Funky Functions

Algebra Level 4

$\large f\left(\frac{1}{1996}\right)+f\left(\frac{2}{1996}\right)+f\left(\frac{3}{1996}\right)+ \ldots+f\left(\frac{1995}{1996}\right)$

Find the value of the summation above if $f\left( x \right) =\frac { { 9 }^{ x } }{ { 9 }^{ x }+3 }$ .

998.5 999 996.5 997 996 999.5 998 997.5

3 solutions

Atul Vaibhav
May 4, 2015

$f(x) = \frac{9^{x}}{9^{x} + 3} , f(1 - x) = \frac{9^{1-x}}{9^{1-x} + 3}$ $f(x) + f(1-x) = \frac{ 9^{x} * 9^{1-x} + 3*9^{x} + 3*9^{1-x} + 9^{x} * 9^{1-x} } {9^{x} * 9^{1-x} + 3*9^{x} + 3*9^{1-x} + 9}$ $= \frac{ 9^{1} + 3*9^{x} + 3*9^{1-x} + 9^{1} } {9^{1} + 3*9^{x} + 3*9^{1-x} + 9} = 1$

Thus, for each 1 <= x <= 997 , we will have f(x) + f(1 - x) = 1 , i.e. a total of 997 lastly, $f(\frac{998}{1996}) = f(\frac{1}{2} ) = \frac{9^\frac{1}{2} } {9^\frac{1}{2} + 3} = \frac{3}{6} =0.5$ Thus, ans = 997 + 0.5 = 997.5

My friend sent me this question, knowing I wouldn't be able to solve it. I'm certain, after seeing the solution, that I mostly understand why it works; however.... I'm confused about the 997. That number, from my perspective, seems to come out of thin air. Would you mind explaining?

Melissa Skrei - 6 years, 1 month ago

since $f(x) + f(1 - x) = 1 ,$ so we have, $f(\frac{1}{1996}) + f(\frac{1995}{1996}) = 1$ $f(\frac{2}{1996}) + f(\frac{1994}{1996}) = 1$ $f(\frac{3}{1996}) + f(\frac{1993}{1996}) = 1$ ..... so on upto $f(\frac{997}{1996}) + f(\frac{999}{1996}) = 1$ , these 997 pairs sum upto 997 .

Atul Vaibhav - 6 years, 1 month ago

Vijaysekhar Chellaboina
May 3, 2015

Easy to see that $f(x)+f(1-x)=1$ . Hence, the summation is simply $997+f\left(\frac{998}{1996}\right)=997+f\left(\frac{1}{2}\right)=997.5.$

Can you please explain the solution?

Ashish Garg - 6 years, 1 month ago

$f(x)=\frac{9^x}{9^x+3}=\frac{9^x}{9^x+9^{1/2}}=\frac{9^{x/2}}{9^{x/2}+9^{(1-x)/2}}$ (I divided numerator and denominator by $9^{x/2}$ .) Now, using that form, find $f(x)+f(1-x)$ .

Akiva Weinberger - 6 years, 1 month ago

Ashish Garg...Its simple man

Atul Vaibhav makes a good way to get f(x)+f(1-x)=1

In the series given we have 1,995 terms i.e., from 1 to 1995 On adding numbers at same position from start and end i.e., f(1/1996)+f(1995/1996)=1 f(2/1996)+f(1994/1996)=1 . . . . f(997/1996)+f(999/1996)=1 (summation of 997 sets) But we miss the term f(998/1996) from symmetry whose value can be judged
as 0.5 So we got 997.5

Dodda Praful - 6 years, 1 month ago

please explain how you obtained the math to get 997 and how you got 1/2 from f(1-x). What I did was subtract x=(1/1996) from x =(1995/1996) and then multiply it by 1996. I got 997.17 which has to be the answer closest to number going forward. I wonder if that is how you obtained 997 as an answer

Ralston Rhoden - 6 years, 1 month ago

David Lee
May 6, 2015

Note: this is Canada 1995 P1. Also in book "The Math Olympian" by Richard Hoshino

Funky Functions

3 solutions

Ashish Garg...Its simple man

Atul Vaibhav makes a good way to get f(x)+f(1-x)=1

0 pending reports