How Does the Derivative Relate to Simple Interest in Accumulation Functions?

[wintermute++]

Homework Statement

Suppose that an amount function $a(t)$ is differentiable and satisfies the property
$a(s + t) = a(s) + a(t) − a(0)$
for all non-negative real numbers $s$ and $t$.

(a) Using the definition of derivative as a limit of a difference quotient, show that $a'(t) = a'(0)$.

(b) Show that $a(t) = 1 + it$ where $i = a(1) − a(0) = a(1) − 1$.

Homework Equations

N/A

The Attempt at a Solution

I do not understand what part b. expects me to do. If $a'(t) = a'(0)$, then I can show that equivalency using the definition of $i$. But, does that really show that $a(t) = 1 + it$? Perhaps the question is poorly worded, and it should read $a(t)$ is a possible solution? Or am I looking at this the wrong way?

 

[pasmith]

Any function of the form $a(t) = Ct + D$ for constants $C$ and $D$ satisfies $a(s + t) = a(s) + a(t) - a(0)$ for all nonnegative $s$ and $t$.

Perhaps the definition of an "amount function" imposes conditions on $a$ which you haven't told us about, for example that $a(0) = 1$.

 

[Stephen Tashi]

$a'(t) = a'(0)$ implies $a'(t)$ is a constant function. You know how to find an antiderivative of a constant function.

 

[LCKurtz]

Homework Statement

Suppose that an amount function $a(t)$ is differentiable and satisfies the property
$a(s + t) = a(s) + a(t) − a(0)$
for all non-negative real numbers $s$ and $t$.

(a) Using the definition of derivative as a limit of a difference quotient, show that $a'(t) = a'(0)$.

(b) Show that $a(t) = 1 + it$ where $i = a(1) − a(0) = a(1) − 1$.

Homework Equations

N/A

The Attempt at a Solution

I do not understand what part b. expects me to do. If $a'(t) = a'(0)$, then I can show that equivalency using the definition of $i$. But, does that really show that $a(t) = 1 + it$? Perhaps the question is poorly worded, and it should read $a(t)$ is a possible solution? Or am I looking at this the wrong way?

The conclusion is false. Try $a(t) = mt$ for any nonzero constant $m$. It satisfies the hypotheses but not the conclusion.

 

[wintermute++]

The textbook writes True, True for the solutions, for whatever that's worth.

My approach was:
Since $a'(t) = a'(0)$, $a(t) = a(0) = 1$. Then $a'(t) = a(1) - 1 = 0 = a'(0)$.

 

[LCKurtz]

The textbook writes True, True for the solutions, for whatever that's worth.

My approach was:
Since $a'(t) = a'(0)$, $a(t) = a(0) = 1$. Then $a'(t) = a(1) - 1 = 0 = a'(0)$.

No. Since $a'(t) = a'(0)$ then $a(t) = ta'(0) + C$, and you aren't given $a(0)=1$.

 

[wintermute++]

No. Since $a'(t) = a'(0)$ then $a(t) = ta'(0) + C$, and you aren't given $a(0)=1$.

My bad. $a(0) = 1$ for accumulation functions.

 

[LCKurtz]

No. Since $a'(t) = a'(0)$ then $a(t) = ta'(0) + C$, and you aren't given $a(0)=1$.

My bad. $a(0) = 1$ for accumulation functions.

Accumulation functions? Who said anything about accumulation functions, whatever they are? Not good to keep secrets when stating a problem...

 

[wintermute++]

Accumulation functions? Who said anything about accumulation functions, whatever they are? Not good to keep secrets when stating a problem...

Miswrote, meant amount function as specified in problem. And sorry, I was lazy and assumed too much of whoever was going to help me.

 

[vela]

Accumulation functions? Who said anything about accumulation functions, whatever they are? Not good to keep secrets when stating a problem...

Apparently the terms amount function and accumulation function come from finance. The accumulation function says how $1 would grow over time. In this problem, the accumulation function $a(t) = 1+it$ corresponds to simple interest. The amount function $A(t)=K a(t)$ is the balance at time $t$ if you start with a principal amount $K$.