Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find the maximum value of $A$ Let $a;b;c>0$ such that $a+b+c=6$. Find the maximum value of $A=a^2bc+a^2+2b^2+2c^2$
WLOG $b\ge c$. I see maximum value of $A=36$ at $(a;b;c)=(2;1;3)$
So i need to prove $A\le 36$. Or I will prove
$(a+b+c)^4\ge 36a^2bc+(a^2+2b^2+2c^2)(a+b+c)^2$
Or $(2a-b-c)(b^3+c^3+a^2b+a^2c+2ab^2+2ac^2-12abc-b^2c-bc^2)\ge 0$
Then Im stuck here, help me solve it.
| Given that $a+b+c=6$ we can substitute $a=6-b-c$ into $A$ to get
$$A=(6-b-c)^2(bc+1)+2(b^2+c^2),$$
where $b,c>0$ and $b+c<6$. Differentiating with respect to $b$ and $c$
yields
$$\frac{\partial A}{\partial b}=-2(6-b-c)(bc+1)+(6-b-c)^2c+4b,$$
$$\frac{\partial A}{\partial b}=-2(6-b-c)(bc+1)+(6-b-c)^2b+4c.$$
If there is a maximum then both derivatives must equal 0 in such a point, and hence
$$0=\frac{\partial A}{\partial b}-\frac{\partial A}{\partial c}=(c-b)((6-b-c)^2-4),$$
so either $b=c$ or $b+c=4$. In the latter case we get $c=4-b$ and so
$$A=2(bc+1)+2(b^2+c^2)=2(b(4-b)+1)+2(b^2+(4-b)^2)=2b^2-8b+34,$$
which does not assume a maximum on $(0,4)$. In the former case where $b=c$ we get
$$A=(6-2b)(b^2+1)+4b^2=-2b^3+10b^2-2b+6,$$
which does not assume a maximum on $(0,3)$. So the maximum does not exist.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the prime factorisation of $6500$ and $1120$, and write down, in factorised form, $\gcd(6500, 1120)$ and $\operatorname{lcm}(6500, 1120)$. (i) Find the prime factorisation of $6500$, and of $1120$.
What is the typical way to go about this? Just using common divisibility rules? That's what I did. I'm not sure if there's a more structured way that I should be doing this, since it could be more difficult depending on the number? The above seem to be easy cases.
$65 \times 100 = 6500$
$13 \times 5 \times 25 \times 4 = 6500$
$13 \times 5 \times 5^2 \times 2^2 = 6500$
$13 \times 5^3 \times 2^2 = 6500$
$1120 = 112 \times 10$
$= 66 \times 2 \times 5 \times 2$
$= 6 \times 11 \times 2^2 \times 5$
$= 3 \times 11 \times 2^3 \times 5$
(ii) Hence write down, in factorised form, $gcd(6500, 1120)$ and $lcm(6500, 1120)$.
For GCD we just selected the highest powers of the numbers that are common to both? So it would be $\gcd(6500, 1120) = 5^3 \times 13 \times 2^3$?
And I think for LCM we take the lowest powers of each number? So it would be $\operatorname{lcm}(6500, 1120) = 5 \times 2^2$?
Thanks for any help.
| One can do the gcd without factoring first. Oh, $$ \operatorname{lcm}(a,b) = \frac{ab}{\gcd(a,b)} $$
$$ \frac{ 6500 }{ 1120 } = 5 + \frac{ 900 }{ 1120 } $$
$$ \frac{ 1120 }{ 900 } = 1 + \frac{ 220 }{ 900 } $$
$$ \frac{ 900 }{ 220 } = 4 + \frac{ 20 }{ 220 } $$
$$ \frac{ 220 }{ 20 } = 11 + \frac{ 0 }{ 20 } $$
Simple continued fraction tableau:
$$
\begin{array}{cccccccccc}
& & 5 & & 1 & & 4 & & 11 & \\
\frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 5 }{ 1 } & & \frac{ 6 }{ 1 } & & \frac{ 29 }{ 5 } & & \frac{ 325 }{ 56 }
\end{array}
$$
$$ $$
$$ 325 \cdot 5 - 56 \cdot 29 = 1 $$
$$ \gcd( 6500, 1120 ) = 20 $$
$$ 6500 \cdot 5 - 1120 \cdot 29 = 20 $$
| {
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"timestamp": "2023-03-29T00:00:00",
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On Lame's Theorem I was trying to study Lame's theorem but I'm a little bit stuck on one specific step... I write here the theorem and the proof:
Lame's theorem:
using the Euclidean algorithm to find the greatest common divisor of two positive integers has number of divisions less than or equal five times the number of decimal digits in the minimum of the two integers.
Proof:
Let $a$ and $b$ be two positive integers where $a > b$.
Applying the Euclidean algorithm to find the greatest common divisor of two integers with $a = r_{0}$ and $b = r_{1}$, we get
$r_{0} = r_{1}q_{1}+r_{2}$, $0≤r_{2}<r_{1}, $
$r_{1} = r_{2}q_{2}+r_{3}$, $0≤r_{3}<r_{2}, $
. . .
$r_{n-2} = r_{n-1}q_{n-1}+r_{n}$, $0≤r_{n}<r_{n-1}, $
$r_{n-1} = r_{n}q_{n}$
Notice that each of the quotients $q_{1}, q_{2}, ..., q_{n-1} $ are all greater than $1$ and $q_{n} ≥ 2$ and this is because $r_{n} < r_{n-1}.$
Thus we have
$r_{n} ≥ 1=f_{2}$,
$r_{n-1} ≥ 2r_{n} ≥ 2f_{2} = f_{3}$,
$r_{n-2} ≥ r_{n-1} + r_{n} ≥ f_{3} + f_{2} = f_{4}$,
$r_{n-3} ≥ r_{n-2} + r_{n-1} ≥ f_{4} + f_{3} = f_{5}$,
...
$r_{2} ≥ r_{3} + r_{4} ≥ f_{n-1} + f_{n-2} = f_{n}$,
$b = r_{1} ≥ r_{2} + r_{3} ≥ f_{n} + f_{n-1} = f_{n+1}$.
Thus notice that $b≥f_{n+1}$.
By a Lemma I don't report here, we have $f_{n+1}>α^{n−1}$ for $n>2$.
As a result, we have $b > α^{n−1}$.
Now notice since $\log_{10} \alpha > \frac{1}{5}$
we see that
$\log_{10}b > (n − 1)/5$.
Thus we have
$(n - 1)< 5 log_{10}b$
Now let $b$ has $k$ decimal digits. As a result, we have $b < 10^k$ and thus $log_{10}b < k$. Hence we conclude that $(n − 1) < 5k$.
Since $k$ is an integer, we conclude that $n ≤ 5k$.
What I really don't understand is just this line:
Now notice since $\log_{10} \alpha > \frac{1}{5}$,
My question is: why $\frac{1}{5}$ has been chosen?
Where does it come from?
Has it been chosen because the theorem says:
[...]less than or equal five times [...]
?
Thank you
| You can play around with things a bit too.
Given $\alpha=\dfrac{1+\sqrt{5}}{2}$, it follows that
\begin{align}
\alpha^2 &= \alpha + 1 \\
\alpha^3 &= \alpha^2 + \alpha \\
&= 2\alpha + 1 \\
\alpha^4 &= 2\alpha^2 + \alpha \\
&= 3\alpha + 2 \\
\alpha^5 &= 3\alpha^2 + 2\alpha \\
&= 5\alpha + 3 \\
&> \dfrac{5+10}{2}+3 &\text{Since $\sqrt 5 >2.$} \\
&> 10
\end{align}
So $\alpha^5 > 10 \implies 5 \log_{10}\alpha > 1 \implies \log_{10}\alpha > \dfrac 15.$
A lot of problems can be solved by assuming that what you find out is useful and then going on to see what happens. He probably chose $\dfrac 15$ because he found, using whatever method, that $\log_{10} \alpha > \dfrac 15$. He then just assumed that that was useful information and proceeded with the rest of the proof. If faith isn't a good enough reason, then just follow through out of curiosity to see what happens. Even if you fail, you now know more that you did before.
| {
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How to evaluate the integral $\int_{0}^{1} \frac{(\log {x})^2}{\sqrt{x^2(1-x)^2+x^2+(1-x)^2}}dx$ $$I=\int_{0}^{1} \frac{(\log {x})^2}{\sqrt{x^2(1-x)^2+x^2+(1-x)^2}}dx$$
My attempt:
$$I=\int_{0}^{1} \frac{(\log {x})^2}{\sqrt{x^2(1-x)^2-2x(1-x)+(x+1-x)^2}}dx$$
$$=\int_{0}^{1} \frac{(\log {x})^2}{\sqrt{(1-x(1-x))^2}}dx=\int_{0}^{1} \frac{(\log {x})^2}{(x^2-x+1)}dx$$
How do I proceed further next step?
| Notice that
$$ I = \frac{1}{2} \int_{0}^{\infty} \frac{\log^2 x}{x^2-x+1} \, dx.$$
To evaluate this integral, we consider the following contour integral
$$ J = \oint_{C}\frac{\operatorname{Log}^3z}{z^2-z+1} \, dz,$$
where $\operatorname{Log}z$ is the complex logarithm with the branch cut $[0,\infty)$ and $C$ is a keyhole contour enclosing this branch cut. A standard argument shows that $J$ reduces to
$$ J
= \int_{0}^{\infty} \frac{\log^3 x - (2\pi i + \log x)^3}{x^2 - x + 1} \, dx.
\tag{1} $$
On the other hand, in the region enclosed by $C$, there are two simple poles of the integrand at $z_1 = e^{\pi i/3}$ and $z_2 = e^{5\pi i/3}$. So by the residue theorem,
$$ J = 2\pi i \sum_{a \in \{z_1, z_2\}} \underset{z=a}{\mathrm{Res}} \, \frac{\operatorname{Log}^3 z}{z^2-z+1} = \frac{248 \pi^4 i}{27\sqrt{3}}. \tag{2} $$
Comparing imaginary parts of $\text{(1)}$ and $\text{(2)}$, we obtain
$$ \frac{248 \pi^4}{27\sqrt{3}} = \operatorname{Im}(J) = \int_{0}^{\infty} \frac{8\pi^3 - 6\pi \log^2 x}{x^2 - x + 1} \, dx $$
and solving this equation with respect to $I$ together with $\int_{0}^{\infty} \frac{dx}{x^2 - x + 1} = \frac{4\pi}{3\sqrt{3}}$ gives
$$ I = \frac{10\pi^3}{81\sqrt{3}}. $$
| {
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Distance between incentre and excentre of a triangle I am unable to get anywhere regarding the distance between the incentre and an excentre of $\triangle ABC$. I do know that it is generalised as:
$$IJ_c=c\sec\left(\frac C2\right)$$
For the incenter $I$ and excenter $J_c$, where side $c$ is opposite $\angle C$.
But how can I prove this? Kindly assist.
| Consider this diagram:
Consider that the distance between $I$ and $E_c$ is simply:
$$d=E_cC-IC\tag{1}$$
And we have:
$$\cos \left(\frac C2\right)=\frac{KC}{E_cC}\implies E_cC=\frac{KC}{\cos(C/2)}\\
\cos\left(\frac C2\right)=\frac{T_bC}{IC}\implies IC=\frac{T_bC}{\cos(C/2)}$$
However, we know that $T_bC=\frac{b+a-c}2$. This makes $IC=\frac{b+a-c}{2\cos (C/2)}$, and from $(1)$, we get:
$$d=\frac{KC}{\cos(C/2)}-\frac{b+a-c}{2\cos(C/2)}\tag{2}$$
However, looking at the diagram, we know that $KC=b+E_cA\sin(A/2)$, because $KC=b+AK$, and $AK$ is easily derivable using law of sines for right triangles. Therefore, $(2)$ becomes:
$$d=\frac{b+E_cA \sin(A/2)}{\cos(C/2)}-\frac{b+a-c}{2\cos(C/2)}\\
\implies d=\frac{b+E_cA\sin(A/2)-\frac{b+a-c}{2}}{\cos (C/2)}$$
However, from the original statement $d=\frac{c}{\cos (C/2)}$, what we only need to show is that:
$$b+E_c A\sin(A/2)-\frac{b+a-c}2=c\tag{3}$$
Polishing $(3)$ we get:
$$b-a+2E_cA\sin(A/2)=c$$
This implies that in order to get $c$, we need:
$$E_cA\sin (A/2)=\frac{a+c-b}{2}\tag{4}$$
From the diagram, consider that $\sin (A/2)$, can be written as:
$$\frac{\sin(C/2)}{IA}=\frac{\sin (A/2)}{IC}\implies \sin (A/2)=\frac{IC \sin (C/2)}{IA}\tag{5}$$
Now, see that $IT_b$ is essentially the inradius, which is equal to:
$$r=IT_b=IC \sin (C/2)$$
Knowing that the inradius $r$ can be written as (which is easily provable):
$$r=\frac{ab\sin C}{a+b+c}$$
Using the value of the updated $(5)$, and substituting it in $(4)$, we get:
$$E_cA\left(\frac{ab\sin C}{(a+b+c)IA}\right)=\frac{a+c-b}2$$
After some, rearrangement, we get:
$$\frac{E_cA}{IA}=\frac{(a+b+c)(a-b+c)}{2ab\sin C}\tag{6}$$
Now, consider the right triangle $E_cAI$ and after some angle-chasing, you will get that $\angle I=\frac{a+c}2$. Notice that simply, $\tan \angle I=\frac{E_cA}{IA}$. Using this information, we get:
$$\tan\left(\frac{A+C}2\right)=\frac{(a+b+c)(a-b+c)}{2ab\sin C} \tag{7}$$
Expanding the RHS, we get:
$$\tan\left(\frac{A+C}2\right)=\frac{a^2-b^2+c^2+2ac}{2ab\sin C} \tag{8}$$
Using the cosine law, we can show that $2ac\cos B=a^2+c^2-b^2$, substituting this in $(8)$ and we get:
$$\tan\left(\frac{A+C}2\right)=\frac{2ac\cos B+2ac}{2ab\sin C}=\frac{c(1+\cos B) }{b\sin C}$$
Now replacing $B=\pi - A-C$, we get:
$$\tan\left(\frac{A+C}2\right)=\frac{c(1-\cos{(A+C)})}{b\sin C}$$
Using the half-angle formula for $\tan$, we get:
$$\frac{\sin (A+C)}{1+\cos(A+C)}=\frac{c(1-\cos{(A+C)})}{b\sin C}\\
b\sin(A+C)\sin C=c(1-\cos^2(A+C))\\
b\sin (A+C)\sin C=c(\sin^2(A+C))\\
\sin(A+C)=\frac{b\sin C}{c}$$
However, using the sine law, we can write $b$ as $b=\frac{c \sin B}{\sin C}$, thus we get:
$$\sin(A+C)=\sin B$$
Which is always true since:
$$B=\pi-A-C\implies \sin(\pi-A-C)=\sin(A+C)\\
Q.E.D.$$
This proof was very long winded, however. There are many ways to write the distance between the excenters and incenters. For example, you have:
$$d=\frac{r_x-r}{\sin(C/2)}$$
For the exradius $r_x$ and in radius $r$. Also, you have $d$:
$$d=\frac{r}{\sin(\frac A2)\cos(\frac{A+C}2)}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the fastest way to estimate the Arc Length of an Ellipse? To estimate the circumference of an ellipse there are some good approximations.
$a$ is the semi-major radius and $b$ is the semi-minor radius.
$$L \approx \pi(a+b) \frac{(64-3d^4)}{(64-16d^2 )},\quad \text{where}\;d = \frac{(a - b)}{(a+b)}$$
Are there any similar formulas to approximate the arc length of an ellipse from $\theta_1$ to $\theta_2$?
If not what are some computationally fast ways to approximate the arc length to within about $1\%$ to $0.1\%$ of $a$?
| I'll assume $\theta_1$ and $\theta_2$ refer to the parametrization
$$ \eqalign{x &= a \cos(\theta)\cr y &= b \sin(\theta)}$$
of the ellipse.
Thus the arc length in question is
$$ L = \int_{\theta_1}^{\theta_2} \sqrt{a^2 \sin^2(\theta) + b^2 \cos^2(\theta)}\; d\theta $$
We want a good approximation of the integrand that is easy to integrate.
It may be best to look at two cases, depending on which of the terms inside the square root is larger.
For $a |\sin(\theta)| \ge b |\cos(\theta)|$, we take
$$\sqrt{a^2 \sin^2(\theta) + b^2 \cos^2(\theta)} = a |\sin(\theta)| \sqrt{1 + \frac{b^2}{a^2} \cot^2(\theta)}$$
and look for a good approximation of $\sqrt{1+t^2}$ for $0 \le t \le 1$.
The best polynomial approximation of degree $3$ for this is approximately
$$ 1.000127929-0.00619431946 \;t+.5478616944\; t^2-.1274538129\; t^3$$
with maximum absolute error $\approx .0001280863448$. Write these coefficients as $c_0, \ldots, c_3$. Thus on the part of the interval where $a |\sin(\theta)| \ge b |\cos(\theta)|$, we can integrate
$$ \pm a \sin(\theta) \left(c_0 + c_1 \frac{b}{a} \cot(\theta) + c_2 \frac{b^2}{a^2} \cot^2(\theta) + \frac{b^3}{a^3} \cot^3(\theta)\right)$$
($+$ on an interval where $\sin(\theta) \ge 0$, $-$ where $\sin(\theta)<0$).
An antiderivative is
$$ \pm\left( - a c_0 \cos(\theta) + b c_1 \sin(\theta) + \frac{b^2}{a} c_2 \left(\cos(\theta)+\ln(\csc(\theta)-\cot(\theta))\right) - \frac{b^3}{a^2} (\csc(\theta)+\sin(\theta))\right)$$
Similarly, for $a |\sin(\theta)| \le b |\cos(\theta)|$ take
$$\sqrt{a^2 \sin^2(\theta) + b^2 \cos^2(\theta)} = b |\cos(\theta)| \sqrt{1 + \frac{a^2}{b^2} \tan^2(\theta)}$$
and integrate
$$\pm b \cos(\theta) \left(c_0 + c_1 \frac{a}{b} \tan(\theta) + c_2 \frac{a^2}{b^2} \tan^2(\theta) + c_3 \frac{a^3}{b^3} \tan^3(\theta) \right)$$
| {
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Prove that $a+b+c\le 0$ $a,b,c$ are integer numbers different from zero and
$$\frac{a}{b+c^2}=\frac{a+c^2}{b}$$
Prove that $a+b+c\le 0$.
I know how to prove that $a+b<0$ but do not know what about $c$.
| Since $c \in \mathbb{Z}$, we have $c \le |c|\le c^2$
$$0=(a+b)c^2+c^4$$
$$a+b+c^2=0$$
$$a+b+c \le a+b+c^2=0$$
| {
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How to find the sum of the the first n elements in the series 1,2,4,5,7,8.... as a function of n? I see I can split it up into two arithmetic series (1,4,7... and 2,5,8...) and add the two sums, but what is the compact notation for the sum of these two series as a function of n?
| Let us first find the formula for even $n$. In this case, $m = n/2$ is a whole number, and moreover the sum of the first $n$ elements of your sequence is the sum of the first $m$ elements of the two arithmetic progressions, that is,
$$
m + 3\frac{m(m-1)}{2} + 2m + 3\frac{m(m-1)}{2} = 3 (m + m(m-1)) = \frac{3n^2}{4}.
$$
If $n$ is odd, then $n-1$ is even and we know that the sum of the first $n-1$ entries is $\frac{3(n-1)^2}{4}$. Moreover, the last entry in the sequence was the $((n-1)/2 + 1)$-st entry of the first arithmetic progression, that is $1,4,7,10, \ldots$. But the value of that entry is $3 ((n-1)/2 + 1) - 2 = 3(n-1)/2 + 1$, so using the formula for the even case, the sum of the first $n$ entries equals
$$
\frac{3(n-1)^2}{4} + 3(n-1)/2 + 1 = \frac{3}{4} (n^2 - 2n + 2n + 1 - 2 + 4/3) = \frac{3n^2}{4} + \frac{1}{4}.
$$
| {
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Given $\sin(t) + \cos(t) = a$, derive an expression in '$a$' for $(\cos(t))^4 + (\sin(t))^4$ I received this question from a student's trigonometry review assignment. After spending an embarrassing amount of time on it, I consulted others and learned that nobody has been able to solve this problem for multiple semesters. I wonder if there is a typo in the statement or if I just haven't been rigorous enough? Here is the question:
$$\text{Given } \sin{t} + \cos{t} = a, \text{ find an equivalent expression for } \sin^4{t} + \cos^4{t} \text{ in terms of } a.$$
Has anybody seen this one before? I tried (among other things) this (which is an approximation of an earlier attempt):
$(\sin{t} + \cos{t})^4$ and using whatever identities I could remember;
$(\sin{t} + \cos{t})^4 = \sin^4{t} + \cos^4(t) + 4\sin{t}\cos^3{t} + 6\sin^2{t}\cos^2{t} + 4\sin^3{t}\cos{t}$
so then
$\begin{align}
\sin^4{t} + \cos^4{t} &= (\sin{t} + \cos{t})^4 - 4\sin{t}\cos^3{t} - 6\sin^2{t}\cos^2{t} - 4\sin^3{t}\cos{t}\\
&= (\sin{t} + \cos{t})^4 - 2\sin{t}\cos{t}(2\cos^2{t} + 3\sin{t}\cos{t} + 2\sin^2{t})\\
&= (\sin{t} + \cos{t})^4 - 2\sin{t}\cos{t}(3\sin{t}\cos{t} + 4)\\
&= a^4 - 2\sin{t}\cos{t}(3\sin{t}\cos{t} + 4)
\end{align}$
Couldn't get further than this, felt like I was overthinking it.
| $$\sin t+\cos t=a\\(\sin t + \cos t)^2=a^2\\2\sin t \cos t=a^2-1
\\(\sin t + \cos t)^4=a^4\\ \sin^4 t + \cos^4 t+4\sin t \cos t (\sin^2 t+\cos^2 t)+6\sin^2 t \cos^2 t=a^4\\ \sin^4 t + \cos^4 t +2(a^2-1)+\frac 32(a^2-1)^2=a^4\\
\sin^4 t+\cos^4 t=a^4-2(a^2-1)-\frac 32(a^2-1)^2$$
| {
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Evaluating $\frac{1}{\sin(2x)} + \frac{1}{\sin(4x)} + \frac{1}{\sin(8x)} + \frac{1}{\sin(16x)}$
Evaluate
$$\dfrac{1}{\sin(2x)} + \dfrac{1}{\sin(4x)} + \dfrac{1}{\sin(8x)} + \dfrac{1}{\sin(16x)}$$
It would be tough for us to solve it using trigonometric identities. There should be strictly an easy trick to proceed.
Rewriting and using trigonometric identities
$$\dfrac{1}{\sin(2x)} + \dfrac{1}{\sin(2x) \cos (2x)} + \dfrac{1}{ 2\big [2\sin (2x)\cos (2x)\cos (4x)\big ]} + \dfrac{1}{\sin(16x)}$$
What am I missing?
Regards
| $$\frac{1}{\sin2x}=\frac{\sin x}{\sin 2x \sin x}=\frac{\sin (2x-x)}{\sin 2x \sin x}=\frac{\sin 2x \cos x - \cos 2x \sin x}{\sin 2x \sin x}=\cot x - \cot2x$$
$$\frac{1}{\sin4x}=\cot 2x - \cot4x$$
$$\frac{1}{\sin8x}=\cot 4x - \cot8x$$
$$\frac{1}{\sin16x}=\cot 8x - \cot16x$$
$$\frac{1}{\sin2x}+\frac{1}{\sin4x}+\frac{1}{\sin8x}+\frac{1}{\sin16x}=\cot x-\cot 16x$$
| {
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find the value of $p(-\pi)$ and $p(0)$? Let $p(x)$ be a polynomial of degree $7$ with real coefficient such that $p(\pi)=\sqrt3.$ and
$$\int_{-\pi}^{\pi}x^{k}p(x)dx=0$$
for $0\leq k \leq 6$. Then find the value of $p(-\pi)$ and $p(0)$?
If I take general polynomial then it is difficult to find and also time consuming so please help to solve. Thanks
| Consider the space $\mathbb{R}_{\le 7}[x]$ of real polynomials of degree at most $7$, equipped with the inner product $\langle f,g\rangle = \int_{-\pi}^\pi fg$.
Your polynomial $p$ satisfies $p \perp \{1, x, \ldots, x^6\}$. We know that a basis for $\mathbb{R}_{\le 7}[x]$ is given by $\{1, x,\ldots, x^7\}$ so applying Gram-Schmidt process on it gives an orthonormal basis
\begin{equation}
\frac1{\sqrt{2\pi}}, \frac{\sqrt{\frac{3}{2}} x}{\pi ^{3/2}},\frac{3 \sqrt{\frac{5}{2}} \left(x^2-\frac{\pi ^2}{3}\right)}{2 \pi ^{5/2}},\frac{5 \sqrt{\frac{7}{2}} \left(x^3-\frac{3 \pi ^2 x}{5}\right)}{2 \pi ^{7/2}},\\
\frac{3 \left(35 x^4-30 \pi ^2 x^2+3 \pi ^4\right)}{8 \sqrt{2} \pi ^{9/2}},\frac{\sqrt{\frac{11}{2}} \left(63 x^5-70 \pi ^2 x^3+15 \pi ^4 x\right)}{8 \pi ^{11/2}},\\\frac{\sqrt{\frac{13}{2}} \left(231 x^6-315 \pi ^2 x^4+105 \pi ^4 x^2-5 \pi ^6\right)}{16 \pi ^{13/2}},\frac{\sqrt{\frac{15}{2}} \left(429 x^7-693 \pi ^2 x^5+315 \pi ^4 x^3-35 \pi ^6 x\right)}{16 \pi ^{15/2}}
\end{equation}
Hence $p(x) = \frac{\lambda}{16 \pi ^{15/2}}\sqrt{\frac{15}{2}} \left(429 x^7-693 \pi ^2 x^5+315 \pi ^4 x^3-35 \pi ^6 x\right)$ for some constant $\lambda \in \mathbb{R}$. Find $\lambda$ by using the condition $p(\pi) = \sqrt{3}$ and you will have found $p$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2910534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $ \lim_{x \to 0} \left( {\frac{1}{x^2}} - {\frac{1} {\sin^2 x} }\right) $ $$\lim_{x\to0}\left({\frac{1}{x^2}}-{\frac{1}{\sin^2x}}\right)$$
Using the L'Hospital Rule I obtained the value $-1/4$, but the answer is given to be $-1/3$. I can't find the mistake. Here's what I did; please point out the mistake.
\begin{align}
\lim_{x\to0}\left({\frac{1}{x^2}}-{\frac{1}{\sin^2x}}\right)&=\lim_{x\to0}\frac{(\sin x+x)(\sin x-x)}{(x\sin x)(x\sin x)} \\[1ex]
&=\lim_{x\to0}\left(\frac{\sin x+x}{x\sin x}\right)\lim_{x\to0}\left(\frac{\sin x-x}{x\sin x}\right) \\[1ex]
&=\lim_{x\to0}\left(\frac{\cos x+1}{\sin x+x\cos x}\right)\lim_{x\to0}\left(\frac{\cos x-1}{\sin x+x\cos x}\right) \\[1ex]
&=\lim_{x\to0}\:(\cos x+1)\,\lim_{x\to0}\left(\frac{\cos x-1}{(\sin x+x\cos x)^2}\right) \\[1ex]
&=\lim_{x\to0}\frac{-\sin x}{(\sin x+x\cos x)(2\cos x-x\sin x)} \\[1ex]
&=-\lim_{x\to0}\left[\frac{1}{1+\cos x\left(\frac{x}{\sin x}\right)}\right]\left(\frac{1}{2\cos x-x\sin x}\right) \\[1ex]
&=-\frac{1}{2}\left[\lim_{x\to0}\,\frac{1}{1+\cos x}\right] \\[1ex]
&=-\frac{1}{4}
\end{align}
| As an alternative by Taylor expansion as $x\to 0$
$$\sin x = x -\frac16x^3 + o(x^3)\implies \sin^2 x = \left(x -\frac16x^3 + o(x^3)\right)^2=x^2-\frac13x^4+o(x^4)$$
we have
$$\frac{1}{x^2} - \frac{1} {\sin^2 x} =\frac{\sin^2 x-x^2}{x^2\sin^2 x}=\frac{x^2-\frac13x^4+o(x^4)-x^2}{x^2\left(x^2-\frac13x^4+o(x^4)\right)}=$$$$=\frac{-\frac13x^4+o(x^4)}{x^4+o(x^4)}=\frac{-\frac13+o(1)}{1+o(1)}\to -\frac13$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2910595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 6
} |
If $\sin^8(x)+\cos^8(x)=48/128$, then find the value of $x$? If $$\sin^8(x)+\cos^8(x)=48/128,$$ then find the value of $x$? I tried this by De Moivre's theorem:
$$(\cos\theta+i\sin\theta)^n=\cos(n\theta)+i\sin(\theta)=e^{in\theta}$$
But could not proceed further please help.
| \begin{align}
\sin^8 x+\cos^8 x&=(\sin^4 x+\cos^4)^2-2\sin^4 x\cos^4 x\\
&=\left[(\sin^2 x+\cos^2 x)^2-2\sin^2 x\cos^2 x\right]^2-\cfrac {\sin^4 2x} 8\\
&=(1-\cfrac {\sin^2 2x}2)^2--\cfrac {\sin^4 2x} 8
\end{align}
Let $t=\sin^2 2x$, we have
$$(1-\cfrac {t}2)^2-\cfrac {t^2}8=\cfrac {48}{128}\Leftrightarrow t^2-8t+5=0,$$ which implies
$$\sin^2 2x=4-\sqrt{11} \hspace{1in}\text{(with the other invalid root discarded)}$$
hence $$ x=\pm \cfrac {\sin^{-1}(\sqrt{4-\sqrt{11}})}2$$or more precisely,
$$x=\pm\cfrac {\sin^{-1}(\sqrt{4-\sqrt{11}})}2+k\pi,\>k\in \mathbb{Z}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2910845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 3
} |
Using geometry to show that $\int_{0}^{b} \sqrt{1-x^2} dx = \frac{1}{2}b\sqrt{1-b^2}+\frac{1}{2}\sin^{-1}b$
Show that for $0\leq b\leq1$,
$$\int_{0}^{b} \sqrt{1-x^2} dx = \frac{1}{2}b\sqrt{1-b^2}+\frac{1}{2}\sin^{-1}b$$
by using geometry.
