The binomial theorem gives the expansion of expressions of the form (a + b)n where n is a positive integer. The general form is:
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
Here,
When n is not an integer, or when we want to expand expressions such as (1 + x)n for small x, we can approximate the expansion using a binomial series. The binomial expansion formula for small x and any real number n is:
(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots(1+x)
For very small x, we can truncate this series after a few terms to obtain an approximation.
How to Use Binomial Expansion for Approximation?
When x is small, higher powers of x become increasingly insignificant, allowing us to approximate the function by using only the first few terms of the binomial expansion. This is particularly useful in applied mathematics, physics, and engineering, where complex expressions need to be simplified for quick calculations.
If x is small in the expression (1 + x)n, we can approximate:
(1 + x)^n \approx 1 + nx
This approximation holds for ∣x∣≪1.
For little larger values of x i.e., |x| < 1, we can use first few terms of expansion such as
(1 + x)^n \approx 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3
When is Binomial Expansion Useful for Approximation?
Binomial expansion is especially useful when the value of the variable is very small, meaning when ∣x∣ is much less than 1. In such cases, the higher powers of x become extremely small and contribute very little to the final result. So, by only considering the first few terms of the expansion, we can quickly approximate the value of expressions like (1 + x)n.
For example:
- Approximating powers of numbers close to 1, such as (1.01)n or (0.99)n.
- Finding quick estimates for square roots or higher roots, like
\sqrt{1.02} or\sqrt[3]{0.98} . - If you want to calculate the future value of an investment using P(1 + r)n for small r:
Solved Questions on Approximation using Binomial Expansion
Question 1: Approximate (1 + 0.02)5 using binomial expansion.
Solution:
Given (1 + 0.02)5, we will expand using the first two terms of the binomial expansion for small x.
The binomial expansion is:
(1 + x)^n \approx 1 + nx
Substitute n = 5 and x = 0.02:
⇒(1 + 0.02)^5 \approx 1 + 5(0.02) = 1 + 0.10 = 1.10 Answer:
(1 + 0.02)^5 \approx 1.10
Example 2: Approximate (1 - 0.01)6 using binomial expansion.
Solution:
For (1 - 0.01)6, we use the binomial expansion formula with x = -0.01 and n = 6.
⇒(1 - 0.01)^6 \approx 1 + 6(-0.01) = 1 - 0.06 = 0.94 Answer:
(1 - 0.01)^6 \approx 0.94
Example 3: Use binomial expansion to approximate
Solution:
We know
\sqrt{1.02} = (1.02)^{1/2} = (1 + 0.02)^{1/2} .
Using binomial expansion, we approximate:(1 + x)^n \approx 1 + \frac{n}{1}x
Substitutingn = \frac{1}{2} and x = 0.02:
⇒\sqrt{1.02} \approx 1 + \frac{1}{2}(0.02) = 1 + 0.01 = 1.01 Answer:
\sqrt{1.02} \approx 1.01
Example 4: Approximate (2.01)4 using binomial expansion.
Solution:
We rewrite 2.01 as (2 + 0.01). Thus, (2.01)4 = (2 + 0.01)4.
Using binomial expansion for (a + x)n, we consider the first two terms:
⇒(2 + 0.01)^4 \approx 2^4 + 4(2^3)(0.01) = 16 + 32(0.01) = 16 + 0.32 = 16.32 Answer:
(2.01)^4 \approx 16.32
Example 5: Approximate (0.98)5 using binomial expansion.
Solution:
We rewrite 0.98 as (1 - 0.02). Thus, (0.98)^5 = (1 - 0.02)^5.
Using the first two terms of the binomial expansion:
⇒(1 - 0.02)^5 \approx 1 + 5(-0.02) = 1 - 0.10 = 0.90 Answer:
(0.98)^5 \approx 0.90
Practice Questions
You can download free worksheet on Approximation using Binomial Expansion for practicing various different questions with their answers from below:
Download Worksheet on Approximation using Binomial Expansion |
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Conclusion
Approximation using binomial expansion is a practical method for simplifying expressions, especially when dealing with small values of x. By truncating the binomial series after a few terms, we can obtain quick approximations for various functions. This technique is widely used in fields such as physics, engineering, and economics, where small perturbations or deviations from a base value are common.
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