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Given a sequence $a_n =\frac{n^2 - 3n + (-1)^n}{3n^2 - 7n + 5}$, prove that $\lim_{n \to \infty} a_{n} = g$. The given Sequence is $a_n = \frac{n^2 - 3n + (-1)^n}{3n^2 - 7n + 5}$.
I showed that the Sequence $a_n$ converges towards a Value $g = \frac{1}{3}$.
How do I determine for each $\epsilon > 0$ an $n_0$ so that: $|a_n - g| < \epsilon \ \forall n > n_0$?
This is what I tried so far:
$\begin{array}{rcl}
\left|a_n - g\right| & < & \epsilon \\
\left|\frac{n^2 - 3n + (-1)^n}{3n^2 - 7n + 5} - \frac{1}{3}\right| & < & \epsilon \\
\left|\frac{3n^2 - 9n + 3 \cdot (-1)^n}{9n^2 - 21n + 15} - \frac{3n^2 - 7n + 5}{9n^2 - 21n + 15}\right| & < & \epsilon \\
\left|\frac{-2n - 5 + 3 \cdot (-1)^n}{9n^2 - 21n + 15}\right| & < & \epsilon \ \text{ Denominator > 0, Numerator < 0 } \forall n \in \mathbf{N} \\
\frac{2n + 5 - 3 \cdot (-1)^n}{9n^2 - 21n + 15} \leq \frac{2n + 5 + 3}{9n^2 - 21n + 15} = \frac{2n + 8}{9n^2 - 21n + 15} & < & \epsilon \\
\end{array}$
Are my Steps correct so far?
How do I proceed from here?
| What you have done is correct. You need $\frac {2n+8} {3n(3n-7)+15} <\epsilon$.
One way to achieve this is to make $3n-7 >\frac 1 {\epsilon}$. $\cdots$ (1).
Now we need $2n+8 <3n+15\epsilon$. For this we need
$n>8-15\epsilon$. Hence $n >8-15\epsilon$ and $n>\frac 1 3 (7+\frac 1 {\epsilon})$ is good enough. Choose $n_0$ to be greater than the maximum of $\frac 1 3 (7+\frac 1 {\epsilon})$ and $8-15\epsilon$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3553643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Proving that $\prod_{k=1}^{n}\tan\left(\frac{\pi k}{2n+1}\right)=\sqrt{2n+1}$ using geometry I have to prove that the following holds. A hint to use complex numbers has been given. I have tried to make a start but not to any result.
$$\prod_{k=1}^{n}\tan\left(\frac{\pi k}{2n+1}\right)=\sqrt{2n+1}$$
My Attempt:
Let us consider $(2n+1)^{\text{th}}$ roots of unity, $z_k=\exp\left(\frac{2k\pi i}{2n+1}\right)$. We can rewrite the product in terms of $\arg(z_k)$ as $\prod_{k=1}^{n}\tan\left(\frac{1}{2}\arg(z_k)\right)$. Or equivalently so, if we consider $(4n+2)^{\text{th}}$ roots of unity, we get this product as $\prod_{k=1}^{n}\tan(\arg(\zeta _k))$, where $\zeta_k= \exp\left(\frac{2k\pi i}{4n+2}\right)$.
I know it can be proved by proving the expression for $\prod\sin(\frac{1}{2}\arg(z_k))$. But I was wondering, is there a way to use telescopic products or purely the geometry of the complex roots of unity to arrive at this
| $\text{We are going to prove that}$
$\displaystyle \prod_{k=1}^{m}\left(\tan \frac{k \pi}{2 m+1}\right)=\sqrt{2 m+1}. \tag*{}$
$\textrm{Considering the roots of the polynomial equation }\\ \text{of degree 2m}$
$\displaystyle (z+1)^{2 m+1}-(z-1)^{2 m+1}=0 \cdots(*)\tag*{} $
$\displaystyle \left(\frac{z-1}{z+1}\right)^{2 m+1}=1 \tag*{} \\$
$\displaystyle \frac{z-1}{z+1} =\cos \frac{2 k \pi}{2 m+1}+i \sin \frac{2 k \pi}{2 m+1} =e^{\frac{2 k \pi}{2 m+1} i}, $
$\text { where } k=1,2, \cdots , 2 m.$
$\textrm{Hence the 2m complex roots of (*) are}$
$ \displaystyle z =\frac{e^{\frac{2 k \pi}{2 m+1} i}+1}{1-e^{\frac{2 k \pi}{2 m+1} i}} =\frac{e^{\frac{k \pi}{2 m+1} i}-e^{-\frac{k \pi}{2 m+1} i}}{e^{-\frac{k\pi}{2 m+1} i}-e^{\frac{k \pi}{2 m+1} i}} \tag*{} $
$\displaystyle \qquad =\frac{2 \cos \left(\frac{k \pi}{2 m+1}\right)}{-2 i\sin \left(\frac{k \pi}{2 m+1}\right)} =i \cot \left(\frac{k \pi}{2 m+1}\right)\tag*{} $
$\displaystyle \because \text { product of roots of }(*) =\frac{\text{constant in (*)}}{\text{leading coefficient of (*)}}\tag*{} $
$\therefore \displaystyle \prod_{k=1} ^{2 m}\left[i \cot \left(\frac{k \pi}{2 m+1}\right)\right]=\frac{2}{2(2 m+1)} \tag*{} \\$
$\displaystyle i^{2 m} \prod_{k=1}^ {2 m}\cot \left(\frac{k \pi}{2 m+1}\right)=\frac{1}{ 2 m+1} \tag*{} $
$\because \displaystyle \cot \frac{(2 m+1-k)\pi}{2 m+1} =-\cot \left(\frac{k \pi}{2 m+1}\right)\tag*{} $
$\therefore\displaystyle (-1)^{m}(-1) ^m \prod_{k=1}^ { m}\left[ \cot ^{2} \frac{k \pi}{2 m+1}\right]=\frac{1}{2 m+1} \tag*{} $
$\displaystyle \left(\prod_{k=1}^{m} \cot \frac{k \pi}{2 m+1}\right)^{2}=\frac{1}{2 m +1 } \tag*{} \\$
$$\displaystyle \quad \prod_{k=1}^{m}\left(\tan \frac{k \pi}{2 m+1}\right)=\sqrt{2 m+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3556048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Prove that $3^{4n-1}-2$ is always a multiple of 5. I am trying to prove the following statement: $f(n) =3^{4n-1}-2 $ is always a multiple of 5, for $n\in \mathbb Z^+$. Using proof by induction:
Base case: $f(1)=25$, which is a multiple of 5 and hence holds for $n=1$.
Assumption step: Assume $f(k) =3^{4k-1}-2 $ is a multiple of of 5.
Considering $n= k+1$: $f(k+1)=3^{4k+3}-2 $
$$=27(3^{4k})-2$$
$$f(k+1)-6f(k)^{**}=27(3^{4k})-2-6(3^{4k-1}-2) $$
$$=27(3^{4k})-2-2(3^{4k}-6)$$
$$=27(3^{4k})-2-2(3^{4k})+12$$
$$=25(3^{4k})+10$$
$$=5(5(3^{4k})+2)$$
**when performing this step, is any multiple of the $n = k$ case allowed to be added or subtracted, or is the arguement not valid for certain values?
| So we need to prove that \begin{equation}
5 | 3^{4 n-1}-2 \quad \forall n \in \mathbb Z^{+}
\end{equation}
For n=1, we have \begin{equation}
5 | 25
\end{equation}. We assume it is true for some $n\in\mathbb Z^+$
Now we prove \begin{equation}
\begin{array}{l}
{3^{4(n+1)-1}-2} \\
{3^{4 n+4-1}-2} \\
\end{array}
\end{equation}
$3^{4} \cdot\left(3^{4 n-1}\right)-2$
\begin{equation}
\begin{aligned}
&3^{4}\left(3^{4 n-1}-2\right)+2 \cdot 3^{4}-2\\
&3^{4}\left(3^{4 n-1}-2\right)+160
\end{aligned}
\end{equation}
\begin{equation}
3^{4 n-1}-2
\end{equation} can be divided by 5 and 160 divided by 5 is 32 so whole is expression can be divided by 5.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3558742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Prove that $\binom{n}{0} + 2\binom{n}{1} + ...(n+1)\binom{n}{n} = (n + 2)2^{n-1}$. How can I prove the following identity?
$$\binom{n}{0} + 2\binom{n}{1} + ...(n+1)\binom{n}{n} = (n +
2)2^{n-1}$$
I thought about differentiating this:
$$(1 + x) ^ n = \sum_{k = 0} ^ {n} \binom{n}{k}x^k$$
and then evaluating it at $x = 1$, but I didn't get to my desired result. I just kept finding that $\displaystyle\sum_{k = 0}^n k \binom{n}{k} = n2^{n-1}$.
| Another way.
Multiply
$(1+x)^n=\sum_{k=0}^n{n\choose k}x^k
$
by $x$
to get
$x(1+x)^{n}=\sum_{k=0}^n{n\choose k}x^{k+1}
$.
Now when you differentiate
you get
$\begin{array}\\
\sum_{k=0}^n(k+1){n\choose k}x^{k}
&=(x(1+x)^{n})'\\
&=(1+x)^n+xn(1+x)^{n-1}\\
&=(1+x)^{n-1}(1+x+nx))\\
&=(1+x)^{n-1}(1+x(n+1))\\
\end{array}
$
Setting
$x=1$ gives
$\sum_{k=0}^n(k+1){n\choose k}
=(n+2)2^{n-1}
$.
More generally,
Multiply
$(1+x)^n=\sum_{k=0}^n{n\choose k}x^k
$
by $x^m$
to get
$x^m(1+x)^{n}=\sum_{k=0}^n{n\choose k}x^{k+m}
$.
Now when you differentiate
you get
$\begin{array}\\
\sum_{k=0}^n(k+m){n\choose k}x^{k+m-1}
&=(x^m(1+x)^{n})'\\
&=mx^{m-1}(1+x)^n+x^mn(1+x)^{n-1}\\
&=x^{m-1}(1+x)^{n-1}(m(1+x)+xn))\\
&=x^{m-1}(1+x)^{n-1}(m+x(n+m))\\
\end{array}
$
Setting
$x=1$ gives
$\sum_{k=0}^n(k+m){n\choose k}
=2^{n-1}(2m+n)
$.
This, of course,
can be gotten by adding
earlier results.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3558942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Area bounded by a hyperbola Find the area "A" of the given graph
Answers should be in terms of a, b and h
My attempt:
(1) Look for an equation in terms of y (or maybe in terms of x since the region is bounded between $y=0$ and $y=h$?)
$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \\
b^2x^2-a^2y^2 = a^2b^2 \\
a^2y^2 = b^2x^2-a^2b^2 \\
y^2 = (\frac{bx}{a})^2-b^2 \\
y = \sqrt{(\frac{bx}{a})^2-b^2}
$
The expression inside the square root looks pretty similar to $\frac{d}{dx}arcsin(x)=\frac{1}{\sqrt{1-x^2}}$... I don't know if/how to make that substitution (if that's what I need to do). After that, I believe I just have to evaluate it from $y=0$ to $y=h$
| Integrate over $y$, it will be easier to work out.
$$A = 2a\int_0^h \sqrt{1+\frac{y^2}{b^2}}dy = 2ab \int_0^{h/b} \sqrt{1+u^2}du $$
The primitive of $\sqrt{1+u^2}$ can be found by substitution. I suggest $u=\sinh t$, then
$$\int \sqrt{1+u^2}du = \int \cosh^2 t dt = \frac{1}{4} \sinh(2t) + \frac{t}{2} = \frac{1}{2} u\sqrt{1+u^2} + \frac{\text{arcsinh}(u)}{2} \; .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3560343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Evaluate $\int_0^{\infty } \frac{\tan ^{-1}\left(\sqrt{a^2+x^2}\right)}{\left(x^2+1\right)\sqrt{a^2+x^2}} \, dx$ J. Borwein's review on experimental mathematics gives the following
$$\int_0^{\infty } \frac{\tan ^{-1}\left(\sqrt{a^2+x^2}\right)}{\left(x^2+1\right)\sqrt{a^2+x^2}} \, dx=\frac{\pi \left(2 \tan ^{-1}\left(\sqrt{a^2-1}\right)-\tan^{-1}\left(\sqrt{a^4-1}\right)\right)}{2 \sqrt{a^2-1}}, \ a>1$$
How can we establish it? Any help will be appreciated.
Update: The original problem has a typo and the current one is easy.
| As suggested by Ali Shather in the comments, we can write the inverse tangent as an integral to obtain
\begin{align}
\int \limits_0^\infty \frac{\arctan\left(\sqrt{a^2+x^2}\right)}{(1+x^2)\sqrt{a^2+x^2}} \, \mathrm{d} x &= \int \limits_0^\infty \int \limits_0^1 \frac{\mathrm{d} y}{1+y^2 (a^2+x^2)} \,\frac{\mathrm{d} x}{1+x^2} \\
&= \int \limits_0^1 \int \limits_0^\infty \frac{\mathrm{d} x}{(1+x^2)(1+y^2(a^2+x^2))} \, \mathrm{d} y \\
&= \int \limits_0^1 \frac{1}{1 + (a^2-1)y^2} \int \limits_0^\infty \left(\frac{1}{1+x^2} - \frac{1}{a^2+\frac{1}{y^2} + x^2}\right) \, \mathrm{d} x \, \mathrm{d} y \\
&= \frac{\pi}{2} \int \limits_0^1 \frac{\mathrm{d} y}{\sqrt{1+a^2 y^2}(y + \sqrt{1+a^2 y^2})} \\&\!\!\!\!\!\!\stackrel{y = \frac{2t}{a(1-t^2)}}{=} \pi \int \limits_0^{\frac{a}{1+\sqrt{1+a^2}}}\frac{\mathrm{d} t}{a(1+t^2)+2t} \\
&= \frac{\pi a}{a^2 - 1} \int \limits_0^{\frac{a}{1+\sqrt{1+a^2}}} \frac{\mathrm{d}t}{1 + \left(\frac{1+at}{\sqrt{a^2 - 1}}\right)^2} \\
&\!\!\!\!\!\!\!\stackrel{\frac{1+at}{\sqrt{a^2 - 1}} = \frac{1}{u}}{=} \frac{\pi}{\sqrt{a^2-1}} \int \limits_{\sqrt{\frac{a^2-1}{a^2+1}}}^{\sqrt{a^2-1}} \frac{\mathrm{d} u}{1+u^2} \\
&= \frac{\pi}{\sqrt{a^2-1}}\left[\arctan\left(\sqrt{a^2-1}\right) - \arctan\left(\sqrt{\frac{a^2-1}{a^2+1}}\right)\right] \\
&= \frac{\pi}{\sqrt{a^2-1}}\left[\arctan\left(\sqrt{a^2-1}\right) - \frac{1}{2}\arctan\left(\sqrt{a^4-1}\right)\right]
\end{align}
for $a > 1$. The last step follows from $\arctan(x) = \frac{1}{2} \arctan\left(\frac{2 x}{1-x^2}\right)$ for $x^2 < 1$. Using the more general argtangent addition formula this result can also be written as
$$ \int \limits_0^\infty \frac{\arctan\left(\sqrt{a^2+x^2}\right)}{(1+x^2)\sqrt{a^2+x^2}} \, \mathrm{d} x = \frac{\pi}{\sqrt{a^2 - 1}} \arctan \left(\frac{\sqrt{a^2-1}}{2 + \sqrt{a^2+1}}\right) \, , \, a > 1 \, .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3562771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 0
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Finding the value of $\sum_{n=1}^\infty\frac{n^2}{(n+1)(n+2)(n+3)(n+4)}$ Problem_
Find the value of $$\sum_{n=1}^\infty\frac{n^2}{(n+1)(n+2)(n+3)(n+4)}$$
It seems like I have to use the partial sum in order to get the exact value. But making it into the partial fractions was not that easy to me because of the numerator.
I tried to segregate the fraction by putting $A, B, C$ and $D$ as:
$$\frac{n^2}{(n+1)(n+2)(n+3)(n+4)}={A\over n+1}+{B\over n+2}+{C\over n+3}+{D\over n+4}$$
By multiplying both hands with $(n+1)(n+2)(n+3)(n+4)$, I could get the values which are $A=1/6, B=-2, C=9/2$ and $D=-8/3$. Unfortunately, there were no significant properties among them.
I've also thought of dividing the numerator into two or more terms such as
$$n^2=n(n+1) - n$$
However, this method also did not give me any hint to solve. Are there other ways to evaluate the series looks like that? Thank you for your interest.
| First add and subtract 1 to the numerator
it becomes $\sum_{n=1}^\infty \frac{(n+4-5)}{(n+2)(n+3)(n+4)} + \sum_{n=1}^\infty\frac{1}{(n+1)(n+2)(n+3)(n+4)}$
$=\sum_{n=1}^\infty\frac{1}{(n+2)(n+3)} -\sum_{n=1}^\infty\frac{5}{(n+2)(n+3)(n+4)}+\sum_{n=1}^\infty\frac{1}{(n+1)(n+2)(n+3)(n+4)}$.(try evaluating this yourself)
$=\frac{1}{3}-\frac{5}{2} \times \frac{1}{12} + \frac{1}{3} \times \frac{1}{24} = \boxed{\frac{5}{36}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3564128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find circumradius of an octagon inscribable into a circle with side lengths $1,1,1,1,3\sqrt2,3\sqrt2,3\sqrt2$ and $3\sqrt2$ An octagon has side lengths $1,1,1,1,3\sqrt2,3\sqrt2,3\sqrt2$ and $3\sqrt2.$ What should be the length of its circumradius?
I tried solving it using elementary geometry, but that was of no use. I tried my hand at complex number geometry as well, but that didn't work.
Can someone provide a formal answer (considering that I am just in my 11th grade)? Is there an elegant approach to this question using complex numbers?
| The circumradius does not depend on the order of the sides. To show this, draw the isosceles triangle from the center to the vertices. The angle subtended by each of the $1$ sides is the same, as is the angle subtended by each of the $3\sqrt 2$ sides. These angles must sum to $\frac \pi 2$. Draw the octagon with the sides alternating.
If $\theta$ is the angle subtended by a side of $1$ and $r$ is the radius of the circle we have $\sin \frac \theta 2 = \frac 1{2r}$. The angle subtended by a side of $1$ and a side of $3\sqrt 2$ must add to $\frac \pi 2$ by symmetry, so the angle subtended by a side of $3\sqrt 2$ is $\frac \pi 2-\theta$. That gives $\sin \left(\frac \pi 4-\frac \theta 2\right)=\frac 3{r\sqrt 2 }$
$$\sin \left(\frac \pi 4-\frac \theta 2\right)=\frac 3{r\sqrt 2 }\\
\sin \frac \pi 4 \cos \frac \theta 2-\cos \frac \pi 4 \sin \frac\theta 2 =\frac 3{r\sqrt 2 }\\
\frac {\sqrt 2}2\cos \frac \theta 2-\frac {\sqrt 2}{4r}=\frac 3{r \sqrt 2}\\
\cos \frac \theta 2-\frac 1{2r}=\frac 3r\\
\cos \frac \theta 2=\frac 7{2r}\\
\left(\frac 7{2r}\right)^2+\left(\frac 1{2r}\right)^2=1\\50=4r^2\\
r=\frac 5{\sqrt 2}$$
Done by hand, checked with Alpha (click exact form on the final result).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3564255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Integral $\int \arcsin \left(\sqrt{\frac{x}{1-x}}\right)dx$ $$\int \arcsin\left(\sqrt{\frac{x}{1-x}}\right)dx$$
I'm trying to solve this integral from GN Berman's Problems on a course of mathematical analysis (Question number 1845)
I tried substituting $x$ for $t^2$:
$$2\int t\arcsin\left(\frac{t}{\sqrt{1-t^2}}\right) dt$$
And then evaluating it by parts, which made it very complicated.
So how do I go about solving this?
| Substitute
$$u = \sqrt{\frac{x}{1-x}} \implies \frac{du}{dx} = \frac{\sqrt{x}}{(1-x)^{\frac{3}{2}}} + \frac{1}{2\sqrt{1-x}\sqrt{x}}$$
Then the integral become:
$$2 \int \frac{u\arcsin(u)}{(u^2 + 1)^2} \,du$$
Integrating by parts $\left(f = \arcsin(u), g'= \frac{u}{(u^2 + 1)^2} \implies f'= \frac{1}{\sqrt{1 - u^2}}, g = -\frac{1}{2(u^2 + 1)}\right)$:
$$2 \int \frac{u\arcsin(u)}{(u^2 + 1)^2} \,du = -\frac{\arcsin u}{2(u^2+1)} - \int \frac{1}{2\sqrt{1-u^2}(u^2+1)} \,du \text{ (1)}$$
We will now solve this integral first:
$$\int \frac{1}{2\sqrt{1-u^2}(u^2+1)} \,du \text{ (2)}$$
Substitute $v = \arcsin(u) \implies du = \cos(v) \,dv$, then:
$$(2) \implies \int \frac{\cos(v)}{2\sqrt{1-\sin(v)^2}(\sin(v)^2+1)} \,dv$$
$$= \frac{1}{2} \int \frac{1}{\sin(v)^2+1} \,dv$$
$$= \frac{1}{2} \int \sec^2(v) \cdot \frac{1}{2\tan(v)^2+1} \,dv \text{ (3)}$$
Substitute $w = \sqrt{2}\tan(v) \implies \frac{dw}{dv} = \sqrt{2}\sec^2(v)$, then:
$$(3) \implies \frac{1}{2\sqrt{2}} \int \frac{1}{w^2+1} \,dw = \frac{1}{2\sqrt{2}} \arctan(w)$$
Undo substitution, we get:
$$\int \frac{1}{2\sqrt{1-u^2}(u^2+1)} \,du = \frac{\arctan(\frac{\sqrt{2}u}{\sqrt{1-u^2}})}{2\sqrt{2}}$$
Plug it back into $(1)$ and undo substitution to get the answer.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $\frac{\sum \sqrt{xy}}{\sqrt{xyz}}-2(\sum \sqrt{x})\ge \sum \frac{\sqrt{xy}}{y}$ Let $x,y,z>0$ such that $x+y+z+2\sqrt{xyz}\le 1$. Prove that $$\frac{\sqrt{xy}+\sqrt{yz}+\sqrt{xz}}{\sqrt{xyz}}-2\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\ge \frac{\sqrt{xy}}{y}+\frac{\sqrt{yz}}{z}+\frac{\sqrt{xz}}{x}$$
My try: Let $$\left(\sqrt{x};\sqrt{y};\sqrt{z}\right)\rightarrow \left(\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}};\frac{b}{\sqrt{\left(c+b\right)\left(b+a\right)}};\frac{c}{\sqrt{\left(a+c\right)\left(b+c\right)}}\right)$$
Or we need to prove $$\sum \frac{\sqrt{\left(a+b\right)\left(a+c\right)}}{a}-2\left(\sum \frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\right)\ge \sum \frac{a\sqrt{b+c}}{b\sqrt{a+c}}$$
I tried to use C-S: $\sqrt{\left(a+b\right)\left(a+c\right)}\ge a+\sqrt{bc}$ and AM-GM but failed. Otherwise we have $cos^2 A+cos ^2 B+ cos ^2 C+2 cos A cos Bcos C=1$ then i let $x=cos^2 A$ but i cant continue.
| The inequality can be written as:
$$\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\geq 2(\sqrt{x}+\sqrt{y}+\sqrt{z})+\frac{\sqrt{x}}{\sqrt{y}}+\frac{\sqrt{y}}{\sqrt{z}}+\frac{\sqrt{z}}{\sqrt{x}}$$
From the condition and using AM-GM:
$$
\begin{aligned}
(1-\sqrt{x})(1+\sqrt{x})&=1-x\\
&\geq y+z+2\sqrt{xyz}\\
&\geq 2\sqrt{yz}+2\sqrt{xyz}\\
&=2\sqrt{yz}(1+\sqrt{x})
\end{aligned}
$$
Therefore $1\geq \sqrt{x}+2\sqrt{yz}$ and similarly $1\geq \sqrt{y}+2\sqrt{zx}, 1\geq \sqrt{z}+2\sqrt{xy}$. Thus:
$$
\begin{aligned}
\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}&\geq
\frac{\sqrt{z}+2\sqrt{xy}}{\sqrt{x}}+\frac{\sqrt{x}+2\sqrt{yz}}{\sqrt{y}}+\frac{\sqrt{y}+2\sqrt{zx}}{\sqrt{z}}
\\
&= 2(\sqrt{x}+\sqrt{y}+\sqrt{z})+\frac{\sqrt{x}}{\sqrt{y}}+\frac{\sqrt{y}}{\sqrt{z}}+\frac{\sqrt{z}}{\sqrt{x}}
\end{aligned}
$$
Equality occurs when $x=y=z=\frac{1}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3565911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the value of $k$ in this question Suppose $x_1$ and $x_2$ are solutions to $x^2+x+k=0$, if $x_1^2+x_1x_2+x_2^2=2k^2$, find the value of $k$.
$x_1+x_2 = -1$
$x_1x_2 = k$
$(x_1+x_2)^2 = x_1^2+2(x_1x_2)+x_2^2=2k^2$
$=\frac {x_1^2}{2}+x_1x_2+\frac {x_2^2}{2}=2k^2=\frac{1}{2}$
$2k^2=\frac{1}{2}$
$k^2=\frac{1}{4}$
$k=\sqrt{\frac{1}{4}}$
Could someone please confirm if this is correct and perhaps identify any errors I may have made?
| $$(x_1^2+x_2^2+x_1x_2)=(x_1+x_2)^2-x_1x_2=1-k=2k^2\implies 2k^2+k-1 \implies k=-1,1/2$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Perron vector of the adjacency matrix of complete bipartite graph $K_{p,q}$. We know that the spectral radius of the adjacency matrix of complete bipartite graph $K_{p,q}$ is $\sqrt{pq}$. Also note that the perron vector corresponding to the spectral radius has positive entries and is of the form $$(a,a,\cdots,a,b,b,\cdots,b)$$ where $a$ is the entry for the partition with vertex set with cardinality $p$ and $b$ for $q$.
Is there any known form how $a$ and $b$ looks like?
| I think that $a= \sqrt{q}$, $b= \sqrt{p}$ works:
\begin{align*}
A_{p,q}v_{\sqrt{pq}} &= \phantom{\sqrt{pq}} \left[\begin{array}{cccccc}
0& \dots & 0 & 1 & \dots& 1 \\
\vdots& & & && \vdots \\
0& \dots & 0 & 1 & \dots& 1 \\
1& \dots & 1 & 0 & \dots& 0 \\
\vdots& & & && \vdots \\
1& \dots & 1 & 0 & \dots& 0 \\
\end{array}\right]
\begin{bmatrix}
\sqrt{q} \\
\vdots \\
\sqrt{q} \\
\sqrt{p} \\
\vdots \\
\sqrt{p} \\
\end{bmatrix} \\
{} &= \phantom{\sqrt{pq}}\begin{bmatrix}
q\sqrt{p} \\
\vdots \\
q\sqrt{p} \\
p\sqrt{q} \\
\vdots \\
p\sqrt{q} \\
\end{bmatrix} \\
{} &= \sqrt{pq} \begin{bmatrix}
\sqrt{q} \\
\vdots \\
\sqrt{q} \\
\sqrt{p} \\
\vdots \\
\sqrt{p} \\
\end{bmatrix} \\
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Convergence In L2 norm Given that
\begin{align*}
&u_n(x) = \frac{x}{\sqrt{x^2+1/n}}&\\
&\text{and }&\\
&v(x) = \left\{ \begin{array}{cc}
\frac{|x|}{x} & \hspace{5mm} x\neq 0 \\
0 & \hspace{5mm} x=0 \\
\end{array} \right.&
\end{align*}
Show that $u_n \to v$ with respect to $L^2$ norm as $n \to \infty$.
My Attempt!
Since $u^{}_n(-x)=-u^{}_n(x)$ and $v(-x)=-v(x)$, and Define
\begin{align*}
&g_n(x) := |u_n^{}(x)-v(x)|&.\\
&\text{Noticing that $g_n(x)$ is even.}&\\
\end{align*}
And
\begin{align*}
&\|u_n(x)-v(x)\|^2 = \|g_n(x)\|^2=\int_{-1}^{1}|g_n(x)|^2dx&\\
&=2\int_{0}^{1}|g_n(x)|^2dx=2\int_{0}^{1}\left|\frac{x}{\sqrt{x^2+(1/n)}}-1\right|^2dx&\\
%&=2\int_{0}^{1}|g_n(x)|^2dx=2\int_{0}^{1}\left|\frac{x}{\sqrt{x^2+(1/n)}}-1\right|^2dx&\\
&=2\int_{0}^{1}\left|\frac{x-\sqrt{x^2+(1/n)}}{\sqrt{x^2+(1/n)}}\right|^2dx&\\
&= 2\int_{0}^{1}\left|\frac{-1/n}{\sqrt{(x^2+1/n)}(x+\sqrt{x^2+(1/n)})}\right|^2dx&\\
&\leq 2\int_{0}^{1}\left|\frac{-1/n}{\sqrt{(x^2+1/n)}}\right|^2dx&\\
&= 2\int_{0}^{1}\frac{1}{n^2} \frac{1}{{(x^2+1/n)}}dx&\\
\end{align*}
The inequality is obtained by dropping the term $x+\sqrt{x^2+1/n}$ from the denominator of preceding term.