This is what I have thus far:
$\int_{0}^{b} \sqrt{1-x^2} dx$
$$=y=\sqrt{1-x^2}$$
$$=y^2=1-x^2$$
$$=x^2+y^2=1$$
From the picture above, to find the area we add the area of the sector and the area of the triangle.
Area of sector: $\frac{θ}{2}r^2$
Area of triangle: $\frac{1}{2}bh$
Side note: $(h=a)$
I have no clue what to do next. Please help.
| The area of the triangle with sides $b$ and $a=\sqrt{1-b^2}$ forming the right angle is
$$
\frac 12b\sqrt{1-b^2}\ .
$$
The angle, marked in black, can be extracted from the triangle with sides $a,b,r=1$, it is $\arcsin (b/1)$. So the sector has area
$$\frac 12\arcsin b\cdot 1^2\ .
$$
Adding the two areas we get the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2913765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find all non-negative integer $n$ that satisfy $f(x+1)+f(x-1)=\sqrt nf(x)$
Find all non-negative integer $n$ that there exists a non-periodic function $f:\Bbb R->\Bbb R\quad f(x+1)+f(x-1)=\sqrt nf(x)\forall x$.
My attempt:
$f(x+2)+f(x)=\sqrt n\cdot f(x+1)$
$\sqrt n \cdot (f(x+1)+f(x+3))=n\cdot f(x+2)$
$f(x+2)+f(x+4)=\sqrt n\cdot f(x+3)$
Therefore $$f(x)+f(x+4)=(n-2)\cdot f(x+2)\tag{*}$$
$n=0\implies f(x)=-f(x+2)\implies f(x)=f(x+4)$
$n=1\implies f(x+2)+f(x)=f(x+1)\,$ and $\,f(x+3)+f(x+1)=f(x+2)\implies f(x)=f(x+3)$
$n=2\quad(*)\implies f(x)+f(x+4)=0\,$ A similar argument hold for $n=0$.
$n=3\quad(*) \implies f(x)+f(x+4)=f(x+2).$ A similar argument hold for $n=1.$
$n=4\quad f(x)=x$ works.
$\forall n>4,f(x)$ can be defined recursively:
$i)\;$ $f(x)=f(\lfloor x\rfloor)\forall x$
$ii)\;$ $f(0)=0,\;f(1)=1$
$iii)\;$ $f(x)+f(x+2)=\sqrt n \cdot f(x+1)\;\forall x\in\Bbb Z$
It's easy to show that this satisfy the riginal functional equation, but we still need to prove it's non-periodic. It's sufficient to prove $iii)$ is monotonic.
$f(x+2)=\sqrt n\cdot f(x+1)-f(x)>2\cdot f(x+1)-f(x)\ge f(x+1)$
Is my proof correct?
Any help appreciated.
| For a nonnegative integer $n$, the roots of the polynomial $t^2-\sqrt{n}\,t+1$ are primitive $k$-th roots of unity for some integer $k>0$ if and only if $n\leq 3$. (Note that $k=4$ for $n=0$, $k=6$ for $n=1$, $k=8$ for $n=2$, and $k=12$ for $n=3$.) This shows that any solution $f:\mathbb{R}\to\mathbb{R}$ to the functional equation
$$f(x+1)+f(x-1)=\sqrt{n}\,f(x)\text{ for all }x\in\mathbb{R}$$
is periodic for each $n=0,1,2,3$. Cesareo gave nonperiodic solutions for integers $n\geq 4$.
Indeed, for $n=0$, we get
$$f(x+4)=-f(x+2)=f(x)\text{ for all }x\in\mathbb{R}\,.$$
For $n=1$, we have
$$\begin{align}f(x+3)&=f(x+2)-f(x+1)
\\&=\big(f(x+1)-f(x)\big)-f(x+1)
\\&=-f(x)\text{ for every }x\in\mathbb{R}\,,\end{align}$$
whence $$f(x+6)=-f(x+3)=f(x)\text{ for all }x\in\mathbb{R}\,.$$
For $n=2$, we have
$$\begin{align}
f(x+4)&=\sqrt{2}\,f(x+3)-f(x+2)\\&=\sqrt{2}\,\big(\sqrt{2}\,f(x+2)-f(x+1)\big)-f(x+2)\\&=f(x+2)-\sqrt{2}\,f(x+1)\\&=-f(x)\text{ for each }x\in\mathbb{R}\,,\end{align}$$
making
$$f(x+8)=-f(x+4)=f(x)\text{ for every }x\in\mathbb{R}\,.$$
Finally, for $n=3$, we have
$$\begin{align}
f(x+6)&=\sqrt{3}\,f(x+5)-f(x+4)\\&=\sqrt{3}\,\big(\sqrt{3}\,f(x+4)-f(x+3)\big)-f(x+4)
\\&=2\,f(x+4)-\sqrt{3}\,f(x+3)\\&=2\,\big(\sqrt{3}\,f(x+3)-f(x+2)\big)-\sqrt{3}\,f(x+3)
\\&=\sqrt{3}\,f(x+3)-2\,f(x+2)\\&=\sqrt{3}\,\big(\sqrt{3}\,f(x+2)-f(x+1)\big)-2\,f(x+2)
\\&=f(x+2)-\sqrt{3}\,f(x+1)=-f(x)
\end{align}$$
for all $x\in\mathbb{R}$. Consequently,
$$f(x+12)=-f(x+6)=f(x)\text{ for all }x\in\mathbb{R}\,.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2917573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Finding solution to differential equation I'm stuck on this problem:
Find the solution of the given initial value problem:
$ty′+4y=t^2−t+5$, $y(1)=7$, $t>0$
When I multiply both sides by $\mu(t)$ I find that $\mu(t) = e^{4t}$
So:
$$
\frac{d}{dt} (\mu y) = \mu t^2 - \mu t + 5 \mu\\
e^{4t} y= \int t^2 e^{4t}-te^{4t}+5e^{4t} dt
$$
This is what I have so far
| For $t y′ + 4y = t^2 − t + 5$ then
\begin{align}
\frac{1}{\mu} \frac{d}{dt} \, ( \mu \, y) &= y' + \frac{\mu'}{\mu} \, y
\end{align}
which leads to
$$ \frac{d}{dt} \ln \mu = \frac{4}{t} = \frac{d}{dt} (4 \ln t) = \frac{d}{dt} \ln(t^4),$$
or $\mu(t) = t^4$. Now,
\begin{align}
y' + \frac{4}{t} \, y &= t - 1 + \frac{5}{t} \\
\frac{1}{t^4} \, \frac{d}{dt} \, (t^4 \, y) &= t - 1 + \frac{5}{t} \\
\frac{d}{dt} \, (t^4 \, y) &= t^5 - t^4 + 5 \, t^3 \\
y(t) &= t^{-4} \, \int^{t} (u^{5} - u^{4} + 5 \, u^{3}) \, du \\
&= \frac{t^2}{6} - \frac{t}{5} + \frac{5}{4} + \frac{c_{0}}{t^4}.
\end{align}
By applying $y(1) = 7$ leads to
$$y(t) = \frac{10 t^6 - 12 t^5 + 75 t^4 + 347}{60 \, t^4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2918297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Integrating $\int {1+\sin x\over x^2+2}dx$ I've been looking for a suitable approach to solve the following integral:
$$\int {1+\sin x\over x^2+2}dx$$
So far, I've tried by substitution and by parts to no avail. I've also attempted the following manipulations:
$$\int {\sin^2x+\cos^2x+\sin x\over x^2+2}dx = \int {\sin x (\sin x+1)+\cos^2 x\over x^2+2}dx $$
(Which appears to be just messing around with it; it got me not closer to any solution!)
And:
$$\begin{align}
\int {1+\sin x\over x^2+2}dx &= \int {(1+\sin x)(1-\sin x)\over (x^2+2)(1-\sin x)}dx \\
&= \int {1-\sin^2 x\over (x^2+2)(1-\sin x)}dx \\
&= \int {\cos^2(x)\over (x^2+2)(1-\sin x)}dx
\end{align}$$
However, I don't feel I'm getting any closer to a form I can solve. From what we are presently learning in our classes, we should be able to solve it using integration by parts, by substitution or by partial fractions.
Could someone point me in the right direction so I can work it out?
Thans in advance.
Artem.
| $$I=\int\frac{1+\sin(x)}{x^2+2}dx=\int\frac{1}{x^2+2}dx+\int\frac{\sin(x)}{x^2+2}dx$$
$$I_1=\int\frac{1}{x^2+2}dx=\int\frac{1}{2\left[(\frac{x}{\sqrt{2}})^2+1\right]}dx$$
$a=\frac{x}{\sqrt{2}}$ so $dx=\sqrt{2}da$
$$I_1=\frac{\sqrt{2}}{2}\int\frac{1}{a^2+1}da=\frac{\sqrt{2}}{2}\arctan\left(\frac{x}{\sqrt{2}}\right)+C$$
now we can move onto the harder second integral:
$$I_2=\int\frac{\sin(x)}{x^2+2}dx=\Im\int\frac{e^{ix}}{x^2+2}dx$$
if we let:
$$I_2(b)=\Im\int\frac{e^{iax}}{x^2+2}dx$$
$$I_2'(b)=\Im\int\frac{ix.e^{iax}}{x^2+2}dx$$
$u=x^2+2$ so $dx=\frac{du}{2x}$
$$I_2'(b)=\Im\frac{i}{2}\int\frac{e^{ia\sqrt{u-2}}}{u}du=\Im\frac{i}{2}\int\frac{1}{u}\sum_{n=0}^{\infty}\frac{(ia\sqrt{u-2})^n}{n!}du$$
and this $\sqrt{u-2}$ could also be expressed as a series to give a double summation which may be solvable.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2919040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Finding the value of a polynomial at certain points $p(x)$ be a polynomial of degree 7 with real coefficients such that $p(π) = √3$ and $$\int_{-π}^{π} x^{k}p(x) = 0, \text{ for} \; 0\leq k \leq 6. $$
I have to find the value of $p(-π)$ and $p(0).$
My initial thoughts were to suppose that $p(x) = a_0 + a_1x + a_2x^2 + \dots + a_7x^7$ and solve it for the coefficients but it was very tedious and like impossible to solve.
Any insight. Thank you.
| Let's start with $$\int_{-\pi}^\pi p(x)dx = 0$$Note that for all of the terms with $x$ having an odd degree, the integral of the function will have that term with even degree, so substituting $\pi$ for $x$ or $-\pi$ will produce the same value. This gives us our key insight: We solely care about the terms with even degree. Hence, the above integral becomes $$2\pi(a_0+\frac{a_2}3\pi^2+\frac{a_4}5\pi^4+\frac{a_6}7\pi^6)=0\to a_0+\frac{a_2}3\pi^2+\frac{a_4}5\pi^4+\frac{a_6}7\pi^6 =0$$We can similarly construct equations with $k=1,2,...6$ to get $$\frac{a_1}3+\frac{a_3}5\pi^2+\frac{a_5}7\pi^4+\frac{a_7}9\pi^6=0$$$$\frac{a_0}3+\frac{a_2}5\pi^2+\frac{a_4}7\pi^4+\frac{a_6}9\pi^6=0$$$$\frac{a_1}5+\frac{a_3}7\pi^2+\frac{a_5}9\pi^4+\frac{a_7}{11}\pi^6=0$$$$\frac{a_0}5+\frac{a_2}7\pi^2+\frac{a_4}9\pi^4+\frac{a_6}{11}\pi^6=0$$$$\frac{a_1}7+\frac{a_3}9\pi^2+\frac{a_5}{11}\pi^4+\frac{a_7}{13}\pi^6=0$$$$\frac{a_0}7+\frac{a_2}9\pi^2+\frac{a_4}{11}\pi^4+\frac{a_6}{13}\pi^6=0$$And finally, we have one equation due to the condition that $p(\pi)=\sqrt{3}$.$$a_0+a_1\pi+a_2\pi^2+a_3\pi^3+a_4\pi^4+a_5\pi^5+a_6\pi^6+a_7\pi^7=\sqrt{3}$$
After solving this system of equations, what you find is that the only coefficients of the polynomial that are non-zero are those of the odd terms! This means that the function is odd, i.e., $p(x)=-p(-x)$. Moreover, since $a_0=0$, we know that $$p(-\pi)=-\sqrt 3$$$$p(0)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2921515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Expressing $\sin(18^{°})$ in algebraic form from $z^5-1$ as starting point I am trying to express $\sin(18°)$ in algebraic form using only complex numbers. I know that when I factor $z^5-1.$ I get an expansion that looks like: $(z-1)(z^4+z^3+z^2+z^1+1)$ The exercise then says substitute for $z+\frac{1}{z}$ in the 'long factor' and then somehow derive $\sin(18°)$. I just have no idea how to I am supposed to do this. But knowing how, could teach me something about complex numbers.
| I guess you realize that $18^\circ$ is $1/20$ of a circle, so that $a=\sin 18^\circ+i\cos 18^\circ$ is a $5$ root of unity. So $a$ must satisfy $z^{5}=1$.
The polynomial $z^{5}-1$ factors and a bit of thought gets you to the point where you are, that $a$ must satisfy $z^4+z^3+z^2+z+1=0$. So if you can find its roots, then one of the real parts will be $\sin 18^\circ.$
Divide the above equation by $z^2.$
$$z^2+z+1+\frac{1}{z}+\frac{1}{z^2}=0.$$
Let $w=z+\frac{1}{z}$ and note the the 2nd and 4th terms equal $w$. Also note that $w^2= z^2+2+\frac{1}{z^2}$, so by adding and subtracting $1$ we make the above
$$w^2+w-1 = 0.$$
Solve this to get
$$w=\frac{-1\pm\sqrt{5}}{2}.$$
For each of these two solutions solve the quadratic equations
$$z+\frac{1}{z} = \frac{-1\pm\sqrt{5}}{2}.$$
Figure out which one is in the 1st quadrant and take its imaginary part.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2925226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Extend $f(x)=\frac{6x^3-11x^2-16x-4}{2x^2-5x-2}$ in a way it is continuous $\forall x \in \mathbb{R}$ Problem
If it is possible to extend $$f(x)=\frac{6x^3-11x^2-16x-4}{2x^2-5x-2}$$ in a way it is continuous $\forall x \in \mathbb{R}$. Extend all $x \not\in X$ where $X$ is domain of $f$. Also define domain $X$.
Attempt to solve
By solving $2x^2-5x-2=0$ we can get $\forall x \not\in X$.
$$ x= \frac{5\pm \sqrt{(-5)^2-4 \cdot 2\cdot(-2)}}{2 \cdot 2} $$
$$ x= \frac{5 \pm \sqrt{41}}{4} $$
Meaning :
$$ X=\mathbb{R}\setminus \{\frac{5-\sqrt{41}}{4},\frac{5+\sqrt{41}}{4}\} $$
In order to $f$ to be continuous $\forall x \
\not\in X$ have to have limits in these undefined points.
$$ \lim_{x \rightarrow \frac{5-\sqrt{41}}{4}} \frac{6x^3-11x^2-16x-4}{2x^2-5x-2}
=\frac{1}{4}(23-3\sqrt{41})$$
and
$$ \lim_{x \rightarrow \frac{5+\sqrt{41}}{4}}\frac{6x^3-11x^2-16x-4}{2x^2-5x-2} = \frac{1}{4}(23+3\sqrt{41})$$
Meaning if i set
$$
\begin{cases}
f(\frac{5-\sqrt{41}}{4})=\frac{1}{4}(23-3\sqrt{41}) \\
f(\frac{5+\sqrt{41}}{4}=\frac{1}{4}(23+3\sqrt{41})
\end{cases}
$$
is continous $\forall x \in \mathbb{R}$
Now only problem is that is this correct solution and if it is how do i solve it more simply than this ? I didn't compute these limits by hand, maybe there is more simpler solution ?
| Just do a polynomial long divison.
$$\frac{6x^3-11x^2-16x-4}{2x^2-5x-2} = 2+3x$$
| {
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"url": "https://math.stackexchange.com/questions/2927664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $z = x+iy$ be a complex number and denote $z^2 = A + Bi$. Solve for $x,y$ in terms of $A,B$. Here, $x, y, A, B$ are all real numbers.
I am just having some problems with the algebra of putting it all together to get a solution.
What I have currently is:
$z^2 = (x+iy)^2 = x^2 + 2xyi - y^2$
This gives $A = x^2 - y^2$ and $B = 2xy$.
I also know that $|z|^2 = |z^2|$, which gives:
$|x+iy|^2 = |A+Bi|$ => $x^2+y^2 = \sqrt{(A^2+B^2)}$.
I know that I must be making some silly error (either in the above calculations or when I try substituting things in) but I keep going in circles with my algebra and not able to get a final answer. How should I proceed from here?
| You already have $x^2- y^2= A$ and $x^2+ y^2= \sqrt{A^2+ B^2}$. Adding those two equations eliminates $y^2$ and gives $2x^2= A+ \sqrt{A^2+ B^2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to know that $2n^3+9n^2+13n+6$ factors into $(n+1)(n+2)(2n+3)$? Apologies in advance, my math is very rusty.
I'm slowly working my way through Schaum's Outline of Discrete Math for some self-study, occasionally filling in large knowledge gaps in my grasp on algebra. In one of the supplementary questions I'm asked to prove (by induction) that:
$$
\sum^n_{i=1}i^2 = \frac{n(n+1)(2n+1)}{6}
$$
In the inductive step I add $(n+1)^2$ to either side and then try to work my way through, multiplying out the factors:
$$
\frac{n(n+1)(2n+1)}{6} + (n+1)^2 \\
= \frac{n(n+1)(2n+1)+6(n+1)^2}{6} \\
= \frac{(n^2+n)(2n+1)+6(n+1)^2}{6} \\
= \frac{2n^3+n^2+2n^2+n+6(n^2+2n+1)}{6} \\
= \frac{2n^3+n^2+2n^2+n+6n^2+12n+6}{6} \\
= \frac{2n^3+9n^2+13n+6}{6}
$$
I think everything up to this point is pretty trivial, but it's around this point that I get a bit lost. I know I need to factor the terms in the 3-degree polynomial in the numerator somehow, but I'm having a hard time with the actual mechanics given the lack of obvious common factors & the fact that 13 is prime.
I'm pretty sure the final result should look something like:
$$
\frac{(n+1)((n+1)+1)(2(n+1)+1)}{6} \\
= \frac{(n+1)(n+2)(2n+3)}{6}
$$
Multiplying the expected result out I can see the two statements are equal and I'm confident that
$$
\frac{2n^3+9n^2+13n+6}{6} = \frac{(n+1)(n+2)(2n+3)}{6}
$$
but I'm just not quite sure how to factor the polynomial myself to arrive at the final result.
Any hints on how to proceed from here, or what I need to be reading up on to get my head around this?
| The Rational Root Theorem tells you that any rational root of the polynomial $2 n^3 + 9 n^2 + 13 n + 6$ has the form $\frac{p}{q}$ for integers $q \mid 2$, $p \mid 6$, leaving 12 possible rational roots. Since all of the coefficients are positive, all of the (real) roots are negative, leaving only 6 possible rational roots, $-6, -3, -2, -\frac{3}{2}, -1, -\frac{1}{2}$.
Substituting, for example, $-1$ (because it's fast to evaluate a polynomial at $-1$---just take the alternating sum of the coefficients) gives $-2 + 9 - 13 + 6 = 0$, so $n = -1$ is a root of the polynomial, and equivalently the polynomial has factor $n + 1$. Dividing the polynomial by $n + 1$ and factoring gives $2 n^2 + 7 n + 6 = (n + 2)(2 n + 3)$, and putting this all together gives the desired factorization:
$$\boxed{2 n^3 + 9 n^2 + 13 n + 6 = (n + 1) (n + 2) (2 n + 3)}.$$
Remark As you probably know, not all polynomials with integer coefficients factor into products of linear terms with integer coefficients as they did in this case. For example, the general formula for $\sum i^4$ analogous to the one in this question contains a factor of $3 n^2 + 3 n - 1$, which is not factorable over the integers.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 0
} |
Aid in identifying an error made whilst evaluating an integral
Evaluate $$I=\int\frac{1}{x^2-8x+8}~dx$$
First, complete the square using the denominator: $$x^2-8x+8=(x-4)^2-8$$
and therefore, let $x=2\sqrt{2}\sec{\theta}+4$, $\therefore dx=2\sqrt{2}\sec{\theta}\tan{\theta}~d\theta$, and hence, we have
$$I=2\sqrt{2}\int\frac{\sec{\theta}\tan{\theta}}{(2\sqrt{2}\sec{\theta}+4-4)^2-8}~d\theta$$ $$=2\sqrt{2}\int\frac{\sec{\theta}\tan{\theta}}{(2\sqrt{2}\sec{\theta})^2-8}~d\theta$$
$$=2\sqrt{2}\int\frac{\sec{\theta}\tan{\theta}}{8\sec^2{\theta}-8}~d\theta$$
$$=\frac{\sqrt{2}}{4}\int\frac{\sec{\theta}\tan{\theta}}{\sec^2\theta-1}~d\theta$$
$$=\frac{\sqrt{2}}{4}\int\frac{\sec{\theta}\tan{\theta}}{\tan^2\theta}~d\theta$$
$$=\frac{\sqrt{2}}{4}\int\frac{\sec{\theta}}{\tan\theta}~d\theta$$
note:$$\frac{\sec{\theta}}{\tan{\theta}}=\frac{\cos{\theta}}{\sin{\theta}}\cdot\frac{1}{\cos{\theta}}=\csc{\theta}$$
therefore,
$$I=-\frac{\sqrt{2}}{4}\ln{|\csc\theta+\cot\theta|}+C$$
now, using the fact that $x=2\sqrt{2}\sec\theta+4$, we have the following identities, based on the definition of the trigonometric functions:
$$\csc\theta=\frac{x-4}{\sqrt{x^2-8x+8}}$$
as well as
$$\cot\theta=\frac{2\sqrt{2}}{\sqrt{x^2-8x+8}}$$
therefore, the final integral is given by:
$$I=-\frac{\sqrt{2}}{4}\ln{\left|\frac{x+2\sqrt{2}-4}{\sqrt{x^2-8x+8}}\right|}+C$$
which, I've been informed, is incorrect. Would anyone be kind enough to help me realize my error? Any responses are appreciated. Thank you.
| Your answer is correct and it's only a matter of rewriting.
Let us consider your integral:
$$I=\int\frac{1}{x^2-8x+8}\ dx.$$
I approached this problem by factoring the denominator instead of completing the square:
$$x^2-8x+8=0.$$
The discriminant is given by $\Delta_x=(-8)^2-4(1)(8)=64-32=32$ such that $\sqrt{\Delta_x}=\sqrt{32}=4\sqrt{2}$.
Solutions for $x$ are given by $$x_{1,2}=\frac{-(-8)\pm4\sqrt{2}}{2(1)}=4\pm2\sqrt{2}.$$
Thus,
$$x^2-8x+8=(x-4-2\sqrt{2})(x-4+2\sqrt{2}).$$
Next I applied partial fraction decomposition on the integrand:
$$\frac{1}{x^2-8x+8}=\frac{A}{x-4-2\sqrt{2}}+\frac{B}{x-4+2\sqrt{2}}.$$
It follows that
$$1=A(x-4+2\sqrt{2})+B(x-4-2\sqrt{2})=(A+B)(x-4)+(A-B)2\sqrt{2}.$$
It can be derived that $A+B=0$ and $(A-B)2\sqrt{2}=1$ which means that $A=\frac{1}{4\sqrt{2}}$ and $B=-\frac{1}{4\sqrt{2}}$.
Thus,
$$\frac{1}{x^2-8x+8}=\frac{1}{4\sqrt{2}}\left(\frac{1}{x-4-2\sqrt{2}}-\frac{1}{x-4+2\sqrt{2}}\right).$$
Substitution into your integral gives
$$I=\frac{1}{4\sqrt{2}}\int\left(\frac{1}{x-4-2\sqrt{2}}-\frac{1}{x-4+2\sqrt{2}}\right) dx.$$
Integration yields
$$I=\frac{1}{4\sqrt{2}}\ln\left|\frac{x-4-2\sqrt{2}}{x-4+2\sqrt{2}}\right|+\text{constant}.$$
Okay, so we have arrived at an expression. Let us rewrite it to yield your expression.
Consider
$$I=-\frac{1}{4\sqrt{2}}\ln\left|\left(\frac{x-4-2\sqrt{2}}{x-4+2\sqrt{2}}\right)^{-1}\right|+\text{constant}.$$
It follows that
$$I=-\frac{1}{4\sqrt{2}}\ln\left|\frac{x-4+2\sqrt{2}}{x-4-2\sqrt{2}}\right|+\text{constant}.$$
Within the argument of the natural logarithm; multiply both the denominator and numerator by $x-4+2\sqrt{2}$:
$$I=-\frac{1}{4\sqrt{2}}\ln\left|\frac{x-4+2\sqrt{2}}{x-4-2\sqrt{2}}\frac{x-4+2\sqrt{2}}{x-4+2\sqrt{2}}\right|+\text{constant}.$$
It follows that
$$I=-\frac{1}{4\sqrt{2}}\ln\left|\frac{(x-4+2\sqrt{2})^2}{x^2-8x+8}\right|+\text{constant}.$$
Write $\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$:
$$I=-\frac{\sqrt{2}}{4}\frac{1}{2}\ln\left|\frac{(x-4+2\sqrt{2})^2}{x^2-8x+8}\right|+\text{constant}.$$
It follows that
$$I=-\frac{\sqrt{2}}{4}\ln\left|\frac{x-4+2\sqrt{2}}{\sqrt{x^2-8x+8}}\right|+\text{constant}.$$
We have obtained your expression.
| {
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"timestamp": "2023-03-29T00:00:00",
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How many 3 -digit numbers can be formed from 0 to 8 whose sum is equal to 20 for repetition and non-repetition? I have a range from $0$ to $8$ . I want to find that how many 3-digit numbers can be formed whose sum is equal to $20$.
