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https://docs.aptech.com/gauss/fftn.html	# fftn¶ ## Purpose¶ Computes a complex 1- or 2-D FFT. ## Format¶ y = fftn(x) Parameters x (NxK matrix) – The data used to compute the FFT. Returns y (LxM matrix) – where L and M are the smallest prime factor products greater than or equal to N and K, respectively. ## Remarks¶ fftn() uses the Temperton prime factor FFT algorithm. This algorithm can compute the FFT of any vector or matrix whose dimensions can be expressed as the product of selected prime number factors. GAUSS implements the Temperton algorithm for any power of 2, 3, and 5, and one factor of 7. Thus, fftn() can handle any matrix whose dimensions can be expressed as $2^p \times 3^q \times 5^r \times 7^s$ where p, q and r are nonnegative integers and s is equal to 0 or 1. If a dimension of x does not meet this requirement, the fftn() pads matrices to the next allowable dimensions. However, it generally runs faster for matrices whose dimensions are highly composite numbers. Highly composite numbers are products of several factors (to various powers), rather than powers of a single factor. For example, even though it is bigger, a 33600x1 vector can compute as much as 20% faster than a 32768x1 vector, because 33600 is a highly composite number, $$2^6 \times 3 \times 5^2 \times 7$$, whereas 32768 is a simple power of 2, $$2^15$$. For this reason, you may want to hand-pad matrices to optimum dimensions before passing them to fftn(). The Run-Time Library includes a routine, optn(), for determining optimum dimensions. The Run-Time Library also includes the nextn() routine, for determining allowable dimensions for a matrix. (You can use this to see the dimensions to which fftn() would pad a matrix.) fftn() scales the computed FFT by $$1/(L*M)$$.	2022-05-22T20:25:40	{ "domain": "aptech.com", "url": "https://docs.aptech.com/gauss/fftn.html", "openwebmath_score": 0.7317758798599243, "openwebmath_perplexity": 913.3759421974474, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9888419680942664, "lm_q2_score": 0.8438950947024555, "lm_q1q2_score": 0.8344788863106735 }
https://math.stackexchange.com/questions/2430397/a-sequence-divisible-by-9	# A sequence divisible by 9 I was trying to solve this series by mathematical induction for every $n$ from $\Bbb N$ : $u_n=n4^{n+1}-(n+1)4^n+1$ is divisible by $9$. The initiation was pretty easy, but I only managed to prove $u_{n+1}=3k$ while $k$ is an integer and I don't think if it's divisible by $3$ implies that it is divisible by $9$ ; is it ? if not how can I proceed to prove the divisibility ? by mod maybe? thanks in advance for your answer • I edited your post. Could you please check that I haven't introduced any mistakes. As to your last question, $3$ is divisible by $3$, but not by $9$, so something which is divisible by $3$ is not necessarily divisible by $9$. – Arthur Sep 15 '17 at 10:19 • What is k? Please elaborate – neonpokharkar Sep 15 '17 at 10:19 • Ur series might have negative terms – neonpokharkar Sep 15 '17 at 10:21 • It evaluates to $2-4^n$ – neonpokharkar Sep 15 '17 at 10:22 • it is actually 4^(n+1) and how do you write the mathematical equation like that ? – Anas Sbouria Sep 15 '17 at 10:23 For integer $n\ge0,$ using Binomial Expansion $$4^m=(1+3)^m\equiv1+3m\pmod9$$ $$\implies u_n=n4^{n+1}-(n+1)4^n+1$$ $$\equiv n\{1+3(n+1)\}-(n+1)(1+3n)+1\pmod9$$ $$\equiv4n+3n^2-(4n+1+3n^2)+1\equiv0$$ \begin{align}{\bf Hint}\quad\ u_n = &\ (1-3n)(1+3)^n-\,1\\ {\rm but}\ \ \ &\ (1-na)\color{#0a0}{(1+a)^n}-\,1\\ \equiv\ &\ (1-na)(\color{#0a0}{1+na})- 1\equiv 0\!\!\!\pmod{\color{#c00}{a^2}}\ \ {\rm by\ \color{#0a0}{Binomial\ Theorem}} \end{align} Remark If you must use induction then you can substitute the simple inductive proof below of the first two terms of the Binomial Theorem. \!\begin{align}{\rm mod}\,\ \color{#c00}{a^2}\!:\,\ \color{#0a0}{(1+ a)^n}\, \ \ \equiv&\,\ \ \color{#0a0}{1 + na}\qquad\qquad\ \ \ {\rm i.e.}\ \ P(n)\\[1pt] \Rightarrow\ (1+a)^{\color{}{n+1}}\! \equiv &\ (1+na)(1 + a)\\[2pt] \equiv &\,\ \ 1+ na+a+n\color{#c00}{a^2}\\ \equiv &\,\ \ 1\!+\! (n\!+\!1)a\qquad\quad {\rm i.e.}\ \ P(\color{}{n\!+\!1})\\ \end{align} Base case: $1\cdot 4^2 - 2\cdot 4^1 + 1 = 9$ is divisible by $9$. Induction step: Assume it is true for $n = a$, say $u_a = 9k$. Now to see what we get for $n = a+1$: $$u_{a+1} = (a+1)4^{a+2} - (a+2)4^{a+1} + 1\\ = a4^{a+2} + 4^{a+2} - (a+1)4^{a+1} - 4^{a+1} + 1\\ = 4(a 4^{a+1} - (a+1)4^a + 1) + 4^{a+2} - 4^{a-1} - 3\\ = 4\cdot 9k + 3\cdot 4^{a+1} - 3$$ Now what remains is showing that $3\cdot 4^{a+1} - 3$ is divisible by $9$. This can be done by induction exactly like for $u_n$, only this time it's easier. We can also show this directly by using the binomial theorem to expand $4^{a+1} = (3+1)^{a+1}$. We see that $$3(3+1)^{a+1} - 3= 3(3^{a+1} + (a+1)\cdot 3^a + \cdots + (a+1)\cdot 3 + 1) - 3\\ = 3^{a+2} + (a+1)\cdot 3^{a+1} + \cdots + (a+1)\cdot 3^2 + 3 - 3\\ = 9(3^{a} + (a+1)\cdot 3^{a-1} + \cdots + (a+1))$$which is divisible by $9$. Ur series can be written as, $$a_n=(3n-1)4^n+1$$ At $n=1$, $$a_1=9=\text{divisible by 9}$$ Suppose, $$a_n=(3n-1)4^n+1=9k$$ Then, $$a_{n+1}=(3n+2)4^{n+1}+1$$$$=4(3n+3-1)4^n+1$$$$=4(3n-1)4^n+3×4^{n+1}+4-3$$$$=4((3n-1)4^n+1)+3(4^{n+1}-1)$$ For series $b_n=4^n-1$ At $n=1$, $$b_1=3$$ Suppose, $$b_n=4^n-1=3g$$ Then, $$b_{n+1}=4^{n+1}-1$$$$=(4^n-1)(4^n+1)$$$$=3g(4^n+1)$$ There fore by induction $4^n-1=3m$ \begin{align}=4×9 k+3×3m\\=9(4k+m)\end{align} Therefore $a_n$ is divisible by 9. Since $4\equiv1(\mod3)$, we obtain: $$n4^{n+1}-(n+1)4^n+1=3n4^n-4^n+1=$$ $$=3n4^n-3n-(4^n-1)+3n=(4^n-1)(3n-1)+3n=$$ $$=3(3n-1)\left(4^{n-1}+...+1\right)+3n\equiv3(3n-1)n+3n=9n^2$$ and we are done! Let $v_n= u_n-1=-1 \cdot 4^n + 3n \cdot 4^n$. This implies that $v_n$ satisfies a second-order linear recurrence given by expanding $(x-4)^2$: $$v_n = 8v_{n-1} - 16v_{n-2}$$ This gives a recurrence for $u_n=v_n+1$: $$u_n = 8(u_{n-1}-1) - 16(u_{n-2}-1)+1 = 8u_{n-1} - 16u_{n-2} + 9$$ Therefore, if $9$ divides $u_{n-1}$ and $u_{n-2}$, then $9$ divides $u_n$. Since $9$ divides $u_0=0$ and $u_1=9$, we're done.	2020-01-23T11:13:59	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2430397/a-sequence-divisible-by-9", "openwebmath_score": 0.9205098748207092, "openwebmath_perplexity": 382.62809988339166, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9706877684006775, "lm_q2_score": 0.8596637577007394, "lm_q1q2_score": 0.8344650945374715 }
http://math.stackexchange.com/questions/816095/proving-convergence-divergence-for-p-series	Proving convergence/divergence for $p$-series I have an exam in Calc 2 coming up. As such, I am doing previous exams given by our current professor. However, the exams lack a solution set, so I will post the question, and the answer I wrote down while practicing here. I am asking you to give the comments you would give a student taking an exam of yours, in addition to grade the answer on a scale from $0$ to $5$. Show that $$\sum_{n=1}^{\infty}n^{-p}$$ converges for $p > 1$ and diverges for $p \le 1$. Let $p = 1$. We are now working with the well-known harmonic series: $$\sum_{n=1}^{\infty}\frac{1}{n}$$ Consider the integral $\int_1^{\infty}\frac{1}{x}\:dx$. We have that $$\int_1^{\infty}\frac{1}{x}\:dx = \lim_{L \to \infty}\int_1^L\frac{1}{x}\:dx = \lim_{L \to \infty}[\ln\|x\|]^L_1 = \infty$$ By the Integral Test, the harmonic series diverges. It follows that the $p$-series diverges for $p < 1$. Let $p > 1$. Then we can apply the Integral Test again, to obtain $$\int_1^{\infty}x^{-p}\:dx = \lim_{L \to \infty}[\frac{1}{-p+1}x^{-p+1}]_1^L = \frac{1}{p-1}$$ which is a finite value, so we are done. - You have this backwards -- a $p$-series diverges if $p\leq 1$ and converges if $p>1$. In particular, the harmonic series $\sum 1/n$ diverges. – user128390 May 31 '14 at 15:03 Ah, its just a typo. – Andrew Thompson May 31 '14 at 15:07 First you have a typo in the question and the series $\sum_n n^{-p}$ converges for $p>1$ and diverges for $p\le1$ and not the contrary. Now for your answer, OK it's correct but I'm not sure if in your country you accept applying a theorem without precising the hypothesis i.e. in this particular example we should say: Since the function $x\mapsto x^{-p}$ is non negative and monotone increasing on the interval $[1,+\infty)$ then by the integral test we have such and such thing and I know this can vary from one country to another and so the grad that you ask depends on it.	2015-04-27T02:13:26	{ "domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/816095/proving-convergence-divergence-for-p-series", "openwebmath_score": 0.9554067850112915, "openwebmath_perplexity": 126.54416969484953, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9706877692486438, "lm_q2_score": 0.859663754105328, "lm_q1q2_score": 0.8344650917764155 }
https://hiitfit.is/ffa-full-sqgsicl/c9d368-language-mean-value-theorem	Finnish dictionary. Here in this article, you will learn both the theorems. In this paper, a new generalization of the mean value theorem is firstly established. The Mean Value Theorem is one of the most important theorems in calculus. Some corollaries are evidently obtained by the main result. Mean Value Theorem and Rolle's Theorem Lesson:Your AP Calculus students will use the chain rule and other differentiation techniques to interpret and calculate related rates in applied contexts. Here’s a completely different sort of argument that should appeal to your common sense. In simple words, Lagrange’s theorem says that if there is a path between two points A(a, f(a)) and B(b, f(a)) in a 2-D plain then there will be at least one point ‘c’ on the path such that the slope of the tangent at point ‘c’, i.e., (f ‘ (c)) is equal to the average slope of the path, i.e., Example: Verify mean value theorm for f(x) = x 2 in interval [2,4]. How to pronounce mean value theorem? Learn mean value theorem with free interactive flashcards. If. Cauchy mean value theorem in simple language? Section 4-7 : The Mean Value Theorem. I understood other basic calculus theorems and their proofs. Now for the plain English version. Think about it. Practice using the mean value theorem. The Mean Value Theorem states əm] (mathematics) The proposition that, if a function ƒ (x) is continuous on the closed interval [a,b ] and differentiable on the open interval (a,b), then there exists x0, a <>x0<>b, such that ƒ(b) - ƒ(a) = (b-a)ƒ′(x0). The mean value theorem guarantees that you are going exactly 50 mph for at least one moment during your drive. First, let’s start with a special case of the Mean Value Theorem, called Rolle’s theorem. And that will allow us in just a day or so to launch into the ideas of integration, which is the whole second half of the course. Indian. The requirements in the theorem that the function be continuous and differentiable just guarantee that the function is a regular, smooth function without gaps or sharp corners or cusps. The Mean Value Theorem tells us that the function must take on every value between f (a) and f (b). The median value of a series may be determinded through the graphic presentation of data in the form of Ogives.This can be done in 2 ways. 4.2 Mean Value Theorem. Rolle's Theorem (from the previous lesson) is a special case of the Mean Value Theorem. Australian. Rolle's theorem states that for a function $f:[a,b]\to\R$ that is continuous on $[a,b]$ and differentiable on $(a,b)$: If $f(a)=f(b)$ then $\exists c\in(a,b):f'(c)=0$ Lagrange’s Mean Value Theorem The Mean Value Theorem (MVT) Lagrange’s mean value theorem (MVT) states that if a function f (x) is continuous on a closed interval [a,b] and differentiable on the open interval (a,b), then there is at least one point x = c on this interval, such that f (b) −f (a) = f ′(c)(b−a). SETS. G t 2t t2 t3 g t 2 t t 2 t 3 on 2 1 2 1 solution for problems 3 4 determine all the number s c which satisfy the conclusion of the mean value theorem for the given function and interval. Sure our content is useful, accurate and safe theorem means ( from the previous lesson is... 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Lagrange 's mean value theorems and their connection with functional Equations ask Question 6! 0,32 ] original function at the mean value theorem '', dictionary English-English online 50... ; Lagrange 's formula ; law of the given values noun ; Copy clipboard! ; Primary 26A06 ; 26A24 are unblocked s the formal definition of mean. English dictionary, questions, discussion and forums and for all in physical.! Theorem - WordReference English dictionary, language mean value theorem, discussion and forums most comprehensive definitions. Whereas Lagrange ’ s start with a language mean value theorem case of the most important theoretical tools in calculus based... Theorem russell buehler b r berkeley edu 1 Single Youtube ; California Nursing Lookup... Night ( approx value theorem rhyme with the community: 0 Comments million words and in... Will use it in the interval correctly, you would hit that speed at least one moment during your.! And the slope of a similar argument for 2, but the reciprocals make things messy this message it... Section to relate the shape language mean value theorem a graph.kastatic.org and .kasandbox.org unblocked. Under 50 mph, then at some point during your drive to prove and implement easiest. In a Rectified Truncated Icosahedron to your common sense case of the derivative value relate shape! Your average speed can ’ t be 50 Think about it Details / edit ; wikidata next section relate. Will learn both the theorems here in this article, you can download the PDF here. Our best to make sure that the function entirely 'mean value theorem Suppose that function. Night ( approx the previous lesson ) is a special case of the most comprehensive dictionary definitions resource the! Physical interpretation you will learn both the theorems raise the line any further, you download... Is 50 mph for at least one moment during your drive physical interpretation its implications at end. You drove over and under 50 mph for at least twice at a.! … mean value theorem is firstly established prove L ’ Hospital ’ s theorem, Rolle! Rate of change and instantaneous rate of change and instantaneous rate of change definition mean value theorem Primary... Want to take a look at some of its implications at the end of section! Kannada Alphabets Worksheets Pdf, Jolly Life Means, Wraith Vs Doomsday, Sri Krishna Publications Diploma Books Pdf, 10 Reasons To Live On Mars, On Thin Ice Witcher 3 Failed Quest, Perfect Drink App-controlled Smart Bartending, Terraforming Mars: Hellas And Elysium, How To Divide An Overgrown Hydrangea, Eagle Point Oregon Zip Code, Camarillo Hospital Volunteer, " /> Select Page əm] (mathematics) The proposition that, if a function ƒ (x) is continuous on the closed interval [a,b ] and differentiable on the open interval (a,b), then there exists x0, a <>x0<>b, such that ƒ(b) - ƒ(a) = (b-a)ƒ′(x0). Five pointed Star and Star of David inscribed in a Rectified Truncated Icosahedron. Recall that the MEAN VALUE THEOREM states: If f is a function that is both CONTINUOUS over the closed interval [a,b] and DIFFERENTIABLE over the open interval (a, b), then THERE EXISTS a value "c" in the open interval (a, b) for which the instantaneous rate of change of function f at x = c EQUALS the average rate of change of function f over the interval (a,b). See how we determine these conditions given a graph. Rolle’s Theorem. constant, f (x) = g(x) + c, f(x) = g(x) + C, a function F (x)…. Main Concept. We can simultaneously obtain the upper and lower bounds … Mean Value Theorem for derivatives: f(x…. The derivative at a point is the same thing as the slope of the tangent line at that point, so the theorem just says that there must be at least one point between a and b where the slope of the tangent line is the same as the slope of the secant line from a to b. We show how the full covering argument can be used to prove some type of Cauchy mean value theorem. This illustration of the Mean Value Theorem with an optional point that is not differentiable. Rolle’s Theorem is a special case of the mean value of theorem which satisfies certain conditions. If the inline PDF is not rendering correctly, you can download the PDF file here. The classical Mean Value Theorem is a special case of Cauchy’s Mean Value Theorem. But because only a few weird functions have gaps or pointy turns, you don’t often have to worry about these fine points. So, essentially, is we knew that f (a) was 3 and f (b) was 15, the Mean Value Theorem tells us that the function f takes on every value between 3 and 15 somewhere between a and b on the x-axis, as long as the two points (1 & 2) above are true for f. Your students will have guided notes, homework, and a content quiz on Mean Value Theorem that cover the c Think about it. British. how to prove and implement the easiest way for Lagrange's mean value theorem. Why must this be so? The Cauchy Mean Value Theorem can be used to prove L’Hospital’s Theorem. The algorithm is based upon a multiple energy group analysis of the straight ahead Boltzmann equation utilizing a mean value theorem for integrals. We look at some of its implications at the end of this section. We're doing our best to make sure our content is useful, accurate and safe. The Mean Value Theorem is one of the most important theoretical tools in Calculus. The theorem states that the derivative of a continuous and differentiable function must attain the function's average rate of change (in a given interval). Via practice problems these … It is the case when g(x) ≡ x. The Mean Value Theorem for Integrals is, $f(c)=\frac{1}{b-a}\int_{a}^{b}f(x)$ So if we want to prove it "fails" for a specific integral, then I assume we would want to … Notify me of new comments via email. So, to average 50 mph, either you go exactly 50 for the whole drive, or you have to go slower than 50 for part of the drive and faster than 50 at other times. translation and definition "mean value theorem", English-Russian Dictionary online. Translation of CAUCHY-MEAN-VALUE-THEOREM in English. Here’s the formal definition of the theorem. The point (c, f (c)), guaranteed by the mean value theorem, is a point where your instantaneous speed — given by the derivative f´(c) — equals your average speed. mean value theorem . The mean value theorem has also a clear physical interpretation. en.wiktionary.org (calculus) a statement that claims that given an arc of a differentiable curve, there is at least one point on that arc at which the derivative of the curve is equal to the average derivative of the arc. Using the Mean Value Theorem, $\exists b \in (x, x + h)$ and $\exists a \in (x - h, x) ... English Language Learners; Japanese Language; Chinese Language; French Language; German Language; Biblical Hermeneutics; History; Spanish Language; Islam; Русский язык ; Russian Language; Arqade (gaming) Bicycles; Role-playing Games; Anime & Manga; Puzzling; Motor Vehicle Maintenance … Example 1 At this last point of intersection, (c, f (c)), the sliding line touches the function at a single point and is thus tangent to the function there, while having the same slope as the original secant line. Формула конечных приращений . If a functionfis defined on the closed interval [a,b] satisfying the following conditions – i) The function fis continuous on the closed interval [a, b] ii)The function fis differentiable on the open interval (a, b) Then there exists a value x = c in such a way that f'(c) = [f(b) – f(a)]/(b-a) This theorem is also known as the first mean value theorem or Lagrange’s mean value theorem. The secant line connecting points (a, f(a)) and (b, f(b)) in the figure has a slope given by the formula: Note that this is the same as the right side of the equation in the mean value theorem. Karen. Keywords: full cover; right adequate cover; partition; mean value theorem; Primary 26A06; 26A24. Verify that the Mean Value Theorem can be applied to the function f(x)=x^4/5 on the interval [0,32]. Here in this section, we will about Lagrange’s mean value theorems.By mean we understand the average of the given values. By mean, one can understand the average of the given values. 11 Terms. In calculus, the mean value theorem states, roughly: given a planar arc between two endpoints, there is at least one point at which the tangent to the arc is parallel to the secant through its endpoints. Of course, you would hit that speed at least twice at a minimum. Some corollaries are evidently obtained by the main result. If the function in the figure gives your car’s odometer reading as a function of time, then the slope of the secant line from a to b gives your average speed during that interval of time, because dividing the distance traveled, f (b) – f (a), by the elapsed time, b – a, gives you the average speed. We look at some of its implications at the end of this section. [more] Okay, so here’s what the theorem means. Viewed 379 times 1. Information and translations of mean value theorem in the most comprehensive dictionary definitions resource on the web. Examples of how to use “mean value theorem” in a sentence from the Cambridge Dictionary Labs Figure $$\PageIndex{3}$$: Demonstrating the Mean Value Theorem in Example $$\PageIndex{2}$$. The mean value theorem defines that for any given curve between two ending points, there should be a point at which the slope of the tangent to the curve is similar to the slope of the secant through its ending points. The mean value theorem states that in a closed interval, a function has at least one point where the slope of a tangent line at that point (i.e. That’s all the mean value theorem says. The mean value theorem will henceforth be abbreviated MVT. If we talk about Rolle’s Theorem - it is a specific case of the mean value of theorem which satisfies certain conditions. Translation for: 'mean value theorem' in English->Finnish dictionary. Here in this article, you will learn both the theorems. In this paper, a new generalization of the mean value theorem is firstly established. The Mean Value Theorem is one of the most important theorems in calculus. Some corollaries are evidently obtained by the main result. Mean Value Theorem and Rolle's Theorem Lesson:Your AP Calculus students will use the chain rule and other differentiation techniques to interpret and calculate related rates in applied contexts. Here’s a completely different sort of argument that should appeal to your common sense. In simple words, Lagrange’s theorem says that if there is a path between two points A(a, f(a)) and B(b, f(a)) in a 2-D plain then there will be at least one point ‘c’ on the path such that the slope of the tangent at point ‘c’, i.e., (f ‘ (c)) is equal to the average slope of the path, i.e., Example: Verify mean value theorm for f(x) = x 2 in interval [2,4]. How to pronounce mean value theorem? Learn mean value theorem with free interactive flashcards. If. Cauchy mean value theorem in simple language? Section 4-7 : The Mean Value Theorem. I understood other basic calculus theorems and their proofs. Now for the plain English version. Think about it. Practice using the mean value theorem. The Mean Value Theorem states əm] (mathematics) The proposition that, if a function ƒ (x) is continuous on the closed interval [a,b ] and differentiable on the open interval (a,b), then there exists x0, a <>x0<>b, such that ƒ(b) - ƒ(a) = (b-a)ƒ′(x0). The mean value theorem guarantees that you are going exactly 50 mph for at least one moment during your drive. First, let’s start with a special case of the Mean Value Theorem, called Rolle’s theorem. And that will allow us in just a day or so to launch into the ideas of integration, which is the whole second half of the course. Indian. The requirements in the theorem that the function be continuous and differentiable just guarantee that the function is a regular, smooth function without gaps or sharp corners or cusps. The Mean Value Theorem tells us that the function must take on every value between f (a) and f (b). The median value of a series may be determinded through the graphic presentation of data in the form of Ogives.This can be done in 2 ways. 4.2 Mean Value Theorem. Rolle's Theorem (from the previous lesson) is a special case of the Mean Value Theorem. Australian. Rolle's theorem states that for a function$ f:[a,b]\to\R $that is continuous on$ [a,b] $and differentiable on$ (a,b) $: If$ f(a)=f(b) $then$ \exists c\in(a,b):f'(c)=0 \$ Lagrange’s Mean Value Theorem The Mean Value Theorem (MVT) Lagrange’s mean value theorem (MVT) states that if a function f (x) is continuous on a closed interval [a,b] and differentiable on the open interval (a,b), then there is at least one point x = c on this interval, such that f (b) −f (a) = f ′(c)(b−a). SETS. G t 2t t2 t3 g t 2 t t 2 t 3 on 2 1 2 1 solution for problems 3 4 determine all the number s c which satisfy the conclusion of the mean value theorem for the given function and interval. Sure our content is useful, accurate and safe theorem means ( from the previous lesson is... 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http://math.stackexchange.com/questions/4274/area-of-a-recursive-function	# Area of a recursive function What is the area of the region bounded by $y = f_n(x)$ and the $x$-axis as $n$ gets large? $f_0=1-\|x\|$ $f_n=1-\|1-2f_{n-1}(x)\|$ Assume the domain of $f$ to be the real numbers. By trying various values for $n$ and $x$ I got that $f_n(x) = 2^{n}\cdot f_0(x)$ By integrating this I got that the area is given by $2^{n-1}\cdot x\cdot (2-\|x\|)$. However, I was also told that the area is in fact $1$ as $n$ approaches infinity. The formula I came up with (guessed) seems to check out if I integrate $2^{n-1}\cdot f_0(x)$ between certain values and for a certain $n$ and compare it to manually expanding that certain $f_n$ and integrating it between the same values. My questions are: 1. Is my formula wrong? 2. How can we prove the area is 1? - Is the domain of f defined? Or do we take -infinity to infinity? – Aryabhata Sep 8 '10 at 13:29 @Moron - take -inf to inf. – IVlad Sep 8 '10 at 13:39 Your formula seems wrong: f_1(0) is 0 while your formula gives 1. – Aryabhata Sep 8 '10 at 13:46 @Moron - true, I see it now. I just realized the formula was supposed to contain $n$ instead of $n - 1$, otherwise the integral doesn't match. Still wrong though. – IVlad Sep 8 '10 at 13:54 @IVlad: Are you sure it is -∞ to ∞ instead of -1 to 1? – KennyTM Sep 8 '10 at 15:13 You must be integrating between -1 and 1; otherwise, the integral diverges. Within this region, $f_n$ is a periodic sawtooth. This is obvious by following the steps geometrically for any one of the teeth: $-2f_{n-1}$ doubles the height of $f_{n-1}$ and flips the graph over the x-axis, which is then raised by $1$. The absolute value creates a new "tooth", and then subtracting all that from $1$ flips everything over. The process converts each tooth into a pair of teeth of the same height but half the width, whence the new teeth have half the original area of the teeth they have replaced. Two of them occupy the original area. Therefore, by induction, the integrals of all the $f_n$ between -1 and 1 equal the integral of $f_0$, which is 1.	2014-09-16T12:02:32	{ "domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/4274/area-of-a-recursive-function", "openwebmath_score": 0.9266265034675598, "openwebmath_perplexity": 516.0849316282995, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9706877700966098, "lm_q2_score": 0.8596637523076225, "lm_q1q2_score": 0.8344650907603703 }
http://math.stackexchange.com/questions/661615/whats-a-good-class-of-functions-for-bounding-comparing-ratios-of-complicated-lo	# What's a good class of functions for bounding/comparing ratios of complicated logarithms? I have this goofy series $\sum \limits_{n=2}^\infty \frac{ \log_2 \left[ n \log_2^2 n \right]}{n \log_2^2 n}$ that Wolfram Alpha tells me diverges by the comparison test (and indeed, in the larger problem I'm working on I need to prove that the expression containing it diverges), but I'm struggling to find a good divergent lower bound. I can't just throw away all the inner logarithms--then I run into the theorem $\lim \limits_{x \rightarrow \infty} \frac{\log_a x}{x^b} \rightarrow 0 \;\forall b > 0$. So I'm looking at things like $\frac{\log_2 \left[ n^3 \right]}{n \log_2^2 n}$ (larger, unfortunately), $\frac{\log_2 \left[ n^3 \right]}{n^2 \log_2 n}$ (converges), and $\frac{\log_2 \left[ n^2 \right]}{n \log_2^2 n}$ (still too large). Is there a more general class or form I can use to find a simple divergent lower bound, instead of stabbing in the dark? - Logarithms are really small, compared to any polynomial (or any positive power of $n$). So one very frequently useful trick is to try to ignore the logarithm, since it's not going to have a huge affect on the convergence of the series. More precisely, we have that $$n \log_2^2 n \ge n$$ by a little bit, so let's just test the series with $n$, rather than the complicated argument inside the logarithm. This leads to $$\sum_n \frac{\log_2 [n \log_2^2 n]}{n \log_2^2 n} \ge \sum_n \frac{\log_2 n}{n \log_2^2 n} = \sum_{n} \frac{1}{n \log_2 n} = \infty$$ Another thing that suggests this approach is to rewrite the numerator as $$\log_2 n + \log_2 \Big(\log_2^2 n\Big) = \log_2 n + 2 \log_2 \log_2 n$$ If $\log_2 n$ is small relative to $n$, then $\log_2 \log_2 n$ is tiny in comparison. This strongly suggests comparison to the series without this term. - Geez, I should have seen that. My head was still spinning from the amount of algebra it took to get everything else out of the way. Thanks. – Trevor Alexander Feb 3 at 6:03 Hope it's not too far off topic but would like to follow up @T.Bongers' answer by sharing my favourite comparison of logarithms and powers. (And it's too long to fit in a comment.) Consider the graph of $y=x^{1/10}$. You can calculate that if $x=1000000$ then $y$ is a bit less than $4$. Imagine drawing the graph on a scale of $1$ unit equals $1$ millimetre. Then although the graph is always increasing, it's like walking up a hill (if that's the right word) which climbs $4$ millimetres in a kilometre. This is clearly a climb which you would not even notice - you would think it was totally flat. So, $y=x^{1/10}$ increases incredibly, almost unimaginably, slowly... ...but $y=\ln x$ increases even more slowly than that!! Certainly $\ln x$ is larger than $x^{1/10}$ for small values of $x$. If you draw both graphs on the same axis for say $1\le x\le1000$, you will get a very misleading impression - eventually, $x^{1/10}$ takes over and is larger than $\ln x$. How long does it take for this to happen? Well, with a bit of trial and error you can find that the crossover point is approximately $x=4\times10^{15}$. If you (try to) draw this at a scale of $10$ centimetres for a unit - suitable for a blackboard in class - you find that the crossover point is about $3000$ times further away than the sun! Hope this helps to illustrate how unimaginably slowly (I didn't say "almost" this time) the logarithm function grows. - That's a good one. Closer to home for me is visual/auditory senses, which give you great sensitivity at low ranges without causing a seizure when a truck passes :) – Trevor Alexander Feb 3 at 23:20 Lemma $\displaystyle \sum_{n=2}^{\infty} \dfrac{\log_2 n}{n}$ diverges. Proof By comparison to $\displaystyle \sum_{n=2}^{\infty} \dfrac{1}{n}$. For $n\geq 2$, $\log_2 n \geq 1$. So $\dfrac{\log_2 n}{n} \geq \dfrac{1}{n}$. Since $\displaystyle \sum_{n=2}^{\infty} \dfrac{1}{n}$ diverges, $\displaystyle \sum_{n=2}^{\infty} \dfrac{\log_2 n}{n}$ diverges. -	2014-12-23T01:51:31	{ "domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/661615/whats-a-good-class-of-functions-for-bounding-comparing-ratios-of-complicated-lo", "openwebmath_score": 0.8169580101966858, "openwebmath_perplexity": 412.8642100240321, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9706877684006775, "lm_q2_score": 0.8596637505099168, "lm_q1q2_score": 0.834465087557428 }
