let $p$ be a given prime.
How to find the number of solutions to $p = a^3+b^3+c^3 - 3abc$ ?
Another question is ; let $w$ be a positive integer. Let $f(w)$ be the number of primes of type prime = $a^3+b^3+c^3 - 3abc$ below $w$. How does the function $f(w)$ behave ? How fast does it grow ? Are those primes of type $A$ $mod$ $B$ for some integers $A$ and $B$ ?
How to deal with this ?
Can this be solved without computing the class number ?
|
We may also prove that there are infinitely many primes when $a+b+c=1$. This answer along with Andre Nicolas' answer and Will Jagy's answer allows us to calculate the exact number of representations $a^3+b^3+c^3-abc=p$ for each prime $p$.
We begin with the factorization $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca).$$ When $a+b+c=1$, we have that $$a^2+b^2+c^2-ab-bc-ca=3a^2+3ab+3b^2-3a-3b+1.$$ Rearranging, this becomes $$\frac{3}{4}\left(2a+b-1\right)^2+\frac{1}{4}\left(3b-1\right)^2.$$ If $x=a+2b-1,$ and $y=a-b,$ then the above is $$\frac{3}{4}\left(x+y\right)^{2}+\frac{1}{4}\left(x-y\right)^{2}=x^{2}+xy+y^{2}.$$ This is the norm form for $\mathbb{Z}\left[\zeta_{3}\right],$ the Eisenstein integers, and we know exactly which primes $p$ can be represented as $$p=x^{2}+xy+y^{2}.$$ To count the number of such representations, see my answer here. Note that this happens if and only if $p\equiv1 \pmod{6}$, and so we see that when $a+b+c=1$, the polynomial $a^3+b^3+c^3-3abc$ represents precisely those primes which are $1$ modulo $6$.
|
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|
Fractions in binary? How would you write a fraction in binary numbers; for example, $1/4$, which is $.25$? I know how to write binary of whole numbers such as 6, being $110$, but how would one write fractions?
|
Not sure how helpful this is but whenever I work with binary I always use a table.
So if you want 0.25 or 15.75 (base 10) in binary:
\begin{array}{r|rrrr|rrrr}
& 8 & 4 & 2 & 1 & \frac{1}{2} & \frac{1}{4} & \frac{1}{8} & \frac{1}{16} \\
& 8 & 4 & 2 & 1 & 0.5 & 0.25 & 0.125 & 0.0625 \\
\hline
0.25 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
15.75 & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 0
\end{array}
Gives 0.01 and 1111.11 respectively.
And another way:
15.375 to binary for example:
\begin{align}
15 - 8 & = 7, & \quad 7 \ge 0 & \to 1 \\
7 - 4 & = 3, & 3 \ge 0 & \to 1 \\
3 - 2 & = 1, & 1 \ge 0 & \to 1 \\
1 - 1 & = 0, & 0 \ge 0 & \to 1 \\
\\
0.375 - 0.5 & = -0.125, & -0.125 \lt 0 & \to 0 \\
0.375 - 0.25 & = 0.125, & 0.125 \ge 0 & \to 1 \\
0.125 - 0.125 & = 0, & 0 \ge 0 & \to 1
\end{align}
Gives 1111.011.
|
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|
Question about Geometry involving angles and lines
The answer is C however if angle ACD is 110 degrees and angle AB is 110 degrees how does it equal 180?
|
Since it's an isosceles triangle, angles $B, C$ (the angles opposite segments $AC, AB$ respectively, which are equal in length) are equal in measure:
$$ \angle B = \angle C$$
We also know the the sum of the measures of the angles of a triangle sum to $180^\circ$.
$$\angle A + \angle B + \angle C = \angle A + 2 \angle C = 180^\circ $$ $$\implies 40^\circ + 2 \angle C = 180^\circ \implies 2\angle C = 180^\circ - 40^\circ = 140^\circ$$
Solve for (the measure of) angle $C$: $$\angle B = \angle C = \dfrac{140^\circ}{2} = 70^\circ$$
Since $\angle C$ and $\angle ACD$ (the unknown angle) are supplementary angles (since $B, C, D$ are colinear: they lie on the same straight line), we know that the sum of the measures of $\angle C$ and $\angle ACD$ is $180^\circ$
$$\angle C + \angle ACD = 180^\circ\implies \angle ACD = 180^\circ - \angle C = 180^\circ - 70^\circ = 110^\circ$$
The answer is, indeed, that the measure of $$\angle ACD = 110^\circ$$
|
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|
Non-induction proof of $2\sqrt{n+1}-2<\sum_{k=1}^{n}{\frac{1}{\sqrt{k}}}<2\sqrt{n}-1$
Prove that $$2\sqrt{n+1}-2<\sum_{k=1}^{n}{\frac{1}{\sqrt{k}}}<2\sqrt{n}-1.$$
After playing around with the sum, I couldn't get anywhere so I proved inequalities by induction. I'm however interested in solutions that don't use induction, if there are some (relatively simple ones, since I'm high-school student).
Also any advice for determining if a sum can be written in "compact" form? For example, $\displaystyle \sum_{k=1}^{n}{(-1)^{k-1}k}$ is actually $\displaystyle-\frac{n}{2}$ for even $n$ and $\displaystyle\frac{n+1}{2}$ for odd $n$.
|
Mean Value Theorem can also be used,
Let $\displaystyle f(x)=\sqrt{x}$
$\displaystyle f'(x)=\frac{1}{2}\frac{1}{\sqrt{x}}$
Using mean value theorem we have:
$\displaystyle \frac{f(n+1)-f(n)}{(n+1)-n}=f'(c)$ for some $c\in(n,n+1)$
$\displaystyle \Rightarrow \frac{\sqrt{n+1}-\sqrt{n}}{1}=\frac{1}{2}\frac{1}{\sqrt{c}}$....(1)
$\displaystyle \frac{1}{\sqrt{n+1}}<\frac{1}{\sqrt{c}}<\frac{1}{\sqrt{n}}$
Using the above ineq. in $(1)$ we have,
$\displaystyle \frac{1}{2\sqrt{n+1}}<\sqrt{n+1}-\sqrt{n}<\frac{1}{2\sqrt{n}}$
Adding the left part of the inequality we have,$\displaystyle\sum_{k=2}^{n}\frac{1}{2\sqrt{k}}<\sum_{k=2}^{n}(\sqrt{k}-\sqrt{k-1})=\sqrt{n}-1$
$\Rightarrow \displaystyle\sum_{k=2}^{n}\frac{1}{\sqrt{k}}<2\sum_{k=2}^{n}(\sqrt{k}-\sqrt{k-1})=2(\sqrt{n}-1)$
$\Rightarrow \displaystyle1+\sum_{k=2}^{n}\frac{1}{\sqrt{k}}<1+2\sum_{k=2}^{n}(\sqrt{k}-\sqrt{k-1})=2\sqrt{n}-2+1=2\sqrt{n}-1$
$\Rightarrow \displaystyle\sum_{k=1}^{n}\frac{1}{\sqrt{k}}<2\sqrt{n}-1$
Similarly adding the right side of the inequality we have,
$\displaystyle\sum_{k=1}^{n}\frac{1}{2\sqrt{k}}>\sum_{k=1}^{n}(\sqrt{k+1}-\sqrt{k})=\sqrt{n+1}-1$
$\Rightarrow \displaystyle\sum_{k=1}^{n}\frac{1}{\sqrt{k}}>2(\sqrt{n+1}-1)$
This completes the proof.
$\displaystyle 2\sqrt{n+1}-2<\sum_{k=1}^{n}{\frac{1}{\sqrt{k}}}<2\sqrt{n}-1.$
|
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|
If $a,b \in \mathbb R$ satisfy $a^2+2ab+2b^2=7,$ then find the largest possible value of $|a-b|$ I came across the following problem that says:
If $a,b \in \mathbb R$ satisfy $a^2+2ab+2b^2=7,$ then the largest possible value of $|a-b|$ is which of the following?
$(1)\sqrt 7$, $(2)\sqrt{7/2}$ , $(3)\sqrt {35}$ $(4)7$
My Attempt: We notice, $(a-b)^2=7-(4ab+b^2)$ and hence $|a-b|=\sqrt {7-(4ab+b^2)}$ and so $|a-b|$ will be maximum whenever $(4ab+b^2)$ will be minimum. But now I am not sure how to progress further hereon.Can someone point me in the right direction? Thanks in advance for your time.
|
As the distance between points is invariant 1,2 under Rotation,
using Rotation of axes by putting $a=x\cos\theta-y\sin\theta$ and $b=x\sin\theta+y\cos\theta$ in $a^2+2ab+2b^2=7$
$(x\cos\theta-y\sin\theta)^2+2(x\cos\theta-y\sin\theta)(x\sin\theta+y\cos\theta)+2(x\sin\theta+y\cos\theta)^2=7$
or, $x^2(\cos^2\theta+2\cos\theta\sin\theta+2\sin^2\theta)$
$+xy(-2\cos\theta\sin\theta+2\cos^2\theta-2\sin^2\theta+4\cos\theta\sin\theta)+y^2(\sin^2\theta-2\cos\theta\sin\theta+2\cos^2\theta)=7$
Applying $\cos2\theta=2\cos^2\theta-1=1-2\sin^2\theta$ and $\sin2\theta=2\cos\theta\sin\theta,$
$x^2(3-\cos2\theta+2\sin2\theta)+2xy(2\cos2\theta+\sin2\theta)+y^2(3+\cos2\theta-2\sin2\theta)=14$
To remove $xy$ term we put $2\cos2\theta+\sin2\theta=0\implies \frac{-\sin2\theta}2=\frac{\cos2\theta}1=\pm\frac1{\sqrt{2^2+1^2}}$
Taking the '$+$' sign and on simplification we get $\frac{x^2}{\frac{7(3+\sqrt5)}2}+\frac{y^2}{\frac{7(3-\sqrt5)}2}=1$
Comparing with $\frac{x^2}{c^2}+\frac{y^2}{d^2}=1,$
the extreme values of $x$ are $\pm\sqrt{\frac{7(3+\sqrt5)}2}=\pm\frac{\sqrt7}2\sqrt{6+2\sqrt5}=\pm\frac{\sqrt7}2\sqrt{(\sqrt5+1)^2}=\pm\frac{\sqrt{35}+\sqrt7}2$
Similarly, the respective extreme values of $y$ are $\pm\frac{\sqrt{35}-\sqrt7}2$
So, maximum value of $|a-b|=\left|\frac{\sqrt{35}+\sqrt7}2-(\frac{\sqrt7-\sqrt{35}}2)\right|=\sqrt{35}$
|
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|
Problem with graphing linear equations Well, I can understand how to graph basic liner equations, for example:
$$y=2x-4$$
The y-intercept would be -4 and the slope would be 2. The coordinates could then be (0,-4)(1, -2)
However, how would I solve a linear equation like this: $$y = \frac{2x}{4}$$
What are the steps to find out the coordinates? The only relationship that I know that can possibly help me is: $$\frac{x}{4}=\frac{1}{4}x$$
|
Strong Hint:
The equation $y = \dfrac{2x}{4}$ is read as the linear relationship described by multiplying the $x$-value by $\dfrac{2}{4}$.
Since the numerator $2$ and the denominator $4$ are both even, we can simplify this fraction, meaning to reduce the fraction to lowest terms. To do that, we divide the numerator and the denominator by their greatest common divisor.
In symbols, the greatest common divisor of two integers $a$ and $b$ is expressed as $\gcd(a, b) = c$ such that $c$ is their greatest common divisor. By order of substitution, we consider $a = 2$ and $b = 4$.
Since $a$ and $b$ are even, their greatest common divisor is a product of $2$, namely itself: $\gcd(2, 4) = 2$
Therefore, we can simplify the fraction $\dfrac{2}{4}$ by dividing the numerator and denominator each by $2$.
Since the numerator is $2$ and the denominator is $4$, we simplify our fraction to $\dfrac{1}{2}$.
Therefore, $y = \dfrac{2x}{4} = \bigg(\dfrac{2}{4}\bigg)x = \dfrac{1}{2}x$.
We cannot simplify the fraction anymore because $\gcd(1, 2) = 1$ and so this fraction is most simplified. It is irreducible. Now, we look back at the general linear equation, which is the general linear relationship of the $x$-value to the $y$-value. This equation is in the following form: $$y = mx + c$$ for which $x$ is the independent variable and $y$ is the dependent variable.
Since $y = \dfrac{1}{2}x$, then by order of substituion, $m = \dfrac12$ and $c = 0$. Therefore, $y = \dfrac12 x + 0$.
What is the value of the slope?
What is the value of the $y$-intercept?
What coordinates would this linear relationship generate?
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\sum_{j = 0}^{n} (-\frac{1}{2})^j = \frac{2^{n+1} + (-1)^n}{3 \times 2^n}$ whenever $n$ is a nonnegative integer. I'm having a really hard time with the algebra in this proof. I'm supposed to use mathematical induction (which is simple enough), but I just don't see how to make the algebra work.
$\sum_{j = 0}^{k} (-\frac{1}{2})^k + (-\frac{1}{2})^{k + 1} = \frac{2^{k+1} + (-1)^k}{3 \times 2^k}+(-\frac{1}{2})^{k + 1}$, by adding $(-\frac{1}{2})^{k + 1}$ to both sides.
