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Find the missing coordinates. Fill in the missing coordinates on the unit circle, represented by the letters.
Using sin and cos, we have a $\sin(45^\circ)$ of $\frac{\sqrt{2}}{2}$ and a $\cos(45^\circ)$ of $\frac{\sqrt{2}}{2}$, and a $\cos(60^\circ)$ of $\frac{1}{2}$, and a $\sin(60^\circ)$ of $\frac{\sqrt{3}}{2}$. The coordinates I need are represented by A, B,C, D, E and F.
The answer is A: $\frac{\sqrt{2}}{2}$, $\frac{\sqrt{2}}{2}$, B: $-\frac{1}{2}$, $\frac{\sqrt{3}}{2}$, C: $-\frac{\sqrt{3}}{2}$, $\frac{1}{2}$, D: $-\frac{\sqrt{2}}{2}$, $-\frac{\sqrt{2}}{2}$; E: $\frac{1}{2}, \frac{\sqrt{3}}{2}$; F: $\frac{\sqrt{3}}{2}$, $-\frac{1}{2}$.
How does one come up with these coordinates? Thank you.
|
Point represented as (value of cos, value of sin)
Point A represent $45^\circ$
$sin45^\circ = \frac{1}{\sqrt2}$,
$cos45^\circ = \frac{1}{\sqrt2}$,
Point B represent $120^\circ$
$sin120^\circ = \frac{\sqrt3}{2}$,
$cos120^\circ = -\frac{1}{\sqrt2}$, [ because in second quadrant]
Point C represent $150^\circ$
$sin150^\circ = \frac{1}{2}$,
$cos 150^\circ = -\frac{\sqrt3}{2}$
Point D is in mid of third quadrant so its like $45^\circ$ but in actual its $225^\circ$.
$sin225^\circ = -\frac{1}{\sqrt2}$, [ because in third quadrant]
$cos225^\circ = -\frac{1}{\sqrt2}$, [ because in third quadrant]
Point E represent $300^\circ$
$sin 300^\circ = \frac{\sqrt3}{2}$,
$cos 300^\circ = \frac{1}{2}$
Point F represent $330^\circ$
$sin 330^\circ = \frac{1}{2}$,
$cos 330^\circ = \frac{\sqrt3}{2}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $P.V. \int^{\infty}_{0} \frac{x^\alpha }{x(x+1)} dx $ where $0 < \alpha <1$ Evaluate
$$P.V. \int^{\infty}_{0} \frac{x^\alpha }{x(x+1)} dx $$
where $0 < \alpha <1$
Thm
Let $P$ and $Q$ be polynomials of degree $m$ and $n$,respectively, where $n \geq m+2$. If $Q(x)\neq 0$. for $Q$ has a zero of order at most 1 at the origin and $f(z)= \frac{z^\alpha P(z)}{Q(z)}$, where $0 < \alpha <1$ then
$$P.V, \int^{\infty}_0 \frac{x^ \alpha P(x)}{Q(x)} dx= \frac{2 \pi i}{1- e^{i \alpha 2 \pi }} \sum^{k}_{j=1} Res [f,z_j] $$
where $z_1,z_2 ,\dots , z_k$ are the nonzero poles of $\frac{P}{Q}$
Attempt
Got that $P(x)=1$ where its degree $m=1$ and $q(x)=x(x+1)$ its degree is $n=1$ so it is not the case that $n \geq m+2$ because $2 \geq 1+2$
|
We assume $0<\alpha<1$. We have
$$
P.V. \int^{\infty}_{0} \frac{x^\alpha }{x(x+1)} dx =\frac{\pi}{\sin(\alpha \pi)}.
$$
Hint. One may prove that
$$
\begin{align}
& \int^{\infty}_{0} \frac{x^\alpha }{x(x+1)} dx
\\\\&=\int^1_{0} \frac{x^\alpha }{x(x+1)} dx+\int^{\infty}_{1} \frac{x^\alpha }{x(x+1)} dx
\\\\&=\int^1_{0} \frac{x^{\alpha-1} }{x+1} dx+\int^0_{1} \frac{x^{\alpha-1} }{1+\frac1x}\cdot \left(- \frac{dx}{x^2}\right)
\\\\&=\int^1_{0} \frac{x^{\alpha-1} }{1+x} dx+\int^1_{0} \frac{x^{-\alpha} }{1+x}dx
\\\\&=\int^1_{0} \frac{x^{\alpha-1} (1-x)}{1-x^2} dx+\int^1_{0} \frac{x^{-\alpha}(1-x) }{1-x^2}dx
\\\\&=\int^1_{0} \frac{x^{\alpha-1} (1-x)}{1-x^2} dx+\int^1_{0} \frac{x^{-\alpha}(1-x)}{1-x^2}dx
\\\\&=\frac12\psi\left(\frac{\alpha+1}2\right)-\frac12\psi\left(\frac{\alpha}2\right)+\frac12\psi\left(1-\frac{\alpha}2\right)-\frac12\psi\left(\frac{1-\alpha}2\right)
\\\\&=\frac{\pi}{\sin(\alpha \pi)}
\end{align}
$$ where we have used the classic integral representation of the digamma function
$$
\int^1_{0} \frac{1-t^{a-1}}{1-t} dt=\psi(a)+\gamma, \quad a>-1,\tag 1
$$ and the properties
$$
\psi(a+1)-\psi(a)=\frac1a,\qquad \psi(a)-\psi(1-a)=-\pi\cot(a\pi).
$$
|
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|
Functions have similar graphs I graphed the two functions $y=x^2\sin\left(\frac{x}{50}\right)$ and $y=\frac{1}{50}x^3$ and I noticed they have extremely similar graphs from $-50 < x <50$. What is the reason for this?
|
Hint. By differentiation, one may prove that
$$
u-\frac{u^3}6\le\sin u\le u,\qquad u \in [0,1]
$$ giving
$$
\frac{x^3}{50}-\frac{x^5}{750000}\le x^2 \sin \left(\frac{x}{50}\right)\le \frac{x^3}{50}, \quad -1<\frac{x}{50}<1
$$ or
$$
\left| x^2 \sin \left(\frac{x}{50}\right)-\frac{x^3}{50}\right|\le \left|\frac{x^5}{750000}\right|, \quad -1<\frac{x}{50}<1.
$$
|
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|
continued fraction for $\sqrt{n^2 − 1}$ Question:
Find the continued fraction for $\sqrt{n^2 − 1}$, where $n \ge 2$ is
an integer.
My attempt:
$n - 1 = \sqrt{n^2} - 1 \lt \sqrt{n^2 − 1} \lt \sqrt{n^2}$
So far, $[n-1; ]$
$\sqrt{n^2 − 1} = n - 1 + \frac{1}{x}$
$\to \frac{1}{x} = \sqrt{n^2 - 1} - (n-1) \to \frac{1}{x^2} = n^2-1-n^2-1+2n-2\times(n-1)\times\sqrt{n^2-1} = 2\times(n-1)\times(1-\sqrt{n^2-1})$
But, it doesn't really help to find continued fraction.
|
Use that $$x = \frac{1}{\sqrt{n^2-1} - (n-1)} = 1+ \frac{\sqrt{n^2-1} - (n-1)}{2(n-1)} = 1 + \frac{1}{2(n-1)x} $$
to conclude that $$\sqrt{n^2-1} = [n-1; \overline{1, 2(n-1)}]$$
|
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|
If $a+b+c+d=0$ and $\{a,b,c,d\}\subset[-1,1]$ so $\sum\limits_{cyc}\sqrt{1+a+b^2}\geq4$ Let $\{a,b,c,d\}\subset[-1,1]$ such that $a+b+c+d=0$. Prove that:
$$\sqrt{1+a+b^2}+\sqrt{1+b+c^2}+\sqrt{1+c+d^2}+\sqrt{1+d+a^2}\geq4$$
I tried Holder and more, but without success.
|
Some thoughts
(For 3-variable problem, see: Prove $\sum_{\mathrm{cyc}} \sqrt{34x^2 + 28y^2 + 7z^2 - xy - 28yz + 41zx} \ge 9x + 9y + 9z$)
It suffices to prove the following:
Problem 1: Let $a, b, c, d \ge -1$ with $a+b+c+d = 0$. Prove that
$$\sqrt{1 + a + b^2} + \sqrt{1 + b + c^2} + \sqrt{1 + c + d^2} + \sqrt{1 + d + a^2} \ge 4.$$
(Thank @tthnew for verifying it by his routine.)
Problem 1 is equivalent to Problem 2 and Problem 3 below.
Problem 2: Let $x, y, z, w \ge 0$ with $x + y + z + w = 4$. Prove that
$$\sqrt{x + (y-1)^2} + \sqrt{y + (z-1)^2} + \sqrt{z + (w-1)^2} + \sqrt{w + (x-1)^2} \ge 4.$$
Problem 3: Let $x, y, z, w \ge 0$. Prove that
$$\sum_{\mathrm{cyc}} \sqrt{5x^2 + 9y^2 + z^2 + w^2 - 2xy + 6xz + 6xw - 6yz - 6yw + 2zw} \ge 4(x+y+z+w).$$
|
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|
Prove by induction that $n^3 < 3^n$ The question is prove by induction that $n^3 < 3^n$ for all $n\ge4$.
The way I have been presented a solution is to consider:
$$\frac{(d+1)^3}{d^3} = (1 + \frac{1}{d})^3 \ge (1.25)^3 = (\frac{5}{4})^3 = \frac{125}{64} <2 < 3$$
Then using this $$(d+1)^3 = d^3 \times \frac{(d+1)^3}{d^3} < 3d^3 < 3 \times 3^d = 3^{d+1}$$ so we have shown the inductive step and hence skipping all the easy parts the above statement is true by induction.
However I don't find this method very intuitive or natural; is there another way to attack this problem?
The approach I wish to take involves starting from $$ 3^{d+1} = 3 \times3^d > 3d^3$$ but then I do not know how show further that $3d^3 > (d+1)^3 $ to complete the inductive step. I have looked around at the proofs related to showing that $2^n > n^2 $ inductively for $n \ge 5$ but cannot relate the proof for that case to this case.
Also, is there a more general method that could be used to solve, say $a^n > n^a $ for $ n \ge k $ for some $k\in \Bbb R$
|
This base case holds because $4^3 < 3^4$.
To show that the inductive step holds, we need to show that:
$(n + 1)^3 < 3^{n + 1}$ holds if $n^3 < 3^n$ holds.
Note that:
$3^{n + 1} = 3 * 3^n > 3n^3$(since $3^n > n^3$ by the inductive hypothesis) > $(n + 1)^3$.
By binomial expansion: $(n + 1)^3 = n^3 + 3n^2 + 3n + 1$.
So:
$3n^3 ≥ (n + 1)^3\Leftrightarrow 3n^3 ≥ n^3 + 3n^2 + 3n + 1 \Leftrightarrow 2n^3 - 3n^2 - 3n - 1 ≥ 0. $
Since $2n^3 - 3n^2 - 3n - 1 = 0$ has a real solution at about $n \approx 2.26$ and $f(3) > 0$, we see that $2n^3 - 3n^2 - 3n - 1 > 0$ holds on the interval $(2.26,\infty)$. Then, because $4^3 < 3^4$, we see that:
$$3^{n + 1} > (n + 1)^3$$ for all $n \geq 4$, which is what we wanted to show.
Hope it helps.
|
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|
prove $(a^3+1)(b^3+1)(c^3+1)\ge 8$ let $a,b,c\ge 0$ and such $a+b+c=3$ show that
$$(a^3+1)(b^3+1)(c^3+1)\ge 8$$
My research:It seem use Holder inequality,so
$$(a^3+1)(b^3+1)(c^3+1)\ge (abc+1)^3$$
Use AM-GM
$$abc\le\dfrac{(a+b+c)^3}{27}=1$$
what? then I think this method is wrong
|
$\sum\limits_{cyc}\left(\ln(a^3+1)-\ln2\right)=\sum\limits_{cyc}\left(\ln(a^3+1)-\ln2-\frac{3}{2}(a-1)\right)$.
Let $f(x)=\ln(x^3+1)-\ln2-\frac{3}{2}(x-1)$.
Easy to see that $\min\limits_{[0,2]}f=0$.
Indeed, $f'(x)=\frac{3x^2}{x^3+1}-\frac{3}{2}=\frac{3(1-x)(x^2-x-1)}{2(x^3+1)}$ and since $x_{max}=\frac{1+\sqrt5}{2}$,
it remains to check that $f(2)>0$, which is true.
Thus, for $\{a,b,c\}\subset[0,2]$ our inequality is proven.
Let $a>2$.
Hence, $\prod\limits_{cyc}(a^3+1)\geq(8+1)\cdot1\cdot1>8$.
Done!
|
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|
What is the remainder left after dividing $1! + 2! + 3! + ... + 100!$ by $5$? I have this question as a homework.
What is the remainder left after dividing $1! + 2! + 3! + \cdots + 100!$ by $5$?
I tried this: I noticed that every $n! \equiv 0 \pmod{5}$ for every $n\geq 5$.
For $n < 5$:
$$\begin{align*}
1! &\equiv 1 \pmod{5} \\
2! &\equiv 2 \pmod{5} \\
3! &\equiv 1 \pmod{5} \\
4! &\equiv 4 \pmod{5} \\
\end{align*}$$
So $1! + 2! + \cdots + 100! \equiv 8 \equiv 3 \pmod{5}$.
Therefore the remainder is $3$.
Am I thinking properly?
Thanks a lot.
|
For the sake of an answer:
YES. You are thinking properly. What you did is nice and correct.
Similar Related questions:
What is the remainder when $1! + 2! + 3! +\cdots+ 1000!$ is divided by $12$?
what is the remainder when $1!+2!+3!+4!+\cdots+45!$ is divided by 47?
|
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|
To prove the inequality:- $\frac{4^m}{2\sqrt{m}}\le\binom{2m}{m}\le\frac{4^m}{\sqrt{2m+1}}$
Problem Statment:-
Prove:-$$\dfrac{4^m}{2\sqrt{m}}\le\binom{2m}{m}\le\dfrac{4^m}{\sqrt{2m+1}}$$
My Attempt:-
We start with $\binom{2m}{m}$ (well that was obvious), to get
$$\binom{2m}{m}=\dfrac{2^m(2m-1)!!}{m!}$$
Now, since $2^m\cdot(2m-1)!!\lt2^m\cdot2^m\cdot m!\implies \dfrac{2^m\cdot(2m-1)!!}{m!}\lt 4^m$
$$\therefore \binom{2m}{m}=\dfrac{2^m(2m-1)!!}{m!}\lt4^m$$
Also, $$2^m\cdot(2m-1)!!\gt2^m\cdot(2m-2)!!\implies 2^m(2m-1)!!\gt2^m\cdot2^{m-1}\cdot(m-1)!\\
\implies \dfrac{2^m\cdot(2m-1)!!}{m!}\gt\dfrac{4^m}{2m}$$
So, all I got to was $$\dfrac{4^m}{2m}\lt\binom{2m}{m}\lt4^m$$
So, if anyone can suggest me some modifications to my proof to arrive at the final result, or just post a whole different non-induction based proof.
|
Taking the product of the ratios of the terms gives
$$
\binom{2n}{n}=\prod_{k=1}^n4\frac{k-1/2}{k}\tag{1}
$$
Bernoulli's Inequality says
$$
\sqrt{\frac{k-1}k}\le\frac{k-1/2}{k}\le\sqrt{\frac{k-1/2}{k+1/2}}\tag{2}
$$
Applying $(2)$ to $(1)$, we get
$$
\frac{4^n}{2\sqrt{n}}\le\binom{2n}{n}\le\frac{4^n}{\sqrt{2n+1}}\tag{3}
$$
In this answer, it is shown that
$$
\frac{4^n}{\sqrt{\pi\left(n+\frac13\right)}}\le\binom{2n}{n}\le\frac{4^n}{\sqrt{\pi\left(n+\frac14\right)}}\tag{4}
$$
which is a much tighter estimate.
|
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|
If $abc+1=0$, verify that' If $abc+1=0$, prove that:
$\frac {1}{1-a-b^{-1}}+\frac {1}{1-b-c^{-1}} +\frac {1}{1-c-a^{-1}}=1$.
My Attempt:
$abc+1=0$
$abc=-1$.
Now,
$$L.H.S=\frac {1}{1-a-b^{-1}}+\frac {1}{1-b-c^{-1}}+\frac {1}{1-c-a^{-1}}$$
$$=\frac{b}{b-ab-1} + \frac {1}{1-b-c^{-1}} + \frac {c^{-1}}{c^{-1}-1-(ca)^{-1}}$$
$$=\frac{b}{b+c^{-1} -1} +\frac{1}{1-b-c^{-1}} +\frac{c^{-1}}{c^{-1}-1+b}$$
What should I do next?
|
$$\dfrac1{1-a-b^{-1}}=\dfrac{bc}{bc-abc-c}=\dfrac{bc}{bc+1-c}$$
$$\dfrac1{1-b-c^{-1}}=\dfrac{-c}{-c+bc+1}$$
As $-abc=1,-a^{-1}=bc$
$$\dfrac1{1-c-a^{-1}}=\dfrac1{1-c+bc}$$
|
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|
Short technique of tackling or another method $\int_{0}^{\infty}{2x\over (x^4+2x^2+1)+\sqrt{x^4+2x^2+1}}dx=\ln{2}$ Prove that,
$$I=\int_{0}^{\infty}{2x\over (x^4+2x^2+1)+\sqrt{x^4+2x^2+1}}dx=\ln{2}$$
I try:
$x^4+2x^2+1=(x^2+1)^2$
$$\int_{0}^{\infty}{2x\over (x^2+1)(x^2+2)}dx$$
Let $u=x^2+1$, $du=2xdx$
$$\int_{1}^{\infty}{1\over u(u+1)}du$$
$$\int_{1}^{\infty}\left({1\over u}-{1\over u+1}\right)du$$
$$I=\ln{1}-\ln{{1\over2}}=\ln{2}$$
Is there another short way of integrating this integral?
|
Note that $$I=\int_{0}^{\infty}\frac{2x}{\left(x^{2}+1\right)\left(x^{2}+2\right)}dx=2\int_{0}^{\infty}\frac{1}{x}\left(\frac{x^{2}}{x^{2}+1}-\frac{\left(x/\sqrt{2}\right)^{2}}{\left(\frac{x}{\sqrt{2}}\right)^{2}+1}\right)dx
$$ hence by the Frullani's theorem we get $$I=-2\log\left(\frac{1}{\sqrt{2}}\right)=\color{red}{\log\left(2\right)}.$$
|
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|
The sum of $\binom{n}{0}\binom{3n}{2n}-\binom{n}{1}\binom{3n-3}{2n-3}+\binom{n}{2}\binom{3n-6}{2n-6}+\cdots \cdots $
The sum of $$\binom{n}{0}\binom{3n}{2n}-\binom{n}{1}\binom{3n-3}{2n-3}+\binom{n}{2}\binom{3n-6}{2n-6}-\cdots \cdots \cdots $$
$\bf{My\; Try::}$ We can write above sum as $$\sum^{n}_{k=0}(-1)^k\binom{n}{k}\binom{3n-3k}{2n-3k} = \sum^{n}_{k=0}(-1)^k\binom{n}{k}\binom{3n-3k}{n}$$
So sum $$ = \sum^{n}_{k=0}(-1)^k\cdot \frac{n!}{k!\cdot (n-k)!}\times \frac{(3n-2k)!}{n!\cdot (2n-3k)!} = \sum^{n}_{k=0}(-1)^k \cdot \frac{(3n-2k)!}{k!\cdot (n-k)!\cdot (2n-3k)!}$$
Now i did not understand how can i solve it after that , help Required, Thanks
|
This also has a very simple algebraic proof. Suppose we seek to
evaluate
$$S_n = \sum_{k=0}^n {n\choose k} (-1)^k {3n-3k\choose n}.$$
Introduce
$${3n-3k\choose n} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2n-3k+1}} \frac{1}{(1-z)^{n+1}} \; dz.$$
We get for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2n+1}} \frac{1}{(1-z)^{n+1}}
\sum_{k=0}^n {n\choose k} (-1)^k z^{3k}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2n+1}} \frac{1}{(1-z)^{n+1}}
(1-z^3)^n
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2n+1}} \frac{1}{1-z}
(1+z+z^2)^n
\; dz.$$
Now residues sum to zero so we have from the poles at zero, one and
infinity
$$S_n - 3^n +
\mathrm{Res}_{z=\infty} \frac{1}{z^{2n+1}} \frac{1}{1-z}
(1+z+z^2)^n = 0.$$
Note that the residue at infinity is
$$-\mathrm{Res}_{z=0} \frac{1}{z^2} z^{2n+1}
\frac{1}{1-1/z} (1+1/z+1/z^2)^n
\\= -\mathrm{Res}_{z=0} z^{2n}
\frac{1}{z-1} (1+1/z+1/z^2)^n
\\ = -\mathrm{Res}_{z=0}
\frac{1}{z-1} (z^2+z+1)^n = 0.$$
This leaves
$$\bbox[5px,border:2px solid #00A000]{S_n = 3^n.}$$
|
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Determine all $a,b,c\in \mathbb{Z}$ for which the equation $(a^2+b^2)x^2-2(b^2+c^2)x-(c^2+a^2)=0$ has rational roots. Determine all $a,b,c\in \mathbb{Z}$ for which the equation $(a^2+b^2)x^2-2(b^2+c^2)x-(c^2+a^2)=0$ has rational roots.. I know $\Delta \ge 0$ and $\sqrt{\Delta}$ must be rational.
|
This is partial solution.
As Χpẘ suggested, we need to determine $a,b,c$ which makes $a^4+b^4+c^4+a^2b^2+a^2c^2+3b^2c^2$ perfect square. We know that $(a^2+b^2+c^2)^2=a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2$ is a square.
We know that $(a,0,0), (0,b,0), (0,0,c)$ is solution. Therefore, let's assume that at most one of $a,b,c$ is $0$. Also, we are dealing with squares, so we can assume that $a,b,c$ is all non-negative. Since if $(x,y,z)$ is a solution, then $(kx,ky,kz)$ is also a solution, we may assume $\gcd(a,b,c)=1$. Also, the equation is symmetric with respect to $b$ and $c$, so we may assume that $b \ge c$.
Case 1: $a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2 = a^4+b^4+c^4+a^2b^2+a^2c^2+3b^2c^2$
If then, $(b^2-a^2)(c^2-a^2)=a^4$. If $a=0$, there is no solution since $b,c$ is nonzero. Similarly, we cannot assume that $b$ or $c$ is zero.
Also if $c \le b < a$, then $0>b^2-a^2>-a^2$ and $0>c^2-a^2>-a^2$, so no solution. If $c \le a \le b$, then LHS is $0$ or negative. Therefore $a<c\le b$.
Let $b^2-a^2=\frac{p}{q}a^2$ and $c^2-a^2=\frac{q}{p}a^2$. Clearly $p \ne q$ and from $b \ge c$, $p > q$. Then we know that both $\frac{p+q}{q}$ and $\frac{p+q}{p}$ are perfect squares, also their reciprocals $\frac{q}{p+q}$ and $\frac{p}{p+q}$. Their sum is exactly $1$ and we know all nontrivial rational solutions to this equation corresponds to Pythagorean triple. Therefore, take any Pythagorean triple $l,m,n$ with $l^2+m^2=n^2$ and $l<m<n$, we have $b^2l^2=a^2n^2$ and $c^2m^2=a^2n^2$. It follows that $a=lm$, $b=mn$, $c=ln$.
Case 2: $a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2 \ne a^4+b^4+c^4+a^2b^2+a^2c^2+3b^2c^2$
This case is really hard... I cannot make any progress worth mentioning.
I made a computer search for the solutions under 1k of Case 2. There are about 760 solutions, and you can see the list and code used here.
def is_square(apositiveint):
x = apositiveint // 2
seen = set([x])
while x * x != apositiveint:
x = (x + (apositiveint // x)) // 2
if x in seen: return False
seen.add(x)
return True
for i in range(0, 1000):
for j in range(0, 1000):
for k in range(j, 1000):
number = i ** 4 + j ** 4 + k ** 4 + 3 * j * j * k * k + i * i * (j * j + k * k)
if number > 2 and not (i == j == 0 or j == k == 0 or k == i == 0) and is_square(number) and (i * i + j * j + k * k) ** 2 != number:
print i, j, k, (number ** 0.5 - (i * i + j * j + k * k)) / 18
https://docs.google.com/spreadsheets/d/1gij-LfwLYKCq5TnYlRtDnbem988Udu6zZU-a7iGMjtY/edit?usp=sharing
|
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|
How to differentiate $y=\ln(x+\sqrt{1+x^2})$? I'm trying to differentiate the equation below but I fear there must have been an error made. I can't seem to reconcile to the correct answer. The problem comes from James Stewart's Calculus Early Transcendentals, 7th Ed., Page 223, Exercise 25.
