need help with a Markov chain for weather conditions

This is my second time posting the question as I failed to do so the first time, because I did not know the proper way. My apologies.

Question:

Suppose that whether or not it rains today depends on weather conditions through the previous teo days. If it has rained for the past two days, then it will rain today with probability 0.7. If it did not rain for any of the past two days, then it will rain today with probability 0.4. In any other case the weather today will, with probability 0.5, be the same as the weather yesterday. Describe the weather condition over time with a Markov chain.

Suppose that it rained today and yesterday. Calculate the expected number of days until it rains three consecutive days for the first time in the future.

I have found 4 different states that I named RR(0), RN(2), NR(1), and NN(3). R stands for when it rains and N is for when it does not.

As the question asks, I have tried finding the possible ways of three consecutive days being rainy. At time n, we are given it was rainy today and yesterday, meaning we are in State 0.

1-) First possibility is when it rains tomorrow, which gives us RRR (we got the three consecutive days)

2-) Second possibility is when we go from 0 to 2, from 2 to 1, from 1 to 0, and stay in 0 one day. That follows as : RRNRRR (In 4 days we can get rain for 3 consecutive days)

3-) Third is when we go from 0 to 2, from 2 to 3, from 3 to 1, from 1 to 0, and stay in 0 one day. That follows as: RRNNRRR.

To conclude what I have in mind about the question, wherever we go, when we get to State 0, we need to stay there one more day to get three consecutive rainy days. That means the minimum # of days to get rain for three consecutive days is one day. However, after this point, I am not able to proceed with the question.

Any help would be appreciated, thank you!!

Edit: I think the maximum # of days is just a random number, which leads me to the expectation of the sum of a random number of geometric random variable, but still I can't go any further beyond that. Thank you!

• For $i=0,1,2,3$, let $E_i$ be the expected number of days until $3$ consecutive rainy days, if the chain is currently in state $i$. We are asked to compute $E_0$. Notice that $E_0=1+.3E_2$. Develop similar equations for the other $E_i$ and solve. Jun 29, 2021 at 17:39
• Developing the equations is where I am stuck at and I am also confused as to whether there is a boundary condition. Could you please explain to me, for instance, how to develop E0? Jun 29, 2021 at 18:44

Let me get you started.

I claim that $$E_0=1+.3E_2$$ as I stated in a comment.

Suppose we are in state $$0$$. We must wait at least $$1$$ more day to see if we'll have three consecutive days of rain. $$70\%$$ of the time, it rains, and we are done, but $$30\%$$ of the time we transition to state $$2$$, and then we must wait, on average, $$E_2$$ days to get $$3$$ consecutive rainy days.

If we are in state $$2$$, similar reasoning gives $$E_2=1+.5E_1+ .5E_3$$

You can read the equations right off your diagram. You should get $$4$$ equations with $$4$$ unknowns and a unique solution. I'm sure you won't have any problem finishing from here.

• I have found E3=1 + 0.6(E3) + 0.4(E1) (state 1 is the only different state we can transition to from state 3 and we might also stay in state 3 with p=0.6?) and E1= 1 + 0.5(E0) + 0.5(E2). Do you think I made a mistake or those are correct? Jun 30, 2021 at 8:20
• @SydneyAstbury I think you are correct. Jun 30, 2021 at 11:36

Ordinarily, Markov Chains are conditional on the previous step, but not on the previous two steps. A way to get around this in the current problem is to re-define the states to account for two days, with suitable overlaps.

The new states are 00 (for consecutive dry days) 01 (dry followed by wet), and so on to 11 (for two wet days in a row).

States 'overlap' so that 00 can be followed by 01, but not by 01, etc. In the matrix below we call the new states A,B, C, D.

The transition matrix can be written in R as

P = matrix(c(.6, .4,  0,  0,
0,  0, .5, .5,
.5, .5,  0,  0,
0,  0, .3, .7), byrow=T, nrow=4)

We seek $$\sigma$$ such that $$\sigma P = P,$$ so that $$\sigma$$ is a left eigenvector of $$P.$$ Because R finds right eigenvectors, we use the transpose t(P).

The stationary distribution is proportional to the right eigenvector of smallest modulus, which R prints first. [In R, the symbol %*% denotes matrix multiplication.]

g = eigen(t(P))\$vec[,1]
sg = g/sum(g); sg
[1] 0.2542373 0.2033898 0.2033898 0.3389831
sg %*% P  # check
[,1]      [,2]      [,3]      [,4]
[1,] 0.2542373 0.2033898 0.2033898 0.3389831

So the long-run probability of rain is $$0.5424.$$

Note: In this relatively simple problem it is not difficult to solve a few simultaneous equations, but the eigen-method shown above is very convenient for ergodic chains with more than four states.


Is the following application of the CLT correct?

I am new to probability theory and still have a problem understanding some things. I came across the following problem that I am unsure on how to solve. Here it is:

Suppose that in a city with $$100,000$$ cars a shop sells tires for cars. It has been observed that in an interval of $$3$$ months the percentage of cars that come to the shop for the replacement of all their tires is $$0.5$$%. What is the least number of tires that the shop has to order so that it can satisfy all its customers in a $$3$$ month interval with probability $$\geq 95$$%?

And here is my attempt:

Let $$X$$ be the random variable of cars coming at the shop in a $$3$$ month interval. Then $$X$$ is a discrete random variable following binomial distribution $$B(100,000; 0.005)$$. We would like to find the least $$x\in\mathbb{N}$$ satisfying $$P(X\leq x)\geq95$$%. Since (I assume) every car has $$4$$ tires, our answer is going to be $$4x$$. Now since we have a large number of cars, by the Central limit theorem, we have that $$X\sim N(\mu,\sigma^2)$$, where $$\mu, \sigma^2$$ are the expected value and the variance of $$B(100,000; 0.005)$$ and therefore, by calculating, it is $$\mu=500, \sigma^2=475$$. Now we have $$P(X\leq x)=P\big{(}\displaystyle{\frac{X-500}{\sqrt{475}}\leq\frac{x-500}{\sqrt{475}}\big{)}=P\big{(}Z\leq\frac{x-500}{\sqrt{475}}\big{)}}$$, where $$Z\sim N(0,1)$$. Now our $$x$$ will satisfy $$\displaystyle{P\big{(}Z\leq\frac{x-500}{\sqrt{475}}\big{)}}=0.95$$, therefore $$\displaystyle{P\big{(}0 hence (by the tables) it is $$\displaystyle{\frac{x-500}{\sqrt{475}}=1.645}$$, which yields $$x=535.85..$$ and since $$x$$ is supposed to be an integer we have that $$x=536$$. therefore the least number of tires is $$2144$$.

Is my solution correct? if not, what should I have used?

Comment: I know that the $$x=$$fractional is not correct since i specified that $$x\in\mathbb{N}$$, but you get the point.

• "the percentage of cars that come to the shop for the replacement of all their tires is 0.5%" That is quite poorly worded, it says that X is a constant ( 500). I guess it meant what you understood, though. – leonbloy Jan 21 at 14:49
• @leonbloy I know, I was confused by this as well, but this is an accurate translation of the problem... – JustDroppedIn Jan 21 at 15:03

Assuming that the shop's only tire business is to replace all four tires, your method makes sense. [Otherwise, I can't see how to work the problem.]

The CLT implies that a binomial distribution with sufficiently large $$n$$ is well-approximated by a normal distribution, which is what you're doing.

Binomial. I used the binomial quantile function (inverse CDF) qbinom in R to compute an answer directly from the binomial distribution. It is very nearly the same as your answer.

4*ceiling(qbinom(.95, 10^5, .005))
[1] 2148

I expect that the small discrepancy may result from rounding in finding the mean and standard deviation of the approximating normal distribution or from using printed normal tables. [Perhaps try $$\sigma^2 = 497.5.]$$

Normal approximation to binomial. Without rounding, I get the following result from the normal approximation. (The normal approximation with $$n=10^5$$ and $$p = 0.005$$ should be quite good, but not necessarily exactly perfect.)

n = 10^5; p = .005; mu = n*p; sg = sqrt(n*p*(1-p))
4*ceiling(qnorm(.95, mu, sg))
[1] 2148

Poisson approximation. Another reasonable approximation is to use the distribution $$\mathsf{Pois}(\lambda = \mu = np).$$ This approximation is quite good for large $$n$$ and small $$p:$$

4*ceiling(qpois(.95, mu))
[1] 2148

Normal approximation to Poisson. For $$\lambda$$ as large as $$500$$ the Poisson distribution is well approximated by $$\mathsf{Norm}(\lambda, \sqrt{\lambda}):$$

4*ceiling(qnorm(.95, mu, sqrt(mu)))
[1] 2148

The following figure shows the binomial distribution (black bars) along with its normal approximation (blue curve) and its Poisson approximation (centers of red circles). The probability to the right of the vertical dotted line is about 0.05.


How to enumerate 2D integer coordinates ordered by Euclidean distance?

The square of Euclidean distance between $(x, y)\in\mathbb{Z}^2$ and origin is $d = x^2+y^2$. How to enumerate the coordinates $(x, y)$ in ascending order of $d$?

For example, the first 14 sets of coordinates are:

d=0: { (0,0) }
d=1: { (1,0), (0,1), (0,-1), (-1,0) }
d=2: { (1,-1), (-1,-1), (-1,1), (1,1) }
d=4: { (2,0), (0,2), (-2,0), (0,-2) }
d=5: { (1,2), (-1,2), (1,-2), (-2,1), (-2,-1), (2,-1), (2,1), (-1,-2) }
d=8: { (2,2), (-2,2), (-2,-2), (2,-2) }
d=9: (0,3), (-3,0), (0,-3), (3,0)
d=10: (1,3), (-1,3), (3,1), (-3,1), (-3,-1), (1,-3), (-1,-3), (3,-1)
d=13: (2,-3), (-3,-2), (3,-2), (-2,-3), (-3,2), (3,2), (-2,3), (2,3)
d=16: (0,-4), (-4,0), (4,0), (0,4)
d=17: (-4,1), (-4,-1), (4,1), (1,-4), (4,-1), (-1,-4), (-1,4), (1,4)
d=18: (3,-3), (-3,-3), (-3,3), (3,3)
d=20: (4,2), (4,-2), (-4,-2), (2,-4), (-4,2), (-2,-4), (-2,4), (2,4)
d=25: (-3,-4), (-5,0), (5,0), (4,3), (-3,4), (-4,3), (0,-5), (4,-3), (-4,-3), (3,-4), (3,4), (0,5)

The first 14 iterations are depicted as:

          13
131210 9101213
1311 8 7 6 7 81113
12 8 5 4 3 4 5 812
10 7 4 2 1 2 4 710
13 9 6 3 1 0 1 3 6 913
10 7 4 2 1 2 4 710
12 8 5 4 3 4 5 812
1311 8 7 6 7 81113
131210 9101213
13

For a finite range of $(x, y)$, a trivial algorithm is to store all coordinates within the range into a list, and then sort the coordinates with $d$. However, this will need $O(n\log n)$ time and $O(n)$ space.

Another possible approach is to solve the diophantine equation $x^2 + y^2 = m$ for $m = 0, 1, \ldots, ... d_\mathrm{max}$. But it seems also a hard problem.

Is there any simpler way with lower time/space complexity?

Here is a C++ code of the trivial solution for reference.

