# Find the smallest positive number k

For any positive number $$n$$, let $$a_n = \sqrt{2+\sqrt{2+{...+\sqrt{2+\sqrt 2}}}}$$ ($$2$$ appear $$n$$) and let $$k$$ is positive number such that $$\displaystyle\frac{1}{k}\leq\frac{3-a_{n+1}}{7-a_n}$$ for any positive number $$n$$, then find the smallest positive number $$k$$.

I have $$a_1=\sqrt 2$$ and $$a_{n+1}=\sqrt{2+a_n}, \forall n \in\mathbb N$$

Consider

$$a_1=\sqrt 2 \lt 2$$

$$a_2=\sqrt{2+a_1}\lt\sqrt{2+2}=2$$

$$a_3=\sqrt{2+a_2}\lt\sqrt{2+2}=2$$

Use Mathematical Induction, I conclude $$\sqrt 2\leq a_n\leq 2,\forall n\in\mathbb N$$

Thus, $$3-a_{n+1}\gt1$$ and $$7-a_{n}\gt 5$$

Since $$k\in\mathbb N$$, I have $$\displaystyle k\geq\frac{7-a_n}{3-a_{n+1}}=\frac{7-a_n}{3-\sqrt{2+a_n}}=3+\sqrt{2+a_n}=3+a_{n+1}$$

Hence, $$3+\sqrt 2\leq 3+a_{n+1}\leq 3+2=5$$

Therefore $$k=5$$

Please check my solution, Is it correct?, Thank you

• $a_n=2\cos\frac{\pi}{2^{n+1}}$. – Riemann Dec 18 '19 at 8:06

You have given a lower & upper bound for $$a_n$$. With this, you've determined a possible value for $$k$$, but you haven't shown it's necessarily the smallest such value of $$k$$.

Instead, note that $$a_{n+1} = \sqrt{2 + a_n}$$. Thus, you have

\begin{aligned} \frac{3-a_{n+1}}{7-a_n} & = \frac{3-\sqrt{2 + a_n}}{7-a_n} \\ & = \frac{(3-\sqrt{2 + a_n})(3 + \sqrt{2 + a_n})}{(7-a_n)(3 + \sqrt{2 + a_n})} \\ & = \frac{9-(2 + a_n)}{(7-a_n)(3 + \sqrt{2 + a_n})} \\ & = \frac{7 - a_n}{(7-a_n)(3 + \sqrt{2 + a_n})} \\ & = \frac{1}{3 + \sqrt{2 + a_n}} \end{aligned}\tag{1}\label{eq1A}

Thus, if $$L$$ is the supremum of $$a_n$$, then $$k = 3 + \sqrt{2 + L}$$. To determine $$L$$, you can prove that $$a_n$$ is a strictly increasing sequence (I'll leave it to you fill in the details, such as what is shown in Solution verification: Prove by induction that $$a_1 = \sqrt{2} , a_{n+1} = \sqrt{2 + a_n}$$ is increasing and bounded by $$2$$), with an upper bound as you've shown, so it must converge to limiting value of its supremum. To determine this value, use

\begin{aligned} L & = \sqrt{2 + L} \\ L^2 & = 2 + L \\ L^2 - L - 2 & = 0 \\ (L - 2)(L + 1) & = 0 \end{aligned}\tag{2}\label{eq2A}

Thus, since $$L \gt 0$$, you have $$L = 2$$, as you surmised. You also have

$$k = 3 + \sqrt{2 + 2} = 5 \tag{3}\label{eq3A}$$

which matches what you got.

• $a_n=2\cos\frac{\pi}{2^{n+1}}$. – Riemann Dec 18 '19 at 8:07


# How to calculate the errors of single and double precision

We consider the initial value problem

$$\left\{\begin{matrix} y'=y &, 0 \leq t \leq 1 \\ y(0)=1 & \end{matrix}\right.$$

We apply the Euler method with $h=\frac{1}{N}$ and huge number of steps $N$ in order to calculate the approximation $y^N$ of the value of the solution $y$ at $t^N, \ y(t^N)=y(1)=e$. At the following table there are, for all $N$, the errors $|\epsilon^N|=|e-y^N|$, when the calculations are done with single and double precision.

$$\begin{matrix} N & |\epsilon^N|\text{ Single-precision } & |\epsilon^N| \text{ Double-precision } \\ - & - & - \\ 100 & 0.13468 \cdot 10^{-1} & 0.13468 \cdot 10^{-1} \\ 200 & 0.67661 \cdot 10^{-2} & 0.67647 \cdot 10^{-2}\\ 400 & 0.33917 \cdot 10^{-2} & 0.33901 \cdot 10^{-2}\\ 800 & 0.16971 \cdot 10^{-2} & 0.16970 \cdot 10^{-2}\\ 1600 & 0.85568 \cdot 10^{-3} & 0.84898 \cdot 10^{-3} \\ \cdots & & \\ 102400 & 0.65088 \cdot 10^{-4} & 0.13273 \cdot 10^{-4} \\ 204800 & 0.21720 \cdot 10^{-3} & 0.66363 \cdot 10^{-5} \\ 409600 & 0.78464 \cdot 10^{-3} & 0.33181 \cdot 10^{-5} \\ 819200 & 0.20955 \cdot 10^{-2} & 0.16590 \cdot 10^{-5} \\ \dots \end{matrix}$$

We notice that the errors of the calculations of double-precision get approximately half. However, in the case of single-precision, for $N>10^5$ the errors increase! Indeed, for a big enough $N$, the errors in our case tend to $1.71828 \dots$.

Could you explain me why the errors, when the calculations are done in single-precision, increase for $N>10^5$ and why they get approximately half when the calculations are done in double-precision?

Also, how can we calculate the error for a given $N$? For example, if we have $N=10^5$ then $\epsilon^N=|e-y^{10^5}|=\left |e- \left( 1+ \frac{1}{10^5} \right)^{10^5} \right |$. How can we calculate the latter, knowing that the zero of the machine is $10^{-6}$ when we have single precision but $10^{-12}$ when we have double precision?

EDIT: It holds that: $$\ln{\left( 1+ \frac{1}{N}\right)}=\sum_{n=1}^{\infty} (-1)^{n+1} \frac{\left( \frac{1}{N}\right)^n}{n}=\frac{1}{N}- \frac{1}{2N^2}+O(N^{-3})$$ Right? If so, then $N \ln{\left( 1+ \frac{1}{N}\right)}=1-\frac{1}{2N}+O(N^{-2})$, right?  If so, then how can we find the difference of the real solution with the approximation when we take into consideration that we have single precision and how when we take into consideration that we have double precision?

• In the case of single-precision, it seems as if truncation error is overcoming the finite-step integration error. – robjohn Mar 15 '15 at 17:59
• Note that IEEE float has the pattern (s,e,m)=(1,8,23) and thus a machine constant $2^{-23}\simeq 0.125·10^{-6}$. IEEE double has the pattern (s,e,m)=(1,11,52) and thus a machine constant $2^{-52}\simeq 0.25·10^{-15}$. – Dr. Lutz Lehmann Mar 15 '15 at 19:05

When $N$ gets large, the size of each step, and thus the size of the change in function value, gets small. You start out with $y(0)=1$, and then you get $$y(h)=1+\epsilon$$ and if $\epsilon$ is small enough, the computer won't be able to handle the difference between $1$ and $1+\epsilon$ with very good precision. This is the source of the increasing error. Since double precision handles this problem better, you get less error.

As for why the error tends to $1.71828\ldots$, if $\epsilon$ is really small, the computer thinks that $y$ doesn't change at all from time step to time step, and therefore thinks that $y$ is a constant function. You're supposed to get $e$ as the final value, so the error is therefore $e-1$.

