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https://www.abragas.com.br/4j94m/997ad1-if-two-triangles-are-equal-in-area%2C-they-are-congruent	[email protected] +55 (21) 2221-6695 if two triangles are equal in area, they are congruent 26 de janeiro de 2021, às 3:11 They can have the same angles but have sides of different lengths. Now, the area of triangle ABC is ABCM/2 and the area of triangle ABD is ABDN/ 2 But CM=DN so these two areas are equal. In the simple case below, the two triangles PQR and LMN are congruent because every corresponding side has the same length, and every corresponding angle has the same measure. (Why? In this case, two triangles are congruent if two sides and one included angle in a given triangle are equal to the corresponding two sides and one included angle in another triangle. find the price at which it was sold. The two triangles have two angles congruent (equal) and the included side between those angles congruent. You can replicate the SSS Postulate using two straight objects -- uncooked spaghetti or plastic stirrers work great. (ii) If two squares have equal areas, they are congruent. #2 That is true, they must both have the same side length to have the same perimeter, therefore they will also have the same area. You now have two triangles, △SAN and △SWA. Definition: Triangles are congruent when all corresponding sides and interior angles are congruent.The triangles will have the same shape and size, but one may be a mirror image of the other. 2 triangles are congruent if they have: exactly the same three sides and SSSstands for "side, side, side" and means that we have two triangles with all three sides equal. If we can show, then, that two triangles are congruent, we will know the following: 1) Their corresponding sides are equal. An included side is the side between two angles. Two triangles are congruent if their corresponding sides are equal in length and their corresponding interior angles are equal in measure. Join now. We could also think this angle right over here. So also, sides d and e must be equal because DEF is isosceles. It can be shown that two triangles having congruent angles (equiangular triangles) are similar, that is, the corresponding sides can be proved to be proportional. I could have one triangle with sides measuring 7 in + 5 in + 8 in (perimeter = 20 in) and another triangle with sides of 6 in + 4 in + 10 in (perimeter = 20 in). Triangles are congruent when all corresponding sides & interior angles are congruent. Only if the two triangles are congruent will they have equal areas. The Angle Side Angle Postulate (ASA) says triangles are congruent if any two angles and their included side are equal in the triangles. Theorem 1 : Hypotenuse-Leg (HL) Theorem. So we have: a=d. Hence all squares are not congruent. ... maths. = True (iii) If two figures have equal areas, they are congruent. Two triangles, ABC and A′B′C′, are similar if and only if corresponding angles have the same measure: this implies that they are similar if and only if the lengths of corresponding sides are proportional. We use the symbol ≅ to show congruence. they have to be congruent to have same perimeters AND area con. …, how much loss and profit computed on his original capital. Technically speaking, that COULD almost be the end of the proof. When triangles are similar, they could be congruent. prove that they are congruent. Move to the next side (in whichever direction you want to move), which will sweep up an included angle. AAS (Angle-Angle-Side): If two pairs of angles of two triangles are equal in measurement, and a pair of corresponding non-included sides are equal in length, then the triangles are congruent. The congruence of two objects is often represented using the symbol "≅". ALL of this is based on a single concept: That the quality that we call "area" is an aspect of dimensional lengths and angles. Light1729 Light1729 If two triangles are congruent then they must overlap each other completely, so, they have exactly equal area. Yes, they are similar. And for sensible cases. Only if the two triangles are congruent will they have equal areas. Furthermore, your question is about congruent triangles and not similar triangles.l. If two triangles have the same area, they must be congruent. (d) if two sides and any angle of one triangle are equal to the corresponding sides and an angle of another triangle, then the triangles are not congruent. You can check polygons like parallelograms, squares and rectangles using these postulates. Log in. e.g. …, nswer then don't give) give me with full steps.don't spam ❌❌❌❌❌. Asked on January 17, 2020 by Premlata Pandagre In a squared sheet, draw two triangles of equal areas such that (i) the triangles are congruent. However, different squares can have sides of different lengths. Two or more triangles that have the same size and shape are called congruent triangles.. Thus, a=d. Two triangles are congruent if their corresponding sides are equal in length and their corresponding interior angles are equal in measure. Asked on December 26, 2019 by Yugandhara Jhunjhunwala. #2 That is true, they must both have the same side length to have the same perimeter, therefore they will also have the same area. All the sides of a square are of equal length. Click hereto get an answer to your question ️ If the area of two similar triangles are equal, prove that they are congruent. Triangles can be considered congruent if following conditions are satisfied. All three triangle congruence statements are generally regarded in the mathematics world as postulates, but some authorities identify them as theorems (able to be proved). (iv) If two triangles are equal in area, they are congruent. The two are not the same. one could look upside-down compared to the other but if they have the same dimensions (i.e. monojmahanta4 monojmahanta4 09.04.2020 Math Secondary School If the area of two triangles are equal. (iii) If two rectangles have equal area, they are congruent. But that does not mean that they have to be congruent. = False (iv) If two triangles are equal in area, they are congruent. That is false..one could have side lengths of 9 and 1, and the other could have side lengths of 3. See the included side between ∠C and ∠A on △CAT? You can specify conditions of storing and accessing cookies in your browser, If the area of two triangles are equal. Testing to see if triangles are congruent involves three postulates, abbreviated SAS, ASA, and SSS. An included angleis an angle formed by two given sides. (ii) If two squares have equal areas, they are congruent. Triangles can be similar or congruent. Ask your question. You can't do it. For example: (See Solving SSS Trianglesto find out more) You may have to rotate one triangle, to make a careful comparison and find corresponding parts. Join now. Why should two congruent squares have the same area? Congruent Triangles. Here, instead of picking two angles, we pick a side and its corresponding side on two triangles. Those are the three magnitudes of plane geometry: length (the sides), angle, and area. Want to see the math tutors near you? Testing to see if triangles are congruent involves three postulates. Corresponding sides and angles mean that the side on one triangle and the side on the other triangle, in the same position, match. Contrapositive of the given statement : If the areas of two traingles are not equal then the triangles are not congruent. For example, these triangles are similar because their angles are congruent. And therefore as congruent shapes have equal lengths and angles they have equal are by definition. So the fact that two triangles have the same area does not necessarily tell us that they are congruent (they could be congruent but that is often not the case) since they can have different measures on their sides or a different size and still have the same area. The triangle which maximises area for a given perimeter is the equilateral triangle with each side a third of the perimeter; the area is then p 2 /√432. (e) There is no AAA congruence criterion. Both the areas would be 9, but the figures would not be congruent. Cut a tiny bit off one, so it is not quite as long as it started out. the same shape and size, or to put it another way, the same angles and side lengths), they are congruent. So I guess the answer is not always. (f) Two circles having same circumference are congruent. Compare them to the corresponding angles on △BUG. Introducing a diagonal into any of those shapes creates two triangles. Definition: Triangles are congruent when all corresponding sides and interior angles are congruent.The triangles will have the same shape and size, but one may be a mirror image of the other. If the area of two similar triangles are equal, prove that they are congruent. The given statement -. So let's just start That triangle BCD is congruent Let’s use congruent triangles first because it requires less additional lines. -If two squares have equal perimeters, then they have equal ...” in Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions. The SAS Postulate says that triangles are congruent if any pair of corresponding sides and their included angle are congruent. Congruent triangles will have completely matching angles and sides. The meaning of congruence in Maths is when two figures are similar to each other based on their shape and size. If the areas of two similar triangles are equal, prove that they are congruent. Put them together. Pick any side of △JOB below. Only if the two triangles are congruent will they have equal areas. Prove that if in two triangles,two angles and the included side of one triangle are equal to two angles and the included side of the other triangle,then two triangles are congruent. the cost price of a flower vase is rs 120 . The diagonals of a rhombus divide it into four triangles of equal area. Converse of the above statement : If the areas of the two triangles are equal, then the triangles are congruent. Congruence is our first way of knowing that magnitudes of the same kind are equal. (iv) If two triangles are equal in area, they are congruent. For example, both of these triangles are isosceles, since they have two equal sides and angles. So go ahead; look at either ∠C and ∠T or ∠A and ∠T on △CAT. So once you realize that three lengths can only make one triangle, you can see that two triangles with their three sides corresponding to each other are identical, or congruent. Congruent triangles are triangles which are identical, aside from orientation. By applying the Side Angle Side Postulate (SAS), you can also be sure your two triangles are congruent. Now you have three sides of a triangle. Divide a square sheet in two triangles of equal area so that they are congruent. If the corresponding angles of two triangles are equal, then they are always congruent. … Using any postulate, you will find that the two created triangles are always congruent. 0 votes . Congruent Triangles. If 2 sides and the included angle of one triangle are equal to the corresponding sides and angle of another triangle, the triangles are congruent. Hence, the congruence of triangles can be evaluated by knowing only three values out of six. If the hypotenuse and one leg of a right triangle are equal to the hypotenuse and one leg of another right triangle, then the two right triangles are congruent. What is the solution set of y=3x+16 and 2y=x+72? So, for equal area, all sides are equal. asked Jul 7 in Congruence of Triangles by Rani01 (52.4k points) congruent triangles; class-7; 0 votes. Two triangles are congruent if they have the same shape and size, but their position or orientation may vary. You may think we rigged this, because we forced you to look at particular angles. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. Think: Two congruent triangles have the same area. ASA (Angle-Side-Angle): If two pairs of angles of two triangles are equal in measurement, and the included sides are equal in length, then the triangles are congruent. Side-Angle-Sideis a rule used to prove whether a given set of triangles are congruent. (if you don't know the a If the areas of two similar triangles are equal, prove that they are congruent. Geometricians prefer more elegant ways to prove congruence. But just to be overly careful, let's compute a/d. $\endgroup$ – Moti Feb 7 '15 at 16:42 add a comment \| Your Answer In the sketch below, we have △CAT and △BUG. State true or false and justify your answer. In most systems of axioms, the three criteria – SAS, SSS and ASA – are established as theorems. (iii) If two figures have equal areas, they are congruent. Since two triangles are similar therefore the ratio of the area is equal to the square of the ratio of its corresponding side a r e a ∆ A B C a r e a ∆ D E F = B C E F 2 = A B D E 2 = A C D F 2 B C E F 2 = A B D E 2 = A C D F 2 = 1 N o w , t a k i n g a n y o n e c a s e 1 = B C E F 2 1 = B C E F E F = B C (ii) If two squares have equal areas, they are congruent. In the simple case below, the two triangles PQR and LMN are congruent because every corresponding side has the same length, and every corresponding angle has the same measure. prove that they are congruent. ... triangles are equal, prove that they are congruent. Those are the three magnitudes of plane geometry: length (the sides), angle, and area. You can compare those three triangle parts to the corresponding parts of △SAN: After working your way through this lesson and giving it some thought, you now are able to recall and apply three triangle congruence postulates, the Side Angle Side Congruence Postulate, Angle Side Angle Congruence Postulate, and the Side Side Side Congruence Postulate. You can think you are clever and switch two sides around, but then all you have is a reflection (a mirror image) of the original. Congruent triangles are triangles having corresponding sides and angles to be equal. https://www.mathsisfun.com › geometry › triangles-congruent.html If the areas of two similar triangles are equal, prove that they are congruent. An included angleis an angle formed by two given sides. But, it is not necessary that triangles having equal area will be congruent to one another they may or may not be congruent. Conditional Statements and Their Converse. For eg., [ example for non-congruent triangles having equal areas] (1) ABC has base BC = 6 cm and height AD = 4 cm, other sides AB =12 cm, CA = 17 cm. Now you will be able to easily solve problems on congruent triangles definition, congruent triangles symbol, congruent triangles Class 8, congruent triangles geometry, congruent t If we can show, then, that two triangles are congruent, we will know the following: 1) Their corresponding sides are equal. A, B and C are three sets, n(A) = 14, n(C) = 13, n(A U BUC) = 28, Local and online. This forces the remaining angle on our △CAT to be: This is because interior angles of triangles add to 180°. ASA (angle, side, angle) ASA stands for “ angle, side, angle ” and means that we have 2 triangles where we know 2 angles and the included side are equal. Similar triangles will have congruent angles but sides of different lengths. are both 90 degrees. Guess what? Congruence is denoted by the symbol ≅. If triangle RST is congruent to triangle WXY and the area of triangle WXY is 20 square inches, then the area of triangle RST is 20 in.². The four triangles are congruent with each other regardless whether they are rotated or flipped. Find n(ABC).9. The given statement - "If two triangles are congruent, then their areas are equal." Learn faster with a math tutor. …, n(A) = n(B), nn Byn(B, C) = 4 and n(An C) = 3. This is the only postulate that does not deal with angles. AAS is equivalent to an ASA condition, by the fact that if any two angles are … If you have two similar triangles, and one pair of corresponding sides are equal, then your two triangles are congruent. Log in. A postulate is a statement presented mathematically that is assumed to be true. Two right triangles can be considered to be congruent, if they satisfy one of the following theorems. To be congruent two triangles must be the same shape and size. asked by priyanka. Congruent triangles sharing a common side. Let's take a look at the three postulates abbreviated ASA, SAS, and SSS. You will see that all the angles and all the sides are congruent in the two triangles, no matter which ones you pick to compare. After you look over this lesson, read the instructions, and take in the video, you will be able to: Get better grades with tutoring from top-rated private tutors. If the hypotenuse and one leg of a right triangle are equal to the hypotenuse and one leg of another right triangle, then the two right triangles are congruent. You can only assemble your triangle in one way, no matter what you do. FALSE. Triangle Congruence Theorems (SSS, SAS, ASA), Congruency of Right Triangles (LA & LL Theorems), Perpendicular Bisector (Definition & Construction), How to Find the Area of a Regular Polygon, Do not worry if some texts call them postulates and some mathematicians call the theorems. "If two triangles are congruent, then their areas are equal." b=e. Pairs - The classic pairs game with simple congruent shapes. So now you have a side SA, an included angle ∠WSA, and a side SW of △SWA. Note that if two angles of one are equal to two angles of the other triangle, the tird angles of the two triangles too will be equal. • SSS (Side Side Side)  if all three corresponding sides are equal in length. 3) They are equal areas. it the shopkeeper sells it at a loss of 10% . TRUE. The legs of each of these isosceles triangles could have any lengths as long as they are equal, but the legs of these two triangles need not be the same. CPCT Rules in Maths. Yes, if two triangles have two congruent angles and two congruent sides then the triangles are guaranteed to be congruent. It is true that all congruent triangles have equal area. Comparing one triangle with another for congruence, they use three postulates. https://www.mathsisfun.com/geometry/triangles-congruent.html Remember that the included angle must be formed by the given two sides for the triangles to be congruent. Illustration of SAS rule: Suppose you have parallelogram SWAN and add diagonal SA. 2) Their corresponding angles are equal. However, they do not have to be congruent in order to be similar. asked Sep 21, 2018 in Class IX Maths by navnit40 ( -4,939 points) That is false..one could have side lengths of 9 and 1, and the other could have side lengths of 3. This will happen when the area is half the multiplication of the sides, or maximum area for such two sides. -If two rectangles have equal areas, then they are congruent. The postulate says you can pick any two angles and their included side. You can now determine if any two triangles are congruent! Find a tutor locally or online. Side-Angle-Sideis a rule used to prove whether a given set of triangles are congruent. Notice that ∠C on △CAT is congruent to ∠B on △BUG, and ∠A on △CAT is congruent to ∠U on △BUG. (g) If two triangles are equal in area, they are congruent. TRUE. Corresponding sides and angles mean that the side on one triangle and the side on the other triangle, in the same position, match. For two triangles to be congruent, the corresponding angles and the sides are to be equal. Perhaps the easiest of the three postulates, Side Side Side Postulate (SSS) says triangles are congruent if three sides of one triangle are congruent to the corresponding sides of the other triangle. Are they congruent? 1. You can only make one triangle (or its reflection) with given sides and angles. Two right triangles can be considered to be congruent, if they satisfy one of the following theorems. Write the following statements in words, farhan sarted his bussiness by investing 75000 in the first year he made a profit of 20% he invested the capital in new business and made loss of 12% prove that they are congruent. If two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the triangles are congruent (Side-Angle-Side or SAS). Converse of the above statement : If the areas of the two triangles are equal, then the triangles are congruent. Since they have to be congruent so let 's compute a/d to get answer... More triangles are congruent not have to be similar triangles by Rani01 ( 52.4k points ) congruent triangles have similar! Be sure your two triangles that ∠C on △CAT together in a different way the. The included side between two angles and two congruent squares have equal areas unit? them... To see if triangles are congruent pick any two angles line SA used. Two straight objects -- uncooked spaghetti or plastic stirrers work great important in the below. At the three magnitudes of the given statement - if if two triangles are equal in area, they are congruent triangles are congruent involves three.. 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What would prove that the two triangles are equal in length like parallelograms squares... A rhombus divide it into four triangles are still congruent at a loss of 10 % because know..., abbreviated SAS, ASA, SAS, SSS and ASA – are as... One pair of corresponding sides are equal, prove that they are congruent in,... They share the side angle side Postulate ( SAS ), angle, and a with. Equal, then your two triangles top-rated professional tutors to prove that they are will!	2021-07-28T07:28:25	{ "domain": "com.br", "url": "https://www.abragas.com.br/4j94m/997ad1-if-two-triangles-are-equal-in-area%2C-they-are-congruent", "openwebmath_score": 0.6212843060493469, "openwebmath_perplexity": 599.1020565360698, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9793540692607816, "lm_q2_score": 0.8670357512127872, "lm_q1q2_score": 0.8491349911448218 }
https://www.physicsforums.com/threads/quick-question-about-integral-of-1-x.300986/	# Quick question about integral of (1/x) 1. ### 2^Oscar 45 I know that the integral of 1/x is ln(x) but if you follow the conventional method of integrating you raise the power and multiply by the new power, and you get an answer that is obviously wrong Would someone mind explaining to me why you cannot integrate this function by normal means? Thanks, Oscar 2. ### jdougherty 25 One way to think of it is the following: If we could find the antiderivative by "normal" means, then we would get $\int \frac{1}{x}~dx = x^{0} = \textrm{constant}$ which is clearly not true from looking at the graph of 1/x Another way to look at it: the reason we can do the reverse power rule usually for two reasons: 1) the fundamental theorem of calculus, and 2) the power rule. In other words, it usually works because we know if we find the function F(x) such that $x^{n}$ is the derivative, then we know the integral is equal to F(x). But the power rule only works when n is not zero, since D[$x^{0}$] = D[1] = 0. 3. ### lurflurf 2,325 This is silly. You are maybe thinking of Integral[x^a]=x^(a+1)/(a+1) (* a!=-1) This cannot hold when a=-1 because it would require division by zero. We can however use limits to extend the result. Integral[x^a]=lim_{a->a} x^(a+1)/(a+1) in particular Integral[x^-1]=lim_{a->-1} x^(a+1)/(a+1)=log(x) Last edited: Mar 20, 2009 4. ### lanedance 3,307 hmmm.. do you mean $$\int x^a = ^{lim}_{b\rightarrow a} \frac{x^{b+1}}{b+1}$$ then $$\int \frac{1}{x}= ^{lim}_{b\rightarrow -1} \frac{x^{b+1}}{b+1} =ln(x)$$ 5. ### HallsofIvy 40,310 Staff Emeritus It is not clear what you are asking. The obvious answer is what lurflurf said. You cannot use the formula $$\int x^n dx= \frac{1}{n+1} x^{n+1}+ C$$ when n= -1 because then you would be dividing by 0. I'm not sure why you think of that as "normal means". 6. ### confinement 192 Think of it this way, the antiderivative of 1/x is the function whose inverse is exactly equal to its own derivative. Let y(x) be the antiderivate of 1/x. Then we have: $$\frac{dy}{dx} = \frac{1}{x}$$ Inverting the Liebniz notation in the way that he intended yields: $$\frac{dx}{dy} = x$$ The last equation says that x' = x, i.e. the function x(y) is equal to its own derivative. This means that x(y) cannot be a polynomial or rational function, since all of those functions change when you differentiate them. It is this perfect property, that the antiderivative of 1/x is the inverse function of the function who is exactly equal to its own derivative, that puts in a class of its own, a special case. 7. ### 2^Oscar 45 Hi, Sorry for being unclear. My question was supposed to ask why you cannot do the following: $$\int x^n dx= \frac{1}{n+1} x^{n+1}+ C$$ for the special case of x-1, I realise that it cannot be done from looking at the graph of y=$$1/x$$ and because the fraction requiring division by 0 cannot be defined. my query was as to why you could not do this for x-1 (which confinement answered). Once again appologies for the lack of clarity in my question, all your responses have cleared some of the clouds from my head :) thanks very much, Oscar 8. ### lurflurf 2,325 Yes, though lim a->a is valid it is confusing to some, I also typoed a into x. Also omiting dx is valid, but confusing to some. Most people denote their favorite base as log, what then is you favorite base if not e? 9. ### lanedance 3,307 yeah its got to be e, though i like the ln just to avoid any confusion Last edited: Mar 21, 2009 10. ### nikkor180 5 Greetings: d/dx [ln(x)] = lim(h-->0) [(ln(x+h) - ln(x)) / h] = lim(h-->0) [(1/h)ln[(x+h)/x] = (1/x) lim(h-->0) [(x/h) ln(1+h/x)] = (1/x) lim(h-->0) [ln(1+h/x)^(x/h)] = (1/x) ln [lim(h-->0) [(1+(h/x))^(x/h)]] {from lim[f(g(x))] = f(lim[g(x)])} = (1/x) ln e = 1/x. Since d/dx [ln(x)] = 1/x, it follows that intgrl[1/x] = ln x. Regards, Rich B. [email protected] 11. ### arildno 12,015 You might prefer the following argument for why the anti-derivative should be ln(x) plus some constant C: 1. We define the arbitrary (non-natural) power of some (positive) variable as follows: $$x^{a}=e^{a\ln(x)}$$ 2. In terms of natural powers, we define the exponential function itself as follows: $$e^{y}=1+\sum_{n=1}^{\infty}\frac{y^{n}}{n!}$$ 3. Let us utilize the following definite integral to prove our point: We have: $$\int_{a}^{b}x^{-1+\epsilon}dx=\frac{1}{\epsilon}(b^{\epsilon}-a^{\epsilon})$$, where a,b,$\epsilon$ are all considered greater than 0. 4. Now, let us utilize definitions from 1, 2 upon the expression in 3. We gain: $$\frac{1}{\epsilon}(b^{\epsilon}-a^{\epsilon})=\sum_{n=1}^{\infty}\frac{\epsilon^{n-1}((\ln(b))^{n}-(\ln(a))^{n}}{n!})$$ 5. Thus, in the limit $\epsilon\to{0}$, all but the first double-term of this sum goes to zero, and we retain: $$\lim_{\epsilon\to{0}}\int_{a}^{b}x^{-1+\epsilon}dx=ln(b)-ln(a)$$ This is in accordance with what we should have! 6. Note that this argument shows that the definite integral of some power of x can be defined as a continuous function of the power variable, even in the "special case" where the power variable has the value -1. Last edited: Mar 21, 2009 12. ### sbcdave 10 I'm new to calculus, but my impression is that an integral should be a function that represents the area under the curve of the function being integrated, which f(x)=ln(x) does not do for f(x)=1/x In the graph of 1/x you can see that from 0 toward infinity the area under the curve would come on quickly and almost stop increasing. Can anyone shed light on what I'm missing. I apologize if this is elementary. I took college algebra 10 years ago, recently took a precalc class at a community college and just finished reading a text book called Brief Calculus. Love math and physics by the way and am glad I found this site. 13. ### checkitagain 99 To 2^Oscar and the rest of this thread's users: --------> $\int{\dfrac{1}{x}}$dx$\ = \ \ln\|x\| \ + \ C$ Other than missing the arbitrary constant, "C," there must be absolute value bars around the "x." $ Edit *$ $\text{Good catch by micromass concerning the missing dx term.}$ Last edited: Dec 21, 2011 14. ### lavinia 1,989 While the answers already given are correct here is a way of looking at it that has always mystified me. 1/x has the remarkable property that the area under it from 1 to some number is the sum of the areas from 1 to any two numbers whose product equals the number. So if x = yz then the area from 1 to x is the sum of the areas from 1 to y and 1 to z. It is easy to show this using the integral definition of area though I would love to see an elementary proof not using calculus. It is not hard to convince yourself that a rational function (quotient of two polynomials) can not have this property. So the logarithm is something new. It's derivative is a rational function but it is not. It is like a portal from the rational to the transcendental. A function that compares two laws of addition and/or multiplication is called a homomorphism. The logarithm is a homomorphism from the positive numbers under multiplication to all of the numbers under addition. The exponential function is the reverse homomorphism taking the numbers under addition to the positive numbers under multiplication. Last edited: Dec 21, 2011 15. ### checkitagain 99 $\text{Not those answers as they concern where the}$ $\text{absolute value symbol is missing, not to mention the "+ C."}$ Please see the post above yours regarding this. 16. ### micromass 18,553 Staff Emeritus You forgot the dx in your post above. So your answer is also incorrect. 17. ### dextercioby 12,304 Somehow (to me at least) related to lavinia's post is the interesting fact discovered by Euler $$\mbox{for x very large} ~ \ln x \simeq \frac{1}{1} + \frac{1}{2} +\frac{1}{3} +...+\frac{1}{x}$$ a result which can be recast rigorously in terms of limits. 18. ### lanedance 3,307 Hey sbcdave -welcome to PF! you're generally best off posting question as new threads, gets less confusing that way, particularly for old threads. The reminann integral (there are other more complex defintions) is interpreted as the area under of a well behaved function With this is mind and considering it as a definite integral (over a given integral where the function is well behaved $$\int_a^bdx \frac{1}{x} = ln(x)\|_a^bdx =ln(b)-ln(a)$$ This is equivalent to the area between the function and the horizontal axis. Notice ln(x) is a a "monotonically" increasing function (ln(b)>ln(a) for all 0<a<b), so it always gives a positive area. I don't totally undertstand your question.. But, at x = 0, f(x) = 1/x is not well defined, so to calculate the area you must use limits. Similarly to deal with taking the integral to infinity, you need to use a limiting process. Things get subtle when you consder lmiting process, so its important to be rigrorous. First lest split the integral into 2 portions, which we can do when the function is well behaved: $$\int_a^bdx f(x) = \int_a^cdx f(x) + \int_c^bdx f(x)$$ In this case lets choose c=1 and the integral becomes $$\lim_{a \to 0^+} \lim_{b \to +\infty} \int_a^b dx \frac{1}{x} = \lim_{a \to 0^+} \int_a^1 dx \frac{1}{x} + \lim_{b \to +\infty}\int_1^bdx \frac{1}{x} = \lim_{a \to 0^+}(ln(1)-ln(a)) + \lim_{b \to +\infty}(ln(b)-ln(1)) = \lim_{a \to 0^+}{-ln(a)} + \lim_{b \to +\infty}ln(b)$$ As both these diverge (become infinite) we actually find there is inifinite area below 1/x in both the intervals (0,1) and (1,inf). In fact in some repsects there is saimialr amount of infinte area in each. Though in both cases the curve compresses against the axis, it doesn't do so quickly enough to result in a finite area - things can get wierd with lmiting processes Hopefully this helped answer your question and didn't confuse things more! can be a nice way to help, learn and keep up with some math skills Mod note: fixed LaTeX Last edited by a moderator: Dec 21, 2011 19. ### sbcdave 10 Thanks Roger But the natural log of anything less than 1 is negative, which seems to say that the area under the curve from 0 to anything less than x=1 on the graph of f(x)=1/x should be a negative area. The graph of f(x)=1/x from x=0 to 1 is above the x-axis and as you say in the next quote is infinite. Example of what I'm struggling with: $$\int_{0.1}^1dx \frac{1}{x}$$ Should be a definable positive integer. Approximately equal to 9(0.1) increments with heights 1/0.9 + 1/0.8 + ... 1/0.1 (left hand method) or approximately equal to 2.829. However, if you do ln(1)-ln(0.1) you get approximately 2.3 (I expected a negative number and need to think about this a little more, there's clearly a flaw in my logic...haha) Sorry. Posting anyways so that you might see where I was coming from. 20. ### sbcdave 10 I was looking at a graph of ln(x) and 1/x together on my ti-83 and assumed that because the ln(x) went negative with x<1 that that meant the area was negative, now I see that because the -y value grows as x gets smaller that your subtracting a larger negative from a negative and end up with a positive. Now I'm a little confused why the left hand method was larger than the ln(x)-ln(x) method instead of smaller though if you have any ideas on that? Thanks again for humoring me once already I'll understand if the thread dies here...lol Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook Have something to add?	2015-04-01T01:08:50	{ "domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/quick-question-about-integral-of-1-x.300986/", "openwebmath_score": 0.8864861130714417, "openwebmath_perplexity": 799.0612908401829, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9793540674530016, "lm_q2_score": 0.8670357477770337, "lm_q1q2_score": 0.8491349862125928 }
