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Is it correct? $\tan x + \cot x \ge 2$ proof The question is to prove $\tan x + \cot x \ge 2$ when $x$ is an acute angel.
This is what I did
$$\begin{align}
\tan x + \cot x &\ge 2\\
\frac{1}{\sin x \cos x} &\ge 2\\
\left(\frac{1}{\sin x \cos x}\right) - 2 &\ge 0\\
\left(\frac{1 - 2\sin x \cos x}{\sin x \cos x}\right) &\ge 0\\
\left(\frac{(\sin x - \cos x)^2}{\sin x \cos x}\right) &\ge 0\\
\left(\frac{(\sin x - \cos x)^2}{\sin x \cos x}\right) &\ge 0\\
\end{align}$$
Both nominator and denominator will never be negative because nominator is powered to two and cosx & sinx are positive when angel is acute.
Is it correct? Is there another way to solve?
|
Since $x$ is an acute angle, $\cot x$ and $\tan x$ are positive.
Therefore, by A.M.-G.M. inequality,
$\frac {\tan x + \cot x}{2} \ge \sqrt{\tan x \times \cot x}$
Now, $\tan x \times \cot x = 1$ (by definition)
Therefore we obtain,
$\tan x + \cot x \ge 2 \sqrt{1}$
And so, $\tan x + \cot x \ge 2 $
Proof complete.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Law of Indices and Quadratic Expressions So I think I need some clarification about the rules for manipulating indices, in particular these two equivalences:
$(x^3)^2 = x^{(3)(2)} = x^6$
$a = a^1$
Take the expression:
$(5+5)^2$, which is equivalent to $(5+5)(5+5) = 100$.
Using the rules for indices above, I would have thought that I could use them to do this: $(5^1+5^1)^2 = 5^{(1)(2)}+5^{(1)(2)} = 5^2+5^2$, but this is obviously wrong as $5^2+5^2 = 50 \neq 100$.
Where am I going wrong with my application of the rules for indices?
|
Your error is here:
$$
(5^1+5^1)^2 \color{red}{\neq} 5^{(1)(2)}+5^{(1)(2)}.
$$
You can easily verify that these two numbers are not the same.
In general, we have $(a+b)^2=a^2+b^2+2ab$, so it is typically the case that $(a+b)^2\neq a^2+b^2$.
This is true:
$$
(5^1+5^1)^2 \color{red}{=} 5^{(1)(2)}+5^{(1)(2)}+2\cdot5^{1}\cdot5^{1}.
$$
Addendum:
Your "rule" $(a+b)^n=a^n+b^n$ does not follow from the two rules you stated in your post.
The two rules you mentioned say nothing about the power of a sum.
To calculate the power of a sum, you need something more.
In the case $n=2$ we get, using the definition $x^2=x\cdot x$, that
$$
(a+b)^2=(a+b)(a+b)=a^2+ab+ba+b^2=a^2+2ab+b^2.
$$
You can get similar formulas for other $n$s in a similar way; see the binomial theorem.
|
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|
Sequence with denominators of products of consecutive Fibonacci numbers I'm trying to figure out a way to solve the value of this:
$$\frac{1}{1\times 2}-\frac{1}{2\times 3}+\frac{1}{3\times 5}-\frac{1}{5\times 8}+\frac{1}{8\times 13}-\dots$$
The only thing I can come up with is a summation involving the $nth$ fibonacci term formula that uses $\phi$. Any other insights?
|
You can do this using partial fractions.
Note that:
$$\begin{align}
\frac{1}{1 * 2} &= \frac{1}{1} + \frac{-1}{2} \\
\frac{1}{2 * 3} &= \frac{-1}{2} + \frac{2}{3} \\
\frac{1}{3 * 5} &= \frac{2}{3} + \frac{-3}{5} \\
\frac{1}{5 * 8} &= \frac{-3}{5} + \frac{5}{8} \\
\frac{1}{8 * 13} &= \frac{5}{8} + \frac{-8}{13}
\end{align}$$
The pattern can be formulated so that this identity can be proven later on
$$
\frac{1}{f_i * f_{i+1}} = (-1)^{i-1}\frac{f_{i-1}}{f_i} + (-1)^i\frac{f_i}{f_{i+1}}.
$$
Then the partial sums of the series are equivalent to
$$\begin{multline}
\frac{1}{1} + \frac{-1}{2} + \frac{1}{2} + \frac{-2}{3} + \frac{2}{3} + \frac{-3}{5} + \frac{3}{5} + \frac{-5}{8} + \frac{5}{8} + \frac{-8}{13} + \cdots + \frac{-f_i}{f_{i+1}} = \\
= 1 - \frac{f_i}{f_{i+1}} = \frac{f_{i+1}}{f_{i+1}} - \frac{f_i}{f_{i+1}} = \frac{f_{i-1}}{f_{i+1}}
\end{multline}$$
|
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|
Surface integral problem: $\iint_S (x^2+y^2)dS$
*
*The problem statement, all variables and given/known data
$\iint_S (x^2+y^2)dS$, $S$is the surface with vector equation $r(u, v)$ = $(2uv, u^2-v^2, u^2+v^2)$, $u^2+v^2 \leq 1$
*Relevant equations
Surface Integral. $\iint_S f(x, y, z)dS = \iint f(r(u, v))\left | r_u \times r_v \right |dA$,
*The attempt at a solution
First, I tried to shift the form of $\iint_S (x^2+y^2)dS$. $x^2+y^2$.
$x^2+y^2$ = $u^4v^4+2u^2v^2+v^4$, and $\left | r_u \times r_v \right |$ = $\sqrt{32v^4+64u^2v^2+32u^4}$
Thus, the initial integral becomes $\iint_S 2^2\sqrt{2}(u^2+v^2)^3 dudv$
I used polar coordinates, as u = rsin$\theta$ and v = $rsin\theta$. $0\leq r\leq 1$, $ 0 \leq \theta \leq 2\pi$. As result, the answer came to be $2^3\sqrt{2}/5*\pi$ but the answer sheet says its zero. Am I missing something?
|
Given parametrization
$$\phi (u,v) = (2uv,{u^2} - {v^2},{u^2} + {v^2})$$
Given function:
$$f(u,v) = {u^2} + {v^2}$$
Using polar coordinates:
$$\begin{gathered}
u = r \cdot cos(\varphi ) \hfill \\
v = r \cdot \sin (\varphi ) \hfill \\
\end{gathered} $$
Transform function:
$$f(r \cdot cos(\varphi ),r \cdot \sin (\varphi )) = {r^2} \cdot co{s^2}(\varphi ) + {r^2} \cdot {\sin ^2}(\varphi ) = {r^2}$$
Transform surface volume element:
$$\begin{gathered}
du = cos(\varphi ) \cdot dr - r\sin (\varphi ) \cdot d\varphi \hfill \\
dv = \sin (\varphi ) \cdot dr + r\cos (\varphi ) \cdot d\varphi \hfill \\
\hfill \\
du \wedge dv = r \cdot dr \wedge d\varphi \hfill \\
\end{gathered} $$
Integrate:
$$\begin{gathered}
\int\limits_{{u^2} + {v^2} \leqslant 1} {f(u,v) \cdot du \wedge dv} = \int\limits_0^{2 \cdot \pi } {\int\limits_0^1 {{r^2} \cdot r \cdot dr \wedge d\varphi } } \hfill \\
\hfill \\
= \int\limits_0^{2 \cdot \pi } {\int\limits_0^1 {{r^3} \cdot dr \cdot d\varphi } } = \int\limits_0^{2\pi } {\frac{1}{4} \cdot d\varphi } \hfill \\
\hfill \\
= \frac{1}{4} \cdot \int\limits_0^{2 \cdot \pi } {d\varphi } = \frac{1}{4} \cdot 2\pi \hfill \\
\hfill \\
= \frac{\pi }{2} \hfill \\
\end{gathered}$$
|
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|
Proving $\lim_{x\rightarrow 1}x^{\frac{1}{n-x-x^2-x^3-\ldots -x^n}}=\frac{1}{\sqrt[k]{e}}$ Proving
$$\lim_{x\rightarrow 1}x^{\left(\frac{1}{n-x-x^2-x^3\cdots -x^n}\right)}=\frac{1}{\sqrt[k]{e}}$$
when $$k=1+2+3+\ldots+n$$
|
Let $S$ be the denominator of the exponent so that $$S = n - x - x^{2} - \cdots - x^{n}$$ We have to calculate the limit $$\lim_{x \to 1}x^{1/S}$$ Let this desired limit be $L$ so that $$\begin{aligned}\log L &= \log\left(\lim_{x \to 1}x^{1/S}\right)\\
&= \lim_{x \to 1}\log x^{1/S}\text{ (by continuity of log)}\\
&= \lim_{x \to 1}\frac{\log x}{S}\\
&= \lim_{x \to 1}\frac{\log x}{x - 1}\cdot\frac{x - 1}{S}\\
&= \lim_{x \to 1}1\cdot\frac{x - 1}{S}\\
&= \lim_{x \to 1}\frac{x - 1}{n - x - x^{2} - \cdots - x^{n}}\\
&= \lim_{x \to 1}\frac{x - 1}{-(x + x^{2} + \cdots + x^{n} - n)}\\
&= \lim_{x \to 1}\frac{x - 1}{-\{(x - 1) + (x^{2} - 1) + \cdots + (x^{n} - 1)\}}\\
&= \lim_{x \to 1}\dfrac{1}{-\left\{1 + \dfrac{x^{2} - 1}{x - 1} + \cdots + \dfrac{x^{n} - 1}{x - 1}\right\}}\\
&= \dfrac{1}{-\left\{1 + 2 + \cdots + n\right\}}\\
&= -\frac{1}{k}\end{aligned}$$ We have used the standard limit $$\lim_{x \to a}\frac{x^{m} - a^{m}}{x - a} = ma^{m - 1}$$ with $a = 1$ and $m = 2, 3, \ldots, n$. It now follows that the desired limit is $$L = e^{-1/k} = \frac{1}{\sqrt[k]{e}}$$
|
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|
Show that $\sin45°+\sin15°=\sin75°$ Steps I took:
1) Finding the value of the left hand side
$$\sin45=\sin\frac { 90 }{ 2 } =\sqrt { \frac { 1-\cos90 }{ 2 } } =\sqrt { \frac { 1 }{ 2 } } =\frac { \sqrt { 2 } }{ 2 } $$
$$\sin15=\sin\frac { 30 }{ 2 } =\sqrt { \frac { 1-\cos30 }{ 2 } } =\sqrt { \frac { 1-\frac { \sqrt { 3 } }{ 2 } }{ 2 } } =\frac { \sqrt { 2-\sqrt { 3 } } }{ 2 } $$
So, $\sin45+\sin15=\frac { \sqrt { 2 } +\sqrt { 2-\sqrt { 3 } } }{ 2 } $
2) Rewriting $\sin75$
$$\sin75=\sin(45+30)=\sin45\cos30+\sin30\cos45$$
$$=\frac { \sqrt { 2 } }{ 2 } \cdot \frac { \sqrt { 3 } }{ 2 } +\frac { 1 }{ 2 } \cdot \frac { \sqrt { 2 } }{ 2 } $$
$$=\frac { \sqrt { 6 } +\sqrt { 2 } }{ 4 } $$
So now I have these two expressions that are obviously equal (when I compute them) but how do I show them as being equal to each other. I assume the question is implying for the resulting expressions to look equal.
|
mathisfun did a nice job of showing you how to reconcile your answers.
Another way of showing that $\sin(45^\circ) + \sin(15^\circ) = \sin(75^\circ)$ is to use the formula
$$\sin\alpha + \sin\beta = 2\sin\left(\frac{\alpha + \beta}{2}\right)\cos\left(\frac{\alpha - \beta}{2}\right)$$
with $\alpha = 45^\circ$ and $\beta = 15^\circ$.
\begin{align*}
\sin(45^\circ) + \sin(15^\circ) & = 2\sin\left(\frac{45^\circ + 15^\circ}{2}\right)\cos\left(\frac{45^\circ - 15^\circ}{2}\right)\\
& = 2\sin\left(\frac{60^\circ}{2}\right)\cos\left(\frac{30^\circ}{2}\right)\\
& = 2\sin(30^\circ)\cos(15^\circ)\\
& = 2 \cdot \frac{1}{2} \cdot \cos(15^\circ)\\
& = \cos(15^\circ)\\
& = \sin(90^\circ - 15^\circ)\\
& = \sin(75^\circ)
\end{align*}
since $\cos\theta = \sin(90^\circ - \theta)$.
|
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|
How can I solve the system of equations? How can I solve the system of equations?
$$\begin{cases}
x^2 y^2+12 x y^3-18 x y-18y^4-4 y^2+27=0,&\\
x^2 y^2-3 x y^3-3 x y+5 y^2=0.
\end{cases}$$
I have not any idea to solve.
|
$$\begin{cases}
x^2 y^2+12 x y^3-18 x y-18y^4-4 y^2+27=0,&\\
x^2 y^2-3 x y^3-3 x y+5 y^2=0.
\end{cases}$$
We subtract the second equation from the first to get
$15xy^3-15xy-18y^4-9y^2+27=0$.
But $15xy^3-15xy=15xy(y^2-1)$, and $-18y^4-9y^2+27=-9(2y^4+y^2-3)=9(-2y^2-3)(y^2-1)$.
Hence $15xy^3-15xy-18y^4-9y^2+27$$=(y^2-1)(15xy-18y^2-27)=3(y^2-1)(5xy-6y^2-9)$. Thus either $y^2=1$ (hence $y=\pm 1$, or $5xy-6y^2-9=0$.
Case 1: $y=\pm 1$.
If $y=1$, Substitute to the second equation to get $x^2-6x+5=0$, so $x=1$ or $x=5$.
If $y=-1$, Substitute to the second equation to get $x^2+6x+5=0$, so $x=-1$ or $x=-5$.
We can get these solutions by factoring.
Case 2: $5xy-6y^2-9=0$
So now we proceed to write $x^2 y^2-3 x y^3-3 x y+5 y^2=0$ in terms of $y$, by changing all $xy$ terms into $\frac{6y^2-9}{5}$.
Thus $x^2y^2-3xy^3-3xy+5y^2=(\frac{6y^2-9}{5})^2-(\frac{6y^2-9}{5})y^2-3(\frac{6y^2-9}{5})+5y^2=0$. This is just a quadratic in $y^2$. But then this quadratic is equal to $3(y^2)^2-14(y^2)+108=0$ (after expanding and regrouping), and it has no real roots as its discriminant, $14^2-4(3)(108)$, is negative.
Thus our only solutions are $\boxed{(x,y)=(\pm 1, \pm 5), (\pm 5, \pm1)}$.
|
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Find the least significant digit Explain why the number $(3^{27}\cdot7^{313}\cdot11^{121})^{1000}$ has 1 as its least significant digit. I know I am supposed to use $\pmod{10}$ but I am not sure how to combine the answers of the individual powers and then raise them to $1000$
|
Notice that $3^3=27\equiv 7\pmod{10}$, $3^4\equiv 21\equiv 1\pmod{10}$. $7^2\equiv 9\pmod{10}$, $7^4\equiv 9^2\equiv 1\pmod{10}$. Also, $11^4\equiv 1^4\equiv 1\pmod{10}$. Now powers distribute over products, so
$$(3^{27}7^{313}11^{121})^{1000}=3^{27000}7^{313000}11^{121000}$$
and since the exponents are all multiples of $4$ we have that each term in the product is congruent to $1$ mod $10$, hence so is the product of all of them.
|
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How to evaluate $\displaystyle\lim_{x\to0^+}\left(\frac{\ln(4^x-3^x)-\ln(4^x-1)}{x}\right)(4^x-1)$? How to evaluate
$$L:=\lim_{x\to0^+}\left(\frac{\ln(4^x-3^x)-\ln(4^x-1)}{x}\right)(4^x-1)$$?
My solution:
$$\begin{align}
L&=\lim_{x \to 0^+} \frac{\ln\left(\frac{4^x-3^x}{4^x-1}\right)}{x}(4^x-1)=\\
&=\lim_{x \to 0^+} \frac{ \ln\left(1-\frac{3^x-1}{4^x-1}\right)}{\frac{3^x-1}{4^x-1}}\times \frac{3^x-1}{4^x-1}\times \frac{4^x-1}{x}
\end{align}$$
How to continue from here?
|
$$\begin{align}L&=\lim_{x \to 0^+} \ln\left(1 - \frac{3^x-1}{4^x-1}\right)\times\frac{4^x-1}{x}=\\
&=\lim_{x \to 0^+} \ln\left(1 - \frac{3^x-1}{4^x-1}\right)\times\lim_{x \to 0^+}\frac{4^x-1}{x}=\\
&=\ln 4 \cdot \lim_{x \to 0^+} \ln\left(1 - \frac{3^x-1}{4^x-1}\right)=\\
&=\ln 4 \cdot \ln \left( \lim_{x \to 0^+}\left[1 - \frac{3^x-1}{4^x-1}\right]\right )=\tag{$\star$}\\
&=\ln 4 \cdot \ln \left(1 - \frac{\ln 3}{\ln 4} \right )\approx-2.17999444
\end{align}$$
In $(\star)$ we used the fact the logarithm is continuous in $1 - \frac{\ln 3}{\ln 4}$, so that we can evaluate the inner limit and then just substitute (there is a theorem about that, if you're curious).
Here is the plot of the function near $0$:
|
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|
Maximum area of a isosceles triangle in a circle with a radius r As said in the title, I'm looking for the maximum area of a isosceles triangle in a circle with a radius $r$.
I've split the isosceles triangle in two, and I solve for the area $A=\frac{bh}{2}$*. I have made my base $x$, and solve for the height by using the Pythagorean theorem of the smaller triangle (seen in picture).
$h=r+\sqrt{r^2-x^2}$
So my the formula, I think, for both triangles should be $A=x(r+\sqrt{r^2-x^2})$
But after I solved for the derivative, when put "$= 0$", and checked on my calculator, I got the maximum to be about $4.3301r$, which differs a lot from my book's answer of $\frac{3\sqrt{3}}{4}r^2$*. Is my formula for the area right? Am I going about this the wrong way, or is it just my derivative that is wrong? Thanks in advance
*
Edited from original post
$A=rx+x\sqrt{r^2-x^2}$
$A'=r+x(\frac{1}{2})(r^2-x^2)^{-\frac{1}{2}}(-2x)+\sqrt{r^2-x^2}$
$r+\sqrt{r^2-x^2}=\frac{x^2}{\sqrt{r^2-x^2}}$
$r\sqrt{r^2-x^2}+(r^2-x^2)=x^2$
$r^2+r\sqrt{r^2-x^2}=2x^2$
$r^2(r^2-x^2)=(2x^2-r^2)^2$
$r^4-r^2x^2=4x^2-4x^2r^2+r^4$
$4x^4=3x^2r^2$
$x=\frac{\sqrt{3}}{2}r$
|
A slightly different perspective:
Use above drawing:
Label apex of isosceles triangle $A$, centre of circle $O$, extend $AO$ to intersect the circle in $C$. Length $AC = 2r$. Pick any point, say, on the circle's left part, call $ B$. Look at $\triangle ABC$.
$\triangle ABC$ is a right triangle (Thales circle)
Draw the height to side $AC$. Let length of height = $x$.
Let footpoint of height on $AC$ be $X$.
Height $x$ divides length $AC$ into $q$ and $ p$, where $q $ the upper part.
Area of $\triangle ABX$ is half the area of the isosceles triangle we are originally looking at.
Altitude rule for $\triangle ABC$ : $x^2 = q×p$.
With $p= 2r - q$ we get: $x^2 = q(2r - q)$.
Area $ \triangle ABX = (1/2) qx = (1/2)q(q(2r - q))^{1/2}$.
Area = $(1/2)(q^3(2r - q)^{1/2}$.
To find extremal points we look at the square of the area function above, factor 1/2 has been dropped, I.e.:
$f(q) = q^3(2r -q)$.
$f'(q) = 6rq^2 - 4q^3 = 0$.
1)$q = 0$, Area = $0$, or
2) $4q = 6r$, $q = (3/2)r$.
Maximum area of original triangle :
AREA = $2(1/2)(q^3(2r-q))^{1/2} $=
$[((3/2)r)^3(2r -(3/2)r)]^{1/2}$ =
$(3/2)r[(3/2)r(r/2)]^{1/2}$ =
$(3/4)r^2√3$.
AREA = $(3/4)√3r^2$.
|
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|
How can I prove the pattern $\sqrt{1 + 155555…5} = 2 \sqrt{3888…89}?$ How can I prove this
$$\sqrt{1+155}=2\sqrt{39}$$
$$\sqrt{1+1555}=2\sqrt{389}$$
$$\sqrt{1+15555}=2\sqrt{3889}$$
$$\sqrt{1+155555}=2\sqrt{38889}$$
|
Well,
$$\sqrt{155+1}=\sqrt{156}=\sqrt{4\times39}=\sqrt{4}\times\sqrt{39}=2\sqrt{39}$$
$$\sqrt{1555+1}=\sqrt{1556}=\sqrt{4\times389}=\sqrt{4}\times\sqrt{389}=2\sqrt{389}$$
And on..
|
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|
How to solve the congruence $ x^4 + x + 3 = 0 \pmod{3^3}$? $$x^4 + x + 3 = 0 \pmod{3^3}$$
I'm not sure how to this, I've tried many times but it never works for me :/ so, I hope someone will help me
|
As the modulus is small, it isn't too hard to just check all the possible values of $x$. For $x = 0, 1, 2, 3$, the calculations are easy enough:
\begin{align*}
x = 0 &\implies x^4 + x + 3 = 3 \not\equiv 0 \bmod 9\\
x = 1 &\implies x^4 + x + 3 = 5 \not\equiv 0 \bmod 9\\
x = 2 &\implies x^4 + x + 3 = 21 \not\equiv 0 \bmod 9\\
x = 3 &\implies x^4 + x + 3 = 87 \not\equiv 0 \bmod 9.
\end{align*}
For $x \geq 4$, the presence of $x^4$ makes calculations slightly less trivial (not only calculating the value of $x^4 + x + 3$, but also determining whether the result is divisible by $9$). One way to proceed is to first work out $x^2 \bmod 9$ then square the result to obtain $x^4 \bmod 9$.
\begin{align*}
x = 4 &\implies x^2 = 16 \equiv 7 \bmod 9 \implies x^4 \equiv 4 \bmod 9\\
x = 5 &\implies x^2 = 25 \equiv 2 \bmod 9 \implies x^4 \equiv 4 \bmod 9\\
x = 6 &\implies x^2 = 36 \equiv 0 \bmod 9 \implies x^4 \equiv 0 \bmod 9\\
x = 7 &\implies x^2 = 49 \equiv 4 \bmod 9 \implies x^4 \equiv 7 \bmod 9\\
x = 8 &\implies x^2 = 64 \equiv 1 \bmod 9 \implies x^4 \equiv 1 \bmod 9.
\end{align*}
With these results at hand, the calculations becomes easier.
\begin{align*}
x = 4 &\implies x^4 + x + 3 \equiv 4 + 4 + 3 \bmod 9 \not\equiv 0 \bmod 9\\
x = 5 &\implies x^4 + x + 3 \equiv 4 + 5 + 3 \bmod 9 \not\equiv 0 \bmod 9\\
x = 6 &\implies x^4 + x + 3 \equiv 0 + 6 + 3 \bmod 9 \equiv 0 \bmod 9\ \ \checkmark\\
x = 7 &\implies x^4 + x + 3 \equiv 7 + 7 + 3 \bmod 9 \not\equiv 0 \bmod 9\\
x = 8 &\implies x^4 + x + 3 \equiv 1 + 8 + 3 \bmod 9 \not\equiv 0 \bmod 9.
