Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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The graphs of $x^2+y^2+6x-24y+72=0$ and $x^2-y^2+6x+16y-46=0$ intersect at four points.Find the sum of the distances of these four points The graphs of $x^2+y^2+6x-24y+72=0$ and $x^2-y^2+6x+16y-46=0$ intersect at four points.Compute the sum of the distances of these four points from the point $(-3,2).$
$x^2+y^2+6x-24y... | Rather than solving the problem by eliminating x we can eliminate y. We will get a parabola which contains all the roots having x axis as directrix and (3,-2) as focus. So, all we need to do is find the sum of ordinates to find the distance. And, then you can eliminate x from equation to get quadratic in y and find sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1702822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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?Where does the integral went wrong $$\int_{-a}^{a}\int_{0}^{b\cdot \sqrt{1-\frac{x^2}{a^2}}}2x^2ydydx=\int_{-a}^{a}2b^2(x^2-x^4/a^2) dx =2b^2(2/3a^2-2a^5/5a^2)=\frac{8}{15}a^3b^2$$
But the actually answer in $\frac{4}{15}a^3b^2$. Which step did I go wrong?
| The first equality should have been $$\int_{-a}^{a}\int_{0}^{b\cdot \sqrt{1-\frac{x^2}{a^2}}}2x^2ydydx=\int_{-a}^{a}b^2(x^2-x^4/a^2)dx$$
since the antiderivative of $2y$ is $y^2$. That's why you get $2$ times the desired result.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Series of squares of n integers - where is the mistake? Given the following two series:
$$1^3 + 2^3 + ... + n^3$$
$$0^3 + 1^3 + .... + (n-1)^3$$
I take the difference vertically of the two:
$$\left(1^3-0^3\right) + \left(2^3-1^3\right) + .... + \left(n^3-(n-1)^3\right)$$
This equals to $n^3$
If I now express this in su... | Note that $\sum_{i=1}^n i = \frac 12n(n{\color{red}+}1)$, hence
\begin{align*}
\frac 13 \left(n^3 + 3\frac{n(n+1)}2 - n\right)
&= \frac 16 n(2n^2 + 3n + 3 - 2)\\
&= \frac 16 n(2n^2 + 3n + 1)\\
&= \frac 16 n(2n+1)(n+1)
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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An elegant way to solve $\frac {\sqrt3 - 1}{\sin x} + \frac {\sqrt3 + 1}{\cos x} = 4\sqrt2 $ The question is to find $x\in\left(0,\frac{\pi}{2}\right)$:
$$\frac {\sqrt3 - 1}{\sin x} + \frac {\sqrt3 + 1}{\cos x} = 4\sqrt2 $$
What I did was to take the $\cos x$ fraction to the right and try to simplify ;
But it lo... | Another approach is to write the equation as
$(\sqrt{3} - 1) \cos x +(\sqrt{3} + 1) \sin x = 4 \sqrt2 \sin x \cos x$
then rearranging gives
$\frac{\sqrt{3} - 1}{2 \sqrt{2}} \cos x +\frac{\sqrt{3} + 1}{2 \sqrt{2}} \sin x = 2 \sin x \cos x$.
Now note that $(\frac{\sqrt{3} - 1}{2 \sqrt{2}})^2 +(\frac{\sqrt{3} + 1}{2 \sqr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Divisibility of $n^4 -n^2$ by 4 (induction proof) We have to show that $$ n^4 -n^2 $$ is divisible by 3 and 4 by mathematical induction
Proving the first case is easy however I do not know how what to do in the inductive step. Thank you.
| Let's prove that
if $n^4-n^2$ is a multiple of both $3$ and $4$,
then so is $(n+1)^4-(n+1)^2$.
Consider
$
((n+1)^4-(n+1)^2)-(n^4-n^2)=4 n^3+6 n^2+2 n
$.
Ignoring $6n^2$, we have $4 n^3+2 n=3n^3+3n+n^3-n=3n^3+3n+6\binom{n+1}{3}$, which is always a multiple of $3$.
Ignoring $4n^3$, we have
$6 n^2+2 n=4n^2+2n(n+1)=4n^2+4... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculate the limit of: $x_n = \frac{\ln(1+\sqrt{n}+\sqrt[3]{n})}{\ln(1 + \sqrt[3]{n} + \sqrt[4]{n})}$, $n \rightarrow \infty$ Is it ok to solve the following problem this way? What I have done is to solve parts of the limit first (that converges to $0$), and then solve the remaining expression? Or is this flawed reaso... | The answer is correct but I would say the working is not quite rigorous. In effect you have said that
$$\lim a=0\ ,\quad \lim c=0$$
implies
$$\lim\frac{a+b}{c+d}=\lim\frac bd\ .$$
It's true in this case since $b,d\to\infty$; it would also be true if $b,d$ have any finite non-zero limits; but it is not always true if $... | {
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If possible, solve $7x^2 - 4x + 1 \equiv 0 $ (mod 11) If possible, solve $7x^2 - 4x + 1 \equiv 0 $ (mod 11)
Not sure if I'm approaching this problem correctly, any help is appreciated.
So far I have:
$7x^2 - 4x + 1 \equiv 0 $ (mod 11)
$21x^2 - 12x + 3 \equiv 0 $ (mod 11)
$-x^2 - x + 3 \equiv 0 $ (mod 11)
$-x^2 - x \eq... | Multiply by $8$ (which is the inverse of $7$ mod $11$) and get: $x^2+x+8=0$ (mod $11$).
Then complete the square. "Half of 1" is the same as "half of 12" (mod 11) so we can rewrite our equation as $(x+6)^2-36+8=0$ (mod $11$) which simplifies to $(x+6)^2=6$ (mod $11$).
So your equation has a solution if and only if 6 is... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Induction for divisibility: $3\mid 12^n -7^n -4^n -1$ I must use mathematical induction to show that
$a_{n} = 12^n −7^n −4^n −1$ is divisible by 3 for all positive integers n.
Assume true for $n=k$
$a_{k} = 12^k -7^k -4^k -1$
Prove true for $n=k+1$
$a_{k} = 12^{k+1} -7^{k+1} -4^{k+1} -1$
$ = (12^k)(12) - (7^k)(7) - (... | Hint: Instead of the last step, I would do this:
\begin{align*} a_{k+1} &= 12^{k+1} -7^{k+1} -4^{k+1} -1 \\
&= 12^k \cdot 12 - 7^k \cdot 7 - 4^k \cdot 4 - 4 + 3 \\
&= 12^k \cdot 8 - 7^k \cdot 3 + 4(12^k-7^k-4^k-1)+3 \end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1711256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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How to integrate $\int\limits_{0}^{\pi/2}\frac{dx}{\cos^3{x}+\sin^3{x}}$? I have$$\int\limits_{0}^{\pi/2}\frac{\text{d}x}{\cos^3{x}+\sin^3{x}}$$
Tangent half-angle substitution gives a fourth-degree polynomial in the denominator that is difficult to factor.
| $$I = \int_{0}^{\pi/2}\frac{dx}{\sqrt{2}\cos\left(x-\frac{\pi}{4}\right)\left(1-\frac{1}{2}\sin(2x)\right)}=\int_{-\pi/4}^{\pi/4}\frac{dx}{\sqrt{2}\cos(x)\left(1-\frac{1}{2}\cos(2x)\right)} $$
hence, through the substitution $x=\arcsin t$:
$$ I = \sqrt{2}\int_{0}^{\pi/4}\frac{dx}{\cos(x)\left(\frac{3}{2}-\cos^2 x\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1712198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
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Derive $\tan(3x)$ in terms of $\tan(x)$ using De Moivre's theorem
Derive the following identity: $$\tan(3x)=\frac{3\tan(x)-\tan^3(x)}{1-3\tan^2(x)}$$
The way I approached the questions is that I first derived $\sin(3x)$ and $\cos(3x)$ because $\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$. Then substituting:
$$\tan... | Don't “normalize” the triplication formulas to only sines and cosines.
We have
$$
\cos3x+i\sin3x=\cos^3x+3i\cos^2x\sin x-3\cos x\sin^2x-i\sin^3x
$$
so
\begin{align}
\sin3x&=3\cos^2x\sin x-\sin^3x \\[6px]
\cos3x&=\cos^3x-3\cos x\sin^2x \\[12px]
\tan3x&=\frac{3\cos^2x\sin x-\sin^3x}{\cos^3x-3\cos x\sin^2x}
\end{align}
No... | {
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Prove the function Prove
$\forall n\in\mathbb{N}: n\ge 1 \rightarrow 2^n\le 2^{n+1}-2^{n-1}-1.$
I did that $2^n \le\ 2^{n+2} - 2^{n}$ and then $2^n < 2^{n+2} - 2^{n-1}$
but have no idea how to add $-1$ in the function and let the $2^n \le\ 2^{n+2} - 2^{n} - 1$
| Use a geometric series to show that $1+2+\cdots +2^n = 2^{n+1}-1$.
Here is an alternative approach:
Write the equation as $2^{n+1} \ge 2^n + 2^{n-1} +1$.
For $n=1$ it is easy to check that $4 \ge 2 + 1 +1$.
Suppose the statement is true for $n$ and multiply across by $2$ to get
$2^{n+2} \ge 2^{n+1} + 2^{n} +2 \ge 2^{n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1716073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the range of $f :R → R$, and $f(x) = x^2 + 6x − 8$ I have this discrete math question I have done completing the square but not sure how to continue. May I get some guide? Thanks!
What is the range of $f :R → R$, and $f(x) = x^2 + 6x − 8$
$f(x)=x^2+6x-8$
$f(x)=(x^2+6x+9)-8-9$
$f(x)=(x+3)^2-17$
| \begin{align}
f(x) & = x^2 + 6x - 8 \\
& = (x^2 + 6x + 9) - 8 - 9 \\
& = (x + 3)^2 - 17 \\
\end{align}
Thus, the range is $[-17, \infty)$, which follows immediately from the fact that $(x + 3)^2 \ge 0$ and that $f(x)$ is not bounded from above.
