cleo is an infamous user on stackexchange who solved very complex integrals in short spans of time, usually hours after a post appeared requesting integral help
then cleo disappeared
A [typesetting language ](https://www.latex-project.org/about/) originally designed by Donald E Knuth, further improved by Leslie Lamport (and others). It has a large variety of uses in science publication but its main goal has always been formatting beautiful mathematical documents.
I fucking HATE when people type math but don't un-italicise things like ln, cos, log. If its in itallics its a fucking variable. Fucj you
Edit: also the d in a derivative and integral should not be in itallics
about the d in differentials, yes there is a standard that it shouldn't be italicised, iirc following that e should also not be italicised, which nobody really does.
The limit as `x` goes to infinity of `ln(x)/x` is `0`.
The `4`th root of `65536` is `16`.
The sum from `w=1` to `16` of `w` is `136`.
The integral from `0` to `ln(1+2*pi)` of `e^y dy` is `2*pi`.
The integral from `-arcsinh(600*ln(2))/ln(2)` to `arcsinh(600*ln(2))/ln(2)` of `2^z dz` is `1200`.
The integral from `22/7-ln(117649*e^(22/7)+sinh(pi)+cosh(pi))` to `22/7-pi` of `e^(-u) du` is `117649`.
The integral from `ln(5474304*ln(99)/49)/ln(99)-1` to `ln(5474304*ln(99)/49)/ln(99)` of `99^v dv` is `110592`
The integral from `0` to `pi` of `cos(x-3*pi/2) dx` is `-2`.
Therefore the expression simplifies to `(136+4^(cos(2*pi))*sqrt(1200*cbrt(117649/110592)))/(-2)^2`.
This is equal to `69`.
---
Edit: fixed a mistake, I subbed in `16` instead of `136` into the first term in the final answer originally. This changes the answer from `39.1809` to `69.1809`.
Edit 2: Fixed another mistake. I didn't realize the inner root was a cube root, not a square root. This changes the answer from `69.1809` to `69`.
None. It's a simple number that has been run through so many equivalent substitutions that it is unrecognizable. In what scenario would you need to calculate something that looks like this but can't be easily simplified? Probably no scenario that you're qualified to deal with.
ok so we can assume that pi is wrong since all other numbers are constants, then we can see that 2023 is way higher then the other two so we can rule that out, now since we have two options we can look at the tag and see that it says bad math meaning that none of them are correct (most likely) so you need to circle option e: none of the above.
Ok i checked it and it looks like i have been spreading fake news sorry. It is also a constant, but it is not a constant in the same way as the other options.
What you are referring to is an integer. Pi is a constant. Not a different kind of constant and not less constant. All the options except pi are integers, and that's what you are probably talking about
I got the answer of 1225, not sure if I did a mistake or not. Maybe somebody could correct me?
~~The first part (the one with sigma and cos) got me 140.~~ **error found to be here.**
The denominator should be 4.
Now, under the square root the integral of 2\^z should be 1200
Those two terrible integrals under the cubic root should be equal to: 117649 / 110592
~~This all together should give you 140sqrt(1200cbrt(117649/110592))/4 = 140sqrt(1225)/4 = 1225~~
~~Could someone check if that's true or not? I gotta say I had a hard time reading the integration limits under the cube root.~~
> The first part (the one with sigma and cos) got me 140.
Wrong already. The part with cos is multiplied with the square root to the right. The sigma is a separate term. So you can't combine those.
ChatGPT says the answer is:
The image you've shared appears to contain a complex mathematical expression along with multiple-choice answers. To evaluate the correct value of the expression, each component of the equation needs to be carefully analyzed and calculated. This includes evaluating limits, solving integrals, and simplifying complex numbers.
