Best Time Saving Mathematics Formulas & Theorems
Formulas are the most important part of mathematics and as we all know one is the backbone of the latter. Considering there are thousands of mathematical formulas to help people develop analytical approach and solve problems easily — there are some that go beyond. Some formulas aren’t just timesaving but those also do wonders. In this article I have collected some of the finest timesaving formulas in mathematics.
The calendar formula
This formula is extremely helpful in finding weekday for a specific date in history. More about this can be found in the article “Calendar Formula: Finding the Weekdays“.
Infinite Summation into Integration
We all know that integration can be interpreted as summation and vice versa. Infinite summations can be easily converted into finite integrals which on solution yields result to the infinite summation. The methods of converting infinite summations is pretty easy and thus the whole process saves a lot of time.
Theory is that if $f$ is a positive increasing function then $f(n) \le \int_n^{n+1} f(x) dx \le f(n+1) $.
Similarly, if $f$ is a positive decreasing function then $f(n) \ge \int_n^{n+1} f(x) dx \ge f(n+1) $.
For increasing $f$, $\sum_{i=0}^{n1} f(i) \le \int_0^{n} f(x) dx \le \sum_{i=0}^{n1} f(i+1) $
or, $\sum_{i=0}^{n} f(i)f(n) \le \int_0^{n} f(x) dx \le \sum_{i=0}^{n} f(i)f(0) $ or $f(0) \le \sum_{i=0}^{n} f(i)\int_0^{n} f(x) dx \le f(n) $.
For a decreasing function, the inequalities are reversed.
Khan Academy has a very good video about converting infinite summations into integral problems.
Integral Equation Magical Formula
This helps in converting a multiple integral equation to a simple linear integral equation.
Consider an integral of order n is given by $ \displaystyle{\int_{\Delta}^{\Box}} f(x) dx^n$
We can prove that $ \displaystyle{\int_{a}^{t}} f(x) dx^n = \displaystyle{\int_{a}^{t}} \dfrac{(tx)^{n1}}{(n1)!} f(x) dx$
This formula can help you solve tedious integral problems in a jiffy. Just compare the parameters and you are good to go!
For example, let’s try to solve $ \int_0^1 x^2 dx^2$
Solution:
$$ \int_0^1 x^2 dx^2$$
$$ = \int_0^1 \dfrac{(1x)^{21}}{(21)!} x^2 dx$$
(Compare, $t=1$)
$ =\int_0^1 (1x) x^2 dx$
$ =\int_0^1 (1x) x^2 dx$
$ =\int_0^1 (x^2x^3) dx =1/12$
Done!
$\sin^m x \cos^n x$ integration formula
To integrate $\sin^m x \cos^n x$ with respect to $x$, we use simple substitution considering if $n$ is odd or $m$ is odd or both are even.

 If n is odd, then ignore what m is and use substitution as $u = \sin x$ or $ du = \cos x \ dx$ and convert the remaining factors of cosine using $\cos^2 x = 1 – \sin^2 x$. This will work even if $m = 0.$
This can be applied to $\int \sin^5 x \cos^3 x \ dx$ like integrals.
 If n is odd, then ignore what m is and use substitution as $u = \sin x$ or $ du = \cos x \ dx$ and convert the remaining factors of cosine using $\cos^2 x = 1 – \sin^2 x$. This will work even if $m = 0.$

 If m is odd and n is even — use substitution as $u = \cos x$ or $du = − \sin x \ dx$ and convert the remaining factors of sine using $\sin^2 x = 1 – \cos^2 x$. This will work if $n = 0.$This can be applied to $\int \sin^3 x \cos^8 x \ dx$ like integrals.
 If both powers are even we reduce the powers using the half angle formulas: $\sin^2 x = \dfrac{1}{2} ( 1 – \cos 2x ) $; and
$\cos^2 x = \dfrac{1}{2} (1 + \cos 2x) $Alternatively, you can switch to powers of sine and cosine using $\cos^2 x+ \sin^2 x = 1$ and use the reduction formulas. Example: $\int \sin^4 x \cos^2 x \ dx$
L’ Hospital’s Rule
According to it if we have an indeterminate form for a limit problem, like 0/0 or $\dfrac{\infty}{\infty}$, then all we need to do is differentiate the numerator and differentiate the denominator and then take the limit. Read more about this on Mathworld.
Liouville & Dirichlet’s Theorem
These theorems are used to easily convert surface or volume integrals into $\Gamma$functions.

 $$ \int \int \int_{V} x^{l1} y^{m1} z^{n1} dx dy ,dz = \frac { \Gamma {(l)} \Gamma {(m)} \Gamma {(n)} }{ \Gamma{(l+m+n+1)} } $$where V is the region given by $ x \ge 0 y \ge 0 z \ge 0 x+y+z \le 1 $ .
 If $ x, y, z$ are all positive such that $ h_1 < (x+y+z) < h_2 $ then $$ \int \int \int_{V} x^{l1} y^{m1} z^{n1} F (x,y,z) dx dy ,dz = \frac { \Gamma {(l)} \Gamma {(m)} \Gamma {(n)} }{ \Gamma{(l+m+n)} } \int_{h_1}^{h_2} F(h) h^{l+m+n1} dh $$
Detailed information about Liouville & Dirichlet’s Theorem can be found here.
Remainder Theorems
There are several formulas in Number Theory on finding divisibility and remainders. As an example, general remainder theorem is a general approach of Euclidean division of polynomials while on the other hand Euler’s Remainder Theorem works as an excellent utility to find divisibility of large numbers.
More to be added soon!
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1 comment add your comment
Nice article sir. I share this article to my brother who loves math all the time.