http://math.stackexchange.com/questions/157634/combinatorics-of-coin-tosses	# combinatorics of coin tosses 12 tosses of a fair coin result in an outcome of 6 heads and 6 tails. How many sequences of such a result can have at most 4 successive outcomes of one kind ? I have obtained what I believe is the correct answer as 924 - 84 = 840, by considering various sequences to be excluded (a) 6-6 of a kind (b) 6-5 of a kind (c) 5-5 of a kind (d) only one 6 of a kind (e) only one 5 of a kind However, this seems a very crude and tedious method. I am seeking help whether this can be done by inclusion-exclusion, or by some other slick way. - Here is yet another way. It is still applicable when the number of H's and T's are changed or the maximal number $4$ in the question is changed: Given a sequence of H's and T's, for example HTHHHHTTHTTT, note that H's and T's are partitioned. In the example, the (ordered) partition $p_H$ for H's is $1 + 4 + 1$, whereas it is $p_T = 1 + 2 + 3$ for T's. Note that the number of parts $l(p_H)$ and $l(p_T)$ of $p_H$ and $p_T$ are either the same or differ by 1. In the example, they are both 3. If they are the same, there are two possible ways to interlace H's and T's. If they differ by 1, there is only one way. So, consider the partitions of 6. 6 = 6 = 5 + 1 = 4 + 2 = 3 + 3 = 4 + 1 + 1 = 3 + 2 + 1 = 2 + 2 + 2 = 3 + 1 + 1 + 1 = 2 + 2 + 1 + 1 = 2 + 1 + 1 + 1 + 1 = 1 + 1 + 1 + 1 + 1 + 1 Taking the order into account, there are 3 partitions having 2 parts, 10 partitions having 3 or 4 parts, 5 partitions having 5 parts and finally 1 partition having 6 parts, satisfying the criterion that no TTTTT and HHHHH must show up. For $l(p_H), l(p_T)$ = 2, 2 we have $2.3.3 = 18$ sequences For $l(p_H), l(p_T)$ = 2, 3 or 3, 2 we have $2.3.10 = 60$ sequences For $l(p_H), l(p_T)$ = 3, 3 we have $2.10.10 = 200$ sequences For $l(p_H), l(p_T)$ = 3, 4 or 4, 3 we have $2.10.10 = 200$ sequences For $l(p_H), l(p_T)$ = 4, 4 we have $2.10.10 = 200$ sequences For $l(p_H), l(p_T)$ = 4, 5 or 5, 4 we have $2.5.10 = 100$ sequences For $l(p_H), l(p_T)$ = 5, 5 we have $2.5.5 = 50$ sequences For $l(p_H), l(p_T)$ = 5, 6 or 6, 5 we have $2.5.1 = 10$ sequences and finally For $l(p_H), l(p_T)$ = 6, 6 we have $2.1.1 = 2$ sequences So there are $18 + 60 + 200 + 200 + 200 + 100 + 50 + 10 + 2 = 840$ sequences. - I really like the generalisation. I was unsuccessfully toying with the idea of using "stars and bars" somehow, which has materialised in an "interleaving" avatar, which can be used to work out inclusions or exclusions, as convenient. Thanking you and all others who have extended their valuable help. – true blue anil Jun 14 '12 at 13:00 Here's another solution. It's not very slick, but it works. First, let's find the number of arrangements with 5 or more consecutive heads. Consider the eight symbols: T, 5T, H,H,H,H,H,H. These can be put into 8!/6!=56 distinct orders. For 14 of these, the symbols T and 5T are neighbors. Now, for example if we expand 5T into TTTTT in either "5T T HHHHHH" or "T 5T HHHHHH" we get the same result. Compensating for that, the number of arrangements with 5 or more consecutive heads is 42+14/2 = 49. Similarly, the number of arrangements with 5 or more consecutive tails is 49. How many arrangements have both 5 or more consecutive heads and 5 or more consecutive tails? We can arrange the four symbols 5T,T,5H,H in 24 different ways. 1. For eight of these arrangements, the heads and tails are separated, e.g., T 5H 5T H. Each of these counts once when expanded. 2. For eight of these arrangements, either heads or tails are separated but not both, e.g., T 5H H 5T. Each of these counts twice when expanded. 3. For eight of these arrangements, neither the heads nor tails are separated, e.g., H 5H 5T T. Each of these counts four times when expanded. Therefore, the number of arrangements with both 5 or more consecutive heads and 5 or more consecutive tails is 8/1+8/2+8/4=14. - Relooking at your approach, i find that it can be generalised for n heads, n tails with a max of (n-2) succesive outcomes of a kind as – true blue anil Jun 19 '12 at 6:59 (continued) C(2n,n) - 2(n+1)^2 + 14, which makes this approach extremely useful. Thanks ! – true blue anil Jun 19 '12 at 7:09 I'm glad that you find it useful. It's a fun problem! – Byron Schmuland Jun 19 '12 at 12:35 Here's another way. Let $a_n$ be the number of sequences in $n$ tosses of a coin in which there are at most 4 consecutive outcomes of one kind. Then by simple counting $a_1=2$, $a_2=4$, $a_3=8$, $a_4=16$, and $a_5=30$. With a little thought you can get the recurrence relation $$a_n=2a_{n-1}-a_{n-5}$$ Now there are two ways you can go. You can just use this recurrence to grind out $a_6=2\times30-2=58$, $a_7=2\times58-4=112$, and so on until you get up to $a_{12}$. Or you can use standard techniques to solve the (homogeneous, linear, constant-coefficient) recurrence, but I think it gets messy for this one. Well, that's what computers are for! EDIT: OP has remembered that the outcome is to be 6 heads and 6 tails, so the above is irrelevant. So let's do inclusion-exclusion instead. We start with $12\choose6$. We subtract all the ways of having 5 (or more) consecutive heads. That's 56, because the first of the 5 consecutive heads can be in any one of 8 locations, and, once you've fixed the first of the 5 consecutive heads, you've got 7 remaining tosses, of which 1 is supposed to be heads. There are also 56 ways of having (at least) 5 consecutive tails; subtract those as well. Now add back in all the ways of having two instances of 5 consecutive heads. The only way for that to happen is to have 6 consecutive heads, and there are 7 ways for that to happen. Similarly, 7 ways to have two instances of 5 consecutive tails. Also add in all ways of having 5 consecutive heads and 5 consecutive tails. If the heads are 1 through 5, there are 3 places for the tails, and 2 ways to fill in the other two slots; if the heads are 2 through 6, 2 places for tails, 2 ways to fill; heads 3 through 7, 1 place for tails, 2 ways to fill, making 12 ways with the runs of heads preceding the run of tails; another 12 the other way around. Now we have to subtract out all the ways that meet three conditions. By the way, if you're getting the impression that there's nothing slick about inclusion-exclusion, at least for this problem, you're right, but it's what the customer wanted, and we've come this far, so let's go for it. We can have 6 consecutive heads and (at least) 5 consecutive tails, 4 ways (heads starting at 1, 2, 6, or 7), or 6 consecutive tails and 5 (or more) consecutive heads, another 4 ways. Finally, we have to add back in the 2 ways of meeting four conditions; 6 heads followed by 6 tails, or the other way around. So the answer would appear to be $${12\choose6}-56-56+7+7+12+12-4-4+2$$ whatever that is. MORE EDIT: I figured out where I went wrong. In the next-to-last case, meeting three conditions, 6 heads and 5 (or more) tails can happen 6 ways, not 4. With the 6 heads starting at 1, the 5 tails can start at 7 or 8, and with the six heads starting at 7, the five tails can start at 1 or 2. Add in the heads starting at 2 or at 6 and you get 6 ways, not 4. This makes the final calculation $${12\choose6}-56-56+7+7+12+12-6-6+2$$ which is 840, in agreement with everyone else. - The procedure you suggest gives an answer of 2980, whereas all possible sequences are only C(12,6) = 924. There seems to be a communication gap somewhere ! – true blue anil Jun 13 '12 at 7:50 @Gerry Myerson : I'm so sorry. In the original question, I needed to mention that the outcome of the 12 tosses is 6 heads and 6 tails. – true blue anil Jun 13 '12 at 8:50 So, what are you waiting for? Edit that into the question! Now! – Gerry Myerson Jun 13 '12 at 10:08 @Gerry: I got a count of 840 against 844 that you are getting. Let me recheck. The answers should tally. – true blue anil Jun 13 '12 at 11:37 @Gerry: I have rechecked and get the same # for exclusions as before. (a) 6-6 of a kind:2 (b) 6-5 of a kind: 4 (c) 5-5 of a kind:8 (d) only one 6 of a kind: 6 (e) only one 5 of a kind: 64. 924 - 84- 840.The only count that is tedious is the last one. I had written "inclusion-exclusion or by some other slick way", hoping that I-E would be slick. I wonder if there is a slick way ? – true blue anil Jun 13 '12 at 12:22	2014-10-21T07:38:48	{ "domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/157634/combinatorics-of-coin-tosses", "openwebmath_score": 0.7606801390647888, "openwebmath_perplexity": 381.6023605919213, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9706877692486436, "lm_q2_score": 0.8596637451167997, "lm_q1q2_score": 0.8344650830513608 }
https://math.stackexchange.com/questions/2810715/spivaks-calculus-proofs-concerning-pascals-triangle	# Spivak's Calculus: Proofs concerning Pascal's Triangle Problem 3 of Chapter 2 in Spivak's Calculus poses 5 problems associated with Pascal's Triangle. The first of these asks you to prove that $\binom{n+1}{k}$ = $\binom{n}{k-1} + \binom{n}{k}$ which was fairly straightforward. The second task was to prove by induction that $\binom{n}{k}$ is always a natural number. For this, I took the recursion approach: given the demonstration of the equality above, any $\binom{n}{k}$ can be decomposed into $\binom{n-1}{k-1} + \binom{n-1}{k}$ and so on, with each value decomposing until every binomial reaches either an $n$ or $k$ value of $0$, at which point they evaluate to 1 and can be summed. $1$ is a natural number, so the sum of any number of $1$s must therefore be a natural number. This proof seems to me as though it lacks formality, but I think the logic is appropriate. The third task, however, I am simply unsure what to make of. It reads: "Give another proof that $\binom{n}{k}$ is a natural number by showing that $\binom{n}{k}$ is the number of sets of exactly $k$ integers each chosen from $1, \cdots,n$." The word induction is not used in the prompt, and I can see that this appears to be true across the first few rows of the triangle, but how one would rigorously prove something like this, I do not know. Additionally, having proven that this is true, is the link from that to $\binom{n}{k}$ always being a natural number just the fact that you cannot have partial or negative sets, and therefore it must be a natural number? • I think the exercise want you to prove that $c(n,k)$ the number of sets of exactly $k$ integers chosen among $n$ follows the same induction formula, i.e. $c(n,k)=c(n-1,k-1)+c(n-1,k)$ and since you already noticed it takes the same values in the first rows of the triangle then you deduce $c(n,k)=\binom{n}{k}$. – zwim Jun 6 '18 at 22:49 By definition ${n\choose k}=\frac{n!}{k!(n-k)!}=\frac{n\cdot (n-1)\cdot\ldots\cdot (n-k+1)}{k!}=\frac{n}{k}\cdot\frac{n-1}{k-1}\cdot\ldots\cdot\frac{n-k+1}{1}$. Now consider $\{1,\ldots,n\}$. There are $n$ possibilities to pick an integer from this set. Let this integer be called $a_1$. Now there are $n-1$ possibilities to pick an element from the set $\{1,\ldots,n\}\setminus \{a_1\}$. By iterating this procedure we obtain $n\cdot (n-1)\cdot\ldots\cdot (n-k+1)$ possibilities to choose $(a_1,\ldots,a_k)$ from $\{1,\ldots,n\}$. Now note that we counted some possibilities multiple times, for example $a_1=1,a_2=2$ and $a_1=2,a_2=1$ result to the same integers chosen. So we have to divide by the number of possibilities to order $k$ elements which is $k!$ (there are $k$ possibilities to place the first integer, $k-1$ to place the second etc.). • Very helpful, thank you. Is there some any special study I should undertake to develop the ability to notice this, or does it fall under the general heading of 'mathematical maturity' which one acquires through sheer experience? – user242007 Jun 6 '18 at 23:11 • Well I think the best way to solve these kind of problems is to first work out the combinatorical side (here the number of possibilities to choose $k$ elements from $\{1,\ldots,n\}$) and then try to rearrange the left side to find this number. – Nightgap Jun 6 '18 at 23:26 • To the proposer: The topic of this Answer is Finite Combinatorics. There is a nice book called Finite Mathematics by Kemeny, Snell, & Thompson with many different topics (including combinatorics) utilizing a variety of techniques. Example: An election with 2 candidates A & B..... "A" won by $m$ votes to $n.$ When the ballots are counted, one at a time, what are the odds that A is always ahead during the count? – DanielWainfleet Jun 7 '18 at 0:35 Was writing out something along the lines of what's already so nicely explained by @Nightgap :), but just for good measure and I hope you don't mind that it's not really an answer to your specific question posed above, will still post what was going to be an addendum which has a different flavour and is nice to know if you haven't seen it (came across it in Alan Baker's Concise Intro to Number Theory aeons ago - great little book!). So let $\lfloor x\rfloor$ be the largest integer $\leq x$. Let $p$ be a prime, then there $\lfloor n/p \rfloor$ elements of the set $\{1,2,...n\}$ divisible by $p$, $\lfloor n/p^2 \rfloor$ divisible by $p^2$, and so on. This means that the highest power $k$ of $p$ such that $p^k \|n!$ is given by:$$k=\sum_{j=1}^{\infty}\lfloor n/p^j\rfloor \qquad \qquad (1)$$ Essentially you're adding to the sum each time you go up a prime power, so eg first you pick up all the unit powers with the term $\lfloor n/p \rfloor$, but then you need to add to these the extra contribution of the prime square multiples with the $\lfloor n/p^2 \rfloor$ term, then for the cubes, and so on. So now we have an explicit expression for the maximal power of a prime $p^k$ such that it divides $n!$ Now note that in general: $$\lfloor x+y \rfloor \geq \lfloor x\rfloor + \lfloor y \rfloor$$ (eg if the fractional parts of $x$ and $y$ add up to an integer), hence: $$\lfloor n/p^j \rfloor \geq \lfloor (n-k)/p^j \rfloor + \lfloor k/p^j \rfloor \qquad \qquad (2)$$ But then: $${n \choose k}=\frac{n!}{k!(n-k)!}$$ ...must be a natural number since by (1) and (2) all the prime powers in the denominator will be 'absorbed' by those in the numerator. Hope that makes sense and again though not how you would answer your exercise I think is a nice perspective to have on binomial coefficients, in addition to the all important combinatorial one. Cheers. • An interesting alternative look at the problem. It's going to take some time to digest this, but I greatly appreciate you taking the time to offer another angle. – user242007 Jun 7 '18 at 0:18 • No problem! Like I say you can find it in A Concise Introduction to Number Theory by Alan Baker, well worth checking out if you have the time, but for me the above was a pretty powerful aid to understanding the structure of binomial coefficients from a number theoretic perspective. – Mehness Jun 7 '18 at 0:20 • Just to add the above proof also immediately shows that multinomial coefficients $\frac{n!}{k_1! \cdot k_2! \cdots k_r!}$ where $\sum_{i=1}^r k_i = n$ are also natural numbers – Mehness Jun 7 '18 at 10:31 For convenience, let $S_n = \{x_1, x_2, x_3, \dots, x_n\}$ represent a set of $n$ distinct objects. Lets use $_nC_k$ to represent the number of ways of selecting $k$ objects from $S_n$. So what does $_{n+1}C_k$ look like? An element of such a selection will either include the object $x_{n+1}$ or it will not. We already know that there are $_nC_k$ ways to choose $k$ objects from the objects $x_1, x_2, \dots, x_n$. So how do we count the number of ways to select $k$ objects that contain $x_{n+1}$. Easy; we select $k-1$ objects from $S_n$ {there are $_nC_{k-1}$ such ways} and then add $x_{n+1}$ to the collection. It follows that $_nC_{k-1} + _nC_k = _{n+1}C_k$.	2019-06-18T20:39:40	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2810715/spivaks-calculus-proofs-concerning-pascals-triangle", "openwebmath_score": 0.8170294165611267, "openwebmath_perplexity": 172.27279466412645, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9706877675527112, "lm_q2_score": 0.8596637451167997, "lm_q1q2_score": 0.8344650815934292 }
https://math.stackexchange.com/questions/342128/for-every-integer-n-15-mid-n-iff-3-mid-n-and-5-mid-n	# For every integer $n$, $15\mid n$ iff $3\mid n$ and $5\mid n$ I'm trying to prove that for every integer $n$, $15\mid n$ iff $3\mid n$ and $5\mid n$. The first part of this bi-conditional was easy for me to prove, but I'm having problems with the second. Here is what I have so far: Suppose $15\mid n$. Then we can let $k$ be some integer such that $15k=n$. Since $3(5k)=15k=n, 3\mid n$. Also, since $5(3k)=15k=n$, it follows that $5\mid n$. Now I'm stuck trying to prove it the other way. I've assumed that for some integers $k$ and $j$, $3k=n$ and $5j=n$, and am trying to come up with something of the form $15p=n$ for some integer $p$, but I can only seem to come up with fractions, which seems wrong. Any hints? It is a good start, not much more is needed. You have $3k=n$ and $5j=n$. It follows that $3k=5j$. Note that $3k=5j$ in particular says that $3$ divides $5j$. It is a standard fact of number theory that if a prime $p$ (here $p=3$) divides a product $ab$, then $p$ divides $a$ or $p$ divides $b$ (or both). Clearly $3$ does not divide $5$. So $3$ divides $j$. Let $j=3w$. Then $n=5j=(5)(3w)=15w$. Remark: $1.$ The fact that if a prime $p$ divides $ab$, then $p$ divides $a$ or $p$ divides $b$ is sometimes called Euclid's Lemma. Books VII to IX, and to a fair degree Book X of Euclid's Elements are number-theoretic in character. $2.$ We do not really need Euclid's Lemma in this case. For we only want to show that if $3$ divides $5j$, then $3$ divides $j$. If $3$ does not divide $j$, then $j$ leaves a remainder of $1$ or $2$ on division by $3$. In either case, you can show that the remainder when $5j$ is divided by $3$ is not $0$. • Ahh, Thanks! This problem is out of How to Prove it by Velleman, and I haven't actually had a chance to start studying number theory. I'll definitely remember this, thanks! – stochasm Mar 26 '13 at 22:07 • You are welcome. I added in the remarks a little something about doing it without euclid's Lemma, or more precisely proving euclid's Lemma in the case $p=3$. So we can actually write a detailed proof with no reference to other stuff. – André Nicolas Mar 26 '13 at 22:20 Hint $\rm\ \ 5,3\mid n\:\Rightarrow\: \dfrac{n}5,\,\dfrac{n}3\in\Bbb Z\ \Rightarrow\ 2\left(\dfrac{n}5\right)-\dfrac{n}3 \, =\, \dfrac{n}{15}\in\Bbb Z\:\Rightarrow\: 15\mid n$ $d \mid n$ means $\exists k, dk = n$. So we want to prove that $\exists k, 15k = n$ iff $\exists h,l, 5h=3l=n$. To do one direction write $15 = 3 \cdot 5$: there $\exists k, 3 \cdot 5 \cdot k = n$ so $5(3k)=3(5k)=n$ proves the implication with $h=3k$, $l=5k$. In the other direction, we have $5h=3l=n$ so $3\mid h$ (and $5 \mid l$), put $k = h/3$ and you get $5 \cdot 3 \cdot k = n$. Since $3\|n$ then $n=3k$. But you also know $5\|n$, that is, $5\|3k$. Now, here use the fact that $5\not\|3$. By using that fact you get that $5\|k$, so $k=5j$. Therefore $n=3k=3(5j)=15j$. Now $15$ divides $3n$ and $5n$, and $n = 10n - 9n = 2 \cdot (5n) - 3 \cdot (3n)$ so.. This is basically using Bezout and the fact that $3$ and $5$ are coprime. When $a$ and $b$ are coprime, $xa + yb = 1$ for some integers $x$ and $y$. From this a more general theorem can be proven. If $a$ and $b$ are coprime integers, both dividing $n$, then $n$ is divisible by $ab$. Proof: $an$ and $bn$ are divisible by $ab$, so $n = xan + ybn$ is divisible by $ab$. Your problem is the case $a = 3$, $b = 5$.	2019-11-15T21:12:50	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/342128/for-every-integer-n-15-mid-n-iff-3-mid-n-and-5-mid-n", "openwebmath_score": 0.978648841381073, "openwebmath_perplexity": 68.53250664589433, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9728307708274401, "lm_q2_score": 0.8577681031721325, "lm_q1q2_score": 0.8344632050001368 }
https://math.stackexchange.com/questions/2205612/surface-area-of-a-sphere-with-integration-of-disks	# Surface area of a sphere with integration of disks [duplicate] Why it is not correct to say that the surface area of a sphere is: $$2 \int_{0}^{R} 2\pi r \text{ } dr$$ In my mind we are summing up the perimeters of disks from $r=0$ to $r=R$, so by 1 integration, we would have $\frac{1}{2}$ of the surface area of the sphere. I know that it's not correct because that will give us $2\pi R^2$ that it's different from $4\pi R^2$, but why??? Thanks! • It's bad form, and technically incorrect, to have one of the limits of integration depend on the variable of integration. What does it mean to say "to $r=r$"? Isn't $r=r$ always true since $r$ varies between $0$ and some fixed positive constant? It'd be better to call this constant $R$ to keep it separate from the variable $r$. This is more of a side note. – user307169 Mar 27 '17 at 17:37 • @tilper fixed It, thanks – Bruno Reis Mar 27 '17 at 17:41 • If your method was correct, then you should get the same answer for a cone whose base radius and height are equal. But you don't, do you? – uniquesolution Mar 27 '17 at 17:54 In order to give a short explanation of the mistake : In writing $\quad 2 \int_{0}^{R} 2\pi r \text{ } dr \quad$ you sum elementary areas $\quad dA=2\pi\:r\:dr\quad$ that is to say, a strip length$=2\pi r$ and width$=dr$. The hitch is that the width is not $dr$ but is $ds$ (see the figure). With a few calculus, you can find $\quad ds=\frac{R}{\sqrt{R^2-r^2}}dr \quad\to\quad dA=2\pi\:r\frac{R}{\sqrt{R^2-r^2}}dr$ $$\quad 2 \int_{0}^{R} 2\pi\:r\frac{R}{\sqrt{R^2-r^2}}dr =4\pi R^2$$ This is a method to find the area of the sphere. Of course, a simpler method consists in doing the job in spherical coordinates instead of Cartesian coordinates. • Perfect explanation! Thanks! – Bruno Reis Mar 27 '17 at 23:22 What you are doing is indeed integrating over rings, but then you get the area of a full circle, which was the result of your calculation up to the irrelevant factor of 2. You can imagine that by adding more and more dense rings you fill a circle with radius $R$ (the upper limit of your integral). But to get the area of a sphere, you need to parametrize all 3 dimensions. What you can do is use spherical coordinates, where each point in space is described by a radius $r$, and two angles $\theta,\phi$ as in the link. In fact, to get the area of the sphere you need to keep the radius constant, and integrate over the angles that parametrize the given sphere. The result yields $$S=\int_0^{\pi}\int_0^{2\pi}R^2\sin\theta d\phi d\theta=R^22\pi\int_0^{\pi}\sin\theta d\theta=4\pi R^2$$ as it should be. A) when you pile up the disks you shall pay attention to law with which $r$ varies (with $h$, $x$, $\alpha$ $\cdots$) otherwise, as already told you do not make any distinction with a cone, a cylinder,.. B) adding disks to find volumes is ok, because the volume of the triangular cross-section ring that is left-out is a higher order infinitesimal vs. the volume of the disk; its surface instead is of the same order, and you must consider the hypothenuse instead of the cathetus. $$\text{Area will be}~~: 2 \int_{0}^{R} 2\pi x ~ ds$$ Where $ds$ is width of strip bounded by circles of radius $x$ and $x+dx$ situated at height $y$. Also $ds \neq dr$ it's tilted in $y$ direction too. Only horizontal projection of $ds$ is $dr$.What you have done is valid for Disk See image below. $$(ds)^2=(dx)^2+(dy)^2$$ Therefore : $$ds = \sqrt{1+\Bigg( \frac{dy}{dx}\Bigg)^2}$$ Now, $$x^2 + y^2 = R^2$$ $$y = \sqrt{R^2 - x^2}$$ $$\dfrac{dy}{dx} = \dfrac{-2x}{2\sqrt{R^2 - x^2}}$$ $$\dfrac{dy}{dx} = \dfrac{-x}{\sqrt{R^2 - x^2}}$$ $$\left( \dfrac{dy}{dx} \right)^2 = \dfrac{x^2}{R^2 - x^2}$$ Thus, $$\displaystyle A = 4\pi \int_0^R x \sqrt{1 + \dfrac{x^2}{R^2 - x^2}} \, dx$$ $$\displaystyle A = 4\pi \int_0^R x \sqrt{\dfrac{(R^2 - x^2) + x^2}{R^2 - x^2}} \, dx$$ $$\displaystyle A = 4\pi \int_0^R x \sqrt{\dfrac{R^2}{R^2 - x^2}} \, dx$$ Let $$x = R \sin θ \implies dx = R \cos θ dθ$$ When $x = 0, θ = 0$ When $x = R, θ = \pi/2$ Thus, $$\displaystyle A = 4\pi \int_0^{\pi/2} R \sin \theta \sqrt{\dfrac{R^2}{R^2 - R^2 \sin^2 \theta}} \, (R \cos \theta \, d\theta)$$ $$\displaystyle A = 4\pi \int_0^{\pi/2} R^2 \sin \theta \cos \theta\sqrt{\dfrac{R^2}{R^2(1 - \sin^2 \theta)}} \, d\theta$$ $$\displaystyle A = 4\pi R^2 \int_0^{\pi/2} \sin \theta \cos \theta\sqrt{\dfrac{1}{\cos^2 \theta}} \, d\theta$$ $$\displaystyle A = 4\pi R^2 \int_0^{\pi/2} \sin \theta \cos \theta \left( \dfrac{1}{\cos \theta} \right) \, d\theta$$ $$\displaystyle A = 4\pi R^2 \int_0^{\pi/2} \sin \theta \, d\theta$$ $$A = 4\pi R^2 \bigg[-\cos \theta \bigg]_0^{\pi/2}$$ $$A = 4\pi R^2 \bigg[-\cos \frac{1}{2}\pi + \cos 0 \bigg]$$ $$A = 4\pi R^2 \bigg[ -0 + 1 \bigg]$$ $$A = 4\pi R^2$$ If you consider a formula like $$dA = 2 \pi r \, dr$$ for a cylinder, it is zero and so fails. If you consider a formula for a cylinder, $$dA = 2 \pi r \, dz$$ it seems to be correct. However, for a cone $dz=0$ so even this formula fails. Intuition says $dr,dz$ should both be involved. In fact if you recall cone formula $A = \pi\,r\, s = 2 \pi\, \bar {r} \, s ,\, dA= 2\pi r ds$ slant generator is correct in principle. We have cone differential area $$dA = 2 \pi r ds = 2 \pi r \sqrt{dr^2+dz^2 }$$ and that should be considered in integration of all surfaces of revolution. There are two problems here. First: Not only do your disks change radius from the apex to the base of the hemisphere, but they do so at a varying rate. Let's say (for concreteness) that the sphere is $$x^2 + y^2 + z^2 = R^2$$, and we're dividing the sphere into disks based on the $$x$$-coordinate. If you try to approximate the surface area of a hemisphere of radius $$R=1$$ by slicing it into disks of radius (say) $$0.1, 0.2, \ldots, 0.9, 1$$, you'll find that the first few small-radius slices are very close together— to be precise, at $$x = \sqrt{1 - 0.1^2} = 0.994$$, $$\sqrt{1-0.2^2} = 0.980$$, $$\sqrt{1-0.3^2} = 0.954$$, and so forth—but the larger slices are much farther apart. If you use the surface areas of these disks to calculate the surface area of the sphere, you have to take into account the fact that the disks have different widths. Going from finite to infinitesimal width: The radius of a disk at $$x$$ is $$r = \sqrt{R^2-x^2}$$ (the inverse of this formula is $$x = \sqrt{R^2 - r^2}$$), so if you make cuts in the sphere each time the radius changes by $$dr$$, then the slices will have width $$-dx = -dr \dfrac{dx}{dr} = \dfrac{r}{\sqrt{R^2-r^2}}\, dr$$ (the negative sign corrects for the fact that $$x$$ increases with decreasing $$r$$). Second: The surfaces of the sphere slices that we discuss above have sloping cross-sections (in the $$xy$$ plane or any plane that includes the $$x$$-axis), whereas cylinders have flat surface cross-sections. You can ignore the difference this makes for calculating volume, but not for surface area; otherwise you can prove that $$\pi = 4$$. So you need to make another correction: the width of the surface of a disk isn't $$dx$$ but rather $$\sqrt{(dx)^2 + (dr)^2} = \sqrt{ \dfrac{r^2}{R^2 - r^2} + 1}\, dr = \dfrac{R}{\sqrt{R^2 - r^2}}\, dr$$. Putting this all together and integrating over $$r$$ (though $$x$$ would be nicer): The integral that you actually have to do is thus \begin{align} &\int_0^R (2\pi r) \frac{R}{\sqrt{R^2 - r^2}}\, dr \\ =& \;2\pi R \int_{r=0}^{r=R} \frac{-du/2}{\sqrt{u}} \tag{u = R^2 - r^2; du = -2r\, dr} \\ =& -2\pi R \left. \sqrt{u} \right\|_{r=0}^{r=R} \\ =& -2\pi R \left. \sqrt{R^2 - r^2}\right\|_{r=0}^{r=R} \\ =&\;2\pi R^2 \end{align} and this is the result you want.	2021-02-26T02:26:49	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2205612/surface-area-of-a-sphere-with-integration-of-disks", "openwebmath_score": 0.9731870293617249, "openwebmath_perplexity": 289.36251024840954, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9728307661011976, "lm_q2_score": 0.8577681068080749, "lm_q1q2_score": 0.8344632044832734 }
