$A(n)$ is an assertion (or proposition) based on $n$.

It's not a term depending on $n$.

Yeah

but see the line

$A(k)$ is the proposition that $1^2 + 2^2 + \dots + (k-1)^2 < \frac{k^3}{3}$.

Start with $A(k)$ and ad $k^{2}$ on both sides

Right

You're doing that to get $1^2 + 2^2 + ... + (k-1)^2 + k^2$ on the LHS.

exactly

but isn't that equal to $A(k+1)$ ?

Nope.

$A(k+1)$ is the proposition that $1^2 + 2^2 + \dots + (k-1)^2 + k^2 < \frac{(k+1)^3}{3}$.

But $ A(k+1) = 1^{2} + 2^{2} + ... + k^{2} < frac{(k+1)^{2}}{3}$

That's exactly where your confusion lies. $A(k)$ isn't equal to anything.

Okay

I'm not getting it

$A(n)$ or whatever you want to call it is the inequality given that depends on $n$.

Can you explain the proposition part

A proposition is merely a statement.

You're proposing it to math. Then math will prove whether it's true or false.

Yeah

okay lemme read it again

So our proposition is that the inequality is true.

Our proposition depends on $n$.

So we need to prove it's true for a set of $n$, which in this case, is the integers.

You're essentially applying the squeeze theorem here.

You've shown that, $1^2 + 2^2 + \dots + (k-1)^2 + k^2 < \dfrac{k^3}{3} + k^2$.

$A(k+1)$ is the proposition that your sum on the LHS is less than $\frac{(k+1)^3}{3}$.

okay

If you show that $\frac{k^3}{3} + k^2 < \frac{(k+1)^3}{3}$, then you've shown that your original LHS is less than $\frac{(k+1)^3}{3}$, which is the proposition that we've called $A(k+1)$.

so basically now we want to show that A(k+1) < (k+1)^3 / 3

for that we just add k^2 on both sides, make it look like a(k+1) and show that this is less than (k+1)^3/3

That's not it. Tell me what $A(k+1)$ is.

I feel that scientists have created ambigious math notations that may not look ambiguous to us humans (at first glance) but could look ambiguous to some aliens.

Then the notation isn't ambiguous.

For only humans live on Earth.

(As far as we know.)

$A(k+1) : 1^{2} + 2^{2} + ... + k^{2} < \frac{(k+1)^{3}}{3}$

No.

$A(k+1)$ isn't equal to anything.

Sorry

That seems ok. Just write it in words next time.

okay

Now this is what we want to show

$A(k+1)$ is the assertion that blah blah is true.

Since we assumed that $A(k)$ is true, we will use it to show that $A(k+1)$ is true?

Exactly.

Why do we need to use $A(k)$?

It's the one before $A(k+1)$

Say an alien comes to our planet Earth. What I'm saying is, 'it' might find our notation so ambiguous (after learning everything about it) that they'd have to read a notation for like a minute to understand it.

Why do we need to assume the assertion that comes directly before $A(k+1)$?

@Huy :-)

@r9m: :-(

Hmm

Because we need to increment by steps of one

If you can answer that, you've understood why induction is so powerful.

I'm not really sure.

You're almost there.

@Huy just kiddin ;) .. how are u doin ? :)

I don't know. :S

If we increment by 1 at a time, this means that we can go on an on forever

If you have assumed a proposition's truth for an arbitrary case, say $n=k$, then have been able to use that assumption to prove the next case's validity, you can establish the proposition's truth for a base case, which will imply the case after that is true, then the one after than, and the one after that.

Exactly.

@r9m: I'm okay. Trying to figure out some ways to make my living room / bedroom look nicer.

@Chris'ssis $\sum\limits_{n=1}^{\infty} \psi^{(1)}(a+n)\psi^{(1)}(b+n)$ looks interesting :-) ..

The answer is 3, @r9m. Rather elementary if you ask me.

@Khallil independent of $a,b$ ?! I don't think so ;) =P

So in mathematical terms, what I'm I doing with $A(k)$ ?

Hahahahaha! I just wanted to feel smurt, @r9m.

And what do we call $A(k) and A(k+1)$

@Khallil ^^'

I know the base case is when n=1

$A(k)$ and $A(k+1)$ are special cases of the general statement (or assertion, or proposition) $A(n)$.

@r9m I think I can do that pretty easily.

Are you comfortable with the words assertion, proposition and statement?

@Chris'ssis :D .. okay !!! :)

I'm comfortable with proposition, which is a statement that is either true or false.

@r9m Piece of cake. :D

Not sure about the 2 others though.

But an Axiom is a statement assumed to be true. Wouldn't that make $A(k)$ an axiom?

@Chris'ssis same strategy as the case $a=b=0$ ? ..

