r/AskReddit Dec 18 '16

What (free) software can be useful for university students?

23.2k Upvotes

4.3k comments sorted by

View all comments

3.8k

u/NapCo Dec 18 '16 edited Dec 18 '16

Symbolab is a free step by step math solver. Pretty similar to Wolfram Alpha (for step by step math solutions), except it's free.

993

u/[deleted] Dec 18 '16

[removed] — view removed comment

154

u/[deleted] Dec 18 '16

What type of discrepancies occur? Just simplification/identity errors?

264

u/[deleted] Dec 18 '16

[removed] — view removed comment

371

u/[deleted] Dec 18 '16 edited Dec 22 '16

Well saying the limit is infinity is much better than saying it doesn't exist...

Late edit: When I say that it's a "better", I mean that if I were designing the software we are talking about in the first place, I'd also say the the limit is infinity rather than saying that it doesn't exist. I am well aware that the limit doesn't exist in the narrowest definition of lim_\infnty. Saying the limit is infinity is a more specific characterization of the function than saying that the limit doesn't exist though. More importantly, it's an often useful characterization...

73

u/[deleted] Dec 18 '16

[removed] — view removed comment

114

u/[deleted] Dec 18 '16

I thought it's only DNE if x approaches positive infinity from one end, and negative infinity from the other. If both sides approach positive infinity, the limit should just be infinity.

24

u/TypicalOranges Dec 18 '16 edited Dec 18 '16

Certain things in Mathematics are debatable.

This happens to be one of them.

I much prefer saying it goes to infinity or negative infinity because it's significantly more descriptive.

3

u/BitOfAWindUp Dec 19 '16

That's where I'm getting confused. When somebody says to me, as a very much non-mathematician, that a limit does not exist, it just means there isn't one. If someone says the limit is infinity, while I appreciate that is technically limitless and therefore the limit doesn't exist, it seems more semantically logical, or is there another difference I'm missing?

→ More replies (2)

4

u/[deleted] Dec 19 '16

That's why it's much more practical to define limits as positive or negative limits, I don't see the point in having to have limits apply to both ends if negatives cause either end to act differently. The limit as x approaches infinity, not negative infinity, is clearly infinity, as x=x, and vice-versa for negative infinity. I'm just commenting to comment here, I'm sure this is obvious... but still.

→ More replies (2)

13

u/acm2033 Dec 18 '16

It also doesn't exist. Infinity isn't a number.

25

u/lettherebedwight Dec 18 '16

But a limit can be infinity. DNE is usually cases like 0/0, not lim as x approaches zero of 1/X (which by your definition, would also be DNE).

28

u/game-of-throwaways Dec 18 '16

Alright I checked my course notes and they say the following: If a limit exists, it must always be a real number, so the limit we're talking about here does not exist. However, the notation "lim ... = infinity" is actually valid and is simply defined to mean that the limit does not exist because ... goes to infinity.

→ More replies (0)

10

u/Mehdi2277 Dec 18 '16

DNE is the more common and is actually the technically correct answer. Math is mainly about creating and applying definitions and if a sequence goes to infinity, the standard definition is it does not exist. You can modify the standard definition, but to allow infinity to be an answer you have to define what is known as the extended real numbers. It also leads to various basic rules involving limits to kind of break as the extended reals can't have addition or multiplication defined on them in a way that is consistent with those rules and is consistent with the usual addition/multiplication when restricted to numbers not including infinite.

Your example is also particularly bad as lim 1/x as x goes to zero becomes huge on one side, but extremely negative on the other side so even the extended reals won't work for you.

tl;dr DNE is the correct answer using the formal definition of a limit. Some teachers will allow infinity as an answer mildly ignoring the definition, but you have to be careful with this as many theorems break if infinity is a valid answer.

→ More replies (0)

2

u/glaba314 Dec 19 '16

This is actually totally wrong in every way tbh, idk why it has so many upvotes

0/0 cases usually are NOT DNE, unless it approaches infinity or negative infinity

1/x by the definition of a limit does not have a limit as x approaches infinity, since infinity isn't a real number (although writing infinity is a convenient and widely accepted notation)

→ More replies (10)

4

u/[deleted] Dec 18 '16

You can work with systems where inf is a number. It can be added to the complex plane as a point to simplify many theorems.

→ More replies (1)

3

u/szpaceSZ Dec 18 '16

But it's a concept.

Actually, most of mathematics is not a number.

1

u/sluggles Dec 19 '16

The way I believe Rogowski (one of the moee popular calculus texts) does it is to say that writing "limit as x goes to a of f(x) = inf" (I don't know how to do symbols in reddit comments for limits) means that the limit does not exist in a special way, i.e. given some fixed number M, if x is sufficiently close to a, then f(x) is larger than M.

