Questions 1 year ago. Important Maths Formula. A vector of unit magnitude is unit vector. Direction Of A Vector Formula. Circumference of a Circle Formula. Linear Approximation Formula. Statistical Significance Formula. Difference of Squares Formula. Regular Square Pyramid Formula. Triangular Pyramid Formula. Arithmetic Sequence Formulas. Volume Of Parallelepiped Formula. Volume Of An Ellipsoid Formula. Isosceles Triangle Formulas. Complex Number Power Formula.
Equilateral Triangle Formulas. Coin Toss Probability Formula. Degree And Radian Measure Formula. Consecutive Integers Formula. Area of Circle Formulas. Frequency Distribution formula. Hypergeometric Distribution formula. I will recommend anyone looking for Business loan to mr benjamin who helped me with Four Million USD loan to startup my business and it's was fast When obtaining a loan from them it was surprising at how easy they were to work with. The process was fast and secure.
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Net, ASP. Maximal entropy means least information content. Assume that the probability that a physical or chemical system is in a state k is pk and that the energy of the state k is Ek. Nature minimizes the free energy X F p1 ,. Find this distribution in general if Ei are given. We see in general that for quadratic f and linear g, we end up with a linear system of equations.
Related to the previous remark is the following observation. It is often possible to reduce the Lagrange problem to a problem without constraint. This is a point of view often taken by economists. Use the method of Lagrange multipliers. A solution x is called an eigenvector. Never mind here that B is not symmetric. Use the later variables. Doctor Manhattan designs a bullet with fixed volume and minimal area. So, simple polygons are allowed.
What island does have the maximal area if the length of the boundary is fixed? This is called the isoperimetric problem. If we look at the problem restricted to polygons with a fixed number n of vertices, then we have a nice finite dimensional Lagrange problem. Problem A: Assume the circumference g x, y of the triangle is 3. Set up the Lagrange equations and solve them. Here is a side problem from good old Euclidean geometry.
Solving the problem to find the n-gon with maximal area is a messy Lagrange problem. It can be done by a computer but there is a more elegant way: Problem C: Use the computation in problem A to show that for a maximal polygon containing vertices Problem D: Conclude that a polygon with n vertices and maximal area must be a regular polygon.
You are on your treasure island G and have two locations A, B in G. The problem to find the shortest connection between A and B can be quite complex in general. An example is when G is bound by a Gosper curve. For the following let us assume that the boundary of G is a convex curve: this means that for any two points A, B in G, the line segment through A, B is contained in G. Problem E: Verify that for a shore triangle, the billiard law of reflection at the boundary holds.
Hint: to see that the incoming angle is the same as the outgoing angle, take a minimal triangle A, B, C, where B is on the island shore, then replace the curve with the tangent curve L at B. Now reflect C at L to get a point C 0.
Verify that the shortest billiard path ABC has the same length than the straight line connecting A with C 0. What polygon with fixed circumference has maximal area? The next time you are cast away on an island, count the number m of mountain peaks, the number s of sinks and the number p of mountain passes.
Make some experiments. If you want to challenge yourself, see whether you can prove the island theorem by deformation. This is probably too hard. Just enjoy the struggle! Assume now that our island is an atoll, a ring shaped reef. First an island with 2 mountain peaks and with 1 mountain pass. Then an island with 3 mountain peaks and 2 mountain passes. The Atafu atoll. Let us look at the one-dimensional case, where we prove things easier. Assume the island is the interval [a, b]. Let m be the number of maxima and s the number of minima sinks.
One-dimensional islands. What is the largest area that such a triangle can have? The result might look a bit strange for a triangle. Similarly find an example with a minimum and an example of a non-Morse function where the critical point is neither a maximum, nor a minimum. Prove this. As already in one dimension, the definition is designed to be independent of an orientation chosen on R.
We are integrating like summing up a spread sheet. Just add up all entries. To justify that the limit exists, we again can use the Heine-Cantor theorem which tells that f is continuous on R if and only if it is uniformly continuous. Because the boundary was assumed to be given by a collection of curves which have finite total arc length L, the number of cubes Qij which intersect the boundary C is bounded by 4Ln a curve of length 1 can maximally touch 4 squares.
Let I be the limsup of In. We rarely evaluate integrals using Riemann sums. Fortunately it is possible to reduce a double integral to single integrals. Is this the same? This is answered with Fubini, which we have already used.