@r9m Yeap.

;-)

@r9m ;-)

@Huy Get a lava lamp.

@MikeMiller: Are you an interior designer?

Kind of, @Sabಠ_ಠ.

But what about statement and assertion?

They're all the same thing in this case. A statement that is either true or false.

aha

We'll go about the induction by assuming that the statement $A(k)$ is true.

@r9m Did you try this version $$\sum\limits_{n=1}^{\infty} \psi^{(1)}(a+n)\psi^{(1)}(b+n) \psi^{(1)}(c+n)$$ ?

or

$$S(k)=\sum\limits_{n=1}^{\infty} \psi^{(k)}(a+n)\psi^{(k)}(b+n) \psi^{(k)}(c+n),\space k\ge 1$$

@Chris'ssis seems you have a nice trick to tackle these summations .. :-)

I can't wrap my head around $A(k+1)$ as a consequence

If I get this right it's all going to be easy.

I was thinking of doing Apostol till part 10 in a month guess it will be hard

@r9m There are more tricks, and sometimes all fail at once. Then I need to create other tools. I always need to invent something to go further. :-)

i'm struggling big-time with rationalizing a nasty limit. in the following latex, the limit is the term on the left, the middle and right terms are what i'm (wanting|hoping) to multiply by, to get rid of the radicals. but i don't know where to start. anyone can tell me how to get started with this beast?

$$\frac{\sqrt{x+2} - \sqrt{3x-2}}{\sqrt{4x+1} - \sqrt{5x-1}} \times \frac{\sqrt{x+2} + \sqrt{3x-2}}{\sqrt{4x+1} + \sqrt{5x-1}} \times \frac{\sqrt{4x+1} + \sqrt{5x-1}}{\sqrt{x+2} + \sqrt{3x-2}}$$

@topper: Where is a limit?

@Chris'ssis okay :) .. but I was wondering if the usual manipulate the series will work for that too ?! .. or do we do it in the triple integral way ?

@Huy looks like this: $$\lim_{x \to 2}\frac{\sqrt{x+2} - \sqrt{3x-2}}{\sqrt{4x+1} - \sqrt{5x-1}}$$

@topper: Why are you multiplying by these two terms? You only need to try to get rid of the square roots in the denominator.

@Huy Haha - why am I chuckling? Because I got myself in a twist when I did that - see math.stackexchange.com/questions/917440/… where once I rationalized the denominator it came out as 2-x, so was still zero when I substituted 2 for x

@r9m I'll give you more details when I write the proof down. My start is based upon the series manipulation.

@Chris'ssis okay :D .. thanks !! :D

@r9m Did you compute that one in $a$ and $b$? I won't ask you to give me the closed form. ;)

I'm making a conscious effort to ask questions on the site before I ask here, so that the next user(s) can benefit from the answers, but I just ran out of ideas this time, despite people trying to help

@Chris'ssis first did it for integer $a,b$'s,(via manipulating the series) and then extending to real $a,b$ was easy :)

@r9m Nice.

@Chris'ssis -_- 'I won't ask you to give me the closed form' ?! sounds like hey kid come here .. I don't want your candy =P

@r9m :-)))))))))

It's my $200^{th}$ consecutive day on SE :P LOL

:D

no @Huy - just a lava lamp advocate

@Huy and anyone else - i'm sorted on the above problem. didn't spot the difference of squares lurking

@Nick whatcha talkin' about?

My sympathies, @r9m

@TedShifrin thank you professor ! :)

Sorry I went away, @Sabಠ_ಠ.

I was watching Hunter X Hunter.

I also read some old chapters of Naruto.

Anywho.

I'll outline it really simply.

We want to prove that the proposition $A(k+1)$ is true by using $A(k)$ which we have assumed to be true to make sure that there's a domino effect starting with $k=1$.

If we can do this, then we can establish that $A(1)$ being true implies $A(2)$ is true and so on and so forth.

We have $A(k)$, the statement that suggests that this inequality is true $$1^2 + 2^2 + \dots + (k-1)^2 < \frac{k^3}{3}$$

Adding $k^2$ to both sides will give you $$1^2 + 2^2 + \dots + (k-1)^2 + k^2 < \frac{k^3}{3} + k^2$$

Np :)

We already know that the proposition $A(k+1)$ tells us that this inequality may or may not be true $$1^2 + 2^2 + \dots + (k-1)^2 + k^2 < \frac{(k+1)^3}{3}$$

Salut, @Sab

Salut @TedShifrin :)

Hi @anon @blue

bah, didn't see daniel's

hi

To prove the validity (truth) of $A(k+1)$, we must use the statement $A(k)$ that's been modified by adding $k^2$ to it's inequality.