• Don't we have to consider the difference between the real solution and the approximation?  If so, then do we have to consider the difference $|y(h)-y^1|$ ? – evinda Mar 15 '15 at 18:17
• @evinda The $\epsilon$ is just $\Delta y$ for the first step. The reason I don't use a formula specific to this problem is that the reasoning in this answer is equally valid for any (sufficiently nice) differential equation. – Arthur Mar 15 '15 at 18:18
• It holds that $t^n=nh=\frac{n}{N}$, right?  So for the first step, doesn't the following hold?  $$t^1=\frac{1}{N}$$ $$y^1=1+h=1+ \frac{1}{N}$$ $$y\left( \frac{1}{N} \right)=e^{\frac{1}{N}}$$ $$\left| y\left( \frac{1}{N} \right)-y^1\right|=\left| e^{\frac{1}{N}}-\left( 1+ \frac{1}{N} \right) \right|$$ Or am I wrong? If it is like that, then how do we calculate the latter? – evinda Mar 15 '15 at 18:27
• $$\left(1+\frac1N\right)^N=e^{N·\ln(1+\frac1N)}=e^{1-\frac 2N+\frac3{N^2}\mp...}=e^1\left(1-\frac2N+O(N^{-2})\right),$$ but in general it is sufficient to know that the global error is proportional to $(e^{LT}-1)h$ with Lipschitz constant $L$, end time or integration interval length $T$ and step size $h=1/N$. – Dr. Lutz Lehmann Mar 15 '15 at 18:40
• It holds that: $$\ln{\left( 1+ \frac{1}{N}\right)}=\sum_{n=1}^{\infty} (-1)^{n+1} \frac{\left( \frac{1}{N}\right)^n}{n}=\frac{1}{N}- \frac{1}{2N^2}+O(N^{-3})$$ Right? If so, then $N \ln{\left( 1+ \frac{1}{N}\right)}=1-\frac{1}{2N}+O(N^{-2})$, right?  If so, then how can we find the difference of the real solution with the approximation when we take into consideration that we have single precision and how when we take into consideration that we have double precision? – evinda Apr 13 '15 at 13:10

The Euler method has local discretization error order 2 and global error order 1. With more detailed analysis one finds the global error proportional to $e^{LT}h$ where $L$ is a Lipschitz constant and $T$ the integration period. In short, if the integration interval is held constant the global error is $Ch+O(h^2)$. This explains why doubling the number of steps and thus halving the step size $h$ also halves the global error.

Each integration step has a random floating point error that is a small multiple $Dμ$ of the machine precision $μ$ for the floating point type. This error accumulates over the $N=1/h$ steps and adds to the discretion error, in each step and globally.

In total this gives an expression for the computational global error like $$Ch+NDμ=Ch+\frac{TD·μ}h=2\sqrt{CDT·μ}+\frac{C}{h}·\left(h-\sqrt{\frac{DT·μ}{C}}\right)^2$$ This is a convex function in $h$ that has a valley at approximately (disregarding the constants) $h≈\sqrt{μ}$ with an error of about the same magnitude $\sqrtμ$.

• For the float type $μ≈ 10^{-7}$ and thus the minimal error is attained for approximately $h≈ 10^{-4}$ to $10^{-3}$.
• For the double type, $μ≈10^{-15}$, so the valley is to be found at step sizes $h≈ 10^{-8}$ to $10^{-7}$.

Thus if you were to continue the table, you would see growing errors also for the double type after reaching about $10^{-7}$ as lowest error. As test, for $h=10^{-9}$ or $N=10^9\simeq 2^{23}·100$ the error should be above $10^{-6}$.

The error progression of the Euler method can be compared to compound interest rate computations where the interest rate is $Lh$ per integration step. This is a consequence of the Gronwall lemma. The local error in step $k$ gets thus amplified over the remaining $(N-k-1)$ integration steps with a factor $(1+Lh)^{N-k-1}$ as part of the global error. The local error is the sum of the discretization error bound by $c·h^2$ and the floating point error bound by $d·μ$ where $c$ contains the magnitude of the system function $f$ and $d$ the magnitude of its derivative resp. the complexity of the evaluation of $f$. Thus the estimate of the global error has the form \begin{align} (c·h^2+d·μ)·\sum_{k=0}^{N-1}(1+Lh)^{N-k-1} &=(c·h^2+d·μ)·\frac{(1+Lh)^N-1}{Lh} \\ &\simeq \left(\frac{c·h}L+\frac{d·μ}{Lh}\right)·(e^{LT}-1) \end{align}

• What do you mean by integration step? – evinda Mar 15 '15 at 19:19
• Moving from $t$ to $t+h$. The solution of differential equations is also called integration. – Dr. Lutz Lehmann Mar 15 '15 at 19:31
• How do we get that if the integration interval is held constant then the global error is $Ch+O(h^2)$? – evinda Mar 15 '15 at 19:53
• This just means that the error is a function of $h$ with a regular root at $h=0$, i.e., $h·g(h)$ with $g(h)=C+O(h)$. Or just a more explicit way to say that the error is of order $O(h)$. – Dr. Lutz Lehmann Mar 15 '15 at 20:05
• 1. The error is $O(h)$ because it is proportional to $e^{LT}h$, or am I wrong?  2. Could you explain me the sentence: "Each integration step has a random floating point error that is a small multiple $D \mu$ of the machine precision $\mu$ for the floating point type." ?  3. How did you get this equality: $$Ch+\frac{TD·μ}h=2\sqrt{CDT·μ}+\frac{C}{h}·\left(h-\sqrt{\frac{DT·μ}{C}}\right)^2$$ ? – evinda Mar 15 '15 at 20:12


# Evaluating $\operatorname{arccos} \frac{2}{\sqrt5} + \operatorname{arccos} \frac{3}{\sqrt{10}}$

Evaluate: $$\operatorname{arccos} \frac{2}{\sqrt5} + \operatorname{arccos} \frac{3}{\sqrt{10}}$$

We let $$\alpha = \operatorname{arccos} \frac{2}{\sqrt5} \qquad \beta = \operatorname{arccos}\frac{3}{\sqrt{10}}$$

Then we have:

\begin{align} \cos(\alpha) = \frac{2}{\sqrt5} &\qquad \cos(\beta) = \frac{3}{\sqrt{10}} \\[4pt] \sin(\alpha) = \frac{1}{\sqrt5} &\qquad \sin(\beta) = \frac{1}{\sqrt{10}} \end{align}

In order to evaluate, we are told, we first determine $\sin(\alpha + \beta)$; we wind up with $1/\sqrt2$, thus we have $\pi/4$.

What I am confused about is why we have to use sin($\alpha + \beta$). For example, if I were to use $\cos(\alpha + \beta)$, I would get the answer $7/(\sqrt{10}\sqrt5)$, which I do not know what to do with. I am having trouble finding out whether there is some kind of pattern to this kind of thing, or did the author just know to use $\sin(\alpha + \beta)$ since he/she checked cos and saw nothing comes out of this?

Any help is much appreciated, thank you

• $7\sqrt{50}$ is $\cos(\alpha-\beta)$, not $\cos(\alpha+\beta)$. – Angina Seng Jul 8 '18 at 4:13
• Actually, you need to do both. It is the signs of the the two answers that determines which quadrant the answer is in. – steven gregory Jul 8 '18 at 4:47

As an alternative approach (which, truthfully, arose from my not reading your question carefully and, essentially, wishing to draw in MS Paint) see the following diagram in which the expression you wish to evaluate is the angle sum $\alpha + \beta$:

Next, let us consider instead $\tan(\alpha + \beta)$ using a tangent identity:

$$\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha \tan\beta} = \frac{1/2 + 1/3}{1 - (1/2)(1/3)} = \frac{5/6}{5/6} = 1$$

Observe $\alpha + \beta \in (0, \pi)$; the unique angle in this interval yielding a tangent of $1$ is $\pi/4$.

• Fantastic answer thank you very much for the insight – Algebra 8 Jul 8 '18 at 16:57

From $\sin(\alpha+\beta)=\frac1{\sqrt2}$, we should expect $|\cos(\alpha+\beta)|=\frac1{\sqrt2}$ from the formula $\sin^2 \theta + \cos^2 \theta = 1$.

\begin{align} \cos(\alpha+\beta)&=\cos(\alpha)\cos(\beta)\color{blue}-\sin(\alpha)\sin(\beta)\\ &=\frac{2}{\sqrt5}\cdot \frac{3}{\sqrt{10}}-\frac1{\sqrt5}\cdot \frac{1}{\sqrt{10}}\\ &=\frac{6}{5\sqrt{2}}-\frac{1}{5\sqrt2}\\ &=\frac{5}{5\sqrt2}\\ &=\frac{1}{\sqrt2} \end{align}

You can directly use the formula $$\cos^{-1}x+\cos^{-1}y=\cos^{-1}(xy-\sqrt{1-x^2}\sqrt{1-y^2})$$ $$x=\frac{2}{\sqrt{5}},y=\frac{3}{\sqrt{10}}$$ $$=\cos^{-1}\left(\left(\frac{2}{\sqrt{5}}\right)\left(\frac{3}{\sqrt{10}}\right)-\sqrt{1-\left(\frac{2}{\sqrt{5}}\right)^2}\sqrt{1-\left(\frac{3}{\sqrt{10}}\right)^2}\right)$$ $$=\cos^{-1}\left(\frac{6}{\sqrt{50}}-\sqrt{1-\frac45}\sqrt{1-\frac{9}{10}}\right)$$ $$=\cos^{-1}\left(\frac{6}{\sqrt{50}}-\sqrt{\frac15}\sqrt{\frac{1}{10}}\right)$$ $$=\cos^{-1}\left(\frac{6}{\sqrt{50}}-\frac{1}{\sqrt{50}}\right)$$ $$=\cos^{-1}\left(\frac{5}{\sqrt{50}}\right)$$ $$\cos^{-1}\frac{2}{\sqrt{5}}+\cos^{-1}\frac{3}{\sqrt{10}}=\cos^{-1}\left(\frac{5}{\sqrt{50}}\right)$$

