Recycle function

Algebra Level pending

g ( a ) = ∣ X − Y ∣ X = remainder when ’a’ is divided by ’a+1’. Y = the quotient when X is divided by a f ( x ) = { 0 ; if x > x 2 1 ; if x < 2 else; f ( g ( x ) ) \large g(a) = |X-Y| \\ \text{X = remainder when 'a' is divided by 'a+1'.} \\ \text{Y = the quotient when X is divided by a} \\ \large f(x) = \begin{cases} 0 \text{; if } x > x^2 \\ 1 \text{; if x < 2} \\ \text{else; } f(g(x)) \end{cases}

The above shows definition of 2 functions

Then find the value of

∑ n = 1 ∞ ( f ( 1 1 0 n ) + f ( n ) − f ( − n ) ) + 1 \large \color{#20A900}{\displaystyle \sum_{n=1}^{\infty} (f(\dfrac{1}{10^n}) + f(n) - f(-n))}+ \color{#3D99F6}{1}


The answer is 1.

Viki Zeta by Viki Zeta , 19, India

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1 solution

Viki Zeta
Sep 23, 2016

g ( x ) = X − Y . X is the remainder when ’x+1’ divides ’x’. So, since x < x + 1 x = 0 × ( x + 1 ) + x ⟹ X = x Y is the quotient when ’x’ divides ’X’ X = x x = X × 1 + 0 ⟹ Y = 1 ∴ g(x) = x - 1 ; x > 0 ______________________________________________________________________________________________________ Now for f(x), x > 2 f ( x ) = f ( g ( x ) ) ; x > 2 f ( x ) = f ( x − 1 ) ; x > 2 So, value for f(x), x > 2, is the predefined value of integer 1-less than the given value. So at one instance, it will reach the inequality : x < 2 ∴ f(x) = -1 ; x > 2 Now, for f(-x), x > 0 − x < 2 ∴ f(-x) = -1 ; x ≥ 0 Now, on evaluating the last case, x > x 2 , this can happen only in case of fractions of type 1 x ; x > 0 ∴ f ( 1 1 0 n ) = 0 ; n > 0 , since square of 1 1 0 n ; n > 0 is always lesser than 1 1 0 n ; n > 0 ________________________________________________________________________________________________________ ∴ ∑ n = 1 ∞ ( f ( 1 1 0 n ) + f ( n ) − f ( − n ) ) + 1 = ∑ n = 1 ∞ ( 0 + − 1 − ( − 1 ) ) + 1 ∑ n = 1 ∞ ( 0 ) + 1 = 1 g(x) = X - Y.\\ \text{X is the remainder when 'x+1' divides 'x'. So, since x < x + 1} \\ x = 0 \times (x+1) + x \implies X = x \\ \text{Y is the quotient when 'x' divides 'X' } \\ \color{#3D99F6}{\fbox{X = x}} \\ x = X\times 1 + 0 \\ \implies \color{#3D99F6}{\fbox{Y = 1}} \\ \therefore \color{#D61F06}{\fbox{g(x) = x - 1}}; x > 0 \\ \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_} \\ \text{Now for f(x), x > 2} \\ f(x) = f(g(x)); x > 2 \\ f(x) = f(x-1); x > 2 \\ \text{So, value for f(x), x > 2, is the predefined value of integer 1-less than the given value.} \\ \text{So at one instance, it will reach the inequality : x < 2} \\ \therefore \color{#D61F06}{\fbox{f(x) = -1}; x > 2} \\ \text{Now, for f(-x), x > 0} \\ -x < 2 \\ \therefore \color{#D61F06}{\fbox{f(-x) = -1}; x \ge 0}\\ \text{Now, on evaluating the last case, } x > x^2 \text{, this can happen only in case of fractions of type } \dfrac{1}{x}; x > 0 \\ \therefore \color{#D61F06}{\boxed{f(\dfrac{1}{10^n}) = 0}; n > 0} \text{, since square of } \dfrac{1}{10^n}; n> 0\text{ is always lesser than } \dfrac{1}{10^n} ; n>0 \\ \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_} \\ \therefore \displaystyle \sum_{n=1}^{\infty} (f(\dfrac{1}{10^n}) + f(n) - f(-n)) + 1 = \displaystyle \sum_{n=1}^{\infty} (0 + -1 - (-1)) + 1\\ \displaystyle \sum_{n=1}^{\infty} (0) + 1 = \color{#3D99F6}{\boxed{ 1 }}

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