数学|MATH1141/MATH1131 Mathematics代写 unsw代写

MATH1131, Mathematics 1A, and MATH1141, Higher Mathematics 1A, are first year courses
taught by the School of Mathematics and Statistics in semester 1, and are each worth six units
of credit. MATH1131 is also taught in semester 2. Students who pass MATH1131 in semester
1 usually continue to study MATH1231, Mathematics 1B, in semester 2. Those students who
pass MATH1141 with a Credit usually continue to study MATH1241, Higher Mathematics 1B,
in semester 2. MATH1231 is also taught in Summer Session. MATH1131 and MATH1231 (or
MATH1141 and MATH1241) are generally specified in Engineering programs, as well as many
Science programs.

这是一份unsw新南威尔斯大学MATH1141/MATH1131的成功案例

数学|MATH1141/MATH1131 Mathematics代写 sydney代写


问题 1.

Next, if $\alpha>\omega$ is a limit ordinal, then according to what we have said before, $\sum_{\xi<\omega^{\alpha}} \xi=\sup {\beta<\alpha} \sum{\xi<\omega^{\sigma}} \xi$, and here in the supremum we can take the supremum for successor ordinals $\beta$ smaller than $\alpha$. Thus, according to what we have just proved, in this case
$$
\sum_{\xi<\omega^{\alpha}} \xi=\sup {\beta<\alpha} \omega^{\beta \cdot 2}=\omega^{\sigma}, $$ where $\sigma=\sup {\beta<\alpha} \beta \cdot 2$. Here if $\alpha$ equals one of the powers of $\omega$, say $\alpha=\omega^{\tau}$, then
$$
\omega^{\tau}=\sup {\beta<\alpha} \beta \leq \sup {\beta<\alpha} \beta \cdot 2 \leq \sup _{\beta<\alpha} \beta \cdot \omega=\omega^{\tau},
$$


证明 .

$$
\beta=\omega^{\xi_{n}} \cdot a_{n}+\cdots+\omega^{\xi_{0}} \cdot\left(a_{0}-1\right)+\delta,
$$
where $\delta<\omega^{\xi_{0}}$, and here
$$
\beta \cdot 2=\omega^{\varepsilon_{n}} \cdot\left(2 a_{n}\right)+\omega^{\xi_{n-1}} \cdot a_{n-1}+\cdots+\omega^{\varepsilon_{0}} \cdot\left(a_{0}-1\right)+\delta,
$$
thus
$$
\sigma=\sup {\beta<\alpha} \beta \cdot 2=\omega^{\xi{n}} \cdot\left(2 a_{n}\right)+\cdots+\omega^{\xi_{0}} \cdot a_{0}=\alpha \cdot 2 .
$$
In summary, the sum in question is equal to $\omega^{2 \alpha-1}$ if $\alpha$ is finite, it equals $\omega^{\alpha}$ if $\alpha$ is a power of $\omega$, and in all other cases it equals $\omega^{\alpha \cdot 2}$.







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MATH1141/MATH1131 COURSE NOTES :

Actually, the method we used for the existence can be easily extended to yield both the existence and unicity of the representation. In fact, if the normal form of $\alpha$ is $\alpha=\omega^{\xi_{n}} \cdot a_{n}+\cdots+\omega^{\varepsilon_{0}} \cdot a_{0}$ and set $\delta_{0}=\xi_{0}$ and choose $\delta_{i}$, $1 \leq i \leq n$, so that $\xi_{i}=\xi_{i-1}+\delta_{i}$, then $\delta_{0} \geq 0$ and $\delta_{i}>0$ for $1 \leq i \leq n$. Now
$$
\alpha=\omega^{\delta_{0}} \cdot a_{0} \cdot\left(\omega^{\delta_{1}}+1\right) \cdot a_{1} \cdots a_{n-1} \cdot\left(\omega^{\delta_{n}}+1\right) \cdot a_{n}
$$
and if $\delta_{0}=\omega^{\gamma_{m}}+\cdots+\omega^{\gamma_{0}}$ with $\gamma_{m} \geq \cdots \geq \gamma_{0}$, then
$$
\alpha=\omega^{\omega^{\gamma m}} \cdots \omega^{\omega v 0} \cdot a_{0} \cdot\left(\omega^{\delta_{1}}+1\right) \cdot a_{1} \cdots a_{n-1} \cdot\left(\omega^{\delta_{n}}+1\right) \cdot a_{n}
$$




















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