# Birth Day of Srinivasa Ramanujam

#### Subhotosh Khan

##### Super Moderator
Staff member

National Mathematics Day ( i.e. Birth Day of Srinivasa Ramanujam )

See This Absolutely Amazing Mathematics Given By Great Mathematician *रामानुजम*
1 x 8 + 1 = 9
12 x 8 + 2 = 98
123 x 8 + 3 = 987
1234 x 8 + 4 = 9876
12345 x 8 + 5 = 98765
123456 x 8 + 6 = 987654
1234567 x 8 + 7 = 9876543
12345678 x 8 + 8 = 98765432
123456789 x 8 + 9 = 987654321

1 x 9 + 2 = 11
12 x 9 + 3 = 111
123 x 9 + 4 = 1111
1234 x 9 + 5 = 11111
12345 x 9 + 6 = 111111
123456 x 9 + 7 = 1111111
1234567 x 9 + 8 = 11111111
12345678 x 9 + 9 = 111111111
123456789 x 9 +10= 1111111111

9 x 9 + 7 = 88
98 x 9 + 6 = 888
987 x 9 + 5 = 8888
9876 x 9 + 4 = 88888
98765 x 9 + 3 = 888888
987654 x 9 + 2 = 8888888
9876543 x 9 + 1 = 88888888
98765432 x 9 + 0 = 888888888

And Look At This Symmetry :

1 x 1 = 1
11 x 11 = 121
111 x 111 = 12321
1111 x 1111 = 1234321
11111 x 11111 = 123454321
111111 x 111111 = 12345654321
1111111 x 1111111 = 1234567654321
11111111 x 11111111 = 123456787654321
111111111 x 111111111 = 12345678987654321

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#### Jomo

##### Elite Member
You failed to show your work. You need to now go back and prove each of those beautiful symmetries (and not by brute force).

#### Jomo

##### Elite Member
Was it Ramanujam who actually proved these symmetries?

#### JeffM

##### Elite Member
According to Hardy, the first set of results he got from Ramanujam consisted of theorems

previously proved by others

never imagined by others but flawed

never imagined by others but sound

#### JeffM

##### Elite Member
@Jomo

Brute force is whole lot easier than algebra.

$$\displaystyle n \text { is an integer } \ge 1.$$

$$\displaystyle f(n) = \sum_{j=1}^n (10 - j) * 10^{(n - j)}.$$

$$\displaystyle g(n) = \sum_{j=1}^n j * 10^{(n - j)}.$$

$$\displaystyle n = 1 \implies f(n) = 9 \text { and } g(n) = 1 \implies f(1) = 8 * g(1) + 1 \implies$$

$$\displaystyle n = 1 \implies f(n) = 8 * g(n) + n.$$

$$\displaystyle \therefore \exists \text { a positive integer } k \text { such that } f(k) = 8g(k) + k.$$

$$\displaystyle f(k) = \sum_{j=1}^k (10 - j) * 10^{(k - j)}$$

$$\displaystyle g(k) = \sum_{j=1}^k j * 10^{(k - j)}.$$

$$\displaystyle f(k + 1) = \sum_{j=1}^{k+1} (10 - j) * 10^{\{(k + 1) - j\}} =$$

$$\displaystyle \left (\sum_{j=1}^k (10 - j) * 10^{\{(k + 1) - j\}} \right ) + \{10 - (k + 1)\} * 10^{\{(k + 1) - (k + 1)\}} =$$

$$\displaystyle 10f(k) + 9 - k = 10\{8g(k) + k\} + 9 - k = 80g(k) + 9(k + 1).$$

$$\displaystyle g(k + 1) = \sum_{j=1}^{k+1} j * 10^{\{(k+1) - j\}} = \left (\sum_{j=1}^k j * 10^{\{(k+1) - j\}} \right ) + k + 1 =$$

$$\displaystyle 10 \left (\sum_{j=1}^k j * 10^{(k - j)} \right ) + k + 1 = 10g(k) + (k + 1).$$

$$\displaystyle \therefore 8g(k + 1) + (k + 1) = 80g(k) + 8(k + 1) + k + 1 = 80g(k) + 9(k + 1).$$

$$\displaystyle f(k + 1) = 80g(k) + 9(k + 1) = 8g(k + 1) + (k + 1) \implies$$

$$\displaystyle f(k + 1) = 8g(k + 1) + k + 1 \implies$$

$$\displaystyle f(n + 1) = 8g(n + 1) + n+ 1 \text { for any positive integer } n.$$ ...edited

And how you generalize that is beyond perplexing.