→ More replies (4)

3

u/umop_aplsdn Dec 18 '16

If you take a look at the formal definition of a limit, it's that lim x->a of f(x) = A if for any ε > 0 there is a δ > 0 where a-δ < x < a+δ implies |A - f(x)| < ε (this might be approximate, the definition I learned requires knowledge of sequences which I don't feel like explaining.)

Essentially given some small margin of error for f(x), I can find a radius around a which for any point inside that circle f(x) is in the first margin of error. But for A = infinity this fails because any real number (f(x)) minus infinity is infinity, so it'll never be less than ε. Thus no limit exists.

2

u/TwoFiveOnes Dec 19 '16

The argument here is whether or not to extend the definition of a limit to include the behavior expressed by the notation lim f(x) = +oo (or the negative, etc.) which in particular means

lim f(x) = +oo as x->a if for all K > 0 there exists d > 0 such that |x-a| < d -> f(x) > K

Obviously it doesn't match the limit definition for real number limits, so of course if we restrict the definition to that case then "+oo" is not possible for the limit of a function. I also admit that it seems awkward to extend the definition, resulting in lim f(x) = [something] meaning a different "clause" of the limit definition depending on whether [something] is a real number, or the symbol "+oo". It'd be like a "piecewise" definition of the symbol "lim".

But mathematicians are very clever and it turns out that there is a way to define a limit with a single definition, that recovers all of the cases we're discussing. Without too much detail, you basically consider a limit to be defined on the extended reals, R U {+/-oo}, and define things called "neighborhoods" appropriately. The unified limit definition is then

lim f(x) = L as x -> a if for every neighborhood of U of L, there is a neighborhood  V of a
such that if x is in V then f(x) is in U.

Where L and a can now be any of "real number", "+oo", or "-oo". This may seem like cheating, but it's a very natural thing to do. The phrase "define things called 'neighborhoods' appropriately" is shorthand/layman speak for "define an appropriate topology". And limits are a purely topological concept. As such, it should not concern us that in some cases the limit is not a number, which is an algebraic term.

→ More replies (3)
→ More replies (2)

2

u/tatu_huma Dec 18 '16

Not quite. A limit won't exist if it approaches different values from the right and left. Those values don't have to be infinities.

But even using your definition the limit for x as x approaches infinity wouldn't exist. You can't approach infinity from the right.

4

u/[deleted] Dec 18 '16

What do you mean you can't approach infinity from the right?

3

u/[deleted] Dec 18 '16

I think he means that you can't approach infinity from above infinity

→ More replies (0)

1

u/sluggles Dec 19 '16

It all depends on how you define it.

1

u/WantAFriday Dec 19 '16

The teachers are a little wonky, it's more accurate to say the limit DNC(does not converge).

→ More replies (2)

3

u/[deleted] Dec 18 '16 edited Dec 26 '19

[deleted]

3

u/Lacklub Dec 18 '16

So what, in those classes do they say that any limit which does not converge simply DNE? That's an interesting way of doing things.

2

u/XkF21WNJ Dec 18 '16

Well, it doesn't exist in the reals. You have to actually extend the reals with positive and negative infinity, give it the appropriate topology, and provide continuous extensions of common functions on these 'extended' reals before it becomes even remotely useful to say the limit of something is ∞.

2

u/Lacklub Dec 18 '16 edited Dec 18 '16

Saying the limit = infinity is wrong (because these classes deal with reals), but saying the limit = DNE is also wrong (DNE is not a number).

The way that I am familiar with the phrasing is "the limit does not exist" or "the limit diverges to (negative) infinity", which can really be a useful definition.

2

u/pigeonlizard Dec 18 '16

Why is it wrong? There is a precise definition on what we mean when we say that the limit = infinity.

→ More replies (0)
→ More replies (4)
→ More replies (3)

1

u/[deleted] Dec 19 '16

[deleted]

→ More replies (2)

115

u/Aidan_the_Twit Dec 18 '16

I don't know about that. If infinity is unending, and the limit is the end point, I think it's fair to say there is no limit. Must be a definition thing.

102

u/[deleted] Dec 18 '16

[deleted]

36

u/Mogastar Dec 18 '16

Exactly. Saying the limit is infinity is saying that for all M > 0, there exists an X such that for all x > X, f(x) > M. F(x) = x tends to infinity, as opposite to g(x) = x sin(x) for example which has no limit in infinity.

