Recent Questions - Mathematics Stack Exchange |
- Interpreting the graphical initial condition of the wave equation.
- Why are these two formulas logically equivalent?
- Two consecutive prime factors in two consecutive composites
- If $f'(x)>f(x)$ for all reals and $f(x_{0})=0$, then $f(x)>0$ for $x>x_{0}$.
- On exercise $1.7.15$ of Guillemin-Pollack
- How is a permutation composed of one cycle the product of disjoint cycles?
- Conditional expectation help
- How to define the circle that this complex equation produces on the complex plane?
- Single variable logistic regression
- Total amount of time spent at a state in asymmetric random walk on $\mathbb{Z}$
- Steps along a diagonal
- How can i evaluate this Dilogarithm Integral $\int_0^1\frac{\text{Li}_2(x)\ln(1+x)}{x}\mathrm dx$?
- Induction to solve Recurrence Equation
- Binomial expansion of x is large
- Square root of the operator on the product space
- Let $p,q,$ and $r$ be constants. One solution to the equation $(x-p)(x-q) = (r-p)(r-q)$ is $x=r$. Find the other soultion in terms of $p,q,$ and $r$.
- Rotating a cone
- Conditional Probability for random variable (E(X))
- How do I get the function and solve the problem using the definition of limit of sum?
- Can you please help me solve this inequality problem? [closed]
- Fibonacci numbers, sum of squares and divisibility
- Can someone explain the integral split in the following?
- Complex analysis with logarithms
- Top Dolbeault cohomology group vs top De Rham cohomology group of a compact complex manifold
- Vinyl Record Arc Length
- Let $A$ be non singular, prove that $\frac{1}{\|A^{-1}\|} = \min_{\|x\| = 1} \|Ax\|$
- Are reduced SVD and truncated SVD the same thing?
- Find coordinates for points on circle given R, 2 Points, and angle or 2 points and center?
- Where does the invariant expression for the exterior derivative come from?
- Prove that intersection of diagonals in trapezium divide parallel segment to equal segments
| Interpreting the graphical initial condition of the wave equation. Posted: 06 Nov 2021 08:43 PM PDT I am solving the wave equation and I am given an initial condition for $\frac{\partial u }{\partial t} |_{t=0}$ as the graph below: How do I describe this mathematically? Is it sufficient to say $\frac{\partial u }{\partial t} |_{t=0} = 1 $ $\forall x \in [-1,1]$?  |
| Why are these two formulas logically equivalent? Posted: 06 Nov 2021 08:42 PM PDT Formula 1: $$(¬P ∧ ¬Q ∧ ¬R) ∨ (¬P ∧ ¬Q ∧ R) ∨ (P ∧ Q ∧ ¬R) ∨ (P ∧ Q ∧ R)$$ Formula 2: $$(¬P ∧ ¬Q) ∨ (P ∧ Q)$$ How are these two formulas logically equivalent? To my understanding when two formulas are logically equivalent, they have identical truth values under all interpretations, these 2 formulas produce completely different truth tables- formula 1 has 3 variables and formula 2 has 2 variables to start off with. I don't understand how they are logically equivalent?  |
| Two consecutive prime factors in two consecutive composites Posted: 06 Nov 2021 08:28 PM PDT Two consecutive composites contain a pair of consecutive primes. If you are given the two primes, how do you find the least composites? The list of these composites begins 8, 9, 14, 21, 65 as the lesser of the pair. If you are given primes 19 and 23, is there a fast way of finding the least pair of composites, in this case $6*19=114$ and $5*23=115$  |
| If $f'(x)>f(x)$ for all reals and $f(x_{0})=0$, then $f(x)>0$ for $x>x_{0}$. Posted: 06 Nov 2021 08:44 PM PDT Someone had asked an interesting question but deleted it after one hour. Since the question is interesting, I re-post it and submit a full solution. Question: Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a differentiable function such that $f'(x)>f(x)$ for all $x\in\mathbb{R}$. Suppose that there exists $x_{0}\in\mathbb{R}$ such that $f(x_{0})=0$. Prove that $f(x)>0$ for all $x>x_{0}$.  |
