Question

given the equation
\[f(x+y)=f(x)+f(y)+x^2y+xy^2\]
and the limit:
\[\lim_{x \to 0} \frac{f(x)}{x}=1\]
a) find f(0)
b) find f'(0)
c) find f'(x)

Answer 1

from the limit i concluded that f'(0)=0
and than happen if f(0)=0
considering that limit to be the f'(0)=lim (f(x)-f(0))/x

Answer 2

is what i did sound fine?

Answer 3

@Kainui

Answer 4

i meant f'(0)=1 not 0

Answer 5

now i'm thinking to let y=x
to get \[f(2x)=2f(x)+2x^3\]
doing the derivative
\[2f'(2x)=2f'(x)+6x^2\]
\[f'(2x)=f'(x)+3x^2\]
not sure where this leads to eheh

Answer 6

hmm i think i didn't get that f(0) correct!

Answer 7

thinking...

Answer 8

this is @freckles 's fav problem!

Answer 9

hehe where is @freckles

Answer 10

i will come back later to think about this
just leave me here your notes :)

Answer 11

\[f(x+y)=f(x)+f(y)+x^2y+xy^2\]
\[\text{ put } x=y \]
\[f(x+x)=f(x)+f(x)+x^3+x^3\]
\[f(2x)=2f(x)+2x^3\]
\[f(2(0))=2f(0)+2(0)^3\]
\[f(0)=2f(0)+0\]
\[f(0)-2f(0)=0\]
\[-1f(0)=0\]
\[f(0)=0\]

But I bet you were trying to skip to the derivative part
\[\lim_{x \rightarrow 0}\frac{f(x)-f(0)}{x-0}=(f(x))'|_{x=0} \\ \text{ and we have } f(0)=0 \\ \text{ so we have } \lim_{x \rightarrow 0} \frac{f(x)}{x}=(f(x))'|_{x=0}\]
But we were given:
\[\lim_{x \rightarrow 0} \frac{f(x)}{x}=1\]
so f'(0)=1
...
now I have to think about the f'(x) again

Answer 12

http://openstudy.com/users/myininaya#/updates/5410d734e4b08b11f1302400
check out what @ganeshie8 wrote towards the end :)

Answer 13

Interesting I used that differently
I found f'(x) but I'm in the phone now I can't post it once I get home I will post it

Answer 14

Eh that's exactly the same question lol
Anyways I will post my f'(x) later

Answer 15

and i think you posted in that question lol

Answer 16

Ahh I remember now xD @xapproaches , here is a copy paste from that thread for computing \(f'(x)\)
:
\[f(x+y)=f(x)+f(y)+x^2y+xy^2\]
\[f(x+y)-f(x) = f(y)+x^2y+xy^2\]

divide through out by \(y\)

\[\dfrac{f(x+y)-f(x)}{y} = \dfrac{f(y)}{y}+x^2+xy\]

take the limit and \(y\to 0\) and stare at left hand side

\[\lim\limits_{y\to 0}~\dfrac{f(x+y)-f(x)}{y} = \lim\limits_{y\to 0} \left[ \dfrac{f(y)}{y}+x^2+xy\right]\]

\[f'(x) = 1 + x^2 + 0\]

Answer 17

eh i didn't remember actually
i was just trying some problems from stewart textbook!
here is what i did
\[\large f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{f(x)+f(h)+x^2h+xh^2-f(x)}{h}\]
\[\large =\lim_{h\to0}\frac{f(h)+x^2h+xh^2}{h}=\lim_{h\to0}[\frac{f(h)}{h}+x^2+hx]\]
\[\large =1+x^2\]
which the same result

Answer 18

i don't why i didn't see that f(0)
even thought i did take y=x hehe

Answer 19

though*

Answer 20

mad skills :p @xapproachesinfinity

Answer 21

damn i feel down that i didn't see that f(0) hehe

Answer 22

Woah, for some reason that method is really quite awesome to use (x+y) and make it (x+h), it just feels so nice.... XD

Answer 23

:)
i will post a different question and how can come up with the solution

Answer 24

I have one question for you, what is f(x) itself? (This isn't too far away I think)

Answer 25

well just take integral yes?

Answer 26

we have f(0)=0 already so we can find C

Answer 27

c=-1 hmm

Answer 28

oops no le me see

Answer 29

c=0 yes

Answer 30

okay let me post it \[f(x)=x+\frac{ x^3}{3}\]

Answer 31

Yeah, =P

Answer 32

:)

Answer 33

somehow you like integrals hehe :P

Answer 34

Integrals are probably the coolest thing I've ever seen in my entire life. Learning solids of revolution was pretty much a spiritual awakening for me, and that is not even (or oddly) an exaggeration.