\end{align*}
So the only solution is $x = 6$.
|
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|
Confusion about a Linear Transformation question. Let $\beta := [M_1, M_2, M_3, M_4]$ be the ordered basis of $R^{2×2}$ defined by:
$$ M_1 := \begin{pmatrix}
1 & 0\\
0 & 0 \end{pmatrix},
M_2 := \begin{pmatrix}
0 & 1\\
0 & 0 \end{pmatrix},
M_3 := \begin{pmatrix}
0 & 0\\
1 & 0 \end{pmatrix},
M_4 := \begin{pmatrix}
0 & 0\\
0 & 1 \end{pmatrix}
$$
Define the linear transformation
$L := A → (A + A^T)$.
a) What is $L \left( \begin{pmatrix}
a & b\\
c & d \end{pmatrix} \right)$?
b) Find the matrix representation of $L$ with respect to the basis $\beta$.
c) Find all solutions to $L(A) = \begin{pmatrix}
4 & 0\\
0 & -6 \end{pmatrix}$.
Does the following look correct?
$L \left( \begin{pmatrix}
a & b\\
c & d \end{pmatrix} \right)$ = $ \begin{pmatrix}
2a & b + c\\
b+ c & 2d \end{pmatrix}$
$Ker(L) = span\{\begin{pmatrix}
0 & c\\
-c & 0 \end{pmatrix}\}$
$range(L) = span\{ \begin{pmatrix}
a & 0\\
0 & 0 \end{pmatrix}, \begin{pmatrix}
0 & s\\
s & 0 \end{pmatrix}, \begin{pmatrix}
0 & 0\\
0 & d \end{pmatrix}\}$ where s denotes the quantity $c + d$
Matrix Representation: $\begin{pmatrix}
2 & 0 & 0 & 0\\
0 & 1 & 1 & 0\\
0 & 1 & 1 & 0\\
0 & 0 & 0 & 2\end{pmatrix} \begin{pmatrix}
a\\
b\\
c\\
d\end{pmatrix}=\begin{pmatrix}
2a\\
b + c\\
b + c\\
2d\end{pmatrix}$
Thanks for the help!
|
$$ \begin{pmatrix}
a & b\\
c & d \end{pmatrix}=a.\begin{pmatrix}
1 & 0\\
0 & 0 \end{pmatrix}+b.\begin{pmatrix}
0 & 1\\
0 & 0 \end{pmatrix}+c.\begin{pmatrix}
0 & 0\\
1 & 0 \end{pmatrix}+d.\begin{pmatrix}
0 & 0\\
0 & 1 \end{pmatrix}$$
Now $L$ is linear and then apply $L$ to each of these four matrices given $L(A)=A+A^T$
|
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|
Prove if $\left\{x_n\right\}$ converges to $2$, then $\left\{\frac{1}{x_n}\right\}$ converges to $\frac{1}{2}$ We know that for all $\varepsilon > 0$, there exists an $N \in \mathbb{N}$ such that $\lvert x_n - 2 \rvert < \epsilon$ for all $n \geq N$, and we want to show that for all $\varepsilon' > 0$, there exists an $N' \in \mathbb{N}$ such that $\left| \frac{1}{x_n} - \frac{1}{2} \right| < \epsilon'$ for all $n \geq N'$.
Let $\varepsilon = \varepsilon' - \frac{3}{2}$. Then by the triangle inequality, we have $$\left| \frac{1}{x_n} - \frac{1}{2} \right| \leq \left| \frac{1}{x_n}\right| + \left| -\frac{1}{2}\right|.$$
Because we proved earlier that $x_n > 1$ for all $n \geq N$, we know that $$\left| \frac{1}{x_n}\right| + \left| -\frac{1}{2}\right| < \left| x_n\right| + \left| -\frac{1}{2}\right|.$$
By the triangle inequality again, we know that $$\left| x_n - \frac{1}{2}\right| \leq \left| x_n\right| + \left| -\frac{1}{2}\right|.$$
Note that $\left| x_n - \frac{1}{2}\right| = \left| (x_n - 2) + (2 - \frac{1}{2})\right|$, so we get $$\left| (x_n - 2) + (2 - \frac{1}{2})\right| \leq \left| x_n - 2\right| + \left| 2 - \frac{1}{2}\right|.$$
Then we have $$\left| x_n - 2\right| + \left|2 - \frac{1}{2}\right| < \varepsilon + \frac{3}{2} = \left(\varepsilon' - \frac{3}{2}\right) + \frac{3}{2} = \varepsilon',$$ so $\frac{1}{x_n} \rightarrow \frac{1}{2}$.
I just realized that I don't know how to make sure that $\varepsilon' - \frac{3}{2}$ is a positive number. How can I do this?
|
Use the Algebraic Limit Theorem for sequences:
*
*$\lim(ca_n)=c\cdot\lim a_n$
*$\lim(a_n+b_n)=\lim a_n + \lim b_n$
*$\lim(a_n b_n)=\lim a_n\cdot\lim b_n$
*$\lim\left(\frac{a_n}{b_n}\right)=\frac{\lim a_n}{\lim b_n}$ provided $b_n \ne 0$ and $\lim b_n \ne 0$
Rule (4) in the backwards direction, with $(a_n) = (1, 1, 1, \dots)$ and $(b_n) \to 2$, will give you what you need. If $\lim a_n = 1$ and $\lim b_n = 2$, then $\lim\left(\frac{1}{b_n}\right) = \frac{1}{2}$.
A proof of the Algebraic Limit Theorem can be found on p. 45 of this book:
Abbott, S. (2001). Understanding analysis. New York, NY: Springer.
|
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|
deriving the sum of $x^n/(n+2)^2$ I am writing a research paper and I have stumbled upon an issue.
I have to evaluate
$$\sum_{n=1}^{\infty} \frac{x^n}{(n+2)^2}$$
Here is what I did:
$$ \sum_{n=1}^{\infty} x^{n-1} = \frac{1}{1-x}$$
$$\sum_{n=1}^{\infty} x^{n+1} = \frac{x^2}{1-x}$$
Integrate once with respect to $x$.
$$\sum_{n=1}^{\infty} \frac{x^{n+2}}{n+2} = \frac{3}{2} - \frac{x^2}{2} - x - \log(1-x)$$
Divide by $x$
$$\sum_{n=1}^{\infty} \frac{x^{n+1}}{n+2} = \frac{3}{2x} - \frac{x}{2} - 1 - \frac{\log(1-x)}{x}$$
Integrate the expression again;
$$\sum_{n=1}^{\infty} \frac{x^{n+2}}{(n+2)^2} = \frac{3\log(x)}{2} - x - \frac{x^2}{4} + Li_2(x)$$
All we have to do is divide by $x^2$
$$\sum_{n=1}^{\infty} \frac{x^n}{(n+2)^2} = \frac{3\log(x)}{2x^2} - \frac{1}{x} - \frac{1}{4} + \frac{Li_2(x)}{x^2} = \frac{4Li_2(x) - x^2 - 4x + 6\log(x)}{4x^2}$$
The issue is WolframAlpha returns it as:
$$\sum_{n=1}^{\infty} \frac{x^n}{(n+2)^2} = \frac{4Li_2(x) - x^2 - 4x}{4x^2}$$
Why is the way I did, wrong? I have an extraneous, extra log term.?
Help is appreciated.
|
Not sure what you did, but all that work seems kinda silly to arrive at your result.
$$\sum_{n=1}^\infty\frac{x^n}{(n+2)^2}=\frac{1}{x^2}\sum_{n=1}^\infty\frac{x^{n+2}}{(n+2)^2}=\frac{1}{x^2}(-x-\frac{x^2}{4}+\sum_{n=1}^\infty\frac{x^n}{n^2})=\frac{1}{x^2}(-x-\frac{x^2}{4}+\text{Li}_2(x))$$
|
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|
Prove that $ ax^2+bx+c=0 $ has at least one root in $(0,1)$ if $10a+12b+15c=0$
If $10a+12b+15c=0$, Prove that $$ ax^2+bx+c=0 $$ has at least one root in $(0,1)$.
Progress
I tried to solve this by Rolle`s theorem ($f'$ has a root between any two roots of $f$), but could not reach the result.
|
Consider $x = \frac{10}{12}$:
$$\begin{align*}
f\left(\frac{10}{12}\right)
&= a\left(\frac{10}{12}\right)^2+b\left(\frac{10}{12}\right)+c\\
&= \frac{10}{144}\left(10a+12b+\frac{144}{10}c\right)\\
&= \frac{10}{144}\left(10a+12b+15c-\frac{6}{10}c\right)\\
&= -\frac{6}{144}c
\end{align*}$$
And consider $x=0$:
$$f(0) = 0^2a + 0b + c = c$$
If $c\ne 0$, by intermediate value theorem, there must be an $x\in\left(0, \frac{10}{12}\right)\subset(0,1)$ which satisfies $f(x)= 0$.
If $c=0$, $\frac{10}{12}\in(0,1)$ is a root.
How I obtained $x = \frac{10}{12}$:
I assume $x = 12k$ and $x^2 = 10k$ for some real number $k$. By solving $10k = (12k)^2$, $k$ is chosen to be the non-zero root $k = \frac{10}{144}$, which appears as the multiplier above. And $x = 12k = \frac{10}{12}$.
|
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|
Prove two identities relating to series Show that:
$$(1)\sum_{n=1}^\infty\ln\left(\cos \frac{x}{2^n}\right)=\ln\frac{\sin x}x$$
$$(2). \sum_{n=1}^\infty\frac1{2^n}\tan \frac{x}{2^n}=\frac1x-\cot x$$
Thank you in advance.
NOTE:
The origial problem (1) I stated is $\sum_{n=1}^\infty\lim_{x\to \infty}\left(\cos \frac{x}{2^n}\right)=\ln\frac{\sin x}x$, which is a mistake as noted by the comments below.
|
For the first one :
$$\sum\limits_{n=1}^{N} \ln \cos \frac{x}{2^n} = \ln \prod\limits_{n=1}^{N} \cos \frac{x}{2^n} = \ln \frac{\sin \frac{x}{2^N}\prod\limits_{n=1}^{N} \cos \frac{x}{2^n}}{\sin \frac{x}{2^N}} = \ln \frac{\sin x}{2^N\sin \frac{x}{2^N}}$$
where, we made use of the identity: $\sin y \cos y = \frac{1}{2}\sin 2y$ repeatedly.
Taking limit as $N \to \infty$, we have $\displaystyle \sum_{n=1}^{\infty}\ln\big(\cos \frac{x}{2^n}\big)=\ln \frac{\sin x}{x}$
For the second one use the identity: $$\tan y = \cot y - 2\cot 2y$$
So, $\displaystyle \sum\limits_{n=1}^{N} \frac{1}{2^n} \tan \frac{x}{2^n} = \sum\limits_{n=1}^{N} \frac{1}{2^n}\left(\cot \frac{x}{2^n} - 2\cot \frac{x}{2^{n-1}} \right) = \frac{1}{2^{N}}\cot \frac{x}{2^N} - \cot x$
Taking the limit as $N \to \infty$,
$$\sum\limits_{n=1}^{\infty} \frac{1}{2^n} \tan \frac{x}{2^n} = \frac{1}{x} - \cot x$$
|
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|
How to compute $\sum_{n\ =\ 1}^{\infty}\arctan\left(\,\frac{3n^{2}}{ 2n^{4} - 1}\,\right)$ I find this problem on facebook group.
$$\mbox{Is it possible to find exact value of}\quad
\sum_{n\ =\ 1}^{\infty}\arctan\left(\,\frac{3n^{2}}{ 2n^{4} - 1}\,\right)\ {\large ?}.
$$
I think this is not telescope sum. And Wolfram Alpha can not find it.
Thank in advances.
|
Let $$\tan x=\frac{1}{n^2} \text{and } \tan y=\frac{1}{2n^2}$$ so we have that $$\tan(x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y}=\frac{\frac{1}{n^2}+\frac{1}{2n^2}}{1-\left(\frac{1}{n^2}\frac{1}{2n^2}\right)}=\frac{3n^2}{2n^4-1}$$ hence we have $$x+y=\arctan\frac{\frac{1}{n^2}+\frac{1}{2n^2}}{1-\left(\frac{1}{n^2}\frac{1}{2n^2}\right)}=\arctan\frac{1}{n^2}+\arctan\frac{1}{2n^2}.$$ From here I think you can do
|
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|
Proving that $\int_0 ^1 \frac{\text{d}s}{\sqrt{1-s^2}}$ converges with no trig functions Let
$$\int_0 ^1 \frac{\text{d}s}{\sqrt{1-s^2}}$$
How to show that it converges with no use of trigonometric functions?
(trivially, it is the anti-derivative of $\sin ^{-1}$ and therfore can be computed directly)
|
Here is a proof in single-variable calculus language. On the interval $0\leq s \leq 1$, we have
$$
\sqrt{1-s^2} = \sqrt{(1+s)(1-s)} \leq \sqrt{2} \sqrt{1-s}
$$
So
$$
\frac{1}{\sqrt{1-s^2}} \geq \frac{1}{\sqrt{2}} \frac{1}{\sqrt{1-s}}
$$
We would like to use the Comparison Test, comparing $\int_0^1 \frac{ds}{\sqrt{1-s}}$ to $\int_0^1 \frac{ds}{\sqrt{1-s}}=\int_0^1 \frac{du}{\sqrt{u}}$. The latter converges by the $p$-test ($p=1/2 < 1$).
Unfortunately, the comparison goes in the wrong direction to apply the Comparison Test. We need to instead bound $\sqrt{1-s^2}$ from below in order to bound $\frac{1}{\sqrt{1-s^2}}$ from above. So assume instead that $\frac{1}{2} \leq s \leq 1$. Then
$$
\sqrt{1-s^2} = \sqrt{(1+s)(1-s)} \geq \sqrt{\frac{3}{2}} \sqrt{1-s}
$$
So
$$
\frac{1}{\sqrt{1-s^2}} \leq \sqrt{\frac{2}{3}} \frac{1}{\sqrt{1-s}}
$$
on this interval.
To put it together, you have
$$
\begin{aligned}
\int_0^1 \frac{ds}{\sqrt{1-s^2}} &= \int_0^{1/2}\frac{ds}{\sqrt{1-s^2}} + \int_{1/2}^1 \frac{ds}{\sqrt{1-s^2}}\\
&\leq \int_0^{1/2}\frac{ds}{\sqrt{1-s^2}} + \sqrt{\frac{2}{3}} \int_{1/2}^1 \frac{ds}{\sqrt{1-s}}
\end{aligned}
$$
The first integral is not improper and the second converges.
|
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|
Find all the prime numbers that satisfy the following conditions There was a brainteaser in the Science Magazine from University of Hong Kong which is as follow:
Find all the prime numbers $p$ such that $\sqrt{\frac{p+7}{9p-1}}$ is rational.
I tried a few numbers and it seems to suggest that $11$ is a suitable candidate.
Can I know the techniques to approach this question?
Thank you.
|
The fact that $p$ is prime is something of a red herring.
Suppose that $$\sqrt{\frac{x+7}{9x-1}}$$
is rational for some positive integer $x$.
Then for some positive integers $k$, $a$, and $b$, we have the equations
$$x+7=ka^2$$
and $$9x-1=kb^2$$
Multiplying the first equation by $9$ and subtracting the second equation from it, we get
$$64=k(9a^2-b^2)=k(3a-b)(3a+b)$$
Now each of $k$, $3a-b$, and $3a+b$ is a positive factor of $64$. As a reminder, $64$ has seven positive factors: $1$, $2$, $4$, $8$, $16$, $32$, and $64$.
Note that the terms $3a-b$ and $3a+b$ add up to be $6a$, so the sum of these terms must be divisible by $6$. Using this criteria, we can quickly reduce the possible pairs $(3a-b,3a+b)$ to
$(2,4)$, $(2,16)$, $(2,64)$, $(4,8)$, $(4,32)$, $(8,16)$, $(8,64)$, and $(16,32)$. However, the products of these pairs is greater than $64$ in all cases except $(2,4)$, $(2,16)$, and $(4,8)$.
Now, solving the resulting equations for $k$, $a$, and $b$, we get the possible solutions $(k,a,b)$ as $(8,1,1)$, $(2,3,7)$, and $(2,2,2)$. These then correspond to solutions for $x$ (we use here $x=ka^2-7$) - giving us $x=8-7=1$, $x=18-7=11$, and $x=8-7=1$. Hence the only positive integers which make
$$\sqrt{\frac{x+7}{9x-1}}$$
rational are $1$ and $11$.
Finally, since we are asked for all prime solutions, we have the only solution as $p=11$.
|
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|
show that $n + \lfloor \sqrt{n} + \frac{1}{2}\rfloor$ is never a perfect square for all positive integers $n$ show that $n + \lfloor \sqrt{n} + \frac{1}{2}\rfloor$ is never a perfect square for all positive integers $n$
I am thinking of a proof by contradiction by assuming $n + \lfloor \sqrt{n} + \frac{1}{2}\rfloor = k^2 \implies \lfloor \sqrt{n} + \frac{1}{2}\rfloor = k^2-n $
|
Let $t=\lfloor\sqrt{n}+\frac{1}{2}\rfloor$, so $t-\frac{1}{2}\le\sqrt{n}<t+\frac{1}{2}$.
If $n+t=k^2$ for some integer $k$,
then $t^2-t+\frac{1}{4}\le n<t^2+t+\frac{1}{4}\implies t^2+\frac{1}{4}\le n+t<t^2+2t+\frac{1}{4}<(t+1)^2\implies$
$t^2<k^2<(t+1)^2$, which gives a contradiction.
|
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|
Equivalence of $\pi$ is the first positive zero of the taylor series for $\sin(x)$ and $\pi/4 = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots$ For $x\in\mathbb{R}$, define $\sin (x) = x - x^3/3!+x^5/5!-\cdots$ and $\pi = 4(1-\frac{1}{3}+\frac{1}{5} -\frac{1}{7}+\cdots)$. Then show that $\sin(\pi/2) = 1$
In the prologue of Real and Complex Analysis by Walter Rudin, pi is defined as the first positive zero of the series defined as $\sin(x)$; I want to check that pi is same as above defined Pi.
|
Here's a detailed outline for how to do this in a fairly standard line of development of trig functions from their power series definitions.
(A "plug in" combinatorial proof would be neat to see as well, of course.)
Step 1. Show the basic trig identities (double angle formulas and $\sin(x)^2 + \cos(x)^2 = 1$) for the taylor series definitions of $\sin(x)$ and $\cos(x)$.
[The double angle formulas are nice to prove using the power series for $e^x$. The second is immediate upon differentiation.]
Step 2. Let $pi$ (not the greek symbol for now) be the first root of $\sin(x)$. From the derivative of $\sin(x)$ at zero and the intermediate value theorem, we get that $\sin(x) > 0$ for $0 < x < pi$.
Step 3. Use trig identities to show that $\sin(2x) = 0$ if and only if $\sin(x) = 0$ or $\cos(x) = 0$. Conclude that $x = pi/2$ is the first time $\cos(x) = 0$.
Step 4. Let $\tan(x) = \frac{\sin(x)}{\cos(x)}$. Show that $\frac{d}{dx} \tan(x) = \frac{1}{\cos(x)^2}$. Conclude that $\tan(x)$ is well-defined, increasing, and unbounded on $(-pi/2,pi/2)$.
Step 5. Define $\arctan(x)$ to be the inverse of $\tan(x): (-pi/2,pi/2) \rightarrow \mathbb{R}$. Show that $\frac{d}{dx}\arctan(x) = \frac{1}{1+x^2}$ from trig identities.
Step 6. Show that $\arctan(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots$ from the series expansion for $\frac{1}{1+x^2}$ and differentiation of power series theorems and the fact that $\tan(0) = 0$. This converges and hence is valid on $(-1,1)$.
Step 7. The series also converges for $x = 1$ and hence (by a theorem on power series convergence), we get $\arctan(1) = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \ldots$. Thus it remains to show $2 \arctan(1) = pi/2$.
Step 8. Show from trig identities that $\tan(x) = \pm 1$ if and only if $\sin(x) = \pm \cos(x)$ if and only if $\cos(2x) = 0$. Conclude that $2 \arctan(1) = pi/2$, as desired (because $2 \arctan(1)$ must be the first zero of $\cos(x)$).
|
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|
Solving an algebraic inequality For any $a$, $b$, and $c$ prove
$$3a^2+3b^2-2b+2a+1>0$$
I tried the following
$$(a+1)^2+(b-1)^2+2(a^2+b^2)-1>0\\
(a+1)^2-1+(b-1)^2+2(a^2+b^2)>0\\
(a+1-1)(a+1+1)+(b-1)^2+2(a^2+b^2)>0\\
(a^2+2a)+(b-1)^2+2(a^2+b^2)>0\\
$$
$a^2$ and $(b-1)^2$ and $2(a^2+b^2)$ are always positive and greater than zero, when $a>0$ our inequality is proved. However when a is negative I can not say $2a$ is always less than or equal to $a^2$. I think this problem should be solved by algebraic identities alone and not by discussing the different signs of $a$? I appreciate your help
|
$f(a) = 3a^2 +2a + (3b^2-2b+1) \to \triangle' = 1^2 - 3(3b^2-2b+1) = -9b^2 +6b - 2 < 0,
\forall b \text{ since } \triangle' = 3^2 -(-9)(-2) = 9 - 18 = -9 < 0 \to f(a) > 0, \forall a$.
|
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|
Evaluation of $\sqrt{\frac12+\sqrt{\frac14+\sqrt{\frac18+\cdots+\sqrt{\frac{1}{2^n}}}}}$ I was just playing around with a calculator, and came to the conclusion that:
$$\sqrt{\frac12+\sqrt{\frac14+\sqrt{\frac18+\cdots+\sqrt{\frac{1}{2^n}}}}} \approx 1.29$$
Now I'm curious. Is it possible to evaluate the exact value of the following?
|
$$y=\sqrt{\frac12+\sqrt{\frac14+\sqrt{\frac18+\cdots+\sqrt{\frac{1}{2^n}}}}}\equiv\sqrt{\frac12+\frac{1}{\sqrt{2}}\left(\sqrt{\frac12+\sqrt{\frac12+\cdots+\sqrt{\frac{1}{2}}}}\right)}$$
$$y=\sqrt{\frac12+\frac{1}{\sqrt{2}}\left(\sqrt{\frac12+\sqrt{\frac12+\cdots+\sqrt{\frac{1}{2}}}}\right)}$$ But the Let the term in bracket be x
therefore:$$x=\sqrt{\frac12+\sqrt{\frac12+\cdots+\sqrt{\frac{1}{2}}}}\equiv\sqrt{\frac12+x}$$
squaring both sides
$$x^2=\frac12+x$$
$$x^2-x-\frac12=0$$solving the equation gives:
$$x=\frac{1+\sqrt{3}}{2}$$
but we have $$y=\sqrt{\frac12+\frac{1}{\sqrt{2}}\left(\sqrt{\frac12+\sqrt{\frac12+\cdots+\sqrt{\frac{1}{2}}}}\right)}=\sqrt{\frac12+\frac{1}{\sqrt{2}}x}$$
Putting the value of x into y gives$$y=\sqrt{\frac12+\frac{1}{\sqrt{2}}\frac{1+\sqrt{3}}{2}}=1.211$$
Therefore $$\sqrt{\frac12+\sqrt{\frac14+\sqrt{\frac18+\cdots+\sqrt{\frac{1}{2^n}}}}}=1.211$$
This is an exact value.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to solve $y=(xy'+2y)^2$? What kind of differential equation is this thing and how to solve it?
$$y=(xy'+2y)^2$$
$$y=x^2y'^2+4xyy'+4y^2$$
|
the differential equation $y = (x\frac{dy}{dx} + 2y)^2$ is singular at $x = 0.$ one has to find the solution away from $x = 0$ and patch it up around
$x = 0$ and $y = 0$ or $y + \frac{1}{4}.$ there is a boundary layer at $x = 0.$
now we will find the outer solution that is valid away from $x = 0.$
$y = (x\frac{dy}{dx} + 2y)^2$ implies $y \ge 0$ so we can make a change of variable $y = u^2, u \ge 0.$ this transforms the differential equations to
$$2xu \frac{du}{dx} + 2u^2 = \pm u , \frac{d(xu)}{dx} = \pm \frac{1}{2}$$ this has the general solution $$ u = \pm \frac{1}{2} + \frac{C}{x}, y = \left( \pm \frac{1}{2} + \frac{C}{x} \right)^2 $$
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|
Can I get a closed-form of $\frac{\zeta(2) }{2}-\frac{\zeta (4)}{2^3}+\frac{\zeta (6)}{2^5}-\frac{\zeta (8)}{2^7}+\cdots$? Can I get a closed-form of
$$\frac{\zeta(2) }{2}-\frac{\zeta (4)}{2^3}+\frac{\zeta (6)}{2^5}-\frac{\zeta (8)}{2^7}+\cdots$$
|
Let $A$ be the series. Then
\begin{align}A &= \sum_{k = 1}^\infty (-1)^k \frac{\zeta(2k)}{2^{2k-1}}\\
&= \sum_{k = 1}^\infty \frac{(-1)^k}{2^{2k-1}} \sum_{n = 1}^\infty \frac{1}{n^{2k}}\\
&= 2\sum_{n = 1}^\infty \sum_{k = 1}^\infty \left(-\frac{1}{4n^2}\right)^k\\
&= -2\sum_{n = 1}^\infty \frac{1}{4n^2+1}\\
& = -\frac{1}{4}\sum_{{n = -\infty \atop n \neq 0}}^\infty \frac{1}{n^2 + \frac{1}{4}}\\
&= \frac{1}{4}\left(\sum_{n = -\infty}^\infty \frac{1}{n^2+\frac{1}{4}} - 4\right)\\
&= \frac{1}{4}\left(\operatorname{Res}_{z = \frac{i}{2}} \frac{\pi \cot(\pi z)}{z^2 + \frac{1}{4}} + \operatorname{Res}_{z = -\frac{i}{2}} \frac{\pi \cot(\pi z)}{z^2 + \frac{1}{4}} - 4\right)\\
&= \frac{1}{4}\left(\frac{\pi\cot\left(\frac{\pi i}{2}\right)}{2\left(\frac{i}{2}\right)} + \frac{\pi \cot\left(-\frac{\pi i}{2}\right)}{2\left(-\frac{i}{2}\right)} - 4\right)\\
&= \frac{1}{4}\left(\pi \coth\left(\frac{\pi}{2}\right) + \pi \coth\left(\frac{\pi} {2}\right) - 4 \right)\\
&=\frac{\pi}{2}\coth\left(\frac{\pi}{2}\right) - 1.
