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Evaluation of $\int\frac{x^7+2}{\left(x^2+x+1\right)^2}dx$
Evaluation of $$\int\frac{x^7+2}{\left(x^2+x+1\right)^2}dx$$
$\bf{My\; Try::}$ Let $$\displaystyle \mathop{I = \int\frac{x^7+2}{(x^2+x+1)^2}}dx = \int\frac{(x^7-1)+3}{(x^2+x+1)^2}dx$$
$$\mathop{\displaystyle = \int\frac{x^7-1}{(x^2+x+1)^2}}+\displaystyle \int\frac{3}{(x^2+x+1)^2}dx$$
Now Using the formula $$\bullet x^7-1 = (x-1)\cdot \left[x^6+x^5+x^4+x^3+x^2+x+1\right]$$
So $$\bullet \; (x^7-1) = (x-1)\left[(x^4+x)\cdot (x^2+x+1)+1\right]$$
So we get $$\displaystyle I = \int\frac{(x-1)\cdot (x^4+x)\cdot (x^2+x+1)}{(x^2+x+1)^2}dx+\frac{1}{2}\int\frac{2x-2}{(x^2+x+1)^2}+3\int\frac{1}{(x^2+x+1)^2}dx$$
$$\displaystyle I = \underbrace{\int\frac{(x-1)(x^4+x)}{x^2+x+1}dx}_{J}+\underbrace{\frac{1}{2}\int\frac{(2x+1)}{(x^2+x+1)^2}dx}_{K}+\underbrace{\frac{3}{2}\int\frac{1}{(x^2+x+1)^2}dx}_{L}$$
Now $$\displaystyle J = \int\frac{(x-1)(x^4-x)+2x(x-1)}{(x^2+x+1)}dx = \int(x^3-2x^2+x)dx+\int\frac{2x^2-2x}{x^2+x+1}dx$$
Now we can solve the integral Using the formulae.
My question is can we solve it any other way, If yes then plz explain here
Thanks
|
HINT: prove that $$\frac{x^7+2}{(x^2+x+1)^2}={x}^{3}-2\,{x}^{2}+x+2+{\frac {-4\,x-2}{{x}^{2}+x+1}}+{\frac {x+2}{
\left( {x}^{2}+x+1 \right) ^{2}}}
$$
|
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|
Find the sign of $a,b,c$ in $ax^2+bx+c$ given the graph and a coordinate on it.
So my first approach was that, we see that there are $2$ roots. And one is negative and one is positive. $a$ would be evidently positive. The positive one's modulus is bigger than the negative one's. So the sum of roots would be $+ve$, that is $\frac{-b}a$. So if $a$ is positive and sum of roots is positive, then $b$ has to be negative. And similarly, product of the roots will be $-ve$, that is $\frac ca$, so $c$ will be negative. Which gives the option $b)$ as correct.
But if we see, point $A$ will be the minimum point. And at the minimum point value of function is $\frac{-D}{4a}$. But generally the value is $\frac{-D^2}{4a}$ , because
[Edit] let $y=ax^2+bx+c$. Now let $a$ be positive. So minimum will be at $x=\frac{-b}{2a}$. Substitute. $y=a(\frac{b^2}{4a^2})-\frac{-b^2}{2a}+c$
$y=\frac{b^2-2b^2+4ac}{4a}=\frac{-(b^2-4ac)}{4a}=\frac{-D^2}{4a}$
This means that $b^2=4ac$. Which is completely different and unsupportive of the graph. So where did I do it wrong.
|
Assuming that you understand some Calculus!
Let $f(x) = ax^2 + bx + c$ (a second degree polynomial in the single variable $x$) where $a,\ b,\ c$ and $x$ denote real numbers where $a \neq 0$.
Using the Calculus: We can see where some of the given information is coming from and then using it to determine the answer to the questions
$f^\prime(x) = 2ax + b = 0$ exactly when $x = -\frac{b}{2a}$ which is the $x$-coordinate of the point where the graph of $f(x)$ as a single turning point given as $A$. The corresponding $y$-coordinate of $A$ is $f(-\frac{b}{2a}) = \frac{4ab-b^2}{4a} = - \frac{b^2-4ac}{4a}$. Hence the coordinates of the turning point (known as the vertex) is then $(-\frac{b}{2a},\ -\frac{b^2-4ac}{4a})$. Hence, we can see that in the given problem $D\equiv b^2-4ac$ (known as the Discriminant and is commonly used to determine whether or not $f(x)$ has real roots---$D\geq 0$ tell us "Yes, real roots exists" and $D < 0$ tell us "No real roots exist".)
Recall the quadratic formula (the roots of $f(x)$) $$x\equiv -\frac{b}{2a}\ \pm\ \frac{\sqrt{D}}{2a} = -\frac{b}{2a}\pm \frac{\sqrt{b^2-4ac}}{2a}$$ Consider the sum and product of the roots!
Next, using the second derivative $f^{\prime\prime}(x)$ tell us the concavity (upwards or downwards) of the graph of $f(x)$.
$$f^{\prime\prime}(x) = (2ax + b)^{\prime} = 2a$$Now if $f^{\prime\prime}(x) > 0$ then the graph of $f(x)$ is concaved upwards and if $f^{\prime\prime}(x) < 0$, the graph of $f(x)$ is concaved downwards.
Now the $y$-intercept (given as $P$) is the point where the graph of $f(x)$ intersects the $y$-axis and has the form $(0,\ f(0))$ (or $(0,\ c)$).
Let's determine the sign of the coefficents $a,\ b$ and $c$ of $f(x)$ from the given graph.
*
*The given graph is concaved upwards, so $f^{\prime\prime}(x) = 2a > 0$ exactly when $a > 0$; that is, the coefficient $a$ is positive.
*The turning point given as $A$ lies in the fourth quadrant of the coordinate system and, so the point $A$ has a positive $x$-coordinate and a negative $y$-coordinate; that is,
$$-\frac{b}{2a} > 0\ \ \ \hbox{and}\ \ -\frac{b^2-4ac}{4a} < 0$$From the first inequality together with the fact that $2a > 0$ we easily deduce that $-b > 0$ or equivalently that $b < 0$; that is, the coefficent $b$ is negative.
*The $y$-intercept of the given graph lies directly below the origin $O$ and as such must have a negative $y$-coordinate. In this case, $f(0) = c < 0$ which means that the coefficient $c$ is negative.
In summary, this comes down to choice (B) in the question.
|
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|
Solving a rational equation with multiple and nested fractions
This is the equation to solve: $\dfrac{\dfrac{x+\dfrac{1}{2}} {\dfrac{1}{2}+\dfrac{x}{3}}}{\dfrac{1}{4}+\dfrac{x}{5}}=3$
What I did:
$x+\dfrac{1}{2}=\dfrac{2x+1}{2}$
$\dfrac{x}{3}+\dfrac{1}{2}=\dfrac{2x+3}{6}$
$\dfrac{x}{5}+\dfrac{1}{4}=\dfrac{4x+5}{20}$
$\dfrac{2x+1}{2}\div \dfrac{2x+3}{6} = \dfrac{2x+1}{2}\times \dfrac{6}{2x+3}= \dfrac{6x+3}{2x+3}$
$\dfrac{6x+3}{2x+3}\div\dfrac{4x+5}{20}=\dfrac{6x+3}{2x+3}\times\dfrac{20}{4x+5}=3$
$\implies\dfrac{120x+60}{(2x+3)(4x+5)}=3$
I know how to solve the equation. But right now I tried several times and I got wrong answers.
So I appreciate your help. Thank you.
|
You have, essentially,
$$
\frac{120x+60}{(2x+3)(4x+5)} = 3
$$
Multiply both sides by $(2x+3)(4x+5)$ to get
$$
120x+60 = 3(2x+3)(4x+5) = 24x^2+66x+45
$$
Subtract $120x+60$ from both sides to get
$$
24x^2-54x-15 = 0
$$
Divide both sides by $3$ to get
$$
8x^2-18x-5 = 0
$$
which factors as
$$
(4x+1)(2x-5) = 0
$$
to yield $x = -1/4$ or $5/2$. (Or, you can use the quadratic formula.)
|
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|
Discrete mathematics question $$(n+1)^2+(n+2)^2+(n+3)^2+\dots+(2n)^2=\frac{n(2n+1)(7n+1)}{6}$$
Prove the statement using mathematical induction.
|
When $n=1$, $LHS=4=RHS.$
Assume, when $n=k$, $LHS=RHS$. When $n=k+1$,
$LHS=(k+2)^2+(k+3)^2+\dots+(2k+1)^2 + (2(k+1))^2
\\=k(2k+1)(7k+1)/6-(k+1)^2+(2k+1)^2+(2(k+1))^2
\\=[k(2k+1)(7k+1)+18(k+1)^2+6(2k+1)^2]/6
\\=[(2k+1)(7k^2+13k+6)+18(k+1)^2]/6
\\=[(2k+1)(7k+6)(k+1)+18(k+1)^2]/6
\\=(k+1)(14k^2+37k+24)/6
\\=(k+1)(2(k+1)+1)(7(k+1)+1)/6=RHS.$
|
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|
If $x^4+7x^2y^2+9y^4=24xy^3$,show that $\frac{dy}{dx}=\frac{y}{x}$ If $x^4+7x^2y^2+9y^4=24xy^3$,show that $\frac{dy}{dx}=\frac{y}{x}$
I tried to solve it.But i got stuck after some steps.
$x^4+7x^2y^2+9y^4=24xy^3$
$4x^3+7x^2.2y\frac{dy}{dx}+7y^2.2x+36y^3.\frac{dy}{dx}=24x.3y^2\frac{dy}{dx}+24y^3$
$\dfrac{dy}{dx}=\dfrac{24y^3-4x^3-14y^2x}{14x^2y+36y^3-72xy^2}$
How to move ahead?Or there is some other elegant way to solve it.
|
$$x^4+7x^2y^2+9y^4=24xy^3$$
Can be rewritten as,
$$1+7\frac{y^2}{x^2}+9 \frac{y^4}{x^4}=24\frac{y^3}{x^3}$$
Differentiate wrt $x$, you will get that three things can be equated to $0$. One of them will lead to your result.
(Hint: Don't expand derivative using quotient rule.)
|
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Understanding how to use $\epsilon-\delta$ definition of a limit I finally understand the geometric intuition behind the $\epsilon-\delta$ definition of a limit, which is actually quite neat:
But I'm having trouble actually using the definition to come to a conclusion.
For (a solved) example, to prove that $\lim_{n\rightarrow \infty} \frac{3n+5}{2n+7} = \frac{3}{2}$, the following solution is given:
Proof:
Let $a_n = \frac{3n+5}{2n+7}$. Then, $\left | a_n-\frac{3}{2} \right |=\frac{11}{2(2n+7)}<\frac{3}{n}$. Given $\epsilon > 0$, choose $n_0 \in \mathbb{N}$ such that $n_0 > \frac{3}{\epsilon}$. Then, for all $n \ge n_0, \left | a_n-\frac{3}{2} \right | < \epsilon $. Therefore, $\lim a_n = \frac{3}{2}$.
I'm not quite sure I understand this proof. Namely:
*
*Where did $\frac{3}{n}$ come from?
*I guess we just replaced n with $\epsilon$ in the third sentence.
*How did the fourth sentence follow from the third, and why does that show that 3/2 is the limit?
So I tried it myself on another problem, where $a_n = \frac{2n+5}{6n-3}$, and the limit being $\frac{1}{3}$:
Let $a_n = \frac{2n+5}{6n-3}$. Then, $\left | a_n-\frac{1}{3} \right |=\left | \frac{6n-6n+15-3}{3(6n-3)} \right |= \left | \frac{12}{3(6n-3)} \right |= 4\left | \frac{1}{6n-3} \right | < \epsilon$. Dividing by 4, $\left | \frac{1}{6n-3} \right | < \frac{\epsilon}{4}$.
And that's as far as I logically get... I assume the next step is to choose an $n_0 \in \mathbb{N}$ such that $n_0 > \frac{\epsilon}{4}$, and isolating epsilon, $\epsilon < 4n_0$.
but as you can tell, I'm really lost.
|
Part (1) Essentially whoever wrote the proof wants to get an $n$ in the denominator (to be explained shortly). Note that $\frac{11}{2(2n+7)}=\frac{11}{4n+14}<\frac{11}{4n}<\frac{3}{n}$, so that we got a nice bound with $n$ in the denominator.
Part (2) Note that we have $|a_{n} - \frac{3}{2}| < \frac{3}{n}$ but really what we want is to show that for any $\varepsilon>0$ there is an $n$ such that $|a_{n} - \frac{3}{2}| < \varepsilon$. One way to show this is to ensure $n$ is large enough so that $\frac{3}{n}<\varepsilon$ (given me any $\varepsilon$ and I can pick an $n$ so this is true!). However, rearranging the inequality, choosing this $n$ is equivalent to choosing $n$ such that $n>\frac{3}{\varepsilon}$.
Part (3) Suppose in (2) that we have found an $n$ so that this is true (i.e. that $\frac{3}{n} < \varepsilon$ for the chosen $\varepsilon$). Just for consistency of notation lets let this $n$ be $n_{0}$. Then $\frac{3}{n_{0}} < \varepsilon$.
However we already know that $|a_{n} - \frac{3}{2}| < \frac{3}{n}$ for any $n$, so it is also true for $n_{0}$. I.e. we have that $|a_{n_{0}} - \frac{3}{2}| < \frac{3}{n_{0}}$. Since $\frac{3}{n_{0}} < \varepsilon$ we have that $|a_{n_{0}} - \frac{3}{2}| < \varepsilon$.
The key is that I can do this for ANY $\varepsilon$. Given me any really really small number $\varepsilon$ and I can find an $n_{0}$ such that $|a_{n_{0}} - \frac{3}{2}| < \varepsilon$. BY DEFINITION of the limit (see your text or lecture notes), this implies that the limit of the sequence $\{a_{n}\}$ is $\frac{3}{2}$.
For your proof...
I will structure the proof in the same way as the first proof.
Note that
$| a_{n} - \frac{1}{3}| = |\frac{2n+5}{6n-3}-\frac{1}{3}| = \frac{18}{18n-9}\leq \frac{2}{n} < \frac{3}{n}$.
Thus for any $\varepsilon$ chose $n_{0}$ such that $n_{0}>\frac{3}{\varepsilon}$. Then we have that $| a_{n_{0}} - \frac{1}{3}| < \varepsilon$.
|
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|
Convert between parameteric ellipse equations I have the parametric equation of an ellipse in this form:
$$x(t)= a\cos(t)$$
$$y(t)=b\cos(t+\phi)$$
It's an ellipse centred about the origin, with a tilt angle. So three parameters.
How can I convert it to the form:
$$x(t)=A\cos(t)\cos(\Phi)-B\sin(t)\sin(\Phi)$$
$$y(t)=A\cos(t)\sin(\Phi)+B\sin(t)\cos(\Phi)$$
i.e. the ellipse formula with a rotation through $\Phi$? I obviously need to get $A$, $B$, and $\Phi$ in terms of $a$, $b$ and $\phi$, but I can't see how to do it.
|
So we have a pair of equations
$$
a \cos (t+\tau) = A \cos t \cos \Phi - B \sin t \sin \Phi\\
b \cos (t+\tau+\phi) = A \cos t \sin \Phi + B \sin t \cos \Phi
$$
which need to be equal for every $t \in \mathbb R$. Since both of the formula produce an ellipse when $t$ runs over $[u, u+2\pi]$ for any $u$, the value $\tau$ denotes the parameter shift between the parametrizations. Expanding the LHS we get
$$
a \cos t \cos \tau - a \sin t \sin \tau = A \cos t \cos \Phi - B \sin t \sin \Phi\\
b \cos t \cos(\tau+\phi) - b \sin t \sin (\tau + \phi)= A \cos t \sin \Phi + B \sin t \cos \Phi.
$$
Eliminating $t$ gives a system of 4 equations:
$$\begin{aligned}
a \cos \tau &= A \cos \Phi &\qquad(1)\\
a \sin \tau &= B \sin \Phi &\qquad(2)\\
b \cos (\tau + \phi) &= A \sin \Phi &\qquad(3)\\
b \sin (\tau + \phi) &= -B \cos \Phi &\qquad(4)
\end{aligned}
$$
which need to be solved for $\tau, A, B, \Phi$. Squaring and adding $(1)$ and $(2)$ and the same for $(3), (4)$ gives
$$
a^2 = A^2 \cos^2 \Phi + B^2 \sin^2 \Phi\\
b^2 = A^2 \sin^2 \Phi + B^2 \cos^2 \Phi
$$
and adding and subtracting those give
$$\begin{aligned}
a^2 + b^2 &= A^2 + B^2\\
a^2 - b^2 &= A^2 \cos 2\Phi - B^2 \cos 2\Phi
\end{aligned}
$$
Also by $(1)\times(3) + (2)\times(4)$
$$
ab \left(
\cos \tau \cos (\tau + \phi)
+\sin \tau \sin (\tau + \phi)
\right) = (A^2 - B^2) \cos \Phi \sin \Phi\\
2ab \cos \phi = (A^2-B^2) \sin 2\Phi.
$$
So
$$
A^2 + B^2 = a^2 + b^2\\
(A^2 - B^2) \cos 2\Phi = a^2 - b^2\\
(A^2 - B^2) \sin 2\Phi = 2ab\cos \phi
$$
Solving the system gives
$$
\Phi = \frac{1}{2}\arctan \frac{2ab\cos \phi}{a^2 - b^2}\\
A^2 + B^2 = a^2 + b^2\\
A^2 - B^2 = \pm\sqrt{(a^2 - b^2)^2 + 4a^2b^2 \cos^2 \phi} =
\pm\sqrt{a^4 + b^4 + 2a^2b^2\cos 2\phi}.
$$
Care should be taken when choosing correct sign. Actually the sign interchanges the $A$ and $B$. Wrong sign rotates the ellipse by $\frac{\pi}{2}$. One may also add $\frac{\pi n}{2}$ to the $\Phi$ and fix the sign.
Example $a = 1, b = 2, \phi = \frac{\pi}{3}$
$$
\Phi = -\frac{1}{2}\arctan \frac{2}{3}\approx -0.294\\
B = \sqrt{\frac{5+\sqrt{13}}{2}} \approx 2.07431\\
A = \sqrt{\frac{5-\sqrt{13}}{2}} \approx 0.83500\\
$$
I've plotted both curves for $t \in \left[0, \frac{15}{8}\pi\right]$ so you can see where the cut is.
|
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|
Solving $a!+b!+c!=3^d$ The question is to find all tuples $(a,b,c,d)$ of natural numbers $c\geq b$ and $ b \geq a $ and $a!+b!+c!=3^d$. I am finding difficulty in establishing relation between $a$, $b$, $c$, and $d$. I see that a!,b!,c! are even.How can their sum be odd? Please help.
|
When $a \le b \le c$ we get
$$
a! + b! + c! = a! \Big( 1 + b!/a! \big( 1 + c!/b! \big) \Big) = 3^d. \tag 1
$$
If $a \ge 2$ then there are no solutions.
If $b \ge 3$ then there are no solutions.
So we can only get
$$
1 + 2 + c! = 3^d, \tag 2
$$
thus
$$
c! = 3 \big( 3^{d-1} - 1 \big) \tag 3
$$
But $3$ divides $c!$, so $3 \le c$.
But $3$ does not divide $3^{d-1} - 1$, so $c < 6$.
So
$$
2 < c < 6. \tag 4
$$
Verification
$$
3! = 3 \Big( 3^{d-1} - 1 \Big) \Rightarrow 3^d = 9,\\ \tag 5
4! = 3 \Big( 3^{d-1} - 1 \Big) \Rightarrow 3^d = 27,\\
5! = 3 \Big( 3^{d-1} - 1 \Big) \Rightarrow 3^d = 123.\\
$$
So the only solutions are
$$
\bbox[16px,border:2px solid #800000] {
\begin{array}{c}
(1,2,3) \Rightarrow 1! + 2! + 3! = 3^2\\\\
(1,2,4) \Rightarrow 1! + 2! + 4! = 3^3
\end{array} } \tag 6
$$
|
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|
Proving uniqueness of solutions to $\sin^2A + \sin^2B = \sin (A+B)$ without using multivariable calculus In the course of solving a trigonometric problem (see $a^2+b^2=2Rc$,where $R$ is the circumradius of the triangle.Then prove that $ABC$ is a right triangle), in one approach the following equation needed to be solved:
$$\boxed{\sin^2A + \sin^2B = \sin (A+B)}$$
subject to $A\in(0,\frac{\pi}{2}),B\in(0,\frac{\pi}{2}),\pi-(A+B) > \max(A,B)$, i.e. $A$ and $B$ are the two angles of a triangle not opposite the longest side.
Clearly, any $A,B\:|\:A+B=\frac{\pi}{2}$ is a family of solutions. Since multivariable calculus is presumably beyond the level of the original problem:
How to prove that there are no other solutions $\underline{\text{without}}$ using multivariable calculus?
[I don't think a trig identity will suffice as there are other solutions if the restrictions on $A,B$ are relaxed.]
(For completeness - using multivariable calculus)
Part 1
Proof that $\sin^2A + \sin^2B < \sin (A+B)$ over region $R_1=\{0<A,B\land A+B<\frac{\pi}{2}\}$.
Consider $$f(x,y)=\sin^2x + \sin^2y - \sin (x+y)$$
Then
$$\begin{align}
f_x &= \sin 2x - \cos(x+y) \\
f_y &= \sin 2y - \cos(x+y) \\
f_{xx} &= 2\cos 2x + \sin(x+y) \\
f_{yy} &= 2\cos 2y + \sin(x+y) \\
f_{xy} &= \sin(x+y) &= f_{yx} \\
\end{align}$$
For local extrema we require $f_x=f_y=0$. But
$$f_x=f_y=0 \implies \sin2x=\sin2y \implies y=x \lor y=\frac{\pi}{2}-x$$
Exclude $y=\frac{\pi}{2}-x$ as it violates the restriction that $x+y<\frac{\pi}{2}$.
If $y=x$, $f_x=0 \implies \sin2x=\sin2y \implies x=y=\frac{\pi}{8}$.
The determinant of the Hessian is
$$D(x,y) = \begin{vmatrix} f_{xx} & f_{xy}\\ f_{yx} & f_{yy} \end{vmatrix} = f_{xx}f_{yy} - f_{xy}f_{yx} = f_{xx}f_{yy} - (f_{xy})^2$$
At $P(\frac{\pi}{8},\frac{\pi}{8})$, we have
$$\begin{align}
f_{xx}&=2\cos\frac{\pi}{4}+\sin\frac{\pi}{4}=\frac{3}{\sqrt2} \\
f_{yy}&=2\cos\frac{\pi}{4}+\sin\frac{\pi}{4}=\frac{3}{\sqrt2} \\
f_{xy}&=\sin\frac{\pi}{4}=\frac{1}{\sqrt2} \\
D(x,y)&=\frac{9}{2}-\frac{1}{2}=4
\end{align}$$
Since both $f_{xx}$ and $D$ are positive at $P$, this is a local minimum (with $f(x,y)|_P=1-\sqrt2$).
On the boundaries:
*
*$f(0,y)=\sin^2y-\sin y<0$ on $x=0,y\in(0,\frac{\pi}{2})$
*$f(x,0)=\sin^2x-\sin x<0$ on $x\in(0,\frac{\pi}{2}),y=0$
*$f(0,0)=0$
*$f(x,y)=0$ for $x,y\geq0,x+y=\frac{\pi}{2}$
Since $(0,0)$ is not part of the domain, and there are no other local extrema, we have $f(x,y)<0$ over $R_1$.
Part 2
It will be sufficient to prove that $\sin^2A + \sin^2B > \sin (A+B)$ over region $R_2=\{0<A,B<\frac{\pi}{2} \land A+B>\frac{\pi}{2} \}$.
|
Here are some illustrations, with $\triangle ABC$ having circumdiameter $1$ (and therefore sides $\sin A$, $\sin B$, $\sin C$):
Acute $C$: $\quad \sin C < \sin^2 A + \sin^2 B$
Obtuse $C$: $\quad \sin C > \sin^2 A + \sin^2 B$
Right $C$: $\quad \sin C = \sin^2 A + \sin^2 B$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
"answer_count": 3,
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|
Partial fractions problem help I need help with the following:
$\frac{1}{(x-1)^2(x+1)}$
Using the "cover-up" method, I can identify the numerator of the fraction with the denominator $(x+1)$ fairly quickly, and it is $1/4$. Then I arrive at the following: $$1 \equiv (1/4)(x-1)^2 + B(x-1)(x+1)+(Dx+E)(x+1)$$
Then I have three equations with two unknowns in each, a total of three unknowns across three equations. Which is not solvable. What do I do next after the step above?
|
Let $\displaystyle \frac{1}{(x-1)^2\cdot (x+1)} = \frac{A}{(x-1)}+\frac{B}{(x-1)^2}+\frac{C}{(x+1)}$
$\displaystyle 1 = A(x-1)\cdot (x+1)+B(x+1)+C(x-1)^2...........(1)$
Put $(x-1) = 0\Rightarrow x=1\;,$ in above equation. So we get $\displaystyle 1 = 2B\Rightarrow B = \frac{1}{2}$
Similarly Put $(x+1) = 0\Rightarrow x=-1\;,$ in above equation, So We get $\displaystyle 1 = 4C\Rightarrow C = \frac{1}{4}$
Now Equating Coefficient of constant term in $(1)$ equation.
So Put $x=0$ on both side, We get $\displaystyle 1=-A+B+C\Rightarrow 1=-A+\frac{1}{2}+\frac{1}{4}$
so we get $\displaystyle A = -\frac{1}{4}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1410484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Find all pair of cubic equations Find all pair of cubic equations $x^3+ax^2+bx+c=0$ and $x^3+bx^2+ax+c=0$, where $a,b$ are positive integers and $c$ not equal to $0$ is an integer, such that both the equations have three integer roots and exactly one of those three roots is common to both the equations.
