How to re-write an equation [duplicate]
This question is an exact duplicate of:
How to solve a difference equation?
1 answer
So I have this equation
$4y_{k+1} = 2y_k$.
If i divide both sides by 4, I get
$$y_{k+1}=frac12y_k$$
However, the equation can be re-written as
so that
$$y_{1}=frac12y_0$$
and
$$y_{2}=frac12 y_1=frac1{2^2}y_0$$
and
$$y_{3}=frac12 y_2=frac1{2^3}y_0$$
$$cdots$$
This is a difference equation, where the initial condition first is y=0 and then y=2.
Can somebody please explain how , from $$y_{k+1}=frac12y_k$$ i can get to the following equations?
differential-equations systems-of-equations
asked Nov 29 at 18:20
GGGG
176
marked as duplicate by MisterRiemann, amWhy, Jyrki Lahtonen, Hans Lundmark, Kevin Long Nov 29 at 22:26
This question was marked as an exact duplicate of an existing question.
|
show 7 more comments
This question is an exact duplicate of:
How to solve a difference equation?
1 answer
So I have this equation
$4y_{k+1} = 2y_k$.
If i divide both sides by 4, I get
$$y_{k+1}=frac12y_k$$
However, the equation can be re-written as
so that
$$y_{1}=frac12y_0$$
and
$$y_{2}=frac12 y_1=frac1{2^2}y_0$$
and
$$y_{3}=frac12 y_2=frac1{2^3}y_0$$
$$cdots$$
This is a difference equation, where the initial condition first is y=0 and then y=2.
Can somebody please explain how , from $$y_{k+1}=frac12y_k$$ i can get to the following equations?
differential-equations systems-of-equations
asked Nov 29 at 18:20
GGGG
176
marked as duplicate by MisterRiemann, amWhy, Jyrki Lahtonen, Hans Lundmark, Kevin Long Nov 29 at 22:26
This question was marked as an exact duplicate of an existing question.
Insert $k=0,1,2,ldots$
– MisterRiemann
Nov 29 at 18:22
If i do so, i get that y1=1 and y2=1 and these are not the answers, what am i doing wrong?
– GGGG
Nov 29 at 18:26
Its a kind of following since i cant understand it really there...
– GGGG
Nov 29 at 18:28
"where the initial condition first is y=0 and then y=2. " I do not understand this question. Do you have a first term. If $y_0 = a$ then this formula is simply $y_k = frac 1{2^k} a$. But you need a term. If the term you are given is $y_m = a$ then the formula is $y_k = frac 1{2^{k-m}}a = 2^{m-k} a$.
– fleablood
Nov 29 at 18:30
the question is : Solve the difference equation: 4yk+1 = 2yk first with the initial condition y0 = 0 and then with the initial condition y0 = 2.
– GGGG
Nov 29 at 18:32
|
show 7 more comments
This question is an exact duplicate of:
How to solve a difference equation?
1 answer
So I have this equation
$4y_{k+1} = 2y_k$.
If i divide both sides by 4, I get
$$y_{k+1}=frac12y_k$$
However, the equation can be re-written as
so that
$$y_{1}=frac12y_0$$
and
$$y_{2}=frac12 y_1=frac1{2^2}y_0$$
and
$$y_{3}=frac12 y_2=frac1{2^3}y_0$$
$$cdots$$
This is a difference equation, where the initial condition first is y=0 and then y=2.
Can somebody please explain how , from $$y_{k+1}=frac12y_k$$ i can get to the following equations?
differential-equations systems-of-equations
asked Nov 29 at 18:20
GGGG
176
This question is an exact duplicate of:
How to solve a difference equation?
1 answer
So I have this equation
$4y_{k+1} = 2y_k$.
If i divide both sides by 4, I get
$$y_{k+1}=frac12y_k$$
However, the equation can be re-written as
so that
$$y_{1}=frac12y_0$$
and
$$y_{2}=frac12 y_1=frac1{2^2}y_0$$
and
$$y_{3}=frac12 y_2=frac1{2^3}y_0$$
$$cdots$$
This is a difference equation, where the initial condition first is y=0 and then y=2.
Can somebody please explain how , from $$y_{k+1}=frac12y_k$$ i can get to the following equations?
This question is an exact duplicate of:
How to solve a difference equation?
1 answer
differential-equations systems-of-equations
differential-equations systems-of-equations
asked Nov 29 at 18:20
GGGG
176
asked Nov 29 at 18:20
GGGG
176
asked Nov 29 at 18:20
GGGG
176
asked Nov 29 at 18:20
GGGG
176
asked Nov 29 at 18:20
GGGG
176
176
marked as duplicate by MisterRiemann, amWhy, Jyrki Lahtonen, Hans Lundmark, Kevin Long Nov 29 at 22:26
This question was marked as an exact duplicate of an existing question.
marked as duplicate by MisterRiemann, amWhy, Jyrki Lahtonen, Hans Lundmark, Kevin Long Nov 29 at 22:26
This question was marked as an exact duplicate of an existing question.
Insert $k=0,1,2,ldots$
– MisterRiemann
Nov 29 at 18:22
If i do so, i get that y1=1 and y2=1 and these are not the answers, what am i doing wrong?
– GGGG
Nov 29 at 18:26
Its a kind of following since i cant understand it really there...
– GGGG
Nov 29 at 18:28
"where the initial condition first is y=0 and then y=2. " I do not understand this question. Do you have a first term. If $y_0 = a$ then this formula is simply $y_k = frac 1{2^k} a$. But you need a term. If the term you are given is $y_m = a$ then the formula is $y_k = frac 1{2^{k-m}}a = 2^{m-k} a$.
