## Saturday, April 22, 2017

## Friday, April 21, 2017

### Bhojpuri lakokti

My grandfather Rameshwar Prasad used to say this

Kose kose gor dhoye

Teen kos par khaye

Jahan jahan man karei

Tahan tahan jaye

The essence is if you wash your feet every mile and eat at every 3 miles then you can go anywhere you feel like. Because washing your feet refreshes you and eating reinvigorate you.

I always feel this is well said.

Kose kose gor dhoye

Teen kos par khaye

Jahan jahan man karei

Tahan tahan jaye

The essence is if you wash your feet every mile and eat at every 3 miles then you can go anywhere you feel like. Because washing your feet refreshes you and eating reinvigorate you.

I always feel this is well said.

### Mohalla clinic in Kamla Nagar, Delhi

Arvind Kejriwal's mohalla clinic is a game changer. I visited the clinic last week and talked to the people there and they seemed very satisfied with the work. The clinic is air conditioned and it was week stock. It has been opened for a month. It's impressive how much if a state government is willing can do for its citizen. I didn't take the camera that time. So today I went to get the picture.

## Wednesday, April 19, 2017

### Sperners Lemma

There are basically two lemmas in Topology which allow us to prove Sperner's Lemma.

Lemma:1 If a close interval [0,1] has finite number of points that divide it into smaller intervals by points either 0 and 1. Then there are odd number of intervals whose end points are different.

Proof: Think of the left most 1 out of all the one. This will form the very first interval of the type (0,1). The next interval if is present will be of type (1,0) and this forces the presence of another interval of type (0,1). So we see that the intervals of type (0,1) occurs at odd positions and since the last interval has to be of type (0,1). It means we will have odd number of intervals.

Lemma:2 Suppose any room of a house has 0,1 and 2 doors. Then the number of dead ends and the outer door have the same parity.

Proof: First thing to notice is that there is a unique path from any given door to another door. Second, there are only 3 such path exist.

1. Outer door to Outer door (let there be m such paths. For each path we have two doors, therefore we have 2m such outer doors).

2. Outer door to Dead end door (let there be n such paths. For each path we have one outer and one dead end door, therefore we have n outer and n dead end doors)

3. Dead end door to Another Dead end door. (let there be p such paths. For each path we have 2 dead end doors. Therefore we have 2p dead end doorss)

Therefore

Number of outer doors = 2m+n

Number of dead end doorss = 2p+n

Thus both have same parity.

Sperners Lemma: It gaurantees the existence of a dead end triangle in any triangulation of a triangle obtained by the following method. The vertices are labelled 1,2 and 3. Then the side containing (1,2) vertex has only 1 and 2 intermediate point and so on. The vertices of triangle inside could be labeled by any of these three numbers.

Proof: Suppose (1,2) and (2,1) are the doors. Then by lemma 2 we have only doors on one side of the triangle and there will be odd number of these. So if we enter through any of the doors either we exit from another door then pairs are taken. So we will always be left by door which is not matched and that mean it will end up going into a triangle with all three different vertices.

Remark: There are only 2 types of triangle which allow a path (1,1,2), (2,2,1) and (1,2,3). All other triangles are inaccessible.

Following two videos explain sperners lemma and using fair division.

This gives a little more explanation to the above. Both complement each other.

Lemma:1 If a close interval [0,1] has finite number of points that divide it into smaller intervals by points either 0 and 1. Then there are odd number of intervals whose end points are different.

Proof: Think of the left most 1 out of all the one. This will form the very first interval of the type (0,1). The next interval if is present will be of type (1,0) and this forces the presence of another interval of type (0,1). So we see that the intervals of type (0,1) occurs at odd positions and since the last interval has to be of type (0,1). It means we will have odd number of intervals.

Lemma:2 Suppose any room of a house has 0,1 and 2 doors. Then the number of dead ends and the outer door have the same parity.

Proof: First thing to notice is that there is a unique path from any given door to another door. Second, there are only 3 such path exist.

