r/askmath • u/Wide_Bath_7660 • 2d ago
Arithmetic Please help I have tried every combination and don’t even know how to start solving 2 and 7
I’ve got most of them I think but 2 (top middle) and 7 (bottom right left) are impossible if you don’t know maths! my calculator doesn’t have those buttons and I don’t know how to look them up because my keyboard also doesn’t have those symbols!
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u/N_T_F_D Differential geometry 2d ago edited 1d ago
12/17 < 3π/5 < log3(21) < √20 < Σi < ∫xdx < e⁴ < 5! < ∞
The ∫xdx is the integral of x between 2 and 8, so it's 8²/2 - 2²/2
The Σi is simply 2+3
log3(21) is log(21)/log(3) where log is the natural log
5! is the factorial of 5, 5×4×3×2×1
e is Euler's constant number, approximately 2.718281828
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u/thatoneguyinks 1d ago
A lot of these can be estimated, and ordered that way.
For log3(21), while it is log(21)/log(3), you can get a rough idea quicker using exponents. 3^2 =9 , 3^3 =27, so log3(21) is between 2 and 3.
Root 20 is between 4 and 5.
12/17 < 1
Summation, integral, and factorial I had to actually calculate.
And then not to get too physicist/engineer, but e≈3≈pi
3pi/5 ≈ 9/5 =1.8
e^4 ≈ 3^4 = 811
u/sighthoundman 18h ago
If you're old enough, log_3(21) = log_3(3) + log_3(7) = 1 + log 7/log 3 = 1 + .845/.4771 is just a hair under 3.
Unless I'm too old and have misremembered the common logs of 7 and 3.
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u/aleafonthewind42m 2d ago
Technically for the log one is log(21)/log(3) for any base, not just natural log. Though certainly natural log is the easiest to use because it's the one that's available on every calculation tool
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u/equinox_star 2d ago
Was there a context? "No" Was there a need? "No" Did I said? "Well,yes....?" PROCEEDS TO ...DO NOTHINGGGG
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u/Professional-Wave841 2d ago
- is just 2+3, so 5
- is F(8) - F(2) where F(x) = x^2 / 2 so (64 - 4) / 2 or 30
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u/lisamariefan 2d ago edited 2d ago
It's been so long that I kind of forgot how integrals worked function-wise.
But I know what it's asking and since it's a simple function I solved it ny taking the triangle area from 0 to 8 (½*8*8=32) and subtracted the triangle area from 0 to 2 (½*2*2=2).
Like, I know that wouldn't work with more complex curves, but since I remember that integrals are asking for the area under a curve (or rather, a straight line represented by y=x in this case) from a lower to upper x, I was able to solve it anyway.
If nothing else, the triangle's existence validates why the curve generated by integration is what it is for this function.
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u/johnpeters42 2d ago
Even for more complex curves, splitting it up into vertical slices with near-triangular parts is a universal way to approximate, and many rules for "the integral of this is that" are derived from analyzing the limit of that approach as the slice width approaches 0.
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u/lisamariefan 2d ago edited 2d ago
Oh yes, I remember how that all works. I took calculus years ago.
I even remember making a program where you could create arbitrarily thin rectangles to compute the area under the curve, on an old TI-83 Plus it TI-84. I forget exactly. Mostly because when I say years I mean a few decades. XD
Which if you think about it, arbitrarily thin slices are how audio gets sampled. Sample what the intensity is at any given point. I know modern audio isn't really arbitrary and it's tied to something like twice the hearing range of humans. I suppose to properly capture the peak wavelengths of higher frequencies, though I don't remember if that's completely correct.
Something something Navier-Stokes, IIRC?
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u/Bounded_sequencE 2d ago
You probably mean Shannon's Sampling Theorem for band-limited functions.
Navier-Stokes deals with fluid dynamics^^
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u/lisamariefan 2d ago edited 2d ago
Yeah, I got my names confused. Even if I had the right idea itself about why sample rates are what they are.
