I am a Mathematics undergraduate and am desperately in love with Mathematics and this project is going to be a fun, tough but most importantly mind expanding ride.

Subscribe to see an new proof each week.

The project beings on August 5th. Watch this space …

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Click on Life in Saudi Arabia

Wooooooowoooooooh! One of the most loved places by me! As we all know its a desert country! Its land connects with Red Sea and Persian (Arabian) Gulf coastlines.

First thing on recalling Golf countries, I remember that camel ride… haha! Yep! Scary but still enjoyed it! It’s home to the Muslim religion’s 2 most sacred mosques: Masjid al-Haram, in Mecca and Medina’s Masjid an-Nabawi, burial site of the prophet Muhammad(SWS)

Map of Saudi Arabia

In the Holy month of Ramadan we get leaves from office and our business at Saudi Arabia. While having causal talk at home, my kids started telling, this Ramadan lets go on long drive. Haha! after thinking for a while we (me & my hubby) agreed and started planning for a **long-drive** as our kids wish :)

My kids ran to bring a pen and a paper to draw a map of our So called long-drive. My home…

View original post 99 more words

Whenever we talk about *understanding* data, we talk about **models**. In statistics, a model is usually some sort of parameterized function – like a probability density function (pdf) or a regression model. But the effectiveness of the model’s outputs will only be as good as its fit to the data. If the characteristics of the data are very different from the assumed model, the bias turns out to be pretty high. So from a probability perspective, how do we quantify this *fit*?

The two entities of interest are – the data , and the parameters . Now consider a function , that returns a number proportional to the degree of ‘fit’ between the two – essentially quantifying their relationship with each other.

There are two practical ways you could deal with this function. If you kept constant and varied the data being analyzed, you would be getting a function whose only argument is the data. The output would basically be a measure of how well your input data satisfies the assumptions made by the model.

But in real life, you rarely know your model with certainty. What you do have, is a bunch of observed data. So shouldn’t we also think of the other way round? Suppose you kept constant, but tried varying instead. Now, what you have got is a function , that computes how well different sets of parameters describe your data (which you have for real).

Mind you, in both cases, the underlying mathematical definition is the same. The input ‘variable’ is what has changed. This is how *probability* and *likelihood* are related. The function is what we call the probability function (or *pdf* for the continuous case), and is called the **likelihood function**. While assumes you know your model and tries to analyze data according to it, keeps the data in perspective while figuring out how well different sets of parameters describe it.

The above definition might make you think that the likelihood is nothing but a rewording of probability. But keeping the data constant, and varying the parameters has huge consequences on the way you interpret the resultant function.

Lets take a simple example. Consider you have a set of different coin tosses, where out of them were , while the others were . Lets say that the coin used for tossing was biased, and the probability of a coming up on it is . In this case,

Now suppose you made coin yourself, so you know . In that case,

On the other hand, lets say you don’t know much about the coin, but you do have a bunch of toss-outcomes from it. You made 10 different tosses, out which 5 were . From this data, you want to measure how likely it is that your guess of is correct. Then,

There is a very, very important distinction between probability and likelihood functions – the value of the probability function sums (or integrates, for continuous data) to 1 over all possible values of the input data. However, the value of the likelihood function *does not* integrate to 1 over all possible combinations of the parameters.

The above statement leads to the second important thing to note: DO NOT interpret the value of a likelihood function, as the probability of the model parameters. If your probability function gave the value of 0.7 (say) for a discrete data point, you could be pretty sure that there would be no other option as likely as it. This is because, the sum of the probabilities of all other point would be equal to 0.3. However, if you got 0.99 as the output of your likelihood function, it wouldn’t necessarily mean that the parameters you gave in are the most likely ones. Some other set of parameters might give 0.999, 0.9999 or something higher.

The only thing you can be sure of, is this: If , then it is more likely that denote the parameters of the underlying model.

The likelihood function is usually denoted as (likelihood of the parameters given the data point ), so we will stick with it from now on. The most common use of likelihood, is to figure out that set of parameters which yields the highest value for it (and thus describes your dataset the best). This method is called **Maximum Likelihood Estimation**. You maximize the value of the likelihood function in a bid to find the optimal parameters for your model. This trick is applied in many areas of data science, such as logistic regression.

Maximum Likelihood Estimation usually involves computing the partial derivative of the likelihood function with respect to the parameters. We generally deal with the log-likelihood (basically the logarithm of the likelihood function) rather than likelihood itself. Since log is a monotonically increasing function, the optimum value of the likelihood function can be calculated using derivatives of log-likelihood as well. The reason we use logarithms, is to make the process of dealing with derivatives easy. Consider the coin-toss example I gave above:

Your Likelihood function for the probability of , given the and , was:

Computing log, we get

To maximise, we will compute the partial derivative with respect to , and equate to zero.

