# Terms of a Sequence

Construct a sequence of interpolating values $Y_n, to,f(1 + \sqrt{10})$, where $f(x) = (1 + X^2)^{-1}$ for $-5 \leq X \leq 5$, as follows: For each n = 1,2, ... ,10, let $h = \frac{10} {n}$

$x_j^n= -5 + jh$, for each j = 0, 1,2, ... ,n. What would my sequence be exactly. I have come up with something but it seems incorrect.

The first term would be : $-5+0*\frac{10} {1}$

The second ther would be : $-5+1*\frac{10} {2}$

And so forth. But I have a feeling that this is incorrect.

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what kind of interpolation are you using? –  robjohn Sep 19 '12 at 14:56
Lagrange interpolation / Neville's Method but that is the easy part. Once I figure out what my x-values are then the rest is quite simple. –  math101 Sep 19 '12 at 15:03

It seems as if you are being asked to use $n$ subintervals (therefore, $n+1$ points) to interpolate the function $f$ on $[-5,5]$. Since the width of the interval is $10$, the width of each subinterval is $h=\frac{10}n$. The mesh for $n$ intervals consists of the points $x_j^n=-5+jh=-5+j\frac{10}n$ for $j=0,1,2,\dots,n$.

For a particular $n$, compute $f(x_j^n)$ for $j=0,1,2,\dots,n$ and interpolate to get $f(1+\sqrt{10})$.

For example, if $n=4$, use the $x$ values $\{-5,-2.5,0,2.5,5\}$ and the $y$ values $\{f(-5),f(-2.5),f(0),f(2.5),f(5)\}$.

The interpolating polynomial is $1-\frac{129}{754}x^2+\frac{2}{377}x^4$ and when that is evaluated at $1+\sqrt{10}$, we get $-\frac1{754}(21+82\sqrt{10})\approx-.3717596394$, which is the fourth value your book gives.

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Ok here is the answer that is in the back of my book: The first 10 terms of the sequences –  math101 Sep 19 '12 at 21:36
The first 10 terms of the sequence are $.038462, .333671, .116605, -.371760, -.0548919, .605935, .190249, -.513353, -.0668173, .448335 \text{ and } f(1+ \sqrt{10})=.0545716$ –  math101 Sep 19 '12 at 21:47
As you can see these 10 terms are in reality the y-values or f(x). They have been plugged into the equation $(1+x^{2})^{-1}$ –  math101 Sep 19 '12 at 21:52
@math101: how are you getting negative values for $(1+x^2)^{-1}$ ? –  robjohn Sep 19 '12 at 22:03
@math101: The answers in the back of the book are for the interpolated values of $f(1+\sqrt{10})$ using $n=1,2,3,\dots,10$. For $n=1$ you are only using $f(-5)=\frac1{26}$ and $f(5)=\frac1{26}$, so the value of $f(1+\sqrt{10})=\frac1{26}$. –  robjohn Sep 19 '12 at 22:20

You want $11$ equally spaced points from $-5$ to $5,$ which are all the integers: $-5, -4, -3, \ldots ,5$. $h$ should be one value, here $1$, which is the spacing of the points. It should be the length of the interval $(10)$ divided by the number of spaces $(10)$. I don't know where you got $h=\frac{10}n.$ So it should be $-5+0*1, -5+1*1, \ldots -5+10*1$

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Your answer seems correct but then it doesn't match the answer I have found unless the answer is incorrect. I wonder –  math101 Sep 19 '12 at 5:19
$h=\frac{10} {n}$ was given in the question –  math101 Sep 19 '12 at 14:18
@math101: that is fine, but it should not be "For each $n=1,2,\ldots ,10$ because $n$ doesn't vary. Maybe this fragment should come before $x_j^n=-5+jh$ –  Ross Millikan Sep 19 '12 at 14:40
For some reason it doesnt match the answers given in my book. For some reason the question was worded unclearly and its hard to put the pieces together –  math101 Sep 19 '12 at 14:52