• Thank you all very much, I am studying for the GRE and the author gave the formula cos($\alpha + \beta$) = cos$\alpha$cos$\beta$ + sin$\alpha$sin$\beta$. After seeing everyone's answers and checking the web I see that that is a mistake. – Algebra 8 Jul 8 '18 at 16:57
• You mean the answer which I have is wrong? – Key Flex Jul 8 '18 at 17:01

Using

$$\operatorname{arccos} \frac{2}{\sqrt5} + \operatorname{arccos} \frac{3}{\sqrt{10}}=\pi-\left(\arcsin\dfrac2{\sqrt5}+\arcsin\frac{3}{\sqrt{10}}\right)$$

as $$\left(\frac{2}{\sqrt5}\right)^2+\left(\frac{3}{\sqrt{10}}\right)^2=\dfrac45+\dfrac9{10}>1$$

$$\arcsin\dfrac2{\sqrt5}+\arcsin\frac{3}{\sqrt{10}}=\pi-\arcsin\left(\dfrac2{\sqrt5}\dfrac1{\sqrt{10}}+\dfrac1{\sqrt5}\dfrac3{\sqrt{10}}\right)=\pi-\arcsin\dfrac1{\sqrt2}=?$$


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Befor do topsort on a graph,graph have to be DAG.Could you please help me how i check whether it is DAG or not? Understnad the logic and implement by your own. The algorithm for the topological sort is as follows: Call dfs(g) for some graph g. The main reason we want to call depth first search is to compute the finish times for each of the vertices. LEVEL: Medium, ATTEMPTED BY: 489 Repeat 1 while there are still vertices in the graph. The topological sort is a simple but useful adaptation of a depth first search. Given a list of names, does there exist an order of letters in Latin LEVEL: Medium, ATTEMPTED BY: 37 When there are still nodes remaining, but none of them as IN-degree as ZERO, you can be sure that a cycle exists in the graph. Programming competitions and contests, programming community. Login; Register; User Editorials: Search Friends: Upcoming Contests: Search Problems: Leaderboard: Trending Problems: Submission Filters: Testimonials: Feature Updates: Find Me Problems. Codeforces. Example Problem Example (Codeforces Round 290 div. But I am not sure if this algorithm is related to topological sort or if should I restructure my work with another point of view. We have avoided using STL algorithms as main purpose of these problems are to improve your coding skills and using in-built algorithms will do no good.. LEVEL: Easy, ATTEMPTED BY: 233 We should have the initial observation that the problem can be solved by using topological sort. CodeChef - A Platform for Aspiring Programmers. if the graph is DAG. For example, another topological sorting of the following graph is “4 5 2 3 1 0”. HackerEarth uses the information that you provide to contact you about relevant content, products, and services. It also helps you to manage and track your programming comepetions training for you and your friends. While the exact order of the items is unknown (i.e. BLACK — Processed. Topological Sort. topological sort: all edges are directed from left to right. The algorithm for the topological sort is as follows: Call dfs(g) for some graph g. The main reason we want to call depth first search is to compute the finish times for each of the vertices. They are related with some condition that one should happen only after other one happened. Search Problems. Problem Name Search Site Tags... OR . Theoretical knowledge of algorithms is important to competitive programmers. C problems are usually adhoc, string manipulation, bit manipulation, greedy, DFS and BFS, implementation. 2.Initialize a queue with indegree zero vertices. 1 Problem A) A list of names are written in lexicographical order, but not in a normal sense. Think how to implement this corner case, rest part is easy though. (Description) Monotonic Queue/Stack. Editorial. Here you will learn and get program for topological sort in C and C++. Given the total number of courses and a list of prerequisite pairs, return the ordering of courses you should take to finish all courses. 1385E Directing Edges (Description) 102006G Is Topo Logical? For example, another topological sorting of the following graph is “4 5 2 3 1 0”. Also go through detailed tutorials to improve your understanding to the topic. 3. We start from this piece, do a BFS, two cakes are called connected if there is a shared edge. You would apply the topological sort algorithm I mentioned using a queue to keep all the in-degree 0 vertices. Also try practice problems to test & improve your skill level. Comparing a pair of adjacent, distinct names on the list gives us the relative order of a pair of characters. All Tracks Algorithms Graphs Topological Sort Problem. An algorithm for solving a problem has to be both correct and efﬁcient, and the core of the problem is often about inventing an efﬁcient algorithm. GREY — In Process 3. 2.Initialize a queue with indegree zero vertices. Analytics. (Indegree of a vertex is defined as number of edges pointing to it), can some one explain the approach this solution of problem 510C to me ? Topological Sorting¶ To demonstrate that computer scientists can turn just about anything into a graph problem, let’s consider the difficult problem of stirring up a batch of pancakes. A topological ordering is possible if and only if the graph has no directed cycles, i.e. GitHub is home to over 50 million developers working together to host and review code, manage projects, and build software together. Programming competitions and contests, programming community. Topological Sorting; graphs If is a DAG then a topological sorting of is a linear ordering of such that for each edge in the DAG, appears before in the linear ordering Solved example of topological sort. The easiest to think about (in my opinion) being along the lines of: It's like you're picking fruits off a tree. Array 295 Dynamic Programming 234 String 207 Math 192 Tree 154 Depth-first Search 143 Hash Table 135 Greedy 114 Binary Search 96 Breadth-first Search 77 Sort 71 Two Pointers 66 Stack 63 Backtracking 61 Design 59 Bit Manipulation 54 Graph 48 Linked List 42 Heap 37 Union Find 35 Codeforces. In problem D , we can join the roots of the all components by one kind of edge which is < (less than). The ordering of the nodes in the array is called a topological ordering. Problem. For topological sort problems,easiest approach is: 1.Store each vertex indegree in an array. Third, Then Write code and submit in the OJ to justify test cases. ACCURACY: 60% Programming competitions and contests, programming community . The recipe is really quite simple: 1 egg, 1 cup of pancake mix, 1 tablespoon oil, and $$3 \over 4$$ cup of milk. The first vertex in topological sorting is always a vertex with in-degree as 0 (a vertex with no incoming edges). Problem link— SPOJ TOPOSORT Topological Sorting /* Harun-or-Rashid CSEDU-23rd Batch */ Select Code #include using namespace std; #define white 0 #define gray 1 #define black 2 … ;), The only programming contests Web 2.0 platform, Educational Codeforces Round 102 (Rated for Div. ACCURACY: 59% Codeforces Round #258 (Div. 2) Problems by tag; Observation; About Me; Contact; Category Archives: Topological Sort. The topological sort is a simple but useful adaptation of a depth first search. R. Rao, CSE 326 5 Posted on May 24, 2014 by sufiantipu111. For example, a topological sorting of the following graph is “5 4 2 3 1 0”. 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# Wavelength of cosine-squared

I am confused. Usually, the wavelength is the x-distance between the tops of two consecutive waves. Here is the graph.

There is only 0.1 m between 2 crests. But the answer counts the wavelength as 0.2 m

This is puzzling. But use the identity $$\cos^2(\theta)=\frac{\cos(2\theta)+1}{2}.$$ The period of $$\cos(2\theta)$$ is $$\pi$$, so the period of $$\cos^2(\theta)$$ is also $$\pi.$$ I think I disagree with the answer in your post. The period is $$0.1$$.
• Yes, there should be more context involved on the question. Still, we should note that the "oscillations" in your consideration would be about the value of $1/2$, and not about zero. Subtle, indeed. – Bruno Anghinoni May 30 at 23:49
• The "oscillations" in the power graph are about the value $R_s/2$. – user52817 May 31 at 0:10


# Proof Checking and input: Generators of $\mathbb{Z}_{pq}$

I'm self-studying abstract algebra (slowly but surely), and I have a question about my answer to the following prompt:

Problem statement:

Show that there are $(q-1)(p-1)$ generators of the group $\mathbb{Z}_{pq}$, where $p$ and $q$ are distinct primes. (Where $\mathbb{Z}_{n}$ is the additive group of integers modulo $n$)

I can use a theorem from my textbook that says:

The integer $r$ generates the group $\mathbb{Z}_n$ iff $$1\le r\lt n \quad\text{and}\quad \gcd(r, n) = 1$$

My attempt at a proof:

Let $p$ and $q$ be distinct primes and $\mathbb{Z}_{pq}$ be the additive group of integers modulo $pq$.
An element $a \in \mathbb{Z}_{pq}$ is a generator of $\mathbb{Z}_{pq}$ iff: $$1\le a\lt pq \quad\text{and}\quad \gcd(a, pq) = 1$$ There are $p-1$ positive multiples of $q$ less than $pq$. Also, there are $q-1$ positive multiples of $p$ less than $pq$. These are the only elements of $\mathbb{Z}_{pq}$ that are not coprime to $pq$. Finally, $0$ is not a generator of $\mathbb{Z}_{pq}$.