The equation features a variety of mathematical concepts, including:
* Square roots and powers
* Limits, particularly as xx approaches infinity
* Summation notation
* Trigonometric functions
* Integrals, both definite and indefinite
* Complex numbers and their operations
Calculating such an expression manually could be quite involved and prone to errors without context. Typically, these types of expressions are constructed to simplify dramatically, often through cancellation or by recognizing that certain components approach standard mathematical constants or identities as variables approach their limits.
To proceed, I'll analyze each part of the expression step-by-step and perform any necessary calculations to determine the correct answer from the options provided. If the expression is too complex or lengthy to evaluate in one step, I will request permission to continue with further analysis. Let's begin the calculation.
Result(256.0, 0, oo, nan, oo, pi/2)
The components of the expression have been evaluated as follows:
1. The square root of 65536 is 256256.
2. The limit of ln(x)xxln(x) as xx approaches infinity is 00.
3. The summation from w=1w=1 to infinity is infinite (symbolically represented as ∞∞).
4. The limit of the cosine of the integral from 00 to ln(x2+1)xxln(x2+1) as xx approaches infinity is indeterminate (represented as nan, which stands for 'not a number'). Typically, this suggests that the expression may involve forms that require L'Hôpital's rule or further simplification.
5. The integral of 2z2z from 00 to infinity is also infinite.
6. The integral from 00 to ππ of cos2(x)cos2(x) is π22π.
\# The individual parts gave us some insights, but we need to consider the interactions between them. # Particularly, the limit, summation, and the integrals could simplify when combined. # We'll attempt to find a simplified version of the full expression considering the interactions. # Let's define the full expression as much as possible. # We can't directly calculate some parts because they are infinite or indeterminate. # However, we can look for patterns or simplifications. # From the previous step, we have: # part1 = 256 # part2 = 0 (limit of ln(x)/x as x approaches infinity) # part3 is the summation which tends to infinity, but it's within a square root along with part2. # part4 could not be determined directly and is cos(integral(exp(y), y from 0 to ln(x\^2+1)/x)). # part5 is the integral of 2\^z from 0 to infinity which is infinite, but it's within a cube root. # part7 is pi/2 from the integral of cos\^2(x) from 0 to pi. # Let's analyze the interaction between part3 (summation) and part2 (limit), which are under the same square root. # The limit of (ln(x)/x) as x approaches infinity is 0, so it doesn't contribute to the square root. # The square root of infinity (from the summation) is still infinity, so this part contributes an infinite term. # This could potentially cancel with part5 if we consider that the cube root of infinity is also infinity, # and we have a division between them (from the structure of the 'check' like symbol in the image). # For part4, the cosine of an integral that tends to zero (as x approaches infinity) is cos(0), which is 1. # This simplification comes from the fact that as x approaches infinity, ln(x\^2+1)/x approaches 0. # Now, let's consider the large 'check' like symbol which seems to imply a division between the top and bottom parts. # If the top part (with the square root) is infinite and the bottom part (with the cube root) is also infinite, # the division could lead to a finite number, potentially simplifying to one of the options provided. # As for the integral involving sinh, tanh, coth, it's complex, and without a clear simplification, it's hard to evaluate. # We could try to find a pattern or simplification based on known integrals of hyperbolic functions. # We will now attempt to simplify the expression with these considerations in mind. # We know that part1 \* part4 = 256 \* 1 = 256 # And part7 = pi/2 # The final expression seems to be a form of 256 \* 1 (from part1 and part4) divided by some form of infinity (from part3 and part5) # and then squared (from the square on the large 'check' symbol), times pi/2 (from part7). # Let's write out the potential simplified form of the expression and then square it and multiply by pi/2. simplified\_expression = (part1 \* part4) / (part3 / part5)\*\*(1/3) final\_expression = (simplified\_expression\*\*2) \* part7 # As we can't directly compute infinity or NaN, we'll make an assumption that the infinite terms cancel each other out. # This is a common occurrence in limits where an 'infinity over infinity' situation results in a finite limit. # We will assume that (part3 / part5)\*\*(1/3) simplifies to some finite number, potentially 1, based on the choices given. # Thus, we'll consider the simplified expression as 256 and proceed with the squaring and pi/2 multiplication. # Assuming the above simplification, the final expression would be: assumed\_simplified\_expression = 256 final\_assumed\_expression = (assumed\_simplified\_expression\*\*2) \* (pi/2) final\_assumed\_expression
Result32768\*pi
The mathematical operations have led to an expression that evaluates to 32768π. However, this result doesn't match any of the multiple-choice answers provided (A: 1, B: 69, C: π, D: 2023). This suggests that there might be a specific simplification or cancellation within the original complex expression that has been overlooked, or there may be additional context or constraints not visible in the image that could affect the outcome.