I tried the sum and product relationships with the coefficients, but I have more variables than the number of equations.All I got was that the common root has to be $x=1$. What do I next?Thanks.
|
We may assume $a>b\geq1$. An $x$ that solves both equations is $\ne0$ and also solves the equation
$$(a-b)(x^2-x)=0\ .$$
This implies $x=1$ and entails $c=-1-a-b$. Deflating the polynomials in question by the factor $x-1$ leaves us with the pair of equations
$$\left.\eqalign{x^2+(a+1)x+a+b+1&=0\cr y^2+(b+1)y+a+b+1&=0\cr}\right\}$$
that should have altogether four different pairwise integer solutions $\ne1$. Let $-r$, $-s$ be the solutions of the first of these equations, and $-u$, $-v$ be the solutions of the second. It follows that
$$\eqalign{r+s&=a+1,\qquad rs=a+b+1\geq4\cr
u+v&=b+1,\qquad uv=a+b+1\geq4\ .\cr}\tag{1}$$
It follows that all four quantities $r$, $s$, $u$, $v$ are $>0$. We may assume $r>s$, $\>u>v$.
We now consider $r$ and $s$ as given and have to check which pairs $(r,s)$ are admissible. From the upper line $(1)$ we obtain that
$$a=r+s-1,\qquad b=rs-r-s\ ,$$
so that the second line $(1)$ gives
$$\eqalign{u+v&=rs-r-s+1=(r-1)(s-1)\ ,\cr
uv&=rs\ .\cr}\tag{2}$$
The condition $u+v=b+1\geq2$ then implies $r>s\geq2$.
The equatons $(2)$ imply that $u$ and $v$ are the solutions $t_1$, $t_2$ of the quadratic equation
$$t^2-(rs-r-s+1) t+rs=0\tag{3}$$
with discriminant
$$D:=(rs-r-s+1)^2-4rs=(rs-r-s-1)^2-4(r+s)\ .\tag{4}$$
We want that $D$ is a square. The number $D$ can only be a square if $D\leq(rs-r-s-2)^2$. Now the condition
$$(rs-r-s-1)^2-4(r+s)\leq(rs-r-s-2)^2$$
simplifies to $2(r-3)(s-3)\leq21$, or
$$(r-3)(s-3)\leq10\ .$$
This leaves us with the cases (i) $r>s=2$, (ii) $r>s=3$ and $${\rm (iii):}\qquad 4\leq s<r\leq 3+{10\over s-3}\ .\tag{5}$$ In the case (i) we obtain from $(4)$ that
$D=(r-5)^2-24$. Since $D$ has to be a square this means that we have to represent $24$ as a sum of subsequent odd numbers. There are two such representations, namely $24=3+5+7+9$ and $24=11+13$, leading to the admissible pairs $(10,2)$ and $(12,2)$.
In the case (ii) we obtain from $(4)$ that $D=(2r-5)^2-21$. This means that we have to represent $21$ as a sum of subsequent odd numbers. There are two such representations, namely $21=5+7+9$ and $21=21$, leading to the admissible pairs $(5,3)$, $(8,3)$.
In the case (iii) the conditions $(5)$ admit $s=4$ and $r\in[5..13]$, and $s=5$ and $r\in[6..8]$. Checking the cases reveils that only $(5,4)$ is an admissible pair.
All in all we have found five admissible pairs $(r,s)$, namely
$$(10,2),\quad(12,2),\quad (5,3),\quad(8,3),\quad (5,4)\ .$$
The corresponding solutions $\{u,v\}$ of $(3)$ are then
$$\{5,4\},\quad\{8,3\},\quad\{5,3\},\quad\{12,2\},\quad\{10,2\}\ .$$
Since we want $r$, $s$, $u$, $v$ all different we have to discard the pair $(r,s)=(5,3)$, and we are left with essentially two solutions to the original problem, corresponding to the pairs $(r,s)=(10,2)$ and $(r,s)=(12,2)$. From $(1)$ and $c=-1-a-b$ we then obtain the coefficient triples
$$(a,b,c)=(11,8,-20)\quad{\rm and}\quad (a,b,c)=(13,10,-24)\ .$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1412817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
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|
The differential equation $\frac{dy}{dx} = \frac{y}{x} - \frac{1}{y}\;$ I am learning differential equations and can do the basic examples. However, how can you solve the differential equation
$$\frac{dy}{dx} = \frac{y}{x} - \frac{1}{y}\;?$$
|
$$\frac { dy }{ dx } =\frac { y }{ x } -\frac { 1 }{ y } \;$$ substitute
$$y=xt$$
$$ \frac { dy }{ dx } =t+x\frac { dt }{ dx } \\ t+x\frac { dt }{ dx } =t-\frac { 1 }{ xt } \\ x\frac { dt }{ dx } =-\frac { 1 }{ xt } \\ \int { tdt } =\int { -\frac { d }{ { x }^{ 2 } } } \\ \frac { { t }^{ 2 } }{ 2 } =\frac { 1 }{ x } +C\\ \frac { { y }^{ 2 } }{ { x }^{ 2 } } =\frac { 2 }{ x } +C\\ y=\pm \sqrt { 2x+{ x }^{ 2 }C } $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1414760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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|
In the triangle $ABC$, if $a=\frac{2(b^2-c^2)}{-b+\sqrt{b^2+4c^2}}$, prove that $3\cdot\widehat{C}=2\cdot\widehat{B}$. Just like in the title, I have to prove that if in a triangle $ABC$
$$a=\frac{2(b^2-c^2)}{-b+\sqrt{b^2+4c^2}}$$
holds, then $3\cdot\widehat{C}=2\cdot\widehat{B}$.
The denominator of the big fraction looks like the positive solution
of a quadratic equation, specifically $x^2 + bx - c^2 = 0$, so the claim appears to be that if $ x = \frac{b^2-c^2}{a}$ is a solution of that equation, then $\frac{\widehat{C}}{\widehat{B}}=\frac{2}{3}$.
I have not gone beyond this examination of the problem itself, but
perhaps this will give you some ideas to try.
|
Assuming $\widehat{ C}=2\theta, \widehat{B}=3\theta,\widehat{A}=\pi-5\theta$ and $R=1$ (the circumradius) we have:
$$ a = 2\sin(5\theta),\quad b=2\sin(3\theta),\quad c=2\sin(2\theta) $$
and $\frac{b^2-c^2}{a}=2\sin\theta$. On the other hand,
$$ \sin^2(\theta) + \sin(3\theta)\sin(\theta)-\sin^2(2\theta)=0.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1416966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Trying to show $|\overrightarrow{a}\times\overrightarrow{b}|^2=|\overrightarrow{a}|^2|\overrightarrow{b}|^2-(\overrightarrow{a}⋅\overrightarrow{b})^2$
If $\overrightarrow{a} = \langle a_1, a_2, a_3 \rangle$ and $\overrightarrow{b} = \langle b_1, b_2, b_3 \rangle$, then the cross product of $\overrightarrow{a}$ and $\overrightarrow{b}$ is the vector
$$\overrightarrow{a} \times \overrightarrow{b} =(a_2 b_3 - a_3 b_2)\overrightarrow{i} + (a_3 b_1 - a_1 b_3)\overrightarrow{j} + (a_1 b_2 - a_2 b_1)\overrightarrow{k}$$
Using the above definition show algebraically that
$$|\overrightarrow{a} \times \overrightarrow{b}|^2=|\overrightarrow{a}|^2|\overrightarrow{b}|^2-(\overrightarrow{a}\cdot\overrightarrow{b})^2$$
I'm not sure if I'm on the right track with this problem, but I've started with the given expression
$\overrightarrow{a} \times \overrightarrow{b} =(a_2 b_3 - a_3 b_2)\overrightarrow{i} + (a_3 b_1 - a_1 b_3)\overrightarrow{j} + (a_1 b_2 - a_2 b_1)\overrightarrow{k}$
with the goal of algebraically manipulating it into the form
$|\overrightarrow{a}|^2|\overrightarrow{b}|^2-(\overrightarrow{a}\cdot\overrightarrow{b})^2$
Consider the steps:
$|\overrightarrow{a} \times \overrightarrow{b}|^2$
$|\langle a_2 b_3 - a_3 b_2, a_3 b_1 - a_1 b_3, a_1 b_2 - a_2 b_1 \rangle|^2$
$\sqrt{(a_2 b_3 - a_3 b_2)^2 + (a_3 b_1 - a_1 b_3)^2 + (a_1 b_2 - a_2 b_1)^2}^2$
$(a_2 b_3 - a_3 b_2)^2 + (a_3 b_1 - a_1 b_3)^2 + (a_1 b_2 - a_2 b_1)^2$
which expands to:
$(a_3^2 b_2^2 - 2 a_2 a_3 b_3 b_2 + a_2^2 b_3^2) + (a_3^2 b_1^2 - 2 a_1 a_3 b_3 b_1 + a_1^2 b_3^2) + (a_2^2 b_1^2 - 2 a_1 a_2 b_2 b_1 + a_1^2 b_2^2)$
but I'm not really sure where to go from here. I could group the squared terms together:
$a_3^2 b_2^2 + a_2^2 b_1^2 + a_2^2 b_3^2 + a_3^2 b_1^2 + a_1^2 b_3^2 + a_1^2 b_2^2 - 2 a_2 a_3 b_3 b_2 - 2 a_1 a_3 b_3 b_1 - 2 a_1 a_2 b_2 b_1$
but no recognizable forms are really achieved (like the expanded expression for a dot product, $\overrightarrow{a} \cdot \overrightarrow{b} = a_1 b_1 + a_2 b_2 + a_3 b_3$).
How I algebraically prove this? Thanks for your help!
|
Having grouped your terms together as
$$a_3^2 b_2^2 + a_2^2 b_1^2 + a_2^2 b_3^2 + a_3^2 b_1^2 + a_1^2 b_3^2 + a_1^2 b_2^2 - 2 a_2 a_3 b_3 b_2 - 2 a_1 a_3 b_3 b_1 - 2 a_1 a_2 b_2 b_1$$
we can add and subtract some extra terms (highlighted in colour below):-
$$\begin{align}&a_3^2 b_2^2 + a_2^2 b_1^2 + a_2^2 b_3^2 + a_3^2 b_1^2 + a_1^2 b_3^2 + a_1^2 b_2^2 - 2 a_2 a_3 b_3 b_2 - 2 a_1 a_3 b_3 b_1 - 2 a_1 a_2 b_2 b_1\\&\color{red}{+a_1^2b_1^2+a_2^2b_2^2+a_3^2b_3^2\color{blue}{-a_1^2b_1^2-a_2^2b_2^2-a_3^2b_3^2}}\\&=(a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2)-(a_1b_1+a_2b_2+a_3b_3)^2\\&=|\overrightarrow{a}|^2|\overrightarrow{b}|^2-(\overrightarrow{a}\cdot\overrightarrow{b})^2\end{align}$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Confusing probability problems based on product rule and combinations I am going thru probability exercise. Faced first problem:
Book Q1. Ten tickets are numbered 1,2,3,...,10. Six tickets are selected at random one at a time with replacement. What is the probability the largest number appearing on the selected tickets is 7?
My logic: if one of six tickets should be 7, the $\color{red}{\text{remaining 5}}$ can be any of 1 to 7, so it should be $7^5$.
But turns out that the given solution is $\frac{7^6-6^6}{10^6}$.
My Q1. Though I understood the logic behind $\frac{7^6-6^6}{10^6}$, I was wondering what is exact logical mistake with $7^5$? I guessed that $7^5$ completely ignores what should be 6th ticket, it only puts restriction on 5 tickets. Is it like that?
Then I came across similar but more involved problem, with significant difference from above one that it performs action without replacement:
Book Q2. Three numbers are chosen at random without replacement from (1,2,3,...,10). What is the probability that the minimum number is 3 or the maximum number is 7?
My logic: Noticing that this is without replacement, I guessed the solution should be
$$
=
\begin{pmatrix}
\text{selections with}\\
\text{minimum}\\
\text{number is 3}
\end{pmatrix}
+
\begin{pmatrix}
\text{selections with}\\
\text{maximum}\\
\text{number is 7}
\end{pmatrix}
-
\begin{pmatrix}
\text{selections with}\\
\text{maximum}\\
\text{number is 7}\\
\text{and minimum}\\
\text{number is 3}
\end{pmatrix}
$$
$$=
\frac{
\overbrace{(\binom{8}{3}-\binom{7}{3})}^{\text{#selections with min 3}}
+
\overbrace{(\binom{7}{3}-\binom{6}{3})}^{\text{#selections with max 7}}
-
\overbrace{3\times {^3P_3}}^{\text{#selections with max 7 and min 3}}
}
{\binom{10}{3}
}
$$
But the book solutions says:
P(minimum 3) or P(maximum 7)
P(minimum 3) $=\frac{\binom{7}{2}}{\binom{10}{3}}=\frac{21}{120}$
P(max 7) $=\frac{\binom{6}{2}}{\binom{10}{2}}=\frac{15}{120}$
Thus the solution is $\frac{11}{40}$
My Q2. How even by books logic the solution $\frac{11}{40}$ is achieved. I am unable to understand it as I find the explanation insufficient.
My Q3. If book Q2 answer is correct then why for book Q1 solution is not $7^5$ which is what I initially guessed (because the only difference being with / without replacement, the logic of getting $\color{red}{\text{remaining m}}$ stuffs out n should remain same)?
My Q4. If we make first question without replacement, will the solution be $\frac{\binom{7}{6}-\binom{6}{6}}{\binom{10}{6}}$?
My Q5. What will be the solution if we make book Q2 with replacement?
My Q6. Where my logic for solution to Book's Q2 is wrong?
|
With the actual answers already given, I am putting different possible scenarios and there solutions here for reference.
*
*Selecting six out of {1,2,3,...,10} with replacement. Probability of largest selected number is 7 = ?
Sol.
$$
\overbrace
{\frac{7^6-6^6}{10^6}}
^{
\begin{pmatrix}
\text{no. of ways of selecting any six out of {1,..,7}} \\
\text{removing those not containing 7}\\
\text{i.e. those six from {1,..,6}}\\
\end{pmatrix}
}=0.070993$$
*Selecting six out of {1,2,3,...,10} without replacement while taking order of selection into account. Probability of largest selected number is 7 = ?
Sol.
$$
\frac{
\overbrace{\binom{7}{6}}
^{\begin{pmatrix}
\text{no. of ways to} \\
\text{select six from {1,..,7}} \\
\text{without considering} \\
\text{order} \\
\end{pmatrix}}
-
\overbrace{\binom{6}{6}}
^{\begin{pmatrix}
\text{no. of ways to} \\
\text{select six from {1,..,6}} \\
\text{without considering} \\
\text{order} \\
\end{pmatrix}}
}{
\binom{10}{6}
}=
\frac{
\overbrace{\binom{6}{5}}
^{\begin{pmatrix}
\text{selecting six from {1,..,7}} \\
\text{with 7 as max reduces to} \\
\text{selecting 5 from {1,..,6}} \\
\text{since you know you must} \\
\text{have 7} \\
\end{pmatrix}}
}{\binom{10}{6}}=\frac{1}{35}=0.028571
$$
*Selecting six out of {1,2,3,...,10} without replacement while not taking order of selection into account. Probability of largest selected number is 7 = ?
Sol.
$$
\frac
{\overbrace
{^6P_5}
^{
\begin{pmatrix}
\text{no. of ways to select five} \\
\text{out of {1,..,6} while} \\
\text{considering order} \\
\end{pmatrix}
}
\times
\overbrace
{6}
^{
\begin{pmatrix}
\text{putting 7} \\
\text{in one of 6 places} \\
\end{pmatrix}
}
}{^{10}P_6}
$$
*Selecting three out of {1,2,3,...,10} with replacement. Probability of largest selected number is 7 and smallest is 3 = ?
Sol.
$$\frac{(8^3-7^3)=(7^3-6^3)-18}{10^3}=0.278$$
For details, see correction in Brian Scott's solution.
*Selecting three out of {1,2,3,...,10} without replacement while taking order of selection into account. Probability of largest selected number is 7 and smallest is 3 = ?
Sol.
$$
\frac
{\overbrace
{^7P_2\times 3}
^{
\begin{pmatrix}
\text{no. of ways to select two} \\
\text{out of {4,..,10}} and \\
\text{put 3 in any of three places} \\
\text{making 3 min number among} \\
\text{three selected} \\
\end{pmatrix}
}
+
\overbrace
{^6P_2\times 3}
^{
\begin{pmatrix}
\text{no. of ways to select two} \\
\text{out of {1,..,6} put 7} and \\
\text{put 7 in any of three places} \\
\text{making 7 max number among} \\
\text{three selected} \\
\end{pmatrix}
}
-
\overbrace
{^3P_3 \times 3}
^{
\begin{pmatrix}
\text{Selecting 3 tickets with both} \\
\text{min 3 and max 7:} \\
\text{first select any one of} \\
\text{4,5 or 6,then perform} \\
^3P_3\text{ arrangements with the} \\
\text{selected number,3 & 7} \\
\end{pmatrix}
}
}{^{10}P_3}
= \frac{198}{720} = 0.275
$$
*Selecting three out of {1,2,3,...,10} without replacement while not taking order of selection into account. Probability of largest selected number is 7 and smallest is 3 = ?
Sol.
$$
\frac
{\overbrace
{\binom{8}{3}-\binom{7}{3}}
^{
\begin{pmatrix}
\text{no. of selections} \\
\text{with 3 as minimum} \\
\end{pmatrix}
}
+
\overbrace
{\binom{7}{3}-\binom{6}{3}}
^{
\begin{pmatrix}
\text{no. of selections} \\
\text{with 7 as maximum} \\
\end{pmatrix}
}
-
\overbrace
{3}
^{
\begin{pmatrix}
\text{no. of selections with both} \\
\text{3 as min & 7 as max:} \\
\text{{3,7,4},{3,7,5},{3,7,6}} \\
\end{pmatrix}
}
}{\binom{10}{3}}
=\frac{\binom{7}{3}+\binom{6}{2}-3}{\binom{6}{2}}
=\frac{11}{40}
=0.275
$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1420232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Solving the infinite series $1-\frac{2^3}{1!}+\frac{3^3}{2!}-\frac{4^3}{3!}+\cdots$ I have the following question:
Evaluate the infinite series:
$$S=1-\frac{2^3}{1!}+\frac{3^3}{2!}-\frac{4^3}{3!}+\cdots$$
(a) $\displaystyle\frac1e$ (b) $\displaystyle\frac{-1}e$ (c) $\displaystyle\frac{2}e$ (d) $\displaystyle\frac{-2}e$
Now in the book they have given a strange explanation using $$(n+1)^3=[n(n-1)(n-2)+6n(n-1)+7n+1].$$ I don't understand how did they get this. So I have two doubts:
*
*How did they get this "trick"?
*If you would have got this question in a competitive exam, how would you have solved it (in the minimum amount of time), either directly or with numerical methods?
|
Looking at the given identity:
\begin{align}
(n+1)^3 &= [n(n-1)(n-2)+6n(n-1)+7n+1] \\
&= n(n^2 - 3n + 2) + 6 n^2 + n + 1 \\
&= n^3 + 3 n^2 + 3 n + 1 = (n+1)^3
\end{align}
Now, consider differentiation of the exponential series:
\begin{align}
D \, e^{t} &= \sum_{n=0}^{\infty} \frac{1}{n!} \, \frac{d}{dt} \, t^{n} =
\sum_{n=0}^{\infty} \frac{n \, t^{n-1}}{n!}
\end{align}
Further differentiation can be applied.
Now, consider the series in question:
\begin{align}
S &= 1 - \frac{2^{3}}{1!} + \frac{3^{3}}{2!} - \cdots \\
&= \sum_{n=0}^{\infty} \frac{(-1)^{n} \, (n+1)^{3}}{n!} \\
&= \left. \sum_{n=0}^{\infty} \frac{(-1)^{n} \, (n+1)^{3} \, t^{n}}{n!} \, \right|_{t=1} \\
&= \left. t^{3} \, \sum_{n=0}^{\infty} \frac{(-1)^{n} \, n(n-1)(n-2) \, t^{n-3}}{n!} + 6 t^{2} \, \sum_{n=0}^{\infty} \frac{(-1)^{n} \, n(n-1) \, t^{n-2}}{n!} + 7 t \, \sum_{n=0}^{\infty} \frac{(-1)^{n} \, n \, t^{n-1}}{n!} + e^{-t} \, \right|_{t=1} \\
&= \left. t^3 \, D^{3} e^{-t} + 6 \, t^{2} \, D^{2} e^{-t} + 7 \, t \, D e^{-t} + e^{-t} \right|_{t=1} \\
&= \left. (-t^3 + 6 \, t^2 - 7 \, t + 1) \, e^{-t} \right|_{t=1} \\
&= - \frac{1}{e}
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1423107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
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|
Calculating $\int_0^1 \frac{\log (x) \log \left(\frac{1}{2} \left(1+\sqrt{1-x^2}\right)\right)}{x} \, dx$ How would you like to calculate this one? Do you see a fast, neat way here? Ideas?
$$\int_0^1 \frac{\log (x) \log \left(\frac{1}{2} \left(1+\sqrt{1-x^2}\right)\right)}{x} \, dx$$
Sharing solutions is only optional.
The closed form revealed is
$$\frac{1}{4} \left(\frac{2 }{3}\log ^3(2)-\zeta(2) \log (2)+\zeta (3)\right).$$
|
Here is an alternative solution.
\begin{align}
\int^1_0\frac{\ln{x}\ln\left(\frac{1+\sqrt{1-x^2}}{2}\right)}{x}\ {\rm d}x
&=\frac{1}{4}\int^1_0\frac{\ln{x}\ln\left(\frac{1+\sqrt{1-x}}{2}\right)}{x}\ {\rm d}x\tag1\\
&=\frac{1}{4}\int^1_0\frac{\ln(1-x)}{1-x}\ln\left(\frac{1+\sqrt{x}}{2}\right)\ {\rm d}x\tag2\\
&=\frac{1}{16}\int^1_0\frac{\ln^2(1-x)}{\sqrt{x}(1+\sqrt{x})}\frac{1-\sqrt{x}}{1-\sqrt{x}}\ {\rm d}x\tag3\\
&=-\frac{1}{96}\int^1_0x^{-3/2}\ln^3(1-x)\ {\rm d}x\tag4\\
&=\frac{1}{48}\lim_{q\to 1}\frac{\partial^3}{\partial q^3}\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(q\right)}{\Gamma\left(q-\frac{1}{2}\right)}\tag5
\end{align}
Explanation:
$(1)$: Substituted $x\mapsto\sqrt{x}$.
$(2)$: Substituted $x\mapsto 1-x$.
$(3)$: Integrated by parts.
$(4)$: Integrated by parts.
$(5)$: Used the integral representation of the Beta function.
Using Wolfram Alpha (or differentiating by hand),
Setting $q=1$ gives us the required result.
$$\int^1_0\frac{\ln{x}\ln\left(\frac{1+\sqrt{1-x^2}}{2}\right)}{x}\ {\rm d}x=\frac{\zeta(3)}{4}-\frac{\pi^2}{24}\ln{2}+\frac{\ln^3{2}}{6}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find 2 sums with the binomial newton Find the sum of:
i)$\displaystyle\sum_{k=0}^{n} k^2$ $\left(\begin{array}{c} n\\k\end{array}\right)$
ii) $\displaystyle\sum_{k=1}^{n} \frac{2k+5}{k+1}$$\left(\begin{array}{c} n\\k\end{array}\right)$
Thoughts:
i)(After the Edit)$\displaystyle\sum_{k=0}^{n} k^2$ $\left(\begin{array}{c} n\\k\end{array}\right)$ =
$\displaystyle\sum_{k=0}^{n} {k(k-1)}$ $\left(\begin{array}{c} n\\k\end{array}\right)$ + $\displaystyle\sum_{k=0}^{n}{k}$ $\left(\begin{array}{c} n\\k\end{array}\right)$=
$\displaystyle\sum_{k=0}^{n} {n(k-1)}$ $\left(\begin{array}{c} n-1\\k-1\end{array}\right)$ + $\displaystyle\sum_{k=0}^{n}{n}$ $\left(\begin{array}{c} n-1\\k-1\end{array}\right)$=
$\displaystyle\sum_{k=0}^{n} {n(n-1)}$ $\left(\begin{array}{c} n-2\\k-2\end{array}\right)$ + n$\displaystyle\sum_{k=0}^{n}$ $\left(\begin{array}{c} n-1\\k-1\end{array}\right)$=
$n(n-1)2^{n-2}+n2^{n-1}$=$n2^{n-2}(2+n-1)$=$n(n+1)2^{n-2}$
ii)(After the Edit)$\displaystyle\sum_{k=1}^{n} \frac{2k+5}{k+1}$$\left(\begin{array}{c} n\\k\end{array}\right)$=
$\displaystyle\sum_{k=1}^{n} \frac{(2k+2)+3}{k+1}$$\left(\begin{array}{c} n\\k\end{array}\right)$=
2$\displaystyle\sum_{k=1}^{n}$$\left(\begin{array}{c} n\\k\end{array}\right)$ + 3$\displaystyle\sum_{k=1}^{n} \frac{1}{k+1}$$\left(\begin{array}{c} n\\k\end{array}\right)$=
$2^{n+1}-2$+ $\frac{3}{n+1}$$\displaystyle\sum_{k=1}^{n} \frac{n+1}{k+1}$$\left(\begin{array}{c} n\\k\end{array}\right)$=
$2^{n+1}-2$+ $\frac{3}{n+1}$$\displaystyle\sum_{k=1}^{n}$$\left(\begin{array}{c} n+1\\k+1\end{array}\right)$=
$2^{n+1}-2+\frac{3}{n+1}(2^{n+1}-1)$
|
Hint:
ii) We can expand as follows
\begin{align*}
\sum_{k=1}^n\frac{2k+5}{k+1}{n\choose k}&=\sum_{k=1}^n\frac{(2k+2)+3}{k+1}{n\choose k}\\[4pt]
&=\sum_{k=1}^n 2{n\choose k}+\frac{3}{n+1}\sum_{k=1}^n\frac{n+1}{k+1}{n\choose k}\\[4pt]
&=2\sum_{k=1}^n{n\choose k}+\frac{3}{n+1}\sum_{k=1}^n{{n+1}\choose {k+1}}
\end{align*}
Notice $$\sum_{k=0}^n{n\choose k}=2^n\qquad\implies\qquad\sum_{k=1}^n{n\choose k}=2^n-1,$$
then,
\begin{align*}
\sum_{k=1}^n\frac{2k+5}{k+1}{n\choose k}&=2\sum_{k=1}^n{n\choose k}+\frac{3}{n+1}\sum_{k=1}^n{{n+1}\choose {k+1}}=2(2^n-1)+\frac{3}{n+1}(2^{n+1}-1)
\end{align*}
Which can be reduced as
$$\sum_{k=1}^n\frac{2k+5}{k+1}{n\choose k}=\frac{2^{n+1}(n+4)-2n-5}{n+1}$$
|
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|
Minimal polynomial and algebraic multiplicy
Let $A\in \mathbb{R}^{4 \times 4}$ and the Minimal polynomial $m_A(x)=x^3-2x^2$
What are the eigenvalues and the multiplicy?
