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Prove $\tan(A+B+Y)=\frac{\tan A+\tan B+\tan Y-\tan A\tan B\tan Y}{1-\tan A \tan B-\tan B\tan Y-\tan Y\tan A}$ I have to prove this most difficult trigonometric identity.
$$\tan(A+B+Y)=\frac{\tan A+\tan B+\tan Y-\tan A\tan B\tan Y}{1-\tan A \tan B-\tan B\tan Y-\tan Y\tan A}.$$
I know
$$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$$
My problem is with the extra $Y$ in this problem. What can I do about I think I know a solution which is to do $\tan(A+B)$ then $\tan(B+Y)$ but I am not sure how to apply it.
| Do $a=\tan A, b= \tan B$ and $y = \tan Y$. Then,
\begin{eqnarray}
\tan (A+B+Y) &=& \dfrac{\tan(A+B) + \tan(Y)}{1-\tan(A+B)y}\\
&=& \dfrac{\dfrac{a+b}{1-ab}+y}{1 - \Big(\dfrac{a+b}{1-ab}\Bigr)y}\\
&=& \dfrac{\dfrac{a+b+(1-ab)y}{1-ab}}{\dfrac{1-ab - (a+b)y}{1-ab}}\\
&=& \dfrac{a+b+y -aby}{1-ab-ay-by}.\\
\end{eqnarray}
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\int_1^3\frac{\ln(x+2)}{x^2+2x+15} \ dx$ Could you please give me a hint on how to compute:
$$
\int_1^3\frac{\ln(x+2)}{x^2+2x+15}dx
$$
Thank you for your help
| First, let's make a change of variables, $x=3 + u$, which maps $(1,3)$ into $(-1,1)$:
$$
\int_1^3 \frac{\log(x+2)}{x^2+x+15}\mathrm{d} x = \int_{-1}^1 \frac{\log(4+u)}{u^2 + 6 u+ 23}\mathrm{d} u
$$
Now use $\log(4+u) = \log(4) + \log\left(1+\frac{u}{4}\right)$:
$$\begin{eqnarray}
\int_1^2 \frac{\log(1+2 u)}{2 u^2+7}\mathrm{d} u &=& \int_{-1}^1 \frac{\log(4)}{u^2 + 6 u+ 23}\mathrm{d} u + \int_{-1}^1 \frac{\log\left(1+\frac{u}{4}\right)}{u^2 + 6 u+ 23}\mathrm{d} u
\end{eqnarray}$$
The first integral is trivially evaluated by completing the squares in the denominator: $u^2+6 u+23 = (u+3)^2 + 14$, giving:
$$
I_0 = \int_{-1}^1 \frac{\log(4)}{u^2 + 6 u+ 23}\mathrm{d} u = \frac{\log(4)}{\sqrt{14}} \arctan\left(\frac{\sqrt{14}}{11} \right) \approx 0.121478
$$
This already gives a good approximation to the correct value ($\approx 0.118650$). The second integral can be done expanding logarithm in a series, and integrating term-wise. The first term:
$$
\Delta_1 = \int_{-1}^1 \frac{u/4}{u^2 + 6 u+ 23}\mathrm{d} u = -\frac{3}{4} \frac{1}{\sqrt{14}} \arctan\left(\frac{\sqrt{14}}{11} \right) + \frac{1}{8} \log\left(\frac{5}{3}\right) \approx -0.001868
$$
The second:
$$
\Delta_2 = -\frac{1}{2} \int_{-1}^1 \frac{(u/4)^2}{u^2 + 6 u+ 23}\mathrm{d} u = -\frac{1}{16} + \frac{5}{32} \frac{1}{\sqrt{14}} \arctan\left(\frac{\sqrt{14}}{11} \right) + \frac{3}{32} \log\left(\frac{5}{3}\right) \approx -0.000918
$$
Combining, $I_0 + \Delta_1+\Delta_2 = 0.118692$ which gives a good approximation.
| {
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Taylor polynomial of $f(x) = 1/(1+\cos x)$ I'm trying to solve a problem from a previous exam. Unfortunately there is no solution for this problem.
So, the problem is:
Calculate the Taylor polynommial (degree $4$) in $x_0 = 0$ of the function:
$$f(x) = \frac{1}{1+\cos(x)}$$
What I tried so far:
*
*calculate all $4$ derivatives
*$1+\cos(x) = 2\cos^2(\frac{x}{2})$ and work with this formula
*$\int\frac{1}{1+\cos(x)}dx = \tan(\frac{x}{2})$ and then use the Taylor series of $\tan(\frac{x}{2})$
*$\frac{1}{1 + \cos(x)} = \frac{1}{1 + \left(1 + \frac{x^2}{2!} + \cdots\right)}$
What do you think, is there a good way to calculate the Taylor polynomial of this function or is there just the hard way (derivatves)?
| $$\cos x=\sum_{k=0}^\infty (-1)^k\frac{x^{2k}}{(2k)!}=1+\cos x= 2-\frac{x^2}{2}+\frac{x^4}{24}-...\Longrightarrow$$
$$\frac{1}{1+\cos x}=\frac{1}{2-\frac{x^2}{2}+\frac{x^4}{24}-...}=\frac{1}{2}\frac{1}{\left[1-\left(\frac{x}{2}\right)^2\right]\left(1-\frac{x^2}{24}+...\right)}=$$
$$=\frac{1}{2}\left(1+\frac{x^2}{4}+\frac{x^4}{8}+...\right)$$by taking the development of $$\frac{1}{1-x^2}$$
| {
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Proving that $\frac{ \cos (A+B)}{ \cos A-\cos B}=-\cot \frac{A-B}{2} \cot \frac{A+B}{2}$ How would I prove the following trig identity?
$$\frac{ \cos (A+B)}{ \cos A-\cos B}=-\cot \frac{A-B}{2} \cot \frac{A+B}{2} $$
My work thus far has been:
$$\dfrac{2\cos\dfrac{A+B}{2} \cos\dfrac{A-B}{2}}{-2\sin\dfrac{A+B}{2} \sin\dfrac{A-B}{2}} =-\cot\dfrac{A+B}{2} \cot\dfrac{A-B}{2} \ .$$
| A couple of hints: use the $\cot$ identity on the RHS to put in terms of $\sin$ and $\cos$, use a product to sum identity, and pay attention to odd and even functions. This should simplify it to an easier expression.
| {
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Legendre symbol, second supplementary law $$\left(\frac{2}{p}\right) = (-1)^{(p^2-1)/8}$$
how did they get the exponent. May be from Gauss lemma, but how.
Suppose we have a = 2 and p = 11. Then n = 3 (6,8,10), but not
$$15 = (11^2-1)/8$$
n is a way to compute Legendre symbols from Gauss lemma:
$$\left(\frac{a}{p}\right) = (-1)^n$$
| Here is a proof from Ireland-Rosen Ch-6 (The key idea is to work in $ \overline{\mathbb{Z}} $, modulo $ p \overline{\mathbb{Z}} $) :
Let $ p $ be an odd prime. We have $ \left( \frac{2}{p} \right) \equiv 2^{\frac{p-1}{2}} (\text{mod } p\mathbb{Z} ) $. Notice we can write $ 2^{\frac{1}{2}} $ as $ 2^{\frac{1}{2}} = 2 \cos \left(\frac{2 \pi}{8} \right) = \zeta + \frac{1}{\zeta} $ (where $ \zeta := e^{\frac{i 2 \pi}{8}} $), and now both $ \zeta, \zeta + \frac{1}{\zeta} \in \overline{\mathbb{Z}} $ (because $\zeta$ satisfies $ X^4 + 1 = 0$, and $ \zeta + \frac{1}{\zeta} $ satisfies $ X^2 - 2 = 0 $).
So we have $ \left( \frac{2}{p} \right) \equiv \left( \zeta + \frac{1}{\zeta} \right)^{p-1} (\text{mod } p\overline{\mathbb{Z}}), $ in $ \overline{\mathbb{Z}} $. Now writing $ \eta := \zeta + \frac{1}{\zeta} ( = 2^{\frac{1}{2}} ) $, we get $ \left( \frac{2}{p} \right) \eta \equiv \eta^p \equiv \zeta^p + \zeta^{-p} \, ( \text{mod } p\overline{\mathbb{Z}})$.
But notice $ \zeta^p + \zeta^{-p} $ is determined by the value of ($ p $ mod $ 8 $), and it is in fact $ 2^{\frac{1}{2}} $ when $ p \equiv \pm 1 \, (\text{mod } 8) $ and $ -2^{\frac{1}{2}} $ when $ p \equiv \pm 3 \, ( \text{mod } 8 ) $. We can encapsulate this casework by writing $ \zeta^p + \zeta^{-p} = (-1)^{\frac{p^2-1}{8}} 2^{\frac{1}{2}} $ (because $ p $ can only be of the form $ 8k \pm 1 $ or $ 8k \pm 3 $, and $ (-1)^{\frac{p^2-1}{8}} $ is $ (-1) $ if former and $ 1 $ if latter).
So finally $ \left( \frac{2}{p} \right) \eta \equiv (-1)^{\frac{p^2-1}{8}} \eta \, (\text{mod }p\overline{\mathbb{Z}}) $. Notice multiplying by $ \eta $ makes both sides integers. Therefore $ \dfrac{\left( \frac{2}{p} \right) 2 - (-1)^{\frac{p^2 - 1}{8}} 2 }{p} \in \overline{\mathbb{Z}} \cap \mathbb{Q} = \mathbb{Z} $, giving us $ \left( \frac{2}{p} \right) = (-1)^{\frac{p^2 -1}{8}} $.
| {
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When a is less than c in $ \int_a^b \frac {dx} {x^4 - c^4} $ $$ \int_a^x \frac {dx} {x^4 - c^4} = \frac {1} {4c^3} \ln \left(\frac {x-c} {x+c} \right)_a^x - \frac {1} {2c^3} \tan^{-1} \Bigl(\frac {x} {c} \Bigr)_a^x $$
$$ =\frac {1} {4c^3} \Bigl[ \ln \Bigl(\frac {x-c} {x+c} \Bigr) - \ln \Bigl(\frac {a-c} {a+c} \Bigr) \Bigr]- \frac {1} {2c^3} \Bigl[\tan^{-1} \Bigl(\frac {x} {c} \Bigr)-\tan^{-1} \Bigl(\frac {a} {c} \Bigr)\Bigr]$$
When $x$ is less than $c$, or when $a$ is less than $c$, the number in the natural log becomes negative. Then, should the general answer have absolute sign instead of parenthesis as below? Is this a more proper or general answer?
$$ =\frac {1} {4c^3} \left[ \ln \left|\frac {x-c} {x+c} \right| - \ln \left| \frac {a-c} {a+c} \right| \right] $$
| Yes. Notice that
$$
\int \frac{dx}{x+c} = \ln \left| x+c \right| + \text{constant}
$$
(where "constant" actually means a piecewise constant function that is constant on $(-\infty,-c)$ and also on $(-c,\infty)$). Similar remarks apply with $x-c$ in place of $x-c$. The partial-fraction decomposition of $1/(x^4-c^4)$ includes both $\text{constant}/(x+c)$ and $\text{constant}/(x-c)$, so those two integrals are where these logarithms come from.
| {
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$\sqrt{x} +y = 4$, $x+ \sqrt{y}= 6$, find the solution $(x,y)$ $\sqrt{x} +y = 4$, $\sqrt{y} +x= 6$, find the solution (x,y). $NOTE$ : $\sqrt{4}+1= 4-1$, $\sqrt{1} +4 =1+4$
| This is basically the method which was suggested in the comments above - turning this into a quartic equation. We will see whether someone suggest a substantially more elegant solution.
$$\sqrt{x}+y=4\\
x+\sqrt{y}=6$$
Using the substitution
$\sqrt{x}=s$ and $\sqrt{y}=t$
we get:
$$s+t^2=4\\
t+s^2=6$$
Which gives
$$s=4-t^2=4-(6-s^2)^2\\
(s^2-6)^2+s-4=0\\
s^4-12s^2+s+32=0$$
It should be possible to solve this as a quartic equation, although it would be quite laborious. You can check what WolframAlpha is able to find out
here
and
here
| {
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Use the given equalities to derive trigonometric functions.
(A) $\sin(-x)=-\sin x$
(B) $\cos(-x)=\cos x$
(C) $\cos(x+y)=\cos x\cos y-\sin x\sin y$
(D) $\sin(x+y)=\sin x\cos y+\cos x\sin y$
Use these equalities to derive the following important trigonometric functions:
f) $\left|\cos\dfrac{x}{2}\right|=\sqrt{\dfrac{1+\cos x}{2}}$
g) $\left|\sin\dfrac{x}{2}\right|=\sqrt{\dfrac{1-\cos x}{2}}$
This is for (f): Since this is a half-angle identity I replace $x$ with $\frac{\pi}{2}$. And I'll use (C). $\cos(\frac{\pi}{2}+\frac{\pi}{2})=\cos\frac{\pi}{2}\cos\frac{\pi}{2}-\sin\frac{\pi}{2}\sin\frac{\pi}{2}\Rightarrow \cos2\frac{\pi}{2}=\cos^2\frac{\pi}{2}-\sin^2\frac{\pi}{2}$
Using power reduction identity of: $\cos^2\theta=\dfrac{1+\cos2\theta}{2}$ yields $\cos2\frac{\pi}{2}=\dfrac{1+\cos2\frac{\pi}{2}}{2}$.
I do not believe this is correct because $\cos^2\theta\ne \cos2\theta$. Please help, but no answers.
| I know from an earlier question of yours that you are familiar with the identities
$$\cos 2w =2\cos^2 w-1=1-2\sin^2 w,\tag{$1$}$$
which can be derived fairly quickly from (C).
(Yes, I have changed the name of the variable. That is deliberate.)
Now let $w=\frac{x}{2}$. Then the identities $(1)$ can be rewritten as
$$\cos x=2\cos^2 \frac{x}{2}-1=1-2\sin^2 \frac{x}{2}.$$
(We are replacing $w$ by $\frac{x}{2}$. So $2w=x$.)
Look first at the identity $\cos x=2\cos^2 \frac{x}{2}-1$. This can be rewritten as
$1+\cos x=2\cos^2 \frac{x}{2}$, and then as $\cos^2\frac{x}{2}=\frac{1+\cos x}{2}$.
Take the square root of both sides. We get
$$\sqrt{\frac{1+\cos x}{2}}=\left|\cos \frac{x}{2}\right|.$$
Here we used the general fact that $\sqrt{a^2}=|a|$.
The other identity is proved the same way. From $\cos x=1-2\sin^2\frac{x}{2}$ we get $2\sin^2\frac{x}{2}=1-\cos x$. Divide both sides by $2$ and take the square root.
| {
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Geometry/Triangles Problem Let $A,B,C$ be a triangle's vertices (FIG.1), and $D$ is a point on $\overline {BC}$. It is given that $\angle B= 2 \angle C $, $\overline{AB}=\overline{DC}$ and also that $\overline{AD}$ is the bisector of angle $A$.. We have to prove that angle $\angle BAC=72°$.
$\hspace{0.54 cm}$
$$\begin{align}\text{FIGURE 1} &\end{align}$$
| As $B=2C \Rightarrow A=\pi-B-C=\pi-3C$ then
$$ 0<B<\pi \Rightarrow 0<2C<\pi \Rightarrow 0<C<\frac{\pi}{2}$$
and $A=\pi-3C$ so, $0<A<\pi \Rightarrow0<\pi - 3C <\pi \Rightarrow0< C< \frac{\pi}{3} $
So, $0< C< \frac{\pi}{3} $
If $∠BAD=x , ∠DAC=x$ and if $∠ADB=y, ∠ADC=\pi-y$
If $AB=c$, $BC=a$, $CA=b \Rightarrow BD=a-c$
Applying the law of sines on triangle $ACD$, $$\frac{c}{\sin x} = \frac{b}{\sin (\pi- y)}\Rightarrow\frac{c}{\sin x} = \frac{b}{\sin y}$$
Applying sine law on triangle ABD, $$\frac{a-c}{\sin x} = \frac{c}{\sin y}$$
Upon division, $$\frac{a-c}{c}=\frac{c}{b}$$
$$\frac{\sin\ A - \sin C}{\sin C}=\frac{\sin C}{\sin B}$$ (applying $a=2R\sin A$ etc.)
$$\frac{\sin\ (\pi - 3C) - \sin C}{\sin C}=\frac{\sin C}{\sin 2C}$$
$$\frac{\sin\ 3C - \sin C}{\sin C}=\frac{\sin C}{\sin 2C}$$
$$(2\sin C \cos2C)\sin2C = (\sin C)^2$$
applying $$\sin2x-\sin2y=2\sin(x-y)\cos(x+y)$$
$$\sin C(2 \cos2C\sin2C - \sin C)=0$$
$$\sin C(\sin 4C - \sin C)=0$$
$$\sin 4C=\sin C \text{ as } 0<C<\pi=>\sin C>0$$
$4C=n\pi+(-1)^nC$ where n is any integer,
If $n$ is even $(=2m)$,say, $4C=2m\pi+C \Rightarrow C=\frac{2m\pi}{3}$,
As $0< C< \frac{\pi}{3} $, so there is no solution if $n$ is even.
If $n$ is odd $(=2m+1)$, say, $4C=(2m+1)\pi-C\Rightarrow C=\frac{(2m+1)\pi}{5}$
$$C=\frac{\pi}{5}\ as\ 0< C< \frac{\pi}{3} $$
$$A=\pi-3C=\frac{2\pi}{5}=72^\circ$$
| {
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Inequality. $a^2+b^2+c^2 \geq a+b+c$ Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that
$a^2+b^2+c^2 \geq a+b+c$.
Thanks
| Let's solve it in an elementary way and start from the fact that:
$$a^2 \ge 2a -1 \tag1$$
$$b^2 \ge 2b-1 \tag2$$
$$c^2 \ge 2c-1 \tag3$$
Then add up $(1)$ $(2)$ $(3)$ and get:
$$a^2+b^2+c^2 \ge a+b+c +a+b+c -3 \tag4$$
By AM-GM we have
$$\frac{a+b+c}{3} \ge (abc)^\frac{1}{3}=1 $$
$$ a+b+c \ge 3 \tag5 $$
Finally, from $(4)$ and $(5)$ we obtain the required inequality:
$$a^2+b^2+c^2 \ge a+b+c +a+b+c -3 \ge a+b+c $$
Q.E.D.
| {
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Inequality. $\frac{1}{16}(a+b+c+d)^3 \geq abc+bcd+cda+dab$ I want to prove the following inequality :
$$\frac{1}{16}(a+b+c+d)^3 \geq abc+bcd+cda+dab, $$
$a,b,c,d \in \mathbb{R}_{+} .$
In my book, at the answers chapter the author uses AM $\geq$ GM, but I haven't any idea how I can use that.
Thanks :)
| As noted by Zarrax, it suffices to show that
$$(a + b + c + d)^3 - 16(abc + bcd + cda + dab) = 4 \sum_{\mathrm{sym}}a^3 + 36 \sum_{\mathrm{sym}} a^2 b - 40 \sum_{\mathrm{sym}}abc$$
is nonnegative for all $a, b, c, d \ge 0$. Here I use the notation $\sum_{\mathrm{sym}}f(a,b ,c, d) := (1/4!) \sum_{\pi \in S_4} f(\pi(a), \pi(b), \pi(c), \pi(d))$ as $\pi$ ranges over all permutations of $\{a, b, c, d\}$.
But by AM-GM, $a^3 + b^3 + c^3 \ge 3 abc$ so that $\sum_{\mathrm{sym}}(a^3 + b^3 + c^3) \ge 3 \sum_{\mathrm{sym}}abc$ or just $\sum_{\mathrm{sym}} a^3 \ge \sum_{\mathrm{sym}} abc$. Similarly, AM-GM also gives $a^2 b + a^2 c + b^2 a + b^2 c + c^2 a + c^2 b \ge 6 abc$, so that $\sum_{\mathrm{sym}} a^2 b \ge \sum_{\mathrm{sym}} abc$. Thus
$$4 \sum_{\mathrm{sym}}a^3 + 36 \sum_{\mathrm{sym}} a^2 b - 40 \sum_{\mathrm{sym}}abc
= 4 \left(\sum_{\mathrm{sym}}a^3 - \sum_{\mathrm{sym}}abc\right) + 36 \left(\sum_{\mathrm{sym}} a^2 b - \sum_{\mathrm{sym}}abc\right) \ge 0.$$
| {
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Summing digits of powers of 2 to get 1 2 4 8 7 5 pattern If you add the decimal digits of multiples of 9,possibly repeatedly, you get 9. For instance, with 18 we have $$18\mapsto 1+8=9$$
and with 909 we have $$ 909\mapsto 9+0+9=18\mapsto 1+8=9.$$ This is proven.
If you do the same with powers of two, you develop a repeating pattern,
1,2,4,8,7,5, even when you descend negative powers. I don't know if this stops eventually, but it seems like there might be a proof for it. Can anyone prove this?
\begin{align*}
.0625\mapsto 6+2+5=13&\mapsto 4 \\
.125&\mapsto 8 \\
.25 &\mapsto 7 \\
.5 &\mapsto 5 \\
1&\mapsto 1 \\
2 &\mapsto 2 \\
4 &\mapsto 4 \\
8&\mapsto 8 \\
16&\mapsto 7 \\
32&\mapsto 5 \\
64\mapsto 10 &\mapsto 1 \\
128 \mapsto 11 &\mapsto 2 \\
256 \mapsto 13 &\mapsto 4 \\
512 &\mapsto 8 \\
1024 &\mapsto 7 \\
2048 \mapsto 14&\mapsto 5 \\
4096\mapsto 19 \mapsto 10 &\mapsto 1 \\
\end{align*}
| We look at $2^n$, where $n$ ranges over the non-negative integers.
The key is the fact that $2^6$ has remainder $1$ on division by $9$. Using congruence notation, we have $2^6\equiv 1\pmod{9}$. Let $n$ be any non-negative integer. We can express $n$ as $6q+r$, where $0\le r\le 5$ ($q$ stands for quotient, $r$ for remainder).
It follows that
$$2^n=2^{6q+r}=(2^6)^q 2^r\equiv (1)^q 2^r\equiv 2^r\pmod{9}.$$
So the remainder when you divide $2^n$ by $9$ depends only on $r$. For $r=0$, $1$, $2$, $3$, $4$, and $5$, these remainders are, as you observed, $1$, $2$, $4$, $8$, $7$, and $5$.
To connect this with sums of (decimal) digits, observe that a decimal number like $6852$ is just $(6)(10^3)+(8)(10^2)+(5)(10^1) +(2)(10^0)$. But for any non-negative integer $k$, we have $10^k\equiv 1\pmod{9}$. So $6852\equiv 6+8+5+2 \pmod{9}$. Thus the remainder when $6852$ is divided by $9$ is the same as the remainder when $6+8+5+2$ is divided by $9$. Asimilar remark holds for any non-negative integer expressed in decimal form.
Since remainders when $2^n$ is divided by $9$ cycle with period $6$, and the "casting out nines" process gives us these remainders, the pattern you observed continues forever.
You extended the pattern to negative exponents. This is an interesting observation that I do not recall seeing before. Express $2^{-n}$ as a decimal, by noting that
$$2^{-n}=\frac{1}{2^n}=\frac{5^n}{10^n}.$$
Then (essentially) you looked at the digit sum (modulo $9$) of $5^n$. Modulo $9$, the numbers $5^n$ cycle with period $6$, for the same reason as with $2^n$.
Now calculate $5^n$ modulo $9$, for $n=0,1, 2, 3, 4, 5$. We get that $5^n$ is congruent in turn to $1$, $5$, $7$, $8$, $4$, and $2$ modulo $9$. So the pattern does indeed continue "backwards."
Remark: For a more general approach, please see Euler's Theorem.