Last edited by a moderator:

#### lex

##### Full Member
@JeffM
Presumably no $$\displaystyle k$$ in your last line of working.

#### JeffM

##### Elite Member
No. Perhaps the great khan will fix it. ................. fixed

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#### JeffM

##### Elite Member
Actually, generalization and recursion are quite clarifying.

$$\displaystyle m \text { and } n \text { are integers such that } m > 0 \text { and } n > 1.$$

$$\displaystyle m = 1 \implies f(m) = n;\\ 1 < m \implies f(m) = (n + 1) * f(m - 1) + \{(n + 1) - m\}.$$
So, for example, if n = 9

$$\displaystyle f(3) = (9 + 1) * f(2) + 9 + 1 - 3 = 10\{(9 + 1) * f(1) + 9 + 1 - 2\} +7 =\\ 10(10 * 9 + 8) + 7 = 10 * (90 + 8)+ 7 = 980 + 7 = 987.$$
$$\displaystyle m = 1 \implies g(m) = 1;\\ 1 < m \implies g(m) = (n + 1) * g(m -1) + m.$$
So, for example, if n = 9

$$\displaystyle g(3) = (9 + 1) * f(2) + 3 = 10\{(9 + 1) * g(1) + 2\} + 3 =\\ 10(10 * 1 + 2) + 3 = 10 * (10 + 2)+ 3 = 120 + 3 = 123.$$
NOTE THAT $$\displaystyle f(1) = n = n - 1 + 1 = \{(n - 1) * 1\} + 1 = (n - 1) * g(1) + 1$$

$$\displaystyle \therefore m = 1 \implies f(m) = (n - 1) * g(m) + m.$$

$$\displaystyle \therefore \exists \text { positive integer } k \text { such that } f(k) = (n - 1) * g(k) + k.$$

$$\displaystyle k \text { is a positive integer} \implies k + 1 > 1. \text { Therefore, by definition,}$$

$$\displaystyle f(k + 1) = (n + 1) * f(k) + (n + 1) - (k + 1) \implies$$

$$\displaystyle f(k + 1) = (n + 1) * f(k) + n - k \implies$$

$$\displaystyle f(k + 1) = (n + 1)\{(n - 1) * g(k) + k\} + n - k \implies$$f(1

$$\displaystyle f(k + 1) = (n^2 - 1) * g(k) + kn + k + n - k \implies$$

$$\displaystyle f(k + 1) = (n^2 - 1) * g(k) + n(k + 1).$$

$$\displaystyle \text {Because } k + 1 > 1,\ g(k + 1) = (n + 1) * g(k) + k + 1.$$

$$\displaystyle \therefore (n - 1) * g(k + 1) = (n^2 - 1) * g(k) + (n - 1)(k + 1) \implies$$

$$\displaystyle (n - 1) * g(k + 1) = (n^2 - 1) * g(k) + n(k + 1) - (k + 1) \implies$$

$$\displaystyle (n - 1) * g(k + 1) + (k + 1)= (n^2 - 1) * g(k) + n(k + 1) = f(k + 1).$$

$$\displaystyle \therefore f(m) = (n - 1) * g(m) + m.$$

There is nothing special about 8 in the original proposition other than that 8 = 9 - 1.

Let's suppose n = 4.

$$\displaystyle f(1) = 4 \text { and } g(1) = 1.\\ f(2) = 5 * 4 + 5 - 2 = 23 \text { and } g(2) = 5 * 1 + 2 = 7.\\ f(3) = 5 * 23 + 5 - 3 = 117 \text { and } g(3) = 5 * 7 + 3 = 38.\\ f(4) = 5 * 117 + 5 - 4 = 586 \text { and } g(4) = 5 * 38 + 4 = 194.$$
Boring.