https://math.stackexchange.com/questions/1156899/proving-a-open-in-v-subseteq-m-metric-space-iff-a-c-cap-v-certain-open	# Proving '$A$ open in $V\subseteq M$ (metric space) iff $A=C\cap V$ (certain open $C$ in $M$)' I want to prove the following: Let $(M,d)$ be a metric space. Let $A\subseteq V\subseteq M$. 1) $A$ is open in $V \Leftrightarrow A = C\cap V$ (for a certain open $C$ in $M$) 2) $A$ is closed in $V \Leftrightarrow A = C\cap V$ (for a certain closed $C$ in $M$) ### Questions: • Could someone check the proof? • 'for a certain open $C$ in $\color{Blue}{M}$.' Would this proof also work for a more specific choice of $C$? Like for a certain open $C$ in $\color{blue}{V}$. I don't really see the added value of choosing $M$ over $V$. • Could some give me some pointers on how to prove $2, \Rightarrow$? ### Proof 1) $\Leftarrow$: Choose $a\in A$. $$\begin{array}{rl} & a \in A = C\cap V\\ \Rightarrow & a \in C\\ \Rightarrow & (\exists r > 0)(B_M(a,r)\subseteq C)\\ \Rightarrow & (\exists r > 0)(B_M(a,r)\cap V \subseteq C\cap V)\\ \Rightarrow & (\exists r> 0) (B_V(a,r)\subseteq A \end{array}$$ $\Rightarrow$: Choose $a\in A$. $$\begin{array}{rl} \Rightarrow & (\exists r_a >0)(B_V(a,r_a) \subseteq A) \end{array}$$ Consider all $a\in A$ then: $$\begin{array}{rl} & A = \bigcup_{a\in A} B_V(a,r_a)\\ \Rightarrow & A = \bigcup_{a\in A} \left[ V\cap B_M(a,r_a)\right]\\ \Rightarrow & A = V\cap\left[ \bigcup_{a\in A} B_M(a,r_a)\right] \end{array}$$ Let $$\left[ \bigcup_{a\in A} B_M(a,r_a)\right] = C$$ which is open as a union of open sets. ### Proof 2) $\Leftarrow$: $$\begin{array}{rrl} & V\setminus A &= V\setminus(C\cap V)\\ \Rightarrow & & = (V\setminus C)\cup (V\setminus V)\\ \Rightarrow && = V\setminus C \end{array}$$ Since $C$ is closed then $V\setminus C$ is open and so is $V\setminus A$. Then $A$ is closed in $V$. $\Rightarrow$: How? So far everything seems right. For the last proof, let $A$ closed in $V$. Then $V\backslash A$ is open in $V$. By the first statement (which you already proved) there exists an open $B$ in $M$ such that $V\backslash A=B\cap V$. Now $A=(M\backslash B)\cap V$. And we know that $C= M\backslash B$ is closed in $M$ because $B$ is open in $M$.	2019-09-23T05:05:28	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1156899/proving-a-open-in-v-subseteq-m-metric-space-iff-a-c-cap-v-certain-open", "openwebmath_score": 0.9542206525802612, "openwebmath_perplexity": 193.85388651152303, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9943580909983967, "lm_q2_score": 0.8539127510928476, "lm_q1q2_score": 0.849095053055873 }
https://math.stackexchange.com/questions/2221033/is-0-an-imaginary-number	# Is $0$ an imaginary number? My question is due to an edit to the Wikipedia article: Imaginary number. The funny thing is, I couldn't find (in three of my old textbooks) a clear definition of an "imaginary number". (Though they were pretty good at defining "imaginary component", etc.) I understand that the number zero lies on both the real and imaginary axes. But is $\it 0$ both a real number and an imaginary number? We know certainly, that there are complex numbers that are neither purely real, nor purely imaginary. But I've always previously considered, that a purely imaginary number had to have a square that is a real and negative number (not just non-positive). Clearly we can (re)define a real number as a complex number with an imaginary component that is zero (meaning that $0$ is a real number), but if one were to define an imaginary number as a complex number with real component zero, then that would also include $0$ among the pure imaginaries. What is the complete and formal definition of an "imaginary number" (outside of the Wikipedia reference or anything derived from it)? • See Complex number : "A complex number whose real part is zero is said to be purely imaginary, whereas a complex number whose imaginary part is zero is a real number." Apr 6 '17 at 15:51 • Here's what wolfram says Apr 6 '17 at 15:52 • I do not think this question should be down voted. It is well edited and clearly there was decent thought put into it. – user416426 Apr 6 '17 at 15:58 • The downvotes are sad. The premise might seem silly, but the question is well-written and clearly thought-out. I like it. Apr 6 '17 at 16:02 • And why not? Mathematics is full of similar cases. For example, the zero function is the unique function that is both even and odd. – MJD Apr 6 '17 at 16:02 The Wikipedia article cites a textbook that manages to confuse the issue further: Purely imaginary (complex) number : A complex number $z = x + iy$ is called a purely imaginary number iff $x=0$ i.e. $R(z) = 0$. Imaginary number : A complex number $z = x + iy$ is said to be an imaginary number if and only if $y \ne 0$ i.e., $I(z) \ne 0$. This is a slightly different usage of the word "imaginary", meaning "non-real": among the complex numbers, those that aren't real we call imaginary, and a further subset of those (with real part $0$) are purely imaginary. Except that by this definition, $0$ is clearly purely imaginary but not imaginary! Anyway, anybody can write a textbook, so I think that the real test is this: does $0$ have the properties we want a (purely) imaginary number to have? I can't (and MSE can't) think of any useful properties of purely imaginary complex numbers $z$ apart from the characterization that $\|e^{z}\| = 1$. But $0$ clearly has this property, so we should consider it purely imaginary. (On the other hand, $0$ has all of the properties a real number should have, being real; so it makes some amount of sense to also say that it's purely imaginary but not imaginary at the same time.) I don't think there is a complete and formal definition of "imaginary number" It's a useful term sometimes. It's an author's responsibility to make clear what he or she means in any particular context where precision matters. If $0$ should count, or not, then the text must say so. Your question shows clearly that you understand the structure of the complex numbers, so you should be able to make sense of any passage you encounter. A complex number z=a+ib where a and b are real numbers is called : 1- purely real , if b=0 ; e.g.- 56,78 ; 2- purely imaginary, if a=0 ,e.g.- 2i, (5/2)i ; 3- imaginary,if b≠ 0 ,e.g.- 2+3i,1-i,5i ; 0 is purely imaginary and purely real but not imaginary.	2022-01-20T15:39:27	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2221033/is-0-an-imaginary-number", "openwebmath_score": 0.8885458111763, "openwebmath_perplexity": 503.67068777645045, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9473810525948927, "lm_q2_score": 0.8962513738264114, "lm_q1q2_score": 0.8490915699252843 }
http://math.stackexchange.com/questions/205778/prove-that-a-language-b-is-regular	# Prove that a language B is regular here is the question I'm dealing with: Let B = {$1^{k}$y\|y $\in$ {0,1}* and y contains at least k 1s, for k $\geq$ 1}. Show that B is a regular language. Can I use the pumping lemma to show that this is a regular language? My intuition tells me that I can't, since I would have to find every string s that satisfies the pumping lemma. Anyways, your suggestions are appreciated. Thank you in advance. - The pumping lemma for regular languages is only a tool for showing that a language is not regular; it cannot be used to show that a language is regular. The two most straightforward ways to demonstrate that a language is regular are (1) to write a regular grammar that generates it, and (2) to design a finite state automaton that recognizes it. In this case, though, it pays to begin by taking a close look at just what words are in $B$. Consider a word $11x$, where $x\in\{0,1\}^$: since we can set $k=1$, such a word is automatically in $B$, since $1x$ certainly contains at least one $1$. A word that begins with $1$ is not in $B$ if and only if it has no other $1$’s; and a word that begins with $0$ is not in $B$. Thus, $$B=\{1x\in\{0,1\}^:x\in\{0,1\}^\text{ and }x\text{ has at least one }1\}\;,$$ the language corresponding to the regular expression $10^1(0\lor 1)^*$. It’s not at all hard to design a regular grammar that generates this language, or a finite state machine that recognizes it. - I just built the DFA and thus proved that B is regular. Thanks so much for confirming my initial assumption. Thanks a lot! – Dave Oct 2 '12 at 3:22 @Dave: You’re welcome! – Brian M. Scott Oct 2 '12 at 3:24 wait a second.. what about string 1101. this is not accepted by B (take k=2). New B would accept it. Am I wrong? (I would be wrong if you could always pick k=1) – Dave Oct 2 '12 at 3:50 @Dave: The way the description of $B$ is worded allows you to choose $k$. Had it said that there is some $k\ge 1$ such that $B=\{1^ky:\text{ etc. }\}$, that would have been a different story, though not much harder. If it meant that you had to use the maximum $k$ allowed by the initial string of $1$’s, it should have said so; that would be very different. – Brian M. Scott Oct 2 '12 at 3:52 @TheNotMe: That’s even better. You’re welcome! – Brian M. Scott Nov 30 '13 at 17:25 show 4 more comments You can never use (directly) the pumping lemma to show a langauge is regular since there are non-regular langauges that satisfy all the lemma properties. In your case the langauge is very simple and you can build a finite state automata to prove regularity -	2014-04-20T22:19:43	{ "domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/205778/prove-that-a-language-b-is-regular", "openwebmath_score": 0.7482568621635437, "openwebmath_perplexity": 195.94850401896198, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9489172673767973, "lm_q2_score": 0.894789473818021, "lm_q1q2_score": 0.8490811823729187 }
https://www.physicsforums.com/threads/change-in-scales-causing-change-in-gradient.947892/	# Change in scales causing change in gradient? ## Main Question or Discussion Point Hello I plotted a graph of velocity against time only to realise that I needed more space on the X axis, so I changed my scale from instead of 2cm= 10 seconds to 2cm =20 seconds. My Y axis remained constant with a scale of 2cm= 10 m/s However, the gradient I got from the first scale was 1.15m/s2 and this even coincides with the data given. But with my new readjusted scale to fit the page I'm getting a gradient of 2.4 m/s2. How can this be? the graph is a straight line so the acceleration is constant according to the data given, am I plotting this wrong?? DrClaude Mentor However, the gradient I got from the first scale was 1.15m/s2 and this even coincides with the data given. How are you calculating that gradient? Dr Claude, I am using the triangle method and then dividing the vertical units by the horizontal units, using two poitns that are NOT given in the data but instead two points from the line of best fit I drew, which is standard practice for this level jedishrfu Mentor It's somewhat hard to determine what you did wrong without seeing any work or a picture of your chart. If you've computed the slope using algebraic means you should get the same value ie $(y1-y0)/(x1-x0)$ for your acceleration (velocity/seconds ie $m/s^2$) If instead you are using the grids on your graph paper then of course you'll get something different unless you adjust for the x length of the grid element. As an example, if you computed it using $(y1-y0)$ in centimeters over $(x1-x0)$ in centimeters then that is wrong. Whereas if you computed it $( (y1-y0).in.cm) * (10m/s.per.cm) )$ over $( ((x1-x0).in.cm) * (20 seconds.per.cm))$ DrClaude Mentor Dr Claude, I am using the triangle method and then dividing the vertical units by the horizontal units Then my guess is that you are not calculating the units correctly. Can you post the plots you did? It may help in figuring where you are going wrong. Ok here are my images with the scales represented above and the data presented to plot orginally... However I realized what I did wrong... I simply forgot to convert thehorizontal units in my new graph the second picture, to suit...it was just one tiny mistake.... how DO I avoid these? The test I'm studying for specifically wants me to do the triangle method, bigger than half the line, etc. BUt theres not much time and I somtimes I just make silly errors, now I feel dumb sorry for this silly post #### Attachments • 26.7 KB Views: 289 • 43.6 KB Views: 293 • 33.1 KB Views: 294 jedishrfu Mentor This is NOT a silly post. We all learn from our mistakes. The most humbling thing is that we learn we make mistakes. The second thing is in reconstructing our reasoning we might determine why we made the mistake. To be fair, you caught your mistake and that is a very hopeful sign that you're really grasping the subject. This mistake will stick with you forever and I'm sure you won't make this same mistake twice. (fingers crossed). You can even pass it on to your kids with some sagely advice. Our mistakes teach us so much and eventually we get to teach them to others. scottdave	2020-04-02T10:39:47	{ "domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/change-in-scales-causing-change-in-gradient.947892/", "openwebmath_score": 0.7141177654266357, "openwebmath_perplexity": 1012.0382805831316, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9489172688214138, "lm_q2_score": 0.8947894597898777, "lm_q1q2_score": 0.849081170353999 }
http://spmd.safoo.de/geometric-series-test-proof.html	## Geometric Series Test Proof Theorem: (Geometric Series Test) If jrj<1, the geometric series P 1 n=0 ar n, where a6= 0, converges with sum a 1 r: If jrj 1, the series diverges. Let s n = be the n th partial sum. In an ecological study of the feeding behavior of birds, the number of hos between flights was counted for several birds. Now, as we have done all the work with the simple arithmetic geometric series, all that remains is to substitute our formula, (Noting that here, the number of terms is n-1). The word "countable" means that you can label. By inspection, it can be difficult to see whether a series will converge or not. Example 13. Definition. The geometric series and the telescoping series make their appearance in this chapter. (Alternating series test) Consider the series. These are both geometric series, so I can sum them using the formula for geometric series: X Series Test says that the series converges. P), then b is the arithmetic mean of a and c. Water authorities identify a threshold geometric mean where beaches or shellfish beds must be closed. The proof by induction method use to proof geometric series can be applied to other progression as well by doing the same step:. Test results for water quality (specifically, fecal coliform bacteria concentrations) are sometimes reported as geometric means. The geometric series converges and has a sum of if. If the limit of a[n] is not zero, or does not exist, then the sum diverges. 4 Ratio test The geometric series leads to a useful test for convergence of the general series X1 n=0 a n= a 0 + a 1 + a 2 + (12) We can make sense of this series again as the limit of the partial sums S n = a 0 + a 1 + + a n as n!1. The Egyptians used this method of finite geometric series mainly to "solve problems dealing with areas of fields and volumes of granaries" but used it for many other uses too, including the pyramids and math problems similar to those one might find on a STAAR test today (see D1, and F1). If we can –nd a function f(x) such that. 1)25 is a geometric progression with a geometric ratio of (1. Now if 1 k=N+2 a k converged, this implies a k!0. usually shown using the integral test. The first proof in Algebra 2! Students learn to derive the formula for the sum of the first n terms of a finite geometric sequence. 6Theorem (Alternating series test): (i) Let be an alternating series such that (ii) Then is convergent. The Mathematics Vision Project: Scott Hendrickson, Joleigh Honey, Barbara Kuehl, Travis Lemon, Janet Sutorius. The nth-term test for divergence. Any such series can be written as ± P (−1)kak with ak ≥ 0 for all k ∈ N. Finally, we give a test which helps us to analyze convergence of an alternating series. Finding Geometry Help is Easy. You can see that this is reasonable by dividing 1 by , or using the the formula for the sum of a geometric series with ratio. Buy ColorBird Geometric Series Tablecloth Diamond Pattern Cotton Linen Dust-Proof Table Cover for Kitchen Dinning Tabletop Linen Decor (Round, 60 Inch, Yellow): Tablecloths - Amazon. Hence, is absolutely convergent, and thus is itself convergent. 1 nm in a pressurized stirred cell (detailed in fig. This task also provides practice in writing and using formulas for arithmetic sequences. Finally, we give a test which helps us to analyze convergence of an alternating series. In partnership with the. In fact, our proof is an extension of the nice result given by Cohen and Knight [2]. These functions would be difficult if not impossible to write down analytically, but there is software to find conformal maps numerically. A SHORT(ER) PROOF OF THE DIVERGENCE OF THE HARMONIC SERIES LEO GOLDMAKHER It is a classical fact that the harmonic series 1+ 1 2 + 1 3 + 1 4 + diverges. Property 1: If \|r\| < 1 then the geometric series converges to. Free math lessons and math homework help from basic math to algebra, geometry and beyond. 1),…, 500(1. You can then "flex" the 3-D hexaflexagon, exposing each of four six-sided faces, one at a time. There are several questions Consider the geometric series: (Sum from k=0 to infinity) of ar^k and consider the repeating decimal. The circle has radius 25 centimetres and the angle of the sector is 280° a) Find the area of the sector of the circle. If L>1 then X a n diverges. The idea is to compare the series witha geometric series witha ratio slightly larger than ½. Determine the number of terms n in each geometric series. This looks to me like an expression of the "ratio test" for convergence of a series. Grappling with the geometric series, geometry formulas or geometric sequence? Our tutors can help. I we see from the graph that because the values of b n are decreasing, the. In fact this series diverges quite slowly. To make a 3-D hexaflexagon, print out the template, cut and fold carefully, then tape (or glue) into shape. If a n diverges, then b n diverges. Great! Think it might be an arithmetic or geometric sequence? If the sequence has a common difference, it's arithmetic. Radius of Convergence. Now, as we have done all the work with the simple arithmetic geometric series, all that remains is to substitute our formula, (Noting that here, the number of terms is n-1). We also consider two speciﬁc. However, our rules of probability allow us to also study random variables that have a countable [but possibly inﬁnite] number of possible values. Tests for convergence: Ratio Test, Root Test and Raabe's Test. So, the sum of the series, which is the limit of. If 1 then 1 for all so does not define a null sequence and the series diverges by the null sequence test. Geometric Proof Video 366. In mathematics, a geometric series is a series with a constant ratio between successive terms. Direct Comparison Test If 0 <= a n <= b n for all n greater than some positive integer N, then the following rules apply: If b n converges, then a n converges. NOTES ON INFINITE SEQUENCES AND SERIES MIGUEL A. Alternating series Theorem (Leibniz’s test) If the sequence {a n} satisﬁes: 0 < a n, and a n+1 6 a n, and a n → 0, then the alternating series P ∞ n=1 (−1) n+1a n converges. This quiz/worksheet combination will test your understanding of the formula to find the sum of the infinite geometric series by providing you with example problems. Suppose the interest rate is loo%, so i = 1. For example, if , Results on geometric series show that the two expressions are equal. The next proof is unique among all known proofs of the infinitude of the set of primes. The geometric series converges, so also does, by Theorem 6. Thus, we will assume that a = 1. Proof: E(X) = xx Pr(X = x) x E(X)x Pr(X = x) x E(X) E(X) Pr(X = x) = E(X) x E(X) Pr(X = x) = E(X) Pr(X E(X)) Example: If X is B100;1=2, then Pr(X 100) = Pr(X 2E(X)) 1 2 Thisisnotaparticularlyusefulestimate. Note, the disk of convergence ends exactly at the singularity z= 1. Get an answer for 'State and prove Raabe's Test. Note that, is not a geometric series. The diﬀerence is that while the Ratio Test for series tells us only that a series converges (ab-solutely), the theorem above tells us that the sequence converges to zero. Does X1 1 2n n3 converge or diverge?. You must use. Test results for water quality (specifically, fecal coliform bacteria concentrations) are sometimes reported as geometric means. Since every converging sequence is bounded, the s n are bounded. Sequences and Series teaches students how to define, notate and interpret different types of series and sequences, such as arithmetic and geometric, and how to use mathematical induction in proofs and on their homework. relate geometric theorems on points, lines, and planes • Logic : Student will use inductive reasoning to draw reasonable conclusions, or deductive reasoning to prove basic theorems, and write conditional statements, converses, inverses and contrapositives. 6Theorem (Alternating series test): (i) Let be an alternating series such that (ii) Then is convergent. The idea is. INDEX proof, 84–86 characteristic polynomial, 292 circle as parametric curve, 259–260 in polar coordinates, 269 closed interval, 11 codomain, 13. Now if 1 k=N+2 a k converged, this implies a k!0. If it converges, then find its sum. Anionic dyes were not used to avoid the potentially confounding. We would like to show you a description here but the site won't allow us. But the integral test easily shows that this series diverges. Proof: Suppose the sequence converges to zero and is monotone decreasing. And hope Now i'm a section of helping you to get a better product. Geometric Sequences - Module 12. then completeness. In these notes we will prove the standard convergence tests and give two tests that aren't in our text. Therefore, the series has bounded partial sums; hence, this sum converges. The reason the test works is that, in the limit, the series looks like a geometric series with ratio L. So, each of the following is geometric. For example, the series. There are several questions Consider the geometric series: (Sum from k=0 to infinity) of ar^k and consider the repeating decimal. A SHORT(ER) PROOF OF THE DIVERGENCE OF THE HARMONIC SERIES LEO GOLDMAKHER It is a classical fact that the harmonic series 1+ 1 2 + 1 3 + 1 4 + diverges. For example, an interesting series which appears in many practical problems in science, engineering, and mathematics is the geometric series + + + + ⋯ where the ⋯ indicates that the series continues indefinitely. THE RATIO TEST We now know how to handle series which we can integrate (the Integral Test), and series which are similar to geometric or p-series (the Comparison Test), but of course there are a. Proof by induction use the basis that if it is true for n = 1 and we assume it is true for n = 1 to some number k and if we can show that it also work for k+1 then we can proof the validity of the geometric progression. Sequences and Series teaches students how to define, notate and interpret different types of series and sequences, such as arithmetic and geometric, and how to use mathematical induction in proofs and on their homework. The nth-term test for divergence. Geometric Proof Video 366. 93Mb) Get help for solving geometric proofs with artificial intelligence methods. If it converges, then find its sum. As an example the geometric series given in the introduction,. Proof: If and are convergent, then it follows from the sum theorem for convergent sequences that is convergent and is valid. This task contains an opportunity to compare the growth of arithmetic and geometric sequences. We begin by giving the following estimate for the partial sum of a p-series: Lemma. Menu Algebra 2 / Sequences and series / Geometric sequences and series A geometric sequence is a sequence of numbers that follows a pattern were the next term is found by multiplying by a constant called the common ratio, r. The series P 1 n=1 1 2 converges (p-series with p= 2 >1). Definition. A geometric series is the sum of terms with a common ratio. For the above proof, using the summation formula to show that the geometric series "expansion" of 0. About This Quiz & Worksheet. But then after the rst N terms the series P nxn is dominated by the geometric series for rjxj, hence converges 8. geometric - definizione, significato, pronuncia audio, sinonimi e più ancora. As a counterexam-ple, few series more clearly illustrate that the convergence of terms. Good GCSE maths result. ALTERNATING SERIES Does an = (−1)nbn or an = (−1)n−1bn, bn ≥ 0? NO Is bn+1 ≤ bn & lim n→∞ YES n = 0? P YES an Converges TELESCOPING SERIES Dosubsequent termscancel out previousterms in the sum? May have to use partial fractions, properties of logarithms, etc. series is a geometric series, our results on geometric series can be used instead. So, each of the following is geometric. Proof: E(X) = xx Pr(X = x) x E(X)x Pr(X = x) x E(X) E(X) Pr(X = x) = E(X) x E(X) Pr(X = x) = E(X) Pr(X E(X)) Example: If X is B100;1=2, then Pr(X 100) = Pr(X 2E(X)) 1 2 Thisisnotaparticularlyusefulestimate. Note: When you talk about an arithmetic sequence, the word arithmetic (in this context) is pronounced air-ith-ME-tic; that is, the accent is on the third syllable. Observe that 1 n2 + 3 < 1 n2 for every n 1. Geometry - MA3110 IC Scope and Sequence Unit Lesson Lesson Objectives Similar Figures Determine if two polygons are similar using dilations. 3 (Part 2) Infinite Geometric Series - Module 12. But if a series converges absolutely, it also converges. Sequences and series, whether they be arithmetic or geometric, have may applications to situations you may not think of as being related to sequences or series. Sequences and Series teaches students how to define, notate and interpret different types of series and sequences, such as arithmetic and geometric, and how to use mathematical induction in proofs and on their homework. It may be one of the most useful tests for convergence. 2 Tests for Convergence Let us determine the convergence or the divergence of a series by comparing it to one whose behavior is already known. The key steps are check cost, condition of pre-order and price recommendation. Its proof is on the separate handout. Geometric Series Convergence. The Limit Superior/Inferior Ratio Test for Series of Complex Numbers. Series with non-negative terms: Comparison test, Cauchy's condensation theorem. Monday, April 30, 2018. Alternating Series test We have the following test for such alternating series: Alternating Series test If the alternating series X1 n=1 ( 1)n 1b n = b 1 b 2 + b 3 b 4 + ::: b n > 0 satis es (i) b n+1 b n for all n (ii) lim n!1 b n = 0 then the series converges. Either the integral test or the Cauchy condensation test shows that the p -series converges for all p > 1 (in which case it is called the over-harmonic series) and diverges for all p ≤ 1. Since , this series diverges. Here, the common ratio (base) is r = sin 2 x , which is always bounded by 1. The sum of two solutions and a scalar multiple of a solution of such a system is again a solution of the system. In Chapter 1, you learned some basic geometric concepts. If it converges, then find its sum. Conditional Convergence; Summary of Tests; Taylor and Maclaurin. of Convergence. 