<![CDATA[Math Is Fun Forum / Formulas]]> 2021-10-02T19:57:46Z FluxBB http://www.mathisfunforum.com/index.php <![CDATA[Formula for Pi using the Nilakantha Series]]> To find a particular term in the series, assuming that the term number is x, replace ∞ with x-1, which works for everything except the first term which is of course 3.

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http://www.mathisfunforum.com/profile.php?id=222180 2021-10-02T19:57:46Z http://www.mathisfunforum.com/viewtopic.php?id=26663&action=new
<![CDATA[Cannot find formula or solution to get answer for feet/inch addition]]> Hi Nickm62388;

Here's my way of tackling these (others may have different methods).

Problem #1
The inches & fraction components in the first length are less than their counterparts in the second length, which adds difficulty to subtracting imperial lengths, particularly those with fractions.

To make the subtraction easier, I start with the following three steps:
1. convert the fraction to the Least Common Denominator (LCD); then
2. adjust the value of the fraction component to make it greater than that of the second length; then
3. adjust the value of the inches component to make it greater than that of the second length.

That leaves a simple subtraction exercise.

Btw, the 'take' and 'add' terminology that I used is like 'borrow' and 'pay', now aka 'regrouping', and is how I've always understood that concept.

Also, I've used symbols ' and " instead of abbreviations 'ft' and 'in' respectively...a common notation practice for these imperial length units. Formula: You could convert both lengths to eighths of an inch, deduct the second from the first, and convert the result to feet, inches and eighths:

That last conversion from eighths is a bit tricky...

Problem #2
Same approach as for Problem #1: Formula: You could convert both angles to seconds, deduct the second from the first, and convert the result to degrees, minutes & seconds:

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http://www.mathisfunforum.com/profile.php?id=40741 2021-07-01T05:29:19Z http://www.mathisfunforum.com/viewtopic.php?id=26416&action=new
<![CDATA[I am confused]]> None of these

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http://www.mathisfunforum.com/profile.php?id=236935 2021-06-02T14:22:47Z http://www.mathisfunforum.com/viewtopic.php?id=24712&action=new
<![CDATA[common mistake]]> Hostible wrote:

Know that -a^2 means -1 x a^2 and not -a x -a.
The reason is that -a^2 has its coefficient as -1
Therefore -a x -a = a^2 and and not -a^2.
This symbol ^ implies exponent.

True but (-a)^2 is totally different.

We know that (-a)^2 = (-a)(-a) = a^2.

Let me see.

-[(a)(a)]

-(a^2)

-a^2

Agree?

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http://www.mathisfunforum.com/profile.php?id=223890 2021-05-15T16:50:47Z http://www.mathisfunforum.com/viewtopic.php?id=19547&action=new
<![CDATA[Elegant Cosine, Heron's and Half Angle Formulae using semiperimeter]]> The Cosine Rule with its squares of sides a , b and c cries out for a fusion with the semiperimeter.

First lets turn the semiperimeter into a few substitutions for later use

Where semiperimeter = s  = ( a + b + c ) / 2

( a + b + c )  = 2 s
( a + b -  c )   =2 ( s – c )
( a -  b + c )   =2 ( s – b )
( b + c -  a )   =2 ( s – a )

Then our Cosine rule

a² = b² +  c² – 2bcCos A

Cos A = ( b² +  c² -  a² ) / 2bc

1 – Cos A = ( 2bc - b² -  c² +  a² ) / 2bc
= ( a – b + c )( a + b - c ) / 2bc
= 2 ( s - b )( s – c ) /  bc

1 + Cos A = ( 2bc + b² +  c² -  a² ) / 2bc
= ( a + b + c )( b + c - a ) / 2bc
= 2 s ( s – a ) /  bc

Sin² A = ( 1 – Cos a )( 1 + Cos A )
= 4 s ( s – a )( s – b )( s - c ) / b²c²

_______________________
Sin A = 2 ( √  s ( s – a )( s – b )( s - c )  ) / bc

Area of Triangle =  ½ . bc Sin A
______________________
=  √  s ( s – a )( s – b )( s - c )   Heron's Area Formula

A much more elegant way of deriving Heron's Formula and there is still the Half angle Formula to go :-

______________
Cos ( A / 2 ) =  √  ( 1 + Cos A ) / 2
______________
=  √  s  (  s – a ) / bc

______________
Sin ( A / 2 ) =  √  ( 1 - Cos A ) / 2
___________________
=  √  ( s – b )(  s – a ) / bc

& Sin A  = 2 Sin ( A / 2 ) Cos ( A / 2 ) gives the same Sin A result as above

Tan ( A / 2 ) = Sin ( A / 2 ) / Cos ( A / 2 )
_______________________
=  √ ( s – b )( s – c ) / s ( s – a )

corrected mistake in half angle formulae

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http://www.mathisfunforum.com/profile.php?id=223802 2021-03-23T14:27:04Z http://www.mathisfunforum.com/viewtopic.php?id=26051&action=new
<![CDATA[Modern Algebra Symbols]]> You have probably solved the problem by now.

Its the geometric symbol for congruent too , and not in open office function editor, which calls the equivalance symbol congruent .