\end{align}
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|
How to find all positive integers $a,b,c,d$ with $a\le\ b\le c$ such that $a!+b!+c!=3^d$ ? How to find all positive integers $a,b,c,d$ with $a\le\ b\le c$ such that $a!+b!+c!=3^d$ ?
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Let $X = a!+b!+c!$.
*
*$a = 1$, otherwise $2 | a! \land 2| b! \land 2| c! \implies 2 | X \implies X \ne 3^d$
*$b \le 2$, otherwise $X \equiv 1 + 0 + 0 \equiv 1 \pmod 3 \implies X \ne 3^d$.
*If $b = 1$, then $1 \le c \le 2$, otherwise, $X \equiv 1 + 1 + 0 \equiv 2 \pmod 3
\implies X \ne 3^d$.
*If $b = 2$, then $2 \le c \le 5$, otherwise $X \equiv 1 + 2 + 0 \equiv 3 \pmod 9$.
Since $X > 3$ in this case, $X \ne 3^d$ once again.
Combine all these, there are only 6 candidates for $(a,b,c)$:
$$(1,1,1), (1,1,2), (1,2,2), (1,2,3), (1,2,4), (1,2,5)$$
By brute force, one can check three of them are solutions:
$$(1,1,1) \leadsto X = 3,\quad (1,2,3) \leadsto X = 9\quad\text{ and }\quad (1,2,4) \leadsto X = 27$$
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|
find x+y if $ (x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=42 $? Let $ (x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=42 $.
How find $x+y$? I tried a number of ways.
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Let's call $\alpha$ the value of $x+\sqrt{x^2+1}$ and $\beta$ the value of $y+\sqrt{y^2+1}$
Then we know that $x=\dfrac{\alpha^2-1}{2\alpha}$ and $y=\dfrac{\beta^2-1}{2\beta}$
We know also that $\alpha\beta=42$
Hence $x+y=\dfrac{\alpha^2-1}{2\alpha}+\dfrac{\beta^2-1}{2\beta}=\dfrac{(\alpha\beta-1)(\alpha+\beta)}{2\alpha\beta}$
$x+y=\dfrac{41}{84}\times(\alpha+\beta)=\dfrac{41}{84}\times(\dfrac{42}{\beta}+\beta)$
$x+y$ can thus take many different values
|
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|
What is $\underbrace{555\cdots555}_{1000\ \text{times}} \ \text{mod} \ 7$ without a calculator
It can be calculated that $\frac{555555}{7} = 79365$. What is the remainder of the number $5555\dots5555$ with a thousand $5$'s, when divided by $7$?
I did the following:
$$\begin{array}
& 5 \ \text{mod} \ 7=& &5 \\
55 \ \text{mod} \ 7= & &6 \\
555 \ \text{mod} \ 7= & &2 \\
5555 \ \text{mod} \ 7= & &4 \\
55555 \ \text{mod} \ 7= & &3 \\
555555 \ \text{mod} \ 7= & &0 \\
5555555 \ \text{mod} \ 7= & &5 \\
55555555 \ \text{mod} \ 7= & &6 \\
555555555 \ \text{mod} \ 7= & &2 \\
5555555555 \ \text{mod} \ 7= & &4 \\
\end{array}$$
It can be seen that the cycle is: $\{5,6,2,4,3,0\}$.
$$\begin{array}
& 1 \ \text{number =} &5 \\
7 \ \text{numbers =} &5 \\
13 \ \text{numbers =} &5 \\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots & \\
985 \ \text{numbers =} &5 \\
991 \ \text{numbers =} &5 \\
997 \ \text{numbers =} &5 \\
998 \ \text{numbers =} &6 \\
999 \ \text{numbers =} &2 \\
\color{red}{1000} \ \color{red}{\text{numbers =}} &\color{red}{4} \\
\end{array}$$
From here, we can conclude that $\underbrace{555\cdots555}_{1000\ \text{times}} \ \text{mod} \ 7 = 4$.
However, I wasn't allowed to use a calculator and solved this in about 12 minutes. Another problem was that there was a time limit of about 5 minutes. My question is: Is there an easier and faster way to solve this?
Thanks a lot in advance!
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Set $C = \displaystyle \sum_{i=0}^{999}5\cdot 10^i$.
Applying the summation formula for a geometric series,
$\quad \displaystyle C = 5 \, (1-10^{1000}) \,(1-10)^{-1}$
Continuing,
$\quad \displaystyle C \equiv 5 \, (1-3\times{3^3}^{333}) \, 5^{-1} \equiv 1 - 3\times(-1)^{333} \equiv 4 \pmod{7}$
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|
Let $C$ be the curve of intersection of the plane $x+y-z=0$ with ellipsoid $\frac{x^2}4+\frac{y^2}5+\frac{z^2}{25}=1$. Let $C$ be the curve of intersection of the plane $x+y-z=0$ and the ellipsoid $$\frac{x^2}4+\frac{y^2}5+\frac{z^2}{25}=1$$ Find the points on $C$ which are farthest and nearest from the origin
When dealing with constraints I tried to consider the function $$F(x,y,z)=x^2+y^2+z^2-\lambda(x+y-z)-\mu\left(\frac{x^2}4+\frac{y^2}5+\frac{z^2}{25}-1\right)$$ However, I cannot solve this equation after differentiating respect to $x,y,z$ because it yields three equations with no common solution.
The system of equations are
$$2x=\lambda+\frac{\mu x}{2}$$
$$2y=\lambda+\frac{2\mu y}{5}$$
$$2z=-\lambda+\frac{2\mu z}{25}$$
How would I approach this problem, thanks.
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Since there's a computer result posted now, I guess I'll describe my approach further. From the set of "Lagrange equations" you describe, we can solve each one for $ \ \lambda \ $ to establish
$$ \lambda \ = \left ( \ 2 \ - \frac{\mu}{2} \right ) \ x \ = \left ( \ 2 \ - \frac{2\mu}{5} \right) \ y \ = \left( \ \frac{2\mu}{25} \ - 2 \right ) \ z \ \ . $$
We can solve the implied pairs of equations for $ \ x \ $ and $ \ y \ $ in terms of $ \ z \ $ ; inserting these into the equation for the "constraint plane" $ \ x \ + \ y \ = \ z \ $ yields
$$ \ \frac{4 \ (\mu \ - \ 25)}{25 \ (4 \ - \ \mu )} \ + \ \frac{\mu \ - \ 25}{5 \ (5 \ - \ \mu )} \ = \ 1 \ \ , $$
leading to the quadratic equation $ \ 34 \mu^2 \ - \ 490 \mu \ + 1500 \ = \ 0 \ $ , for which the solutions are $ \ \mu \ = \ 10 \ $ and $ \ \mu \ = \ \frac{75}{17} \ $ .
The first solution, $ \ \mu \ = \ 10 \ $ , used in the proportionality relation among the coordinates gives us $ \ x \ = \ \frac{2}{5} z \ $ and $ \ y \ = \ \frac{3}{5} z \ $ . We can now use these in the equation for the ellipsoidal contraint surface to find
$$ z^2 \ = \ \frac{125}{19} \ \ \Rightarrow \ \ x \ = \ \pm 2 \sqrt{\frac{5}{19}} \ \ , \ \ y \ = \ \pm 3 \sqrt{\frac{5}{19}} \ \ , \ \ z \ = \ \pm 5 \sqrt{\frac{5}{19}} \ \ . $$
Similarly, the second solution, $ \ \mu \ = \ \frac{75}{17} \ $ , produces the proportions $ \ x \ = \ 8 z \ $ and $ \ y \ = \ -7 z \ $ for the points on the intersection curve found from
$$ z^2 \ = \ \frac{100}{2584} \ = \ \frac{25}{646} \ \ \Rightarrow \ \ x \ = \ \pm \frac{40}{\sqrt{646}} \ \ , \ \ y \ = \ \mp \frac{35}{\sqrt{646}} \ \ , \ \ z \ = \ \pm \frac{5}{\sqrt{646}} \ \ . $$
The first pair represents the maximal-distance points with
$$ s^2 \ = \ (2^2 \ + \ 3^2 \ + \ 5^2 ) \ \cdot \ \frac{5}{19} \ = \ 10 \ \ $$
and the second pair are the minimal-distance points with
$$ s^2 \ = \ (8^2 \ + \ [-7]^2 \ + \ 1^2 ) \ \cdot \ \frac{25}{646} \ = \ \frac{75}{17} \ \ \approx \ \ 4.4118 \ \ . $$ [These are confirmed by 111's Sage output.]
We see that these coordinate sets do satisfy the constraint $ \ x \ + \ y \ = \ z \ $ , which also explains why there are just two points for each solution in $ \ \mu \ $ . [The number $ \ 646 \ $ factors as $ \ 17 \cdot 19 \ $ , which clarifies the results shown in the computer results.]
We would expect the points of minimal and maximal distance from the origin to be arranged with some sort of symmetry, since the ellipsoid is centered on the origin with its axes parallel to the coordinate axes and the plane passes through the origin and is symmetrical about the plane $ \ x \ = \ y \ $ . (We should not expect the coordinate values to be too "pretty", since the proportionality of the ellipsoid's axes are 2 : √5 : 5 .)
Here is a graph of the geometrical arrangement, showing one minimal-distance and one maximal-distance point; the partners of each pair are not visible.
|
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|
Probability for a word to start with $\text{2,0,0,4}$ Let $A= \left\{ 2,2,4,4,0,0,0,0\right\}$. We arrange those $8$ numbers randomly. What is the probability to get a sequence starting with $2,0,0,4$?
The answer is:$$\frac{2\cdot 4 \cdot 3 \cdot 2 \cdot 4!}{8!}$$
The denominator is simple, there are $8!$ permutations. Can you explain the numerator?
Thanks.
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The letters that are left to be distributed are {2, 4, 0, 0}. These can be arranged in $A=\binom{4}{2}\times\binom{2}{1}=\frac{2\times4!}{2!\times2!}$ different ways.
The total amount of ways those 8 numbers can be arranged is not 8!, but $B=\binom{8}{4}\times\binom{4}{2}=\frac{8!\times4!}{4!\times4!\times2!\times2!}$. The chance is then $\frac{A}{B}=\frac{2\times4!\times4!}{8!}=\frac{2\times4\times3\times2\times4!}{8!}$
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Prove that $\int_{-\infty}^{+\infty} \frac{1}{x^4+x^2+1}dx = \frac{\pi}{\sqrt{3}}$ Good evening everyone,
how can I prove that
$$\int_{-\infty}^{+\infty} \frac{1}{x^4+x^2+1}dx = \frac{\pi}{\sqrt{3}}\;?$$
Well, I know that $\displaystyle\frac{1}{x^4+x^2+1} $ is an even function and the interval $(-\infty,+\infty)$ is symmetric about zero, so
$$\int_{-\infty}^{+\infty} \frac{1}{x^4+x^2+1}dx = 2\int_0^\infty \frac{1}{x^4+x^2+1}dx.$$
Then I use the partial fraction:
$$2\int_0^\infty \frac{1}{x^4+x^2+1}dx= 2\int_0^\infty \left( \frac{1-x}{2(x^2-x+1)} + \frac{x+1}{2(x^2+x+1)} \right)dx.$$
So that's all. What's next step?
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The next step is to write, for example, $$\frac{x+1}{x^2+x+1} = \frac{1}{2} \cdot \frac{2x+1}{x^2+x+1} + \frac{1}{2} \cdot \frac{1}{x^2+x+1}$$ from which we then have $$\int \frac{x+1}{x^2+x+1} \, dx = \frac{1}{2} \log\left|x^2+x+1\right| + \frac{1}{2} \int \frac{dx}{(x+1/2)^2+(\sqrt{3}/2)^2},$$ and the second integral is, after an appropriate substitution, expressible as an inverse tangent. A analogous approach applies to the other term in your original expression.
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|
testing for the convergence of a series I have to check if the following series converges. I thought that I have to use comparison test, but it did not lead me anywhere.
$$\sum_{n=2}^{\infty}\frac{ln^5(2n^7+13)+10sin(n)}{nln^6(n^{7/8}+2n^{1/2}-1)ln(ln(n+(-1)^n))}$$
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Note that for sufficiently large $n$
$$\frac{\ln^5{(2n^7+13)+10\sin(n)}}{\ln^5{(n^\frac{7}{8}}+2\sqrt{n}-1)} > \frac{\ln^5{(2n^7+13)-10}}{\ln^5{(n^\frac{7}{8}}+2\sqrt{n}-1)} > 1.$$
Also
$$ \lim_{n \to \infty}\frac{n^{\frac{7}{8}}+2\sqrt{n}-1}{n} = 0,$$
and for sufficiently large $n$ we have
$$ \ln(n^{\frac{7}{8}}+2\sqrt{n}-1) < \ln n.$$
Since, $ n+ (-1)^n \leqslant n+1$, we have $\ln \ln (n+ (-1)^n) \leqslant \ln \ln (n+1)$.
Thus, for sufficiently large $n$,
$$\frac{\ln^5{(2n^7+13)+10\sin(n)}}{n\cdot \ln^6{(n^\frac{7}{8}}+2\sqrt{n}-1)\cdot\ln{\ln{(n+(-1)^n}})} > \frac{1}{n \cdot \ln n \cdot \ln \ln (n+1)} \\ > \frac{1}{(n+1) \cdot \ln (n+1) \cdot \ln \ln (n+1)} .$$
The series with terms given by the RHS of the above inequality diverges by the integral test, since
$$\int_2^{\infty}\frac{dx}{(x+1) \cdot \ln (x+1) \cdot \ln \ln (x+1) }= \lim_{x \to \infty}[\ln \ln \ln(x+1)- \ln \ln \ln (3)] = \infty.$$
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How to proof that $\sum_{i=1}^{2^n} 1/i \ge 1+n/2$ I had troubles trying to prove that for every $n\ge1$
$$\sum_{i=1}^{2^n}\frac1i\ge 1+\frac n2$$
Can you give me a hint about the induction proof or show me in detail how can I prove it? I would appreaciate any help. Thanks!
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$$\sum_{i=1}^{2^{k+1}}\frac{1}{i} = \sum_{i=1}^{2^{k}}\frac{1}{i}+\sum_{i=2^k+1}^{2^{k+1}}\frac{1}{i} \\ = \sum_{i=1}^{2^{k}}\frac{1}{i}+\left(\frac{1}{2^k+1}+\frac{1}{2^k+1} \ldots +\frac{1}{2^{k+1}-1}+ \frac{1}{2^{k+1}}\right) \\ \geq \sum_{i=1}^{2^{k}}\frac{1}{i}+\left(\frac{1}{2^{k+1}}+\frac{1}{2^{k+1}} \ldots +\frac{1}{2^{k+1}}+ \frac{1}{2^{k+1}}\right)$$ The "greater than or equal to" is true because you are making the denominators larger. Next, you know you are adding up $2^k$ quantities of $\frac{1}{2^{k+1}}$ since $2^{k+1}-2^{k} = 2^k(2-1) = 2^k$. Hence $$\sum_{i=1}^{2^{k}}\frac{1}{i}+\left(\frac{1}{2^{k+1}}+\frac{1}{2^{k+1}} \ldots +\frac{1}{2^{k+1}}+ \frac{1}{2^{k+1}}\right) = \sum_{i=1}^{2^{k}}\frac{1}{i}+2^k\left(\frac{1}{2^{k+1}}\right) \\ = \sum_{i=1}^{2^{k}}\frac{1}{i}+\frac{1}{2}$$ Now apply the induction hypothesis and do some algebra to get your result!
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Is there a closed-form of $ \sum_{n=0}^{\infty }\frac{(-1)^n}{(2n+1)(2n+2)(2n+3)(2n+4)}$ Is there a closed-form of $$\sum_{n=0}^{\infty }\frac{(-1)^n}{(2n+1)(2n+2)(2n+3)(2n+4)}$$
Thanks for any help
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To give another approach:
The sum can be calculated using the observation that $$\int_0^1\left(1-x^{\frac{1}{n}}\right)^kdx=\frac{1}{\binom{n+k}{n}}$$
which is related to the beta function. Using this we get:
$$\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)(2n+2)(2n+3)(2n+4)}=\frac{1}{4!}\sum_{n=0}^{\infty}\frac{(-1)^n}{\binom{2n+4}{4}}=\frac{1}{4!}\sum_{n=0}^{\infty}(-1)^n\int_0^1\left(1-x^{\frac{1}{4}}\right)^{2n}dx=\frac{1}{4!}\int_0^1\sum_{n=0}^{\infty}(-1)^n\left(1-x^{\frac{1}{4}}\right)^{2n}dx=\frac{1}{4!}\int_0^1\frac{1}{1+\left(1-x^{\frac{1}{4}}\right)^2}dx=\frac{1}{4!}\int_0^1\frac{4x^3}{1+\left(1-x\right)^2}dx=\frac{1}{3!}\int_0^1\frac{(1-x)^3}{1+x^2}dx=\frac{1}{3!}\int_0^1\frac{1-3x+3x^2-x^3}{1+x^2}dx=\frac{1}{6}\left[\arctan(x)-\frac32\ln\left(1+x^2\right)+3x-3\arctan(x)-\frac{x^2}{2}+\frac{1}{2}\ln\left(1+x^2\right)\right]_0^1=\frac{1}{6}\left(-\frac{\pi}{2}-\ln(2)+\frac{5}{2}\right)$$
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|
Find joint likelihood function of observations $x_1, x_2, \ldots, x_n$ and $y_1, y_2, \ldots, y_m$
Let $x_1,\ldots,x_n$ be observations from a normal distribution with mean $0$ and s.d $s_1$.
Similarly let $y_1,\ldots,y_m$ be observations from a normal distribution with mean $0$ and s.d $s_2$.
Find the combined likelihood.
I've written down the joint likelihood of the $x$ observations multiplied by the joint likelihood of the $y$ observations. I don't think this is correct however. Many thanks
|
That is correct if $(x_1,\ldots,x_n)$ is independent of $(y_1,\ldots,y_n)$.
For the following to be valid, one must assume all $n+m$ observations are independent. One has
\begin{align}
L(s_1,s_2) & = \prod_{i=1}^n \left(\frac1{\sqrt{2\pi}\,s_1} \exp\left(\frac{-1}2 \left( \frac{x_i}{s_1} \right)^2 \right)\right) \prod_{j=1}^n \left(\frac1{\sqrt{2\pi}\,s_2} \exp\left(\frac{-1}2 \left( \frac{y_j}{s_2} \right)^2 \right)\right) \\[10pt]
& = \frac 1 {(2\pi)^{n+m} s_1^n s_2^m} \exp\left( \frac{-1}2 \left( \frac 1 {s_1^2} \sum_{i=1}^n x_i^2 + \frac 1 {s_2^2} \sum_{j=1}^m y_j^2 \right) \right),
\end{align}
so
$$
\ell(s_1,s_2) = -n\log s_1 - m \log s_2 -\frac 1 2\left( \frac A{s_1^2} + \frac B {s_2^2} \right) + \text{constant}
$$
where $A$ and $B$ are of course the sums of squares of the $x$s and $y$s.
PS in response to comments: Under the null hypothesis of equal variances, the log-likelihood is
$$
\ell(s) = -(n+m)\log s -\frac 1 2 \left( \frac{A+B}{s^2} \right).
$$
Then we have
$$
\ell'(s) = \frac{-(n+m)}s + \frac{A+B}{s^3} = \frac{A+B-s^2(n+m)}{s^3} \begin{cases} >0 & \text{if }s^2<\frac{A+B}{n+m}, \\[10pt]
= 0 & \text{if }s^2=\frac{A+B}{n+m}, \\[10pt]
<0 & \text{if }s^2>\frac{A+B}{n+m}. \end{cases}
$$
Therefore the MLE under the null hypothesis is $\widehat{s}= \sqrt{\frac{A+B}{n+m}}$. We have
$$
\ell(\widehat{s}) = -(n+m)\log\widehat{s} - \frac 1 2 \frac{A+B}{\widehat{s}^2} = -\frac{n+m}2 \log\frac{A+B}{n+m} - \frac 1 2 \frac{A+B}{\left(\frac{A+B}{n+m}\right)}. \tag 1
$$
Under the alternative hypothesis we have $\widehat{s}_1=\sqrt{A/n}$ and $\widehat{s}_2=\sqrt{B/m}$, and then
\begin{align}
\ell(\widehat{s}_1,\widehat{s}_2) & = -n\log\widehat{s}_1-m\log\widehat{s}_2 -\frac 1 2 \left( \frac{A}{\widehat{s}_1^2} + \frac{B}{\widehat{s}_2^2}\right) \\[10pt]
& = -\frac n 2 \log\frac A n - \frac m 2 \log \frac B m - \frac 1 2\left(\frac A {A/n} + \frac B {B/m} \right) . \tag 2
\end{align}
Subtracting $(2)$ from $(1)$ we get the logarithm of the likelihood ratio:
$$
-\frac{n+m} 2 \log\frac{A+B}{n+m} + \frac n 2 \log\frac A n + \frac m 2 \log \frac B m + \text{terms not depending on $A$ and $B$}.
$$
It is easy (if a bit tedious) to show that this is a monotone function of
$$
\frac{A/n}{B/m}
$$
and that has an F-distribution with $n$ and $m$ degrees of freedom. You reject if this is either too big or too small.
|
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|
Investigate convergence of the series Investigate convergence of the series:
$$\left( \frac{n^2+3n+10}{n^2+5n+17} \right)^{n^2 (\sqrt{n+1}-\sqrt{n-1})}$$
It should be solvable with simple manipulations with the formula, i guess, but how to do that?
|
Note that
$$
\sqrt{n+1}-\sqrt{n-1}=\frac2{\sqrt{n+1}+\sqrt{n-1}}\ge\frac1{\sqrt{n+1}}
$$
and by cross-multiplication, for $n\ge20$,
$$
\frac{n^2+3n+10}{n^2+5n+17}\le\frac{n}{n+2}
$$
Therefore, using Bernoulli's Inequality,
$$
\begin{align}
\left(\frac{n^2+3n+10}{n^2+5n+17}\right)^{n^2(\sqrt{n+1}-\sqrt{n-1})}
&\le\left(\frac1{1+\frac2n}\right)^{n^2/\sqrt{n+1}}\\
&\le\frac1{1+\frac{2n}{\sqrt{n+1}}}\\[4pt]
&\to0
\end{align}
$$
Upon reading the question again, I am not certain whether you were wanting to know whether the terms above converge or whether their sum converges. Using the inequality
$$
1+x\le e^x
$$
we get that
$$
\begin{align}
\left(\frac{n^2+3n+10}{n^2+5n+17}\right)^{n^2(\sqrt{n+1}-\sqrt{n-1})}
&\le\left(1-\frac2{n+2}\right)^{n^2/\sqrt{n+1}}\\
&\le e^{-\frac{2n^2}{(n+2)\sqrt{n+1}}}\\[9pt]
&\le e^{-2\sqrt{n-5}}
\end{align}
$$
since $\frac{2n^2}{(n+2)\sqrt{n+1}}\ge2\sqrt{n-5}$. Thus, this term decays faster than any power of $n$. Therefore, even the sum converges.
|
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|
Matrices, determinants, and applications to identities involving Fibonacci numbers Preamble
It is well known that since:
$$
\begin{pmatrix}
F_{n+1} \\
F_n \\
\end{pmatrix} =
\begin{pmatrix}
1 & 1 \\
1 & 0 \\
\end{pmatrix}
\begin{pmatrix}
F_n & F_{n-1} \\
\end{pmatrix}
$$
it is valid that:
$$
\begin{pmatrix}
1 & 1 \\
1 & 0 \\
\end{pmatrix}^n =
\begin{pmatrix}
F_{n+1} & F_n \\
F_n & F_{n-1} \\
\end{pmatrix}
$$
By calculating determinants, it immediately follows that:
$$F_{n-1}F_{n+1} - {F_n}^2
=\det\left[\begin{matrix}F_{n+1}&F_n\\F_n&F_{n-1}\end{matrix}\right]
=\det\left[\begin{matrix}1&1\\1&0\end{matrix}\right]^n
=\left(\det\left[\begin{matrix}1&1\\1&0\end{matrix}\right]\right)^n
=(-1)^n$$
or
$$F_{n-1}F_{n+1} - {F_n}^2=(-1)^n$$
This is known as Cassini's identity.
