Closed Form Expressions and Solutions
A closed-form expression is a mathematical expression or a formula that uses only basic, well-known functions — like addition, multiplication, roots, exponents, and logarithms — written out in a finite number of steps.
In a closed form solution the answer is expressed as a precise symbolic expression using standard mathematical operations — no iterations involved. The answer is exact in form and can be written down completely.
Consider a linear equation $ax+b=0$ where $a\ne 0, b \in \mathbb{R}$ or a quadratic equation $ax^2+bx+c=0$ where $a\ne 0, b,c \in \mathbb{R}$. These equations can be solved using so called Close-Form expressions.
For linear, $x=-\frac{b}{a}$ and for quadratic, it is $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$
Quick Examples
If $f(x) = 2x – 6 = 0$. The solution is $x = 3$
If $f(x) = x^2 – 5x + 6 = 0$. Using the above closed form we get $x = 2$ or $x = 3$.
Alos it can be noted that both solutions are exact integers
Closed form is not limited to polynomials.
Consider $\log_2 x = 3$. The solution is $x = 2^3 = 8$, exactly.
In general, if $\log_a x = b$ where $a > 0, a \ne 1; x>0; b \in \mathbb{R}$, then the solution is $x = a^b$ — a closed form in terms of $a$ and $b$.
These are exact symbolic answers, derived directly from the structure of the equation.
Now consider
- $e^x = 7$ has the closed form solution $x = \ln 7$
- $f(x) = x^2 – 2 = 0$. The closed form solution is $x = \sqrt{2}$.
As a symbol, this is exact — it identifies precisely which number solves the equation.
But their numerical values $\ln 7 = 1.94591015 \ldots \ldots ,\sqrt{2} = 1.41421356\ldots$, are irrational and non-terminating.
The closed form solutions are exact but their numerical solutions are not exactly an integer or terminating rationals like $x=\frac{1}{2}=0.5$, the solution for $2x=1$.
This distinction is important. Closed form is a statement about symbolic completeness. Whether that symbol evaluates to an integer, terminating rationals etc depends entirely on the nature of the solution.
Now consider $f(x) = x + \ln x – 5$. There is no algebraic manipulation, no standard formula, no symbolic way that “isolates” $x$ on one side. No closed form solution exists.
Approximations — Rounding and Significant Digits
Once we move from symbolic expressions to numerical computation, approximation may be unavoidable. Two principal sources of error arise: rounding and significant digit truncation.
Rounding
Rounding is the process of replacing a number with a nearby value that fits within the representation being used.
The most familiar rule is round-half-up: if the digit being dropped is $5$ or above, the preceding digit is increased by one. So $2.35$ rounded to one decimal place gives $2.4$, and $2.45$ rounded to one decimal place gives $2.5$.
But this rule introduces a systematic bias — over many calculations, values ending in exactly $5$ are always pushed upward, causing accumulated error to drift in one direction.
A common correction is banker’s rounding, also called round-half-to-even: when the dropped portion is exactly halfway, the result is rounded to whichever neighbouring value has an even last digit. So $2.5$ rounds to $2$, and $3.5$ rounds to $4$, and $4.5$ rounds to $4$, and $5.5$ rounds to $6$.
Over a large number of operations, half the half-way cases round up and half round down, and the bias cancels out. This is standard practice in financial computation and in many numerical libraries precisely because of this property.
But regardless of the rounding rule used, rounding is a departure from the true value. The error introduced at each step may be tiny, but it does not disappear — it carries forward into every subsequent calculation that uses the rounded value.
Significant Digits
Significant digits refer to the meaningful digits in a number — the digits that carry actual information about its precision. The question of how many significant digits to keep is not merely a matter of display; it determines what information is retained and what is permanently discarded.
Consider the number:
$12.53453753434787476476010005\ldots$
This number has digits running on without end. In any real computation, we may decide how many significant digits to work with.
With $4$ significant digits, we keep $12.53$ and discard everything after — the retained value is $12.53$. The discarded portion begins at $453753\ldots$ — already in the fourth decimal place. The absolute error introduced is $0.00453753\ldots$, which looks small, but as a fraction of the last retained digit it is not negligible.
With $5$ significant digits, we keep $12.535$. The retained value is $12.535$ and the discarded portion begins at $3753\ldots$ The absolute error is now $0.000453753\ldots$, ten times smaller than before.