There are $pq$ elements in $\mathbb{Z}_{pq}$, so the number of generators is: \begin{align} pq - (p-1) - (q-1) - 1 &= pq -p -q + 1 \\ &= p(q-1) - (q - 1) \\ &= (q-1)(p-1) \end{align}

My questions:

• Obviously, if there's a flaw in the proof, I'd like to know. :) Aside from that...

• When I assert "these are the only elements of $\mathbb{Z}_{pq}$ that are not coprime to $pq$," do I need to further show that this is true? It is patently obvious to me, but I know that just claiming "this is obvious" is not a valid method of proof.

• I'd like to have some input on the style/format of my proof. Is there a better (read: more formal/traditional) way to phrase something in this proof, or is there a format that I'm not following? As I'm self-teaching, I don't want to learn bad habits...

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On your second point, the general rule that I like to follow is: If you need to ask whether something has to be justified then it doesn't hurt to actually justify it. People find different things to be obvious and what's obvious to you may not be obvious to me. – EuYu Sep 20 '13 at 1:22

I don't see anything wrong with your proof.

For your second question, you could expand a bit on that statement, perhaps saying something like "since $1$,$p$, $q$ and $pq$ are the only divisors $pq$, any integer $n$not divisible by $p$ or $a$ must have $\gcd(n,pq)=1$, and be coprime to $pq$", but in situations like this, where a moment's though and writing down a few definitions will give a proof, omitting it is usually safe.

In writing proofs, I have found the following idea helpful: a proof is really intended to convince someone (a reader, a teacher, yourself) that a theorem is true. If, after reading your proof and spending a little time thinking about each step, this person could still doubt that your theorem is true, then you should add more. Otherwise, your safe. This reasoning, however, is very audience-dependant. For some audiences, the following would be an acceptable proof of your theorem:

The generators of $\Bbb{Z}_{pq}$ correspond with integers less than and coprime to $pq$. Since these are counted by Euler's $\phi(n)$, $\phi(n)$ is multiplicative, and, for any prime $p$, $\phi(p) = p-1$, the number of generators of $\Bbb{Z}_{pq}$ is $\phi(pq)=\phi(p)\phi(q) = (p-1)(q-1)$.

For lots of other audiences, it would not be.

Remember, a proof doesn't always have to be written down, finalized, and perfect. It can be more of a conversation.

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# Prove Two Functions are Equal?

Given two generic functions $$x(t)$$ and $$y(t)$$ I want to prove that $$x(t) = y(t)$$.

To do so, I take the derivative, which turn out to be: $$\dot x(t) = Ax(t) + B$$ $$\dot y(t) = Ay(t) + B$$ where $$A$$ and $$B$$ are the same in both derivatives.

Is this sufficient to say both functions are equal?

The reason I ask is that, in general, $$x(t)$$ and $$y(t)$$ could have vastly different forms (ie. $$x(t)$$ could be the resultant of a complicated integral $$y(t)$$ or something similar). Because of this, I am wondering if I have to go through the trouble of reducing $$x(t)$$ to $$y(t)$$ or vice-versa.

• Are $A,B$ constants or functions of $x,y,t$ ? – Yves Daoust Mar 4 '20 at 13:52
• @YvesDaoust they are constants – Clark Mar 4 '20 at 15:38

They both satisfy the linear first order differential equation:

$$\begin{equation*} f'(t) = A f(t) + B \end{equation*}$$

You need at least one other condition, say prove that $$x(t_0) = y(t_0)$$ for some value $$t_0$$, or perhaps the same derivative at a point. Note that the solution is

$$\begin{equation*} f(t) = c e^{A t} - B/A \end{equation*}$$

here $$c$$ is an unknown constant, to be determined by other conditions.

• do you mean $x(t_0) = y(t_0)$? So if I can prove both functions have the same value at one point, they should have the same value at all points? (ie. from existence and uniqueness?) – Clark Mar 4 '20 at 13:13
• @Clark, yes; and fixed. – vonbrand Mar 4 '20 at 13:15
• Awesome, thanks! It's all coming back – Clark Mar 4 '20 at 13:18

If $$x(t)$$ and $$y(t)$$ are $$C^1$$ functions such that $$x(t_0) = y(t_0)$$ for some $$t_0$$ furthermore $$x'(t) = y'(t) = f(t)$$ for some function (Lipschitz-continuous) $$f$$, then according to the uniqueness of the solution of a differential equation you have $$x(t)=y(t)$$ for all $$t\geq t_0$$.

By subtraction,

$$\dot x-\dot y=\dot{(x-y)}=A(x-y)$$ so that the two given functions can differ by a solution of this equation, given by

$$x-y=Ce^{At}.$$

• This doesn't contradict the other answers right? Let's say $x(t_0) = y(t_0) = 0$ where $t_0 = 0$ for simplicity. This would then give $C = 0$ and, therefore $x(t) = y(t)$ correct? – Clark Mar 4 '20 at 15:47
• @Clark: absolutely. My goal was to show how to solve the question without integrating the original equations and focusing on the difference. (Though the benefit is tiny.) – Yves Daoust Mar 4 '20 at 15:50