I mean, a single simple integral allows you to take a circle, imagine that it has an infinite number of smaller rings inside of it. But then you actually get to add up all those circles of "0" width inside together, multiplied by their circumferences (just baby's length*width) to get the area of the circle.

Then people come along and say they invented with axioms, what a bunch of cheap liars. Not only that, but they're too afraid to say the truth that division by zero is defined, it's just such a sensitive subject we have to call it "calculus" as if it's separate... Ok now I'm rambling but yeah, somehow I like integrals and no one else seems to realize how incredible they truly are! haha.

Answer 35

haha, that's deep ^^

Answer 36

why do you call them cheap though, didn't they do a lot of awesome stuff hehe

Answer 37

Depends on who we're talking about lol

Answer 38

idk who you were talking about heheh, at any rate must be some mathematicians lol

Answer 39

I mean I'm not saying axioms can't be helpful, I just think people believe that math comes from axioms. It's useful for creating a giant system based on pure logic so that you can compare it to similar mathematical structures. But I don't believe the set theory construction of calculus is actually calculus.

I'm not really talking about anyone in particular anyways haha, like everyone has their own way and if they like/need/want/use axioms let them. I just don't haha.

Answer 40

so you believe 5/0 is defined
if so then how do you define it?

Answer 41

I never said that.

Answer 42

" Not only that, but they're too afraid to say the truth that division by zero is defined,
"
oh you were talking about 0/0

Answer 43

maybe

Answer 44

well i must admit axioms are not so convincing sometimes
or perhaps i just don't got it hehe

Answer 45

well if you define 0/0 then you need to define 5/0 as well?

Answer 46

not really
0/0 is indeterminate form
some cases you can define it and some cases you can't

Answer 47

I think anyways

Answer 48

Yeah, I am not saying division by zero is defined for all values, just stuff like 0/0. Basically all of calculus is just dividing 0 by 0 and doing stuff that's scary. It depends on what the context is, does what you're doing with the division by zero have some meaning behind it or is it worthless nonsense?

Answer 49

ok well that was the only thing that scared me of your whole paragraph

Answer 50

I don't see any reason why things have to have definitions at all in mathematics. If you do something and what you get out of it is a useful result and the interpretation you get out of it gives you something else useful, why fight it? I am just not interested in rigor at all, and usually proofs are not interesting to me because of several really stupid reasons that I will spare you lol.

Answer 51

Are you a mathematician or an engineer?

Answer 52

or i mean I guess you can be both

Answer 53

Neither haha.

Answer 54

like i met an engineer they say they didn't have to care about proofs just results
and you remind me of her

Answer 55

Well there's a difference between why an engineer doesn't like proofs and why I don't like proofs.

Answer 56

hehe math with no proofs is just not working no?

Answer 57

I am kind of being misunderstood here when I say these things, but I'm not sure I am really interested in making the point I'm trying to say here haha. Let's just say I like "good" proofs.

Answer 58

Kainui real analysis is all about rigor. It starts with construction of rational numbers and builds all the way up to integrals. It is more like language of mathematics, defining each term and proving every little thing along the way. I have fallen in love with real analysis because of its rigor despite my 10 year brain damaging engineering experience lol

Answer 59

you're talking about real analysis when you talk about quality of proofs

Answer 60

Yeah but isn't real analysis kind of fake compared to complex analysis? I just can't get into it, I feel like there's kind of a lot of broken arguments going on in real analysis since they are depriving themselves of a fuller picture.

Where do axioms come from? They are also kind of redundant, which bothers me, it just doesn't feel right or like real mathematics at all. All the definitions in analysis seem just arbitrary to me and not natural in any way, it just makes me too uncomfortable to study.

When I think back to when I took real analysis we were told to prove things like that there are an infinite number of integers, and use epsilon delta proofs which seems to just tuck infinities away rather than actually dealing with them. It just feels like a big meaningless game to me.

Answer 61

I like both, but i guess i like real analysis more as it is my first love xD

Answer 62

you guys are mathematicians :)

Answer 63

In my opinion you are what you do everyday. And I do math every single day, so in that sense I am a mathematician I'll agree haha. @xapproachesinfinity

Answer 64

hehe
everyday without exceptions
yeah i would agree no less if you live with math like that