1. Outer door to Outer door (let there be m such paths. For each path we have two doors, therefore we have 2m such outer doors).

2. Outer door to Dead end door (let there be n such paths. For each path we have one outer and one dead end door, therefore we have n outer and n dead end doors)

3. Dead end door to Another Dead end door. (let there be p such paths. For each path we have 2 dead end doors. Therefore we have 2p dead end doorss)

Therefore

Number of outer doors = 2m+n

Number of dead end doorss = 2p+n

Thus both have same parity.

Sperners Lemma: It gaurantees the existence of a dead end triangle in any triangulation of a triangle obtained by the following method. The vertices are labelled 1,2 and 3. Then the side containing (1,2) vertex has only 1 and 2 intermediate point and so on. The vertices of triangle inside could be labeled by any of these three numbers.

Proof: Suppose (1,2) and (2,1) are the doors. Then by lemma 2 we have only doors on one side of the triangle and there will be odd number of these. So if we enter through any of the doors either we exit from another door then pairs are taken. So we will always be left by door which is not matched and that mean it will end up going into a triangle with all three different vertices.

Remark: There are only 2 types of triangle which allow a path (1,1,2), (2,2,1) and (1,2,3). All other triangles are inaccessible.

Following two videos explain sperners lemma and using fair division.

This gives a little more explanation to the above. Both complement each other.

## Tuesday, April 18, 2017

### Shaking hands with Arvind Kejriwal

I am a big fan of Arvind Kejriwal. He is the rare politician who does what he says. To me he is bigger than Gandhi. He has brought tangible changes to the lives of the people and has shown what good intentions can do. So when I heard that he was coming to Kayla Nagar. I reached there and had the fortune of seeing him in person. Here are some photographs.

## Tuesday, March 07, 2017

### Harmonic series without number 9 is convergent

There are tons of proofs that harmonic series diverges. What if we have a harmonic series where we get rid of all the numbers which contains $9$, will it still converge.
For single digit there are $8$ numbers which do not contain number $9$.
Single digit numbers $8$
Two digit numbers $8 \times 9 $
Three digit numbers $8 \times 9 \times 9 = 8\cdot 9^2$
4 digit numbers $latex 8\cdot 9^3$
...
Also notice that largest fraction in harmonic series
Two digits $\frac{1}{10}$
Three digits $\frac{1}{100}$
Four digits $\frac{1}{1000}$
So if we replace the terms of harmonic with these fractions we will get an upper bound on the harmonic series devoid of number $9$
It is $\frac{1}{1}\cdot 8+\frac{1}{10}\cdot 8 \cdot 9+\frac{1}{100}\cdot 8 \cdot 9^2+\cdots+..$ and
$\Rightarrow 8\cdot \frac{1}{1}+8\cdot \frac{9}{10}+8\cdot \frac{9^2}{10^2}+8\cdot \frac{9^3}{10^3}+\cdots$
This is a geometric series and the sum is
$\frac{8}{1-\frac{9}{10}} = 80$.
Thus the geometric series with common ratio of $\frac{9}{10}$ is convergent.

Labels: 9, Convergence, Harmonic series

## Tuesday, February 21, 2017

### At least one of 3 consecutive odd numbers is a multiple of 3

The way to approach this problem is to show that there will always be a number among three consecutive odd numbers that is divisible by $3$.
Let three consecutive odd numbers be $2n+1,2n+3,2n+5$.
If $2n+1$ is a multiple of $3$ then we are done.
Else the remainder when divided by $3$ is either $1$ or $2$. Suppose its $1$ then it means $2x$ is divisible by $3$ which means $2x+3$ is divisible by $3$.
Or suppose the remainder is $2$, which means $2x-1$ is divisible by $3$ which means $2x+2$ and $2x+5$ are divisible by $3$. Hence proved.