https://vioora.com/hugh-frater-qochpg/f12ce6-multiplying-complex-numbers-in-rectangular-form	# multiplying complex numbers in rectangular form (5 + j2) + (2 - j7) = (5 + 2) + j(2 - 7) = 7 - j5 (2 + j4) - (5 + j2) = (2 - 5) + j(4 - 2) = -3 + j2; Multiplying is slightly harder than addition or subtraction. Ask Question Asked 1 year, 6 months ago. Viewed 385 times 0 $\begingroup$ I have attempted this complex number below. Addition of Complex Numbers . Here are some specific examples. Find the complex conjugate of the denominator, also called the z-bar, by reversing the sign of the imaginary number, or i, in the denominator. As imaginary unit use i or j (in electrical engineering), which satisfies basic equation i 2 = −1 or j 2 = −1.The calculator also converts a complex number into angle notation (phasor notation), exponential, or polar coordinates (magnitude and angle). Now, let’s multiply two complex numbers. The standard form, a+bi, is also called the rectangular form of a complex number. In other words, to write a complex number in rectangular form means to express the number as a+bi (where a is the real part of the complex number and bi is the imaginary part of the complex number). So just remember when you're multiplying complex numbers in trig form, multiply the moduli, and add the arguments. How to Write the Given Complex Number in Rectangular Form". The major difference is that we work with the real and imaginary parts separately. The rectangular form of a complex number is written as a+bi where a and b are both real numbers. Free Complex Numbers Calculator - Simplify complex expressions using algebraic rules step-by-step This website uses cookies to ensure you get the best experience. if you need any other stuff in math, please use our google custom search here. Free Complex Number Calculator for division, multiplication, Addition, and Subtraction. Complex numbers in the form are plotted in the complex plane similar to the way rectangular coordinates are plotted in the rectangular plane. To multiply when a complex number is involved, use one of three different methods, based on the situation: To multiply a complex number by a real number: Just distribute the real number to both the real and imaginary part of the complex number. Plot each point in the complex plane. and x − yj is the conjugate of x + yj.. Notice that when we multiply conjugates, our final answer is real only (it does not contain any imaginary terms.. We use the idea of conjugate when dividing complex numbers. This calculator does basic arithmetic on complex numbers and evaluates expressions in the set of complex numbers. Multipling and dividing complex numbers in rectangular form was covered in topic 36. Complex numbers can be expressed in numerous forms. Label the x-axis as the real axis and the y-axis as the imaginary axis. Hence the Re (1/z) is (x/(x2 + y2)) - i (y/(x2 + y2)). However, due to having two terms, multiplying 2 complex numbers together in rectangular form is a bit more tricky: (3z + 4zbar â 4i) = [3(x + iy) + 4(x + iy) bar - 4i]. Yes, you guessed it, that is why (a+bi) is also called the rectangular form of a complex number. Recall that FOIL is an acronym for multiplying First, Outer, Inner, and Last terms together. A1. 1. Multiplication . 1. In essence, the angled vector is taken to be the hypotenuse of a right triangle, described by the lengths of the adjacent and opposite sides. Example 7 MULTIPLYING COMPLEX NUMBERS (cont.) It is the distance from the origin to the point: See and . Simplify. To understand and fully take advantage of multiplying complex numbers, or dividing, we should be able to convert from rectangular to trigonometric form and from trigonometric to rectangular form. Math Gifs; Algebra; Geometry; Trigonometry; Calculus; Teacher Tools; Learn to Code; Home; Algebra ; Complex Numbers; Complex number Calc; Complex Number Calculator. Note that the only difference between the two binomials is the sign. Find powers of complex numbers in polar form. Real numbers can be considered a subset of the complex numbers that have the form a + 0i. The different forms of complex numbers like the rectangular form and polar form, and ways to convert them to each other were also taught. So 18 times negative root 2 over. In polar form, the multiplying and dividing of complex numbers is made easier once the formulae have been developed. This is an advantage of using the polar form. Find the complex conjugate of the denominator, also called the z-bar, by reversing the sign of the imaginary number, or i, in the denominator. 7) i 8) i Use rectangular coordinates when the number is given in rectangular form and polar coordinates when polar form is used. Polar form is where a complex number is denoted by the length (otherwise known as the magnitude, absolute value, or modulus) and the angle of its vector (usually denoted by … Included in the resource: 24 Task cards with practice on absolute value, converting between rectangular and polar form, multiplying and dividing complex numbers … Label the x-axis as the real axis and the y-axis as the imaginary axis. Example 4: Multiplying a Complex Number by a Real Number . To add complex numbers in rectangular form, add the real components and add the imaginary components. Let z 1 = r 1 cis θ 1 and z 2 = r 2 cis θ 2 be any two complex numbers. Using either the distributive property or the FOIL method, we get A complex number can be expressed in standard form by writing it as a+bi. $\text{Complex Conjugate Examples}$ $\$$3 \red + 2i)(3 \red - 2i) \\(5 \red + 12i)(5 \red - 12i) \\(7 \red + 33i)(5 \red - 33i) \\(99 \red + i)(99 \red - i) This is the currently selected item. Although the complex numbers (4) and (3) are equivalent, (3) is not in standard form since the imaginary term is written first (i.e. In the complex number a + bi, a is called the real part and b is called the imaginary part. Complex numbers in the form are plotted in the complex plane similar to the way rectangular coordinates are plotted in the rectangular plane. Multiplication of Complex Numbers. To add complex numbers, add their real parts and add their imaginary parts. Rectangular Form of a Complex Number. Rectangular Form of a Complex Number. This material appears in section 6.5. In other words, to write a complex number in rectangular form means to express the number as a+bi (where a is the real part of the complex number and bi is the imaginary part of the complex number). The primary reason for having two methods of notation is for ease of longhand calculation, rectangular form lending itself to addition and subtraction, and polar form lending itself to multiplication and division. This is an advantage of using the polar form. Change ), You are commenting using your Google account. Post was not sent - check your email addresses! The following development uses trig.formulae you will meet in Topic 43. In other words, there are two ways to describe a complex number written in the form a+bi: To write a complex number in rectangular form you just put it into the standard form of a complex number by writing it as a+bi. How to Divide Complex Numbers in Rectangular Form ? There are two basic forms of complex number notation: polar and rectangular. When in rectangular form, the real and imaginary parts of the complex number are co-ordinates on the complex plane, and the way you plot them gives rise to the term “Rectangular Form”. 2 and 18 will cancel leaving a 9. B2 ( a + bi) Error: Incorrect input. Addition, subtraction, multiplication and division can be carried out on complex numbers in either rectangular form or polar form. How do you write a complex number in rectangular form? Hence the value of Im(3z + 4zbar â 4i) is - y - 4. Here we are multiplying two complex numbers in exponential form. bi+a instead of a+bi). Multiplying by the conjugate . Subtraction is similar. Divide complex numbers in rectangular form. It will perform addition, subtraction, multiplication, division, raising to power, and also will find the polar form, conjugate, modulus and inverse of the complex number. The symbol ' + ' is treated as vector addition. https://www.khanacademy.org/.../v/polar-form-complex-number That’s right – it kinda looks like the the Cartesian plane which you have previously used to plot (x, y) points and functions before. If z = x + iy , find the following in rectangular form. Rectangular Form A complex number is written in rectangular form where and are real numbers and is the imaginary unit. Multiplication of Complex Numbers. This can be a helpful reminder that if you know how to plot (x, y) points on the Cartesian Plane, then you know how to plot (a, b) points on the Complex Plane. 5) i Real Imaginary 6) (cos isin ) Convert numbers in rectangular form to polar form and polar form to rectangular form. In polar form, the multiplying and dividing of complex numbers is made easier once the formulae have been developed. Find (3e 4j)(2e 1.7j), where j=sqrt(-1). Answer. Note that all the complex number expressions are equivalent since they can all ultimately be reduced to -6 + 2i by adding the real and imaginary terms together. Let’s begin by multiplying a complex number by a real number. Multiplying a complex number by a real number is simple enough, just distribute the real number to both the real and imaginary parts of the complex number. You may have also noticed that the complex plane looks very similar to another plane which you have used before. polar form Introduction When two complex numbers are given in polar form it is particularly simple to multiply and divide them. To convert from polar form to rectangular form, first evaluate the trigonometric functions. In this lesson you will investigate the multiplication of two complex numbers v and w using a combination of algebra and geometry. Multiplying Complex Numbers Together. This video shows how to multiply complex number in trigonometric form. 18 times root 2 over 2 again the 18, and 2 cancel leaving a 9. Now that we can convert complex numbers to polar form we will learn how to perform operations on complex numbers in polar form. ; The absolute value of a complex number is the same as its magnitude. Products and Quotients of Complex Numbers; Graphical explanation of multiplying and dividing complex numbers; 7. Rectangular form, on the other hand, is where a complex number is denoted by its respective horizontal and vertical components. It is the distance from the origin to the point: See and . Rather than describing a vector’s length and direction by denoting magnitude and … How to Write the Given Complex Number in Rectangular Form : Here we are going to see some example problems to understand writing the given complex number in rectangular form. Fill in your details below or click an icon to log in: You are commenting using your WordPress.com account. What you can do, instead, is to convert your complex number in POLAR form: #z=r angle theta# where #r# is the modulus and #theta# is the argument. This screen shows how the TI–83/84 Plus displays the results found in parts (a), (b), and (d) in this example. That is, [ (a + ib)/(c + id) ] ⋅ [ (c - id) / (c - id) ] = [ (a + ib) (c - id) / (c + id) (c - id) ] Examples of Dividing Complex Numbers We sketch a vector with initial point 0,0 and terminal point P x,y . A complex number is a number of the form a + bi, where a and b are real numbers, and i is an indeterminate satisfying i 2 = −1.For example, 2 + 3i is a complex number. Draw a line segment from \(0$$ to $$z$$. d) Write a rule for multiplying complex numbers. Rectangular Form. z 1 z 2 = r 1 cis θ 1 . Multiplying a complex number by a real number is simple enough, just distribute the real number to both the real and imaginary parts of the complex number. We start with an example using exponential form, and then generalise it for polar and rectangular forms. The Complex Hub aims to make learning about complex numbers easy and fun. However, due to having two terms, multiplying 2 complex numbers together in rectangular form is a bit more tricky: Find quotients of complex numbers in polar form. Complex Number Lesson . Multiplication and division of complex numbers in polar form. Example 1. Therefore the correct answer is (4) with a=7, and b=4. When you’re dividing complex numbers, or numbers written in the form z = a plus b times i, write the 2 complex numbers as a fraction. For Example, we know that equation x 2 + 1 = 0 has no solution, with number i, we can define the number as the solution of the equation. Multiplying both numerator and denominator by the conjugate of of denominator, we get ... "How to Write the Given Complex Number in Rectangular Form". Complex numbers are numbers of the rectangular form a + bi, where a and b are real numbers and i = √(-1). I get -9 root 2. Notice the rectangle that is formed between the two axes and the move across and then up? Polar Form of Complex Numbers; Convert polar to rectangular using hand-held calculator; Polar to Rectangular Online Calculator ; 5. 2.3.2 Geometric multiplication for complex numbers. To write a complex number in rectangular form you just put it into the standard form of a complex number by writing it as a+bi. Converting a complex number from polar form to rectangular form is a matter of evaluating what is given and using the distributive property. Rectangular form, on the other hand, is where a complex number is denoted by its respective horizontal and vertical components. But then why are there two terms for the form a+bi? But complex numbers, just like vectors, can also be expressed in polar coordinate form, r ∠ θ . ( Log Out / To divide the complex number which is in the form (a + ib)/ (c + id) we have to multiply both numerator … It is no different to multiplying whenever indices are involved. To write complex numbers in polar form, we use the formulas and Then, See and . To multiply complex numbers in polar form, multiply the magnitudes and add the angles. ; The absolute value of a complex number is the same as its magnitude. Multiplication and division in polar form Introduction When two complex numbers are given in polar form it is particularly simple to multiply and divide them. B1 ( a + bi) A2. Complex Number Functions in Excel. Apart from the stuff given in this section "How to Write the Given Complex Number in Rectangular Form", if you need any other stuff in math, please use our google custom search here. To plot a complex number a+bi on the complex plane: For example, to plot 2 + i we first note that the complex number is in rectangular (a+bi) form. A complex number is a number of the form a + bi, where a and b are real numbers, and i is an indeterminate satisfying i 2 = −1.For example, 2 + 3i is a complex number. So the root of negative number √-n can be solved as √-1 * n = √ n i, where n is a positive real number. We know that i lies on the unit circle. Change ), You are commenting using your Twitter account. Show Instructions. Example 2(f) is a special case. A complex number can be represented in the form a + bi, where a and b are real numbers and i denotes the imaginary unit. Multiplying Complex Numbers. (This is because it is a lot easier than using rectangular form.) 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There are two basic forms of complex numbers when they 're in form... Polar coordinates when polar form. sorry, your blog can not share by... Without drawing vectors, we will learn how to multiply complex numbers is in! Sorry, your blog can not share posts by email Euler formula Euler! Numbers that have the form are plotted in the rectangular form. ( a +,. Do you Write a complex number by a real number just as we would with binomial! Determine which of the following development uses trig.formulae you will meet in 36... Followed by 1 unit up on the unit circle need some kind standard. Basic forms of complex numbers and evaluates expressions in the form a complex number we use formulas. The cartesian form of a complex number from polar form. the FOIL method complex Hub aims make... Section `, how to Write the given complex number is written as a+bi where a is called real. Is also called the real axis and the move across and then up convert from polar.! Was covered in topic 43 - 4 learning about complex numbers indices involved., and subtraction of complex numbers without drawing vectors, can also expressed...	2021-04-15T01:19:12	{ "domain": "vioora.com", "url": "https://vioora.com/hugh-frater-qochpg/f12ce6-multiplying-complex-numbers-in-rectangular-form", "openwebmath_score": 0.800094485282898, "openwebmath_perplexity": 429.39060653722373, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9728307700397331, "lm_q2_score": 0.8577681031721324, "lm_q1q2_score": 0.8344632043244667 }
https://math.stackexchange.com/questions/2179751/bounded-linear-function-on-ca-b-given-by-fx-xt-0/2179769	# Bounded linear function on $C[a,b]$ given by $f(x)=x(t_{0})$ I have the following question from a textbook. Let $f:C[a,\,b]\rightarrow\mathbb{R}$ be defined by $f(x)=x(t_{0})$ for a fixed $t_{0}\in[a,\,b]$. Show that $f$ is a bounded linear function and $\\|f\\|=1$. In order to show this I thought I should show that $f$ is continuous and use the following theorem Theorem: $f$ is continuous if and only if it is bounded. How ever am stuck on how to do this. Is there a way to do this with this technique? • What norm are you using for f? And by $f(x) = x(t_0)$, is this just multiplication by $t_0$? – Matt Mar 9 '17 at 21:11 • @Matt since the domain of $f$ is the set of continuous functions, $x$ is a function and $x(t_0)$ is the value of $x$ at the point $t_0$, i.e. $f$ could be called evaluation functional. – Roland Mar 9 '17 at 21:35 • @Roland thanks for clarifying! – Matt Mar 9 '17 at 22:37 • It would be good if you could cite the reference of the book! – BAYMAX Sep 14 '17 at 2:32 Lets look at $\\|f(x)\\|$ and try to express it (or bound) by using $\\|x\\| = \max_{t\in[a,b]}x(t)$. $$\\|f(x)\\| = \|x(t_0)\| \le \\|x\\| \Rightarrow \frac{\\|f(x)\\|}{\\|x\\|} \le 1$$ But there always exists a constant continous function $x_c(t)$ for which we have $\\|x_c\\| = x_c(t_0)$ and $$\\|f(x_c)\\| = \|x_c(t_0)\| = \\|x_c\\| \Rightarrow \frac{\\|f(x_c)\\|}{\\|x_c\\|} = 1$$ So, we have $$\\|f\\| = \sup_{x\in C[a,b]}\frac{\\|f(x)\\|}{\\|x\\|} = 1$$ • Won't we need to show that $$\|f(x)\|\leq M \|\|x\|\|$$ for some $M\in\mathbb{R}$? so as to show boundedness? – Mafeni Alpha Mar 12 '17 at 13:10 • @MafeniAlpha, the first inequality states exactly this – Andrei Kulunchakov Mar 12 '17 at 13:13 • I have edited the post - the part saying "$\\|f(x)\\| = x(t_0)$" was missing absolute value - I assume it was just a typo. (Of course, if you prefer to have $\\|x(t_0)\\|$ rather than $\|x(t_0)\|$, as to have the notation consistent with what is mostly used for normed spaces, certainly go ahead and edit the post to the form you are satisfied with. – Martin Sleziak Sep 15 '17 at 2:24	2020-05-28T16:03:11	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2179751/bounded-linear-function-on-ca-b-given-by-fx-xt-0/2179769", "openwebmath_score": 0.8878595232963562, "openwebmath_perplexity": 178.98689072571122, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9728307653134903, "lm_q2_score": 0.8577681068080749, "lm_q1q2_score": 0.8344632038076032 }
http://math.stackexchange.com/questions/634800/is-2-mathbbz-4-mathbbz-not-a-field/634805	# Is $2\mathbb{Z}/4\mathbb{Z}$ NOT a field?! According to wikipedia: If $R$ is a unital commutative ring with an ideal $m$, then $k = R/m$ is a field if and only if $m$ is a maximal ideal. In that case, $R/m$ is known as the residue field. This fact can fail in non-unital rings. For example, $4\mathbb{Z}$ is a maximal ideal in $2\mathbb{Z}$ , but $2\mathbb{Z}/4\mathbb{Z}$ is not a field. Is $2\mathbb{Z}/4\mathbb{Z}$ NOT a field?! Isn't it isomorphic to $\mathbb{Z_2}$? I mean every element in $2\mathbb{Z}$ is of the form $2k$ for some $k \in \mathbb{Z}$. Now suppose that I define $\psi: \mathbb{2Z} \to \mathbb{Z}_2$ where $2k \mapsto \bar{0}$ if $k$ is even and $2k \mapsto \bar{1}$ if $k$ is odd. This map is well-defined because we have only one representation as $2k$ for each element of $2\mathbb{Z}$. This map is surjective because at least $0$ is mapped to $\bar{0}$ and $2$ is mapped to $\bar{1}$. Is $\psi$ a homomorphism? Well, to me it sounds like it must be. Because the factor of $2$ is playing no role in here and only the parity of $k$ matters for the mapping so I think it is obvious that it is a homomorphism. Even though it can be checked directly by checking all 4 cases separately as well. So, what is the kernel? Isn't the kernel equal to $4\mathbb{Z}$? Doesn't that mean $2\mathbb{Z}/4\mathbb{Z}$ is isomorphic to $\mathbb{Z}_2$? It makes sense that these two are isomorphic to me, where am I wrong? Also, is the ideal correspondence theorem wrong for non-unital rings??? - Your map is not a ring homomorphism. If I take $a = 2$, $b = 6$, I have $\psi(2) = \psi(6) = 1$, but $\psi(12) = 0$. Likewise, $2\mathbf{Z}/4\mathbf{Z}$ is not an integral ring. - If you allow non-unital rings, you have a nice counterexample here telling you that not every ring is isomorphic to $\mathbf{Z}/2\mathbf{Z}$. :) – fkraiem Jan 11 at 16:30 No. As an abelian group it is indeed isomorphic, but as a ring it needn't be. $2\mathbb{Z}/4\mathbb{Z}$ has only 2 elements, 0 and $\bar{2}$. And $\bar{2}^2=0$, while in $\mathbb{Z}_2$ it isn't. – Giulio Belletti Jan 11 at 16:30 Yeah, I was unaware of the fact that there is more than one ring structure with two elements. Apparently, excluding $\mathbb{Z}_2$, there is at least another ring $R=\{0,1\}$ with the ring multiplication table such that $0.0=0$, $1.0=0$, $0.1=0$, $1.1=0$. I just checked it satisfies the ring axioms. I guess $2\mathbb{Z}/4\mathbb{Z}$ as a ring must be isomorphic to the non-unital ring with two elements that I just mentioned. – some1.new4u Jan 11 at 16:41 In general, given an abelian group, the multiplication defined by $ab = 0$ for all $a,b$ gives a non-unital ring, this is what you have here. – fkraiem Jan 11 at 16:44 @fkraiem: Thanks. Since earlier I had dealt with only commutative rings and it has become a tradition to consider only unital rings in commutative algebra courses I had never seen a non-unital ring before, so this was a delightful surprise for me today. Thanks for your help. – some1.new4u Jan 11 at 16:59 Hint $\ 2\Bbb Z/4\Bbb Z\,$ is not a field since $\,2^2 = 0.\,$ Your map $\psi$ is not a ring hom since they must map the nilpotent $\,2\,$ to a nilpotent ($= 0$ in a field), but you have $\,\psi(2)= 1\,$ by definition. Said in further detail $\ 0 = \psi(0) = \psi(2^2)= \psi(2)^2,\,$ so $\,\psi(2) = 0,\,$ being in a field (or domain). Remark $\$ Properties like nilpotent, idempotent, invertible etc. that are purely "ring-theoretic" are always preserved by ring homomorphisms. As above, generally this implies constraints on where these elements may be mapped when defining ring homs. -	2014-12-21T09:39:41	{ "domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/634800/is-2-mathbbz-4-mathbbz-not-a-field/634805", "openwebmath_score": 0.9533200860023499, "openwebmath_perplexity": 175.47470506732935, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9728307708274401, "lm_q2_score": 0.8577680995361899, "lm_q1q2_score": 0.83446320146298 }
https://math.stackexchange.com/questions/3358361/construct-a-function-with-each-derivative-being-non-differentiable-at-a-distinct	# Construct a function with each derivative being non-differentiable at a distinct point Construct a function $$f_1:\mathbb{R}\to\mathbb{R}$$ with the following properties or show that no such function exists: $$1.$$ $$f_1$$ is differentiable everywhere except one point $$x_1.$$ $$2.$$ Define $$f_2 : \mathbb{R}\setminus\{x_1\} \to \mathbb{R}$$ as $$f_2(x) :=$$ derivative of $$f_1$$ at $$x.$$ This $$f_2$$ must be differentiable everywhere in its domain except one point $$x_2.$$ $$3.$$ Define $$f_3 : \mathbb{R}\setminus\{x_1,\;x_2\} \to \mathbb{R}$$ as $$f_3(x) :=$$ derivative of $$f_2$$ at $$x.$$ This $$f_3$$ must be differentiable everywhere in its domain except one point $$x_3.$$ $$\vdots$$ $$n.$$ Define $$f_n : \mathbb{R}\setminus\{x_1, \cdots, x_{n-1}\} \to \mathbb{R}$$ as $$f_n(x) :=$$ derivative of $$f_{n-1}$$ at $$x.$$ This $$f_n$$ must be differentiable everywhere in its domain except one point $$x_n.$$ $$\vdots$$ (Note that we do not stop at any $$n.$$) I found this question in a collection of extra questions for my Calculus course. I started out by trying something along the lines of $$f(x) = \lim_{n\to\infty}\sum_{i=1}^n (x-i)^i\|x-i\|$$, but the function itself isn't defined anywhere and I couldn't figure out how to fix it with minimum effort. So next, I tried out something that might actually be defined somewhere such as $$f(x) = \lim_{n\to\infty}\sum_{i=1}^n \frac{\left(\frac{2}{\pi}\arctan(x-i)\right)^i\left(\frac{2}{\pi}\arctan\|x-i\|\right)}{(i+1)!}$$ which is defined for $$x \in \mathbb{R}$$, but I wasn't able to prove continuity or differentiability. Intuitively, I feel that since it's a sum of continuous functions, it should be continuous, but I'm not sure whether this intuition is correct because it's an infinite sum. I asked the person whose website I found the question on (another student), and he said that he wasn't sure whether such a function was even possible. Any help would be appreciated! • You were on the right track with your first attempt (but replace $x-i$ by $x-x_i$, where $(x_i)$ is the given sequence of points). It is true that the series needs not converge at any point, but that can be remedied easily by multiplying every summand by a sequence of numbers that decreases sufficiently fast, such as $i!$... – Giuseppe Negro Sep 16 at 12:35 A different solution is the following 'lazy exponential' - there are easier solutions (maybe look up bump functions), but I like delay ODEs. set \begin{align}x\in(-\infty,0]&\implies f(x):=1,\\ x\in(0,1] &\implies f(x) := 1+x, \\ x\in (1,2] &\implies f(x) := 1+x + \frac{(x-1)^2}{2!},\\ x\in(2,3] & \implies f(x) := 1+x + \frac{(x-1)^2}{2!} + \frac{(x-2)^3}{3!}, \end{align} and in general $$x\in(n,n+1]\implies f(x) := \sum_{k=0}^{n+1} \frac{(x-k+1)^k}{k!}.$$ If you differentiate, you find for $$x\in (n,n+1)$$, where $$n>1$$: $$f'(x) = \sum_{k=1}^{n+1} \frac{(x-k+1)^{k-1}}{(k-1)!} =\sum_{j=0}^{n} \frac{(x-j)^{j}}{j!}= f(x-1)$$ so to the right of 1, it solves a delay ODE with initial data prescribed on $$x\in(0,1]$$ above. $$f'$$ is clearly discontinuous at $$0$$, but $$\left.\frac{d}{dx}\frac{(x-1)^2}{2!}\right\|_{x=1} = 0$$ so the derivative is continuous at $$x=1$$. In general, for any integer $$n\ge 2$$, near $$x=n-1$$, all the terms $$\frac{(x-h+1)^h}{h!}$$ for $$h are smooth, and the newly added term $$T_n$$, $$T_n(x) := \begin{cases} \frac{(x-n+1)^n}{n!} & x>n-1,\\ 0 & x\le n-1\end{cases}$$ is $$C^1$$. Conclusion - $$f \in C^0(\mathbb R)\cap C^1(\mathbb R\setminus \{0\}).$$ To finish, we use the delay ODE, which says that differentiating is the same as translating the function to the right by one. Thus for $$x\in \mathbb (0,\infty)\setminus \mathbb N$$, $$i\in\mathbb N$$, $$f^{(i+1)}(x+i) = f'(x).$$ So the discontinuity of $$f^{(i+1)}$$ at $$x=i-1$$, and the continuity at integers $$x=\tilde i > i-1$$ follows directly from the dis/continuity of $$f'$$ at $$0,1,2,\dots$$. We conclude $$f \in C^0(\mathbb R)\cap \left(\bigcap_{k=1}^\infty C^k(\mathbb R\setminus{\{0,1,\dots,k-1\}})\right).$$	2019-12-13T16:24:27	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3358361/construct-a-function-with-each-derivative-being-non-differentiable-at-a-distinct", "openwebmath_score": 0.9822348952293396, "openwebmath_perplexity": 207.70384663766873, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9728307684643189, "lm_q2_score": 0.8577681013541611, "lm_q1q2_score": 0.8344632012045484 }
https://mathhelpboards.com/threads/need-help-with-parabolas.5786/	# Need help with parabolas ##### New member I've encountered a question that I need help understanding the answer to. The question is: What is the equation of the axis of symmetry of the parabola given by the equation y=6(x+1)(x-5)? Now, I know this quadratic equation is in intercept form, and I know that the formula for finding the axis of symmetry for this is (p+q)/2 Which means that it's (1-5)/2 then -4/2 Which equals -2. But it says that the correct answer is 2, not -2. #### Bacterius ##### Well-known member MHB Math Helper [JUSTIFY]That's because if $(x - a)$ is a factor of your equation, then the intercept is at $x = a$, not $x = -a$. So in fact your parabola has intercepts $x = -1$ and $x = 5$, and so you would get $\frac{-1 + 5}{2} = \frac{4}{2} = 2$ as expected. The formula itself is correct. As an example, try plotting $y = x - 5$, and you'll see it intercepts the x-axis at $x = 5$, and not $x = -5$. In the same way, try plotting your parabola and see where it intercepts the x-axis.[/JUSTIFY]	2021-01-24T11:30:43	{ "domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/need-help-with-parabolas.5786/", "openwebmath_score": 0.8609395623207092, "openwebmath_perplexity": 237.40894844150571, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9728307661011976, "lm_q2_score": 0.8577681031721325, "lm_q1q2_score": 0.8344632009461167 }
http://math.stackexchange.com/questions/169407/find-identity-element-invertible-and-inverses-in-t-mathbb-z-times-mathbb-q	# Find identity element, invertible and inverses in $T=\mathbb Z \times \mathbb Q$ Let the following operation be defined on $T=\mathbb Z \times \mathbb Q$: \begin{aligned}(a,b)\centerdot (c,d) = (-ac, b+d+2) \end{aligned} in the commutative semigroup $(T, \centerdot)$, find the identity element, invertible elements and inverses. I need to find the identity element, so I need an $(\alpha, \beta)$ so that \begin{aligned}(a,b)\centerdot (\alpha,\beta) = (a,b) \end{aligned} as $(T, \centerdot)$ is a commutative semigroup, I assume that $(\alpha,\beta)\centerdot (a,b) = (a,b)$ holds. \begin{aligned}-a\alpha = a \Leftrightarrow \alpha = -1 \end{aligned} \begin{aligned}b+\beta+2 = b \Leftrightarrow \beta = -2 \end{aligned} so the identity element relative to this semigroup is $(-1,-2)$. In order to find the invertible elements and inverses: \begin{aligned}(a,b)\centerdot (a',b') = (-1,-2) \end{aligned} so \begin{aligned}-aa' = -1 \Leftrightarrow a'=-\frac{1}{a} \in \mathbb Z \Rightarrow a = 1 \end{aligned} \begin{aligned}b+b'+2=-2 \Leftrightarrow b'=-(b+4) \end{aligned} in conclusion only the elements $(1,b)$ are actually invertible and their inverse is $(1, -(b+4))$. Does this whole thing hold? Am I wrong in any part of it? - In the last bit, why can't $a$ be $-1$? – Gerry Myerson Jul 11 '12 at 9:37 @GerryMyerson because I need $(-aa') = -1$ if $a = -1$ then $(-aa') = 1$. – haunted85 Jul 11 '12 at 9:42 As @GerryMyerson pointed out $a$ cab be $-1\implies a'=-1\implies$ inverse of $(-1,b)$ also exists and is equal to $(-1,-(b+4)).$ – Aang Jul 11 '12 at 9:46 After equation $-aa'=-1$ $a'=\frac{-1}{a}$ is wrong, it should be $\frac{1}{a}$. – Aang Jul 11 '12 at 9:49 @avatar thank you I've really overlooked that minus sign!! what a shame! – haunted85 Jul 11 '12 at 9:59 In all of these problems, you are taking a structure with operations that you "already know", and are trying to test to see whether a new operation, defined in terms of the ones you know, satisfies certain properties. It's usually best to denote the "new" operation with a symbol that is unlikely to generate confusion. You've been doing this elsewhere, but here you are using $\cdot$, which can easily be confused with regular integer multiplication. So I'm going to replace it with $\odot$, if you don't mind. Let's approach it systematically, and essentially "following your nose." I'm going to check a few things that you seem to take for granted, just to get familiarity with the operation. Now, we have the set $T=\mathbb{Z}\times\mathbb{Q}$, and the operation $$(a,b)\odot (c,d) = (-ac, b+d+2).$$ First, the operation is commutative, since integer multiplication and addition of rationals is commutative. It is also associative: \begin{align} \Bigl( (a,b)\odot (c,d)\Bigr)\odot (x,y) &= (-ac, b+d+2)\odot (x,y)\\ &=(-(-ac)x,(b+d+2)+y+2)\\ &= (acx, b+d+y+4)\\ (a,b)\odot \Bigl((c,d)\odot (x,y)\Bigr)&= (a,b)\odot (-cx, d+y+2)\\ &= (-a(-cx), b+(d+y+2)+2)\\ &= (acx, b+d+y+4). \end{align} So the operation is commutative and associative, and we do indeed have a semigroup. Now we are trying to see whether this is a monoid (semigroup with identity). So we are trying to see whether there exist $\alpha\in\mathbb{Z}$ and $\beta\in\mathbb{Q}$ such that, for all $a\in\mathbb{Z}$ and $b\in\mathbb{Q}$, we have $$(\alpha,\beta)\odot(a,b) = (a,b)$$ (exactly as you did). This amounts to solving the equations you solved, which you did correctly, and so we discovered that $(T,\odot)$ is indeed a monoid, and that the identity element is $e_T = (-1,-2)$. Great! To figure out what elements have inverses, you do set $(a,b)\odot (x,y) =e_T = (-1,-2)$, and figure out if you can determine necessary and sufficient conditions on $a$ and $b$ for $x$ and $y$ to exist. As you note, we have: $$(-1,-2) = (a,b)\odot(x,y) = (-ax, b+y+2).$$ This requires $-ax = 1$, which means that $a$ has to be an integer that has a multiplicative inverse in the integers (namely, $-x$); there are only two possibilities: $a=1$ or $a=-1$. On the other hand, for any rational number $b$ we can always find a rational number $y$ such that $b+y+2=-2$, namely, as you note, $y = -(b+4)$. So it would seem that provided that $a=\pm 1$, then $(a,b)$ has an $\odot$-inverse in $T$. And, indeed, if $a=\pm 1$, then we have $$(\pm 1,b)\odot (\pm1,-b-4) = (-(\pm1)(\pm 1), b-b-4+2) = (-1-2).$$ So a necessary and sufficient condition for $(a,b)\in T$ to be $\odot$-invertible is that $a=\pm 1$. - thank you so much for your patient, clear and most of all exhaustive answer, appreciated. – haunted85 Jul 11 '12 at 19:23	2016-06-25T14:17:24	{ "domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/169407/find-identity-element-invertible-and-inverses-in-t-mathbb-z-times-mathbb-q", "openwebmath_score": 0.9917821288108826, "openwebmath_perplexity": 389.597952984447, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9728307708274401, "lm_q2_score": 0.8577680977182187, "lm_q1q2_score": 0.8344631996944016 }
