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4 years ago

What set of transformations is performed on triangle ABC to form triangle A′B′C′?

Mathematics

1 answer:

Simora [160]4 years ago

4 0

Answer: A translation 5 units down followed by a 180-degree counterclockwise rotation about the origin .

Step-by-step explanation:

From the given figure, the coordinates of ΔABC are A(-3,4), B(-3,1), C(-2,1) and the coordinates of ΔA'B'C' are A'(3,1), B'(3,4), C'(2,4).

When, a translation of 5 units down is applied to ΔABC, the coordinates of the image will be

Then applying 180° counterclockwise rotation about the origin, the coordinates of the image will be :-

which are the coordinates of ΔA'B'C'.

Hence, the set of transformations is performed on triangle ABC to form triangle A’B’C’ is " A translation 5 units down followed by a 180-degree counterclockwise rotation about the origin ".

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$\displaystyle\lim_{n\to\infty}\left(k!+\frac{(k+1)!}{1!}+\cdots+\frac{(k+n)!}{n!}\right)=\lim_{n\to\infty}\dfrac{\displaystyle\sum_{i=0}^n\frac{(k+i)!}{i!}}{n^{k+1}}=\lim_{n\to\infty}\frac{a_n}{b_n}$

By the Stolz-Cesaro theorem, this limit exists if

$\displaystyle\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$

also exists, and the limits would be equal. The theorem requires that $b_n$ be strictly monotone and divergent, which is the case since $k\in\mathbb N$ .

You have

$a_{n+1}-a_n=\displaystyle\sum_{i=0}^{n+1}\frac{(k+i)!}{i!}-\sum_{i=0}^n\frac{(k+i)!}{i!}=\frac{(k+n+1)!}{(n+1)!}$

so we're left with computing

$\displaystyle\lim_{n\to\infty}\frac{(k+n+1)!}{(n+1)!\left((n+1)^{k+1}-n^{k+1}\right)}$

This can be done with the help of Stirling's approximation, which says that for large $n$ , $n!\sim\sqrt{2\pi n}\left(\dfrac ne\right)^n$ . By this reasoning our limit is

$\displaystyle\lim_{n\to\infty}\frac{\sqrt{2\pi(k+n+1)}\left(\dfrac{k+n+1}e\right)^{k+n+1}}{\sqrt{2\pi(n+1)}\left(\dfrac{n+1}e\right)^{n+1}\left((n+1)^{k+1}-n^{k+1}\right)}$

Let's examine this limit in parts. First,

$\dfrac{\sqrt{2\pi(k+n+1)}}{\sqrt{2\pi(n+1)}}=\sqrt{\dfrac{k+n+1}{n+1}}=\sqrt{1+\dfrac k{n+1}}$

As $n\to\infty$ , this term approaches 1.

Next,

$\dfrac{\left(\dfrac{k+n+1}e\right)^{k+n+1}}{\left(\dfrac{n+1}e\right)^{n+1}}=(k+n+1)^k\left(\dfrac{k+n+1}{n+1}\right)^{n+1}=e^{-k}(k+n+1)^k\left(1+\dfrac k{n+1}\right)^{n+1}$

The term on the right approaches $e^k$ , cancelling the $e^{-k}$ . So we're left with

$\displaystyle\lim_{n\to\infty}\frac{(k+n+1)^k}{(n+1)^{k+1}-n^{k+1}}$

Expand the numerator and denominator, and just examine the first few leading terms and their coefficients.

$\displaystyle\frac{(k+n+1)^k}{(n+1)^{k+1}-n^{k+1}}=\frac{n^k+\cdots+(k+1)^k}{n^{k+1}+(k+1)n^k+\cdots+1+n^{k+1}}=\frac{n^k+\cdots+(k+1)^k}{(k+1)n^k+\cdots+1}$

Divide through the numerator and denominator by $n^k$ :

$\dfrac{n^k+\cdots+(k+1)^k}{(k+1)n^k+\cdots+1}=\dfrac{1+\cdots+\left(\frac{k+1}n\right)^k}{(k+1)+\cdots+\frac1{n^k}}$

So you can see that, by comparison, we have

$\displaystyle\lim_{n\to\infty}\frac{(k+n+1)^k}{(n+1)^{k+1}-n^{k+1}}=\lim_{n\to\infty}\frac1{k+1}=\frac1{k+1}$

so this is the value of the limit.

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