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Harder 3U question (1 Viewer)
- Thread starter apollo1
- Start date
bobthebuilder94
Member
I'd like to know how you did ii if thats possible 
{hopefully that isn't the one you can't get}
{hopefully that isn't the one you can't get}
apollo1
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f(x) is greater than or equal to zero (which can be deduced from the fact that a1 a2 ... are all greater than zero)
therefore the discriminant is less than or equal to zero (positive definite), hence you solve for b^2-4ac less than or equal to zero with the expression found in (i).
therefore the discriminant is less than or equal to zero (positive definite), hence you solve for b^2-4ac less than or equal to zero with the expression found in (i).
bobthebuilder94
Member
hey @hup, was wondering if you could post how to do ii, still have no clue where the sigma comes in from :S
From ii)
]\geq \frac{1}{n}([\sum_{k=1}^{n}(a_{k})])^{2})
now, let
sub this in:
^{2}]\geq \frac{1}{n}([\sum_{k=1}^{n}(2k-1)])^{2})
Expand each series out as follows:
-1)^{2}+(2(2)-1)^2+...+(2n-1)^2\geq \frac{1}{n}([(2(1)-1)+(2(2)-1)+...+(2n-1)])^{2})
This simplifies to:
^{2}+(3)^2+...+(2n-1)^2\geq \frac{1}{n}([(1)+(3)+...+(2n-1)])^{2})
Now the RHS bracket is just a simple AP and turns to be:
^{2}+(3)^2+...+(2n-1)^2\geq \frac{1}{n}([\frac{n}{2}(1+(2n-1)])^{2}\: \: using\: \: S_{n}=\frac{n}{2}(a+l))
This simplifies to:
^{2}+(3)^2+...+(2n-1)^2\geq \frac{1}{n}(n^{2})^{2})
which then gives us our first result:
^{2}+(3)^2+...+(2n-1)^2\geq n^{3})
For the second inequality let:
^2)
Using the inequality from part ii) this becomes:
^{4}]\geq \frac{1}{n}([\sum_{k=1}^{n}(2k-1)^{2}])^{2})
Expanding the LHS gives:
^{4}\geq \frac{1}{n}([\sum_{k=1}^{n}(2k-1)^{2}])^{2})
using the result from the first of this question we get:
^{2}\geq n^{3})
Subbing this in to the new inequality gives:
^{4}\geq \frac{1}{n}(n^{3})^{2})
which gives us our desired result:
^{4}\geq n^{5})
now, let
sub this in:
Expand each series out as follows:
This simplifies to:
Now the RHS bracket is just a simple AP and turns to be:
This simplifies to:
which then gives us our first result:
For the second inequality let:
Using the inequality from part ii) this becomes:
Expanding the LHS gives:
using the result from the first of this question we get:
Subbing this in to the new inequality gives:
which gives us our desired result:
apollo1
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cool. tnks for the solution.From ii)
now, let
sub this in:
Expand each series out as follows:
This simplifies to:
Now the RHS bracket is just a simple AP and turns to be:
This simplifies to:
which then gives us our first result:
For the second inequality let:
Using the inequality from part ii) this becomes:
Expanding the LHS gives:
using the result from the first of this question we get:
Subbing this in to the new inequality gives:
which gives us our desired result:
Hey the sigma is just the expansion of the f(x). So;
Now as the equation is always above the x-axis or has a double root (ie, it just touches the axis), then the discriminant will be