Non-Repetition of numbers:
$$
5+7+8=20 \\
$$
Repetition of numbers:
$$
4+8+8=20, \\
6+7+7=20, \\
8+6+6=20,
$$
I want to know the number of ways that I can represent 20 using the sum of 3- positive integers. In repetition case, there is only 1 way which is mentioned above but in non-repetition case, we can also do it on paper but I want the general solution.
| The problem can be transformed into a couple of related problems as follows:
Problem 1: Consider the non-repetition problem with no restriction on range. Here we are to find natural numbers $0\leq a<b<c$ such that:
$$a+b+c=20$$
Given appropriate $x,y,z\geq0$ we could reformulate this in the following way:
$$
\begin{align}
a&=x \\
b&=x+1+y\\
c&=x+1+y+1+z
\end{align}
$$
so the equation becomes
$$
3x+2(1+y)+1+z=20 \\
\Updownarrow\\
3x+2y+z=17
$$
Constraint 1: To make use of the results from problem 1, we need to identify the cases where $c\leq 8$. For this to be the case, we must have:
$$
c=x+1+y+1+z\leq 8 \\
\Updownarrow \\
x+y+z \leq 6
$$
Solutions 1: In order to search for solutions to problem 1 subject to constraint 1, we can start from a particular solution:
$$
(x,y,z)=(5,1,0) \implies (a,b,c) = (5,7,8)
$$
and try adding neutral vectors (vectors with $3x+2y+z=0$) such as $(-2,3,0)$ and $(0,-1,2)$ to this in order to produce further solutions. Indeed
$$
(5,1,0)+(-2,3,0)=(3,4,0)\implies(a,b,c)=(3,8,9)
$$
would have been another solution, had it not been for the constraint $x+y+z\leq 6$. Since adding $(-2,3,0)$ or $(0,-1,2)$ increases $x+y+z$ by one and since $(x,y,z)=(5,1,0)$ already has sum $6$, soon one realizes that this is the only solution. Note that $(-2,3,0)$ and $(0,-1,2)$ form a basis for the nullspace of the problem since the nullspace defined by $3x+2y+z=0$ must be two-dimensional.
Problem 2: Consider the repetition case $0\leq a\leq b\leq c$. This can be described via:
$$
\begin{align}
a &= x \\
b &= x+y \\
c &= x+y+z
\end{align}
$$
so we have:
$$
3x+2y+z=20
$$
Constraint 2: Here we have:
$$
c=x+y+z\leq 8
$$
Solutions 2: Again we start from a particular solution:
$$
(x,y,z) = (6,1,0) \implies(a,b,c)=(6,7,7)
$$
Note that $x+y+z=7$ so this is a valid solution. But as soon as we add $(-2,3,0)$ or $(0,-1,2)$ we increase the sum $x+y+z$ by one, so we have very little wiggle room. In fact we can add one of those exactly once, so the only two other solutions must be:
$$
\begin{align}
(6,1,0)+(-2,3,0) &= (4,4,0) \implies &&(a,b,c)=(4,8,8) \\
(6,1,0)+(0,-1,2) &= (6,0,2) \implies &&(a,b,c)=(6,6,8)
\end{align}
$$
I hope this makes sense!
| {
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Proving identity $\frac{1}{x+iy}=\frac{x}{x^2+y^2}-i\frac{y}{x^2+y^2}$ Below are some of the identities provided early in Needhams "Visual Complex analysis"
I want to verify the identity in my title. From the identities given before it, I could only write
$$\frac{1}{x+iy}=\frac{1}{x^2+y^2}\big(-\arctan\frac{y}{x}\big)$$
Not sure how to derive that $\big(-\arctan\frac{y}{x}\big)=x-iy$
thanks
| $$\frac{1}{x+iy} \frac{x-iy}{x-iy} = \frac{x-iy}{(x+iy)(x-iy)} = \frac{x-iy}{x^2-i^2y^2} = \frac{x-iy}{x^2+y^2} = \frac{x}{x^2+y^2} - i\frac{y}{x^2+y^2}$$ as required
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Prove that if $x^3 + 3x + 3$ is irrational then $x$ is irrational, by proving the contrapositive I don't understand how to do this considering that the contrapositive of $x^3$ is irrational. For example $2$ to the cube root is irrational, but I am trying to prove that is is rational
| A more pedestrian proof (following @Mike's first step):
Suppose $x$ is rational, and the quotient of two integers $x = \frac{a}{b}.$ Then
\begin{align}
f(x)
&= x^3 + 3x + 3 \\[8pt]
&= \left(\frac{a}{b}\right)^3 + 3\frac{a}{b} + 3\frac{1}{1} \\[8pt]
&= \frac{a^3}{b^3} + 3\frac{a}{b}\frac{b^2}{b^2} + 3\frac{1}{1}\frac{b^3}{b^3} \\[8pt]
&= \frac{a^3}{b^3} + 3\frac{ab^2}{b^3} + \frac{3b^3}{b^3} \\[8pt]
&= \frac{a^3 + 3ab^2 + 3b^3}{b^3}
\end{align}
which is a quotient of two integers (namely, $a^3 + 3ab^2 + 3b^2$ and $b^3$). Hence $f(x)$ is a rational as well.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2941328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $\frac{\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{2}-\sqrt{2+\sqrt{3}}}=-3-2\sqrt{3}$
Show that $$\frac{\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{2}-\sqrt{2+\sqrt{3}}}=-3-2\sqrt{3}.$$
My attempt: I could find a way to develop the LHS by un-nesting the double radical, figuring out that
$$\sqrt{2+\sqrt{3}}=\frac{\sqrt{6}+\sqrt{2}}{2}\ \ (1)$$
By substituting (1) in the LHS of the original expression,
it is straightforward to show that it is equal to the RHS.
But my problem is on how to show without the un-nesting, by another approach. I've used a standard approach of
multiplying both terms by $\sqrt{2}+\sqrt{2+\sqrt{3}}$, leading to
$$\frac{4+\sqrt{3}+2\sqrt{4+2\sqrt{3}}}{-\sqrt{3}}\Leftrightarrow \frac{(4+\sqrt{3}+2\sqrt{4+2\sqrt{3}})\sqrt{3}}{-3}.$$
But I was not able to show that this last expression is equal to RHS by this approach.
Hints and answers not using my first un-nesting approach (if possible) will be appreciated. Sorry if this is a duplicate.
| You need to denest $\sqrt{2+\sqrt{3}}$ at some point, but this can be deferred to the last step. Let $s=\sqrt{2}$ and $r=\sqrt{3}$. Then
\begin{align}
\frac{s+\sqrt{s^2+r}}{s-\sqrt{s^2+r}} +3+2r
&=\frac{1+\sqrt{1+r/2}}{1-\sqrt{1+r/2}} +3+2r\\
&=\frac{1+\sqrt{1+r/2}}{1-\sqrt{1+r/2}}\times\frac{1+\sqrt{1+r/2}}{1+\sqrt{1+r/2}} +3+2r\\
&=\frac{2+r/2+2\sqrt{1+r/2}}{-r/2}+3+2r\\
&=\frac{-(r+1)+\sqrt{4+2r}}{-r/2}
\end{align}
and it remains to show that $\sqrt{4+2r}=r+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2941552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Determining the minimum value of the function $y = x + 2\sqrt{x^2 - \sqrt{2}x + 1}$ I am curious whether there is an algebraic verification for $y = x + 2\sqrt{x^2 - \sqrt{2}x + 1}$ having its minimum value of $\sqrt{2 + \sqrt{3}}$ at $\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{6}}$. I have been told the graph of it is that of a hyperbola.
| Scale and shift
$$\begin{align}u&=\sqrt{2}x-1&v&=\tfrac{1}{\sqrt{2}}y-\tfrac{1}{2}\text{.}
\end{align}$$
Then it is just as well to minimize
$$v=\tfrac{u}{2}+\sqrt{u^2+1}\text{.}$$
with respect to $u$.
Use the stereographic projection
$$u=\frac{2t}{t^2-1}\text{.}$$
Then it is just as well to minimize
$$v=1+\frac{t+2}{t^2-1}=1+\frac{3}{2(t-1)}-\frac{1}{2(t+1)}$$
over $t^2>1\text{.}$
Use the Cayley transform
$$s=\frac{t-1}{t+1}\text{.}$$
Then it is just as well to minimize
$$v=\frac{3}{4s}+\frac{s}{4}$$
over $s>0$. Rescale
$$\begin{align}w&=2\frac{v}{\sqrt{3}}&r&=\frac{s}{\sqrt{3}}\text{.}\end{align}$$
Then it is just as well to minimize
$$w=\frac{r+\tfrac{1}{r}}{2}$$
over $r>0$.
But by the arithmetic-geometric mean inequality,
$$w\geq 1$$
with equality iff $r=1$. Retracing the steps gives
$$\begin{align}x&=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{6}}&y&=\frac{\sqrt{6}}{2}+\frac{\sqrt{2}}{2}\text{.}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2942263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 3
} |
Prove all derivatives of $f(x)=\frac{1}{1+x}$ by induction Problem
Prove all derivatives of:
$$ f(x)=\frac{1}{1+x} $$
by induction.
Attempt to solve
I compute few derivatives of $f(x)$ so that i can form general expression for induction hypothesis. I compute all derivatives utilizing formula:
$$ \frac{d}{dx}x^n=nx^{n-1} $$
First 4 derivatives are:
$$ f'(x)=(-1)\cdot(1+x)^{-2}\cdot 1 = -\frac{1}{(1+x)^2} $$
$$ f''(x)=(-1)(-2)(1+x)^{-3}\cdot 1 = \frac{2}{(1+x)^3} $$
$$ f'''(x)=(-1)(-2)(-3)(1+x)^{-4}\cdot 1 = -\frac{6}{(1+x)^4} $$
$$ f''''(x)=(-1)(-2)(-3)(-4)(1+x)^{-5} \cdot 1 = \frac{24}{(1+x)^5} $$
Observe that $(-1)(-2)(-3)(-4)\dots (-n)$ can be generalized with:
$$ (-1)(-2)(-3)(-4)\dots(-n) = (-1)^n\cdot n! $$
Expression follows factorial of $n$ except every other value is positive and every other is negative. If i multiply it by $(-1)^n$ it is positive when $n \mod 2 = 0$ and negative when $n \mod 2 \neq 0$.
Rest of the expression can be generalized as:
$$ (1+x)^{-n-1} = (1+x)^{-(n+1)}=\frac{1}{(1+x)^{n+1}} $$
Combining these gives formula in analytic form:
$$ f(n)= \frac{(-1)^n \cdot n!}{(1+x)^{n+1}} $$
I can form induction hypothesis such that:
$$ \frac{d^n}{dx^n}\frac{1}{1+x} = \frac{(-1)^n\cdot n!}{(1+x)^{n+1}} $$
Induction proof
Base case
Base case when $n=0$:
$$ \frac{d^0}{dx^0}\frac{1}{1+x}=\frac{1}{1+x}=\frac{(-1)^0\cdot 0!}{(1+x)^{0+1}} $$
Induction step
$$ \frac{d^n}{dx^n}\frac{1}{1+x} =_{\text{ind.hyp}} \frac{(-1)^n\cdot n!}{(1+x)^{n+1+1}} $$
$$ \frac{d^n}{dx^n}\frac{1}{1+x} = \frac{(-1)^n\cdot n!}{(1+x)^{n+2}} $$
Now the problem is that formula i used for derivation can only be used recursively. I believe this is correct notation for $n$:th derivative but computing one is only defined recursively with formula i used:
$$ \frac{d}{dx} x^n = nx^{n-1} $$
Which is not defined for case:
$$ \frac{d^n}{dx^n}x^n = \text{
undefined} $$
The idea is to show that this recursion can be expressed in analytical form and it is valid for all $n\in \mathbb{Z}+$ by induction. Problem is i don't know how do you express this in recursive form and how do you get from recursion formula to the analytical one.
| Your hypothesis is
$$\frac{d^n}{dx^n}\frac1{1+x}=\frac{(-1)^nn!}{(1+x)^{n+1}}$$
We want to show that
$$\frac{d^{n+1}}{dx^{n+1}}\frac1{1+x}=\frac{(-1)^{n+1}(n+1)!}{(1+x)^{n+2}}$$
Let's verify:
\begin{align}
\frac{d^{n+1}}{dx^{n+1}}\frac1{1+x} &=\frac{d}{dx}\left(\frac{d^n}{dx^n}\frac1{1+x} \right)\\
&=\frac{d}{dx}\left(\frac{(-1)^nn!}{(1+x)^{n+1}} \right) \\
&=(-1)^n(n!) \frac{d}{dx}(1+x)^{-(n+1)} \\
&= (-1)^n(n!) (-(n+1)) (1+x)^{-(n+2)}\\
&=\frac{(-1)^{n+1}(n+1)!}{(1+x)^{n+2}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2942357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find the equation in spherical coordinates of $x^2 + y^2 – z^2 = 4$. Find the equation in spherical coordinates of $x^2 + y^2 – z^2 = 4$.
$$\begin{align}
x^2 + y^2 &= r^2\sin^2(\theta)\\
z^2 &= r^2 \cos(\theta) \\
x^2 + y^2 - z^2&=r^2(\sin^2(\theta) - \cos^2(\theta)) = 4
\end{align}$$
Thus, $$-r^2(\cos(2\theta)) = 4$$
Is this right?
| $x^2 +y^2-z^2=4 \implies \rho ^2-2z^2=4$$
$$ \implies \rho ^2(1-2\cos ^2 ( \phi ))=4$$
$$\implies \rho ^2(\cos 2( \phi ))=4 \implies \rho ^2=4\sec (2\phi)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2943176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Finding the complex square roots of a complex number without a calculator
The complex number $z$ is given by $z = -1 + (4 \sqrt{3})i$
The question asks you to find the two complex roots of this number in the form $z = a + bi$ where $a$ and $b$ are real and exact without using a calculator.
So far I have attempted to use the pattern $z = (a+bi)^2$, and the subsequent expansion $z = a^2 + 2abi - b^2$. Equating $a^2 - b^2 = -1$, and $2abi = (4\sqrt{3})i$, but have not been able to find $a$ and $b$ through simultaneous equations.
How can I find $a$ and $b$ without a calculator?
| Observe that $\|z\| = \sqrt{(-1)^2+(4\sqrt3)^2} = \sqrt{49} = 7$. Therefore the root of $z$ will have length $\sqrt 7$, so $a^2+b^2=7$. Combine this with $a^2-b^2=-1$ to get $a$ and $b$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 5
} |
Recurrence $a_j = \frac{n}{n-j} + \frac{j}{n-j} a_{j-1}$ Let $n\in \mathbb{N}$. How to solve the recurrence
$$a_j = \frac{n}{n-j} + \frac{j}{n-j} a_{j-1}$$
for $1\leq j <n$, and $a_0=1$?
I calculated it for some $n$s:
$n=2: [1, 3]$
$n=3: [1, 2, 7]$
$n=4: [1, \frac{5}{3}, \frac{11}{3}, 15]$
$n=5: [1, \frac{3}{2}, \frac{8}{3}, \frac{13}{2}, 31]$
$n=6: [1, \frac{7}{5}, \frac{11}{5}, \frac{21}{5}, \frac{57}{5}, 63]$
$n=7: [1, \frac{4}{3}, \frac{29}{15}, \frac{16}{5}, \frac{33}{5}, 20, 127]$
It looks like $a_{n-1} = 2^n - 1$.
I tried to use generating function $A(x) = \sum_{k=0}^{n-1} a_k x^k$ and got
$$(n-x)A(x) - 2xA'(x) = n\frac{1-x^n}{1-x}-(n-1)a_{n-1}x^{n-1} - a_{n-1}x^n$$
but don't know how to solve this or if it's correct (it's highly probable I made a mistake somewhere).
| From the recurrence relation, it is easy to check that
$$ \binom{n-1}{j} a_j - \binom{n-1}{j-1}a_{j-1} = \binom{n}{j}. $$
Therefore
$$ a_j = \frac{\sum_{k=0}^{j} \binom{n}{k}}{\binom{n-1}{j}}. $$
Addendum. By a different method, one can also prove that
$$ a_j = n \int_{0}^{1} (1-u)^{n-1-j} (1+u)^j \, du = \sum_{k=0}^{j} \binom{j}{k} (-1)^{j-k}2^k \frac{n}{n-k}. $$
| {
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Formalization/Verification of a beginner combinatorics problem I have the following task:
To build a computer chip, 4 non-distinct components are needed. Of 12 existing components, 2 are defective. Given that 4 components are chosen, what are the probabilities of the following combinations?
*
*No defective components
*Exactly one defective component
*Exactly two defective components
Progress so far:
The general approach I use is to consider the number of non-defective components over the total number of components for repeated decrements of each set.
*
*$\frac{10}{12} \cdot \frac{9}{11} \cdot \frac{8}{10} \cdot \frac{7}{9} \approx 0.4242 $
For the next question, I sum the probabilities of each case, depending on if the defective component is chosen last, second last, etc.
*$\frac{10}{12} \cdot \frac{9}{11} \cdot \frac{8}{10} \cdot \frac{2}{9} \ + \frac{10}{12} \cdot \frac{9}{11} \cdot \frac{2}{10} \cdot \frac{8}{9} \ + \frac{10}{12} \cdot \frac{2}{11} \cdot \frac{9}{10} \cdot \frac{8}{9} \ + \frac{2}{12} \cdot \frac{10}{11} \cdot \frac{9}{10} \cdot \frac{8}{9} $
I could employ a similar intuitive approach for 3. , however I feel like this is messy and not particularly rigorous. How do I answer these questions formally?
| The probability on $k\in\{0,1,2\}$ defectives among the $4$ chosen is:$$\frac{\binom4k\binom8{2-k}}{\binom{12}2}$$
Hypergeometric distribution is used.
Actually this approach turns things around: there are $4$ chosen items and $8$ non-chosen items. Selecting $2$ items (the defective ones) what is the probability that $k$ of them will come from the chosen ones?
| {
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Limit of $\sin(\pi \sqrt{4n^2+n})$ We can notice that $\sqrt{4n^2 + n} = \sqrt{4n^2(1+ \frac{1}{4n})} = 2n\sqrt{1+\frac{1}{4n}}$. Therefore
$$\lim_{n \to \infty} \sin (2n\pi \sqrt{1+ \frac{1}{4n}}) \text{ will be an even number}$$
Because the square root becomes $1$ and we end up with an even number: $\sin(\text{even number})$
And sine of an even number is $0$. But apparently that is not the right answer.
| Consider that
$$\sin(\pi\sqrt{4n^2+n})=\sin(\pi\sqrt{4n^2+n}-2\pi n)=\sin\left(\frac{\pi n}{\sqrt{4n^2+n}+2n}\right)\to\sin\frac\pi4.$$
| {
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Solution of $X^2+Y^2=Z^2$ under condition $X+Y+Z=1000$ As we know before the general solution of the equation $x^2+y^2=z^2$ in the number theory is as follows:
$x=\pm(a^2-b^2)c$, $y=\pm 2abc$ and $z=\pm(a^2+b^2)c$.
I want to know how to find a solution for the equation $x^2+y^2=z^2$ under condition $x+y+z=1000$?
A hint would be greatly appreciated. Thanks, Payam
| If the solutions you want have to be integer, you can simply add up the formulas for $x, y, z$ under the condition $x^2+y^2=z^2$. For example, let $x=(a^2-b^2)c, y=2abc, z=(a^2+b^2)c$, then $x+y+z=2ac(a+b)=1000\rightarrow ac(a+b)=500=2^2\cdot5^3 $. Since $a,b,c$ need to be integer, we can let $a=2,b=3,c=50$, then we have a solution $x=-250,y=600,z=650$. Other integer solutions can be found in the same way.
Actually, if you allow real solutions, the equations will have infinite solutions. Just let $x=zcos\theta,y=zsin\theta$, this satisfies the first equation. And then, the second equation becomes $z(cos\theta+sin\theta+1)=1000$. Obviously, it has infintely many real solutions.
| {
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Prove $\ 1<\frac{1}{n+1}+\frac{1}{n+2}+ ... + \frac{1}{3n+1}<2 $ using Cauchy-Schwarz How do I prove this inequality
$$\ 1<\frac{1}{n+1}+\frac{1}{n+2}+ ... + \frac{1}{3n+1}<2 $$
I've tried to prove that $$\ \frac{1}{n+1}+\frac{1}{n+2}+ ... + \frac{1}{3n+1} \ $$ is less than $$\ \frac{1}{1}+\frac{1}{2}+ ... + \frac{1}{n} \ $$ and that this sum is either less than $2$ or equal to $2$ using C-S.
| By C-S
$$\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{2n+2}+...+\frac{1}{3n}+\frac{1}{3n+1}=$$
$$=\left(\frac{1}{n+1}+\frac{1}{3n+1}\right)+\left(\frac{1}{n+2}+\frac{1}{3n}\right)+...+\left(\frac{1}{2n}+\frac{1}{2n+2}\right)+\frac{1}{2n+1}>$$
$$>\frac{(1+1)^2}{4n+2}+\frac{(1+1)^2}{4n+2}+...+\frac{(1+1)^2}{4n+2}+\frac{1}{2n+1}=1.$$
The right inequality we can prove by the similar way:
$$\sum_{k=1}^{2n+1}\frac{1}{n+k}=\sum_{k=1}^n\left(\frac{1}{n+k}+\frac{1}{3n+2-k}\right)+\frac{1}{2n+1}=$$
$$=\sum_{k=1}^n\frac{4n+2}{(n+k)(3n+2-k)}+\frac{1}{2n+1}<\sum_{k=1}^n\frac{4}{3n}+\frac{1}{3}=\frac{5}{3}<2.$$
| {
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Proof of Bellard's formula I'm reading Bellard's proof for his eponymous formula computing pi digits, and I can't get past the first line.
Given that:
*
*$\displaystyle-\ln(1-x) = \sum_{n=1}^\infty \frac{x^n}{n}$ for $|x| < 1$,
*$\arctan(y) = \Im[-\ln(1-iy)]$ for any real $y$
(I didn't know the last one but Wolfram Alpha agrees, so okay), the author gets
$$\arctan\left(\frac{1}{a-1}\right) = \Im\left[-\ln\left(1- \frac{1+i}{a}\right) \right] = \sum_{n=0}^\infty \frac{(-1)^n2^{2n}}{a^{4n+3}}\left( \frac{a^2}{4n+1} + \frac{2a}{4n+2} + \frac{2}{4n+3}\right)$$
In this equation, I don't even get the first equality, and assuming it to be correct, I don't see how to derive the second equality... I've tried expanding using Newton's theorem and expanding the binomial coefficients, but it did not work.
Any ideas?
EDIT
A friend nearly found the solution for the second equality (up to some mysterious sign error), and the first equality is actually invalid when $a \in (0, 1)$, according to Wolfram Alpha: on this interval, we rather have $\arctan\left(\frac{1}{a-1}\right) = \Im\left[-\ln\left(1- \frac{1+i}{a}\right) \right] \color{red}{+ \pi}$ .
So I still don't know why $\arctan\left(\frac{1}{a-1}\right) = \Im\left[-\ln\left(1- \frac{1+i}{a}\right) \right]$ for $a \notin (0,1)$, but given that Bellard uses $a = 2$, let's assume it's fine. However I'm interested if someone could find the solution to this problem.
Now, from $\Im\left[-\ln\left(1- \frac{1+i}{a}\right) \right]$ to $\displaystyle\sum_{n=0}^\infty \frac{(-1)^n2^{2n}}{a^{4n+3}}\left( \frac{a^2}{4n+1} + \frac{2a}{4n+2} + \frac{2}{4n+3}\right)$:
$$\begin{align*}
-\ln\left(1 - \frac{1+i}{a}\right) &= \sum_{k=1}^\infty \frac{1}{k}\left(\frac{1+i}{a}\right)^k \\
&= \sum_{n=0}^\infty \left[ \frac{1}{4n+1} \left( \frac{1+i}{a}\right)^{4n+1} + \dotso + \frac{1}{4n+4} \left( \frac{1+i}{a}\right)^{4n+4}\right]\\
&= \sum_{n=0}^\infty\left( \frac{1+i}{a}\right)^{4n+1}\left[ \frac{1}{4n+1}+ \dotso + \frac{1}{4n+4}\left(\frac{1+i}{a}\right)^3 \right]
\end{align*}$$
We remark that $1 + i = \sqrt{2} e^{i\pi / 4}$ so :
$$\begin{align*}
-\ln\left(1 - \frac{1+i}{a}\right) &= \sum_{n=0}^\infty \frac{\left( \sqrt{2} e^{i \pi/4}\right)^{4n+1}}{a^{4n+1}}\left[ \frac{1}{4n+1}+ \dotso + \frac{1}{4n+4}\frac{\left( \sqrt{2} e^{i \pi/4}\right)^3}{a^3} \right]\\
&=\sum_{n=0}^\infty \frac{2^{2n}(-1)^n}{a^{4n+1}}\left[ \frac{\left( \sqrt{2} e^{i \pi/4}\right)}{4n+1}+ \dotso + \frac{1}{4n+4}\frac{\left( \sqrt{2} e^{i \pi/4}\right)^4}{a^3} \right]\\
\end{align*}$$
Taking the imaginary part, knowing $\sqrt{2} e^{i \pi/4} = 1+i$ and $(1+i)^2 = 2i$, $(1+i)^3 = 2 - 2i$ and $(1+i)^4 = -2$:
$$\begin{align*}
\Im \left[-\ln\left(1 - \frac{1+i}{a}\right)\right] &= \sum_{n=0}^\infty \frac{2^{2n}(-1)^n}{a^{4n+1}}\left[ \frac{1}{4n+1}+ \frac{2}{4n+2}\frac{1}{a} - \frac{2}{4n+3} \frac{1}{a^2}\right]\\
&= \sum_{n=0}^\infty \frac{2^{2n}(-1)^n}{a^{4n+3}}\left[ \frac{a^2}{4n+1}+ \frac{2a}{4n+2} \color{red}- \frac{2}{4n+3}\right]\\
\end{align*}$$
So everything is good, except for the last minus sign...
| 1. The $\operatorname{arctan}$ formulas.
Consider what it means for a complex number $u$ to be a logarithm of $z$: essentially, that $e^u = z$.
Since $e^u = e^{\Re u + i \Im u} = e^{\Re u} e^{i\Im u} = r e^{i \theta}$ you see that pretty much by definition, $\Im \ln(z) = \Im u = \theta$ is the angular component of $z$ in the complex plane.
Now, take the point $z = 1-iy$, is it an elementary trigonometric exercise to check that the angle forme between the real axis and $z$ is $\theta$ such that $\tan\theta = -y$. By the above remark this means exactly that $\Im \ln(1-iy) = \arctan(-y)$.
Since $\arctan$ is an odd function, this means that $-\Im \ln(1-iy) = \arctan(y)$.
The very same argument works for $z = 1 - (1+i)/a$. Indeed, this time we have
$$\begin{align*}
\tan \theta
& = \frac{\Im z}{\Re z} \\
& = \frac{-1/a}{1 - 1/a} \\
& = \frac{-1}{a - 1}
\end{align*}
$$
And therefore, $-\Im\ln(1 - (1+i)/a) = \arctan(1/(a-1))$.
2. The summation formula
You have it right, except that $(1+i)^3 = 2i(1+i) = -2 + 2i$. And not $2 - 2i$ as you wrote, which was the source of this erroneous sign.
Well done! For additional fun formulas and the rationale behind them (in French), you can have a look at this pdf from the olden days.
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Why isn't Legendre's Conjecture resolved by the work done by Nair and Hanson in relation to Least Common Multiples? I recently discovered the work done by Hanson and Nair. As I worked through Hanson's proof, an argument occurred to me regarding Legendre's Conjecture.
Clearly, my argument is wrong. It is too simple not to have been discovered in the time that these proofs have been available.
