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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
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• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|微积分代写Calculus代写|Approximations with fractions

Approximations with fractions are as simple as approximating the numerator and denominator separately and then dividing. You may already use this procedure in the real-number setting; for instance,
$$\frac{\sqrt{2}+3}{5-\sqrt{7}} \approx \frac{4.4142}{2.3542} \approx 1.8750 \text {. }$$
Of coursé, we could entè the entire fraction into thé calculator all at once, rather than calculate the numerator and denominator separately before dividing, in order to reduce roundoff error. This is good practice; but behind the scenes, the calculator is actually calculating the numerator and denominator separately before dividing, just to a greater precision than it shows on the screen. It is still an approximation of the type demonstrated.
Example 14 Approximate $\frac{2+\omega-\omega^2}{6+\sqrt{\omega}}$.
Solution We approximate the numerator and denominator separately. The numerator is on the real level, so we throw away the infinitesimals, giving $2+\omega-\omega^2 \approx 2$. The same is true for the denominator, and $6+\sqrt{\omega} \approx 6$. Therefore,
$$\frac{2+\omega-\omega^2}{6+\sqrt{\omega}} \approx \frac{2}{6}=\frac{1}{3} .$$
The numerator and denominator do not need to be on the same level for the procedure to work; we still approximate the numerator and denominator separately.
Example 15 Approximate $\frac{\Omega^2+3}{2 \omega+\omega^4}$.
Solution The numerator is on the $\Omega^2$ level and the denominator is on the $\omega$ level. Throwing away the lower-level terms in each numerator and denominator yields
$$\frac{\Omega^2+3}{2 \omega+\omega^4} \approx \frac{\Omega^2}{2 \omega}=\frac{1}{2} \Omega^2 \cdot \Omega=\frac{1}{2} \Omega^3 .$$

## 数学代写|微积分代写Calculus代写|Absolute value approximations

Recall that the absolute value of a positive number is that number, whercas the absolute valuc of a ncgative number changes the sign of the number. Therefore, evaluating an absolute value requires us to know whether the number is positive or negative. Because $\omega$ is a positive infinitesimal,
$$|\omega|=\omega .$$
Because $-\omega$ is negative,
$$|-\omega|=-(-\omega)=\omega .$$
The sign changes to make the result positive. But what about more complicated hyperreals?

Consider $\left|\omega-\omega^2\right|$. To evaluate the absolute value, we need to know whether the number inside is positive or negative, and a first glance shows that $\omega$ is positive whereas $-\omega^2$ is negative. So which is it? Because $\omega-\omega^2 \approx \omega$, the number must be positive (it is approximately the number, and $\left|\omega-\omega^2\right|=\omega-\omega^2$.
Example 18 Perform the arithmetic: $|38-15 \omega|$.
Solution Since $38-15 \omega \approx 38$, the number inside is positive and
$$|38-15 \omega|=38-15 \omega .$$
If the number inside is negative, its sign is changed by taking the negative of what’s inside: $|-5|=-(-5)=5$.
Example 19 Perform the arithmetic: $|38-15 \Omega|$.
Solution Since $38-15 \Omega \approx-15 \Omega$, the number is negative. Therefore,
$$|38-15 \Omega|=-(38-15 \Omega)=-38+15 \Omega \text {. }$$

# 微积分代考

## 数学代写|微积分代写Calculus代写|Approximations with fractions

$$\frac{\sqrt{2}+3}{5-\sqrt{7}} \approx \frac{4.4142}{2.3542} \approx 1.8750 .$$

$$\frac{2+\omega-\omega^2}{6+\sqrt{\omega}} \approx \frac{2}{6}=\frac{1}{3} .$$

$$\frac{\Omega^2+3}{2 \omega+\omega^4} \approx \frac{\Omega^2}{2 \omega}=\frac{1}{2} \Omega^2 \cdot \Omega=\frac{1}{2} \Omega^3 .$$

## 数学代写|微积分代写Calculus代写|Absolute value approximations

$$|\omega|=\omega \text {. }$$

$$|-\omega|=-(-\omega)=\omega \text {. }$$

$$|38-15 \omega|=38-15 \omega .$$

$$|38-15 \Omega|=-(38-15 \Omega)=-38+15 \Omega$$

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