While this is the general approach to finding ranges of quadratic functio... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Finding limit of $ \lim \limits_{x,y \to 0,0}{(1 + x^2 y^2)}^{-\frac{1}{x^2 + y^2}}$ Here is my limit:
$$ \lim \limits_{x,y \to 0,0}{(1 + x^2 y^2)}^{-\frac{1}{x^2 + y^2}}$$
I have learned two methods. One where we replace y with for example $y = kx $ (because $y = y_0 + k(x - x_0)$ and $y_0 = 0, x_0 = 0$). Or with $x ... | We can use your idea of setting $x= r \cos \theta$, $y=r \sin \theta$ and the limit becomes
$$\lim_{r \rightarrow 0} \left(1+\frac{r^4 \sin^2 2\theta}{4} \right)^{-\frac{1}{r^2}}$$
For a given $r$, the maximum and minimum values of this function are
$1$ and $\left( 1+\frac{r^4}{4} \right)^{-\frac{1}{r^2}}$ obtained ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1718699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Trigonometric substitution $\tan{\frac{x}{2}}=t$. What is $\cos{x}$ then? For example, the integral is:
$$\int \frac{\sin{x}}{3\sin{x}+4\cos{x}}dx$$
And we use the substitution: $\tan{\frac{x}{2}}=t$
Now, to get $\cos{x}$ in terms of $\tan\frac{x}{2}$, I first expressed $\cos^2\frac{x}{2}$ and $\sin^2\frac{x}{2}$ in te... | These are well-known formulae :
$$\sin x =\frac{2t}{1+t^2},\qquad \cos x =\frac{1-t^2}{1+t^2},\qquad \tan x =\frac{2t}{1-t^2}\quad(t\not\equiv\pm\frac\pi4\mod\pi).$$
A geometric proof of the formulae:
Consider the unit circle and a line passing through the point $(-1,0)$which has equation $y=t(x+1)$. Let $M$ be the s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1719033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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integrate $\int \frac{\tan^4x}{4}\cos^3x$
$$\int \frac{\tan^4x}{4}\cos^3x$$
$$\int \frac{\tan^4x}{4}\cos^3x=\frac{1}{4}\int \frac{\sin^4x}{\cos^4x}\cos^3x=\frac{1}{4}\int\frac{\sin^4x}{\cos x}=\frac{1}{4}\int\frac{\sin^2x\cdot\sin^2x}{\cos x}=\frac{1}{4}\int\frac{\sin^2x\cdot(1-\cos^2x)}{\cos x}=\frac{1}{4}\int \frac... | No, you've made a mistake in your fifth step! Solve it like this:
$$\int\frac{\tan^4(x)\cos^3(x)}{4}\space\text{d}x=\frac{1}{4}\int\tan^4(x)\cos^3(x)\space\text{d}x=\frac{1}{4}\int\tan(x)\sin^3(x)\space\text{d}x=$$
Substitute $u=\sin(x)$ and $\text{d}u=\cos(x)\space\text{d}x$:
$$\frac{1}{4}\int-\frac{u^4}{u^2-1}\spac... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to compute such a limit? Knowing $f(x,y) = 2x^2 +3y^2 -7x +15y$, one simply proves $$|f(x,y)|\leq 5(x^2+y^2)+22 \sqrt{x^2 + y^2}$$ How can I use this info to compute
$$ \lim_{(x,y)\to(0,0)} \frac{f(x,y) - 2(x^2+y^2)^{1/4}}{(x^2+y^2)^{1/4}}\;\;\; ?$$
Thanks!
| This can be done quite easily if you convert to polar coordinates.
We convert
$$lim_{x, y \to (0, 0)} \frac{2x^2 + 3y^2 - 7x + 15y - 2(x^2 + y^2)^{\frac{1}{4}}}{(x^2 + y^2)^{\frac{1}{4}}}$$
turns into
$$lim_{r \to 0} \>\> \frac{2r^2\cos^2 \theta + 3r^2 \sin^2 \theta - 7r \cos \theta + 15 r \sin \theta - 2 \sqrt{r}}{\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Let $F=\mathbf Z_2$ and let $f(x)=x^3+x+1 \in F[x]$ and let a be a zero of $f(x)$ in some extension of $F$ a) How many elements does $F(a)$ have?
b) Express $a^5, a^{-2}, a^{100}$ in the form $c_2a^2+c_1a+c_0$
I know that $F(a)$ has 8 elements since $f(x)$ has no zeros in $\mathbf{Z_2}$ and by a theorem that states :
$... | If $a$ is a zero of $x^3+x+1$, then $a^3=-a-1=a+1$, since we're working mod $2$.
Then, $a^4=a\cdot a^3=a(a+1)=a^2+a$.
Moreover, $a(a^2+1)=1$ and so $a^{-1}=a^2+1$ and $a^{-2}=(a^2+1)^2=a^4+1=a^2+a+1$.
Now, $a$ is a non-zero element in a field of $8$ elements. Hence, by Lagrange's theorem applied to the multiplicative g... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Exponential equation on the set of real numbers Solve the following equation on the set of real numbers:
$8^x+27^x+2·30^x+54^x+60^x=12^x+18^x+20^x+24^x+45^x+90^x$
$x=1; x=0; x=-1$ are trivial solutions, but I'm stuck with proving that there are no others...
| Assuming we are looking for real valued solutions for $x$:
Start by writing all the terms on the left hand side, with zero on the other side, and write $a=2^x$, $b=3^x$ and $c=5^x$
The equation then becomes $$a^3+b^3+2abc+ab^3+a^2bc-a^2b-ab^2-a^2c-a^3b-b^2c-ab^2c=0$$
When $b$ is replaced by $a$ the polynomial is identi... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Sums involving floor function I am looking for a direct formula for this sum
$$\sum_{k=0}^n \lfloor{\sqrt{n+k}}\rfloor\lfloor{\sqrt{k}}\rfloor$$
Or a method to efficiently compute the sum for large n
| Here's an exact $O(\sqrt{n})$ formula:
$$\left\lfloor \sqrt{n}\right\rfloor \left((n+1) \left\lfloor \sqrt{n}\right\rfloor +\frac{1}{3} \left(-\left(\left\lfloor \sqrt{n}\right\rfloor +1\right)^3+\frac{3}{2} \left(\left\lfloor \sqrt{n}\right\rfloor +1\right)^2+\frac{1}{2} \left(-\left\lfloor \sqrt{n}\right\rfloor -1\r... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find formula for $\frac{1}{\sqrt 1}+ \frac{1}{\sqrt 2}+\cdots+\frac{1}{\sqrt n}$ I have the series:
$$\frac{1}{\sqrt 1}+ \frac{1}{\sqrt 2}+\cdots+\frac{1}{\sqrt n}$$
I find hard to generalize into one formula, any explanation would be helpful.
| Analogous to Euler-Mascheroni Constant we have:
$$1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}=
2\sqrt{n}-1.4603545088\ldots+\frac{1}{2\sqrt{n}}-\frac{1}{24\sqrt{n^{3}}}+O\left( \frac{1}{\sqrt{n^{7}}} \right)$$
where $\displaystyle
\lim_{n\to \infty} \left( 2\sqrt{n}-1-\frac{1}{\sqrt{2}}-\frac{1}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1727376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\bigg\lfloor\frac{m}{p^k}\bigg\rfloor-\bigg\lfloor\frac{n}{p^k}\bigg\rfloor-\bigg\lfloor\frac{m-n}{p^k}\bigg\rfloor$ equal to zero or one
Prove that $\bigg\lfloor\frac{m}{p^k}\bigg\rfloor-\bigg\lfloor\frac{n}{p^k}\bigg\rfloor-\bigg\lfloor\frac{m-n}{p^k}\bigg\rfloor$ equal to zero or one for all $k,m,n\in \... | Note that
$$
\bigg\lfloor\frac{m}{p^k}\bigg\rfloor\le \frac{m}{p^k}<\bigg\lfloor\frac{m}{p^k}\bigg\rfloor+1\quad\text{and}\quad \bigg\lfloor\frac{n}{p^k}\bigg\rfloor\le \frac{n}{p^k}<\bigg\lfloor\frac{n}{p^k}\bigg\rfloor+1.
$$
Hence,
$$
\frac{m}{p^k}-1-\frac{n}{p^k}<\bigg\lfloor\frac{m}{p^k}\bigg\rfloor-\bigg\lfloor\fr... | {
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"url": "https://math.stackexchange.com/questions/1728994",
"timestamp": "2023-03-29T00:00:00",
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Integrate $\int_0^4 \frac{x}{\sqrt{16 + 5x}} \, dx$ I'm trying to solve
$$
\int_0^4 \frac{x}{\sqrt{16 + 5x}} \, dx
$$
using the substitution rule. The substitution rule, as far as I know it, reads
Let $g^\prime$ be a continuous function on $[a,b]$ whose range is an interval $I$, and let $f$ be continuous on $I$. The... | $$\int_{0}^{4}\frac{x}{\sqrt{16+5x}}\space\text{d}x=$$
Substitute $u=16+5x$ and $\text{d}u=5\space\text{d}x$.
This gives a new lower bound $u=16+5\cdot0=16$ and upper bound $u=16+5\cdot4=36$:
$$\frac{1}{5}\int_{16}^{36}\frac{u-16}{5\sqrt{u}}\space\text{d}u=$$
$$\frac{1}{5}\int_{16}^{36}\left[\frac{\sqrt{u}}{5}-\frac{... | {
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Prove the convergence and find the sum of series $\sum\limits_{n=1}^{\infty}\left(n^3\sin\frac{\pi}{3^n}\right)$. We know that $0<\sin\frac{\pi}{3^n}\le\frac{\sqrt 3}{2},\forall n\ge 1$. How to find the boundary for $n^3\sin\frac{\pi}{3^n}$ (how to use comparison test here)?
I tried using the ratio test, but the limit ... | Since $\sin x \le x$ for $x\ge 0$ and $0\le \frac{\pi}{3^n}\le \pi$,
$$
0\le n^3 \sin\frac{\pi}{3^n} \le \frac{n^3 \pi}{3^n}
$$
and $\sum_{n=1}^{\infty}\frac{n^3 \pi}{3^n}$ converges by ratio test. By comparison test, given series converges.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 0
} |
Is there integral or series for $\sqrt{10}-\frac{4^4}{3^4}$ (to prove the inequality)? Both of these numbers are bad approximations for $\pi$, but they turn out to be much closer together:
$$\sqrt{10}-\frac{4^4}{3^4}=0.00178$$
Since there is a lot of questions here about integrals and series which prove such close ineq... | Well, this is the same as $\sqrt{2}\sqrt{5} - \frac{2^8}{3^4}$. So, you can use tools from calculus to show that this function is positive at x = 2:
$f(x) = \sqrt{x}\sqrt{5} - \frac{x^8}{3^4}$
First of all, its zero is at:
$\frac{x^8}{\sqrt{x}} = 3^4 \sqrt{5} $
This gives us:
$x= 3^{(8/15)} 5^{(1/15)}$
This is slightly... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1730287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
how do I prove this by induction? (recursion) The terms are given recursively:
$P_0=3$
$P_1=7$
and $P_n = 3P_{n-1}-2P_{n-2}$ for $n\ge2$
What should I assume and what step proves that $P_n=2^{n+2}-1$ is a closed form of the sequence.
Suppose $n_0=1$ and the base cases are $0$ and $1$.
