How do you find an equation for this function that is not recursive?

#f(n)=-2f(n-1)+(-1)^n(n-1)#

Thanks in advance!

3 Answers
Mar 30, 2018

Answer:

#f(n) = (-1)^n(2^n(k+1)-n-1)#

Explanation:

I will assume that this is a function of integers, since otherwise there are many possible variations.

Given:

#f(n) = -2f(n-1)+(-1)^n(n-1)#

Let:

#g(n) = (-1)^n f(n)#

Then:

#g(n) = (-1)^n f(n)#

#color(white)(g(n)) = (-1)^n (-2f(n-1)+(-1)^n(n-1))#

#color(white)(g(n)) = (-1)^n (-2(-1)^(n-1)g(n-1)+(-1)^n(n-1))#

#color(white)(g(n)) = 2g(n-1)+(n-1)#

Let:

#k = g(0)#

Then #g(0)#, #g(1)#, #g(2)#,... form a sequence:

#k#, #2k#, #4k+1#, #8k+4#, #16k+11#, #32k+26#,...

Putting #k=0#, this sequence becomes:

#0, 0, 1, 4, 11, 26,...#

This has differences:

#0, 1, 3, 7, 15,...#

recognisable as #2^n-1#

Let us compare the previous sequence with #2^n#:

#1, 2, 4, 8, 16, 32,...#

A matching formula is:

#2^n-n-1#

So the general formula for #g(n)# can be written:

#g(n) = 2^n(k+1)-n-1#

So the formula for #f(n)# can be written:

#f(n) = (-1)^n(2^n(k+1)-n-1)#

Mar 31, 2018

Answer:

See below.

Explanation:

This is a linear difference equation. It can be solved as

#f(n) = f_h(n)+f_p(n)# with

#f_h(n)+2f_h(n-1) = 0#
#f_p(n)+2f_p(n-1)=(-1)^n(n-1)#

To solve #f_h(n)# we make

#f(n) = a^n# and after substitution

#a^n+2a^(n-1) = 0 rArr a = -2 rArr f_h(n) = (-1)^n 2^n#

for the particular we make #f_p(n) = c_n (-1)^n 2^n# and after substitution

#c_n(-1)^n2^n+2c_(n-1)(-1)^(n-1)2^(n-1) = (-1)^n(n-1)# or

#c_n-c_(n-1) = (n-1)2^-n# or

#c_n = c_(n-1) + (n-1)2^-n# or

#c_n = c_0 + sum_(k=1)^n (k-1)2^-k#

and finally

#f(n) = (c_0 + sum_(k=1)^n (k-1)2^-k)(-1)^n2^n#

This formulation can be simplified a lot but we let this task to the reader.

NOTE

#sum_(k=1)^n(k-1)x^k = x^2 sum_(k=0)^(n-1)x^k = x^2d/dx((x^n-1)/(x-1))# then

#sum_(k=1)^n(k-1)x^k =(x (x + (n (x-1) - x) x^n))/(x-1)^2#

now making #x = 2^-1# we obtain

#sum_(k=1)^n(k-1)2^-k =1-(n+1)2^-n#

Apr 1, 2018

Answer:

# f(n) = (-1)^n[2^n(f(0)+1)-n-1]#

Explanation:

The equation

#f(n) = -2f(n-1)+(-1)^n(n-1)#

can be rewritten in the form

#f(n)+(-1)^n(n+1) = -2f(n-1)+(-1)^n{(n-1)+(n+1)}#
#qquad qquad = -2(f(n-1)+(-1)^(n-1) n)#

This shows that the function, #F(n)#, defined by

#F(n) equiv f(n)+(-1)^n(n+1)#

obeys

#F(n) = -2F(n-1)#

The obvious solution for this equation is

#F(n) =(-2)^nF(0)#

Since #F(0) = f(0)+1#, we have

#F(n) equiv f(n)+(-1)^n(n+1) = (-2)^n(f(0)+1)#

and so

#f(n) = (-2)^n(f(0)+1) - (-1)^n(n+1)#
#qquad = (-1)^n[2^n(f(0)+1)-n-1]#