I find both Cassini's identity and this proof exceptionally attractive.
Now, there are other identities that resemble Cassini's identity:
$$F_{n-2}F_{n-1}F_{n+3} - {F_n}^3=(-1)^nF_{n-3}$$
$$F_{n+2}F_{n+1}F_{n-3} - {F_n}^3=(-1)^nF_{n+3}$$
$${F_{n-3}}{F_{n+1}}^2-{F_{n-2}}^2{F_{n+3}}=4 (-1)^n{F_{n}}$$
$${F_{n-1}}^2{F_{n+1}}^2-{F_{n-2}}^2{F_{n+2}}^2=4 (-1)^n{F_{n}}^2$$
$$F_{n-2}F_{n-1}F_{n+1}F_{n+2} - {F_n}^4=-1$$
Question
Is there a proof of five identities above that relies on matrices, the proof that would be attractive and similar to the mentioned proof
of Cassini's identity?
Also, is there any other Fibonacci identity that follows from a suitable correspondant matrix equation?
Trivia
Not directly related to any regular (meaning having the form of an equality) Fibonacci identity, I found also some fairly surprising matrix identities involving Fibonacci numbers, and among them the strangest is:
$$\pmatrix{
3&6&-3&-1\\
1&0&0&0\\
0&1&0&0\\
0&0&1&0
}^n\pmatrix{
2197&512&125&27\\
512&125&27&8\\
125&27&8&1\\
27&8&1&1
}=\pmatrix{
F_{n+7}^3&F_{n+6}^3&F_{n+5}^3&F_{n+4}^3\\
F_{n+6}^3&F_{n+5}^3&F_{n+4}^3&F_{n+3}^3\\
F_{n+5}^3&F_{n+4}^3&F_{n+3}^3&F_{n+2}^3\\
F_{n+4}^3&F_{n+3}^3&F_{n+2}^3&F_{n+1}^3\\
}
$$
Here, one more unusual fact: $8$, $27$, and $125$ are also cubes of a Fibonacci number.
|
The "strange matrix identity" at the bottom is just a linear recurrence among the cubed Fibonacci numbers:
$$F_{n+4}^3=3F_{n+3}^3+6F_{n+2}^3-3F_{n+1}^3-F_n^3$$
which looks somewhat less formidable than the matrix equation.
To derive this, we start with Binet's formula
$$ F_n = \frac{\phi^n-(-\phi^{-1})^n)}{\sqrt 5} $$
and cube it to get
$$F_n^3=\frac{(\phi^3)^n-3(-\phi)^n+3(\phi^{-1})^n-(-\phi^{-3})^n}{5\sqrt5}$$
The polynomial whose roots are the numbers raised to the $n$th power here is
$$(x−\phi^3)(x+\phi)(x−\phi^{-1})(x+\phi^{-3})=x^4−3x^3−6x^2+3x+1$$
which supplies the coefficients of the above recurrence.
In a similar way one can derive $$F^2_{n+3}=2F_{n+2}^2+2F_{n+1}^2-F_n^2$$ and its associated matrix form
$$ \begin{pmatrix}2 & 2& -1\\1&0&0\\0&1&0\end{pmatrix}^n
\begin{pmatrix}1\\1\\0\end{pmatrix} =
\begin{pmatrix}F_{n+2}^2\\F_{n+1}^2\\F_n^2\end{pmatrix} $$
See A055870 for the corresponding coefficients for higher powers of Fibonacci numbers.
|
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|
How to find $ \binom {1}{k} + \binom {2}{k} + \binom{3}{k} + ... + \binom{n}{k} $
Find $$ \binom {1}{k} + \binom{2}{k} + \binom{3}{k} + ... + \binom {n}{k} $$ if $0 \le k \le n$
Any method for solving this problem? I've not achieved anything so far. Thanks in advance!
|
For a polynom $P(x)=a_mx^m +\dots +a_1x+a_0$ denote $[x^k]P(x)=a_k$.
Notice that for $k\le i \le n$
$$\dbinom{i}{k}=[x^k](1+x)^i$$
hence
\begin{align}
\sum_{i=1}^n\dbinom{i}{k}&=\sum_{i=k}^n\dbinom{i}{k}\\
&=\sum_{i=1}^n[x^k](1+x)^i\\
&=[x^k]\sum_{i=k}^n(1+x)^i\\
&=[x^k]\frac{(1+x)^{n+1}-(1+x)^k}{(1+x)-1}\\
&=[x^k]\frac{(1+x)^{n+1}-(1+x)^k}{x}\\
&=[x^{k+1}]\left((1+x)^{n+1}-(1+x)^k\right)\\
&=[x^{k+1}](1+x)^{n+1}+[x^{k+1}](1+x)^{k}\\
&=\dbinom{n+1}{k+1}+0\\
&=\dbinom{n+1}{k+1}
\end{align}
|
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|
Prove that $\cos(2a) + \cos(2b) + \cos(2c) \geq -\frac{3}{2}$ for angles of a triangle Let the three internal angles of a triangle are $a,b,c$. Prove that
$$\cos(2a) + \cos(2b) + \cos(2c) \geq -\frac{3}{2}.$$
I'm looking for an elementary, geometric proof. So avoid derivatives and tools from optimalization theory if it's possible.
|
If you are interested in an algebraic proof, mine goes as follows:
$\cos{2A} + \cos{2B} = 2 \cos{(A+B)} \cos{(A-B)} = %
2 \cos{(\pi - C)} \cos{(A-B)} = \\-2 \cos{C} \cos{(A-B)}$
Hence,
\begin{eqnarray*}
& \cos{2A} + \cos{2B} + \cos{2C} &\geq& - \dfrac{3}{2} \\
\iff & -2 \cos{C} \cos{(A-B)} + \cos{2C} &\geq& - \dfrac{3}{2} \\
\iff & -2 \cos{C} \cos{(A-B)} + 2\cos^2{C} - 1 &\geq& - \dfrac{3}{2} \\
\iff & 4\cos^2{C} - 4\cos{(A-B)} \cos{C} + 1 &\geq& 0
\end{eqnarray*}
We prove the last inequality with a quadratic function.
Consider $f(t) = 4t^2 - 4 \cos{\phi} \; t + 1$, $t, \phi \in \mathbb{R}$. The descriminant is given by $D = 16(\cos^2{\phi} - 1) \leq 0$. Hence $f(t) \geq 0$ for all $t, \phi \in \mathbb{R}$. The result is now obvious if we let $t = \cos{C}$ and $\phi = A - B$.
Also, $f(t) = 0$ if and only if $\phi = n\pi$, $n \in \mathbb{Z}$, which is translated to $A = B$ for triangle $ABC$. In this case, $t = 1/2$, which means that $\cos{C} = 1/2 \implies C = \frac{\pi}{3}$. Thus, equality holds if and only if $A = B = C = \frac{\pi}{3}$. i.e. if and only if the triangle is equilateral.
|
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|
Is there a trick to prove that $(x^4+y^4)^{1/2} \leq x^2+y^2$? $(x^{4}+y^{4})^{1/2} \leq x^{2}+y^{2}$
I tried multiplying the original by 1 = $\displaystyle \frac{(x^{4}+y^{4})^{1/2}}{(x^{4}+y^{4})^{1/2}}$, but that just brings me back to the original.
Ive looked up sqrt and exponent laws and can't find anything helpful. What am I missing?
|
$$
\begin{align}
(x^2 + y^2)^2 &= x^4 + 2x^2y^2 + y^4\\
&= (x^4 + y^4) + 2x^2y^2
\end{align}
$$
Now compare $(x^4 + y^4)^{\frac{1}{2}}$ with $\left[(x^4 + y^4) + 2x^2y^2\right]^{\frac{1}{2}}$, knowing that for all $x, y$, $2x^2y^2\geq 0$.
|
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|
Pythagorean triples So I am given that $65 = 1^2 + 8^2 = 7^2 + 4^2$ , how can I use this observation to find two Pythagorean triangles with hypotenuse of 65.
I know that I need to find integers $a$ and $b$ such that $a^2 + b^2 = 65^2$, but I don't understand how to derive them from that observation.
Here is my attempt.
$65^2 = (8^2+1^2)(7^2+4^2) = 8^27^2 + 1^24^2 + 1^27^2 + 8^24^2 = (8\cdot7)^2 + 4^2 + 7^2 + (8\cdot4)^2$ but now I am stuck here, any suggestions!
|
Hint If $m,n$ are integers then $(m^2-n^2, 2mn, m^2+n^2)$ is a Pytagorean triple.
|
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|
Laurent series expansion, can one simplify this? I have to expand $f(z)=\frac{z-1}{(z^2+1)z}$ in an annulus $R(i,1,2)$.
$$f(z)=\frac{1}{z-i}\frac{1}{z+i}-\frac{1}{z-i}\Big(\frac{i}{z+i}-\frac{i}{z}\Big)$$
$$\frac{1}{z-i}\frac{1}{z+i}=\frac{1}{z-i}\cdot\frac{1}{2i}\cdot\frac{1}{1-(-(z-i)/2i)}=\sum_{n=0}^{\infty}(-1)^n (\frac{1}{2i})^{n+1}(z-i)^n$$
$$\frac{i}{z+i}=\frac{1}{2}\cdot \frac{1}{1-(-(z-i)/2i)}=\sum_{n=0}^{\infty}(-1)^n (\frac{1}{2i})^{n}\frac{1}{2}(z-i)^n$$
$$\frac{i}{z}=\frac{i}{z-i}\cdot\frac{1}{1-(-i/(z-i))}=\sum_{n=0}^{\infty}(-1)^n i^{n+1}(\frac{1}{z-i})^{n+1}$$
Finally we got:
$$f(z)=\sum_{n=0}^{\infty}(-1)^n (\frac{1}{2i})^{n+1}(z-i)^n-\sum_{n=0}^{\infty}(-1)^n (\frac{1}{2i})^{n+1}(z-i)^{n-1}+\sum_{n=0}^{\infty}(-1)^n i^{n+1}(\frac{1}{z-i})^{n+2}=\frac{1}{2i}\cdot\frac{1}{z-i}+\sum_{n=1}^{\infty}\Big( (-1)^{n-1}(\frac{1}{2i})^{n}-(-1)^n (\frac{1}{2i})^{n+1}\Big)(z-i)^{n-1}+\sum_{n=0}^{\infty}(-1)^n i^{n+1}(\frac{1}{z-i})^{n+2}$$
My questions are: is this expansion correct? And can one simplify it?
|
From partial fractions, we have that
$$
f(z) = \frac{-1}{z} + \frac{1}{2}\frac{1+i}{z+i} + \frac{1}{2}\frac{1-i}{z-i}
$$
For the Laurent series in annulus, I obtained:
\begin{align}
\frac{-1}{z} + \frac{1}{2}\frac{1+i}{z+i} + \frac{1}{2}\frac{1-i}{z-i}
&= \frac{-1}{z-i}\frac{1}{1+\bigl(\frac{i}{z-i}\bigr)} + \frac{1}{4i}\frac{1+i}{1+\bigl(\frac{z -i}{2i}\bigr)}+\frac{1}{2}\frac{1-i}{z-i}\\
&= \frac{1}{2}\frac{1-i}{z-i}+\sum_{n=0}^{\infty}\biggl[i^n\frac{(-1)^{n+1}}{(z-i)^{n+1}}+\frac{1+i}{4i}\Bigl(\frac{i}{2}\Bigr)^n(z-i)^n\biggr]
\end{align}
I checked the results on Mathematica as well and it checked out. However, Wolfram online wouldn't compute it, so I attached a screen shot of my results.
|
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|
How do I find the coefficient of $x^2$ in this polynomial, given its value at three points? We are given the following data about a polynomial $P(x)$ of unknown degree: $P(0) = 2$, $P(1) = −1$, $P(2) = 4$.
Determine the coefficient of $x^2$ in $P(x)$ if all the third-order differences are $1$.
|
Since all third order differences are $1$ we know that this polynomial is of degree $3$ and leading coëfficiënt $\frac{1}{6}$.
$P(0) = 2$, so we know $P(x) = \frac{1}{6} x^3 + b x^2 + cx + 2$.
Because $P(1) = -1$ and $P(2) = 4$ we know that :
$ \frac{1}{6} + b + c + 2 = -1$
$ \frac{8}{6} + 4b + 2c + 2 = 4$.
The solution is $b = \frac {7}{2}$ and $c= \frac {-20} {3}$.
|
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|
Prove that if $({x+\sqrt{x^2+1}})({y+\sqrt{y^2+1}})=1$ then $x+y=0$ Let
$$\left({x+\sqrt{x^2+1}}\right)\left({y+\sqrt{y^2+1}}\right)=1$$
Prove that $x+y=0$.
This is my solution:
Let
$$a=x+\sqrt{x^2+1}$$
and
$$b=y+\sqrt{y^2+1}$$
Then $x=\dfrac{a^2-1}{2a}$ and $y=\dfrac{b^2-1}{2b}$. Now $ab=1\implies b=\dfrac1a$. Then I replaced $x$ and $y$:
$$x+y=\dfrac{a^2-1}{2a}+\dfrac{b^2-1}{2b}=\dfrac{a^2-1}{2a}+\dfrac{\dfrac{1}{a^2}-1}{\dfrac{2}{a}}=0$$
This solution is absolutely different from solution in my book. Is my solution mathematically correct? Did I assumed something that may not be true?
|
Hint: Let $x=\sinh a$ and $y=\sinh b$. Then, using the fact that $\cosh^2u-\sinh^2u=1$ and $\sinh u+\cosh u=e^u$, we arrive at $e^{a+b}=1\iff a+b=0$, assuming a and b are reals. Since $\sinh u$, just like $\sin u$, is an odd function, the proof is complete.
|
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|
Real solutions of $x^n + y^n = (x+y)^n$ I have to find all real solutions of the following equation:
$x^n + y^n = (x+y)^n$
Clearly for $n = 1$, the equation holds for every $x,y$ real numbers.
If $n$ is greater or equal to $2$, we do binomial expansion on RHS and from $x^n + y^n - (x+y)^n = 0 $ it follows that the roots are either $x=0$ or $y=0$.
Am I missing out on something or these are the only solutions?
|
First, if $x=0$ or $y=0$ then the equation is satisfied.
Second, if $x>0$ and $y>0$ then because by binomial expansion
$$ (x+y)^n = x^n + y^n + \text{strictly positive terms}$$
the equation can never be satisfied. On the other hand, if $x<0$ and $y<0$ then the equation is equivalent to $(-x)^n+(-y)^n=(-x-y)^n$ where all terms are positive so this has no solutions by the previous case.
So we are left with the case where, without loss of generality $x>0>y$, or if we use $-y$ instead of $y$ as a variable, the case
$$ x^n + (-y)^n = (x-y)^n, \text{ where } x,y>0.$$
If $n$ is even then all terms are positive and clearly $x^n + (-y)^n = x^n+y^n > (x-y)^n = |x-y|^n$, because either $x>|x-y|$ or $y>|x-y|$. If $n$ is odd then, depending on the sign of $x-y$ the equation becomes
$$ x^n + (y-x)^n = y^n \text{ or } y^n + (x-y)^n = x^n,$$
with all three terms positive. This is itself of the form $X^n+Y^n = (X+Y)^n$ so the only solutions are given by $y=x$ (which means $x=-y$ in the original equation), $x=0$ and $y=0$.
|
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|
A problem of inequality Let $a_1, a_2, a_3$; $b_1, b_2, b_3$; $c_1, c_2, c_3$; $d_1, d_2, d_3$ be all real numbers.
We need to show that
$$\begin{align}(a_1b_1c_1d_1 + a_2b_2c_2d_2 &+ a_3b_3c_3d_3)^4\\ &\leq (a_1^4+a_2^4+a_3^4)(b_1^4+b_2^4+b_3^4)(c_1^4+c_2^4+c_3^4)(d_1^4+d_2^4+d_3^4)\end{align}$$
I have used Cauchy-Schwarz inequality on $(a_1,a_2,a_3)$ and $(A_1,A_2,A_3)$ where $A_i=b_ic_id_i$ and get the following
$$(a_1A_1+a_2A_2+a_3A_3)< (a_1^2+a_3^2+a_3^2)^{1/2}(A_1^2+A_3^2+A_3^2)^{1/2}$$ Next my aim was to assume $B_i=c_id_i$ and go forward. But this does not help and fails therefore.
|
Your inequality is an application of the Hölder's Inequality.
More specifically,
$$(a_1^4 + a_2^4 + a_3^4)^\frac{1}{4}\cdots(d_1^4 + d_2^4 + d_3^4)^\frac{1}{4} \ge \left[ a_1^{\left(4 \cdot \frac{1}{4}\right)}b_1^{\left(4 \cdot \frac{1}{4}\right)}c_1^{\left(4 \cdot \frac{1}{4}\right)}d_1^{\left(4 \cdot \frac{1}{4}\right)} + \cdots + a_4^{\left(4 \cdot \frac{1}{4}\right)}b_4^{\left(4 \cdot \frac{1}{4}\right)}c_4^{\left(4 \cdot \frac{1}{4}\right)}d_4^{\left(4 \cdot \frac{1}{4}\right)}\right]$$
$$\iff (a_1^4 + a_2^4 + a_3^4)\cdots(d_1^4 + d_2^4 + d_3^4) \ge (a_1b_1c_1d_1 + \cdots + a_4b_4c_4d_4)^4$$
|
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|
Eigenvalues of operator on $S_n$'s group algebra Take the group algebra of the symmetric group $S_n$ (or equivalently consider $S_n$'s regular representation) - I guess over $\mathbb{C}$.
If $e_{i,j} \in S_n$ denotes the element which swaps only the pair $i$ and $j$, consider the linear operator
$$
O = (e_{1,2} + e_{2, 3} + \cdots + e_{n-1, n} + e_{n,1})
$$
which acts on the group algebra.
The question is, for generic $n$, what are the eigenvalues of this operator?
For example, in the easy case of $S_2$, the operator $O$ is $e_{1,2} + e_{2,1} = 2e_{1, 2}$, the group algebra is two-dimensional, and $O$ has eigenvectors $1_{S_2} + e_{1, 2}$ and $1_{S_2} - e_{1, 2}$ with eigenvalues $2, -2$ respectively.
|
For $n=4$, the eigenvalues are $4, -4, 0, 0, 0, 0, 0, 0, 2, 2, 2, 2, 2, 2, 2, 2, -2, -2, -2, -2, -2, -2, -2, -2$.
For $n=5$, the eigenvalues are the roots of
$(x - 5) \cdot (x + 5) \cdot (x - 1)^5 \cdot (x + 1)^5 \cdot x^{24} \cdot (x^2 - 5)^6 \cdot (x^2 - 5\cdot x + 5)^8 \cdot (x^2 + 5\cdot x + 5)^8 \cdot (x^2 - 2\cdot x - 4)^{10} \cdot (x^2 + 2\cdot x - 4)^{10}$
(courtesy of Sage cell server), so don't get your hopes up for rational spectra. These are not the Young-Jucys-Murphy elements.
For the sake of reference, here is my crappy Sage code:
n = 5
QSn = SymmetricGroupAlgebra(QQ, n)
sn = list(Permutations(n))
eln = QSn(Permutation([n]+range(2, n)+[1]))
for i in range(1, n):
eln += QSn(Permutation(range(1, i) + [i+1, i] + range(i+2, n+1)))
mx = []
for p in sn:
elf = eln * QSn(p)
mx.append([elf.coefficient(q) for q in sn])
m = matrix(QQ, mx)
print m.characteristic_polynomial().factor()
For $n=6$, degree-$3$ factors appear, so don't hope for anything semi-nice either.
|
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|
Solve the recurrence $T(n) = 2T(n-1) + n, T(1)=1, n\geq 2$ This question has been already solved here, I just want to figure out why I'm not being able to solve it using my method. Here's what I did -
$T(n)=2T(n-1)+n$
$T(n-1)=2T(n-2)+(n-1)$
$\therefore \,T(n)=2\{2T(n-2)+(n-1)\}+n$
$T(n)=4T(n-2)+3n-2$
$T(n-2)=2T(n-3)+(n-2)$
$\therefore T(n)=4\{2T(n-3)+(n-2)\}+3n-2$
$T(n)=8T(n-3)+7n-10$
$
\\.
\\.
\\.
\\T(n)=2^kT(n-k)+(2^k-1)n-c
\\\therefore T(n)=2^{n-1}T(1)+(2^{n-1}-1)n-c
\\T(n)=2^{n-1}+(2^{n-1}-1)n-c
$
This is where I'm stuck. I've seen other methods at the link above but I want to keep it simple by using induction. How should I proceed?
The answer is supposed to be : $2^{n+1}-n-2$
|
look for a particular solution of the form $T(n) = an + b$ where $a,b$ need to be determined. we need $an + b = 2(an - a + b) + n$ equating the coefficient of $n$ gives you $a = -2$ and then $b = 2b -a$ so $b = a + -2$ and $T(n) = -n-2$ is a particular solutions. the solution to the homogeneous equation is $T(n) = C2^{n}$ and the general solution is $$T(n) = C2^n - n - 2 $$ fix the $C$ to satisfy $T(1) = 1$
$\bf edit: $
we will need $1*2 + 2*2^2 + 3*2^3 + \cdots + (n-2)*2^{n-2} = (n-3)2^{n-1}+2$
here is your method taking from where you sopped
$\begin{align}T(n) &= 2^{n-1}T(1) + (2^{n-1} - 1)n -\left(1*2 + 2*2^2 + 3*2^3 + \cdots+ (n-2)2^{n-2} \right)\\
&=2^{n-1} + (2^{n-1} - 1)n -\left( (n-3)2^{n-1}+2 \right)\\
&=2^{n+1}-n -2
\end{align}$
|
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|
Prove that the equation $a^2+b^2=c^2+3$ has infinitely many integer solutions $(a,b,c)$. Prove that the equation $a^2+b^2=c^2+3$ has infinitely many integer solutions $(a,b,c)$.