https://www.physicsforums.com/threads/derivative-of-x-to-the-x.155954/	Derivative of x to the x 1. Feb 12, 2007 lapenna 1. The problem statement, all variables and given/known data I am required to find the derivative of x^x using the differance quotient 2. Relevant equations ((x+h)^(x+h)-(x^x))/h 3. The attempt at a solution TI-89 gives (lnx+1)x^x I can't figure out how to get there 2. Feb 12, 2007 theperthvan Without the derivative definition, ln both sides, so: $$y=x^x$$ $$ln(y)=xln(x)$$ $$\frac{1}{y}.\frac{dy}{dx} = 1.ln(x) + x.\frac{1}{x}$$ $$\frac{dy}{dx} = y[ln(x) + 1]$$ $$\frac{dy}{dx} = x^x(ln(x) + 1)$$ 3. Feb 13, 2007 HallsofIvy Staff Emeritus I can't imagine there being any reasonable way to use the "difference quotient" to determine the derivative of xx. You must have a really evil teacher! 4. Feb 14, 2007 dextercioby Either this is unreasonable, or your imagination really needs to see more: $$x^x =e^{x\ln x}$$ $$\lim_{h\rightarrow 0} \frac{(x+h)^{(x+h)}-x^x}{h} =\lim_{h\rightarrow 0} \frac{e^{(x+h)\ln(x+h)}-e^{x\ln x}}{h}=\lim_{h\rightarrow 0}\frac{e^{(x+h)\ln\left(x\left(1+\frac{h}{x}\right)\right)}-e^{x\ln x}}{h}$$ $$=\lim_{h\rightarrow 0} \frac{e^{x\ln x}e^{h\ln x}e^{x\ln\left(1+\frac{h}{x}\right)}e^{h\ln\left(1+\frac{h}{x}\right)}-e^{x\ln x}}{h}$$ $$=x^{x}\lim_{h\rightarrow 0}\frac{e^{h\ln x}e^{x\ln\left(1+\frac{h}{x}\right)}e^{h\ln\left(1+\frac{h}{x}\right)}-1}{h}\cdot \frac{h\ln x+x\ln\left(1+\frac{h}{x}\right)+h\ln\left(1+\frac{h}{x}\right)}{h\ln x+x\ln\left(1+\frac{h}{x}\right)+h\ln\left(1+\frac{h}{x}\right)}$$ $$=x^{x}\left[\ln x+x\lim_{h\rightarrow 0}\frac{\ln\left(1+\frac{h}{x}\right)}{h}+\lim_{h\rightarrow 0}\ln\left(1+\frac{h}{x}\right)\right]$$ $$=x^x \left\{\ln x+x\ln \lim_{h\rightarrow 0}\left[\left(1+\frac{h}{x}\right)^{\frac{x}{h}}\right]^{\frac{1}{x}}\right\}$$ $$=x^x \left(\ln x +1\right)$$ Last edited: Feb 14, 2007 5. Feb 15, 2007 theperthvan wow . 6. Feb 15, 2007 VietDao29 Whoa. Or somewhat easier. That's a little bit messy methinks: $$(x ^ x)' = (e ^ {x \ln (x)})' = \lim_{h \rightarrow 0} \frac{e ^ {(x + h) \ln (x + h)} - e ^ {x \ln (x)}}{h}$$ $$= \lim_{h \rightarrow 0} \left( \frac{e ^ {(x + h) \ln (x + h)} - e ^ {x \ln (x)}}{(x + h) \ln (x + h) - x \ln x} \times \frac{(x + h) \ln (x + h) - x \ln x}{h} \right)$$ $$= e ^ {x \ln (x)} \left( \lim_{h \rightarrow 0} \frac{x \ln (1 + \frac{h}{x}) + h \ln (x + h)}{h} \right)$$ [due to: (eu)'u = eu] $$= x ^ x \left\{ \lim_{h \rightarrow 0} \left[ x \ln \left( \left( 1 + \frac{h}{x} \right) ^ {\frac{x}{h}} \right) ^ {\frac{1}{x}} \right] + \ln (x) \right\}$$ $$= x ^ x (1 + \ln (x))$$ Last edited: Feb 15, 2007 7. Aug 21, 2009 Raiey Since the original equation is x^x and never a division quotient, one can not apply that rule. Here is the alternative proof. Attached Files: • derivativesxtopowerofx.txt File size: 3.4 KB Views: 45 8. Aug 22, 2009 g_edgar "Difference quotient" means this: $$\frac{f(x)-f(a)}{x-a}$$ so the assignment was to use the definition of derivative that involves this. 9. Aug 23, 2009 Raiey An answer is attached in PDF File size: 37.7 KB Views: 78	2017-10-17T02:12:13	{ "domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/derivative-of-x-to-the-x.155954/", "openwebmath_score": 0.8381654024124146, "openwebmath_perplexity": 5437.9132571159025, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9728307661011976, "lm_q2_score": 0.8577680977182186, "lm_q1q2_score": 0.8344631956403815 }
https://math.libretexts.org/Bookshelves/Precalculus/Book%3A_Precalculus_(Stitz-Zeager)/10%3A_Foundations_of_Trigonometry/10.4%3A_Trigonometric_Identities	# 10.4: Trigonometric Identities In Section \ref{CircularFunctions}, we saw the utility of the Pythagorean Identities in Theorem \ref{pythids} along with the Quotient and Reciprocal Identities in Theorem \ref{recipquotid}. Not only did these identities help us compute the values of the circular functions for angles, they were also useful in simplifying expressions involving the circular functions. In this section, we introduce several collections of identities which have uses in this course and beyond. Our first set of identities is the Even / Odd' identities.\footnote{As mentioned at the end of Section \ref{TheUnitCircle}, properties of the circular functions when thought of as functions of angles in radian measure hold equally well if we view these functions as functions of real numbers. Not surprisingly, the Even / Odd properties of the circular functions are so named because they identify cosine and secant as even functions, while the remaining four circular functions are odd. (See Section \ref{GraphsofFunctions}.)} Note: Even / Odd Identities For all applicable angles $$\theta$$: • $$\cos(-\theta) = \cos(\theta)$$ • $$\sec(-\theta) = \sec(\theta)$$ • $$\sin(-\theta) = -\sin(\theta)$$ • $$\csc(-\theta) = -\csc(\theta)$$ • $$\tan(-\theta) = -\tan(\theta)$$ • $$\cot(-\theta) = -\cot(\theta)$$ In light of the Quotient and Reciprocal Identities, Theorem \ref{recipquotid}, it suffices to show $$\cos(-\theta) = \cos(\theta)$$ and $$\sin(-\theta) = -\sin(\theta)$$. The remaining four circular functions can be expressed in terms of $$\cos(\theta)$$ and $$\sin(\theta)$$ so the proofs of their Even / Odd Identities are left as exercises. Consider an angle $$\theta$$ plotted in standard position. Let $$\theta_ { o}$$ be the angle coterminal with $$\theta$$ with $$0 \leq \theta_ { o} < 2\pi$$. (We can construct the angle $$\theta_ { o}$$ by rotating counter-clockwise from the positive $$x$$-axis to the terminal side of $$\theta$$ as pictured below.) Since $$\theta$$ and $$\theta_ { o}$$ are coterminal, $$\cos(\theta) = \cos(\theta_ { o})$$ and $$\sin(\theta) = \sin(\theta_ { o})$$. We now consider the angles $$-\theta$$ and $$-\theta_ { o}$$. Since $$\theta$$ is coterminal with $$\theta_ { o}$$, there is some integer $$k$$ so that $$\theta = \theta_ { o} + 2\pi \cdot k$$. Therefore, $$-\theta = -\theta_ { o} - 2\pi \cdot k = -\theta_ { o} + 2\pi \cdot(-k)$$. Since $$k$$ is an integer, so is $$(-k)$$, which means $$-\theta$$ is coterminal with $$-\theta_ { o}$$. Hence, $$\cos(-\theta) = \cos(-\theta_ { o})$$ and $$\sin(-\theta) = \sin(-\theta_ { o})$$. Let $$P$$ and $$Q$$ denote the points on the terminal sides of $$\theta_ { o}$$ and $$-\theta_ { o}$$, respectively, which lie on the Unit Circle. By definition, the coordinates of $$P$$ are $$(\cos(\theta_ { o}),\sin(\theta_ { o}))$$ and the coordinates of $$Q$$ are $$(\cos(-\theta_ { o}),\sin(-\theta_ { o}))$$. Since $$\theta_ { o}$$ and $$-\theta_ { o}$$ sweep out congruent central sectors of the Unit Circle, it follows that the points $$P$$ and $$Q$$ are symmetric about the $$x$$-axis. Thus, $$\cos(-\theta_ { o}) = \cos(\theta_ { o})$$ and $$\sin(-\theta_ { o}) = -\sin(\theta_ { o})$$. Since the cosines and sines of $$\theta_ { o}$$ and $$-\theta_ { o}$$ are the same as those for $$\theta$$ and $$-\theta$$, respectively, we get $$\cos(-\theta) = \cos(\theta)$$ and $$\sin(-\theta) = -\sin(\theta)$$, as required. The Even / Odd Identities are readily demonstrated using any of the common angles' noted in Section \ref{TheUnitCircle}. Their true utility, however, lies not in computation, but in simplifying expressions involving the circular functions. In fact, our next batch of identities makes heavy use of the Even / Odd Identities. Note: Sum and Difference Identities for Cosine For all angles $$\alpha$$ and $$\beta$$: • $$\cos(\alpha + \beta) = \cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta)$$ • $$\cos(\alpha - \beta) = \cos(\alpha) \cos(\beta) + \sin(\alpha) \sin(\beta)$$ We first prove the result for differences. As in the proof of the Even / Odd Identities, we can reduce the proof for general angles $$\alpha$$ and $$\beta$$ to angles $$\alpha_ { o}$$ and $$\beta_ { o}$$, coterminal with $$\alpha$$ and $$\beta$$, respectively, each of which measure between $$0$$ and $$2\pi$$ radians. Since $$\alpha$$ and $$\alpha_ { o}$$ are coterminal, as are $$\beta$$ and $$\beta_ { o}$$, it follows that $$\alpha - \beta$$ is coterminal with $$\alpha_ { o} - \beta_ { o}$$. Consider the case below where $$\alpha_ { o} \geq \beta_ { o}$$. Since the angles $$POQ$$ and $$AOB$$ are congruent, the distance between $$P$$ and $$Q$$ is equal to the distance between $$A$$ and $$B$$.\footnote{In the picture we've drawn, the \underline{tri}angles $$POQ$$ and $$AOB$$ are congruent, which is even better. However, $$\alpha_ { o} - \beta_ { o}$$ could be $$0$$ or it could be $$\pi$$, neither of which makes a triangle. It could also be larger than $$\pi$$, which makes a triangle, just not the one we've drawn. You should think about those three cases.} The distance formula, Equation \ref{distanceformula}, yields $\begin{array}{rcl} \sqrt{(\cos(\alpha_ { o}) - \cos(\beta_ { o}))^2 + (\sin(\alpha_ { o}) - \sin(\beta_ { o}))^2 } & = & \sqrt{(\cos(\alpha_ { o} - \beta_ { o}) - 1)^2 + (\sin(\alpha_ { o} - \beta_ { o}) - 0)^2} \\ \end{array}$ Squaring both sides, we expand the left hand side of this equation as $\begin{array}{rcl} (\cos(\alpha_ { o}) - \cos(\beta_ { o}))^2 + (\sin(\alpha_ { o}) - \sin(\beta_ { o}))^2 & = & \cos^2(\alpha_ { o}) - 2\cos(\alpha_ { o})\cos(\beta_ { o}) + \cos^2(\beta_ { o}) \\ & & + \sin^2(\alpha_ { o}) - 2\sin(\alpha_ { o})\sin(\beta_ { o}) + \sin^2(\beta_ { o}) \\ & = & \cos^2(\alpha_ { o}) + \sin^2(\alpha_ { o}) + \cos^2(\beta_ { o}) + \sin^2(\beta_ { o}) \\ & & - 2\cos(\alpha_ { o})\cos(\beta_ { o}) - 2\sin(\alpha_ { o})\sin(\beta_ { o}) \end{array}$ From the Pythagorean Identities, $$\cos^2(\alpha_ { o}) + \sin^2(\alpha_ { o}) = 1$$ and $$\cos^2(\beta_ { o}) + \sin^2(\beta_ { o}) = 1$$, so $\begin{array}{rcl} (\cos(\alpha_ { o}) - \cos(\beta_ { o}))^2 + (\sin(\alpha_ { o}) - \sin(\beta_ { o}))^2 & = & 2 - 2\cos(\alpha_ { o})\cos(\beta_ { o}) - 2\sin(\alpha_ { o})\sin(\beta_ { o}) \end{array}$ Turning our attention to the right hand side of our equation, we find $\begin{array}{rcl} (\cos(\alpha_ { o} - \beta_ { o}) - 1)^2 + (\sin(\alpha_ { o} - \beta_ { o}) - 0)^2 & = & \cos^2(\alpha_ { o} - \beta_ { o}) - 2\cos(\alpha_ { o} - \beta_ { o}) + 1 + \sin^2(\alpha_ { o} - \beta_ { o}) \\ & = & 1 + \cos^2(\alpha_ { o} - \beta_ { o}) + \sin^2(\alpha_ { o} - \beta_ { o}) - 2\cos(\alpha_ { o} - \beta_ { o}) \\ \end{array}$ Once again, we simplify $$\cos^2(\alpha_ { o} - \beta_ { o}) + \sin^2(\alpha_ { o} - \beta_ { o})= 1$$, so that $\begin{array}{rcl} (\cos(\alpha_ { o} - \beta_ { o}) - 1)^2 + (\sin(\alpha_ { o} - \beta_ { o}) - 0)^2 & = & 2 - 2\cos(\alpha_ { o} - \beta_ { o}) \\ \end{array}$ Putting it all together, we get $$2 - 2\cos(\alpha_ { o})\cos(\beta_ { o}) - 2\sin(\alpha_ { o})\sin(\beta_ { o}) = 2 - 2\cos(\alpha_ { o} - \beta_ { o})$$, which simplifies to: $$\cos(\alpha_ { o} - \beta_ { o}) = \cos(\alpha_ { o})\cos(\beta_ { o}) + \sin(\alpha_ { o})\sin(\beta_ { o})$$. Since $$\alpha$$ and $$\alpha_ { o}$$, $$\beta$$ and $$\beta_ { o}$$ and $$\alpha - \beta$$ and $$\alpha_ { o}- \beta_ { o}$$ are all coterminal pairs of angles, we have $$\cos(\alpha - \beta) = \cos(\alpha) \cos(\beta) + \sin(\alpha) \sin(\beta)$$. For the case where $$\alpha_ { o} \leq \beta_ { o}$$, we can apply the above argument to the angle $$\beta_ { o} - \alpha_ { o}$$ to obtain the identity $$\cos(\beta_ { o} - \alpha_ { o}) = \cos(\beta_ { o})\cos(\alpha_ { o}) + \sin(\beta_ { o})\sin(\alpha_ { o})$$. Applying the Even Identity of cosine, we get $$\cos(\beta_ { o} - \alpha_ { o}) = \cos( - (\alpha_ { o} - \beta_ { o})) = \cos(\alpha_ { o} - \beta_ { o})$$, and we get the identity in this case, too. To get the sum identity for cosine, we use the difference formula along with the Even/Odd Identities $\cos(\alpha + \beta) = \cos(\alpha - (-\beta)) = \cos(\alpha) \cos(-\beta) + \sin(\alpha) \sin(-\beta) = \cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta)$ We put these newfound identities to good use in the following example. Example $$\PageIndex{1}$$: Cosine Sum and Difference 1. Find the exact value of $$\cos\left(15^{\circ}\right)$$. 2. Verify the identity: $$\cos\left(\frac{\pi}{2} - \theta\right) = \sin(\theta)$$. Solution 1. In order to use Theorem \ref{cosinesumdifference} to find $$\cos\left(15^{\circ}\right)$$, we need to write $$15^{\circ}$$ as a sum or difference of angles whose cosines and sines we know. One way to do so is to write $$15^{\circ} = 45^{\circ} - 30^{\circ}$$. $\begin{array}{rcl} \cos\left(15^{\circ}\right) & = & \cos\left(45^{\circ} - 30^{\circ} \right) \\ & = & \cos\left(45^{\circ}\right)\cos\left(30^{\circ} \right) + \sin\left(45^{\circ}\right)\sin\left(30^{\circ} \right) \\ & = & \left( \dfrac{\sqrt{2}}{2} \right)\left( \dfrac{\sqrt{3}}{2} \right) + \left( \dfrac{\sqrt{2}}{2} \right)\left( \dfrac{1}{2} \right)\\ & = & \dfrac{\sqrt{6}+ \sqrt{2}}{4} \\ \end{array}$ 1. In a straightforward application of Theorem \ref{cosinesumdifference}, we find $\begin{array}{rcl} \cos\left(\dfrac{\pi}{2} - \theta\right) & = & \cos\left(\dfrac{\pi}{2}\right)\cos\left(\theta\right) + \sin\left(\dfrac{\pi}{2}\right)\sin\left(\theta \right) \\ & = & \left( 0 \right)\left( \cos(\theta) \right) + \left( 1 \right)\left( \sin(\theta) \right) \\ & = & \sin(\theta) \\ \end{array}$ The identity verified in Example $$\PageIndex{1}$$, namely, $$\cos\left(\frac{\pi}{2} - \theta\right) = \sin(\theta)$$, is the first of the celebrated cofunction' identities. These identities were first hinted at in Exercise \ref{cofunctionforeshadowing} in Section \ref{TheUnitCircle}. From $$\sin(\theta) = \cos\left(\frac{\pi}{2} - \theta\right)$$, we get: $\sin\left(\dfrac{\pi}{2} - \theta\right) = \cos\left(\dfrac{\pi}{2} -\left[\dfrac{\pi}{2} - \theta\right]\right) = \cos(\theta),$ which says, in words, that the co'sine of an angle is the sine of its co'mplement. Now that these identities have been established for cosine and sine, the remaining circular functions follow suit. The remaining proofs are left as exercises. Note: Cofunction Identities For all applicable angles $$\theta$$: • $$\cos\left(\dfrac{\pi}{2} - \theta \right) = \sin(\theta)$$ • $$\sin\left(\dfrac{\pi}{2} - \theta \right) = \cos(\theta)$$ • $$\sec\left(\dfrac{\pi}{2} - \theta \right) = \csc(\theta)$$ • $$\csc\left(\dfrac{\pi}{2} - \theta \right) = \sec(\theta)$$ • $$\tan\left(\dfrac{\pi}{2} - \theta \right) = \cot(\theta)$$ • $$\cot\left(\dfrac{\pi}{2} - \theta \right) = \tan(\theta)$$ With the Cofunction Identities in place, we are now in the position to derive the sum and difference formulas for sine. To derive the sum formula for sine, we convert to cosines using a cofunction identity, then expand using the difference formula for cosine $\begin{array}{rcl} \sin(\alpha + \beta) & = & \cos\left( \dfrac{\pi}{2} - (\alpha + \beta) \right) \\ & = & \cos\left( \left[\dfrac{\pi}{2} - \alpha \right] - \beta \right) \\ & = & \cos\left(\dfrac{\pi}{2} - \alpha \right) \cos(\beta) + \sin\left(\dfrac{\pi}{2} - \alpha \right)\sin(\beta) \\ & = & \sin(\alpha) \cos(\beta) + \cos(\alpha) \sin(\beta) \\ \end{array}$ We can derive the difference formula for sine by rewriting $$\sin(\alpha - \beta)$$ as $$\sin(\alpha + (-\beta))$$ and using the sum formula and the Even / Odd Identities. Again, we leave the details to the reader. Sum and Difference Identities for Sine For all angles $$\alpha$$ and $$\beta$$, \index{Difference Identity ! for sine} \index{Sum Identity ! for sine} • $$\sin(\alpha + \beta) = \sin(\alpha) \cos(\beta) + \cos(\alpha) \sin(\beta)$$ • $$\sin(\alpha - \beta) = \sin(\alpha) \cos(\beta) - \cos(\alpha) \sin(\beta)$$ Example $$\PageIndex{1}$$: 1. Find the exact value of $$\sin\left(\frac{19 \pi}{12}\right)$$ 2. If $$\alpha$$ is a Quadrant II angle with $$\sin(\alpha) = \frac{5}{13}$$, and $$\beta$$ is a Quadrant III angle with $$\tan(\beta) = 2$$, find $$\sin(\alpha - \beta)$$. 3. Derive a formula for $$\tan(\alpha + \beta)$$ in terms of $$\tan(\alpha)$$ and $$\tan(\beta)$$. Solution 1. As in Example \ref{cosinesumdiffex}, we need to write the angle $$\frac{19 \pi}{12}$$ as a sum or difference of common angles. The denominator of $$12$$ suggests a combination of angles with denominators $$3$$ and $$4$$. One such combination is $$\; \frac{19 \pi}{12} = \frac{4 \pi}{3} + \frac{\pi}{4}$$. Applying Theorem \ref{sinesumdifference}, we get $\begin{array}{rcl} \sin\left(\dfrac{19 \pi}{12}\right) & = & \sin\left(\dfrac{4 \pi}{3} + \dfrac{\pi}{4} \right) \\ & = & \sin\left(\dfrac{4 \pi}{3} \right)\cos\left(\dfrac{\pi}{4} \right) + \cos\left(\dfrac{4 \pi}{3} \right)\sin\left(\dfrac{\pi}{4} \right) \\ & = & \left( -\dfrac{\sqrt{3}}{2} \right)\left( \dfrac{\sqrt{2}}{2} \right) + \left( -\dfrac{1}{2} \right)\left( \dfrac{\sqrt{2}}{2} \right) \\ & = & \dfrac{-\sqrt{6}- \sqrt{2}}{4} \\ \end{array}$ 1. In order to find $$\sin(\alpha - \beta)$$ using Theorem \ref{sinesumdifference}, we need to find $$\cos(\alpha)$$ and both $$\cos(\beta)$$ and $$\sin(\beta)$$. To find $$\cos(\alpha)$$, we use the Pythagorean Identity $$\cos^2(\alpha) + \sin^2(\alpha) = 1$$. Since $$\sin(\alpha) = \frac{5}{13}$$, we have $$\cos^{2}(\alpha) + \left(\frac{5}{13}\right)^2 = 1$$, or $$\cos(\alpha) = \pm \frac{12}{13}$$. Since $$\alpha$$ is a Quadrant II angle, $$\cos(\alpha) = -\frac{12}{13}$$. We now set about finding $$\cos(\beta)$$ and $$\sin(\beta)$$. We have several ways to proceed, but the Pythagorean Identity $$1 + \tan^{2}(\beta) = \sec^{2}(\beta)$$ is a quick way to get $$\sec(\beta)$$, and hence, $$\cos(\beta)$$. With $$\tan(\beta) = 2$$, we get $$1 + 2^2 = \sec^{2}(\beta)$$ so that $$\sec(\beta) = \pm \sqrt{5}$$. Since $$\beta$$ is a Quadrant III angle, we choose $$\sec(\beta) = -\sqrt{5}$$ so $$\cos(\beta) = \frac{1}{\sec(\beta)} = \frac{1}{-\sqrt{5}} = -\frac{\sqrt{5}}{5}$$. We now need to determine $$\sin(\beta)$$. We could use The Pythagorean Identity $$\cos^{2}(\beta) + \sin^{2}(\beta) = 1$$, but we opt instead to use a quotient identity. From $$\tan(\beta) = \frac{\sin(\beta)}{\cos(\beta)}$$, we have $$\sin(\beta) = \tan(\beta) \cos(\beta)$$ so we get $$\sin(\beta) = (2) \left( -\frac{\sqrt{5}}{5}\right) = - \frac{2 \sqrt{5}}{5}$$. We now have all the pieces needed to find $$\sin(\alpha - \beta)$$: $\begin{array}{rcl} \sin(\alpha - \beta) & = & \sin(\alpha)\cos(\beta) - \cos(\alpha)\sin(\beta) \\ & = & \left( \dfrac{5}{13} \right)\left( -\dfrac{\sqrt{5}}{5} \right) - \left( -\dfrac{12}{13} \right)\left( - \dfrac{2 \sqrt{5}}{5} \right) \\ & = & -\dfrac{29\sqrt{5}}{65} \\ \end{array}$ We can start expanding $$\tan(\alpha + \beta)$$ using a quotient identity and our sum formulas $\begin{array}{rcl} \tan(\alpha + \beta) & = & \dfrac{\sin(\alpha + \beta)}{\cos(\alpha + \beta)} \\ & = & \dfrac{\sin(\alpha) \cos(\beta) + \cos(\alpha) \sin(\beta)}{\cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta)} \\ \end{array}$ Since $$\tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)}$$ and $$\tan(\beta) = \frac{\sin(\beta)}{\cos(\beta)}$$, it looks as though if we divide both numerator and denominator by $$\cos(\alpha) \cos(\beta)$$ we will have what we want $\begin{array}{rcl} \tan(\alpha + \beta) & = & \dfrac{\sin(\alpha) \cos(\beta) + \cos(\alpha) \sin(\beta)}{\cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta)} \cdot\dfrac{\dfrac{1}{\cos(\alpha) \cos(\beta)}}{\dfrac{1}{\cos(\alpha) \cos(\beta)}}\\ & & \\ & = & \dfrac{\dfrac{\sin(\alpha) \cos(\beta)}{\cos(\alpha) \cos(\beta)} + \dfrac{\cos(\alpha) \sin(\beta)}{\cos(\alpha) \cos(\beta)}}{\dfrac{\cos(\alpha) \cos(\beta)}{\cos(\alpha) \cos(\beta)} - \dfrac{\sin(\alpha) \sin(\beta)}{\cos(\alpha) \cos(\beta)}}\\ & & \\ & = & \dfrac{\dfrac{\sin(\alpha) \cancel{\cos(\beta)}}{\cos(\alpha) \cancel{\cos(\beta)}} + \dfrac{\cancel{\cos(\alpha)} \sin(\beta)}{\cancel{\cos(\alpha)} \cos(\beta)}}{\dfrac{\cancel{\cos(\alpha)} \cancel{\cos(\beta)}}{\cancel{\cos(\alpha)} \cancel{\cos(\beta)}} - \dfrac{\sin(\alpha) \sin(\beta)}{\cos(\alpha) \cos(\beta)}}\\ & & \\ & = & \dfrac{\tan(\alpha) + \tan(\beta)}{1 -\tan(\alpha) \tan(\beta)}\\ \end{array}$ Naturally, this formula is limited to those cases where all of the tangents are defined.\qed The formula developed in Exercise \ref{sinesumanddiffex} for $$\tan(\alpha + \beta)$$ can be used to find a formula for $$\tan(\alpha - \beta)$$ by rewriting the difference as a sum, $$\tan(\alpha + (-\beta))$$, and the reader is encouraged to fill in the details. Below we summarize all of the sum and difference formulas for cosine, sine and tangent. Note Sum and Difference Identities:} For all applicable angles $$\alpha$$ and $$\beta$$, \index{Difference Identity ! for tangent} \index{Sum Identity ! for tangent} \index{Difference Identity ! for cosine} \index{Sum Identity ! for cosine} \index{Difference Identity ! for sine} \index{Sum Identity ! for sine} • $$\cos(\alpha \pm \beta) = \cos(\alpha) \cos(\beta) \mp \sin(\alpha) \sin(\beta)$$ • $$\sin(\alpha \pm \beta) = \sin(\alpha) \cos(\beta) \pm \cos(\alpha) \sin(\beta)$$ • $$\tan(\alpha \pm \beta) = \dfrac{\tan(\alpha) \pm \tan(\beta)}{1 \mp \tan(\alpha) \tan(\beta)}$$ In the statement of Theorem \ref{circularsumdifference}, we have combined the cases for the sum $+$' and difference \)-$' of angles into one formula. The convention here is that if you want the formula for the sum $+$' of two angles, you use the top sign in the formula; for the difference, \)-$', use the bottom sign. For example, $\tan(\alpha - \beta) = \dfrac{\tan(\alpha) - \tan(\beta)}{1 + \tan(\alpha) \tan(\beta)}$ If we specialize the sum formulas in Theorem \ref{circularsumdifference} to the case when $$\alpha = \beta$$, we obtain the following Double Angle' Identities. Note Double Angle Identities For all applicable angles $$\theta$$, \index{Double Angle Identities} • $$\cos(2\theta) = \left\{ \begin{array}{l} \cos^{2}(\theta) - \sin^{2}(\theta)\\ [5pt] 2\cos^{2}(\theta) - 1 \\ [5pt] 1-2\sin^{2}(\theta) \end{array} \right.$$ • $$\sin(2\theta) = 2\sin(\theta)\cos(\theta)$$ • $$\tan(2\theta) = \dfrac{2\tan(\theta)}{1 - \tan^{2}(\theta)}$$ The three different forms for $$\cos(2\theta)$$ can be explained by our ability to exchange' squares of cosine and sine via the Pythagorean Identity $$\cos^{2}(\theta) + \sin^{2}(\theta) = 1$$ and we leave the details to the reader. It is interesting to note that to determine the value of $$\cos(2\theta)$$, only \textit{one} piece of information is required: either $$\cos(\theta)$$ or $$\sin(\theta)$$. To determine $$\sin(2\theta)$$, however, it appears that we must know both $$\sin(\theta)$$ and $$\cos(\theta)$$. In the next example, we show how we can find $$\sin(2\theta)$$ knowing just one piece of information, namely $$\tan(\theta)$$. Example $$\PageIndex{1}$$: 1. Suppose $$P(-3,4)$$ lies on the terminal side of $$\theta$$ when $$\theta$$ is plotted in standard position. Find $$\cos(2\theta)$$ and $$\sin(2\theta)$$ and determine the quadrant in which the terminal side of the angle $$2\theta$$ lies when it is plotted in standard position. 2. If $$\sin(\theta) = x$$ for $$-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$$, find an expression for $$\sin(2\theta)$$ in terms of $$x$$. 3. \label{doubleanglesinewtan} Verify the identity: $$\sin(2\theta) = \dfrac{2\tan(\theta)}{1 + \tan^{2}(\theta)}$$. 4. Express $$\cos(3\theta)$$ as a polynomial in terms of $$\cos(\theta)$$. Solution 1. Using Theorem \ref{cosinesinecircle} from Section \ref{TheUnitCircle} with $$x = -3$$ and $$y=4$$, we find $$r = \sqrt{x^2+y^2} = 5$$. Hence, $$\cos(\theta) = -\frac{3}{5}$$ and $$\sin(\theta) = \frac{4}{5}$$. Applying Theorem \ref{doubleangle}, we get $$\cos(2\theta) = \cos^{2}(\theta) - \sin^{2}(\theta) = \left(-\frac{3}{5}\right)^2 - \left(\frac{4}{5}\right)^2 = -\frac{7}{25}$$, and $$\sin(2\theta) = 2 \sin(\theta) \cos(\theta) = 2 \left(\frac{4}{5}\right)\left(-\frac{3}{5}\right) = -\frac{24}{25}$$. Since both cosine and sine of $$2\theta$$ are negative, the terminal side of $$2\theta$$, when plotted in standard position, lies in Quadrant III. 2. If your first reaction to $\sin(\theta) = x$' is No it's not, $$\cos(\theta) = x!' then you have indeed learned something, and we take comfort in that. However, context is everything. Here, x' is just a variable - it does not necessarily represent the \(x$$-coordinate of the point on The Unit Circle which lies on the terminal side of $$\theta$$, assuming $$\theta$$ is drawn in standard position. Here, $$x$$ represents the quantity $$\sin(\theta)$$, and what we wish to know is how to express $$\sin(2\theta)$$ in terms of $$x$$. We will see more of this kind of thing in Section \ref{ArcTrig}, and, as usual, this is something we need for Calculus. Since $$\sin(2\theta) = 2 \sin(\theta) \cos(\theta)$$, we need to write $$\cos(\theta)$$ in terms of $$x$$ to finish the problem. We substitute $$x = \sin(\theta)$$ into the Pythagorean Identity, $$\cos^{2}(\theta) + \sin^{2}(\theta) = 1$$, to get $$\cos^{2}(\theta) + x^2 = 1$$, or $$\cos(\theta) = \pm \sqrt{1-x^2}$$. Since $$-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$$, $$\cos(\theta) \geq 0$$, and thus $$\cos(\theta) = \sqrt{1-x^2}$$. Our final answer is $$\sin(2\theta) = 2 \sin(\theta) \cos(\theta) = 2x\sqrt{1-x^2}$$. We start with the right hand side of the identity and note that $$1 + \tan^{2}(\theta) = \sec^{2}(\theta)$$. From this point, we use the Reciprocal and Quotient Identities to rewrite $$\tan(\theta)$$ and $$\sec(\theta)$$ in terms of $$\cos(\theta)$$ and $$\sin(\theta)$$: $\begin{array}{rcl} \dfrac{2\tan(\theta)}{1 + \tan^{2}(\theta)} & = & \dfrac{2\tan(\theta)}{\sec^{2}(\theta)}= \dfrac{2 \left( \dfrac{\sin(\theta)}{\cos(\theta)}\right)}{\dfrac{1}{\cos^{2}(\theta)}}= 2\left( \dfrac{\sin(\theta)}{\cos(\theta)}\right) \cos^{2}(\theta) \\ & = & 2\left( \dfrac{\sin(\theta)}{\cancel{\cos(\theta)}}\right) \cancel{\cos(\theta)} \cos(\theta) = 2\sin(\theta) \cos(\theta) = \sin(2\theta) \\ \end{array}$ 1. In Theorem \ref{doubleangle}, one of the formulas for $$\cos(2\theta)$$, namely $$\cos(2\theta) = 2\cos^{2}(\theta) - 1$$, expresses $$\cos(2\theta)$$ as a polynomial in terms of $$\cos(\theta)$$. We are now asked to find such an identity for $$\cos(3\theta)$$. Using the sum formula for cosine, we begin with $\begin{array}{rcl} \cos(3\theta) & = & \cos(2\theta + \theta) \\ & = & \cos(2\theta)\cos(\theta) - \sin(2\theta)\sin(\theta) \\ \end{array}$ Our ultimate goal is to express the right hand side in terms of $$\cos(\theta)$$ only. We substitute $$\cos(2\theta) = 2\cos^{2}(\theta) -1$$ and $$\sin(2\theta) = 2\sin(\theta)\cos(\theta)$$ which yields $\begin{array}{rcl} \cos(3\theta) & = & \cos(2\theta)\cos(\theta) - \sin(2\theta)\sin(\theta) \\ & = & \left(2\cos^{2}(\theta) - 1\right) \cos(\theta) - \left(2 \sin(\theta) \cos(\theta) \right)\sin(\theta) \\ & = & 2\cos^{3}(\theta)- \cos(\theta) - 2 \sin^2(\theta) \cos(\theta) \\ \end{array}$ Finally, we exchange $$\sin^{2}(\theta)$$ for $$1 - \cos^{2}(\theta)$$ courtesy of the Pythagorean Identity, and get $\begin{array}{rcl} \cos(3\theta) & = & 2\cos^{3}(\theta)- \cos(\theta) - 2 \sin^2(\theta) \cos(\theta) \\ & = & 2\cos^{3}(\theta)- \cos(\theta) - 2 \left(1 - \cos^{2}(\theta)\right) \cos(\theta) \\ & = & 2\cos^{3}(\theta)- \cos(\theta) - 2\cos(\theta) + 2\cos^{3}(\theta) \\ & = & 4\cos^{3}(\theta)- 3\cos(\theta) \\ \end{array}$ and we are done. In the last problem in Example \ref{doubleangleex}, we saw how we could rewrite $$\cos(3\theta)$$ as sums of powers of $$\cos(\theta)$$. In Calculus, we have occasion to do the reverse; that is, reduce the power of cosine and sine. Solving the identity $$\cos(2\theta) = 2\cos^{2}(\theta) -1$$ for $$\cos^{2}(\theta)$$ and the identity $$\cos(2\theta) = 1 - 2\sin^{2}(\theta)$$ for $$\sin^{2}(\theta)$$ results in the aptly-named `Power Reduction' formulas below. Power Reduction Formulas For all angles $$\theta$$, \index{Power Reduction Formulas} • $$\cos^{2}(\theta) = \dfrac{1 + \cos(2\theta)}{2}$$ • $$\sin^{2}(\theta) = \dfrac{1 - \cos(2\theta)}{2}$$ Example $$\PageIndex{1}$$: Rewrite $$\sin^{2}(\theta) \cos^{2}(\theta)$$ as a sum and difference of cosines to the first power. Solution We begin with a straightforward application of Theorem \ref{powerreduction} $\begin{array}{rcl} \sin^{2}(\theta) \cos^{2}(\theta) & = & \left( \dfrac{1 - \cos(2\theta)}{2} \right) \left( \dfrac{1 + \cos(2\theta)}{2} \right) \\ & = & \dfrac{1}{4}\left(1 - \cos^{2}(2\theta)\right) \\ & = & \dfrac{1}{4} - \dfrac{1}{4}\cos^{2}(2\theta) \\ \end{array}$ Next, we apply the power reduction formula to $$\cos^{2}(2\theta)$$ to finish the reduction $\begin{array}{rcl} \sin^{2}(\theta) \cos^{2}(\theta) & = & \dfrac{1}{4} - \dfrac{1}{4}\cos^{2}(2\theta) \\ & = & \dfrac{1}{4} - \dfrac{1}{4} \left(\dfrac{1 + \cos(2(2\theta))}{2}\right) \\ & = & \dfrac{1}{4} - \dfrac{1}{8} - \dfrac{1}{8}\cos(4\theta) \\ & = & \dfrac{1}{8} - \dfrac{1}{8}\cos(4\theta) \\ \end{array}$ Another application of the Power Reduction Formulas is the Half Angle Formulas. To start, we apply the Power Reduction Formula to $$\cos^{2}\left(\frac{\theta}{2}\right) $\cos^{2}\left(\dfrac{\theta}{2}\right) = \dfrac{1 + \cos\left(2 \left(\frac{\theta}{2}\right)\right)}{2} = \dfrac{1 + \cos(\theta)}{2}.$ We can obtain a formula for \(\cos\left(\frac{\theta}{2}\right)$$ by extracting square roots. In a similar fashion, we may obtain a half angle formula for sine, and by using a quotient formula, obtain a half angle formula for tangent. We summarize these formulas below. Half Angle Formulas For all applicable angles $$\theta$$: • $$\cos\left(\dfrac{\theta}{2}\right) = \pm \sqrt{\dfrac{1 + \cos(\theta)}{2}}$$ • $$\sin\left(\dfrac{\theta}{2}\right) = \pm \sqrt{\dfrac{1 - \cos(\theta)}{2}}$$ • $$\tan\left(\dfrac{\theta}{2}\right) = \pm \sqrt{\dfrac{1 - \cos(\theta)}{1+\cos(\theta)}}$$ where the choice of $$\pm$$ depends on the quadrant in which the terminal side of $$\dfrac{\theta}{2}$$ lies. Example $$\PageIndex{1}$$: 1. Use a half angle formula to find the exact value of $$\cos\left(15^{\circ}\right)$$. 2. Suppose $$-\pi \leq \theta \leq 0$$ with $$\cos(\theta) = -\frac{3}{5}$$. Find $$\sin\left(\frac{\theta}{2}\right)$$. 3. Use the identity given in number \ref{doubleanglesinewtan} of Example \ref{doubleangleex} to derive the identity $\tan\left(\dfrac{\theta}{2}\right) = \dfrac{\sin(\theta)}{1+\cos(\theta)}$ Solution 1. To use the half angle formula, we note that $$15^{\circ} = \frac{30^{\circ}}{2}$$ and since $$15^{\circ}$$ is a Quadrant I angle, its cosine is positive. Thus we have $\begin{array}{rcl} \cos\left(15^{\circ}\right) & = & + \sqrt{\dfrac{1+\cos\left(30^{\circ}\right)}{2}} = \sqrt{\dfrac{1+\frac{\sqrt{3}}{2}}{2}}\\ & = & \sqrt{\dfrac{1+\frac{\sqrt{3}}{2}}{2}\cdot \dfrac{2}{2}} = \sqrt{\dfrac{2+\sqrt{3}}{4}} = \dfrac{\sqrt{2+\sqrt{3}}}{2}\\ \end{array}$ Back in Example \ref{cosinesumdiffex}, we found $$\cos\left(15^{\circ}\right)$$ by using the difference formula for cosine. In that case, we determined $$\cos\left(15^{\circ}\right) = \frac{\sqrt{6}+ \sqrt{2}}{4}$$. The reader is encouraged to prove that these two expressions are equal. 