Using we get,

Solving, we get the intuitive result:

In most cases, when you compute likelihood, you would be dealing with a bunch of data points , rather than a single one. The likelihood of with respect to the data-set then gets defined as follows:

Using log, the overall likelihood becomes:

]]>**Rule*** :*–

Write the 0 and 9 with one minus the no of the 3’s given.Then to left of 0 increment it by 1. To left of 0 write two 8’s. ( p-1) no of 1’s / 0 /(p-1) no of 8’s /9

Lets look at the following examples:-

**Example**** 1:-**

333^2 =

Here p=3

Therefore, 333^2 =

= (3-1) no of 1’s / 0/(3-1) no of 8’s /9

= 11/0/88/9

**Example 2*** *:-

33^2 =

Here p=2

1/0/8/9

Therefore ,33^2 =1089

**Example3 :** –

33333^2 =

Looks like it is difficult ?

Here we go with same logic….

p=5

Therefore, (1111) /0/(8888)/9

33333^2 =1111088889

Bring your notebook 1 for submission. Also bring Notebook 2 tomorrow.

Regards,

Pallavi

]]>My requirements included:

- The formulae should all be of a similar form.
- Where there is a change of dimension, formulae should be given both in terms of the source and the target dimensons.
- No formula should have a surd (root) in the denominator.
- The terms in a surd should have reduced factors. (So, in particular, any integer under a square root should be square-free.)

I use the symbols:

a = length of an edge

r = inradius, i.e. the radius of the insphere

ρ (rho) = midradius, i.e. the radius of the midsphere

R = circumradius, i.e. the radius of the circumsphere

S = total surface area of polyhedron

C = volume or capacity of polyhedron

The tables were generated with the help of *Maxima* and *Python*. Much of the work to produce the final forms required hand calculation. The final tables have been checked by computer for correctness.

It is possible that these formulae are not the simplest possible, if I have missed an available simplification.

For example, it happens that:

(a + b√c)² = h + k·√c

where

h = a² + b²c

k = 2ab

so

h + k·√c = (a + b√c)²

where, solving for a and b,

a = ±√( (h ± √(h²−k²c)) / 2 )

b = ±√( (h ± √(h²−k²c)) / 2c )

and so it is possible that a form such as

√( h + k·√c )

might simplify to

a + b√c

for certain integers c, h and k.

My formulae include consideration for the following spheres:

- The insphere is the largest that fits inside the polyhedron; it touches the centres of the faces.
- The midsphere passes through the midpoints of the edges.
- The circumsphere is the smallest that fits around the polyhedron; it passes through the vertices.

The formulae are for the radius of each of these spheres. I do not calculate the volumes or surface areas of these spheres, as those are related directly to the radii by the formulae

Area = 2τ rad²

Vol = 2τ rad³ / 3

and

rad = √( Area / 2τ )

rad = ∛( 3 Vol / 2τ )

(where π=τ/2)

In addition, I do not calculate the areas of individual faces. Those are easily obtained by dividing the total surface area S by the number of faces.

The tables are also available in PDF format.

given a = edge |
tetrahedron (4) |
cube / hexahedron (6) |
octahedron (8) |
dodecahedron (12) |
icosahedron (20) |
---|---|---|---|---|---|

a = edge |
a | a | a | a | a |

r = inradius |
a·√6/12 ≈ 0.204124⁺ · a |
a·1/2 ≈ 0.5 · a |
a·√6/6 ≈ 0.408248⁺ · a |
a·√(10·(25+11·√5))/20 ≈ 1.113516⁺ · a |
a·√(6·(7+3·√5))/12 ≈ 0.755761⁺ · a |

ρ = midradius |
a·√2/4 ≈ 0.353553⁺ · a |
a·√2/2 ≈ 0.707107¯ · a |
a·1/2 ≈ 0.5 · a |
a·(3+√5)/4 ≈ 1.309017¯ · a |
a·(1+√5)/4 ≈ 0.809017¯ · a |

R = circumradius |
a·√6/4 ≈ 0.612372⁺ · a |
a·√3/2 ≈ 0.866025⁺ · a |
a·√2/2 ≈ 0.707107¯ · a |
a·√(6·(3+√5))/4 ≈ 1.401259¯ · a |
a·√(2·(5+√5))/4 ≈ 0.951057¯ · a |

S = surface area |
a²·√3 ≈ 1.732051¯ · a² (a·∜3)² ≈ (1.316074⁺ · a)² |
a²·6 ≈ 6.0 · a² (a·√6)² ≈ (2.449490¯ · a)² |
a²·2·√3 ≈ 3.464102¯ · a² (a·∜12)² ≈ (1.861210¯ · a)² |
a²·3·√(5·(5+2·√5)) ≈ 20.645729¯ · a² (a·∜(45·(5+2·√5)))² ≈ (4.543757⁺ · a)² |
a²·5·√3 ≈ 8.660254⁺ · a² (a·∜75)² ≈ (2.942831¯ · a)² |