I would greatly appreciate it if someone could help me understand where my reasoning is wrong.
(1) Nair proved:
For any $n \in N^*$, we have:
$$\text{lcm}({n\choose 1},2{n\choose 2},\dots,n{n\choose n})=\text{lcm}(1,2,\dots,n)$$
(2) So it follows:
$$\text{lcm}(1,2,\dots,(n^2+n)) > \left(\frac{n^2+n}{2}\right){{n^2+n}\choose{\frac{n^2+n}{2}}}$$
(3) It is straight forward to show for $n \ge 4$
$${{2n}\choose n} > \frac{4^n}{n}$$
(4) So it follows:
$$\text{lcm}(1,2,\dots,(n^2+n)) > \left(\frac{n^2+n}{2}\right)\frac{4^{n^2+n}}{n^2+n} = \frac{4^{n^2+n}}{2}$$
(5) Hanson proved:
Let $B(n)$ denote the least common multiple of the integers $1,\dots,n$. Then
$$B(n) < 3^n$$
(6) So, it follows:
$$\text{lcm}(1,2,\dots,n^2) < 3^{n^2}$$
(7) Let $R(x,y) = \dfrac{\text{lcm}(1,2,\dots,x)}{\text{lcm}(1,2,\dots,y)}$
(7) So that:
$$R(n^2+n,n^2) > \frac{\frac{4^{n^2+n}}{2}}{3^{n^2}}=\frac{4^n}{2}\left(\frac{4}{3}\right)^{n^2} > 4^n$$
(8) Assume that there is no prime between $n^2$ and $n^2+n$.
(9) So, the value of $R(n^2+n,n^2)$ must be equal to the product of the relative powers of primes $p^a$ such that $n^2 < p^a \le n^2+n$.
Note: I say "relative" since for $a \ge 2$, $R(p^a,p^a-1) = p$
(10) If $p$ is a prime such that $n^2 < p^a \le (n^2+n)$, it follows that $p \le \lfloor\sqrt{n^2+n}\rfloor = n$
(11) If $n^2 < p^a \le (n^2+n)$, then $p^{a+1} = p(p^a) > pn^2$
(12) It is well known that $\prod\limits_{p \le n} p < 4^n$ where $p$ is a prime.
(13) So that, $R(n^2+n,n^2) < 4^{\pi(n)}$ where $\pi(n)$ is the prime counting function.
(14) It is well known for $n\ge8$ that $\pi(n) \le \frac{n}{2}$
(15) But then we have a contradiction since:
$$R(n^2+n,n^2) \le 4^{n/2}$$
(16) So, we reject our assumption that there is no prime between $n^2$ and $n^2+n$.
|
(4) So it follows:$$\text{lcm}(1,2,\dots,(n^2+n)) > \left(\frac{n^2+n}{2}\right)\frac{4^{n^2+n}}{n^2+n} = \frac{4^{n^2+n}}{2}$$
This is incorrect. It should be
$$\text{lcm}(1,2,\dots,(n^2+n)) > \left(\frac{n^2+n}{2}\right)\frac{4^{(n^2+n)/2}}{(n^2+n)/2}=4^{(n^2+n)/2}$$
| {
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Find the smallest positive integer $X$ such that $478^{870} \equiv X \ (\text{mod} \ 273)$ Appreciate if one could advise if my solution is correct. Here is my attempt of the problem:
Since $(273, 478) =1,$ by Euler's theorem, $478^{\phi(273)}=478^{144} \equiv 1 \ \ (\text{mod} \ 273) \implies 478^{864} \equiv 1 \ \ (\text{mod} \ 273).$
Next, $478^{2} \equiv 22 \ \ (\text{mod} \ 39) \implies 478^{6} \equiv 22^{3} \equiv 1 \ \ (\text{mod} \ 39)$ and $478^{\phi(7)} = 478^{6} \equiv 1 \ \ (\text{mod} \ 7) $
Hence, $478^{6} \equiv 1 \ \ (\text{mod} \ 273)$ and $478^{864+6}= 478^{870} \equiv 1 \ \ (\text{mod} \ 273)$
| Your idea to use Chinese Remainder Theorem is good, but we can apply it more cleanly. We have $273 = 3\cdot 7\cdot 13,$ so we localize at each prime:
$$X\equiv 478^{870} \equiv 1^{870} \equiv 1 \pmod{3}$$
$$X\equiv 478^{870} \equiv 2^{6\cdot 145} \equiv 1 \pmod{7}$$
$$X\equiv 478^{870} \equiv (-3)^{6+12\cdot 72} \equiv 3^6 \equiv 27^2 \equiv 1^2 \equiv 1 \pmod{13}$$
By Chinese Remainder Theorem $X\equiv 1 \pmod{273}.$
| {
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solve for $x$: $(x^4-13x^2+36)^4+|x^2+x-6|+\sqrt{x^3-7x+6}=0$ There is an equation that I think it is complicated ,a little!
$$(x^4-13x^2+36)^4+|x^2+x-6|+\sqrt{x^3-7x+6}=0$$
Actually we must solve for $x$ here.
I want you to hint me how can I simplify the equation and solve it.
| Hint:
The first expression is biquadratic and
$$x^4-13x^2+36=(x^2-4)(x^2-25).$$
The second is quadratic,
$$x^2+x-6=(x-2)(x+3).$$
The third is cubic and by inspection $x=1$ is a root. So
$$x^3-7x+6=(x-1)(x^2+x-6)=(x-1)(x-2)(x+3).$$
Now you need to find the common roots.
| {
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Solve $\sqrt{\frac{4-x}x}+\sqrt{\frac{x-4}{x+1}}=2-\sqrt{x^2-12}$ The equation is
$$\sqrt{\frac{4-x}x}+\sqrt{\frac{x-4}{x+1}}=2-\sqrt{x^2-12}$$
I tried squaring both left side and right side then bringing them to same numerator but got lost from there ... any ideas of how should this be solved?
I got to:
$$2\cdot\sqrt{-\frac{(x-4)^2}{x^2+x}}+4\cdot\sqrt{x^2-12}=x^2-8-\frac{4-x}{x^2+x}$$
| Since there are three square roots you will need to square the whole equation at least three times. This is not the best way since it produces false solutions every single time and also the complexity of the equation will increase every time.
Therefore observe the numerator within the square roots of LHS and the square roots on the right
$$\sqrt{\frac{4-x}x}+\sqrt{\frac{x-4}{x+1}}=2-\sqrt{x^2-12}$$
Note that for $x=4$ the LHS will be zero hence both numerators will become zero and the denominator will be defined. So we get
$$0=2-\sqrt{4^2-12}\Rightarrow 0=2-\sqrt{16-12}\Rightarrow 0=0$$
And therefore $x_1=4$ is a solution. Also it is the only real solution (according to WolframAlpha). The other two are given by $x_{2/3}= -0.889727 \pm 0.079797 i$ but honestly speaking I guess you only can get them by squaring over and over again eliminating the false solutions in the end.
| {
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Find the value of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
Suppose $(a, b, c)\in\Bbb R^3$, $a,b,c$ are all nonzero, and we have $
\sqrt{a+b}+\sqrt{b+c}=\sqrt{c+a}$. Find the value of $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}.$$
Here is my attempt:
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{bc+ac+ab}{abc}.$$
I am having trouble in figuring out the best approach to simplify $
\sqrt{a+b}+\sqrt{b+c}=\sqrt{c+a}$
so that I can find the value of $
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ . Hope somebody has an idea.
| Just to do it the hard way.
Let $m = a+b$ and $n = b+c$ and $k = a + c$.
$\sqrt m + \sqrt n = \sqrt k$ so $k = m + n + 2\sqrt mn$ and
$\frac 1a + \frac 1b + \frac 1c = \frac 2{m - n +k} + \frac 2{n-k + m} + \frac 2{k-m + n}$
$=2(\frac {(n-k+m)(k-m + n) + (m-n+k)(k-m+n)+ (m-n+k)(n-k+m)}{(m-n+k)(n-k+m)(k-m+n)})$
$n-k +m = -2\sqrt{mn};$ $k-m+n = 2(n + \sqrt{mn});$ and $m-n+k = 2(m + \sqrt{mn})$
so we have
$\frac 1a + \frac 1b + \frac 1c=\frac {2*4}{8} \frac {-\sqrt{mn}(n+\sqrt{mn}) + (m+\sqrt{mn})(n+\sqrt{mn}) -\sqrt{mn}(m+\sqrt{mn})}{-\sqrt{mn}(n+\sqrt{mn}) (m + \sqrt{mn})}$
$=\frac {-n\sqrt{mn} - mn + mn +n\sqrt{mn} + m\sqrt{mn} + mn - m\sqrt{mn} - mn}{nm\sqrt{nm} + m^2n + mn^2 + mn\sqrt{mn}}$
$= \frac {0}{nm\sqrt{nm} + m^2n + mn^2 + mn\sqrt{mn}} = 0$
====
The "other" hard way to do it:
$\sqrt{a + b} + \sqrt{b + c} = \sqrt{a+c}$. So it seems reasonable we can attempt to solve $b$ in terms of $a $ and $c$
$(a + b) + (b+c) + 2\sqrt{a+b}\sqrt{b+c}= a + c$ so
$\sqrt{a+b}\sqrt{b+c}= - b$ (so $b \le 0$)
So $(a+ b)(b+c) = b^2$ and $ab + b^2 +ac + bc = b^2$ so $b(a+c) = -ac$
If $a + c \ne 0$ then $b = \frac {-ac}{a+c}$.
If $a +c = 0$ then $ac = 0$ and either $a$ or $c$ equal $0$ and we were told they were non-zero.
So $b = \frac {-ac}{a+ c}$.
So $\frac 1a + \frac 1b + \frac 1c = \frac 1a - \frac {a+c}{ac} + \frac 1c= \frac c{ac} - \frac {a+c}{ac}+ \frac a{ac} = \frac {a -(a+c) +c}{ac} $ and .... well, now it just falls out. It's almost ridiculous.
....
But it's interesting to note that $b$ is negative and $a$ and $c$ are positive.
It makes one wonder just how obvious it should be that $\sqrt{a - \frac 1{\frac 1a + \frac 1c}} + \sqrt{c - \frac 1{\frac 1a + \frac 1c}} =\sqrt{a + c}$?
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Seeking methods to solve: $\int_{0}^{1} \frac{1}{1 + \arctan(x)} \:dx$ I've been playing with the following definite integral and was wondering if anyone knew of any methods to solve?
$$I = \int_{0}^{1} \frac{1}{1 + \arctan(x)} \:dx$$
| The approach I took:
First let $u = \tan(x)$ to yield:
$$I = \int_{0}^{\frac{\pi}{4}} \frac{\tan^2(u) + 1}{1 + u} \:du$$
Unfortunately I had no luck with my usual tactics, so I decided to use the Taylor series of $\tan^2(u)$ at $u = 0$ which I was greatly helped with here.
We find that:
$$ \tan^{2}(u) + 1 = \sum^{\infty}_{n=1} \frac{B_{2n} (-4)^n \left(1-4^n\right)\left(2n - 1\right)}{(2n)!} u^{2n-2} $$
Which holds for $|u| < \frac{\pi}{2}$. As the domain of the integral is within that, we can use this expansion (I believe).
Hence, we arrive at,
$$I = \int_{0}^{\frac{\pi}{4}} \frac{\sum_{n = 1}^{\infty} C_n u^{2n - 2}}{1 + u} \:du = \sum_{n = 1}^{\infty}C_n\int_{0}^{\frac{\pi}{4}} \frac{u^{2n - 2}}{1 + u} \:du = \sum_{n = 1}^{\infty}C_n\: F_n $$
Where
$$ C_n = \frac{B_{2n} (-4)^n \left(1-4^n\right)\left(2n - 1\right)}{(2n)!} ,\qquad F_n = \int_{0}^{\frac{\pi}{4}} \frac{u^{2n - 2}}{u + 1} \: du$$
Taking the Taylor expansion for $\frac{1}{1 + u}$ at $u = 0$ (as given here) we can evaluate $F_n$:
\begin{align}
F_n &= \int_{0}^{\frac{\pi}{4}} \frac{u^{2n - 2}}{u + 1} \: du \\
&= \left[\ln|u + 1| + \sum_{k = 1}^{2n - 2} \left(-1\right)^k u^k \right]_{0}^{\frac{\pi}{4}} \\
&= \ln\left|\frac{\pi}{4} + 1 \right| + \sum_{k = 1}^{2n - 2} \left(-1\right)^k \frac{\pi^k}{4^kk}
\end{align}
From which we arrive at:
\begin{align}
I = \sum_{n = 1}^{\infty} C_n\:F_n &= \sum_{n = 1}^{\infty} C_n \left[\ln\left|\frac{\pi}{4} + 1 \right| + \sum_{k = 1}^{2n - 2} \left(-1\right)^k \frac{\pi^k}{4^kk} \right] \\
&= \ln\left|\frac{\pi}{4} + 1 \right| \sum_{n = 1}^{\infty} C_n + \sum_{n = 1}^{\infty}\sum_{k = 1}^{2n - 2} C_n \left(-1\right)^k \frac{\pi^k}{4^kk}
\end{align}
Recall that
$$ \tan^{2}(u) + 1 = \sum^{\infty}_{n=1} \frac{B_{2n} (-4)^n \left(1-4^n\right)\left(2n - 1\right)}{(2n)!} u^{2n-2} = \sum_{n = 1}^{\infty} C_n u^{2n - 2} $$
Hence,
$$ \tan^{2}(1) + 1 = \sec^{2}(1) = \sum_{n = 1}^{\infty} C_n$$
Thus,
\begin{align}
I &= \ln\left|\frac{\pi}{4} + 1 \right| \sum_{n = 1}^{\infty} C_n + \sum_{n = 1}^{\infty}\sum_{k = 1}^{2n - 2} C_n \left(-1\right)^k \frac{\pi^k}{4^kk} \\
&= \ln\left|\frac{\pi}{4} + 1 \right|\sec^2(1) + \sum_{n = 1}^{\infty}\sum_{k = 1}^{2n -2}\frac{\left(-1\right)^kB_{2n} (-4)^n \left(1-4^n\right)\left(2n - 1\right)\pi^k}{(2n)!4^kk}
\end{align}
However at this stage, I'm lost as to how this could be simplified.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2971586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Can't figure out $O(n \log n)$ divide and conquer algorithm For an $n$ that is a power of $2$, the $n × n$ Weirdo matrix $W_n$ is defined as follows. For
$n = 1, W_1 = [1]$. For $n > 1$, $W_n$ is defined inductively by
$W_n =
\left[
\begin{matrix}
W_\frac{n}{2} & -W_\frac{n}{2} \\
I_\frac{n}{2} & W_\frac{n}{2} \\
\end{matrix}
\right]$
where $I_k$ denotes the $k × k$ identity matrix. For example,
$W_2 =
\left[
\begin{matrix}
1 & −1 \\
1 & 1 \\
\end{matrix}
\right]$,
$W_4 =
\left[
\begin{matrix}
1 & −1 & −1 & 1 \\
1 & 1 & −1 & −1 \\
1 & 0 & 1 & −1 \\
0 & 1 & 1 & 1 \\
\end{matrix}
\right]$,
$W_8 =
\left[
\begin{matrix}
1 & −1 & −1 & 1 & −1 & 1 & 1 & −1 \\
1 & 1 & −1 & −1 & −1 & −1 & 1 & 1 \\
1 & 0 & 1 & −1 & −1 & 0 & −1 & 1 \\
0 & 1 & 1 & 1 & 0 & −1 & −1 & −1 \\
1 & 0 & 0 & 0 & 1 & −1 & −1 & 1 \\
0 & 1 & 0 & 0 & 1 & 1 & −1 & −1 \\
0 & 0 & 1 & 0 & 1 & 0 & 1 & −1 \\
0 & 0 & 0 & 1 & 0 & 1 & 1 & 1 \\
\end{matrix}
\right]$
How would I find a $O(nlogn)$ algorithm that calculates the product $W_n · \bar{x}$, where $\bar{x}$ is a vector of
length $n$ and $n$ is a power of $2$?
| I'm going to change indicies so that $W_n$ is a $2^n \times 2^n$ matrix. Let's assume that $x = (a, b)$ is a vector of length $2^n$, split up into two subvectors $a, b$ each of length $2^{n-1}$. Then we have that
$$ W_n x = \begin{pmatrix} W_{n-1} & -W_{n-1} \\ I_{n-1} & W_{n-1} \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} W_{n-1} a - W_{n-1} b \\ a + W_{n-1} b \end{pmatrix} $$
Let $f(n)$ be the time needed to evaluate the product $W_n x$ for $x$ a vector of length $2^n$. Then the above computation shows that to compute $W_n x$, we need to compute the two products $W_{n-1} a$ and $W_{n-1} b$, and then do a linear amount of work to combine them. This is the same recurrence relation as merge-sort, and so we get an $O(n \log n)$ running time.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2972937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
how to find a, b that satisfy $\displaystyle \lim_{x \to 0} \frac{e^{-2x} -\frac{1+ax}{1+bx}}{x^2}=0$ How can I find those $a$ and $b$ in $\displaystyle \lim_{x \to 0} \frac{e^{-2x} -\frac{1+ax}{1+bx}}{x^2}=0$ ?
[my attempt] Since the denominator is $x^2$, it will be $0$. And $e^{-2x}$ is 1 so I get $1-\frac{1+ax}{1+bx} = 0$. I get $a=b$ but Im not sure this is right. Am I taking a wrong way?
| The given limit condition is equivalent to $$\lim_{x\to 0}\frac{(1+bx)-(1+ax)e^{2x}}{x^2}=0\tag{1}$$ We need to make use of the following limits $$\lim_{x\to 0}\frac{e^x-1}{x}=1,\lim_{x\to 0}\frac{e^x-1-x}{x^2}=\frac{1}{2}\tag{2}$$ The first one is standard and the second one is derived from first one via L'Hospital's Rule.
By splitting the numerator in $(1)$ appropriately we can rewrite $(1)$ as $$\lim_{x\to 0}\frac{1+2x-e^{2x}}{(2x)^2}\cdot 4+\frac{b-a-2} {x} -a\cdot\frac{e^{2x}-1} {2x}\cdot 2=0$$ And using limits in $(2)$ we see that the above is equivalent to $$\lim_{x\to 0}\frac{b-a-2}{x}=2+2a$$ Thus we have $b-a=2$ and $2+2a=0$ so that $a=-1,b=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2973041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Evaluate $\lim_{n\to \infty}\left(\frac{1}{\sqrt{n(n+1)}}+\frac{1}{\sqrt{(n+1)(n+2)}}+\cdots+\frac{1}{\sqrt{(2n-1)2n}}\right)$ I want to evaluate $$\lim_{n\to \infty}\left(\frac{1}{\sqrt{n(n+1)}}+\frac{1}{\sqrt{(n+1)(n+2)}}+\cdots+\frac{1}{\sqrt{(2n-1)2n}}\right).$$
I have computed
\begin{align*}
a_{n+1}-a_n & = \frac{1}{\sqrt{(2n+1)(2n+2)}}-\frac{1}{\sqrt{n(n+1)}}\\
& = \frac{1}{\sqrt{n+1}}\left( \frac{1}{\sqrt{2(2n+1)}}-\frac{1}{\sqrt{n}} \right).
\end{align*}
Now I'm stuck.
| Hint: $$\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}<\frac{1}{\sqrt{n(n+1)}}+\frac{1}{\sqrt{(n+1)(n+2)}}+\cdots+\frac{1}{\sqrt{(2n-1)(2n)}}$$
$$<\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n-1}$$ and use $\lim_{n \rightarrow \infty} (\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n})=\ln 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2973156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Computational Complexity of Euclidean Algorithm for Polynomials Let us assume that the two polynomials that we have are degree $n$ polynomials. The naive Euclidean Algorithm for univariate polynomial does $O(n)$ divisions and each division takes $O(n^2)$. So shouldn't the naive Euclidean algorithm run for $O(n^3)$ time? But I see in Wikipedia that the algorithm runs for $O(n^2)$. I am not sure what I am missing.
| Here is an example, see what you think
$$ \left( 6 x^{5} + 5 x^{4} + 4 x^{3} + 3 x^{2} + 2 x + 1 \right) $$
$$ \left( x^{5} + 2 x^{4} + 3 x^{3} + 4 x^{2} + 5 x + 6 \right) $$
$$ \left( 6 x^{5} + 5 x^{4} + 4 x^{3} + 3 x^{2} + 2 x + 1 \right) = \left( x^{5} + 2 x^{4} + 3 x^{3} + 4 x^{2} + 5 x + 6 \right) \cdot \color{magenta}{ \left( 6 \right) } + \left( - 7 x^{4} - 14 x^{3} - 21 x^{2} - 28 x - 35 \right) $$
$$ \left( x^{5} + 2 x^{4} + 3 x^{3} + 4 x^{2} + 5 x + 6 \right) = \left( - 7 x^{4} - 14 x^{3} - 21 x^{2} - 28 x - 35 \right) \cdot \color{magenta}{ \left( \frac{ - x }{ 7 } \right) } + \left( 6 \right) $$
$$ \left( - 7 x^{4} - 14 x^{3} - 21 x^{2} - 28 x - 35 \right) = \left( 6 \right) \cdot \color{magenta}{ \left( \frac{ - 7 x^{4} - 14 x^{3} - 21 x^{2} - 28 x - 35 }{ 6 } \right) } + \left( 0 \right) $$
$$ \frac{ 0}{1} $$
$$ \frac{ 1}{0} $$
$$ \color{magenta}{ \left( 6 \right) } \Longrightarrow \Longrightarrow \frac{ \left( 6 \right) }{ \left( 1 \right) } $$
$$ \color{magenta}{ \left( \frac{ - x }{ 7 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ - 6 x + 7 }{ 7 } \right) }{ \left( \frac{ - x }{ 7 } \right) } $$
$$ \color{magenta}{ \left( \frac{ - 7 x^{4} - 14 x^{3} - 21 x^{2} - 28 x - 35 }{ 6 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ 6 x^{5} + 5 x^{4} + 4 x^{3} + 3 x^{2} + 2 x + 1 }{ 6 } \right) }{ \left( \frac{ x^{5} + 2 x^{4} + 3 x^{3} + 4 x^{2} + 5 x + 6 }{ 6 } \right) } $$
$$ \left( 6 x^{5} + 5 x^{4} + 4 x^{3} + 3 x^{2} + 2 x + 1 \right) \left( \frac{ - x }{ 42 } \right) - \left( x^{5} + 2 x^{4} + 3 x^{3} + 4 x^{2} + 5 x + 6 \right) \left( \frac{ - 6 x + 7 }{ 42 } \right) = \left( -1 \right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2974177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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If $ \int_{0}^{1} \frac{\ln x}{1-x^2} dx = -\frac{π^2}{\lambda} $ find $\lambda$ given that $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{π^2}{6} $
If $$\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{π^2}{6} $$ then $$ \int_{0}^{1} \frac{\ln x}{1-x^2} dx = -\frac{π^2}{\lambda} $$ then the value of $\lambda$ equals?
My attempt- I tried using integrating by parts to the integral. As a result, I'm left with $$-\frac{1}{2} \int_{0}^{1} \frac{\ln\left(\frac{1+x}{1-x}\right)}{x} dx $$ Now I'm stuck in here! I don't know how to move on from here! Or maybe integrating by parts was a bad option? If it is, please guide me to a solution or Please help me on how to continue from here!
Any help would be appreciated.
| Use the expansion
$$\ln\frac{1+x}{1-x}=2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots\right)$$
then
\begin{align}
-\frac{1}{2} \int_{0}^{1} \frac{\ln\frac{1+x}{1-x}}{x} dx
&= -\frac{1}{2} \int_{0}^{1} \dfrac1x2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots\right) dx \\
&= -\left(1+\frac{1}{3^2}+\frac{1}{5^2}+\cdots\right) \\
&= \color{blue}{-\dfrac{\pi^2}{8}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2974841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Functional equation - Cyclic Substitutions Please help solve the below functional equation for a function $f: \mathbb R \rightarrow \mathbb R$:
\begin{align}
&f(-x) = -f(x) , \text{ and } f(x+1) = f(x) + 1, \text{ and } f\left(\frac 1x\right) = \frac{f(x)}{x^2} \\
&\text{ for all } x \in \mathbb R \text{ and } x \ne 0 .
\end{align}
I know this will be solved by cyclic substitutions, but I'm unable to figure out the exact working. Can someone explain step wise?
| For $x\neq0$ and $x\neq1$ we obtain: $$f(x)=x^2f\left(\frac{1}{x}\right)=x^2f\left(\frac{1}{x}-1+1\right)=x^2\left(f\left(\frac{1}{x}-1\right)+1\right)=$$
$$=x^2+x^2f\left(\frac{1-x}{x}\right)=x^2+x^2\cdot\frac{f\left(\frac{x}{1-x}\right)}{\frac{x^2}{(x-1)^2}}=x^2+(x-1)^2f\left(\frac{x}{1-x}\right)=$$
$$=x^2+(x-1)^2f\left(\frac{1}{1-x}-1\right)=x^2+(x-1)^2\left(f\left(\frac{1}{1-x}\right)-1\right)=$$
$$=x^2-(x-1)^2+(x-1)^2\cdot\frac{f(1-x)}{(1-x)^2}=2x-1-f(x-1).$$
Thus, $$f(x)+f(x-1)=2x-1$$ or
$$f(x)+f(x+1)=2x+1,$$ which gives
$$f(x)+f(x)+1=2x+1$$ or
$$f(x)=x.$$
Now, show that $f(0)=0$ and $f(1)=1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2975898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Arranging 5 pairs of brothers in ten tables Suppose I have $10$ people composed of $5$ pairs of brothers. They are seated randomly around $5$ tables. Each table has only two places.
What is the possibility that exactly $3$ pairs of brothers will be sitting at the same table?
$$\\ \frac{ {5 \choose 3} \cdot 3\cdot 2\cdot {4 \choose 1}{2 \choose 1}}{{10 \choose 2,2,2,2,2} / 5!} $$
What I thought was picking $\ 3 $ pairs out of the $\ 5 $ then seating each of them at a different table. Therefore, $\ {5 \choose 3} \cdot 3 \cdot 2 \cdot 1$ and then I'm left with $\ 4 $ people which is two pairs and picking one of them $\ {4 \choose 1} \cdot {2 \choose 1} $, because after picking one of the people, I have only two options left for the second one and none for the pairing the last two.
The sample space I thought would be $\ {10 \choose 2,2,2,2,2} / 5! $ because it's the number of ways to divide $10$ people into $5$ groups of $2$ with no difference between the groups.
I really don't know what I am doing wrong here.
| Let's take as our sample space the number of ways of placing two people at each table, which is
$$\binom{10}{2, 2, 2, 2, 2} = \binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}$$
For the favorable cases, choose which three of the five tables will receive a pair of brothers. Choose a pair of brothers to sit at each of those tables. That leaves four people and two tables. Choose which of those two tables the youngest of those four people sits at. One of the other three people will be his brother. To ensure that there are exactly three pairs of brothers sitting together, we must choose one of the other two people to sit with the youngest person remaining. The final two people must sit together at the remaining table. Hence, the number of favorable cases is
$$\binom{5}{3} \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 2$$
Thus, the probability that exactly three pairs of brothers will be seated together at the five tables if they are seated randomly is
$$\frac{\dbinom{5}{3} \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 2}{\dbinom{10}{2}\dbinom{8}{2}\dbinom{6}{2}\dbinom{4}{2}\dbinom{2}{2}}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding $B^{225}$ without many computations I have that $B$ is a 4x4 matrix. $B-5I=\begin{pmatrix} -2 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 \\ 3 & 0 & -3 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$. The question asks to find $B^{225}$ without performing many computations, leaving the answer as a product of 3 matrices (univerted matrices are acceptable).