13 GEOMETRIC SERIES, POWER SERIES, RATIO TEST 3 13. N ∞ (−1) j. For example, an interesting series which appears in many practical problems in science, engineering, and mathematics is the geometric series + + + + ⋯ where the ⋯ indicates that the series continues indefinitely. Prove the convergence of the geometric series using $\epsilon$, N definition $\begingroup$ What I gave is the proof you Uniform Convergence using Abel's test. To solve, determine the value of the cumulative distribution function (cdf) for the geometric distribution at x equal to 3. usually shown using the integral test. We've already looked at these. Students may accept the formula for the sum of an infinite geometric series given that $$\left\| r \right\| < 1$$, and they may even understand the proof of this formula; but they usually are not shown (informally) that the defining feature of a convergent infinite series is that the limit of the series is the limit of its sequence of partial. 19981 THE GEOMETRIC SERIES IN CALCULUS 37--. 1 (A convergent series) A boy is given a chocolate, and he decides that each day he is going to eat. geometric series: Consider the following sequence of numbers: 2, 8, 32, 128, … This is called a geometric progression with a geometric ratio of four. The idea of the ratio test is that we compute the absolute value of that ratio, and. For instance the series 1+3x+9x2+27x3+81x4+ +3kxk+. Proof To prove Property 1, assume that and choose such that By the definition of the limit of a sequence, there exists some such that for all Therefore, you can write the following inequalities. Then is a null sequence, so is a null sequence (by Theorem 7. We must now compute its sum. The Egyptians used this method of finite geometric series mainly to "solve problems dealing with areas of fields and volumes of granaries" but used it for many other uses too, including the pyramids and math problems similar to those one might find on a STAAR test today (see D1, and F1). Print Working with Geometric Sequences Worksheet Recognize a geometric sequence and the quiz allows me to test their knowledge on whatever subject in social studies I am teaching at the. We’ve already looked at these. An infinite geometric series does not converge on a number. You have to to look the same items to test cost as it sometimes will help you in purchasing Design Toscano Set Of 2 Marble Obelisks Geometric Sculptures. Intro to Geometric Sequences. If a n;b n 0 and a n b n for all n, then if the series X1 n=1 b n converges then the series X1 n=1 a n converges. A note about the geometric series Before we get into today's primary topic, I have to clear up a little detail about the geometric series. It's pretty simple; we're really just asking whether the n th term of a series converges to zero, but the divergence test has some important limitations that we should get out of the way right away. Infinite series and the biggest maths problem of them all: One famous series is the Riemann zeta function, which is involved in one of the biggest open problems in maths: the Riemann hypothesis. The term r is the common ratio, and a is the first term of the series. Most importantly, it also tells you the areas which need your attention. Test for Divergence: If lim → ≠0, then diverges. Where a 1 = the first term, a 2 = the second term, and so on a n = the last term (or the n th term) and a m = any term before the last term. Rather than write down the proof, let me just motivate why this theorem is true. for some constants a and r. Then, once you get an explicit formula for f ( x ), you can plug in x = π/3. Another way of looking at that is to ask for an average number of trials before the first occurrence of the event. The Harmonic Series Diverges Again and Again∗ Steven J. Finally, we give a test which helps us to analyze convergence of an alternating series. P), then b is the arithmetic mean of a and c. The Ratio Test Proof (1): If 0 ˆ<1, we can apply the previous theorem to see P 1 n=1 ja nj converges. A proof is a mathematical argument used to verify the truth of a statement. The geometric series diverges if. (Notethatin thissectionwewill sometimesbeginourseriesat n 0 and sometimesbegin them at n 1. For this question you will write a two-column proof of the first part of the Overlapping Angle Theorem. Proof of 1 (if L < 1, then the series converges) Proof of 2 (if L > 1, then the series diverges) Proof of 1 (if L < 1, then the series converges) Our aim here is to compare the given series with a convergent geometric series (we will be using a comparison test). The common ratio of the series is positive. The key steps are check cost, condition of pre-order and price recommendation. Determine whether the series converges or diverges. ALTERNATING SERIES Does an = (−1)nbn or an = (−1)n−1bn, bn ≥ 0? NO Is bn+1 ≤ bn & lim n→∞ YES n = 0? P YES an Converges TELESCOPING SERIES Dosubsequent termscancel out previousterms in the sum? May have to use partial fractions, properties of logarithms, etc. ap calculus bc college board calculus nth term and geometric series test fun worksheet. N ∞ (−1) j. Complete the Lesson Practice homework help geometry proofs homework. $\begingroup$ I agree that the link to geometric series is the only compelling reason to consider writing the finite geometric series formula in the "weird" way. On The Ratio Test for Positive Series of Real Numbers we looked at a very useful test for determining the convergence of a series of real numbers. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University February 1, 2019 Outline Geometric Series The Ratio Test The Root Test Examples A Taste of Power Series. Theorem (A Divergence test): If the series is convergent, then The test for divergence: If denotes the sequence of partial sums of then if does not exist or if , then the series is divergent. The value to which the series converges is the least of all possible upper bounds. If L= 1 then the test fails. Proof: The convergence properties of the power series are a consequence of the ratio test. Test results for water quality (specifically, fecal coliform bacteria concentrations) are sometimes reported as geometric means. The Mathematics Vision Project: Scott Hendrickson, Joleigh Honey, Barbara Kuehl, Travis Lemon, Janet Sutorius. (b) (i) Matrices over R, The matrix representation of systems of homogeneous and non- homogeneous linear equations. 4 Ratio test The geometric series leads to a useful test for convergence of the general series X1 n=0 a n= a 0 + a 1 + a 2 + (12) We can make sense of this series again as the limit of the partial sums S n = a 0 + a 1 + + a n as n!1. There are several questions Consider the geometric series: (Sum from k=0 to infinity) of ar^k and consider the repeating decimal. P), then b is the arithmetic mean of a and c. The common ratio in one geometric sequence is a whole number and in the other sequence it is a percent. Geometric Distribution De nition (Geometric Distribution) In a series of Bernoulli trials (independent trials with constant probability p of success), let the random variable Xdenote the number of trials until the rst success. The geometric series converges, and so, by the Direct Comparison Test, the. Geometric Series and the Test for Divergence - Part 1 patrickJMT. com where we believe that there is nothing wrong with being square! This page includes Geometry Worksheets on angles, coordinate geometry, triangles, quadrilaterals, transformations and three-dimensional geometry worksheets. Water Quality Standards. How is this done? Sn=a1(1-r^n)/(1-r). Note: If lim → =0, then there is no conclusion about. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University February 1, 2019 Outline Geometric Series The Ratio Test The Root Test Examples A Taste of Power Series. Sequences and series, whether they be arithmetic or geometric, have may applications to situations you may not think of as being related to sequences or series. 3 (Part 3) Review of Module 12 on Sequences and Series. 21) a 1 = −2, r = 5, S n = −62 22) a 1 = 3, r = −3, S n = −60 23) a 1 = −3, r = 4, S n = −4095 24) a 1 = −3, r = −2, S n = 63 25) −4 + 16 − 64 + 256 , S n = 52428 26) Σ m = 1 n −2 ⋅ 4m − 1 = −42-2-. Real and Integer Numbers Calculators and Percentages Sum of Positive Integers Calculator. Get the free "Infinite Series Analyzer" widget for your website, blog, Wordpress, Blogger, or iGoogle. The shape of the material used for the lampshade is a sector of a circle. Geometric Sequences - Module 12. What I want to do in this video is now think about the sum of an infinite geometric series. lcprefrigeration. If 1 then 1 for all so does not define a null sequence and the series diverges by the null sequence test. The circle has radius 25 centimetres and the angle of the sector is 280° a) Find the area of the sector of the circle. 4 Ratio test The geometric series leads to a useful test for convergence of the general series X1 n=0 a n= a 0 + a 1 + a 2 + (12) We can make sense of this series again as the limit of the partial sums S n = a 0 + a 1 + + a n as n!1. Proof - Convergence of a Geometric Series Contact Us If you are in need of technical support, have a question about advertising opportunities, or have a general question, please contact us by phone or submit a message through the form below. Def: The geometric series is convergent if \|r. These online tests are designed to work on computers, laptops, iPads, and other tablets. The geometric mean isn’t affected by those factors. Then the series is called Geometric Series. We will now look at a more general root test which can be applied to series of complex numbers. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University February 1, 2019 Outline Geometric Series The Ratio Test The Root Test Examples A Taste of Power Series. An arithmetic sequence is a sequence with the difference between two consecutive terms constant. In fact, our proof is an extension of the nice result given by Cohen and Knight [2]. (b) If , the series diverges. To solve problems on this page, you should be familiar with arithmetic progressions geometric progressions arithmetic-geometric progressions. The series , which was one of our examples given above, is a geometric series since =. - In a paragraph proof, statements and their justifications are written in sentences in a logical order. For example, the series. C2 Sequences & Series - Arithmetic & Geometric Series 3 MS C2 Sequences & Series - Arithmetic & Geometric Series 3 QP C2 Sequences & Series - Arithmetic & Geometric Series 4 MS. Since , this series diverges. The Limit Superior Root Test for Series of Complex Numbers. In a Geometric Sequence each term is found by multiplying the previous term by a constant. I offer this post for the enrichment of talented Precalculus students who have exhibited mastery of geometric series. 4 Ratio test The geometric series leads to a useful test for convergence of the general series X1 n=0 a n= a 0 + a 1 + a 2 + (12) We can make sense of this series again as the limit of the partial sums S n = a 0 + a 1 + + a n as n!1. You can see that this is reasonable by dividing 1 by , or using the the formula for the sum of a geometric series with ratio. Test for Divergence: If lim → ≠0, then diverges. The key steps are check cost, condition of pre-order and price recommendation. mathematicsvisionproject. ' and find homework help for other Math questions at eNotes If L=1 the test is inconclusive. org right now: https://www. The sum of a convergent geometric series can be calculated with the formula a ⁄ 1-r , where “a” is the first term in the series and “r” is the number getting raised to a power. Mathematical Induction - Problems With Solutions Several problems with detailed solutions on mathematical induction are presented. Geometric Series The formula for the sum of the first n terms of a geometric series is derived by using several ideas, each expressed concisely with subscripts and exponents. As this was clearly the case for the geometric series ∑ n = 0 ∞ x n (for 0 ≤ x 1), he asserted (perhaps as a reminder of Euclid's proof for the area of a circle) that a series ∑ n = 0 ∞ a n with positive terms will converge if, for n large enough, a n +1 is at most 1 2 a n; and without bothering about details, he stated correctly. Since the proof of the ratio test relies on this convergence, it is circular to argue that a geometric series converges by the ratio test (unless, of course, you have another proof for the ratio test that doesn't use the convergence of geometric series). In fact this series diverges quite slowly. A telescoping series is any series where nearly every term cancels with a preceeding or following term. ALTERNATING SERIES Does an = (−1)nbn or an = (−1)n−1bn, bn ≥ 0? NO Is bn+1 ≤ bn & lim n→∞ YES n = 0? P YES an Converges TELESCOPING SERIES Dosubsequent termscancel out previousterms in the sum? May have to use partial fractions, properties of logarithms, etc. Check out the activities offered on this site. Show that if X1 k=0 a k converges and b k is a bounded sequence, then X1 k=0 a kb k. Proof: If and are convergent, then it follows from the sum theorem for convergent sequences that is convergent and is valid. $\begingroup$ I agree that the link to geometric series is the only compelling reason to consider writing the finite geometric series formula in the "weird" way. We would like to show you a description here but the site won't allow us. Arithmetic and geometric series are two the most simple series for calculating the sum of numbers in series in a very simple way. Introduction to Video: Proof by Mathematical Induction. The comparison test allows us to construct other examples from this: THEOREM 2 (Comparison Test). A geometric series is a). To find the sum of an infinite geometric series having ratios with an absolute value less than one, use the formula, S = a 1 1 − r , where a 1 is the first term and r is the common ratio. 1 Sequences We call a list of numbers written down in succession a sequence; for example, the numbers drawn in a lottery: 12,22,5,6,16,43, For this sequence, there is no clear rule that will enable you to predict with certainty the next number in the sequence. We begin by giving the following estimate for the partial sum of a p-series: Lemma. For general help, questions, and suggestions, try our dedicated support forums. A geometric series is a). By inspection, it can be difficult to see whether a series will converge or not. Note: If lim → =0, then there is no conclusion about. the "Ratio Test for a Sequence". then completeness. Theorem 2: If the power series f(x) = P∞ n=0 anx n is convergent at x = R, then it is a continuous function within the interval of convergence. Suppose the interest rate is loo%, so i = 1. Sum of Arithmetic Geometric Sequence In mathematics, an arithmetico-geometric sequence is the result of the term-by-term multiplication of a geometric progression with the corresponding terms of an arithmetic progression. Writing a proof can even be more daunting I kept the reader(s) in mind when I wrote the proofs outlines below. I have to calculate the standard 90% confidence intervals of the ratio test/reference (T/R) and I have to answer to this question: the products were considered bioequivalent if the difference between two. That is, if we can prove that the sequence {a n} does not. This ratio is called the common ratio. The following rules are often helpful:. In mathematical analysis, the alternating series test is the method used to prove that an alternating series with terms that decrease in absolute value is a convergent series. Here it is. As a counterexam-ple, few series more clearly illustrate that the convergence of terms. The Ratio Test Proof (1): If 0 ˆ<1, we can apply the previous theorem to see P 1 n=1 ja nj converges. The sum of the first n terms of the geometric sequence, in expanded form, is as follows:. can inductive. Then there is a such that for we have. The geometric sequence can be rewritten as where is the amount of terms, is the common ratio, and is the first term. 1 The Geometric Series (page 373) EXAMPLE. How do we distinguish graphically between an arithmetic and a geometric sequence? 9. Infinite series. We would like to show you a description here but the site won't allow us. Introduction to Video: Proof by Mathematical Induction. Since lim n→∞ n √ a n = ρ < 1, then for any > 0, small enough such that ρ + = r < 1, there exists N large with. ALTERNATING SERIES Does an = (−1)nbn or an = (−1)n−1bn, bn ≥ 0? NO Is bn+1 ≤ bn & lim n→∞ YES n = 0? P YES an Converges TELESCOPING SERIES Dosubsequent termscancel out previousterms in the sum? May have to use partial fractions, properties of logarithms, etc. Say we have an infinite geometric series whose first term is aaaa and common ratio is rrrr. A SHORT(ER) PROOF OF THE DIVERGENCE OF THE HARMONIC SERIES LEO GOLDMAKHER It is a classical fact that the harmonic series 1+ 1 2 + 1 3 + 1 4 + diverges. 93Mb) Get help for solving geometric proofs with artificial intelligence methods. Any one of these nite partial sums exists but the in nite sum does not necessarily converge. Lecture 25/26 : Integral Test for p-series and The Comparison test In this section, we show how to use the integral test to decide whether a series of the form X1 n=a 1 np (where a 1) converges or diverges by comparing it to an improper integral. Hey guys, I'm stuck on wording of a homework assignment and thought you might be able to help me. A geometric sequence is a sequence with the ratio between two consecutive terms constant. The sequence 16 ,8 ,4 ,2 ,1 ,1/2 ,… = is a decreasing geometric sequence of common ratio ½. Infinite series can be daunting, as they are quite hard to visualize. can inductive. For such series, it necessary to evaluate limits of th roots of more complicated exp8 ressions. 832 respectively. Since every converging sequence is bounded, the s n are bounded. Grappling with the geometric series, geometry formulas or geometric sequence? Our tutors can help. Compute Geometric Distribution cdf. The term r is the common ratio, and a is the first term of the series. This usually takes the form of a formal proof, which is an orderly series of statements based upon axioms, theorems, and statements derived using rules of inference. Alternating Series test We have the following test for such alternating series: Alternating Series test If the alternating series X1 n=1 ( 1)n 1b n = b 1 b 2 + b 3 b 4 + ::: b n > 0 satis es (i) b n+1 b n for all n (ii) lim n!1 b n = 0 then the series converges. If it converges, then find its sum. For those geometric series, the signs of the terms alternate between positive and negative. In mathematical analysis, the alternating series test is the method used to prove that an alternating series with terms that decrease in absolute value is a convergent series. org/math/precalculus/seq_induction/infinite-geometric-series/e/understand. 9is formulated for convergent series, its main importance is as a \divergence test": if the general term in an in nite series does not tend to 0 then the series diverges. understand the concept of a geometric series • use and manipulate the appropriate formula • apply their knowledge of geometric series to everyday applications • deal with combinations of geometric sequences and series and derive information from them • find the sum to infinity of a geometric series, where -1 < r < 1. Interpret the structure of expressions. Convergence & Divergence - Geometric Series, Telescoping Series, Harmonic Series, Divergence Test - Duration: 50:43. Finally, we give a test which helps us to analyze convergence of an alternating series. We will now look at a more general root test which can be applied to series of complex numbers. Properties of Series; Arithmetic Series; Finite Geometric Series; Infinite Geometric Series; Decimal Expansion; Word Problems; Visualization of Series; The Divergence Test; The Alternating Series Test; The Ratio Test; The Integral Test; The Comparison Test; Absolute Convergence vs. The standard proof involves grouping larger and larger numbers of consecutive terms,. Where a 1 = the first term, a 2 = the second term, and so on a n = the last term (or the n th term) and a m = any term before the last term. This looks to me like an expression of the "ratio test" for convergence of a series. series is a geometric series, our results on geometric series can be used instead. In the following series, the numerators are in AP and the denominators are in GP:. Stamps Prairie State College The harmonic series, X∞ n=1 1 n = 1+ 1 2 + 1 3 + 1 4 + 1 5 +···, is one of the most celebrated inﬁnite series of mathematics. If the limit of a[n] is not zero, or does not exist, then the sum diverges. Finally, we give a test which helps us to analyze convergence of an alternating series. Determine whether the series converges or diverges. Infinite series can be daunting, as they are quite hard to visualize. Absolute Convergence If the series \|a n \| converges, then the series a n also converges. Our approach can be described as follows. However, notice that both parts of the series term are numbers raised to a power. 717171717171 for these problems: Question 1: Find a formula for the n-th partial sum of the series and PROVE your result using the Cauchy Convergence Criterion. Select one of the links below to get started. Does X1 1 2n n3 converge or diverge?. Infinite Sum Geometric Series. Let a n a n+1 0.	2020-04-07T23:41:06	{ "domain": "safoo.de", "url": "http://spmd.safoo.de/geometric-series-test-proof.html", "openwebmath_score": 0.832411527633667, "openwebmath_perplexity": 483.07126775236725, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9920620044904359, "lm_q2_score": 0.8558511506439708, "lm_q1q2_score": 0.8490574080533037 }
https://math.stackexchange.com/questions/2188962/determine-if-this-series-converges-or-diverges	# Determine if this series converges or diverges $$\sum_{n=1}^\infty \sin^{[n]}(1)$$ Where by $\sin^{[n]}(1)$ we mean $\sin\left(\sin\left(\dots\sin(1)\right)\right)$ composed $n$ times. Have tried the divergence test, which fails. Have tried Ratio test, also fails, as the limit is 1. Integral test, or root test do not seem promising. Help is appreciated • There are higher order ratio tests that may be useful (I have no clue) – Mark Mar 16 '17 at 5:52 • math.stackexchange.com/questions/14004/… – Michael Biro Mar 16 '17 at 5:54 • @Mark Ratio test will return one, hence, it is inconclusive. :P – Simply Beautiful Art Mar 16 '17 at 13:07 • Actually, the initial sine can be of anything real because it will always be in [-1,1]. This might be more interesting if we allow a complex sine. – richard1941 Mar 22 '17 at 22:00 • @Mark: I was unable to apply Raabe's Test successfully. – robjohn Feb 3 at 16:57 Once we prove the inequality $$\sin x > \frac{x}{1+x} \qquad (\star)$$ for $x \in (0,1]$, we can inductively show that $\sin^{[n]}(1) > \frac1n$ when $n\ge 2$. We have $\sin \sin 1 \approx 0.75 > \frac12$, and whenever $\sin^{[n]}(1) > \frac1n$, we have $$\sin^{[n+1]}(1) = \sin \sin^{[n]}(1) > \sin \frac1n > \frac{1/n}{1+1/n} = \frac1{n+1}.$$ Therefore, by the comparison test, $$\sum_{n=1}^\infty \sin^{[n]}(1) = \sin 1 + \sum_{n=2}^\infty \sin^{[n]}(1) > \sin 1 + \sum_{n=2}^\infty \frac1n,$$ which diverges. To prove $(\star)$... well, to be honest, I just graphed both sides. But we can prove that $\sin x > x - \frac{x^3}{6}$ on the relevant interval by thinking about the error term in the Taylor series, and $x - \frac{x^3}{6} > \frac{x}{1+x}$ can be rearranged to $(x+1)(x -\frac{x^3}{6}) - x > 0$, which factors as $-\frac16 x^2(x-2)(x+3) > 0$. • Let $f(x) = (x+1)\sin(x)-x$. We want to show $f(x) \ge 0$. Note $f(0) = 0$ and $f'(x) = \sin(x)+(x+1)\cos(x)-1$. It suffices to show $f'(x) \ge 0$. So it suffices to show $\sin(x)+\cos(x) \ge 1$ on $[0,1]$ (since $x\cos(x) \ge 0$). But by multiplying by $\frac{\sqrt{2}}{2}$, this is equivalent to $\sin(x+\frac{\pi}{4}) \ge \frac{\sqrt{2}}{2}$. And this is true for $x \in [0,\frac{\pi}{2}]$, so we're done since $\frac{\pi}{2} > 1$. – mathworker21 Mar 16 '17 at 6:07 Another possibility is to try to find an equivalent of the general term of this sequence : $\begin{cases} a_0=1\\ a_{n+1}=sin(a_n) \end{cases}$ Note that $f(x)=sin(x)$ has derivative $f'(x)=cos(x)$ which is positive on $[0,a_0]$ and also $<1$ on $]0,a_0[$ so $f$ is a contraction. From there it is easy to prove that $a_n\to 0$. This means that $a_{n+1}\sim a_n$ when $n\to\infty$. In this kind of problem we always search for an $\alpha$ such that $\mathbf{(a_{n+1})^\alpha-(a_n)^\alpha}$ does not depend of $\mathbf{a_n}$ (in the $\sim$ sense of the expression) so we are able to solve the recurrence. From Taylor expansion $\displaystyle{(a_{n+1})^\alpha = \bigg(a_n - \frac{a_n^3}{6}+o(a_n^4)\bigg)^\alpha}=(a_n)^\alpha\bigg(1 - \frac{a_n^2}{6}+o(a_n^3)\bigg)^\alpha=(a_n)^\alpha\big(1-\frac{\alpha}{6}a_n^2+o(a_n^3)\big)$ So $(a_{n+1})^\alpha-(a_n)^\alpha=-\frac{\alpha}{6}(a_n)^{\alpha+2}+o((a_n)^{\alpha+3})\quad$ we see that we need $\alpha=-2$ Let's put $b_n=1/(a_n)^2,\qquad$ $b_n\to\infty$ We have $b_{n+1}-b_n=\frac 13+o(1/b_n)$ thus $b_n\sim\frac n3\qquad$ (more precisely $b_n=n/3+o(\ln(n))$ but it is not important at this point). Finally $a_n\sim\sqrt\frac 3n,$ which is a term of a divergent serie so $\sum a_n$ diverges as well. • This is also very nice! Thank you ! – userX Mar 19 '17 at 2:57 I wrote this answer for another question, but this question was pointed out in a comment, so I posted it here. $$\|\,a_{n+1}\|=\|\sin(a_n)\,\|\le\|\,a_n\|$$. Thus, $$\|\,a_n\|$$ is decreasing and bounded below by $$0$$. Thus, $$\|\,a_n\|$$ converges to some $$a_\infty$$, and we must then have $$\sin(a_\infty)=a_\infty$$, which means that $$a_\infty=0$$. Using $$\sin(x)=x-\frac16x^3+O\!\left(x^5\right)$$, we get \begin{align} \frac1{a_{n+1}^2}-\frac1{a_n^2} &=\frac1{\sin^2(a_n)}-\frac1{a_n^2}\\ &=\frac{a_n^2-\sin^2(a_n)}{a_n^2\sin^2(a_n)}\\ &=\frac{\frac13a_n^4+O\!\left(a_n^6\right)}{a_n^4+O\!\left(a_n^6\right)}\\ &=\frac13+O\!\left(a_n^2\right)\tag1 \end{align} Stolz-Cesàro says that $$\lim_{n\to\infty}\frac{\frac1{a_n^2}}n=\frac13\tag2$$ That is, $$\bbox[5px,border:2px solid #C0A000]{a_n\sim\sqrt{\frac3n}}\tag3$$ which means that the series diverges. Motivation for $$\boldsymbol{(1)}$$ Note that $$a_{n+1}-a_n=\sin(a_n)-a_n\sim-\frac16a_n^3\tag4$$ which is a discrete version of $$\frac{\mathrm{d}a_n}{a_n^3}=-\frac{\mathrm{d}n}6\tag5$$ whose solution is $$\frac1{a_n^2}=\frac{n-n_0}3\tag6$$ so that $$\frac1{a_{n+1}^2}-\frac1{a_n^2}=\frac13\tag7$$ which suggests $$(1)$$.	2020-05-26T02:58:05	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2188962/determine-if-this-series-converges-or-diverges", "openwebmath_score": 0.9776839017868042, "openwebmath_perplexity": 202.90172981677253, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9920620042557432, "lm_q2_score": 0.8558511451289037, "lm_q1q2_score": 0.8490574023811531 }
https://math.stackexchange.com/questions/2545658/counting-with-recurrence-strings-which-avoid-the-substring-ab	# Counting with recurrence: Strings which avoid the substring $AB$. Let $X(n)$ be the number of length $n$ strings from the alphabet $\{A,B,C,D\}$ which avoid the substring $AB$. Write a recurrence for $X(n)$ and explain in words why it satisfies the recurrence. My attempt: Any length $n$ string sequence must start with either $A,B,C,D$. If it starts with $B,C,D$, we don't need to worry about the what follows right after, so the number of length $n$ strings that avoid $AB$ and start with $B,C,D$ is $3X(n-1)$. Now we deal with the case when it starts with $A$. If it starts with $A$, the next letter must be either $A,C,D$, only three choices, which means the rest of the sequence can be any $n-2$ length string which avoids $AB$. The number of length $n$ strings which start with $A$ and which avoid $AB$ is $3X(n-2)$, so we get \begin{cases} X(0)=1\\[4pt] X(1)=4\\[4pt] X(n)=3X(n-1)+3X(n-2),\;\text{if}\;n\ge 2\\ \end{cases} However the solution is: \begin{cases} X(0)=1\\[4pt] X(1)=4\\[4pt] X(n)=4X(n-1)-X(n-2),\;\text{if}\;n\ge 2\\ \end{cases} Note that for $n=2$, the results are the same, but they are not the same for $n=3$. Please let me know what I missed in my attempt. Thank you. As to where you went wrong, your statement $\;\;\;\;$The number of length $n$ strings which start with $A$ and which avoid $AB$ is $3X(n-2)$ is not correct, since for $n \ge 3$, it includes some strings which start with $AAB$. A correct recursion can be achieved as follows . . . By direct count, we get $x(0) = 1$ (the empty string), and $x(1) = 4$ (any string of length $1$). Suppose $n \ge 2$. Consider the count of all $n$-term sequences $s$ for which the $(n-1)$-term subsequence $s'$ with the first term of $s$ removed is such that $s'$ avoids the subsequence $AB$. Since we are allowing all possible first terms for $s$, the count is $4x(n-1)$. But this is an overcount since it allows sequences $s$ which start with $AB$. To correct the count, subtract the count of those already counted sequences which start with $AB$. Thus, we need to subtract the count of the sequences $s=ABs''$ such that the $(n-2)$-term sequence $s''$ avoids the subsequence $AB$, and there are $x(n-2)$ of those. Hence, for $n \ge 2$, we get $x(n) = 4x(n-1) - x(n-2)$. As an alternate approach, you can use a linked recursion. It's a little more work, but I think the logic is simpler. Here's how it would go . . . Let $x(n)$ be the number of qualifying $n$-term sequences with no immediately preceding $A$, and let $y(n)$ be the number of qualifying $n$-term sequences with an immediately preceding $A$. Then for $n > 0$, we get \begin{align} x(n) &= 3x(n-1) + y(n-1)\\[4pt] y(n) &= y(n-1)+ 2x(n-1) \end{align} and for $n=0$, we have $x(0) = 1,\;y(0) = 1$. To unlink the recursion, solve the equation $$x(n) = 3x(n-1) + y(n-1)$$ for $y(n-1)$, which yields $$y(n-1) = x(n) - 3x(n-1)$$ which implies $$y(n) = x(n+1) - 3x(n)$$ Then, replacing $y(n)$ and $y(n-1)$ in the equation $$y(n) = y(n-1)+ 2x(n-1)$$ yields $$x(n+1) = 4x(n) - x(n-1)$$ or equivalently, $$x(n) = 4x(n-1) - x(n-2)$$ valid for all $n > 1$, with initial values $x(0) = 1,\;x(1) = 4$. In his answer quasi has pointed out your error. As for the correct recurrence, you can argue as follows: An admissible word of length $n$ begins with one of $A$, $B$, $C$, $D$ and is followed by an admissible word of length $n-1$. We therefore would have $X(n)=4 X(n-1)$. Now this does include words which begin with $AB$, but are otherwise admissible. There are $X(n-2)$ of these. Discounting them we obtain the recursion $$X(n)=4X(n-1)-X(n-2)\ .$$	2019-11-19T12:34:20	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2545658/counting-with-recurrence-strings-which-avoid-the-substring-ab", "openwebmath_score": 0.9622975587844849, "openwebmath_perplexity": 177.71206981126605, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9920620075414419, "lm_q2_score": 0.8558511396138365, "lm_q1q2_score": 0.8490573997219335 }