My attempt:
$(a+1)(a-1)+(b+1)(b-1)=c^2+1$
This form didn't help so I thought of $\mod 3$, but that didn't help either. Please help. Thank you.
|
In order that an integer $n$ can be represented as $c^2-b^2$, it suffices that $n\not\equiv{2}\pmod{4}$. In such a case $n$ can be written as the product of two divisors with the same parity, $n=pq$, and we can take $c=\frac{p+q}{2},d=\frac{p-q}{2}$. So it is sufficient to prove that for an infinite number of integers $a$, $a^2-3\not\equiv 2\pmod{4}$. This happens every time $a$ is even.
For instance, take $a=10$. We can write $n=a^2-3=97$ as $1\cdot 97$, hence
$$ (a,b,c) = (10,48,49) $$
is a solution of $a^2-3=c^2-b^2$.
|
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What values of $a$ and $b$ does this system have infinitely many solutions? As a disclosure, this question is more for me to confirm that I did my work correctly. More specifically, the "solution" provided to me claims there are two values of $a$ and $b$ that yield infinite solutions, but I found only one.
That said, this is the problem:
For what values of $a$ and $b$ does this system have infinitely many solutions
\begin{equation} A =
\begin{pmatrix}
a & 0 & b & |2 \\
a & 2 & a & |b \\
b & 2 & a & |a \\
\end{pmatrix}
\end{equation}
I reduced the matrix to
\begin{equation} A =
\begin{pmatrix}
a & 0 & b & |2 \\
0 & 2 & a-b & |b-2 \\
0 & 0 & b-\frac{b^{2}}{a} & |2-b+a-2\frac{b^{2}}{a} \\
\end{pmatrix}
\end{equation}
For this system to have infinite solutions, then one row must be $0$. In order for that to happen, I determined that $b=0$, $a=-2$.
However, the solution I have claims $b=0$, $a=-2$, or $a = b \neq 0$. I don't see how the latter is possible based on my reduction. Either I did something wrong, or the solution is wrong. Any insight would be appreciated.
|
You can check easily if the solution is wrong. (It is not.) Put $a=b$. Then you have:
\begin{equation} A =
\begin{pmatrix}
a & 0 & a & |2 \\
a & 2 & a & |a \\
a & 2 & a & |a \\
\end{pmatrix}
\end{equation}
Reducing to (III - I) and (II - I)
\begin{equation} A =
\begin{pmatrix}
a & 0 & a & |2 \\
0 & 2 & 0 & |a-2 \\
0 & 0 & 0 & |0 \\
\end{pmatrix}
\end{equation}
For $a\neq 0$ this has infinitely many solutions. (For $a=0$ it has none.)
Thus there is a problem in your computation.
|
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|
How to find $\int_0^{1/4}\frac{1}{x\sqrt{1-4x}}\ln\left({\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}}}\right)dx$ Let $H_n$ be the harmonic series. I want to find the value of $A=\displaystyle\sum_{n=0}^\infty \binom{2n}{n}\left(\frac{1}{4}\right)^n\frac{H_n}{n} $.
From this paper : https://cs.uwaterloo.ca/journals/JIS/VOL15/Boyadzhiev/boyadzhiev6.pdf
, I found that $$\sum_{n=0}^\infty \binom{2n}{n}H_n x^n=\frac{1}{\sqrt{1-4x}}\ln\left({\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}}}\right)=f(x)$$
So I get $A=\displaystyle\int_0^{1/4}\frac{f(x)}{x}dx$.
How to find the exact value of
$$\int_0^{1/4}\frac{1}{x\sqrt{1-4x}}\ln\left({\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}}}\right)dx$$.
Thank in advances.
|
Set $x=1$ in
$$
\sum_{n=1}^\infty \frac{{2n\choose n}}{4^n}\frac{H_n}{n} x^n=2\operatorname{Li}_2\left(\frac{1-\sqrt{1-x}}{1+\sqrt{1-x}}\right)$$
we get
$$\sum_{n=1}^\infty \frac{{2n\choose n}}{4^n}\frac{H_n}{n}=2\operatorname{Li}_2(1)=2\zeta(2)$$
For a different result, pick $x=-1$,
$$\sum_{n=1}^\infty (-1)^n\frac{{2n\choose n}}{4^n}\frac{H_n}{n}=2\operatorname{Li}_2\left(\frac{1-\sqrt{2}}{1+\sqrt{2}}\right)=2\operatorname{Li}_2(2\sqrt{2}-3).$$
|
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|
Surface Area by Integration $$2\pi\int_{3}^6\left(\frac{1}{3}x^\frac{3}{2}-x^\frac{1}{2}\right)\left(1+\left(\frac{1}{2}x^\frac{1}{2}-\frac{1}{2}x^\frac{-1}{2}\right)^2\right)^\frac{1}{2}dx$$
I've managed to simplify this down to the equation below (not sure if it'll help), but I still can't integrate it.
Please help.
$$\frac{\pi}{3}\int_{3}^6(x^4-4x^3-2x^2+12x+9)^\frac{1}{2}dx$$
PS. The answer you should get is 9$\pi$.
|
You did a good bit of the hard work yourself, but these other computations may help:
\begin{align}
\frac{\pi}{3}\int_3^6\sqrt{x^4-4x^3-2x^2+12x+9}\;dx &= \frac{\pi}{3}\int_3^6 (x^2-2x-3)\;dx\tag{simplify}\\[1em]
&= \frac{\pi}{3}\int_3^6 x^2\;dx - \frac{2\pi}{3}\int_3^6x\;dx -\pi\int_3^61\;dx\\[1em]
&= \left.\frac{\pi x^3}{9}\right|_3^6-\frac{2\pi}{3}\int_3^6x\;dx-\pi\int_3^61\;dx\tag{FTC}\\[1em]
&= 21\pi-\frac{2\pi}{3}\int_3^6x\;dx-\pi\int_3^61\;dx\\[1em]
&= 21\pi+\left.\left(-\frac{\pi x^2}{3}\right)\right|_3^6-\pi\int_3^61\;dx\tag{FTC}\\[1em]
&= 12\pi-\pi\int_3^61\;dx\\[1em]
&= 12\pi+\left.(-\pi x)\right|_3^6\\[1em]
&= 9\pi.
\end{align}
Hence, you can see the integral evaluates to $9\pi$, as desired.
|
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|
Solving $\cos x + 3^{1/2} \sin x = 1$ for $0\leq x \leq 360^{\circ}$ $\cos x + \sqrt3 \sin x = 1$
Not sure what the step would be to get an answer.
|
If we square both sides of the equation:
$$
\Big(\cos(x) + \sqrt{3}\sin(x)\Big)^2 = 1^2 \\
\cos^2(x) + 3\sin^2(x) + 2\sqrt{3}\sin(x)\cos(x) = 1 \\
\cos^2(x) + \sin^2(x) + 2\sin^2(x) + 2\sqrt{3}\sin(x)\cos(x) = 1 \\
1 + 2\sin^2(x) + 2\sqrt{3}\sin(x)\cos(x) = 1 \\
2\sin^2(x) + 2\sqrt{3}\sin(x)\cos(x) = 0 \\
\sin(x)\Big(\sin(x) + \sqrt{3}\cos(x)\Big) = 0 \\
$$
So either $\sin(x) = 0$ or:
$$
\sin(x) + \sqrt{3}\cos(x) = 0 \\
\sin(x) = -\sqrt{3}\cos(x) \\
\tan(x) = -\sqrt{3} \\
$$
Then for what angles $x$ does $\sin(x) = 0$ or $\tan(x) = -\sqrt{3}$?
|
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|
A level Integration: $\int\frac{x^3}{\sqrt{x^2-1}}dx$ Using the substitution $x=\cosh (t)$ or otherwise, find
$$\int\frac{x^3}{\sqrt{x^2-1}}dx$$
The correct answer is apparently
$$\frac{1}{3}\sqrt{x^2-1}(x^2+2)$$
I seem to have gone very wrong somewhere; my answer is way off, can someone explain how to get this answer to me.
Thanks.
My working:
$$\int\frac{\cosh^3t}{\sinh^2t}dt$$
$$u=\sinh t$$
$$\int\frac{1+u^2}{u^2}du$$
$$\frac{-1}{u}+u$$
$$\frac{-1}{\sinh t}+\sinh t$$
$$\frac{-1}{\sqrt{x^2-1}}+\sqrt{x^2-1}$$
^my working, I'm pretty sure this is very wrong though.
Edit: I've spotted my error. On the first line it should be
$$\int \cosh^3t \, dt$$
not
$$\int\frac{\cosh^3t}{\sinh^2t}dt$$
|
Let $\displaystyle u=x^2, dv=\frac{x}{\sqrt{x^2-1}}dx$ $\;\;$and $\;\;du=2xdx, v=\sqrt{x^2-1}$ to get
$\displaystyle\int\frac{x^3}{\sqrt{x^2-1}}\;dx=x^2\sqrt{x^2-1}-\int2x\sqrt{x^2-1}\;dx=x^2\sqrt{x^2-1}-\frac{2}{3}(x^2-1)^{3/2}+C$
Alternatively, let $u=\sqrt{x^2-1}, \;\;du\displaystyle=\frac{x}{\sqrt{x^2-1}}dx, \;\;x^2=u^2+1$ to get
$\displaystyle\int\left(u^2+1\right)du=\frac{1}{3}u^3+u+C=\frac{1}{3}(x^2-1)^{3/2}+\sqrt{x^2-1}+C$
|
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|
Proof that the harmonic series diverges (without improper integrals)
Show that $$\sum\limits_{k=1}^n \frac{1}{k} \geq \log(n)$$ Use this to deduce that the series $$\sum\limits_{k=1}^\infty \frac{1}{k}$$ diverges.
Hint: Use the estimate $$\frac{1}{k} \geq \int_k^{k+1} \frac{1}{x}\,\mathrm{d}x$$
Note that all instances of "$\log$" mean the natural logarithm, the inverse of the exponential. After doing the integration, I've concluded that $\frac{1}{k} \geq \log(1+\frac{1}{k})>-\log(k)$
I suspect I'm supposed to use the hint to arrive at the first inequality, and then since $\log(n)$ can become arbitrarily large as $n$ increases, so must the harmonic series (essentially the comparison test).
I'm having trouble making this jump. Hints are appreciated, but please no solutions.
|
we can use the fact that $$\dfrac{1}{k-1} + \dfrac{1}{k}+\dfrac{1}{k+1} = \dfrac{1}{k}+\dfrac{2k}{k^2 - 1} > \dfrac{1}{k}+\dfrac{2}{k} = \dfrac{3}{k}\text{ for } k \ge 2$$ to show that $s=1 + \frac12 + \frac13 + \cdots$ cannot be finite, therefore the harmonic series must diverge.
$\begin{align}
s &= \dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3} +\dfrac{1}{4} + \dfrac{1}{5}+ \cdots \\
&=1+\left(\dfrac{1}{2} + \dfrac{1}{3}+\dfrac{1}{4} \right)+
\left(\dfrac{1}{5} + \dfrac{1}{6}+\dfrac{1}{7} \right) +
\left(\dfrac{1}{8} + \dfrac{1}{9}+\dfrac{1}{10} \right) +
\left(\dfrac{1}{11} + \dfrac{1}{12}+\dfrac{1}{13} \right) + \cdots \\
&> 1+3\left(\frac13 + \frac16+\frac19+\frac1{12} \cdots\right) =
1+1 +\frac12 + \frac13+\frac14 + \cdots\\
&=1 + s
\end{align}$
we have $ > s+1$, therefore $s$ cannot be a finite number.
|
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|
Number of paths in 3D coordinates A cute problem which is an extension of a well-known counting problem:
Find the number of paths of length $12$ from $(0,0,0)$ to $(4,4,4)$ passing through adjacent lattice points (for two ajacent points in the path, the distance between them is $1$), which avoid $(1,1,1), (2,2,2), (3,3,3)$.
|
We can use inclusion exclusion. let $f(S)$ be the number of paths that pass through all the elements of $S$. Then we want:
$\color{blue}{f(\emptyset)}-\color{red}{f((1,1,1))}-\color{green}{f((2,2,2))}-\color{purple}{f((3,3,3))}+\color{blue}{f((1,1,1),(2,2,2))}+\color{red}{f((1,1,1),(2,2,2))}+\color{green}{f((1,1,1),(3,3,3))}+\color{purple}{f((1,1,1),(3,3,3))}-\color{blue}{f((1,1,1),(2,2,2),(3,3,3)}$
Calculating each one we get:
$\color{blue}{\binom{12}{4,4,4}}-\color{red}{\binom{3}{1,1,1}\binom{9}{3,3,3}}-\color{green}{\binom{6}{2,2,2}\binom{6}{2,2,2}}-\color{purple}{\binom{9}{3,3,3}\binom{3}{1,1,1}}+\color{grey}{3\binom{3}{1,1,1}\binom{3}{1,1,1}\binom{6}{2,2,2}}-\color{blue}{\binom{3}{1,1,1}\binom{3}{1,1,1}\binom{3}{1,1,1}\binom{3}{1,1,1}}$
|
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|
Factorizing $(x+a)(x+b)(x+c)$ I was solving questions related to polynomial factorization. I have learnt the remainder and factor theorems, and some basic identities.
There was a question like this one:
$$p(x)=x^3+8x^2+19x+12$$
The method we have been taught is to first find the factors of the last number, here it is $12$, so the factors are $\pm1$, $\pm2$, $\pm3$, $\pm4$, $\pm6$, $\pm12$. Then we have to substitute each of these values for $x$ till $p(x)=0$. Then, using factor theorem, we get a factor. For example, if $p(2)=0$, $x-2$ is a factor. Then, we use long division to get the product of the other two factors. Finally, we split the middle term and factor that expression.
I noticed that all of the answers were of the form $(x+a)(x+b)(x+c)$, which is equal to $x^3+x^2(a+b+c)+x(ab+bc+ca)+abc$. So, for the given question, if you find three integers such that $a+b+c=8$ (coefficient of $x^2$) and $abc=12$, can the factorization be simply written as $(x+a)(x+b)(x+c)$? In this case, it is $1+3+4=8$ and $1\times3\times4=12$, so factorization is $(x+1)(x+3)(x+4)$
I tried explaining this to my teacher, but she said that we must use the prescribed method of the NCERT book only.
So, is my method right? (I think so, but I'm not 100% sure) Also, is it necessary to confirm that the coefficient of $x$ is $ab+bc+ca$, or will that always hold true?
|
Since $$P(-1)=(-1)^3+8(-1)^2+19(-1)+12=0$$
Therefore $x+1$ is one of the factors. After dividing $p(x)$ by $x+1$ we have
$$x^3+8x^2+19x+12=(x+1)(x^2+7x+12)$$
Since $$x^2+7x+12=(x+3)(x+4)$$ We are going to have
$$p(x)=x^3+8x^2+19x+12=(x+1)(x+3)(x+4)$$
|
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|
Finding Smallest x and y to Satisfy Equation Find the smallest natural numbers $x$ and $y$ such that $$7^2x=5^3y$$
I'm unsure how to proceed with this question. Could someone explain the process for determining the answer?
Added from the comments:
$x=5^3$ would satisfy $5^3$ divides $7^2x$
|
First consider left side. $5^3$ divides right side, so $5^3$ must also divides left side. Now $7^2$ and $5^3$ are relatively prime, so $5^3$ occurs in $x$, in other word:
$$x=5^3 \cdot x'$$
For some natural $x'$. $x'$ is natural, so $x' \geq 1$, so:
$$x=5^3 \cdot x' \geq 5^3$$
The same way we get:
$$y \geq 7^2$$
But if we put $x=5^3$ and $y=7^2$ the equation is satisfied, so $x,y$ are the smallest possible solutions.
|
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|
Let $a,b,c,d>0$ and $a+b+c+d=1$. Prove that $\frac{abc}{1+bc}+\frac{bcd}{1+cd}+\frac{cda}{1+ad}+\frac{dab}{1+ab}\le \frac{1}{17}$ Let $a,b,c,d>0$ and $a+b+c+d=1$. Prove that $\dfrac{abc}{1+bc}+\dfrac{bcd}{1+cd}+\dfrac{cda}{1+ad}+\dfrac{dab}{1+ab}\le \dfrac{1}{17}$
My attempt:
I figured out that if each of the element could be like $\dfrac{abc}{1+bc}\le \dfrac{1}{68}$ then we would be done.
From a little manipulation we get, $\dfrac{1}{bc}+1\le 68a$ , $\dfrac{1}{cd}+1\le 68b$ , $\dfrac{1}{da}+1\le 68c$ , $\dfrac{1}{ab}+1\le 68d$. Summing then we get,
$\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{cd}+\dfrac{1}{da}\le 64$
But, I suppose my first assumption could be wrong, if not then please help me further and if so then please help with the solution. Thank you.
|
Here is my approach. I am stuck at a place but I hope this is a good method and I don't reach a dead end. The LHS is equivalent to $\frac{a}{1/bc+1}+\frac{b}{1/cd+1}+\frac{c}{1/da+1}+\frac{d}{1/ab+1}$. From AM-GM, we have $1+1/bc>2/\sqrt{bc}$. Inverse the inequality with inverting the inequality sign and then multiply by $a$ and write the other 3 terms. Then add all the terms. Denote our original expression as P. We have $P<a\sqrt{bc}+b\sqrt{cd}+c\sqrt{da}+d\sqrt{ab}/2$. Which itself is less than $ab+bd+ab+ac+bc+bd+cd+ac/4$. Now I am working to prove this less than 1/17, hope this helps.
|
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|
Prove the limit of a sequence using definition Find the limit of the following sequence and justify it using the definition:
$$x_n = \frac{1}{n}sin\frac{nπ}{4}+\frac{n}{2n^2+cosn} $$
The definition:
X is the limit of the sequence $x_n$ if the following condition holds:
For each real number Ɛ > 0, there exists a natural number N such that, for every natural number n > N, we have $|x_n - x|$ < Ɛ.
I guessed the limit should be 0 and tried to evaluate $|x_n - 0|$ in many ways but I can't get it into a form where it's easy to see what n must be.
|
You are correct, the limit is zero. To prove it by definition, let $\varepsilon > 0$ and set $N > \frac{2}{\epsilon}$. If $n > N$, then
\begin{align}|x_n - 0| &= \left|\frac{1}{n}\sin \frac{n \pi}{4} + \frac{n}{2n^2 + \cos n}\right|\\
& \le \frac{1}{n}\left|\sin \frac{n\pi}{4}\right| + \left|\frac{n}{2n^2 + \cos n}\right|\\
& \le \frac{1}{n} + \frac{n}{2n^2 - 1} \\
& \le \frac{1}{n} + \frac{n}{n^2} \quad (\text{since $n^2 - 1 \ge 0$})\\
& = \frac{2}{n} < \frac{2}{N}\\
& < \epsilon
\end{align}
|
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"url": "https://math.stackexchange.com/questions/1138185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Calculate integral using beta and gamma functions I have to calculate the following integral using beta and gamma functions:
$$
\int\limits_0^1 \frac{x\,dx}{(2-x)\cdot \sqrt[3]{x^2(1-x)}}
$$
I came up with this terrible solution. Firstly, let's break it into two parts:
$$
\int\limits_0^1 \frac{(x-2)\,dx}{(2-x)\cdot \sqrt[3]{x^2(1-x)}} + \int\limits_0^1 \frac{2\,dx}{(2-x)\cdot \sqrt[3]{x^2(1-x)}}
$$
The first one is $-B\left(\frac 13,\frac 23\right)$. The second one can be simplified with substitution $x = 1 - \frac 1t$ to
$$
2\int\limits_{-\infty}^0 \frac{dt}{(t+1)(t-1)^\frac 23}
$$
But it's too unwieldy in my opinion. Furthermore, it's not so easy to evaluate $B\left(\frac 13,\frac 23\right)$. Is there any easier solution?
|
For the second integral, make the substitution $ \displaystyle x = \frac{u}{1+u}$.
Then
$$ \begin{align} 2 \int_{0}^{1} \frac{dx}{(2-x) \sqrt[3]{x^2(1-x)}} &= 2 \int_{0}^{\infty} \frac{du}{(2+u)u^{2/3}} \\ &= \int_{0}^{\infty} \frac{du}{\left(1+ \frac{u}{2}\right)u^{2/3}} \\ &= \frac{2}{2^{2/3}} \int_{0}^{\infty} \frac{1}{(1+t)t^{2/3}} \, dt \\ &= \frac{2}{2^{2/3}} \int_{0}^{\infty} \frac{t^{1/3-1}}{(1+t)^{1/3+2/3}} \, dt \\ &= \frac{2}{2^{2/3}} B \left(\frac{1}{3}, \frac{2}{3} \right) \tag{1} \\&= \frac{2}{2^{2/3}} \frac{\Gamma \left(\frac{1}{3} \right) \Gamma \left(1- \frac{1}{3} \right)}{\Gamma(1)} \\ &= \frac{2}{2^{2/3}} \frac{2 \pi}{\sqrt{3}} \tag{2} \\ &= \frac{2^{4/3} \pi}{\sqrt{3}}. \end{align}$$
$(1)$ https://en.wikipedia.org/wiki/Beta_function#Properties
$(2)$ https://en.wikipedia.org/wiki/Gamma_function#General
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Solution of equation $[1+\frac{x}{b}]e^{-x/b}=z$ Can we solve this equation
$$\left(1+\frac{x}{b}\right)e^{-x/b}=z$$
We have to determine value of $x$ in term of $z$.
Problem occur while calculating the following integral.
$$\frac{a}{b^{2}}\int_{0}^{\infty}(1-(1+\frac{x}{b})e^{-\frac{x}{b}})^{a-1}x^{r+1}e^{-\frac{x}{b}}dx $$
and subtituting
$$(1+\frac{x}{b})e^{-\frac{x}{b}}=z$$
|
Your original problem is
$f(a, b, r)
=\frac{a}{b^{2}}\int_{0}^{\infty}(1-(1+\frac{x}{b})e^{-\frac{x}{b}})^{a-1}x^{r+1}e^{-\frac{x}{b}}dx
$.
I'll play with it and see what I get.
Letting
$x/b = y$,
so
$dx = b dy$,
$f(a, b, r)
=\frac{a}{b}\int_{0}^{\infty}(1-(1+y)e^{-y})^{a-1}(by)^{r+1}e^{-y}dy
=ab^r\int_{0}^{\infty}(1-(1+y)e^{-y})^{a-1}y^{r+1}e^{-y}dy
$.
Since $e^y > 1+y$
for $y ? 0$,
$(1+y)e^{-y}
< 1$
so we could expand
$(1-(1+y)e^{-y})^{a-1}
$
with the binomial theorem.
This gives
$(1-(1+y)e^{-y})^{a-1}
=\sum_{n=0}^{\infty}
(-1)^n
\binom{a-1}{n}
((1+y)e^{-y})^n
$.