I think this book has a mistake
pr... | Note that $2^{0+2}-1=2^{2}-1=4-1=3$ and $2^{1+2}-1=2^{3}-1=8-1=7$,
so that the base case holds. Suppose now that the statement holds
for $P_{n-1}$ and $P_{n-2}$ and note that
\begin{align*}
P_{n} & =3P_{n-1}-2P_{n-2}\\
& =3\left(2^{n+1}-1\right)-2\left(2^{n}-1\right)\\
& =3\cdot2^{n+1}-3-2^{n+1}+2\\
& =2\cdot2^{n+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1730998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int_0^1 \sqrt{x^2+4x+1}\,dx$
Evaluate $\displaystyle\int_0^1 \sqrt{x^2+4x+1}\,dx$
I tried simplify by doing this:
$$\int_0^1 \sqrt{x^2+4x+1}\:dx=\displaystyle\int_0^1 \sqrt{(x+2)^2-3}\:dx$$
Then, by letting $t=x+2$, $dt=dx$
$$\int_0^1 \sqrt{t^2-3}\:dt=\displaystyle\int_0^1 \sqrt{(t-\sqrt 3)(t+\sqrt 3)}\:dt... | Using the Euler substitution $\sqrt{x^2+4x+1}=t-x$, we obtain
$\displaystyle x=\frac{t^2-1}{2(t+2)}$ and $\displaystyle dx=\frac{t^2+4t+1}{2(t+2)^2}dt$,
so $\displaystyle\int_0^1\sqrt{x^2+4x+1}dx=\int_1^{1+\sqrt{6}}\left(t-\frac{t^2-1}{2(t+2)}\right)\frac{t^2+4t+1}{2(t+2)^2}dt=\frac{1}{4}\int_1^{1+\sqrt{6}}\frac{(t^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1731891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Integration of a $\int \frac{v^2+1}{v^3-v^2+v+1}\,dv$ While solving the DE $$\frac{dy}{dx}=\frac{y^2-x^2}{y^2+x^2}$$ with the initial substitution $y=vx$ I got stuck in the integration of :
$$\int \frac{v^2+1}{v^3-v^2+v+1}\,dv$$
I don't know how to proceed further. Kindly help.
| The main difficulty is that your cubic doesn't have nice roots. Lets call them $v_r$, $v_c$ and $\bar{v_c}$ and let $v_c=a+bi$. I.e.
$$v^3-v^2+v+1=(v-v_r)(v-v_c)(v-\bar{v_c})$$
$$=(v-v_r)(v^2-2av+a^2+b^2)=(v-v_r)((v-a)^2+b^2)$$
Hence we can write:
$$\frac{v^2+1}{v^3-v^2+v+1}=\frac{A}{v-v_r}+\frac{Bv+C}{(v-a)^2+b^2}$$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1734103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Algebraic shortcut to find $a^n + b ^n$? Recently, I found this problem online:
Given $a+b=1$ and $a^2+b^2=2$, find $a^7+b^7$.
Although I could've solved it by substituting the first equation into the second and then using the quadratic formula; the way the question was set up, I suspected that there was a shortcut. ... | Note that
$$a^{n+1}+b^{n+1}=(a^n+b^n)(a+b)-ab(a^{n-1}+b^{n-1}).\tag{1}$$
Since $(a+b)^2=1$, and $a^2+b^2=2$, we have $2ab=-1$ and therefore $ab=-\frac{1}{2}$.
It follows from (1) that
$$a^{n+1}+b^{n+1}=(a^n+b^n)(1)+\frac{1}{2}(a^{n-1}+b^{n-1}).$$
Let $f(k)=a^k+b^k$. We have obtained the recurrence
$$f(n+1)=f(n)+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1736581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Trigonometric Indentities $\dfrac{1}{\cos x}+ \dfrac{2\cos x}{\cos 2x}$
If $x=\dfrac{\pi}{7}$ Prove that above expression is equal to $4$.
I worked a few steps and reached here:
$\dfrac{(4\cos^2 x -1)}{\cos x (2\cos^2 x-1)}$
Not able to proceed after this!!!
| $$F=\dfrac{\cos2x+2\cos^2x}{\cos x\cos2x}=\dfrac{3-4\sin^2x}{\cos x\cos2x}$$
If $\sin x\ne0,$
$$F=\dfrac{\sin x(3-4\sin^2x)}{\sin x\cos x\cos2x}=\frac{2\sin3x}{\sin2x\cos2x}=4\cdot\dfrac{\sin3x}{\sin4x}$$
Now $\sin4x=\sin3x$
if $4x=n\pi+(-1)^n3x$ where $n$ is any integer
If $n$ is odd $=2m+1,$(say), $7x=(2n+1)\pi, x=\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1737060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Determine the real numbers $x, y, z, t$ satisfying an equation Be non-zero real numbers $a, b, c$ so $a + b + c \neq 0$ and $ab + bc + ca = 0$.
Determine the real numbers $x, y, z, t$ knowing that
$$\frac{\frac{a^2 + b^2 + c^2}{2}+x^2+y^2+z^2}{a+b+c}= t+ \sqrt{ax+by+cz-t(a+b+c)}.$$
I tried converting equality data but ... |
is even possible to determine the numbers x, y, z, t?
Yes, it's possible.
Let $x^2+y^2+z^2=P,a+b+c=Q$.
Since
$$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)=a^2+b^2+c^2$$
we have
$$\frac{\frac{a^2 + b^2 + c^2}{2}+x^2+y^2+z^2}{a+b+c}= t+ \sqrt{ax+by+cz-t(a+b+c)}$$
$$\Rightarrow\quad \frac{\frac{Q^2}{2}+P}{Q}= t+ \sqrt{ax+by+cz-Qt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1737406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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$\forall x\in R,$find the range of the function $f(x)=\cos x(\sin x+\sqrt{\sin^2x+\sin^2\alpha});\alpha\in[0,\pi]$ $\forall x\in R,$find the range of the function $f(x)=\cos x(\sin x+\sqrt{\sin^2x+\sin^2\alpha});\alpha\in[0,\pi]$
$f(x)=\cos x(\sin x+\sqrt{\sin^2x+\sin^2\alpha});\alpha\in[0,\pi]$
$f'(x)=\cos x(\cos x+\... | Let $$y=\cos x\left[\sin x+\sqrt{\sin^2 x+\sin^2 a}\right] = \sin x\cdot \cos x+\cos x\cdot \sqrt{\sin^2 x+\sin^2 a}$$
Now Using $\bf{Cauchy\; Schwartz\; Inequality}$
We get $$(\sin^2 x+\cos ^2 x)\cdot \left[\cos^2 x+\sin^2 x+\sin^2 a\right]\geq \left(\sin x\cdot \cos x+\cos x\cdot \sqrt{\sin^2 x+\sin^2 a}\right)^2$$
S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1740147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
curve of $(x^2+y^2)^2=2(x^2-y^2)$ The diagram shows the curve $(x^2+y^2)^2=2(x^2-y^2)$ and one of its maximum points $M$. Find the coordinates of $M$.
My attempt.
Differentiate the equation and I got
$\frac{dy}{dx}=-\frac{x(x^2+y^2-1)}{y(x^2+y^2+1)}$
How should I proceed?
| I think you take the top part set it to zero $x(x^2+y^2-1)=0$, since thats how you get the critical numbers. Then you have :
$x=0$ or $y=\sqrt{1-x^2}$
plug y back into original formula and solve to get something like
$x=\pm \sqrt{\frac{3}{4}}$ so max would be at one of those $x$ values.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1740632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
find the rate of change $f(x) = 4\sin^3 x$ when $x = \frac{5\pi}{6}$ find the rate of change $f(x) = 4\sin^3 x$ when $x = \frac{5\pi}{6}$
To find the rate of change I need to find $\frac{dy}{dx}$ using the chain rule $h'(x) = g'(f(x)).f'(x)$
$g'(f(x)) = 12\sin^2 x$
$f'(x) = \cos x$
$h'(x) = 12\sin^2 x \cos x$
After thi... | Trig by Reference Triangles:
The angle $\frac{5 \pi}{6}$ in radians will be given by $150^o$. This simple conversion can be done by remembering that $180^o = \pi$ $rad $.
To solve for
$h'(x)=12 sin^2x cosx$
$h'( \frac{ 5 \pi}{6} )=12 sin^2 ( \frac{5 \pi}{6}) cos( \frac{5 \pi}{6} )$
we can use right-angled triangles of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1746358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Multiple Angle formulas, alternate forms Relatively simple question, that might not be simple to answer:
I have noticed that there are ways of expressing every double angle formula of a given trigonometric function using only that function except for $\sin$ and $\csc$. That is,
$\sin2\theta=2\sin\theta\cos\theta=?$
$\c... | A formula using no trig functions other than the sine, and
without the ambiguity of symbols such as $\pm$, is
$$
\sin(2\theta) = \begin{cases}
2\sin\theta \sqrt{1-\sin^2\theta} &
\text{if $2n\pi-\frac\pi2 \leq \theta \leq 2n\pi+\frac\pi2$
for some integer $n$,} \\
-2\sin\theta \sqrt{1-\sin^2\theta} & \text{otherwise}.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1747209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove that this system of linear equations generates $\left| \left( \begin{smallmatrix} 1/2 \\ n \end{smallmatrix} \right) \right|$ as a solution? This infinite system of linear equations:
$$ \begin{array}( 2x_1=1 \\ 3x_1+4x_2=2 \\ 4x_1+5x_2+6x_3=3 \\ \cdots \end{array} $$
In other words, this is particular case of a s... | Alright, I figured it out. First, we write the recurrence relation for the absolute value of the binomial coefficients:
$$x_m= \left| \left( \begin{matrix} \frac{1}{2} \\ m \end{matrix} \right) \right|=\frac{1/2 (1-1/2)(2-1/2)\cdots (m-1-1/2)}{m!}=\frac{m-3/2}{m}x_{m-1}$$
$$x_1=\frac{1}{2}$$
Next, we work with our syst... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1747688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Limit $\lim_{n\to\infty}\sum_{k=0}^n\frac{n}{n^2+k}$ I came across this problem that I'm supposed to be able to solve in under 5 minutes (for a competition).
$$\lim_{n\to \infty} \sum_{k=0}^n \frac {n}{n^2+k}$$
I tried solving this for small sums, $\sum_{k=0}^2 \frac {n}{n^2+k}$, $\sum_{k=0}^3 \frac {n}{n^2+k} $ and m... | $\displaystyle \sum_{k=0}^n \frac {n}{n^2+k} = \frac {n}{n^2+0} + \frac{n}{n^2+1} +.......... +\frac{n}{n^2+n} $
$\displaystyle = \frac {n(n^2+1)(n^2+2)\cdots(n^2+n)+n(n^2)(n^2+2)......}{n^2(n^2+1)(n^2+2)....(n^2+n)} = \frac {n.n^{2n}(n+1)+}{n^2\cdot n^{2n}+}$
$\displaystyle \lim_{n\rightarrow \infty} \sum_{k=0}^n \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1748429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Pade approximant for the function $\sqrt{1+x}$ I'm doing the followiwng exercise:
The objective is to obtain an approximation for the square root of any given number using the expression
$$\sqrt{1+x}=f(x)\cdot\sqrt{1+g(x)}$$
where g(x) is an infinitesimal. If we choose $f(x)$ as an approximation of $\sqrt{1+x}$, then ... | Using the fact that $\sqrt{1+x}=1+\frac x{1+\sqrt{1+x}}$, we get
$$
\begin{align}
\sqrt{1+x}
&=1+\cfrac x{2+\cfrac x{2+\cfrac x{2+\cfrac x{2+\cdots}}}}
\end{align}
$$
Which gives the approximants
$$
\begin{align}
\color{#C00000}{1+\frac12x}&=1+\frac12x\\
\color{#00A000}{\frac{1+\frac34x}{1+\frac14x}}&=1+\frac12x-\frac1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1750968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
How to evaluate this limit using Taylor expansions? I am trying to evaluate this limit:
$\lim_{x \to 0} \dfrac{(\sin x -x)(\cos x -1)}{x(e^x-1)^4}$
I know that I need to use Taylor expansions for $\sin x -x$, $\cos x -1$ and $e^x-1$. I also realise that all of these are just their regular Taylor expansions with their f... | Turns out to just be a foolish mistake. When I factored $x$ out of the denominator I should have raised it to 4. That would be an $x^5$ on the bottom:
$\lim_{x \to 0} \dfrac{x^3(-\frac{1}{3!}+\frac{x^2}{5!}...)x^2(-\frac{9}{2!}+\frac{81x^2}{4!}...)}{x^5({1}+\frac{x}{2!}+\frac{x^2}{3!}+\frac{x^3}{4!}...)^4}$
Now I can s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1752176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Find the limit $\lim_{x\rightarrow 0} \frac{\sin x - x}{\tan x - x}$ Find the limit if it exists. (this exercise is taken from Calculus - The Classic Edition by Swokowski Chapter 10, section 1, no. 9, p.498)
$$\lim_{x\rightarrow 0} \frac{\sin x - x}{\tan x - x}$$
since the limit is $0/0$ therefore, we use L'Hopital's ... | For $x\to 0$ we have $\sin x\sim x-\frac{x^3}{3!} $ and $\tan x\sim x+\frac{x^3}{3} $
then
$$
\frac{\sin x- x}{\tan x-x}\sim \frac{-\frac{x^3}{3!}}{+\frac{x^3}{3}}=-\frac{1}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1752715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 4
} |
n tends to infinity $\displaystyle\lim_{n\to\infty}
{\left(
\frac{\sqrt {n^2+n} - 1}{n}
\right)}^{
\left(2 \sqrt{n^2+n} - 1
\right)}$
Sorry for the bad format of the question I don't know much latex
I tried rationalization of the power and the base but was not able to get the answer.