With $6$ significant digits, we keep $12.5346$. The retained value is $12.5346$ and the discarded portion begins at $53\ldots$ The absolute error is $0.0000453\ldots$
Each additional significant digit reduces the absolute error by roughly a factor of ten. But the digits being discarded are real — they are part of the number. Once discarded, they are gone from the computation.
Now consider a number less than one:
$0.000078534637281900045\ldots$
Here the leading zeros after the decimal point are not significant — they merely locate the decimal point. The significant digits begin at $7$. With $4$ significant digits, the retained value is $0.00007853$ and everything from $4637\ldots$ onward is discarded.
With $5$ significant digits, we retain $0.000078535$ (rounding the $4$ up because the next digit is $6$). The precision gained at each step is the same in relative terms.
Now consider a negative number:
$-12.53453753434787476476010005\ldots$
The sign does not affect the count of significant digits. With $4$ significant digits, the retained value is $-12.53$ and the same truncation applies as before.
With $5$ significant digits, $-12.535$. The error introduced has the same magnitude as in the positive case — the sign is preserved but does not change what is lost.
The deeper issue with significant digits is what happens when truncated values are used in further calculations. Suppose $a = 12.53453753\ldots$ is retained as $12.53$ and $b = 0.00453753\ldots$ is retained as $0.004538$.
If we compute $a – b$, the true answer is $12.53000000\ldots$ but the computed answer using the retained values is $12.53 – 0.004538 = 12.525462$. The subtraction has introduced an error that is larger in relative terms than either of the individual truncations.
This phenomenon is one of the most consequential effects of significant digit truncation in numerical computation.
Analytical software, platforms, programming environments usually have a provision / function to handle these rounding / siginificance digits operations. The impact can be seen in this R example
A Note on Asymptotic Behaviour
Asymptotic behaviour describes how a function behaves as its argument moves toward an extreme — as $x \to \infty$, as $x \to -\infty$, or as $x$ approaches some special point where the function changes character fundamentally.
Consider $e^x$.
As $x \to \infty$, $e^x$ grows without bound and does so faster than any polynomial — faster than $x^{100}$, faster than $x^{1000}$.
As $x \to -\infty$, $e^x$ decays toward zero, approaching it from above but never reaching it. The horizontal axis is an asymptote — the function gets arbitrarily close but never touches it.
Now consider $\ln x$.
As $x \to \infty$, $\ln x$ also grows without bound, but extraordinarily slowly — slower than $x^{0.001}$, slower than any positive power of $x$ however small.
This slowness has practical consequences: quantities that grow logarithmically are well-behaved even when the underlying variable becomes very large.
As $x \to 0^+$, $\ln x \to -\infty$ — the function blows up downward, with a vertical asymptote at $x = 0$.
This connects directly to the question of approximations. Any polynomial approximation to a function is constructed at a specific point and is trustworthy only within a neighbourhood of that point.
As $x$ moves away, terms that were small at the centre begin to grow, and eventually the approximation departs from the true function.
Asymptotic thinking makes this precise: it describes how quickly the neglected terms grow as $x$ drifts, and therefore how far the approximation can be trusted. The Taylor series — discussed next — is fundamentally an asymptotic question.
Taylor’s Theorem
Taylor’s theorem formalises the idea that any sufficiently smooth function $f(x)$ can be approximated near a chosen point $a$ by a polynomial built entirely from the value and successive derivatives of $f$ at that single point.
The Taylor expansion of $f(x)$ around the point $a$, is $f(a) + f'(a)(x-a) + \dfrac{f”(a)}{2!}(x-a)^2 + \dfrac{f”'(a)}{3!}(x-a)^3 + \dfrac{f^{(4)}(a)}{4!}(x-a)^4 + \cdots$
And carried to the fourth order, is:
$f(x) \approx f(a) + f'(a)(x-a) + \dfrac{f”(a)}{2!}(x-a)^2 + \dfrac{f”'(a)}{3!}(x-a)^3 + \dfrac{f^{(4)}(a)}{4!}(x-a)^4 $
Each term adds one more layer of local information about the shape of $f$ at the point $a$.
The zeroth term $f(a)$ is simply the value of the function at $a$ — a constant approximation, valid only at the point itself.
