MHB Finding a5 in an+1 Sequence Using a4 and 1/2

[brunette15]

I have been given the following sequence: a_n+1 = a_n + (1/2)ⁿ⁺¹, n>=0 with a₀ = 1.

I am trying to express a₅ in terms of 1/2(a₄) .

I have started by writing out the first few terms however i am still struggling with getting a₅.

Can anyone please help me out with this?
Thanks in advance!

 

[MarkFL]

What I would do here is find the closed form for $a_n$ first. Can you identify the homogeneous solution $h_n$ and the form of the particular solution $p_n$? If so, can you then find the particular solution and then by the principle of superposition give the general solution, and then use the given initial value to obtain the solution satisfying all given conditions?

Once you do this, then expressing $a_5$ as a function of $\frac{1}{2}a_4$ is straightforward. :D

 

[brunette15]

What I would do here is find the closed form for $a_n$ first. Can you identify the homogeneous solution $h_n$ and the form of the particular solution $p_n$? If so, can you then find the particular solution and then by the principle of superposition give the general solution, and then use the given initial value to obtain the solution satisfying all given conditions?

Once you do this, then expressing $a_5$ as a function of $\frac{1}{2}a_4$ is straightforward. :D

Hi, thanks for the reply. I kind of see where you are heading with this, i am a bit unsure what you mean by closed form of an though?

 

[MarkFL]

Hi, thanks for the reply. I kind of see where you are heading with this, i am a bit unsure what you mean by closed form of an though?

When we find the closed form for the solution to a difference equation, we find (in this case) $a_n$ as a function of $n$. For example, if given:

$$a_n=a_{n-1}+n$$ where $a_1=1$, then the closed form is:

$$a_n=\frac{n(n+1)}{2}$$

 

[MarkFL]

I will walk you through the process for finding the closed form for the given recursion, or difference equation:

$$a_{n+1}=a_{n}+\left(\frac{1}{2}\right)^{n+1}$$ where $a_0=1$.

The associated homogeneous equation is:

$$a_{n+1}-a_{n}=0$$

Now, we see the characteristic equation is:

$$r-1=0\implies r=1$$

Thus we know the homogeneous solution is given by:

$$h_n=c_1\cdot1^n=c_1$$

Inspection of the original equation tells us the particular solution must have the form:

$$p_n=A\left(\frac{1}{2}\right)^n$$

We can now use the method of undetermined coefficnets to find $A$. Substituting this solution into the original equation, we obtain (after rearranging a bit):

$$p_{n+1}-p_{n}=\left(\frac{1}{2}\right)^{n+1}$$

Now, using the form of $p_n$, we get:

$$A\left(\frac{1}{2}\right)^{n+1}-A\left(\frac{1}{2}\right)^n=\left(\frac{1}{2}\right)^{n+1}$$

Multiplying though by $2^{n+1}\ne0$, there results:

$$A-2A=1\implies A=-1$$

And so, using the principle of superposition, we obtain the general solution:

$$a_n=h_n+p_n=c_1-\frac{1}{2^n}$$

Now, all that's left to do is find the parameter $c_1$ using the given initial value:

$$a_0=c_1-\frac{1}{2^0}=1\implies c_1=2$$

Hence, the closed form can be written as:

$$a_n=2-2^{-n}$$

So, can you now write $a_5$ as a function of $\dfrac{1}{2}a_4$?

 

[brunette15]

I will walk you through the process for finding the closed form for the given recursion, or difference equation:

$$a_{n+1}=a_{n}+\left(\frac{1}{2}\right)^{n+1}$$ where $a_0=1$.

The associated homogeneous equation is:

$$a_{n+1}-a_{n}=0$$

Now, we see the characteristic equation is:

$$r-1=0\implies r=1$$

Thus we know the homogeneous solution is given by:

$$h_n=c_1\cdot1^n=c_1$$

Inspection of the original equation tells us the particular solution must have the form:

$$p_n=A\left(\frac{1}{2}\right)^n$$

We can now use the method of undetermined coefficnets to find $A$. Substituting this solution into the original equation, we obtain (after rearranging a bit):

$$p_{n+1}-p_{n}=\left(\frac{1}{2}\right)^{n+1}$$

Now, using the form of $p_n$, we get:

$$A\left(\frac{1}{2}\right)^{n+1}-A\left(\frac{1}{2}\right)^n=\left(\frac{1}{2}\right)^{n+1}$$

Multiplying though by $2^{n+1}\ne0$, there results:

$$A-2A=1\implies A=-1$$

And so, using the principle of superposition, we obtain the general solution:

$$a_n=h_n+p_n=c_1-\frac{1}{2^n}$$

Now, all that's left to do is find the parameter $c_1$ using the given initial value:

$$a_0=c_1-\frac{1}{2^0}=1\implies c_1=2$$

Hence, the closed form can be written as:

$$a_n=2-2^{-n}$$

So, can you now write $a_5$ as a function of $\dfrac{1}{2}a_4$?

Thankyou so much!

 

[MarkFL]

To finish up the problem, what I did was write:

$$a_5=2-2^{-5}$$

$$\frac{1}{2}a_4=\frac{1}{2}\left(2-2^{-4}\right)=1-2^{-5}$$

Now, if we solve both equations for $2^{-5}$ and equate the results, we obtain:

$$2^{-5}=2-a_5=1-\frac{1}{2}a_4$$

And from this, by solving for $a_5$, we get:

$$a_5=\frac{1}{2}a_4+1$$

Another way to proceed would be to begin with the given recursion and compute the following values:

$$a_1=\frac{3}{2},\,a_2=\frac{7}{4},\,a_3=\frac{15}{8},\,a_4=\frac{31}{16}$$

Then use:

$$a_5=a_4+\frac{1}{32}=\frac{1}{2}a_4+\frac{1}{2}\cdot\frac{31}{16}+\frac{1}{32}=\frac{1}{2}a_4+1$$

 

[Evgeny.Makarov]

It is also possible to say
\[
a_{n+1}=a_n+\frac{1}{2^{n+1}}\implies a_5=a_4+\frac{1}{2^5}=2\frac{a_4}{2}+\frac{1}{2^5}.
\]