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Proving that $\frac{\csc\theta}{\cot\theta}-\frac{\cot\theta}{\csc\theta}=\tan\theta\sin\theta$
Prove $\dfrac{\csc\theta}{\cot\theta}-\dfrac{\cot\theta}{\csc\theta}=\tan\theta\sin\theta$
So, LS=
$$\dfrac{\csc\theta}{\cot\theta}-\dfrac{\cot\theta}{\csc\theta}$$
$$\left(\dfrac{1}{\sin\theta}\cdot \dfrac{\tan\theta}{1}\right)-\left(\dfrac{1}{\tan\theta}\cdot \dfrac{\sin\theta}{1}\right)$$
$$\dfrac{\tan\theta}{\sin\theta}-\dfrac{\sin\theta}{\tan\theta}$$
Now, considering the fact that I must have a common denominator to subtract, would this be correct:
$$\dfrac{\tan^2\theta}{\tan\theta\sin\theta}-\dfrac{\sin^2\theta}{\tan\theta\sin\theta}\Rightarrow \dfrac{\tan^2\theta-\sin^2\theta}{\tan\theta\sin\theta}$$
I feel like I'm close to the answer because the denominator is the RS of the OP. Please help. Do not give me the answer.
| Or you could:
$$
\dfrac{\csc\theta}{\cot\theta}-\dfrac{\cot\theta}{\csc\theta} = \frac{\sin \theta}{\sin \theta \cos\theta} - \frac{\cos\theta\sin\theta}{\sin\theta}.
$$
Then cancel terms. Then common denominator. Then...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/185205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 2
} |
Proving $n^4 + 4 n^2 + 11$ is $16k\,$ for odd $n$
if $n$ is an odd integer, prove that $n^4 + 4 n^2 + 11$ is of the form
$16 k$.
And I went something like:
$$\begin{align*}
n^4 +4 n^2 +11
&= n^4 + 4 n^2 + 16 -5 \\
&= ( n^4 +4 n^2 -5) + 16 \\
&= ( n^2 +5 ) ( n^2-1) +16
\end{align*}$$
So, now we have to prove that the product of $( n^2 +5 )$ and $( n^2-1)$ is a multiple of 16.
But, how can we do this?
If anybody has any idea of how I can improve my solution, please share it here.
Edit updated to include the necessary hypothesis that $n$ is odd.
| If $2|k=>16|n^4$ and $4|n^2=>16|(n^4+4n^2)=>n^4+4n^2+11≡11\pmod{16}$
Else
$n$ is odd$=2k+1$(say), $n^2=(2k+1)^2=8\cdot\frac{k(k+1)}{2}+1≡1\pmod{8}=>8|(n^2-1)$
(i)So, $n^4+4n^2+11=(n^2-1)^2+6(n^2-1)+16≡0\pmod{16}$ if $n$ is odd.
(ii)When $n$ is odd, $2|(n^2+1)$ and $8|(n^2-1)$(already proved) $=>2\cdot8|(n^2-1)\cdot(n^2+1)=>16|(n^4-1) $
(iii)When $n$ is odd, $n^2≡1\pmod{8}=1+8m$(say),
So, $n^4=(n^2)^2=(1+8m)^2=1+16m+64m^2≡1\pmod{16}$
So using (ii) or (iii), $n^4≡1\pmod{16}$ and $n^2≡1\pmod{8}=>4n^2≡4\pmod{32}$ if $n$ is odd,
So, $n^4+4n^2+11≡1+4+11\pmod{16}≡0\pmod{16}$ if $n$ is odd.
Alternatively, using Carmichael Function, $\lambda(16)=\frac{\phi(16)}{2}=4$ and $\lambda(8)=\frac{\phi(8)}{2}=2$
So, $n^4≡1\pmod{16}$ and $n^2≡1\pmod{8}=>4n^2≡4\pmod{32}$ if $(16,n)=1$ i.e., $n$ is odd,
So, $n^4+4n^2+11≡0\pmod{16}$ if $n$ is odd(like (ii)).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/187033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
Prove that $\frac{2}{1+\tan A}+\frac{2}{1+\tan B}+\frac{2}{1+\tan C} \le 3(\sqrt{3}-1)$ let $ABC$ be an acute triangle with all angles greater than $45^o$
Prove that
$$\frac{2}{1+\tan A}+\frac{2}{1+\tan B}+\frac{2}{1+\tan C} \le 3(\sqrt{3}-1)$$
I let $\tan A=a$, $\tan B=b$, $\tan C=c$ with $a+b+c=abc$ then the inequality equivalent to $\frac{2}{1+a}+\frac{2}{1+b}+\frac{2}{1+c} \le 3(\sqrt{3}-1)$ but it's not work.
I don't known what to do. Algebra or trigonometric?
| What about Calculus:
$f(x)=\frac{1}{1+\tan(x)}$ has a positive second derivative on $(\frac{\pi}{4}, \frac{\pi}{2})$. This solves the problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/187938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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If $a$, $a+2$ and $a+4$ are prime numbers then, how can one prove that there is only one solution for $a$? If $a$, $a+2$ and $a+4$ are prime numbers then, how can one prove that there is only one solution for $a$?
when, $a=3$
we have, $a+2=5$ and $a+4=7$
| You have observed that $a=3$ is a possible solution. Now assume $a>3$. What forms can $a$ take? Since $a$ is prime it should be of the form $3k+2$ or $3k+1$ where $k$ is a positive integer. If it was of the form $3k+2$ then $a+4=3k+2+4=3(k+2)$ will not be prime since $k+2 >1 $. If it was of the form $3k+1$ then $a+2=3k+1+2=3(k+1)$ will not be a prime since $k+1>1 $. Therefore, $k=3$ is the only answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/189179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 8,
"answer_id": 7
} |
Generalization of $\frac{x^n - y^n}{x - y} = x^{n - 1} + yx^{n - 2} + \ldots + y^{n - 1}$ I thought about a generalization for the formula
$$\frac{x^n - y^n}{x - y} = x^{n - 1} + yx^{n - 2} + \ldots + y^{n - 1}$$
It can be written as
$$\frac{x^n - y^n}{x - y} = x^{n - 1} + yx^{n - 2} + \ldots + y^{n - 1} = \sum_{i + j = n - 1}x^iy^j$$
So we would like to generalize:
$$\sum_{i_1 + i_2 + i_3 + ... +i_k = n - 1}{x_1}^{i_1}{x_2}^{i_2}{x_3}^{i_3} \cdots {x_k}^{i_k}$$
For example"
$$\sum_{i + j + k = n - 1}{x}^{i}{y}^{j}{z}^{k} = \sum_{k=0}^{n - 1}{z}^{k}
\sum_{i + j = n - k - 1}{x}^{i}{y}^{j} = \sum_{k=0}^{n - 1}{z}^{k}
\frac{x^{n - k} - y^{n - k}}{x -y} = \frac{1}{x-y} \left(\frac{x^{n + 1} - z^{n + 1}}{x - z} - \frac{y^{n + 1} - z^{n + 1}}{y - z}\right)$$
It seems that the generalized expression is the divided difference of $x^{n + k - 2}$ in the points $x_1, x_2, \ldots, x_k$.
Does anyone have an idea how to prove it?
| I think that this formula is what you are looking for. If $\mathbf x = (x_1,\dotsc,x_r)$, then
$$ \sum_{|I|=n} \mathbf{x}^I = \sum_i\frac{x_i^{n}}{\prod_{j\neq i}(1-\frac{x_j}{x_i})}. $$
With 1 variable, it gives
$$ x^n = x^n, $$
for two,
$$ \sum_{i+j = n} x^i y^j = \frac{x^{n+1}}{x-y} + \frac{y^{n+1}}{y-x}, $$
and for three, if gives
$$ \sum_{i+j+k=n} x^i y^j z^k= \frac{x^{n+2}}{(x-y)(x-z)} + \frac{y^{n+2}}{(y-x)(y-z)} + \frac{z^{n+2}}{(z-x)(z-y)}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/189225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 0
} |
When does $\lfloor (n-1)x \rfloor + \lfloor x \rfloor = \lfloor nx \rfloor$? I am trying to find the conditions under which $\lfloor(n-1)x\rfloor + \lfloor x \rfloor = \lfloor nx \rfloor$. The trivial case is whenever $x \in \mathbb{Z}$. If $n = 2$, then $x - \lfloor x \rfloor \lt \frac{1}{2}$.
As $n$ increases however, the pattern becomes more complicated. For example, for $n = 3$, if $x - \lfloor x \rfloor \lt \frac{1}{3}$ then $\lfloor 3x \rfloor = 3\lfloor x \rfloor$, and $\lfloor 2x \rfloor = 2\lfloor x \rfloor$ so the identity is satisfied. However, if $ \frac{1}{2} \lt x - \lfloor x \rfloor \lt \frac{2}{3}$, then the identity is also satisfied.
I tried seeing whether Hermite's identity can help but I can't see an obvious way of applying it.
| Let $m=\lfloor x\rfloor$ and $\alpha=x-m$. Then
$$\lfloor (n-1)x \rfloor + \lfloor x \rfloor =\lfloor(n-1)m+(n-1)\alpha\rfloor+m=nm+\lfloor(n-1)\alpha\rfloor\;,$$
and $$\lfloor nx \rfloor=\lfloor nm+n\alpha\rfloor=nm+\lfloor n\alpha\rfloor\;,$$
so $\lfloor (n-1)x \rfloor + \lfloor x \rfloor = \lfloor nx \rfloor$ iff $\lfloor(n-1)\alpha\rfloor=\lfloor n\alpha\rfloor$, i.e., iff there is an integer $k$ such that $$k\le(n-1)\alpha<n\alpha<k+1\;.\tag{1}$$
$(1)$ is equivalent to $$\frac{k}{n-1}\le\alpha<\frac{k+1}n\;,$$
so we want the non-empty intervals of the form $$\left[\frac{k}{n-1},\frac{k+1}n\right),\quad k=0,\dots,n-1\;.$$
Now $\frac{k}{n-1}<\frac{k+1}n$ iff $kn<(k+1)(n-1)=kn+n-k-1$ iff $n>k+1$, so
$$\lfloor (n-1)x \rfloor + \lfloor x \rfloor = \lfloor nx \rfloor\quad\text{iff}\quad x-\lfloor x\rfloor\in\bigcup_{k=0}^{n-2}\left[\frac{k}{n-1},\frac{k+1}n\right)\;.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/190605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Find a positive number $\delta$? Find a positive number $\delta$ such that when $|x-1|<\delta$, then $|x^2-1| < 0.45$
Part 2:
Find the LARGEST number $\lambda$ with the property that when $|x-1|< \lambda$, then $|x^2-1| < 0.45$
I don't understand this at all. Help?
| Suppose $0 < \epsilon < 1$. Consider the case $x^2 - 1 \ge 0$ first, i.e., $x \in (-\infty, -1] \cup [1, \infty)$. In this case, the inequality $|x^2 - 1| < \epsilon$ becomes
$$
x^2 - 1 - \epsilon < 0.
$$
Factor this as
$$
(x - \sqrt{1 + \epsilon})(x + \sqrt{1 + \epsilon}) < 0.
$$
This is true when $x \in (-\sqrt{1 + \epsilon}, \sqrt{1 + \epsilon})$. So we get $x \in (-\sqrt{1 + \epsilon}, -1] \cup [1, \sqrt{1 + \epsilon})$ for this case.
The other case is when $x^2 - 1 \le 0$, i.e., $x \in [-1, 1]$, the inequality $|x^2 - 1| < \epsilon$ becomes
$$
x^2 - 1 + \epsilon > 0.
$$
Factor this as
$$
(x - \sqrt{1 - \epsilon})(x + \sqrt{1 - \epsilon}) > 0.
$$
This is true when $x \in (-\infty, -\sqrt{1 - \epsilon}) \cup (\sqrt{1 - \epsilon}, \infty)$, so we get $x \in [-1, -\sqrt{1 - \epsilon}) \cup (\sqrt{1 - \epsilon}, 1]$ for this case. Combining them both, we get
$$
x \in (-\sqrt{1 + \epsilon}, -\sqrt{1 - \epsilon}) \cup (\sqrt{1 - \epsilon}, \sqrt{1 + \epsilon}).
$$
So to pick $\lambda$, make $\lambda = 1 - \sqrt{1 - \epsilon}$. This is because $1 - \sqrt{1 - \epsilon} > \sqrt{1 + \epsilon} - 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/191112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Proving the inequality $a^2+b^2+c^2+ab+bc+ca\ge6$ Given that $a$, $b$, $c$ are non-negative real numbers such that $a+b+c=3$, how can we prove that:
$a^2+b^2+c^2+ab+bc+ca\ge6$
| By the Cauchy-Schwarz inequality,
$$6=1*(a+b)+1*(b+c)+1*(a+c)\leq\sqrt{1^2+1^2+1^2}\sqrt{(a+b)^2+(b+c)^2+(a+c)^2}$$
In other words
$$(a+b)^2+(b+c)^2+(a+c)^2\geq 12$$
which is the same as the desired inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/193140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
} |
A function $f(x)$ on the real domain and real constants $a$ and $b\neq 0$ for which $f(x)-f(x-\delta)+a+bx^2=0$ for some real $\delta\neq 0$ Is there a function $f(x)$ on the real domain and real constants $a$ and $b\neq 0$ for which the following is true:
$$f(x)-f(x-\delta)+a+bx^2=0$$
for some real $\delta\neq 0$?
EDIT: I missed a very important constraint in the original posting of this question: $b\neq 0$. Sorry!
| $f(x)-f(x-\delta)+a+bx^2=0$
$f(x)-f(x-\delta)=-bx^2-a$
In fact this belongs to a functional equation of the form http://eqworld.ipmnet.ru/en/solutions/fe/fe1108.pdf.
The general solution of this functional equation is $f(x)=\Theta(x)+f_p(x)$ , where $\Theta(x)$ is an arbitrary periodic function with period $\delta$
Luckily we can find $f_p(x)$ by method of undetermined coefficients:
Let $f_p(x)=Ax^3+Bx^2+Cx$ ,
Then $f_p(x-\delta)=A(x-\delta)^3+B(x-\delta)^2+C(x-\delta)=Ax^3-3A\delta x^2+3A\delta^2x-A\delta^3+Bx^2-2B\delta x+B\delta^2+Cx-C\delta=Ax^3+(B-3\delta A)x^2+(3\delta^2A-2\delta B+C)x-\delta^3A+\delta^2B-\delta C$
$\therefore Ax^3+Bx^2+Cx-(Ax^3+(B-3\delta A)x^2+(3\delta^2A-2\delta B+C)x-\delta^3A+\delta^2B-\delta C)\equiv-bx^2-a$
$3\delta Ax^2+(2\delta B-3\delta^2A)x+\delta^3A-\delta^2B+\delta C\equiv-bx^2-a$
$\therefore\begin{cases}3\delta A=-b\\2\delta B-3\delta^2A=0\\\delta^3A-\delta^2B+\delta C=-a\end{cases}$
$\begin{cases}A=-\dfrac{b}{3\delta}\\B=-\dfrac{b}{2}\\C=-\dfrac{b\delta}{6}-\dfrac{a}{\delta}\end{cases}$
$\therefore f(x)=\Theta(x)-\dfrac{bx^3}{3\delta}-\dfrac{bx^2}{2}-\dfrac{b\delta x}{6}-\dfrac{ax}{\delta}$ , where $\Theta(x)$ is an arbitrary periodic function with period $\delta$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/194855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Finding $\alpha$ for $\sin(4 \alpha + \frac{\pi}{6}) = \sin (2\alpha + \frac{\pi}{5})$ I try to find $\alpha$ for $\sin(4 \alpha + \frac{\pi}{6}) = \sin (2\alpha + \frac{\pi}{5})$.
Left side:
$$\sin(4 \alpha + \frac{\pi}{6}) =$$
$$= \sin4\alpha \times \cos \frac{\pi}{6} + \cos 4\alpha \times \sin\frac{\pi}{6} =$$
$$= \sin4\alpha \times \frac{\sqrt{3}}{2} + \cos 4\alpha \times \frac{1}{2} = $$
$$= 2\sin2\alpha \times \cos2\alpha \times \frac{\sqrt{3}}{2} + (\cos^2 2\alpha - \sin^2 2\alpha) \times \frac{1}{2} = $$
$$= 2\sin\alpha \times \cos \alpha \times \cos2\alpha \times \sqrt{3} + \frac{\cos^2 2\alpha - \sin^2 2\alpha}{2} = $$
$$= 2\sin\alpha \times \cos \alpha \times (\cos^2\alpha - \sin^2\alpha) \times \sqrt{3} + \frac{(\cos^2 \alpha - \sin^2\alpha)(\cos^2 \alpha - \sin^2\alpha) - \sin^2 2\alpha}{2} = $$
$$= 2\sqrt{3}\sin\alpha \cos \alpha (\cos^2\alpha - \sin^2\alpha) + \frac{(\cos^2 \alpha - \sin^2\alpha)(\cos^2 \alpha - \sin^2\alpha) - (2\sin\alpha \times \cos\alpha)(2\sin\alpha \times \cos\alpha)}{2} = $$
$$= 2\sqrt{3}\sin\alpha \cos \alpha (\cos^2\alpha - \sin^2\alpha) + \frac{(\cos^2 \alpha - \sin^2\alpha)(\cos^2 \alpha - \sin^2\alpha)}{2} - 2\sin^2\alpha \times \cos^2\alpha = $$
$$= 2 \times \sqrt{3}\sin\alpha \cos \alpha (\cos^2\alpha - \sin^2\alpha) + \frac{\cos^4 \alpha - (\sin^2\alpha \times \cos^2\alpha) - (\sin^2\alpha \times \cos^2\alpha) + \sin^4 \alpha}{2} - 2\sin^2\alpha \times \cos^2\alpha = $$
$$= 2 \sqrt{3}\sin\alpha \cos \alpha (\cos^2\alpha - \sin^2\alpha) + \frac{(\cos^2\alpha - \sin^2\alpha)^2}{2} - 2\sin^2\alpha \times \cos^2\alpha = $$
$$= 2\sqrt{3}(\sin\alpha \times \cos^3 \alpha - \sin^3\alpha \times \cos\alpha) + \frac{(1 - 2\sin^2\alpha)^2}{2} - 2\sin^2\alpha \times \cos^2\alpha = $$
$$= 2\sqrt{3}(\sin\alpha \times \cos^3 \alpha - \sin^3\alpha \times \cos\alpha) + \frac{1}{2}-2\sin^2\alpha+2\sin^4\alpha - 2\sin^2\alpha \times \cos^2\alpha = $$
$$= 2\sin\alpha(\sqrt{3}\times \cos^3 \alpha - \sqrt{3}\sin^2\alpha \times \cos\alpha + \frac{1}{4\sin\alpha}-\sin\alpha + \sin^3\alpha - \sin\alpha \times \cos^2\alpha) = $$
Right side:
$$\sin(2\alpha + \frac{\pi}{5}) = \sin2x \times \cos\frac{\pi}{5} + \cos2\alpha \times \sin\frac{\pi}{5} = $$
$$= 2\sin\alpha \cos\alpha \times \cos\frac{\pi}{5} + (\cos^2\alpha - \sin^2\alpha)\sin\frac{\pi}{5} = $$
$$= 2\sin\alpha \cos\alpha \times \cos\frac{\pi}{5} + (1 - 2\sin^2\alpha)\sin\frac{\pi}{5} = $$
$$= 2\sin\alpha \cos\alpha \times \cos\frac{\pi}{5} + \sin\frac{\pi}{5} - 2\sin^2\alpha \times \sin\frac{\pi}{5} = $$
$$= 2\sin\alpha(\cos\alpha \times \cos\frac{\pi}{5} + \frac{\sin\frac{\pi}{5}}{2\sin\alpha} - \sin\alpha \times \sin\frac{\pi}{5})$$
I can't get any further on either. Bringing them together I get:
$$2\sin\alpha(\sqrt{3}\times \cos^3 \alpha - \sqrt{3}\sin^2\alpha \times \cos\alpha + \frac{1}{4\sin\alpha}-\sin\alpha + \sin^3\alpha - \sin\alpha \times \cos^2\alpha) = 2\sin\alpha(\cos\alpha \times \cos\frac{\pi}{5} + \frac{\sin\frac{\pi}{5}}{2\sin\alpha} - \sin\alpha \times \sin\frac{\pi}{5})$$
$$\iff \sqrt{3}\times \cos^3 \alpha - \sqrt{3}\sin^2\alpha \times \cos\alpha + \frac{1}{4\sin\alpha}-\sin\alpha + \sin^3\alpha - \sin\alpha \times \cos^2\alpha = \cos\alpha \times \cos\frac{\pi}{5} + \frac{\sin\frac{\pi}{5}}{2\sin\alpha} - \sin\alpha \times \sin\frac{\pi}{5}$$
Where do I go from here? Om am I already dead wrong?
| $\sin(4\alpha+\frac{\pi}{6})=\sin(2\alpha+\frac{\pi}{5})$
$\sin(4\alpha+\frac{\pi}{6})-\sin(2\alpha+\frac{\pi}{5}) = 0$
Consider the formula: $\sin\alpha-\sin\beta=2\cos\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}$,
and $A\cdot B=0\Leftrightarrow A=0\vee B=0$
$2\cos\frac{6\alpha+\frac{11\pi}{30}}{2}\sin\frac{2\alpha-\frac{\pi}{30}}{2}=0$
$\cos\frac{6\alpha+\frac{11\pi}{30}}{2}\sin\frac{2\alpha-\frac{\pi}{30}}{2}=0$
$\cos\frac{6\alpha+\frac{11\pi}{30}}{2}=0\vee\sin\frac{2\alpha-\frac{\pi}{30}}{2}=0$
$\frac{6\alpha+\frac{11\pi}{30}}{2}=m\Rightarrow\cos m=0\Rightarrow m=\frac{\pi}{2}+k\pi$
Hence we have:
$\frac{6\alpha+\frac{11\pi}{30}}{2}=\frac{\pi}{2}+k\pi\Rightarrow\alpha=\frac{k\pi}{3}+\frac{19\pi}{180}$, $k\in Z$
$\frac{2\alpha-\frac{\pi}{30}}{2}=n\Rightarrow \sin n=0\Rightarrow n=k\pi$
Hence we have:
$\frac{2\alpha-\frac{\pi}{30}}{2}=k\pi\Rightarrow\alpha=k\pi+3$, $k\in Z$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/195229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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How I can prove that the sequence $\sqrt{2} , \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}$ converges to 2?
Prove that the sequence $\sqrt{2} , \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}} \ $ converges to $2$.
My attempt
I proved that the sequence is increasing and bounded by $2$, can anyone help me show that the sequence converges to $2$?
Thanks for your help.
| We have
$$\sqrt{2}=2^{\frac{1}{2}}.$$
Exciting, no?
We also have
$$\sqrt{2\sqrt{2}}=2^{\frac{1}{2}+\frac{1}{4}},$$
and
$$\sqrt{2\sqrt{2\sqrt{2}}}=2^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}}.$$
For the next term, we take the previous term, multiply by $2$, getting exponent of $2$ equal to $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}$, then take the square root, getting exponent of $2$ equal to $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}$.
The pattern continues, since the "next term" is always obtained by the same process.
It is well-known that the series $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots$ converges to $1$. Since the function $f(x)=2^x$ is continuous, our original sequence converges to $2^1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/200416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "56",
"answer_count": 7,
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Solve $(\sqrt{5+2\sqrt{6}})^{x}+(\sqrt{5-2\sqrt{6}})^{x}=10$.
