zeebobDD
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http://www.boredofstudies.org/courses/maths/3u/1205396875_2006_Mathematics_Extension_1_Notes.pdf can someone help me with Question3? ty
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Mathemaatical Inducction (1 Viewer)
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I'm on my iPhone ATM, but question 3 should be
Done using sum of rectangles under curve and using
The bigger and lower rectangles to bound it. I'll show
You later if someone else doesn't
But if you had to do it using induction it is easiest
to separate the inequality and do induction twice
and then conclude that inequality
Done using sum of rectangles under curve and using
The bigger and lower rectangles to bound it. I'll show
You later if someone else doesn't
But if you had to do it using induction it is easiest
to separate the inequality and do induction twice
and then conclude that inequality
Last edited:
deterministic
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Consider each part of the inequality separately. When we add 
to all parts of inequality, the left part of the inequality becomes:
}+1}{\sqrt{n+1}} \\ &= \frac{\sqrt{n^2+n}+1}{\sqrt{n+1}} \\ &\geq \frac{\sqrt{n^2}+1}{\sqrt{n+1}} \\&=\frac{n+1}{\sqrt{n+1}}\\ &=\sqrt{n+1})
Where the third line is due to the fact
for 
The right hand part of inequality becomes:
}+1}{\sqrt{n+1}} -1\\ &= \frac{2\sqrt{n^2+n}+1}{\sqrt{n+1}} -1\\ &\leq \frac{2\sqrt{n^2+n+\frac{1}{4}}+1}{\sqrt{n+1}}-1 \\&=\frac{2\sqrt{(n+\frac{1}{2})^2}+1}{\sqrt{n+1}}-1\\ &=\frac{2(n+\frac{1}{2})+1}{\sqrt{n+1}}-1 \\&= \frac{2(n+1)}{\sqrt{n+1}}-1 \\ &= 2\sqrt{n+1}-1)
It follows that:
Where the third line is due to the fact
The right hand part of inequality becomes:
It follows that:
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I'll go straight to the inductive proof step for the LHS of the inequality
First we establish a result as follows
}+1 > n+1\\\\$Divide both sides by $\sqrt{n+1}\\\\\therefore\sqrt{n}+\dfrac{1}{\sqrt{n+1}} > \sqrt{n+1})
Now to the induction step
Hence we can claim that
However, the result holds for n = 1 as an equality and so subsequent integers of n will lead to strict inequality, hence the result as a weak inequality is true by induction i.e.
First we establish a result as follows
Now to the induction step
Hence we can claim that
However, the result holds for n = 1 as an equality and so subsequent integers of n will lead to strict inequality, hence the result as a weak inequality is true by induction i.e.
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