Much as before, the idea is that in your post, you surround equations with special characters, and MathJax will convert the contained text into an equation for you. There's two types of equation that you can typeset:
\( y=x^2 \)
. Note that we do not support $ signs as most LaTeX users would be familiar with, since this occurs too frequently in text.
\[ y = \int f(x) dx \]
, which we note is exactly what one would type in a usual LaTeX document.
For reference, the old guide is still available and has a number of useful examples for those getting started.
Finally, please note that for legacy posts, the old [math] [/math]
tags will still continue to work and these will display equations as inline. However it's likely that older posts may look different to the way that they did before.
So I'll start off:
1) http://linuxfreak87.googlepages.com/
1) Covers a lot of stuff, Maths and some physics.
2) http://mathworld.wolfram.com/
2) Amazing maths resource, lot of advanced stuff.
3) Basic and advanced maths here. Good tutorials.
4) Again more good tutorias and weekly challanges.
5) http://mathforum.org/dr.math/
5) LOTS of question solutions here, examples too. This one has helped me a lot in the past and still does
6) http://www.ics.uci.edu/~eppstein/junkyard/
6) Lots of fun geometry, useful stuff and interesting stuff here.
7) http://en.wikipedia.org/wiki/Category:Mathematics
7) As always Wikipedia is a great resource for one and all.
8) http://www.research.att.com/~njas/sequences/Seis.html
8) If your interest is number sequences this is the place to go. Useful for research.
9) Equations, equations and yes you guessed - MORE equations. Very useful resorce for reference.
10) http://home.att.net/~numericana/
10) Lots of interesting stuff and some other useful links too.
11) http://www.mcs.surrey.ac.uk/Personal/R.Knott/
11) Lots of interesting stuff, the mysteries of the Fibonacci Numbers etc.
12) http://integrals.wolfram.com/
12) Very useful too, online integral solver!
13) Maths in music, what next?
If you have more to add pease share them
Cheers,
Ryan Jones
]]>a^{3} = 345 / n^{2}
Is there any way to calculate da/dt?
]]>
This issue is offtopic in a thread about electrons so I have started a new thread for discussion.
https://www.scienceforums.net/topic/115443-electrons-how-do-they-work/?tab=comments#comment-1062243
Quote1 hour ago, studiot said:Indeed, which is why I based my figure smack in the middle of the range that can can currently be found.
This range is very large, larger than the range of atomic sizes for instance.
Furthermore zero is a finite number.
Finite is often used to mean a non-zero number
https://en.wikipedia.org/wiki/Finite_number
"In mathematical parlance, a value other than infinite or infinitesimal values and distinct from the value 0"
I prefer the definitions of
Dedekind : not infinite
Russell : Able to be counted using a terminating sequence of natural numbers.
Consider the equation
x2 - 2x + 1= 0
Is the difference between the two roots of this quadratic or infinite?
]]>
This statement is where we could use my own new type of maths I am trying to bring into secondary education through my company "Hydra." I have spent a lot of time planning new approached to maths, at secondary and tertiary levels, so, please bear with me and observe where the strengths are of my new system? This is just an example of how it can be applied, of course.
[9H] * [6K] * [2X] * [1Q] = [108A]. This is because the powers can be applied to the symbols and then taken as '[X1]'. This leads to a simple set of symbols to multiply.
Then, we take the [108A] divided into the symbols on the left that we recognize, being [H], [X] and [K], coming to [HXK] = [108].
This leaves the [DG] being equal to 'the left over bits.' This means we can say [HKX] = [108] and [DG] = [X], so, that means the sum on the left equals [HK2X] = [108].
Or, did I make a mistake somewhere... I am in a heck of a hurry!
]]>This would be [infinity] * [2n] = [3] * [2] = [6] * [x]
Then, [5] * [2] = [10] * [x]
Then, [7] * [2] = [14] * [x]
So, the answer is double prime times by [x].
Or,
[N] * [2] *[X].
]]>I have had the urge for years so if anyone could help that would be great.
]]>
I am interested to know in the context of intrinsic curvature but feel I need to get this concept well understood first.