I want to show that the right side is equal to:
$\frac{2^{k+1+1} + (-1)^{k+1}}{3 \times 2^{k+1}}$
Thank you!
|
Try writing your sum
$$\sum_{j = 0}^{k} \left(-\frac{1}{2}\right)^k + \left(-\frac{1}{2}\right)^{k + 1} $$
$$ =\frac{2^{k+1} + (-1)^k}{3 \times 2^k} + \frac{(-1)^{k+1}}{2^{k+1}} $$
$$ = \frac{2^{k+2} + 2(-1)^k}{3 \times 2^{k+1}} + \frac{(-1)^{k+1}}{2^{k+1}} \;$$
$$ = \frac{2^{k+2} + 2(-1)^k}{3 \times 2^{k+1}} + \frac{3(-1)^{k+1}}{3\times 2^{k+1}}$$
$$ = \quad\quad?$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find Maclaurin series of $(\sin(x^3))^{1/3}$ How do I find Maclaurin series for the function:
$$\sqrt[3]{\sin(x^3)}$$
The answer should be:
$$ x - \frac {x^7}{18} - \frac {{x}^{13}}{3240} + o(x^{13})$$
I tried:
$$\sin x = x - \frac {x^3}{3!} + \frac {x^5}{5!} - \frac {x^7}{7!} + ...$$
So, I changed $x$ to $x^3$ to get:
$$\sin(x^3) = x^3 - \frac {x^9}{3!} + \frac {x^{15}}{5!} - \frac {x^{21}}{7!} + ...$$
But, I'm stuck when it comes to power of 1/3:
$$\sqrt[3]{x^3 - \frac {x^9}{3!} + \frac {x^{15}}{5!} - \frac {x^{21}}{7!} + ...} = a_0+a_1x+a_2x^2+a_3x^3+...$$
|
If you factor out $x^3$ from cubic root you'll get
$$
x\sqrt[3]{1-\left (\frac{x^6}{3!}-\frac{x^{12}}{5!} +o(x^{12})\right )}
$$
Now, use power series expansion for cubic root
$$
\sqrt[3]{1-x} = 1-\frac x3-\frac {x^2}9+o(x^2) \\
\sqrt[3]{1-\left (\frac{x^6}{3!}-\frac{x^{12}}{5!} +o(x^{12})\right )} = 1-\frac 13\left(\frac{x^6}{3!}-\frac{x^{12}}{5!} \right)-\frac 19\left(\frac {x^{12}}{3!3!}+o(x^{12}) \right ) = \\
\\= 1-\frac {x^6}{3\cdot 3!}+\left ( \frac 1{3\cdot5!}-\frac 1{9\cdot 3! \cdot 3!}\right)x^{12}+o(x^{12}) = 1-\frac {x^6}{18}-\frac {x^{12}}{3240}+o(x^{12})
$$
After multiplying to $x$ you'll get your answer.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integrating :$\int\sqrt{\sin x} \cos^{\frac{3}{2}}x dx$ How to integrate :
$$\int\sqrt{\sin x} \cos^{\frac{3}{2}}x dx$$
|
Let's make a change of variables $u = \sin^2(x)$. Formally, $\sqrt{\sin(x)} = u^{1/4}$, $\cos^{3/2}(x) = (1-u)^{3/4}$, and $\mathrm{d}x = \frac{\mathrm{d}u}{2 \sqrt{u} \sqrt{1-u}}$.
Thus:
$$
\int \sqrt{\sin( x)} \cos^{3/2}( x) \, \mathrm{d}x = \frac{1}{2}\int u^{-1/4} (1-u)^{1/4} \mathrm{d} u
$$
In another answer of mine I show how to use differentiation properties of the Gauss's hypergeometric function ${}_2F_1$ to evaluate:
$$
\int \left(1-u\right)^a u^b \mathrm{d}u = \frac{u^{b+1}}{b+1} {}_2 F_1\left( \left. \begin{array}{cc} -a & b+1 \cr &b+2& \end{array} \right| u \right) +\color\gray{\text{const.}}
$$
Using the above for $b=-1/4$ and $a=1/4$:
$$
\int u^{-1/4} (1-u)^{1/4} \mathrm{d} u = \frac{4}{3} u^{3/4} \cdot {}_2 F_1\left( \left. \begin{array}{cc} -1/4 & 3/4 \cr &7/4& \end{array} \right| u \right) +\color\gray{\text{const.}}
$$
Recombining we get:
$$
\int \sqrt{\sin( x)} \cos^{3/2}( x) \, \mathrm{d}x = \frac{2}{3} \sin^{3/2}(x) \cdot {}_2 F_1\left( \left. \begin{array}{cc} -1/4 & 3/4 \cr &7/4& \end{array} \right| \sin^2(x)\right) +\color\gray{\text{const.}} \tag{$\ast$}
$$
Since we use formal operation, like $\sqrt{\sin(x)} = \sqrt{\sqrt{u}} \stackrel{?}{=} u^{1/4}$ we should differentiate $(\ast)$ to check the result. Differentiating we get:
$$
\frac{\mathrm{d}}{\mathrm{d} x} \left( \frac{2}{3} \sin^{3/2}(x) \cdot {}_2 F_1\left( \left. \begin{array}{cc} -1/4 & 3/4 \cr &7/4& \end{array} \right| \sin^2(x)\right) \right) = \sqrt{\sin(x)} \cos(x) \left(\cos^2(x)\right)^{1/4}
$$
The above is different from the original integrand by a factor of $\frac{(\cos^2(x))^{1/4}}{\sqrt{\cos(x)}}$ which is a differential constant (whose fourth power simplifies to 1, and which equals 1 where $\cos(x)>0$), and hence we can adjust the $(\ast)$ by simply dividing over it, giving:
$$
\int \sqrt{\sin( x)} \cos^{3/2}( x) \, \mathrm{d}x = \frac{2}{3} \frac{\sqrt{\cos(x)} \, \sin^{3/2}(x)}{(\cos^2(x))^{1/4}} \cdot {}_2 F_1\left( \left. \begin{array}{cc} -1/4 & 3/4 \cr &7/4& \end{array} \right| \sin^2(x)\right)+\color\gray{\text{const.}}
$$
which makes us realize that the constant of integration means a differential constant.
|
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|
Need to prove $\frac{3}{5}(2^{\frac{1}{3}}-1)\le\int_0^1\frac{x^4}{(1+x^6)^{\frac{2}{3}}}dx\le1$ I need to show that
$$\frac{3}{5}(2^{\frac{1}{3}}-1)\le\int_0^1\frac{x^4}{(1+x^6)^{\frac{2}{3}}}dx\le1$$
I just know that if in $[a,b]$, $f(x)\le g(x)\le h(x)$, then
$$\int_a^bf(x)dx\le\int_a^bg(x)dx\le\int_a^bh(x)dx$$
How do I get a function like $f$ and $h$ in the range $[0,1]$, for the function $g(x)=\frac{x^4}{(1+x^6)^{\frac{2}{3}}}$? I mean, is there an approach without solving backwards?
|
Consider left part first. You need to get integral there, something like $f(1)-f(0)$
$$
\frac 35 \left ( 2^{\frac 13}-1\right ) = \frac 35 \left ( (1+\fbox 1)^{\frac 13}-(1+\fbox 0)^{\frac 13}\right ) =\frac 35 \left . \left ( 1+x^n\right )^{\frac 13} \right |_0^1
$$
Now, you can imagine that values inside of the boxes are values of almost any power of $x$, since $$1^n = 1 \\
0^n = 0$$
So $f(x) = \frac 35 \left (1+x^n \right )^{\frac 13}$, and therefore $f'(x) = \frac n5 \left ( 1+x^n\right )^{-\frac 23}x^{n-1}$
To choose which power should be there, just pay attention to the integral given. It has $x^4$ in enumerator, and $(1+x^6)$ in denominator, so power could be either 5 (to match enumerator after differentiating) or 6 (to match denominator).
$$
n = 5 \\
f(x) = \frac 35 \left (1+x^5 \right )^{\frac 13} \\
f'(x) = \frac {x^4}{\left ( 1+x^5\right)^{\frac 23}}
$$
Now, all you have to prove is
$$
\frac{x^4}{\left (1+x^5 \right)^{\frac 23}} \le \frac {x^4}{\left (1+x^6 \right)^{\frac 23}},\quad \text{for}\ \forall x \in [0,1]
$$
which is equivalent to
$$
1 + x^6 \le 1 + x^5
$$
or
$$
x \le 1
$$
which is true.
If you choose to match denominator
$$
n = 6 \\
f(x) = \frac 35 \left ( 1+x^6\right )^{\frac 13} \\
f'(x) = \frac 65 \left ( 1+x^6\right)^{-\frac 23} x^5
$$
You have to prove that
$$
\frac {6 x^5}{5\left ( 1+x^6\right)^{\frac 23}} \le \frac {x^4}{\left (1+x^6 \right)^{\frac 23}},\quad \text{for}\ \forall x \in [0,1]
$$
which is equivalent to
$$
6x \le 5
$$
which is not true for all $x \in [0,1]$, so it's better to stick with $n = 5$.
Right part is easier, I can put some notes if it's necessary.
|
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|
How do I completely solve the equation $z^4 - 2z^3 + 9z^2 - 14z + 14 = 0$ where there is a root with the real part of $1$. I would please like some help with solving the following equation:
$$z^4 - 2z^3 + 9z^2 - 14z + 14 = 0$$
All I know about the equation is that there is a root with the real part of $1$.
My approach has been to factor out the root: $1 + yi$ and divide the equation with it. The following calculations (using the division algorithm) is quite cumbersome, and I wonder if there is any better way of doing this?
Thank you kindly for your help!
|
As the coefficients of the different powers of $z$ are real,
if one of the four roots is $1+yi,$ the other must be its conjugate $1-yi$ .
If the other two roots are $x,w,$
then using Vieta's Formulas for the coefficient of $z^3$, $1+yi+1-yi+x+w=2\implies w=-x$
So, $$(z-x)(z+x)\{z-(1+yi)\}\{z-(1-yi)\}=0$$
$$\implies (z^2-x^2)(z^2-2z+1+y^2)=0$$
$$\implies z^4-2z^3+z^2(1+y^2-x^2)+2x^2z-x^2(1+y^2)=0$$
Now compare the coefficients of the different powers of $z$
So, $2x^2=-14,x^2=-7,x=\pm \sqrt{-7}=\pm \sqrt7i$ and $1+y^2-x^2=9\implies 1+y^2+7=9,y^2=1$
So, the roots are $\pm \sqrt7i,1\pm i$
|
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|
Rewrite so that the denominator does not have any root expressions: $\frac{\sqrt[3]{49} +\sqrt[3]{7x} + \sqrt[3]{x^2}}{\sqrt[3]{x} -\sqrt[3]{7}}$ I am struggling with rewriting the following so that the denominator does not have any root expressions:
$$\frac{\sqrt[3]{49} +\sqrt[3]{7x} + \sqrt[3]{x^2}}{\sqrt[3]{x} -\sqrt[3]{7}}$$
I guess I should start with the denominator and try to get rid of the cube root expressions. But I cannot really get how one would do that easily. Is there another way to solve this problem?
Thank you kindly for your help!
|
Hint:$a^3-b^3=(a-b)(a^2+ab+b^2)$
Let $a=x^{1/3},b=7^{1/3}$
$x-7=a^3-b^3=(a-b)(a^2+ab+b^2)=(x^{1/3}-7^{1/3})(7^{2/3}+(7x)^{1/3}+x^{2/3})$
$\displaystyle \frac{(7^{2/3}+(7x)^{1/3}+x^{2/3})}{(x^{1/3}-7^{1/3})}=\frac{(7^{2/3}+(7x)^{1/3}+x^{2/3})}{(x^{1/3}-7^{1/3})}\frac{(7^{2/3}+(7x)^{1/3}+x^{2/3})}{(7^{2/3}+(7x)^{1/3}+x^{2/3})}=\frac{(7^{2/3}+(7x)^{1/3}+x^{2/3})^2}{(x-7)}$
|
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|
Factorization of cyclic polynomial
Factorize $$a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)$$
Since this is a cyclic polynomial, factors are also cyclic
$$f(a) = a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)$$
$$f(b) = b(b^2-c^2)+b(c^2-b^2)+c(b^2-b^2) = 0 \Rightarrow a-b$$
is a factor of the given expression. Therefore, other factors are $(b-c)$ and $(c-a)$. The given expression may have a coefficient a constant factor which is nonzero. Let it be $m$.
$$\therefore a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2) = m(a-b)(b-c)(c-a)$$
Please guide further on how to find this coefficient.
|
\begin{align}
a (b^2-c^2)+b (c^2-a^2)+c (a^2-b^2)
&=ab^2-ac^2+bc^2-ba^2+ca^2-cb^2\\
&=a^2c-a^2b+ab^2-ac^2+bc^2-cb^2\\
&=a^2 (c-b)-a (c^2-b^2)+bc (c-b)\\
&=(c-b)(a^2-ac-ab+bc)\\
&=(c-b)(bc-ab-ac-a^2)\\
&=(c-b)(b (c-a)-a (c-a))\\
&=(c-b)(b-a)(c-a)\\
&=(a-b)(b-c)(c-a)
\end{align}
|
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|
Obtaining the pdf of the sum of two iid random variables with pdf $f_{X}(x)= \sqrt{\frac{1}{2\pi x}e^{-x/2}}$ X1, and X2 are independent with the pdf: $f_{X}(x)= \sqrt{\frac{1}{2\pi x}e^{-x/2}}$ defined for x>0
Y=X1+X2
What is the pdf of Y?