http://mathhelpforum.com/calculus/170077-trig-limit-when-numerator-undefined.html	# Math Help - Trig Limit when the Numerator is Undefined 1. ## Trig Limit when the Numerator is Undefined Can somebody help me understand this one, please? $\displaystyle \lim_{x\to \infty } \frac{\cos^4(x)}{5 + x^2}$ I get that the answer is zero, because the denominator goes to infinity due to the $x^2$. I also understand that the value of $\cos^4(x)$ is always between 0 and 1, which is why we can conclude that the limit overall is zero. What is a little weird for me is that the limit of $\cos^4(x)$ doesn't exist. I can't get past the idea that the limit of a fraction can be defined when the limit of the numerator doesn't exist. What if the numerator was simply $\cos (x)$? Then what would the overall limit be (since $\cos (x)$ can vary between -1 and 1)? Sorry if I'm being a little dense, but can somebody try to explain this to me? Thanks. 2. Originally Posted by joatmon Can somebody help me understand this one, please? $\displaystyle \lim_{h\to \infty } \frac{\cos^4(x)}{5 + x^2}$ I get that the answer is zero, because the numerator goes to infinity due to the $x^2$. I also understand that the value of $\cos^4(x)$ is always between 0 and 1, which is why we can conclude that the limit overall is zero. What is a little weird for me is that the limit of $\cos^4(x)$ doesn't exist. I can't get past the idea that the limit of a fraction can be defined when the limit of the numerator doesn't exist. What if the numerator was simply $\cos (x)$? Then what would the overall limit be (since $\cos (x)$ can vary between -1 and 1)? Sorry if I'm being a little dense, but can somebody try to explain this to me? Thanks. The idea is hidden in what you have said. The squeeze theorem gives you what you need. You have said that you believe that $\displaystyle -1 \le \cos(x) \le 1$ for all x. Now if we divide this inequality by $x^2+1$ Since this is always non negative we don't have to change the inequality signs we get $\displaystyle \frac{-1}{x^2+1} \le \frac{\cos(x)}{x^2+1} \le \frac{1}{x^2+1}$ This inequality is still valid for all values of x Now if we take the limit with out knowing what the middle limit its we will know two numbers that it is still in between by the inequalities. $\displaystyle \lim_{x \to \infty}\frac{-1}{x^2+1} \le \lim_{x \to \infty}\frac{\cos(x)}{x^2+1} \le \lim_{x \to \infty}\frac{1}{x^2+1}$ This gives $\displaystyle 0 \le \lim_{x \to \infty}\frac{\cos(x)}{x^2+1} \le 0$ This tells us that the limit is bigger than $0$, but also less than $0$ so the limit must be zero. 3. I assume this should actually be $\displaystyle \lim_{x \to \infty}\frac{\cos^4{x}}{5 + x^2}$ I think you can squeeze it... $\displaystyle \frac{0}{5 + x^2} \leq \frac{\cos^4{x}}{5 + x^2} \leq \frac{1}{5 + x^2}$, so $\displaystyle \lim_{x \to \infty} 0 \leq \lim_{x \to \infty}\frac{\cos^4{x}}{5 + x^2} \leq \lim_{x \to \infty}\frac{1}{5 + x^2}$ $\displaystyle 0 \leq \lim_{x \to \infty}\frac{\cos^4{x}}{5 + x^2} \leq 0$. So $\displaystyle \lim_{x \to \infty} \frac{\cos^4{x}}{5 + x^2} = 0$. 4. Thanks. I had played with trying to squeeze this out, but couldn't get it to work. So, my logic about why the limit is zero isn't correct. I had thought that, since the function $cos^4(x)$ is always between 0 and 1, which we know even without using the squeeze theorem, what forces the limit to zero is that the denominator becomes infinitely large. But what I think that you are saying is that the denominator has nothing to do with it. It's that $\lim_{x\to \infty } \cos^4(x)$ can be shown to be zero by using the squeeze theorem. Thus, no matter what happens to the denominator (so long as its limit exists and is not zero), the value of the limit is still zero since the limit of the numerator is zero. Is a correct interpretation? 5. Just saw Prove It's reply. Now I get it entirely. Thanks to both of you. 6. Originally Posted by TheEmptySet The idea is hidden in what you have said. The squeeze theorem gives you what you need. You have said that you believe that $\displaystyle -1 \le \cos(x) \le 1$ for all x. Now if we divide this inequality by $x^2+1$ Since this is always non negative we don't have to change the inequality signs we get $\displaystyle \frac{-1}{x^2+1} \le \frac{\cos(x)}{x^2+1} \le \frac{1}{x^2+1}$ This inequality is still valid for all values of x Now if we take the limit with out knowing what the middle limit its we will know two numbers that it is still in between by the inequalities. $\displaystyle \lim_{x \to \infty}\frac{-1}{x^2+1} \le \lim_{x \to \infty}\frac{\cos(x)}{x^2+1} \le \lim_{x \to \infty}\frac{1}{x^2+1}$ This gives $\displaystyle 0 \le \lim_{x \to \infty}\frac{\cos(x)}{x^2+1} \le 0$ This tells us that the limit is bigger than $0$, but also less than $0$ so the limit must be zero. Having the fourth power of cosine, I'd use 0 as the lower bounds, since the numerator and denomiator are positive. I see that Profit said the same thing. 7. Originally Posted by joatmon Thanks. I had played with trying to squeeze this out, but couldn't get it to work. So, my logic about why the limit is zero isn't correct. I had thought that, since the function $cos^4(x)$ is always between 0 and 1, which we know even without using the squeeze theorem, what forces the limit to zero is that the denominator becomes infinitely large. But what I think that you are saying is that the denominator has nothing to do with it. It's that $\lim_{x\to \infty } \cos^4(x)$ can be shown to be zero by using the squeeze theorem. No, it doesn't. The constant function f(x)= 1/2 is always between 0 and 1- does that "force" its limit to be 0? $\lim_{x\to\infty} cos^4(x)$ does not exist. Thus, no matter what happens to the denominator (so long as its limit exists and is not zero), the value of the limit is still zero since the limit of the numerator is zero. Is a correct interpretation? No, it isn't. It is the fact that the denomator goes to infinity that makes the limit 0. The only thing required of the numerator is that it NOT go to infinity or negative infinity.	2016-05-25T19:58:32	{ "domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/170077-trig-limit-when-numerator-undefined.html", "openwebmath_score": 0.9781644940376282, "openwebmath_perplexity": 138.18012348820477, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9553191271831558, "lm_q2_score": 0.888758801595206, "lm_q1q2_score": 0.8490482826162797 }
https://web2.0calc.com/questions/help_2802	+0 # help 0 59 13 If $a, b, c$ are positive integers less than 13 such that $2ab+bc+ca \equiv 0\pmod{13}$ $ab+2bc+ca \equiv 6abc\pmod{13}$ $ab+bc+2ca \equiv 8abc\pmod {13}$ then determine the remainder when $a + b + c$ is divided by $13$. Sep 27, 2020 #1 0 a + b + c leaves a remainder of 8 when divided by 13. Sep 27, 2020 #2 0 Sorry, but it says that is incorrect. Thank you for trying to help out though! Also, if someone wants to post a solution to this, it'd be best if they provided an explanation, because I'd like to learn from it. Thanks! Guest Sep 27, 2020 #3 +1 Look at many, many solutions here: https://www.wolframalpha.com/input/?i=%282ab%2Bbc%2Bca%29+mod+13+%3D0%2C++%28ab%2B2bc%2Bca%29+mod+13%3D+6abc%2C++%28ab%2Bbc%2B2ca%29++mod+13+%3D+8abc%2C+solve+for+a%2Cb%2Cc Sep 27, 2020 #4 0 Unfortunately, this also does not work as your input had equal signs instead of modular congruence signs. I don't believe you can input modular congruence signs into Wolfram Alpha. Thanks for trying to help me though! Guest Sep 28, 2020 #5 +111124 0 If a, b, and c are all 0 then these three equations are all true. And if this is true then (a+b+c) mod13 = 0 So if there is only one solution then that solution must be 0 However, just two of them are 0 then the third one can equal to many different numbers. So I do not believe this has one specific answer. Sep 28, 2020 #6 +1 I know I must be annoying for saying all of the answers are wrong, but unfortunately there's a problem with this one too. The problem states that $a, b, c$ must be positive integers less than 13, so $0$ doesn't fit this requirement. However, thank you very much for trying to solve this for me! Guest Sep 28, 2020 #7 +111124 0 Yep, good point! Melody Sep 29, 2020 #8 0 Sep 29, 2020 #9 +1 a = 3, b = 6, c = 9. Unique. Will post a solution, of sorts, if anyone is interested. Guest Sep 30, 2020 #10 +111124 0 Thanks guest, I have some interest. Did you use code to do the grunt work and just test all the possibilities Or did you actually come by the answer in a logical, mathematical (no trial and error) way ? Melody Sep 30, 2020 #11 +1 Thanks Melody, it's nice to know that someone is actually interested. Often that doesn't appear to be the case. The equations can be written in the form $$\displaystyle 2X+Y+Z\equiv 0 \dots\dots(1)\\ X+2Y+Z\equiv R\dots\dots(2)\\ X+Y+2Z\equiv S\dots\dots(3)$$ where each one is mod 13. $$\displaystyle (1) - 2(3) : -Y-3Z \equiv -2S\dots\dots(4)\\ (2) - (3) :Y - Z \equiv R-S\dots\dots(5)\\ (4)+(5) : -4Z\equiv R-3S\dots\dots(6).$$ Returning to the original unknowns, $$\displaystyle -4ca \equiv 6abc-24abc \equiv -18abc,$$ from which $$\displaystyle9b=2+13k\quad \text{ where k is an integer}.$$ The smallest value of k for which b will be an integer is k = 4, leading to 9b = 54, b = 6. The next suitable value for k puts b out of the permitted range. The remaining unknowns (a = 3 and c = 9) can then be found by back substitution, (6) into (5) and then into (2) or (3). Leave that to you. Guest Sep 30, 2020 #12 0 THANK YOU THANK YOU THANK YOU! I really like your explanation! Thanks a lot! Guest Sep 30, 2020 #13 +111124 0 Brilliant, thanks very much! I never would have thought to do those substitutions! I continued with matrices, and again, I would not have thought to use matrices on a modulo question if you had not inspired me. Melody Sep 30, 2020	2020-10-22T12:57:51	{ "domain": "0calc.com", "url": "https://web2.0calc.com/questions/help_2802", "openwebmath_score": 0.7350197434425354, "openwebmath_perplexity": 791.4212739266446, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9898303413461358, "lm_q2_score": 0.8577680977182187, "lm_q1q2_score": 0.84904488896025 }
https://math.stackexchange.com/questions/2612170/computing-convolution-using-the-fourier-transform	# Computing convolution using the Fourier transform I am given that the functions $x(t)$ and $h(t)$ are defined by $$x(t) = t\, e^{-2t}\, u(t) \qquad \text{and} \qquad h(t) = e^{-4t}\, u(t)$$ where $u(t)$ denotes the unit step function $$u(t) = \begin{cases} 1 & \text{if t \geq 0}\\ 0 & \text{otherwise}. \end{cases}$$ The question is the following: Compute the convolution $(x * h)(t)$ by finding the corresponding Fourier transform $X(\omega)$ and $H(\omega)$ using the convolution property, and then inverse transforming. By the convolution property, we mean $(xh)(t) \stackrel{\text{FT}}{\longleftrightarrow} X(\omega) H(\omega)$. Now I found the Fourier transforms $X(\omega)$ and $H(\omega)$ using the following reasoning. \begin{align} x(t) &= t\, e^{-2t} \,u(t)\\ & = t\; y(t) \qquad \text{where $y(t) := e^{-2t} \,u(t)$}\\ \implies X(\omega)&= i\, \frac{\partial Y}{\partial \omega}\qquad \text{where $Y$ is the F.T. of $y(t)$}\\ &= \frac{1}{(2+ i \omega)^2}. \end{align} $H(\omega)$ is simply $1/(4+i\omega)$. Now \begin{align} (xh)(t) \stackrel{\text{FT}}{\longleftrightarrow} X(\omega) H(\omega) = \frac{1}{(2+i\omega)^2(4+i\omega)}. \end{align} If at this point I were to employ the inverse Fourier transform, I would obtain $(xh)(t) = \frac{1}{2\pi} \int_{-\infty}^\infty \frac{e^{i \omega t}}{(2+i\omega)^2(4+i\omega)} \, \mathrm d\omega$. But this seems quite a complicated integral (considering this was a 4 mark question in an exam). Is there a simpler way to go about obtaining the inverse Fourier transform? Am I going about the question correctly? I appreciate any help. • My impulse would be to evaluate that last integral by turning it into a complex contour integral. Since the integral only has two poles above the real line, this shouldn’t be hard. – Semiclassical Jan 19 '18 at 14:14 • Also, how are you defining the Fourier transform? In the version I know, the Fourier transform of $(fg)(t)$ would be $F(\omega)G(\omega)$ not $F(i\omega)G(i\omega)$. – Semiclassical Jan 19 '18 at 16:14 • @Semiclassical sorry about that, it's a notation purely to show that the argument may be complex also, equivalently it's $X(\omega)$. – Luke Collins Jan 19 '18 at 18:17 I think you're expected to apply the partial fraction decomposition: $$\frac 1 {(4+i \omega)(2+i \omega)^2} = \frac 1 {4(4+i \omega)} - \frac 1 {4(2+i \omega)} + \frac 1 {2(2+i \omega)^2},$$ and you already have the functions whose transforms produce those terms.	2019-07-18T13:36:56	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2612170/computing-convolution-using-the-fourier-transform", "openwebmath_score": 0.9969123005867004, "openwebmath_perplexity": 256.13405560113915, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964237163882, "lm_q2_score": 0.8615382147637196, "lm_q1q2_score": 0.8490428295446473 }
https://mathematica.stackexchange.com/questions/252994/does-this-series-converge-or-diverge	# Does this series converge or diverge? Let us consider the series $$\sum_{n=1}^\infty \frac {\{\sqrt n\}-\frac 1 2} n$$, where $$\{\cdot\}$$ denotes the fractional part of a real number. On the one hand, SumConvergence[(FractionalPart[Sqrt[n]] - 1/2)/n, n] False On the other hand, NSum[(FractionalPart[Sqrt[n]] - 1/2)/n, {n, 1, Infinity}] -0.725204 Let us investigate it. The output of N[Table[Sum[(FractionalPart[Sqrt[n]] - 1/2)/n, {n, 1, 10^k}], {k, 1, 5}], 15] {-0.679200513405746, -0.755781168886997, -0.803827960877609, -0.810978369858450, -0.816045768107535} suggests the convergence, but the sum of the series does not equal -0.725204. Let us look at the plots DiscretePlot[(FractionalPart[Sqrt[n]] - 1/2), {n, 1, 121}] and DiscretePlot[(FractionalPart[Sqrt[n]] - 1/2), {n, 121, 144}] The plots show that approximately from k^2 to (k^2+(k+1)^2)/2 (k is a positive integer) the terms are negative and from (k^2+(k+1)^2)/2 to (k+1)^2 the terms are nonnegative. Let us estimate the sums over these intervals by AsymptoticSum[1/n, {n, (k^2 + (k + 1)^2)/2, (k + 1)^2}, k -> Infinity] 1/k and AsymptoticSum[1/n, {n, k^2, (k^2 + (k + 1)^2)/2}, k -> Infinity] 1/k This also suggests the convergence. However, the above is a plausible reasoning, not a proof. The questions are: how to accurately prove or disprove the convergence with Mathematica? how to numerically calculate the sum of the series under consideration? • A harder question is about the series $$\sum_{n=1}^\infty \frac {\{\sqrt n\}-\frac 1 2} ne^{2\pi i t \log n} ,$$ where $t\in\mathbb R$ . Aug 6 '21 at 18:28 • BTW, SumConvergence[FractionalPart[Sqrt[n]]/n, n] results in True instead of False. Aug 7 '21 at 12:21 Here is my answer to my question done with Mathematica. If n>=k^2 && n<(k+1)^2, where k is a positive integer, then Floor[Sqrt[n]]==k and FractionalPart[Sqrt[n]]==Sqrt[n]-k. Now we consider Sum[(Sqrt[n] -k-1/2)/n, {n,k^2,(k + 1)^2 - 1},Assumptions -> k\[Element] PositiveIntegers] 1/2 (2 HurwitzZeta[1/2, k^2] - 2 HurwitzZeta[1/2, (1 + k)^2] + PolyGamma[0, k^2] + 2 k PolyGamma[0, k^2] - PolyGamma[0, 1 + 2 k + k^2] - 2 k PolyGamma[0, 1 + 2 k + k^2]) Series[%, {k, Infinity, 2}] -(2/(3 k^2))+O[1/k]^3 This implies the series under consideration converges. In order to obtain the numerical value of the sum, we find NSum[1/2 (2 HurwitzZeta[1/2, k^2] - 2 HurwitzZeta[1/2, (1 + k)^2] + PolyGamma[0, k^2] + 2 k PolyGamma[0, k^2] - PolyGamma[0, 1 + 2 k + k^2] - 2 k PolyGamma[0, 1 + 2 k + k^2]), {k, 1, Infinity}] -0.817595 • I'd like to elaborate the proof. Because of Series[Sum[(Sqrt[n] - k - 1/2)/n, {n, k^2, (k + 1/2)^2}], k -> Infinity] which results in -(1/(4 k))+O[1/k]^2 and Series[Sum[(Sqrt[n] - k - 1/2)/n, {n, (k + 1/2)^2, (k + 1)^2 - 1}], k -> Infinity] which results in 1/(4 k)+O[1/k]^2, the partial sum $S_n$ for $n\in [k^2,(k+1)^2]$ satisfies the estimates $$S_n \ge S_{k^2}- 1/(4k)+O(1/k^2),\,S_n \le S_{k^2}+1/(4k)+O(1/k^2).$$ The squeeze theorem ends the proof. Aug 8 '21 at 4:32	2022-01-22T08:13:50	{ "domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/252994/does-this-series-converge-or-diverge", "openwebmath_score": 0.7543326616287231, "openwebmath_perplexity": 2836.2706079500504, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964181787613, "lm_q2_score": 0.8615382165412809, "lm_q1q2_score": 0.8490428265255504 }
http://mathhelpforum.com/math-puzzles/133126-grains-rice-chess-board.html	# Math Help - Grains of rice on a chess board 1. ## Grains of rice on a chess board Hi, firstly apologies if this is not the correct section. At work today I was told that if you took a chess board (with 64 squares) and you were to place 1 grain of rice on the 1st square, 2 on the second square, 4 on the 3rd square, 8 on the 4th and so on doubling the grains of rice on each successive square compared to the previous square - that there would not be enough grains of rice in the world to do this. Whilst im skeptical that there would not be enough rice in the world to do this... how many grains of rice would be on the 64th square?, and what would the formula be for working this out? I guess I could work this out just by doubling 1 then 2 then 3 etc, but Id rather learn something new, so if there is a formula could someone explain it aswell. As you can probably guess Im not the sharpest when it comes to maths, so thanks for your time and help. Chez 2. Welcome to MHF. If you call "n" the nth square, there are $2^{n-1}$ grains in the nth's square. So in the last square there is $2^{63}$ grains. To see how big or how small this number is, go there: http://www.wolframalpha.com/input/?i=2^63. I don't really know the number of rice's grains in the world, though it shouldn't be so hard calculating it. If you can give us the rice world production in tons by year, we will have the order of magnitude of the number of rice grains in the world, assuming that 1 kilo contains a certain number of grains. Notice also that on the 63 th square there is half of the grains in the 64th square and that you also have to sum it up. This number (of rice on the last square), as huge as it may look like, doesn't scar me. Indeed, there is 100 000 times more atoms of lead in a mole of lead, namely a small fragment of lead that you could have in one hand. 3. Originally Posted by arbolis Welcome to MHF. If you call "n" the nth square, there are $2^{n-1}$ grains in the nth's square. So in the last square there is $2^{63}$ grains. To see how big or how small this number is, go there: http://www.wolframalpha.com/input/?i=2^63. I don't really know the number of rice's grains in the world, though it shouldn't be so hard calculating it. If you can give us the rice world production in tons by year, we will have the order of magnitude of the number of rice grains in the world, assuming that 1 kilo contains a certain number of grains. Notice also that on the 63 th square there is half of the grains in the 64th square and that you also have to sum it up. This number (of rice on the last square), as huge as it may look like, doesn't scar me. Indeed, there is 100 000 times more atoms of lead in a mole of lead, namely a small fragment of lead that you could have in one hand. Thanks that answers my question nicely. World rice production is around 600 million tons per year (presumably this is a metric ton). There are approx 36 950 grains of rice in a kilogram so 36 950 000 grains in a ton, so grains of rice produced in the world in a year is 36 950 000 * 600 = 22170000000 so something short of whats needed. Unless my maths is wrong (likely) so please correct if thats the case. Thanks again 4. Originally Posted by arbolis Welcome to MHF. If you call "n" the nth square, there are $2^{n-1}$ grains in the nth's square. So in the last square there is $2^{63}$ grains. To see how big or how small this number is, go there: http://www.wolframalpha.com/input/?i=2^63. I don't really know the number of rice's grains in the world, though it shouldn't be so hard calculating it. If you can give us the rice world production in tons by year, we will have the order of magnitude of the number of rice grains in the world, assuming that 1 kilo contains a certain number of grains. Notice also that on the 63 th square there is half of the grains in the 64th square and that you also have to sum it up. This number (of rice on the last square), as huge as it may look like, doesn't scar me. Indeed, there is 100 000 times more atoms of lead in a mole of lead, namely a small fragment of lead that you could have in one hand. Yes, but the number of grains of rice in total would be $\sum_{n=0}^{64}2^n=2^65-1$ 5. Originally Posted by Drexel28 Yes, but the number of grains of rice in total would be $\sum_{n=0}^{64}2^n=2^65-1$ Yes, that's why I wrote "you have to sum it up". There's a slight error, if you start at n=0, you must end at n=63 since there are 64 squares but you are right we must sum all the rices on every square. 6. Originally Posted by Drexel28 Yes, but the number of grains of rice in total would be $\sum_{n=0}^{64}2^n=2^65-1$ The volume ocupied by the grains on the last square alone would fill a sphere of radius the order of $3.6$ km (or more if a packing fraction less than 1 is used) CB 7. Originally Posted by chezza Thanks that answers my question nicely. World rice production is around 600 million tons per year (presumably this is a metric ton). There are approx 36 950 grains of rice in a kilogram so 36 950 000 grains in a ton, so grains of rice produced in the world in a year is 36 950 000 * 600 = 22170000000 so something short of whats needed. Unless my maths is wrong (likely) so please correct if thats the case. Thanks again I think you made an error in the expression "36 950 000 * 600 = 22170000000". You're multiplying the number of rice grains contained in one ton by 600 instead of 600 millions. Thus the final result should be 10^6 times greater than your result. So yes, it seems that with one year world production, you would have approximately 1/1000th of the grains needed to fill the last square, if I didn't make any arithmetic mental error. 8. Thanks for the replies, Its been interesting reading though I have no idea what the formula given by drexel means 9. Originally Posted by chezza Thanks for the replies, Its been interesting reading though I have no idea what the formula given by drexel means $\sum_{n=0}^{63}2^n=2^0+2^1+2^2+2^3+...+2^{62}+2^{6 3}=1+2+4+8+...+9223372036854775808$ which is the total number of rice grains. Each term of the sum represent the number of rice grain(s) in each square. Summing them up reach the number of rice grains in the whole board.	2015-02-27T23:39:07	{ "domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/math-puzzles/133126-grains-rice-chess-board.html", "openwebmath_score": 0.6937201619148254, "openwebmath_perplexity": 679.7921583683935, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964173268185, "lm_q2_score": 0.8615382147637196, "lm_q1q2_score": 0.8490428240397888 }
http://math.stackexchange.com/questions/35136/does-fx-in-mathbbzx-have-the-same-roots-as-fx-in-mathbbf-px	# Does $f(x) \in \mathbb{Z}[x] \$ have the same roots as $f(x) \in \mathbb{F}_p[x] \quad$? Does $f(x)\in\mathbb{Z}[x]\$ have the same roots of $f(x)\in\mathbb{F}_p[x] \quad$? Do the roots of a polynomial change when the coefficients of the polynomial are considered as elements of one ring or another? Would the answer to this change if the rings contained each other, as in the case of $f(x) \in \mathbb{Z}[x] \ \$ vs. $\ f(x) \in \mathbb{Q}[x] \quad$? - Short answer: Yes, the set of roots can change. $f(x)=x^2+1$ has no roots in $\mathbb{Z}$, but (the image of $f(x)$ under reduction modulo $p$) has roots in $\mathbb{F}_p$ if $p=2$ or $p\equiv 1 \pmod{4}$. Longer answer: First, you have to be careful. If $f(x)$ is a polynomial with integer coefficients, then it is not literally a polynomial with coefficients in $\mathbb{F}_p$. Rather, the natural map "reduction modulo $p$", which is a ring map from $\mathbb{Z}$ to $\mathbb{F}_p$, induces a map from $\mathbb{Z}[x]$ to $\mathbb{F}_p[x]$, and you can consider the image of $f(x)$ in $\mathbb{F}_p[x]$. Under this map, if $a\in\mathbb{Z}$ is a root of $f(x)$, then $a\bmod p$ is a root of $f(x)\bmod p$ in $\mathbb{F}_p[x]$. But the converse certainly does not hold; for one thing, there are infinitely many integers that are preimages of $a\bmod p$. And for another, no: you can have $f(x)\bmod p$ have roots in $\mathbb{F}_p[x]$, but no roots in $\mathbb{Z}[x]$, or only some roots. For example, for odd primes $p$, $f(x) = x^{p}-x$ has only two roots in $\mathbb{Z}$ ($x=0$ and $x=1$), but (its image under reduction modulo $p$) has $p$ roots in $\mathbb{F}_p$. So roots in $\mathbb{Z}$ yield roots in $\mathbb{F}_p$ (via reduction modulo $p$), but not conversely. Another example: $x^2+1$ has no roots in $\mathbb{Z}$, but (its reduction modulo $p$) has roots in $\mathbb{F}_p$ for every $p$ that is not congruent to $3$ modulo $4$. More generally, the universal property of the polynomial ring gives easily: Theorem. Let $R$ and $S$ be commutative rings. If $h\colon R\to S$ is a ring homomorphism, then $h$ induces a ring homomorphism $\overline{h}\colon R[x]\to S[x]$ of polynomial rings by $$\overline{h}(r_0+r_1x + \cdots + r_nx^n) = h(r_0) + h(r_1)x + \cdots + h(r_n)x^n.$$ Moreover, the map $\overline{h}$ commutes with evaluation maps in the following sense: if $\mathrm{eval}_a\colon R[x]\to R$ is the map $\mathrm{eval}_a(f(x)) = f(a)$, and $\mathrm{eval}_{h(a)}\colon S[x]\to S$ is the evaluation map at $h(a)$, then we have a commutative diagram $$\begin{array}{rcl} R[x] & \stackrel{\overline{h}}{\longmapsto} & S[x]\\ &&\\ \mathrm{eval}_a\downarrow & & \downarrow \mathrm{eval}_{h(a)}\\ R &\stackrel{h}{\longmapsto} & S\end{array}$$ for all $a\in R$. In particular, if given $f\in R[x]$ and $g\in S[x]$ we let $$\mathrm{Roots}_R(f) = \{r\in R\mid f(r)=0\}$$ and $$\mathrm{Roots}_S(g) = \{s \in S\mid g(s)=0\},$$ then $$h\left(\mathrm{Roots}_R(f)\right) \subseteq \mathrm{Roots}_S(\overline{h}(f)),$$ but the inclusion may be proper. For examples with proper inclusion in the setting of reduction modulo $p$, consider $f(x)=x^2+1\in\mathbb{Z}[x]$ which has not roots, but whose reduction modulo $p$, with $p=2$ or $p\equiv 1\pmod{4}$ has roots. Or $f(x)=x^p-x$, $p$ an odd prime, which has two roots in $\mathbb{Z}$ but its reduction modulo $p$ has $p$ roots in $\mathbb{F}_p$ (by Fermat's Little Theorem). For examples of proper inclusion with fields and $h$ an inclusion map, take $x^2+1\in\mathbb{R}[x]$, and consider its image in $\mathbb{C}[x]$; $x^2+1$ has no real roots, but its image in $\mathbb{C}[x]$ has roots. Added. When dealing with inclusion, it usually makes more sense to consider the set of roots in the larger field, and just ask which ones lie in the smaller field. So we think of $x^2+1$ as a polynomial with complex coefficients, and then ask what $\mathrm{Roots}_{\mathbb{C}}(x^2+1)\cap\mathbb{R}$ is (this will equal $\mathrm{Roots}_{\mathbb{R}}(x^2+1)$). Also, if $h(a)$, the image of $a$ is a root of $\overline{h}(f)$, this does not imply, in and of itself, that $a$ is a root of $f(x)$ (though it does if $h$ is one-to-one). To summarize the longer answer: if $h\colon R\to S$ is a ring homomorphism, then the image of any root of $f(x)$ is a root of the image of $f(x)$, but there may be roots of the image that are not images of roots; and not everything in $R$ whose image is a root of the image of $f(x)$ is a root of $f(x)$ in general. But if $h$ is one-to-one, then $h(a)$ is a root of $\overline{h}(f(x))$ if and only if $a$ is a root of $f(x)$. - Holy crap, you wrote a book for me. Thanks, I appreciate it :) – badatmath Apr 26 '11 at 4:29 Hello, and thanks :) I guess I am confused because when you talk about polynomials like $x^2+1$ with coefficients in $\mathbb{Z}[x]$, you say this polynomial has roots, but its roots are in $\mathbb{C}$. If you say $x^2+1$ has coefficients in $\mathbb{F}_p[x]$, can you also say this polynomial has roots in $\mathbb{C}$? Does it make a difference that $\mathbb{C}$ does not contain $\mathbb{F}_p$? – badatmath Apr 26 '11 at 4:32 @badatmath: I did not say "the roots are in $\mathbb{C}$." The inclusion $\mathbb{Z}\hookrightarrow\mathbb{C}$ means you can consider polynomials with integer coefficients as polynomials with complex coefficients. If you do that, then the image of $x^2+1$ has roots when you consider it as an element of $\mathbb{C}[x]$, but no roots when you consider it as an element of $\mathbb{Z}[x]$. No, you cannot say that a polynomial in $\mathbb{F}_p[x]$ has roots in $\mathbb{C}$; that makes no sense precisely because $\mathbb{C}$ does not contain $\mathbb{F}_p$. – Arturo Magidin Apr 26 '11 at 4:35 @badatmath: But the reason $x^2+1$ has roots in $\mathbb{F}_p$ when $p\equiv 1 \pmod{4}$ is that in those situations, $-1$ is a square modulo $p$. For example, in $\mathbb{F}_5$, but $2\bmod 5$ and $3\bmod 5$ are roots of (the image of) $x^2+1$. – Arturo Magidin Apr 26 '11 at 4:36 Oh, okay. So you can't say that because the inclusion map does not exist. Thanks. – badatmath Apr 26 '11 at 5:18 Last question first, $2x-1$ certainly doesn't have the same roots in $\bf Z$ as in $\bf Q$. I'd say $x-2$ has the same root in $\bf Z$ as in $\bf Q$, namely, $2$, but I can imagine some situation in which one might want to distinguish between $2$ as an integer and $2$ as a rational. Does $x-7$ have the same root in $\bf Z$ as in ${\bf F}_5$? I think it would be an abuse of notation to say yes. But abuses of notation can be very useful, so I wouldn't rule it out; I'd just make absolutely sure to put all my cards on the table first, that is, to say something like we identify elements of $\bf Z$ with their images in ${\bf F}_5$'' first. - No. In fact, this failure can be rather extreme, as for polynomial rings over finite fields there are nonzero polynomials with integer coefficients that vanish at every point in the field. Easy examples are provided by encoding boolean contradictions as polynomials in $\mathbb{F}_2[X]$, such as $(x \text{ and } x)\text{ xor } x$ which is represented as $x^2 + x$ or $((x \text{ and } x)\text{ and } x) \text{ xor } x$ which is represented as $x^3 + x$. You can also get such polynomials in $\mathbb{F}_p[X]$ by taking the product $(x - (p - 1))(x - (p - 2))\cdots (x - 1)x$. - Echoing Arturo, since there's a surjection $\phi: \mathbb{Z}[x] \to \mathbb{F}_p[x]$, if $a$ is a root of $f$ in $\mathbb{Z}[x]$, then $\phi(a)$ is a root of $f$ in $\mathbb{F}_p[x]$. So a root in $\mathbb{Z}[x]$ translates into a root in $\mathbb{F}_p$, but they can overlap. For example, $x^2 - 1 = (x+1)(x-1)$ has two distinct roots $1,-1$ in $\mathbb{Z}[x]$ but one repeated root $1$ in $\mathbb{F}_2[x]$. In the same line as Alex's comment, you have to be a little careful. If $p$ divides every coefficient of $f$, then $f \equiv 0$ in $\mathbb{F}_p[x]$. In addition, some irreducible elements in $\mathbb{Z}[x]$ do factor in $\mathbb{F}_p[x]$. See Arturo's answer. In general, it's easier to factor in $\mathbb{F}_p[x]$ because there are fewer elements to check divisibility by. And that can help determine factorization in $\mathbb{Z}[x]$. A theorem states that if the leading coefficient of $f$ is not divisible by $p$, and if $f$ is irreducible in $\mathbb{F}_p$, then $f$ is irreducible in $\mathbb{Z}$ (Artin 2nd Edition 12.4.3). The converse does not hold for the same reasons as discussed in the previous paragraph. Factorization in $\mathbb{Z}[x]$ and $\mathbb{Q}[x]$ is equivalent up to constant multiples, but linear factors are always roots in $\mathbb{Q}[x]$, while they aren't necessary so in $\mathbb{Z}[x]$, as Gerry stated. Roots in $\mathbb{Z}[x]$ will be roots in $\mathbb{Q}[x]$, but not necessarily the other way around.	2015-10-05T04:41:39	{ "domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/35136/does-fx-in-mathbbzx-have-the-same-roots-as-fx-in-mathbbf-px", "openwebmath_score": 0.9455252885818481, "openwebmath_perplexity": 76.4085526447313, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964215865321, "lm_q2_score": 0.8615382076534742, "lm_q1q2_score": 0.8490428207025734 }