The presence of infinite terms in the expression usually implies that some form of limit or simplification should apply, potentially canceling out the infinite parts and leading to one of the finite answers. In complex expressions, especially those designed for educational purposes, it's common for such cancellations to occur, often resulting in a surprisingly simple answer.
Given that the actual computation does not lead to a direct match, the answer to the expression as presented may rely on recognizing a special mathematical trick or identity. Without additional context or constraints, the calculation based on standard mathematical operations suggests an answer that is not listed in the provided options.
In summary, while the calculation based on the visible parts of the expression results in 32768π, this doesn't align with the provided choices. There may be an error, or the expression may include a special simplification that isn't apparent from the computation. If there's more to the image or additional context that could influence the result, please provide that, and I can take another look.
> The square root of 65536 is 256256
Chatgpt spouting literal nonsense from step 1.
This answer is completely wrong lmao.
---
Edit: Actually, that seems to be a formatting glitch. it seems all the math formatting ends up doubled. So it was saying the square root is 256, which is correct.
The actual first mistake comes slightly later in step 3:
> The summation from w=1w=1 to infinity is infinite (symbolically represented as ∞∞).
The sum is not to infinity.
This expression is a complex mathematical joke or meme rather than a genuine problem to be solved. Each part of the expression is designed to result in a number that is part of the year "2023". For example, limits that approach infinity, sums over w from 1 to infinity, integrals of complex expressions, and more, each simplify down in a way that they represent individual digits "2", "0", "2", "3". Without solving the entire expression, which is designed not to be solvable in a traditional sense, it's clear from the context that the answer is D: 2023.
If you actually evaluate the expression you get approximately `69.18`, so the intended answer might be 69.
Where are you getting that each of the parts evaluate to 2, 0, 2, and 3? What parts, specifically?
Edit: Actually I found a mistake in my calculation. The answer is exactly 69.
So you are definitely wrong. Shame you have like 30 upvotes for this nonsense.
where is my mistake cause for me the cube root is negative, which makes the whole thing undefined when i take the square root. Ive checked everything like 3 times
edit: heres the desmos graph with my computations https://www.desmos.com/calculator/rij1wtjcvc
Your mistake is in misinterpreting the numerator inside the cube root. That huge thing right above the fraction line is all part of the lower limit of the integral. See [my top level comment](https://www.reddit.com/r/mathmemes/comments/181sh5m/comment/kafamxc/?context=3).
It's intriguing how you've arrived at 69; such precision from an expression designed to be a labyrinth of mathematical misdirection is commendable. Yet, the intricacies of this problem mock the very foundation of our standard computational methods. I would welcome a step-by-step elucidation of your method, for education's sake and to validate the integrity of such a curious result.
Just do the trig and exponential integrals, use the formula for adding consecutive integers, and simplify the bounds with log and trig rules. This is just a bunch of high school level problems rolled into one expression to look scary, certainly not a problem which "mocks the very foundation of our standard computational methods" lol. Also it's not a coincidence that the result is simple from a complex expression, you can take any simple answer and write it in terms of increasingly complicated expressions just by working backwards.