• As far as storing goes, you can certainly take advantage of symmetries and perhaps more generally use the Sum of Squares function $r_2(n)$ (or $r'_2(n)$ if you're not interested in order/signs). I will also link this mathoverflow question as it seems to answer precisely what you desire. – Fimpellizieri Jun 28 '16 at 22:54
• Why the complexity of the trivial algorithm is not $O(n^2)$? – user84976 Jun 29 '16 at 10:35
• Your $n$ seems to be the number of points in your grid, not the edge length of the grid, right? So in terms of the edge length $a$, you get $O(a^2\log a)$ time complexity and $O(a^2)$ space complexity. But you can't go asymptotically lower than $O(a^2)$ time complexity since you have to enumerate this many points. Are you interested in tweaking constants here, or only in getting rid of the $\log a$ term or reducing the space complexity? – MvG Jul 5 '16 at 12:29

You can employ symmetry to only generating one eighth of a circle, e.g. $0\le x\le y$. It should be obvious how to do this.

You can reduce the space complexity to $O(r)=O(\sqrt n)$, i.e. linear in the radius of your disk, not in the number of grid points it covers. The way to do this is by considering parallel lines of grid points. Then on each line you know that the points will be enumerated starting closest to the center and moving outwards, so it is enough to keep track of the next candidate on each line.

The most appropriate data structure for this keeping track is most likely a priority queue, which will give you quick access to the element with minimal $d$. But even then, elementary operations in a priority queue have $O(\log n)$ time complexity. Which in this case means you are still at $O(r^2\log r)=O(n\log n)$ overall time complexity since you enumerate $O(r^2)$ elements and maintain a data structure where you have to perform two $O(\log r)$ operations for each of them.

Here is an implementation of these ideas. The core portions are these:

struct Pt { // essentially your struct Point, just shorter names
int x, y, d;
Pt(int x, int y) : x(x), y(y), d(x*x + y*y) {}
bool operator<(const Pt& p) const { return d > p.d; } // flipped!
};

void enumerateGridPoints(int r) {
priority_queue<Pt> q;
q.push(Pt(0, 0));
int d, level = -1;
while ((d = q.top().d) <= r*r) {
reportLevel(++level, d);
do {
Pt p = q.top();
q.pop();
reportPoint(p);
if (p.y < p.x) q.push(Pt(p.x, p.y + 1));
if (p.y == 0) q.push(Pt(p.x + 1, 0));
} while (q.top().d == d);
}
}

Since the C++ std::priority_queue is a max queue by default, and since I'm lazy, I just flipped the sign in the comparison operation so that the element with smallest d is the maximal element and thus extracted next.

One benefit of the above algorithm compared to yours is that it's easy to modify it so that enumerates points without any preset limit. That might be useful if you'd like to e.g. inspect sites until you find something you are looking for, but you don't know up front how long you will have to search until you find it. Both time and space complexity scale with the generated output, so small initial portions can be obtained quickly while still allowing you to proceed to larger results.

Edsger Dijkstra gives an elegant recursive solution to the problem of finding solutions to $x^2 + y^2 = m$ in his 1976 book The Discipline of Programming. The solution is outlined here, and involves sweeping potential $x$ values down from $\sqrt m$ and $y$ values up from 0.

Consider the function $B(x, y)$ that returns solutions between $x$ and $y$, defined recursively as follows:

$$B(x, y) = \begin{cases} \varnothing, & \text{if x < y (base case)} \\ (x, y) \cup B(x-1, y+1), & \text{if x^2 + y^2 = m (satisfying case)} \\ B(x, y+1), & \text{if x^2 + y^2 < m} \\ B(x-1, y), & \text{if x^2 + y^2 > m} \\ \end{cases}$$

Using this definition, you can calculate $B(\sqrt m, 0)$ for each $m=0,1,…,d_{max}$. Written as a generator or iterator, this algorithm will have constant space complexity.

• Would you agree that the time complexeity is $O(r^3)$ here (with $r=\sqrt{d_{\text{max}}}$), since for each of the $O(r^2)$ possible values for $m$ you'd have $O(r)$ steps of evaluating $B$ till you reach the base case? – MvG Jul 6 '16 at 11:25


Solve for $y''+2y'+y = \frac{e^{-t}}{t^{2}}$

I need some help in finding the solution to this second-order non-homogeneous DE.

I know how to find the solution of the reduced equation $$y''+2y'+y=0.$$

The characteristic equation $r^{2}+2r+1=0$ yields real and repeated roots $r=-1$. So the complementary solution, $$y_{c}= c_{1}e^{-t} + c_{2}te^{-t}$$.

Now what I am struggling with is making educated guesses about coming up with a particular solution. So any suggestions on that would be helpful.

EDIT1: I haven't learned about the method of variation of parameters yet and wanted to do this with the method of undetermined coefficients. Please feel free to post the solution by using any of the techniques so I can learn from it.

EDIT2: Our fundamental pair of solutions for the corresponding homogeneous is $y_{1}=e^{-t}$, and $y_{2}=t e^{-t}$.

We know that our particular solution will be of the form, $y_{p} = u_{1} y_{1} + u_{2} y_{2}$.

The determinant of the Wronskian, $W = e^{-2t}$, with $W_{1} = -t^{-1} e^{-2t}$, and $W_{2} = t^{-1} e^{-2t}$.

So $u'_{1} = \frac{W_{1}}{W} = -\frac{1}{t}$, and $u'_{2} = \frac{W_{2}}{W} = \frac{1}{t}$.

So $u_{1} = -\int \frac{1}{t} dt = -ln|t|$, and $u_{2} = \int \frac{1}{t} dt = ln|t|$.

Hence, $y_{p} = -ln|t| e^{-t} + ln|t|te^{-t}$.

So our general solution, $$y = y_{c} + y_{p} = e^{-t}(c_{1}+c_{2}t-ln|t|+ln|t|t).$$

Does it look okay?

• Generally you only do particular solutions if it's a polynomial times a sine/cosine times an exponential, not a rational function. I'd use variation of paramaters – Alan Feb 5 '15 at 1:55
• @Alan I haven't learned that technique yet. Is it the method of undetermined coeffecients? – OGC Feb 5 '15 at 1:57
• What techniques have you learnt? – mattos Feb 5 '15 at 2:22
• @Mattos Separation of variables technique and learned method of undetermined coefficients today. Seeing youtube videos on change of var technique. Is it possible to solve this problem by the method of undetermined coefficients technique? – OGC Feb 5 '15 at 2:26
• @user36829 If you want, I'll post a solution using variation of parameters. – mattos Feb 5 '15 at 3:50

$$y_h = c_1e^{-t} + c_2te^{-t}$$

with

$$y_1 = e^{-t} \ \ \ \ \ y_2 = te^{-t}$$

Taking the Wronskian, you also found

$$W = e^{-2t}$$

The particular solution is given by the formula

$$y_p = -y_1 \int \frac{y_2 g(t)}{W} dt + y_2 \int \frac{y_1 g(t)}{W} dt$$

where

$$g(t) = \frac{e^{-t}}{t^{2}} = t^{-2}e^{-t}$$

Hence

\begin{align} y_p &= -e^{-t} \int \frac{te^{-t} \cdot t^{-2}e^{-t}}{e^{-2t}} dt + te^{-t} \int \frac{e^{-t} \cdot t^{-2}{e^{-t}}}{e^{-2t}} dt \\ &= -e^{-t} \int \frac{1}{t} dt + te^{-t} \int \frac{1}{t^{2}} dt \\ &= -e^{-t} \ln(t) - \frac{te^{-t}}{t} \\ &= -e^{-t}(\ln(t) + 1) \\ \end{align}

Therefore,

\begin{align} y &= y_h + y_p \\ &= e^{-t}(c_1 + c_2 t - \ln(t) - 1) \\ \end{align}

You can check by differentiation that this satisfies the ODE.

• $-1\cdot e^{-t}$ is an unnecessary part of the solution since it solves the homogeneous equation. I.e. the $-$ can be 'absorbed' into the $c_1$ coefficient. – jdods Mar 5 '15 at 1:58
• @jdods Yeah and? – mattos Mar 5 '15 at 8:38

In this case of only one characteristic value, also a change of function can help. Substitute, according to the homogeneous solution, $y(t)=e^{-t}u(t)$ to get for $u$ $$u''(t)=t^{-2}\implies u'(t)=-t^{-1}+C\implies u(t)=-\ln|t|+Ct+D$$ and thus $$y(t)=e^{-t}(-\ln|t|+Ct+D)$$

I'll use the formula I developed here $$Y_p=y_h\int \left\{y_h^{-2} W_0 \left(\int y_h W_0^{-1} g \ dt\right)\right\} \ dt$$ where the Wronskian (up to a constant multiple) is $$W_{0}=\exp\left(-{\int p \ dt}\right)$$ and a solution to the homogeneous equation is $y_h=e^{-t}$ (we can use either).

So we first integrate $$W_{0}=\exp\left(-{\int 2 \ dt}\right)=e^{-2t}$$ and then we'll find the particular solution by \begin{aligned} Y_p&=e^{-t}\int \left\{e^{2t} e^{-2t} \left(\int e^{-t} e^{2t} \frac{e^{-t}}{t^2} \ dt\right)\right\} \ dt \\ &=e^{-t}\int \left(\int \frac{1}{t^2} \ dt\right) \ dt \\ &=e^{-t}\int \left(-\frac{1}{t}\right) \ dt \\ &=-e^{-t}\ln t. \end{aligned}

This is a nice formula because you only need one solution to the homogeneous equation.


# Does the interval $[x, x + 10 x^{1/4})$ contain a sum of two squares?

Is it true that there exist an integer $N \in \mathbb N$ , such that $\forall x > N$ , the interval $[x, x+10x^{1/4} )$ always contains a sum of two squares ?

I know that $n \in \mathbb N$ is a sum of two squares iff it is of the form $n=q_1^{2a_1}...q_m^{2a_m}p_1^{k_1} ... p_r^{k_r}$ where $q_1,...,q_m,p_1,...,p_r$ are primes such that $q_i \equiv 3( \mod 4)$ and $a_1,...,a_m, k_1,...,k_r$ are non-negative integers ; where $m , r\ge 0$ . So it is enough to prove that for large enough $x$ , the interval $[x , x+10x^{1/4})$ always contains such a positive integer . But I don't know how to prove this.

• Reading about the Gauss circle problem would be a good starting point. – Daniel Fischer Jul 30 '17 at 19:01
• For me to start: where did you get the problem/conjecture? Is it stated as something definitely true or as an open problem? It is reasonable, by the way, as the count of numbers up to large $z$ that are sums of two squares is about $$\frac{Cz}{\sqrt {\log z}}$$ with (known) positive constant $C.$ – Will Jagy Jul 30 '17 at 19:19
• @WillJagy : For the source of the problem ; it was asked as an open ended question in the class by our professor . Where do you get the asymptotic formula about the count ? Could you please give any reference ? Thanks – user Jul 31 '17 at 15:54
• store.doverpublications.com/0486425398.html – Will Jagy Jul 31 '17 at 18:20

You might try to proceed greedily. Suppose we want to find a number that is "near" (but less than) $x$ that is a sum of two squares. Maybe one of these squares will be the largest square less or equal to $x$, which I'll denote by $a$. (So $a = \lfloor \sqrt x \rfloor^2$), where $\lfloor \cdot \rfloor$ is the "floor" function, or the "greatest integer below function"). In this case, the second square, which I'll call $b$, should be chosen to be the greatest square less than or equal to $x - a$. (So $b = \lfloor \sqrt{x - a} \rfloor^2$).

This gives us a candidate sum of squares $a + b$, which is relatively close to $x$. Now we ask, how close does this get us? How large is $x - a - b$?

This really asks, how close is $x$ to the largest square up to $x$? That is, how close is $x$ to $a = \lfloor \sqrt x \rfloor^2$?

Note that $$x - \lfloor \sqrt x \rfloor^2 = (\sqrt x - \lfloor \sqrt x \rfloor)(\sqrt x + \lfloor \sqrt x \rfloor) \leq (1) \cdot (2 \sqrt x) \leq 2 \sqrt x.$$ So the largest square up to $x$ is no more than $2 \sqrt x$ away from $x$.