Please differentiate $y=\ln(x+\sqrt{1+x^2})$
My Answer:
Differentiate using the natural log rule:
$$y'=\left(\frac{1}{x+(1+x^2)^{1/2}}\right)\cdot\left(x+(1+x^2)^{1/2}\right)'$$
Now to differentiate the second term, note the chain rule applied and then simplification:
$$\left(x+(1+x^2)^{1/2}\right)'=1+\frac{1}{2}\cdot(1+x^2)^{-1/2}\cdot(2x)$$
$$1+\frac{1}{2}\cdot(1+x^2)^{-1/2}\cdot(2x)=1+\frac{x}{(1+x^2)^{1/2}}$$
Our expression is now:
$$y'=\left(\frac{1}{x+(1+x^2)^{1/2}}\right)\cdot\left(1+\frac{x}{(1+x^2)^{1/2}}\right)$$
Distribute the left term across the two right terms for my result:
$$y'=\left(\frac{1}{x+(1+x^2)^{1/2}}\right)+\left(\frac{x}{\left(x+(1+x^2)^{1/2}\right)\left(1+x^2\right)^{1/2}}\right)$$
$$y'=\left(\frac{1}{x+(1+x^2)^{1/2}}\right)+\left(\frac{x}{\left(x(1+x^2)^{1/2}\right)+(1+x^2)^{1}}\right)$$
At this point I can see that if I simplify further by adding the fractions I'll still have too many terms, and it will get awfully messy. The answer per the book (below) has far fewer terms than mine. I'd just like to know where I've gone wrong in my algebra. Thank you for your help. Here's the correct answer:
$$y'=\frac{1}{\sqrt{1+x^2}}$$
|
You wrote:
Distribute the left term across the two right terms
That is the same mistake that we see when we consider simplifying $\dfrac 3 4\times \dfrac 5 3.$ One can multiply:
$$
\frac{3\times 5}{4\times 3} = \frac{15}{12}
$$
but that is not a good way to simplify. Usually one should cancel before multiplying:$\require{cancel}$
$$
\frac{\bcancel 3} 4 \times \frac 5 {\bcancel 3}.
$$
Do the same thing here: cancel before multiplying:
\begin{align}
\left(\frac{1}{x+(1+x^2)^{1/2}}\right)\cdot\left(1+\frac{x}{(1+x^2)^{1/2}}\right) & = \left(\frac{1}{x+(1+x^2)^{1/2}}\right)\cdot\left( \frac{ (1+x^2)^{1/2} + x}{(1+x^2)^{1/2}} \right) \\[10pt]
& =\left(\frac 1 {\cancel{x+(1+x^2)^{1/2}}}\right)\cdot\left( \frac{ \cancel{ (1+x^2)^{1/2} + x}}{(1+x^2)^{1/2}} \right)
\end{align}
|
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|
Prove $10^{n+1}+3\cdot 10^n+5$ is divisible by $9$? How do I prove that an integer of the form $10^{n+1}+3\cdot 10^{n}+5$ is divisible by $9$ for $n\geq 1$?I tried proving it by induction and could prove it for the Base case n=1. But got stuck while proving the general case. Any help on this ? Thanks.
|
Let $S(n)$ be the statement: $10^{n+1}+3\cdot{10^{n}}+5$ is divisible by $9$
Basis step: $S(1)$:
$\Rightarrow 10^{(1)+1}+3\cdot{10^{(1)}}+5=10^{2}+30+5$
$\hspace{45.5 mm}=100+35$
$\hspace{45.5 mm}=135$, which is divisible by $9$
Inductive step:
Assume $S(k)$ is true, i.e. assume that $10^{k+1}+3\cdot{10^{k}}+5$ is divisible by $9$
$\hspace{59 mm} \Rightarrow 10^{n+1}+3\cdot{10^{n}}+5=9A$
$\hspace{59 mm} \Rightarrow 10\cdot{10^{n}}+3\cdot{10^{n}}+5=9A$
$\hspace{59 mm} \Rightarrow 13\cdot{10^{n}}+5=9A$
$\hspace{59 mm} \Rightarrow 10^{n}=\dfrac{9A-5}{13}$
Then, $S(k+1)$: $10^{(k+1)+1}+3\cdot{10^{(k+1)}}+5$
$\hspace{23.5 mm} =10^{k+2}+3\cdot{10}\cdot{10^{k}}+5$
$\hspace{23.5 mm} =100\cdot{10^{k}}+30\cdot{10^{k}}+5$
$\hspace{23.5 mm} =130\cdot{10^{k}}+5$
$\hspace{23.5 mm} =130\cdot{\bigg(\dfrac{9A-5}{13}\bigg)}+5$
$\hspace{23.5 mm} =10\cdot{(9A-5)}+5$
$\hspace{23.5 mm} =90A-50+5$
$\hspace{23.5 mm} =90A-45$
$\hspace{23.5 mm} =9\hspace{1 mm}(10A-5)$, which is divisible by $9$
So, $S(k+1)$ is true whenever $S(k)$ is true.
Therefore, $10^{n+1}+3\cdot{10^{n}}+5$ is divisible by $9$.
|
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|
Simplifying $9^{3/4}$, I get $3\sqrt[4]{9}$, but that's not the answer. Why? I am trying to simplify:
$9^\frac{3}{4}$
So this is what I did:
$9^\frac{3}{4} = \sqrt[4]{9^3}$
$\sqrt[4]{3*3*3*3*3*3}$
$3\sqrt[4]{3*3}$
$3\sqrt[4]{9}$
$3\sqrt[4]{3^2}$
I don't see how I can simplify this even more, however the answer I provided is incorrect. How can I simplify this even more?
|
We know that $9=3^2$ .So, $$\sqrt [4]{9^3} =\sqrt [4]{(3^2)^3} =\sqrt [4]{3^2*3^2*3^2} =\sqrt {3*3*3*3*3*3} $$ After this you have proceeded correctly. You can simplify the last step as: $$ 3\sqrt [4]{3^2} =3\times 3^{2/4} =3\times 3^{1/2} =3\sqrt {3} $$ Hope it helps.
|
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|
The inequality $a^cb^d(c+d)^{c+d}\le c^cd^d(a+b)^{c+d}$
The inequality $a^cb^d(c+d)^{c+d}\le c^cd^d(a+b)^{c+d}$ for $a,b,c,d>0$
The inequality is equivalent to:
$$\displaystyle\frac{a^cb^d}{(a+b)^{c+d}}\le\frac{c^cd^d}{(c+d)^{c+d}}$$
and the right side should be at least $\left(\frac{1}{2}\right)^{c+d}$ if the rearrangement inequality is true also for reals.
So it would be sufficient if left side is below $\left(\frac{1}{2}\right)^{c+d}$, but this seems to be true that if only $a,b,c,d\gt1$.
can you help
|
Since $f(x)=x\ln x$ is a convex function, by Jensen we obtain:
$$\frac{a}{a+b}\left(\frac{c}{a}\ln\frac{c}{a}\right)+\frac{b}{a+b}\left(\frac{d}{b}\ln\frac{d}{b}\right)\geq\left(\frac{a}{a+b}\cdot\frac{c}{a}+\frac{b}{a+b}\cdot\frac{d}{b}\right)\ln\left(\frac{a}{a+b}\cdot\frac{c}{a}+\frac{b}{a+b}\cdot\frac{d}{b}\right)$$ or
$$\left(\frac{c}{a}\right)^c\left(\frac{d}{b}\right)^d\geq\left(\frac{c+d}{a+b}\right)^{c+d},$$
which is your inequality.
Done!
|
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|
Find the range of the function $f(x) = \sqrt{\tan^{-1}x+1}+\sqrt{1-\tan^{-1}x}$ Problem :
Find the range of the function $f(x) = \sqrt{\tan^{-1}x+1}+\sqrt{1-\tan^{-1}x}$
My approach :
Let $\tan^{-1}x =t $
$y = \sqrt{t+1}+\sqrt{1-t}$
Squaring both sides we get :
$y^2= t+1+1-t +2\sqrt{(t+1)(1-t)}$
$\Rightarrow y^2= 2+2\sqrt{(t+1)(1-t)}$
Now how to get the range of this function, please guide, will be of great help , thanks.
|
Note that $f(x)$ is even, so we may find extrema in either $x\geq 0$ or $x \leq 0$.
Also, we can find the domain of the function to be $[\tan(1), \tan(1)]$.
Now, consider $f'(x)$ for x $\geq0$:
$$f'(x)=\left(\dfrac{-1}{\sqrt{1-\arctan(x)}}+\dfrac{1}{\sqrt{1+\arctan(x)}}\right)\times\dfrac{1}{2(1+x^2)}$$
$f'(x)< 0$ for $x > 0$. Therefore f is decreasing for $x > 0$.
Extrema occurs at endpoints of the interval.
Therefore the range is $[\sqrt{2},2]$
|
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|
Another variant of vandermonde's identity: $\binom{r}{m+k} \binom{s}{n+k}$ It can be shown that $$\sum_{k}\binom{r}{m+k} \binom{s}{n+k} = \binom{r+s}{r-m+n}$$ using either lattice paths or manipulation of the binomial coefficients in the identity. How so? I've played with this for hours, and the only worthwhile offering (which is not that hard to guess) I have is that on the lattice of dimension
$$(r-m+n) \times (s-n+m)$$
which represents the RHS binomial coefficient,
the lattice paths represented by $\binom{r}{m+k}$ lie on it. In particular, their right-most corners lie on one line.
|
Observe that we certainly get all non-zero values when $k\ge -m-n$
so we may write
$$\sum_k {r\choose m+k} {s\choose n+k}
= \sum_{k\ge -m-n} {r\choose m+k} {s\choose n+k}
\\ = \sum_{k\ge 0} {r\choose k-n} {s\choose k-m}
= \sum_{k\ge 0} {r\choose r+n-k} {s\choose s+m-k}.$$
Now put
$${r\choose r+n-k} =
\int_{|z|=\epsilon}
\frac{1}{z^{r+n-k+1}} (1+z)^r \; dz$$
and
$${s\choose s+m-k} =
\int_{|w|=\gamma}
\frac{1}{w^{s+m-k+1}} (1+w)^s \; dw$$
so that we obtain for the sum
$$\int_{|z|=\epsilon}
\frac{1}{z^{r+n+1}} (1+z)^r
\int_{|w|=\gamma}
\frac{1}{w^{s+m+1}} (1+w)^s
\sum_{k\ge 0} z^k w^k
\; dw\; dz
\\ = \int_{|z|=\epsilon}
\frac{1}{z^{r+n+1}} (1+z)^r
\int_{|w|=\gamma}
\frac{1}{w^{s+m+1}} (1+w)^s
\frac{1}{1-wz}
\; dw\; dz.$$
This converges for $|zw|\lt 1.$ Now in the inner integral residues sum
to zero. We obtain the desired sum for the residue from the pole at
$w=0.$ Write for the residue from the pole at $w=1/z$
$$-\int_{|z|=\epsilon}
\frac{1}{z^{r+n+2}} (1+z)^r
\int_{|w|=\gamma}
\frac{1}{w^{s+m+1}} (1+w)^s
\frac{1}{w-1/z}
\; dw\; dz$$
for a contribution of
$$-\int_{|z|=\epsilon}
\frac{1}{z^{r+n+2}} (1+z)^r
z^{s+m+1} \frac{(1+z)^s}{z^s} \; dz
= -\int_{|z|=\epsilon}
\frac{1}{z^{r+n-m+1}} (1+z)^{r+s}
\; dz
\\ = -{r+s\choose r-m+n}.$$
For the residue at infinity we get
$$\mathrm{Res}_{w=\infty} \frac{1}{w^{s+m+1}} (1+w)^s
\frac{1}{1-wz}
\\ = -\mathrm{Res}_{w=0} \frac{1}{w^2} w^{s+m+1}
\frac{(1+w)^s}{w^s} \frac{1}{1-z/w}
= -\mathrm{Res}_{w=0} w^{m} \frac{(1+w)^s}{w-z} = 0.$$
Hence we have the closed form
$$\bbox[5px,border:2px solid #00A000]{
{r+s\choose r-m+n} = {r+s\choose s-n+m}.}$$
|
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|
Choosing one type of ball without replacement. Suppose I have $9$ balls, among which $3$ are green and $6$ are red. What is the probability that a ball randomly chosen is green?
It is $\dfrac{3}{9}=\dfrac{1}{3}$.
If three balls are randomly chosen without replacement, then what is the probability that the three balls are green?
Is it $\dfrac{3}{9}\times\left\{\dfrac{2}{8}+\dfrac{3}{8}\right\}\times\left\{\dfrac{1}{7}+\dfrac{2}{7}+\dfrac{3}{7}\right\}=\dfrac{90}{504}$?
But in hypergeometric distribution formula, it is
$$f(X=3)=\dfrac{\binom{3}{3}\binom{9-3}{3-3}}{\binom{9}{3}}=\dfrac{1}{84}.$$
|
The answer is given by:
$$
P = \frac{3}{9} \cdot \frac{2}{8}\cdot \frac{1}{7} = \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{1}{7} = \frac{1}{84}
$$
|
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|
Show that:$\sum\limits_{n=1}^{\infty}{n\over (4n^2-1)(16n^2-1)}={1\over 12}(1-\ln{2})$ Show that
$$\sum_{n=1}^{\infty}{n\over (4n^2-1)(16n^2-1)}={1\over 12}(1-\ln{2})$$
My try:
We split into partial decomposition
$$n={A\over 2n-1}+{B\over 2n+1}+{C\over 4n-1}+{D\over 4n+1}$$
Setting $n={1\over 2}$, ${-1\over2}$ we have $A={1\over3}$ and $B={-1\over 3}$
Finding C and D is a bit tedious
I wonder what is the closed form for
$$\sum_{n=1}^{\infty}{1\over an+b}=F(a,b)?$$
This way is not a good approach. Can anyone help me with a better approach to tackle this problem? Thank you.
|
$\left(\frac{1}{an+b}\right)_{\substack{n\in\mathbb{N}\\an+b\neq0}}$ is not summable if $a\neq0$ (this has the same behavior as the harmonic series). Thus $\sum\frac{1}{an+b}$ diverges, and your $F(a,b)$ is not well-defined.
However you can sum up to some integer $N$, and use an asymptotic development of:
$$\sum_{n=1}^{N}\frac{1}{an+b}=\frac{1}{a}\sum_{n=1}^{N}\frac{1}{n+\frac{b}{a}}$$
This can be done typically with a comparison with an integral.
UPD: Actually you just need to know an asymptotic estimation of the harmonic series:
$$H_N=\sum_{n=1}^N\frac{1}{n}=\log(n)+\gamma+o(1)$$
Because:
$$\sum_{n=1}^{N}\frac{1}{n+\lfloor{\frac{b}{a}}\rfloor+1}<\sum_{n=1}^{N}\frac{1}{n+\frac{b}{a}}\le\sum_{n=1}^{N}\frac{1}{n+\lfloor\frac{b}{a}\rfloor}$$
Thus:
$$\sum_{n=1+\lfloor{\frac{b}{a}}\rfloor+1}^{N+\lfloor{\frac{b}{a}}\rfloor+1}\frac{1}{n}<\sum_{n=1}^{N}\frac{1}{n+\frac{b}{a}}\le\sum_{n=1+\lfloor{\frac{b}{a}}\rfloor}^{N+\lfloor{\frac{b}{a}}\rfloor}\frac{1}{n}$$
i.e.
$$H_{N+\lfloor{\frac{b}{a}}\rfloor+1}-H_{1+\lfloor{\frac{b}{a}}\rfloor}<\sum_{n=1}^{N}\frac{1}{n+\frac{b}{a}}\le H_{N+\lfloor\frac{b}{a}\rfloor}-H_{\lfloor\frac{b}{a}\rfloor}$$
Ok, let's write it.
$$\sum_{n=1}^{N}\frac{1}{an+b}=\frac{1}{a}\sum_{n=1}^{N}\frac{1}{n+\frac{b}{a}}$$
According to our previous estimations: $\sum_{n=1}^{N}\frac{1}{an+b}=\frac{1}{a}\log(N)+O(1)$. But actually we need a $o(1)$ precision. So let $\epsilon_N=\sum_{n=1}^{N}\frac{1}{an+b}-\frac{1}{a}\log(N)$.
$$\epsilon_{N}-\epsilon_{N-1}=\frac{1}{aN+b}-\frac{1}{a}(\log(N)-\log(N-1))=\frac{1}{aN+b}+\frac{1}{a}\log\left(1-\frac{1}{N}\right)$$
So, using asymptotic estimations of $\log(1+x)$ and $\frac{1}{1+x}$ in $x=0$:
$$\epsilon_{N}-\epsilon_{N-1} = \frac{1}{aN}\left(1-\frac{b}{aN}+o\left(\frac{1}{N}\right)\right)+\frac{1}{a}\left(-\frac{1}{N}-\frac{1}{2N^2}+o\left(\frac{1}{N^2}\right)\right)
\sim \frac{C}{N^2}$$
with $C=-\frac{b}{a^2}-\frac{1}{2a}$
Summing equivalents there exists $C'$ s.t. $\epsilon_N = C' - \frac{C}{N}+o\left(\frac{1}{N}\right)$
Now we can solve the original problem, using the partial decomposition-form and adding our estimations.
|
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|
System of $3$ nonlinear equations
Find positive solutions to the following:
\begin{align*}
x^2+y^2+xy=&1\\
y^2+z^2+yz=&3\\
z^2+x^2+xz=&4
\end{align*}
I simplified and got $x+y+z=\sqrt{7}$ and $x^2+y^2+xy=1$. How do I continue?
|
Through the cosine theorem, the problem can be stated in the following way:
In a triangle $ABC$ with side lengths $1,\sqrt{3},2$, what are the
distances of the Fermat point $F$ from the vertices of $ABC$?
Well, such a triangle is a right triangle, since $1^2+\sqrt{3}^2=2^2$, and $FA+FB+FC$ is the length of the Steiner net of $ABC$. It follows that:
$$ x+y+z = \sqrt{1^2+\sqrt{3}^2-2\sqrt{3}\cos 150^\circ}=\sqrt{7}$$
To find $x,y,z$, i.e. the lenghts of $FA,FB,FC$, it is enough to embed this construction in the plane by setting, for instance, $A=(0,0), B=(1,0), C=(0,2)$, then finding the coordinates of $F$ by intersecting a couple of lines. As an alternative, since the trilinear coordinates of $F$ are given by
$$ F = \left[\frac{1}{\sin(A+\pi/3)};\frac{1}{\sin(B+\pi/3)};\frac{1}{\sin(C+\pi/3)}\right]$$
with the previous assumptions we have
$$ F=\left[1;\frac{\sqrt{5}}{2+\sqrt{3}};\frac{\sqrt{5}}{1+2\sqrt{3}}\right]$$
and the tripolar coordinates of $F$ are $[1,2,4]$.
On the other hand, once we get $x+y+z=\sqrt{7}$ our job is done, since the original system of equations implies:
$$\left\{\begin{array}{rcr}(x-z)(x+y+z)&=&-2\\(y-x)(x+y+z)&=&-1\\(z-y)(x+y+z)&=&3 \end{array}\right. $$
or:
$$\left\{\begin{array}{rcr}z-x&=&\frac{2}{\sqrt{7}}\\x-y&=&\frac{1}{\sqrt{7}}\\z-y&=&\frac{3}{\sqrt{7}} \end{array}\right. $$
so that $(x,y,z)=\left(\tau,\tau-\frac{1}{\sqrt{7}},\tau+\frac{2}{\sqrt{7}}\right)$.
By solving the original system with respect to $\tau$, we get $\tau=\frac{2}{\sqrt{7}}$. The other solutions (non-geometrical, associated with negative lengths) can be derived in a similar way.
|
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|
Show that $\sqrt{4 + 2\sqrt{3}} - \sqrt{3}$ is rational.
Show that $\sqrt{4 + 2\sqrt{3}} - \sqrt{3}$ is rational.
I've tried to attempt algebra on this problem. I noticed that there is some kind of nesting effect when trying to solve this. Please help me to understand how to attempt to denest this number.
Any help would be greatly appreciated.
|
Method 1: Consider this denesting algorithm:
Denested square roots: Given a radical of the form $\sqrt{X\pm {Y}}$ with $X,Y\in\mathbb{R}$ and $X>Y$, we have a possible simplification as$$\sqrt{X\pm Y}=\sqrt{\dfrac {X+\sqrt{X^2-Y^2}}2}\pm\sqrt{\dfrac {X-\sqrt{X^2-Y^2}}2}\tag1$$
Using $(1)$ on $\sqrt{4+2\sqrt3}$, we have$$\sqrt{4+2\sqrt3}=\sqrt{\dfrac {4+2}2}+\sqrt{\dfrac {4-2}2}=\sqrt3+1\tag2$$
So the original expression becomes$$\color{brown}{\sqrt{4+2\sqrt3}}-\sqrt3=\color{brown}{\sqrt3+1}-\sqrt3=1$$
Which is rational.
Method 2: If you don't like $(1)$ and think it's too complicated, I present you an alternative method. Simply set $\sqrt{4+2\sqrt{3}}-\sqrt3$ equal to a variable and simplify!
Here, we have$$\begin{align*} & \sqrt{4+2\sqrt3}-\sqrt3=\alpha\\ & \sqrt{4+2\sqrt3}=\alpha+\sqrt3\\ & 4+2\sqrt3=(\alpha+\sqrt3)^2\\ & 4+2\sqrt3=\alpha^2+3+2\alpha\sqrt3\end{align*}$$
To solve for $\alpha$, we have $2\alpha\sqrt3=2\sqrt3\implies\alpha=1$. Checking with the other half, $\alpha^2+3=1+3=4$ holds. Hence,$$\sqrt{4+2\sqrt3}-\sqrt3=1$$
|
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"url": "https://math.stackexchange.com/questions/2076737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 4
}
|
Linear transformation matrix of vector space $\mathbb R^{2x2}$
Select a basis B of a vector space $\mathbb R^{2x2}$ and for linear transformation $f:\mathbb R^{2x2}→\mathbb R^{2x2}$ given by $f(X) = \begin{pmatrix}1 & 0 \\ 1 & 0 \end{pmatrix}X+\begin{pmatrix}1 & 1 \\ 1 & 1 \end{pmatrix}X^T$ compute the matrix relative to the base B.
I selected canonical basis $B=\{\begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix},\begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix},\begin{pmatrix}0 & 0 \\ 1 & 0 \end{pmatrix},\begin{pmatrix}0 & 0 \\ 0 & 1 \end{pmatrix}\}$
Then I computed transformation of each vector in the basis according to the task.
Got $f(B)=\{\begin{pmatrix}2 & 0 \\ 2 & 0 \end{pmatrix},\begin{pmatrix}1 & 1 \\ 1 & 1 \end{pmatrix},\begin{pmatrix}0 & 1 \\ 0 & 1 \end{pmatrix},\begin{pmatrix}0 & 1 \\ 0 & 1 \end{pmatrix}\}$, if I'm not wrong.
What should I do now?
|
If it's indeed the case that
$$f\begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix} = 2 \begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix} + 2 \begin{pmatrix}0 & 0 \\ 1 & 0 \end{pmatrix}$$
(I didn't check your calculations), then this would tell us that the first row of the matrix is $2, 0, 2, 0$, and so forth.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2078951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
sum of series $\frac{n}{1\cdot 2 \cdot 3}+\frac{n-1}{2\cdot 3\cdot 4}+\frac{n-2}{3\cdot 4 \cdot 5}+\cdots \cdots n$ terms sum of series $\displaystyle \frac{n}{1\cdot 2 \cdot 3}+\frac{n-1}{2\cdot 3\cdot 4}+\frac{n-2}{3\cdot 4 \cdot 5}+\cdots \cdots n$ terms
assuming $\displaystyle S_{n} =\frac{n}{1\cdot 2 \cdot 3}+\frac{n-1}{2\cdot 3\cdot 4}+\frac{n-2}{3\cdot 4 \cdot 5}+\cdots +\frac{n-(n-1)}{n\cdot n+1 \cdot n+2}$
$\displaystyle S_{n} = \sum^{n-1}_{r=0}\frac{n-r}{(r+1)\cdot (r+2) \cdot (r+3)}$
wan, t be able to solve after that , could some help me with this
|
Hint:
$\dfrac{n-r}{(r+1)(r+2)(r+3)}=\dfrac{n+1-(r+1)}{(r+1)(r+2)(r+3)}$
$=(n+1)\cdot\dfrac1{(r+1)(r+2)(r+3)}-\dfrac1{(r+2)(r+3)}$
Now $\dfrac1{(r+2)(r+3)}=\dfrac{r+3-(r+2)}{(r+2)(r+3)}=?$
See Telescoping series
Again, $\dfrac2{(r+1)(r+2)(r+3)}=\dfrac{r+3-(r+1)}{(r+1)(r+2)(r+3)}=\dfrac1{(r+2)(r+1)}-\dfrac1{(r+2)(r+3)}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2079081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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|
Help find closed form for:$\sum_{n=0}^{\infty}{(-1)^n\left({{\pi\over 2}}\right)^{2n}\over (2n+k)!}=F(k)$ What is the closed form for
$$\sum_{n=0}^{\infty}{(-1)^n\left({{\pi\over 2}}\right)^{2n}\over (2n+k)!}=F(k)?$$
My try:
I have found a few values of $F(k)$, but was unable to find a closed form for it.