C = volume |
a³·√2/12 ≈ 0.117851⁺ · a³ (a·⁶√41472/12)³ ≈ (0.490280⁺ · a)³ |
a³ | a³·√2/3 ≈ 0.471405¯ · a³ (a·⁶√162/3)³ ≈ (0.778272¯ · a)³ |
a³·(15+7·√5)/4 ≈ 7.663119¯ · a³ (a·∛(16·(15+7·√5))/4)³ ≈ (1.971523⁺ · a)³ |
a³·5·(3+√5)/12 ≈ 2.181695¯ · a³ (a·∛(720·(3+√5))/12)³ ≈ (1.296974⁺ · a)³ |

given r = inradius |
tetrahedron (4) |
cube / hexahedron (6) |
octahedron (8) |
dodecahedron (12) |
icosahedron (20) |
---|---|---|---|---|---|

a = edge |
r·2·√6 ≈ 4.898979⁺ · r |
r·2 ≈ 2.0 · r |
r·√6 ≈ 2.449490¯ · r |
r·√(2·(25−11·√5)) ≈ 0.898056¯ · r |
r·√(6·(7−3·√5)) ≈ 1.323169⁺ · r |

r = inradius |
r | r | r | r | r |

ρ = midradius |
r·√3 ≈ 1.732051¯ · r |
r·√2 ≈ 1.414214¯ · r |
r·√6/2 ≈ 1.224745¯ · r |
r·√(2·(5−√5))/2 ≈ 1.175571¯ · r |
r·√(6·(3−√5))/2 ≈ 1.070466⁺ · r |

R = circumradius |
r·3 ≈ 3.0 · r |
r·√3 ≈ 1.732051¯ · r |
r·√3 ≈ 1.732051¯ · r |
r·√(3·(5−2·√5)) ≈ 1.258409¯ · r |
r·√(3·(5−2·√5)) ≈ 1.258409¯ · r |

S = surface area |
r²·24·√3 ≈ 41.569219⁺ · r² (r·2·∜108)² ≈ (6.447420¯ · r)² |
r²·24 ≈ 24.0 · r² (r·2·√6)² ≈ (4.898979⁺ · r)² |
r²·12·√3 ≈ 20.784610¯ · r² (r·2·∜27)² ≈ (4.559014⁺ · r)² |
r²·30·√(2·(65−29·√5)) ≈ 16.650873⁺ · r² (r·∜(1800·(65−29·√5)))² ≈ (4.080548⁺ · r)² |
r²·30·√(6·(47−21·√5)) ≈ 15.162168⁺ · r² (r·∜(5400·(47−21·√5)))² ≈ (3.893863¯ · r)² |

C = volume |
r³·8·√3 ≈ 13.856406⁺ · r³ (r·⁶√192)³ ≈ (2.401874¯ · r)³ |
r³·8 ≈ 8.0 · r³ (r·2)³ ≈ (2.0 · r)³ |
r³·4·√3 ≈ 6.928203⁺ · r³ (r·⁶√48)³ ≈ (1.906369¯ · r)³ |
r³·10·√(2·(65−29·√5)) ≈ 5.550291⁺ · r³ (r·⁶√(200·(65−29·√5)))³ ≈ (1.770538¯ · r)³ |
r³·10·√(6·(47−21·√5)) ≈ 5.054056⁺ · r³ (r·⁶√(600·(47−21·√5)))³ ≈ (1.716116⁺ · r)³ |

given ρ = midradius |
tetrahedron (4) |
cube / hexahedron (6) |
octahedron (8) |
dodecahedron (12) |
icosahedron (20) |
---|---|---|---|---|---|

a = edge |
ρ·2·√2 ≈ 2.828427⁺ · ρ |
ρ·√2 ≈ 1.414214¯ · ρ |
ρ·2 ≈ 2.0 · ρ |
ρ·(3−√5) ≈ 0.763932⁺ · ρ |
ρ·(√5−1) ≈ 1.236068¯ · ρ |

r = inradius |
ρ·√3/3 ≈ 0.577350⁺ · ρ |
ρ·√2/2 ≈ 0.707107¯ · ρ |
ρ·√6/3 ≈ 0.816497¯ · ρ |
ρ·√(10·(5+√5))/10 ≈ 0.850651¯ · ρ |
ρ·√(6·(3+√5))/6 ≈ 0.934172⁺ · ρ |

ρ = midradius |
ρ | ρ | ρ | ρ | ρ |

R = circumradius |
ρ·√3 ≈ 1.732051¯ · ρ |
ρ·√6/2 ≈ 1.224745¯ · ρ |
ρ·√2 ≈ 1.414214¯ · ρ |
ρ·√(6·(3−√5))/2 ≈ 1.070466⁺ · ρ |
ρ·√(2·(5−√5))/2 ≈ 1.175571¯ · ρ |

S = surface area |
ρ²·8·√3 ≈ 13.856406⁺ · ρ² (ρ·2·∜12)² ≈ (3.722419⁺ · ρ)² |
ρ²·12 ≈ 12.0 · ρ² (ρ·2·√3)² ≈ (3.464102¯ · ρ)² |
ρ²·8·√3 ≈ 13.856406⁺ · ρ² (ρ·2·∜12)² ≈ (3.722419⁺ · ρ)² |
ρ²·6·√(10·(25−11·√5)) ≈ 12.048685¯ · ρ² (ρ·∜(360·(25−11·√5)))² ≈ (3.471122¯ · ρ)² |
ρ²·10·√(6·(7−3·√5)) ≈ 13.231691¯ · ρ² (ρ·∜(600·(7−3·√5)))² ≈ (3.637539⁺ · ρ)² |