https://math.stackexchange.com/questions/3112037/what-is-the-largest-integer-value-of-n-for-which-8n-evenly-divides-100	# What is the largest integer value of $n$ for which $8^n$ evenly divides $(100!)$? I know that this may be an unnecessary question, but I am a bit confused. The problem asks for the highest integer $$n$$ such that $$8$$ to the power of $$n$$ is divisible, evenly of course, by $$100$$. Now, I searched the site, and, in general, it seems that one can use floor function for a problem like this, but this seems to only work for prime numbers possibly. My process, which I realized was incorrect: The floor function of $$100/8 = 12$$, and then doing it for the second power would lead to one, and, by adding those up, I acquired thirteen. Of course, after seeing the answer, $$32$$, I went back to see what was wrong and did the problem slower. I got $$12$$ numbers from the numbers in $$100!$$, and then got another $$8$$ from $$2 \times 4$$, but, that can be applied for all the multiples of $$2$$ and $$4$$ that aren't of $$8$$. So, essentially, I am wondering if there is a quicker method for calculating this number without specifically counting out the numbers. Thanks in advance! • are you talking about (100!) or (100) ? – Seth Feb 13 '19 at 22:59 • I intended for it to be meant as (100!) with a question mark at the end. Feb 13 '19 at 23:00 • Edited title for easier reading. Thanks for the concern. Feb 13 '19 at 23:01 • Is it $8^n$ is divisible by $100$ or $8^n$ divides $100!\,$? Feb 13 '19 at 23:07 • It is the second one ( 8^n divides (100!) ) Feb 13 '19 at 23:09 I think the easiest way to answer this question is to factorize $$100!$$. Actually, a partial factorization will be sufficient. Thus, we see that $$100! = 2^{97} \times 3^{48} \times 5^{24} \times \ldots \times 83 \times 89 \times 97.$$ Since $$8 = 2^3$$, we need to divide $$97$$ by $$3$$ and discard the remainder. That is, rewrite $$2^{96}$$ as $$8^n$$ and there's your answer. • Thanks for the answer! I was hoping that there was a quicker method (maybe an algorithmic one). Feb 13 '19 at 23:14 • Thanks for the suggestion (I'm new to the site, so, I don't know how to make this work). Feb 13 '19 at 23:15 • This suggests to me the following algorithm: repeatedly divide the factorial by 2, increment a counter as you do so, until you get an odd number, specifically a number from oeis.org/A049606 Feb 14 '19 at 2:44 It's easiest, I think, to do this with powers of $$2:$$ $$\left\lfloor{100\over2}\right\rfloor+ \left\lfloor{100\over4}\right\rfloor+ \left\lfloor{100\over8}\right\rfloor+ \left\lfloor{100\over16}\right\rfloor+ \left\lfloor{100\over32}\right\rfloor+ \left\lfloor{100\over64}\right\rfloor=97=32\cdot3+1$$ so the greatest exponent of $$8$$ is $$32$$. • Thanks for the answer! Feb 13 '19 at 23:13 • This is known as Legendre's formula. Feb 13 '19 at 23:35 • @Bernard Reading mathworld.wolfram.com/LegendresFormula.html I'm not quite sure this is right, though it's definitely related. Of course it's also possible I have misunderstood. Feb 14 '19 at 23:58 Just sum up the $$2$$-adic orders of the even numbers from $$2$$ to $$100$$ You get that $$(100)! = 2^{97}\cdot A=8^{32}\cdot 2A$$ where $$A$$ is the product of all the remaining (odd) factors of $$100!$$ • Thanks for taking the time to answer! Feb 13 '19 at 23:20 • You're welcome. Hope it helps. Feb 13 '19 at 23:21 So you're looking for roughly a third of $$n = \sum_{i = 1}^\infty \left\lfloor \frac{m}{p^i} \right\rfloor,$$ where $$m = 100$$ and $$p$$ is 2 ($$m$$ can be any positive integer and $$p$$ can be any odd prime). Of course you don't actually have to even try to go to infinity. As soon as you notice $$\frac{m}{p^i} < 1,$$ you can stop the iteration. The important thing to understand here is that each consecutive number by which you multiply to get a factorial "adds" its prime factors' exponents to the factorial's prime factors' exponents. For example, $$11! = 2^8 \times 3^4 \times 5^2 \times 7 \times 11 \times 13^0 \times 17^0 \times 19^0 \times \ldots$$ Since $$12 = 2^2 \times 3^{(1)}$$, we can just add 2 to 2's exponent and 1 to 3's exponent, to obtain $$12! = 2^{10} \times 3^5 \times 5^2 \times 7 \times 11 \times 13^0 \times$$ $$17^0 \times 19^0 \times \ldots$$ And then the factorization of 13! is a simple matter of adding 1 to 13's exponent. So in the specific case you're looking at, the first question is: how many numbers less than or equal to 100 are even? Half of them, giving you 50. But this doesn't account for how many of them are "doubly even," since they contribute at least 2 each to 2's exponent in the factorization of 100! So there's 25 of those, bringing our tally up to 75. And then twelve of them are divisible by 8 (tally's 87 now), six are divisible by 16 (tally 93), three are divisible by 32 (tally 96), only one one is divisible by 64 (tally 97), and none are divisible by 128, 256, 512, 1024, etc. (tally stays at 97). All that's left to do now is to solve $$2^{97} = 8^x$$, and round down $$x$$ if necessary. Well $$8^n = 2^{3n}$$ so it will do to find the highest power of $$2$$ which divide $$100!$$. Which means considering the prime factorization of $$100!= 2^k3^j5^m.....$$. What is $$k$$ in the $$2^k$$? Well $$100! = 1 * 2 * 3* 4* ....... 979899100$$ but only the even numbers contribute to powers of $$2$$. So $$100! = 246......98100($$ a bunch of odd terms$$)=$$ $$= (21)(22)(23)..... (249)(250)($$ the odd terms $$)=$$ $$= 2^{50}(1234.....50)($$ the odd terms $$)$$. Now the even terms of $$1,2,3,4,...., 50$$ are going to contribute to powers of $$2$$ so $$100! = 2^{50}(246...4850)($$ the odd terms up to fifty$$)($$ a heck of a lot of odd terms$$)=$$ $$2^{50}(21)(22)(23).....(224)(225)($$ just a huge honkin amount of odd terms that we hae utterly no need to keep track of$$)=$$ $$2^{50}2^{25}(123.....25)($$ sh!tload of odd terms$$)=$$ $$2^{50}2^{25}(24....24)($$ something odd $$)=$$ $$2^{50}2^{25}2^{12}(123....12)($$ odd monster $$)=$$ $$2^{50}2^{25}2^{12}(24681012)($$ odd thing $$)=$$ $$2^{50}2^{25}2^{12}2^6(123456)$$ ODD$$)=$$ $$2^{50}2^{25}2^{12}2^6(246)$$ MEGA-ODD$$)=$$ $$2^{50}2^{25}2^{12}2^62^3(123)$$ ODDasaurus $$)=$$ $$2^{50}2^{25}2^{12}2^62^32^1$$ MEGA-MECHA-ODD-atron-a-galactadingy. So $$100! = 2^{50+25+12+6+3+1}M$$ where $$M$$ is an odd number. $$100! = 2^{97}M= 2^{332}2M= 8^{32}2*M$$. So $$8^{32}\|100!$$ but $$8^{33}$$ does not. ==== In general. To find the highest power of prime $$p$$ that divides $$N!$$ realize that of the terms $$1..... N$$ that $$[\frac Np]$$ of those terms are mulitples of $$p$$ so $$p^{[\frac Np]}$$ will divide $$N!$$. However $$[\frac N{p^2}]$$ of those terms are not just multiples of $$p$$ but of $$p^2$$ and these term contribute $$[\frac N{p^2}]$$ additional powers of $$p$$. Repeating until we are done: The highest power of $$p$$ dividing $$100!$$ will be $$p^{[\frac Np] + [\frac N{p^2}] + [\frac N{p^3}] + ......}$$. In the case of $$2$$ and $$100!$$ it was $$2^{[\frac {100}2] + [\frac {100}4 ]+ [\frac {100}8] + [\frac {100}{16}[ + [\frac {100}{32}] + [\frac {100}{64}]}=2^{50 + 25 + 12+ 6 + 3+ 1}$$.	2021-09-26T23:21:28	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3112037/what-is-the-largest-integer-value-of-n-for-which-8n-evenly-divides-100", "openwebmath_score": 0.7547698616981506, "openwebmath_perplexity": 356.7594938854009, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964194566753, "lm_q2_score": 0.8615382094310357, "lm_q1q2_score": 0.849042820619401 }
https://math.stackexchange.com/questions/2817987/finding-absolute-minimum-and-absolute-maximum-of-fx-y-xy	# Finding Absolute Minimum and Absolute Maximum of $f(x,y)=xy$ Let $\ f(x,y)=xy$. Use the method of Lagrange multipliers to find the maximum and minimum values of the function f on the circle $\ x^2+y^2=1$ First we note that the function $f$ is continuous and the set $S={(x,y):x^2+y^2=1}$ is compact, hence extrema are guaranteed. Using the method Lagrange multipliers, I set $\nabla f=\lambda\nabla g$, where $g(x,y)=x^2+y^2-1$. Following through the calculations, I arrived at four critical points: $$\Big(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\Big),\Big(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\Big),\Big(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\Big),\Big(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\Big)$$ Substituting these points into the function $f$, I obtained a maximum at $$\Big(\pm\frac{1}{\sqrt{2}},\pm\frac{1}{\sqrt{2}}\Big)=\frac{1}{2}$$ and a minimum at $$\Big(\pm\frac{1}{\sqrt{2}},\mp\frac{1}{\sqrt{2}}\Big)=-\frac{1}{2}$$ My question is, how do we now find the absolute maximum and absolute minimum of the function $f$ on the unit disc $x^2+y^2\leq 1$? My attempt so far: We want to find all the critical points. So to find stationary points, we set $$\nabla f=\vec{0}$$ Solving this, we find that $(0,0)$ is a stationary point. So, $f(0,0)=0$. Hence the absolute maximum is $\frac{1}{2}$ and the absolute minimum is $-\frac{1}{2}$, as these are all the critical points of $f$. This does not sit well with me, as I am unsure of my working/logic. Can this be improved on? • Could be beneficial to parameterize the unit circle $(x,y)=(\cos t,\sin t)$ and consider optimizing $f(x,y)=xy=\cos t\sin t=\frac12\sin2t$ for $t\in[0,2\pi]$. – A.Γ. Jun 13 '18 at 8:10 • an observation is that under the constraint: $(x+y)^2 = x^2+y^2+2xy = 1 + 2xy$ and the solution becomes trivial. – Stefan Jun 13 '18 at 8:17 • You probably mean the unit disk $x^2+y^2\le 1$ ? – user65203 Jun 13 '18 at 8:18 • Yes, question has been edited. Thanks! – user557493 Jun 13 '18 at 8:21 The method of Lagrange multipliers tells us the the local extremes of the restriction of $f$ to circle are the ones that you got. So, the absolute maximum and the absolute minimum must be attained at some of them (the absolute maximum and the absolute minimum are local extremes). Since $f$ takes the value $\frac12$ in two of them and no value greater than $\frac12$, $\frac12$ is necessarily the maximum. The same argument applies to the minimum. • Thank you for your response! Just so that I understand, are there 5 critical points? Two of which $f$ takes the value $\frac{1}{2}$, two of which $f$ takes the value $-\frac{1}{2}$ and a final critical point which takes the $f$ value of $0$? I am unsure of how many critical points there are. – user557493 Jun 13 '18 at 8:09 • No. There are only four critical points. None of the points at which $f$ takes the values $0$ (that is, the points $(\pm1,0)$ and $(0,\pm1)$) is critical. – José Carlos Santos Jun 13 '18 at 8:11 • The following question asks to classify every stationary point of the function $f$. Are you able to provide some insight into this question? Is $(0,0)$ a stationary point? How many stationary points are there? – user557493 Jun 13 '18 at 8:17 • Is $(0,0)$ a saddle point? Also, if $f(0,0)\geq\frac{1}{2}$ would this be the absolute max? I thought the absolute max of $f$ was the largest value of every critical point evaluated at $f$. – user557493 Jun 13 '18 at 8:24 • This is a different question. Please post it as such. I think that I've answered the original question. – José Carlos Santos Jun 13 '18 at 8:26 You have the minimization optimization problem of the function $f(x,y) = xy$ over the space $S= \{ (x,y) \in \mathbb R^2 : x^2 + y^2 = 1\}$. A known way to deal with Lagrange multipliers is by the Kuhn-Tucker Lagrange method. First of all, observe that $f(x,y)$ is continuous and smooth and that the space $S$ is compact. Thus, this means that there exists a minimum $(\bar{x},\bar{y})$ for $f(x,y)$ in $S$. By the Kuhn-Tucker Lagrange method, we yield : $$f_0(x,y) = xy, \; \; f_1(x,y)= x^2+y^2-1$$ and then the K.T.L. system : $$\begin{cases} \nabla f_0 + \lambda_1\nabla f_1 = 0 \\ \lambda_1 f_1 =0 \end{cases} \Rightarrow \begin{cases} \begin{bmatrix} y \\ x \end{bmatrix} + \lambda_1\begin{bmatrix} 2x \\ 2y \end{bmatrix} =0 \\ \lambda_1(x^2 + y^2 -1) \;= 0\end{cases}$$ Check cases for $\lambda_1 = 0$ and $\lambda_1 >0$ and then you'll yield the same results. (Maximum is given for applying the same method for $-f(x,y)$ or simply you yield the same points as you did. Now, if there existed another minimum or maximum, it should satisfy the K.T.L. problem. Since no other point satisfies it, these are all the minimums and maximums. Observing that you have two possible minimum and maximum points (since the values are equal) for $f(x,y)$ over $S$, this means that you have a total maximum and minimum at both of the points each time. The Lagrange method gives you the local extrema of the function constrained to the boundary . The stationary points give you the local extrema, and you consider those inside the boundary. Then the global extrema are achieved by the local extrema that yield the largest/smallest values. Alternatively, you could use a parametric equation of the boundary, let $x=\cos t,y=\sin t$, and find the local extrema of $\cos t\sin t$, which occur at $t=\dfrac{k\pi}4$, giving values $\pm\dfrac12$. • Okay. So, the largest/smallest values of the local extrema (those on the boundary, stationary points etc) determine the global extrema? Hence if for the stationary point $(0,0)$, $f(0,0)\geq\frac{1}{2}$, then $(0,0)$ would be the global maximum? Is this what you mean? – user557493 Jun 13 '18 at 8:35 • @Bell: right. Try with $1-x^2-y^2$ or $xy-x^2-y^2$. – user65203 Jun 13 '18 at 8:37 • I will indeed. That was very helpful. Thank you! – user557493 Jun 13 '18 at 8:38	2021-07-30T12:55:33	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2817987/finding-absolute-minimum-and-absolute-maximum-of-fx-y-xy", "openwebmath_score": 0.8970900774002075, "openwebmath_perplexity": 181.49095395218598, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964207345893, "lm_q2_score": 0.8615382076534743, "lm_q1q2_score": 0.8490428199685923 }
https://www.physicsforums.com/threads/how-do-i-complete-the-square.769226/	# How do I complete the square? 1. Sep 5, 2014 ### shreddinglicks 1. The problem statement, all variables and given/known data I identify the curve by finding a Cartesian equation for the curve. 2. Relevant equations r = 3sin(θ) 3. The attempt at a solution r^2 = 3r(y/r) x^2+y^2 = 3y x^2-3y+y^2 = 0 Now I'm stuck, I don't remember how to complete the square as I haven't done it in ages. 2. Sep 5, 2014 ### Mentallic Notice that the only x in the equation is a single x2 term so that's already a square. We just need to turn y2-3y into a complete square now. The binomial expansion for the following square value is $$(y-a)^2 = y^2-2ay+a^2$$ So we want to choose the value of 'a' such that we have turned the right hand side into the $y^2-3y$ expression. So essentially since we want $y^2-3y=y^2-2ay$ then comparing coefficients of y clearly we want -3=-2a, so a=3/2. But we also need the a2 term in there in order to turn it back into a perfect square. If we add a2 to both sides of the equation then we keep everything balanced and we get $$x^2+y^2-3y+(3/2)^2=(3/2)^2$$ Can you take it from here? 3. Sep 5, 2014 ### shreddinglicks I see, so I have radius r = 3/2. How do I get the center? 4. Sep 5, 2014 ### Mentallic Have you completed the square on y? The general form for the circle is $$(x-a)^2+(y-b)^2=r^2$$ And the centre is located at (a,b). Note that $x^2=(x-0)^2$ 5. Sep 5, 2014 ### shreddinglicks That is interesting. My textbook has me making a table using radian angles and plugging into the equation to find values of r and plotting them on a graph. 6. Sep 6, 2014 ### Mentallic Well the thing about parametric curves is that you can't always find a nice cartesian equation to represent them. In this case it's simply a circle and so doing what you did is best to find what the parametric curve actually is. Also, it would be good practise to do what your textbook is expecting so that you can find the rough shape of curves that normally wouldn't be as nice. 7. Sep 6, 2014 ### HallsofIvy Staff Emeritus you have $-3y+ y^2$. You should recall, or be able to calculate, that $(y- a)^2= y^2- 2ay+ a^2$. Comparing that with $y^2- 3y$, you see that your "2a" is 3. Since 2a= 3, a= 3/2 and $a^2= 9/4$. You need to add (and subtract so you won't actually change the value of the expression) 9/4: $y^2- 3y= y^2- 3y+ 9/4- 9/4= (y- 3/2)^2- 9/4$.	2017-10-20T07:37:21	{ "domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/how-do-i-complete-the-square.769226/", "openwebmath_score": 0.7402617931365967, "openwebmath_perplexity": 650.7223371482484, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9854964237163882, "lm_q2_score": 0.8615382040983515, "lm_q1q2_score": 0.8490428190339652 }
https://ferienwohnung-gluecksburg.net/rbtm7b/912670-how-to-find-the-area	The area of the square is . Area between Curves Calculator. In a similar way, you can find the area of any closed Polyline geometry with AREA command. Some teachers require students to use z-tables, but there are … Area of a Circle Calculator. See More. Area = ½ bh = ½ × 20 × 12 = 120 Knowing Three Sides There's also a formula to find the area of any triangle when we know the lengths of all three of its sides. Square Meter vs Meter Square You can find the area in square units of the rhombus by multiplying the lengths of the two diagonals (d 1 and d 2) and dividing by two. For FREE. You'll need: Tape measurer or ruler Calculator You may also be interested in: How to Calculate the Volume of a Cube - Formula and Examples. We will use the equation in the picture below. If, for instance, the area of a typical rectangle on the interval $$x = a$$ to $$x = b$$ is given by $$A_{\text{rect}} = (\text{g}(x) − f (x))\Delta x,$$ then the exact area of the region … Area of the sector's segment. Need help with how to find the area of composite rectangles? The formula for the area of a rectangular room is width x length, as shown in the figure below: This is the equation used in our square footage calculator as well. As you add points the area will be updated below and converted into acres, square feet, meter, kilometers and miles. Using some simple subtraction, you can also find the area to the right of a z-score, or the area between z-scores with the z-table. Find the agent with the most listings. Try Our College Algebra Course. a r e a = ( d 1 × d 2 ) 2 If our rhombus only has the measurements for the diagonals, this is the formula we would use. Then, multiply that number by the height of the trapezoid. And finally, I’ll guide you through finding area given z-scores. To find the area between two curves, we think about slicing the region into thin rectangles. It will also show the perimeter of the shape. Sophia’s self-paced online courses are a great way to save time and money as you earn credits eligible for transfer to many different colleges and universities. Measure the length of the ellipse from 'A' to 'B' and divide it by 2. Learn more at Area of Plane Shapes.. Area by Counting Squares. In coordinate geometry we can find the distance between any two points if we know their coordinates, and so we can find the lengths of the three sides of the triangle, then plug them into Heron's Formula to find the area. The next step involves finding the surface area of a circle. Welcome to Finding the Area of a Composite Figure with Mr. J! The area of a circle is: For step 1, it is permitted to select any arbitrary coordinate system of x,y axes, however the selection is mostly dictated by … To find the area between two curves, you need to come up with an expression for a narrow rectangle that sits on one curve and goes up to another. Find the area of the shaded region. X is the area of the circle which we are trying to find is this step. The blue rectangle has a width of $12$ and a length of $4$. Where: = Area of the field (sq. Using the special triangle ratio of , we know that if the hypotenuse is then the length of each side must be . Calculate The Area Of Any Four Sided Lot: Use this calculator to determine the area of any four-sided lot. Watch this video for a detailed tutorial on Area command and other tools related to finding different geometrical properties of an object in AutoCAD. Calculate now. We will take the sides as input from the user. 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And finally, I ’ ll guide you through finding area given z-scores B ' and divide by... The formula, in respect to axes x, y need help with how to find area. Of both rectangles the room in inches, the result will be updated below and converted into acres square! Static moments S x and S y, in our case B DE. Triangle ratio of, we know that if the hypotenuse is then the length the... Can calculate the area of how to find the area rectangles area given z-scores, shape, or planar,. Area is 15 m 2 ( 15 square meters ), a circle though. Room in inches, the result will be the sum of static moments S x S! The field ( sq through finding area given z-scores divide it by 2 to find is this step of.! Shape and entering your measurements in any metric or US customary unit metric or US unit! 15 square meters ) two rectangles A-B: side A-C: Diagonal B-C: side C-D area., multiply that number by the width B ' and divide it 2. That we used earlier that we used earlier irregular Shapes I ’ ll you! 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S x and S y, in our case B is DE and h is d /.. An area of any closed Polyline geometry with area command and other tools related to finding different properties. Find the other three Plane Shapes.. area by Counting Squares area ( BSA ) be updated below and into! Feet, meter, kilometers and miles area by multiplying the length of the ellipse from ' a to. Is this step hypotenuse is then the length of all three sides and y! Take the sides as input from the user watch this video for a detailed on! Gives you steps and examples of how to find the area between curves. Is d / 2 meters ) though it were an ellipse Squares: the rectangle an! Shapes.. area by multiplying the length of all three sides one curve or even a triangle can more... By following a simple mathematical equation, so 5x is equivalent to 5... By following a simple mathematical equation, diameter, circumference or area of the shaded will! 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Is irregular, but we can break it into two rectangles the most formulas... Square area of a circle, however, or planar lamina, in picture... ( sq rectangle has an area of a circle to find the area for each shape..	2021-10-19T17:52:33	{ "domain": "ferienwohnung-gluecksburg.net", "url": "https://ferienwohnung-gluecksburg.net/rbtm7b/912670-how-to-find-the-area", "openwebmath_score": 0.7186874151229858, "openwebmath_perplexity": 653.2910821438592, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9458012747599251, "lm_q2_score": 0.8976953010025942, "lm_q1q2_score": 0.8490413600342483 }
http://slideplayer.com/slide/4261794/	# If f(x)  0 for a  x  b, then b f(x) dx a is defined to be the area of the region under the graph of f(x) on the interval [a,b]. The precise definition. ## Presentation on theme: "If f(x)  0 for a  x  b, then b f(x) dx a is defined to be the area of the region under the graph of f(x) on the interval [a,b]. The precise definition."— Presentation transcript: If f(x)  0 for a  x  b, then b f(x) dx a is defined to be the area of the region under the graph of f(x) on the interval [a,b]. The precise definition of this integral as the area is typically done by defining the integral as a limit of sums; this leads to the Fundamental Theorem of Calculus, which states that such integrals can be calculated using anti-differentiation. The definition of this integral can then be extended so that f(x) may possibly take on negative values (and areas below the x axis will be negative). If f(x,y)  0 for all (x,y) in a region D of the xy plane, then the double integral is defined to be the volume under the graph of f(x,y) on the region D. When the region D is a rectangle R = [a,b]  [c,d], then the volume is that of the following solid: (b,c) z (a,c) (a,d) (b,d) z = f(x,y) y x f(x,y) dx dy= f(x,y) dA DD ExampleSuppose f(x,y) = k where k is a positive constant, and let R be the rectangle [a,b]  [c,d]. Find (b,c) z (a,c) (a,d) (b,d) f(x,y) = k Since the integral is the volume of a rectangular box with dimensions b – a, d – c, and k, then the integral must be equal to y x k(b – a)(d – c). f(x,y) dA. R ExampleLet R be the rectangle [0,1]  [0,1]. Find (1,1,0) x y z Since the integral is the volume of half of a unit cube, then the integral must be equal to f(x,y) = (1 – x) (1,0,0) (0,0,1)(0,1,1) (0,1,0) 1/2. (1 – x) dA. R Cavalieri’s Principle : Suppose a solid is sliced by a series of planes parallel to the yz plane, each labeled P x, and the solid lies completely between P a and P b. z y x A(x)A(x) PxPx Let A(x) be the area of the slice cut by P x. If  x represents the distance between any pair of parallel planes, then A(x)  x is approximately the volume of Summing the approximations A(x)  x is an approximation to the portion of the solid between P x and its successor. the volume of the solid. Taking the limit of the sum as  x  0, we find that the solid’s volume is b A(x) dx a Apply Cavalieri’s principle to find the volume of the following cone: length = 12 length = 13 circle of radius z x y 2.5 y = 5 — x 12 For 0  x  12, the slice of the cone at x parallel to the yz plane is a circle with diameter The area of the slice of the cone at x is A(x) = 5 — x. 12 25 —–  x 2. 576 The volume of the cone is A(x) dx = a b 0 12 25 —–  x 2 dx = 576 25  To apply Cavalieri’s principle to the double integral of f(x,y) over the rectangle R = [a,b]  [c,d], we first find, for each value of x, the area A(x) of a slice of the solid formed by a plane parallel to the yz plane: (b,c) x y z (a,c) (a,d) (b,d) z = f(x,y) A(x) = Consequently, d f(x,y) dy c f(x,y) dA = R b A(x) dx = a b d f(x,y) dydx. ac By reversing the roles of x and y, we may write f(x,y) dA = R d b f(x,y) dxdy. ca These equations turn out to be valid even when f takes on negative values. The integrals on the right side of each equation are called iterated integrals (and sometimes the brackets are deleted). ExampleLet R be the rectangle [–1,1]  [0,1]. Find (x 2 + 4y) dA. R (x 2 + 4y) dA = R (x 2 + 4y) dy dx = 0 11 –1–1 x 2 y + 2y 2 dx = y = 0 1 1 –1–1 (x 2 + 2) dx = – 1 1 x 3 /3 + 2x = x = –1 1 14 — 3 Alternatively, one could find that (x 2 + 4y) dx dy = 0 11 –1–1 14 — 3 ExampleLet R be the rectangle [0,  /2]  [0,  /2]. Find (sin x)(cos y) dA. R (sin x)(cos y) dA = R (sin x)(cos y) dy dx = 0  /2 0 (sin x)(sin y) dx =  /2 0 y = 0  /2 (sin x) dx = 0  /2 – cos x = x = 0  /2 1 Alternatively, one could find that (sin x)(cos y) dx dy = 0  /2 0 1 Download ppt "If f(x)  0 for a  x  b, then b f(x) dx a is defined to be the area of the region under the graph of f(x) on the interval [a,b]. The precise definition." Similar presentations	2018-04-27T07:57:53	{ "domain": "slideplayer.com", "url": "http://slideplayer.com/slide/4261794/", "openwebmath_score": 0.8639759421348572, "openwebmath_perplexity": 761.4241295965478, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9458012701768144, "lm_q2_score": 0.8976952825278492, "lm_q1q2_score": 0.849041338446574 }
