Can you find the real root of this polynomial using only SPM Coordinate Geometry skills?





Tips: a straight line, slope, multiple 2-point lines, root occurs @ y = 0

This post has been edited by Critical_Fallacy: Dec 19 2013, 04:20 PM

Hi RED-HAIR-SHANKS & crazywing26,

Please listen to v1n0d's advice because he is an experienced math teacher. Perhaps I should make myself clear. If a Row can be simplified, it means the terms share a common divisor. When the diagonal element is normalized, the entire row is divided by the value of the diagonal element.

In fact, Gauss Elimination procedure does not require the diagonal elements to be normalized. Some students are not comfortable dealing with fractions. As long as you can reduce the entries below the diagonal elements to zeroes, then you solve the linear system. To avoid normalization, you can find the least common multiple between the diagonal element and the entries below it, so that you can perform the elimination.

For example, because a11 = 3, and a21 = 2, and the least common multiple is 6, therefore ERO {R2' = 3*R2 - 2*R1} will reduce a21 =0, without dealing with fractions. The is a price to pay: Be extra careful when doing the arithmetic calculations for the TWO Rows!



This post has been edited by Critical_Fallacy: Dec 19 2013, 09:11 PM

Hi RED-HAIR-SHANKS,





The idea behind using Coordinate Geometry to find the real root, of a curve is to exploit the slope formula of a straight line.

Since you know the real root of f(x) lies at the interval [4, 5], you can pick any two points within the interval, such that the lower bound , and the upper bound .

From these two points, and , a straight line can be drawn, where it intercepts x-axis. The x value that intercepts x-axis is closer to the real root, .

This post has been edited by Critical_Fallacy: Dec 19 2013, 09:40 PM

Differentiation (Newton's method) is not required although it can be very efficient. I wouldn't train you how to solve if it cannot be solved using only SPM Coordinate Geometry. Like I told you, pick 2 points within [4, 5], and with good judgment, I'd pick xL = 4.00 and xU = 4.25. To complete the info of points, find f(4.00) and f(4.25). Now you can find the slope (m) of the straight line between these two points.



With the slope and one of the points, you can find the equation of the straight line. To find the x-intercept of the straight line, just let y = 0. The x-intercept must be within the interval and is getting closer to the real root. Now, check if you understand the following algorithm (procedure).



It is convenient to tabulate your calculations using a table. I'll show the first 2 iterations and you can do the rest. In reality, sometimes you will encounter non-differentiable functions. So, you cannot use Newton's method to solve it. I discover this method independently when I was Form 4 (no one taught me Newton's method). My purpose is to show you how powerful SPM Add Maths can be. Of course, this requires Higher Order Thinking Skills (HOTS)!



This post has been edited by Critical_Fallacy: Dec 20 2013, 01:59 AM

Your answer is correct, but you used Newton's method. Perhaps this graphical representation of Coordinate Geometry will give you the better idea. As you can see, the x-intercept is getting closer to the true root of the curve. By the way, this SIMPLE root-finding technique is known as the False Position method.

Consider the definition of absolute function of |x − 3|:



It is not difficult to see that



This means that no matter how close x gets to 3, there will be both positive and negative x-values that yield f(x) = 1 or f(x) = −1. This is clearly illustrated on the graph of the function:



Because approaches a different number from the right side of 3 than it approaches from the left side, the limit does NOT exist.

where the Jacobian matrix is given by



Do you mean finding a transformation from a region S in the uv-plane, to a region R in the xy-plane?

Wow! This is Matrix Calculus. To be honest, I'm not familiar with Fisher matrix, but a close examination shows that the Fisher information is closely related to the Maximum Likelihood and the Fisher Information Matrix (FIM), can be defined as I in the following equation:



where the likelihood function of parameter θ is defined as a discrete probability distribution P depending on a parameter θ:



For N parameters in the model, so that θ is a N×1 vector , the Fisher information takes the form of an N×N matrix as shown below.



To integrate a matrix of this size N, all we do is integrate the individual entries, which can be quite tedious. If you are familiar with the computational software MATLAB, you may consider using the commands mvnrfish (without missing data) or ecmnfish (with missing data) in MATLAB Financial Toolbox to compute a Fisher information matrix based on current maximum likelihood or least-squares parameter estimates.

You can post up to seven images in a single post.

We define the length with respect to the Riemann metric g of a piecewise continuously differentiable curve represented by a map



by the formula using Einstein convention



We see immediately that the Euler-Lagrange equation for the Riemann metric will be a pain because of the square root in the Lagrangian. But then, this problem has a surprisingly simple solution, which, at first, cannot possibly seem right: that is, simply omit the square root! Thus, you probably want to consider the functional:



To justify this, recall a condition is called parametrization by arc length, that if the functional Sg indeed has a minimum in the space of continuously differentiable curves with fixed boundary points A, B, then the minimum curve also minimizes the functional sg, and furthermore is parametrized by arc length! Therefore,



The equality arises if and only if is constant in t.