Solve $(\sqrt{5+2\sqrt{6}})^{x}+(\sqrt{5-2\sqrt{6}})^{x}=10$
I square the both sides and get $(5+2\sqrt{6})^{x}+(5-2\sqrt{6})^{x}=98$. But I don't know how to carry on. Please help. Thank you.
| Nice solution, just for clarification
$\left(5+2\sqrt{6}\right)^{\frac{x}{2}}\cdot \left(5-2\sqrt{6}\right)^{\frac{x}{2}}=\left[\left(5+2\sqrt{6}\right)\cdot \left(5-2\sqrt{6}\right)\right]^{\frac{x}{2}}=\left[5^2-\left(2\sqrt{6}\right)^2\right]^{\frac{x}{2}}=\left(25-24\right)^{\frac{x}{2}}=1$
and so for $t=\left(\sqrt{5+2\sqrt{6}}\right)^x$......
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/202078",
"timestamp": "2023-03-29T00:00:00",
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understanding definition of limit for x_n= [(n+1)/n]^0.5; where x_n is the n-th term in the sequence. $x_n= \sqrt{\frac{n+1}{n}}$where $x_n$ is the $n^{th}$ term in the sequence.
Now, $x_n \rightarrow 1$ as $n\rightarrow\infty$.
Then applying definition of limit: $$\left|\sqrt{\frac{n+1}{n}}- 1 \right|= \frac{\frac{n+1}{n}-1}{\sqrt{\frac{n+1}{n}}+1}<\frac{1}{2n}<\epsilon$$
whenever $n>N=1/(2\epsilon)$.
-The difficulty for me is inability to understand how $1/(2n)$ was deduced.
Thank you for assistance and apologies for poor notations
the equation in the code should read after absolute value equation = {[(n+1)/n]-1}/{([(n+1)/n]^0.5)+1}<1/2n< epsilon
| I see that since $n(n+1)>n$ then $\sqrt{n(n+1)}+n>2n$ and so $$\bigg|\frac{1}{\sqrt{n(n+1)}+n}\bigg|<\frac{1}{2n}$$ But, $$\bigg|\frac{1}{\sqrt{n(n+1)}+n}\bigg|=\bigg|\frac{\frac{n+1}{n}-1}{\sqrt{\frac{n+1}{n}}+1}\bigg|$$ Therefore $$\bigg|\sqrt{\frac{n+1}{n}}-1\bigg|=\bigg|\frac{\frac{n+1}{n}-1}{\sqrt{\frac{n+1}{n}}+1}\bigg|=\bigg|\frac{1}{\sqrt{n(n+1)}+n}\bigg|<\frac{1}{2n}$$
| {
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How to derive the equation for x in a quadratic equation?
Possible Duplicate:
Why can ALL quadratic equations be solved by the quadratic formula?
How to derive this:
$x = \frac{-b + {\sqrt{b^2 + 4ac}}}{2a}$
From this:
$ax^2 + bx + c = 0$
I know this may be a little elementary :)
| The usual approach is to complete the square:
$$\begin{align*}
ax^2+bx+c&=a\left(x^2+\frac{b}ax+\frac{c}a\right)\\
&=a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}+\frac{c}a\right)\\
&=a\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a}+c\;,
\end{align*}$$
which equals $0$ if and only if $$a\left(x+\frac{b}{2a}\right)^2=\frac{b^2}{4a}-c=\frac{b^2-4ac}{4a}\;.$$ Now divide both sides by $a$ to get
$$\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}\;,$$ and take square roots:
$$x+\frac{b}{2a}=\pm\sqrt{\frac{b^2-4ac}{4a^2}}=\pm\frac1{2a}\sqrt{b^2-4ac}\;.$$ Hence
$$x=-\frac{b}{2a}\pm\frac1{2a}\sqrt{b^2-4ac}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\;.$$
| {
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Level curves on ellipsoid Let $a,b,c>0$ with $a\leq b\leq c$. Let $E$ be the ellipsoid determined by $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=1$$
Is there a function $f:E\rightarrow \mathbb{R}$ such that $f\in C^{\infty}(E)$ and all level curves of $f$ are circles?
Thanks
| Generalizing @Rahul's solution (but, as we'll see, not-generalizing that solution), consider the plane, through the origin, with normal $(p,q,r)$ such that $p^2+q^2+r^2=1$. Of course, obtaining circular level curves in the case $p^2+q^2 = 0$ requires $a=b$ and is easily dispatched; similarly, $q^2+r^2 = 0$ and $r^2 + p^2 = 0$; so, from here on, we assume that at most one of $p$, $q$, $r$ is zero.
The plane is spanned by orthogonal unit vectors
$$\mathbf{u} := \frac{(q,-p,0)}{\sqrt{p^2+q^2}} \qquad \mathbf{v} := \frac{(-pr,-qr,p^2+q^2)}{\sqrt{p^2+q^2}}$$
The point $u \mathbf{u} + v \mathbf{v}$ lies on the ellipse iff
$$\begin{align}
1 &= \frac{(uq-vpr)^2}{a^2} + \frac{(-up-vqr)^2}{b^2} + \frac{v^2(p^2+q^2)^2}{c^2} \\ &= u^2 \left( \frac{q^2}{a^2}+\frac{p^2}{b^2} \right) + 2uvpqr\frac{a^2-b^2}{a^2b^2} + v^2 \left(\frac{p^2r^2}{a^2}+\frac{q^2r^2}{b^2}+\frac{(p^2+q^2)^2}{c^2}\right) \\
&=: A u^2 + 2 B u v + C v^2
\end{align}$$
For the parameterized curve to be a circle, we must have $B = 0$ and $A=C$.
For $B=0$, we have either that $a=b$, or that (at most) one of $p$, $q$, $r$ is zero. Together, $a=b$ and $A=C$ (and $p^2+q^2+r^2=1$) imply $a=c$, so that the ellipsoid is in fact a sphere. On the other hand, $pqr=0$ and $A=C$ allow us to either solve for $p$, $q$, $r$ or derive other constraints on $a$, $b$, $c$:
Case $p=0$: We have, from $A=C$, that $\frac{1}{a^2} = \frac{r^2}{b^2}+\frac{q^2}{c^2}$, so that
$$q^2 a^2(c^2-b^2) = -c^2(b^2-a^2) \qquad r^2 a^2(c^2-b^2)= b^2(c^2-a^2)$$
Since $c^2-b^2$, $b^2-a^2$, and $c^2-a^2$ are all non-negative, the $q$ equation requires $a=b=c$ and we have another sphere.
Case $q=0$: Here, $\frac{1}{b^2} = \frac{r^2}{a^2}+\frac{p^2}{c^2}$, which is the same as the above case, with $p \to q \to r \to p$ and $a\to b\to c \to a$. Thus,
$$r^2 b^2(c^2-a^2) = a^2(c^2-b^2) \qquad p^2 b^2(c^2-a^2)= c^2(b^2-a^2)$$
which allows solutions for $p$ and $r$ even for non-spheres. (This is @Rahul's case.)
Case $r=0$: Here, $\frac{q^2}{a^2}+\frac{p^2}{b^2}=\frac{1}{c^2}$, which is the same as the above case under the mentioned cycling of variables, so that
$$p^2 c^2(b^2-a^2) = b^2(c^2-a^2) \qquad q^2 c^2(b^2-a^2)= -a^2(c^2-b^2)$$
Again, the $q$ equation implies that the ellipse is a sphere.
Consequently, @Rahul's circles are the only ones for non-spherical $E$.
Note. The fact that the three cases are cyclically related suggests that there should be an insightful, cyclically-symmetric way of writing the equation $A=C$, but I'm not seeing it.
| {
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Arithmetic Mean & Geometric Mean Question 1: if the arithmetic mean of two numbers is twice of their geometric mean, their ratio of sum of numbers to the difference of numbers equals?
Question 2: if the quadratic equation:
$(b^2+c^2)x2-2(a+b)cx+(c^2+a^2)=0$
has equal roots then?what is its AP & GP?
Question 3: If the expansion of:
$(1+x)^{50}$
let S be the sum of the coefficient of the odd power of x, then S will be?
Please help with this problems in brief.
-Thanks.
| Question 1:
Arithmetic mean: $\frac{a+b}{2}$
Geometric mean: $\sqrt{ab}$
So the condition becomes $\frac{a+b}{2}=2\sqrt{ab}$. Square both sides to get
$$
\frac{a^2+2ab+b^2}{4}=4ab
$$
which, assuming $b\ne0$, results in
$$
\left(\frac ab\right)^2-14\frac ab+1=0
$$
and
$$
\frac ab=7\pm4\sqrt{3}
$$
Then we can compute
$$
\frac{a+b}{a-b}=\frac{\frac ab+1}{\frac ab-1}=\frac{4\pm2\sqrt{3}}{3\pm2\sqrt{3}}\frac{3\mp2\sqrt{3}}{3\mp2\sqrt{3}}=\pm\frac{2\sqrt{3}}{3}
$$
Question 3:
The sum of all the coefficients is $(1+1)^{50}=2^{50}$
The sum of the even coefficients minus the sum of the odd coefficients is $(1-1)^{50}=0^{50}$
The sum of the odd coefficients is $\frac12(2^{50}-0^{50})=2^{49}$.
| {
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sum of the series
could anyone tell me how to calculate these sums?I am not finding any usual way to calculate them.
| 5.7:
$$\begin{align*}
1+\frac{3}{4}+\frac{3\cdot5}{4\cdot8}+\cdots
&=1+\sum_{n=1}^{\infty}\frac{3\cdot5\cdots(2n+1)}{4\cdot8\cdots4n}\\
&=1+\sum_{n=1}^{\infty}\frac{3\cdot5\cdots(2n+1)}{2\cdot4\cdots2n}(\frac{\sqrt{2}}{2})^{2n}
\end{align*}$$
Set $f(x)=\sum_{n=1}^{\infty}\frac{3\cdot5\cdots(2n-1)}{2\cdot4\cdots2n}x^{2n+1}$, then $1+\frac{3}{4}+\frac{3\cdot5}{4\cdot8}+\cdots=1+f^{\prime}(\frac{\sqrt{2}}{2})$.
$$\begin{align*}
f^{\prime}(x)&=\sum_{n=1}^{\infty}\frac{3\cdot5\cdots(2n+1)}{2\cdot4\cdots2n}x^{2n}\\
&=\sum_{n=1}^{\infty}\frac{3\cdot5\cdots(2n-1)\cdot(2n+1)}{2\cdot4\cdots2n}x^{2n}\\
&=\sum_{n=1}^{\infty}\frac{3\cdot5\cdots(2n-1)\cdot2n}{2\cdot4\cdots(2n-2)\cdot2n}x^{2n}+\sum_{n=1}^{\infty}\frac{3\cdot5\cdots(2n-1)}{2\cdot4\cdots2n}x^{2n}\\
&=x^{2}(1+\sum_{n=2}^{\infty}\frac{3\cdot5\cdots(2n-1)}{2\cdot4\cdots(2n-2)}x^{2n-2})+\frac{1}{x}f(x)\\
&=x^{2}(1+f^{\prime}(x))+\frac{1}{x}f(x)
\end{align*}$$
Set $g(x)=f(x)+x$, then $g^{\prime}(x)=x^{2}g^{\prime}(x)+\frac{1}{x}g(x)$, by calculation:
$$g^{\prime}(x)=\frac{1}{x(1-x^{2})}g(x)$$
$$(\frac{\sqrt{1-x^{2}}}{x}g(x))^{\prime}=0$$
$$\frac{\sqrt{1-x^{2}}}{x}g(x)=c$$
$$g(x)=c\frac{x}{\sqrt{1-x^{2}}}$$
$$g^{\prime}(x)=c\frac{1}{\sqrt{1-x^{2}}^{\frac{3}{2}}}$$
as $g^{\prime}(0)=f^{\prime}(0)+1=1$, so $c=1$, $1+\frac{3}{4}+\frac{3\cdot5}{4\cdot8}+\cdots=1+f^{\prime}(\frac{\sqrt{2}}{2})=g^{\prime}(\frac{\sqrt{2}}{2})=2\sqrt{2}$.
| {
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Can any integer can be written as the sum of 8 integer cubes? Can anyone offer an elementary proof why: $$\forall n \in {\mathbb Z} \space \exists a, b, c, d, e, f, g, h \in {\mathbb Z}$$such that$$n=a^3+b^3+c^3+d^3+e^3+f^3+g^3+h^3{\rm ?}$$
In other words, why every integer is the sum of eight cubes integers.
My first thought was that since the gaps between $p^3$ and $(p+1)^3$ increase with $p,$ this would only hold for small integers. This is wrong; can anyone tell me why?
N.B. This is different to Waring's problem because $n,a,b,c...$ dot not have to be natural numbers, so we can have $27=3^3+(-1)^3+(-1)^3+(-1)^3+(-1)^3$
| Allowing negative cubes, five suffice. It is suspected that four suffice but this is an open problem. I will see if I can find the argument for five, it is just one or two explicit formulas. By the way, these are called the "Easier" Waring problems.
Can't find them at this point, so here is section D5 form Richard K. Guy, Unsolved Problems in Number Theory. Note that he is allowing cubes to be positive, negative, or zero:
Is every number the sum of four cubes? This has been proved for all
numbers except possibly those of the form $9n \pm 4$
Part 1: $$ 6n = (n+1)^3 + (n-1)^3 - n^3 - n^3 $$
Part 2:
$$ 6n - 2 = n^3 +(n+2)^3 - (n+1)^3 - (n+1)^3 - 2^3 $$
$$ 6n-1 = (n+1)^3 + (n-1)^3 - n^3 - n^3 - 1^3 $$
$$ 6n = (n+1)^3 + (n-1)^3 - n^3 - n^3 $$
$$ 6n+1 = (n+1)^3 + (n-1)^3 - n^3 - n^3 + 1^3 $$
$$ 6n + 2 = n^3 +(n-2)^3 - (n-1)^3 - (n-1)^3 + 2^3 $$
$$ 6n + 3 = (n-3)^3 +(n-5)^3 - (n-4)^3 - (n-4)^3 + 3^3 $$
| {
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$16^{18}+16^{18}+16^{18}+16^{18}+16^{18}=4^x$ without a calculator How to solve the following equation without using calculator
$$16^{18}+16^{18}+16^{18}+16^{18}+16^{18}=4^x$$
| $16^{18}+16^{18}+16^{18}+16^{18}+16^{18}$
$=5\times 16^{18}$
$= 5\times {4^2}^{18} $
$= 5\times 4^{2\times 18} = 5\times 4^{36} = 4^x$
$ 5\times 4^{36} = 4^{\log_4{5}} \times 4^{36} $
$= 4^{\log_4{5} +36}$
$\rightarrow x= \log_4{5} +36$
| {
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Evaluating $\frac{1}{2\pi}\int_{0}^{2\pi}\frac{1}{1-2t\cos\theta +t^2}d\theta$ I need solve this integral, and I tried various methods of solving and did not get it. The integral is:
$$\frac{1}{2\pi}\int_{0}^{2\pi}\frac{1}{1-2t\cos\theta +t^2}d\theta,$$
where $t$ is a positive integer.
| $$\frac{1}{1-2t\cos \theta +t^2}=\frac 1{1+t^2-2t\frac{(1-\tan^2\frac{\theta}2)}{(1+\tan^2\frac{\theta}2)}}$$
$$=\frac{\sec^2\frac{\theta}2}{(1+t^2)(1+\tan^2\frac{\theta}2)-2t(1-\tan^2\frac{\theta}2)}=\frac{\sec^2\frac{\theta}2}{(1-t)^2+\tan^2\frac{\theta}2(1+t)^2}$$
$$=\frac1{(1+t)^2}\frac{\sec^2\frac{\theta}2}{(\frac{1-t}{1+t})^2+\tan^2\frac{\theta}2}$$
If $f(\theta)=\frac{1}{1-2t\cos \theta +t^2}, f(2\pi-\theta)=f(\theta)$,
So, $\int_{0}^{2\pi}f(\theta)d\theta=2\int_{0}^{\pi}f(\theta)d\theta$
$$I=\frac{1}{2\pi}\int_{0}^{2\pi}\frac{1}{1-2t\cos\theta +t^2}d\theta
=\frac{1}{\pi}\int_{0}^{\pi}\frac{1}{1-2t\cos\theta +t^2}d\theta,$$
$$=\frac{1}{\pi(1+t)^2}\int_{0}^{\pi}\frac{\sec^2\frac{\theta}2}{(\frac{1-t}{1+t})^2+\tan^2\frac{\theta}2}d\theta$$
Now we can put $z=\tan \frac{\theta}2$ in the given problem, if $\theta=0,z=0$ and if $\theta=\pi,z=∞$ and $dz=\frac{\sec^2\frac{\theta}{2}d\theta}{2}$
So, $$I=\frac{1}{\pi(1+t)^2}\int_{0}^{∞}\frac{2dz}{(\frac{1-t}{1+t})^2+z^2}$$
$$=\frac{2}{\pi(1-t^2)} \tan^{-1}{\frac{(1+t)z}{1-t}} \mid_{0}^{∞}$$
At $z=0, \tan^{-1}{\frac{(1+t)z}{1-t}}=0$ if $t \ne 1$
At $z=∞, \tan^{-1}{\frac{(1+t)z}{1-t}}=\frac{\pi}2$ if $ \frac{1+t}{1-t}>0 $ or if $ \frac{(1+t)(1-t)}{(1-t)^2}>0$ or if $1-t^2>0$ or if $-1< t< 1$
At $z=∞, \tan^{-1}{\frac{(1+t)z}{1-t}}=-\frac{\pi}2$ if $t>1$ or $t<-1$
| {
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Common tangent to two circles
Find the equations of the common tangents to the 2 circles:
$$(x - 2)^2 + y^2 = 9$$
and
$$(x - 5)^2 + (y - 4)^2 = 4.$$
I've tried to set the equation to be $y = ax+b$, substitute this into the 2 equations and set the discriminant to zero, we then get a simultaneous quadratic equations. But they are really difficult to solve. So is there any simpler way to do this? Thank you.
| For the circles $$(x - 2)^2 + y^2 = 9\text{ and } (x - 5)^2 + (y - 4)^2 = 4,$$
using Article $180$ of The elements of coordinate geometry by Loney,
Case $1:$ If $T_2$ is the point dividing internally the line joining the centres
in the ratio $2:3$ then its coordinates are $$\left(\frac{2\cdot2+5\cdot3}{2+3},\frac{2\cdot0+4\cdot3}{2+3}\right)=\left(\frac{19}5,\frac{12}5\right)$$
Now, the equation of any line passing through $T_2\left(\frac{19}5,\frac{12}5\right)$ is $$\frac{y-\frac{12}5}{x-\frac{19}5}=m\implies 5mx-5y+12-19m=0$$ where $m$ is the gradient
If it is the tangent of the circle, the distance of this line from the centre of each circle will be equal to the radius of the respective circle.
$$\implies \left|\frac{5m\cdot 2-5\cdot0+12-19m}{\sqrt{(5m)^2+(-5)^2}}\right|=3\implies |12-9m|=15\sqrt{m^2+1}$$
$$\text{On squaring and re-arrangement, } 16m^2+24m+9=0\implies m=-\frac34$$
So, there will be only one crossed common tangent
Case $2:$ If $T_1$ is the point dividing externally the line joining the centres
in the ratio $2:3$ then its coordinates are $$\left(\frac{2\cdot2-5\cdot3}{2-3},\frac{2\cdot0-4\cdot3}{2-3}\right)=\left(11,12\right)$$
Now, the equation of any line passing through $T_1\left(11,12 \right)$ is $$\frac{y-12}{x-11}=n\implies nx-y+12-11n=0$$ where $n$ is the gradient
Can you follow the method used in Case $1$ to find the two values of $n,$ which implies there will be two simple common tangents
| {
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Least Square Solutions
Find the closest point or points to b $= (-1,2)^T$ that lie on
the line $x + y = 0$ and also find the line $2x + y = 0$.
The question is obviously elementary but I do not know how to set it up properly to find the points. I know for $x + y = 0$ we have a slope of -1 going through the origin and hence the closest point there is $(-\frac32, \frac32)$, but how can I set it up so that I can find that point. Also, I already know how to do it using differentiation, I only want to know how to do it algebraically using $Ax = b$ form.
| Note that $x + y = 0$ describes a slope in the direction of $\left( \begin{matrix} 1 \\ -1 \end{matrix} \right)$. We want to find the point $c \left( \begin{matrix} 1 \\ -1 \end{matrix} \right)$ on this line such that $\left( \left( \begin{matrix} -1 \\ 2 \end{matrix} \right) - c \left( \begin{matrix} 1 \\ -1 \end{matrix} \right) \right) \cdot \left( \begin{matrix} 1 \\ -1 \end{matrix} \right) = 0$. Solving for $c \left( \begin{matrix} 1 \\ -1 \end{matrix} \right)$ we get:
$\frac{\left( \begin{matrix} -1 \\ 2 \end{matrix} \right) \cdot \left( \begin{matrix} 1 \\ -1 \end{matrix} \right)}{\left( \begin{matrix} 1 \\ -1 \end{matrix} \right) \cdot \left( \begin{matrix} 1 \\ -1 \end{matrix} \right)} \cdot \left( \begin{matrix} 1 \\ -1 \end{matrix} \right) = \left( \begin{matrix} -\frac{3}{2} \\ \frac{3}{2} \end{matrix} \right)$
| {
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Find $ord_m b^2$ if $ord_m a = 10$ and $ab\equiv 1\pmod m$ If $ab \equiv 1 \pmod {m}$ and if $ord_ma=10$, find $ord_mb^2$.
Could somebody give me a hint? What I know is that $ab \equiv 1 \pmod {m}$ can be used when finding the multiplicative inverse. Would this mean that $10$ is the multiplicative inverse, so $b=10$? If so, what does this tell me, and how does it help me find $ord_mb^2$?
Thanks.
| $\newcommand{\ord}{\operatorname{ord}}$No, $\ord_ma=10$ says that $a^{10}\equiv 1\pmod m$, but if $1\le n < 10$, then $a^n\not\equiv 1\pmod m$. That is, $10$ is the smallest positive integer $n$ such that $a^n\equiv 1\pmod m$. Similarly, $\ord_mb^2$ is the smallest positive integer $n$ such that $\left(b^2\right)^n\equiv 1\pmod m$, i.e., such that $b^{2n}\equiv 1\pmod m$.
You know that $ab\equiv 1\pmod m$. Clearly this implies that $(ab)^n\equiv 1\pmod m$ for all $n$, and in particular, $(ab)^{10}\equiv 1\pmod m$. But $1\equiv (ab)^{10}=a^{10}b^{10}\equiv 1\cdot b^{10}\pmod m$, so $b^{10}\equiv 1\pmod m$, and therefore $\left(b^2\right)^5\equiv 1\pmod m$. Thus, $\ord_mb^2\le 5$.
Is it possible that $\ord_mb^2<5$? No: it’s a basic theorem that if $c^n\equiv 1\pmod m$, then $\ord_mc\mid n$. We know that $\left(b^2\right)^5\equiv 1\pmod m$, so $\ord_mb^2\mid 5$. $5$ is prime, so $\ord_mb^2$ must be either $1$ or $5$. If it were $1$, that would mean that $b^2\equiv 1\pmod m$. What would that tell you about $a^2$? And why is that impossible?