| On exercise $1.7.15$ of Guillemin-Pollack Posted: 06 Nov 2021 08:24 PM PDT $\mathbf {The \ Problem \ is}:$ If $X$ is an embedded submanifold in $\mathbb R^N$ ,show there exists a linear map $L : \mathbb R^N \to \mathbb R$ such that $L|_X$ is a Morse map . $\mathbf {My \ approach}:$ Exc $14$ is a special case with $X=S^{N-1}$ and $L=π_N$ , so I think for general case we can take $L=π_N$ $=> x\in X$ is critical point iff $T_xX= \mathbb R^{N-1}×\{0\}$ but after that I can't approach . A hint is appreciated, thanks in advance .  |
| How is a permutation composed of one cycle the product of disjoint cycles? Posted: 06 Nov 2021 08:21 PM PDT I'm reading through an abstract algebra textbook, and one of the exercises is as follows:
I tried to figure it out on my own, but I got stuck on the special case in which a permutation is a single cycle (e.g. $(1 2 3)$ in $S_3$), since it cannot be disjoint with any other cycle in that permutation group, and you need two elements to perform multiplication. I thought maybe it would be disjoint from the empty set, but then multiplication is not well defined. I ended up finding a proof here, but it doesn't seem to consider the special case. I also tried looking this question up online, but I can't find any examples of it being asked before. The closest thing I can find is this wiki page, which just says that a cycle is a permutation. So, my question is as follows: Is a permutation composed of one cycle a product of disjoint cycles even if there is no multiplication? If so, how?  |
| Posted: 06 Nov 2021 08:00 PM PDT Let U be a continuous random variable with the following pdf: $f_{U}{(t)} = \frac{1}{4}1_{[-2\_0)}(t) + \frac{1}{4}(1-\frac{t}{4})1_{[0\_4]}(t)$ Define V = $U^{2}*1\{U > \frac{1}{2}\}$ Find E(U | V near 9) Ive managed to reduce this to $\frac{E(U*1\{V \:near \:3^2\})}{P(V \:near\: 3^2)}$. I am not sure if at this point I can just replace the V near 9 with U = 3. Any help would be appreciated.  |
| How to define the circle that this complex equation produces on the complex plane? Posted: 06 Nov 2021 07:58 PM PDT I have an equation as follows: $\left( C_1 - \frac{C_2}{X} \right) \left(C_3 - \frac{0.75}{X}\right)^*=0$ where: $'*'$ represents the complex conjugate and $C_1$, $C_2$, $C_3$ are constants that are complex numbers, and $X$ is the complex number I am solving for. I have kept the constants general, but I can provide the numbers if required. I am told that if you plot $\frac{C_2}{X}$ on the complex plane it produces a circle, but I am having trouble how to define it. I can easily find one point on the circle by setting the left-most bracket expression to zero and solving for X. I can find another point on the circle by setting the right-most bracket expression to zero and solving for X. After that point, I am really stumped. I cannot define a unique circle with two points. Any help would be greatly appreciated. I've been stuck on this for weeks.  |
| Single variable logistic regression Posted: 06 Nov 2021 07:57 PM PDT I am trying to write out the code for using Newton-Raphson to solve for a coefficient and intercept in a single variable logistic regression. However in both the math and in the code I see that the Hessian matrix is not invertible. Is Newton-Raphson not applicable to single variable logistic regression or am I just doing something wrong? If it is not applicable is there anything that converges faster than GD/SGD that I can use? An external library does not fit the use case.  |
| Total amount of time spent at a state in asymmetric random walk on $\mathbb{Z}$ Posted: 06 Nov 2021 07:50 PM PDT Let $S(A)$ be the total amount of time spent in $A\subset \mathbb{Z}$ by a discrete-time asymmetric random walk on $\mathbb{Z}$. The transition probabilities are $P(i,i+1)=a,P(i,i-1)=b,P(i,i)=1-a-b$.
