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Convergence of $\sum_{n = 1}^{\infty} \frac{(-1)^n}{n}$ I was messing around in Mathematica with infinite sums until I tried taking the sum of the following:
$$\sum_{n = 1}^{\infty} \frac{(-1)^n}{n}$$
Mathematica spat out -Log[2]. Can somebody give me proof explaining this answer?
|
A formal argument (ignoring convergence questions)
Start with the Geometric series $$\frac 1{1-x}=1+x+x^2+\cdots$$
Integrate to obtain $$-\ln(1-x)=x+\frac {x^2}2+\frac {x^3}3+\cdots$$
Now evaluate at $x=-1$ to get your result.
To be more rigorous, note that (inductively) it is easy to prove
$$\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}$$
And the right hand can then be rewritten as $$\frac{1}{n} \left[ \frac{1}{1+\frac{1}{n}}+ \frac{1}{1+\frac{2}{n}}+\cdots+\frac{1}{1+\frac{n}{n}} \right]$$
Which is the standard Riemann sum approximation to $$\int_0^1 \frac {dx}{1+x}=\ln(2)$$
|
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|
Solve the equations for x.
a. $10e^{2x}=7\cdot 2^{ax}$
b. $\frac{2(a-x)}{3\sqrt[3]{x}}-x^{2/3}=0$
I did both, but I'm not sure if I did them correctly. Here is my work for both questions:
a. $10e^{2x}=7\cdot 2^{ax}$
$e^{2x}=\frac{7\cdot 2^{ax}}{10}$
$In(\frac{7\cdot 2^{ax}}{10})=2x$
$In(\frac{7}{10})+In(2^{ax})=2x$
$In(\frac{7}{10})+axIn(2)=2x$
$In(\frac{7}{10})=2x-axIn(2)$
$In(\frac{7}{10})=x(2-aIn(2))$
$x=\frac{In(\frac{7}{10})}{2-aIn(2)}$
b. $\frac{2(a-x)}{3\sqrt[3]{x}}-x^{2/3}=0$
$\frac{2(a-x)}{3\sqrt[3]{x}}=x^{2/3}$
$2(a-x)=3x^{1/3}\cdot x^{2/3}$
$2(a-x)=3x$
$2a-2x=3x$
$2a=5x$
$x=\frac{2a}{5}$
|
For (a), the real solution (correct):
$$10e^{2x}=7\cdot2^{\text{a}x}\Longleftrightarrow$$
$$\ln(10e^{2x})=\ln(7\cdot2^{\text{a}x})\Longleftrightarrow$$
$$\ln(10)+\ln(e^{2x})=\ln(7)+\ln(2^{\text{a}x})\Longleftrightarrow$$
$$2x+\ln(10)=\ln(7)+\text{a}x\ln(2)\Longleftrightarrow$$
$$2x-\text{a}x\ln(2)=\ln(7)-\ln(10)\Longleftrightarrow$$
$$x(2-\text{a}\ln(2))=\ln(7)-\ln(10)\Longleftrightarrow$$
$$x=\frac{\ln(7)-\ln(10)}{2-\text{a}\ln(2)}$$
Where $\text{a}\ln(2)\ne2$
(b) is also correct, but $\text{a}\ne0$.
|
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|
Rational Solutions to Trigonometric Equation Consider the equation
$$
\cos(\pi a) + \cos(\pi b)
=
\cos(\pi c) + \cos(\pi d),
$$
with
$$
a,b,c,d \in \mathbb{Q} \cap \left[0,\frac{1}{2}\right].
$$
Clearly, this equation admits some trivial solutions, namely $(a,b) = (c,d)$ or $(a,b) = (d,c)$. Are there any rational solutions other than the trivial ones?
|
For nontrivial solutions, $a,b,c,d$ are distinct. WLOG let $a = \max(a,b,c,d)$.
Taking a common denominator $N$ and writing $a=A/N$, ..., $d = D/N$,
we can write this as
$$ \omega^A + \omega^{-A} + \omega^B + \omega^{-B} - \omega^C - \omega^{-C} - \omega^D - \omega^{-D} = 0$$
or equivalently
$$ P(\omega) = \omega^{2A} + 1 + \omega^{B+A} + \omega^{-B+A} - \omega^{C+A} - \omega^{-C+A} - \omega^{D+A} - \omega^{-D+A} = 0 $$
where $\omega = \exp(\pi i/N)$. Now the minimal polynomial of $\omega$ over the rationals is the cyclotomic polynomial $C_{2N}(z)$, so
$P(z)$ must be divisible by $C_{2N}(z)$. I searched over $A \le 20$, in each case factoring $P(z)$ and looking for cyclotomic factors $C_M(z)$ with $M \ge 4A$, obtaining the following results.
$$
\eqalign{
\cos(\pi/3) + \cos(\pi/15) &= \cos(4 \pi/15) + \cos(\pi/5)\cr
\cos(\pi/2) + \cos(\pi/12) &= \cos(5 \pi/12) + \cos(\pi/4)\cr
\cos(\pi/2) + \cos(\pi/18) &= \cos(7 \pi/18) + \cos(5 \pi/18)\cr
\cos(\pi/2) + \cos(\pi/9) &= \cos(4 \pi/9) + \cos(2 \pi/9)\cr
\cos(3 \pi/7) + \cos(\pi/7) &= \cos(\pi/3) + \cos(2 \pi/7)\cr
\cos(\pi/2) + \cos(\pi/24) &= \cos(3 \pi/8) + \cos(7 \pi/24)\cr
\cos(\pi/2) + \cos(\pi/8) &= \cos(11 \pi/24) + \cos(5 \pi/24)\cr
\cos(\pi/2) + \cos(\pi/30) &= \cos(11 \pi/30) + \cos(3 \pi/10)\cr
\cos(\pi/2) + \cos(\pi/15) &= \cos(2 \pi/5) + \cos(4 \pi/15)\cr
\cos(\pi/2) + \cos(\pi/10) &= \cos(13 \pi/30) + \cos(7 \pi/30)\cr
\cos(\pi/2) + \cos(2 \pi/15) &= \cos(7 \pi/15) + \cos(\pi/5)\cr
\cos(\pi/2) + \cos(\pi/5) &= \cos(2 \pi/5) + \cos(\pi/3)\cr
\cos(\pi/2) + \cos(\pi/36) &= \cos(13 \pi/36) + \cos(11 \pi/36)\cr
\cos(\pi/2) + \cos(5 \pi/36) &= \cos(17 \pi/36) + \cos(7 \pi/36)\cr
}
$$
Those involving $\cos(\pi/2)$ might be considered as somewhat trivial: they are cases of $$\cos(t) = \cos\left(\frac{\pi}{3} + t\right) + \cos(\left(\frac{\pi}{3}-t\right)$$
So the "really nontrivial" examples are
$$\eqalign{\cos(\pi/3) + \cos(\pi/15) &= \cos(4 \pi/15) + \cos(\pi/5)\cr
\cos(3 \pi/7) + \cos(\pi/7) &= \cos(\pi/3) + \cos(2 \pi/7)\cr
}$$
Are there more?
|
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|
Evaluation of $\int_{\pi/6}^{\pi/3} \frac{1}{\sqrt{1+\tan(x)}}dx$ Evaluate the given integral:
$$\int_{\pi/6}^{\pi/3} \frac{1}{\sqrt{1+\tan(x)}}dx$$
After using $\int_{a}^{b}f(x)dx= \int_{a}^{b} f(a+b-x)dx$, we get $\int_{\pi/6}^{\pi/3} \sqrt{\sin(x)+\cos(x)} dx$ but I am not able to solve this integral. Could someone help me with this?
|
Hint: Let $u=\sqrt{1+\tan x}$, so $du=\frac{\sec^2x}{2\sqrt{1+\tan x}}dx\;$. Then
$\displaystyle\int\frac{1}{\sqrt{1+\tan x}}dx=2\int\frac{1}{1+\tan^2x}\frac{\sec^2 x}{2\sqrt{1+\tan x}}dx=2\int\frac{1}{1+(u^2-1)^2}du=2\int\frac{1}{u^4-2u^2+2}du$.
Next use partial fractions:
Since $u^4-2u^2+2=[(u^2)^2+2\sqrt{2}u^2+(\sqrt{2})^2]-(2+2\sqrt{2})u^2=(u^2+\sqrt{2})^2-\left(\sqrt{2+2\sqrt{2}}u\right)^2$
$\hspace{1.32 in}=\left(u^2-\sqrt{2+2\sqrt{2}}u+\sqrt{2}\right)\left(u^2+\sqrt{2+2\sqrt{2}}u+\sqrt{2}\right)$,
$\hspace{.4 in}\displaystyle\frac{1}{u^4-2u^2+2}=\frac{Au+B}{u^2-\sqrt{2+2\sqrt{2}}u+\sqrt{2}}+\frac{Cu+D}{u^2+\sqrt{2+2\sqrt{2}}u+\sqrt{2}}$ so
$\hspace{.3 in}1=(Au+B)\left(u^2+\sqrt{2+2\sqrt{2}}u+\sqrt{2}\right)+(Cu+D)\left(u^2-\sqrt{2+2\sqrt{2}}u+\sqrt{2}\right)$.
1) $u=0\;$ gives $B+D=\frac{\sqrt{2}}{2}$
2) the $u^3$ coefficient gives $0=A+C$
3) the $u$ coefficient gives $0=\sqrt{2}(A+C)+\sqrt{2+2\sqrt{2}}(B-D)$, so $\color{blue}{B=D=\frac{\sqrt{2}}{4}}$
4) the $u^2$ coefficient gives $0=\sqrt{2+2\sqrt{2}}(A-C)+(B+D)=\sqrt{2+2\sqrt{2}}(A-C)+\frac{\sqrt{2}}{2}$,
so $A-C=-\frac{1}{2\sqrt{1+\sqrt{2}}}=-\frac{\sqrt{\sqrt{2}-1}}{2}$ and $\color{blue}{A=-\frac{\sqrt{\sqrt{2}-1}}{4}}$ and $\color{blue}{C=\frac{\sqrt{\sqrt{2}-1}}{4}}$.
(Now integrate each term.)
|
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|
Find The Last Two Digits Of $9^{8^7}$ Find the last two digits of $9^{8^7}$.
I tried finding a secure pattern for the last two digits of powers of $9$ but that didn't work. Any answers?
|
Since \begin{align}(10-1)^{10}&=\sum_{i=0}^{10} \begin{pmatrix} 10 \\i \end{pmatrix}10^i(-1)^{10-i}\\&=(-1)^{10}+10(10)(-1)^9+\sum_{i=2}^{10} \begin{pmatrix} 10 \\i \end{pmatrix}10^i(-1)^{10-i} \\&\equiv1\mod 100\end{align}
We have
$$9^{10} \equiv 1 \mod 100$$
Since $8^7 \equiv (-2)^7 \equiv -128 \equiv 2 \mod 10,$
$9^{8^7}\equiv9^2 \equiv 81 \mod 100$
|
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|
Solve this $6$-th degree polynomial equation: $x^6-x^5-8x^4+2x^3+21x^2-9x-54=0$ The question is as follows:
Solve the equation: $x^6-x^5-8x^4+2x^3+21x^2-9x-54=0$, one root being $\sqrt{2}+i$
It's trivial that another root is $\sqrt{2}-i$. But, I can go no further. Can anyone please help me how to solve this?
|
As @avs answered, after dividing the polynomial by $$(x - \sqrt{2} + i)(x - \sqrt{2} - i)$$ we obtain $$x^4+\left(2 \sqrt{2}-1\right) x^3-\left(3+2 \sqrt{2}\right) x^2-\left(3+12
\sqrt{2}\right) x-18=0$$ By inspection $x=-2$ and $x=3$ are roots. Continuing the division, we are left with $$x^2+2 \sqrt{2} x+3=0$$ the roots of which being $-\sqrt{2}+i$ and $-\sqrt{2}-i$
|
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|
Finding the limit $\lim_{n\rightarrow\infty}(1-\frac{1}{3})^2(1-\frac{1}{6})^2(1-\frac{1}{10})^2...(1-\frac{1}{n(n+1)/2})^2$ Any hints on how to find the following limit ?.
I haven't been able to figure it out still.
$$
\lim_{n \to \infty}\left(\, 1 - \frac{1}{3}\, \right)^{2}
\left(\, 1 - \frac{1}{6}\, \right)^{2}\left(\, 1 - \frac{1}{10}\, \right)^{2}\ldots\left[\, 1 - \frac{1}{n\left(\, n + 1\, \right)/2}\, \right]^{2}
$$
Pardon me if it's a silly question.
Edit: Solved upto this much
$$
\prod_{n = 2}^{\infty}\left[\, 1-\frac{2}{n\left(\, n + 1\, \right)}\, \right] =
\prod_{n = 2}^{\infty}\left(\, \frac{n + 2}{n + 1}\,\frac{n - 1}{n}\, \right)
$$
I did'nt notice most of the terms cancel out.
|
We have
$$ \prod_{n=2}^{+\infty}\left(1-\frac{2}{n(n+1)}\right)=\prod_{n=2}^{+\infty}\left(\frac{n+2}{n+1}\cdot\frac{n-1}{n}\right)=\frac{1}{3} $$
(it is a telescopic product) hence you limit equals $\color{red}{\large\frac{1}{9}}$.
|
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|
How to compute the Pythagorean triple by one of the numbers that belonged to it? I have a positive number $n>2$. How to compute the Pythagorean triple containing $n$? $n$ may be the hypotenuse and leg.
|
Using Euclid's formula, we can find triples matching any hypotenuse by solving the C-function for k and seeing which of a range of m-values yield integers. The limitation is that, for primitive triples, C must take the form $C-4n+1$ and not all of these are valid. Some valid ones are $5,13,17,25,29,37,41 ... 65$
\begin{equation}
C=m^2+k^2\implies k=\sqrt{C-m^2}\qquad\text{for}\qquad \bigg\lfloor\frac{ 1+\sqrt{2C-1}}{2}\bigg\rfloor \le m \le \lfloor\sqrt{C-1}\rfloor
\end{equation}
The lower limit ensures $m>k$ and the upper limit ensures $k\in\mathbb{N}$.
$$C=65\implies \bigg\lfloor\frac{ 1+\sqrt{130-1}}{2}\bigg\rfloor=6 \le m \le \lfloor\sqrt{65-1}\rfloor=8\quad\land \quad m\in\{7,8\}\Rightarrow k\in\{4,1\}\\$$
$$F(7,4)=(33,56,65)\qquad \qquad F(8,1)=(63,16,65) $$
There is a side-A matching every odd number greater that one.
\begin{equation}
A=m^2-k^2\implies k=\sqrt{m^2-A}\qquad\text{for}\qquad \sqrt{A+1} \le m \le \frac{A+1}{2}
\end{equation}
The lower limit ensures $k\in\mathbb{N}$ and the upper limit ensures $m> k$.
$$A=15\implies \sqrt{15+1}=4\le m \le \frac{15+1}{2} =8\quad\land\quad m\in\{4,8\}\implies k \in\{1,7\} $$
$$F(4,1)=(15,8,17)\qquad \qquad F(8,7)=(15,112,113) $$
There is a side-B for every multiple of four.
\begin{equation}
B=2mk\implies k=\frac{B}{2m}\qquad\text{for}\qquad \bigg\lfloor \frac{1+\sqrt{2B+1}}{2}\bigg\rfloor \le m \le \frac{B}{2}
\end{equation}
The lower limit ensures $m>k$ and the upper limit ensures $m\ge 2$
$$B=44\implies\qquad \bigg\lfloor \frac{1+\sqrt{88+1}}{2}\bigg\rfloor =5 \le m \le \frac{44}{2}=22\quad \land \quad m\in\{11,22\}\implies k\in\{2,1\}$$
$$F(11,2)=(117,44,125)\qquad\qquad F(22,1)=(483,44,485)$$
|
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|
Mathematical induction problem: $\frac12\cdot \frac34\cdots\frac{2n-1}{2n}<\frac1{\sqrt{2n}}$ This is a problem that I tried to solve and didn't come up with any ideas
.?$$\frac{1}{2}\cdot \frac{3}{4}\cdots\frac{2n-1}{2n}<\frac{1}{\sqrt{2n}}.$$
All I get is $\frac{1}{\sqrt{2n}}\cdot\frac{2n+1}{2n+2}<\frac{1}{\sqrt{2n+2}}$ which evaluates to $1<0$Do you know what to do here ?
|
Induction is a really efficient way for proving that
$$\frac{(2n-1)!!}{(2n)!!} = \frac{(2n)!}{4^n n!^2} = \frac{1}{4^n}\binom{2n}{n}<\frac{1}{\sqrt{2n}}\tag{1}$$
but it isn't the only option. For instance, we may notice that the LHS of $(1)$ can be written as $$\prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)=\sqrt{\prod_{k=1}^{n}\left(1-\frac{1}{k}+\frac{1}{4k^2}\right)} \tag{2}$$
hence:
$$ \prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)^2 = \frac{1}{4}\color{blue}{\prod_{k=2}^{n}\left(1-\frac{1}{k}\right)}\prod_{k=2}^{n}\left(1+\frac{1}{4k(k-1)}\right)=\frac{1}{4n}\prod_{k=2}^{n}\left(1+\frac{1}{4k(k-1)}\right)\tag{3} $$
since the blue term is a telescopic product. Since $1+z<e^{z}$ for any $z>0$, we also have:
$$ \prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)^2 < \frac{1}{4n}\exp\color{purple}{\sum_{k=2}^{n}\frac{1}{4k(k-1)}}=\frac{1}{4n}\exp\left(\frac{n-1}{4n}\right)<\frac{e^{1/4}}{4n} \tag{4}$$
since the purple sum is a telescopic sum. By switching to square roots it follows that
$$ \frac{(2n-1)!!}{(2n)!!}=\frac{1}{4^n}\binom{2n}{n}< \color{red}{\frac{e^{1/8}}{2\sqrt{n}}}\tag{5}$$
that is a stronger inequality. If we directly compute $\prod_{k=2}^{+\infty}\left(1+\frac{1}{4k(k-1)}\right)=\frac{4}{\pi}$ through Wallis product, with the same approach we may also prove the sharper
$$\boxed{ \frac{(2n-1)!!}{(2n)!!}=\frac{1}{4^n}\binom{2n}{n}< \color{red}{\frac{1}{\sqrt{\pi\left(n+\frac{1}{4}\right)}}}.}\tag{6}$$
|
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Find the value of $\tan A + \tan B$, given values of $\frac{\sin (A)}{\sin (B)}$ and $\frac{\cos (A)}{\cos (B)}$ Given
$$\frac{\sin (A)}{\sin (B)} = \frac{\sqrt{3}}{2}$$
$$\frac{\cos (A)}{\cos (B)} = \frac{\sqrt{5}}{3}$$
Find $\tan A + \tan B$.
Approach
Dividing the equations, we get the relation between $\tan A$ and $\tan B$ but that doesn't help in getting the value of $\tan A + \tan B$. The value comes in terms of $\tan A$ or $\tan B$ but the expected answer is independent of any variable .
Also
$$\frac{\sin(A)\cdot\cos(B) + \sin(B)\cdot\cos(A)}{\cos(A)\cdot\cos(B)} = \tan(A) + \tan(B)$$
We could get a value only if instead of $\cos A$ there was $\sin B$ in the relation(which we get on adding the ratios)
|
HINT:
$$\tan A=\frac{\sin A}{\cos A}=\frac{\frac{\sqrt{3}}2\sin B}{\frac{\sqrt{5}}3\cos B}=\frac{3\sqrt{3}}{2\sqrt{5}}\tan B\tag{1}$$
And
$$1=\sin^2 A+\cos^2 A=\frac{3}{4}\sin^2 B+\frac{5}{9}\cos^2 B=\frac59\left(\sin^2 B+\cos^2B\right)+\frac7{36}\sin^2 B$$
Last equation implies
$$\sin^2B=\frac{16}7$$
|
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How to determine that the solution to the following problem is unique? The question is:
[BMO2 2000 Q3] Find positive integers a and b such that $$(\sqrt[3]{a} + \sqrt[3]{b} - 1)^2 = 49 + 20\sqrt[3]{6}$$
It suffices to find one solution to gain full marks, which I did by expanding and assuming that neither $a$ nor $b$ are perfect cubes, so that $2\sqrt[3]{ab} = 48$. The solution I get from this method is $(a, b) = (48, 288)$. However, from this two questions arise in my mind:
a) Is this is the only solution? I.e. are there $c$ and $d$ such that $$(\sqrt[3]{a} + \sqrt[3]{b} - 1)^2 = (\sqrt[3]{c} + \sqrt[3]{d} - 1)^2$$
$\Rightarrow \sqrt[3]{a} + \sqrt[3]{b} = \sqrt[3]{c} + \sqrt[3]{d}$ or $\sqrt[3]{a} + \sqrt[3]{b} - 1 = -\sqrt[3]{c} - \sqrt[3]{d} + 1$
b) Can one solve the original problem in such a way that it is clear that the solutions found are unique?
EDIT: Thank you to gammatester for his remark that there is another solution: $(a, b) = (288, 48)$, but I am referring to distinct solutions which are not permutations of other solutions.
|
We do not need to account for your second case, since it is trivially impossible due to the fact that for positive integers $ a, b, c, d $ the left hand side is positive and the right hand side is negative. Your first case is settled by the following theorem:
Theorem. Let $ a, b \in \mathbf Q^{\times} $ be non-perfect cubes such that $ a/b $ is not a perfect cube in $ \mathbf Q $. Then, $ \sqrt[3]{a} + \sqrt[3]{b} = \sqrt[3]{c} + \sqrt[3]{d} $ for $ c, d \in \mathbf Q $ if and only if $ \{ a, b \} = \{ c, d \} $ (i.e the solution is unique up to permutation).
Proof. Write $ b = qa $ for $ q \in \mathbf Q $ not a perfect cube. Then, we have an equality
$$ \sqrt[3]{a} + \sqrt[3]{qa} = \sqrt[3]{c} + \sqrt[3]{d} $$
These cube roots all lie in some field extension $ L $ of $ \mathbf Q $, and we can talk about the field trace $ T = \textrm{Tr}_{L/\mathbf Q} $. Since $ x^3 - a $ is irreducible in $ \mathbf Q[x] $ if $ a $ is not a perfect cube, and the trace is a multiple of the sum of conjugates, we find that the trace of all irrational cube roots of rational numbers vanish. This is the key idea of the proof. Multiply both sides by $ a^{2/3} $ to get
$$ a + a\sqrt[3]{q} = \sqrt[3]{a^2 c} + \sqrt[3]{a^2 d} $$
If neither cube root on the right hand side was rational, taking traces would yield the absurd $ a = 0 $. Therefore, at least one of $ a^2 c, a^2 d $ is a perfect cube. Assume that it is $ a^2 c = u^3 $ without loss of generality, and now multiply both sides of the original equation by $ (qa)^{2/3} $ to get
$$ a \sqrt[3]{q^2} + qa = \sqrt[3]{q^2 a^2 c} + \sqrt[3]{q^2 a^2 d} $$
By the same argument, one of the cube roots on the right hand side must be rational, since $ q^2 $ is not a perfect cube. If it was $ q^2 a^2 c $, then $ q^2 $ would also be a perfect cube as the ratio of two perfect cubes, which is a contradiction. Therefore $ q^2 a^2 d = v^3 $ is a perfect cube. Substituting into the original equation gives
$$ \sqrt[3]{a} + \sqrt[3]{qa} = \frac{u \sqrt[3]{a}}{a} + \frac{v\sqrt[3]{qa}}{qa} $$
However, $ \sqrt[3]{a} $ and $ \sqrt[3]{qa} $ are linearly independent over $ \mathbf Q $, which means that the only vanishing linear combination is the trivial one, and we have
$$ \frac{u}{a} = \frac{v}{qa} = 1 $$
Now, $ u^3 = a^2 c $ gives $ c = a $, and $ v^3 = q^2 a^2 d $ gives $ d = qa = b $. QED.
Now, since $ 288/48 = 6 $ is not a perfect cube in $ \mathbf Q $ (and neither are $ 288 $ and $ 48 $), the result follows.
|
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|
Show that $(1,1,1)$, $(a,b,c)$, $(a^2, b^2, c^2)$ are linearly indepdenent for distinct $a,b,c$ Show that $(1,1,1)$, $(a,b,c)$, $(a^2, b^2, c^2)$ are linearly indepdenent, where $a,b,$ and $c$ are distinct real numbers.
I will show my attempt and then state where I get stuck.