I am little hesitant to drop this term because it depends on $n$.
Can any body tell me if I am doing it right so far?
| No, since
$$ \frac{1}{x + \sqrt{x^2 + 1/n}} > 1 $$
for $n$ large and $x \in [0, 1]$ small. I think the easiest proof is probably by the dominated convergence theorem, as $u_n \rightarrow v$ pointwise and $|u_n| \leq 1$. But if you want to proceed in this way, you can. Since
$$ \frac{1}{x + \sqrt{x^2 + 1/n}} \leq \frac{1}{\sqrt{x^2 + 1/n}}, $$
using Mathematica I find:
$$ ||g_n(x)||^2 \leq \frac{2}{n^2}\int_0^1 \frac{dx}{(x^2+1/n)^2} \xrightarrow{n\to\infty} 0.$$
Although that integral, which I have omitted if you want to evaluate it yourself, seems rather complicated.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to solve modular equivalence with polynomial I'm finding the solutions for the modular equation $x^2 -3x+2 \equiv 0\pmod{14}$. This is what I've done so far:
\begin{align}
0 & \equiv x^2 -3x+2 \pmod{14}\\
& \equiv (x-1)(x-2) \pmod{14} \\
\end{align}
This implies that two of the solutions are the polynomial's usual roots: 1, 2.
This is where I get lost. I realize I should be using the following fact to find the other solutions, but don't know how to approach it.$$(x-1)(x-2) \overset{14}{\equiv} 0 \Leftrightarrow 14|(x-1)(x-2)$$ The provided answer says that 8,9 are the other solutions. How would I go about finding these?
| If $14|(x-1)(x-2)$, we cannot conclude $14|x-1$ or $14|x-2$, because $14$ is not prime.
But $14=2\times7$, so $14|(x-1)(x-2)$ means $2|(x-1)(x-2)$ and $7|(x-1)(x-2)$.
Since $2$ and $7$ are prime, this means ($2|x-1 $ or $2|x-2$) and ($7|x-1$ or $7|x-2$).
Thus, there are four solutions (mod $14$). Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does $\int_{0}^{\infty} \frac{2x +3}{\sqrt {x^3 + 2x + 5}} \,dx $ converge? Does the following integral converge? I will post my solution, but I am unsure if it is true.
$$\int_{0}^{\infty} \frac{2x +3}{\sqrt {x^3 + 2x + 5}} \,dx $$
My solution
Let
$$ g(x) = \frac{2x}{\sqrt {x^3}}$$
$$ f(x) = \frac{2x +3}{\sqrt {x^3 + 2x + 5}} $$
Then $$\lim_{k \to \infty} \frac{f(x)}{g(x)} = 1$$
Therefore whatever one does, so does the other.
$$ \int_{0}^{\infty} g(x) = \int_{0}^{\infty} \frac{2}{\sqrt {x}} = +\infty $$
Therefore g(x) diverges, thus
$$\int_{0}^{\infty} f(x) = \int_{0}^{\infty} \frac{2x +3}{\sqrt {x^3 + 2x + 5}} \,dx = + \infty$$
diverges too
| Perhaps redundant:
Let $x \ge 2$.
$\dfrac{x}{\sqrt{3x^3}}\lt \dfrac{2x+3}{\sqrt{x^3+2x+5}}$, or
$0<g(x):=(1/√3)x^{-1/2} <$
$\dfrac{2x+3}{\sqrt{x^3+2x+5}}=:f(x);$
$\int_{2}^{\infty} g(x)dx$ diverges, so does $\int_{2}^{\infty}f(x)dx$ (Monotonicity of integral).
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
solve for $(\cos x)(\cos2x)(\cos3x)=\frac{1}{4}$ solve for $(\cos x)(\cos2x)(\cos3x)=\frac{1}{4}$. what is the general solution for $x$.
I wrote $\cos2x$ as $2\cos^2x-1$ and $\cos3x$ as $4\cos^3x-3\cos x$. Then the expression gets reduced to $\cos x(2\cos^2x-1)(4\cos^3x-3\cos x)=\frac{1}{4}$. substituted $\cos x=t$ and got
$t(2t^2-1)(4t^3-3t)=\frac{1}{4}$. how do i proceed further?
| $$4\cos x\cos3x\cos2x=2\cos2x(\cos2x+\cos4x)=2\cos2x(\cos2x+2\cos^22x-1)$$
$$\implies0=4\cos^32x+2\cos^22x-2\cos2x-1$$
$$=2\cos^22x(2\cos2x+1)-(2\cos2x+1)$$
$$=(2\cos2x+1)(2\cos^22x-1)$$
If $2\cos2x+1=0,$
$$\cos2x=-\cos\dfrac\pi3=\cos\left(\pi-\dfrac\pi3\right)$$
Else $0=2\cos^22x-1=\cos4x,4x=(2n+1)\dfrac\pi2$ where $n$ is any integer
| {
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"url": "https://math.stackexchange.com/questions/3570048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Given that $3^{15a} = 5^{5b} = 15^{3c}$, show that $5ab-bc-3ac=0$ Given that $3^{15a} = 5^{5b} = 15^{3c}$, show that $5ab-bc-3ac=0$
The only thing I can do is:
$3^{5a} = 5^{b} = 15^{\frac{3}{5}c}$
and then i am stuck, I figure that there must be a relationship between 3 and 5? should i utilise $5ab-bc-3ac=0$?
| To get the variables directly take logarithms. (Doesn't matter which base).
$\log 3^{15a} = \log 5^{5b} = \log 15^{3c}$ so
$15a \log 3 = 5b \log 5 = 3c \log 15$.
So flip a coin and choose which one we should express the others in. I pick ... $b$.
$a = \frac {\log 5}{3\log 3} b$ and $c = \frac {5\log 5}{3\log 15} b$
So
$5ab - bc -3ac =$
$\frac {5\log 5}{3\log 3}b^2 - \frac {5\log 5}{3\log 15}b^2 - \frac {15(\log 5)^2}{9\log 3\log 15} b^2 = $
$b^2 (\frac {5\log 5}{3\log 3} - \frac {5\log 5}{3\log 15} - \frac {15(\log 5)^2}{9\log 3\log 15} )$
Which reduces to proving $\frac {5\log 5}{3\log 3} - \frac {5\log 5}{3\log 15} - \frac {15(\log 5)^2}{9\log 3\log 15} = 0$
Which requires comfort with logs.
$\frac {5\log 5}{3\log 3} - \frac {5\log 5}{3\log 15} - \frac {15(\log 5)^2}{9\log 3\log 15}=$
$\frac {5\log 5\log 15 - 5\log 5\log 3}{3\log 3 \log 15}-\frac{5(\log 5)^2}{3\log 3\log 15}=$
$\frac {5\log 5\log 15 - 5\log 5\log 3-5(\log 5)^2}{3\log 3\log 15}=$
$\frac {5[\log 5(\log 3 + \log 5) -\log 5\log 3 -(\log 5)^2] }{3\log 3\log 15}=$
$\frac {5[\log 5\log 3 + (\log 5)^2 - \log 5\log3 - (\log 5)^2]}{3\log 3 \log 15}=$
$\frac {5\cdot 0}{3\log 3 \log 15}=0$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find $f(x)$ such that: $ f'(x) + f(x^2) = 2x + 1 $ How to find the function $f(x)$ that is derivable on $\mathbb{R}$ and satisfies the equation:
$$ f'(x) + f(x^2) = 2x + 1 \text{ } \text{ } \forall x \in \mathbb{R}$$
My attempt:
Substitute $-x$ for $x$ in the equation, we have a system of 2 equations :
\begin{cases}
f'(x) + f(x^2) &= 2x + 1\\
f'(-x) + f(x^2) &= -2x + 1
\end{cases}
Take the difference of them, we have:
$$ f'(x) - f'(-x) =4x $$
Integrate both sides:
$$ f(x) + f(x) = 2x^2 + C_1 $$
Therefore, we have:
$$ f(x) = x^2 + C_2 $$
However, this method results in an invalid solution to the original equation. I wonder whether there is another way to solve this problem or why my solution is wrong.
Thanks in advance.
| We can write a power series for $f(x)$ as $$f(x) = \sum_{n=0}^\infty a_nx^n$$
Then we get that $$f'(x) + f(x^2) = \sum_{n=0}^{\infty}(n+1)a_{n+1}x^{n} + \sum_{n=0}^{\infty} a_nx^{2n} = 2x+1$$
Breaking this into odd and even powers of $x$, this can be rewritten as $$\sum_{n=0}^{\infty}((2n+1)a_{2n+1} + a_n)x^{2n} + \sum_{n=0}^{\infty}(2n+2)a_{2n+2}x^{2n+1} = 2x+1$$
From the second sum, we can get that $(2n+2)a_{2n+2} = 0 \to a_{2n+2} = 0$ for $n \ge 1$ and $a_2 = 1$. Then from the first sum, we have these relations.
$$a_1 + a_0 = 1$$
$$3a_3 + a_1 = 0$$
$$5a_5 + a_2 = 0$$
$$7a_7 + a_3 = 0$$
$$9a_9 + a_4 = 0$$
$$...$$
Because $a_4, a_6, a_8, a_{10}, ...$ are $0$, it is also true that $a_9, a_{13}, a_{17}, a_{21}, ..$ are also $0$. Because those are $0$, it is also true that $a_{19}, a_{27}, a_{35}, ...$ are also $0$. By induction, the only ones that are not $0$ are the powers of $x$ where the exponent is either of the form $2^n-1, n \ge 0$ or $3 \cdot 2^n - 1, n \ge 0$.
From $a_2 = 1$, we can get that $a_5 = -1/5$, which leads to $a_{11} = 1/55$. In general, $$a_{3 \cdot 2^n - 1} = \frac{(-1)^n}{\prod_{k=1}^{n} (3 \cdot 2^k - 1)}$$
Similarly, given $a_1$, we can get that $a_0 = 1-a_1$, and $$a_{2^n-1} = \frac{(-1)^{n+1}}{\prod_{k=1}^n (2^k - 1)} a_1 $$
From this, we have that $$f(x) = (1-a_1) + x^2 + \sum_{n=1}^{\infty} \frac{(-1)^n}{\prod_{k=1}^{n} (3 \cdot 2^k - 1)} x^{{3 \cdot 2^n - 1}} + a_1\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{\prod_{k=1}^n (2^k - 1)} x^{2^n-1}$$
$a_1$ can be anything. Note that the required functional equation only holds true for $-1 < x < 1$ for my $f(x)$, as it only converges for $-1 < x < 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Trig substitution for $\sqrt{9-x^2}$ I have an integral that trig substitution could be used to simplify.
$$ \int\frac{x^3dx}{\sqrt{9-x^2}} $$
The first step is where I'm not certain I have it correct. I know that, say, $\sin \theta = \sqrt{1-cos^2 \theta}$, but is it correct in this case $3\sin \theta = \sqrt{9 - (3\cos \theta)^2}$?
Setting then $x = 3\cos \theta; dx = -3\sin \theta d\theta$
$$-\int \frac{(3\cos\theta)^3}{3\sin\theta}3\sin\theta d\theta$$
$$-27\int\cos^3\theta d\theta$$
$$-27\int(1-\sin ^2\theta)\cos \theta d\theta$$
Substituting again, $u=\sin \theta; du=\cos \theta d\theta$
$$-27\int(1-u^2)du $$
$$-27u + 9u^3 + C$$
$$-27\sin \theta + 9 \sin^3 \theta + C$$
$$-9\sqrt{9-x^2} + 3\sin\theta\cdot 3\sin\theta\cdot \sin \theta + C$$
$$-9\sqrt{9-x^2} + (\sqrt{9-x^2})^2 \cdot \frac{\sqrt{9-x^2}}{3} + C$$
$$-9\sqrt{9-x^2} + \frac{1}{3}(9-x^2)(9-x^2)^{\frac{1}{2}} + C$$
$$-9\sqrt{9-x^2} + \frac{1}{3}(9-x^2)^\frac{3}{2} + C $$
I guess I have more doubts that I've done the algebra correctly than the substitution, but in any case I'm not getting the correct answer. Have I calculated correctly? Is the answer simplified completely?
EDIT
Answer needed to be simplified further:
$$-9\sqrt{9-x^2} + \frac{1}{3}(\sqrt{9-x^2}^2 \sqrt{9-x^2}) + C$$
$$-9\sqrt{9-x^2} + \frac{1}{3}((9-x^2)\sqrt{9-x^2}) + C$$
$$\sqrt{9-x^2} \left (-9 + \frac{1}{3}(9-x^2) \right ) + C$$
$$\sqrt{9-x^2} \left (-6 - \frac{x^2}{3} \right ) + C$$
$$ \bbox[5px,border:2px solid red]
{
- \left ( 6+ \frac{x^2}{3} \right ) \sqrt{9-x^2}
}
$$
This is the answer the assignment was looking for.
| You work is correct, You may further simplify
$$I=-9\sqrt{9-x^2}+\frac{1}{3}\left(\sqrt{9-x^2}\right)^3 +C= \sqrt{9-x^2} \left (-9+\frac{9-x^2}{3}\right)+C$$
$$\implies I=-\frac{1}{3}\sqrt{9-x^2}~~(18+x^2)+C,$$
which is the final correct answer.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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prove that: $\sqrt{2}=\frac{F_n^2+F_{n+1}^2+F_{n+2}^2}{\sqrt{F_n^4+F_{n+1}^4+F_{n+2}^4}}$ Pythagoras's constant in Fibonacci number!
How do I show that?
$$\sqrt{2}=\frac{F_n^2+F_{n+1}^2+F_{n+2}^2}{\sqrt{F_n^4+F_{n+1}^4+F_{n+2}^4}}\tag1$$
Where $F_n$ is Fibonacci sequence.
$$2(F_n^4+F_{n+1}^4+F_{n+2}^4)^2=F_n^2+F_{n+1}^2+F_{n+2}^2$$
$$2(F_n^8+F_{n+1}^8+F_{n+2}^8+2F_n^2F_{n+1}^2+2F_n^2F_{n+2}^2+F_{n+1}^2F_{n+2}^2)=F_n^2+F_{n+1}^2+F_{n+2}^2$$
This is getting too messy. I hope there is an easy way of proving this formula.
| Using the same notation as Matteo $x=F_{n}$ and $y=F_{n+1}$ we have \begin{eqnarray}F_n^4+F_{n+1}^4+F_{n+2}^4&=&x^4+y^4+(x+y)^4 \\
&=& 2x^4+4x^3y+6x^2y^2+4xy^3+2y^4\\
&=&2(x^4+2x^3y+3x^2y^2+2xy^3+y^4)\\
&=&2(x^2+y^2+xy)^2
\end{eqnarray}
On the other hand we have \begin{eqnarray}F_n^2+F_{n+1}^2+F_{n+2}^2&=& x^2+y^2+(x+y)^2\\ &=& 2(x^2+xy+y^2)\end{eqnarray} and we are done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Isosceles triangle $ABC$ with an inside point $M$, find $\angle BMC$ We have an isosceles $\triangle ABC, AC=BC, \measuredangle ACB=40^\circ$ and a point $M$ such that $\measuredangle MAB=30^\circ$, $\measuredangle MBA=50^\circ$.
Find $\measuredangle BMC$.
Starting with $\angle ABC=\angle BAC=70^\circ \Rightarrow \angle CBM=20 ^\circ$. Let us construct the equilateral $\triangle ABH$. If we look at $\triangle ACH, \angle ACH=20^\circ$ and $\angle CAH=10^\circ$. Can we show $\triangle AHC \cong CHB$? Any other ideas?
| Here's a trigonometric approach. Let $\angle BCM=\varphi\Rightarrow \angle ACM=40^{\circ}-\varphi$. Apply the law of sines in $\triangle AMC$ and $\triangle BMC$:
$$\frac{AC}{CM}=\frac{\sin(80^{\circ}-\varphi)}{\sin(40^\circ)} \\ \frac{BC}{CM}=\frac{\sin(20^{\circ}+\varphi)}{\sin(20^\circ)} $$
Since $AC=BC$, the two ratios with the sines are equal. We have $\sin(40^\circ)=2\sin(20^\circ)\cos(20^\circ)$, so
$$\frac{\sin(80^{\circ}-\varphi)}{2\cos(20^\circ)}=\sin(20^\circ+\varphi) \Leftrightarrow \\ \sin(80^{\circ}-\varphi)=2\sin(20^\circ+\varphi)\cos(20^\circ)$$ Then use the sum-product identities:
$$\sin(80^{\circ}-\varphi)=\sin(\varphi)+\sin(\varphi+40^\circ) \Leftrightarrow \\
\sin(\varphi)=\sin(80^{\circ}-\varphi)-\sin(\varphi+40^\circ) \Leftrightarrow \\
\sin(\varphi)=2\sin(20^\circ-\varphi)\cos(60^\circ) \Leftrightarrow \\
\sin(\varphi)=\sin(20^\circ-\varphi) $$
Since $0<\varphi<40^{\circ}$, the last equality implies $\varphi=20^\circ-\varphi\Leftrightarrow \varphi=10^{\circ}$, and we find $\angle BMC=150^{\circ}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $(x^2-5x+4)(y^2+y+1)<2y$ for all real $y$, then $x$ belongs to the interval $(2,b)$, then $b$ can be? If $(x^2-5x+4)(y^2+y+1)<2y$ for all real $y$, then $x$ belongs to the interval $(2,b)$, then $b$ can be?
$$y^2(x^2-5x+4)+y(x^2-5x+2)+(x^2-5x+4)<0$$
As it is true for all real y, hence $D<0$
$$(x^2-5x+2)^2-4(x^2-5x+4)^2<0$$
Let $x^2-5x=u$
$$(u+2)^2-4(u+4)^2<0$$
$$u^2+4+4u-4(u^2+16+8u)<0$$
$$-3u^2-28u-60<0$$
$$3u^2+28u+60>0$$
$$3u^2+18u+10u+60>0$$
$$(3u+10)(u+6)>0$$
Back substituting u
$$(3x^2-15x+10)(x^2-5x+6)>0$$
$$\left(x-\dfrac{15+\sqrt{225-120}}{6}\right)\left(x-\dfrac{15-\sqrt{225-120}}{6}\right)(x-2)(x-3)>0$$
$$x\in\left(-\infty,\dfrac{15+\sqrt{105}}{6}\right)\cup(2,3)\cup\left(\dfrac{15+\sqrt{105}}{6},\infty\right)$$
But in the question it is given $x\in(2,b)$, what mistake am I making here.
| It is only necessary that the discriminant is less than $0$; not sufficient. We also need that the leading coefficient is $<0$ (otherwise the quadratic is $>0$ for all $y$). Thus, we also require
$$x^2-5x+4<0\implies 1<x<4.$$
Intersecting this with what you got gives the interval $(2,3)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Compute $\left[\begin{smallmatrix}1-a & a \\ b & 1-b\end{smallmatrix}\right]^n$ Compute $\begin{bmatrix}1-a & a \\ b & 1-b\end{bmatrix}^n$, where the power of $n\in\mathbb N$ denotes multiplying the matrix by itself $n$ times; $a,b\in[0,1]$.
Edit:
I considered using induction, computed the desired matrix:
$$\begin{bmatrix}(1-a)^2+ab & a(2-a-b) \\ b(2-a-b) & (1-b)^2+ab\end{bmatrix}$$ and $$\begin{bmatrix}(1-a)^3+ab(1-a)+ba(2-a-b) & a(1-a)^2+a^2b+a^2(2-a-b) \\ (1-a)b(2-a-b)+b(1-b)^2+ab^2 & ab(2-a-b)+(1-b)^3+ab(1-b)\end{bmatrix}$$ for $n=2$ and $3$, respectively. However, I fail to see a pattern here that can be used as an induction hypothesis.
| Note that the characteristic polynomial of $A$ is given by
$$\begin{align}
\chi_A(\lambda) &= \lambda^2 - (2-a-b)\lambda + 1-a-b \\
&= \left( \lambda - \left( 1 - \frac{a+b}{2} \right) \right)^2 - \left( \frac{a+b}{2} \right)^2
\end{align}$$
so the eigenvalues of $A$ are $1-\dfrac{a+b}{2} \pm \dfrac{a+b}{2}$, i.e. $1$ and $1-a-b$.
Use these to find an invertible matrix $P$ such that $A = P \begin{pmatrix} 1 & 0 \\ 0 & 1-a-b \end{pmatrix} P^{-1}$; the columns of $P$ are eigenvectors of $A$.
When you've done this, you'll have
$$A^n = P \begin{pmatrix} 1 & 0 \\ 0 & 1-a-b \end{pmatrix}^n P^{-1} = P \begin{pmatrix} 1 & 0 \\ 0 & (1-a-b)^n \end{pmatrix} P^{-1}$$
and then expanding the matrix product gives a closed-form expression for $A^n$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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On the decomposition of $1$ as the sum of Egyptian fractions with odd denominators Suppose that we decompose $1$ as a sum of Egyptian fractions with odd denominators.
I noticed (from a cursory view) that the fraction
$$\frac{1}{3}$$
appears in each of such decompositions.
Questions
Must the fraction $1/3$ appear in each such decomposition? Is it possible to prove this? Or is there a counterexample?
| $1 = \frac{1}{5} + \frac{1}{7} + \frac{1}{9} + \frac{1}{11} + \frac{1}{13} + \frac{1}{15} + \frac{1}{17} + \frac{1}{19} + \frac{1}{21} + \frac{1}{23} + \frac{1}{25} + \frac{1}{27} + \frac{1}{33} + \frac{1}{611} + \frac{1}{265525} + \frac{1}{97544139723} + \frac{1}{8457652617058141652925}
$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Stuck in a finite summation in a physics problem.r³/(r-1)² Actually it was not a hard problem as seen from physics point of view, just some application of Coulomb's law (multiple charge system) but it finally ended up as the summation of:-
$$
\sum_{r=2}^n \frac{(r+1)^3}{r^2}
$$
I tried to open things up but it finally took me to summation of $\frac{1}{r^2}$ and $\frac{1}{r}$, and I know they can be solved but it's too much for my level I guess (as I am a JEE aspirant). Is there any other way to solve the summation?
| We have:
$$\frac{(r+1)^3}{r^2}=\frac{r^3+3r^2+3r+1}{r^2}=r+3+\frac{3}{r}+\frac{1}{r^2}$$
Thus:
$$\sum_{r=2}^n \frac{(r+1)^3}{r^2}=\sum_{r=2}^n \bigg( r+3+\frac{3}{r} \frac{1}{r^2} \bigg)$$
$$\sum_{r=2}^n \frac{(r+1)^3}{r^2}=\sum_{r=2}^n r +\sum_{r=2}^n 3 +3 \sum_{r=2}^n \frac{1}{r} + \sum_{r=2}^n \frac{1}{r^2}$$
$$\sum_{r=2}^n \frac{(r+1)^3}{r^2}=\bigg( \frac{n(n+1)}{2}-1 \bigg) +3(n-1) +3 \sum_{r=2}^n \frac{1}{r} + \sum_{r=2}^n \frac{1}{r^2}$$
$$\sum_{r=2}^n \frac{(r+1)^3}{r^2}=\frac{n^2+7n-8}{2}+(3H_n-3)+(H_n^{(2)}-1)=\frac{n^2+7n-16}{2}+3H_n+H_n^{(2)}$$
Now, it is not possible to get the exact values for $H_n$ and $H_n^{(2)}$, but for large $n$, you can approximate:
$$\lim_{n \to \infty} H_n - \ln{n} = \gamma = 0.577 \ldots \implies H_n \approx \ln{n}+ \gamma$$
$$\lim_{n \to \infty} H_n - \frac{\pi^2}{6}=0 \implies H_n^{(2)} \approx \frac{\pi^2}{6}$$
Hence, we have:
$$\sum_{r=2}^n \frac{(r+1)^3}{r^2} \approx \frac{n^2+7n-16}{2}+3 \ln{n} + 3 \gamma + \frac{\pi^2}{6}$$
As $n$ gets really large, the difference between the LHS and RHS tends to $0$. This shows that we have a really good approximation.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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On the integral $\int_0^{\sqrt{2}/2} \frac{\arctan \sqrt{1-2t^2}}{1+t^2} \, \mathrm{d}t$ I'm having a difficult time evaluating the integral
$$\mathcal{J} = \int_0^{\sqrt{2}/2} \frac{\arctan \sqrt{1-2t^2}}{1+t^2} \, \mathrm{d}t$$
This is integral arose after simplifying the integral $\displaystyle \int_{0}^{\pi/4 } \arctan \sqrt{\frac{1-\tan^2 x}{2}} \, \mathrm{d}x$;
\begin{align*}
\require{cancel.js}
\int_{0}^{\pi/4} \arctan \sqrt{\frac{1-\tan^2 t}{2}}\, \mathrm{d}t &\overset{1-\tan^2 t \mapsto 2t^2}{=\! =\! =\! =\! =\! =\!=\!=\!} \int_{0}^{\sqrt{2}/2} \frac{t \arctan t}{\sqrt{1-2t^2} \left ( 1-t^2 \right )} \, \mathrm{d}t \\
&=\cancelto{0}{\left [ - \arctan \sqrt{1-2t^2} \arctan t \right ]_0^{\sqrt{2}/2}} + \int_{0}^{\sqrt{2}/2} \frac{\arctan \sqrt{1-2t^2}}{1+t^2} \, \mathrm{d}t
\end{align*}
My main guess is that differentiation under the integral sign is the way to go here. Any ideas?
| This answer is based on Feynman's trick. Put
\begin{equation*}
I(a) = \int_{0}^{\pi/4}\arctan\left(a\sqrt{\dfrac{1-\tan^2 x}{2}}\right)\, dx .
\end{equation*}
Then
\begin{gather*}
I'(a) = \int_{0}^{\pi/4}\dfrac{1}{1+a^2\dfrac{1-\tan^2 x}{2}}\cdot \sqrt{\dfrac{1-\tan^2 x}{2}} \, dx = \\[2ex]\int_{0}^{\pi/4}\dfrac{1}{1+a^2\dfrac{\cos 2x}{1+\cos 2x}}\cdot \sqrt{\dfrac{\cos 2x}{1+\cos 2x}} \, dx =
[y=\cos 2x]\\[2ex]
= \dfrac{1}{2}\int_{0}^{1}\dfrac{1}{1+(a^2+1)y}\cdot\sqrt{\dfrac{y}{1-y}}\,dy= \left[z=\sqrt{\dfrac{y}{1-y}}\right] =\\[2ex]
\dfrac{1}{2}\int_{-\infty}^{\infty}\dfrac{z^2}{(1+(a^2+2)z^2)(z^2+1)}\, dz = [\mbox{ residue calculus }]=\\[2ex]
\dfrac{\pi}{2}\left(\dfrac{1}{a^2+1}-\dfrac{1}{(a^2+1)\sqrt{a^2+2}}\right)
\end{gather*}
Finally we get
\begin{gather*}
I(1)=I(1)-I(0)=\int_{0}^{1}I'(a)\, da =\dfrac{\pi}{2}\left[\arctan a -\arctan\dfrac{a}{\sqrt{a^2+2}}\right]_{0}^{1} =\dfrac{\pi^2}{24}.
\end{gather*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the locus of the complex number from the given data If $|\sqrt 2 z-3+2i|=|z||\sin (\pi/4+\arg z_1) + \cos (3\pi/4-\arg z_1)|$, where $z_1=1+\frac{1}{\sqrt 3} i$, then the locus of $z$ is ...
$$\arg~z_1 =\frac{\pi}{6}$$
So RHS of the equation is $\;|z|\dfrac{1}{\sqrt 2}$.
Then
$\;|\sqrt 2 z-3+2i|=|z|\dfrac{1}{\sqrt 2}$.