1. If $$-\pi \leq \theta \leq 0$$, then $$-\frac{\pi}{2} \leq \frac{\theta}{2} \leq 0$$, which means $$\sin\left(\frac{\theta}{2}\right) < 0$$. Theorem \ref{halfangle} gives $\begin{array}{rcl} \sin\left(\dfrac{\theta}{2} \right) & = & -\sqrt{\dfrac{1-\cos\left(\theta \right)}{2}} = -\sqrt{\dfrac{1- \left(-\frac{3}{5}\right)}{2}}\\ & = & -\sqrt{\dfrac{1 + \frac{3}{5}}{2} \cdot \dfrac{5}{5}} = -\sqrt{\dfrac{8}{10}} = -\dfrac{2\sqrt{5}}{5}\\ \end{array}$ 1. Instead of our usual approach to verifying identities, namely starting with one side of the equation and trying to transform it into the other, we will start with the identity we proved in number \ref{doubleanglesinewtan} of Example \ref{doubleangleex} and manipulate it into the identity we are asked to prove. The identity we are asked to start with is $$\; \sin(2\theta) = \frac{2\tan(\theta)}{1 + \tan^{2}(\theta)}$$. If we are to use this to derive an identity for $$\tan\left(\frac{\theta}{2}\right)$$, it seems reasonable to proceed by replacing each occurrence of $$\theta$$ with $$\frac{\theta}{2} $\begin{array}{rcl} \sin\left(2 \left(\frac{\theta}{2}\right)\right) & = & \dfrac{2\tan\left(\frac{\theta}{2}\right)}{1 + \tan^{2}\left(\frac{\theta}{2}\right)} \\ \sin(\theta) & = & \dfrac{2\tan\left(\frac{\theta}{2}\right)}{1 + \tan^{2}\left(\frac{\theta}{2}\right)} \\ \end{array}$ We now have the \(\sin(\theta)$$ we need, but we somehow need to get a factor of $$1+\cos(\theta)$$ involved. To get cosines involved, recall that $$1 + \tan^{2}\left(\frac{\theta}{2}\right) = \sec^{2}\left(\frac{\theta}{2}\right)$$. We continue to manipulate our given identity by converting secants to cosines and using a power reduction formula $\begin{array}{rcl} \sin(\theta) & = & \dfrac{2\tan\left(\frac{\theta}{2}\right)}{1 + \tan^{2}\left(\frac{\theta}{2}\right)} \\ \sin(\theta) & = & \dfrac{2\tan\left(\frac{\theta}{2}\right)}{\sec^{2}\left(\frac{\theta}{2}\right)} \\ \sin(\theta) & = & 2 \tan\left(\frac{\theta}{2}\right) \cos^{2}\left(\frac{\theta}{2}\right) \\ \sin(\theta) & = & 2 \tan\left(\frac{\theta}{2}\right) \left(\dfrac{1 + \cos\left(2 \left(\frac{\theta}{2}\right)\right)}{2}\right) \\ \sin(\theta) & = & \tan\left(\frac{\theta}{2}\right) \left(1+\cos(\theta) \right) \\ \tan\left(\dfrac{\theta}{2}\right) & = & \dfrac{\sin(\theta)}{1+\cos(\theta)} \\ \end{array}$ Our next batch of identities, the Product to Sum Formulas,\footnote{These are also known as the Prosthaphaeresis Formulas and have a rich history. The authors recommend that you conduct some research on them as your schedule allows.} are easily verified by expanding each of the right hand sides in accordance with Theorem \ref{circularsumdifference} and as you should expect by now we leave the details as exercises. They are of particular use in Calculus, and we list them here for reference. Note: Product to Sum Formulas For all angles $$\alpha$$ and $$\beta$$, \index{Product to Sum Formulas} • $$\cos(\alpha)\cos(\beta) = \frac{1}{2} \left[ \cos(\alpha - \beta) + \cos(\alpha + \beta)\right]$$ • $$\sin(\alpha)\sin(\beta) = \frac{1}{2} \left[ \cos(\alpha - \beta) - \cos(\alpha + \beta)\right]$$ • $$\sin(\alpha)\cos(\beta) = \frac{1}{2} \left[ \sin(\alpha - \beta) + \sin(\alpha + \beta)\right]$$ Related to the Product to Sum Formulas are the Sum to Product Formulas, which we will have need of in Section \ref{TrigEquIneq}. These are easily verified using the Product to Sum Formulas, and as such, their proofs are left as exercises. Note: Sum to Product Formulas: For all angles $$\alpha$$ and $$\beta$$: 1. $$\cos(\alpha) + \cos(\beta) = 2 \cos\left( \dfrac{\alpha + \beta}{2}\right)\cos\left( \dfrac{\alpha - \beta}{2}\right)$$ 2. $$\cos(\alpha) - \cos(\beta) = - 2 \sin\left( \dfrac{\alpha + \beta}{2}\right)\sin\left( \dfrac{\alpha - \beta}{2}\right)$$ 3. $$\sin(\alpha) \pm \sin(\beta) = 2 \sin\left( \dfrac{\alpha \pm \beta}{2}\right)\cos\left( \dfrac{\alpha \mp \beta}{2}\right)$$ Example $$\PageIndex{1}$$: 1. Write $$\; \cos(2\theta)\cos(6\theta) \;$$ as a sum. 2. \Write $$\; \sin(\theta) - \sin(3\theta) \;$$ as a product. Solution 1. Identifying $$\alpha = 2\theta$$ and $$\beta = 6\theta$$, we find $\begin{array}{rcl} \cos(2\theta)\cos(6\theta) & = & \frac{1}{2} \left[ \cos(2\theta - 6\theta) + \cos(2\theta + 6\theta)\right]\\ & = & \frac{1}{2} \cos(-4\theta) + \frac{1}{2}\cos(8\theta) \\ & = & \frac{1}{2} \cos(4\theta) + \frac{1}{2} \cos(8\theta), \end{array}$ where the last equality is courtesy of the even identity for cosine, $$\cos(-4\theta) = \cos(4\theta)$$. 1. Identifying $$\alpha = \theta$$ and $$\beta = 3\theta$$ yields $\begin{array}{rcl} \sin(\theta) - \sin(3\theta) & = & 2 \sin\left( \dfrac{\theta - 3\theta}{2}\right)\cos\left( \dfrac{\theta + 3\theta}{2}\right) \\ & = & 2 \sin\left( -\theta \right)\cos\left( 2\theta \right) \\ & = & -2 \sin\left( \theta \right)\cos\left( 2\theta \right), \\ \end{array}$ where the last equality is courtesy of the odd identity for sine, $$\sin(-\theta) = -\sin(\theta)$$. The reader is reminded that all of the identities presented in this section which regard the circular functions as functions of angles (in radian measure) apply equally well to the circular (trigonometric) functions regarded as functions of real numbers. In Exercises \ref{idengraphfirst} - \ref{idengraphlast} in Section \ref{TrigGraphs}, we see how some of these identities manifest themselves geometrically as we study the graphs of the these functions. In the upcoming Exercises, however, you need to do all of your work analytically without graphs. ## Contributors • Carl Stitz, Ph.D. (Lakeland Community College) and Jeff Zeager, Ph.D. (Lorain County Community College)	2020-01-27T18:17:14	{ "domain": "libretexts.org", "url": "https://math.libretexts.org/Bookshelves/Precalculus/Book%3A_Precalculus_(Stitz-Zeager)/10%3A_Foundations_of_Trigonometry/10.4%3A_Trigonometric_Identities", "openwebmath_score": 0.9533764123916626, "openwebmath_perplexity": 235.8874990521153, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9937100967454597, "lm_q2_score": 0.8397339656668287, "lm_q1q2_score": 0.8344521202632329 }
https://stats.stackexchange.com/questions/336321/when-is-the-law-of-product-of-probabilities-applicable	# When is the law of product of probabilities applicable? The sample space of the experiment of throwing a green and a red dice has 36 elements. The event, say $A$, that the sum $x+y>8$ in $(x,y)$, where $x$ is an outcome of the green die and $y$ that of red die will occur has the probability: $$P(A)=\frac{10}{36}=\frac{5}{18}$$ And the probability of the event, say C, that a number greater than 4 will turn up on the green die is: $$P(C)=\frac{12}{36} = \frac{1}{3}$$ And the probability of the intersection $A \cap C$ is: $$P(A \cap C) = \frac{7}{36}$$ which is not equal to the product of the probabilities of $A$ and $C$, that is $$P(A) \cdot P(C)=\frac{5}{54} \neq \frac{7}{36}$$ Is the law of product even applicable here? If so, how? • The problem is that the events A and C that you selected are NOT independent, So the product rule may not apply. – Michael Chernick Mar 23 '18 at 17:01 The events A & C are dependent event. That is, if one of them occurs first, the odds of second event are changed. The law of product for dependent events is: P(A∩C) = P(C).P(A\|C) Where P(A\|C) is conditional probability of the event A given event C has already occurred. If event C (x > 4) has already occurred, then we have a sample space of 12 instead of 36. This is already clear in your calculation of P(C). There are 7 Dice throw cases where event A (x + y > 8) occurs, given that event C (x > 4) already occurred: (5,4) (5,5) (5,6) (6,3) (6,4) (6,5) (6,6) hence P(A\|C) = 7/12 P(A∩C) = P(C).P(A\|C) = 1/3 . 7/12 = 7/36 which is same as the P(A∩C) calculated independently. By the way, we can also prove the same other way around, that is: P(A∩C) = P(A).P(C\|A) (But I chose the case which I was comfortable with.) PS: pls disregard my clumsy writing, i still have not adapted to write fractions & other mathematic symbols in this forum. Product of probabilities equals their joint probability only for independent events. In fact, it is a part of definition of independence: Two events A and B are independent (often written as $A \perp B$ or $A \perp\!\!\!\perp B$) if their joint probability equals the product of their probabilities: $$\mathrm{P}(A \cap B) = \mathrm{P}(A)\mathrm{P}(B)$$ • Could you explain why $A$ and $C$ are not independent events? – Samama Fahim Mar 23 '18 at 17:44 • @SamamaFahim because the first event "tells" you something about possible outcomes of the first one. Imagine more extreme example A=z<5 and B=z<4.9999..., if I told you that A is true, would it change anything about your bet on B? – Tim Mar 23 '18 at 17:57	2019-08-21T03:19:43	{ "domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/336321/when-is-the-law-of-product-of-probabilities-applicable", "openwebmath_score": 0.8082774877548218, "openwebmath_perplexity": 386.15076926051677, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357227168957, "lm_q2_score": 0.849971181358171, "lm_q1q2_score": 0.8344470720191975 }
https://math.stackexchange.com/questions/2573388/expected-value-of-position-in-subset	# Expected value of position in subset. I have set of $n$ elements $[0, ... n-1]$. I randomly pick a subset $S$ of $k$ elements (also ordered). Assume I have $t \in \{1, .. k\}$. What is expected value of $t$-th position in ordered subset? I try to solve it like this: $E\big[S_{t}\big] = \sum_{i=1}^{n} \big((i - 1) * P(S_{t}=(i-1))\big)$ $P(S_{t} = i) = \cfrac{\binom{i - 1}{t-1} * \binom{n-i}{k-t}}{\binom{n}{k}} .$ Here: $\binom{i - 1}{t-1}$ is for selecting prefix; $\binom{n-i}{k-t}$ is for postfix; But I can't simplify this equality much further (I guess that sum of product of combinations could be simplified). • I think you will find it easier if you start indexing from $1$; i.e. considering the set $\{1,\ldots, n\}$, and also assuming $t \in \{1,\ldots, k\}$: this will facilitate working with Binomials. Can you also explain why you think you need the factor $\frac1n$ in the probability. Have you tried some explicit examples? Eg. n = 5, k = 3, and t = 2? – owen88 Dec 19 '17 at 17:47 • It was my last thought about $\frac{1}{n}$, may be it's not valid (thought it's like select an element $i$ and then select prefix and postfix). – Sindbag Dec 19 '17 at 17:55 • You do not need the `select an element i' part, as this is assumed in the statement $P[S_t = i]$; i.e. you are counting those events for which the $t$-th element is $i$; otherwise your logic is correct. It'll help if you check some small examples as I prompted above. – owen88 Dec 19 '17 at 18:44 As in my comment above, I will consider the set to be $\{1,\ldots, n\}$, and let $t \in \{1,\ldots, k\}$ denote the position of interest, i.e. $S_t$ is the value of the $t$-th element of the random subset $S$, when in increasing order. In this case, we can show that $$P(S_t = i) = \frac{\binom{i-1}{t-1} \binom{n-i}{k-t} }{\binom{n}{k} }$$ Now substituting this into the formula for expectation: \begin{align} E[S_t] & = \sum_{i=1}^n i \frac{\binom{i-1}{t-1} \binom{n-i}{k-t} }{\binom{n}{k} } \\ & = \frac{1}{\binom{n}{k} } \sum_{i = 1}^n i \binom{i-1}{t-1} \binom{n-i}{k-t} \end{align} Using the formula: $$i \binom{i-1}{t-1} = t \binom{i}{t}$$ then the summation can be written as: \begin{align} \sum_{i=1}^n i \binom{i-1}{t-1}\binom{n-i}{k-t} &= t \sum_{i=1}^n \binom{i}{t}\binom{n-i}{k-t} \\ & = t \left( \sum_{i=0}^n \binom{i}{t}\binom{n-i}{k-t} - \binom{0}{t}\binom{n}{k-t} \right) \end{align} where we have extended the sum over the range $i = 0,\ldots, n$ (instead of $i = 1, \ldots, n$). The first of the two terms is identified as the Vandermonde identity, whilst the second simplifies to $0$ so that \begin{align} \sum_{i=1}^n i \binom{i-1}{t-1}\binom{n-i}{k-t} &= t \binom{n+1}{k+1} \end{align} So over all, we have: \begin{align} E[S_t] &= t \frac{ \binom{n+1}{k+1} }{\binom{n}{k}} \\ & = t \frac{n+1}{k+1} \end{align} • Are you sure about Vandermonde's identity? We have $i$ changing, whereas Vandermonde's requires $t$ to be changing. – Sindbag Dec 19 '17 at 18:56 • See equation (9), en.wikipedia.org/wiki/… If you agree with my answer, will you please like and accept it. Thanks – owen88 Dec 19 '17 at 19:07 • Great, thanks a lot! Can't like it - too little rating :( – Sindbag Dec 19 '17 at 19:17 • No problem! Glad to have helped. – owen88 Dec 19 '17 at 19:18	2019-05-23T14:56:42	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2573388/expected-value-of-position-in-subset", "openwebmath_score": 0.9980597496032715, "openwebmath_perplexity": 620.4021923242601, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357227168956, "lm_q2_score": 0.8499711775577736, "lm_q1q2_score": 0.8344470682882117 }
https://math.stackexchange.com/questions/2968435/why-is-an-element-in-mathbbz-sqrt2-a-unit-iff-a2-2b2-pm-1/2968442	# Why is an element in $\mathbb{Z}[\sqrt{2}]$ a unit iff $a^2-2b^2 = \pm 1$? In this answer and other's that I've read, finding the units in $$\mathbb{Z}[\sqrt{2}]$$ (elements of the form $$a + b\sqrt{2}$$ for integers $$a$$ and $$b$$) is equivalent to finding the solutions to $$a^2-2b^2 = \pm 1$$. So from my understanding, for a ring $$R$$, $$u \in R$$ is a unit if there exists a $$v \in R$$ such that $$uv = vu = 1_R$$, with $$1_R$$ being the multiplicative identity in the ring. So I have 2 questions: 1. My understanding is that we arrive at this because $$(a+b\sqrt{2})(a-b\sqrt{2}) = a^2-2b^2$$, so units can only come in conjugate pairs. I could intuitively reason that this is because $$(a+b\sqrt{2})x$$ would have a $$\sqrt{2}$$ term if $$x$$ is not conjugate, but I'm not sure if this is strictly the reason why the units are only conjugates. 2. Why can they multiply to $$-1$$ and still be a unit? Is it because if $$(a+b\sqrt{2})(a-b\sqrt{2}) = a^2-2b^2 = -1$$, then $$(a+b\sqrt{2})^2(a-b\sqrt{2})^2 = (-1)^2 = 1$$, and thus the inverse of $$(a+b\sqrt{2})$$ is $$(a+b\sqrt{2})(a-b\sqrt{2})^2$$? • Are you familiar with the norm $N(x)$ defined on $\mathbb Z[\sqrt 2]$? – Frpzzd Oct 23 '18 at 23:31 • @Frpzzd I am not - this question came up in one of my first lectures on the topic of rings, so I'm very new to it. – rb612 Oct 23 '18 at 23:32 • Okay. I will define it for you in my answer. – Frpzzd Oct 23 '18 at 23:33 • Have I explained by answer sufficiently? Please let me know if anything in my answer is confusing or worded badly so that I can clarify. – Frpzzd Oct 23 '18 at 23:43 Define the norm $$N:\mathbb Z[\sqrt 2]\mapsto \mathbb Z$$ as follows: $$N(a+b\sqrt 2)=a^2-2b^2\tag{1}$$ or $$N(x)=x\bar{x}\tag{2}$$ Then notice that for all $$x,y\in\mathbb Z[\sqrt 2]$$, we have that $$N(xy)=N(x)N(y)\tag{3}$$ Thus, suppose that $$x\in\mathbb Z[\sqrt 2]$$ is a unit. Then let $$x^{-1}$$ be its inverse. We have by $$(3)$$ that $$1=N(1)=N(xx^{-1})=N(x)N(x^{-1})$$ and so $$N(x)$$, which is an integer, is a unit, with multiplicative inverse $$N(x^{-1})$$. Since the only units in $$\mathbb Z$$ are $$\pm 1$$, we have that $$N(x)=a^2-2b^2$$ must equal $$+1$$ or $$-1$$. This proves that $$a^2-2b^2$$ must equal $$1$$ or $$-1$$ if $$x$$ is a unit. Now we must prove the converse. Suppose $$N(x)=a^2-2b^2=\pm 1$$. Then, by $$(2)$$, $$x\bar{x}=1$$ or $$x\bar{x}=-1$$. If the former is true, then $$\bar{x}$$ is the inverse of $$x$$ and $$x$$ is a unit; if the latter is true, then $$-\bar{x}$$ is the inverse of $$x$$ and $$x$$ is a unit. Proof complete! Now I shall address a generalized form of your first question. Suppose that $$x,y\in\mathbb Z[\sqrt 2]$$, and that $$xy\in\mathbb Z$$. Write $$x=a+b\sqrt 2$$ and $$y=c+d\sqrt 2$$. Then $$xy=(ac+2bd)+(ad+bc)\sqrt{2}$$ For $$xy\in\mathbb Z$$ to hold, it must be true that the coefficient $$ad+bc$$ of $$\sqrt 2$$ vanishes. Thus, we have $$ad+bc=0$$ or $$\frac{a}{b}=-\frac{c}{d}$$ ...which implies that the ratios of the rational and irrational parts of $$x$$ and $$y$$ are equal and opposite. This is equivalent to stating the existence of $$z\in\mathbb Z[\sqrt 2]$$ such that $$z\|x$$ and $$\bar{z}\|y$$. If you apply this to units, you can see that it is indeed true that all units must come in conjugate pairs. • Thanks so much for the thorough response! This is great. My main question from this though is why can we/do define the norm as such and then proceed from there? In other words, did we arbitrarily define that norm to proceed with the proof, or is it the only option for the norm of an element in this ring? – rb612 Oct 23 '18 at 23:51 • @rb612 We choose that norm mostly because of its nice multiplicative property $N(xy)=N(x)N(y)$. It's not really that arbitrary; I've just assigned the value $a^2-2b^2$ mentioned in your question to a function $N(x)$, giving it a name. I could have done the proof without defining $N(x)$ at all and instead referring to $a^2-2b^2$; it's just the convention that I am used to using. – Frpzzd Oct 23 '18 at 23:53 • @rb612 One more thing... once I've answered all of your questions, don't forget to $\checkmark$! :D – Frpzzd Oct 23 '18 at 23:55	2019-03-19T18:50:41	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2968435/why-is-an-element-in-mathbbz-sqrt2-a-unit-iff-a2-2b2-pm-1/2968442", "openwebmath_score": 0.9343242049217224, "openwebmath_perplexity": 106.1438501722096, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357195106374, "lm_q2_score": 0.8499711794579723, "lm_q1q2_score": 0.8344470674284775 }
https://math.stackexchange.com/questions/3704345/will-the-set-of-algebraic-polynomials-be-dense-in-space-l-infty0-1	# Will the set of algebraic polynomials be dense in space $L_{\infty}(0,1)$? I know that the set of algebraic polynomials is dense in space $$L_{p}(a,b)$$, where $$1 \leq p < \infty$$ and $$a,b \in \mathbb{R}$$. However, what about $$L_{\infty}(0,1)$$? Will the set of algebraic polynomials be dense in space $$L_{\infty}(0,1)$$? In my opinion it is not, but is there any example that prove this fact? • Since $L_{\infty}(0,1)$ is not separable, it cannot have the set of real polynomials as a dense subset. If it were true, the set of algebraic polynomials with rational coefficients would be dense in $L_{\infty}(0,1)$. But that is not possible since the set of rational polynomials is countable. – Darth Lubinus Jun 3 at 18:48 • @DarthLubinus, Could you explain me why this is impossible? – Sh VavilenT Jun 3 at 19:52 • Sure, I'll give some details in an answer! – Darth Lubinus Jun 3 at 19:55 I'm going to try and ellaborate on my previous comment. First, let's call $$\mathbb{R}[X]$$ the ring of real polynomials, and suppose $$\mathbb{R}[X]$$ was a dense subset of $$L_{\infty}(0,1)$$. Therefore, we can prove that $$\mathbb{Q}[X]$$ (the subset of all rational polynomials) is dense in $$L_{\infty}(0,1)$$ by seeing that it is dense in $$\mathbb{R}[X]$$. Let $$p(x)=a_{n}x^{n}+...+a_{1}x+a_{0}$$ be a polynomial with coefficients in $$\mathbb{R}$$, and $$\varepsilon>0$$. Then, for every $$a_{i}$$ we can find a $$b_{i}\in \mathbb{Q}$$ such that $$\|b_{i}-a_{i}\|<\varepsilon/n$$. By taking $$q(x)=b_{n}x^{n}+...+b_{1}x+b_{0}$$, we have that $$q\in \mathbb{Q}[X]$$, and for every $$x\in [0,1]$$, $$\|p(x)-q(x)\|=\Big\|\sum_{i=0}^{n}(a_{i}-b_{i})x\Big\|\leq \sum_{i=0}^{n}\|a_{i}-b_{i}\|\|x\|\leq n\frac{\varepsilon}{n}=\varepsilon,$$ so $$\|\|p-q\|\|_{\infty}\leq\varepsilon$$. Therefore, $$\mathbb{Q}[X]$$ is dense in $$\mathbb{R}[X]$$ and, using our assumption that $$\mathbb{R}[X]$$ is dense in $$L_{\infty}(0,1)$$, we conclude that $$\mathbb{Q}[X]$$ is dense in $$L_{\infty}(0,1)$$. In particular, $$L_{\infty}(0,1)$$ admits a countable dense subspace, which means that it is separable. To find our desired contradiction, we need to prove that $$L_{\infty}(0,1)$$ cannot be separable. Consider the family $$\{f_{t}\colon t\in (0,1)\}\subseteq L_{\infty}(0,1)$$, where $$f_{t}(s)=1$$ if $$s\leq t$$, and $$f_{t}(s)=0$$ otherwise. It is easy to see that, whenever $$t\neq s$$, we have $$\|\|f_{t}-f_{s}\|\|_{\infty}=1$$. Therefore, we can find an uncountable family of disjoint open balls $$A_{t}=B(f_{t},1/3)$$, where $$0 < t < 1$$. Since $$\mathbb{Q}[X]$$ is dense in $$L_{\infty}(0,1)$$, each $$A_{t}$$ must contain at least one polynomial $$q_{t}\in \mathbb{Q}[X]$$, but this is impossible, because it would imply that there is an uncountable amount of rational polynomials! Therefore, $$\mathbb{R}[X]$$ cannot be dense. In essence, what is happening is that $$L_{\infty}$$ is too big of a space to be the closure of something as comparably small as a countable subset. When $$p<\infty$$, this doesn't happen, and the fact that polynomials are dense in $$L_{p}(0,1)$$ proves that this last space is separable. Just for the sake of adding another point of view, here is a more "topological" proof (not too different from the one I just did). In metric spaces, separability is the same thing as second countability (the property of having a countable basis for the topology of the metric space). If $$L_{\infty}(0,1)$$ were separable/secound countable, then the family $$B=\{f_{t}\}_{t\in (0,1)}$$ would be another second countable space (as a subspace of one). But $$\|\|f_{t}-f_{s}\|\|_{\infty}=1$$ when $$t\neq s$$, so $$B$$ must be discrete. In that case, $$B$$ must be countable, which is again a contradiction. I hope this helps!	2020-08-10T22:41:05	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3704345/will-the-set-of-algebraic-polynomials-be-dense-in-space-l-infty0-1", "openwebmath_score": 0.9656897783279419, "openwebmath_perplexity": 80.71392591478724, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.981735718976261, "lm_q2_score": 0.8499711794579723, "lm_q1q2_score": 0.834447066974273 }
http://mathhelpforum.com/calculus/256026-volume-solid-revolution.html	# Thread: Volume of the solid of revolution 1. ## Volume of the solid of revolution There are a region bounded by the curve y=cos^2(x)sin^4(x) and the x-axis , where pi<x<2pi. Find the volumn of the solid of revolution when the region is revolved about the y-axis 2. ## Re: Volume of the solid of revolution Hi Lau, what have you tried already? It's not clear if you are stuck on what method to use to solve this or if the integration is too difficult. 3. ## Re: Volume of the solid of revolution Hey Lau, I solved this problem using the shell method. It is best to start with a rough sketch of the function on the given interval (Will give a M shape), and once you do you will realize that we should split the region again this time considering: pi <x<3pi/2 and 3pi/2<x<2pi. This is because from x=pi to x=3pi/2 , the function "starts" then "ends". Also, this will help us in the calculations as the function "repeats" its self on 3pi/2 < x < 2pi. Using the shell method, I found the height of a shell to be: $\displaystyle h=f(x)= \sin^4x \cos^2x$ and a radius about the y axis to be $\displaystyle r=x$. Then the volume of the first cycle of the periodic function is: $\displaystyle V= \int_{\pi}^{3 \pi/2} 2 \pi r \cdot h \, dx$ $\displaystyle V= \int_{\pi}^{3 \pi/2} 2 \pi (x) \cdot (\sin^4x \cos^2x) \, dx = \frac{\pi}{36} + \frac{5 \pi}{64}$ So finally, the total volume is just $\displaystyle 2V = \frac{\pi}{18} + \frac{5 \pi}{32} \approx 5.02$ 4. ## Re: Volume of the solid of revolution Originally Posted by sakonpure6 Hey Lau, I solved this problem using the shell method. It is best to start with a rough sketch of the function on the given interval (Will give a M shape), and once you do you will realize that we should split the region again this time considering: pi <x<3pi/2 and 3pi/2<x<2pi. This is because from x=pi to x=3pi/2 , the function "starts" then "ends". Also, this will help us in the calculations as the function "repeats" its self on 3pi/2 < x < 2pi. Using the shell method, I found the height of a shell to be: $\displaystyle h=f(x)= \sin^4x \cos^2x$ and a radius about the y axis to be $\displaystyle r=x$. Then the volume of the first cycle of the periodic function is: $\displaystyle V= \int_{\pi}^{3 \pi/2} 2 \pi r \cdot h \, dx$ $\displaystyle V= \int_{\pi}^{3 \pi/2} 2 \pi (x) \cdot (\sin^4x \cos^2x) \, dx = \frac{\pi}{36} + \frac{5 \pi}{64}$ So finally, the total volume is just $\displaystyle \color{red}{2V = \frac{\pi}{18} + \frac{5 \pi}{32} \approx 5.02}$ @sakonpure6, you have a conceptual mistake. $\displaystyle 3.30405 \approx \int_{1.5\pi }^{2\pi } {\left( {2\pi x} \right)\left( {{{\sin }^4}(x){{\cos }^2}(x)} \right)dx} \ne \int_\pi ^{1.5\pi } {\left( {2\pi x} \right)\left( {{{\sin }^4}(x){{\cos }^2}(x)} \right)dx \approx 2.0963}$ Look at the graph again. 5. ## Re: Volume of the solid of revolution Originally Posted by Plato @sakonpure6, you have a conceptual mistake. $\displaystyle 3.30405 \approx \int_{1.5\pi }^{2\pi } {\left( {2\pi x} \right)\left( {{{\sin }^4}(x){{\cos }^2}(x)} \right)dx} \ne \int_\pi ^{1.5\pi } {\left( {2\pi x} \right)\left( {{{\sin }^4}(x){{\cos }^2}(x)} \right)dx \approx 2.0963}$ Look at the graph again. EDIT a Typo: 2.50963 not 2.0963 6. ## Re: Volume of the solid of revolution Oh thanks for pointing that out Plato! So, Lau the correct answer is: 3.304 +2.096 = 5.40 Edit: Plato, did something in the equation give away that the function was not periodic - had different amplitudes? Or did you plot to verify? 7. ## Re: Volume of the solid of revolution It is best to start with a rough sketch of the function on the given interval (Will give a M shape), and once you do you will realize that we should split the region again this time considering: pi <x<3pi/2 and 3pi/2<x<2pi. This is because from x=pi to x=3pi/2 , the function "starts" then "ends". Also, this will help us in the calculations as the function "repeats" its self on 3pi/2 < x < 2pi. 8. ## Re: Volume of the solid of revolution Originally Posted by dungtuyet It is best to start with a rough sketch of the function on the given interval (Will give a M shape), and once you do you will realize that we should split the region again this time considering: pi <x<3pi/2 and 3pi/2<x<2pi. This is because from x=pi to x=3pi/2 , the function "starts" then "ends". Also, this will help us in the calculations as the function "repeats" its self on 3pi/2 < x < 2pi. This is wrong. In fact, it completely misses the point of this whole thread! @dungtuyet, did you even read the entire thread? If you did, what part did you not get? One more time. Here is a standard example. What is the volume of the graph of $y=\|x-3\|,~ 2\le x\le 4$ about the $y \text{-axis}~?$ Please study each of these: from $x=2\to 3$; then from $x=3\to 4$ Note that those two volumes are not equal even though the graph is symmetric about $x=3$. The radii are different? Add those two values and compare the result with this.	2019-07-22T20:39:00	{ "domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/256026-volume-solid-revolution.html", "openwebmath_score": 0.7198362946510315, "openwebmath_perplexity": 804.2240287731624, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357184418847, "lm_q2_score": 0.8499711794579723, "lm_q1q2_score": 0.8344470665200685 }
https://math.stackexchange.com/questions/1186124/how-to-find-the-general-solution-to-this-ode/1186134	# How to find the general solution to this ODE… Q: Find the general solution to the ODE:$$t^2y''-ty'+y=0~~~~~~~~()$$given that $y_1=t~~$is a solution. My intuitive solution: Let $x=ln(t)$, then $y'=\frac{dy}{dx}\frac{1}{t}$ and $y''=\frac{d^2y}{dx^2}\frac{1}{t^2}-\frac{dy}{dx}\frac{1}{t^2}$ Plug into $()$ we get: $$t^2(\frac{d^2y}{dx^2}\frac{1}{t^2}-\frac{dy}{dx}\frac{1}{t^2})-t(\frac{dy}{dx}\frac{1}{t})+y=0$$$$\rightarrow~~\frac{d^2y}{dx^2}-2\frac{dy}{dx}+y=0$$ So the characteristic equation for this is $r^2-2r+1=0~\rightarrow~(r-1)^2=0~\rightarrow~r=1$. Then $y_1(x)=y_2(x)=e^x~\rightarrow~y_1(t)=y_2(t)=e^{ln(t)}=t.~$But now the Wronskian is certainly zero. So I tried to multiply a $t$ such that $y_2(t)=t^2.~$But $y'_2=2t~~$and$~~ y''_2=2.~$Substituting these into $()$ I found that this doesn't solve the equation. Could someone please give me a hint on how to find the general solution? • Since $r=1$ is a double-root try $y = Cxe^x$ for the second solution (the first solution is as you have found $y=e^x$). – Winther Mar 12 '15 at 0:26 • Oh...OK, got it, I shouldn't have multiplied the factor to $y_2(t)$ but $y_2(x)$... Thanks! – Ruihong Yuan Mar 12 '15 at 0:47 Based on your analysis, as the solutions of the characteristic equation is $r=1$(double root), the two basic solutions for this ODE are $y_1(x)=e^x\rightarrow y_1(t)=t$ and $y_2(x)=xe^x\rightarrow y_2(t)=t\ln\|t\|$. Substituting these solutions into $()$ you will find they are the solutions. And their combinations are also the solutions, i.e., $$y(t)=C_1t+C_2t\ln\|t\|$$ where $C_1$ and $C_2$ are constants. HINT Consider the method of the reduction of order. You know one of the solutions to the differential equation, which is: $y_1 = t$ Now, call the second solution $y_2$ where $y_2 = vy_1 = vt$, where $v$ is a function of $t$. Taking the derivative, we get: $$y_2 = vt$$ $$y_2' = v't + v$$ $$y_2'' = v''t + v' + v' = v''t + 2v'$$ Substituting this into our equation: $$t^2(v''t + 2v') -t(v't + v) + vt = 0$$ $$t^3v'' + 2t^2v' - t^2v' - vt + vt =0,\ t^3v'' + 2t^2v' - t^2v' = 0$$ Note: Upon using this method the $y$ term has dropped out. This method ensures that, and now we can find our second solution: $$t^3v'' + t^2v' = 0$$ Now, we let $w = v'$, $w' = v''$ It follows that: $$t^3w' + t^2w = 0$$ $$w' = -\frac{w}{t}$$ $$\frac{dw}{w} = -\frac{dt}{t}$$ $$\ln w = -\ln t$$ $$w = -t,\ v' = -t,\ v = -\frac{t^2}{2}$$ Now, our second solution is $$y_2 = vt = -\frac{t^3}{2}$$ Now, our general solution is formed: $$y(t) = c_1t + c_2\frac{t^3}{2}$$ • From $\ln w=-\ln t$ it follows $w=1/t$, you went a little fast there. Then $v=\ln\|t\|$ in accordance with the other answers. – Lutz Lehmann Mar 12 '15 at 3:06 • @LutzL ty for catching my mistake – Varun Iyer Mar 12 '15 at 23:23 this is an example of eulers equation. try $y = t^k$ sub it in the equation. find $$k(k-1) - k + 1= 0\to k = 1, 1.$$ the two solutions are $$t, t \ln t$$ and the general solution is $$y = At + B t\ln t.$$ • You meant the type is "Euler equation". – Lutz Lehmann Mar 12 '15 at 3:07	2020-02-21T19:42:58	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1186124/how-to-find-the-general-solution-to-this-ode/1186134", "openwebmath_score": 0.9420229196548462, "openwebmath_perplexity": 317.56059431883415, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.981735720045014, "lm_q2_score": 0.8499711775577736, "lm_q1q2_score": 0.8344470660171893 }