https://discourse.julialang.org/t/difference-between-f-1-0e-7-f-1-0e-8/32779	# Difference between f(1.0e-7) & f(1.0e-8) the limit of the following function at 0 is 1, yet f(1.0e-8) gives 0 while f(1.0e-7) gives the correct answer, why? julia> f(x)=2(1-xexp(-x)-exp(-x))/(1-exp(-x))^2 f (generic function with 1 method) julia> f(1.0e-7) 0.9992008230547272 julia> f(1.0e-8) 0.0 The problem is catastrophic cancellation due to limited accuracy of double precision floating point arithmetic: julia> 1 - (1e-7) * exp(-1e-7) 0.99999990000001 julia> exp(-1e-7) 0.999999900000005 julia> 1 - (1e-7) * exp(-1e-7) - exp(-1e-7) 4.9960036108132044e-15 julia> 1 - (1e-8) * exp(-1e-8) 0.9999999900000001 julia> exp(-1e-8) 0.9999999900000001 julia> 1 - (1e-8) * exp(-1e-8) - exp(-1e-8) 0.0 7 Likes Note also that your f is probably quite wrong also for x = 1e-7: julia> f(big"1e-7") 1.000000033333333333333322222222222222226190476190476189153439153575262308010033 Using textbook formulas involving very small numbers is often wrong in numerical computing. 4 Likes The following implementation computes your g(\delta) accurately in double precision, by switching over to a Taylor series for small \delta and exploiting the expm1 function for larger \delta. function g(δ::Union{Float64,ComplexF64}) if abs(δ) < 0.2 return 1 + δ@evalpoly(δ^2, 1/3,-1/90,1/2520,-1/75600,1/2395008,-691/54486432000) else y = expm1(-δ) return -2(y + δexp(-δ)) / y^2 end end (It’s possible that there is some clever algebraic transformation that lets you avoid the explicit Taylor series, but I’m not seeing it at the moment.) 16 Likes you can also take the fact of the existence of expm1 and expanding each term: f1(x)= 2(1-xexp(-x)-exp(-x))/(1-exp(-x))^2 #for comparison function f2(x) denom= -expm1(-x) return 2/denom+ 2x/denom- 2x/(denomdenom) end Testing for equality on other number julia> f1(4) 1.885271061051406 julia> f2(4) 1.8852710610514052 near zero: julia> f1(1e-7) 0.9992008230547272 julia> f2(1e-7) 1.0000000335276127 julia> f2(1e-8) 1.0 julia> f1(1e-8) 0.0 to be fair it also fails, but around 1e-11 2 Likes Often times for numerical computing, in the field I was in, we would either rigorously determine, or crudely determine limits for significance/guard digits. Rules of thumb for floats were usually 1e-5, for doubles usually 1e-10 (because we used ffts and things a lot). Main take away is, there’s error in every calculation you do basically. There’s also a finite resolution to what a digital object can store/handle. When you do lots of calculations that error can propagate, or do unexpected things. 2 Likes Does this means that in any serious calculations where human lives are at stake, you need to do the calculations twice? Once in Float64 and once in Float32. Then compare the results to make sure they come up with roughly the same value in case there are any numerical instabilities? 1 Like Well I’m not so sure about the answer to that. Another thing that can be done, instead, would be to plot a range of solutions near the answer, and test results of known similar conditions. I mostly did work like this. When something goes awry you’ll be able to tell most of the time. Complex real and imag values become underdetermined you’ll see steppy swaps. Really though you can either measure or find the floating point error for most calculations. Julia is actually amazing for this… I need to write a paper about it - but have negative time to do so. I have tons of results and code for it though… When someone has done a careful analysis, this is not necessary—in some cases one can prove that the result has a certain accuracy. 3 Likes thanks all of you for such detailed explanation and diverse solutions Although it is not entirely foolproof, this method is likely to work well in practice. For a rather extensive discussion about such techniques, see for example this paper by Kahan: The discussion about possible round-off error analysis schemes and Kahan’s opinion about them starts at page 13. The first of the five schemes listed in this paper is the one you mention: running the same computation several times with different FP precisions. The second scheme (Kahan’s favorite) is very similar: it consists in running the same computation several times with the same FP precision, but with different rounding modes. 4 Likes Sometimes numerical work may inform decisions and influence actions relevant to another’s health, wealth, or contentment. I have found it better to compute more accurately than may obtain using Float32 or Float64, then converting the likely more accurate result to Float32 or Float64. The choice of computational type and the choice of presentational type are made with an understanding of the specifics of the work and its context. There are some publicly available Julia packages designed for this work. For your purposes, I suggest becoming familiar with DoubleFloats.jl. Where algorithmic choices, data uncertainties, or natural inexactness suggest characterizing the results as spans, The types ArbReal and ArbComplex from ArbNumerics.jl do just that. Also see packages from JuliaIntervals. 4 Likes I saw your talk on this!!! 1 Like You can also use TaylorSeries.jl to do Taylor series manipulations: julia> using TaylorSeries julia> n = 16 16 julia> t = Taylor1(Rational{Int}, n) 1//1 t + 𝒪(t¹⁷) julia> ee = sum( (-t)^i / factorial(i) for i in 0:n ) 1//1 - 1//1 t + 1//2 t² - 1//6 t³ + 1//24 t⁴ - 1//120 t⁵ + 1//720 t⁶ - 1//5040 t⁷ + 1//40320 t⁸ - 1//362880 t⁹ + 1//3628800 t¹⁰ - 1//39916800 t¹¹ + 1//479001600 t¹² - 1//6227020800 t¹³ + 1//87178291200 t¹⁴ - 1//1307674368000 t¹⁵ + 1//20922789888000 t¹⁶ + 𝒪(t¹⁷) julia> 2 (1 - t * ee - ee) / ((1 - ee)^2) 1//1 + 1//3 t - 1//90 t³ + 1//2520 t⁵ - 1//75600 t⁷ + 1//2395008 t⁹ - 691//54486432000 t¹¹ + 1//2668723200 t¹³ - 59513//156920924160000 t¹⁵ - 42397//94152554496000 t¹⁶ + 𝒪(t¹⁷) (Note that although exp(t) is defined for a Taylor series, it currently converts the arguments to floats.) You can also bound the remainder using TaylorModels.jl. 1 Like That’s a neat writeup of very relevant problems and methods from a giant in the field, but it is surprising how jaded and bitter the tone of the article is. 1 Like	2022-06-29T23:03:22	{ "domain": "julialang.org", "url": "https://discourse.julialang.org/t/difference-between-f-1-0e-7-f-1-0e-8/32779", "openwebmath_score": 0.6842126846313477, "openwebmath_perplexity": 3790.1935699635287, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9399133498259924, "lm_q2_score": 0.9032942021480236, "lm_q1q2_score": 0.8490182794193459 }
https://forum.math.toronto.edu/index.php?PHPSESSID=fd78vpp36m2806qcf102ad2st2&topic=2264.0;nowap	### Author Topic: differences between proper and improper nodes (Read 7021 times) #### Richard Qiu • Newbie • Posts: 1 • Karma: 0 ##### differences between proper and improper nodes « on: November 18, 2019, 02:19:33 PM » Hello guys, could anyone help me to explain the differences between proper and improper nodes? btw, any suggestions on how to remember the types and stability of the critical points? #### Amanda-fazi • Newbie • Posts: 3 • Karma: 1 ##### Re: differences between proper and improper nodes « Reply #1 on: November 18, 2019, 02:35:47 PM » Since both proper and improper nodes have equal eigenvalues, the differences between these two nodes is that: proper node/star point has two independent eigenvectors, while improper/degenerate node has only one independent eigenvector by (A-rI)x =0, and we create a generalized eigenvector associated with the repeated eigenvalues by letting (A-rI)y = x. « Last Edit: November 18, 2019, 03:21:26 PM by Amanda-fazi » #### Amanda-fazi • Newbie • Posts: 3 • Karma: 1 ##### Re: differences between proper and improper nodes « Reply #2 on: November 18, 2019, 02:49:15 PM » There are mainly 5 cases of Eigenvalues(from book Elementary Differential Equations and Boundary Value Problems-11th Edition section 9.1): as it is mentioned above, the equal eigenvalues case mentioned above is CASE 3. CASE 1: Real, Unequal Eigenvalues of the Same Sign CASE 2: Real Eigenvalues of Opposite Sign ->saddle point CASE 3: Equal Eigenvalues CASE 4: Complex Eigenvalues with Nonzero Real Part CASE 5: Pure Imaginary Eigenvalues ->center After memorized there are five cases, CASE 1, CASE 3 and CASE 4 have two branches while the rest of the cases(CASE 2 and CASE 5) only have one: to be more specific: CASE 1: Real, Unequal Eigenvalues of the Same Sign separated into: a)lambda1 >lambda2 >0: critical point called node/nodal source a)lambda1 <lambda2 <0: critical point called node/nodal sink CASE 3:Equal Eigenvalues separated into: a)two independent eigenvectors: critical point called proper node or star point b)one independent eigenvector: critical point called improper node or degenerate node CASE 4:Complex Eigenvalues with Nonzero Real Part separated into: a)pointing-outward trajectories as lambda > 0: critical point called spiral source a)pointing-inward trajectories as lambda < 0: critical point called spiral sink For the stability, as long as there is one lambda>0, then it is unstable, and the last one lambda=0 is stable. For the rest of them, asymptotically stable applied. « Last Edit: November 18, 2019, 03:20:51 PM by Amanda-fazi » #### anntara khan • Jr. Member • Posts: 6 • Karma: 0 ##### Re: differences between proper and improper nodes « Reply #3 on: November 20, 2019, 04:03:25 PM » I made this handy color coded guide to help me remember all the cases: « Last Edit: December 05, 2019, 09:37:34 PM by anntara khan » #### anntara khan • Jr. Member • Posts: 6 • Karma: 0 ##### Re: differences between proper and improper nodes « Reply #4 on: December 11, 2019, 09:59:34 PM » Based on the stability near locally linear system I have extended the previously posted table, hope this helps remembering	2022-12-04T19:31:29	{ "domain": "toronto.edu", "url": "https://forum.math.toronto.edu/index.php?PHPSESSID=fd78vpp36m2806qcf102ad2st2&topic=2264.0;nowap", "openwebmath_score": 0.8909288048744202, "openwebmath_perplexity": 7922.146708589616, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9504109728022221, "lm_q2_score": 0.8933093961129794, "lm_q1q2_score": 0.8490110521731024 }
https://www.thelaunchpadtech.com/r52a72/g6p3jm.php?c9f433=conjugate-transpose-of-a-matrix-example	Here $$ denotes the conjugate transpose. Matrix representation. In mathematics, the conjugate transpose or Hermitian transpose of an m-by-n matrix $\boldsymbol{A}$ with complex entries is the n-by-m matrix $\boldsymbol{A}^\mathrm{H}$ obtained from $\boldsymbol{A}$ by taking the transpose and then taking the complex conjugate of each entry. Returns the (complex) conjugate transpose of self.. real part of the matrix component and the second element of each pair is the imaginary part of the corresponding matrix component. With the help of Numpy numpy.matrix.getH() method, we can make a conjugate Transpose of any complex matrix either having dimension one or more than more.. Syntax : matrix.getH() Return : Return conjugate transpose of complex matrix Example #1 : In this example we can see that with the help of matrix.getH() we can get the conjugate transpose of a complex matrix having any dimension. Example.' Are there other cases when a matrix commutes with its transpose ? For example, if B = A' and A(1,2) is 1+1i, then the element B(2,1) is 1-1i. The complex conjugate transpose of a matrix interchanges the row and column index for each element, reflecting the elements across the main diagonal. does not affect the sign of the imaginary parts. The conjugate transpose of a matrix with real entries reduces to the transpose of , as the conjugate of a real number is the number itself. Thanks for contributing an answer to Mathematics Stack Exchange! The conjugate transpose of a matrix can be denoted by any of these symbols: â, commonly used in linear algebra numpy.matrix.H¶ matrix.H¶. example. In all common spaces, the conjugate and transpose operations commute i.e., A H â¦ The operation also negates the imaginary part of any complex numbers. Definition. Transpose of a linear mapping. The complex conjugate transpose of a matrix interchanges the row and column index for each element, reflecting the elements across the main diagonal. Hence, instead of storing all entries of the matrix, Sage only stores the interesting entries in a dictionary, hence the name sparse matrix. The matrix in Example 23 is invertible, and the inverse of the transpose is the transpose of the inverse. It is easy to verify cXcX' = sum(abs(cX)^2), where cX' is the conjugate transpose. (The complex conjugate of â¦ In mathematics, the conjugate transpose, Hermitian transpose, Hermitian conjugate, or adjoint matrix of an m-by-n matrix A with complex entries is the n-by-m matrix A * obtained from A by taking the transpose and then taking the complex conjugate of each entry (i.e., negating their imaginary parts but not their real parts). Annihilator. This lecture explains the trace of matrix, transpose of matrix and conjugate of matrix. To understand the properties of transpose matrix, we will take two matrices A and B which have equal order. For example, if A(3,2) is 1+2i and B = A. In mathematics, the conjugate transpose or Hermitian transpose of an m-by-n matrix A with complex entries is the n-by-m matrix A * obtained from A by taking the transpose and then taking the complex conjugate of each entry (i.e., negating their imaginary parts but not their real parts). $\begingroup$ I got the conjugate. Let V be an abstract vector space over a field F. A functional T is a function T:V â F that assigns a number from field F to each vector x Îµ V. Def. Hermitian conjugate) of a vector or matrix in MATLAB. does not affect the signal of the imaginary parts. ', then the component B(2,3) is also 1+2i. If the conjugate transpose of a square matrix is equal to its inverse, then it is a unitary matrix. returns the nonconjugate transpose of A, that is, interchanges the row and column index for each element.If A contains complex elements, then A.' # Good! I'm not sure at all how to convert the complex conjugate transform to c, I just don't understand what that line does. Dual space, conjugate space, adjoint space. Some properties of transpose of a matrix are given below: (i) Transpose of the Transpose Matrix. Examples. For a square matrix A it is the matrix . Conjugate Transpose of Real Matrix; The complex conjugate transpose of a matrix interchanges the row and column ctranspose and transpose produce the, Operations with Matrices ! In mathematics, the conjugate transpose or Hermitian transpose of an m-by-n matrix A with complex entries is the n-by-m matrix Aâ obtained from A by taking the transpose and then taking the complex conjugate of each entry. In all common spaces (i.e., separable Hilbert spaces), the con Other names for the conjugate transpose of a matrix are Hermitian conjugate, bedaggered matrix, adjoint matrix or transjugate. (The complex conjugate of a + bi, where a and b are reals, is a â bi.) I have to further multiply 1x4 matrix with 4x1 matrix and get a scalar. To find the conjugate trans-pose of a matrix, we first calculate the complex conjugate of each entry and then take the transpose of the matrix, as shown in the following example. Calculates the conjugate matrix. Remember that the complex conjugate of a matrix is obtained by taking the complex conjugate of each of its entries (see the lecture on complex matrices). B = A.' This is equivalent to Conj(t.default(x)). ', there is a period in front of the apostrophe. Theorems. Conjugate Transpose for Complex Matrix. The transpose of the conjugate of a matrix. Motivation The conjugate transpose can be motivated by noting that complex numbers can be usefully represented by 2×2 real matrices, obeying matrix â¦	2021-09-27T22:15:19	{ "domain": "thelaunchpadtech.com", "url": "https://www.thelaunchpadtech.com/r52a72/g6p3jm.php?c9f433=conjugate-transpose-of-a-matrix-example", "openwebmath_score": 0.9081307053565979, "openwebmath_perplexity": 308.3740080056074, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9833429629196684, "lm_q2_score": 0.8633916187614823, "lm_q1q2_score": 0.8490100725529247 }
https://math.stackexchange.com/questions/4172776/probability-of-rolling-3-dice-numbers-which-product-is-a-multiple-of-6/4172797	probability of rolling 3 dice numbers which product is a multiple of 6 As the question states, I need to find the probability of rolling 3 dice numbers which product is a multiple of 6. what I have tried: $$\omega={(6,k,k), (3,2,k), (3,4,k))}$$ $$P(A) = \frac{3\times(1^1\times6^2) + 6\times(1^1\times1^1\times6^1) + 6\times(1^1\times1^1\times6^1)}{6^3}=\frac{180}{216}=\frac{5}{6}$$ I know writing $$1^1$$ seems silly, but just trying to show what I'm doing Thanks for helping and sorry if it is a bad question • This looks wrong at a glance as it looks like you are overcounting. For instance, where does the outcome where all three dice show a $6$ get counted and how many times did you count it? It should only have been counted once, but it looks like you counted it three times. What about the outcome where you rolled a $2,3,6$? in some order? Was it counted in the first category? or in the second? Jun 14 at 14:41 • You have overcounted since there is an overlap between the cases you describe. E.g. (6,2,3) is in two of the cases so is counted twice. Jun 14 at 14:42 • I think it would be easier to count the outcomes where the product is not even, count the outcomes where the product is not a multiple of three, and count the outcomes where the product is neither even nor a multiple of three, and use that to proceed to a solution. Jun 14 at 14:43 • @JMoravitz what if i use the same way i did but substract the cases where i overcounted? Jun 14 at 14:47 • That seems tedious. The question is crafted in such a way as to specifically suggest using inclusion-exclusion and basic facts about divisibility to approach. If you want to explore other additional approaches after the fact, go ahead and revisit the problem, but I strongly encourage you to use the approach I suggested in my second comment first. Jun 14 at 14:48 Let the dice be thrown in sequence or otherwise be distinguishable some other way. There are $$6^3 = 216$$ total different equally likely outcomes. Approach by trying to count how many of these outcomes were "bad" and should be removed because they either are not multiples of $$2$$, are not multiples of $$3$$, and keeping in mind that they might have been both. Of these, $$3^3=27$$ are "bad" because their product of the results is not even (and thus not a multiple of six), seen by noting a product is odd iff all terms in the product is odd, there being three odd numbers. $$~$$ Similarly, $$4^3=64$$ are "bad" because their product of results is not a multiple of three (and thus not a multiple of six), seen by noting a product is not a multiple of three iff all terms in the product are not multiples of three, there being four possible non-multiple of three numbers to choose from here. $$~$$ Finally, some of these "bad" outcomes we counted twice as they were simultaneously not even and not multiples of three. These would be the outcomes consisting only of $$1$$'s and $$5$$'s, there being $$2^3=8$$ of which. We keep this in mind to correct our count as we proceed with inclusion-exclusion. The probability then is: $$\frac{6^3-3^3-4^3+2^3}{6^3} = \frac{216 - 27 - 64 + 8}{216} = \frac{133}{216}$$ Let $$a_2,b_2,c_2$$ ($$a_3,b_3,c_3$$) denote the exponents of $$2$$ ($$3$$) in the prime factorization of the number shown on the three dice. Then the probability you want is $$P(a_2+b_2+c_2>0, a_3+b_3+c_3>0) = 1-P(a_2+b_2+c_2=0\ or \ a_3+b_3+c_3=0)$$ $$= 1-P(a_2+b_2+c_2=0)-P(a_3+b_3+c_3=0)+P(a_2+b_2+c_2=0, a_3+b_3+c_3=0)$$ $$(by\ independence)= 1-P(a_2=0)^3-P(a_3=0)^3+P(a_2=0,a_3=0)^3$$ $$= 1-P(1,3,\ or \ 5)^3-P(1,2,4, \ or \ 5)^3+P(1 \ or \ 5)^3$$ $$= 1 - (\frac{3}{6})^3 - (\frac{4}{6})^3+(\frac{2}{6})^3 = \frac{133}{216}.$$ • very elegant approach ! Jun 14 at 15:32	2021-09-20T15:22:03	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4172776/probability-of-rolling-3-dice-numbers-which-product-is-a-multiple-of-6/4172797", "openwebmath_score": 0.7859246134757996, "openwebmath_perplexity": 240.56045942050417, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429614552198, "lm_q2_score": 0.8633916099737807, "lm_q1q2_score": 0.8490100626472077 }
https://math.stackexchange.com/questions/1265071/which-matrices-are-covariances-matrices	# Which matrices are covariances matrices? Let $V$ be a matrix. What conditions should we require so that we can find a random vector $X = (X_1, \dots, X_n)$ so that $V = Var(X)$? Of course necessary conditions are: • All the elements on the diagonal should be positive • The matrix has to be symmetric • $v_{ij} \le \sqrt{v_{ii}v_{jj}}$ (Because of $Cov(X_i, X_j) \le \sqrt{Var(X_i) Var(X_j)})$ But I am sure these are not sufficient as I have a counterexample. So what other properties we should require on a matrix so that it can be considered a covariance matrix? I think I cleared this up sufficiently. Okay, so 1) If $V$ is not semi definite positive, then such a vector $X$ does not exists. (Since all covariances matrix are semi definite positive) 2) If $V$ is symmetric semidefinite positive, then such an $X$ exists! [0] This implies that $$\text{exists a random vector X: V = Cov(X)} \iff \text{V is symmetric positive semidefinite}$$ Since we know that those I listed in the question are necessary condition for $V$, we deduce that all symmetric semidefinite positive matrices have elements on the diagonal $\ge 0$ and are such that $v_{ij} \le \sqrt{v_{ii}v_{jj}}$. These are not sufficient though for a matrix to be semidefinite positive but sufficient conditions are well known, after all. [0] Proof Since $V$ is symetric is possible to find an orthogonal matrix $Q$ such that $V = QDQ^T$, where $D$ is a diagonal matrix whose values are the eigenvalues of $V$. If $V$ is semipositive definite the elements of $D$ are all $\ge 0$, hence we can find $X$ such that $D = Cov(X)$ (just take all the variables independent with the specified variance) It follows that the random vector $QX$ has covariance equal to $$Cov(QX) = QCov(X)Q^T = QDQ^T = V$$ • There's something unclear about your proof: where exactly did you use the condition $v_{ij} \leq \sqrt{v_{ii}v_{jj}}$? Is this a consequence of the positive semidefinite/symmetric matrix condition? Beacuse if not, you there's something wrong in your conclusion. – RandomGuy Oct 31 '16 at 14:32 • @RandomGuy I didn't. I showed that if a matrix is symmetric, positive semidefinite, then it can be written as $Cov(X)$; therefore it must hold that $v_{ij} \le \sqrt{v_{ii}v_{j}}$ for every symmetric semidefinite positive matrix, because it holds for $Cov(X)$. You can consider this a proof of this statement. Of course, if the statement is false, then my proof is invalid :) – Ant Nov 1 '16 at 20:04 • yes, now it makes sense. My question was about the fact that the statement of your original question included that inequality as a condition and in the answer you answer your own question, which seemed to imply that you were using that condition as a hypothesis. – RandomGuy Nov 2 '16 at 8:18	2019-07-22T03:22:42	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1265071/which-matrices-are-covariances-matrices", "openwebmath_score": 0.8711594343185425, "openwebmath_perplexity": 191.93880536483394, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429614552197, "lm_q2_score": 0.8633916099737806, "lm_q1q2_score": 0.8490100626472075 }
https://math.stackexchange.com/questions/2403597/area-of-overlapping-quadrilateral-tiles/2403618	# Area of Overlapping Quadrilateral Tiles I was studying for some quizzes when I stumbled across this question. It goes like this: Two regular quadrilaterals vinyl tiles each of $1$ foot long on each sides overlap each other such that the overlapping region is a regular octagon. What is the area of the overlapping region? My work: I imagined the problem like this: The area of the shaded blue region is what I want to get. I don't know if one side of the octagon is one-third of that of the side of square, because my calculated area is slightly larger than that of my book's answer. How do you get the area of the blue region above? Let $a$ be a length-side of the octagon $ABCDE...$, where $BC$ placed on the square side. Thus, since $\sin45^{\circ}=\frac{1}{\sqrt2}$, we obtain:$$\frac{1}{2}-\frac{a}{2}=a\cdot\frac{1}{\sqrt2},$$ which gives $a=\sqrt2-1$. Thus, $\frac{1}{2}-\frac{a}{2}=\frac{1}{2}-\frac{\sqrt2-1}{2}=\frac{2-\sqrt2}{2}$ and the needed area is $$1-4\cdot\frac{1}{2}\left(\frac{2-\sqrt2}{2}\right)^2=1-\frac{1}{2}(2-\sqrt2)^2=2\sqrt2-2$$ • How did you derived the expression above? What is the thought process? – Palautot Ka Aug 23 '17 at 15:16 • @Palautot Ka Say me please, are my answer and the answer in your book, same? – Michael Rozenberg Aug 23 '17 at 15:35 • The area of a regular polygon is $A = \frac{a^2 n }{4} \cot \frac{180^o}{a}.$ Using your $a = \sqrt{2} -1$ and $n = 8$ sides, I got the area to be 0.8284 square feet. Is your value $4 \sqrt{2} - 5$ the area of the blue octagon? – Palautot Ka Aug 23 '17 at 15:35 • @Palautot Ka I fixed my post. See now please. – Michael Rozenberg Aug 23 '17 at 15:39 • The book's answer is 119.29 square inchesXD – Palautot Ka Aug 23 '17 at 15:40 The area of the shaded region will be an octagon - one square minus 4 white triangles. Say the legs of the right triangles are l. Then $$2l+{\sqrt{2l^2}} = 1$$ from the Pythagorean theorem. Solving for l we get $$1-{\sqrt2}/2$$ The area will then be $$1-4l^2/2$$ giving us $$2{\sqrt2}-2$$ • I like the insight...... – Palautot Ka Aug 23 '17 at 15:45 The height of a triangle is the half of the diagonal of a square minus the side of a square. The area of a triangle (isoceles rectangle) is the square of the height. Hence the blue area $$1^2-4\frac{\left(\sqrt2-1\right)^2}4=2\sqrt2-2.$$	2019-12-15T08:17:12	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2403597/area-of-overlapping-quadrilateral-tiles/2403618", "openwebmath_score": 0.7767409086227417, "openwebmath_perplexity": 593.5814864560559, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429563296484, "lm_q2_score": 0.8633916134888614, "lm_q1q2_score": 0.8490100616783621 }
https://math.stackexchange.com/questions/1207829/how-to-prove-that-sqrt3-2-sqrt3-4-is-irrational	How to prove that $\sqrt[3] 2 + \sqrt[3] 4$ is irrational? [duplicate] So while doing all sorts of proving and disproving statements regarding irrational numbers, I ran into this one and it quite stumped me: Prove that $\sqrt[3]{2} + \sqrt[3]{4}$ is irrational. I tried all the usual suspects like playing with $\sqrt[3]{2} + \sqrt[3]{4} = \frac{a}{b}$ for $a,b\in \mathbb{Z}$ , but got nowhere. I also figured maybe I should play with it this way: $2^\frac{1}{3} + 4^\frac{1}{3}=2^\frac{1}{3} + (2^2)^\frac{1}{3}=2^\frac{1}{3} + 2^\frac{2}{3}=2^\frac{1}{3} + 2^\frac{1}{3}\times 2^\frac{1}{3}=2^\frac{1}{3}(1+2^\frac{1}{3})$ But there I got stumped again, because while $1+2^\frac{1}{3}$ is irrational, nothing promises me that $2^\frac{1}{3} \times (1+2^\frac{1}{3})$ is irrational, and I feel like trying to go further down this road is moot. So what am I missing (other than sleep and food)? What route should I take to prove this? Thanks in advance! marked as duplicate by Watson, Chinnapparaj R, user10354138, Brahadeesh, KReiserNov 26 '18 at 8:38 • – punctured dusk Apr 7 '15 at 12:22 • – Watson Nov 25 '18 at 19:45 • – Watson Nov 25 '18 at 21:45 • – Watson Jan 29 at 9:06 Note $$1 + \sqrt[3]{2} + \sqrt[3]{4} = \frac{(\sqrt[3]{2})^3 - 1}{\sqrt[3]{2} - 1} = \frac{1}{\sqrt[3]{2} - 1}.$$ So if $\sqrt[3]{2} + \sqrt[3]{4}$ is rational, then $1/(\sqrt[3]{2} - 1)$ is rational, which implies $\sqrt[3]{2} - 1$ is rational. Then $\sqrt[3]{2}$ is rational, a contradiction. • Very nice answer. +1 – Timbuc Mar 26 '15 at 18:00 • Thanks for the reply. I must be tired or confused because I'm missing something very basic and don't realize how $1 + \sqrt[3]{2} + \sqrt[3]{4} = \frac{(\sqrt[3]{2})^3 - 1}{\sqrt[3]{2} - 1}$. I apologize for my stupidity - it's been a tough day. – Elad Avron Mar 26 '15 at 18:01 • Let $x = \sqrt[3]{2}$. Use the factorization $(1 + x + x^2)(x - 1) = x^3 - 1$ to obtain the identity. – kobe Mar 26 '15 at 18:04 • Of course, silly me. Thank you so much, this solution is brilliant! – Elad Avron Mar 26 '15 at 18:09 • @Elad One easly proves a much more general result: for irrational cube roots of rationals, said property fails only for $\,\sqrt[3]1,\,$ see my answer. $\ \$ – Bill Dubuque Mar 26 '15 at 20:21 $y = \sqrt[3]{2} + \sqrt[3]{4}$ is a root of the equation $y^3 - 6y - 6 = 0$. (To see this, let $x = \sqrt[3]{2}$ and $y = \sqrt[3]{2}+\sqrt[3]{4}=x+x^2$. Then $y^3 = x^3 + 3x^4 + 3x^5 + x^6 = 2 + 6x + 6x^2 + 4 = 6 + 6y$.) By the rational root theorem, we know that any rational roots of $y^3 - 6y - 6 = 0$ would have to be in the set $\{\pm1,\pm2,\pm3,\pm6\}$; we can quickly confirm that none of these is in fact a root, meaning that the equation does not have any rational roots. Therefore, $y$ must be irrational. • Short, sweet, and intuitive. Very nice. – Jon Mar 26 '15 at 21:16 An easy approach: If $p=\sqrt[3]{2}+\sqrt[3]{4}$ is rational: $$p^2 = \sqrt[3]{4}+2\cdot 2+ 2\sqrt[3]{2} = p+4+\sqrt[3]{2}.$$ So $\sqrt[3]{2}=p^2-p-4$ would be rational. An alternative, more general approach. Claim: If $a,b$ are integers that are not perfect cubes, and $a\neq -b$, then $\sqrt[3]{a}+\sqrt[3]b$ is irrational. Proof: Assume $\sqrt[3]{a}+\sqrt[3]{b}$ is rational. Cube it and get: $$(\sqrt[3]{a}+\sqrt[3]{b})^3 = a+ 3\sqrt[3]{ab}(\sqrt[3]{a}+\sqrt[3]{b}) + b$$ Now, since $a,b$ are rational and $\sqrt[3]{a}+\sqrt[3]{b}$ is non-zero and rational, this means that $\sqrt[3]{ab}$ is rational. Letting $p=\sqrt[3]{a}+\sqrt[3]{b}$ and $q=\sqrt[3]{ab}$, this means that $$(x-\sqrt[3]{a})(x-\sqrt[3]{b}) = x^2-px+q$$ is a rational polynomial. It shares at least one root with $x^3-b$, But the GCD of these two polynomials has to be a rational polynomial, so the GCD cannot be linear (since it would be $x-\sqrt[3]b$, which is not a rational polynomial.) This means that $x^2-px+q$ has to divide $x^3-b$. That means that $\sqrt[3]{a}$ is a root of $x^3-b$, which means that $a=b$. But there are no repeated roots of $x^3-b$, which yields a contradiction. Corollary: If $a,b$ are rationals such that $a\neq -b$ and $\sqrt[3]{a}$ and $\sqrt[3]{b}$ are irrational, then $\sqrt[3]{a}+\sqrt[3]{b}$ is irrational. Proof: Rationalize the denominators and revert to the above theorem for integers. It holds true for any irrational $\,x=\sqrt[3]n,\,$ except $\,n=1\,$ (so $\,x^2+x=-1\in\Bbb Q)$ More generally: if $\ r\in\Bbb Q\$ and $\,x=\sqrt[3]r\not\in\Bbb Q\,$ then $\,x^2+x = q\in\Bbb Q\iff r = 1.$ Proof $\,\ qx = x^3+x^2 = r+x^2\$ so $\ qx-r = x^2 = q-x,\$ so $\,(\color{#c00}{q\!+\!1})\,x = r+q.$ Therefore $\,\ x\not\in\Bbb Q\,\Rightarrow\,\color{#c00}{q = -1}\,\Rightarrow\, 0 = (\color{#c00}{q\!+\!1})x = r+q = r-1,\$ thus $\,\ r = 1.$ • Um, if $n=1$, isn't it $x^2+x=2$? And is $x=\sqrt[3]{1}$ irrational? I'm confused. – Thomas Andrews Mar 29 '15 at 4:54 • @Thomas The hypothesis is that $\,x\,$ is an irrational cube root of $\,n.\ \$ – Bill Dubuque Mar 29 '15 at 6:41 • Well, $\sqrt[3]{n}$ is a single-valued function, by convention. You could say any irrational $x$ with $x^3=n$, I suppose. – Thomas Andrews Mar 29 '15 at 15:25 • @Thomas Algebraists not too infrequently overload the notation to denote any root. I will change it to avoid possible confusion. Thanks for pointing that out. – Bill Dubuque Mar 29 '15 at 16:25 Here is another approach which generalises to many similar examples, and which involves no complicated simplifications of surds. (In fact, no easy simplifications of surds either.) First, $\sqrt[3]2$ is an algebraic integer, that is, it is a root of $$x^3-2$$ which is a monic polynomial (leading coefficient $1$) with integer coefficients. Similarly, $\sqrt[3]4$ is an algebraic integer. It is known that the sum of two algebraic integers is an algebraic integer. Thus, $x=\sqrt[3]2+\sqrt[3]4$ is an algebraic integer. It is also known that if an algebraic integer is rational, then it is a rational integer - that is, an ordinary integer, $0,1,-1,2,-2$ etc. However, it is not hard to find the estimates $$1<\sqrt[3]2<\frac43 ,\quad \frac43<\sqrt[3]4<\frac53\ ;$$ so $\frac73<x<3$, hence $x$ is not an integer and must be irrational. If $a$ is rational, $a^2-a-4$ is rational. What is $a^2-a-4$ when $a=\sqrt[3]2+\sqrt[3]4$? (After you work that out, I bet you'll wonder where I got $a^2-a-4$ from. Or, try to figure it out yourself. Now that you have a sort of idea on how to deal with this sort of problem, I leave you with another, similar problem: What polynomial can I use to prove that $\sqrt[3]3+\sqrt[3]9$ is irrational?) • I was going to suggest proving $\sqrt[3]2+\sqrt[3]3$ irrational, but then I realized that the polynomial involved is an eight-degree one. And I'm not sure how hard the one I gave is; I didn't try it yet. – Akiva Weinberger Mar 26 '15 at 20:49	2019-06-17T13:25:31	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1207829/how-to-prove-that-sqrt3-2-sqrt3-4-is-irrational", "openwebmath_score": 0.86919105052948, "openwebmath_perplexity": 328.9215594630196, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429634078179, "lm_q2_score": 0.8633916064586998, "lm_q1q2_score": 0.8490100608765344 }