For example must a mathematical "surface" in a 4-D space be 2-dimensional (like a skin) or is it 3-dimensional (like a volume)?
If it is 3-dimensional,what defines it as a surface?
]]>
This is my final work on the Prime product problem.
I know it is just x^2 * y^2 = PNP^2
However the terms would just cancel out. Instead I have decided to let x^2 equal a pattern of x and PNP. So I just substituted the equation which is more complex and will not equal the right side of the equation for x^2.
In calculus where you have a complex derivative where you let du/dx equal a portion of the derivative so you can understand and simplify the manipulation of the integral. I am instead taking a more complex pattern and leaving it so x^2 does not cancel x squared. By doing this I hope it solves the pattern.
So if you could solve this polynomial equation you would solve the factorization problem.
If you couldn’t solve the polynomial? Well you could just write an algorithm that plugged in Prime numbers from smallest to largest. And because the polynomial is set up to find PNP you would get a feel for the range x was in. I mean, that this time when you try a number how far away the computed value is from PNP is significant.
So if this works it is faster than using division to factor.
But of course I await any disagreeing opinions. I now this problem gets that. But it was my final attempt before moving on to a different problem to pursue.
(((((x^2*PNP^4 + 2*PNP^2 * x^5) + x^8)/ PNP^4 ) - ((1 - x^2/(2*PNP)))) * ((PNP^2/x^2 ))) == PNP^2 Above is the pattern of x^2 * y^2 = PNP^2 It is not to be simplified yet x put and tested in that place. It is faster than division since the equation approaches PNP as the proper x is used. Smallest to largest Prime numbers are to be used. PNP = 85 x = 5 (((((x^2*PNP^4 + 2*PNP^2 * x^5) + x^8)/ PNP^4 ) - ((1 - x^2/(2*PNP)))) * ((PNP^2/x^2 ))) 85 5 4179323/578 N[4179323/578, 14] Sqrt[7230.66262975778546712802768166089965397924`14.] 85.033303062728 ((((x^2*PNP^4 + 2*PNP^2 * x^5) + x^8)/ PNP^4 ) - ((1 - x^2/(2*PNP)))) 4179323/167042 N[4179323/167042, 14] 25.019593874594 ((PNP^2/x^2)) 289 Questions to ask: Is it unique to factors of PNP or does it just give a decimal to all real numbers? Does it work for other values of PNP and x? Is it faster than factoring (recursion)?. Is it just x = x and as so not a useful pattern? Can the error be programmed? Verify then post. That is what I need to do. But I actually believe there is a pattern here. The question is does it work for all PNP and x values. I have many patterns. Some answer some of the questions. But I haven't found a polynomial I can solve after these questions have been answered. For example if this worked, I would need to solve the given equation. And this proves to be challenging. PNP = 85 x = 3 (((((x^2*PNP^4 + 2*PNP^2 * x^5) + x^8)/ PNP^4 ) - ((1 - x^2/(2*PNP)))) * ((PNP^2/x^2 ))) 85 3 847772947/130050 N[847772947/130050, 14] 6518.8231218762 Sqrt[6518.8231218762] 80.739229138481]]>
Now let's say you want there to be a chance of a red marble appearing instead of a green one. In order to make this happen, you must define a separate 'mean time to happen' for the red marbles. But you do not want to change the overall chance of any marble to appear; the combined probability of a red or green marble appearing must be the same as the previous probability of a green one appearing.
Aim at, say, a fifteen percent probability of a red marble appearing instead of a green one, with the current 'mean time to happen' being one thousand seconds. What 'mean times to happen' would you have to assign to both green and red?
]]>Is there any way to calculate from those numbers that the semi-major axis is 26534.9039306654?
]]>
[math]\phi^{(\displaystyle\frac{\pi + \phi}{2})}= \pi[/math]
which manages to roughly approximate [math]\pi[/math]. I then found if you did
[math]\phi^{(\displaystyle\frac{\pi + \phi}{x})}= \pi[/math] with [math]x = 2.000811416[/math],
the equation exactly reached [math]\pi[/math]. But [math]x = 2.000811416[/math] seems too random to me, is there any connection between [math]\pi and \phi[/math] that would produce [math]x = 2.000811416[/math]?