Here is the Fubini theorem: Figure 1. Integrating over a region via a Riemann integral. A double integral is a signed volume. Parts where f RR Rb We know from single variable calculus that a f x dx is the signed area under R b R f x the curve of f. Note that as we have defined the integrals, the equivalence would be wrong if f x is negative somewhere.
It is the double integral which is the correct notion of area. RR The integral R f x, y dxdy can be interpreted as the signed volume under the graph of f above the region R. Find the area of a disc of radius a. We will next The hoof solid was considered by Archimedes already.
Cover S with cubes Qij as in the last lecture. Applying in every direction, Taylor with remainder, we see Z i j i j 1 Mn F. R 1For the C 1 case, see J. Schwartz, Mathematical Monthly 61, , or P. While we usually could ignore talking about orientation, it is evident here that the integrals considered so far, we do not care about the orientation of the space. If the change of coordinates switches the orientation, the resulting integral does not change.
Coordinate change. Rd Rb In higher dimensions, one can then apply the derivative to anti-symmetric tensors. The trouble also continues in the new coordinate system and it is even more dramatic. Here is a famous open problem about coordinate changes. It is called the Jacobian conjecture. It deals with polynomial coordinate changes, where x u, v and y u, v are polynomials in u, v.
One knows that if the conjecture is false, then there exists a counter example with integer polynomials and Jacobian determinant 1. The conjecture is open since at least Problem: Here is RR a famous problem.
Solution: this problem looks difficult 2 at first as we can not integrate with respect to x or y. Find E[x], E[x2 ], E[x3 ] and E[x4 ]. What is hot now is 22 spinner with 23 bearings! To keep our bearings, we do not count the bearings.
It can produce butterflies in your stomach but there are some tricks to do that fast. Relax with the Math 22 fidget spinner for example! The math 22 spinner and the butterfly. Find an example for the Jacobian conjecture where both polynomials are not linear!
In this seminar we look a bit around in the literature and collect problem solving strategies. We have seen already a few methods: Already seen principles 1. Induction Theorem on unique row reduced echelon form 2. Contradiction Clairaut theorem 3. Deformation Hopf Umlaufsatz 4. Invariant Sum of Morse indices on island We will introduce a few more principles and tips and take the opportunity to introduce a bit the literature.
We only look at 4 books. If you read and absorb this book, you immediately get measurably stronger in math. Still after more than 70 years, it is the best. Understand the problem: unknowns, data, draw figure.
Devise a plan: similar or related problem? Carry out the plan: check each step. Examine the solution: can other problems be solved as such? But it is amazing to see the power in a method. Why is it powerful? Because if one sees a harder problem the first time, one is totally lost. Proof: if not, then the problem was easy Where do we start? This is where it is good already to have a guide telling you: well, just first start to understand the problem.
The problem is featured even on the cover of some later editions of the book. An here is another problem from Polya, slightly reformulated.
At which rate is the water level rising if the water depth is z meters? Like Polya, also Tao has proven new important theorems many as a single author and so got some street cred. Here are some problems from his book: Problem C: An integer n has the same last digit than n5. We have done the case of the sum of the first n squares in a practice exam. Tao does not give a formal list of strategies, but explains in an example on page 4 the following principles.
Consider special, extreme or degenerate cases. Solve a simplified version of the problem c. Formulate a conjecture d. Derive intermediate steps which would get it. Reformulate, especially try contraposition. Examine solutions of similar problems g. Generalize the problem The book of Perkins analyses skillfully the mechanisms of break through ideas. It destills the following mechanism for break through ideas.
It captures it pretty well, since problems which are solved quickly rarely cover new ground. Perkins 1. Long search. Work for years or decades. Little apparent progress. Many failures. A precipitating event. Maybe external circumstances. A cognitive snap. Usually in a flash. Flesh it out. Try to solve it yourself and also keep track on how you pursue the task to solve the problem.
Problem F: Someone brings an old coin to a museum director and offers it for sale. The coin is stamped B. Instead of considering the purchase, the museum director calls the police. If this was too easy experiments show that some people can answer it very quickly. For others it takes longer , try this one, also from Perkins: Problem G: You are driving a jeep through the Sahara desert. You encounter someone lying face down in the sand, dead.
