blocks|key|2659812|text|一种选择是，当n趋于无穷时，取(f(n)+%2B+g(n))+/+f(n)的极限。如果它收敛到有限的非零值，则f(n)+%2B+g(n)+=Θ(f(n))。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2659813|假设f(+n+)对于足够大的n是非零的，那么在这个极限中，上述比率是|2659814|(f(n)+%2B+g(n))+/+f(n)
=+f(n)+/+f(n)+%2B+g(n)+/+f(n)
=1%2B+g(n)+/+f(n)。|blockquote|2659815|因此，取n到无穷远的极限，上面的表达式收敛到1，因为这个比率等于零(+g(n)是o(f(n))的意思)。|2659816|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|N|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|O|8|@]|9|@]|A|$]]]|J|$]]

One option would be to take the limit of (f(n) + g(n)) / f(n) as n tends toward infinity. If this converges to a finite, nonzero value, then f(n) + g(n) = &Theta;(f(n)).

Assuming that f(n) is nonzero for sufficiently large n, the above ratio, in the limit, is

<blockquote>
 (f(n) + g(n)) / f(n)
 
 = f(n) / f(n) + g(n) / f(n)
 
 = 1 + g(n) / f(n).
</blockquote>

Therefore, taking the limit as n goes to infinity, the above expression converges to 1 because the ratio goes to zero (this is what it means for g(n) to be o(f(n)).

blocks|key|167484|text|到目前一切尚好。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|167485|接下来，回顾一下，在最好的情况下，0+<=+g(n)；这将使您在g(n)+%2B+f(n)上有一个下限。|offset|length|style|CODE|167486|entityMap^0|0|H|9|W|B|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@$D|L|E|M|F|G]|$D|N|E|O|F|G]]|9|@]|A|$]]|$1|H|3|-4|5|6|7|P|8|@]|9|@]|A|$]]]|I|$]]

So far so good.

For the next step, recall that in the best case, <code>0 &lt;= g(n)</code>; this should get you a lower bound on <code>g(n) + f(n)</code>.

blocks|key|2569381|text|在我们开始之前，让我们首先说明小o和大Theta符号是什么意思：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2569382|小o表示法
形式上，g(n)+=+o(f(n))+(或g(n)+∈+o(f(n)))持有足够大的n，这意味着对于每一个正常数ε，都存在一个常数N，使得
N>N+(%2B)所有n>N(%2B)|blockquote|offset|length|style|CODE|2569383|来自notation。|2569384|大Θ表示法
h(n)+=+Θ(f(n))表示存在正常数k_1、k_2和N，因此对于n+>+N，k_1+·+%7Cf(n)%7C和k_2+·+%7Cf(n)%7C分别是%7Ch(n)%7C上的上界和下界。
所有n>N(+k_1+)n>N+(%2B%2B)的k_2·x(N)_f(N)_x(N)_(N)_(N)(N)(N)_(N)_(N)|2569385|来自https://www.khanacademy.org/computing/computer-science/algorithms/asymptotic-notation/a/big-big-theta-notation。|2569386|Given:+g(n)+∈+o(f(n))|BOLD|2569387|因此，在我们的例子中，对于每一个ε>0，我们都可以找到一些常量的N，比如(%2B)，我们的函数g(n)和f(n)。因此，对于n>N，我们有|2569388|%7Cg(n)%7C+≤+ε*%7Cf(n)%7C,+for+some+ε>0,+for+all+n>N

Choose+a+constant+ε+<+1+(recall,+the+above+holds+for+all+ε+>+0),+
with+accompanied+constant+N.+
Then+the+following+holds+for+all+n>N