| Let $y = \displaystyle\lim_{n\to\infty}
{\left(
\frac{\sqrt {n^2+n} - 1}{n}
\right)}^{2 \sqrt{n^2+n} - 1}$
and let $g(n) = \sqrt{n^2 + n}$
Then $g'(n) = \dfrac{2n+1}{2g(n)}$ and
\begin{align}
\ln(y)
&= (2\sqrt{n^2+n} - 1) \ln{ \left( \frac{\sqrt{n^2+n} - 1}{n} \right)} \\
&= (2g(n) - 1) \ln{ \left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1754965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Ellipse and chord length There is a analytic geometry problem:
In the ellipse $\frac{x^2}{4}+y^2=1$, segment $AB$ is a chord and $AB=\sqrt{3}$, find the maximum and minimum area of $\triangle AOB$.
My progress
Assume $A(x_1,y_1),B(x_2,y_2)$, we have
$$ \left\{
\begin{array}{}
(x_1-x_2)^2+(y_1-y_2)^2=3 \\
\frac{x_1^... |
It likely is a conditional extremum problem and I don't know what to do next.
I don't know what to do next either.
So, let us take another approach.
If the line $AB$ is parallel to $y$-axis, we can easily see that the area of the triangle is $\sqrt 3/2$.
If the line $AB$ is not parallel to $y$-axis, we can set $AB : ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1756245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Test for the convergence of the sequence $S_n =\frac1n \left(1 + \frac{1}{2} + \frac{1}{3} + \cdots+ \frac{1}{n}\right)$ $$S_n =\frac1n \left(1 + \frac{1}{2} + \frac{1}{3} + \cdots+ \frac{1}{n}\right)$$
Show the convergence of $S_n$ (the method of difference more preferably)
I just began treating sequences in school, a... | Avoiding appeals to logarithms:
Writing $H_n = 1 +\frac{1}{2}+\cdots + \frac{1}{n}$, note that for $n>M$:
$$H_n=1 +\frac{1}{2}+\cdots + \frac{1}{n}=1 +\frac{1}{2}+\cdots + \frac{1}{m}+\cdots+\frac{1}{n}=H_M+\frac{1}{M+1}+\cdots+\frac{1}{n}$$
Now, each of the terms after $H_M$ is $<\frac{1}{M}$, whence $H_n<H_M+\frac{n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1757859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Can this puzzle be solved without brute force?
Consider positive integers $a$ and $b$, where $a \ge b$ and the sum $\frac{a+1}{b}+\frac{b+1}{a}$ is also an integer. Find the sum of all $a$ values less than $1000$ that meet this criteria.
For example, if $a=3$ and $b=2$, then
$$\frac{a+1}{b}+\frac{b+1}{a} = \frac{3+1}... | We use a now-standard technique called Vieta jumping, which has a fascinating and surprisingly recent history.
Suppose $a^2 + a + b^2 + b = kab$ for some integer $k$. Then, writing this as a polynomial in $a$ with integer coefficients:
$$a^2 - (kb-1) a + (b^2+b) = 0,$$
we see that if $(a,b)$ is one of the roots of the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1758896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
is it possible to choose points on the graph of $y = x^2$ to form vertices of an equilateral triangle? is it possible to choose points on the graph of $y = x^2$ to form vertices of an equilateral triangle $\Delta ABC$?where three ponit not $(0,0)$ .and find $(S_{\Delta ABC})_{min}$
| Summary
*
*Are there other points which can form equilateral triangle?
The answer is yes. In fact, there are infinitely many of them.
The centroids of the triangles lie on another parabola $y = 9x^2 + 2$.
*The minimal area is $3\sqrt{3}$. achieved by the equilateral triangle with vertices $(0,0)$, $(\pm\sqrt{3},3)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1763529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Evaluate the $\lim_{x \to \ -\infty} (x + \sqrt{x^2 + 2x})$ Evaluate : $$\lim_{x \to \ -\infty} (x + \sqrt{x^2 + 2x})$$
I've tried some basic algebraic manipulation to get it into a form where I can apply L'Hopital's Rule, but it's still going to be indeterminate form.
This is what I've done so far
\begin{align}
... | This is based on @Battani's answer but with a more in-depth explanation
\begin{align}
\lim _{ x\to -\infty } \left( \frac { -2x }{ x-\sqrt { x^{ 2 }+2x } } \right) &= \lim _{ x\rightarrow -\infty }\left( \frac { -2x }{ x-\sqrt { { x }^{ 2 }\left( 1+\frac { 2 }{ x } \right) } } \right) \\ \\
&\text{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1766582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
How to evaluate $\lim_{c \rightarrow \infty} \int_{-c}^c f(x)dx$ I'm trying to evaluate:
$$\lim_{c \rightarrow \infty} \int_{-c}^c \frac{1+x}{1+x^2}dx$$
but I don't understand how to evaluate
$$\lim_{c \rightarrow \infty} \int_{-c}^c f(x)dx$$
How?
| Hint. Observe that
$$
\int_{-c}^c \frac{1+x}{1+x^2}dx=\int_{-c}^c \frac{1}{1+x^2}dx=2\int_0^c \frac{1}{1+x^2}dx=2\arctan(c).
$$ Can you take it from here?
Some details. We have
$$
\begin{align}
\color{blue}{\int_{-c}^c \frac{x}{1+x^2}dx}&=\int_{-c}^0 \frac{x}{1+x^2}dx+\int_0^c \frac{x}{1+x^2}dx
\\\\&=\int_{c}^0 \frac{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1767151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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How do I solve the following differential equation (It's not seperable)? I'm trying to Solve the following equation: find the solution $y:(-1,1) \rightarrow \mathbb{R}$ of $y'=\dfrac{y}{1-x^2}+x$?
It is not separable and I have no other Tools to solve it.
| Subtract $ \frac {y(x)}{1-x^2} $ from both sides :
$ \frac {dy(x)}{dx} + \frac {y(x)}{x^2-1} = - \frac {-x^3 + x}{x^2-1}$
Let $ m(x) = e^{\int \frac {1}{x^2-1} dx } = \frac {\sqrt{1-x}}{\sqrt{x+1}} $
Then multiplay both sides by m(x) and substitute : $ \frac {\sqrt{1-x}}{\sqrt{x+1}(x^2-1)} = \frac {d}{dx} \frac {\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1768299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $ 1 + \frac{\sqrt{x+3}}{1+\sqrt{1-x}} = x + \frac{\sqrt{2x+2}}{1+\sqrt{2-2x}} $
Solve for $x \in \mathbb{R}$
$$ 1 + \dfrac{\sqrt{x+3}}{1+\sqrt{1-x}} = x + \dfrac{\sqrt{2x+2}}{1+\sqrt{2-2x}} $$
I tried some substitutions and squaring but that didn't help. I also tried to use inequalities as done in my previous p... | Given : $ 1 + \dfrac{\sqrt{x+3}}{1+\sqrt{1-x}} = x + \dfrac{\sqrt{2x+2}} {1+\sqrt{2-2x}} $
Let $\alpha = x+3 $ ; $\beta = 2x+2$ ; $\left(\beta - \alpha\right) = x-1$
$\implies 1+ \dfrac{\sqrt{\alpha}}{1+\sqrt{4-\alpha}} = x + \dfrac{\sqrt{\beta}} {1+\sqrt{4-\beta}} $
$\implies \alpha + \dfrac{\sqrt{\alpha}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1770721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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} |
Solving a mixed radical and quadratic equation
Solve for $x \in \mathbb{R}$
$$4x^2(x+2) +3(2x^2-4x-3)\sqrt{4x+3} +6x = 0$$
I tried taking square by isolating the radical, but the resultant equation couldn't be solved.
Any help will be appreciated.
Thanks.
| Let $X=4x+3$. Then,
$$3(2x^2-X)\sqrt{X}=-2x(2x^2+X)$$
Squaring the both sides gives
$$9(2x^2-X)^2X=4x^2(2x^2+X)^2,$$
i.e.
$$4x^4X-4x^2X^2+9X^3-16x^6+16x^4X-36x^2X^2=0$$
$$X(4x^4-4x^2X+9X^2)-4x^2(4x^4-4x^2X+9X^2)=0$$
$$(X-4x^2)(4x^4-4x^2X+9X^2)=0$$
since $9X^2=X^2+8X^2$
$$(X-4x^2)(4x^4-4x^2X+X^2+8X^2)=0$$
to have
$$(X-4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1770825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
if $x^y=y^x$ show that $x+y>2e$
Let $0<x<y$, such that
$$x^y=y^x$$
show that
$$x+y>2e$$
Since $$y\ln{x}=x\ln{y}\Longrightarrow \dfrac{\ln{y}}{y}=\dfrac{\ln{x}}{x}$$
Let
$$f(x)=\dfrac{\ln{x}}{x}\Longrightarrow f'(x)=\dfrac{1-\ln{x}}{x^2}$$
If $0<x<e$ then $f'(x)>0$, if $x>e$,then $f'(x)<0$
$\Longleftrightarrow... | As MathematicianByMistake answer,we want prove
$$g(r)=r^{\frac{1}{r-1}}+r^{\frac{r}{r-1}}=(x+1)^{\frac{1}{x}}+(x+1)^{\frac{x+1}{x}}=(x+1)^{\frac{1}{x}}(x+2),x>0$$
since
$$g'(x)=(x+1)^{\frac{1}{x}-1}\cdot\dfrac{x(x^2+2x+2)-(x^2+3x+2)\ln{(x+1)}}{x^2}$$
we only prove
$$h(x)=x(x^2+2x+2)-(x^2+3x+2)\ln{(x+1)}>0,x>0$$
since
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1771348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 3,
"answer_id": 1
} |
Use induction to prove that $2^n \gt n^3$ for every integer $n \ge 10$. Use induction to prove that $2^n \gt n^3$ for every integer $n \ge 10$.
