Accumulation function + unique property

[MasterNewbie]

I apologize for the vague title, but I'm not really sure how to describe the problem. I am also not sure if this belongs here or in financial math (the crux of the problem is calculus-based so I'm posting it here).

The problem in its entirety:
Suppose an accumulation function a(t) is differentiable and satisfies
a(s + t) = a(s) + a(t) - a(0)
for all non-negative reals s and t.
a) Using the definition of derivative, show that a'(t) = a'(0).
b) Show that a(t) = 1 + i*t, where i = a(1) - a(0) = a(1) - 1

Of note, all accumulation functions a(t) have the property a(0) = 1.

I finished part a, and began with part b.
a'(t) = a'(0) = i (a'(t) evaluated at t = 0 is a constant, which I am calling i)
Integrating it,
a(t) = i*t + c
We know a(0) = 1 = i*0 + c
implies c = 1
So a(t) = 1 + i*t.

I am stuck at showing that i = a(1) - a(0).
I tried doing a(2) = a(1+1) = a(1) + a(1) - a(0) but the resulting equation is dependent, and anything of a similar vein ends up being dependent.

 

[Ishuda]

I apologize for the vague title, but I'm not really sure how to describe the problem. I am also not sure if this belongs here or in financial math (the crux of the problem is calculus-based so I'm posting it here).

The problem in its entirety:
Suppose an accumulation function a(t) is differentiable and satisfies
a(s + t) = a(s) + a(t) - a(0)
for all non-negative reals s and t.
a) Using the definition of derivative, show that a'(t) = a'(0).
b) Show that a(t) = 1 + i*t, where i = a(1) - a(0) = a(1) - 1

Of note, all accumulation functions a(t) have the property a(0) = 1.

I finished part a, and began with part b.
a'(t) = a'(0) = i (a'(t) evaluated at t = 0 is a constant, which I am calling i)
Integrating it,
a(t) = i*t + c
We know a(0) = 1 = i*0 + c
implies c = 1
So a(t) = 1 + i*t.

I am stuck at showing that i = a(1) - a(0).
I tried doing a(2) = a(1+1) = a(1) + a(1) - a(0) but the resulting equation is dependent, and anything of a similar vein ends up being dependent.

Any linear function f can be written as
f(t) = i * t + f(0)
and satisfies
f(s+t) = f(s) + f(t) - f(0)
irregardless of the slope i or the value of f(0) [the a(0) in your problem].

However, to find the slope of a linear function, one might consider a(1)-a(0) where
a(t) = i t + a(0) = i t + 1
[from the boundary conditions, i.e. a(0)=1]

 

[MasterNewbie]

Calculating for the slope using the slope formula yields i.
a(1) = i + 1
a(0) = 1

[i + 1 - 1]/[1 - 0] = i.

However, taking note of
a(1) = i + 1 = i + a(0)
i = a(1) - 1 = a(1) - a(0)

Thank you, I don't think I would've stumbled upon that without trying a(1) to calculate the slope.