On the slight chance you understood what i said, do you know where [math]x = 2.000811416[/math] can be derived from?
Cheers,
Rob
]]>everything dean has shown was known at the time godel did his proof but no one meantioned any of it
http://gamahucherpress.yellowgum.com/books/philosophy/GODEL5.pdf
look
godel used the 2nd ed of PM he says
“A. Whitehead and B. Russell, Principia Mathematica, 2nd edition, Cambridge 1925. In particular, we also reckon among the axioms of PM the axiom of infinity (in the form: there exist denumerably many individuals), and the axioms of reducibility and of choice (for all types)”
note he says he is going to use AR
but
Russell following wittgenstien took it out of the 2nd ed due to it being invalid
godel would have know that
russell and wittgenstien new godel used it but said nothing
ramsey points out AR is invalid before godel did his proof
godel would have know ramseys arguments
ramsey would have known godel used AR but said nothing
Ramsey says
Such an axiom has no place in mathematics, and anything which cannot be
proved without using it cannot be regarded as proved at all.
This axiom there is no reason to suppose true; and if it were true, this
would be a happy accident and not a logical necessity, for it is not a
tautology. (THE FOUNDATIONS OF MATHEMATICS* (1925) by F. P. RAMSEY
every one knew AR was invalid
they all knew godel used it
but nooooooooooooo one said -or has said anything for 76 years untill dean
the theorem is a fraud the way godel presents it in his proof it is crap
]]>The motion graph should look something like in the picture. It may look like sqrt(x) at first but this isn't the case, I am wondering if there is some function that actually reaches a slope of 0 relatively quickly (or a VERY tiny slope) ?
Thanks!
]]>
Let all abstract numbers be defined exactly as concrete numbers.
Concrete number: A numerical quantity with a corresponding unit.
Let the corresponding unit exist as an abstract dimension notated with the use of (_).
Let the length and width of all dimensional units remain abstract and undeclared.
Let the dimensional unit be equal in quantity to the numerical quantity it corresponds to.
Let all numerical quantities inhabit their corresponding abstract dimensional units.
Let zero be assigned a single dimensional unit.
Classic Isomorphic
0 = (0) = (0,_) = (0,0_)
1 = (1) = (1,_) = (1,1_)
2 = (2) = (2,_,_) = (2,2_)
3 = (3) = (3,_,_,_) = (3,3_)
(-1) = (-1) = (-1,_) = (-1,1_)
(-2) = (-2) = (-2,_,_) = (-2,2_)
(-3) = (-3) = (-3,_,_,_) = (-3,3_)
Therefore:
Any classic number (n) = isomorphic (n) = (n,n_).
Where (_) is defined as a dimensional unit, the quantity of which corresponds to a given numerical quantity.
Where (n) is defined as the numerical quantity separate from the dimensional unit.
Where (n_) is defined as the dimensional unit separate from the numerical quantity, and equal in quantity to the numerical quantity it corresponds to.
Let addition and subtraction exist without change. Except regarding notation: (a+b = c: a+0 = a: a-0 = a: 0+0 = 0: 0-0 = 0).
In any binary expression of multiplication let one number (n) represent only a numerical quantity or (n), let the other number (n) represent only a quantity of dimensional unit equal in quantity to the number it corresponds to, or (n_).
In any binary expression of division let the numerator (n) always exist as a numerical quantity or (n), let the denominator (n) always exist as a dimensional unit quantity equal in quantity to the number it corresponds to, or (n_). Therefore, in all cases of binary division (n/n): (n) is notated as (n/n_).
Let multiplication be defined as the placing of a given numerical quantity, with addition, equally into each given quantity of dimensional unit. Then all numerical quantities in all dimensional units are added.
Let division be defined as the placing of a given numerical quantity, with subtraction, equally into each given quantity of dimensional unit. Then all numerical quantities in all dimensional units are subtracted except one.
In all binary operations of multiplication containing a number (0) and a non-zero number (n), the notation of the number (0) as (0) or as (0_), will dictate the notations of the binary non-zero number (n) in the operation.