My thinking is that this problem may have to do with diagonalizing $B$ based on eigenvectors. $\lambda=5$ is clearly an eigenvalue because $det(B-5I)=0$, and I also reasoned that possible eigenvectors could be $\begin{pmatrix} 3 \\ 0 \\ -2 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} -3 \\ 0 \\ 2 \\ 0 \end{pmatrix}$, just based on the way the matrix multiplication would work to yield linear combinations of these eigenvectors. So for diagonalization we need to bases $\alpha=...$ and $\beta=\left\{\begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix},\begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix},\begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix},\begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}\right\}$. I believe these eigenvectors should belong in $\alpha$ but I don't know which other eigenvectors belong or how to find them.
Even beyond that point, I don't quite understand the meaning of this question and how we are supposed to make the logical leap to $B^{225}$. What am I missing?
| Let $A=B-5I$. Then $A^2=-5A$. Therefore $A^n=(-5)^{n-1}A$ for $n\ge1$.
Then
$$B^n=(5I+A)^n=5^n I+\sum_{k=1}^n {n\choose k}5^{n-k}A^k
=5^n I+\sum_{k=1}^n (-1)^{k-1}{n\choose k}5^{n-1}A
=5^n I+5^{n-1}A.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find the Max, Min,Sup and inf of $A = \left \{ \frac{1}{3^\frac{x}{2}+3^\frac{2}{x}} \right \}$ I need to find the min, max, sup and inf of $$A = \left \{ \frac{1}{3^\frac{x}{2}+3^\frac{2}{x}} , x>0, x \in \mathbb{R} \right \}$$
My attempt:
$3^\frac{x}{2} + 3^\frac{2}{x} \geqslant 2\sqrt{3^{2 \left | x \right |}} = 2 \times {3^x}$ By the A.M G.M inequality. Also:
$$ 0 \leqslant \frac{1}{3^\frac{x}{2}+3^\frac{2}{x}} \leq \frac{1}{2 \times 3^x} $$
I'm a little stuck now and I would appreciate some assistance.
| Consider the function $f(x)=\frac{1}{3^\frac{x}{2}+3^\frac{2}{x}}$ defined on $(0,\infty).$
*
*Clearly, $f(x)>0$ for any $x>0.$ Since $\lim_{x \to 0^+}f(x)=0\quad$ and $\quad\lim_{x \to \infty}f(x)=0,$ $$Inf A=0 \quad \text{and A doesn't have a minimum}.$$
*If $x \in (0,2),$ then $3^\frac{x}{2}<3^\frac{2}{x}$ and $$f(x)<\frac{1}{2 \cdot 3^{x\over 2}}<\frac{1}{6}$$
If $x \in (2,\infty),$ then $3^\frac{2}{x}<3^\frac{x}{2}$ and $$f(x)<\frac{1}{2 \cdot 3^{2\over x}}<\frac{1}{6}$$
Thus $1\over 6$ is un upper-bound of $A.$ Since $f(2)={1\over 6},$ this is the maximum of $A.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding $\int_{0}^{\pi} 2\pi\frac{R^2(z-R\cos\theta)\sin\theta }{\sqrt{(R^2+z^2-2zR\cos\theta)^3}}\,d\theta$ I am doing some physics problem and in order to solve it I need to solve the following integral:
$$ \int_{0}^{\pi} 2\pi\frac{R^2(z-R\cos\theta)\sin\theta }{\sqrt{(R^2+z^2-2zR\cos\theta)^3}}\,d\theta $$ where $z>R$.
Considering physics of the problem I expect to get $\dfrac{4\pi R^2}{z^2}$ (although that can be wrong). I tried to do substitution $u=R^2+z^2-2zR\cos\theta$, but I did not succeed to get the solution. Also i tried to convert that real integral into complex integral to solve it, but also I did not succeed to get the solution. So my question is how to solve that integral?
Thank you for any help!
| The change of variables $t = \frac{2 z R}{z^2+R^2} \cos(\theta)$ yields
\begin{align} f(z,R) &\equiv 2\pi R^2 \int \limits_{0}^{\pi} \frac{z-R\cos(\theta)}{(z^2+R^2-2zR\cos\theta)^{3/2}} \, \sin(\theta) \mathrm{d} \theta = \frac{\pi R}{\sqrt{z^2+R^2}} \int \limits_{-\frac{2 z R}{z^2+R^2}}^{\frac{2 z R}{z^2+R^2}} \frac{1-\frac{z^2+R^2}{2 z^2} t}{(1-t)^{3/2}} \, \mathrm{d} t \\
&= \frac{\pi R}{\sqrt{z^2+R^2}} \left[\frac{2}{\sqrt{1-t}} - \frac{z^2+R^2}{z^2} \frac{2-t}{\sqrt{1-t}}\right]_{t=-\frac{2 z R}{z^2+R^2}}^{t=\frac{2 z R}{z^2+R^2}} \, ,
\end{align}
where the second integral can be computed via integration by parts, for example. Now plug in the limits and after a few lines of algebra you should indeed arrive at
$$ f(z,R) = 4 \pi \frac{R^2}{z^2} \, . $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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For every natural number $m$, $2^{2m}-1$ is divisible by $3$ Can anyone tell me if I missed any small details in this proof?
(basis) $2^{2(1)}-1 = 3*1$ which is divisible by 3.
(induction) Fix $m>1$ so $p<m$. Hence for every natural number $p,$ $2^{2p}-1=3p.$ then for every natural number $(p+1),$ $2^{2(p+1)}-1=3(p+1)$ then
$2^{2(p+1)}-1=3(p+1)$
$2*2^{2p}-1=3(p+1)$
$2*2^{2p}-1=3(p+1)$
$2(3p+1)-1=3(p+1)$ by induction hypothesis
$2(3p+1)+2^2-3=3(p+1)$
$6p+3=3(p+1)$
$3(p+1)=3(p+1)$
Thus, the statement is true for the natural number n whenever it is true for any natural number less than n.
| $2^{2m} - 1 = (2^m)^2 - 1 = (2^m - 1)(2^m + 1); \tag 1$
of the three consecutive integers
$2^m - 1, \; 2^m, \; 2^m + 1 \tag 2$
precisely one of the must be divisible by $3$ (this is true of any three consecutive integers); clearly
$3 \not \mid 2^m; \tag 3$
therefore either
$3 \mid 2^m - 1 \; \text{or} \; 3 \mid 2^m + 1, \tag 4$
which implies
$3 \mid (2^m - 1)(2^m + 1) = 2^{2m} - 1. \tag 5$
Doesn't seem to use induction, does it? But to be thorough, we need to prove that $3$ divides precisely one of any three consecutive positive integers . . . and induction might help here! For example, restricting ourselves to the naturals $\Bbb N$, we see that $3$ divides precisely one out of $1, \; 2, \; 3$; now suppose for some $k$, $3$ divides precisely one of $k - 2$, $k - 1$, $k$; now consider the triplet $k - 1$, $k$, $k + 1$; if $3 \not \mid k - 2$, it must by hypothesis divide one of $k - 1$, $k$; but then $3 \not \mid k + 1$, lest it also divide $k - 2 = (k + 1) - 3$, which we have rejected. So $3$ divides exactly one of the triplet $k - 1$, $k$, $k + 1$; and if $3 \mid k - 2$, we have $3 \mid (k - 2) + 3 = k + 1$, so once again $3$ is a factor of precisely one of $k - 1$, $k$, $k + 1$ and our little induction is done.
I find this method a little easier to get my head around, although I commend Bram28's suggestion for a proof, and his explanation, as well.
Note Added in Edit, Thursday 1 November 2018, 6:44 PM PST: OK, so I fooled around with it a little more and found another proof which doesn't invoke the "three consecutive naturals" business, and I think is likely what Bram28 was getting at in his closing sentences; the argument performs induction directly $2^{2k} - 1$; suppose
$3 \mid 2^{2k} - 1; \tag 6$
then
$\exists q \in \Bbb N, \; 2^{2k} - 1 = 3q; \tag 7$
thus,
$2^{2k} = 3q + 1; \tag 8$
now,
$2^{2(k + 1)} = 2^{2k + 2} = 2^{2k}2^2 = 4 \cdot 2^{2k} = 4(3q + 1) = 12q + 4, \tag 9$
whence
$2^{2(k + 1)} - 1 = 12q + 3 = 3(4q + 1), \tag{10}$
and that does it, does it not? $OE\Delta$ End of Note.
| {
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Prove that $f$ is additive if $f(x)f(x-y)+f(y)f(x+y)= f(x)^2+f(y)^2$
Say $f:\mathbb{R}\to \mathbb{R}$ is non-constant such that $$f(x)f(x-y)+f(y)f(x+y)= f(x)^2+f(y)^2$$
Prove that $f(x+y)=f(x)+f(y)$.
If we put $a=f(0)$ and $y=0$ we get $$f(x)^2+af(x)= f(x)^2+a^2 \implies af(x) = a^2$$
If $a\ne 0$ then $f(x)=a$ is constant function which can not be, so $a=0$. Now if we put $x=y$ we get $$f(x)f(2x)=2f(x)^2$$
From here I have no more idea what to do.
Edit after Lulu's comment: If we put also $y=-x$ we get $$2f(x)^2= f(x)f(2x) = f(x)^2+f(-x)^2\implies \boxed{f(x)^2=f(-x)^2}$$
| Clearly any constant solution is the zero function. So assume that $f$ is nonzero. Let $P(x,y)$ denote the given assertion. We have $f(0)=0$ by $P(x,0)$. Take $f(x)\neq 0$, then $P(x,x)$ implies $f(2x)=2f(x)$ and following an easy induction $f(kx)=kf(x)$ for all $k\in \mathbb N.$ If $f(x_0)=0$ then $P(x_0,-x_0)$ gives $f(-x_0)=0.$ So for all $x,$ $f$ is odd. Now $P(x,-y)$ and $P(x,y)$ implies $$f(x-y)(f(x)+f(y))=f(x+y)(f(x)-f(y)).$$ If $f(x)+f(y)\neq 0$ then substituting value of $f(x-y)$ in original equation gives $f$ is additive. If $f(x)+f(y)=0$ but $f(x+y)\neq 0$, then $f(x)=f(y)=0,$ and so $P(x+y,y)$ gives $f(x+y)=0,$ absurd.
| {
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Integral of $\int \frac{-x^2+2x-3}{x^3-x^2+x-1}dx$ I have this simple integral: $\int \frac{-x^2+2x-3}{x^3-x^2+x-1}dx$ and I can't come up with the correct answer.
Here's what I did:
I found the roots for $x^3-x^2+x-1$ so I could do partial fractions:
$\frac{-x^2+2x-3}{x^3-x^2+x-1}=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}=\frac{A(x^2+1)+(Bx+C)(x-1)}{(x-1)(x^2+1)}$
I then found the values for $A$, $B$ and $C$:
$x^2(A+B)+x(-B+C)+A-C=-x^2+2x-3$
$
\left\{
\begin{array}{c}
A+B=-1 \\
-B+C=2 \\
A-C=-3
\end{array}
\right.
$
$A=-1$, $B=0$, $C=2$
So now back again with the integral:
$\int \frac{-1}{x-1}dx+\int \frac{2}{x^2+1}dx=-\int \frac{1}{x-1}dx + 2 \int \frac{1}{x^2+1}dx=-\ln|x-1|+2|x^2+1|=2\ln|\frac{x^2+1}{x-1}|+C$
But why the correct answer is $2 \arctan(x)-\ln|x-1|+C$? Where does the $arctan$ come from?
| Your computation is almost correct, but you made a mistake here.
Note that$\int\dfrac{2}{x^2+1}=2\arctan(x)$, but not $\int\dfrac{2}{x^2+1}\ne2|x^2+1|$
| {
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Proving $5 \mid (n^5-n)$ for all $n \in \mathbb{Z}^+$
Prove for all $n \in \mathbb{Z}^+$ that $5 \mid (n^5-n)$
My proof
Basis step: Since $5 \mid (1^5-1) \iff 5 \mid 0$ and $5 \mid 0$ is true, the statement is true for $n=1$.
Inductive step: Assume the statement is true for $n = k$; that is, assume that $5 \mid (k^5-k)$ is true. Then there is $m \in \mathbb{Z}^+$ such that $$k^5 - k = 5m.$$ We must show that this statement is true for $n = k+1$, i.e. show that there is $\ell \in \mathbb{Z}^+$ such that $$(k+1)^5 - (k+1) = 5\ell.$$ Note that $(k+1)^5 - (k+1)$ expands as $k^5 + 5k^4 + 10k^3 + 10k^2 + 4k$.
We can try to find a polynomial $P(x)$ such that $$(k^5-5) + P(x) =(k+1)^5 - (k+1) = k^5 + 5k^4 + 10k^3 + 10k^2 + 4k$$ so as to try to add $P(x)$ to both sides of the assumption. We find that
$$\begin{align}P(x) &= k^5 + 5k^4 + 10k^3 + 10k^2 + 4k - (k^5-5) \\
&=5k^4 + 10k^3 + 10k^2 + 5k \end{align}$$ and we can also observe that since $k\in\mathbb{Z}^+$, we have that $\frac{1}{5}P(x) = k^4 + 2k^3 + 2k^2 + k$ is a positive integer. Thus we add this to both sides of our assumption $$ \begin{align}
k^5 - k &= 5m \\
(k^2 - k) + P(x) &= 5m + P(x) \\
(k+1)^5 - (k+1) &= 5\left(m + \tfrac{1}{5}P(x)\right)
\end{align}.$$ Since $\frac{1}{5}P(x),m \in \mathbb{Z}^+$, it follows that $m + \tfrac{1}{5}P(x) \in \mathbb{Z}^+$. Thus $5 \left| \big[ (k+1)^5 - (k+1) \big] \right.$
By PMI, $5 \mid (n^5-n)$ for all $n \in \mathbb{Z}^+$.
My questions
*
*Is this proof valid?
*What other ways can this be proved by induction? The polynomial expansions took a while to deal with, so I was wondering if there are any alternate methods. (Just FYI, I only have college first-year-level knowledge)
| An alternative proof:
Note that we can write $n^{5} - n$ as follows:
$$n^{5} - n = 5\left[6{n\choose 2} + 30{n\choose 3} + 48{n\choose 4} + 24{n\choose 5}\right].$$
This is very clearly a multiple of $5$ (the expression is $5$ times another integer), so we are done.
Here, ${n\choose k} = {\frac{n!}{k! (n-k)!}}$
| {
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Evaluate a series with binomial coefficients
If
$$y=\frac{3}{5}+\frac{1\cdot 3}{2!}(\frac{2}{5})^2+\frac{1\cdot 3\cdot 5}{3!}(\frac{2}{5})^3+\dots$$
Then find the value 0f $y^2+2y$.
My approach is as follow
$$y-\frac{1}{5}=\frac{2}{5}+\frac{1\cdot 3}{2!}(\frac{2}{5})^2+\frac{1\cdot 3\cdot 5}{3!}(\frac{2}{5})^3+\dots$$
Let $x=\frac{2}{5}$, $Y=y-\frac{1}{5}$. Then
$$Y=x+\frac{1\cdot 3}{2!}(x)^2+\frac{1\cdot 3\cdot 5}{3!}(x)^3+\dots$$
I am not able to proceed from here.
| Hint. Recall the definition of double factorial,
$$1\cdot 3\cdots (2k-1)=\frac{k!}{2^k}\binom{2k}{k}.$$
Hence
$$Y=x+\frac{1\cdot 3}{2!}(x)^2+\frac{1\cdot 3 \cdot 5}{3!}(x)^3+\dots=
\sum_{k=1}^{\infty}\binom{2k}{k}(x/2)^k.$$
Now take a look at How to show that $1 \over \sqrt{1 - 4x} $ generate $\sum_{n=0}^\infty \binom{2n}{n}x^n $ .
| {
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Extending the question: Is $S$ a basis for $P^3$?
Let $S = \{4 - t, t^3, 6t^2, 3t + t^3, -1 + 4t\}$. Is $S$ a basis for $P^3$?
Polynomial Equation:
0t3 + 0t2 - t + 4 = 0
t3 + 0t2 + 0t + 0 = 0
0t3 + 6t2 + 0t + 0 = 0
t3 + 0t2 + 3t + 0 = 0
0t3 + 0t2 + 4t - 1 = 0
Transform into an augmented matrix:
\begin{bmatrix}0&0&-1&4&0\\1&0&0&0&0\\0&6&0&0&0\\1&0&3&0&0\\0&0&4&-1&0\end{bmatrix}
RREF:
\begin{bmatrix}1&0&0&0&0\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&0\end{bmatrix}
I answered the question saying that S is not a basis for p3 because the all zero bottom row means that there is a free variable within the equation, making the overall system linearly dependent and therefore not a basis.
To push my understanding, can I further say that the vector (-1 + 4t) is the only dependent variable and if it was not in S, then S would be a basis of p3?
Also, the presence of the free variable would mean that the dependent vector would be a non-unique combination of ANY of the other vectors found within S?
| $P^3$ is 4 dimensional with a natural basis, $1, t, t^2, t^3$. In this basis, we can represent $S$ as the set of columns of
\begin{pmatrix}
4 & 0 & 0 & 0 & -1 \\
-1 & 0 & 0 & 3 & 4 \\
0 & 0 & 6 & 0 & 0 \\
0 & 1 & 0 & 1 & 0 \\
\end{pmatrix}
This row reduces to
\begin{pmatrix}
1 & 0 & 0 & 0 & -\frac{1}{4} \\
0 & 1 & 0 & 0 & -\frac{5}{4} \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & \frac{5}{4} \\
\end{pmatrix}
Row reducing gives us a basis for the relations between columns (the null space). Specifically, we have one relation
$$ (4 - t) + 5 (t^3) - 5(3t + t^3) + 4(-1 + 4t) = 0 $$
I can solve this equation for any element of $S$ except $6t^2$. This means that when I remove $4-t$ or $t^3$ or $3t+t^3$ or $-1 + 4t$ from $S$, I get a basis (because I no longer have the above relation and because I have a way of writing the removed vector in terms of the remaining vectors [using the above relation]).
| {
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Prove that $\sqrt[3]{\frac{1}{9}}+\sqrt[3]{-\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}$ is a root for $x^3+\sqrt[3]{6}x^2-1$
Prove that
$$c=\sqrt[3]{\frac{1}{9}}+\sqrt[3]{-\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}$$
is a root for
$$P(x)=x^3+\sqrt[3]{6}x^2-1$$
Source: list of problems for math contest preparation.
I have no clue on how to approach the problem. Most surely not by direct substitution, but I'm not seeing how. Some hint will be appreciated.
| Simplify $c$ before substituting to $P(x)=x^3+\sqrt[3]{6}x^2-1$:
$$c=\sqrt[3]{\frac{1}{9}}+\sqrt[3]{-\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}=\frac{1-\sqrt[3]{2}+\sqrt[3]{2^2}}{\sqrt[3]{9}}=\frac{1+2}{\sqrt[3]{9}(1+\sqrt[3]{2})}=\frac{\sqrt[3]{3}}{1+\sqrt[3]{2}};\\
\begin{align}P\left(\frac{\sqrt[3]{3}}{1+\sqrt[3]{2}}\right)&=\frac{3}{(1+\sqrt[3]{2})^3}+\sqrt[3]{6}\cdot \frac{\sqrt[3]{9}}{(1+\sqrt[3]{2})^2}-1=\\
&=\frac{3+3\sqrt[3]{2}(1+\sqrt[3]{2})-(1+\sqrt[3]{2})^3}{(1+\sqrt[3]{2})^3}=\\
&=\frac{\color{red}3+\color{blue}{3\sqrt[3]{2}}+3\sqrt[3]{4}-\color{red}1-\color{blue}{3\sqrt[3]{2}}-3\sqrt[3]{4}-\color{red}2}{(1+\sqrt[3]{2})^3}=\\
&=0.\end{align}$$
| {
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Solution to differential equation system and solution to its conversion into 2nd order differential equation Following is the differential equation system with IVP
$\vec{x}'=\small\begin{pmatrix}3&-9\\4&-3\end{pmatrix}\vec{x},
\vec{x}(0)=\small\begin{pmatrix}2\\-4\end{pmatrix}$ The particular solution to this differential equation system is given below.
$\vec{x}(t)=\frac23\small\begin{pmatrix}3\cos{(3\sqrt{3}t)}\\\cos{(3\sqrt{3}t)}+\sqrt{3}\sin{(3\sqrt{3}t)}\end{pmatrix}+\frac{14}{3\sqrt{3}}\small\begin{pmatrix}3\sin{(3\sqrt{3}t)}\\\sin{(3\sqrt{3}t)}-\sqrt{3}\cos{(3\sqrt{3}t)}\end{pmatrix}...(1)$
When i converted this differential equation system into 2nd order differential equation, I got $y"+13y'-7y=0, y(0)=2,y'(0)=-4$
Now, the particular solution to this 2nd order equation is $-\frac{2\sqrt{6}}{3}\sin{(\sqrt{6}t)}+2\cos{(\sqrt{6}t)}...(2)$
Now why there is a difference between these two solutions namely (1) and (2)
| That is not the second order equation I get. The system is:
$$\left\{
\begin{matrix}
\dot x=3x-9y \\
\dot y=4x-3y \\
\end{matrix}
\right.$$
From the second equation we get
$$\ddot y=4\dot x-3\dot y\tag 1$$
and
$$4x=\dot y+3y\tag 2$$
Substituting the value for $\dot x$ from the first equation into $(1)$
$$\ddot y=12x-36y-3\dot y\tag 3$$
And substituting $x$ from $(2)$ into $(3)$
$$\ddot y=3\dot y+9y-36y-3\dot y$$
$$\ddot y+27y=0$$
It has as solution:
$$y=c_1\sin\left(3\sqrt{3}t\right)+c_2\cos\left(3\sqrt{3}t\right)$$
Further, I get $\dot y(0)=20$ as $\dot y(t)=4x(t)-3y(t)$ and hence $\dot y(0)=4·2-3(-4)$
The equation is different and, waiting for the initial values to apply, the argument for the sine and cosine is the same as for the solution for the system.
| {
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Expectation of total scores when rolling a die until the score is not 6 An unbiased six-sided die 1, 2, ..., 6 is used during a game, where if one player scores a 6, then he rolls the die again. The player continues playing the game until he scores another value which is NOT 6.
Find the expected value of the total score on one turn.
My working:
$\mathbb{E}[X]=\sum_{k=1}^{\infty}kP(X=k).$
where, $P(X=k) = \frac{1}{6}^{k-1}\frac{5}{6}$
So, $\mathbb{E}[X]=\sum_{k=1}^{\infty}kP(X=k)$ $= \frac{5}{6}\sum_{k=1}^{2}k\frac{1}{6}^{k-1} = \frac{5}{6}(1+2(\frac{1}{6})) = \frac{10}{9}$
EDITED
So, $\mathbb{E}[X]=\sum_{k=1}^{\infty}kP(X=k)$ $= \frac{5}{6}\sum_{k=1}^{\infty}k\frac{1}{6}^{k-1} = \frac{5}{6}\frac{1}{(1-\frac{1}{6})^{2}} = \frac{6}{5}$
But on the answer sheet, it is $\frac{21}{5}$
I would appreciate if someone can point out my mistake. Thanks!
| The expected sum after $k$ rolls is not $k$ but $6(k-1)+3$ indeed if we stop at first roll it means that we get $1,2,3,4,5$ and the average sum will the mean i.e. $3$, if we rolled twice means that the first roll we get $6$ and now one of the values $1,2,3,4,5$ and so on. It means that the exact formula is
\begin{align}
\mathbb{E}[X]&=\sum_{k=1}^{\infty}(6(k-1)+3)P(X=k)=\frac{5}{6}\sum_{k=1}^{\infty}(6(k-1)+3)(\frac{1}{6})^{k-1}=\\
& = 5\sum_{k=1}^{\infty}(k-1)(\frac{1}{6})^{k-1}+\frac{5}{2}\sum_{k=1}^{\infty}(\frac{1}{6})^{k-1}=\\
&=\frac{6}{5}+3=\frac{21}{5}
\end{align}
| {
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Indefinite integral of $\frac{\sqrt{ 25x^2 - 4}}{x}$ Right off the bat i factored a 4 from the radicand to get it into a form such that i can leverage ${\tan^2(\theta)}$ = ${\sec^2(\theta)} - 1$
Then i set $\frac{25}{4}$${x^2}$ = ${\sec^2(\theta)}$
${x}$ = $\frac{2}{5}\sec(\theta)$
${dx}$ = $\frac{2}{5} \sec(\theta)\tan(\theta) \ d \theta$
Now i went ahead and substituted into the integrand, after simplification and evaluation of the new integral we get
$2\tan\theta - 2\theta$
Now to sub back in terms of x by using a right triangle $2\tan(\theta$) = $\sqrt{ 25x^2 - 4}$
(Note: on the right triangle, Opposite = $\sqrt{ 25x^2 - 4}$, Adjacent = $2$, hypotenuse = $5x$)
However for $2\theta$ i made the resubstitution that
${x}$ = $\frac{2}{5}(\sec\theta)$
so to solve for $\theta$
$\operatorname {arcsec} \frac{5}{2}$${x}$ = ($\theta$)
However that was the wrong resubstitution for ($\theta$), the correct one was
($\theta$) = ${\arccos(\frac{2}{5x})}$
How did $\arccos$ even get into the picture? we already had ${x} = \frac{2}{5}(\sec(\theta))$ so intuitively i just solved for theta. I dont understand how its $\arccos$ and not $\operatorname {arcsec}$, can someone help me make sense of the last part? pls halp
| Generally, seeing $\sqrt{x^2 - a^2}$ or something of the like begs for the substitution of $x = a\cos(u)$, since $\cos^2(x) - 1 = \sin^2(x)$. (Or perhaps the hyperbolic version; similar process, I'll leave that calculation up to you if you prefer.) So, notice, by factoring out $\sqrt{25}=5$ from the numerator,
$$\mathcal I := \int \frac{\sqrt{25x^2 - 4}} x dx = 5 \int \frac{\sqrt{x^2 - (2/5)^2}}{x}dx$$
Let $x = 2/5 \cdot \cos(u)$. Then $dx = -2/5 \cdot \sin(u)du$. Notice the radicand becomes
$$x^2 - (2/5)^2 = (2/5)^2 (\cos(u)^2 - 1) = (2 \sin(u) / 5)^2$$
Thus,
$$\mathcal I = 5 \int \frac 2 5 \frac{\sin(u)}{\cos(u)} \cdot \frac{-2}{5} \sin(u)du$$
Some simplification gets us to
$$\mathcal I = \frac{-4}{5} \int \frac{\sin^2(u)}{\cos(u)}du = \frac{-4}{5} \int \frac{1 - \cos^2(u)}{\cos(u)}du = \frac{-4}{5}\left( \int \sec(u)du-\int\cos(u)du \right)$$
These remaining integrals are fairly elementary. You can see this for a list of various integrals. Most relevant,
$$\begin{align}
\int \cos(u)du &= \sin(u) + C\\
\int \sec(u)du &= \ln|\sec(u)+\tan(u)|+C=2 \text{ arctanh} \left( \tan \frac u 2 \right)+C
\end{align}$$
Everything that remains is just back-substitution, simplification, and a bit of algebra, which I'll leave to you.
| {
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$a,b,c,d,e$ are zeroes of $6x^5+5x^4+4x^3+3x^2+2x+1$ find $(a+1)(b+1)(c+1)(d+1)(e+1)$
If $a,b,c,d,e$ are zeroes of the polynomial
$$6x^5+5x^4+4x^3+3x^2+2x+1$$
find the value of $(a+1)(b+1)(c+1)(d+1)(e+1)$.
According to me in order to solve this problem one should first factorize the given polynomial in the form of:
$$(x-a)(x-b)(x-c)(x-d)(x-e)$$
and equate the two and then compare the coefficients to get the following equations:
$(a+b+c+...)=-5$
$(ab+ac+ad+bd+.....)=4$
$(abc+abd+abe+...)=-3$
$(abcd+abce+abde+.....)=2$
$abcde=-1$
Now, adding all the equations and with some factorization we get:
$$(a+1)(b+1)(c+1)(d+1)(e+1)=-2$$
but my book says the answer is $0.5$.