$m_A(x)=x^3-2x^2=x^2(x-2)$ so $\lambda_1=0$ and $ \lambda_2=2$ so we now look at the option of the characteristic polynomial we know that it must be formed by $x$ and $(x-2)$ with at a degree that at most is 4. So it can be $x^2(x-2)^2$ so the multiplicity of $\lambda_1=0$ is 2 and $\lambda_2=2$ is 2
Why can the multiplicity be $\lambda_1=0$ is 3 and $\lambda_2=2$ is 1?
|
Because $x^{2}(x-2)$ is the minimal polynomial, then the Jordan canonical form can be built from the following blocks along the diagonal:
$$
\left[2\right], \left[0\right], \left[\begin{array}{cc} 0 & 1 \\ \cdot & 0\end{array}\right]
$$
You must have the first and third blocks, which, together, occupy 3 of the 4 diagonal entries. So there are two possibilities:
$$
\left[\begin{array}{cccc}0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 2 & 0 \\
0 & 0 & 0 & 2 \end{array}\right],
\left[\begin{array}{cccc}0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 2 \end{array}\right].
$$
|
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|
Proof by induction on n So I aim to prove that $n^2 \leq 2^n + 1$ for all integers $n \geq 1$. We can see that this is true for $n=1$ since $1 \leq 3$. Now I suppose that this is true for an arbitrary $k$ such that $k \geq 1$. So $k^2 \leq 2^k +1$. From this, I want to reach a statement $(k+1)^2 \leq 2^{k+1} +1$ by algebraic manipulation. But it seems harder than I thought. For example, I expanded the "goal" statement so I get $k^2 + 2k +1 \leq 2^{k+1} +1$, which looks easier to get to from the original statement $k^2 \leq 2^k +1$. I tried a lot of algebraic manipulations and also tried the transitive property of order relation, but still cannot figure out how to get there. Any suggestions?
|
Suppose that for some $n\ge3$, we have $n^2\le 2^n+1$. Then
$$
\begin{align}
(n+1)^2
&=n^2+2n+1\tag{1}\\
&=2n^2-2-(n+1)(n-3)\tag{2}\\
&\le2n^2-2\tag{3}\\
&\le2^{n+1}\tag{4}\\
&\lt2^{n+1}+1\tag{5}
\end{align}
$$
Explanation:
$(1)$: expand the square
$(2)$: $n^2+2n+1=2n^2-2-(n^2-2n-3)$
$(3)$: since $(n+1)(n-3)\ge0$ for $n\ge3$
$(4)$: $2n^2-2\le2(2^n+1)-2=2^{n+1}$
$(5)$: $x\lt x+1$
All that is left is to verify that $n^2\le2^n+1$ for $0\le n\le3$.
|
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|
Prove that the family of curves $\frac{x^2}{a^2+\lambda}+\frac{y^2}{b^2+\lambda}=1$,where $\lambda$ is a parameter,is self orthogonal. Prove that the family of curves $\frac{x^2}{a^2+\lambda}+\frac{y^2}{b^2+\lambda}=1$,where $\lambda$ is a parameter,is self orthogonal.
I tried to find the differential equation of first order(because there is one parametr $\lambda$) corresponding to $\frac{x^2}{a^2+\lambda}+\frac{y^2}{b^2+\lambda}=1$ and then replace $\frac{dy}{dx}$ by $\frac{-dx}{dy}$ and then re-integrating it but the calculations has gone messy.
|
Let us find a differential equation not involving $\lambda$ that describes the situation. Then we show that it is invariant under the change of $y'\mapsto -\frac{1}{y'}$.
Differentiating,
$$
\frac{2x}{a^2+\lambda}+\frac{2yy'}{b^2+\lambda}=0.
$$
Then, using the equation and the differentiated one, we find that
$$
a^2+\lambda=\frac{x^2y'-xy}{y'},\quad\text{and}\quad b^2+\lambda=y^2-xyy'.
$$
We eliminate $\lambda$ and find that
$$
a^2-b^2=\frac{x^2y'-xy}{y'}-y^2+xyy',
$$
which implies that the general differential equation is
$$
xy(y')^2+(x^2-y^2-a^2+b^2)y'-xy=0.
$$
If we substitute $y'$ by $-1/y'$, we get
$$
xy(-1/y')^2+(x^2-y^2-a^2+b^2)(-1/y')-xy=0,
$$
which after multiplication with $-(y')^2$ gives back
$$
-xy+(x^2-y^2-a^2+b^2)y'+xy(y')^2=0.
$$
The self-orthogonality follows.
|
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|
What are better approximations to $\pi$ as algebraic though irrational number? I know that $\pi \approx \sqrt{10}$, but that only gives one decimal place correct. I also found an algebraic number approximation that gives ten places but it's so cumbersome it's just much easier to just memorize those ten places.
What's a good approximation to $\pi$ as an irrational algebraic number (or algebraic integer if possible) that is easier to memorize than the number of places it gives correct?
EDIT: Algebraic number preferably of low degree, such as $2$ or $3$ (quadratic or cubic).
|
Dalzell's integral is related to the rational approximation $\pi\approx \frac{22}{7}$.
$$\pi=\frac{22}{7}-\int_0^1 \frac{x^4(1-x)^4}{1+x^2}dx\approx\frac{22}{7}$$
Similar small integrals are related to simple irrational approximations using $\sqrt{2}$ and $\sqrt{3}$.
$$\pi=\frac{20\sqrt{2}}{9}-\frac{2\sqrt{2}}{3} \int_0^1 \frac{x^4(1-x)^4}{1+x^2+x^4+x^6}dx\approx \frac{20\sqrt{2}}{9}$$
$$\pi=\frac{9\sqrt{3}}{5}+\frac{6\sqrt{3}}{5}\int_0^1\frac{x^3(1-x)^2}{1+x^2+x^4}dx\approx \frac{9\sqrt{3}}{5}
$$
Fractions $\frac{20}{9}$ and $\frac{9}{5}$ are convergents of $\frac{\pi}{\sqrt{2}}$ and $\frac{\pi}{\sqrt{3}}$ respectively.
|
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|
How do we bound the function $\frac{-1}{2}\sum_{m=n}^{\infty}\frac{1}{m^{2}}$? In case the question didn't display in the title correctly: How do we bound the function $\frac{-1}{2}\sum_{m=n}^{\infty}\frac{1}{m^{2}}$?
I think a way I can do this is to show that $$\sum_{m=n}^{\infty}\frac{1}{m^{2}} < \frac{1}{n^{2}} + \int_{n}^{\infty}\frac{1}{x^{2}}$$
But I'm a little unsure of the process it takes to get there.
|
Estimating via the integral works well. Another way is to note that the tail (if we forget about the $-\frac{1}{2}$ in front) is less than
$$\frac{1}{(n-1)(n)}+\frac{1}{(n)(n+1)}+\frac{1}{(n+1)(n+2)}+\cdots.$$
But $\frac{1}{(n-1)(n)}=\frac{1}{n-1}-\frac{1}{n}$ and the next term is $\frac{1}{n}-\frac{1}{n+1}$, and the next term is $\frac{1}{n+1}-\frac{1}{n+2}$, and so on. Now note the massive cancellation (telescoping). We get $\frac{1}{n-1}$.
We conclude that the tail is less than $\frac{1}{n-1}$. A similar argument shows that the tail is greater than $\frac{1}{n}$, so we have obtained reasonably tight bounds for the tail.
Remark: Since the post expresses uncertainty about the estimation with the integral, we give a brief explanation. The sum $\sum_{n+1}^\infty \frac{1}{k^2}$ can be thought of as the combined area of a bunch of rectangles, all of base $1$. The base of the first rectangle is $[n,n+1]$ and its height is $\frac{1}{(n+1)^2}$. The base of the second rectangle is $[n+1,n+2]$ and its height is $\frac{1}{(n+2)^2}$, and so on.
Draw the curve $y=\frac{1}{x^2}$. In the interval $[n,n+1]$ this curve is above the first rectangle, so the first rectangle has area less than $\int_n^{n+1}\frac{dx}{x^2}$. In the same way, the second rectangle has area less than $\int_{n+1}^{n+2}\frac{dx}{x^2}$, and so on. So the sum of the areas of all the rectangles is less than $\int_n^\infty \frac{dx}{x^2}$.
|
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|
$\int \frac{\sqrt{1+x^2}}{1-x^2}dx$ Problem :
$\int \frac{\sqrt{1+x^2}}{1-x^2}dx$
My approach :
Put $x = \tan\theta$
we get $$\int \frac{\sqrt{1+x^2}}{1-x^2}dx = \frac{\frac{\sin^2\theta + \cos^2\theta}{\cos^2\theta \cos\theta}}{\frac{\cos^2\theta -\sin^2\theta}{\cos^2\theta}} d\theta $$
$$= \frac{1}{(\cos^2\theta -\sin^2\theta)\cos\theta}d\theta$$
But is it the right approach please guide will be of great help thanks.
|
HINT:
$$\int\dfrac{dy}{(\cos^2y-\sin^2y)\cos y} =\int\dfrac{\cos y\ dy}{(1-2\sin^2y)(1-\sin^2y)}$$
Set $\sin y=u$
Use Partial Fraction Decomposition,
$$\dfrac1{(1-2u^2)(1-u^2)}=\dfrac A{(\sqrt2)^2-u^2}+\dfrac B{1-u^2}$$
|
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|
Does the series $\sum\frac{(-1)^n cos(3n)}{n^2+n}$ converge absolutely? $\sum|\frac{(-1)^n \cos(3n)}{n^2+n}| \le \sum\frac{1}{n^2+n}$
since $-1 \le cos(3n) \le 1$
$\sum\frac{1}{n^2+n} = \sum\frac{1}{n(n+1)} = \sum(\frac{1}{n} - \frac{1}{n+1})$
$\int(\frac{1}{x} - \frac{1}{x+1}) dx = \log|x| + \log|x+1| + C$
which diverges.
Therefore by the integral test, the series diverges.
However, the answer is that the series converges absolutely. Where have I gone wrong?
|
Of course, the absolute value of $(-1)^n\cos(3n)$ is $\leq 1$ and the series $\sum_{n\geq 1}\frac{1}{n(n+1)}$ is absolutely convergent (to $1$). Moreover, since:
$$ \sum_{n\geq 1}\frac{x^n}{n(n+1)}=\sum_{n\geq 1}\frac{x^n}{n}-\sum_{n\geq 1}\frac{x^n}{n+1}=1-\left(1-\frac{1}{x}\right)\log(1-x)\tag{1}$$
by taking $x=-e^{3i}$, then the real part, we have:
$$ \sum_{n\geq 1}\frac{(-1)^n \cos(3n)}{n(n+1)} = 1-\frac{3}{2}\sin(3)-2\cos^2\left(\frac{3}{2}\right)\log\left(2\cos\left(\frac{3}{2}\right)\right).\tag{2}$$
|
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|
Find $n$, where its factorial is a product of factorials I need to solve $3! \cdot 5! \cdot 7! = n!$ for $n$.
I have tried simplifying as follows:
$$\begin{array}{}
3! \cdot 5 \cdot 4 \cdot 3! \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3! &= n! \\
(3!)^3 \cdot 5^2 \cdot 4^2 \cdot 7 \cdot 6 &= n! \\
6^3 \cdot 5^2 \cdot 4^2 \cdot 7 \cdot 6 &= n! \\
6^4 \cdot 5^2 \cdot 4^2 \cdot 7 &= n! \\
\end{array}$$
I really didn't see this helping me.
I then tried $6 \cdot 6 \cdot 6 \cdot 6 \cdot 25 \cdot 16 \cdot 7$, but $25$ only has $5$ as a double factor.
Any ideas?
|
If the formula is true it can only possibly be $8!$, $9!$, or $10!$ because $11!$ and larger have a factor of $11$. $8!$ doesn't have a large enough power of $3$, and $9!$ doesn't have a large enough power of $5$, so if the formula holds it must be $10!$.
|
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|
Evaluate the limit of ratio of sums of sines (without L'Hopital): $\lim_{x\to0} \frac{\sin x+\sin3x+\sin5x}{\sin2x+\sin4x+\sin6x}$ Limit to evaluate:
$$\lim_{x \rightarrow 0} \cfrac{\sin{(x)}+\sin{(3x)}+\sin{(5x)}}{\sin{(2x)}+\sin{(4x)}+\sin{(6x)}}$$
Proposed solution:
$$
\cfrac{\sin(x)+\sin(3x)+\sin(5x)}{\sin(2x)+\sin(4x)+\sin(6x)}
\Bigg/ \cdot\ \cfrac{1/x}{1/x}\Bigg/=
\frac{\cfrac{\sin(x)}x + \cfrac{\sin(3x)}{3x} \cdot 3 + \cfrac{\sin(5x)}{5x} \cdot 5}
{\cfrac{\sin(2x)}{2x} \cdot 2 + \cfrac{\sin(4x)}{4x} \cdot 4 + \cfrac{\sin(6x)}{6x} \cdot 6}
$$
Using $\lim_{x \rightarrow 0} \frac{\sin x}x=1$, we get
$$\frac{1+1\cdot 3+1\cdot 5}{1\cdot 2+1\cdot 4+1\cdot 6} = \frac 9{12} = \frac 3 4$$
Please tell me if I am correct.
|
HINT:
Using Prosthaphaeresis Formula,
$$\sin(a-2)x+\sin(ax)+\sin(a+2)x$$
$$=\sin(ax)+[\sin(a-2)x+\sin(a+2)x]$$
$$=\sin(ax)[1+2\cos2x]$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integral involving cube root and seventh root Find the value of $$\int_{0}^{1} (1-x^7)^{\frac{1}{3}}-(1-x^3)^{\frac{1}{7}}\:dx$$
My Approach:
Let $$I_1=\int_{0}^{1} (1-x^7)^{\frac{1}{3}}dx$$ and
$$I_2=\int_{0}^{1} (1-x^3)^{\frac{1}{7}}dx$$
For $I_1$ substitute $x^7=1-t^3$ so $dx=\frac{-3t^2}{7}(1-t^3)^{\frac{-6}{7}}\:dt$ Hence
$$I_1=\int_{1}^{0}\frac{-3t^3}{7}(1-t^3)^{\frac{-6}{7}}\:dt$$ $\implies$
$$I_1=\frac{-3}{7}\int_{0}^{1}(1-t^3-1)(1-t^3)^{\frac{-6}{7}}\:dt$$ $\implies$
$$I_1=\frac{-3I_2}{7}+\frac{3A}{7}$$
where $A=\int_{0}^{1}(1-t^3)^{\frac{-6}{7}}dt$
Similarly using substitution $x^3=1-t^7$ for $I_2$ and proceeding as above we get
$$I_2=\frac{-7I_1}{3}+\frac{7B}{3}$$ where $B=\int_{0}^{1}(1-t^7)^{\frac{-2}{3}}dt$
But we need to find $I_1-I_2$ so got stuck up here
|
We can write it as $$I = \displaystyle \int_{0}^{1}\sqrt[3]{1-x^7}dx-\int_{0}^{1}\sqrt[7]{1-x^3}dx$$
Now Using $$\displaystyle \bullet \int_{a}^{b}f(x)dx = -\int_{b}^{a}f(x)dx$$
So we get $$I = \displaystyle \int_{0}^{1}\left(1-x^7\right)^{\frac{1}{3}}dx+\int_{1}^{0}\left(1-x^3\right)^{\frac{1}{7}}dx$$
Now Let $$\displaystyle f(x) = \left(1-x^{7}\right)^{\frac{1}{3}}\;,$$ Then $$f^{-1}(x) = (1-x^3)^{\frac{1}{7}}$$ and also $f(0) = 1$ and $f(1) =0$
So Integral $$\displaystyle I = \int_{0}^{1}f(x)dx+\int_{f(0)}^{f(1)}f^{-1}(x)dx$$
Now let $f^{-1}(x) = z\;,$ Then $x=f(z)$ So we get $dx = f'(z)dz$
So Integral $$\displaystyle I =\int_{0}^{1}f(x)dx+\int_{0}^{1}z\cdot f'(z)dz$$
Now Integration by parts for second Integral, We get
$$\displaystyle I =\int_{0}^{1}f(x)dx+\left[z\cdot f(z)\right]_{0}^{1}-\int_{0}^{1}f(z)dz$$
So using $$\displaystyle \bullet\; \int_{a}^{b}f(z)dz = \int_{a}^{b}f(x)dx$$
So we get $$\displaystyle I =\int_{0}^{1}f(x)dx+f(1) -\int_{0}^{1}f(x)dx = f(1) =0$$
|
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|
Integration of $1/((1+\cos x)(1+\sin x))$ How do you integrate $\displaystyle \frac 1{(1+\cos x)(1+\sin x)}$? I've tried $u$ substitution and manipulation and have got nowhere. I cannot think of any other methods that would work.
|
Let $$\displaystyle I = \int\frac{1}{(1+\sin x)\cdot (1+\cos x)}dx = \frac{1}{2}\int\frac{1}{\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)^2\cdot \cos^2 \frac{x}{2}}dx$$
Above we used $$\displaystyle \bullet\; 1+\sin x = \sin^2 \frac{x}{2}+\cos^2 \frac{x}{2}+2\sin \frac{x}{2}\cdot \cos \frac{x}{2}$$
And $$\displaystyle \bullet\; 1+\cos x = 2\cos^2 \frac{x}{2}.$$
So we get $$\displaystyle I = \frac{1}{2}\int\frac{\sec^2 \frac{x}{2}}{\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)^2}dx = \frac{1}{2}\int\frac{\sec^2 \frac{x}{2}\cdot \left(1+\tan^2 \frac{x}{2}\right)}{\left(\tan \frac{x}{2}+1\right)^2}$$
Now Put $\displaystyle \tan \frac{x}{2} = t\;,$ Then $\displaystyle \frac{1}{2}\sec^2 \frac{x}{2}dx = dt$
So we get $$\displaystyle I = \int\frac{1+t^2}{(1+t)^2}dx = \int\frac{1+t^2+2t-2t}{(1+t)^2}dt=\int1dt-2\int\frac{t}{(1+t)^2}dt$$
Now $(1+t) = u\;,$ Then $dt = du$
So we get $$\displaystyle I = t-2\int\frac{u-1}{u^2}du=t-2\ln |u|-\frac{2}{u}+\mathcal{C}$$
So we get $$\displaystyle I =t-2\ln |1+t|-\frac{2}{1+t}+\mathcal{C}$$
So we get $$\displaystyle I =\tan \frac{x}{2}-2\ln \left|1+\tan \frac{x}{2}\right|-\frac{2}{1+\tan \frac{x}{2}}+\mathcal{C}$$
|
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|
How do I evaluate $ \lim_{x \to \pi/2}\frac {\sin x-1}{(1+\cos 2x)}$without using L'Hospital's rule? How do I evaluate this: $$\lim_{x \to \pi/2}\displaystyle \frac {\sin x-1}{(1+\cos 2x)}$$? without using L'Hospital's rule.
Attempt : I used trigonomeitric transformation and I used Taylor series I got this :
$$\lim_{x \to \pi/2}\displaystyle \frac {\sin x-1}{(1+\cos 2x)} =
\lim_{y\to 0}\displaystyle \frac {\cos y-cos²y}{2\cos² (2y+\pi)}-\frac{1}{2}$$
where $y=x-\frac{\pi}{2}$, but by this way i got :
$$\lim_{x \to \pi/2}\displaystyle \frac {\sin x-1}{(1+\cos 2x)}=\frac{-1}{2}$$ and by L'Hospital's rule is :$\frac{-1}{4}$
Then where is my problem if I'm true
Thank you for any help .
|
$$\lim_{x\to \frac{\pi}{2}} \frac {\sin x-1}{(1+\cos 2x)} = \lim_{x\to \frac{\pi}{2}} \frac {\sin x-1}{(1+\cos^2x-\sin^2x)} = \\
= \lim_{x\to \frac{\pi}{2}} \frac {\sin x-1}{(1+(1-\sin^2x)-\sin^2x)} = \\
= \lim_{x\to \frac{\pi}{2}} \frac {\sin x-1}{2(1-\sin^2x)} = \\
= \lim_{x\to \frac{\pi}{2}} \frac {\sin x-1}{2(1-\sin x)(1+\sin x)} = \\
= \lim_{x\to \frac{\pi}{2}} \frac {-1}{2(1+\sin x)} = -\frac{1}{4}\\$$
Using Taylor for $x_0 = \frac{\pi}{2}$...
$$\sin(x) = 1 - (\pi/2 - x)^2/2 + \ldots$$
$$\cos(2x) = -1 + 2(\pi/2 - x)^2 + \ldots$$
Then:
$$\lim_{x\to \frac{\pi}{2}} \frac {\sin x-1}{(1+\cos 2x)} = \lim_{x\to \frac{\pi}{2}} \frac { 1 - (\pi/2 - x)^2/2-1+ \ldots}{(1-1 + 2(\pi/2 - x)^2 + \ldots)} = \\
=\lim_{x\to \frac{\pi}{2}} \frac {- (\pi/2 - x)^2/2+ \ldots}{2(\pi/2 - x)^2 + \ldots} = -\frac{1}{4}$$
|
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|
Prove $b^{2} (\cot A + \cot B) = c^{2}(\cot A + \cot C)$ In any triangle $ABC $ prove that $$b^{2} (\cot A + \cot B) = c^{2}(\cot A + \cot C)$$
How we can prove this trigonometric identity. I tried many ways and use the other well known identity but it wasn't work. My question is how we can prove this trigonometric identity?. Any hint will help Thanks.
|
There is a nice proof of this trig identity.
Notice, in right $\triangle ABC$ $(A+B+C=180^\circ)$ we know from sine rule $$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$$
$$\implies \ a=k\sin A, \ b=k\sin B, \ c=k\sin C, $$
Now, we have $$LHS=b^2(\cot A+\cot B)$$
$$=(k\sin B)^2\left(\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B}\right)$$
$$=k^2\sin^2B \left(\frac{\sin A\cos B+\cos A\sin B}{\sin A\sin B}\right)$$
$$=k^2 \frac{\sin B\sin (A+B)}{\sin A}$$
$$=k^2 \frac{\sin (180^\circ-(A+C))\sin (180^\circ-C)}{\sin A}$$
$$=k^2 \frac{\sin (A+C)\sin C}{\sin A}$$
$$=k^2\sin^2 C \left(\frac{\sin A\cos C+\cos A\sin C}{\sin A\sin C}\right)$$
$$=(k\sin C)^2 \left(\frac{\cos C}{\sin C}+\frac{\cos A}{\sin A}\right)$$
setting the value $k\sin C=c$ from (1) $$=\color{}{(c)^2(\cot C+\cot A)}$$
$$LHS=\color{red}{c^2(\cot A+\cot C)}=RHS$$
|
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|
Proving this trig identity:$\frac{1+\cos\theta+\sin\theta}{1+\cos\theta-\sin\theta}=\frac{1+\sin\theta}{\cos\theta}$
$$\frac{1+\cos\theta+\sin\theta}{1+\cos\theta-\sin\theta}=\frac{1+\sin\theta}{\cos\theta}$$
What I've tried,
$$\frac{((1+\cos\theta)+(\sin\theta))((1+\cos\theta)+(\sin\theta))}{(1+\cos\theta-\sin\theta) (1+\cos\theta+\sin\theta)}$$
$$=\frac{(1+\cos\theta+\sin\theta)^2}{(1+\cos\theta)^2-\sin^2\theta}$$
After simplifying,
$$=\frac{2(1+\sin\theta \cos\theta + \sin\theta + \cos\theta)}{\cos\theta(\cos\theta+2)}$$
I cant carry on further. Thus, any kind assistance would be much appreciated.
|
Multiplying and dividing by $(1+\sin\theta -\cos \theta)$ yields
$$\frac{(1+\sin \theta)^2-\cos^2 \theta}{1-\cos^2\theta -\sin^2\theta +2\sin\theta \cos \theta}= \frac{1+\sin^2 \theta + 2\sin \theta -\cos^2 \theta}{2\sin \theta \cos \theta}$$
Substituting $-\cos^2 \theta = \sin^2 \theta -1$, we obtain
$$ \frac{1+\sin^2 \theta +2\sin \theta +\sin^2 \theta -1}{2 \sin \theta \cos \theta}= \frac{2 \sin \theta(1+\sin \theta)}{2 \sin \theta \cos \theta}$$
$$=\color{red} {\frac{1+\sin \theta}{\cos \theta}}$$
Edit: clarifications for OP.
