Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Are these polynomials irreducible in $\mathbb{Q}[x]$? $$x^3+x^2+x+1$$
$$x^5+x^3+x^2+1$$
$$x^5+x^3+x+1$$
I've tried applying Eisenstein criterion but I can't figure it out.
Thanks in advance.
| $$x^3+x^2+x+1=(x+1)(x^2+1)$$
$$x^5+x^3+x^2+1=(x^2+1)(x^3+1)$$
For $f(x)=x^5+x^3+x+1$, we have $f(2)=43$, which is prime, so by Cohn's criterion, $f(x)$ is irreducible over $\mathbb{Z}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $p(x)=ax^3 -2x^2 +bx+c$, find $a, b$ and $c$ if $p(0)=12$, $p(-1)=3$ and $p(2)=36$ If $p(x)=ax^3 -2x^2 +bx+c$, find $a, b$ and $c$ if $p(0)=12$, $p(-1)=3$ and $p(2)=36$
Can someone please teach me how to do this question thanks!
| This is really a problem in basic linear algebra. If we simply plug in the three given values of the independent variable $x$, we obtain the three equations for $a$, $b$, $c$:
$p(0) = 12$ becomes
$c = 12$;
$p(-1) = 3$ becomes
$-a -b +c = 5$;
$p(2) = 36$ becomes
$8a + 2b + c = 44$;
looks like we get $c$ for free! Plug... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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smallest $k$ s.t. $(x+y)^2\leq k(x^2-xy+y^2)$ I would appreciate if somebody could help me with the following problem
Q: Find $K$?
$$(x+y)^2\leq k(x^2-xy+y^2)$$.
where $\forall x,y\in \mathbb{R}$
| $$(x+y)^2\leq k(x^2-xy+y^2)\iff (k-1)x^2-(2+k)xy+(k-1)y^2\geq0$$
using the inequality
$$xy\leq \frac{1}{2}(x^2+y^2)$$
we have
$$(k-1)x^2-(2+k)xy+(k-1)y^2\geq (k-1)x^2-\frac{2+k}{2}(x^2+y^2)+(k-1)y^2\\
=(\frac{k}{2}-2)(x^2+y^2)\geq 0\iff k\geq 4$$
and if we take $x=y=1$ we find that $4\leq k$.
So $k= 4$ answers the que... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $\alpha +\beta = \dfrac{\pi}{4}$ prove that $(1 + \tan\alpha)(1 + \tan\beta) = 2$
If $\alpha +\beta = \dfrac{\pi}{4}$ prove that $(1 + \tan\alpha)(1 + \tan\beta) = 2$
I have had a few ideas about this:
If $\alpha +\beta = \dfrac{\pi}{4}$ then $\tan(\alpha +\beta) = \tan(\dfrac{\pi}{4}) = 1$
We also know that $\tan... | You were almost there:
$$\begin{align}
1 &= \frac{\tan \alpha + \tan\beta}{1 - \tan\alpha \tan\beta}\\
1 - \tan\alpha\tan\beta &= \tan\alpha + \tan\beta\\
2 &= 1 + \tan\alpha + \tan\beta + \tan\alpha\tan\beta\\
2 &= (1+\tan\alpha)(1+\tan\beta)
\end{align}$$
where each equation is equivalent to the preceding/following, ... | {
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Show that $\frac {\sin(3x)}{ \sin x} + \frac {\cos(3x)}{ \cos x} = 4\cos(2x)$ Show that
$$\frac{\sin(3x)}{\sin x} + \frac{\cos(3x)}{\cos x} = 4\cos(2x).$$
| HINT:
$\sin (3x) = 3\sin x - 4 \sin^3 x$
and
$\cos(3x) = 4\cos^3x - 3\cos x$
| {
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Evaluating the limit of a sequence given by recurrence relation $a_1=\sqrt2$, $a_{n+1}=\sqrt{2+a_n}$. Is my solution correct? Problem
The sequence $(a_n)_{n=1}^\infty$ is given by recurrence relation:
*
*$a_1=\sqrt2$,
*$a_{n+1}=\sqrt{2+a_n}$.
Evaluate the limit $\lim_{n\to\infty} a_n$.
Solution
*
*Show that th... | Looks great. Here is a fun trick I've seen to answer this question.
Using the half angle formula, notice the following:
$$\cos\left(\frac{\pi}{4}\right)=\frac{1}{2}\sqrt 2\\\cos\left(\frac{\pi}{8}\right)=\sqrt{\frac{1}{2}(1+\frac{1}{2}\sqrt 2)}=\frac{1}{2}\sqrt{2+\sqrt 2}\\\cos\left(\frac{\pi}{16}\right)=\sqrt{\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/449592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 2,
"answer_id": 1
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Show that $\sin2\alpha\cos\alpha+\cos2\alpha\sin\alpha = \sin4\alpha\cos\alpha - \cos4\alpha\sin\alpha$
Show that $\sin2\alpha\cos\alpha+\cos2\alpha\sin\alpha = \sin4\alpha\cos\alpha - \cos4\alpha\sin\alpha$
I know that $\sin2\alpha = 2\sin\alpha\cos\alpha$
so
$$\sin2\alpha\cos\alpha=2\sin\alpha\cos^2\alpha$$
and $\... | You're making this way harder than it needs to be. One is $\sin{(2 \alpha + \alpha)}$. The other is $\sin{(4 \alpha-\alpha)}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Maximum and minimum function on circle Find minimum and maximum value of function
$f(x,y) = 3x+4y + |x-y|$
on circle
$$\left\{ (x,y): x^2+y^2 = 1 \right\}$$
I used polar coordinate system. So I have $x = \cos t$ and $y=\sin t$ where $ t \in [0, 2 \pi)$.
Then i exploited definition of absolute function and i got:
$$h(t... | We have $\frac{\partial{f}}{\partial{x}}=\frac{x-y}{\sqrt{(x-y)^2}}+3$ and $\frac{\partial{f}}{\partial{y}}=\frac{y-x}{\sqrt{(x-y)^2}}+4$, i.e., $\nabla f=(\frac{x-y}{\sqrt{(x-y)^2}}+3, \frac{y-x}{\sqrt{(x-y)^2}}+4)$. Let $\phi(x,y)=x^2+y^2$ $\Rightarrow$ $\nabla \phi= 2(x,y)$. Now, using Lagrange multipliers we have $... | {
"language": "en",
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How to get value of $a^2b^2+b^2c^2+c^2a^2$ If $a+b+c=0$ and $a^2+b^2+c^2=36$, what is the value of $a^2b^2+b^2c^2+c^2a^2$?
| We'll work with what we have until we get what we want:
$$0 = (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2 ac = 36 + 2(ab + bc + ac)$$
And therefore we can conclude that
$$
- 18 = ab + bc + ac
$$
This in turn means that
$$
324 = (ab + bc + ac) ^2 = a^2b^2 + b^2 c^2 + a^2 c^2 + 2(a^2bc + ab^2 c + abc^2) \\\\
=a^2b^... | {
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Conversion of sum of series into product form: $\sum_{n=1}^\infty \frac1{n(n+1)} = \frac12 \prod_{n=2}^\infty \left( 1+\frac{1}{n^2-1} \right)$ Show that the following series and product are equivalent:
$$
\sum_{n=1}^\infty \left[ \dfrac{1}{n(n+1)} \right] = \dfrac{1}{2} \prod_{n=2}^\infty \left[ 1+\dfrac{1}{n^2-1} \r... | Hint: First, simplify the fractions to get
$$\prod\limits_{n = 2}^\infty 1 + \frac{1}{n^2 - 1} = \prod\limits_{n = 2}^\infty \frac{n^2}{n^2 - 1}$$
Denote the partial product by $P_n$, and take a logarithm:
$$\log{P_N} = \log{\prod\limits_{n = 2}^N \frac{n^2}{n^2 - 1}} = \sum\limits_{n = 2}^N \log{\frac{n^2}{n^2 - 1}}$$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that 360 divides (a-2)(a-1)a.a.(a+1)(a+2) there's a question which asks to prove that
360 | a2(a2-1)(a2-4)
I attempted it in the following manner.
a2(a2-1)(a2-4) = (a-2)(a-1)(a)(a+1)(a+2)(a)
The first 5 terms represent the product of 5 consecutive terms. Hence,
One of them will have a factor of 5
One of them will... | We know from here or here, the product $n$ consecutive integers is divisible by $n!$ for integer $n>0$
$$(a-2)(a-1)a\cdot a(a+1)(a+2)$$
$$=(a-2)(a-1)a(a+3-3)(a+1)(a+2)$$
$$=\underbrace{(a-2)(a-1)a(a+1)(a+2)(a+3)}_{\text{ the product of } 6\text{ consecutive integers }}-3\underbrace{(a-2)(a-1)a(a+1)(a+2)}_{\text{ the ... | {
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"timestamp": "2023-03-29T00:00:00",
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Need help with a number theory question I can't solve this question I got from a Math Olympiad past paper:
Find all integers $a$ such that $\frac{a^2+4}{2a+1}$ is also an integer
I know $a$ can be $0$ and $-1$ but I can't ascertain if there are other values $a$ can take.
| André's answer is pretty much the best way to do this. A more naïve approach is as follows:
Suppose $\frac{a^2+4}{2a+1} = k\in\mathbb{Z}$. Then:
$a^2 - 2ka + (4-k) = 0$
This is a quadratic equation with solutions:
$a = \frac{2k \pm \sqrt{4k^2 - 4(4-k)}}{2}= k \pm\sqrt{k^2+k-4}$
For $a$ to be an integer we thus need $k^... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Geometry problem on finding the diameter of a circle $AB$ is the diameter of a circle. Tangents $AD$ and $BC$ are drawn so that $AC$ and $BD$ intersect at a point on the circle. If $|AD|=a$ and $|BC|=b$, $(a \neq b)$ then find the diameter of the circle.
|
Lets say AC and BD intersect at E , so at right angles as diameter subtends right angle at circumference .
Using pythagorous theorem
$$ \displaystyle\boxed{(diameter)^2 = AE^2 + BE^2} $$
So now we should calculate AE and BE to get d
as $$CE \times CA = CB^2 ---> 1$$
and $$DE \times DB = DA^2 ---> 2$$
and as the tr... | {
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"url": "https://math.stackexchange.com/questions/463467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Does this hold for three numbers If $a\ge b\ge c\ge0$, does it hold that $\sqrt[3]{\left(a-b+c\right)^{2}}\ge\sqrt[3]{a^{2}}-\sqrt[3]{b^{2}}+\sqrt[3]{c^{2}}$?
