UTM offers Mathematics ? I didn't know that . Thanks .

Here, these are my workings, but I think they're incomplete:


My problem arouse when I am unable to obtain zero in all three of the entries that is above the leading diagonal(above all the 1's). I'm in deadlock.

Whether Shanks simplifies Rows 1 & 2 or not, it doesn't really matter, because the ultimate objective is to normalize all diagonal elements. It does not change the fact that Shanks still can reduce the augmented matrix to row echelon form within the minimum 6 STEPS in Gauss Elimination of a 3-by-3 matrix. before performing the back-substitution.

can I ask how to solve this? =( Sorry I'm that bad at self learning .....


Thanks =D

To integrate a matrix of this size N, all we do is integrate the individual entries, which can be quite tedious.

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!

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.

5?

A standard way to solve this type of problem in STPM, taking the logarithm at both sides:



Hence, the least value of n is 5.

Hi Flame Haze, RED-HAIR-SHANKS, maximR, crazywing26 and v1n0d,

Can you simplify this equation?

Critical_Fallacy and v1n0d

Look at what I found !

http://eprints.usm.my/view/subjects/

Oh . But it'd be better if you do what you love . 

Assume that the equation system
x*2 + sxy + y*2 - 1 = 0;
x*2 + y*2 - s*2 + 3 = 0
defi…ne x and y implicitly as differentiable functions of s.
(a) Differentiate the system (i.e. fi…nd the differentials) and fi…nd the values of x'(s) =dx/ds and y'(s)= dy/ds when x=0, y=1 and s=2

Can someone give me an insight of this?

Perhaps you can study this example. Please check for any mistake.

From the question that I get, yeah it's just sxy.

Any insight from you?

I reckon it'll hit the acorn. When the archer aims horizontally and shoots an arrow straight at an acorn, the arrow will move slightly downwards in a curve due to some gravitational attraction towards the Earth. At the same time, the acorn will fall downwards too when Scrat drops the acorn. And if we ignore the air drag, the arrow will surely hit the acorn(assume that the acorn falls with an acceleration that is tantamount to the gravitational attraction, where a=g=9.8ms^-2).

This is just my estimation and bits of basic knowledge from my Physics SPM, I'd love to see some more detailed and explicit explanation.

Aren't both x and y functions of s?

x=x(s), y=y(s)?

How could you differentiate sxy and only treat one as a function of s?

Luckily v1n0d and studyboy are here!

Sorry, I'm bad at discrete math. Can you check this working?

You're right, and there is a condition imposed by the initial velocity as mentioned by v1n0d.
If there were no gravity, the arrow would fly straight to Scrat and the acorn. Since gravity gives the dropped acorn and the released arrow the same constant acceleration downward, they each fall the same vertical distance below the positions they would have had with no gravity. Thus, the arrow ends up hitting the acorn no matter what the initial velocity** of the arrow. The higher the velocity of the arrow, the sooner they meet and the shorter the vertical distance that the acorn falls before being hit.



** Assuming the arrow is fired from ground level, the initial velocity of the arrow must be higher than the minimum velocity given by:



where hmax is the height of the acorn from ground before it is dropped, xmin is the horizontal distance of the archer away from Scrat / acorn, and tmax is the time of the acorn strikes the ground after it is released by Scrat.