Question

can someone help me with the attached induction question?

Answer 1

@ganeshie8 @mathmale @satellite73

Answer 2

So you have proved for the base case right?

Answer 3

Sorry I don't know what I should prove fro the base case

Answer 4

for*

Answer 5

prove (1+x)^n >= 1+nx for n=1 where x>=-1

Answer 6

well (1 + x) ^1 = (1 + (1) x)

Answer 7

and that means the inequality holds since both sides are the same when n=1

Answer 8

and that is obviously true for all x, so it's true for x>= -1

Answer 9

now assume for integer k you have (1+x)^k >=1+kx for x>=-1

then show from that (1+x)^(k+1) >=1+(k+1)x follows

Answer 10

now assume the condition is true for all n >=1 to N

Answer 11

so is that the inductive hypothesis?

Answer 12

the inductive hyp is the assumption part

Answer 13

yes

and to proceed I would write two relations (assuming the induction hypothesis)
(1+x) >= 1+x
(1+x)^n >= (1+nx)
and multiply them

Answer 14

and I omitted positive
when I said integer
should be positive integer
assume for some positive integer k*

Answer 15

and make note that (1+x) is 0 or positive for all x>=-1
so you know the relation does not change sign

Answer 16

I'm a little lost, how come you need to multiply them?

Answer 17

\[(1+x)^{k+1} \ge 1+(k+1)x \text{ is what we want to show} \\ (1+x)^{k+1}=(1+x)^{k}(1+x).... \text{ use the inductive hypothesis }\]

Answer 18

so the (1+x)^k(1+x) simplifies to 1 + (k + 1)x?

Answer 19

no that doesn't simplify to that

Answer 20

do you remember the inductive hypothesis ?

Answer 21

(1+x)^k >=1+kx for x>=-1?

Answer 22

\[\text{ assume for some positive integer } k \text{ we have } (1+x)^{k} \ge 1+kx \text{ for } x \ge -1\]

Answer 23

so we use the inductive step? like what we know?

Answer 24

we want to use this hypothesis here:
\[(1+x)^{k+1}=(1+x)^k (1+x) \ge ...\]

Answer 25

well (1 + x)^k >= 1 + kx so do we substitute it in?

Answer 26

can you show me what you mean

Answer 27

like (1 + x)^(k+1) = (1+ kx) (1+x)

Answer 28

well the symbol between those two things should be =>

Answer 29

\[(1+x)^{k+1}=\color{blue}{(1+x)^k} (1+x) \color{blue}{\ge (1+kx)(1+x)} \\ \text{ where we used the inductive hyp I put in blue }\]

Answer 30

now let's try to remember what we are trying to show
we are trying to show
\[(1+x)^{k+1} \ge 1+(k+1)x\]

Answer 31

do you have any ideas on what we can do with (1+kx)(1+x) to shoe this is greater than 1+(k+1)x

Answer 32

greater than/equal to*

Answer 33

can we multiply (1+kx)(1+x) together?

Answer 34

let's try that

Answer 35

so what is the product of (1+kx) and (1+x)

Answer 36

1+x+kx+kx^2

Answer 37

cool and you know x+kx can be written as (k+1)x

Answer 38

\[1+(k+1)x+kx^2 \]

Answer 39

now again we are trying to show this is greater than or equal to 1+(k+1)x

Answer 40

x^2 is always zero or positive
and k is a positive integer
kx^2 is therefore positive or zero

Answer 41

so greater or equal

Answer 42

yes 1+(k+1)x+kx^2>= 1+(k+1)x

Answer 43

because of kx^2 being positive or zero

Answer 44

okay is that the proof then?

Answer 45

I'm not sure how we're supposed to write it all out

Answer 46

yep

proving the base case (check)
stating the inductive hypothesis (check)
proving the inductive statement using the inductive hypothesis (check)

put at the end something like: therefore this concludes (1+x)^n >=1+kn for all real number x>=-1 and any positive integer k

Answer 47

alrighty

Answer 48

you have any conclusions
your proof should really have 4 main parts: base case, statement of inductive hypothesis and I like to write what I want to show so you could just say inductive statement, then proof of the inductive statement, and finally the conclusion

Answer 49

Base case I might write:
\[\text{ Base case } n=1: \\ (1+x)^1 \ge 1+1 x \\ 1+x \ge 1+x \\ 1+x=1+x \\ \text{ So the base case holds } \]

Answer 50

so like write what we want and what we have?

Answer 51

\[\text{ Inductive statement } \\ \text{ If }\color{red}{\text{ for all real } x \ge -1 \text{ and positive integer }k \text{ we have } (1+x)^k \ge 1+kx } \\ \text{ then } \color{blue}{\text{we have } (1+x)^{k+1} \ge 1+(k+1)x }.\]
The inductive hypothesis is in red.
What we want to show is in blue using the the inductive hypothesis.

Answer 52

So you can say:
\[\text{ Assume for all real } x \ge -1 \text{ and positive integer } k \text{ we have } (1+x)^{k} \ge 1+kx \\ \text{ then start your next line with } (1+x)^{k+1}.... \\ \text{ this where you are trying \to show this is bigger than} 1+(k+1)x\]

Answer 53

\[(1+x)^{k+1} =(1+x)^k (1+x) \ge (1+kx)(1+x) \\ =1+x+kx+kx^2=1+(k+1)x+kx^2 \ge 1+(k+1)x\]

Answer 54

And then finally the last line of your proof is showing your conclusion

Answer 55

alright that makes sense now, thank you so much! :)

Answer 56

np