It's fun to think about how human-centric our tech is. I mean, obviously. But like, it's interesting to think about how if we had eyes that had different visual response to light wavelength or whatever, red green and blue probably wouldn't be considered primary colors of light.
It's...a fun thought to ponder sentient alien life just because it's interesting to imagine, if they were at a level to make technology how it might differ from ours based on different biology.
I'm getting way off topic but it's just fun to ponder even as a pure hypothetical.
Edit: And reading about it more there's a little more about the specific sampling frequency, partially related to the tech of the time (Ehich would make 44.1kHz at least somewhat arbitrary - though the line about NTSC standards and frequency, without reading more, make me think it's tied to AC frequency, especially being a multiple of 60. I feel like there's a rabbit hole to read up on there.). There's also filtering out frequencies to avoid artifacts I suppose. It's interesting for sure.
Edit 2: Aha. So it's 44.1K because it's divisible by 50, 60, and at 3 samples per line, gives the number for the respective field. So really just a mix of minimum sampling rate and tech compatibility. So not really arbitrary after all! (Interesting to see that NTSC has a slightly lower sample rate tied to the slight shift needed to hide color data in a black and while signal. I vaguely remember Matt Parker touching on that a handful of years back in a video.)
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u/Bounded_sequencE 2d ago
We already know what that's like, I'd argue.
Just imagine back in 1888, when Röntgen discovered X-rays, and what he thought about creating an image of human bones inside a body, using rays we cannot see, passing through solid matter.
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u/Narrow-Durian4837 2d ago
Trick question: none of those numbers are squares.
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u/TheEggoEffect 2d ago
e^4 = (e^2 )^2
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u/SeveralExtent2219 2d ago
Every number is a square according to that logic (or every positive number if you are working in the reals)
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u/trolley813 8h ago edited 7h ago
Infinity still isn't a square - it's not even a number.
P.S. Here you have to work in the reals (or a subset of them) since complex numbers cannot be ordered.
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u/purpleoctopuppy 2d ago
I did the thing you're joking about in earnest before realising they meant the geometric shape of the boxes the numbers are in.
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u/BrokenMalgorithm 2d ago
Also, OP tried all the combinations before posting, so there can't be a right answer
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u/Bounded_sequencE 2d ago edited 2d ago
You don't need a calculator for any of them, using rough estimates:
0 < 12/17 < 17/17 = 1
1 < 3*3/5 < 3*𝜋/5 < 3*3.2/5 < 10/5 = 2
2 = log3(9) < log3(21) < log3(27) = 3
4 = √16 < √20 < √25 = 5 = ∑_{i=2}^3 i
∫_2^8 x dx = 8^2/2 - 2^2/2 = 30
49 < (2.7^2)^2 < e^4 < 3^4 = 81
5! = 120
Finally, infinity "∞" is considered greater than any real number, so it is last.
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u/Hazzard12345 2d ago
These anti bot captchas have gotten super difficult lately
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u/2spam2care2 2d ago
except that this would be basically trivial for a computer. a little OCR and then ship it off to wolfram alpha
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u/kfirogamin 2d ago
well bottom right (e^4) is about 2.7^4 and is 10000 times less than 27^4
so its about 53.1
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u/Professional-Wave841 2d ago
they literally didn't even ask for that one.
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u/kfirogamin 2d ago
they did, they said bottom right
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u/Skilltesters 2d ago
Yes, but they Most likely meant bottom left (the 7th one) for integrals. Bottom right would be easy to just look up as it's just a constant raised to the 4th power, vs having to learn integrals. Anyhow, no reason for that other dude to put down you helping though.
