Best Time Saving Mathematics Formulas & Theorems
Here’s the deal: mathematics has thousands of formulas, but only a handful will genuinely change how fast you solve problems. After years of teaching and competitive math, I’ve compiled the formulas that consistently save the most time. These aren’t just shortcuts—they’re problem-solving weapons.
Let me walk you through each one with practical examples.
The Calendar Formula
Ever needed to figure out what day of the week a historical date fell on? This formula handles it instantly—no calendar lookup required.
The calendar formula lets you calculate the weekday for any date in history using modular arithmetic. It accounts for leap years, century adjustments, and month codes to output a number from 0 to 6 (representing Sunday through Saturday).
I’ve written a detailed breakdown with worked examples in my article on Calendar Formula: Finding the Week-days. Trust me—once you master this, you’ll never need to Google “what day was July 4, 1776” again.
Converting Infinite Summations to Integrals
This technique is a game-changer. Instead of wrestling with infinite series term by term, you convert them into finite integrals that are often much easier to evaluate.
The core principle relies on bounding partial sums with integrals. Here’s how it works:
For a positive increasing function \( f \):
$$f(n) \le \int_n^{n+1} f(x) \, dx \le f(n+1)$$
For a positive decreasing function \( f \):
$$f(n) \ge \int_n^{n+1} f(x) \, dx \ge f(n+1)$$
Summing these inequalities across the range gives us powerful bounds. For an increasing \( f \):
$$\sum_{i=0}^{n-1} f(i) \le \int_0^{n} f(x) \, dx \le \sum_{i=0}^{n-1} f(i+1)$$
Which rearranges to:
$$f(0) \le \sum_{i=0}^{n} f(i) – \int_0^{n} f(x) \, dx \le f(n)$$
For decreasing functions, flip all the inequality signs.
Khan Academy explains this beautifully in the following video:
The Cauchy Formula for Repeated Integration
This is my personal favorite. It converts a tedious \( n \)-fold iterated integral into a single integral. The time savings are massive.
The formula: If you have an integral of order \( n \):
$$\int_{a}^{t} \int_{a}^{t_1} \cdots \int_{a}^{t_{n-1}} f(x) \, dx \, dt_{n-1} \cdots dt_1 = \int_{a}^{t} \frac{(t-x)^{n-1}}{(n-1)!} f(x) \, dx$$
In compact notation: \( \displaystyle\int_{a}^{t} f(x) \, dx^n = \int_{a}^{t} \frac{(t-x)^{n-1}}{(n-1)!} f(x) \, dx \)
Let me show you how powerful this is with a worked example.
Example: Evaluate \( \displaystyle\int_0^1 x^2 \, dx^2 \)
Solution:
Here we have \( a = 0 \), \( t = 1 \), \( n = 2 \), and \( f(x) = x^2 \). Applying the formula:
$$\int_0^1 x^2 \, dx^2 = \int_0^1 \frac{(1-x)^{2-1}}{(2-1)!} x^2 \, dx$$
$$= \int_0^1 (1-x) x^2 \, dx$$
$$= \int_0^1 (x^2 – x^3) \, dx$$
$$= \left[ \frac{x^3}{3} – \frac{x^4}{4} \right]_0^1 = \frac{1}{3} – \frac{1}{4} = \frac{1}{12}$$
What would have required two nested integrations took just one. The kicker? This works for any order \( n \).
For more on integral equations, see my series starting with Integral Equations: Definitions and Types.
Integration of \( \sin^m x \cos^n x \)
Integrating products of sine and cosine powers comes up constantly. Here’s the decision framework I use every time:
Case 1: \( n \) is odd (cosine has odd power)
Substitute \( u = \sin x \), so \( du = \cos x \, dx \). Convert remaining cosine factors using \( \cos^2 x = 1 – \sin^2 x \).
This works even when \( m = 0 \). Use it for integrals like \( \int \sin^5 x \cos^3 x \, dx \).
Case 2: \( m \) is odd, \( n \) is even (sine has odd power)
Substitute \( u = \cos x \), so \( du = -\sin x \, dx \). Convert remaining sine factors using \( \sin^2 x = 1 – \cos^2 x \).
This works even when \( n = 0 \). Use it for integrals like \( \int \sin^3 x \cos^8 x \, dx \).
Case 3: Both powers are even
Use the half-angle formulas to reduce powers:
$$\sin^2 x = \frac{1}{2}(1 – \cos 2x)$$
$$\cos^2 x = \frac{1}{2}(1 + \cos 2x)$$
Alternatively, use the identity \( \cos^2 x + \sin^2 x = 1 \) to manipulate powers, then apply reduction formulas. Example: \( \int \sin^4 x \cos^2 x \, dx \).
Quick Reference Table:
| Condition | Substitution | Identity Used |
|---|---|---|
| \( n \) odd | \( u = \sin x \) | \( \cos^2 x = 1 – \sin^2 x \) |
| \( m \) odd, \( n \) even | \( u = \cos x \) | \( \sin^2 x = 1 – \cos^2 x \) |
| Both even | None (reduce powers) | Half-angle formulas |
L’Hôpital’s Rule
When you hit an indeterminate form like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), L’Hôpital’s Rule is your escape hatch.
The rule: If \( \lim_{x \to a} \frac{f(x)}{g(x)} \) gives an indeterminate form, then:
$$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$$
provided the limit on the right exists (or is \( \pm\infty \)).
Pro tip: You can apply the rule repeatedly if you keep getting indeterminate forms. But here’s the catch—always verify you have an indeterminate form before applying. I’ve seen students blindly differentiate when the limit was perfectly evaluable.
