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Add initials proof

so this is kind of a fun proof by induction example because it's not your standard algebraic one it also has multiple base cases so the first part of this problem we're going to show or demonstrate that it works for base cases so normally just says for the base case but in our case thank you so we have lots of base cases in our case I'd be N equals 1 2 & 3 so we want to say okay is it true that T sub 1 is less than 2 to the N and is it true or starting that to the end 2 to the 1 and is it true Akali not equal signs less than sign is it true that T 2 is less than 2 squared and is it true that T 3 is less than 2 cubed and in fact it is because we know that all of these are 1 because that's what we're showing right here we're given that all those values are 1 so indeed 1 is less than 2 1 is also less than 4 and 1 is also less than eighths are actually totally fine the next thing that we want to do is we want to assume it works for K but actually we also want to show so if we're gonna start at say N equals 4 okay we can't just assume that it works for 4 we also have to assume that it works for 3 and 2 as well because remember the idea is we're going to move up and say okay well if it works for 1 2 & 3 then it's got to work for 4 all right well it works for a 2 3 and forth and it's got to work for 5 and it works for 2 3 or 3 4 & 5 it's got to work for 6 so the only way that we can show that it works for K is we also have to assume that we've got to back and that's because we have three total initial conditions or base cases so we want to assume that it works for case we have to assume that T sub K is less than 2 to the K we want to assume that it works for K minus 1 and want to assume that works for K minus two see this is where it gets kind of fun okay so we want to assume that all of those are true now our goal is to show that it works for K plus 1 all right so again we've assumed we're going to assume that it works for 1 2 & 3 and show that it still works for 4 ok so our goal is to show that T sub K plus 1 is less than 2 to the k plus 1 okay so we need to go back and we need to look at our original our recall this recursive definition so our original recursive definition is that wherever we are is equal to the sum of the three previous terms so we're going to start with that we're going to say ok so we have T sub k plus 1 that's going to be equal the sum of the two of the three previous terms K minus 1 and K minus 2 so all of these have to be less than or equal to their counterparts and this is where our assumptions come in so this one has to be less than to the K so the sums of all of these can I guess what I'm saying is if I sum all these up some still has to be less than the sum of those so that's kind of what we're doing here so 2 K minus 1 and then 2 K minus 2 so I can go and put in equality pull out the 2 to the K I've got 1 plus and that's going to be 1/2 right because it's going to beat them already this will be 2 to the negative 1 and then 2 to the negative 2 also known as 1 plus 1/2 plus 1/4 and if I'm going to be all messy about it fourths I have 4 plus 2 plus 1 that's going to be 7 so 2 to the K over 7/4 okay now we're going to do a bit of a thinking step it's kind of silly and again when you're dealing with these proof by induction it's not always just straight cord outdoor where you've got to be a little squirrely especially when you have these less than or less than or equal to signs or inequalities in general so remembering back to what our goal is our goal is to have something in the term 2 k plus 1 so I'm thinking about that the way that I can get to a 2 k plus 1 is to multiply something by 2 now if I want that to be multiplied by 2 the only way to do 2 times something is 7/4 is I could say well 2 times 3 and 1/2 for its is 7/4 I mean that's dumb but it's true so then I'm able to take this 2 here and make that K plus 1 times 3/5 over poor yeah so that's fun other people would call that 7/8 so if it helps you to write it as 7 back I think it helps you to write that as 7/8 I would recommend doing that so 2 k plus 1 times 7 eighths now again looking at our goal our goal is to say that it's less than or equal to 2 k plus 1 and while 7/8 of something is certainly less than the whole of something so 7/8 of this is definitely less than that by itself so I'm actually able to come in here and make that less than sign and I'll go ahead and make it a little bit more dramatic by putting it on its own line so that you get it otherwise it just kind of fades into the background but again the idea that 7/8 of something is clearly less than the entire thing of something but yes so we were able to show getting from the base case and from the recursive definition so it was a little bit different cuz we didn't have a summation to perturb we had a recursive definition that we just kind of messed with and so we got to where we needed to go