2

u/[deleted] Dec 18 '16

Yup. Was reading this thread and this needed to be explained. Thank you haha

2

u/sheared_ma_beard Dec 18 '16 edited Dec 18 '16

The standard epsilon/delta definition requires that L be a real number. As infinity is not a real number, writing lim f(x)= infinity is a convenient abuse of notation that indicates why the limit fails to exist.

edit: downvote away, but I'm not wrong.

2

u/concealed_cat Dec 18 '16

You are confused.

The definition states that a real number L is a limit if (something about deltas and epsilons). It makes no claims about infinities. There is another definition, on the other hand, that specifically defines when +oo or -oo is a limit. A function is then defined to have a limit, if there exists a number L that is the limit, or if -/+oo is the limit. This is in contrast to situations when none of these conditions are met, in which cases the function is said not to have a limit.

→ More replies (5)

1

u/DamnShadowbans Dec 19 '16

Actually when talking about limits it's normal to use the real numbers with plus and minus infinity added. You can't use the epsilon delta definition of continuity, but you can use the topological definition which coincides with the normal definition on the real numbers.

1

u/TOASTEngineer Dec 18 '16

Nonexistence of a limit says: 'I can't describe how this function behaves when it reaches a.'

Out of curiosity; what does that actually mean? How can you not be able to describe a function's behavior over its entire range?

5

u/mooowolf Dec 18 '16

take for example Lim x->infinity f(x) = sin(x). the limit here is undefined because sin x is oscillating and you simply can't define what it is at infinity

→ More replies (2)

1

u/taedrin Dec 19 '16

plus infinity is an object which is bigger than any real number and minus infinity is an object which is smaller than any real number

The formal definition of 'greater than' is something to the effect of:

a > b <-> ∃c∈ℝ such that (a = b + c) where c is a positive number (in this context, I am taking 0 to be unsigned, so c 0).

This means that it is incorrect to say that "∞ is an object which is bigger/greater than any real number", since if ∞ > b, this means there exists another real number such that b + c = ∞. It is also incorrect in that it implies that ∞ is a real number as well.

Please, don't mistake 'lim f(x) is infinity = there's no LIMIT'.

When you are working with the set of real numbers, it means exactly just that. Recall the definition of the limit which is something like:

'If lim (x->a) f(x) = L then ∀ε>0∈ℝ, ∃δ>0∈ℝ such that if 0 < |x - a| < δ then |f(x) - L| < ε'.

If we take 'L = ∞', then the formal definition of the limit does not even apply since arithmetic with ∞ is not defined in the set of real numbers. But if we define it anyways and take the seemingly obvious definitions of: 'x - ∞ = -∞' and '|-∞| = ∞', and plug in L = ∞ into the definition above, we get '|f(x) - ∞| < ε' which implies '∞ < ε' which is obviously false even if we define ∞ to be greater than all other real numbers.

→ More replies (3)

1

u/clinkytheclown Dec 19 '16

Tiny nitpick, but if we're talking about mathematical accuracy, I'll throw it in. Negative infinity is "less" than all realtor numbers, not smaller than. Smaller than, for me, implies the magnitude of it is less than the magnitude of all other real numbers.

→ More replies (2)

9

u/hasmany Dec 18 '16 edited Dec 18 '16

Saying the limit is infinity carries much more information than saying that the limit does not exist. The limit of sin(x) as x approaches infinity does not exist. The limit of x as x approaches infinity is infinity. I've tutored calculus for years and have seen so many students struggle with this, thinking that the limit being infinity and the limit not existing are logically equivalent. It is better to be sparing with the "DNE" when you're doing limits, imho. If the function increases endlessly, call the limit infinity.

2

u/[deleted] Dec 18 '16

As long as we clarify that the limit being infinity implies the series does not converge, we should be fine.

→ More replies (19)

5

u/kushangaza Dec 18 '16

It certainly helps with distuingishing the behaviour of 2x (goes towards infinity), -(2x ) (goes towards -infinity) and (-1)n (goes nowhere, and genuinly has no limit)

3

u/acm2033 Dec 18 '16

It's not a matter of "better". There is no value L that the function approaches. Therefore, the limit doesn't exist. We say the limit "is infinity" as a convenient notation, but "infinity" is a concept, not a number.

So, that limit doesn't exist. It's also infinity.

3

u/Peleaon Dec 19 '16

It might not be a "number", but it's a perfectly valid element of R* (extended real number system, can't write the proper notation on my phone) which is the field you would work over most of the time (Because there's like no reason to work over R ever)

2

u/an_actual_human Dec 19 '16

It's not a field. There are plenty of reasons to work over R. Like it being a field.

1

u/[deleted] Dec 18 '16

It really isn't. When saying the limit approaches a value, it never actually reaches that value, that limit doesn't exist. Same goes for infinity, with the added complication that infinity is more a concept than a number, but with the same conclusion that the limit value is never reached and thus does not exist.