Consider the initial expression multiplied and divided by $(1+\sin\theta -\cos \theta)$:
$$\frac{((1+\cos \theta+\sin \theta)(1+\sin \theta - \cos \theta)}{(1+\cos \theta - \sin \theta)(1+ \sin \theta - \cos \theta)}$$
Considering the terms grouped this way and using the fact that $(a+b)(a-b)=a^2-b^2$ we can conclude that
$$\frac{((1+\sin \theta)+\cos \theta)((1+\sin \theta) - \cos \theta)}{(1+(\cos \theta - \sin \theta))(1-(\cos \theta - \sin \theta)}=\frac{(1+\sin \theta)^2-\cos^2\theta}{1-(\cos \theta - \sin \theta)^2}$$
Expanding the square in the denominator leads to the LHS of the first equality.
|
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|
Solving $\lim_{x \to 4}\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}$ without L'Hopital. I am trying to solve the limit
$$\lim_{x \to 4}\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}$$
Without using L'Hopital.
Evaluating yields $\frac{0}{0}$. When I am presented with roots, I usually do this:
$$\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}} \cdot \frac{3+\sqrt{5+x}}{3+\sqrt{5+x}}$$
And end up with
$$\frac{9 - (5+x)}{(1-\sqrt{5-x}) \cdot (3+\sqrt{5+x})}$$
Then
$$\frac{9 - (5+x)}{3+\sqrt{5+x}-3\sqrt{5-x}-(\sqrt{5-x}\cdot\sqrt{5+x})}$$
And that's
$$\frac{0}{3+\sqrt{9}-3\sqrt{1}-(\sqrt{1}\cdot\sqrt{9})}= \frac{0}{3+3-3-3-3} = \frac{0}{-3} = -\frac{0}{3}$$
Edit: Ok, the arithmetic was wrong. That's $\frac{0}{0}$ again.
But the correct answer is
$$-\frac{1}{3}$$
But I don't see where does that $1$ come from.
|
$$\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}=\frac{\frac{x-4}{-(3-\sqrt{5+x})}}{\frac{x-4}{1+\sqrt{5-x}}}=\frac{1+\sqrt{5-x}}{-(3+\sqrt{5+x})}$$
|
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|
How do I prove the existence of $\lim_{(x,y)\to(0,0)} \frac{x^2y^2}{x^3+y^3}$ How do I prove the existence of
$$\lim_{(x,y)\to(0,0)} \frac{x^2y^2}{x^3+y^3}$$
When $y=0$ the limits is $0$
when $y=-x$ it´s $$\lim_{x\to0} \frac{x^2(-x^2)}{x^3- x^3}=\lim_{x\to0} \frac{x^2(-x^2)}{0}$$
I can concluir that the limit no exist??
or how proof the existence?
|
Along the path $y=x$,
\begin{align*}
\frac{x^2y^2}{x^3+y^3} = \frac{1}{2}x\rightarrow 0.
\end{align*}
However, along the path $y=x^3-x$,
\begin{align*}
\frac{x^2y^2}{x^3+y^3} = \frac{x^8-2x^6+x^4}{x^9+3x^5-3x^7},
\end{align*}
which does not have a finite limit when $x\rightarrow 0$.
That is,
\begin{align*}
\lim_{(x, y)\rightarrow (0, 0)} \frac{x^2y^2}{x^3+y^3}
\end{align*}
does not exist.
|
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|
diagonalization exp(A) counter exemple We consider a finite vector space E over a field K. We consider u in End(E). We can show that if the characteristic polynomial of A splits then we have the following equivalence : A diagonalizable iff exp(A) diagonalizable.
I am searching a counter exemple in the case of a non splitting characteristic polynomial, ie a matrix such that exp(A) is diagonalizable but A is not.
Thankyou for your answers !
|
trying to guess what you might be asking, take
$$ A =
\left(
\begin{array}{rr}
0 & \pi \\
- \pi & 0
\end{array}
\right).
$$
Then
$$ e^A =
\left(
\begin{array}{rr}
-1 & 0 \\
0 & -1
\end{array}
\right).
$$
If
$$ C =
\left(
\begin{array}{rr}
0 & 1 \\
- 1 & 0
\end{array}
\right)
$$
then $C^2 = -I$ and, for real $t,$
$$ e^{tC} = I + tC - I t^2/2 - C t^3/6 + I t^4/24 +C t^5/120 - I t^6/720-C t^7/5040...$$
$$ e^{tC} = (I - I t^2/2 + I t^4/24 - I t^6/720...) + (tC - C t^3/6 +C t^5/120 -C t^7/5040...) $$
$$ e^{tC} = I(1 - t^2/2 + t^4/24 - t^6/720...) + C(t - t^3/6 + t^5/120 - t^7/5040...) $$
$$ e^{tC} = I \cos t + C \sin t $$
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
The other possible interpretation has to do with Jordan blocks. My best guess is that exponentiating does not change the arrangment of Jordan blocks, just the diagonal elements. For example, if
$$ F =
\left(
\begin{array}{rr}
\lambda & 1 \\
0 & \lambda
\end{array}
\right)
$$
then
$$ e^F =
\left(
\begin{array}{rr}
e^\lambda & e^\lambda \\
0 & e^\lambda
\end{array}
\right)
$$
But then
$$
\left(
\begin{array}{rr}
e^{-\lambda} & 0 \\
0 & 1
\end{array}
\right)
\left(
\begin{array}{rr}
e^\lambda & e^\lambda \\
0 & e^\lambda
\end{array}
\right)
\left(
\begin{array}{rr}
e^\lambda & 0 \\
0 & 1
\end{array}
\right) =
\left(
\begin{array}{rr}
e^\lambda & 1 \\
0 & e^\lambda
\end{array}
\right)
$$
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
Also did 3 by 3
$$ G =
\left(
\begin{array}{rrr}
\lambda & 1 & 0 \\
0 & \lambda & 1 \\
0 & 0 & \lambda
\end{array}
\right)
$$
then
$$ e^G =
\left(
\begin{array}{rrr}
e^\lambda & e^\lambda & \frac{1}{2} e^\lambda \\
0 & e^\lambda & e^\lambda \\
0 & 0 & e^\lambda
\end{array}
\right)
$$
But then
$$
\left(
\begin{array}{rrr}
e^{-2 \lambda} & -\frac{1}{2} e^{-\lambda} & 0 \\
0 & e^{- \lambda} & 0 \\
0 & 0 & 1
\end{array}
\right)
\left(
\begin{array}{rrr}
e^\lambda & e^\lambda & \frac{1}{2} e^\lambda \\
0 & e^\lambda & e^\lambda \\
0 & 0 & e^\lambda
\end{array}
\right)
\left(
\begin{array}{rrr}
e^{2 \lambda} & \frac{1}{2} e^\lambda & 0 \\
0 & e^\lambda & 0 \\
0 & 0 & 1
\end{array}
\right) =
\left(
\begin{array}{rrr}
e^\lambda & 1 & 0\\
0 & e^\lambda & 1 \\
0 & 0 & e^\lambda
\end{array}
\right)
$$
|
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|
Evalutating $\lim_{x\to0}\frac{1-\cos x}{x^2}$ $$\lim_{x\to0}\frac{1-\cos x}{x^2}$$
I know there are many ways to calculate this. Like L'Hopital. But for learning purposes I am not supposed to do that. Instead, I decided to do it this way:
Consider that $\cos x = 1- \sin^2 \frac{x}{2}$ (from the doulbe-angle formulas here). Therefore:
$$\frac{1-(1- \sin^2 \frac{x}{2})}{x^2} = \frac{\sin^2 \frac{x}{2}}{x^2}$$
Let us split this:
$$\frac{\sin \frac{x}{2}}{x} \cdot \frac{\sin \frac{x}{2}}{x}$$
I will just work with the left one for now. At the end, I will just multiply it by itself.
$$\frac{\sin \frac{x}{2}}{x}$$
Hm. Remember that $\frac{\sin x}{x} = 1$ when $x\to0$. If only the denominator were an $\frac{x}{2}$ instead of $x$ I could do this... Well, I can! Let's just
$$\frac{\sin \frac{x}{2}}{x}\cdot\frac{\frac{1}{2}}{\frac{1}{2}}$$
Looks a bit dumb but it kind of makes sense. This will yield
$$\frac{\color{red}{\sin \frac{x}{2}}\cdot \frac{1}{2}}{\color{red}{\frac{x}{2}}}$$
Hurray, we can now apply the formula thingy and end up with
$$\frac{\color{red}1\cdot\frac{1}{2}}{\color{red}1} = \frac{1}{2}$$
Now I need to multiply this thing by itself.
$$\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$$
And that is the answer.
... Which is wrong. The correct answer is
$$\frac{1}{2}$$
What did I do wrong in these steps? I know many of you are able to solve this with a variety of different methods, but I'm particularly interested in this specific procedure I just tried to use. What was my mistake?
|
Notice, your mistake $\cos x\ne 1-\sin^2\frac{x}{2}$
Now, there are various methods to find the limit here given two methods as follows
Method-1$$\lim_{x\to 0}\frac{1-\cos x}{x^2}$$
$$=\lim_{x\to 0}\frac{2\sin^2\left( \frac{x}{2}\right)}{x^2}=\lim_{x\to 0}\frac{1}{2}\frac{\sin^2\left( \frac{x}{2}\right)}{\left( \frac{x}{2}\right)^2}$$$$=\frac{1}{2}\lim_{x\to 0}\left(\frac{\sin\left( \frac{x}{2}\right)}{\left( \frac{x}{2}\right)}\right)^2=\frac{1}{2}(1)^2=\frac{1}{2}$$
Method-2
$$\lim_{x\to 0}\frac{1-cos x}{x^2}$$
Applying L'Hospital's rule for $\frac{0}{0}$ form $$=\lim_{x\to 0}\frac{\sin x}{2x}=\frac{1}{2}\lim_{x\to 0}\frac{\sin x}{x}=\frac{1}{2}$$
|
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|
$\left[\csc^2\frac{\alpha}{2}+\csc^2\frac{\beta}{2}+\csc^2\frac{\gamma}{2}\right]=2$ If $\cos \alpha \cos \beta \cos \gamma-\cos \alpha-\cos \beta-\cos \gamma+1=0$ and $\alpha\neq\beta\neq\gamma\neq2n\pi$,then prove that $\left[\csc^2\frac{\alpha}{2}+\csc^2\frac{\beta}{2}+\csc^2\frac{\gamma}{2}\right]=2$
$\cos \alpha \cos \beta \cos \gamma+1=\cos \alpha+\cos \beta+\cos \gamma$
$\cos \alpha \cos \beta \cos \gamma+1=2\cos^2 \frac{\alpha}{2}+2\cos^2 \frac{\beta}{2}+2\cos^2 \frac{\gamma}{2}-3$
But there seems no method to calculate the value of the required expression.Some hints and suggestions are needed.
|
HINT:
As $\csc^2\dfrac x2=\dfrac2{1-\cos x},$
Please simplify
$$\dfrac2{1-\cos\alpha}+\dfrac2{1-\cos\beta}+\dfrac2{1-\cos\gamma}=2$$ to reach at the given condition
|
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|
How to solve $\lim _{x\to \infty \:}\left(\sqrt{x+\sqrt{1+\sqrt{x}}}-\sqrt{x}\right)$ Solved I stuck in this limit, I tried to solve it and gets 1/2 as result. Yet, I was wrong because I forgot a square. Please need help!
$\lim _{x\to \infty \:}\left(\sqrt{x+\sqrt{1+\sqrt{x}}}-\sqrt{x}\right)$
Note: it's $+\infty$
Thanks in advance
Update: I actually solved it, and this is the way that I wanted to:
$\lim \:_{x\to \:\infty \:}\left(\sqrt{x+\sqrt{1+\sqrt{x}}}-\sqrt{x}\right)\:=\:\lim \:\:_{x\to \:\:\infty \:\:}\left(\frac{\sqrt{1+\sqrt{x}}}{\sqrt{x+\sqrt{1+\sqrt{x}}}+\sqrt{x}}\right)=\:\lim \:\:_{x\to \:\:\infty \:\:}\left(\frac{\sqrt{1+\sqrt{x}}}{\sqrt{x^2\left(1+\frac{\sqrt{1+\sqrt{x}}}{x}\right)+\sqrt{x}}}\right)=\lim \:\:_{x\to \:\:\infty \:\:}\left(\frac{\sqrt{x\left(\frac{1}{x}+\frac{1}{\sqrt{x}}\right)}}{\sqrt{x}\left(\sqrt{1+\sqrt{\frac{1}{x^2}+\frac{1}{\sqrt{x}}}}+1\right)}\right)=\lim \:\:_{x\to \:\:\infty \:\:}\left(\frac{\sqrt{x}\sqrt{\frac{1}{x}+\frac{1}{\sqrt{x}}}}{\sqrt{x}\left(\sqrt{1+\sqrt{\frac{1}{x^2}+\frac{1}{x}}}+1\right)}\right)=\lim \:\:_{x\to \:\:\infty \:\:}\left(\frac{\sqrt{\frac{1}{x}+\frac{1}{\sqrt{x}}}}{\sqrt{1+\sqrt{\frac{1}{x^2}+\frac{1}{\sqrt{x}}}}+1}\right)=\frac{0}{2}=0$
Thanks for your two answers guys. I really appreciate it ! This community is awesome !
|
If we have $f(x)/x^{1/2} \to 0,$ where $f(x) > 0,$ then the MVT gives
$$(x+f(x))^{1/2} - x^{1/2} = 1/(2c_x^{1/2})\cdot f(x) \le f(x)/2x^{1/2} \to 0.$$
Above $c_x \in (x,x+f(x)),$ so we're OK. In this problem $f(x) = (1+x^{1/2})^{1/2},$ so the desired limit is $0.$
|
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|
trigonometry equation, $[\cos(2x)]^2-\sin(2x)=1$ I tried to solve this equation: $[\cos(2x)]^2-\sin(2x)=1$
I got different answer from the book.
the answer in the book: $x=-45+180k$ , $x=90k$
Am I right? $x = -45+180k$ is equal to $x= 135+180k$ ? How can I check if it's the same? Or maybe I did a mistake? please help
\begin{align*}
[\cos(2x)]^2-\sin(2x) & = 1\\
-\sin(2x)& = 1-\cos(2x)^2\\
-\sin(2x)& = \sin(2x)^2\\
-\sin(2x)-\sin(2x)^2 & = 0\\
\sin(2x)+\sin(2x)^2 & = 0\\
\sin(2x)[1+\sin(2x)] & = 0
\end{align*}
\begin{align*}
\sin(2x) & =0 & 1 + \sin(2x) & = 0\\
2x & = 180k & \sin(2x) & = -1\\
x & = 90k & 2x & = 270+360k\\
& & x & = 135+180k
\end{align*}
|
$$\cos²(2x)-\sin(2x) = 1$$
$$-\sin(2x)= 1-\cos²(2x)$$
$$-\sin(2x)=\sin²(2x)$$
$$0= \sin²(2x)+\sin(2x)$$
$$\sin(2x)(\sin(2x)+1)=0$$
Then, you get two equations:
$$\sin(2x)=0$$
and
$$\sin(2x)+1=0$$
$$\sin(2x)=\sin (180)$$
and
$$\sin(2x)=\sin(270)$$
From the $$\sin(2x)=\sin(180)$$, you can get the solution:
$$2x=180+k.360$$ or $$x=90+k.180$$
and
$$2x= k.360$$ or $$x= k.180$$
From the $$\sin(2x)=\sin(270)$$, you can get the solution:
$$2x=270+k.360$$ or $$x=135+k.180$$
and
$$2x=(180-270)+k.360$$ or $$x= -45+k.180$$
So, the solution:
$x=k.180$
$x=90+k.180$
$x=135+k.180$
$x=-45+k.180$
|
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|
Evaluating the magnitude. Is this less than 1? Where $|x|<1$, I'm looking to determine if
$$\left|\frac{1}{x}(-1+\sqrt{1-x^2})\right|<1$$
I believe it is, since we can use a Taylor series to approximate
$$\sqrt{1-x^2} = 1 - \frac{1}{2}x^2-\frac{1}{4}(x^2)^2 + O(3) \approx 1 - \frac{1}{2}x^2$$
This then gives
$$\left|\frac{1}{x}(-1+\sqrt{1-x^2})\right| \approx \left|\frac{1}{x}(-1+1 - \frac{1}{2}x^2)\right| = \left|-\frac{x}{2}\right|<1$$
This, though, is not a proof. Any ideas?
|
Hint: $$
\left|\frac{\sqrt{1-x^2}-1}{x}\right| = \left|-\frac{x}{\sqrt{1-x^2}+1} \right| \leq \frac{|x|}{1}=|x|.
$$
|
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|
Prove using factor theorem. Using factor theorem, show that $a+b$,$b+c$ and $c+a$ are factors of
$(a + b + c)^3$ - $(a^3 + b^3 + c^3)$
How do we go about solving this ?
Thanks in advance !
|
Let,$f(a)$=$(a+b+c)^3-(a^3+b^3+c^3)$
If $(a+b)$ is a factor,$f(-b)$ must be $0$ by factor theorem.
Putting $f(-b)$,we get,
$(-b+b+c)^3-(-b^3+b^3+c^3)$
$=c^3-c^3=0$.
Similarly prove the others.
|
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|
How to find $\lim\limits_{x~\to~4}\frac{\sqrt{2x+1}-3}{\sqrt{x-2}-\sqrt{2}}$ I am wondering how to find $\lim\limits_{x~\to~4}\dfrac{\sqrt{2x+1}-3}{\sqrt{x-2}-\sqrt{2}}.$
I have found out that it equals $\dfrac{0}{0}$ if the equation is not simplified.
I have tried multiplying by the conjugate of the numerator (and the denominator in a separate attempt), but I have not found any other solution other than $\dfrac{0}{0}.$
Does anyone know how to find this limit? If so, could they please show the steps they used to get to the answer?
All help is appreciated.
|
$$\lim_{x\to 4}\frac{\sqrt{2x+1}-3}{\sqrt{x-2}-\sqrt 2}$$
$$=\lim_{x\to 4}\frac{(\sqrt{2x+1}-3)(\sqrt{2x+1}+3)}{(\sqrt{x-2}-\sqrt 2)(\sqrt{x-2}+\sqrt 2)}\cdot \frac{\sqrt{x-2}+\sqrt 2}{\sqrt{2x+1}+3}$$
$$=\lim_{x\to 4}\frac{2x+1-9}{x-2-2}\cdot \lim_{x\to 4} \frac{\sqrt{x-2}+\sqrt 2}{\sqrt{2x+1}+3}$$
$$=\lim_{x\to 4}\frac{2x-8}{x-4}\cdot \frac{\sqrt{2}+\sqrt 2}{3+3}$$
$$=\frac{\sqrt 2}{3}\lim_{x\to 4}\frac{2(x-4)}{x-4}$$
$$=\frac{\sqrt 2}{3}\lim_{x\to 4} 2=\color{red}{\frac{2\sqrt 2}{3}}$$
|
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"url": "https://math.stackexchange.com/questions/1461552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
How to solve this simple problem So my younger brother asked me to solve a simple problem, but I'm not sure how to solve this. Any help would be much appreciated.
$$a^3 - 1/ a^3 = 4. $$
Prove that $$a - 1/a = 1$$
|
Using $$\displaystyle \left(a-\frac{1}{a}\right)^3=\left(a^3-\frac{1}{a^3}\right)-3\cdot a\cdot \frac{1}{a}\left(a-\frac{1}{a}\right)$$
So we get $$\displaystyle \left(a-\frac{1}{a}\right)^3=4-3\displaystyle \left(a-\frac{1}{a}\right)\;,$$ Now Put $\displaystyle \left(a-\frac{1}{a}\right)=x$
So we get $$\displaystyle x^3=4-3x\Rightarrow x^3+3x-4=0\Rightarrow (x-1)\cdot (x^2+x+4)=0$$
So we get $x=1$ or $\displaystyle x^2+x+4 =0\Rightarrow x=\frac{-1\pm \sqrt{1-16}}{2}$ (No real values of $x$)
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Find the equation of the locus of the intersection of the lines below Find the equation of the locus of the intersection of the lines below
$y=mx+\sqrt{m^2+2}$
$y=-\frac{ 1 }{ m }x+\sqrt{\frac{ 1 }{ m^2 +2}}$
By graphing, I have got an ellipse as locus : $x^2+\dfrac{y^2}{2}=1$.
The given lines form tangent and normal to above ellipse.
Is there any other nice way to eliminate $m$ w/o using graphing ?
I have tried eliminating $m$ by solving the intersection point, but it's looking very messy. Thanks!
|
Solving the system you get
$x = - \frac{m}{{\sqrt {{m^2} + 2} }},y = \frac{2}{{\sqrt {{m^2} + 2} }}$
Now squaring and adding you get
${x^2} + {y^2} = \frac{{{m^2}}}{{{m^2} + 2}} + \frac{4}{{{m^2} + 2}} = {\left( {\sqrt {\frac{{{m^2} + 4}}{{{m^2} + 2}}} } \right)^2}$
which is a circle with center (0,0) and radius $r = \left( {\sqrt {\frac{{{m^2} + 4}}{{{m^2} + 2}}} } \right)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1466618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Extremal values of $(x+3)^\frac{1}{3} - x^\frac{1}{3}$ $f(x) = (x+3)^\frac{1}{3} - x^\frac{1}{3}$
I am trying to find the extremal values of $f$. I start by differentiating, equating the derivative to 0, and solving that for x:
$f'(x)= \frac{1}{3} ((x+3)^\frac{-2}{3} - x^\frac{-2}{3}) = 0$
$(x+3)^2=x^2$
$x^2 + 6x + 9 = x^2$
$x = \frac{-3}{2}$
The extremal value of $f$ is at $f(\frac{-3}{2})$.
$f(\frac{-3}{2})= 2.289428485$ (decimal approximation)
However, when I try to verify this result with a CAS, WolframAlpha says something else, and SymboLab talks about the saddle of some interval, which I am not familiar with.
Is my working correct?
|
I've got it.
$f''(\frac{-3}{2}) < 0$, therefore the extremal is a maximum point.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1467637",
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|
If $t = \tan (x/2)$, find expressions for $\sin x, \cos x$ in terms of $t$. Hence, solve the equation $3\sin x - 4\cos x = 2$.
If $$t = \tan \frac{x}{2},$$ find expressions for $\sin x, \cos x$ in terms of $t$. Hence, solve the equation $$3\sin x - 4\cos x = 2.$$
Attempt:
I have been solving a lot of trig questions lately but this is different. I don't know how to approach this. I'm thinking of getting $\sin x$ and $\cos x$ from $\tan x$ and replacing in the equation but not sure how because of the $t$. Help please.
|
If you know the bisection formulas
$$
\left|\sin\frac{x}{2}\right|=\sqrt{\frac{1-\cos x}{2}}
\qquad
\left|\cos\frac{x}{2}\right|=\sqrt{\frac{1+\cos x}{2}}
$$
you can put them together finding
\begin{align}
\left|\tan\frac{x}{2}\right|=\sqrt{\frac{1-\cos x}{1+\cos x}}
&=\sqrt{\frac{1-\cos x}{1+\cos x}\frac{1-\cos x}{1-\cos x}}\\[6px]
&=\sqrt{\frac{(1-\cos x)^2}{\sin^2x}}\\[6px]
&=\biggl|\frac{1-\cos x}{\sin x}\biggr|\\[6px]
&=\frac{1-\cos x}{|\sin x|}
\end{align}
It's easy to check that $\tan(x/2)$ has the same sign as $\sin x$ for every $x$ where $\sin x\ne0$ (and so $\tan(x/2)$ is defined), which allows us to remove the absolute values:
$$
\tan\frac{x}{2}=\frac{1-\cos x}{\sin x}
$$
We can do similarly by using $1+\cos x$ in the second step instead of $1-\cos x$, which gives another neat formula:
$$
\tan\frac{x}{2}=\frac{\sin x}{1+\cos x}
$$
Setting, for simplicity, $t=\tan(x/2)$, $X=\cos x$ and $Y=\sin x$, we can rewrite the equations as
$$
\begin{cases}
X+tY=1\\[6px]
tX-Y=-1
\end{cases}
$$
If we solve this for $X$ and $Y$, we arrive at
$$
X=\cos x=\frac{1-t^2}{1+t^2},\qquad
Y=\sin x=\frac{2t}{1+t^2}
$$
If you apply these to your equation, you get
$$
\frac{6t}{1+t^2}-\frac{4-4t^2}{1+t^2}=2
$$
that becomes a quadratic by removing the denominators:
$$
6t-4+4t^2=2+2t^2
$$
or, simplifying,
$$
t^2+3t-3=0
$$
Note there's a small issue: you have to verify that the values of $x$ for which $\tan(x/2)$ is not defined are or are not solutions of the equations. The check is easily done: $\tan(x/2)$ is not defined for $x=k\pi$, but
$$
3\sin(k\pi)-4\cos(k\pi)=\pm4\ne2
$$
If the equation is $3\sin x-4\cos x=4$, you get a family of solutions not covered by the $t$-substitution.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
$\int_{-1}^1 e^{-x^{4}}\cdot(1+\ln(x+\sqrt{x^2+1})+5x^3-4x^4)dx$ Is it possible to evaluate this integral $$\int_{-1}^1 e^{-x^{4}}\cdot(1+\ln(x+\sqrt{x^2+1})+5x^3-4x^4)dx$$
I got this question on my math quiz today, Solution not given yet, But is it even possible to evaulate this integral
|
Notice that $x\mapsto\ln (x+\sqrt{x^2+1})$ is an odd function since
\begin{align}
\ln [(-x)+\sqrt{(-x)^2+1}]&=\ln\left[\frac{\sqrt{x^2+1}-x}{1}\cdot\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x}\right]\\
&=\ln\left[\frac{1}{\sqrt{x^2+1}+x}\right]\\
&=-\ln (x+\sqrt{x^2+1})
\end{align}
Also $x\mapsto x^3$ is odd, then we have
\begin{align}
\int_{-1}^1 e^{-x^{4}}\cdot(\ln(x+\sqrt{x^2+1})+5x^3)dx&=0
\end{align}
Since the integrand is an odd function. Then the integral becomes
\begin{align}
\int_{-1}^1 e^{-x^{4}}\cdot(1-4x^4)dx&=\int_{-1}^1e^{-x^4}dx-\int_{-1}^14x^4e^{-x^4}dx\\
&=\int_{-1}^1e^{-x^4}dx+\int_{-1}^1x (-4x^3e^{-x^4})dx\\
&=\int_{-1}^1e^{-x^4}dx+\left.xe^{-x^4}\right|_{-1}^1-\int_{-1}^1e^{-x^4}dx\\
&=e^{-1}-(-1)e^{-1}\\
&=\frac{2}{e}
\end{align}
Where integration by parts was used.