Thanks for any help.
| Consider one case of $a = b \ge c \ge 0$. Both sides are equal, and the inequality holds.
Now, partial differentiate $\sqrt[3]{(a-b+c)^2} - \sqrt[3]{a^2} + \sqrt[3]{b^2} - \sqrt[3]{c^2}$ with respect to a.
$$\frac{2}{3}(a-b+c)^{-\frac{1}{3}} - \frac{2}{3}a^{-\frac{1}{3}}$$
$$=\frac{2}{3\sqrt[3]{a-b+c}} - \frac{2}{3\sq... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that if $d\cdot e| d(d+1)+e\cdot e$ then $d\cdot (d+1)+e\cdot e=3de$ Prove that if $d\cdot e| d(d+1)+e\cdot e$ then $d\cdot (d+1)+e\cdot e=3de$ where $d$ and $e$ are positive integers.
| We can simplify the problem a bit. Let $\gamma = \gcd(d,e)$ and write $d = \gamma\delta,\; e = \gamma\varepsilon$. Then the condition becomes
$$\gamma^2\delta\varepsilon \mid \gamma\delta(\gamma\delta+1) + \gamma^2\varepsilon^2\tag{1}$$
and we see that we must have $\gamma \mid \delta$. So write $d = \gamma^2\kappa$ to... | {
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"timestamp": "2023-03-29T00:00:00",
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Some identities with the Riemann zeta function Can someone either help derive or give a reference to the identities in Appendix B, page 27 of this, http://arxiv.org/pdf/1111.6290v2.pdf
Here is a reproduction of Appendix B from Klebanov, Pufu, Sachdev and Safdi's $2012$ preprint (v2) 'Renyi Entropies for Free Field Theo... | The first identity is essentially the functional equation for the Hurwitz Zeta Function. You can find a proof and associated results here, where the identity appears on page 2 of the pdf (pg 1916) as equation (2) where the $F$ series is defined on the previous page.
| {
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"timestamp": "2023-03-29T00:00:00",
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$a+b+c+d=4n$; where $a,b,c$ lie between $0$ and $n $ and $d$ lies between $0$ and $2n.$ In the examination the max marks for each of three papers is $n$ and the fourth paper is $2n$. find the number of ways in which a candidate can gets $3n$ marks ?
| Rather than applying inclusion-exclusion, you can let generating functions do the work for you. The possible (integral) marks that can be obtained for a single paper with maximum mark$~m$ is given by the polynomial $1+X+\cdots+X^m=\frac{1-X^{m+1}}{1-X}$. You want to get the coefficient of $X^{4n}$ (or maybe of $X^{3n}$... | {
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"timestamp": "2023-03-29T00:00:00",
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How to expand $\left(1+px+qx^2\right)^{-2}$ The coefficients of $x$ and $x^2$ in the expansion of $\left(1+px+qx^2\right)^{-2}$ in ascending powers of $x$ are 4 and 14 respectively. Find the values of $p$ and $q$.
How should you go about solving this?
| In general
$$(1+y)^{-2} = \sum_{k=0}^{\infty} (-1)^k (k+1) \, y^k$$
so that
$$(1+p x+q x^2)^{-2} = 1 - 2 (p x+q x^2) + 3 (p x + q x^2)^2 - \cdots$$
Out to $O(x^2)$:
$$(1+p x+q x^2)^{-2} = 1 -2 p x + (3 p^2 - 2 q) x^2 + \cdots$$
$$-2 p = 4 \implies p=-2$$
$$3 p^2-2 q=14 \implies -2 q=2 \implies q = -1$$
You can also con... | {
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Determine the minimum of $a^2 + b^2$ if $a,b\in\mathbb{R}$ are such that $x^4 + ax^3 + bx^2 + ax + 1 = 0$ has at least one real solution I just wanted the solution, a hint or a start to the following question.
Determine the minimum of $a^2 + b^2$ if $a$ and $b$ are real numbers for which
the equation
$$x^4 + ax^3 + bx^... | As has already been observed, if $c$ is a root of your equation, so is $c^{-1}$. So suppose your equation has roots $c,d,c^{-1},d^{-1}$.
We have
$$a=-(c+c^{-1} + d + d^{-1})$$
and
$$b=cc^{-1}+ cd+cd^{-1} +c^{-1}d +c^{-1}d^{-1} + dd^{-1}$$
$$=2+(c+c^{-1})(d+d^{-1})$$
Now if $c\in \mathbb{R}$, we must also have $d+d^{-1}... | {
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"timestamp": "2023-03-29T00:00:00",
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Integral using Euler's substitution Question:
Solve this:
$$\displaystyle \int \frac {2x-\sqrt {4x^2-x+1} } {x-1} \,dx$$
Our solution:
$$16 \ln|t+4|-\ln|t+0.25|+C$$ when $t=\sqrt {4x^2-x+1}-2x$ (using Euler's formula)
But wolfram's solution was totally different. We were wondering if these two are equal.
We found this... | Tools that might help for a direct check if the two solutions are equal up to a constant ... and perhaps useful in the future.
*
*We have that the expression $\sqrt{4x^2-x+1}-2x$ can be re-written as $\frac{(\sqrt{4x^2-x+1}-2x)(\sqrt{4x^2-x+1}+2x)}{\sqrt{4x^2-x+1}-2x}=\frac{(4x^2-x+1)-4x^2}{\sqrt{4x^2-x+1}+2x}$.
*... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/475688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Forget about the sin and cos functions, show that $(x-x^3/3!+x^5/5!-x^7/7!+...)^2+ (1-x^2/2!+x^4/4!-x^6/6!+...)^2=1$. Forget about the $\sin$ and $\cos$ functions, are there possibly some brilliant way to show that
$$\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots\right)^2+
\left(1-\frac{x^2}{2!}+\frac{x^4... | If we denote
$$f(x)=\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots\right)^2+
\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots\right)^2$$
then by differentiation term by term we see that $f'(x)=0$ so $f(x)=f(0)=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/477353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
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Finding minima and maxima of $\sin^2x \cos^2x$ I am trying to find the turning points of $\sin^2x \cos^2x$ in the range $0<x<\pi$.
So far I have:
$\begin{align} f'(x) & =\cos^2x \cdot 2\sin x \cdot \cos x + \sin^2x \cdot 2\cos x \cdot -\sin x \\
& = 2 \cos^3x \sin x - 2\sin^3x \cos x \\
& = 2 \sin x \cos x (\cos^2x - \... | One might broadly interpret "algebraically" to mean "without calculus." This is actually quite doable, and in my opinion less complicated than dealing with messy trigonometric equations.
Note the following identities:
$f(x) = \sin^2 x \cos^2 x = \sin^2 x - \sin^4 x = \sin^2 x (1-\sin^2 x) \ge 0$
Equality holds when $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/477636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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Sum of the series $1+\frac{1\cdot 3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+\cdots$ Decide if the sum of the series
$$1+\frac{1\cdot3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+ \cdots$$
is: (i) $\infty$, (ii) $1$, (iii) $2$, (iv) $4$.
| Note that
$$\binom{n}{r} = \frac{n(n-1)(n-2)\dots(n-r+1)}{1\cdot2\cdot3\cdots r}$$
and so
$$\binom{1/2}{r} = \frac{\frac12 \cdot \frac{-1}2 \cdot \frac{-3}{2} \cdot \frac{-5}{2} \cdots \frac{-(2r-3)}{2}}{1\cdot2\cdot3\cdots r} = \frac{(-1)^{r-1}1 \cdot 3 \cdot 5 \cdots (2r-3)}{2^r r!}$$
Now, in this problem, each term ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/479610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 8,
"answer_id": 0
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A problem on limit It seems
$\lim_{n \rightarrow \infty} \sum_{r=1}^{n-1}\binom{n}{r-1}
\sum_{j=0}^{n-r+1} (-1)^j \binom{n-r+1}{j} \frac{1}{r+j+1}$
is $\frac{1}{2}$. Here's a plot of the function for $n \leq 300$:
But I can not prove this. Any hints?
| $\begin{align}
S(n)
&=\sum_{r=1}^{n-1}\binom{n}{r-1}
\sum_{j=0}^{n-r+1} (-1)^j \binom{n-r+1}{j} \frac{1}{r+j+1}\\
&= \sum_{r=0}^{n-2}\binom{n}{r}
\sum_{j=0}^{n-r} (-1)^j \binom{n-r}{j} \frac{1}{r+j+2}\\
\text{(replace }r \text{ by } n-r)\quad&= \sum_{r=2}^{n}\binom{n}{r}
\sum_{j=0}^{r} (-1)^j \binom{r}{j} \frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/480160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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} |
Showing Whether a Sequence is Bounded Above or Not I am trying to solve the following problem about a sequence:
Consider the sequence ${a_n}$ where $a_n = 1 + \frac{1}{1 \cdot 3} + \frac {1}{1 \cdot 3 \cdot 5} + \frac {1}{1 \cdot 3 \cdot 5 \cdot 7} + ... + \frac{1}{1 \cdot 3 \cdot ... \cdot (2n-1)}.$ Decide whether ${a... | If you want a direct proof that doesn't use geometric series outright, note that if you consider the terms in $a_n$ shifted forward one position and then compare to the terms in $a_{n+1}$, you get $a_{n+1} \leq 1 + a_n/3$. Also note $a_1 = 1 < 3/2$. If $a_n < 3/2$ then $a_{n+1} \leq 1 + a_n/3 < 1 + 1/2 = 3/2$ as wel... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/480266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Roots of biquadratic equation This question also was a part of my today's maths olympiad paper:
If squares of the roots of $x^4 + bx^2 + cx + d = 0$ are $\alpha, \beta, \gamma, \delta$ then prove that:
$64\alpha\beta\gamma\delta - [4\Sigma \alpha\beta - (\Sigma \alpha)^2]^2 = 0$
I found value of $\alpha\beta\gamma\delt... | Complete solution of the problem:
Given that α,β,γ & δ are squares of the roots of $x^4+bx^2+cx+d=0$ hence $\sqrt{α},\sqrt{β},\sqrt{γ}$ & $\sqrt{δ}$ are the roots of the given equation.