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Frequency of a trigonometric function - Where is my mistake? I need to find the frequency of the following trigonometric function.$$y=\sin^4(x)+\cos^4(x)$$
The "answers" section says the answer is: $$F_y=\frac{\pi}{2}$$
This is what i did:
Finding $\sin(x)^4$ frequency (I'll call it F1):
$$\cos(2x)=1-\sin^2(x)$$
$$\sin^2(x)=\frac{1-\cos(2x)}{2}$$
$$\sin^4(x)=\frac{\cos^2(2x)-2\cos(2x)+1}{4}=\frac{cos^2(2x)+4\sin^2(x)-1}{4}$$
Finding $\cos(2x)^2$ frequency:
$$\cos(4x)=2\cos^2(2x)-1$$
$$\cos^2(2x)=\frac{\cos(4x)+1}{2}$$
$$f_1=\frac{2\pi}{4}=\frac{\pi}{2}$$
Finding $\sin(x)^2$ frequency:
$$\cos(2x)=1-2\sin^2(x)$$
$$\sin^2(x)=\frac{1-\cos(2x)}{2}$$
$$f_2=\frac{2\pi}{2}=\pi$$
$$F_1: \frac{f_1}{f_2}=\frac{\frac{\pi}{2}}{\pi}=\frac{1}{2}$$
$$F_1=\frac{\pi}{2}\times2=\pi$$
Finding $cos(x)^4$ frequency (I'll call it F2):
$$\cos(2x)=2\cos^2(x)-1$$
$$\cos^2(x)=\frac{\cos(2x)+1}{2}$$
$$\cos^4(x)=\frac{\cos^2(2x)+2\cos(2x)+1}{4}$$
Finding $\cos(2x)$ frequency (we already have $\cos(2x)^2$ frequency - f1):
$$f_3=\frac{2\pi}{2}=\pi$$
$$F_2: \frac{f_1}{f_3}=\frac{\frac{\pi}{2}}{\pi}=\frac{1}{2}$$
$$F_2=\frac{\pi}{2}\times2=\pi$$
Finding $y$'s frequency:
$$F_y: \frac{F_1}{F_2}=\frac{\pi}{\pi}=\frac{1}{1}$$
$$F_y=\pi\times1=\pi$$
| $$\sin^4\left(x+\frac{\pi}{2}\right)=\left(\sin x\cos\left(\frac{\pi}{2}\right)+\sin\left(\frac{\pi}{2}\right)\cos x\right)^4=\cos^4x$$
$$\cos^4\left(x+\frac{\pi}{2}\right)=\left(\cos x\cos\left(\frac{\pi}{2}\right)-\sin x\sin\left(\frac{\pi}{2}\right)\right)^4=(-\sin x)^4=\sin^4 x$$
Thus, the period of $\,\sin^4 x+\cos^4x\,$ indeed is not more than $\,\pi/2\,$
| {
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Expected Number of Successes in a Sample
$200$ calculators are ordered and of those $200$, $20$ are broken. $10$ calculators are selected at random. Calculate the expected value of broken calculators in the selection.
Solution:
Chance of broken calculator: $\dfrac{1}{10}$.
Do I need to calculate the odds of $0$ - $10$ calculators being broken, multiply the probabilities with the respective $0$ - $10$ and add those together?
For example:
\begin{align}
{10 \choose 0} \cdot \left(\frac{1}{10}\right)^0 \cdot \left(\frac{9}{10}\right)^{10} &= 0.3487 \tag{0}, \\
{10 \choose 1} \cdot \left(\frac{1}{10}\right)^1 \cdot \left(\frac{9}{10}\right)^9 &= 0.3874 \tag{1}, \\
\vdots & \\
{10 \choose 10} \cdot \left(\frac{1}{10}\right)^{10} \cdot \left(\frac{9}{10}\right)^0 &= 1 \cdot 10^{-10}. \tag{10}
\end{align}
Adding them together:
\begin{align}
E[X] = (0) \cdot 0 + (1) \cdot 1 + \ldots + (10) \cdot 10.
\end{align}
Or is it enough to calculate
\begin{align}
E[X] &= (1+2+3+4+5+6+7+8+9+10) \cdot \left(\frac{1}{10}\right).
\end{align}
Thanks.
| (See note below for the correct answer.)
Let $X_k = 1$ if the $k$th calculator is broken, and $0$ otherwise.
Assuming independence of 'brokenness', we have the probability of the $k$th calculator being broken is $P(X_k = 1) = \frac{20}{200} = \frac{1}{10}$.
Then the expected number of broken calculators is
$$E (\sum_{k=1}^{10} X_k) = \sum_{k=1}^{10}1 P(X_k) = 10 \frac{1}{10} = 1$$
NOTE: The above relies on the randomness of selection, not of any characteristic of 'brokenness'. Hence the reasoning is incorrect, even if the answer is correct. Here is a correct answer with correct reasoning.
Let $J = \{1,...,200\}$ represent the calculators. Let $B \subset J$, with $|B| = 20$, represent the (indices of) the broken calculators. Let $n = 10$ represent the number of calculators selected.
Our sample space is $\Omega = \{I \subset J | |I| = n \}$. Assuming the sample is selected uniformly, we have the probability $P(I) = \frac{1}{|\Omega|}$ for all $I \in \Omega$.
As above, let $X_k = 1$ if the $k$th calculator is broken, and $0$ otherwise, that is $X_k = 1_{B}(k)$. Obviously, we have $\sum_{k \in J} X_k = |B|$. Let $I \in \Omega$, then $\sum_{k \in I} X_k$ represents the number of broken calculators in the sample $I$.
We wish to compute $E (\sum_{k \in I} X_k)$, and since the distribution is uniform, we have $E (\sum_{k \in I} X_k) = \sum_{I \in \Omega} \sum_{k \in I} X_k \frac{1}{|\Omega|}$.
To simplify this expression, count the number of times a given $X_k$ occurs in the above summation. There are a total of $n|\Omega|$ '$X_*$s, so any particular $X_k$ occurs $\frac{n|\Omega|}{|J|}$ times. Consequently we have
$$E (\sum_{k \in I} X_k) = \sum_{I \in \Omega} \sum_{k \in I} X_k \frac{1}{|\Omega|} = \frac{n|\Omega|}{|J|} \frac{1}{|\Omega|} \sum_{k \in J} X_k = \frac{n|B|}{|J|} = \frac{10\cdot 20}{200} = 1$$
| {
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How many distinct real root does the equation $x^{4}-x^{3}\cdot\sin(x)-x^{2}\cdot\cos(x)=0 $ have? How many distinct real root does the equation $x^{4}-x^{3} \cdot \sin(x)-x^{2} \cdot \cos(x)=0 $ have?
Is there any quick solution(less than 2 minutes)?
| You can factor out $x^2$ to get $x^2(x^2-x \sin x - \cos x ) = 0$. So, clearly, there is a root of multiplicity $2$ at $x=0$. Letting $f(x) = x^2-x \sin x - \cos x $, we note that $f$ is even, $f(0) <0$, $\lim_{x\to\infty} f(x) = \infty$ and $f'(x)> 0$ when $x>0$. Consequently $f$ has one real root in $(0,\infty)$.
So the answer is 3 distinct real roots.
| {
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Iterated Integral requiring u-substitution This is an iterated integral. I have tried solving it several times using u-substition, but I am not getting the correct answer. My latest result is (4/15)(10^(5/2)-33). Obviously something is off, but what? Could you please show me some steps and your final answer so I can work it out on my own and make sure I get it right?
Calculate the iterated integral:
$$
\int_0^3\int_0^1 4xy\sqrt{x^2+y^2}~dy~dx
$$
My latest attempt:
u=x^2+y^2 _ du=2ydy _ @y=1, u=x^2+1 _ @y=0, u=x^2
int(0 to 3)[int(x^2 to x^2+1) 2xu^(1/2)du]dx = int(0, 3)[4/3 * xu^(3/2) for u= x^2 to u=x^2+1]dx = 4/3 * int(0 to 3) [x(x^2+1)^(3/2) - x(x^2)^(3/2)]dx
= 4/3 int(0 to 3) [x(x^2+1)^(3/2)]dx - 4/3 int(0 to 3) [x^4]dx
v= x^2+1 _ dv=2xdx _ @x=3, v=10 _ @x=0, v=1
4/3 int(1 to 10) [1/2 * v^(3/2)]dv - 4/3 int(0 to 3) [x^4]dx
= 2/3[2/5 * v^(5/2) for v=1 to v=10] - 4/3[1/5 * x^5 for x=0 to x=3]
=4/15(10^(5/2) - 1) - 4/15 (32 - 0) = 4/15(10^(5/2) - 33)
| You could start with letting
$$u = \sqrt {{x^2} + {y^2}}$$
so that
$$4\int {xy\sqrt {{x^2} + {y^2}} dy = \frac{4}{3}x{{({x^2} + {y^2})}^{3/2}}}.$$
| {
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Reason behind particular induction step I'm wondering why and how you get some of these steps in an Induction proof.
Okay, so the question is to prove the following statement by induction:
$1 - x + x^2 - x^3$ $+ ... +$ $x^{2n-2} = \frac{x^{2n-1}+1}{x+1}$
The first step that sort of confuses me is having a base case of $n=2$. Why would you start out with that, and not $n=1$? Maybe it's just trivial to show the $n=1$ base case.
Thes next step that also confuses me would be when we show that this is true for $n = k+1$
We write it out as:
$1 - x + x^2 - x^3$ $+ ... +$ $x^{2k-2} - x^{2k-1} + x^{2k} = \frac{x^{2k-1}+1}{x+1} - x^{2k-1} + x^{2k}$
I don't understand how two terms were added ($- x^{2k-1} + x^{2k}$)
Any sort of explanation of clarification would be helpful! Thanks.
| Let $P(n)$ be the statement
$$1-x+x^2-x^3+\ldots+x^{2n-2}=\frac{x^{2n-1}+1}{x+1}\;;\tag{1}$$
you want to prove that $P(n)$ is true for all $n\ge 2$. (I’ll come back to the choice of $2$ later.) For the induction step you assume $P(k)$ for some $k\ge 2$ and try to prove $P(k+1)$. Let’s see what those statements really are. $P(k)$ is
$$1-x+x^2-x^3+\ldots+x^{2k-2}=\frac{x^{2k-1}+1}{x+1}\;,$$
obtained by substituting $k$ for $n$ in $(1)$. $P(k+1)$ is obtained similarly, by substituting $k+1$ for $n$ in $(1)$, so it’s
$$1-x+x^2-x^3+\ldots+x^{2(k+1)-2}=\frac{x^{2(k+1)-1}+1}{x+1}\;.$$
This can be simplified. First, the righthand side is clearly $$\frac{x^{2k+1}+1}{x+1}\;.$$ The lefthand side is $1-x+x^2-x^3+\ldots+x^{2k}$; if you display a couple more terms, working backwards from the end, you’ll see that it’s
$$1-x+x^2-x^3+\ldots+x^{2k-2}-x^{2k-1}+x^{2k}\;,$$
which is $$\Big(1-x+x^2-x^3+\ldots+x^{2k-2}\Big)-x^{2k-1}+x^{2k}\;.$$ Thus, $P(k+1)$ is the statement
$$1-x+x^2-x^3+\ldots+x^{2k-2}-x^{2k-1}+x^{2k}=\frac{x^{2k+1}+1}{x+1}\;,$$
which can be usefully parenthesized as
$$\Big(1-x+x^2-x^3+\ldots+x^{2k-2}\Big)-x^{2k-1}+x^{2k}=\frac{x^{2k+1}+1}{x+1}\;.\tag{2}$$
This is useful because $P(k)$, the induction hypothesis, lets us replace
$$1-x+x^2-x^3+\ldots+x^{2k-2}$$
by $\dfrac{2^{2k-1}+1}{x+1}$ in $(2)$ to see that $P(k+1)$ is equivalent to the claim that
$$\frac{2^{2k-1}+1}{x+1}-x^{2k-1}+x^{2k}=\frac{x^{2k+1}+1}{x+1}\;,$$
which is easily verified by a little elementary algebra.
I don’t know why the induction was started at $n=2$; presumably the theorem was stated that way, as the assertion that $P(n)$ holds for every $n\ge 2$. It could just as well have been started at $n=1$: since $2\cdot1-2=0$, $P(1)$ is
$$1=\frac{x^1+1}{x+1}=1\;,$$
which is certainly true.
| {
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$\sum k! = 1! +2! +3! + \cdots + n!$ ,is there a generic formula for this? I came across a question where I needed to find the sum of the factorials of the first $n$ numbers. So I was wondering if there is any generic formula for this?
Like there is a generic formula for the series:
$$ 1 + 2 + 3 + 4 + \cdots + n = \frac{n(n+1)}{2} $$
or
$$ 1^{2} + 2^{2} + 3^{2} + 4^{2} + \cdots + n^{2} = \frac{n(n+1)(2n + 1)}{6} $$
Is there is any formula for:
$$ 1! +2! +3! + 4! + \cdots + n! $$
and
$$ {1!}^2 +{2!}^2 +{3!}^2 + \cdots + {n!}^2 $$?
Thanks in advance.
If not, is there any research on making this type of formula?
Because I am interested.
| In addition to the special functions given by J.M., an asymptotic expansion can be computed
$$
\begin{align}
\sum_{k=0}^n k!
&=n!\left(\frac11+\frac1n+\frac1{n(n-1)}+\frac1{n(n-1)(n-2)}+\dots\right)\\
&=n!\left(1+\frac1n+\frac1{n^2}+\frac2{n^3}+\frac5{n^4}+\frac{15}{n^5}+O\left(\frac1{n^6}\right)\right)\\
&=\sqrt{2\pi n}\frac{n^n}{e^n}\left(1+\frac{13}{12n}+\frac{313}{288n^2}+\frac{108041}{51840n^3}+\frac{12857717}{2488320n^4}+O\left(\frac1{n^5}\right)\right)
\end{align}
$$
As with most asymptotic expansions, the series does not converge, and cannot be used to get an exact answer, but it gives a good approximation.
Edit: I forgot to give
$$
\begin{align}
\sum_{k=0}^nk!^2
&=n!^2\left(\frac11+\frac1{n^2}+\frac1{n^2(n-1)^2}+\frac1{n^2(n-1)^2(n-2)^2}+\dots\right)\\
&=n!^2\left(1+\frac1{n^2}+\frac1{n^4}+\frac2{n^5}+\frac4{n^6}+\frac{10}{n^7}+O\left(\frac1{n^8}\right)\right)\\
&=2\pi\frac{n^{2n+1}}{e^{2n}}\left(1+\frac1{6n}+\frac{73}{72n^2}+\frac{1049}{6480n^3}+\frac{157541}{155520n^4}+O\left(\frac1{n^5}\right)\right)
\end{align}
$$
| {
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How to calculate the maximum value of: $\frac{25x}{x^2+1600x+640000}$? Wolfram says it's 800, but how to calculate it?
$$
\frac{25x}{x^2+1600x+640000}
$$
| Let $$\frac{25x}{(x^2+1600x+640000)}=y$$
or $$x^2y+x(1600y-25)+640000y=0$$
This is a quadratic equation in $x$.
As $x$ is real, the discriminant $(1600y-25)^2-4\cdot y\cdot 640000y\ge 0$
On simplification, $-128y+1\ge 0\implies 128y\le 1\implies y\le \frac1 {128}$
So, the maximum value of $y=\frac{25x}{(x^2+1600x+640000)}$ is $\frac1 {128}$
The value of $x$ for the maximum value of $y$ is $-\frac{1600y-25}{2y}$ where $y=\frac1 {128}$,
so $x$ will be $\frac{25-1600\cdot \frac 2{128} }{\frac2{128}}=\frac{25\cdot 128-1600}2=800$
Observe that $y$ does not have any lower limit\minimum value.
This approach can be applied to the expression like $\frac{ax^2+bx+c}{Ax^2+Bx+C}$
Reference: Minimum value of given expression
| {
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Area enclosed by the curve $x^2+y^2=|x|+|y|$.
Find the area enclosed by the curve $x^2+y^2=|x|+|y|$ on the coordinate plane.
I have no idea how this curve looks like, therefore can't find the area. Please help. Thank you.
| Wolfram|Alpha shows you how the curve looks like. To obtain the area, first note that by symmetry we can concentrate on the first quadrant. There the equation is equivalent to $x^2 + y^2 = x+y\iff y^2 - y + (x^2-x) = 0$, so
\[
y_{1,2} = \frac 12 \pm \sqrt{\frac 14 - (x^2-x)}
\]
The randicand is nonnegative for $x^2 - x \le \frac 14$, that is $0 \le x \le \frac 12(1 + \sqrt 2)$ (note that we are in the first quadrant, hence $x \ge 0$). $y_2$ is positive if moreover $x^2-x \ge 0$, that is $x \ge 1$. So the area is
\begin{align*}
A &= 4 \left(\int_0^1 \left(\frac 12 + \sqrt{\frac 14 - (x^2-x)}\right)\,dx
+ \int_1^{\frac 12(1 +\sqrt 2)} 2\sqrt{\frac 14 - (x^2 - x)}\, dx\right)
\end{align*}
| {
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How to find the roots of $x^4 +1$ I'm trying to find the roots of $x^4+1$. I've already found in this site solutions for polynomials like this $x^n+a$, where $a$ is a negative term. I don't remember how to solve an equation when $a$ is a positive term as the equation above.
Thanks
| We have:
$$\begin{align}x^4+1 =0 \\
x^4+1+2x^2-2x^2 =0 \\
(x^2+1)^2-(\sqrt{2}x)^2 =0 \\
(x^2+1+\sqrt{2}x)(x^2+1-\sqrt{2}x) =0 \end{align}$$
Then on solving these factors by quadratic formula, we will have:
$$x= \frac{-1}{\sqrt{2}}+\frac{i}{\sqrt{2}} , \frac{-1}{\sqrt{2}}-\frac{i}{\sqrt{2}}, \frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}, \frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}$$
| {
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Fibonacci conjecture: $(F_{n+5})^2 - (F_n)^2 = 3((F_{n+3})^2 - (F_{n+2})^2) + 8 F_{n+2} F_{n+3} $. So this is the question I have
The Fibonacci sequence is a recurrence system given by $$F_1 = 1, \ F_2 = 1, \ F_{n+2} = F_{n+1} + F_n \qquad (n = 1, 2, 3, \ldots).$$ This question concerns the following conjecture: $$(F_{n+5})^2 - (F_n)^2 = 3((F_{n+3})^2 - (F_{n+2})^2) + 8 F_{n+2} F_{n+3} \qquad (n = 1, 2, 3, \ldots).$$ (a) Confirm that the conjecture is true when $n = 6$.
(b) Prove that the conjecture is true for all integers $n \geq 1$.
But when I go through the solution I don't understand one line (separated and marked).
(b) Let $n$ be any integer such that $n \geq 1$.
$$\begin{align}
\left( F_{n+5} \right)^2 - \left( F_n \right)^2
&= \left( F_{n+5} - F_n \right) \left( F_{n+5} + F_n \right) \\
&= \left( F_{n+4} + F_{n+3} - F_{n+2} + F_{n+1} \right) \left( F_{n+4} + F_{n+3} + F_{n+2} - F_{n+1} \right) \\
\\
&= \left( \color{red}{2} F_{n+3} + F_{n+1} \right) \left( 2 F_{n+3} + 2 F_{n+2} - F_{n+1} \right) \tag{*} \\
\\
&= \left( 2 F_{n+3} - F_{n+3} - F_{n+2} \right) \left( 2 F_{n+3} + 2 F_{n+2} - F_{n+3} + F_{n+2} \right) \\
&= \left( 3 F_{n+3} - F_{n+2} \right) \left( F_{n+3} + 3 F_{n+2} \right) \\
&= 3 \left( \left( F_{n+3} \right)^2 - \left( F_{n+2} \right)^2 \right) + 8 F_{n+2} F_{n+3}
\end{align}$$
Hence, the conjecture is true for all integers $n \geq 1$
In the marked line, where does the red $\color{red}{2}$ come from?
| Use $F_{n+4}-F_{n+2}=F_{n+3}$ in the first factor and $F_{n+4}=F_{n+3}+F_{n+2}$ in the second factor to obtain the asterisked line from the preceding one.
| {
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Showing $\lim_{n\rightarrow\infty}\sqrt[3]{n^3+n^2}-\sqrt[3]{n^3+1}\rightarrow\frac{1}{3}$ $$\lim_{n\rightarrow\infty}\sqrt[3]{n^3+n^2}-\sqrt[3]{n^3+1}\rightarrow\frac{1}{3}$$
I tried to say we can erase the $1$ from the equation, as it's a constant. But I don't know how to do the rest without running into this mistake: $$\lim_{n\rightarrow\infty}\sqrt[3]{n^3+n^2}-n=\frac{\sqrt[3]{\frac{n^3}{n^3}+\frac{n^2}{n^3}}-\frac{n}{n}}{\frac{1}{n}}=\frac{1-1}{0}$$
| You should use that $a^3-b^3=(a-b)(a^2+ab+b^2)$. Take $a=\sqrt[3]{n^3+n^2}$, $b=\sqrt[3]{n^3+1}$ and then multiply your expression by $(a^2+ab+b^2)/(a^2+ab+b^2)$. Then use the trick you are trying to use.
| {
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"timestamp": "2023-03-29T00:00:00",
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Geometry Prove - two perpendicular lines in a circle In a circle of radius r, two lines (AB and CD) are perpendicular to each other and meet at X.
Show that:
| a^2 + y^2 = r^2\\
B: b^2 + y^2 = r^2 \\
C: x^2 + c^2 = r^2 \\
D: x^2 + d^2 = r^2
$$
From $A$ and $B$ one can find that $a = -b$, and analogously $c = -d$. Then
$XA^2 + XB^2 + XC^2 + XD^2 = (a-x)^2 + (b-x)^2 + (c-y)^2 + (d-y)^2 = \\
a^2+b^2+c^2+d^2-2x(a+b)-2y(c+d)+2x^2+2y^2 = a^2+b^2+c^2+d^2+2x^2+2y^2$
which is sum of all LHS for the equations $A-D$ above and consequently $4r^2$
| {
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Trying to find $\det (B)$ Let $A=\left(\begin{matrix}1&1&-1\\-1&1&1\\1&-1&1\end{matrix} \right)$,and ${A}^{T}B{\left( \cfrac{1}{2}{A}^{T}\right)}^{T}-8{A}^{-1}B=I$, How to compute $\left|B \right|$?。
| Consider the permutation matrix $$P = \pmatrix{0 & 1 & 0\cr 0 & 0 & 1\cr 1 & 0 & 0\cr}$$
Note that $P^3 = I$ and $A = I + P - P^2$. Since $A P^2 = P^2 + I - P = -A + 2 I$, we see
that $A^{-1} = (P^2 + I)/2$, while $A^T = I + P^2 - P$. It seems reasonable, then, that a solution $B$ might also be a linear combination of $I$, $P$ and $P^2$. In fact, if
$B = a I + b P + c P^2$ we get
$$ \eqalign{\frac{1}{2} A^T B A &- 8 A^{-1} B = \frac{1}{2} (I + P^2 - P)(a I + b P + c P^2)(I + P - P^2) - 4 (P^2 + I) (aI + b P + cP^2)\cr
&= \frac{-5 a - 9 b-c}{2} I + \frac{-a-5b-9c}{2} P + \frac{-9 a - b - 5 c}{2} P^2} $$
For this to be $I$, we just have to solve the $3 \times 3$ system
$$ \eqalign{ 5a + 9 b + c &= -2\cr
a + 5 b + 9 c &= 0\cr
9 a + b + 5 c &= 0\cr}$$
obtaining $a = -2/45, b = -19/90, c = 11/90$.
Now note that the eigenvalues of $P$ are the three cube roots of $1$, say $1, \omega, \omega^2$; the eigenvalues of $B = a I + b P + c P^2$ are then $a+b+c$, $a+b\omega + c \omega^2$ and $a + b\omega^2 + c \omega$, so (using $1 + \omega + \omega^2 = 0$)
$$\eqalign{\det(B) &= (a+b+c)(a+b\omega + c \omega^2)(a + b \omega^2 + c \omega)\cr
&= a^3 + b^3 + c^3 - 3 a b c\cr} $$
Substituting the values of $a,b,c$ gives us the answer $-1/90$.
EDIT: Note that this shows that there is a $B$ with determinant $-1/90$, but not that this is the only solution.
The calculation can be shortened: if $B$ is of the form $aI + bP + cP^2$, then it commutes with $A$. So we can simplify the equation to
$$ \left(\frac{A^T A}{2} - 8 A^{-1}\right) B = I$$
Thus $B = \left(\dfrac{A^T A}{2} - 8 A^{-1}\right)^{-1}$, and
$\det B = 1/\det \left(\dfrac{A^T A}{2} - 8 A^{-1}\right)$.