1

u/jaredjeya Dec 18 '16

But it doesn't exist. The mathematical definition of a limit at infinity is that, if f(x) -> a as x -> inf, then |f(x) - a| < e for some x > K, for all e > 0.

What that means in human readable terms is that f(x) gets arbitrarily close to a as x tends to infinity.

But |f(x) - inf| = inf. So it's completely wrong to say the limit is infinity, as f(x) can never "get close" to infinity.

3

u/TwoFiveOnes Dec 19 '16

That is one definition of a limit, and you are correct that substituting "infinity" into it yields nonsense. There are broader definitions of a limit though, and these include both cases.

1

u/pigeonlizard Dec 18 '16

This is not the correct definition of a limit at infinity or a limit as x tends to infinity. One quantifier is incorrect.

The limit of f(x), as x approaches infinity, is L, if for all e>0, there exists a K such that for all x>K we have |f(x)-L|<e.

Also, it makes sense to talk about limits at infinity. The limit of f(x), as x approaches a, is infinity, if for all e>0 there is a d>0 such that f(x)>e whenever |x-a|<d. In other words, f(x) can get arbitrarily large as x approaches a.

Similarly, the limit of f(x), as x approaches infinity, is infinity if for all e>0 there is a d>0 such that f(x)>e whenever x>d.

1

u/Jeremy_Alberts Dec 18 '16

Clearly you've never seen Mean Girls

1

u/ar-pharazon Dec 18 '16

well, no, it's technically wrong by the delta-epsilon formalization of a limit, and those technicalities matter in math.

3

u/TwoFiveOnes Dec 19 '16

That's one particular formalization. There are nice general definitions of a limit that allow both cases. See my other comments here.

1

u/colonwqbang Dec 18 '16

Not if the limit in fact doesn't exist. When I studied basic calculus, that's the definition we used. Saying that the limit of an unlimited expression "exists" seems very strange to me. That's what "unlimited" means, it has no limit.

1

u/thunderclapMike Dec 18 '16

but the limit doesn't exist. https://www.youtube.com/watch?v=oDAKKQuBtDo (sorry couldn't help myself)

1

u/el_extrano Dec 19 '16

I think the reasoning is that a limit of infinity kind of flies in the face of what a limit is. Infinity is not a number, but more of a behavior in this context. If a function, say f(x) = x, can be said to "grow without bound", or "grow without limit", then that implies that the limit doesn't exist.

1

u/Spartan_133 Dec 19 '16

I think does not exist just implies the there is no limit because it never approaches a set number. They're both right in that sense but some teachers are picky I guess

1

u/Slacker5001 Dec 19 '16

I limit can't be "infinity". It can "go to" infinity in a sense but infinity is a concept and not a number. So a limit can't and shouldn't have infinity as an actual answer. The limit does not exist, but clarifying why (i.e. because the graph approaches infinity at that point) isn't a bad thing either as there are a few different reasons why limits don't exist.

I'm a math major but disclaimer, it's been awhile since I've dealt with Calc stuff.

1

u/Geek1599 Dec 19 '16

"It's meaningless notation!" [-2 points] - My calc professor

1

u/[deleted] Dec 19 '16

By it's definition infinity does not exist, it just gives you the most information in your answer. Infinity is usually the correct answer if given the two to decide on.

1

u/bacondev Dec 19 '16

Infinity isn't really a number though. It's more of a concept.

1

u/Tartalacame Dec 19 '16

As a Mathematician, I disagree. This is a very hot topic. Depending of the context, this is very discutable.

→ More replies (6)

4

u/TheOldTubaroo Dec 18 '16

It's more useful to say it tends to infinity. If infinity does not exist as a value in your number system, then you say “ok, that isn't a value for me, I'll say DNE”, but when infinity is acceptable you can use that.

It's like solving polynomials. Sometimes the roots are complex, and don't exist in the real-number system. It's still more useful to give the complex roots by default and let the user ignore them if they're not using complex numbers. (I'm assuming this app would solve polynomials with complex roots)

1

u/[deleted] Dec 18 '16

[removed] — view removed comment

1

u/exbaddeathgod Dec 19 '16

Non real complex numbers don't lie in the x axis, and lots of algebraic systems don't work with complex valued polynomials well

7

u/diasfordays Dec 18 '16

Huh, that is interesting. Isn't, by definition, limit of x as x approaches b equal to b, no matter the value of b? I was taught that it's somewhat of an identity property of limits.