$F(0)=0$
$F(1)={2\over \pi}$
$F(2)=\left({2\over \pi}\right)^2$
$F(3)=\left({2\over \pi}\right)^2-\left({2\over \pi}\right)^3$
$F(4)={1\over 2}\left({2\over \pi}\right)^2-\left({2\over \pi}\right)^4$
$F(5)={1\over 6}\left({2\over \pi}\right)^2-\left({2\over \pi}\right)^4+\left({2\over \pi}\right)^5$
|
The sum can also be expressed in terms of the Regularized Incomplete Gamma function ($Q(a,z)$).
In fact, premised that for a general function of a integer $f(k)$ we have
$$
\begin{gathered}
\frac{{\left( {i^{\,k} + \left( { - i} \right)^{\,k} } \right)}}
{2}f(k) = \left( {\frac{{e^{\,i\,k\frac{\pi }
{2}} + e^{\, - \,i\,k\frac{\pi }
{2}} }}
{2}} \right)f(k) = \cos \left( {k\frac{\pi }
{2}} \right)f(k) = \hfill \\
= \left[ {k = 2j} \right]\left( { - 1} \right)^{\,j} f(2j)\quad \left| {\;k,j\; \in \;\;\mathbb{Z}\,} \right. \hfill \\
\end{gathered}
$$
and that the Lower Incomplete Gamma function can be expressed as:
$$
\begin{gathered}
\gamma (s,z) = \int_{\,0\,}^{\,z\,} {t^{\,s - 1} \;e^{\, - \,t} dt} = \hfill \\
= z^{\,s} \;e^{\,\, - z} \;\Gamma (s)\;\sum\limits_{0\, \leqslant \,k} {\frac{{z^{\,k} }}
{{\Gamma (s + k + 1)}}} = \frac{{z^{\,s} \;e^{\,\, - z} }}
{s}\sum\limits_{0\, \leqslant \,k} {\frac{{z^{\,k} }}
{{\left( {s + 1} \right)^{\,\overline {\,k\,} } \,}}} = \hfill \\
= z^{\,s} \;e^{\,\, - z} \sum\limits_{0\, \leqslant \,k} {\frac{{z^{\,k} }}
{{s^{\,\overline {\,k + 1\,} } \,}}} = z^{\,s} \sum\limits_{0\, \leqslant \,j} {\frac{{\left( { - 1} \right)^{\,j} }}
{{\,\left( {s + j} \right)}}\frac{{z^{\,j} }}
{{j!}}} \hfill \\
\end{gathered}
$$
we can then write
$$
\begin{gathered}
F(x,m) = \sum\limits_{0\, \leqslant \,k} {\left( { - 1} \right)^{\,k} \frac{{x^{\,2k} }}
{{\left( {2k + m} \right)!}}} = \hfill \\
= \frac{1}
{2}\left( {\sum\limits_{0\, \leqslant \,k} {\frac{{\left( {i\,x} \right)^{\,k} }}
{{\left( {k + m} \right)!}}} + \sum\limits_{0\, \leqslant \,k} {\frac{{\left( { - \,i\,x} \right)^{\,k} }}
{{\left( {k + m} \right)!}}} } \right) = \hfill \\
= \frac{1}
{{2\,\Gamma (m)}}\left( {\frac{{e^{\,\,i\,x} \,\gamma (m,i\,x)}}
{{\left( {i\,x} \right)^{\,m} }} + \frac{{e^{\, - \,i\,x} \,\gamma (m, - i\,x)}}
{{\left( { - i\,x} \right)^{\,m} }}} \right) = \hfill \\
= \frac{1}
{{\left| x \right|^{\,\,m} }}\,\operatorname{Re} \left( {e^{\,\,i\,x} e^{\,\, - \,i\,m\,\left( {sign(x)\pi /2} \right)} \,\frac{{\gamma (m,i\,x)}}
{{\Gamma (m)}}} \right) = \hfill \\
= \frac{1}
{{\left| x \right|^{\,\,m} }}\,\operatorname{Re} \left( {e^{\,\,i\,x} e^{\,\, - \,i\,m\,\left( {sign(x)\pi /2} \right)} \,\left( {1 - Q(m,i\,x)} \right)} \right) \hfill \\
\end{gathered}
$$
So, for $x=\pi /2$ we get
$$
\begin{gathered}
F(\pi /2,m) = \sum\limits_{0\, \leqslant \,k} {\left( { - 1} \right)^{\,k} \frac{{\left( {\pi /2} \right)^{\,2k} }}
{{\left( {2k + m} \right)!}}} = \hfill \\
= \left( {\frac{2}
{\pi }} \right)^{\,\,m} \,\operatorname{Re} \left( {e^{\,\, - \,i\,\left( {m - 1} \right)\,\left( {\pi /2} \right)} \,\frac{{\gamma (m,i\,\pi /2)}}
{{\Gamma (m)}}} \right) = \hfill \\
= \left( {\frac{2}
{\pi }} \right)^{\,\,m} \,\operatorname{Re} \left( {e^{\,\, - \,i\,\left( {m - 1} \right)\,\left( {\pi /2} \right)} \,\left( {1 - Q(m,i\,\pi /2)} \right)} \right) \hfill \\
\end{gathered}
$$
example with $m=5$
$$
\begin{gathered}
F(\pi /2,5) = \sum\limits_{0\, \leqslant \,k} {\left( { - 1} \right)^{\,k} \frac{{\left( {\pi /2} \right)^{\,2k} }}
{{\left( {2k + 5} \right)!}}} = \hfill \\
= \left( {\frac{2}
{\pi }} \right)^{\,\,5} \,\operatorname{Re} \left( {e^{\,\, - \,i\,4\,\left( {\pi /2} \right)} \,\left( {1 - Q(5,i\,\pi /2)} \right)} \right) = \hfill \\
= \left( {\frac{2}
{\pi }} \right)^{\,\,5} \,\operatorname{Re} \,\left( {1 - \frac{1}
{{24}}\left( {12\,\pi - \frac{1}
{2}\pi ^{\,3} + i\left( {3\pi ^{\,2} - \frac{1}
{{16}}\pi ^{\,4} - 24} \right)} \right)} \right) = \hfill \\
= \left( {\frac{2}
{\pi }} \right)^{\,\,5} \left( {1 - \,\frac{\pi }
{2} + \frac{1}
{{48}}\pi ^{\,3} } \right) = \left( {\frac{2}
{\pi }} \right)^{\,\,5} - \,\left( {\frac{2}
{\pi }} \right)^{\,\,4} + \frac{1}
{6}\left( {\frac{2}
{\pi }} \right)^{\,2} \hfill \\
\end{gathered}
$$
which matches with the value given by Zaid Alyafeai,
as well as with the fomula indicated by Igor Rivin
$$
\begin{gathered}
F(\pi /2,5) = \sum\limits_{0\, \leqslant \,k} {\left( { - 1} \right)^{\,k} \frac{{\left( {\pi /2} \right)^{\,2k} }}
{{\left( {2k + 5} \right)!}}} = \left( {\frac{2}
{\pi }} \right)^{\,\,5} - \,\left( {\frac{2}
{\pi }} \right)^{\,\,4} + \frac{1}
{6}\left( {\frac{2}
{\pi }} \right)^{\,2} = \hfill \\
= 0.007860176 \cdots = \hfill \\
= \frac{1}
{{5!}}{}_1F_2 \left( {1\;;\;\frac{5}
{2} + \frac{1}
{2},\;\frac{5}
{2} + 1\;;\; - \frac{1}
{4}\left( {\frac{\pi }
{2}} \right)^{\,2} } \right) \hfill \\
\end{gathered}
$$
and concerning the latter, a computer calculation over multiple values of $m$ and $x$ shows a full match.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2080167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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|
Find the sum of this infinite series
Sum the series $$\frac{1^2}{2^2}+\frac{1^2\cdot3^2}{2^2\cdot4^2}+\frac{1^2\cdot3^2\cdot5^2}{2^2\cdot4^2\cdot6^2}+\ldots$$
Someone please help me in finding sum of this infinite series. I have never found the sum of this kind of series.
|
The function $${}_2 F_1(1/2,1/2;1;z) = 1 + \sum_{n=1}^{\infty} \frac{((1/2)(3/2)...(1/2 + n - 1))^2}{n!^2} z^{n} = 1 + \sum_{n=1}^{\infty} \Big( \frac{1 \cdot 3 \cdot ... \cdot (2n-1)}{2 \cdot 4 \cdot ... \cdot (2n)} \Big)^2 z^{n}$$
can be expressed as an elliptic integral $$\frac{2}{\pi} \int_0^{\pi/2} \frac{1}{\sqrt{1 - z \sin^2(\theta)}} \, \mathrm{d}\theta$$ whenever $|z| < 1.$
From the integral representation it is easy to see that diverges as $z$ approaches $1$. For fun, the alternating series with $z = -1$ can be evaluated as $$1 -\frac{1^2}{2^2} + \frac{1^2 \cdot 3^2}{2^2 \cdot 4^2} \pm ... = \frac{\Gamma(1/4)^2}{(2\pi)^{3/2}} \approx 0.8346$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2082626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Relation between inverse tangent and inverse secant I've been working on the following integral
$$\int\frac{\sqrt{x^2-9}}{x^3}\,dx,$$
where the assumption is that $x\ge3$. I used the trigonometric substitution $x=3\sec\theta$,which means that $0\le\theta<\pi/2$. Then, $dx=3\sec\theta\tan\theta\,dx$, and after a large number of steps I achieved the correct answer:
$$\int\frac{\sqrt{x^2-9}}{x^3}\,dx=\frac16\sec^{-1}\frac{x}{3}-\frac{\sqrt{x^2-9}}{2x^2}+C$$
I was able to check my answer using Mathematica.
expr = D[1/6 ArcSec[x/3] - Sqrt[x^2 - 9]/(2 x^2), x];
Assuming[x >= 3, FullSimplify[expr]]
Which returned the correct response:
Sqrt[-9 + x^2]/x^3
Mathematica returns the following answer:
Integrate[Sqrt[x^2 - 9]/x^3, x, Assumptions -> x >= 3]
-(Sqrt[-9 + x^2]/(2 x^2)) - 1/6 ArcTan[3/Sqrt[-9 + x^2]]
Which I can write to make more clear.
$$-\frac16\tan^{-1}\frac{3}{\sqrt{x^2-9}}-\frac{\sqrt{x^2-9}}{2x^2}+D$$
Now, you can see that part of my answer is there, but here is my question. How can I show that
$$\frac16\sec^{-1}\frac{x}{3}\qquad\text{is equal to}\qquad -\frac16\tan^{-1}\frac{3}{\sqrt{x^2-9}}$$
plus some arbitrary constant? What identities can I use? Also, can anyone share the best web page for inverse trig identities?
Update: I'd like to thank everyone for their help. The Trivial Solution's suggestion gave me:
$$\theta=\sec^{-1}\frac{x}{3}=\tan^{-1}\frac{\sqrt{x^2-9}}{3}$$
Then the following identity came to mind:
$$\tan^{-1}x+\tan^{-1}\frac1x=\frac{\pi}{2}$$
So I could write:
\begin{align*}
\frac16\sec^{-1}\frac{x}{3}-\frac{\sqrt{x^2-9}}{2x^2}
&=\frac16\tan^{-1}\frac{\sqrt{x^2-9}}{3}-\frac{\sqrt{x^2-9}}{2x^2}\\
&=\frac16\left(\frac{\pi}{2}-\tan^{-1}\frac{3}{\sqrt{x^2-9}}\right)-\frac{\sqrt{x^2-9}}{2x^2}\\
&=\frac{\pi}{12}-\frac16\tan^{-1}\frac{3}{\sqrt{x^2-9}}-\frac{\sqrt{x^2-9}}{2x^2}
\end{align*}
Using Olivier's and Miko's thoughts, I produced this plot in Mathematica.
Plot[{1/6 ArcSec[x/3] - Sqrt[x^2 - 9]/(
2 x^2), -(1/6) ArcTan[3/Sqrt[x^2 - 9]] - Sqrt[x^2 - 9]/(
2 x^2)}, {x, -6, 6},
Ticks -> {Automatic, {-\[Pi]/12, \[Pi]/12}}]
Which shows that the two answers differ by $\pi/12$, but only for $x>3$.
|
Since
$$ \sec[..]^2 = 1+ \tan[..]^2 $$
we directly have identities
$$ \sec^{-1}x = \tan^{-1}\sqrt {x^2 - 1} $$
and
$$ \tan ^{-1} x= \sec^{-1}\sqrt {1 + x^2} $$
Also plot between $ \sec^{-1}...,\, \tan^{-1}. $
Note the $ \pi/4,\pi $ intercept on $y$, and period respy.Proper sign to be taken.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2082980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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|
The sum of series with natural logarithm: $\sum_{n=1}^\infty \ln\left(\frac{n(n+2)}{(n+1)^2}\right)$ Calculate the sum of series:
$$\sum_{n=1}^\infty \ln\left(\frac{n(n+2)}{(n+1)^2}\right)$$
I tried to spread this logarithm, but I'm not seeing any method for this exercise.
|
In another, more straight, way:
$$
\begin{gathered}
\sum\limits_{1\, \leqslant \,n} {\ln \left( {\frac{{n\left( {n + 2} \right)}}
{{\left( {n + 1} \right)^{\,2} }}} \right)} = \ln \left( {\prod\limits_{1\, \leqslant \,n} {\frac{{n\left( {n + 2} \right)}}
{{\left( {n + 1} \right)^{\,2} }}} } \right) = \hfill \\
= \ln \left( {\frac{{\prod\limits_{1\, \leqslant \,n} n }}
{{\prod\limits_{1\, \leqslant \,n} {\left( {n + 1} \right)} }}\;\frac{{\prod\limits_{1\, \leqslant \,n} {\left( {n + 2} \right)} }}
{{\prod\limits_{1\, \leqslant \,n} {\left( {n + 1} \right)} }}} \right) = \ln \left( {\frac{{\prod\limits_{1\, \leqslant \,n} n }}
{{\prod\limits_{2\, \leqslant \,n} n }}\;\frac{{\prod\limits_{3\, \leqslant \,n} n }}
{{\prod\limits_{2\, \leqslant \,n} n }}} \right) = \hfill \\
= \ln \left( {1\;\frac{1}
{2}} \right) = \ln \left( {\frac{1}
{2}} \right) \hfill \\
\end{gathered}
$$
Rewriting the above in more rigorous terms, we have
$$
\begin{gathered}
\sum\limits_{1\, \leqslant \,n\, \leqslant \,q} {\ln \left( {\frac{{n\left( {n + 2} \right)}}
{{\left( {n + 1} \right)^{\,2} }}} \right)} = \ln \left( {\prod\limits_{1\, \leqslant \,n\, \leqslant \,q} {\frac{{n\left( {n + 2} \right)}}
{{\left( {n + 1} \right)^{\,2} }}} } \right) = \hfill \\
= \ln \left( {\frac{{\prod\limits_{1\, \leqslant \,n\, \leqslant \,q} n }}
{{\prod\limits_{1\, \leqslant \,n\, \leqslant \,q} {\left( {n + 1} \right)} }}\;\frac{{\prod\limits_{1\, \leqslant \,n\, \leqslant \,q} {\left( {n + 2} \right)} }}
{{\prod\limits_{1\, \leqslant \,n\, \leqslant \,q} {\left( {n + 1} \right)} }}} \right) = \ln \left( {\frac{{\prod\limits_{1\, \leqslant \,n\, \leqslant \,q} n }}
{{\prod\limits_{2\, \leqslant \,n\, \leqslant \,q + 1} n }}\;\frac{{\prod\limits_{3\, \leqslant \,n\, \leqslant \,q + 2} n }}
{{\prod\limits_{2\, \leqslant \,n\, \leqslant \,q + 1} n }}} \right) = \hfill \\
= \ln \left( {\frac{1}
{{q + 1}}\;\frac{{q + 2}}
{2}} \right) \hfill \\
\end{gathered}
$$
and therefore
$$
\mathop {\lim }\limits_{q\, \to \,\infty } \sum\limits_{1\, \leqslant \,n\, \leqslant \,q} {\ln \left( {\frac{{n\left( {n + 2} \right)}}
{{\left( {n + 1} \right)^{\,2} }}} \right)} = \mathop {\lim }\limits_{q\, \to \,\infty } \ln \left( {\frac{1}
{2}\;\frac{{q + 2}}
{{q + 1}}} \right) = \ln \left( {\frac{1}
{2}\mathop {\lim }\limits_{q\, \to \,\infty } \;\frac{{q + 2}}
{{q + 1}}} \right) = \ln \left( {\frac{1}
{2}} \right)
$$
|
{
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"url": "https://math.stackexchange.com/questions/2083217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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|
Integer solutions to nonlinear system of equations $(x+1)^2+y^2 = (x+2)^2+z^2$ and $(x+2)^2+z^2 = (x+3)^2+w^2$
Do there exist integers $x,y,z,w$ that satisfy \begin{align*}(x+1)^2+y^2 &= (x+2)^2+z^2\\(x+2)^2+z^2 &= (x+3)^2+w^2?\end{align*}
I was thinking about trying to show by contradiction that no such integers exist. The first equation gives $y^2 = 2x+3+z^2$ while the second gives $z^2 = 2x+5+w^2$. How can we find a contradiction from here?
|
$-1^2 + 0^2 = 0^2 + 1^2 = 1^2 + 0^2$ if you want a trivial example including negative and zero integers.
So $x = -2$ etc.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that there do not exist integers that satisfy the system
Prove that there do not exist integers $x,y,z,w,t,$ that satisfy \begin{align*}(x+1)^2+y^2 &= (x+2)^2+z^2\\(x+2)^2+z^2 &= (x+3)^2+w^2\\(x+3)^2+w^2 &= (x+4)^2+t^2.\end{align*}
I thought about using a modular arithmetic argument. The given system is equivalent to \begin{align*}y^2 &= 2x+3+z^2\\z^2 &= 2x+5+w^2\\w^2 &= 2x+7+t^2.\end{align*} Thus $y^2 = 6x+15+t^2$. How can we find a contradiction from this?
|
Let's look at the equations modulo $8$. Notice that the only squares modulo $8$ are $0,1$ and $4$.
Case $(1)$: $x \equiv 0,4 \pmod 8$
Then we have the system
\begin{align*}
y^2&\equiv z^2+3\\
z^2&\equiv w^2+5\\
w^2 &\equiv t^2+7\end{align*}
The first equation implies $y^2\equiv 4$ and $z^2\equiv 1$. This implies $w^2 \equiv 4$ via the second equation. But then the third equation implies becomes $t^2 \equiv 5$, which cannot be.
Case $(2)$: $x \equiv 2,6 \pmod 8$
Then we have the system
\begin{align*}
y^2&\equiv z^2+7\\
z^2&\equiv w^2+1\\\
w^2 &\equiv t^2+3\end{align*}
The first equation implies $y^2\equiv 0$ and $z^2\equiv 1$. This implies $w^2 \equiv 0$ via the second equation. But then the third equation implies becomes $t^2 \equiv 5$, which cannot be.
Case $(3)$: $x \equiv 1,5 \pmod 8$
Then we have the system
\begin{align*}
y^2&\equiv z^2+5\\
z^2&\equiv w^2-1\\
w^2 &\equiv t^2+1\end{align*}
The first equation implies $y^2\equiv 1$ and $z^2 \equiv 4$, but then the second equation cannot be satisfied.
Case $(4)$: $x \equiv 3,7 \pmod 8$
Then we have the system
\begin{align*}
y^2&\equiv z^2+1\\
z^2&\equiv w^2+3\\
w^2 &\equiv t^2+5\end{align*}
The first equation implies $y^2\equiv 1$ and $z^2\equiv 0$, but then the second cannot be satisfied, and this concludes our proof.
|
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"url": "https://math.stackexchange.com/questions/2085316",
"timestamp": "2023-03-29T00:00:00",
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|
Weird Integration problem: $\int_{-2}^{2} \frac{x^2}{1+5^x}dx$ $\int_{-2}^{2} \frac{x^2}{1+5^x}dx$
I am stuck at the first step and have tried replacing $5^x$ with $e^{\ln(5^x)}$ but nothing simplifies out in the end.
Any hints how I should proceed?
|
$$I=\int_{-2}^{2} \frac{x^2}{1+5^x}dx$$
Let $y=-x$, then:
$$I=\int_{-2}^{2} \frac{y^2}{1+5^{-y}}dy=\int_{-2}^{2} \frac{y^25^y}{1+5^y}dy=\int_{-2}^{2} \frac{x^25^x}{1+5^x}dx$$
So:
$$I+I=\int_{-2}^{2} \frac{x^2}{1+5^x}dx+\int_{-2}^{2} \frac{x^25^x}{1+5^x}dx=\int_{-2}^2x^2\,dx=2\int_0^2x^2\,dx=2\left(\frac{2^3}{3}\right)=\frac{16}{3}$$
So, the integral is equal to $\dfrac{8}{3}$.
Edit: A clarification on the changing of limits under $y=-x$:
$$\int_{x=-2}^{x=2}f(x)\,dx=\int_{-y=-2}^{-y=2}f(-y)(-\,dy)=-\int_{y=2}^{y=-2}f(-y)\,dy=\int_{y=-2}^{y=2}f(-y)\,dy$$
$$=\int_{-2}^{2}f(-y)\,dy$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2085415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Given $x^2-y^2+2x+2y \geq 2xy+1$ and $y^2-x^2=\frac{1}{3}$ with the condition $x,y > 0$ find $\frac{x}{y}$ Had this question popup while studying for a test but unsure how to go about solving it:
Given $x^2-y^2+2x+2y \geq 2xy+1$ and $y^2-x^2=\frac{1}{3}$ with the condition $x,y > 0$ find $\frac{x}{y}$
Initial thoughts: It feels like $\frac{x}{y}$ doesn't have an actual value but a region but not completely sure.
This is my attempt at finding out $\frac{x}{y}$:
From the first condition
$$x^2-y^2+2x+2y \geq 2xy+1$$
$$ -(y^2-x^2)+2(x+y) \geq 2xy+1$$
$$ -\frac{1}{3}+2(x+y) \geq 2xy+1$$
$$ x+y-xy \geq \frac{2}{3} $$
$$ y(1-x) \geq \frac{2}{3}-x$$
$$ y(1-x) \geq \frac{2-3x}{3}$$
$$ y \geq \frac{2-3x}{3(1-x)} ~~~~\text{if x<1}$$
$$ y \leq \frac{2-3x}{3(1-x)} ~~~~\text{if x>1}$$
But stuck now :(
|
First, lets solve for $y$ in $y^2-x^2=\frac{1}{3}$. Since $y>0$, we find $$y=\sqrt{\frac{1+3x^2}{3}}.$$ Now define $g(x,y)=x^2-y^2+2x+2y-2xy-1$. As you noted, we can substitute to get $$g(x,y)=-\frac{1}{3}+2x+2y-2xy-1.$$ However, we can also get rid of $y$ with the previous condition: define $$f(x)=g(x,\sqrt{\frac{1+3x^2}{3}})=-\frac{1}{3}+2x+2\sqrt{\frac{1+3x^2}{3}}-2x\sqrt{\frac{1+3x^2}{3}}-1.$$ Now, the question is for what values of $x$ is $f(x)\geq 0$? Using mathematica and the fact that $x>0$, we find that the roots are $$x_1=\frac{1-\sqrt{\frac{7-4\sqrt{2}}{3}}}{2},\ x_2=\frac{1+\sqrt{\frac{7+4\sqrt{2}}{3}}}{2}.$$ Now, substituting into $$\frac{x}{y}=\frac{x}{\sqrt{\frac{1+3x^2}{3}}}$$ we get that $$\frac{x}{y}\in [\frac{3-\sqrt{21-12\sqrt{2}}}{\sqrt{6(7-2\sqrt{2}-\sqrt{21-12\sqrt{2}})}},\frac{3+\sqrt{21+12\sqrt{2}}}{\sqrt{6(7+2\sqrt{2}+\sqrt{21+12\sqrt{2}})}}]\approx[0.2754,0.9353]$$
|
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|
Hint in integration $\int\frac{x^{2}}{\left(x\cos x-\sin x \right )\left( x\sin x+\cos x \right )}\,\mathrm{d}x$ In the following integration
$$\int \frac{x^{2}}{\left ( x\cos x-\sin x \right )\left ( x\sin x+\cos x \right )}\, \mathrm{d}x$$
I tried alot. But does not get any proper start.
Can anybody provide me a hint.
|
Notice
$$x^{2}=x^2\left ( \sin^2x+\cos^2x \right )$$
then
\begin{align*}
\int \frac{x^{2}}{\left ( x\cos x-\sin x \right )\left ( x\sin x+\cos x \right )}\, \mathrm{d}x&=\int \frac{x^2\left ( \sin^2x+\cos^2x \right )}{\left ( x\cos x-\sin x \right )\left ( x\sin x+\cos x \right )}\, \mathrm{d}x\\
&=\int \frac{x\cos x}{x\sin x+\cos x}\, \mathrm{d}x+\int \frac{x\sin x}{x\cos x-\sin x}\, \mathrm{d}x\\
&=\ln\left | x\sin x+\cos x \right |-\ln\left | x\cos x-\sin x \right |+C
\end{align*}
|
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|
Calculate $A \cdot B$ and say if it's equal to $B \cdot A$ (matrixes)
Calculate $A \cdot B$ and $B \cdot A$ for $A= \begin{pmatrix} 1 & 1\\
1 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & -1\\ 1 & 1
\end{pmatrix}$
Is $A \cdot B=B \cdot A$?
