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:

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

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

 

[MarkFL]

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

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

The associated homogeneous equation is:

\(\displaystyle a_{n+1}-a_{n}=0\)

Now, we see the characteristic equation is:

\(\displaystyle r-1=0\implies r=1\)

Thus we know the homogeneous solution is given by:

\(\displaystyle h_n=c_1\cdot1^n=c_1\)

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

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

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):

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

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

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

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

\(\displaystyle A-2A=1\implies A=-1\)

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

\(\displaystyle a_n=h_n+p_n=c_1-\frac{1}{2^n}\)

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

\(\displaystyle a_0=c_1-\frac{1}{2^0}=1\implies c_1=2\)

Hence, the closed form can be written as:

\(\displaystyle a_n=2-2^{-n}\)

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:

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

The associated homogeneous equation is:

\(\displaystyle a_{n+1}-a_{n}=0\)

Now, we see the characteristic equation is:

\(\displaystyle r-1=0\implies r=1\)

Thus we know the homogeneous solution is given by:

\(\displaystyle h_n=c_1\cdot1^n=c_1\)

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

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

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):

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

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

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

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

\(\displaystyle A-2A=1\implies A=-1\)

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

\(\displaystyle a_n=h_n+p_n=c_1-\frac{1}{2^n}\)

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

\(\displaystyle a_0=c_1-\frac{1}{2^0}=1\implies c_1=2\)

Hence, the closed form can be written as:

\(\displaystyle a_n=2-2^{-n}\)

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:

\(\displaystyle a_5=2-2^{-5}\)

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

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

\(\displaystyle 2^{-5}=2-a_5=1-\frac{1}{2}a_4\)

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

\(\displaystyle a_5=\frac{1}{2}a_4+1\)

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

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

Then use:

\(\displaystyle a_5=a_4+\frac{1}{32}=\frac{1}{2}a_4+\frac{1}{2}\cdot\frac{31}{16}+\frac{1}{32}=\frac{1}{2}a_4+1\)

 

[Evgeny.Makarov]

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