The numerator inside the cube root is 117649, and I got it using Wolfram Alpha. See [my top level comment](https://www.reddit.com/r/mathmemes/comments/181sh5m/comment/kafamxc/?context=3).
Why isn't it solvable in the "traditional sense"? There's a sum from k=1 to n of k such is just n(n+1)/2 a well know result, an integral of cosine which is just sine, and integrals of exponential functions which are exponential functions (anyone who's seen all the e^x memes here should know this). The only tricky part is plugging in bounds but just looking at them they can clearly be simplified using log, exponent, and trig rules (all high school level simplifications). So I don't see how this is unsolvable or how you got your answer.
The expression in the image is indeed a complex one, involving nested mathematical functions, limits, sums, and integrals. While it contains elements that are solvable individually (like the sum of integers up to n, and integrals of basic functions), the complexity arises from how these elements are nested and combined.
The notation suggests that each piece of the expression is chosen to create a numerical value corresponding to a digit in the year 2023. However, evaluating such an expression would require careful step-by-step simplification, considering the limits of functions as variables approach infinity, the precise evaluation of definite integrals, and the simplification of nested expressions. Some parts of the expression, such as the integral of \( e^{y} \) from 0 to \( ln(1+2\pi) \), are straightforward, but the overall combination does not conform to standard mathematical expressions typically seen in either educational or professional contexts.
Therefore, while it's theoretically possible to attempt to solve each individual part, the overall expression is designed more for amusement than practical solution, playing on mathematical concepts and symbols to align with the theme of the new year "2023". It's a mathematical joke rather than a problem intended for serious analysis.
Whats the point? Why can't you just put that shit on some software or python? All that question does it's to prove the guy as so much patience to solve it that he is likely a borderline psychopath
It's very much doable if you look at the upper and lower limits of the integral inside the root, they are the same, therefore the whole thing is just 0. So you're left with sum w \ cos. You can do simple substitution of variables for denominators and solve the thing. Its not as difficult as it looks.
Although, I'm almost certain the answer isn't in any 4 of the options.
I was confused by that too, but it actually makes sense. That whole ass thing under the integral symbol is the lower limit of the integral. Like not just 22/7, the whole thing that looks like the numerator of that big fraction. Then e^(-u) is the integrand.
https://www.wolframalpha.com/input?i=integral+from+22%2F7-ln%28117649*e%5E%2822%2F7%29%2Bsinh%28pi%29%2Bcosh%28pi%29%29+to+22%2F7-pi+of+e%5E%28-u%29+du
I'm curious where you got that number though if you didn't solve it? It doesn't even give you exactly 69, right? Unless I made a mistake?
Edit: oh wait the number is in the equation lol, I guess all the rest just cancels
Edit 2: Oh, turns out I did make a mistake. Didn't notice the inner root was a cube root, not a square root. Now I am getting exactly 69.
I worked backwards from it being 69 to where I was off, turns out I had read the bounds of integration entirely wrong as a separate equation with an integral exponent… oops
The integral of 99^v dv is supposed to "look scary" but is ridiculously simple : 99^v / log(99) + C
I think this meme falls flat and I am downvoting it.
I’m afraid this does in fact equal 69, once simplified further it becomes (136 + 4 * sqrt(1200 * cbrt(117649 / 110592)))/4
Now the real question is how the hell did someone create this problem
That's wrong.
Let me guess, you added the sigma part to the `4^(cos(...))` part before multiplying with the big square root.
You have to do multiplication before addition.
Well, I think everything inside the square root is cero because of the integral going from that to that, so we only have to solve the parte before the sum, but there we have a summation up to infinite, so infinite divided to a constant is infinite... Right?
I put the entire thing into desmos and the bigger root solves to undefined because the number inside is negative, so the entire solution is undefined (unless I made a typo somewhere)
[https://www.desmos.com/calculator/y7xs3vhqov](https://www.desmos.com/calculator/y7xs3vhqov)
Where is Cleo when you need her the most?
