MHB On induction as you kindly requested (detailed)

  • Thread starter Thread starter sarrah1
  • Start date Start date
  • Tags Tags
    Induction
AI Thread Summary
The discussion revolves around proving an inequality for a sequence defined by a recurrence relation using mathematical induction. The user presents an initial inequality and attempts to derive a general form for any n, but encounters difficulties in maintaining the correct structure during the induction step. Clarifications are sought regarding the definitions of norms and the correct application of the recurrence relation. The conversation highlights the importance of correctly substituting terms and maintaining consistency in notation throughout the proof process. The user expresses gratitude for the assistance in resolving these mathematical challenges.
sarrah1
Messages
55
Reaction score
0
Hi Evgeny

Thank you very much for your reply.

I will try to be more clear this time and in detail. The reason why it was not as clear, as it's lengthy. I have the following

$|{y}_{n}(x)|\le\int_{a}^{b} \,|k(x,s)| |{y}_{n-1}(s)| ds+\int_{a}^{x} \,|g(x,s)| |{y}_{n-1}(s)| ds$ , $x\in[a,b]$ $a,b$ real ... (1)

By defining operator $K$

$Ky=\int_{a}^{b} \,k(x,s) y(s) ds$

I define norms

$||K||={max}_{a<x<b}\int_{a}^{b} \,|k(x,s)|ds $ , $M={max}_{a<x<b}|g(x,s)|$ , $||y||={max}_{a<x<b}|y(x|$

From (1) $|{y}_{1}|\le||{y}_{0} ||K||+||{y}_{0}||M(x-a)$

Another iteration gives

$|{y}_{2}|\le||{y}_{0}|| {||K||}^{2}+||{y}_{0}|| ||K||M(x-a)+||{y}_{0}|| M||K|| (b-a)+||{y}_{0}||{M}^{2}\frac{{(x-a)}^{2}}{2!}$

In general for any $n$

$|{y}_{n}(x)|\le\left\{||K||\left(\sum_{j=0}^{n-1}{(||K||+M(b-a))}^{n-j-1}{M}^{j}\frac{{(x-a)}^{j}}{j!}\right)+{M}^{n}\frac{{(x-a)}^{n}}{n!}\right\}||{y}_{0}||$ ... (2)

MY PROBLEM STARTS HERE:

Instead of writing the above inequality for $n=3,4,$ etc.. I want to apply mathematical induction to prove (2). Assuming (2) is valid for $n$ I obtain using one further iteration

$|{y}_{n+1}(x)|\le\left\{||K||\left(\sum_{j=0}^{n-1}{(||K||+M(b-a))}^{n-j-1}{M}^{j}\left(||K||\frac{{(b-a)}^{j}}{j!}+M\frac{{(x-a)}^{j+1}}{j+1!}\right)\right)+{M}^{n}\left(||K||\frac{{(b-a)}^{n}}{n!}+M\frac{{(x-a)}^{n+1}}{n+1!}\right)\right\}||{y}_{0}||$ ... (3)

But since $(x-a)\le(b-a)$

$|{y}_{n+1}(x)|\le\left\{||K||\left(\sum_{j=0}^{n-1}{(||K||+M(b-a))}^{n-j}{M}^{j}\frac{{(b-a)}^{j}}{j!}\right)+{M}^{n}\left(||K||\frac{{(b-a)}^{n}}{n!}+M\frac{{(x-a)}^{n+1}}{n+1!}\right)\right\}||{y}_{0}||$ ... (4)

$\le\left\{||K||\left(\sum_{j=0}^{n}{(||K||+M(b-a))}^{n-j}{M}^{j}\frac{{{(b-a)}^{j}}}{j!}\right)+{M}^{n+1}\frac{{(x-a)}^{n+1}}{n+1!}\right\}||{y}_{0}||$ ... (5)

So whereas I need the form (2) in $x$ to carry the integration of ${(x-a)}^{j}$ , I ended up with (5) in which $(b-a)$ replaced $(x-a)$ in the first inside bracket. usually I must end in induction to exactly the same form except $n -> n+1$. Yet form (5) is really what I need except that I cannot start with $b-a$ instead because when integrated doesn't give $\frac{{(x-a)}^{n}}{n!}$ which what I need

I am really very grateful to sort out where is the problem
Sarrah
 
Physics news on Phys.org
You did not write how $y_{n+1}$ is obtained from $y_n$.

sarrah said:
$M=\max_{a<x<b}|g(x,s)|$
This expression depends on $s$.

sarrah said:
From (1) $|{y}_{1}|\le||{y}_{0} ||K||+||{y}_{0}||M(x-a)$
Is it correct that $|{y}_{n+1}|\le\|{y}_{n} \|K+\|{y}_{n}\|M(x-a)$ for all $n$? Do you mean $\|y_n\|$ instead of $|y_n|$? Also, how come $x$ is not used in the left-hand side of the inequality, but it is used in the right-hand side? Do you mean $|{y}_{1}(x)|\le||{y}_{0} \|K+\|{y}_{0}\|M(x-a)$ for all $x\in[a,b]$?
 