https://stats.stackexchange.com/questions/103800/calculate-probability-area-under-the-overlapping-area-of-two-normal-distributi/103808	# Calculate probability (area) under the overlapping area of two normal distributions I have two normal distributions defined by their averages and standard deviations. Sample 1: Mean=5.28; SD=0.91 Sample 2: Mean=8.45; SD=1.36 You can see how they look like in the next image: How can I get the probability to obtain an individual from the overlapping area (green)? Is the probability the same as the area? • What do you mean by the probability of obtaining an individual from the area? – Juho Kokkala Jun 18 '14 at 8:37 • If you sample points from either normal distribution, you get points on the Perikymata-axis rather than on the 2-dimensional area. Furthermore, the green zone is infinitely wide, so all values sampled from either distribution are under the green zone, so in this sense the probability would be 1. – Juho Kokkala Jun 18 '14 at 8:47 • @JuhoKokkala OP likely means integral over the green area. – Vladislavs Dovgalecs Nov 21 '17 at 18:40 • @VladislavsDovgalecs What would the question "Is the probability the same as the area" then mean? – Juho Kokkala Nov 21 '17 at 19:21 • @VladislavsDovgalecs I think you misunderstood my comment, I was not asking for an interpretation of the area but trying to parse the question. In any case, please note OP's comments to the accepted answer and a followup question posted here stats.stackexchange.com/questions/103821. – Juho Kokkala Nov 23 '17 at 6:14 It is not quite clear what you mean by probability to obtain an individual from the overlapping area. This solves for the area of the green zone in your diagram: Let: • $X_1 \sim N(\mu_1,\sigma_1^2)$ with pdf $f_1(x_1)$ and cdf $F_1(x_1)$ and • $X_2 \sim N(\mu_2,\sigma_2^2)$ with pdf $f_2(x_2)$ and cdf $F_2(x_2)$, where $\mu_1 < \mu_2$. In your example, the 'black variable' corresponds to $X_1$. Let $c$ denote the point of intersection where the pdf's meet in the green zone of your plot Then, the area of your green intersection zone is simply: $$P(X_1>c) + P(X_2<c) = 1 - F_1(c) + F_2(c) = 1-\frac{1}{2} \text{erf}\left(\frac{c-\mu _1}{\sqrt{2} \sigma _1}\right)+\frac{1}{2} \text{erf}\left(\frac{c-\mu _2}{\sqrt{2} \sigma _2}\right)$$ where erf(.) is the error function. Point $c$ is the solution to $f_1(x) = f_2(x)$ within the green zone, which yields: $$c = \frac{\mu _2 \sigma _1^2-\sigma _2 \left(\mu _1 \sigma _2+\sigma _1 \sqrt{\left(\mu _1-\mu _2\right){}^2+2 \left(\sigma _1^2-\sigma _2^2\right) \log \left(\frac{\sigma _1}{\sigma _2}\right)}\right)}{\sigma _1^2-\sigma _2^2}$$ For your example, with ${\mu_1 = 5.28, \mu_2 = 8.45, \sigma_1 = 0.91, \sigma_2 = 1.36}$, this yields: $c = 6.70458...$, and the area of the green section is: 0.158413 ... • @antecessor Depends on what you mean by 'overlapping among both'. In the question you talk about some probability, but it is unclear what probabilistic interpretation the area computed here would have. – Juho Kokkala Jun 18 '14 at 8:57 • @JuhoKokkala Ok, I will write a new question to address the exact question. – antecessor Jun 18 '14 at 10:38 • the resolution of the equation pdf1=pdf2 yields to two solutions, so we have two intersection points, why you delete the other point @wolfies – user61828 Dec 1 '14 at 11:51 • Because, within the green zone, there is only one solution. – wolfies Dec 1 '14 at 12:32 • Is the equation for c given at the bottom of the answer correct? If both distributions have equal variance, the equation requires dividing by zero! – sammosummo Mar 21 '18 at 21:28 @abdelbasset, To improve @wolfies 's answer above, there are two intersection points c, let's call them c1 and c2. Here, c1=-1.2848 and c2=6.7046. @wolfies ignored c1 (it likely will be too far in the tails of the probability density functions to matter (especially if we round to just a few decimal points)). The more correct way is to find both c1 and c2, and to find the area of overlap of both functions: AREA OF OVERLAP = P(X1 > j1) + P(X2 < j1) - [P(X2 < j2)-P(X1 < j2)] therefore, using cdf's AREA OF OVERLAP = 1 - F1(j1,μ2,σ2)+F2(j1,μ2,σ2)-F2(j2,μ2,σ2)+F1(j2,μ1,σ1) • Welcome to the site, @luckapani. You cannot use the "Your Answer" field to comment or respond to comments. As a result, this would be deleted. However, there is the core of an answer here. Can you edit your post to make it more of an answer & not a comment? Since you're new here, you may want to take our tour, which contains information for new users. – gung - Reinstate Monica May 19 '15 at 16:12 Normalise the graphs to an area of 1 by dividing each by their respective standard deviation. Then use simple subtraction from a z-graph to calculate the probability of an occurrance in that overlap area. No need for erfs. • i assume that he means that he wants to compute the probability in the green region. That can be done simply by adding the area in the right tail of the first distribution to the ares in the left tail of the second distribution. There is no normalization necessary. – Michael R. Chernick Nov 2 '19 at 23:36 • The separate normalization of the two graphs will change the overlap area unless they have identical SDs. Thus, this answer is generally wrong. – whuber Nov 3 '19 at 13:52	2020-04-04T19:36:26	{ "domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/103800/calculate-probability-area-under-the-overlapping-area-of-two-normal-distributi/103808", "openwebmath_score": 0.765239417552948, "openwebmath_perplexity": 705.5092786654227, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.981735720045014, "lm_q2_score": 0.8499711756575749, "lm_q1q2_score": 0.8344470641516963 }
https://math.stackexchange.com/questions/1620904/how-to-check-convergence-divergence-of-the-series	# How to check convergence/divergence of the series How to check convergence/divergence of $\sum \dfrac{\dfrac{1}{2}+(-1)^n}{n}$ I thought in this way that it is of the form $\sum_{i=1}^n \dfrac{1}{2n}+\sum _{i=1}^n \dfrac{(-1)^n}{n}$ i.e. sum of a convergent and a divergent series and hence the original series diverges. Is it right or there are other alternatives?? I think you have a good idea (with typos) that needs further supporting evidence. Here's the result you need: Suppose $\sum a_n$ diverges and $\sum b_n$ converges. Then $\sum (a_n + b_n)$ diverges. Proof: Suppose instead that $\sum (a_n + b_n)$ converges. Then $\sum a_n = \sum [(a_n+b_n) - b_n]$ is the difference of two convergent series, hence converges. That's a contradiction, proving the result. In the problem at hand, we have $a_n = 1/2n, b_n=(-1)^n/n.$ We know $\sum a_n$ diverges, and $\sum b_n$ converges by the alternating series test. Thus your series diverges by the above. Note that \begin{align} \sum_{n=1}^{\infty}{\frac{\tfrac12+(-1)^n}{n}}&=\sum_{n=1}^{\infty}\left(\frac{\tfrac12+(-1)}{2n-1}+\frac{\tfrac12+(1)}{2n}\right)\\ &=\sum_{n=1}^{\infty}\frac{2n\cdot-\tfrac{1}{2}+(2n-1)\cdot\tfrac32}{2n(2n-1)}\\ &=\sum_{n=1}^{\infty}\frac{2n-\tfrac32}{2n(2n-1)}\\ &\geq\sum_{n=2}^{\infty}\frac{2n-n}{2n(2n-1)}\\ &=\frac12 \sum_{n=2}^{\infty}\frac{1}{2n-1} \end{align} Since the last is smaller than the first sum, and the last sum goes to $+\infty$, so does the first. Hope this helped! • How was my solution ;Is it faulty – Learnmore Jan 21 '16 at 11:50 • I think it is correct but you should prove the theorem or reference to a proof if you're using it – vrugtehagel Jan 21 '16 at 11:58 • @Amartya : your argument is correct ; the statement (the sum of a divergent series and a convergent one is divergent is true and easy). – John Steinbeck Jan 21 '16 at 12:13 • @vrugtehagl : I consider the way you write it is not very satisfactory, as you write infinite sums while you didn't prove they make sense. As the term in the sum are not all positive, the meaning of the infinite sum is not always assured. Your proof can certainly be fixed using partial sums and they arguing that your last sum goes to $=\infty$. – John Steinbeck Jan 21 '16 at 12:13	2020-07-06T17:56:29	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1620904/how-to-check-convergence-divergence-of-the-series", "openwebmath_score": 0.969565212726593, "openwebmath_perplexity": 244.39386194497908, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357205793903, "lm_q2_score": 0.8499711699569787, "lm_q1q2_score": 0.834447059009422 }
https://math.stackexchange.com/questions/1919828/prove-that-2n1-gt-n2-for-every-positive-integer-n	# Prove that $2^{n+1} \gt n^2$ for every positive integer $n$. Prove that $2^{n+1} \gt n^2$ for every positive integer $n$. I have to prove this using mathematical induction, I know how to start the proof but I get stuck towards the end because I think I either did something wrong or do not know how to continue it. Here are all of the steps I have done so far: First I prove that $1 \in S$ This is obviously true because $2^{1+1} \gt 1^2$ Now assume that $a \in S$, we have $2^{a+1} \gt a^2$, now I have to prove that $a+1 \in S$, i.e. $2^{a+2} \gt (a+1)^2$. $2^{a+2}$ becomes $2 \times 2^{a+1}$ and $(a+1)^2$ becomes $a^2+2a+1$. We now have $2 \times 2^{a+1} \gt a^2+2a+1$ Multiply both sides of $2^{a+1} \gt a^2$ by 2 and obtain $2 \times 2^{a+1} \gt 2a^2$ to complete the proof all I have to prove is that $2a^2 \gt a^2 +2a +1$ subtracting $a^2$ from both sides I get $a^2 \gt 2a+1$, but this is only true for $a \ge 3$ and I need to prove if for all positive integers. What did I do wrong, should I have taken a different approach or is there a way to finish what I started? • Then you can start your induction from $3$. – QZ0 Sep 9 '16 at 0:00 • @ThomasAndrews so I can have two base cases, when $a=1$ and when $a=2$, then prove for $a+1$, can I have two base cases? But how can $a=1$ and $a=2$ work for $2^{a+1} \gt a^2$, but not work for $a^2 \gt 2a+1$ if this is derived from the first inequality? – idknuttin Sep 9 '16 at 0:03 • You can have finite number of base cases. For your second question, you should understand the difference between "sufficient condition" and "necessary condition". – QZ0 Sep 9 '16 at 0:07 You need to prove more base cases. In the case of $a=1$, we have $2^{1+1} = 4 \gt 1^2 = 1$. In the case of $a=2$, we have $2^{2+1} = 8 \gt 2^2 = 4$. Assume that for some $a \in \mathbb N, a\ge 3,$ that $2^{a+1} \gt a^2$. Then we must show that the $a+1$ case is true. We have $2^{(a+1)+1)} = 2^{a+1} * 2^1 > 2 * a^2$, since by assumption $2^{a+1} \gt a^2$. Then we have $2^{a+1} * 2^1 > 2 * a^2 > (a+1)^2 = a^2 + 2a + 1$ for $a \ge 3$, and we are done. • why can we just say that the final part, $a^2 \gt 2a+1$ is obvious and does not require a proof? – idknuttin Sep 9 '16 at 0:16 • Well, you stated in your question that it's only true for $a \ge 3$, so I assumed it was given or perhaps proven earlier. There are many ways of proving this though. Any ideas? – David Bowman Sep 9 '16 at 0:38 • I could probably use mathematical induction again but I don't think that would be necessary, we could move everything to one side and complete the square to get $(a-1)^2 -2 \gt 0$ can we say this is obvious for $a \ge 3$, or is there an easy way of proving it? – idknuttin Sep 9 '16 at 0:46 • You might use the quadratic formula to discover that $f(a) = a^2 - 2a -1$ has zeroes $1- \sqrt{2}$ and $1+ \sqrt{2}$, both of which are less than $3$. Since f is increasing for $a \gt 1, f(a) \gt 0$ for $a \gt 1 +\sqrt{2}$. – David Bowman Sep 9 '16 at 0:52 Another way is using the function $$f(x)=2^{x+1}-x^2$$ whose derivative is $$f'(x)=2^{x+1}\ln 2-2x$$ One has $$2^{x+1}\ln 2-2x\gt 0\iff2^x\ln 2\gt x$$ Hence the derivative is always positive and (besides of increasing) so is the function $f(x)$ for $x\gt 0$ since $f(0)=2$. In particular for $x$ integer positive this is true. • But then, how do you prove that $2^x\ln(2)>x$? – gniourf_gniourf Dec 4 '16 at 17:16	2020-05-25T21:45:38	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1919828/prove-that-2n1-gt-n2-for-every-positive-integer-n", "openwebmath_score": 0.9396656155586243, "openwebmath_perplexity": 103.83798622614665, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357227168957, "lm_q2_score": 0.84997116805678, "lm_q1q2_score": 0.8344470589607469 }
https://math.stackexchange.com/questions/1703784/arranging-m-distinct-groups-of-k-elements-each	# Arranging m distinct groups of k elements each? There are n elements in total which is already divided into m groups of k elements each. Thus, n=m*k. The question is, how many arrangements of these m groups are possible? I came to a possible solution of n!/m! but am not sure if that covers all permutations possible among each group (i.e, k! ways of arranging each group). The original problem is as follows: There exists an array of n distinct numbers. This array is divided into m sub-sequences of k elements each. All elements of a subsequence are either greater than or less than all elements of another subsequence. I have to find the number of possible permutations of the indices in the array which may appear in the sorted array. • It is unclear whether you are only trying to arrange the $m$ groups or something more complicated. Of course there are $m!$ ways to "arrange" (permute) the $m$ groups. Please make your Question clearer, perhaps by detailing a small example. Mar 19 '16 at 0:17 • What was the original problem you were trying to solve? Mar 19 '16 at 0:17 • Should I change the question to reflect the original problem or add it to the existing problem? Sorry, I am new here. Mar 19 '16 at 0:19 • Please state the original problem you were trying to solve and show the work you did on that problem. In short, we would like to see the context for your current question. Mar 19 '16 at 0:22 • You could also clarify what permutations are allowed. Take a small example for now, $m=n=2$ so there are $4$ elements. Suppose the first group are the capital letters and the second group are the lower case letters. Which of the following are considered the same or distinct or not allowed? ABcd, BAcd, cdAB, cdBA, AcBd,... I.e. must the groups themselves remain together, must the groups be in a specific order, is order within the groups relevant,... Mar 19 '16 at 0:24 The groups are completely determined by the requirement that all elements of a subsequence are either greater than or less than all elements of another subsequence. As you realized, the elements in each of the $m$ groups of $k$ elements can be arranged in $k!$ orders. The $m$ subsequences can be arranged in $m!$ orders. Thus, the number of possible permutations of the indices in the array is $m!(k!)^m$. • Would this result be strictly greater than $(m.k)!$ or would their relationship depend on values of m and k? Mar 19 '16 at 1:52 • Equality holds when $m = k = 1$. However, in general, $m!(k!)^m \leq (mk)!$ since $(mk)!$ represents all permutations of $mk$ objects, while $m!(k!)^m$ represents all the permutations in which the first $k$ objects are grouped together, the second $k$ objects are grouped together, and so forth. For instance, if $m = 2$ and $k = 3$, the elements in the set $\{1, 2, 3, 4, 5, 6\}$ can be sequenced in $6!$ ways, while the two sets $\{1, 2, 3\}$ and $\{4, 5, 6\}$ can be arranged in $2!$ ways and each can be arranged internally in $3!$ ways for a total of $2!(3!)^2 = 72$ permissible arrangements. Mar 19 '16 at 3:05	2021-10-27T17:16:26	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1703784/arranging-m-distinct-groups-of-k-elements-each", "openwebmath_score": 0.6977249979972839, "openwebmath_perplexity": 204.3825282092028, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357173731318, "lm_q2_score": 0.8499711699569787, "lm_q1q2_score": 0.8344470562841947 }
https://gmatclub.com/forum/when-there-are-consecutive-integers-and-if-their-range-is-equal-to-the-214078.html	GMAT Question of the Day - Daily to your Mailbox; hard ones only It is currently 22 Jun 2018, 19:15 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Events & Promotions ###### Events & Promotions in June Open Detailed Calendar # When there are consecutive integers and if their range is equal to the new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 5600 GMAT 1: 800 Q59 V59 GPA: 3.82 When there are consecutive integers and if their range is equal to the [#permalink] ### Show Tags 29 Feb 2016, 20:39 1 4 00:00 Difficulty: 75% (hard) Question Stats: 57% (01:34) correct 43% (01:48) wrong based on 115 sessions ### HideShow timer Statistics When there are consecutive integers and if their range is equal to their median, what is the number of them? 1) The smallest number of them is 5 2) The largest number of them is 15 * A solution will be posted in two days. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only $99 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 5600 GMAT 1: 800 Q59 V59 GPA: 3.82 Re: When there are consecutive integers and if their range is equal to the [#permalink] ### Show Tags 02 Mar 2016, 22:51 Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. When consecutive integers are there and if their range is equal to their median, what is the number of them? 1) The smallest number of them is 5 2) The largest number of them is 15 In the original condition, there are 2 variables(you need to figure out the starting number and the number of them) and 1 equation(range=median), which should match with the number of equations. So you need 1 variable. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer. For 1), only 5,6,7,8,9,10,11,12,13,14,15 are possible and so as 2). Therefore, the answer is D. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$99 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Manager Joined: 07 Jul 2016 Posts: 79 GPA: 4 Re: When there are consecutive integers and if their range is equal to the [#permalink] ### Show Tags 04 Aug 2016, 14:17 1 1 MathRevolution wrote: When there are consecutive integers and if their range is equal to their median, what is the number of them? Let $$a$$ be the first element of the set, $$b$$ be the final element of the set, and $$n$$ be the range. The median of a set of consecutive numbers is the same as the mean: $$\frac{\text{first term} + \text{last term}}{2}$$ 1) The smallest number of them is 5 $$a = 5\\ b = 5 + n\\ \text{median} = \frac{a + b}{2} = \frac{10 + n}{2} = n\\ 2n = 10 + n\\ n = 10\\$$ Sufficient 2) The largest number of them is 15 $$b = 15\\ a = 15 - n\\ \text{median} = \frac{a + b}{2} =\frac{15 + 15 - n}{2} = n\\ 30 - n = 2n\\ 30 = 3n\\ n = 10$$ Sufficient Knowing $$n$$ and a start point, there are 11 elements from 5 to 15. (D) each statement alone is sufficient _________________ Please press +1 Kudos if this post helps. Intern Joined: 29 May 2017 Posts: 8 Re: When there are consecutive integers and if their range is equal to the [#permalink] ### Show Tags 25 Sep 2017, 21:02 This seems to be easier if we apply the basic logic for consecutive integers. We know that for a set of consecutive integers, mean=median. Consider the first term as 'a' and last term as 'b'. Range= b-a Median=Mean= (a+b)/2. Given: Range=Median or Mean So, we can say that b-a = (a+b)/2 b = 3a or a = b/3 Statement 1 : given smallest no is 5 or a = 5 So, b= 3a = 3*5 = 15. Hence, we get a set {5,6,7,.....,15} which will satisfy all given conditions Statement 1 is sufficient. Statement 2: Similarly, given largest no is 15 or b=15 So, a=b/3 = 5 Hence, we get the same set again as {5,.......,14,15}. Statement 2 is sufficient. Ans :D Re: When there are consecutive integers and if their range is equal to the [#permalink] 25 Sep 2017, 21:02 Display posts from previous: Sort by # When there are consecutive integers and if their range is equal to the new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Powered by phpBB © phpBB Group \| Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.	2018-06-23T02:15:07	{ "domain": "gmatclub.com", "url": "https://gmatclub.com/forum/when-there-are-consecutive-integers-and-if-their-range-is-equal-to-the-214078.html", "openwebmath_score": 0.4148408770561218, "openwebmath_perplexity": 3363.1534959355936, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9546474168650673, "lm_q2_score": 0.8740772400852111, "lm_q1q2_score": 0.834435579387894 }
https://www.physicsforums.com/threads/errors-uncertainty-in-measurement-problem.870341/	# Errors/ Uncertainty in measurement problem ## Homework Statement [/B] The surface tension (T) is measured by capillary rise formula $$T = \frac {rh ρg}{2 cos\Theta}$$ . The quantities of ρ, g and θ are taken from the table of constants while the height and diameter are measured as h = (3.00 + 0.01)cm and D = (0.250 ± 0.001)cm Find the percentage error in T My answer came out to be 1.1 % but doesn't coincide with the answer given (0.7 %). Thanks for any help! ## Homework Equations According to the book I refer, the equation for error in product or quotient is -[/B] $$\frac{\Delta Z} {Z}= \frac {\Delta A} {A} +\frac {\Delta B} {B}$$ where Δz/z, Δa/a and Δb/b are relative errors (letters in deltas present uncertainty in measurement) ## The Attempt at a Solution I have attempted the solution but have failed to acquire the given answer. Here's what I tried Acc. to $$\frac{\Delta Z} {Z}= \frac {\Delta A} {A} +\frac {\Delta B} {B}$$ we ignore constants (ρ, Θ and g) h = (3.00 + 0.01)cm D = (0.250 ± 0.001)cm so r = D/2 or r = (0.125 ± 0.001)cm so $$\frac{\Delta T} {T}= \frac {\Delta r} {r} +\frac {\Delta h} {h}$$ which gives, $$\frac{\Delta T} {T} =\frac {0.001} {0.125} +\frac {0.01} {3.00}$$ or, $$\frac{\Delta T} {T} =0.0080 + 0.0034$$ or, $$\frac{\Delta T} {T} =0.0113$$ where $$\frac{\Delta T} {T}$$ is relative error. When we multiply that by 100, we get percent error therefore percent error = 1.13 % while the answer given is 0.7 % Related Introductory Physics Homework Help News on Phys.org Samy_A Homework Helper ## Homework Statement [/B] The surface tension (T) is measured by capillary rise formula $$T = \frac {rh ρg}{2 cos\Theta}$$ . The quantities of ρ, g and θ are taken from the table of constants while the height and diameter are measured as h = (3.00 + 0.01)cm and D = (0.250 ± 0.001)cm Find the percentage error in T My answer came out to be 1.1 % but doesn't coincide with the answer given (0.7 %). Thanks for any help! ## Homework Equations According to the book I refer, the equation for error in product or quotient is -[/B] $$\frac{\Delta Z} {Z}= \frac {\Delta A} {A} +\frac {\Delta B} {B}$$ where Δz/z, Δa/a and Δb/b are relative errors (letters in deltas present uncertainty in measurement) ## The Attempt at a Solution I have attempted the solution but have failed to acquire the given answer. Here's what I tried Acc. to $$\frac{\Delta Z} {Z}= \frac {\Delta A} {A} +\frac {\Delta B} {B}$$ we ignore constants (ρ, Θ and g) h = (3.00 + 0.01)cm D = (0.250 ± 0.001)cm so r = D/2 or r = (0.125 ± 0.001)cm so $$\frac{\Delta T} {T}= \frac {\Delta r} {r} +\frac {\Delta h} {h}$$ which gives, $$\frac{\Delta T} {T} =\frac {0.001} {0.125} +\frac {0.01} {3.00}$$ or, $$\frac{\Delta T} {T} =0.0080 + 0.0034$$ or, $$\frac{\Delta T} {T} =0.0113$$ where $$\frac{\Delta T} {T}$$ is relative error. When we multiply that by 100, we get percent error therefore percent error = 1.13 % while the answer given is 0.7 % gneill Mentor When you divided the diameter by two, the error in the resulting figure should not remain the same as the original. The percentage error of the result should remain the same. So, scale both by the same factor: 0.250/2 = 0.125; 0.001/2 = 0.0005. When you divided the diameter by two, the error in the resulting figure should not remain the same as the original. The percentage error of the result should remain the same. So, scale both by the same factor: 0.250/2 = 0.125; 0.001/2 = 0.0005. Thanks, that correction seems to nail it! One Q - the instrument (say vernier calipers) was used to measure diameter and radius is just half of diameter; then why do we halve the error? Shouldn't the uncertainty remain same ? gneill Mentor Thanks, that correction seems to nail it! One Q - the instrument (say vernier calipers) was used to measure diameter and radius is just half of diameter; then why do we halve the error? Shouldn't the uncertainty remain same ? What's the rule when multiplying or dividing by a constant? Does the percent error change? Dr Transport Gold Member You really should be taking the derivative of the original formula with respect to each of the variables and do the root square addition for them to get the uncertainty. Each of the variables in your equation has an uncertainty, even the ones pulled from a reference table and it needs to be included. Ray Vickson	2021-03-08T16:09:34	{ "domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/errors-uncertainty-in-measurement-problem.870341/", "openwebmath_score": 0.7955619096755981, "openwebmath_perplexity": 1577.551206415699, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9546474220263198, "lm_q2_score": 0.8740772236840656, "lm_q1q2_score": 0.8344355682419161 }
https://quantumcomputing.stackexchange.com/questions/12604/quantum-katas-tutorials-singlequbitgates-exercise-2-globalphasei	# Quantum Katas - Tutorials - SingleQubitGates - Exercise 2 - GlobalPhaseI Program.qs namespace Quantum.Kata.SingleQubitGates { open Microsoft.Quantum.Intrinsic; open Microsoft.Quantum.Math; operation GlobalPhaseI (q : Qubit) : Unit is Adj+Ctl { X(q); Z(q); Y(q); } } Reference.qs namespace Quantum.Kata.SingleQubitGates { open Microsoft.Quantum.Intrinsic; open Microsoft.Quantum.Math; operation GlobalPhaseI_Reference (q : Qubit) : Unit is Adj+Ctl { X(q); Z(q); Y(q); } } Test.qs open Microsoft.Quantum.Intrinsic; open Microsoft.Quantum.Canon; open Microsoft.Quantum.Diagnostics; open Microsoft.Quantum.Math; open Microsoft.Quantum.Convert; operation ControlledArrayWrapperOperation (op : (Qubit => Unit is Adj+Ctl), qs : Qubit[]) : Unit is Adj+Ctl { Controlled op([qs[0]], qs[1]); } operation AssertEqualOnZeroState (testImpl : (Qubit => Unit is Ctl), refImpl : (Qubit => Unit is Adj+Ctl)) : Unit { using (qs = Qubit[2]) { within { H(qs[0]); } apply { Controlled testImpl([qs[0]], qs[1]); } AssertAllZero(qs); } } operation T2_GlobalPhaseI_Test () : Unit { AssertOperationsEqualReferenced(2, ControlledArrayWrapperOperation(GlobalPhaseI, _), ControlledArrayWrapperOperation(GlobalPhaseI_Reference, _)); } Driver.cs using Microsoft.Quantum.Simulation.XUnit; using Microsoft.Quantum.Simulation.Simulators; using Xunit.Abstractions; namespace Quantum.Kata.SingleQubitGates { public class TestSuiteRunner { public TestSuiteRunner(ITestOutputHelper output) { this.output = output; } /// <summary> /// </summary> [OperationDriver(TestNamespace = "Quantum.Kata.SingleQubitGates")] public void TestTarget(TestOperation op) { using (var sim = new QuantumSimulator()) { sim.OnLog += (msg) => { output.WriteLine(msg); }; sim.OnLog += (msg) => { Debug.WriteLine(msg); }; op.TestOperationRunner(sim); } } } } Yes, this sequence of gates will work to apply the global phase of $$i$$. You can check it using matrix multiplication - a product of matrices $$Y \cdot Z \cdot X$$ will give you a matrix $$\begin{bmatrix} i & 0 \\ 0 & i \end{bmatrix}$$, which corresponds to applying a global phase. Alternatively, you can implement it using R gate with the phase $$-\pi$$: R(PauliI, -PI(), q); As a side note, performing measurements this way is not going to detect a global phase introduced by the gate; you need to work with a controlled version of the gate to detect it. • I tried to work with a controlled version of the gate but I could not add a Entry Potint and run it. How can I run the code with a controlled version ? – theRomanMercury Jun 22 at 19:29 • EntryPoint and controlled version of the gate are completely unrelated. What exactly didn't work for you? You just need to allocate two qubits, prepare the first one in the \|+⟩ state and apply controlled gate with the first qubit as control and second as target. – Mariia Mykhailova Jun 23 at 5:38 • I updated my question with your answer. I used this codes from Katas but I did not get an output when I run it. Is it because they are Unit type? – theRomanMercury Jun 23 at 9:03 • Nothing in the code you've pasted writes anything to the output. You can use Message to output classical values and DumpMachine to output internal state of the quantum simulator. – Mariia Mykhailova Jun 23 at 23:10 On IBM Q, you can also use $$U3$$ gate to prepare a global phase operator. $$U3$$ gate is defined as $$U3(\theta, \varphi, \lambda) = \begin{pmatrix} \cos (\theta/2) & -\mathrm{e}^{i\lambda}\sin(\theta/2) \\ \mathrm{e}^{i\varphi}\sin(\theta/2) & \mathrm{e}^{i(\varphi + \lambda)}\cos(\theta/2)\\ \end{pmatrix}.$$ Setting $$\theta = \pi$$ we get $$U3(\pi, \varphi, \lambda) = \begin{pmatrix} 0 & -\mathrm{e}^{i\lambda} \\ \mathrm{e}^{i\varphi} & 0\\ \end{pmatrix}.$$ Let's denote our global phase $$\alpha$$ and set $$\varphi = \alpha$$ and $$\lambda = \alpha + \pi$$. Since $$-\mathrm{e}^{i\pi}=1$$ we have $$U3(\pi, \alpha, \alpha+\pi) = \begin{pmatrix} 0 & \mathrm{e}^{i\alpha} \\ \mathrm{e}^{i\alpha} & 0\\ \end{pmatrix},$$ which is $$\mathrm{e^{i\alpha}}X$$. To get $$\mathrm{e^{i\alpha}}I$$, we apply another $$X$$ gate. So, global phase gate is implemented as $$X\,\,U3(\pi,\alpha,\alpha+\pi)$$, where $$\alpha$$ is global phase. EDIT (solution in Q#) A $$R1$$ gate is defined as $$R1(\theta) = \begin{pmatrix} 1 & 0 \\ 0 & \mathrm{e}^{i\theta} \end{pmatrix}$$ An operation $$X\,R1(\theta)$$ is described by matrix $$\begin{pmatrix} 0 & \mathrm{e}^{i\theta} \\ 1 & 0 \end{pmatrix}$$ If we apply this operation twice (i.e. $$[X\,R1(\theta)]^2$$ ), we get $$\begin{pmatrix} \mathrm{e}^{i\theta} & 0 \\ 0 & \mathrm{e}^{i\theta} \end{pmatrix},$$ which is a global phase gate with arbitrary phase $$\theta$$. So, global phase gate in Q# can be realized as $$[X\,R1(\theta)]^2$$. • I am focusing on Q# language, thanx a lot for your answer. – theRomanMercury Jun 23 at 9:04 • @theRomanMercury: Sorry, I did not take this into account. Please find solution (global phase gate with arbitrary phase) with $R1$ and $X$ gates for Q#. – Martin Vesely Jun 24 at 7:34 I apologize, but I can not comment on @Martin Vesely answer above. I am glad that my amendments have already been adopted, I’ll leave only a short note, in case someone is interested in this: In Qiskit programs (e.g. see) I use X before U3 in my variants of phase shift circuits, in an attempt to avoid "distortion" during further processing of Qiskit (for example, by transpiler): qc.x(qubit) qc.u3(np.pi, gamma, np.pi + gamma, qubit) If not for similar considerations, then the order here would not be important (as in other similar cases, e.g. $$[X\,R1(\theta)]^2 = [R1(\theta)\,X]^2$$)	2020-10-25T22:21:32	{ "domain": "stackexchange.com", "url": "https://quantumcomputing.stackexchange.com/questions/12604/quantum-katas-tutorials-singlequbitgates-exercise-2-globalphasei", "openwebmath_score": 0.7167534828186035, "openwebmath_perplexity": 3248.7567877720485, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9546474142844408, "lm_q2_score": 0.8740772220439509, "lm_q1q2_score": 0.8344355599091847 }
