How do you interpolate values on a time series to get a function (equation)?

Question

I have following time series:

1   10000
2   
3
4   20000
5   
6   
7
8   
9   
10  
11  
12
13  
14  
15  
16  200000

And I need to interpolate those values to get the rest of them. How do I do that?

Answer 1

From the comments, it sounds like what you want is a best-fit (as opposed to exact-fit) exponential function. The usual approach to this would be as follows:

Note that if $x,y$ values followed (exactly) the relationship $y = ae^{bx}$, then we would have $$ y = ae^{bx} \implies \log(y) = \ln(a) + bx \leadsto z = a_0 + bx. $$ In other words, the variables $x$ and $z = \ln(y)$ satisfy the relationship $a_0 + bx$ for some numbers $a_0,b$. These numbers would have to satisfy the equation $$ \pmatrix{1&1\\1&4\\1&16} \pmatrix{a_0\\b} = \pmatrix{\ln(10000)\\ \ln(20000)\\ \ln(200000)}. $$ As it turns out, this equation has no exact solution. However, we can obtain the least-squares solution by solving the associated equation $$ \pmatrix{1&1\\1&4\\1&16}^T\pmatrix{1&1\\1&4\\1&16} \pmatrix{a_0\\b} = \pmatrix{1&1\\1&4\\1&16}^T\pmatrix{\ln(10000)\\ \ln(20000)\\ \ln(200000)}. $$ Solving this equation leads to the answer $a_0 = 9.0576, b = 0.1975$. The associated value $a$ is given by $$ \ln(a) = a_0 \implies a = e^{a_0} = 8583.5, $$ which leads to the best-fit model $$ y = 8583.5 \cdot e^{0.1975 \cdot x}. $$