We just have to calculate $\det \left(\dfrac{A^T A}{2} - 8 A^{-1}\right) = -90$ to say $\det(B)=-1/90$.
| {
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"url": "https://math.stackexchange.com/questions/241116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the fixed points of the two systems What are the fixed points of the two non linear systems below?
\begin{align}
x(t)& = x(3-x-2y)\\
y(t)& = y(2-x-y)
\end{align}
I know that $(0,0)$, $(0,2)$, $(3,0)$, $(1,1)$ are the fixed points. However, I don't understand some of the steps to obtain this. Any help will be appreciated.
| Unless I missed something in the question, the fixed points are given by solutions of $x = x(3-x-2y)$, $y=y(2-x-y)$.
If $x=0$, the first equation is trivially satisfied, and the second gives $y = y(2-y)$, or in other words $y(y-1) = 0$. Hence $(0,0)$ and $(0,1)$ are solutions.
If $y=0$, the second equation is trivially satisfied, and the first gives $x=x(3-x)$ or in other words $x(2-x) = 0$. Hence $(0,0)$ and $(2,0)$ are solutions.
Now suppose $x\neq 0$ and $y\neq 0$. Divide the equations by $x,y$ respectively to get $1 = 3-x-2y$, $1 = 2-x-y$, or equivalently $x+2y-2=0$ and $1-x-y=0$. Adding the first equation to two times the second gives $x=0$ which cannot be, hence there are no solutions with $x\neq 0$ and $y\neq 0$.
Hence the only solutions are $(0,0), (0,1),(2,0)$.
Addendum: Following Hans comments below, I will look for stationary points of the original system. This means solving $0 = x(3-x-2y)$, $0=y(2-x-y)$ instead.
The same approach works: If $x=0$, then the first equation is trivially satisfied, and the second becomes $0=y(2-y)$. The solutions are $(0,0), (0,2)$.
If $y=0$, then the second equation is trivially satisfied, and the first becomes $0 = x(3-x)$. The solutions are $(0,0), (3,0)$.
Now suppose $x\neq 0$ and $y\neq 0$. Divide the equations by $x,y$ respectively to get $3-x-2y=0$, $2-x-y = 0$. Subtracting the first equation from the second yields $y=1$ from which it follows that the solution is $(1,1)$.
Hence the solutions are $(0,0), (3,0), (0,2), (1,1)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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What are $2222^{5555}+5555^{2222} \pmod 7$ and $9^{2n+1}+8^{n+2} \pmod{73}$? Tell me hint for solve :
1) $ 2222^{5555}+5555^{2222} \equiv \mathord? \pmod 7$
2) $ 9^{2n+1}+8^{n+2} \equiv \mathord ?\pmod{73}$
thank you.
| (2)I was trying to figure out how the problem came to being.
Observe that $73=8\cdot9+1$
Replacing $8$ with $a$ we get $a(a+1)+1=a^2+a+1$
Now as the root of $a^2+a+1=0$ are the imaginary cube roots of $1,$
let $\omega$ be one of the imaginary cube roots of $1.$
So, $\omega^3=1\implies \omega^{3n}=1$
$\omega^{3n+2m+n}=\omega^{2m+n}$
Now,$\omega^{3n+2m+n}=(\omega^2)^{m+2n}=\{-(1+\omega)\}^{m+2n}=(-1)^m (1+\omega)^{m+2n}$
So, $\omega^{2m+n}=(-1)^m (1+\omega)^{m+2n}$
So, $\omega$ is a root of $x^{2m+n}-(-1)^m (1+x)^{m+2n}=0$
$\implies(x-\omega)\mid \{x^{2m+n}-(-1)^m (1+x)^{m+2n}\}$
Similarity, $\omega^2$ is a root,$\implies (x-\omega^2)\mid \{x^{2m+n}-(-1)^m (1+x)^{m+2n}\}$
But we know, $(x-\omega)(x-\omega^2)=x^2+x+1$
So, $$x^{2m+n}-(-1)^m (1+x)^{m+2n}\equiv0\pmod{x^2+x+1}$$
Once derived, it can be verified using Congruence like following:
As $1+x\equiv-x^2\pmod{x^2+x+1}$ and $x^3\equiv1\pmod{x^2+x+1},$
$(1+x)^{m+2n}\equiv(-x^2)^{m+2n}\pmod{x^2+x+1}\equiv(-1)^{m+2n} x^{2m+4n}\equiv(-1)^m x^{2m+n} (x^3)^n$
or, $(1+x)^{m+2n}\equiv (-1)^m x^{2m+n}\pmod{x^2+x+1}$
or, $(1+x)^{m+2n}(-1)^m\equiv x^{2m+n}$
Here in this problem, $x=8,m=1$
(1)$$2222\equiv3\pmod 7\implies 2222^5\equiv 3^5\pmod 7\equiv3^3\cdot3^2\equiv(-1)9\equiv5$$
$$5555\equiv4\pmod 77\implies 5555^2\equiv4^2\pmod7\equiv 2$$
So, $$2222^5+5555^2\equiv5+2\pmod 7\equiv 0$$
But , $$(2222^5+5555^2)\mid\{(2222^5)^{1111}+(5555^2) ^{1111} \} $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Integer solutions to $\sqrt{a} + \sqrt{b} = \sqrt{c}$ An example I can see using basic algebra is $\sqrt{2} + \sqrt{8} = \sqrt{18}$, but is there a general method to find integer solutions to the problem? Another question: say you are given the value of $c$; can you find the values of $a$ and $b$?
| Let $c=d n^2$ with $n\ge 0$ and with $d \gt 0$ square-free, i.e. $d$ has no repeated prime factors. $d$ is the product of the individual prime factors of $c$ which occur an odd number of times.
Then $a=d k^2$ and $b=d (n-k)^2$ with $0 \le k \le n$ are the only solutions to $\sqrt a + \sqrt b = \sqrt c$, and there are $n+1$ such pairs.
This is similar to N.S.'s solution, except for the restriction on $d$.
$18 = 2^1 \times 3^2$ so $d=2$ and thus $n=3$. So there are $n+1=4$ solutions to $\sqrt a + \sqrt b = \sqrt{18}$ namely
*
*$\sqrt{0^2 \times 2} + \sqrt{3^2 \times 2} = \sqrt{3^2 \times 2}$ i.e.$\sqrt{0} + \sqrt{18} = \sqrt{18}$
*$\sqrt{1^2 \times 2} + \sqrt{2^2 \times 2} = \sqrt{3^2 \times 2}$ i.e.$\sqrt{2} + \sqrt{8} = \sqrt{18}$
*$\sqrt{2^2 \times 2} + \sqrt{1^2 \times 2} = \sqrt{3^2 \times 2}$ i.e.$\sqrt{8} + \sqrt{2} = \sqrt{18}$
*$\sqrt{3^2 \times 2} + \sqrt{0^2 \times 2} = \sqrt{3^2 \times 2}$ i.e.$\sqrt{18} + \sqrt{0} = \sqrt{18}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
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Factor $1+x+x^2+x^3+...+x^{14}$ In a previous task, I was asked to factor $1+x+x^2+x^3$ for $x \in \mathbb{R}$, which I accomplished by solving
$1+x+x^2+x^3 = 0 \to $
$1+x(1+x+x^2) = 0 \to $
$x(1+x+x^2) = -1$
which has a solution $x = -1$, and thus I knew $(x+1)$ was a factor. A bit of guesswork gave me $(x+1)(x^2+1)$.
Now I'm asked to factor $1+x+x^2+x^3+...+x^{14}$ for $x \in \mathbb{R}$ and I'm a bit stuck. Again, we have the implication $x(1+x+x^2+...+x^{13}) = -1$, for which $x=-1$ is a solution, so again we have a factor $x+1$. But now I cannot apply guesswork to determining the rest of the factors, so I feel there is some kind of conclusion I can draw about the powers (perhaps their parity) to solve this problem?
| Note that the polynomial has 15 terms, so try grouping it in 5 groups of 3:
$$ 1 + x + x^2 + \cdots + x^{14} =\\ =(1+x+x^2) + x^3(1+x+x^2)+ x^6(1+x+x^2) + x^9(1+x+x^2)+ x^{12}(1+x+x^2)$$
and then factor out $(1+x+x^2)$
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate the sum of series $\sum\limits_{i=0}^{n-1} i2^i$
Possible Duplicate:
How to compute the formula $\sum \limits_{r=1}^d r \cdot 2^r$?
How can I calculate precise value of that series: $\sum\limits_{i=0}^{n-1} i2^i$ ?
So far, I tried to differentiate the $ \sum\limits_{i=0}^{n} 2^i = 2^{i-1} - 1 $ series, but by result $2^n(n+2)$ isn't correct according to Wolfram ($2^n (n-2)+2$).
| $$\begin{array}{rll}
S &=1\cdot2^1+&2\cdot2^2+3\cdot2^3+\cdots+(n-2)\cdot2^{n-2}+(n-1)\cdot2^{n-1} \\
2S &= &1\cdot2^2+2\cdot2^3+\cdots+(n-3)\cdot2^{n-2}+(n-2)\cdot2^{n-1}+(n-1)\cdot2^{n}
\end{array}$$
Subtracting,
$$S-2S=1\cdot2^1+(2-1)\cdot2^2+\cdots+\{(n-2)-(n-3)\}\cdot2^{n-2}+\{(n-1)-(n-2)\}\cdot2^{n-1}-(n-1)2^n=(2^1+2^2+\cdots+2^{n-1})-(n-1)2^n=2\left(\frac{2^{n-1}-1}{2-1}\right)-(n-1)2^n=2^n\{1-(n-1)\}-2$$
So, $S=2+2^n(n-2)$
Refernce: Arithmetico-geometric series
| {
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Inequality. $(a^2+bc)(b^2+ca)(c^2+ab) \geq abc(a+b)(b+c)(c+a)$ Let $a,b,c$ be three real positive(strictly) numbers. Prove that:
$$(a^2+bc)(b^2+ca)(c^2+ab) \geq abc(a+b)(b+c)(c+a).$$
I tried :
$$abc\left(a+\frac{bc}{a}\right)\left(b+\frac{ca}{b}\right)\left(c+\frac{ab}{c}\right)\geq abc(a+b)(b+c)(c+a) $$
and now I want to try to prove that for example $$a+\frac{bc}{a} \geq a+b$$
but I don't know if is is a good idea.
Thanks:)
| Here's another approach that is less intensive. Show that
$$ (a^2 + bc) ( b^2 + ca) \geq ab(c+b)(c+a) $$
This is equivalent to
$$ c (a-b)^2 (a+b) \geq 0 $$
Multiply the 3 cyclic versions of the first inequality, and you get your conclusion. Equality holds if and only if $a=b=c$.
Note: The factorization of the second line is easily guessed, by observing the equality holds when $ c = 0, a = b, a = - b$.
| {
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If $\gcd(a,b)=1$, then $\gcd(a+b,a^2 -ab+b^2)=1$ or $3$. Hint: $a^2 -ab +b^2 = (a+b)^2 -3ab.$
I know we can say that there exists an $x,y$ such that $ax + by = 1$. So in this case,
$(a+b)x + ((a+b)^2 -3ab)y =1.$
I thought setting $x = (a+b)$ and $y = -1$ would help but that gives me $3ab =1.$ Any suggestions?
| $\gcd(a+b, a^2 - ab + b^2) = \gcd(a+b, (a+b)^2 - 3ab) = \gcd(a+b, 3ab)$
Since $\gcd(a,b) = 1$ hence $\gcd(a+b,a) = 1$ and $\gcd(a+b, b)$ = 1.
Thus, $\gcd(a+b, 3ab) = \gcd (a+b, 3) $ by applying the Lemma below.
Hence, $\gcd(a+b, a^2 -ab + b^2) = \gcd(a+b, 3)=1$ or 3. (This is very simple to evaluate for given values.)
Lemma If $\gcd(m,n) = 1$, then $\gcd (m, np) = \gcd(m,p)$.
Proof: This is obvious when you consider GCD via prime factorization.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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simplify summation of factorial (random walk) I suspect that the expression
$$\sum_{n=0}^N \frac{(N-2n)^2}{n!(N-n)!}$$
simplifies to
$$\frac{2^N}{(N-1)!}$$
But I cannot find the intermediate steps. Can someone give me a hint how I can deduce this result?
(The expression comes up when calculating the average final position after a random walk of $N$ steps.)
| $$
\begin{align}
\sum_{n=0}^N\frac{(N-2n)^2}{n!(N-n)!}
&=
\sum_{n=0}^N\frac{((N-n)-n)^2}{n!(N-n)!}
\\
&=
\sum_{n=0}^N\frac{(N-n)^2}{n!(N-n)!}+\sum_{n=0}^N\frac{n^2}{n!(N-n)!}-2\sum_{n=0}^N\frac{(N-n)n}{n!(N-n)!}
\\
&=
\sum_{n=0}^N\frac{n^2}{n!(N-n)!}+\sum_{n=0}^N\frac{n^2}{n!(N-n)!}-2\sum_{n=1}^{N-1}\frac1{(n-1)!(N-n-1)!}
\\
&=
2\sum_{n=0}^N\frac{n^2}{n!(N-n)!}-2\sum_{n=0}^{N-2}\frac1{n!(N-2-n)!}\;.
\end{align}
$$
Now consider
$$
\sum_{n=0}^m\binom mnq^n=(1+q)^m
$$
and thus
$$
\sum_{n=0}^m\frac{q^n}{n!(m-n)!}=\frac{(1+q)^m}{m!}\;.
$$
Applying $q\partial/\partial q$ twice yields
$$
\sum_{n=0}^m\frac{n^2q^n}{n!(m-n)!}=\frac{(1+q)^{m-1}}{(m-1)!}+q^2\frac{(1+q)^{m-2}}{(m-2)!}\;.
$$
Now setting $q=1$ yields
$$
\sum_{n=0}^N\frac{(N-2n)^2}{n!(N-n)!}=2\left(\frac{2^{N-1}}{(N-1)!}+\frac{2^{N-2}}{(N-2)!}\right)-2\frac{2^{N-2}}{(N-2)!}=\frac{2^N}{(N-1)!}\;.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the value of : $\lim_{n\to\infty}[(n+1)\int_{0}^{1}x^{n}\ln(1+x)dx]$. I am stuck on the following problem:
Find the value of : $$\lim_{n\to\infty}[(n+1)\int_{0}^{1}x^{n}\ln(1+x)dx].$$
My attempts: Let $$I_{n}= \lim_{n\to\infty}[(n+1)\int_{0}^{1}x^{n}\ln(1+x)dx]=\lim_{n\to\infty}[\ln(2)-\int_{0}^{1}\frac{x^{n+1}}{1+x}dx]$$ and now i can not progress. Please help. Thanks in advance for your time
| You can also use power series. Here's how Euler probably would have done it. I say that because the proof as I present it here is a bit informal but it can be patched up easily. We do term by term integration and then later interchange a limit and an infinite sum. Start with
$\frac{1}{1-x}=1+x+x^2+x^3+x^4+\cdots$
Replace $x$ with $-x$
$\frac{1}{1+x}=1-x+x^2-x^3+x^4-\cdots$
Integrate term by term (we will use this log series again at the bottom)
$\ln(1+x)=\int_0^x \frac{dt}{1+t}=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots$
Multiply by $x^n$
$x^n\ln(1+x)=x^{n+1}-\frac{x^{n+2}}{2}+\frac{x^{n+3}}{3}-\frac{x^{n+4}}{4}+\cdots$
Integrate it term by term again
$\int_0^1 x^n\ln(1+x)=\frac{1}{n+2}-\frac{1}{2(n+3)}+\frac{1}{3(n+4)}-\frac{1}{4(n+5)}+\cdots$
Multiply by $(n+1)$
$(n+1)\int_0^1 x^n\ln(1+x)=\frac{n+1}{n+2}-\frac{n+1}{2(n+3)}+\frac{n+1}{3(n+4)}-\frac{n+1}{4(n+5)}+\cdots$
Now take the limit as $n$ tends to infinity
$$\lim_{n\rightarrow\infty} (n+1)\int_0^1 x^n\ln(1+x)=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\cdots$$
which is the alternating harmonic series. In case you don't remember that this converges to $\ln(2)$ then go back to that natural log series and plug in $x=1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Can someone show me why this factorization is true? $$x^n - y^n = (x - y)(x^{n-1} + x^{n-2}y + \dots + xy^{n-2} + y^{n-1})$$
Can someone perhaps even use long division to show me how this factorization works? I honestly don't see anyway to "memorize this". I like to see some basic intuition behind this
| Since you said above that you already believe that the result is true, that makes me think you really want to know how to come up with the result (if it had not been presented to you).
My answer would be: play with several examples!
That is, sharpen your pencil and get busy factoring out $x^n-y^n$ for $n=1,2,3,\dots$ using long division or any other tools at your disposal. Then look at the pattern formed by the coefficients as well as the powers of $x$ and $y$:
\begin{array}{c|c}
n & x^n-y^n \\\hline
1 & x-y \\
2 & (x-y) (x+y) \\
3 & (x-y) \left(x^2+y x+y^2\right) \\
4 & (x-y) \left(x^3+y x^2+y^2 x+y^3\right) \\
5 & (x-y) \left(x^4+y x^3+y^2 x^2+y^3 x+y^4\right) \\
6 & (x-y) \left(x^5+y x^4+y^2 x^3+y^3 x^2+y^4 x+y^5\right) \\
7 & (x-y) \left(x^6+y x^5+y^2 x^4+y^3 x^3+y^4 x^2+y^5 x+y^6\right) \\
8 & (x-y) \left(x^7+y x^6+y^2 x^5+y^3 x^4+y^4 x^3+y^5 x^2+y^6 x+y^7\right) \\
9 & (x-y) \left(x^8+y x^7+y^2 x^6+y^3 x^5+y^4 x^4+y^5 x^3+y^6 x^2+y^7 x+y^8\right) \\
10 & (x-y) \left(x^9+y x^8+y^2 x^7+y^3 x^6+y^4 x^5+y^5 x^4+y^6 x^3+y^7 x^2+y^8 x+y^9\right) \\
\end{array}
Hopefully this would lead you to conjecture that the formula you gave is indeed true for all $n$. Then to prove this, you would proceed by induction.
Hope that helps.
| {
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"timestamp": "2023-03-29T00:00:00",
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Indefinite integral involving arctan How should I evaluate this definite integral? I am unable to figure out how to start.
$$\int \tan^{-1} \left(1 + x + x^{2}\right) dx $$
| Proceed by integration by parts, we get
\begin{align*}
\int \arctan \left( 1 + x + x^2 \right) dx & = x \arctan \left(1+x+x^2 \right) - \int \dfrac{x(2x+1)}{(x^2+x+1)^2+1} dx
\end{align*}
Now note that $$\left(x^2+x+1 \right)^2+1 = \left(x^2+1 \right) \left(x^2+2x+2 \right)$$
Hence, $$\dfrac{x(2x+1)}{(x^2+x+1)^2+1} = \dfrac{x}{x^2+1} - \dfrac{x}{x^2 + 2x+2}$$
Hence,
$$\int \dfrac{x(2x+1)}{(x^2+x+1)^2+1} dx = \int \dfrac{x dx}{x^2+1} - \int \dfrac{x dx}{x^2 + 2x+2}$$
$$\int \dfrac{x dx}{x^2+1} = \dfrac12 \log \left(1+x^2 \right)$$
$$\int \dfrac{x dx}{x^2 + 2x+2} = \int \dfrac{\left(x + 1 \right) dx}{\left( x+1 \right)^2+1} - \int \dfrac{dx}{\left( x+1 \right)^2+1} = \dfrac12 \log \left((x+1)^2+1\right) - \arctan(x+1)$$
Putting all this together, we get that
$$x \arctan \left( 1+x+x^2 \right) - \dfrac12 \log \left( 1+x^2\right) + \dfrac12 \log \left((x+1)^2+1\right) - \arctan(x+1)$$
Rearranging gives us
$$x \arctan \left( 1+x+x^2 \right) - \arctan(x+1) + \dfrac12 \log \left(\dfrac{(x+1)^2+1}{x^2+1}\right) + \text{ constant}$$
| {
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Do there exist any odd prime powers that can be represented as $n^4+4^n$? Well, I wrote up a solution on it, but according to the place I found the problem, it isn't quite correct. Ah, I'm simply hoping someone will point out where I got wrong.
Now, let, $n^4+4^n = p^k$, where $p$ is an odd prime and $k$ is a positive integer.
Further, $p^k \equiv 1 \pmod 2$. Therefore, $n^4 + 4^n \equiv n^4 \equiv 1 \pmod 2$, and so $n \equiv 1 \pmod 2$. $n$ must be odd.
Okay, now let, $n = 2m +1$. Substituting it in $n^4+4^n$ and using the Sophie Germain inequality, we have,
$$n^4+4\cdot4^{2m} = n^4 + 4(2^m)^4 = (n^2 + 2^n+2^{m+1}\,n)(n^2 + 2^n-2^{m+1}\,n) = p^k$$
Now, as $p^k$ can only be factorized into smaller powers of $p$, let $n^2 + 2^n+2^{m+1}\,n = p^i$, and let $n^2 + 2^n-2^{m+1}\,n = p^j$ where $i+j= k$, obviously, and $i>j$.
Now consider this:
$$\begin{align}
p^i - p^j & \equiv 0\\
2\cdot2^{m+1}\,n = 2^{m+2}\,n &\equiv 0 \pmod p
\end{align}$$
But, as $p$ is odd, $\gcd(p, 2) = 1$, so $n \equiv 0 \pmod p$.
Now look at this:
$$\begin{align}
p^i + p^j &\equiv 0 \\
2(n^2 + 2^n) &\equiv 0 \\
n^2 + 2^n &\equiv 0 \pmod p
\end{align}$$
But we just established that $n \equiv 0 \pmod p$, so $2^n \equiv 0 \pmod p$.
Therefore, let $2^n = jp$ for some integer $j$.
Now, $2^n$ is its own prime factorization, which is unique according to the Fundamental Theorem of Arithmetic and does not include $p$.
Therefore, the above statement is an impossibility!
There exist no such $p$ and $n$, and no odd prime powers can be written as $n^4+4^n$.
Ah, well, that's it.
Sorry for the tediousness of it. I've still no clue how to use $\LaTeX$.
Thank you everybody,
Cheers.
| For odd $n>1$ the expression is never prime, it can be factored into a product where each factor is the sum of two squares.
$$[((2^m)+n)^2 +(2^m)^2][(2^m)^2 +((2^m)^2-n)^2]$$
where $m = (n-1)/2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/261925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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incircle and circumcircle Given a triangle $\Delta ABC$ with circumcenter $O$, and incenter $I$.
If $∠ABC = 45°$, $OI = d$, and $c − b = d√2$,
Find the value of $\sin A$.