Let's check: does (n - 1) * g(m) + m = f(m).

$$\displaystyle 3 * 1 + 1 = 4. \ \checkmark \\ 3 * 7 + 2 = 23. \ \checkmark \\ 3 * 38 + 3 = 117. \ \checkmark \\ 3 * 194 + 4 = 586. \ \checkmark$$

It checks, but there is nothing WOWish about it. But lets do it with numerals to base 4 + 1 = 5

$$\displaystyle f(1_5) = 4_5 \text { and } g(1) = 1_5.$$

$$\displaystyle f(2_5) = 43_5 \text { and } g(2) = 12_5.$$

$$\displaystyle f(3_5) = 432_5 \text { and } g(3) = 123_5.$$

$$\displaystyle f(4_5) = 4321_5 \text { and } g(4) = 1234_5.$$

lex

#### lex

##### Full Member
@JeffM
Ramanujan lives!
You've really gotten into this.

#### lex

##### Full Member
@JeffM
I'll share a couple of proofs of your original one, but you've gone way beyond this now.

You want to prove:

$$\displaystyle \left(\sum_{j=1}^{n}{8j\times{10}^{n-j\ }}\right)+n=\left(\sum_{j=1}^{n}{\left(10-j\right)\times{10}^{n-j\ }}\right)$$

which is equivalent to proving that:

$$\displaystyle \left(\sum_{j=1}^{n}{\left(10-9j\right)\times{10}^{n-j\ }}\right)=n$$

This can be easily done using mathematical 'induction' on $$\displaystyle n$$:

Clearly it is true for $$\displaystyle n=1$$: LHS=1, RHS=1
Assume true for $$\displaystyle n$$, then prove:

$$\displaystyle \left(\sum_{j=1}^{n+1}{\left(10-9j\right)\times{10}^{n+1-j\ }}\right)=n+1$$

LHS =$$\displaystyle \text{ }10\sum_{j=1}^{n}{\left(10-9j\right)\times{10}^{n-j\ }}+\left(10-9\left(n+1\right)\right){10}^0$$

$$\displaystyle \qquad=10n+10-9n-9\qquad$$(using the inductive hypothesis)

$$\displaystyle \qquad =n+1\qquad=\qquad$$RHS $$\displaystyle \qquad\qquad$$

True for $$\displaystyle n=1$$, true for $$\displaystyle n\rightarrow$$ true for $$\displaystyle n+1$$, therefore by principle of mathematical induction it is true for all $$\displaystyle n\in\mathbb{Z}^+$$.

#### lex

##### Full Member
@JeffM
Another alternative:
You want to prove:

$$\displaystyle \left(\sum_{j=1}^{n}{8j\times{10}^{n-j\ }}\right)+n=\left(\sum_{j=1}^{n}{\left(10-j\right)\times{10}^{n-j\ }}\right)$$

which is equivalent to proving that:
$$\displaystyle \left(\sum_{j=1}^{n}{\left(10-9j\right)\times{10}^{n-j\ }}\right)=n$$

This can be done directly using:

(1)$$\displaystyle \qquad\sum_{j=1}^{n}x^{j\ }=\frac{x\left(1-x^n\right)}{1-x}$$
and (2)$$\displaystyle \qquad\sum_{j=1}^{n}{jx^{j\ }}=x\frac{d}{dx}\left(\frac{\left(1-x^{n+1}\right)}{1-x}\right)=\frac{x\left(nx^{n+1}-\left(n+1\right)x^n+1\right)}{\left(1-x\right)^2}$$
With $$\displaystyle x=\frac{1}{10}$$
$$\displaystyle \left(1\right)\rightarrow\sum_{j=1}^{n}{10}^{-j\ }=\frac{1}{9}\left(1-{10}^{-n}\right)$$
$$\displaystyle \left(2\right)\rightarrow\sum_{j=1}^{n}{j10}^{\left(n-j\right)} ={10}^n\sum_{j=1}^{n}{j10}^{-j}=\frac{1}{81}\left(-9n-10+{10}^{n+1}\right)$$

$$\displaystyle \therefore\sum_{j=1}^{n}{\left(10-9j\right)\times{10}^{n-j\ }}=\frac{{10}^{n+1}}{9}\left(1-{10}^{-n}\right)-\frac{9}{81}\left(-9n-10+{10}^{n+1}\right)$$
$$\displaystyle =\frac{{10}^{n+1}}{9}-\frac{10}{9}+n+\frac{10}{9}-\frac{{10}^{n+1}}{9}$$

$$\displaystyle =n$$

#### JeffM

##### Elite Member
@JeffM
Ramanujan lives!
You've really gotten into this.
That is very flattering. Thank you. But of course it is not true. I just walked along paths made by Ramanjum.

#### lex

##### Full Member
@JeffM
Too modest again - looks like a lot of original thinking.