This is what I did so far:
$$f_{Y}(y) = \int_{0}^{\infty } f_{X_{1}}(y-x_{2})f_{X_{2}}(x_{2})dx_{2}$$
$$ = \int_{0}^{\infty } \sqrt{\frac{1}{2\pi (y-x_{2})}e^{-(y-x_{2})/2}} \sqrt{\frac{1}{2\pi x_{2}}e^{-x_{2}/2}} dx_{2}$$
$$=\int_{0}^{\infty } \sqrt{\frac{1}{2\pi (y-x_{2})}e^{(-y+x_{2})/2}} \sqrt{\frac{1}{2\pi x_{2}}e^{-x_{2}/2}} dx_{2}$$
$$=\int_{0}^{\infty } \sqrt{\frac{1}{2\pi (y-x_{2})}e^{(-y+x_{2})/2} \frac{1}{2\pi x_{2}}e^{-x_{2}/2}} dx_{2}$$
$$=\int_{0}^{\infty } \sqrt{\frac{1}{2\pi (y-x_{2})} \frac{1}{2\pi x_{2}}e^{-y/2}} dx_{2}$$
$$=\frac{1}{2\pi} \sqrt{e^{-y/2}} \int_{0}^{\infty } \sqrt{\frac{1}{ (y-x_{2})x_{2}} } dx_{2}$$
$$=\frac{1}{2\pi} \sqrt{e^{-y/2}} \int_{0}^{\infty } \sqrt{\frac{1}{ yx_{2}-(x_{2})^{2}} } dx_{2}$$
At this point I wasn't sure how to proceed with the integral, so I went to wolfram alpha. According to it, this integral does not converge, so not sure what to do now.
|
Effectively $x_2$ only goes to $y$, not to $\infty$. (The density is $0$ at negatives.)
For the work that remains, complete the square. After making the right substitution, you should end up needing to find $\int\frac{du}{\sqrt{1-u^2}}$, which is $\arcsin u +C$.
|
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|
Find of $\int e^x \cos(2x) dx$ I did the following.
Using the LIATE rule:
$$\begin{align*}
u &=& \cos(2x)\\
u\prime &=& -2 \sin(2x)\\
v &=& e^x\\
v\prime&=&e^x
\end{align*}$$
We get:
$$\int e^x \cos(2x)dx = e^x \cos(2x) +2 \int e^x \sin(2x)dx $$
Now we do the second part.
$$\begin{align*}
u &=& \sin(2x)\\
u\prime &=& 2 \cos(2x)\\
v &=& e^x\\
v\prime&=&e^x
\end{align*}$$
We get:
$$\int e^x \sin(2x)dx = e^x \sin(2x) -2 \int e^x \cos(2x)dx$$
Putting it together we get:
$\begin{align*}
\int e^x \cos(2x)dx &=& e^x \cos(2x) +2 \int e^x \sin(2x)dx \\
&=&e^x \cos(2x) + 2[e^x \sin(2x) -2 \int e^x \cos(2x)dx]\\
&=&e^x \cos(2x) + 2e^x \sin(2x) -4 \int e^x \cos(2x)dx
\end{align*}$
$\begin{align*}
\int e^x \cos(2x)dx &=&e^x \cos(2x) + 2e^x \sin(2x) -4 \int e^x \cos(2x)dx\\
5\int e^x \cos(2x)dx &=&e^x(\cos(2x) + 2 \sin(2x))\\
\int e^x \cos(2x)dx &=&\frac{e^x( \cos(2x) + 2 \sin(2x))}{5}
\end{align*}$
I am not sure if this is right, if it is right, is there a better way of doing this.
|
We know that $\cos(2x)$ is the real part of $e^{2ix}$, so we'll calculate $\int e^xe^{2ix}dx $, then we take its real part. We have
$$\int e^xe^{2ix}dx=\frac{1}{2i+1}e^{(2i+1)x}+C=\frac{e^x}{5}(1-2i)(\cos(2x)+i\sin(2x))+C\\=\frac{1}{5}(e^x\cos2x+2e^x\sin2x)+C+i(\text{imaginary part}).$$
Now, you can conclude.
|
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|
Please, help me to find where is a mistake in the solutions of the equation. I have this equation and I will be very thankful to anyone who can provide me any help with the one discrepancy in my solution and the solution from the self-learning website:
$$
\frac{1+\tan(x) + \tan^2(x) + ... + \tan^n(x) + ...}{1-\tan(x) + \tan^2(x) - ... +(-1)^n\tan^n(x) ...}= 1+\sin(2x)
$$
My solution
I solved it in not too graceful way. First I found the roots for this equation:
$$
\require{cancel}
\begin{align}
&\boxed{\frac{1+\tan(x)}{1-\tan(x)}= 1+\sin(2x)} \\ \\
&\frac{(1+\tan(x))\cancel{(1-\tan(x))}}{\cancel{1-\tan(x)}}= (1+\sin(2x))(1-\tan(x))\\ \\
&\cancel{1}+\tan(x)=\cancel{1} - \tan(x) + \sin(2x) - \sin(2x)\tan(x) \\
&\frac{\sin(2x)}{\tan(x)} - \frac{\sin(2x)\cancel{\tan(x)}}{\cancel{\tan(x)}} -\frac{2\cancel{\tan(x)}}{\cancel{\tan(x)}} = 0\\ \\
&\bbox[Beige]{\boxed{\sin(2\alpha)=2\sin(\alpha)\cos(\alpha)}} \\ \\
&\frac{\cancel{2}\cancel{\sin(x)}\cos(x)\cos(x)}{\cancel{\sin(x)}} - \cancel{2}\sin(x)\cos(x) - \cancel{2} = 0 \\ \\
&\cos^2(x) - \sin(x)\cos(x) -1 = 0\\ \\
&\bbox[Beige]{\boxed{\sin^2(\alpha) + \cos^2(\alpha) = 1}} \\ \\
&\cancel{\cos^2(x)} - \sin(x)\cos(x) - \sin^2(x) - \cancel{\cos^2(x)} = 0\\ \\
&\sin^2(x) + \sin(x)\cos(x) = 0 \\ \\
&\sin(x)(\sin(x) + \cos(x)) = 0 \\ \\
&\sin(x) = 0 \implies \boxed{x_1=\pi k ,\space k \in \mathbb{Z}} \\ \\
&\sin(x) + \cos(x) = 0 \implies \tan(x) = -1 \implies \boxed{x_2=\frac{\pi}{4}\left(4m - 1\right) ,\space m \in \mathbb{Z}}
\end{align}
$$
Also I checked that there were no parasite and missing roots, cause I've divided the equation by $\tan(x)$ and multiplied it by $1-\tan(x)$. There were no such ones cause $\tan(x)$ cannot assume $1$, when $\sin(x)=0$ or $\tan(x) = -1$, and the missing roots with $\tan(x) = 0$ are a subset of $x_1$.
Then I checked which of these roots fitted the next equation:
$$
\frac{1+\tan(x) + \tan^2(x)}{1-\tan(x) + \tan^2(x)}= 1+\sin(2x)
$$
And it turned out that only $x_1$ did. Then my logic was: if there were another roots for the next steps of this sequence, e.g. $\frac{1+\tan(x) + \tan^2(x) + \tan^3(x)}{1-\tan(x) + \tan^2(x) - \tan^3(x)}= 1+\sin(2x)$ they would not fit the first step while $x_1$ would fit all steps. So the only solution is $x_1 = \pi k ,\space k \in \mathbb{Z}$
Possibly the wrong solution with my correction. You can see the original one at the www.bymath.com
It is obvious that the fraction numerator and denominator are geometrical sequences (progressions) with the common ratios $\tan(x)$ and $-\tan(x)$ correspondingly. Note, that here $|\tan(x)| < 1$, otherwise the left-hand side expression is meaningless. Therefore it is possible to transform the fraction numerator and denominator by the sum formula for the unboundedly decreasing geometric sequence (progression) $\bbox[Beige]{\boxed{S = \frac{b_1}{1-q}}}$, where $b$ is the first member of a sequence and $q$ in a ratio.
$$
\begin{align}
&\frac{1}{1 - \tan(x)} : \frac{1}{1 - (-\tan(x))} = 1 + \sin(2x) \\
&\frac{1}{1 - \tan(x)} * (1 + \tan(x)) = 1 + \sin(2x) \\
&\bbox[Beige]{\boxed{\sin(\alpha) = \frac{2\tan(\frac{\alpha}{2})}{1+\tan^2(\frac{\alpha}{2})}}} \\
&\frac{1 + \tan(x)}{1 - \tan(x)} - 1 - \frac{2\tan(x)}{1+\tan^2(x)} = 0\\ \\
&\frac{(1 + \tan(x))(1+\tan^2(x)) - (1 - \tan(x))(1+\tan^2(x)) - 2\tan(x)(1 - \tan(x))}{(1 - \tan(x))(1+\tan^2(x))} = 0\\ \\
&\frac{1 + \tan^2 x + \tan x + \tan^3 x -(1 + \tan^2 x - \tan x - \tan^3 x ) -2\tan x + 2\tan^2 x}{(1 - \tan x)(1+\tan^2 x)} =0 \\
\end{align}
$$
So here is the point where this second solution is wrong. The right is this:
$$
\begin{align}
&\frac{\cancel{1} + \cancel{\tan^2 x} + \cancel{\tan x} + \tan^3 x -\cancel{1} - \cancel{\tan^2 x} + \cancel{\tan x} + \tan^3 x -\cancel{2\tan x} + 2\tan^2 x}{(1 - \tan x)(1+\tan^2 x)} =0 \\
&\frac{2\tan^3 x + 2\tan^2 x}{(1 - \tan x)(1+\tan^2 x)} =0 \implies \tan x = 0, \ \tan x = -1 \\
\end{align}
$$
Whereas the author of the site got this: $\frac{4\tan^3 x}{(1 - \tan x)(1+\tan^2 x)} =0 \implies \tan x = 0$ and lost one root $x_2=\frac{\pi}{4}\left(4m - 1\right) ,\space m \in \mathbb{Z}$
ACTUALLY QUESTION
So when I looked to the second solution I was pretty confused with this:
in that solution it is said that unless $|tan(x)| < 1$ the left-hand side of the equation in meaningless. Why, can somebody explain, please? I suppose that this is not true cause:
*
*while the transformation with the sum formula was correct, the solution itself was wrong
*the right solution gives us two roots one of which does not fit all possible steps of the sequences, hence the numerator and denominator are not a unboundedly decreasing geometric sequence (progression).
Is my conclusion correct?
My sincerest appreciation.
|
If $|\tan x|\ge 1$ both numerator and denominator of the left-hand side are not finite, because the geometric series
$$N=1+\tan(x) + \tan^2(x) + ... + \tan^n(x) + ...$$
and
$$D=1-\tan(x) + \tan^2(x) - ... +(-1)^n\tan^n(x)+...$$
don't converge. If $|\tan x|< 1$, we have
$$N=\lim_{n\rightarrow \infty }\frac{\tan ^{n}x-1}{\tan x-1}=\frac{0-1}{\tan x-1}=\frac{1}{1-\tan x}$$
and
$$D=\lim_{n\rightarrow \infty }\frac{(-\tan x)^{n}-1}{(-\tan x)-1}=\frac{0-1}{-\tan x-1}=\frac{1}{1+\tan x};$$
and the given equation is equivalent to
$$\begin{eqnarray*}
\frac{1+\tan x}{1-\tan x} &=&1+\frac{2\tan x}{1+\tan ^{2}x} \\
\frac{1+\tan x}{1-\tan x}&=&\frac{(1+\tan ^{2}x)^{2}}{1+\tan ^{2}x},\qquad |\tan x|< 1,\tag{1}
\end{eqnarray*}$$
because
$$\sin 2x=\frac{2\tan x}{1+\tan ^{2}x}.$$
To solve equation $(1)$ we can observe that it is equivalent to
$$\frac{\left( 1+\tan x\right) ( 1+\tan ^{2}x) }{(1-\tan
x) \left( 1+\tan ^{2}x\right) }=\frac{(1+\tan x)^{2}\left( 1-\tan
x\right) }{\left( 1-\tan x\right) \left( 1+\tan ^{2}x\right) },\qquad |\tan x|< 1,$$
and to
$$\begin{eqnarray*}
\left( 1+\tan x\right) ( 1+\tan ^{2}x) &=&(1+\tan x)^{2}\left(
1-\tan x\right) \\
&\Leftrightarrow &1+\tan ^{2}x=(1+\tan x)\left( 1-\tan x\right) \\
&\Leftrightarrow &1+\tan ^{2}x=1-\tan ^{2}x \\
&\Leftrightarrow &\tan ^{2}x=-\tan ^{2}x \\
&\Leftrightarrow &2\tan ^{2}x=0 \\
&\Leftrightarrow &\tan x=0,
\end{eqnarray*}$$
whose single solution is $x_1=k\pi,k\in\mathbb{Z}$.
|
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|
equation $\displaystyle a_{n+1}.x^2-2x\sqrt{a^2_{1}+a^2_{2}+.......+a^2_{n}+a^2_{n+1}}+\left(a_{1}+a_{2}+.......+a_{n}\right) = 0$ has real roots The Largest positive integer $n$ such that for all real no. $a_{1},a_{2},.......,a_{n},a_{n+1}.$ The equation
$\displaystyle a_{n+1}.x^2-2x\sqrt{a^2_{1}+a^2_{2}+.......+a^2_{n}+a^2_{n+1}}+\left(a_{1}+a_{2}+.......+a_{n}\right) = 0$ has real roots.