I know that this would be transient but I'm confused about the total time spent part and how to incorporate that into the expectation calculation? For example, if the total time spent is $1$, I don't know how to find the corresponding probability.  |
| Posted: 06 Nov 2021 07:49 PM PDT I was thinking about this recently, on a walk around the neighborhood... I have a square with 1 unit on each side and I want to go from the bottom-left corner to the top-right corner. If I go along the perimeter it will take me two units: one unit up and one unit right. If I go along the diagonal it will take me sqrt(2) units (thanks Pythagoras). But if I divide the diagonal into steps, the answer never approaches sqrt(2). E.g. 1/2 up, 1/2 right, 1/2 up, 1/2 right equals two units. But also 1/1000 up, 1/1000 right...etc. will also equal two units. I can't seem to reconcile this. Why does it never approach sqrt(2) as the step size gets infinitesimally small?  |
| How can i evaluate this Dilogarithm Integral $\int_0^1\frac{\text{Li}_2(x)\ln(1+x)}{x}\mathrm dx$? Posted: 06 Nov 2021 08:31 PM PDT Well, I've been trying to solve the following integral: \begin{equation*} \int_0^1\frac{\text{Li}_3(x)}{1+x}\mathrm dx, \end{equation*} where by integration by parts, making $u=\text{Li}_3(x)$ and $\mathrm dv=\frac1{1+x}$, i got: \begin{align*} \int_0^1\frac{\text{Li}_3(x)}{1+x}\mathrm dx=\ln(2)\zeta(3)-\int_0^1\frac{\text{Li}_2(x)\ln(1+x)}{x}\mathrm dx.\tag{1} \end{align*} Particularly, I was curious about the integral on the right side of (1), so I proceeded naively to apply integration by parts again. This time, doing $u=\text{Li}_2(x)\ln(1+x)$ and $\mathrm dv=\frac1x$: (note: the choice $u=\text{Li}_2(x)$ and $\mathrm dv=\frac{\ln(1+x)}{x}$ it has a certain symmetry with the integral I want to calculate) $$\int_0^1\frac{\text{Li}_2(x)\ln(1+x)}{x}\mathrm dx=\int_0^1\frac{\ln(x)\ln(1-x)\ln(1+x)}{x}\mathrm dx-\int_0^1\frac{\text{Li}_2(x)\ln(x)}{1+x}\mathrm dx.$$ In order not to extend this post, I briefly know that it is possible to calculate the first integral above, but the second is another problem I don't know how to deal with. There is the possibility of using generator functions for the integral of my question, but I'm trying to avoid that, another thing is that I can still accept this type of solution.  |
| Induction to solve Recurrence Equation Posted: 06 Nov 2021 08:44 PM PDT Show that: $B(n, k) = \binom{n+k}{n} $ is the solution to the following recurrence equation: $B(n, k) = B(n − 1, k) + B(n, k − 1)$ $for$ $n > 0$ $k > 0$ $B(n, 0) = 1$ $B(0, k) = 1$ Now, I don't want people to do my homework, but I am looking for guidance. I calculate the base step, which it's easy. $B(n, k) = \binom{n+k}{n} $ $B(n, k) = \frac{(n+k)!}{n!(n + k - n)!}$ $B(n, k) = \frac{(n+k)!}{n! * k!}$ $B(n, 0) = 1 = \frac{(n+0)!}{n! * 0!} = \frac{n!}{n!} = 1$ Now, I need to use the inductive hypothesis to prove that: $B(n, k) = \binom{n+k}{n} $ Now I know I have to use the inductive step: $n + 1$ and maybe $k + 1$ to do the proof, but I don't know how to continue from here. Any help is really appreciated.  |
| Binomial expansion of x is large Posted: 06 Nov 2021 07:41 PM PDT I have the function: $f(x) = \frac{5}{(1+x^2)(2+x)}$ How do I find the first n terms in the expansion of f(x) if $x$ is large? I have no issue in finding it if x is small - we convert it into partial fraction first, then expand as usual. What is the difference between if x is large and x is small anyway? P.S. I am new here, still figuring out how to use math jax  |
