A difficult sequence

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santo3vong

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This is a calculus text in our school. I'm stuck at question 2. Find the coefficient of x2
 
First i use this sequence to discribe it
Screenshot_2020-02-29-12-25-45-780_com.android.chrome.jpg
And don't know how to do next. Can someone help me
 
It will help us if you can translate the problem to English; I can guess what it asks, but am not entirely sure what it is that is repeated n times.

Your work is a good start, but the third line is wrong, and the fourth doesn't follow from it; you probably didn't type what you meant, or else didn't check what you wrote.

For the next step, I'd like to know what methods you have learned for working with recursive sequences.
 
Dr.Peterson said:
It will help us if you can translate the problem to English; I can guess what it asks, but am not entirely sure what it is that is repeated n times.

Your work is a good start, but the third line is wrong, and the fourth doesn't follow from it; you probably didn't type what you meant, or else didn't check what you wrote.

For the next step, I'd like to know what methods you have learned for working with recursive sequences.
Click to expand...
\[\begin{align}
& \text{i }\!\!'\!\!\text{ m sorry that yesterday i type it on the android and it }\!\!'\!\!\text{ s so difficult to type all the problems}\text{.} \\
& \text{Now on my laptop}\text{, i will type them all}\text{.} \\
& \\
& \text{i }\!\!'\!\!\text{ m stuck at problem 2}\text{, 3 and 4} \\
& \\
& Question\text{ 2: Consider the expression}\text{.} \\
& {{\text{(}...{{\text{(((x-2}{{\text{)}}^{2}}-2{{)}^{2}}-2)}^{2}}-...2)}^{2}}\text{ (square n times)} \\
& \text{Find the coefficient of }{{\text{x}}^{2}}. \\
& \\
& Question3:\text{Proof that the limit below converges:} \\
& \underset{B\to \infty }{\mathop{\lim }}\,\int\limits_{0}^{B}{\sin (x)\sin ({{x}^{2}})dx} \\
\end{align}\]
 
#3 Why would it converge? It shows no signs of doing so.
 
Dr.Peterson said:
It will help us if you can translate the problem to English; I can guess what it asks, but am not entirely sure what it is that is repeated n times.

Your work is a good start, but the third line is wrong, and the fourth doesn't follow from it; you probably didn't type what you meant, or else didn't check what you wrote.

For the next step, I'd like to know what methods you have learned for working with recursive sequences.
Click to expand...
f.gif
 
I too am surprised that the integral diverges.
What have you tried? Please post the work you have done so how and we can give you some hints.
 
Look at the following and post your results. Hopefully we will see pattern for the coefficient for x^2.
n=1: (x-2)^2
n=2: ((x-2)^2-2)^2
n=3: (((x-2)^2-2)^2-2)^2
 
Jomo said:
Look at the following and post your results. Hopefully we will see pattern for the coefficient for x^2.
n=1: (x-2)^2
n=2: ((x-2)^2-2)^2
n=3: (((x-2)^2-2)^2-2)^2
Click to expand...
I do something upper, please check it out and maybe you have an idea
 
I'm trying to make a joint recursion of the coefficients of [MATH]1[/MATH], [MATH]x[/MATH], and [MATH]x^2[/MATH], using the fact that only these three are needed to determine what they are for the next step. That is, if we call the nth polynomial [MATH]P_n(x) = a_n + b_nx + c_n x^2 + ...[/MATH], then [MATH]P_{n+1}(x) = ((a_n-2) + b_nx + c_n x^2 + ...)^2[/MATH]. The sequences [MATH]a_n[/MATH] and [MATH]b_n[/MATH] turn out to be simple; [MATH]c_n[/MATH], which we want, will be just a little more complicated.
 
santo3vong said:
I do something upper, please check it out and maybe you have an idea
Click to expand...
The work will give you a quadratic equation which you can solve. However it will not be for n-terms. It will be for infinite terms.

Maybe there are easier ways but try what I am saying above. If you can see a pattern, then we can try to prove that result.
 
Dr.Peterson said:
I'm trying to make a joint recursion of the coefficients of [MATH]1[/MATH], [MATH]x[/MATH], and [MATH]x^2[/MATH], using the fact that only these three are needed to determine what they are for the next step. That is, if we call the nth polynomial [MATH]P_n(x) = a_n + b_nx + c_n x^2 + ...[/MATH], then [MATH]P_{n+1}(x) = ((a_n-2) + b_nx + c_n x^2 + ...)^2[/MATH]. The sequences [MATH]a_n[/MATH] and [MATH]b_n[/MATH] turn out to be simple; [MATH]c_n[/MATH], which we want, will be just a little more complicated.
Click to expand...
Ah my teacher has tought it and it calls "phương pháp hàm sinh"
 
Oh i solve it out. I’ve changed my thought and it effects surprisingly. I won’t use recursion equation here. By treating

2.gif

We will have

\begin{align}

& {{u}_{2}}={{\left( {{u}_{1}}-2 \right)}^{2}}={{\left( {{\left( a+\frac{1}{a} \right)}^{2}}-2 \right)}^{2}}={{\left( {{a}^{2}}+\frac{1}{{{a}^{2}}}+2-2 \right)}^{2}}={{\left( {{a}^{2}}+\frac{1}{{{a}^{2}}} \right)}^{2}} \\

& {{u}_{3}}={{\left( {{u}_{2}}-2 \right)}^{2}}={{\left( {{a}^{4}}-\frac{1}{{{a}^{4}}} \right)}^{2}}={{\left( {{a}^{{{2}^{2}}}}-\frac{1}{{{a}^{{{2}^{2}}}}} \right)}^{2}} \\

& \Rightarrow {{u}_{n}}={{\left( {{a}^{{{2}^{n-1}}}}+\frac{1}{{{a}^{{{2}^{n-1}}}}} \right)}^{2}}\left( 1 \right) \\

\end{align}.