## Bisection Method

1) compute a sequence of increasingly accurate estimates of the root. In this paper we extend the traditional recursive bisection stan-dard cell placement tool Feng Shui to directly consider mixed block designs. This is a visual demonstration of finding the root of an equation $$f(x) = 0$$ on an interval using the Bisection Method. You begin with two initial approximations p 0 and p 1 which bracket the root and have f p 0 f p 1 < 0. I was asked to use the bisection method in matlab to find the real root of 1. Numerically solve F(X)=LN(X)-1/X=0 by forming a convergent, fixed point iteration, other than Newton's, starting from X(1)=EXP(1). Tabular Example of Bisection Method Numerical Computation. Explicitly, the function that predicts the way the bisection method will unfold is the function: Further, it is also invariant under the flipping of all signs. Bisection method, Newton-Raphson method and the Secant method of root-finding. Assume f(x) is an arbitrary function of x as it is shown in Fig. b] that contains a root (We can use the property sign of f(a) ≠ sign of f(b) to find such an initial interval). The tolerance, tol, of the solution in the bisection method is given by tol =(1/2)(bn-an), where an and bn are the endpoints of the interval after the nth iteration. Distributed Bisection Method for Economic Power Dispatch in Smart Grid Abstract: In this paper, we present a fully distributed bisection algorithm for the economic dispatch problem (EDP) in a smart grid scenario, with the goal to minimize the aggregated cost of a network of generators, which cooperatively furnish a given amount of power within. Le Stack Exchange Network Stack Exchange network consists of 175 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. ***** *****MATLAB CODE ***** x = linspace(0, 2*pi, 100); y = sin(x); plot(x, y, ’*r’);. After reading this chapter, you should be able to: 1. It is a bit difficult to apply bisection method to a non-deterministic function. Bisection method is a popular root finding method of mathematics and numerical methods. 84070158) ≈ 0. 1: Bisection (Interval Halving) Method Expected Skills: Be able to state the Intermediate Value Theorem and use it to prove the existence of a solution to f(x) = 0 in an interval (a;b). Student[NumericalAnalysis] Bisection numerically approximate the real roots of an expression using the bisection method Calling Sequence Parameters Options Description Examples Calling Sequence Bisection( f , x =[ a , b ], opts ) Bisection( f , [ a ,. Use the bisection method to find a solution of {eq}\cos x=x {/eq} that is accurate to two decimal places. To use this module, we should import it using − This method is same as insort() method. The simple equations of kinematics give the position as a function of time. Let’s start with a method which is mostly used to search for values in arrays of every size, Bisection. Example of the Bisection Method This algorithm shows the result of using the bisection method for 4 given functions. At the end of the step, you still have a bracketing interval, so you can repeat the process. However, instead of simply dividing the region in two, a linear interpolation is used to obtain a new point which is (hopefully, but not necessarily) closer to the root than the equivalent estimate for the bisection method. $299 vinyl cutter to start your home business - Duration: 17:41. Algorithm is quite simple and robust, only requirement is that initial search interval must encapsulates the actual root. NET (or, how to make an async method synchronous) Bisection method; Execute an instance method of Object and call in its block instance methods of another object; get URL Params (2 methods) Rake Migrate (newest method) order/format of params in method definition; XML Load methods; Kohana helper method for Askimet. Bisection Method Example - Polynomial • If limits of 0 to 10 are selected, the solution converges to x = 4 Engineering Computation: An Introduction Using MATLAB and Excel 21. Assume f(x) is an arbitrary function of x as it is shown in Fig. Use the bisection method to approximate this solution to within 0. Pros of Bisection Method 1. Given a function f(x) and an interval which might contain a root, perform a predetermined number of iterations using the bisection method. The convergence is linear, slow but steady. 1 word related to bisection: division. Octave / MATLAB. It is Fault Free (Generally). 01 and |f(1. Example: Solving x − sin(x) = 0 using bisection method • At every step, the function sin(x) needs to be evaluated ⇒ truncation errors • Round-off errors occur with every operation of (+, −, ×, ÷) and accumulate along the way • The bisection process cannot go on forever; has to stop at a finite number of iterations ⇒ further errors. (c) Use Newton’s method to evaluate the same root as in (b). Bisection Method - Half-interval Search This code calculates roots of continuous functions within a given interval and uses the Bisection method. It is a very simple and robust method, but it is also rather slow. (a) The smallest positive root of x = 1+ :3cos( x ) Let f (x ) = 1+ :3cos( x ) x. Newton and Raphson used ideas of the Calculus to generalize this ancient method to find the zeros of an arbitrary equation Their underlying idea is the approximation of the graph of the function f ( x ) by the tangent lines, which we discussed in detail in the previous pages. Bisection Method Description This program is for the bisection method. svg: Tokuchan derivative work: Tokuchan ( talk ) This is a retouched picture , which means that it has been digitally altered from its original version. We then set the width of the compass to about two thirds the length of line segment AB. The main way Bisection fails is if the root is a double root; i. xl xu Bisection algorithm. The red curve shows the function f and the blue lines are the secants. One of your comments says you are creating an object to round values to 6 places, but you are not creating an object there. This Demonstration shows the steps of the bisection root-finding method for a set of functions. Let a = 0 and b = 1. It is a bit difficult to apply bisection method to a non-deterministic function. Bisection method is a closed bracket method and requires two initial guesses. In this paper, we have presented a new method for computing the best-fitted rectangle for closed regions using their boundary points. The theory is kept to a minimum commensurate with comprehensive coverage of the subject and it contains abundant worked examples which provide easy understanding through a clear and concise theoretical treatment. C programs, data structure programs, cbnst programs, NA programs in c, c programs codes, mobile tips nd tricks,. Philippe B. ***** *****MATLAB CODE ***** x = linspace(0, 2*pi, 100); y = sin(x); plot(x, y, ’*r’);. Assumptions We will assume that the function f(x) is continuous. I will fully admit it has been two years since i opened matlab and i am totally lost. Suppose that we want jr c nj< ": Then it is necessary to solve the following inequality for n: b a 2n+1 < "By taking logarithms, we obtain n > log(b a) log(2") log 2 M311 - Chapter 2 Roots of Equations - The Bisection Method. Figure 1: The graphs of y=x (black) and y=\cos x (blue) intersect. Bisection Method of Solving a Nonlinear Equation. Bisection Method Using C. Assume f(x) is an arbitrary function of x as it is shown in Fig. which proves the global convergence of the method. Brackets are unions of similar simplexes. Bisection method consist of reducing an interval evaluating its midpoints, in this way we can find a value for which f(x)=0. The bisection method is used to solve transcendental equations. The bisection algorithm is then applied recursively to the sub-interval where the sign change occurs. In fact, the common proof of the Intermediate Value Theorem uses the Bisection method. The bisection method in mathematics is a root-finding method that repeatedly bisects an interval and then selects a subinterval in which a root must lie for further processing. Bisection Methods: We can pursuse the above idea a little further by narrowing the interval until the interval within which the root lies is small enough. Calculus textbook. The bisection method is discussed in Chapter 9 as a way to solve equations in one unknown that cannot be solved symbolically. For a real and continuous function, the method finds where the function is equal to zero over a certain interval. The IVT states that suppose you have a segment (between points a and b, inclusive) of a continuous function , and that function crosses a horizontal line. At which point, things got better. Learn via an example, the bisection method of finding roots of a nonlinear equation of the form f(x)=0. Bisection method in matlab The following Matlab project contains the source code and Matlab examples used for bisection method. Algorithm is quite simple and robust, only requirement is that initial search interval must encapsulates the actual root. Newton- Raphson method. It provides a convenient command line inter-. The bisection method applied to sin(x) starting with the interval [1, 5]. It is also known as Binary Search or Half Interval or Bolzano Method. For example, suppose that we would like to solve the simple equation 2 x = 5. This method, also known as binary chopping or half-interval method, relies on the fact that if f(x) is real and continuous in the interval a < x < b , and f(a) and f(b) are of opposite signs, that is,. A few steps of the bisection method applied over the starting range [a 1;b 1]. 84070742] and sin(40. This process involves ﬁnding a root, or solution, of an equation of the form f(x) = 0 for a given function f. Always Convergent. the bisection method. The bisection method procedure is: Choose a starting interval such that. THE BISECTION METHOD AND LOCATING ROOTS. Table of Contents 1 - The interval-halving (bisection) method, Java/OOP style 2 - The interval halving method written in a slightly more functional style 3 - The same 'halveTheInterval' function in a completely FP style After writing the code first in what I’d call a “Java style,” I then. Otherwise, the Intermediate Value Theorem is used to determine whether the root lies on the subinterval$(a_n, p_n)$or the subinterval$(p_n, b_n)$. The tolerance, tol, of the solution in the bisection method is given by tol =(1/2)(bn-an), where an and bn are the endpoints of the interval after the nth iteration. This method is closed bracket type, requiring two initial guesses. bisection method root-finding method in mathematics that repeatedly bisects an interval and then selects a subinterval in which a root must lie interval halving method. Bisection Method is one of the simplest, reliable, easy to implement and convergence guarenteed method for finding real root of non-linear equations. The bisection method in mathematics is a root-finding method that repeatedly bisects an interval and then selects a subinterval in which a root must lie for further processing. 4 comments I am using equation f(x): x^ 3-2x-5 = x*x*x-2*x-5 (a=2 and b=3). The bisection method is a method used to find the roots of a function. This is achieved by selecting two points A and B on that interval. In mathematics , the bisection method is a root-finding method that applies to any continuous functions for which one knows two values with opposite signs. BISECTION METHOD Bisection method is the simplest among all the numerical schemes to solve the transcendental equations. The bisection method is a root-finding method based on simple iterations. 001, m = 100) Arguments f. Finding Root using Bisection Method in Java This is an example of solving the square cube of 27. The method assumes that we start with two values of z that bracket a root: z1 (to the left) and z2 (to the right), say. The bisection method requires two points aand bthat have a root between them, and Newton’s method requires one. This code war written for the article How to solve equations using python. It is assumed that f(a)f(b) <0. The bisection method in mathematics is a root finding method which repeatedly bisects an interval and then selects a subinterval in which a root must lie for further processing. Given an initial. Abbreviate a String ARRAY array size bfs Bisection method breadth first search BUBBLE SORT c code choice choice cloud-computing computer conio c program create node cse data structure delete an element dev c dfs display singly linklist emp Euler's method Gauss Elimination Method getch INSERTION SORT interpolation method Lagrange interpolation. Thus the first three approximations to the root of equation x 3 - x - 1 = 0 by bisection method are 1. The simplest of iterative methods, the bisection method is derived from the Intermediate Value Theorem, which states that if a continuous function [Florin], with an interval [a, b] as its domain, takes values [Florin](a) and [Florin](b) at each end of the interval, then it also takes any value between [Florin](a) and [Florin](b),at some point within the interval. The simplest way to solve an algebraic equation of the form g(z) = 0, for some function g is known as bisection. Bisection method is a popular root finding method of mathematics and numerical methods. Now at the very end, we want to output the result, and this is a function. Just like any other numerical method bisection method is also an iterative method, so it is advised to tabulate values at each iteration. Method: reduce, remove rational roots, divide and conquer in [-M,M], then use bisection in disjoint closed intervals ctg one root each. The setup of the bisection method is about doing a specific task in Excel. Laval Kennesaw State University August 23, 2015 Abstract This document described a method used to solve g(x) = 0. OF TECH & SCI. Select a Web Site. (here in my code i don't know why the loop doesn't work as it should. Consider a root finding method called Bisection Bracketing Methods • If f(x) is real and continuous in [xl,xu], and f(xl)f(xu)<0, then there exist at least one root within (xl, xu). The bisect algorithm is used to find the position in the list, where the data can be inserted to keep the list sorted. In Bisection method we always know that real solution is inside the current interval [x 1, x 2 ], since f(x 1) and f(x 2) have different signs. Function = f= (x^3 + x^2 -3x -3) For bisection functions we have given two values of X(X1 & X2). Matlab Build-in Function. any help would be appreciated. The variable f is the function formula with the variable being x. (b) Use the bisection method to evaluate one root of your choice. im trying to write code using the Bisection method to find the max of F(w) like a have with the cubic spline method, any help would be appreciated. Be able to apply the Bisection (Interval Halving) Method to approximate a solution to f(x) = 0. Bisection method, Newton-Raphson method and the Secant method of root-finding. I can speak English, Hindi and a little French. The algorithm is. sign ( f ( a )) == np. If a function changes sign over an interval, the function value at the midpoint is evaluated. The theorem is demonstrated in Figure 2. Getting root of an equation by Bisection Method through C programming language. follow the algorithm of the bisection method of solving a nonlinear equation, 2. 