In all cases of a binary expression where the notation is not given for the number (0), the numerical quantity (0) is notated for (0), and the dimensional quantity (n_) is notated for (n).
Therefore: (n*0 = n_*0 = 0).
Let exponents and logarithms exist without change. Except regarding notation: (a^b = c).
Assertion:
All binary operations of multiplication and division remain unchanged except binary operations involving the number (0). As well as defining division by the number (0) as an operation of a given numerical quantity (n) into the dimensional unit quantity (0_).
Multiplication
Classic
2*3 = 6
Isomorphic
2*(_,_,_) = 6
Where:
Classic (2): is the numerical quantity.
Classic (3): is the dimensional unit quantity.
(_,_,_): the dimensional unit quantity of the number (3).
(2,2,2): the numerical quantity (2) added equally into all dimensional unit quantities.
(2+2+2 = 6): the numerical quantity (2) added equally into all dimensional unit quantities, then all numerical quantities in all dimensional unit quantities are added.
Therefore:
2*(_,_,_) = 6
Or,
3*(_,_) = 6
Where:
Classic (2): is the dimensional unit quantity.
Classic (3): is the numerical quantity.
(_,_): the dimensional unit quantity of the number (2).
(3,3): the numerical quantity (3) added equally into all dimensional unit quantities.
(3+3 = 6): the numerical quantity (3) added equally into all dimensional unit quantities, then all numerical quantities in all dimensional unit quantities are added.
Therefore:
3*(_,_) = 6
Classic
2*0 = 0
Isomorphic
2*(_) = 2
Where:
Classic (2): is the numerical quantity.
Classic (0): is the dimensional unit quantity.
(_): the dimensional unit quantity of the number (0).
(2): the numerical quantity (2) added equally into all dimensional unit quantities.
(2): the numerical quantity (2) added equally into all dimensional unit quantities, then all numerical quantities in all dimensional unit quantities are added.
Therefore:
2*(_) = 2
Or,
0*(_,_) = 0
Where:
Classic (0): is the numerical quantity.
Classic (2): is the dimensional unit quantity.
(_,_): the dimensional unit quantity of the number (2).
(0,0): the numerical quantity (0) added equally into all dimensional unit quantities.
(0+0 = 0): The numerical quantity (0) added equally into all dimensional unit quantities, then all numerical quantities in all dimensional unit quantities are added.
Therefore:
0*(_,_) = 0
Classic
0*0 = 0
Isomorphic
0*(_) = 0
(_): the dimensional unit quantity of the number (0).
(0): the numerical quantity of (0) added equally into all dimensional unit quantities.
(0): the numerical quantity of (0) added equally into all dimensional unit quantities, then all numerical quantities in all dimensional unit quantities are added.
Therefore:
0*(_) = 0
Therefore, the product of binary multiplication by the number (0) with a non-zero number, is relative to the number (0) declared as a numerical quantity or as a dimensional unity quantity in the binary expression.
Isomorphic expressions containing variables.
Where: (n) =/= 0
n*(0_) = n = (0_)*n
n*(_) = n = (_)*n
n_*0 = 0 = 0*n_
Division
Classic
6/2 = 3
Isomorphic
6/(_,_) = 3
Where:
Classic (6): is the numerical quantity.
Classic (2): is the dimensional unit quantity.
(_,_): the dimensional unit quantity of the number (2).
(3,3): the numerical quantity (6) subtracted equally into all dimensional unit quantities.
(3): the numerical quantity (6) subtracted equally into all dimensional unit quantities, then all numerical quantities in all dimensional unit quantities are subtracted except one.
Therefore:
6/(_,_) = 3
Classic
1/4 = .25
Isomorphic
1/(_,_,_,_) = .25
Where:
Classic (1): is the numerical quantity.
Classic (4): is the dimensional unit quantity.
(_,_,_,_): the dimensional unit quantity of the number (4).
(.25,.25,.25,.25): the numerical quantity (1) subtracted equally into all dimensional unit quantities.