My Try:: If Given equation has real roots. Then its Discriminant $\geq 0$
So $\displaystyle 4\left(a^2_{1}+a^2_{2}+.......+a^2_{n}+a^2_{n+1}\right)-4.a_{n+1}.\left(a_{1}+a_{2}+.......+a_{n}\right)\geq 0$
So $\left(a^2_{1}+a^2_{2}+.......+a^2_{n}+a^2_{n+1}\right)-a_{n+1}.\left(a_{1}+a_{2}+.......+a_{n}\right)\geq 0$
after that How can i solve it,plz explain me Thanks
|
Let $x=a_{n+1}$ then the last inequality is $x^2-xT+S \ge 0$ where $T=a_1+a_2+\cdots+a_n$ and $S=a_1^2+a_2^2+\cdots+a_n^2$. This requires its own discriminant $\Delta=T^2-4s\le 0$. As you try increasing the number of $a_i$'s you see at 4 elements the inequality can be made to work but not after that. Here is how it turns out after a bit of algebra
for $n=1$ we have
$-\Delta=3a_1^2\ge 0$.
for $n=2$ we have
$-\Delta=2(a_1^2+a_2^2)+(a_1-a_2)^2 \ge 0$.
for $n=3$ we have
$-\Delta=1(a_1^2+a_2^2+a_3^2)+(a_1-a_2)^2 +(a_2-a_3)^2+(a_3-a_1)^2\ge 0$.
for $n=4$ we have
$-\Delta=0(a_1^2+a_2^2+a_3^2+a_4^2) \\ +(a_1-a_2)^2+(a_1-a_3)^2+(a_1-a_4)^2 +(a_2-a_3)^2+(a_2-a_4)^2+(a_3-a_4)^2\ge 0$.
for $n=5$ the inequality does not work for all possible $a_i$'s. Use the suggestion of @vonbrand, above, of setting all of them equal to say $t$ to see $-\Delta=4(5t^2)-(5t)^2=-5t^2 <0 $ for $t\ne 0$.
|
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|
values of basic trig function, given values in terms of trig functions Get the function value if $\displaystyle \sin x = \frac{2}{\sqrt{13}}$ and $\displaystyle \cos x = -\frac{3}{\sqrt{13}}$ and the function value is going to be calculated from $\displaystyle \sin\left(2x-\frac{7\pi}{6}\right)$.
|
$$\sin(2x-\frac{7\pi}{6})=\sin2x\cos(7\pi/6)-\cos2x\sin(7\pi/6)=$$
$$=2\sin x\cos x\cos(7\pi/6)-(\cos^2x-\sin^2x)\sin(7\pi/6)=$$
$$=2(2/\sqrt 13)(-3/\sqrt13)\cos(7\pi/6)-(4/13-9/13)\sin(7\pi/6)=$$
$$=-12/13\cos(7\pi/6)+5/13\sin(7\pi/6)=$$
$$=-12/13(-\sqrt 3/2)+5/13(-1/2)=$$
$$=12\sqrt 3/26-5/26=(12\sqrt 3-5)/26$$
|
{
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|
Partial Fraction Decomposition $\frac{2x^2+3x+3}{(x+1)(x^2+1)}$ How would one go about decomposing this fraction?
$$
2x^2+3x+3\over
(x+1)(x^2+1)
$$
Here is what I have so far:
$$
{2x^2+3x+3\over
(x+1)(x^2+1)
} =
{ A\over x+1 } + { Bx+C\over x^2+1} $$
$$
{2x^2+3x+3\over
(x+1)(x^2+1)
} =
{ A(x^2+1) } + (Bx+C)(x+1) $$
I know how to solve for A - it's 1. But how do I get B and C?
|
$$
\begin{align}
2x^2 + 3x +3 &= A(x^2 + 1) + (Bx+C)(x+1)\\
&= x^2(A+B) + x(B+C) + A+C
\end{align}
$$
By comparing the corresponding co-efficients :
$$
A+B = 2;\\
B+C = 3;\\
C+A = 3;
$$
By solving, $A = 1$, $B = 1$, $C = 2$.
|
{
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|
How to find $E(f(f(f(\ldots f(x)))$ I have a random function $f(x)$ which returns one of the integers in the range $[0, x-1]$ with equal probability and $f(0) = 0$.
What is the expected value $E(f(f(f(\ldots f(x)))$ ($n$-times $f(x)$)? The answer should be a function of $x$ and $n$.
|
I fail to find a general pattern, but here are some results.
Write $E_n(x) = E(f^n(x))$.
It is easily seen that $E_n(x) = 0$ if $x \leq n$ and $E_n(n+1) = \frac{1}{(n+1)!}$.
Furthermore, if $x \geq 1$, we have
$$
E_1(x) = \frac{x-1}{x} + \dots + \frac{1}{x} = \frac{x-1}{2}
$$
$$
E_2(x) = \frac{1}{x}\sum_{y=1}^{x-1}\frac{y-1}{2} = \frac{1}{4}\frac{(x-1)(x-2)}{x} = \frac{1}{4}\left(x-3+\frac{2}{x}\right)
$$
$$
E_3(x) = \frac{1}{4x} \sum_{y=1}^{x-1}(y-3) + \frac{1}{2x}\sum_{y=1}^{x-1}\frac{1}{y} = \frac{(x-1)(x-4)}{8x} + \frac{H_{x-1}}{2x}
$$
The appearance of the harmonic sum $H_x$ makes me think that no simplification will be possible in general.
|
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|
Does there exists a polynomial $Q(x,y)$ such that $x-1=Q(x^2-1,x^3-1)$. I have a question: Does there exists a polynomial $Q(x,y)$ such that $x-1=Q(x^2-1,x^3-1)$.
I did as following: Let $Q(x,y)=\sum\limits_{k=0}^n\sum\limits_{i+j=k}a_{ij}x^iy^j$. Then I could find some coefficients: the coefficient of $(x^2-1)$ is $0$,coefficient of $x^3-1$ is $-\frac13$, ...
Now I get stuck. Is the proposition true? How to prove it?
|
The polynomials $x^2-1,x^3-1,(x^2-1)^2,(x^2-1)(x^3-1),(x^2-1)^3,(x^3-1)^2,(x^2-1)^2(x^3-1),(x^2-1)^4,(x^2-1)(x^3-1)^2$ are $9$ in number and all of degree at most $8$. I suspect they span the $8$-dimensional space of polynomials of degree at most $8$ vanishing at $x=1$. If that's right, then $x-1$ will be a linear combination of them. Worth a try, I think.
|
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|
Integers with 15 divisors (from brilliant.org) How many integers from 1 to 19999 have exactly 15 divisors?
Note: This is a past question from brilliant.org.
|
If $n = p_1^{a_1}p_2^{a_2} \ldots p_k^{a_k}$, then the number of divisors is given by $d(n) = (1+a_1)(1+a_2)\cdots(1+a_k)$. We are given that $d(n) = 15$. From this, show that there cannot be $3$ distinct primes diving $n$. Hence, $n=p_1^{a_1}$ or $n=p_1^{a_1} p_2^{a_2}$.
$1$. If $n=p_1^{a_1}$, then $(1+a_1) = 15 \implies a_1 = 14$. And since $n \leq 19999$, we have $p_1=2$.
$2$. If $n=p_1^{a_1}p_2^{a_2}$, (assume $a_1 \geq a_2$) then $(1+a_1)(1+a_2) = 15$ gives us $$a_1 = 4,a_2 = 2$$
$a_1 = 4$, then $p_1 \in \{2,3,5,7\}$, since $13^4 > 20000$ and $11^4 > \dfrac{20000}{2^2}$.
*
*$a_1 = 4$, $p_1=2$ gives us $p_2 \neq 2$ and $p_2^2 < \dfrac{20000}{2^4} \implies p_2 \leq 31$. Hence, number of options is $10$.
*$a_1 = 4$, $p_1=3$ gives us $p_2 \neq 3$ and $p_2^2 < \dfrac{20000}{3^4} \implies p_2 \leq 13$. Hence, number of options is $5$.
*$a_1 = 4$, $p_1=5$ gives us $p_2 \neq 5$ and $p_2^2 < \dfrac{20000}{5^4} \implies p_2 \leq 5$. Hence, number of options is $2$.
*$a_1 = 4$, $p_1=7$ gives us $p_2 \neq 7$ and $p_2^2 < \dfrac{20000}{7^4} \implies p_2 \leq 2$. Hence, number of options is $1$.
Hence, if $n$ has only one distinct prime divisor, then there is only one $n$ and if $n$ has two distinct prime divisor, then there are $10+5+2+1 = 18$ such $n$'s. Hence, the total number of $n$'s less than $20000$, with $15$ divisors is $19$.
|
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|
Showing $(x,y)$ pairs exist for $\sqrt{\quad\mathstrut}$ If we were to show that there exists infinitely many $(x,y)$ pairs in $\mathbb{Q}^2$ for which both $\sqrt{x^2+y^4}$ and $\sqrt{x^4+y^2}$ are rational. If the power root for $x$ and $y$ vary but never the same, how can we prove that we can always represent the outcome as a rational number? One way to do it would be to use the Pell equation. Is there any other ways?
|
Let $x=y=t^2-1$. Then $x^2+y^3=x^3+y^2=t^2(t^2-1)^2$ so both squareroots are $t(t^2-1)$, which is rational provided $t$ is rational. This is infinitely many pairs $(x,y).$
With a little work and Pell's equation, one can find pairs $(x,y)$ of form $(2t,t)$ which work, if it's a concern that $x=y$ in the above example.
EDIT I just noticed the OP needs distinct $x,y$. With $x=2t$ and $y=t$ we have
$$x^2+y^3=t^2(t+4),\ \ x^3+y^2=t^2(8t+1).$$
Thus we want $t+4=a^2, 8t+1=b^2.$ Solving the first and putting it into the second gives the Pell type equation
$$[1]\ \ b^2-8a^2=-31.$$
This has a solution $(b,a)=(13,5),$ while the associated Pell equation $b^2-8a^2=1$ has a solution $(b,a)=(3,1)$. So in the usual Pell fashion, infinitely many solutions to [1] may be obtained by multiplying out the expression
$$(13+5\sqrt{8})\cdot (3+\sqrt{8})^n$$
and putting it in the form $b+a\sqrt{8}$. Then such $a,b$ must be substituted back into one of $t+4=a^2,\ 8t+1=b^2$ to get the $t$ for which $(x,y)=(2t,t)$ gives rational squareroots.
|
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|
Please help me to find the value of $ABCDE$ (step by step) $ABCD\times E = DCBA$ with $A,B,C,D$, and $E$ distinct decimal digits (and $ABCD$ representing the concatenation of those digits). How can I find the value each of them?
|
2178*4 = 8712
Consider $E$ cannot be 1. otherwise $D=A$.If $E\ge 5$, then $A=1$, $D=E$, contradiction!
So $E\ge 2$, thus we have
*
*$E = 2$, then $A$ can only be $1,2,3,4$, since the product should be
4-digit number.
$DE$ has end digit as $A$, then $A$ is even, $A$ can be 2,4, since $A\neq E$, therefore $A=4$. Then $D = 7$($D\neq E$), but we look at the first digit of $\overline{DCBA}$, we should know that $D\ge E*A = 8$, contradiction!
*$E=3$, then $A$ can be $1,2,3$, and we exclude $1,3$ for the same reason. $A=2$, thus $D = 4$, But we expect $D\ge E*A = 6$.
*$E=4$, thus $A = 2$, $DE$ has end digit as $2$, thus $D=8$ or $D=3$, and $D\ge E*A =8$, THUS $D=8$.