Adding $f'(a)(x-a)$ gives a linear approximation — the tangent line to $f$ at $a$. This is the best straight line fit to $f$ near $a$.
Adding $\dfrac{f”(a)}{2!}(x-a)^2$ introduces curvature. The second derivative measures how fast the slope itself is changing, and this term bends the approximation to match the local curvature of $f$ at $a$.
Adding $\dfrac{f”'(a)}{3!}(x-a)^3$ captures the asymmetry — the way $f$ leans differently to the left and right of $a$. This is what the third derivative, sometimes called the rate of change of curvature, contributes.
Adding $\dfrac{f^{(4)}(a)}{4!}(x-a)^4$ refines further, capturing the rate at which the asymmetry itself changes.
Every piece of information used — the value, the slope, the curvature, the lean, the rate of change of lean — is drawn entirely from $f$ at the single point $a$.
Differentiation is a local operation, measuring behaviour at a point. Taylor’s theorem says that if we know $f$ and all its derivatives at one point well enough, we can reconstruct the behaviour of $f$ away from that point, term by term, to any desired order.
The error in stopping at $n$ terms is not left vague. Taylor’s theorem provides it an exact form — the Lagrange remainder:
$R_n(x) = \dfrac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$
for some $c$ lying between $a$ and $x$. The error is controlled by the size of the $(n+1)$-th derivative of $f$, multiplied by $(x-a)^{n+1}$ — which shrinks as $x$ stays close to $a$ — and divided by $(n+1)!$ — which grows very rapidly with $n$.
Together, these factors mean that for well-behaved functions, including more terms drives the error down rapidly, especially near $a$. The error is given a precise mathematical shape, and that shape is itself a function of the derivatives of $f$.
Examples Function Speed
$f(x) = e^x$ around $a = 0$
The exponential function has a property: every derivative of $e^x$ is $e^x$ itself.
That is,
$f(x) = e^x$
$f'(x) = e^x$
$f”(x) = e^x$, and so on for every order.
At $a = 0$, every derivative evaluates to $e^0 = 1$.
The Taylor expansion is:
$e^x = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dfrac{x^4}{4!} + \cdots$
To first order: $e^x \approx 1 + x$
To second order: $e^x \approx 1 + x + \dfrac{x^2}{2}$
To third order: $e^x \approx 1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6}$
To fourth order: $e^x \approx 1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6} + \dfrac{x^4}{24}$
For small $x$, the first order approximation $1 + x$ is useful.
The remainder after stopping at order $n$ is:
$R_n(x) = \dfrac{e^c}{(n+1)!} x^{n+1}$
for some $c$ between $0$ and $x$. Since all derivatives of $e^x$ are $e^x$ itself, the $(n+1)$-th derivative evaluated at $c$ is simply $e^c$. For bounded $x$, $e^c$ is bounded, and the $(n+1)!$ in the denominator grows without bound as $n$ increases. The error therefore shrinks to zero for any fixed $x$, no matter how large.
$f(x) = \ln(1+x)$ around $a = 0$
Here $f(x) = \ln(1+x)$.
Computing the derivatives:
$f(x) = \ln(1+x)$, so $f(0) = 0$
$f'(x) = \dfrac{1}{1+x}$, so $f'(0) = 1$
$f”(x) = -\dfrac{1}{(1+x)^2}$, so $f”(0) = -1$
$f”'(x) = \dfrac{2}{(1+x)^3}$, so $f”'(0) = 2$
$f^{(4)}(x) = -\dfrac{6}{(1+x)^4}$, so $f^{(4)}(0) = -6$
Substituting into the Taylor formula:
$\ln(1+x) = x – \dfrac{x^2}{2} + \dfrac{x^3}{3} – \dfrac{x^4}{4} + \cdots$
To first order: $\ln(1+x) \approx x$
To second order: $\ln(1+x) \approx x – \dfrac{x^2}{2}$
To third order: $\ln(1+x) \approx x – \dfrac{x^2}{2} + \dfrac{x^3}{3}$
To fourth order: $\ln(1+x) \approx x – \dfrac{x^2}{2} + \dfrac{x^3}{3} – \dfrac{x^4}{4}$
The alternating signs — positive, negative, positive, negative — reflect the alternating signs of the derivatives at $0$.
Each successive term corrects the overshoot of the previous one, pulling the approximation alternately above and below the true value before converging.