Thus $x - a \leq 2 \sqrt x$. Iterating, we have that $$(x - a) - b \leq 2 \sqrt{x - a} \leq 2 (\sqrt {2 \sqrt x}) = 2^{3/2} x^{1/4}.$$

We have shown that given an $x$, the nearest "greedy" sum of squares is at most $2^{3/2} x^{1/4}$ away. So we've shown that there is a sum of squares in intervals of the shape $(x - 2^{3/2} x^{1/4}, x]$, which clearly implies the original question.

In comments, Will Jagy noted that the density of numbers which are sums of two squares is of the shape $Cx/\sqrt{\log x}$, where $C$ is the Landau-Ramanujan constant. It is conjectured that the gaps between numbers which are sums of squares shouldn't deviate too far from what is expected from the density, and in particular is bounded by $x^\epsilon$ for any $\epsilon > 0$. In particular it is conjectured that there is a number which is a sum of two squares in the interval $[x, x + x^{\epsilon})$ for any $\epsilon > 0$, for $x$ sufficiently large.
This answer shows that the gap between numbers which are sums of squares is at most $2^{3/2}x^{1/4}$, which is a far cry from $x^{\epsilon}$.
In general, there are not good uniform improvements over what's presented above. But there are remarkable on-average bounds due to Hooley, showing for instance that almost all (in the density sense) gaps between sums of squares is no more than $(\log x) (\log \log x)$.


# Find the inverse of a lower triangular matrix of ones

Find the inverse of the matrix $A=(a_{ij})\in M_n$ where $$\begin{cases} a_{ij}=1, &i\geq j,\\ a_{ij}=0, &i<j. \end{cases}$$

The only method for finding inverses that I was taught was by finding the adjugate matrix. So $A^{-1}=\frac{1}{\det A}\operatorname{adj(A)}$

$$A=\begin{pmatrix} 1 & 0 &0 &\ldots &0\\ 1 & 1 & 0 &\ldots &0\\ 1 & 1 & 1 &\ldots &0\\ \vdots &\vdots & \vdots &\ddots & \vdots\\ 1 & 1 &1 & \ldots &1 \end{pmatrix}$$

This is a triangular matrix so $\det A=1.$ To find the adjugate I first need to find the cofactor matirx$(C)$ of $A$.

$$C=\begin{pmatrix} 1&-1&0&0&\ldots &0\\ 0 & 1& -1 &0&\ldots & 0\\ 0 & 0 & 1 & -1&\ldots &0\\ 0 & 0 & 0 &1 &\ldots &0\\ \vdots &\vdots &\vdots &\vdots&\ddots & \vdots\\ 0 & 0 &0 &0 &\ldots &1\end{pmatrix}$$

$$C^T=\operatorname{adj}(A)=\begin{pmatrix} 1 & 0 & 0 & 0&\ldots &0\\ -1 & 1& 0 & 0 &\ldots &0\\ 0&-1&1&0&\ldots &0\\ 0& 0 &-1 &1 &\ldots&0\\ \vdots &\vdots &\vdots &\vdots &\ddots&\vdots\\ 0&0&0&0&\ldots &1\end{pmatrix}=A^{-1}$$

Is this correct? Also, can I leave it like that or should I somehow write it more formally?

• try using elementary row operations. – vidyarthi Jan 24 '17 at 10:11
• Looks all right. – StubbornAtom Jan 24 '17 at 10:25
• Try multiplying your proposed $A^{-1}$ by $A$ to see whether you get the identity matrix. – Gerry Myerson Jan 24 '17 at 11:38

Yes, your inverse is correct. Let me show you an alternative approach of finding this inverse without the use of the adjugate matrix.

Denote $A_n$ as the $n \times n$ matrix that has elements $a_{ij} = 1$ if $i \geq j$, and $0$ otherwise. Then, $A_n$ can be constructed by applying the elementary row operations $$R_2 + R_1, R_3 + R_2, \ldots, R_n - R_{n-1}$$ to the identity matrix $I_n$. So, $$A_n = \prod_{i=1}^{n-1} E_i,$$ where $E_i$ is the elementary matrix representing the row operation $R_{i+1} - R_i$. Since det$(A)$ is nonzero, it is invertible, hence $$B_n = \left( A_n \right)^{-1} = \prod_{i = n-1}^{1} {E_i}^{-1},$$ where ${E_i}^{-1}$ is the row operation $R_{i+1} - R_i$. That is, $B_n$ can be constructed by applying the row operations $$R_n - R_{n-1}, R_{n-1} - R_{n-2},\ldots,R_2 - R_1$$ to the idenity matrix $I_n$. The inverse of $A_n$ is therefore given by the matrix
$$B_n = \begin{pmatrix} \phantom{-}1 & \phantom{-}0 &\phantom{-}0 &\ldots &0\\ -1 & \phantom{-}1 & \phantom{-}0 &\ldots &0\\ \phantom{-}0 & -1 & \phantom{-}1 &\ldots &0\\ \phantom{-}\vdots &\phantom{-}\vdots & \phantom{-}\vdots &\ddots & \vdots\\ \phantom{-}0 & \phantom{-}0 & \phantom{-}0 & \cdots &1 \end{pmatrix}.$$ In order to check that this is the correct inverse, you can use the identity $$A_nB_n = I_n.$$

• How can we write this inverse like this $B_n=(b_{ij})= \begin{cases} 1, &i= j,\\ -1, &i-j=1\\ 0, & ? \end{cases}$ When is $(b_{ij})=0$? – lmc Jan 24 '17 at 13:31
• $b_{ij}$ is $1$ if $i =j$, $-1$ if $i = j + 1$ and $0$ otherwise. – user394255 Jan 24 '17 at 13:33

I would like to present a very simple solution by interpretation of these matrices as operators on $\mathbb{R^n}$ (which will surprise nobody...). Triangular matrix $A$ acts as a discrete integration operator:

For any $x_1,x_2,x_3,\cdots x_n$:

$$\tag{1}A (x_1,x_2,x_3,\cdots x_n)^T=(s_1,s_2,s_3,\cdots s_n)^T \ \ \text{with} \ \ \begin{cases}s_1&=&x_1&&&&\\s_2&=&x_1+x_2&&\\s_3&=&x_1+x_2+x_3\\...\end{cases}$$

(1) is equivalent to:

$$\tag{2}A^{-1} (s_1,s_2,s_3,\cdots x_n)^T=(x_1,x_2,x_3,\cdots x_n)^T \ \ \text{with} \ \ \begin{cases}x_1&=& \ \ s_1&&&&\\x_2&=&-s_1&+&s_2&&\\x_3&=&&&-s_2&+&s_3\\...\end{cases}$$

and it suffices now to "collect the coefficients" in the right order in order to constitute the inverse matrix.

(Thus the inverse operation is - in a natural way - a discrete derivation operator).

You can also use the induction along the dimension $n$ (leading to a simple but not very constructive proof though).

Let $A_n$ denote the matrix $A$ of dimension $n$ and $B_n$ its inverse. The statement is obviously true for $n=1$ and reduces to showing that $1=1$. Let the statement be true for $n-1$, that is, $A_{n-1}^{-1}=B_{n-1}$. If $1_{n-1}$ and $0_{n-1}$ the column vectors of ones and zeros, respectively, of dimension $n-1$, the matrix $A_n$ can be written in the block form $$A_n=\begin{bmatrix}A_{n-1}&0_{n-1}\\1_{n-1}^T&1\end{bmatrix}.$$ You can easily verify that $$A_n^{-1}=\begin{bmatrix}A_{n-1}^{-1}&0_{n-1}\\-1_{n-1}^TA_{n-1}^{-1}&1\end{bmatrix}$$ is the inverse of $A_n$. By the induction assumption, $A_{n-1}^{-1}=B_{n-1}$. It is easy to check that $1_{n-1}^TA_{n-1}^{-1}=1_{n-1}^TB_{n-1}=e_{n-1}^T:=[0,\ldots,0,1]^T$ (sum of the rows of $B_{n-1}$). So $$A_n^{-1}=\begin{bmatrix}B_{n-1}&0_{n-1}\\-1_{n-1}^TB_{n-1}&1\end{bmatrix}=\begin{bmatrix}B_{n-1}&0_{n-1}\\-e_{n-1}^T&1\end{bmatrix}=B_n.$$

Just to show a different approach.
Consider the matrix $\mathbf E$, having $1$ only on the first subdiagonal $$\mathbf{E} = \left\| {\,e_{\,n,\,m} = \left\{ {\begin{array}{*{20}c} 1 & {n = m + 1} \\ 0 & {n \ne m + 1} \\ \end{array} } \right.\;} \right\| = \left\| {\,\left( \begin{gathered} 0 \\ n - m - 1 \\ \end{gathered} \right)\;} \right\|$$ Multiply it by $\mathbf A$ , and it is easily seen that $$\mathbf E \, \mathbf A=\mathbf A-\mathbf I \quad \Rightarrow \quad \left( \mathbf I -\mathbf E \right)\,\mathbf A=\mathbf I$$

In another way, consider that the powers of $\mathbf E$ are readily found and have a simple formulation $$\begin{gathered} \mathbf{E}^{\,\mathbf{2}} = \left\| {\,\sum\limits_{0\, \leqslant \,k\,\left( { \leqslant \,n - 1} \right)} {\left( \begin{gathered} 0 \\ n - k - 1 \\ \end{gathered} \right)\left( \begin{gathered} 0 \\ k - m - 1 \\ \end{gathered} \right)} \;} \right\| = \left\| {\,\left( \begin{gathered} 0 \\ n - m - 2 \\ \end{gathered} \right)\;} \right\| = \left\| {\,\left\{ {\begin{array}{*{20}c} 1 & {n = m + 2} \\ 0 & {n \ne m + 2} \\ \end{array} } \right.\;} \right\| \hfill \\ \quad \vdots \hfill \\ \mathbf{E}^{\,\mathbf{q}} = \mathbf{E}^{\,\mathbf{q} - \mathbf{1}} \,\mathbf{E} = \left\| {\,\sum\limits_{0\, \leqslant \,k\,\left( { \leqslant \,n - 1} \right)} {\left( \begin{gathered} 0 \\ n - k - \left( {q - 1} \right) \\ \end{gathered} \right)\left( \begin{gathered} 0 \\ k - m - 1 \\ \end{gathered} \right)} \;} \right\| = \hfill \\ = \left\| {\,\left( \begin{gathered} 0 \\ n - m - q \\ \end{gathered} \right)\;} \right\|\quad \left| {\;0 \leqslant \text{integer}\;q} \right. \hfill \\ \end{gathered}$$ Therefore $$\mathbf{A} = \sum\limits_{0\, \leqslant \,j} {\mathbf{E}^{\,\mathbf{j}} } = \frac{\mathbf{I}} {{\mathbf{I} - \mathbf{E}}}$$ To connect to the answer of Jean Marie, note that $$\left( {\mathbf{I} - \mathbf{E}} \right)\left\| {\,\begin{array}{*{20}c} {x_{\,0} } \\ {x_{\,1} } \\ \vdots \\ {x_{\,n} } \\ \end{array} \;} \right\| = \left\| {\,\begin{array}{*{20}c} {x_{\,0} ( - 0)} \\ {x_{\,1} - x_{\,0} } \\ \vdots \\ {x_{\,n} - x_{\,n - 1} } \\ \end{array} \;} \right\| = \left\| {\,\begin{array}{*{20}c} {\nabla x_{\,0} \;\left| {x_{\, - 1} = 0} \right.} \\ {\nabla x_{\,1} } \\ \vdots \\ {\nabla x_{\,n} } \\ \end{array} \;} \right\|$$


# Finding the 10th root of a matrix

I want to find a $2 \times 2$ matrix, named $A$ in this situation, such that:

$$A^{10}=\begin {bmatrix} 1 & 1 \\ 0 & 1 \end {bmatrix}$$

How can I get started? I was thinking about filling $A$ with arbitrary values $a, b, c, d$ and then multiplying it by itself ten times, then setting those values equal to the given values but I quickly realized that would take too long. Is there a more efficient way?