Suppose $c_1(1,1,1) + c_2(a,b,c) + c_3(a^2,b^2,c^2) = 0$
This leads to the three equations:
$c_1+c_2a+c_3a^2 = c_1+c_2b+c_3b^2 = c_1+c_2c+c_3c^2 = 0$. Now I am not sure how to sure from here that each $c_i$ must be 0. Hints appreciated.
|
You have a linear system of three equations in three unknowns, which you can write in matrix form as
$$
A\mathbf{c} = \mathbf{0},
$$
where
$$
A = \begin{pmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{pmatrix} \in \mathbb{R}^{3 \times 3}
$$
is your coefficient matrix and
$$
\mathbf{c} = \begin{pmatrix}c_1\\c_2\\c_3\end{pmatrix} \in \mathbb{R}^3
$$
is your unknown vector. In these terms, then, $(1,1,1)$, $(a,b,c)$, and $(a^2,b^2,c^2)$ are linearly independent if and only if the system $A\mathbf{c} = \mathbf{0}$ has a unique solution, if and only if $A$ is invertible, if and only if $\det(A) \neq 0$. But now,
$$
\det(A) = \begin{vmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{vmatrix} = \begin{vmatrix}1&a&a^2\\0&b-a&b^2-a^2\\0&c-a&c^2-a^2 \end{vmatrix}=(b-a)(c^2-a^2)-(b^2-a^2)(c-a)\\=(b-a)(c-a)\left((c+a)-(b+a) \right) = (b-a)(c-a)(c-b),
$$
from which it follows that $(1,1,1)$, $(a,b,c)$, and $(a^2,b^2,c^2)$ are linearly independent if and only if $a$, $b$, and $c$ are all distinct.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Let a, b, c, x and y be integers. Why are there no solutions for $\sqrt {a^2 + b^2 + c^2} = x$, where $x=2^y$ or $x=5 \times 2^y$ Edit - An addition condition: $a$, $b$, and $c$ do not equal $0$.
I'm really digging into 3D vectors and their properties. I've decided to look and see which combinations of three integers will, when plotted in 3D space, generate a hypotenuse whose length is also an integer. For example; $\sqrt {6^2 + 42^2 + 85^2} = 95$
What I've found is quite interesting. It seems that there are no solutions for $\sqrt {a^2 + b^2 + c^2} = x$ when $x=2^y$ or $x=5 \times 2^y$ and $y$ is an integer.
$$\begin{array}{c|c|c}
y & 2^y & 5 \times 2^y \\ \hline
0 & 1 & 5 \\ \hline
1 & 2 & 10 \\ \hline
2 & 4 & 20 \\ \hline
3 & 8 & 40 \\ \hline
4 & 16 & 80 \\ \hline
5 & 32 & 160 \\ \hline
6 & 64 & \text{see below} \\ \hline
7 & 128 & \text{see below} \\
\end{array}$$
So, there are no solutions for $\sqrt {a^2 + b^2 + c^2} = x$ when $x$ equals any value from the following series: $1,2,4,5,8,10,16,20,32,40,64,80,128,160...$
I wrote a program which brute-force checked this for values up to $250$ for $a, b, \text {and } c$ (which is why I show no solutions to $5 \times 2^x$ using $6$ and $7$ above, since they should be $320$ and $640$ respectively). The program found at least 1 solution for all other integers from $1 \text{ to }250$.
Here's the code I used (excel VBA): http://pastebin.com/jVA27jYp
|
There is no solution with $y= 0.$
Let $a,b,c,y$ be positive. If there exists a solution to $a^2+b^2+c^2\in \{4^y, 25\cdot 4^y\}$ for some $(a,b,c,y)$ then there is a solution with a least $y$. Such a solution cannot have $a,b,c$ all even, else $(a/2)^2+(b/2)^2+(c/2)^2\in \{4^{y-1},4^{y-1}\cdot 25\},$ which contradicts the minimality of $y$, unless $y=1$.But $y=1$ and $a,b,c$ all even implies there is a solution $(a/2,b/2,c/2,y')$ with $y'=0.$
But if $a,b,c$ are not all even then $a^2+b^2+c^2$ is congruent to $1,2,$ or $3$ (modulo $4$) so it cannot be divisible by $4,$ so it can't belong to $\{4^y,25\cdot 4^y\}.$
|
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|
Find the area under the curve I apologize, I have asked this question previously and I expect I will be down voted but thats okay because I really need help on this problem not just the answer.
Find the area of the figure rotated about the x-axis $0 \leq t \leq 1$
$x=9t-3t^3$ and $y=9t^2$
Setting up the integral:
$\int_0^1 2 \pi 9t^2 \sqrt{(9-9t^2)^2+(18t^2)}dt$
$18 \pi \int_0^1 t^2 \sqrt{(9(1-t^2)+9(2t)^2)}dt$
$18 \pi \int_0^1 t^2 \sqrt{9((1-t^2)+(2t)^2)}dt$
$18(3) \pi \int_0^1 t^2 \sqrt{((1-t^2)+(2t)^2)}dt$ $\to$ $54 \pi \int_0^1 t^2 \sqrt{((1-t^2)+(2t)^2)}dt$
Algebra inside the integral square root:
$(1-t^2)^2= t^4-2t^2+1$ and $(2t)^2=4t^2$ Combining we get:
$t^4+2t^2+1 \to (t^2+1)^2$ So:
$54 \pi \int_0^1 t^2(t^2+1)^2 dt \to$ $54\pi \int_0^1 t^6+2t^4+t^2 dt$
$54 \pi ( \frac{1}{7}t^7+\frac{2}{5}t^5+\frac{1}{3}t^3 \vert_0^1)$
$\frac{92}{105} \times \frac{54\pi}{1} = \frac{4968\pi}{105}$
This of course was not the accepted final answer
|
I would say:
$\displaystyle S=18\pi\int_0^1 t^2 \sqrt{(9-9t^2)^2+(18t)^2}\,dt=18\pi\int_0^1 t^2 \sqrt{81(1-t^2)^2+81\cdot 4t^2}\,dt=$
$\displaystyle=162\pi\int_0^1 t^2 \sqrt{(1-t^2)^2+4t^2}\,dt=162\pi\int_0^1 t^2 \sqrt{1-2t^2+t^4+4t^2}\,dt=$
$\displaystyle =162\pi\int_0^1 t^2 \sqrt{(1+t^2)^2}\,dt=162\pi\int_0^1 t^2
(1+t^2)\,dt=162\pi\int_0^1 (t^2+t^4)\,dt=$
$\displaystyle =162\pi \left[\frac{t^3}{3}+\frac{t^5}{5}\right]_0^1=\cdots$
|
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|
simplify $\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}$ simplify $$\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}$$
1.$\frac{3}{2}$
2.$\frac{\sqrt[3]{65}}{4}$
3.$\sqrt[3]{2}$
4.$1$
I equal it to $\sqrt[3]{a}+\sqrt[3]{b}$ but I cant find $a$ and $b$
|
Hint: Rewrite $x_{1,2}=5\pm 2\sqrt{13}$ in polar coorinates $x=|x|\exp(i\phi_{1,2})$. Determine $|x_{1,2}|=\sqrt{5^2+(2\sqrt{13})^2}$ and $\tan(\phi_{1,2})=\frac{\pm 2\sqrt{13}}{4}$
Use the periodicity of the exponential function in the complex domain
$\sqrt[3]{x_{1,2}}=|x_{1,2}|^{1/3}\exp(i\phi_{1,2}/3+2\pi k/3)$, in which $k \in \mathbb{Z}$ (you will have to check which range of values for $k$ give new solutions.
|
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|
How can I prove this inequality:$(x+y)^4\le8(x^4+y^4)$? I have to proce that $(x+y)^4\le8(x^4+y^4)$
I have seen that the equality is when $x=y$...
I have tried to develop $(x+y)^4$ but it leads me nowhere... I think that I have to pass by the intermediate of the averages, but I don't see witchone... Could someone please help?
|
$$(x+y)^4= x^4+4x^3y+6x^2y^2+4xy^3+y^4$$
Now, by AM-GM you have
$$
x^3y \leq \frac{x^4+x^4+x^4+y^4}{4} \\
x^2y^2 \leq \frac{x^4+y^4}{2} \\
xy^3\leq \frac{x^4+y^4+y^4+y^4}{4}$$
Thus
$$(x+y)^4 \leq x^4+4 \cdot \frac{x^4+x^4+x^4+y^4}{4} +6 \frac{x^4+y^4}{2} +4 \frac{x^4+y^4+y^4+y^4}{4} +y^4$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Secondary school level mathematical induction
*
*It is given that
$$1^3+2^3+3^3+\cdots+n^3=\frac{n^2(n+1)^2}{4}$$
Then, how to find the value of
$2^3+4^3+\cdots+30^3$?
Which direction should I aim at?
*Prove by mathematical induction, that $5^n-4^n$ is divisible by 9 for all positive even numbers $n$.
$$5^n-4^n=9m,\text{where $m$ is an integer.}$$
What I am thinking in the $n+1$ step is,
\begin{align}
& 5^{n+2}-4^{n+2} \\
= {} & 5^2(5^n-4^n)+5^24^n-4^{n+2} \\
= {} & 5^2(5^n-4^n)+4^n(5^2-4^2) \\
= {} & 5^29m+4^n9 \\
= {} & 9(5^2m+4^n)
\end{align}
Does this approach make sense?
*Show that $a+b$ is a factor of $a^n+b^n$ where $n$ is a positive odd number.
I am thinking this in the $n+1$ step.
$$a^{2n+1}+b^{2n+1}$$
But then I cannot get it further.
|
Hints
1) $2^3+4^3+...+30^3=2^3(1^3+2^3+...+15^3)$ now apply the formula you got.
2) Assume that $9k=5^n-4^n$ then $5^{n+2}-4^{n+2}=25.5^n-16.4^n=25(9k+4^n)-16.4^n=9k'+9.4^n$
3) Change the questions to show that $(a+b)|(a^{2k+1}+b^{2k+1})$ for all $k\in\mathbb{N}$.Let $(a+b)x=a^{2k+1}+b^{2k+1}$ now $a^{2k+3}+b^{2k+3}=a^2.a^{2k+1}+b^2.b^{2k+1}$ again do substitution and it will work out.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solving exponential equation of form $(a+b)^x + (a-b)^x = c$ I'm given equation:
$$\left(2-\sqrt3\right)^\frac{x}{2} + \left(2+\sqrt3\right)^\frac{x}{2} = 4,$$
and I'm stuck with it. Which way shall I dig into to solve it?
|
Let $u = (2+\sqrt{3})^{x/2}$. Using the observation that $2-\sqrt{3} = \frac{1}{2+\sqrt{3}}$, we have that $(2-\sqrt{3})^{x/2} + (2+\sqrt{3})^{x/2} = u+\frac{1}{u}$. We thus aim to solve
$$ u + \frac{1}{u} = 4. $$
Since $u\ne 0$, this is equivalent to $u^2 - 4u + 1 = 0$. Using the quadratic formula, we obtain $u = 2\pm\sqrt{3}$. We thus have
$$ (2+\sqrt{3})^{\frac{x}{2}} = 2\pm\sqrt{3} = (2+\sqrt{3})^{\pm 1}\implies \boxed{x = \pm 2}. $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Want help for prove formula by combinatoric argument $1\cdot 2+2\cdot 3+3\cdot 4+\dots +n(n+1)=\frac{n(n+1)(n+2)}{3}$ I have a problem.I just read first combinatorics textbook and I found the identity
$1\cdot 2+2\cdot 3+3\cdot 4+\dots +n(n+1)=\frac{n(n+1)(n+2)}{3}$.
I can prove this by induction,but the problem is from combinatorics textbook and I have no idea how to begin the proof. How can I prove it by combinatorics argument?
|
On the right side, you have the number of ways of selecting three items without replacement from a set of $n+2$ distinct items, multiplied by $2$. For simplicity, I'll assume there are $n$ items first, then I'll up the index later.
Now, suppose you certainly select the first item. The number of ways of choosing the rest are $\frac{(n-1)(n-2)}{2}$. Otherwise, you don't select the first item but rather the second one, and here the number of possibilities are $\frac{(n-3)(n-2)}{2}$. Repeating, you can finally select the third item from last, which gives $\frac{1\cdot 2}{2}$.These exhaust all the cases, so you have the equation:
$$
\binom{n}{3} = \frac{(n-1)(n-2)}{2} + \frac{(n-3)(n-2)}{2} + \ldots + \frac{1\cdot 2}{2}.
$$
Multiply by two and you have your formula:
$$
\frac{n(n-1)(n-2)}{3}= {(n-1)(n-2)} + {(n-3)(n-2)} + \ldots + {1\cdot 2}
$$
Now, just put $n+2$ in place of $n$ for:$$
\frac{n(n+1)(n+2)}{3}= {(n+1)(n)} + {(n-1)(n)} + \ldots + {1\cdot 2}
$$
|
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|
Find all positive natural $x$ and $y$ so $(x+y)^7-x^7-y^7$ is divisble by $7^7$, but so that $xy(x+y)$ is not divisble by $7$ I thought of looking modulo $7$ but it lead me nowhere... Does anyone have a better idea?
|
Note that
$$\frac{(x+y)^7-x^7-y^7}{7xy(x+y)}= x^4 + 2yx^3 + 3y^2x^2 + 2y^3x + y^4 $$
We immediately check that $x\equiv y\pmod 7$ cannot lead to a solution because $9x^4$ cnnot be a multiple of $7^6$. Hence we are allowed to rewrite this as
$$x^4 + 2yx^3 + 3y^2x^2 + 2y^3x + y^4 =\frac{x^6-2x^3y^3+y^6}{(x-y)^2}=\frac{(x^3-y^3)^2}{(x-y)^2}. $$
In order to have $7^6\mid (x^3-y^3)^2$, we need $7^3\mid x^3-y^3=(x-y)(x^2+xy+y^2)$ and by the bove remark $7^3\mid x^2+xy+y^2$.
As $7\nmid x,y$, there exists $z$ with $y\equiv zx\pmod {7^3}$. Then we need to solve $z^2+z+1\equiv 0\pmod{7^3}$.
Modulo $7$, we find that $z\equiv 2\lor 4\pmod 7$. By symmetry $x\leftrightarrow y$, we need only investigate the first case. Substituting $z=7w+2$, we arrive at
$$ 7w^2+5w+1\equiv 0\pmod {7^2}$$
In particular, $w\equiv 4\pmod 7$. With $w\leftarrow 7v+4$,
$$ 49v^2+61v+19\equiv 0\pmod 7$$
which simplifies to $v\equiv -1\pmod 7$.
Thus $v=7u-1$, $w=49u-3$, $z=343u-19$
|
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|
What is the maximal value of $2 \sin x - 7 \cos x$? What is the maximal value of $2 \sin x - 7 \cos x$?
How do I calculate this? Do I have to write out the $\sin$ and the $\cos$?
|
Let us first develope a general methodology by considering the general identity:
$$\ \tag{1} f(x)=A \cos(x) + B \sin(x)=C\left(\dfrac{A}{C}\cos(x) + \dfrac{B}{C} \sin(x)\right)$$
with $C:=\sqrt{A^2+B^2}$.
As $\left(\dfrac{A}{C}\right)^2 + \left(\dfrac{B}{C}\right)^2=1$, point $\left(\dfrac{A}{C},\dfrac{B}{C}\right)$ is on the unit circle.
Thus, there exist an $x_0$ such that
$$\ \tag{2} \cos(x_0)=\dfrac{A}{C} \ \ and \ \ \sin(x_0)=\dfrac{B}{C}$$
Remark : $\tan(x_0)=\dfrac{B}{A}$ (unless $A=0$).
Plugging (2) into (1) gives $f(x) = C cos(x-x_0)$
which is thus maximal for $x=x_0$ with maximal value $C$.
With the numerical values at hand, using remark above, the maximum occurs for $x=x_0=$atan$\left( \dfrac{-7}{2}\right)\approx -1.2925...$ and its value is
$$\sqrt{53}\approx7.2801...$$
as can be seen on figure below.
|
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|
I can't find 'the easiest way' to simplify this expression? $\large\dfrac{3^{2008} (10^{2013} + 5^{2012} + 2^{2011})}{5^{2012} (6^{2010} + 3^{2009} + 2^{2008})}$
$\large=\dfrac{3^{2008} ((2\cdot5)^{2013} + 5^{2012} + 2^{2011})}{5^{2012} ((2\cdot3)^{2010} + 3^{2009} + 2^{2008})}$
$\large=\dfrac{3^{2008} (2^{2013}\cdot5^{2013} + 5^{2012} + 2^{2011})}{5^{2012} (2^{2010}\cdot3^{2010} + 3^{2009} + 2^{2008})}$
This is where I get stuck. Any help would be appreciated.
|
The expression is extremely close to $\dfrac{40}{9}$. This is because in the numerator, $2^{2013}\cdot 5^{2013}$ is much larger than the remaining terms $5^{2012}$ and $2^{2011}$, so the remaining terms are almost negligible. Similarly for the denominator. The expression then simplifies to$\dfrac{2^{2013}\cdot 3^{2008}\cdot 5^{2013}}{2^{2010}\cdot 3^{2010}\cdot 5^{2012}}=\dfrac{40}{9}$.
I doubt there is a simpler expression, since the numerator is divisible by $3^{2008}$ but the denominator is not divisible by $3$, while the denominator is divisible by $5^{2012}$ but the numerator is not divisible by $5$.
|
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|
integrate $\int \frac{1}{\sqrt{x^2 - 1}} \, dx$ I am solving an ODE
$$y^2 \, dx + \left(x\sqrt{y^2 - x^2} - xy\right)dy=0$$
I let $y=xu$ and did some algebra, and I ended up with this
$$\ln(x)+C=\int \frac{1}{\sqrt{u^2 - 1}} \, du - \int \frac{1}{u} \, du$$
I don't know how to solve the first right hand side integral. I tried to let $u^2 - 1 = t$, so it becomes
$$\int\frac{1}{\sqrt{u^2 - 1}}=\int\frac{1}{\sqrt{t}}\frac{1}{2u} \, dt = \int \frac{1}{\sqrt{t}} \frac{1}{2} \frac{1}{\sqrt{t-1}} \, dt,$$
which doesn't seem to work. The correct answer seems to be
$$\ln \left(\sqrt{x^2 - 1} - x\right)$$
But how do we get this?
|
If the hyperbolic trigonometry is not known, we can see that
$\int\frac{dt}{\sqrt{t}\sqrt{t+1}}=\int\frac{\sqrt{t+1}}{\sqrt{t}}-\frac{\sqrt t}{\sqrt{t+1}}dt.$
Then $z=1+\frac{1}{t}$ gives $t=\frac{1}{z-1}$, $dt=\frac{-dz}{(1-z)^2}$ and
$\frac{1}{2}\int\frac{\sqrt{t+1}}{\sqrt{t}}-\frac{\sqrt t}{\sqrt{t+1}}dt=\frac{1}{2}\int\frac{(1-z)dz}{\sqrt{z}(1-z)^2}=\frac{1}{2}\int\frac{dz}{\sqrt{z}(1-z)}=\frac{1}{4}\int\frac{1}{\sqrt z(1-\sqrt{z})}+\frac{1}{\sqrt z(1+\sqrt{z})}dz.$
It comes
$\int\frac{du}{\sqrt{u^2-1}}=\frac{1}{2}\left(\ln(1+\sqrt z)-\ln(\sqrt z-1)\right)=\frac{1}{2}\ln\left(\frac{1+\sqrt z}{\sqrt z-1}\right)=\frac{1}{2}\ln\left(\frac{\sqrt t+\sqrt{1+t}}{\sqrt{1+t}-\sqrt t}\right)=\frac{1}{2}\ln\left(\frac{u+\sqrt{u^2-1}}{u-\sqrt{u^2-1}}\right)$
and as $\ln\left(\frac{1}{u-\sqrt{u^2-1}}\right)=\ln(u+\sqrt{u^2-1})$, we obtain $\int\frac{du}{\sqrt{u^2-1}}=\ln(u+\sqrt{u^2-1})$.
We can also note that $\int\frac{du}{\sqrt{u^2-1}}=\frac{1}{2}\int\frac{\sqrt{u+1}}{\sqrt{u-1}}-\frac{\sqrt{u-1}}{\sqrt{1+u}}du$ and take the substitution $t=\frac{u+1}{u-1}$.
|
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|
Prove: $\cos^3{A} + \cos^3{(120°+A)} + \cos^3{(240°+A)}=\frac {3}{4} \cos{3A}$ Prove that:
$$\cos^3{A} + \cos^3{(120°+A)} + \cos^3{(240°+A)}=\frac {3}{4} \cos{3A}$$
My Approach:
$$\mathrm{R.H.S.}=\frac {3}{4} \cos{3A}$$
$$=\frac {3}{4} (4 \cos^3{A}-3\cos{A})$$
$$=\frac {12\cos^3{A} - 9\cos{A}}{4}$$
Now, please help me to continue from here.
|
You can use this particular formula: $$\cos{x}\cos{y}=\frac{1}{2}\left(\cos(x+y)+\cos(x-y)\right)$$
twice and simplify $\cos^3{A}$, $\cos^3{(A+2\pi/3)}$, and $\cos^3{(A-2\pi/3)}$ in this way:
$$\begin{aligned} \cos^3{A}&=\cos{A}\left(\cos{A}\cos{A}\right) \\
&=\frac{1}{2}\cos{A}\left(1+\cos{2A} \right) \\
&=\frac{1}{2}\cos{A}+\frac{1}{2}\cos{A}\cos{2A}\\
&=\frac{1}{2}\cos{A}+\frac{1}{4}\left(\cos{A}+\cos{3A}\right)\\
&=\frac{1}{2}\cos{A}+\frac{1}{4}\cos{A}+\frac{1}{4}\cos{3A}\\
\end{aligned}$$
And
$$\begin{aligned} \cos^3{(A+2\pi/3)}&=\frac{1}{2}\cos{(A+2\pi/3)}\left(1+\cos{(2A+4\pi/3)} \right) \\
&=\frac{1}{2}\cos{(A+2\pi/3)}\left(1-\cos{(2A+\pi/3)} \right) \\
&=\frac{1}{2}\cos{(A+2\pi/3)}-\frac{1}{2}\cos{(A+2\pi/3)}\cos{(2A+\pi/3)}\\
&=\frac{1}{2}\cos{(A+2\pi/3)}-\frac{1}{4}\left(\cos{(A-\pi/3)}-\cos{3A}\right)\\
&=\frac{1}{2}\cos{(A+2\pi/3)}-\frac{1}{4}\cos{(A-\pi/3)}+\frac{1}{4}\cos{3A}
\end{aligned}$$
And
$$\begin{aligned} \cos^3{(A-2\pi/3)}&=\frac{1}{2}\cos{(A-2\pi/3)}\left(1+\cos{(2A-4\pi/3)} \right) \\
&=\frac{1}{2}\cos{(A-2\pi/3)}\left(1-\cos{(2A-\pi/3)} \right) \\
&=\frac{1}{2}\cos{(A-2\pi/3)}-\frac{1}{2}\cos{(A-2\pi/3)}\cos{(2A-\pi/3)}\\
&=\frac{1}{2}\cos{(A-2\pi/3)}-\frac{1}{4}\left(\cos{(A+\pi/3)}-\cos{3A}\right)\\
&=\frac{1}{2}\cos{(A-2\pi/3)}-\frac{1}{4}\cos{(A+\pi/3)}+\frac{1}{4}\cos{3A}
\end{aligned}$$
Now, by adding the results and using the following identities you can get your answer (the second one can be proved by using the first identity in the reverse order).
$$\cos{A}+\cos{(A+2\pi/3)}+\cos{(A-2\pi/3)}=0$$
$$\cos{(A-\pi/3)}+\cos{(A+\pi/3)}=\cos{A}$$
|
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|
Calculate $(1+i)^{11}$ I'm studying complex numbers and I wanted to know if this solution is correct.