This is a task from an old exam and I'd like to know if I did it correctly?
I will only ask for $A \cdot B$ because the other is done the same way.
| 1 -1
A*B | 1 1
---------------------
1 1 | c_11 c_12
1 1 | c_21 c_22
$c_{11}= 1 \cdot 1+1 \cdot 1=2$
$c_{12}= 1 \cdot (-1)+1 \cdot 1=0$
$c_{22}= 1 \cdot 1+1 \cdot 1=2$
$c_{22}= 1 \cdot (-1) +1 \cdot 1=0$
Thus, $A \cdot B = \begin{pmatrix} 2 & 0\\
2 & 0 \end{pmatrix}$
Is it correct? I would do $B \cdot A$ this way too and then check if they are equal to complete the task.
|
Your calculation is correct.
Notice that if you were not required to calculate $B \cdot A$ then you would not have needed to calculate it entirely in order to get that $A \cdot B \neq B \cdot A$.
|
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|
What must be the simplest proof of the sum of first $n$ natural numbers? I was studying sequence and series and used the formula many times $$1+2+3+\cdots +n=\frac{n(n+1)}{2}$$ I want its proof.
Thanks for any help.
|
I'm not sure how simple this gets, but I still think it's worth noting that this can be applied to higher powers such as $1^2+2^2+\cdots n^2$ or $1^3+2^3+\cdots n^3$.
Solving by the use of Indeterminate Coefficients:
Assume the series$$1+2+3+4+5\ldots+n\tag1$$
Is equal to the infinite series$$1+2+3+4+5+\ldots+n=A+Bn+Cn^2+Dn^3+En^4+\ldots\&c\tag2$$
If we 'replace' $n$ with $n+1$, we get$$1+2+3+4+\ldots+(n+1)=A+B(n+1)+C(n+1)^2+\ldots\&c\tag3$$
And subtracting $(3)-(2)$, gives$$\begin{align*} & n+1=B+C(2n+1)\tag4\\n & +1=2Cn+(B+C)\tag5\end{align*}$$
Therefore, $C=\dfrac 12,B=\dfrac 12,A=0$ and $(1)$ becomes$$1+2+3+4+5\ldots+n=\dfrac n2+\dfrac {n^2}2=\dfrac {n(n+1)}{2}$$
|
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|
Let $f(x) = x^3-\frac{3}{2}x^2+x+\frac{1}{4}.$ Then the value of $ \int^{3/4}_{1/4}f(f(x))\mathrm dx$ If $\displaystyle f(x) = x^3-\frac{3}{2}x^2+x+\frac{1}{4}.$ then the value of $\displaystyle \int^{\frac{3}{4}}_{\frac{1}{4}}f(f(x))\mathrm dx$
$\displaystyle \int^{\frac{3}{4}}_{\frac{1}{4}}(f(x))^3-1.5(f(x))^2+f(x)+0.25 \mathrm dx$
could some help me with this, thanks
|
You can use the property given here:
$$\color{blue}{\int_a^b f(x)dx=\int_a^{(a+b)/2}[f(x)+f(a+b-x)]dx \Rightarrow \\ \int_a^b f(f(x))dx=\int_a^{(a+b)/2}[f(f(x))+f(f(a+b-x))]dx};\\
\int_{1/4}^{3/4} f(f(x))dx=\int_{1/4}^{1/2}[f(f(x))+f(f(1-x))]dx=\\
\int_{1/4}^{1/2}[f(f(x))+f(f(1-x))]dx=\\
\int_{1/4}^{1/2}[1]dx=\frac14,$$
because:
$$f(f(x))=x^3-\frac32x^2+x+\frac14;\\
f(f(1-x))=(1-x)^3-\frac32(1-x)^2+(1-x)+\frac14=
-x^3+\frac32x^2-x+\frac34.$$
Also, the property mentioned by RedFloyd is given in the above source:
$$\color{blue}{\int_a^b f(x)dx=\int_a^b f(a+b-x)dx \Rightarrow \\
\int_a^b f(f(x))dx=\int_a^b f(f(a+b-x))dx}.$$
|
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|
$1+8k$ is a quadratic residue modulo $2^n$
Prove that $1+8k$, for all nonnegative integers $k$, is a quadratic residue modulo $2^n$, where $n$ is a positive integer.
For $n = 5$, we have $0,1,9,17,25$, which are all quadratic residues modulo $2^5$. How do we prove this in general?
|
We can assume $n\ge 4$, because otherwise it is trivial.
Fact 1: Two odd numbers have the same square modulo $2^n$ if and only if they are equal or opposite modulo $2^{n-1}$.
Proof: If $(2s+1)^2\equiv (2t+1)^2\pmod {2^n}$ then $s^2+s\equiv t^2+t\pmod {2^{n-2}}$ and
$$(s-t)(s+t+1)\equiv 0\pmod{2^{n-2}}$$
Only one of these factors is even, so we have $s\equiv t$ or $s+t\equiv -1 \pmod{2^{n-2}}$ and the fact follows.
An immediate consequence is that each odd square modulo $2^n$ has exactly four square roots.
Then there are exactly $2^{n-3}$ odd quadratic residues modulo $2^n$, a fourth of the odd residues.
On the other hand,
Fact 2: If $r$ is odd and $m\not\equiv 1\pmod 8$ then $r$ is not a quadratic residue modulo $2^n$.
Proof: Suppose the contrary. If $r\equiv m^2\pmod{2^n}$, we have that $2^n$ divides $(m^2-1)-(r-1)$. Since $8$ divides $m^2-1$, it also divides $r-1$, a contradiction.
To finish the proof, note that the set of odd numbers from $0$ to $2^n$ can be divided in $2^{n-2}$ packs of $4$ numbers each. These packs can be of the form $\{8k+1,8k+3,8k+5, 8k+7\}$. Among these four numbers only $8k+1$ can be a square. Since a fourth of the odd residues are quadratic, $8k+1$ must be a square for each $k$.
|
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|
Show that if $2+2\sqrt{28n^2+1}$ is an integer then it must be perfect square. As written in title, I want to prove that
If $n$ is an integer, show that if $2+2\sqrt{28n^2+1}$ is an integer than it must be perfect square.
I m struggling in making a start . Please help.
|
Here's another answer which uses the Pell equation.
As mentioned above, in order for $2 + 2\sqrt{28y^2+1}$ to be an integer, it is necessary and sufficient that $28y^2 + 1$ is a perfect square.
Hence we must look at the solutions to Pell's equation $x^2 - 28y^2 = 1$.
All solutions to this equation are of the form $x_n+y_n\sqrt{28} = \pm\left(127+24\sqrt{28}\right)^n, n \in \mathbb{Z}$.
This means $x_n = \pm\frac{1}{2}\left[\left(127+24\sqrt{28}\right)^n+\left(127-24\sqrt{28}\right)^n\right]$.
Also note that $2 + 2\sqrt{28y_n^2 + 1} = 2+2\sqrt{x_n^2} = 2+2|x_n| = 2 + \left(127+24\sqrt{28}\right)^n + \left(127-24\sqrt{28}\right)^n$.
But $127 + 24\sqrt{28} = \left(8+3\sqrt{7}\right)^2$ and $\left(8+3\sqrt{7}\right)\left(8-3\sqrt{7}\right) = 1$ so we finally obtain:
$$2+2\sqrt{28y_n^2+1} = \left(8+3\sqrt{7}\right)^{2n} + 2\left(8+3\sqrt{7}\right)^n\left(8-3\sqrt{7}\right)^n+\left(8-3\sqrt{7}\right)^{2n} = \left[\left(8+3\sqrt{7}\right)^n+\left(8-3\sqrt{7}\right)^n\right]^2$$
which is the square of an integer (you can check with the binomial theorem).
|
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|
Prove conjecture $a_{n+1}>a_{n}$ if $a_{n+1}=a+\frac{n}{a_{n}}$ Let sequence $\{a_{n}\}$ such $a_{1}=a>0$,and
$$a_{n+1}=a+\dfrac{n}{a_{n}}$$
I used the software to find this following conjecture :
if $n>\dfrac{4}{a^3}$,we have
$$a_{n+1}>a_{n}$$
|
Here is an answer which proves that the conjecture holds for $a > 1$. For the general case I couldn't show it but I present some ideas.
The proof follows by induction.
We want $a_{n+1} > a_n$. So we need (see the calculations by Han de Bruijn):
$$
a_n < \frac{a}{2} + \sqrt{\left(\frac{a}{2}\right)^2+n} = U(n)
$$
So this constitutes an upper bound $U(n)$ for all $a_n$. Since this must hold for all $a_n$, we can ask for a condition for this to hold for $a_{n+1}$. We get
$$
a + \frac{n}{a_n} = a_{n+1} < \frac{a}{2} + \sqrt{\left(\frac{a}{2}\right)^2+n+1} = U(n+1)
$$
which gives (SOME MORE STEPS ADDED HERE by request)
$$
a_{n} > \frac{n}{- {a} + U(n+1) } = \frac{n}{n+1} U(n+1)
$$
The latter equality can be verified by clearing denominators:
$$
n+ 1 = {- {a} U(n+1) } + (U(n+1))^2
$$
inserting $U(n+1)$
$$
n+ 1 = - {a} \left[\frac{a}{2} + \sqrt{\left(\frac{a}{2}\right)^2+n+1} \right] + \left[\frac{a}{2} + \sqrt{\left(\frac{a}{2}\right)^2+n+1} \right]^2
$$
and expanding the brackets:
$$
n+ 1 = - \frac{a^2}{2} - a \sqrt{\left(\frac{a}{2}\right)^2+n+1} + \frac{a^2}{4} + a \sqrt{\left(\frac{a}{2}\right)^2+n+1} + \left(\frac{a}{2}\right)^2+n+1
$$
which clearly is an identity. (END EDIT)
Summarizing, we have
$$
a_n > \frac{n}{n+1} \left[\frac{a}{2} + \sqrt{\left(\frac{a}{2}\right)^2+n+1} \right] = \frac{n}{n+1} U(n+1) = L(n)
$$
So this constitutes a lower bound $L(n)$ for all $a_n$. By induction, if for some $N$, $L(N) < a_N < U(N)$, then these lower and upper bounds are necessary conditions for all further $n > N$ in order that $a_{n+1} > a_n$.
Now we give a (further) sufficient condition for that to hold. The idea is to show that for any $a_n$ with $L(n) < a_n < U(n)$, also $L(n+1) < a_{n+1} < U(n+1)$ holds. Since $a_{n+1} = a + \frac{n}{a_n}$, we have that 1) the highest $a_{n+1}$ will be obtained from the lowest $a_{n} = L(n)$ and 2)
the lowest $a_{n+1}$ will be obtained from the highest $a_{n} = U(n)$.
So we need for 1)
$$
a + \frac{n}{L(n)} \leq U(n+1)
$$
and for 2)
$$
a + \frac{n}{U(n)} \geq L(n+1)
$$
Establishing this, we have, by induction, that all $a_n$ will always stay within the limits $L(n) < a_n < U(n)$ and hence we have proved the OP's claim.
For 1): Inserting $L(n)$ and $U(n+1)$ gives
$$
a + \frac{n}{\frac{n}{n+1} U(n+1)} \leq U(n+1)
$$
Solving this for $U(n+1)$ gives
$$
U(n+1) \geq \frac{a}{2} + \sqrt{\left(\frac{a}{2}\right)^2+n+1}
$$
which holds by the definition of $U(n)$.
For 2): Inserting $L(n+1)$ and $U(n)$ gives
$$
a + \frac{n}{U(n)} \geq \frac{n+1}{n+2} U(n+2)
$$
The LHS equals $U(n)$ (clear by inserting it). Then we need
$$
{U(n)} \geq \frac{n+1}{n+2} U(n+2)
$$
Inserting $U(n)$ and doing the algebra gives
$$
a \sqrt{\left(\frac{a}{2}\right)^2+n} > 1 - \frac{a^2}{2}
$$
Clearly, for $a > 1$ this holds. For $a < 1$ we can square and get
$$
a^2 \left(\frac{a}{2}\right)^2+ n a^2 > \frac{a^4}{4} + 1 - a^2
$$
or
$$
n > \frac{1}{a^2} -1
$$
Let's first look at the case $a > 1$. This gives for $n=1$ (start of induction) that $ \frac12 \left[\frac{a}{2} + \sqrt{\left(\frac{a}{2}\right)^2+2} \right] = L(1) < a_1 < U(1) = \frac{a}{2} + \sqrt{\left(\frac{a}{2}\right)^2+1}$
Since $a_1 = a$ by the task description, we must have that
$ \frac12 \left[\frac{a}{2} + \sqrt{\left(\frac{a}{2}\right)^2+2} \right] < a < \frac{a}{2} + \sqrt{\left(\frac{a}{2}\right)^2+1}$
The right inequality is obvious and the left inequality holds for $a > 1$. This establishes my first claim that the OP's conjecture holds for $a > 1$.
As for $a <1$, in this case we have that the initial condition $a_1=a$ violates the lower bound $L(1)$. Further, our sufficient (conservative) condition will only be satisfied for $
n > \frac{1}{a^2} -1
$.
If we require, as in the original setting, $n > \frac{4}{a^3}$, then for $a <1$, the condition $
n > \frac{1}{a^2} -1
$ will be satisfied anyway. So the program of proof could be:
show that, if we start with $a_1 = a$, and take $N = \frac{4}{a^3}$ many iteration steps $a_{n+1} = a + n/a_n$, then $L(N) < a_N < U(N)$. From this the general proof would follow.
|
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|
Prove or disprove $|x+y+z|\ge\sqrt{3}$ Let $x,y,z\in R$, such $|x|\neq|y|\neq|z|$,and $|x|,|y|,|z|>1$ and $xy+yz+xz=-1$
prove or disprove
$$|x+y+z|\ge\sqrt{3}$$
I try
$$(x-y)^2+(y-z)^2+(z-x)^2\ge 0$$
so $$|x+y+z|^2\ge 3(xy+yz+xz)=-3$$
why?
|
Counterexample:
Suppose $x = 1.02, y = -1.0001, z = \frac {-1-xy}{x+y} = \frac {0.020102}{0.0199} \approx 1.01015$
$|x|,|y|,|z| > 1$
and
$xy + yz + xz = -1$
and $|x+y+z| = 1.03 < \sqrt 3$
Since $|x|, |y|, |z| > 1, x^2 + y^2 + z^2 > 3$
$(x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy + xz + yz) \ge 0\\
(x+y+z)^2 \ge 3 -2 \ge 0\\
(x+y+z)^2 \ge 1\\
|x+y+z| \ge 1$
|
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|
For $x+y=1$, show that $x^4+y^4\ge \frac{1}{8}$ As in the title. Let $x,y$ be two real numbers such that $x+y=1$. Prove that $x^4+y^4\ge \frac{1}{8}$.
Any hints? Basically, the only method I am aware of is plugging $y=1-x$ into the inequality and investigating the extrema of the function, but I don't think it's the best method. I'm looking for a cleverer way to prove that inequality.
|
Also a Cauchy-Schwarz approach works:
\begin{align*}
1 = x + y \le \sqrt{(x^2+y^2)(1^2+1^2)} \implies x^2 + y^2 \ge \frac{1}{2} \\
\frac{1}{2} \le x^2 + y^2 \le \sqrt{(x^4 + y^4)(1^2 + 1^2)} \implies x^4 + y^4 \ge \frac{1}{8}
\end{align*}
|
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|
Show the following equality Basically I want to show the following:
$$\sqrt{2}\ |z|\geq\ |\operatorname{Re}z| + |\operatorname{Im}z|$$
So what I did is the following:
Let $z = a + bi$
Consider the following:
$$2|z|^2 = 2a^2 + 2b^2 = a^2 + b^2 +a^2+b^2$$
Since $(a-b)^2\geq0$, hence $a^2+b^2\geq 2ab$
Thus $2|z|^2 \geq a^2+b^2+2ab = (a+b)^2$
Hence $\sqrt{2}|z| \geq a + b$ but $a = |\operatorname{Re}z|~,~ b = |\operatorname{Im}z|$
Did I make a mistake somewhere, if yes I would appreciate it if it could be pointed out and perhaps provide some guideline on how to prove this.
|
It suffices to consider $x, y > 0$. Other case simply changes the sign of $x,y$ and of $\theta$. Thus: $\dfrac{|\text{Re(z)}|}{|z|} = \dfrac{x}{\sqrt{x^2+y^2}}= \sin \theta, \dfrac{\text{Im(z)}}{|z|}= \dfrac{y}{\sqrt{x^2+y^2}} = \cos \theta, \theta \in \left(0,\dfrac{\pi}{2}\right)$. Thus you prove: $\sin \theta + \cos \theta \le \sqrt{2}$, which is true since $\sin \theta + \cos \theta = \sqrt{2}\sin\left(\theta+\dfrac{\pi}{4}\right)\le \sqrt{2}$ .
|
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|
Number of cubes We have $X$ cubes with $8000\le X\le10000$.
We have built columns with $2×2$ bases, leaving 2 cubes. We have also built columns with $3×3$ and $5×5$ bases, leaving 4 cubes in these cases.
How can we calculate the number of cubes?
I have created the equations
$$n\equiv2\bmod4$$
$$n\equiv4\bmod9$$
$$n\equiv4\bmod25$$
but I am not sure how to proceed in calculating the right number. What is the best way to calculate it? Thanks for help.
|
The Chinese Remainder is used for calculating a number $n$ in this case that when divided by $4$ has remainder $2$, by $9$ remainder $4$ and by $25$ remainder $4$.
Then $lcm(4,9,25) = 900$ and we need $a+b+c \equiv r \pmod{900}$
Let's start calculating $a=2\cdot9\cdot25\cdot t$ where $t_1 \equiv (9\cdot25)^{-1} \pmod 4$.
Then $t_1 \equiv 225^1 \equiv 1 \pmod 4$ and $a=450$
For $b=4\cdot4\cdot25\cdot t_2$ where $t_2 \equiv (4\cdot25)^{-1} \pmod 9$.
Then $t_2 \equiv 1 \equiv 100^5 \pmod 9$ and $b=400$
For $c=4\cdot4\cdot9\cdot t_3$ where $t_3 \equiv (4\cdot9)^{-1} \pmod 9$
Then $t_3 \equiv 16 \equiv 36^{19} \pmod{25}$ and $c=2304$
Thus $454 \equiv a+b+c \equiv 3154 \pmod{900}$ so $n=454$ but we aren't done yet.
Since you want that $n$ satisfies $8000<n<10000$ then $n=8554$ or $n=9454$
|
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|
Decomposition of this partial function I came across this
$$\int \frac{dx}{x(x^2+1)^2}$$
in "Method of partial functions" in my Calculus I book.
The thing is that, however, decomposing $\frac{1}{x(x^2+1)^2}$ would take long in this way:
$$\frac{1}{x(x^2+1)^2}= \frac{A}{x}+\frac{Bx+C}{x^2+1}+\frac{Dx+E}{(x^2+1)^2}$$
So is there a quicker or a more practical way that I can use?
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Let's take the decomposition you give and see how quick we can go
$$\frac{1}{x(x^2+1)^2}= \frac{A}{x}+\frac{Bx+C}{x^2+1}+\frac{Dx+E}{(x^2+1)^2}$$
Multiplying by $x$ the two sides and making $x=0$ one gets $A=1$.
Multiplying by $(x^2+1)^2$ and making $x=i$ we get $-i=Di+E$ and this means $D=-1$ and $E=0$ . Now rewrite the decomposition
$$\frac{1}{x(x^2+1)^2}= \frac{1}{x}+\frac{Bx+C}{x^2+1}-\frac{x}{(x^2+1)^2}$$
This is equivalent to
$$\begin{align}{Bx+C\over x^2+1}=&{1\over x(x^2+1)^2}-{1\over x}+{x\over (x^2+1)^2}\\=&{(x^2+1)(1-1-x^2)\over x(x^2+1)^2}\\=&-{x\over x^2+1}\end{align}$$
Looks pretty quick to me
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Find $\int_{0}^{\infty }\cos \left (x \right )\sin \left (x^{2} \right )\mathrm{d}x$ How to prove
$$\int_{0}^{\infty }\cos \left (x \right )\sin \left (x^{2} \right )dx=\frac{1}{2}\sqrt{\frac{\pi }{2}}\left ( \cos\frac{1}{4}-\sin\frac{1}{4} \right )$$
any hint ?Thanks!
|
\begin{align*}
\int_{0}^{\infty }\cos x\sin x^2\, \mathrm{d}x&=\frac{1}{4}\int_{-\infty}^{\infty }\left [ \sin\left ( x^{2}+x \right )+\sin\left ( x^{2}-x \right ) \right ]\mathrm{d}x\tag1\\
&=\frac{1}{4}\int_{-\infty}^{\infty }\left [ \sin\left ( \left ( x+\frac{1}{2} \right )^{2}-\frac{1}{4} \right )+\sin\left ( \left ( x-\frac{1}{2} \right )^{2}-\frac{1}{4} \right ) \right ]\mathrm{d}x\\
&=\frac{1}{2}\int_{-\infty}^{\infty }\sin\left ( x^{2}-\frac{1}{4} \right )\mathrm{d}x\\
&=\frac{1}{2}\cos\frac{1}{4}\int_{-\infty}^{\infty }\sin x^{2}\, \mathrm{d}x-\frac{1}{2}\sin\frac{1}{4}\int_{-\infty}^{\infty }\cos x^2\, \mathrm{d}x\tag2
\end{align*}
The use the Fresnel integral , we will get the answer as wanted.
$(1)$ : $\sin\alpha \cos\beta =\dfrac{1}{2}\left [ \sin\left ( \alpha +\beta \right )+\sin\left ( \alpha -\beta \right ) \right ]$
$(2)$ : $\sin\left ( \alpha -\beta \right )=\sin\alpha \cos\beta -\cos\alpha \sin\beta $
Use the same way we can get
$$\int_{0}^{\infty }\cos x\cos x^2\, \mathrm{d}x=\frac{1}{2}\sqrt{\frac{\pi }{2}}\left ( \cos\frac{1}{4}+\sin\frac{1}{4} \right )$$
|
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How solve the given logarithmic problem Let
$$\log_{12}(18) = a$$
Then $$\log_{24}(16)$$
is equal to what in terms of a?
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We have: $12^a = 18 \implies 2^{2a}\cdot 3^a = 2\cdot 3^2\implies 2a\ln 2+ a\ln 3 = \ln 2+2\ln 3$. Thus $b = \log_{24}16\implies 24^b = 16 = 2^4\implies 2^{3b}\cdot 3^b = 2^4\implies 3b\ln 2+b\ln 3=4\ln2\implies 3b+b\dfrac{\ln3}{\ln2}=4$. You can divide by $\ln2$ the first equation, and solve for $\dfrac{\ln3}{\ln2}$, then substitute it into the second equation to solve for $b$ in term of $a$.
|
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Proof that $\frac{a}{a+3b+3c}+\frac{b}{b+3a+3c}+\frac{c}{c+3a+3b} \ge \frac{3}{7}$ for all $a,b,c > 0$ So I am trtying to proof that $\frac{a}{a+3b+3c}+\frac{b}{b+3a+3c}+\frac{c}{c+3a+3b} \ge \frac{3}{7}$ for all $a,b,c > 0$. First I tried with Cauchy–Schwarz inequality but got nowhere.
Now I am trying to find a convex function, so I can use jensen's inequality, but I can't come up with one which works.. Has anyone an idea?
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moving $$\frac{3}{7}$$ to the left and clearing the denominators we get
$$3\,{a}^{3}-{a}^{2}b-{a}^{2}c-a{b}^{2}-3\,abc-a{c}^{2}+3\,{b}^{3}-{b}^{
2}c-b{c}^{2}+3\,{c}^{3}
\geq 0$$ and this is equivalent to $$a^3+b^3+c^3-3abc+(a-b)(a^2-b^2)+(b^2-c^2)(b-c)+(a^2-c^2)(a-c)\geq 0$$
which is clear
(the first Terms are AM-GM)
|
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Prove that the largest power of $2$ dividing $(n+1)(n+2) \cdots (an)$ is greater than$2^{(a-1)n}$
Let $a,n$ be positive integers. Find all $a$ such that for some $n$ the largest power of $2$ dividing $(n+1)(n+2) \cdots (an)$ is greater than $2^{(a-1)n}$.
Since I thought there were no such $a$, I thought about proving this by contradiction. That is, assume that $2^{(a-1)n+1}$ divides $(n+1)(n+2) \cdots (an)$. How can we get to a contradiction from here?
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There are $(a-1)n$ factors in the original expression.
$\lfloor \frac {(a-1)n + 1}{2}\rfloor$ are divisible by $2.$
$\lfloor\frac {(a-1)n + 1}{4}\rfloor$ are divisible by $4$.