(.25): the numerical quantity (1) subtracted equally into all dimensional unit quantities, then all numerical quantities in all dimensional unit quantities are subtracted except one.
Therefore:
1/(_,_,_,_) = .25
Classic
0/2 = 0
Isomorphic
0/(_,_) = 0
Where:
Classic (0): is the numerical quantity.
Classic (2): is the dimensional unit quantity.
(_,_): the dimensional unit quantity of the number (2).
(0,0): the numerical quantity (0) subtracted equally into all dimensional unit quantities.
(0): the numerical quantity (0) subtracted equally into all dimensional unit quantities, then all numerical quantities in all dimensional unit quantities are subtracted except one.
Therefore:
0/(_,_) = 0
Classic
2/0 = undefined
Isomorphic
2/(_) = 2
Where:
Classic (2): is the numerical quantity.
Classic (0): is the dimensional unit quantity.
(_): the dimensional unit quantity of the number (0).
(2): the numerical quantity (2) is subtracted equally into all dimensional unit quantities.
(2): the numerical quantity (2) is subtracted equally into all dimensional unit quantities, then all numerical quantities in all dimensional unit quantities are subtracted except one.
Therefore:
2/(_) = 2
Classic
0/0 = undefined
Isomorphic
0/(_) = 0
Where:
Classic numerator (0): is the numerical quantity.
Classic denominator (0): is the dimensional unit quantity.
(_): the dimensional unit quantity of the number (0).
(0): the numerical quantity (0) subtracted equally into all dimensional unit quantities.
(0): the numerical quantity (0) subtracted equally into all dimensional unit quantities, then all numerical quantities in all dimensional unit quantities are subtracted except one.
Therefore:
0/(_) = 0
Isomorphic expressions containing variables.
Where (n) =/= 0
n/(0_)= n
n/(_) = n
0/(n_) = 0
Therefore, division by zero is expressible as a quotient. By definition of division, the numerical quantity (0) can never exist as a divisor. Only the dimensional unit quantity of the number (0) or (_), or (0_) may exist as a divisor.
Therefore, all division is defined as a specific operation of a given numerical quantity into a given dimensional unit quantity. So that division by zero is defined as a given numerical quantity operated into the dimensional unit quantity of the number (0).
Assertion:
The defining of abstract numbers and the operations of multiplication and division as given above will allow for a mathematical construct in which it is possible to define division by zero. It will also do so in such a manner as to not contradict any given field axiom.
*As all operations of addition and subtraction exist without change only the field axioms regarding multiplication will be addressed*
Field Axioms
Associative: (ab)c = a(bc)
Commutative: ab = ba
Distributive: (a+b)c = ac+bc
Identity: a*1 = a = 1*a
Inverses: a*a^(-1) = 1 = a^(-1) * a: if a =/= 0
For the field axioms to hold, the defining of special operations for binary multiplication of the number (0) on the number (n) must be considered. In these special cases alone, binary expressions of multiplication may exist without a unique numerical quantity and a unique dimensional unit quantity.
Allow that: (0*0 = 0)
As the numerical quantity of the number (0) can be added to the numerical quantity of the number (0): But cannot yield a product containing a dimensional unit quantity.
Allow that: (0_*0_ = 0_)
As the dimensional unit quantity of the number (0) can be added to the dimensional unit quantity of the number (0): But cannot yield a product containing a numerical quantity.
Where any number (0) exists as undefined in a binary expression of multiplication:
(0*0 = 0): (0*0_ = 0): (0*0 = 0)
Therefore:
(n+0 = n): (n+0_ = n): (n+0 = n)
Where (n) =/= (0): and (0) exists as undefined in a binary expression:
(n*0) = (n*0) = (n_*0) = 0
Associative
(ab)c = a(bc)
Isomorphic equations.
(a*b)c = a(b*c)
Let: a = 1, b = 2, c = 0: 0 (is a numerical quantity for use in all binary expressions)
(1_*2)0 = 1(2_*0)
2_*0 = 1*0
0 = 1_*0
0 = 0
Let: a = 1, b = 2, c = 0: 0_ (is a dimensional quantity for use in all binary expressions)
(1_*2)0 = 1(2*0_)
2*0_ = 1*2_
2 = 2
Continued isomorphic examples of the associative axiom.