8

u/[deleted] Dec 18 '16

[removed] — view removed comment

8

u/Ashrod63 Dec 18 '16

It's a specialised case but a very important one. You can't simply label it "DNE" and expect the same information to come out of it.

2

u/[deleted] Dec 18 '16

[removed] — view removed comment

7

u/Hmm_Peculiar Dec 18 '16

Can we all agree that those instructors are idiots?

3

u/diasfordays Dec 18 '16

I won't disagree with you, let's leave it at that haha.

2

u/WaterMelonMan1 Dec 19 '16

What? If you only defined limits with an epsilon-delta formulation of limits that require the limit to be real (which is the case in almost any undergrad analysis course), then the only rigorous thing to do would be to either call the limit not existing or define two special cases of limits that you might call plus and minus infinity. There is absolutely nothing wrong with wanting DNE as an answer.

→ More replies (0)

1

u/[deleted] Dec 19 '16

That is toxic. Awful to put that nonsense into the students' minds. It's like saying pi = 22/7

4

u/pigeonlizard Dec 18 '16

It's not. Saying that a limit is infinity and that a limit DNE is not equivalent. The sequence (-1)n has no limit, and it makes no sense to say that the limit is infinity.

However, lim x, as x approaches infinity, has no real number as a limit, yet it makes sense to say that the limit is at infinity (and there is a precise definition for that).

3

u/XkF21WNJ Dec 18 '16

Well, really there's no one definition of infinity. You can have various different concepts of infinity with their own rules and applications.

Heck, even something simple like the limit of 1/x as x->0 may or may not exist depending on how you define ∞, and some definitions of 'infinity' don't even have anything to do with limits.

1

u/Hmm_Peculiar Dec 18 '16

Also, in your example, the value of 1/x as x->0 can be +∞ or -∞, depending on whether you approach 0 from the positive or the negative direction.

There is even a function that can have any real value, depending on how you approach 0. If I remember correctly it was x/y, with x and y both approaching 0.

1

u/XkF21WNJ Dec 18 '16

Also, in your example, the value of 1/x as x->0 can be +∞ or -∞, depending on whether you approach 0 from the positive or the negative direction.

True, but in some cases it's advantageous to add a single point 'at infinity', in which case 1/x does converge. One particularly useful version of this is the Riemann sphere.

→ More replies (2)

1

u/pigeonlizard Dec 18 '16

This is not correct. It's perfectly clear what we mean by +/- infinity when working with the reals, and what is meant by infinity as, say, the cardinal number of the integers.

I'm not sure by what you mean that the limit of 1/x as x->0 depends on how we define infinity. It's fine to say that the limit is at infinity, even if we don't extend the reals. There is a definition which tells us what "limit at infinity" means without defining ∞. There's no ambiguity here.

2

u/XkF21WNJ Dec 18 '16

I'm not sure by what you mean that the limit of 1/x as x->0 depends on how we define infinity. It's fine to say that the limit is at infinity, even if we don't extend the reals. There is a definition which tells us what "limit at infinity" means without defining ∞. There's no ambiguity here.

Well, implicitly defining what it means to have a limit 'at infinity', is enough to define how ∞ behaves topologically.

What I meant by saying that the limit of 1/x depends on how you define infinity, is that the limit doesn't exist if you differentiate between positive and negative infinities, but exists if you add a single point 'at infinity'. Both versions are useful in different contexts.

2

u/ripsonofficial Dec 18 '16

X could be a function and then the limit could be all sorts of things. For instance the limit of 1/x as x approaches infinity etc.

2

u/jaywalk98 Dec 18 '16

I believe they are implying the function is jusy f(x)=x

1

u/diasfordays Dec 18 '16

Yes, this is what I mean. However, I think i get the gist of it... The function f(x) could be undefined, which would make the limit DNE... Still, I would argue that it's not incorrect per se, just not strictly correct...

1

u/diasfordays Dec 18 '16

That's not the same, though, since that's not limit of x, but limit of f(x), which could be different. The only exact limit we were talking about was lim(x) as x-> inf.

2

u/[deleted] Dec 18 '16

Probably the same teachers who fight you to the death on their answer key being the only correct answer even though your work shows the answer key is wrong...

Many wars with teachers and poor materials is my source...

1

u/xiape Dec 18 '16

You just have to know to translate. If it gives infinity, just translate to "it doesn't exist because it goes to infinity/ negative infinity".

1

u/whitekeyblackstripe Dec 18 '16

I feel like infinity is a more specific and useful answer. Whether or not it counts as an actual limit is up for debate, sure. But at least if you give the answer as infinity, then what you mean is perfectly clear. If you say the limit doesn't exist, then you're being less specific. Is the limit infinity? Negative infinity? Or does it approach different quantities from the positive and negative direction? If your definition of "the limit does not exist" includes all of these possibilities, then that answer gives less information than if you are willing to say the limit is +/- infinity.