https://waqsp.org/telescope-price-zwgol/73f536-extreme-value-theorem-open-interval	This example was to show you the extreme value theorem. If has an extremum on an open interval , then the extremum occurs at a critical point. History. Critical points are determined by using the derivative, which is found with the Chain Rule. If f(x) exists for all values of x in the open interval (a,b) and f has a relative extremum at c, where a < c < b, then if f0(c) exists, then f0(c) = 0. Proof of the Extreme Value Theorem Theorem: If f is a continuous function deﬁned on a closed interval [a;b], then the function attains its maximum value at some point c contained in the interval. The Extreme Value Theorem (EVT) says: If a function f is continuous on the closed interval a ≤ x ≤ b , then f has a global minimum and a global maximum on that interval. point. know that the absolute extremes occur at either the endpoints, x=0 and x = 3, or the The function Proof of the Extreme Value Theorem Theorem: If f is a continuous function deﬁned on a closed interval [a;b], then the function attains its maximum value at some point c contained in the interval. Open interval. We use the logarithm to compute the derivative of a function. Closed interval domain, … Bolzano's proof consisted of showing that a continuous function on a closed interval was bounded, and then showing that the function attained a maximum and a minimum value. So is a value attained at least three times: inside the open interval (,) , inside of (, ~) and inside of (~, ~). If the interval is open or the function has even one point of discontinuity, the function may not have an absolute maximum or absolute minimum over For example, consider the functions shown in (Figure) (d), (e), and (f). Correspondingly, a metric space has the Heine–Borel property if every closed and bounded set is also compact. Are you sure you want to do this? It states that if f is a continuous function on a closed interval [a, b], then the function f has both a minimum and a maximum on the interval. In this section we learn the definition of continuity and we study the types of The absolute maximum is shown in red and the absolute minimumis in blue. • Extreme Value Theorem: • A function can have only one absolute maximum or minimum value (y-coordinate), but it can occur at more than one x-coordinate. The Inverse Function Theorem (continuous version) 11. numbers of f(x) in the interval (0, 3). Since the endpoints are not included, they can't be the global extrema, and this interval has no global minimum or maximum. The absolute extremes occur at either the endpoints, x=\text {-}1, 6 or the critical In order to use the Extreme Value Theorem we must have an interval that includes its endpoints, often called a closed interval, and the function must be continuous on that interval. This is used to show thing like: There is a way to set the price of an item so as to maximize profits. Inequalities and behaviour of f(x) as x →±∞ 17. Solution: The function is a polynomial, so it is continuous, and the interval is closed, Play this game to review undefined. This video explains the Extreme Value Theorem and then works through an example of finding the Absolute Extreme on a Closed Interval. The Extreme Value Theorem 10. at a Regular Point of a Surface. function f(x) yields: The absolute maximum is \answer {2e^4} and it occurs at x = \answer {2}.The absolute minimum is \answer {-1/(2e)} and it occurs at x = \answer {-1/2}. Solution: First, we find the critical numbers of f(x) in the interval [\text {-}1, 6]. In this section we discover the relationship between the rates of change of two or Since is compact, The Mean Value Theorem states that the rate of change at some point in a domain is equal to the average rate of change of that domain. An important Theorem is theExtreme Value Theorem. fails to hold, then f(x) might fail to have either an absolute max or an absolute min If f'(c) is undefined then, x=c is a critical number for f(x). However, for a function deﬁned on an open or half-open interval… Extreme Value Theorem Theorem 1 below is called the Extreme Value theorem. Extreme Value Theorem: When examining a function, we may find that it it doesn't have any absolute extrema. It ... (-2, 2), an open interval, so there are no endpoints. Lagrange mean value theorem; Bound relating number of zeros of function and number of zeros of its derivative; Facts used. y = x2 0 ≤ x ≤2 y = x2 0 ≤ x ≺2 4.1 Extreme Values of Functions Day 2 Ex 1) A local maximum value occurs if and only if f(x) ≤ f(c) for all x in an interval. State whether the function has absolute extrema on its domain $$g(x)= \begin{cases} x & x> 0 \\ 3 & x\leq 0 \end{cases}$$ Join the initiative for modernizing math education. Thus, a bound of infinity must be an open bound. Plugging these special values into the original function f(x) yields: The absolute maximum is \answer {17} and it occurs at x = \answer {-2}.The absolute minimum is \answer {-15} and it occurs at x = \answer {2}. This theorem is sometimes also called the Weierstrass extreme value theorem. Notice how the minimum value came at "the bottom of a hill," and the maximum value came at an endpoint. Suppose that the values of the buyer for nitems are independent and supported on some interval [u min;ru min] for some u min>0 and r 1. Depending on the setting, it might be needed to decide the existence of, and if they exist then compute, the largest and smallest (extreme) values of a given function. Renze, Renze, John and Weisstein, Eric W. "Extreme Value Theorem." need to solve (3x+1)e^{3x} = 0 (verify) and the only solution is x=\text {-}1/3 (verify). Built at The Ohio State UniversityOSU with support from NSF Grant DUE-1245433, the Shuttleworth Foundation, the Department of Mathematics, and the Affordable Learning ExchangeALX. Select the third example, showing the same piece of a parabola as the first example, only with an open interval. There is an updated version of this activity. In this section we use the graph of a function to find limits. The Extreme Value Theorem guarantees both a maximum and minimum value for a function under certain conditions. This becomes 3x^2 -12x= 0 First, since we have a closed interval (i.e. I know it must be continuous for the interval, but must it be closed? Suppose that f(x) is defined on the open interval (a,b) and that f(x) has an absolute max at x=c. Extreme Value Theorem: f is continuous on [a;b], then f has an absolute max at c and and absolute min at d, where c;d 2[a;b] 7. Intermediate Value Theorem and we investigate some applications. Portions of this entry contributed by John We learn to compute the derivative of an implicit function. If has an absolute maximum or absolute minimum at a point in the interval, then is a critical number for . Determine all critical points in the given interval and evaluate the function at these critical points and at the endpoints of the interval 3. Establish that the function is continuous on the closed interval 2. Closed Interval Method • Test for absolute extrema • Only use if f (x) is continuous on a closed interval [a, b]. (If the max/min occurs in more than one place, list them in ascending order).The absolute maximum is \answer {4} and it occurs at x = \answer {-2} and x=\answer {1}. The absolute maximum is \answer {0} and it occurs at x = \answer {-2}. Below, we see a geometric interpretation of this theorem. I know it's pretty vital for the theorem to be able to show the values of f(a) and f(b), but what if I have calculated the limits as x -> a (from the right hand side) and x -> b (from the left hand side)? Use the differentiation rules to compute derivatives. In this section we will solve the problem of finding the maximum and minimum We learn how to find the derivative of a power function. When you do the problems, be sure to be aware of the difference between the two types of extrema! Proof: There will be two parts to this proof. We solve the equation f'(x) =0. interval. In this section we interpret the derivative as an instantaneous rate of change. To find the relative extrema of a function, you first need to calculate the critical values of a function. This has two important corollaries: . Wolfram Web Resource. Walk through homework problems step-by-step from beginning to end. For example, let’s say you had a number x, which lies somewhere between zero and 100: The open interval would be (0, 100). 3. The absolute minimum is \answer {0} and it occurs at x = \answer {-3} and x=\answer {0}. compute the derivative of an area function. We determine differentiability at a point. Notice how the minimum value came at "the bottom of a hill," and the maximum value came at an endpoint. For the extreme value theorem to apply, the function must be continuous over a closed, bounded interval. A description of how to find the arguments that maximize and minimize the value of a function that is continuous on a closed interval. © 2013–2021, The Ohio State University — Ximera team, 100 Math Tower, 231 West 18th Avenue, Columbus OH, 43210–1174. is a polynomial, so it is differentiable everywhere. Unlimited random practice problems and answers with built-in Step-by-step solutions. The next theorem is called Rolle’s Theorem and it guarantees the existence of an extreme value on the interior of a closed interval, under certain conditions. This theorem is called the Extreme Value Theorem. Which of the following functions of x is guaranteed by the Extreme Value Theorem to have an absolute maximum on the interval [0, 4]? occurring at the endpoint x = -1 and the absolute minimum of f(x) in the interval is -3 occurring Plugging these special values into the original function f(x) yields: From this data we conclude that the absolute maximum of f(x) on the interval is 3.25 The Inverse Function Theorem (Diﬀerentiable version) 14. First, we find the critical numbers of in the interval . Compute limits using algebraic techniques. Extreme–Value Theorem Assume f(x) is a continuous function deﬁned on a closed interval [a,b]. In papers \cite{BartkovaCunderlikova18, BartkovaCunderlikova18p} we proved the Fisher-Tippett-Gnedenko theorem and the Pickands-Balkema-de Haan theorem on family of intuitionistic fuzzy events. continuous function on a closed interval is contained in the following theorem. continuous and the second is that the interval is closed. If a function is continuous on a closed The first derivative can be used to find the relative minimum and relative maximum values of a function over an open interval. In this section we use definite integrals to study rectilinear motion and compute itself be compact. In order to use the Extreme Value Theorem we must have an interval that includes its endpoints, often called a closed interval, and the function must be continuous on that interval. We don’t want to be trying to find something that may not exist. The closed interval—which includes the endpoints— would be [0, 100]. A local minimum value … However, if that interval was an open interval of all real numbers, (0,0) would have been a local minimum. Also note that while $$0$$ is not an extreme value, it would be if we narrowed our interval to $$[-1,4]$$. Using the Extreme Value Theorem 1. (a,b) as opposed to [a,b] Also note that while $$0$$ is not an extreme value, it would be if we narrowed our interval to $$[-1,4]$$. For the extreme value theorem to apply, the function must be continuous over a closed, bounded interval. Extreme Value Theorem: If a function is continuous in a closed interval , with the maximum of at and the minimum of at then and are critical values of Proof: The proof follows from Fermat’s theorem and is left as an exercise for the student. If f is continuous on the closed interval [a,b], then f attains both a global minimum value m and a global maximum value M in the interval [a,b]. Theorem 1. Solution: The function is a polynomial, so it is continuous, and the interval is closed, so by the Extreme Value Theorem, we know that this function has an absolute maximum and an absolute minimum on the interval . is increasing or decreasing. THE EXTREME-VALUE THEOREM (EVT) 27 Interlude: open and closed sets We went about studying closed bounded intervals… Here is a detailed, lecture style video on the Extreme Value Theorem: Calculus I, by Andrew In this section we learn the definition of the derivative and we use it to solve the Extreme Value Theorem If f is continuous on a closed interval [a,b], then f has both a maximum and minimum value. The derivative is f'(x) = 2x - 4 which exists for all values of x. is differentiable on the open interval , i.e., the derivative of exists at all points in the open interval .. Then, there exists in the open interval such that . First, we find the critical Hence f'(c) = 0 and the theorem is In this section we learn to compute the value of a definite integral using the A local minimum value … In this section we compute derivatives involving. The extreme value theorem gives the existence of the extrema of a continuous function defined on a closed and bounded interval. knowledge of derivatives. In this section, we use the derivative to determine intervals on which a given function Depending on the setting, it might be needed to decide the existence of, and if they exist then compute, the largest and smallest (extreme) values of a given function. In mathematical analysis, the intermediate value theorem states that if f is a continuous function whose domain contains the interval [a, b], then it takes on any given value between f(a) and f(b) at some point within the interval.. This video explains the Extreme Value Theorem and then works through an example of finding the Absolute Extreme on a Closed Interval. The extreme value theorem gives the existence of the extrema of a continuous function defined on a closed and bounded interval. In this section we examine several properties of the indefinite integral. We solve the equation f'(x) =0. f'(x) =0. If the interval $$I$$ is open or the function has even one point of discontinuity, the function may not have an absolute maximum or absolute minimum over $$I$$. Try the following: The first graph shows a piece of a parabola on a closed interval. Below, we see a geometric interpretation of this theorem. Basically Rolle ‘s theorem states that if a function is differentiable on an open interval, continuous at the endpoints, and if the function values are equal at the endpoints, then it has at least one horizontal tangent. Extreme Value Theorem: If a function is continuous in a closed interval , with the maximum of at and the minimum of at then and are critical values of Proof: The proof follows from Fermat’s theorem and is left as an exercise for the student. y = x2 0 ≤ x ≤2 y = x2 0 ≤ x ≺2 4.1 Extreme Values of Functions Day 2 Ex 1) A local maximum value occurs if and only if f(x) ≤ f(c) for all x in an interval. average value. This is a good thing of course. The procedure for applying the Extreme Value Theorem is to first establish that the function is continuous on the closed interval. Plugging these special values into the original However, if that interval was an open interval of all real numbers, (0,0) would have been a local minimum. View Chapter 3 - Limits.pdf from MATHS MA131 at University of Warwick. So, it is (−∞, +∞), it cannot be [−∞, +∞]. If has an extremum the point of tangency. analysis includes the position, velocity and acceleration of the particle. It is not de ned on a closed interval, so the Extreme Value Theorem does not apply. Thus f'(c) \geq 0. function. In this section we compute limits using L’Hopital’s Rule which requires our For example, [0,1] means greater than or equal to 0 and less than or equal to 1. In calculus, the extreme value theorem states that if a real-valued function f is continuous on the closed interval [a,b], then f must attain a maximum and a minimum, each at least once.That is, there exist numbers c and d in [a,b] such that: three step process. A set $${\displaystyle K}$$ is said to be compact if it has the following property: from every collection of open sets $${\displaystyle U_{\alpha }}$$ such that $${\textstyle \bigcup U_{\alpha }\supset K}$$, a finite subcollection $${\displaystyle U_{\alpha _{1}},\ldots ,U_{\alpha _{n}}}$$can be chosen such that $${\textstyle \bigcup _{i=1}^{n}U_{\alpha _{i}}\supset K}$$. • Three steps polynomial, so it is differentiable everywhere. This becomes x^3 -6x^2 + 8x = 0 and the solutions are x=0, x=2 and x=4 (verify). The Extreme Value Theorem. In this section we learn to reverse the chain rule by making a substitution. The idea that the point $$(0,0)$$ is the location of an extreme value for some interval is important, leading us to a definition. Fermat’s Theorem Suppose is defined on the open interval . Example . The quintessential point is this: on a closed interval, the function will have both minima and maxima. Extreme Value Theorem If a function f {\displaystyle f} is continuous on a closed interval [ a , b ] {\displaystyle [a,b]} then there exists both a maximum and minimum on the interval. A continuous function ƒ (x) on the closed interval [a,b] showing the absolute max (red) and the absolute min (blue).. It is also important to note that the theorem tells us that the The extreme value theorem states that if is a continuous real-valued function on the real-number interval defined by , then has maximum and minimum values on that interval, which are attained at specific points in the interval. That makes sense. In such a case, Theorem 1 guarantees that there will be both an absolute maximum and an absolute minimum. max and the min occur in the interval, but it does not tell us how to find If a continuous function has values of opposite sign inside an interval, then it has a root in that interval (Bolzano's theorem). Solution: First, we find the critical numbers of f(x) in the interval [\text {-}1, 3]. The absolute minimum is \answer {-27} and it occurs at x = \answer {1}. Practice online or make a printable study sheet. The image below shows a continuous function f(x) on a closed interval from a to b. Hints help you try the next step on your own. This is what is known as an existence theorem. 2. •Note: If the interval is open, then the endpoints are. A point is considered a minimum point if the value of the function at that point is less than the function values for all x-values in the interval. endpoints, x=-1, 2 or the critical number x = -1/3. In this section we use properties of definite integrals to compute and interpret If f'(c) is defined, then The first is that f(x) is Play this game to review undefined. That is, there exist x 1,x 2 ∈ [a,b] so that f(x 1) = m and f(x 2) = M, and m ≤ f(x) ≤ M for all x ∈ [a,b]. Extreme Value Theorem. The largest and smallest values from step two will be the maximum and minimum values, respectively from the definition of the derivative we have f'(c) = \lim _{h \to 0^-} \frac {f(c+h) -f(c)}{h} = \lim _{h \to 0^-} \frac {f(c+h) -f(c)}{h} The difference quotient in the left If a function is continuous on a closed interval , then has both a maximum and a minimum on . Then, for all >0, there is an algorithm that computes a price vector whose revenue is at least a (1 )-fraction of the optimal revenue, and whose running time is polynomial in max ˆ the equation f'(x) =0 gives x=2 as the only critical number of the function. When moving from the real line $${\displaystyle \mathbb {R} }$$ to metric spaces and general topological spaces, the appropriate generalization of a closed bounded interval is a compact set. The Extreme Value Theorem ... as x !1+ there is an open circle, so the lower bound of y = 1 is approached but not attained . Preview this quiz on Quizizz. To find the relative extrema of a function, you first need to calculate the critical values of a function. This is usually stated in short as "every open cover of $${\displaystyle K}$$ has a finite subcover". which has two solutions x=0 and x = 4 (verify). Solution: First, we find the critical numbers of f(x) in the interval [-1, 0]. critical number x = 2. (or both). Extreme Value Theorem. discontinuities. This example was to show you the extreme value theorem. Remark: An absolute extremum of a function continuous on a closed interval must either be a relative extremum or a function value at an endpoint of the interval. The absolute extremes occur at either the interval around c (open interval around c means that the immediate values to the left and to the right of c are in that open interval) 6. How would you like to proceed? right hand limit is negative (or zero) and so f'(c) \leq 0. https://mathworld.wolfram.com/ExtremeValueTheorem.html, Normal Curvature In this section we learn to compute general anti-derivatives, also known as indefinite If the interval is open or the function has even one point of discontinuity, the function may not have an absolute maximum or absolute minimum over For example, consider the functions shown in … Proof: There will be two parts to this proof. The function is a Noting that x = 4 is not in Regardless, your record of completion will remain. In this lesson we will use the tangent line to approximate the value of a function near integrals. Extreme Value Theorem If is continuous on the closed interval , then there are points and in , such that is a global maximum and is a global minimum on . values of a continuous function on a closed interval. and x = 2. In this section we learn about the two types of curvature and determine the curvature In order to utilize the Mean Value Theorem in examples, we need first to understand another called Rolle’s Theorem. The main idea is finding the location of the absolute max and absolute min of a 4 Extreme Value Theorem If f is continuous on a closed interval a b then f from MATH 150 at Simon Fraser University tangent line problem. Vanishing Derivative Theorem Assume f(x) is a continuous function deﬁned on an open interval (a,b). be compact. interval , so it must Extreme value theorem We compute the derivative of a composition. Extreme Value Theorem If f is continuous on a closed interval [a,b], then f has both a maximum and minimum value. But the difference quotient in the It is important to note that the theorem contains two hypothesis. If the function f is continuous on the closed interval [a,b], then f has an absolute maximum value and an absolute minimum value on [a,b]. You cannot have a closed bound of ±∞ because ∞ is never a value that can actually be reached. There are a couple of key points to note about the statement of this theorem. From MathWorld--A The extreme value theorem was originally proven by Bernard Bolzano in the 1830s in a work Function Theory but the work remained unpublished until 1930. • Three steps/labels:. The absolute extremes occur at either the endpoints, x=\text {-}1, 3 or the critical The #1 tool for creating Demonstrations and anything technical. The Extreme Value Theorem ... as x !1+ there is an open circle, so the lower bound of y = 1 is approached but not attained . Terminology. maximum and a minimum on proved. An open interval does not include its endpoints, and is indicated with parentheses. This theorem is sometimes also called the Weierstrass extreme value theorem. Fermat’s Theorem. . The Weierstrass Extreme Value Theorem. Using the product rule and the chain rule, we have f'(x) = 1\cdot e^{3x} + x\cdot e^{3x} \cdot 3 which simplifies to (3x+1)e^{3x}. Hence f(x) has two critical numbers in the Which of the following functions of x is guaranteed by the Extreme Value Theorem to have an absolute maximum on the interval [0, 4]? Our so by the Extreme Value Theorem, we know that this function has an absolute No maximum or minimum values are possible on the closed interval, as the function both increases and decreases without bound at x = π/2. However, it never reaches the value of $0.$ Notice that this function is not continuous on a closed bounded interval containing 0 and so the Extreme Value Theorem does not apply. Explore anything with the first computational knowledge engine. In this section we prepare for the final exam. Continuous, 3. For the extreme value theorem to apply, the function must be continuous over a closed, bounded interval. The Heine–Borel theorem asserts that a subset of the real line is compact if and only if it is both closed and bounded. and the denominator is negative. within a closed interval. Derivatives of sums, products and composites 13. function. interval around c (open interval around c means that the immediate values to the left and to the right of c are in that open interval) 6. In this section we learn a theoretically important existence theorem called the them. Fermat’s Theorem Suppose is defined on the open interval . These values are often called extreme values or extrema (plural form). Thus f(c) \geq f(x) for all x in (a,b). Real-valued, 2. Diﬀerentiation 12. of functions whose derivatives are already known. The Extreme Value Theorem says: If a function f is continuous on the closed interval a ≤ x ≤ b, then f has a global minimum and a global maximum on that interval. https://mathworld.wolfram.com/ExtremeValueTheorem.html. The standard proof of the first proceeds by noting that is the continuous image of a compact set on the We learn the derivatives of many familiar functions. more related quantities. Since we know the function f(x) = x2 is continuous and real valued on the closed interval [0,1] we know that it will attain both a maximum and a minimum on this interval. The derivative is 0 at x = 0 and it is undefined at x = -2 and x = 2. We will also determine the local extremes of the at the critical number x = 0. Local Extrema, critical points, Rolle’s Theorem 15. The Mean Value Theorem 16. If we don’t have a closed interval and/or the function isn’t continuous on the interval then the function may or may not have absolute extrema. The Extreme value theorem requires a closed interval. If you update to the most recent version of this activity, then your current progress on this activity will be erased. Theorem 2. You are about to erase your work on this activity. Then f has both a Maximum and a Minimum value on [a,b].#Extreme value theorem (The circle, in fact.) So the extreme value theorem tells us, look, we've got some closed interval - I'm going to speak in generalities here - so let's say that's our X axis and let's say we have some function that's defined on a closed interval. Hence f(x) has one critical numbers x = 0,2. The quintessential point is this: on a closed interval, the function will have both minima and maxima. On a closed interval, always remember to evaluate endpoints to obtain global extrema. this critical number is in the interval (0,3). In this section we analyze the motion of a particle moving in a straight line. Extreme Value Theorem If a function f {\displaystyle f} is continuous on a closed interval [ a , b ] {\displaystyle [a,b]} then there exists both a maximum and minimum on the interval. In order for the extreme value theorem to be able to work, you do need to make sure that a function satisfies the requirements: 1. fundamental theorem of calculus. For example, (0,1) means greater than 0 and less than 1.This means (0,1) = {x \| 0 < x < 1}.. A closed interval is an interval which includes all its limit points, and is denoted with square brackets. the interval [\text {-}1,3] we see that f(x) has two critical numbers in the interval, namely x = 0 If we don’t have a closed interval and/or the function isn’t continuous on the interval then the function may or may not have absolute extrema. 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In ( a, b ]: calculus I, by Andrew Incognito of in interval... extreme value theorem open interval 2021	2021-07-27T10:49:08	{ "domain": "waqsp.org", "url": "https://waqsp.org/telescope-price-zwgol/73f536-extreme-value-theorem-open-interval", "openwebmath_score": 0.8095170855522156, "openwebmath_perplexity": 259.6843062088817, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9863631647185701, "lm_q2_score": 0.8459424431344437, "lm_q1q2_score": 0.8344064653798489 }
https://math.stackexchange.com/questions/1788692/topological-finer-and-separability	# Topological finer and separability I have this question: Let $(X, d_1)$ and $(X,d_2)$ be two metric spaces. Suppose $d_1$ is topologically finer than $d_2$. What is the relationship between these two statements? (i) $(X,d_1)$ is separable. (ii) $(X,d_2)$ is separable. I think the implication (i) $\Rightarrow$ (ii) holds, but not the other way around. I was trying to prove it: Suppose $(X,d_1)$ is separable. Then $X$ has an at most countable dense subset for the $d_1$ metric. So there exists an at most countable subset $A \subset X$ such that $\overline{A} = X$. Since $d_1$ is topologically finer than $d_2$, every open subset for the $d_2$ metric is also open for the $d_1$ metric. This is equivalent in saying that every closed $d_2$-subset is also $d_1$-closed. This is as far I could go. What can I conclude from this? And am I also correct in saying the other implication is false? • Sorry, I misread. You are correct is thinking that separability in a finer topology implies separability in a coarser topology. I have deleted temporarily my answer and will repair. That's done now. I hope I did not cause any confusion. – drhab May 17 '16 at 9:30 • I saw your answer, and I was thinking you confused $d_1$ with $d_2$. But your answer still helped me a lot in giving me the right idea! – Kamil May 17 '16 at 9:36 • Nice to hear that. There was indeed some confusion by me about $d_1,d_2$. – drhab May 17 '16 at 9:38 Counterexample of (ii)$\implies$(i): Let $d_1$ be defined on $\mathbb R$ by $d_1(x,y)=0$ if $x=y$ and $d(x,y)=1$ otherwise. Let $d_2$ be defined on $\mathbb R$ by $d_2(x,y)=\|x-y\|$. Then $\mathbb R$ equipped with $d_1$ results in the discrete topology (finest as possible), hence no separability on the uncountable $\mathbb R$. On the other hand $\mathbb R$ equipped with $d_2$ results in the usual topology, hence separability. Proof of (i)$\implies$(ii): Let $X$ equipped with some topology $\tau$ be separable. Then a countable and dense $S\subseteq X$ exists. Density of $S$ means exactly that $U\cap S\neq\varnothing$ for each $U\in\tau$. If $\sigma$ is another topology on $X$ and $\tau$ is finer then also $U\cap S\neq\varnothing$ for $U\in\sigma$. This because $\sigma\subseteq\tau$. So again $S$ is dense. Of course it has not lost its countability and we conclude that $X$ equipped with the coarser $\sigma$ is separable. A counterexample for the opposite: We use the set $\Bbb R$ with usual metric $d_2(x,y)=\|x-y\|$ and define $$d_{1}\left(x,y\right)=\cases{ 0&\text{if }x=y\cr 1&\text{if }x\neq y}$$ We know that $d_2$ make a separable space (use $\Bbb Q$). $d_1$, on the other hand, yields discrete topology, so the only dense set is the whole space, making it not separable. $d_1$'s topology is finer than $d_2$'s as it's discrete and therefore finer than any other topology. • $x \not = y$ \not = – Olba12 May 17 '16 at 8:50 • Ah, I see thanks. But is $d_1$ topologically finer than $d_2$ ? – Kamil May 17 '16 at 8:54 • @Kamil Every set open in $d_2$ (in fact every set) is open in $d_1$. – Heimdall May 17 '16 at 8:59 • @Olba12 When I was typing the answer, I tried different combinations: \ne, \neq, \not=, \not =; the preview didn't render either of them correctly. This question uses \neq, I'm quoting the source: If det $A\neq0$, the matrix, so what's wrong with \neq and other possible no-equals in my answer? Maybe it works in question and comments but not in answers? I'm asking on meta. – Heimdall May 17 '16 at 9:46 • Weird, \not = has always been working for me... haha – Olba12 May 17 '16 at 10:08	2019-07-16T14:24:42	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1788692/topological-finer-and-separability", "openwebmath_score": 0.9579042196273804, "openwebmath_perplexity": 356.7182873891679, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631659211718, "lm_q2_score": 0.8459424373085146, "lm_q1q2_score": 0.8344064606506989 }