– fleablood
Nov 29 at 18:30
the question is : Solve the difference equation: 4yk+1 = 2yk first with the initial condition y0 = 0 and then with the initial condition y0 = 2.
– GGGG
Nov 29 at 18:32
|
show 7 more comments
Insert $k=0,1,2,ldots$
– MisterRiemann
Nov 29 at 18:22
If i do so, i get that y1=1 and y2=1 and these are not the answers, what am i doing wrong?
– GGGG
Nov 29 at 18:26
Its a kind of following since i cant understand it really there...
– GGGG
Nov 29 at 18:28
"where the initial condition first is y=0 and then y=2. " I do not understand this question. Do you have a first term. If $y_0 = a$ then this formula is simply $y_k = frac 1{2^k} a$. But you need a term. If the term you are given is $y_m = a$ then the formula is $y_k = frac 1{2^{k-m}}a = 2^{m-k} a$.
– fleablood
Nov 29 at 18:30
the question is : Solve the difference equation: 4yk+1 = 2yk first with the initial condition y0 = 0 and then with the initial condition y0 = 2.
– GGGG
Nov 29 at 18:32
Insert $k=0,1,2,ldots$
– MisterRiemann
Nov 29 at 18:22
Insert $k=0,1,2,ldots$
– MisterRiemann
Nov 29 at 18:22
If i do so, i get that y1=1 and y2=1 and these are not the answers, what am i doing wrong?
– GGGG
Nov 29 at 18:26
If i do so, i get that y1=1 and y2=1 and these are not the answers, what am i doing wrong?
– GGGG
Nov 29 at 18:26
Its a kind of following since i cant understand it really there...
– GGGG
Nov 29 at 18:28
Its a kind of following since i cant understand it really there...
– GGGG
Nov 29 at 18:28
"where the initial condition first is y=0 and then y=2. " I do not understand this question. Do you have a first term. If $y_0 = a$ then this formula is simply $y_k = frac 1{2^k} a$. But you need a term. If the term you are given is $y_m = a$ then the formula is $y_k = frac 1{2^{k-m}}a = 2^{m-k} a$.
– fleablood
Nov 29 at 18:30
"where the initial condition first is y=0 and then y=2. " I do not understand this question. Do you have a first term. If $y_0 = a$ then this formula is simply $y_k = frac 1{2^k} a$. But you need a term. If the term you are given is $y_m = a$ then the formula is $y_k = frac 1{2^{k-m}}a = 2^{m-k} a$.
– fleablood
Nov 29 at 18:30
the question is : Solve the difference equation: 4yk+1 = 2yk first with the initial condition y0 = 0 and then with the initial condition y0 = 2.
– GGGG
Nov 29 at 18:32
the question is : Solve the difference equation: 4yk+1 = 2yk first with the initial condition y0 = 0 and then with the initial condition y0 = 2.
– GGGG
Nov 29 at 18:32
|
show 7 more comments
1 Answer
1
active
oldest
votes
Notice that $$y_n={1over 2}y_{n-1}={1over 2^2}y_{n-2}={1over 2^3}y_{n-3}=cdots ={1over 2^k}y_{n-k}$$let $k=n$ therefore $$y_n={1over 2^n}y_0$$
answered Nov 29 at 18:45
Mostafa Ayaz
13.7k3836
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Notice that $$y_n={1over 2}y_{n-1}={1over 2^2}y_{n-2}={1over 2^3}y_{n-3}=cdots ={1over 2^k}y_{n-k}$$let $k=n$ therefore $$y_n={1over 2^n}y_0$$
answered Nov 29 at 18:45
Mostafa Ayaz
13.7k3836
add a comment |
Notice that $$y_n={1over 2}y_{n-1}={1over 2^2}y_{n-2}={1over 2^3}y_{n-3}=cdots ={1over 2^k}y_{n-k}$$let $k=n$ therefore $$y_n={1over 2^n}y_0$$
answered Nov 29 at 18:45
Mostafa Ayaz
13.7k3836
add a comment |
Notice that $$y_n={1over 2}y_{n-1}={1over 2^2}y_{n-2}={1over 2^3}y_{n-3}=cdots ={1over 2^k}y_{n-k}$$let $k=n$ therefore $$y_n={1over 2^n}y_0$$
answered Nov 29 at 18:45
Mostafa Ayaz
13.7k3836
Notice that $$y_n={1over 2}y_{n-1}={1over 2^2}y_{n-2}={1over 2^3}y_{n-3}=cdots ={1over 2^k}y_{n-k}$$let $k=n$ therefore $$y_n={1over 2^n}y_0$$
answered Nov 29 at 18:45
Mostafa Ayaz
13.7k3836
answered Nov 29 at 18:45
Mostafa Ayaz
13.7k3836
answered Nov 29 at 18:45
Mostafa Ayaz
13.7k3836
answered Nov 29 at 18:45
Mostafa Ayaz
13.7k3836
13.7k3836
add a comment |
add a comment |
Insert $k=0,1,2,ldots$
– MisterRiemann
Nov 29 at 18:22
If i do so, i get that y1=1 and y2=1 and these are not the answers, what am i doing wrong?
– GGGG
Nov 29 at 18:26
Its a kind of following since i cant understand it really there...
– GGGG
Nov 29 at 18:28
"where the initial condition first is y=0 and then y=2. " I do not understand this question. Do you have a first term. If $y_0 = a$ then this formula is simply $y_k = frac 1{2^k} a$. But you need a term. If the term you are given is $y_m = a$ then the formula is $y_k = frac 1{2^{k-m}}a = 2^{m-k} a$.
– fleablood
Nov 29 at 18:30
the question is : Solve the difference equation: 4yk+1 = 2yk first with the initial condition y0 = 0 and then with the initial condition y0 = 2.
– GGGG
Nov 29 at 18:32