I've thought of using the fact that $d^2 = R(R-2r)$ where $R$ is the circumradius and $r$ is the inradius. But I didn't know where to apply it.
| keep going:$\ (4+2\sqrt{2})x^3-(2+2\sqrt{2})x^2-(2+3\sqrt{2})x+(3+\sqrt{2})=4x^3-2\sqrt{2}x^2+2\sqrt{2}x^3-2x^2-3\sqrt{2}x+3-2x+\sqrt{2}=2\sqrt{2}x^2*(\sqrt{2}x-1)+2x^2(\sqrt{2}x-1)-3(\sqrt{2}x-1)-(\sqrt{2})(\sqrt{2}x-1)=(\sqrt{2}x-1)[(2\sqrt{2}+2)x^2-(3+\sqrt{2})]$=0
$\ x=\sqrt{2}/2$
or
$\ x^2=(3+\sqrt{2})/(2\sqrt{2}+2)$ ie $\ x=\sqrt{(2\sqrt{2}-1)/2} $
now you can know $\sin A$
| {
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Showing that: $(\frac{a}{b+c})^2+(\frac{b}{a+c})^2+(\frac{c}{a+b})^2+\frac{10abc}{(a+b)(b+c)(c+a)}\ge 2$ Let a;b;c>0. Prove:
$$\left(\frac{a}{b+c}\right)^2+\left(\frac{b}{a+c}\right)^2+\left(\frac{c}{a+b}\right)^2+\frac{10abc}{(a+b)(b+c)(c+a)}\geq 2$$
I think
$$\frac{2a}{b+c}=x;\frac{2b}{c+a}=y;\frac{2c}{a+b}=z$$
We have: $xy+yz+zx+xyz=4$
$$(\frac{x}{2})^2+(\frac{y}{2})^2+(\frac{z}{2})^2+\frac{10xyz}{8} \ge 2$$
$\Leftrightarrow x^2+y^2+z^2+5xyz \ge 8$
$\Leftrightarrow x^2+y^2+z^2-5(xy+yz+zx) +12 \ge 0$
deadlock
Can you help me? Thank you very much
| You are almost there. You have $x^{2}+y^{2}+z^{2}\ge xy+yz+xz$ by matching squares. So $$x^{2}+y^{2}+z^{2}-5(xy+yz+zx)+12\ge 12-4(xy+yz+zx)\ge 0$$ In other words you need to show $xy+yz+zx\le 3$. Now by Cauchy-Schwarz the left hand side only obtain maximum of $x=ky, y=kz,z=kx$. So we have either $x=y=z$ or $x=-y, y=-z,z=-x$. The second case is impossible since $a,b,c> 0$. Therefore the maximum is obtained when $x=y=z=1$.
| {
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$N \equiv 3 (\textrm{mod } 4)$ and Collatz conjecture Can the Collatz conjecture also be interpreted as behaviour, transformation of number of form $N\equiv 3(\textrm{mod }4)$ to the form of $N\equiv 1(\textrm{mod }4)$
Because integers of the form $N\equiv 3(\textrm{mod }4)$ can be later always only ones be divided by $2$. Number of form $N\equiv 1(\textrm{mod }4)$ have much more diversity.
$3,7,11,15,19,23,29,35, \ldots, N \equiv 3(\textrm{mod }4) $(distance $4$ between numbers)
using $(3n+1)/2$ or applying $+2,+4,+6,\ldots$ turns into
$5,11,17,23,29,35,41,47,\ldots, N \equiv 1 (\textrm{mod }4)$ or $3$ (50%) (distance $6$ between numbers)
again using $(3n+1)/2$ or applying $+3,+6,+9$ turns into
$8,17,26,35,44,53,62,71,\ldots$ (distance 9 between numbers), $N(\textrm{mod }4)$ is various.
| To add to @coffemath approach: we can separate the class of odd numbers a,b in the following way such that $b=(3a+1)/2^A$ where $A$ is the value which makes b an odd integer:
$$ \begin{array} {r|c|l||r|c|l}
& k \ge 0& & && k\gt 0\\
A & a=& b= & & A & a= & b= \\
\hline \\
0 & 2^0 \cdot 2 k +0 & k & & 1 & 2^1 \cdot 2 k -1 & 6 k - 1 \\
2 & 2^2 \cdot 2 k +1 & 6 k + 1 & &3 & 2^3 \cdot 2 k -3 & 6 k - 1 \\
4 & 2^4 \cdot 2 k +5 & 6 k + 1 & &5 & 2^5 \cdot 2 k -11 & 6 k - 1 \\
6 & 2^6 \cdot 2 k +21 & 6 k + 1 & &7 & 2^7 \cdot 2 k -43 & 6 k - 1 \\
&\cdots & \cdots & && \cdots & \cdots \\
A & 2^A \cdot 2 k + r & 6 k + 1 & &A & 2^A \cdot 2 k -s & 6 k - 1 \\
\end{array} $$
where the residues in the columns below the a are dependent on A by
$$r = { 2^A -1 \over 4-1} \qquad s ={ 2^A +1 \over 4-1} $$
Of course, the entry at $A=0$ should not occur in this table, if we look only at odd a but leave it there for completeness/smoothness of the table.
(That this table covers all odd positive integers is a nice exercise, left to the reader...)
[added:]
The known "trivial" cycle is that where a=1 in the category A=2 where k=0 such that $$ 2^2 \cdot 2 k + 1 \to 6 k + 1 \quad||_{k=0}\quad 2^2 \cdot 2 \cdot 0+1 \to 1 $$
If we look also at the negative odd integers we have the cycle $-1 \to -1 $ at A=1, k=0 such that
$$ 2^1 \cdot 2 k - 1 \to 6 k - 1 \quad||_{k=0}\quad 2^1 \cdot 2 \cdot 0 -1 \to -1 $$
and the cycle $-5 \to -7 \to -5 $ by $A_1=1$, $k_1=-1$ and $A_2=2$, $k_2=-1$ :
$$ 2^1 \cdot 2 k - 1 \to 6 k - 1 \quad ||_{k=-1} \quad -5 \to -7 \\
2^2 \cdot 2 k + 1 \to 6 k + 1 \quad||_{k=-1}\quad -7 \to -5 \\ $$
The other known cycle in the negative integers is longer and not shown here.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Range of a function $f(x)$
As x varies over all real numbers, the range of the function $$f(x)=\frac{x^2-3x+4}{x^2+3x+4}$$ is (1) $[\frac{1}{7},7]$, (2) $[-\frac{1}{7},7]$, (3) $[-7,7]$ (4) $(-\infty,\frac{1}{7})\bigcup(7,\infty)$.
Trial:$$\begin{align} \frac{x^2-3x+4}{x^2+3x+4} &=\frac{(x-\frac{3}{2})^2+\frac{7}{4}}{(x+\frac{3}{2})^2+\frac{7}{4}} >0 \end{align}$$ So,(1) will be the right answer. But how I show that $\frac{1}{7} \leq f(x) \leq 7$. Please help.
Note: Originally it was stated that
$$f(x)=\frac{x^2-3x+1}{x^2+3x+1},$$
and this is what the initial answers were based on.
| Answer to modified question: Note that the line $y=1$ is an asymptote to the curve $y=f(x)$.
Calculate $f'(x)$. It turns out to be $\frac{6x^2-24}{(x^2+3x+4)^2}$. So $f(x)$ is increasing from $-\infty$ to $-2$, then decreasing until $2$, then increasing.
For negative $x$, the curve is above the asymptote, and for positive $x$ it is below the asymptote.
The maximum value (at $x=-2$) is $7$, and the minimum value (at $x=2$) is $\frac{1}{7}$. So the range is $[1/7,7]$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Continued fraction of a square root If I want to find the continued fraction of $\sqrt{n}$ how do I know which number to use for $a_0$? Is there a way to do it without using a calculator or anything like that? What's the general algorithm for computing it? I tried to read the wiki article but was overwhelmed and lost. I tried Googling but couldn't find a website that actually explained this question.
If anyone has a good site that answers these questions either, please let me know. Thanks!
| Here the easiest method to generate continued fraction for any square (or more) root. Lets take $\sqrt{5}$:
$$\sqrt{5} \approx 2,2360679775...$$
$$\sqrt{5} = 2 + 0,2360679775$$
So $2=\left \lfloor \sqrt{5} \right \rfloor$
$$\sqrt{5} =2 + x$$ with $x=0,2360679775$
$$\sqrt{5}^{2} = (x + 2)^{2} \Rightarrow 5= x^2 + 4x + 4$$
$$1= x^2 + 4x$$
$$\frac{1}{x}= x + 4 \Rightarrow x = \frac{1}{4+x}$$
So $x = \frac{1}{4+\frac{1}{4+\frac{1}{4+\frac{1}{4+\ddots}}}}$
Finally, we got: $\sqrt{5} = 2 + \frac{1}{4+\frac{1}{4+\frac{1}{4+\frac{1}{4+\ddots}}}}$.
So for any $x$, $\sqrt{x} = \left \lfloor \sqrt{x} \right \rfloor + \frac{x-\left \lfloor \sqrt{x} \right \rfloor^{2}}{2\left \lfloor \sqrt{x} \right \rfloor+\frac{x-\left \lfloor \sqrt{x} \right \rfloor^{2}}{2\left \lfloor \sqrt{x} \right \rfloor+\frac{x-\left \lfloor \sqrt{x} \right \rfloor^{2}}{2\left \lfloor \sqrt{x} \right \rfloor + \ddots}}}$
So to answer, $a_0 =\left \lfloor \sqrt{x} \right \rfloor$ (because any periodic continued fraction is smaller than 1).
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculating the integral $\int_0^{\infty}{\frac{\ln x}{1+x^n}}$ using complex analysis I need to calculate $\int_0^{\infty}{\frac{\ln x}{1+x^n}}$ $,n\geq2$ using complex analysis. I probably need to use the Residue Theorem.
I use the function $f(z)={\frac{\ln z}{1+z^n}}$ in the normal branch.
I've tried to use this contour.
Where $\theta$ is an angle so that only $z_{0}$ will be in the domain (I hope this is clear from the drawing)
I estimated $|\int_{\Gamma_{R}}|$ and $|\int_{\Gamma_{\epsilon}}|$ and showed that they must tend to $0$ when $\epsilon \rightarrow0$ and $R \rightarrow\infty$. (Is it true?)
However I'm having trouble calculating $\int_{\Gamma_{2}}$ . Does it have something to do with choosing the "right" $\theta$?
Any ideas?
Thanks!
UPDATE
After Christopher's comment I chose $\theta=\frac{2\pi}{n}$ which gives, after the paramatrization $\Gamma (t) = te^{\frac{2\pi i}{n}}$, $t\in(\epsilon,R)$:
$$\int_{\Gamma_{2}}{\frac{\ln z}{1+z^n}dz} = \int_{\epsilon}^{R}{\frac{\ln (te^\frac{2\pi i}{n})}{1+t^n}e^\frac{2\pi i}{n}dt} =
e^\frac{2\pi i}{n}\int_{\epsilon}^{R}{\frac{\ln t}{1+t^n}dt} +
ie^\frac{2\pi i}{n}\int_{\epsilon}^{R}{\frac{\frac{2\pi}{n}}{1+t^n}dt} =$$
$$
=
e^\frac{2\pi i}{n}\int_{0}^{\infty}{\frac{\ln t}{1+t^n}dt} +
ie^\frac{2\pi i}{n}\int_{0}^{\infty}{\frac{\frac{2\pi}{n}}{1+t^n}dt}
$$
But I have no idea how to deal with the second integral.
| This is NOT an answer to the original question. This is a justification of the final identity in Peter's answer. This was requested in chat, but too long to put there or in a comment box.
@Peter, all you need is recurrence and reflection for the trigamma function. Taking $z = 1/2n,$ first we find
$$ \psi^{(1)} \left( 1 + \frac{1}{2n} \right) = \psi^{(1)} \left( \frac{1}{2n} \right) - \frac{1}{z^2} = \psi^{(1)} \left( \frac{1}{2n} \right) -4 n^2. $$
Next
$$ \psi^{(1)} \left( 1 - \left( 1 + \frac{1}{2n} \right) \right) + \psi^{(1)} \left( 1 + \frac{1}{2n} \right) = \pi^2 \csc^2 \frac{\pi}{2n}, $$ so
$$ \psi^{(1)} \left( - \frac{1}{2n} \right) + \psi^{(1)} \left( \frac{1}{2n} \right) - 4 n^2 = \pi^2 \csc^2 \frac{\pi}{2n}, $$ or
$$ \psi^{(1)} \left( - \frac{1}{2n} \right) + \psi^{(1)} \left( \frac{1}{2n} \right) = 4 n^2 + \pi^2 \csc^2 \frac{\pi}{2n}. $$
Then
$$ \psi^{(1)} \left( \frac{1}{2} - \frac{1}{2n} \right) + \psi^{(1)} \left( \frac{1}{2} + \frac{1}{2n} \right) = \pi^2 \csc^2 \left(\frac{\pi}{2} - \frac{\pi}{2n} \right) = \pi^2 \sec^2 \left( \frac{\pi}{2n} \right). $$
| {
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$\lim_{x \to 0}\frac{|x|\sin \left(\frac{1}{3 \sqrt{x}}\right)}{\sqrt{x^4+4x^2+7}}$
Find $$\lim_{x \to 0}\frac{|x|\sin \left(\frac{1}{3 \sqrt{x}}\right)}{\sqrt{x^4+4x^2+7}}$$
I know that $\lim_{x \to 0} \frac{\sin x}{x}=1$ But here $\sin \left(\frac{1}{3 \sqrt{x}}\right)$ is given when $x \to 0$. Need help.
| The limit of the denominator is $\sqrt7$ so we just need to tame the numerator . Observe
$$\left|\left|x\right|\sin\frac{1}{3\sqrt{x}}\right|=\left|x\right|\left|\sin\frac{1}{3\sqrt{x}}\right|\le \left|x\right|\cdot 1$$
What does this tell you?
Also note that $x\to 0^+$ as $\sqrt x$ must be defined
| {
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$\epsilon$-$\delta$ proof that $f(x) = x \sin(1/x)$, $x \ne 0$, is continuous I'm doing an exercise that asks me to prove that $f$ is continuous using a $\epsilon$-$\delta$ proof. I have that
$$
f(x) = \begin{cases}
x\cdot \sin \frac1x,&x\neq 0
\\
0,&x = 0
\end{cases}
$$
I've already managed to show this property for $x=0$. How can I show it for $x \ne 0$, also using a $\epsilon$-$\delta$ proof?
Thank you very much.
| Let $a\neq 0$. By the triangle inequality,
$$\begin{array}{ccc}
\left|f(x)-f(a)\right| &=& \left|x\sin \frac1x-a\sin \frac1a\right| \\
&=& \left|x\sin \frac 1x-a\sin \frac 1x+a\sin \frac 1x-a\sin \frac1a\right| \\
&\le& \left|x-a\right|\left|\sin \frac 1x\right|+a\left|\sin \frac 1x-\sin \frac1a\right| \\
&<& \delta+a\left|\sin \frac 1x-\sin \frac1a\right|
\end{array}$$
It all comes down to bounding the second term. By the trigonometric identity
\begin{equation}\sin \alpha-\sin \beta=2\sin \frac{\alpha-\beta}2\cos \frac{\alpha+\beta}2\end{equation}
we have
$$\begin{array}{ccc}
\left|\sin \frac 1x-\sin \frac1a\right| &=&
\left|2\sin \frac{\frac1x-\frac1a}{2}\cos\frac{\frac1x+\frac1a}{2} \right| \\
&=& 2\left|\sin \frac{x-a}{2xa}\cos\frac{x+a}{2xa} \right| \\
&\le& 2\left|\sin \frac{x-a}{2xa}\right|\end{array}$$
Because $\left|\sin \alpha\right|\le \alpha$,
\begin{equation}2\left|\sin \frac{x-a}{2xa}\right|\le 2\left|\frac{x-a}{2xa}\right|=\frac{\left|x-a\right|}{\left|x\right|\left|a\right|}\end{equation}
As $\left|x-a\right|<\delta\implies \left|x\right|>\left|a\right|-\delta$, the situation is simplified if we choose $\delta<\frac{\left|a\right|}2$. Then,
\begin{equation}\left|x-a\right|<\delta\implies \left|x\right|>\left|a\right|-\delta>\frac{\left|a\right|}2\implies \frac1{\left|x\right|}<\frac{2}{\left|a\right|}\end{equation}
and so
\begin{equation}\left|\sin \frac 1x-\sin \frac1a\right|\le\frac{\left|x-a\right|}{\left|x\right|\left|a\right|}<\frac{2\delta}{a^2}\end{equation}
I belive you can finish this off.
| {
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Is there any trick to evaluate this integral?
Possible Duplicate:
Please help me to evaluate $\int\frac{dx}{1+x^{2n}}$.
Is there any trick to evaluate
$$\int_{-\infty}^\infty \frac{{\rm d} x}{x^{2n}+1}?$$
| This is just a try ...
Since the function is even, we have $2 \int_{0}^\infty \frac{1}{x^{2n}+1}dx$
$$ \int_0^1 \frac{1}{x^{2n}+1} dx + \int_1^\infty\frac{1}{x^{2n}+1} dx \\
= \int_0^1 \sum_{k=0}^\infty (-1)^k (x)^{2kn} + \int_1^\infty \frac{1}{x^{2n}} \sum_{k=0}^\infty (-1)^kx^{-2nk} \\
= \sum_{k=0}^\infty (-1)^k \left[\frac{(x)^{2kn+1}}{2kn+1}\right]_0^1 + \sum_{k=0}^{\infty}(-1)^k \left[ \frac{x^{-2nk-2n+1}}{-2nk-2n+1}\right]_1^\infty \\
$$
So, we have $ $
$$=\sum_{k=0}^\infty(-1)^k \frac{1}{2kn+1} + \sum_{k=0}^\infty (-1)^k \frac{1}{2nk+2n-1} \\
=\sum_{k=0}^\infty (-1)^k \left( \frac{1}{2kn+1} + \frac{1}{2n(k+1) - 1}\right)$$
$$ \int_{-\infty}^\infty \frac{{\rm d} x}{x^{2n}+1}? = 2 \sum_{k=0}^\infty (-1)^k \left( \frac{1}{2kn+1} + \frac{1}{2n(k+1) - 1}\right)$$
Or on complex analysis this is just special case of this problem.
| {
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"timestamp": "2023-03-29T00:00:00",
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An integration to first order I am having some trouble evaluating an integral -- involving taking an approximation. It would be great if someone could help me.
I wish to evaluate
$$\int_0^\pi {\cos\theta\cos \left[\omega t-{\omega \over c}(a^2+r^2-2rb\cos \theta)^{1\over 2}\right]\over (a^2+r^2-2rb\cos \theta)^{1\over 2}}d\theta$$
for the case where $a\gg r$ and $b=a\sin \phi$, to first order in $1\over a$.
The answer is supposed to be
$$\pi r^2\omega(\sin \phi)[\sin(\omega t-{\omega\over c}a)]\over 2a$$but I can't see how to get there...
There must be some Taylor expansion required. I suppose we could write $$(a^2+r^2-2rb\cos \theta)^{1\over 2}\approx a\left(1-{rb\over a^2}\cos\theta\right)$$
but when next?
Ok, if we ignore the given answer -- assuming it is wrong, what should I be getting?
For the argument of the $\cos \left[\omega t-{\omega \over c}(a^2+r^2-2rb\cos \theta)^{1\over 2}\right]$, suppose I Taylor expand $\cos$. It would go something like $$1-{1\over 2}\left[\omega t-{\omega \over c}(a^2+r^2-2rb\cos \theta)^{1\over 2}\right]^2+...$$
However, note that the higher order terms involve terms of $\left(1\over a\right)^n$, with values of $n$ of the non-negligible magnitude. So that can't be the way to do it?
Please help! Thanks.
| The supposed answer is incorrect. Up to first order, we have $$\int_0^\pi\frac{\cos\theta\cos\left(\omega t-\frac\omega c\sqrt{a^2+r^2-2rb\cos\theta}\right)}{a\sqrt{a^2+r^2-2rb\cos\theta}}\,d\theta\approx\frac{\pi rb\omega\sin(\omega a/c-\omega t)}{2ca^2}.$$
Proof: Write $\sqrt{a^2+r^2-2rb\cos\theta}=a\sqrt{1+(r^2-2rb\cos\theta)/a^2}$, which allows us to use $$(1+x)^{\pm1/2}=1\pm\frac{x^2}2+\mathcal O(x^4)$$ for $|x|<1$ (since $a\gg r$). Then\begin{align}I&=\int_0^\pi\frac{\cos\theta\cos\left(\omega t-\frac\omega c\cdot a\sqrt{1+(r^2-2rb\cos\theta)/a^2}\right)}{a\sqrt{1+(r^2-2rb\cos\theta)/a^2}}\,d\theta\\&=\int_0^\pi\left(\frac1a+\mathcal O\left(\frac1{a^3}\right)\right)\cos\theta\cos\left(\omega t-\frac\omega c\left(a+\frac{r^2-2rb\cos\theta}{2a}+\mathcal O\left(\frac1{a^3}\right)\right)\right)\,d\theta\\&\approx\int_0^\pi\frac{\cos\theta}a\cos\left(\omega t-\frac\omega c\left(a+\frac{r^2-2rb\cos\theta}{2a}\right)\right)\,d\theta\end{align} up to first order in terms of $1/a$. The cosine addition identity yields \begin{align}I&\approx\small\int_0^\pi\frac{\cos\theta}a\left[\cos\left(\omega t-\frac\omega ca\right)\cos\left(\frac\omega c\frac{r^2-2rb\cos\theta}{2a}\right)+\sin\left(\omega t-\frac\omega ca\right)\sin\left(\frac\omega c\frac{r^2-2rb\cos\theta}{2a}\right)\right]\\&\approx\int_0^\pi\frac{\cos\theta}a\left[\cos\left(\omega t-\frac\omega ca\right)+\sin\left(\omega t-\frac\omega ca\right)\cdot\frac\omega c\frac{r^2-2rb\cos\theta}{2a}\right]\,d\theta\end{align} again up to first order, since $\cos x=1+\mathcal O(x^2)$ and $\sin x=x+\mathcal O(x^3)$. Thus, \begin{align}I&\approx\frac{\omega\sin(\omega t-\omega a/c)}{2ca^2}\int_0^\pi(r^2-2rb\cos\theta)\cos\theta\,d\theta=-\frac{\pi rb\omega\sin(\omega t-\omega a/c)}{2ca^2},\end{align} which fits reasonably well when $a\gg r$.
| {
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Calculate $\int\frac{1}{(x+1)\sqrt{1+x^2}} dx $ How can I calculate the following integral :
$$\int\frac{1}{(x+1)\sqrt{1+x^2}} dx $$
I try to write the integral like :
$$\int\frac{1+x-x}{(x+1)\sqrt{1+x^2}}=\int\frac{1}{\sqrt{1+x^2}}-\int\frac{x}{(x+1)\sqrt{1+x^2}}=\int\frac{1}{\sqrt{1+x^2}}-\int\frac{(\sqrt{x^2+1})'}{(x+1)}$$
but still nothing .
thanks :)
| This is how I would go, term $\sqrt{1+x^2}$ makes me go for $x=\sinh(u)$, which will convert to $\cosh(u)$ and cancel with the one brought by $du$.
Then I convert to exponential form ($t=e^u$) to get a rational fraction.