| Square root of the operator on the product space Posted: 06 Nov 2021 07:51 PM PDT Consider the space $\mathbb R \oplus L^2(\mathbb R)$ and define an operator on it $$ \Psi = \left( \begin{matrix} A&B^*\\B&D \end{matrix} \right) $$ where $A\in \mathbb R$, $B \in L^2(\mathbb R)$, $C$ is a bounded linear operator on $L^2(\mathbb R)$ and $B^*(x) = \langle B,x \rangle$. I am looking for the square root of this operator such that $$ QQ=\Psi $$ IdeaFormally, for the numbers $A, B, D\in\mathbb R$ the 2x2 matrix square root can be written $$ R=\frac{1}{t}\left(\begin{array}{cc} A+s & B \\ B & D+s \end{array}\right) $$ where $\tau = A+D$, $\delta = AD-B^2$, $s =\pm \sqrt{\delta}$ and $t=\pm \sqrt{\tau + 2s}$. However, in above case, $B$ and $D$ are not the numbers anymore. I was wondering, however, if one could use the analogy (perhaps via spectral decomposition to write operator as multiplicative one) to get the square root of $\Psi$. Many thanks.  |
| Posted: 06 Nov 2021 08:07 PM PDT Let $p,q,$ and $r$ be constants. One solution to the equation $(x-p)(x-q) = (r-p)(r-q)$ is $x=r$. Find the other soultion in terms of $p,q,$ and $r$. The quadratic in $x$ is $x^2-(p+q)x-pq$. This means the sum of roots $= p+q$. Does that mean either $p =r$ or $q=r$?  |
| Posted: 06 Nov 2021 07:41 PM PDT I learned that, given a point $(x,y)$ in the plane, $\sigma_{\phi}(x,y) = (x\cos(\phi)-y\sin(phi),x\sin(\phi)+x\cos(\phi))$ is the point corresponding to rotating $(x,y)$ by an angle $\phi$ counter-clockwise. This can be used for finding the equation of a rotated parabola, for example. I thought about the formula and realized that, writing $(x,y)$ in polar form $(r\cos(\varphi), r\sin(\varphi))$, it is clear that the rotated point is $(r\cos(\varphi+\phi)), r\sin(\varphi+\phi)$. Applying the formulas for the sine and the cosine of the sum of two angles, I obtained the expression of $\sigma_{\phi}(x,y)$. EDIT: the inspiration of this questions is that I'd like to rotate a cone so its axis becomes another line (for example, so a 'tilted cone' becomes a vertical one). Hence the title of the question. Now, I'd like to know how to rotate a figure in space so that the $z$ axis becomes a certain line of my election. I thought that using spherical coordinates would be the natural approach. I spherical coordinates, $$x = r\sin(\phi)\cos(\theta)\\ y = r\sin(\phi)\sin(\theta)\\ z = r\cos(\phi) $$ being $\theta$ the polar angle and $\phi$ the angle with the $z$ axis. In my case, I want to add $\phi_0$ to the angle $\phi$. Applying the formulas of the sine and cosine of the sum of angles, I got that the point $(x',y',z')$ corresponding to rotating a point $(x,y,z)$ is $$ x' = x\cos(\phi_0)+z\sin(\phi_0)\cos(\theta) \\ y' = y\cos(\phi_0)+z\sin(\phi_0)\sin(\theta)\\ z' = z\cos(\phi_0)-r\sin(phi_1)\sin(\phi) $$ The angles $\theta$ and $\phi$ should be found from $(x,y,z)$. Is my approach correct? Is there an easier way to do this? I would really appreciate if someone could point me in the right direction.  |
| Conditional Probability for random variable (E(X)) Posted: 06 Nov 2021 07:59 PM PDT Need help with part b and c. We deal a hand of two cards (drawn at random) from a standard 52-card deck. Let the random variable X represent the number of aces in this hand. Compute the following: (a) E(X) (b) E(X| the ace of ♠ is in this hand ) (c) E(X| at least one ace is in this hand ). For part a, $x=0,1,2$ using the formula $^nCr=n!/(r!(n-r)!)$ $$P(x=0)= (4C0 *48C2)/(52C2) = 0.851$$ $$P(x=1)= (4C1 *48C1)/(52C2) = 0.145$$ $$P(x=2)= (4C2 *48C0)/(52C2) = 0.0045$$ $$E(X)= Σ(x*p(xi)=0.154$$ Where I'm running into issues with b and c. How would this need to be step up so that it can be solved.  |