Then we solve the other two unknowns by solving an equation:
$8000+400B+40C+32 = 8000+100C+10B+2 $,
which is
$39B+3 = 6C$, where we can see $B\le 1$, since $C\le 9$.
we have $(B,C) = (1,7)$, ONLY.
|
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|
How to integrate $\int \frac 1 {\begin{pmatrix}\frac1 3 t - 8\end{pmatrix}^5} \,dt$ by substitution? I am required to solve this question via integration by an appropriate substitution. This is the first time I am doing it with no help on the substitution part. Hence, could somebody help me check this working?
$$
\begin{align*}
\int \frac 1 {\begin{pmatrix}\frac1 3 t - 8\end{pmatrix}^5} \,dt
= 3\int \frac 1 {\begin{pmatrix}\frac1 3 t - 8\end{pmatrix}^5} \,\frac 1 3dt
\end{align*}
$$
Consider the substitution $u = \begin{pmatrix}\frac 1 3 t -8\end{pmatrix}$ and so $du = \frac 1 3 dt$ and perform the substitution
$$
\begin{align*}
3\int \frac 1 {\begin{pmatrix}\frac1 3 t - 8\end{pmatrix}^5} \,\frac 1 3dt
&= 3\int\frac 1 {u^5} \, du
\\&= -\frac 3 4 u^{-4} + C
\\&= -\frac 3 4 \begin{pmatrix}\frac 1 3 t -8\end{pmatrix}^{-4} +C
\end{align*}
$$
|
Thank you to all who have helped, my proposed solution has been verified and is correct.
|
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|
Definite Integral $\int_0^{\pi/2} \frac{\log(\cos x)}{x^2+\log^2(\cos x)}dx = \frac{\pi}{2}\left(1-\frac{1}{\log 2}\right)$ I want to prove that
$$\int_0^{\pi/2} \frac{\log(\cos x)}{x^2+\log^2(\cos x)}dx = \frac{\pi}{2}\left(1-\frac{1}{\log 2}\right)$$
|
Consider $$f(z) = \frac{1}{\log(1-iz)} \frac{1}{1+z^{2}}$$ where the branch cut for $\log (1-iz)$ runs down the imaginary axis from $z=-i$.
Then integrate around a contour that consists of the line segment $[-R,R]$ and the upper half of the circle $|z|=R$, and is indented around the simple pole at the origin.
Since $ z f(z) ->0 $ uniformly as $R \to \infty$, $\displaystyle \int f(z) \ dz$ vanishes along the upper half of $|z|=R$ as $ R \to \infty$.
So we have
$$ \begin{align} \text{PV} \int_{-\infty}^{\infty} \frac{1}{\log(1-ix)} \frac{dx}{1+x^{2}} &= \text{PV} \int_{-\infty}^{\infty} \frac{1}{\frac{1}{2}\log(1+x^{2}) - i\arctan x} \frac{dx}{1+x^{2}} \\ &= \text{PV} \int_{-\infty}^{\infty}\frac{\frac{1}{2} \log(1+x^{2})+ i \arctan x }{\frac{1}{4} \log^{2}(1+x^{2})+ \arctan^{2} x} \frac{dx}{1+x^{2}} \\ &= 2 \pi i \ \text{Res}[f(z), i] + \pi i \ \text{Res}[f(z),0] \\ &= 2 \pi i \left(\frac{1}{2i \log 2} \right) + \pi i \left( -\frac{1}{i} \right) \\ &= \pi \left( \frac{1}{\log 2} - 1\right). \end{align}$$
Equating the real parts on both sides of the equation,
$$ \int_{-\infty}^{\infty} \frac{\frac{1}{2} \log(1+x^{2})}{\frac{1}{4} \log^{2}(1+x^{2}) + \arctan^{2} x} \frac{dx}{1+x^{2}} = \pi \left(\frac{1}{\log 2} -1 \right). $$
Now let $x= \tan u$.
Then
$$ \begin{align} \int_{-\pi /2}^{\pi /2} \frac{\frac{1}{2} \log(\sec^{2}u)}{\frac{1}{4} \log^{2}(\sec^{2}u) + u^{2}} du &= \int_{-\pi/2}^{\pi /2} \frac{\log (\sec u)}{\log^{2}(\sec u)+u^{2}} \ du \\ &= - \int_{-\pi/2}^{\pi/2} \frac{\log(\cos u)}{\log^{2}(\sec u) + u^{2}} \ du \\ &= \pi \left(\frac{1}{\log 2} -1 \right) \end{align}$$
which implies
$$ \int_{0}^{\pi/2} \frac{\log(\cos u)}{u^{2} + \log^{2}(\cos u)} \ du = \frac{\pi}{2} \left(1- \frac{1}{\log 2} \right).$$
|
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|
Prove that: $(1+a_1)(1+a_2)...(1+a_n) \le 1 + S_n + (S_n)^2/2! + ... + (S_n)^n/n!$ I am currently working on this problem from Hardy's Course of Pure Mathematics and have gotten stuck near the end. I was wondering if someone could help me determine what to go next.
Question
If $a_1, a_2, ...,a_n$ are all positive and $S_n=a_1+a_2+...+a_n$ then:
$(1+a_1)(1+a_2)...(1+a_n) \le 1 + S_n + \dfrac{(S_n)^2}{2!} + ... + \dfrac{(S_n)^n}{n!}$
My Attempt
Proof by induction:
I had first shown that it was true for $n=1$ and $n=2$.
Now suppose that it is true for n. Then it must also be true for $n+1$
$(1+a_1)(1+a_2)...(1+a_n)(1+a_{n+1}) \le 1 + S_n + \dfrac{(S_n)^2}{2!} + ... + \dfrac{(S_n)^n}{n!} + \dfrac{(S_{n})^{n+1}}{(n+1)!}$
Define: $(1+a_1)(1+a_2)...(1+a_n)=x$ and $1 + S_n + \dfrac{(S_n)^2}{2!} + ... + \dfrac{(S_n)^n}{n!}=y$
Then: $x+xa_{n+1} \le y+\dfrac{(S_{n})^{n+1}}{(n+1)!}$
By the induction assumption, we know that $x \le y$.
What I can't figure out
From this point, what I think the next natural step would be is to show that
$xa_{n+1} \le \dfrac{(S_{n})^{n+1}}{(n+1)!}$
I know this rearranges to:
$xa_{n+1} \le \dfrac{(S_{n})^{n}}{n!} \times \dfrac{S_n}{(n+1)} $
And feel there may be something I can do here, but haven't been able to figure anything out.
|
Continue the proof using MI. Assume P(n) is true.
Let $y=a_{n+1}, S=S_n$
Let $f(y)=1+(S+y)+\frac{(S+y)^2}{2!}+...+\frac{(S+y)^{(n+1)}}{(n+1)!}-(1+a_1)(1+a_2)...(1+a_n)(1+y) \quad y>0$
$f'(y)=1+(S+y)+\frac{(S+y)^2}{2!}+...+\frac{(S+y)^{n}}{n!}-(1+a_1)(1+a_2)...(1+a_n)$
$ > 1+S+\frac{S^2}{2!}+...+\frac{S^{n}}{n!}-(1+a_1)(1+a_2)...(1+a_n) \geq 0 $
$f(0) = 1+S+\frac{S^2}{2!}+...+\frac{S^{(k+1)}}{(k+1)!}-(1+a_1)(1+a_2)...(1+a_n)$
$> 1+S+\frac{S^2}{2!}+...+\frac{S^{n}}{n!}-(1+x_1)(1+x_2)...(1+x_n) \geq 0$
Since f is differentiable w.r.t y, thus $f(y) \geq f(0) \geq 0$ for $y>0$
Rewrite $(S+y)$ as $S_{n+1}$ and $y$ as $a_{n+1}$
$1+S_{n+1}+\frac{S_{n+1}^2}{2!}+...+\frac{S_{n+1}^{(n+1)}}{(n+1)!} \geq (1+a_1)(1+a_2)...(1+a_n)(1+a_{n+1})$
Then P(n+1) is also true.
|
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"url": "https://math.stackexchange.com/questions/326721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
Range of $\frac{1}{2\cos x-1}$ How can we find the range of $$f(x) =\frac{1}{2\cos x-1}$$
Since range of $\cos x$ can be given as : $-1 \leq \cos x \leq 1$
therefore we can proceed as :$$\begin{array}{rcl}
-2 \leq & 2\cos x & \leq 2 \\
-2-1 \leq & 2\cos x -1 & \leq 2-1\\
-3 \leq & 2\cos x -1 & \leq 1 \\
\frac{-1}{3} \leq & \frac{1}{2\cos x-1} & \leq 1
\end{array}$$
Is it the range? Please suggest and guide.
|
No, you are fine down to $-3 \le 2\cos x -1 \le 1$, but you need to change the sense of the inequality when you invert: Think of a positive $x \le 1$, then $\frac 1x \ge 1$. The idea is that if the denominator is small (and you have trapped it near $0$) the inverse is large.
|
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"url": "https://math.stackexchange.com/questions/328493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
How to evaluate $\int_0^{\pi/2}x^2\ln(\sin x)\ln(\cos x)\ \mathrm dx$ Find the value of
$$I=\displaystyle\int_0^{\pi/2}x^2\ln(\sin x)\ln(\cos x)\ \mathrm dx$$
We have the information that
$$J=\displaystyle\int_0^{\pi/2}x\ln(\sin x)\ln(\cos x)\ \mathrm dx=\dfrac{\pi^2}{8}\ln^2(2)-\dfrac{\pi^4}{192}$$
|
Presented below is a complete solution that evaluates the integral to the following close-form
\begin{align}
&\int_0^{\pi/2}x^2\ln(\sin x)\ln(\cos x)\ dx\\
=& -\frac\pi2 \text{Li}_4(\frac12)-\frac{3\pi}8\ln2\ \zeta(3)
+\frac{\pi^5}{320}+\frac{\pi^3}{16}\ln^22-\frac\pi{48}\ln^42\tag1
\end{align}
To derive (1), evaluate the integrals below sequentially, with $\int_0^{\pi/2}\ln^2(2\sin x)dx=\frac{\pi^3}{24}$ and $\int_0^{\pi/2}x^2 \ln^2(2\cos x) dx=\frac{11\pi^5}{1440}$
\begin{align}
&\int_0^{\pi/2}x^2
\overset{x\to\frac\pi2-x}{\ln^2(2\sin x)}\ dx
=\pi\int_0^{\pi/2}x
\ln^2(2\sin x)\ dx-\frac{\pi^5}{360}\\
\\
&\int_0^{\pi/2}x^2
\ln^2(2\sin 2x)\overset{2x\to x}{dx} \\
=& \ \frac18\int_0^{\pi/2}(\pi^2-2\pi x+2x^2 )
\ln^2(2\sin x)\ dx=\frac{13\pi^5}{2880}\\
\\
&\int_0^{\pi/2}x^2\ln(2\sin x)\ln(2\cos x)\ dx\\
=& \ \frac12 \int_0^{\pi/2}[x^2
\ln^2(2\sin 2x)-x^2 \ln^2(2\sin x)-x^2 \ln^2(2\cos x)]\ dx\\
=& - \frac{\pi^5}{5760}-\frac\pi2 \int_0^{\pi/2}x
\ln^2(2\sin x)\ dx\
\end{align}
Writing out both sides of the last integral and utilizing below to yield the close-form (1)
\begin{align}
\int_0^{\pi/2}x\ln(2\sin x)dx=\frac7{16}\zeta(3),\>\>\>
\int_0^{\pi/2}x^2\ln(2\sin x)dx=\frac{3\pi}{16}\zeta(3)\\
\end{align}
$$\int_0^{\pi/2}x\ln^2(\sin x)dx
= \text{Li}_4(\frac12)
-\frac{19\pi^4}{2880}+\frac{\pi^2}{12}\ln^22+\frac1{24}\ln^42
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/330057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "60",
"answer_count": 6,
"answer_id": 5
}
|
Computing $\int \frac{6x}{\sqrt{x^2+4x+8}} dx$ I am trying to compute an indefinite integral. Thanks!
$$\int \frac{6x}{\sqrt{x^2+4x+8}} dx.$$
|
$$\begin{align}\int \dfrac{6x}{\sqrt{x^2 + 4x + 8}}\, dx\;
& = \;\int \dfrac{6x+12 -12}{\sqrt{x^2 + 4x + 8}} \,dx \\ \\ \\ \\
& = \int \dfrac {6(x + 2) - 12}{\sqrt {\color{blue}{\bf (x^2+ 4x + 4) + 4}}}\,dx
\end{align}$$
split the integral into the difference of two integrals:
$$ = \;\int \dfrac {6(x + 2)}{\sqrt {(x^2 + 4x + 8}}\,dx\;\; -\;\; 12\int \dfrac{1}{\sqrt{\color{blue}{\bf (x+2)^2 + 2^2}}}\, dx $$
Now, for the left hand integral,
*
*let $u = x^2 + 4x + 8 \implies du = 2x + 4 = 2(x + 2) \,dx,\implies\; 3du = 6(x+2)\,dx$. Substitute: you'll have an integral of the form $\;3\int u^{-1/2} \,du$
For the right-hand integral,
*
*let $\;v = x+2\;\implies\;dv = dx.$
*And as suggested, you'll want to use an appropriate trigonometric
substitution for the right-hand integral: in particular, find the
promising hyperbolic trig substitution.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
$ \sqrt{2-\sqrt{2}} $ simplified So I have a (nested?) square root $ \sqrt{2-\sqrt{2}} $. I know that $ \sqrt{2-\sqrt{3}} = \frac{\sqrt{6}-\sqrt{2}}{2} $. I know how to turn the simplified version into the complex one, but not vice versa. What is $ \sqrt{2-\sqrt{2}} $ simplified in this fashion and what are the steps?