All that remains to be done is to show that $$ \frac{k^3}{3} + k^2 < \frac{(k+1)^3}{3}$$

If we can do that, then we can use our last three inequalities to show that $$1^2 + 2^2 + \dots + (k-1)^2 < \frac{k^3}{3} + k^2 < \frac{(k+1)^3}{3} $$

It's just like having $a<b$ and $b<c$.

It's clear that you can combine these inequalities to show that $a<b<c$, or more succinctly, $a<c$.

Back to school, @anon? Anything good?

I teach intermediate and college algebra now.

That's all there is, @Sabಠ_ಠ. I'm off now. It's getting really late where I am, now.

just had first lecture today

Oh wow. Don't lose patience.

I'm saving this @Khalil :D THanks a lot :) Hopefully by tomorrow I'll be up and ready to do any proof by induction :D

Great explanation ;)

They hire undergrads? Or have you been promoted?

I hope I helped. It took me far longer than you can imagine, to grasp induction, @Sabಠ_ಠ. I wasn't taught it well at all. I was just taught to remember a bunch of steps.

Same

I have to learn most(if not all) of calculus by myself

We're all exemplary teachers, @Khallil

(Also, it's Kha**ll**il. I won't get pinged if you forget one of the **l**s. You could just write @Kha to make it easier on yourself!)

THey focus too much on questions and not much on concepts :(

ahh sorry @Khallil :P

On MSE, @TedShifrin?

No, all math teachers, @Khallil:)

@Ted I got a brand new Spivak which I now proudly own :P Too bad Apostol is wayy to expensive will use the library's one :P

Next after Soivak you can proudly own my multivariable/linear book, @Sab :D

Hopefully yeah :D I see your book pop up all the time among the best :D

I dunno about that ...

ohh you shouuld :O

I was on a forum last time and your book was among the maths books that should be on any mathematicians shelf :3

I wouldn't go as far as saying that, @Ted. I've had some pretty hopeless math teachers. They'd only teach to the exams and wouldn't enrich my learning in any way.

@Khallil: I was being exceedingly sarcastic.

I like the style of Apostol and Spivak. They make you feel b**thurt but when you get it it's so awesome :D

Sorry! My sarcasmometer is broken, @TedShifrin.

LOL

@Khallil these are my teachers currently

So you know how I feel.

All too well, @Sabಠ_ಠ.

Luckily MSE exists :)

I have a great debt to TSR as well.

UK?

@TedShifrin yes, I think I am the second undergrad to be able to teach

Plenty of my students want to be spoonfed and memorize. Suffice to say, they aren't fond of me ...

That's why I do bad at exams.

I can't memorize

@anon: That is a supreme honor. They're paying you like a TA?

I need to understand first and then I get realistic that calculus can't be learned in 6-7 months

now I got barely 2 months for the exams and I need to work a lot

Yep. It's this site, @Sabಠ_ಠ.

So you did A-levels?

@TedShifrin 15/hr, I think the others get that too

Hrs of preparation and office hours, too?

Yep, @Sabಠ_ಠ.

A-levels is so overrated. Got A in maths got in uni failed my first test.

I agree. They're way too watered down.

@TedShifrin I don't officially have office hours - we are associated with a Math Lab which they can come to for help

Yep I still blame it till date. Feels like I learned nothing in highschool

Are you from the UK, or did you do the international A-Level papers, @Sabಠ_ಠ?

Do they pay you prep time?

I did the international one.

Hello @Ted, @anon

Rehi @Mike

@TedShifrin I think they pay on the assumption it takes an hour to prepare an hour of lecture, although our class notes are for the most part given to us

Ah, I see. Which exam board did you do math (and further math?) with? Was it Edexcel, @Sabಠ_ಠ?

I didn't notice a prehi, @Ted

CIE(the easiest IMO)

Ok, that's not unfair.

I didn't do further and I regret it.

there was @Mike

Well, prehi to you again.

@TedShifrin I am okay with limiting the conceptual background and theory behind math in low-level classes - most of the students have no interest in math and won't see any higher math courses, and we don't have time to cover theory (it also hurts grades because of information overload). if the theory behind math will help the student learn and remember facts and formulas, I'll try to add it in though.

So, I got a comment on a text I'm writing that since for $u \in U$ and $v \in U$, i have that $u + v \not\in U$ , then I can't call $U$ a space, but it must be called a set. I don't really have any problems with calling it a set (there is just one occurrence in the text), but I've never really heard anyone refer to the "trial set" in mathematics, only the "trial space". I tried to find exact definitions of spaces and sets, but it doesn't seem all that clear to me.

Most people do, @Sabಠ_ಠ. Honestly, it only takes about a month to go over all of further.