Let: a = 1, b = 0: 0, c = 0: 0
(1_*0)0 = 1(0*0)
0*0 = 1_*0
0 = 1_*0
0 = 0
Let: a = 1, b = 0: 0_, c = 0: 0_
(1*0_)0 = 1(0_*0_)
1*0_ = 1*0_
1 = 1
Let: a = 1, b = 0: 0, c = 0: 0_
(1_*0)0 = 1(0*0_)
0*0_ = 1*0
0 = 1_*0
0 = 0
Let: a = 1, b = 0: 0_, c = 0: 0
(1*0_)0 = 1(0_*0)
1_*0 = 1*0
0 = 1_*0
0 = 0
Therefore, the associative axiom still holds as true.
Commutative
a*b = b*a
Isomorphic equations.
a*b = b*a
Let: a = 2: 2, b = 3: 3_
2*(_,_,_) = (_,_,_)*2
2*3_ = 3_*2
6 = 6
Let: a = 2: 2_, b = 3: 3
3*(_,_) = (_,_)*3
3*2_ = 2_*3
6 = 6
Continued isomorphic examples of the commutative axiom.
If (a) = 0: 0
0*b_ = b_*0
0 = 0
If (a) = 0: 0_
0_*b = b*0_
b = b
If (b) = 0: 0
a_ *0 = 0*a_
0 = 0
If (b) = 0: 0_
a*0_ = 0_*a
a = a
Therefore, the commutative axiom still holds true.
Distributive
(a+b)c = a*c+b*c
Isomorphic equations.
(a+b)c = a*c+b*c
Let: a = 1, b = 2, c = 0: 0
(1+2)0 = 1_*0+2_*0
3_*0 = 0+0
0 = 0
Let: a = 1, b = 2, c = 0: 0_
(1+2)0 = 1*0_+2*0_
3*0_ = 1+2
3 = 3
Continued isomorphic examples of the distributive axiom.
Let: a = n, b = 0: 0, c = 0: 0
(n+0)0 = n_*0+0*0
n_*0 = 0+0
0 = 0
Let: a = n, b = 0: 0_, c = 0: 0_
(n+0)0 = n*0_+0_*0_
n*0_ = n+0_
n = n
Let: a = n, b = 0: 0, c = 0: 0_
(n+0)0 = n*0_+0*0_
n*0_ = n+0
n = n
Let: a = n, b = 0: 0_, c = 0: 0
(n+0)0 = n_*0+0_*0
n_*0 = 0+0
0 = 0
Therefore, the distributive axiom still holds as true.
Identity
a*1 = a = 1*a
Isomorphic
a*1 = a = 1*a
For the identity axiom to hold: (a) =/= (0)
Where (a) = 0: the operations of (0) by the multiplicative identity (1) is given previously in the text.
Where (a) =/= 0: All binary expressions not involving zero exist without change.
Therefore, except regarding the number (0), the identity axiom still holds as true.
Inverses
a*a^(-1) = 1 = a^(-1) * a: if a =/= 0
Isomorphic
a*a^(-1) = 1 = a^(-1) * a: if a =/= 0
As all binary expressions not involving zero exist without change, the inverse axiom holds as true.
Where (a) = 0: the number (0) remains without a multiplicative inverse.
The dimensional unit quantity of the number (0): (_), or (0_), cannot be considered the multiplicative inverse of the number (1). By definition the multiplicative inverse of the number (1) must be a numerical quantity. Therefore, the numerical quantity (1) remains the only multiplicative inverse for the number (1).
Therefore, all field axioms continue to exist as true.
Examples as to the validity for the necessity of Numerus “Numerans-Numeratus”.
1. Provides for a mathematical construct in which it is possible to define division by zero.
2. As division by zero is defined, any slope formula expressing division by zero is definable. Therefore, the slope of a formula expressing division by zero can be expressed as “vertical”.