$\lfloor\frac {(a-1)n + 1}{2^i}\rfloor$ are divisible by $2^i$
The largest power of $2$ that divides $(n+1)\cdots(an) = 2^{\left(\lfloor \frac {(a-1)n + 1}{2}\rfloor + \lfloor \frac {(a-1)n + 1}{2^2}\rfloor \cdots +\lfloor \frac {(a-1)n + 1}{2^i}\rfloor\cdots+\lfloor \frac {(a-1)n + 1}{2^{k}}\rfloor\right)}$
Where k is the largest integer $\le \log_2 ((a-1)n + 1) $
$\sum_\limits{i=1}^{k} \lfloor \frac {(a-1)n + 1}{2^i} \rfloor < ((a-1)n + 1)\sum_\limits {i=1}^{\infty} \frac 1{2^i}\\\sum_\limits {i=1}^{\infty} \frac 1{2^i} = 1\\
\sum_\limits{i=1}^{k } \lfloor \frac {(a-1)n + 1}{2^i} \rfloor < (a-1)n + 1\\
\sum_\limits{i=1}^{k } \lfloor \frac {(a-1)n + 1}{2^i} \rfloor \le (a-1)n$
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if $f(x)+2f(\frac{1}{x})=2x^2$ what is $f(\sqrt{2})$ Question
if $f(x)+2f(\frac{1}{x})=2x^2$ what is $f(\sqrt{2})$
My steps
I tried to plug in $\sqrt{2}$ into the equation but that didn't get me anywhere because then i would have $2f(\frac{1}{\sqrt{2}})$ in the way. I was wondering on how to solve this equation?
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If $f(x)+2f(\frac{1}{x})=2x^2$, then by substitution $f(\frac{1}{x})+2f(x)=\frac{2}{x^2}$ and so $f(\frac{1}{x})=\frac{2}{x^2}-2f(x)$. Substituting $f(\frac{1}{x})$ into the first equation yields $f(x)+2[\frac{2}{x^2}-2f(x)]=2x^2$. Solve for $f(x)$: $f(x)=\frac{2}{3}[\frac{2}{x^2}-x^2]$. So $f(\sqrt{2})=-\frac{2}{3}$.
|
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How to compute the sum $\sum_{n=0}^\infty \tfrac{n^2}{2^n}$?
How to find this sum :
$\sum_{n=0}^\infty \dfrac{n^2}{2^n}$
$\sum_{n=0}^\infty \dfrac{n^2}{2^n}=\dfrac{1}{2}+\dfrac{4}{4}+\dfrac{9}{8}+\dfrac{16}{16}+\dfrac{25}{32}+\dfrac{36}{64}+\dfrac{49}{128}+\dots$
Now $\sum_{n=0}^\infty \dfrac{n}{2^n}\leqslant \sum_{n=0}^\infty \dfrac{n^2}{2^n}$
And I know that $\sum_{n=0}^\infty \dfrac{n}{2^n}=2$.
But how to find this sum ? I am confused.Please give some hints.
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If $-1< x < 1$, we have, by differentiation and adding :
$$\begin{align} \sum_{n=0}^\infty x^n & = \frac 1{1-x} \implies & \\ \sum_{n=1}^\infty nx^{n-1} & = \frac 1{(1-x)^2} \implies \\ \sum_{n=1}^\infty nx^n & = \frac x{(1-x)^2} \implies & \small \\ \sum_{n=1}^\infty n^2x^{n-1} & = \frac {1+x}{(1-x)^3} \implies & \small \\ \sum_{n=1}^\infty n^2x^n & = \frac {x(1+x)}{(1-x)^3} \end{align}$$
Put $x=\frac{1}{2}$. We are done.
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Finding two possible values of $z^2 + z + 1$ given that $z$ is one of the three cube roots of unity. Factorise $z^3 - 1 $.
If $z$ is one of the three cube roots of unity, find the two possible values of $z^2 + z + 1$.
Factorising gives you :
$(z - 1)(z^2 + z + 1) = 0$ since $z$ is one of the three cube roots of unity.
z is complex so $z \neq 1$ so $z-1 \neq 0$
Hence, $z^2 + z + 1 = 0$
Where do I go from here?
Any help is appreciated!!
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You are done. The two possible values of $z^2+z+1$ are $0$ for $z$ a third root of unity different from $1$, as you have computed, and $1+1+1=3$ for $z=1$.
|
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Help for series calculation $\sum_{n\ge1} \frac{1}{4n^3-n}$ I want to find the following series.
$$\sum_{n\ge1} \frac{1}{4n^3-n}$$
So, fisrt of all, patial fraction : $$\frac{1}{4n^3-n}=\frac{1}{2n+1}+\frac{1}{2n-1}-\frac{1}{n}.$$
Next, I consider the geometric series $$\sum_{n\ge1} x^{2n-2}+x^{2n}-x^{n-1}=\frac{1}{1-x^2}+\frac{x^2}{1-x^2}-\frac{1}{1-x}\tag{1}$$ where ${-1\lt x\lt1}$.
Then, I integrate both side of (1) from $0$ to $1$:
$$\sum_{n\ge1}\frac{1}{2n-1}+\frac{1}{2n+1}-\frac{1}{n}=\int_0^1 \frac{x^2-x}{1-x^2}dx$$
Finally, I get the value that is equal to $ln(2)-1$ by improper integral but it's less than $0$. What did I do wrong? All help is appreciated.
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You can do this way. It is easier. Let
$$ f(x)=\sum_{n=1}^\infty\frac{1}{4n^3-n}x^{2n+1}. $$
Then $f(1)=\sum_{n=1}^\infty\frac{1}{4n^3-n}$ and
$$ f'(x)=\sum_{n=1}^\infty\frac{1}{(2n-1)n}x^{2n}, f''(x)=2\sum_{n=1}^\infty\frac{1}{(2n-1)}x^{2n-1}, f'''(x)=2\sum_{n=1}^\infty x^{2n-2}=\frac2{1-x^2}$$
and hence
$$ f(1)=\int_0^1\int_0^s\int_0^t\frac2{1-x^2}dxdtds=2\ln2-1.$$
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How many triples satisfy $ab + bc + ca = 2 + abc $
$a^2 + b^2 + c^2 - \frac{a^3 + b^3 + c^3 - 3abc}{a+b+c} = 2 + abc$
How many triples $(a,b,c)$ satisfies the statement? Here $a,b,c > 1$.
It is easy to simplify the statement to
$$ab + bc + ca = 2 + abc.$$
But now how to proceed I don't know. I think this is somehow related to stars and bars theorem but I don't know how to convert this to that problem. Any hint will be helpful.
Source this is a problem from BdMO 2015 Dhaka regional.
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If $a$, $b$, and $c$ are required to only be positive integers and some of them is $1$, then we have a unique solution $(a,b,c)=(1,1,1)$. For solutions with $a,b,c>1$, note that
$$(a-1)(b-1)(c-1)=abc-bc-ca-ab+a+b+c-1=a+b+c-3\,.$$
Set $x:=a-1$, $y:=b-1$, and $z:=c-1$. Therefore,
$$xyz=x+y+z\,.$$
Without loss of generality, suppose that $x\leq y\leq z$ (noting that they are positive integers). Now, we have
$$y+z=x(yz-1)\geq yz-1\,,\text{or }(y-1)(z-1)\leq 2\,.$$
If $(y-1)(z-1)=0$, then $y=1$, making $x=1$ as well, so $$z=xyz=x+y+z=2+z\,$$ which is absurd. If $(y-1)(z-1)=1$, then $y=2$ and $z=2$, so $$4x=xyz=x+y+z=x+4\,,$$ leading to $x\notin\mathbb{Z}$, which is a contradiction. Thus, $(y-1)(z-1)=2$, and so $y=2$ and $z=3$. Ergo, $$6x=xyz=x+y+z=x+5\,,\text{ or }x=1\,.$$
Consequently, $(a,b,c)=(2,3,4)$ is the only solution, up to permutation.
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Find all possible positive integer $n$ such that $3^{n-1} + 5^{n-1} \mid 3^n + 5^n $
Find all possible positive integers $n$ such that $3^{n-1} + 5^{n-1} \mid 3^n + 5^n $.
I don't know how to start with. Any hint or full solution will be helpful.
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If $n= 1$ then we have $3^{n-1} + 5^{n-1} = 1+ 1 = 2|8 = 3^{n}+5^{n}$. So that's one solution.
Assume $n > 1$.
$3^n + 5^n = 3*3^{n-1} + 5*5^{n-1} = 3*3^{n-1} + 3*5^{n-1} + 2*5^{n-1} = 3(3^{n-1} + 5^{n-1}) + 2*5^{n-1}$.
So $3^{n-1} + 5^{n-1}|2*5^{n-1}$
If $p$ is prime and $p|3^{n-1} + 5^{n-1}$ then $p|2*5^{n-1}$ then $p = 5$ or $p= 2$
If $p = 5$ then $p\not |3^{n-1} + 5^{n-1}$. So $p = 2$ and $3^{n-1} + 5^{n-1} = 2^k$ but $2^k| 2*5^{n-1}$ means $k = 1$ so $3^{n-1} + 5^{n-1} = 2$. Which can only happen if $n = 1$.
So $n =1$ is only solution.
|
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How to find $\lim_{x\to1}\frac{\sqrt[3]{x+7}+2\sqrt{3x+1}-6}{\sqrt[4]{x}-1}$ without using L'Hospital Rule? Compute the following limit. I've tried using l'Hospital. And it work the result was $\dfrac{7}{3}$. But how can I do this without this rule? I am trying to help I friend who hasn't done derivatives yet.
$$\lim_{x\to1}\dfrac{\sqrt[3]{x+7}+2\sqrt{3x+1}-6}{\sqrt[4]{x}-1}$$
|
Hint:
$$\lim_{x\to1}\dfrac{\sqrt[3]{x+7}+2\sqrt{3x+1}-6}{\sqrt[4]{x}-1}=\\
\lim_{x\to1}\dfrac{\sqrt[3]{x+7}-2}{\sqrt[4]{x}-1}+\lim_{x\to1}\dfrac{2\sqrt{3x+1}-4}{\sqrt[4]{x}-1}\\
\lim_{x\to1}\dfrac{\sqrt[3]{x+7}-2}{\sqrt[4]{x}-1}\times\dfrac{(\sqrt[3]{x+7}+2)(\sqrt[4]{x}+1)}{(\sqrt[3]{x+7}+2)(\sqrt[4]{x}+1)}\dfrac{\sqrt{x}+1}{\sqrt{x}+1}\\+\lim_{x\to1}\dfrac{2(\sqrt{3x+1}-2)}{\sqrt[4]{x}-1}\times
\dfrac{(\sqrt{3x+1}+2)(\sqrt[4]{x}+1)}{(\sqrt{3x+1}-2)(\sqrt[4]{x}+1)}\times\dfrac{\sqrt{x}+1}{\sqrt{x}+1}\\$$
|
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Field with 2 elements Let $ A $ be a ring with $ x^{2}+y^{2}+z^{2}=xy+yz+zx+xyz+1,\forall x,y,z\in A^{*} $.
Prove that $ A $ is a field with 2 elements.
If we put $ x=y=z=1 $ we obtain that $ 1+1=0 $.
If we put $y=z=1 $ we have that $ x^2=x $, which means $ A $ is a boolean ring.
That's all I did so far.
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$x^2 + x^2 + x^2 = x*x + x*x + x*x + x*x *x + 1 =x^2 + x^2 +x^2 + x^3 + 1$ so $x^3 + 1 = 0$ and $x^3 = -1$
$x^2 + x^2 + 1 = x*x + x*1 + 1*x + x*x *1 = x^2 + x + x + x^2 + 1$ so $x + x =0$ and$x = -x $ and therefore $x^3 = -1 = 1$
$x^2 = x^2 + ( 1-1) = x^2 + 1 + 1 = x^2 + 1^2 + 1^2 = x*1 + 1*1 + 1*x + x*1*1 + 1=x+x + x +1 + 1 = x$
So $1=x^3 = x(x^2) = xx = x^2 = x$
So $x =1$ for all $x \ne 0$.
Any ring with only two elements is a field.
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Explain why $(a−b)^2 = a^2 −b^2$ if and only if $b = 0$ or $b = a$. This is a question out of "Precalculus: A Prelude to Calculus" second edition by Sheldon Axler. on page 19 problem number 54.
The problem is Explain why $(a−b)^2 = a^2 −b^2 $ if and only if $b = 0$ or $b = a$.
So I started by expanding $(a−b)^2$ to $(a−b)^2 = (a-b)(a-b) = a^2 -2ab +b^2$. To Prove that $(a−b)^2 = a^2 −b^2 $ if b = 0 I substituted b with zero both in the expanded expression and the original simplified and I got $(a−b)^2 = (a-0)^2 = (a-0)(a-0) = a^2 - a(0)-a(0)+0^2 = a^2$ and the same with $a^2 -2ab +b^2$ which resulted in $a^2 - 2a(0) + 0^2 = 2a$ or if I do not substite the $b^2$ I end up with $a^2 + b^2$. That's what I got when I try to prove the expression true for $b=0$.
As for the part where $b=a$, $(a−b)^2 = (a-b)(a-b) = a^2-2ab+b^2$, if a and b are equal, let $a=b=x$ and I substite $a^2-2ab+b^2 = x^2-2(x)(x) + x^2 = x^2-2x^2+x^2 = 1-2+1=0$ I do not see where any of this can be reduced to $a^2-b^2$ unless that equals zero......I do see where it holds but I do not see how would a solution writting out look.After typing this it seems a lot clearer but I just can't see how to phrase a "solution".
P.S: This is my first time asking a question here so whatever I did wrong I am sorry in advance and appreciate the feedback.
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You have to prove two things:
1) If $b = 0$ or $b = a$ then $(a-b)^2 = a^2 - b^2$.
And
2) If $(a-b)^2 = a^2 - b^2$ then either $b = 0 $ or $b = a$.
To prove 1: we do what you did correctly:
If $b = 0$ then $(a - b)^2 = (a-0)^2 = a^2$. An $a^2 - b^2 = a^2 - 0^2 = a^2 - 0 = a^2$. So $(a - b)^2= a^2 - b^2$.
If $b =a$ then $(a- b)^2 = (a-a)^2 = 0^2 = 0$. and $a^2 - b^2 = a^2 - b^2 = 0$. SO $(a- b)^2 = a^2 - b^2$.
That was the easy part.
To prove 2)
$(a-b)^2 = a^2 - b^2$ means
$(a -b)^2 = a^2 -2ab + b^2 = a^2 - b^2$ so
$a^2 - 2ab + b^2 + b^2 = a^2 - b^2 + b^2$ so
$a^2 - 2ab + 2b^2 = a^2$
$a^2 - 2ab + 2b^2- a^2 = a^2 - a^2$
$-2ab + 2b^2 = 0$
$2b(a- b) = 0$
Now if $mn = 0$ either $m = 0$ or $n=0$; that should be a basic fact you know.
So either $2b = 0$ or $a-b = 0$.
If $2b = 0$ then $b = 0$. If $a-b = 0$ then $a = b$.
So if $(a-b)^2 = a^2 -b^2$ then either $b =0$ or $b= a$.
|
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Approximation of a summation by an integral I am going to approximate $\sum_{i=0}^{n-1}(\frac{n}{n-i})^{\frac{1}{\beta -1}}$ by $\int_{0}^{n-1}(\frac{n}{n-x})^{\frac{1}{\beta -1}}dx$, such that $n$ is sufficiently large.
*
*Is the above approximation true?
*If the above approximation is true, by which theorem or method (like Newton's method) it can be holds? What is the error of approximation?
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If we write $s = 1/(\beta-1)$, your sum and integral can be re-written as
$$ \sum_{i=0}^{n-1} \left( \frac{n}{n-i} \right)^s = n^s \sum_{k=1}^n \frac{1}{k^s}, \qquad \int_{0}^{n-1} \left( \frac{n}{n-x} \right)^s \, dx = n^s \int_{1}^{n} \frac{dx}{x^s}. $$
In this case, the Euler-Maclaurin formula provides a way of estimating the difference within $\mathcal{O}(n^{-K})$ for any prescribed exponent $K$. To see this, let $\tilde{B}_k(x) = B_k(x - \lfloor x \rfloor)$ be the periodic Bernoulli polynomials. Then
\begin{align*}
\int_{1}^{n} \frac{dx}{x^s}
&= \left( \int_{[1,n]} \frac{d\lfloor x \rfloor}{x^s} + \int_{[1,n]} \frac{d\tilde{B}_1(x)}{x^s} \right) \\
&= \sum_{k=1}^{n} \frac{1}{k^s} + \int_{[1,n]} \frac{d\tilde{B}_1(x)}{x^s} \\
&= \sum_{k=1}^{n} \frac{1}{k^s} + \left[ \frac{\tilde{B}_1(x)}{x^s} \right]_{1^-}^{n} + s \int_{1}^{n} \frac{\tilde{B}_1(x)}{x^{s+1}} \, dx \\
&= \sum_{k=1}^{n} \frac{1}{k^s} - \frac{1 + n^{-s}}{2} + s \int_{1}^{n} \frac{\tilde{B}_1(x)}{x^{s+1}} \, dx
\end{align*}
However, there is an issue with this form. Indeed, if we keep using $[1, n]$ as the domain of integration, the error term of the Euler-Maclaurin formula never vanishes as $n \to \infty$. This is because for each fixed $K$,
$$ \int_{1}^{n} \frac{\tilde{B}_1(x)}{x^{s+K}} \, dx = \Theta(1) \qquad \text{as } n \to \infty $$
To resolve issue, let us assume $s > 0$ and we split the last integral as the difference of two:
\begin{align*}
\int_{1}^{n} \frac{dx}{x^s}
&= \sum_{k=1}^{n} \frac{1}{k^s} -\frac{1 + n^{-s}}{2} + s \int_{1}^{\infty} \frac{\tilde{B}_1(x)}{x^{s+1}} \, dx - s \int_{n}^{\infty} \frac{\tilde{B}_1(x)}{x^{s+1}} \, dx
\end{align*}
We note that integration by parts easily checks the asymptotics $\int_{n}^{\infty} \frac{\tilde{B}_K(x)}{x^{s+K}} \, dx = \mathcal{O}(n^{-s-K})$ for each fixed $K$. Then letting $n \to \infty$ together with the extra assumption $s > 1$, this yields
\begin{align*}
s \int_{1}^{\infty} \frac{\tilde{B}_1(x)}{x^{s+1}} \, dx
= \frac{1}{2} + \frac{1}{s-1} - \zeta(s).
\end{align*}
This continues to hold for $s > 0$ by the principle of analytic continuation. Plugging this back and simplifying in terms of the sum,
\begin{align*}
\sum_{k=1}^{n} \frac{1}{k^s}
&= \int_{1}^{n} \frac{dx}{x^s} + \frac{1 + n^{-s}}{2} + s \left( \zeta(s) - \frac{1}{2} - \frac{1}{s-1} \right) + s \int_{n}^{\infty} \frac{\tilde{B}_1(x)}{x^{s+1}} \, dx.
\end{align*}
Then we can continue the procedure to extract more terms: for each fixed $K$,
\begin{align*}
\sum_{k=1}^{n} \frac{1}{k^s}
&= \int_{1}^{n} \frac{dx}{x^s} + \frac{1 + n^{-s}}{2} + s \left( \zeta(s) - \frac{1}{2} - \frac{1}{s-1} \right)\\
&\qquad - \sum_{j=2}^{K} \frac{\Gamma(s+j-1)}{\Gamma(s)} \frac{B_j}{j!} \frac{1}{n^{s+j-1}} + \underbrace{\frac{\Gamma(s+K)}{K!\Gamma(s)} \int_{n}^{\infty} \frac{\tilde{B}_K(x)}{x^{s+K}} \, dx}_{\mathcal{O}(n^{-s-K-1})}
\end{align*}
as $n \to \infty$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2114574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
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|
Show that the equation $x^2+3y^2+4yz-6x+8y+8=0$ becomes a surface. Show that the equation $x^2+3y^2+4yz-6x+8y+8=0$ becomes a surface generated by the movement of a line and explicite its rectilinear generatrices.
I have tried to make its matrix, knowing that $a_{11}=1$, $a_{12}=3$...$a_{00}=8$.. But I don't think that's the way to solve it because the matrix that i will get will have a lot of 0's..
|
If you want to make a matrix do it like this.
$\mathbf x^T \begin {bmatrix}
1 &0 &0\\
0 &3 &2\\
0 &2 &0\\ \end{bmatrix}\mathbf x + \begin {bmatrix} -6&8&0\end{bmatrix} \mathbf x = 0$
Since that matrix is symmetric is is diagonalizable with ortho-normal basis.
$\mathbf x^t P^T D P \mathbf x + BP^TP\mathbf x = -8\\
\mathbf u = P \mathbf x$
$P = \begin {bmatrix}
1 &0 &0\\
0 &\frac 2{\sqrt5} &-\frac 1{\sqrt5}\\
0 &\frac 1{\sqrt5} &\frac 2{\sqrt5}\\ \end{bmatrix}$
$D = \begin {bmatrix}
1 &0 &0\\
0 &4 &0\\
0 &0 &-1\\ \end{bmatrix}$
$u_1^2 + 4u_2^2 - u_3^2 - 6u_1 + \frac {16}{\sqrt5} u_2 +\frac {8}{\sqrt5} u_3=0\\
(u_1-3)^2 + 4(u_2+\frac2{\sqrt 5})^2 - (u_3- \frac 4{\sqrt 5})^2 = 1
$
That is a hyperboliod of one sheet.
$(x-3)^2 + 4(\frac 2{\sqrt5} y - \frac 1{\sqrt5} z+\frac2{\sqrt 5})^2 - (\frac 1{\sqrt5} y + \frac 2{\sqrt5}z- \frac 4{\sqrt 5})^2 = 1
$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2115379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Show that determinant of $\small\begin{pmatrix}2 & 2 & 8\\ 3& 2 & 3 \\ 4 & 5 & 6\end{pmatrix}$ is divisible by $19$ Using that the numbers
$228,
323$
and
$456$
are
divisible
by
$19$.
Show
that
the
determinant of matrix
$\begin{pmatrix}2 & 2 & 8\\ 3& 2 & 3 \\ 4 & 5 & 6\end{pmatrix}$
is
divisible
by
$19$.
|
There's an alternative and maybe easier way to prove the result, with Gaussian elimination.
Consider the matrix with coefficients over $\mathbb{Z}/19\mathbb{Z}$; the inverse of $2$ is $10$, so
\begin{align}
\begin{bmatrix}
2 & 2 & 8 \\
3 & 2 & 3 \\
4 & 5 & 6
\end{bmatrix}
&\to
\begin{bmatrix}
1 & 1 & 4 \\
3 & 2 & 3 \\
4 & 5 & 6
\end{bmatrix}
&&R_1\gets 10R_1
\\
&\to
\begin{bmatrix}
1 & 1 & 4 \\
0 & -1 & -9 \\
0 & 1 & 9
\end{bmatrix}
&&\begin{aligned}R_2\gets R_2-2R_1\\R_3\gets R_3-4R_1\end{aligned}
\end{align}
which clearly has rank $2$, so its determinant is $0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2116555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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|
Find the limit: $\lim_{x\to 0} \frac{8^x-7^x}{6^x-5^x}$ Find the limit:
$$\lim_{x\to 0} \frac{8^x-7^x}{6^x-5^x}$$
I really have no idea what to do here, since I obviously can't plug in $x=0$ nor divide by highest power...Any help is appreciated!
|
By using hint of @dxix we get
$$\lim _{ x\to 0 } \frac { 8^{ x }-7^{ x } }{ 6^{ x }-5^{ x } } =\lim _{ x\to 0 } \frac { { 7 }^{ x }\left[ { \left( \frac { 8 }{ 7 } \right) }^{ x }-1 \right] }{ { 5 }^{ x }\left[ { \left( \frac { 6 }{ 5 } \right) }^{ x }-1 \right] } =\lim _{ x\to 0 } \frac { { 7 }^{ x }\frac { \left[ { \left( \frac { 8 }{ 7 } \right) }^{ x }-1 \right] }{ x } }{ { 5 }^{ x }\frac { \left[ { \left( \frac { 6 }{ 5 } \right) }^{ x }-1 \right] }{ x } } =\\ =\frac { \ln { \left( 8/7 \right) } }{ \ln { \left( 6/5 \right) } } $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2118307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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|
How to show that $\sum_{k} (-1)^k{{a+b}\choose{a+k}}{{b+c}\choose{b+k}}{{c+a}\choose{c+k}} = \frac{(a+b+c)!}{a!b!c!}$ How to show that
$$\sum_{k} (-1)^k{{a+b}\choose{a+k}}{{b+c}\choose{b+k}}{{c+a}\choose{c+k}} = \frac{(a+b+c)!}{a!b!c!}$$
|
Here is an answer based upon Short Proofs of Saalschütz's and Dixon's theorems by Ira Gessel and Dennis Stanton.
The answer is provided in three steps
*
*Step 1: Relationship between coefficients of the constant term of a bivariate Laurent series and a transformed of it.
*Step 2: Application of this relationship to a specific Laurent series with the coefficient of the constant term being strongly related with OPs binomial identity.
*Step 3: Substitutions to reveal the symmetry of the binomial identity and to transform it finally to OPs stated binomial identity.