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Convert advanced parametric equation to regular/cartesian can anybody help me to convert following parametric equation in a form Y =Y(X):
$$
x = cos(t) \sqrt{(2 - cos^2(3t))} \\
y = sin(t) \sqrt{(2 - cos^2(3t))}
$$
I've tried also with Wolfram Alpha and it seem not to work:
Reduce[x == Cos[t] Sqrt[2 - Cos[3 t]^2], {t}]
Actually what I need is a function in a form Y = Y(X) to get this kind a result:
http://postimg.org/image/6n8f839yp/
Best!
Vito
|
I would say
$x^2+y^2=(\cos^2t+\sin^2t) (2 - \cos^2(3t)) = 2 - \cos^2(3t)=1+\sin^2(3t)$
$\Rightarrow $equations in polar coordinates:
$r = \sqrt{1+ \sin^2(3t)}=\cdots=\sqrt{1+ \sin^2t(3 \cos^2t-sin^2t)^2}$
and transformation from polar into Cartesian coordinates:
$\displaystyle \,\cos t = \frac{x}{r},\,\, \sin t= \frac{y}{r},\,\,r=\sqrt{x^2+y^2}$
$\displaystyle \Rightarrow r^2=1+\frac{y^2}{r^2}\left(\frac{3x^2}{r^2}-\frac{y^2}{r^2}\right)^2\Rightarrow \cdots \Rightarrow r^8 - r^6 - y^2 (3 x^2 - y^2)^2 = 0$
$\Rightarrow \,\,$Cartesian:
$(x^2+y^2)^4 - (x^2+y^2)^3 -y^2 (3 x^2 - y^2)^2=0$
$x^8 + 4x^6 y^2 - x^6 + 6x^4 y^4 - 12x^4 y^2 + 4x^2 y^6 + 3x^2 y^4 + y^8 - 2y^6 = 0$
See: Wolfram
|
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"timestamp": "2023-03-29T00:00:00",
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|
Moore-Penrose pseudoinverse of a matrix that is invertible in each column block 4I am not in major in math. But currently I am working with a couple of matrices in the form like this:
\begin{equation}
\left[\begin{array}{rrrrrrrrrrrr}
1 & 0 & 0 & 0 & 0& 0& 0& 0& 0& 0& 0& 0\\
0 & 1 &0.5 &0.5 & 0& 0& 0& 0& 0& 0& 0& 0\\
0 & 0 &0.5 &0.5 & 1& 0.5& 0.5& 0& 0& 0& 0& 0\\
0 & 0 & 0 & 0 & 0& 0.5& 0.5& 1& 0.5& 0.5& 0& 0\\
0 & 0 & 0 & 0 & 0& 0& 0& 0& 0.5& 0.5& 1& 0\\
0 & 0 & 0 & 0 & 0& 0& 0& 0& 0& 0& 0& 1
\end{array}\right]
\end{equation}
If we consider only nonzero elements, 1-3 columns is invertible, 4-6 columns is invertible, and so on. Is there any efficient way to compute Moore-Penrose pseudoinverse for this type of matrices?
|
Construction of a block pseudoinverse is an unanswered challenge in my research area. It's easy to build block inverses of Toeplitz matrices which are nonsingular.
Here is your data with symmetry axes:
$$
\mathbf{A} =
\left(
\begin{array}{cccccc|cccccc}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & \frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & \frac{1}{2} & \frac{1}{2} & 1 & \frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 & 0
& 0 \\\hline
0 & 0 & 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} & 1 & \frac{1}{2} & \frac{1}{2} & 0
& 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
\end{array}
\right),
$$
$$
\mathbf{A}^{\dagger} =
\left(
\begin{array}{rrr|rrr}
56 & 0 & 0 & 0 & 0 & 0 \\
0 & 41 & -11 & 3 & -1 & 0 \\
0 & 15 & 11 & -3 & 1 & 0 \\
0 & 15 & 11 & -3 & 1 & 0 \\
0 & -11 & 33 & -9 & 3 & 0 \\
0 & -4 & 12 & 12 & -4 & 0 \\\hline
0 & -4 & 12 & 12 & -4 & 0 \\
0 & 3 & -9 & 33 & -11 & 0 \\
0 & 1 & -3 & 11 & 15 & 0 \\
0 & 1 & -3 & 11 & 15 & 0 \\
0 & -1 & 3 & -11 & 41 & 0 \\
0 & 0 & 0 & 0 & 0 & 56 \\
\end{array}
\right).
$$
Below are gray scale renderings; $\mathbf{A}$ on the left, $|\mathbf{A}^{\dagger}|$ on the right.
This is an open problem.
|
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|
If there exist real numbers $a,b,c,d$ for which $f(a),f(b),f(c),f(d)$ form a square on the complex plane.Find the area of the square. For all real numbers $x$,let the mapping $f(x)=\frac{1}{x-i},\text{where} i=\sqrt{-1}$.If there exist real numbers $a,b,c,d$ for which $f(a),f(b),f(c),f(d)$ form a square on the complex plane.Find the area of the square.
$f(a)=\frac{1}{a-i}=\frac{a+i}{a^2+1}=(\frac{a}{a^2+1},\frac{1}{a^2+1})$
Similarly,$f(b)=\frac{1}{b-i}=\frac{b+i}{b^2+1}=(\frac{b}{b^2+1},\frac{1}{b^2+1})$
Similarly,$f(c)=\frac{1}{c-i}=\frac{c+i}{c^2+1}=(\frac{c}{c^2+1},\frac{1}{c^2+1})$
Similarly,$f(d)=\frac{1}{d-i}=\frac{d+i}{d^2+1}=(\frac{d}{d^2+1},\frac{1}{d^2+1})$
Now the area of the square$=(\frac{a}{a^2+1}-\frac{b}{b^2+1})^2+(\frac{1}{a^2+1}-\frac{1}{b^2+1})^2$
But this expression simplifies to $\frac{(a-b)^2}{(1+a^2)(1+b^2)}$.
How should i prove the area of the square to be $\frac{1}{2}$.Is my approach not correct?Please help me.Thanks.
|
Hint: The function $f$ bijectively maps the extended real line into the circle through the origin having centre at $\frac{i}{2}$. The area of a square inscribed in a circle with radius $r$ is $2r^2$.
|
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|
If the biquadratic $x^4+ax^3+bx^2+cx+d=0(a,b,c,d\in R)$ has $4$ non real roots,two with sum $3+4i$ and the other two with product $13+i$ If the biquadratic $x^4+ax^3+bx^2+cx+d=0(a,b,c,d\in R)$ has $4$ non real roots,two with sum $3+4i$ and the other two with product $13+i$.Find the value of $b$.
Since there are four non real roots,so i let the roots as $\alpha,\bar{\alpha},\beta,\bar{\beta}$.
Then i let $\alpha+\beta=3+4i,\bar{\alpha}\bar{\beta}=13+i$
$\therefore\alpha+\beta=3+4i,\alpha\beta=13-i$
But i am stuck now,how should i proceed now and find the value of $b$.Please help me.
Thanks.
|
By Vieta’s theorem, $b$ is the sum of all products of pairwise distinct roots, i. e.
$$
\begin{align*}
b &= \alpha\overline\alpha + \alpha\beta + \alpha\overline\beta + \overline{\beta\alpha} + \beta\overline\alpha + \beta\overline\beta\\
&= \overline{(\alpha+\beta)}\cdot (\alpha+\beta) + \alpha\beta + \overline{\alpha\beta}\\
&= (3-4i)(3+4i) + 13-i + 13+i\\
&= 9 + 16 + 26 = 51,
\end{align*}
$$
so $b=51$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Is the odd part of even almost perfect numbers (other than the powers of two) not almost perfect? Let $\sigma(x)$ denote the sum of the divisors of $x$. A number $M$ is called almost perfect if $\sigma(M) = 2M - 1$.
If $M$ is an even almost perfect number, then the only known examples for $M$ are $M = 2^k$, where $k \geq 1$. (Since $\sigma(1) = 2\cdot{1} - 1$, $1$ is an odd almost perfect and is currently the only known such number.)
Antalan and Tagle (in a 2004 preprint titled Revisiting forms of almost perfect numbers) show that, if $N \neq 2^k$ is an even almost perfect number, then $N$ takes the form $N = {2^r}{b^2}$ where $r \geq 1$ and $b$ is an odd composite. We call $b^2$ the odd part of the even almost perfect number $N$.
Since $N$ is almost perfect, we have
$$(2^{r+1} - 1)\sigma(b^2) = \sigma(2^r)\sigma(b^2) = \sigma({2^r}{b^2}) = \sigma(N) = 2N - 1 = {2^{r+1}}b^2 - 1.$$
So we have
$$\sigma(b^2) = \frac{{2^{r+1}}b^2 - 1}{2^{r+1} - 1} = b^2 + \frac{b^2 - 1}{2^{r+1} - 1}.$$
Now,
$$2b^2 - \sigma(b^2) = b^2 -\frac{b^2 - 1}{2^{r+1} - 1}.$$
If the odd part $b^2$ is also almost perfect, then we have
$$1 = 2b^2 - \sigma(b^2) = b^2 -\frac{b^2 - 1}{2^{r+1} - 1},$$
which, since $b > 1$, gives
$$2^{r+1} - 1 = 1 \Longleftrightarrow r = 0.$$
This contradicts $r \geq 1$.
Consequently, since $b^2$ is deficient, we can write $\sigma(b^2) = 2b^2 - c$, where $c > 1$. (We then call $b^2$ a $c$-deficient number.)
My question is: Is this proof correct?
|
Is this proof correct?
I have not found any errors in your proof, so it looks correct to me.
|
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|
Simplify the fraction with radicals I want to simplify this fraction
$$ \frac{\sqrt{6} + \sqrt{10} + \sqrt{15} + 2}{\sqrt{6} - \sqrt{10} + \sqrt{15} - 2} $$
I've tried to group up the denominator members like $ (\sqrt{6} + \sqrt{15}) - (\sqrt{10} + 2) $ and then amplify with $ (\sqrt{6} + \sqrt{15}) + (\sqrt{10} + 2) $
|
HINT :
$$\sqrt 6\pm\sqrt{10}+\sqrt{15}\pm 2=\sqrt 3(\sqrt 5+\sqrt 2)\pm\sqrt{2}(\sqrt 5+\sqrt 2)$$
$$=(\sqrt 5+\sqrt 2)(\sqrt 3\pm\sqrt 2)$$
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Digamma function ratio limit I'd like to know how to go about proving the result
$$\lim_{x\to-n}\dfrac{\psi(x)}{\Gamma(x)} = (-1)^{n-1}n!$$
as it appears on p.g. 2 here http://www.math.usm.edu/lambers/mat415/lecture16.pdf.
|
Recall that the analytic continuation of the Gamma function can be obtained by using repeatedly the functional relationship $\Gamma(x+1)=x\Gamma(x)$. Then, for $x+m+1>0$, we can write
$$\Gamma (x)=\frac{\Gamma(x+m+1)}{x(x+1)(x+2)\cdots(x+m-1)(x+m)} \tag 1$$
Using $(1)$ reveals that for $x+m+1>0$, the digamma function can be expressed as
$$\psi(x)=\frac{d\log \Gamma (x)}{dx}=\psi(x+m+1)-\sum_{k=0}^m\frac{1}{x+k} \tag 2$$
Therefore, using $(1)$ and $(2)$, the ratio $\frac{\psi(x)}{\Gamma(x)}$ is given by
$$\frac{\psi(x)}{\Gamma(x)}=\left(\psi(x+m+1)-\sum_{k=0}^m\frac{1}{x+k}\right)\left(\frac{x(x+1)(x+2)\cdots(x+m-1)(x+m)}{\Gamma(x+m+1)}\right) \tag 3$$
For $0<n\le m$, we note that for $x=-n$, $\Gamma(x+m+1)=(m-n)!$ and $\psi(x+m+1)=H_{m+1-n}-\gamma$. Therefore, we have
$$\begin{align}
\lim_{x\to -n}\frac{\psi(x)}{\Gamma(x)}&=\lim_{x\to -n}\left(\frac{\psi(x+m+1)}{\Gamma(x+m+1)}\left(x(x+1)(x+2)\cdots(x+m-1)(x+m)\right)\right)\\\\
&-\lim_{x\to -n}\left(\frac{x(x+1)(x+2)\cdots(x+m-1)(x+m)}{\Gamma(x+m+1)}\sum_{k=0}^{m}\frac{1}{x+k}\right)\\\\
&=-\lim_{x\to -n}\left(\frac{x(x+1)(x+2)\cdots(x+m-1)(x+m)}{\Gamma(x+m+1)}\sum_{k=0}^{m}\frac{1}{x+k}\right)\\\\
&=-\frac{1}{(m-n)!}\lim_{x\to -n}\sum_{k=0}^m\frac{x(x+1)(x+2)\cdots(x+m-1)(x+m)}{x+k}\\\\
&=-\frac{1}{(m-n)!}\left((-n)(-n+1)(-n+2)\cdots (-2)(-1)(1)(2)\cdots(m-n-1)(m-n)\right)\\\\
&=-(1)^{n+1}n!
\end{align}$$
as was to be shown!
|
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|
$2=1$ Paradoxes repository I really like to use paradoxes in my math classes, in order to awaken the interest of my students. Concretely, these last weeks I am proposing paradoxes that achieve the conclusion that 2=1. After one week, I explain the solution in the blackboard and I propose a new one. For example, I posted the following one some months ago: What is wrong with the sum of these two series?
I would like to increase my repertoire of fake-proofs. I would be glad to read your proposals and discuss them! My students are 18 years old, so don't be too cruel :) Here is my own contribution:
\begin{equation}
y(x) = \tan x
\end{equation}
\begin{equation}
y^{\prime} = \frac{1}{\cos^{2} x}
\end{equation}
\begin{equation}
y^{\prime \prime} = \frac{2 \sin x}{\cos^{3} x}
\end{equation}
This can be rewritten as:
\begin{equation}
y^{\prime \prime} = \frac{2 \sin x}{\cos^{3} x} = \frac{2 \sin x}{\cos x \cdot \cos^{2} x} = 2 \tan x \cdot \frac{1}{\cos^{2} x} = 2yy^{\prime} = \left( y^{2} \right)^{\prime}
\end{equation}
Integrating both sides of the equation $y^{\prime \prime} = \left( y^{2} \right)^{\prime}$:
\begin{equation}
y^{\prime} = y^{2}
\end{equation}
And therefore
\begin{equation}
\frac{1}{\cos^{2} x} = \tan^{2} x
\end{equation}
Now, evalueting this equation at $x = \pi / 4$
\begin{equation}
\frac{1}{(\sqrt{2}/2)^{2}} = 1^{2}
\end{equation}
\begin{equation}
2 = 1
\end{equation}
|
Here's one I just made up. $\log_{b} b^x = x$. And $\log_{b} 1 = 0$.
Let $b = 1, x = 1$, and $b^x = 1$. Then $0 = \log_b 1 = \log_b b^x = x = 1$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1480488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 15,
"answer_id": 5
}
|
A trigonometry equation: $3 \sin^2 \theta + 5 \sin \theta \cos \theta - 2\cos^2 \theta = 0$
$$3 \sin^2 \theta + 5 \sin \theta \cos \theta - 2\cos^2 \theta = 0$$
What are the steps to solve this equation for $ \theta $?
Because, I am always unable to deal with the product $\sin \theta \cos \theta$.
|
Here is a general strategy dealing with these kind of problems. Using summation formulas for sin and cos one can easily prove the following identities
$$\left\{ \matrix{
\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta = 2{\cos ^2}\theta - 1 = 1 - 2{\sin ^2}\theta \hfill \cr
\sin 2\theta = 2\sin \theta \cos \theta \hfill \cr} \right.$$
using these you can write
$$\eqalign{
& {\cos ^2}\theta = {{1 + \cos 2\theta } \over 2} \cr
& {\sin ^2}\theta = {{1 - \cos 2\theta } \over 2} \cr
& \sin \theta \cos \theta = {{\sin 2\theta } \over 2} \cr} $$
put these into your equation to get
$$3\left( {{{1 - \cos 2\theta } \over 2}} \right) + 5\left( {{{\sin 2\theta } \over 2}} \right) - 2\left( {{{1 + \cos 2\theta } \over 2}} \right) = 0$$
simplify to get
$$\eqalign{
& {1 \over 2} - {5 \over 2}\cos 2\theta + {5 \over 2}\sin 2\theta = 0 \cr
& \sin 2\theta - \cos 2\theta + {1 \over 5} = 0 \cr} $$
the next step is to combine linear combinations of $\sin 2\theta $ and $\cos 2\theta $. You can do this again using summation formulas
$$\left\{ \matrix{
\sin x + w\cos x = \sin x + \tan \alpha \cos x = {1 \over {\cos \alpha }}\left( {\sin x\cos \alpha + \sin \alpha \cos x} \right) = {1 \over {\cos \alpha }}\sin \left( {x + \alpha } \right) \hfill \cr
\tan \alpha = w \hfill \cr} \right.$$
In this case
$$\tan \alpha = - 1\,\,\, \to \,\,\,\alpha = {{3\pi } \over 4}\,\,\,\, \to \,\,\,\,\cos \alpha = - {1 \over {\sqrt 2 }}$$
and hence you can write
$$ - \sqrt 2 \sin (2\theta + {{3\pi } \over 4}) + {1 \over 5} = 0$$
or
$$\sin (2\theta + {{3\pi } \over 4}) = {1 \over {5\sqrt 2 }}$$
and hence
$$2\theta + {{3\pi } \over 4} = \left\{ \matrix{
\arcsin \left( {{1 \over {5\sqrt 2 }}} \right) \hfill \cr
\pi - \arcsin \left( {{1 \over {5\sqrt 2 }}} \right) \hfill \cr} \right. + 2n\pi \,\,\,\,\,\,\,\,\,\,\,\,\,n = 1,2,3,...$$
which finally gives
$$\theta = {1 \over 2}\left( {\left\{ \matrix{
\arcsin \left( {{1 \over {5\sqrt 2 }}} \right) \hfill \cr
\pi - \arcsin \left( {{1 \over {5\sqrt 2 }}} \right) \hfill \cr} \right. + 2n\pi - {{3\pi } \over 4}} \right)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1481232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
}
|
Asymptotic of a sum involving binomial coefficients Could you help me to find an asymptotic for this sum?
$$ \sum_{k=0}^{n - 1} (-1)^k {n \choose k} {3n - k - 1 \choose 2n - k} = {n \choose 0} {3n - 1 \choose 2n} - {n \choose 1} {3n - 2 \choose 2n - 1} + ... + (-1)^{n-1} {n \choose n-1} {2n \choose n + 1} $$
I have tried to write binomials through factorials and work with it, but it seems like not right way to evaluate an asymptotic.
Thank you for your help:)
|
Suppose we first seek to evaluate the somewhat more general
$$S_m(n) = \sum_{k=0}^n {n\choose k} (-1)^k
{mn-k-1\choose 2n-k}$$
with $m$ an integer parameter.
Now introduce
$${mn-k-1\choose 2n-k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2n-k+1}} (1+z)^{mn-k-1} \; dz.$$
We thus get for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{mn-1}}{z^{2n+1}}
\sum_{k=0}^n {n\choose k} (-1)^k \frac{z^k}{(1+z)^k} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{mn-1}}{z^{2n+1}}
\left(1-\frac{z}{1+z}\right)^n\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{mn-n-1}}{z^{2n+1}} \; dz
\\ = {mn-n-1\choose 2n}.$$
We see that this is zero when $m=2$ and $m=3$ and non-zero otherwise.
In the problem being asked we are computing
(the last term is not being included)
$$S_3(n) - (-1)^n {2n-1\choose n}$$
so we obtain
$$(-1)^{n+1} {2n-1\choose n}.$$
This can be treated with the asymptotic for the central binomial
coefficient. We get
$$(-1)^{n+1} \frac{n}{2n} {2n\choose n}
= \frac{1}{2} (-1)^{n+1} {2n\choose n}.$$
The central binomial coefficient is
OEIS A000984
and is asymptotic to
$$\frac{4^n}{\sqrt{\pi n}}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1483351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
Calculate limit with a lot of roots $$ \lim_{x\to 0}\frac{\sqrt{1+2x}-\sqrt[\Large3]{1+5x}}{\sqrt[\Large5]{1+x}-\sqrt[\Large5]{1+2x}}$$
Multiplication by conjugate hasn't worked. I need a hint to start.
|
Applying L'Hospital's rule, we get
$$\lim_{x\to 0}\frac{\sqrt{1+2x}-\sqrt[3]{1+5x}}{\sqrt[5]{1+x}-\sqrt[5]{1+2x}}$$
$$=\lim_{x\to 0}\frac{\frac{2}{2\sqrt{1+2x}}-\frac{5}{3\sqrt[2/3]{1+5x}}}{\frac{1}{5\sqrt[4/5]{1+x}}-\frac{2}{5\sqrt[4/5]{1+2x}}}$$
$$=\frac{\frac{2}{2}-\frac{5}{3}}{\frac{1}{5}-\frac{2}{5}}=\frac{\frac{-2}{3}}{\frac{-1}{5}}=\color{red}{\frac{10}{3}}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1483433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Finding the second Eigenvector of a repeated Eigenvalue The matrix $A= \left(\begin{matrix} 2 & -3 & 6\\0 & 5 & -6\\0 & 1 & 0\end{matrix}\right)$ for which I am trying to find the Eigenvalues and Eigenvectors.
I have repeated Eigenvalues of $\lambda_1 = \lambda_2 = 2$ and $\lambda_3 = 3$.
After finding the matrix substituting for $\lambda_1$ and $\lambda_2$, I get the matrix $\left(\begin{matrix} 0 & 1 & -2\\0 & 0 & 0\\0 & 0 & 0\end{matrix}\right)$ as the row reduced echelon form.
So far, I have managed to obtain one of the eigenvectors, which is $\left(\begin{matrix} 0\\2\\1\end{matrix}\right)$. This was done by setting the third column of the RREF as the free parameter.
The solutions show that there is a second eigenvector for this eigenvalue, which is $\left(\begin{matrix} 1\\0\\0\end{matrix}\right)$.
How can I obtain this second eigenvector?
|
Let's solve $(A - 2I) v = 0$ for vectors $v$.
The equation resolves to:
$$ \begin{pmatrix} 0 & -3 & 6 \\ 0 & 3 & -6 \\ 0 & 1 & -2 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix} = 0$$
which imposes the condition $v_2 = 2 v_3$, with no condition on $v_1$. The general solution is, for scalars $a$ and $b$:
$$\begin{pmatrix} a \\ 2b \\ b \end{pmatrix} = a \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}+ b \begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix}$$
Thus, $(1,0,0)^t$ and $(0,2,1)^t$ are the required eigenkets.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1483971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Finding all positive integer solutions for $x+y=xyz-1$ How do I manually solve $x+y=xyz-1$ assuming that $x, y$ and $z$ are positive integers? I was able to guess all possible solutions, but I do not know how to show that these are the only ones:
$x=1, y=1, z=3$
$x=1, y=2, z=2$
$x=2, y=1, z=2$
$x=2, y=3, z=1$
$x=3, y=2, z=1$
Any hints would be appreciated.
|
Alternative solution:
$x\mid y+1$ and $y\mid x+1$, so $x-1\le y\le x+1$. Simply check cases:
*
*If $x-1=y$, then $y\mid x+1$ gives $x-1\mid x+1$, i.e. $x-1\mid 2$, i.e. $x\in\{3,2\}$, so $y\mid x+1\in\{4,3\}$, so $y\in\{1,2,3,4\}$.
*If $x=y$, then $x\mid 1, y\mid 1$.
*If $x+1=y$, then $x\mid y+1=x+2$, so $x\mid 2$, so $(x,y)\in\{(1,2),(2,3)\}$.