For a deeper dive, check out Wolfram MathWorld’s treatment of L’Hôpital’s Rule.
Dirichlet’s Theorem and Liouville’s Extension
These theorems are lifesavers when you need to evaluate multiple integrals over specific regions. They convert messy volume or surface integrals directly into Gamma functions.
Dirichlet’s Theorem: For the region \( V \) where \( x \ge 0 \), \( y \ge 0 \), \( z \ge 0 \), and \( x + y + z \le 1 \):
$$\iiint_{V} x^{l-1} y^{m-1} z^{n-1} \, dx \, dy \, dz = \frac{\Gamma(l) \Gamma(m) \Gamma(n)}{\Gamma(l+m+n+1)}$$
Liouville’s Extension: For positive \( x, y, z \) with \( h_1 < x + y + z < h_2 \):
$$\iiint_{V} x^{l-1} y^{m-1} z^{n-1} F(x+y+z) \, dx \, dy \, dz = \frac{\Gamma(l) \Gamma(m) \Gamma(n)}{\Gamma(l+m+n)} \int_{h_1}^{h_2} F(h) h^{l+m+n-1} \, dh$$
The beauty here is that complicated triple integrals reduce to products of Gamma functions—which have known values for many arguments (remember, \( \Gamma(n) = (n-1)! \) for positive integers).
I’ve written a comprehensive guide on Dirichlet’s Theorem and Liouville’s Extension with more worked examples.
Remainder Theorems in Number Theory
Number theory offers several powerful tools for divisibility and remainder calculations. Two stand out:
The Polynomial Remainder Theorem: When you divide a polynomial \( p(x) \) by \( (x – a) \), the remainder equals \( p(a) \). Simple, but surprisingly useful for factoring and root-finding.
Euler’s Theorem: If \( \gcd(a, n) = 1 \), then:
$$a^{\phi(n)} \equiv 1 \pmod{n}$$
where \( \phi(n) \) is Euler’s totient function (the count of integers less than \( n \) that are coprime to \( n \)).
This is incredibly powerful for finding remainders of large powers. For example, finding \( 7^{100} \mod 10 \) becomes trivial: since \( \phi(10) = 4 \) and \( \gcd(7, 10) = 1 \), we have \( 7^4 \equiv 1 \pmod{10} \), so \( 7^{100} = (7^4)^{25} \equiv 1^{25} = 1 \pmod{10} \).
For more, see the Wikipedia article on Euler’s Theorem.
Frequently Asked Questions
When should I use the summation-to-integral conversion?
Use it when you need to estimate or bound an infinite series, or when the series doesn’t have a closed-form sum but the corresponding integral does. It’s especially useful for proving convergence and finding asymptotic behavior of partial sums.
What is the Cauchy formula for repeated integration?
The Cauchy formula converts n nested integrals into a single integral: ∫ₐᵗ f(x) dxⁿ = ∫ₐᵗ [(t-x)^(n-1)/(n-1)!] f(x) dx. This dramatically simplifies calculations involving repeated integration, reducing computational effort from n integration steps to just one.
How do I know which substitution to use for sin^m x cos^n x integrals?
Check the powers: if the cosine power (n) is odd, substitute u = sin x. If the sine power (m) is odd and cosine power is even, substitute u = cos x. If both powers are even, use half-angle formulas to reduce the powers first.
Can L’Hôpital’s Rule be applied multiple times?
Yes, you can apply L’Hôpital’s Rule repeatedly as long as each application still results in an indeterminate form (0/0 or ∞/∞). However, always verify the indeterminate form exists before each application—blindly differentiating can lead to incorrect answers.
What is the Gamma function and why does it appear in Dirichlet’s theorem?
The Gamma function Γ(n) extends the factorial to non-integer values, with Γ(n) = (n-1)! for positive integers. It appears in Dirichlet’s theorem because the multiple integrals over simplex regions naturally decompose into products of Beta functions, which are ratios of Gamma functions.
What’s the difference between Euler’s theorem and Fermat’s Little Theorem?
Fermat’s Little Theorem is a special case of Euler’s theorem. Fermat states that a^(p-1) ≡ 1 (mod p) when p is prime and gcd(a,p) = 1. Euler generalizes this to any modulus n: a^φ(n) ≡ 1 (mod n) when gcd(a,n) = 1, where φ(n) is Euler’s totient function.
How accurate is the calendar formula for historical dates?
The calendar formula is exact for dates in the Gregorian calendar (post-1582 for most of Europe, later for other regions). For dates before the Gregorian reform, you need to account for the Julian calendar, which has different leap year rules. The formula includes adjustments for century years and leap years.
Why do half-angle formulas help with even-powered trig integrals?
When both sine and cosine have even powers, simple u-substitution doesn’t work because you can’t separate out a single factor. Half-angle formulas convert sin²x and cos²x into expressions involving cos(2x), reducing the power and often making the integral directly computable or amenable to further reduction.
What are the limitations of L’Hôpital’s Rule?
L’Hôpital’s Rule only applies to 0/0 and ∞/∞ indeterminate forms. Other forms like 0·∞, ∞-∞, 0⁰, 1^∞, and ∞⁰ must be algebraically manipulated into 0/0 or ∞/∞ first. Also, the rule fails if the derivative of the denominator is zero at the limit point.
Can Dirichlet’s theorem be extended to more than three variables?
Yes, Dirichlet’s theorem generalizes to any number of variables. For n variables x₁, x₂, …, xₙ with powers l₁-1, l₂-1, …, lₙ-1 over the simplex where all variables are non-negative and sum to at most 1, the integral equals Γ(l₁)Γ(l₂)…Γ(lₙ)/Γ(l₁+l₂+…+lₙ+1).