Thus, we have
sum of the roots
$$=\sqrt{α}+ \sqrt{β}+\sqrt{γ}+ \sqrt{δ}=\frac{-B}{A}=\frac{0}{1}=0$$ &
sum of the products of two r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/481306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Compute $ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx$ Compute $$ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx$$
Of course we have $$ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx = 2 \int_{0}^{\infty} \frac{x^2}{(1+x^2)^2} dx = 2 \int_{0}^{\infty} \left( \frac{x}{1+x^2} \right) ^2 dx = \lim_{ A \to \inft... | HINT:
$$I=\int\frac{x^2dx}{(x^2+1)^2}=\int x\cdot \frac x{(x^2+1)^2}dx$$
Integrating by parts,
$$I=x\int \frac x{(x^2+1)^2}dx-\int\left(\frac{d(x)}{dx}\cdot \frac x{(x^2+1)^2}dx\right)dx$$
$$\text{As }\int \frac x{(x^2+1)^2}dx=\frac12\int\frac{d(x^2+1)}{(x^2+1)^2}=-\frac1{2(x^2+1)}$$
$$I=-x\cdot\frac1{2(x^2+1)}+\int \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/483180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 3
} |
How do I write this sum in summation notation? $$\sum_{n=?}^\infty \left(\frac{x^n}{?}\right) = \frac{x^0}{1} + \frac{x^1}{x^2 -1}+\frac{x^2}{x^4 - x^2 +1}+\frac{x^3}{x^6 -x^4 + x^2 -1}+\frac{x^4}{x^8-x^6 +x^4 - x^2 +1}+\cdots$$
I am pretty sure I have the numerator of the summation, $x^n$ correct, but don't know how t... | Once you realise that the denominators are sums of powers of $-x^2$, this becomes easy. You have
$$
\sum_{n=0}^\infty\frac{x^n}{\sum_{i=0}^n(-x^2)^i}.
$$
Actually that denominator is a geometric series, that can be written explicitly without summation (provided that $X=-x^2\neq1$) using
$$
\sum_{i=0}^nX^i=\frac{1-X... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/483327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
How to show $\sqrt{4+2\sqrt{3}}-\sqrt{3} = 1$ I start with $x=\sqrt{4+2\sqrt{3}}-\sqrt{3}$, then
$\begin{align*}
x +\sqrt{3} &= \sqrt{4+2\sqrt{3}}\\
(x +\sqrt{3})^2 &= (\sqrt{4+2\sqrt{3}})^2\\
x^2 + (2\sqrt{3})x + 3 &= 4+ 2\sqrt{3}\\
x^2 + (2\sqrt{3})x - 1 - 2\sqrt{3} &= 0
\end{align*}$
So I have shown that there is so... | Hint: $$4+2\sqrt{3} = 1+2\sqrt{3}+\sqrt{3}^2 = (1+\sqrt{3})^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/483398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
} |
IMO 1988, problem 6 In 1988, IMO presented a problem, to prove that $k$ must be a square if $a^2+b^2=k(1+ab)$, for positive integers $a$, $b$ and $k$.
I am wondering about the solutions, not obvious from the proof. Beside the trivial solutions a or $b=0$ or 1 with $k=0$ or $1$, an obvious solution is $a=b^3$ so that t... | Supposing that
$$ \frac{a^2 + b^2}{ab + 1} = k$$ then
$$ a^2 - a(kb) + (b^2 - k) = 0 $$
So using quadratic formula gives: $$ a = \frac{kb \pm \sqrt{k^2b^2 -4(b^2 -k)}}{2}$$
The solutions are when $k = b^2 $ and thus $a= kb = b^3$
So $$ b = 1 , a = 1 , k= 1$$
$$b = 2 , a = 8, k = 4$$
$$ b= 3, a= 27, k = 9 $$ and so on... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/483771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 7,
"answer_id": 2
} |
Left and Right Ideal Generated by Two Matrices. Let $R= {\rm Mat}_2(\Bbb R)$ be the ring (with $1$) of $2\times2$-matrices with entries in $\Bbb R$. Let $$M = \left\{\begin{pmatrix}1&0 \\0&0\end{pmatrix},\begin{pmatrix}0&1\\0&0\end{pmatrix}\right\},$$ i.e. a set of two matrices. What are the left and right ideal genera... | Sorry, there are some issues here caused by your choice of notation. By multiplying the same $a,b,c,d$ matrix with both generators, you've overlooked that there's nothing wrong with multiplying the generators with two different matrices and adding, which will produce many more possiblities.
In the left ideal generated ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/483839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\lim_{x\to0} \frac{6}{x}-\frac{42}{x^2+7x}$ Evaluate $\lim_{x\to0} \frac{6}{x}-\frac{42}{x^2+7x}$
I'm I want to say that you cross multiply to get the same denominator, but I could be wrong.
Please Help!!
| First, we find the common denominator:
$\require{cancel}$
$$\frac ab - \frac c{bd} = \frac{ad - c}{bd}$$
Applying the "general rule" in your case:
$$\begin{align}
\frac 6x - \frac{42}{x^2 + 7x} & = \frac 6x - \frac{42}{x(x + 7)} \\ \\
& = \frac{6(x + 7) - 42}{x(x + 7)} \\ \\ & = \frac{6x}{x(x+7)} \\ \\ & = \frac{6\col... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/485185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Proving Muirhead-like inequalities Let $T_{m,n,p}(x,y,z)=\sum_{Sym} x^m y^n z^p$.
For $x,y,z>0$, prove
$2T_{6,3,0}(x,y,z)+T_{3,3,3}(x,y,z)+3T_{4,4,1}(x,y,z)\geq 6T_{5,2,2}(x,y,z)$.
I tried to prove that by using AM-GM inequality, without success.
Is there a general way to prove these "Muirhead-like" inequalities?
| This is a very old question, but I'll answer it anyway.
We should prove that:
$$ \sum_{sym}2x^6 y^3 + x^3 y^3 z^3 + 3 x^4 y^4 z^1 - 6 x^5 y^2 z^2 \ge 0$$
If we divide the inequality by $x^3 y^3 z^3$, we get:
$$ \sum_{sym} 2 \frac{x^3}{z^3} + 1 + 3 \frac{x y}{z^2} - 6 \frac{x^2}{y z} \ge 0$$
Apply the following subsitu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/485272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Expression of an Integer as a Power of 2 and an Odd Number (Chartrand Ex 5.4.2[a])
Let $n$ be a positive integer. Show that every integer $m$ with $ 1 \leq m \leq 2n $ can be expressed as $2^pk$, where $p$ is a nonnegative integer and $k$ is an odd integer with $1 \leq k < 2n$.
I wrote out some $m$ to try to conceive... | I personally found this problem easier to do if you consolidate case 2 and 3 into one case and then prove by contradiction:
Assume m is even and that m cannot be written in the form $2^p \bullet k$ where $p$ is a nonnegative integer and k is an odd integer with $1 \le k \le 2n$. Thus, $m \neq 2^p \bullet k$. We can re... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/486395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
} |
Series of modified Bessel functions There is a known identity to evaluate a sum of the form
$$\sum_{n\geq1} \rho^n I_n(\omega) $$
Where $\rho>0$, $\omega >0$ and $I_n$ is the modified Bessel function of the first kind.
???
Thanks.
| $$ S = \sum_{n=1}^{\infty} x^nI_n(y) = \sum_{n=1}^{\infty} x^n \sum_{k = 0}^{\infty} \frac{\left(\frac{y}{2} \right)^{2k + n}}{k!\Gamma(n+k+1)}$$
$$ = \sum_{n=1}^{\infty} \left( \frac{xy}{2} \right)^n \sum_{k=0}^{\infty} \frac{\left( \frac{y^2}{4} \right)^k }{k!(n+k)!} = \sum_{n=1}^{\infty} \frac{\left( \frac{xy}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/488462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
find this limit $\lim_{x\to0^{+}}\frac{\tan{(\tan{x})}-\tan{(\sin{x})}}{\tan{x}-\sin{x}}$ find the limit.
$$\lim_{x\to0^{+}}\dfrac{\tan{(\tan{x})}-\tan{(\sin{x})}}{\tan{x}-\sin{x}}$$
my try:
$$\tan{x}=x+\dfrac{1}{3}x^3+o(x^3),\sin{x}=x-\dfrac{1}{6}x^3+o(x^3)$$
so
$$\tan{(\tan{x})}=\tan{x}+\dfrac{1}{3}(\tan{x})^3+o(\t... | I think this identity is useful here:
$$\tan(a-b)=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}$$ and the fact that $\tan(a)\sim a$ while $a\sim 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/490470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Prove $3^n \ge n^3$ by induction Yep, prove $3^n \ge n^3$, $n \in \mathbb{N}$.
I can do this myself, but can't figure out any kind of "beautiful" way to do it.
The way I do it is:
Assume $3^n \ge n^3$
Now,
$(n+1)^3 = n^3 + 3n^2 + 3n + 1$,
and $\forall{} n \ge 3$,
$3n^2 \le n^3, \,\, 3n + 1 \le n^3$
Which finally giv... | Basis:
$n = 1$
$$3^1 \ge 1^3 \implies 3 \ge 1\text{, which is true}$$
Inductive hypothesis
Let $n=k$ and also let this inequality hold:
$$3^k \ge k^3$$
Inductive step
We'll prove that it also holds when $n=k+1$
$$3^{k+1} \ge (k+1)^3$$
$$3^k \cdot 3 \ge k^3 + 3k^2 + 3k + 1$$
$$3^k + 3^k + 3^k \ge k^3 + 3k^2 + 3k + 1$$
N... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/490983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
How to evaluate the limit $\lim\limits_{x \to 1} \left(\frac{2}{1-x^2} - \frac{3}{1-x^3}\right) $, and others? $$ \lim_{x \to 1} \left( \frac{2}{1-x^2} - \frac{3}{1-x^3} \right)$$
In my opinion the function is not defined at $ x = 1 $ but somehow when I look at the graph, it's continuous and there is no break. I learne... | Consider the expression
$$\lim_{x \to 1}
\left(
\frac1{1-x}
-
\frac1{1-x}
\right)
$$
This is "obviously" zero,
even though both terms blow up
as $x \to 1$,
because the singularities cancel.
The same thing happens here.
The basic identity needed here is
$1-x^n
=(1-x)(1+x+x^2+...+x^{n-1})
$.
