Factorials of the Plane!

Calculus Level 1

Let x x be a real number satisfying x ! + ( x − 1 ) ! = 3 x ( x − 1 ) ! x! + (x-1)! = 3x(x-1)! , find the value of 4 x 4x .

To clarify, x ! = Γ ( x + 1 ) x! = \Gamma(x + 1) , where Γ ( ⋅ ) \Gamma(\cdot) denotes the gamma function . In other words, even if x x is not an integer, x ! x! can be computed.


The answer is 2.

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Nanayaranaraknas Vahdam by Nanayaranaraknas Vahdam , 22, India

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6 solutions

x ! + ( x − 1 ) ! = 3 x ( x − 1 ) ! x! + (x-1)! = 3x(x-1)!

x ! = ( 3 x − 1 ) ( x − 1 ) ! x! = (3x-1)(x-1)!

x ! ( x − 1 ) ! \frac{x!}{(x-1)!} = = 3 x − 1 3x-1

x = 3 x − 1 x=3x-1 → \rightarrow x = 1 2 x=\boxed{\frac{1}{2}}

Thus, 4 x = 2 4x=\boxed{2}

9 Helpful 0 Interesting 1 Brilliant 0 Confused

x(x-1)!+(x-1)!=3x(x-1)!, x+1=3x, 2x=1, x=1/2, 4x = 2. Simplest solution. a suggestion to you all "NEVER TRY TO OVERCOMLICATE THINGS, A DOT IS A DOT

Kushagra Sahni - 7 years, 1 month ago

Too easy! But is there 1/2!?

dada haha - 7 years ago

If x^2015/2015+x^2014/2014+x^2013/2013+⋯+x+1=0, find the number of real roots.

rama rao - 7 years, 1 month ago

If a1,a2,a3,b1,b2,b3 are the values of n for which ∑ (r=0)^(n-1)▒x^2r is divisible by ∑ (r=0)^(n-1)▒x^r then the triangle having vertices (a1, b1), (a2, b2) and (a3, b3) cannot be Isosceles triangle A right angled isosceles triangle A right angled triangle An equilateral triangle

rama rao - 7 years, 1 month ago

The question has error because 0.5! Doesn't exist. So...

Varun Kotharkar - 5 years, 10 months ago

If x = 1 2 x = \boxed { \frac { 1 } { 2 } } , then 1 2 ! + ( 1 2 − 1 ) ! = ( 3 2 ) ( 1 2 − 1 ) ! \frac { 1 } { 2 } ! + ( \frac { 1 } { 2 } - 1 )! = ( \frac { 3 } { 2 } ) ( \frac { 1 } { 2 } - 1 ) ! .

But ( − 1 2 ) ! ( - \frac { 1 } { 2 } ) ! is not defined. ( ( − 1 2 ) ! ( - \frac { 1 } { 2 } ) ! is undefined. ).

. . - 3 months, 3 weeks ago

But i think there isn't 1 2 ! \frac{1}{2}!

Reynan Henry - 7 years, 1 month ago

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And there isn't any − 1 2 ! -\frac{1}{2}!

Reynan Henry - 7 years, 1 month ago

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Actually it is defined by the Gamma function

Nanayaranaraknas Vahdam - 7 years, 1 month ago
Ahmed Obaiedallah
May 30, 2015

x ! + ( x − 1 ) ! = 3 x ( x − 1 ) ! x!+(x-1)!=3x(x-1)!

n o t e : x ( x − 1 ) ! = x ! note: x(x-1)!=x!

x ! x ( x − 1 ) ! + ( x − 1 ) ! x ( x − 1 ) ! = 3 \frac {x!}{x(x-1)!} +\frac{(x-1)!}{x(x-1)!}=3

x ! x ! + ( x − 1 ) ! x ( x − 1 ) ! = 3 \frac {x!}{x!} +\frac{(x-1)!}{x(x-1)!}=3

1 + 1 x = 3 1+\frac{1}{x}=3

2 x = 1 , 4 x = 2 2x=1, \boxed{4x=2}

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Bala Vidyadharan
May 18, 2015

x!+(x-1)!=3x(x-1)! x.(x-1)!+(x-1)!=3x(x-1)! Taking (x-1)! common (x-1)![x+1]=3x(x-1)! Cancelling (x-1)! on both sides x+1=3x 3x-x=1 2x=1 x=1/2

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Edwin Gray
Jul 14, 2018

Let x! = x (x -1)!. Then the equation becomes x (x - 1)! + (x - 1)! = 3x*(x - 1)!. Dividing by (x - 1)!, x + 1 = 3x, 1 =2x, and 2 = 4x. Ed Gray

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Sundeep Prasad
Aug 7, 2015

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Dipak Prajapati
Jul 5, 2015

Same solution as Nanayaranaraknas

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