$\begin{align}\require{cancel}\int\frac {dx}{(x+1)\sqrt{1+x^2}}
&=\int\frac {\cancel{\cosh(u)}du}{(\sinh(u)+1)\cancel{\sqrt{1+\sinh(u)^2}}}
=\int\frac {du}{\sinh(u)+1}\\\\
&=\int\frac {dt/t}{(t-1/t)/2+1}
=\int\frac {2\,dt}{t^2+2t-1}
=\int\frac {2\,dt}{(t+1)^2-2}\\\\
&=C-\sqrt{2}\tanh^{-1}\Big(\tfrac{t+1}{\sqrt{2}}\Big)\end{align}$
The substitutions proposed appear quite naturally, though the final result (especially after back substitution) is not as appealing as the one proposed by Quanto...
| {
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"url": "https://math.stackexchange.com/questions/279526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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} |
Cyclotomic polynomial simplification I try to work out what $\Phi_{12}(z)$ is:
By the fundamental theorem of arithmetic:
$$\Phi_{12}(z)=(z-\xi^0)(z-\xi^1)(z-\xi^2)(z-\xi^3)(z-\xi^4)(z-\xi^5)(z-\xi^6)\\(z-\xi^7)(z-\xi^8)(z-\xi^9)(z-\xi^{10})(z-\xi^{11})$$
$$\Phi_{12}(z)=(z-\xi^0)(z-\xi^1)(z-\xi^2)(z-\xi^3)(z-\xi^4)(z-\xi^5)\\
(z+\xi^0)(z+\xi^1)(z+\xi^2)(z+\xi^3)(z+\xi^4)(z+\xi^5)
$$
$$\Phi_{12}(z)=(z^2-\xi^{0\cdot 2})(z^2-\xi^{1\cdot 2})(z^2-\xi^{2\cdot 2})(z^2-\xi^{3\cdot 2})(z^2-\xi^{4\cdot 2})(z^2-\xi^{5\cdot 2})$$
$$\Phi_{12}(z)=(z^2-\xi^{0})(z^2-\xi^{2})(z^2-\xi^{4})(z^2-\xi^{6})(z^2-\xi^{8})(z^2-\xi^{10})$$
$$\Phi_{12}(z)=(z^2-\xi^{0})(z^2-\xi^{2})(z^2-\xi^{4})(z^2+\xi^{0})(z^2+\xi^{2})(z^2+\xi^{4})$$
$$\Phi_{12}(z)=(z^4-\xi^{0\cdot 2})(z^4-\xi^{2\cdot 2})(z^4-\xi^{4\cdot 2})$$
$$\Phi_{12}(z)=(z^4-\xi^{0})(z^4-\xi^{4})(z^4-\xi^{8})$$
$$\Phi_{12}(z)=(z^4-\xi^{0})(z^8-z^4(\xi^{4}+\xi^{8})+\xi^{4}\xi^{8})$$
$$\Phi_{12}(z)=(z^4-1)(z^8-z^4+1)$$
Why is $\Phi _{12} (z)$ given as $z^4-z^2+1$?
(Or, more precisely, what is wrong with my method (or am I not noticing their equivalence)?).
| I suppose $\xi$ is a primitive $12^{th}$ root of unity.
The polynomial $\Phi_{12}(x)$ is the product $\prod(z-\xi^k)$ over the primitive $12^{th}$ roots of unity (i.e. for $k=1,5,7,11$).
What you are calculating is, $\displaystyle{\prod_{k=0}^{11}(z-\xi^k)}$, the product over all $12^{th}$ roots of unity which is just the polynomial $z^{12}-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/280890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How to determine the sign of : I've got some trouble: I don't know how to determine the sign of :
$$\frac{4t^2- \sqrt t}{t}$$
for $t > 0$.
Thank you in advance
| Since $\sqrt{t}$ is only defined when $t \ge 0$, we can only consider inputs from that range.
Now we have to consider the sign of both the numerator and the denominator. The fraction will be positive whenever the sign of $4t^2 - \sqrt{t}$ matches that of $t$. Clearly though, $t$ is only positive when $t \ge 0$, so we see that the sign of the numerator determines the sign of the fraction.
Consider the equation $4t^2 - \sqrt{t} = 0.$ We can solve this as follows:
$$4t^2 = \sqrt{t}$$
$$(4t^2)^2 = t$$
$$16t^4 = t$$
$$16t^4 - t = 0$$
$$t(16t^3 - 1) = 0.$$
So the numerator equals zero whenever $t = 0$ (which is not on our interval) or $16t^3 - 1 = 0$. Solving this last equation we see that the numerator equals zero when $t = (\frac{1}{16})^\frac{1}{3}$.
We now pick a point larger than this value, say t = 1. Plugging in, we see that $4(1)^2 - \sqrt{1} = 4 - 1 = 3 > 0.$ Thus, whenever $t > (\frac{1}{16})^\frac{1}{3}$, the numerator, and the fraction, are positive. Picking a point between $0$ and $(\frac{1}{16})^\frac{1}{3}$ will reveal that the numerator is negative on the interval $(0, (\frac{1}{16})^\frac{1}{3}).$ For instance, $t = \sqrt{\frac{1}{16}}.$ Plugging in this value to the numerator gives
$$4(\sqrt{\frac{1}{16}})^2 - \sqrt{\sqrt{\frac{1}{16}}} = $$
$$4(\frac{1}{16}) - (\frac{1}{16})^{\frac{1}{4}} = $$
$$ \frac{4}{16} - \frac{1}{2} < 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/282794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Inequality: $(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27$
Let be $a,b,c \geq 0$ such that: $a^2+b^2+c^2=3$.
Prove that:
$$(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27.$$
I try to apply $GM \leq AM$ for $x=a^3+a+1$, $y=b^3+b+1,z=c^3+c+1$ and
$$\displaystyle \sqrt[3]{xyz} \leq \frac{x+y+z}{3}$$ but still nothing.
Thanks :-)
| I have a new answer. We will apply the Cebysev's inequality.
$$(a^3+a+1)(b^3+b+1)\leq3(a^3b^3+ab+1)$$
So:
$$\left[(a^3+a+1)(b^3+b+1)(c^3+c+1)\right]^{2} \leq 27(a^3b^3+ab+1)(b^3c^3+cb+1)(c^3a^3+ac+1).$$
Now, we will prove that:
$$27(a^3b^3+ab+1)(b^3c^3+cb+1)(c^3a^3+ac+1) \leq 27^{2}.$$ Equivalent with:
$$(a^3b^3+ab+1)(b^3c^3+cb+1)(c^3a^3+ac+1) \leq 27.$$
Now we use $AM \geq GM$ :
$$(a^3b^3+ab+1)(b^3c^3+cb+1)(c^3a^3+ac+1) \leq \left(\frac{a^3b^3+b^3c^3+a^3c^3+ab+bc+ca+3}{3}\right)^{3}.$$
Now we will apply the following inequality to obtain the desired result:
$$a^3b^3+b^3c^3+a^3c^3 \leq \frac{(a^2+b^2+c^2)^{3}}{9}=3,$$
$$ab+bc+ca \leq a^2+b^2+c^2=3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/283895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "43",
"answer_count": 8,
"answer_id": 3
} |
Integral of $\frac{\sqrt{x^2+1}}{x}$ So I have to do the integration of $\frac{\sqrt{x^2+1}}{x}$. Give me a hint. What should I replace? Should I do it with integration by parts?
| You could use the substitution
$$
\begin{align}
&&u&=x+\sqrt{x^2+1},&x&=\frac{u^2-1}{2u} \\
&&\sqrt{x^2+1}&=\frac{u^2+1}{2u},&dx&=\frac{u^2+1}{2u^2}du
\end{align}
$$
to get (unless I did some mistake)
$$
\int\frac{\sqrt{x^2+1}}{x}dx=
\int\frac{\frac{u^2+1}{2u}}{\frac{u^2-1}{2u}}\frac{u^2+1}{2u^2}du=
\int\frac{(u^2+1)^2}{2u^2(u^2-1)}du=
\int\frac{(u^2-1)^2+4u^2}{2u^2(u^2-1)}du=
$$
$$
=\int\left(\frac{(u^2-1)^2}{2u^2(u^2-1)}+\frac{4u^2}{2u^2(u^2-1)}\right)\,du=
\int\left(\frac{u^2-1}{2u^2}+\frac{2}{u^2-1}\right)\,du=
$$
$$
=\frac{u}{2}+\frac{1}{2u}+\ln|u-1|-\ln|u+1|+C.
$$
Now just do the backsubstitution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/286118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Determinant of an $n\times n$ matrix with 5's on the diagonal and 2's on the superdiagonal and subdiagonal
Possible Duplicate:
Special determinant formula for a specific matrix
How to find $\det A_n$ as a function of $n$?
$$A_n=\begin{pmatrix} 5&2 &0& 0 & \ldots & 0\\
2& 5& 2& 0 & \ldots & 0\\
0 &2& 5 &2 & \ldots & 0\\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots\\
0 &0& 0& 0 & \ldots & 5\end{pmatrix}$$
I tried to develop Laplace and mathematical induction, but I don't know how to do it.
| Example:
$$
\left|\begin{matrix}
5&2&0&0\\
2&5&2&0\\
0&2&5&2\\
0&0&2&5
\end{matrix}\right|
=5\left|\begin{matrix}
5&2&0\\
2&5&2\\
0&2&5
\end{matrix}\right|
-2\left|\begin{matrix}
2&2&0\\
0&5&2\\
0&2&5
\end{matrix}\right|
=5\left|\begin{matrix}
5&2&0\\
2&5&2\\
0&2&5
\end{matrix}\right|
-2^2\left|\begin{matrix}
5&2\\
2&5
\end{matrix}\right|.
$$
The first equality is obtained by Laplace expansion along the first row, while the second one is obtained by Laplace expansion along the first column. So you get a recurrence relation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/286198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Power Series and Radius of Convergence
Determine for following Power Series in $\mathbb{C}$ the radius of
Convergence.
a) $\sum _{ n=0 }^{ \infty }{ (2+\sqrt { n } )^{ n }z^{ n } } $
b) $\sum _{ n=1 }^{ \infty }{ (1-\frac { 1 }{ n } )^{ n }z^{ n } } $
c) $\sum _{ n=0 }^{ \infty }{ n!n^{ -n }z^{ n } } $
For the radius of convergence, on has the formula: $ r=\frac { 1
> }{ L } ,\quad wo\quad L:=\lim _{ n\rightarrow \infty }{ sup\sqrt [ n
> ]{ \left| { a }_{ n } \right| } }$
Or the simpler formula: $r=\lim _{ n\rightarrow \infty }{ \left|
> \frac { { a }_{ n } }{ { a }_{ n+1 } } \right| }$
Hear are the solutions:
a) ${ a }_{ n }:=(2+\sqrt { n } )^{ n }, \text{for} \ \ n\in\mathbb{N} $ we have
$\sqrt [ n ]{ \left| { a }_{ n } \right| } =\sqrt [ n ]{ \left| 2+\sqrt { n } \right| ^{ n } } =\left| 2+\sqrt { n } \right| $ and so $\lim _{ n\rightarrow \infty }sup{ \sqrt [ n ]{ \left| { a }_{ n } \right| } } =\infty $ and the Convergence Radius is $r=\frac { 1 }{ \lim _{ n\rightarrow \infty } sup{ \sqrt [ n ]{ \left| { a }_{ n } \right| } } } =0$
b) ${ a }_{ n }:=(1-\frac { 1 }{ n } )^{ n }$ for $n\in \mathbb{N}$ we have $\sqrt [ n ]{ \left| { a }_{ n } \right| } =\sqrt [ n ]{ \left| 1-\frac { 1 }{ n } \right| ^{ n } } =1-\frac { 1 }{ n } $ and so $\lim _{ n\rightarrow \infty }{ sup\sqrt [ n ]{ \left| { a }_{ n } \right| } } =1$ and the Convergence Radius is $r=\frac { 1 }{ \lim _{ n\rightarrow \infty } sup{ \sqrt [ n ]{ \left| { a }_{ n } \right| } } } =1$
c) $a_{n}:=n!n^{-n} \neq0$ for all $n\in\mathbb{N}$ and $\left| \frac { a_{ { n } } }{ a_{ { n }+1 } } \right| =\frac { n!(n+1)^{ { n+1 } } }{ n^{ n }(n+1)! } =\frac { 1 }{ n+1 } \frac { (n+1)^{ n+1 } }{ n^{ n } } =(\frac { n+1 }{ n } )^{ n }=(1+\frac { 1 }{ n } )^{ n }$ we know that $\lim _{ n\rightarrow \infty }{ (1+\frac { 1 }{ n } ) } ^{ n }=e$ so the Convergence Radius is $r=\lim _{ n\rightarrow \infty }{ \left| \frac { a_{ { n } } }{ a_{ { n+1 } } } \right| } =e$
| For a), the series diverges $\forall \, z \ne 0$ because $\lim_{n \rightarrow \infty} (2 + \sqrt{n}) = \infty$.
For b)
$$\lim_{n \rightarrow \infty} \left ( 1-\frac{1}{n} \right )^n = \frac{1}{e}$$
so by the comparison test with a geometric series, the radius of convergence is $1$, i.e. the series converges only when |z| < 1$.
For c), use Stirling's approximation and the root test to show that the series converges $\forall \, |z| < e$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/288231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove $|\frac{1}{n} - \frac{1}{x}| \le \frac{1}{n^2}$ Prove $|\frac{1}{n} - \frac{1}{x}| \le \frac{1}{n^2}$ for all $n \le x \le n+1$, using the mean value theorem applied to $f(x) = \frac{1}{x}$
Immediately, I can recognize some components of the mean value theorem. $\frac{1}{n^2}$ likely comes from the slope of $f$ at $n$, and the $|\frac{1}{n} - \frac{1}{x}|$ expression likely comes from the secant expression $\frac{\frac{1}{n} - \frac{1}{x}}{x}$. But I cannot figure out how they fit together in the end.
| Assume $n>0$.
By mean value theorem, there exists $c\in(n,n+1)$ such that
$$\frac{1}{n}-\frac{1}{x}=\frac{1}{c^2}(x-n)$$
Since $c>n$, we get $c^2>n^2$. So
$$\left|\frac{1}{n}-\frac{1}{x}\right|<\frac{x-n}{n^2}\le\frac{1}{n^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/288580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Finding $\int^1_0 \frac{\log(1+x)}{x}dx$ without series expansion I was trying to evaluate $$\int^1_0 \frac{\log(1+x)}{x}dx.$$
I expanded $\log(1+x) $ as
$x -\frac{x^2}{2}... $ and got the answer. I would like to know if there is any way to do it without series expanding.
| With $\int^1_0 \frac{\ln(1+x)}{x}dx
= 2\int^1_0 \frac{\ln(1+x^2)}{x}dx
= 3\int^1_0 \frac{\ln(1+x^3)}{x}dx $
\begin{align}
&\int^1_0 \frac{\ln(1+x)}{x}dx\\= &\ \frac67\int^1_0 {\frac{\ln\frac{(1+x^2)(1+x)} {(1+x^3)}}{x} } dx
=\frac67\int^1_0 \frac{\ln\frac{1+x^2}{1-x+x^2}}{x}dx \\= &\ \frac67\int_0^1\int_0^{\frac\pi6}
\frac{2\cos t }{1-2x\sin t + x^2}dt\ dx
=\frac67\int_0^{\frac\pi6}\left(\frac\pi2+t\right)dt
=\frac{\pi^2}{12}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/288830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 5,
"answer_id": 1
} |
Find $\lim_{x \rightarrow 0} \frac{e^{\sin x} - e^x}{\sin^3 2x}$ I have to find $\displaystyle\lim_{x \rightarrow 0} \frac{e^{\sin x} - e^x}{\sin^3 2x}$ using Taylor polynomials.
Here's what I've done so far:
*
*$e^x = 1 + x + \frac{1}{2} x^2 + \frac{1}{6} x^3 + o(x^3)$
*$\sin x = x - \frac{1}{6} x^3 + o(x^4)$
*$e^{\sin x} = 1 + x - \frac{1}{6} x^3 + o(x^4)$
*$\sin 2x = 2x + o(x^2)$
*$\sin^3 2x = 8x^3 + o(x^6)$
Therefore I can rewrite my limit as: $\displaystyle\lim_{x \rightarrow 0} \frac{-\frac{1}{3} x^3 - \frac{1}{2} x^2 + o(x^3)}{8x^3 + o(x^6)}$
In this form, the limit appers to be $-\frac{1}{24}$, but the correct result is $-\frac{1}{48}$.
Could you tell me what I'm doing wrong?
| Let's rewrite the limit as
$$\displaystyle\lim_{x \rightarrow 0} e^x\times\lim_{x \rightarrow 0}\frac{e^{\sin x-x} - 1}{\sin x-x}\times\lim_{x \rightarrow 0}\left(\frac{2x}{\sin 2x}\right)^3\times \lim_{x \rightarrow 0}\frac{\sin x-x}{8x^3}=\lim_{x \rightarrow 0}\frac{x-x^3/6+O(x^5)-x}{8x^3}=-\frac{1}{48}$$
Q.E.D.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Minimum of $\left| \sin x- 1\right| + \left|\sin x- 2\right| + \left| \sin x -3\right| + \left| \sin x+1\right|$ I would like to know the minimum value of $$\left| \sin x- 1\right| + \left|\sin x- 2\right| + \left| \sin x -3\right| + \left| \sin x+1\right|$$
for $x \in \mathbb{R}$.
| Note that $-1\le\sin x\le1$ for all real $x$. This allows you to rewrite each of the $4$ summands without the absolute value bars. For example: $\sin x-3\leq 1-3=-2<0$, so $|\sin x-3|=3-\sin x$; and $\sin x+1\geq -1+1=0$, so $|\sin x+1|=\sin x+1$. Simplify the resulting expression, and then use the fact that $\sin x$ ranges from $-1$ to $1$ (inclusive) to minimize it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/289185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 4
} |
Equation of a line in homogenous coordinates given 2 points in affine coordinates So if I have 2 points $A$ and $B$ such that $F(A) = (1; a, a^3)$, and $F(B) = (1; b, b^3)$. how do I find the equation of this line in homogeneous coordinates?
So I know how to get a line the "normal" way. If I take the points $a$ and $b$ and represent them as regular Cartesian coordinates $A = (a, a^3), B = (b, b^3)$ and then say a line equation is $y = mx + b$. I could find the slope to be $(b^3-a^3)/(b-a)$ and plug that in for $m$ and then find '$b$' by plugging in the coordinates for $A$ and $B$.
But then I wouldn't know how to take that $y = mx+b$ and turn it into homogeneous coordinates. I don't think this is necessarily how I'm supposed to go about it.
| In planar geometry the line passing through two points A and B in homogeneous coordinates is found from their vector cross product:
$$ \begin{pmatrix} C_0 \\ C_1 \\ C_2 \end{pmatrix} = \begin{pmatrix} A_0 \\ A_1 \\ A_2 \end{pmatrix} \times \begin{pmatrix} B_0 \\ B_1 \\ B_2 \end{pmatrix} = \begin{pmatrix} A_1 B_2 - A_2 B_1 \\ A_2 B_0 - A_0 B_2 \\ A_0 B_1 - A_1 B_0 \end{pmatrix} $$
The equation for the line is
$$ C_0 + C_1 x + C_2 y = 0 $$
Example
Two points A and B have coordinates $A=(1,7,2)$ and $B=(1,3,-2)$. The line that joints them is
$$ C = \begin{pmatrix} 1 \\ 7 \\ 2 \end{pmatrix} \times \begin{pmatrix} 1 \\ 3 \\ -2 \end{pmatrix} = \begin{pmatrix} -20 \\ 4 \\ -4 \end{pmatrix}$$
with the equation $$ (-20)+(4)x+(-4)y=0 \\ x-y-5 = 0 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/290315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find five positive integers whose reciprocals sum to $1$ Find a positive integer solution $(x,y,z,a,b)$ for which
$$\frac{1}{x}+ \frac{1}{y} + \frac{1}{z} + \frac{1}{a} + \frac{1}{b} = 1\;.$$
Is your answer the only solution? If so, show why.
I was surprised that a teacher would assign this kind of problem to a 5th grade child. (I'm a college student tutor) This girl goes to a private school in a wealthy neighborhood.
Please avoid the trivial $x=y=z=a=b=5$. Try looking for a solution where $ x \neq y \neq z \neq a \neq b$ or if not, look for one where one variable equals to another, but explain your reasoning. The girl was covering "unit fractions" in her class.
| There are many ways to obtain many solutions. Here is one systematic way to obtain solutions.
First look at a class of solutions such that $x \leq y \leq z \leq a \leq b \leq 10$ (other solution can be obtained as permutation of these).
This implies that $2 \leq x \leq 5$. So now start with $x=2$. This now means that $3 \leq y \leq 8$. Choose $y=3$. This now means that $7 \leq z \leq 10$. You will quickly find that no solution exists such that $x \leq y \leq z \leq a \leq b \leq 10$. Then choose $y=4$. Again you will find that no solution exists such that $x \leq y \leq z \leq a \leq b \leq 10$.
Going through this you will find that if we want $x=2$, and $x \leq y \leq z \leq a \leq b \leq 10$, then $$[2, 5, 10, 10, 10];
[2, 6, 9, 9, 9];
[2, 8, 8, 8, 8]$$ are the only solutions with $x=2$ and $x \leq y \leq z \leq a \leq b \leq 10$.
Now set $x=3$ and get bounds for the remaining variables to see that $$
[3, 3, 9, 9, 9];
[3, 4, 6, 8, 8];
[3, 5, 5, 6, 10];
[3, 6, 6, 6, 6]; $$ are the only solutions with $x=3$ and $x \leq y \leq z \leq a \leq b \leq 10$.
Here are the solutions such that $x \leq y \leq z \leq a \leq b \leq 10$
$$[x,y,z,a,b] \in \{[2, 5, 10, 10, 10];
[2, 6, 9, 9, 9];
[2, 8, 8, 8, 8];
[3, 3, 9, 9, 9];
[3, 4, 6, 8, 8];
[3, 5, 5, 6, 10];\\
[3, 6, 6, 6, 6];
[4, 4, 4, 8, 8];
[4, 4, 5, 5, 10];
[4, 4, 6, 6, 6];
[5, 5, 5, 5, 5]; \}
$$
There are $114$ distinct solutions i.e. without permutations such that
$$x \leq y \leq z \leq a \leq b \leq 100$$ and can be found here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/290435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "373",
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"answer_id": 8
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How to find real roots? $4 ^{x} +6 ^{x ^{2}} =5 ^{x} +5 ^{x ^{2}}$ How to find real roots?
$$4 ^{x} +6 ^{x ^{2}} =5 ^{x} +5 ^{x ^{2}}$$
I try LMVT but vary difficult
| The following argument is admittedly on the intuitive side, using concavity mixed with "flatter than" comparisons.
If $x<0$ then since $x^2>0$ we have
$$5^x-4^x<0<6^{x^2}-5^{x^2}.$$
If $0<x<1$ then in fact $0<x^2<x<1$, and the functions $u^x$ and $u^{x^2}$ are both concave down, with $u^{x^2}$ being flatter between 5 and 6 than is $u^x$ there. So
$$6^{x^2}-5^{x^2}<6^x-5^x<5^x-4^x.$$
Finally if $x>1$ then in fact $1<x<x^2$ and the functions $u^x$ and $u^{x^2}$ are both concave up, with $u^x$ being flatter between 5 and 6 than is $u^{x^2}$ there. So
$$6^{x^2}-5^{x^2}>6^x-5^x>5^x-4^x.$$
If the above (admittedly vague) arguments are right, we have ruled out all possibilities other than $x=0,1$ for the equation $5^x-4^x=6^{x^2}-5^{x^2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/293323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Combinatorics. Please explain me how to do it. In how many different ways three persons A, B, C having 6, 7 and 8 one rupee coins respectively can donate Rs.10 collectively? This isn't a homework question. Please explain me the steps. Thank you!
| If $A$ gives $a$ coins, clearly, $0\le a\le 6$
and $B+C=10-a$
Now, $0\le B\le 7\implies 0\le 10-a-C\le 7\implies 3-a\le C\le 10-a $
Also, $0\le C\le 8\implies$ max $(3-a,0)\le C\le $ min$(10-a,8)$
If $a=0,$ max $(3,0)\le C\le $ min$(10,8)\implies 3\le C\le 8$ so $C$ can assume $8-3+1=6$ values.
Similarly, for $a=1,2,3,4,5,6;$ $ C$ can assume $7,8,8,7,6,5$ values
respectively.
So, the number possible combinations are $6+7+8+8+7+6+5=47$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/293610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Square root of surds: $\sqrt{12+2\sqrt{6}}$? I got this question
Find the square root of $12+2\sqrt{6}$ expressing your answer in the form $\sqrt{m}+\sqrt{n}$.
I have no idea what this means and how to go about it.
| Let $x = \sqrt{12 + 2\sqrt{6}} = \sqrt{n} + \sqrt{m}$. Then $x^2 = 12 + 2\sqrt 6 = n + m + 2 \sqrt{nm}$.