There are no tracks anywhere around. There has been no wind for days to destroy tracks. What do you find? The book of Posamentier and Krulik is more intended for the teacher and less for the research mathematician. It goes through the following principles Posamentier-Krulik 1. Reason logically 3. Work backwards 5. Consider extreme cases 7. Organize data 9. Account all possibilities 2.
Recognize patterns 4. Adopt different view 6. Solve simpler problems 8. Make a picture Experiment, guess and test Linear Algebra and Vector Analysis The teacher wants every student to change place and move to a seat to the left, right, front or left.
It it possible? In which cases is it possible? Once you have an idea, prove the statement. Homework Ives, I met a man with seven wives, Each wife had seven sacks, Each sack had seven cats, Each cat had seven kits: Kits, cats, sacks, and wives, How many were there going to St.
The rhyme was inspired by one of the oldest problems texts in math, the Rhind Papyrus. What am I? There is one hole 0 left. The task is to reorder a scrambled puzzle so that all numbers are in order and 0 at the very bottom right.
The player can switch 0 with a neighboring piece. Sam Loyd suggested to start with stone 14 and 15 switched. Prove that one can not win the prize. A solid is aRRR finite union of such basic solids. Solids in R3 are sets which are unions of solids bound by smooth surfaces.
The second solid appears in homework There are two basic strategies to compute the the first is to slice the RR R b integral: region up along a line like the z-axis then form a R z f x, y, z dxdydz. To get R 1 RR the volume of a cone for example, integrate 0 [ R z 1dxdy]dz.
A second reduction is to see the solid sandwiched between two graphs of a function on a region RR R h x,y R, then form R [ g x,y f x, y, z dz]dxdy. In the cone case, we have for R the disc p of radiusp 1.
Burgers and fries! The first reduces to a single integral, the second to a double integral. For cylindrical coordinates, the situation is the same as for polar coordinates. RRR RRR 2 To find E 0 0 0 We will see some details in class. What is its volume? Find its volume. For example, if f is a density on the surface then RR this S f dS is the mass.
Again, we have to stress that in this integral, the orientation of the surface is irrelevant. The distortion factor dr is always non-negative. The proof is the same as seen in the two-dimensional change of variable situation. We will talk about determinants in detail later but if A is in row reduced echelon form then AT A is the 1Unfortunately, scalar integrals are often placed close to the integration of differential forms like volume forms.
The later are of different nature and use an integration theory in which spaces come with orientation. So far, if we replace r u, v with r v, u gives the same result like area or mass.
Linear Algebra and Vector Analysis identity matrix and the determinant is 1, agreeing with the volume. If a column of A is added to another column, then this does change det B T B. The only row reduction step which affects the dr is the scaling.
But that is completely in sync what happens with the volume. The last theorem covers everything we have seen and we ever need to know when integrating scalar functions over manifolds.
R Examples The surface area is infinite. The 6-sphere has maximal surface area The volume of the ball is only 0. The surface area of the sphere for example is only 0. Compare with a n-unit cube of volume 1 and a boundary surface area 2n. High dimensional spheres and balls are tiny! Tunnels connecting different parts of space-time appear frequently in science fiction. Find the total surface area of the lantern. In this seminar we have the honor to have Archimedes as a special guest.
The actual Archimedes did not know about the interview. How does it work? Quantum space-time produces sometimes tiny wormhole constellations in which a wave function can be trapped. By harvesting many of those trapped waves, we can rebuild and interact with an object or person from a previous time.
It is enough time however for a short interview. We take the opportunity and ask him about his theorems. Math 22a: What a pleasure to have you here. It must be a dream. Math 22a: yes, it is also a hot spot for science, but there are many now. We are eager to learn a bit about your proof expertise. Math 22a: What result of yours do you consider the most important one? Archimedes: Definitely the formula for the volume of the sphere! Math 22a: Why?
Archimedes: It was much harder to get this than the circumference of the circle or the surface area of the sphere. It was also harder to test the result experimentally. Math 22a: How did you measure? Archimedes: We build wood models of cylinders, cones and spheres of the same base radius and height and measured their volume ratios.
Problem A: Explain how Archimedes can using wooden models measure their volumes. Given a cylinder C, a cone O and a sphere S of base length 1. Math 22a: Was the comparison of the sphere with the complement of a cone in the cylinder historically the first proof?
Archimedes: The relation had been conjectured before. Problem B: Explain why slicing the unit sphere at height z gives the same area as a ring of radius 1 in which a hole of size z has been has been drilled.