We consider formal Laurent series
\begin{align*}
A(x,y)=\sum_{m,l}a_{m,l}x^my^l
\end{align*}
in which for only finitely many pairs $(m,l)$ the coefficients $a_{m,l}\ne 0$ with $m<0$ or $l<0$. Let $[x^n]$ denote the coefficient of $x^n$ in a series. We start with a useful observation
Step 1: Let $A(x,y)$ be a formal Laurent series. The following is valid
\begin{align*}
[x^0y^0]A\left(\frac{x}{1+y},\frac{y}{1+x}\right)=[x^0y^0]\frac{1}{1-xy}A(x,y)\tag{1}
\end{align*}
By linearity it is sufficient to prove (1) by considering $A(x,y)=x^ly^m$ with $l,m\in \mathbb{Z}$. We have to show
\begin{align*}
[x^0y^0]\frac{x^l}{(1+y)^l}\cdot\frac{y^m}{(1+x)^m}=[x^0y^0]\frac{x^ly^m}{1-xy}\tag{2}
\end{align*}
We show the LHS of (2) fulfills
\begin{align*}
[x^0y^0]\frac{x^l}{(1+y)^l}\cdot\frac{y^m}{(1+x)^m}=
\begin{cases}
1&\qquad\text{if }l=m\leq 0\\
0&\qquad\text{otherwise}
\end{cases}\tag{3}
\end{align*}
*
*If $l>0$ or $m>0$ the result in (3) is $0$ since the binomial expansion of $(1+y)^{-l}(1+x)^{-m}$ has no negative powers.
*If $l=-r,m=-s$ with $r,s\geq 0$ we note that
\begin{align*}
[x^0y^0]\frac{(1+x)^s(1+y)^r}{x^ry^s}&=[x^ry^s](1+x)^s(1+y)^r\\
&=\binom{s}{r}\binom{r}{s}=
\begin{cases}
1&\qquad\text{if }r=s\\
0&\qquad\text{if }r\neq s
\end{cases}
\end{align*}
and (3) follows.
We look at the RHS of (2) and observe
\begin{align*}
[x^0y^0]\frac{x^ly^m}{1-xy}=[x^{-l}y^{-m}]\sum_{j\geq 0}(xy)^j
=\begin{cases}
1&\qquad\text{if }l=m\leq 0\\
0&\qquad\text{otherwise}
\end{cases}
\end{align*}
which is the same as (3) and the claim (2) follows.
Step 2: We now apply (1) to the Laurent series
\begin{align*}
f(x,y)=\frac{(x-y)^n}{x^ly^m(1-xy)^n}\qquad\qquad\qquad l,m\in\mathbb{Z},n\in\mathbb{N}
\end{align*}
We obtain
\begin{align*}
&f\left(\frac{x}{1+y},\frac{y}{1+x}\right)\\
&\qquad=\left(\frac{x}{1+y}-\frac{y}{1+x}\right)^n
\cdot\frac{(1+y)^l}{x^l}\cdot\frac{(1+x)^m}{y^m}\cdot\left(1-\frac{xy}{(1+x)(1+y)}\right)^{-n}\\
&\qquad=\frac{\left((x-y)+(x^2-y^2)\right)^n}{(1+x)^n(1+y)^n}\cdot
\frac{(1+y)^l}{x^l}\frac{(1+x)^m}{y^m}\cdot
\frac{(1+x)^n(1+y)^n}{(1+x+y)^n}\\
&\qquad=\frac{(1+x)^m(1+y)^l(x-y)^n}{x^ly^m}\tag{4}
\end{align*}
on the other hand we get
\begin{align*}
\frac{1}{1-xy}f(x,y)=\frac{(x-y)^n}{x^ly^m(1-xy)^{n+1}}\tag{5}
\end{align*}
equating (4) and (5) we obtain according to (2)
\begin{align*}
[x^0y^0]\frac{(1+x)^m(1+y)^l(x-y)^n}{x^ly^m}=[x^0y^0]\frac{(x-y)^n}{x^ly^m(1-xy)^{n+1}}\tag{6}
\end{align*}
The LHS of (6) gives
\begin{align*}
[x^0y^0]&\frac{(1+x)^m(1+y)^l(x-y)^n}{x^ly^m}\\
&=[x^ly^m]\sum_{k}\binom{n}{k}(-y)^kx^{n-k}(1+x)^m(1+y)^l\\
&=\sum_{k}\binom{n}{k}(-1)^k[x^{l-n+k}y^{m-k}]
\sum_{i=0}^m\binom{m}{i}x^i\sum_{j=0}^l\binom{l}{j}y^j\\
&=\sum_{k}(-1)^k\binom{n}{k}\binom{m}{l-n+k}\binom{l}{l-m+k}
\end{align*}
In order to consider the RHS of (6), we do a little trick. We set $x=xz$ and $y=yz$ and note that
\begin{align*}
[x^0y^0]f(x,y)&=[x^0y^0z^0]f(xz,yz)\\
&=[x^0y^0z^0]\frac{(xz-yz)^n}{(xz)^l(yz)^m(1-z^2xy)^{n+1}}\\
&=[x^0y^0z^0]\frac{(x-y)^n}{x^ly^mz^{l+m-n}(1-z^2xy)^{n+1}}\tag{7}
\end{align*}
We observe that (7) is zero if $l+m-n$ is odd. If $l+m-n=2r$ is even, we obtain
\begin{align*}
[x^0y^0z^0]&\frac{(x-y)^n}{x^ly^mz^{l+m-n}(1-z^2xy)^{n+1}}\\
&=[x^ly^mz^{2r}](x-y)^n\sum_{j\geq 0}\binom{-(n+1)}{j}(-z^2xy)^j\\
&=[x^ly^mz^{2r}](x-y)^n\sum_{j\geq 0}\binom{n+j}{j}(z^2xy)^j\\
&=\binom{n+r}{r}[x^{l-r}y^{m-r}]\sum_{j=0}^nx^k(-y)^{n-k}\\
&=\binom{n+r}{r}\binom{n}{l-r}(-1)^{m-r}\tag{8}
\end{align*}
With focus on (8) at which $l+m-n=2r$ is even we conclude
\begin{align*}
\sum_{k}(-1)^k\binom{n}{k}\binom{m}{l-n+k}\binom{l}{l-m+k}
=\binom{n+r}{r}\binom{n}{l-r}(-1)^{m-r}\tag{9}
\end{align*}
Step 3: We can now put more symmetry into (9) by setting
\begin{align*}
n&=a+b\\
m&=c+a\\
l&=b+c
\end{align*}
We obtain
\begin{align*}
\sum_{k}(-1)^k\binom{a+b}{k}\binom{c+a}{c-a+k}\binom{b+c}{b-a+k}=(-1)^{a}\binom{a+b+c}{a+b}\binom{a+b}{a}
\end{align*}
Finally multiplying this identity with $(-1)^a$ and shifting the index $k\rightarrow k+a$ we obtain
\begin{align*}
\sum_{k}(-1)^k\binom{a+b}{a+k}\binom{b+c}{b+k}\binom{c+a}{c+k}&=\binom{a+b+c}{a+b}\binom{a+b}{a}\\
&=\frac{(a+b+c)!}{a!b!c!}
\end{align*}
and the claim follows.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "7",
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|
Is the following limit finite ....? I would like to see some clue for the following problem:
Let $a_1=1$ and $a_n=1+\frac{1}{a_1}+\cdots+\frac{1}{a_{n-1}}$, $n>1$. Find
$$
\lim_{n\to\infty}\left(a_n-\sqrt{2n}\right).
$$
|
Quick and dirty proof.
Assume for a moment that the sequence $\,a_n\,$ is a continuous
and differentiable function $\,a(n)\,$ of $\,n\,$ as is suggested in the picture below.
Then consider the following sequence. Fasten your seatbelts!
$$
a_{n+1} = a_n + \frac{1}{a_n} \\
\frac{a_{n+1} - a_n}{1} = \frac{1}{a_n} \\
\frac{a(n+dn) - a(n)}{dn} = \frac{1}{a(n)} \\
a'(n)\,a(n) = 1 \\ \frac{da^2(n)}{dn} = 2 \\
a^2(n) = 2n+C
$$
[Scaling argument deleted. I see no way to get it consistent, let it be rigorous. See edits of this post]
Now we would like to have $\,C=0$ ; and fortunately there is a boundary condition, for $\,n=2$ , that fits the bill : $\,C = a_2^2 - 2\cdot 2 = 0$ . So:
$\;a_n = \sqrt{2n}$ .
The larger $n$ , the better all of the approximations. A much neater way to express the end-result is:
$$
\large \boxed{\lim_{n\to\infty} \frac{a_n}{\sqrt{2n}} = 1}
$$
BONUS. Lemma:
$$
a_{n+1} = a_n + \frac{1}{a_n} \quad \Longrightarrow \quad
a_{n-1}^2 - a_na_{n-1} + 1 = 0 \quad \Longrightarrow \quad a_{n-1} = \frac{a_n}{2} + \sqrt{\left(\frac{a_n}{2}\right)^2-1}
$$
Now compute (the discretization of) the second derivative, assuming again that $\,a_n\,$ is large:
$$
a''(n) = a_{n+1} - 2a_n + a_{n-1} = a_n + \frac{1}{a_n} - 2a_n + \frac{a_n}{2} +
\frac{a_n}{2}\sqrt{1-\frac{1}{(a_n/2)^2}} \approx \\ \frac{1}{a_n} - \frac{a_n}{2} + \frac{a_n}{2}
\left[1-\frac{1}{2}\frac{1}{(a_n/2)^2} - \frac{1}{8}\left(\frac{1}{(a_n/2)^2}\right)^2\right] = -\frac{1}{a_n^3}
$$
Which means that the discrete function $\,a_n\,$ for large $\,n\,$ is actually very smooth , when seen as a continuous
and differentiable $\,a(n)$ . This is an even stronger motivation for the above treatment.It is noticed that the "true" derivatives exhibit the same structure as the discretizations:
$$
\left(\sqrt{2x}\right)'' = \left(\frac{1}{\sqrt{2x}}\right)' = -\frac{1}{\left(\sqrt{2x}\right)^3}
$$
There are no coincidences in mathematics.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "12",
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|
In right triangle $ABC$ ( $BC$ is hypotenuse),$D$ is a point on $BC$.Calculate $AB$ given that...
In right triangle $ABC$ ( $BC$ is hypotenuse),$D$ is a point on $BC$.Calculate $AB$ given that: $AD$=5,$BD=3$,$CD=6$.
|
Let $\angle{ADC}$ be $\alpha$. Then by the cosine theorem
$b^2=25+36-60\cos\alpha$,
$c^2=25+9-30\cos(180^{\circ}-\alpha)=34+30\cos\alpha$.
This gives
$b^2+2c^2=129$.
Since $b^2+c^2=9^2=81$, it gives $c^2=48.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2120648",
"timestamp": "2023-03-29T00:00:00",
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|
Compute $\int_0^3 \frac{x^4}{x^4+(x-3)^4} \,dx$ $$\int_0^3 \frac{x^4}{x^4+(x-3)^4} \,dx$$
Then the question ask me to change it into
$$\int_0^3 \frac{(x-3)^4}{x^4+(x-3)^4} \,dx$$
Then how to evaluate it
if let $$u=x-3$$
$$du=dx$$
$$x^4 =(u+3)^4$$
$$\int_3^0 \frac{u^4}{(u+3)^4+u^4} \,du$$
it is still the same
so what should i do?
|
Let,
$$I=\int_0^3 \frac{x^4}{x^4+(x-3)^4} \,dx$$
Let $x=3-u$ and change dummy variable back to $x$.
$$I=\int_{0}^{3} \frac{(3-x)^4}{x^4+(x-3)^4} dx$$
$$=\int_{0}^{3} \frac{(x-3)^4}{x^4+(x-3)^4} dx$$
Add this to the first form of $I$.
$$2I=\int_{0}^{3} 1 dx$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2123553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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|
Sum of the combinatorial series $$\frac{\binom{b+i}{1}}{\binom{i}{1}} + \frac{\binom{b+i}{2}}{\binom{i}{2}} + \frac{\binom{b+i}{3}}{\binom{i}{3}} +... + \frac{\binom{b+i}{i}}{\binom{i}{i}} $$
where $b \in \mathbb{R}$ and $b\in [-1, 1]$ .
|
Proof of $ \displaystyle\enspace \sum\limits_{j=1}^i \binom{b+i}{j}\binom{i}{j}^{-1} = \frac{b+i}{1-b}-\frac{b}{1-b} \binom{b+i}{i} \enspace$ by induction for $b\ne 1$.
$i:=1$ : $\displaystyle \enspace \sum\limits_{j=1}^1 \binom{b+1}{j}\binom{1}{j}^{-1}=\frac{b+1}{1-b}-\frac{b}{1-b} \binom{b+1}{1} \enspace $ which is true because of $b+1=b+1$
$i\to i+1$ :
We have $\displaystyle \enspace \sum\limits_{j=1}^{i+1} \binom{b+i+1}{j}\binom{i+1}{j}^{-1} =\frac{b+i+1}{i+1}(1+\sum\limits_{j=1}^i \binom{b+i}{j}\binom{i}{j}^{-1}) \enspace $ and therefore have to proof $\displaystyle \enspace \frac{b+i+1}{1-b}-\frac{b}{1-b} \binom{b+i+1}{i+1} =\frac{b+i+1}{i+1}(1+\frac{b+i}{1-b}-\frac{b}{1-b} \binom{b+i}{i}) $ .
But elementary transformations show immediately, that this is true.
A note for $\enspace b=1$ :
$ \displaystyle\enspace \sum\limits_{j=1}^i \binom{1+i}{j}\binom{i}{j}^{-1}=(i+1)H_i \enspace $ where $H_i$ is the $i^{th}$ partial sum of the harmonic series.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2124986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
How do I solve $(x+1)^5 +36x+36 = 13(x+1)^3$? I tried $$(x+1)^5 + 36 x + 36 = 13 (x +1)^3\\
(x+1)^5 + 36(x+1) = 13 (x +1)^3\\
(x+1)^4 +36 = 13 (x+1)^2
$$
But, don't understand how to solve further. Can somebody show step by step please. Thanks!
|
Continue from your last step.
Put $(x + 1)^2 = z$
We have,
$z^2 + 36 = 13z$
$z^2 - 13z + 36 = 0$
$z^2 - 9z - 4z + 36 = 0$
$z(z - 9) - 4(z - 9) = 0$
$(z - 9)(z - 4) = 0$
When z = 9
$(x + 1)^2 = 9$
$(x + 1) = \pm 3$
x = 2 or -4
When z = 4
$(x + 1)^2 = 4$
$(x + 1) = \pm 2$
x = 1 or -3
x = -4, -3, 1, 2
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2125300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
How does this simplify to 1? I am working on this differetiation problem:
$ \frac{d}{dx}x(1-\frac{2}{x})$
and I am currently stuck at this point:
$1\cdot \left(1-\frac{2}{x}\right)+\frac{2}{x^2}x$
Symbolab tells me this simplifies to $1$ but I do not understand how. I am under the impression that;
$1\cdot \left(1-\frac{2}{x}\right)+\frac{2}{x^2}x \equiv 1- 2x^{-x^2}-2^{-x}$
|
Do not get confused by fractions and exponents.
You should remember that $\frac {k}{n} = kn^{-1}$ and not $k^{-n} $ and that $\frac {k}{n} l = kln^{-1}$ and not $kl^{-n} $.
We have $$1\times (1-\frac {2}{x}) = 1-\frac {2}{x} \tag {1}$$ and then $$\frac {2}{x^2}x = \frac {2}{x^2} \times x = \frac {2}{x} \tag {2} $$
What do we get by adding $(1)$ and $(2)$? The result is $1-\frac {2}{x} + \frac {2}{x} = 1$. Hope it helps.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Prove that $\sum\limits_{n=1}^{\infty} \frac{n}{3^n} = \frac{3}{4}$ How can you derive that
$$ \sum\limits_{n=1}^{\infty} \frac{n}{3^n} = \frac{3}{4} \, ?$$
I suspect some clever use of the geometric series will do, but I don't know how.
|
$$\sum_{n=1}^{\infty} \frac{n}{3^n} = \sum_{n=1}^{\infty}\sum_{k=n}^{\infty}\frac{1}{3^n}\frac{1}{3^{k-n}}=\\
=\sum_{n=1}^{\infty}\frac{\left(\frac{1}{3}\right)^n}{1-\frac{1}{3}}=\frac{3}{2}\sum_{n=1}^{\infty}\left(\frac{1}{3}\right)^n =\\
= \frac{3}{2}\frac{\frac{1}{3}}{1-\frac{1}{3}} = \frac{3}{2}\frac{1}{2}=\frac{3}{4}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2131479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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|
Proving the inequality $(a^2-ab+b^2)(c^2-ac+a^2)(b^2-bc+c^2) \le 12$. This is a follow up question to my previous post "Inequalities of expressions completely symmetric in their variables". An answer provided a counterexample to me reasoning: under the constraints $a,b,c\in\Bbb{R}^+$ and $a+b+c=3$,
$$
(a^2-ab+b^2)(c^2-ac+a^2)(b^2-bc+c^2) \le 12.
$$
I demanded a proof for this inequality, however since it was an entirely different question, I felt the need for a new post.
|
Let $a\geq b\geq c\geq0$.
Hence,
$$\prod_{cyc}(a^2-ab+b^2)\leq(a^2-ab+b^2)a^2b^2=((a+b)^2-3ab)a^2b^2\leq(9-3ab)a^2b^2=$$
$$=12(3-ab)\cdot\frac{ab}{2}\cdot\frac{ab}{2}\leq12\left(\frac{3-ab+\frac{ab}{2}+\frac{ab}{2}}{3}\right)^3=12$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2132401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
How to do this question that talks about dependency of x Let $x > 0$. Prove that the value of the following expression doesn't depend on x
$$\int_{0}^{x} \frac{1}{1+t^2} dt + \int_{0}^{\frac{1}{x}} \frac{1}{1+t^2}dt$$
Attempt:
Left: f'(x) = $\frac{1}{1+x^2}$
Right: f'(x) = $\frac{1}{1+(\frac{1}{x})^2} -\frac{1}{x^2}$
$= \frac{1}{(1+\frac{1}{x^2})} - \frac{1}{x^2}$
$=\frac{x^2}{1+x^2} - \frac{1}{x^2}$
$=\frac{x^4 - x^2 - 1}{1+x^2}$
Yeah I don't know what I am doing, I tried to remove the integral but failed miserably
|
Using the $u$ sub $t' = t^{-1}$, we see $dt'= -(t')^2dt$ that
\begin{align}
\int^{\frac{1}{x}}_0\frac{1}{1+t^2}\ dt = -\int^x_\infty \frac{1}{1+(t')^{-2}} \frac{dt'}{(t')^2}= \int^\infty_x \frac{1}{1+(t')^2}\ dt' = \frac{\pi}{2} - \int^x_0\frac{1}{1+(t')^2}\ dt'.
\end{align}
But we see that
\begin{align}
\frac{\pi}{2} - \int^x_0\frac{1}{1+(t')^2}\ dt'= \int^x_0\frac{1}{1+(t')^2}\ dt'
\end{align}
provided
\begin{align}
\int^x_0\frac{1}{1+(t')^2}\ dt' = \frac{\pi}{4}.
\end{align}
Thus, your claim is incorrect since equality holds only when $x=1$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the maximum and minimum value of $\cos^2x-6\sin(x)\cos(x)+3\sin^2x+2$ Find the maximum and minimum value of $$\cos^2x-6\sin(x)\cos(x)+3\sin^2x+2$$.
i simplified and reach to expression as follows :
$5 + 2\sin(x)[\sin(x)-3\cos(x)]$. How do i go from here?
Thanks
|
$$\cos^2x-6\sin(x)\cos(x)+3\sin^2x+2$$
$$-3\sin(2x)+2\dfrac{1-\cos2x}{2}+3$$
$$-3\sin(2x)-\cos(2x)+4$$
now use
$$|a\sin\alpha+b\cos\alpha|\leq\sqrt{a^2+b^2}$$
|
{
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|
Evaluate the integral $\int \frac{x^2(x-2)}{(x-1)^2}dx$
Find $$\int \frac{x^2(x-2)}{(x-1)^2} dx .$$
My attempt:
$$\int \frac{x^2(x-2)}{(x-1)^2}dx = \int \frac{x^3-2x^2}{x^2-2x+1}dx $$
By applying polynomial division, it follows that
$$\frac{x^3-2x^2}{x^2-2x+1} = x + \frac{-x}{x^2-2x+1}$$
Hence $$\int \frac{x^3-2x^2}{x^2-2x+1}dx = \int \left(x + \frac{-x}{x^2-2x+1}\right) dx =\int x \,dx + \int \frac{-x}{x^2-2x+1} dx \\ = \frac{x^2}{2} + C + \int \frac{-x}{x^2-2x+1} dx $$ Now using substitution $u:= x^2-2x+1$ and $du = (2x-2)\,dx $ we get $dx= \frac{du}{2x+2}$.
Substituting dx in the integral:
$$\frac{x^2}{2} + C + \int \frac{-x}{u} \frac{1}{2x-2} du =\frac{x^2}{2} + C + \int \frac{-x}{u(2x-2)} du $$
I am stuck here. I do not see how using substitution has, or could have helped solve the problem. I am aware that there are other techniques for solving an integral, but I have been only taught substitution and would like to solve the problem accordingly.
Thanks
|
Your last integral, $\int \frac{-x}{u(2x-2)} du$, doesn't make me happy.
You have both $x$ and $u$ in the integral. The point of the substitution is to get rid of the $x$'s and leave the integral in terms of $u$.
I'll take it from the last integral in terms of $x$, everything looks fine before and it goes astray after that.
$\int \frac{-x}{x^2-2x+1} dx = \int \frac{-x}{(x-1)^2} dx$. Let $u=x-1$ therefore $du=dx$ and $-x=-1-u$, then $\int \frac{-x}{(x-1)^2} dx=\int \frac{-1-u}{u^2} du=\int \frac{-1}{u^2} du -\int \frac{1}{u} du= \frac{1}{u}-\ln u=\frac{1}{x-1} - \ln {(x-1)}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Permute order of summation change the sum : $\sum_{n=1}^\infty \frac{(-1)^n}{n}$. In a course of analysis it's written that if $\sum_{k=1}^\infty x_k$ converge but it's not absolutely convergent, then changing any order of summation will change the sum.
For example, if I consider $$\sum_{n=1}^\infty \frac{(-1)^n}{n}=\ln(2).$$
What will be $-\frac{1}{2}+1+\sum_{n=3}^\infty \frac{(-1)^n}{n}$? To me it should be $$-\frac{1}{2}+1+\ln(2)-1+\frac{1}{2}=\ln(2),$$
so may be it's only by changing an infinite number of element of the sum no? If we change just a finite number, it should be the same, no?
|
You are right. Changing finite numbers wont change the sum.
Here is a nice example: Set $S: =\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n}$. Then
\begin{align*}
\sum_{n=1}^{\infty}\left(\frac{1}{2n-1}-\frac{1}{4n-2}-\frac{1}{4n}\right)
\end{align*}
is a rearrangement of $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n}$ and since
\begin{align*}
\frac{1}{2n-1}-\frac{1}{4n-2}-\frac{1}{4n} = \frac{1}{4n-2}-\frac{1}{4n}=\frac{1}{2}\left(\frac{1}{2n-1} -\frac{1}{2n}\right)
\end{align*}
for all $n\in\mathbf N$ we have
$$\sum_{n=1}^{\infty}\frac{1}{2n-1}-\frac{1}{4n-2}-\frac{1}{4n} =\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{2n-1} -\frac{1}{2n}=\frac{1}{2}S.$$
This is just a special case of the Riemann series theorem since $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n}$ is obviously conditional convergent.
|
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"url": "https://math.stackexchange.com/questions/2140637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
To prove that prove that $cos^8 \theta sec^6 \alpha , \frac{1}{2 } ,sin^8 \theta cosec^6 \alpha $ are in A.P If $\cos^4 \theta \sec^2 \alpha , \frac{1}{2 } ,\sin^4 \theta \csc^2 \alpha $ are in A.P ,
then prove that $\cos^8 \theta \sec^6 \alpha , \frac{1}{2 } ,\sin^8 \theta \csc^6 \alpha $ are in A.P
Now i have reached upto
$1=\cos^4 \theta \sec^2 \alpha + \sin^4 \theta \csc^2 \alpha$.
By completing square i have $(\sin^2 \theta \cot\alpha - \cos^2 \theta \tan \alpha)^2=0$ so i get $\tan \theta= \pm \tan \alpha$
How do i proceed?