|
this is a general way to do it:
$\sqrt{2-\sqrt{2}}=\sqrt x+\sqrt y $
then $2-\sqrt 2=2 \sqrt{xy}+x+y$
and then make the rational parts equal to the rational parts as such:
$2=x+y$
now we know $y=2-x$. substitute this into the irrational part:
$-\sqrt 2 =2\sqrt{(x)}\sqrt{(2-x)}$ square both sides to get
$2=4(x)(2-x) \rightarrow 2=8x-4x^2$
However the soulutions to this quadratic equation are $1-\frac{1}{\sqrt{2}}$ and $1+\frac{1}{\sqrt{2}}$
so $ \sqrt{ 2-\sqrt{2}}=\sqrt{1-\frac{1}{\sqrt{2}}} +\sqrt{1+\frac{1}{\sqrt{2}}}$
In other words there is no way to denest your radical.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/333185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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|
Solving recursion with generating function I am trying to solve a recursion with generating function, but somehow I ended up with mess.....
$$y_n=y_{n-1}-2y_{n-2}+4^{n-2}, y_0=2,y_1=1 $$
\begin{eqnarray*}
g(x)&=&y_0+y_1x+\sum_2^{\infty}(y_{n-1}-2y_{n-2}+4^{n-2})x^n\\
&=&2+x+\sum^{\infty}_2y_{n-1}x^n-2\sum_2^{\infty}y_{n-2}x^n+\sum_2^{\infty}4^{n-2}x^n\\
&=&2+x+x\sum_1^{\infty}y_{n-1}x^{n-1}-2x^2\sum_0^{\infty}y_{n-2}x^{n-2}+\frac{1}{4^2}\sum^{\infty}_{2}(4x)^{n}\\
&=&2+x+x(g(x)-1)-2x^2g(x)+\frac{1}{4^2}\left(\frac{1}{1-4x}-1-4x\right)\\
g(x)(1-x+2x^2)&=&2+\frac{1}{4^2}\frac{1}{1-4x}-\frac{1}{4^2}-\frac{x}{4}\\
g(x)&=& \frac{x^2-8x+2}{(1-4x)(1-x+2x^2)}
\end{eqnarray*}
How do I go from here to get $y_n$ as a complete solution, and also I noticed that $(1-x+2x^2)$ has imaginary roots, what does it mean? no solution?
|
Use the technique in Wilf's "generatingfunctionology" Define the generating function $A(z) = \sum_{n \ge 0} y_n z^n$, and write:
$$
y_{n + 2} = y_{n + 1} - 2 y_n + 4^n \quad y_0 = 2, y_1 = 1
$$
Using properties of generating functions:
$$
\frac{A(z) - y_0 - y_1 z}{z^2}
= \frac{A(z) - y_0}{z} - 2 A(z) + \frac{1}{1 - 4 z}
$$
Solving for $A(z)$, reducing to partial fractions:
$$
A(z) = \frac{17z - 27}{14 (1 - z + 2 z^2)} + \frac{1}{14} \frac{1}{1 - 4 z}
$$
The first term's denominator splits into:
$$
\left( 1 - \frac{1 + i \sqrt{7}}{2} z \right)
\left(1 - \frac{1 - i \sqrt{7}}{2} z \right)
$$
This gets quite ugly, and sadly the magnitude of those is also 4. This gives a solution that fluctuates wildly. The full solution is of the form:
$$
y_n = \alpha \left(\frac{1 + i \sqrt{7}}{2}\right)^n
+ \overline{\alpha} \left(\frac{1 - i \sqrt{7}}{2}\right)^n
+ \frac{4^n}{14}
$$
for some complex constant $\alpha$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/335289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Prove $\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6$
If $a,b,c$ are non-negative numbers and $a+b+c=3$, prove that:
$$\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6.$$
Here's what I've tried:
Using Cauchy-Schawrz I proved that:
$$(3a + b^3)(3 + 1) \ge (3\sqrt{a} + \sqrt{b^3})^2$$
$$\sqrt{(3a + b^3)(4)} \ge 3\sqrt{a} + \sqrt{b^3}$$
$$\sqrt{(3a + b^3)} \ge \frac{3\sqrt{a} + \sqrt{b^3}}{2}$$
Also I get:
$$\sqrt{(3b + c^3)} \ge \frac{3\sqrt{b} + \sqrt{c^3}}{2}$$
$$\sqrt{(3c + a^3)} \ge \frac{3\sqrt{c} + \sqrt{a^3}}{2}$$
If I add add 3 inequalities I get:
$$\sqrt{(3a + b^3)} + \sqrt{(3b + c^3)} + \sqrt{(3c + a^3)} \ge \frac{3\sqrt{a} + \sqrt{b^3}}{2} + \frac{3\sqrt{b} + \sqrt{c^3}}{2} + \frac{3\sqrt{c} + \sqrt{a^3}}{2}$$
Now i need to prove that:
$$\frac{3\sqrt{a} + \sqrt{a^3}}{2} + \frac{3\sqrt{b} + \sqrt{b^3}}{2} + \frac{3\sqrt{c} + \sqrt{c^3}}{2} \ge 6 = 2(a+b+c)$$
It's enough now to prove that:
$$\frac{3\sqrt{a} + \sqrt{a^3}}{2} \ge b+c = 3-a$$
$$\frac{3\sqrt{b} + \sqrt{b^3}}{2} \ge a+c = 3-b$$
$$\frac{3\sqrt{c} + \sqrt{c^3}}{2} \ge b+a = 3-c$$
All three inequalities are of the form:
$$\frac{3\sqrt{x} + \sqrt{x^3}}{2} \ge 3-x$$
$$3\sqrt{x} + \sqrt{x^3} \ge 6-2x$$
$$(3\sqrt{x} + \sqrt{x^3})^2 \ge (6-2x)^2$$
$$9x + x^3 + 6x^2 \ge 36 - 24x + 4x^2$$
$$x^3 + 2x^2 + 33x - 36 \ge 0$$
$$(x-1)(x^2 + 3x + 33) \ge 0$$
Case 1:
$$(x-1) \ge 0 \ \ \ \ \text{ for any }\ x \geq 1$$
$$(x^2 + 3x + 33) \ge 0 \ \ \ \ \text{ for any x in R} $$
Case 2:
$$0 \ge (x-1) \ \ \ \ \text{ for any }\ 1 \geq x$$
$$0 \ge (x^2 + 3x + 33) \ \ \ \ \text{there are no solutions in R} $$
This proves that for $$x \geq 1$$ $$(x-1)(x^2 + 3x + 33) \ge 0$$ is true and so it is
$$\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6$$, but a, b, c can be every non-negative number. I proved it's true for $$a,b,c \geq 1$$, but i can't for $$a,b,c \geq 0$$
|
Proof by KaiRain.
By squaring and collecting in terms, the original inequality is equivalent with:
\begin{align*} \sum_{cyc} \sqrt{ \left(3a+b^3 \right) \left(3b+c^3 \right)} \ge \frac{3 \left(a+b \right) \left(b+c \right) \left(c+a \right)}{2} \end{align*}
We will prove that:
\begin{align} \sqrt{ \left(3a+b^3 \right) \left(3b+c^3 \right)} \ge \frac{3ab^2+b^2c}{2}+c^2a+abc \ \ \ (1)\end{align}
holds true for all non-negative reals $a,b,c$ such that $a+b+c=3$. Indeed, this inequality can be proved by C-S and AM-GM in a rather tricky way. Following are the details:
\begin{align*}3= \frac{ \left(a+b+c \right)^2}{3} = ab+bc+ca + \frac{ \left(b+c-2a \right)^2 +3\left(b-c \right)^2 }{12} \ge ab+bc+ca+ \frac{ \left(b-c \right)^2}{4} \end{align*}
Using the above estimation, it turns out:
\begin{align*} \sqrt{ \left(b^3+3a \right) \left(c^3+3b \right)} & \ge \sqrt{ b^3+a \left(ab+bc+ca \right) } \ \sqrt{c^3 + b \left(ab+bc+ca \right) +\frac{b \left(b-c \right)^2}{4}} \\ & = \sqrt{b^3+a^2b+ca^2+abc} \ \sqrt{ \frac{ b\left(b+c \right)^2}{4}+ab^2+c^3+abc} \\ & \ge \frac{b^2\left(b+c \right)}{2}+\sqrt{a^3b^3}+c^2a+abc\\ &= \frac{ b^3+\sqrt{a^3b^3}+\sqrt{a^3b^3} }{2} +\frac{b^2c}{2}+c^2a+abc\\ &\ge \frac{3ab^2}{2} +\frac{b^2c}{2}+c^2a+abc.\end{align*}
Now, by adding two similar inequalities with $(1)$ and taking a note that:
\begin{align*} \sum_{cyc} \left(\frac{3ab^2+b^2c}{2}+c^2a+abc \right) = \frac{3 \left(a+b \right) \left(b+c \right) \left(c+a \right)}{2} \end{align*}
the result follows. Hence, the proof is completed.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "17",
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|
Solve the equation by putting new variable I want to solve the equation
$$x^2+(2 x+3) \sqrt{3 x^2+6 x+2}=6 x+5.$$
I tried. Put $t = \sqrt{3 x^2+6 x+2}.$ We have
$$x^2+(2 x+3)t - 6x - 5 = 0.$$
Discriminant of this equation (with unknown $x$) is $56-36t+4t^2$. It is not a square number.
|
Just put it like this:
$(2 x + 3) \sqrt{(3 x^2 + 6 x + 2)} = (6 x + 5 - x^2)$
then square both sides:
$(2 x + 3)^2 (3 x^2 + 6 x + 2) = (6 x + 5 - x^2)^2$
and collect the terms:
$(-1 + x (5 + x)) (7 + x (17 + 11 x))=0$
The answer is straightforward. Will this help?:)
P.S.: the solutions are
$\left\{x\to \frac{1}{22} \left(-17-i \sqrt{19}\right)\right\},\left\{x\to \frac{1}{22} \left(-17+i \sqrt{19}\right)\right\},\left\{x\to \frac{1}{2} \left(-5-\sqrt{29}\right)\right\},\left\{x\to \frac{1}{2} \left(\sqrt{29}-5\right)\right\}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/336548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Why $\sum_{k=1}^{\infty} \frac{k}{2^k} = 2$? Can you please explain why
$$
\sum_{k=1}^{\infty} \dfrac{k}{2^k} =
\dfrac{1}{2} +\dfrac{ 2}{4} + \dfrac{3}{8}+ \dfrac{4}{16} +\dfrac{5}{32} + \dots =
2
$$
I know $1 + 2 + 3 + ... + n = \dfrac{n(n+1)}{2}$
|
Start with the geometric series
$$s(x) = \sum_{k=0}^{\infty} x^k = \frac{1}{1-x}$$
Then
$$x \frac{d}{dx} s(x) = \sum_{k=1}^{\infty} k x^k = \frac{x}{(1-x)^2}$$
Your case has $x=1/2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/337937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
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|
Evaluate $\lim\limits_{n\rightarrow \infty}\left\{\frac{1^3}{n^4}+\frac{2^3}{n^4}+\frac{3^3}{n^4}+\dots +\frac{n^3}{n^4}\right\}$
*
*Evaluate $\displaystyle\lim_{n\rightarrow \infty}\left\{\frac{1^3}{n^4}+\frac{2^3}{n^4}+\frac{3^3}{n^4}+\dots +\frac{n^3}{n^4}\right\}$.
*Examine whether $x^{1/x}$ possesses a maximum or minimum and determine the same.
|
It is happens that there is an attractive closed form for $1^3+2^3+\cdots+n^3$, which can readily be proved by induction:
$$1^3+2^3+3^3+\cdots +n^3=\left(\frac{n(n+1)}{2}\right)^2.$$
Divide by $n^4$. Fairly quickly we find that the desired limit is $\dfrac{1}{4}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/338121",
"timestamp": "2023-03-29T00:00:00",
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|
generating functions / combinatorics Calculate number of solutions of the following equations:
$$ x_1 + x_2 + x_3 + x_4 = 15 $$
where $ 0 \le x_i < i + 4 $
I try to solve it using generating functions/enumerators :
$$ (1+x+x^2+x^3+x^4)(1+x+x^2+x^3+x^4+x^5)(1+x+x^2+x^3+x^4+x^5+x^6)(1+x+x^2+x^3+x^4+x^5+x^6+x^7)$$
and take coefficient near $15$. But I do not know how to quickly calculate it. Maybe there exists any faster way?
|
To take a slightly different route from i707107's solution, once you obtain the product of rational functions, you can use the identity:
and
$$
\frac{1}{(1-x)^n} = 1 + \binom{1 + n -1}{1}x + \binom{2 + n -1}{2}x^2 + \dots + \binom{r + n -1}{r}x^r + \dots.
$$
to expand the last term of the simplified product (2nd line) that i707107 has written. Then you would simply take product of the first $4$ polynomials, which isn't that bad since they're just $2$ terms each. Then you'd find the coefficient of the products that give $x^{15}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/339230",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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|
Given the graphic sequence of a simple graph, how to construct the adjacency matrix? Suppose I have the graphic sequence (7,6,6,6,5,4,4,2), how do I get the adjacency matrix out of it?
|
The graph has $8$ vertices, and $\tfrac{1}{2}(7+6+6+6+5+4+4+2)=20$ edges by the Handshaking Lemma. We can exhaustively generate the non-isomorphic $8$-vertex $20$-edge graphs with vertex degrees between $2$ and $7$ using geng which comes with nauty using:
geng 8 20:20 -d2 -D7
I wrote a script that filters out the ones that don't have degree sequence (7,6,6,6,5,4,4,2) leaving two remaining. The two graphs are as follows:
I've marked the vertices with their degrees, and their corresponding adjacency matrices are given below.