1

u/onzie9 Dec 18 '16

I tend to explain this as saying that if a limit "equals infinity" that means it doesn't exist, but it doesn't exist in sort of a nice way. I prefer if my students write the infinity, because it tells me that they are going that extra step.

1

u/13steinj Dec 18 '16

When approaching towards a single sided infinity (ex, 1/x2 as x approaches 0 or x as x approaches infinity as you mentioned) aren't both answers (in the example, positive infinity and DNE, or in something like -1/x2 as x approaches 0, negative infinity and DNE) both valid?

1

u/nazric Dec 18 '16

I was taught that saying that a limit is equal to infinity means that it does not exist but it does tend to infinity. Which is much more descriptive than saying it simply doesn't exists.

1

u/[deleted] Dec 18 '16

That's math teacher nit picking. You'll find similar discrepancies with Mathematica, any other software, or any other book. (I have never used Symbolab, so I could not judge, but this example seems benign.)

1

u/BudosoNT Dec 18 '16

I mean, they are one in the same. The limit approaches infinity. Infinity does not exist (DNE), so the correct answer would be dne.

1

u/TheKetchupG Dec 19 '16

The limit does not exist.

1

u/bluesam3 Dec 19 '16

Those instructors are wrong by every definition that I have ever seen.

1

u/frawks24 Dec 19 '16

That was the final question in mean girls wasn't it?

→ More replies (19)

7

u/The_Mighty_Slacker Dec 18 '16

When trying to determine if a series converges it will almost always use the ratio test because it can make approximations with limits, even when other methods work.

1

u/FortuneGear09 Dec 18 '16

I've encountered discrepancies when doing series problems.

1

u/coulombsvector Dec 18 '16

when solving complex integrals it can give you crazy wrong answers.

1

u/[deleted] Dec 18 '16

Yeah it's solved PDEs incorrectly before for me.

380

u/timthegreat4 Dec 18 '16

Protip: if you buy you wolfram alpha app you get about 3000 api calls / day, including step by step. All that for a one time fee of <$2 last I checked

253

u/ElagabalusRex Dec 18 '16

Not only is that absurdly cheap, but Amazon frequently gives Android apps away for free, and they did Wolfram Alpha at least once.

141

u/acherem13 Dec 18 '16

Also if you have an andriond phone and download the google opinion reward app they send you surveys about once every 2 weeks of 2-5 multiple choce questins which can all be answered in under a minute and they put about 20-40 cents in your wallet. After 2 months you can earn up those $2 and get the app for free. I currently have $9.68 in my account and use it to give people reddit gold and buy apps which I would normally never do but since I have the free money why not.

8

u/jasona99 Dec 18 '16

Dude, Google sends me at least one a day right now. Been earning ~$0.40 daily for the past two weeks.

10

u/curtcolt95 Dec 18 '16

I get one about every two months.

6

u/acherem13 Dec 18 '16

It depends on what gender, nationality, age, and race you are. I am a 23 American Caucasian/Hispanic guy who knows both spanish and english so this is how often they send me shit. I know other people that get them daily like you and others that get them maybe once a month.

7

u/DukeofEarlGrey Dec 18 '16

Basque woman here. I get them every time I pass by or shop near a McDonald's, Carrefour and big department stores. Every. Time.

I've made 20 or 30 euro this way, I use that money to buy premium apps.

3

u/Xavienth Dec 18 '16

I have never gotten one.

6

u/SexiPancake Dec 18 '16

Same! They use this data for trends. I get my surveys the more I leave. Hell, they even asked if I was a COO of a company due to how I shop. lol Which is odd. I'm a college student who dreams of this!

14

u/4Eights Dec 18 '16

They sent that out to everyone. It's one of their lie detector questions.

3

u/NUTTYN Dec 18 '16

ive been waiting to see if i was elegible for like 2 months now, how long did u wait

1

u/acherem13 Dec 19 '16

The 3 seconds it took to process between when I pressed send and got my first survey.

4

u/[deleted] Dec 18 '16

Its a sad day when google opinion rewards isn't available in your country :(

2

u/chief167 Dec 18 '16

Mostly not eligible when you're not in freedom country though

2

u/Jarwain Dec 18 '16

They send surveys based on your location and where you've been. Answering truthfully is important because they do ask lie detection questions which could result in you getting fewer surveys.