++++ε(%7Cg(n)%7C+%2B+%7Cf(n)%7C)+≤+2%7Cf(n)%7C+≤+2(%7Cg(n)%7C+%2B+%7Cf(n)%7C)+≤+4*%7Cf(n)%7C++++(*)|code-block|syntax|javascript|2569389|去掉(*)中最左边的不等式，除以2，我们有：|2569390|%7Cf(n)%7C+≤+%7Cg(n)%7C+%2B+%7Cf(n)%7C+≤+2*%7Cf(n)%7C,+n>N++++++++++++++++++++++++++++(**)+|2569391|我们看到，这就是用(%2B%2B)表示的用常量k_1+=+1、k_2+=+2和h(n)+=+g(n)%2Bf(n)表示的大Θ表示法。因此|2569392|(**)+=>+g(n)+%2B+f(n)+is+in+Θ(f(n))|2569393|我们已经证明了g(n)+∈+o(f(n))意味着(g(n)+%2B+f(n))+∈+Θ(f(n))。|2569394|entityMap|0|LINK|mutability|MUTABLE|url|https://en.wikipedia.org/wiki/Big_O_notation#Little-o_notation|1|https://www.khanacademy.org/computing/computer-science/algorithms/asymptotic-notation/a/big-big-theta-notation^0|0|A|E|R|E|1C|1|1Q|1|1Z|1|0|2|8|0|0|6|E|R|3|V|3|Z|1|15|5|1B|C|1O|C|23|6|0|2|32|1|0|0|6|7|E|0|G|3|W|1|10|3|19|4|1E|4|1O|3|0|0|2|3|0|0|9|4|J|7|R|7|Z|G|0|0|7|E|O|N|0^^$0|@$1|2|3|4|5|6|7|1I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|1J|8|@$E|1K|F|1L|G|H]|$E|1M|F|1N|G|H]|$E|1O|F|1P|G|H]|$E|1Q|F|1R|G|H]|$E|1S|F|1T|G|H]]|9|@]|A|$]]|$1|I|3|J|5|6|7|1U|8|@]|9|@$E|1V|F|1W|1|1X]]|A|$]]|$1|K|3|L|5|D|7|1Y|8|@$E|1Z|F|20|G|H]|$E|21|F|22|G|H]|$E|23|F|24|G|H]|$E|25|F|26|G|H]|$E|27|F|28|G|H]|$E|29|F|2A|G|H]|$E|2B|F|2C|G|H]|$E|2D|F|2E|G|H]]|9|@]|A|$]]|$1|M|3|N|5|6|7|2F|8|@]|9|@$E|2G|F|2H|1|2I]]|A|$]]|$1|O|3|P|5|6|7|2J|8|@$E|2K|F|2L|G|Q]|$E|2M|F|2N|G|H]]|9|@]|A|$]]|$1|R|3|S|5|6|7|2O|8|@$E|2P|F|2Q|G|H]|$E|2R|F|2S|G|H]|$E|2T|F|2U|G|H]|$E|2V|F|2W|G|H]|$E|2X|F|2Y|G|H]|$E|2Z|F|30|G|H]]|9|@]|A|$]]|$1|T|3|U|5|V|7|31|8|@]|9|@]|A|$W|X]]|$1|Y|3|Z|5|6|7|32|8|@$E|33|F|34|G|H]]|9|@]|A|$]]|$1|10|3|11|5|V|7|35|8|@]|9|@]|A|$W|X]]|$1|12|3|13|5|6|7|36|8|@$E|37|F|38|G|H]|$E|39|F|3A|G|H]|$E|3B|F|3C|G|H]|$E|3D|F|3E|G|H]]|9|@]|A|$]]|$1|14|3|15|5|V|7|3F|8|@]|9|@]|A|$W|X]]|$1|16|3|17|5|6|7|3G|8|@$E|3H|F|3I|G|H]|$E|3J|F|3K|G|H]]|9|@]|A|$]]|$1|18|3|-4|5|6|7|3L|8|@]|9|@]|A|$]]]|19|$1A|$5|1B|1C|1D|A|$1E|1F]]|1G|$5|1B|1C|1D|A|$1E|1H]]]]

Before we begin, lets first state what little-o and Big-Theta notations means:

<blockquote>
 Little-o notation
 
 Formally, that <code>g(n) = o(f(n))</code> (or <code>g(n) ∈ o(f(n))</code>) holds for
 sufficiently large <code>n</code> means that for every positive constant <code>ε</code>
 there exists a constant <code>N</code> such that

<pre><code>|g(n)| ≤ ε*|f(n)|, for all n &gt; N (+)
</code></pre>
</blockquote>

From <a href="https://en.wikipedia.org/wiki/Big_O_notation#Little-o_notation" rel="nofollow">https://en.wikipedia.org/wiki/Big_O_notation#Little-o_notation</a>.