$\begin{array}{|cccccccc|} \hline 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 1 & 1 &
1 & 1 \\ 1 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 & 0 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 & 1 & 0 & 1 & 1 \\ 0 & 1 & 1 & 1 &
1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 0 \\ \hline \end{array}
\quad\quad\quad
\begin{array}{|cccccccc|} \hline 0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 & 0 & 1 &
1 & 1 \\ 0 & 0 & 1 & 0 & 1 & 1 & 1 & 1 \\ 1 & 1 & 0 & 1 & 0 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 & 1 & 0 & 1 & 1 \\ 0 & 1 & 1 & 1 &
1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 0 \\ \hline \end{array}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to find the value of $\sin{\dfrac{\pi}{14}}+6\sin^2{\dfrac{\pi}{14}}-8\sin^4{\dfrac{\pi}{14}}$
Determine
$$
\sin\left(\pi \over 14\right) + 6\sin^{2}\left(\pi \over 14\right)
-8\sin^{4}\left(\pi \over 14\right)
$$
My idea: Let $\displaystyle{\sin\left(\pi \over 14\right)} = x$.
|
A non-trivial result is that $\sin \dfrac{\pi}{14}$ is a root to the equation $^{\color{blue}{(**)}}$
$$1 - 4x - 4x^2 + 8x^3 = 0. \tag{1}$$
If you believe this, then you can simply apply polynomial division to obtain
$$x + 6x^2 - 8x^4 = -\left(x + \frac{1}{2}\right)(8x^3 - 4x^2 - 4x + 1) + \frac{1}{2}.$$
So then, filling in $x = \sin \dfrac{\pi}{14}$ gives you the answer:
$$\boxed{\sin{\dfrac{\pi}{14}}+6\sin^2{\dfrac{\pi}{14}}-8\sin^4{\dfrac{\pi}{14}} = \dfrac{1}{2}.}$$
$^{\color{blue}{(**)}}$ For a proof that $\sin \dfrac{\pi}{14}$ is indeed a root of equation $(1)$, you need some patience and careful bookkeeping. Let us write $c_k = \cos \dfrac{k \pi}{14}$ and $s_k = \sin \dfrac{k \pi}{14}$. Then, using the triple angle formula for the cosine, we have:
$$c_3 = 4 c_1^3 - 3c_1 = 4 c_1 (1 - s_1^2) - 3c_1 = c_1 (1 - 4 s_1^2).$$
On the other hand, using the double angle formula for the sine twice, we also have:
$$c_3 = s_4 = 2s_2 c_2 = 4 s_1 c_1 (1 - 2s_1^2) = c_1 (4 s_1 - 8 s_1^2)).$$
Equating these two expressions and dividing by $c_1 \neq 0$, we get
$$1 - 4s_1^2 = 4s_1 - 8s_1^3.$$
Bringing all terms to one side, this leads to
$$1 - 4s_1 - 4s_1^2 + 8s_1^3 = 0.$$
|
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|
Sum of $\sum_{n=0}^\infty \frac{(x+2)^{n+2}}{3^n} $ Calculate the sum of the next series and for which values of $x$ it converges:
$$\sum_{n=0}^\infty \frac{(x+2)^{n+2}}{3^n}$$
I used D'Alembert and found that the limit is less than 1, so: $-5 < x < 1$ (because the fraction must be less than 1).
and then I assigned the values: $x=-5$ and $x=1$ in the series and got:
for $x=-5$ and $x=1$, it diverges.
then the series converges in the range of $(-5,1)$, $R=3$ and the center point is for $x=2$.
Please let know if there is a mistake and find the sum.
|
Your sum is equal to
$$
\sum_{n=0}^\infty\frac{(x+2)^{n+2}}{3^n}=(x+2)^2\sum_{n=0}^\infty\left(\frac{x+2}{3}\right)^n,
$$
which you recognize as a geometric series $\sum_n q^n$ with sum $1/(1-q)$. Thus,
$$
\sum_{n=0}^\infty\frac{(x+2)^{n+2}}{3^n}=(x+2)^2\frac{1}{1-\frac{x+2}{3}} =\frac{3(x+2)^2}{1-x}
$$
as long as $|(x+2)/3|<1$.
If $|(x+2)/3|\geq 1$, the sum is not convergent.
|
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|
Find the value of $\alpha $ given $2\sin\theta -\sqrt 5 \cos \theta \equiv - 3\cos (\theta + \alpha )$ Given:
$$2\sin\theta -\sqrt 5 \cos \theta \equiv - 3\cos (\theta + \alpha ),$$
where $$0 <\alpha < 90^\circ, $$
find $α.$
The issue I have with this question is the $-3$ on the right hand side, it really complicates things and I don't know how to deal with it.
When I usually come across equations of the form $$a\sin \theta + b\cos \theta $$ it's a relatively straight forward process of converting them into another equation of the form
$$\sqrt {{a^2} + {b^2}} \left({a \over {\sqrt {{a^2} + {b^2}} }}\sin \theta - {b \over {\sqrt {{a^2} + {b^2}} }}\cos \theta \right)$$
where $$\cos \alpha = {a \over {\sqrt {{a^2} + {b^2}} }}$$
and $$\sin \alpha = {b \over {\sqrt {{a^2} + {b^2}} }}$$
(or some other variant of this).
I cant wrap my head around this, specifically. If
$$2\sin\theta -\sqrt 5 \cos \theta \equiv - 3\cos (\theta + \alpha ),$$
this must mean the square root expression prefixing the cosine addition identity must be $-3$, as
$$\sqrt {{a^2} + {b^2}} \cos(\theta + \alpha ) \equiv - 3\cos(\theta + \alpha ).$$
I am aware the square root operation outputs negative values I just dont know how to deal with that in this instance.
I hope this doesn't read too convoluted, thank you.
|
$$\Rightarrow {\rm{2sin}}\theta {\rm{ - }}\sqrt 5 \cos \theta = - 3\cos (\theta + \alpha )$$
$$\Rightarrow -\frac{2}{3}\sin \theta+\frac{\sqrt 5 }{3}\cos \theta= \cos (\theta + \alpha ) $$
$$\Rightarrow -\frac{2}{3}\sin \theta+\frac{\sqrt 5 }{3}\cos \theta= \cos \theta \cos \alpha - \sin \theta \sin \alpha$$
$$\Rightarrow \frac{\sqrt 5 }{3}\cos \theta -\frac{2}{3}\sin \theta = \cos \theta \cos \alpha - \sin \theta \sin \alpha$$
Equating both sides and considering $0<\alpha < 90^0$ i.e. both $\cos \alpha$ and $\sin \alpha$ lie on the first quadrant.
$$\cos \alpha = \frac{\sqrt 5 }{3}\tag1$$
$$\sin \alpha = \frac{2}{3}\tag2$$
Dividing $(2)$ by $(1)$
$$\frac{\sin \alpha}{\cos \alpha} =\frac{\frac{2}{3}}{\frac{\sqrt 5 }{3}}=\frac{2}{sqrt(5)}$$
$$\tan \alpha = \frac{2}{\sqrt5}$$
$$\alpha = tan^{-1}\frac{2}{\sqrt5}$$
Note, the reason you went wrong because of your formulation,
Note
$$\sqrt {{a^2} + {b^2}} \left({a \over {\sqrt {{a^2} + {b^2}} }}\sin \theta - {b \over {\sqrt {{a^2} + {b^2}} }}\cos \theta \right) \ne \sqrt {{a^2} + {b^2}} \cos(\theta + \alpha )$$
But rather
$$-\sqrt {{a^2} + {b^2}} \left({b \over {\sqrt {{a^2} + {b^2}} }}\cos - {a \over {\sqrt {{a^2} + {b^2}} }}\sin \theta \theta \right) = - \sqrt {{a^2} + {b^2}} \cos(\theta + \alpha )$$
Which will derive to
$$-\sqrt {{a^2} + {b^2}} \cos(\theta + \alpha ) = - 3\cos(\theta + \alpha ).$$
|
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|
If $x \neq 0,y \neq 0,$ then $x^2+xy+y^2$ is ..... I came across the following problem that says:
If $x \neq 0,y \neq 0,$ then $x^2+xy+y^2$ is
1.Always positive
2.Always negative
3.zero
4.Sometimes positive and sometimes negative.
I have to determine which of the aforementioned options is right.
Now since $x \neq 0,y \neq 0$, so $ x^2+xy+y^2=(x-y)^2+3xy > 0$,if $x,y$ are of same sign. But if $x,y$ are of different sign,I am not sure about the conclusion.
Can someone point me in the right direction? Thanks in advance for your time.
|
$x^2 + xy + y^2$ is a quadratic form and can be written
$$x^2 + xy + y^2 = \begin{bmatrix} x & y\end{bmatrix}\begin{bmatrix} 1 & \frac{1}{2} \\ \frac{1}{2} & 1 \end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix}$$
A matrix $\mathbf{A}$ is positive-definite if $\mathbf{z'}\mathbf{A}\mathbf{z}>0 $ for all $\mathbf{z}\neq 0$
Show that $\begin{bmatrix} 1 & \frac{1}{2} \\ \frac{1}{2} & 1 \end{bmatrix}$ is a positive-definite matrix and you will have a very nice solution to your problem
|
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|
Generating function for the solutions of equation $ 2x_1 + 4x_2 = n$ Let's denote $h_n$ as the number of soulutions of the following equation:
$$ 2x_1 + 4x_2 = n$$
where $x_i \in \mathbb N$.
Find generating function of the sequence $h_n$ and calculate $h_{2000}$.
I've found the generating function:
$$\frac{1}{1-x^2}\cdot \frac{1}{1-x^4},$$
but I don't know how to expand it now. Any ideas?
|
As you observed, if the generating function for $h_n$ is $$f(x) = h_0 + h_1x + h_2x^2 + h_3x^3 + \dots + h_nx^n + \dots,$$
then $$f(x) = (1 + x^2 + x^4 + x^6 + \dots)(1 + x^4 + x^8 + x^{12} + \dots) = \frac{1}{1-x^2}\frac{1}{1-x^4}$$
To actually get the coefficients of $x^n$ in the above, we resort to partial fractions. The denominator of the above can be factored into irreducible factors, using the identity $1-y^2 = (1+y)(1-y)$, as $(1+x)(1-x)(1+x^2)(1+x)(1-x) = (1+x)^2(1-x)^2(1+x^2)$. Therefore, by the general theory of partial fractions, $f(x)$ can be written as
$$f(x) = \frac{A}{1+x\vphantom{(1+x)^2}} + \frac{B}{(1+x)^2} + \frac{C}{1-x\vphantom{(1-x)^2}} + \frac{D}{(1-x)^2} + \frac{Ex + F}{1+x^2} + \frac{Gx + H}{(1+x^2)^2}$$
where $A, B, C, D, E, F, G, H$ are constants. Using various painful tricks it's possible to determine the constants, but because I'm not the one doing this as homework, I'll just turn to Wolfram Alpha which says that
$$f(x) = \frac{1}{4(1+x)} + \frac{1}{8(1+x)^2} + \frac{1}{4(1-x)} + \frac{1}{8(1-x)^2} + \frac{1}{4(1+x^2)}.$$
Here the five terms are, respectively,
$\frac14 \sum_n{(-1)^n x^n}$ and
$\frac18 \sum_n {(-1)^n (n+1)x^n}$ and
$\frac14 \sum_n x^n$ and
$\frac18 \sum_n (n+1)x^n$ and
$\frac14 \sum_n{(-1)^n x^{2n}}$,
so $$h_{2000} = \frac14 (-1)^{2000} + \frac18 (-1)^{2000}2001 + \frac14 + \frac18 2001 + \frac14 (-1)^{1000} = 501.$$
[That's the answer, but as you can see the whole thing is a quite painful process that I wouldn't wish on my worst enemy, which is why I keep saying that generating functions are not the best way to solve these counting problems, despite the dazzle of the first step where you get some cute expression for the generating function.]
Edit: Just for contrast, the solution without generating functions: Clearly $n$ must be even, so we're counting solutions to $x + 2y = n/2$ in the nonnegative integers. From $0 \le 2y \le n/2$ we have $0 \le y \le \left\lfloor \frac{n/2}{2} \right\rfloor$, and for each such $y$ there is a unique $x = n/2 - 2y$ as solution. So the number of solutions (for even $n$) is $1 + \left\lfloor \frac{n/2}{2} \right\rfloor$, which for $n = 2000$ is $1 + \left\lfloor \frac{1000}{2} \right\rfloor = 501$.