Using this,
$\begin{align}
\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/492643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Compute $\int_0^n \left[\frac x {x+1} + \frac x {2x+4} + \frac x {3x+9} + \cdots\right] \, dx$ for $x>0$ I want to compute $\displaystyle \int_0^n \left[\frac x {x+1} + \frac x {2x+4} + \frac x {3x+9} + \cdots \right] \, dx$ for $x>0$
My attempt: The integrand can be written as a sum of $\displaystyle f_k(x)=\frac x{kx... | We have
\begin{align}
f(x) & = \sum_{k=1}^{\infty} \dfrac{x}{kx+k^2} = \sum_{k=1}^{\infty} \left(\dfrac1k - \dfrac1{k+x} \right) = \sum_{k=1}^{\infty} \int_0^1 \left(t^{k-1} - t^{k+x-1}\right)dt\\
& = \int_0^1 (1-t^x) \left(\sum_{k=1}^{\infty} t^{k-1} \right) dt = \int_0^1 \dfrac{1-t^x}{1-t} dt
\end{align}
We now have
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/493696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How prove two segments are congruent If $\angle ACB=2\angle ABC=4\angle BAD$,and $D$ is on $BC$, such $CG\parallel AB$, $CG=CD$, $HG \parallel AC$.
show that
$$BD=HG$$
My idea: let $\angle BAD=x$, then $\angle ABC=2x$, $\angle ACB=4x$.
Lemma 1：in $\Delta ABC,AB=c,AC=b,BC=a$ if $\angle A=2\angle B$,then $a^2=b(b+c... | The statement can be proved by applying the sine theorem four times.
If $\widehat{BAD}=\theta,\widehat{ABC}=2\theta,\widehat{ACB}=4\theta$, then $$\widehat{DAC}=\pi-7\theta,\widehat{DCG}=2\theta,\widehat{ADC}=3\theta,\widehat{DHG}=7\theta,\widehat{CDG}=\widehat{CGD}=\frac{\pi}{2}-\theta,\\ \widehat{HDG}=\frac{\pi}{2}-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/495112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Different Ways of Integrating $3\sin x\cos x$ I am asking this question for my son who is in (equivalent) twelfth
grade and I failed to answer his query.
When he tries to integrate $3\sin x\cos x$, he finds that
this can be done in at least following three ways.
And these three ways do not produce equivalent results.
O... | All three answers are correct provided you add a constant to each one of those.
Because from the very definition of integration, it is the area under the curve, so it requires bounds to give a unique value.
You can't evaluate the value of an indefinite integral without including constant.
And I am sure that in the exam... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/495159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
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Prove limit of $\frac1{n+1}+\frac1{n+2}+\ldots+\frac1{n+n}$ exists and lies between $0$ and $1$. Prove limit of $\displaystyle \frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{n+n}$ exists and lies between $0$ and $1$.
So far I have $\displaystyle \lim_{n\rightarrow\infty}\sum_{k=1}^n\frac{1}{k+n}=L$ for some $L>0$.
Then gi... | It's a monotone rising sequence: if $$s_n = \frac{1}{n+1} + ... + \frac{1}{2n},$$ then we have $$s_{n+1} - s_n = \frac{1}{2n+2} + \frac{1}{2n+1} - \frac{1}{n+1} = \frac{1}{2n+1} - \frac{1}{2n+2} > 0.$$
It is bounded below by $0$; and it's bounded above by $1$, because $$s_n = \frac{1}{n+1} + ... + \frac{1}{2n} < \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/498040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
} |
Mathematics max and min $x$ and $y$ values of an ellipse I am trying to find the max and min x and y values of an ellipse. The ellipse is given by $$x^2-xy+y^2=3$$
So I need to find the max points, so I need the derivative of the function.
I got $$\dfrac{dy}{dx} = \dfrac{(y-2x)}{(2y-x)}$$
So I then thought that I shoul... | (originally posted on gamedev)
The general equation of an ellipse centered at the origin is:
$$(ax+by)^2+(cx+dy)^2=r^2$$
Expand w.r.t $x$ and $y$:
$$(a^2+c^2)x^2+(b^2+d^2)y^2+2(ab+cd)xy=r^2$$
An horizontal line has equation $y=k$. Either it intersects with the ellipse in two points, or it intersects in only one point (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/498622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Improving bound on $\sqrt{2 \sqrt{3 \sqrt{4 \ldots}}}$ An old challenge problem I saw asked to prove that $\sqrt{2 \sqrt{3 \sqrt{4 \ldots}}} < 3$. A simple calculation shows the actual value seems to be around $2.8$, which is pretty close to $3$ but leaves a gap. Can someone find $C < 3$ and prove $\sqrt{2 \sqrt{3 \sq... | Put $K=\log C$. The problem is then to found a tight upper bound for:
$$ K = \sum_{n=1}^{+\infty}\frac{\log(1+n)}{2^n}. $$
Note that:
$$ K/2=K-K/2 = \sum_{n=1}^{+\infty}\frac{\log\left(1+\frac{1}{n}\right)}{2^n}. $$
By writing $K/4$ as $K/2-K/4$ just like above, we get:
$$\frac{K}{4}=\frac{\log 2}{2}-\sum_{n=1}^{+\inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/498774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 3
} |
When is $2^n \pm 1$ a perfect power Is there an easy way of showing that $2^n \pm 1$ is never a perfect power, except for $2^3 + 1 = 3^2 $?
I know that Catalan's conjecture (or Mihăilescu's theorem) gives the result directly, but I'm hopefully for a more elementary method.
I can show that it is never a square, except... | (This proof was completed with an insightful comment from Gottfried. I'm placing it as an answer so that it is readily seen that an answer exists, as opposed to just leaving it in the question.)
Suppose we have $ 2^n \pm 1 = x^k$ for some positive integers $x, k$. Cases $n=1, 2, 3$ are easily dealt with. $n=3$ yields t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/500616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "36",
"answer_count": 3,
"answer_id": 2
} |
Calculus limit homework problem $$ \lim_{n→\infty} \frac1 n \left(\left(a + \frac 1 n\right)^2 + \left(a + \frac 2 n\right)^2 + ... + \left(a + \frac{n-1}{n}\right)^2\right)$$
$$ \text{hint: }\ 1^2 + 2^2 + ... + n^2 = \frac{n(n+1)(2n+1)}{6} $$
I can't figure out how to find this problem's limit. Does anybody have ide... | Ignore the limit for now. We will find the expression first, and then evaluate the limit.
$$\frac{1}{n}\left((a^2+2a/n+1/n^2)+(a^2+4a/n+4/n^2) + \ldots + (a^2 + (k) 2a/n + k^2/n^2) + \ldots + (a^2+(n-1)2a/n) + (n-1)^2/n^2)\right).$$
This simplifies to $$\frac{1}{n}((n-1)a^2 + 2a/n (1 + 2 + \ldots + n - 1) + 1/n^2 (1^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/503814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove $\sqrt{s_n+1} = \frac{1}{2}(1+\sqrt{5})$ This is to prove how the limit of $s_n$ converges to $\frac{1}{2}(1+\sqrt{5})$.
Assume: $s_1 = 1$; for $n \geq 1$, $s_{n+1} = \sqrt{s_n + 1}$.
How to prove this converges to $\frac{1}{2}(1+\sqrt{5})$?
| Argue by induction:
*
*The terms of the sequence are all positive.
*They are all bounded above (by $3$): The base case is clear, and if $s_n<3$, then $s_{n+1}=\sqrt{s_n+1}<\sqrt{3+1}=2<3$.
*The sequence is increasing: $s_2=\sqrt2>1=s_1$, and if $s_n<s_{n+1}$, then $s_n+1<s_{n+1}+1$, so $s_{n+1}=\sqrt{s_n+1}<\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/504427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Taylor Series Expansion for $\sin^2(\omega t)$ What are the first few terms for the Taylor Series Expansion for $\sin^2(\omega t)$? $(\omega$=$2\pi f$)
If you could show some working, that would be helpful
| Sometimes a little sledgehammering is good practice. I prefer Andrés solution, which is elegant, but this can also be a good exam question to test various Taylor expansion skills by insisting that no short cuts be used.
Applying basic Taylor expansion
Consider $f(x) = \sin^2 x$:
\begin{align*}
f'(x) &= 2\sin x\cos x =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/505360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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How to find the type of triangle when given the ratio of it's sides?
Q.The sides of a triangle are in ratio 4 : 6 : 7, then the triangle is:
(A) acute angled
(B) obtuse angled
(C) right angled
(D) impossible
It's definitely not (C) right-angled since $7^2 ≠ 6^2+4^2$
Is it possible to use trigonometry here ev... | Let $ABC$ be a triangle with sides $a,b,c$.
If $c^2=a^2+b^2$, then the angle at $C$ is a right angle.
If $c^2\lt a^2+b^2$, then the angle at $C$ is acute.
If $c^2\gt a^2+b^2$, then the angle at $C$ is obtuse.
We can think of these facts as coming from the Cosine Law
$$c^2=a^2+b^2-2ab\cos(\angle C).$$
Remark: But we don... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/507175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
minimal polynomial of a matrix with some unknown entries Question is to prove that :
characteristic and minimal polynomial of $ \left( \begin{array}{cccc}
0 & 0 & c \\
1 & 0 & b \\
0 & 1 & a \end{array} \right) $ is $x^3-ax^2-bx-c$.
what i have done so far is :
characteristic polynomial of a matrix $A$ is given by $\... | It is very easy to see what the minimal polynomial is for this matrix, even without knowing let alone factoring the characteristic polynomial. The argument even works for matrices of this type (called companion matrices) of arbitrary size $n$. Since $A\cdot e_i=e_{i+1}$ for $0<i<n$, where $e_i$ is the standard basis ve... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
If $ k = \sqrt{n\cdot (n+1)\cdot(n+2)\cdot (n+3)}$. Then $\lfloor k \rfloor =$ If $ k = \sqrt{n\cdot (n+1)\cdot(n+2)\cdot (n+3)}$, where $n\in \mathbb{N}$. Then $\lfloor k \rfloor = $
$\underline{\bf{My\; Try}}::$ We can write the expression $n\cdot (n+1)\cdot(n+2)\cdot (n+3) = (n^2+3n).(n^2+3n+2)$
$ = (n^2+3n)^2+2\cdo... | We want to determine
$\lfloor v \rfloor$,
where
$v = \sqrt{n\cdot (n+1)\cdot(n+2)\cdot (n+3)}$.
Let $x = n+3/2$,
so
$x^2 = n^2+3n+9/4$.