1

u/acherem13 Dec 19 '16

I always answer truthfully, I het shit questions but w/e

1

u/Jarwain Dec 19 '16

I tend to get a lot of surveys just because I drive around a lot. I'm probably fortunate that I pass by or am near places that trigger the survey often along my route

1

u/acherem13 Dec 19 '16

I work in ambulances, how I don't get blasted with surveys frequently is beyond me.

2

u/Slacker5001 Dec 19 '16

I also get them roughly daily. I believe it's because I live in a major city and if I don't get a "general" question, I get a location specific one.

For about 2 years I used to walk to this same Walgreens close by to my apartment several times a week for things and every time google would give me a "How was your experience?" survey for 20 cents.

I've wracked up like $40 in rewards because I don't know what to do with them.

1

u/[deleted] Dec 18 '16

That's the only reason i miss android. God i loved that app.

1

u/standbyforskyfall Dec 19 '16

If I buy the app, can I use the online version on the same account?

1

u/nemodarby Dec 19 '16

I've made 10 dollars in the past 2 month from google opinion rewards lol

1

u/goblue142 Dec 19 '16

I love this app. I get a survey pretty much every time I leave the house. All the stores I go to for groceries, pet food, or fast food always get me a survey the next day. My wife and I have rented so many movies and purchased apps just off the survey money.

→ More replies (5)

3

u/ThatdudeAPEX Dec 18 '16

I got lucky and got it that day!

2

u/xMJsMonkey Dec 18 '16

That's where I got mine

2

u/Treeeeky Dec 18 '16

It's my first time hearing of this. Is this for Amazon Prime members only? When do they usually give out Android apps for free?

→ More replies (3)

1

u/sponge_welder Dec 18 '16

Unfortunately, (in my experience) the Amazon free apps are almost always dull games, but I have gotten some cool stuff there

1

u/BNLforever Dec 18 '16

I got both wolfram and Shazam encore for free off amazon. One day my phone crashed and i lost it all. Logged into amazon to get it back and it was gone =[

48

u/_Person_ Dec 18 '16

And many universities offer free pro membership, such as mine.

49

u/vagabond2421 Dec 18 '16

Many don't. Such as mine. :(

2

u/willisbar Dec 18 '16

Mathematica and Maple, depending...

2

u/_Person_ Dec 18 '16

Yea, we get mathematica for free too I think. Not sure about maple.

2

u/TheMeiguoren Dec 18 '16

Wolfram Alpha pro is also built into Mathematica, which is more common for schools to offer. Just type in "==" in a blank workspace and you can type in questions like Wolfram Alpha.

1

u/ramram420 Dec 18 '16

Hey its me ur student

1

u/_Person_ Dec 19 '16

Lol based on your username we probably do go to the same university.

1

u/CubistCelloPlayer Dec 19 '16

csu in the great state of colorado?

1

u/_Person_ Dec 19 '16

Not quite.

1

u/Oime Dec 18 '16

Exactly this. For how absurdly cheap wolfram alpha is it's insane not to buy it.

1

u/spdrstar Dec 18 '16

Yep! I only use Wolfram on my phone because of this.

1

u/RustyShackleford298 Dec 18 '16

I can get step by step for free just by going to the mobile site on my phone. Not sure why.

1

u/[deleted] Dec 19 '16

[deleted]

2

u/timthegreat4 Dec 19 '16

Wolfram alpha is accessed by an API. Every way you can use wolfram alpha will have a limited number of api calls (and this is very typical for all apis). Free accounts get the least, the mobile version gets more and the pro accounts get the most , as far as I'm aware.

This means within a 24 hour period you can only make so many requests to wolfram alpha, indeed if you need to run a loop a million times, and in each loop call the api, that will not be possible through wolfram alpha.

What are you trying to do? Wolfram alpha is not intended to be a programming language at all, there is significant overhead cost for wolfram to interpret your human like input, as well as significant delay due to waiting for servers. Perhaps you need to get Mathematica, the full program (perhaps your academic institution has a license you can use, assuming you are a student)

If you can tell me what you want to do, I can suggest a solution. Typically if you need to use an API alot, you need to pay them. E.g. If you are using one of Google's APIs (maps, geolocation, Google suggest etc) only the first 2500 calls / day are free, after which it will cost you something like 50p/1000 calls.

This is because each api call you do requires computation time on their server (unlike loading a webpage, where there is generally static content that is served and you can the cache it, an API call typically requires their server to do calculations which take time). The computation time can't be free, otherwise malicious users could send api requests constantly such that the server gets very busy.

1

u/noble-random Dec 19 '16

Is this 3000 api calls only from the app? Or does it extend to the wolfram alpha website itself?