The best stop gap solution would be to screen shot your formula and edit in a photoshop type editor.

I use gimp which is a free editor and just as powerful .

If you then search Google images for ' congruent symbol ' you can copy the one you prefer to your clipboard and paste it in.

There are a lot of formula editors on the market all have limitations but you are using the best of the bunch I think.

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http://www.mathisfunforum.com/profile.php?id=223802 2021-03-20T21:42:23Z http://www.mathisfunforum.com/viewtopic.php?id=24434&action=new
<![CDATA[Primality test and easy factorization]]> Hugo28036 wrote:

This is an algorithm that test if one number is prime and if the number it´s not prime returns the biggest factor of that number.

Are you going to keep it a secret and profit off of it or are you going to enlighten our poor souls as to what it is? ]]>
http://www.mathisfunforum.com/profile.php?id=188456 2020-09-11T23:46:17Z http://www.mathisfunforum.com/viewtopic.php?id=25864&action=new
<![CDATA[Proof of the Twin primes Conjecture and Goldbach's conjecture]]> Proof of the Twin primes Conjecture and Goldbach's conjecture
We can find infinite prime numbers with the separation we want and we can express every even number as the sum of two prime numbers.
Proof of the Twin primes Conjecture and Goldbach's conjecture

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http://www.mathisfunforum.com/profile.php?id=222504 2020-08-15T06:13:34Z http://www.mathisfunforum.com/viewtopic.php?id=25831&action=new
<![CDATA[Sum of first n natural numbers - fourth, fifth, and sixth powers.]]> Hi,

.

We know

Sum of first n natural numbers is

Sum of squares of first n natural numbers is

Sum of cubes of first n natural numbers is

In a similar manner, this is additional information.

Sum of fourth powers of n natural numbers is

Sum of fifth powers of n natural numbers is

Sum of sixth powers of n natural numbers is

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http://www.mathisfunforum.com/profile.php?id=682 2020-07-16T05:00:41Z http://www.mathisfunforum.com/viewtopic.php?id=25783&action=new
<![CDATA[Do mental multiplication quickly]]> [deleted, please post if you need this]

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http://www.mathisfunforum.com/profile.php?id=222180 2020-06-29T18:04:47Z http://www.mathisfunforum.com/viewtopic.php?id=25746&action=new
<![CDATA[Decimal to Fraction]]> Hi,

Since edited.

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http://www.mathisfunforum.com/profile.php?id=682 2019-09-06T03:12:10Z http://www.mathisfunforum.com/viewtopic.php?id=25099&action=new
<![CDATA[Solid Geometry Formulas]]> irspow wrote:

or...

A = s²/(4tan[180/n]) where s is length of side and n is number of sides (for those who don't want to use pi to calculate the area of a polygon) Unfortunately, trigonometry tends to turn more people off than π.

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http://www.mathisfunforum.com/profile.php?id=212031 2019-04-18T08:08:29Z http://www.mathisfunforum.com/viewtopic.php?id=3292&action=new
<![CDATA[Tests For Divisibility]]> Here’s a YouTube video on tests of divisibility and how to apply them to some math problems.

[list=*]
[/list]

I found the solution to finding all palindromic four-digit powers particularly interesting.

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http://www.mathisfunforum.com/profile.php?id=217821 2018-07-24T22:21:40Z http://www.mathisfunforum.com/viewtopic.php?id=13168&action=new
<![CDATA[Trigonometry Formulas]]> ganesh wrote:

(Angles are given in degrees, 90 degrees, 180 degrees etc.)

I.
Sin(-θ)=-Sinθ
Cos(-θ) = Cosθ
tan(-θ) = -tanθ
cot(-θ) = -cotθ
sec(-θ) = secθ
cosec(-θ)= - cosecθ

II.
sin(90-θ) = cosθ
cos(90-θ) = sinθ
tan(90-θ) = cotθ
cot(90-θ) = tanθ
sec(90-θ) = cosecθ
cosec(90-θ) = secθ

III.
sin(90+θ) = cosθ
cos(90+θ) = -sinθ
tan(90+θ) = -cotθ
cot(90+θ) = -tanθ
sec(90+θ) = -cosecθ
cosec(90+θ) = secθ

IV.
sin(180-θ) = sinθ
cos(180-θ) = -cosθ
tan(180-θ) = -tanθ
cot(180-θ) = cotθ
sec(180-θ) = -secθ
cosec(180-θ) = cosecθ

V.
sin(180+θ) = -sinθ
cos(180+θ) = -cosθ
tan(180+θ) = tanθ
cot(180+θ) = cotθ
sec(180+θ) = -secθ
cosec(180+θ) = -cosecθ

cot(180-θ)=-cotθ
since cot(180-θ)=1/tan(180-θ)

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http://www.mathisfunforum.com/profile.php?id=219080 2018-04-27T08:48:10Z http://www.mathisfunforum.com/viewtopic.php?id=3290&action=new
<![CDATA[Vector Formulas]]> ganesh wrote:

Where is the supposed "c" in the Left Hand Side?

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http://www.mathisfunforum.com/profile.php?id=212031 2018-01-27T07:09:29Z http://www.mathisfunforum.com/viewtopic.php?id=3289&action=new