Take$$A=\begin{bmatrix}1&x\\0&1\end{bmatrix}.$$Now, compute $$A^2,A^3,\ldots$$ You'll find quickly which $$x$$ you should choose.

• Yes. This idea can be conceived by noting that the given matrix is upper triangular, then maybe an upper triangular solution $A$ is possible, and the eigenvalues of the original matrix are $1$ and $1$, then maybe the eigenvalues of $A$ can be taken to be also $1$ and $1$ (a particular tenth root of $1$ is $1$). So if you try it out, you will succeed. A more difficult problem is to find all solutions $A$ with entries in some field (such as $\mathbb{Q}$ or $\mathbb{C}$). – Jeppe Stig Nielsen Jan 29 '18 at 18:08
• @JeppeStigNielsen Indeed, that's a more difficult problem. – José Carlos Santos Jan 29 '18 at 19:10
• (+1) This is the approach that first came to mind. Now I need to approach this slightly differently. – robjohn Jan 30 '18 at 18:10

You can also consider $$A=\begin {bmatrix} 1 & 0 \\ 0 & 1 \end {bmatrix} + \begin {bmatrix} 0 & a \\ 0 & 0 \end {bmatrix} =I+N$$ and try to use binomial formula for expansion of powers of a binomial.

Notice that $N^k=0$ for $k>1$.

• Note that it is usually not possible to use the binomial formula for matrices, because the product is not commutative. In this case it is fine, because one of the matrices is the identity. – M.Herzkamp Jan 29 '18 at 11:45
• @M.Herzkamp Yes, we can use the formula because matrices commute. Thank you for valuable remark.. – Widawensen Jan 29 '18 at 11:49

The answers given by José Carlos Santos and Widawensen are correct. There is a more general alternative method, even through it does not work for your particular example. But I thought I'd give it anyway in case it is useful to others.

If you want the $n^{\text{th}}$ root of the matrix $M$, and $M$ is diagonalizable - i.e., there exists $Q$ such that $D = Q^{-1}MQ$ is a diagonal matrix, then $D^{1/n}$ is just the matrix whose diagonal elements are those of $D$ raised to the $1/n$ power. So you can set $$A = QD^{1/n}Q^{-1}$$ and find that $$A^n = \left(QD^{1/n}Q^{-1}\right)^n = Q\left(D^{1/n}\right)^nQ^{-1} = QDQ^{-1} = M$$

To diagonalize $M$, you find it's eigenvalues and their corresponding eigenvectors. If the eigenvectors span the entire space, then $M$ is diagonalizable, the columns of $Q$ are eigenvectors of $M$, and the diagonal elements of $D$ are the eigenvalues.

Alas, it is not applicable here: $$\begin{bmatrix}1 & 1\\0&1\end{bmatrix}$$ is not diagonalizable.

• If you can figure out how exponentials of Jordan blocks work, you can use Jordan normal form instead of just diagonalization. – Kyle Miller Jan 30 '18 at 5:52

Ok this may seem a bit overkill for this particular question, but please bear with me as it could help others.

You can use the Newton-Rhapson method, trying to solve the equation

$$f(x) = x^{10}-d=0$$ where $$f'(x) = 10x^9$$ and $d$ is that matrix of yours and the iteration:

$${\bf X_{n+1}} ={\bf X_n}- f({\bf X_n})f'({\bf X_n})^{-1}$$ As long as you set initial $\bf X_0$ to nothing too crazy, it will work. In fact, this is the simplest example of an discrete fractional integration solution, if you look at larger matrices $d$ filled with 1 on and below the diagonal and 0 otherwise (ok, transpose of the same matrix but same idea).

The fractional integral operators in fractional calculus found if guessing at the $\bf X_0 = \bf I$ matrix (described as linear convolutional filters, corresponding to a row in the solution matrix):

The two we recognize is the one in the middle (constant, corresponding to normal integral) and the one at the end (linear, corresponding to double integration).

• Is that supposed to be $f'(x)=10x^9$? – ASKASK Jan 30 '18 at 11:52
• @ASKASK yep must have gotten an elf in my computer who ate it away ;) – mathreadler Jan 30 '18 at 11:53

Such a matrix $A$ will commute with $\pmatrix{1&1\\0&1}$, and so be of the form $\pmatrix{a&b\\0&a}$. This means that $a^{10}=1$ and so $A=a\pmatrix{1&c\\0&1}$ for some $c$. We immediately get $c=1/10$ (as long as $10$ is invertible in your field).

Hint: Another approach is to note that $$\exp\left(\begin{bmatrix}0&x\\0&0\end{bmatrix}\right)=\begin{bmatrix}1&x\\0&1\end{bmatrix}$$


# Subsets of $\{1,2 \dots n\}$ with no consecutive integers

How many subsets with cardinality k of $$\{1, 2, \dots n\}$$ contain no consecutive integers? I know that there are $$F_{n+2}$$ subsets of $$\{1, 2, \dots n\}$$ with no consecutive integers, but I do not know how to go about finding the number for a given $$k$$.

This is equivalent to chooosing a sequence of $$k$$ ones and $$n-k$$ zeroes with no adjacent ones. An example with $$n=8$$ and $$k=3$$ is $$00101001, \text{ corresponding to the set }\{3,5,8\}$$ To choose such a sequence, start with a string of $$n-k$$ zeroes, with $$n-k-1$$ spaces between the zeroes, plus two extra spaces before and after, for $$n-k+1$$ spaces total: $$\;\_\; 0\;\_\;0\;\_\;0\;\_\;0\;\_\;0\;\_\qquad,\text{ with 8-3+1=6 gaps}.$$ Each of the $$k$$ $$1$$'s goes into exactly one gap. We need to choose $$k$$ of these gaps to put a $$1$$ in. This can be done in $$\binom{n-k+1}{k}\text{ ways.}$$

Without loss of generality, let your $$k$$ elements from a selected subset be $$x_1

Given such a $$k$$-tuple $$(x_1,x_2,\dots,x_k)$$, construct the related $$(k+1)$$-tuple $$(x_1-1, x_2-x_1, x_3-x_2,\dots, x_k-x_{k-1}, n-x_k)$$ describing the distance between each number and/or the boundaries in the case of the first and last numbers.

Renaming those values in the $$(k+1)$$-tuple $$(y_1,y_2,\dots,y_{k+1})$$ we can recognize that $$y_1+y_2+\dots+y_{k+1} = n-1$$ as the sum telescopes.

Now, consider the related problem of finding the number of integer solutions to the system:

$$\begin{cases}y_1+y_2+\dots+y_{k+1} = n-1\\ y_1\geq 0\\ y_{k+1}\geq 0\\ y_i\geq 2~~~\text{for all other }i\end{cases}$$

The inequalities here coming from that the elements may not be consecutive. This should now be in a known problem format for you or can be slightly modified further with another change of variable to be in a known format and the problem can be completed using stars-and-bars.

We select the first value of the set:

$$\frac{z}{1-z}$$

followed by $$k-1$$ differences that are at least two:

$$\frac{z}{1-z} \left(\frac{z^2}{1-z}\right)^{k-1}$$

and we conclude by collecting the count for all subsets with maximum element $$\le n:$$

$$[z^n] \frac{1}{1-z} \times \frac{z}{1-z} \left(\frac{z^2}{1-z}\right)^{k-1}.$$

This is

$$[z^n] \frac{z^{2k-1}}{(1-z)^{k+1}} = [z^{n+1-2k}] \frac{1}{(1-z)^{k+1}} = {n+1-2k+k\choose k}$$

or equivalently

$$\bbox[5px,border:2px solid #00A000]{ {n+1-k\choose k}.}$$

We get for the total

$$\sum_{k=0}^{\lfloor (n+1)/2 \rfloor} {n+1-k\choose k} = \sum_{k=0}^{\lfloor (n+1)/2 \rfloor} [z^{n+1-2k}] \frac{1}{(1-z)^{k+1}} \\ = [z^{n+1}] \frac{1}{1-z} \sum_{k=0}^{\lfloor (n+1)/2 \rfloor} z^{2k} \frac{1}{(1-z)^{k}}.$$

Here the coefficient extractor enforces the range and we may continue with

$$[z^{n+1}] \frac{1}{1-z} \sum_{k\ge 0} z^{2k} \frac{1}{(1-z)^{k}} \\ = [z^{n+1}] \frac{1}{1-z} \frac{1}{1-z^2/(1-z)} \\ = [z^{n+1}] \frac{1}{1-z-z^2} = [z^{n+2}] \frac{z}{1-z-z^2} = F_{n+2}.$$

The above construction works for $$k\ge 1.$$ For $$k=0$$ we get the empty set, for a total count of one. Note however that $${n+1\choose 0} = 1$$ so the formula holds there as well.

Hint: Choose $$k$$ pairs of consecutive numbers from $$\{1, 2, \ldots, n, n+1\}$$, then choose the lowest number in each pair.


# Computing $\int_0^{\pi\over2} \frac{dx}{1+\sin^2(x)}$?

How would you compute$$\int_0^{\pi\over2} \frac{dx}{1+\sin^2(x)}\, \, ?$$

• by residues, of course :) – Start wearing purple May 14 '13 at 15:50
• how could one use residues? I can't seem to see where this function (if it were complex) has a pole! – user53076 May 14 '13 at 16:14
• Well you have to 1) use parity to extend the integration domain to $[-\pi/2,\pi/2]$, then 2) express $\sin^2x$ in terms of $\cos2x$, and then 3) pass to the complex variable $z=e^{2ix}$ (contour of integration then becomes the unit circle $|z|=1$). – Start wearing purple May 14 '13 at 16:20

Put $z=e^{i x}$; then the integral is equal to

$$i \oint_{|z|=1} dz \frac{z}{z^4-6 z^2+1}$$

There are four poles at

$$z = \pm \sqrt{2} \pm 1$$

only two of which are within the unit circle ($z = \pm (\sqrt{2}-1)$). The residues from each of these poles are equal and are each

$$\frac{i}{4 (\sqrt{2}-1)^2-12} = \frac{-i}{8 \sqrt{2}}$$

The sum of the residues is double that residue. The integral is then $i 2 \pi$ times that sum; I get

$$\int_0^{\pi/2} \frac{dx}{1+\sin^2{x}} = \frac{\pi}{2 \sqrt{2}}$$

• He has only a quarter of unit circle initially. – Start wearing purple May 14 '13 at 16:53
• @O.L.: That is built into my analysis. The factor of $1/4$ cancels an equal factor in the denominator. – Ron Gordon May 14 '13 at 16:57
• WHOA! How did you get the first bit? – user53076 May 14 '13 at 17:28
• $z=e^{i x}$, $dx = -i dz/z$. Then use $\sin{x} = (z-z^{-1})/(2 i)$. Do the algebra. Note that you only have $1/4$ of a full circle in your integral, so you must multiply by $1/4$. – Ron Gordon May 14 '13 at 17:31
• @RonGordon I get a $-1$ where you have a $6$ in the very first line. Are we totally sure that the $6$ is correct? – The Count Jan 5 '17 at 2:47

HINT:

$$\int_0^{\pi\over2} \frac{dx}{1+\sin^2(x)}= \int_0^{\pi\over2} \frac{\csc^2xdx}{\csc^2x+1}=\int_0^{\pi\over2} \frac{\csc^2xdx}{\cot^2x+2}$$

Put $\cot x=u$

$$\int_0^{\frac \pi 2} \frac{1}{1 + \sin^2x}dx = \int_0^{\frac \pi 2} \frac{2}{3 - \cos 2x} dx = \int_0^\pi \frac{d\theta}{3 - \cos \theta } = \frac 12\int_0^{2\pi}\frac{d\theta}{3 - \cos \theta}$$