My method:
If $n = 10$, $2^n \gt n^3$ where $2^{10} \gt 10^3$ which is equivalent to $1024 \gt 1000$, which holds for $n = 10$. $2^k \gt k^3$.
$2^{k + 1} \gt (k + 1)^3$
$2^{k ... | We do the induction step, perhaps somewhat along the lines that you had considered.
Let $k\ge 10$, and suppose that $2^k\gt k^3$. We want to show that $2^{k+1}\gt (k+1)^3$.
We have
$$2^{k+1}=2\cdot 2^k\gt k^3+k^3.$$
It will now be enough to show that $k^3+k^3\gt (k+1)^3$, or equivalently that $(k+1)^3\lt k^3+k^3$.
We ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1772026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find the orthogonal projection of the given vector on the given subspace $W$ of the inner product space $V$. $V=P(R)$ with the inner product $\langle f(x),g(x) \rangle$=$\int_0^1 f(t)g(t )dt$, $h(x)=4+3x-2x^2$ and $W=P_1(R)={\{1,x}\}$.
I don't know how to do this the question. All I know is that it has somethin... | The inner product structure of your vector space $V$ is $$\langle f | g
\rangle = \int_0^1 f(x)g(x) dx$$
To project a vector
$$h(x) = 4 + 3x - 2x^2$$
on the subspace $W$ of the vector space $V$, you just add the projections of $h$ on each of the basis vectors of the subspace.
In this case, since
$$W = P_1 = \left\{ 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1773834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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Prove that $a^2b+b^2c+c^2a \leqslant 3$ for $a,b,c >0$ with $a^ab^bc^c=1$
Let $a,b,c >0$ and $a^ab^bc^c=1$. Prove that
$$a^2b+b^2c+c^2a \leqslant 3.$$
I don't even know what to do with the condition $a^ab^bc^c=1$. At first I think $x^x>1$, but I was wrong. This inequality is true, following by the verification from M... | Remarks: @Wiley gave a very nice proof. Here we give an alternative proof.
Using the well-known inequality $a^2b + b^2c + c^2a \le \frac{4}{27}(a + b + c)^3 - abc$, it suffices to prove that
$$\frac{4}{27}(a + b + c)^3 - abc \le 3. \tag{1}$$
The condition $a^a b^b c^c = 1$ is written as $$a\ln a + b\ln b + c\ln c = 0.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1774794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 5,
"answer_id": 3
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A rational function integration Evaluate $$\int \frac{3x^2+1}{(x^2-1)^3}dx$$ I tried breaking the numerator in terms of the denominator but it didn't help much. I also tried a few substitutions buy most of them were useless. Please give some hints. Thanks.
| Let $$I = \int\frac{3x^2+1}{(x^2-1)^3}dx = \int\frac{3x^2+1}{x^{\frac{3}{2}}\left(x^{\frac{3}{2}}-x^{-\frac{1}{2}}\right)^3}dx$$
So $$I = \int\frac{3x^{\frac{1}{2}}+x^{-\frac{3}{2}}}{\left(x^{\frac{3}{2}}-x^{-\frac{1}{2}}\right)^3}dx\;,$$ Now Put $\left(x^{\frac{3}{2}}-x^{-\frac{1}{2}}\right)=t\;,$ Then $\left(3x^{\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1776283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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How to differentiate this fraction $\frac{2}{x^2+3^3}$? $\frac{2}{(x^2+3)^3}$.
I have ${dy}/{dx}$ x 2 x ${x^2+3^3}$ - 2 x ${dy}/{dx}$ x ${x^2+3^3}$ over $({x^2+3)^6}$
And then simplifying to $-12x^5 + 36x^2$ over $({x^2+3)^6}$
I'm not sure if this is right.
| $$\frac {d}{dx} \frac 2{(x^2+3)^3}=\frac {d}{dx} (2(x^2+3)^{-3})=2\frac {d}{dx} (x^2+3)^{-3}=2\times(-3)(x^2+3)^{-4}2x=-\frac {12x}{(x^2+3)^{4}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1776669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
If $x \geq C$, where $C > 0$ is a constant, then what is the least upper bound for $\dfrac{2x}{x + 1}$? The title says it all.
Since
$$f(x) = \dfrac{2x}{x + 1} = 2\left(1 - \dfrac{1}{x + 1}\right),$$
then because $x \geq C$ where $C > 0$, an upper bound is given by
$$\dfrac{2x}{x + 1} < 2.$$
Note that the lower bound i... | \begin{align*}
x&\ge C \\
f(x) = \dfrac{2x}{x + 1} &= 2\left(1 - \dfrac{1}{x + 1}\right)\\
x&\in [C,\infty)\\
x+1&\in[C+1, \infty)\\
\frac{1}{x+1}&\in\Bigg(0,\frac{1}{C+1}\Bigg]\\
1-\frac{1}{x+1}&\in \Bigg[1-\frac{1}{C+1}, 1\Bigg)\\
\dfrac{2x}{x + 1} = 2\left(1 - \dfrac{1}{x + 1}\right)&\in\Bigg[\frac{2C}{C+1}, 2\Bigg)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1776908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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How to prove that $(x^2+y)(y^2+z)(z^2+x)+2xyz \leqslant 10$, where $x,y,z \in \mathbb{R}^+$ and $x^2+y^2+z^2+xyz=4$ $x,y,z \in \mathbb{R}^+$ and $x^2+y^2+z^2+xyz=4$, prove
$$(x^2+y)(y^2+z)(z^2+x)+2xyz \leqslant 10$$
I try several trig substitutions but feel hopeless with the cyclic term here. The condition $x^2+y^2+z^2... | $$ x^2+y^2+z^2+xyz=4;\space\space x,y,z>0\qquad (1)\\ (x^2+y)(y^2+z)(z^2+x)+2xyz \le 10\qquad (2)$$
$(1)\Rightarrow 0<x,y,z\le 2 \text{ and}\space x=2\iff y=z=0$
If $x=2$ then $(2)$ is trivially verified.
Put $x=2-\epsilon$ where $0\le \epsilon\le 2$ so $(1)$ becomes
$$y^2+z^2+(2-\epsilon)yz=\epsilon(4-\epsilon)\qquad ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1777259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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When is $\frac{a}{n} = n^a$ true? Title says it all really, I am trying to figure out if theres a situation where
$\frac{a}{n} = n^a$ is true or if this is impossible.
This is not realy from somewhere, just for the sake of curiosity.
| $\frac an = n^a \implies$
$a = n^{a+1} \implies $
$1 = \log_a a = \log_a n^{a+1} = (a +1 ) \log_a n \implies$
$\frac 1{a+1} = \log_a n \implies$
$n = a^{\frac 1{a+1}}$
So for all positive $a$, $n$ is solvable. As $0/0$ (as well as $0^0$) are undefined $n = a = 0$ is not a solution (despite $0 = 0^1$).
For $a < 0$ it... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1781442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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For real numbers $a,b,c$ calculate the value of: $\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}$ if we have... For real numbers $a,b,c$ we have: $a+b+c=11$ and $\frac1{a+b}+\frac1{b+c}+\frac1{c+a}=\frac{13}{17}$, calculate the value of: $\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}$
I think we should use a trick to solve thi... | Hint:
As $(a+b+c) ( \frac1{a+b}+\frac1{b+c}+\frac1{c+a})= 3+\frac{a}{b+c} +\frac{b}{c+a} + \frac{c}{a+b}$
Now applying the values given in question we get $\frac{a}{b+c} +\frac{b}{c+a} + \frac{c}{a+b}=(11 \cdot \frac{13}{17})- 3=\frac{92}{17}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1783523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Prove that $4n+2=x^2+y^2+z^2$ for some odd $x,y$ and even $z$ Show that for all $n\in \mathbb{N}$, exists $x,y,z \in \mathbb{N}$, such that $x,y$ are odd and $z$ is even, such that $4n+2=x^2+y^2+z^2$.
I started by using the fact that every natural number has a decomposition to a sum of 4 square, and tried to prove that... | If x and y are odd while z is even, then you can set $x = 2a+1, y = 2b+1, z = 2c$
$4n+2 = (2a+1)^2+(2b+1)^2+(2c)^2 = 4a^2 + 4a+1+4b^2+4b+1+4c^2 = 4(a^2+a+b^2+b+c^2)+2$
Now you're trying to prove that every possible $n$ can be described as $n = a^2+a+b^2+b+c^2 = a(a+1) + b(b+1) + c^2$
Note that $a(a+1)+b(b+1) \equiv 0 (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1784123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How does one solve this recurrence relation? We have the following recursive system:
$$
\begin{cases}
& a_{n+1}=-2a_n -4b_n\\
& b_{n+1}=4a_n +6b_n\\
& a_0=1, b_0=0
\end{cases}
$$
and the 2005 mid-exam wants me to calculate answer of $ \frac{a_{20}}{a_{20}+b_{20}} $.
Do you have any idea how to solve this recursive equ... | We can write the recurrence relation in matrix form
$$\begin{bmatrix} a_{k+1}\\ b_{k+1}\end{bmatrix} = \begin{bmatrix}-2 & -4\\ 4 & 6\end{bmatrix} \begin{bmatrix} a_{k}\\ b_{k}\end{bmatrix}$$
Hence,
$$\begin{bmatrix} a_{n}\\ b_{n}\end{bmatrix} = \begin{bmatrix}-2 & -4\\ 4 & 6\end{bmatrix}^n \begin{bmatrix} a_{0}\\ b_{0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1786760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 4,
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Determine all integral solutions of $a^2+b^2+c^2=a^2b^2$ I started like this :
$a^2+c^2=b^2(a^2-1)\\c^2 +1=(a^2-1)(b^2-1)$
but it's leads to nowhere.
can you help please ?
| Suppose that this Diophantine equation has an integral solution $(a,b,c)\neq(0,0,0)$. Then, $a\neq 0$ and $b\neq 0$, whence $a^2-1\geq 0$ and $b^2-1\geq 0$. If $a$ or $b$ is even, then $a^2-1\equiv 3\pmod{4}$ or $b^2-1\equiv3\pmod{4}$. Hence, $\left(a^2-1\right)\left(b^2-1\right)=c^2+1$ is divisible by a prime natura... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1787444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Edwards Differential Calculus for Beginners, 1896. Chapter 4, Question 82 This question asks for an evaluation of an infinite series assuming only knowledge of basic differential calculus. I couldn't figure it out, but user 'Dr. MV' gave me a hint which was sufficient for me to find the answer. I hope this makes the qu... | We have:$$\frac{2^nx^{2^n-1}+2^{n+1}x^{2^{n+1}-1}}{1+x^{2^n}+x^{2^{n+1}}}-\frac{2^nx^{2^n-1}-2^{n+1}x^{2^{n+1}-1}}{1-x^{2^n}+x^{2^{n+1}}}=\frac{2^{n+1}x^{2^{n+1}-1}+2^{n+2}x^{2^{n+2}-1}}{1+x^{2^{n+1}}+x^{2^{n+2}}}$$
So if we add and substract $$\frac{1+2x}{1+x+x^2}$$ at the lefthand side of the sum, the subtracted term... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Inverse of almost quadratic functions Hi all first question here,
I'm currently trying to invert the following two (two since the $\pm$ sign) functions:
$y = x^2 \pm \sqrt{(ax + b \sin{\theta})^2 + (b\cos{\theta})^2}$
Where a,b, $\theta$ are real and positive.
