DSA for Friday, 25-July-2025
Series for sin(x) and cos(x)
We start with presenting the series for sin(x) and cos(x)
Maclaurin Series for sin(x):
\($$
\sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n \cdot x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots
$$\)
Maclaurin Series for cos(x):
\($$
\cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n \cdot x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots
$$
\)
We will try to learn how to do this in tiny steps.
This is how:
Natural Numbers:
Make a Program to print and sum these series.1,2,3,4,5,6,…
1,3,5,7,…
0,2,4,6,8,…Alternating Signs:
Make a Program to print and sum this series.
1,-1,1,-1,1,-1,,,Alternating Sign Powers increase by 2:
Make a Program to print and sum these series.1,-2,3,-4,5,-6,…
2 - 8 + 32 - 128 + 512 - 2048
1 - 4 + 16 - 64 + 256 - 1024\($$ 2 - 8 + 32 - 128 + 512 - 2048 $$ \)\($$ x - x^3 + x^5 - x^7 + \cdots $$ \)Factorials:\(\[ n! = 1 \times 2 \times 3 \times \cdots \times n \] \)
\(\[
\begin{aligned}
0! &= 1 \quad \text{(by definition)} \\
1! &= 1 \\
2! &= 1 \times 2 = 2 \\
3! &= 1 \times 2 \times 3 = 6 \\
4! &= 1 \times 2 \times 3 \times 4 = 24 \\
5! &= 1 \times 2 \times 3 \times 4 \times 5 = 120 \\
6! &= 1 \times 2 \times 3 \times 4 \times 5 \times 6 = 720 \\
7! &= 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 = 5040
\end{aligned}
\]
\)
Alternating Signs Factorials with a difference of 2:
0!−2!+4!−6!=1−2+24−720
1!−3!+5!−7!=1−6+120−5040
Please print and sum both.Get Sin x and Cos x now.
\($$ \sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n \cdot x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots $$ \)
\($$
\cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n \cdot x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots
$$
\)
Please leave your solutions and suggestions in the comments
Solution of question 2
#include <stdio.h>
int main() {
int n, i, term, sum = 0;
// Ask user for number of terms
printf("Enter the number of terms: ");
scanf("%d", &n);
printf("Series: ");
for(i = 1; i <= n; i++) {
// Term is 1 for odd positions, -1 for even positions
if(i % 2 == 1)
term = 1;
else
term = -1;
printf("%d", term);
if(i != n) {
printf(", ");
}
sum += term;
}
printf("\nSum of the series: %d\n", sum);
return 0;
}
#include <stdio.h>
int main() {
int n, i;
int sum_natural = 0, sum_odd = 0, sum_even = 0;
printf("Enter the number of terms: ");
scanf("%d", &n);
// Natural numbers
printf("\nNatural Numbers Series: ");
for(i = 1; i <= n; i++) {
printf("%d ", i);
sum_natural += i;
}
printf("\nSum of Natural Numbers: %d\n", sum_natural);
// Odd numbers
printf("\nOdd Numbers Series: ");
for(i = 0; i < n; i++) {
int odd = 2 * i + 1;
printf("%d ", odd);
sum_odd += odd;
}
printf("\nSum of Odd Numbers: %d\n", sum_odd);
// Even numbers
printf("\nEven Numbers Series: ");
for(i = 0; i < n; i++) {
int even = 2 * i;
printf("%d ", even);
sum_even += even;
}
printf("\nSum of Even Numbers: %d\n", sum_even);
return 0;
}