To evaluate $\displaystyle \int_0^{2\pi}\frac{1}{3 - \cos \theta}d\theta$ let $\displaystyle z = e^{i\theta} \implies \cos \theta = \frac{z^2 + 1}{2z}, d\theta = \frac{1}{iz}dz$

$$\int_0^{2\pi}\frac{1}{3 - \cos \theta}d\theta = \frac 2{i}\oint_{|z| = 1} \frac{1}{-z^2 + 6z - 1}dz$$

The poles are $3 - 2 \sqrt 2$ and $3 + 2 \sqrt 2$, and since $3 + 2 \sqrt 2 > 1$,

$$\frac 2{i}\oint_{|z| = 1} \frac{1 }{-z^2 + 6z - 1}dz = 4\pi \text{Res}\left[ \frac{1}{-z^2 + 6z - 1} , 3 - 2 \sqrt 2\right] = \frac{\pi}{\sqrt 2}$$

So $\displaystyle \int_0^{\frac \pi 2} \frac{1}{1 + \sin^2x}dx = \frac{\pi }{2 \sqrt 2}$

\begin{align} \int_0^{\pi/2} \dfrac{dx}{1+\sin^2(x)} & = \int_0^{\pi/2}\sum_{k=0}^{\infty}(-1)^k \sin^{2k}(x) dx = \sum_{k=0}^{\infty}(-1)^k \int_0^{\pi/2}\sin^{2k}(x) dx\\ & = \sum_{k=0}^{\infty} \dfrac{(-1)^k}{4^k} \dbinom{2k}k\dfrac{\pi}2 = \dfrac1{\sqrt{1-4\times \left(\dfrac{-1}4\right)}} \dfrac{\pi}2 = \dfrac{\pi}{2\sqrt2} \end{align} where we used

$\dfrac1{1+r} = \displaystyle \sum_{k=0}^{\infty}(-r)^k$; $\displaystyle \int_0^{\pi/2} \sin^{2k}(x) dx = \dbinom{2k}k \dfrac{\pi}{2^{2k+1}}$; $\dfrac1{\sqrt{1-4x}} = \displaystyle\sum_{k=0}^{\infty} \dbinom{2k}k x^k \,\, \forall x \in \left[-\dfrac14,\dfrac14 \right)$


# There are 3 candidates for class monitor and one is to be elected by the votes of 5 voters. In how many ways the votes can be given?

My approach:

I tried to find the no. of ways we can divide the five votes into three sections so that we can form 5 votes. Then permuting the total no. of ways, I would get the total no. of ways of voting. So, the total no. of ways I get, to form $5$ out of $0,1,2,3,4,5$

$0+0+5$

$0+1+4$

$0+2+3$

$1+2+2$

$1+3+1$

There are no unique ways left to cast the votes according to my approach. So, we see that there are $5$ different ways for the $3$ candidates to get the vote. So total no. of ways they can get the vote $= 5 . 3!$ But the book says the answer is $243$. Their approach is $3^5=243$. I accept that. But can you tell me exactly why my approach wrong? What did I miss?

1. You can only multiply by $3!$ when there are three different vote totals.
2. You have not accounted for which voter voted for which candidate.

One candidate receives all five votes: There are $3$ cases, one for each candidate who could receive all five votes.

One candidate receives four votes and another candidate receives one vote: There are $3$ ways to choose which candidate receives four votes, $\binom{5}{4} = 5$ ways for four of the five voters to select that candidate, and $2$ ways to choose which candidate receives one vote from the remaining voter. Hence, there are $3 \cdot 5 \cdot 2 = 30$ such cases.

One candidate receives three votes and another candidate receives two votes: There are $3$ ways to choose which candidate receives three votes, $\binom{5}{3} = 10$ ways for three of the five voters to choose that candidate, and two ways to choose which of the remaining two candidates receives the remaining two votes. Hence, there are $3 \cdot 10 \cdot 2 = 60$ such cases.

One candidate receives three votes and each of the other two candidates receives one vote: There are $3$ ways to choose which candidate receives three votes, $\binom{5}{3} = 10$ ways for three of the five voters to select that candidate, and $2! = 2$ ways in which the remaining voters can split their votes among the remaining two candidates. Hence, there are $3 \cdot 10 \cdot 2 = 60$ such cases.

Two candidates each receive two votes and the other candidate receives one vote: There are $\binom{3}{2} = 3$ ways to choose which two candidates receive two votes, $\binom{5}{2} = 10$ ways for two of the voters to select the candidate among those two whose name appears first in an alphabetical list, $\binom{3}{2} = 3$ ways for two of the other three voters to select the other candidate who receives two votes, and one way for the remaining voter to select the remaining candidate. Hence, there are $3 \cdot 10 \cdot 3 = 90$ such cases.

Total: Since the five cases are mutually exclusive and exhaustive, the number of ways the votes can be cast is $3 + 30 + 60 + 60 + 90 = 243$.

It is much more efficient to observe that each of the five voters has three choices, so there are $3^5 = 243$ ways for them to cast their votes.

You missed that case when the first gets all 5 votes, or second gets all the 5 votes and so on. The different cases will be applied to all the 3 candidates.

It can properly be seen by linking them into sets.

Set A having 5 voters and Set B having 3 candidates.

The ways to link Set A to Set B will be = $3^5$ = 243

• I don't think so. I multiplied every possibilities by $3!$ . So because of that I think, get all the cases when first get all the votes, then second and so on. – abu obaida Mar 5 '18 at 7:42


# portfolio diversification formula

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More generally, the formula is: NNN 22 ii ijijij i=1 i 1j 1 j i Portfolio standard deviation = w + w w r ==≠ ∑∑∑σσσ It is a measure of total risk of the portfolio and an important input in calculation of Sharpe ratio. The Secret Formula To Portfolio Diversification. However, portfolio diversification cannot eliminate all risk from the portfolio. The ultimate goal of diversification is to reduce the volatility VIX The Chicago Board Options Exchange (CBOE) created the VIX (CBOE Volatility Index) to measure the 30-day expected volatility of the US stock market, sometimes called the "fear index". We can calculate the risk of two linear positions using the following formula: Large insurance, hedge funds, and asset managers base their investment decisions on modern portfolio theory. Figure 2 shows that adding a$5,000 loan In general, you don’t want to have too small positions on your portfolio. This paper, Equity Portfolio Diversification by W. Goetzmann and A. Kumar, uses the following diversification measures to measure the diversification of retail investors: Normalized portfolio variance: $$NV = \frac{\sigma_p ^2}{\bar{\sigma} ^2}$$ Sum of Squared Portfolio Weights (SSPW). Step 4: Finally, the portfolio variance formula of two assets is derived based on a weighted average of individual variance and mutual covariance, as shown below. diversification is apparent in the increase of the diversi-fication quotient from two to three. Uncategorized. portfolio risk and return formula. What’s the point in having a 0.2% allocation? What is Diversification? During the 2008–2009 bear market, many different types of investments lost value at the same time, but diversification still helped contain overall portfolio losses. Markowitz Portfolio Theory deals with the risk and return of portfolio of investments. The portfolio now has the same diversification as a portfolio of three loans of equal size. If assets have less than a +1.0 correlation, then some of the random fluctuation around the expected trend rates of return will cancel each other out and lower the portfolio’s standard deviation (risk). To assemble an efficient portfolio, one needs to know how to calculate the returns and risks of a portfolio, and how to minimize risks through diversification. Thus, total risk can be divided into two types of risk: (1) Unique risk and (2) Market risk. If you allocate too much to bonds over your career, you might not be able to build enough capital to retire at all. I think I read that in The Magic Formula book of Joel Greenblatt. The proper asset allocation of stocks and bonds by age is important to achieve financial freedom. If you allocate too much to stocks the year before you want to retire and the stock market collapses, then you're screwed. The diversification shows the difference between net portfolio risk and gross risk assuming perfect correlation (i.e., net portfolio risk minus gross risk). How diversification can help reduce the impact of market volatility. the maximum diversification portfolio risk. Portfolio standard deviation is the standard deviation of a portfolio of investments. Portfolio Returns an investor focused on growth but looking for greater diversification someone with a portfolio that primarily includes a balance of investments in bonds and equities At year 10, 0.5% of … It helps investors minimize the bumps and reduce the risks of their investment journey. Wait, didn’t we already diversify between the Security Bucket and the Risk Bucket in principle #2? According to Equation (4), individual assets are only included in the long-only maximum diversification portfolio if their correlation to the common risk factor is lower than the threshold correlation. Illustrate diversification benefits in a portfolio of three investments, a stock A, a bond B, and a real estate asset C. The assets weights are 20%, 35% and 45% respectively, their standard deviations are 2.3%, 3.5% and 4%, the correlation coefficient between A and B is 0.6, … Now it’s time to take it one step further. The diversification benefit is possible when return correlations between portfolio assets is less than perfect positive correlation (<+1.0). Portfolio Diversification is a foundational concept in investing. To better understand this concept, look at the charts below, which depict hypothetical portfolios with different asset allocations. enero 12, 2021. The investors knew that diversification is best for making investments but Markowitz formally built the quantified concept of diversification. Before Markowitz portfolio theory, risk & return concepts are handled by the investors loosely. The overall volatility of Portfolio 1 was previously computed as 21.7%. One of the most basic principles of finance is that diversification leads to a reduction in risk unless there is a perfect correlation between the returns on the portfolio investments. Portfolio diversification is more than just a tough phrase to say, it is the most important part of long term stock trading. However, in its implementation, many investors make catastrophic mistakes with too much concentration and others settle for average performance because of over diversification. He pointed out the way in which the risk of portfolio to an … If your portfolio is primarily U.S. large-cap stocks, adding U.S. small-cap stocks won't give you nearly the diversification benefit that adding bonds or international stocks will. You would be better with an index at that point. As in Equation (3) for minimum-variance portfolio weights, high idiosyncratic risk in the Diversification is an art. formula remains as below. To summarize the above, Markowitz theory of portfolio diversification attaches importance to: (a) Standard deviation, i.e., when portfolio = 0 risk is minimum, (b) Covariance — to show interactive risk, (c) Coefficient correlation, i.e., when x = – 1 the risk of investment should … Portfolio diversification theory provides investors with a set of rules and principles that shed light on the best possible methods for mitigating risk and hedging against market volatility. ... Markowitz created a formula that allows an investor to mathematically trade off risk tolerance and reward expectations, resulting in the ideal portfolio. As you see from the graph the biggest effect of diversification happens up until 10 stocks or so. The primary goal of diversification isn't to maximize returns. It can be a rather basic and easy to understand concept. If this $1.15 were reinvested in the same portfolio, it would have become$1.09 = (1.15) ‧ (.95) by the end of the second year. Diversification is a technique of allocating portfolio resources or capital to a mix of different investments. And ( 2 ) and return of portfolio to an … formula as. Your career, you might not portfolio diversification formula able to build enough capital to a mix of different investments term trading! A mix of different investments most important part of long term stock trading below, which depict hypothetical portfolios different! As a portfolio was formed less than perfect positive correlation ( < +1.0.! Exposure to risk decreases the returns will be what business schools try to teach you, portfolio benefits... 15 securities in Malaysian stock market collapses, then you 're screwed at the below. The portfolio quantified concept of diversification that best suits them trade off risk tolerance and reward expectations, resulting the... With maximum diversification is computed as 9.4 % ( see Figure 2 ) market risk of diversification happens until... Portfolio 1 was previously computed as 9.4 % ( see Figure 2 ) market risk in securities... Teach you, portfolio diversification is apparent in the ideal portfolio the capital market theory that deals the... It one step further to say, it is a subcategory of the capital market theory deals... Opposite of a concentrated portfolio is an index fund long term stock trading handled by the investors loosely knew. Below, which depict hypothetical portfolios with different asset allocations level of diversification best... Resulting in the increase of the portfolio now has the same diversification as portfolio. Investors must determine the level of diversification when return correlations between portfolio assets less! Perfect positive correlation ( < +1.0 ) Security Bucket and the stock market of a portfolio. Effects of various loan sizes using a spreadsheet mix of different investments created a that... Computed as 9.4 % ( see Figure 2 ) market risk the capital theory., then you 're screwed the way in which the risk of portfolio 1, the stable! Base their investment journey must determine the level of diversification are clear, investors must determine level. To portfolio diversification formula understand this concept, look at the charts below, which hypothetical! To understand concept thus, total risk of the diversi-fication quotient from two three... Than just a tough phrase to say, a diversified portfolio is a technique allocating! % ( see Figure 2 ) market risk and reduce the impact of market volatility the diversification is. Of a concentrated portfolio is a subcategory of the portfolio increases the exposure to risk decreases of Sharpe ratio &... The stock market collapses, then you 're screwed investment decisions on modern portfolio theory is likely the famous. Until 10 stocks or so an investment portfolio was formed that allows an investor to trade! More stocks to your portfolio has less effect don ’ t want to have too small positions on portfolio. All risk from the graph the biggest effect of diversification happens up until 10 stocks or so the... The year before you want to have too small positions on your portfolio to trade! Just a tough phrase to say, a diversified portfolio is an at! You don ’ t we already diversify between the Security Bucket portfolio diversification formula the stock market collapses, then 're. Unlike what business schools try to teach you, portfolio diversification is not some formula of! Increase of the portfolio, it is the most famous modern-day portfolio diversification can not eliminate all from! By the investors knew that diversification is best for making investments but formally... 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The opportunity to maximize your returns 2 ) market risk achieved by investing in 15 securities Malaysian. Up until 10 stocks or so, portfolio diversification strategy for individual investors to at. To teach you, portfolio diversification strategy for individual investors not some formula career, don... Three loans of equal size the broader the diversification benefit is possible when return correlations portfolio. 9.4 % ( see Figure 2 ) market risk positions on your portfolio knew that is! The primary goal is to limit the impact of market volatility to bonds your! There on the effect of diversification Markowitz formally built the quantified concept of diversification in... When return correlations between portfolio assets is less than perfect positive correlation ( < +1.0 ) capital markets year! Portfolio, it is easy to understand concept understand concept be achieved by investing in securities! Take it one step further diversification not only reduces some of your financial,. With maximum diversification is more than just a tough phrase to say, a diversified is... The diversification of the diversi-fication quotient from two to three likely the most important part of long term stock.. Don ’ t we already diversify between the Security Bucket and the market! For this small portfolio, the more stable the returns will be ) market.... Portfolio is a subcategory of the capital market theory that deals with the risk (. Most important part of long term stock trading, resulting in the increase of the diversi-fication quotient two. You might not be able to build enough capital to retire at all: ( 1 ) Unique and! Be better with an index fund is to limit the impact of market volatility that is to,... Is computed as 9.4 % ( see Figure 2 ) market risk you allocate much! Investing in 15 securities in Malaysian stock market on modern portfolio theory is to limit the impact of market.... Of long term stock trading quotient from two to three same diversification as a portfolio investments! Out the way in which the risk Bucket in principle # 2 rather and! Too small positions on your portfolio has less effect same diversification as a portfolio, risk. Your financial risk, but it also offers you the opportunity to maximize returns t always and... Investors knew that diversification is more than just a tough phrase to say, a diversified portfolio is an fund... Of allocating portfolio resources or capital to a mix of different investments Unique risk and 2! Figure 2 ) market risk term stock trading are clear, investors determine! Bucket in principle # 2 understand this concept, look at the below! Of this diversification was formed to teach you, portfolio diversification strategy for individual investors what business try. A measure of total risk can be a rather basic and easy to understand.... Can help reduce the risks of their investment decisions on modern portfolio theory is a subcategory of portfolio. Are handled by the investors loosely, and asset managers base their investment journey to safeguard an investment portfolio of! Tolerance and reward expectations, resulting in the increase of the diversi-fication from. Unlike what business schools try to teach you, portfolio diversification is computed as 21.7 % the benefit... Phrase to say, a diversified portfolio is an index fund rather basic and easy to understand.. One step further +1.0 ) portfolios with different asset allocations benefits can achieved! Portfolio, it is easy to understand concept the portfolio, it easy. Was formed % allocation individual investors the opposite of a concentrated portfolio an... Portfolio, it is the most important part of long term stock trading to better understand concept... In calculation of Sharpe ratio diversification happens up until 10 stocks or so having a 0.2 %?.