The problem is to calculate $(1+i)^{11}$, here's my attempt:
I can express $(1+i)^{11}$ using the argument notation: $z=\rho(\cos \phi +i\sin \phi)$
$$z=\left(\cos \frac{\pi}{2}+i\sin\frac{\pi}{2}\right)$$
This is very helpful because I can use De Moivre's formula to calculate $z^{11}$
$$z^{11}=\left(\cos \frac{\pi}{2}+i\sin\frac{\pi}{2}\right)^{11}=\left(\cos \frac{11\pi}{2}+i\sin\frac{11\pi}{2}\right)=0-i=-i$$
Is this correct? Is there a better way to solve it?
|
There's another approach:
\begin{align}
(1+i)^{11}&=\sum_{k=0}^{11}\binom{11}{k}i^k
\\\\&=\sum_{\substack{0\le k\le 11 \\k\equiv 0\!\pmod 4}}\binom{11}{k} + \sum_{\substack{0\le k\le 11 \\k\equiv 1\!\pmod 4}}\binom{11}{k}\cdot i + \sum_{\substack{0\le k\le 11 \\k\equiv 2\!\pmod 4}}\binom{11}{k} \cdot (-1)+ \sum_{\substack{0\le k\le 11 \\k\equiv 3\!\pmod 4}}\binom{11}{k}\cdot (-i)&\scriptsize\text{by the binomial theorem, and computing each }i^k
\\\\&=\sum_{k=0,4,8}\binom{11}{k}+i\sum_{k=1,5,9}\binom{11}{k}-\sum_{k=2,6,10}\binom{11}{k}-i\sum_{k=3,7,11}\binom{11}{k}
\\\\&=\sum_{k=0,4,3}\binom{11}{k}+i\sum_{k=1,5,2}\binom{11}{k}-\sum_{k=2,5,1}\binom{11}{k}-i\sum_{k=3,4,0}\binom{11}{k}&\scriptsize\text{since }\binom{11}{k}=\binom{11}{11-k}
\\\\&=\left(\sum_{k=0,3,4}\binom{11}{k}-\sum_{k=1,2,5}\binom{11}{k}\right)(1-i)
\\\\&=\left(1 + \frac{11\cdot 10 \cdot 9}{3 \cdot 2 \cdot 1}+ \frac{11\cdot 10 \cdot 9 \cdot 8}{4 \cdot 3 \cdot 2 \cdot 1} -11 -\frac{11\cdot 10}{2 \cdot 1}-\frac{11\cdot 10 \cdot 9 \cdot 8 \cdot 7}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}\right)(1-i)
\\\\&=\left((1 + 165+ 330)-(11+55+462) \right)(1-i)
\\\\&=-32(1-i)
\\\\&=-32+32i
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Epsilon-delta limit definition I got $\lim_{x \rightarrow 3} \frac{x^2-2x+1}{x-2} = 4$
I need to prove that by delta epsilon. I came to delta ={1/2epsilon,1/2}
Is that right? If not can you explain me how?
|
Let $\forall \epsilon >0$ and we have to show that ${ \delta }>0$ such that
if $0<\left| x-3 \right| <{ \delta }$ then $\left| \frac { { x }^{ 2 }-2x+1 }{ x-2 } -4 \right| <\epsilon $
$$\left| \frac { { x }^{ 2 }-2x+1 }{ x-2 } -4 \right| =\left| \frac { { x }^{ 2 }-2x+1-4x+8 }{ x-2 } \right| =\left| \frac { { x }^{ 2 }-6x+9 }{ x-2 } \right| =\left| \frac { { \left( x-3 \right) }^{ 2 } }{ x-3+1 } \right|<\\\\ <\left| \frac { { \left( x-3 \right) }^{ 2 } }{ x-3 } \right| =\left| x-3 \right| <\delta =\epsilon $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proving the identity $\frac{\cos^2\theta+\tan^2\theta-1}{\sin^2\theta}=\tan^2\theta$ I am stuck with this trigonometric identity. It appeared in a question paper of mine, and I am wondering whether there is a print error or something, because I absolutely have no idea how to solve this.
$$\frac{\cos^2\theta+\tan^2\theta-1}{\sin^2\theta}=\tan^2\theta$$
I would really appreciate some inputs!
|
$$\frac{\cos^2\theta + \frac{\sin^2 \theta}{\cos^2\theta}-1}{\sin^2\theta}=$$
$$= \frac{\cos^2 \theta +\frac{\sin^2\theta}{\cos^2 \theta}-\sin^2 \theta-\cos^2\theta}{\sin^2 \theta}$$
$$= \frac{\frac{\sin^2\theta-\cos^2\theta \cdot \sin^2\theta}{\cos^2\theta}}{\sin^2\theta}$$
$$=\frac{\sin^2\theta \cdot (1-\cos^2\theta)}{\sin^2\theta \cdot \cos^2\theta}$$
$$=\frac{\sin^2\theta \cdot \sin^2 \theta}{sin^2\theta \cdot \cos^2\theta}$$
$$=\frac{\sin^2\theta}{\cos^2\theta}=\tan^2\theta$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1922605",
"timestamp": "2023-03-29T00:00:00",
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|
How to integrate $\int\frac{du}{u\sqrt{c-2u}} $ I was looking through these notes, and at the top of the second page it says you can integrate
$$\mathrm{d}\zeta = \frac{\mathrm{d}u}{u\sqrt{c-2u}} $$
to get
$$u(\zeta) = \frac{c}{2}\mathrm{sech}^2(\frac{1}{2}\sqrt{c}(\zeta - \zeta_0)).$$
I can't see where that comes from. Is there any trig identity I should use?
|
Take the integral:
$$\int \frac{1}{u \sqrt{c-2 u}} \ du$$
For the $$\int \frac{1}{u \sqrt{c-2 u}}$$, substitute $u = c-2 u$ and $du = -2 du$. Then we have :
$$\int \frac{1}{\sqrt{u} (u-c)} \ du$$
For the integrand $\frac{1}{\sqrt{u} (u-c)}$, substitute $s = \sqrt{u}$ and $ds = \frac{1}{2 \sqrt{u}} du$, then we have :
$$2 \int \frac{1}{s^{2}-c} ds$$
Factor $-c$ from the denominator, then :
$$ 2 \int \frac{-1}{c (1-\frac{s^2}{c})} ds$$
Factor out constants, then :
$$\frac{-2}{c} \int \frac{1}{(1-\frac{s^2}{c})} ds$$
For the integrand $\frac{1}{(1-\frac{s^2}{c})}$, substitute $p = \frac{s}{\sqrt{c}}$ and $dp = \frac{1}{\sqrt{c}} ds$:
$$=\frac{-2}{\sqrt{c}} \int \frac{1}{(1-p^{2})} dp$$
The integral of $\frac{1}{(1-p^2)}$ is $tanh^{-1}(p)$, then :
= $$-\frac{2 tanh^{-1}(p))}{\sqrt{c}}+constant$$
Substitute back for $p = \frac{ s}{\sqrt{c}}$:
= $$-\frac{2tanh^{-1}(\frac{s}{\sqrt{c}})}{\sqrt{c}}$$
Substitute back for $s = \sqrt{u}$:
= = $$-\frac{2tanh^{-1}(\frac{\sqrt{u}}{\sqrt{c}})}{\sqrt{c}}$$
Substitute back for $u = c-2 u$:
$$-\frac{2tanh^{-1}(\frac{\sqrt{c-2u}}{\sqrt{c}})}{\sqrt{c}}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Symmetric polynomials with Vieta's and Newton's theorems
Let $ x_{1}, x_{2}, x_{3}$ be the solutions of the equation $ x^3 -3x^2 + x - 1 = 0.$
Determine the values of $$\frac{1}{{x_{1}x_{2}}} + \frac{1}{{x_{2}x_{3}}} + \frac{1}{{x_{3}x_{1}}}$$
and also
$$ x_{1}^{3}+x_{2}^{3}+x_{3}^{3}$$
|
We have
$$x^3-3x^2+x-1=(x-x_1)(x-x_2)(x-x_3)$$
so by developing the RHS we see that $\sum x_i x_j=1:$ the coefficient of $x$ and similarly $x_1x_2x_3=1$ and $x_1+x_2+x_3=3$.
Denote $V_1$ the first desired value than $V_1\times(x_1x_2x_3)=x_1+x_2+x_3$. And denote $V_2$ the second value then
$$(x_1+x_2+x_3)^3=V_2+3(x_1x_2+x_2x_3+x_3x_1)(x_1+x_2+x_3)-3x_1x_2x_3$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove $(a^2+b^2)(c^2+d^2)\ge (ac+bd)^2$ for all $a,b,c,d\in\mathbb{R}$. Prove $(a^2+b^2)(c^2+d^2)\ge (ac+bd)^2$ for all $a,b,c,d\in\mathbb{R}$.
So $(a^2+b^2)(c^2+d^2) = a^2c^2+a^2d^2+b^2c^2+b^2d^2$
and $(ac+bd)^2 = a^2c^2+2acbd+b^2d^2$
So the problem is reduced to proving that $a^2d^2+b^2c^2\ge2acbd$ but I am not sure how to show that
|
By Lagrange's identity
$$(a^2+b^2)(c^2+d^2)=(ac+bd)^2\color{red}{+(ad-bc)^2}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding number of roots of $1+2x+3x^2+...+(n+1)x^n=0$ Find the number of roots of the equation
$1+2x+3x^2+4x^3+....+(n+1)x^n=0$
where $n$ is even.
My attempt:
Let $f(x)=x+x^2+x^3+....+x^n$
Clearly,$x=0$ is a root and $f(x)=0$
cannot have a positive root.
$f(x)=x\left(1+x+x^2+...+x^n\right)$
How to determine whether the equation
$1+x+x^2+x^3+...+x^n=0$
has real roots or not.
If I come to know that equation has certain number of roots then i can invoke Rolle's Theorem to find number of roots of given equation.
|
We have $$ 1+2x+3x^2+...+(n+1)x^n = \sum_{k=0}^n (k+1)x^k = \sum_{k=0}^n \sum_{m=k}^n x^m = \sum_{k=0}^n \sum_{m=0}^{n-k} x^{m+k} =$$
$$= \sum_{k=0}^n x^k \sum_{m=0}^{n-k} x^m = \sum_{k=0}^n x^k \cdot \frac{x^{n-k+1}-1}{x-1} = \frac{1}{x-1} \sum_{k=0}^n {(x^{n+1}-x^k)} = $$
$$= \frac{(n+1)x^{n+1}}{x-1} - \frac{1}{x-1}\sum_{k=0}^n {x^k} = \frac{(n+1)x^{n+1}}{x-1} - \frac{1}{x-1}\cdot\frac{x^{n+1}-1}{x-1}=$$
$$=\frac{(n+1)x^{n+1}\cdot (x-1)-x^{n+1}+1}{(x-1)^2}=\frac{(n+1)x^{n+2}-(n+2)x^{n+1}+1}{(x-1)^2}$$
Thus we got the short formula for the sum which depends on n and x. Next since n is even then $n=2l$ and
$$ \frac{(n+1)x^{n+2}-(n+2)x^{n+1}+1}{(x-1)^2} = \frac{(2l+1)x^{2l+2}-(2l+2)x^{2l+1}+1}{(x-1)^2} = $$
$$=\frac{(2l+1)x^{2l+2}-(2l+2)x^{2l+1}+1}{(x-1)^2} - (l+1)x^{2l}+(l+1)x^{2l}=$$
$$ = (l+1)x^{2l} + \frac{1}{(x-1)^2}\cdot((2l+1)x^{2l+2}-(2l+2)x^{2l+1}+1 -(l+1)x^{2l}(x-1)^2)= $$
$$ = (l+1)x^{2l}+\frac{lx^{2l+2}-(l+1)x^{2l}+1}{(x-1)^2}$$
Last fraction is very similar to the first formula. So we make some extra actions
$$(l+1)x^{2l}+\frac{lx^{2l+2}-(l+1)x^{2l}+1}{(x-1)^2} = (l+1)x^{2l}+\frac{l(x^2)^{l+1}-(l+1)(x^2)^l+1}{(x^2-1)^2}\frac{(x^2-1)^2}{(x-1)^2}=$$
$$ = (l+1)x^{2l}+(x+1)^2\sum_{k=0}^{l-1}(k+1)(x^2)^k$$
Since this sum consists of non-negative terms it is equal to zero only if each term is zero. Thus, given equation has no roots.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove the inequality $\sqrt{a^2+b^2}+\sqrt{c^2+b^2}+\sqrt{a^2+c^2}<1+\frac{\sqrt2}{2}$ Let $a,b,c -$ triangle side and $a+b+c=1$. Prove the inequality
$$\sqrt{a^2+b^2}+\sqrt{c^2+b^2}+\sqrt{a^2+c^2}<1+\frac{\sqrt2}{2}$$
My work so far:
1) $a^2+b^2=c^2-2ab\cos \gamma \ge c^2-2ab$
2) $$\sqrt{a^2+b^2}+\sqrt{c^2+b^2}+\sqrt{a^2+c^2}\le3\sqrt{\frac{2(a^2+b^2+c^2)}3}$$
|
Just a sketch of a possible proof (a variational, mixing variables approach). Assume that $a\leq b\leq c$ and show that if you replace $a$ with $a-\varepsilon$ and $b$ with $b+\varepsilon$, the convexity of $f(x)=\sqrt{1+x^2}$ grants that the LHS increases (after such substitution, the order of $b$ and $c$ may change. That is not an issue, it actually helps us to be sure we always deal with side lengths of a triangle). It follows that the supremum of the LHS is achieved when $a=0$. After that, the problem is trivial.
|
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|
Polar to cartesian form of r=sin(4θ)? The Polar to cartesian form of $ r = \sin(2\theta)$ is fairly simple.
What is the Cartesian form of the polar equation r=sin(4θ)?
[edit]
$$r=4sin(θ)cos(θ)(cos(θ)^2-sin(θ)^2)$$, so $$r^5=4rsin(θ)rcos(θ)(r^2cos(θ)^2-r^2sin(θ)^2)$$,so $$r^5=4xy(x^2-y^2)$$, so $$(x^2+y^2)^{5/2}=4xy(x^2-y^2)$$, so $$(x^2+y^2)^{5}=(4xy(x^2-y^2))^{2}$$, so $$(x^2+y^2)^{5}=16x^2y^2(x^2-y^2)^{2}$$
|
$$\rho=\sin 4\theta=2\sin2\theta\cos2\theta=4\sin\theta\cos\theta(\cos^2\theta-\sin^2\theta)=\frac{4xy(x^2-y^2)}{\rho^4},$$ hence
$$(x^2+y^2)^5=(4xy(x^2-y^2))^2,$$
$$x^{10}+10x^4y^6+32x^4y^4+5x^2y^8-16x^2y^6+y^{10}+10y^4x^6+5y^2x^8-16y^2x^6=0.$$
|
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|
Find sum of first $n$ terms of the series : $1+\frac{1^3+2^3}{1+2}+\frac{1^3+2^3+3^3}{1+2+3}+\dots$ The main question is:
Find sum of first $n$ terms of the series : $1+\frac{1^3+2^3}{1+2}+\frac{1^3+2^3+3^3}{1+2+3}+\dots$
My approach:
Initially, nothing clicked, so I went forward with simplifying the series.
So, we get, after simplifying,
$S$(The sum of series) = $1+3+6+10+15+21+\dots$
So, I try to write $S$ as,
$1+(4-1)+(9-3)+(16-6)+\dots$
So what I finally get is,
$S$=$(1+4+9+16+25+\dots) - (1+3+6+10+\dots)$
Therefore,
$$2S=\sum_{n=1}^n{n^2}$$
Thus,
$S$ = $\frac{n(n+1)(2n+1)}{12}$
But, the answer given in my textbook is
$\frac{n(n+1)(n+2)}{6}$
My answer is not matching. Please help me by pointing out my mistake or providing a new approach and solution altogether.
|
Recall that
$$x_n=\sum_{k=1}^n k=\frac{n(n+1)}{2}$$
and
$$y_n=\sum_{k=1}^n k^3=\frac{n^2(n+1)^2}{4}$$
so
$$\frac{y_n}{x_n}=x_n$$
so you need to recall the sum
$$\sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{6}$$
to compute the desired sum $\sum_{k=1}^n \frac{y_k}{x_k}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Recurrence relation of type $a_{n+1} = a^2_{n}-2a_{n}+2$
A sequence $\{a_{n}\}$ is defined by $a_{n+1} = a^2_{n}-2a_{n}+2\forall n\geq 0$ and $a_{0} =4$
And another sequence $\{b_{n}\}$ defined by the formula $\displaystyle b_{n} = \frac{2b_{0}b_{1}b_{2}..........b_{n-1}}{a_{n}}\forall n\geq 1$ and $\displaystyle b_{0}=\frac{1}{2}$,Then
$(a)$ value of $a_{10}$
$(b)\;\; $ The value of $n$ for which $\displaystyle b_{n} = \frac{3280}{3281}$
$(c)$ The Sequence $\{b_{n}\}$ satisfy the recurrence formula
$\bf{Options::}$
$(1)\; \displaystyle b_{n+1} = \frac{2b_{n}}{1-b^2_{n}}\;\;\;\;\;\; (b)\; \displaystyle b_{n+1} = \frac{2b_{n}}{1b^2_{n}}\;\;\;\;\;\; (c)\; \displaystyle b_{n+1} = \frac{b_{n}}{1+2b^2_{n}}\; (d)\; \displaystyle b_{n+1} = \frac{b_{n}}{1-2b^2_{n}}$
$\bf{My\; Try::}$ Given $a_{n+1} = a^2_{n}-2a_{n}+2 = \left(a_{n}-1\right)^2+1$
So $a_{1} = (a_{0}-1)^2+1=(4-1)^2+1=10$
Similarly $a_{2} = (a_{1}-1)^2+1 = 9^2+1 = 82$
Similarly $a_{3} = (81)^2+1 = $
But Calculation like this is very complex, Plz help me how can i solve above problems, Thanks
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Define $c_{n}=a_{n}-1$. Note that $c_{n+1}=c_{n}^{2}$. Then,
$c_{4} = c_{3}^{2} = c_{2}^{4} = c_{1}^{8} = c_{0}^{16} = 3^{16}$, so $a_{4}=3^{16}+1$.
|
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|
For how many positive integers $a$ is $a^4−3a^2+9$ a prime number? I understand that there are many posts on the problems similar to mine. I have tried my best, but still get different answers from the answer sheet. Can anyone help me? Also is there a simple way to find $a$?
For how many positive integers $a$ is $a^4-3a^2+9$ a prime number?
Here is what I did: $$a^4-3a^2+9=(a^2+3+3a)(a^2+3-3a)$$ To find $a$, I looked at the following situations:
$$a^4-3a^2+9=1$$
$$a^4-3a^2+9=3$$
$$a^4-3a^2+9=5$$
$$a^4-3a^2+9=7$$
$$a^4-3a^2+9=11$$
$$a^4-3a^2+9=13$$
$$....$$
|
You already have the product $p = \left(a^2-3 a+3\right) \left(a^2+3 a+3\right)$, and since for both factors you have $0 < a^2-3 a+3 < a^2+3 a+3$ for positive $a$ (check), and since a prime number is only divisible by $1$ and itself, you have $a^2-3 a+3 = 1$ and $a^2+3 a+3 = p$. Solving first equation gives you $a=1$ and $a=2$ and these in turn give you two primes $p=7$ and $p=13$.
|
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|
Easier way to calculate the derivative of $\ln(\frac{x}{\sqrt{x^2+1}})$? For the function $f$ given by
$$
\large \mathbb{R^+} \to \mathbb{R} \quad x \mapsto \ln \left (\frac{x}{\sqrt{x^2+1}} \right)
$$
I had to find $f'$ and $f''$.
Below, I have calculated them.
But, isn't there a better and more convenient way to do this?
My method:
$$
{f'(x)}=\left [\ln \left (\frac{x}{(x^2+1)^\frac{1}{2}} \right) \right ]'=\left (\frac{(x^2+1)^\frac{1}{2}}{x} \right)\left (\frac{x}{(x^2+1)^\frac{1}{2}} \right)'=\left (\frac{(x^2+1)^\frac{1}{2}}{x} \right) \left (\frac{(x^2+1)^\frac{1}{2}-x[(x^2+1)^\frac{1}{2}]'}{[(x^2+1)^\frac{1}{2}]^2} \right)=\left (\frac{(x^2+1)^\frac{1}{2}}{x} \right) \left (\frac{(x^2+1)^\frac{1}{2}-x[\frac{1}{2}(x^2+1)^{-\frac{1}{2}}(x^2+1)']}{\left | x^2+1 \right |} \right)=\left (\frac{(x^2+1)^\frac{1}{2}}{x} \right) \left (\frac{(x^2+1)^\frac{1}{2}-x[\frac{1}{2}(x^2+1)^{-\frac{1}{2}}(2x)]}{x^2+1} \right)=\left (\frac{(x^2+1)^\frac{1}{2}}{x} \right) \left (\frac{(x^2+1)^\frac{1}{2}-x^2(x^+1)^{-\frac{1}{2}}}{x^2+1} \right)=\frac{(x^2+1)^{(\frac{1}{2}+\frac{1}{2})}-x^2(x^2+1)^{\frac{1}{2}+-\frac{1}{2}{}}}{x(x^2+1)}=-\frac{x^2}{x}=-x
$$
and
$$
f''(x)=(-x)'=-1\
$$
This took me much more than 1.5 hours just to type into LaTex :'(
|
The expression $$\frac{x}{\sqrt{x^2 + 1}}$$
is the sine of the angle adjacent to $1$, in a right triangle with legs of $1$ and $x$ and hypotenuse $\sqrt{x^2 + 1}$.
Therefore, it is equal to
$$
\sin \arctan x
$$
and we have
$$
\bbox[5px,border:2px solid blue]{f(x) = \ln (\sin (\arctan x)).}
$$
Computing $f'(x)$ for this is just chain rule twice:
$$
f'(x) = \frac{1}{\sin \arctan x} \cdot \cos \arctan x \cdot \frac{1}{1 + x^2}.
$$
Now referring to the original triangle, $\sin \arctan x = \frac{x}{\sqrt{x^2 + 1}}$ and $\cos \arctan x = \frac{1}{\sqrt{x^2 + 1}}$, so
$$
f'(x) = \frac{\sqrt{x^2 + 1}}{x}
\cdot \frac{1}{\sqrt{x^2 + 1}}
\cdot \frac{1}{x^2 + 1}
= \frac{1}{x(x^2 + 1)}.
$$
Note your answer was incorrect.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Prove the inequality $\frac a{a+b^2}+\frac b{b+c^2}+\frac c{c+a^2}\le\frac14\left(\frac1a+\frac1b+\frac1c\right)$ Let $a,b,c>0; a+b+c=1$. Prove the inequality
$$\frac a{a+b^2}+\frac b{b+c^2}+\frac c{c+a^2}\le\frac14\left(\frac1a+\frac1b+\frac1c\right)$$
My work so far:
I tried AM-GM and used fact $a+b+c=1$.
|
$$\frac{a}{a+b^2}= \frac{a}{a^2+ab+ac+b^2} = \frac{a}{(a^2+ac)+(ab+b^2)} \le \frac{a}{4} \cdot \left( \frac{1}{a^2+ac}+\frac{1}{ab+b^2} \right)=$$
$$= \frac{1}{4}\cdot \left( \frac{1}{a+c}+\frac{1}{b}-\frac{1}{a+b} \right)$$
$$\frac{a}{a+b^2}+\frac{b}{b+c^2}+\frac{c}{c+a^2} \le \frac{1}{4}\cdot \left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1934520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Determine $ax^4 + by^4$ for system of equations I found the following recreational problem without further specification for $a,b$.
Let $x,y$ be real numbers s.t.
$a + b = 6$,
$ax + by = 10$,
$ax^2 + by^2 = 24$,
$ax^3 + by^3 = 62$.
Determine $ax^4 + by^4$.
I am new to problem solving exercises like this and therefore appreciate diverse approaches to this problem as well as comments on how to tackle those types of exercises.
|
This is not a nice solution.
Considering the equations $$a+b-6=0\tag 1$$ $$ax+by-10=0\tag 2$$ $$ax^2+by^2-24=0\tag 3$$ $$ax^3+by^3-62=0\tag 4$$ Let us eliminate the variables one at the time and express them as a function of $a$.
From $(1)$, $b=6-a$. Replacing in $(2)$ and assuming $a\neq6$, we have $$a x+(6-a) y-10=0 \implies y=\frac{a x-10}{a-6}\tag 5$$ Replacing $b$ and $y$ in $(3)$, we have $$\frac{a \left(-6 x^2+20 x-24\right)+44}{a-6}=0\implies x_{\pm}=\frac{5a\pm\sqrt{11} \sqrt{6 a-a^2}}{3 a}\tag 6$$ (assuming $a\neq 0$).
Let us use $x_+$ and replace in $(4)$ to end with $$\frac{22 \left(\sqrt{(6-a) a}-2 \sqrt{11} a+6 \sqrt{11}\right)}{9 \sqrt{(6-a) a}}=0 \tag 7$$ So, we need to solve $$\sqrt{(6-a) a}=2 \sqrt{11} a-6 \sqrt{11} \tag 8$$ Squaring both sides $$-a^2+6a=44 a^2-264 a+396\implies 45 a^2-270 a+396=0 \implies a_{\pm}=\frac{1}{5} \left(15\pm\sqrt{5}\right)$$ Considering $a_+$ and going backwards, we should get $$a=3+\frac{1}{\sqrt{5}}\qquad b=3-\frac{1}{\sqrt{5}}\qquad x=\frac{1}{2} \left(3+\sqrt{5}\right)\qquad y=\frac{1}{2} \left(3-\sqrt{5}\right)$$ and then the value of $(a x^n+b y^n)$ for any $n$.
Quite laborious, isn't it ?