Math 22a: Do you remember the precise moment, when the discovery stuck? Math 22a: This discovery must have occurred after you got the circle circumference computed. How difficult was the later? Archimedes: also this needed some time. It emerged pretty early that the circumference is somehow proportional to the radius. The measurement of the constant was then a bit trickier even so it remained open what fraction it is.
I did that using the following picture. How can you make this precise as in reality the circular sector does not have the same area as the triangle. The circle proof. It was exciting to see a solid bound partly by round parts to have a rational volume, which is 2 third of the height.
Problem C: a Take a hoof of height 1 and cut in triangular pieces which are obtained if y is constant. Math 22a: Also very impressive is your computation of the surface area of the sphere by relating it with the surface area of a cylinder. What was the intuition there? Archimedes: Actually, a drawing which is accurate enough shows this pretty well. As both situations have circular symmetry, we only need to understand what happens with the lengths on a sphere when it is projected on the cylinder.
There are similar triangles. Take a stick of some length and place it onto the sphere pointing to the north pole. As it gets closer to the pole and its height-length is one half of the actual length. In the sphere case, the factor one-half is applied to the radius. In the cylinder case it is applied to the height. Problem C: Explain this in more modern terms. We have a unit sphere and a cylinder of radius 1. Math 22a. A last question: What is a function in mathematics?
We also have rules which tell how to compute rates of change. This is cool. Let me see: does this also work for the area and circumference of a disc?
Go ahead. Yes, that is correct. Math 22a Thank you very much for the interview. It will inspire us for the second midterm exam. Maybe you can visit and take the exam on Tuesday or review on Sunday? Find some. What are the formulas for the volumes and surface areas? It is seen in the picture above to the right. Please explain. Functions dance upwards. Prove that n is odd. Problem 28P. Which one? Complete the formula. We have seen two major types each a three capital letter acronym. Which type is it?
Only pick points A-J. Linear Algebra and Vector Analysis y Does f have a global maximum? Find the minimum of f on this constraint using the Lagrange method! We start all together at AM. You can already fill out your name in the box above. Please stow away any other material and electronic devices. Remember the honor code. Except for problems There is also space on the back of each page. Archimedes sends his good luck wishes. He just sent us his selfie. Oh well, these celebrities!
Assuming f is a Morse function, how many saddle points are there? We think of F as a force field. Even so F and r are column vectors, we write in this lecture C [F1 x ,. A line integral in the plane and a line integral in space. Figure 1 shows the situation. We go more against the field than with the field. We think of f as the potential. Is every vector field F a gradient field? Proof: this is a consequence of the Clairaut theorem.
It can not be a gradient field. Differentiate f with respect to y and compare fy with Q. Solve for C y. Here is a direct formula for the potential. Let Cxy be the straight line path which goes from 0, 0 to x, y.
Proof: By the fundamental theorem of line integral, we can replace Cxy by a path [t, 0] Rgoing from 0, 0 toR x, 0 and then with R x[x, t] to x, y. Examples R If f x, y, z is a tempera ature and r t the path of a fly in the room, then f r t is the temperature, which the fly experiences at the point r t at time t. The change of temperature for the fly is d f r t. A device which implements a non-gradient force field is called a perpetual motion machine.
It realizes a force field for which the energy gain is positive along some closed loop. The first law of thermodynamics forbids the existence of such a machine. We will look at examples in the seminar. Linear Algebra and Vector Analysis Figure 3. The vortex vector field has a singularity at 0, 0. All the curl is concentrated at 0, 0. C Problem The result should depend on a and b. Humans have dreamed about this for centuries. There is no mathematical proof that such a machine can not exist.
It is an experimental fact that all isolated physical process we know preserve energy. So, how come we can harvest energy from the wind force for example? Wind energy is driven by external sources, in particular the solar energy which heats up different parts of the earth surface. The sun energy comes from nuclear processes, mainly the fusion process.
It is a nice sport to come up with machines which seem to work or then to analyze a given machine which has been constructed and to find why it fails.
Our first machine is a circular pipe which is half filled with water. On the side without water, the gravitational force pulls a wooden ball down. On the water side, the buoyancy force pulls the ball up. Valves are in place so that the water stays in place. Problem A: Analyse the pipe machine. You can assume that operating the valves uses arbitrary little energy and when opening one of the valves, the water stays in place.
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