Integrating term-by-term,
$\begin{array}\\
f(a, b, r)
&=ab^r\int_{0}^{\infty}(1-(1+y)e^{-y})^{a-1}y^{r+1}e^{-y}dy\\
&=ab^r\int_{0}^{\infty}\sum_{n=0}^{\infty}
(-1)^n
\binom{a-1}{n}
((1+y)e^{-y})^ny^{r+1}e^{-y}dy\\
&=ab^r\sum_{n=0}^{\infty}(-1)^n
\binom{a-1}{n}
\int_{0}^{\infty}
((1+y)e^{-y})^ny^{r+1}e^{-y}dy\\
\end{array}
$
Looking at the inside integral,
$\begin{array}\\
\int_{0}^{\infty}
((1+y)e^{-y})^ny^{r+1}e^{-y}dy
&=\int_{0}^{\infty}
(1+y)^ne^{-ny}y^{r+1}e^{-y}dy\\
&=\int_{0}^{\infty}
\sum_{k=0}^n \binom{n}{k}y^ke^{-(n+1)y}y^{r+1}dy\\
&=\sum_{k=0}^n \binom{n}{k}\int_{0}^{\infty}
e^{-(n+1)y}y^{k+r+1}dy\\
&=\sum_{k=0}^n \binom{n}{k}\int_{0}^{\infty}
e^{-z}\left(\frac{z}{n+1}\right)^{k+r+1}\frac{dz}{n+1}\quad (z = (n+1)y)\\
&=\sum_{k=0}^n \binom{n}{k}\frac{1}{(n+1)^{k+r+2}}\int_{0}^{\infty}
e^{-z}z^{k+r+1}dz\\
&=\sum_{k=0}^n \binom{n}{k}\frac{(k+r+1)!}{(n+1)^{k+r+2}}\\
&=\frac1{(n+1)^{r+2}}\sum_{k=0}^n \binom{n}{k}\frac{(k+r+1)!}{(n+1)^{k}}\\
\end{array}
$
There might be something
that can be done if
this is inserted into the sum above,
but I'll leave it at this.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find matrix from Eigenvectors and Eigenvalues A matrix $A$ has eigenvectors
$v_1 = \left(
\begin{array}{c}
2 \\
1 \\
\end{array}
\right)$
$v_2 = \left(
\begin{array}{c}
1 \\
-1 \\
\end{array}
\right)$
with corresponding eigenvalues $\lambda_1$= 2 and $\lambda_2$= -3, respectively.
Determine Ab for the vector b = $
\left(
\begin{array}{c}
1 \\
1 \\
\end{array}
\right)$
I know how to find eigenvalues and eigenvectors from a given matrix A, but not this one,
the vector A is a 2x1 matrix, determinant does not exist here, so how to find the matrix A as stated in the question?
|
By definition of eigenvalue and eigenvector, we have
$$\tag{1}A\left(
\begin{array}{c}
2 \\
1 \\
\end{array}
\right)=2\left(
\begin{array}{c}
2 \\
1 \\
\end{array}
\right)\mbox{ and }A\left(
\begin{array}{c}
1 \\
-1 \\
\end{array}
\right)=-3\left(
\begin{array}{c}
1 \\
-1 \\
\end{array}
\right).$$
Now, since
$$\left(
\begin{array}{c}
1 \\
1 \\
\end{array}
\right)=\frac{2}{3}\left(
\begin{array}{c}
2 \\
1 \\
\end{array}
\right)-\frac{1}{3}\left(
\begin{array}{c}
1 \\
-1 \\
\end{array}
\right),$$
we have
$$A\left(
\begin{array}{c}
1 \\
1 \\
\end{array}
\right)=\frac{2}{3}A\left(
\begin{array}{c}
2 \\
1 \\
\end{array}
\right)-\frac{1}{3}A\left(
\begin{array}{c}
1 \\
-1 \\
\end{array}
\right)=....\mbox{(using $(1)$)}$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
What are all the twin primes $p$ and $q = p + 2$ for which $pq - 2$ is also prime? It seems that $p = 3$ and $q = 5$ ($pq - 2 = 13$) are the only solutions. However I'm having a difficult time proving this.
I have that all primes can be represented as $3k + 1$ or $3k + 2$ so if $p = 3k + 2$ then $q = 3k + 4$ and
\begin{align*}
pq - 2 &= (3k + 2)(3k + 4) - 2\\
&= 9k^2 + 18k + 8 - 2\\
&= 9k^2 + 18k + 6\\
&= 3(3k^2 + 6k + 2) \text{ which can't be prime}
\end{align*}
So that seems promising. But when I let $p = 3k + 1$ and $q = 3k + 3$ I get
\begin{align*}
pq -2 &= (3k + 1)(3k + 3) - 2\\
&= 9k^2 + 12k + 3 - 1\\
&= 9k^2 + 12k + 2,
\end{align*}
which could very well be prime. So what do I do? It appears, though, that the only solution is $p = 3$.
|
Hint $\,\ {\rm mod\ 3}\!:\ p\not\equiv \color{#c00}0\,\Rightarrow\, \color{#0a0}q\,(pq\!-\!2) = \overbrace{(p\,+\,2)}^{\large 0\ {\rm if}\ p\,\equiv\,\color{#c00}{\bf 1}\ }\,\overbrace{(p\,(p+2)-2)}^{\large\ 0\ {\rm if}\ p\,\equiv\,\color{#c00} 2}\equiv 0 $
hence we conclude $\ 3\nmid \color{#0a0}q\,\Rightarrow\,3\mid pq-2$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1142961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Why does $\int\frac{1}{2x+1}dx=\frac{1}{2}\ln|2x+1|+C$? The way I am thinking is as follows:
$$\int\frac{1}{2x+1}\,dx = \int\frac{1}{2}\frac{1}{x+\frac{1}{2}}\,dx = \frac{1}{2}\int\frac{1}{x+\frac{1}{2}}\,dx = \frac{1}{2}\ln\left|x+\frac{1}{2}\right|+C$$
However, the textbook answer is $\frac{1}{2}\ln|2x+1|+C$. Where is my thinking wrong?
|
Let us given
$$ \int{1\over2x+1}dx$$
Let us do the following substitution
$$ u = 2x+1 $$
So,
$$ du = 2dx $$
Finally,
$$ \int{1\over2x+1}dx = \frac12\int{1\over{u}}du = \frac12\ln\lvert{u}\rvert+C=\frac12\ln\lvert{2x+1}\rvert+C$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1148261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Finding the value of the infinite sum $1 -\frac{1}{4} + \frac{1}{7} - \frac{1}{10} + \frac{1}{13} - \frac{1}{16} + \frac{1}{19} + ... $ Can anyone help me to find what is the value of $1 -\frac{1}{4} + \frac{1}{7} - \frac{1}{10} + \frac{1}{13} - \frac{1}{16} + \frac{1}{19} + ... $ when it tends to infinity
The first i wanna find the pattern but it seems do not have any unique pattern can anyone help me?
|
Hint. Observe that the general term of your series is
$$\frac{(-1)^n}{3n+1}. $$
Then you may write
$$
\begin{align}
\sum_{n=0}^{\infty}\frac{(-1)^n}{3n+1}&=\sum_{n=0}^{\infty}(-1)^n\int_0^1 x^{3n} dx\\\\
&=\int_0^1 \sum_{n=0}^{\infty}(-1)^nx^{3n} dx\\\\
&=\int_0^1 \frac{1}{1+x^3} dx\\\\
&=\int_0^1 \left(\frac{1}{3 (1+x)}+\frac{2-x}{3 \left(1-x+x^2\right)}\right) dx\\\\
&=\frac{1}{3}\ln 2+\frac{\pi\sqrt{3}}{9}
\end{align}
$$
Some details.
$$
\begin{align}
\int_0^1 \frac{1}{1+x^3} dx&=\frac{1}{3}\int_0^1\!\frac{dx}{(1+x)}+\frac{1}{3}\int_0^1\!\frac{2-x}{(x-1/2)^2+3/4} dx\\\\
&=\frac{1}{3}\ln 2+\frac{1}{3}\int_{-1/2}^{1/2}\!\frac{3/2-u}{u^2+3/4} du \quad (u=x-1/2, \,dx=du)\\\\
&=\frac{1}{3}\ln 2+\frac{1}{3}\int_{-1/2}^{1/2}\!\frac{3/2}{u^2+3/4} du\\\\
&=\frac{1}{3}\ln 2+\int_{0}^{1/2}\!\frac{1}{u^2+3/4} du\\\\
&=\frac{1}{3}\ln 2+\left.\frac{2}{\sqrt{3}}\arctan \left( \frac{2}{\sqrt{3}}u\right)\right|_{0}^{1/2} \\\\
&=\frac{1}{3}\ln 2+\frac{\pi\sqrt{3}}{9}
\end{align}
$$ where we have used $\displaystyle \arctan \!\left(\! \frac{1}{\sqrt{3}}\!\right)\!=\frac{\pi}{6}$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1148862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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|
Area of triangle bounded by line and degenerate "crossed lines" conic The question is
Show that the two lines given by
$$(A^2 - 3B^2)x^2 + 8ABxy +(B^2 - 3A^2)y^2=0$$
and the line given by $$Ax+By+C=0$$ determine an equilateral triangle of area $$\frac{C^2}{\sqrt{3}\;(A^2+B^2)}$$
I tried factorizing the pair of straight lines into two straight lines. But that seemed to be not really simple. And I doubt I was proceeding rightly. It would be great if anyone helps.
|
$$( A^2 - 3 B^2 ) x^2 + 8 A B x y + ( B^2 - 3 A^2 ) y^2 = 0 $$
are a pair of straight lines through the the origin making angle $ \pi/3 $ or $2 \pi/3 $ with each other at the origin, just like two sides along diameter of circum-circle of a regular hexagon.
They are factorisable into two straight lines as it is easy to verify that for two cases in particular :
$$ A = 1, B=0 \rightarrow x^2 - 3 y^2 = 0 ; A = 0, B=1 \rightarrow y^2 - 3 x^2 = 0 $$
and, in general,
$$ (A^2-3 B^2) x^2 - 8 A B x y + ( B^2 - 3 A^2 ) y^2 = 0 $$ has slopes
$$ y = m_1 x , y = m_2 x ; \,( m_1, m_2) = \frac {4\, A B \pm \sqrt{3}(A^2 +B^2 )}{ (A^2 -B^2 )} $$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Summation inductional proof: $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2}<2$ Having the following inequality
$$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2}<2$$
To prove it for all natural numbers is it enough to show that:
$\frac{1}{(n+1)^2}-\frac{1}{n^2} <2$ or $\frac{1}{(n+1)^2}<2-\frac{1}{n^2} $
|
By using some telescopic sums, we can be even more accurate. We may notice that:
$$ \frac{1}{n^2}-\frac{1}{n(n-1)} = -\frac{1}{n^2(n-1)} $$
and if $n>1$:
$$ \frac{1}{n^2(n-1)}-\frac{1}{(n-1)n(n+1)}=\frac{1}{(n-1)n^2(n+1)} $$
so:
$$ \sum_{n=1}^{N}\frac{1}{n^2} = 1+\sum_{n=2}^{N}\frac{1}{n(n-1)}-\sum_{n=2}^{N}\frac{1}{(n-1)n(n+1)}-\sum_{n=2}^{N}\frac{1}{(n-1)n^2(n+1)}$$
or:
$$\begin{eqnarray*} \sum_{n=1}^{N}\frac{1}{n^2} &=& 2-\frac{1}{N}-\frac{N^2+N-2}{4N(N+1)}-\sum_{n=2}^{N}\frac{1}{(n-1)n^2(n+1)}\\&=&\frac{7}{4}-\frac{2N+1}{2N(1+N)}-\sum_{n=2}^{N}\frac{1}{(n-1)n^2(n+1)}\\&\leq&\color{red}{\frac{7}{4}}-\frac{1}{N+1}.\end{eqnarray*}$$
The telescopic approach (or the Euler's acceleration method) also leads to:
$$\forall N\geq 2,\qquad \sum_{n=1}^{N}\frac{3}{n^2\binom{2n}{n}}\leq \zeta(2) \leq \frac{1}{N^2\binom{2N}{N}}+\sum_{n=1}^{N}\frac{3}{n^2\binom{2n}{n}}$$
so, by just choosing $N=3$,
$$ \zeta(2)\leq\color{red}{\frac{593}{360}} $$
and the approximation is accurate up to two figures.
|
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"timestamp": "2023-03-29T00:00:00",
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|
The Cauchy product of $\sum_{n=0}^\infty{(-1)^{n+1}\over n+1}$ with itself. In Apostol's Mathematical Analysis there is the following exercise. Show that the Cauchy product of
$$\sum_{n=0}^\infty{(-1)^{n+1}\over n+1}$$
with itself is the series
$$2\sum_{n=1}^\infty{(-1)^{n+1}\over n+1}\left(1+\frac12+\cdots+\frac1n\right).$$
This doesn't seem to be true though. Writing out the terms,
$$\left(\sum_{n=0}^\infty{(-1)^{n+1}\over n+1}\right)\left(\sum_{n=0}^\infty{(-1)^{n+1}\over n+1}\right)=\frac11-\left(\frac11\frac12+\frac12\frac11\right)+\left(\frac11\frac13+\frac12\frac12+\frac13\frac11\right)+\cdots$$
and
$$2\sum_{n=1}^\infty{(-1)^{n+1}\over n+1}\sum_{k=1}^n\frac1k=2\left(-\frac12+\frac13\left(1+\frac12\right)-\frac14\left(1+\frac12+\frac13\right)+\cdots\right).$$
I can see that in the first sum, all terms of the form $\frac1n\frac1m$ with $m\ne n$ are doubled, but the terms $\frac1{n^2}$ don't seem to be accounted for. It would be nice if the sum of the missing terms was $0$, but in fact what we seems to have is
$$\begin{align}
\left(\sum_{n=0}^\infty{(-1)^{n+1}\over n+1}\right)\left(\sum_{n=0}^\infty{(-1)^{n+1}\over n+1}\right) & =2\sum_{n=1}^\infty{(-1)^{n+1}\over n+1}\sum_{k=1}^n\frac1k+\sum_{n=1}^\infty\frac1{n^2}\\
& =2\sum_{n=1}^\infty{(-1)^{n+1}\over n+1}\sum_{k=1}^n\frac1k+{\pi^2\over 6}.
\end{align}$$
Any help is appreciated.
|
The Cauchy product will be
$$\sum_{n = 0}^\infty \sum_{k = 0}^n \frac{(-1)^{k+1}}{k+1} \frac{(-1)^{n-k+1}}{n-k+1} = \sum_{n = 0}^\infty \sum_{k = 0}^n \frac{(-1)^n}{(k+1)(n-k+1)}.$$
Now
\begin{align}\sum_{k = 0}^n \frac{(-1)^n}{(k+1)(n-k+1)} &= (-1)^n \sum_{k = 0}^n \frac{1}{n+2}\left(\frac{1}{n-k+1} + \frac{1}{k+1}\right)\\
&= \frac{(-1)^n}{n+2}\left\{\sum_{k = 0}^n \frac{1}{n-k+1} + \sum_{k = 0}^n \frac{1}{k+1}\right\}\\
&= \frac{(-1)^n}{n+2}\left\{\sum_{j = 0}^n \frac{1}{j+1} + \sum_{k = 0}^n \frac{1}{k+1}\right\} \quad (j = n - k)\\
&= \frac{2(-1)^n}{n+2}\sum_{k = 0}^n \frac{1}{k+1}\\
\end{align}
So the product is reduced to
$$2\sum_{n = 0}^\infty \frac{(-1)^n}{n+2}\sum_{k = 0}^n \frac{1}{k+1} = 2\sum_{n = 1}^\infty \frac{(-1)^{n+1}}{n+1}\left(1 + \frac{1}{2} + \cdots + \frac{1}{n}\right).$$
|
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|
showing certain vertices form an equilateral triangle Suppose $a,b,c$ lie in the unit circle of the complex plane and satisfy $a+b+c = 0$. Then, $a,b,c$ form the vertices of an equilateral triangle.
Try
So, I want to show that $|b-a|=|c-a|=|b-c|$. We are given $|a| = |b| = |c| = 1$ and $a+b+c=0$. Since $|z| = z \overline{z} $, we have
$$ (b-a)( \overline{b}-\overline{a} ) = (c-a)( \overline{c} - \overline{a} )$$
which simplifies to
$$ |b| - b \overline{a} - a \overline{b} - |a| = |c| - c \overline{a} - a \overline{c} - |a| $$
Hence,
$$ c \overline{a} + a \overline{c} - b \overline{a} - a \overline{b} = 0$$
$$ a( \overline{c} - \overline{b} ) + \overline{a} ( c-b) = 0 $$
Here I am stuck. Perhaps this is the wrong approach?
|
Without loss of generality you can assume that $a=1$, because the centre of mass of the three points, the origin, is invariant under their rotation about it. This implies that $c = \overline b$, and that the real component of $b$ and $c$ is $-1/2$.
|
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|
How do I calculate: $\int \frac{dx}{3\sin^2 x+5\cos^2x}?$ How can I calculate this integral ?
$$\int \frac{dx}{3\sin^2 x+5\cos^2x}=\text{?}$$
Thank you! I've tried using universal substitution but the result was too complicated to be somehow integrated. Can you please give me a useful hint ?
|
At first substitute: $$\int \frac{dx}{3\sin^2\left(x\right)+5\cos^2\left(x\right)}dx = \int \frac{1}{3\sin^2\left(\arctan\left(u\right)\right) + 5\cos^2\left(\arctan\left(u\right)\right)}\frac{1}{1+u^2}du$$ where $x=\arctan\left(u\right)$ and $dx=\frac{1}{1+u^2}du$.
Then we can write this as $$\int \frac{1}{\left(u^2+1\right)\left(\frac{3u^2}{u^2+1}+\frac{5}{u^2+1}\right)}du=\int \frac{1}{3u^2+5}$$
Lets substitue again, where $u=\frac{\sqrt{5}}{\sqrt{3}}v$ and thus $du=\sqrt{\frac{5}{3}}dv$: $$\int \frac{1}{3\left(\frac{\sqrt{5}}{\sqrt{3}}v\right)^2+5}\sqrt{\frac{5}{3}}dv=\int \frac{1}{\sqrt{15}\left(v^2+1\right)}dv=\frac{1}{\sqrt{15}}\int \frac{1}{v^2+1}dv$$
As we know this relation, we can directly write $\frac{1}{\sqrt{15}}\arctan\left(v\right)$, then going the substitutions backwards: $$\frac{\arctan\left(\sqrt{\frac{3}{5}}\tan\left(x\right)\right)}{\sqrt{15}}$$
Thus: $$\int \frac{dx}{3\sin^2\left(x\right)+5\cos^2\left(x\right)}dx = \frac{\arctan\left(\sqrt{\frac{3}{5}}\tan\left(x\right)\right)}{\sqrt{15}} + C$$
|
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|
How to factorise $x^4 - 3x^3 + 2$, so as to compute the limit of a quotient? Question:
Find the limit: $$\lim_{x \to 1}\frac{x^4 - 3x^3 + 2}{x^3 -5x^2+3x+1}$$
The denominator can be simplified to: $$(x-1)(x^2+x)$$
However, I am unable to factor the numerator in a proper manner (so that $(x-1)$ will cancel out)
I know upon graphing that the limit is $5\over4$. What should I do here?
Note: To be done without the use of L'Hospital Rule
|
Using the Euclidean division we get
$$\begin{array}\\x^4-3x^3+2&\Bigg|&x-1\\
-(x^4-x^3)&\Bigg|&x^3\\
=-2x^3+2&\Bigg|&-2x^2\\
-(-2x^3+2x^2)\\
=-2x^2+2&\Bigg|&-2x\\-(-2x^2+2x)\\=-2x+2&\Bigg|&-2
\end{array}$$
so we find that
$$x^4-3x^3+2=(x-1)(x^3-2x^2-2x-2)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$y''+y'^{2}+y=0$ equation solution How would you solve this differential equation $y''+y'^{2}+y=0$? I can't apply the ansatz method (or more formally apply the characteristic polynomial method). Thanks
|
Consider $y=ax^2+bx+c,$ then we have $y'=2ax+b, (y')^2=4a^2x^2+4abx+b^2, y''=2a$ which leads to
$$c=-2a-b^2\\
-ax^2-bx=4a^2x^2+4abx\\
-ax-b=4a^2x+4ab\\
-ax-b=4a(ax+b)\\
a=-\frac 14\text{ or }ax+b=0$$
$b$ is a free variable when $a=-\frac 14$, leading to a set of solutions $y=-\frac 14x^2+bx+\frac 12-b^2$. When $ax+b=0$ we get the "set" of solutions $a=b=c=0$. Double checking with the original for $a=-\frac 14$, we get
$$y''+(y')^2+y=0\\
=-\frac 12+\frac 14x^2-bx+b^2-\frac 14x^2+bx+\frac 12-b^2$$
which reduces to $0$ as required.
|
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|
When is the number of $N$'s factors $1 + \sqrt{N}$? (Answer: Only $N = 4$ and $N = 16$.)
The following question arose in a course for pre-service and in-service elementary school teachers:
For what $N \in \mathbb{N}$ is it the case that the number of $N$'s factors is $1 + \sqrt{N}$?
The initial observations were that this holds for $N = 4$ and $N = 16$, with $1 + \sqrt{4} = 3$ and $1 + \sqrt{16} = 5$ factors, respectively. (The factors of $4: 1, 2, 4$; the factors of $16: 1, 2, 4, 8, 16$.)
Noting $\sqrt{N} \in \mathbb{N}$ requires all primes in the prime factorization of $N$ to be raised to even powers, I tried a few individual cases using the standard technique for enumerating the number of factors.
For example:
If we have a prime squared $N = p^2$, then the number of factors is $2+1 = 3$.
Moreover, $1 + \sqrt{N} = 1 + p$. We now seek a prime $p$ for which $1 + p = 3$; so $p=2$ is the unique solution. In a similar vein, consider $p^4$ has $4+1 = 5$ factors, and square root $p^2$; so we need a prime $p$ for which $1+p^2 = 5$. Again $p = 2$ works, so we recover the aforementioned cases $4$ and $16$ that were initially found by just exploring.
I checked a few more examples, but did not find any other $N$ that worked.
An example that did not work: Consider the product of distinct primes squared $N = p^2 q^2$. Then the number of factors is $(2+1)(2+1) = 9$, and $1 + \sqrt{N} = 1 + pq$. We now seek distinct primes $p$ and $q$ for which we have $1 + pq = 9$, i.e., $pq = 8$; but no such primes exist, since $8 = 2^3$. So there is no admissible $N$ with this multiplicative structure.
One can work out other, individual examples with reasonable celerity: $p^2 q^4$ has $15$ factors, and equating this to one plus its square root gives $pq^2 = 14$; again, no distinct primes $p$ and $q$ exist to satisfy this condition, since $14 = 2 \cdot 7$.
I imagine there is a reasonably short (and totally elementary - even if not in the sense of "elementary school") approach to this problem in generality. I expect that there are no solutions other than $4$ and $16$, but do not see a quick way to argue as much.
For the sake of clarity, here is the question stated again:
For what $N \in \mathbb{N}$ is it the case that the number of $N$'s factors is $1 + \sqrt{N}$?
|
Just to rephrase Robert Israel's answer into more palatable language for the level at which I am presently teaching:
There are two functions to think about here.
The first is the divisor function, $N \rightarrow d(N) := \#\{N$'s distinct divisors$\}$.
The second is the function $N \rightarrow 1 + \sqrt{N}$.
The issue with finding (necessarily square) integers $N$ for which $d(N)$ and $1+\sqrt{N}$ are equal is that the latter grows very quickly; in fact, we have equality iff $N \in \{2^2, 2^4\}$.
For example, consider $5^2$: Then $d(5^2) = 2+1 = 3$, and $1+\sqrt{5^2} = 6$, where $6 > 3$.
Similarly, consider $7^4$: Then $d(7^4) = 4+1 = 5$, and $1 + \sqrt{7^4} = 50$, where $50 > 5$.
So if we consider a number like $N = 5^2 \cdot 7^4$, then $d(N) = 3 \cdot 5 = 15$, but $1 + \sqrt{N} = 246$.
In this way, one comes to see that we will almost always have $d(N) < 1 + \sqrt{N}$ by a fair bit.
The three exceptions to this (noting again that we want $N$ to be a perfect square so that its square root is an integer) are $2^2, 2^4,$ and $3^2$. Even with $N = 3^4$ we find $d(N) = 5 < 10 = 1 + \sqrt{N}$.
So we examine $N$ equal to each of $2^2, 2^4, 3^2, 2^2 \cdot 3^2,$ and $2^4 \cdot 3^2$.
(RI eliminates $3^2$ using a separate parity argument.)
The cases of $2^2$ and $2^4$ satisfy the desired equality; the others do not.
In particular:
$d(2^2) = 3 = 1 + \sqrt{2^2}$ and $d(2^4) = 5 = 1 + \sqrt{2^4}$.