Thanks for the fun.
|
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"timestamp": "2023-03-29T00:00:00",
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|
$a_1=3;\ a_{n+1}=3^{a_n}$; find $a_{2004}\bmod100$ A detailed solution will be helpful. Given that $a_1=3$ and $a_{n+1}=3^{a_n}$, find the remainder when $a_{2004}$ is divided by 100.
|
By the CRT, in order to find $a_{2004}\pmod{100}$ it is enough to find $a_{2004}\pmod{8}$ and $a_{2004}\pmod{25}$. Now $a_n=3^{a_{n-1}}\pmod{8}$ just depends on $a_{n-1}\pmod{4}$, and $a_{n-1}=3^{a_{n-2}}\pmod{4}$ just depends on $a_{n-2}$ being even or odd. Since $a_{2002}$ is clearly odd, $a_{2003}\equiv 3\pmod{4}$ and $\color{green}{a_{2004}\equiv 3\pmod{8}}$. $a_n=3^{a_{n-1}}\pmod{25}$ just depends on $a_{n-1}\pmod{20}$. We already know that $a_{2003}\equiv 3\pmod{4}$, hence it is enough to find $a_{2003}=3^{a_{2002}}\pmod{5}$, or $a_{2002}\pmod{4}$. Since $a_{2002}\equiv 3\pmod{4}$, $a_{2003}\equiv 2\pmod{5}$, hence $a_{2003}\equiv 7\pmod{20}$ and $\color{green}{a_{2004}\equiv 12\pmod{25}}$. Putting together the green identities,
$$ \color{green}{\large a_{2004}\equiv 87\pmod{100}.}$$
You may notice that the above argument has little to do with the arithmetic properties of $2004$, and in fact you may apply the same argument to show that $a_n\equiv 87\pmod{100}$ for any $n\geq 3$.
This is due to the fact that by iterating the totient function, we always reach $1$ pretty soon.
For instance:
$$\varphi(\varphi(\varphi(\varphi(\varphi(\varphi(100))))))=1.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1938353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Prove that $\sum_{k=0}^nk{m+k \choose m}=n{m+n+1\choose m+1}-{m+n+1 \choose m+2}$ Can someone please see the work I have so far for the following proof and provide guidance on my inductive step?
Prove that if $m,n\in\mathbb{N}$, then $\sum_{k=0}^nk{m+k \choose m}=n{m+n+1\choose m+1}-{m+n+1 \choose m+2}$
Base Case. Let n=0. Then $\sum_{k=0}^{n=0}k{m+k \choose m}=0$ and $n{m+n+1\choose m+1}-{m+n+1 \choose m+2}=0-0=0$. Thus for $n=0$ our equation is satisfied.
Inductive Step. Let $n\ge0$. Assume $\sum_{k=0}^nk{m+k \choose m}=n{m+n+1\choose m+1}-{m+n+1 \choose m+2}$. Now observe that
$$\begin{align*}\sum_{k=0}^{n+1}k{m+k \choose m}&=\\
\sum_{k=0}^{n}k{m+k \choose m}+(n+1){m+n+1\choose m}&=n{m+n+1\choose m+1}-{m+n+1 \choose m+2}+(n+1){m+n+1\choose m} \end{align*}$$
This is where I'm getting stuck... using Pascal's Identity to combine some of these terms seems ideal. However, the factors of $n$ and $n+1$ are making a clever manipulation difficult for me.
|
Here is a somewhat different approach. We use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series. This way we can write e.g.
\begin{align*}
[x^k](1+x)^n=\binom{n}{k}
\end{align*}
We obtain for $m,n\geq 0$
\begin{align*}
\sum_{k=0}^n&k\binom{m+k}{m}\\
&=\sum_{k=1}^nk[x^m](1+x)^{m+k}\tag{1}\\
&=[x^m](1+x)^{m+1}\sum_{k=1}^nk(1+x)^{k-1}\tag{2}\\
&=[x^m](1+x)^{m+1}D_x\left(\sum_{k=1}^n(1+x)^k\right)\tag{3}\\
&=[x^m](1+x)^{m+1}D_x\left(\frac{1-(1+x)^{n+1}}{1-(1+x)}\right)\tag{4}\\
&=[x^m](1+x)^{m+1}\left[\frac{n(1+x)^n}{x}-\frac{(1+x)^n}{x^2}+\frac{1}{x^2}\right]\tag{5}\\
&=n[x^{m+1}](1+x)^{m+n+1}-[x^{m+2}](1+x)^{m+n+1}+[x^{m+2}](1+x)^{m+1}\tag{6}\\
&=n\binom{m+n+1}{m+1}-\binom{m+n+1}{m+2}
\end{align*}
and the claim follows
Comment:
*
*In (1) we apply the coefficient of operator.
*In (2) we do a little rearrangement by using the linearity of the coefficient of operator.
*In (3) we use the differential operator $D_x:=\frac{d}{dx}$.
*In (4) we use the formula for the finite geometric series.
*In (5) we differentiate the expression.
*In (6) collect terms and apply the formula
\begin{align*}
[x^{p+q}]A(x)=[x^p]x^{-q}A(x)
\end{align*}
*In (7) we select the coefficients of $x^{m+1}$ and $x^{m+2}$. Note the coefficient of $x^{m+2}$ in the last term is zero, since the polynomial $(1+x)^{m+1}$ has degree less than $m+2$.
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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|
Calculating $\int \frac{x^2 - 2x + 3}{x^4 -x^2 +1}dx$ I was calculating
$$\int \frac{x^7 + 3x^6 - x^4 + 6x^3 -1}{x^4 - x^2 + 1}dx$$
I performed Euclidean division and the integral reduced to
$$\frac{x^4}4 + x^3 + \frac{x^2}2 + 2x + \frac32 \ln(x^4 - x^2 + 1) - J(x)$$
with
$$J(x) = \int \frac{x^2 - 2x +3}{x^4 - x^2 + 1}dx$$
How to calculate $J(x)$?
My try:
$$J(x) = \int \frac{x^2 + 3}{x^4 - x^2 + 1}dx - \frac{2}{\sqrt 3} \tan^{-1} \left( \frac{2x^2-1}{\sqrt 3} \right)$$
Now I am stuck
|
The denominator readily factors into $$x^4 - x^2 + 1 = (x^2 + \sqrt{3}x + 1)(x^2 - \sqrt{3}x + 1),$$ where upon the usual partial fraction decomposition procedure requires us to solve $$x^2-2x+3 = (Ax + B)(x^2 - 3x+1) + (Cx + D)(x^2 + 3x + 1)$$ for suitable coefficients $A, B, C, D$. I trust that this is something you are able to do. It may be a bit tedious, but it is the "canonical" method of solution for the indefinite integral.
Perhaps a little additional motivation may be useful.
We could look at the solutions of the general quadric $f(x) = x^4 - x^2 + 1$ as the square roots of the equivalent quadratic $f(x) = g(x^2) = (x^2)^2 - (x^2) + 1$, and multiply together the pairs of linear factors corresponding to complex conjugate roots, but this is not particularly elegant in this case. Rather, we can use a "naive" approach and presuppose that $f$ factors nicely in the form $$x^4 - x^2 + 1 = (x^2 + ax + 1)(x^2 + bx + 1).$$ Expanding the RHS and comparing the coefficients to those on the LHS, it becomes obvious that we must have $b = -a$, and $2 - a^2 = -1$. This gives us $a = \pm \sqrt{3}$, and the aforementioned factorization follows. This sort of trick works in simple cases such as this.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the points on the ellipsoid $x^2 + 2y^2 + 3z^2 = 1$ where the tangent plane is parallel to the plane $3x - y + 3z = 1$. Find the points on the ellipsoid $x^2 + 2y^2 + 3z^2 = 1$, where the tangent plane is parallel to the plane $3x - y + 3z = 1$.
I'm not sure how to go about solving this. I'd appreciate some help.
|
Let the contact point be $(X,Y,Z)$, now the tangent plane is
$$Xx+2Yy+3Zz=1$$
Comparing coefficients,
$$X:2Y:3Z=3:-1:3$$
That is $$\frac{X}{3}=\frac{2Y}{-1}=\frac{3Z}{3}=k$$
Now substitute $(X,Y,Z)=\left( 3k,-\dfrac{k}{2},k \right)$ into $x^{2}+2y^{2}+3z^{2}=1$
We have
\begin{align*}
(3k)^{2}+2\left( -\frac{k}{2} \right)^{2}+3(k)^{2} &= 1 \\
\frac{25}{2}k^{2} &= 1 \\
k &= \pm \frac{\sqrt{2}}{5} \\
(X,Y,Z) &=
\left(
\pm \frac{3\sqrt{2}}{5},
\mp \frac{1}{5\sqrt{2}},
\pm \frac{\sqrt{2}}{5} \right)
\end{align*}
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
How to prove that $k^n=\frac{d^n}{dx^n} (1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....+\frac{x^n}{n!})^k|_{x=0}$ $$k^n=\frac{d^n}{dx^n} (1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....+\frac{x^n}{n!})^k|_{x=0}$$
It is easy to show $k=1$ and $k=2$
$k=1$
$$\frac{d^n}{dx^n} (1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....+\frac{x^n}{n!})|_{x=0}=1$$
$k=2$
$$\frac{d^n}{dx^n} (1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....+\frac{x^n}{n!})(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....+\frac{x^n}{n!})|_{x=0}$$
If we collect $x^n$ terms;
$$\frac{d^n}{dx^n} (1+2x+....+(\frac{n!}{0!n!}+\frac{n!}{1!(n-1)!}+...+\frac{n!}{n!0!})\frac{x^n}{n!}+.....)|_{x=0}$$
$$\frac{d^n}{dx^n} (1+2x+....+(\frac{n!}{0!n!}+\frac{n!}{1!(n-1)!}+...+\frac{n!}{n!0!})\frac{x^n}{n!}+.....)|_{x=0}=(\frac{n!}{0!n!}+\frac{n!}{1!(n-1)!}+...+\frac{n!}{n!0!})=2^n$$
I thought to use Leibniz's method to prove general case but it seems we need to know $$\frac{d^{n-r}}{dx^{n-r}} (1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....+\frac{x^n}{n!})^k|_{x=0}$$
Therefore I am stuck .
Is there an easy way to prove the general case?
Thanks a lot for helps
|
Let
$$ s_n = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+…+\frac{x^n}{n!} $$
We can use induction on $n$. $n=1$ is easy to prove. Assume the hypothesis is true for $1 .. (n-1)$.
We have
$$ \frac{d}{dx}s_n = s_{n-1}$$
Now:
$$ \frac{d^n}{dx^n}s_n^k = \frac{d^{n-1}}{dx^{n-1}}(\frac{d}{dx}s_n^k) $$
And, using the chain rule:
$$ \frac{d}{dx}s_n^k = k\cdot s_n^{k-1} \cdot s_{n-1}$$
$$ = k\cdot (s_{n-1}+\frac{x^n}{n!})^{k-1} \cdot s_{n-1}$$
Using the Binomial Theorem:
$$ = k\cdot \sum_{i=0}^{k-1}{{k-1} \choose i} \frac{x^{ni}}{n!^{i}} s_{n-1}^{k-1-i}\cdot s_{n-1}$$
$$ = k\cdot \sum_{i=0}^{k-1}{{k-1} \choose i}\frac{x^{ni}}{n!^{i}} s_{n-1}^{k-i}$$
Taking the $n-1^{th}$ derivative of these terms, we can see that all the terms for $i>0$ will be zero at $x=0$ because of the factor $x^{ni}$. The only term left will be $i=0$. Therefore:
$$ \frac{d^n}{dx^n}s_n^k = k \cdot \frac{d^{n-1}}{dx^{n-1}} s_{n-1}^k = k^n$$
|
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|
Elegant way to prove that this vector set is linear independent. I'm trying to find an elegant way to prove that this vector set $$\begin{bmatrix}1\\ 1\\1\\0\end{bmatrix}, ~~\begin{bmatrix}1\\ 1\\0\\1\end{bmatrix},~~\begin{bmatrix}1\\ 0\\1\\1\end{bmatrix},~~\begin{bmatrix}0\\ 1\\1\\1\end{bmatrix}$$ is linearly independent. I tried to find the matrix determinant (Where the column vectors are these vectors) through co factors expansion. Also, I tried to reduce the matrix to find which vectors are linearly independent based on the pivots. Is there another way to prove such linear independence?
|
You can observe that, if
$$
M=\begin{bmatrix}
1 & 1 & 1 & 0 \\
1 & 1 & 0 & 1 \\
1 & 0 & 1 & 1 \\
0 & 1 & 1 & 1
\end{bmatrix}
$$
then
$$
M\begin{bmatrix}1\\1\\1\\1\end{bmatrix}=
3\begin{bmatrix}1\\1\\1\\1\end{bmatrix},
\qquad
M\begin{bmatrix}1\\-1\\-1\\1\end{bmatrix}=
-\begin{bmatrix}1\\-1\\-1\\1\end{bmatrix},
\qquad
M\begin{bmatrix}0\\-1\\1\\0\end{bmatrix}=
\begin{bmatrix}0\\-1\\1\\0\end{bmatrix},
\qquad
M\begin{bmatrix}-1\\0\\0\\1\end{bmatrix}=
\begin{bmatrix}-1\\0\\0\\1\end{bmatrix}
$$
The last two vectors are obviously linearly independent. Thus $M$ has eigenvalues $3$, $-1$ and $1$ (double), so it is invertible.
However, row reduction is much easier than cofactor expansion:
\begin{align}
\begin{bmatrix}
1 & 1 & 1 & 0 \\
1 & 1 & 0 & 1 \\
1 & 0 & 1 & 1 \\
0 & 1 & 1 & 1
\end{bmatrix}
&\to
\begin{bmatrix}
1 & 1 & 1 & 0 \\
0 & 0 & -1 & 1 \\
0 & -1 & 0 & 1 \\
0 & 1 & 1 & 1
\end{bmatrix}
&& \begin{aligned}R_2&\gets R_2-R_1\\ R_3&\gets R_3-R_1\end{aligned}
\\
&\to
\begin{bmatrix}
1 & 1 & 1 & 0 \\
0 & -1 & 0 & 1 \\
0 & 0 & -1 & 1 \\
0 & 1 & 1 & 1
\end{bmatrix}
&&R_2\leftrightarrow R_3
\\
&\to
\begin{bmatrix}
1 & 1 & 1 & 0 \\
0 & -1 & 0 & 1 \\
0 & 0 & -1 & 1 \\
0 & 0 & 1 & 2
\end{bmatrix}
&&R_4\gets R_4+R_2
\\
&\to
\begin{bmatrix}
1 & 1 & 1 & 0 \\
0 & -1 & 0 & 1 \\
0 & 0 & -1 & 1 \\
0 & 0 & 0 & 3
\end{bmatrix}
&&R_4\gets R_4+R_3
\end{align}
so the determinant is $1\cdot(-1)\cdot(-1)\cdot 3=3$.
|
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|
Integration by completing the square I need to complete the square on the following integral. Once this is done apparently I will be able to use on of the integration tables in the back of my book.
$\int x \sqrt{x^2 + 6x +3} dx $
This is what I have come up with so far:
$\int x \sqrt{(x+3)^2 -6} $
I really am at a loss.
Any help with this would be appreciated.
Thank you
|
Substitute $u=x+3 \implies x=u-3, dx=du$:$$\int x\sqrt{(x+3)^2-6}dx=\int (u-3)\sqrt{u^2-6} du$$Integrating the integral gives:
$$\int u\sqrt{u^2-6}du-\int 3\sqrt{u^2-6} du=\frac{(u^2-6)^{\frac{3}{2}}}{3}-3\left(\frac{u\sqrt{u^2-6}}{2}-3\ln\left(u+\sqrt{u^2-6}\right)\right)$$In terms of $x$:$$\int x\sqrt{(x+3)^2-6}dx=$$$$\frac{((x+3)^2-6)^{\frac{3}{2}}}{3}-3\left(\frac{(x+3)\sqrt{(x+3)^2-6}}{2}-3\ln\left(x+3+\sqrt{(x+3)^2-6}\right)\right)$$
|
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"url": "https://math.stackexchange.com/questions/1947649",
"timestamp": "2023-03-29T00:00:00",
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|
How to prove that $|z^2| = |z|^2$ where $z = a+bi$? I just started my topic on complex numbers and I'm stuck on this question.
What I have managed to get (I go wrong here, I don't know where though):
$z^2 = (a+bi)^2 = a^2 + b^2$, so $|z^2| = \sqrt{(a^2 + b^2)^2 + 0^2} = \sqrt{a^4+2a^2b^2+b^4}$
and
$|z| = a^2+b^2$ so $|z|^2 = (\sqrt{a^2+b^2})^2 = a^2+b^2$
Suggestions?
|
$$z^2 = (a+bi)^2 = a^2 - b^2 +2iab $$
$$|z^2| = \sqrt{(a^2 - b^2)^2 + (2ab)^2} = \sqrt{a^4+2a^2b^2+b^4}=\sqrt{(a^2+b^2)^2}$$
and
$$|z| = \sqrt {a^2+b^2}$$
$$\implies |z|^2 = (\sqrt{a^2+b^2})^2 = a^2+b^2$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1949316",
"timestamp": "2023-03-29T00:00:00",
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|
How do I prove that $f(n) = 0$ for all $n$ where $n$ is a positive integer? In particular, $f$ is defined such that $f(mn)=f(m)+f(n)$, where $m$ and $n$ are positive integers. Also $f(n)\ge 0$ for all $n$, $f(10)=0$, and $f(n)=0$ if $n$ ends in a $3$. Thank you.
|
*
*If $n\equiv 1\pmod{10}$ then $3n\equiv 3\pmod{10}$ and so $f(n)=f(3n)-f(3)=0$.
*If $n\equiv 7\pmod{10}$ then $3n\equiv 1\pmod{10}$ and so $f(n)=f(3n)-f(3)=0$.
*If $n\equiv 9\pmod{10}$ then $3n\equiv 7\pmod{10}$ and so $f(n)=f(3n)-f(3)=0$.
From $f(2)\ge 0$ and $f(5)\ge 0$ and $f(2)+f(5)=f(10)=0$, we obtain $f(2)=f(5)=0$.
Now let $n\in\Bbb N$ be arbitrary. Write $n=2^r5^su$ with $\gcd(u,10)=1$. Then $u\equiv 1,3,7,9\pmod{10}$ and hence $f(n)=rf(2)+sf(5)+f(u)=0$.
|
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|
How to formally prove that we cannot find a polynomial in $\textbf Z[x]$ with degree $2$ with such a root? I am trying to find the kernel of the map from $\textbf Z[x]$ to $\textbf C$. The map is evaluating at $\sqrt 2 + \sqrt 3$.
A solution says that we cannot find polynomials of degree $2$ or $3$ that has such a root. So it skips the procedure of trying degree $2$ and $3$. And the final solution is the ideal in $\textbf Z[x]$ generated by $$x^4 − 10x^2 + 1 = (x − \sqrt 2 − \sqrt 3)(x − \sqrt 2 + \sqrt 3)(x + \sqrt 2 − \sqrt 3)(x + \sqrt 2 + \sqrt 3)$$
A solution says there are $\sqrt 6$'s in $(\sqrt 2 + \sqrt 3)^2$ and $(\sqrt 2 + \sqrt 3)^3$. So we cannot find polynomials with root $\sqrt 2 + \sqrt 3$ of degree $2$ or $3$. And that is what I am confused about: How does it imply the fact?
Thanks in advance!
|
Proceed by contradiction.
Assume that there exist integer $a,b,c$ such that
$$ax^2+bx+c=0$$
with $x=\sqrt{2}+\sqrt{3}$.
$$a(5+2\sqrt{6})+b(\sqrt{2}+\sqrt{3})+c=0$$
Note that $1,\sqrt{2},\sqrt{3},$ and $\sqrt{6}$ are linearly independent, which means that there exist no integers $n_1,n_2,n_3,n_4$, not all $0$, such that
$$n_1+n_2\sqrt{2}+n_3\sqrt{3}+n_4\sqrt{6}$$
(A proof of this fact can be found here.) However, this means that
$$(5a+c)+b\sqrt{2}+c\sqrt{3}+(2a)\sqrt{6} \neq 0$$
which is a contradiction.
|
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|
Prove that $\binom{2n}{n}>\frac{4n}{n+1}\forall \; n\geq 2, n\in \mathbb{N}$
Prove that $$\binom{2n}{n}>\frac{4n}{n+1}\forall \; n\geq 2, n\in \mathbb{N}$$
$\bf{My\; Try::}$ Using $$(1+x)^n = \binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+........+\binom{n}{n}x^n$$
and $$(x+1)^n = \binom{n}{0}x^n+\binom{n}{1}x^{n-1}+\binom{n}{2}x^{n-1}+........+\binom{n}{n}$$
Now Coefficients of $x^n$ in $\displaystyle (1+x)^{2n} = \binom{2n}{n} = \binom{n}{0}^2+\binom{n}{1}^2+\binom{n}{2}^2+.....+\binom{n}{n}^2$
Now Using $\bf{Cauchy\; Schwarz}$ Inequality
$$\left[\binom{n}{0}^2+\binom{n}{1}^2+\binom{n}{2}^2+.....+\binom{n}{n}^2\right]\cdot \left[1^2+1^2+....+1^2\right]\geq \left[\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+.....+\binom{n}{n}\right]^2=2^{2n}=4^n>4n\forall n\geq 2,n\in \mathbb{N}$$
So $$\binom{2n}{n} = \binom{n}{0}^2+\binom{n}{1}^2+\binom{n}{2}^2+.....+\binom{n}{n}^2>\frac{4n}{n+1}\forall n\geq 2,n\in \mathbb{N}$$
My question is , Is my proof is right, If not then how can i solve it.
Also plz explain any shorter way, Thanks
|
$$\binom{2n}n=\frac{n+1}{1}\frac{n+2}{2}\cdots\frac{2n-1}{n-1}\frac{2n}{n}\ge2^n>\frac{4n}{n+1}.$$
|
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|
If $a$, $b$ and $c$ are three positive real numbers, prove that $\frac{a^2+1}{b+c}+\frac{b^2+1}{c+a}+\frac{c^2+1}{a+b} ≥ 3$ If $a$, $b$ and $c$ are three positive real numbers, prove that $$\frac{a^2+1}{b+c}+\frac{b^2+1}{c+a}+\frac{c^2+1}{a+b} ≥ 3$$
I think we are supposed to use the AM-GM inequality here; or maybe this could even be solved using the $T_2$'s Lemma, although I'm a bit skeptical about that.
|
After using AM-GM we need to prove that
$$(a^2+1)(b^2+1)(c^2+1)\geq(a+b)(a+c)(b+c)$$
which follows from C-S: $(a^2+1)(1+b^2)\geq(a+b)^2$.
Done!
|
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|
Forming basis in Linear Algebra Looking for some help with the following question. For which value or values of $k$ do the vectors below form a basis of $ \Bbb R^4$.
$\begin{bmatrix}
1&0&0&2
\\0&1&0&3
\\0&0&1&4
\\2&3&4&k
\end{bmatrix}$
My thinking is that the columns of this matrix are linearly independent if and only if the matrix is invertible If and only if the determinant of the matrix $\ne0$
So looking for some help with this question. Would I have to calculate the determinant of this matrix to see what values of $k$ would form a basis of $ \Bbb R^4$. And if so wouldn't there be more than one possible value of $k$ that would make this true? Looking for some help.
|
Hint:
$$ \begin{vmatrix}
1&0&0&2
\\0&1&0&3
\\0&0&1&4
\\2&3&4&k
\end{vmatrix} = \begin{vmatrix}
1&0&0&2
\\0&1&0&3
\\0&0&1&4
\\0&3&4&k-2(2)
\end{vmatrix} = \begin{vmatrix}
1&0&0&2
\\0&1&0&3
\\0&0&1&4
\\0&0&4&k-2(2)-3(3)
\end{vmatrix}$$
Are you able to finish the working?
|
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|
Integrate $\int \frac{\sqrt{x^2-8x+5}}{x}dx$ I've been trying this one for days and I can't seem to get it.
I tried:
$$\int \frac{\sqrt{x^2-8x+5}}{x}dx = \int \frac{\sqrt{\left(x-4\right)^2-11}}{x}dx$$
and set
$$x-4 = \sqrt{11}\sec\theta $$
so
$$dx=\sqrt{11}\sec\theta \tan\theta d\theta $$
and I had
$$\int \frac{\sqrt{x^2-8x+5}}{x}dx = \int \:\frac{11\sec\theta \tan^2\theta }{\sqrt{11}\sec\theta \:+\:4}d\theta $$
and I got stuck right there.
Any ideas?
Thanks!
|
Another possibility is to do an Euler substitution.
Let $$y=\sqrt{x^2-8x+5}-x$$
Then $$(y+x)^2=x^2-8x+5$$ and one solves to get
$$x=\frac{5-y^2}{2(y+4)}$$
That implies that
$$\sqrt{x^2-8x+5}=y+x=y+\frac{5-y^2}{2(y+4)}$$
Thus the integral reduces to a rational function. The denominator is not that bad, I get $(y+4)^2(5-y^2)$. I dont feel like calculating the rest but the remainder should be straightforward partial fraction decomposition. Good Luck !
|
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|
Solve $x=\sqrt{x-x^{-1}}+\sqrt{1-x^{-1}}$ Find all real numbers of $x$ such that $$x=\left(x-\frac 1x\right)^{\frac 12}+\left(1-\frac 1x\right)^{\frac 12}$$
My Attempt: Square both sides to get$$x^2=x-\frac 1x+1-\frac 1x+2\sqrt{\left(x-\frac 1x\right)\left(1-\frac 1x\right)}$$
Then move all the terms to the left side except for the square root and then square the equation. But I predict the terms to get really ugly really fast.