I tried banging my head against the wall and using Maple, ... | $$y-x^2=\pm\sqrt{(ax+b\sin\theta)^2+(b\cos\theta)^2}$$
$$(y-x^2)^2=(ax+b\sin\theta)^2+(b\cos\theta)^2$$
$$y^2-2x^2y+x^4=ax^2+2ab\sin\theta x+b^2(\sin^2\theta+\cos^2\theta)$$
$sin^2\theta+\cos^2\theta=1$
$$y^2-2x^2y+x^4=ax^2+2ab\sin\theta x+b^2$$
$$x^4-2(y+a)x^2-2ab\sin\theta x+y^2-b^2=0$$
$$x^4+kx^2+l=mx^2+2ab\sin\thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1792114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\sum\limits_{cyc}\left(\frac{a^4}{a^3+b^3}\right)^{\frac34} \geqslant \frac{a^{\frac34}+b^{\frac34}+c^{\frac34}}{2^{\frac34}}$ When $a,b,c > 0$, prove
$$\left(\frac{a^4}{a^3+b^3}\right)^{\frac34}+\left(\frac{b^4}{b^3+c^3}\right)^ {\frac34}+\left(\frac{c^4}{c^3+a^3}\right)^{\frac34} \geqslant \frac{a^{\frac34}+b^... | It's not a proof, but I think it can help.
We can rewrite this inequality in the following form.
$$\sum_{cyc}\frac{a^4}{\sqrt[4]{\left(\frac{a^4+b^4}{2}\right)^3}}\geq a+b+c,$$
where $a$, $b$ and $c$ are positives.
Now, since for $(a,b,c)=(1.98,0.89,1.38)$
$$\sum_{cyc}\frac{a^4}{\sqrt[4]{\left(\frac{a^4+b^4}{2}\right)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1793680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 2
} |
How do I find the cdf of $X_1 + X_2$? $X_1$ uniform $(0,1)$ and $X_2$ uniform $(0,2)$
$$
\begin{cases}
f(x_1,x_2) = \frac{1}{2}, &\quad \mbox{for} \ 0<x_1<1, 0<x_2<2 \\
0, & \quad \mbox{otherwise}
\end{cases}
$$
The density of $X_1+X_2$:
$$
f(x_1,x_2) =
\begin{cases}
\displaystyle \frac{z}{2} & \quad 0<z<... | The PDF of $Y = X_1 + X_2$ is the convolution of the PDFs of $X_1$ and $X_2$
$$f_Y (y) = \begin{cases} \dfrac{y}{2} & \text{if } y \in [0,1]\\\\ \dfrac{1}{2} & \text{if } y \in [1,2]\\\\ \dfrac{3-y}{2} & \text{if } y \in [2,3]\\\\ 0 & \text{otherwise}\end{cases}$$
Integrating, we obtain the CDF
$$F_Y (y) = \begin{cases... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1793934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Sum of Specific Convergent Series Let L be the sum of the following alternating convergent series $$L = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}$$
Now consider the rearrangement
$$1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}+... = \sum_{n=1}^{\infty}\left(\frac{1}{4n... | Multiply the given relation by a factor of $\frac{1}{2}$ and add it to the original sequence. Now we get:
$$L = 1 - \frac{1}{2} + \frac 13 - \frac 14 + \frac 15 - \frac 16 + \frac 17 - \frac 18 + ...$$
$$\frac 12 L = 0 + \frac 12 + 0 - \frac 14 + 0 + \frac 16 + 0 - \frac 18+ ...$$
$$\frac 32 L = 1 + 0 + \frac 13 - \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1795962",
"timestamp": "2023-03-29T00:00:00",
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Prove $x+y^2+z^3 \geqslant x^2y+y^2z+z^2x$ for $xy+yz+zx=1$
Let $x,y,z \geqslant 0$ and $xy+yz+zx=1$. Prove that
$$x+y^2+z^3 \geqslant x^2y+y^2z+z^2x.$$
What I try:
$$x+y^2+z^3 \geqslant x^2y+y^2z+z^2x$$
$$\Leftrightarrow x(xy+yz+zx)+y^2+z^3- x^2y-y^2z-z^2x \geqslant 0$$
$$\Leftrightarrow \left(x^2z+\frac{z^3}{4}-z^2... | Alternative proof:
Using $xy + zx \le 1$ and $\sqrt{yz} \le 1$, we have
\begin{align*}
&x + y^2 + z^3 - x^2 y - y^2z - z^2x \\
\ge\,& x(xy + zx) + y^2\sqrt{yz} + z^3 - x^2 y - y^2z - z^2x\\
=\,& x^2z + y^2\sqrt{yz} + z^3 - y^2z - z^2x \\
=\,& \left(x^2z - z^2x + \frac14 z^3\right) + \frac34 z^3 + y^2\sqrt{yz} - y^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1796337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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What am I doing wrong in calculating the following limit? $$\lim_{x\to-2} \frac{x+2}{\sqrt{6+x}-2}=\lim_{x\to-2} \frac{1+2/x}{\sqrt{(6/x^2)+(1/x)}-2/x^2}$$ Dividing numerator and denominator by $x \neq0$
$$\frac{1+2/-2}{\sqrt{(6/4)+(1/-2)}-2/4}=\frac{0}{1/2}=0$$ but the limit is $4$ according to Wolfram Alpha?
| You say you divide by $x$, but that's not what you do in the denominator; it would be:
$$\lim_{x\to-2} \frac{x+2}{\sqrt{6+x}-2}
=\lim_{x\to-2} \frac{1+\tfrac{2}{x}}{\tfrac{\sqrt{6+x}}{x}-\tfrac{2}{x}} =\lim_{x\to-2} \frac{1+\tfrac{2}{x}}{-\sqrt{\tfrac{6}{x^2}+\tfrac{1}{x}}-\tfrac{2}{x}} $$
A better approach:
$$\begin{a... | {
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"url": "https://math.stackexchange.com/questions/1796440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Which exponents r>0 is the limit finite I am trying to find values of $r>0$ such that $\lim\limits_{n\rightarrow \infty} \sum\limits_{k=1}^{n^2}\frac{n^{r-1}}{n^r+k^r}$ is finite. I have tried to use integral methods for this limit such as $\sum\limits_{k=1}^{n^2}\frac{n^{r-1}}{n^r+k^r}=\sum\limits_{k=1}^{n}\frac{n^{r-... | In short: diverges for $r\in [0,1]$, converges for $r>1$.
*
*First case (simpler): $r\in(0,1)$:
$$
\sum_{k=1}^{n^2}\frac{n^{r-1}}{n^r+k^r} = \frac{1}{n}\sum_{k=1}^{n^2}\frac{n^{r}}{n^r+k^r}
\geq \frac{1}{n}\sum_{k=1}^{n^2}\frac{n^{r}}{n^r+n^{2r}}
\geq \frac{1}{n}\sum_{k=1}^{n^2}\frac{n^{r}}{n^{2r}+n^{2r}}
= \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1798985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show $x^2 + y^2 + 1 = 0 \pmod m$, iff $\,m \pmod 4 \ne 0$.
Show that $x^2 + y^2 + 1 = 0$ $\pmod m$ has solutions iff $\,m \pmod 4
\ne 0$.
I know hot to show that this equation has solutions if m = p It's easy to show "$=>$", but I'm completery stucked with the opposite direction.
| If $m \equiv 0\pmod 4$ it is simple to prove there is no solution, test all possible pairs $(x,y)$. If $2m \equiv 2\pmod 4$ then a solution can be constructed by the chinese remainder theorem from the solution of the odd $m$ and the solution $x=0,y=1$ modulo $2$. So we assume that $m$ is odd.
If $m$ is a prime then $\f... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Show that the series converges uniformly Show that $\sum_{n=1}^{\infty} \sin \left(\dfrac{x}{n}\right)^2$ converges uniformly on $[a,-a]$ , $a\in \mathbb{R}$.
My attempt:
Using Taylor's formula, we have:$$ \sin\left(\dfrac{x}{n}\right)^2 = \dfrac{x^2}{n^2} + o\left( \dfrac{x^2}{n^2}\right)$$
Therefore,
$$\sin\left(\d... | Since your $x$ is bounded, there exists an $N$ such that
\begin{equation}
0<\frac{x^2}{n^2}\leq \frac{\pi}{2}
\end{equation}
for all $n\geq N$ and $x\in[-a,a]$. Now you can use the inequality $0\leq \sin x\leq x $ for $0\leq x\leq \frac{\pi}{2}$.
| {
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"source": "stackexchange",
"question_score": "1",
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proof that $\sum_{n=1}^{\infty}\frac{n}{n^2+2n+1}$ diverges, by comparsion I need to prove that
$$\sum_{n=1}^{\infty}\frac{n}{n^2+2n+1}$$
diverges by comparsion.
The way I did was to use
$$\frac{n}{n^2+2n+1}>\frac{n}{n^2+2n^2+n^2} = \frac{n}{4n^2} = \frac{1}{4n}$$
which diverges. Can I do that? Because $$2n+1<2n^2+n... | $n/(n^2+2 n+1)=1/(n+2 +1/n^2)\geq 1/(n+3)$, and $\sum_n 1/(n+3)$ diverges.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solution of an initial value problem (MCQ) (CSIR DEC 2015) The solution of the initial value problem
$ (x-y) u_{x} + (y-x-u) u_{y} = u $
with the initial condition $u(x,0) = 1$ satisfies
*
*$ u^2(x-y+u) + (y-x-u) = 0$
*$ u^2(x+y+u) + (y-x-u) = 0$
*$ u^2(x-y+u) - (x+y+u) = 0$
*$ u^2(y-x+u) + (x+y-u) = 0$
This is... | Second characteristic curve :
$\frac{dx}{x-y} = \frac{dy}{y-x-u} = \frac{du}{u} =\frac{dx-dy+du}{(x-y)-(y-x-u)+u} = \frac{d(x-y+u)}{2(x-y+u)} = \frac{du}{u} $
$\frac{1}{2}\ln|x-y+u|-\ln|u|=$constant $\quad\to\quad \frac{x-y+u}{u^2}=C_2$
The implicit solution is $\quad x+y+u=F\left(\frac{x-y+u}{u^2}\right) \quad $any d... | {
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"source": "stackexchange",
"question_score": "3",
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The best notation for this identity involving pentagonal numbers $\omega(n)$ and the $3x+1$ map Let the $3x+1$ map
$$ f(n) =
\begin{cases}
3n+1 & \text {if $n$ is odd} \\
\frac{n}{2} & \text {if $n$ is even}
\end{cases} .$$
Now we read the Wikipedia's page for the Collatz problem, also known by several name... | At first some notes: We usually regard the closed formula
\begin{align*}
\omega(n)=\frac{3n^2+n}{2}\qquad\qquad\qquad\qquad\qquad n\geq 1\tag{1}
\end{align*}
as simpler than the summation formula
\begin{align*}
\omega(n)=\sum_{k=0}^{n-1}(3k+1)\qquad\qquad\qquad\qquad\quad n\geq 1\tag{2}
\end{align*}
So, if we want to r... | {
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Prove $\sqrt[2]{x+y}+\sqrt[3]{y+z}+\sqrt[4]{z+x} <4$
$x,y,z \geqslant 0$ and $x^2+y^2+z^2+xyz=4$, prove
$$\sqrt[2]{x+y}+\sqrt[3]{y+z}+\sqrt[4]{z+x} <4$$
A natural though is that from the condition $x^2+y^2+z^2+xyz=4$, I tried a trig substitutions $x=2\cdot \cos A$, $y=2\cdot \cos B$ and $z=2\cdot \cos C$, where $A,... | Remark: My second proof is given in https://artofproblemsolving.com/community/c6h2483725p20868007.