C = volume |
ρ³·8/3 ≈ 2.666667¯ · ρ³ (ρ·2·∛9/3)³ ≈ (1.386723¯ · ρ)³ |
ρ³·2·√2 ≈ 2.828427⁺ · ρ³ (ρ·√2)³ ≈ (1.414214¯ · ρ)³ |
ρ³·8·√2/3 ≈ 3.771236⁺ · ρ³ (ρ·2·⁶√162/3)³ ≈ (1.556543⁺ · ρ)³ |
ρ³·2·(3·√5−5) ≈ 3.416408¯ · ρ³ (ρ·∛(2·(3·√5−5)))³ ≈ (1.506110¯ · ρ)³ |
ρ³·10·(√5−1)/3 ≈ 4.120227¯ · ρ³ (ρ·∛(90·(√5−1))/3)³ ≈ (1.603148⁺ · ρ)³ |

given R = circumradius |
tetrahedron (4) |
cube / hexahedron (6) |
octahedron (8) |
dodecahedron (12) |
icosahedron (20) |
---|---|---|---|---|---|

a = edge |
R·2·√6/3 ≈ 1.632993⁺ · R |
R·2·√3/3 ≈ 1.154701¯ · R |
R·√2 ≈ 1.414214¯ · R |
R·√(6·(3−√5))/3 ≈ 0.713644⁺ · R |
R·√(10·(5−√5))/5 ≈ 1.051462⁺ · R |

r = inradius |
R·1/3 ≈ 0.333333⁺ · R |
R·√3/3 ≈ 0.577350⁺ · R |
R·√3/3 ≈ 0.577350⁺ · R |
R·√(15·(5+2·√5))/15 ≈ 0.794654⁺ · R |
R·√(15·(5+2·√5))/15 ≈ 0.794654⁺ · R |

ρ = midradius |
R·√3/3 ≈ 0.577350⁺ · R |
R·√6/3 ≈ 0.816497¯ · R |
R·√2/2 ≈ 0.707107¯ · R |
R·√(6·(3+√5))/6 ≈ 0.934172⁺ · R |
R·√(10·(5+√5))/10 ≈ 0.850651¯ · R |

R = circumradius |
R | R | R | R | R |

S = surface area |
R²·8·√3/3 ≈ 4.618802⁺ · R² (R·2·∜108/3)² ≈ (2.149140¯ · R)² |
R²·8 ≈ 8.0 · R² (R·2·√2)² ≈ (2.828427⁺ · R)² |
R²·4·√3 ≈ 6.928203⁺ · R² (R·2·∜3)² ≈ (2.632148⁺ · R)² |
R²·2·√(10·(5−√5)) ≈ 10.514622⁺ · R² (R·∜(40·(5−√5)))² ≈ (3.242626¯ · R)² |
R²·2·√(30·(3−√5)) ≈ 9.574541⁺ · R² (R·∜(120·(3−√5)))² ≈ (3.094276¯ · R)² |

C = volume |
R³·8·√3/27 ≈ 0.513200⁺ · R³ (R·2·⁶√3/3)³ ≈ (0.800625¯ · R)³ |
R³·8·√3/9 ≈ 1.539601¯ · R³ (R·2·⁶√27/3)³ ≈ (1.154701¯ · R)³ |
R³·4/3 ≈ 1.333333⁺ · R³ (R·∛36/3)³ ≈ (1.100642⁺ · R)³ |
R³·2·√(30·(3+√5))/9 ≈ 2.785164¯ · R³ (R·⁶√(1080·(3+√5))/3)³ ≈ (1.406966¯ · R)³ |
R³·2·√(2·(5+√5))/3 ≈ 2.536151¯ · R³ (R·⁶√(648·(5+√5))/3)³ ≈ (1.363719⁺ · R)³ |

given S = surface area |
tetrahedron (4) |
cube / hexahedron (6) |
octahedron (8) |
dodecahedron (12) |
icosahedron (20) |
---|---|---|---|---|---|

a = edge |
√S·∜27/3 ≈ 0.759836¯ · √S √(S·√3/3) ≈ √(0.577350⁺ · S) |
√S·√6/6 ≈ 0.408248⁺ · √S √(S·1/6) ≈ √(0.166667¯ · S) |
√S·∜1728/12 ≈ 0.537285¯ · √S √(S·√3/6) ≈ √(0.288675⁺ · S) |
√S·∜(225·(5−2·√5))/15 ≈ 0.220082⁺ · √S √(S·√(5−2·√5)/15) ≈ √(0.048436⁺ · S) |
√S·∜675/15 ≈ 0.339809¯ · √S √(S·√3/15) ≈ √(0.115470⁺ · S) |