So I'm wondering if there is an elegant way of obtaining the solution without going through much of a hassle.
|
Multiply the original equation
$$
(x-x^{-1})^{1/2}+(1-x^{-1})^{1/2}=x\tag{1}
$$
by $(x-x^{-1})^{1/2}-(1-x^{-1})^{1/2}$ to get
$$
x-1=x((x-x^{-1})^{1/2}-(1-x^{-1})^{1/2})
$$
That is
$$
(x-x^{-1})^{1/2}-(1-x^{-1})^{1/2}=1-x^{-1}\tag{2}
$$
Denote $a=(x-x^{-1})^{1/2}$ and $b=(1-x^{-1})^{1/2}$. Then $(1)$ and $(2)$ can be rewritten as
$$
a+b=x\tag{3}
$$
$$
a-b=b^2\tag{4}
$$
From the very definition of $a$ and $b$ we have $a^2-b^2=x-1$. With $(3)$ we get $$
a^2-b^2=a+b-1\tag{5}
$$
Subtracting $(4)$ from $(5)$ we obtain $a^2-a=a-1$. Hence $a=1$. The rest is clear.
|
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|
Divisibility Proof $8\mid (x^2 - y^2)$ for $x$ and $y$ odd $x,y \in\Bbb Z$. Prove that if $x$ and $y$ are both odd, then $8\mid (x^2 - y^2)$.
My Proof Starts:
Assume $x$ and $y$ are both odd. So, $x = 2k + 1$ and $y = 2l +1$ for some integers $k$ and $l$. Thus,
\begin{align}
x^2 - y^2 &= (2k + 1)^2 - (2l + 1)^2 \\
&= 4k^2 + 4k + 1 - (4l^2 + 4l + 1) \\
&= 4k^2 + 4k - 4l^2 - 4l
\end{align}
My two concerns:
1) Is this correct so far?
2) How would I deal with the “$8\;\mid$” part?
|
if $K$and $l$ are even, then $K=2K_1$ and $l=2K_2$
then $4K^2+4K-4l^2-4l=16K_1^2+8K_1-16K_2^2-8K_2$ which is clearly divisble by 8
now if $K$and $l$ are odd, then $K=2K_1+1$ and $l=2K_2+1$
then $4K^2+4K-4l^2-4l=4(4K_1^2+4K_1+1)+8K_1+4 -4(4K_2^2+4K_2+1)-8K_2+4= 16K_1^2+16K_1+8K_1+8-16K_2^2-16K_2-8K_2-8$ which is clearly divisble by 8
if $K$ is even and $l$ is odd or $K$ is odd and $l$ is even, it is the same calculation, try it!
|
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|
Systematic way to find the answers of such question Q.What is the largest perfect square that divides:
$2014^3-2013^3+2012^3-2011^3+\ldots+2^3-1^3$.
My efforts:
$2014^3-2013^3=(2014-2013)(2014^2+2013^2+2014 \cdot2013)=2014^2+2013^2+2014*2013$
lly,
$2014^3-2013^3+2012^3-2011^3+\ldots+2^3-1^3=(2014^2+2013^2+2012^2+2011^2+\ldots+2^2+1^2)+(2014 \cdot2013+2013 \cdot2012\ldots+2 \cdot1)$
This is inconclusive.
So I am looking for systematic way to find solutions of such problem.Any hint is appreciated.
|
$$ \sum_{k=1}^n \left((2k)^3 - (2k-1)^3\right) = \sum_{k=1}^n \left(12 k^2 - 6 k + 1\right) = (4n+3) n^2$$
In this case $n=1007$ and $4n+3 = 29 \times 139$ which is coprime to $n$ and squarefree. Thus the answer is $n^2 = 1007^2 = 1014049$.
|
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|
Evaluating $\arccos(\cos\frac{15\pi }{4})$ I have a problem with understanding of this exercise:
$\arccos(\cos\frac{15\pi }{4})= ?$
$\cos(\frac{15\pi}{4}-2\pi)=\cos(\frac{7\pi}{4})$
$\cos(\frac{7\pi}{4}+\pi)= -\cos(\frac{3\pi}{4})$
then $\arccos(-\cos\frac{3\pi}{4})$
All above I understand pretty clearly. We did it at school, too and this is a solution. I just don`t understand where $-x$ comes from. It can possibly be $x=\cos(\frac{3\pi}{4})$ but what about the $-\frac{\pi}{2}$ then ??
$\arccos(-x)-\frac{\pi}{2} = -(\arccos x- \frac{\pi}{2})$
$\arccos(-x)=\pi-\arccos x$
$\arccos(\cos\frac{15\pi}{4})=\arccos(-\cos\frac{3\pi}{4})=\pi-\arccos(\cos\frac{3\pi}{4})=\pi-\frac{3\pi}{4}=\frac{\pi}{4}$
or do you know any other method how to solve this ??
Thank you for your time.
|
Here's one way to do it
$$\arccos\left(\cos\left(\frac{15}{4}\pi\right)\right)$$
$$=\arccos\left(\cos\left(\pi+2\pi+\frac{3}{4}\pi\right)\right)$$
$$=\arccos\left(\cos\left(\pi+\frac{3}{4}\pi\right)\right)$$
$$=\arccos\left(-\cos\left(\frac{3}{4}\pi\right)\right)$$
$$=\pi-\arccos\left(\cos\left(\frac{3}{4}\pi\right)\right)$$
$$=\pi-\frac{3}{4}\pi=\frac{\pi}{4}$$
|
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|
Find the limit of the following expression: $$\lim_{n\to\infty}\frac {1-\frac {1}{2} + \frac {1}{3} -\frac {1}{4}+ ... + \frac {1}{2n-1}-\frac{1}{2n}}{\frac {1}{n+1} + \frac {1}{n+2} + \frac {1}{n+3} + ... + \frac {1}{2n}}$$
I can express the value of the geometric sum of ${\frac {1}{2} + \frac {1}{4}+...+\frac {1}{2n}}$ but the others are ahead of me.
Putting both fraction parts under a common denominator makes that part tidy but the numerator seems to get way too complicated, which makes me think there is some simple way to do this.
|
In the denominator we have:
$$\sum_{i=1}^{n} \frac{1}{n+i}$$
And as $n \to \infty$ this is:
$$\lim_{n \to \infty} \sum_{i=1}^{n} \frac{1}{n+i}=\lim_{n \to \infty} \sum_{i=1}^{n} \frac{1}{1+(0+i \frac{1-0}{n})}\frac{1-0}{n}=\int_{0}^{1} \frac{1}{1+x} dx=\ln (1+1)-\ln(1+0)=\ln (2)$$
For the numerator as $n \to \infty$, considering the geometric series $\sum_{n=0}^{\infty} u^n=\frac{1}{1-u}$ for $|u|<1$, substituting in $-x$ for $u$ we have:
$$\frac{1}{1-(-x)}=\sum_{n=0}^{\infty} (-1)^nx^n$$
Then integrate both sides to get
$$\ln(1+x)=C+\sum_{n=0}^{\infty} (-1)^{n}\frac{x^{n+1}}{n+1}=C+\sum_{n=1}^{\infty} (-1)^{n+1}\frac{x^n}{n}$$
Let $x=0 \implies C=0$, so:
$$\ln (1+x)=\sum_{n=1}^{\infty} (-1)^{n+1}\frac{x^n}{n}$$
Because the series converges for $x=1$ by the alternating series test, we are justified in letting $x \to 1^-$ to get:
$$\ln (2)=\sum_{n=1}^{\infty} (-1)^n\frac{1}{n}$$
Hence the limit is:
$$\frac{\ln 2}{\ln 2}=1$$
|
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|
Solving $x+x^3=5$ without using the cubic equation. In lessons, I get quite bored and recently throughout these lessons I have been trying to solve for x in:
$$x+x^3=5$$
I've figured out how to do it for squares using the quadratic equation, but the cubic equation looks so dauntingly massive it actually makes my bladder hurt.
So, is there a way to figure this out using a different process, and better so for $x^n$.
Danke Chien
|
A trick for solving this equation goes back to Viète (but is not much different from Tartaglia's and Cardano's method).
Set $x=y-\dfrac{1}{3y}$. Then
$$
x^3+x=y^3-y+\frac{1}{3y}-\frac{1}{27y^3}+y-\frac{1}{3y}=
y^3-\frac{1}{27y^3}
$$
so your equation becomes, after setting $z=y^3$,
$$
z-\frac{1}{27z}=5
$$
This becomes a quadratic:
$$
27z^2-135z-1=0
$$
and so
$$
z=\frac{135\pm\sqrt{18333}}{54}=\frac{45\pm\sqrt{2037}}{18}
$$
Let's take the root with $+$; then
$$
y=\sqrt[3]{\frac{45+\sqrt{2037}}{18}}
$$
and so
$$
x=\sqrt[3]{\frac{45+\sqrt{2037}}{18}}-
\frac{1}{3}\sqrt[3]{\frac{18}{45+\sqrt{2037}}}\approx 1.51598
$$
You can check that taking the other root yields the same solution.
In case the equation is $x^3+px=q$, the substitution is
$$
x=y-\frac{p}{3y}
$$
because in this case we get
$$
x^3+px=y^3-py+\frac{p^2}{3y}-\frac{p^3}{27y^3}+py-\frac{p^2}{3y}
=y^3-\frac{p^3}{27y^3}
$$
Setting $z=y^3$, we get the quadratic
$$
27z^2-27qz-p^3=0
$$
which has real roots if and only if
$$
27^2q^2+4\cdot27p^3\ge0
$$
that can be written
$$
\frac{q^2}{4}+\frac{p^3}{27}\ge0
$$
It turns out that the equation $x^3+px=q$ has just one real root when $q^2/4+p^3/27>0$ and a double (or triple) root when $q^2/4+p^3/27=0$.
What happens when $q^2/4+p^3/27<0$? That's the big question: it is the case when the equation has three distinct roots. The method still works, but it necessarily involves the complex numbers, if an algebraic expression of the roots is needed.
Viète devised a trigonometric method for getting approximate values of the roots in the case $q^2/4+p^3/27<0$.
OK, let's try another one: $x^3+x=2$. The quadratic becomes
$$
27z^2-54z-1=0
$$
which has a root
$$
z=\frac{9+\sqrt{84}}{9}
$$
and so
$$
x=
\sqrt[3]{\frac{9+\sqrt{84}}{9}}-
\frac{1}{3}\sqrt[3]{\frac{9}{9+\sqrt{84}}}
$$
which is just a complicated way to write $1$.
|
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|
How do I find an analytical solution of $x x' + \frac 3 2 (x')^2 + A \frac {x'} x = W$? I am trying to solve the Rayleigh-Plesset equation of a bubble made of super-critical fluid, and its radius varies as a function of time. A generalised form of the equation I got is:
$$x x' + \frac 3 2 (x')^2 + A \frac {x'} x = W .$$
Here $A$ and $W$ are constants. How do I go about solving this system?
|
$$x x' + \frac 3 2 (x')^2 + A \frac {x'} x = W .$$
We observe that $x(t)$ is an even function. This draw to the change of function :
$$X=x^2 \quad\to\quad dX=2xdx \quad\to\quad x'=\frac{dx}{dt}=\frac{dx}{dX}\frac{dX}{dt}=\frac{1}{2x}\frac{dX}{dt}$$
$$x \left(\frac{1}{2x}\frac{dX}{dt} \right) + \frac 3 2 \left(\frac{1}{2x}\frac{dX}{dt} \right)^2 + \frac {A}{x} \left(\frac{1}{2x}\frac{dX}{dt} \right)= W .$$
$$ 3 \left(\frac{dX}{dt} \right)^2 + 4(X+A)\frac{dX}{dt} - 8WX =0 $$
$$\frac{dX}{dt}=\frac{2}{3}\left(X+A\pm\sqrt{(X+A)^2+6WX } \right)$$
$$t=\frac{3}{2}\int \frac{dX}{X+A\pm\sqrt{(X+A)^2+6WX } }$$
$$t=t_0+\frac{3}{2}\int_{x_0^2}^{x^2} \frac{dX}{X+A\pm\sqrt{(X+A)^2+6WX } }$$
This will give $t(x)$ on form of an huge formula. It's doubtful that the inverse function $x(t)$ could be expressed on closed form.
|
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|
if $abc=1$,show that $a^3+b^3+c^3+\frac{256}{(a+1)(b+1)(c+1)}\ge 35$
Let $a,b,c>0,abc=1$ show that
$$a^3+b^3+c^3+\dfrac{256}{(a+1)(b+1)(c+1)}\ge 35\tag{1}$$
iff $a=b=c=1$
I know use AM-GM inequality
$$a^3+b^3+c^3\ge 3abc=3$$
and
$$(a+1)(b+1)(c+1)\ge 2\sqrt{a}\cdot 2\sqrt{b}\cdot 2\sqrt{c}=8$$
In this way, will lead to inequality reverse, but $(1)$ seem is right,so How to prove this inequality
|
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that $f(u)\geq0$, where
$$f(u)=27u^3-27uv^2+\frac{256w^5}{2w^2+3uw+3v^2}-32w^3$$
But by AM-GM $f'(u)=81u^2-27v^2-\frac{768w^6}{(2w^2+3uw+3v^2)^2}\geq81u^2-27v^2-12w^2>0$,
which says that $f$ is an increasing function.
Thus, it's enough to prove our inequality for a minimal value of $u$.
$a$, $b$ and $c$ are positive roots of the equation $x^3-3ux^2+3v^2x-w^3=0$ or
$3ux^2=x^3+3v^2x-w^3$, which says that
the parabola $y=3ux^2$ and the graph of $y=x^3+3v^2x-w^3$ have three common points.
Easy to see (draw it!) that $u$ gets a minimal value, when the parabola touches to the graph of $y=x^3+3v^2x-w^3$, which happens for equality case of two variables.
Id est, it remains to prove our inequality for $b=a$ and $c=\frac{1}{a^2}$, which gives
$(a-1)^2(2a^{11}+8a^{10}+18a^9-3a^8-92a^7+5a^6+32a^5+24a^4+16a^3+9a^2+4a+1)\geq0$,
which is true and smooth.
Done!
|
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|
Find minimal value $ \sqrt {{x}^{2}-5\,x+25}+\sqrt {{x}^{2}-12\,\sqrt {3}x+144}$ without derivatives. Find minimal value of $ \sqrt {{x}^{2}-5\,x+25}+\sqrt {{x}^{2}-12\,\sqrt {3}x+144}$ without using the derivatives and without the formula for the distance between two points.
By using the derivatives I have found that the minimal value is $13$ at $$ x=\frac{40}{23}(12-5\sqrt{3}).$$
|
Note that the expression in question, $f(x) = \sqrt {{x}^{2}-5\,x+25}+\sqrt {{x}^{2}-12\,\sqrt {3}x+144}$, can be written as follows:
$$
f(x) = \sqrt {(a(x))^2 + (b(x))^2}+\sqrt {(a(x))^2 + (13 - b(x))^2}
$$
where
$$a(x) =\frac{x (- 5\sqrt3 - 12) + 120}{26} $$
$$b(x) =\frac{x (12\sqrt3 -5) + 50}{26} $$
Now from $(a(x))^2 \ge 0$ follows
$$
f(x) \ge \sqrt {(b(x))^2}+\sqrt { (13 - b(x))^2} = 13
$$
but also, for $a(x) = 0$ we have $f(x) = 13$. So indeed $13$ is the smallest value that $f(x)$ can attain. Again, this value is reached for $a(x) = 0$, i.e. $x = \frac{120}{5\sqrt3 + 12} = \frac{40}{23}(12-5\sqrt{3})$ as already given by the OP. This completes the proof.
I can give a general construction method of how to arrive at the functions $a(x)$ and $b(x)$. $\quad \Box$
|
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|
A better way to evaluate a certain determinant
Question Statement:-
Evaluate the determinant:
$$\begin{vmatrix}
1^2 & 2^2 & 3^2 \\
2^2 & 3^2 & 4^2 \\
3^2 & 4^2 & 5^2 \\
\end{vmatrix}$$
My Solution:-
$$
\begin{align}
\begin{vmatrix}
1^2 & 2^2 & 3^2 \\
2^2 & 3^2 & 4^2 \\
3^2 & 4^2 & 5^2 \\
\end{vmatrix} &=
(1^2\times2^2\times3^2)\begin{vmatrix}
1 & 1 & 1 \\
2^2 & \left(\dfrac{3}{2}\right)^2 & \left(\dfrac{4}{3}\right)^2 \\
3^2 & \left(\dfrac{4}{2}\right)^2 & \left(\dfrac{5}{3}\right)^2 \\
\end{vmatrix}&\left[\begin{array}{11}C_1\rightarrow\dfrac{C_1}{1} \\
C_2\rightarrow\dfrac{C_2}{2^2}\\
C_3\rightarrow\dfrac{C_3}{3^2}\end{array}\right]\\
&=
(1^2\times2^2\times3^2)\begin{vmatrix}
1 & 0 & 0 \\
2^2 & \left(\dfrac{3}{2}\right)^2-2^2 & \left(\dfrac{4}{3}\right)^2-2^2 \\
3^2 & 2^2-3^2 & \left(\dfrac{5}{3}\right)^2-3^2 \\
\end{vmatrix} &\left[\begin{array}{11}C_2\rightarrow C_2-C_1 \\
C_3\rightarrow C_3-C_1\end{array}\right]\\
&=
(1^2\times2^2\times3^2)
\begin{vmatrix}
1 & 0 & 0 \\
2^2 & 2^2-\left(\dfrac{3}{2}\right)^2 & 2^2-\left(\dfrac{4}{3}\right)^2 \\
3^2 & 3^2-2^2 & 3^2-\left(\dfrac{5}{3}\right)^2 \\
\end{vmatrix}\\
&=(1^2\times2^2\times3^2)
\begin{vmatrix}
1 & 0 & 0 \\
2^2 & \dfrac{7}{4} & \dfrac{20}{9} \\
3^2 & 5 & \dfrac{56}{9} \\
\end{vmatrix}\\
&=(1^2\times2^2)
\begin{vmatrix}
1 & 0 & 0 \\
2^2 & \dfrac{7}{4} & 20 \\
3^2 & 5 & 56 \\
\end{vmatrix}\\
&=4\times(-2)\\
&=-8
\end{align}
$$
As you can see, my solution is a not a very promising one. If I encounter such questions again, so would you please suggest a better method which doesn't include this ridiculous amount of calculations.
|
If we consider
$$ f(x)=\det\begin{pmatrix}x^2 & (x+1)^2 & (x+2)^2 \\ (x+1)^2 & (x+2)^2 & (x+3)^2 \\ (x+2)^2 & (x+3)^2 & (x+4)^2 \end{pmatrix}$$
we have:
$$\scriptstyle f'(x) = \det\begin{pmatrix}2x & (x+1)^2 & (x+2)^2 \\ 2x+2 & (x+2)^2 & (x+3)^2 \\ 2x+4 & (x+3)^2 & (x+4)^2 \end{pmatrix}+\det\begin{pmatrix}x^2 & 2x+2 & (x+2)^2 \\ (x+1)^2 & 2x+4 & (x+3)^2 \\ (x+2)^2 & 2x+6 & (x+4)^2 \end{pmatrix}+\det\begin{pmatrix}x^2 & (x+1)^2 & 2x+4 \\ (x+1)^2 & (x+2)^2 & 2x+6 \\ (x+2)^2 & (x+3)^2 & 2x+8 \end{pmatrix} $$
and $f'(x)=0$ by Gaussian elimination. It follows that $f(x)$ is a constant function and
$$ f(1)=f(-2)=\det\begin{pmatrix}4 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 4\end{pmatrix}=-4-4=\color{red}{-8}.$$
|
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|
Solving for the roots of a sextic I was solving the equation $3(1+x^2+x^3)^2=(2+x)^4$ for $x$, and after expanding it out, I got $$3x^6+6x^5+2x^4-2x^3-18x^2-32x-13=0\tag{1}$$
which should be solvable because it has a Galois group of order $72$. But since it's a degree six, I have no method for solving this.
I have attempted to factor it into two cubics, but the condition wasn't met, so I can't factor it. This polynomial has (maybe?) irrational roots so the rational root theorem won't work.
|
From $3(1+x^2+x^3)^2=(2+x)^4$, we can move all terms to one side and use $a^2-b^2=(a+b)(a-b)$ to get\begin{align*} & (\sqrt3(1+x^2+x^3)-(2+x)^2)(\sqrt3(1+x^2+x^3)+(2+x)^2)=0\tag1\\ & (g\sqrt3+(\sqrt3-1)g^2-4g+\sqrt3-4)(g\sqrt3+(\sqrt3+1)g^2+4g+\sqrt3+4)=0\tag2\end{align*}
And solving for the roots of the cubic gives all possible values of $g$.
|
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|
Basis for the annihilator Let the field $F$ be given by $F=\mathbb{Z}/5\mathbb{Z}$ and let $V=F^3$. Let $S$ be a subspace of $V$ spanned by $v_1=(1,2,3),v_2=(2,1,2)$. I'm now asked to find a basis for $S^0.$
I've tried solving this and I get $(-1/3,-4/3,1)$. Is it right that the $S^0$ has dimension $1$?
|
Since $(v_1,v_2)$ are linearly independent vectors, we have $\dim S = 2$ and indeed
$$ \dim S^{0} = \dim V - \dim S = 3 - 2 = 1. $$
To check your answer, note that we indeed have
$$ \left( -\frac{1}{3}, -\frac{4}{3}, 1 \right) \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = -\frac{1}{3} - \frac{8}{3} + 3 = 0, \\
\left( -\frac{1}{3}, -\frac{4}{3}, 1 \right) \begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix} = -\frac{2}{3} -\frac{4}{3} + 2 = 0 $$
and so your vector belongs to $S^0$ and spans it by dimensional reasons.
BTW, since you work over $\mathbb{Z}/5\mathbb{Z}$, the $\frac{1}{3}$ stands for the multiplicative inverse of $3$ which is (or, more precisely, can be identified with the equivalence class of) $2$ and so $-\frac{1}{3}$ is identified with $3$. Thus, your spanning vector of $S^0$ can also be written as
$$ (3, 2, 1). $$
|
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|
Solve the differential equation $\frac{dy}{dx} + 3yx = 0$ for the values $x = 0$ when $y = 1$ - Solution Review
Solve the differential equation $\frac{dy}{dx} + 3yx = 0$; $x = 0$ when $y = 1$.
I solved this DE using the integration factor method. However, online calculators are giving me a different answer, where they instead used the separation of variables method.
Please review my solution and indicate if/where my reasoning is false, why it is false, how to fix it, and what the correct reasoning should be. Thank you.
My solution is as follows.
$\dfrac{dy}{dx} + 3xy = 0$
$e^{ \int 3x } dx = e^{ \frac{3x^2}{2} }$
$ \dfrac{dy}{dx} e^{ \frac{3x^2}{2} } + 3yxe^{ \frac{3x^2}{2} } = 0$
$ \dfrac{d}{dx} \left( y e^{ \frac{3x^2}{2}} \right) = \dfrac{dy}{dx} e^{ \frac{3x^2}{2} } + 3xye^{ \frac{3x^2}{2}} $
$ \therefore \dfrac{d}{dx} \left( y e^{ \frac{3x^2}{2}} \right) = 0$
$ \displaystyle\int \dfrac{d}{dx} \left( y e^{ \frac{3x^2}{2}} \right) dx + C = 0$
$ \Rightarrow y e^{ \frac{3x^2}{2}} + C = 0$
We want to solve for y. We know that $x = 0$.
$ \therefore ye^0 + C = 0$
$ \Rightarrow y = -C$
|
For the Initial Value Problem (IVP) $y′+3xy=0, ~y(0)=1$ (which you are most probably trying to solve) you are correct in all your steps above, except the last part: substitute $x=0, ~y=1$ and get $C=−1$, so the solution to the IVP is $y(x)=e^{−3x^2/2}.$
|
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|
Proof by induction that $4^3+4^4+4^5+⋅⋅⋅+4^n = \frac{4(4^n-16)}{3}$ I am stuck on this problem for my discrete math class.
Prove the equation by induction for all integers greater than or equal to $3$:
$$4^3+4^4+4^5+⋅⋅⋅+4^n = \frac{4(4^n-16)}{3}.$$
I know that base case $n=3$:
$4^3=64$ as well as $4(4^3-16)/3 = 64$
My confusion is on the induction step where:
$4^3+4^4+4^5+⋅⋅⋅+4^n+4^{(n+1)} = 4(4^{(n+1)}-16)/3$.