Where am I going wrong and what is the actual method to solve this, with the correct answer?
| $6(x-a)(x-b)(x-c)(x-d)(x-e) = 6x^5 + 5x^4 + 4x^3 + 3x^2 + 2x + 1$
Plugging x = -1
$6(-1-a)(-1-b)(-1-c)(-1-d)(-1-e) = -6(a+1)(b+1)(c+1)(d+1)(e+1) = -6 + 5 - 4 + 3 - 2 + 1 = -3\\
(a+1)(b+1)(c+1)(d+1)(e+1) = \frac 12$
| {
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Easy way of tackling $\sqrt{x^2}+x=\sqrt{5}$ Solve for x
$$\sqrt{x^2}+x=\sqrt{5}\tag1$$
$$\sin(\sqrt{x^2}+x)=\sin({\sqrt{5}})\tag2$$
$$\sin(\sqrt{x^2})\cos(x)+\cos(\sqrt{x^2})\sin(x)=\sin(\sqrt{5})\tag3$$
$$\sin(\sqrt{x^2})+\cos(\sqrt{x^2})\tan(x)=\sin(\sqrt{5})\sec(x)\tag4$$
$$\sin(\sqrt{x^2})+\cos(\sqrt{x^2})\sqrt{\sec^2(x)-1}=\sin(\sqrt{5})\sec(x)\tag5$$
finally after all the hard work $(5)$ reveals
$$x+x=\sqrt{5}\tag6$$
$$x=\frac{\sqrt{5}}{2}\tag7$$
Is there another easy way solving this $(1)?$
| Use cases. Certainly $x \neq 0.$
Case I: $x > 0.$ Then $\sqrt{x^2} + x = 2x.$ Thus $x = \frac{\sqrt{5}}{2}.$
Case II: $x < 0.$. Then $\sqrt{x^2} = -x$, but then this says $ 0 = \sqrt{5},$ contradiction. Hence we must have the only solution $x = \frac{\sqrt{5}}{2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2990656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving $\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac{1}{2}$
Prove the identity
$$8\cos^4 \theta -4\cos^3 \theta-8\cos^2 \theta+3\cos \theta +1=\cos4\theta-\cos3\theta$$
If $7\theta $ is a multiple of $2\pi,$ Show that $\cos4\theta=\cos3\theta$ and deduce,
$$\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac{1}{2}$$
My Work
I was able to prove identity using half angle formula and $\cos3\theta $ expansion.
Since
$$7\theta=2n\pi$$ $$4\theta=2n\pi-3\theta$$
$$\therefore \cos4\theta=\cos3\theta$$
I cannot prove the final part.
Please help me. Thanks in advance.
| $\cos\frac{0\pi}{7}, \cos\frac{2\pi}{7}, \cos\frac{4\pi}{7}, \cos\frac{6\pi}{7}$ are distinct roots of the fourth order polynomial $$P(x)=8x^4-4x^3-8x^2+3x+1$$
So $P(x)$ can be re-written $$P(x)=8\left(x-\cos\frac{0\pi}{7}\right)\left(x-\cos\frac{2\pi}{7}\right)\left(x-\cos\frac{4\pi}{7}\right)\left(x-\cos\frac{6\pi}{7}\right)$$
Therefore looking at $x^3$ coefficient gives $$\cos\frac{0\pi}{7}+\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=\frac{4}{8}=\frac{1}{2}$$
Actually, for any $n>=2$, $$P_n(x)=T_n(x)-T_{n-1}(x)=2^{n-1}x^n-2^{n-2}x^{n-1}+...$$
Where $T_n$ is the nth Chebyshev polynomial.
So $$P_n(cos(x))=cos(nx)-cos((n-1)x)$$And Likewise, $$\sum_{k=0}^{n-1} cos\frac{2k\pi}{2n-1} = \frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2991542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Number of garlands which could be formed by using $3$ flowers of same type and $12$ flowers of other type. Question : Find number of garlands which could be formed by using $3$ flowers of same type and $12$ flowers of other type.
By garland I mean a perfectly circular one .
All $3$ flowers are alike of one type, all other $12$ are alike of different type.
An attempt at the solution.
Firstly we arrange all the $3$ of one type flowers then arrange the other ones, now after arranging the $3$ my circle got divided into $3$ parts and the sum of the flowers in that part has to be $12$.
Let the three parts be $a, b$, and $c$, so
$$a+b+c=12$$
Now the number of integral solutions (without considering the order as flowers are alike) of this equation will be my answer it should be $14\choose 2$.
However, it is wrong. Somebody help me out.
| The numbers that you are after are called compositions (with 0s allowed, rotated such that the largest number comes first, and are not of the form a + b + a with a and b distinct). The compositions of 12 into 3 parts are:
12 + 0 + 0
11 + 1 + 0
11 + 0 + 1
10 + 2 + 0
10 + 1 + 1
10 + 0 + 2
9 + 3 + 0
9 + 2 + 1
9 + 1 + 2
9 + 0 + 3
8 + 4 + 0
8 + 3 + 1
8 + 2 + 2
8 + 1 + 3
8 + 0 + 4
7 + 5 + 0
7 + 4 + 1
7 + 3 + 2
7 + 2 + 3
7 + 1 + 4
7 + 0 + 5
6 + 6 + 0
6 + 5 + 1
6 + 4 + 2
6 + 3 + 3
6 + 2 + 4
6 + 1 + 5
5 + 5 + 2
5 + 4 + 3
5 + 3 + 4
4 + 4 + 4
So there are 31 of these.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2995262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that the following inequality holds when $x>0$ $\require{cancel}$
Show that the following inequality holds for $x>0$
$$1+\frac{x}{2}-\frac{x^2}{8}<\sqrt{x+1}<1+\frac{x}{2}.$$
I proceeded as follows
$$\sqrt{x+1}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16 (\xi+1)^{5/2}},\quad \xi\in[0,x]$$
and substituing in the inequality yields
$$1+\frac{x}{2}-\frac{x^2}{8}<1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16 (\xi+1)^{5/2}}<1+\frac{x}{2}$$
which is equivalent to
$$\begin{cases}
\cancel{1+\frac{x}{2}-\frac{x^2}{8}}\cancel{<1+\frac{x}{2}-\frac{x^2}{8}}+\frac{x^3}{16 (\xi+1)^{5/2}}\quad(1)\\
\cancel{1+\frac{x}{2}}-\frac{x^2}{8}+\frac{x^3}{16 (\xi+1)^{5/2}}<\cancel{1+\frac{x}{2}}\qquad\,\,\,\,\,\, (2)
\end{cases}$$
Inequality $(1)$ holds for the values given by the problem statement. Instead for $(2)$
$$\frac{x^3}{16 (\xi+1)^{5/2}}<\frac{x^2}{8}$$
But $$\displaystyle{\max_{0\leq\xi\leq x}\Bigg\{\frac{1}{16 (\xi+1)^{5/2}}\Bigg\}}=\displaystyle{\min_{0\leq\xi\leq x}{\Big\{16 (\xi+1)^{5/2}}\Big\}}$$
which occurs at $\xi=0$, and thus, for $(2)$, it is left to prove that
$$\frac{x^3}{16}<\frac{x^2}{8}$$
but this happens for $x<0\,\vee\,0<x<2$.
Instead, if I use $\xi=x$ then inequality $(2)$ becomes
$$\frac{x^3}{16 (x+1)^{5/2}}<\frac{x^2}{8}$$
which holds for $-1<x<0\,\vee\,x>0$, and thus coincides with the restriction given by the problem.
Would it be correct to take $\xi=x$ rather than $0$? Is this approach correct at all?
| The right inequality:
We need to prove that:
$$1+x<\left(1+\frac{x}{2}\right)^2$$ or
$$1+x<1+x+\frac{x^2}{4}.$$
The left inequality.
Let $x+1=t^2,$ where $t>1$.
Thus, we need to prove that:
$$1+\frac{t^2-1}{2}-\frac{(t^2-1)^2}{8}<t$$ or
$$(t-1)^3(t+3)>0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2997403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Given $3^x = 12$ find $(\sqrt{4/3})^\frac{1}{x-2}$ Given $3^x = 12$ find $$\left(\sqrt{\dfrac{4}{3}}\right)^\dfrac{1}{x-2}$$ in simple form.
I've faced this problem, in a high school book, but failed to solve.
I've tried to calculate
*
*$3^{x-2} = \dfrac{12}{9}$
*$x -2 = \log_3 \dfrac{12}{9}$
*$\dfrac{1}{(x -2)} = \dfrac{ \log 3} { \log (\frac43)}$
However, replacing $\dfrac{1}{(x -2)}$ on the power doesn't help for me.
$$\left(\dfrac{4}{3}\right)^\dfrac{1}{\dfrac{2 \log 3} { \log (\frac43)} } = \left(\dfrac{4}{3}\right)^\dfrac{\log \left(\frac{4}{3}\right)}{2 \log 3 } = \left(\dfrac{4}{3}\right)^\dfrac{2 \log 2 - \log 3}{2 \log 3 } $$
Any hint?
| Let $$y= (\sqrt {4/3})^{\frac {1}{x-2}}$$
We have $$\ln y = \frac {\ln (4/3)}{2(x-2)}$$
Since $$x-2 = \frac {\ln (4/3)}{\ln 3}$$
We get $$\ln y = \ln \sqrt 3 $$ which implies $$ y=\sqrt 3 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3001578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Fast way to solve $(s-2i)^2 (s+2i)^2$ $(s-2i)^2 (s+2i)^2=$
$(s^2-4si-4) (s^2+4si-4)=$
$s^4+4s^3i-4s^2-4s^3i+16s^2+16si-4s^2-16si+16 =$
$(s^4-8s^2+16) =$
$(s^2+4)^2$
Is there a quicker way to see that $(s-2i)^2 (s+2i)^2= (s^2+4)^2$
| We have
$$(s-2i)^2 (s+2i)^2=[(s-2i) (s+2i)]^2$$
and recall that $(A-B)(A+B)=A^2-B^2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What am I doing wrong finding $\lim_{x\to 0} \left( \frac{1+x\cdot2^x}{1+x\cdot3^x} \right)^{1/x^2}$? It has an answer here, but I'd like to know where my solution went wrong.
$$\lim_{x\to 0} \left( \frac{1+x\cdot2^x}{1+x\cdot3^x} \right)^{\frac{1}{x^2}} $$
$$\lim_{x\to 0} \left( \frac{1+x\cdot2^x +x\cdot 3^x-x\cdot 3^x}{1+x\cdot3^x} \right)^\frac{1}{x^2} $$
$$\lim_{x\to 0} \left( 1 + \frac{x\cdot2^x-x\cdot 3^x}{1+x\cdot3^x} \right)^\frac{1}{x^2} $$
$$\lim_{x\to 0} \left( 1 + \frac{x\cdot2^x-x\cdot 3^x}{1+x\cdot3^x} \right)^{\frac{1}{x^2}\cdot \frac{1+x\cdot 3^x}{x\cdot2^x-x\cdot 3^x}\cdot\frac{x\cdot2^x-x\cdot 3^x}{1+x\cdot3^x}} $$
$$\lim_{x\to 0} e^{\frac{1}{x^2}\cdot \frac{x\cdot2^x-x\cdot 3^x}{1+x\cdot3^x}} $$
$$\lim_{x\to 0} e^{\frac{1}{x}\cdot \frac{2^x-3^x}{1+x\cdot3^x}} $$
$$\lim_{x\to 0} e^{\frac{1}{x}\cdot \frac{(1+1)^x-(2+1)^x}{1+x\cdot3^x}} $$
$$\lim_{x\to 0} e^{\frac{1}{x}\cdot \frac{1+x+o(x)-1-x2-o(x)}{1+x\cdot3^x}} $$
$$\lim_{x\to 0} e^{\frac{-1}{1+x\cdot3^x}} $$
$$e^{-1}$$
The answer in the book is $\frac{2}{3}$.
| You want to evaluate:
$$\lim_{x\to 0}\frac{1}{x}\cdot \frac{2^x-3^x}{1+x\cdot3^x}=\lim_{x\to 0}\frac{1}{x}\cdot \frac{(1+x\color{red}{\ln 2}+o(x^2))-(1+x\color{red}{\ln 3}+o(x^2))}{1+x\cdot3^x}=\\
\lim_{x\to 0}\frac{\ln (2/3)+o(x^2)}{1+x\cdot3^x}=\ln \frac23.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3003184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
$\frac{|a|}{|b-c|} + \frac{|b|}{|c-a|} + \frac{|c|}{|b-a|} \geq 2$ If $a, b, c$ are distinct real numbers then you demonstrate that:
$$ S=\frac{|a|}{|b-c|} + \frac{|b|}{|c-a|} + \frac{|c|}{|b-a|} \geq 2.$$
Using inequality $ |x-y|\leq |x|+|y|$ we showed that $ S >\frac{3}{2}.$
For $b = 2a, c = 3a, S=5,$ that is, the sum has values greater than 2, but I have not been able to prove this.
| Without loss of generality we can assume that $b$ and $c$ have the same sign. Then $|b|=|b-c|+|c|$. Also, $|c-a|\le |c|+|a|$ and $|a-b| \le |a|+|b-c|+|c|$. Therefore
$$
\frac{|a|}{|b-c|}+\frac{|b|}{|c-a|}+\frac{|c|}{|a-b|} \ge \frac{|a|}{|b-c|} + \frac{|b-c|+|c|}{|c|+|a|} + \frac{|c|}{|a|+|b-c|+|c|} = (*).$$
Denote $x=|a|$, $y=|b-c|$, $z=|c|$. Then, by Schwarz and by $x^2+y^2 \ge 2xy$, we get
\begin{align*}
(*) & = \frac{x}{y}+\frac{y+z}{z+x}+\frac{z}{x+y+z} \\
&\ge \frac{(x+(y+z)+z)^2}{xy+(y+z)(z+x)+z(x+y+z)} \\
&= \frac{x^2+y^2+4z^2+2xy+4yz+4zx}{2z^2+2xy+2yz+2zx} \\
&\ge \frac{2xy+4z^2+2xy+4yz+4zx}{2z^2+2xy+2yz+2zx} \\
&=2.\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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An AMM-like integral $\int_0^1\frac{\arctan x}x\ln\frac{(1+x^2)^3}{(1+x)^2}dx$
How can we evaluate $$I=\int_0^1\frac{\arctan x}x\ln\frac{(1+x^2)^3}{(1+x)^2}dx=0?$$
I tried substitution $x=\frac{1-t}{1+t}$ and got
$$I=\int_0^1\frac{2 \ln \frac{2 (t^2+1)^3}{(t+1)^4} \arctan \frac{t-1}{t+1}}{t^2-1}dt\\
=\int_0^1\frac{2 \ln \frac{2 (t^2+1)^3}{(t+1)^4} (\arctan t-\frac\pi4)}{t^2-1}dt$$
I'm able to evaluate $$\int_0^1\frac{\ln \frac{2 (t^2+1)^3}{(t+1)^4}}{t^2-1}dt$$
But I have no idea where to start with the rest one.
| Through the dilogarithm/trilogarithm machinery it can be shown that
$$ \int_{0}^{1}\frac{\log(1+i x)\log(1+x)}{x}\,dx=\\\frac{\pi K}{2}-\frac{9i\pi^3}{64}+3iK\log(2)-\frac{3\pi i}{16}\log^2(2)+\frac{5\pi^2}{32}\log(2)-\frac{\log^3(2)}{8}-\frac{69}{16}\zeta(3)+6\,\text{Li}_3\left(\tfrac{1+i}{2}\right) $$
$$ \int_{0}^{1}\frac{\log^2(1+i x)}{x}\,dx=\\
-\frac{\pi K}{2}-\frac{3i\pi^3}{64}+iK\log(2)-\frac{\pi i}{16}\log^2(2)+\frac{5\pi^2}{96}\log(2)-\frac{\log^3(2)}{24}-\frac{3}{16}\zeta(3)+2\,\text{Li}_3\left(\tfrac{1+i}{2}\right) $$
$$ \int_{0}^{1}\frac{\log(1+ix)\log(1-ix)}{x}\,dx= \frac{\pi K}{2}-\frac{27}{32}\zeta(3)$$
hence the claim follows by $\arctan x=\text{Im}\,\log(1+ix)$ and $\log(1+x^2)=\log(1+ix)+\log(1-ix)$.
| {
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"url": "https://math.stackexchange.com/questions/3006106",
"timestamp": "2023-03-29T00:00:00",
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Limit $ a_{n+1} = \frac{1}{3}(2a_n + \frac{5}{a_n^{2}}), a_1 > 0 $
Let $ a_{n+1} = \frac{1}{3}(2a_n + \frac{5}{a_n^{2}}), a_1 > 0 $. Show the sequence converge and find it's limit.
I did the following:
We see that $a_n$ is positive for every n obviously (considering $a_1 > 0$). Then:
$$ a_{n+1} - a_n = \frac{2a_n^{3} + 5}{3a_n^{2}} - a_n = \frac{5 - a_n^{3}}{3a_n^{2}}$$
We see that for each $a_n > ^3\sqrt{5}$ the sequence is decreasing.
That means the proof collapses to prove $a_n > ^3\sqrt{5}$ in order for $a_n$ to be decreasing. Suppose $ a_n > ^3\sqrt{5}$ then obviously $a_{n+1} = \frac{2a_n^{3} + 5}{3a_n^{2}} > \frac{2(\sqrt[3]{5})^{3} + 5}{3(\sqrt[3]{5})^{2}} = \sqrt[3]{5}$ so we deduced $a_{n+1} > ^\sqrt[3]{5} $.
(It's important to note $a_1$ might be less then $\sqrt[3]{5}$, hence the sequence will decrease from the second element.)
So we observed $a_n$ is always positive and decreasig hence it's bounded below so it's bounded. As for it's limit:
$$\begin{align} L = \lim_{n \to \infty} a_{n+1} = \lim_{n \to \infty} \frac{1}{3}(2a_n + \frac{5}{a_n^{2}}) = \frac{1}{3}(2L + \frac{5}{L^{2}}) = \sqrt[3]{5}\end{align}$$
I'm not sure as of the correctness of the solution. I'm aware that it's not a contradiction because it only implies it's in the epsilon-environment of $\sqrt[3]{5}$, or rather in this case $(\sqrt[3]{5}, \sqrt[3]{5}+\epsilon)$.
Is it correct?
EDIT: If anyone encounter a question in the likes, to sum it up you obviously should show the sequence is montonic and bounded. In this case for example we can show it using AM-GM inequality to see the sequence is bounded by $\sqrt[3]{5}$ and then in the difference of two consequective elements we see the difference is negative due to every element being greater then the third root of 5. From here we know the sequence is bounded below and decreasing hence it's bounded. If we plug in the limit of the sequence to itself we will see it's the third root of five. And that's it.
| First observe that
$$
a_{n+1}=\frac{1}{3}\left(2a_n+\frac{5}{a_n^2}\right)=\frac{1}{3}\left(a_n+a_n+\frac{5}{a_n^2}\right)\ge \left(a_n\cdot a_n\cdot \frac{5}{a_n^2}\right)^{1/3}=5^{1/3}.
$$
Hence, $a_n\ge 5^{1/3}$, for all $n>1$.
Next observe that
$$
f(x)=\frac{1}{3}\left(2x+\frac{5}{x^2}\right)<x,
$$
for all $x$ in $[5^{1/3},\infty)$, since
$$
x-f(x)=\frac{1}{3}\left(x-\frac{5}{x^2}\right)=\frac{x^3-5}{3x^2}\ge 0.
$$
Hence
$$
5^{1/3}\le a_{n+1}=f(a_n)\le a_n, \quad \text{for all $n>1$.}
$$
Thus $\{a_n\}$ is bounded and decreasing, and therefore it is convergent.
Also, if $a_n\to a$, then, as $f$ is continuous in $(0,\infty)$, then
$$
a=\lim a_n=\lim a_{n+1}=\lim f(a_n)=\lim f(a)
$$
and hence the limit $a$ satisfies
$$
a=f(a)=\frac{1}{3}\left(2a+\frac{5}{a^2}\right)
$$
and thus $a^3=5$, which implies that $a_n\to 5^{1/3}$.
| {
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"url": "https://math.stackexchange.com/questions/3007963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Limit of $\lim_{x \to 0} (\cot (2x)\cot (\frac{\pi }{2}-x))$ (No L'Hôpital)
$\lim_{x \to 0} (\cot (2x)\cot (\frac{\pi }{2}-x))$
I can't get to the end of this limit. Here is what I worked out:
\begin{align*}
& \lim_{x \to 0} \frac{\cos 2x}{\sin 2x}\cdot\frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )}
\lim_{x \to 0}\frac{\frac{\cos2x }{2x}}{\frac{\sin 2x}{2x}}\cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )}
= \lim_{x \to 0} \frac{\cos 2x}{2x} \cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )} \\
= & \lim_{x \to 0} \frac{{\cos^2 (x)}-{sin^2 (x)}}{2x}\cdot\frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )}
= \lim_{x \to 0} \left(\frac{\cos^2(x)}{2x}-\frac{sin^2 x}{2x}\right)\cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )} \\
= & \lim_{x \to 0} \left(\frac{1-\sin^2 x}{2x}-\frac{sin x}{2}\right)\cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )}
= \lim_{x \to 0} \left(\frac{1}{2x}-\frac{\sin^2 x}{2x}-\frac{sin x}{2}\right) \cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )} \\
= & \lim_{x \to 0} \left(\frac{1}{2x}-\frac{\sin x}{2}-\frac{sin x}{2}\right)\cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )}
= \lim_{x \to 0} \left(\frac{1}{2x}-2\frac{\sin x}{2}\right)\cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )} \\
= & \lim_{x \to 0} \left(\frac{1}{2x}-\sin x\right)\cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )}
\end{align*}
Here is where I can't seem to complete the limit, the 2x in the denominator is giving me a hard time and I don't know how to get rid of it. Any help would be appreciated. (In previous questions I got a really hard time because of my lack of context, I hope this one follows the rules of the site. I tried.)
| Hint:
$$\lim_{x \to 0} \cot (2x)\cot (\frac{\pi }{2}-x)=\lim_{x \to 0} \cot (2x)\tan x=\lim_{x \to 0} \dfrac{\cos2x}{\sin2x}\dfrac{\sin x}{\cos x}=\lim_{x \to 0} \dfrac{\cos2x}{1}\dfrac{2x}{\sin2x}\dfrac{\sin x}{x}\dfrac{1}{\cos x}\dfrac12$$
| {
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"url": "https://math.stackexchange.com/questions/3011543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Product of the Riemann sum Hi I want to prove that for all $x,y\in R$, the following holds
$$(\sum_{n=0}^{\infty}\frac{x^n}{n!})(\sum_{n=0}^{\infty}\frac{y^n}{n!})=\sum_{n=0}^{\infty}\frac{(x+y)^n}{n!}$$
without using $e$ nor $\sum_{n=0}^\infty \frac{x^n}{n!}=\lim_{n\rightarrow\infty}(1+\frac{x}{n})^n$
What I know:
$$LHS = 1\sum_{n=0}^{\infty}\frac{y^n}{n!}+ x\sum_{n=0}^{\infty}\frac{y^n}{n!} +\frac{x^2}{2!}\sum_{n=0}^{\infty}\frac{y^n}{n!}+\cdots$$
$$=\ \ 1\ \ \ +\ y\ \ \ +\ \frac{y^2}{2!}\ \ \ +\ \frac{y^3}{3!}\ \ \ +\ \cdots\ \ $$
$$\ +\ x\ \ +\ xy\ \ +\ x\frac{y^2}{2!}\ \ +\ x \frac{y^3}{3!}\ \ +\ \ \cdots$$
$$\ +\frac{x^2}{2!}+\frac{x^2}{2!}y+\frac{x^2}{2!}\frac{y^2}{2!}+\frac{x^2}{2!} \frac{y^3}{3!}+\ \cdots$$
$$\vdots\hspace{6cm}$$
And if we see the expression above as a matrix with $a_{1,1}=1, \ a_{1,2}=y, \ a_{2,1}=x,$ etc, it's clear that it equals:
$$= a_{1,1} + (a_{1,2}+a_{2,1}) + (a_{1,3}+a_{2,2}+a_{3,1}) + \cdots$$
$$=1 + (x+y) + \frac{(x+y)^2}{2}+\cdots=RHS\hspace{1.5cm}$$
But this seems really not rigorous, so I was wondering if there's a way to rigorously prove the equation, and if I'm not on the right path, what could be a way to approach it.
| First, prove that $\sum_{n\ge 0} \frac {x^n}{n!}$ and $\sum_{n\ge 0} \frac{y^n}{n!}$ converge absolutely by using the Ratio Test and compute their Cauchy Product:
$$
\begin{align}
\left(\sum_{n\ge 0} \frac {x^n}{n!}\right)\left(\sum_{n\ge 0} \frac {y^n}{n!}\right)&=\sum_{n\ge 0} c_n = \sum_{n\ge 0} \sum_{i=0}^n \frac{x^i}{i!}\frac{y^{n-i}}{(n-i)!} \\ &=\sum_{n\ge 0} \sum_{i=0}^n \frac 1{n!}\binom{n}{i}x^iy^{n-i}=\sum_{n\ge 0}\frac 1{n!}(x+y)^n
\end{align}
$$
Where we used the Binomial Theorem in the last equality.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Ellipse on a Circular Cylinder in Cylindrical Coordinates Is there an equation in cylindrical coordinates for an ellipse (tilted at some angle) on the surface of a right circular cylinder of radius r? For simplicity, I envision the cylinder to be coincident with the x-axis.
I am aware that the cylinder could be "unwrapped" into a plane, which would result in the ellipse becoming a sine curve. I am just not sure how that information ties into cylindrical coordinates.
EDIT: I have realized that I am looking for a parametric equation. For an ellipse on the surface of a cylinder of radius r which has a certain angle of inclination, is there a parameterization where I can calculate the axial coordinate seperately from the azimuth angle for a given t?
Thank you.
| As stated before assume the (normalized) equation of the plane is
$$ \frac{a x + b y + c z}{\sqrt{a^2+b^2+c^2}} = d $$
and the parametric equation of the cylinder
$$ \pmatrix{x & y & z}= f(\varphi,z) = \pmatrix{ R \cos\varphi, R\sin\varphi, z} $$
Where the two intersect you have your ellipse in cartesian coordinates
$$ z(\varphi) = \frac{d \sqrt{a^2+b^2+c^2}-R (a \cos\varphi+b \sin\varphi) }{c} $$
or
$$\vec{r}_{\rm curve}(\varphi) = \pmatrix{x\\y\\z} = \pmatrix{ R \cos\varphi \\ R \sin\varphi \\ \frac{d \sqrt{a^2+b^2+c^2}-R (a \cos\varphi+b \sin\varphi) }{c} } $$
Now let's find the properties of this ellipse.
The center of the ellipse is at $\vec{r}_{\rm cen} = \pmatrix{0 & 0 & \frac{d \sqrt{a^2+b^2+c^2}}{c}} $
The ellipse in polar coordinates is
$$ r(\varphi) = \| \vec{r}_{\rm curve}-\vec{r}_{\rm cen} \| = \sqrt{R^2 + \frac{R^2}{c^2} \left( \frac{a^2+b^2}{2} + a b \sin(2 \varphi) + \frac{a^2-b^2}{2} \cos(2\varphi) \right)} $$
This allows us to find the major and minor radii
$$ \begin{aligned}
r_{\rm major} & = R \frac{ \sqrt{a^2+b^2+c^2}}{c} \\
r_{\rm minor} & = R
\end{aligned} $$
The principal axes of the ellipse are on
$$ \varphi = \frac{1}{2} {\rm atan}\left( \frac{2 a b}{a^2-b^2} \right) + n \frac{\pi}{2} $$
| {
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An Inconsistency in Numerical Approximation Consider the expression
$$
10^5 - \frac{10^{10}}{1+10^5}.
$$
Using the elementary properties of fractions we can evaluate the expression as
$$
10^5 - \frac{10^{10}}{1+10^5} = \frac{10^5 + 10^{10} - 10^{10}}{1+10^5} = \frac{10^5}{1+10^5}\approx 1.
$$
Note that the approximation $10^5+1 \approx 10^5$ is used in the last step. Now suppose we use the same approximation, but apply it before we perform the subtraction. We get
$$
10^5 - \frac{10^{10}}{1+10^5} \approx 10^5 - \frac{10^{10}}{10^5} = 0.
$$
The same logic works for
$$
10^p - \frac{10^{2p}}{1+10^p}
$$
for arbitrary large $p$, so it cannot be simply an issue with the accuracy of the approximation.