r = inradius |
√S·∜12/12 ≈ 0.155101¯ · √S √(S·√3/72) ≈ √(0.024056⁺ · S) |
√S·√6/12 ≈ 0.204124⁺ · √S √(S·1/24) ≈ √(0.041667¯ · S) |
√S·∜3/6 ≈ 0.219346¯ · √S √(S·√3/36) ≈ √(0.048113¯ · S) |
√S·∜(360·(65+29·√5))/60 ≈ 0.245065⁺ · √S √(S·√(10·(65+29·√5))/600) ≈ √(0.060057¯ · S) |
√S·∜(600·(47+21·√5))/60 ≈ 0.256814⁺ · √S √(S·√(6·(47+21·√5))/360) ≈ √(0.065954¯ · S) |

ρ = midradius |
√S·∜108/12 ≈ 0.268642⁺ · √S √(S·√3/24) ≈ √(0.072169¯ · S) |
√S·√12/12 ≈ 0.288675⁺ · √S √(S·1/12) ≈ √(0.083333⁺ · S) |
√S·∜108/12 ≈ 0.268642⁺ · √S √(S·√3/24) ≈ √(0.072169¯ · S) |
√S·∜(1800·(25+11·√5))/60 ≈ 0.288091⁺ · √S √(S·√(2·(25+11·√5))/120) ≈ √(0.082997¯ · S) |
√S·∜(5400·(7+3·√5))/60 ≈ 0.274911⁺ · √S √(S·√(6·(7+3·√5))/120) ≈ √(0.075576⁺ · S) |

R = circumradius |
√S·∜12/4 ≈ 0.465302⁺ · √S √(S·√3/8) ≈ √(0.216506⁺ · S) |
√S·√2/4 ≈ 0.353553⁺ · √S √(S·1/8) ≈ √(0.125 · S) |
√S·∜27/6 ≈ 0.379918¯ · √S √(S·√3/12) ≈ √(0.144338¯ · S) |
√S·∜(200·(5+√5))/20 ≈ 0.308392⁺ · √S √(S·√(2·(5+√5))/40) ≈ √(0.095106¯ · S) |
√S·∜(27000·(3+√5))/60 ≈ 0.323177⁺ · √S √(S·√(30·(3+√5))/120) ≈ √(0.104444¯ · S) |

S = surface area |
S | S | S | S | S |

C = volume |
√S³·∜12/36 ≈ 0.051700⁺ · √S³ √(S·⁶√243/18)³ ≈ √(0.138781¯ · S)³ |
√S³·√6/36 ≈ 0.068041⁺ · √S³ √(S·1/6)³ ≈ √(0.166667¯ · S)³ |
√S³·∜3/18 ≈ 0.073115⁺ · √S³ √(S·⁶√972/18)³ ≈ √(0.174853¯ · S)³ |
√S³·∜(360·(65+29·√5))/180 ≈ 0.081688⁺ · √S³ √(S·⁶√(250·(65+29·√5))/30)³ ≈ √(0.188267¯ · S)³ |
√S³·∜(600·(47+21·√5))/180 ≈ 0.085605¯ · √S³ √(S·⁶√(303750·(47+21·√5))/90)³ ≈ √(0.194237⁺ · S)³ |

given C = volume |
tetrahedron (4) |
cube / hexahedron (6) |
octahedron (8) |
dodecahedron (12) |
icosahedron (20) |
---|---|---|---|---|---|

a = edge |
∛C·⁶√72 ≈ 2.039649¯ · ∛C ∛(C·6·√2) ≈ ∛(8.485281⁺ · C) |
∛C | ∛C·⁶√288/2 ≈ 1.284898⁺ · ∛C ∛(C·3·√2/2) ≈ ∛(2.121320⁺ · C) |
∛C·∛(25·(7·√5−15))/5 ≈ 0.507222⁺ · ∛C ∛(C·(7·√5−15)/5) ≈ ∛(0.130495⁺ · C) |
∛C·∛(75·(3−√5))/5 ≈ 0.771025⁺ · ∛C ∛(C·3·(3−√5)/5) ≈ ∛(0.458359⁺ · C) |

r = inradius |
∛C·⁶√243/6 ≈ 0.416342¯ · ∛C ∛(C·√3/24) ≈ ∛(0.072169¯ · C) |
∛C·1/2 ≈ 0.5 · ∛C ∛(C·1/8) ≈ ∛(0.125 · C) |
∛C·⁶√972/6 ≈ 0.524558¯ · ∛C ∛(C·√3/12) ≈ ∛(0.144338¯ · C) |
∛C·⁶√(250·(65+29·√5))/10 ≈ 0.564800⁺ · ∛C ∛(C·√(10·(65+29·√5))/200) ≈ ∛(0.180171¯ · C) |
∛C·⁶√(303750·(47+21·√5))/30 ≈ 0.582711⁺ · ∛C ∛(C·√(6·(47+21·√5))/120) ≈ ∛(0.197861¯ · C) |