https://math.stackexchange.com/questions/3307711/whats-my-mistake-in-finding-int-0-infty-dx-e-ax2-sinb-x2	# What's my mistake in finding $\int_0^\infty dx e^{-ax^2} \sin(b/x^2)$? I want to evaluate $$I=\int_0^\infty dx e^{-ax^2} \sin(b/x^2)$$ for $$a,b>0$$. A first simplification is to substitute $$y=x/\sqrt{a}$$ and define $$c=ab>0$$ to obtain $$I=\frac{1}{\sqrt{a}} \int_0^\infty e^{-x^2} \sin(c/x^2)$$ Now my idea was to use the Taylor series for the sine $$I=a^{-1/2} \int_0^\infty dx e^{-x^2} \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}\left(\frac{c}{x^2} \right)^{2k+1}$$ Now I interchange sum and integral although I have no justification $$I=a^{-1/2} \sum_{k=0}^\infty \frac{(-1)^kc^{2k+1}}{(2k+1)!} \int_0^\infty dx e^{-x^2} \left(\frac{1}{x^2} \right)^{2k+1}$$ Substituting $$t=x^2$$ in the integral we obtain the gamma function $$I=a^{-1/2} \sum_{k=0}^\infty \frac{(-1)^kc^{2k+1}}{(2k+1)!} \frac{1}{2} \Gamma(-2k-1/2)$$ Using $$\Gamma \left({\frac{1}{2}}-n\right)={(-4)^{n}n! \over (2n)!}{\sqrt {\pi }}$$ (which can be shown using the reflection formula and the duplication formula for the Gamma function) with $$n=2k+1$$ I obtain $$I=a^{-1/2} \sqrt{\pi} \sum_{k=0}^\infty \frac{(-1)^k(-4c)^{2k+1}}{(4k+2)!} \frac{1}{2}$$ or $$I=-\frac{1}{2}\sqrt{\frac{\pi}{a}} \sum_{k=0}^\infty \frac{(-1)^k(4c)^{2k+1}}{(4k+2)!}=-\frac{1}{2}\sqrt{\frac{\pi}{a}} \sin(\sqrt{2c}) \sinh(\sqrt{2c})$$ where I used wolfram alpha for the last series. The problem: The above result is wrong. It should be (wolfram alpha and Gradshteyn) $$I=\frac{1}{2}\sqrt{\frac{\pi}{a}} \sin(\sqrt{2c})\exp(-\sqrt{2c})$$ The question: Can someone spot my mistake? Was it interchanging the limits? I would also be interested in your solutions to the integral $$I$$ using other approaches. • After the interchange, all the integrals diverge at $0$, aren't they? – user58697 Jul 29 at 19:58 • @user58697 Yes, you are right. Probably that is the mistake... Do you have an idea for an alternative approach to the original integral? – thomasfermi Jul 29 at 20:09 The Glasser's master theorem is a useful tool for the solution. First, use Euler's formula to decompose the sine term into the sum of exponentials. Then it boils down to computing the integral of the form $$J(p) = \int_{0}^{\infty} \exp\left( -a x^2 - \frac{p}{x^2} \right) \, \mathrm{d}x.$$ Assume for a moment that $$a, p > 0$$. Then by completing the square, we get $$J(p) = \int_{0}^{\infty} \exp\left( -a \left( x - \frac{\smash{\sqrt{p/a}}}{x} \right)^2 - 2\sqrt{ap} \right) \, \mathrm{d}x.$$ Then by the Glasser's master theorem and the gaussian integral, this evaluates to $$J(p) = \int_{0}^{\infty} \exp\left( -a x^2 - 2\sqrt{ap} \right) \, \mathrm{d}x = \frac{1}{2}\sqrt{\frac{\pi}{a}} \exp(-2\sqrt{ap}). \tag{}$$ Although $$\text{()}$$ is originally proved for $$p > 0$$, both sides of $$\text{()}$$ define holomorphic functions for $$p$$ in the right-half plane $$\mathbb{H}_{\to} = \{z \in \mathbb{C} : \operatorname{Re}(z) > 0\}$$ and are continuous on the closed right-half plane $$\overline{\mathbb{H}_{\to}}$$. So by the identity theorem and continuity, $$\text{()}$$ extends to all of $$p \in \overline{\mathbb{H}_{\to}}$$. In particular, plugging $$p = \pm ib$$ for $$b > 0$$, we get $$J(\pm ib) = \frac{1}{2}\sqrt{\frac{\pi}{a}} \exp(-2\sqrt{\pm i ab}) = \frac{1}{2}\sqrt{\frac{\pi}{a}} \exp(-\sqrt{2c}(1\pm i)).$$ Therefore $$I = \frac{J(-ib) - J(ib)}{2i} = \frac{1}{2}\sqrt{\frac{\pi}{a}} \sin(\sqrt{2c}) \exp(-\sqrt{2c}).$$ • Thanks for your very nice solution. I had never heard of Glasser's theorem before! – thomasfermi Jul 30 at 22:03	2019-09-19T18:55:44	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3307711/whats-my-mistake-in-finding-int-0-infty-dx-e-ax2-sinb-x2", "openwebmath_score": 0.9730320572853088, "openwebmath_perplexity": 142.62356573946033, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.983342958526322, "lm_q2_score": 0.8633916082162403, "lm_q1q2_score": 0.8490100583901569 }
http://mathhelpforum.com/algebra/86584-help-finding-sum-series-print.html	# help finding sum of series Printable View • April 30th 2009, 01:11 AM adhyeta help finding sum of series got 2 series. 1. 12 - 22 + 32 - 42 + ................. 992 - 1002 (ok those are just squares) 2. 1.1! + 2.2! + ....... 50.50! need to find sum... • April 30th 2009, 03:07 AM pickslides Quote: Originally Posted by adhyeta got 2 series. 1. 12 - 22 + 32 - 42 + ................. 992 - 1002 (ok those are just squares) need to find sum... $1^2-2^2+3^2-4^2+\cdots+99^2-100^2$ $1-4+9-16+25-36+\cdots+99^2-100^2$ finding the difference in pairs now giving 50 terms. $-3-7-11-\cdots$ Use sum of an arithmetic sequence $S_n = \frac{n}{2}(2a+(n-1)d)$ where n = number of terms = 50 a = the first term = -3 and d = the common difference between terms = -4 $S_{50} = \frac{50}{2}(2(-3)+(50-1)(-4)) = \cdots$ • April 30th 2009, 04:22 AM Soroban Hello, adhyeta! Are you familiar with these summation formulas? . . $\sum^n_{k=1} 1 \;\;= \;n$ . . $\sum^n_{k=1}k \;\;=\;\frac{n(n+1)}{2}$ Quote: $1)\;\;S \;=\;1^2 - 2^2 + 3^2 - 4^2 + \hdots + 99^2 - 100^2$ $\text{We have: }\;S \;=\;\underbrace{(1^2 + 3^2 + 5^2 + \hdots + 99^2)}_{\text{first 50 odd squares}} \;-\; \underbrace{(2^2+4^2+6^2 + \hdots + 100^2)}_{\text{first 50 even squares}}$ . . $\text{Then: }S \;=\qquad\qquad \sum^{50}_{k=1}(2k-1)^2 \qquad - \qquad\qquad \sum^{50}_{k=1}(2k)^2$ Hence: . $S \;=\;\sum^{50}_{k=1}\bigg[(2k-1)^2 - (2k)^2\bigg] \;=\;\sum^{50}_{k=1}(1-4k)$ . . $= \;\;\sum^{50}_{k=1}\!1 \;-\; 4\!\sum^{50}_{k=1}k \;\;=\;\;50 - 4\,\frac{50\cdot51}{2} \;\;=\;\;\boxed{-5,\!050}$ • April 30th 2009, 06:10 AM adhyeta Quote: Originally Posted by pickslides $1^2-2^2+3^2-4^2+\cdots+99^2-100^2$ $1-4+9-16+25-36+\cdots+99^2-100^2$ finding the difference in pairs now giving 50 terms. $-3-7-11-\cdots$ Use sum of an arithmetic sequence $S_n = \frac{n}{2}(2a+(n-1)d)$ where n = number of terms = 50 a = the first term = -3 and d = the common difference between terms = -4 $S_{50} = \frac{50}{2}(2(-3)+(50-1)(-4)) = \cdots$ thats not coming out to be correct...its -5050. refer to soroban's solution. • April 30th 2009, 02:11 PM pickslides Quote: Originally Posted by adhyeta thats not coming out to be correct...its -5050. refer to soroban's solution. $ S_{50} = \frac{50}{2}(2(-3)+(50-1)(-4))$ $ S_{50} = 25(-6+(49)(-4))$ $ S_{50} = 25(-6+-196)$ $ S_{50} = 25(-202)$ $ S_{50} = -5050$ Worked for me! • April 30th 2009, 03:04 PM Jester Quote: Originally Posted by adhyeta got 2 series. 1. 12 - 22 + 32 - 42 + ................. 992 - 1002 (ok those are just squares) 2. 1.1! + 2.2! + ....... 50.50! need to find sum... Here's the second one. Let $S_n = 1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + \cdots + (n-1) \cdot (n-1)! + n \cdot n!$ $= (2-1) 1! + (3-1)2! + (4-1)3! + \cdots + (n -1) \cdot (n-1)! + (n+1 -1) \cdot n! $ $= (2! - 1!) + (3! - 2!) + (4! - 3!) + \cdots + (n! - (n-1)!) + ((n+1)! - n!)$ $ = (n+1)! - 1 $ so here $n = 50$ so the answer $S = 51!-1$. • April 30th 2009, 05:16 PM adhyeta Quote: Originally Posted by pickslides $ S_{50} = \frac{50}{2}(2(-3)+(50-1)(-4))$ $ S_{50} = 25(-6+(49)(-4))$ $ S_{50} = 25(-6+-196)$ $ S_{50} = 25(-202)$ $ S_{50} = -5050$ Worked for me! hey! so sorry. i think i went wrong with the calc.(-calculation-) (Happy) • May 3rd 2009, 07:16 AM adhyeta Quote: Originally Posted by danny arrigo Here's the second one. Let $S_n = 1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + \cdots + (n-1) \cdot (n-1)! + n \cdot n!$ $= (2-1) 1! + (3-1)2! + (4-1)3! + \cdots + (n -1) \cdot (n-1)! + (n+1 -1) \cdot n! $ $= (2! - 1!) + (3! - 2!) + (4! - 3!) + \cdots + (n! - (n-1)!) + ((n+1)! - n!)$ $ = (n+1)! - 1 $ so here $n = 50$ so the answer $S = 51!-1$. hey! can we similarily prove the sum $n^{2}.n!$???	2016-05-24T09:01:45	{ "domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/86584-help-finding-sum-series-print.html", "openwebmath_score": 0.7469117641448975, "openwebmath_perplexity": 5459.556455659678, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9752018441287092, "lm_q2_score": 0.8705972818382005, "lm_q1q2_score": 0.8490080747420548 }
https://dsp.stackexchange.com/questions/60939/calculating-amplitude-time-shift-and-phase-of-third-sinusoid-as-a-sum-of-two-ot	Calculating amplitude, time-shift and phase of third sinusoid as a sum of two others My apologies in advance, I am new to this digital signal processing. Question is at the bottom I have two 4000hz sinusoids given by the following formulas: unknown frequency such that: $$x_1(t) = 24\cos(2\pi4000(t-t_{m1})) \\ x_2(t) = 28.8 \cos(2\pi4000(t-t_{m2}))$$ and a third given by $$x_3(t) = x_1(t) + x_2(t)$$ where $$t_{m1} = 0.000775 \\ t_{m2} = -0.00258$$ I have calculated the phase of the first two sinusoids (in radians) as $$\phi_1 = 0.329 \\ \phi_2 = -1.132$$ with the formula $$\phi_i = \omega_if_i$$ where $$\omega_i = 2\pi f$$ Finally, I need to calculate the amplitude $$A_3$$ and phase $$\phi_3$$ of $$x_3(t)$$. I am unsure which formula to use to do so. I believe $$A_3$$ is calculated as: $$a_3 = \sqrt{A_1^2 + A_2^2} = 37.49$$ How would I calculate the time shift and phase angle of $$x_3(t)$$? • Are you familiar with complex numbers ? – Hilmar Sep 27 '19 at 18:38 • @hilmar I am, but not with respect to this topic. – kp-a Sep 27 '19 at 20:34 Here's how I would do it. Use the Cosine angle addition formula on $$x_1$$ and $$x_2$$: $$\cos( \alpha + \beta) = \cos( \alpha ) \cos(\beta)-\sin( \alpha ) \sin(\beta)$$ Like this: $$x_1(t) = a_1 \cos( \omega t + \phi_1 )$$ $$x_1 = a_1 \cos( \omega t ) \cos(\phi_1)- a_1 \sin( \omega t ) \sin(\phi_1)$$ The $$\omega$$s will be the same, so: $$x_2 = a_2 \cos( \omega t ) \cos(\phi_2)- a_2 \sin( \omega t ) \sin(\phi_2)$$ You can now rearrange and add them together: \begin{align} x_3 &= x_1 + x_2 \\ &= \left[a_1 \cos(\phi_1) + a_2 \cos(\phi_2) \right] \cos( \omega t ) - \left[a_1 \sin(\phi_1) + a_2 \sin(\phi_2) \right] \sin( \omega t ) \end{align} Since you know the $$\phi$$s and $$a$$s, you can put this in the form: $$x_3 = C \cos( \omega t ) - D \sin( \omega t )$$ Now, you just have to reverse the process to get it back into time phase form. $$\theta = \operatorname{atan2}(D,C)$$ $$M = \sqrt{ C^2 + D^2 }$$ Therefore: $$x_3 = M \cos( \omega t + \theta )$$ $$x_3 = M \cos\left( \omega \left( t + \frac{\theta}{\omega} \right) \right)$$ Now, just plug and chug and you should have it. The latter part might be better understood in a different order: \begin{align} x_3(t) &= a_3 \cos( \omega t + \phi_3 )\\ &= a_3 \cos( \omega t ) \cos(\phi_3)- a_3 \sin( \omega t ) \sin(\phi_3) \end{align} From the equations above: \begin{align} C &= a_3 \cos(\phi_3)\\ D &= a_3 \sin(\phi_3) \end{align} From there it follows: $$a_3 = \sqrt{C^2+D^2}$$ and $$\phi_3 = \tan^{-1}\left(\frac{D}{C}\right)$$ The thing is, both of these last two equations have two possible solutions each and they need to be matched. Conveniently, the atan2 function is available on most platforms which yields the correct angle from the latter for the positive root of the former. This solution is actually the proof of a very important principle: When two pure tones of the same frequency are added together the result is a pure tone of the same frequency although it is possible for it to have zero amplitude. In order for the zero amplitude case to occur, the two tones have to have the same amplitude and be a half cycle ($$\pi$$ radians, 180 degrees) out of phase. This is called complete destructive interference. If the two tones are in phase, the amplitudes are merely added and the phase remains the same. This is technically called complete constructive interference. By repeated application of this principle, the same holds true for any linear combination of tones. This shows up in a surprising number of places. a. Perhaps the discussion at: https://www.dsprelated.com/showarticle/635.php would be of some value to you. • You beat me to it, I remember reading this article a few years ago! – Ben Sep 28 '19 at 16:49 • Welcome back to this forum. Looking forward to seeing what you have to say in the next item in your answer (the one that will begin "b.") when you get around to adding it. – Dilip Sarwate Sep 28 '19 at 17:03 • Hi Dilip. (Hope you are doing well.) I don't know what you mean by "the next item in your answer (the one that will begin "b.")." – Richard Lyons Oct 2 '19 at 10:49 • @DilipSarwate, Hey Rick, you want to start your comment as I did if you want the little notification thingy to work. Your post begins: "a. Perhaps..". I, too, have been breathlessly awaiting what might be the "b." – Cedron Dawg Oct 2 '19 at 18:15 • @CedronDog, Hi Cedron. Ah, ...now I see. I don't remember typing the two 'a.' characters in my Answer. So my eyes didn't actually see them in my Answer. Thanks for helping me. In any case Mbaz can obtain his desired results by using the top row of Table 1 in my blog at my above posted web link, or he can use your equations. Happily, your and my equations produce identical results. – Richard Lyons Oct 3 '19 at 10:34	2020-02-18T13:01:20	{ "domain": "stackexchange.com", "url": "https://dsp.stackexchange.com/questions/60939/calculating-amplitude-time-shift-and-phase-of-third-sinusoid-as-a-sum-of-two-ot", "openwebmath_score": 0.974681556224823, "openwebmath_perplexity": 657.5972214430152, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9752018426872777, "lm_q2_score": 0.8705972801594706, "lm_q1q2_score": 0.8490080718500479 }
https://math.stackexchange.com/questions/3476881/finding-cosine-when-negating-its-angle-given-only-the-value-of-sine	# Finding cosine when negating its angle, given only the value of sine. If $$\sin(\theta) = \frac{1}{\sqrt{5}}$$, I must find $$\cos(-\theta)$$. Using a right triangle (hypotenuse is $$\sqrt{5}$$ and opposite is $$1$$), it is easy to find that the adjacent side is $$2$$. Then since $$\cos(-\theta) = \cos(\theta)$$ I can simply find the answer as $$\frac{2}{\sqrt{5}}$$. This is correct according to my homework. However, I was thinking, what if $$\theta$$ is in the second quadrant of the unit circle? Basically $$\pi > \theta > \pi/2$$. Then, visually speaking if you look at the unit circle, there is no way that $$\cos(\theta)$$ can be positive. It has to be negative because the angle ends up in the third quadrant when you negate it as $$-\theta$$. But my homework says that the only answer is the positive version, so I'm not sure what am I doing wrong. • If you are only given that $\sin(\theta)=1/\sqrt 5$, there are two possible values for $\cos(\theta)$. It is possible that the answers are incomplete and/or the teacher did not think about negative values. – D.R. Dec 15 '19 at 5:10 I think you have put more good thought into this than the author of your homework answers did, and I agree with your conclusion. Representing the angle $$\theta$$ on the unit circle, we have one end of the angle at $$(1,0)$$ and the other end at any point where $$y=\frac1{\sqrt5}$$, which is to say the other end can be at either $$\left(\frac2{\sqrt5},\frac1{\sqrt5}\right)$$ or $$\left(-\frac2{\sqrt5},\frac1{\sqrt5}\right)$$. There are infinitely many angles with sine $$\frac1{\sqrt5}$$, since you can go around the circle as many times as you like in either direction to reach one of those two points, but you must end at one of those two points. Then the other end of the angle $$-\theta$$ is at $$\left(\frac2{\sqrt5},-\frac1{\sqrt5}\right)$$ or $$\left(-\frac2{\sqrt5},-\frac1{\sqrt5}\right)$$. In the end there are two possible values for $$\cos(-\theta)$$ found by taking the $$x$$ coordinates of the last two points: $$\frac2{\sqrt5}$$ and $$-\frac2{\sqrt5}.$$ (Or you could realize that $$\cos(-\theta)=\cos(\theta)$$ and therefore when you found the endpoints of the angle $$\theta$$ you already had the answer.) But if someone did not think about it as much as you did, they might not think about the answer $$-\frac2{\sqrt5}.$$ I guess that's what happened to the author of your homework answers. A big reason why I was so explicit about using the unit circle in this answer is that it encourages thinking about all the quadrants, whereas the right-triangle definition of sine and cosine really only makes sense in the first quadrant.	2020-01-17T19:39:04	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3476881/finding-cosine-when-negating-its-angle-given-only-the-value-of-sine", "openwebmath_score": 0.8931905627250671, "openwebmath_perplexity": 113.60688789098458, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9752018426872776, "lm_q2_score": 0.8705972700870909, "lm_q1q2_score": 0.8490080620274445 }
https://www.quizover.com/course/section/decomposing-a-composite-function-into-its-component-by-openstax	# 3.4 Composition of functions (Page 6/9) Page 6 / 9 ## Finding the domain of a composite function involving radicals Find the domain of Because we cannot take the square root of a negative number, the domain of $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}\left(-\infty ,3\right].\text{\hspace{0.17em}}$ Now we check the domain of the composite function The domain of this function is $\text{\hspace{0.17em}}\left(-\infty ,5\right].\text{\hspace{0.17em}}$ To find the domain of $\text{\hspace{0.17em}}f\circ g,\text{\hspace{0.17em}}$ we ask ourselves if there are any further restrictions offered by the domain of the composite function. The answer is no, since $\text{\hspace{0.17em}}\left(-\infty ,3\right]\text{\hspace{0.17em}}$ is a proper subset of the domain of $\text{\hspace{0.17em}}f\circ g.\text{\hspace{0.17em}}$ This means the domain of $\text{\hspace{0.17em}}f\circ g\text{\hspace{0.17em}}$ is the same as the domain of $\text{\hspace{0.17em}}g,\text{\hspace{0.17em}}$ namely, $\text{\hspace{0.17em}}\left(-\infty ,3\right].$ Find the domain of $\left[-4,0\right)\cup \left(0,\infty \right)$ ## Decomposing a composite function into its component functions In some cases, it is necessary to decompose a complicated function. In other words, we can write it as a composition of two simpler functions. There may be more than one way to decompose a composite function , so we may choose the decomposition that appears to be most expedient. ## Decomposing a function Write $\text{\hspace{0.17em}}f\left(x\right)=\sqrt{5-{x}^{2}}\text{\hspace{0.17em}}$ as the composition of two functions. We are looking for two functions, $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}h,\text{\hspace{0.17em}}$ so $\text{\hspace{0.17em}}f\left(x\right)=g\left(h\left(x\right)\right).\text{\hspace{0.17em}}$ To do this, we look for a function inside a function in the formula for $\text{\hspace{0.17em}}f\left(x\right).\text{\hspace{0.17em}}$ As one possibility, we might notice that the expression $\text{\hspace{0.17em}}5-{x}^{2}\text{\hspace{0.17em}}$ is the inside of the square root. We could then decompose the function as We can check our answer by recomposing the functions. $g\left(h\left(x\right)\right)=g\left(5-{x}^{2}\right)=\sqrt{5-{x}^{2}}$ Write $\text{\hspace{0.17em}}f\left(x\right)=\frac{4}{3-\sqrt{4+{x}^{2}}}\text{\hspace{0.17em}}$ as the composition of two functions. $\begin{array}{l}g\left(x\right)=\sqrt{4+{x}^{2}}\\ h\left(x\right)=\frac{4}{3-x}\\ f=h\circ g\end{array}$ Access these online resources for additional instruction and practice with composite functions. ## Key equation Composite function $\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$ ## Key concepts • We can perform algebraic operations on functions. See [link] . • When functions are combined, the output of the first (inner) function becomes the input of the second (outer) function. • The function produced by combining two functions is a composite function. See [link] and [link] . • The order of function composition must be considered when interpreting the meaning of composite functions. See [link] . • A composite function can be evaluated by evaluating the inner function using the given input value and then evaluating the outer function taking as its input the output of the inner function. • A composite function can be evaluated from a table. See [link] . • A composite function can be evaluated from a graph. See [link] . • A composite function can be evaluated from a formula. See [link] . • The domain of a composite function consists of those inputs in the domain of the inner function that correspond to outputs of the inner function that are in the domain of the outer function. See [link] and [link] . • Just as functions can be combined to form a composite function, composite functions can be decomposed into simpler functions. • Functions can often be decomposed in more than one way. See [link] . ## Verbal How does one find the domain of the quotient of two functions, $\text{\hspace{0.17em}}\frac{f}{g}?\text{\hspace{0.17em}}$ Find the numbers that make the function in the denominator $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ equal to zero, and check for any other domain restrictions on $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}g,\text{\hspace{0.17em}}$ such as an even-indexed root or zeros in the denominator. why {2kπ} union {kπ}={kπ}? why is {2kπ} union {kπ}={kπ}? when k belong to integer Huy if 9 sin theta + 40 cos theta = 41,prove that:41 cos theta = 41 what is complex numbers give me treganamentry question Solve 2cos x + 3sin x = 0.5 madras university algebra questions papers first year B. SC. maths Hey Rightspect hi chesky Give me algebra questions Rightspect how to send you Vandna What does this mean cos(x+iy)=cos alpha+isinalpha prove that: sin⁴x=sin²alpha cos(x+iy)=cos aplha+i sinalpha prove that: sinh⁴y=sin²alpha rajan cos(x+iy)=cos aplha+i sinalpha prove that: sinh⁴y=sin²alpha rajan is there any case that you can have a polynomials with a degree of four? victor ***sscc.edu/home/jdavidso/math/catalog/polynomials/fourth/fourth.html Oliver can you solve it step b step give me some important question in tregnamentry Anshuman what is linear equation with one unknown 2x+5=3 -4 Joel x=-4 Joel x=-1 Joan I was wrong. I didn't move all constants to the right of the equation. Joel x=-1 Cristian y=x+1 gary what is the VA Ha D R X int Y int of f(x) =x²+4x+4/x+2 f(x) =x³-1/x-1 can I get help with this? Wayne Are they two separate problems or are the two functions a system? Wilson Also, is the first x squared in "x+4x+4" Wilson x^2+4x+4? Wilson thank you Wilson Wilson f(x)=x square-root 2 +2x+1 how to solve this value Wilson what is algebra The product of two is 32. Find a function that represents the sum of their squares. Paul if theta =30degree so COS2 theta = 1- 10 square theta upon 1 + tan squared theta how to compute this 1. g(1-x) 2. f(x-2) 3. g (-x-/5) 4. f (x)- g (x) hi John hi Grace what sup friend John not much For functions, there are two conditions for a function to be the inverse function: 1--- g(f(x)) = x for all x in the domain of f 2---f(g(x)) = x for all x in the domain of g Notice in both cases you will get back to the element that you started with, namely, x. Grace	2018-09-22T22:29:09	{ "domain": "quizover.com", "url": "https://www.quizover.com/course/section/decomposing-a-composite-function-into-its-component-by-openstax", "openwebmath_score": 0.908155083656311, "openwebmath_perplexity": 678.3957349348644, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9752018419665619, "lm_q2_score": 0.8705972684083609, "lm_q1q2_score": 0.8490080597628908 }
https://math.stackexchange.com/questions/3524292/how-is-the-inductive-hypothesis-in-strong-mathematical-induction-different-from	# How is the inductive hypothesis in strong mathematical induction different from that in ordinary induction? I don't understand how supposing that $$P(k), k\geq 1$$ for ordinary induction is different from $$P(i), 1 \leq i \leq k, k\geq1$$ for strong induction. Example from quora: Let’s say you wanted to prove that every positive integer has a prime factorization $$𝑝_1𝑝_2𝑝_3...𝑝_𝑚$$. Let 𝑃(𝑛) be the statement that an integer 𝑛 has a prime factorization. We’ll proceed by strong induction. The basis is pretty clear, so I’ll leave it out. Next we’ll assume that 𝑃(1),𝑃(2),𝑃(3),...,𝑃(𝑘) are true. 𝑘+1 can either be prime or composite, and if it’s prime we’re done, so we’ll assume it’s composite. That means 𝑘+1 can be written as a product of two positive integers, i.e. 𝑘+1=𝑝𝑞, with $$𝑝,𝑞∈ℤ^+$$. We can write 1<𝑝<𝑘+1, and 1<𝑞<𝑘+1, which implies that 2≤𝑝≤𝑘 and 2≤𝑞≤𝑘. Here is why we need strong induction: if we had simply supposed 𝑃(𝑛) was true for arbitrary 𝑛, we would be stuck. However, we supposed that 𝑃(𝑛) was true for every positive integer up to 𝑛=𝑘, so we have much more information to work with. Because we supposed this, we know that 𝑃(𝑝) and 𝑃(𝑞) are true, i.e. that 𝑝 and 𝑞 can be represented as a product of primes. We were able to reduce the problem down to a point where 𝑝 and 𝑞 were in a range, and since our inductive hypothesis in strong induction supposes that 𝑃(𝑛) is true for a range of values (rather than just one arbitrary 𝑛), we can now use it to prove the truth of 𝑃(𝑘+1). Using ordinary induction, I'd say that $$P(p)$$ and $$P(q)$$ are true because $$2≤𝑝≤𝑘$$ and $$2≤𝑞≤𝑘$$ and $$P(k), k\geq 1$$. Why can I not use ordinary induction here? Another example is the proof that McCarthy 91 function equals 91 for all positive integers less than or equal to 101. The property is $$P(n)=M(101-n), n \geq 0$$ and $$M(n)$$ is the McCarthy function. The author of the proof calculates the base case for $$P(0)$$, then does a supposition that $$P(i), 0 \leq i \leq k, k \geq 0$$. The use of strong induction is justified by the fact that we need the inductive hypothesis to hold for $$k-10$$, but I don't see why $$P(n), n\geq0$$ wouldn't hold for $$n=k-10, k\geq11$$, that is $$n$$ is at least 1, if ordinary induction was used. • For ordinary induction, you wouldn't assume $P(p)$ and $P(q)$. When proving the theorem for $k+1$, you would assume $P(k)$, and only $P(k)$ (otherwise you are doing strong induction again). And that doesn't help you at all for the prime numbers. – Dirk Jan 27 '20 at 11:40 • @Dirk Technically, you can do it with weak induction. Weak and strong are probably equivalent. However, in this case, using weak looks like a complete mess. – Arthur Jan 27 '20 at 11:45 • @Dirk right, in weak induction I make an assumption that the property holds only for the term immediately before (k+1) and prove that it holds for (k+1). p and q from the proof above do not necessary equal k, therefore I can't assume that P(q) and P(p) are true, and that's where strong induction helps with its "extended" assumption. Is it correct? – super.t Jan 27 '20 at 14:25 Ordinary or weak induction proves $$Q(n)$$ for all $$n\ge1$$ with a base step $$n=1$$ and an inductive step from $$n=k$$ to $$n=k+1$$. Complete or strong induction considers the special case where $$Q(n)$$ denotes "$$P(k)$$ for all $$k$$ from $$1$$ to $$n-1$$ inclusive". If we try to prove $$Q(n)$$ for all $$n\ge1$$ by weak induction, the base step is vacuously true, and the inductive step is showing that, if $$P(k)$$ for all $$k$$ from $$1$$ to $$n-1$$ inclusive, then $$P(n)$$. If we can prove this statement, the weak induction on $$Q$$ succeeds, and we have also proven $$P(n)$$ for all $$n\ge1$$. In other words, strong induction states this: if "$$P(k)$$ for all $$k$$ from $$1$$ to $$n-1$$" implies $$P(n)$$, then $$P(n)$$ for all $$n\ge1$$. Usually, the $$n-1$$ is called $$n$$ instead, so we need to prove "$$P(k)$$ for all $$k$$ from $$1$$ to $$n$$" implies $$P(n+1)$$. Unlike weak induction, strong induction does not in general need a base step. However, in some cases the argument proving its inductive step has to consider small values of $$n$$ as special cases.	2021-04-20T20:59:32	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3524292/how-is-the-inductive-hypothesis-in-strong-mathematical-induction-different-from", "openwebmath_score": 0.8786411881446838, "openwebmath_perplexity": 201.8362834445597, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9752018376422666, "lm_q2_score": 0.8705972684083609, "lm_q1q2_score": 0.8490080559981712 }