Is there an easy explanation of what's going on here?
| It is simply an issue of accuracy of approximation. Let me write $x = 10^p$. Then your expression is
$$ x - \frac{x^2}{1+x}$$
Note that $$\frac{x^2}{1+x} = \frac{x}{1/x + 1} = x (1 - 1/x + O(1/x^2)) = x - 1 + O(1/x)$$
so that
$$ x - \frac{x^2}{1+x} = x - (x - 1 + O(1/x)) = 1 + O(1/x)$$
In your second calculation you only evaluated $x^2/(1+x)$ to within $O(1)$, not $O(1/x)$, so naturally you have an error at the end that is $O(1)$.
| {
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Find sum of series $ \sum_{n=1}^{\infty} (n\cdot \ln \frac{2n+1}{2n-1} - 1) $ how can I find sum of series $ \sum_{n=1}^{\infty} (n\cdot \ln \frac{2n+1}{2n-1} - 1) $?
It is so weird for me because I put this to Mathematica and it tells me that sum does not converge...
Let consider sum no to infinity, but to n
$$ \sum_{k=1}^{n} (k\cdot \ln \frac{2k+1}{2k-1} - 1) =$$
$$ ln \frac{3}{1}\cdot \left(\frac{5}{3}\right)^2 \cdot...\cdot \left(\frac{2n+1}{2n-1}\right)^n - n = ln \frac{1}{1}\cdot \frac{1}{3}\cdot \frac{1}{5}\cdot ... \frac{1}{2n-1} - n $$
but $$ n = ln e^n $$
so
it will be $$ln\frac{1}{e^n} \cdot \frac{1}{1}\cdot \frac{1}{3}\cdot \frac{1}{5}\cdot ... \frac{1}{2n-1}$$
So the limit of it is $-\infty$
Have I done this well or I missed sth?
| We have that
$$\sum_{n=1}^{N} \left(n\cdot \ln \frac{2n+1}{2n-1} - 1\right)=\sum_{n=1}^{N} \left(n\cdot \ln (2n+1)-n\ln (2n-1) - 1\right)=$$
$$=(1\cdot \ln 3-1\cdot \ln1-1)+(2\cdot \ln 5-2\cdot \ln3-1)+(3\cdot \ln 7-3\cdot \ln5-1)+\ldots=$$
$$=-\ln(3\cdot 5\cdot 7\cdot \ldots\cdot (2N-1))+N\cdot\ln(2N+1)-N=$$$$=-\ln\left(\frac{(2N)!}{2^NN!}\right)+N\cdot\ln(2N+1)-N=$$
$$=\ln\left(\frac{(2^N)^2N!N^N}{(2N)!e^N}\right)+N\cdot\ln\left(1+\frac1{2N}\right)$$
and by Stirling's approximation $N!\sim \sqrt{2\pi N}\left(\frac{N}{e}\right)^N$
$$\frac{(2^N)^2N!N^N}{(2N!)e^N}\sim\frac{(2^N)^2N^N}{e^N}\frac{\sqrt{2\pi N}}{\sqrt{4\pi N}}\frac{N^Ne^{2N}}{e^N4^NN^{2N}}=\frac{1}{\sqrt 2}$$
| {
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Ways for Two Dice to Sum to 8: $x_1 +x_2 =8$ I am trying to determine the number of ways two dice can be rolled to have a sum of 8. This question can be easily brute-forced, but I want to use a technique that can be applied to similar questions that cannot be brute-forced so easily.
My solution was as follows,
$$x_1 +x_2 = 8, x_1, x_2 \geq 1$$
$$x_1+1+x_2+1 = 8 \rightarrow x_1+x_2 =6$$
Now I can use the stars and bars method.
So $C(6 +(2-1), (2-1))$ should be the answer, which is equal to $7$. But when I brute-forced this question, I only found $5$ solutions. What did I do wrong?
Normally if I wanted to find, say for example, the number of integer solutions to $x_1+x_2+x_3 = 12, x_i \geq 0$, I would apply the same method and do $C(12+(3-1), (3-1))$. What am I doing wrong?
| Your sum count is wrong because you are including the cases where some of the $x_i>6.$ For example, $7+3+2=12,$ but you don't want to count that case.
Summary:
The number of ways of getting the sum $S$ from rolling $k$ fair $n$-sided dice labeled $1$ to $n$ is:
$$\sum_{i=0}^{k}(-1)^i\binom{k}{i}\binom{S-in-1}{k-1}$$
You have to be careful evalating this, because $\binom{m}{j}$ is not a polynomial in $m$ for fixed $j$, since it is $0$ when $m<j.$
A trick to make it easier is that you can replace any $S$ with $(n+1)k-S$ and get the same count. So you can always use $S\leq \frac{(n+1)k}{2}$, in which case you can ignore terms with $i>\frac{k+1}{2}$ since then $\binom{S-in-1}{k-1}=0.$
Longer:
This is a case for either generating functions or an inclusion-exclusion argument.
The generating approach applied to the sum of two fair six-sided dice is to compute the coefficients of $x^8$ in:
$$(x+x^2+x^3+\cdots + x^6)^2$$
or the coefficient of $x^6$ in:
$$(1+x+\cdots+x^5)^2$$
Now, this might seem harder, but it isn't, because $1+x+\cdots+x^5=\frac{1-x^6}{1-x}$ So you are seeking the coefficient if $x^6$ in:
$$(1-x^6)^2\frac{1}{(1-x)^2}\tag{1}=(1-2x^6+x^{12})\frac{1}{(1-x)^2}$$
And we have the general result:
$$\frac{1}{(1-x)^{k}}=\sum_{m=0}^{\infty} \binom{m+k-1}{k-1}x^m \tag{2}$$
So when $k=2$ you get:
$$\frac{1}{(1-x)^2}=\sum_{m=0}^{\infty}(m+1)x^m$$
Multiplying by $(1-x^6)^2=1-2x^6+x^{12}$ this means the coefficient of $x^j$ in $(1)$ is:
$$\binom{j+1}{1}-2\binom{j-5}{1}+\binom{j-11}{1}$$
We can't replace $\binom{n}{1}$ with $n$ because we need to $\binom{n}1=0$ when $n<1.$ The term $\binom{j-11}{1}$ is only ever really used to "zero out" the cases where $\binom{j+1}{1}-2\binom{j-5}{1}$ is negative.
When seeking the number of ways to get a sum of $8$, we use $j=8-2=6$ this is $7-2=5.$
When the sum we are seeking, $S,$ is between $2$ and $7$, inclusive, this formula gives us a count of:
$$S-1$$
because the term $\binom{S-7}{1}=0$ when $S\leq 7.$
and when $7<S\leq 12$ this number of ways is:
$$S-1 - 2(S-7)=13-S$$
When $S>12$ all the sum is:
$$(S-1)-2(S-7)+(S-13)=0$$
And when $S<2$ all the terms are zero.
More generally, if you are rolling a set of $k$ $n$-sided dice, you get,via $(2)$, then the number of ways to get a total of $S$ is the coefficient of $x^{S-k}$ in:
$$(1-x^n)^k\sum_{m=0}^{\infty}\binom{m+k-1}{k-1}x^m$$
Thus we get that the number of ways to get a sum $S$ from $k$ $n$-sided dice is: $$\sum_{i=0}^{k}(-1)^i\binom{k}{i}\binom{S-in-1}{k-1}$$
You can also prove this formula using inclusion-exclusion, rather than using generating functions.
Example: when throwing three eight-sided dice, the number of ways to get a sum of $S$ is:
$$\binom{S-1}{2}-3\binom{S-9}{2}+3\binom{S-17}{2}-\binom{S-25}{2}$$
Which yields:
$$\begin{cases}
0&S<3\\
\frac{(S-1)(S-2)}{2}&3\leq S<11\\
27S-S^2-134&11\leq S<19\\
\frac{(25-S)(26-S)}{2}&19\leq S\leq 24\\
0&25\leq S
\end{cases}
$$
You also have, in general, the value for $S$ and $(n+1)k-S$ should be equal. This means that you can always solve for $S$ by using at most the first $\frac{k+1}{2}$ terms of the sum, with the other terms zero.
When dealing with differently-sided dice, you get the same kind of result, but the numerator is different.
For example, a four- and six-sided die rolled together gives you:
$$x^2(1-x^4)(1-x^6)\sum_{m=0}^{\infty}\binom{m+1}{1}x^m$$
Then the coefficient of $x^S$ in this is:
$$\binom{S-1}{1}-\binom{S-5}{1}-\binom{S-7}{1}+\binom{S-11}{1}$$
| {
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Use complete induction to prove that $a_n < 2^n$ for every integer $n \geq 2$ Define the sequence of integers $a_0, a_1, a_2, \cdots$ as follows
$$ a_i =
\begin{cases}
i+1 & 0 \leq i \leq 2 \\
a_{i-1} + a_{i-2} + 2a_{i-3} & i > 2 \\
\end{cases}
$$
Use complete induction to prove that $a_n < 2^n$ for every integer $n \geq 2$
I will prove $a_n < 2^n, \forall n \geq 2$ by using complete induction
Base Case: Three cases $n = 2, 3, 4$
let $n = 2$
$a_{n} = a_{2} = 2 + 1 = 3 < 4 = 2^2 = 2^n$ By definition, and holds
let $n = 3$
$a_{n} = a_{3} = 3 + 2 + 1 = 6 < 8 = 2^3 = 2^n$ By definition, and holds
let $n = 4$
$a_{n} = a_{4} = 4 + 3 + 2 = 9 < 16 = 2^4 = 2^n$ By definition, and holds
Inductive step: let $n > 4$. Suppose $a_j < 2^j$ whenever $2 \leq j < n$. [Inductive hypothesis]
What to prove: $a_n < 2^n$
$a_{n} = a_{i-1} + a_{i-2} + 2a_{i-3}$ [By definition]
$< 2^{n-1} + 2^{n-2} + 2^{n-3+1}$ [By Inductive hypothesis]
$= 2^{n-1} + 2^{n-2} + 2^{n-2}$
$= 2^{n-1} + 2^{n-2 + 1}$
$= 2^{n-1 + 1}$
$= 2^n$
as wanted.
Would this be correct?
| The second part is fine. But the base case is not just the $n=2$ case. You should have checked it for $n=0$, and $n=1$ too.
| {
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Prove a geometric sequence a, b, c from the arithmetic progression $1/(b-a)$, $1/2b$, $1/(b-c)$ The given task is:
The following forms an arithmetic sequence: $$\frac{1}{b-a}, \frac{1}{2b}, \frac{1}{b-c}.$$
Show, that $a, b, c$ forms an geometric sequence.
It's easily enough to understand that $$ \frac{1}{2b}-\frac{1}{b-a}=\frac{1}{b-c}-\frac{1}{2b} \iff \frac{a+b}{a-b}=\frac{b+c}{b-c}$$ and that I need to prove that $$\frac{b}{a}=\frac{c}{b},$$ but between the two I just make it more and more complicated.
| Taking it from where you left off, use cross-products and simplify
$$(a+b)(b-c) = (a-b)(b+c) \iff \color{blue}{ab}-ac+b^2\color{green}{-bc} = \color{blue}{ab}+ac-b^2\color{green}{-bc}$$
$$2b^2 = 2ac \iff b^2 = ac \iff \frac{b}{a} = \frac{c}{b}$$
| {
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Finding $\lim\limits_{n→∞}n^3(\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt2)$ What is$$\lim_{n→∞}n^3(\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt2)?$$So it is$$\lim_{n→∞}\frac{n^3(\sqrt{n^2+\sqrt{n^4+1}})^2-(n\sqrt{2})^2}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}=\lim_{n→∞}\frac{n^3(n^2+\sqrt{n^4+1}-2n^2)}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}.$$
I do not know what to do next, because my resuts is $∞$ but the answer from book is $\dfrac{1}{4\sqrt{2}}$.
| HINT
You only need to apply the trick twice, indeed we have that
$$\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt{2}=(\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt{2})\cdot\frac{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}=$$$$=\frac{n^2+\sqrt{n^4+1}-2n^2}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}$$
and
$$\frac{\sqrt{n^4+1}-n^2}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}=\frac{\sqrt{n^4+1}-n^2}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}\cdot \frac{\sqrt{n^4+1}+n^2}{\sqrt{n^4+1}+n^2}=$$$$=\frac{1}{(\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2})(\sqrt{n^4+1}+n^2)}$$
Can you conclude form here?
| {
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Solving the Diophantine equation $y^2 = x^4+x+ 2$ for $x,y \in \mathbb{Z}$ I want to solve the Diophantine equation $y^2 = x^4+x+ 2$ for $x,y \in \mathbb{Z}$.
I already found 4 solutions: $(x,y) = (1,\pm2)$ and $(x,y)=(-2,\pm4)$. It can probably be solved using some factorization argument, but I don't know how.
| If $x\geq -1$, then $$x^4<y^2<(x^2+2)^2\,.$$
Therefore, for $x\geq -1$, there exists a solution iff $x^4+x+2=(x^2+1)^2$. The only integer root of the last equation is $x=1$, yielding the solutions $(x,y)=(1,\pm2)$.
If $x\leq -3$, then $$(x^2-2)^2<y^2<x^4\,.$$
Therefore, for $x\leq -3$, there exists a solution iff $x^4+x+2=(x^2-1)^2$. The last equation does not have an integer root, whence there does not exist a solution in this case.
If $x=-2$, then we have the solutions $(x,y)=(-2,\pm 4)$. Therefore, there are only four solutions $(x,y)\in\mathbb{Z}\times\mathbb{Z}$ to $y^2=x^4+x+2$, which are $(1,\pm2)$ and $(-2,\pm4)$.
| {
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Prove $\sum\limits_{n=1}^{\infty} a_{f(n)} = \dfrac{3}{2} \ln(2) $ Where $a_n = \dfrac{(-1)^{n+1}} { n}$ and $f:\mathbb{N}\to \mathbb{N} $ is a bijection which for $k \in \mathbb{N}\setminus \{0\}$ is given by: $$f(3k+1)= 4k+1$$ $$f(3k+2)= 4k+3$$ $$f(3k+3)= 2k+2.$$
Could some please help me get started on this proof? I'm having extra difficulty seeing where $f$ is constructed from (I've looked at various proofs of convergence of alternating harmonic series and am still confused). And then how to set up $a_{3n}$ based on this definition (again confused and didn't find that much step-by-step help on google). Thank you!!
| Euler's constant:
The sequence
$$\gamma_n := 1+\frac 12 + \frac 13 + \cdots + \frac 1n - \ln(n)$$
converges to a finite limit $\gamma$, where $\gamma$ is a constant:
$$\lim_{n \rightarrow \infty} \gamma_n = \gamma$$
The subsequence lemma:
The sequence $b_n$ converges to a limit $L$ if and only if all of the subsequences $b_{kn}$, $b_{kn+1}$, $b_{kn+2}$, $\dots$, $b_{kn+k-1}$ converge and have the same limit $L$.
Solution:
Define the sequence of partial sums
$$S_k = \sum_{n=1}^ka_{f(n)} = 1+\frac 13 - \frac 12 + \frac 15 + \frac 17 - \frac 14 + \cdots + a_{f(k)}$$
Let us first look at the subsequence $T_k := S_{3k}$.
\begin{align}
T_k = S_{3k} = & \bigg(1+\frac 13 - \frac 12\bigg) + \bigg(\frac 15 + \frac 17 - \frac 14\bigg) + \cdots + \bigg(\frac{1}{4k-3}+\frac{1}{4k-1}-\frac{1}{2k}\bigg) \\
= & \bigg(1+\frac 13 + \frac 15 + \cdots + \frac{1}{4k-1} \bigg)-\bigg(\frac 12 + \frac 14 + \frac 16 + \cdots + \frac{1}{2k} \bigg) \\
= & \bigg[\bigg(1+\frac 12 + \frac 13 + \cdots + \frac{1}{4k-1}\bigg) - \bigg(\frac 12 + \frac 14 + \frac 16 + \cdots + \frac{1}{4k-2}\bigg) \bigg] \\
& - \bigg(\frac 12 + \frac 14 + \frac 16 + \cdots + \frac{1}{2k} \bigg) \\
= & \bigg(1+\frac 12 + \frac 13 + \cdots + \frac{1}{4k-1}\bigg) - \frac 12 \bigg(1+\frac 12 + \frac 13 + \cdots + \frac{1}{2k-1}\bigg) \\
& -\frac 12 \bigg(1+\frac 12 + \frac 13 + \cdots + \frac{1}{k}\bigg) \\
= & [\gamma_{4k-1}+\ln(4k-1)]-\frac 12 [\gamma_{2k-1}+\ln(2k-1)] - \frac 12 [\gamma_{k}+\ln(k)] \\
= & \gamma_{4k-1}-\frac 12 \gamma_{2k-1} - \frac 12 \gamma_{k} + \frac 12 \ln\frac{(4k-1)^2}{(2k-1)k} \\
\rightarrow & \gamma - \frac 12 \gamma - \frac 12 \gamma + \frac 12 \ln (8) \\
= & {\color{red}{\frac32}}\ln (2)
\end{align}
Thus, $\lim_{k \rightarrow \infty}S_{3k} = {\color{red}{\frac32}}\ln (2)$. Of course, this is not enough to show that $\lim_{k\rightarrow \infty}S_k ={\color{red}{\frac32}} \ln (2)$ (take the Grandi series as a counter-example). We also need to show that $\lim_{k\rightarrow \infty}S_{3k+1} = \lim_{k\rightarrow \infty}S_{3k+2} = {\color{red}{\frac32}}\ln (2)$.
This is easily done with
$$\lim_{k\rightarrow \infty}S_{3k+1} = \lim_{k\rightarrow \infty}\bigg(S_{3k}+\frac{1}{4k+1} \bigg) = \lim_{k\rightarrow \infty}S_{3k}+\lim_{k\rightarrow \infty}\bigg(\frac{1}{4k+1} \bigg) = {\color{red}{\frac32}}\ln (2)+0 = {\color{red}{\frac32}}\ln (2)$$
and similarly,
$$\lim_{k\rightarrow \infty}S_{3k+2} = \lim_{k\rightarrow \infty}\bigg(S_{3k}+\frac{1}{4k+1}+\frac{1}{4k+3} \bigg) = {\color{red}{\frac32}}\ln (2)+0+0 = {\color{red}{\frac32}}\ln (2)$$
It follows that
$$\sum_{n=1}^\infty a_{f(n)} = \lim_{k \rightarrow \infty} S_k = {\color{red}{\frac32}}\ln (2)$$
| {
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Half of Vandermonde's Identity We know Vandermonde's Identity, which states
$\sum_{k=0}^{r}{m \choose k}{n \choose r-k}={m+n \choose r}$.
Does anyone know what happens if we walk bigger steps with k? Say we skip all the odd ks, is something like
$\sum_{k=0}^{r/2}{m \choose 2k}{n \choose r-2k}=\frac{1}{2} {m+n \choose r}$
or at least
$\sum_{k=0}^{r/2}{m \choose 2k}{n \choose r-2k}=\Theta \left( \frac{1}{2} {m+n \choose r}\right)$
true?
Maybe someone here has even some general insight on other step widths?
Thank you!
| We derive a binomial identity which shows the deviation of OPs sum from $\frac{1}{2}\binom{m+n}{r}$. It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series. This way we can write for instance
\begin{align*}
\binom{n}{k}=[z^k](1+z)^n\tag{1}
\end{align*}
We assume wlog $n\geq m$ and obtain
\begin{align*}
\color{blue}{\sum_{k=0}^{r/2}}&\color{blue}{\binom{m}{2k}\binom{n}{r-2k}}\\
&=\sum_{k\geq 0}\binom{m}{2k}[z^{r-2k}](1+z)^n\tag{2}\\
&=[z^r](1+z)^n\sum_{k\geq 0}\binom{m}{2k}z^{2k}\tag{3}\\
&=[z^r](1+z)^n\frac{1}{2}\left((1+z)^m+(1-z)^m\right)\tag{4}\\
&=\frac{1}{2}[z^r](1+z)^{m+n}+\frac{1}{2}[z^r](1+z)^n(1-z)^m\\
&=\frac{1}{2}\binom{m+n}{r}+\frac{1}{2}[z^r](1-z^2)^m(1+z)^{n-m}\tag{5}\\
&=\frac{1}{2}\binom{m+n}{r}+\frac{1}{2}[z^r]\sum_{k=0}^{r/2}\binom{m}{k}(-1)^kz^{2k}(1+z)^{n-m}\\
&=\frac{1}{2}\binom{m+n}{r}+\frac{1}{2}\sum_{k=0}^{r/2}\binom{m}{k}(-1)^k[z^{r-2k}](1+z)^{n-m}\tag{6}\\
&\,\,\color{blue}{=\frac{1}{2}\binom{m+n}{r}+\frac{1}{2}\sum_{k=0}^{r/2}\binom{m}{k}\binom{n-m}{r-2k}(-1)^k}\tag{7}
\end{align*}
Comment:
*
*In (2) we apply the coefficient of operator as indicated in (1) and we set the upper limit of the sum to $\infty$ without changing anything since we are adding zeros only.
*In (3) we use the linearity of the coefficient of operator.
*In (4) we write the sum as polynomial in closed form.
*In (5) we select the coefficient of $z^r$ of the left polynomial and we rewrite the other polynomial keeping in mind that $n\geq m$.
*In (6) we use the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.
*In (7) we select the coefficient of $z^{r-2k}$.
| {
"language": "en",
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"question_score": "5",
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} |
Find the sum of these variables. Five real numbers $a_1, a_2, a_3, a_4\;\text{and}\; a_5\;$ are such that
$$\sqrt{a_1- 1} + 2\sqrt{a_2- 4}+3\sqrt{a_3- 9}
+4\sqrt{a_4- 16} + 5 \sqrt{a_4- 25} =\frac{a_1+a_2+a_3+a_4+a_5}{2}.$$
Find $a_1+a_2+a_3+a_4+a_5.$
Thanks for checking this out!
| Define$$b_1=\sqrt{a_1-1}\\b_2=\sqrt{a_2-4}\\b_3=\sqrt{a_3-9}\\b_4=\sqrt{a_4-16}\\b_5=\sqrt{a_5-25}\\$$therefore by substitution we have $$b_1+2b_2+3b_3+4b_4+5b_5={b_1^2+b_2^2+b_3^2+b_4^2+b_5^2\over 2}+{55\over 2}$$or$$2b_1+4b_2+6b_3+8b_4+10b_5={b_1^2+b_2^2+b_3^2+b_4^2+b_5^2}+{55}$$by rearranging we obtain$$(b_1-1)^2+(b_2-2)^2+(b_3-3)^2+(b_4-4)^2+(b_5-5)^2=0$$therefore $$b_i=i\quad ,\quad i=1,2,3,4,5$$and we have $$a_i=2i^2$$which yields to $$\sum a_i=110$$
| {
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} |
$\int_{-\infty}^{\infty}\frac{\mathrm{d}x}{ax^2+bx+c}=\pi$ similar identities I recently found that
$$\int_{-\infty}^{\infty}\frac{\mathrm{d}x}{ax^2+bx+c}=\pi$$
iff
$$b^2-4ac=-4.$$
I found it by integrating
$$I=\int_{-\infty}^{\infty}\frac{\mathrm{d}x}{ax^2+bx+c}.$$
If the reciprocal of the function is to be integrated over the entire real line, then the function must not have any real zeros. This implies that $$b^2-4ac<0$$
With this in mind, we complete the square:
$$I=\int_{-\infty}^{\infty}\frac{\mathrm{d}x}{a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a}}.$$
Then setting $g=c-\frac{b^2}{4a}$, and $x+\frac{b}{2a}=\sqrt{\frac{g}{a}}\tan u$,
$$I=\sqrt{\frac{g}{a}}\int_{-\pi/2}^{\pi/2}\frac{\sec^2u\ \mathrm{d}u}{g\tan^2u+g}=\frac{\pi}{\sqrt{ag}}.$$
So our identity holds for
$$ag=1.$$
And with a little algebra,
$$b^2-4ac=-4.$$
So my question is, are there any other similar identities involving other famous constants? Cheers!
| First if we consider the integral:
\begin{equation}
I(a) = \int_{-\infty}^{\infty}\frac{1}{x^2 + a^2}\:dx = \frac{\pi}{a}
\end{equation}
Then we can see:
\begin{equation}
I(a) + cI(b) = \left[\int_{-\infty}^{\infty}\frac{1}{x^2 + a^2}\:dx + c\int_{-\infty}^{\infty}\frac{1}{x^2 + b^2}\:dx \right] = \frac{\pi}{a} + c\frac{\pi}{b} = \pi\left[\frac{1}{a} + c\frac{1}{b} \right]
\end{equation}
We can see that this can be expanded to any number of $I(a)$ terms and any sequence of $c_i$ values, i.e.
\begin{equation}
\sum_{i = 1}^{\infty} (-1)^{i + 1} \frac{1}{i + 1}I(i) = \pi \sum_{i = 1}^{\infty} \frac{(-1)^{i + 1}}{i + 1} = \pi\ln(2)
\end{equation}
Thus,
\begin{equation}
\sum_{i = 1}^{\infty} \int_{-\infty}^{\infty} \frac{(-1)^{i + 1}}{x^2 + \left(i + 1\right)^2} \:dx = \pi \sum_{i = 1}^{\infty} \frac{(-1)^{i + 1}}{i + 1} = \pi\ln(2)
\end{equation}
Or as another example,
\begin{equation}
\sum_{i = 1}^{\infty} \int_{-\infty}^{\infty} \frac{1}{x^2 + i^4} \:dx = \frac{ \pi^3}{6}
\end{equation}
What's even better (IMO) is that we can take each integral spoken to and apply (through Glasser Master Theorem)
$$x = t - \sum_{i = 1}^{n - 1}\frac{\left| d_{i}\right|}{ t - e_{i}}$$
Where $d_i, e_i \in \mathbb{R}$ and $n \in N$ that the value of the integrals remain unchanged!!
As an addendum to the example:
\begin{equation}
\sum_{i = 1}^{\infty} \int_{-\infty}^{\infty} \frac{(-1)^{i + 1}}{x^2 + \left(i + 1\right)^2} \:dx = \pi \sum_{i = 1}^{\infty} \frac{(-1)^{i + 1}}{i + 1} = \pi\ln(2)
\end{equation}
If we call on this integral as addressed here we see that
\begin{equation}
2\int_{0}^{\frac{\pi}{2}} \frac{x}{\tan(x)} \:dx = 2\sum_{i = 1}^{\infty} \int_{-\infty}^{\infty} \frac{(-1)^{i + 1}}{x^2 + \left(i + 1\right)^2} \:dx =\pi\ln(2)
\end{equation}
Or,
\begin{equation}
\int_{0}^{\frac{\pi}{2}} \frac{x}{\tan(x)} \:dx = \sum_{i = 1}^{\infty} \int_{-\infty}^{\infty} \frac{(-1)^{i + 1}}{x^2 + \left(i + 1\right)^2} \:dx = \frac{\pi\ln(2)}{2}
\end{equation}
| {
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Show that $T: K\to K,$ defined by $Tx=(0,x^2_1,x^2_2,x^2_3,\cdots)$ is not non-expansive Let $X = l_\infty$ (the space of sequences of real numbers which are bounded). Let $K=\{x\in l_\infty:\Vert x \Vert_\infty\leq 1\}.$ Define \begin{align} T:& K\to K \\&x\mapsto Tx=(0,x^2_1,x^2_2,x^2_3,\cdots)\end{align}
I want to show that $T$ is not non-expansive.