Then
$\begin{align}
v^2
&=(x-3/2)(x-1/2)(x+1/2)(x+3/2)\\
&=(x^2-(3/2)^2)(x^2-(1/2)^2)\\
&=x^4-5x^2/2+9/16\\
&=(x^2-5/4)^2-25/16+9/16\\
&=((n^2+3n+9/4)-5/4)^2-1\\
&=(n^2+3n+1)^2-1\\
\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/508512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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How find the maximum of $\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{z}{1+z^2}$ let $x,y,z$ are postive numbers,and such
$$xy+zx+yz=1$$
find the maxum of
$$\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}+\dfrac{z}{1+z^2}$$
my try
note
$$x^2+1=x^2+xy+yz+xz=(x+y)(x+z)$$
$$y^2+1=(y+x)(y+z)$$
$$z^2+1=(z+x)(z+y)$$
$$\dfrac{1}{1+x^2}+\dfrac{1}{... | Trigonometric substitution seems easier here. We may set $x = \tan\frac{A}{2}, y = \tan\frac{B}{2}, z = \tan\frac{C}{2}$, where $A, B, C$ are angles of a triangle, so that the constraint is satisfied. Then we have to find the maximum of
$$f = \dfrac{1}{1+\tan^2\frac{A}{2}}+\dfrac{1}{1+\tan^2\frac{B}{2}}+\dfrac{\tan\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/508582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
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How find this equation $\prod\left(x+\frac{1}{2x}-1\right)=\prod\left(1-\frac{zx}{y}\right)$ let $x,y,z\in(0,1)$, find the pairs of $(x,y,z)$ such
$$\left(x+\dfrac{1}{2x}-1\right)\left(y+\dfrac{1}{2y}-1\right)\left(z+\dfrac{1}{2z}-1\right)=\left(1-\dfrac{xy}{z}\right)\left(1-\dfrac{yz}{x}\right)\left(1-\dfrac{zx}{y}\ri... | The only $(x,y,z)$ is $(\frac12,\frac12,\frac12)$.
For all other $x,y,z \in (0,1)$ the left-hand side is larger.
We prove this by showing that if we fix the average $a=\frac13(x+y+z)$ then
(i) the left-hand side is minimized when $x=y=z=a$, while
(ii) the right-hand side is maximized when $x=y=z=a$.
Hence it is enough ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
Show $\lim_{x\to3} \frac{2x+3}{4x-9}=3$ by the definition of the limit. Show $\lim_{x\to3} \frac{2x+3}{4x-9}=3$ by the definition of the limit.
The definition of the limit: if for any given $\epsilon>0$ $\exists\delta>0$ such that if $x\in D(f(x))$ and $0<|x-c|<\delta$, then $|f(x)-L|<\epsilon$ where $L$ is the limit.
... | We have that
$$|f(x)-3|=\left|\frac{2x+3}{4x-9}-3\right|=\left|\frac{10(x-3)}{4x-9}\right|=\frac{10}{|4x-9|}|x-3|.$$
We want to simplify the denominator, ideally getting something like
$$|f(x)-3|<(...)|x-3|.$$
Restrict $x$ to a range around $3$ where $|4x-9|$ is bounded below, like $x\in(\frac{5}{2},\frac{7}{2})$. When... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Inequality....(RMO $1994$...question $8$) If $a$, $b$, $c$ are positive real numbers such that $a+b+c = 1$, prove that
$$(1+a)(1+b)(1+c) ≥ 8(1-a)(1-b)(1-c)\text{.}$$
| We can easily prove the inequality: $(a+b)(b+c)(c+a)\ge 8abc$.
Replacing $a$ by $(1-a)$, $b$ by $(1-b)$ and $c$ by $(1-c)$ we get:
$$(2-(a+b))(2-(c+b))(2-(c+a))\geq 8(1-a)(1-b)(1-c)$$
but $a + b + c= 1$ therefore:
$$a+b=1-c,~ b+c=1-a, ~\text{and}~ c+a=1-b$$
substituting these values in LHS we get :
$$(1+a)(1+b)(1+c)\ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/511970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
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Use mathematical induction to prove that 9 divides $n^3 + (n + 1)^3 + (n + 2)^3$; Looking for explanation, I already have the solution. I have the solution for this but I get lost at the end, here's what I have so far.
basis $n = 0$; $9 \mid 0^3 + (0 + 1)^3 + (0 + 2)^2 ?$
$9 \mid 1 + 8$ = true
Induction: Assum... | We will show that $(n+1)^3+(n+2)^3+(n+3)^3$ is divisible by $9$. The following compuation shows that it is indeed so:
$(n+1)^3+(n+2)^3+(n+3)^3= n^3+(n+1)^3+(n+2)^3+9(n^2+3n+3)$
since the sum of first 3 terms on the RHS is divisible by $9$ by assumption.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/512072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Check whether $36^{36}+41^{41}$ a multiple of $77$ Let $$a=\frac{36^{36}+41^{41}}{77}.$$
Is $a$ an integer?
I know that:
*
*The last digit ofr $41^{41}$ is $1$.
*The last digit of $36^{36}=6^{72}$ is $6$.
How can I use this information to answer my question, or do I need another thing to help me?
| Let $n=36^{36}+41^{41}$. Three elementary steps:
*
*What happens modulo $7$: Note that $36\equiv 1\pmod{7}$ and that $41\equiv -1\pmod{7}$ hence $n \equiv 1^{36}+(-1)^{41}=1+(-1)=0\pmod{7}$.
*What happens modulo $11$: Note that $36\equiv 3\pmod{11}$ and that $41\equiv -3\pmod{11}$ hence $n\equiv 3^{36}+(-3)^{41}=3... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How prove this inequality $a^2+b^2+c^2+d^2\ge abcd$ let $a,b,c,d$ are positive numbers,and such
$$2(a+b+c+d)\ge abcd$$
show that
$$a^2+b^2+c^2+d^2\ge abcd$$
My try:if $a,b,c,d\le 16$,then we have
$$a^2+b^2+c^2+d^2\ge 4\sqrt{abcd}$$
| You are on the right track. So if, $abcd\le 16,$ then $4\sqrt{abcd}\ge abcd.$ What if $abcd\ge 16?$ Well, then you need another estimate for $a^2+b^2+c^2+d^2.$ The most natural one is $$a^2+b^2+c^2+d^2\ge \frac{(a+b+c+d)^2}{4}\ge \frac{(abcd)^2}{16}\ge abcd. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/513067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How to prove $\sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-\cdots}}}}}}} = \frac{2+\sqrt 5 +\sqrt{15-6\sqrt 5}}{2}$ Ramanujan stated this radical in his lost notebook:
$$\sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-\cdots}}}}}}} = \frac{2+\sqrt 5 +\sqrt{15-6\sqrt 5}}{2}$$
I don't have any idea on... | The correct period has length 4, namely (+,+,+,-)
$$x_1=\small+\sqrt{5+\sqrt{5+\sqrt{5\color{red}-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5\color{red}-\cdots}}}}}}} = \frac{2+\sqrt 5 +\sqrt{15-6\sqrt 5}}{2}=2.7472\dots$$
The other roots of the quartic in $x$ are given by the patterns $\small(+,+,-,+),\; (+,-,+,+),\; (-,+,+,+)$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/515253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "51",
"answer_count": 2,
"answer_id": 1
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A conjectural closed form for $\sum\limits_{n=0}^\infty\frac{n!\,(2n)!}{(3n+2)!}$ Let
$$S=\sum\limits_{n=0}^\infty\frac{n!\,(2n)!}{(3n+2)!},\tag1$$
its numeric value is approximately $S \approx 0.517977853388534047...$${}^{[more\ digits]}$
$S$ can be represented in terms of the generalized hypergeometric function:
$$S=... | partial answer
Use the method of LINK
Since
$$
\int_0^1 t^n (1-t)^{2n}\;dt = \frac{n!(2n)!}{(3n+1)!} ,
$$
If we write
$$
S(x) = \sum_{n=0}^\infty \frac{n!(2n)!}{(3n+2)!}\;x^{3n+2}
$$
then
$$
S(x) = \int_0^1 \left(\sum_{n=0}^\infty t^n(1-t)^{2n} \frac{x^{3n+2}}{3n+2}\right)dt
$$
and $S = S(1)$.
But the derivative with ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "50",
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"answer_id": 2
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If both the sum and sum of squares of two rationals are integers, the two rationals are integers too There are two rational numbers $\alpha, \beta$ such that $\alpha + \beta,\ \alpha^2 + \beta^2$ are both integers. Prove that $\alpha, \beta$ are integers.
I started off by assuming that $\alpha = \frac{a}{b}, \beta = \f... | Hint: If $\alpha+\beta=k$ and $\alpha=\dfrac{a}{b}$, then $\alpha^2+(k-\alpha)^2-k^2=2\alpha^2-2\alpha k=\dfrac{2a}{b}\left(\dfrac{a}{b}-k\right)$ must be integer.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the equations of the tangent line(s) that are parallel to the line $y= -4x +3$ Find the equations of the tangent line(s) to $\;f(x) = x^3 -4x^2 + 2\;$ that are parallel to the line $y= -4x +3$.
I know the slope is -4 and $f'(x)= 3x^2-8x.$
| if $f'(x)=-4$ then the tangent line to that point will be parallel to your line.
Set $f'(x)=-4$ then solved for $x$.
$$-4=3x^2-8x$$
$$3x^2-8x+4=0$$
$$x=\dfrac{8\pm\sqrt{(-8)^2-4(3)(4)}}{2(3)}$$
$$x=\dfrac{8\pm\sqrt{64-48}}{6}$$
$$x=\dfrac{8\pm4}{6}$$
$$x=\dfrac{8+4}{6}\text{ or }\dfrac{8-4}{6}$$
$$x=2\text{ or }\dfrac2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Given $f(x+1)=x^2-3x+2$, how can I find $f(x)$? Given $f(x+1)=x^2-3x+2$, how can I find $f(x)$?
| Let $y=x+1$. We then have $x=y-1$. Hence,
\begin{align}
f(y) & = f(x+1) = x^2 -3x+2 = (y-1)^2 - 3(y-1)+2\\
& = y^2-2y+1-3y+3+2 = y^2-5y+6
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The sum of three square roots bounded below by $\sqrt{82}$
Let $a,b,c >0$ and $a+b+c \le 1$. Prove: $$\sqrt{a^2+1/a^2}+\sqrt{b^2+1/b^2}+\sqrt{c^2+1/c^2} \ge \sqrt{82}$$
Progress: I tried to have 3 vectors $(a,1/a)$, $(b,1/b)$ and $(c,1/c)$, played around with the vector inequalities but no success.
| Assume first $a+b+c=1$.