| How do I get the function and solve the problem using the definition of limit of sum? Posted: 06 Nov 2021 07:42 PM PDT Express the following as a definite integral $\int_0^1f(x)\,dx$ and find its exact value. $$\lim_{n\to\infty}\sum_{i=1}^n\frac1{\sqrt n\sqrt{n+i}}$$ Hi, have some doubts regarding this question on how to get the definite integral using the definition of limit of sum, converting the Riemann sum to a definite integral , the process or steps in doing so from the limits equation above.  |
| Can you please help me solve this inequality problem? [closed] Posted: 06 Nov 2021 07:53 PM PDT Prove that: $3(x^2 + y²+ xy) (y^2+z^2+yz) (z^2+x^2+ zx) \geq (x+ y+z)^2(xy+ yz+zx)^2.$  |
| Fibonacci numbers, sum of squares and divisibility Posted: 06 Nov 2021 08:11 PM PDT Let $(F_n)$ be the Fibonacci sequence (i.e. $F_{n+2} = F_{n+1}+F_n$ with $F_{-1}=1$ and $F_0=0$). Consider $a,b \in \mathbb{Z}_{\ge 1}$ with $a \le b$, and $D=1+a^2+b^2$, such that $a$ and $b$ divide $D$. Question: Must the ordered pair $(a,b)$ be equal to $(F_{2n-1},F_{2n+1})$ for some $n \ge 0$? Remark: It was checked (by SageMath) for $b<10^5$. Note that $F_{2n\pm1}$ divides $1+ F_{2n-1}^2 + F_{2n+1}^2$ thanks to the following identity: Proposition: $1+ F_{2n-1}^2 + F_{2n+1}^2 = 3F_{2n-1}F_{2n+1}$. Cassini's identity: $F_{n-1}F_{n+1} = (-1)^n + F_n^2$.  |
| Can someone explain the integral split in the following? Posted: 06 Nov 2021 08:19 PM PDT Let $\mathbb{F} \subset \mathbb{G}$ where $\mathbb{G}$ is the filtration progressively enlarged by the filtration $\mathbb{F}$ and let $\tau$ be a $\mathbb{G}$-stopping time. Define $B_t:= \exp (\int_0^t r_s ds)$ where $r$ is $\mathbb{G}$-adapted and let $A$ be a $\mathbb{G}$-adapted process with finite variation, can some one explain the following equality? $$B_\tau\mathbb{E} \bigg[\int_{[\tau,T]} (B_s)^{-1} dA_s \bigg| \mathcal{G}_\tau \bigg]= A_\tau - A_{\tau-} + B_\tau\mathbb{E} \bigg[\int_{(\tau,T]} (B_s)^{-1} dA_s \bigg| \mathcal{G}_\tau \bigg]$$ How does the right hand side change if $r$ and\or $A$ were $\mathbb{F}$-adapted?  |
| Complex analysis with logarithms Posted: 06 Nov 2021 07:42 PM PDT Question: If $\Im z_{1} > 0$ and $\Im z_{2} > 0$, does $Log(z_{1}z_{2}) = Log z_{1} + Log z_{2}$? If yes, prove the statement. If no, give a counterexample. My approach: Suppose that $\Im z_{1} > 0$ and $\Im z_{2} > 0$, then $z_{1} = r_{1}\exp i \theta_{1}$ and $z_{2} = r_{2}\exp i \theta_{2}$ where $0<\theta_{1}<\pi$ and $0<\theta_{2}<\pi$ Thus, $0<\theta_{1} + \theta_{2}<2\pi$ Thus, $Log(z_{1}z_{2}) = Log[(r_{1}r_{2})\exp i (\theta_{1} + \theta_{2})] = Log(r_{1}r_{2}) + i (\theta_{1} + \theta_{2})$ What should I do next to prove that the statement is true? I'm not sure if it is true but cannot find a counterargument. Thank you so much for your help! P/S: The principal value of $log z$ is $Log z$. Please do not edit when you don't know the actual terms.  |
| Top Dolbeault cohomology group vs top De Rham cohomology group of a compact complex manifold Posted: 06 Nov 2021 08:09 PM PDT I cannot reconcile two simple facts for a compact complex manifold $X$ with $dim_{\mathbb{C}}=n$. Let $A^{p,q}(X)$ be differential forms of degree $(p,q)$ on $X$. One the one hand, $$H^{2n}_{DR} (X)= A^{n,n}(X)/d\left(A^{n-1,n}(X) + A^{n,n-1}(X)\right) = A^{n,n}(X)/\left(\partial(A^{n-1,n}(X)) + \bar{\partial}(A^{n,n-1})\right)$$ and this group is isomorphic to $\mathbb{C}$ via integration of forms and Stokes' theorem. On the other hand, $$H^n(X,K_X)\stackrel{Dolbeault}{====} A^{n,n}(X)/ \bar{\partial}(A^{n,n-1}(X))$$ is also said to be isomorphic to $\mathbb{C}$ via integration and before Serre's duality is established (p. 135 of Voisin's "Hodge Theory and Complex Algebraic Geometry I"; in fact, the above fact is used to prove Serre's duality in the book.) Without the $\partial\bar{\partial}$-lemma (that requires $X$ to be Kahler) I don't see how to show that $$\partial \omega=d \omega, \quad \omega \in A^{n-1, n}(X)$$ is zero in Dolbeault cohomology (since it clearly integrates to zero by Stokes' theorem). And yet it should be true by Serre's duality that does not require the Kahler condition...  |