Labels: Consecutive odd numbers, Number theory, Prime Number

### Variance of Uniform Distribution

Let the corresponding probabilities are $\frac{1}{n+1}$ for all at the points $\{0,1,2,\cdots,n+1\}$
The general formula is $\sigma^2 = E[x^2]-(E[X])^2$
Let's first calculate $E[X]=0\cdot \frac{1}{n+1}+1\cdot \frac{1}{n+1}+\cdots+n\cdot \frac{1}{n+1}=\frac{0+1+2+\cdots+n}{n+1}=\frac{n(n+1)}{2(n+1)}=\frac{n}{2}$
Now lets calculate $E[X]=0^2\cdot \frac{1}{n+1}+1^2\cdot \frac{1}{n+1}+\cdots+n^2\cdot \frac{1}{n+1}=\frac{0^2+1^2+2^2+\cdots+n^2}{n+1}=\frac{n(n+1)(2n+1)}{6(n+1)}=\frac{n(2n+1)}{6}$
Therefore $\sigma^2 = \frac{n(2n+1)}{6}-\left ( \frac{n}{2}\right )^2 $
$\Rightarrow \frac{n(2n+1)}{6}-\frac{n^2}{4}$
$\Rightarrow \frac{2n(2n+1)-3n^2}{12}$
$\Rightarrow \frac{4n^2+2n-3n^2}{12}$
$\Rightarrow \frac{n^2+2n}{12}=\frac{n(n+2)}{12}$
Now suppose we have same $n+1$ terms shifted from $a$ to $b$ in that case the variance becomes $\frac{(b-a)(b-a+2)}{12}$

Labels: Uniform Distribution, Variance

## Wednesday, January 04, 2017

### Six deceptive problems that no one can solve

This is the title of the article which appeared here

The problems are

1. Twin Prime Conjecture

2. The Moving Sofa Problem

3. The Collatz Conjecture

4. The Beal Conjecture

5. The Inscribed Square Problem

6. Goldbach Conjecture

I am aware of the problems number 1,3 and 6. So I need to investigate the other 3. Which are Moving Sofa, Beal Conjecture and the Inscribed Square Problem.

The problems are

1. Twin Prime Conjecture

2. The Moving Sofa Problem

3. The Collatz Conjecture

4. The Beal Conjecture

5. The Inscribed Square Problem

6. Goldbach Conjecture

I am aware of the problems number 1,3 and 6. So I need to investigate the other 3. Which are Moving Sofa, Beal Conjecture and the Inscribed Square Problem.

Labels: famous unsolved problems

### How many triangular numbers are also Fibonacci Numbers

I heard this from Bruce Edwards that they are only 5 and these are 1,1,3,21, and 55.
I wrote a program in Sagemath to verify that. Here it is
Here is the program with its output

def fibo(n):

fib_1 = 1

fib_2 = 1

if n == 1 or n== 2:

return 1

fib_n=0

for i in range(3,n+1):

fib_n = fib_1+fib_2

fib_1 = fib_2

fib_2 = fib_n

return fib_n

def triNumList(n):

a=[]

for i in range(1,n):

b=i*(i+1)/2

a.append(b)

return a

def fiboList(n):

a=[]

for i in range(1,n):

a.append(fibo(i))

return a

def TriFiboCompare(n):

f = fiboList(n)

t = triNumList(n)

print(f)

print(t)

for i in f:

if i in t:

print("Match",i)

TriFiboCompare(100)