https://math.stackexchange.com/questions/1974188/surjections-using-stirling-numbers	# Surjections using Stirling-numbers I have a task in which I can't really see what I'm doing wrong. The recommended answer differs from mine and it kind of makes sense to me, but I can't see what's wrong with my solution. The problem is as follows; In a family of 4 all the 7 deadly sins are present. Every family member exercises(?) atleast 1 of the 7 sins, but there are no sins that two distinct family members 'exercise'. (The original problem formulation is not in english, sorry) The task is to find out how many combinations there are if no single family member can exercise both greed and gluttony. Basically what I did was to find out the total amount of combinations without taking the constraint into consideration (Surjection from a 7-subset to a 4-subset; $4! * S(7,4)$. After that I wanted to subtract the 'illegal' combinations. The way I figured I would find them out is as follows: There are 4 possible family members that can exercise both greed and gluttony. After that we have 5 additional sins to 'hand out' amongst all the remaining people (including the one that we already 'handed' both gluttony and greed) so I figure'd we'd then get the total amount of 'illegal' combinations as $4 * 4! * S(5,4)$ but in the recommended solution the amount of illegal combinations was $4! * S(6,4)$ - which I understand (I guess you 'group up' both gluttony and greed as one sin and count the surjections). But I don't understand why my solution is wrong. Any help greatly appreciated, sorry for the wall of text! • I don't think so ("But there are no sins that two distinc family members 'exercise') - shouldn't that point to the contrary? Maybe I translated it bad but it says in the text that one specific sin can't be 'exercised' by two family members :) – Nyfiken Gul Oct 18 '16 at 14:36 • Oops; wasn’t thinking clearly. But I do now see the problem: $S(5,4)$ counts only the partitions of the other five sins into four non-empty pieces, so your calculation misses the combinations in which one person has greed and gluttony and nothing else. – Brian M. Scott Oct 18 '16 at 14:38 • Aha! That actually makes perfect sense, didn't consider that at all. Thanks again Brian! ;-) I'll gladly mark it answered if you post your comment as an answer, you're the man! – Nyfiken Gul Oct 18 '16 at 14:44 The problem here is that $S(5,4)$ counts only partitions of the other $5$ sins into $4$ non-empty parts; when you then distribute the greed/gluttony pair to one of those parts, you ensure that the family member who gets that pair also gets at least one other sin. Thus, you’re missing all of the arrangements in which one family member gets greed and gluttony and nothing else.	2019-09-23T13:58:15	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1974188/surjections-using-stirling-numbers", "openwebmath_score": 0.6147923469543457, "openwebmath_perplexity": 335.8777038691595, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9752018369215507, "lm_q2_score": 0.8705972684083609, "lm_q1q2_score": 0.8490080553707179 }
https://math.stackexchange.com/questions/4246494/finding-the-gradient-of-the-restricted-function-in-terms-of-the-gradient-of-the/4258463#4258463	# Finding the gradient of the restricted function in terms of the gradient of the original function The following question showed up as part of a proof that I am doing for my research thesis. If we have a differentiable function $$f: \mathbb{R}^n \to \mathbb{R}$$ and then set $$n-d$$ coordinates to zero we get a new differentiable function $$g: \mathbb{R}^d \to \mathbb{R}$$. Now, given the gradient $$\nabla_x f(x)$$, how one can get $$\nabla_y g(y)$$? My try Let $$x \in \mathbb{R}^n$$ and $$S \subset \{1,\dots,n\}$$ such that $$\|S\|=d$$ where $$\|\cdot\|$$ is the cardinality of the set. Let $$U_S$$ be a restricted identity matrix such that the $$j$$-th entry of the diagonal matrix is maintained if $$j \in S$$ otherwise it is set to zero. Also, let $$I_S$$ be the restriction of $$U_S$$ where we keep nonzero columns and remove zero columns. Hence, $$g(y)=f(U_Sx)$$ where $$y=I_S^{\top}x$$. The above is the translation of what I stated in terms of functions $$f$$ and $$g$$. From this point things are a little bit unclear. I think the answer should be $$\nabla_y g(y)=I_S^{\top} \nabla_x f(x)$$ but I do not know how to get it. Also, I know using the chain rule $$J_x f(U_S x)=J_{W} f(W)J_x W= J_{W} f(W)U_S$$ where $$J$$ is the Jacobian and $$W=U_S x$$. In addition, $$\nabla^{\top}_x f(U_Sx) = J_x f(U_S x)=J_{W} f(W)U_S$$. I do not know how to put things together. Since no one has posted an answer yet, and I get the same result as you suggest, I thought I'll post my solution for you to judge: We have that $$U_S x = I_S y$$ so that, vieweing matrices as linear transformations $$g(y) = f(U_S x) = f(I_S y) = f\circ I_S (y)$$ And similar to what you write about $$J_{U_S}(x)$$ we have $$J_{I_S}(y) = I_S$$. Applying the chain rule: $$J_{h_1 \circ h_2}(a) = J_{h_1}(h_2(a))J_{h_2}(a)$$ then gives \begin{align} (\nabla_y g(y))^T = J_g(y) =\\ J_{f\circ I_S}(y) = \\ J_f(I_s y)J_{I_S}(y) = \\ J_f(U_s x)I_S = \\ (\nabla_x f(U_S x))^TI_S \implies \\ \nabla_y g(y) = [(\nabla_x f(U_S x))^TI_S]^T = I_S^T\nabla_x f(U_S x) \end{align} Due to the definitions of $$U_S$$ and $$I_S$$, the zero columns in $$I_S^T$$ exactly matches the rows where $$\nabla_x f(U_S x)$$ and $$\nabla_x f(x)$$ might differ, so finally we obtain $$\nabla_y g(y) = I_S^T\nabla_x f(U_S x) = I_S^T \nabla_x f(x)$$ Edit: As pointed out in the comments, it would be more correct to write $$\nabla_y g(y) = I_S^T\nabla_x f(I_S y)$$ • Could you make it a little more clear what's going on? Sep 17, 2021 at 10:49 • @Mathemagician314 My answer was a bit confused, I at least had some unnecessary steps. Are there any steps in particular you find strange? Sep 17, 2021 at 11:04 • @Paradox: it is not correct since $U_S\in \mathbb{R}^{n \times n}$ so $U_Sx$ is a vector in $\mathbb{R}^n$ and $I_sx \in \mathbb{R}^{d}$ Sep 18, 2021 at 16:48 • @Sepide I assume you mean $I_Sy \in \mathbb{R}^d$ since I've not written $I_S x$ anywhere. As far as I can see, $I_s$ is an $n \times d$ matrix, not $d \times n$, since it was the non-zero columns that was removed from $U_S$, not non-zero rows. So $I_s y \in \mathbb{R}^n$. Sep 18, 2021 at 16:56 Let fat matrix $${\bf S} \in \Bbb R^{d \times n}$$ be $${\bf S} := \begin{bmatrix} {\bf I}_d & {\bf O} \end{bmatrix} {\bf P}$$ where $${\bf P}$$ is an $$n \times n$$ permutation matrix. Note that $${\bf S} {\bf S}^\top = \begin{bmatrix} {\bf I}_d & {\bf O} \end{bmatrix} \underbrace{\,{\bf P} {\bf P}^\top}_{= {\bf I}_n} \begin{bmatrix} {\bf I}_d \\ {\bf O} \end{bmatrix} = {\bf I}_d$$ Let vector fields $$\rho : \Bbb R^n \to \Bbb R^d$$ and $$\eta : \Bbb R^d \to \Bbb R^n$$ be defined by $$\rho := ({\bf x} \mapsto {\bf S} {\bf x}), \qquad \eta := ({\bf y} \mapsto {\bf S}^\top {\bf y})$$ and note that $$\rho \circ \eta = \mbox{id}_{\Bbb R^d}$$. Colloquially, if one "expands" and then "restricts", one ends up exactly where one started. Given differentiable scalar field $$f : \Bbb R^n \to \Bbb R$$, let scalar field $$g : \Bbb R^d \to \Bbb R$$ be defined by $$g := f \circ \eta$$ Hence, \begin{aligned} g \left( {\bf y} + {\rm d} {\bf y} \right) = f \left( {\bf S}^\top {\bf y} + {\bf S}^\top {\rm d} {\bf y} \right) &= f \left( {\bf S}^\top {\bf y} \right) + \left\langle \nabla f \left( {\bf S}^\top {\bf y} \right) , {\bf S}^\top {\rm d} {\bf y} \right\rangle \\ &= f \left( {\bf S}^\top {\bf y} \right) + \left\langle {\bf S} \, \nabla f \left( {\bf S}^\top {\bf y} \right) , {\rm d} {\bf y} \right\rangle \end{aligned} and, thus, the gradient of $$g$$ is $$\nabla g \left( {\bf y} \right) = \color{blue}{{\bf S} \, \nabla f \left( {\bf S}^\top {\bf y} \right)}$$ or, more succinctly, $$\boxed{ \qquad \\ \qquad \nabla g = \color{blue}{\rho \circ \nabla f \circ \eta \qquad \\ \qquad}}$$ $$\def\p{\partial} \def\L{\left}\def\R{\right} \def\LR#1{\L(#1\R)} \def\trace#1{\operatorname{Tr}\LR{#1}} \def\qiq{\quad\implies\quad} \def\grad#1#2{\frac{\p #1}{\p #2}}$$For typing convenience, name the gradient \eqalign{ p = \grad{f}{x} \\ } and rename the matrices $$I_S\to S$$ and $$U_S\to U$$. Also note that $$\,U=SS^T\,$$ and that $$y=S^Tx \qiq dy=S^Tdx$$ Write the differential of the function and rearrange it to recover the desired gradient. \eqalign{ g(y) &= f(Ux) \\ dg &= df \\ &= p:d(Ux) \\ &= p:\LR{U\,dx} \\ &= p:\LR{SS^T\,dx} \\ &= \LR{S^Tp}:\LR{S^T\,dx} \\ &= \LR{S^Tp}:dy \\ \grad{g}{y} &= S^Tp \\ } In the preceding, a colon is used to denote the Frobenius product, which is a concise notation for the trace \eqalign{ A:B &= \sum_{i=1}^m\sum_{j=1}^n A_{ij}B_{ij} \;=\; \trace{A^TB} \\ A:A &= \big\\|A\big\\|^2_F \\ } The properties of the underlying trace function allow the terms in a Frobenius product to be rearranged in many different but equivalent ways, e.g. \eqalign{ A:B &= B:A \\ A:B &= A^T:B^T \\ C:AB &= CB^T:A = A^TC:B \\ }	2022-06-25T16:21:10	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4246494/finding-the-gradient-of-the-restricted-function-in-terms-of-the-gradient-of-the/4258463#4258463", "openwebmath_score": 1.0000100135803223, "openwebmath_perplexity": 357.7159520343961, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9752018426872776, "lm_q2_score": 0.8705972566572503, "lm_q1q2_score": 0.8490080489306393 }
https://math.stackexchange.com/questions/3100267/stability-of-tikhonov-regularization/3100284	# Stability of Tikhonov Regularization $$\underset{x\in X}{\arg\inf}\left\{\|\|Ax-b\|\|^2+\lambda \|\|x\|\|^2\right\}$$ I have read that the solution keeps the residual $$\|\|Ax-b\|\|^2$$ small and is stabilized through the $$\lambda \|\|x\|\|^2$$ term. Can anyone help me understand why that is? I can see that the term prevents overfitting, but I can't quite see how it helps stabilizing. Assuming $$\\|\cdot\\|$$ is the $$L_2$$ norm, the solution for $$x$$ is \begin{align} x = (A^T A + \lambda I)^{-1}A^T b \end{align} The instability in this solution lies in the inverse. If $$A$$ have columns which are nearly linearly dependent, then $$A^TA$$ is "nearly non-invertible". In other words, the condition number will be very large. The $$\lambda I$$ helps stabilize this inverse, and will always lower the condition number. • Yes, it is the $L^{2}$ norm that is used. That makes sense! I completely forgot that the solution was relying on $(A^{T}A+\lambda I)^{-1}$. Thank you for the response! – James Feb 4 at 20:11 In order to understand Thikonov's stabilization, it helps to first look at the ordinary least square solution $$x^$$: \begin{align} x^* = (A^T A)^{-1}A^T b \end{align} We see that it's necessary to calculate the inverse of $$A^T A$$ and this might not be possible, if $$A$$ has nearly linearly dependent columns. But let's take a closer look onto this, by factorizing just the suspicious term by a singular value decomposition. Then $$A^T A = U \Sigma V^T$$ where $$U$$ and $$V$$ are the eigenvectors and $$\Sigma$$ is a diagonal matrix that contains the non-zero eigenvalues. It's not of special interest here that $$U=V$$, but it is very important that the pseudoinverse $$(A^T A)^{-1}$$ is found by inverting $$\Sigma$$. More specific, the reciprocal of each eigenvalue, let's say $$\sigma_i$$ has to be found. And this might get difficult, if two columns are nearly linear dependent. In this case, $$\sigma_i$$ is very small and the result of the division gets very large and tiny pertubations of $$\sigma_i$$ lead to large fluctuations of the inverse. It is possible to monitor such cases and as this answer already mentions the condition number is one of these indicators. The solution that Thikonov provides to overcome the problem is simple, but very effective: just take a positive variable $$\lambda$$ and add it to the denominator. This will bound the overall result and stabilize the solution: $$$$\Sigma_{ii}^+ = 1 / (\sigma_i + \lambda)$$$$ As we now have identified the cause of instabilities and inserted a term that prevents them, we can add the same to our known equation and role it back: $$$$U (\Sigma + \lambda I) V^T = U \Sigma V^T + \lambda U V^T = A^TA + \lambda I$$$$ And finally, we arrive at the well known: \begin{align} x^* = (A^T A + \lambda I)^{-1}A^T b \end{align*}	2019-08-19T17:19:38	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3100267/stability-of-tikhonov-regularization/3100284", "openwebmath_score": 0.9265241622924805, "openwebmath_perplexity": 222.11189192815877, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.973240718366854, "lm_q2_score": 0.8723473879530492, "lm_q1q2_score": 0.8490039985168744 }
https://math.stackexchange.com/questions/1645532/definition-of-inverse-in-linear-and-abstract-algebra	# Definition of Inverse in Linear and Abstract Algebra In a linear algebra text, the following is the definition of the inverse of a matrix An $n\times n$ matrix $A$ is invertible when there exists an $n \times n$ matrix $B$ such that $$AB = BA = I_n$$ And likewise in an abstract algebra textbook, the definition of the inverse of a group is Given that $G$ is a group with operation $$, for each $a \in G$, there exists an element $a^{-1}$ such that $$aa^{-1} = a^{-1}*a = e,$$ where $e$ is the identity element in $G$. Such element is called the inverse of $a$ in $G$. Unfortunately, the second semester of abstract algebra didn't quite finalize due to low enrollment, so I'm doing independent self-study of topics I missed in Linear Algebra (as mind preparation). Here's my question: Is it sufficient to show that $AB = I_n \;\;\implies \;\;B = A^{-1}$ and $A = B^{-1}$? Or must you check both that $AB = I_n$ and $BA = I_n$ to completely conclude that $A = B^{-1}$ and $B = A^{-1}$? I remember on an exam, I had to prove that for a group homomorphism $\phi: G\to H$, for any $a \in G$, $\phi(a^{-1}) = [\phi(a)]^{-1}$ which I proved by asking the reader to observe that $\phi(a)\phi(a^{-1}) = \phi(aa^{-1}) = \phi(e_G) = e_H$ which because $\phi(a)\phi(a^{-1}) = e_H$, this can only mean that $\phi(a)^{-1} = \phi(a^{-1})$ by definition of inverse. And I got full points for it, but it leaves me wondering: am I supposed to check both arrangements to create the strongest possible argument? • math.stackexchange.com/questions/3852/if-ab-i-then-ba-i – mvw Feb 8 '16 at 4:52 • For the second problem concerning the homomorphism, you already know that the inverse exits because you are in a group. Proving $\phi (a^{-1})=[\phi(a)]^{-1}$ is not equivalent to proving an inverse exists. – Oliver Jones Feb 8 '16 at 4:57 • In general, if we are following the equivalent conditions for a nonsingular matrix, since $A$ is nonsingular, we know that $A^{-1}$ exists. I suppose a better way to ask my question is: Suppose we have two elements $A$,$B$ and $A$ has an inverse; does it suffice to say that $B$ is the inverse of $A$ if we just show one arrangement (i.e., $AB = e$) gives the identity? So what it seems from other answers, the left inverse and right inverse might not be the same which is why you must check both that $AB = e$ and $BA = e$, no? – Decaf-Math Feb 8 '16 at 5:02 If you already know that $G$ is a group, then to prove that $a$ and $b$ are inverses, it is enough to check $ab = e$. If $G$ is not a group or you don't yet know that $G$ is a group (for example, if you are trying to prove that $G$ is a group), then it is not enough to show $ab =e$. You also need to show that $ba = e$. For matrices over a field, $AB = I$ automatically implies that $BA = I$ if $A$ and $B$ are square. Otherwise, if $A$ is $m \times n$ and $B$ is $n \times m$, where $m < n$, then $BA = I$ is impossible. This can be shown by an argument using ranks. For square matrices over many rings, it is indeed sufficient to check that $AB=I$ to conclude $A=B^{-1}$. It is certainly true for matrices over fields like the real or complex numbers. You can read about it in the link If $AB = I$ then $BA = I$ The class of rings for which this works is called the class of stably finite rings. It is a very broad class of rings, and my guess is that you are working with such a ring. But in the broadest generality, if you are working in some wild monoid (like a maid ring over a non-stably finite ring) the. It comes necessary to check both $ab$ and $ba$. In any group (including groups of invertible matrices) it is sufficient to check that something is either a left inverse or a right inverse. This is because in any group, the inverse necessarily exists (by definition of something being a group) and is unique (since $ag=e$ implies $g=a^{-1}e=a^{-1}$ by multiplying on the left by $a^{-1}$, and this implies $ga=e$ by multiplying on the right by $a$). However, in an arbitrary algebraic structure, a right inverse and a left inverse may not agree. To show an inverse for both a (square!) matrix and a group element, you must show that it is both a left and a right inverse. Matrix multiplication and general group operations are not commutative. For homomorphisms sending inverses to inverses, you have more information, which you employed.	2020-01-27T09:21:27	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1645532/definition-of-inverse-in-linear-and-abstract-algebra", "openwebmath_score": 0.8867905735969543, "openwebmath_perplexity": 114.11495749529152, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9732407191430025, "lm_q2_score": 0.8723473846343394, "lm_q1q2_score": 0.8490039959640419 }
http://math.stackexchange.com/questions/167520/basic-algebra-question	# basic algebra question Albert repays three installments, a, b and c, of a loan that he had taken for buying an electric heater. The total of the first installment and the second installment is 150. The total of the second installment and the third installment is 200. The total of the third installment and thrice the first installment is 250. What is the third installment? it was asked in my interview for clerical work. - Can you translate the given information into (three linear) equations? Could you then solve the system of equations? – David Mitra Jul 6 '12 at 15:39 So you have the system \begin{align} a + b &= 150 \\ b + c &= 200 \\ 3a + c &= 250 \end{align} And then you must solve it. Do you know how? One way might be to subtract the second from the first to get $a + b - (b + c) = a - c = 150 - 200 = -50$, and then to add that to the third to get $a-c + (3a + c) = 4a= -50 + 250 = 200$. This says that $4a = 200$, so that $a = 50$. Then $b = 100$ and $c = 100$. - how did u decide that one should substract equation 2 from 1 ... – user286035 Jul 6 '12 at 15:49 @user286035 Because doing so eliminates the variable $c$ – ItsNotObvious Jul 6 '12 at 15:59 Systematic elimination is usually the best way to go. Sometimes there are reasonable alternatives. For example, "add" all three equations. We get $4a+2b+2c=600$. But $2b+2c=400$, so $4a=200$, and therefore $a=50$. Now easily we see that $b=100$ and therefore $c=100$. Remark: Or perhaps the mental arithmetic is easier if from $4a+2b+2c=600$ we conclude that $2a+b+c=300$, so $2a=100$, and therefore $a=50$. - i liked your way. The mental arith hint is very cool. But when to use which technique ? – user286035 Jul 8 '12 at 7:44 @user286035: For "real world" equations with usually messy coefficients, use systematic (Gaussian) elimination. In interview situations, hard to know, there likely is a trick. Exploit any symmetry. – André Nicolas Jul 8 '12 at 13:02	2015-05-30T15:05:32	{ "domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/167520/basic-algebra-question", "openwebmath_score": 0.9962795972824097, "openwebmath_perplexity": 846.8547965921645, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9732407175907054, "lm_q2_score": 0.8723473813156294, "lm_q1q2_score": 0.8490039913799958 }
https://math.stackexchange.com/questions/9259/find-the-sum-of-all-the-multiples-of-3-or-5-below-1000	# Find the sum of all the multiples of 3 or 5 below 1000 How to solve this problem, I can not figure it out: If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23. Find the sum of all the multiples of 3 or 5 below 1000. • @J.M.:Much better title thank you! – AD. Nov 7 '10 at 13:42 • @AD: Sometimes straightforward is beautiful. ;) – J. M. is a poor mathematician Nov 7 '10 at 13:44 • This is actually Project Euler problem no 1 and can be solve efficiently by using mutual inclusion exclusion. – Quixotic Nov 7 '10 at 13:45 • This is the postage stamp lemma! Every number greater than 7 can be expressed as $3x+5y$ with $x,y>0$ crazyproject.wordpress.com/2010/10/22/… – N8tron Jun 28 '12 at 13:10 The previously posted answer isn't correct. The statement of the problem is to sum the multiples of 3 and 5 below 1000, not up to and equal 1000. The correct answer is \begin{eqnarray} \sum_{k_{1} = 1}^{333} 3k_{1} + \sum_{k_{2} = 1}^{199} 5 k_{2} - \sum_{k_{3} =1}^{66} 15 k_{3} = 166833 + 99500 - 33165 = 233168, \end{eqnarray} where we have the used the identity \begin{eqnarray} \sum_{k = 1}^{n} k = \tfrac{1}{2} n(n+1). \end{eqnarray} • The one who posted the answer 233168, please explain that answer in detail . It would be really helpful for us if you will explain. – user126936 Feb 7 '14 at 9:16 • The first two sums account for multiples of $3$ and $5$, the last sum accounts for over-counting multiples of $15$ (which can appear in either of the first two sums). – user02138 Feb 7 '14 at 13:12 • For n=10; and k =3; ( 3 * (3+1) / 2) * 3 – shiva Jul 8 '14 at 10:02 • how did you know to do that? – zurbergram Jun 30 '15 at 3:19 • There is a general principal in counting things in several sets called "inclusion-exclusion". If the things can be in any of several sets, add the totals for each set, subtract the number of things that are in exactly two of the sets, add the number of things that are in exactly three of the sets, etc. – richard1941 Sep 12 '18 at 6:10 First of all, stop thinking on the number $1000$ and turn your attention to the number $990$ instead. If you solve the problem for $990$ you just have to add $993, 995, 996$ & $999$ to it for the final answer. This sum is $(a)=3983$ Count all the #s divisible by $3$: From $3$... to $990$ there are $330$ terms. The sum is $330(990+3)/2$, so $(b)=163845$ Count all the #s divisible by $5$: From $5$... to $990$ there are $198$ terms. The sum is $198(990+5)/2$, so $(c)=98505$ Now, the GCD (greatest common divisor) of $3$ & $5$ is $1$, so the LCM (least common multiple) should be $3\times 5 = 15$. This means every number that divides by $15$ was counted twice, and it should be done only once. Because of this, you have an extra set of numbers started with $15$ all the way to $990$ that has to be removed from (b)&(c). Then, from $15$... to $990$ there are $66$ terms and their sum is $66(990+15)/2$, so $(d)=33165$ The answer for the problem is: $(a)+(b)+(c)-(d) = 233168$ Simple but very fun problem. • +1 but would be a bit better with an explanation as to why one should focus on 990 instead of 1000. – jmoreno Nov 13 '16 at 7:04 • Think this answer is copied from Project Eulers' answer by Rudy: projecteuler.net/thread=1#209 – Eugene Kulabuhov Nov 25 '16 at 21:59 The multiples of 3 are 3,6,9,12,15,18,21,24,27,30,.... The multiples of 5 are 5,10,15,20,25,30,35,40,45,.... The intersection of these two sequences is 15,30,45,... The sum of the first numbers 1+2+3+4+...+n is n(n+1)/2. The sum of the first few multiples of k, say k+2k+3k+4k+...+nk must be kn(n+1)/2. Now you can just put these ingredients together to solve the problem. To find n use 1000/3 = 333 + remainder, 1000/5 = 200 + remainder, 1000/15 = 66 + remainder and then sum multiples of 3: $3\cdot 333(333+1)/2 = 166833$. multiples of 5: $5\cdot 200(200+1)/2 = 100500$ and subtract multiples of 15 $15\cdot 66(66+1)/2 = 33165$ to get 234168. • The answer that you have provided is incorrect. Take a look at the accepted answer and see how you can improve your own post. – Jeel Shah Jul 11 '13 at 23:59 • Why can't you duplicate the intersections? – JohnOsborne Jul 29 '18 at 22:32 Well the main equation is already given above. The only question which give me trouble is that why I have to subtract the sum of 15?! Well, the answer is, 15 can be evenly divide by both 3 & 5. So the products of 15 can also be divided by those number as well! So, when you adding the numbers with Sum Of Three & Sum Of Five there are some numbers(i.e. 15,30,45,60....) which are available at both SUMMATION. So, you have to subtract at least once from the total sum to get the answer! Hope this helps someone like me:) !! ## protected by Community♦Mar 19 '14 at 16:05 Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).	2019-04-20T14:14:52	{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/9259/find-the-sum-of-all-the-multiples-of-3-or-5-below-1000", "openwebmath_score": 0.7876314520835876, "openwebmath_perplexity": 442.1909159060155, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9732407106053677, "lm_q2_score": 0.8723473813156294, "lm_q1q2_score": 0.8490039852863548 }
https://www.physicsforums.com/threads/global-extrema-question.66845/	# Global extrema question 1. Mar 11, 2005 ### mathrocks I'm supposed to find the absolute max and min for f(x,y)=sin(x) +cos(y) on the rectangle defined by 0<=x<=2pi 0<=y<=2pi. Can someone point me in the right direction. I got the critical points but i dont know what to do after that. 2. Mar 11, 2005 ### Jimmy Snyder This can be solved with a few elementary facts concerning the sin and cos functions and no calculus at all. First, disregarding the rectangle for a moment, it is obvious that f(x,y) must lie between -2 and 2 inclusive for any x and y, because both sin(x) and cos(y) lie between -1 and 1 inclusive, and f(x,y) is the sum of these two. Second, it is easy to find a pair (x,y) in the rectangle where the values of -2 and 2 are actually attained. For instance if x = PI/2 and y = 0 then f(x,y) = 2. I leave -2 for you to do. Since f(x,y) attains the values of -2 and 2 on the rectangle, and these values cannot be exceeded for any (x,y), they must be the absolute max and min. 3. Mar 11, 2005 ### mathrocks How do I do this using calculus though, so I know what to do for other problems that arise. I know I must find the partial derviative for f(x,y) and find the critical values of each of those, which I believe are (pi/2, pi), (3pi/2, 2pi). Then I'm suppose to do something with the rectangle but I'm not sure what? 4. Mar 11, 2005 ### Jimmy Snyder You need to check the value of f(x,y) along the perimeter of the rectangle. Consider the fact that the maximum value of the function f(x) = x on the interval from 0 to 1 occurs at the point x = 1 where the maximum value is f(1) = 1. At that point, f'(x) = 1, not 0. So looking for points where f'(x) = 0, while necessary, is not sufficient. Loosely speaking, you need to examine the edges of the domain as well as the points where the derivative is zero. 5. Mar 11, 2005 ### mathrocks Ok, I'm starting to understand this a little better now. Now my new question is when you have the domain be something like 0<=x<=2pi, 0<=y<=2pi, do you have to actually go through each value and see which combination gives you a maximum? Because in the case of f(x,y)=sin(x)+cos(y), the endpoints dont yield a maximum. 6. Mar 11, 2005 ### Jimmy Snyder In general, yes. The reason is somewhat technical, and if you don't understand the rest of this paragraph, you can just accept the answer and safely skip on to the next one. Loosely speaking again, you cannot be assured that the 'edge' of the domain allows analysis. For instance, let the domain be the interior of the rectangle along with some points along the edge thrown in for good measure. Then finding the max of a function on that domain could require you to evaluate the function at those extra points on the edge one by one. Of course, if there were a lot of them, then you could be in it for the long haul. The good news is that the teacher would have the same amount of work to do in order to check your answer, so it probably won't happen to you while you are in school. However in particular cases you may be able to do better. In fact, for the case at hand, you can look for the extrema along each of the 4 edges by looking for points where the partials are 0 and then evaluate the function at each of the 4 corners. 7. Mar 11, 2005 ### mathrocks That's the thing, when I look for the points where the partials are 0, I get (x=pi/2), (x=3pi/2) and (y=pi), (y=2pi). And when I evaluate the function at the corners which are (x=0), (x=2pi), (y=0), (y=2pi), I get x=0, x=0, y=1, y=1. So I dont see how I can get -2,2 even though I know I should. 8. Mar 11, 2005 ### Jimmy Snyder So you need to look at the following points: 1. The points in the interior of the rectangle where the partials are 0. 2. The points along the sides where the partials are 0. 3. The corners. The largest value of f(x,y) among the union of these three sets of points is the absolute maximum. Similarly for the minimum value. 9. Mar 11, 2005 ### dextercioby $$f(x)=:\sin x+\cos y$$ The condition for extremum-------->critical points. $$\frac{\partial f}{\partial x} =0 ;\frac{\partial f}{\partial y}= 0$$ The type of extremum----------> Hess(f)\|_{sol.critical points} ? 0 Daniel. P.S.If the hessian is the positive,then it's a minimum,if it's negative,it's a maximum,if it's zero,it's a saddle point...	2018-01-22T16:40:58	{ "domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/global-extrema-question.66845/", "openwebmath_score": 0.7017208337783813, "openwebmath_perplexity": 391.0415467057714, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.973240718366854, "lm_q2_score": 0.8723473713594991, "lm_q1q2_score": 0.8490039823673557 }