How should I proceed?
| Continue with $|\sqrt 2 z-3+2i|=\frac{1}{\sqrt 2}|z|$ to get,
$$(\sqrt 2 z-3+2i) (\sqrt 2 \bar z-3-2i) =\frac{1}{2}|z|^2$$
or,
$$3|z|^2 -2\sqrt2 (3-2i) \bar z -2\sqrt2 (3+2i) z +26=0$$
which can be written in the form of a circle, i.e.
$$\bigg| z - \frac{2\sqrt2}3(3-2i) \bigg|^2 = \frac{26}9$$
Thus, the locus is a circle with the center at $\ \frac{2\sqrt2}3(3-2i)$ and the radius $\frac{\sqrt{26}}3$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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For non-negative reals $x,y$ such that $x+y\le 4$ prove that $y(x-3)(y-3) \le 3(4-y)$ For non-negative reals $x,y$ such that $x+y\le 4$ prove that $y(x-3)(y-3) \le 3(4-y)$
ATTEMPT
We transform the equation into $xy^2+12y-3xy-3y^2\le 12$
I noticed that for $x=0, y=2$ equality is achieved, but I am lost here.
| I am interested in elementary solution
$$
\begin{aligned}
(y-2)^{2}&\geq 0\\
y^{2}-4y+4&\geq 0\\
4-y&\geq y(3-y)\\
\\
x&\geq 0\\
3&\geq 3-x
\end{aligned}
$$
Only one of $y(3-y)$ and $3-x$ can be negative, thus $3(4-y)\geq y(3-x)(3-y)$. from the inequalities, equality is when $y=2$ and $x=0$
| {
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In triangle $ABC$, $AA_1$, $BB_1$, $CC_1$ divide sides in ratio of $1: 2$ and meet at $M$, $K$, $L$. Find area relation of $KLM$ and $ABC$
Points $A_1$, $B_1$, $C_1$ divide sides $BC$, $CA$, $AB$ equilateral triangle $ABC$ in a ratio of $1: 2$.
The line segments $AA_1$, $BB_1$, $CC_1$ determine the triangle $KLM$.
Is the triangle $KLM$ also an equilateral side? In what relation are the area of triangle $KLM$ the area of triangle$ ABC$?
My attempt:
I can see that $KLM$ is an equilateral triangle. But, why the fraction $\frac{1}{7}$ is the area ratio of the triangle $KLM$ to the triangle $ABC$?
| Here is a proof without using Menelaus's thm. Let $A_2$, $B_2$, and $C_2$ be the midpoints of $CA_1$, $AB_1$, and $BC_1$, resp. Draw $MA_1$, $KA_2$, $KB_1$, $LB_2$, $LC_1$, $LC_2$, $MA_2$, $KB_2$, and $LC_2$. In this proof, $[\mathcal{P}]$ is the area of a polygon $\mathcal{P}$.
Due to symmetry $\triangle KLM$ is equilateral. Therefore $\angle C_1LA=\angle KLM=60^\circ$. Since $AC_1=AB_2$, $\angle C_1B_2A=60^\circ$. Therefore $\square AC_1LB_2$ is cyclic so we see that $\angle B_2LA=\angle B_2C_1A=60^\circ$. That is $\angle KLB_2=180^\circ-\angle KLM-\angle B_2LA=60^\circ$.
By symmetry $\angle CKB_1=\angle C_1LA=60^\circ$ too. Hence $\angle CKB_1=\angle KLB_2$, so $KB_1\parallel LB_2$. Since $CB_1=B_1B_2$, we get $LK=KC$.
Turn $\triangle KB_1C$ around $C$ in the counterclockwise direction until $CB_1$ coincides with $CA_2$. Suppose $K_c$ is the image of $K$ under this transformation. Then $\triangle CKK_c$ is equilateral with side length $KC=KL$. This shows that
$$[CB_1KA_2]=[CKK']=[KML].$$
Now $[KB_1B_2]=[KB_1C]$ because $\triangle KB_1B_2$ and $\triangle KB_1C$ share the same altitude from $K$ and have the same base legth $B_1B_2=B_1C$. Furthermore $\triangle KLB_2\cong \triangle CKA_2$ because $KL=CK$, $\angle KLB_2=60^\circ = \angle CKA_2$, and $LB_2=KB_2$ (by symmetry). Therefore
$$[KB_1B_2L]=[KB_1B_2]+[KLB_2]=[KB_1C]+[CKA_2]=[CB_1KA_2].$$
By symmetry, we have
$$[AC_1LB_2]=[BA_1MC_2]=[CB_1KA_2]$$
and
$$[MA_1A_2K]=[KB_1B_2L]=[LC_1C_2M].$$
Therefore $[KLM]=\frac{1}{7}[ABC]$.
Here is another argument why $\triangle KML$ is equilateral (which is not just saying "due to symmetry"). Note that $\triangle ABA_1$ is obtained by a rotation of $\triangle CAC_1$ about the center of $\triangle ABC$ by $60^\circ$ in the counterclockwise direction. Therefore $CC_1$ and $AA_1$ make an angle of $60^\circ$. Hence $\angle KLM=60^\circ$. By the same argument $\angle LMK=60^\circ$ and $\angle MKL=60^\circ$ too.
If you want to use a Menelaus argument, you can do it like this. From Menelaus's thm (on $\triangle ABA_1$ with the traversal $C_1LC$)
$$\frac{AC_1}{C_1B}\cdot\frac{BC}{CA_1}\cdot \frac{A_1L}{LA}=-1.$$
Here the length ratios are signed (the ratio of lengths measured in the same direction is positive, and the ratio of lengths measured in the opposite directions is negative). That is
$$\frac{1}{2}\cdot\left(-\frac{3}{2}\right)\cdot\frac{AL_1}{LA}=-1\implies \frac{AL_1}{LA}=\frac{4}{3}.$$
Using Menelaus's thm on $\triangle AA_1C$ with traversal $BMB_1$ yields
$$\frac{AM}{MA_1}\cdot \frac{A_1B}{BC}\cdot\frac{CB_1}{B_1A}=-1.$$
Hence
$$\frac{AM}{MA_1}\cdot\left(-\frac{1}{3}\right)\cdot\frac{1}{2}=-1\implies \frac{A_1M}{MA}=\frac{1}{6}.$$
This shows that $A_1M:ML:LA=1:3:3$.
We have
$$\frac{[KLM]}{[LA_1C]}=\frac{LM}{LA_1}\cdot \frac{LK}{LC}.$$
By symmetry $\frac{LK}{LC}=\frac{ML}{MA}=\frac12$, so
$$\frac{[KLM]}{[LA_1C]}=\frac{3}{3+1}\cdot\frac{1}{2}=\frac{3}{8}.$$
Similarly
$$\frac{[LA_1C]}{[C_1BC]}=\frac{CA_1}{CB}\cdot\frac{CL}{CC_1}=\frac{2}{3}\cdot\frac{3+3}{1+3+3}=\frac{4}{7}.$$
Finally
$$\frac{[C_1BC]}{[ABC]}=\frac{BC_1}{BA}=\frac{2}{3}.$$
Therefore
$$\frac{[KLM]}{[ABC]}=\frac{[KLM]}{[LA_1C]}\cdot \frac{[LA_1C]}{[C_1BC]}\cdot \frac{[C_1BC]}{[ABC]}=\frac{3}{8}\cdot\frac{4}{7}\cdot\frac{2}{3}=\frac17.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Convergence of $S(x,y)=\sum_{n=1}^{\infty}\left(\frac{x+2n+1}{(x+2n+1)^2+y^2}+\frac{x+2n}{(x+2n)^2+y^2}-\frac{x+n}{(x+n)^2+y^2}\right)$ I know that the following sum is convergent and equal to $\ln2$ for all $x\not\in\mathbb{Z}$
$$\sum_{n=0}^{\infty}\left(\frac{1}{x+2n+1}+\frac{1}{x+2n}-\frac{1}{x+n}\right)-\ln2=0$$
Can I somehow quickly prove that the following sum is also convergent for $x\not\in\mathbb{Z}$ and $y\in\mathbb{R}$ ?
$$S(x,y)=\sum_{n=0}^{\infty}\left(\frac{x+2n+1}{(x+2n+1)^2+y^2}+\frac{x+2n}{(x+2n)^2+y^2}-\frac{x+n}{(x+n)^2+y^2}\right)$$
| According to Wolfy,
each term is
$
\dfrac{\left(8 n^4 x - 4 n^4 + 20 n^3 x^2 - 4 n^3 x + 12 n^3 y^2 - 2 n^3 + 18 n^2 x^3 + 3 n^2 x^2 + 18 n^2 x y^2 - 3 n^2 x + 9 n^2 y^2 + 7 n x^4 + 4 n x^3 + 10 n x^2 y^2 - n x^2 + 8 n x y^2 + 3 n y^4 + n y^2 + x^5 + x^4 + 2 x^3 y^2 + 2 x^2 y^2 + x y^4 + y^4\right)}
{(n^2 + 2 n x + x^2 + y^2) (4 n^2 + 4 n x + x^2 + y^2) (4 n^2 + 4 n x + 4 n + x^2 + 2 x + y^2 + 1)}
=\dfrac{\left(n^4(8 x - 4) + n^3(20 x^2 - 4 x + 12 y^2 - 2)\\
+ n^2(18 x^3 + 3 x^2 + 18 x y^2 - 3 x + 9 y^2 )\\
+ n(7 x^4 + 4 x^3 + 10 x^2 y^2 - x^2 + 8 x y^2 + 3 y^4)\\ + y^2 + x^5 + x^4 + 2 x^3 y^2 + 2 x^2 y^2 + x y^4 + y^4\right)}
{\left((n^2 + 2 n x + x^2 + y^2) \\(4 n^2 + 4 n x + x^2 + y^2) \\(4 n^2 + 4 n x + 4 n + x^2 + 2 x + y^2 + 1)\right)}
$.
Since the highest power of $n$
in the numerator is
$n^4$
and the denominator has
$n^6$,
the terms are like
$\dfrac1{n^2}$
and the sum of these converge.
| {
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"timestamp": "2023-03-29T00:00:00",
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Given that $a+b+c=0$, show that $2(a^4+b^4+c^4)$ is a perfect square Given that $a+b+c=0$. Show that: $2(a^4+b^4+c^4)$ is a perfect square
MY ATTEMPTS:
I found that when $a+b+c=0$, $a^3+b^3+c^3=3abc$
So I did:
$(a^3+b^3+c^3)(a+b+c)$ -- $a^4+b^4+c^4=-(a^3c+a^3b+ab^3+b^3c+c^3a+c^3b)$
And then I tried to substitute $a^4+b^4+c^4$, but I found nothing that I thought relevant to the question
| I don't pretend that my solution is shorter, but for such questions, I try to avoid "redicovering all from scratch" ; I systematically apply the Newton-Girard identities one finds here, in particular for the present case :
$$p_4=\underbrace{e_1^4-4e_1 e_2+4e_1 e_3}_{= \ 0 \ \text{because} \ e_1= \ 0}+2e_2^2-\underbrace{4e_4}_{= \ 0}\tag{1}$$
with notations
*
*$p_k$ for the sum of $k$-th powers and
*$e_k$ for the coefficient of $(-1)^k x^k$ in the expansion $(x-a)(x-b)(x-c)...$.
Consequence of (1): $2p_4=(2 e_2)^2$ is indeed a perfect square ; otherwise said :
$$2(a^4+b^4+c^4)=(2(ab+bc+ca))^2\tag{2}$$
Of course, the previous expression can be given different forms ; for example, variable $c$ can be eliminated by using relationship $c=-a-b$.
| {
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Prove by mathematical induction that $3^n>2n^3$ I'm having trouble with this question:
"Prove by mathematical induction that for all integers $n\ge 6$, $3^n>2n^3$".
I got to $P(k)=2k^3<3^k$ and $P(k+1)=2(k+1)^3<3^{k+1}=2k^3+6k^2+6k+2<3^k*3$,
but I dont know how I can get $P(k+1)$ from $P(K)$...
Thanks
| $P(k)$ is the statement $ 2n^3<3^n $. Do not write "$P(k)=.....$"; $P(k)$ is not a mathematical value.
If we assume that then we have $2n^3 < 3^n$
So $2(n+1)^3 = 2n^3 + 6n^2 + 6n + 1$ And we have $2n^3 < 3^n$ so
$2(n+1)^3 =2n^3 + 6n^2 + 6n + 1< 3^n + 6n^2 + 6n + 1$
And $n\ge 6$ so $6n^2 \le n*n^2 =n^3$ and $6n+1 < 6n+n < 6n*n=6n^2 < n*n^3 < n^3$.
So
$2(n+1)^3 =2n^3 + 6n^2 + 6n + 1< 3^n + (6n^2) + (6n + 1)$
$< 3^n + n^3 + n^3 = 3^n + 2n^3 \le 3^n + 3^n < 3^n + 3^n + 3^n$
$< 3*3^n = 3^{n+1}$.
So $2(n+1)^3 < 3^{n+1}$ and so the statement $P(k+1)$ is true.
| {
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How to find $\int \frac {(1+x^2)(2+x^2)}{(x \cos x+\sin x)^4}dx$? I came across this integral while studying indefinite integrals.
So far I have had many unsuccessful attempts which include - trying by parts - but I could not find a way to proceed with it.
I even tried dividing both numerator and denominator by $x^4$ to yield $$\int \frac {(1+x^{-2})(1+2x^{-2})}{( \cos x+ \frac \sin x)^4}dx.$$ But I am still not able to move forward.
I even tried to cheat a little bit by taking derivative of options, still nothing.
And many more..
So the problem still stands. Can someone tell me how to proceed?
(Note: this is a problem from very elementary calculus course so no contour integrals, no multivariable and such stuff, however I think differentiation under integration would be fine. Also it would help if answers are one of those present in options (see image).)
Edit: Thanks to comments now I know answer is C but I was wondering if anyone could show me a straightforward way to do it thanks!
Edit 2 : turns out B is also correct.
| Let $f_n(x) = \left( \frac {\cos x - x \sin x}{x\cos x + \sin x} \right)^n$ and note that,
$$\frac{df_n(x)}{dx}
= -\frac{n(\cos x - x \sin x)^{n-1}(x^2+2)}{(x\cos x + \sin x)^{n+1}}$$
Then, for $n=1$ and $n=3$, we have respectively,
$$\frac{df_1(x)}{dx}
= -\frac{x^2+2}{(x\cos x + \sin x)^{2}}$$
$$\frac{df_3(x)}{dx}=- \frac{3(\cos x - x \sin x)^2(x^2+2)}{(x\cos x + \sin x)^{4}}$$
which leads to
$$\frac {(1+x^2)(2+x^2)}{(x \cos x+\sin x)^4} = -\frac{d}{dx}\left(\frac13f_3(x)+f_1(x)\right)$$
Thus,
$$\begin{align}
& \int \frac {(1+x^2)(2+x^2)}{(x \cos x+\sin x)^4}dx \\
& =-\frac13f_3(x)-f_1(x) \\
& =-\frac13 \left( \frac {\cos x - x \sin x}{x\cos x + \sin x} \right)^3
- \frac {\cos x - x \sin x}{x\cos x + \sin x} \\
& =-\frac13 \left( \frac {1- x \tan x}{x + \tan x} \right)^3
- \frac {1- x \tan x}{x + \tan x} + C\\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3610320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
What are the integer solutions to $a^{b^2} = b^a$ with $a, b \ge 2$ I saw this in quora.
What are all the
integer solutions to
$a^{b^2} = b^a$
with $a, b \ge 2$?
Solutions I have found so far:
$a = 2^4 = 16, b = 2,
a^{b^2}
= 2^{4\cdot 4}
=2^{16},
b^a = 2^{16}
$.
$a = 3^3, b = 3,
a^{b^2} = 3^{3\cdot 9}
=3^{27},
b^a = 3^{3^3}
=3^{27}
$.
In the general case,
$a$ and $b$ have the
same set of prime divisors,
so let
$a
=\prod_P p_i^{a_i}$,
$b
=\prod_P p_i^{b_i}
$
with each
$a_i \ge 1, b_i \ge 1$.
$b^a
=b^{\prod p_i^{a_i}}
=(\prod p_j^{b_j})^{\prod p_i^{a_i}}
=\prod p_j^{b_j\prod p_i^{a_i}}
$
$a^{b^2}
=a^{\prod p_i^{2b_i}}
=(\prod p_j^{a_j})^{\prod p_i^{2b_i}}
=\prod p_j^{a_j\prod p_i^{2b_i}}
$
Therefore,
for each $p_j$,
$b_j\prod p_i^{a_i}
=a_j\prod p_i^{2b_i}
$.
I haven't gotten any further than this.
I conjecture that
there are no other solutions.
| Render $a=tb^2$. Then
$(tb^2)^{b^2}=b^{tb^2}$
$tb^2=b^t$ ($b$ is assumed nonzero)
$t=b^{t-2}$ ($t$ is nonzero bdcause $a$ is assumed nonzero)
If $b$ is to be a positive integer and $t$ is rational, then $t$ will be a positive integer or $b$ will be a perfect power. But if $t>4$ then $b=t^{1/(t-2)}$ will lie strictly between $1$ and $2$, and $b\ge 4$ requires $t\ge 4^{t-2}$ which also excludes $t>4$. We are forced to a finite set of candidates for $t$, namely $\{1,2,3,4\}$, from which there can be only finitely many solutions for $b$ and thus for $a=tb^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3611719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How can I solve this geometry problem without trigonometry?
Let $ABC$ be a triangle with $D$ on the side $AC$ such that $\angle DBC=42^\circ$ and $\angle DCB=84^\circ$. If $AD = BC$, find $x = \angle DAB$.
It's supposed to be solved with constructions, but I couldn't figure out. See the trigonometric version here.
|
Here is a mostly geometric solution. First of all, let $C'$ be another point on $AC$ such that $BC'=BC$. Observe that the triangle $BC'D$ has $\angle C'BD=30^\circ$ and $\angle BDC'=\angle BDC=54^\circ$. Thus, by the Law of Sines on the triangle $BC'D$, we obtain
$$\frac{C'D}{C'B}=\frac{\sin(\angle C'BD)}{\sin(\angle BDC')}=\frac{\sin(30^\circ)}{\sin(54^\circ)}\,.$$
Since $\sin(54^\circ)=\cos(36^\circ)=\dfrac{1+\sqrt{5}}{4}$ (a proof is given here) and $\sin(30^\circ)=\dfrac{1}{2}$, we conclude that
$$\frac{C'D}{C'B}=\frac{-1+\sqrt{5}}{2}\,.$$
From $AD=BC=BC'$, we see that
$$\begin{align}\frac{C'A}{C'B}&=\frac{AD+C'D}{BC'}=1+\frac{CD'}{BC'}\\&=1+\frac{-1+\sqrt{5}}{2}
=\frac{1+\sqrt{5}}{2}=\frac{C'B}{C'D}\,.\end{align}$$
Hence, $C'A\cdot C'D=(C'B)^2$. This shows that $C'B$ is a tangent to the circumcircle of the triangle $ADB$. Therefore,
$$x=\angle BAC=\angle BAC'=\angle C'BD=30^\circ\,.$$
Here is a proof that $\dfrac{C'D}{C'B}=\dfrac{-1+\sqrt{5}}{2}$ without using trigonometry (and thereby, proving that $\cos(36^\circ)=\dfrac{1+\sqrt{5}}{4}$). Let $J$ be the reflection of $C'$ with respect to $BD$. Therefore, $BC'J$ is an equilateral triangle. If $r:=\dfrac{C'D}{C'B}$, then $r=\dfrac{C'D}{C'J}$.
Draw a regular pentagon $C'JMLK$ so that $D$ is an interior point of this pentagon. The thick line segments $JB$, $BC'$, $C'J$, $JM$, $ML$, $LK$, and $KC'$ are easily seen to be of the same length. Note that $D$ is on the diagonal $C'M$ of the pentagon. The triangle $DMJ$ is also an isosceles triangle with $DM=MJ$. As $MJ=BC'$, we conclude that, in fact, $M=A$.
Thus, as $C'DJ$ and $C'JM$ are similar triangles,
$$r=\frac{C'D}{C'J}=\frac{C'J}{C'M}=\frac{C'J}{C'D+DM}\,.$$
(At this stage, since $A=M$, it already follows that $C'B^2=C'J^2=C'D\cdot CM=C'D\cdot C'A$, establishing that $C'B$ is a tangent to the circumcircle of the triangle $ADB$. Hence, knowing the exact value of $r$ is unnecessary.) Thus, from $DM=MJ=C'J$, we get
$$r=\frac{C'J}{C'D+C'J}=\frac{1}{\frac{C'D}{C'J}+1}=\frac{1}{r+1}\,.$$
That is, $r^2+r-1=0$, or $r=\dfrac{-1\pm\sqrt{5}}{2}$. As $r>0$, we get $r=\dfrac{-1+\sqrt{5}}{2}$, as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3611833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
Show that $1+2^n+2^{2n}$ is divisible by $7$, when $n$ is not a multiple of $3$ Problem taken from a paper on mathematical induction by Gerardo Con Diaz. Although it doesn't look like anything special, I have spent a considerable amount of time trying to crack this, with no luck.
Show that $1+2^n+2^{2n}$ is divisible by $7$, when $n$ is not a multiple of 3.
| $n=3k+1$, $1+2^{3k+1}+2^{2(3k+1)}$, $2^3=1$ mod $7$ implies that
$1+2^{3k+1}+2^{2(3k+1)}=1+2+4$ mod $7$.
$n=3k+2$, $1+2^{3k+2}+2^{2(3k+2)}=1+4+16$ mod $7$ and $1+4+16=0$ mod $7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3613766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
If a,b,c be roots of $2x^3+x^2+x-1=0$ show that some expression is equal to 16. If $a,b,c$ are roots of $2x^3+x^2+x-1=0,$ show that
$$\bigg(\frac{1}{b^3}+\frac{1}{c^3}-\frac{1}{a^3}\bigg)\bigg(\frac{1}{c^3}+\frac{1}{a^3}-\frac{1}{b^3}\bigg)\bigg(\frac{1}{a^3}+\frac{1}{b^3}-\frac{1}{c^3}\bigg)=16$$
My attempt:
Let $\frac{1}{a}=p,\frac{1}{b}=q,\frac{1}{c}=r$
$p+q+r=1$
$pqr=2$
$$pq+qr+rp=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}$$
$$=\frac{ab+bc+b^2}{(abc)^2}$$
$$=4\bigg(\frac{1}{pq}+\frac{1}{qr}+\frac{1}{q^2}\bigg)$$
$$pq+qr+rp=4\bigg(\frac{pq+qr+rp}{pq^2r}\bigg)$$
$$pq^2r=4$$
$$\implies q=2 \implies b=\frac{1}{2}$$
So p, r are roots of $x^2+x+1=0$
$\implies p^3=q^3=1$
But this condition gives a different value of the required expression, so what am I doing wrong? Please tell me the right solution.
| $$2x^3+x^2+x-1=0`~~~(1)$$
$a,b, c$ are roots of the cubic, let $$s=\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{a^3b^3+b^3c^3+c^3a^3}{(abc)^3}=0$$ Let one root od the new cubic be $$y=-\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=s-\frac{2}{a^3}=s-\frac{2}{x^3} \implies x=(\frac{2}{-y})^{1/3}~~~~(2)$$
From (1) we can write:
$$(2x^3+1)^3=(x^2+x)^3 \implies (2x^3+1)^3=x^6+x^3+3x^3(x^2+x)$$
$$\implies (2x^3+1)^3-[x^6+x^3+3x^3(-3x^3-1)] \implies 8x^9++17x^6+8x^3+1=0~~~(3)$$
Putting this transformation (2) in (3), we a cubic in $y$ as
$$y^3-16y^2+68y-64=0~~~~(4)$$ The required expression is nothing but $y_1 y_2 y_3$ and its value from (4) is $64$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3615742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Prove that $\mathbb{Q}[\sqrt{3}, \sqrt{5}] = \mathbb{Q}[\sqrt{3}+ \sqrt{5}]$ Prove that $\mathbb{Q}[\sqrt{3}, \sqrt{5}] = \mathbb{Q}[\sqrt{3}+ \sqrt{5}]$
Take $x \in \mathbb{Q}[\sqrt{3}+ \sqrt{5}]. x = a_x + b_x( \sqrt{3} + \sqrt{5}) = a_x + b_x\sqrt{3} + b_x\sqrt{5} \in \mathbb{Q}[\sqrt{3}, \sqrt{5}] \Rightarrow \mathbb{Q}[\sqrt{3}+ \sqrt{5}] \subset \mathbb{Q}[\sqrt{3}, \sqrt{5}] $
I am having trouble proving the other inclusion.
| While it is possible to show the other inclusion, I think it is far easier to show that the dimensions of each field extension (as vector spaces over $\mathbb{Q}$) are the same (which is $4$ in this case). Then since one vector space is contained within the other, they must be the same if they have the same dimension. Can you handle this from here?
If you wanted to show the other inclusion direction, then you might argue as follows. Note that $(\sqrt 3 + \sqrt 5)^2 = 3 + 2 \sqrt{15} + 5$, and thus $\sqrt{15} \in \mathbb{Q}[\sqrt 3 + \sqrt 5]$. And
$$\begin{align}
(\sqrt 3 + \sqrt 5 + \sqrt{15})^2 &= 3 + 5 + 15 + 2 \sqrt{15} + 2 \sqrt{45} + 2 \sqrt{75} \\&= 23 + 2 \sqrt{15} + 6 \sqrt{5} + 10 \sqrt{3}.
\end{align}$$
Subtracting $23 + 2 \sqrt{15} + 6(\sqrt{3} + \sqrt{5})$ shows that $4 \sqrt{3} \in \mathbb{Q}[\sqrt{3} + \sqrt{5}]$, and thus $\sqrt{3} \in \mathbb{Q}[\sqrt{3} + \sqrt{5}]$. (And similarly for $\sqrt 5$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3617442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Verifying that Kummer hypergeometric function is a solution to $xy''+(b-x)y'-ay=0$ The following second order differential equation (see YouTube link)
$$xy''+(b-x)y'-ay=0$$
has two solutions, one of them resmenble Kummer function of the first kind:
$$y=M(a,b,x)=\sum_{n=0}^\infty \frac{a^{(n)}x^n}{b^{(n)}n!},$$
where $a^{(n)}$ are rising factorial of $a$.
While verifying that that function is the solution to the differential equation, I got this:
$$n(n-1)+nb-nx-ax=0$$
I am wondering if there is something wrong. Actually, I thought that all terms will cancel each other.
Thanks.
| $\def\d{\mathrm{d}}$If $y(x) = \sum\limits_{n = 0}^∞ \dfrac{a^{(n)} x^n}{b^{(n)} n!} = 1 + \sum\limits_{n = 1}^∞ \dfrac{a^{(n)} x^n}{b^{(n)} n!}$, then$$
ay(x) = a + \sum_{n = 1}^∞ \frac{a · a^{(n)} x^n}{b^{(n)} n!},
$$\begin{gather*}
(b - x) y'(x) = (b - x) \sum_{n = 1}^∞ \frac{a^{(n)} nx^{n - 1}}{b^{(n)} n!} = \sum_{n = 1}^∞ b · \frac{a^{(n)} nx^{n - 1}}{b^{(n)} n!} - \sum_{n = 1}^∞ x · \frac{a^{(n)} nx^{n - 1}}{b^{(n)} n!}\\
= \sum_{n = 1}^∞ \frac{a^{(n)} x^{n - 1}}{(b + 1)^{(n - 1)} (n - 1)!} - \sum_{n = 1}^∞ \frac{a^{(n)} x^n}{b^{(n)} (n - 1)!} = a + \sum_{n = 1}^∞ \frac{a^{(n + 1)} x^n}{(b + 1)^{(n)} n!} - \sum_{n = 1}^∞ \frac{a^{(n)} x^n}{b^{(n)} (n - 1)!},
\end{gather*}\begin{gather*}
xy''(x) = \sum_{n = 2}^∞ x · \frac{a^{(n)} n(n - 1)x^{n - 2}}{b^{(n)} n!} = \sum_{n = 2}^∞ \frac{a^{(n)} x^{n - 1}}{b^{(n)} (n - 2)!} = \sum_{n = 1}^∞ \frac{a^{(n + 1)} x^n}{b^{(n + 1)} (n - 1)!},
\end{gather*}
and\begin{align*}
&\mathrel{\phantom{=}}{} xy''(x) + (b - x)y'(x) - ay(x)\\
&= {\small \sum_{n = 1}^∞ \frac{a^{(n + 1)} x^n}{b^{(n + 1)} (n - 1)!} + \left( a + \sum_{n = 1}^∞ \frac{a^{(n + 1)} x^n}{(b + 1)^{(n)} n!} - \sum_{n = 1}^∞ \frac{a^{(n)} x^n}{b^{(n)} (n - 1)!} \right) - \left( a + \sum_{n = 1}^∞ \frac{a · a^{(n)} x^n}{b^{(n)} n!} \right)}\\
&= \sum_{n = 1}^∞ \left( \frac{a^{(n + 1)}}{b^{(n + 1)} (n - 1)!} + \frac{a^{(n + 1)}}{(b + 1)^{(n)} n!} - \frac{a^{(n)}}{b^{(n)} (n - 1)!} - \frac{a · a^{(n)}}{b^{(n)} n!} \right) x^n\\
&= \sum_{n = 1}^∞ \bigl( (a + n)n + (a + n)b - (b + n)n - a(b + n) \bigr) \frac{a^{(n)}}{b^{(n + 1)} n!} x^n = 0.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3618386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Precalculus linear combination problem What is the maximum possible length of the vector resulting from the following linear combination?
$$
\frac{1}{\| \mathbf{v_1} \|} \,\mathbf{v_1} + \frac{1}{\|\mathbf{v_2}\|} \,\mathbf{v_2} + \cdots + \frac{1}{\| \mathbf{v_n} \|} \,\mathbf{v_n}
$$
| Let $\mathbf{u_i} = \frac{1}{\| \mathbf{v_i} \|} \,\mathbf{v_i}$, then $\| \mathbf{u_i} \| = 1.$
$
\Big\|\frac{1}{\| \mathbf{v_1} \|} \,\mathbf{v_1} + \frac{1}{\|\mathbf{v_2}\|} \,\mathbf{v_2} + \cdots + \frac{1}{\| \mathbf{v_n} \|} \,\mathbf{v_n}\Big\| =
\| \mathbf{u_1} + \mathbf{u_2} + \dots + \mathbf{u_n}\| \le \| \mathbf{u_1}\| + \|\mathbf{u_2}\| + \dots + \|\mathbf{u_n}\| = n.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3622452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Area and center location of an ellipse generated by the intersection of an ellipsoid and plane I am working on a model that requires that I know the area and center coordinates of the ellipse that is created by the intersection of an ellipsoid and a plane.
