Expand Expressions In Ascending Powers Of X: A Detailed Guide

Hey guys! Today, we're diving deep into the fascinating world of expanding expressions in ascending powers of x. This is a crucial topic in mathematics, especially in calculus and series expansions. We'll tackle two challenging problems step-by-step, making sure you grasp every concept. So, buckle up and let's get started!

Expanding $6 \sqrt{1-2x} - \frac{1}{1+4x}$ in Ascending Powers of $x$ Up to $x^3$

Okay, let's break down this first part. We need to expand the expression $6 \sqrt{1-2x} - \frac{1}{1+4x}$ into a series of ascending powers of x, but only up to the term x³. This means we're looking for an expression of the form a + bx + cx² + dx³, where a, b, c, and d are constants. The process involves using the binomial theorem or, more specifically, the binomial series expansion, which is a powerful tool for handling expressions like these. It's like having a mathematical Swiss Army knife – super versatile! The binomial series expansion allows us to approximate functions using polynomials, which is incredibly useful in various fields, from physics to engineering.

Step 1: Expand $6 \sqrt{1-2x}$

First, let's focus on expanding $6 \sqrt{1-2x}$. We can rewrite this as $6(1-2x)^{\frac{1}{2}}$. Now, we can apply the binomial series expansion, which states:

$(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + ...$

In our case, we have $(1-2x)^{\frac{1}{2}}$, so we can think of x in the formula as -2x and n as \frac{1}{2}. Plugging these values into the binomial series formula, we get:

$(1-2x)^{\frac{1}{2}} = 1 + \frac{1}{2}(-2x) + \frac{\frac{1}{2}(\frac{1}{2}-1)}{2!}(-2x)^2 + \frac{\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)}{3!}(-2x)^3 + ...$

Let's simplify this step by step. First, we have:

$1 + \frac{1}{2}(-2x) = 1 - x$

Next, let's calculate the x² term:

$\frac{\frac{1}{2}(\frac{1}{2}-1)}{2!}(-2x)^2 = \frac{\frac{1}{2}(-\frac{1}{2})}{2}(4x^2) = -\frac{1}{2}x^2$

Now, let's find the x³ term:

$\frac{\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)}{3!}(-2x)^3 = \frac{\frac{1}{2}(-\frac{1}{2})(-\frac{3}{2})}{6}(-8x^3) = -x^3$

So, the expansion of $(1-2x)^{\frac{1}{2}}$ up to the x³ term is:

$1 - x - \frac{1}{2}x^2 - x^3$

Remember, we need to multiply this by 6, so:

$6(1-2x)^{\frac{1}{2}} = 6(1 - x - \frac{1}{2}x^2 - x^3) = 6 - 6x - 3x^2 - 6x^3$

Step 2: Expand $\frac{1}{1+4x}$

Now, let's tackle the second part: expanding $\frac{1}{1+4x}$. We can rewrite this as $(1+4x)^{-1}$. Again, we'll use the binomial series expansion. This time, n is -1, and what we previously called x in the formula, we now call 4x. So, plugging these into our trusty binomial series expansion formula:

$(1+4x)^{-1} = 1 + (-1)(4x) + \frac{(-1)(-1-1)}{2!}(4x)^2 + \frac{(-1)(-1-1)(-1-2)}{3!}(4x)^3 + ...$

Let's simplify each term:

The constant term is simply 1.

