# Proof of Fundamental Theorem of Calculus

• The quadrature of all figures follow from the inverse method of tangents, and thus the whole science of sums and quadratures can be reduced to analysis, a thing nobody even had any hopes of before. (Leibniz)
• Knowing thus the Algorithm of this calculus, which I call Differential Calculus, all differential equations can be solved by a common method. (Leibniz

Let us now study the effect of the time step in solution of the basic IVP

• $\dot u(t)=f(t),\quad\mbox{for }t>0,\quad u(0)=u^0$,

by Forward Euler time stepping

• $u(ndt+dt)=u(ndt)+f(ndt)dt,\quad n=0,1,2,...$

We compare taking one step with time step $dt$ with two steps of time step $\frac{dt}{2}$, for a given $n$:

• $u(ndt+dt)-\bar u({ndt+dt)}=f(ndt)dt - (f(ndt)+f(ndt+\frac{dt}{2}))\frac{dt}{2}$
• $=(f(ndt)-f(ndt+\frac{dt}{2}))\frac{dt}{2}$,

where $\bar u$ is computed with time step $\frac{dt}{2}$, and we assume that the same intial value for $t=ndt$ is used so that $\bar u(ndt)=u(ndt)$. Assuming that $f(t)$ is Lipschitz continuous with Lipschitz constant $L$, we then find that

• $\vert u(ndt +dt)-\bar u(ndt +dt)\vert \le \frac{L}{4} dt^2$.

Summing now the contributions from all time steps with $n=0,1,2,...,N$, where $T=(N+1)dt$ is a final time, we get using that $\sum_{n=0}^Ndt=T$,

• $\vert u(T)-\bar u(T)\vert \le \frac{LT}{4}\, dt$,

where thus $u(T)$ is computed with time step $dt$ and $\bar u(T)$ with time step $\frac{dt}{2}$.

Repeating the argument with successively refined times step $\frac{dt}{4},\frac{dt}{8},...$, we get

• $\vert u(T)-\bar u(T)\vert \le \frac{LT}{2}\, dt$

for the difference between $u(T)$ computed with time step $dt$ and $\bar u(T)$ computes with vanishingly small time step, since

• $\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...< \frac{1}{2}$.

The fundamental step in the proof of the Fundamental Theorem.

We have now proved the Fundamental Theorem of Calculus:

Theorem If $f:[0,T]\rightarrow R$ is Lipschitz continuous, then the function $u(t)=\int_0^tf(s)\, ds$ defined by Forward Euler time-stepping with vanishing time step, solves the IVP: $\dot u(t)=f(t)$ for $t\in (0,T)$, $u(0)=0$.

# Understanding the Fundamental Theorem

The proof shows what it means to understand the Fundamental Theorem of Calculus:  This is to realize that (letting $k$ denote a finite time step and $dt$ a vanishingly small step)

• $u(T)=\int_0^Tf(t)\, dt\approx \sum_{n=0}^Nf(nk)k\quad\mbox{if }T=(N+1)k$,

as a consequence of

• $u((n+1)k)\approx u(nk)+f(nk)k,\quad\mbox{or}\quad\frac{u((nk+k)-u(nk)}{k}\approx f(nk)$,

where the sum is referred to as a Riemann sum, with the following bound for the difference

• $\vert \int_0^Tf(t)\, dt-\sum_{n=0}^Nf(nk)k\vert \le \frac{LTk}{2}\quad\mbox{with}T=(N+1)k$,

assuming $f:[0,T]\rightarrow R$ is Lipschitz continuous with Lipschitz constant $L$.

In other words, understanding the integral $u(t)=\int_0^tf(s)\, ds$ of a function $f:[0,T]\rightarrow R$, means to understand that:

1. $u(t)=\int_0^tf(s)\, ds$ is determined by Riemann sums with vanishingly small step size, as the solution to the IVP $\dot u(t)=f(t)$, $u(0)=0$,
2. the difference between two Riemann sums with mesh size $k$ and $\frac{k}{2}$, is bounded by $Lk$ (or more precisely by $\frac{L}{4}k$).

# Even Better Understanding

As a serious student, you now probably ask: In precisely what sense the differential equation $\dot u(t)=f(t)$ is satisfied by an Euler Forward solution $u(t)$ with time step $k$? It certainly is so constructed, but can we get a direct verification? One way to do this is to associate a continuous piecewise linear function determined by the values $u(nk)$ at the discrete time levels $nk$,again denoted by $u(t)$. We then have on each interval $(nk,(n+1)k)$, by the definition of $u(t)$:

• $\dot u(t)=\frac{u((n+1)k)-u(nk)}{k}=f(nk)$,

from which we conclude that

• $\vert \dot u(t)-f(t)\vert\le \vert f(nk)-f(t)\vert\le Lk\quad\mbox{for }t\in ((n+1)k,nk)$.

We can thus say that $u(t)$ satisfies the differential equation $\dot u(t)=f(t)$ for all $t$ with a precision of $Lk$. In other words, the residual $\dot u(t)-f(t)$ is smaller than $Lk$.

We have now understood the Fundamental Theorem even better, right?

We shall see below that extending a function defined on a discrete set of points to a continuous piecewise linear function, is a central aspect of approximation in general and of the Finite Element Method FEM in particular.

• What is fundamental about the Fundamental Theorem?
• Why is $\frac{d}{dt}\int_0^tf(s)\, ds=f(t)$? (compare with last argument)
• What is the effect of finite precision computation according to Constructive Calculus in Finite Precision? (see below)
• What is the Riemann sum error using the Trapezoidal Rule ?

Hint:

• $\int_0^{t+dt}f(s)\, ds-\int_0^tf(s)\, ds=\int_t^{t+dt}f(s)\, ds = f(t)dt\pm \frac{L}{2}dt^2$.

The sad result of not understanding the mathematics of Archimedes.

# Fundamental Theorem in Finite Precision

Lipschitz continuity in the presence of finite precision can be defined as follows:  A real-valued function $f(t)$ of a real variable $t$ is Lipschitz continuous with Lipschitz constant $L$ in finite precision $\epsilon > 0$, if for all $t$ and $\Delta t$

• $\vert f(t+\Delta t) - f(t)\vert\le L\vert \Delta t+ \epsilon\vert$.

We see that here $dt$ will effectively be bounded below by $\epsilon$. With this extension of the concept of Lipschitz continuity to finite precision, the first step of the above proof takes the form

• $\vert u(T)-\bar u(T)\vert \le \frac{LT}{4}\, (dt+\epsilon)$.

In the second step, the repetition with successively refined times step $\frac{dt}{4},\frac{dt}{8},...$, is performed until $\frac{dt}{2^N} <\epsilon$ for some natural number $N$, which gives

• $\vert u(T)-\bar u(T)\vert \le \frac{LT}{2}dt+N\frac{LT}{4}\epsilon$

for the difference between $u(T)$ computed with time step $dt$ and $\bar u(T)$ computed with time step $dt \ge \epsilon$. Here $N$ is the 2-logarithm of $\frac{\epsilon}{dt}$ and thus is a constant of moderate size (not large). In essence, we thus obtain the previous estimate with $dt$ replaced by $dt +\epsilon$ and $\epsilon$ appears as a lower bound of the time step $dt$

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