Further Maths - The Method of Differences
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What is the requirement of a series for the method of differences to be applicable?
The general term, $u _ r$ of a series can be expressed in the form $f(r) - f(r+1)$.
If the general term of a series $u _ r$ can be expressed as $f(r) - f(r + 1)$, how could you write the series?
For a general series $\sum^n _ {r = 1} (f(r) - f(r + 1))$, what is $u _ 1$?
For a general series $\sum^n _ {r = 1} (f(r) - f(r + 1))$, what is $u _ 2$?
For a general series $\sum^n _ {r = 1} (f(r) - f(r + 1))$, what is $u _ r$?
For a general series $\sum^n _ {r = 1} (f(r) - f(r + 1))$, what are the first and last few terms of the series?
\[\begin{align*} +\ f(1) &- f(2) \\ +\\ f(2) &- f(3) \\ +\ f(3) &- f(4) \\ \dots \\ +\ f(n) &- f(n + 1) \end{align*}\]
What does this cancel down to?
2021-01-28
If you’re doing a method of differences question with three partial fractions, what is probably true that means they cancel out?
The fractions along the diagonals add up to something that is cancelled out.
\[\frac{1}{3} - \frac{1}{2(n + 1)}\]
What is the value of this expression as $n \to \infty$?
If asked to find the limit of a series after a Method of Differences question, should you combine the fractions or leave them seperated?
Leave them seperated.
\[\frac{1}{2(n + 1)}\]
How would you write what this is equal to in an exam?
As $n \to \infty$
\[\frac{1}{2(n+1)} \to 0\]2022-05-15
\[\frac{1}{1} + \frac{1}{3} - \frac{1}{2}\]
\[\frac{1}{2} + \frac{1}{6} - \frac{1}{4}\]
\[\frac{1}{3} + \frac{1}{9} - \frac{1}{8}\]
How would you notice these fractions cancelling out for the method of differences?
They cancel along the diagonal.