Problem 8 If $f(x+1)+f(x-1)=2 f(x)$ and ... [FREE SOLUTION]

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Rejaul Makshud

$Math Studyset 91影视 Explanations$ Math

2019 Edition

Chapter 1: Problem 8

If $f(x+1)+f(x-1)=2 f(x)$ and $f(0)=0$ then find $f(n), n \in N .$

Short Answer

Expert verified

After performing all the steps, it is found that $f(n) = 0$ for all $n \in N$. This is the simplest form of the function.

Step by step solution

Unfolding the recurrence relation

First, it is given that $f(0)=0$. Now, try to find $f(1)$, which can be obtained by setting $x=0$. This will give $f(1)+f(-1)=2 f(0)$. Replacing $f(0)=0$, we get $f(1)= -f(-1)$

Find the value of $f(-1)$

Now, to find the value of $f(-1)$, replace $x$ by $-1$ in the original functional relation. We get $f(0)+f(-2)=2 f(-1)$. Using $f(0)=0$, we are left with $f(-2)=-2 f(-1)$. But due to the property derived in Step 1, we can replace $-f(-1)$ by $f(1)$. So we have $f(-2)=-2 f(1)$

Note the pattern and generalize

Observe that we have three equations up to now, namely $f(1)=-f(-1)$, $f(-1)=f(1)$, and $f(-2)=-2 f(1)$. We can see that whether n is positive or negative, always $f(n)= (-1)^n * n * f(1)$. This can serve as the general formula for $f(n)$ for all $n \in N$

Normalizing the function

Inserting $x=1$ to the original functional relation we get $f(2)+f(0)=2 f(1)$. Then by substituting the given $f(0)=0$ and the derived expression for $f(2)= -2 f(1)$ we obtain on simplifying, $f(1)=0$.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Recurrence Relations

Recurrence relations form a crucial part of functional equations, allowing us to express sequences or functions based on previous terms. In our exercise, we use the relation $f(x+1) + f(x-1) = 2f(x)$ to find expressions for unknown terms. This relation tells us how each term in the sequence can be derived from its neighbors.

To use this concept effectively:

Start with any given initial values, such as $f(0) = 0$.
Using the recurrence formula, express further terms. When $x = 0$, the equation simplifies to $f(1) = -f(-1)$.
Substitute back to find other terms, like $f(-1)$, $f(2)$, etc.

By unfolding the relation, we systematically expose more terms, eventually leading to a general expression.

Mathematical Induction

Mathematical induction is a proof technique to confirm the validity of a proposition for all natural numbers by first proving it for an initial value and then for a general step. In our problem, though not explicitly used, it informs the way we validate patterns noticed in recurrence relations.

For practical steps using induction:

Verify the proposition for a base case: here, $n=0$.
Assume the expression holds for $n=k$.
Prove it holds for $n=k+1$, using the relation to extend from $f(k)$ to $f(k+1)$.

While unfolding terms and spotting patterns frequently relates to induction, it helps reassure that each step logically follows, suggesting a reliable formula.

General Formula Deduction

The deduction of a general formula allows us to represent an entire sequence with a single expression. In this exercise, recognizing the pattern across derived terms helps deduce that $f(n) = (-1)^n \cdot n \cdot f(1)$ serves as a general formula.

To deduce such formulas:

Identify any consistent patterns from specific cases, such as the sign and magnitude pattern.
Formulate a general expression capturing these findings.
Verify with initial conditions or logical examination.

Here, once it was set $f(1)=0$ through normalization, the constant part of the formula maintained coherence across terms. This step finalizes the function's description over all relevant natural numbers.

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91影视

If \(f(x+1)+f(x-1)=2 f(x)\) and \(f(0)=0\) then find \(f(n), n \in N .\)

Short Answer

Step by step solution

Unfolding the recurrence relation

Find the value of \(f(-1)\)

Note the pattern and generalize

Normalizing the function

Key Concepts

Recurrence Relations

Mathematical Induction

General Formula Deduction

One App. One Place for Learning.

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