2004 Judith Koeller The bisection method can be used to approximate a solution p to an equation f(p)=0 where f(x) is a continuous function. Mujahid Islam Md. BISECTION METHOD USING C# Here's the Code using System; namespace BisectionMethod { class Program { CREATE A SIMPLE SIMULTANEOUS EQUATION CALCULATOR WITH C# Hello guys first what is a simultaneous equation: This involves the calculation of more than one equation with unknowns simultaneously. The problem is that it seems like the teachers recommended solution to the task isn't quite right. Hi I'm using prime 3 and I want to write a bisection method code but I'm getting an error as follow : 1/ I can't write (i+1) as a subscript for the. What are synonyms for bisection?. Hi, my code doesn't seem to continue beyond the first iteration of the bisection method in my loop. The two most well-known algorithms for root-finding are the bisection method and Newton’s method. Rafiqul Islam Khaza Fahmida Akter 2. Assumptions We will assume that the function f(x) is continuous. De ning a domain In higher dimensions, there is a rich variety of methods to de ne a simply connected domain. This is a visual demonstration of finding the root of an equation $$f(x) = 0$$ on an interval using the Bisection Method. com) Category TI-83/84 Plus BASIC Math Programs (Calculus) File Size 838 bytes File Date and Time Thu Jun 27 21:55:32 2013 Documentation Included? Yes. 84070158, 40. Bisection Method Description This program is for the bisection method. I think there is a need for an improvement, e. We start with a line segment AB. For a given function as a string, lower and upper bounds, number of iterations and tolerance Bisection Method is computed. Find more Mathematics widgets in Wolfram|Alpha. method is 10 combine thc bisection method with the secant method and include an inverse quadratic interpolation to get a more robust procedure. The bisection method is discussed in Chapter 9 as a way to solve equations in one unknown that cannot be solved symbolically. Given a function f(x) and an interval which might contain a root, perform a predetermined number of iterations using the bisection method. The bisection method is probably the simplest root-finding method imaginable. The bisection method is used to solve transcendental equations. Any zero-finding method (Bisection Method, False Position Method, Newton-Raphson, etc. xl xu Bisection algorithm. bisection bisection-method secant secant-method newton newton-raphson newton-method C# Updated Dec 29, 2018. Consider a root finding method called Bisection Bracketing Methods • If f(x) is real and continuous in [xl,xu], and f(xl)f(xu)<0, then there exist at least one root within (xl, xu). It also makes a graph available of the iterates. This module provides support for maintaining a list in sorted order without having to sort the list after each insertion. In mathematics , the bisection method is a root-finding algorithm which repeatedly divides an interval in half and then selects the subinterval in which a root exists. The bisection method guarantees a root (or singularity) and is used to limit the changes in position estimated by the Newton-Raphson method when the linear assumption is poor. 001 using the bisection method. Bisection method is a popular root finding method of mathematics and numerical methods. In this tutorial you will get program for bisection method in C and C++. The IVT states that suppose you have a segment (between points a and b, inclusive) of a continuous function , and that function crosses a horizontal line. Matlab Build-in Function. The Bisection Method is a numerical method for estimating the roots of a polynomial f(x). a b 3 Regula falsi Consider the ﬁgure in which the root lies between a and b. We will study three diﬀerent methods 1 the bisection method 2 Newton’s method 3 secant method and give a general theory for one-point iteration methods. It is a very simple and robust method, but it is also relatively slow. The problem was the calling of the function. This is intended as a summary and supplementary material to the required textbook. 2 Estimate how many iterations will be needed in order to approximate this root with an accuracy of ε=0. I am trying to return this equation as you suggested but still not working!. xl xu Bisection algorithm. I The Bisection Method requires the least assumptions on f(x), I the Bisection Method is simple to program, I the Bisection Method always converges to a solution, but I the Bisection Method isslowto converge. Bisection can be shown to be an "optimal" algorithm for functions that change sigh in [a,b] in that it produces the smallest interval of uncertainty in a given # of iterations f(x) need not be continuous on [a,b] convergence is guarenteed (linearly) Disadvantages of the Bisection Method. The graph of this equation is given in the figure. The bisection method is discussed in Chapter 9 as a way to solve equations in one unknown that cannot be solved symbolically. So let's take a look at how we can implement this. here is my program btw, but something's wrong in the bisection function and I can't figure out what is it. Mathematica. If the function values at points A and B have opposite signs…. Lecture Material. Bisection Method. Figure 1: The graphs of y=x (black) and y=\cos x (blue) intersect. thanks Code:. After reading this chapter, you should be able to: 1. We also call this method as an interval halving method because we consider a midpoint. We stay with our original. i just tried my vba code for bisection method but it doesn't work. Show that 𝑓𝑥=𝑥3+4𝑥2−10= 0has a root in [1, 2], and use the Bisection method to determine an approximation to the root that is accurate to at least within 10−4. And as I mentioned last time, this was the state of the art until the 17th century. The method is based on the following theorem. It is used in cases where it is known that. By testing different. The Bisection method is a numerical method which finds approximate solutions to polynomial equations with the use of midpoints. Select a Web Site. Equations don't have to become very complicated before symbolic solution methods give out. I dream of being a Theoretical Physicist one day. SECANT METHODS Convergence If we can begin with a good choice x 0, then Newton’s method will converge to x rapidly. The bisection method requires two points aand bthat have a root between them, and Newton’s method requires one. a b 3 Regula falsi Consider the ﬁgure in which the root lies between a and b. Bisection Method 1- Flowchart. (a) Bisection Method: This is one of the simplest and reliable iterative methods for the solution of nonlinear equation. 1) compute a sequence of increasingly accurate estimates of the root. In mathematics , the bisection method is a root-finding algorithm which repeatedly divides an interval in half and then selects the subinterval in which a root exists. It is Fault Free (Generally). If your calculator can solve equations numerically, it most likely uses a combination of the Bisection Method and the Newton-Raphson Method. Author jamespatewilliamsjr Posted on July 29, 2018 July 29, 2018 Categories C# Computer Programming Language, Root Finding Algorithms Tags Bisection Method, Brent's Method, C#, Computer Science, Newton's Method, Numerical Analysis, Regula Falsi Leave a comment on Root Finding Algorithms by James Pate Williams, BA, BS, MSwE, PhD. Numerical Analysis: Root Solving with Bisection Method and Newton’s Method. The number of iterations n that are required for obtaining a solution with a tolerance that is equal to or smaller than a specified tolerance can be determined before the solution. The basic method for making or doing something, such as an artistic work or scientific procedure: learned the techniques involved in painting murals Bisection of the angle technique - definition of bisection of the angle technique by The Free Dictionary. Bisection Method. The theorem is demonstrated in Figure 2. As a result, f(x) is approximated by a secant line through. any help would be appreciated. BISECTION METHOD. Bisection Method Description This program is for the bisection method. It is a very simple and robust method, but it is also rather slow. The most straightforward root-finding method. If the function values at points A and B have opposite signs…. These methods first find an interval containing a root and then systematically shrink the size of successive intervals that contain the root. 3 The bisection method converges very slowly 4 The bisection method cannot detect multiple roots Exercise 2: Consider the nonlinear equation ex −x−2=0. It was observed that the Bisection method. For the PASSFAIL method, the measure must pass for one limit and fail for the other limit. CGN 3421 - Computer Methods Gurley Numerical Methods Lecture 6 - Optimization page 107 of 111 Single Variable - Golden Section Search Optimization Method Similar to the bisection method • Define an interval with a single answer (unique maximum) inside the range sign of the curvature does not change in the given range. The Bisection Method is a successive approximation method that narrows down an interval that contains a root of the function f(x) The Bisection Method is given an initial interval [a. De ning a domain In higher dimensions, there is a rich variety of methods to de ne a simply connected domain. m" So create new. Apply the bisection method over a "large" interval. Use the bisection method to locate % a zero of the function f(x) = x sin(x) - 1. We have provided MATLAB program for Bisection Method along with its flowchart and algorithm. False Position or Regular Falsi method uses not only in deciding. 1 Show there is a root αin the interval (1,2). The routine assumes that an interval [a,b] is known, over which the function f(x) is continuous, and for which f(a) and f(b) are of opposite sign. The algorithm is iterative. The simplest way to solve an algebraic equation of the form g(z) = 0, for some function g is known as bisection. While the subject itself is quite interesting, the programming environment being used in the lab is Turbo C, a DOS based IDE which has been abandoned a long time ago. Brent's method combines the bisection method, secant method, and the method of inverse quadratic interpolation. If that is the case, you could save that data to an array and plot that array when you exit the loop like. (b) Use the bisection method to evaluate one root of your choice. -Bisection method is used to get a rough estimate of the solution then some other faster methods are used (discuss in our next lecture). Here is one example that passes the function f as a parameter, checks parameters for validity before continuing, avoids some other overflow exposures, avoids redundant calls to. Any zero-finding method (Bisection Method, False Position Method, Newton-Raphson, etc. The method is also called the interval halving method, the binary search method, or the dichotomy method. Mujahid Islam Md. The bisection method is far more efficient than algorithms which involve a search over frequencies, and of course the usual problems associated with such methods (such as determining how fine the search should be) do not arise. A genetic approach using direct representation of solutions for the parallel task scheduling problem. In the ﬁrst iteration of bisection method, the approximation lies at the small circle. If a n and b n are satistfy equation (3) then b n − a n ≤ b −a 2n, for n ≥ 1 where b1 = b,a1 = a. 5 Bisection Method (cont’d) •It always converge to the true root (but be careful about the following) •f(x L) * f(x U) < 0 is true if the interval has odd number of roots, not necessarily one root. PROGRAM(Simple Version):. The method assumes that we start with two values of z that bracket a root: z1 (to the left) and z2 (to the right), say. The setup of the bisection method is about doing a specific task in Excel. However, instead of simply dividing the region in two, a linear interpolation is used to obtain a new point which is (hopefully, but not necessarily) closer to the root than the equivalent estimate for the bisection method. Using C program for bisection method is one of the simplest computer programming approach to find the solution of nonlinear equations. Bisection method. The bisection algorithm is then applied recursively to the sub-interval where the sign change occurs. 0 (193 ratings) Course Ratings are calculated from individual students’ ratings and a variety of other signals, like age of rating and reliability, to ensure that they reflect course quality fairly and accurately. Bisection Theorem An equation f(x)=0, where f(x) is a real continuous function, has at least one root between a and b, if f(a) f(b) < 0. Select a Web Site. The bisection method is discussed in Chapter 9 as a way to solve equations in one unknown that cannot be solved symbolically. Be able to apply the Bisection (Interval Halving) Method to approximate a solution to f(x) = 0. The bisection method is a simple root-finding method. ) exa_myfpi. If a function changes sign over an interval, the function value at the midpoint is evaluated. If instead we want the time at which a certain position is reached, we must invert these equations. 1) in terms of all five of the desirable attributes. Description: Given a closed interval [a,b] on which f changes sign, we divide the interval in half and note that f must change sign on either the right or the left half (or be zero at the midpoint of [a,b]. I followed the same steps for a different equation with just tVec and it worked. 3 The bisection method converges very slowly 4 The bisection method cannot detect multiple roots Exercise 2: Consider the nonlinear equation ex −x−2=0. The convergence is linear, slow but steady. It is a very simple and robust method, but it is also relatively slow. Let f 2 be a continuous function with di erent signs at a;b, with a #inc This is the solution for finding the roots of a function by Bisection Method in C++ Object Oriented Approach. We will study three diﬀerent methods 1 the bisection method 2 Newton's method 3 secant method and give a general theory for one-point iteration methods. follow the algorithm of the bisection method of solving a nonlinear equation, 2. Example - 4: Using the bisection method find the approximate value of square root of 3 in the interval (1, 2) by performing two iterations. Formerly, engineers built external drivers to submit multiple parameterized Star-Hspice jobs, with each job exploring a region of the operating envelope of the circuit. What is Bisection Method? It is an iterative method based on a well known theorem which states that if f(x) be a continuous function in a closed interval [a,b] and f(a)f(b)<0, then there exists at least one real root of the equation f(x)=0, between a and b. Sample C program that uses Bisection Method in mathematics. By the Intermediate Value Theorem, there exists p in with. Pick starting points, precision and method. Compute where is the midpoint. Plot the function sin(x) on [0;2ˇ]. In fact, the common proof of the Intermediate Value Theorem uses the Bisection method. This tutorial explores a simple numerical method for finding the root of an equation: the bisection method. The secant method avoids this issue by using a nite di erence to approximate the derivative. Use the bisection method to approximate this solution to within 0. 9 Abeam is loaded as shown in Fig. What are synonyms for bisection?. 725 option price). I dream of being a Theoretical Physicist one day. 1 word related to bisection: division. Bisection method is used for finding root of the function in given interval. Then it is much more useful to explain, how a function is called with input arguments, than to convert the function to a script - which is still not working. Suppose we want to solve the equation f(x)=0,where f is a continuous function. Root is found by repeatedly bisecting an interval. In case, you are interested to look at the comparison between bisection method (adopted by Mibian Library) and my code please have look at screenshot of results obtained :-As you can see, bisection method didn’t converge well (to$13. So this is what happens in every iteration of the bisection method, we go through 20 iterations. sign ( f ( a )) == np. Suppose we know the two points of an interval and , where , and. The convergence of the bisection method is very slow. This can be achieved if we joint the coordinates (a,f(a)) and (b. Now at the very end, we want to output the result, and this is a function. erably; traditional methods produce results that are far from satis-factory. Bisection Method: The idea of the bisection method is based on the fact that a function will change sign when it passes through zero. Convergence • Theorem Suppose function 𝑓(𝑥) is continuous on [ , ], and 𝑓 ∙𝑓 <0. The bisection method is a simple root-finding method. The bisection method is probably the simplest root-finding method imaginable. if a and b are two. Methods for finding roots are iterative and try to find an approximate root $$x$$ that fulfills $$|f(x)| \leq \epsilon$$, where $$\epsilon$$ is a small number referred later as tolerance. To find a root very accurately Bisection Method is used in Mathematics. In a nutshell, the former is slow but robust and the latter is fast but not robust. This will open a new tab with the resource page in our marketplace. 01, and therefore we chose b = 1. The main way Bisection fails is if the root is a double root; i. Suppose we know the two points of an interval and , where , and. Since the line joining both these points on a graph of x vs f(x), must pass through a point, such that f(x)=0. Bisection is the division of a given curve, figure, or interval into two equal parts (halves). % % Enter the starting endpoints for [a,b] in a and b % % Enter the tolerance in delta. Bisection Method. conventional methods like Newton-Raphson method (N-R), Regula Falsi method (R-F) & Bisection method (BIS). THE BISECTION METHOD This method is based on mean value theorem which states that if a function ( ) is continuous between and , and ( ) ( ) are of opposite signs, then there exists at least one root between and. Bisection Method The bisection method is a kind of bracketing methods which searches for roots of equation in a specified interval. Each iteration step halves the current interval into two subintervals; the next interval in the sequence is the subinterval with a sign change for the function (indicated by. What is Bisection Method? The method is also called the interval halving method, the binary search method or the dichotomy method. Use this tag for questions related to the bisection method, which is a root-finding method that repeatedly bisects an interval and then selects a subinterval in which a root must lie for further processing. 2 Using the Bisection Method to Prove the Intermediate Value Theorem Now suppose that fis continuous on [a;b], f(a) <0 and f(b) >0. So it is dependent on. Advantages of the Bisection method. It also makes a graph available of the iterates.


