A set of rational numbers 2 (correct version)

Calculus Level 3

Lets say you have two natural numbers x x and n n with n ≠ 1 n\neq 1 .

Lets say you also have a set with all multiples of n n until x ⋅ n x \cdot n :

M = { 1 ⋅ n ; 2 ⋅ n ; 3 ⋅ n ; ⋯ ; x ⋅ n } M=\left\{ 1\cdot n;2\cdot n;3\cdot n;\cdots ;x\cdot n \right\}

Lets say you have another set with the powers of n n until n x { n }^{ x } :

N = { n 1 ; n 2 ; n 3 ; ⋯ ; n x } N=\left\{ { n }^{ 1 };{ n }^{ 2 };{ n }^{ 3 };\cdots ;{ n }^{ x } \right\}

Now you store every possible fraction, where the numerator is an element of M M and the denominator is an element of N N in a third set Q Q . Q Q can hold multiple times the same value. For example, when n = 2 n=2 and x ≥ 2 x\ge 2 , Q Q holds 2 2 \frac { 2 }{ 2 } and 4 4 \frac { 4 }{ 4 } even both fractions have the same value.

Q ˉ \bar { Q } is the arithmetic mean of Q Q . f ( n ) = lim ⁡ x → ∞ Q ˉ f\left( n \right) =\lim _{ x\rightarrow \infty }{ \bar { Q } } . Which one is correct?

Here you can find the first problem.

For those, who wonder why I reuploaded this problem after a few minutes. There was a typing error in the correct solution.

f ( n ) = 1 n − 1 f\left( n \right) =\frac { 1 }{ n-1 } f ( n ) = n 2 ⋅ ( n − 1 ) f\left( n \right) =\frac { n }{ 2\cdot \left( n-1 \right) } f ( n ) = 1 2 ⋅ ( n − 1 ) f\left( n \right) =\frac { 1 }{ 2\cdot \left( n-1 \right) } f ( n ) = 0 f\left( n \right) =0 not enought information f ( n ) = n n − 1 f\left( n \right) =\frac { n }{ n-1 }

CodeCrafter 1 by CodeCrafter 1 , 17, Germany

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1 solution

CodeCrafter 1
May 12, 2019

From the first problem we know:

Q ˉ = ( ∑ i = 1 x n ⋅ i ) ⋅ ( ∑ j = 1 x 1 n j ) x 2 \bar { Q } =\frac { \left( \sum _{ i=1 }^{ x }{ n\cdot i } \right) \cdot \left( \sum _{ j=1 }^{ x }{ \frac { 1 }{ { n }^{ j } } } \right) }{ { x }^{ 2 } }

We also know that:

∑ i = 1 x n ⋅ i = n ⋅ x ⋅ ( x + 1 ) 2 \sum _{ i=1 }^{ x }{ n\cdot i } =n\cdot \frac { x\cdot \left( x+1 \right) }{ 2 }

∑ j = 1 x 1 n j = n x − 1 n x ⋅ ( n − 1 ) \sum _{ j=1 }^{ x }{ \frac { 1 }{ { n }^{ j } } } =\frac { { n }^{ x }-1 }{ { n }^{ x }\cdot \left( n-1 \right) }

Thus Q ˉ = n ⋅ x ⋅ ( x + 1 ) ⋅ ( n x − 1 ) 2 ⋅ x 2 ⋅ n x ⋅ ( n − 1 ) = x ⋅ n x − x + n x − 1 2 ⋅ x ⋅ n x − 1 ⋅ ( n − 1 ) = x ⋅ n x − 1 ⋅ ( n − 1 n x + n x − 1 x ⋅ n x − 1 ) 2 ⋅ x ⋅ n x − 1 ⋅ ( n − 1 ) = n − 1 n x + n x − 1 x ⋅ n x − 1 2 ⋅ ( n − 1 ) \bar { Q } =\frac { n\cdot x\cdot \left( x+1 \right) \cdot \left( { n }^{ x }-1 \right) }{ 2\cdot { x }^{ 2 }\cdot { n }^{ x }\cdot \left( n-1 \right) } =\frac { x\cdot { n }^{ x }-x+{ n }^{ x }-1 }{ 2\cdot x\cdot { n }^{ x-1 }\cdot \left( n-1 \right) } =\frac { x\cdot { n }^{ x-1 }\cdot \left( n-\frac { 1 }{ { n }^{ x } } +\frac { n }{ x } -\frac { 1 }{ x\cdot { n }^{ x-1 } } \right) }{ 2\cdot x\cdot { n }^{ x-1 }\cdot \left( n-1 \right) } =\frac { n-\frac { 1 }{ { n }^{ x } } +\frac { n }{ x } -\frac { 1 }{ x\cdot { n }^{ x-1 } } }{ 2\cdot \left( n-1 \right) }