Let $x=(a,1/a)$, $y=(b,1/b)$, $z=(1/c)$. Then the triangle inequality for the Euclidean norm tells us that
$$
\|x+y+z\|\leq\|x\|+\|y\|+\|z\|.
$$
This looks like
$$
\sqrt{1+\left(\frac1a+\frac1b+\frac1c\right)^2}\leq\sqrt{a^2+\frac1{a^2}}+\sqrt{b^2+\frac1{b^2}}+\sqrt{c^2+\frac1{c^2}}.
$$
So to p... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 1
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How do you factor $x^3 - 8 = 0$? I factored $x^3 - 8 = 0$ and I only got $x = 2$, but the answer said it's $x=2$ and $x=-1 \pm \sqrt{3}i$? How do you get $x=-1 \pm\sqrt{3}i$?
| Let $x = y-1$.
Then $x^3 - 8 = y^3 - 3y^2 + 3y - 9 = (y-3)(y^2 + 3)$.
Setting this last expression equal to zero, we find all three roots:
One is $y = 3$ and the other two are $y = \pm i\sqrt{3}$.
Since $x = y-1$, we conclude the roots are $x = 2$ and $x = \pm i\sqrt{3} - 1$.
"QED"
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How prove this $a+b\le 1+\sqrt{2}$ let $0<c\le b\le 1\le a$, and such $a^2+b^2+c^2=3$, show that
$a+b\le 1+\sqrt{2}$
My try: let $ c^2=3-(a^2+b^2)\le b$
| As Sun stated, given a valid tuple of $(a,b,c)$, replace it with $(A, b, 0)$ where $ A^2 = a^2 + c^2$. Observe that $a + b \leq A + b$, hence it remains to show that $ A + b \leq 1 + \sqrt{2}$.
Squaring this, we need to show that $ A^2 + 2Ab + b^2 \leq 3 + 2 \sqrt{2}$ or that $AB \leq \sqrt{2}$.
But since $A^2 + 2b^2 \... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Proof by Math Induction I have 3 math induction proofs I have been struggling with for a while. I understand how to do summation proofs but these ones, I can't find a general pattern to solve. Please help.
1) $D(n) = {n(n-3) \over2}$ for all $n \ge 3$ This is for the n diagonals of a polygon.
2)$\binom{n}{0} + \binom... | I assume that $D(n)$ is the number of diagonals in an $n$-sided convex polygon. Descriptively: each vertex of the polygon can be connected to $n-3$ other (connecting it with itself or its neighbors will not produce diagonals). That way, you make each diagonal twice, hence the formula.
If you want to use induction, then... | {
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Finding the remainder when a polynomial is divided by a product of numbers whose remainders are known We have a polynomial $f(x)$ with rational roots that leaves remainders $15, 2x + 1$ when divided by the polynomials $x - 3, (x-1)^2$ respectively. What is the remainder when $f(x)$ is divided by $(x - 3)(x-1)^2$?
From... | Write $$f(x)=A(x-3)(x-1)^2+B(x-1)^2+C(x-3)+D$$ where $A$ is an integer polynomial or constant
$\implies 15=f(3)=4B+D\implies D=15-4B$
$\implies f(x)=A(x-3)(x-1)^2+B(x-1)^2+C(x-3)+15-4B$
Now, $f(x)\equiv C(x-3)+15-4B\pmod{(x-1)^2}\equiv Cx+15-4B-3C $
But $f(x)\equiv 2x+1\pmod{(x-1)^2}$
$\implies C=2$ and $15-4B-3C=1$ an... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Characteristic polynomial of a matrix is monic? Given a $n \times n$ matrix A, I need to show that its characteristic polynomial, defined as $P_A (x) = det (xI-A)$ is monic. I am trying induction. But no clue after induction hypothesis.
| For $n=1$,
\begin{equation}
P_A(x)=det(xI_1-A)=|s-a_{11}|=s-a_{11}.
\end{equation}
So, $P_A(x)$ is monic for $n=1$.
Assume $P_A(x)$ is monic for $n=k$, i.e.
\begin{equation}
P_A(x)=det(xI_k-A)=\begin{vmatrix}
(x-a_{11})&-a_{12}&...&-a_{1k}\\
-a_{21}&(x-a_{22})&...&-a_{2k}\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/525707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 5
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How to solve this first-order ODE $\frac{dy}{dx}=\frac{y^6-2x^2}{2x^2y+2y^3-y}$? solve this ODE equation
$$\dfrac{dy}{dx}=\dfrac{y^6-2x^2}{2x^2y+2y^3-y}$$
My try:
$$\dfrac{y~dy}{dx}=\dfrac{y^6-2x^2}{2x^2+2y^2-1}$$
let
$$u=y^2$$
then
$$\dfrac{du}{dx}=\dfrac{2u^3-4x^2}{2x^2+2u-1}$$
then I can't work. Thank yo... | You did it correctly.
Now for $\dfrac{du}{dx}=\dfrac{2u^3-4x^2}{2x^2+2u-1}$ ,
$(2u+2x^2-1)\dfrac{du}{dx}=2u^3-4x^2$
Let $v=u+x^2-\dfrac{1}{2}$ ,
Then $u=v-x^2+\dfrac{1}{2}$
$\dfrac{du}{dx}=\dfrac{dv}{dx}-2x$
$\therefore2v\left(\dfrac{dv}{dx}-2x\right)=2\left(v-x^2+\dfrac{1}{2}\right)^3-4x^2$
$2v\dfrac{dv}{dx}-4xv=2v^3-... | {
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"timestamp": "2023-03-29T00:00:00",
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How to find the $f^{-1}(x)$ of $f(x)=x^{3}-12x+\frac{48}{x}-\frac{64}{x^{3}}$ It is a question from a quiz.
The following is the whole question.
Let
\begin{eqnarray}
\\f(x)=x^{3}-12x+\frac{48}{x}-\frac{64}{x^{3}} , \space x\in (-\infty ,0),
\end{eqnarray}
find $f^{-1}(x)$. Hint : $f(x)$ can be written in the form,... | for $x \in (-\infty,0)$ the inverse function has the form
$$
f^{-1}(x)=\frac{1}{2}\left( \sqrt[3]{x}-\sqrt{\sqrt[3]{x^2}+16} \right)
$$
As mentioned above we have
$y = \left(x - \dfrac4x\right)^3 \Longrightarrow x^2 - x\sqrt[3]{y} - 4 = 0$.
By solving the equation we get
$$
x=\frac{1}{2}\left( \sqrt[3]{y}-\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/528477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 3
} |
Prove that there are no positive integers $a,b,c,d$ such that $a^2 + b^2 = c^2$ and $a^2 - b^2 = d^2$ Prove that there are no positive integers $a,b,c,d$ such that $a^2 + b^2 = c^2$ and $a^2 - b^2 = d^2$?
Could you show me the proof?
| Dividing by $(a,b)$, we can assume that $(a,b)=1$.
Multiplying the two equations yields
$$
a^4-b^4=c^2d^2\tag{1}
$$
whereby
$$
b^4+(cd)^2=a^4\tag{2}
$$
This answer characterizes all primitive Pythagorean Triples; that is, there must be a pair of positive integers $(m,n)$ so that $(m,n)=1$ and $m+n$ is odd so that
$$
x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/530003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Find the limit of $\sum\limits_{k=1}^n\left(\sqrt{1+\frac{k}{n^2}}-1\right)$
$$\lim_{n\rightarrow\infty}\sum_{k=1}^n\left(\sqrt{1+\frac{k}{n^2}}-1\right)$$
Note that $\forall x\ge 0, \sqrt{x}-1\le\sqrt{1+x}-1\le x$
Then
$$\sum_{k=1}^n\left(\sqrt{\frac{k}{n^2}}-1\right)\le S_n\le \sum_{k=1}^n\frac{k}{n^2}=\frac{1}{2}-... | For every $x$ in $(0,1)$, $\frac12x-x^2\leqslant\sqrt{1+x}-1\leqslant\frac12x$ hence
$$
\frac1{2n^2}T_n-\frac1{n^4}R_n\leqslant S_n\leqslant\frac1{2n^2}T_n,\qquad T_n=\sum_{k=1}^nk,\quad R_n=\sum_{k=1}^nk^2.
$$
One knows that $T_n=\frac12n(n+1)$ and $R_n\leqslant\sum\limits_{k=1}^nn^2=n^3$ hence
$$
\frac{n+1}{4n}-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/532404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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Proving any product of four consecutive integers is one less than a perfect square Prove or disprove that : Any product of four consecutive integers is one less than a
perfect square.
OK so I start with $n(n+1)(n+2)(n+3)$ which can be rewritten $n(n+3)(n+1)(n+2)$
After multiplying we get $(n^2 + 3n)(n^2 + 3n + 2)$
How ... | Observe that the product $p$ of four consecutive integers can be written as $p=(x-\frac{3}{2})(x-\frac{1}{2})(x+\frac{1}{2})(x+\frac{3}{2})$ where $x=n+\frac{1}{2}$ for some integer $n$. Then $p=(x^2-\frac{9}{4})(x^2-\frac{1}{4}) = (x^2-\frac{5}{4}+1)(x^2-\frac{5}{4}-1) = (x^2-\frac{5}{4})^2-1$. It remains to show that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/532737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 1
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What is the tens digit of $3^{100}$? Is there a general formula to calculate the n-th digit of any big number?
| We would like to calculate $3^{100}\pmod{100}$. This will tell us the last two digits of $3^{100}$, which includes the tens digit.