There, KaiRain@AoPS gave a nice proof.
Proof without using derivative
We split into three cases:
*
*$y+z \le 1$ and $z+x \le 1$: We have $x + y \le 2$.
Thus, we have $\mathrm{LHS} \le \sqrt{2} + 1 + 1 < 4$. The inequali... | {
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"source": "stackexchange",
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Choosing Functions for the Squeeze Theorem
Evaluate $$\lim_{n\to \infty}\dfrac{1}{\sqrt{n^2}}+\dfrac{1}{\sqrt{n^2+1}}+...+\dfrac{1}{\sqrt{n^2+2n}}$$
$$$$
I came across the the question on this site itself but had a few doubts on the given solution. As I do not yet have 50 reputation points, I cannot comment over ther... | Remember that $a<b\implies a\le b$. This is because $a\le b$ means $a<b$ or $a=b$ as you already noted in the comment.
If you are still confused, recall that "$p\implies q$" means $q$ is true whenever $p$ is true. And when $a<b$ is true, $a<b$ or $a=b$ is true. Hence, $a<b\implies a\le b$.
For example, is "$0\le 1$" ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Solve the differential equation $\frac{dx}{dt}=-\frac{1}{5}\sqrt{(x+1)}$ Solve the differential equation $$\frac{dx}{dt}=-\frac{1}{5}\sqrt{(x+1)}$$ given that $x=80$ when $t=0$. Give your answer in the form $x=f(t)$.
\begin{align}
\frac{dx}{dt} & = -\frac{1}{5}\sqrt{(x+1)} \\
\frac{dt}{dx} & = \frac{-5}{\sqrt{(x+1)}} \... | You didn't put $+C$ in the right place. You should have $C-\frac{t}{10}=\sqrt{x+1}$. Solve for $C$ here, then isolate $x$.
| {
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How to find $L = \int_0^1 \frac{dx}{1+{x^8}}$ Let $L = \displaystyle \int_0^1 \frac{dx}{1+{x^8}}$ . Then
*
*$L < 1$
*$L > 1$
*$L < \frac{\pi}{4}$
*$L > \frac{\pi}{4}$
I got some idea from this video link. But got stuck while evaluating the second integral.
Please help!!
Thanks in advance!
| $\forall x \in [0,1], \frac{1}{1+x^8}<1$, so $L<1$.
$\forall x \in [0,1], \frac{1}{1+x^8}>\frac{1}{1+x^2}$, so $L>\int_0^1 \frac{dx}{1+{x^2}}=\frac{\pi}{4}$
How to evaluate this integral :
$$\begin{align} \int \frac{1}{1+x^8}dx &= \int (1+x^8)^{-1}dx \\&= \int_0^x(1+t^8)^{-1}dt +a \\&= \int_0^{x^8} (1+t)^{-1}d(t^{1/8}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Please remind me of how this technique works ... We had a high school mathematics teacher who taught us a cool technique that I've forgotten. It can be used, for example, for developing a formula for the sum of squares for the first "n" integers. You start by making a column for Sn, and then determine the differences u... | *
*backgrounded on the fact that the difference of two consecutive squares is an odd number:
$(n+1)^2-n^2=2n+1$
the difference of two consecutive odd numbers is always $2$ thus the self-saying technique.
So when we proceed to find a square $n^2$ we sum up all odd numbers from 1 to $2n-1$ which is an arithmetic seque... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find $\max\int_{a}^{b}\left (\frac{3}{4}-x-x^2 \right )\,dx$ over all possible values of $a$ and $b$, $(aI tried finding the maxima of $f(x)=\frac{3}{4}-x-x^2$ by taking the derivative and so on and use the fact that $\displaystyle\int_{a}^{b}f(x)\,dx \leq M(b-a)$ where $M$ is the global maximum, but then the ma... | Given $$f(a,b) = \int_{a}^{b}\left(\frac{3}{4}-x-x^2\right)dx$$
Do the integration to get:
$$f(a,b) = \frac{a^3}{3}+\frac{a^2}{2}-\frac{3 a}{4}-\frac{b^3}{3}-\frac{b^2}{2}+\frac{3 b}{4} $$
Diff $f(a,b)$ with w.r.t $a$ and set to $0$
$$a^2+a-\frac{3}{4} =0$$
Solve. We get these $2$ critical points:
$$a = - \frac{3}{2}... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "7",
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Show that $1^7+7^7+13^7+19^7+23^7\equiv{0}\pmod{63}$ Show that $1^7+7^7+13^7+19^7+23^7\equiv{0}\pmod{63}$
According to Fermat's theorem:
$$1^7+7^7+13^7+19^7+23^7\equiv{1+7+13+19+23}\pmod{7}\equiv{63}\pmod{7}\equiv{0}\pmod{7}$$
Now we need to show: $1^7+7^7+13^7+19^7+23^7\equiv{0}\pmod{9}$ , but how??
| Since OP demands a proof without Euler's theorem, below is one:
First we denote the sum as $S$ and modulo $9:$
$$S\equiv1^7+(-2)^7+4^7+1^7+5^7\pmod9.$$
Since $(-2)^3\equiv1\pmod9,$ we find that $(-2)^7\equiv-2\pmod9,$ so that $$S\equiv4^7+5^7\pmod9.$$
But $5\equiv-4\pmod9,$ so $$S\equiv4^7+(-4)^7\equiv0\pmod9.$$
Hope... | {
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"timestamp": "2023-03-29T00:00:00",
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Using a Taylor polynomial to approximate $\cos(\frac14)$ with an error no more than $10^{-12}$ So the Lagrange remainder is given by:
$$R_n(x)=\frac{f^{n+1}(c)}{(n+1)!}(x-a)^{n+1}.$$
We want $\cos(\frac14)$ and we can do it around $a=0$. We know that $f^{n+1}$ is either $\pm\cos x$ or $\pm\sin x$, which are both bounde... | You want the error to be less than $10^{-12}$, so you want $|R_n(x)|<\frac{1}{10^{12}}$. You know that
$$|R_n(x)|\leq\frac{1}{(n+1)!}\left(\frac{1}{4}\right)^n,$$
so you want to find $n$ such that
$$\frac{1}{(n+1)!}\left(\frac{1}{4}\right)^n<\frac{1}{10^{12}},$$
so that you can combine the inequalities to get $|R_n(x)|... | {
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If $\sin \theta+\cos\theta+\tan\theta+\cot\theta+\sec\theta+\csc\theta=7$, then $\sin 2\theta$ is a root of $x^2 -44x +36=0$ My own bonafide attempt.
$$ 0<\theta<\pi/2$$
and $$\sin\theta+\cos\theta+\tan\theta+\cot\theta+\sec\theta+\csc\theta=7$$
then show that $\sin 2\theta$ is a root of the equation $$x^2 -44x +3... | $$\sin(\theta)+\cos(\theta)+\tan(\theta)+\cot(\theta)+\sec(\theta)+\csc(\theta)=7$$
$$\sin(\theta)+\cos(\theta)+\frac{\sin(\theta)}{\cos(\theta)}+\frac{\cos(\theta)}{\sin(\theta)}+\frac{1}{\cos(\theta)}+\frac{1}{\sin(\theta)}=7$$
$$\sin^2\theta\cos\theta+\sin\theta\cos^2\theta+\sin^2\theta+\cos^2\theta+\sin\theta+\cos\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814187",
"timestamp": "2023-03-29T00:00:00",
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Integration - Find the area under the curve. Im not sure how to do this at all. Need help walking through the steps on how to get the answer. So below is the question. I need any and all help.
Find the area under the curve:
$$ y=2\sin(3x-\pi/3)$$
between $x=0$ and $x=\pi/2$, and give your answer to $4$ decimal places.
| I assume by "area under the curve," you mean area between the graph of $y=f(x)$ and the $x$-axis. The area under the curve between $x=x_1$ and $x=x_2$ is the definite integral of the function $f(x)$ with bounds $x_1$ and $x_2$.
Here is a picture of this area from Wolfram Alpha (note that when the function goes below t... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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For $f(x)=\frac {1}{x}$ and $g(x)=\sqrt{x-4}$, find the domain of the composite function $g\circ f(x)$.
For $f(x)=\frac {1}{x}$ and $g(x)=\sqrt{x-4}$, find the domain of the composite function $g\circ f(x)$.
My Attempt
Here, $$f(x)=\frac {1}{x}$$
$$g(x)=\sqrt{x-4}$$
Now,
$$g\circ f(x)=\sqrt{\frac {1-4x}{x}}$$
How can... | We have $$g\circ f(x)=\sqrt{f(x)-4}=\sqrt{\frac{1}{x}-4}$$
This function is defined when $x\neq 0$ because of the $\frac{1}{x}$ and when $\frac{1}{x}-4\geq 0$ because of the $\sqrt{\quad}$. Now, if $x<0$ then $\frac{1}{x}<0$ and so $\frac{1}{x}-4<0$. Hence, we need $x>0$. Finally if $x>0$, then
$$\frac{1}{x}-4\geq 0 \... | {
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"timestamp": "2023-03-29T00:00:00",
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First integrals for solving system of ODEs Assume a problem $$
\begin{cases}
\frac{\mathrm{dx}}{\mathrm{dt}} = \frac{y}{x-y}, \\[2ex]
\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{x}{x-y}.
\end{cases}$$
Additionally, $x = x(t)$ and $y=y(t)$.
