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Home/ Questions/Q 3485896
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Editorial Team
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Editorial Team
Asked: May 18, 20262026-05-18T10:57:47+00:00 2026-05-18T10:57:47+00:00

Big oh notation says that all g(n) are an element c.f(n), O(g(n)) for some

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Big oh notation says that all g(n) are an element c.f(n), O(g(n)) for some constant c.

I have always wondered and never really understood why we need this arbitrary constant to multiply with the bounding function f(n) to get our bounds?

Also how does one decide what number this constant should be?

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  1. Editorial Team
    Editorial Team
    2026-05-18T10:57:48+00:00Added an answer on May 18, 2026 at 10:57 am

    The constant itself doesn’t characterize the limiting behavior of the f(n) compared to g(n).

    It is used for the mathematical definition, which enforces the existence of a constant M such that

    alt text

    If such a constant exists then you can state that f(x) is an O(g(x)), and this is the usual notation when analyzing algorithms, you just don’t care about which is the constant but just the complexity of operations itself. The constant is able make that disequation correct by ensuring that M|g(x)| is an upper bound of f(x).

    How to find that constant depends on f(x) and g(x) and it is the mathematical point that must be proved to ensure that f(x) has a g(x) big-o so there’s not a general rule. Look at this example.

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