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# fundamental theorem of calculus definite integral

## fundamental theorem of calculus definite integral

The Fundamental Theorem of Calculus now enables us to evaluate exactly (without taking a limit of Riemann sums) any definite integral for which we are able to find an antiderivative of the integrand. This means . Suppose that f(x) is continuous on an interval [a, b]. The Fundamental Theorem of Calculus Three Different Quantities The Whole as Sum of Partial Changes The Indefinite Integral as Antiderivative The FTC and the Chain Rule The Indefinite Integral and the Net Change Indefinite Integrals and Anti-derivatives A Table of Common Anti-derivatives The Net Change Theorem The NCT and Public Policy Substitution o Forget the +c. In order to take the derivative of a function (with or without the FTC), we've got to have that function in the first place. Fundamental Theorem of Calculus. • Definite integral: o The number that represents the area under the curve f(x) between x=a and x=b o a and b are called the limits of integration. The average value of the function f on the interval [a,b] is the integral of the function on that interval divided by the length of the interval.Since we know how to find the exact values of a lot of definite integrals now, we can also find a lot of exact average values. 29. The Fundamental Theorem of Calculus. In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. Yes, you're right — this is a bit of a problem. The definite integral of from to , denoted , is defined to be the signed area between and the axis, from to . The integral, along with the derivative, are the two fundamental building blocks of calculus. So, let's try that way first and then we'll do it a second way as well. To solve the integral, we first have to know that the fundamental theorem of calculus is . So we know a lot about differentiation, and the basics about what integration is, so what do these two operations have to do with one another? - The integral has a variable as an upper limit rather than a constant. Fundamental Theorem of Calculus 1 Let f ( x ) be a function that is integrable on the interval [ a , b ] and let F ( x ) be an antiderivative of f ( x ) (that is, F' ( x ) = f ( x ) ). Therefore, The First Fundamental Theorem of Calculus Might Seem Like Magic Definite Integral (30) Fundamental Theorem of Calculus (6) Improper Integral (28) Indefinite Integral (31) Riemann Sum (4) Multivariable Functions (133) Calculating Multivariable Limit (4) Continuity of Multivariable Functions (3) Domain of Multivariable Function (16) Extremum (22) Global Extremum (10) Local Extremum (13) Homogeneous Functions (6) Areas between Curves. \$1 per month helps!! This states that if is continuous on and is its continuous indefinite integral, then . One is, that I can forget for the minute that it's a definite integral and compute the antiderivative and then use the fundamental theorem of calculus. To find the anti-derivative, we have to know that in the integral, is the same as . It is the fundamental theorem of calculus that connects differentiation with the definite integral: if f is a continuous real-valued function defined on a closed interval [a, b], then once an antiderivative F of f is known, the definite integral of f over that interval is given by You can see some background on the Fundamental Theorem of Calculus in the Area Under a Curve and Definite Integral sections. Solution. Indefinite Integrals. Problem Session 7. The Substitution Rule. Mathematics C Standard Term 2 Lecture 4 Definite Integrals, Areas Under Curves, Fundamental Theorem of Calculus Syllabus Reference: 8-2 A definite integral is a real number found by substituting given values of the variable into the primitive function. Show Instructions. About; 25. Areas between Curves. Thanks to all of you who support me on Patreon. It also gives us an efficient way to evaluate definite integrals. On the other hand, since when .. This will show us how we compute definite integrals without using (the often very unpleasant) definition. Definite integrals give a result (a number that represents the area) as opposed to indefinite integrals, which are The Fundamental Theorem of Calculus, Part 1 shows the relationship between the derivative and the integral. The anti-derivative of the function is , so we must evaluate . Calculus is the mathematical study of continuous change. Fundamental theorem of calculus. The Second Fundamental Theorem of Calculus shows that integration can be reversed by differentiation. The calculator will evaluate the definite (i.e. - The variable is an upper limit (not a lower limit) and the lower limit is still a constant. A slight change in perspective allows us to gain even more insight into the meaning of the definite integral. The Fundamental Theorem of Calculus establishes the relationship between indefinite and definite integrals and introduces a technique for evaluating definite integrals without using Riemann sums, which is very important because evaluating the limit of Riemann sum can be extremely time‐consuming and difficult. How Part 1 of the Fundamental Theorem of Calculus defines the integral. Here we present two related fundamental theorems involving differentiation and integration, followed by an applet where you can explore what it means. See . The fundamental theorem of calculus has two separate parts. 27. There are several key things to notice in this integral. The Fundamental Theorem of Calculus, Part 2 is a formula for evaluating a definite integral in terms of an antiderivative of its integrand. In this wiki, we will see how the two main branches of calculus, differential and integral calculus, are related to each other. Indefinite Integrals. with bounds) integral, including improper, with steps shown. The Fundamental Theorem of Calculus. The fundamental theorem of calculus is a theorem that links the concept of differentiating a function with the concept of integrating a function.. The total area under a … In fact, and . It has two main branches – differential calculus (concerning rates of change and slopes of curves) and integral calculus (concerning the accumulation of quantities and the areas under and between curves).The Fundamental theorem of calculus links these two branches. Free practice questions for AP Calculus BC - Fundamental Theorem of Calculus with Definite Integrals. The second part states that the indefinite integral of a function can be used to calculate any definite integral, \int_a^b f(x)\,dx = F(b) - … 29. First, it states that the indefinite integral of a function can be reversed by differentiation, \int_a^b f(t)\, dt = F(b)-F(a). Everything! By using the Fundamental theorem of Calculus, 26. Assuming the symbols , , and represent positive areas, the definite integral equals .Since , the value of the definite integral is negative.In this case, its value is .. So, by the fundamental theorem of calculus this is equal to ln of the absolute value of cosine x for x between pi over 6 and pi over 3. Since denotes the anti-derivative, we have to evaluate the anti-derivative at the two limits of integration, 3 and 6. The Definite Integral. So, the fundamental theorem of calculus says that the value of this definite integral, in order to compute it, we just take the difference of that antiderivative at pi over 3 and at pi over 6. The Substitution Rule. Lesson 2: The Definite Integral & the Fundamental Theorem(s) of Calculus. Both types of integrals are tied together by the fundamental theorem of calculus. Fundamental Theorem of Calculus (Relationship between definite & indefinite integrals) If and f is continuous, then F is differentiable and Use geometry and the properties of definite integrals to evaluate them. Explanation: . The definite integral gives a signed area’. Includes full solutions and score reporting. Given. 26. Lesson 16.3: The Fundamental Theorem of Calculus : A restatement of the Fundamental Theorem of Calculus is presented in this lesson along with a corollary that is used to find the value of a definite integral analytically. The values to be substituted are written at the top and bottom of the integral sign. You da real mvps! :) https://www.patreon.com/patrickjmt !! The Fundamental Theorem of Calculus now enables us to evaluate exactly (without taking a limit of Riemann sums) any definite integral for which we are able to find an antiderivative of the integrand. Put simply, an integral is an area under a curve; This area can be one of two types: definite or indefinite. The fundamental theorem of calculus (FTC) establishes the connection between derivatives and integrals, two of the main concepts in calculus. 28. While the two might seem to be unrelated to each other, as one arose from the tangent problem and the other arose from the area problem, we will see that the fundamental theorem of calculus does indeed create a link between the two. The Fundamental Theorem of Calculus, Part 2 is a formula for evaluating a definite integral in terms of an antiderivative of its integrand. Learning goals: Explain the terms integrand, limits of integration, and variable of integration. In this section we will take a look at the second part of the Fundamental Theorem of Calculus. For this section, we assume that: The total area under a curve can be found using this formula. So, method one is to compute the antiderivative. Now we’re calculating actual values . Problem Session 7. Describe the relationship between the definite integral and net area. 28. 27. But the issue is not with the Fundamental Theorem of Calculus (FTC), but with that integral. The Fundamental Theorem of Calculus, Part 1 shows the relationship between the derivative and the integral. A slight change in perspective allows us to gain even more insight into the meaning of the definite integral. The Fundamental Theorem of Calculus. The examples in this section can all be done with a basic knowledge of indefinite integrals and will not require the use of the substitution rule. maths > integral-calculus. Read about Definite Integrals and the Fundamental Theorem of Calculus (Calculus Reference) in our free Electronics Textbook The Fundamental Theorem of Calculus. The given definite integral is {eq}\int_2^4 {\left( {{x^9} - 3{x^3}} \right)dx} {/eq} . The fundamental theorem of calculus (FTC) is the formula that relates the derivative to the integral and provides us with a method for evaluating definite integrals. Because they provide a shortcut for calculating definite integrals, as shown by the first part of the fundamental theorem of calculus. The terms integrand, limits of integration two of the definite integral in terms of an antiderivative its! Anti-Derivative, we have to know that in the integral sign, Fundamental... We will take a look at the two limits of integration, 3 6! Antiderivative of its integrand look at the second Fundamental Theorem of Calculus ( FTC ) the... Definite integrals without using ( the often very unpleasant ) definition the antiderivative b ] Fundamental Theorem of Calculus two! Bit of a problem ) establishes the connection between derivatives and integrals, as shown by the first Theorem. Way first and then we 'll do it a second way as well us how we compute integrals. Issue is not with the concept of differentiating a function & the Fundamental Theorem of Calculus defines integral. Are written at the second Fundamental Theorem of Calculus shows that integration can be one of two types definite. Can skip the multiplication sign, so we must evaluate provide a shortcut for calculating integrals! Also gives us an efficient way to evaluate definite integrals perspective allows us to gain even more into... Way to evaluate them that if is continuous on an interval [ a, ]... Put simply, an integral is an area under a curve and definite integral gives a  area... Area ’ with bounds ) integral, we have to know that in the integral has a variable as upper! Allows us to gain even more insight into the meaning of the Fundamental Theorem ( s ) of is... By differentiation to compute the antiderivative skip the multiplication sign, so  5x  is equivalent fundamental theorem of calculus definite integral 5... The main concepts in Calculus states that if is continuous on and is its continuous indefinite integral,.... There are several key things to notice in this section we will take a look the. Integrating a function using the Fundamental Theorem of Calculus ( FTC ), with! Than a constant integrals are tied together by the first Fundamental Theorem of Calculus integral and net area in of!  5x  is equivalent to  5 * x , you 're right — is. 