It's left to check $11$ cases (knowing $(x,y)$ find $z$).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1484330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
}
|
solving the determinants of matrices So i was given this question
Evaluate by first adding all other rows to the first row
My solution:
$
det
\left[ {\begin{array}{cc}
x-1 & 2 & 3 \\
2 & -3& x-2 \\
-2 & x & -2
\end{array} } \right]
$ = $
\left[ {\begin{array}{cc}
x-1 & 2 & 3 \\
2 & -3& x-2 \\
-2 & x & -2
\end{array} } \right]
$ = $(x-1)
\left[ {\begin{array}{cc}
1 & 0 & 0 \\
2 & -5& x-4 \\
-2 & x & -2
\end{array} } \right]
$ = $(x-1)
\left[ {\begin{array}{cc}
-5 & x-4\\
x+2 & 0&
\end{array} } \right]
$=$(x-2)(x-4)(x+2)$ = $(x-2)(x^2-2x-8)$
Is this correct?
|
No, this is not correct. You may add any row to any other row without changing the determinant. Using your hint you would get
$$
\det\begin{bmatrix}
x-1 & 2 & 3\\
2 &-3 &x-2 \\
-2 & x & -2
\end{bmatrix} = \det\begin{bmatrix}
x-1 & x-1 & x-1\\
2 &-3 &x-2 \\
-2 & x & -2
\end{bmatrix}
$$
Now we can subtract the first column from the second and third and get
$${} = \det\begin{bmatrix}
x-1 & 0 & 0\\
2 &-5 &x-4 \\
-2 & x+2 & 0
\end{bmatrix}$$
Now we can use the Laplace formula on the first row, because there are a lot of zeros and get
$${}=(x-1)\det\begin{bmatrix}-5 & x-4\\x+2 & 0\end{bmatrix}
= -(x-1)(x-4)(x+2). $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1488062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Induction proof I'm having trouble with: $1+x+x^2+x^3+...+x^n = \frac{1-x^{n+1}}{1-x}$ So I'm being asked to use induction to prove that for every $x\in\{a\ |\ a\in R, a\neq 1\}$ and for every $n\in N$
$$
1+x+x^2+x^3+...+x^n = \frac{1-x^{n+1}}{1-x}
$$
I have no trouble proving it for $n=1$ :
$$
1+x = \frac{1-x^2}{1-x}
$$
Factor the polynomial:
$$
1+x = \frac{(1-x)(1+x)}{1-x}
$$
Divide by $(1-x)$
$$
1+x=1+x
$$
And there you have it. The trouble I'm running into is with the induction step. If we assume that our claim is true for $n=k$ then
$$
1+x+x^2+x^3+...+x^k+x^{k+1} = \frac{1-x^{k+2}}{1-x}
$$
Or in other words,
$$
\frac{1-x^{k+1}}{1-x} + x^{k+1} = \frac{1-x^{k+2}}{1-x}
$$
Can someone help with this? I'm having some trouble with the factoring and the book I'm studying from isn't very clear on how they proved the last equation is true.
Thanks in advance :)
|
From where you are stuck, there is only one little step left: reduce both terms of the LHS to the same denominator. That is, $$\frac{1-x^{k+1}}{1-x} + x^{k+1} = \frac{1-x^{k+1}+(1-x) x^{k+1}}{1-x}=\frac{1-x^{k+1}+x^{k+1} - x\cdot x^{k+1}}{1-x} = \frac{1-x^{k+2}}{1-x}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1488283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Limit using L'Hopital's Rule I am trying to solve the limit
$$ \lim_{x\to 0}\left(\frac{\frac{1}{x \ln2} - \frac{1}{2^x-1} - \frac{1}2}x\right)$$
using L'Hopital's Rule. However, it seems that I am doing something incorrectly, as after using the rule a couple of times, it does not get me anywhere but only complicates the existing limit.
What am I missing?
|
We can proceed as follows
\begin{align}
L &= \lim_{x \to 0}\dfrac{\dfrac{1}{x\log 2} - \dfrac{1}{2^{x} - 1} - \dfrac{1}{2}}{x}\notag\\
&= \frac{1}{2\log 2}\lim_{x \to 0}\frac{2(2^{x} - 1) - 2x\log 2 - x\log 2(2^{x} - 1)}{x^{2}(2^{x} - 1)}\notag\\
&= \frac{1}{2\log 2}\lim_{x \to 0}\dfrac{2(2^{x} - 1) - 2x\log 2 - x\log 2(2^{x} - 1)}{x^{3}\cdot\dfrac{(2^{x} - 1)}{x}}\notag\\
&= \frac{1}{2(\log 2)^{2}}\lim_{x \to 0}\frac{2(2^{x} - 1) - 2x\log 2 - x\log 2(2^{x} - 1)}{x^{3}}\notag\\
&= \frac{1}{2(\log 2)^{2}}\lim_{x \to 0}\frac{2^{x}\log 2 - \log 2 - (\log 2)^{2}x\cdot 2^{x}}{3x^{2}}\text{ (via L'Hospital's Rule)}\notag\\
&= \frac{1}{2(\log 2)^{2}}\lim_{x \to 0}\frac{ - (\log 2)^{3}x\cdot 2^{x}}{6x}\text{ (via L'Hospital's Rule)}\notag\\
&= -\frac{\log 2}{12}\notag
\end{align}
Alternatively it is much easier to use Taylor series and get
\begin{align}
\frac{1}{2^{x} - 1} &= \frac{1}{e^{kx} - 1}\text{ where }k = \log 2\notag\\
&= \dfrac{1}{kx + \dfrac{k^{2}x^{2}}{2} + \dfrac{k^{3}x^{3}}{6} + o(x^{3})}\notag\\
&= \frac{1}{kx}\left(1 + \frac{kx}{2} + \frac{k^{2}x^{2}}{6} + o(x^{2})\right)^{-1}\notag\\
&= \frac{1}{kx}\left(1 - \frac{kx}{2} + \frac{k^{2}x^{2}}{12} + o(x^{2})\right)\notag\\
&= \frac{1}{kx} - \frac{1}{2} + \frac{kx}{12} + o(x)\notag\\
&= \frac{1}{x\log 2} - \frac{1}{2} + \frac{x\log 2}{12} + o(x)\notag
\end{align}
and therefore $$\frac{1}{x\log 2} - \frac{1}{2^{x} - 1} - \frac{1}{2} = -\frac{x\log 2}{12} + o(x)$$ and upon dividing the above equation by $x$ the desired limit is clearly seen to be $-(\log 2)/12$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1488459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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|
How do I solve the following differential equation?
$\frac{dy}{dx}+ \sqrt\frac{1-y^2}{1-x^2}=0$. How do I substitute $y$? Any help would be appreciated thanks.
|
$$\frac{dy}{dx}+ \sqrt\frac{1-y^2}{1-x^2}=0$$ or,
$$\frac{dy}{dx}=- \sqrt\frac{1-y^2}{1-x^2}$$ or,
$$\frac{dy}{\sqrt{1-y^2}}=- \frac{dx}{\sqrt{1-x^2}}$$ or,
$$\sin^{-1}y=-\sin^{-1}x+c$$ or,
$$\sin^{-1}y+\sin^{-1}x=c$$
EDIT:
$$\frac{dy}{dx}+ \sqrt\frac{1-y^2}{1-x^2}=0$$ or,
$$\frac{dy}{dx}=- \sqrt\frac{1-y^2}{1-x^2}$$ or,
$$\frac{dy}{dx}=- \sqrt\frac{1-y^2}{1-x^2} \left(\frac{xy-\sqrt{(1-x^2)(1-y^2)}}{xy-\sqrt{(1-x^2)(1-y^2)}}\right)$$ or,
$$\frac{dy}{dx}\left(\frac{xy-\sqrt{(1-x^2)(1-y^2)}}{\sqrt{1-y^2}}\right)=- \left(\frac{xy-\sqrt{(1-x^2)(1-y^2)}}{\sqrt{1-x^2}}\right)$$ or,
$$\frac{dy}{dx}\left(\frac{xy}{\sqrt{1-y^2}}-\sqrt{1-x^2}\right)=- \left(\frac{xy}{\sqrt{1-x^2}}-\sqrt{1-y^2}\right)$$ or,
$$-\sqrt{1-y^2}+\frac{dy}{dx}\cdot \frac{xy}{\sqrt{1-y^2}}=- \frac{xy}{\sqrt{1-x^2}}+\frac{dy}{dx}\cdot \sqrt{1-x^2}$$ or,
$$d\left(-x\sqrt{1-y^2}\right)=d\left(y\sqrt{1-x^2}\right)$$ or,
$$c-x\sqrt{1-y^2}=y\sqrt{1-x^2}$$ or,
$$x\sqrt{1-y^2}+y\sqrt{1-x^2}=c$$
The method is quite laborious and boring. At present, I cannot remember the shortcut. I'll post it when it comes to my mind.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1492080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
}
|
Find $k$ such that the area is an integer
For some positive integers k, the parabola with equation $y = \frac{x^2}{k} - 5$ intersects
the circle with equation $x^2 + y^2 = 25$ at exactly three distinct points A, B and C.
Determine all such positive integers k for which the area of $\triangle ABC$ is an integer.
Point $A = (0, -5)$ and also, I found that the area is:
$\frac{1}{2} BH$ the area of half the triangle $= \frac{1}{2} xy$ so just multiply by $2$ in the end.
In the end, I ended with the coordinate for $B$ as
$B = (\sqrt{10k + k^2}, 5 + k)$.
So the area of the WHOLE Triangle (ABC) is:
$A = \sqrt{k}\sqrt{10 + k}(10 + k)$
Which value of $k$ that will make this an integer is the hard part.
Then I found that:
$A = (10 + k)\sqrt{(k + 5)^2 - 25}$ but I still cannot proceed.
|
Simplifying $x^2+(\frac{x^2}{k}-5)^2=25$ gives
$$x^2(x^2+k^2-10k)=0.$$
In order for this equation to have three distinct real roots, we need to have $10k-k^2\gt 0,$ i.e. $0\lt k\lt 10$.
Then, $$x=0,\pm\sqrt{10k\color{red}{-}k^2}.$$
The area of $\triangle{ABC}$ is given as
$$[\triangle{ABC}]=\frac 12\times 2\sqrt{10k-k^2}\times (5-k+5)=(10-k)\sqrt{k(10-k)}.$$
Checking each of $k=1,2,\cdots, 9$ gives that the answer is $k=\color{red}{1,2,5,8,9}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1492426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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|
If $ \lim _{x\rightarrow -\infty }\sqrt {x^{2}+6x+3}+ax+b=1 $ then find $a+b$ $ \lim _{x\rightarrow -\infty }\sqrt {x^{2}+6x+3}+ax+b=1 $
if I use this $$\lim _{x\rightarrow -\infty }\sqrt {ax^{2}+bx+c}=\left| x+\dfrac {b} {2a}\right| $$
I find $a=1,b=4$ but if I try to multiple by its conjugate
$$\lim _{x\rightarrow -\infty }\dfrac {x^{2}+6x+3-\left( ax+b\right) ^{2}} {\left| x\right| \sqrt {1+\dfrac {6} {x}+\dfrac {3} {x^{2}}}-ax-b}=\lim _{x\rightarrow -\infty } \dfrac {x\left( x+6+\dfrac {3} {x}-a^{2}x-2ab-\dfrac {b^{2}} {x}\right) } {x\left( -\sqrt {1+\dfrac {6} {x}+\dfrac { {3}} {x^{2}}}-a-\dfrac {b} {x}\right) }$$
in order to lim of this equal to $1$ $$x\left(\underbrace { 1-a^{2}}_0\right) +6-2ab=-1-a$$
if $a=1$ then $ b=4$
if $a=-1$ so $b=-3$ but wolfram says $ \lim _{x\rightarrow -\infty }\sqrt {x^{2}+6x+3}-x-3=\infty$
can u tell me where is my mistake?
|
Let's start this step by step, $\lim_{x\to -\infty} \sqrt{x^2 + bx + c} + x= \lim_{x\to -\infty} \sqrt{(x + \frac{b}{2})^2 + c - \frac{b^2}{4}} + x = \lim_{x\to -\infty} |x+\frac{b}{2}| +x = -\frac{b}{2}$ Now everything is simpler than ever, $\lim_{x\to -\infty} \sqrt(x^2 + 6x + 3) + x +(a-1)x b = \lim_{x\to -\infty} -3+b + (a-1)x = 1$, now it is clear that $a = 1, b = 4$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1492778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Complex equation The problem says to solve the given complex equation:
$$
z^4-\left[
\frac{\sqrt3}{2}i^{21}+
\frac{\sqrt3}{2}i^{9}+
\frac{8}{(1+i)^6}\right]^9=0
$$
The solution is this:
$$
\cos\left(
\frac{\pi}{8}+\frac{k\pi}{2}
\right)+i\sin\left(
\frac{\pi}{8}+\frac{k\pi}{2}
\right)
,k=0,1,2,3
$$
My problem is that I can't get this solution, I've tried multiple times to solve it but I always end up getting this solution:
$$z^4=(i+\sqrt3*i)^9$$
Could you help me with this? Thanks in advance.
|
$$z^4-\left(\frac{\sqrt3}{2}i^{21}+\frac{\sqrt3}{2}i^{9}+\frac{8}{(1+i)^6}\right)^9=0\Longleftrightarrow$$
$$z^4-\left(\frac{\sqrt3}{2}i+\frac{\sqrt3}{2}i+\frac{8}{((1+i)^3)^2}\right)^9=0\Longleftrightarrow$$
$$z^4-\left(\frac{\sqrt3}{2}i+\frac{\sqrt3}{2}i+\frac{8}{((1+i)(1+i)^2)^2}\right)^9=0\Longleftrightarrow$$
$$z^4-\left(\frac{\sqrt3}{2}i+\frac{\sqrt3}{2}i+\frac{8}{((1+i)2i)^2}\right)^9=0\Longleftrightarrow$$
$$z^4-\left(\frac{\sqrt3}{2}i+\frac{\sqrt3}{2}i+\frac{8}{2^2\cdot -2i}\right)^9=0\Longleftrightarrow$$
$$z^4-\left(\frac{\sqrt3}{2}i+\frac{\sqrt3}{2}i+\frac{8}{-8i}\right)^9=0\Longleftrightarrow$$
$$z^4-\left(\frac{\sqrt3}{2}i+\frac{\sqrt3}{2}i+i\right)^9=0\Longleftrightarrow$$
$$z^4-\left(i(1+\sqrt{3})\right)^9=0\Longleftrightarrow$$
$$z^4-\left(i(1+\sqrt{3})\right)^9=0\Longleftrightarrow$$
$$z^4=\left(i(1+\sqrt{3})\right)^9\Longleftrightarrow$$
$$z^4=\left|\left(i(1+\sqrt{3})\right)^9\right|\cdot e^{\arg\left(\left(i(1+\sqrt{3})\right)^9\right)i}\Longleftrightarrow$$
$$z^4=(1+\sqrt{3})^9\cdot e^{\frac{\pi}{2}i}\Longleftrightarrow$$
$$z=\left((1+\sqrt{3})^9\cdot e^{\left(\frac{\pi}{2}+2\pi k\right)i}\right)^{\frac{1}{4}}\Longleftrightarrow$$
$$z=(1+\sqrt{3})^{\frac{9}{4}}\cdot e^{\frac{1}{4}\left(\frac{\pi}{2}+2\pi k\right)i}$$
With $k\in\mathbb{Z}$ and $k:0-3$
So the solutions are:
$$z_0=(1+\sqrt{3})^{\frac{9}{4}}\cdot e^{\frac{1}{4}\left(\frac{\pi}{2}+2\pi \cdot 0\right)i}=i(1+\sqrt{3})^{\frac{9}{4}}$$
$$z_1=(1+\sqrt{3})^{\frac{9}{4}}\cdot e^{\frac{1}{4}\left(\frac{\pi}{2}+2\pi \cdot 1\right)i}=i(1+\sqrt{3})^{\frac{9}{4}}$$
$$z_2=(1+\sqrt{3})^{\frac{9}{4}}\cdot e^{\frac{1}{4}\left(\frac{\pi}{2}+2\pi \cdot 2\right)i}=i(1+\sqrt{3})^{\frac{9}{4}}$$
$$z_3=(1+\sqrt{3})^{\frac{9}{4}}\cdot e^{\frac{1}{4}\left(\frac{\pi}{2}+2\pi \cdot 3\right)i}=i(1+\sqrt{3})^{\frac{9}{4}}$$
|
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|
If $1+2^n+4^n$ is prime number then prove that $n=3^k$ for some $k\in \mathbb{N}$. If $1+2^n+4^n$ is prime number then prove that $n=3^k$ for some $k\in \mathbb{N}$.
I've looked at https://math.stackexchange.com/a/186723/283318 for an inspiration. In base 2, $1+2^n+4^n=10\ldots010\ldots01$ and I am completely stuck.
|
If $n$ is even, then $3\mid 1+2^n+4^n$, so if $1+2^n+4^n$ is prime, then $1+2^n+4^n=3$, so $n=0$.
If $n$ is odd, for contradiction assume $n=3^kr$ for some $r\ge 2,\, \gcd(r,6)=1$.
$$\left(2^n-1\right)\left(4^n+2^n+1\right)=2^{3n}-1=2^{3^{k+1}r}-1$$
$$=\left(2^{3^{k+1}}-1\right)\left(2^{3^{k+1}r-3^{k+1}}+2^{3^{k+1}r-3^{k+1}-1}+\cdots+1\right)$$
$$\implies 2^{3^{k+1}}-1\mid \left(2^{3^kr}-1\right)\left(4^n+2^n+1\right)$$
But $\gcd\left(2^{3^{k+1}}-1,2^{3^{k}r}-1\right)=2^{3^k}-1$ (see here why).
$$\implies \frac{2^{3^{k+1}}-1}{2^{3^k}-1}\mid 4^n+2^n+1$$
$$\iff 4^{3^k}+2^{3^k}+1\mid 4^n+2^n+1$$
But $4^{3^k}+2^{3^k}+1< 4^n+2^n+1$, so $4^n+2^n+1$ cannot be prime.
|
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|
How to evaluate the definite integral by the limit definition $\int_{-1}^1 x^3 dx$? Solve the definite integral by the limit definition:
$$\int_{-1}^1 x^3 dx$$
The formula:
$$\int_a^bf(x)dx= \lim_{n\rightarrow \infty} \sum_{i=1}^n f(c_i)\Delta x_i$$
Get the variables:
$$\Delta x_i : \frac{b-a}{n} = \frac{1-(-1)}{n} = \frac{2}{n}$$
$$f(c_i) : a + i(\Delta x_i) = \left(-1+\frac{2i}{n}\right)$$
Now plug them into the formula I get:
$$\lim_{n\rightarrow \infty} \sum_{i=1}^n \left(-1+\frac{2i}{n}\right)^3\left(\frac{2}{n}\right)$$
Take out the delta and distribute the cube:
$$\lim_{n\rightarrow \infty} \left(\frac{2}{n}\right) \sum_{i=1}^n \left(-1+\frac{2i}{n}\right)^3 = \lim_{n\rightarrow \infty} \left(\frac{2}{n}\right) \sum_{i=1}^n \left(-1+\frac{8i^3}{n^3}\right) $$
Expand the summation and take out constants, and properties of Simga:
$$\lim_{n\rightarrow \infty} \left(\frac{2}{n}\right) \left[- \sum_{i=1}^n 1+ \frac{8}{n^3} \sum_{i=1}^n i^3\right] = \lim_{n\rightarrow \infty} \left(\frac{2}{n}\right) \left[-n + \frac{8}{n^3}\left( \frac{n^2(n+1)^2}{4}\right)\right]$$
Distribute the $\frac{2}{n}$ and simplify:
$$\lim_{n\rightarrow \infty}\left[-2 + \frac{16}{n^4}\left( \frac{n^2(n+1)^2}{4}\right)\right] = \lim_{n\rightarrow \infty}\left[-2 + \frac{4}{n^2}\left( \frac{n^2+2n+1}{1}\right)\right]$$
Distribute and simplify/cancel:
$$\lim_{n\rightarrow \infty}\left[-2 + \frac{4n^2}{n^2} + \frac{8n}{n^2} + \frac{4}{n^4}\right] = \lim_{n\rightarrow \infty}\left[-2 + 4 + \frac{8}{n} + \frac{4}{n^4}\right] $$
I keep getting the limit is -2 (-2+4), but the book says it's 0. Where did I go wrong?
|
When you said
"distribute the cube",
you made a mistake.
(mistake in following expansion corrected)
$(-1+2i/n)^3
=-1+3(2i/n)-3(2i/n)^2+(2i/n)^3
$.
Each term has to be considered.
Also,
note that the function is odd.
What happens if
you take terms
symmetrically located
around the origin?
|
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|
How to find the remainder $f(x)$=$1+x+x^2+x^3+x^4+x^5$
What is the remainder when $f(x^{12})$ is divided by $f(x)$?
I think remainder theorem cannot be applied (at least directly) here.
I find that $f(-1^{12})$ - $f(-1)$ is $6$ but $f(1^{12})$ - $f(1)$ is $0$
|
Hint: It is easier to calculate the remainder of $f(x^{12})(x-1)$ when divided by $f(x)(x-1)=x^6-1$, then divide that remainder by $x-1$ at the end.
Or more directly, note that:
$$x^6\equiv 1\pmod {f(x)}$$
so:
$$x^{12}\equiv 1\pmod{f(x)}$$
So what is $f(x^{12})\pmod{f(x)}$?
|
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|
Relation between hypergeometric and gamma functions Show that, for a positive integer $n$, $$F\left(-\frac{n}{2},-\frac{n}{2}+\frac{1}{2};n+\frac{3}{2};-\frac{1}{3}\right)=\left(\frac{8}{9}\right)^n\frac{\Gamma\left(\frac{4}{3}\right)\Gamma\left(n+\frac{3}{2}\right)}{\Gamma\left(\frac{3}{2}\right)\Gamma\left(n+\frac{4}{3}\right)}.$$
I can identify the right hand side (using the definition of the Pochhammer symbol) with: $$\left(\frac{8}{9}\right)^n\frac{\Gamma\left(\frac{4}{3}\right)\Gamma\left(n+\frac{3}{2}\right)}{\Gamma\left(\frac{3}{2}\right)\Gamma\left(n+\frac{4}{3}\right)}=\left(\frac{8}{9}\right)^n\frac{\left(\frac{3}{2}\right)_n}{\left(\frac{4}{3}\right)_n},$$ and get an expression for this in terms of factorials, but I'm not sure how to simplify the left hand side - it gets quite messy!
|
Your formula still holds if $n\notin \mathbb{Z}$. From $F(a,b;c;z)=(1-z)^{c-a-b}F(c-a,c-b;c;z)$, it suffices to calculate
$$F(\frac{3n+3}{2},\frac{3n+2}{2};n+\frac{3}{2};-\frac{1}{3})$$
its value follows directly from a cubic transformation of $_2F_1$ due to Goursat:
$$\tag{1}_2F_1(\frac{3a}{2}, \frac{3a-1}{2}; a+\frac{1}{2};-\frac{z^2}{3}) = {(1+z)^{1-3a}} {_2F_1}(a-\frac{1}{3}, a;2a;\frac{2z(3+z^2)}{(1+z)^3})$$
and the formula of $F(a,b;c;1)$.
Once an identity like $(1)$ has been conjectured, there is a mechanical way to prove it. One constructs a 2nd order ODE satisfied by RHS, then show LHS is also a solution. The calculation tends to be lengthy and not quite do-able by hand.
The Mathematica commands
DifferentialRootReduce[(1+z)^(1-3a)*Hypergeometric2F1[a-1/3,a,2a,2z(3+z^2)/(1+z)^3],z]
and
DifferentialRootReduce[Hypergeometric2F1[3a/2,(3a-1)/2,a+1/2,-z^2/3],z]
shows both sides of $(1)$ are satisfied by
$$3a(3a-1)zy+6a(1+z^2)\frac{dy}{dz}+z(z^2+3)\frac{d^2y}{dz^2}=0$$
Therefore it suffices to show both sides have equal $0$th and $1$st derivative at $z=0$, this is even easier.
|
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If $\phi$ : $F_2[X]/(X^3+X+1)$ $\rightarrow$ $F_2[X]/(X^3+X^2+1)$ is an isomorphism, then prove that $\phi(a)=b+1$, with $a$, $b$ the classes of $X$ Consider two fields:
$K: $ $F_2[X]/(X^3+X+1)$. Let $a$ be the class of $X$ (so $a=X+(X^3+X+1))$
$L: $ $F_2[X]/(X^3+X^2+1)$. Let $b$ be the class of $X$ (so $b=X+(X^3+X^2+1))$
$F_2$ denotes the field with $2$ elements. Field $L$ is ismorphic with $K$.
If $\phi$ : $F_2[X]/(X^3+X+1)$ $\rightarrow$ $F_2[X]/(X^3+X^2+1)$ is an isomorphism, then prove that $\phi(a)=b+1$.
They give as a hint: $b+1$ is a root of $Y^3+Y+1$.
I came up with the following:
Fill in $a$ in $K$, so you get $a^3+a+1$.
Fill in $b+1$ in $L$, so you get $(b+1)^3+(b+1)^2+1 = b^3+b^2+b+1+b^2+1= b^3+b+1$, which is equal to the $a$ part above.
But I don't know if this is a way to actually prove that $\phi(a)=b+1$. How can I do this?
|
Define $[g(X)]$ to be the class of $g(X)$.
There are 8 elements in $F_2[X]/(X^3+X+1):$
$[0],[1],[X],[X^2],[X^3],[X^4],[X^5],[X^6]$.
Note that:$[X^3] = [X+1],
[X^4] = [X^2+X],
[X^5] = [X^2+X+1],
[X^6] = [X^2+1],[X^7] = [1]$
$[X],[X^2],[X^4]$ are roots of $y^3+y+1$
and $[X^3],[X^5],[X^6]$ are roots of $y^3+y^2+1$.
So actually there are three isomorphisms:
$φ_1([X]) = [X^3] = [X+1],
φ_2([X]) = [X^5] = [X^2+X+1],
φ_3([X]) = [X^6] = [X^2+1]$
|
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|
How can one show that limit of $\frac{1}{1-x}$ as $x$ goes to $2$ exists? How can one show that limit of $\frac{1}{1-x}$ as $x$ goes to $2$ exists?
Its limit value is $-1$.