Attempt
Multiply the first equation by $x$ and the second one by $y$. Subtract the l... | Don't forget $x^2-y^2=(x+y)(x-y)$,
\begin{align*}
\frac{(x+y)(C-t)}{2} &= C_{0} \\
x+y &= \frac{2C_{0}}{C-t}
\end{align*}
Hence,
$$\left \{
\begin{align*}
x &= \frac{C_{0}}{C-t}+\frac{C-t}{2} \\
y &= \frac{C_{0}}{C-t}+\frac{t-C}{2} \\
\end{align*} \right.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1816984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\sum_{n=0}^{\infty}{2^n(n^2-n\pi+1)(n^2+n-1)\over (2n+1)(2n+3){2n\choose n}}=1$ $$S=\sum_{n=0}^{\infty}{2^n(n^2-n\pi+1)(n^2+n-1)\over (2n+1)(2n+3){2n\choose n}}=1$$
Beta function
$${1\over {2n\choose n}}=n\int_{0}^{1}x^{n-1}(1-x)^ndx$$
$$\sum_{n=0}^{\infty}{2^n\over {2n\choose n}}=\int_{0}^{1}{2(1-t)\over [1-2t(... | Through Euler's beta function you may prove that:
$$ \forall x\in(0,4), \qquad f(x)=\sum_{n\geq 0}\frac{x^n}{(2n+1)\binom{2n}{n}}=\frac{4\arcsin\left(\frac{\sqrt{x}}{2}\right)}{\sqrt{x(4-x)}} \tag{1}$$
(have also a look at this question), then by computing $\int x^2 f(x^2)\,dx$ we get:
$$ \forall x\in(0,2),\quad g(x)=\... | {
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Find the equation of the sphere which touches the sphere $x^2+y^2+z^2+2x-6y+1=0$ at $(1,2,-2)$ and pass through $(1,-1,0)$ Find the equation of the sphere which touches the sphere $x^2+y^2+z^2+2x-6y+1=0$ at $(1,2,-2)$ and pass through $(1,-1,0)$
My Attempt:
Let the equation of the sphere be $x^2+y^2+z^2+2ux+2vy+2wz+d=... | As sphere touch each other at $(1,2,-2)$ so distance between their centres=$r_1+r_2$ and distance between centre of sphere and point $(1,-1,0)$ is the radius of sphere(at that point we have a tangent). Now you have $4$ equations and $4$ unknowns you can now solve them
| {
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Using AM-GM inequality to prove prove that $$x^4 + y^4 + z^4 \geq xyz(x+y+z)$$
This AM-GM inequalities are seriously stumping me. I'd appreciate a full proof and explanation and hints for proving other inequalities like this. Thanks.
| The inequality is equivalent to ( by expanding RHS)
$$ x^4 + y^4 +z^4 \ge x^2yz + y^2xz + z^2xy$$
By AM-GM
$$ x^4+x^4 +y^4 +z^4 \ge 4(x^8y^4z^4)^\frac{1}{4} = 4x^2yz$$
$$ y^4+y^4 +x^4 +z^4 \ge 4(y^8x^4z^4)^\frac{1}{4} = 4y^2xz$$
$$ z^4+z^4 +x^4 +y^4 \ge 4(z^8x^4y^4)^\frac{1}{4} = 4z^2xy$$
Adding them up we get
$$4(x^4... | {
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"answer_id": 2
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USA $2011$ contest inequality problem, proving $\frac{ab+1}{(a+b)^2}+\frac{bc+1}{(b+c)^2}+\frac{ca+1}{(c+a)^2}\ge 3$, under given condition.
If $a^2+b^2+c^2+(a+b+c)^2\le4$, then $$\frac{ab+1}{(a+b)^2}+\frac{bc+1}{(b+c)^2}+\frac{ca+1}{(c+a)^2}\ge 3.$$
My attempt:
From the given criteria, one can easily obtain that $$(... | For positives $a$, $b$ and $c$ by AM-GM we obtain:
$2\sum\limits_{cyc}\frac{1+ab}{(a+b)^2}\geq\sum\limits_{cyc}\frac{a^2+b^2+c^2+ab+ac+bc+2ab}{(a+b)^2}=3+\sum\limits_{cyc}\frac{(c+a)(c+b)}{(a+b)^2}\geq3+3=6$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1820337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Showing $\int_{0}^{1}{x^{2n}-x\over 1+x}\cdot{dx\over \ln{x}}=\ln\left({2\over \pi}\cdot{(2n)!!\over (2n-1)!!}\right)$ Integrate
$$I=\int_{0}^{1}{x^{2n}-x\over 1+x}\cdot{dx\over \ln{x}}=\ln\left({2\over \pi}\cdot{(2n)!!\over (2n-1)!!}\right)\tag1$$
$${x^{2n}-x\over 1+x}=\sum_{k=0}^{\infty}(-1)^k(x^{2n}-x)x^k\tag2$$
S... | Enforce the substitution $x\to e^{-x}$ to write
$$\begin{align}
I(n)&=\int_0^1 \frac{x^{2n}-x}{1+x}\frac{1}{\log(x)}\,dx\\\\
&=\int_0^\infty \frac{e^{-x}-e^{-2nx}}{x}\frac{e^{-x}}{1+e^{-x}}\,dx\\\\
&=\sum_{k=0}^\infty (-1)^k \int_0^\infty \frac{e^{-(k+2)x}-e^{-(k+2n+1)x}}{x}\\\\
&=\sum_{k=0}^\infty (-1)^k \log\left(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1821080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to put derivative of composition in Jacobian matrix? Here are two functions:
$f\left(u,v\right)=u^{2}+3v^{2}$
$g\left(x,y\right)=\begin{pmatrix} e^{x}\cos y \\ e^{x}\sin y \end{pmatrix} $
I need to make Jacobian matrix of $f\circ g$. I found derivative of their composition:
$\frac{d\left(f\circ g\right) }{d\left(... | Using the chain rule instead:
\begin{align*}D(f\circ g)(x,y)& =\color{red}{Df(g(x,y))}\cdot\color{blue}{ Dg(x,y)}\\
& = \color{red}{\begin{pmatrix} 2u&6v \end{pmatrix}\circ(g(x,y))}\cdot \color{blue}{ \begin{pmatrix}e^x\cos y & -e^x\sin y \\ e^x\sin y & e^x\cos y\end{pmatrix}}\\
& =\color{red}{ \begin{pmatrix} 2e^x\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1822008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Equation involving the Jacobi symbol: $\left( \frac {-6} p \right) = 1$? I have to determine the values of $p \in \{0, \dots, 23 \}$ such that $\left( \frac {-6} p \right) = 1$.
I have that:
$$\left( \frac {-6} p \right) = \left( \frac 2 p \right) \left( \frac {-3} p \right)$$
and I know that $\left( \frac 2 p \right) ... | I saw this the other day as well. Call this a side note, the primes $1,7 \pmod {24}$ are all represented by $x^2 + 6 y^2.$ It is easy to see that if $p = x^2 + 6 y^2$ in integers, then $(-6|p) = 1.$
$$ 1, 7, 31, 73, 79, 97, 103, 127, 151, 193,
199,
$$
the primes $5,11 \pmod {2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1822204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the area of $S=\{(x,y)|\rm{\exists ~}\theta,\beta,x=\sin^2{\theta}+\sin{\beta},y=\cos^2{\theta}+\cos{\beta}\}$ Let $S$ be the domain defined by $$S=\{(x,y)|\rm{\exists ~}\theta,\beta,x=\sin^2{\theta}+\sin{\beta},y=\cos^2{\theta}+\cos{\beta}\}$$
find the area of $S$
This is middle school problem,so I think it can b... | From $(x - \sin^2\theta)^2 + (y - \cos^2\theta)^2 = 1$ we get that $S$ is a set of circles with radius $1$ and centers on the points of $(\sin^2\theta, \cos^2\theta)$, i.e. on the segmet from point $(1,0)$ to point $(0,1)$ (see figure below).
So the area of $S$ is sum of areas of two semicircles and area of rectngle wi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1826004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\lim\limits_{x\to \infty} a\sqrt{x+1}+b\sqrt{x+2}+c\sqrt{x+3}=0$ if and only if $ a+b+c=0$ Prove that
$$ \displaystyle \lim_{x\to\infty } \left({a\sqrt{x+1}+b\sqrt{x+2}+c\sqrt{x+3}}\right)=0$$
$$\text{if and only if}$$
$$ a+b+c=0.$$. I tried to prove that if $a+b+c=0$, the limit is $0$ first, but after ge... | The limit is bounded if a+b+c = 0
verify:
$\lim_\limits{x \to \infty} \sqrt{x+k} - \sqrt {x} = 0$
$\lim_\limits{x \to \infty} a\sqrt{x+1} -a\sqrt{x}+ b\sqrt{x+2} -b\sqrt{x} + c\sqrt{x+3} -c\sqrt{x}= 0$
$\lim_\limits{x \to \infty} (a\sqrt{x+1} + b\sqrt{x+2} + c\sqrt{x+3}) = \lim_\limits{x \to \infty}(a+b+c) \sqrt{x}$
$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1827190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
Is function invertible? Reflection on the unit circle:
Let $E=\mathbb R ^{2} - \left\{0,0\right\} $ be perforated plane and $f: E \mapsto E$ defined by $f\left(x,y\right)=\left(\frac{x}{x^{2}+y^{2} } , \frac{y}{x^{2}+y^{2} } \right)
$
Show with Jacobian matrix that $f$ is in all points local invertible. Show that... | See the comment on the question for local invertibility.
Now, we prove that $f$ is invertible by showing that $f$ is one-one and onto. Given $(x_1, y_1), (x_2,y_2) \in \Bbb R^2 - \{(0,0)\}$, we have:
$$f(x_1,y_1) = f(x_2, y_2) \implies \begin{cases} \frac{x_1}{x_1 ^2 + y_1^2} = \frac{x_2}{x_2 ^2 + y_2^2} \\ \frac{y_1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1830158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
The solution of equation $4+6+8+10+\cdots +x =270$ is 15. The solution of equation
$4+6+8+10+\cdots +x =270$
is $x=15$.
How can I prove it?
I ve tried the geometric sequence but I cannot figure out the pattern.
| $$4+6+8+10+\cdots +x=270\\ \frac { 4+x }{ 2 } \left( \frac { x-4 }{ 2 } +1 \right) =270\\ \frac { \left( 4+x \right) }{ 2 } \frac { \left( x-2 \right) }{ 2 } =270\\ \left( x-2 \right) \left( x+4 \right) =1080\\ { x }^{ 2 }+2x-1088=0\\ x=-1\pm \sqrt { 1089 } \\ x=32$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1830968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
problem proving: $(1+q)(1+q^2)(1+q^4)...(1+q^{{2}^{n}}) = \frac{1-q^{{2}^{n+1}}}{1-q}$ I'm trying to prove this, and it is really frustrating, because it seems a really easy problem to prove, however, I'm having a little problem with exponents:
$$(1+q)(1+q^2)(1+q^4)...(1+q^{{2}^{n}}) = \frac{1-q^{{2}^{n+1}}}{1-q}$$
Hyp... | Define $f(q)=(1+q)(1+q^2)(1+q⁴)...(1+q^{{2}^{n}})$,
We calculate
\begin{align*}
(1-q)f(q)&= (1-q)(1+q)(1+q^2)(1+q⁴)...(1+q^{{2}^{n}}) \\
&=(1-q^2)(1+q^2)(1+q^{2^2})...(1+q^{{2}^{n}})\\
&=(1-q^{2^2})(1+q^{2^2})...(1+q^{{2}^{n}})\\
&\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \vdots \\
&=1-q^{2^{n+1}}
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1831623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
induction clarification about the step $n+1$ Suppose i need to prove that $\frac{1}{2^2}+\frac{1}{3^2}...+\frac{1}{n^2}<1-\frac{1}{n}$
So in the step of $n+1$, the right side becomes $<1-\frac{1}{n+1}$ or is it: $<1-\frac{1}{n}-\frac{1}{n+1}$? i guess it's the first one but why? i mean in the left side it's in additio... | Your induction should be as follows: assume that this property holds for all $k≤n$. Then your inductive step for $n+1$ is: $$\frac{1}{3^2} +\ldots +\frac{1}{n^2}+\frac{1}{(n+1)^2}<\frac{1}{2^2}+\frac{1}{3^2}+\ldots +\frac{1}{n^2}$$ This holds as you can verify for yourself that $\frac{1}{2^2}>\frac{1}{(n+1)^2}$ for any... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1832551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
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