## i have two array x=[1 6 8 9],y=[3 6 9 14],then i plot (x,y) in matlab. i have no linear equation of x,y. i have only above array of x and y. if value of x is 4 then what is the value of y. is it possible using logic?

Asked by vipul utsav

### vipul utsav (view profile)

on 29 Nov 2012
Accepted Answer by Matt Fig

### Matt Fig (view profile)

i have two array x=[1 6 8 9],y=[3 6 9 14],then i plot (x,y) in matlab. i have no linear equation of x,y. i have only above array of x and y. if value of x is 4 then what is the value of y.

is it possible using logic?

PK

### PK (view profile)

on 29 Nov 2012

please be clear with you question ,, what exactly do u require or to execute?

vipul utsav

### vipul utsav (view profile)

on 29 Nov 2012

'interp1' this is inbuilt function ,so i just want logic behind it

Azzi Abdelmalek

### Azzi Abdelmalek (view profile)

on 29 Nov 2012

What result are you expecting?

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### Matt Fig (view profile)

Answer by Matt Fig

### Matt Fig (view profile)

on 29 Nov 2012
```x=[1 6 8 9];
y=[3 6 9 14];
interp1(x,y,4,'linear')
```

Matt Fig

### Matt Fig (view profile)

on 29 Nov 2012

Use the same approach, simply with the new values. You can use any of the interpolating methods available int INTERP1. Have a look at the help:

```help interp1
```
vipul utsav

### vipul utsav (view profile)

on 29 Nov 2012

yes i read help,but for 11 inter11 doesn't work. and for 4 it is work. 11 is outside of maximum value in x.

Matt Fig

### Matt Fig (view profile)

on 29 Nov 2012

If you read the help, you would see there is an extrapolation option. When you want to look at values outside the range, you are extrapolating. You should also know that extrapolating is usually something one does not want to do for real applications.... especially for so sparsely populated data as this!

```interp1(x,y,11,'linear','extrap')
```

### Jan Simon (view profile)

Answer by Jan Simon

### Jan Simon (view profile)

on 29 Nov 2012
Edited by Jan Simon

### Jan Simon (view profile)

on 29 Nov 2012

There is an infinite number of possible values: You can e.g. define an infinite number of polynomials through these points. And there are much more functions than only polynomials.

Therefore this problem cannot be solved, when there are no further restrictions.

One workaround would be to choose 19. This is not really serious, but fast and it solves the question as good as any other value also. The choice of 19 is explained in Carl E. Liderholm: "Mathematics Made Difficult", 1972:

```Each sequence of natural number can be continued by 19.
```