3. Allows for division by zero in a field, without contradicting the field axioms.
4. Allows dimensional analysis to define division by zero with “actual concrete numbers”, within the confines of its own system. The possibility of which was previously unexplored, the application of which is applicable to physics.
5. Therefore, physics, semantics, philosophy and mathematics can be considered to be unified to an extent. As all abstract numbers have been shown to exist and function, exactly as concrete numbers. Therefore, the unification of abstract and concrete principles, both in mathematics and in physics.
]]>
1) The number of letters in Hamlet (or characters if you want it to include spacing, punctuation etc., but excluding capitalization)
2) The average time it takes someone to type one letter, or in other words, words per minute. We must be given some leeway here because we must agree upon whether the monkey is frantically mashing the button with its fingers (not whole hands, because then the probability would always be 0) or it is typing at a rate of an average human.
3) The number of accepted buttons we will give the monkey (or words in the alphabet, depending on what we want to do)
I think given these variables, the calculation should be easy. Of course, this assumes that all keys/letters have an equal probability of being hit
Another thing we can do is calculate the probability of the monkey writing Hamlet on the first attempt. This has one less variable (2 is excluded) so it is an even easier calculation. I'm sure I could find the answer out by googling, but it is more fun this way.
So, does anyone care to add the necessary information?
]]>
Copyrighted©PiyushGoel
^{}
3^{2} + 4^{2 }= 5^{2}
9 + 16 = 25
1.(3/4)^{2} + (4/4)^{2} = (5/4)^{2}
0.75^{2} + 1^{2} = 1.25^{2}
0.5625 + 1 = 1.5625
2.(3/5)^{2} + (4/5)^{2} = (5/5)^{2}
0.6^{2} + 0.8^{2} = 1^{2}
0.36 + 0.64 = 1
O.K. Fellas, let’s notice a difference here.
3.(3/3)^2 +(4/3)^2 = (5/3)^2
1^2 + 1.3^2 = 1.6^2
1 + 1.69 = 2.56
Difference => 2.69 – 2.56 = 0.13
Accuracy =>0.13/2.69=.048
4.(3/3)^2 +(4/3)^2 = (5/3)^2
1^2 + 1.33^2 = 1.66^2
1 + 1.7689 = 2.7556
Difference => 2.7689 – 2.7556 = 0.133
Accuracy => 0.133/2.7689 = 0.0048
5.(3/3)^2 +(4/3)^2 = (5/3)^2
1^2 + 1.333^2 = 1.666^2
1 + 1.776889 = 2.775556
Difference => 2.776889 – 2.775556 = 0.001333
Accuracy => 0.001333/2.776889 = 0.00048
6.(3/3)^2 +(4/3)^2 = (5/3)^2
1^2 + 1.3333^2 = 1.6666^2
1 + 1.77768889 = 2.77755556
Difference => 2.77768889 – 2.77755556 = 0.00013333
Accuracy => 0.0001333/2.77768889 = 0.000048
7.(3/3)^2 +(4/3)^2 = (5/3)^2
1^2 + 1.33333^2 = 1.66666^2
1 + 1.7777688889 = 2.7777555556
Difference => 2.7777688889 – 2.7777555556 = 0.000013333
Accuracy => 0.000013333/2.7777688889 = 0.0000048
8.(3/3)^2 +(4/3)^2 = (5/3)^2
1^2 + 1.333333^2 = 1.666666^2
1 + 1.777776888889 = 2.777775555556
Difference => 2.77777688889 – 2.777775555556 = 0.0000013333
Accuracy => 0.0000013333/2.777776888889 = 0.00000048
So, as you can see its pretty much visible here that irrespective of the precision there is in division, the presence of digits 4 and 8 in the Accuracy Factor is mesmeric to look. This may come under Number Pattern Theory where the operation eventually leads to same numerical digits though quantitatively they are different.
Author: Piyush Goel
I could go on and on, but I'll go onto explain some of the basic syntax of LaTeX.
Syntax
Functions & General Syntax
Basically put, if you want to write a math equation in LaTeX, you just write it. If you wanted f(x) = 3, then bung that between to math tags and you're done, producing [math]f(x)=3[/math]. Don't worry about extra spaces or carriage returns, because in general LaTeX will ignore them. It does get a little more complex than this, but don't worry about that for now. Remember that any letters you type in will be presumed to be some kind of variable and hence will be italicised.