Specifically, for the location of the center of the ellipse I want to know the coordinates of this ellipse in Cartesian coordinates.
The general equations for the ellipsoid and plane I have started with are:
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$$
$$m(x-x_0) + n(y-y_0) + k(z-z_0) = 0$$
I am writing the general forms of these equations because actually I need to be able to solve this for a number of different ellipses and planes with different orientations.
One specific case that I would like to use this solution for first is the case of an ellipsoid that is defined by:
$$a = 7, b = 5, c = 6$$
and a plane defined by:
$$(y+b)tan(\theta)-z + \frac{1}{2} = 0$$
Where $\theta$ is the desired angle of the plane, in this case $\theta=30^o$ is perfectly fine (arbitrary example).
Please note that "b" is the same b used in the equation of the ellipsoid.
I looked at some of the other threads that have asked about plane ellipsoid intersections. However, since I specifically need to calculate the area of the ellipse generated by this intersection and the location of its center, when I tried to use a parametric solution, I had difficulty doing this once I had the parametric equations.
I would really love to learn how to solve this problem so I can include it into my model. Help is greatly appreciated.
Thank you!
-Christian
| Without loss of generality, you can translate the ellipsoid to origin, rotate it so that its semiaxes are parallel to the Cartesian coordinate axes, and reorder the axes so that the $z$ component of the intersection plane normal in the rotated coordinates has the largest magnitude. Then, the points on the ellipsoid fulfill
$$\frac{x^2}{r_x^2} + \frac{y^2}{r_y^2} + \frac{z^2}{r_z^2} = 1 \tag{1}\label{AC1}$$
where $r_x$, $r_y$, and $r_z$ are the ellipsoid semiaxes.
Define the plane using its normal $(n_x, n_y, n_z)$ and signed distance from origin $n_d$, i.e. point $(x, y, z)$ is on the plane if and only if
$$x n_x + y n_y + z n_z = n_d \tag{2a}\label{AC2a}$$
Solving this for $z$ yields
$$z = \frac{n_d - x n_x - y n_y}{n_z} \tag{2b}\label{AC2b}$$
(We reorder the axes so that $\lvert n_z \rvert \ge \lvert n_y \rvert$, $\lvert n_z \rvert \ge \lvert n_x \rvert$, to avoid division by zero, and in numerical computations, for best numerical stability.)
Substituting $\eqref{AC2b}$ into $\eqref{AC1}$ yields general quadratic form
$$a x^2 + b x y + c y^2 + d x + e y + f = 0$$
where (assuming $n_z r_z \ne 0$, both sides multiplied by $n_z^2 r_z^2$)
$$\left\lbrace ~ \begin{aligned}
a &= n_x^2 + n_z^2 r_z^2 / r_x^2 \\
b &= 2 n_x n_y \\
c &= n_y^2 + n_z^2 r_z^2 / r_y^2 \\
d &= -2 n_x n_d \\
e &= -2 n_y n_d \\
f &= n_d^2 - n_z^2 r_z^2 \\
\end{aligned} \right . \tag{3}\label{AC3}$$
Using this answer by Osmund Francis and the Wikipedia Ellipse article, we can express the following:
The discriminant $D$ (note sign convention!) is
$$D = b^2 - 4 a c$$
where the intersection is an ellipse if and only if $D \lt 0$.
Center of the ellipse is at $(x_0, y_0)$,
$$\left\lbrace ~ \begin{aligned}
\displaystyle x_0 &= \frac{2 c d - b e}{D} \\
\displaystyle y_0 &= \frac{2 a e - b d}{D} \\
\end{aligned} \right.$$
Semimajor axis length $r_+$ and semiminor axis length $r_-$ are
$$\begin{aligned}
r_+ &= \frac{1}{-D}\sqrt{ 2 ( a e^2 + c d^2 - b d e + f D ) ( a + c + \sqrt{ b^2 + ( a - c )^2 } ) } \\
r_- &= \frac{1}{-D}\sqrt{ 2 ( a e^2 + c d^2 - b d e + f D ) ( a + c - \sqrt{ b^2 + ( a - c )^2 } ) } \\
\end{aligned}$$
and the area $A$ of the ellipse is
$$A = \pi ~ r_+ ~ r_-$$
The angle $\theta$ between $x$ axis and major axis is
$$\theta = \begin{cases}
\operatorname{atan2}\left(c - a - \sqrt{b^2 + (a-c)^2}, b\right), & b \ne 0 \\
0, & b = 0, \quad a \lt c \\
90^o, & b = 0, \quad a \gt c \\
0, & b = 0, \quad a = c, \quad \lvert d \rvert \ge \lvert f \rvert \\
90^o, & b = 0, \quad a = c, \quad \lvert d \rvert \lt \lvert f \rvert \\
\end{cases}$$
Note that the major axis intersects the ellipse at $(x_{+1}, y_{+1})$ and $(x_{+2}, y_{+2})$,
$$\left\lbrace ~ \begin{aligned}
x_{+1} &= x_0 + r_+ ~ \cos\theta \\
y_{+1} &= y_0 + r_+ ~ \sin\theta \\
\end{aligned} \right ., \quad \left\lbrace ~ \begin{aligned}
x_{+2} &= x_0 - r_+ ~ \cos\theta \\
y_{+2} &= y_0 - r_+ ~ \sin\theta \\
\end{aligned} \right .$$
and similarly the minor axis intersects the ellipse at $(x_{-1}, y_{-1})$ and $(x_{-2}, y_{-2})$,
$$\left\lbrace ~ \begin{aligned}
x_{-1} &= x_0 + r_- ~ \cos\theta \\
y_{-1} &= y_0 + r_- ~ \sin\theta \\
\end{aligned} \right ., \quad \left\lbrace ~ \begin{aligned}
x_{-2} &= x_0 - r_- ~ \cos\theta \\
y_{-2} &= y_0 - r_- ~ \sin\theta \\
\end{aligned} \right .$$
Finally, using angle parameter $\varphi$, we can parametrize the ellipse as $\bigr(x(\varphi), y(\varphi)\bigr)$, $0 \le \varphi \le 360^o$,
$$\left\lbrace ~ \begin{aligned}
x(\varphi) &= x_0 + r_+ \cos\theta \cos\varphi - r_- \sin\theta \sin\varphi \\
y(\varphi) &= y_0 + r_+ \sin\theta \cos\varphi + r_- \cos\theta \sin\varphi \\
\end{aligned} \right .$$
which is just $x = r_+ \cos \varphi$, $y = r_- \sin\varphi$, rotated by $\theta$, and translated to $(x_0, y_0)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Fastest way to solve $x^3\equiv x \pmod{105}$ $$x^3\equiv x \pmod{105}$$
I'm trying to solve this equation. Here's what I tried so far:
$$x^3\equiv x \pmod{105} \iff x^2\equiv 1 \pmod{105}$$
Then, applying the Chinese remainder theorem, I got the system:
$$\cases{x^2 \equiv 1 \pmod{5}\\x^2 \equiv 1 \pmod{7}\\x^2 \equiv 1 \pmod{3}}$$
With the following solutions:
$$\cases{x \equiv \pm1 \pmod{5}\\x \equiv \pm1 \pmod{7}\\x \equiv \pm1 \pmod{3}}$$
At this point, I think I need to pretty much solve these eight systems:
$$\cases{x \equiv 1 \pmod{5}\\x \equiv 1 \pmod{7}\\x \equiv 1 \pmod{3}}
\cases{x \equiv 1 \pmod{5}\\x \equiv 1 \pmod{7}\\x \equiv -1 \pmod{3}}
\cases{x \equiv 1 \pmod{5}\\x \equiv -1 \pmod{7}\\x \equiv 1 \pmod{3}}
\cases{x \equiv -1 \pmod{5}\\x \equiv 1 \pmod{7}\\x \equiv 1 \pmod{3}}$$$$
\cases{x \equiv -1 \pmod{5}\\x \equiv 1 \pmod{7}\\x \equiv -1 \pmod{3}}
\cases{x \equiv -1 \pmod{5}\\x \equiv -1 \pmod{7}\\x \equiv 1 \pmod{3}}
\cases{x \equiv 1 \pmod{5}\\x \equiv -1 \pmod{7}\\x \equiv -1 \pmod{3}}
\cases{x \equiv -1 \pmod{5}\\x \equiv -1 \pmod{7}\\x \equiv -1 \pmod{3}}$$
Here's how I solved the first one:
Considering the first two equations, we get:
$$x=5k+1=7h+1$$ from which $k = 7+7y, h = 5+5y$, with $y \in \mathbb{Z}$. Therefore, $$x=36+35y\iff x\equiv1\pmod{35}$$
Adding in the third equation, we have that $36+35y = 1+3 w$, from which $x = 1281 + 35w \iff x \equiv1\pmod{105}$.
However, this one seems like a really tedious method as I'd have to do the same calculations for seven more systems. Is there anything I'm missing? Is there a faster way to do this?
| As $x^3-x=(x-1)x(x+1)$ is a product of three consecutive integers
$3$ must divide $x^3-x$
So, we need $$x^3\equiv x\pmod{5\cdot7}$$
If $(x-1)x(x+1)\equiv0\pmod 5$
$\implies x\equiv0\ \ \ \ (1), x\equiv-1\ \ \ \ (2), x\equiv1\pmod5\ \ \ \ (3)$
Similarly, $x\equiv0\ \ \ \ (4), x\equiv-1\ \ \ \ (5), x\equiv1\pmod7\ \ \ \ (6)$
Now apply CRT on $(1),(4); (1),(5);(1),(6);(2),(4); (2),(5);(2),(6)$
| {
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"question_score": "2",
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Question about $\lim\limits_{x \to2} \frac{\sqrt[3]{x} - \sqrt[3]{2}}{\sqrt{x} - \sqrt{2}}$ $\lim\limits_{x \to2} \frac{\sqrt[3]{x} - \sqrt[3]{2}}{\sqrt{x} - \sqrt{2}}$
Let $x=t^6$, then it becomes $\lim\frac{t^2 - (\sqrt[6]{2})^2}{t^3 - (\sqrt[6]{2})^3}$=$\lim\frac{(t+\sqrt[6]{2})(t-\sqrt[6]{2})}{(t-\sqrt[6]{2})(t^2+t\sqrt[6]{2}+(\sqrt[6]{2})^2}$=$\lim\frac{(t+\sqrt[6]{2})}{(t^2+t\sqrt[6]{2}+(\sqrt[6]{2})^2}$
But when $x\to 2$, t can go to $\sqrt[6]{2}$ or -$\sqrt[6]{2}$, which gives two different limits, where I was wrong? Thanks!
| Use the (not so) popular algebraic limit formula $$\lim_{x\to a} \frac{x^n-a^n} {x-a} =na^{n-1}$$ Dividing both the numerator and denominator of the expression under limit by $x-2$ we can see that the desired limit equals $$\left. \frac{1}{3}\cdot 2^{-2/3}\middle/\frac{1}{2}\cdot 2^{-1/2}\right.=\frac {2^{5/6}}{3}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A question about a function $q(b)$ User Mathphile invented the function $\ q(b)\ $ which is defined as the sum of the values $ \lfloor \frac bj \rfloor$ for $j=1,\cdots ,b$ without duplicates.
Example : The values for $b=11$ are $[11, 5, 3, 2, 2, 1, 1, 1, 1, 1, 1]$ , without duplicates we have $[1, 2, 3, 5, 11]$ , the sum is $22$ , hence $q(11)=22$
In PARI/GP $q(b)$ can be calculated with
f(b)=vecsum(Set(vector(b,j,b\j)))
I have two conjectures :
*
*$q(b)$ is strictly increasing
*For $b\ge 2$ : $q(b) = q(b-1) + 1$ holds , if and only if $b$ is an odd prime.
$q(1)=1$ , $q(2)=3$, hence the claim is true for $b=2$
If $b>2$, we have $\lfloor \frac{b}{b-1} \rfloor=1$ , hence the fraction for $j=b$ is a duplicate, hence does not prolong the set.
If $b$ is an odd prime, the fractions for $j=2,\cdots, b-1$ coincide for $b-1$ and $b$ , since $\frac{b}{j}$ is never an integer. For $j=1$, the fraction increases from $b-1$ to $b$. Hence the sum increases by $1$. This shows one direction of the claim.
It remains to show that the sum increases by more than $1$ in the case of a composite number , which would also show that $q(b)$ is strictly increasing. How can I do that ?
| If $b$ is composite integer then for $ 1\leq j\leq [\sqrt{b}]$ we have that $[\frac{b}{j}] > [\frac{b}{j+1}]$ because $\frac{b}{j}-\frac{b}{j+1}\geq 1$ which is true for all $1 \leq j \leq [-0.5+0.5 \sqrt{1+4b}] \leq [\sqrt{b}]$, so for $j=1,2,3,\cdots, [\sqrt{b}]$ and we have that $[\frac{b}{j}]\geq [\frac{b-1}{j}]$ and its strictly bigger when $j|b$ and we have at least two cases $j=1,a$ when $b$ is composite integer.
Let $ [\sqrt{b}]< j \leq b-1$ since $[\frac{b}{b-1}]=1 = [\frac{b}{b}]$ for all $b\geq 3$
If $b|j$ then $[\frac{b}{j}] > [\frac{b-1}{j}]$ if $\frac{b}{j}$ is repeated for the $q(b)$ series and $\frac{b}{j}-1$ is new for $q(b-1)$ this might make $q(b-1)>q(b)$, but we will prove that if this is the case then $[\frac{b}{j+1}]= [\frac{b-1}{j+1}] = \frac{b}{j}-1$ which is to prove that $ \frac{b}{j} > \frac{b}{j+1},\frac{b-1}{j+1} \geq \frac{b}{j}-1$ which is true for all $ j \geq \sqrt{b} $ and we are done.
Note : we always have $j+1$ if we extended the sum for infinity and knowing that $[\frac{b}{j}]=[\frac{b-1}{j}]=0 $ for $j\geq b+1$( to make the proof more rigorous)
| {
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"question_score": "3",
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Show that $\int \frac{dx}{(a+b\cos x)^n}=\frac{A\sin x}{(a+b\cos x)^{n-1}}+B\int {dx \over (a+b\cos x)^{n-1}}+C\int {dx \over (a+b\cos x)^{n-2}}$ I've been trying to develop this integral in parts twice, but I can't come up with anything meaningful,
Could you give me some hints on how to develop this exercise?
Prove that
$$\int \frac{dx}{(a+b\cos x)^n}=\frac{A\sin x}{(a+b\cos x)^{n-1}}+B\int {dx \over (a+b\cos x)^{n-1}}+C\int {dx \over (a+b\cos x)^{n-2}}, \hspace{1cm} (|a|\neq |b|). $$
and determine the coefficients $A,B$ and $C$, if $n$ is a natural number greater than the unit
| Hint
By differentiating the equality, we should prove $$\frac{1}{(a+b\cos x)^n}={d\over dx}\frac{A\sin x}{(a+b\cos x)^{n-1}}+{B \over (a+b\cos x)^{n-1}}+{C \over (a+b\cos x)^{n-2}}, \hspace{1cm}$$ with $|a|\neq |b|$. Also $${d\over dx}\frac{A\sin x}{(a+b\cos x)^{n-1}}{={A\cos x(a+b\cos x)^{n-1}+(n-1)Ab\sin^2 x(a+b\cos x)^{n-2}\over (a+b\cos x)^{2n-2}
}
\\=
{A\cos x(a+b\cos x)+(n-1)Ab\sin^2 x\over (a+b\cos x)^{n}}
\\={Aa\cos x+Ab\cos^2 x+(n-1)Ab\sin^2 x\over (a+b\cos x)^{n}}
}$$
The task of finding proper values $A,B,C$ such that $$\frac{1}{(a+b\cos x)^n}-{B \over (a+b\cos x)^{n-1}}-{C \over (a+b\cos x)^{n-2}}={Aa\cos x+Ab\cos^2 x+(n-1)Ab\sin^2 x\over (a+b\cos x)^{n}}$$should be easy.
| {
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"source": "stackexchange",
"question_score": "2",
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What is $\frac{a b \sin x}{2\sqrt{(a^2 + b^2 + 2 ab \cos x ) \cdot (a+b - \sqrt{a^2+b^2+ 2 ab \cos x})}}$ for $x \rightarrow 0$? I am trying to find the limit
$\lim_{x \rightarrow0} \frac{a b \sin x}{2\sqrt{(a^2 + b^2 + 2 ab \cos x ) \cdot (a+b - \sqrt{a^2+b^2+ 2 ab \cos x})}}$
for $a,b \in \rm{I\!R}_{+}$. Applying L'Hospital's rule leads to
$\lim_{x \rightarrow0}\frac{\cos x \cdot \sqrt{(a^2 + b^2 + 2 ab \cos x ) \cdot (a+b - \sqrt{a^2+b^2+ 2 ab \cos x})}}{-2 \sin x \cdot (a+b - \sqrt{a^2+b^2+ 2 ab \cos x}) + \frac{\sin x \cdot (a^2 + b^2 + 2 ab \cos x )}{\sqrt{a^2+b^2+ 2 ab \cos x}} }$.
However, this remains with both, cosine and sine. Maybe one could use a trigonometric identity, which I cannot find.
| Assume $u=\sqrt{a^+b^2+2ab\cos x} $ so that $u\to a+b$ as $x\to 0$. The expression under limit is $$\frac{ab\sin x} {2u\sqrt{a+b-u}}$$ whose limit is same as that of $$\frac{ab} {2(a+b)}\frac{x}{\sqrt{a+b-u}}$$ Next we can observe that $$\frac{a+b-u} {x^2}=\frac{(a+b) ^2-u^2}{x^2(a+b+u)}=\frac{2ab}{a+b+u}\cdot\frac{1-\cos x}{x^2}\to\frac{ab}{2(a+b)} $$ It follows that the desired limit is $$\frac{ab} {2(a+b)}\sqrt {\frac{2(a+b)}{ab}}=\sqrt{\frac{ab}{2(a+b)}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3640785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\lim_{x \to 0} \frac{\sin\left(a+2x\right)-2\sin\left(a+x\right)+\sin\left(a\right)}{x^{2}}$ without L'Hôpital This limit is one of the "Problems Plus" from Stewart Calculus:
$$\lim_{x \to 0} \frac{\sin\left(a+2x\right)-2\sin\left(a+x\right)+\sin\left(a\right)}{x^{2}}$$
Note that the limit is of the indeterminate form $\frac{0}{0}$. The problem appears several chapters before L'Hôpital's rule is discussed, so I assume there is a solution without using L'Hopital.
Looking at a graph, the local behavior of the function near $0$ appears to be $-\sin(a+x)$, which, of course, suggests a limit of $-\sin(a)$. Using L'Hôpital's rule twice confirms this guess:
$$\begin{align}
& \lim_{x \to 0} \frac{\frac{d}{dx} [\sin\left(a+2x\right)-2\sin\left(a+x\right)+\sin\left(a\right)]}{ \frac{d}{dx} x^{2}} \\
& = \lim_{x \to 0} \frac{2\cos(a+2x)-2\cos(a+x)}{2x} \\
& = \lim_{x \to 0} \frac{-4\sin(a+2x)+2\sin(a+x)}{2} \\
& = -\sin(a).
\end{align}$$
Can anyone give a hint or solution for evaluating this limit without L'Hôpital?
| Without heavy machinery, only that $\lim_{x \to 0} \sin x / x = 1$, you can use the sum-to-product formulas, as follows.
\begin{align*}
\sin(a+2x) - 2 \sin(a+x) + \sin a & = \sin(a+2x) - \sin(a+x) + \sin a - \sin(a+x) \\
& = 2 \sin(x/2) \cos(a + 3x/2) - 2 \sin (x/2) \cos(a + x/2) \\
& = 2 \sin(x/2) \left (\cos(a + 3x/2) - \cos(a + x/2) \right ) \\
& = - 4 \sin(x/2) \sin(a+x) \sin(x/2) \\
& = - 4 \sin^2(x/2) \sin(a+x).
\end{align*}
Then
$$
\lim_{x \to 0} \frac{\sin^2(x/2)}{x^2} = \frac14 \lim_{x \to 0} \frac{\sin^2(x/2)}{(x/2)^2} = \frac14 \lim_{y \to 0} \left ( \frac{\sin y}{y} \right )^2 = \frac14,
$$
and finally the limit of the expression you are looking at is indeed
$$
-4 \times \frac14 \times \sin a = - \sin a.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3643583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Floor functions in a set of numbers Find the number of distinct numbers in the list
$$\left\lfloor \frac{1^2}{1000} \right\rfloor, \ \left\lfloor \frac{2^2}{1000} \right\rfloor, \ \left\lfloor \frac{3^2}{1000} \right\rfloor, \ \dots, \ \left\lfloor \frac{1000^2}{1000} \right\rfloor.$$
The floor of all numbers till $\left\lfloor\frac{32^2}{1000}\right\rfloor$ are all $0.$ Assuming this goes in cycles of $32,$ there will be $31$ cycles of the $32,$ making there $\boxed{31}$ distinct numbers in this list. Is this the right track?
| Consider $\lfloor \frac{x^2}{1000} \rfloor = k$ for some integer $k$. This means:
$$k\leq \frac{x^2}{1000}<k+1$$
$$1000k\leq x^2 \leq 1000k+999$$
It is clear that when $x\geq 500$, $(x+1)^2-x^2=2x+1>1000$ so each value of $\lfloor\frac{x^2}{1000}\rfloor$ is distinct. That's $\fbox{501}$ distinct values from $x=500,501,\cdots,1000$
For $x < 500$ i.e. $k < 250$, all $k$ can be satisfied. $k$ ranges from $0$ to $249$ here, so $\fbox{250}$ distinct values here.
Therefore the answer is $250+501=751$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the polynomial $P(x)$ s.t. for any $a,b\in\Bbb C$ s.t. $a^2+b^2=ab$, $P(a+b)=6\big(P(a)+P(b)\big)+15a^2b^2(a+b)$.
Suppose that $P(x)$ is a polynomial such that
$$P(a+b)=6\big(P(a)+P(b)\big)+15a^2b^2(a+b)$$
for any complex numbers $a$ and $b$ satisfying $a^2+b^2=ab$, then find the polynomial $P(x)$.
I don't know how to even approach this one, it seems very complex. Your help is appreciated!
| In this solution, $P$ is assumed to be an entire function (which is more generalised than just being a polynomial). Let $\omega_1,\omega_2$ be roots of $x^2-x+1=0$.
We note that when $a=\omega_jt$ and $b=t$, we have $a^2+b^2=ab$, so that
$$P\big((\omega_j+1)t\big)-6\big(P(\omega_j t)-6P(t)\big)-15\omega_j^2(\omega_j+1)t^5=0\tag{1}$$
for $j=1,2$.
Suppose that
$$P(x)=\sum_{k=0}^\infty p_kx^k.$$
If $k\neq 5$, then the coefficient of $t^k$ of the LHS of $(1)$ is $p_k\cdot f_k(\omega_j)$, where
$$f_k(x)=(x+1)^k-6(x^k+1).$$
If $p_k\ne 0$, then $f_k(\omega_j)=0$. For $k\geq 5$, we have
$$\big|(\omega_j+1)^k\big|\geq |\omega_j+1|^k=\sqrt{3}^k=9\sqrt{3}$$
and
$$\big|6(\omega_j^k+1)\big|\leq 6\big(|\omega_j|^k+1\big)=6\cdot 2=12.$$
Thus
$$\big|(\omega_j+1)^k-6(\omega_j^k+1)\big|\geq \big|(\omega_j+1)^k\big|-\big|6(\omega_j^k+1)\big|\geq 9\sqrt{3}-12>0\,.$$
For $k=0,1,2,3,4$, we can easily verify by hand that
$$(\omega_j+1)^k-6(\omega_j^k+1)\ne 0.$$
For $k=0,1$, this is trivial. For $k=2$, we note that $(x+1)^2-6(x^2+1)=-5x^2+2x-5$ is not divisible by $x^2-x+1$. For $k=3$, we note that $\omega_j^3=-1$ and $(\omega_j+1)^3\ne 0$. For $k=4$, note that $$(\omega_j+1)^4=\big((\omega_j+1)^2\big)^2=(3\omega_j)^2=9\omega_j^2$$ and $$\omega_j^4=-\omega_j,$$ so $$(\omega_j+1)^4-6(\omega_j^4+1)=9\omega_j^2+6\omega_j-6\ne0.$$
Thus $\omega_j$ is not a root of $f_k(x)$ for any integer $k\geq 0$. This is a contradiction, so $p_k=0$ for $k\ne 5$.
The coefficient of $t^5$ in $(1)$ is
$$p_5(\omega_j+1)^5-6(\omega_j^5+1)-15\omega_j^2(\omega_j+1).\tag{2}$$
Noting that
$$(x+1)^5-6(x^5+1)-15x^2(x+1)=-5(x+1)(x^2-x+1)^5,$$
we conclude that
$$(\omega_j+1)^5-6(\omega_j^5+1)-15\omega_j^2(\omega_j+1)=0.\tag{3}$$
Subtract $(3)$ from $(2)$ to get
$$(p_5-1)(\omega_j+1)^5=0.$$
Therefore $p_5=1$, so $P(x)=x^5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3647743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Linear Algebra matrix calculation help please A = $$
\begin{pmatrix}
2 & 1 \\
3 & 4 \\
\end{pmatrix}
$$
How can i find matrix B s.t B^3 = A?
| You can define $B$ as
$$B = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$
And solve the equation following equation system:
$$B^3 = \begin{bmatrix} a & b \\ c & d \end{bmatrix}*\begin{bmatrix} a & b \\ c & d \end{bmatrix}*\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix},$$
what equals to
$$\begin{bmatrix} a^3+2abc+bdc & a^2b+b^2c+abd+bd^2 \\ a^2c+acd+c^2d+d^2c & abc+2bcd+d^3 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}$$
And solving that 4 equation system will give you the solution.
| {
"language": "en",
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"source": "stackexchange",
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Prove that for all positives $a, b$ and $c$, $(\sum_{cyc}\frac{c + a}{b})^2 \ge 4(\sum_{cyc}ca)(\sum_{cyc}\frac{1}{b^2})$.
Prove that for all positives $a, b$ and $c$, $$\left(\frac{b + c}{a} + \frac{c + a}{b} + \frac{a + b}{c}\right)^2 \ge 4(bc + ca + ab) \cdot \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right)$$
Let $ca + ab = m$, $ab + bc = n$ and $bc + ca = p$, we have that $$\left(\frac{m}{a^2} + \frac{n}{b^2} + \frac{p}{c^2}\right)^2 \ge 2(m + n + p) \cdot \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right)$$
$$\iff \left(\frac{m}{a^2} + \frac{n}{b^2} + \frac{p}{c^2} - 1\right)^2 \ge 2 \cdot \sum_{cyc((m, n, p), (a, b, c))}\left[n \cdot \left(\frac{1}{c^2} + \frac{1}{a^2}\right)\right] + 1$$
Expanding $\displaystyle \sum_{cyc((m, n, p), (a, b, c))}\left[n \cdot \left(\frac{1}{c^2} + \frac{1}{a^2}\right)\right]$ gives $$2 \cdot \sum_{cyc}\frac{ca}{b^2} + \left(\frac{b + c}{a} + \frac{c + a}{b} + \frac{a + b}{c}\right)$$
Let $\dfrac{b + c}{a} = m'$, $\dfrac{c + a}{b} = n'$ and $\dfrac{a + b}{c} = p'$, we have that $$(m' + n' + p' - 1)^2 \ge 2 \cdot \left[2 \cdot \sum_{cyc}\frac{ca}{b^2} + (m' + n' + p')\right] + 1$$
Moreover, $$(m')^2 + (n')^2 + (p')^2 = \sum_{cyc}\left[\left(\frac{c + a}{b}\right)^2\right] \ge 2 \cdot \sum_{cyc}\frac{ca}{b^2}$$
$$\implies (m' + n' + p' - 1)^2 \ge 2 \cdot \left[(m')^2 + (n')^2 + (p')^2 + m' + n' + p'\right] + 1$$
$$\iff -[(m')^2 + (n')^2 + (p')^2] + 2(m'n' + n'p' + p'm') - 4(m' + n' + p') \ge 0$$, which is definitely not correct.
