Confused about proving the base case

[potent bison]

How does the assumption that T(n) <= c for n <= 2 help with the base case?

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[slate thorn]

It establishes the best case for the proof of the inductive hypothesis.

The assumption provides a concrete bound for T(n) when n is small (specifically, for n = 1 and n = 2). This allows us to explicitly verify that the function behaves as expected for these initial values.

[potent bison]

but it says assume T <= c instead of >=

we are provin lower bound so shouldnt it have been T >= c

[slate thorn]

Finishing this though, by assuming T(n) <= c for n <= 2, we can focus on proving the inductive step for larger values of n. This assumption ensures that our proof starts from a known point, giving us a solid foundation to build on.

The predicate P(n) is defined as T(n) >= cn^log_2^3.

By making sure the base case holds (via the assumption), we can then just use mathematical induction to show that P(n) holds for all n.

[slate thorn]

Yes, the assumption ( T(n) \leq c ) for ( n \leq 2 \ serves as a base case to establish that the function is bounded at small values, providing a foundation for proving the lower bound ( T(n) \geq cn^{\log_2 3} ) for larger ( n ).

[olive holly]

Stratos
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[slate thorn]

So again, this makes sure of a smooth transition in the inductive proof.

[potent bison]

Im ngl i feel like u explained the why not how coz i dont get it

im saying how does that assumption help prove the base case

[slate thorn]

The assumption is bounded and allows us to start our proof from a known point, making it easier to demonstrate the lower bound for larger n.

[potent bison]

right so for this we woulda be starting for n > 2

but im still not understanding why that specific assumption just lets us skip base cases without explaining it

[slate thorn]

Think of it this way. So, the assumption ( T(n) \leq c ) for ( n \leq 2 ) means we can confidently state that the function does not exceed a certain value for small inputs, which simplifies your analysis. Since the behavior of ( T(n) ) is known for these base cases, we can focus on proving the lower bound for larger ( n ) without needing to verify the base cases individually.

[olive holly]

Stratos

[slate thorn]

This allows you to use mathematical induction effectively!

[potent bison]

ohhhhhhhg

and when provin for omega i would say that its true for smaller inputs?

wait what idk acc

but for omega how would the assumption work

[slate thorn]

For reiteration, the assumption ( T(n) \leq c ) for ( n \leq 2 ) shows that the function is bounded for small inputs, allowing us to focus on proving the lower bound for larger ( n ).

[olive holly]

Stratos

[slate thorn]

Now, on the otherhand, for proving ( \Omega ), we establish that the function meets the lower bound as ( n ) increases, using the behavior from the base cases to support the argument.

[olive holly]

Stratos

[potent bison]

wait huhhhh

i thought we said that the big o doesnt work for smaller values

[slate thorn]

in this context

we focus on the growth of the function for larger n

the assumption for small values helps us establish a baseline but when proving omega, we primarily demonstrate that T(n) meets or exceeds the lower bound as n increases

keep in mind

regardless of its behavior for small inputs

does that make sense @potent bison

[potent bison]

so for omega we focus on larger n

oh wait im so sorry for being slow i meant but for big O

we were talking about omega previously i forgot

[slate thorn]

You're all good.

Just keep in mind that for ( O ), the assumption about small inputs helps establish that the function doesn't exceed a certain value, but we primarily focus on showing that ( T(n) ) grows no faster than the upper bound for larger ( n )

[olive holly]

Stratos

[potent bison]

one more question why did they pick 2?

for the assumption

because I am trynna do another question in a similiar manner but not sure about the assumption