<blockquote>
 Big-Θ notation
 
 <code>h(n) = Θ(f(n))</code> means there exists positive constants <code>k_1</code>, <code>k_2</code>
 and <code>N</code>, such that <code>k_1 · |f(n)|</code> and <code>k_2 · |f(n)|</code> is an upper bound
 and lower bound on on <code>|h(n)|</code>, respectively, for <code>n &gt; N</code>, i.e.

<pre><code>k_1 · |f(n)| ≤ |h(n)| ≤ k_2 · |f(n)|, for all n &gt; N (++)
</code></pre>
</blockquote>

From <a href="https://www.khanacademy.org/computing/computer-science/algorithms/asymptotic-notation/a/big-big-theta-notation" rel="nofollow">https://www.khanacademy.org/computing/computer-science/algorithms/asymptotic-notation/a/big-big-theta-notation</a>.

<hr>

Given: <code>g(n) ∈ o(f(n))</code>

Hence, in our case, for every <code>ε&gt;0</code> we can find some constant <code>N</code> such that <code>(+)</code>, for our functions <code>g(n)</code> and <code>f(n)</code>. Hence, for <code>n&gt;N</code>, we have

<pre><code>|g(n)| ≤ ε*|f(n)|, for some ε&gt;0, for all n&gt;N

Choose a constant ε &lt; 1 (recall, the above holds for all ε &gt; 0), 
with accompanied constant N. 
Then the following holds for all n&gt;N

 ε(|g(n)| + |f(n)|) ≤ 2|f(n)| ≤ 2(|g(n)| + |f(n)|) ≤ 4*|f(n)| (*)
</code></pre>

Stripping away the left-most inequality in <code>(*)</code> and dividing by 2, we have:

<pre><code>|f(n)| ≤ |g(n)| + |f(n)| ≤ 2*|f(n)|, n&gt;N (**) 
</code></pre>

We see that this is the very definition Big-Θ notation, as presented in <code>(++)</code>, with constants <code>k_1 = 1</code>, <code>k_2 = 2</code> and <code>h(n) = g(n)+f(n)</code>. Hence

<pre><code>(**) =&gt; g(n) + f(n) is in Θ(f(n))
</code></pre>

Ans we have shown that <code>g(n) ∈ o(f(n))</code> implies <code>(g(n) + f(n)) ∈ Θ(f(n))</code>.

So I'm struggling with proving (or disproving) the above question. I feel like it is true, but I'm not sure how to show it. 

Again, the question is if g(n) is o(f(n)), then f(n) + g(n) is Theta(f(n))

Note, that is a little-o, not a big-o!!! 

So far, I've managed to (easily) show that: 

g(n) = o(f(n)) -> g(n) &lt; c*f(n)

Then g(n) + f(n) &lt; (c+1)*f(n) -> (g(n) + f(n)) = O(f(n))

However, for showing Big Omega, I'm not sure what to do there. 

Am I going about this right? 

EDIT: Everyone provided great help, but I could only mark one. THANK YOU.

Proving if g(n) is o(f(n)), then f(n) + g(n) is Theta(f(n))

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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因此，我很难证明(或反驳)上述问题。我觉得这是真的，但我不知道如何表现出来。同样，如果g(n)是o( f(n) )，那么f(N)+ g(n)就是Theta(f(n))。注意，这是一个小-o，而不是一个大-o！到目前为止，我已经(很容易)证明了：g(n) = o(f(n)) -> g(n) < c*f(n)然后g(n) ...

问证明g(n)是o( f(n) )，则f(N)+ g(n)是Theta(f(n))
EN

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问证明g(n)是o( f(n) )，则f(N)+ g(n)是Theta(f(n))EN