Another attempt, let $(0 <) \ a \le b \le c \implies ab \le ca \le bc \iff ca + ab \le ab + bc \le bc + ca$
$\iff m \le n \le p$ and $a^2 \le b^2 \le c^2 \iff \dfrac{1}{a^2} \ge \dfrac{1}{b^2} \ge \dfrac{1}{c^2}$.
By the Chebyshev inequality, we have that $$3 \cdot \left(\frac{m}{a^2} + \frac{n}{b^2} + \frac{p}{c^2}\right) \le (m + n + p) \cdot \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right)$$
Any help would be appreciated.
| A NOTE
The given inequality
$$
\left(\sum_{cyc}\frac{a+b}{c}\right)^2\geq 4\left(\sum_{cyc}ab\right)\left(\sum_{cyc}\frac{1}{a^2}\right)\textrm{, }a,b,c>0\tag 1
$$
is written as
$$
\left(\sum_{cyc}ab(a+b)\right)^2\geq4\left(\sum_{cyc}ab\right)\left(\sum_{cyc}a^2b^2\right).\tag 2
$$
Hence if $s=a+b+c$ and $p=abc$ and $s_h=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$, then
$$
\left(\sum_{cyc}ab(a+b)\right)^2=p^2(ss_h-3)^2
$$
and
$$
4\left(\sum_{cyc}ab\right)\left(\sum_{cyc}a^2b^2\right)=4p^3s_h(s_h^2-2s/p)
$$
Hence (2) becomes
$$
9+2ss_h+s^2s_h^2-4ps_h^3\geq 0.\tag 3
$$
Inequality (3) is a equivalent to
$$
9\leq ss_h\leq \rho\left(\frac{4p}{s^3}\right),
$$
where $\rho=\rho(t)$ is the real root of
$$
t \rho^3=9+2 \rho+\rho^2.
$$
Hence given positive numbers $a,b,c$ we have
$$
9\leq (a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\leq x,
$$
where $x$ is the real root of
$$
9+2x+x^2=\frac{4abc}{(a+b+c)^3}x^3.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Solving $\displaystyle \int_{-\tan^{-1}\sqrt{4a^2-1}}^{\tan^{-1}\sqrt{4a^2-1}}\log(2\cos \theta)d\theta$ $$\int_{-\tan^{-1}\sqrt{4a^2-1}}^{\tan^{-1}\sqrt{4a^2-1}}\int_{\frac{1}{a}}^{2\cos\theta}\frac{1}{x}dxd\theta=\int_{-\tan^{-1}\sqrt{4a^2-1}}^{\tan^{-1}\sqrt{4a^2-1}}(\log(2\cos \theta)\,\mathrm +\log(a))d\theta=$$
$$=2\log(a)(\tan^{-1}\sqrt{4a^2-1})+\int_{-\tan^{-1}\sqrt{4a^2-1}}^{\tan^{-1}\sqrt{4a^2-1}}\log(2\cos \theta)d\theta$$
I want to to know if when $a\rightarrow \infty$ the original integral converges and I don't know how to solve the last integral and it seems its difficult (maybe its necesary use further techniques).
So maybe a better way its use inequalites and find an easier integral but I cant find it, so can you help me please?
| Note that, for $a\to \infty$, $\tan^{-1}\sqrt{2a^2-1} \to \frac\pi2$. Then, the Integral
$$\int_{-\frac\pi2}^{\frac\pi2}\log(2\cos \theta)d\theta=0$$
However, the first term $2\log(a)(\tan^{-1}\sqrt{4a^2-1})$
diverges; so does the original integral.
————-
Edit:
\begin{align}
I&=\int_0^{\frac{\pi}{2}}\log(2\cos x)dx=\int_0^{\frac{\pi}{2}}\log(2\sin x)dx\\
&=\int_0^{\frac{\pi}{2}}\log(2\cos\frac{x}{2})dx+\int_0^{\frac{\pi}{2}}\log(2\sin\frac{x}{2})dx\\
&=2\int_0^{\frac{\pi}{4}}\log(2\cos x)dx+2\int_0^{\frac{\pi}{4}}\log(2\sin x)dx\\
&=2\int_0^{\frac{\pi}{4}}\log(2\cos x)dx+2\int^{\frac\pi2}_{\frac{\pi}{4}}\log(2\cos x)dx\\
&=2\int_0^{\frac{\pi}{2}}\log(2\cos x)dx=2I \implies I=0
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\frac{x^{2}}{(x-y)^{2}}+\frac{y^{2}}{(y-z)^{2}}+\frac{z^{2}}{(z-x)^{2}} \geq 1$ Question -
Let $x, y, z$ be distinct real numbers. Prove that
$$
\frac{x^{2}}{(x-y)^{2}}+\frac{y^{2}}{(y-z)^{2}}+\frac{z^{2}}{(z-x)^{2}} \geq 1
$$
My work -
first i apply directly C-S and after simplification i have to prove that
$4(xy+yz+zx)>x^2+y^2+z^2$ which i am not able to prove ..
then multiplying $x^2,y^2,z^2$ to numerators and denominators of corresponding fractions respectively i again apply C-S and this time we have to prove that after simplification
$(xy)^2+(yz)^2+(zx)^2 + 2x^3y+2y^3z+2z^3x > 0$ which again i fail to prove ...
I try to do with other inequalities but none of them working.
any help will be helpful
thankyou
| Let $a = \dfrac{x}{x-y}, b = \dfrac{y}{y-z}, c = \dfrac{z}{z-x}$, and note
$a+b+c = ab+bc+ca+1$. We need to show $a^2+b^2+c^2 \geqslant 1$. But this is now just equivalent to $(a+b+c-1)^2 \geqslant 0$
Alternately, the original inequality $\iff \dfrac{(x^2y+y^2z+z^2x-3xyz)^2}{(x-y)^2(y-z)^2(z-x)^2}\geqslant 0$
| {
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Let $x, y \in \mathbb R$ such that $x^2+y^2=2x-2y+2$. Find the largest possible value of $x^2+y^2$
Let $x, y \in \mathbb R$ s. t . $x^2+y^2=2x-2y+2$. Find the largest possible value of $x^2+y^2-\sqrt{32}$
I know this is a duplicate of another question, but that question has solutions involving calculus and geometry, while I want a solution relying on algebra and basic inequalities only to solve this problem.
| Yet another approach: Let $S=\{(x,y) : x^2+y^2=2x-2y+2\}$ be the algebraic curve of the constraint and let $A=\{(x,y): x+y=0\}$ be the anti-diagonal. We have the basic inequality: $(x-y)^2 \leq 2(x^2+y^2)$ with equality iff $(x,y)\in A$. Thus, for points in $S$ we have:
$$ (x-y)^2 \leq 4(x-y) + 4 $$
with equality iff $(x,y)\in A$ (in addition to being in $S$).
Equivalently, $(x-y-2)^2 \leq 8$ or $2-2\sqrt{2} \leq x-y \leq 2+2\sqrt{2}$, with the two extremal cases corresponding to $S\cap A$. Evaluating on $S$ we get the upper bound:
$$ x^2+y^2 -\sqrt{32} \leq 2(2+2\sqrt{2}) + 2 - 4\sqrt{2} = 6$$
with equality being attained at the unique intersection of $x-y=2+2\sqrt{2}$ and $x+y=0$, i.e. $x=-y=1+\sqrt{2}$
| {
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How to compute $\int_0^{\frac{\pi}{2}}\frac{\arctan(\sqrt{\tan(x)})}{\tan(x)}dx$ I have been asked to compute the integral $$\int_0^{\frac{\pi}{2}}\frac{\arctan(\sqrt{\tan(x)})}{\tan(x)}dx$$
I have been told that it converges, and we only need its value.
I tried the substitutions $ u=\tan(x)$, $\;\; v^2=\tan(x) $ .
I thought about by parts integration, but it becomes more complicated.
I thank you in advance for an idea .
| $$\begin{align}
\int_{0}^{\frac{\pi}{2}}\frac{\arctan\left(\sqrt{\tan\left(x\right)}\right)}{\tan\left(x\right)}dx &= \frac{1}{2}\int_{0}^{\pi}\frac{\arctan\left(\sqrt{\tan\left(\frac{x}{2}\right)}\right)}{\tan\left(\frac{x}{2}\right)}dx
\\&\stackrel{(1)}{=}\int_{0}^{\infty}\frac{\arctan\left(\sqrt{t}\right)}{t\left(1+t^{2}\right)}dt
\\&=2\int_{0}^{\infty}\frac{\arctan u}{u\left(1+u^{4}\right)}du
\\&\stackrel{(2)}{=}2\int_{0}^{\infty}\int_{0}^{1}\frac{1}{1+u^{2}t^{2}}\frac{1}{1+u^{4}}dtdu
\\&\stackrel{(3)}{=}2\int_{0}^{1}\int_{0}^{\infty}\frac{1}{1+u^{2}t^{2}}\frac{1}{1+u^{4}}dudt
\\&\stackrel{(4)}{=}2\int_{0}^{1}\frac{1}{1+t^{4}}\int_{0}^{\infty}\left(\frac{1}{1+u^{4}}+\frac{t^{4}}{t^{2}u^{2}+1}-\frac{t^{2}u^{2}}{1+u^{4}}\right)dudt
\\&=2\int_{0}^{1}\frac{1}{1+t^{4}}\left(\frac{\pi}{2\sqrt{2}}+\frac{\pi t^{3}}{2}-\frac{\pi t^{2}}{2\sqrt{2}}\right)dt
\\&= \frac{\pi}{\sqrt{2}}\int_{0}^{1}\left(\frac{1-t^{2}}{1+t^{4}}\right)dt+\pi\int_{0}^{1}\frac{t^{3}}{1+t^{4}}dt
\\&=\frac{\pi}{\sqrt{2}}\frac{\operatorname{arccoth}\left(\sqrt{2}\right)}{\sqrt{2}}+\frac{\pi\log\left(2\right)}{4}
\\&=\frac{\pi}{4}\log\left(6+4\sqrt{2}\right) = \frac{\pi}{2}\log\left(2+\sqrt{2}\right)
\end{align}$$
Where:
$(1)$ is the Weirstrass Subsitution
$(2)$ is using the fact $\frac{\arctan{x}}{x} = \int_0^1 \frac{du}{1+x^2 u^2} $
$(3)$ is interchanging integrals via Fubini's Theorem (our integrand is positive and decays rapidly)
$(4)$ is a partial fraction decomposition
| {
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Prove $\cos 20^\circ \cos 40^\circ \cos 80^\circ= \frac18$ geometrically I understand the wizardry that $\cos 20^\circ \cos 40^\circ \cos 80^\circ=\dfrac18$
Proving it isn't that hard. Taking the left hand side and multiplying it up and down by $\sin 20$ yields:
\begin{align}
& \dfrac{\sin 20^\circ}{\sin 20^\circ} \cdot \cos 20^\circ \cos 40^\circ \cos 80^\circ \\
=&\dfrac{\sin 40^\circ}{2\sin 20^\circ} \cdot \cos 40^\circ \cos 80^\circ \\ =& \dfrac{\sin 80^\circ}{4\sin 20^\circ} \cdot \cos 80^\circ \\ =& \dfrac{\sin 160^\circ}{8\sin 20^\circ}=\dfrac{\sin 20^\circ}{8\sin 20^\circ}=\boxed{\dfrac{1}{8}}
\end{align}
My question is, is there a geometric proof of this, or the corollary $\sin 10 \sin 50 \sin 70^\circ$? Inspired by the possibility from this other indentity
I'm hoping that it's possible. I tried doing a similar construction as in the link that didn't go so well, mainly because of my repeated use of the cosine rule.
Is my quest achievable? Thanks for any guidance and advice.
|
Configure the angles in the diagram. Then
\begin{align}
& \triangle ABE: \>\>\> \cos 20 = \frac{\frac y2}x = \frac y{2x}\\
& \triangle BGE: \>\>\> \cos 40 = \frac{\frac x2}1 = \frac x2\\
& \triangle ABC: \>\>\> \cos 80 = \frac{\frac12}y = \frac1{2y} \\
\end{align}
Thus,
$$\cos 20\cos 40 \cos 80 = \frac y{2x}\frac x2\frac1{2y} = \frac18$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3658066",
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Prove that $\sqrt{\frac{2 a^{2}+b c}{a^{2}+2 b c}}+\sqrt{\frac{2 b^{2}+c a}{b^{2}+2 c a}}+\sqrt{\frac{2 c^{2}+a b}{c^{2}+2 a b}} \geq 2 \sqrt{2}$ Question -
Prove that for all non-negative real numbers a,b, c, we have
$$
\sqrt{\frac{2 a^{2}+b c}{a^{2}+2 b c}}+\sqrt{\frac{2 b^{2}+c a}{b^{2}+2 c a}}+\sqrt{\frac{2 c^{2}+a b}{c^{2}+2 a b}} \geq 2 \sqrt{2}
$$
My work -
we may assume that $a b c=1$
The problem becomes
$$
\sqrt{\frac{2 x+1}{x+2}}+\sqrt{\frac{2 y+1}{y+2}}+\sqrt{\frac{2 z+1}{z+2}} \geq 2 \sqrt{2}
$$
where $x=a^{3}, y=b^{3}, z=c^{3}$
now i did not know where to go from here ...i tried all classic inequalities like chebyshev,re-arangement but none of them working .
can anyone solve this using classic inequalities
any help will be appreciated
thank you
| By C-S twice we obtain: $$\sum_{cyc}\sqrt{\frac{2a^2+bc}{a^2+2bc}}-2\sqrt2=\sum_{cyc}\frac{\sqrt{(2a^2+bc)(a^2+2bc)}}{a^2+2bc}-2\sqrt2\geq$$
$$\geq\sum_{cyc}\frac{\sqrt2(a^2+bc)}{a^2+2bc}-2\sqrt2=\sqrt2\left(\sum_{cyc}\left(\frac{a^2+bc}{a^2+2bc}-\frac{1}{2}\right)-\frac{1}{2}\right)=$$
$$=\sqrt2\left(\sum_{cyc}\frac{a^2}{2(a^2+2bc)}-\frac{1}{2}\right)\geq \sqrt2\left(\frac{(a+b+c)^2}{2\sum\limits_{cyc}(a^2+2bc)}-\frac{1}{2}\right)=0.$$
| {
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Finding $\frac{c\sin(A-B)}{a^2-b^2}-\frac{b\sin(C-A)}{c^2-a^2}$
If $a, b$ and $c$ (all distinct) are the sides of a triangle ABC opposite to the angles $A, B$ and $C$, respectively, then $\frac{c\sin(A-B)}{a^2-b^2}-\frac{b\sin(C-A)}{c^2-a^2}$ is equal to $?$
By opening, $\sin(A-B)$ as $\sin A\cos B-\cos A\sin B$ and then $\sin A$ as $\frac{a}{2R}$ and $\cos A$ as $\frac{b^2+c^2-a^2}{2bc}$, I am able to get the answer as zero. But I am looking for a shorter solution, maybe by putting values of angles and sides.
My first instinct was to assume the triangle to be equilateral. But the question invalidates that case. Then I thought of a right angled triangle with pythogorean triplet as $3,4,5$. But here, I don't know other two angles.
I wonder if I could just keep a dummy triangle handy whose sides and angles I know, which I could quickly use to solve such questions. Any help please? Thanks.
| Assume that $a>b$.
There exists a point $D$ on line segment $AB$ such that $CD=b$, Produce $CD$ to meet the circumscribed circle of $\Delta ABC$ at point $E$. (See the figure below.)
Since $AC=CD$, $\angle CDA=\angle CAD=A$. Hence, $\angle BCD=A-B$.
Since $\angle BAE$ and $\angle BCE$ are angles in the same segment, $\angle BAE=\angle BCE=A-B$.
Applying the power theorem on the circle with center $C$ and radius $b$, we have $AB\cdot BD=a^2-b^2$. Since $AB=c$, $BD=\dfrac{a^2-b^2}{c}$.
Note that $\Delta BDE$ and $\Delta CDA$ are similar. We have $BE=BD=\dfrac{a^2-b^2}{c}$.
Applying the sine formula on $\Delta ABE$, we have $\dfrac{BE}{\sin\angle BAE}=2R$, where $R$ is the radius of the circumscribed circle of $\Delta ABC$. Hence, we have $\dfrac{c\sin(A-B)}{a^2-b^2}=\dfrac{1}{2R}$.
If $b>a$ instead, then we have $\dfrac{c\sin(B-A)}{b^2-a^2}=\dfrac{1}{2R}$. But $\dfrac{c\sin(A-B)}{a^2-b^2}$ and $\dfrac{c\sin(B-A)}{b^2-a^2}$ are indeed equal.
Similarly, we have $\dfrac{b\sin(C-A)}{c^2-a^2}=\dfrac{1}{2R}$. Thus, we have $\dfrac{c\sin(A-B)}{a^2-b^2}-\dfrac{b\sin(C-A)}{c^2-a^2}=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does $\sum _{n=1}^{\infty }\sum _{i=1}^{n }\frac{\left(-1\right)^n}{i\cdot n}$ have a finite value? If so, evaluate its closed form. Does $\sum _{n=1}^{\infty }\sum _{i=1}^{n }\frac{\left(-1\right)^n}{i\cdot n}$ have a finite value? If so, evaluate its closed form.
I'm pretty sure its related to $\begin{array}{l}\zeta \left(2\right)=\frac{\pi ^2}{6}\end{array}$
But i am having trouble converting this into a form that can be further manipulated.
Thanks ☺☺☺
| Consider $$\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}...\infty\right)^2=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}...\infty+2\left(1\left(-\frac{1}{2}\right)+1\left(\frac{1}{3}\right)+...\ +-\frac{1}{2}\left(\frac{1}{3}\right)+-\frac{1}{2}\left(-\frac{1}{4}\right)...\infty\right)$$
The numbers in the brackets are the sum of products of unique pairs of all numbers.*(-1)^(sum of both numbers)
We can rearrange it so that the denominators are in a sequence.
Let $$x=-1\left(-\frac{1}{2}\right)+1\left(\frac{1}{3}\right)+...\ +-\frac{1}{2}\left(\frac{1}{3}\right)+-\frac{1}{2}\left(-\frac{1}{4}\right)...\infty $$
$$=\left(-\frac{1}{1\cdot 2}-\frac{1}{2\cdot 3}-\frac{1}{3\cdot 4}...\infty \right)+\left(\left(\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 8}...\infty \right)+\left(\frac{1}{2.4}+\frac{1}{4.6}+...\right)\right)-\left(\frac{1}{1\cdot 4}+...and\ so\ on\right)$$
Now each term telescopes,
$$T_n=\frac{1}{n}\sum _{i=1}^n\frac{1}{i}$$
This is equivalent to the nth term in the question, So the sum is
$$\begin{array}{l}\frac{\left(\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}...\infty \right)^2-\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}...\infty \right)\right)}{2}=\frac{\left(\ln \left(2\right)\right)^2}{2}-\frac{\pi ^2}{12}\end{array}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Integral $ \int_{-\infty}^{+\infty} \frac{ \sin^2(\sqrt{(x-a)^2 + b^2}\,\,t)}{(x-a)^2 + b^2} dx$ I am trying to solve the following definite integral:
$$ \int_{-\infty}^{+\infty} \frac{ \sin^2(\sqrt{(x-a)^2 + b^2}\,\,t)}{(x-a)^2 + b^2} dx$$
with $a$ and $b$ being real constants. Notice that the integrand is non-negative for all x , with a peak around $x=a$ for small values of $b/a$. So the integration does give a finite value.
To go about this, I tried substituting $(x-a)^2 + b^2 = y^2$
This gives me $dx \,(x - a) = y \, dy$ whence:
$$ \int_{?}^{+\infty} \frac{ \sin^2(|y|\,\,t)}{y} \frac{1}{\sqrt{y^2 - b^2}} dy$$
While the upper limit of y transforms to $+\infty$, clearly the lower limit is not $-\infty$ (even if it is $-\infty$ the integrand is odd so it evaluates to zero which cannot be true).
This suggests its a case of bad substitution. The only better alternative I can think of is to substitute $x-a = y$, this gives me:
$$ \int_{-\infty}^{+\infty} \frac{ \sin^2(\sqrt{y^2 + b^2}\,\,t)}{y^2 + b^2} dy $$
Any thoughts on how I can get an expression for this integral ?
Thanks for your time!
| Thanks Gary. Your solution was quite impressive.
Let me complicate the integral to a level higher and see if we can proceed similarly to get an analytical solution. I now write the integral to be solved as:
$$ \int_{-\infty}^{+\infty} \frac{\sin^2 \sqrt{(x^2-a^2)^2 + b^4 }\, t}{(x^2-a^2)^2 + b^4 } \, \, dx $$
You will observe that for $x \approx a$ we can write $x^2 - a^2 \approx 2 a (x - a)^2 $ thus reducing this integral to a form similar to what you have solved previously. However, I now want to arrive at a solution without making such an approximation.
I proceed now similar to what you had done previously.
I attempted to re-write the integral by replacing $\left(\frac{x^2- a^2}{b^2} \right)^2 = w^2 $ - however the even integrand will no longer continue to be even with this substitution. So I continue without such a replacement.
$$ \frac{2}{b^4} \int_{0}^{+\infty} \frac{\sin^2 b^2 t \sqrt{\left(\frac{x^2-a^2}{b^2} \right)^2 +1}}{ \left(\frac{x^2-a^2}{b^2} \right)^2 +1 } \,\,dx $$
Differentiating with respect to time, this can be shown to reduce to:
$$ \frac{2}{b^2} \int_{0}^{+\infty} \frac{\sin 2 b^2 t \sqrt{\left(\frac{x^2-a^2}{b^2} \right)^2 +1}}{ \sqrt{ \left(\frac{x^2-a^2}{b^2} \right)^2 +1 } } \,\,dx $$
Now comes the hard part: I attempt replacing $\sqrt{\left(\frac{x^2-a^2}{b^2} \right)^2 +1 } = u $. This integral will be equivalent to:
$$2 b^2 \int_{\sqrt{\frac{a^4}{b^4}+1}}^{+\infty} \frac{\sin 2 b^2 u t }{ \sqrt{u^2 - 1} } \,\, \frac{1 }{\sqrt{a^2 + b^2 \sqrt{u^2 - 1} }} du $$ At this point I have no idea how to proceed. Any ideas ?
Thanks!
| {
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Can the expression $a^2 + b^2 - c^2$ be factored as a product of two quaternions, where $a$, $b$, $c$ are real numbers? I'm trying to factor the expression
$$a^2+b^2-c^2$$
as a product of two quaternions, where $a,b,c$ are reals. Can anyone give me the answer?
I think it can't be done.
I started with multiplying
$$(Aa,Bb,Cc,Dd)(Ea,Fb,Gc,Hd)=\\
(a^2 A E - b^2 B F - c^2 C G - d^2 D H,\\
a b B E + a b B F - c d D G + c d C H, \\
a c C E + b d D F + a c A G - b d B H, \\
a d D E - b c C F + b c B G + a d A H)$$
For some real numbers $A, B, \dotsc, H$ , where I use the convention $(A,B,C,D)=A +Bi +Cj+Dk$. Since we want the result to be real for all $a,b,c,d$ we can write
$$D E + A H = 0\\
C F - B G = 0\\
C E + A G =0\\
D F -B H = 0\\
B E + A F = 0\\
D G - C H = 0$$
What remains is to set the real part to be equal to the desired result.
For example if we choose the polynomial $a^2+b^2+c^2$ we get
$$a^2+b^2+c^2=(0,a,b,c)(0,-a,-b,-c).$$
However, my question was to find the factorization for $$a^2+b^2-c^2.$$ In that case the system doesn't have any real solutions for $A,B,\dotsc,H$.
I suspect therefore that all the polynomials, where one sign differs from the others, e.g.
$$a^2-b^2+d^2\\
b^2+c^2-d^2$$
will also not be factorisable.
Can anyone confirm this?
| It cannot be written as the product of two $\mathbb{H}$-linear combinations of variables $a,b,c$. (We assume the formal variables commute with all quaternion scalars, of course.)
Suppose it were, say $(ap+q)(ar+s)$ where $p,q\in\mathbb{H}$ and $r,s$ are $\mathbb{H}$-linear combinations of $b$ and $c$. The leading term (with respect to $a$) would then be $pra^2$ which must be $a^2$ and so $pr=1$. We may factor the coefficients of $a$ out as $(a+qp^{-1})pr(a+r^{-1}s)=(a+u)(a+v)$. Now the middle term is $(u+v)a$, which must be $0$ as in the expression $a^2+b^2-c^2$, so $v=-u$ and we get $(a+u)(a-u)=a^2-u^2$. Thus, we want some combination of $b$ and $c$ (that is, $u$) which squares to $b^2-c^2$.
Writing $(xb+yc)^2=b^2-c^2$, we see $x^2=1$ so $x=\pm1$. But then the middle term is $\pm2ybc$ which can only be $0$ if $y=0$ in which case we don't see $-c^2$ as the final term.
It is however factorizable in the split quaternions. Then $a^2+b^2-c^2-d^2$ is the product of the split quaternion $a+bi+cj+dk$ with its conjugate $a-bi-cj-dk$.
| {
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Euclidean geometry : find out the area.
${\angle}ABC=45°$, $AD{\bot}AE$, $AD=AE$, $AC{\bot}BE$, $F$ is the intersection of $AC$ and $BE$, $BF=9$, so what is the area of ${\square}ABCE$?
| Let $F(0,0), A(0,a), B(-b,0), C(0,-c), E(e,0)$, then $D=tB+(1-t)C=
(-tb,-(1-t)c)$ for some $t$ ($t\in(0;1)$ if we believe the drawing, but I won't use it).
The desired area is then $\frac12(ae+ec+bc+ba)$.
We have:
$AD\perp AE\Leftrightarrow (D-A).(E-A)=0 \Leftrightarrow (-tb,-(1-t)c-a).(e,-a)=0 \Leftrightarrow -tbe+(1-t)ca+a^2=0$
$|AD|=|AE|\Leftrightarrow (D-A)^2=(E-A)^2 \Leftrightarrow t^2b^2+(1-t)^2c^2+a^2+2(1-t)ca=a^2+e^2$
$AC\perp BE$ ve have already used when introducing the coordinate system
$BF=9\Leftrightarrow b=9$
and $\angle ABC=45^\circ \Leftrightarrow (A-B).(C-B)=\frac{\sqrt{2}}{2}|A-B|\cdot|C-B|\Rightarrow ((a,b).(b,-c))^2=\frac{1}{2}(a^2+b^2)(b^2+c^2)
\Leftrightarrow 2(ab-bc)^2=(a^2+b^2)(b^2+c^2)$
Then we wolframalpha this thing $\begin{cases}
-tbe+(1-t)ca+a^2=0\\
t^2b^2+(1-t)^2c^2+a^2+2(1-t)ca=a^2+e^2\\
b=9\\
2(ab-bc)^2=(a^2+b^2)(b^2+c^2)
\end{cases}$
and see there's one condition missing as the desired area $S=a^2 - \frac{2916}{a - 9} - 243$ can be made arbitrary large if we take $a$ close enough to $9$.
| {
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Determining the period of $\sin 2x +\sin\frac{x}{2}$ without using the LCM of periods
I tried to calculate period of function described as:
$$y=\sin 2x +\sin\frac{x}{2}$$
but without using LCM of periods.
From definition of periodic function we have:
$$\begin{align}
0 &= \sin(2x+2T)+\sin\left(\frac{x+t}{2}\right)-\sin2x-\sin\left(\frac{x}{2}\right) \\[4pt]
&=2\sin\left(\frac{5x+5T}{4}\right)\cos\left(\frac{x+T}{4}\right)-2\sin\left(\frac{5x}{4}\right)\cos\left(\frac{3x}{4}\right)
\end{align}$$
I cannot do further more.
| Let $$\sin2(x+T)+\sin\frac{x+T}{2}=\sin2x+\sin\frac{x}{2},$$ where $T>0$.
