Here are my points  $x = 0, y = 10000 \\ x = 1, y = 10421.716 \\ x = 2, y = 10882.302  \\ x = 3, y = 11385.341\\ x = 4, y = 11934.746 \\x = 5, y = 12534.791 \\x = 6, y = 13190.142 \\ x = 7, y = 13905.899\\ x = 8, y = 14687.628$  Here are my values.   The initial value for $y$ is $10000$. As $x$ value increase by $1$, $y$ is multiplied by $1.8^{0.15}$ and subtracted by $500$.  How can I make natural exponential equation with these points? or is it impossible to make natural exponential equation with this?
How do I make exponential equation with points?  Here are my points:  x = 0, y = 0 x = 1, y = 4.75000 x = 2, y = 9.26250 x = 3, y = 13.54937 x = 4, y = 17.62191   this is same as   x = 0, y = 0 x = 1, y = (0 + 5) * 0.95  x = 2, y = ((5 * 0.95) + 5) * 0.95 x = 3, y = (((5 * 0.95) + 5) * 0.95 + 5)) * 0.95 x = 4, y = ((((5 * 0.95) + 5) * 0.95 + 5)) * 0.95 + 5) ) 0.95   as x value increase by 1, y value is added by 5 and multiplied by 0.95 i tried to make an equation for this but i couldn't  also i tried to use natural exponential but the initial value is 0 so it didn't work  please help me
I'd like to do something like this:  &gt;&gt; foo = @() functionCall1() functionCall2()   So that when I said:  &gt;&gt; foo()   It would execute functionCall1() and then execute functionCall2().  (I feel that I need something like the )  EDIT:  functionCall1 and functionCall2 are not necessarily functions that return values. Prove by Mathematical Induction: $1(1!) + 2(2!) + \cdot \cdot \cdot +n(n!) = (n+1)!-1$ Prove by Mathematical Induction . . .  $1(1!) + 2(2!) + \cdot \cdot \cdot +n(n!) = (n+1)!-1$  I tried solving it, but I got stuck near the end . . .  a. Basis Step:      $(1)(1!) = (1+1)!-1$      $1 = (2\cdot1)-1$      $1 = 1 \checkmark$   b. Inductive Hypothesis     $1(1!) + 2(2!) + \cdot \cdot \cdot +k(k!) = (k+1)!-1$   Prove k+1 is true.     $1(1!) + 2(2!) + \cdot \cdot \cdot +(k+1)(k+1)! = (k+2)!-1$      $\big[RHS\big]$      $(k+2)!-1 = (k+2)(k+1)k!-1$      $\big[LHS\big]$      $=\underbrace{1(1!) + 2(2!) + \cdot \cdot \cdot + k(k+1)!} + (k+1)(k+1)!$ (Explicit Last Step)      $= \underbrace{(k+1)!-1}+(k+1)(k+1)!$ (Inductive Hypothesis Substitution)      $= (k+1)!-1 + (k+1)(k+1)k!$      $= (k+1)k!-1 + (k+1)^{2}k!$   My [LHS] looks nothing like my [RHS] did I do something wrong?  EDIT:     $ = (k+1)k! + (k+1)^2k! -1 $      $ = (k+1)(k!)(1 + (k+1))-1$      $ = (k+1)(k!)(k+2)-1 = (k+2)(k+1)k!-1$ Which queue for an unmarried EU/non-EU couple travelling through UK Immigration Control together? How to properly secure a wireless home network I'm soon going to be setting up a wireless network for me and some buddies and (hopefully) making it secure. 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Possible Duplicate:             I want to add some shell scripts in my launcher as shortcuts..  They are located deep inside and i have to navigate to it every time to execute them...  I tried  link where the 2nd answers seems to solvee it but as said in comments neither can i find any option to add a launcher..  I guess it existed in 11.04 with a "+" button on launcher but i use 12.04.  So how do i get this working?? And i need those commands everytime for sure..! Are there any official specifications regarding the torque for camera mounts? I've been wondering why I've never seen any official specifications of the maximum allowed torque that can be exerted upon lens mounts. It seems to me like a very useful specification, but I can't seem to find it from any manufacturer.  indicates that there might be no specifications for Nikon mounts. Are there really no manufacturer that publish these specification of their mounts? 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