# X0 value for Newton's forward interpolation formula

1. Oct 27, 2015

### Raghav Gupta

If x values are: 10, 20, 30, 40, 50
corresponding y values are:20, 65, 180, 390, 505
then what is y value at x=25.
I was having an argument with maths sir. I was saying to take x0 as 10 to get the answer as 107.14, which is correct application of formula,
but our maths sir said that we should take x0 as 20, since it is closest value to 25.
According to me in formula we should take first value incase if it is not backward interpolation formula as otherwise then we are missing then the 4th difference in difference table.

2. Oct 27, 2015

### andrewkirk

According to Wolfram, you should use 20 as your base, because the interpolation point should not be more than one step away from the base. That is, you are trying to estimate $f(x_0+a)$ and it is required that $a\in[0,1]$. Dividing the x values by 10 to give steps of 1, we find that the only value of $x_0$ that can give $a$ in that range is $x_0=2, a=0.5$, which corresponds to using the original value of 20 as a base.

You are right that doing that loses the ability to use the 4th diff. But that is more than made up for by the fact that your first difference will be more accurate because you start three times closer to the interpolation point.

3. Oct 28, 2015

### Raghav Gupta

But when we will compute the interpolating polynomial, and then give value as 25, definitely the answer would come taking 10 as the base.

4. Oct 28, 2015

### andrewkirk

I'm afraid I don't understand what you are saying.

If you are saying that it's possible to do the problem using 10 as the base, you're right.

It's just that in general it will be expected to be more accurate if you use a base that's as close as possible to, but not greater than, the interpolation point. In this case that base is x=20.

5. Oct 28, 2015

### Raghav Gupta

Can you explain why to take base as 20 and why it would be more accurate?
Can you give a graph or equation related to these data points?

6. Oct 28, 2015

### andrewkirk

The higher powers of $\Delta$ in the formula are used to compensate for curvature in the function. They correct for the fact that, the further one moves away from the base point, the less accurate the linear approximation, which only uses the first power of $\Delta$, becomes.

If the function can be expressed as a power series (this is called being an 'analytic' function), the formula will give an exactly correct answer, no matter how far we are away from the base point, provided we use all of the infinitely many powers of $\Delta$. Since we can't do that, the approximation will be imperfect, so the curvature adjustment will never be complete.

If we use 10 as base point we have four powers of $\Delta$ to use, compared to three powers if we use 20 as base. That extra power needs to compensate for the curvature that occurs between points 10 and 20. But it's only partial compensation. To fully compensate we'd need all the infinitely many terms. So instead we take all of that curvature out of the equation by moving the base point closer to the interpolation point.

This is a verbal explanation, intended to give intuition. I'm confident a formal mathematical explanation could be given, but it would be long to write and long to read, so I haven't done one.

7. Oct 28, 2015

### Raghav Gupta

Suppose we initially don't know for what value of x we have to find y value
and we find y in terms of x by the given data points.
Now put x=25 in equation, then definitely a different value would come for y, isn't it? Since we have taken base as 10.

8. Oct 28, 2015

### andrewkirk

If you don't know what your interpolation value will be then, other things being equal, you should choose the base to be the one that gives you the greatest number of available powers of $\Delta$. But that's a different question from the one posed in the OP.

9. Oct 28, 2015

### Raghav Gupta

Yes,suppose now we have calculated the polynomial in terms of x. Now we give the x value as 25. Is this correct application or not?

Also in this problem instead of Newton interpolation formula should we apply Central difference formula?