(a) Show that the binomial expansion of $(4+5 x)^{\frac{1}{2}}$ in ascending powers of $x$, up to and including the term in $x^2$ is $2+\frac{5}{4} x+k x^2$, giving the value of the constant $k$ as a simplified fraction.
b) (i) Use the expansion from part (a), with $x=\frac{1}{10}$, to find an approximate value for $\sqrt{2}$ Give your answer in the form $\frac{p}{q}$ where $p$ and $q$ are integers.
(ii) Explain why substituting $x=\frac{1}{10}$ into this binomial expansion leads to a valid approximation.
Answer (2)
Find the binomial expansion of ( 4 + 5 x ) 2 1 up to the x 2 term and determine that k = − 64 25 .
Substitute x = 10 1 into the expansion to get an approximate value of 256 543 for 2 3 .
Calculate 2 = 256 543 3 = 543 768 .
Verify that the approximation is valid since ∣ 4 5 x ∣ < 1 when x = 10 1 , and the final answer is 543 768 .
Explanation
Problem Analysis We are given the binomial expansion of ( 4 + 5 x ) 2 1 and asked to find the constant k in the x 2 term. Then, we will use this expansion to approximate 2 by substituting x = 10 1 into the expansion. Finally, we need to explain why this substitution leads to a valid approximation.
Applying Binomial Theorem First, we rewrite the given expression using the binomial theorem: ( 4 + 5 x ) 2 1 = 4 2 1 ( 1 + 4 5 x ) 2 1 = 2 ( 1 + 4 5 x ) 2 1 Now, we apply the binomial expansion formula: ( 1 + y ) n = 1 + n y + 2 ! n ( n − 1 ) y 2 + ... , where y = 4 5 x and n = 2 1 .
Finding k Expanding the expression up to the x 2 term, we have: 2 [ 1 + 2 1 ( 4 5 x ) + 2 ! 2 1 ( 2 1 − 1 ) ( 4 5 x ) 2 + ... ] Simplifying the expression, we get: 2 [ 1 + 8 5 x + 2 2 1 ( − 2 1 ) ( 16 25 x 2 ) + ... ] 2 [ 1 + 8 5 x − 8 1 ⋅ 16 25 x 2 + ... ] 2 + 4 5 x − 64 25 x 2 + ... Thus, k = − 64 25 .
Approximating sqrt(2) Now, we use the expansion with x = 10 1 to find an approximate value for 2 . Substituting x = 10 1 into the expansion, we have: 2 + 4 5 ( 10 1 ) − 64 25 ( 10 1 ) 2 = 2 + 40 5 − 6400 25 = 2 + 8 1 − 256 1 = 256 512 + 32 − 1 = 256 543 So, the approximate value is 256 543 .
Finding sqrt(2) Since ( 4 + 5 x ) 2 1 = ( 4 + 5 ( 10 1 ) ) 2 1 = ( 4 + 2 1 ) 2 1 = ( 2 9 ) 2 1 = 2 3 , we have 2 = 256 543 3 = 543 3 ⋅ 256 = 543 768 .
Validity of Approximation The binomial expansion is valid when ∣ 4 5 x ∣ < 1 . Substituting x = 10 1 , we get ∣ 4 5 ⋅ 10 1 ∣ = ∣ 8 1 ∣ < 1 . Since ∣ 8 1 ∣ < 1 , the approximation is valid because x = 10 1 lies within the radius of convergence of the binomial series.
Final Answer Therefore, the value of k is − 64 25 , the approximate value for 2 is 543 768 , and the approximation is valid because ∣ 4 5 x ∣ < 1 .
Examples
Binomial expansions are used in various fields such as physics and engineering to approximate complex expressions. For example, in physics, when dealing with velocities much smaller than the speed of light, the relativistic kinetic energy formula can be approximated using a binomial expansion. This simplifies calculations and provides accurate results in many practical scenarios. Similarly, in engineering, binomial expansions can be used to approximate the behavior of systems with small perturbations, making analysis more tractable.
The binomial expansion of ( 4 + 5 x ) 2 1 up to x 2 is 2 + 4 5 x − 32 25 x 2 , where k = − 32 25 . Substituting x = 10 1 gives an approximate value for 2 as 256 543 , which is valid since the series converges. Overall, this method allows for a practical approximation of square roots using binomial expansion techniques.
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