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Well it's true if $-\lambda$ is not in the spectrum of $X^TX$: \begin{align*} (X^TX+\lambda I)^{-1}X^TX+\lambda(X^TX+\lambda I)^{-1}&=(X^TX+\lambda I)^{-1}X^TX+(X^TX+\lambda I)^{-1}\cdot\lambda I\\ &=(X^TX+\lambda I)^{-1}(X^TX+\lambda I) \\ &=I. \end{align*} Then subtracting $\lambda(X^TX+\lambda I)^{-1}$ from both sides, we obtain the result. ...

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tl;dr This is a well known problem with several suggested solutions each of which can be characterized as fair or unfair depending on your definition of "fair". Some have formulas, others are algorithms that work from the data. Here are some of many links to the literature, focused on the application of your problem to the apportionment of seats in ...

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Read about Least Square Fitting. I think that this is what you need. To sum it up briefly, I'd say the you should consider which type of function you expect the approximation would be (linear, polinomial, etc.) and then, using LSF, you should approximate this function with the observations you have. Hope you'll find it helpful.

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Except for a=0.25, all the curves look pretty linear. You might try a standard linear least squares fit of each data set to get y = mx+b and see what m and b look like as functions of a. Do you have a model of what the curves should look like? If you do, I think you should try to fit that type of curve to each data set.

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At year 1, Let T(1) be the Tuition and there is a probability that the government cancels the tuition in which case you would have paid only 10,000 at the start. At year 4, Let T(4) be the Tuition and there is a probability that the government cancels the tuition in which case you would have paid $$\frac{1.02^4-1}{1.02-1}$$ In a similar argument T(8) ...

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Consider the statement If I look at an individual particle entering the chamber at time $0$, it is travelling $Ce^{−At}$ metres per second after $t$ seconds. The constants $C$ and $A$ are the same for every particle. If $t=0$, the speed would be $C\cdot e^{−A*0}=C\cdot e^0=C$. Therefore, $v_i=C$ (Thanks @Semiclassical for pointing this out before). So ...

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