Master Method Equasion

[HL777]

Hey there,

so I'm reading this book about algorithms and there's a simplifying operation performed on an equasion that I can't follow.

Basically how do you go from

a^j * c * (n ÷ b^j)^d

to

c * n^d * (a ÷ b^d)^j

I've checked it with a number example, this is legitimate, the equasions are equal, I just don't see the steps that would take you from one to the other.

Any help is appreciated!

 

[Otis]

… Basically how do you go from
a^j * c * (n ÷ b^j)^d
to
c * n^d * (a ÷ b^d)^j …

Hi HL. Basically, we use commutative, associative, and exponent properties from algebra. Here's one way.

\(\displaystyle a^j \cdot c \cdot \Big(\frac{n}{b^j}\Big)^d\)

\(\displaystyle a^j \cdot c \cdot \frac{n^d}{b^{j \cdot d}}\)

\(\displaystyle \frac{a^j \cdot c \cdot n^d}{b^{j \cdot d}}\)

\(\displaystyle \frac{c \cdot n^d \cdot a^j}{b^{j \cdot d}}\)

\(\displaystyle c \cdot n^d \cdot \frac{a^j}{b^{j \cdot d}}\)

\(\displaystyle c \cdot n^d \cdot \Big(\frac{a}{b^d}\Big)^j\)

?

 

[Deleted member 4993]

Hi HL. Basically, we use commutative, associative, and exponent properties from algebra. Here's one way.

\(\displaystyle a^j \cdot c \cdot \Big(\frac{n}{b^j}\Big)^d\)

\(\displaystyle a^j \cdot c \cdot \frac{n^d}{b^{j \cdot d}}\)

\(\displaystyle \frac{a^j \cdot c \cdot n^d}{b^{j \cdot d}}\)

\(\displaystyle \frac{c \cdot n^d \cdot a^j}{b^{j \cdot d}}\)

\(\displaystyle c \cdot n^d \cdot \frac{a^j}{b^{j \cdot d}}\)

\(\displaystyle c \cdot n^d \cdot \Big(\frac{a}{b^d}\Big)^j\)

?

Otis,

you posted a Xerox-ready solution and OP did not show a line of work! Do you think the OP learned anything from that? I hope!!

 

[Steven G]

Hey there,

so I'm reading this book about algorithms and there's a simplifying operation performed on an equasion that I can't follow.

Basically how do you go from

a^j * c * (n ÷ b^j)^d

to

c * n^d * (a ÷ b^d)^j

I've checked it with a number example, this is legitimate, the equasions are equal, I just don't see the steps that would take you from one to the other.

Any help is appreciated!

Trying the identity with a few numbers that work is not enough to conclude that you have an identity.

Example. Suppose you think that x^y means x*y, which is wrong. Now 2^2 does equal 2*2 and 1^1 does equal 1*1. So can you conclude that x^y is just x*y? NO! Why not? Since 2^3 = 2*2*2=8 and 8 does NOT equal 2*3.

It is enough to come up with a counter example to show that the identity, say x^y=x*y, is false. It is not enough to show that for a few examples that x^y=x*y to conclude that for all numbers that x^y=x*y

 

[lev888]

It is enough to come up with a counter example to show that the identity, say x^y=x*y, is true.

You mean false?

 

[Otis]

… you posted a Xerox-ready solution and OP did not show a line of work! Do you think the OP learned anything from that?…

Hi Subhotosh! I don't read the op as a request for homework help. I think HR is reading a book, on their own.

Were I to speculate, I'd say the OP has a good chance of learning something, as they seem to be engaged in the material they're reading. (That's why I made the effort to provide lesson links.)

If it turns out that HR's question is actually a problem from a school assignment and they turn in my work as their own, then I botched it.

?

 

[HL777]

Thanks for the reply!

This is no homework and I did indeed learn something from it

 

[HL777]

Trying the identity with a few numbers that work is not enough to conclude that you have an identity.

You're absolutely right, if I wanted to be scientific with this this would not suffice for any kind of proof.

In reality I trust the book enough to think that the odds of such specific equasions being the same by some accident is pretty low.

 

[Steven G]

You're absolutely right, if I wanted to be scientific with this this would not suffice for any kind of proof.

In reality I trust the book enough to think that the odds of such specific equasions being the same by some accident is pretty low.

OK, I support what you say.