Solving Complex Numerical Expressions and Understanding Negative Exponents

Whether you are a student, a mathematician, or an enthusiast, understanding complex numerical expressions and negative exponents can be quite challenging. In this article, we will break down these concepts and provide step-by-step solutions to specific numerical expressions. We will also explore the application of these concepts in more advanced mathematical problems such as integrals and derivatives.

Understanding Negative Exponents

A negative exponent means that the base of the exponent is in the denominator of a fraction, and the exponent is positive. For example, ( b^{-n} ) is equivalent to ( frac{1}{b^n} ), where ( n ) is a positive integer.

Solving Numerical Expressions

Let us start with the following numerical expressions as examples:

Part (b)

Given the expression:
[ 200^{-1} - 100 cdot 100^{-1} - 200 ]
We can rewrite each term with positive exponents by using the definition of negative exponents:

[ frac{1}{200} - 100 cdot frac{1}{100} - 200 ]

Further simplification yields:

[ frac{1}{200} - frac{2}{200} - 200 -frac{1}{200} - 200 -frac{1 200 cdot 200}{200} -frac{1 40000}{200} -frac{40001}{200} ]

The final simplified form of this expression is:

[ boxed{-299frac{197}{200}} ]

Part (c)

Let us consider the numerical expression:
[ left(frac{3}{5}right)^2 - left(frac{5}{2}right)^{-1} - left(frac{4}{3}right)^{-2}left(frac{1}{4}right)^{-2} ]
We can rewrite each term using the properties of negative exponents:

[ frac{9}{25} - frac{2}{5} - frac{9}{16} cdot 16 ]

Now, we convert each term to have a common denominator of 400:

[ frac{144}{400} - frac{160}{400} - frac{225}{400} -frac{241}{400} ]

The final simplified form of this expression is:

[ boxed{-frac{241}{400}} ]

Advanced Mathematical Concepts

Understanding the application of these concepts in calculus can be crucial for solving more advanced problems. Consider the following problem involving an integral and an exponential function:

We are given the expression:
[ e^{-x} f(x) 2 int_{0}^{x} sqrt{t^4 1} , dt ]
This implies:
[ f(x) e^x left[ 2 int_{0}^{x} sqrt{t^4 1} , dt right] ]
Taking ( f^{-1} ) on both sides, we obtain:
[ x f^{-1} left[ 2 e^x e^x int_{0}^{x} sqrt{t^4 1} , dt right] ]
Differentiating both sides, we use the chain rule and the integral theorem stated as follows:

[ frac{d}{dx} left[ int_{a(x)}^{b(x)} f(x) , dx right] f(b(x)) cdot b'(x) - f(a(x)) cdot a'(x) ]

Letting ( x 0 ) into the equation, we get:

[ 1 left[ f^{-1} left( 2 right) right]^{-1} cdot 201 ]

This simplifies to:

[ left[ f^{-1} left( 2 right) right]^{-1} frac{1}{3} ]

Thus, we have:

[ left[ f^{-1} left( 2 right) right] 3 ]

The solution to this problem showcases an application of both negative exponents and integral calculus.

Conclusion

Understanding negative exponents and complex numerical expressions is essential for tackling advanced mathematical problems. By breaking down each term and applying the appropriate rules, we can solve even the most challenging expressions. Mastering these concepts opens up a world of possibilities in mathematics and its applications.