# To show for following sequence $\lim_{n \to \infty} a_n = 0$ where $a_n$ = $1.3.5 … (2n-1)\over 2.4.6…(2n)$

How can I show

$\lim_{n \to \infty} a_n = 0$

$a_n = {1.3.5 ... (2n-1)\over 2.4.6...(2n)}$

I have shown that $a_n$ is monotonically decreasing. I thought to shown sequence is bounded from below then it automatically would converge and hence my question will be solve. But I'm unable to show its boundedness... Or there maybe another method to prove this. Thanks

• Clarification request: what do you mean by $1.3.5...$? Are these numbers being added? – Mike Pierce Nov 17 '14 at 8:14
• I think is multiplication of odd numbers. Isn't $\pi$ in the answer? I vaguely remember a result like this from Euler. – ReverseFlow Nov 17 '14 at 8:15
• It is certainly bounded below, since all terms are positive, so it converges. But that does not show it converges to $0$. The Stirling approximation will show convergence to $0$, but one can do it with less machinery. – André Nicolas Nov 17 '14 at 8:21
• Try to prove using induction that:$$a_n\lt\frac{1}{\sqrt{3n+1}}$$ – pointer Nov 17 '14 at 8:26
• That the limit is zero is intuitively obvious, seeing as $a_n=\binom{2n}n/4^n$is the probability of getting exactly $n$ heads in $2n$ independent tosses of a fair coin. – bof Nov 17 '14 at 8:30

Multiply the numerator by the denominator; so $$a_n=\frac{1\times3\times5\times ... \times(2n-1)}{ 2\times4\times6\times...\times(2n)}=\frac{1\times2\times3\times ... \times(2n)}{\Big(2\times4\times6\times...\times(2n)\Big)^2}=\frac{(2n)!}{4^n(n!)^2}$$ If now you use Stirling approximation $$m! \approx \sqrt{2 \pi } e^{-m} m^{m+\frac{1}{2}}$$ and then $$a_n \approx \frac{1}{\sqrt{\pi n} }$$ A more detailed approach would show for the asymptotic behavior $$a_n \approx \frac{1}{\sqrt{\pi n} }\Big(1-\frac{1}{8n}\Big)$$