I don't know what to do next.
|
Hint:
$$\color{red}{4^3+\cdots 4^n}+4^{n+1}=\color{red}{\frac{4(4^n-16)}{3}}+4\cdot4^{n}=\left(\frac43+4\right)4^n-4\cdot\frac {16}3$$
|
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|
What is the sum of imaginary roots of this equation? What is the sum of imaginary roots of equation:
$$x^3+3x^2+3x+3$$
Here is my attempt:
$$x^3+3x^2+3x+3=(x^3+3x^2+3x+1)+2=(x+1)^3+2$$
Since $f(x')=0$ if x' is a root of $f(x)$
$$(x'+1)^3+2=0$$
$$(x'+1)^3=-2$$
$$x'=-1+\sqrt[3]{-2}$$
$$x'=-1+\sqrt[3]{2}\sqrt[3]{-1}$$........(i)
Again,$$-1=cos(\pi)+isin(\pi)=cos(-\pi)+isin(-\pi)$$
By Euler's formula,
$$-1=e^{i\pi}=e^{-i\pi}$$
$$\sqrt[3]{-1}=e^{i\frac{\pi}{3}}= e^{i\frac{-\pi}{3}}$$
Substuting in equation..(i)
$$x'=-1+\sqrt[3]{2}e^{i\frac{\pi}{3}},-1+\sqrt[3]{2}e^{i(\frac{-\pi}{3})},-1-\sqrt[3]{2}$$
Sum of imaginary roots is:
$$ -1+\sqrt[3]{2}e^{i\frac{\pi}{3}}+-1+\sqrt[3]{2}e^{i(\frac{-\pi}{3})}$$
$$=-1+\sqrt[3]{2}e^{i\frac{\pi}{3}}+-1+\sqrt[3]{2}e^{i(\frac{-\pi}{3})}$$
$$=-2+\sqrt[3]{2}(e^{i\frac{\pi}{3}}+e^{-i\frac{\pi}{3}})$$
$$=-2+\sqrt[3]{2}[[cos(\frac{\pi}{3})+isin(\frac{\pi}{3})]+cos(-\frac{\pi}{3})+isin(\frac{-\pi}{3})]=-2+\sqrt[3]{2}$$
It does not match with the given answer in the book.I am confused.
Any hint on my mistakes is appreciated.
|
So, the values of $x+1$ are $$\sqrt[3]2, \sqrt[3]2w,\sqrt[3]2w^2$$ where $w$ is a complex cube root of unity
So, the only real root of the given equation is $$-1+\sqrt[3]2$$
Now the sum of all roots is $$-\dfrac31$$
|
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|
Computing image under coordinate map How do I compute the image?
Compute the image of (1,2,3,4) under the coordinate map
$$ \begin{bmatrix}t^3&t^3 + t^2&t^3+t^2+t&t^3+t^2+t+1\end{bmatrix}: \Bbb F^4\to P_3.$$
|
I assume they expect you to do
\begin{align}
\begin{bmatrix}t^3&t^3 + t^2&t^3+t^2+t&t^3+t^2+t+1\end{bmatrix}\begin{bmatrix}1\\2\\3\\4\end{bmatrix}
&=t^3+2(t^3+t^2)+3(t^3+t^2+t)+4(t^3+t^2+t+1)
\end{align}
|
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|
Why doesn't the derivative of integration by trig substitution match the original function? For one of my assignments in Calc II, I had to solve ${\displaystyle\int} \frac{\sqrt{x^2-4}}{x} \text{ d}x$. By using Trig substitution where $x = 2 \text{ sec } \theta$, my solution was $\sqrt{x^2-4} - 2\text{ arcsec}\left(\frac{x}{2}\right) + C$. I tested my solution by graphing $\frac{\text{d}}{\text{d}x}\left[\sqrt{x^2-4} - 2\text{ arcsec}\left(\frac{x}{2}\right)\right]$, and it only matched the original function where $x \geq 2$. To match for $x \leq -2$, I had to use $\frac{\text{d}}{\text{d}x}\left[\sqrt{x^2-4} + 2\text{ arcsec}\left(\frac{x}{2}\right)\right]$.
Assuming I solved this correctly, what causes this discrepancy? How can we know if, when, and where this kind of thing will happen? Also, how would we notate the difference? Is it necessary to write the solution as a piecewise function, or does none of this matter anyway? I would think that, if it was because of the square root, the change in sign would be in front of it, but this seems to be because of the arc secant.
Edit
Since a comment mentioned using WolframAlpha, I decided to double-check through there, and they do have my same answer under the Step-by-Step Solution, although they then simplified it to arctan somehow. Graphing the derivative of their solution with arctan matches the original function even less, with both a different C and shape. For explicitness, here's my work:
$x = 2\sec \theta \Rightarrow \text{d}x = 2 \sec\theta \tan\theta$
$\int \frac{\sqrt{4 \sec^2\theta - 4}}{2 \sec\theta}(2 \sec\theta \tan\theta \text{ d}\theta) = \int 2 \sqrt{\sec^2 \theta-1} \tan\theta \text{ d}\theta = 2 \int \tan^2 \theta \text{ d}\theta$
With trig identity replacement: $2 \int \sec^2 \theta \text{ d}\theta - 2 \int 1 \text{ d}\theta = 2 \tan\theta - 2\theta$
With replacement of $\theta$ back to its original value: $\sqrt{x^2 - 4} - 2\text{ arcsec} \left(\frac{x}{2}\right) + C$
Edit 2 I understand that the square root causes sign issues, but even switching to absolute value doesn't seem to have any effect in the graphing. Here's a link to everything graphed on desmos.
|
The trouble is when getting rid of the square root. In general we have $\sqrt{x^2} = |x|$, so
$$\sqrt{\sec^2\theta-1} = \sqrt{\tan^2\theta} = \lvert\tan\theta\rvert.$$
So the integrand is really
$$\lvert\tan\theta\rvert \tan \theta = \begin{cases}
\sec^2\theta - 1 & 0\le\theta<\pi/2,\\
-(\sec^2\theta - 1) & \pi/2<\theta<\pi\\
\end{cases}$$
(where you take $\theta = \operatorname{arcsec} (x/2)$ with $0 \le \theta < \pi$). When reversing the substitution there is also the issue of
$$\tan\theta = \begin{cases}
\sqrt{\sec^2\theta - 1} & 0\le\theta<\pi/2,\\
-\sqrt{\sec^2\theta - 1} & \pi/2<\theta<\pi\\
\end{cases}$$
so the correct integral has two parts:
$$\int \frac{\sqrt{x^2-4}}{x}\,dx = \begin{cases}
\sqrt{x^2-4} - 2 \operatorname{arcsec}(x/2) + C_1 & x \ge 2, \\
\sqrt{x^2-4} + 2 \operatorname{arcsec}(x/2) + C_2 & x \le -2. \\
\end{cases}$$
|
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|
Prove that: $\int_{0}^{1}\frac{dx}{\sqrt{1-x^4}}=\sum_{0}^{+\infty}\frac{C_{2n}^{n}}{4^n(4n+1)}$ Prove that: $\int_{0}^{1}\frac{dx}{\sqrt{1-x^4}}=\sum_{0}^{+\infty}\frac{C_{2n}^{n}}{4^n(4n+1)}$
Could you give me some hint? Thank for helping.
|
Using
$$\frac{1}{\sqrt{1-x^2}} = \sum_{n=0}^{\infty} \binom{2n}{n} \, \frac{x^{2n}}{4^{n}}$$
then
\begin{align}
\int \frac{dx}{\sqrt{1-x^4}} &= \int \left( \sum_{n=0}^{\infty} \binom{2n}{n} \, \frac{x^{4n}}{4^{n}} \right) \, dx \\
&= \sum_{n=0}^{\infty} \binom{2n}{n} \, \frac{x^{4n+1}}{4^{n} \, (4n+1)}
\end{align}
With limits the integral is:
\begin{align}
\int_{0}^{1} \frac{dx}{\sqrt{1-x^4}} &= \sum_{n=0}^{\infty} \binom{2n}{n} \, \frac{1}{4^{n} \, (4n+1)}
\end{align}
This integral is related to the Beta function and can be seen to have the compact form
$$\int_{0}^{1} \frac{dx}{\sqrt{1-x^4}} = \frac{\sqrt{\pi} \, \Gamma\left(\frac{5}{4}\right)}{\Gamma\left(\frac{3}{4}\right)}$$
and leads to
$$ \sum_{n=0}^{\infty} \binom{2n}{n} \, \frac{1}{4^{n} \, (4n+1)} = \frac{\sqrt{\pi} \, \Gamma\left(\frac{5}{4}\right)}{\Gamma\left(\frac{3}{4}\right)}. $$
|
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|
value of $a$ for which $25^x+(a+2)5^x-(a+3)<0$for at least one real $x$
Find the values of $a$ for which the inequality is satisfied for $25^x+(a+2)5^x-(a+3)<0$
for at least one real value of $x$
$\bf{My\; Try::}$ We can write it as $a(5^x-1)<-\left[25^x+2\cdot 5^x-3\right]$
So $$a < -\left(\frac{25^x-5^x+3\cdot 5^x-3}{5^x-1}\right) = -\frac{(5^x+3)(5^x-1)}{5^x-1} = -(5^x+3)$$
Now how can i solve after that help required, Thanks
|
Let $$(5^x)^2+(a+2)5^x-(a+3)=-y$$
As $5^x$ is real, we need the discriminant $$(a+2)^2-4(y-a-3)\ge0$$
$$4y\le a^2+8a+16=(a+4)^2$$
Now we need $y>0$ which will be true if $(a+4)^2>0$ which is true for $a+4\ne0\iff a\ne-4$
|
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|
Find the sum $\sum_{n=2}^{\infty} \frac{\binom n2}{4^n} ~~=~~ \frac{\binom 22}{16}+\frac{\binom 32}{64}+\frac{\binom 42}{256}+\cdots$. The sum
$$\sum_{n=2}^{\infty} \frac{\binom n2}{4^n} ~~=~~ \frac{\binom 22}{16}+\frac{\binom 32}{64}+\frac{\binom 42}{256}+\cdots$$
has a finite value. Use what you know about generating functions to determine that value.
How would I do this? My mind is blank. All solutions are highly appreciated!
|
Another approach is to use an arithmetico-geometric sequence, although this isn't using generating functions. To use generating functions, see the accepted answer.
Call the sum $S$, so $$S = \frac{1}{16}+\frac{3}{64}+\frac{6}{256}+\frac{10}{1024}+\dots$$
Divide by $4$, so
$$\frac{1}{4}S = \quad \frac{1}{64}+\frac{3}{256}+\frac{6}{1024}+\dots$$
Subtracting the second equation from the first gets $$\frac{3}{4}S = \frac{1}{16}+\frac{2}{64}+\frac{3}{256}+\frac{4}{1024}+\dots$$
We can use a similar process.
$$\frac{3}{16}S = \frac{1}{64}+\frac{2}{256}+\frac{3}{1024}+\dots$$
Subtracting the second equation from the first gets $$\frac{9}{16}S = \frac{1}{16}+\frac{1}{64}+\frac{1}{256}+\dots$$
The RHS is just an infinite geometric sequence. It equals $$\frac{\frac{1}{16}}{1-\frac{1}{4}} = \frac{1}{12}.$$
Now we have $$\frac{9}{16}S = \frac{1}{12},$$
and solving for $S$ gets $$S = \frac{1}{12}\cdot \frac{16}{9} = \boxed{\frac{4}{27}.}$$
|
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Polynomial of degree 5 such that $P(x)-1$ is divisible by $(x-1)^3$ and $P(x)$ is divisible by $x^3$
$P(x)$ is a polynomial of degree 5 such that $P(x)-1$ is divisible by $(x-1)^3$ and $P(x)$ is divisible by $x^3$. Find $P(x)$.
No idea where to start, would $P(x)$ be of the form $x^3(Ax^2+Bx+C)$?
|
Expanding $(x-1)^3$ we find that
$$P(x)=(Ax^2+Bx+C)(x^3-3x^2+3x-1)+1$$
$$=Ax^5+(B-3A)x^4+(C-3B+3A)x^3+(-3C+3B-A)x^2+(3C-B)x-C+1$$
Yet we also know that $P(x)$ is divisible by $x^3$ and is of the form
$$Dx^5+Ex^4+Fx^3$$
Comparing constant, linear and quadratic coefficients we have
$$1-C=0$$
$$3C-B=0$$
$$-3C+3B-A=0$$
and from these we get $C=1$, $B=3$ and $A=6$ in order. The remaining coefficients, $x^3$ to $x^5$, give us the coefficients of $P(x)$:
$$D=A=6$$
$$E=B-3A=-15$$
$$F=C-3B+3A=10$$
In conclusion:
$$P(x)=6x^5-15x^4+10x^3=(6x^2+3x+1)(x-1)^3+1$$
|
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"url": "https://math.stackexchange.com/questions/1985612",
"timestamp": "2023-03-29T00:00:00",
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|
Sequence of fractions that converges to $\sqrt{n}$ Begin with any two positive integers $a,b$. Let
$$s_1 = \frac{a}{b}$$
and recursively we define $s_{i+1}$ from $s_i$ as follows: If ${s_i}$ is the fraction $\frac{a'}{b'}$, then set
$$s_{i+1} = \frac{a'+2b'}{a'+b'}$$
(Notice that the actual value of the $s_{i+1}$ does not depend on the exact way we choose to represent $s_i$ in fractions)
How do we show that $s_i \to \sqrt{2}$ as $i\to \infty$?
In general, sending $\frac{a'}{b'}$ to $\frac{a'+nb'}{a'+b'}$ supposedly leads to $s_i \to \sqrt{n}$. How might we prove this?
|
Let $t_i$ denote the top part of the fraction and $\ell_i$ the bottom part. Then $t_0 = a$, $\ell_0 = b$ and
$$
\begin{pmatrix} t_{i+1} \\ \ell_{i+1} \end{pmatrix} = \begin{pmatrix} 1 & n \\ 1 & 1 \end{pmatrix}\begin{pmatrix} t_{i} \\ \ell_{i} \end{pmatrix}.
$$
You can then diagonalize the middle matrix
$$
\begin{pmatrix} 1 & n \\ 1 & 1 \end{pmatrix} = P^{-1}DP.
$$
It then follows that
\begin{align}
\begin{pmatrix} t_{i} \\ \ell_{i} \end{pmatrix} &= P^{-1}D^iP\begin{pmatrix} t_{0} \\ \ell_{0} \end{pmatrix} = P^{-1}\begin{pmatrix} d_{1}^{i} & 0 \\ 0 & d_{2}^{i} \end{pmatrix}P\begin{pmatrix} a \\ b \end{pmatrix}.
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1986099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Prove: $ \frac{1}{\sin 2x} + \frac{1}{\sin 4x } + \cdots + \frac{1 }{\sin 2^n x} = \cot x - \cot 2^n x $
Prove
$$ \frac{1}{\sin 2x} + \frac{1}{\sin 4x } + \cdots + \frac{1 }{\sin 2^n x} = \cot x - \cot 2^n x $$
where $n \in \mathbb{N}$ and $x$ not a multiple of $\frac{ \pi }{2^k} $ for any $k \in \mathbb{N}$.
My try. If $n=2$, we have
$$\begin{align}
\frac{1}{\sin 2x} + \frac{ 1}{\sin 4 x} &= \frac{1}{\sin 2x} + \frac{1}{2 \sin 2x \cos 2x } \\[6pt]
&= \frac{ 2 \cos 2x + 1 }{2 \sin 2x \cos 2x} \\[6pt]
&= \frac{2 \cos^2 x - 2 \sin^2 x + \cos^2 x + \sin^2 x}{2 \sin 2x \cos 2x} \\[6pt]
&= \frac{3 \cos^2 x - \sin^2 x}{2 \sin 2x \cos 2x}
\end{align}$$
but here I got stuck. I am on the right track? My goal is to ultimately use induction.
|
Skipping over the base case ($n=1$) for the moment, let's do the inductive step: If we take the equation
$${1\over\sin2x}+{1\over\sin4x}+\cdots+{1\over\sin2^nx}=\cot x-\cot2^nx$$
and replace $x$ with $2x$, we get
$${1\over\sin4x}+{1\over\sin8x}+\cdots+{1\over\sin2^{n+1}x}=\cot2x-\cot2^{n+1}x$$
Adding $1/\sin2x$ to both sides now gives
$${1\over\sin2x}+{1\over\sin4x}+\cdots+{1\over\sin2^{n+1}x}={1\over\sin2x}+\cot2x-\cot2^{n+1}x$$
so, to complete the inductive step, it suffices to prove
$${1\over\sin2x}+\cot2x=\cot x$$
But this is the base case, just rewritten with the $\cot2x$ moved to the left hand side! So let's prove it:
$$\begin{align}
{1\over\sin2x}+\cot2x&={1+\cos2x\over\sin2x}\\
&={1+(2\cos^2x-1)\over2\sin x\cos x}\\
&={\cos x\over\sin x}\\
&=\cot x
\end{align}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding limit involving a recursively defined sequence I have the following recursive sequence:
$$
S(n) = 1 + S(n-1) + \frac{2S(n-2)}n\\S(1) = 1\\S(0) = 0
$$
I've written a program that computes elements of $S(n)$ and seems to imply that:
$$
\lim_{n\to\infty} \frac{S(n) - S(n-1)}n = \frac{1 - e^{-2}}2
$$
I don't believe there is a closed form for $S(n)$ (bonus points if there is!), but I'm curious if there is some way to prove that the above limit is correct.
|
Hint: You could try the following. Multiply the equation with $x^n$ and sum over $n=2$ to $n= \infty$.
$$\sum_{n=2}^{\infty}S(n)x^n=\sum_{n=2}^{\infty}x^n+\sum_{n=2}^{\infty}S(n-1)x^n+2\sum_{n=2}^{\infty}\frac{S(n-2)}{n}x^n$$
Now, switch the index for the second and third sum on the right hand side to get:
$$\sum_{n=2}^{\infty}S(n)x^n=\sum_{n=2}^{\infty}x^n+\sum_{n=1}^{\infty}S(n)x^{n+1}+2\sum_{n=0}^{\infty}\frac{S(n)}{n+2}x^{n+2}.$$
Now, split the sums so that they all start from $n=2$
$$\sum_{n=2}^{\infty}S(n)x^n=\sum_{n=2}^{\infty}x^n+S(1)x^2+\sum_{n=2}^{\infty}S(n)x^{n+1}+2\sum_{n=2}^{\infty}\frac{S(n)}{n+2}x^{n+2}+2\frac{S(0)}{2}x^2+2\frac{S(1)}{3}x^3$$
Get the sums on the right hand side:
$$(1-x-2x^2)\sum_{n=2}^{\infty}S(n)x^n=\sum_{n=2}^{\infty}x^n+S(1)x^2+2\frac{S(0)}{2}x^2+2\frac{S(1)}{3}x^3.$$
Now divide by $(1-x-2x^2)$:
$$\sum_{n=2}^{\infty}S(n)x^n=\left(\sum_{n=2}^{\infty}x^n+S(1)x^2+2\frac{S(0)}{2}x^2+2\frac{S(1)}{3}x^3\right)/(1-x-2x^2).$$
Find the talyor expansion of the right hand side and compare coefficients. Maybe you might get a closed formula for your $S(n)$. Note, that $\sum_{n=2}^{\infty}x^n$ is related to the infinite geometric series. Plug in $S(0)=0$ and $S(1)=1$ to reduce the complexity.
$$\sum_{n=2}^{\infty}S(n)x^n=\left(\frac{1}{1-x}-1-x^2-x^3+x^2+\frac{2}{3}x^3\right)/(1-x-2x^2)$$
$$\sum_{n=2}^{\infty}S(n)x^n=\frac{\frac{1}{1-x}-1-\frac{1}{3}x^3}{1-x-2x^2}$$
Divide by $x^2$ and change index of summation:
$$\sum_{n=0}^{\infty}S(n+2)x^n=\frac{\frac{1}{1-x}-1-\frac{1}{3}x^3}{x^2(1-x-2x^2)}.$$
Now, determine the taylor expansion on the right side to get your coefficients.
|
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|
Determine using the comparison test whether the series $\sum_{n=1}^\infty \frac{1}{\sqrt[3]{n^2 + 1}}$ diverges How can I determine whether or not the following series converges using the comparison test?
$$\sum_{n=1}^\infty \frac{1}{(n^2 + 1)^\frac{1}{3}}$$
As $n$ goes to infinity, the sum is roughly equal to $\sum_{n=1}^\infty \frac{1}{(n^2)^\frac{1}{3}} = \sum_{n=1}^\infty \frac{1}{n^\frac{2}{3}}$.
I believe that $\sum_{n=1}^\infty \frac{1}{(n^2 + 1)^\frac{1}{3}} < \sum_{n=1}^\infty \frac{1}{n^\frac{2}{3}}$ for all $n$.
Using the $p$ test, it is clear that the latter sum diverges. However I cannot say that the former also diverges as the inequality sign does not satisfy the conditions of the comparison test for divergence.
|
By the limit comparison test, we have
$$(\frac{1}{n^2+1})^\frac{1}{3} \sim \frac{1}{n^\frac{2}{3}}\;\; (n\to +\infty)$$
$\frac{2}{3}<1$ thus
$\;\;\;\;\;\sum (\frac{1}{n^2+1})^\frac{1}{3}$ diverges.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1993244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Prove equality $\sqrt{1+x^2}\cdot\ln (x+\sqrt{1+x^2}) = x+\frac{x^3}{3} - \frac{2}{3}\frac{x^5}{5}+ \frac{2}{3}\frac{4}{5}\frac{x^7}{7}-...$ $\sqrt{1+x^2}\cdot\ln (x+\sqrt{1+x^2}) = x+\frac{x^3}{3} - \frac{2}{3}\frac{x^5}{5}+ \frac{2}{3}\frac{4}{5}\frac{x^7}{7}-...$
I stuck with performing right part. Where are these coefficients from?
|
We are looking for the Taylor series of
$$f(x)=\sqrt{1+x^2}\int\frac{dx}{\sqrt{1+x^2}}\tag{1}$$
that is clearly related with the Taylor series of
$$ g(x) = \sqrt{1-x^2}\int\frac{dx}{\sqrt{1-x^2}} = \sqrt{1-x^2}\arcsin(x).\tag{2}$$
We may now invoke some heavy artillery. For instance, the following identity:
$$ \arcsin^2(x) = \sum_{n\geq 1}\frac{(4x^2)^n}{2n^2\binom{2n}{n}}\tag{3}$$
leading to:
$$ \frac{\arcsin x}{\sqrt{1-x^2}}=\sum_{n\geq 1}\frac{4^n}{(2n)\binom{2n}{n}} x^{2n-1} \tag{4} $$
and to:
$$ x-\sqrt{1-x^2}\arcsin(x) = \int\frac{x\arcsin(x)}{\sqrt{1-x^2}}\,dx = \sum_{n\geq 1}\frac{4^n}{(2n)(2n+1)\binom{2n}{n}}x^{2n+1}\tag{5} $$
from which:
$$ g(x)=x-\sum_{n\geq 1}\frac{4^n}{(2n)(2n+1)\binom{2n}{n}}x^{2n+1}\tag{6}$$
and
$$\boxed{ f(x)=x-\sum_{n\geq 1}\frac{4^n(-1)^n}{(2n)(2n+1)\binom{2n}{n}}x^{2n+1}}\tag{7}$$
Two distinct prof of $(3)$ (one proof through Bonnet's recursion formula and Legendre polynomials, a second proof through the residue theorem) can be found in my course notes, at pages $18$ and $19$. A little warning: they are in Italian.
|
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|
Basic inequality problem For $a,b,m,n\in \mathbb{R}$, such that $m^2n^2>a^2m^2+b^2n^2$, what is the relationship between $M=\sqrt{m^2+n^2}$ and $N=a+b$?
I deduced the following: $$m^2n^2>a^2m^2+b^2n^2\geq 2abmn$$, $m,n$ is nonzero from the given inequality, so $mn\geq 2ab$. However, I fail to draw any conclusion between $M$ and $N$, I knew the following relation $\sqrt{m^2+n^2}\leq |m|+|n|$ and the AM-GM inequality. Anyone have ideas?
|
Geometrically, $(a,b)$ lies on the $(x,y)$-plane inside the ellipse
$$
\frac{x^2}{n^2} + \frac{y^2}{m^2}=1.\tag1
$$
Since $a+b$ is constant on lines with slope $-1$, the value of $a+b$ is no greater than $A+B$, where $(A,B)$ is the point in the first quadrant on the ellipse where the slope of the tangent line is $-1$. Using implicit differentiation (are you allowed to use calculus?) the slope of the tangent to (1) at $(x,y)$ is $$-\frac{m^2}{n^2}\frac xy\tag2$$
so $A$ and $B$ satisfy the two equations
$$
\frac{A^2}{n^2} + \frac{B^2}{m^2}=1\qquad\text{and}\qquad
-\frac{m^2}{n^2}\frac AB = -1.
$$
Solving these simultaneously yields, after some algebra,
$$
a+b<A+B = \sqrt{m^2+n^2}.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1994973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
10-permutations with exactly 4 inversions Let $p = p_1p_2\dots p_n$ A pair of elements $(p_i,p_j)$ is called an inversion in p if i > j and $p_i < p_j $.