The x term is:

$(-1)(4x) = -4x$

The x² term is:

$\frac{(-1)(-2)}{2}(16x^2) = 16x^2$

The x³ term is:

$\frac{(-1)(-2)(-3)}{6}(64x^3) = -64x^3$

So, the expansion of $(1+4x)^{-1}$ up to the x³ term is:

$1 - 4x + 16x^2 - 64x^3$

Step 3: Combine the Expansions

Now, we need to subtract the second expansion from the first:

$(6 - 6x - 3x^2 - 6x^3) - (1 - 4x + 16x^2 - 64x^3)$

Combine like terms:

$6 - 1 = 5$

$-6x - (-4x) = -2x$

$-3x^2 - 16x^2 = -19x^2$

$-6x^3 - (-64x^3) = 58x^3$

Therefore, the expansion of $6 \sqrt{1-2x} - \frac{1}{1+4x}$ up to the x³ term is:

$5 - 2x - 19x^2 + 58x^3$

Step 4: Determine the Range of Valid Values

Now, for the crucial part: the range of values for which this expansion is valid. Remember, the binomial series expansion is valid when |x| < 1. In our case, we have two binomial expansions: $(1-2x)^{\frac{1}{2}}$ and $(1+4x)^{-1}$.

For $(1-2x)^{\frac{1}{2}}$, we need |-2x| < 1, which simplifies to |x| < \frac{1}{2}.

For $(1+4x)^{-1}$, we need |4x| < 1, which simplifies to |x| < \frac{1}{4}.

To ensure both expansions are valid, we need to take the stricter condition, which is |x| < \frac{1}{4}. This means the range of valid values is $-\frac{1}{4} < x < \frac{1}{4}$.

Expanding $\left(1+\frac{x}{8}\right)^{-\frac{1}{2}}$ in Ascending Powers of $x$

Alright, let's move on to the second part of the problem! We need to expand the expression $\left(1+\frac{x}{8}\right)^{-\frac{1}{2}}$ in ascending powers of x. This is another perfect opportunity to use the binomial series expansion. Guys, this formula is truly your best friend in these situations. It saves so much time and effort compared to other methods. In this case, our "x" in the formula is actually \frac{x}{8}, and n is -\frac{1}{2}. Let's plug these values into the binomial series expansion formula:

$\left(1+\frac{x}{8}\right)^{-\frac{1}{2}} = 1 + (-\frac{1}{2})(\frac{x}{8}) + \frac{(-\frac{1}{2})(-\frac{1}{2}-1)}{2!}(\frac{x}{8})^2 + \frac{(-\frac{1}{2})(-\frac{1}{2}-1)(-\frac{1}{2}-2)}{3!}(\frac{x}{8})^3 + ...$

Step 1: Simplify the Terms

Let's break down each term and simplify. The constant term is, again, just 1.

The x term is:

$(-\frac{1}{2})(\frac{x}{8}) = -\frac{x}{16}$

The x² term is:

$\frac{(-\frac{1}{2})(-\frac{3}{2})}{2}(\frac{x^2}{64}) = \frac{\frac{3}{4}}{2}(\frac{x^2}{64}) = \frac{3x^2}{512}$

The x³ term is:

$\frac{(-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})}{6}(\frac{x^3}{512}) = \frac{-\frac{15}{8}}{6}(\frac{x^3}{512}) = -\frac{5x^3}{16384}$

Step 2: Combine the Terms

Now, let's put it all together. The expansion of $\left(1+\frac{x}{8}\right)^{-\frac{1}{2}}$ in ascending powers of x up to the x³ term is:

$1 - \frac{x}{16} + \frac{3x^2}{512} - \frac{5x^3}{16384} + ...$

Step 3: Determine the Range of Valid Values

Finally, let's find the range of values for which this expansion is valid. For the binomial series expansion to be valid, we need |\frac{x}{8}| < 1. This simplifies to |x| < 8. Therefore, the range of valid values is -8 < x < 8.

Conclusion

And there you have it, guys! We've successfully expanded two challenging expressions in ascending powers of x and determined their ranges of validity. Remember, the binomial series expansion is a powerful tool, but it's crucial to understand its conditions for validity. Practice makes perfect, so keep tackling these types of problems, and you'll become a pro in no time! This whole exercise really highlights the beauty and power of series expansions in approximating complex functions. It's like having a mathematical lens that lets you zoom in and understand the behavior of functions in a specific range. Keep exploring, keep learning, and most importantly, keep having fun with math!