Thus, for $x=0$ we obtain:
$$\sin2T+\sin\frac{T}{2}=0$$ and for $x=2\pi$ we obtain:
$$\sin2T-\sin\frac{T}{2}=0,$$
which gives $$\sin2T=\sin\frac{T}{2}=0$$ or
$$T=2\pi k,$$ where $k$ is a positive integer number.
We see that $k=1$ is not valid, but $k=2$ gives the answer:
$$T=4\pi.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3672859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $z^2+z+2=0$, then $z^2 + \frac{4}{z^2} = -3$
If $z^2+z+2=0$, for $z\in\mathbb C$, demonstrate:
$$z^2 + \frac{4}{z^2} = -3$$
$z^2 = - 2 - z$, but it didn't help me.
Is there any other elegant solution?
| $$z^2+\frac{4}{z^2}+3$$
$$=\frac{z^4+3z^2+4}{z^2}$$
$$=\frac{(z^2+z+2)(z^2-z+2)}{z^2}$$
$$=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3674164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove this inequality with $xyz=1$ let $x,y,z>0$ and such $xyz=1$,show that
$$f(x)+f(y)+f(z)\le\dfrac{1}{8}$$
where $f(x)=\dfrac{x}{2x^{x+1}+11x^2+10x+1}$
I try use this $2x^x\ge x^2+1$,so we have
$$2x^{x+1}+11x^2+10x+1\ge x^3+11x^2+11x+1=(x+1)(x^2+10+1)$$
It need to prove
$$\sum_{cyc}\dfrac{x}{(x+1)(x^2+10x+1)}\le\dfrac{1}{8},$$where $xyz=1$
then I can't
| The TL method helps!
Since $$x^x\geq\frac{x^3-x^2+x+1}{2},$$ it's enough to prove that
$$\sum_{cyc}\frac{x}{x^4-x^3+12x^2+11x+1}\leq\frac{1}{8}$$ or
$$\sum_{cyc}\left(\frac{1}{24}-\frac{x}{x^4-x^3+12x^2+11x+1}\right)\geq0$$ or
$$\sum_{cyc}\left(\frac{1}{24}-\frac{x}{x^4-x^3+12x^2+11x+1}-\frac{1}{48}\ln{x}\right)\geq0.$$
Now, prove that for any $0<x<6$ we have $$\frac{1}{24}-\frac{x}{x^4-x^3+12x^2+11x+1}-\frac{1}{48}\ln{x}\geq0,$$
which says that our inequality is proven for $\max\{x,y,z\}<6.$
Let $x\geq6$.
Now, we see that for any $x>0$ we have $$\frac{x}{x^4-x^3+12x^2+11x+1}\leq\frac{1}{17}$$ and for any $x\geq6$ we have $$\frac{x}{x^4-x^3+12x^2+11x+1}\leq\frac{6}{6^4-6^3+12\cdot6^2+11\cdot6+1}\leq\frac{6}{1579}.$$
Id est,
$$\sum_{cyc}\frac{x}{x^4-x^3+12x^2+11x+1}\leq\frac{6}{1579}+\frac{2}{17}<\frac{1}{8}$$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3674817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Convergence of the recursive sequence $x_n=\frac{x_{n-1}}2+\frac3{x_{n-1}},n\ge 1$ Consider the iterative scheme :
$x_n=\frac{x_{n-1}}2+\frac3{x_{n-1}},n\ge 1$
with the initial point $x_0 \gt 0$. Then the sequence $\{x_n\}$
$(a)$ converges only if $x_o \lt 3$
$(b)$converges for any $x_0$
$(c)$does not converge for any $x_0$
$(d)$converges only if $x_0\gt 1$
My attempt :
Since $x_0\gt 0$ ,we have $x_n\gt 0, \forall n\in \mathbb{N}$
Now , $x_n-x_{n-1}=\frac{x_{n-1}^2+6-2x_{n-1}^2}{x_{n-1}}=\frac{6-x_{n-1}^2}{2x_{n-1}}$
$\Rightarrow x_n\lt ,\gt x_{n-1}$ according as $x_{n-1}\gt ,\lt \sqrt{6}$.
Again,by the given recurrence relation ,
$x_{n-1}^2-2x_nx_{n-1}+6=0$
This is quadratic in $x_{n-1}$ and to have a real solution
$4x_{n}^2-24\ge 0$ i.e $x_n\ge \sqrt6$ or $x_{n}\le -\sqrt6$
The latter is not possible, so $x_n\gt \sqrt6$
What does this mean?? Does this mean that even if $x_0$ be any value greater than zero and smaller than $\sqrt6$, the sequence is ultimately monotone decreasing converging to $\sqrt6$ (limit obtained from the relation)? So, should my answer be $(b)$?
Please help me with your ideas. Thanks for your time.
| There is a nice substitution and technique applied here. It enables us to find a closed form of $x_n$ from which the correct alternative can be easily derived. Given
\begin{align*}
x_n=\frac{x_{n-1}}{2}+\frac{3}{x_{n-1}}\qquad\qquad n\geq 1
\end{align*}
we substitute
$\color{blue}{x_n=\sqrt{6}\,y_n}$
and we obtain
\begin{align*}
y_n&=\frac{y_{n-1}}{2}+\frac{1}{2y_{n-1}}\\
&=\frac{y_{n-1}^2+1}{2y_{n-1}}\tag{1}
\end{align*}
From (1) we get
\begin{align*}
y_n-1&=\frac{\left(y_{n-1}-1\right)^2}{2y_{n-1}}\\
y_n+1&=\frac{\left(y_{n-1}+1\right)^2}{2y_{n-1}}\\
\frac{y_n-1}{y_n+1}&=\left(\frac{y_{n-1}-1}{y_{n-1}+1}\right)^2=\cdots=\underbrace{\left(\frac{y_0-1}{y_0+1}\right)^{2^n}}_{C_n}\tag{2}
\end{align*}
We can now calculate $y_n$ from (2) and obtain a closed form of $x_n$ this way:
\begin{align*}
y_{n}-1&=C_n\left(y_n+1\right)\\
y_n\left(1-C_n\right)&=1+C_n\\
y_n&=\frac{1+C_n}{1-C_n}\qquad\Rightarrow\qquad
\color{blue}{x_n=\sqrt{6}\,\frac{1+\left(\frac{x_0-\sqrt{6}}{x_0+\sqrt{6}}\right)^{2^n}}
{1-\left(\frac{x_0-\sqrt{6}}{x_0+\sqrt{6}}\right)^{2^n}}\qquad n\geq 0}\tag{3}
\end{align*}
We conclude from (3)
\begin{align*}
\left\{x_n\right\}_{n\geq 0}\ \mathrm{convergent}\quad&\Leftrightarrow\quad\left|\frac{x_0-\sqrt{6}}{x_0+\sqrt{6}}\right|<1\\
&\Leftrightarrow\quad\left|x_0-\sqrt{6}\right|<\left|x_0-\left(-\sqrt{6}\right)\right|\\
&\,\,\color{blue}{\Leftrightarrow\quad x_0>0}
\end{align*}
and (b) is the correct answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3676016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Finding pdfs of $\frac1{X^2}$ and $\frac{1}{4}\left(\frac1{X^2}+\frac1{W^2}\right)$ where $X,W$ are independent $N(0,1)$ $X,W$ are independent random variables, both $N(0,1)$, i.e. $f_X(x)=f_W(x)= \frac{1}{{\sqrt{2\pi}}}e^{-\frac{x^2}{2}}$.
*
*Find PDF of $Y:=\frac{1}{X^2}$
*Find PDF of $\frac{1}{4}\left(\frac{1}{X^2}+\frac{1}{W^2}\right)$
For the first case $a)$ I have:
\begin{eqnarray*}
F_Y(y)=P(Y \leq y)=P(\frac{1}{X^2}\leq y) \:=
\end{eqnarray*}
\begin{Bmatrix}P(\frac{1}{X^2}\leq y) \: y<0
& \\ P(\frac{1}{X^2} \leq y), \: y>0
&
\end{Bmatrix}
\begin{Bmatrix} 0, \: y<0
& \\ P(\frac{1}{X^2} \geq y), \: y>0
&
\end{Bmatrix}
$$ P(X^2 \geq \frac{1}{y})= P(X \leq -\frac{1}{\sqrt{y}} \cup X\geq \frac{1}{\sqrt{y}})=$$
$$ = P(X\leq -\frac{1}{\sqrt{y}}) + 1 - P(X\leq \frac{1}{\sqrt{y}}))= $$
$$= F_X(-\frac{1}{\sqrt{y}}) +1 - F_X(\frac{1}{\sqrt{y}}) $$
we have therefore:
$$ F_Y(y)= \left\{\begin{matrix}F_X(-\frac{1}{\sqrt{y}}) +1 - F_X(\frac{1}{\sqrt{y}}), \: y>0
& \\ 0, \: y<0
&
\end{matrix}\right. $$
for $y>0$ we have:
$$ f_Y(y) = f_X(-\frac{1}{\sqrt{y}}))*(-\frac{1}{\sqrt{y}}))´ - f_X(\frac{1}{\sqrt{y}}))*(\frac{1}{\sqrt{y}}))´ $$
$$ = f_X(-\frac{1}{\sqrt{y}}))*\frac{y^{\frac{-3}{2}}}{2} - f_X(\frac{1}{\sqrt{y}}))*(-\frac{y^{\frac{-3}{2}}}{2})$$
$$ =\frac{y^{\frac{-3}{2}}}{2}*(f_X(-\frac{1}{\sqrt{y}})) - f_X(\frac{1}{\sqrt{y}}))) $$
$$ = \frac{1}{2} * y^{\frac{-3}{2}}(\frac{1}{\sqrt{2\pi }}e^{-\frac{1}{2y}} + \frac{1}{\sqrt{2\pi }}e^{-\frac{1}{2y}}) = y^{\frac{-3}{2}}*\frac{1}{\sqrt{2\pi }}*e^{-\frac{1}{2y}}$$
For the second part, there is a hint that we should use polar coordinates and that the substitution $t=\cot(2 \theta)$ could be helpful.
So since we have the PDF of both X and W (if what I have in (a) is correct), I tried using the convolution formula and calculating the integral, but I haven't been very successful, so I'd appreciate some help.
| Your work for the first problem is correct.
Let $\Phi(\cdot)$ and $\phi(\cdot)$ be the cdf and pdf of standard normal distribution, as usual.
Since $X$ is standard normal, you have for every $y>0$,
$$P\left(\frac1{X^2}\le y\right)=\cdots=1-\Phi\left(\frac1{\sqrt y}\right)+\Phi\left(-\frac1{\sqrt y}\right)=2\left(1-\Phi\left(\frac1{\sqrt y}\right)\right)$$
So the pdf of $Y=\frac1{X^2}$ must be
$$f_Y(y)=\frac1{y^{3/2}}\phi\left(\frac1{\sqrt y}\right)=\frac1{\sqrt{2\pi}y^{3/2}}e^{-1/2y}\mathbf1_{y>0}$$
You can also use change of variables to directly say
\begin{align}
f_Y(y)&=\phi\left(\frac1{\sqrt y}\right)\left|\frac{\mathrm d}{\mathrm dy}\left(\frac1{\sqrt y}\right)\right|+\phi\left(\frac{-1}{\sqrt y}\right)\left|\frac{\mathrm d}{\mathrm dy}\left(\frac{-1}{\sqrt y}\right)\right|
\\&=2\phi\left(\frac1{\sqrt y}\right)\left|\frac{\mathrm d}{\mathrm dy}\left(\frac1{\sqrt y}\right)\right|
\\&=\frac1{y^{3/2}}\phi\left(\frac1{\sqrt y}\right)
\end{align}
For the second problem, transform to polar coordinates $(X,W)\mapsto (R,\Theta)$ such that $X=R\cos\Theta$ and $W=R\sin\Theta$. Then
$$Z=\frac14\left(\frac1{X^2}+\frac1{W^2}\right)=\frac{X^2+W^2}{4X^2W^2}=\frac1{(R\sin(2\Theta))^2}$$
All you have to do now is show that $R\sin(2\Theta)$ also has a standard normal distribution, so that $$Z\stackrel{d}=\frac1{X^2}=Y$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3676992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Are there formulas for $\sin\left(\frac{\theta}{3}\right)$ and $\cos\left(\frac{\theta}{3}\right)$? So far, I have seen the half angle formula (i.e.)
$$\sin^2\left(\frac{\theta}{2}\right) = \frac{1-\cos\theta}{2}$$
Is there a way to find $\sin\left(\frac{\theta}{3}\right)$ and $\cos\left(\frac{\theta}{3}\right)$ in terms of $\sin\theta$ and/or $\cos\theta$ (i.e. a closed form solution)?
| There is no analytic formula for $\sin\frac{\theta}{3}$ & $\cos\frac{\theta}{3}$ . But the value of $\sin\frac{\theta}{3}$ & $\cos\frac{\theta}{3}$ can be derived in terms of $\sin\theta $ & $\cos\theta$ using triple angle formula (which may also give complex roots) as follows
$$\sin3x=3\sin x-4\sin^3x$$
setting $3x=\theta$ or $x=\frac{\theta}{3}$ in above trig. identity,
$$\sin\theta=3\sin \frac{\theta}{3}-4\sin^3\frac{\theta}{3}$$$$\boxed{4\sin^3\frac{\theta}{3}-3\sin \frac{\theta}{3}+\sin\theta=0}$$
Similarly,
$$\cos3x=4\cos^3x-3\cos x$$
Setting $3x=\theta$ or $x=\frac{\theta}{3}$ in above trig. identity,
$$\boxed{4\cos^3\frac{\theta}{3}-3\cos \frac{\theta}{3}-\cos\theta=0}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3683165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Probability when putting balls from one box into another and then draw one ball from it I am struggling with the following problem:
Consider two boxes with several blue and red colored balls in.
In the right box there are 4 red and 5 blue balls.
In the left one are 7 red and 3 blue balls.
Now you blindly draw 4 balls from the right box and put them into the left one.
What is the probability of getting one red ball when you grab into the left box?
This has for sure something to do with conditional probability but in my opinion the answer depends on how many red balls are transfered into the left box.
Case 1: all of the 4 tranfered balls are blue.
So the left box contains 7 red and 7 blue balls. The probability to get a red one is $\frac{1}{2}$. The probability that the transfered balls are all blue is $\frac{5}{9} \cdot \frac{4}{8} \cdot \frac{3}{7} \cdot \frac{2}{6} = \frac{5}{126}$.
But how am I able to calculate P('red ball' | 'all transf. balls are blue')? By the Kolmogorov definition it is $\dfrac{P(\text{'red ball' } \cap \text{'all transf. balls are blue'})}{\frac{5}{126}}$ but surely P('red ball' $\cap$ 'all transf. balls are blue') is not simply $\frac{1}{2} \cdot \frac{5}{126}$.
Futhermore, do I need to differentiate between these different cases? The question sounds like the probability is always the same.
| Let $R$ be the event that after the transfer, you draw a red ball. Let $T(n)$ be the event that you transfer $n$ red balls from the right bin to the left bin (and $4-n$ blue balls).
Then the probability you are looking for is:
$$P(R) = P(R|T(0))P(T(0))+P(R|T(1))P(T(1))+P(R|T(2))P(T(2))+P(R|T(3))P(T(3))+P(R|T(4))P(T(4))$$
Since $P(R|T(n)) = \dfrac{P(R\cap T(n))}{T(n)}$, this can be rewritten as:
$$P(R) = P(R\cap T(0))+P(R\cap T(1))+P(R\cap T(2))+P(R\cap T(3))+P(R\cap T(4))$$
So, calculating this out, we have:
$$\begin{align*}P(R)& = \dfrac{7}{14}\cdot \dfrac{\dbinom{5}{4}}{\dbinom{9}{4}} + \dfrac{8}{14}\cdot \dfrac{\dbinom{5}{3}\dbinom{4}{1}}{\dbinom{9}{4}}+\dfrac{9}{14}\cdot \dfrac{\dbinom{5}{2}\dbinom{4}{2}}{\dbinom{9}{4}}+\dfrac{10}{14}\cdot \dfrac{\dbinom{5}{1}\dbinom{4}{3}}{\dbinom{9}{4}}+\dfrac{11}{14}\cdot \dfrac{1}{\dbinom{9}{4}} \\ & = \dfrac{7\cdot 5+8\cdot 10\cdot 4+9\cdot 10\cdot 6 + 10\cdot 5\cdot 4 + 11\cdot 1}{14\cdot 126} \\ & = \dfrac{79}{126}\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3683506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the radius of the circle given in the picture below. This is the image of the question. I am not able to get how to find the radius. Please help with that.
This is my try. I can't proceed now after it.
Thanks
|
$$PT^2=PC\cdot PA\implies PT^2=2\cdot 5=10$$
Using Pythagorean theorem in right $\Delta PTO$, $PO=\sqrt{OT^2+PT^2}=\sqrt{R^2+10}$
Using cosine rule in $\Delta ABP$, $$\cos\angle APB=\frac{PB^2+PA^2-AB^2}{2(PB)(PA)}=\frac{(\sqrt{R^2+10}+R)^2+5^2-(\sqrt{11})^2}{2(\sqrt{R^2+10}+R)(5)}\tag1$$
Similarly, using cosine rule in $\Delta AOP$, $$\cos\angle APB=\frac{PO^2+PA^2-OA^2}{2(PO)(PA)}=\frac{(\sqrt{R^2+10})^2+5^2-(R)^2}{2(\sqrt{R^2+10})(5)}\tag2$$
Equating the values of $\cos \angle APB$ from (1) & (2),
$$\frac{(\sqrt{R^2+10}+R)^2+5^2-(\sqrt{11})^2}{2(\sqrt{R^2+10}+R)(5)}=\frac{(\sqrt{R^2+10})^2+5^2-(R)^2}{2(\sqrt{R^2+10})(5)}$$
$$\frac{2R\sqrt{R^2+10}+2R^2+24}{\sqrt{R^2+10}+R}=\frac{35}{\sqrt{R^2+10}}$$
$$(11-2R^2)\sqrt{R^2+10}=2R^3-15R$$
$$(11-2R^2)^2(\sqrt{R^2+10})^2=(2R^3-15R)^2$$
$$28R^4-272R^2+605=0$$
$$R^2=\frac{-(-272)\pm\sqrt{(-272)^2-4\cdot 28\cdot605}}{2\cdot 56}=\frac{68\pm\sqrt{389}}{14}$$
$$R=\sqrt{\frac{68+\sqrt{389}}{14}}\approx 2.50318487$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3687003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Spivak calculus chapter 1 consequence of property 9 Spivak state's that one of the consequence of the distributive property is being able to solve multiplication of Arabic numerals.
1 3
x 2 4
_______
3 1 2
is arranged as:
$13 . 24$
$=13 . (2 . 10 + 4)$
$= 13 . 2 . 10 + 13 . 4$
$= 26 . 10 + 52$
$= 312$
and
1 3
x 4
_____
5 2
is arranged as:
$=13 . 4$
$=(1 . 10 + 3) . 4$
$=1 . 10 . 4 + 3 . 4$
$=4 . 10 + 12$
$=4 . 10 + 1 . 10 + 2$
$=(4 + 1) . 10 + 2$
$=5 . 10 + 2$
$=52$
so why isn't $4 . 10 + 12$ directly written as $52$ like in the first problem?
| He is showing us how did he obtain $13.4=52$ in the first working using $P9$ property explicitly.
\begin{align}13\cdot 24 &= 13 \cdot (2\cdot 10 + 4) \\
&=13 \cdot 2 \cdot 10 + \color{blue}{13 \cdot 4} \\
&= 26 \cdot 10 + \color{blue}{52}
\end{align}
The multiplication $13\cdot 4=52$ uses $P9$ also:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3687714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Technique for simplifying, e.g. $\sqrt{ 8 - 4\sqrt{3}}$ to $\sqrt{6} - \sqrt{2}$ How to find the square root of an irrational expression, to simplify that root. e.g.:
$$
\sqrt{ 8 - 4\sqrt{3} } = \sqrt{6} - \sqrt{2}
$$
Easy to verify:
\begin{align}
(\sqrt{6} - \sqrt{2})^2 = 6 - 2\sqrt{12} +2
= 8 - 4 \sqrt{3}
\end{align}
But how to work it out in the first place? I feel there's a standard technique (Completing-the-square? Quadratic formula?), but don't recall it or what it's called...
BTW: this came up in verifying equivalence of different calculations of $\cos{75°}$ (the above divided by $4$), as $\cos{\frac{90°+60°}{2}}$ vs $\cos{(45°+30°)}$, from 3Blue1Brown's lockdown video on complex numbers and trigonometry.
| Note the denesting formula
$$ \sqrt{a-\sqrt c} = \sqrt{\frac{a+\sqrt {a^2-c}}2}- \sqrt{\frac{a-\sqrt {a^2-c}}2}
$$
which can be verified by squaring both sides, and apply it to
$$\sqrt{8-4\sqrt3}=2\cdot \sqrt{2-\sqrt3}= 2\left(\sqrt{\frac32} -\sqrt{\frac12}\right)=\sqrt6-\sqrt2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3689064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Why does Stolz- Cesaro fail to evaluate the limit of $\dfrac{n + n^2 + n^3 + n^4 + \ldots + n^n}{1^n + 2^n + 3^n + 4^n + \ldots +n^n}$, I need to find the limit of the sequence
$\dfrac{n + n^2 + n^3 + n^4 + \ldots + n^n}{1^n + 2^n + 3^n + 4^n + \ldots +n^n}$,
My strategy is to use Stolz's Cesaro theorem for this sequence.
Now, the numerator is given by :
$x_r = n^1+ n^2 +n^3 + \ldots +n^r$, so $x_{n+1} - x_{n} = n^{n+1}$
Similarly for denominator
$y_r = 1^n + 2^n + 3^n +\ldots +r^n$, so $y_{n+1}- y_{n} = (n+1)^n$
Using Stolz Cesaro, this limit is equivalent to
$\displaystyle \lim \dfrac{n^{n+1}}{(n + 1)^n}$, which diverges to $ +\infty$,
However ans given to me is $\dfrac{e-1}{e}$, Can anyone tell where is the error in my solution ?
Thanks.
| Note that, as mentioned in the comments below, your computation of the ratio is incorrect. Regardless, the hypothesis of Stolz-Cesaro assumes that the limit $\lim_{n \to \infty} \frac{a_{n+1} - a_n}{b_{n+1} - b_n}$ exists. If it doesn't exist, it does not imply that the original limit does not exist.
A better way to approach is to write it as follows:
$$
\frac{n + n^2 + \cdots + n^n}{1^n + 2^n + \cdots +n^n} = \frac{n^{-(n-1)} + n^{-(n-2)} + \cdots + n^{-1} + 1}{\left(\frac{1}{n}\right)^n + \left(\frac{2}{n}\right)^n + \cdots + \left(\frac{n-1}{n}\right)^n + 1}
$$
As $n \to \infty$, clearly the numerator $\to 1$. For the denominator, see this.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3695175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Maximize $\sum\limits_{k =1}^n x_k (1 - x_k)^2$ Given problem for maximizing
\begin{align}
&\sum_{k =1}^n x_k (1 - x_k)^2\rightarrow \max\\
&\sum_{k =1}^n x_k = 1,\\
&x_k \ge 0, \; \forall k \in 1:n.
\end{align}
My attempt: first of all i tried AM-GM, or we can just say, that $x_k (1 - x_k)^2 \le x_kx_k^2 =x_k^3$, but now we got just sum of cubes. Can we say, that $\sum\limits_{k =1}^n x_k^3 \le \sum\limits_{k =1}^n x_k,$ because $x_i \le 1$ and get our maximum - 1?
It looks very easy, but i suppose i'm wrong
| For $x_1=x_2=...=x_n=\frac{1}{n}$ we get a value $\frac{(n-1)^2}{n^2}.$
We'll prove that it's a maximal value.
Indeed,
\begin{align}
\frac{(n-1)^2}{n^2}-\sum_{k=1}^nx_k(1-x_k)^2&=\sum_{k=1}^n\left(\frac{(n-1)^2}{n^3}-x_k(1-x_k)^2\right)\\
&=\frac{1}{n^3}\sum_{k=1}^n(1-nx_k)(n^2x_k^2-n(2n-1)x_k+(n-1)^2)
\end{align}
\begin{align}
&=\frac{1}{n^3}\sum_{k=1}^n\left((1-nx_k)(n^2x_k^2-n(2n-1)x_k+(n-1)^2)-(1-nx_k)(n^2-4n+3)\right)\\
&=\frac{1}{n^3}\sum_{k=1}^n(1-nx_k)^2(2n-2-nx_k)\geq0\text{ for any } n\geq2.
\end{align}
For $n=1$ we have $x_1=1$ and $$x_1(1-x_1)^2=0=\frac{(n-1)^2}{n^2},$$
which says that $\frac{(n-1)^2}{n^2}$ is the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3696327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Use differentiation under integral sign to prove $\int_{0}^{\infty} e^{-(x^2+\frac {a^2}{x^2})b^2} dx=\frac {\sqrt {\pi}}{2b} e^{-2ab^2}$ Using differentiation under integral sign, prove that $$\int_{0}^{\infty} e^{-\Big(x^2+\frac {a^2}{x^2}\Big)b^2} dx=\frac {\sqrt {\pi}}{2b}e^{-2ab^2}$$
My Attempt:
Let,
$$F(a)= \int_{0}^{\infty} e^{-\Big(x^2+\frac {a^2}{x^2}\Big)b^2} dx$$
Differentiate with respect to $a$
$$\frac {dF(a)}{da}=\int_{0}^{\infty} e^{-\Big(x^2+\frac {a^2}{x^2}\Big)b^2} \Bigg(-\frac {2ab^2}{x^2}\Bigg) dx$$
Put $\frac {a}{x}=y$
$$\frac {dF(a)}{da}=-2b^2 \int_{0}^{\infty} e^{-\Big(\frac {a^2}{y^2}+y^2\Big)b^2} dy$$
How to proceed after this?
| Proceed by averaging the two expressions as follows
\begin{align}
F’(a)&= -b^2 \int^{\infty}_{0} e^{-\Big(x^2+\frac {a^2}{x^2} \Big)b^2} \left(1+\frac a{x^2}\right)dy\\
&=- b^2 e^{-2ab^2}\int^{\infty}_{0} e^{-\left(x-\frac {a}{x} \right)^2 b^2} d\left(x-\frac a{x}\right)\\
&=- b \>e^{-2ab^2}\int^{\infty}_{-\infty} e^{-u^2} du
= -\sqrt{\pi}b \>e^{-2ab^2}
\end{align}
Then
$$F(a)= -\int_a^\infty F’(s)ds= \sqrt{\pi}b \int_a^\infty e^{-2sb^2}ds
= \frac {\sqrt {\pi}}{2b} e^{-2ab^2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3696598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solution for the roots of $x^4+x^2+1=0$ Is this solution to find the roots of $x^4+x^2+1=0$ correct?
$x^4+x^2+1=0$
$x^4+2x^2+1-x^2=0$
$(x^2+1)^2-x^2=0$
$[(x^2+1)-x][(x^2+1)+x]=0$
$(x^2-x+1)(x^2+x+1)=0$
For this equation to be true, either $(x^2-x+1)=0$ or/and $(x^2+x+1)=0$.
Using the quadratic formula, I got
$x=\frac{1\pm{\sqrt{3}i}}{2}$
and
$x=\frac{-1\pm{\sqrt{3}i}}{2}$
Are these values of x under the set of complex numbers the roots of $x^4+x^2+1=0$?
To answer this, I tried to check my answers using a computing website. According to the website, here are the solutions:
| Given equation
\begin{eqnarray}
x^4+x^2+1=0
\end{eqnarray}
Suppose $x^2=y$ then $y^2+y+1=0$
using quadratic formula we have
\begin{eqnarray}
y=\frac{-1\pm \sqrt{-3}}{2}\\
\end{eqnarray}
\begin{eqnarray}
Thus~~~~~~~~ x^2=\frac{-1\pm \sqrt{-3}}{2}\\
\Rightarrow ~~x=\pm\sqrt\frac{-1\pm \sqrt{-3}}{2}
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3703345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Using the Maclaurin series for $\frac{1}{1-x}$ to find $\frac{x}{1+x^2}$ Suppose I know the Maclaurin series for $$\frac{1}{1-x}=1+x+x^2+x^3+...= \sum_{n=0}^{\infty}x^n \tag{1}$$
then I can find the Maclaurin series for $\frac{1}{(1-x)^2}$ by the substitution $x\to x(2-x)$, which is obtained by solving the following equation for $x$:
$$1-x=(1-u)^2$$ $$x=u(2-u)$$
and replacing $u$ by and $x$, and $=$ by $\to$ , which leads to
$$\frac{1}{(1-x)^2}=1+x(2-x)+x^2(2-x)^2+x^3(2-x)^3+... \tag{2}$$
Succumbing to the same 'substitution'approach with $\frac{x}{1+x^2}$, I put
$$\frac{1}{1-x}=\frac{u}{1+u^2}$$ $$x=1-\frac{1}{u}-u$$
so with $x \to 1-\frac{1}{x}-x$
I should have:
$$\frac{x}{1+x^2}=1+\left(1-\frac{1}{x}-x\right)+\left(1-\frac{1}{x}-x\right)^2+\left(1-\frac{1}{x}-x\right)^3+...\tag{3}$$
but this does not seem correct (at least according to my Desmos graph for $|x|<1$).