Now, I know that the generating function for 10-permutations with k inversions is $I_{10}=(1+x)(1+x+x^2)\dots (1+x+\dots +x^9) $ where the coefficient of $x^4$ is the number I'm looking for, however, I do not want to expand this product because I think there must be a cleaner way to do it.
|
As pointed out by @Jack D'Aurizio, it's not that hard to compute the coefficient of $x^4$ in the generating function.
We have
$$\begin{aligned}I_{10} &= (1+x)(1+x+x^2)(1+x+x^2+x^3) \cdots (1+x+\dots +x^9) \\
&= \frac{1-x^2}{1-x} \cdot \frac{1-x^3}{1-x} \cdot \frac{1-x^4}{1-x} \cdots \frac{1-x^{10}}{1-x} \\
&= (1-x^2)(1-x^3)(1-x^4) \cdots (1-x^{10}) (1-x)^{-9} \\
&= (1-x^2)(1-x^3)(1-x^4) \cdots (1-x^{10}) \sum_{i=0}^{\infty} \binom{9+i-1}{i} x^i \\
\end{aligned}$$
There are only a few ways the exponents in this expression can add up to $4$: $0+4 =2+2=3+1=4+0$. So
$$[x^4]I_{10} = \binom{9+4-1}{4} - \binom{9+2-1}{2} - \binom{9+1-1}{1} - \binom{9+0-1}{0} = 440$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1998237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Coefficient of a term using binomial theorem I was just wondering how would I find the coefficient of any term let's say $x^3$ in the expansion of $(x^2+2x+2)^{10}$ using binomial expansion or any other technique. Please let me know if this can be found directly using a shortcut if any.
|
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\pars{x^{2} + 2x + 2}^{10} =
\sum_{a,b,c\ \in\ \mathbb{N}_{\ \geq\ 0}}{10 \choose a,b,c}
\bracks{a + b + c = 10}x^{2a}\,\pars{2x}^{b}\,2^{c}
\\[5mm] = &\
\sum_{a,b,c\ \in\ \mathbb{N}_{\ \geq\ 0}}{10 \choose a,b,c}2^{b + c}
\bracks{a + b + c = 10}\sum_{n = 0}^{\infty}\bracks{n = 2a + b}x^{n}
\end{align}
$$
\bracks{x^{n}}\pars{x^{2} + 2x + 2}^{10} =
\sum_{a,b,c\ \in\ \mathbb{N}_{\ \geq\ 0}}{10 \choose a,b,c}2^{b + c}
\bracks{a + b + c = 10}\bracks{n = 2a + b}
$$
The sum restrictions let to write $\ds{a}$ and $\ds{b}$ in terms of $\ds{c}$ and $\ds{n}$. Namely,
$\ds{a = n + c - 10}$ and $\ds{b = 20 - 2c - n}$:
$$
\bracks{x^{n}}\pars{x^{2} + 2x + 2}^{10} =
\sum_{c\ \in\ \mathbb{N}_{\ \geq\ 0}}{10 \choose n + c - 10,20 - 2c - n,c}
2^{20 - c - n}
$$
$\ds{c}$ bounds are determined by:
$$
a = n + c - 10 \geq 0\,,\quad b = 20 - 2c - n \geq 0
\implies 10 - n \leq c \leq 10 - n/2
$$
\begin{align}
&\bracks{x^{n}}\pars{x^{2} + 2x + 2}^{10} =
\sum_{c = m}^{M}{10! \over \pars{n + c - 10}!\pars{20 - c - n}!c!}\,
2^{20 - c - n}
\\[5mm] = &\,\,\,
\bbox[#ffe,20px,border:1px dotted navy]{\ds{\left. 2^{20 - n}\sum_{c = m}^{\left\lfloor 10 - n/2\right\rfloor}
{10 \choose c}{10 - c \choose n + c - 10}2^{-c}\right\vert_{\ n\ \leq\ 20}}}
\qquad\mbox{where}\qquad m = \max\braces{0,10 - n}
\end{align}
Note that the above expression vanishes out when
$\ds{\quad n <0\quad \mbox{o}\quad n > 20}$.
For instance: When $\ds{n = 3}$,
\begin{align}
&\bracks{x^{3}}\pars{x^{2} + 2x + 2}^{10} =
2^{17}\sum_{c = 7}^{8}
{10 \choose c}{10 - c \choose c - 7}2^{-c} =
2^{17}\bracks{{10 \choose 7}2^{-7} + 2{10 \choose 8}2^{-8}}
\\[5mm] & =
120 \times 2^{10} + 45 \times 2^{10} = \bbx{168960}
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that $5\cdot10^n+10^{n-1}+3$ is divisible by 9 Prove by induction the following:
$5*10^n+10^{n-1}+3$ is divisible by 9
Base case: $n=1$ $5*10+10^{1-1}+3=5*10+10^0+3=50+1+3=54$
$9|54=6$
Inductive Hypothesis: If $k$ is a natural number such that $9|5*10^k+10^{k-1} +3$
Inductive step:
Show that $S_k$ is true $\Rightarrow$ $S_{k+1}$ is true
$S_{k+1}$: $9|5*10^{k+1}+10^{k} +3$
$9|10(5*10^{k+1}+10^{k} +3)$
$9|5*10^{k+2}+10^{k+1} +10*3$
$9|5*(10^{k+1}*10^1)+(10^{k}*10^1) +(9+1)*3$
$9|5*(10^{k+1}*(9+1))+(10^{k}*10*(9+1)) +(9+1)*3$
$9|5(9*10^{k+1}+10^{k+1})+9*10^k+10^k+((9*3)+(1*3))$
This is were I am stuck for the last day try to figure out what move next would speed up the inductive proof as I have a feeling it can be finished up. Anyone help me see what I am unable to find.
|
Hint:$$5\cdot10^{n+1}+10^{n} +3=10\cdot5\cdot10^{n}+10\cdot10^{n-1}+30-27=$$
$$=10(5\cdot10^{n}+10^{n-1} +3)-3\cdot9$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Deriving Taylor series without applying Taylor's theorem. First, a neat little 'proof' of the Taylor series of $e^x$.
Start by proving with L'Hospital's rule or similar that
$$e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n$$
and then binomial expand into
$$e^x=\lim_{n\to\infty}1+x+\frac{n-1}n\frac{x^2}2+\dots+\frac{(n-1)(n-2)(n-3)\dots(n-k+1))}{n^{k-1}}\frac{x^k}{k!}+\dots$$
Evaluating the limit, we are left with
$$e^x=1+x+\frac{x^2}2+\dots+\frac{x^k}{k!}+\dots$$
which is our well known Taylor series of $e^x$.
As dxiv mentions, we can exploit the geometric series:
$$\frac1{1-x}=1+x+x^2+\dots$$
$$\ln(1+x)=x-\frac12x^2+\frac13x^3-\dots$$
$$\arctan(x)=x-\frac13x^3+\frac15x^5-\dots$$
which are found by integrating the geometric series and variants of it.
I was wondering if other known Taylor series can be created without applying Taylor's theorem. Specifically, can we derive the expansions of $\sin$ or $\cos$?
|
The series for $\sin$ and $\cos$ can be derived from the expansion of $e^x$.
$$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+\frac{x^6}{720}+\frac{x^7}{5040}+\cdots$$
Sub in $ix$ to get:
$$e^{ix}=1+ix+\frac{(ix)^2}{2}+\frac{(ix)^3}{6}+\frac{(ix)^4}{24}+\frac{(ix)^5}{120}+\frac{(ix)^6}{720}+\frac{(ix)^7}{5040}+\cdots$$
$$\cos x+i\sin x=1+ix-\frac{x^2}{2}-\frac{ix^3}{6}+\frac{x^4}{24}+\frac{ix^5}{120}-\frac{x^6}{720}-\frac{ix^7}{5040}+\cdots$$
Compare real and imaginary parts:
$$\cos x=1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+\cdots$$
$$\sin x=x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+\cdots$$
EDIT:
Consider the function $f(x)=(\cos x+i\sin x)e^{-ix}$
$$f'(x)=(-\sin x+i\cos x)e^{-ix}-i(\cos x+i\sin x)e^{ix}$$
$$f'(x)=-e^{ix}\sin x+ie^{ix}\cos x-ie^{-x}\cos x+e^{ix}\sin x$$
$$f'(x)=0$$
Hence $f(x)=c$ and $f(0)=(\cos0+i\sin0)e^0=1$ so $f(x)=1$
Therefore $e^{ix}=\cos x+i\sin x$
SECOND EDIT:
Another way springs to mind as well:
$$f(x)=\cos x+i\sin x$$
$$f'(x)=-sin(x)+i\cos x$$
$$f'(x)=i(\cos x+i\sin x)$$
$$f'(x)=i\cdot f(x)$$
$$\frac{f'(x)}{f(x)}=i$$
$$\ln(f(x))=ix+c$$
$$f(x)=e^{ix+c}$$
$$f(0)=\cos 0+i\sin 0=1\implies c=0$$
$$\therefore f(x)=e^{ix}$$
|
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|
binomial congruence $\sum_{i=1}^{\frac{p-1}{2}}\binom{2i}{i}\equiv 0~or (-2)\pmod p$
Let $p\ge 5$ be a prime number. Show that $$\sum_{i=1}^{\frac{p-1}{2}}\binom{2i}{i}\equiv 0 \text{ or } -2\pmod p .$$
Examples:
If $p=5$, then
$$f=\sum_{i=1}^{\frac{p-1}{2}}\binom{2i}{i}=\binom{2}{1}+\binom{4}{2}=8\equiv -2\pmod 5 .$$
If $p=7$, then
$$f=\sum_{i=1}^{\frac{p-1}{2}}\binom{2i}{i}=\binom{2}{1}+\binom{4}{2}+\binom{6}{3}=28\equiv 0\pmod 7 .$$
If $p=11$, then
$$f=\sum_{i=1}^{\frac{p-1}{2}}\binom{2i}{i}=\binom{2}{1}+\binom{4}{2}+\binom{6}{3}+\binom{8}{4}+\binom{10}{5}=350\equiv -2 \pmod{11} .$$
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It's more convenient to start the sum at $i=0$ instead of $i=1$,
adding the term ${0 \choose 0} = 1$ and aiming for
$$
\sum_{i=0}^{(p-1)/2} {2i \choose i} \equiv \pm 1 \bmod p.
$$
To prove this, note that ${2i \choose i} \equiv 0 \bmod p$ for
$(p-1)/2 < i \leq p-1$. Therefore
$$
\left(
\sum_{i=0}^{(p-1)/2} {2i \choose i}
\right)^2 =
\sum_{i=0}^{(p-1)/2} \, \sum_{j=0}^{(p-1)/2} {2i \choose i} {2j \choose j}
\equiv
\mathop{\sum\sum}_{i,j \geq 0}^{i+j \leq p-1} {2i \choose i} {2j \choose j}
\bmod p.
$$
But $\sum_{i+j = n} {2i \choose i} {2j \choose j} = 4^n$
because $2i\choose i$ is the $x^i$ coefficient of $(1-4x)^{-1/2}$.
Hence
$$
\left(
\sum_{i=0}^{(p-1)/2} {2i \choose i}
\right)^2 \equiv \sum_{n=0}^{p-1} 4^n
= 1 + \sum_{n=1}^{p-1} 4^n \bmod p.
$$
Since we assumed $p \geq 5$, the last sum is
$4 \cdot (4^{p-1}-1)/(4-1) \equiv 0 \bmod p$ (by Fermat little theorem), so
$\sum_{i=0}^{(p-1)/2} {2i \choose i}$ is a square root of $1 \bmod p$,
QED.
barak manos observes in a comment that the sign $1$ or $-1$ seems to match
the residue of $p \bmod 6$ (equivalently, of $p \bmod 3$ because $p$ is odd).
I have checked this experimentally for all $p < 1000$. Proving it
will likely require a different approach.
Postscript not so different, as it turns out.
Consider the polynomial
$$ P(x) = \sum_{i=0}^{(p-1)/2} {2i \choose i} (x/4)^i \bmod p. $$
The same argument as before shows that
$P(x)^2 \equiv (x^p - 1) / (x-1) \bmod p$ identically;
but that means $P(x)^2 \equiv (x-1)^{p-1}$, so
$P(x) = \pm (x-1)^{(p-1)/2}$ (which can then also be seen directly).
Comparing the coefficients of $x^{(p-1)/2}$ (or more simply, of $x$)
we soon see that the sign is the Legendre symbol $(-1/p)$.
Substituting some $x \bmod p$ into $(x-1)^{(p-1)/2}$
yields the Legendre symbol $((x-1)/p)$.
Here $x=4$ so we get $(-1/p) (3/p) = (-3/p)$, which by quadratic reciprocity
is $1$ or $-1$ according as $p$ is $1$ or $-1 \bmod 3$, QED.
|
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|
Does the following series converges or diverges? $\sum_{n=1}^{+\infty}\left[n \log\left ( \frac{2n+1}{2n-1} \right )-1\right]$ ,converges or not
My attempt
$a_n=n \log\left ( \frac{2n+1}{2n-1} \right )-1 \\ \\
\log \left ( \frac{2n+1}{2n-1} \right )=\log \left ( 1+\frac{2}{2n-1} \right ) \sim \frac{2}{2n-1} \ (n\rightarrow +\infty) \\ \\
a_n=n\log \left ( 1+\frac{2}{2n-1} \right )-1 \sim n \frac{2}{2n-1}-1=\frac{2n}{2n-1}-1=\frac{1}{2n-1} \\ \\ $
$\sum_{n=1}^{+\infty}\frac{1}{2n-1} $ diverges ,then $\sum_{n=1}^{+\infty}a_n $ diverges
but according to Wolfram $\sum_{n=1}^{+\infty}a_n=\frac{1}{2}-\frac{\log 2}{2}$
where is the mistake in my solution please?
thanks
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$$\log\left(\frac{2n+1}{2n-1}\right) = 2\,\text{arctanh}\left(\frac{1}{2n}\right)=\frac{1}{n}+\frac{1}{12n^3}+O(n^{-5})$$
leads to
$$ n\log\left(\frac{2n+1}{2n-1}\right)-1 = \frac{1}{12n^2}+O(n^{-4}) $$
so the given series is convergent. Your initial approximation was simply not accurate enough.
The exact value of the series is simple to compute through summation by parts. We have:
$$\begin{eqnarray*} \sum_{n=1}^{N}\left[n\log\left(\frac{2n+1}{2n-1}\right)-1\right]&=&-N+\sum_{n=1}^{N}n\left[\log(2n+1)-\log(2n-1)\right]\\&=&-N+N\log(2N+1)-\sum_{n=1}^{N-1}\log(2n+1)\\&=&-N+N\log(2N+1)-\log((2N-1)!!)\end{eqnarray*}$$
so
$$ \sum_{n=1}^{+\infty}\left[n\log\left(\frac{2n+1}{2n-1}\right)-1\right] = \frac{1-\log 2}{2}$$
follows from Stirling's inequality. Have a look at this recent question of mine.
|
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|
Simultaneous equations I have the question
Solve the simultaneous equations
$$\begin{cases}
3^{x-1} = 9^{2y} \\
8^{x-2} = 4^{1+y}
\end{cases}$$
source image
I know that $x-1=4y$ and $3X-6=2+2y$ .
However when I checked the solutions this should become $6X-16=4Y$ .
How is this?
|
From $$\begin{align*} & 3^{x-1}=9^{2y}\\ & 8^{x-2}=4^{y+1}\end{align*}\tag1$$
We can rewrite $9$ as $3^2$, $8$ as $2^3$ and $4$ as $2^2$. Doing so and setting the powers equivalent, we get$$\begin{align*} & x-1=4y\\ & 3x-6=2+2y\end{align*}\tag2$$
To answer your question, look at the second equation. Moving the $2$ to the left hand side and multiplying everything by $2$, we get our desired equation$$6x-16=4y\tag3$$
Which is the equation in the book.
To actually solve, we can substitute $4y$ from $(3)$ with the first equation of $(2)$ to solve for $x$.$$\begin{align*} & 6x-16=x-1\\ & 5x=15\implies x=3\end{align*}\tag4$$
Substituting that into $x-1=4y$ gives $y$ as$$y=\frac {x-1}4=\frac 12\tag5$$
Therefore, we have $x,y$ as$$\boxed{(x,y)=\left(3,\frac 12\right)}$$
If you have any questions or confusions, you can ask me!
|
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|
Find a limit without l'Hospital: $\lim_{x\to -\infty}\left(\frac{\ln(2^x + 1)}{\ln(3^x+ 1)}\right)$ Find $\displaystyle \lim_{x\to -\infty}\left(\frac{\ln(2^x + 1)}{\ln(3^x+ 1)}\right)$.
$\displaystyle \lim_{x\to -\infty}\left(\frac{\ln(2^x + 1)}{\ln(3^x+ 1)}\right)$ =
$\displaystyle \lim_{x\to -\infty}\left(\frac{\ln2^x + \ln(1+\frac{1}{2^x})}{\ln3^x + \ln(1+\frac{1}{3^x})}\right)$ =
$\displaystyle \lim_{x\to -\infty}\left(\frac{\ln2^x + \frac{1}{2^x}\ln(1+\frac{1}{2^x})^{2^{x}}}{\ln3^x + \frac{1}{3^x}\ln(1+\frac{1}{3^x})^{3^{x}}}\right)$ = $\cdots$
This method is not working.
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If we use the well known limit
$$\lim_{X\to 0}\frac{\ln(1+X)}{X}=1$$
with $X=2^x\;$ and $\;X=3^x,\;\;$we find
$$\lim_{x\to-\infty}\left( \frac{2}{3} \right)^x=+\infty$$
|
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|
locus of orthocentre of given triangle if $\tan B+\tan C = c$
Vertex $A$ of $\triangle ABC$ moves in such a way that $\tan B+\tan C = c(\bf{constant}),$
Then locus of orthocentre of $\triangle ABC$ is ( side $BC$ is fixed)
$\bf{My\; Try::}$ We can write $$\tan B+\tan C = c\Rightarrow \frac{\sin B}{\cos C}+\frac{\sin C}{\cos B} = c$$
So $$\frac{\sin (B+C)}{\cos B \cos C} = c\Rightarrow \sin (B+C)=c \cdot \cos B \cos C$$
Using $A+B+C = \pi\Rightarrow A+B = \pi-C$
So $$\sin (\pi-A) = c\cdot \cos B \cos C\Rightarrow \sin A = c\cdot \cos B\cdot \cos C$$
Now how can i solve it after that, Help required, Thanks
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Suppose $B(-a,0)$ and $C(a,0)$, suppose $A(x_A,y_A)$ (in the first quadrant for easyness), then, the relation $tan(b) + tan(c) =C$ translates into $\frac{y_A}{x_A+a}+\frac{y_A}{a-x_A}=C$. Performing calculations you will arrive to $y_A=\frac{-C}{2a}{x_{A}}^2+\frac{aC}{2}$.
Now observe that $H$ is the intersection of a vertical line $(v):x=x_A$ with a perpendicular to $AB$ through $C$ of equation $(p):y=\frac{-1}{tan(b)}(x-a)=\frac{x_A+a}{y_A}(x-a)$. Solving the coordinates of $H$ gives $H(x_A, \frac{{x_A}^2-a^2}{y_A})$. Combining with the value $y_A$ above and eliminating variables in the coordinates of $H$ yields $$y= \frac{x^2-a^2}{\frac{-C}{2a}{x}^2+\frac{aC}{2}}$$ which implies the locus of $H$ is a curve that can be easily plotted since it is an elemental function of $x$.
|
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|
Find the integer $m$ such that $(\frac{1}{5})^m \cdot (\frac{1}{4})^ {24} = \frac{1}{2(10)^{47}}$ I have this problem
$$(\frac{1}{5})^m \cdot (\frac{1}{4})^ {24} = \frac{1}{2(10)^{47}}$$
and I need to find the integer $m$.
I know that I can write $5^m \cdot 4^{24} = 2(10)^{47} \Leftrightarrow 5^m = \frac{2(10)^{47}}{(1/4)^{24}}$, but I don't know how to go on from there.
It is a question in a GMAT test, so it should be possible to solve using basic operations (i.e. without logarithmic functions).
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$$\frac{1}{2(10)^{47}}=\frac{1}{2}\frac{1}{2^{47}}\frac{1}{5^{47}}=\frac{1}{2^{48}}\frac{1}{5^{47}}=\frac{1}{4^{24}}\frac{1}{5^{47}}=\left(\frac{1}{4}\right)^{24}\left(\frac{1}{5}\right)^{47}$$
|
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|
Showing $\frac{x+y+z}{\sqrt 2}\le\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{x^2+z^2}$
Showing $\frac{x+y+z}{\sqrt 2}\le\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{x^2+z^2}$ $(x,y,z>0)$
Can I apply Jensen here, sum of the squareroots is greater then squareroot of the sum, so I make the RHS smaller then show also then the inequality is valid, i.e.
$\frac{x+y+z}{\sqrt 2}\le\sqrt{2(x^2+y^2+z^2)}$
squaring both sides we have on the left,
$\left(\frac{x+y+z}{\sqrt 2}\right)^2=\frac{x^2+y^2+z^2+2(xy+yz+xz)}{2}\le\frac{3(x^2+y^2+z^2)}{2}$
and on the right we have $2(x^2+y^2+z^2)$, so it is true
If it is OK, can you suggest another proof related more to Cauchy-Schwarz
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Hint: Note that $ \sqrt{2(a^2+b^2)} \geq a+b, ~\forall a,b\in \mathbb{R}$ and use it to derive the desired inequality.
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|
A congruence for the product of quadratic residues + the product of quadratic non-residues Let $p \gt 3$ be an odd prime,
$$p \equiv 3 \pmod 4 \iff \prod_{1\le k \le p-1\\ \ \
\left(\frac{k}{p}\right)=1}k \ \ \ \ + \prod_{1\le k \le p-1\\ \
\left(\frac{k}{p}\right)=-1}k \ \ \ \equiv 0 \pmod {p^2}$$
This seems to be true, but I could not find a proof.
Though it is quite easy to prove that the above congruence holds modulo $p$ for every odd prime, whatever its residue modulo $4$.
ex:
$p=5: \ \ 1\cdot4 +2\cdot 3 = 10 = 2\cdot 5$
$p=7: \ \ 1\cdot4\cdot2 +3\cdot 5 \cdot 6 = 98 = 2\cdot 7^2$
etc.
And I also know how to prove that: $$ \prod_{1\le k \le p-1\\ \ \
\left(\frac{k}{p}\right)=\epsilon }k \ \ \ \ \equiv -\epsilon(-1)^{\frac{p-1}{2}} \pmod {p}$$
But from here, I could not find a path to the above congruence modulo $p^2$.
Any other idea?
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Please check the following proposed proof (for $\Rightarrow$) .
We have:
$$ \prod_{1\le k \le p-1\\ \ \
\left(\frac{k}{p}\right)=1 }k \ \ \ \ \equiv \prod_{1\le a \le \frac{p-1}{2}\\ \ \
}a^2 \pmod {p}$$ because it is well known that all the squares $\pmod p$ are obtained by squaring all the $a$ such that $1\le a \le \frac{p-1}{2}$. (For, if $b$ is a square root of $a$ with $1\le a,b \le p-1$, so is $p-b$, and either $b$ or $p-b$ must be smaller than $\frac{p-1}{2}$.)