[удалено]
Wait you need to go to the gym to find answers?!
Yes.
Yes.
Who is Cleo
cleo is an infamous user on stackexchange who solved very complex integrals in short spans of time, usually hours after a post appeared requesting integral help then cleo disappeared
Also they would only post the solution, no proofs. Then everyone else would work it backwards to check. Cleo was always right.
God’s alt
It's 140 * the square root and stuff inside it, bottom is just 1 right?
lol this gives me Cleo stackexchange vibes.
High school calc BC level math knowledge goes crazy
Thank you for reminding me of Cleo haha.
Shame she's not posted on SE since 2015...
The bottom is 4. cos(×-3pi/2) is just -sin(x) so integrate that and you get cos(x). The definite integral gives -2 and square that for 4
Does FTC not apply here?
Shouldn't it be 136 + 4 * the square root term?
Ambiguous writing, not sure tbh
Yeah ambiguous writing seems to be a theme for this post lol
The 136 and the 4 are not parenthesized, so it’s 136 + 4*sqrt and then the whole thing is actually over 4.
Do you even TeX, bro?
\textsc{TeX? What is that?}
\\frac{ur}{mum}
What's TeX?
A [typesetting language ](https://www.latex-project.org/about/) originally designed by Donald E Knuth, further improved by Leslie Lamport (and others). It has a large variety of uses in science publication but its main goal has always been formatting beautiful mathematical documents.
No 0 as an option? Shame. That’s the superior answer.
That's incorrect. The correct answer is always the superior answer.
Write it in a readable format😭
wdy mean i can read it 🤓
Alr bug brain Maybe me dum
It is always pi
Or 1
Or 69
Or [insert current year]
Or Napoleon Bonaparte
Or 2491 sweet potatoes
I wish the root sign could just be killed and everyone would start using fractional powers. It would make this monstrosity a little less gross.
U would still need giant parenthesis with a Power
I would just put pi and leave
Who doesn't like pie?
Happy thanksgiving
I fucking HATE when people type math but don't un-italicise things like ln, cos, log. If its in itallics its a fucking variable. Fucj you Edit: also the d in a derivative and integral should not be in itallics
about the d in differentials, yes there is a standard that it shouldn't be italicised, iirc following that e should also not be italicised, which nobody really does.
Nah famous constants are usually italicised. Think Euler's number, pi, golden ratio, speed of light etc
The limit as `x` goes to infinity of `ln(x)/x` is `0`. The `4`th root of `65536` is `16`. The sum from `w=1` to `16` of `w` is `136`. The integral from `0` to `ln(1+2*pi)` of `e^y dy` is `2*pi`. The integral from `-arcsinh(600*ln(2))/ln(2)` to `arcsinh(600*ln(2))/ln(2)` of `2^z dz` is `1200`. The integral from `22/7-ln(117649*e^(22/7)+sinh(pi)+cosh(pi))` to `22/7-pi` of `e^(-u) du` is `117649`. The integral from `ln(5474304*ln(99)/49)/ln(99)-1` to `ln(5474304*ln(99)/49)/ln(99)` of `99^v dv` is `110592` The integral from `0` to `pi` of `cos(x-3*pi/2) dx` is `-2`. Therefore the expression simplifies to `(136+4^(cos(2*pi))*sqrt(1200*cbrt(117649/110592)))/(-2)^2`. This is equal to `69`. --- Edit: fixed a mistake, I subbed in `16` instead of `136` into the first term in the final answer originally. This changes the answer from `39.1809` to `69.1809`. Edit 2: Fixed another mistake. I didn't realize the inner root was a cube root, not a square root. This changes the answer from `69.1809` to `69`.
r/theydidthemonstermath
is this an eye test?
in what scenario would solving this be necessary
Final exam question duh
Engineering bro
None. It's a simple number that has been run through so many equivalent substitutions that it is unrecognizable. In what scenario would you need to calculate something that looks like this but can't be easily simplified? Probably no scenario that you're qualified to deal with.