| Posted: 06 Nov 2021 08:12 PM PDT Right now the I am using this formula to calculate the arc length: $$L=\int_a^b\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}d\theta$$ The grooves on the vinyl replicate an Archimedean spiral given in the form of $r= \lambda θ$, where $\lambda $ is a constant. I am having trouble determining the exact number of complete turns and what the value for spacing between each turn should be. Currently, the vinyl record I am using is $754~\text{s}$ long and is played at $40~\text{rpm}$ . Can someone help me formulate an equation? I am not quite sure where to go from here?  |
| Let $A$ be non singular, prove that $\frac{1}{\|A^{-1}\|} = \min_{\|x\| = 1} \|Ax\|$ Posted: 06 Nov 2021 07:57 PM PDT Let $A$ be non singular, prove that $\frac{1}{\|A^{-1}\|} = \min_{\|x\| = 1} \|Ax\|$ I know that $\|A\| = \sup_{\|x\| = 1} \|Ax\|$ but I have no idea how to proceed from here.  |
| Are reduced SVD and truncated SVD the same thing? Posted: 06 Nov 2021 08:22 PM PDT Truncated SVD: http://scikit-learn.org/stable/modules/generated/sklearn.decomposition.TruncatedSVD.html Reduced SVD, I thought this is essentially the same thing, and it appears to be actually more commonly called this way. If you could provide reference, that'll be great.  |
| Find coordinates for points on circle given R, 2 Points, and angle or 2 points and center? Posted: 06 Nov 2021 08:02 PM PDT I would like to find coordinates for points on a circle given:
Ultimately, I would like to write a formula in excel to calculate points on circle to stake out coordinates for surveying. Thanks again for your help. Please let me know if I can provide any more information or rewrite question using better terminology.  |
| Where does the invariant expression for the exterior derivative come from? Posted: 06 Nov 2021 08:24 PM PDT So I've just spent about four TeXed pages (plus about a dozen TeXed pages of discarded work) proving the identity \begin{align*} d \omega(\zeta_1, \ldots, \zeta_{k+1}) &= \sum_{i=1}^{k+1} (-1)^{i-1} \zeta_i \cdot \omega(\zeta_1, \ldots, \hat{\zeta_i}, \ldots, \zeta_{k+1}) \\ &\qquad + \sum_{1 \leq i < j \leq k + 1} (-1)^{i+j} \omega([\zeta_i, \zeta_j], \zeta_1, \ldots, \hat{\zeta_i}, \ldots, \hat{\zeta_j}, \ldots, \zeta_n). \end{align*} Obviously this is not how this identity was discovered. But I don't see any explanation written down anywhere -- just things along the lines of "it can be laboriously checked that the right hand side is in fact linear in $\zeta_1 \wedge \cdots \wedge \zeta_{k+1}$, and then the identity can be verified on an appropriate basis." Questions:
Regarding the first question: In particular, I notice that, on a basis, the first sum gives the desired results and the second sum vanishes; it's simply the case that neither sum is actually a linear function of $\zeta_1 \wedge \cdots \wedge \zeta_{k+1}$, but combining the two gives a linear map. Is there some way to regard the second sum as the result of some "correction" process applied to the first that makes it linear? (Compare for instance situations where you average over a group of symmetries to get an invariant map, for instance.)  |
| Prove that intersection of diagonals in trapezium divide parallel segment to equal segments Posted: 06 Nov 2021 07:55 PM PDT I've got exercise to do as en exercise to my school leaving exam and I have no idea how to prove it:
I'm sorry for my poor translation but I'm not good at math english yet.  |
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