ρ = midradius |
∛C·∛3/2 ≈ 0.721125¯ · ∛C ∛(C·3/8) ≈ ∛(0.375 · C) |
∛C·√2/2 ≈ 0.707107¯ · ∛C ∛(C·√2/4) ≈ ∛(0.353553⁺ · C) |
∛C·⁶√288/4 ≈ 0.642449⁺ · ∛C ∛(C·3·√2/16) ≈ ∛(0.265165⁺ · C) |
∛C·∛(25·(5+3·√5))/10 ≈ 0.663962⁺ · ∛C ∛(C·(5+3·√5)/40) ≈ ∛(0.292705⁺ · C) |
∛C·∛(75·(1+√5))/10 ≈ 0.623773¯ · ∛C ∛(C·3·(1+√5)/40) ≈ ∛(0.242705⁺ · C) |

R = circumradius |
∛C·⁶√15552/4 ≈ 1.249025¯ · ∛C ∛(C·9·√3/8) ≈ ∛(1.948557⁺ · C) |
∛C·√3/2 ≈ 0.866025⁺ · ∛C ∛(C·3·√3/8) ≈ ∛(0.649519⁺ · C) |
∛C·∛6/2 ≈ 0.908560⁺ · ∛C ∛(C·3/4) ≈ ∛(0.75 · C) |
∛C·⁶√(168750·(3−√5))/10 ≈ 0.710749⁺ · ∛C ∛(C·3·√(30·(3−√5))/40) ≈ ∛(0.359045⁺ · C) |
∛C·⁶√(56250·(5−√5))/10 ≈ 0.733289¯ · ∛C ∛(C·3·√(10·(5−√5))/40) ≈ ∛(0.394298⁺ · C) |

S = surface area |
∛C²·6·⁶√3 ≈ 7.205622¯ · ∛C² ∛(C·6·∜108)² ≈ ∛(19.342259¯ · C)² |
∛C²·6 ≈ 6.0 · ∛C² ∛(C·6·√6)² ≈ ∛(14.696938⁺ · C)² |
∛C²·⁶√34992 ≈ 5.719106¯ · ∛C² ∛(C·6·∜27)² ≈ ∛(13.677042⁺ · C)² |
∛C²·3·⁶√(200·(65−29·√5)) ≈ 5.311614¯ · ∛C² ∛(C·3·∜(1800·(65−29·√5)))² ≈ ∛(12.241644⁺ · C)² |
∛C²·3·⁶√(600·(47−21·√5)) ≈ 5.148349¯ · ∛C² ∛(C·3·∜(5400·(47−21·√5)))² ≈ ∛(11.681589¯ · C)² |

C = volume |
C | C | C | C | C |

The Princeton Review GRE books are pretty helpful (found in the reference section of any good library), but they seem to me to focus more on ways to game the test than they do re-instilling a thorough understanding of the mathematical processes and ideas behind the questions. This may work well for many people, but it was leaving me feeling lost and confused at times since there are* so many* kinds of gaming techniques that it’s hard to remember them all, especially if you don’t feel you have a good grasp of the kind of problem you’re solving to begin with. I expect this is the same for many of you as well: my memory won’t hold onto a fact or idea unless it fits into some larger idea or system. If I don’t understand the *why*, I just can’t seem to remember the *what*.

So I was feeling pretty stressed out by the feeling of working very hard but gaining too little, and decided I needed to back up and get a good solid grasp of the basic concepts again. The company that creates and administers the GRE has a list of Khan Academy lessons and practice sessions that pertain to the test posted on their website. They are so well designed, so well-explained, and they’re free, hooray! I feel so much better now about the progress I’m making, and re-discovering the fun of basic and intermediate algebra. Once I had gotten the hang of it, it had always seemed more like games than work to me!

So thank you from the bottom of my heart, Sal Khan and the good people at Khan Academy, you are the best. And yes, I will donate to your Indigogo campaign to fund courses on American government.

*Ordinary Philosophy and its Traveling Philosophy / History of Ideas series is a labor of love and ad-free, entirely supported by patrons and readers like you. Please offer your support today!*

Squaring of Two Digit Number

Don’t Know , So what , I know the Other way

Squaring of Two Digit Number

Rule:

1) Find the reverse of a number.

2) Find the square of reverse of number.

3) Find the difference of square of digits.

4) Add it to 1st two digits of step2. And subtract

result of Step3 from last 2 digits.

5) Atlast result is obatined.

Let’s take an example to understand in much

better way.

Consider squaring of 58.

52 ^2 =???

25 ^2

Ending in 5

Therefore 25 ^2 =

=2 (2 +1)/25

=625.

Now 5^2 -2^2 =25 -4

=21.

25^2 = 06 / 25

Add 21 to 06 and subtract 21 from 25.

We get 27 /04

Therefore , 52^2 =2704.

*Example 2 *

62^2 =??

26^2 = 0676

Now 6^2 -2^2 = 36-4 =32

Now add 32 to 06 and subtract 32 from 76.