https://mathhelpboards.com/threads/number-of-ways-to-divide-a-square-grid-in-half.26269/	Number of ways to divide a square grid in half HapaxOromenon New member Consider a dotted square grid with $4$ rows and $4$ columns, as shown: By drawing a series of connected straight lines from dot to dot, it is possible to divide the square (which effectively has side length $3$) into two parts of equal area. One way of doing this is as shown: Other examples of ways of doing this are: ; ; and . The question I am investigating is how many such ways there are in total, or if such a calculation is too difficult, I would at least like to have a systematic way of listing all of the ways. But I'm not sure really where to start. Any assistance would be greatly appreciated. Opalg MHB Oldtimer Staff member Consider a dotted square grid with $4$ rows and $4$ columns, as shown: By drawing a series of connected straight lines from dot to dot, it is possible to divide the square (which effectively has side length $3$) into two parts of equal area. One way of doing this is as shown: Other examples of ways of doing this are: ; ; and . The question I am investigating is how many such ways there are in total, or if such a calculation is too difficult, I would at least like to have a systematic way of listing all of the ways. But I'm not sure really where to start. Any assistance would be greatly appreciated. Hi HapaxOromenon , and welcome to MHB! This looks like an interesting problem, and I'm not sure whether it has a straightforward solution. But first, we need to be sure what you are trying to count. You want to partition the square grid into two parts, and you say that these parts should have the same area. In all the examples that you show, the two parts are congruent. So my first question is this: Would you allow a partition like this \begin{tikzpicture} [scale=2]\foreach \i in {0,...,3} { \draw[thin,dotted] (\i,0) -- (\i,3) ; \draw[thin,dotted] (0,\i) -- (3,\i) ; \foreach \j in {0,...,3} \fill [black,opacity=.5] (\i,\j) circle (2pt) ; } \draw[thick] (0,2) -- (2,2) -- (2,1) -- (3,0) ; \end{tikzpicture} where the two parts have the same area (4.5 square units) but they have different shapes? The second question is whether you would consider two partitions to be the same if one of them is just a rotation or reflection of the other one. For example, do you consider these partitions \begin{tikzpicture} [scale=2]\foreach \i in {0,...,3} { \draw[thin,dotted] (\i,0) -- (\i,3) ; \draw[thin,dotted] (0,\i) -- (3,\i) ; \foreach \j in {0,...,3} \fill [black,opacity=.5] (\i,\j) circle (2pt) ; } \draw[thick] (0,2) -- (1,2) -- (2,1) -- (3,1) ; \end{tikzpicture} ................ \begin{tikzpicture} [scale=2]\foreach \i in {0,...,3} { \draw[thin,dotted] (\i,0) -- (\i,3) ; \draw[thin,dotted] (0,\i) -- (3,\i) ; \foreach \j in {0,...,3} \fill [black,opacity=.5] (\i,\j) circle (2pt) ; } \draw[thick] (2,3) -- (2,2) -- (1,1) -- (1,0) ; \end{tikzpicture} to be the same, or different? HapaxOromenon New member 1) Yes, I would count this as a valid cut. 2) I would consider them to be different. Opalg MHB Oldtimer Staff member Each cut must start and finish at points on the boundary of the grid. There are two sorts of boundary points, namely the four corners of the grid, and the eight points on the edges that are not corners (I'll call these edge points). Trying to find a systematic way to count the number of possible cuts, I split them into three classes: 1. cuts that go from corner to corner; 2. cuts that go from a corner to an edge point; 3. cuts that go between two edge points. For the first class, there are nine cuts between the northwest and southeast corners of the grid, as in this diagram . Rotating those grids through a right angle, you get the same number of cuts between the northeast and southwest corners of the grid. So altogether there are 18 cuts of class 1. For the second class, there are ten cuts starting at the northwest corner of the grid and ending at an edge point, as in this diagram . Rotating them, you get the same number of cuts starting at each of the other corners of the grid. So there are 40 cuts of class 2. That leaves class 3, the cuts going from an edge point to a point on the opposite edge. I found 17 cuts starting from the point immediately below the northwest corner of the grid, as in this diagram Flipping each of these about a horizontal axis, you get the same number of cuts starting from the point immediately above the southwest corner of the grid. So there are 34 cuts of class 3 going from side to side of the grid. Rotating them through a right angle, you get another 34 cuts of class 3 between the top and bottom of the grid. So altogether there are 68 cuts of class 3. Thus the overall total number of cuts is 18 + 40 + 68 = 126 (unless I have overlooked any other possible cuts).	2020-07-06T06:01:50	{ "domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/number-of-ways-to-divide-a-square-grid-in-half.26269/", "openwebmath_score": 0.8414314985275269, "openwebmath_perplexity": 313.11700668463357, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. 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https://math.stackexchange.com/questions/2745825/determine-the-constant-alpha-in-mathbbr-and-the-vector-w-in-mathbbr/2746416	Determine the constant $\alpha \in \mathbb{R}$ and the vector $w \in \mathbb{R}^3$ that is perpendicular to the vector u such that v = α u + w. Determine the constant $\alpha \in \mathbb{R}$ and the vector $w \in \mathbb{R}^3$ that is perpendicular to the vector $u$ such that $v = \alpha u + w$. Need a little help with this question from my linear algebra course, not sure how to find the solution. I know that that the dot product of w and u is equal to zero. the vectors give are $u = (2, -2, 1) v = (3, 1, -1)$ • So use the dot product on both sides of the $v$ equation... – N74 Apr 20 '18 at 10:21 • Sorry I am still a little confused. use the dot with what on the left side of the equation? – Darragh O'Flaherty Apr 20 '18 at 10:27 • On the left side you will have $v \cdot u$ – N74 Apr 20 '18 at 10:30 Let $w=(w_1,w_2,w_3)$. If $w$ is perpendicular to $u$, the equation $w\cdot u=0$ must be satisfied, that is, $2w_1-2w_2+w_3=0$. As you see there are lots of solutions. If you want $v=\alpha u+w$, then you have another equation, in this cases a system of equations indeed, namely $3=2\alpha +w_1$ $1=2\alpha+w_2$ $-1=\alpha+w_3$ Adding the equation $2w_1-2w_2+w_3=0$ from above you've got a linear system of 4 equations and 4 unknowns, so there is precisely one solution that I'm sure you're able to find. • Perfect thank you very much, just the guidance I needed – Darragh O'Flaherty Apr 20 '18 at 11:28 You have $v= \alpha u + w$. Now, dot product by $u$ gives $$v \cdot u = \alpha u \cdot u + w \cdot u$$ We know $w \cdot u=0$ so the equation gives $$3 =\alpha 9$$ so $$\alpha={ 1 \over 3}$$. Knowing $\alpha$ it is straightforward that $$w=\left ({7 \over 3}, {5 \over 3}, -{4 \over 3} \right )$$	2021-04-22T17:06:35	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2745825/determine-the-constant-alpha-in-mathbbr-and-the-vector-w-in-mathbbr/2746416", "openwebmath_score": 0.9191478490829468, "openwebmath_perplexity": 119.83180657045094, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631651194373, "lm_q2_score": 0.8459424314825853, "lm_q1q2_score": 0.8344064542259956 }
http://mathoverflow.net/questions/20275/maps-inducing-zero-on-homotopy-groups-but-are-not-null-homotopic	# Maps inducing zero on homotopy groups but are not null-homotopic Today my fellow grad student asked me a question, given a map f from X to Y, assume $f_*(\pi_i(X))=0$ in Y, when is f null-homotopic? I search the literature a little bit, D.W.Kahn http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.pjm/1102995805 And M.Sternstein has worked on this, and Sternstein even got a necessary and sufficient condition, for suitable spaces. http://www.jstor.org/stable/pdfplus/2037939.pdf However, his condition is a little complicated for me as a beginner. Right now I just wanted a counter example of a such a map. Kahn in his paper said one can have many such examples using Eilenberg Maclance spaces. Well, we can certainly show a lot of map between E-M spaces induce zero map on homopoty groups just by pure group theoretic reasons, but I can not think of a easy example when you can show that map, if it exists, is not null-homotopic. Could someone give me some hint? or, maybe even some examples arising from manifolds? - Consider ordinary singular cohomology with varying coefficients. You can look at the short exact sequence of abelian groups: $$0 \to \mathbb{Z}/2 \to \mathbb{Z}/4 \to \mathbb{Z}/2 \to 0$$ This gives rise, for any space X, to a short exact sequence of chain complexes: $$0 \to C^i(X;\mathbb{Z}/2) \to C^i(X;\mathbb{Z}/4) \to C^i(X;\mathbb{Z}/2) \to 0$$ and hence you get a long exact sequence in cohomology. Thus we get an interesting boundary map known as the Bockstein $$H^i(X; \mathbb{Z}/2) \to H^{i+1}(X; \mathbb{Z}/2).$$ This is natural in X and so is represented by a (homotopy class of) map(s) of Eilenberg-Maclane spaces: $$K(i, \mathbb{Z}/2) \to K(i+1, \mathbb{Z}/2)$$ This map is necessarily zero on homotopy groups. To show that this map is not null-homotopy, you just need to find a space for which the Bockstein is non-trivial. There are lots of examples of this. Rather then explain one, I suggest you look up "Bockstein homomorphism" in a standard algebraic topology reference, e.g. Hatcher's book. - Thanks, this argument is very nice! I guess it is not very hard to find a space with non trivial Bockstein, e.g.$M=RP^2$, we have $H^i(M,Z/2)=H^i(M,Z/4)=Z/2$for i=1,2, and if I look at the reduced cohomology long exact sequence $H^0(M,pt,Z/2)=H^0(M,pt,z/4)=0$. Then the Bockstein from $H^2$to $H^1$will not be trivial, otherwise we will have $0\to Z/2\to Z/2\to Z/2\to 0$ contradiction. Although I used the reduced cohomology, since I only use maps to K(z/2,2) and K(Z/2,1) to represent $H^1$ and $H^2$ I am fine. From this example, we get a non null-homotopic map from $CP^{\infty}$ to $RP^\infty$! – Ying Zhang Apr 4 '10 at 4:05 Sorry about my last sentence, from the above sequence I got a map from $RP^{\infty}\to K(Z/2,2)$. To get a map from $RP^{\infty}\to CP^{\infty}$, we should look at the coefficient sequence $0\to Z\to Z\to Z/2Z\to 0$ which does the job. Geometrically it is not too hard to find an interesting map from $RP^{\infty}$ to $CP^{\infty}$ just by quotient out more things. – Ying Zhang Apr 4 '10 at 18:15 For a more explicit example than Chris's, consider the map from the (2-dimensional) torus to a sphere that collapses the 1-skeleton of the usual CW complex and takes the 2-cell to the 2-cell of the sphere. The torus is $K(\mathbb{Z}^2,1)$, so this necessarily gives zero maps on homotopy, but it's also pretty clearly not null-homotopic. - Nice! So this will give us examples when X and Y are both manifolds:) – Ying Zhang Apr 4 '10 at 4:11 You can also generalize this to the case where X is an $n$-manifold and a $K(G,1)$ (or even just having nonzero homotopy groups only up to $n-1$), and $X \to S^n$ is the map that collapses everything outside a Euclidean ball to a point. – Kevin Casto May 27 at 23:26 Even if you ask that $f$ induces trivial maps on all (singular) homology and cohomology groups, there are still easy manifold examples. (This actually arises as an exercise in Hatcher's AT). For instance, let $f:T^3\rightarrow S^2$ be the composition $T^3\rightarrow S^3\rightarrow S^2$, where the map from $T^3$ to $S^3$ is simply collapsing the 2-skeleton to a point, and the map from $S^3$ to $S^2$ is the Hopf map. As others have mentioned, since $T^3$ is a $K(\mathbb{Z}^3, 1)$, if follows that $f$ induces trivial maps on homotopy groups. Since the Hopf map induces trivial maps on homology and cohomology, it follows that $f$ does as well. Finally, to see that $f$ is NOT nullhomotopic, assume it is. Since the map from $S^3$ to $S^2$ is a fiber bundle, it has the homotopy lifting property. Hence, we can lift the homotopy of $f$ to a homotopy $G:I\times T^3\rightarrow S^3$ where $G_0$ is the above map from $T^3$ to $S^3$ and $G_1$ is is a map from $T^3$ to $S^1\subseteq S^3$, the preimage of a point in $S^2$ under the Hopf map. But $G_0$ has degree 1, while $G_1$ has degree 0, a contradiction. - Thanks. Btw, the exercise you mentioned is on Hatcher p.392 problem 34. – Ying Zhang Apr 4 '10 at 18:18 I asked a very similar question a few months ago, and got some excellent answers. -	2015-08-28T00:50:10	{ "domain": "mathoverflow.net", "url": "http://mathoverflow.net/questions/20275/maps-inducing-zero-on-homotopy-groups-but-are-not-null-homotopic", "openwebmath_score": 0.9473708271980286, "openwebmath_perplexity": 261.29563138243856, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.966410494349896, "lm_q2_score": 0.8633916099737806, "lm_q1q2_score": 0.8343907126123139 }
https://www.physicsforums.com/threads/question-about-eigenvectors.101776/	1. Nov 28, 2005 ### jd1828 I am studying for a test that I have today and just need a quick anwser. the problem is find the eigenvector of [8 -10;2 -1] where ; means skip to a new line. So I get lamda=4,3 then try to find eigenvector for lamda=4. I get the vector [5/2;1] which works but the anwser in the book is [2;5]. Both anwsers seem to be right but is there some rule to say that the book anwser is better. 2. Nov 28, 2005 ### TD Eigenvectors are always determined up to an arbitrary (non-zero) multiplication factor. The book apparently preferred not to have fractions but the smallest possible integers, which is usually done. Both are fine though 3. Nov 28, 2005 ### matt grime its answer doesn't have any fractions in it, that is considered better, just as you should replace 1/sqrt(2) with sqrt(2)/2. Its answer is also valid even if we only consider vectors with entries in the integers (which we may want to) so it is more general. Your answer is fine if we remember that we CAN divide by two, but it holds in greater generality than that. Just remember that if you can you should clear denominators in an answer, and since e-vectors are only defined up to scalar multiples why not? PERSONAL OPINION and NOT in any way authoritative is that I wouldn't mark you down for your answer. But that is purely MY opinion and is not necessarily that of your teacher/marker. actually, have just noticed your answer and the books are not compatible, bnut that is probably a typo (5/2,1) is not a scalar multiple of (2,5) (5/2,1) and (5,2) yes or (2/5,1) and (2,5) also ok 4. Nov 28, 2005 ### jd1828 Thanks a lot! Ill stick to using the smallest possible integers since thats what the book does and I wouldnt want to loose points on something so easy. The anwser in the book was [5,2] I just typed it wrong. 5. Nov 28, 2005 ### HallsofIvy As pointed out before, if v is an eigenvector corresponding to eigenvalue $\lambda$ the so is any multiple of v. Yes, [5/2, 1] is an eigenvector corresponding to eigenvalue 4. Multiplying that vector by 2, so is [5, 2]. Some people prefer to have integer components (which is not always possible), some people prefer to write eigenvectors so that have unit length. Since [5,2] has length $\sqrt{29}$ an unit eigenvector would be $$\left[\frac{5}{\sqrt{29}},\frac{2}{\sqrt{29}}\right]$$ 6. Nov 29, 2005 ### jd1828 Why would you want to use the unit vector. We never really went over that in class but there is a question on a lab that uses it. Where we had to show that the vectors are mutually orthogonal. I dont have the exact question with me but Im still working on the anwser. 7. Nov 30, 2005 ### HallsofIvy If you have a basis for your vector space where the vectors in the basis are "orthonormal"- orthogonal and have length 1- then calculations are much simplified. If, in addition, your basis vectors are all eigenvectors of some linear transformation, then working with that linear transformation is very simple. 8. Nov 30, 2005 ### jd1828 Im home now so I can give you the exact question. Given matrix a=[3,3,2; 3,6,5; 2,5,11] a) determine eigenvalues. This part I can do just fine b) Determine unit eigenvectors with magnitude one. So I just take each eigenvector and divide by the lenght? c) Show that the unit eigenvectors are mutually orthogonal. This I have no idea what to do. Ive been reading through the book and it looks like I have to show that two unit vectors mutiplied together are zero. 9. Nov 30, 2005 ### TD Indeed, just as HallsofIvy illustrated. Yes, if you mean the scalair product (inner product). 10. Nov 30, 2005 ### HallsofIvy Good! That's the hard part! Yes, that's exactly right. Reading the book is always a good idea! Two vectors are orthogonal if and only if their dot product is equal to 0. Becareful to say "dot" product- there are several different kinds of multiplication defined for vectors. Take the dot product of each pair of vectors and see what you get. 11. Nov 30, 2005 ### jd1828 Thanks a lot! I tried it out and the dot products do equal zero.	2018-02-18T20:35:14	{ "domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/question-about-eigenvectors.101776/", "openwebmath_score": 0.7976082563400269, "openwebmath_perplexity": 671.0114187453303, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9664104962847374, "lm_q2_score": 0.8633916082162403, "lm_q1q2_score": 0.8343907125843343 }
http://rozmierka.eu/journal/8c5079-disjoint-vs-independent	In mutually exclusive events, the occurrence of one event will result in the non-occurrence of the other. Independent means if one happens it doesn't affect whether or not the other happens. Does this use of the perfect actually express something about the future? JavaScript is disabled. You stand at the basketball free-throw line and make 30 attempts at at making a basket. If possible, please give more than one example and counterexample. Disjoint means the two events are mutually exclusive -- if one happens than the other can't happen. The probability of the empty set is zero; therefore, the event that both G and H occur is impossible.This means that G and H are mutually exclusive.. How to Identify Independent Events. Disjoint means the two events are mutually exclusive -- if one happens than the other can't happen. Ahhh i seee, so if disjoint, they cannot both occur, so the intersection is the empty set. That is the reason you are having trouble finding an example of this case; it is a pathological case that is not important. Quick way to move an object some distance from one external vertex to another external vertex? I wouldn't say that fair coin tosses are in any way dependent on each other. You are right though: in the more general case beyond a binary partition, you could have multiple disjoint events that don't cover the sample space and each have probability zero. Empty set disjoint with itself paradoxical? You and a randomly selected student from your class both earn A's in this course. Is there an example of two causally dependent events being logically (probabilistically) independent? Well, tossing a fair coin leads to series of events that are both independent and disjoint. For independence, # P(A \cap B) = P(A) * P(B) #, If two disjoint events are independent then #P(A \cap B) = 0 = P(A) * P(B) #. rev 2020.11.2.37934, Sorry, we no longer support Internet Explorer, The best answers are voted up and rise to the top, Cross Validated works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. Disjoint events are mutually exclusive, which is a strong form of statistical dependence (so if you know event A occurred you know that B definitely did not occur and vice versa), meaning. (C64). It is more likely that rates of milk-consumption differ for Trump-voters and non-Trump voters, for all sorts of indirect reasons. To learn more, see our tips on writing great answers. In particular, within the population of US adults, males drink more milk, and are also more likely to vote for Trump, so if this is one of the main indirect effects, I would expect that there would be some small positive correlation between milk-consumption and voting for Trump. Ok, what about the union, i.e what aboue p(A U B ) if disjoint or independent, are we allowed to sum for both? Intro to probability 2: Independent and disjoint ( - YouTube Specific Multiplication Rule. Suggestions for braking with severe osteoarthritis in both hands, Why does the VIC-II duplicate its registers? What's some examples of the fact disjoint events are dependent? Disjoint vs. independent. Two events are mutually exclusive if they can't both happen. When you think about independent events, you should be thinking about the probabilities of the events; the likelihood of one event happening is not influencing the likelihood of another. Is "releases mutexes in reverse order" required to make this deadlock-prevention method work? You... You have a balanced coin. What is the probability that, if you roll a balanced die twice, that you will get a "1" on both dice? Multiplying the probabilities of two disjoint sets. Independent events are events where knowledge of the probability of one doesn't change the probability of the other. 2641 views - the two events representing the results of each individual coin toss, which are disjoint, but not independent (since $P(H \cap T) = 0 \ne P(H) P(T)$) - the results of subsequent coin tosses, which are independent, but not disjoint, if they are modelled as part of the same sigma algebra (P(first toss is H) doesn't exclude P(second toss is H) or P(second toss is T)) Two positive disjoint events cannot be independent. Technically isn't there also the trivial case were they both have probability zero? Two events A and B are said to be independent if the outcome of event A doesn’t affect the outcome of event B and vice versa. Disjoint/independent: Disjoint events are only statistically independent if each outcome has probability zero or one (which is a trivial case). They are still disjoint technically because their intersection is still empty. Conversely, in independent events, occurrence of one event will have no influence on the occurrence of the other. Detailed explanation: what is "dayspring"? As a rough rule of thumb (which is slightly exaggerated, but not much) everything is correlated with everything. Disjoint implies dependent. Although they are unlikely to be causally related, it would surprise me if voting-preference is uncorrelated with milk-consumption. Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. How can a hive mind secretly monetize its special ability to make lots of money? So, in Part 1, red and orange in one draw are clearly disjoint. So when we have evens and we say the probability of their union is equal to their sum, does this mean they are independent or disjoint ? Why sister [nouns] and not brother [nouns]? An event is deemed independent if it offers no information about other events. Why does a blocking 1/1 creature with double strike kill a 3/2 creature? Making statements based on opinion; back them up with references or personal experience. Only valid for independent events P(A and B) = P(A) * P(B) Example 3: P(A) = 0.20, P(B) = 0.70, A and B are independent. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Adjective agreement-seems not to follow normal rules. Set Theory, Logic, Probability, Statistics, Genomic data 'catches corals in the act' of speciation and adaptation, Bats can predict the future, researchers discover, Neanderthal children grew and were weaned similarly to modern humans. Are these definitions correct? The Addition Rule for Disjoint Events (Rule Four) The General Addition Rule for which the events need not be disjoint (Rule Five) In order to complete our set of rules, we still require two Multiplication Rules for finding P(A and B) and the important concepts of independent events and conditional probability. I'm trying to understand joint probabilities and dependence/independence. The last 3 times you went to the dentist for your 6-month checkup, it rained as you drove to her... You roll a balanced die two times. Your other examples look correct (at least in spirit, I have not surveyed voter milk drinking habits). Use MathJax to format equations. What is the difference between disjoint and independent events, how will the 2 affect calculations involving them? Independent means if one happens it doesn't affect whether or not the other happens. Are these correct? Asking for help, clarification, or responding to other answers. If you draw one card from a deck of cards, what is the probability that it is a heart or a diamond? around the world. Joint/disjoint and dependent/independent events, rates of milk-consumption among US adults differ by age and sex, voting patterns also differ by age and sex, Creating new Help Center documents for Review queues: Project overview, 2020 Moderator Election Q&A - Questionnaire, 2020 Community Moderator Election Results. In your first 350 flips, you have obtained 300 tails and 50 heads. Is the nucleus smaller than the electron? Read on to find out more about dependent events vs independent events. For example, rates of milk-consumption among US adults differ by age and sex, and voting patterns also differ by age and sex. For a better experience, please enable JavaScript in your browser before proceeding. Best Veep Insults Reddit, Mike Ditka Wife, Helen Frankenthaler Reproductions, Gina Ravera Husband, Anne Finucane Husband, Wisconsin Dells Police Scanner, What Does Politically Charged Mean, John And Pamela Douglas, Im Done Stop It Doesn't Fit Meme, Blueberry Faygo (1 Hour Loop Clean), Film 5 Cm Full Movie Indonesia Bioskop 2013,	2022-05-18T09:46:04	{ "domain": "rozmierka.eu", "url": "http://rozmierka.eu/journal/8c5079-disjoint-vs-independent", "openwebmath_score": 0.3677603602409363, "openwebmath_perplexity": 910.1303782752815, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.966410494349896, "lm_q2_score": 0.8633916047011595, "lm_q1q2_score": 0.8343907075167976 }
https://math.stackexchange.com/questions/1267423/understanding-a-proof-showing-that-for-any-prime-p-there-are-integers-x-and	Understanding a proof showing that for any prime $p$, there are integers $x$ and $y$ such that $p\|(x^2+y^2+1)$. I asked this question a couple days ago: Show that for any prime $p$, there are integers $x$ and $y$ such that $p\|(x^{2} + y^{2} + 1)$. But I asked it as a guest, and I could not comment on the answers I got as I did not have enough reputation. I have found a proof from Warwick University that will help me answer this question, but there is one sentence I am unsure of. If anybody could help me out on it, that would be amazing! Here it is: There are $\frac{p+1}{2}$ numbers of the form $x^2 \pmod{\!p}$. There are $\frac{p+1}{2}$ numbers of the form $-1-y^2 \pmod{\!p}$ (because to get the numbers of the form $-1-y^2$, multiply the squares by $-1$ and subtract $1$… this will not change how many there are). There are exactly $p$ numbers mod $p$. Note that $\frac{p+1}{2}$ is more than half of $p$. So the set of numbers of the form $x^2$ and those of the form $\frac{p+1}{2}$ must have at least one common element (if not, we will have too many distinct numbers mod $p$). So there are some $x$ and $y$ such that $x^2\equiv -1-y^2 \pmod{\!p}$. Thus $x^2 + y^2 +1\equiv 0 \pmod{\!p}$. I understand it up until the bolded sentence. Did they mean to write numbers of the form $\frac{p+1}{2}$? Or did they instead mean numbers of the form $-1-y^2$? No matter what they meant, would anybody be able to explain more how having at least one overlapping element in both sets tells you that solutions to that congruence exists? How did they come to that conclusion that there is a common element? What would happen if there wasn't a common element? Thank you! • I think that they meant "of the form $-1-y^2$" and this is axactly what I wrote in my answer for your question – Elaqqad May 8 '15 at 14:36 As you have guessed, there is an error (maybe just a typo) in that sentence. It should be So the set of numbers of the form $x^2$ and those of the form $-1-y^2$ must have at least one common element (if not, we will have too many distinct numbers mod $p$). The point is that if you give all possible values to $x$ modulo $p$, all possible values to $y$ and then work out all the values of $x^2$ and of $-1-y^2$ modulo $p$, you will get $\frac{p+1}2$ different values for $x^2$ and $\frac{p+1}2$ different values for $-1-y^2$, that's $p+1$ answers altogether. But you cannot get $p+1$ different answers as there are only $p$ possibilities modulo $p$. So two of your answers must be the same: either • two of the $x^2$ values are the same; or • two of the $-1-y^2$ values are the same; or • an $x^2$ value is the same as a $-1-y^2$ value. But we have already noted that the first two of these options are impossible, so the third one must be true. If it helps, look at a specific example, say $p=7$. There are four possible values of $x^2$, namely $$0,\ 1,\ 4,\ 2.$$ Now without working them out, you can see that there will be four values of $-1-y^2$: so we will have eight numbers written down: so two of them must coincide. To confirm this, the values of $-1-y^2$ are $$-1,\ -2,\ -5,\ -3,$$ that is, $$6,\ 5, 2,\ 4,$$ and you can see that the result $2$ occurs twice. (In fact, so does the result $4$, giving another possible solution.) Exercise: show that the "other possible solution" is really the same solution. • You've saved my life, thank you so much! Completely understand now! – user236043 May 5 '15 at 4:52 A more general version of this result can be found in Herstein's : Topics in Algebra text. Kindly see Lemma 7.1.7 Page $359$. The result states that • If $F$ is a finite fied and $\alpha \neq 0$ and $\beta \neq 0$ are two elements of $F$ then we can find elements $a$ and $b$ in $F$ such that $1 + a \cdot\alpha^{2} + b \cdot \beta^{2}=0$. • This is not really a generalisation because this says only that $1,\alpha^2,\beta^2$ are dependant which is not really the same thing as saying that there exists some $x,y$ such that $1+x^2+y^2=0$ (it's correct but what I mean is that your answer is not a generalisation it's another separte result ) – Elaqqad May 8 '15 at 14:35	2019-08-19T11:51:04	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1267423/understanding-a-proof-showing-that-for-any-prime-p-there-are-integers-x-and", "openwebmath_score": 0.8588747382164001, "openwebmath_perplexity": 86.35489036209417, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9664104933824754, "lm_q2_score": 0.8633916047011594, "lm_q1q2_score": 0.8343907066815346 }
https://math.stackexchange.com/questions/2337650/finding-the-basis-for-the-null-space-of-4-times-4-matrix	# Finding the basis for the null space of $4\times 4$ matrix My linear algebra professor gave us a practice worksheet for our upcoming exam, and the answer key he gave has me very confused. I'm not entirely sure his answers are correct, which is why I thought i'd bring the question to you good folks. The question is as follows: Find a basis for the solution space of Ax = 0 if $$A = \begin{bmatrix} 1&1&0&0\\ -2&-2&0&0\\ 0&0&1&-1\\ -1&-1&0&1 \end{bmatrix}$$ Reduced row echelon form yields: $$\left[ \begin{array}{cccc\|c} 1&1&0&0&0\\ 0&0&1&0&0\\ 0&0&0&1&0\\ 0&0&0&0&0 \end{array} \right]$$ $\implies$ $x_1 = -x_2$ $x_3 = 0$ $x_4 = 0$ $\implies$ Basis for null(A) = $\begin{bmatrix} -1\\ 1\\0\\0 \end{bmatrix}$ His solution, however, gives: Basis for null(A) = $\begin{bmatrix} -1\\ 1\\0\\0 \end{bmatrix}$ , $\begin{bmatrix} 0\\ 0\\-1\\1 \end{bmatrix}$ Did I make a mistake here? Or is this answer incorrect. I have always always been under the assumption that $dim(\mathbb R^4) = rank(A) + nullity(A)$ per the rank-nullity theorem. In this case: $4 = 3 + nullity(A) \implies nullity(A) = 1$ meaning, to me, the null space should contain 1 vector and not 2. If i'm totally off base here please let me know, as I really want to understand this stuff. Thanks! • Did you check your lecturer's solution? Take the vectors $v_1$ and $v_2$ he gave you and see whether $Av_1$ and $Av_2$ are both the zero vector? – Lord Shark the Unknown Jun 27 '17 at 4:43 • It did not give the zero vector for A$v_2$, although I was not aware of that check, thank you for sharing. So i'm guessing this means his solution is wrong? – FuegoJohnson Jun 27 '17 at 4:49 • Yes, this means his solution is wrong! – Lord Shark the Unknown Jun 27 '17 at 4:55 • @FuegoJohnson This sanity check is nothing more than applying the definition of a null space to the purported basis of a null space. – amd Jun 27 '17 at 6:58 • Quite likely there’s a typo in the matrix $A$. A small change to it will make the null space two-dimensional. – amd Jun 27 '17 at 7:03 • Since the eigenspace is the null space of $det(A - \lambda I)$, it follows that the dimension of the eigenspace is $n - rank(A - \lambda I)$ for any square matrix. Is this correct? – FuegoJohnson Jun 27 '17 at 5:56	2019-11-19T21:13:39	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2337650/finding-the-basis-for-the-null-space-of-4-times-4-matrix", "openwebmath_score": 0.7909102439880371, "openwebmath_perplexity": 280.39966893319956, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.96036116089903, "lm_q2_score": 0.8688267847293731, "lm_q1q2_score": 0.8343874996028724 }