However:
$d(3^2) = 3 < 4 = 1 + \sqrt{3^2}$;
$d(2^2 \cdot 3^2) = 9 > 7 = 1 + \sqrt{2^2 \cdot 3^2}$; and
$d(2^4 \cdot 3^2) = 15 > 13 = 1 + \sqrt{2^4 \cdot 3^2}$.
(For completeness, we might note that $d(1) = 1 < 2 = 1 + \sqrt{1}$.)
From here: Any other choice of $N$ will yield $d(N) < 1 + \sqrt{N}$.
QED
|
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|
Proof $\frac{a_n^2+a_{n+1}^2+1}{a_{n}a_{n+1}} $ is constant I would appreciate if somebody could help me with the following problem:
Question: Defined by $a_{1} =1,a_{2}=2$ and $a_n a_{n+2}=a_{n+1}^2+1(n\geq 1)$
Proof. $\frac{a_n^2+a_{n+1}^2+1}{a_{n}a_{n+1}} $ is constant for $n\geq 1$
I tried (Mathematical Induction) but couldn’t get it that way.
|
The formula $$\frac{a_n^2+a^2_{n+1}+1}{a_na_{n+2}}=3$$ holds for $n=1,2$. Assume it is true for $k$ and show it is true for $k+1$. Here is another approach:
$$
\frac{a_{k+1}^2+a^2_{k+2}+1}{a_{k+1}a_{k+2}}\cdot\frac{a_{k}}{a_{k}}
=\frac{a_{k+1}^2+a^2_{k+2}+1}{a_{k}a_{k+2}}\cdot\frac{a_{k}}{a_{k+1}}
=\frac{a_{k+1}^2+a^2_{k+2}+1}{a_{k+1}^2+1}\cdot\frac{a_{k}}{a_{k+1}}
$$
$$
=\frac{a_k}{a_{k+1}}\left(\frac{a_{k+1}^2+1}{a_{k+1}^2+1}+\frac{a_{k+2}^2}{a_{k+1}^2+1}\right)
=\frac{a_k}{a_{k+1}}\left(1+\frac{a_{k+2}^2}{a_k a_{k+2}}\right)
=\frac{a_k+a_{k+2}}{a_{k+1}}.
$$
So now we need to show that is constant in $k$. See solution above by NikolajK.
|
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|
Explaining that $1 \cdot 3 \cdot 5 \dotsm (2n+1) = 1 \cdot 3 \cdot 5 \dotsm (2n-1)(2n+1)$ I have a few students that are having trouble understanding that
$$1 \cdot 3 \cdot 5 \dotsm (2n+1) = 1 \cdot 3 \cdot 5 \dotsm (2n-1)(2n+1),$$
specifically that
$$\frac{1 \cdot 3 \cdot 5 \dotsm (2n+1)}{1 \cdot 3 \cdot 5 \dotsm (2n-1)} = 2n+1.$$
I've tried explaining it a few different ways, but it didn't seem to go over very well. What would be a good way to explain this?
|
I would suggest including more terms at the end of the products. That makes it easier to see the pattern and to see why it's just $(2n+1)$ that's left at the end.
$$
\require{cancel}
\begin{align}
\frac{1 \cdot 3 \cdot 5 \dotsm (2n+1)}{1 \cdot 3 \cdot 5 \dotsm (2n-1)} &=
\frac{1 \cdot 3 \cdot 5 \dotsm (2n-5)(2n-3)(2n-1)(2n+1)}{1 \cdot 3 \cdot 5 \dotsm (2n-5)(2n-3)(2n-1)\quad\quad\quad} \\
&=\frac{\cancel{1 \cdot 3 \cdot 5} \dotsm \cancel{(2n-5)}\cancel{(2n-3)}\cancel{(2n-1)}(2n+1)}{\cancel{1 \cdot 3 \cdot 5} \dotsm \cancel{(2n-5)}\cancel{(2n-3)}\cancel{(2n-1)}\quad\quad\quad} \\
&=2n+1
\end{align}$$
It might also simply be a matter of spacing: students might get confused by seeing the $2n+1$ directly above the $2n-1$. I've tried to format the above to emphasize that the $2n+1$ is an extra factor that doesn't appear in the bottom.
Finally, it could also be helpful to pick a small $n$ and work out the product concretely.
|
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|
How to prove $x^3-y^3 = (x-y)(x^2+xy+y^2)$ without expand the right side? I can prove that $x^3-y^3 = (x-y)(x^2+xy+y^2)$ by expanding the right side.
*
*$x^3-y^3 = (x-y)x^2 + (x-y)(xy) + (x-y)y^2$
*$\implies x^3 - x^2y + x^2y -xy^2 + xy^2 - y^3$
*$\implies x^3 - y^3$
I was wondering what are other ways to prove that $x^3-y^3 = (x-y)(x^2+xy+y^2)$
|
You can rearrange it into two possible ways.
Show, by left side, that
$$\frac{x^3-y^3}{x-y} = x^2+xy+y^2,$$
or
$$\frac{x^3-y^3}{x^2+xy+y^2} = x-y.$$
You may read about "Long Division of Polynomials". See also LINK for knowing the process.
|
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|
Solve $16x^{-3}=-2$
Solve $16x^{-3}=-2$.
My working:
\begin{align}
16x^{-3}&=-2\\
\frac{1}{16x^{3}}&=-2\\
\frac{16x^3}{16x^3}&=-32x^3\\
1&=-32x^{3}\\
-32x^{3}&=1\\
-32x&=\sqrt[3]{1}\\
-32x&=1\\
x&=\frac{-1}{32}
\end{align}
Is this right? What have I done wrong?
|
Obviously you could check that $x=-\frac1{32}$ doesn't solve $16x^{-3} = -2$.
Instead you can solve as follows
$$\begin{align*}
16 x^{-3} & = -2 & \text{reciprocal } a=b \Leftrightarrow \frac1a = \frac1b \\
\Leftrightarrow \frac1{16} x^3 & = -\frac12 & \text{multiply by } 16\\
\Leftrightarrow x^3 & = -8 & \text{take cube root}\\
\Leftrightarrow x & = \sqrt[3]{-8} = -2
\end{align*}$$
|
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|
Critical numbers of the function: $x\sqrt{5-x}$ Let f(x) = $$\displaystyle f(x) = x\sqrt{5-x} $$
On the interval: [-6,4]
Critical numbers are the the values of x in the domain of f for which f'(x) = 0 or f'(x) is undefined.
Derivative of the function:
$$ \frac{1}{2} \cdot x (5-x)^{\frac{-1}{2}} \cdot -1$$
$$ \frac {\frac{-x}{2}}{\sqrt{5-x}}$$
$f'(x) = 0$, when $x = 0, $ and is undefined when x= 5
Plugging in the roots of the derivative function and the end points of the interval into the original function:
\begin{align*}
f(0) & = 0\\
f(5) & = 0\\
f(-6) & = -6\sqrt{11}\\
f(4) & = 4 \cdot 1 = 4
\end{align*}
So why is the 4 not the absolute maximum value?
p.s. I assumed the first term goes to zero when taking a derivative by the product rule. I confused d/dx x = 1, with any number d/dd = 0
|
Your derivative is wrong. Use the product rule to separate the $x$ from the $\sqrt{5-x}$.
$$\frac{d}{dx}\left(x\sqrt{5-x} \right)
=x\frac{d}{dx}\left(\sqrt{5-x} \right)+\sqrt{5-x}\frac{d}{dx}\left(x \right)$$
$$=x\frac{1}{2\sqrt{5-x}}(-1)+\sqrt{5-x}\cdot 1$$
$$=\frac{-x}{2\sqrt{5-x}}+\frac{2(5-x)}{2\sqrt{5-x}}$$
$$=\frac{10-3x}{2\sqrt{5-x}}$$
Setting this equal to zero, you get
$$x=\frac{10}3$$
This is undefined only at $x\ge 5$, which is not in your interval of interest.
You should be able to continue from here.
|
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|
chain rule derivative This is an derivatives Problem.
find the tangent to $y=\sqrt{x^2-x+5\;}\;$ at $x=5$.
$y = \boxed{\;?\;}$
What I did was first find y by plugin x into the equation. The answer is $y = 5$. Then I found the derivative of both $x^2-x+5$ and $\sqrt{x^2-x+5\;}\;$ the answers for those are $2x-1$ and $\tfrac 1 2 (x)^{-1/2}$.
From here I know I have to find the answers by plugin in $5$ to $\tfrac 1 2 (x^2-x+5)^{-1/2} \cdot (2(x)-1)$ to find $m$.
After $y= mx +b$ to find $b$.
and help anyone Im on my last chance and really need the help.
|
You're almost there. You have applied the chain rule to get the derivative. You just have to substitute at the point: $x:=5$.
$\begin{align}
y &= \sqrt{x^2-x+5}
\\[2ex]
{y}_x' & = {(x^2-x+5)}_x' \cdot {(u^{1/2})}_u'\mid_{u:=(x^2-x+5)}
\\[1ex] &= (2x-1) \cdot \tfrac 1 2 (x^2-x+5)^{-1/2}
\\[2ex]
{y}_x'|_{x:=5} & = (2\cdot 5 - 1)\cdot \tfrac 1 2(5^5 -5+5)^{-1/2}
\\[1ex] & = \frac 9 {10}
\end{align}$
Then since $y|_{x:=5} = \sqrt{5^2-5+5} = 5$ the tangent is $y_t -5 = \frac 9{10} (x_t-5)$
$$y_t = \frac 9 {10} x_t + \frac 1 2$$
|
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|
Finding the interval after substitution Given this problem
$$8\cdot 3^{\sqrt{x}+\sqrt[4]{x}}+9^{\sqrt[4]{x}+1}\geq 9^{\sqrt{x}}$$
$$8\cdot 3^{\sqrt{x}+\sqrt[4]{x}}+3^{2\sqrt[4]{x}+2}\geq 3^{2\sqrt{x}}\\8\cdot 3^{\sqrt{x}+\sqrt[4]{x}-2\sqrt{x}}+3^{2\sqrt[4]{x}+2-2\sqrt{x}}\geq 1\\8\cdot 3^{\sqrt[4]{x}-\sqrt{x}}+3^{2\sqrt[4]{x}-2\sqrt{x}+2}\geq 1\\8\cdot 3^{\sqrt[4]{x}-\sqrt{x}}+9\cdot 3^{2\sqrt[4]{x}-2\sqrt{x}}\geq 1$$
After simplifying I get $8\cdot 3^{\sqrt[4]{x}-\sqrt{x}}+9\cdot 3^{2\sqrt[4]{x}-2\sqrt{x}}\geq 1$ now putting $t=3^{\sqrt[4]{x}-\sqrt{x}}$ we get $8t+9t^2\geq 1$ and solving we get $t\in (-\infty,-1)\cup(\frac{1}{9},+\infty)$ since $t$ is exponential function with positive base then $t>0$ hence we're only looking at the interval $(\frac{1}{9},\infty)$.Now I have no idea how to find the interval for $x$,I've tried substituting $\sqrt[4]{x}-\sqrt{x}=-2$ and I get $x=16$ and substituting $\sqrt[4]{x}-\sqrt{x}\to \infty$ which is impossible so I have no idea what to do now.
|
you already have
$t=3^{\sqrt[4]{x}-\sqrt{x}} \ge \dfrac{1}{9} \iff \sqrt[4]{x}-\sqrt{x} \ge -2 $
$p= \sqrt[4]{x} \ge 0, \implies p-p^2\ge -2 \iff -1\le p \le2 \cap p\ge0 \implies 0 \le p \le 2 \implies 0\le x \le 16 $
|
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|
Equation solving; three expressions as squares If positive integer $x,y$ satisfy,
$$2x^2+x=3y^2+y$$ then $x-y,2(x+y)+1,3(x+y)+1$ are perfect squares.
I somehow managed to prove $x-y$ is a perfect square but couldn't prove the others.
|
$2x^2+x=3y^2+y\Longrightarrow (x-y)(3x+3y+1)=x^2$. Now observe if a prime $p$ divides both the factors, $x-y$ and $3x+3y+1$ on the LHS, then $p^2$ divides $x^2$ and so $p$ divides $x$.
Now, $p$ divides $x$ and $p$ divides $x-y$ implies $p$ divides $y$. But this is a contradiction as $p$ divides $3x+3y+1$. So we can conclude that the two factors on LHS are co prime. Now if two co prime numbers multiply to yield a perfect square, then each of them should be perfect squares.
Now observe again that the given equation is equivalent to $(x-y)(2x+2y+1)=y^2$. A similar reasoning follows.
|
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|
Given the $ x+y+z =3$ Prove that $ x^2+y^2+z^2 \geq 3xyz$ Given the $ x+y+z =3$ and x, y and z are positive numbers. How to prove that $ x^2+y^2+z^2 \geq 3xyz$.
I tried many methods but I failed.
I did the AM-HM in-equality, but failed.
$\frac{\frac{x}{yz} + \frac{y}{zx} + \frac{z}{xy}}{3} \geq \frac{3}{\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}} $.
The denominator of RHS of the equation must have, minimum value. So it boils down to
$\frac{x^2y^2+y^2z^2+z^2x^2}{xyz} \leq 3 $. Don't know how to proceed after that.
Taking partial derivative in range $(0,3)$ for the function $(x^2+y^2+z^2)$ also didn't give me the required insight.
Thankyou.
|
Using AM-GM and Cauchy-Schwarz inequalities, For $x,y,z > 0 \Rightarrow 3 = x+y+z \geq 3\sqrt[3]{xyz} \Rightarrow 1 = 1^3 \geq xyz \Rightarrow 3^2 = (x+y+z)^2 \leq 3(x^2+y^2+z^2) \Rightarrow 3xyz \leq 3 \leq x^2+y^2+z^2 \Rightarrow 3xyz \leq x^2+y^2+z^2$
|
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|
Calculating the limit of a sequence using Stirling's approximation I have the following limit:
$$
\lim_{n \to \infty} \frac{(2n)!\sqrt{n}}{n!^24^n}
$$
Now, in order to get somewhere further in calculating this limit, I used Stirling's approximation that led me to the following:
$$
\lim_{n \to \infty} \left[\frac{(2n)!}{(2n)^{2n}e^{-2n}\sqrt{4\pi n}} \times \frac{(2n)^{2n}ne^{-2n}\sqrt{4\pi}}{n!^24^n} \right]= \lim_{n \to \infty} \frac{(2n)^{2n}ne^{-2n}\sqrt{4\pi}}{n!^24^n}
$$
But am having trouble getting anywhere further on simplifying this. What is a direction I can take from this point on? Or should I have done a different approach perhaps? Any insights/hints are much welcome.
|
Take:
$$
lim_{n \to \infty} \frac{(2n)!\sqrt{n}}{n!^24^n} = lim_{n \to \infty} \frac{(2n)!}{n!^2} \cdot \frac{\sqrt{n}}{4^n}
$$
Recall Stirling's Approximation:
$$n! \sim \sqrt{2\pi n}\left(\frac n e\right)^n$$
Whereby:
$$(2n)! \sim \sqrt{2\pi 2n}\left(\frac {2n} e\right)^{2n} = \sqrt{4\pi n}\left(\frac {2n} e\right)^{2n} = \sqrt{4}\sqrt{\pi n}\left(\frac {2n} e\right)^{2n} = 2\sqrt{\pi n}\left(\frac {2n} e\right)^{2n}$$
And:
$$n!^2 \sim \left(\sqrt{2\pi n}\left(\frac n e\right)^n\right)^2 = \left(\sqrt{2\pi n}\right)^2 \cdot \left(\left(\frac n e\right)^n\right)^2
=2\pi n \cdot \left(\frac n e\right)^{2n}
$$
So:
$$
lim_{n \to \infty} \frac{(2n)!}{n!^2} \cdot \frac{\sqrt{n}}{4^n}
=
\frac{2\sqrt{\pi n}\left(\frac {2n} e\right)^{2n}}
{2\pi n \cdot \left(\frac n e\right)^{2n}}
\cdot
\frac{\sqrt{n}}{4^n}
=
\frac{\sqrt{\pi n}\left(\frac {2n} e\right)^{2n}}
{\pi n \cdot \left(\frac n e\right)^{2n}}
\cdot
\frac{\sqrt{n}}{4^n}
=\frac{\left(\frac {2n} e\right)^{2n}}
{\left(\frac n e\right)^{2n}}
\cdot
\frac{\sqrt{\pi n}\sqrt{n}}{\pi n 4^n}
$$
Continuing:
$$
\frac{\left(\frac {2n} e\right)^{2n}}
{\left(\frac n e\right)^{2n}}
\cdot
\frac{\sqrt{\pi n}\sqrt{n}}{\pi n 4^n}
= \frac{\left(\frac {2n} e\right)^{2n}}
{\left(\frac n e\right)^{2n}}
\cdot
\frac{\sqrt{\pi}\sqrt{n}\sqrt{n}}{\pi n 4^n}
=
\frac{\left(\frac {2n} e\right)^{2n}}
{\left(\frac n e\right)^{2n}}
\cdot
\frac{\sqrt{\pi}n}{\pi n 4^n}
=
\frac{\left(\frac {2n} e\right)^{2n}}
{\left(\frac n e\right)^{2n}}
\cdot
\frac{\sqrt{\pi}}{\pi 4^n}
$$
Continuing:
$$
\frac{\left(\frac {2n} e\right)^{2n}}
{\left(\frac n e\right)^{2n}}
\cdot
\frac{\sqrt{\pi}}{\pi 4^n}
= \left( \frac{\frac{2n}e}{\frac n e} \right)^{2n}
\cdot
\frac{\sqrt{\pi}}{\pi 4^n}
=\left( \frac{2n}e\cdot\frac e n \right)^{2n}
\cdot
\frac{\sqrt{\pi}}{\pi 4^n}
= 2^{2n}\cdot
\frac{\sqrt{\pi}}{\pi 4^n}
$$
Now we note that $4=2^2$, so $4^n= (2^2)^n = 2^{2n}$, whereby:
$$
2^{2n}\cdot
\frac{\sqrt{\pi}}{\pi 4^n}
=
2^{2n}\cdot
\frac{\sqrt{\pi}}{\pi 2^{2n}}
= \frac{\sqrt{\pi}2^{2n}}{\pi 2^{2n}}
= \frac {\sqrt \pi}{\pi}
= \pi^{1/2} \cdot \pi^{-1} = \pi^{-1/2} = \frac 1 {\sqrt{\pi}}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
The sum of the squares is less than or equal to the square of the sums for all n. I am trying to understand this proof. Rather an important part of the proof. I have already shown this is true for $n=2$ and am assuming the $a_n$ case is true.
$$(a_1^2+a_2^2+...+a_n^2) \le (a_1+a_2+...+a_n)^2$$
Want to show that
$$(a_1^2+a_2^2+...+a_n^2 + a_{n+1}^2) \le (a_1+a_2+...+a_n+a_{n+1})^2$$
$=$
$$(a_1^2+a_2^2+...a_n^2) + a_{n+1}^2 \le ((a_1+a_2+...a_n)+(a_{n+1}))^2$$
$=$
$$(a_1^2+a_2^2+...+a_n^2 + a_{n+1}^2) \le (a_1+a_2+...+a_n)^2+2(a_1+a_2+...+a_n)(a_{n+1})+(a_{n+1}) ^2$$ and here is the part I am not understanding. For some reason the proof moves some of the terms over and I cannot identify what is being replaced or why. My guess is that the terms that moves are the ${n+1}$ terms. But, I am not certain.
$$a_1^2+a_n^2+a_{n+1}^2...+2(a_1+a_2+...a_n)(a_{n+1}) \le (a_1+a_2+...a_n)^2$$
|
inductive step:
the claim being correct for
$$(a_1^2+a_2^2+...+a_n^2) \le (a_1+a_2+...+a_n)^2$$
implies
$$(a_1^2+a_2^2+...+a_n^2+a_{n+1}^2) \le (a_1+a_2+...+a_n+a_{n+1})^2$$
Proof
\begin{align}
a_1^2+a_2^2+...+a_n^2+a_{n+1}^2 &=(a_1^2+a_2^2+...+a_n^2)+a_{n+1}^2\\
& \leq (a_1+a_2+...+a_n)^2+a_{n+1}^2 \mbox{ (using the assumption)}\\
&=y^2+a_{n+1}^2 \mbox{ (rewriting } y=a_1+a_2+...+a_n)\\
& \leq (y+a_{n+1})^2 \mbox{ (using: } a^2+b^2\leq (a+b)^2)\\
&=(a_1+a_2+...+a_n+a_{n+1})^2 \mbox{ (plug back for } y)
\end{align}
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
How to solve $\int\frac{dx}{\sqrt{4-x^2}}$ with trigonometric substitution? The integral
$$\int\frac{dx}{\sqrt{4-x^2}}$$
I've found the variable
$$x=2\sin\theta$$
$$x^2=4\sin^2\theta$$
$$dx=2\cos\theta\,d\theta$$
Which gave me by substitution
$$\int\frac{2\cos\theta}{4-\sqrt{4\sin^2\theta}}\,d\theta$$
$$\int\frac{\cos\theta}{2-\sin\theta}\,d\theta$$
$$\frac{1}{2}\int \cos\theta \,d\theta-\int\frac{\cos\theta}{\sin\theta}\,d\theta$$
$$\frac{1}{2}\int \cos\theta \,d\theta-\int\cot\theta \,d\theta$$
$$= \frac{\sin\theta}{2}-\ln|\sin\theta| + C$$
Now if I look at the expected answer, it should be
$$\arcsin\left(\frac{x}{2}\right)+C$$
What am I missing ?
|
Do you need to use trigonometric substitution? Because you can solve it easier using regular substitution. First:
$$\int{\frac{dx}{\sqrt{4(1-\frac{x^2}{4})}}}=\frac{1}{2}\int{\frac{dx}{\sqrt{1-(\frac{x}{2})^2}}}$$
We use substitution $\frac{x}{2}=t ; \frac{dx}{2}=dt$.
Now we have:
$$\int{\frac{dt}{\sqrt{1-t^2}}}=arcsint=arcsin\frac{x}{2}+c$$
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Compute the infinite product $\prod\limits_{n=2}^\infty \left(1+\frac{1}{2^n-2}\right)$ I am trying to compute the infinite product
$$
\prod\limits_{n=2}^\infty \left(1+\frac{1}{2^n-2}\right) .
$$
Wolfram Alpha says the result is $2$, but I can't seem to figure out why.
|
Write
$$1 + \frac{1}{2^n - 2} = \frac{2^n - 1}{2^n - 2} = \frac{1}{2} \frac{2^n - 1}{2^{n-1} - 1}$$
so by telescoping,
$$\prod_{k = 2}^N \left(1 + \frac{1}{2^n - 2}\right) = \frac{1}{2^{N-1}}(2^N - 1) = 2 - \frac{1}{2^{N-1}}.$$
Since $\frac{1}{2^{N-1}} \to 0$ as $N \to \infty$, we have
$$\prod_{n = 2}^\infty \left(1 + \frac{1}{2^n-2}\right) = 2.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1183416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Solve $ab = 2(a+b)$ using modular arithmetic
I have the following equation which I want to solve where $a,b \in \mathbb{N}$:
\begin{align*}ab &= 2(a+b) \\ \iff ab &= 2a + 2b \\ \iff ab - 2a &= 2b \\ \iff a(b-2) &= 2b \\ \iff a &= \frac{2b}{b-2}\end{align*}
Therefore I must solve for $a,b$ such that all pairs $\displaystyle(a,b) = \bigg(\frac{2b}{b-2},b\bigg)$ where $a,b\in \mathbb{N}$
I must now find, using modulo properties, the values of $a,b \in \mathbb{N}$ which satisfies the equation.
Clearly $(0,0)$ is a solution, but $0 \not\in \mathbb{N}$. We cannot have $b=1$ since then $a<1$ and $b =2$ is a horizontal asymptote.
Notice, however that $ \displaystyle a = \frac{2b}{b-2} = 2 + \frac{4}{b-2}$. Thus for $a$ to be an integer, we require $(b-2)|4$.
Hence we must have that $4 \equiv 0 \mod{(b-2)}$.