We get

62^2 =38/44

Therefore, 62^2=3844.

U prilogu se nalaze riješeni zadaci (Matematika A razina) sa državne mature, ljeto 2016.god.

U prvom prilogu se nalazi prvih 15 zadataka dok u drugom ostali zadaci.

Ukoliko ima kakvih nejasnoća pošaljite mail na eugen.sunic@live.com

Link na maturu, Matematika A razina 2016, ljeto: https://www.scribd.com/doc/316830479/MAT-A-razina-pitanja

**English**

In the attachment below you can find solved problems from the 2016 (summer) math exam, level A.

Students across the country must pass this exam in order to inscribe them-self to a desired faculty.

If you have any question, reffer to eugen.sunic@live.com

Exam link, Mathematics A level, 2016, summer: https://www.scribd.com/doc/316830479/MAT-A-razina-pitanja

Riješeni zadaci s postupkom, 1.dio/solved problems with steps, first part:

MAT A, viša razina, ljeto 2016.god. 1.dio

Riješeni zadaci s postupkom, 2.dio/solved problems with steps, first part:

]]>Mathematics Worksheet Busters: Grade 2.

Language: English

ISBN-10: 1480725781

ISBN-13: 978-1480725782

Software credit: I used *Stella 4d* to make this, and you can find that program at http://www.software3d.com/Stella.php, with a free trial download available there.

In formal meaning of education it is linked with SCHOOLING! Main process of education will be fulfilled by school/college or any other education institution. At the end of an academic year or at the end of a particular course of education, after passing the examination, one will be considered as a GRADUATE! Clearly, Difference between education and graduation!

So, education is a bridge between man and human being! Education at schools/ universities helps in preparing oneself to get a good job which is the need of life, which will give a standard living.

During schooling, what are the main concepts comes in ones life?

- Dealing with attendance
- class notes
- homework
- laboratory works [record writing, observations, submission of records on date etc]
- assignments
- studies
- exams
- parental pressure
- and love life!

These all together becomes an unsolved problem for students. Actually period of teenage brings rapid changes in individuals physical, mental, moral, spiritual, sexual and social outlook. In this period he/she learn new things.

We have discussed ,much about academic progress in How to get good grades! & The process you need to follow to learn & make adjustment to learn. As we mentioned above,the problems of dealing with attendance, class notes, homework, laboratory works, assignments, studies and exams can be solved according to How to get good grades! and Forgetfulness/ Forgetting/ Difficulty in Remembering the concepts which is studied earlier! [click on the links for more]

What is remaining now? **LOVE LIFE! A biggest confusion! LOVE or EDUCATION! Is it necessary to choose one among these? Can we live by choosing any one of these? Are you stable in life to get in to these? **It is here where parental pressure starts! Parents start worrying about children future and it is totally acceptable.They have the right to worry for you, if not who will worry and pressurize for your goodness!

What is love? In teenage due to rapid development in his/her physic, one will feel like becoming independent, hatred feeling towards those who advice them, feeling guilty or shame for little things, not able to distinguish the real life concepts! It is a period where a child moves from *later childhood *to *adulthod. * Period of maturation! So, it is common to get attracted towards opposite sex, **DO YOU THINK THIS IS CALLED LOVE? **NO! You are absolutely not!

Love is a strong feeling of affection. If you are also feeling same then, let us go little more deeper! You may ask like is it enough if we choose only one among love and education! No! Both love and education/carrier are equally important in life! From an idealistic point of view it shouldn’t be a choice.It’s two sides of the same coin. **But being a student/teenager is it right time to fall for someone? You may ask as “do you think love happens at the right stage of life?” **This is also right!

Yes!There’s no age for love. No right time. It’ll happen when it’s supposed to and you’re supposed to continue living your life the same way. It’s all about finding the right balance. You need to be sure about your feeling, words and life decisions. Love is also important thing in life. But unfortunately one cannot live on love and fresh air. We need basic things to live and that’s where education/schooling/career takes first place than LOVE!

Be practical! Choose yourself! Love yourself! Your education/career is very precious and valuable. Your career is the one that will feed you. Dont be hasty in making decisions. Love can never feed you, forever. If you believe your partner is good/perfect for you, you have long way to go, frankly and honestly, make your life stable, unless don’t fall for someone.

As a teenager you have lot more to enjoy your life than falling in LOVE! Yo! Love yourself and love your life. I guess looking after better career opportunities would be a better option. Be practical!

Best luck!

Basically, we homomorphisms are special maps that satisfy a certain property. This is rather superficial and abstract at the moment, but we will later see how homomorphisms preserve the structure of groups. Anyhow, this post will be dedicated to developing the fundamentals of homomorphisms and completing some exercises to consolidate our understanding.