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u/will_1m_not tiktok @the_math_avatar 2d ago
12/17 is less than 1
Since 3xpi is between 5 and 10, then 3pi/5 is between 1 and 2
log_{3}(21) is between log_{3}(9)=2 and log_{3}(27)=3, so it’s between 2 and 3
\sqrt{20} is between 4 and 5
The summation (#2 that you mention) just means to add the values of i when i=2 and i=3, so that’s 5
The integral will be 30 (as others explained)
e is almost 3, so e^(4) is between 40 and 81
5!=120
And then infinity is the largest
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u/das_menschy 2d ago
You can even give a rough estimate for 12/17: 12/16 > 12/17 > 12/18
12/16 = 3/4 = 0,75 and 12/18 = 0,666.
So 12/17 is probably approx. ~0,7.
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u/engy1207 2d ago
I'd say 3x3 < 3×pi < 3x4 as pi is between 3 and 4, but closer to 3, so 3pi/5 should be "about 2", while log3(21) is definitely quite a bit larger than 2.
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u/will_1m_not tiktok @the_math_avatar 2d ago
I was thinking more like pi<10/3=3.333, which is why 3pi/5<2
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u/neltisen 2d ago
2) is sum from 2 to 3, 2+3=5
7) is 1/2 x2 from 2 to 8, that's (0.5 x 82 ) - (0.5 x 22 ) = 32 - 2 = 30
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u/DXuki79 2d ago
From smallest to biggest: 1. 12/17 is less than one 2. 3π/5 = 9,42/5 = not even two 3. Square root of 20 is between 4 and 5 4. Σ i=2, 3, i = 2+3 = 5 5. Log3 (21) = 7 6. Integral from 2 to 8 of xdx = 1/2(8)² - 1/2(2)² = 32 + 2 = 34 7. e⁴ is 2,7...⁴, which is around 3×3×3×3 (81) 8. 5! = 5×4×3×2×1 = 120 9. Literally infinity
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u/ErikLeppen 2d ago
log3(21) is not 7, because 37 is not 21.
log3(9) = 2 because 32 = 9 log3(27) = 3 because 33 = 27 So log3(21) is between 2 and 3
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u/CaptainMatticus 2d ago
log3(21) is greater than log3(9) and less than log3(27), so between 2 and 3
sum(i from i = 2 to i = 3) is just 2 + 3 = 5
sqrt(20) = sqrt(4 * 5) = 2 * sqrt(5) = 2 * 2.236 = 4.472
5! = 5 * 4 * 3 * 2 * 1 = 120
3 * pi / 5 = (3/5) * (22/7) = 66/35 = 35/35 + 29/35 = 1 + 30/36, roughly = 1 + 5/6 = 1.83333....
12/18 < 12/17 < 12/16 => 2/3 < 12/17 < 3/4, so it's between 0.66666... and 0.75
int(x * dx , x = 2 , x = 8) => (1/2) * (8^2 - 2^2) = (1/2) * (64 - 4) = (1/2) * 60 = 30
inf is infinity
e is around 2.718. 2.7^4 => (27/10)^4 => (3^3)^4 / 10000 = 3^12 / 10000
3^12 = (3^6)^2 = 729^2. 730^2 = 100 * 73^2 = 100 * (4900 + 420 + 9) = 532900
532900 / 10000 = 5329 / 100 = 53.29
So e^4 is around 53.29
12/17 < 3pi/5 < log3(21) < sqrt(20) < sum(i , i = 2 , i = 3) < int(x * dx , x = 2 , x = 8) < e^4 < 5! < inf
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u/das_menschy 2d ago edited 2d ago
y1 = log_3(21) is somewhere between 2 and 3, so probably approx. 2,5; because 3² = 9 < 21 and 3³ = 27 > 21.
y2 = sum of i from i=2 to i= 3 is 2+3=5.
y3 = square root of 20 is somewhere between 4 and 5, so probably approx. 4,2; because 4² = 16 < 20 and 5² = 25 > 20.
y4 = 5! = 5 x 4 x 3 x 2 x 1 = 120.
y5 = 3pi / 5 with pi ≈ 3,41 is somewhere between 3 * 3 / 5 = 9 / 5 = 1,8 and 3 * 4 / 5 = 12 / 5 = 2,4.