Evgeny.Makarov said:
You did not write how $y_{n+1}$ is obtained from $y_n$.

This expression depends on $s$.

Is it correct that $|{y}_{n+1}|\le\|{y}_{n} \|K+\|{y}_{n}\|M(x-a)$ for all $n$? Do you mean $\|y_n\|$ instead of $|y_n|$? Also, how come $x$ is not used in the left-hand side of the inequality, but it is used in the right-hand side? Do you mean $|{y}_{1}(x)|\le||{y}_{0} \|K+\|{y}_{0}\|M(x-a)$ for all $x\in[a,b]$?

Thank you very much Evgeny for your prompt reply. I did some typing mistakes


$M={max}_{a\le s,x\le b}|g(x,s)|$

I have the recurrence

${y}_{n+1}(x)=\int_{a}^{b} \,k(x,s) {y}_{n}(s)ds +\int_{a}^{x} \,g(x,s) {y}_{n}(s) ds$ (i)

True $|{y}_{n+1}(x)|\le\|{y}_{n} \|K+\|{y}_{n}\|M(x-a)$ , but I WILL NOT use it since I need to integrate again the term $(x-a)$.

It is true that
$|{y}_{1}(x)|\le||{y}_{0} \|K+\|{y}_{0}\|M(x-a)$ , for all $x\in[a,b]$ (ii)
because
$|{y}_{1}(x)|\le\int_{a}^{b} \,|k(x,s)| |{y}_{0}(x)|ds +\int_{a}^{x} \,|g(x,s)| |{y}_{0}(x)|ds$
$\le{max}_{a \le s\le b}|{y}_{0}(s)|\int_{a}^{b} \,|k(x,s)|ds +{max}_{a \le s\le x}|{y}_{0}(s)|\int_{a}^{x} \,|g(x,s)|ds$
i.e. $|{y}_{1}(x)|\le||{y}_{0}||. ||K||+||{y}_{0}|| M(x-a)$

Upon substituting for $|{y}_{1}(s)|$ inside (i) I obtain $|{y}_{2}(x)|$ etc..
many thanks
Sarrah
 
sarrah said:
Thank you very much Evgeny for your prompt reply. I did some typing mistakes


$M={max}_{a\le s,x\le b}|g(x,s)|$

I have the recurrence

${y}_{n+1}(x)=\int_{a}^{b} \,k(x,s) {y}_{n}(s)ds +\int_{a}^{x} \,g(x,s) {y}_{n}(s) ds$ (i)

True $|{y}_{n+1}(x)|\le\|{y}_{n} \|K+\|{y}_{n}\|M(x-a)$ , but I WILL NOT use it since I need to integrate again the term $(x-a)$.

It is true that
$|{y}_{1}(x)|\le||{y}_{0} \|K+\|{y}_{0}\|M(x-a)$ , for all $x\in[a,b]$ (ii)
because
$|{y}_{1}(x)|\le\int_{a}^{b} \,|k(x,s)| |{y}_{0}(x)|ds +\int_{a}^{x} \,|g(x,s)| |{y}_{0}(x)|ds$
$\le{max}_{a \le s\le b}|{y}_{0}(s)|\int_{a}^{b} \,|k(x,s)|ds +{max}_{a \le s\le x}|{y}_{0}(s)|\int_{a}^{x} \,|g(x,s)|ds$
i.e. $|{y}_{1}(x)|\le||{y}_{0}||. ||K||+||{y}_{0}|| M(x-a)$

Upon substituting for $|{y}_{1}(s)|$ inside (i) I obtain $|{y}_{2}(x)|$ etc..
many thanks
Sarrah

sorry I discovered a mistake, it's ${y}_{0}(s)$ and not ${y}_{0}(x)$ inside integral

$|{y}_{1}(x)|\le\int_{a}^{b} \,|k(x,s)| |{y}_{0}(s)|ds +\int_{a}^{x} \,|g(x,s)| |{y}_{0}(s)|ds$

thanks
sarrah
 
Back
Top