I think you should encourage them to understand where things come from, @anon ... Not formal proof, but things that help it make sense.

@MikaelÖhman what is a trial space? what do you want "trial set" to mean?

yep @Khallil My friends who did it do really well in the first year maths

Common phrase in finite element analysis, it's all the space/set of functions for the approximative solution to the PDE

@Khallil true. Put it the time get the results.

for example, our class notes don't mention the fact that the distance formula is from the pythagorean theorem, or why the midpoint formula is what it is.

@Khallil how come you're not in uni yet? :O

Gap year. Going at the end of the month though, @Sabಠ_ಠ!

Hah nice. I took a gap year too :)

How did you spend the year, @Sabಠ_ಠ?

I programmed a lot

Oh, nice! Did you have to do a programming module at university, @Sabಠ_ಠ?

I started with no programming background and went on to build a whole ecommerce platform

What's an ecommerce platform, @Sabಠ_ಠ?

@Khallil I'm actually getting a double major in CS and maths :)

@Khallil e-commerce a platform which allows you to buy online :)

@anon: They'll need Pythagoras in other contexts, so making that connection is good. Midpoint of line segment in $\Bbb R$ as average should precede two-D, I think.

That sounds awesome. Double major!

Does that mean you're raking in the cash, @Sabಠ_ಠ? ;-)

Nop :p but I could :p

I did it to learn and in the process I learned so much I can;t believe it myself thinking about it at times

Yeah double major, our university forces us to do 2 majors

But for the first year we have to do 4 full courses and I'm doing applied maths, stats and physics along with math and cs

seems like 5 but credit-wise they count 4 in all :)

Next year I'll do some computer engineering more maths, stats and cs

and finally I'll do cs and maths

In US standards it would be like getting a degree in maths and cs and 2 minors in stats and com eng.

Oh, that's pretty cool. You'll be really versatile!

Can someone find a duplicate?

Just need to survive though :P

Very true!

@MJD: Binomial theorem?

I'm putting 8 hours of maths everyday from tomorrow

the rest of time will be split among the others

@ted your question contains no verb.

I not verbs.

@MJD

close enough :P

I was hoping to find one that was exactly the same, not merely similar, not a generalization.

Beggars can't be choosers.

I kid. Look hard enough and I'm sure you'll find one.

@MJD There's a pretty interesting combinatorial interpretation.

I've seen 'em pop up everywhere.

@BalarkaSen that's what I was looking for.

jinx

Googol is not helping

@Ted I hope that post didn't look like I was disparaging the assignment. That wasn't intended.

Hi

Yo!

@Khallil how are you doing

Very well, thank you. How are you, @nablablah? ^_^

OK

I might be off now. It's getting late. Later than late.

Bye bye

cya :)

See ya later, guys!

^_^

cheers

@MJD $2^n$ is the number of ways one can pick up one elt from $\{0,1\}$ from $n$ boxes. More simply, it's the number of ways to pick up anything at all from $n$ boxes (the zero being interpreted as picking nothing). However, $\binom{n}{k}$ is the number of ways of picking up $k$ stuffs out of $n$ boxes. The result follows.

Just a note - various people have helped me with my struggles the last few days, well the material's "learnt" and now I'm just doing questions. So far done around 25 limit questions, and already feeling more comfortable, and bizarrely, even starting to feel pangs of fun / enjoyment! Who knew???

And limits are my nemesis. When I get to derivatives, investigating functions, matrices, I'm much more comfortable

@topper that's what math does to you

you'll get used to limits soon

@BalarkaSen I really had my arse kicked the last few weeks, and I know I'm not stupid. But I love it when it starts to come together!

I know I'm stupid :(

@nablablah You know something wrong.

Hi Daniel

@nablablah Well, we all are and aren't. All relative. :)

I know I am stupid though. Everyone does.

In fact, I think I am not even stupid.

I kind of like being the dumbest person in the room, though

Learning new stuff all day

general question - do people believe it's a good idea to copy down the given equation(s) or whatever at the start of an answer? just to internalize it?

Yes @topper: Get used to writing complete mathematical sentences, verbs and all.

.

@Chris'ssis This almost choked me to death :O .. how did they find that function $\displaystyle \pi \cot(\pi z) \Big(\psi(-z) + \gamma \Big) \psi_{1}(-z)$ ?!

see creasson's posts on page 2 and 3 .. he did the alternating sums as well :O

What is the name for a continuous function with no endpoints?

@Balarka you are in the I&S forum right ? do you know by any chance who creasson is ?

@r9m who?

@BalarkaSen integralsandseries.prophpbb.com/…

nuh-huh

I know this seems like a really easy problem, but I need an exactly value as an answer. Can anyone clarify g(3) for me? i.imgur.com/Mxk7n9A.png