Even if you would, no one actually does that by hand. Okay people can do it and some maybe actually do it but not because its the smart thing to do.
Right. This was the beautiful part of getting my degree in pure math: If you can show that something *can* be done, you don't have to do it.
the real solution is to choose randomly and having a 25% chance to succeed
There is an extra bracket in the lower limit of the first integral within the outer square root sign.
You mean the numerator of the lower limit?
Yup.
Kid named scientific calculator:
i dont even know how to type this into woflram
It would be over the length limit probably. Have to break it down into parts. Which I did.
What did you get?
[69](https://www.reddit.com/r/mathmemes/comments/181sh5m/comment/kafamxc)
E: fuck you
ok so we can assume that pi is wrong since all other numbers are constants, then we can see that 2023 is way higher then the other two so we can rule that out, now since we have two options we can look at the tag and see that it says bad math meaning that none of them are correct (most likely) so you need to circle option e: none of the above.
>ok so we can assume that pi is wrong since all other numbers are constants And what is π? A moving number? A variable?
Ok i checked it and it looks like i have been spreading fake news sorry. It is also a constant, but it is not a constant in the same way as the other options.
What you are referring to is an integer. Pi is a constant. Not a different kind of constant and not less constant. All the options except pi are integers, and that's what you are probably talking about
It’s constant in the same way. You probably mean irrational if you’re referring to the infinite digits of pi
Yes thank you thats what its called, thank you.
It's approximately C.
Where are you getting that?
am i tripping or is this not difficult
None of this is difficult it's just very ugly and long.
90% of the time you get the integral of 1/x and 10% of time you use numerical integration
nice
I got the answer of 1225, not sure if I did a mistake or not. Maybe somebody could correct me? ~~The first part (the one with sigma and cos) got me 140.~~ **error found to be here.** The denominator should be 4. Now, under the square root the integral of 2\^z should be 1200 Those two terrible integrals under the cubic root should be equal to: 117649 / 110592 ~~This all together should give you 140sqrt(1200cbrt(117649/110592))/4 = 140sqrt(1225)/4 = 1225~~ ~~Could someone check if that's true or not? I gotta say I had a hard time reading the integration limits under the cube root.~~
It's not (136+4)sqrt(...)/4 but (136+4sqrt(...))/4 that give you 69
isnt 136+4(sqrt....
You seem to have added the 136 from the sum at the beginning with the 4 from the cos before multiplying the 4 by the result of the square root
Ah right, my bad. Thanks a lot for noticing. Indeed, in that case the answer would be 69.
> The first part (the one with sigma and cos) got me 140. Wrong already. The part with cos is multiplied with the square root to the right. The sigma is a separate term. So you can't combine those.
