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May 23

quadratic function

Please Help!! How do I write a bio-poem for the quadratic function? Thanks! arn4utoo

(in case anyone was wondering what a bio-poem is: [1], [2]). We're here to help with math, not to write poetry for you. All I can suggest is to look at Quadratic function and use the special properties described in the article in a way that's similar to the usual bio-poem format. nadav (talk) 03:45, 23 May 2007 (UTC)[reply]

what is criteria for mathmatical oprator

Respected sir, i am confused about the use of mathematical operators in certain formula, i want to know about criteria for a operator i.e. how do we choose one specific operator.

For example in given formula

W=m.g

here weight is function of mass, we can write this as

W=f(m)

where f is function of W.
Then my question is how do we choose only multiplication as a operator why not addition, subtraction, division.
Is there any specific criteria? Please, explain.

We'll need more information to understand your question. In the formula you gave, multiplication is used for physical—not mathematical—reasons (see Newton's second law). There is no one general reason for the choice of operations used in equations. nadav (talk) 07:35, 23 May 2007 (UTC)[reply]

W=f(m). Don't you mean f is a function of mass?Zain Ebrahim 08:51, 23 May 2007 (UTC)[reply]

A very strange question... Maybe you'll benefit from reading multiplication, division, addition and subtraction in order to understand how they differ and in which situations they are used. 213.48.15.234 09:55, 23 May 2007 (UTC)[reply]

I'm not sure if this is going to help but the operators are used based on the definition of the thing you're trying to quantify. So if you look at Newton's second law or Weight then you'd understand why we say W=mg and not (for example) W=m/g. Zain Ebrahim 10:58, 23 May 2007 (UTC)[reply]

(after edit conflict) I think the first response by Nadav1 is more to the point than the next two. The above equation, giving weight as a function of mass and gravitational acceleration, can be found in our article Weight (in the section "Conversion between weight (force) and mass"), and in Force (twice even; once in the section "Examples", and once on the section "Quantitative definition"). As the latter article duly points out, this is true by definition: weight is by definition the magnitude of the force of an object acted upon by gravity, so, using the definition of force F = ma from Newton's second law, replacing F by W and setting a = g, we have W = mg. (For nitpickers: I know this is not the "official" definition of force, but a simplified one for constant mass; for the sake of simplicity I've also omitted the vector aspect). Now this is not really a satisfactory answer to the rationale.

For one thing, mass and acceleration have different dimensions. Adding mass and acceleration is meaningless. Ask yourself: what would it mean to add 10.7 liters to 1.6 kilometers? OK, but why is the law not F = ma², for example? The answer is not mathematical. This is a law of physics, and the laws of physics are those formulas and equations that have proven to give a good quantitative description of the observed behaviour of things in reality. If you want to send a rocket to the Moon and you use Newton's laws to calculate its trajectory, you have a chance it may arrive. If you replace these laws by a randomly chosen other set of laws, no way it will get there. --Lambiam ^Talk 11:01, 23 May 2007 (UTC)[reply]

Actually, it can't be F = ma², because again you have a dimensional problem - this time, you're trying to say that a force (measured in Newtons, or kilograms metres per second per second) is equal to a mass times an acceleration squared, in units of kilograms metres metres per second per second per second per second (of course you'd actually say something like "kilograms metres squared per second to the fourth power" but this shows the difference more clearly). Then again, there could always be some kind of fundamental constant with appropriate dimensions to make it work out. See dimensional analysis to see how you can go from knowing which variables may influence each other to a valid (though not necessarily correct) formula relating them. Confusing Manifestation 22:45, 23 May 2007 (UTC)[reply]

In some other universe, with different laws of nature, the dimension of force could be ML²T⁻⁴. Before Newton formulated his law, he did not know force was supposed to have the same dimension as the newton. (The last sentence is also true in the alternative universe, in which 1 newton is 1 kg m² s^-4.) --Lambiam ^Talk 22:59, 23 May 2007 (UTC)[reply]

Huh ? Since force is defined as rate of change of momentum and momentum is defined as the product of mass and velocity, how can force have any other dimension than MLT^-2 - unless we are changing definitions, in which case everything is wibble. Gandalf61 21:48, 24 May 2007 (UTC)[reply]

Yeah, from what I hear "force" is an arbitrary concept that is not really used in more advanced physics (e.g. Lagrangian mechanics), which focuses more on energy and potential. But this question is better served by the science ref desk. nadav (talk) 06:52, 25 May 2007 (UTC)[reply]

A bit confused...

How can I find the height of a pyramid if I know the base area and volume?

This is what I have: "I would subtract the area of the base from the area of the triangular sides. Next I would... Here I stop."

This is what someone mentioned to me: "It includes division and multiplication by 3."

Colud someone tell me where all this ties in?? Thanks, IP 20:53, 23 May 2007 (UTC)

There's a formula relating the height, area of the base, and volume of a pyramid. It made quite an impression on me in Grade 8 that the volume of a flat-bottomed object that comes to a point on top is equal to a third of the area of the base multiplied by the height. iames 21:13, 23 May 2007 (UTC)[reply]

Let us call the volume

V

, the base area

A

and the height

h

. The relation between these quantities is

V={\frac {1}{3}}Ah

.

You can find the height from that equation. Do you know how to do that? —Bromskloss 21:16, 23 May 2007 (UTC)[reply]

This formula can be found in the article Pyramid (geometry). --Lambiam ^Talk 21:53, 23 May 2007 (UTC)[reply]

H = V · 3 ÷ B? I believe that is it. I ran it through many tests and it came out right. IP 16:47, 24 May 2007 (UTC)

Simple algebraic manipulation.

V={\frac {1}{3}}Ah\to 3V=Ah\to Ah=3V\to h=3{\frac {V}{A}}

. Yes you are correct. Obscurans 03:19, 25 May 2007 (UTC)[reply]

Square Roots mod n

How can I find all the square roots of a number x mod n. Is it easier if x=a² and n=ab for two numbers a and b? Thanks *Max* 23:25, 23 May 2007 (UTC)[reply]

If you can read PostScript files, you'll find an algorithm at http://zoo.cs.yale.edu/classes/cs460/Spring98/quadr.ps. (I haven't checked it, but this should be a reliable source.) --Lambiam ^Talk 23:37, 23 May 2007 (UTC) On further inspection, this seems only to cover the case that n is prime. 23:41, 23 May 2007 (UTC)[reply]

IIRC, you want to look at Euler's criterion and Legendre symbol.

(e/c) For prime moduli, the problem is relatively easy and there a number of algorithms on offer. When n is composite, then it is equivalent to the integer factorization problem since we need the factorization of n. A friendly introduction to anything you ever wanted to know about square roots, including modular square roots, is Ezra Brown's "Square Roots From 1; 24, 51, 10 to Dan Shanks," published in The College Mathematics Journal but also available at [3]. It explains the Shanks-Tonelli algorithm, one of the algorithms for finding the root when n is prime. (also see quadratic residue) nadav (talk) 03:16, 24 May 2007 (UTC)[reply]