We proceed as follows: $3^4 = 81$, so:
$$\begin{align}
3^8 & \equiv (81)^2 \equiv 61 \pmod{100} \\
3^{16} & \equiv (61)^2 \equiv 21\pmod{100} \\
3^{24} & \equiv 61\cdot 21 \equiv 81\pmod{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/533175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Is i true that $x+y\le(2x^2+2y^2)^{1/2}\le...\le(2^{p-1}x^p+2^{p-1}y^p)^{1/p}$ Is it true that,
$$
x+y\le(2x^2+2y^2)^{1/2}\le...\le(2^{p-1}x^p+2^{p-1}y^p)^{1/p}
$$
How to prove or disprove it?
| The first link is OK. $x+y \le |x+y|$, and then note we have
$$|x+y| \le (2x^2+2y^2)^{1/2} \tag{1}$$
iff $(x-y)^2 \ge 0$, after squaring both (nonnegative) sides of $(1),$ rearranging and factoring. So this gives the first inequality of the list
$x+y \le (2x^2+2y^2)^{1/2}.$
The second inequality (if I see the pattern) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/533523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Matrix Equation $A^3-3A=\begin{pmatrix}-7 & -9\\ 3 & 2\end{pmatrix}$ How can I solve in $\mathcal{M}_{2}(\mathbb{Z})$ the equation $$A^3-3A=\begin{pmatrix}-7 & -9\\ 3 & 2\end{pmatrix}?$$
I try to use $$A^2-Tr(A)A+detA\cdot I_2=O_2$$ but I don't still obtain anything.
thanks.
| $$
A=\pmatrix{{2}&{3}\\{-1}&{-1}}.$$
Note that $a^3+b^3=(a+b)^3-3ab(a+b)$.
Let $a,b$ be eigenvalues of $A$. Then $trA=a+b$, $detA=ab$, and they are integers.
Let $X=trA$.
We see from $A(A^2-3I)$ has determinant $13$, we have $detA | 13$.
From the identity in the beginning, we have $X^3-3(ab+1)X+5=0$ , which we can o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/537095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 0
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Prove : $\frac{1}{a}+\frac{1}{b}-\frac{1}{c}< \frac{1}{abc}$ $a;b;c>0$ such that $a^2+b^2+c^2=\frac{5}{3}$. Prove :
$\frac{1}{a}+\frac{1}{b}-\frac{1}{c}< \frac{1}{abc}$
| $$
c(a+b)\leq \frac{c^2+(a+b)^2}{2}=\frac{a^2+b^2+c^2}{2}+ab=\frac 56+ab
$$
which is the rearrangement of your inequality and even the stronger one.
$$
c(a+b)\leq \frac 56+ab \implies \\
\frac 1a+\frac 1b\leq \frac 5{6abc}+\frac 1c\implies
\frac 1a+\frac 1b-\frac 1c\leq \frac 5{6abc}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/538234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving the recurrence $T(n) = 2T\left(\frac{n}{2}\right) + \frac{n}{2}\log(n)$ Please help me solve the recurrence
$$ T(n) = 2T\left(\frac{n}{2}\right) + \frac{n}{2}\log(n) $$
| Here is an exact solution of a related recurrence that has the same complexity as yours. We take $T(0)=0$ and the recurrence is
$$T(n) = 2 T(\lfloor n/2 \rfloor) + \frac{1}{2} n \; (\lfloor\log_2 n\rfloor+1).$$
Let the binary digits of $n$ be given by
$$n = \sum_{k=0}^{\lfloor\log_2 n\rfloor} d_k 2^k.$$
Then the exact ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/538463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How find this integral $I=\int_{0}^{\infty}\frac{\log{\cos^2{x}}}{1+e^{2x}}dx$ find the value
$$I=\int_{0}^{\infty}\dfrac{\log{\cos^2{x}}}{1+e^{2x}}dx$$
My try: let $$e^{2x}=u\Longrightarrow x=\dfrac{1}{2}\log{u}$$
then
$$I=\int_{1}^{\infty}\dfrac{\log{(\cos^2{(\dfrac{1}{2}\log{u})})}}{2u(1+u)}du$$
then I can't.Thank y... | Referring to the identity
$$ \log|\cos x| = - \log 2 + \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \cos (2nx) $$
(which can be easily obtained by considering the real part of $\log (1 + e^{2ix})$), we find that
\begin{align*}
I&= \int_{0}^{\infty} \frac{2}{1+e^{2x}} \left\{ - \log 2 + \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/539160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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How to derive this second derivative using the quotient rule? If a given first derivative is: $\ {dy \over dx} = {-48x \over (x^2+12)^2} $
What are the steps using the quotient rule to derive the second derivative: $\ {d^2y \over dx^2} = {-144(4-x^2) \over (x^2+12)^3} $
My Steps:
\begin{align*}
{d^2y \over dx^2}
&= {-4... | We can factor out the constant term to make life easier (and just multiply by that $-48$ at the end of our calculation) as:
$$-48 \dfrac{x}{(x^2+12)^2}$$
This makes it easier to use the quotient and chain rule (I will assume you know these).
The derivative of $\ {dy \over dx} = {-48x \over (x^2+12)^2} $, using the quot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/541696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Calculate $a^3+b^3+c^3+d^3$ for the real roots of $x^4+2x^3-3x^2-3x+2$ $a,b,c,d \in \mathbb{R}$ are the real roots of $x^4+2x^3-3x^2-3x+2$.
Calculate $a^3+b^3+c^3+d^3$.
With approximation i found out, that $a^3+b^3+c^3+d^3 = -17$, but how can I proof that without calculating the roots exactly?
Cheers
| Express $a^3 + b^3 + c^3 + d^3$ in terms of elementary symmetric polynomials and use Vieta's formulas.
Also, here is an alternative approach for the fans of linear algebra. Consider a matrix
$$
A = \left(\matrix{0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -2 & 3 & 3 & -2}\right).
$$
A trivial check shows that it... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Show that $\sum_{n=0}^\infty r^n e^{i n \theta} = \frac{1- r\cos(\theta)+i r \sin(\theta)}{1+r^2-2r\cos(\theta)}$ Show that $$\sum_{n=0}^\infty r^n e^{i n \theta} = \frac{1- r\cos(\theta)+i r \sin(\theta)}{1+r^2-2r\cos(\theta)},$$ where $0\leq r <1$.
Using this, prove that $\sum_{n=0}^\infty r^n \cos(n\theta)$ and $\su... | The first sum is a geometric series
$$\sum_{n=0}^{\infty}{r^ne^{in\theta}} = \sum_{n=0}^{\infty}(re^{i\theta})^n$$
$$=\frac{1}{1-re^{i\theta}} = \frac{1}{1-r\cos\theta -ir\sin{\theta}}\cdot\frac{1-r\cos\theta + ir\sin\theta}{1-r\cos\theta + ir\sin\theta} =\frac{1-r\cos\theta + ir\sin\theta}{1+r^2-2r\cos\theta}$$
For t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/548150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Calculate $\sum\limits_{k=0}^{\infty}\frac{1}{{2k \choose k}}$ Calculate $$\sum \limits_{k=0}^{\infty}\frac{1}{{2k \choose k}}$$
I use software to complete the series is $\frac{2}{27} \left(18+\sqrt{3} \pi \right)$
I have no idea about it. :|
| Consider the function
$$f(x) = \frac{\arcsin{x}}{\sqrt{1-x^2}}$$
$f(x)$ has a Maclurin expansion as follows:
$$f(x) = \sum_{n=0}^{\infty} \frac{2^{2 n}}{\displaystyle (2 n+1) \binom{2 n}{n}} x^{2 n+1}$$
Differentiating, we get
$$f'(x) = \frac{x \, \arcsin{x}}{(1-x^2)^{3/2}} + \frac{1}{1-x^2} = \sum_{n=0}^{\infty} \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/548563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 3,
"answer_id": 1
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Find the Area of the ellipse Given $$\frac{x^2}{a^2} + \frac{y^2}{b^2}=1$$
where $a>0$, $b>0$
I tried to make $y$ the subject from the equation of the ellipse and integrate from $0$ to $a$. Then multiply by $4$ since there are $4$ quadrants.
$$Area=4\int^a_0\left(b^2-x^2\left(\frac{b^2}{a^2}\right)\right)^\frac{1}{2}dx... |
In order to find the the area inside the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, we can use the transformation $(x,y)\rightarrow(\frac{bx}{a},y)$ to change the ellipse into a circle. Since the lengths in the $x$-direction are changed by a factor $b/a$, and the lengths in the $y$-direction remain the same, the are... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/548876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 1
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Let $a,b,c \in \mathbb{R^+}$, does this inequality holds $\frac{a}{na + kb} + \frac{b}{nb+kc} + \frac{c}{nc + ka} \ge \frac{3}{k+n}$? Does the following statement/inequality holds for $a,b,c \in \mathbb{R^+}$?
$$\frac{a}{na + kb} + \frac{b}{nb+kc} + \frac{c}{nc + ka} \ge \frac{3}{k+n}$$
I've been thinking for hours an... | If $k\ge 2n$ then by CS, the inequality is true as proved by @medicu. If $k<2n$, strictly, then letting $a\to 0$, we have
$$\frac{b}{nb+kc} \ge \frac{3}{k+n}-\frac{1}{n}$$
But this is not true because the right hand side is positive, while we can make the left hand side goes to zero when $c\to \infty$.
So the necessar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/549149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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How to find $f$ if $f(f(x))=\frac{x+1}{x+2}$ let $f:\mathbb R\to \mathbb R$,and such
$$f(f(x))=\dfrac{x+1}{x+2}$$
Find the $f(x)$
My try
I found $f(x)=\dfrac{1}{x+1}$
because when $f(x)=\dfrac{1}{x+1}$,then
$$f(f(x))=f\left(\dfrac{1}{x+1}\right)=\dfrac{1}{\dfrac{1}{x+1}+1}=\dfrac{x+1}{x+2}$$
so $f(x)=\dfrac{1}... | You can easily check that if we define
$$f_{A} = \frac{ax+b}{cx+d} \quad \text{for} \quad A = \begin{pmatrix}a & b \\ c & d \end{pmatrix}, $$
then $f_{A} \circ f_{B} = f_{AB}$. Thus any matrix $A$ satisfying
$$ A^{2} = k \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}, \quad \text{for some} \ k \neq 0 $$
gives rise to a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/549897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 6,
"answer_id": 3
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Prove that $\frac{n^2+(-1)^nn+2}{7n^2+3}$ converges to $\frac{1}{7}$ I want to show that $\frac{n^2+(-1)^nn+2}{7n^2+3}$ converges to $\frac{1}{7}$ using the definition of convergence.
Skratch work:
I need $\mid\frac{n^2+(-1)^nn+2}{7n^2+3}-\frac{1}{7}\mid<\epsilon$.