The solution without generating functions is not always this simple, but I think this is a good example of how going down the generating functions route can be a bad idea if you want exact numbers (as opposed to asymptotic estimates, say).
|
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|
Help please on complex polynomials I wanted to know if there's any good approaches to these questions
a)By considering $z^9-1$ as a difference of two cubes, write $1+z+z^2+z^3+z^4+z^5+z^6+z^7+z^8$
as a product of two real factors one of which is a quadratic.
b) Solve $z^9-1=0$ and hence write down the 6 solutions of $z^6+z^3+1=0$
c)By letting $y=z+\frac{1}{z}$ and dividing $z^6+z^3+1=0$ by $z^3$, deduce that:
$cos\frac{2\pi}{9}+cos\frac{4\pi}{9}+cos\frac{8\pi}{9}=0$
Any help would be greatly appreciated.
|
(a)
\begin{align*}
(z - 1)(z^8 + z^7 + \ldots + 1) = z^9 - 1 & = (z^3 - 1)(z^6 + z^3 + 1) \\
& = (z - 1)(z^2 + z + 1)(z^6 + z^3 + 1) \\
\therefore \qquad z^8 + z^7 + \ldots + 1 & = (z^2 + z + 1)(z^6 + z^3 + 1).
\end{align*}
(b)
The roots of $z^9 - 1$ are $r_n = e^{\frac{2\pi}9ni}$, for $n = 0, \ldots, 8$. Since $r_0 = 1$ is the root of $z - 1$, the other 8 roots are precisely the roots of $z^8 + z^7 + \ldots + 1 = (z^2 + z + 1)(z^6 + z^3 + 1)$. The quadratic $z^2 + z + 1$ has roots $\frac{-1 \pm \sqrt{3}i}{2}$, which are $r_3$ and $r_6$. Therefore, the 6 roots of $z^6 + z^3 + 1$ are $r_1, r_2, r_4, r_5, r_7, r_8$.
(c)
\begin{align*}
y^3
& = z^3 + 3z + 3\frac 1z + \frac 1{z^3} \\
y^3 - 3y + 1 & = z^3 + 1 + \frac 1{z^3} = \frac{1}{z^3}(z^6 + z^3 + 1) \\
& = \frac{1}{z^3}(z - r_1)(z - r_2)(z - r_4)(z - r_5)(z - r_7)(z - r_8).
\end{align*}
The roots of $y^3 - 3y + 1$ are $r_i + \frac{1}{r_i}$, but some $r_i$ give the same root because $r_i = \frac{1}{r_{9-i}}$. More precisely, the three roots are $y_1 = r_1 + \frac{1}{r_1} = r_8 + \frac{1}{r_8}$, $y_2 = r_2 + \frac{1}{r_2} = r_7 + \frac{1}{r_7}$, and $y_3 = r_4 + \frac{1}{r_4} = r_5 + \frac{1}{r_5}$.
We also know that the sum of the roots of $y^3 - 3y + 1$ is $0$ because the coefficient in front of $y^2$ is $0$. This means
\begin{align*}
y_1 + y_2 + y_3 = r_1 + \frac 1{r_1} + r_2 + \frac 1{r_2} + r_4 + \frac 1{r_4} & = 0 \\
2\left(\cos\frac{2\pi}9 + \cos\frac{4\pi}9 + \cos\frac{8\pi}9\right) & = 0
\end{align*}
|
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|
Simplify $\sum_{i=0}^n (i+1)\binom ni$ Simplifying this expression$$1\cdot\binom{n}{0}+ 2\cdot\binom{n}{1}+3\cdot\binom{n}{2}+ \cdots+(n+1)\cdot\binom{n}{n}= ?$$
$$\text{Hint: } \binom{n}{k}= \frac{n}{k}\cdot\binom{n-1}{k-1} $$
|
We have
$$\sum_{k=0}^n\dbinom{n}k x^k = (1+x)^n$$
Hence, we get that
$$\sum_{k=0}^n\dbinom{n}k x^{k+1} = x(1+x)^n$$
Differentiating the above, we get
$$\sum_{k=0}^n(k+1) \dbinom{n}k x^{k} = \dfrac{d(x(1+x)^n)}{dx}$$
Now set $x=1$ to get the answer.
$$\sum_{k=0}^n(k+1) \dbinom{n}k x^{k} = \dfrac{d(x(1+x)^n)}{dx} = (1+x)^{n-1} \left(nx+x+1\right)$$ Setting $x=1$, we get that $$\sum_{k=0}^n(k+1) \dbinom{n}k x^{k} = (n+2) 2^{n-1}$$
|
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|
Really Confused on a surface area integral can't seem to finish the integral off. Basically the question asks to compute $\int \int_{S} ( x^{2}+y^{2}) dA$ where S is the portion of the sphere $x^{2} + y^{2}+ z^{2}= 4$ and $z \in [1,2]$ we start with a chnage of variables
$x=x $
$y=y$
$ z= 2 \cdot(4-(x^{2} + y^{2}))^{1/2}$
$Det(u,v)= \begin{bmatrix}
i & j& k \\
1 & 0 & \frac {-x}{(4-(x^{2} + y^{2}))^{1/2}} \\
0 & 1 & \frac {-y}{(4-(x^{2} + y^{2}))^{1/2}} \\
\end{bmatrix}=(\frac {x}{(4-(x^{2} + y^{2}))^{1/2}})i + (\frac {y}{(4-(x^{2} + y^{2}))^{1/2}})j + k$
$dA=(\frac {x^{2}+y^{2}}{(4-(x^{2} + y^{2}))} +1)^{1/2}$
$\int \int_{S} (\frac {(x^{2}+y^{2})^{3}}{(4-(x^{2} + y^{2}))}+(x^{2}+y^{2})^{2})^{1/2}$
Projecting when z=1 and z=2 we have $x^{2} + y^{2}= 4-1$ $\to r= 0,(3)^{1/2}$
going to polar we have:
$(\frac {r^{6}}{(4-r^{2})}+r^{4})^{1/2}rdrd\theta=(\frac {4r^{4}}{(4-r^{2})})^{1/2}rdrd\theta$
my problem is $2\int^{2\pi}_{0} \int^{(3)^{1/2}}_{0} (\frac {4r^{4}}{(4-r^{2})})^{1/2}rdrd\theta$ there is no nice way i can think of to integrate this. it can also be written as:
$2\int^{2\pi}_{0} \int^{(3)^{1/2}}_{0} \frac {2r^{2}}{((4-r^{2}))^{1/2}}rdrd\theta$
|
Sometimes cylindrical coordinates is at least as easy when there is axial symmetry. The integral to be evaluated becomes
$$2 \pi \int_1^2 dz \, r(z)^3 \sqrt{1+\left ( \frac{d r}{dz} \right)^2} $$
where $r(z) = \sqrt{4 - z^2}$ and $r$ is the distance from the axis to the sphere. This reduces to, upon evaluation of the terms in the integrand
$$4 \pi \int_1^2 dz \: (4-z^2) = \frac{20 \pi}{3}$$
|
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|
Constructing a degree 4 rational polynomial satisfying $f(\sqrt{2}+\sqrt{3}) = 0$ Goal: Find $f \in \mathbb{Q}[x]$ such that $f(\sqrt{2}+\sqrt{3}) = 0$.
A direct approach is to look at the following
$$
\begin{align}
(\sqrt{2}+\sqrt{3})^2 &= 5+2\sqrt{6} \\
(\sqrt{2}+\sqrt{3})^4 &= (5+2\sqrt{6})^2 = 49+20\sqrt{6} \\
\end{align}
$$
Putting those together gives
$$
-1 + 10(\sqrt{2}+\sqrt{3})^2 - (\sqrt{2}+\sqrt{3})^4 = 0,
$$
so $f(x) = -1 + 10x^2 - x^4$ satisfies $f(\sqrt{2}+\sqrt{3}) = 0$.
Is there a more mechanical approach? Perhaps not entirely mechanical, but something more abstract.
|
A 'mechanical' approach follows. Let $x=\sqrt{2}+\sqrt{3}$. Then $x^2=5+2\sqrt{6}$ which means $x^2-5=2\sqrt{6}$. Now $$(x^2-5)^2=24\Longrightarrow(x^2-5)^2-24=0.$$ By construction, one of the roots of $f(x)=(x^2-5)^2-24$ is $\sqrt{2}+\sqrt{3}$.
|
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|
How do I evaluate $\lim_{n \rightarrow \infty} \frac{1\cdot2+2\cdot3 +3\cdot4 +4\cdot5+\ldots}{n^3}$ How to find this limit : $$\lim_{n \rightarrow \infty} \frac{1\cdot2+2\cdot3 +3\cdot4 +4\cdot5+\ldots+n(n+1)}{n^3}$$ As, if we look this limit problem viz. $\lim_{x \rightarrow \infty} \frac{1+2+3+\ldots+n}{n^2}$ then we take the sum of numerator which is sum of first n natural numbers and we can write :
$$\lim_{x \rightarrow \infty} \frac{n(n+1)}{2n^2}$$ which gives after simplification :
$$ \frac{1}{2} $$ as other terms contain $\frac{1}{x}$ etc. and becomes zero.
|
Or we can apply Stolz–Cesàro theorem to obtain $$\lim_{n \rightarrow \infty} \frac{1\cdot2+2\cdot3 +3\cdot4 +4\cdot5+\ldots+n(n+1)}{n^3}=\lim_{n \to \infty}\frac{(n+2)(n+1)}{(n+1)^3-n^3}=\lim_{n\to \infty}\frac{(n+2)(n+1)}{3n^2+3n+1}=\frac{1}{3} $$
|
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|
Solving a 2nd order differential equation by the Frobenius method Can you, please, help me to solve this equation:
$$(x+1)^2y''+(x+1)y'-y=0$$
Here, for me the problem is, I am finding relationship among 3 members: $a_n, a_{n+1}, a_{n+2}$, not between 2 members: $a_n$ and $a_{n+1}$
I would like to solve it using the Frobenius method.
|
Let $y=\sum\limits_{n=0}^\infty a_n(x+1)^{n+r}$ ,
Then $y'=\sum\limits_{n=0}^\infty(n+r)a_n(x+1)^{n+r-1}$
$y''=\sum\limits_{n=0}^\infty(n+r)(n+r-1)a_n(x+1)^{n+r-2}$
$\therefore(x+1)^2\sum\limits_{n=0}^\infty(n+r)(n+r-1)a_n(x+1)^{n+r-2}+(x+1)\sum\limits_{n=0}^\infty(n+r)a_n(x+1)^{n+r-1}-\sum\limits_{n=0}^\infty a_n(x+1)^{n+r}=0$
$\sum\limits_{n=0}^\infty(n+r)(n+r-1)a_n(x+1)^{n+r}+\sum\limits_{n=0}^\infty(n+r)a_n(x+1)^{n+r}-\sum\limits_{n=0}^\infty a_n(x+1)^{n+r}=0$
$\sum\limits_{n=0}^\infty((n+r)^2-1)a_n(x+1)^{n+r}=0$
Since there are not any indical equations present, that means $r$ can be chosen as any complex number.
However, in fact, take $r=1$ or $r=-1$ will bring the above equation most simplified.
Moreover, in fact, we can find all groups of the linearly independent solutions by just taking $r=-1$ :
$\sum\limits_{n=0}^\infty((n-1)^2-1)a_n(x+1)^{n-1}=0$
$\sum\limits_{n=0}^\infty n(n-2)a_n(x+1)^{n-1}=0$
$\therefore n(n-2)a_n=0$
$\therefore\begin{cases}a_0=a_0\\a_2=a_2\\a_n=0~\forall n\in\mathbb{N}\setminus\{2\}\end{cases}$
Hence $y=\dfrac{C_1}{x+1}+C_2(x+1)$
|
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|
Find $(a, b)$ such that $\lim_{x \to 0} \frac{ax -1 + e^{bx}}{x^2} = 1$ Find $(a, b)$ such that $\lim_{x \to 0} \frac{ax -1 + e^{bx}}{x^2} = 1$
I am able to find $b = \pm \sqrt{2}$ using L'Hopitals Rule, but unable to do anything for $a$.
|
As the limit in question is of the indeterminate form "$\frac{0}{0}$", we use L'Hôpital's rule and consider the limit $$\lim_{x \to 0} \frac{a +be^{bx}}{2x}.$$ Note that $\lim_{x\to 0}a+be^{bx} = a+be^0 = a+b$ and $\lim_{x\to 0} 2x = 0$. The only way the overall limit can be $1$ is if we have a limit in the indeterminate form "$\frac{0}{0}$". For this to occur, we need $a = - b$. Then, using L'Hôpital's rule once more, we consider the limit $$\lim_{x\to 0}\frac{b^2e^{bx}}{2} = \frac{b^2}{2}.$$ As we want this limit to be $1$, we find that $b = \pm\sqrt{2}$ and hence $a = \mp\sqrt{2}$.
That is, $$\lim_{x\to 0}\frac{a-1+e^{bx}}{x^2} = 1$$ when $(a, b) = (\sqrt{2}, -\sqrt{2})$ or $(a, b) = (-\sqrt{2}, \sqrt{2})$.
|
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"url": "https://math.stackexchange.com/questions/366384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
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.