Since and are groups, they both must have their own identity elements and respectfully. There will exist some element so that (through ), it is mapped to the identity element (we will soon show why there must always exist at least one non-empty element). We define this set of particular elements as the kernel of , which is denoted as . The precise definition is therefore:

As quoted from Wikipedia “*the kernel of a homomorphism measures the degree at which the homomorphism fails to be injective*.” If the kernel were to have multiple elements, then there exists multiple domain values that map to the same element (the identity ). Therefore, if there is only one element in the kernel then is an injection. However, as the number of elements in the kernel increases, the map becomes ‘less’ of an injection (even though they aren’t injection at all at this point) since multiple points in the range map to a single range element.

Since we have a concept revolving around the domain, perhaps we should fabricate another addressing the range:

Notice how the range is not necessarily the entire set . If the range does encompass the entirety of , then must be a surjection since every element in the range () can be traced back to at least one element from the domain ().

With all this in mind, we can introduce a special type of homomorphism: an isomorphism:

We will consider isomorphism in greater detail in another post but for now, definition 4 is all we need to know about isomorphic maps. Let’s try a quick example:

Example 1:For the map where , determine whether or not is a homomorphism and if so find the kernel and range and deduce if is an isomorphism as well.

First, let . Therefore . However, and so . Therefore, is an isomorphism by definition 1. Since is a homomorphism, the map must have a kernel. We must first identity the identity element in the mapped group () and deduce the element from the domain () that maps to the said identity element via . The identity element under the binary operation is 1 and since , the kernel must be . Finally, the range is all of the mapped elements that come from , which is all the positive real numbers. Therefore .

Notice that is also an injection since every element in the range can only be mapped back to one element in the domain. The kernel also has only one element, which supports this. Additionally, is a surjection since the range is equal to the mapped group, which implies that every element in the codomain is in the range. Therefore is a bijection and since is a homomorphism, is also an isomorphism.

It was mentioned before that every homomorphism has a non-empty kernel. In fact, every homomorphism has the domain’s identity in its kernel:

To prove this, let . Since is a homomorphism, we can write . But and so . must be an element of , which is a group and so there must exist an inverse (here the -1 is denoting the elemental inverse, *not* function inverse). If we ‘multiply’ the left sides of the equation by , we get . However, must be associative and so . Therefore, and so . We will always have the identity element in the kernel if is a homomorphism!

We have another elegant property that all homomorphisms exhibit:

If we let then . Therefore, . From theorem 1 and the fact that is an homomorphism, . By the definition of inverse, and must be inverses and so , which proves theorem 2.

When we introduced the kernel, we mentioned how if the kernel has only one element (which we now know to be ), then it is possible that is an injection. However, if there are multiple elements in the kernel then there is no way is an injection. Interestingly, knowing that is the only element in the kernel is enough to prove that is an injection. We can prove this by first considering the ‘forwards’ case. If is an injection and since then no other element in can map to . Therefore, the kernel must only contain .

Now, let’s consider the ‘backwards’ case. Let us say we are given that the kernel of only has . Suppose that where and . Now consider . Since then by theorem 2. Therefore . But we said that the kernel only contains and so . Multiplying on the right on both sides by , which contradicts . Therefore, implies that and so is an injection. Therefore, in example 1 when we deduced that the kernel has only 0 as an element, we can immediately deduce that is an injection.

Moving on to our last theorem:

This is easily provable using the the general subgroup test. We will start with the kernel, which is a subset of since it is defined as the elements from that maps to the identity element in . Also, since is always true, the kernel is non-empty. Now suppose we and are in the kernel so that . Since is a homomorphism, , where we used the fact that is a group and so the inverse of must exist. By theorem 2, . Therefore , which proves that .

Onto the range. is a subset of since it is defined as the elements mapped by from to . Also since , the range will always contain and so is non-empty. Suppose that where and where . Consider by theorem 2. Since is a homomorphism, for some element since is a group and must be closed. Therefore , which proves that .

Now here are three more examples taken from Haese’s:

Example 2:For the map where , determine whether or not is a homomorphism and if so find the kernel and range and deduce if is an isomorphism as well.

Example 3:For the map where , determine whether or not is a homomorphism and if so find the kernel and range and deduce if is an isomorphism as well.

Example 4:Let be the additive group of polynomials of degree of . Consider the map where . Show that is a homomorphism and determine the kernel and range. Is an isomorphism?

Below are the solutions. Starting with example 2, let . Then . However, and so and so is a homomorphism. The identity element for is just 0. Therefore, any integer multiple of 5 will map to the identity element and so . The range is just the integer modulo-5 . The kernel has multiple elements besides and so is not an injection but is a surjection. Therefore is a homomorphism but not isomorphic.

We apply the same methods for example 3. Let and so . However , which is not . Therefore is not a homomorphism.

Finally on the last example. Let and let us consider and since the derivative of a sum is the sum of derivatives (in other words, the derivative is a linear operator) and so . Therefore, is a homomorphism. Now the identity element for must be the 0 polynomial and the derivative of any constant is 0. Therefore . The range is just since it doesn’t matter how many times we differentiate any polynomial , we will always get a polynomial in . Therefore . Since the kernel has multiple (in fact, infinite) number of elements, can not be an injection and so is not a bijection. Therefore, is not an isomorphism.

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