So I take $\frac{n^2+(-1)^nn+2}{7n^2+3}-\frac{1}{7}=\f... |
you can choose n in better way
you can use sup of a(n) or a bit more to find n>N ,which satisfy that relation
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/550604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How prove $\sum\frac{1}{2(x+1)^2+1}\ge\frac{1}{3}$ let $x,y,z>0$ and such $xyz=1$ show that
$$\dfrac{1}{2(x+1)^2+1}+\dfrac{1}{2(y+1)^2+1}+\dfrac{1}{2(z+1)^2+1}\ge\dfrac{1}{3}$$
My try: I will find a value of the $k$ such
$$\dfrac{1}{2(x+1)^2+1}\ge\dfrac{1}{9}+k\ln{x}$$
note $\ln{x}+\ln{y}+\ln{z}=0$,so
$$\sum_{cyc}\dfra... | Hint:
Use Tringnometric transformation such that
x=sinu, y=secu and z = cotu
For any value of u you can find x,y,z >0 such that xyz = 1 by definition.
For a value of u = pi/6, x=.5, y = 1.154 and z = 1.732
Further, I computed LHS, for u = pi/4 and its value is .3365
Just to double check, I computed LHS for u = pi/3 an... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Function range problem $(sin^2\theta +\sin\theta -1)/(sin^2\theta -\sin\theta+ 2)$ .
It is asked to find the range of this function.
Here i was assumed a variable
$z=sin\theta$ so that i get
$-1<=z<=1 $
and then i obtained a quadratic equation as :-
$(y-1)z^2 -(y+1)z + 2y +1 =0$ from here i need some help or if ther... | Let $f(x)=\frac{\sin^2 x+\sin x-1}{\sin^2 x-\sin x+2}$
Then $g(z):=f(\sin^{-1} z)=\frac{z^2+z-1}{z^2-z+2}$ $$g'(x)=\frac{(2x+1)(x^2-x+2)-(2x-1)(x^2+x-1)}{(x^2-x+2)^2}=\frac{-2x^2+6x+1}{(x^2-x+2)^2}$$
In order to find maxima and minima, set $g'(x)=0$: $-2x^2+6x+1=0$. Thus $x=\frac{3\pm\sqrt{11}}{2}$. Note that $\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/555587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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problem on solving equations with three variables Find all positive real numbers $x,y,z$ which satisfy the following equations simultaneously.
$x^3+y^3+z^3=x+y+z$
$x^2+y^2+z^2=xyz$
| This is an answer that I am writing to give an alternative, brainless, path for these kind of problems, as an answer using means will soon appear (has already appeared).
Check the details as I have to go now.
First we use Newton's identities to write everything in terms of the elementary symmetric polynomials.
Denote
... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1} $ Prove that $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1} $
Thanks in advance, my professor asked us to this a couple weeks ago, but I was enable to get to the right answer.
Good luck!
Here is what I got up to;
$\frac{(n+1)!}{(n-r)!(r+1)!} = \frac{(n)!... | Even though it seems a little far-fetched I will use the Binomial Theorem. The definition of number $\binom{n}{r}$ has a reason to come to exist in the development of $(x + y)^n$. So I think the natural and instructive. For all $x,y\in\mathbb{R}$ we have,
\begin{array}{rrl}
\hspace{2cm}&( x+y)^{n+1}= & (x+y)\cdot ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Elementary proof that $\pi < \sqrt{5} + 1$ I wanted to show that
$$ \frac{\pi}{4\phi} < \frac{1}{2} $$
Where $\phi$ is the golden ratio. I have confirmed the results numerically, and by
simple algebra the inequality simplifies down to
$$
\pi < \sqrt{5} + 1
$$
This is a weaker relation than what was shown here. Prove t... | Just use that $\pi ^2 \leq \frac{61}{6}$ which comes from the series representation of $\frac{\pi^2}{6}$.
Then
$
\begin{align}
\pi^2 <5+1+2\cdot \sqrt5&\Leftarrow\\
\frac{61}{6} <6+2\cdot \sqrt5&\Leftrightarrow\\
\frac{5^2}{6}<2\sqrt 5&\Leftrightarrow\\
\frac{5^3}{36}<4\Leftrightarrow\\
125<144
\end{align}
$
$$$$
Inde... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/560315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 1
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Find a polynomial $p$ of degree $3$ if its value in $4$ points is given Find a polynomial $p$ of degree $3$ such that
\begin{align*}
p(−4) &= −142, \\
p(1) &= −2, \\
p(−5) &= −242, \\
p(4) &= 10.
\end{align*}
Then use your polynomial to approximate $p(2)$.
\begin{align*}
p(x) &= ?, \\
p(2) &= ?.
\end{align*}
I can't fi... | Hint: Find constants $a$, $b$, $c$ and $d$ such that the polynomial
$$a(x-1)(x+5)(x-4)+b(x+4)(x+5)(x-4)+c(x+4)(x-1)(x-4)+d(x+4)(x-1)(x-5)$$
takes on the desired values.
For example, to compute $a$, we want $a(-4-1)(-4+5)(-4-4)=-142$.
| {
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"url": "https://math.stackexchange.com/questions/560649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Differentiation of $2\arccos \left(\sqrt{\frac{a-x}{a-b}}\right)$ Okay so the question is:
Show that the function
$$2\arccos \left(\sqrt{\dfrac{a-x}{a-b}}\right)$$
is equal to
$$\frac{1}{\sqrt{(a-x)(x-b)}} .$$
I started by changing the arccosine into inverse cosine, then attempted to apply chain rule but I di... | $$\dfrac{d}{du} 2\arccos u = - 2\dfrac{1}{\sqrt{1 - u^2}} ~du$$
See the Proof Wiki for a proof of this.
In this problem, we have, $u = \sqrt{\dfrac{a-x}{a-b}}$, and we need to find $dx$, so we have:
$$ \dfrac{d}{dx} \left(\sqrt{\dfrac{a-x}{a-b}} \right) = -\dfrac{\sqrt{\dfrac{a-x}{a-b}}}{2 (a-x)} = -\dfrac{1}{2 \sqrt{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/561667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Arithmetic sequence of tangent values I have two angles $A_1, A_2 > 0$, and $A_1+A_2 < \pi$, is it possible to find an $A_0$ such that
$$\tan(A_0),\ \tan(A_0+A_1),\ \tan(A_0+A_1+A_2)$$
forms an arithmetic sequence on the same continuous range of tangent?
I have been looking for a formula for sum of tangents (not tangen... | (Resolving my own question)
Let the first term $\tan(A_0) = a$, term difference be $d$, then
$$\begin{align*}
\cot A_1 &= \cot(A_0+A_1-A_0)\\
&= \frac{1+\tan(A_0+A_1)\tan A_0}{\tan(A_0+A_1)-\tan A_0}\\
&= \frac{1+(a+d)a}{d}\\
\cot A_2 &= \cot(A_0+A_1+A_2-A_0-A_1)\\
&= \frac{1+\tan(A_0+A_1+A_2)\tan(A_0+A_1)}{\tan(A_0+A_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/562028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to prove this inequality $(a^2+bc^4)(b^2+ca^4)(c^2+ab^4) \leq 64$? Question:
If $a,b,c$ are nonnegative real numbers such that $a+b+c=3,$ then
$$(a^2+bc^4)(b^2+ca^4)(c^2+ab^4) \leq 64$$
My try: I found the equality holds only if $(a,b,c)=(2,0,1)$ or all of its permutations.
But I can't prove this inequality it. ... | → → The three dimensional problem can be simplified to a two dimensional problem by introducing
(again
and again)
suitable triangle coordinates:
$$
\left[ \begin{array}{c} a \\ b \\ c \end{array} \right] =
\left[ \begin{array}{c} 3 \\ 0 \\ 0 \end{array} \right] +
\left[ \begin{array}{c} -3 \\ 3 \\ 0 \end{array... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/565217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 2,
"answer_id": 0
} |
Inequality: $ab^2+bc^2+ca^2 \le 4$, when $a+b+c=3$. Let $a,b,c $ are non-negative real numbers, and $a+b+c=3$.
How to prove inequality
$$
ab^2+bc^2+ca^2\le 4.\tag{*}
$$
In other words, if $a,b,c$ are non-negative real numbers, then how to prove inequality
$$
27(ab^2+bc^2+ca^2)\le 4(a+b+c)^3.\tag {**}
$$
$\color{gray}... | I prove this stronger inequality,
With loss of out,let $a=\min{(a,b,c)}$
\begin{align*}&4(a+b+c)^3-27(a^2b+b^2c+c^2a+abc)\\
&=9a(a^2+b^2+c^2-ab-bc-ac)+(4b+c-5a)(a+b-2c)^2\ge 0
\end{align*}
other nice methods:
with out loss of let $b=mid{(a,b,c)}$,then $(b-a)(b-c)\le 0$,so
$$a^2b+b^2c+c^2a+abc\le b(a^2+c^2+ac)+abc=b(a+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/566768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Someone who is calculating $43434343^2$. The answer is $18865ab151841649$, where the two digits $a$ and $b$ were lost. So I used congruence $\bmod 9$ and $\bmod 11$. First, $43434343^2 \pmod{11}$ is $5$ and I did the answer $18865ab15184169 \pmod{11}$ and got $-b+a+3 \pmod{11}$. I also did $\bmod 9$ and got $a+b+67 \pm... | $\let\cong\equiv$
So you know that
$$43434343^2 \cong 5 \pmod{11}$$
and
$$43434343^2 \cong 1 \pmod{9}$$
and you found that
$$\begin{aligned}
18865ab15184169 &\cong a-b+3 \pmod{11}, \\
18865ab15184169 &\cong a+b+67 \pmod{9}.
\end{aligned}$$
You want $18865ab15184169$ to be equal to $43434343^2$, so the congruence classe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/566871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find $\lim _{n\rightarrow \infty }\frac {na^{2}_{n}+1}{\sum ^{n}_{k=1}\left( 1+2+3+\ldots +k\right) }$ Define $\left\{ a_{n}\right\} $ is an arithmetic sequence that all terms are positive integers. If $a_{10}-a_{1}=225$, find $$\lim _{n\rightarrow \infty }\dfrac {na^{2}_{n}+1}{\sum\limits^{n}_{k=1}\left( 1+2+3+\ldots ... | HINT:
If $d$ is the common difference, $a_n=a_1+(n-1)d\implies a_n-a_1=(n-1)d,$
putting $n=10,225=9d\implies d=25\implies a_n=a_1+(n-1)25=25n+a_1-25$
$$na_n^2+1=n(25n+a_1-25)^2+1=625n^3+50n^2(a_1-25)+n(a_1-25)^2+1=625n^3+O(n^2)$$
As $1+2+\cdots+k-1+k=\frac{k^2+k}2$
$$\sum_{1\le k\le n}(1+2+\cdots+k-1+k)=\frac{\sum_{1\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/568410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
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