We also have functions to display more complex things like matrices and fractions, and they have the syntax of having a \ before them, usually followed by some kind of argument. For example, \sin will produce the function sin and \frac{num}{denom} will produce a fraction with a specified numerator and denominator. More on these later.
Also remember that LaTeX is case sensitive, so \sigma is NOT the same as \Sigma.
Subscripts and Superscripts
This is perhaps one of the easiest things to do in LaTeX, and one of the most useful. Let's, for the sake of argument, say you wanted to write x^{2}. Then you'd write x^{2}, producing [math]x^2[/math]. Notice that you don't necessarily need the { and } in cases where you only have 1 thing in the index, for example x^2. But it does care if you want to write something like [math]x^{3x+2}[/math]. Subscripts are done similarly, but you use the _ operator instead of ^. If you want both subscript and superscript, then use the syntax x^{2}_{1} - which is equivalent to x_{1}^{2}.
Fractions and functions
As I've mentioned, fractions are generated by using the function \frac{num}{denom}. For example:
[math]\frac{1}{3}[/math]
[math]\frac{7}{x^2}[/math]
If you want smaller fractions, you can use \tfrac, to produce things like [math]\tfrac{1}{2}[/math] which will fit into a line nicely without having to seperate it.
LaTeX has some nice in-built functions like \sin, \cos, etc. I'm not going to write them all down here, but I'll point you to a website at the end of the document that contains them. Likewise, you can write symbols (such as infinity by using \infty) and Greek letters (e.g. \phi, \Sigma, \sigma, etc)
Bracketing
You can get all your usual brackets just by typing them straight in; for instance, (, |, [, etc. However, sometimes they won't be the right size, especially if you want to write something like (1/2)^{n}. You can get around this by using the \left and \right commands, and then placing your favourite brackets after them. For instance, to write (1/2)^{n}, we have:
[math]\left( \frac{1}{2} \right)^{n}[/math]
Integrals, Summations and Limits
Integrals can be produced by using \int, summations by \sum and limits by \lim. You can put limits on them all in the right places by using the normal subscript/superscript commands. For instance:
[math]\int_a^b x^2 \,dx[/math]
[math]\lim_{n\to\infty} \frac{1}{n} = 0[/math]
[math]\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}.[/math]
Summary
There's a lot more things you can do with LaTeX, and I'll try to add to this as time goes by. Have a look at:
http://www.maths.tcd.ie/~dwilkins/LaTeXPrimer/'>http://www.maths.tcd...ns/LaTeXPrimer/ - the LaTeX primer
http://omega.albany.edu:8008/Symbols.html'>http://omega.albany....08/Symbols.html - some symbols that you might find useful.
If you have any questions about the system, send me a PM and I'll try to help
Cheers.
The question is: given a binary string that contains 11 digits what is the probability that there is exactly four 1s in the number. I can solve half of it, I know that 11 digits means 2^11 or 2048 total numbers, but I can't figure out how to find out how many contain exactly four 1s through a formula.
]]>Lot of mathematicians have proved Pythagoras theorem in their own ways. If you google it you will indeed found hundred of ways.
Meanwhile I was also sure that maybe one day I could find something new out of this incredible Pythagoras theorem and Recently I got something which I would like to share with you.
To Prove: Deriving the equation of area of trapezium using Arcs
Proof: There is a triangle ABC with sides a b and c as shown in the figure.
Now,
Area of ∆ BCEG = Area of ∆ BDC +Area of ⌂ DCEF + Area of ∆ EFG
c^2=ac/2+ Area of ⌂ DCEF + (c-b) c/2
(2c^2– ac –c^2+ bc )/2=Area of ⌂ DCEF
(c^2– ac+ bc )/2=Area of ⌂ DCEF
c(c– a+ b)/2=Area of ⌂ DCEF
Area of ⌂ DCEF=BC(DE+CF)/2
Copyrighted©PiyushGoel