Question

Question about sequences

Answer 1

It's part a) .

Answer 2

I can show the base case and assume the case where n=k .

Answer 3

Someone want to help me do the 3rd part of the induction step?

Answer 4

hmm, i cant say ive ever come across induction in this manner ....

Answer 5

Here let me post a link to a similar question.

Answer 6

http://facultypages.morris.umn.edu/~mcquarrb/teachingarchive/M1102/Practice/11.1.pdf

Answer 7

Scroll down to page 4.

Answer 8

I just don't get why they could assume sqrt(2ak) is a(k+1)

Answer 9

you looking for bounded or for induction?

Answer 10

Well I'm supposed to show it's bounded by induction.

Answer 11

then lets work that, since i understand better now

Answer 12

Like bounded above by 5 using induction.

Answer 13

given:

\[x_{n+1}=\sqrt{15+2x_n}\]
is it true that x_1 <= 5?
\[1<= 5\]its true for the basis statment

Answer 14

\[\text{ Maybe \it might be easier \to try \to show } x_{n+1}^2<25\]

Answer 15

Yep. I got that.

Answer 16

Maybe? I've always done sequences without ever needing to square it though.

Answer 17

change n to k and assume its true

then determine if its true for x_{k+1} which has already been defined

Answer 18

actually that seems to work out perfectly using that square statement and also the statement before I squared it :)

Answer 19

And I should be less than or equal up there

Answer 20

Okay I'll do it out myself :) . I really want to see the "regular" way though.

Answer 21

And yeah, I follow you upto there @amistre64 .

Answer 22

\[x_{k+1}:=\sqrt{15+2x_k}\le5\]
solve for x_k

\[\sqrt{15+2x_k}\le5\]
\[15+2x_k\le25\]
\[2x_k\le10\]
\[x_k\le5\]

Answer 23

Weird... I wonder how I never noticed that :P .

Answer 24

I guess after doing so much induction for the past 2 years it never came to me. I can also see how this relates to @myininaya 's idea.

Answer 25

i dint actually induct there ... start with x_k

Answer 26

\[x_k\le5\]
assume this is true, and now build this into x_k+1
\[2x_k\le2(5)\]
\[15+2x_k\le15+2(5)\]
\[\sqrt{15+2x_k}\le\sqrt{15+2(5)}\]
\[x_{k+1}\le\sqrt{15+2(5)}\]

Answer 27

i have a bad tendency to work induction backwards lol ... but sometimes backwards gives you a good idea on how to work it forwards

Answer 28

Well if I substitute n=k+1 into n+1 would it not be n+2 instead?

Answer 29

\[x_{k+1}=\sqrt{15+2x_k} \le 5 \\x_{k+1}^2=15+2x_k \le 25 \\ \text{ Now we need \to show } x^2_{k+2} \le 25 \\ x^2_{k+2}=15+2x_{k+1} \le 15+2(5)=25 \\ x_{k+2} \le 5 \]

Answer 30

Yeah. What she said.

Answer 31

So sorry I can't give both of you medals!

Answer 32

induction is about form

Answer 33

Amistre doesn't need anymore.

Answer 34

He has like a million already.

Answer 35

i waste all my medals on fish bait anyways :)

Answer 36

You can borrow my for fish bait to as long as you catch my a big one.

Answer 37

me*

Answer 38

since we know x_k <=5 for some k (namely k=1)

then we need to show that the next term fits the same form by building off of x_k

Answer 39

Ahh I see it now. Thank you :) . THe next parts asks why I know it converges (Monotonic convergence theorem) and the limit of that function (Very easy) .

Answer 40

Also you can give amistre the medal. He smells better so he deserves it.

Answer 41

pizza and roses ... pizza and roses lol

Answer 42

I gave it to you :( .

Answer 43

I swear I would be at 99 if I came here more often. So few people can do Calc III here :P .

Answer 44

so few ask about calc iii here :/

Answer 45

True but so many people haven't done Calc II for years here so when it does get asked few people can answer :P .

Answer 46

polpak was good at it :) wonder whatever happened to him

Answer 47

i gotta get, so yall have fun ;)

Answer 48

You too!