MY TRIAL
$T$ is not non-expansive if for all $k\in [0,1]$, $\Vert Tx-Ty \Vert \geq k\Vert x-y \Vert $. (Note: I am not sure of this definition). Now, take $x=\left(\frac{3}{4},\frac{3}{4},\frac{3}{4},\cdots \right)$ and $y=\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\cdots\right).$ Then,
\begin{align} \Vert Tx-Ty \Vert &=\Vert T\left(\frac{3}{4},\frac{3}{4},\frac{3}{4},\cdots \right)-T\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\cdots\right)\Vert\\&=\Vert \left(0,\frac{9}{16},\frac{9}{16},\frac{9}{16},\cdots\right)-\left(0,\frac{1}{4},\frac{1}{4},\frac{1}{4},\cdots\right)\Vert \\&=\sup \{ 0,\left|\frac{9}{16}-\frac{1}{4}\right|,\left|\frac{9}{16}-\frac{1}{4}\right|\cdots\}\\&=\frac{5}{4}\sup \{ 0,\left|\frac{3}{4}-\frac{1}{2}\right|,\left|\frac{3}{4}-\frac{1}{2}\right|\cdots\}\\&\geq\sup \{ \left|\frac{3}{4}-\frac{1}{2}\right|,\left|\frac{3}{4}-\frac{1}{2}\right|\cdots\}\\&=\Vert x-y \Vert.\end{align}
I am not sure of my work. Kindly check please? If I'm wrong, then kindly provide a way-out. Thanks!
| The only problem I see is the part where you write
$$
\frac{5}{4}\sup \{ 0,\left|\frac{3}{4}-\frac{1}{2}\right|,\left|\frac{3}{4}-\frac{1}{2}\right|\cdots\} \geq\sup \{ \left|\frac{3}{4}-\frac{1}{2}\right|,\left|\frac{3}{4}-\frac{1}{2}\right|\cdots\},
$$
the sign $\ge$ should be $>$ instead because in order to show that $T$ is not nonexpansive we need to show that $\exists x,y\in K$ such that
$$
||Tx-Ty||>||x-y||.
$$
| {
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Prove that $\sum_{k=0}^{2n} \binom {2n+k}{k} \binom{2n}{k} \frac{(-1)^k}{2^k} \frac{1}{k+1} = 0. $
Let $n$ be a positive integer. Prove that
$$
\sum_{k=0}^{2n} \binom {2n+k}{k} \binom{2n}{k} \frac{(-1)^k}{2^k} \frac{1}{k+1} = 0.
$$
I am trying to solve this by using induction on $n$. I have proven the sum to be zero in the case $n=1$. Assuming that the sum is zero for $n=m$ ($m$ is a positive integer), how do I prove that it implies that the sum is zero for $n=m+1$?
Can I get some hints?
| Let's approach the sum through the Hypergeometric Function.
To this purpose let's rewrite it as
$$
\eqalign{
& S(n) = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,2n} \right)} {\binom{2n+k}{k} \binom{2n}{k}
{{\left( { - 1} \right)^{\,k} } \over {2^{\,k} \left( {k + 1} \right)}}} = \cr
& = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,2n} \right)} {\binom{2n+k}{k} \binom{2n}{k}
{1 \over {\left( {k + 1} \right)}}\left( { - {1 \over 2}} \right)^{\,k} } = \cr
& = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,2n} \right)} {t_{\,k} \left( { - {1 \over 2}} \right)^{\,k} } \cr}
$$
The $t_k$ are in the following ratio
$$
\eqalign{
& t_{\,0} = 1 \cr
& {{t_{\,k + 1} } \over {t_{\,k} }} = \cr
& = {{\left( {2n + k + 1} \right)!} \over {\left( {k + 1} \right)!\left( {k + 1} \right)!\left( {2n - k - 1} \right)!\left( {k + 2} \right)}}
{{k!k!\left( {2n - k} \right)!\left( {k + 1} \right)} \over {\left( {2n + k} \right)!}} = \cr
& = - {{\left( {k + 2n + 1} \right)\left( {k - 2n} \right)} \over {\left( {k + 2} \right)}}{1 \over {\left( {k + 1} \right)}} \cr}
$$
so the sum can be expressed as
$$
\eqalign{
& S(n) = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,2n} \right)} {\binom{2n+k}{k} \binom{2n}{k}
{{\left( { - 1} \right)^{\,k} } \over {2^{\,k} \left( {k + 1} \right)}}} = \cr
& = {}_2F_{\,1} \left( {\left. {\matrix{ {2n + 1,\; - 2n} \cr 2 \cr } \,} \right|1/2} \right) \cr}
$$
For $n=0$ this gives
$$
S(0) = {}_2F_{\,1} \left( {\left. {\matrix{ {1,\;0} \cr 2 \cr } \,} \right|1/2} \right) = 1
$$
while for $0<n$ we have
$$
\eqalign{
& {}_2F_{\,1} \left( {\left. {\matrix{ {2n + 1,\; - 2n} \cr 2 \cr } \,} \right|1/2} \right)\quad \left| {\;0 < n} \right.\quad = \cr
& = {{\Gamma \left( 2 \right)} \over {\Gamma \left( {2n + 1} \right)\Gamma \left( { - 2n} \right)}}\sum\limits_{0\, \le \,k\,}
{{{\Gamma \left( {2n + 1 + k} \right)\Gamma \left( { - 2n + k} \right)} \over {\Gamma \left( {2 + k} \right)}}} {1 \over {2^{\,k} k!}} \cr}
$$
Note that we can arrive to the same result by expressing the binomials through the Gamma function and
performing some algebraic simplifications.
To the fraction outside the sum we can apply the Reflection formula for the Gamma function,
which in the inverted form is valid all over the complex field
$$
{1 \over {\Gamma \left( {z + 1} \right)\,\Gamma \left( { - z} \right)}} = - {{\sin \left( {\pi \,z} \right)} \over \pi }\quad \left| {\;\forall z \in \mathbb C} \right.
$$
then clearly
$$
S(n)\quad \left| {\;0 < n \in Z} \right. = \sin \left( {2\pi \,n} \right) \cdot \left( \cdots \right) = 0
$$
| {
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"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Prove the inequality $\ln {(1+\frac{1}{x})}> \frac{2}{2x+1}$ Prove the inequality $$\ln {(1+\frac{1}{x})}> \frac{2}{2x+1}$$
for $x>0$.
My attempt: Let $$f(x)=\ln {(1+\frac{1}{x})}-\frac{2}{2x+1}$$
Then $$f'(x)=-\frac{1}{x(x+1)}+\frac{4}{(2x+1)^2}$$
$$f''(x)=\frac{1}{x^2}-\frac{1}{(x+1)^2}-\frac{8}{(2x+1)^3}>0$$
Then the function $f$ is convex. There exists a minimal point $x_0$ such that $f(x)\geq f(x_0)$. However, there's no critical point $x_0$ such that $f'(x_0)=0$, and $\lim_{x \rightarrow \infty} \sup {f'(x)}=0$. Then I want to show that $f(x)>0$, how do I continue my proof?
I have been trying another approach using Cauchy's MVT by letting
$$f(x)=\ln {x}$$ $$g(x)=\frac{1}{2x+1}$$ such that $$\frac{f(x+1)-f(x)}{g(x+1)-g(x)}=\frac{f'(c)}{g'(c)}$$ where $c \in (x,x+1)$ but failed.
As what I did is
$$\ln {(1+\frac{1}{x})}=\frac{1}{c} \cdot \frac{(2c+1)^2}{2} \cdot \frac {2}{(2x+1)(2x+3)}$$
I can't simply do the inequality
$$\frac{1}{c} \cdot \frac{(2c+1)^2}{2}>\frac{1}{x} \cdot \frac{(2x+1)^2}{2}$$
as $c>x$ because $\frac{1}{c} < \frac{1}{x}$ but $\frac{(2c+1)^2}{2} > \frac{(2x+1)^2}{2}$.
Edited: Of course, I know that $$\ln {(1+ \frac{1}{x})}>\frac{x}{1+x}$$
for $x>-1$. I just need to prove that $$\frac{x}{x+1}>\frac{2}{2x+1}$$ But I hope to find out another approach using calculus method.
| With your approach:
$$
f'(x)=-\frac{1}{x(x+1)}+\frac{4}{(2x+1)^2} = \frac{-1}{x(x+1)(2x+1)^2} < 0
$$
so that $f$ is strictly decreasing, and therefore
$$
f(x) > \lim_{t\to \infty} f(t) = 0 \, .
$$
Or you substitute $y = 1/x$ and consider
$$
g(y) = \ln (1+y) - \frac{2y}{2+y}
$$
with $g(0) = 0$ and
$$
g'(y) = \frac{y^2}{(y+1)(y+2)^2} > 0 \, .
$$
Remark: The estimate $\ln {(1+ \frac{1}{x})}>\frac{x}{1+x}$ is not good enough because $\frac{x}{x+1}<\frac{2}{2x+1}$ for small $x$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How would I go about solving for $x$ in $\frac{(x-a)\sqrt{x-a}+(x-b)\sqrt{x-b}}{\sqrt{x-a}+\sqrt{x-b}}=a-b$? The question
This is a homework question. Given the following, I am to solve for $x$ in terms of $a$ and $b$:
$$\frac{(x-a)\sqrt{x-a}+(x-b)\sqrt{x-b}}{\sqrt{x-a}+\sqrt{x-b}}=a-b;a>b.$$
My attempt
Although I see the pattern of multiple occurrences of $(x-a)$, $(x-b)$ I can't see any way to simplify the fraction further, so I go on to simplify the expression by multiplying by $\sqrt{x-a}+\sqrt{x-b}$:
$$\begin{align*}
(x-a)\sqrt{x-a}+(x-b)\sqrt{x-b}&=(a-b)(\sqrt{x-a}+\sqrt{x-b})\\
&=a\sqrt{x-a}+a\sqrt{x-b}-b\sqrt{x-a}-b\sqrt{x-b}
\end{align*}$$
Now I have the following:
$$(x-a)\sqrt{x-a}+(x-b)\sqrt{x-b}=a\sqrt{x-a}+a\sqrt{x-b}-b\sqrt{x-a}-b\sqrt{x-b}$$
Simplifying the RHS as I was out of ideas at that point:
$$x\sqrt{x-a}-a\sqrt{x-a}+x\sqrt{x-b}-b\sqrt{x-b}=a\sqrt{x-a}+a\sqrt{x-b}-b\sqrt{x-a}-b\sqrt{x-b}$$
I noticed that all one of the common factors $\sqrt{x-a},\sqrt{x-b}$ so I tried to isolate them and factor them out -- that is, all $\sqrt{x-b}$ terms on one side and $\sqrt{x-a}$ terms on the other.
$$\sqrt{x-b}(x-a)=\sqrt{x-a}(2a-b-x)$$
I tried to then square both sides, but that led to quite a mess.
$$(x-b)(x^2-2ax+a^2)=(x-a)(4a^2-4ab+2bx-4ax+b^2+x^2)$$
I'm afraid to even begin trying to simplifying this. I'm convinced I'm going about it in the wrong way.
The $a>b$ hint is interesting, but I have no clue what implication it may have here.
I think the $(x-a)\sqrt {x-a}$ patterns may mean something, perhaps I could do something with $a\sqrt a=\sqrt{a^3}$, but at this point it is probably a dead end.
I appreciate any help.
| Hint: Write your equation in the form
$$\sqrt{x-a}(x+b-2a)=\sqrt{x-b}(a-x)$$ and square it. We get in the case of $$a>b$$ $$x=a$$ or $$x=\frac{1}{3}(4a-b)$$
| {
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Every positive power of $5$ appears in the last digits of bigger power of $5$ Problem. Show that for every positive integer $n$, there is an integer $N > n$ such that the number $5^n$ appears as the last few digits $5^N$. For example, if $n = 3$, we have $5^3 = 125$ and $5^5 = 3125$, so $N = 5$ would satisfy.
This is a problem from the worksheet of the class Putnam Seminar at CMU. Please give hints towards the right direction and not the full solutions. Thanks!!
| Fix $n$.
The cases $n \le 3$ can be handled directly.
We now assume $n > 3$.
Let $m = \lceil n \log_{10} 5 \rceil$ be the smallest integer such that $10^m > 5^n$. For $n > 3$, we have $m < n$.
You want to find $N$ such that $5^N = k 10^m + 5^n$ for some integer $k$. Dividing both sides by $5^n$ yields $$5^{N-n} - 1 = k 2^m/ 5^{n-m}.$$
Thus if you show you can find a large integer $q$ such that $5^q - 1$ is divisible by $2^m$ then you can choose $N$ and $k$ appropriately to conclude the proof.
Base case: For $m=2$ we have $5^1 - 1$ divisible by $2^m$. Inductive step: if $5^q-1$ is divisible by $2^m$, then $5^{2q}-1 = (5^q-1)(5^q+1)$ is divisible by $2^{m+1}$.
| {
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"source": "stackexchange",
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How to factor this quadratic expression? A bit confused on how to factor $2x^2 + 5x − 3 = 0$. Firstly, I multiplied $a \cdot c$, so $2(-3)=-6$, however couldn't find two numbers that will add up to $5$. Then, I thought of the factors could be $6(-1)=-6$, and $6+-1=5$. However, shouldn't the $a$ and $c$ values always be multiplied together, (the $(2)(-3)$, rather than $6(-1)$?
When following the factors $-1$ and $6$. I have
$(2x^2-1x)(6x-3)$
$x(2x-1)+3(2x-1)$
Is this correct, if not; what is the best way to solve a leading coefficient when factoring?
| So we can actually generalize this. Say we have the polynomial
$$p(x)=ax^2+bx+c$$
Fact:
$$p(x)=\bigg(ax+\frac{b-\sqrt{b^2-4ac}}{2}\bigg)\bigg(ax+\frac{b+\sqrt{b^2-4ac}}{2}\bigg)$$
Proof:
Let's assume the existence of three real numbers $r_1$, $r_2$, and $e_1$ such that
$$ax^2+bx+c=e_1(x-r_1)(x-r_2)$$
If we expand the product on the right hand side and then compare coefficients,
$$ax^2+bx+c=e_1x^2-e_1(r_1+r_2)x+e_1r_1r_2$$
we get a system of equations
$$e_1=a\\-e_1(r_1+r_2)=b\\e_1r_1r_2=c$$
Evidently, we get $e_1=a$ for free. So we update our system of equations:
$$r_1+r_2=-\frac{b}{a}\\r_1r_2=\frac{c}{a}$$
We can solve each equation for $r_2$:
$$r_2=-\frac{b}{a}-r_1\\r_2=\frac{c}{ar_1}$$
So we can set the two equations equal to each-other:
$$-\frac{b}{a}-r_1=\frac{c}{ar_1}$$
$$r_1+\frac{c}{ar_1}=-\frac{b}{a}$$
multiplying both sides by $ar_1$,
$$ar_1^2+br_1=-c$$
Then we add $\frac{b^2}{4a}$ to both sides:
$$ar_1^2+br_1+\frac{b^2}{4a}=\frac{b^2}{4a}-c$$
Then we note that
$$a(r_1+b/2a)^2=ar_1^2+br_1+\frac{b^2}{4a}$$
So we plug it in:
$$a(r_1+b/2a)^2=\frac{b^2}{4a}-c$$
$$a(r_1+b/2a)^2=\frac{b^2-4ac}{4a}$$
$$(r_1+b/2a)^2=\frac{b^2-4ac}{4a^2}$$
$$r_1+b/2a=\sqrt{\frac{b^2-4ac}{4a^2}}$$
$$r_1+b/2a=\frac{\sqrt{b^2-4ac}}{\sqrt{4a^2}}$$
$$r_1+b/2a=\frac{\sqrt{b^2-4ac}}{2a}$$
$$r_1=\frac{-b+\sqrt{b^2-4ac}}{2a}$$
And since we know that
$$r_2=-\frac{b}{a}-r_1$$
We know that
$$r_2=-\frac{b}{a}-\bigg(\frac{-b+\sqrt{b^2-4ac}}{2a}\bigg)$$
$$r_2=-\frac{2b}{2a}+\frac{b-\sqrt{b^2-4ac}}{2a}$$
$$r_2=\frac{b-2b-\sqrt{b^2-4ac}}{2a}$$
$$r_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$$
And by definition,
$$ax^2+bx+c=a\bigg(x+\frac{b-\sqrt{b^2-4ac}}{2a}\bigg)\bigg(x+\frac{b+\sqrt{b^2-4ac}}{2a}\bigg)$$
$$ax^2+bx+c=\bigg(ax+\frac{b-\sqrt{b^2-4ac}}{2}\bigg)\bigg(ax+\frac{b+\sqrt{b^2-4ac}}{2}\bigg)$$
And with that our proof is complete :)
| {
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"url": "https://math.stackexchange.com/questions/3049315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Given $\left(x + \sqrt{1+y^2}\right)\left(y + \sqrt{1+x^2}\right) = 1$, prove $\left(x + \sqrt{1+x^2}\right)\left(y + \sqrt{1+y^2}\right) = 1$.
Let $x$ and $y$ be real numbers such that
$$\left(x + \sqrt{1+y^2}\right)\left(y + \sqrt{1+x^2}\right) = 1$$
Prove that
$$\left(x + \sqrt{1+x^2}\right)\left(y + \sqrt{1+y^2}\right) = 1$$
| Hint :
By the hypothesis of the problem, do some algebra and show that :
$$\left(x + \sqrt{1+y^2}\right)\left(y + \sqrt{1+x^2}\right) = 1 \Leftrightarrow \dots \Leftrightarrow y = -x$$
Now, substitute $y=-x$ on the expression $\left(x + \sqrt{1+x^2}\right)\left(y + \sqrt{1+y^2}\right)$ and see what happens.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $p$ is prime, then $x^2 +5y^2 = p \iff p\equiv 1,9 $ mod $(20)$. Let $p\neq 2,5$ be prime. I wish to show that: $x^2 +5y^2 = p \Leftrightarrow p\equiv 1,9 $ mod $(20)$.
I proved to $\Rightarrow$ part, means $x^2 +5y^2=p \Rightarrow p\equiv 1,9 $ mod $(20)$.
For $\Leftarrow$ , $p\equiv 1,9(20) \Rightarrow p\equiv 1(4)$ , $p\equiv1 ,4 (5)$ thus $(\frac{4}{p})=1,(\frac{-1}{p}) =1$ (using legendre symbols) , also $(\frac{5}{p})=_{p\equiv1(4)}(\frac{p}{5})$ and $p\equiv1(5)$ so $(\frac{5}{p})=1$ , so $(\frac{-20}{p})=(\frac{5}{p})(\frac{4}{p})(\frac{-1}{p}) = 1$. So $-20$ is a quadratic residue mod $p$.
Yet I don't succeed to go on from this point (I don't know even if its possible to do so).
| Let $p\equiv1$ or $9\pmod{20}$. Then $-5$ is a quadratic residue modulo $p$,
so there are integers $r$ and $s$ with $$r^2+ps=-5.$$
This means that the integer quadratic form
$$f(X,Y)=pX^2+2rXY+sY^2$$
has discriminant $-20$ and is also positive definite. Then $p$
is represented by $f$. Now $f$ is equivalent to a reduced form. There
are two reduced forms of discriminant $-20$: $g_1(X,Y)=X^2+5Y^2$
and $g_2(X,Y)=2X^2+2XY+3Y^2$. But $g_2$ cannot represent any number
congruent to $1$ or $9$ modulo $20$. Therefore $g_1$ represents $p$.
| {
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"url": "https://math.stackexchange.com/questions/3057465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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a problem on complex numbers Let $w\neq 1$ and $w^{13} = 1$.
If $a = w+ w^3 + w^4 + w^{-4} + w^{-3} + w^{-1}$ and $b = w^2+ w^5 + w^6 + w^{-6} + w^{-5} + w^{-2}$, then the quadratic equation whose roots are $a$ and $b$ is ... ?
I got $w=\cos(\frac{2\pi}{13})+i\sin(\frac{2\pi}{13})$
And then I found $a$ and $b$ in trigonometric form. But when I multiplied them to get the product of roots it gets very difficult. How to solve it?
| $$ a^2 + a = \frac{ w^{16} + 2w^{15} + w^{14} + 2w^{13} + 3w^{12} + 3w^{11} + 3w^{10} + 3w^9 + 6w^8 + 3w^7 + 3w^6 + 3w^5 + 3w^4 + 2w^3 + w^2 + 2w + 1}{w^8} $$
This is not impressive without
$$ w^{16} + 2w^{15} + w^{14} + 2w^{13} = w^3+2w^2+w+2. $$ Therefore
$$ a^2 + a = \frac{ 3w^{12} + 3w^{11} + 3w^{10} + 3w^9 + 6w^8 + 3w^7 + 3w^6 + 3w^5 + 3w^4 + 3w^3 + 3w^2 + 3w + 3}{w^8} $$
The one coefficient out of line is $6 w^8 / w^8,$ so we need to subtract 3
$$ a^2 + a -3 = \frac{ 3w^{12} + 3w^{11} + 3w^{10} + 3w^9 + 3w^8 + 3w^7 + 3w^6 + 3w^5 + 3w^4 + 3w^3 + 3w^2 + 3w + 3}{w^8} $$
$$ a^2 + a -3 = \frac{ 3 \left( w^{12} + w^{11} + w^{10} + w^9 + w^8 + w^7 + w^6 + w^5 + w^4 + w^3 + w^2 + w + 1 \right)}{w^8} $$
and
$$ a^2 + a - 3 = 0 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3060694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\frac{1}{2}(x+\frac{a}{x}) \ge \sqrt{a}$ Let x and a be real numbers > 0. Prove that $\frac{1}{2}(x+\frac{a}{x}) \ge \sqrt{a}$
My idea is that I'm going to use $a>b \iff a^2>b^2$ since we are only dealing with postive real numbers we won't run into problems with the root.
$\frac{1}{2}(x+\frac{a}{x}) \ge \sqrt{a} \iff (\frac{1}{2}(x+\frac{a}{x}))^2 = \frac{1}{4}(x+\frac{a}{x})(x+\frac{a}{x})= \frac{1}{4}(x^2+2a+\frac{a^2}{x^2}) \ge \sqrt{a}^2=a$
And from that we get:
$\frac{1}{4}x^2 +\frac{1}{2}a+\frac{a^2}{4x^2} \ge a$
I don't really know how to proceed from here. What about looking at different possibilities for the values of a and x?
Let x>a. Then:
$\frac{1}{4}x^2 +\frac{1}{2}a+\frac{a^2}{4x^2} \ge \frac{1}{4}a^2 +\frac{1}{2}a+\frac{a^2}{4x^2} \ge a \iff \frac{1}{4}a^2x^2 +\frac{1}{2}ax^2+\frac{a^2}{4} \ge ax^2$
And now I don't know how to prcoeed from here and doubt that my approach is correct. Can someone please help me out here? Tanks in advance.
| $(\sqrt{x}-\sqrt{a/x})^2\geq0\Rightarrow x+a/x-2\sqrt{a}\geq 0\Rightarrow 1/2(x+a/x)\geq \sqrt{a}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3061160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Trying to simplify $\frac{\sqrt{8}}{1-\sqrt{3x}}$ to be $\frac{2\sqrt{2}+2\sqrt{6x}}{1-3x}$ I am asked to simplify $\frac{\sqrt{8}}{1-\sqrt{3x}}$. The solution is provided as $\frac{2\sqrt{2}+2\sqrt{6x}}{1-3x}$ and I am unable to arrive at this. I was able to arrive at $\frac{1+2\sqrt{2}\sqrt{3x}}{1-3x}$
Here is my working:
$\frac{\sqrt{8}}{1-\sqrt{3x}}$ = $\frac{\sqrt{8}}{1-\sqrt{3x}}$ * $\frac{1+\sqrt{3x}}{1+\sqrt{3x}}$ = $\frac{1+\sqrt{8}\sqrt{3x}}{1-3x}$ = $\frac{1+\sqrt{2}\sqrt{2}\sqrt{2}\sqrt{3x}}{1-3x}$ = $\frac{1+2\sqrt{2}\sqrt{3x}}{1-3x}$
Is $\frac{1+2\sqrt{2}\sqrt{3x}}{1-3x}$ correct and part of the way? How can I arrive at the provided solution $\frac{2\sqrt{2}+2\sqrt{6x}}{1-3x}$?
| $$\frac{\sqrt{8}}{1-\sqrt{3x}} = \frac{\sqrt{8}}{1-\sqrt{3x}} \cdot \frac{1+\sqrt{3x}}{1+\sqrt{3x}} = \color{blue}{\frac{\sqrt{8}\cdot\left({1+\sqrt{3x}}\right)}{1-3x}} = \frac{\sqrt{8}+\sqrt{24x}}{1-3x}$$
$$= \frac{\sqrt{2^3}+\sqrt{2^3\cdot3x}}{1-3x} = \frac{2\sqrt{2}+2\sqrt{6x}}{1-3x}$$
Notice the step I highlighted in blue, which is where you made an error. You have to multiply $\sqrt{8}$ to $\left(1+\sqrt{3x}\right)$ completely. You multiplied it by only $\sqrt{3x}$ and added $1$, which wasn’t correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3062999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Non-homogenous differential equation help
*a) Given $x = e^u$ and $$x^2\frac{d^2y}{dx^2} - 4x \frac{dy}{dx} + 6y = 12$$ show that $$\frac{d^2y}{du^2} - 5 \frac{dy}{du} + 6y = 12$$
I was able to do this:
$x = e^u$
$$\frac{dx}{du} = e^u = x $$
$$\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} = \frac{1}{x}\frac{dy}{du}$$
$$\frac{d^2y}{dx^2} = -\frac{1}{x^2}\frac{dy}{du} + \frac{1}{x}\frac{d^2y}{du^2}\frac{du}{dx} = -\frac{1}{x^2}\frac{dy}{du} + \frac{1}{x^2}\frac{d^2y}{du^2}$$
$$x^2\left(-\frac{1}{x^2}\frac{dy}{du} + \frac{1}{x^2}\frac{d^2y}{du^2}\right) -4x\left(\frac{1}{x}\frac{dy}{du}\right) + 6y = 12 $$
$$-\frac{dy}{du} + \frac{d^2y}{du^2} - 4\frac{dy}{du} +6y = 12 $$
$$\frac{d^2y}{du^2} - 5\frac{dy}{du} + 6y = 12 $$
part b is where I got stuck:
b) Hence solve the equation $$x^2\frac{d^2y}{dx^2} - 4x\frac{dy}{dx} + 6y = 12$$ given that $y(1) = 7$ and $y(2) = 14$
$$\frac{d^2y}{du^2} - 5\frac{dy}{du} + 6y = 12 $$
auxiliary equation: $\lambda^2 - 5λ + 6 = 0 $
Hence $\lambda = 3, 2$
Complementary function: $$y = Ae^{3u} + Be^{2u}$$
particular integral in the form $y = c$
$$\frac{dy}{dx} = 0 = \frac{d^2y}{dx^2}$$
$$6C = 12 $$
$$C = 2$$
general solution: $$y = Ae^{3u} + Be^{2u} + 2 $$
where I began to get stuck:
replacing $u = \ln x$ means you get left with
$$y = Ax^3 + Bx^2 + 2$$
$$\frac{dy}{dx} = 3Ax^2 + 2Bx $$
$$\frac{d^2y}{dx^2} = 6Ax + 2B $$
my teacher told me to "substitute this into the solution of the DE and just apply the conditions"
(this is where I think I may me wrong, I think I might have to equate the differential equation to 0 after substituting)
through substituting these values into the differential equation you get:
$$x^2 (6Ax + 2B) - 5(3Ax^2 + 2Bx) + 6 (Ax^3 + Bx^2 + 2) = 12 $$
$$ 6Ax^3 + 2Bx^2 -15Ax^2 - 10Bx + 6Ax^3 + 6Bx^2 + 12 = 12 $$
$$ 12Ax^3 + 8Bx^2 -15Ax^2 - 10Bx = 0 $$
This may resolve very simply, but I didn't want to try without knowing I was doing the correct thing.
| You're correct that the general solution is
$$ y(x) = Ax^3 + Bx^2 + 2 $$
Using the given boundary conditions we have
\begin{align}
y(1) &= A + B + 2 = 7 \\
y(2) &= 8A + 4B + 2 = 14
\end{align}
or
\begin{align}
A + B &= 5\\
8A + 4B &= 12
\end{align}
Solving the above system gives $A=-2$, $B=7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3064380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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