Something off 'The margin'!

Number Theory Level 3

f ( x ) = a 1 + a x + b 1 + b x − c 1 + c x \large f(x) = \frac{a}{1+ax} + \frac{b}{1+bx} - \frac{c}{1+cx}

Let f ( x ) f(x) be a function as defined above where a , b a, \ b and c c are arbitrary non-negative integral constants and a b c ≠ 0 abc \neq 0 . Which of the following results are possible?

I. f ( 0 ) f(0) can be zero.

II. f ( 1 ) ( 0 ) f^{(1)} (0) can be zero.

III. f ( 2 ) ( 0 ) f^{(2)} (0) can be zero.

IV. f ( 3 ) ( 0 ) f^{(3)} (0) can be zero.

Notations: f ( n ) ( a ) f^{(n)}(a) denotes the n t h n^{th} derivative of f ( x ) f(x) at x = a x=a .

Follow up problem:

What conclusion do you deduce for f ( n ) ( 0 ) f^{(n)}(0) for non-negative integral values of n n (note that f ( 0 ) ( 0 ) f^{(0)}(0) is the same as f ( 0 ) f(0) )? Does this problem have trivial solutions? If yes, then what are they?

I only II only II and IV I, II, III and IV I and III I, II and III III and IV I and II

Tapas Mazumdar by Tapas Mazumdar , 20, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Viki Zeta
Oct 12, 2016

f ( x ) = a 1 + a x + b 1 + b x − c 1 + c x x = 0 f ( 0 ) = a + b − c a b c ≠ 0 ⟹ a ≠ 0 , b ≠ 0 , c ≠ 0 Take the case, a ≠ c , b ≠ c , a + b = c f ( 0 ) = c − c = 0 So, f(x) = 0 can be true for some integer ’a’, ’b’ and ’c’ f ′ ( x ) = d d x f ( x ) = d d x a 1 + a x + d d x b 1 + b x − d d x c 1 + c x = − a 2 ( 1 + a x ) 2 − b 2 ( 1 + b x ) 2 + c 2 ( 1 + c x ) 2 f ′ ( 0 ) = c 2 − a 2 − b 2 = c 2 − ( a 2 + b 2 ) Take the case, c 2 = a 2 + b 2 , f ′ ( 0 ) = c 2 − c 2 = 0 So, f’(x)=0 can be true for some integer ’a’, ’b’, and ’c’ f ′ ′ ( x ) = a 3 ( 1 + a x ) 3 + b 3 ( 1 + b x ) 3 − c 3 ( 1 + c x ) 3 f ′ ′ ( 0 ) = a 3 + b 3 − c 3 This is only possible if a 3 + b 3 = c 3 , But according to Fermat’s last theorem There is no integer n ≥ 3 ; such that a n + b n = c n The other two options follows the same condition. f(x) = \dfrac{a}{1+ax} + \dfrac{b}{1+bx} - \dfrac{c}{1+cx} \\ \boxed{x=0} \\ f(0) = a + b - c \\ abc \ne 0 \\ \implies a \ne 0, b\ne 0, c \ne 0 \\ \text{Take the case, } a \ne c, b \ne c, a + b = c \\ f(0) = c - c = 0 \\ \text{So, f(x) = 0 can be true for some integer 'a', 'b' and 'c'} \\ f'(x) = \dfrac{d}{dx} f(x) = \dfrac{d}{dx}\dfrac{a}{1+ax} + \dfrac{d}{dx}\dfrac{b}{1+bx} - \dfrac{d}{dx}\dfrac{c}{1+cx} \\ = -\dfrac{a^2}{(1+ax)^2} -\dfrac{b^2}{(1+bx)^2} + \dfrac{c^2}{(1+cx)^2} \\ f'(0) = c^2 - a^2 - b^2 = c^2 - (a^2 + b^2) \\ \text{Take the case, } c^2 = a^2 + b^2, \\ f'(0) = c^2 - c^2 = 0 \\ \text{So, f'(x)=0 can be true for some integer 'a', 'b', and 'c'}\\ f''(x) = \dfrac{a^3}{(1+ax)^3} +\dfrac{b^3}{(1+bx)^3} - \dfrac{c^3}{(1+cx)^3} \\ f''(0) = a^3 + b^3 - c^3 \\ \text{This is only possible if } a^3 + b^3 = c^3 \text{, But according to Fermat's last theorem} \\ \text{There is no integer } n \ge 3 \text{; such that} \\ a^n + b^n = c^n \\ \text{The other two options follows the same condition.}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

If a+b = c, and a^2+b^2 = c^2. Then ab =0? Which means both equations can be true separately, but not simultaneously?

Siva Bathula - 4 years, 8 months ago

Log in to reply

a + b = 0 a+b=0 is for first case. You should not use it in next case

Viki Zeta - 4 years, 8 months ago

Log in to reply

Sorry a + b = c a+b=c

Viki Zeta - 4 years, 8 months ago

Magnificent! The exact method I was looking for. If you wish you might as well include the answer to my followup question in your solution as well.

Tapas Mazumdar - 4 years, 8 months ago

If both 1 and 2 are true, meaning, f(0) 'can be zero' and first derivative 'can also be zero' at 0. (1) gives a + b = c and (2) a^2 + b^2 = c^2. But if both (1) and (2) are true simultaneously, squaring (1) would give a.b = 0. Meaning either of a or b is zero. Which means only either (1) or (2) can be true, but not both.

Siva Bathula - 4 years, 8 months ago

Log in to reply

Yes. Separately they are both true. You shouldn't follow the result of one to the other. That's why both I and II when checked are true.

Tapas Mazumdar - 4 years, 8 months ago

Think about it. If you were to check for f ( 0 ) f(0) and for f ( 1 ) ( 0 ) f^{(1)} (0) , would you consider the result of one for the other? And my problem said 'arbitrary' values for a , b a,\ b and c c , which means they can take up any necessary values to satisfy the f ( 0 ) = 0 f(0) = 0 and f ( 1 ) ( 0 ) = 0 f^{(1)} (0) = 0 , after which, by Fermat's Last Theorem, there are no possible positive integral solutions for a , b a, \ b and c c .

Tapas Mazumdar - 4 years, 8 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...