And notice that

\begin{align}

& {{u}_{1}}={{\left( a+\frac{1}{a} \right)}^{2}}\Rightarrow \sqrt{{{u}_{1}}}=a+\frac{1}{a}\Rightarrow a\sqrt{{{u}_{1}}}={{a}^{2}}+1\Rightarrow {{a}^{2}}-a\sqrt{{{u}_{1}}}+1=0 \\

& \Delta ={{b}^{2}}-4ac={{u}_{1}}-4 \\

& \Rightarrow {{a}_{1}}=\frac{\sqrt{{{u}_{1}}}+\sqrt{{{u}_{1}}-4}}{2}\,\,\,\,and\,\,\,{{a}_{2}}=\frac{\sqrt{{{u}_{1}}}-\sqrt{{{u}_{1}}-4}}{2}\, \\

\end{align}

And notice that

3.gif

And place the value to (1), we have

\begin{align}

& {{u}_{n}}={{\left( {{\left( \frac{\sqrt{{{u}_{1}}}+\sqrt{{{u}_{1}}-4}}{2} \right)}^{{{2}^{n-1}}}}+{{\left( \frac{\sqrt{{{u}_{1}}}-\sqrt{{{u}_{1}}-4}}{2} \right)}^{{{2}^{n-1}}}} \right)}^{2}} \\

& ={{\left( {{\left( \frac{\sqrt{x}+\sqrt{x-4}}{2} \right)}^{{{2}^{n-1}}}}+{{\left( \frac{\sqrt{x}-\sqrt{x-4}}{2} \right)}^{{{2}^{n-1}}}} \right)}^{2}} \\

\end{align}

And I still cannot point the coefficient out because the square root, can you help me?
 
Last edited:
santo3vong said:
Ah my teacher has tought it and it calls "phương pháp hàm sinh"
Click to expand...
Dr.Peterson said:
I'm trying to make a joint recursion of the coefficients of [MATH]1[/MATH], [MATH]x[/MATH], and [MATH]x^2[/MATH], using the fact that only these three are needed to determine what they are for the next step. That is, if we call the nth polynomial [MATH]P_n(x) = a_n + b_nx + c_n x^2 + ...[/MATH], then [MATH]P_{n+1}(x) = ((a_n-2) + b_nx + c_n x^2 + ...)^2[/MATH]. The sequences [MATH]a_n[/MATH] and [MATH]b_n[/MATH] turn out to be simple; [MATH]c_n[/MATH], which we want, will be just a little more complicated.
Click to expand...
Jomo said:
The work will give you a quadratic equation which you can solve. However it will not be for n-terms. It will be for infinite terms.

Maybe there are easier ways but try what I am saying above. If you can see a pattern, then we can try to prove that result.
Click to expand...

Oh i solve it out. I’ve changed my thought and it effects surprisingly. I won’t use recursion equation here. By treating

View attachment 16939

We will have

\begin{align}

& {{u}_{2}}={{\left( {{u}_{1}}-2 \right)}^{2}}={{\left( {{\left( a+\frac{1}{a} \right)}^{2}}-2 \right)}^{2}}={{\left( {{a}^{2}}+\frac{1}{{{a}^{2}}}+2-2 \right)}^{2}}={{\left( {{a}^{2}}+\frac{1}{{{a}^{2}}} \right)}^{2}} \\

& {{u}_{3}}={{\left( {{u}_{2}}-2 \right)}^{2}}={{\left( {{a}^{4}}-\frac{1}{{{a}^{4}}} \right)}^{2}}={{\left( {{a}^{{{2}^{2}}}}-\frac{1}{{{a}^{{{2}^{2}}}}} \right)}^{2}} \\

& \Rightarrow {{u}_{n}}={{\left( {{a}^{{{2}^{n-1}}}}+\frac{1}{{{a}^{{{2}^{n-1}}}}} \right)}^{2}}\left( 1 \right) \\

\end{align}.

And notice that

\begin{align}

& {{u}_{1}}={{\left( a+\frac{1}{a} \right)}^{2}}\Rightarrow \sqrt{{{u}_{1}}}=a+\frac{1}{a}\Rightarrow a\sqrt{{{u}_{1}}}={{a}^{2}}+1\Rightarrow {{a}^{2}}-a\sqrt{{{u}_{1}}}+1=0 \\

& \Delta ={{b}^{2}}-4ac={{u}_{1}}-4 \\

& \Rightarrow {{a}_{1}}=\frac{\sqrt{{{u}_{1}}}+\sqrt{{{u}_{1}}-4}}{2}\,\,\,\,and\,\,\,{{a}_{2}}=\frac{\sqrt{{{u}_{1}}}-\sqrt{{{u}_{1}}-4}}{2}\, \\

\end{align}

And notice that

View attachment 16940

And place the value to (1), we have

\begin{align}

& {{u}_{n}}={{\left( {{\left( \frac{\sqrt{{{u}_{1}}}+\sqrt{{{u}_{1}}-4}}{2} \right)}^{{{2}^{n-1}}}}+{{\left( \frac{\sqrt{{{u}_{1}}}-\sqrt{{{u}_{1}}-4}}{2} \right)}^{{{2}^{n-1}}}} \right)}^{2}} \\

& ={{\left( {{\left( \frac{\sqrt{x}+\sqrt{x-4}}{2} \right)}^{{{2}^{n-1}}}}+{{\left( \frac{\sqrt{x}-\sqrt{x-4}}{2} \right)}^{{{2}^{n-1}}}} \right)}^{2}} \\

\end{align}

And I still cannot point the coefficient out because the square root, can you help me?
 
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