And finally:

Q ˉ = lim ⁡ x → ∞ n − 1 n x + n x − 1 x ⋅ n x − 1 2 ⋅ ( n − 1 ) = n 2 ⋅ ( n − 1 ) \bar { Q } =\lim _{ x\rightarrow \infty }{ \frac { n-\frac { 1 }{ { n }^{ x } } +\frac { n }{ x } -\frac { 1 }{ x\cdot { n }^{ x-1 } } }{ 2\cdot \left( n-1 \right) } } =\boxed { \frac { n }{ 2\cdot \left( n-1 \right) } }

3 Helpful 0 Interesting 1 Brilliant 0 Confused

For those, who want a proof that ∑ j = 1 x 1 n j = n x − 1 n x ⋅ ( n − 1 ) \sum _{ j=1 }^{ x }{ \frac { 1 }{ { n }^{ j } } } =\frac { { n }^{ x }-1 }{ { n }^{ x }\cdot \left( n-1 \right) } is true:

∑ j = 1 x 1 n j = 1 n 1 + 1 n 2 + 1 n 3 + ⋯ + 1 n x = m \sum _{ j=1 }^{ x }{ \frac { 1 }{ { n }^{ j } } } =\frac { 1 }{ { n }^{ 1 } } +\frac { 1 }{ { n }^{ 2 } } +\frac { 1 }{ { n }^{ 3 } } +\cdots +\frac { 1 }{ { n }^{ x } } =m

Then n ⋅ m = 1 + 1 n 1 + 1 n 2 + 1 n 3 + ⋯ + 1 n x − 1 n\cdot m=1+\frac { 1 }{ { n }^{ 1 } } +\frac { 1 }{ { n }^{ 2 } } +\frac { 1 }{ { n }^{ 3 } } +\cdots +\frac { 1 }{ { n }^{ x-1 } } .

So: n ⋅ m − m = m ⋅ ( n − 1 ) = 1 + 1 n 1 + 1 n 2 + 1 n 3 + ⋯ + 1 n x − 1 − ( 1 n 1 + 1 n 2 + 1 n 3 + ⋯ + 1 n x ) = 1 − 1 n x n\cdot m-m=m\cdot \left( n-1 \right) =1+\frac { 1 }{ { n }^{ 1 } } +\frac { 1 }{ { n }^{ 2 } } +\frac { 1 }{ { n }^{ 3 } } +\cdots +\frac { 1 }{ { n }^{ x-1 } } -\left( \frac { 1 }{ { n }^{ 1 } } +\frac { 1 }{ { n }^{ 2 } } +\frac { 1 }{ { n }^{ 3 } } +\cdots +\frac { 1 }{ { n }^{ x } } \right) =1-\frac { 1 }{ { n }^{ x } }

and finally:

m = 1 n − 1 − 1 n x ⋅ ( n − 1 ) = n x − 1 n x ⋅ ( n − 1 ) m=\frac { 1 }{ n-1 } -\frac { 1 }{ { n }^{ x }\cdot \left( n-1 \right) } =\boxed { \frac { { n }^{ x }-1 }{ { n }^{ x }\cdot \left( n-1 \right) } }

CodeCrafter 1 - 2 years, 1 month ago

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