• In first step how have u written in denominatorof last inequality as $4^{n}(n!)^{2}$ .. – godonichia Nov 17 '14 at 12:11
• Factor $2$ inside the brackets of denominator and square. – Claude Leibovici Nov 17 '14 at 17:21
• @ClaudeLeibovici in stirlings aproximstion as you have mentioned in order to find formula for 2m! I just have to replace m by 2m in above formula .if i do this then im not getting to answer, thanks – Jessica Gtb Dec 25 '14 at 12:20
• @JessicaGtb. I don't know whee you had a problem since if you use Stirling, you should get $$(2n)! \approx \sqrt{\pi } 2^{2 m+1} e^{-2 m} m^{2 m+\frac{1}{2}}$$ Post if you still have problems. Cheers :-) – Claude Leibovici Dec 26 '14 at 7:22
• How did you get that expression for 2n! .it is not same when you replace n by 2n in formula you mentioned in answer – Jessica Gtb Dec 26 '14 at 12:29

Using $\boldsymbol{1+x\le e^x}$

Since $1+x\le e^x$ for all $x\in\mathbb{R}$, \begin{align} a_n &=\prod_{k=1}^n\frac{2k-1}{2k}\\ &=\prod_{k=1}^n\left(1-\frac1{2k}\right)\\ &\le\prod_{k=1}^ne^{-\large\frac1{2k}}\\[3pt] &=e^{-\frac12H_n}\tag{1} \end{align} where $H_n$ are the Harmonic Numbers. Since the Harmonic Series diverges, $(1)$ tends to $0$.

Using Gautschi's Inequality \begin{align} a_n &=\prod_{k=1}^n\frac{2k-1}{2k}\\ &=\prod_{k=1}^n\frac{k-\frac12}{k}\\ &=\frac{\Gamma\!\left(n+\frac12\right)}{\sqrt\pi\,\Gamma(n+1)}\\[6pt] &\le\frac1{\sqrt{\pi n}}\tag{2} \end{align}

Commentary

Gautschi's Inequality also says that \begin{align} a_n &=\frac{\Gamma\!\left(n+\frac12\right)}{\sqrt\pi\,\Gamma(n+1)}\\ &\ge\frac1{\sqrt{\pi\left(n+\frac12\right)}}\tag{3} \end{align} Therefore, $$\lim_{n\to\infty}\sqrt{n}\,a_n=\frac1{\sqrt\pi}\tag{4}$$ Putting the second line of $(1)$ and the limit in $(4)$ together, we can show that $$\gamma+\sum_{n=2}^\infty\frac{\zeta(n)}{n2^{n-1}}=\log(\pi)\tag{5}$$ where $\gamma$ is the Euler-Mascheroni Constant and $\zeta$ is the Riemann Zeta Function.

Let's prove using induction that $$a_n\le\frac{1}{\sqrt{3n+1}}.$$ For $n=1$ it is true. Now we just need to prove that$$\frac{(2n+1)^2}{(2n+2)^2}\le\frac{3n+1}{3n+4}$$or $$(4n^2+4n+1)(3n+4)\le(4n^2+8n+4)(3n+1)$$or$$12n^3+28n^2+19n+4\le12n^3+28n^2+20n+4.$$

• If we are not given the limit it converges to then how would we go about it – godonichia Nov 17 '14 at 8:51
• I don't know. Maybe we could find asymtothics using Stirling's formula, then try to find this inequality. – pointer Nov 17 '14 at 9:08

Your sequence can be rewritten as $$a_n=\frac{1\cdot3\cdot...\cdot(2n-1)}{2\cdot4\cdot...\cdot2n}=\frac{1\cdot2\cdot3\cdot...\cdot(2n-1)\cdot2n}{(2\cdot4\cdot...\cdot2n)^2}=\frac{(2n)!}{2^{2n}(n!)^2}$$ Using Stirling's approximation we get $$a_n=\frac{(2n)!}{2^{2n}(n!)^2}\sim\frac{1}{2^{2n}}\cdot\frac{\sqrt{2\pi 2n}\cdot(\frac{2n}{e})^{2n}}{(\sqrt{2\pi n}(\frac{n}{e})^n)^2}=\frac{1}{2^{2n}}\cdot\frac{\sqrt{2}\cdot2^{2n}}{\sqrt{2\pi n}}=\frac{\sqrt{2}}{\sqrt{2\pi n}}\to 0$$ as $n\to\infty$.

• i havent studied stirlings approximation yet . Can u show me another way which uses sequence series concepts – godonichia Nov 17 '14 at 8:35


# Math Help - Help needed for a probablilty question

1. ## Help needed for a probablilty question

Ok so this is really bugging me, because I thought i finally understood it, but then my book is giving me different answers to what I'm getting.
Anyways, heres the problem, can anyone answer it and show working , so I can learn. Thanks

Given P(R|S) = 0.5, P(R|S') = 0.4 and P(S) = 0.6 find:

(a) P(R) (b) P(S|R) (c) P(S'|R) (d) P(S'|R')

2. P(R)
= P(R(S+S'))
= P(R1)
= P(RS)+P(RS')
= P(R|S)P(S)+P(R|S')P(S')
= P(R|S)P(S)+P(R|S')(1-P(S))

P(RS)
= P(R|S)P(S)
= P(S|R)P(R)

Now you know P(R) and P(RS), so you know P(RS')
P(RS')
= P(S'|R)P(R)

P(S)
= ...
= P(S|R)P(R)+P(S|R')(1-P(R))
gives
P(SR')
= P(S|R')P(R')
= P(S|R')(1-P(R))
Using this,
P(1)
= P((R+R')(S+S'))
= P(RS) + P(RS')+P(R'S)+P(R'S')
gives
P(R'S')
= P(S'|R')P(R')

3. Originally Posted by kamicool17
Given P(R|S) = 0.5, P(R|S') = 0.4 and P(S) = 0.6 find:
(a) P(R) (b) P(S|R) (c) P(S'|R) (d) P(S'|R')
$P(RS)=P(R|S)P(S)$
$P(RS')=P(R|S')P(S')$
$P(R)=P(RS)+P(RS')$

Now finish.

4. Hello, kamicool17!

Another approach . . . (much longer, though)

Given: . $P(R|S) = 0.5,\;\;P(R|S') = 0.4,\;\;P(S) = 0.6$

Find: . $(a)\;P(R) \qquad (b)\;P(S|R)\qquad (c)\;P(S'|R) \qquad (d)\;P(S'|R')$

We have this table:

. . $\begin{array}{c|| c|c||c}
& S & S' & \text{Total } \\ \hline \hline
R & {\color{blue}(a)} & {\color{blue}(c)} & {\color{blue}(e)} \\ \hline
R' & {\color{blue}(b)} & {\color{blue}(d)} & {\color{blue}(f)} \\ \hline \hline
\text{Total} & 0.60 & 0.40 & 1.00 \end{array}$

We have: . $P(S) = 0.60 \quad\Rightarrow\quad P(S') = 0.40$
So we can fill in the bottom row.

We have: . $P(R|S) = 0.5$

Then: . $\frac{P(R \wedge S)}{P(S)} \;=\;0.5 \quad\Rightarrow\quad \frac{P(R \wedge S)}{0.6} \:=\:0.5
$

Hence: . $P(R \wedge S) \,=\,0.30\;{\color{blue}(a)} \quad\Rightarrow\quad P(R' \wedge S) \:=\:0.30\;{\color{blue}(b)}$

. . $\begin{array}{c|| c|c||c}
& S & S' & \text{Total } \\ \hline \hline
R & 0.30 & (c) & (e) \\ \hline
R' & 0.30 & (d) & (f) \\ \hline \hline
\text{Total} & 0.60 & 0.40 & 1.00 \end{array}$

We have: . $P(R|S') = 0.4$

Then: . $\frac{P(R \wedge S')}{P(S')} \:=\:0.4 \quad\Rightarrow\quad \frac{P(R \wedge S')}{0.4} \:=\:0.4$

Hence: . $P(R \wedge S') \:=\:0.16\;{\color{blue}(c)} \quad\Rightarrow\quad P(R' \wedge S') \:=\:0.24\;{\color{blue}(d)}$

Also: . $P(R) \:=\:0.46 \;{\color{blue}(e)}\quad\Rightarrow\quad P(R') \:=\:0.54\;{\color{blue}(f)}$

And we have completed the table:

. . $\begin{array}{c|| c|c||c}
& S & S' & \text{Total } \\ \hline \hline
R & 0.30 & 0.16 & 0.46 \\ \hline
R' & 0.30 & 0.24 & 0.54 \\ \hline \hline
\text{Total} & 0.60 & 0.40 & 1.00 \end{array}$

You can now answer the questions . . .