Find $n$ and $m$ such that $n + m = 12$ and $nm = 6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/294032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Determinant of block matrix Here is a determinant of a $(k+m) \times (k+m)$ Block matrix.
\begin{align}
D=\begin{vmatrix}
a_{11} &a_{12} & \cdots & a_{1k} &0 &\cdots &0 \\
a_{21}& a_{22}& \cdots & a_{2k} & 0 &\cdots &0 \\
\vdots& \vdots & & \vdots & \vdots & &\vdots\\
a_{k1} & a_{k2} & \cdots & a_{kk} & 0 &\cdots & 0\\
c_{11}& c_{12} & \cdots& c_{1k} & b_{11} & \cdots & b_{1m}\\
\vdots& \vdots & & \vdots & b_{21}&\cdots & b_{2m}\\
c_{m1}& c_{m2} & \cdots & c_{mk} & b_{m1}& \cdots & b_{mm}
\end{vmatrix}
\end{align}
If I have got a determinant $$D_1=
\begin{vmatrix}
0 &\cdots &0&a_{11} &a_{12} & \cdots & a_{1k} \\
0 &\cdots &0 &a_{21}& a_{22}& \cdots & a_{2k} \\
\vdots& & \vdots & \vdots & \vdots & &\vdots\\
0 &\cdots & 0&a_{k1} & a_{k2} & \cdots & a_{kk} \\
b_{11} & \cdots & b_{1m}& c_{11}& c_{12} & \cdots& c_{1k}\\
\vdots& & \vdots & \vdots & \vdots& & \vdots\\
b_{m1}& \cdots & b_{mm}&c_{m1}& c_{m2} & \cdots & c_{mk}
\end{vmatrix}
$$
Then $D_1$ is equal to $(-1)^{k \times m}$$D$.
I know that the existence of the factor -1 is due to the interchange of 2 row, but i have a question on that $k \times m$.In my book,it said that i have to do $k\times m$ times row operations to transform $D_1$ into $D$.However,i thought only k times is needed for $D_1$ transform into $D$.
If i have done 1 times row operation for $D_1$
$$D_1=
\begin{vmatrix}
0 &\cdots &0&a_{11} &a_{12} & \cdots & a_{1k} \\
0 &\cdots &0 &a_{21}& a_{22}& \cdots & a_{2k} \\
\vdots& & \vdots & \vdots & \vdots & &\vdots\\
0 &\cdots & 0&a_{k1} & a_{k2} & \cdots & a_{kk} \\
b_{11} & \cdots & b_{1m}& c_{11}& c_{12} & \cdots& c_{1k}\\
\vdots& & \vdots & \vdots & \vdots& & \vdots\\
b_{m1}& \cdots & b_{mm}&c_{m1}& c_{m2} & \cdots & c_{mk}
\end{vmatrix}
=\begin{vmatrix}
a_{11} &\cdots &0&0 &a_{12} & \cdots & a_{1k} \\
a_{21} &\cdots &0 &0& a_{22}& \cdots & a_{2k} \\
\vdots& & \vdots & \vdots & \vdots & &\vdots\\
a_{k1} &\cdots & 0& 0& a_{k2} & \cdots & a_{kk} \\
c_{11} & \cdots & b_{1m}&b_{11} & c_{12} & \cdots& c_{1k}\\
\vdots& & \vdots & \vdots & \vdots& & \vdots\\
c_{m1}& \cdots & b_{mm}&b_{m1}& c_{m2} & \cdots & c_{mk}
\end{vmatrix}
$$
Correct me if i have made any mistakes
| You can think of it in this way: take column $k+1$ (the first one containing $b$s) and swap it with column $k$. Then swap column $k$ with column $k-1$, etc, until you have "bubbled" the original column $k+1$ all the way to the left, after a total of $k$ swaps.
Repeating this process for all $m$ columns on the right requires a total of $km$ swaps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/297288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Overlapping spheres Say you have two spheres that are partially overlapping. How would I find the volume of the portion of one of the spheres that is not overlapping with the other based on how far apart the two spheres are and the spheres' individual radii?
| Assume that the 2 spheres have equal radii. The volume of the intersection is given by
$$V_I = 2 \pi \int_{-a}^{-d/2} dx \: (a^2-x^2)=\frac{4 \pi}{3} a^3-\pi d \left (a^2-\frac{d^2}{12}\right)$$
where $a$ is the radius of each sphere and $d$ is the separation between the centers of the spheres. So the volume in a sphere outside of the intersection is
$$\pi d \left (a^2-\frac{d^2}{12}\right)$$
For the general case, assume that the spheres have radii $a$ and $b>a$. The geometry of the intersection is a lens of thickness $d$, the thickness of the lens surface of radius $a$ being
$$t_a = \frac{a^2-(b-d)^2}{2 d}$$
and that of the lens surface of radius $b$ is
$$t_b = \frac{b^2-(a-d)^2}{2 d}$$
The volume of the lens is then
$$\begin{align}V_I &= \pi \int_{a-t_a}^a dx \: (a^2-x^2) + \pi \int_{-b}^{-b+t_b} dx \: (b^2-x^2)\\ &=\frac{\pi}{12 d}(a+b-d)^2 \left(2 d (a+b)+9 (b-a)^2+d^2\right) \end{align}$$
To get the volume outside of this lens in either sphere, subtract $V_I$ from either $\frac{4 \pi}{3} a^3$ or $\frac{4 \pi}{3} b^3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/297751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Proof that $(1+1/x)^x$ is monotonic increasing How does one prove that $(1+\frac{1}{x})^x$ is monotonic increasing for any $x \in [1,\infty)$?
Thanks a million!
| $$f(x) = \left ( 1+ \frac{1}{x} \right )^x$$
$$\log{f(x)} = x \log{\left( 1+ \frac{1}{x} \right )}$$
$$\frac{d}{dx}\log{f(x)} = \log{\left( 1+ \frac{1}{x} \right )} + \frac{x}{1+\frac{1}{x}}\left (\frac{-1}{x^2} \right ) = \log{\left( 1+ \frac{1}{x} \right )} - \frac{1/x}{1+ \frac{1}{x}}$$
or
$$\frac{d}{dx}\log{f(x)} = \log{\left( 1+ \frac{1}{x} \right )} + \frac{1}{1+ \frac{1}{x}} - 1 $$
Because $y \log{y} \ge y-1 \; \forall y \ge 1$, we may say that, $\forall x \ge 1$:
$$\frac{d}{dx}\log{f(x)} =\log{\left( 1+ \frac{1}{x} \right )} + \frac{1}{1+ \frac{1}{x}} - 1 \ge 0$$
and $f(x)$ is monotone increasing for all $x \ge 1$, as $\log$ is monotone increasing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/297916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\int\frac{\cos x}{\sin x + \cos x}\,\text{d}x$. I'm trying to figure out how to take this indefinite integral:
$$ \int\frac{\cos x}{\sin x + \cos x}\,\text{d}x.$$
I tried simplifying and rearranging it, and this is the best I got:
$$\int\frac{1}{\tan x + 1 }\,\text{d}x.$$
But I still can't figure out how to integrate from there. I know that it's integrable, as Wolfram Alpha indicates that the integral is $ \frac{1}{2}\big(x+\ln{(\sin x + \cos x)}\big)+C$, but I can't figure out the steps to deriving it. Does anyone know how to evaluate this integral?
| $\displaystyle \int \frac{1}{1+ \tan x} \ dx $
$ \displaystyle = \int \frac{1}{1+u} \frac{1}{1+u^{2}} \ du$ (let $u = \tan x$)
$ \displaystyle = \frac{1}{2} \int \left( \frac{1}{1+u} + \frac{1-u}{1+u^{2}} \right) \ du$
$ \displaystyle = \frac{1}{2} \int \left( \frac{1}{1+u} - \frac{u}{1+u^{2}} + \frac{1}{1+u^{2}} \right) \ du $
$ \displaystyle =\frac{1}{2} \left(\ln(1+u) - \frac{1}{2} \ln(1+u^{2}) + \arctan u \right) + C$
$ \displaystyle =\frac{1}{2} \left(\ln(1+u) - \ln(\sqrt{1+u^{2}}) + \arctan u \right) + C$
$ \displaystyle = \frac{1}{2} \left( \ln(1+\tan x) - \ln (\sqrt{1+\tan^{2}x}) + x \right) + C$
$ \displaystyle = \frac{1}{2} \left( \ln(1+\tan x) - \ln (\sec x) + x \right) + C$
$ \displaystyle = \frac{1}{2} \Big( \ln \big(\frac{1+ \tan x}{\sec x}\big) + x \Big) + C$
$ \displaystyle = \frac{1}{2} \left( \ln (\cos x + \sin x ) + x \right) + C $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/298461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Finding limit of a function ($0$ appears in denominator) This is the function:
$$f(x)= \frac{x^3 + 3x^2+2x}{x^2-2x-8} $$
So I tried to find this limit:
$$\lim_{x\to -2}\frac{x^3 + 3x^2+2x}{x^2-2x-8}=\left[\frac{-8-12-4}{4+4-8}=\frac{-24}{0}\right]= \lim_{x\to -2}\frac{x(x-1)(x-2)}{(x-4)(x+2)}$$
This isn't a case where numerator and denominator, after factorization, have some elements which can reduce a "problematic" element... I need to get rid of $(x+2)$ which makes the fraction explode but I don't know how.
| I assume you meant $$f(x)=\frac{x^3+3x^2+2x}{x^2-2x-8}$$
Then $\lim_{x\to2-}f(x)$ is a indeterminate form or type $\frac{0}{0}$, therefore we apply l'Hopital rule and receive
$$\lim_{x\to2-}f(x)=\lim_{x\to2-}\frac{3x^2+6x+2}{2x-2}=\frac{3\cdot4-2\cdot6+2}{-2\cdot2-2}=\frac{2}{-6}=-\frac{1}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/299055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to find the exact value of $ \cos(36^\circ) $? The problem reads as follows:
Noting that $t=\frac{\pi}{5}$ satisfies $3t=\pi-2t$, find the exact value of
$$\cos(36^\circ)$$
it says that you may find useful the following identities:
$$\cos^2 t+\sin^2 t = 1,\\
\sin 2t = 2\sin t\cos t,\\
\sin 3t = 3\sin t - 4\sin^3 t.
$$
Do I have to do a system of linear equations in function of ..what? $t$? $\cos$?
Thanks in advance :)
| To find $\cos{\pi/5}$, note that
$$\sin{(3 \pi/5)} = \sin{(2 \pi/5)}$$
and
$$\sin{(3 \pi/5)} = 3 \sin{(\pi/5)} - 4 \sin^3{(\pi/5)} = 2 \sin{(\pi/5)} \cos{(\pi/5)}$$
Thus
$$2 \cos{(\pi/5)} = 3 - 4 \sin^2{(\pi/5)} = 4 \cos^2{(\pi/5)} - 1$$
Let $y=\cos{(\pi/5)}$. Then
$$4 y^2-2 y-1=0 \implies y = \frac{1 + \sqrt{5}}{4}$$
because $y>0$. Thus, $\cos{(\pi/5)} = (1+\sqrt{5})/4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/300081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
} |
Prove by Mathematical Induction: $1(1!) + 2(2!) + \cdot \cdot \cdot +n(n!) = (n+1)!-1$ Prove by Mathematical Induction . . .
$1(1!) + 2(2!) + \cdot \cdot \cdot +n(n!) = (n+1)!-1$
I tried solving it, but I got stuck near the end . . .
a. Basis Step:
$(1)(1!) = (1+1)!-1$
$1 = (2\cdot1)-1$
$1 = 1 \checkmark$
b. Inductive Hypothesis
$1(1!) + 2(2!) + \cdot \cdot \cdot +k(k!) = (k+1)!-1$
Prove k+1 is true.
$1(1!) + 2(2!) + \cdot \cdot \cdot +(k+1)(k+1)! = (k+2)!-1$
$\big[RHS\big]$
$(k+2)!-1 = (k+2)(k+1)k!-1$
$\big[LHS\big]$
$=\underbrace{1(1!) + 2(2!) + \cdot \cdot \cdot + k(k+1)!} + (k+1)(k+1)!$ (Explicit Last Step)
$= \underbrace{(k+1)!-1}+(k+1)(k+1)!$ (Inductive Hypothesis Substitution)
$= (k+1)!-1 + (k+1)(k+1)k!$
$= (k+1)k!-1 + (k+1)^{2}k!$
My [LHS] looks nothing like my [RHS] did I do something wrong?
EDIT:
$ = (k+1)k! + (k+1)^2k! -1 $
$ = (k+1)(k!)(1 + (k+1))-1$
$ = (k+1)(k!)(k+2)-1 = (k+2)(k+1)k!-1$
| Your LHS may not look much like your RHS yet, but that's because you haven't finished getting it into the simplest possible form. You have $(k+1)k! - 1 + (k+1)^2 k!$. You're looking to get something minus $1$, so that's somewhat promising. Now what factors do the other two terms (the ones involving $k$) have in common?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/301615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
If $a$ and $b$ are two roots of $x^4 + x^3 - 1 = 0$ prove that $ab$ is a root of $x^6 + x^4 + x^3 - x^2 - 1$. If $a$ and $b$ are two roots of $x^4 + x^3 - 1 = 0$ prove that $ab$ is a root of $x^6 + x^4 + x^3 - x^2 - 1$.
Students and I are unsure how to go about this problem. Also will this be a problem I can solve and prove in front of a class in 20 minutes?
| The resultant
$$h(z)=\text{Res}( \ \text{Res} ( \ z-xy, \ x^4+x^3-1, \ x), \ y^4+y^3-1, \ y)$$
is a $16^{th}$degree polynomial in $z$ with roots $\{a_ia_j:i,j=1,2,3,4\}$, where $a_1,a_2,a_3,a_4$ are the roots of $x^4+x^3-1$.
By expanding the above resultant we get that the polynomial $h(z)$ is $$h(z)= \left( {z}^{4}-{z}^{3}-2\,{z}^{2}+1 \right) \left( {z}^{6}+{z}^{4}+{z}^{3}-{z}^{2}-1 \right)^2.$$
The roots of the polynomial $\left( {z}^{4}-{z}^{3}-2\,{z}^{2}+1 \right)$ are $a_i^2$ and the roots of $\left( {z}^{6}+{z}^{4}+{z}^{3}-{z}^{2}-1 \right)$ are $a_ia_j$ with $i<j$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/301823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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The remainder of $1^2+3^2+5^2+7^2+\cdots+1013^2$ divided by $8$ How to find the remainder of $1^2+3^2+5^2+7^2+\cdots+1013^2$ divided by $8$
| Of course, you could just sum the series:
$$\sum_{k=1}^n (2 k-1)^2 = \frac{4 n (n+1)(n-1)}{3} + n$$
In this case, $n=507$, and you want $507 (4 \cdot 169 \cdot 506 + 1) \mod{8}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/304041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
How can I find the possible values that $\gcd(a+b,a^2+b^2)$ can take, if $\gcd(a,b)=1$ If $\gcd(a,b)=1$, how can I find the values that $\gcd(a+b,a^2+b^2)$ can possibly take? I can't find a way to use any of the elemental divisibility and gcd theorems to find them.
| We have $a^2 + b^2 - a(a+b) = b^2 - ab = -b (a-b)$ and $a^2 + b^2 - b(a+b) = a^2 - ba = a (a-b)$.
So if $d$ divides both $a+b$ and $a^2+b^2$, then $d$ divides $$\gcd(a (a-b), b (a-b)) = \gcd(a, b) (a-b) = a - b.$$
So $d$ divides $a+b + a - b = 2a$ and $a+b - (a - b) = 2b$.
So $d$ divides $2\gcd(a,b)=2$.
So the possibilities for the $\gcd$ appear to be $1$ and $2$, and both clearly occur.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/307545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 4
} |
Calculation of polynomial $g(x)$ satisfies $x\cdot g(x+1)=(x-3)\cdot g(x)$
If a polynomial $g(x)$ satisfies $x\cdot g(x+1)=(x-3)\cdot g(x)$ for all $x$, and $g(3)=6$, then $g(25)=$?
My try: $x\cdot g(x+1)=(x-3)\cdot g(x)$,
Put $x=3$, we get $g(4)=0$, means $(x-4)$ is a factor of $g(x)$.
Similarly put $x=0$. We get $g(0)=0$, means $x$ is a factor of $g(x)$.
This means $g(x)=x\cdot (x-4)h(x)$, where $h(x)$ is a polynomial.
Then how can I calculate it?
Thanks.
| A repeated use of $g(x+1)=\dfrac{x-3}x g(x)$ gives
$$\begin{array} gg(25)&=&\dfrac{21}{24} g(24)\\ &=&\dfrac{22}{24}\dfrac{21}{23} g(23)\\&=&\cdots \\ &=&\dfrac{21}{24}\dfrac{20}{23}\cdots \frac{1}{4} g(3)\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/307739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Values of $a$ for which $(a+4)x^2-2ax+2a-6 <0$ for all $x \in R$ How can we find all values of $a$ for which the inequality $(a+4)x^2-2ax+2a-6 <0$ is satisfied for all $x \in R$?
For the given condition, $D >0$, therefore $ (-2a)^2-4(2a-6)(a+4) >0$. Solving for $a$, I get $(a+6)(a-4) <0$, but the answer is $(-\infty, -6]$ which is not my answer.
| Let $f(x)=(a+4)x^2-2ax+2a-6$. Then $f(x)<0$ for all real $x$ precisely when $a<-4$ and $(2a)^2-4(a+4)(2a-6)<0$. The last inequality reduces to $(a+6)(a-4)>0$, so that $a<-6$ or $a>4$. Putting these together, we have the condition $a<-6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/307956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Geodesic on a surface of revolution using Christoffel Symbols. I have the following problem:
For a function $f:[a,b]\rightarrow \mathbb{R}_{>0}$ and for the open set $U=\{(u_1,u_2)\vert\: a<u_1<u_2, 0\leq u_2<2\pi\}$ consider the (local) surface of revolution $M$ obtained as the image of $\sigma:U\rightarrow U'\subset\mathbb{R}^3$ where
\begin{equation}
\sigma(u_1,u_2) = \left(f(u_1)cos(u_2),f(u_1)sin(u_2),u_1\right)
\end{equation}
For a constant $0\leq c<2\pi$, shot that the meridian curve $\gamma(t)=\sigma(u_1(t),c)$ is a geodesic in $M$.
The way I have tried to solve this as follows:
Recall that $\gamma(t)=\sigma(\gamma_1(t),\gamma_2(t)) = \sigma(u_1(t),c)$ and consider the geodesic equations:
\begin{equation}
\frac{d^2\gamma_k}{dt^2} + \sum_{i,j=1,2}\Gamma_{ij}^k \frac{d\gamma_i}{dt}\frac{d\gamma_j}{dt} = \frac{d^2\gamma_k}{dt^2} + \Gamma_{1,1}^k\left(\frac{d\gamma_1}{dt}\right)^2 + 2\Gamma_{1,2}^k \frac{d\gamma_1}{dt}\frac{d\gamma_2}{dt} + \Gamma_{2,2}^k\left(\frac{d\gamma_2}{dt}\right)^2 = 0
\end{equation}
which should be satisfied for $k=1,2$ if $\gamma(t)$ is a geodesic ($\Gamma_{ij}^k$ are the Christoffel Symbols). Since $\frac{d\gamma_2}{dt} = 0$, the equations above reduce to
\begin{equation}
\frac{d^2\gamma_k}{dt^2} + \Gamma_{1,1}^k\left(\frac{d\gamma_1}{dt}\right)^2 = 0
\end{equation}
Computing $\Gamma_{1,1}^1$ we get:
\begin{equation}
\Gamma_{1,1}^1 =\frac{1}{2}\left(g^{-1}\right)^{1,1}\frac{\partial g_{1,1}}{\partial u_1}= \frac{\frac{\partial f(u_1)}{\partial u_1}\frac{\partial^2f(u_1)}{\partial u_1^2}}{\left(\frac{\partial f(u_1)}{\partial u_1}\right)^2 + 1}
\end{equation}
and $\Gamma_{1,1}^2 = 0$.
The first fundamental form is:
\begin{equation}
(g_{ij}) = \begin{pmatrix}
f'(u_1)^2+1 & 0 \\
0 & f(u_1)^2
\end{pmatrix} = \begin{pmatrix}
\frac{1}{f'(u_1)^2+1} & 0 \\
0 & \frac{1}{f(u_1)^2}
\end{pmatrix}^{-1}.
\end{equation}
This does not seem to work out, and I was wondering why? What mistake have I made?
Cheers!
| Thank you THW, although I ended up doing it another way:
First of all, lets compute the first fundemental form and its inverse:
\begin{equation}
(g_{ij}) = \begin{pmatrix}
e & f \\
f & g
\end{pmatrix} = \begin{pmatrix}
f'(u_1)^2+1 & 0 \\
0 & f(u_1)^2
\end{pmatrix} = \begin{pmatrix}
\frac{1}{f'(u_1)^2+1} & 0 \\
0 & \frac{1}{f(u_1)^2}
\end{pmatrix}^{-1}.
\end{equation}
For $\gamma(t)$ to a geodesic, we only need to show that the geodesic curvature of $\gamma(t)$ is zero. Beltrami’s formula for geodesic curvature states that the geodesic curvature $\kappa_g$ can expressed as:
\begin{align*}
\kappa_g &= [\Gamma_{1,1}^2\left(\frac{d\gamma_1}{dt}\right)^3 + \left(2\Gamma_{1,2}^2 - \Gamma_{1,1}^1\right)\left(\frac{d\gamma_1}{dt}\right)^2\frac{d\gamma_2}{dt} + \left(\Gamma_{2,2}^2 -
2\Gamma_{1,2}^1\right)\frac{d\gamma_1}{dt}\left({d\gamma_2}{dt}\right)^2 \\
&\quad\quad\quad - \Gamma_{2,2}^1 \left(\frac{d\gamma_2}{dt}\right)^3 + \frac{d\gamma_1}{dt}\frac{d^2\gamma_2}{dt^2} - \frac{d^2\gamma_1}{dt^2}\frac{d\gamma_1}{dt}]\sqrt{eg-f^2}
\end{align*}
Since $\frac{d\gamma_2}{dt} = 0$, the expression reduces to
\begin{equation}
\Gamma_{1,1}^2\left(\frac{du_1}{dt}\right)^3\sqrt{eg}
\end{equation}
Computing $\Gamma_{1,1}^2$:
\begin{equation}
\Gamma_{1,1}^2 = \left(g^{-1}\right)^{2,2}\left(\frac{\partial g_{1,2}}{\partial u_1} + \frac{\partial g_{2,1}}{\partial u_1} - \frac{\partial g_{2,2}}{\partial u_2}\right) = 0
\end{equation}
Thus the geodiesic curvature $\kappa_g$ of $\gamma(t)$ is zero and $\gamma(t)$ is therefore a geodesic.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/309069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
A few definite integrals a) $\displaystyle \int_{-1}^0 x \sqrt{2 - 5x}dx. $
I think I should use some change of variables here, but it didn't work the way I tried to use it.
b) $\displaystyle \int_{0}^{2} \frac{1}{(x + 2)(x + 3)}dx.$
| Another way is by a rationalizing substitution:
$$
\begin{align}
u & = \sqrt{2-5x} \\[8pt]
u^2 & = 2 - 5x \\[8pt]
2u\,du & = -5\,dx \\[8pt]
\frac{-2}{5} u \, du & = dx \\[8pt]
x & = \frac{2-u^2}{5}
\end{align}
$$
When $x=-1$ then $u=\sqrt{7}$ and when $x=0$ then $u=\sqrt{2}$. So
$$
\int_{-1}^0 x \sqrt{2-5x} \, dx = \int_\sqrt{7}^\sqrt{2} \frac{2-u^2}{5}\cdot u\left(\frac{-2}{5} u\, du \right) = \frac{4}{25}\int_\sqrt{2}^\sqrt{7}(2u^2-u^4)\,du
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/309392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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