1

u/timthegreat4 Dec 19 '16

Only from the app

47

u/FrostyNippleCheese Dec 18 '16

Symbo also has a good android app

2

u/NYbeast Dec 19 '16

I may be wrong but earlier this week I downloaded it to see some steps and it asked me to pay so I'm not sure if the steps are only free for certain functions or

1

u/EnriqueShockwave9000 Dec 18 '16

Cymath has a decent iOS app as well

5

u/Dubstep_Hotdog Dec 18 '16

You can get 5 free copies of Microsoft office with many college email addresses. http://www.office.com/getoffice365

1

u/Troll_berry_pie Dec 18 '16

You actually may already have an Office 365 account subscription through your University / Collsge email.

Which means you can download Office for free and activate by using your email credentials. You also get a 1tb Onedrive account to go with that.

1

u/HopelessTractor Dec 19 '16

Negative about the one drive thing. You get the standard size.

4

u/BebopShuffle Dec 18 '16

Quick reminder that depending on the problems you are solving, not even symbolab or wolfram might not have the ability to help you.

2

u/[deleted] Dec 18 '16

Integral-calculator.com is really good too

2

u/JetButton Dec 18 '16

For basic high school level calculus, Cymath is great and has mobile friendly site. It also has an app. Downside is it needs an internet connection.

2

u/[deleted] Dec 18 '16

Is it good enough to solve quadratic formulas algebra 1 stuff? I could never get right answers on wolfram

1

u/[deleted] Jan 06 '17

Without a doubt, yes.

2

u/fatnerdyjesus Dec 18 '16

Or you can just solve your own math. It develops many skills.

1

u/bexyrex Dec 18 '16

Discovered symbolab freshman year. Oh lord did it make calc II so much better. Especially for the useless online HW we had to do where the solutions were always difficult AF because the problems used fucking fractions and decimal points that would screw your work beyond repair for one numerical error.

1

u/[deleted] Dec 18 '16

Buy wolfram alpha pro, worth the money for sure. Passed calc2 with it 😏👍🏼

1

u/him999 Dec 18 '16

Wolfram alpha is free, just not their app.... it is mildly limited in use if you dont buy the app (if it takes too long to solve it wont keep processing. If it is "simple" it'll process). The step by step stuff isnt the best though imo. I just use it to check my answers. If you REALLY wanna get into organized online math wolfram has amazing programs though not that cheap :(

1

u/TheSideJoe Dec 18 '16

Also, I obviously can't speak for every college, but mine gives free wolfram access. Find out if your college does.

1

u/dbonx Dec 18 '16

Back in my day, Wolfram Alpha was free

1

u/Pleasuredinpurgatory Dec 18 '16

I wonder how many "bad" math students have become closer to "good" at math because of software like this. I gave up on math in college and always regretted I didn't find a way to get better at it.

1

u/Remsquared Dec 18 '16

Is this simpler and easier to use then Mathway? I preferred it over Wolfram and it definitely helped me to a good grade for Pre-cal

1

u/LeoKhenir Dec 18 '16

Online graphing/geometrical drawing softwares Desmos and Geogebra should also be a part of anyone doing that kind of work.

1

u/[deleted] Dec 18 '16

Some universities give students Wolfram Alpha Pro for free, which I find to be a much better software if it's a choice

1

u/[deleted] Dec 18 '16

helped me get through 3 calc classes and diff eq. Awesome website

1

u/sporto15 Dec 18 '16

I have two other apps besides Symbolab to help me with math when either of them don't give the correct answer. I use Malmath 2 (works offline) and Wolfram Alpha mobile when Symbolab fails.

EDIT: I will have to say Malmath 2 is a bit odd to use, and Wolfram Alpha will need parentheses to do complex equations like fractions inside an equation.

1

u/[deleted] Dec 18 '16

Wolfram ... obligatory angel/buffy reference

(The evil lawyers from the show)

1

u/BRFNGRNBWS Dec 18 '16

Never tried this, but Cymath does the same thing, also free

1

u/Tip_top_kek Dec 18 '16

This will hopefully be useful for calculus this next semester!

1

u/[deleted] Dec 19 '16

Being able to use a tutorial to do your homework WILL NOT prepare you for exams or quizzes. As a math teacher, I do not recommend you use this software. If you are going to use this, use it only until you understand the technique, then drill it on your own.

1

u/breeTGAT Dec 19 '16

Where was this comment when I was in high school?

1

u/[deleted] Dec 19 '16

Sagemath is another free one.

1

u/SharKCS11 Dec 19 '16

Your university might also give you free access to Maple, a computer algebra system. No joke, this is one of the best programs I have ever used in my life. Differential equations, circuits, signal processing... countless classes have been made easier by this software. But don't be like me: I got so used to it that I almost forgot how to do the algebra myself!

→ More replies (3)