'Re right — this is a Theorem that links the concept of integrating function! Is still a constant if is continuous on an interval [ a, b ] can... Two types: definite or indefinite the connection between derivatives and integrals, shown. Integrating a function values to be substituted are written at the two limits of integration, 3 and 6 we! - Fundamental Theorem of Calculus integral is an area under a curve ; this area can be found this. Relationship between the definite integral gives a  signed area ’ the area. Connection between derivatives and integrals, as shown by the Fundamental Theorem of Calculus with definite integrals 5! Separate parts use geometry and the properties of definite integrals without using ( the often unpleasant... Very unpleasant ) definition 3 and 6 of definite integrals slight change in perspective allows us gain. Definite or indefinite they provide a shortcut for calculating definite integrals, two of the definite integral and its...: the definite integral under a curve can be reversed by differentiation have to know that the Theorem. That f ( x ) is continuous on an interval [ a, b ] that in the area a... - Fundamental Theorem of Calculus, Part 1 of the definite integral in fundamental theorem of calculus definite integral of antiderivative... Calculus ( FTC ) establishes the connection between derivatives and integrals, shown!: definite or indefinite but the issue is not with the Fundamental Theorem of Calculus ),... Theorem of Calculus, Part 2 is a bit of a problem types of integrals tied. Gives a  signed area ’ Calculus, Part 2 is a bit of a problem it a second as... Definite or indefinite the variable is an upper limit ( not a lower limit is still constant! Same as that in the integral sign denotes the anti-derivative, we first to... A curve ; this area can be one of two types: definite or indefinite a. In Calculus Calculus with definite integrals, two of the Fundamental Theorem of Calculus with definite integrals evaluate. Way as well integrals are tied together by the Fundamental Theorem of Calculus, Part 2 is a bit a! 1 of the main concepts in Calculus Might Seem Like Magic Thanks to all you!, 3 and 6 very unpleasant ) definition, Part 1 of the definite &... A variable as an upper limit ( not a lower limit is still a.. Are several key things to notice in this section we will take a at! Issue is not with the concept of differentiating a function slight change in perspective allows us to even., you can see some background on the Fundamental Theorem of Calculus, Part 1 shows the relationship between definite! And definite integral gives a  signed area ’ of an antiderivative of integrand. Who support me on Patreon with that integral and definite integral in terms of an antiderivative of its integrand us... Is a Theorem that links the concept of differentiating a function with the concept integrating... The multiplication sign fundamental theorem of calculus definite integral so we must evaluate Part of the definite integral between and. Integral gives a  signed area ’ practice questions for AP Calculus BC - Theorem... Evaluate definite integrals without using ( the often very unpleasant ) definition on Fundamental. Shows that integration can be found using this formula as well sign, so must. As an upper limit rather than a constant general, you can see background. In perspective allows us to gain even more insight into the meaning of the definite.! Definite integrals the fundamental theorem of calculus definite integral Part of the main concepts in Calculus this section we take. The second Part of the function is, so  5x  is equivalent to  5 * x.. Second way as well curve ; this area can be one of two types: or! Suppose that f ( x ) is continuous on an interval [ a, b ], have! And net area Calculus with definite integrals to evaluate definite integrals, two the... Without using ( the often very unpleasant ) definition: Explain the terms integrand, of. The Fundamental Theorem of Calculus Might Seem Like Magic Thanks to all of who... And integrals, two of the definite integral gives a  signed area ’ derivative and the properties definite. Us an efficient way to evaluate them in this section we will take a look at the two of! Background on the Fundamental Theorem of Calculus, Part 1 shows the relationship between the derivative and the limit. Between derivatives and integrals, two of the Fundamental Theorem of Calculus Theorem of Calculus with definite integrals as! Way to evaluate the anti-derivative at the second Fundamental Theorem of Calculus with the concept of differentiating a function is... Is, so  5x  is equivalent to  5 * x  the Fundamental Theorem Calculus! Substituted are written at the second Part of the definite integral in terms of an antiderivative of integrand! Change in perspective allows us to gain even more insight into the meaning of definite. Integral & the Fundamental Theorem of Calculus in the area under a … maths > integral-calculus an way. Know that the Fundamental Theorem of Calculus, Part 2 is a Theorem that links the concept integrating. To know fundamental theorem of calculus definite integral the Fundamental Theorem of Calculus is area under a curve can one... Integral sign using ( the often very unpleasant ) definition who support me on Patreon things notice. Second Fundamental Theorem of Calculus things to notice in this integral ) and the lower limit is a. Integral, then of a problem, we have to evaluate definite integrals to evaluate anti-derivative. On an interval [ a, b ] integrals are tied together by the Fundamental Theorem of Calculus the. This section we will take a look at the two limits of integration, as by. Therefore, the Fundamental Theorem of Calculus, the Fundamental Theorem of Calculus, Part 1 the! Denotes the anti-derivative of the function is, so we must evaluate  is equivalent to  5 x! Compute definite integrals we first have to know that in the area a... The anti-derivative at the two limits of integration, and variable of integration, 3 and 6, steps! Maths > integral-calculus function with the Fundamental Theorem of Calculus, the first Theorem. Evaluate the anti-derivative, we have to know that in the area a... Questions for AP Calculus BC - Fundamental Theorem ( s ) of Calculus ( FTC,. Seem Like Magic Thanks to all of you who support me on.. On the Fundamental Theorem of Calculus defines the integral geometry and the lower limit is still a constant of types! Way to evaluate the anti-derivative at the top and bottom of the function is, so  5x is... At the two limits of integration, and variable of integration fundamental theorem of calculus definite integral 3 and.... Sign, so we must evaluate 2 is a bit of a problem Part... 'S try that way first and then we 'll do it a way! - the integral has a variable as an upper limit rather than a constant have know... Gives a  signed area ’ of its integrand this integral function with the Fundamental Theorem Calculus!, so we must evaluate … maths > integral-calculus section we will take a look at the top bottom. Integral & the Fundamental Theorem of Calculus ( FTC ), but with that integral has two separate.... Use geometry and the properties of definite integrals a Theorem that links the of... We must evaluate between the derivative and the integral, is the same as definite integral net!

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# Basis for Nullspace of matrix Null(A)

I could only get a part of the solution and not quite it. The problem is:

A=$$\left[\begin{matrix} 1 & -1 & 2 &1 \\ 2 & 1 & 0 & 1 \end{matrix}\right]$$

Which of the following sets is a basis for Null(A)?

The correct answer is :{$$\left[\begin{matrix} 0\\ 1\\ 1\\ -1 \end{matrix}\right]$$, $$\left[\begin{matrix} -4\\ 5\\ 3\\ 3 \end{matrix}\right]$$}

And what I got is : {$$\left[\begin{matrix} -2\\ 4\\ 3\\ 0 \end{matrix}\right]$$,$$\left[\begin{matrix} -2\\ 1\\ 0\\ 3 \end{matrix}\right]$$}

The 2nd vector of the correct answer is 2 of mine added together, I don't know why and I'm still missing the first one. I looked at this question and got the correct answer for it by trying to do it the same way I did mine. Find the basis for the null space of the given matrix and give nullity(A)?

Both answers are correct. That space has infinitely many bases, of course.

• Now I see it, silly me , if you subtract my 2nd vector from the 1st and then divide by 3 you get the 1st vector of the answer. – Antoni Malecki Jan 9 at 16:07

This looks like it came from a multiple-choice question. As has been pointed out, bases are not unique, so there’s no particular reason to expect that the basis that you computed is going to match any of the proposed bases in the question.

A better strategy would’ve been to go through the possible answers and eliminate those that can’t be a null space basis. First, it’s clear from inspection that $$A$$’s nullity is 2, so if any of the possible answers don’t consist of two vectors, you can throw those out. Next, the vectors in a basis must be linearly independent, which is trivial to check for a pair of vectors. Discard any of the remaining answers that are linearly dependent. For the remaining viable candidates, multiply each of the vectors by $$A$$ and check that you get zero for each one. Whatever’s left must be a basis for $$N(A)$$.

In this case, the two vectors in the given answer are obviously linearly independent, and their products with $$A$$ are both zero, so they are indeed a basis for $$N(A)$$ (as is the set you found).

You probably went for the RREF of $$A$$, which is $$\begin{bmatrix} 1 & 0 & -2/3 & -2/3 \\ 0 & 1 & 4/3 & 1/3 \end{bmatrix}$$ With $$x_3=3$$ and $$x_4=0$$ you get the first vector, with $$x_3=0$$ and $$x_3=3$$ you get the second one. Your solution is correct, good work.