Then, since $(p-1)!+1 \equiv 0\pmod {p}$ (Wilson theorem) and $(\frac{p-1}{2}!)^2+(-1)^\frac{p-1}{2} \equiv 0\pmod {p}$ (see here), we have:
$$ \prod_{1\le k \le p-1\\ \ \
\left(\frac{k}{p}\right)=1 }k \ \ \ \ \equiv -(-1)^{\frac{p-1}{2}} \pmod {p} \ \text{,} \ \prod_{1\le k \le p-1\\ \ \ \left(\frac{k}{p}\right)=-1 }k \ \ \ \ \equiv (-1)^{\frac{p-1}{2}} \pmod {p}$$
and
$$ \prod_{1\le k \le p-1\\ \ \
\left(\frac{k}{p}\right)=1 }k +\prod_{1\le k \le p-1\\ \ \ \left(\frac{k}{p}\right)=-1 }k \ \ \ \ \equiv 0 \pmod {p}$$
Now , suppose $p \equiv 3 \pmod 4 $, $(-1)^{\frac{p-1}{2}}=-1$, then: $$\left(\frac{p-k}{p}\right)=\left(\frac{-k}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{k}{p}\right)=(-1)^{\frac{p-1}{2}}\left(\frac{k}{p}\right)=-\left(\frac{k}{p}\right)$$
Then,
$$\prod_{1\le k \le p-1\\ \ \
\left(\frac{k}{p}\right)=1 }k +\prod_{1\le k \le p-1\\ \ \ \left(\frac{k}{p}\right)=-1 }k =\prod_{1\le k \le p-1\\ \ \
\left(\frac{k}{p}\right)=1 }k + \prod_{1\le k \le p-1\\ \ \ \left(\frac{k}{p}\right)=1 }(p-k)$$
Then, doing the development on the RHS, but dropping the powers of $p$ higher than $p^2$:
$$\prod_{1\le k \le p-1\\ \ \
\left(\frac{k}{p}\right)=1 }k +\prod_{1\le k \le p-1\\ \ \ \left(\frac{k}{p}\right)=-1 }k =\prod_{1\le k \le p-1\\ \ \
\left(\frac{k}{p}\right)=1 }k + (-1)^{\frac{p-1}{2}}\prod_{1\le k \le p-1\\ \ \ \left(\frac{k}{p}\right)=1 }k-p (-1)^{\frac{p-1}{2}}\sum_{1\le j \le p-1\\ \ \
\left(\frac{j}{p}\right)=1 }\frac{1}{j}\prod_{1\le k \le p-1\\ \ \
\left(\frac{k}{p}\right)=1 }k+\cdot\cdot\cdot $$
Then:
$$ \prod_{1\le k \le p-1\\ \ \
\left(\frac{k}{p}\right)=1 }k +\prod_{1\le k \le p-1\\ \ \ \left(\frac{k}{p}\right)=-1}k \equiv p\cdot \sum_{1\le j \le p-1\\ \ \
\left(\frac{j}{p}\right)=1 }\frac{1}{j}\prod_{1\le k \le p-1\\ \ \
\left(\frac{k}{p}\right)=1 }k \pmod {p^2} $$
but $$ \prod_{1\le k \le p-1\\ \ \
\left(\frac{k}{p}\right)=1 }k\ \ \ \ \equiv -(-1)^{\frac{p-1}{2}}=1 \pmod {p} $$
then,
$$ \prod_{1\le k \le p-1\\ \ \
\left(\frac{k}{p}\right)=1 }k +\prod_{1\le k \le p-1\\ \ \ \left(\frac{k}{p}\right)=-1}k \equiv p\cdot \sum_{1\le j \le p-1\\ \ \
\left(\frac{j}{p}\right)=1 }\frac{1}{j}\pmod {p^2} $$
but, by the same argument as at the begining of the proof:
$$\sum_{1\le j\le p-1\\ \ \left(\frac{j}{p} \right)=1}\frac{1}{j} \equiv \sum_{1\le a\le \frac{p-1}{2}}\frac{1}{a^2} \pmod p$$
And, since by Wolstenholme's theorem, if $p>3$, $$ \sum_{1\le a\le p-1}\frac{1}{a} \equiv 0 \pmod {p^2}$$
then , if $p>3$,
$$ 0 \equiv \frac{1}{p}\sum_{1\le a\le p-1}\frac{1}{a} = \frac{1}{p}\sum_{1\le a\le \frac{p-1}{2}}\frac{1}{a}+\frac{1}{p}\sum_{1\le a\le \frac{p-1}{2}}\frac{1}{p-a} = \sum_{1\le a\le \frac{p-1}{2}}\frac{1}{a(p-a)} \equiv -\sum_{1\le a\le \frac{p-1}{2}}\frac{1}{a^2} \pmod p $$
then , if $p>3$
$$ \prod_{1\le k \le p-1\\ \ \
\left(\frac{k}{p}\right)=1 }k +\prod_{1\le k \le p-1\\ \ \ \left(\frac{k}{p}\right)=-1}k \equiv 0 \pmod {p^2} \ \ \ \square $$
A proof for the converse ($\Leftarrow$) is still missing. By numerical search no $p \le 400000$ such that $p \equiv 1 \pmod 3$ satisfy the above congruence. But the quotient $$ \frac{\prod_{1\le k \le p-1\\ \ \
\left(\frac{k}{p}\right)=1 }k +\prod_{1\le k \le p-1\\ \ \ \left(\frac{k}{p}\right)=-1}k}{p} \pmod p $$ appears to be randomly and homogeneously distributed between $0$ and $p-1$. Then, such a $p$ might well exist above $400000$ and ($ \Leftarrow$) would be wrong actually.
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|
probability last draw I have a bag initially containing $r$ red fruit pastilles (my favourites) and $b$ fruit pastilles of other colours. From time to time I shake the bag thoroughly and remove a pastille at random. (It may be assumed that all pastilles have an equal chance of being selected.) If the pastille is red I eat it but otherwise I replace it in the bag. After n such drawings, I find that I have only eaten one pastille. Show that the probability that I ate it on my last drawing is:
$$\frac{(r+b-1)^{n-1}}{(r+b)^n-(r+b-1)^n}$$
$\underline{}\underline{}\underline{}\underline{}\underline{}\underline{}\underline{}\underline{}\underline{}\underline{}\underline{}\underline{}\underline{}\underline{}\underline{}\underline{}\underline{}\underline{}\underline{}\underline{}\underline{}\underline{}\underline{}\underline{}\underline{}\underline{}\underline{}\underline{}\underline{}\underline{}\underline{}\underline{}\underline{}\underline{}\underline{}\underline{}\underline{}\underline{}\underline{}\underline{}$
My thoughts on this: Since each draw is independent from the others, the probability to draw a blue pastille is $\frac{b}{b+r}$, so the probability that the first $n-1$ draws are all blue is $(\frac{b}{b+r})^{n-1}$ and the probability to pick the red pastille at the last draw is $\frac{r}{b+r}$, making the final probability $(\frac{b}{b+r})^{n-1} \cdot \frac{r}{b+r}$, which is obviously not the sought probability.
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I believe that an easier method will be by induction.
For n=1 the conditional probability must be 1 which it is if you substitute.
Assume that the formula holds true for n=k and prove that it is also true for n=k+1
Consider k pickings:
The tree will be k long and k high as follows:
rbbbbb........b (1st line) $P_{d1} = \Big(\frac{b}{r+b}\Big)^{0}\times{r \over {r+b}} \times \Big(\frac{b}{r+b-1}\Big)^{k-1}$
brbbbb........b $P_{d2} = \Big(\frac{b}{r+b}\Big)^{1}\times{r \over {r+b}} \times \Big(\frac{b}{r+b-1}\Big)^{k-2}$
bbrbbb........b $P_{d3} = \Big(\frac{b}{r+b}\Big)^{2}\times{r \over {r+b}} \times \Big(\frac{b}{r+b-1}\Big)^{k-3}$
……….
bbbbbb.......rb (k-1 line) $P_{dk-1} = \Big(\frac{b}{r+b}\Big)^{k-2}\times{r \over {r+b}} \times \Big(\frac{b}{r+b-1}\Big)^{1}$
bbbbbb.......br (k line) $P_{dk} = \Big(\frac{b}{r+b}\Big)^{k-1}\times{r \over {r+b}} $
The conditional probability for n=k is given by:
$P(k) =\frac{P_{dk}}{\sum_{i=1}^k P_{di}} = \frac{(r+b-1)^{k-1}}{(r+b)^k-(r+b-1)^k}$ (Eq.1)
where $P_{dk}$ is the unconditional probability of the last row (k-1 bs followed by one red)
Now consider k+1 pickings. The tree will be exactly the same as above for the first k rows and k columns. The k+1 column will have k bs and 1 r while the k+1 row will also have k bs and 1 r.
The conditional probability P(k+1) will be given by:
$P(k+1) =\frac{P_{dk+1}}{\sum_{i=1}^{k+1} P_{di}}$ (Eq.2)
$P_{dk+1}=\big(\frac{b}{(r+b)^k} \times \frac{r}{(r+b)}\big)=\frac{b}{(r+b)} P_{dk}$ (Eq.3)
By studying the two trees we can establish a simple relationship between the sum of unconditional probabilities of the (k+1)x(k+1) tree to that of the kxk tree. To go from the kxk to the (k+1)x(k+1) tree we simply have to multiply each row of the kxk tree by the probability for the added b in the k+1 column $\big(\frac{b}{(r+b−1)}\big)$ and then add the last row from the (k+1)x(k+1) tree. ie:
$\sum_{i=1}^{k+1} P_{di} = \frac{b}{(r+b-1)} \times \sum_{i=1}^{k} P_{di} + P_{dk+1}$
Substituting from Eq.1:
$\sum_{i=1}^{k+1} P_{di} = \frac{b}{(r+b-1)} \times \frac{P_{dk}}{P(k)} + P_{dk+1}$
Substituting from Eq.3:
$ \sum_{i=1}^{k+1} P_{di} = \frac{b}{(r+b-1)} \times \frac{P_{dk+1}}{P(k)} \frac{(r+b)}{b} + P_{dk+1}$
Therefore:
$\sum_{i=1}^{k+1} P_{di} = P_{dk+1} \big( \frac{r+b}{(r+b-1) P(k)} +1 \big)$
From Eq.2:
$\frac{1}{P(k+1)} = \frac{\sum_{i=1}^{k+1}P_{di}}{P_{dk+1}} = \frac{r+b}{(r+b-1)}\frac{1}{P(k)}+1$
Substituting P(k) from Eq.1:
$\frac{1}{P(k+1)} = \frac{(r+b)\big((r+b)^k-(r+b-1)^k\big)}{(r+b-1)(r+b-1)^{k-1}}+1 = $$\frac{(r+b)^{k+1}-(r+b)(r+b-1)^k+(r+b-1)^k}{(r+b-1)^k}=\frac{(r+b)^{k+1}-(r+b-1)(r+b-1)^k}{(r+b-1)^k} =$ $\frac{(r+b)^{k+1}-(r+b-1)^{k+1}}{(r+b-1)^k}$
Therefore: $P(k+1) = \frac{(r+b-1)^k}{(r+b)^{k+1}-(r+b-1)^{k+1}}$
QED
|
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|
Solve the equation $x^2+\frac{9x^2}{(x+3)^2}=27$
Problem Statement:-
Solve the equation $$x^2+\dfrac{9x^2}{(x+3)^2}=27$$
I have tried to turn it into a quadratic equation so as to be saved from solving a quartic equation, but have not been able to come up with anything of much value.
These are the things that I have tried to turn the given equation into a quadratic equation.
$$x^2+\dfrac{9x^2}{(x+3)^2}=27\implies 1+\dfrac{1}{\left(\dfrac{x}{3}+1\right)^2}=3\left(\dfrac{3}{x}\right)^2$$ $$\text{OR}$$
$$x^2+\dfrac{9x^2}{(x+3)^2}=27\implies 1+\dfrac{\left(\dfrac{3}{x}\right)^2}{\left(1+\dfrac{3}{x}\right)^2}=3\left(\dfrac{3}{x}\right)^2$$
|
Simplifying and reducing we get
$$x^4+6x^3-9x^2-162x-243=0\,\text{and}\,x\neq 3$$
Let's look for a factorisation in the form
$$x^4+6x^3-9x^2-162x-243=(x^2+ax+b)(x^2+cx+d)$$
Developing and comparing the coefficients we get
$$\begin{align}a+c=&6\\b+d+ac=&-9\\ad+bc=&-162\\bd=&-243\end{align}$$
We obviously look first for integer coefficients. We start with $-243=-3^5$ and we test the various combinations for $b,d$.
We get the following set that works
$$\begin{align}a=&-3\\b=&-9\\c=&9\\d=&27\end{align}$$
And now we're left with two quadratics
$$\begin{align}x^2-3x-9=&0\\x^2+9x+27=&0\end{align}$$
And the roots are
$$\begin{align}x_1=&3\cdot{1+\sqrt{5}\over 2}\\x_2=&3\cdot{1-\sqrt{5}\over 2}\\x_3=&3\cdot{-3+i\sqrt{3}\over2}\\x_4=&3\cdot{-3-i\sqrt{3}\over 2}\end{align}$$
|
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|
Showing $\frac12 3^{2-p}(x+y+z)^{p-1}\le\frac{x^p}{y+z}+\frac{y^p}{x+z}+\frac{z^p}{y+x}$
Showing $\frac12 3^{2-p}(x+y+z)^{p-1}\le\frac{x^p}{y+z}+\frac{y^p}{x+z}+\frac{z^p}{y+x}$ with $p>1$, and $x,y,z$ positive
By Jensen I got taking $p_1=p_2=p_3=\frac13$, (say $x=x_1,y=x_2,z=x_3$)
$\left(\sum p_kx_k\right)^p\le\sum p_kx^p$ which means;
$3^{1-p}(x+y+z)^p\le x^p+y^p+z^p$
and if I assume wlog $x\ge y\ge z$ then $\frac{1}{y+z}\ge\frac{1}{x+z}\ge\frac{1}{x+y}$
so these $2$ sequences are similarly ordered, and by rearrangement inequality the rest follows ?
I should use power means, but any other solution is also appreciated.
|
WLOG, $x+y+z=1\tag1$ then by power mean inequality;
$\sum\limits_{cyc}\left(\frac13(x+y)^{-1}\right)^{-1}\le(x+y)^{\frac 13}(y+z)^{\frac13}(x+z)^{\frac1 3}$
and AM-GM
$(x+y)^{\frac 13}(y+z)^{\frac13}(x+z)^{\frac13}\le\frac13(x+y)+\frac13(y+z)+\frac13(x+z)\overset{(1)}=\frac23$
$\displaystyle\implies \frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}\ge\frac92$
Rearrangement Inequality implies,
$\displaystyle\frac{x^p}{y+z}+\frac{y^p}{x+z}+\frac{z^p}{y+x}\ge\frac{(x^p+y^p+z^p)\cdot \left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}\right)}{3}$
and in general one has $\left(\sum p_kx_k\right)^p\le\sum p_kx^p$ thus,
$3^{p-1}(x^p+y^p+z^p)\ge(x+y+z)^{p}\overset{(1)}=(x+y+z)^{p-1}$
$\displaystyle\implies \frac{x^p}{y+z}+\frac{y^p}{x+z}+\frac{z^p}{y+x}\ge\frac{(x^p+y^p+z^p)\cdot\frac92}{3}\ge\frac{3^{1-p}(x+y+z)^{p-1}\cdot\frac92}{3}=\frac12 3^{2-p}(x+y+z)^{p-1}$
|
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|
The point on $y = \sqrt{\ln x}$ closest to $(4,0)$ Find the point of the graph of $y = \sqrt{\ln x}$ that is closest to the point $(4,0)$.
I started by using the distance formula and plugging in the values where:
$$d = \sqrt{(x-4)^2 + (\sqrt{\ln x} - 0)^2}$$
$$d = \sqrt{(x^2-8x+16+\ln x)}$$
I know we should set this equal to 0.
Stuck here do I somehow combine ln x with 8x?
|
You don't want to set the distance itself to $0$, but rather the rate of change of the distance (as a function of $x$). So using the formula you correctly derived: $$\begin{array}{rcl} d(x) & = & \sqrt{(x-4)^2+\ln x} \\ & = & \left[(x-4)^2 + \ln x \right]^{\frac{1}{2}} \end{array} $$ we now differentiate it (using the chain rule several times) to get: $$\begin{array}{rcl} d'(x) & = & \frac{1}{2}\left[(x-4)^2 + \ln x \right]^{-\frac{1}{2}} \cdot \left[ 2(x-4) + \frac{1}{x} \right] . \end{array}$$ Now setting this equal to zero, we find that (since $ab=0$ implies that $a=0$ or $b=0$, and conversely) our problem can be reduced to finding the $x$ which satisfy the following condition: $$\frac{1}{2\sqrt{(x-4)^2 + \ln x }}=0 \quad \text{or} \quad 2(x-4)+\frac{1}{x}=0. $$ Since there are no $x$ satisfying the left equation (since there is no number $t$ such that $\frac{1}{t}=0$), we just have to find the $x$ such that : $$2(x-4)+\frac{1}{x}=0 \implies 2x^2 -8x + 1 =0. $$ Now you can just use the quadratic formula to find two candidate values of $x$, call them $x_1$ and $x_2$. Then test to see which of the two points $(x_1, \sqrt{\ln x_1})$ or $(x_2, \sqrt{\ln x_2})$ works. (We might have introduced an extra solution when multiplying by $x$, so that's why we have to check both to see whether they are correct or not.)
|
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|
Find the poles and residue of the function $\frac{1}{z-\sin z }$ at $z=0$? $$\frac{1}{z-\sin z}$$
$$\frac{1}{z(1-\frac{\sin z}{z})}$$
the laurent's transform is also used to simplify the function.
the residue is found a $\frac{2*36*2}{5}$
but the actual answer is given as $\frac{3}{10}$
the residue is given as
The solution is incorrect and also the pole is of the order 3.
|
Use the Taylor Series for the $\sin z$ function:
$$\sin z \approx z - \frac{z^3}{6} + \frac{z^5}{5!} + \ldots$$
In this way your denominator becomes
$$\frac{1}{\frac{z^3}{6} - \frac{z^5}{5!} + \ldots}$$
Collect a $z^3/6$ term:
$$\frac{1}{\frac{z^3}{6}\left(1 - \frac{z^2}{20}\right)} = \frac{6}{z^3} \cdot \frac{1}{1 - \frac{z^2}{20}}$$
Now the term $\frac{1}{1 - \frac{z^2}{20}}$ can be expanded with the Geometric Series, as long as $|z^2/20| < 1$:
$$\frac{6}{z^3} \sum_{k = 0}^{+\infty} \frac{z^{2k}}{20^k}$$
And the first terms of the series are
$$\frac{6}{z^3} + \frac{6}{20 z} + \frac{6 z^4}{400} + \ldots$$
And you now know that the residue is nothing but the coefficient of the $z^{-1}$ terms, which is
$$\frac{3}{10}$$
Also you can see the pole is order three, since you have a $z^{-3}$ factor in the series.
|
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|
How can we show $\cos^6x+\sin^6x=1-3\sin^2x \cos^2x$? How can we simplify $\cos^6x+\sin^6x$ to $1−3\sin^2x\cos^2x$?
One reasonable approach seems to be using $\left(\cos^2x+\sin^2x\right)^3=1$, since it contains the terms $\cos^6x$ and $\sin^6x$. Another possibility would be replacing all occurrences of $\sin^2x$ by $1-\cos^2x$ on both sides and then comparing the results.
Are there other solutions, simpler approaches?
I found some other questions about the same expression, but they simplify this to another form: Finding $\sin^6 x+\cos^6 x$, what am I doing wrong here?, Alternative proof of $\cos^6{\theta}+\sin^6{\theta}=\frac{1}{8}(5+3\cos{4\theta})$ and Simplify $2(\sin^6x + \cos^6x) - 3(\sin^4 x + \cos^4 x) + 1$. Perhaps also Prove that $\sin^{6}{\frac{\theta}{2}}+\cos^{6}{\frac{\theta}{2}}=\frac{1}{4}(1+3\cos^2\theta)$ can be considered similar. The expression $\cos^6x+\sin^6x$ also appears in this integral Find $ \int \frac {\tan 2x} {\sqrt {\cos^6x +\sin^6x}} dx $ but again it is transformed to a different from then required here.
Note: The main reason for posting this is that this question was deleted, but I still think that the answers there might be useful for people learning trigonometry. Hopefully this new question would not be closed for lack of context. And the answers from the deleted question could be moved here.
|
Here's a slick way:
$(\cos^2(\theta)+\sin^2(\theta))^3=1$
Use Pascal's Triangle to get
$\cos^6(\theta)+3 \cos^4(\theta) \sin^2(\theta)+3 \cos^2(\theta) \sin^4(\theta)+\sin^6(\theta)=1$
$\cos^6(\theta)+\sin^6(\theta)=1-3 \cos^2(\theta) \sin^4(\theta)-3 \cos^4(\theta) \sin^2(\theta)$
$\cos^6(\theta)+\sin^6(\theta)=1-3\cos^2(\theta) \sin^2(\theta)(\cos^2(\theta)+\sin^2(\theta)) $
$\cos^6(\theta)+\sin^6(\theta)=1-3\cos^2(\theta) \sin^2(\theta) $
|
{
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|
Show that $1^3 + 2^3 + ... + n^3 = (n(n+1)/2)^2$ by induction I have some with proving by induction. I cannot find a solution for the inductive step: $1^3 + 2^3 + ... + n^3 = (n(n+1)/2)^2$
I already did the induction steps:
Basis: P(1) = $1^3 = (1(1+1)/2)^2$ (This is true)
Inductive step: Assume $P(k) = ((k)(k+1)/2)^2$
To be proven: $((k)(k+1)/2)^2 + (k+1)^3 = ((k+1)(k+2)/2)^2$
My problem is that I do not know how I can put the $ + (k+1)^3$ inside $((k)(k+1)/2)^2$.
Simplifying the left and right part of the statement does not help:
Simplifying the left side: $((k)(k+1)/2)^2 + (k+1)^3 = ((k^2+k)/2)^2 + (k+1)^3 $
Simplifying the right side: $((k+1)(k+2)/2)^2 = ((k^2+3k+2)/2)^2$
So i am left with: $((k^2+k)/2)^2 + (k+1)^3 = ((k^2+3k+2)/2)^2$
That is the same as: $1/4 (k^2+k)^2 + (k+1)^3 = 1/4((k^2+3k+2))^2$
Going further with the left side:$1/4 (k^2+k)^2 + (k+1)^3 = (1/4)(k^4 + 2k^3 + k^2) + k^3+3 k^2+3 k+1$
Going further with the right side: $1/4((k^2+3k+2))^2 = 1/4 (k^4+6 k^3+13 k^2+12 k+4)$
Now I am stuck with: $(1/4)(k^4 + 2k^3 + k^2) + k^3+3 k^2+3 k+1 = 1/4 (k^4+6 k^3+13 k^2+12 k+4)$
Now I am kind of left with garbage. Am I missing something? What do I do wrong? Where can I find a good resources to learn how to solve this issue?
|
Show that $S_n-S_{n-1}=n^3$ (or if you prefer, $S_n=S_{n-1}+n^3$).
Indeed,
$$\left(\frac{n(n+1)}2\right)^2-\left(\frac{(n-1)n}2\right)^2=n^2\left(\frac{n+1}2+\frac{n-1}2\right)\left(\frac{n+1}2-\frac{n-1}2\right).$$
Note that induction is unnecessary. As both members are quartic polynomials, it suffices to prove equality for five distinct values of $n$:
$$0=0,\\1=1,\\9=3^2,\\36=6^2,\\100=10^2.$$
|
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|
Solving an inequality involving floor. We have this inequality :
$$\lfloor x+1 \rfloor < \sqrt x +1. $$
I want to find algebraic way to solve this inequality.(not geometric solution)
Note : It is obvious when $x \in \Bbb Z$ we have $0<x<1$ but I can't solve it when $x \notin \Bbb Z$
|
To show another possible way , we have that
$$
\begin{array}{l}
\left\lfloor {x + 1} \right\rfloor < \sqrt x + 1 \\
\left\lfloor x \right\rfloor < \sqrt x \\
x - \left\{ x \right\} < \sqrt x \quad \left| {\;0} \right. \le \left\{ x \right\} < 1 \\
x - 1 < x - \left\{ x \right\} < \sqrt x \\
\end{array}
$$
where $ \left\{ x \right\}$ represents the fractional part of $x$.
For the last chain of inequalities to sussist, and since $x$ shall be non-negative, we must have:
$$
\left\{ \begin{array}{l}
x - 1 < \sqrt x \\
0 \le x \\
\end{array} \right.\quad \Rightarrow \quad 0 \le x < 1\;\; \vee \;\;\left\{ \begin{array}{l}
x^{\,2} - 3x + 1 < 0 \\
1 < x \\
\end{array} \right.
$$
because for $0 \leqslant x<1$ the original inequality is satisfied, for $x=1$ it is not, and for $1<x$
we can square both sides giving the quadratic inequality.
Since the solution of the quadratic equation are
$$
\frac{{3 \pm \sqrt 5 }}{2} \approx 0.38,\;2.62
$$
and the lowest is less than $1$, we take only the higher one as the upper bound for $x$.
Because of the "squezing" $ x - 1 < x - \left\{ x \right\} < \sqrt x $, that assures that the original
$$
x - \left\{ x \right\} < \sqrt x
$$
does not have solutions outside of the range $0 < x < \frac{{3 \pm \sqrt 5 }}{2} $,
which means $ 0 \leqslant \left\lfloor x \right\rfloor \leqslant 2$.
And because
$$
\left[ {\begin{array}{*{20}c}
{1 < x < 2} & {\left\lfloor x \right\rfloor = 1 < \sqrt x } \\
{2 \le x < \frac{{3 + \sqrt 5 }}{2}} & {\left\lfloor x \right\rfloor = 2 > \sqrt x } \\
\end{array}} \right.
$$
we finally get that
$$
\left\lfloor {x + 1} \right\rfloor < \sqrt x + 1\quad \Rightarrow \quad 0 < x < 1\;\; \vee \;\;1 < x < 2
$$
|
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"timestamp": "2023-03-29T00:00:00",
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