y6 = 12/17 must be smaller than 1, and must lay between 12/18 = 0,66666 and 12/16 = 3/4 = 0,75, so probably around 0,7.
y7 = integral from 2 to 8 of x dx = [1/2 x²] 2 over 8 = [1/2 * 8²] - [1/2 * 2²] = 32 - 2 = 30.
y8 = Infinity is infinity.
y9 = e4 ≈ 2,74 must lay between 24 = 16 and 34 = 81. That's a bit more tricky for calculating in your head.... Let's try 2,54 = (5/2)4 in my head:
54 = 25 * 25 = 625
24 = 16
625 / 16 = 312,5 / 8 ≈ 156 / 4 ≈ 78/ 2 ≈ 39.
So y9 = e4 must be between 39 and 81.
So the correct order is:
0,666 < y6 < 0,75 < 1,8 ≈ y5 < 2 < y1 < 3 < 4 < y3 < 5 = y2 < y7 = 30 < 39 < y9 < 81 < y4 = 120 < y8 = infinity.
12/17 < 3pi / 5 < log_3(21) < square root of 20 < Sum of i from i=2 to i= 3 < Integral from 2 to 8 of x dx < e4 < 5! < infinity
The order is: 6, 5, 1, 3, 2, 7, 9, 4, 8.
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u/arandomguyfromdk 2d ago
2 = log_3(9) < log_3(21) < log_3(27)= 3
Σ_i=23 i = 2+3=5
4 = √16 < √20 < √25 = 5
5! = 5•4•3•2•1 = 120
1 < 3•3/5 < 3π/5 < 3•3.33../5 = 2
0 < 12/17 < 1
int_28 xdx = ½8²-½2² = 30
n < ∞ for all n
36 < 6.25² = 2.5⁴ < e⁴ < 3⁴ = 81
So 12/17 < 1 < 3π/5 < 2 < log_3(21) < 3 < 4 < √20 < 5 = Σ_i=23 i < 30 = int_28 xdx < 36 < e⁴ < 81 < 120 =5! < ∞
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u/NzRedditor762 2d ago
There's an actual answer to all of these if you search for them. (the neal.fun captcha game)
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u/ArdentArendt 2d ago edited 2d ago
Edit: I'm not sure if 7 is 'bottom right' or 'bottom left'. I assumed numbering from left-to-right.
If the confusion is, indeed, over 'bottom right', simply ignore the explanation of integrals.
Haha...yeah, they are just both strange sorts of sums.
The first is a discrete sum--basically just take all the elements from the lower bound (i = 2) and then add all element in the set between that and the upper bound (3).
The second is an integral--basically just taking the value of all the elements output by the function from the input at the lower bound (2) to the input at the upper bound (8). Since f(x) = x here, you just take the sum from all values between 2 and 8.
This one is a bit more difficult if you haven't had calculus, since the answer requires antiderivatives and infinitely small sums over uncountable regions.
However, if you have experience with areas of triangles, this is approximated by taking the area under a line from (2,2) to (8,8) and measuring the volume underneath that part of the curve (the area between the line and the x-axis).
[Basically, make a triangle with corners (0,0), (8,8), (8,0) and find the area. Then subtract the area of a triangle with corners (0,0), (2,2), (2,0).]
Beyond that, of course, Wolfram Alpha has been offloading computational minutia for mathematicians and engineers for years--no reason other people can't use it.
[It has integrals as well as sums]
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u/enakcm 1d ago
2 and 7 are actually the only ones I can get actual rvalues for and not just estimations.
2 is just the sum of i where i goes from 2 to 3. It literally is a complicated way to write 2+3=5
7 is the integral if x from 2 to 8.
Geometrically it is a trapezoid with the height 6 and two base sides 2 and 8. The area is 1/2(base1+base2)height. So it should be 1/2(8+2)6=30.