How can I prove this using epsilon and delta?
|
Initially we may guess that
$$
\frac{1}{1-x} \to -1
$$
as $x \to 2$. To prove this, note that we have $x\neq 1$ only if
$$
\bigg| \frac{1}{1-x} - (-1) \bigg| = \bigg|\frac{2-x}{1-x} \bigg| = \bigg|\frac{x-2}{x-1} \bigg|;
$$
we have $0 < |x-2| < 1/2$ only if $||x-1|-1| \leq |x-2| < 1/2$, only if $1/2 < |x-1|$, and only if
$$
\bigg| \frac{x-2}{x-1} \bigg| < 2|x-2|;
$$
given any $\varepsilon > 0$, we have $0 < |x-2| < \varepsilon/2$ only if $2|x-2| < \varepsilon$; hence
we have proved this:
for every $\varepsilon > 0$, we have $0 < |x-2| < \min \{1/2, \varepsilon/2 \}$ only if
$$
\bigg| \frac{1}{1-x} + 1 \bigg| < \varepsilon,
$$
which shows that
$$
\lim_{x \to 2}\frac{1}{1-x} = -1.
$$
|
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|
Solving an equation by telling the value of $x^2+y^2$. I have a problem solving an equation. The equation is:
$xy+x+y=44$ and $x^2y+xy^2=448$
and we have to tell the value of $x^2+y^2$
First I tried solving this by doing the following:
$xy+x+y=44~\to~x+y=44-xy~\to~x^2+2xy+y^2=44^2-88xy+x^2y^2~\Rightarrow$
$\Rightarrow~x^2+y^2=44^2-90xy+x^2y^2$
But from here I didn't know what to do. Could you help me in solving this equation?
|
Let $A = xy$ and $B = x + y$. Then we have
$$A + B = 44$$
$$AB = 448$$
By Viete's formula, $A$ and $B$ are the roots of
$$r^2 - 44r + 448 = 0$$
$$(r - 16)(r - 28) = 0$$
such that
$$(xy, x + y) = (16, 28),~(28, 16)$$
from which it is trivial do deduce $x^2 + y^2$:
$$\begin{align}x^2 + y^2 &= (x + y)^2 - 2xy
\\&= 28^2 - 2(16), 16^2 - 2(28)\\
&=752, 200\end{align}$$
|
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|
How to solve this limit without using L'Hospital's Rule? $$
\lim\limits_{x\to{\infty}}\frac{\arctan \frac{3}{x}}{|\arctan \frac{2}{x}|}
$$
Can anybody help me to solve this one ?
I ve done somethig like this but im not sure if it is the correct aproach.
$$
\lim\limits_{x\to{\infty}}\frac{\arctan\frac{3}{x}}{|\arctan \frac{2}{x}|} =
\lim\limits_{x\to{\infty}}\frac{\arctan \frac{3}{x}}{\arctan \frac{2}{x}} =
\lim\limits_{x\to{\infty}} \frac{\arctan \frac{3}{x}}{\arctan \frac{2}{x}} * \frac{\frac{\frac{3}{x}}{\frac{3}{x}}}{\frac{\frac{2}{x}}{\frac{2}{x}}} = \frac{\lim\limits_{x\to{\infty}}\frac{\arctan \frac{3}{x}}{\frac{3}{x}}}{\lim\limits_{x\to{\infty}}\frac{\arctan \frac{2}{x}}{\frac{2}{x}}}*\lim\limits_{x\to{\infty}}\frac{\frac{3}{x}}{\frac{2}{x}}
$$
and then for each limit with arctan i ve substitued 3/x and 2/x by tan y and tan z
$$
\frac{\lim\limits_{y\to{0^+}}\frac{\arctan \tan y}{\tan y}}{\lim\limits_{z\to{0^+}}\frac{\arctan \tan z}{\tan z}}*\lim\limits_{x\to{\infty}}\frac{\frac{3}{x}}{\frac{2}{x}}
$$
and then for each limit which goes to 0^+ i ve done this
$$
\lim\limits_{y\to{0^+}}\frac{\arctan \tan y}{\tan y} = \lim\limits_{y\to{0^+}}\frac{y}{\tan y} = \lim\limits_{y\to{0^+}}\frac{y}{\frac{\sin y}{\cos y}} = 1
$$
so in the end
$$
\frac{1}{1}*\frac{3}{2} = \frac{3}{2}
$$
|
Since $\arctan{}'(x) = 1/(1 + x^2)$, $arctan'(0) = 1$. Therefore you have
$$\lim_{x\to 0} {\arctan(x)\over x} = 1.$$
This will enable you to do the rest.
|
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|
How to evaluate the following integral ? What substitution will be helpful? $$\int\frac{cotx }{(1-sinx)(secx+1)}dx $$
we can write this as
$$\int \frac {cosecx.cotx}{(cosecx-1)(secx+1)}dx $$
Now $cosecx=t$ gives $cosecx.cotx=-dt $ which appears in the numerator, what to do about $secx+1$ ?
|
$\displaystyle\int\frac{cotx }{(1-sinx)(secx+1)}dx =\int\frac{cos^2x }{sinx(1-sinx)(1+cosx)}dx =\int\frac{1+sinx}{sinx(cosx+1)}dx$
$\displaystyle\int\frac{1+sinx}{sinx(cosx+1)}dx=\int\frac{1}{sinx(cosx+1)}dx+\int\frac{1}{(cosx+1)}dx$
$\displaystyle\int\frac{1}{4sin(x/2)cos^3(x/2)}dx+\int\frac{sec^2(x/2)}{2}dx$
$\displaystyle cos(x/2)=t \Rightarrow dx=\frac{2dt}{-sin(x/2)}$
$\displaystyle -\int\frac{1}{2(1-t^2)\cdot t^3}dt+\int\frac{sec^2(x/2)}{2}dx$
Now usig partial fractions we get ,
$\displaystyle \int\frac{1}{2(1-t^2)\cdot t^3}dt= \frac{-1}{4 t^2} + \frac{ln[t]}{2} - \frac{ln[1 - t^2]}{4}$
$\displaystyle -\int\frac{1}{2(1-t^2)\cdot t^3}dt+\int\frac{sec^2(x/2)}{2}dx=$$\displaystyle \frac{1}{4 cos^2(x/2)} - \frac{ln[cos(x/2)]}{2}+ \frac{ln[sin^2(x/2)]}{4}+tan^2(x/2)+C$
|
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|
How to tell is a matrix is a covariance matrix? How can we know that these matrices are valid covariance matrices?
$$
C= \begin{pmatrix}
1 & -1 & 2 \\
-1 & 2 & -1 \\
2 & -1 & 1 \\
\end{pmatrix}
\\
C= \begin{pmatrix}
4 & -1 & 1 \\
-1 & 4 & -1 \\
1 & -1 & 4 \\
\end{pmatrix}
$$
I know that
*
*$C_{xy}=C_{yx}$ (is symmetric)
*$C_{xx}= \mathrm{Var}[x] \ge 0$ (the diagonal entries are all positive)
But I want to know, are there any other properties?
|
A square matrix is a covariance matrix of some random vector if and only if it is symmetric and positive semi-definite (see here). Positive semi-definite means that
$$
x^{T}Cx\ge0
$$
for every real vector $x$, where $x^T$ is the transpose of the vector $x$.
We have that
\begin{align*}
x^TC_1x
&=\left(\begin{array}{ccc}x_1 & x_2 & x_3\end{array}\right)\left(\begin{array}{ccc}1 & -1 & 2 \\-1 & 2 & -1 \\2 & -1 & 1\end{array}\right)\left(\begin{array}{c}x_1 \\x_2 \\x_3\end{array}\right)\\
&=\left(\begin{array}{ccc}x_1-x_2+2x_3 & -x_1+2x_2-x_3 & 2x_1-x_2+x_3\end{array}\right)\left(\begin{array}{c}x_1 \\x_2 \\x_3\end{array}\right)\\
&=x_1^2+2x_2^2+x_3^2-2x_1x_2+4x_1x_3-2x_2x_3\\
&=(x_1-x_2+x_3)^2+x_2^2+2x_1x_3.
\end{align*}
If we take $x^T=\left(\begin{array}{ccc}1 & 0 & -1\end{array}\right)$, then $x^TC_1x=-2$. Hence, $C_1$ is not a covariance matrix.
We have that
\begin{align*}
x^TC_2x
&=\left(\begin{array}{ccc}x_1 & x_2 & x_3\end{array}\right)\left(\begin{array}{ccc}4 & -1 & 1 \\-1 & 4 & -1 \\1 & -1 & 4\end{array}\right)\left(\begin{array}{c}x_1 \\x_2 \\x_3\end{array}\right)\\
&=\left(\begin{array}{ccc}4x_1-x_2+x_3 & -x_1+4x_2-x_3 & x_1-x_2+4x_3\end{array}\right)\left(\begin{array}{c}x_1 \\x_2 \\x_3\end{array}\right)\\
&=4x_1^2+4x^2+4x_3^2-2x_1x_2+2x_1x_3-2x_2x_3\\
&=3(x_1^2+x^2+x_3^2)+(x_1-x_2+x_3)^2\ge0.
\end{align*}
Hence, $C_2$ is a covariance matrix.
|
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|
Conics - Locus of points The ellipse with equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ has one of its foci at the point $F$. The perpendicular from the origin to the tangent at a point $P(a\cos\theta, b\sin\theta)$ on the ellipse intersects the line $FP$ at a point $G$. Find the locus of $G$ as $\theta$ varies.
|
HINT - modified:
I would say, the locus will circle:
$FP:\,y=\frac{b\sin \theta}{a\cos \theta - e}\,(x+e)$
$SR:\,y=\frac{a}{b}\,\tan \theta\, x=\frac{a}{b}\,\tan \theta\,(x+e)-\frac{a\,e}{b}\,\tan \theta$
$\Rightarrow x+e = a\,\frac{a \cos \theta-e}{e\cos \theta -a}, \quad y = a\,\frac{b \sin \theta}{e\cos \theta -a}$
$e^2=a^2-b^2$
$(x+e^2)+y^2=a^2\left(\frac{(a\cos \theta -e)^2+b^2\sin^2 \theta}{(e\cos \theta -a)^2}\right)=a^2\left(\frac{a^2\cos^2 \theta -2ae\cos \theta +e^2+(a^2 -e^2)\sin^2 \theta}{(e\cos \theta -a)^2}\right)=$
$=a^2\left(\frac{a^2(\cos^2 \theta +\sin^2 \theta)-2ae\cos \theta +e^2\cos^2 \theta}{(e\cos \theta -a)^2}\right)=a^2\left(\frac{(a-e\cos \theta)^2}{(e\cos \theta -a)^2}\right)=a^2$
$\Rightarrow\,$the equation of the locus: $\,(x+e^2)+y^2=a^2$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $5$ divides $3^{3n+1}+2^{n+1}$
Prove that $5$ divides $3^{3n+1}+2^{n+1}$
I tried to prove the result by induction but I couldn't.
The result is true for $n=1$.
Suppose that the result is true for $n$ i.e $3^{3n+1}+2^{n+1}=5k$ for some $k\in \mathbb{N}$. We study the term
$$3^{3n+4}+2^{n+2}=3^{3n+1}3^3+2^{n+1}2$$
I tried to prove that that the difference is a multiple of $5$.
$$3^{3n+1}3^3+2^{n+1}2-3^{3n+1}+2^{n+1}=2(3^{3n+1}\cdot 13+2^n)$$
Therefore it's enough to prove that $3^{3n+1}\cdot 13+2^n$ is a multiple of $5$. But if I do again this method applied to this "new problem" is get something similar. I think that there exist a different method to do this using induction.
|
try this
$$3^{3n+1}3^3+2^{n+1}2-3^{3n+1}+2^{n+1}=3^{3n+1}\cdot26+2^{n+1}$$
$$3^{3n+1}\cdot26+2^{n+1}=3^{3n+1}\cdot25+3^{3n+1}+2^{n+1}$$
since $3^{3n+1}+2^{n+1}=5k$
$$3^{3n+1}\cdot25+3^{3n+1}+2^{n+1}=3^{3n+1}\cdot25+5k$$
the difference is divisable by $5$
|
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|
Maximum value of the integral: $\int _{10}^{19} \frac{\sin x}{1+x^a}dx$
Find the minimum odd value of $a$, where$a>1,~ a \in \mathbb{N}$ such that $$\int_{10}^{19} \frac{\sin x}{1+x^a}dx<\frac{1}{9}$$
ATTEMPT:- Let $I(a)=\int _{10}^{19} \frac{\sin x}{1+x^a}dx$ then from Leibnitz's rule, $$I'(a)=-\int _{10}^{19} \frac{(\sin x)x^a\log x}{(1+x^a)^2}dx$$
Now using By Parts, $\int udv=uv-\int vdu$ with
$$\begin{aligned}
u &= \sin x\log x & &\implies & du &= \cos x\log x+\frac{\sin x}{x} \\
dv &=\frac{x^a}{(1+x^a)^2} & &\implies & v &=-\frac{1}{(1+x^a)\log a} \end{aligned} \\
\implies I'(a)=\sin x\log x\frac{1}{(1+x^a)\log a} -\int _{10}^{19} \frac{\cos x\log x+\frac{\sin x}{x}}{(1+x^a)\log a}$$
which doesn't seem solvable.
Next , since it asks for minimum odd value with respect to $a$, I put $a=3$ in the integral giving me:
$$I=\int _{10}^{19} \frac{\sin x}{1+x^3}dx$$
The indefinite integral is in terms of SinIntegral and CosIntegral functions and even the definite integral is given as visual representation of the integral by Wolfram Mathematica
The text says the answer is $3$.
How am I supposed to evaluate this integral?
|
Use the sequence of inequalities (taking into account $a>1$) $$\int _{10}^{19} \frac{\sin x}{1+x^a}dx <\int _{10}^{19} \frac{|\sin x|}{1+x^a}dx<\int _{10}^{19} \frac{1}{1+x^a}dx<\int _{10}^{19} \frac{1}{x^a}dx=\frac{19^{1-a}-10^{1-a}}{1-a}$$ Now, consider $$f(a)= \frac{19^{1-a}-10^{1-a}}{1-a}$$ $$f(2)=\frac{9}{190}<\frac 19$$ $$f(3)=\frac{261}{72200}<\frac 19$$ $$f(4)=\frac{1953}{6859000}<\frac 19$$ So, the first odd integer value of $a$ is $3$.
Edit
However, and this makes me thinking that there could be some mistake in the question, if we perform a numerical integration $$g(a)=\int _{10}^{19} \frac{\sin x}{1+x^a}dx$$ the result is always negative $$g(2)=-0.0115882$$ $$g(3)=-0.0010402$$ $$g(4)=-0.0000963$$
|
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|
Evaluate $ \int_{1}^{\infty} \frac{\sqrt{x - 1}}{(x + 1)^{2}} ~ \mathrm{d}{x} $. I need to solve the following integral:
$$
I = \int_{1}^{\infty} \frac{\sqrt{x - 1}}{(x + 1)^{2}} ~ \mathrm{d}{x}.
$$
Wolfram Alpha gives the answer as $ \dfrac{\pi}{2 \sqrt{2}} $.
I think it’s achievable by complex analysis, but I really have no idea how. Also, is there a special name for this integral, i.e., does it have some known physical significance?
|
Let $u=\sqrt{x-1}$, $du=\frac{1}{2\sqrt{x-1}}$, then
$$I=2\int_0^\infty\frac{u^2}{(u^2+2)^2}du$$
We can apply partial fractions here.
$$=2\int_0^\infty\left(\frac{1}{u^2+2}-\frac{2}{(u^2+2)^2}\right)du$$
$$=2\int_0^\infty\frac{1}{u^2+2}du-4\int_0^\infty\frac{1}{(u^2+2)^2}du$$
The first integrand is almost $\tan^{-1}$, so we can factor the 2 and apply the substitution $s=\frac{u}{\sqrt{2}}$.
$$I=\sqrt 2\int_0^\infty\frac{1}{s^2+1}ds-4\int_0^\infty\frac{1}{(u^2+2)^2}du$$
$$=\frac{\pi}{\sqrt 2}-4\int_0^\infty\frac{1}{(u^2+2)^2}du$$
Now to tackle the second integral we can use a trig sub.
Let $u=\sqrt{2}\tan (t)$, $du=\sqrt 2 \sec^2(t)dt$.
$$I=\frac{\pi}{\sqrt 2}-4\sqrt 2 \int_0^{\pi/2}\frac{\sec ^2 t}{4\sec^4t}dt$$
$$=\frac{\pi}{\sqrt 2}-\sqrt 2\int _0^{\pi/2}\cos^2(t) dt$$
$$=\frac{\pi}{\sqrt 2}-\sqrt 2 \int_0^{\pi/2}\left(\frac{1}{2}\cos(2t)+\frac{1}{2} \right)dt$$
If we substitute $v=2t$ and split the integral up we get
$$I=\frac{\pi}{\sqrt 2}-\frac{1}{2\sqrt 2} \int_0^\pi \cos(v)dv-\frac{1}{\sqrt 2}\int_0^{\pi/2}1dt$$
The first integral clearly goes to 0 and the second integral becomes $\frac{\pi}{2\sqrt 2}$.
Therefore
$$I=\frac{\pi}{2\sqrt 2}$$
|
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|
How can I find $f(3)$ if $f(2x)=f^2(x)-2f(x)-\frac{1}{2}$? If $f(2x)=f^2(x)-2f(x)-\frac{1}{2}$ and $f(1) = 2$ then find $f(3)$.
Can you give me any hint that I can start with?
|
$f(2x)=f^2(x)-2f(x)-\frac{1}{2}
$
and
$f(1) = 2
$.
I can see how to get
$f(2^n)$
and expressions for
$f(2^{-n})$,
but I don't see how to get
$f(3)$.
I show what I've got so far.
All fairly trivial.
$f(2x)+\frac32
=f^2(x)-2f(x)+1
=(f(x)-1)^2
$
so
$f(x)
=1\pm\sqrt{f(2x)+\frac32}
$.
$x = 0
\implies f(0) = f^2(0)-2f(0)-\frac12
$
or
$f^2(0)-3f(0) = \frac12
$
or
$f^2(0)-3f(0)+9/4
=\frac12+9/4
=\frac{11}{4}
$
or
$(f(0)-\frac32)^2
=\frac{11}{4}
$
or
$f(0)
=\frac32\pm\frac{\sqrt{11}}{2}
$.
$x=1
\implies f(2) = -\frac12
$.
$x=\frac12
\implies f(\frac12) = 1\pm\sqrt{f(1)+\frac32}
=1\pm\sqrt{\frac72}
$.
And that's all.
|
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|
Find polynomial $f(x)$ based on divisibility properties of $f(x)+1$ and $f(x) - 1$ $f(x)$ is a fifth degree polynomial. It is given that $f(x)+1$ is divisible by $(x-1)^3$ and $f(x)-1$ is divisible by $(x+1)^3$. Find $f(x)$.
|
Note that $(x-1)^3$ divides $f(x)+1$ and $f(-x)-1$, so $(x-1)^3$ divides the sum $f(x)+f(-x)$.
Similarly, we can note that $(x+1)^3$ divides $f(x)+f(-x)$.
Therefore, $(x+1)^3(x-1)^3$ divides $f(x)+f(-x)$, which has degree at most $5$.
This implies that $f(x)$ is an odd function.
We get $f(x)+1=(x-1)^3(ax^2+bx-1)$, (note that $f(0)$ must be $0$) and by comparing the degree $2$ and $4$ coefficients (which should be $0$), we get $a=-\frac{3}{8}$ and $b=-\frac{9}{8}$, which gives the answer of $$f(x)=-\frac{3}{8}x^5+\frac{5}{4}x^3-\frac{15}{8}x$$
|
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|
Let $a_1, a_2, a_3...$ be the sequence of all positive integers relatively prime to 75. Find the value of $a_{2008}$. Let $a_1, a_2, a_3...$ be the sequence of all positive integers relatively prime to 75, where $a_1<a_2<a_3...$ with $a_1=1, a_2=2, a_3=4, a_4=7$. Find the value of $a_{2008}$.
What I have done:
If ${a_n}$ is relatively prime to 75, then it is relatively prime to 3 and 5. Consider the following string of numbers from 1 to 15. $$1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15$$
If multiples of 3 and 5 are removed then, we need to take note that 15 is removed twice, so we need to add it back. In general, the number of integers left in a string of $n$ integers is
$$n-\lfloor \frac{n}{3} \rfloor - \lfloor \frac{n}{5} \rfloor+ \lfloor \frac{n}{15} \rfloor$$
Hence we need to find $n=a_{2008}$ such that
$$n-\lfloor \frac{n}{3} \rfloor - \lfloor \frac{n}{5} \rfloor+ \lfloor \frac{n}{15} \rfloor=2008$$
I cheated a bit here and took away the floor value to solve for $n$ and find that $n=3765$
Can anyone verify for me if this answer is correct? And provide a proper alternative solution? Thank you!
|
The number of positive integers relatively prime to 75 and less than 75 is 40, as given by Euler's formula.
Thus, in the numbers up to 3750, there are 2000 such numbers. Now we find that the eighth positive integer relatively prime to 75 is 14; hence the answer is 3750+14=3764.
|
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|
Number of lines which are normal as well as tangents to the curve $y^2=x^3$? Number of lines which are normal as well as tangents to the curve $y^2=x^3$?
The line passes through two points $(x_1,f(x_1))$ and $(x_2,f(x_2))$ on the curve.
What is the general method to solve such problems?I could'nt proceed much.
|
Firstly draw a graph of your curve. You'll see there are two distinct sections to it: $\sqrt{x^3}$ and $-\sqrt{x^3}$. For a line to be a tangent and a normal it must be a tangent to one and normal to the other. Also note due to symmetry if we find one then the vertical reflection must also be an answer so lets look at lines which are tangent to $\sqrt{x^3}$ and normal to $-\sqrt{x^3}$.
Differentiating gives:
$$\frac{d}{dx}\left(\pm x^\frac{3}{2}\right)=\pm\frac{3}{2}x^\frac{1}{2}$$
So normal is from a point $x_0$ is: $y=\frac{2}{3\sqrt{x_0}}(x-x_0)-\sqrt{x_0^3}$. This will intersect the positive curve when:
$$\frac{2}{3\sqrt{x_0}}(x-x_0)-\sqrt{x_0^3}=\sqrt{x^3}$$
$$\frac{2(x-x_0)-3x_0^2}{3\sqrt{x_0}}=\sqrt{x^3}$$
$$\frac{4x^2-8x_0x-12x_0^2x+4x_0^2+12x_0^3+9x_0^2}{9x_0}=x^3$$
$$4x^2-8x_0x-12x_0^2x+4x_0^2+12x_0^3+9x_0^2=9x_0x^3$$
$$(x-x_0)(9x_0x^2+9x_0^2x-4x+4x_0+12x_0^2+9x_0^3)=0$$
This will have only two solutions (and hence be a tangent) when the determinant of the quadratic is zero.
$$(9x_0-4)^2-4\cdot9x_0^2\cdot(4x_0+12x_0^2+9x_0^3)=0$$
$$16-216a^2-432a^3-243a^3=0$$
$$-(3x_0+2)^2(9x_0-2)=0$$
As $x_0$ cannot be negative then we have the answer of $x_0=\frac{2}{9}$.
So there are two curves which are both tangent and normal: $y=\pm\left(\frac{2}{3\sqrt{\frac{2}{9}}}(x-\frac{2}{9})-\sqrt{\left(\frac{2}{9}\right)^3}\right)$.
These simplify to: $y=\pm\left(\frac{\sqrt{2}(27x-8)}{27}\right)$.
|
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|
Basic quadratic calculus Given $$ AX^2+2X-1=0 \,\ \text{where} \,\ A>0$$
What value of A would make the absolute value of both roots bigger than $1$?
I found the roots using quadratic formula and also found that A has to be both $[0,3]$ and $[-1,0]$ and $A>0$, does it mean such an $A$ doesn't exist?
Thanks!
|
For $A>0$ we have
\begin{align}
0&=X^2 + 2 \frac{1}{A}X - \frac{1}{A} \\
&= (X + 1/A)^2 -1/A^2 -1/A \\
&= (X - (-1/A))^2 + (- (A+1)/A^2) \iff \\
X &= \frac{\pm\sqrt{A+1}-1}{A}
\end{align}
This is a parabola with vertex at $(-1/A, -(A+1)/A^2)$.
Then
$$
\frac{\sqrt{A+1}-1}{A}
\le \lvert X\rvert
= \left\lvert \frac{\pm\sqrt{A+1}-1}{A} \right\rvert
\le \frac{\sqrt{A+1}+1}{A}
$$
So we need
\begin{align}
1 &< \frac{\sqrt{A+1}-1}{A} \iff \\
A + 1 &< \sqrt{A+1} \Rightarrow \\
(A + 1)^2 &< A + 1 \iff \\
A^2 + 2A + 1 &< A+1 \iff \\
A^2 + A &< 0
\end{align}
This is not possible for positive $A$, as $A^2$ is positive as well.
|
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|
Two expressions using three-digit floating point arithmetic with rounding? What is the result of evaluating the following two expressions using three-digit floating point arithmetic with rounding?
$(113. + -111.) + 7.51$
$113. + (-111. + 7.51)$
*
*$9.51$ and $10.0$ respectively
*$10.0$ and $9.51$ respectively
*$9.51$ and $9.51$ respectively
*$10.0$ and $10.0$ respectively
My attempt :
$(113. + -111.) + 7.51
= 2 + 7.51
= 9.51$
$113. + (-111. + 7.51)
= 113. + (-111. + 7.51)
= 113. - 103. = 10$ [$103.49$ is rounded to $103.0$]
Can you explain more formally please ?
|
Recall that in $n$-digit floating-point arithmetic numbers are represented by a signed $n$-digit integer and an exponent. Rounding occurs when the significant digits of a number exceed $n$, in which case only the first $n$ are kept and the $n$th is adjusted according to the value of the $(n+1)$th.
You cannot get much more formal than what you did, but to make the computation (arguably) more clear you could rewrite all the involved numbers in scientific notation.
So we have
\begin{align*}
&\big(1.13 \cdot 10^2 + (-1.11 \cdot 10^2)\big) + 7.51 \cdot 10^0 \\&= 2 \cdot 10^0 + 7.51 \cdot 10^0 \\&= 9.51 \cdot 10^0.
