Question

calculus@michele_laino @satellite73

Answer 1

@Michele_Laino @satellite73

Answer 2

\[\frac{d}{dx}f(2x)=2f'(2x)\]

Answer 3

how do you know?

Answer 4

chain rule, it is always true

Answer 5

like \[\frac{d}{dx}\sin(2x)=2\cos(2x)\]

Answer 6

okay.

Answer 7

so you know \[2f'(2x)=f'(x)\] since they tell you they are equal

Answer 8

yes, ok

Answer 9

And if f'(1) = 1....then can you substitute that?

Answer 10

yes

Answer 11

you almost had it there

Answer 12

2f'(2x) = 1
f'(2x) = 1/2

Answer 13

\[2f'(2\times 1)=f'(1)=1\]

Answer 14

you should replace x by 1 in both spots

Answer 15

f'(2) = 1/2

Answer 16

looks good to me

Answer 17

Wow, thanks! I didn't realize it would be so easy. Thanks!

Answer 18

lol everything is easy when you know how

Answer 19

btw can you think of a function \(f\) where \[f'(2x)=f'(x)\]?

Answer 20

hint: is is a pretty common function, but not a polynomial, or rational function

Answer 21

hmmm....x^0

Answer 22

?

Answer 23

\(x^0=1\)

Answer 24

i suppose a constant function is one example, but i was looking for an example of a function that is not constant

Answer 25

1/x

?

Answer 26

no, nvm. you said it wasn't rational

Answer 27

idk, what is it?

Answer 28

ask @freckles

Answer 29

\[u(2x)-u(x)=0 \\ u(x)=C \cdot n^{x } \\ u(2x)=C \cdot n^{2x} \\ \\ C n^{2x}-C n^{x}=0 \\ \text{ assume } C \neq 0, n \neq 0 \\ n^{2x}-n^{x}=0 \\ n^{x}(n^x-1)=0 \\ n^x=0 \text{ or } n^x=1 \\ \implies n=1 \\ \\ \text{ so } u(x)=C \\ f'(x)=C \\ \text{ since I replaced } f'(x) \text{ with } u(x) \\ f(x)=Cx+D \]
even though I said C cannot be 0
C=0 does also work

anyways this is a linear function and satellite asked for a non-polynomial
and for some reason I cannot think that way

lol I been playing with recurrence relations too much
I could instead try to use a generating function to solve u(2x)-u(x)=0 and see if I get a different result

Answer 30

did my picture help?

Answer 31

oh log function

Answer 32

that one will work
i am not sure any others will

Answer 33

I'm not seeing which log function for f satisfies f'(2x)=f'(x)
:(

Answer 34

did you mean
\[(f(2x))'=f'(x)\]
@satellite73 \[2 f'(2x)=f'(x) \\ f'(x)=\frac{k}{x} \\ f''(2x)=\frac{k}{2x} \\ 2 \cdot \frac{k}{2x}=\frac{k}{x} \text{ is true } \\ f'(x)=\frac{k}{x} \implies f(x)= k \ln(x)+C\]

Answer 35

oops that '' ignore that should be a single mark

Answer 36

in the link it says \[\frac{d}{dx}f(2x)=f'(x)\]

Answer 37

yes but I have been trying to solve f'(2x)=f'(x) forever now and get log somehow from it :p

Answer 38

so since \[\log(2x)=\log(2)+\log(x)\] you have \[\log'(2x)=\log'(x)\]

Answer 39

oh i see

Answer 40

you like to abuse things I see

Answer 41

i do

Answer 42

if i had clue one as to how to solve a differential equation, maybe i could prove that that is the only solution

Answer 43

\[2 f'(2x)=f'(x)\]
i was thing about treating it similar to a recurrence relation and see what happens

Answer 44

ooh i see

Answer 45

so if \[2f(2x)=f(x)\] you have \[f(x)=\frac{k}{x}\]right?

Answer 46

yes

Answer 47

so it is the log in every case

Answer 48

learn something new every day

Answer 49

i learned something else just now