Thanks
|
You have done rightly. Now we can proceed as follows: $$\tan \theta = \pm \tan \alpha \Rightarrow \tan^2 \theta = \tan^2 \alpha \Rightarrow \sec^2 \theta -1 =\sec^2 \alpha -1 \Rightarrow \sec^2 \theta = \sec^2 \alpha \Leftrightarrow \sec^6 \theta = \sec^6 \alpha \tag{1}$$
Also, $$\tan^2 \theta = \tan^2 \alpha \Rightarrow \cot^2 \theta = \cot^2 \alpha \Rightarrow \csc^2 \theta -1 = \csc^2 \alpha -1 \Rightarrow \csc^2 \theta = \csc^2 \alpha \Leftrightarrow \csc^6 \theta = \csc^6 \alpha \tag {2}$$
Hope you can take it from here by using $(1)$ and $(2)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2142174",
"timestamp": "2023-03-29T00:00:00",
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|
Is there a quicker way to evaluate $\int_{0}^{\infty} \frac{1-x^2}{x^4+3x^2+1}\ dx$? The integral is: $$\int_{0}^{\infty} \frac{1-x^2}{x^4+3x^2+1}\ dx$$
My procedure:
$$4\int_{0}^{\infty} \frac{1-x^2}{(2x^2+3)^2-5}\ dx=4\int_{0}^{\infty} \frac{1-x^2}{(2x^2+3-\sqrt5)(2x^2+3+\sqrt5)}\ dx$$
$$$$Using partial fractions to separate the integral, we obtain:
$$\frac1{\sqrt5}\int_{0}^{\infty} \left(\frac{5-\sqrt5}{2x^2+3-\sqrt5}-\frac{5+\sqrt5}{2x^2+3+\sqrt5}\right)\ dx$$
$$=\frac{5-\sqrt5}{\sqrt5}\int_{0}^{\infty} \frac1{2x^2+3-\sqrt5}\ dx-\frac{5+\sqrt5}{\sqrt5}\int_{0}^{\infty}\frac1{2x^2+3+\sqrt5}\ dx$$
$$$$We need write the fractions in the form $\frac1{u^2+1}$:
$$=\frac{5-\sqrt5}{\sqrt5}\int_{0}^{\infty} \frac1{(3-\sqrt5)\left(\frac{2x^2}{3-\sqrt5}+1\right)}\ dx-\frac{5+\sqrt5}{\sqrt5}\int_{0}^{\infty}\frac1{(3+\sqrt5)\left(\frac{2x^2}{3+\sqrt5}+1\right)}\ dx$$
$$=\frac{5-\sqrt5}{\sqrt5(3-\sqrt5)}\int_{0}^{\infty} \frac1{\left(\frac{\sqrt2x}{\sqrt{3-\sqrt5}}\right)^2+1}\ dx-\frac{5+\sqrt5}{\sqrt5(3+\sqrt5)}\int_{0}^{\infty}\frac1{\left(\frac{\sqrt2x}{\sqrt{3+\sqrt5}}\right)^2+1}\ dx$$
$$\frac{5-\sqrt5}{\sqrt{10(3-\sqrt5)}}\int_{0}^{\infty} \frac{\sqrt2}{\sqrt{3-\sqrt5}\left(\left(\frac{\sqrt2x}{\sqrt{3-\sqrt5}}\right)^2+1\right)}\ dx-\frac{5+\sqrt5}{\sqrt{10(3+\sqrt5)}}\int_{0}^{\infty}\frac{\sqrt2}{\sqrt{3+\sqrt5}\left(\left(\frac{\sqrt2x}{\sqrt{3+\sqrt5}}\right)^2+1\right)}\ dx$$
$$$$Evaluating the indefinite integrals, we obtain:
$$\arctan\left(\frac{\sqrt2x}{\sqrt{3-\sqrt5}}\right)-\arctan\left(\frac{\sqrt2x}{\sqrt{3+\sqrt5}}\right)+C$$
$$$$Evaluating the improper integral:
$$\lim_{z \to \infty}\left[\arctan\left(\frac{\sqrt2x}{\sqrt{3-\sqrt5}}\right)-\arctan\left(\frac{\sqrt2x}{\sqrt{3+\sqrt5}}\right)\right]_0^z$$
$$=\lim_{z \to \infty}\left(\arctan\left(\frac{\sqrt2z}{\sqrt{3-\sqrt5}}\right)-\arctan\left(\frac{\sqrt2z}{\sqrt{3+\sqrt5}}\right)\right)$$
$$=\lim_{z \to \infty}\left(\arctan\left(\infty\right)-\arctan\left(\infty\right)\right)=\frac{\pi}2-\frac{\pi}2=0$$
The way that I evalauted the integral is pretty long and has a lot of calculations. Is there an easier way?
|
Define $$f(x) = \frac{1-x^2}{1+3x^2+x^4}$$ so that $$f(x^{-1}) = -\frac{x^2(1-x^2)}{1+3x^2+x^4}.$$ Since $$\int_{x=0}^\infty f(x) \, dx = \int_{x=0}^1 f(x) \, dx + \int_{x=1}^\infty f(x) \, dx,$$ the transformation $$x = u^{-1}, \quad dx = -u^{-2} \, du$$ gives $$\int_{x=1}^\infty f(x) \, dx = \int_{u=1}^0 -\frac{u^2(1-u^2)}{1+3u^2+u^4} (-u^{-2}) \, du = -\int_{u=0}^1 f(u) \, du.$$ Thus, we immediately obtain cancellation and the integral of $f$ on $(0,\infty)$ is $0$.
For the indefinite integral, we write $$f(x) = \frac{x^{-2} - 1}{x^{-2} + 3 + x^2} = \frac{x^{-2} - 1}{1 + (x + x^{-1})^2}.$$ This suggests the substitution $$u = x + x^{-1}, \quad du = (1 - x^{-2}) \, dx,$$ giving $$\int f(x) \, dx = \int \frac{-du}{1+u^2} = -\tan^{-1} u + C = -\tan^{-1}\left(x + \frac{1}{x}\right) + C.$$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Linear Equations system The system
$$\begin{cases}x-y+3z=-5\\5x+2y-6z=\alpha \\2x-y+\alpha z = -6 \end{cases}$$
for which $\alpha$ values the linear equation system:
*
*has no solution
*has one solution
*has more than one solution
I started to do Gauss elimination on it, but i have no idea what i am looking for and how to approach this, I'm stuck with the Gauss elimination.
My work so far:
\begin{align}
\left(\begin{array}{rrr|r}
1 & -1 & 3 & -5 \\
5 & 2 & 6 & \alpha \\
2 & -1 & \alpha & -6
\end{array}\right)
&\leadsto \left(\begin{array}{rrr|r}
1 & -1 & 3 & -5 \\
0 & 7 & -9 & \alpha + 25 \\
2 & -1 & \alpha & -6
\end{array}\right) \\
&\leadsto \left(\begin{array}{rrr|r}
1 & -1 & 3 & -5 \\
0 & 7 & -9 & \alpha+25 \\
0 & 1 & \alpha-6 & 4
\end{array}\right) \\
&\leadsto \left(\begin{array}{rrr|r}
1 & 0 & \alpha + 3 & -1 \\
0 & 7 & -9 & \alpha+25 \\
0 & 1 & \alpha-6 & 4
\end{array}\right) \\
\end{align}
|
Start with the matrix and data vector
$$
\mathbf{A} =
\left[
\begin{array}{rrr}
1 & -1 & 3 \\
5 & 2 & -6 \\
2 & -1 & \alpha \\
\end{array}
\right], \ \alpha \in \mathbb{C}, \quad
%
b =
\left[
\begin{array}{r}
-5 \\
\alpha \\
-6
\end{array}
\right]
$$
The reduced row eschelon form is
$$
\mathbf{E}_{A} =
\left[
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right].
$$
The matrix $\mathbf{A}$ has full rank $m=n=\rho=3$. The column vectors span $\mathbb{C}^{3}$.
Select option 2: a solution will always exist, and the solution will be unique. In fact it is
$$
\mathbf{A} x = b \qquad \Rightarrow \qquad x =
\frac{1}{7}
\left[
\begin{array}{r}
\alpha - 10 \\
\alpha + 22 \\
-1
\end{array}
\right] .
$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Solving and proving inequalities? If $a,b$ and $c$ are positive real numbers, how do I prove that:
$$\frac{a^3}{b^2-bc+c^2}+
\frac{b^3}{c^2-ca+a^2} + \frac{c^3}{a^2-ab+b^2} \geq 3 \cdot
\frac{ab+bc+ca}{a+b+c}.$$
and when is equality?
Are there general techniques to solve these symmetric and or cyclic inequalities?
|
Lemma 1: If $a,b$ and $c$ are positive real numbers, then $\frac{a^3}{b^2-bc+c^2}+\frac{b^3}{a^2-ac+c^2}+\frac{c^3}{a^2-ab+b^2}\geq a+b+c.$
Proof: Proving the inequality $\frac{a^3}{b^2-bc+c^2}+\frac{b^3}{a^2-ac+c^2}+\frac{c^3}{a^2-ab+b^2}\geq a+b+c$ $\blacksquare$
Lemma 2: If $a,b$ and $c$ are positive real numbers, then $a^2 + b^2 + c^2 ≥ ab + bc + ac.$
Proof: Is this correct method to prove that $a^2 + b^2 + c^2 ≥ ab + bc + ac$, when $a,b,c \geq 0$? $\blacksquare$
Claim: If $a,b$ and $c$ are positive real numbers, then $a+b+c ≥ 3 \cdot
\frac{ab+bc+ca}{a+b+c}.$
Proof: $a+b+c ≥ 3 \cdot
\frac{ab+bc+ca}{a+b+c}$
$\iff a^3+b^3+c^3 \geq 3abc$
$\iff (a+b+c) \cdot (a^2 + b^2 + c^2 - ab - bc - ac) \geq 0$
If $a+b+c=0$ then the above is trivially true.
Thus if $a+b+c>0$ we need to show that $a^2 + b^2 + c^2 ≥ ab + bc + ac$, which is true by Lemma 2. $\blacksquare$
Combining the Claim and Lemma 1 we prove your inequality that is $$ \frac{a^3}{b^2-bc+c^2}+
\frac{b^3}{c^2-ca+a^2} + \frac{c^3}{a^2-ab+b^2} \geq 3 \cdot
\frac{ab+bc+ca}{a+b+c}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\int_{-1}^{1}e^{-\frac{1}{1-x^2}}dx$, can it be computed? Is there a way to compute $\int^{1}_{-1} e^{-\frac{1}{1-x^2}}dx$ ?
I have tried a few change of variables and also to write down $\frac{1}{1-x^2} = \frac{1}{2} ( \frac{1}{1-x} + \frac{1}{1+x})$ But I didn't get anything so far.
Edit: changed $e^{\frac{1}{1-x^2}}$ to $e^{-\frac{1}{1-x^2}}$
|
$$\int_{-1}^{1}\exp\left(\frac{1}{x^2-1}\right)\,dx = 2\int_{0}^{1}\exp\left(\frac{1}{x^2-1}\right)\,dx = \int_{0}^{1}\exp\left(\frac{1}{x-1}\right)\frac{dx}{\sqrt{x}}$$
equals:
$$ \int_{0}^{1}\exp\left(-\frac{1}{x}\right)\frac{dx}{\sqrt{1-x}} = \int_{1}^{+\infty}\frac{e^{-x}}{x^{3/2}\sqrt{x-1}}\,dx=\frac{1}{e}\int_{0}^{+\infty}\frac{e^{-x}\,dx}{(x+1)^{3/2}\sqrt{x}} $$
or:
$$ \frac{2}{e}\int_{0}^{+\infty}\frac{e^{-x^2}}{(x^2+1)^{3/2}}\,dx = \frac{\sqrt{\pi}}{e}\,U\left(\frac{1}{2},0,1\right)\approx 0.443993816168$$
where $U$ is Tricomi's confluent hypergeometric function. An accurate and simple upper bound is given by the Cauchy-Schwarz inequality:
$$ \int_{0}^{+\infty}\frac{e^{-x^2}}{(x^2+1)^{3/2}}\,dx \leq \sqrt{\int_{0}^{+\infty}e^{-2x^2}\,dx\int_{0}^{+\infty}\frac{dx}{(x^2+1)^3}}$$
gives that the original integral is $\leq \frac{\sqrt{3}}{2e}\cdot\left(\frac{\pi}{2}\right)^{3/4}$.
|
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|
Find the remainder of $\frac{3^{11}-1}{2}$ divided by $9$ I have the following question:
Find the remainder of $$\frac{3^{11}-1}{2}$$ divided by $9$
I tried to reformat the question:
$$\frac{3^{11} -1 } {2} \times \frac{1}{9} = \frac{3^{11}-1}{18}$$
Since $3^2 = 9$
$$\frac{3^2(3^9) -1}{3^2 \times2}$$
I don't know where to go next. Anyway, this is one of my many attempts to solve this question, and most of them ends with a complicated solution. I don't want to use modular arithmetic for this question. A hint or anything will help me.
|
$\frac {3^{11} - 1}{2} = \frac {3^{11} - 1}{3-1}=$
$3^{10} + 3^9 + .... + 3^2 + 3 + 1=$
$9(3^8 + ..... + 1) + 4$
so the remainder is $4$.
..or...
$\frac {3^{11}-1}2 = 3^{11}*\frac 12 - \frac 12 \equiv k \mod 9$
$ 3^{11}-1 \equiv 2k \mod 9$
$2k + 1 \equiv 0 \equiv 9 \mod 9$
$2k \equiv 8 \mod 9$
$k \equiv 4 \mod 9$.
|
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probability of discrete random variable In a nuclear reaction, a particle can either separate into two pieces,
or not separate, with respective probabilities $2/3$ and $1/3$. Knowing that
pieces behave like new independent particles, find the law of
probability and mean of the number of particles obtained after two reactions from of a single particle.
Let $X$ be the random variable which designates the number of pieces obtained after two reactions. It is clear that $X\in \{1, 2, 3, 4\}$ and the probability distribution of this variable
${(K, P (X = k)) / 1 ≤ k ≤ 4}$
And he came after that and give me the probability of each case.
$P(k=1) = 3/27$
$P(k=2) = 8/27$
$P(k=3) = 8/27$
$P(k=4) = 8/27$
Can anyone help me to prove this solution?
|
You start with only one particle, let's call it $A$. The set of your particles at the beginning is $$\{A\}.$$
After the first reaction, the particle $A$ can produce another particle $B$, or not. Then we can have the following particle sets:
$$\begin{cases}
\{A\} & ~\text{with probability}~\frac{1}{3}\\
\{A,B\} & ~\text{with probability}~\frac{2}{3}\\
\end{cases}$$
After the second reaction, we can have various scenario:
*
*we start from $\{A\}$ and we don't produce anything with probability $\frac{1}{3}$, obtaining $\{A\}$;
*we start from $\{A\}$ and we produce a new particle $C$ with probability $\frac{2}{3}$, obtaining $\{A,C\}$.
*we start from $\{A,B\}$ and $A$ produce a new particle $D$ with probability $\frac{2}{3}$, while $B$ produce a new particle $E$ with probability $\frac{2}{3}$ obtaining $\{A,B,D,E\}$ (total probability $\left(\frac{2}{3} \cdot \frac{2}{3}\right)$).
*we start from $\{A,B\}$ and $A$ produce a new particle $F$ with probability $\frac{2}{3}$, while $B$ does not produce anything with probability $\frac{1}{3}$ obtaining $\{A,B,F\}$ (total probability $\left(\frac{2}{3} \cdot \frac{1}{3}\right)$).
*we start from $\{A,B\}$ and $A$ does not produce anything with probability $\frac{1}{3}$, while $B$ produce the particle $G$ with probability $\frac{2}{3}$ obtaining $\{A,B,G\}$ (total probability $\left(\frac{1}{3} \cdot \frac{2}{3}\right)$).
*we start from $\{A,B\}$ and $A$ does not produce anything with probability $\frac{1}{3}$, while $B$ does not produce anything with probability $\frac{1}{3}$ obtaining $\{A,B\}$ (total probability $\left(\frac{1}{3} \cdot \frac{1}{3}\right)$).
Summaryzing:
$$\begin{array}{cccc}
\text{Beginning} & \text{After first reaction} & \text{After second reaction} & \text{Probability}\\
\{A\} & \{A\} & \{A\} & \frac{1}{3} \cdot \frac{1}{3} = \frac{3}{27}\\
\{A\} & \{A\} & \{A,C\} & \frac{1}{3} \cdot \frac{2}{3} = \frac{6}{27}\\
\{A\} & \{A,B\} & \{A,B,D,E\} & \frac{2}{3} \cdot \left(\frac{2}{3} \cdot \frac{2}{3}\right) = \frac{8}{27} \\
\{A\} & \{A,B\} & \{A,B,F\} & \frac{2}{3} \cdot \left(\frac{2}{3} \cdot \frac{1}{3}\right) = \frac{4}{27}\\
\{A\} & \{A,B\} & \{A,B,G\} & \frac{2}{3} \cdot \left(\frac{1}{3} \cdot \frac{2}{3}\right) = \frac{4}{27}\\
\{A\} & \{A,B\} & \{A,B\} & \frac{2}{3} \cdot \left(\frac{1}{3} \cdot \frac{1}{3}\right) = \frac{2}{27}\\
\end{array}.$$
Now, you have only one case where at the end you have only one particle. This happens with probability $$p(k=1)=\frac{3}{27}.$$
Further, you have two cases where at the end you have two particles ($\{A,C\}$ and $\{A,B\}$. Summing up, you get:
$$p(k=2) = \frac{6}{27} + \frac{2}{27} = \frac{8}{27}.$$
You have two cases where at the end you have three particles ($\{A,B,F\}$ and $\{A,B,G\}$), Summing up:
$$p(k=3) = \frac{4}{27} + \frac{4}{27} = \frac{8}{27}.$$
Finally, in only one case you have four particles ($\{A,B,D,E\}$), with probability $$p(k=4) = \frac{8}{27}.$$
|
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Smallest positive integral value of $a$ such that ${\sin}^2 x+a\cos x+{a}^2>1+\cos x$ holds for all real $x$
If the inequality $${\sin}^2 x+a\cos x+{a}^2>1+\cos x$$ holds for all $x \in \Bbb R$ then what's the smallest positive integral value of $a$?
Here's my approach to the problem $$\cos^2 x+(1-a)\cos x-a^2<0$$
Let us consider this as a quadratic form respect to $a$.
Applying the quadratic formula $a=\frac{-\cos x\pm\sqrt{5\cos^2 x+4\cos x}}2 $
and substituting $\cos x$ with $1$ and $-1$
we get 3 values of where the graph should touch the x axis $-2,0,1$
How should I proceed now?
|
$$
\cos^2{x}+(1-a)\cos{x}-a^2\,\lt0 \quad\&\quad \left|\,\cos{x}\,\right|\,\le1 \\[6mm]
-1\le\,\cos{x}=\frac12\left(\,-(1-a)\pm\sqrt{(1-a)^2+4a^2}\,\right)\,\le+1 \\
-1-a\,\le\,\pm\sqrt{5a^2-2a+1}\,\le\,+3-a \\[6mm]
\begin{align}
&\text{For}\,\colon\,\,\,\pm\sqrt{5a^2-2a+1}\,\ge\,-1-a \implies 5a^2-2a+1=a^2+2a+1 \\
&\implies 4a^2-4a=a^2-a=0\implies a=0,\,a=1\quad\color{red}{a\in[0,1]} \\[6mm]
&\text{For}\,\colon\,\,\,\pm\sqrt{5a^2-2a+1}\,\le\,+3-a \implies 5a^2-2a+1=a^2-6a+9 \\
&\implies 4a^2+4a-8=a^2+a-2=0\implies a=-2,\,a=1\quad\color{blue}{a\in[-2,1]} \\[6mm]
\end{align}
$$
Hence, for the inequality $\,\{\cos^2{x}+(1-a)\cos{x}-a^2\,\lt0\}\,$ to hold for all $\,x\in\mathbb{R}\,$ and the operator without equal (less than only), Then:
$$ \boxed{ \quad \color{blue}{a \in (-\infty,\,-2)\,\wedge\,(1,\infty) } \quad } $$
And for $\,a\in\mathbb{N}^{+}\implies\color{red}{a=2}\,$
|
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|
For $a, b, c$ is the length of three sides of a triangle. Prove that $\left|\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}\right|<\frac{1}{8}$ For $a, b, c$ is the length of three sides of a triangle. Prove that $$\left|\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}\right|<\frac{1}{8}$$
|
Let $a=y+z$, $b=x+z$ and $c=x+y$.
Hence, $x$, $y$ and $z$ be positives and we need to prove that
$$(2x+y+z)(2y+x+z)(2z+x+y)\geq8\sum_{cyc}(y-x)(2x+y+z)(2y+x+z)$$ or
$$\sum_{cyc}\left(2x^3+15x^2y-x^2z+\frac{16}{3}xyz\right)\geq0,$$
which is obvious because $x^3+y^3+z^3\geq x^2z+y^2x+z^2y$ by Rearrangement.
|
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|
For $x,y,z>0$. Minimize $P=\frac{1}{4x^{2}-yz+2}+\frac{1}{4y^{2}-zx+2}+\frac{1}{4z^{2}-xy+2}$ For $x,y,z>0$ and $xy+yz+xz=1$, minimize $$P=\frac{1}{4x^{2}-yz+2}+\frac{1}{4y^{2}-zx+2}+\frac{1}{4z^{2}-xy+2}$$
My try: Let $xy=a; yz=b;zx=c \Rightarrow a+b+c=1$
$\Rightarrow x^2=\frac{ac}{b};y^2=\frac{ab}{c};z^2=\frac{bc}{a}$
Hence $$P=\sum \frac{1}{\frac{4ac}{b}-b+2}=\sum \frac{1}{\frac{4ac}{b}-b+2(a+b+c)}=\sum \frac{1}{\frac{4ac}{b}+2a+b+2c}=\sum \frac{b}{4ac+2ab+b^2+2bc}$$
$$P=\sum \frac{b}{(2a+b)(2c+b)} \geq \sum \frac{4b}{(2a+2b+2c)^2}=\sum \frac{b}{(a+b+c)^2}=1$$
And i need new method ?
|
You proved that $P \ge 1$.
And for $x=y=z=\frac{1}{\sqrt{3}}$ you have
$$P=\frac{1}{4x^{2}-yz+2}+\frac{1}{4y^{2}-zx+2}+\frac{1}{4z^{2}-xy+2}= \frac{1}{3} + \frac{1}{3}+\frac{1}{3}=1.$$ Hence the minimum of $P$ for positive values of $x,y,z$ is one.
|
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|
Evaluating some integrals I need to evaluate two kind of similar integrals
The first one:
$$\lim _{n\to \infty }\int _0^{\frac{\pi }{3}}\:\frac{\sin ^n\left(x\right)}{\sin ^n\left(x\right)+\cos ^n\left(x\right)}dx$$
The second one:
$$\int _0^{2\pi }\:\frac{x\sin ^{100}\left(x\right)}{\sin ^{100}\left(x\right)+\cos ^{100}\left(x\right)}dx$$
For the first one, I got to this form
$$\lim _{n\to \infty }\int _0^{\frac{\pi }{3}}\:\frac{\sin ^n\left(x\right)}{\sin ^n\left(x\right)+\cos ^n\left(x\right)}dx = \frac{\pi }{3}-\int _0^{\frac{\pi }{3}}\:\frac{\cos ^n\left(x\right)dx}{\sin \:^n\left(x\right)+\cos \:^n\left(x\right)}$$
Now, I would make a substitution $u = \frac{\pi}{3}-x$ but it seems it doesn't help in this situation.
I have put the second integral here because I think it should have the same solving concept.
|
For the first try substituting $x=\frac{2}{3}u$ and divide through by $\sin^n(\frac{2}{3}x)$ top and bottom.
So now we have,
$$\frac{2}{3} \lim_{n \to \infty} \int_{0}^{\frac{\pi}{2}} \frac{1}{1+\cot^n (\frac{2}{3}x)} dx $$
Now for $\frac{2}{3}x \in (0,\frac{\pi}{4})$ we have $\cot \frac{2}{3} x>1$ so there the integrand tends to $0$. This is when $x \in (0,\frac{3\pi}{8})$. Whereas for $x \in (\frac{3\pi}{8},\frac{3\pi}{4})$ we have $\cot \frac{2}{3} x \in (0,1)$ so there the integrand tends to $1$ and thus we have,
$$=\frac{2}{3} \int_{\frac{3}{8}\pi}^{\frac{\pi}{2}} 1 dx$$
$$=\color{red}{\frac{\pi}{12}}$$
For the second integral with $x=u+\pi$ we have,
$$I=\int_{-\pi}^{\pi} (x+\pi)\frac{\sin^{100} (x)}{\sin^{100} (x)+\cos^{100} (x)} dx$$
Notice that $x\frac{\sin^{100} (x)}{\sin^{100} (x)+\cos^{100} (x)}$ is odd.
$$\color{blue}{I}=\color{blue}{\pi \int_{-\pi}^{\pi} \frac{\sin^{100} (x)}{\sin^{100} (x)+\cos^{100} (x)} dx}$$
$$=\pi \int_{-\pi}^{\pi} \frac{\sin^{100} (x)+\cos^2(x)-\cos^2(x)}{\sin^{100} (x)+\cos^{100} (x)}$$
$$=\pi(2\pi-\int_{-\pi}^{\pi}\frac{\cos^{100} (x)}{\sin^{100} (x)+\cos^{100} (x)} dx)$$
$$=2\pi^2-\color{blue}{\pi\int_{-\pi}^{\pi}\frac{\cos^{100} (x)}{\sin^{100} (x)+\cos^{100} (x)}} dx$$
But,
$$\color{blue}{I}=\pi \int_{-\pi}^{\pi}\frac{\cos^{100} (x)}{\sin^{100} (x)+\cos^{100} (x)} dx$$
$$=\pi \int_{-\pi}^{\pi} \frac{\sin^{100} (x)}{\sin^{100} (x)+\cos^{100} (x)} dx$$
Follows from $x=\frac{\pi}{2}-u$ then $u-\frac{\pi}{2}=v$.