[1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169, 63245986, 102334155, 165580141, 267914296, 433494437, 701408733, 1134903170, 1836311903, 2971215073, 4807526976, 7778742049, 12586269025, 20365011074, 32951280099, 53316291173, 86267571272, 139583862445, 225851433717, 365435296162, 591286729879, 956722026041, 1548008755920, 2504730781961, 4052739537881, 6557470319842, 10610209857723, 17167680177565, 27777890035288, 44945570212853, 72723460248141, 117669030460994, 190392490709135, 308061521170129, 498454011879264, 806515533049393, 1304969544928657, 2111485077978050, 3416454622906707, 5527939700884757, 8944394323791464, 14472334024676221, 23416728348467685, 37889062373143906, 61305790721611591, 99194853094755497, 160500643816367088, 259695496911122585, 420196140727489673, 679891637638612258, 1100087778366101931, 1779979416004714189, 2880067194370816120, 4660046610375530309, 7540113804746346429, 12200160415121876738, 19740274219868223167, 31940434634990099905, 51680708854858323072, 83621143489848422977, 135301852344706746049, 218922995834555169026] [1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 406, 435, 465, 496, 528, 561, 595, 630, 666, 703, 741, 780, 820, 861, 903, 946, 990, 1035, 1081, 1128, 1176, 1225, 1275, 1326, 1378, 1431, 1485, 1540, 1596, 1653, 1711, 1770, 1830, 1891, 1953, 2016, 2080, 2145, 2211, 2278, 2346, 2415, 2485, 2556, 2628, 2701, 2775, 2850, 2926, 3003, 3081, 3160, 3240, 3321, 3403, 3486, 3570, 3655, 3741, 3828, 3916, 4005, 4095, 4186, 4278, 4371, 4465, 4560, 4656, 4753, 4851, 4950] ('Match', 1) ('Match', 1) ('Match', 3) ('Match', 21) ('Match', 55)

def fibo(n):

fib_1 = 1

fib_2 = 1

if n == 1 or n== 2:

return 1

fib_n=0

for i in range(3,n+1):

fib_n = fib_1+fib_2

fib_1 = fib_2

fib_2 = fib_n

return fib_n

def triNumList(n):

a=[]

for i in range(1,n):

b=i*(i+1)/2

a.append(b)

return a

def fiboList(n):

a=[]

for i in range(1,n):

a.append(fibo(i))

return a

def TriFiboCompare(n):

f = fiboList(n)

t = triNumList(n)

print(f)

print(t)

for i in f:

if i in t:

print("Match",i)

TriFiboCompare(100)

[1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169, 63245986, 102334155, 165580141, 267914296, 433494437, 701408733, 1134903170, 1836311903, 2971215073, 4807526976, 7778742049, 12586269025, 20365011074, 32951280099, 53316291173, 86267571272, 139583862445, 225851433717, 365435296162, 591286729879, 956722026041, 1548008755920, 2504730781961, 4052739537881, 6557470319842, 10610209857723, 17167680177565, 27777890035288, 44945570212853, 72723460248141, 117669030460994, 190392490709135, 308061521170129, 498454011879264, 806515533049393, 1304969544928657, 2111485077978050, 3416454622906707, 5527939700884757, 8944394323791464, 14472334024676221, 23416728348467685, 37889062373143906, 61305790721611591, 99194853094755497, 160500643816367088, 259695496911122585, 420196140727489673, 679891637638612258, 1100087778366101931, 1779979416004714189, 2880067194370816120, 4660046610375530309, 7540113804746346429, 12200160415121876738, 19740274219868223167, 31940434634990099905, 51680708854858323072, 83621143489848422977, 135301852344706746049, 218922995834555169026] [1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 406, 435, 465, 496, 528, 561, 595, 630, 666, 703, 741, 780, 820, 861, 903, 946, 990, 1035, 1081, 1128, 1176, 1225, 1275, 1326, 1378, 1431, 1485, 1540, 1596, 1653, 1711, 1770, 1830, 1891, 1953, 2016, 2080, 2145, 2211, 2278, 2346, 2415, 2485, 2556, 2628, 2701, 2775, 2850, 2926, 3003, 3081, 3160, 3240, 3321, 3403, 3486, 3570, 3655, 3741, 3828, 3916, 4005, 4095, 4186, 4278, 4371, 4465, 4560, 4656, 4753, 4851, 4950] ('Match', 1) ('Match', 1) ('Match', 3) ('Match', 21) ('Match', 55)

Labels: fibonacci, sagemath, triangular numbers