Math Problem Statement

The image shows a mathematics problem related to the expansion of (1+5x)^3/5 in ascending powers of xx up to and including the term in x2x2. The problem is divided into three parts:

Solution

Thank you for the clarification. Let's address the problem, focusing on the expansion of (1+5x)3/5(1 + 5x)^{3/5} in ascending powers of xx up to and including the term in x2x^2. Here's how you would approach each part of the problem:

Part (a): Expansion of (1+5x)3/5(1 + 5x)^{3/5}

To expand (1+5x)3/5(1 + 5x)^{3/5} using the binomial theorem, we use the general formula for the binomial expansion:

(1+ax)n=1+n(ax)+n(n1)2!(ax)2+n(n1)(n2)3!(ax)3+(1 + ax)^n = 1 + n(ax) + \frac{n(n-1)}{2!}(ax)^2 + \frac{n(n-1)(n-2)}{3!}(ax)^3 + \cdots

Here, n=35n = \frac{3}{5} and a=5a = 5. We expand up to the term in x2x^2:

(1+5x)3/5=1+35(5x)+35(351)2!(5x)2+(1 + 5x)^{3/5} = 1 + \frac{3}{5}(5x) + \frac{\frac{3}{5} \left(\frac{3}{5} - 1\right)}{2!} (5x)^2 + \cdots

Simplifying the expression:

=1+3x+35(25)225x2= 1 + 3x + \frac{\frac{3}{5} \left(\frac{-2}{5}\right)}{2} \cdot 25x^2

=1+3x65×5x2= 1 + 3x - \frac{6}{5} \times 5x^2

=1+3x65×5x2= 1 + 3x - \frac{6}{5} \times 5x^2

=1+3x65×25x2=1+3x6x2= 1 + 3x - \frac{6}{5} \times 25x^2 = 1 + 3x - 6x^2

So the expansion up to x2x^2 is:

(1+5x)3/5=1+3x6x2(1 + 5x)^{3/5} = 1 + 3x - 6x^2

Part (b): Approximate Value

Now, to use the expansion to find an approximate value for a function at a specific value of xx, you would substitute the value of xx into the expansion. For example, if you needed to find the approximate value of (1+5x)3/5(1 + 5x)^{3/5} at x=0.01x = 0.01, you would substitute x=0.01x = 0.01 into the expression:

1+3(0.01)6(0.01)2=1+0.030.0006=1.02941 + 3(0.01) - 6(0.01)^2 = 1 + 0.03 - 0.0006 = 1.0294

Part (c): Percentage Error

If you have a specific actual value of the function, say at x=0.01x = 0.01, you would compare it to the approximation found in part (b) to calculate the percentage error:

Percentage Error=Actual ValueApproximate ValueActual Value×100%\text{Percentage Error} = \left| \frac{\text{Actual Value} - \text{Approximate Value}}{\text{Actual Value}} \right| \times 100\%

For example, if the actual value of (1+5×0.01)3/5(1 + 5 \times 0.01)^{3/5} is known, you can plug it in to find the percentage error.

Would you like further details or have any specific values you'd like to evaluate in part (b) or (c)?


Here are some related questions to deepen understanding:

  1. How is the binomial expansion formula derived for fractional exponents?
  2. What are the limitations of using a truncated binomial expansion for approximation?
  3. Why does the expansion yield better approximations when xx is small?
  4. How does the sign of the coefficient in the expansion affect the approximation?
  5. What happens if we expand (1+5x)3/5(1 + 5x)^{3/5} beyond the x2x^2 term?
  6. Can the binomial expansion be used for negative values of xx in this problem?
  7. How does the expansion change if xx is a complex number?
  8. What role does the binomial expansion play in solving differential equations?

Tip: Always consider the error introduced by truncating a series expansion. Higher-order terms can have significant effects, especially for larger values of xx.

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Math Problem Analysis

Mathematical Concepts

Binomial Theorem
Fractional Exponents
Series Expansion

Formulas

Binomial theorem expansion formula

Theorems

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Suitable Grade Level

Advanced High School