Question

For fun. Prove by induction:
\[\frac{d}{dx} |x^n|=nx^{n-2}|x| \text{ for all odd integer } n \ge 1\]

Answer 1

*

Answer 2

ermm can i use something else but not induction :P

Answer 3

Induction is the preferred method.
But I welcome all responses.

Answer 4

\(\frac{d}{dx} |x^n|=nx^{n-2}|x| \text{ for all odd integer } n \ge 1 \)

ok it goes well when n=1
lets this be true for k ( let k be odd)
\(\frac{d}{dx} |x^k|=kx^{k-2}|x| \)
we need to show for k+2
\(\frac{d}{dx} |x^k+2|= \frac{d}{dx} |x^k x^2|=\frac{d}{dx} |x^k|| x^2|... \text{ since x^2>0 }\\=(kx^{k-2}|x|)x^2+2x(x^k)=2x^{k+1}+kx^k|x|\\ = \)

Answer 5

hmm something went wrong with my induction xD

Answer 6

maybe i need to show for
2k+1 not any k

Answer 7

Yeah do
show for k=1
then assume true it is true for 2k+1
then show it is true for 2k+3 since 2k+3 is the next odd

Answer 8

hey your proof works just fine except for a silly mistake in the last line

Answer 9

i thought so .

Answer 10

|x^k|

Answer 11

oh yeah k, k+2 works too
where k is odd

Answer 12

i'll tell u why it works .
its induction any order set might work :P

Answer 13

ordered*

Answer 14

\(\frac{d}{dx} |x^{k+2}|= \frac{d}{dx} |x^k x^2|=\frac{d}{dx} |x^k|| x^2|... \text{ since x^2>0 }\\=(kx^{k-2}|x|)x^2+2x \color{Red}{|}x^k\color{Red}{|} = kx^k|x| + 2x^k |x| ~~\color{Red}{\star}\\ =(k+2)x^k|x| ~~ \blacksquare \)

Answer 15

oh yes

Answer 16

\(\color{Red}{*}\) : when k is odd we have \(|x^k| = x^{k-1}|x|\)

Answer 17

yes yes i put practices thats why i got confused a bit

Answer 18

nice question

Answer 19

It comes from @idku 's questions for today sorta.

Answer 20

was gonna say something about induction but na forget it :P

Answer 21

What were you going to say earlier before I said induction is the preferred method? Were you just going to do it using the chain rule and all that jazz?

Answer 22

yes

Answer 23

but then realized its hmm so clear in that either hmm

Answer 24

i should say also not either right ?

Answer 25

you are saying 'but then I realized its hmm so clear in that chain rule way' ?

Answer 26

yes

Answer 27

that is what I translated what you said if you are asking about translations

Answer 28

but i do not know if either would be the correct word

Answer 29

haha nvm leave it

Answer 30

i think also is the right one since induction is also clear :P

Answer 31

I think I misunderstood what you said @marki . I'm sorry.

Answer 32

;)

Answer 33

I like induction more as i am born biased to induction proofs xD
using chain rule :
\[\frac{d}{dx}(\left|x^{2k+1}\right|) = \frac{d}{dx}(x^{2k}\cdot
\left|x\right|) = 2kx^{2k-1}|x| + x^{2k}\frac{|x|}{x} = (2k+1)x^{2k-1}|x| \]

Answer 34

this chainrule thing also requires induction to say that it holds for all k i guess

Answer 35

We weren't talking about English/grammar at all. You were saying either way works.

Answer 36

ganesh since u used 2k+1 then no need

Answer 37

basically all proofs touch induction in someway hmm
i could be wrong though with that statement..

Answer 38

k can be anything the result would odd either way

Answer 39

now this is what i meant to say about induction >.<
it came up after WOP which means to prove for natural
in specific this set
{1,.....,k,k+1}
why we need k and k+1
k for odd\even k+1 foe even\odd

Answer 40

we can choose to induct on odd integers and define the set with least positive integer accordingly right ?

Answer 41

so need to be careful , working with induction sometimes is dumb
usual they ask u to do it to make u practice more
it does not give proper proof , its only for generalization

Answer 42

yeah u can do anything , but whats the point >.<

Answer 43

induction works when you have a formula.. it wont help you with deriving the formula though

Answer 44

yes , means for generalization

Answer 45

Exactly! generalization(induction) is a valid proof technique.

I am not happy about this statement :
`it does not give proper proof , its only for generalization`

Answer 46

hmm ur free to not be happy :P
im a fan of wop , i love Archimedean property and glade they made induction of it and happy with D.A

Answer 47

feeling bored @myininaya any other questions :|

Answer 48

not at the moment :(
@ganeshie8 is full of them usually

Answer 49

I really loved that one question with that one answer
that mod 9 question you asked
\[5^{2015}+4^{2015} \\ (5-9)^{2015}+4^{2015} \\ (-4)^{2015}+4^{2015} \\ -4^{2015}+4^{2015} \\ 0\]

Answer 50

haha yeah clever when power is odd

Answer 51

that was an answer by @mathmath333 or whatever his name is

Answer 52

yes i remember

Answer 53

i have to go
be back for more math fun later

Answer 54

ok take care bbye !