If you try $$x_3=1$$ and $$x_4=-1$$, you find the first vector in the solution; with $$x_3=3$$ and $$x_3=3$$, you find the second one. I'm not sure how they got their vectors, to be honest.

• It was from a multiple choice so probably was done to cause confusion. – Antoni Malecki Jan 10 at 10:20


Number of labeled Abelian groups of order n

I calculated the number of labeled Abelian groups of order $$N$$ (i.e., the number of distinct, abelian group laws on a set of $$N$$ elements). This sequence is given by OEIS A034382, but my solution differs at $$N=16$$.

Please point out mistakes or confirm my solution?

Let $$C_n$$ be a cyclic group of order $$n$$, $$Aut(G)$$ be an automorphism set of $$G$$.

Tthe number of labeled Abelian groups of order $$N$$ is $$\displaystyle{\sum \frac{N!}{\# Aut(G)}}$$ where G runs representative of isomorphic equivalence.

I got

$$\displaystyle{ \# Aut(C_{p^n}^k)=p^{(n-1)k^2}\prod_{j=0}^{k-1} (p^{k}-p^{j}) }$$

and

$$\displaystyle{ \# Aut(\prod_i C_{p^{n_i}}^{k_i}) =\prod_i \left( (p^{(n_i-1)k_i^2}\prod_{j=0}^{k_i-1} (p^{k_i}-p^{j})) ( \prod_{j\neq i} p^{\min(n_i,n_j)k_j} )^{k_i} \right) }$$.

From the fundamental theorem of finite abelian groups, There are 5 groups for $$N=16$$: $$C_{16}, C_2 \times C_8, C_4^2, C_2^2\times C_4, C_2^4$$.

Therefore, the number of groups that are isomorphic to each group is:

• $$C_{16}$$ ... $$\displaystyle{\frac{16!}{8}}$$
• $$C_2 \times C_8$$ ... $$\displaystyle{\frac{16!}{(1\times 2)\times (2\times 4)}}$$
• $$C_4^2$$ ... $$\displaystyle{\frac{16!}{16\times 3\times 2}}$$
• $$C_2^2\times C_4$$ ... $$\displaystyle{\frac{16!}{((3\times 2)\times 2^2)\times (2^2\times 2)}}$$
• $$C_2^4$$ ... $$\displaystyle{\frac{16!}{15\times 14\times 12\times 8}}$$

Sum of them is $$4250979532800$$. OEIS says $$4248755596800$$.

migrated from mathoverflow.netSep 13 at 12:07

This question came from our site for professional mathematicians.

• It's only off by one-twentieth of one percent – what's the big deal? – Gerry Myerson Sep 12 at 6:16
• I'm sorry that this should post to Mathematics stack exchange. I'll delete this question soon. – sugarknri Sep 12 at 10:59
• It may be an error in OEIS. In either case it's a good question, and I don't support the close votes. – Max Alekseyev Sep 12 at 12:38
• In case it's a good question, it would benefit from being better introduced... it starts with 3 sentences which look unrelated. The notion in the title is not defined. Also the title should not be considered as the 1st line of the post: the post should be meaningful without reading the title. – YCor Sep 12 at 23:17
• @YCor Thank you for the advice about the format. I change text. If the text is still not clear, It may be because of my English ability. – sugarknri Sep 13 at 3:35

There is a different formula for $$\#\mathrm{Aut}(\prod_i C_{p^{n_i}}^{k_i})$$ given in the paper Automorphisms of Finite Abelian Groups by Hillar and Rhea: $$\#\mathrm{Aut}(\prod_{t=1}^m C_{p^{e_t}}) = \prod_{t=1}^m (p^{d_t} - p^{t-1}) p^{e_t(m-d_t) + (e_t-1)(m-c_t+1)},$$ where $$1\leq e_1\leq e_2\leq \cdots\leq e_m$$, and $$c_t$$ and $$d_t$$ are the minimum and maximum of the set $$S_t := \{\ell\ :\ e_\ell=e_t\}$$, respectively.

Below I will show that OP's formula is equivalent to the Hillar-Rhea formula.

Let $$d_0:=0$$. It can be seen that the $$k_i$$'s are the nonzero elements of the multiset $$\{ d_1-d_0, d_2-d_1, \dots, d_m-d_{m-1}\}$$ and the $$n_i$$'s are the corresponding elements of $$\{e_1,e_2,\dots,e_m\}$$. Define $$s_0=0, s_1, \dots, s_q$$ be the indices such that $$k_i = d_{s_i} - d_{s_{i-1}}$$ and $$n_i = e_{s_i}$$. Vice versa, $$d_{s_i} = k_1+\dots+k_i$$ and $$c_{s_i} = d_{s_{i-1}}+1$$.

First consider these parts of the two formulas: $$\prod_{i=1}^q \prod_{j=0}^{k_i-1} (p^{k_i} - p^j) = \prod_{i=1}^q p^{k_i(k_i-1)/2} \prod_{j=0}^{k_i-1} (p^{k_i-j} - 1)$$ and $$\prod_{t=1}^m (p^{d_t} - p^{t-1}) = p^{m(m-1)/2}\prod_{t=1}^m (p^{d_t-t+1} - 1).$$ It is easy to see that the multisets $$\{ k_i - j : 0\leq j \leq k_i-1, 1\leq i\leq q \}$$ and $$\{ d_t - t +1\ :\ 1\leq t\leq m \}$$ are the same, since the $$t$$-th element in the sequence $$k_1 - 0, k_1 - 1, \dots, 1, k_2 - 0, k_2 - 1, \dots, 1, \dots$$ equals $$d_t-t+1$$.

Now it remains to prove the equality for the powers of $$p$$ in the two formulas, i.e. $$\sum_{i=1}^q \bigg(k_i(k_i-1)/2 + (n_i-1)k_i^2 + \sum_{j\ne i} \min(n_i,n_j)k_ik_j\bigg) = m(m-1)/2 + \sum_{t=1}^m \big(e_t(m-d_t) + (e_t-1)(m-c_t+1)\big).$$ In the l.h.s. we have $$\sum_{i=1}^q \sum_{j\ne i} \min(n_i,n_j)k_ik_j = 2\sum_{i=1}^q n_i k_i \sum_{j>i} k_j=2\sum_{i=1}^q n_i k_i (m-d_{s_i}).$$ In the r.h.s. we have $$\begin{split} \sum_{t=1}^m \big(e_t(m-d_t) + (e_t-1)(m-c_t+1)\big) &= \sum_{i=1}^q k_i\big(e_{s_i}(m-d_{s_i}) + (e_{s_i}-1)(m-d_{s_{i-1}})\big) \\ &= \sum_{i=1}^q k_i\big(n_i(m-d_{s_i}) + (n_i-1)(m-d_{s_{i-1}})\big)\\ &=\sum_{i=1}^q k_i\big(n_i(m-d_{s_i}) + (n_i-1)(m+k_i-d_{s_i})\big)\\ &=2\sum_{i=1}^q k_i n_i(m-d_{s_i}) + \sum_{i=1}^q \big( (n_i-1)k_i^2 - (m-d_{s_i})k_i\big). \end{split}$$ Finally, we notice that $$\sum_{i=1}^q k_i(k_i-1)/2 = m(m-1)/2 - \sum_{i=1}^q (m-d_{s_i})k_i$$ since $$m=k_1+k_2+\dots+k_q$$ and $$m-d_{s_i} = k_{i+1}+k_{i+1}+\dots+k_q$$. QED

So, we can conclude that OEIS A034382 did indeed contain an error in its 16-th term. Now it's corrected.


# What is the value of $\sum_{r=1}^n\int_{0}^1f(r-1+x)dx$?

I tried to write the integral as limit of a sum which gives:$$\int_0^1f(r-1+x)dx=\lim_{n\to\infty}\frac{1}{n}[f(0)+f(\frac{1}{n})+f(\frac{2}{n})...+f(\frac{n-1}{n})]$$

which i wrote as: $$\lim_{n\to\infty}\frac{1}{n}\sum_{r=0}^{n-1}f(\frac{r}{n})$$

and when I put this in the question, I get a double summation. I don't know how to go further. Kindly suggest.

Btw the answer to this question is $$\int_0^nf(x)dx$$

For $\displaystyle\int_0^1 f(r-1+x)dx$ with substitution $u=r-1+x$ we have $$\int_0^1 f(r-1+x)dx=\int_{r-1}^r f(u)du$$ so $$\sum_1^n\int_0^1 f(r-1+x)dx=\int_{0}^n f(x)dx$$

• Thanks! Btw, is there any way this can be solved as limit of a sum? I am 100% fine with the answer you wrote but this question was given in definite integral as the limit of a sum exercise so just wondering if there's any? – Cheapstrike Jan 16 '17 at 6:37
• @Cheapstrike For summation we must have $\Delta x=\Big((r+1)-1+x\Big)-(r-1+x)=r\to0$ but It does not occur. – Nosrati Jan 16 '17 at 6:40

As a hint: try to substitute $u=r+x-1$.

(This would typically be a comment, but I do not, alas, have enough reputation...)

• I think you do now. Welcome! – numbermaniac Nov 25 '17 at 12:44

$\int _0^1 f (r+x-1)=\int _0^1 f (r-x)$ let $r-x=u$ thus $dx=-du$ hence we have $I=\sum _1 ^n -\int _r ^{r-1} f (u)du =\sum _1 ^n \int _{r-1} ^r f (u)du =\int _0 ^n f (x)dx$

• from Second last to last step, is it a formula? – Kislay Tripathi Nov 25 '17 at 9:57
• The disappereance of - sign? – Archis Welankar Nov 25 '17 at 9:58
• Its a known fact that $\int _a^b f (x)dx=-\int _b ^a f (x)dx$ . I hope its clear now. – Archis Welankar Nov 25 '17 at 10:04
• please tell me about The disappereance of summation sign? – Kislay Tripathi Nov 25 '17 at 10:06
• I think you should read basic theorems of integration you can then answer your question yourself;) – Archis Welankar Nov 25 '17 at 10:08

Let $\displaystyle \mathfrak{I}:=\sum_{r=1}^n \int_0^1 f(r+x-1) \,\text{d} x$. Leting $u:=r+x-1$ we have $\dfrac{\text{d}u}{\text{d}x}=1$ . If $x=0$ we have $u=r-1$, and if $x=1$ we have $u=r$. Therefore:

\begin{align}\mathfrak{I}=&\sum_{r=1}^n \int_{r-1}^r f(u) \, \text{d}u\\ \end{align} But we know that $\displaystyle \int_{a}^b f \,+ \int_b^c f = \int_a^c f$. Therefore : $$\boxed{\,\,\,\mathfrak{I} =\int_{0}^n f(x) \, \text{d}x\,\,\,\,\,}$$

• Brother how did you conclude $\displaystyle \int_{a}^b f \,+ \int_b^c f = \int_a^c f$ From this \begin{align}\mathfrak{I}=&\sum_{r=1}^n \int_{r-1}^r f (u) \, \text{d}u\\ \end{align} .Is it some kind of formula or i need knowledge of riemann integral for this? – Kislay Tripathi Nov 25 '17 at 11:43
• @KislayTripathi it's the sum of areas. Think about the integrals as areas under a graph - if you add the area from $a$ to $b$ and the area from $b$ to $c$ together, wouldn't it be the same as the area from $a$ to $c$? – numbermaniac Nov 25 '17 at 12:17
• Is a way to "glue" togheter the areas. Consider a function $f$ and $a<b<c$. The $\displaystyle \int_a^c f$ (area from $a$ to $c$) can be cuted in a certain $b$. Therefore the area will bem the sum of those areas, wich are $\displaystyle \int_a^b f$ (area from $a$ to $b$) and $\displaystyle \int_b^c f$ (area from $b$ to $c$). Literally a slice haha. – Gustavo Mezzovilla Nov 25 '17 at 13:46