So,
$$I=2\pi^2-I$$
$$I=\color{red}{\pi^2}$$
In fact everything I wrote for this will work for $n$ an even positive integer instead of $100$.
|
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|
Quadratic equation find all the real values of $x$ Find all real values of $x$ such that
$\sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x$
I tried sq both sides by taking 1 in RHS but it didn't worked out well...
|
$x\geq1$ and $1$ is not root.
Hence we can rewrite our equation in the following form:
$$x-1=x\left(\sqrt{x-\frac{1}{x}}-\sqrt{1-\frac{1}{x}}\right)$$ or
$$\sqrt{x-\frac{1}{x}}-\sqrt{1-\frac{1}{x}}=1-\frac{1}{x}$$ and with the given we obtain
$$2\sqrt{x-\frac{1}{x}}=x+1-\frac{1}{x}$$ or
$$x-\frac{1}{x}-2\sqrt{x-\frac{1}{x}}+1=0$$ or
$$\left(\sqrt{x-\frac{1}{x}}-1\right)^2=0$$ or
$$x^2-x-1=0,$$
which gives $x=\frac{1+\sqrt5}{2}$.
|
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|
Simplifying Algebraic Expression Under A Square Root This is part of a problem for calculating the length of a curve. Unfortunately, I'm stuck on a pretty basic algebra concept.
Solutions for my problem say that:
$\sqrt{\frac{1}{2t} + 1 + \frac{t}{2}} = \frac{\sqrt{t} + \frac{1}{\sqrt{t}}}{\sqrt{2}}$
I cannot understand how they got this at all.
The most I can simplify the original expression is as
$\sqrt{\frac{1}{2}(\frac{1}{t} + 2 + t)} = \sqrt{1/2}\sqrt{\frac{1}{t} + 2 + t} = \frac{1}{\sqrt{2}}\sqrt{\frac{1}{t} + 2 + t}$
I can't see how they get simplify the algebraic expression under the square root.
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Put over a common denominator: $$\sqrt{\frac{1}{2t} + 1 + \frac{t}{2}} =\sqrt{\frac{1 + 2t + t^2}{2t}}$$
Factor the top term: $$\sqrt{\frac{(1+t)^2}{2t}}$$
Take the square root: $$\frac {1+t}{\sqrt{2t}}$$
Divide top and bottom by $\sqrt t$: $$\frac {\frac 1{\sqrt{t}} + \sqrt{t}}{\sqrt 2}$$
|
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|
How do I solve derivative of $(2x^2+x+3)(5x+7)$ using product rule. I've to find the derivative of
$$(2x^2+x+3)(5x+7)$$
Using product rule I get
$$(2x^2+x+3)×5+5×(2x^2+x+3)\\20x^2+10x+30$$
Which is wrong. Please help, where I went wrong.
|
$y=(2x^2+x+3)(5x+7) $ then
$$y'=(2x^2+x+3)'(5x+7)+(2x^2+x+3)(5x+7)'\\
=(4x+1)(5x+7)+5(2x^2+x+3) $$
|
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If $\sin A+\sin^2 A=1$ and $a\cos^{12} A+b\cos^{8} A+c\cos^{6} A-1=0$ If $\sin A+\sin^2 A=1$ and $a\cos^{12} A+b\cos^{8} A+c\cos^{6} A-1=0$.
Find the value of $2b + \dfrac {c}{a}$.
My Attempt:
$$\sin A+\sin^2 A=1$$
$$\sin A + 1 - \cos^2 A=1$$
$$\sin A=\cos^2 A$$
Now,
$$a(\cos^2 A)^{6}+b(\cos^2 A)^{4}+c(\cos^2 A)^{3}=1$$
$$a\sin^6 A+ b\sin^4 A+c\sin^3 A=1$$
How do I proceed further?
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HINT:
$$\cos^4A=(\cos^2A)^2=\cdots=1-\cos^2A\iff\cos^4A+\cos^2A-1=0$$
Divide $a\cos^{12}A+b\cos^8A+c\cos^8A-1$ by $\cos^4A+\cos^2A-1$
|
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How do I can simplify the below intersection? let $S$ be a sphere defined by this equation :$ \left(x-\frac{4}{3}\right)^2+\left(y-\frac{1}{3}\right)^2+\left(z-\frac{5}{3}\right)^2=\frac{25}{36}$ and $(p)$ the plane defined by the following equation :$2x+y-2z+4=0$ ,
My question here is : How do I can get the intersection between $(p)$ and $(S)$? By the way I want to get the center and the radius of the circle which it is the geometrical intersection of $(p)$ and $(S)$ .
Attempt : I took $y=2(z-x-2)$ from the equation of $(p)$ and by substitution in $(S)$ I have got this complicated form
|
\begin{align*}
\text{Radius of the sphere} &= \frac{5}{6} \\
\text{Distance of the centre from the plane} &=
\frac{2\left( \frac{4}{3} \right)+
\left( \frac{1}{3} \right)-
2\left( \frac{5}{3} \right)+4}
{\sqrt{2^2+1^2+2^2}} \\
&= \frac{11}{3} \\
&> \frac{5}{6}
\end{align*}
Hence, no intersections between plan and sphere.
|
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|
Find a complex number $w$ such that $w^2=-\sqrt{3} - i$ This is a problem in my undergrad foundations class.
\begin{equation}
w^2=-\sqrt{3} - i \\w=(-\sqrt{3}-i)^{\frac{1}{2}} \\w=\sqrt{2}\bigg(\cos\frac{5\pi}{12}+i\sin\frac{5\pi}{12}\bigg)
\end{equation}
So I get to here and the next step is
\begin{equation}
\sqrt{2}\bigg(\cos\bigg(\frac{5\pi}{12}+\pi\bigg)+i\sin\bigg(\frac{5\pi}{12}+\pi\bigg)\bigg)
\end{equation}
and this is equal to
\begin{equation}
w(\cos\pi+i\sin\pi)=-w
\end{equation}
Can someone help me with understanding the last two steps. Thanks
|
$\omega^2 = $$2(-\frac {\sqrt 3}{2} + i \frac 12)\\
2(\cos \frac {7\pi}{6} + i \sin \frac {7\pi}{6})$
$\omega = \sqrt 2(\cos \frac {7\pi}{12} + i \sin \frac {7\pi}{12})$
And here you are....
When we take the square root of the square of something, there are always two solutions. For example $x^2 = 9 \implies x = \pm 3$
We can put in the $\pm symbol here, but that technique breaks down with higher roots.
A more general approach...
$\omega^2 =2(\cos (\frac {7\pi}{6}+2n\pi) + i \sin (\frac {7\pi}{6}+2n\pi))$
Takes advantage of the period nature of the trig functions.
$\omega =\sqrt 2(\cos (\frac {7\pi}{12}+n\pi) + i \sin (\frac {7\pi}{12}+n\pi))$
And now we can look at just the solutions where the argument is in $[0,2\pi)$
$\omega =\sqrt 2(\cos \frac {7\pi}{12} + i \sin \frac {7\pi}{12}),\sqrt 2(\cos (\frac {7\pi}{12}+\pi) + i \sin (\frac {7\pi}{12}+\pi))$
|
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Solving $DEF+FEF=GHH$, $KLM+KLM=NKL$, $ABC+ABC+ABC=BBB$ She visits third class and is $8$ years old (you can imagine how ashamed I felt when I said so to her). I helped her with lots of maths stuff today already but this is very unknowable for me. Sorry it's in German but I have translated it :)
It's saying "Each letter represents a digit. Determine them". First question, what is "them"? The letters I guess?
How shall I determine them when they are unknown? Or is it simply $A=1, B=2, C=3, D=4, E=5, F=6, G=7, H=8, K=11, L=12, M=13, N=14$?
Alright...
With this we gave a) the first try:
It doesn't seem to make sense to set $A=1, B=2, ...$
Or we did something wrong.. Any ideas how this could be solved? :s
|
For the 3rd sum:Let the carry from the 1st column to the 2nd column be $d$, and let the carry from the 2nd column to the 3rd column be $e$.
The second column implies that $3B+d $ is equal to $B$ plus a multiple of $10,$ so $2B+d$ is a multiple of $10$. So $d$ is even, so $d=0$ or $d=2$.
If $d=0$ then $2B=2B+d$ is a multiple of $10$ so $B=0$ or $B=5$. Now $B=0$ gives the "trivial" solution $A=B=C=0$ (because $3\cdot ABC=BBB=000=0$.) But if $d=0$ and $B=5$ then $3C=B+10d=B=5$, which is impossible.
So a no-trivial solution must have $d=2.$ The $2B+2=2B+d$ is a multiple of $10$, which requires $B=4$ or $B=9$.
But if $d=2$ and $B=9$ then $3C=10d+B=20+9=29,$ which is impossible.
With $d=2$ and $B=4$ we have $3C=10d+B=20+4=24$, giving $C=8.$ Finally $3B+d=12+2=14$, giving the carry $e=1.$ And $3A+1=3A+e=B=4,$ giving $A=1$. (Solution:$148+148+148=444$).
8-year-olds who can solve this on their own would be rare.
|
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Is my proof consider to be correct? Problem: Let $a,b ∈ \mathbb N$, prove that at least one of the following $ab, a+b, a-b$ is evenly divisible by 3.
My solution:
Case 1: If $a_{mod}3=0 \text{ or } b_{mod}3=0$ then we can say that $a = 3k,\ ab=3kb$ which is evenly divisible by three
Case 2: If $a_{mod}3=b_{mod}3$ then $a=3k_1+a_{mod}3,\\b=3k_2+b_{mod}3,\\a-b=3k_1+a_{mod}-(3k_2+b_{mod}3)\\a-b=3k_1+a_{mod} -3k_2-b_{mod}3\\\text{since }a_{mod}3=b_{mod}3\\a-b=3k_1-3k_2\\a-b=3(k_1-k_2)$
in this case we are prove that $a-b$ is evenly divisible by three.
Case 3: If $a_{mod}3$ doesn't equal to $b_{mod}3$, if this stands for true, there are two combinations
$a_{mod}3=1, b_{mod}3=2 \text{ and}\\a_{mod}3=2, b_{mod}3=1\\
\text{if this stands for true we can prove that } a+b \text{ is divisible by three}\\
a+b=3k_1+a_{mod}3+3k_2+b_{mod}3, \text{since } a_{mod}3+b_{mod}3=3\\
a+b=3k_1+3k_2+3\\
a+b=3\cdot(k_1+k_2+1)$ we proved for all cases.
Is my proof correct?
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Seems correct, but if you're allowed to use Fermat's little theorem, you could go like this :
If $a$ or $b$ are divisible by $3$ then it is obvious that $ab$ is divisible by $3$.
If neither $a$ or $b$ are divisible by $3$, then let's take a look at $(a-b)$ and $(a+b)$. Take their product : $(a-b)(a+b) = a^2 - b^2$ and rewrite it as : $(a^2-1) - (b^2-1)$.
By Fermat's LT, both $a^2-1$ and $b^2-1$ are divisible by $3$, so is their difference. This means that $(a-b)(a+b)$ is divisible by $3$. Therefore, either $a-b$ or $a+b$ is divisible by 3.
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Why is $ab=(a-b)(a+b)$ false when $a,b$ are coprime? I could not find this, and do not have a background in number theory so sorry if this sounds trivial.
If $a$ and $b$ are coprime integers, why is the statement
$$ab=(a-b)(a+b)$$
a contradiction?
|
it becomes $$ a^2 - ab - b^2 = 0. $$ If $b \neq 0,$ divide through by $b^2$ and introduce $r = \frac{a}{b},$ this gives
$$ r^2 - r - 1 = 0 $$ So $r$ is either the Golden ratio
$$ \frac{1 + \sqrt 5}{2} $$ or negative its reciprocal. Put briefly, irrational, contradicting $a,b$ being integers.
It is also possible to use anisotropy in $\mathbb Q_5,$ which is probably what they want. Well, why not. Take
$$ a^2 - ab - b^2 \equiv 0 \pmod {25} $$ and ASSUME $\gcd(a,b) =1.$
The number $4$ is invertible $\pmod 5 $ and $\pmod {25},$ so we change nothing demanding
$$ 4 a^2 - 4ab - 4 b^2 \equiv 0 \pmod {25}, $$
$$ \left( 2a -b \right)^2 - 5 b^2 \equiv 0 \pmod {25}. $$
It follows that $(2a-b)$ is divisible by $5,$ and then $\left( 2a -b \right)^2$ is divisible by $25.$ In turn, this means $5 b^2$ is divisible by $25,$ so $b^2$ is divisble by $5,$ finally $b$ itself is divisible by $5.$ As $(2a-b)$ is also divisible by $5,$ we find that $a$ is divisible by $5.$ So far, both $a,b$ are divisible by $5.$ This means $5 | \gcd(a,b)$ and $\gcd(a,b) \neq 1$
This is the method I like to show that there is, in fact, no solution of $a^2 - ab - b^2 = 0$ in integers unless both $a,b$ are zero. The lemma that you need to really, really believe, is that, should there be a solution with at least one variable nonzero, then we can divide through by the gcd of those original $a,b$ numbers, divide both by that, resulting in a solution in coprime integers. Our proof of impossibility for coprime integers thus extends to a proof of impossibility at least one variable nonzero, finally to any nonzero solution. The quadratic form name for this argument is "anisotropy." We say that the quadratic form $a^2 - ab - b^2 $ is not isotropic in the $5$-adic numbers $\mathbb Q_5$
It is also true that the quadratic form $a^2 - ab - b^2 $ is not isotropic in the $2$-adic numbers $\mathbb Q_2.$ This is posted in some earlier answers. One may check, there are just four cases, $a^2 - ab - b^2 $ is odd unless both $a,b$ are even. Thus is was enough to use $\pmod 2,$ for this prime it was not even necessary to use $\pmod 4.$
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|
Proving these binomial sums Need help proving with these two:
1)
$$\sum_{k=0}^m (-1)^k \binom{n}{k}= (-1)^m \binom{n - 1}{m}$$
2)
$$\sum_{k=0}^n \binom{n}{k} \binom{k}{m}= \binom{n}{m} 2^{n-m}$$
I tried using these two properties below, but got nowhere. $$\sum_{k=0}^n (-1)^k \binom{n}{k} = 0$$
$$ \sum_{k=0}^n \binom{n}{k} = 2^n$$
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For the first equality, it can be easily proved by induction on $m$. When $m = 0$, the LHS is $1$ and the RHS is also $1$, thus the equality holds. For $m \geq 1$, we have
\begin{align}
\sum_{k=0}^m (-1)^k \binom{n}{k} &= \sum_{k=0}^{m-1}(-1)^k\binom{n}{k} + (-1)^m\binom{n}{m} \\
&= (-1)^{m-1}\binom{n-1}{m-1} + (-1)^{m}\binom{n}{m}\\
&= (-1)^m\left(\binom{n}{m} - \binom{n-1}{m-1}\right) \\
&= (-1)^m \binom{n-1}{m}
\end{align}
For the second equality, we have
\begin{align}
\sum_{k=0}^n \binom{n}{k}\binom{k}{m} &= \sum_{k=0}^n \binom{n}{m} \binom{n-m}{n-k} \\
&= \binom{n}{m} \sum_{k=0}^n \binom{n-m}{n-k} \\
&= \binom{n}{m} 2^{n-m}
\end{align}
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|
Find minimum of $y=\sqrt{-x^2+4x+12}-\sqrt{-x^2+2x+3}$
Find minimum of $y=\sqrt{-x^2+4x+12}-\sqrt{-x^2+2x+3}$
My work so far:
1) $$y =\frac{2x+9}{\sqrt{-x^2+4x+12}+\sqrt{-x^2+2x+3}} $$
2) I used a derivative and found the answer ($y=\sqrt3$ at $x=0$). Is there any other way?
|
$$y=\sqrt{-x^2+4x+12}-\sqrt{-x^2+2x+3}$$
$$\implies y=\sqrt{16-4-x^2+4x}-\sqrt{4-1-x^2+2x}$$
$$\implies y=\sqrt{16-(4+x^2-4x)}-\sqrt{4-(1+x^2-2x)}$$
$$\implies y=\sqrt{16-(2-x)^2}-\sqrt{4-(1-x)^2}$$
From the first radical, we have that $$-4 \le 2-x\le 4 \implies -2 \le x\le 6$$
And from the second radical, we have that $$-2 \le 1-x\le 2 \implies -1 \le x\le 3$$
Net domain will be $[-1,3]$.
Further $y(-1)=\sqrt7$ , $y(0)=\sqrt{12}-\sqrt3$ , $y(1)=\sqrt{15}-2$, $y(2)=4-\sqrt{3}$ and $y(3)=0$
See if this helps.
|
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|
Prove that $b_n = b_{n+100}$
Let $b_n$ denote the units digit of $\displaystyle\sum_{a=1}^n a^a$. Prove that $b_n = b_{n+100}$.
I tried rewriting the sum, but didn't see how to prove the equality. For example, if $n = 178$ we have $$\displaystyle\sum_{a=1}^{178} a^a = (1^1+2^2+3^3+\cdots+78^{78})+(79^{79}+80^{80}+81^{81}+\cdots+178^{178})$$ and so we need to show that $79^{79}+80^{80}+81^{81}+\cdots+178^{178} \equiv 0 \pmod{10}$. Is there an easier way to prove the general result?
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We can reduce the numbers to the residues modulo $10$, i.e:
$$79^{79} + 80^{80} + \cdots + 178^{178} \equiv 9^{79} + 0^{80} + \cdots + 8^{178} \pmod {10}$$
Now note that for each residues we have exactly $10$ numbers such that it appears in the base of the exponent. Group them and take one group. Now notice as their parity is equal the sum in the group is always even, so hence divisible by $2$. Now it's enough to prove it's divisible by $5$. Now if the base of the exponent is divisible by $5$ we're done, so assume it's not. Hence they are coprime. Now note that we can reduce the exponents modulo $4$, as $a^{4} \equiv 1 \pmod 5$ by Fermat's Theorem and as $\frac{4}{(4,10)} = 2$ the exponents will repeat in cycles of two, so we can group them in $5$ groups having same remainder, hence the sum is divisible by $5$.
For a better illustration of the last step check this:
$$79^{79} + 89^{89} + 99^{99} + 109^{109} + 119^{119} + 129^{129} + 139^{139} + 149^{149} + 159^{159} + 169^{169} $$
$$\equiv 9^{79} + 9^{89} + 9^{99} + 9^{109} + 9^{119} + 9^{129} + 9^{139} + 9^{149} + 9^{159} + 9^{169} $$
$$\equiv 9^{3} + 9^{1} + 9^{3} + 9^{1} + 9^{3} + 9^{1} + 9^{3} + 9^{1} + 9^{3} + 9^{1} \equiv 5(9^3 + 9^1) \equiv 0 \pmod 5$$
|
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|
Evaluate the sum of series $\sum_{k=1}^{+\infty}(-1)^{k-1}\frac{1}{k(4k^2-1)}$ Evaluate the sum of series
$$\sum_{k=1}^{+\infty}(-1)^{k-1}\frac{1}{k(4k^2-1)}$$
I have tried two methods:
1) using power series
2) using partial sums
but I can't find the sum.
1) Using power series:
$$\sum_{k=1}^{+\infty}(-1)^{k-1}\frac{1}{k(4k^2-1)}x^{k(4k^2-1)}$$
$$f(x)=\sum_{k=1}^{+\infty}(-1)^{k-1}\frac{1}{k(4k^2-1)}x^{k(4k^2-1)}$$
After derivation:
$$f'(x)=\sum_{k=1}^{+\infty}(-1)^{k-1}x^{4k^3-k-1}$$
The problem here is that:
$$\sum_{k=1}^{+\infty}(-1)^{k-1}x^{4k^3-k-1}=x^2-x^{29}+x^{104}-...$$
Is it possible to find the closed form for the last series?
2) Using partial sums:
$$S_n=\sum_{k=0}^{n}(-1)^{k}\frac{1}{(k+1)(4(k+1)^2-1)}$$
Now, using the formula:
$$S_n+a_{n+1}=a_0+\sum_{k=1}^{n+1}(-1)^{k}\frac{1}{(k+1)(4(k+1)^2-1)}\Rightarrow$$
$$S_n+(-1)^{n+1}\frac{1}{(k+2)(4(k+2)^2-1)}=\frac{1}{3}+\sum_{k=1}^{n+1}(-1)^{k}\frac{1}{(k+1)(4(k+1)^2-1)}$$
$$S_n=\frac{1}{3}-\frac{1}{30}+...+(-1)^{n}\frac{1}{(n+1)(4(n+1)^2-1)}$$
$$\sum_{k=1}^{n+1}(-1)^{k}\frac{1}{(k+1)(4(k+1)^2-1)}=T_n=-\frac{1}{30}+...+(-1)^{n+1}\frac{1}{(n+2)(4(n+2)^2-1)}$$
$$T_n=S_n-\frac{1}{3}+(-1)^{n+1}\frac{1}{(n+2)(4(n+2)^2-1)}$$
Going back to the formula $$S_n+a_{n+1}=a_0+\sum_{k=1}^{n+1}(-1)^{k}\frac{1}{(k+1)(4(k+1)^2-1)}$$
we have that $S_n$ cancels, so we can't determine partial sums using this method?
$$S_n+(-1)^{n+1}\frac{1}{(k+2)(4(k+2)^2-1)}=\frac{1}{3}+T_n$$
$$S_n+(-1)^{n+1}\frac{1}{(k+2)(4(k+2)^2-1)}=\frac{1}{3}+S_n-\frac{1}{3}+(-1)^{n+1}\frac{1}{(n+2)(4(n+2)^2-1)}$$
Question: How to find the sum of this series?
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Observe that checking for absolute convergence we get
$$\frac1{k(4k^2-1)}\le\frac1{k^2}\;\implies\;\text{since}\;\;\sum_{k=1}^\infty\frac1{k^2}$$
converges so does the series of the left side. What can you then deduce?
|
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|
solving a recurrence relation - finding the general solution Solve the recurrence relation:
$u_{n+2} = 2u_{n+1}-u_n$
$u_0 = 1 $ and $u_1 = 4$
My calculations:
I have calculated that the characteristic equation is: $t^2-2t+1 = 0$ so the roots are $r_1=1$ and $r_2=1$
here is where I am stuck. The answer says that the general solution is: $u_n=(A+Bn)1^n$ But how do I know and come to that conclusion?
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$\begin{bmatrix}u_{n+2}\\u_{n+1}\end{bmatrix} = \begin{bmatrix} 2&-1\\1&0\end{bmatrix}\begin{bmatrix}u_{n+1}\\u_{n}\end{bmatrix}$
$\mathbf u_n = B^n \mathbf u_0$
Unfortunately B is not diagonalizable.
$\lambda^2 - 2\lambda + 1 = 0\\(\lambda-1)^2$
Choose $v_1$ such that
$(B-\lambda I)v_1 = 0\\
v_1 = \begin{bmatrix}1\\1\end{bmatrix}$
Choose $v_2$ such that
$(B-I)v_2 = v_1$
$v_2 = \begin{bmatrix}1\\0\end{bmatrix}$
$B\begin{bmatrix}1&1\\1&0\end{bmatrix} = \begin{bmatrix}1&1\\1&0\end{bmatrix}\begin{bmatrix}1&1\\0&1\end{bmatrix}\\
B =
\begin{bmatrix}1&1\\1&0\end{bmatrix}\begin{bmatrix}1&1\\0&1\end{bmatrix}\begin{bmatrix}0&1\\1&-1\end{bmatrix}\\
B^n = \begin{bmatrix}1&1\\1&0\end{bmatrix}\begin{bmatrix}1&1\\0&1\end{bmatrix}^n\begin{bmatrix}0&1\\1&-1\end{bmatrix}= \begin{bmatrix}1&1\\1&0\end{bmatrix}\begin{bmatrix}1&n\\0&1\end{bmatrix}\begin{bmatrix}0&1\\1&-1\end{bmatrix}$
$\begin{bmatrix}u_{n+1}\\u_{n}\end{bmatrix} =$$ \begin{bmatrix}1&1\\1&0\end{bmatrix}\begin{bmatrix}1&n\\0&1\end{bmatrix}\begin{bmatrix}0&1\\1&-1\end{bmatrix}\begin{bmatrix}4\\1\end{bmatrix}\\
\begin{bmatrix}1&1\\1&0\end{bmatrix}\begin{bmatrix}1&n\\0&1\end{bmatrix}\begin{bmatrix}1\\3\end{bmatrix}\\
\begin{bmatrix}1&1\\1&0\end{bmatrix}\begin{bmatrix}1+3n\\3\end{bmatrix}\\
\begin{bmatrix}4+3n\\1+3n\end{bmatrix}\\$
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.