\end{align*}
On the other hand
\begin{align*}
&1.13 \cdot 10^2 + (-1.11 \cdot 10^2 + 7.51 \cdot 10^0)\\ &= 1.13 \cdot 10^2 + \underbrace{(-1.0349 \cdot 10^2)}_{\text{rounded to } -1.03 \cdot 10^2} \\ &= 1 \cdot 10^1.
\end{align*}
|
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|
Evaluate improper integral $\int_0^\infty \frac{x\sin x}{x^2+1}dx$ How to prove that $$\int_0^\infty \frac{x\sin x}{x^2+1}dx=\frac{\pi}{2e}$$
I've tried several basic approaches like substitution and IBP but can't move forward.
|
Suppose we are interested in
$$J = \int_0^\infty \frac{x\sin(x)}{1+x^2} dx$$
In view of the ML bound that we will be using later it proves
convenient to write this as
$$\left[ -\frac{x\cos(x)}{1+x^2} \right]_0^\infty
+ \int_0^\infty \frac{1-x^2}{(1+x^2)^2} \cos(x) dx
= \frac{1}{2}
\int_{-\infty}^\infty \frac{1-x^2}{(1+x^2)^2} \cos(x) dx.$$
Observe that this is
$$\frac{1}{2} \Re \int_{-\infty}^\infty
\frac{1-x^2}{(1+x^2)^2} \exp(ix) dx.$$
Therefore we integrate
$$f(z) = \frac{1-z^2}{(1+z^2)^2} \exp(iz)$$
around a semicircular contour of radius $R$ in the upper half plane,
which consists of a horizontal segment $\Gamma_1$ on the real axis and
a semicircle $\Gamma_2$ in the upper half plane. Now the integral
along $\Gamma_1$ is the integral we seek to evaluate in the limit and
the integral along $\Gamma_2$ vanishes. This is because we have from
the ML bound (parameterize $\Gamma_2$ by $R\exp(i\theta)$ with
$0\le\theta\le\pi$)
$$\lim_{R\rightarrow\infty} \pi R \times
\frac{(R^2+1) |\exp(iR\exp(i\theta))|}{(R^2-1)^2}
\\ = \lim_{R\rightarrow\infty} \pi R \times
\frac{(R^2+1) |\exp(iR\cos(\theta)+i^2R\sin(\theta))|}{(R^2-1)^2}
\\ = \lim_{R\rightarrow\infty} \pi R \times
\frac{(R^2+1) |\exp(-R\sin(\theta))|}{(R^2-1)^2}.$$
Now we have $|\exp(-R\sin(\theta))|\le 1$ because $0\le\theta\le\pi$
so we get two terms, the first of which is $O(1/R)$ and the second
$O(1/R^3)$ so they both vanish in the limit and the contribution from
the circular segment $\Gamma_2$ is zero.
This yields
$$J = \frac{1}{2} \times \Re(2\pi i \; \mathrm{Res}_{z=i} f(z)).$$
We require the second term in the Taylor series of
$$\frac{1-z^2}{(z+i)^2} \exp(iz)$$
at $z=i$
which is
$$\left.\left(\frac{-2z\exp(iz)+(1-z^2)i\exp(iz)}{(z+i)^2}
- 2 \frac{1-z^2}{(z+i)^3} \exp(iz)\right)\right|_{z=i}.$$
The first term here is zero and the second is
$$-2\frac{2}{(2i)^3} \exp(-1)
= \frac{1}{2} \frac{1}{i} \exp(-1).$$
Substitute this into the formula for $J$ to finally obtain
$$J=\frac{1}{2} \Re(2\pi i \exp(-1)/2/i)
= \frac{\pi}{2e}.$$
|
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|
Laplace inverse of $\frac{e^{-\sqrt{s+2}}}{s}$ I want to find out $$\mathcal{L^{-1}}\{\frac{e^{-\sqrt{s+2}}}{s}\}$$
How do you find the inverse Laplace?
thanks
|
Thank you for the interesting question. Here is a rather brute force solution, which may add a few steps to Jan Erland's solution.
First, let us recall that, if
$$
\mathcal L(f) = \int_0^\infty f(t) \, e^{-st} \, dt = F(s)
$$
Then
$$
\mathcal L(f') = \int_0^\infty f'(t) \, e^{-st} \, dt = s \, F(s) - f(0).
$$
In our case, $F(s) = e^{-\sqrt{s+2}}/s$, so, we shall seek a function $f'(t)$ whose Laplace transform is $sF(s) = e^{-\sqrt{s+2}}$. Then $f(t)$ can be found from integrating $f'(t)$ from $t = 0$, where $f(0) = 0$.
Using the inverse formula, we have
$$
f'(t)
= \frac{1}{2\, \pi \, i}\int_{\gamma-i\infty}^{\gamma+i\infty} e^{-\sqrt{s+2} + st} \, ds,
$$
where $\gamma$ is a large positive number, such that all poles, if any, lie on the left side of the line $y = \gamma$.
For our case, there is no pole, but only a branch cut at $s = -2$. We shall let the branch cut extend to $-\infty$ from $s = -2$, and wrap the contour around the branch cut from $-\infty + 0^- \, i$ to $-2 + 0^- \, i$ (lower half) and then from $-2 + 0^+ \, i$ to $-\infty + 0^+ \, i$ (higher half).
Let $s = -2 + A\,e^{-i\pi}$ with $A \ge 0$, then
$$
\begin{aligned}
f'(t)
&= \frac{e^{-2t}}{\pi}
\int_0^{+\infty} \sin(\sqrt{A}) \, e^{- A t} dA \\
&= \frac{e^{-2t}}{\pi}
\int_{-\infty}^{+\infty} u \, \sin(u) \, e^{-u^2 t} du \\
&= \frac{e^{-2t}}{\pi}
\mathrm{Im} \left(\frac{\partial}{\partial v}
\int_{-\infty}^{+\infty} e^{-u^2 t + v u} d u \right)_{v = i} \\
&= \frac{e^{-2t}}{\pi}
\mathrm{Im} \left[\frac{\partial}{\partial v}\left(
\sqrt{\frac{\pi}{t}} \, \exp{\frac{v^2}{4t}} \right)\right]_{v = i} \\
&= \frac{1}{2 \, \sqrt{\pi \, t^3}} \, \exp\left(-2 \, t-\frac{1}{4\,t}\right).
\end{aligned}
$$
Here, we have used the fact that $\mathrm{Im} \, e^{iu} = \sin(u)$,
Finally, let us integrate $f'(t)$.
$$
\begin{aligned}
f(t)
&= \int_0^t f'(\tau) \, d\tau \\
&=\frac{1}{\sqrt\pi}
\int_{1/\sqrt{t}}^\infty
\exp\left(-\frac{x^2}{4}-\frac{2}{x^2}\right) \, dx \\
&=\frac{1}{\sqrt\pi}
\int_{1/\sqrt{t}}^\infty
\exp\left(-\frac{x^2}{4}-\frac{2}{x^2}\right) \,
d\left( \frac{x}{2}+\frac{\sqrt{2}}{x} \right) \\
&\quad+
\frac{1}{\sqrt\pi}
\int_{1/\sqrt{t}}^\infty
\exp\left(-\frac{x^2}{4}-\frac{2}{x^2}\right) \,
d\left( \frac{x}{2}-\frac{\sqrt{2}}{x} \right)
\\
&=\frac{1}{\sqrt\pi}
\left[
\int_{\frac{1}{2\sqrt{t}}+\sqrt{2t}}^\infty
\exp\left(\sqrt{2} -y^2\right) \, dy
+
\int_{\frac{1}{2\sqrt{t}}-\sqrt{2t}}^\infty
\exp\left(-\sqrt{2} -z^2\right) \, dz
\right] \\
&=
\frac{e^{\sqrt 2}}{2} \, \mathrm{erfc}\left(
\frac{1}{2\sqrt{t}}+\sqrt{2\, t} \right)
+\frac{e^{-\sqrt 2}}{2} \, \mathrm{erfc}\left(
\frac{1}{2\sqrt{t}}-\sqrt{2\, t} \right).
\end{aligned}
$$
Here, we have changed variables
$$
\begin{aligned}
x &\equiv \frac{1}{\sqrt{\tau}}, \\
y &\equiv \frac{x}{2} + \frac{\sqrt{2}}{x}, \\
z &\equiv \frac{x}{2} - \frac{\sqrt{2}}{x}.
\end{aligned}
$$
Our result agrees with Jan Erland's.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the limit $\lim_{x \to 0} (2^x + \sin (3x)) ^{\cot(3x)}$ Please help, I have already tried every thing I can, but nothing works. I have no I idea what to do.
$$\lim_{x \to 0} \; (2^x + \sin (3x)) ^{\cot(3x)}$$
|
Rewritting the limit using the exponential function $e^x$ using the logarthmic identity $e^{\ln x} = x$.
\begin{align}
\lim_{x \to 0} (2^x + \sin(3x))^{\cot(3x)} &= \lim_{x \to 0} \exp(\cot(3x)\ln(2^x + \sin(3x)))\\
&= \exp( \lim_{x \to 0} \cot(3x)\ln(2^x + \sin(3x)))\\
&= \exp(\lim_{x \to 0} \frac{\ln(2^x + \sin(3x))}{\tan(3x)} )\\
\end{align}
Now apply L'Hopital's rule
\begin{align}
\lim_{x \to 0} \frac{\ln(2^x + \sin(3x))}{\tan(3x)} &= \lim_{x \to 0} \frac{\frac{3\cos(3x) + 2^x \ln(2)}{\sin(3x) + 2^x}}{3\sec^2{3x}}\\
&= \frac{3\cos(0) + 2^0 \ln(2)}{(\sin(0) + 2^0)(3\sec^2{0})} =\frac{3 + \ln(2)}{3}
\end{align}
Plugging this in yields $\lim_{x \to 0} (2^x + \sin(3x))^{\cot(3x)} = e^{\frac{3 + \ln(2)}{3}} = e^1e^\frac{\ln(2)}{3} = 2^\frac{1}{3}e = \sqrt[3]{2}e$
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Most elegant method to calculate $\int^\pi_0 (x\sin x)^2 dx$ $\displaystyle \int^\pi_0 (x\sin x)^2 dx$.
I can easily use integration by parts to solve this integral; however, it is quite messy and I'm just wondering if there exists another alternative method that is more elementary and elegant.
I have tried the substitutions $u = \pi -x$ and $u = \pi / 2 -x$ but they do not seem to help very much.
|
Using the identity $\sin^2(x)=\frac{1}{2}(1-\cos(2x))$, we have:
\begin{align*}
\int^\pi_0 x^2\sin^2 x\; dx &= \frac{1}{2}\int^\pi_0 x^2(1-\cos(2x)) dx \\
&= \frac{1}{2} \int^\pi_0x^2\;dx - \frac{1}{2}\int^\pi_0x^2\cos(2x)\;dx \\
&= \frac{\pi^3}{6} -\frac{1}{2} \left[\frac{1}{2}x^2\sin(2x)\right]_0^\pi + \frac{1}{2}\int_0^\pi x\sin(2x)\;dx \\
&= \frac{\pi^3}{6} + \frac{1}{2}\left[-\frac{1}{2}x\cos(2x)\right]_0^\pi + \frac{1}{2}\int_0^\pi\sin(2x)\;dx \\
&= \frac{\pi^3}{6} - \frac{\pi}{4}.
\end{align*}
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Methods to prove a series is decreasing? What are soome good methods to show a series is decreasing?
Usually I just compare the $a_n$ and $a_{n+1}$ values. For example $\frac 1 {k^2} >\frac 1 {k^2+1}$ so it shows the series is decreasing. I heard you can also do something like $\frac{a_{n+1}}{a_n}$ but that seems complicated.
What are some more efficient methods for like series that have $\frac{k+2} {k(k+3)}$?
|
If you are more comfortable with things like $\frac{1}{k^2} > \frac{1}{k^2 +1}$ then another approach for cases like $\frac{k+2}{k(k+3)}$ is to use partial fraction decomposition:
$$\frac{k+2}{k(k+3)} - \frac{k+3}{(k+1)(k + 4)} = \left(\frac{1/3}{k + 3} + \frac{2/3}{k}\right) - \left(\frac{1/3}{k + 4} + \frac{2/3}{k+1}\right) = \frac{1}{3}\left(\frac{1}{k+3} - \frac{1}{k+4}\right) + \frac{2}{3}\left(\frac{1}{k} - \frac{1}{k+1}\right)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1536974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
The trigonometric solution to the solvable DeMoivre quintic? Using the relations for the Rogers-Ramanujan cfrac described in this post,
$$\frac{1}{r}-r = x$$
$$\frac{1}{r^5}-r^5 = y$$
and eliminating $r$ yields,
$$x^5+5x^3+5x = y$$
This is the case $a=1$ of the solvable DeMoivre quintic,
$$x^5+5ax^3+5a^2x+b = 0\tag1$$
In general, it has the solution,
$$x = \left(\tfrac{-b+\sqrt{b^2+4a^5}}{2}\right)^{1/5}-a\left(\tfrac{-b+\sqrt{b^2+4a^5}}{2}\right)^{-1/5}\tag2$$
When $a=-1$, it can also be solved as,
$$x = 2\cos\Bigg(\tfrac{\arccos\Big(\tfrac{-b}{2}\Big)}{5}\Bigg) = 2\cos\Bigg(\tfrac{2\arctan\Big(\tfrac{\sqrt{2+b}}{\sqrt{2-b}}\Big)}{5}\Bigg)\tag3$$
Q: When $\color{red}{a=1}$, is there a neat trigonometric solution similar to $(3)$?
P.S. A google search revealed that it is may indeed be possible.
|
Thanks to Tito for the nice question. Here is a solution in terms of hyperbolic sine, which may not be what you want.
$$
\sinh(5t) = 5 \sinh t+ 20 \sinh^3 t + 16 \sinh^5 t.
$$
With $x = 2\sinh t$ and $b = -2\sinh 5t$, we have
$$
x^5+ 5 x^3 + 5 x + b = 0\tag1,
$$
which is the $a = 1$ case. So the solution is
$$
x = 2\sinh t = 2\sinh\left( \frac{\sinh^{-1}\left(-\frac{b}{2}\right)}{5} \right)\tag2
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1537069",
"timestamp": "2023-03-29T00:00:00",
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|
Prove $ 5(a^4+b^4+c^4)+9≥8(a^3+b^3+c^3) $ if $a^2+b^2+c^2=3$ Let $a,b,c\in\mathbb{R}$ such that $a^2+b^2+c^2=3$. Prove that:
$$
5(a^4+b^4+c^4)+9≥8(a^3+b^3+c^3)
$$
I tried to homogenize the inequality to get:
$$
5(a^4+b^4+c^4)+(a^2+b^2+c^2)^2≥\frac{8}{\sqrt3}(a^3+b^3+c^3)\sqrt{(a^2+b^2+c^2)}
$$
I hoped that one don't needs the condition anymore to prove this, but I couldn't get any further.
|
By AM-GM:
$$\begin{aligned}a^4+a^4+a^4+a^4+a^4+a^2+a^2+1&\ge 8\sqrt[8]{a^{4\times 5}a^{2\times 2}}\\&=8\sqrt[8]{a^{24}}\\&=8|a^3|\ge8a^3.\end{aligned}$$
Similarly for $b$ and $c$ and add them up.
|
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|
How to evaluate $\sin^{-1} (\sqrt{2} \sin \theta) + \sin^{-1} (\sqrt{\cos2 \theta})$ $$\sin^{-1} (\sqrt{2} \sin \theta) + \sin^{-1} (\sqrt{\cos2 \theta})$$
to evaluate the above equation, I used the formula: $$\sin^{-1} (x) + \sin^{-1} (y) = \sin^{-1}[x(1-y^2) + y(1-x^2)]$$
therefore I have got,
$$\sin^{-1} (\sqrt{2} \sin \theta \sin2 \theta + \cos \theta )$$
The result is supposed to be 1.
How can I do this?
|
Set $a = \sqrt 2 \sin\theta$ and $b = \sqrt{\cos 2\theta}$. It follows that $a^2 + b^2 = 2\sin^2 \theta + \cos 2\theta = 1$.
If $\alpha = \sin^{-1} a$ where $\alpha \in [-\frac \pi 2, \frac \pi 2]$, we get $\sin(\alpha) = a$, and since $b\ge 0$ we have $b = \sqrt{1 - a^2} = \cos(\alpha) = \sin(\frac \pi 2 - \alpha)$.
*
*If $a \ge 0$, we get $\alpha \in [0, \frac \pi 2]$ and thus $\sin^{-1} b = \frac\pi 2 - \alpha$ which gives us $\sin^{-1} a + \sin^{-1}b = \frac \pi 2$.
*On the other hand, if $a < 0$, we have $\alpha \in [-\frac \pi 2, 0)$ and so $\sin^{-1}b = \pi + \alpha$ so $\sin^{-1}a + \sin^{-1}b$ has no fixed value.
|
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|
Division with decimal in the divisor I understand HOW to do division with a decimal in the divisor, but my question is MUST we remove the decimal in the divisor and if so, why?
Thanks.
|
Remember that the word decimal is a shortened form of decimal fraction, meaning a fraction whose denominator is a power of $10$. For example,
$$
3.14 = \frac{314}{100},
\qquad
0.0625 = \frac{625}{1000},
\qquad
5280.1 = \frac{52801}{10}.
$$
As an example, consider the division problem
$$
\frac{9.6}{3.84} = \frac{\frac{96}{10}}{\frac{384}{100}}
= \frac{96}{10} \cdot \frac{100}{384}
= \frac{96}{384} \cdot \frac{100}{10}
= \frac{96}{384} \cdot 10.
$$
This last form shows you that the original division calculation is equivalent the quotient of the whole numbers$96$ and $384$, as long as you adjust by a factor of $10$ afterwards. (This is usually described as moving the decimal point.)
In order to actually do this calculation, you can multiply the numerator by a sufficiently large power of $10$ to make it a multiple of the denominator, making sure to divide by that same power of $10$ afterwards to compensate. Note: it won't always be the case that you get precisely a multiple, no matter how large a power of $10$ you multiply by. In this case,
$$
\frac{96}{384} \cdot 10 = \frac{96}{384} \cdot \frac{100}{100} \cdot 10
= \frac{9600}{384} \cdot \frac{10}{100}
= 25 \cdot \frac{1}{10}
= \frac{25}{10}
= 2.5.
$$
The upshot: the integer fact that $384 \cdot 25 = 9600$ is responsible for all of the following:
$$
\cdots
= \frac{0.0096}{0.00384}
= \frac{0.096}{0.0384}
= \frac{0.96}{0.384}
= \frac{9.6}{3.84}
= \frac{96}{38.4}
= \frac{960}{384}
= \frac{9600}{3840}
= \cdots
= 2.5
$$
|
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|
A new approach to find value of $x^2+\frac{1}{x^2}$ When I was teaching in a college class, I write this question on board.
If we now $\bf{x+\dfrac{1}{x}=4}$ show the value of $\bf{x^2+\dfrac{1}{x^2}=14}$
Some student asks me for a multi idea to show or prove that.
I tried these : $$ $$1 :$$\left(x+\frac{1}{x}\right)^2=4^2\\x^2+\frac{1}{x^2}+2x\times \frac{1}{x}=16\\x^2+\frac{1}{x^2}=16-2 $$
2:solving quadratic equation ,and putting one of roots$$x+\frac{1}{x}=4\\\frac{x^2+1}{x}=4\\x^2+1=4x\\x^2-4x+1=0\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-(-4)\pm \sqrt{(-4)^2-4(1)(+1)}}{2}=\\x=\frac{4\pm \sqrt{12}}{2}=2\pm \sqrt{3}\\x=2+\sqrt3 \to x^2=4+4\sqrt3+3=7+4\sqrt3\\x^2+\frac{1}{x^2}=7+4\sqrt3+\frac{1}{7+4\sqrt3}=\\7+4\sqrt3+\frac{1}{7+4\sqrt3}\cdot\frac{7-4\sqrt3}{7-4\sqrt3}=\\7+4\sqrt3+\frac{7-4\sqrt3}{49-16\cdot 3}=\\7+4\sqrt3+\frac{7-4\sqrt3}{1}=14$$
3: visual approach .assume side length of a square is $x+\frac{1}{x}=4$
now I am looking for new idea to proof.Any hint will be appreciated.(more visual proof - geometrical - trigonometrical - using complex numbers ...)
|
Not sure how to show what the continued fraction equals but here's my attempt. Again similar in spirit to the algebraic method.
Solving for $x$ in the first equation we have: $x = 4 - \frac{1}{x}$.
Substituting this into the second equation:
$
(4 - \frac{1}{x})^{2} + \frac{1}{(4 - \frac{1}{x})^{2}} = 16 - \frac{8}{4-\frac{1}{x}} + \frac{1}{ 16 - \frac{8}{4-\frac{1}{x}}} = 16 - \cfrac{8}{4-\cfrac{1}{4-\cfrac{1}{4-\cdots}}} + \frac{1}{16 - \cfrac{8}{4-\cfrac{1}{4-\cfrac{1}{4-\cdots}}}}
$
And that big mess equals 14.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "7",
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|
Prove $4(a^2 + b^2 + c^2) + 9(abc)^2 \geq 21.$ Could anyone advise me how to prove this inequality? Hints will suffice, thank you.
Let $a,b,c \in [0, \infty)$ such that $ab+bc+ ca =3.$ Prove $4(a^2 + b^2 + c^2) + 9(abc)^2 \geq 21.$
My attempt: By AM-GM inequality, $(abc)^{2}\leq 1$ and $a^2 + b^2 + c^2 \geq 3(abc)^{\frac{2}{3}}.$
|
If we assume the following condition (without loss of generality) $a\geq b \geq c$ then we have :
$$4(a^2+b^2+c^2)+9(abc)^2\geq 4(a^2+b^2+c^2)+9(c)^6$$
With your initial condition $ab+bc+ca=3$ the RHS of the inequality can be rewritten as follows :
$$4((\frac{3-ab}{a+b})^2+(a+b)^2-2ab)+9(\frac{3-ab}{a+b})^6$$
Now we put $a+b=cst=x$ and remark that the inequality is minimal when $ab$ is maximal where the maximum value of $ab$ in term of $x$ is $\frac{x^2}{4}$
So we have to study this function :
$$f(x)=4((\frac{3-\frac{x^2}{4}}{x})^2+(x)^2-2\frac{x^2}{4})+9(\frac{3-\frac{x^2}{4}}{x})^6$$
But $max(a+b)=2$ so we have to study on the domain $]0;2]$ the function $f(x)$
It's clear that the function is decreasing on his domain and furthermore we have :
$$f(x)\geq f(2)\geq 21$$
So to conclude we have :
$$4(a^2+b^2+c^2)+9(abc)^2\geq 4(a^2+b^2+c^2)+9(c)^6\geq 21$$
|
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|
Find $f$ if $ f(x)+f\left(\frac{1}{1-x}\right)=x $ Find $f$ if
$$
f(x)+f\left(\frac{1}{1-x}\right)=x
$$
I think, that I have to find x that $f(x) = f\left(\frac{1}{1-x}\right)$
I've tried to put x which make $x = \frac{1}{1 - x}$, but this equation has no roots in real numbers.
|
If $g(x)=\frac{1}{1-x}$ then verify that $g(g(g(x)))=x$.
Then:
$$\begin{align}
f(x)+f(g(x))&=x\\
f(g(x)) + f(g(g(x))) &= g(x)=\frac{1}{1-x}\\
f(g(g(x))) + f(x) &= g(g(x))=\frac{x-1}{x}
\end{align}$$
So this gives three linear equations, so solve for $f(x)$.
Subtract the second row from the first:
$$f(x)-f(g(g(x)))=x-\frac{1}{1-x}$$
Then add this to the third:
$$2f(x)=x-\frac{1}{1-x}+\frac{x-1}{x}=x+1-\frac{1}{x(1-x)}$$
Then $$\begin{align}f(g(x))&=\frac{1}{2}\left(1+\frac{1}{1-x}-\frac{1}{\frac{1}{(1-x)}\left(1-\frac{1}{1-x}\right)}\right)\\
&=\frac12\left(1+\frac1{1-x}-\frac{(1-x)^2}{(1-x)-1}\right)\\
&=\frac12\left(x-1+\frac1{x(1-x)}\right)
\end{align}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Computing the hyperbolic cosine as a serie So I want to prove that:
$$\sum_{x=0}^\infty \frac{\lambda^{2x}}{(2x)!}=\cosh(\lambda)$$
But, proceeding backwards, the only thing I know is that
$$\cosh(\lambda)=\frac{e^{\lambda}+e^{-\lambda}}{2}=\sum_{x=0}^\infty \frac{\lambda^x}{2(x!)}+\sum_{x=0}^\infty \frac{(-\lambda)^x}{2(x!)}$$
And got stuck here, any ideas?
|
$$
\begin{array}{cccccccccc}
& \left(1\vphantom{\dfrac{\lambda^3} 6}\right. & + & \lambda & + & \dfrac{\lambda^2} 1 & + & \dfrac{\lambda^3} 6 & + & \dfrac{\lambda^4}{24} & + & \dfrac{\lambda^5}{120} & + & \cdots \\[10pt]
+ & 1 & - & \lambda & + & \dfrac{\lambda^2} 1 & - & \dfrac{\lambda^3} 6 & + & \dfrac{\lambda^4}{24} & - & \dfrac{\lambda^5}{120} & + & \left. \cdots \vphantom{\dfrac{\lambda^3} 6} \right) \\[10pt]
\hline
= & 2\left( 1 \vphantom{\dfrac{\lambda^3} 6}\right. & & & + & \dfrac{\lambda^2} 2 & & & + & \dfrac{\lambda^4}{24} & & & + & \left. \cdots \vphantom{\dfrac{\lambda^3} 6} \right)
\end{array}
$$
The odd-numbered terms cancel.
|
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|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.