Critical Scientist

Archimedes (287–212 b.c.) was the greatest mathematician of the ancient world. He was born in Syracuse, a Greek colony on Sicily, a generation after Euclid. One of his many discoveries is the Law of the Lever. He famously said, “Give me a place to stand and a fulcrum for my lever, and I can lift the earth.”



Renowned as a mechanical genius for his many engineering inventions, he designed pulleys for lifting heavy ships and the spiral screw for transporting water to higher levels. He is said to have used parabolic mirrors to concentrate the rays of the sun to set fire to Roman ships attacking Syracuse.



King Hieron II of Syracuse once suspected a goldsmith of keeping part of the gold intended for the king’s crown and replacing it with an equal amount of silver. The king asked Archimedes for advice. While in deep thought at a public bath, Archimedes discovered the solution to the king’s problem when he noticed that his body’s volume was the same as the volume of water it displaced from the tub. Using this insight he was able to measure the volume of each crown, and so determine which was the denser, all-gold crown. As the story is told, he ran home naked, shouting “Eureka, eureka!” (“I have found it, I have found it!”) This incident attests to his enormous powers of concentration.



In spite of his engineering prowess, Archimedes was most proud of his mathematical discoveries. These include the formulas for the volume of a sphere, and the surface area of a sphere, and a careful analysis of the properties of parabolas and other conics.

This post has been edited by Critical_Fallacy: Dec 22 2013, 06:27 PM

It probably be pretty tedious for me to type the proof here. But it doesn't matter because the concept is about applying Jensen’s inequality to a Lebesgue-integrable function, for .

When p = 2, it becomes , which is quadratically integrable. Applying the special case of Jensen’s inequality using Measure-theoretic approach, you can show that



For more information, you can find various articles online on spaces, Jensen’s inequality, Hölder’s inequality, and Minkowski’s inequality.

From the Electric Potential V due to two point Charges q, satisfying the condition V = 0,



it can be shown that





From the figure, we can find the Electric Field in the x-direction at a point on the x-axis when V = 0, by applying the formula:



P.S. Your answers are correct. But my approach is more elegant.

This post has been edited by Critical_Fallacy: Dec 23 2013, 01:31 PM

Topping up on VengenZ statement, if you are evaluating the potentials along the x-axis, these neutral “points” (V = 0) are not really discrete points in a two-particle configuration in the x-direction. In fact, the electric potential was found to be zero for all values of y at specific x positions. These neutral positions can be determined fairly easily once you understand the concept as shown graphically below.



Using the formula for finding the Electric Potential V due to two point Charges q,



it can be shown that

.

If we set



it follows that

.

Similarly, using algebraic manipulations, you can determine the position of the neutral point, if it is located at the LEFT side of the positively-charged particle, .

You were not alone!

Hi RED-HAIR-SHANKS, maximR, crazywing26 and iChronicles,

Now it is a good time to test your understanding on SPM/STPM Series.

Can you find the sum of the following series if n is a positive integer?

Could you circle the part that you don't particularly understand? It helps me to narrow down my focus on your cognitive problem in the Electricity topic. Thanks!

Thanks for pointing out! It would be meaningless if the sum to infinity because the series diverges. Just find the formula of the sum up to a finite n positive integer.

This post has been edited by Critical_Fallacy: Dec 24 2013, 12:22 AM

Your basic concept is correct, except the “center” position. “Center” means the point is equally distant from the positively charged particle and from the negatively charged particle . The electric potential is zero ONLY when at the center of an electric dipole (a pair of electric charges of equal magnitude but of opposite sign or polarity).

Refer to the figure on Post #399 or #396, now do this for me:

(1) Find the electric potential at the point β (far right) due to a positively charged particle , separated by a distance .

(2) Find the electric potential at the point β (far right) due to a negatively charged particle , separated by a distance .

(3) Find the net electric potential at the point β (far right) due to both charged particles.

(4) Equate the net electric potential V = 0, and then simplify the equation in terms of ratio rA / rB.

This post has been edited by Critical_Fallacy: Dec 24 2013, 12:34 AM

You are good!

Solving physics problems demand concentration. Find Electric Potential V, NOT Electric Field E.

How about your answer for No.4?

You can use the Coulomb's constant, for simplification.

This post has been edited by Critical_Fallacy: Dec 24 2013, 01:00 AM

Hi delsoo,

There is a mistake.

(1) Find the electric potential at the point β (far right) due to a positively charged particle , separated by a distance .

You can use the Coulomb's constant, for simplification.



This post has been edited by Critical_Fallacy: Dec 24 2013, 01:05 AM