Can someone please explain what my conceptual errors are?
| $$\frac{x}{1+x^2}=\frac{1}{1-(\color{red}{1-x-\frac 1x})}$$
$$=\frac{1}{1-u}$$
When $ x \to 0 $, $ u =\color{red}{1-x-\frac 1x} $ goes to $ \infty $, so you are not allowed to repalce $ X $ by $ u $ in the expansion
$$\frac{1}{1-X}=1+X+X^2+..$$
In the first part, $ u=x(2-x)$ goes to zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3703913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Prove $(a^2+b^2+c^2)^3 \geqq 9(a^3+b^3+c^3)$ For $a,b,c>0; abc=1.$ Prove$:$ $$(a^2+b^2+c^2)^3 \geqq 9(a^3+b^3+c^3)$$
My proof by SOS is ugly and hard if without computer$:$
$$\left( {a}^{2}+{b}^{2}+{c}^{2} \right) ^{3}-9\,abc \left( {a}^{3}+{b} ^{3}+{c}^{3} \right)$$
$$=\frac{1}{8}\, \left( b-c \right) ^{6}+{\frac {117\, \left( b+c \right) ^{4} \left( b+c-2\,a \right) ^{2}}{1024}}+{\frac {3\,{a}^{2} \left( 40\,{a }^{2}+7\,{b}^{2}+14\,bc+7\,{c}^{2} \right) \left( b-c \right) ^{2}}{ 32}}$$
$$+{\frac {3\, \left( b+c \right) ^{2} \left( 3\,a-2\,b-2\,c \right) ^{2} \left( b-c \right) ^{2}}{32}}+\frac{3}{16}\, \left( a+2\,b+2\,c \right) \left( 4\,a+b+c \right) \left( b-c \right) ^{4}$$
$$+{\frac { \left( 16\,{a}^{2}+24\,ab+24\,ac+11\,{b}^{2}+22\,bc+11\,{c}^{ 2} \right) \left( 4\,a-b-c \right) ^{2} \left( b+c-2\,a \right) ^{2} }{1024}} \geqq 0$$
I think$,$ $uvw$ is the best way here but it's not concordant for student in The Secondary School.
Also$,$ BW helps here, but not is nice, I think.
So I wanna nice solution for it! Thanks for a real lot!
| Because $(a+b+c)(ab+bc+ca) \geqslant 9abc,$ so we will prove stronger inequality
$$(a^2+b^2+c^2)^3 \geqslant (a+b+c)(ab+bc+ca)(a^3+b^3+c^3).$$
or
$$(a^2+b^2+c^2-ab-bc-ca)^2\sum (a^2+bc)+ \frac{ab+bc+ca}{2} \sum a^2(b-c)^2 \geqslant 0.$$
Done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3704808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Fourier series for function $f(x) = \arccos (\cos x), \ \ x\in ]-\pi, \pi[$ $f(x) = \arccos (\cos x), \ \ x\in [-\pi; \pi]$
Here is what I did:
$$a_0 = \frac{1}{2\pi} \int^{\pi}_{-\pi} \arccos (\cos x)dx = \frac{\pi}{2} \\ a_n = \frac{1}{\pi} \int^{\pi}_{-\pi} \arccos (\cos x) \cos \left(\frac{\pi n x}{\pi} \right)dx = \frac{1}{\pi}\left(\left( x\cos(n x)\right|^{\pi}_{-\pi} - \frac{1}{n}\int^{\pi}_{-\pi} \sin nx \ dx\right) = -2\pi \\ b_n = \frac{1}{\pi} \int^{\pi}_{-\pi} \arccos (\cos x) \sin \left(\frac{\pi n x}{\pi} \right)dx = \frac{1}{\pi} \left(x \sin nx |^{\pi}_{-\pi} + \frac{1}{n} \int^{\pi}_{-\pi} \cos nx dx \right)= 0 \\ f(x) = \frac{\pi}{4}-2\pi \sum^{\infty}_{n=1} \cos\left(\frac{2\pi kx }{2\pi}\right)$$
Is it correct?
| I am a bit puzzled as to why you write $f(x) = \arccos(\cos x))$. This is identical to $f(x) = \lvert x \rvert$ which seems to me a simpler form, and I think is the form you used to evaluate the integrals.
Then the calculation for $a_0 = \pi / 2$ is fine. $b_n = 0$ because $f(x) \sin nx$ is an odd function. For $a_n$ you can say,
\begin{align}
a_n &= \frac{1}{\pi} \int_{-\pi}^\pi \lvert x\rvert \cos nx ~ dx \\
&=\frac{2}{\pi}\int_0^\pi x \cos nx ~ dx \\
&\hspace{1cm} \small \text{(because the integrand is an even function)}\\
&= \frac{2}{\pi} \Bigg\{\Big[ \frac{1}{n}x\sin nx \Big]_0^\pi - \int_0^\pi \frac{1}{n}\sin nx ~ dx\Bigg\} \\
&= \frac{2}{\pi}\Big[ \frac{1}{n^2} \cos nx \Big]_0^\pi \\
&= -\frac{2}{\pi n^2}\Big( 1 -(-1)^n \Big)
\end{align}
That means by my reckoning your calculation of $a_n$ is incorrect. It should be $a_n = 0$ for even $n$ and $-\dfrac{4}{\pi n^2}$ for odd $n$. Thus
$$ f(x) = \frac{\pi}{2} - \frac{4}{\pi} \Bigg( \cos x+ \frac{1}{3^2}\cos 3x + \frac{1}{5^2}\cos 5x + \cdots \Bigg)$$
Incidentally, you can use this Fourier series to solve the Basel Problem, evaluate $\displaystyle \zeta (2) = S = \sum_{n\geqslant 1} \frac{1}{n^2}$.
To do so use pointwise convergence for $f(x)$ when $x=0$ (it is continuous and piecewise smooth). That gives a formula for the sum of odd terms in $S$. The even terms in $S$ sum simply to $\frac{1}{4} S$ and the result, $\frac{\pi^2}{6}, $ follows with a little manipulation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3705572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the minimun of $MN+\frac{3}{5}MP$, $MN$ and $MP$ is two sides of a quadrilateral. In a quadrilateral $OPMN$ ，$\angle NOP=90^\circ$,$ON=1$,$OP=3$, and $M$ satisfy $\vec{MO}\cdot\vec{MP}=4$, find the minimum of
$MP+\frac{3}{5}MN$
I choose the vertex $O$ of $OPMN$ as the origin of the coordinate system, and ON as the $x$-axis, $OP$ as the $y$-axis.
so $N(1,0)$ and $P(0,-3)$ and assume $M(x,y)$ then $\vec{MO}=(-x,-y)$ $\vec{MP}=(-x,-3-y)$. then $$x^2+3y+y^2=4\Rightarrow x^2+\left(y+\frac{3}{2}\right)^2=\left(\frac{5}{2}\right)^2$$
I know trail of point $M$ is a circle centered at $(0,-\frac{3}{2})$.
but I have no idea to go to next. and I use $x=\frac{5}{2}\cos
\theta, y=\frac{5}{2}\sin\theta -\frac{3}{2}$ and we have $$MP+\frac{3}{5}MN =f(\theta)= \sqrt{\frac{17}{2}+\frac{15}{2}\sin
\theta} + \frac{3}{5}\sqrt{\frac{19}{2}-5\cos\theta-\frac{15}{2}\sin\theta}$$
I use Mathematica to find the numerical minimum of $f(\theta)$. It is about $$f_{\min}(\theta)\approx f(-1.45162)=3.45245$$
it seems when $M$ near the bottom of the circle, $MP+\frac{3}{5}MN$ take the minimum . Please help me .thanks very much. :)
| Now, $f'(\theta)=0$ gives $$(17\cos\theta-3\sin\theta-5)(15\sin2\theta+10\cos2\theta-6\sin\theta-16\cos\theta)=0$$ and
$$17\cos\theta-3\sin\theta-5=0$$ gives a minimal value for
$$\frac{17}{\sqrt{298}}\cos\theta-\frac{3}{\sqrt{298}}\sin\theta=\frac{5}{\sqrt{298}},$$
which gives $$\theta_{min}=-\arccos\frac{17}{\sqrt{298}}-\arccos\frac{5}{\sqrt{298}}.$$
Id est, $$\min{f}=f\left(\theta_{min}\right)=\sqrt{\frac{85}{2}\cos\theta_{min}-4}+\frac{3}{5}\sqrt{22-\frac{95}{2}\cos\theta_{min}}=$$
$$=\frac{85-3\sqrt{273}}{2\sqrt{298}}+\frac{3\left(57+5\sqrt{273}\right)}{10\sqrt{298}}=\frac{85\cdot5+3\cdot57}{10\sqrt{298}}=\frac{\sqrt{298}}{5}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3706588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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If $\frac{t^2-x^2}{(t^2+x^2)^2}+\frac{(1-x)^2-t^2}{((1-x)^2+t^2)^2}=0$ for real $t$ and $0
$$\frac{t^2-x^2}{(t^2+x^2)^2}+\frac{(1-x)^2-t^2}{((1-x)^2+t^2)^2}=0$$
where $0<x<1$ and $t\in\mathbb{R}$. Prove that $x=1/2$.
It is evident that $x=1/2$ satisfies the above equation. Please help.
| Well, define $y \in \mathbb{R}$ such that $x=1-y$ (noting that also $0<y<1$), we get
$$\frac{t^2-x^2}{(t^2+x^2)^2}+\frac{(1-x)^2-t^2}{((1-x)^2+t^2)^2}=0$$
$$\implies \frac{t^2-(1-y)^2}{(t^2+(1-y)^2)^2}+\frac{y^2-t^2}{(y^2+t^2)^2}=0$$
Note that the first equation implies
$$\frac{t^2-x^2}{(t^2+x^2)^2}=\frac{t^2-(1-x)^2}{((1-x)^2+t^2)^2}$$
While the second implies
$$\frac{t^2-(1-y)^2}{(t^2+(1-y)^2)^2}=\frac{t^2-y^2}{(y^2+t^2)^2}$$
So, in fact, the two equations are equivalent, which means $y=x \implies 1-x=x \implies x=1/2$
Edit: I figured out that what I proved was if $a$ is a solution (for $x$) then $1-a$ is also a solution. This means that $x=1/2$ for sure is a solution for all $t$.
But when I tried graphing the equation on desmos and added a slider for $t$, I get that as $t$ ranges over all reals (or $t^2$ ranges over positive reals), $x$ and $1-x$ vary with it. For example, $t^2 = \frac{1}{6}\left(\sqrt{601} - 13\right) \implies x \in \{-2,1/2,3\}$
Note that $x=1/2$ is the general solution, and $x=3$ is a specific solution and if $3$ is a solution then $1-3=-2$ is also a solution.
If we want $x=4$ as a solution we have $x=1-4=-3$ is also a solution and $t^2=\frac{1}{6}\left(\sqrt{2353} - 25\right)$ satisfies the equation. So, there's no point of solving for a specific $x$ if $x$ just varies depending on $t^2$ over the reals (and of course $1-x$ accordingly), but for all $t$, $x=1/2$ is certainly a solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3708072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Combinations series: $\frac{{n \choose 1}(n-1)^3+{n \choose 3}(n-3)^3+\ldots}{n^2(n+3)\cdot 2^n}$
Evaluate $\frac{{n \choose 1}(n-1)^3+{n \choose 3}(n-3)^3+\ldots}{n^2(n+3)\cdot 2^n}$ for $n=10$.
Attempt: I'll deal with the case n being even, as we need to evaluate for n=10.
the numerator is
$${n \choose 1}(n-1)^3+{n \choose 3}(n-3)^3+\ldots$$
$$=\sum_{r=odd} {n \choose r}(n-r)^3$$(not sure if this is a correct notation).
$$=\sum_{r=odd}{n \choose n-r}r^3=\sum_{r=odd} {n \choose r}r^3$$(parity being same as n is even and r is odd, although I don't think this matters much).
Using the identity ${n \choose r}=\frac{n}{r} {n-1 \choose r-1}$ repeatedly in following steps,
$$=n\sum_{r=even} {n-1 \choose r-1}r^2$$
$$=[n(n-1)](1+\sum_{r=odd} {n-2 \choose r-2}[(r-2+3)+\frac{1}{r-1}]$$
$$=[n(n-1)](1+(n-2)\sum_{r=even}{n-3 \choose r-3}+3\sum_{r=odd}{n-2 \choose r-2}+\frac{1}{n-1} \sum_{r=even}{n-1 \choose r-1} -1)$$
$$=[n(n-1)]((n-2)\cdot 2^{n-4} +3\cdot 2^{n-4}+\frac{2^{n-2}}{n-1}$$
This simplfies to $n \cdot 2^{n-4} (n^2+7n-4)$.
Which is incorrect. The answer for $n=10$ (numerator/denominator is given as $\frac{1}{16}$).
Where am I going wrong?
Also the hint given for this problem was "expand $\frac{(e^x+1)^n - (e^x-1)^n}{2}$ in two different ways". I didn't quite understand this approach?
Could someone please explain this approach and any other approach also?
| Here's an alternative approach that does not depend on the hint.
Because $$\frac{1+(-1)^k}{2}=\begin{cases}1&\text{if $k$ is even}\\0&\text{if $k$ is odd}\end{cases}$$
we have $$\sum_{k\ge 0} a_{2k} = \sum_{k\ge 0} a_k \frac{1+(-1)^k}{2}.$$
Now take $a_k=\binom{n}{k+1}(k+1)^3$ to obtain
\begin{align}
&\sum_{k\ge 0} \binom{n}{2k+1}(2k+1)^3 \\
&= \sum_{k\ge 0} \binom{n}{k+1}(k+1)^3 \frac{1+(-1)^k}{2} \\
&= \sum_{k\ge 0} \frac{n}{k+1}\binom{n-1}{k}(k+1)^3 \frac{1+(-1)^k}{2} \\
&= n\sum_{k\ge 0} \binom{n-1}{k}(k+1)^2 \frac{1+(-1)^k}{2} \\
&= n\sum_{k\ge 0} \binom{n-1}{k}\left(2\binom{k}{2}+3k+1\right) \frac{1+(-1)^k}{2} \\
&= n\sum_{k\ge 0} \left(2\binom{n-1}{2}\binom{n-3}{k-2}+3(n-1)\binom{n-2}{k-1}+\binom{n-1}{k}\right) \frac{1+(-1)^k}{2} \\
&= \frac{n}{2}\left(2\binom{n-1}{2}\sum_{k\ge 0}\binom{n-3}{k-2}+3(n-1)\sum_{k\ge 0}\binom{n-2}{k-1}+\sum_{k\ge 0}\binom{n-1}{k}\right) \\
&+ \frac{n}{2}\left(2\binom{n-1}{2}\sum_{k\ge 0} \binom{n-3}{k-2}(-1)^k+3(n-1)\sum_{k\ge 0} \binom{n-2}{k-1}(-1)^k+\sum_{k\ge 0} \binom{n-1}{k}(-1)^k\right) \\
&= \frac{n}{2}\left(2\binom{n-1}{2}2^{n-3}+3(n-1)2^{n-2}+2^{n-1}\right) \\
&+ \frac{n}{2}\left(2\binom{n-1}{2}(1-1)^{n-3}+3(n-1)(1-1)^{n-2}+(1-1)^{n-1}\right) \\
&= 2^{n-4} n \left(2\binom{n-1}{2}+6(n-1)+4\right) \\
&+ \frac{n}{2}\left(2\binom{n-1}{2}[n=3]+3(n-1)[n=2]+[n=1]\right) \\
&= 2^{n-4} n^2 (n+3) + 3[n=3]+3[n=2]+\frac{1}{2}[n=1]
\end{align}
So the fraction for even $n \ge 4$ is
$$\frac{2^{n-4} n^2 (n+3)}{2^n n^2 (n+3)} = \frac{1}{16}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3709836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Explain the value of $(0,0)$ for $f(x,y)=1+x\frac{x^2-y^2}{x^2+y^2}$
Let $f:\mathbb R^2\to\mathbb R$ (where $\mathbb R$ is the real numbers) continuous in $\mathbb R^2$, such that
$$f(x,y)=1+x\frac{x^2-y^2}{x^2+y^2}$$ for $(x,y)\neq(0,0)$
Justify the value of $f(0,0)$.
The function is continuous in its domain, and $f(0,0)$ is defined by another equation. What else do I have to justify?
| We have: $\left|x\cdot \dfrac{x^2-y^2}{x^2+y^2}\right| \le |x|\cdot \left|\dfrac{x^2}{x^2+y^2}+\dfrac{y^2}{x^2+y^2}\right| = |x| \implies \displaystyle \lim_{(x,y) \to (0,0)} f(x,y) = 1+0 = 1 \implies f(0,0) = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3710340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Calcuating an Integral via Residues I am trying to solve the following integral:
$$\int_0^\infty\frac{dx*x}{(x^2+1)(2+x)}$$
What I have done is an analytical extenstion to the upper half plane only by recognising that the above integral is just 1/2 the integral over the entire real line, then only considered poles that exist in the upper half.
I then found the corresponding residues of the poles and used the resiude theorem.
My final answer is:
$$(\frac{1}{10} - \frac{i}{5}) \pi$$
However, using mathematica I find that the actual answer should be:
$$\frac{1}{10}(\pi + Log[16])$$
What am I doing wrong? Any ideas?
| Consider $\displaystyle \oint_C \frac{z \, \textrm{Log }z \, dz}{(z^2+1)(z+2)}$ where $C$ is the keyhole contour that goes from $\epsilon$ to $R$ then around a large circle of radius $R$ counterclockwise, from $R$ to $\epsilon$ (below the positive real axis) and around a circle of radius $\epsilon$ clockwise. There are three poles within the contour at $\{-2,i,-i \}$.
The integral is equal to $2\pi i$ times the sum of the residuals at the poles inside the contour.
The integral around the keyhole is equal to the sum of the integrals along the curves making up the keyhole contour. On the circles of radius $R$ and radius $\epsilon$, the integrals approach zero in the limit.
Taking the limit as $\epsilon\rightarrow 0$ and $R\rightarrow \infty$, and showing only the non-zero terms that remain, we have
$$\lim_{\epsilon \rightarrow 0 , R \rightarrow \infty} \left[ \int_\epsilon^R \frac{x \log x \, dx}{(x^2+1)(x+2)} - \int_\epsilon^R \frac{x (\log x+2\pi i)\,dx}{(x^2+1)(x+2)}\right]= 2\pi i \sum_{z\in \textrm{poles}} \textrm{Res }\left[ \frac{z \, \textrm{Log }z}{(z^2+1)(z+2)}\right].$$
The left-hand side is
$$-2\pi i \int_0^\infty \frac{x \, dx}{(x^2+1)(x+2)}$$
The sum of the residues (multiplied by $2\pi i$) is also purely imaginary, so we obtain a real expression for the integral.
$$ \int_0^\infty \frac{x \, dx}{(x^2+1)(x+2)} =
- \left. \text{Res } \frac{z\,\text{Log }z}{(z^2+1)(z+2)}\right|_{z=i}
- \left. \text{Res } \frac{z\,\text{Log }z}{(z^2+1)(z+2)}\right|_{z=-i}
- \left. \text{Res } \frac{z\,\text{Log }z}{(z^2+1)(z+2)}\right|_{z=-2}.$$
$$ \int_0^\infty \frac{x \, dx}{(x^2+1)(x+2)} =
- \frac{ i (\frac{i\pi}{2})}{2i(2+i)}
- \frac{ (-i) (\frac{3\pi i}{2})}{(-2i)(2-i)}
+ \frac{2(\log 2 + \pi i)}{5}.$$
After simplifying, we get
$$\int_0^\infty \frac{x \, dx}{(x^2+1)(x+2)}=\frac{\pi}{10} + \frac{2 \ln 2}{5}.$$
This agrees with the result you quoted from Mathematica.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3710485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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When the quotient $\frac {13n^2+n}{2n+2}$ is an integer? Find the nonzero values of integer $n$ for which the quotient $\frac {13n^2+n}{2n+2}$ is an integer?
My Attempt
I assumed $(2n+2 )| (13n^2+n)$ implies existence of integer $k$ such that $$13n^2+n=k(2n+2)$$
$\implies\ 13n^2+(1-2k)n-2k=0$
$\implies\ n=\frac{(2k-1)\pm\sqrt{(1-2k)^2+104}}{26}$
But this is taking me nowhere.
| You made a mistake in the last line of your post. The $n^2$ shouldn’t be there. You need: $(1-2k)^2+104=h^2\implies h^2-(1-2k)^2=104\implies (h+1-2k)(h-1+2k)=104=1\cdot 104= 2\cdot 52 = 4\cdot 26= 8\cdot 13= (-1)(-104) = (-2)(-52) = (-4)(-26) = (-8)(-13)$. Those are the possibilities for them.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3714351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Rearranging these formulas The entropy of mixing formula is $\Delta S_{\mathrm{mix}} = R(x_1\ln(x_1) + x_2\ln(x_2))$ where $x$ is the mole fraction $x_1 = \frac{n_1}{n_1+n_2}$ and $x_2 = \frac{n_2}{n_2+n_1}$ where $n$ is the number of moles and the subscript denotes which compound it is. I am asked to express this entropy of mixing formula in terms of the mass fraction which is $y_1 = \frac{m_1}{m_1+m_2}$ where $m$ is the mass of the substances. now $m_1 = n_1M_1$ and $m_2 = n_2M_2$ where $M$ is the molar mass, I have tried to rearrange them a lot and could not manage to express $x$ in terms of $y$. Could anyone help me please?
| Given
\begin{align}
\Delta S_{\mathrm{mix}}
&=
R(x_1\ln(x_1) + x_2\ln(x_2))
\tag{1}\label{1}
,\\
x_1 &= \frac{n_1}{n_1+n_2}
\tag{2}\label{2}
,\\
x_2 &= \frac{n_2}{n_1+n_2}
\tag{3}\label{3}
,\\
y_1 &= \frac{m_1}{m_1+m_2}
\tag{4}\label{4}
,\\
m_1 &= n_1M_1
\tag{5}\label{5}
,\\
m_2 &= n_2M_2
\tag{6}\label{6}
.
\end{align}
From \eqref{4}-\eqref{6}
we have
\begin{align}
y_1 &= \frac1{1+\displaystyle\frac{n_2}{n_1}\cdot \frac{M_2}{M_1}}
\tag{7}\label{7}
,
\end{align}
from \eqref{2}-\eqref{3} follows
\begin{align}
x_1+x_2&=1
\tag{8}\label{8}
,\\
\frac{n_2}{n_1}
&=
\frac{x_2}{x_1}
=\frac{1-x_1}{x_1}
\tag{9}\label{9}
,
\end{align}
so \eqref{9} combined with \eqref{7}, \eqref{8} gives
the desired:
\begin{align}
y_1 &= \frac1{1+\frac{x_2}{x_1}\cdot \tfrac{M_2}{M_1}}
\tag{10}\label{10}
,\\
y_1 &= \frac1{1+\frac{1-x_1}{x_1}\cdot \tfrac{M_2}{M_1}}
\tag{11}\label{11}
,\\
y_1 &= \frac1{1+\left(\frac1{x_1}-1\right)\cdot \tfrac{M_2}{M_1}}
\tag{12}\label{12}
,\\
\frac1{y_1} &= {1+\left(\frac1{x_1}-1\right)\cdot \tfrac{M_2}{M_1}}
\tag{13}\label{13}
,\\
\frac1{y_1}-1 &= {\left(\frac1{x_1}-1\right)\cdot \tfrac{M_2}{M_1}}
\tag{14}\label{14}
,\\
\left(\frac1{y_1}-1\right)\cdot \tfrac{M_1}{M_2} &= \frac1{x_1}-1
\tag{15}\label{15}
,\\
1+\left(\frac1{y_1}-1\right)\cdot \tfrac{M_1}{M_2} &= \frac1{x_1}
\tag{16}\label{16}
,
\end{align}
\begin{align}
x_1 &= \frac 1{1+\displaystyle\frac{1-y_1}{y_1}\cdot \tfrac{M_1}{M_2}}
\tag{17}\label{17}
,\\
x_2 &= \frac 1{1+\displaystyle\frac{y_1}{1-y_1}\cdot \tfrac{M_2}{M_1}}
\tag{18}\label{18}
.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3714828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
$\frac{3x+1}{x+1}+\frac{3y+1}{y+1}+\frac{3z+1}{z+1} \le \frac{9}{2}$ I'm having trouble proving that for any $x,y,z>0$ such that $x+y+z=1$ the following inequality is true:
$\frac{3x+1}{x+1}+\frac{3y+1}{y+1}+\frac{3z+1}{z+1} \le \frac{9}{2}$
It seems to me that Jensen's inequality could do the trick, but I'm having trouble finding the right function and the right arguments. Any help is appreciated.
| Another way.
We need to prove that
$$\sum_{cyc}\left(\frac{3}{2}-\frac{3x+1}{x+1}\right)\geq0$$ or
$$\sum_{cyc}\frac{1-3x}{x+1}\geq0$$ or
$$\sum_{cyc}\left(\frac{1-3x}{x+1}+\frac{3}{4}(3x-1)\right)\geq0$$ or
$$\sum_{cyc}(3x-1)\left(\frac{3}{4}-\frac{1}{x+1}\right)\geq0$$ or
$$\sum_{cyc}\frac{(3x-1)^2}{x+1}\geq0$$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3717128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Given polynomial $P(x) = x^2 + ax + b$ and that there only exists one $c$ such that $P^2(c) = c$. Calculate the minimum value of $a + b + c$.
Given polynomial $P(x) = x^2 + ax + b$. Knowing that there only exists one value of real $c$ such that $P^2(c) = c$, calculate the minimum value of $a + b + c$.
Notation: $P^2(x) = (P \circ P)(x)$.
We have that $P^2(c) = c \iff (c^2 + ac + b)^2 + a(c^2 + ac + b) + b = c$
$$\iff [c^2 + (a - 1)c + b][c^2 + (a + 1)c + a + b + 1] = 0$$
For the above equation to have only one solution, $$(a - 1)^2 - 4b, (a + 1)^2 - 4(a + b + 1) \le 0 \implies b \ge \frac{(1 - a)^2}{4}$$
$$\implies \left[c^2 + (a - 1)c + \frac{(1 - a)^2}{4}\right] \cdot \left[c^2 + (a + 1)c + a + \frac{(1 - a)^2}{4} + 1\right] \ge 0$$
$$\implies (a + 2c - 1)^2[(a + 2c + 1)^2 + 4] \ge 0$$
It could be implied that 'the only solution' is $c = \dfrac{1 - a}{2}$, where $b = \dfrac{(1 - a)^2}{4}$.
$$a + b + c = a + \frac{1 - a}{2} + \frac{(1 - a)^2}{4} = \frac{a^2 + 3}{4} \ge \frac{3}{4}$$
Is the above solution correct? If not, could you provide with one? Thanks for your attention.
| Just a less computational argument for the first part of the solution: if $P(x) = x$ had no real solutions, then $P(x)>x$ for all real $x$ and hence $P(P(x)) > P(x) > x$, contradicting the existence of $c$. Any solution of $P(x) = x$ is also a solution of $P(P(x)) =x$, hence $P(x) = x$ must have exactly one solution, $c$. That implies $P(x) = x+(x-c)^2$.
Therefore $a=1-2c$, $b= c^2$ and $a+b+c = c^2-c+1 = (c-\frac{1}{2})^2 + \frac{3}{4} \geq \frac{3}{4}$. Equality holds for $c=\frac{1}{2}$, $a=0$, and $b = \frac{1}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3718840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the sum of the series with terms given by ${T}_{r}=\frac{r}{(r+1)(r+3)(r+4)}$ The given series has general term as $${T}_{r}=\frac{r}{(r+1)(r+3)(r+4)}$$
I have tried to approach this problem by making a telescopic series as follows, but I end up cancelling $r$ in the numerator, $$\frac{1}{(r+1)(r+3)}-\frac{1}{(r+3)(r+4)}=\frac{3}{(r+1)(r+3)(r+4)}$$ Please provide an alternate approach to form telescopic series.
| In general you can guess (in general more hope) a decomposition $\frac{r}{(r+1)(r+3)(3+4)} = \frac{a}{r+1}+\frac{b}{r+3} + \frac{c}{r+4}$ for certain $a,b,c \in \mathbb{R}$.