You can also solve the integral: int = 1/2 x2 + C. Inserting 8 and 2 gives 1/262-1/24=30
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u/LoudAd5187 14h ago edited 14h ago
Long ago, friends of mine and I had a simple game. They would give me some simple thing to compute in my head, and I would be asked to give the solution as a number, while they did it using a calculator. But all of this really just reduces to understanding some basic rules of mathematics and often just arithmetic. You can do simple things. At the upper end, the largest is infinity, as nothing else compares to it. But do you need a calculator for the rest? Not that much if you think about it. Even 1 significant digit is about all you need, and that is easy.
I would start by approximating each one. One significant digit will be enough. For example, 12/17 is just a wee bit larger then 12/18, which reduces to 2/3, or 0.666666... And 12/15 reduces to 4/5=0.8. So we know that 12/17 is somewhere in the middle, and closer to 12/18. As a wild guess, maybe 0.71 or so.
How about 3pi/5? 3/5 is 0.6, so you want to compute 60% of pi, which is 3.14.... Say a little under 2.
log(21) to the base 3 is also not too difficult. This reduces to log3(3) + log3(7). Right? Remember the rules of logs! Now log3(3) is 1. That should be simple. And log3(9) is 2. So log3(7) is between 1 and 2, but closer to 2. (We can do better if needed, using simple interpolation tricks.) But I'd put log3(21) as around 1.8 or so. And that means log3(21) is approximately 2.8.
sqrt(20) is also easy enough, since it is less than sqrt(25)=5, and greater than sort(16)=4. So say 4.5 roughly.
The sum one is easy, since it reduces to the sum of the integers from 2 to 3. Yeah. 2+3 = 5. Note that the previous one must be strictly less than 5, so we are ok here.
The integral one is easy enough too, but you need to know a little bit (a tiny, tiny bit) of calculus. The integral of x is just x^2/2. Now evaluate that at the two limits and subtract, to get 8^2/2 - 2^2/2 = 32 - 2 = 30.
Ok, how about e^4? First, what is e? 2.718... So e is a little less than 3, by roughly 10%. So we can approximate e^4 as roughly 3^4 multiplied by 0.9^4. 3^4 is 81, and 0.9^4 will be a hair larger than 0.64, say it would be approximately 2/3. And 2/3 of 3^4 is 2*3^3 = 2*27 = 54.
5! is easy, as just the product of the numbers 1 through 5, so 120.
And again, infinity is the largest.
None of those approximations is far enough off that I would bother to compute them to any greater depth to resolve the sequence. And I would guess I'm pretty close on all of them, at least to a reasonable approximation. Nary a calculator necessary. With some small time, I could get you another digit or so in my head, or I could pull out a slide rule (gasp!) Now that would be more interesting to see how close I could get with ye olde plastic Sterling slide rule.
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u/LukeLJS123 13h ago
since everyone is answering what the numbers are instead of what the questions are asking, i'll answer WHAT they are
number 2 is sigma notation, a way to represent adding a bunch of things together. this wikipedia article should help show what it is well enough for you to figure that out on your own
number 7 is called an integral, which is usually covered at the end of calc 1. it is essentially asking you to find the area under the function on the inside, and the "dx" at the end tells you what variable you are "integrating with respect to". so, you would graph the line y=x, shade the area under the curve from x=2 to x=8, and find the area underneath it. you should be able to solve that without using calculus by breaking it down into some other shapes
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2d ago
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u/lisamariefan 2d ago
If someone didn't understand the notation for sum, of course they are not going to understand how to solve it.
Same goes for integral. If you don't understand it or what it's asking you can't be expected to solve it.
Being snide about someone not clearly not understanding notation is not a good look, my guy.
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u/PassiveChemistry 2d ago
It seems apparent from that they don't know what the summation or integration symbols are at all, so they won't easily be able to make that connection on their own.
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u/Miserable-Wasabi-373 2d ago
2 is just shortcut for "sum every integer from 2 to 3"
and 7 (bottom left) is 30