Thus we have the following equations:
$b-2 = 1 \implies b = 3 \implies a = 3$
$b-2 = 2 \implies b = 4 \implies a = 4$
$b-2 = 4 \implies b = 6 \implies a = 3$
Thus our solutions are given by $$\{ (6,3), (4,4), (3,6) \} $$
Is the steps and reasoning used correct?
|
HINT
$$a =\dfrac{2b}{b-2} = 2 + \dfrac{4}{b-2}$$
That means, for $a$ to be an integer we must have $(b-2) | 4$
So $b-2$ has to be a factor of $4$.
|
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|
Show that $\int_1^{\infty } \frac{(\ln x)^2}{x^2+x+1} \, dx = \frac{8 \pi ^3}{81 \sqrt{3}}$ I have found myself faced with evaluating the following integral: $$\int_1^{\infty } \frac{(\ln x)^2}{x^2+x+1} \, dx. $$
Mathematica gives a closed form of $8 \pi ^3/(81 \sqrt{3})$, but I have no idea how to arrive at this closed form. I've tried playing around with some methods from complex analysis, but I haven't had much luck (it has been a while). Does anyone have any ideas? Thanks in advance!
|
It can be observed that $x^{2} + x + 1 = (x-a)(x-b)$ where $a = e^{2\pi i/3}$ and $b = e^{-2\pi i/3}$. Now
\begin{align}
I &= \int_{1}^{\infty} \frac{ (\ln(x))^{2} }{ (x-a)(x-b) } \, dx
= \frac{1}{a-b} \, \int_{1}^{\infty} \left( \frac{1}{x-a} - \frac{1}{x-b} \right) \, (\ln(x))^{2} \, dx.
\end{align}
From Wolfram Alpha the integral
\begin{align}
\int \frac{ (\ln(x))^{2} }{ x - a } dx = -2 Li_{3}\left( \frac{x}{a} \right) +2 \log(x) \, Li_{2} \left( \frac{x}{a} \right) + \log^{2}(x) \log\left( 1-\frac{x}{a} \right)
\end{align}
for which the integral in question becomes
\begin{align}
I &= \left[ \frac{-2}{a-b} \left(Li_{3}\left( \frac{x}{a} \right) - Li_{3}\left(\frac{x}{b} \right) \right) + \frac{2}{a-b} \log(x) \, \left(Li_{2} \left( \frac{x}{a} \right) - Li_{2}\left( \frac{x}{b} \right) \right) + \frac{1}{a-b} \, \log^{2}(x) \log\left( \frac{a-x}{b-x} \right) \right]_{1}^{\infty} \\
&= \frac{-2}{a-b} \left[ Li_{3}\left( \frac{1}{a} \right) - Li_{3}\left(\frac{1}{b} \right) \right].
\end{align}
This can then be seen as
\begin{align}
I &= \frac{-2i}{\sqrt{3}} \left[ Li_{3}\left( e^{2\pi i/3} \right) - Li_{3}\left( e^{- 2\pi i/3} \right) \right] \\
&= \frac{-2 i}{\sqrt{3}} \cdot \frac{4 \pi^{3} i}{81} = \frac{8 \pi^{3}}{81 \sqrt{3}}.
\end{align}
\begin{align}
\int_{1}^{\infty} \frac{ (\ln(x))^{2} }{ x^{2} + x + 1 } \, dx = \frac{8 \pi^{3}}{81 \sqrt{3}}.
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "10",
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|
Multivariable taylor polynomial $$f(x, y) = e^{2x+xy+y^2}$$
Find the 2nd order taylor polynomial to the above function about (0,0)
The formula is:
$$P(x,y)=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)+\frac 12[f_{xx}(x-a)^2+2f_{xy}(x-a)(y-b)+f_{yy}(y-b)^2]$$
$$f_x=e^{2x+xy+y^2}(2+y)$$
$$f_y=e^{2x+xy+y^2}(x+2y)$$
$$f_{xx}=e^{2x+xy+y^2}(2+y)^2$$
$$f_{yy}=e^{2x+xy+y^2}(x+2y)^2+2e^{2x+xy+y^2}$$
$$f_{xy}=e^{2x+xy+y^2}(2+y)(x+2y)+e^{2x+xy+y^2}$$
But I still get the wrong answer. What I am doing wrong?
|
We have:
$$f(x, y) = e^{2x+xy+y^2}$$
Finding partials and evaluating them at the point $(a, b) = (0,0)$, yields:
*
*$f_x(x, y) = (y+2) e^{x y+2 x+y^2} \implies f_x(0,0) = 2$
*$f_y(x, y) = e^{x y+2 x+y^2} (x+2 y) \implies f_y(0,0) = 0$
*$f_{xx}(x, y) = (y+2)^2 e^{x y+2 x+y^2} \implies f_{xx}(0,0) = 4$
*$f_{xy}(x, y) = (y+2) e^{x y+2 x+y^2} (x+2 y)+e^{x y+2 x+y^2} \implies f_{xy}(0,0) = 1$
*$f_{yy}(x, y) = e^{x y+2 x+y^2} (x+2 y)^2+2 e^{x y+2 x+y^2} \implies f_{yy}(0,0) = 2 $
Next, we have:
$P(x,y)=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)+\dfrac 12\left[f_{xx}(a,b)(x-a)^2+2f_{xy}(a,b)(x-a)(y-b)+f_{yy}(a,b)(y-b)^2\right]$
So, we get:
$$P(x, y) = 1 + 2 x + 0 +\frac 12[4 x^2 + 2(1) x y + 2 y^2] = 1 + 2x + 2x^2 + xy + y^2$$
Notice that the book forgot to divide $\frac 12 (4)$ and that is the source of their error. Your answer is correct.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Differentiate $\,y = 9x^2 \sin x \tan x:$ Did I Solve This Correctly? I'm posting my initial work up to this point.
Criticism welcomed!
Using the formula $(fgh)' = f'gh+fg'h + fgh'$, differentiate$$y = 9x^2\sin x \tan x$$
$$\begin{align} y' &= 9\frac d{dx}(x^2)\sin x \tan x + 9x^2 \frac d{dx} (\sin x) \tan x + 9x^2\sin x \frac d{dx}(\tan x)\\ \\
& = 9(2x\sin x \tan x + 9x^2-\cos x \tan x + 9x^2\sin x \sec^2x\\ \\
&=9\Big(2x\sin x \tan x + x^2 -\cos x \tan x + x^2\sin x \sec^2 x\Big)
\end{align}$$
Have I done it correctly up and until this point?
|
\begin{aligned}
\frac{d}{dx}(9x^2 \sin(x) \tan(x)) & = \frac{d}{dx}[9x^2]\sin(x) \tan(x) + 9x^2 \frac{d}{dx}[\sin(x)]\tan(x)+9x^2 \sin(x) \frac{d}{dx}[\tan(x)] & \\
& = 9(2x) \sin(x) \tan(x)+9x^2 \cos(x) \tan(x) + 9x^2 \sin(x) \sec^2(x)&
\end{aligned}
and $\cos(x)\tan(x) = \sin(x)$, so:
\begin{aligned}
\frac{d}{dx}(9x^2 \sin(x) \tan(x)) & = \color{red}{9(2x) \sin(x) \tan(x)+9x^2 \cos(x) \tan(x) + 9x^2 \sin(x) \sec^2(x)} & \\
& = 9(2x) \sin(x) \tan(x)+9x^2 \sin(x) + 9x^2 \sin(x) \sec^2(x) & \\
& = 9x\sin(x)[2\tan(x) + x + x\sec^2(x)]&
\end{aligned}
So the derivative is:
$$\boxed{9x\sin(x)[2\tan(x) + x + x\sec^2(x)]}$$
I noticed these errors in your work: In the $\color{red}{\mathrm{red}}$ line, you wrote a minus sign. I suspect you might have thought that $\frac{d}{dx}[\sin(x)] = -\cos(x)$. In your final line, you probably misinterpreted that minus sign no longer as a negative, but a difference:
$$9[2x \sin(x) \tan(x) + \boxed{x^2 -\cos(x) \tan(x)} + x^2 \sin(x) \sec^2(x)],$$
the $\boxed{\mathrm{boxed}}$ part should actually be
$$x^2 \cos(x)\tan(x).$$
|
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|
Prove that $\displaystyle\tanh^{-1}(x)=\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$ for $-1Prove that $\displaystyle\tanh^{-1}(x)=\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$ for $-1<x<1$.
So far I have got $y=\tanh^{-1}(x)\Longleftrightarrow\tanh(y)=x$ .
Differentiating, $\displaystyle\tanh^{-1}(y)=x \Rightarrow \text{sech}^2(y)\cdot y'=1\Rightarrow y'=\frac{1}{\text{sech}^2(y)}=\frac{1}{\tanh^2(y)-1}=\frac{1}{x^2-1}$.
Stucked here.
|
The hyperbolic arctangent of $x$, $y=\operatorname{arctanh} x$, is the solution of $\tanh y=x$, i.e.
$$ \tanh y=\frac{e^y-e^{-y}}{e^y+e^{-y}}=\frac{e^{2y}-1}{e^{2y}+1}=x,\tag{1}$$
leading to:
$$ e^{2y} = \frac{x+1}{x-1}\tag{2}$$
so that:
$$\operatorname{arctanh} x = y = \frac{1}{2}\log e^{2y} = \frac{1}{2}\,\log\left(\frac{1+x}{1-x}\right).\tag{3}$$
As an alternative, since the hyperbolic arctangent is the inverse function of the hyperbolic tangent,
$$\frac{d}{dz}\operatorname{arctanh} z = \frac{1}{1-z^2} = \frac{1}{2}\left(\frac{1}{1+z}+\frac{1}{1-z}\right)=\frac{1}{2}\frac{d}{dz}\log\left(\frac{1+z}{1-z}\right) \tag{4}$$
and we just have to check that the RHS and LHS of $(3)$ match in $x=0$.
|
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|
Prove: If $a$, $b$, and $c$ are consecutive integers such that $a< b < c $ then $a^3 + b^3 \neq c^3$.
Prove: If $a$, $b$, and $c$ are consecutive integers such that $a< b < c $ then $a^3 + b^3 \neq c^3$.
My Attempt: I start with direct proof.
Let $a,b,c$ be consecutive integers and $a< b < c $, there exists a integer $k$ such that $a=k, b=k+1, c=k+2$. Then $a^3+b^3=k^3+(k+1)^3=k^3+(k^3+3k^2+3k+1)$ and $c^3=(k+2)^3=k^3+6k^2+12k+8=(k^3+3k^2+3k+1)+(3k^2+9k+7)$. Hence, we have $k^3+(k^3+3k^2+3k+1)\neq (k^3+3k^2+3k+1)+(3k^2+9k+7)$ which implies $a^3+b^3\neq c^3$.
Does my proof valid ? And, is use Contradiction easier? If not, can anyone give me a hit or suggestion ?
Thanks
|
$(k-1)^3+k^3=(k+1)^3\\\implies k^3=(k+1)^3-(k-1)^3\\\implies k^3=(k+1-k+1)(k^2+2k+1+k^2-1+k^2-2k+1)\\\implies k^3=2(3k^2+2)$
Let $p$ be a prime that divides $k$.
If $p$ be odd then $k^3=2(3k^2+2)\implies p\mid3k^2+2\implies p\mid 2$
Hence $k$ can't have an odd prime divisor. Therefore $k=2^t$. $(?)$
Now assume that $t>0\implies t\ge 1$
Then we have,
$2^{3t}=2(3\cdot2^{2t}+2)\implies 2^{3t-1}=3\cdot 2^{2t}+2\implies2^{3t-2}=3\cdot 2^{2t-1}+1\implies t=\dfrac{1}{2}\ (?)$
Which implies $t=0$. But putting $t=0$ the equation is not satisfied. Hence the conclusion follows.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Infinite sequence series. Limit If $0<x<1$ and
$$A_n=\frac{x}{1-x^2}+\frac{x^2}{1-x^4}+\ldots+\frac{x^{2^n}}{1-x^{2^{n+1}}},$$
then $\lim_{n\to\infty} A_n$ is
$$\text{a) }\ \dfrac{x}{1-x} \qquad\qquad \text{b) }\ \frac{1}{1-x} \qquad\qquad \text{c) }\ \frac{1}{1+x} \qquad\qquad \text{d) }\ \frac{x}{1+x}$$
How to do this? Not able to convert in any standard form.
|
Using the factorization $1 - x^{2^{k+1}} = (1 - x^{2^k})(1 + x^{2^k})$ we decompose
\begin{align}&\frac{x}{1 - x^2} + \frac{x^2}{1 - x^4} + \cdots + \frac{x^{2^n}}{1 - x^{2^n}}\\
&= \left(\frac{1}{1 - x} - \frac{1}{1 - x^2}\right) + \left(\frac{1}{1 - x^2} - \frac{1}{1 - x^4}\right) + \cdots + \left(\frac{1}{1 - x^{2^n}} - \frac{1}{1 - x^{2^{n+1}}}\right)\\
&=\frac{1}{1 - x} - \frac{1}{1 - x^{2^{n+1}}}.
\end{align}
Since $0 < x < 1$, $x^{2^{n+1}} \to 0$, so the last expression converges to
$$\frac{1}{1 - x} - 1 = \frac{x}{1 - x}.$$
Therefore, the answer is a).
|
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|
Express $\sqrt{3}\sin\theta - \cos\theta$ as: $a\cos (\theta + \alpha) $ Express $\sqrt{3}\sin\theta - \cos\theta$ as: $a\cos (\theta + \alpha) $
Can someone please explain to me how to go about doing this?
|
$% Predefined Typography
\newcommand{\paren} [1]{\left({#1}\right)}
\newcommand{\polar}[2] {\left(#1 ~\angle~ #2\right)}
$
$$\sqrt{3}\sin(\theta) - \cos(\theta)$$
First write as $\cos() + \cos()$:
$$\sqrt{3}\sin(\theta) - \cos(\theta) = \sqrt{3}\cos\paren{\theta - \frac{\pi}2} + \cos(\theta + \pi)$$
Then add (polar coordinates) $\polar{\sqrt{3}}{-\frac{\pi}{2}} + \polar{1}{\pi}$ to get:
$$\polar{\sqrt{3}}{-\frac{\pi}{2}} + \polar{1}{\pi} = \polar{2}{-\frac{2\pi}{3}}$$
So
$$\sqrt{3}\cos\paren{\theta - \frac{\pi}2} + \cos(\theta + \pi) = 2\cos\paren{\theta - \frac{2\pi}{3}}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Is there another way to solve this integral? My way to solve this integral. I wonder is there another way to solve it as it's very long for me.
$$\int_{0}^{\pi}\frac{1-\sin (x)}{\sin (x)+1}dx$$
Let $$u=\tan (\frac{x}{2})$$
$$du=\frac{1}{2}\sec ^2(\frac{x}{2})dx $$
By Weierstrass Substitution
$$\sin (x)=\frac{2u}{u^2+1}$$
$$\cos (x)=\frac{1-u^2}{u^2+1}$$
$$dx=\frac{2du}{u^2+1}$$
$$=\int_{0}^{\infty }\frac{2(1-\frac{2u}{u^2+1})}{(u^2+1)(\frac{2u}{u^2+1}+1)}du$$
$$=\int_{0}^{\infty }\frac{2(u-1)^2}{u^4+2u^3+2u^2+2u+1}du $$
$$=2\int_{0}^{\infty }\frac{(u-1)^2}{u^4+2u^3+2u^2+2u+1}du $$
$$=2\int_{0}^{\infty }\frac{(u-1)^2}{(u+1)^2(u^2+1)}du $$
$$=2\int_{0}^{\infty }(\frac{2}{(u+1)^2}-\frac{1}{u^2+1})du $$
$$=-2\int_{0}^{\infty }\frac{1}{u^2+1}du+4\int_{0}^{\infty}\frac{1}{(u+1)^2}du $$
$$\lim_{b\rightarrow \infty }\left | (-2\tan^{-1}(u)) \right |_{0}^{b}+4\int_{0}^{\infty}\frac{1}{(u+1)^2}du$$
$$=(\lim_{b\rightarrow \infty}-2\tan^{-1}(b))+4\int_{0}^{\infty}\frac{1}{(u+1)^2}du$$
$$=-\pi+4\int_{0}^{\infty}\frac{1}{(u+1)^2}du$$
Let $$s=u+1$$
$$ds=du$$
$$=-\pi+4\int_{1}^{\infty}\frac{1}{s^2}ds$$
$$=-\pi+\lim_{b\rightarrow \infty}\left | (-\frac{4}{s}) \right |_{1}^{b}$$
$$=-\pi+(\lim_{b\rightarrow \infty} -\frac{4}{b}) +4$$
$$=4-\pi$$
$$\approx 0.85841$$
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$\big[$Another (standard) method$\big]$
When we have integrals of the form: $\displaystyle \int \frac{dx}{a+b\sin (x) + c\cos (x)} $ we can perform the substitution: $t=\tan (x/2)$. Then, we have: $\displaystyle \sin(x) = \frac{2t}{1+t^2} $ , $\displaystyle \cos (x) = \frac{1-t^2}{1+t^2} $ , $\displaystyle dx= \frac{2dt}{1+t^2}$.
First we must remove the sine term from the numerator.
$\displaystyle \int_0^{\pi} \frac{1-\sin(x) -1 + 1}{\sin(x) + 1} dx = \int_0^{\pi} \frac{2}{\sin(x)+1} -1 dx$
Now we substitute. Also @$x=0 \rightarrow t=0$ and @$x=\pi \rightarrow t=\infty$. Thus:
$\displaystyle \int_0^{\infty} \Bigg( \frac{2}{\frac{2t}{1+t^2} +1} -1 \Bigg) \cdot \frac{2}{1+t^2} dt = \int_0^{\infty} \frac{4}{2t+1+t^2} - \frac{2}{1+t^2} dt = $
$\displaystyle = \int_0^{\infty} \frac{4}{(t+1)^2} dt - 2\tan^{-1}(t)\big|_0^{\infty} = 4 \left[ -\frac{1}{t+1} \right]_0^{\infty} -2\frac{\pi}{2} = -4 \left( \frac{1}{\infty} - \frac{1}{0+1} \right) -\pi = 4-\pi $
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1194139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 9,
"answer_id": 6
}
|
Various evaluations of the series $\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}$ I recently ran into this series:
$$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}$$
Of course this is just a special case of the Beta Dirichlet Function , for $s=3$.
I had given the following solution:
$$\begin{aligned}
1-\frac{1}{3^3}+\frac{1}{5^3}-\cdots &=\sum_{n=0}^{\infty}\frac{(-1)^n}{\left ( 2n+1 \right )^3} \\
&\overset{(*)}{=} \left ( 1+\frac{1}{5^3}+\frac{1}{9^3}+\cdots \right )-\left ( \frac{1}{3^3}+\frac{1}{7^3}+\frac{1}{11^3}+\cdots \right )\\
&=\sum_{n=0}^{\infty}\frac{1}{\left ( 4n+1 \right )^3} \; -\sum_{n=0}^{\infty}\frac{1}{\left ( 4n+3 \right )^3} \\
&= -\frac{1}{2\cdot 4^3}\psi^{(2)}\left ( \frac{1}{4} \right )+\frac{1}{2\cdot 4^3}\psi^{(2)}\left ( \frac{3}{4} \right )=\frac{1}{2\cdot 4^3}\left [ \psi^{(2)}\left ( 1-\frac{1}{4} \right )-\psi^{(2)}\left ( \frac{1}{4} \right ) \right ]\\
&=\frac{1}{2\cdot 4^3}\left [ 2\pi^3 \cot \frac{\pi}{4} \csc^2 \frac{\pi}{4} \right ] \\
&=\frac{\pi^3 \cot \frac{\pi}{4}\csc^2 \frac{\pi}{4}}{4^3}=\frac{\pi^3}{32}
\end{aligned}$$
where I used polygamma identities and made use of the absolute convergence of the series at $(*)$ in order to re-arrange the terms.
Any other approach using Fourier Series, or contour integration around a square, if that is possible?
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We begin with an integral representation:$$S:=\sum_{n\ge0}\frac{(-1)^n}{(2n+1)^3}=\frac12\int_0^\infty\frac{x^2e^{-x}}{1+e^{-2x}}dx.$$The integrand is even, so$$S=\frac14\int_{-\infty}^\infty\frac{x^2e^{-x}}{1+e^{-2x}}dx.$$Or with $e^{-x}=\tan t$,$$S=\frac14\int_0^{\pi/2}\ln^2(\tan t)dt.$$Since $\int_0^{\pi/2}\tan^{2k-1}tdt=\frac{\pi}{2}\csc(k\pi)$,$$S=\frac{\pi}{32}\left.\left(\partial_k^2\csc(k\pi)\right)\right|_{k=\frac12}=\frac{\pi^3}{32}\left(\csc\frac{\pi}{2}\right)\left(\cot^2\frac{\pi}{2}+\csc^2\frac{\pi}{2}\right)=\frac{\pi^3}{32}.$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1195285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
}
|
Proving $\tan^210^\circ+\tan^250^\circ+\tan^270^\circ=9$ I need help proving the following identity.
$$\tan^210^\circ+\tan^250^\circ+\tan^270^\circ=9$$
I am not sure if it is even true.
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If $\tan3y=\tan30^\circ$
$3y=n180^\circ+30^\circ$ where $n$ is any integer
$y=60^\circ n+10^\circ$ where $n=0,1,2$
For $n=2,y=130^\circ,\tan130^\circ=\tan(180^\circ-50^\circ)=-\tan50^\circ$
Now $\tan3y=\dfrac{3\tan y-\tan^3y}{1-3\tan^2y}$
and consequently, $\dfrac{3\tan y-\tan^3y}{1-3\tan^2y}=\dfrac1{\sqrt3}$ as $\tan30^\circ=\dfrac1{\sqrt3}$
Rearrange to form a cubic equation in $\tan y$ where $y=60^\circ n+10^\circ$ where $n=0,1,2$
We need $\tan^210^\circ+\tan^250^\circ+\tan^270^\circ$
$=(\tan10^\circ)^2+(-\tan50^\circ)^2+(\tan70^\circ)^2$
$=[\tan10^\circ+(-\tan50)+\tan70^\circ]^2$
$-2[\tan10^\circ(-\tan50)+(-\tan50)\tan70^\circ+\tan70^\circ\tan10^\circ]$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1198248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Prove that $f(x) = \sqrt{x^2 + x}$ where $x \in [0, +\infty)$ is uniformly continuous Prove that $f(x) = \sqrt{x^2 + x}$ where $x \in [0, +\infty)$ is uniformly continuous.
So lets take:
${\mid \sqrt{x^2+x} - \sqrt{y^2+y} \mid}^2 \, \leqslant \,\, {\mid \sqrt{x^2+x} - \sqrt{y^2+y} \mid}{\mid \sqrt{x^2+x} + \sqrt{y^2+y} \mid} \,\, = \,\, {\mid x^2+x-y^2-y \mid} \,\, = \,\, {\mid (x+y)(x-y)+(x-y) \mid} \,\, = \,\, {\mid (x-y)(x+y+1) \mid } < {\epsilon}^2 \Rightarrow \,\, {\mid \sqrt{x^2+x} - \sqrt{y^2+y} \mid} < \epsilon$
So now we know that if we take $\delta = {\epsilon}^2$ the condition for uniform continuity of this function will be met because
${\mid x - y \mid} < (\delta = {\epsilon}^2) \Rightarrow {\mid f(x)-f(y) \mid} < \epsilon$
Is this proof valid? Or I miss something?
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Let's work with $f$ on an interval $[a,\infty[$, with $a>0$.
$$f'(x) = \frac{2x + 1}{2\sqrt{x^2 + x}} \le \frac{2x + 1}{2x} = 1 + \frac{1}{2x} \le 1 + \frac{1}{2a} $$
Now using MVT, $f$ becomes $(1 + \frac{1}{2a})$-Lipschitz on $[a,\infty[$. But Lipschitz $\implies$ Uniformly continuous.
For $[0,a]$, use the fact that this is a closed interval in $\mathbb R$, and that $f$ is continuous on this interval.
Thus $f$ is uniformly continuous on $[0,a]$ and $[a,\infty[$ each, hence on their union as well.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1198508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"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.