So what you have to do is to solve the system given by $ \frac{r}{(r+1)(r+3)(3+4)} = \frac{a}{r+1}+\frac{b}{r+3} + \frac{c}{r+4} = \frac{a(r+3)(r+4) + b(r+1)(r+4) + c(r+1)(r+3)}{(r+1)(r+3)(r+4)}$.
Basically you will have to impose that the leading coefficient of $r^{2}$ will be $0$, as well as the costant term, and the one of $r$ equals to $1$.
This traslates into a linear system. If my calculus are correct we get $a(r^{2}+7r+12) + b(r^{2}+5r+4)+ c(r^{2}+4r+3) = r^{2}(a+b+c) + r(7a+5b+4c)+12a+4b+3c$. The system now becomes $\begin{cases}a+b+c = 0 \\ 7a+5b+4c = 1 \\ 12a+4b+3c = 0\end{cases}$. I'm more comfort evaluating the system as $\begin{pmatrix} 1 & 1 & 1 \ 7 & 5 & 4 \ 12 & 4 & 3\end{pmatrix}\begin{pmatrix} a\ b \ c\end{pmatrix} = \begin{pmatrix} 0\ 1 \ 0\end{pmatrix}
Since the matrix is invertible (the determinant should be nonzero if I'm not mistaken) it surely exists $a,b,c \in \mathbb{R}$ as required, so you can proceed in solving the system to determine them which will link to other answers given.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3722271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to Prove $\int_{0}^{\infty}\frac {1}{x^8+x^4+1}dx=\frac{π}{2\sqrt{3}}$ Question:- Prove that
$\int_{0}^{\infty}\frac {1}{x^8+x^4+1}dx=\frac{π}{2\sqrt{3}}$
On factoring the denominator we get,
$\int_{0}^{\infty}\frac {1}{(x^4+x^2+1)(x^4-x^2+1)}dx$
Partial fraction of the integrand contains big terms with their long integral.So i didn't proceed with partial fraction.I'm unable figure out any other method.I think that there might be some other method for evaluation of this definite integral since its value is
$\frac{π}{2\sqrt{3}}$.
Does anyone have nice way to solve it!
| Use $x^8+x^4+1=(x^4+x^2+1)(x^4-x^2+1)$ to partial decompose
\begin{align} \int_{0}^{\infty}\frac {1}{x^8+x^4+1}dx
&=\frac12\int_{0}^{\infty}\frac {x^2+1}{x^4+x^2+1}dx
- \frac12\int_{0}^{\infty}\frac {x^2-1}{x^4-x^2+1}dx\\
&= \frac12\int_{0}^{\infty}\frac {d(1-\frac1x)}{(x-\frac1x)^2+3}
- \frac12\int_{0}^{\infty}\frac {d(1+\frac1x)}{(x+\frac1x)^2-3}\\
&=\frac12\cdot \frac{π}{\sqrt{3}}+\frac12\cdot0= \frac{π}{2\sqrt{3}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3728939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Diophantine equation : $6^m+2^n+2=x^2$
Find $m,n,x\in\mathbb{N}$ such that $6^m+2^n+2=x^2$.
My first approach is to show that for $m,n\geq2$, there exist no solution for $x$ by using modulo $4$.
Case $1$ : $m=1$, $x^2=2^n+8$.
As $n\geq1\implies2\mid RHS\implies2\mid x^2\implies4\mid x^2\implies4\mid LHS\implies 4\mid 2^n\implies n\geq 2$.
The equation can be reduced into $2+2^{n-2}=\bar x^2$ where $2\bar x=x$.
If $n-2\geq2$, $LHS\equiv2$ and $RHS\equiv0,1\mod4$. Therefore $n-2<2\implies n\leq3$.
Checking for $2\leq n\leq3$, we have $m=1,n=3,x=4$ as a solution.
Case $2$ : $n=1$, $x^2=6^m+4$.
$m=1$ is not a solution, therefore $m\geq2\implies 4\mid LHS\implies2\mid x$.
The equation can be reduced into $2^{m-2}3^m+1=\bar x^2$ where $2\bar x=x$.
I do not know how to solve the problem after this step. Any hints or solution is appreciated.
| $\textbf{Hint:}$Write the last equality as,
$2^{m-2}3^m=(\bar x-1)(\bar x+1)$.
Here, $gcd(\bar x+1,\bar x-1)=2$ unless $m=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3731058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integrate $\int \frac{x^2-1}{x^3 \sqrt{2x^4-2x^2+1}} \mathop{dx}$ Evaluate $$\int \frac{x^2-1}{x^3 \sqrt{2x^4-2x^2+1}} \mathop{dx}$$
I tried $u=\sqrt{2x^4-2x^2+1}$, $u=\dfrac{1}{x}$ and $u=\sqrt{x}$ but none of these worked.
A friend gave this and said it's from IIT JEE and answer is $$\dfrac{\sqrt{2x^4-2x^2+1}}{2x^2}+C$$
I would like a hint or suggestion.
| $$\int \frac{x^2-1}{x^3 \sqrt{2x^4-2x^2+1}} \mathop{dx}=\int \dfrac{1-\dfrac{1}{x^2}}{x^3 \sqrt{2-\dfrac{2}{x^2}+\dfrac{1}{x^4}}}dx$$$$=\int \frac{\left(1-\dfrac{1}{x^2}\right)dx}{x^3 \sqrt{\left(\dfrac{1}{x^2}-1\right)^2+1}}$$
$$=\frac14\int \frac{2\left(\dfrac{1}{x^2}-1\right)\left(\dfrac{-2}{x^3}\right)dx}{ \sqrt{\left(\dfrac{1}{x^2}-1\right)^2+1}}$$
$$=\frac14\int \frac{d\left(\left(\dfrac{1}{x^2}-1\right)^2+1\right)}{ \sqrt{\left(\dfrac{1}{x^2}-1\right)^2+1}}$$
$$=\frac14\cdot 2 \sqrt{\left(\dfrac{1}{x^2}-1\right)^2+1}\ +C$$
$$=\color{blue}{\frac{1}{2x^2}\sqrt{2x^4-2x^2+1}}\ +C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3731783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\frac{1}{n}+\ln {n}<1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}<1+\ln{n}$ for $n\ge2$. How to prove the following bound:
$\frac{1}{n}+\ln {n}<1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}<1+\ln{n}$ for $n\ge2$.
My Attempt:
We have:$$\int_1^n\frac1xdx=\ln n$$So,$$\sum_{i=1}^n\frac{1}{i+1}\le\int_1^n\frac1xdx=\ln n\le\sum_{i=1}^n\frac1i$$
Adding $1$ on both sides to get the inequality: $1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}<1+\ln{n}$
How Can I prove the inequality given on the LHS? Also, in this case: Why is the sum $\sum_{x=1}^n\frac{1}{x}$ more than the value of the integral $\int_1^{n}\frac{1}{x}dx$? Thanks for the help.
| Since $e^x>1+x$ for all $x>0$, we have for $k\geq 1$ that
$$
e^{\frac{1}{k}} > 1 + \frac{1}{k} = \frac{{k + 1}}{k}.
$$
Thus
$$
e^{\sum\limits_{k = 1}^{n - 1} {\frac{1}{k}} } > \prod\limits_{k = 1}^{n - 1} {\frac{{k + 1}}{k}} = n,
$$
$$
\sum\limits_{k = 1}^{n - 1} {\frac{1}{k}} > \log n,
$$
$$
\sum\limits_{k = 1}^n {\frac{1}{k}} > \log n + \frac{1}{n}.
$$
To prove the upper bound note that
$$
\sum\limits_{k = 1}^n {\frac{1}{k}} = 1 + \sum\limits_{k = 2}^n {\frac{1}{k}} < 1 + \sum\limits_{k = 2}^n {\int_{k - 1}^k {\frac{{dt}}{t}} } = 1 + \int_1^n {\frac{{dt}}{t}} = 1 + \log n.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3735002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Residue of $f(z)=\frac{1}{z^2(e^z-1)}$ I want to find the residue of the function $f(z)=\frac{1}{z^2(e^z-1)}$. I have tried something that I am 99 precent sure about, but still I would appreciate some feedback. What I have done is that:
$$\frac{1}{z^2(e^z-1)}=\frac{1}{z^2}\cdot\frac{1}{1+z+\frac{z}{2}+o(z^2)-1}=\frac{1}{z^2}\frac{1}{z+\frac{z}{2}+o(z^2)}$$
$$=\frac{1}{z^2}\cdot\frac{1}{1+\frac{z}{2}+o(z)}=\frac{1}{z^2}\cdot\frac{1}{1-(-\frac{z}{2}+o(z))}$$
$$=\frac{1}{z^2}\cdot[1+(-\frac{z}{2}+o(z))+(-\frac{z}{2}+o(z))^2+\cdots]=\frac{1}{z^2}[1-\frac{z}{2}+\text{greater powers of }z]$$ $$=\frac{1}{z^2}-\frac{1}{2z}+\text{greater powers of }z$$
$$\Downarrow$$
$$\operatorname{Res}(f,0)=-\frac{1}{2}$$
| The value found is not correct. You need to expand the exponential at order $3$:
\begin{align}
\frac{1}{z^2(e^z-1)}&=\frac{1}{z^2\bigl(z+\tfrac{z^2}2+\tfrac{z^3}{6}+o(z^3)\bigr)}=\frac{1}{z^3\bigl(1+\tfrac{z}2+\tfrac{z^2}{6}+o(z^2)\bigr)}\\
&=\frac{1}{z^3}\Bigl(1-\Bigl(\frac{z}2+\frac{z^2}6\Bigr)+\Bigl(\frac{z}2+\frac{z^2}6\Bigr)^2+o(z^2)\Bigr)\\
&=\frac{1}{z^3}\Bigl(1-\Bigl(\frac{z}2+\frac{z^2}6\Bigr)+\frac{z^2}4+o(z^2)\Bigr) \\&=\frac{1}{z^3}\Bigl(1-\frac{z}2+\frac{z^2}{12}+o(z^2)\Bigr)=\frac{1}{z^3}-\frac1{2z^2}+\frac{1}{12z}+o\Bigl(\frac1{z}\Bigr)
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/3735299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplify $\frac{4\cos ^2\left(2x\right)-4\cos ^2\left(x\right)+3\sin ^2\left(x\right)}{4\sin ^2\left(x\right)-\sin ^2\left(2x\right)}$
Simplify:
$$\frac{4\cos ^2\left(2x\right)-4\cos ^2\left(x\right)+3\sin ^2\left(x\right)}{4\sin ^2\left(x\right)-\sin ^2\left(2x\right)}$$
After the substitution as $\cos(x)=a$ and $\sin(x)=b$, $(a^2+b^2=1)$, the expression becomes
$$\frac{4(a^2-b^2)^2-4a^2+3b^2}{4b^2-4a^2b^2}=\frac{4a^4-8a^2b^2+4b^4-4a^2+3b^2}{4b^4}=\bigg(\frac{a^2}{b^2}-1\bigg)^2-\frac{4a^2-3b^2}{4b^4}$$But I don't think I got anything useful... Any help is appreciated.
| So we have :
$$\frac{4\cos ^2\left(2x\right)-4\cos ^2\left(x\right)+3\sin ^2\left(x\right)}{4\sin ^2\left(x\right)-\sin ^2\left(2x\right)}$$
First of all, by using double-angles formulas we can get this:
$$\cos ^2\left(x\right) = \frac{1+\cos\left(2x\right)}{2}$$
$$\sin ^2\left(x\right) = \frac{1-\cos\left(2x\right)}{2}$$
So
$$4\cos ^2\left(2x\right)-4\cos ^2\left(x\right)+3\sin ^2\left(x\right) = \frac{8\cos ^2\left(2x\right)-7\cos \left(2x\right)-1}{2}$$ which equals to $$ \frac{(cos\left(2x\right) -1)(8\cos \left(2x\right)+1)}{2}$$
Lets look at the denominator $${4\sin ^2\left(x\right)-\sin ^2\left(2x\right) = \cos^2 \left(2x\right) - 2\cos \left(2x\right) + 1}$$ Since:
$$4\sin ^2\left(x\right) = 2 -2\cos \left(2x\right) $$
$$\sin^2\left(2x\right) = 1 - \cos^2 \left(2x\right)$$
So we get:
$$(\cos^2 \left(2x\right) - 2\cos \left(2x\right) + 1) = (\cos \left(2x\right) - 1)^2$$
Combining these equations, our first equation becomes :
$$ \frac{(cos\left(2x\right) -1)(8\cos \left(2x\right)+1)}{2(\cos \left(2x\right) - 1)^2}$$
Which can be reduced to:
$$\frac{8\cos \left(2x\right)+1}{2\cos \left(2x\right) - 2}$$
By using half angle formulas, we have :
$$8\cos\left(2x\right) = 8 - 16\sin^2\left(x\right)$$
$$2\cos\left(2x\right) = 2 - 4\sin^2\left(x\right)$$
So our final result is : $$\frac{8\cos \left(2x\right)+1}{2\cos \left(2x\right) - 2 } = \frac{16\sin^2\left(x\right) - 9}{4\sin^2\left(x\right)}$$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
} |
How to calculate the integral $I\left(a,b,c\right)=\int_{a}^{b}exp\left(u^{2}\right)\times\mbox{erfi}\left(\sqrt{\frac{u}{c}}\right)du$? I want to compute the following integral depending on a,b and c all strictly positive real numbers:
$$I\left(a,b,c\right)=\int_{a}^{b}exp\left(u^{2}\right)\times\mbox{erfi}\left(\sqrt{\frac{u}{c}}\right)du$$
I tend to use the 'Brute force' solution that is to rely on the integral series of $exp(x)$ and $erfi(x)$. This gives this development:
$$I\left(a,b,c\right)=\int_{a}^{b}\sum_{k=0}^{+\infty}\frac{\left(u^{2k}\right)}{k!}\times\mbox{erfi}\left(\sqrt{\frac{u}{c}}\right)du$$
$$\Leftrightarrow I\left(a,b,c\right)=\sum_{k=0}^{+\infty}\frac{1}{k!}\int_{a}^{b}u^{2k}\times\mbox{erfi}\left(\sqrt{\frac{u}{c}}\right)du$$
As:$$\mbox{erfi}\left(x\right)=\frac{2}{\sqrt{\pi}}\sum_{j=0}^{+\infty}\frac{x^{2j+1}}{\left(2j+1\right)j!}$$
this gives:
$$I\left(a,b,c\right)=\sum_{k=0}^{+\infty}\frac{1}{k!}\int_{a}^{b}u^{2k}\times\frac{2}{\sqrt{\pi}}\sum_{j=0}^{+\infty}\frac{\frac{u}{c}^{\frac{2j+1}{2}}}{\left(2j+1\right)j!}du$$
$$\Leftrightarrow I\left(a,b,c\right)=\frac{2}{\sqrt{\pi}}\sum_{k=0}^{+\infty}\frac{1}{k!}\sum_{j=0}^{+\infty}c^{-\frac{2j+1}{2}}\int_{a}^{b}\frac{u^{\frac{4k+2j+1}{2}}}{\left(2j+1\right)j!}du$$
The inner integral evaluates like:
$$\int_{a}^{b}u^{\frac{4k+2j+1}{2}}du=\left[\frac{u^{\frac{4k+2j+3}{2}}}{\frac{4k+2j+3}{2}}\right]_{a}^{b}$$
$$\Leftrightarrow\int_{a}^{b}u^{\frac{4k+2j+1}{2}}du=\frac{2}{4k+2j+3}\left[b^{\frac{4k+2j+3}{2}}-a^{\frac{4k+2j+3}{2}}\right]$$
This leads to:
$$I\left(a,b,c\right)=\frac{2}{\sqrt{\pi}}\sum_{k=0}^{+\infty}\frac{1}{k!}\sum_{j=0}^{+\infty}\frac{2c^{-\frac{2j+1}{2}}}{(4k+2j+3)\left(2j+1\right)j!}\left[b^{\frac{4k+2j+3}{2}}-a^{\frac{4k+2j+3}{2}}\right]$$
I assume this to be right but is there a more elegant solution than this double sum to the infinite ?
EDIT: i correct the question to state that a and b are strictly positive real numbers, and also the missing term in the final result.
| Let's begin by an expansion of the $\mbox{erfi}(x)$
function:
$$I\left(a,b,c\right)=\int_{a}^{b}exp\left(u^{2}\right)\times\mbox{erfi}\left(\sqrt{\frac{u}{c}}\right)du$$
$$\Leftrightarrow I\left(a,b,c\right)=\int_{a}^{b}exp\left(u^{2}\right)\times\frac{2}{\sqrt{\pi}}\sum_{j=0}^{+\infty}\frac{\sqrt{\frac{u}{c}}^{2j+1}}{\left(2j+1\right)j!}du$$
$$\Leftrightarrow I\left(a,b,c\right)=\frac{2}{\sqrt{\pi}}\sum_{j=0}^{+\infty}\frac{1}{\left(2j+1\right)j!}\int_{a}^{b}e^{u^{2}}\sqrt{\frac{u}{c}}^{2j+1}du$$
$$\Leftrightarrow I\left(a,b,c\right)=\frac{2}{\sqrt{\pi}}\sum_{j=0}^{+\infty}\frac{c^{-\frac{\left(2j+1\right)}{2}}}{\left(2j+1\right)j!}\int_{a}^{b}e^{u^{2}}u^{\frac{2j+1}{2}}du$$
$$\Leftrightarrow I\left(a,b,c\right)=\frac{2}{\sqrt{\pi}}\sum_{j=0}^{+\infty}\frac{c^{-\frac{\left(2j+1\right)}{2}}}{\left(2j+1\right)j!}\int_{a}^{b}e^{u^{2}}u^{\frac{2j+1}{2}}du$$
Then we need to compute the value of the integral $\int_{a}^{b}e^{u^{2}}u^{\frac{2j+1}{2}}du$ .
First let's split the integral in two new ones:
$$\int_{a}^{b}e^{u^{2}}u^{\frac{2j+1}{2}}du=\int_{a}^{+\infty}e^{u^{2}}u^{\frac{2j+1}{2}}du-\int_{b}^{+\infty}e^{u^{2}}u^{\frac{2j+1}{2}}du$$
Let $x=\frac{u^{2}}{a^{2}}$
, then: $u=a\sqrt{x}$ $\Rightarrow dx=\frac{2u}{a^{2}}du=\frac{2\sqrt{x}}{a}du$
so that: $$\int_{a}^{+\infty}e^{u^{2}}u^{\frac{2j+1}{2}}du=\int_{1}^{+\infty}e^{a^{2}x}\left(a\sqrt{x}\right)^{\frac{2j+1}{2}}\frac{a}{2\sqrt{x}}dx$$
$$=\frac{a^{\frac{2j+3}{2}}}{2}\int_{1}^{+\infty}e^{a^{2}x}x^{\frac{2j-1}{4}}dx$$
This last integral is nothing less than the expression of the Generalized Exponential Integral function $E_{n}\left(y\right)$ such that:
$$E_{n}\left(y\right)=\int_{1}^{+\infty}\frac{e^{-yt}}{t^{n}}dt=\int_{1}^{+\infty}\frac{e^{-yt}}{t^{n}}dt$$
with $n=\frac{1-2j}{4}$ and $y=-a^{2}$.
Then, we have:
$$\int_{a}^{b}e^{u^{2}}u^{\frac{2j+1}{2}}du=\frac{a^{\frac{2j+3}{2}}}{2}E_{\frac{1-2j}{4}}\left(-a^{2}\right)-\frac{b^{\frac{2j+3}{2}}}{2}E_{\frac{1-2j}{4}}\left(-b^{2}\right)$$
so that in the end, we have:
$$I\left(a,b,c\right)=\frac{2}{\sqrt{\pi}}\sum_{j=0}^{+\infty}\frac{c^{-\frac{\left(2j+1\right)}{2}}}{\left(2j+1\right)j!}\left(\frac{a^{\frac{2j+3}{2}}}{2}E_{\frac{1-2j}{4}}\left(-a^{2}\right)-\frac{b^{\frac{2j+3}{2}}}{2}E_{\frac{1-2j}{4}}\left(-b^{2}\right)\right)$$
Note that this last result may also be expressed in terms of incomplete Gamma function as $$E_{n}\left(y\right)=y^{n-1}\Gamma\left(1-n,y\right)$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3736601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the sum of all positive integers $n$ such that when $1^3+2^3+3^3 +\dots+ n^3$ is divided by $n+5$ the remainder is $17.$
Find the sum of all positive integers $n$ such that when $1^3+2^3+3^3 +\dots+ n^3$ is divided by $n+5$ the remainder is $17.$
Letting $k= n+5$ we get that $1^3+2^3+3^3 +\dots+ (k-5)^3 \equiv 17 \text{ (mod $k$)}$.
Knowing the sum of cubes formula we get that ($\frac{(k-5)(k-4)}{2})^2\equiv 17 \text{ (mod $k$)}$.
From here I'm not sure how I should continue. What would be my options?
| $\frac{1}{4}(k-5)^2(k-4)^2$ = $\frac{1}{4}(k^4-18k^3+121k^2-360k+400)$=$pk+17$ for some integer $p$. So, $\frac{1}{4}(k^4-18k^3+121k^2-360k+332)$ = $pk$ => $k$ should divide $332$ . So we get possible values of $k$ as $1,83,166,332$. As $k>4$, $k=1$ is not possible. Similarly for $k=332$, $(k^3-18k^2+121k-360+1)$ is clearly not divisible by $4$. So, $p=\frac{1}{4}(k^3-18k^2+121k-360+1) $ isn't an int, which is not possible. When $k=83$, it is easy to see that $p=\frac{1}{4}(k^3-18k^2+121k-360+4)$ is an integer(Why?). With a bit of clever reasoning you can verify that $k =166$ also fits the bill. So, $n= 78 \space and \space 161$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3737447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
prove $\sum\cos^3{A}+64\prod\cos^3{A}\ge\frac{1}{2}$ In every acute-angled triangle $ABC$,show that
$$(\cos{A})^3+(\cos{B})^3+(\cos{C})^3+64(\cos{A})^3(\cos{B})^3(\cos{C})^3\ge\dfrac{1}{2}$$
I want use Schur inequality
$$x^3+y^3+z^3+3xyz\ge xy(y+z)+yz(y+z)+zx(z+x)$$
then we have
$$x^3+y^3+z^3+6xyz\ge (x+y+z)(xy+yz+zx)$$
But I can't use this to prove my question
and I use this post methods links also can't solve my problem,use $AM-GM $ inequality$$\cos^3{A}+\dfrac{\cos{A}}{4}\ge\cos^2{A}$$
so
$$LHS\ge \sum_{cyc}\cos^2{A}-\dfrac{1}{4}\sum_{cyc}\cos{A}+64\prod_{cyc}\cos^3{A}$$use
$$\cos^2{A}+\cos^2{B}+\cos^2{C}+2\cos{A}\cos{B}\cos{C}=1$$
it must to prove
$$\frac{1}{2}+64\cos^3{A}\cos^3{B}\cos^3{C}\ge 2\cos{A}\cos{B}\cos{C}+\dfrac{1}{4}(\cos{A}+\cos{B}+\cos{C})$$
| Another way.
Let $a^2+b^2-c^2=z$, $a^2+c^2-b^2=y$ and $b^2+c^2-a^2=x$.
Thus, $x$, $y$ and $z$ are positives, $\cos\alpha=\frac{x}{\sqrt{(x+y)(x+z)}},$ $\cos\beta=\frac{y}{\sqrt{(x+y)(y+z)}}$, $\cos\gamma=\frac{z}{\sqrt{(x+z)(y+z)}}$
and we need to prove that
$$2\sum_{cyc}\frac{x^3}{\sqrt{(x+y)^3(x+z)^3}}+\frac{128x^3y^3z^3}{\prod\limits_{cyc}(x+y)^3}\geq1.$$
Now, by AM-GM $$\sum_{cyc}\frac{x^3}{\sqrt{(x+y)^3(x+z)^3}}=\sum_{cyc}\frac{2x^3}{2(x+y)(x+z)\sqrt{(x+y)(x+z)}}\geq$$
$$\geq \sum_{cyc}\frac{2x^3}{(x+y)(x+z)(2x+y+z)}.$$
Id est, it's enough to prove that:
$$ \sum_{cyc}\frac{4x^3}{(x+y)(x+z)(2x+y+z)}+\frac{128x^3y^3z^3}{\prod\limits_{cyc}(x+y)^3}\geq1,$$ which is obvious by BW (https://artofproblemsolving.com/community/c6h522084 ).
By the way, a full expanding gives $$\sum_{sym}(2x^9y^3+7x^8y^4+10x^7y^5+5x^6y^6)+$$
$$+xyz\sum_{sym}(6x^8y+28x^7y^2+30x^6y^3-8x^5y^4)+$$
$$+x^2y^2z^2\sum_{sym}(21x^6+40x^5y-154x^4y^2-158x^3y^3)+$$
$$+x^3y^3z^3\sum_{sym}(-8x^3+58x^2y+121xyz)\geq0,$$
which is not so trivial.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3737759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
} |
Doubt about how to compute $\sum_{n=1}^{2019}{\frac{1+2+3+4...+n}{1^3+2^3+3^3...+2019^3}}$ In the book I am reading, I have encountered the following sum.
$$S = \sum_{n=1}^{2019}{\frac{1+2+3+4...+n}{1^3+2^3+3^3...+2019^3}}$$
From here, I factored the denominator since it does not seem to be dependent on $n,$ and I rewrote the expression as
$$\frac{1}{1^3+2^3+3^3+ \cdots +2019^3} \sum_{n=1}^{2019}{1+2+3+4+ \cdots +n}=\frac{1}{1^3+2^3+3^3 + \cdots +2019^3} \sum_{n=1}^{2019}{\sum_{k=1}^n}k$$
From here, I proceeded to compute the sum since there is an equation for the sum of the first $n$ terms, and then, I solved the sum of this equation. Also, I applied the equation of the sum of the first $n^3$ terms. Finally, my result was wrong. My question is regarding my initial approach to the problem since everything else is very simple. I have thought about other ways to solve this, and I can not find it. The official answer is
$$\frac{2019}{2010}.$$
Give this problem a try, and if you arrive at the solution, I invite you to help me understand it.
| Let's replace $2019$ by $T$. In the numerator:
$$ \sum_{n=1}^T \sum_{k=1}^n k = \sum_{n=1}^T \frac{n(n+1)}{2} = \frac{(T+1)^3 - (T+1)}{6}$$
In the denominator:
$$ \sum_{k=1}^T k^3 = \frac{(T+1)^4 - 2 (T+1)^3 + (T+1)^2}{4} $$
So:
$$ \frac{ \sum_{n=1}^T \sum_{k=1}^n k }{\sum_{k=1}^T k^3} = \frac{4}{6} \frac{(T+1)^3 - (T+1)}{(T+1)^4 - 2 (T+1)^3 + (T+1)^2} = \frac{2 (T+2)}{3 T (T+1)}$$
For $T=2019$ this is $\dfrac{2021}{3030 \cdot 2019}$. The official answer is wrong.
Or maybe the question was supposed to be
$$ \frac{1}{2}\sum_{n=1}^{2019} \frac{1 + 2 + \ldots + n}{1^3 + 2^3 + \ldots + n^3} $$
which would work out to $\dfrac{2019}{2020}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3740187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
prove formula by induction I need to prove by induction that $\sum_{n=2}^{m} \frac{1}{n^2 - 1} = \frac{1}{2}(1+\frac{1}{2}-\frac{1}{m}-\frac{1}{m+1})$.
I've seen some similar questions about the convergence of the infinite series, however none about the finite case, I've tried in several ways to factorize or add things in order the find the case m+1 but without success.
| Here is an explicit evaluation which may be useful. This is focused on the $m+1$ portion of the inductive proof:
$P\left(m+1\right)=\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{m+1}-\frac{1}{m+2}\right)$
$P\left(m+1\right)=\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{m+1}\right)-\frac{1}{2}\frac{1}{m+2}$
In order for the inductive step to be valid, the above expression needs to be equal to:
$
P\left(m\right)+\frac{1}{\left(m+1\right)^{2}-1}=P\left(m\right)+\frac{1}{m^{2}+2m}
$
where $P\left(m\right)=\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{m}-\frac{1}{m+1}\right)$ is assumed true.
We thus require that:
$
\frac{1}{2}\frac{1}{m+2}=\frac{1}{2m+4}=\frac{1}{2m}-\frac{1}{m^{2}+2m}
$
where the first term on the right completes the expression for $P\left(m\right)$ and the second term is the new increment of $\frac{1}{n^{2}-1}$ for $m+1$.
Cross-multiplying the right hand side of the above yields:
$
\frac{1}{2m+4}=\frac{m^{2}}{2m^{3}+4m^{2}} \checkmark
$
I hope this helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3740597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
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