Regression with inverse independent variable Let's suppose I have a $N$-vector $Y$ of dependent variables, and an $N$-vector $X$ of independent variable. When $Y$ is plotted against $\frac{1}{X}$, I see that there is a linear relationship (upward trend) between the two. Now, this also means that there is a linear downward trend between $Y$ and $X$.
Now, if I run the regression: $Y = \beta * X + \epsilon$
and get the fitted value $\hat{Y} = \hat{\beta}X$
Then I run the regression: $Y = \alpha * \frac{1}{X} + \epsilon$ and get the fitted value  $\tilde{Y} = \hat{\alpha} \frac{1}{X}$
Will the two predicted values, $\hat{Y}$and $\tilde{Y}$ be approximately equal?
 A: I see no reason for them to be "approximately equal" in general -- but what exactly do you mean by approximately equal?
Here's a toy example:
library(ggplot2)
n <- 10^3
df <- data.frame(x=runif(n, min=1, max=2))
df$y <- 5 / df$x + rnorm(n)
p <- (ggplot(df, aes(x=x, y=y)) +
      geom_point() +
      geom_smooth(method="lm", formula=y ~ 0 + x) +  # Blue, OP's y hat
      geom_smooth(method="lm", formula=y ~ 0 + I(x^-1), color="red"))  # Red, OP's y tilde
p

The picture:

The "blue" model would do much better if it were allowed to have an intercept (i.e. constant) term...
A: 
 When Y is plotted against $\frac{1}{X}$, I see that there is a linear relationship (upward trend) between the two. Now, this also means that there is a linear downward trend between Y and X

The last sentence is wrong: there is a downward trend, but it is by no means linear:
 
I used a $f(x) = \frac{1}{x}$ as function plus a bit of noise on $Y$. As you can see, while plotting $Y$ against $\frac{1}{X}$ yields a linear behaviour, $Y$ against $X$ is far from linear. 
(@whuber points out that the $Y$ against $\frac{1}{X}$ plot doesn't look homoscedastic. I think it appears to have higher variance for low $Y$ because the much higher case density leads to larger range which is essentially what we perceive. Actually, the data is homoscedastic: I used Y = 1 / X + rnorm (length (X), sd = 0.1) to generate the data, so no dependency on the size of $X$.) 
So in general the relationship is very much non-linear. That is, unless your range of $X$ is so narrow that you can approximate $\frac{d \frac{1}{x}}{dx} = - \frac{1}{x^2} \approx const.$ Here's an example: 
 
Bottomline: 


*

*In general, it is very hard to approximate a  $\frac{1}{X}$-type function by a linear or polynomial function. And without offset term you'll never get a reasonable approximation. 

*If the $X$ interval is narrow enough to allow a linear approximation, you'll anyways not be able from the data to guess the relation should be $\frac{1}{X}$ and not linear ($X$).