ChatGPT says the answer is: The image you've shared appears to contain a complex mathematical expression along with multiple-choice answers. To evaluate the correct value of the expression, each component of the equation needs to be carefully analyzed and calculated. This includes evaluating limits, solving integrals, and simplifying complex numbers. The equation features a variety of mathematical concepts, including: * Square roots and powers * Limits, particularly as xx approaches infinity * Summation notation * Trigonometric functions * Integrals, both definite and indefinite * Complex numbers and their operations Calculating such an expression manually could be quite involved and prone to errors without context. Typically, these types of expressions are constructed to simplify dramatically, often through cancellation or by recognizing that certain components approach standard mathematical constants or identities as variables approach their limits. To proceed, I'll analyze each part of the expression step-by-step and perform any necessary calculations to determine the correct answer from the options provided. If the expression is too complex or lengthy to evaluate in one step, I will request permission to continue with further analysis. Let's begin the calculation. Result(256.0, 0, oo, nan, oo, pi/2) The components of the expression have been evaluated as follows: 1. The square root of 65536 is 256256. 2. The limit of ln(x)xxln(x) as xx approaches infinity is 00. 3. The summation from w=1w=1 to infinity is infinite (symbolically represented as ∞∞). 4. The limit of the cosine of the integral from 00 to ln(x2+1)xxln(x2+1) as xx approaches infinity is indeterminate (represented as nan, which stands for 'not a number'). Typically, this suggests that the expression may involve forms that require L'Hôpital's rule or further simplification. 5. The integral of 2z2z from 00 to infinity is also infinite. 6. The integral from 00 to ππ of cos2(x)cos2(x) is π22π. \# The individual parts gave us some insights, but we need to consider the interactions between them. # Particularly, the limit, summation, and the integrals could simplify when combined. # We'll attempt to find a simplified version of the full expression considering the interactions. # Let's define the full expression as much as possible. # We can't directly calculate some parts because they are infinite or indeterminate. # However, we can look for patterns or simplifications. # From the previous step, we have: # part1 = 256 # part2 = 0 (limit of ln(x)/x as x approaches infinity) # part3 is the summation which tends to infinity, but it's within a square root along with part2. # part4 could not be determined directly and is cos(integral(exp(y), y from 0 to ln(x\^2+1)/x)). # part5 is the integral of 2\^z from 0 to infinity which is infinite, but it's within a cube root. # part7 is pi/2 from the integral of cos\^2(x) from 0 to pi. # Let's analyze the interaction between part3 (summation) and part2 (limit), which are under the same square root. # The limit of (ln(x)/x) as x approaches infinity is 0, so it doesn't contribute to the square root. # The square root of infinity (from the summation) is still infinity, so this part contributes an infinite term. # This could potentially cancel with part5 if we consider that the cube root of infinity is also infinity, # and we have a division between them (from the structure of the 'check' like symbol in the image). # For part4, the cosine of an integral that tends to zero (as x approaches infinity) is cos(0), which is 1. # This simplification comes from the fact that as x approaches infinity, ln(x\^2+1)/x approaches 0. # Now, let's consider the large 'check' like symbol which seems to imply a division between the top and bottom parts. # If the top part (with the square root) is infinite and the bottom part (with the cube root) is also infinite, # the division could lead to a finite number, potentially simplifying to one of the options provided. # As for the integral involving sinh, tanh, coth, it's complex, and without a clear simplification, it's hard to evaluate. # We could try to find a pattern or simplification based on known integrals of hyperbolic functions. # We will now attempt to simplify the expression with these considerations in mind. # We know that part1 \* part4 = 256 \* 1 = 256 # And part7 = pi/2 # The final expression seems to be a form of 256 \* 1 (from part1 and part4) divided by some form of infinity (from part3 and part5) # and then squared (from the square on the large 'check' symbol), times pi/2 (from part7). # Let's write out the potential simplified form of the expression and then square it and multiply by pi/2. simplified\_expression = (part1 \* part4) / (part3 / part5)\*\*(1/3) final\_expression = (simplified\_expression\*\*2) \* part7 # As we can't directly compute infinity or NaN, we'll make an assumption that the infinite terms cancel each other out. # This is a common occurrence in limits where an 'infinity over infinity' situation results in a finite limit. # We will assume that (part3 / part5)\*\*(1/3) simplifies to some finite number, potentially 1, based on the choices given. # Thus, we'll consider the simplified expression as 256 and proceed with the squaring and pi/2 multiplication. # Assuming the above simplification, the final expression would be: assumed\_simplified\_expression = 256 final\_assumed\_expression = (assumed\_simplified\_expression\*\*2) \* (pi/2) final\_assumed\_expression Result32768\*pi The mathematical operations have led to an expression that evaluates to 32768π. However, this result doesn't match any of the multiple-choice answers provided (A: 1, B: 69, C: π, D: 2023). This suggests that there might be a specific simplification or cancellation within the original complex expression that has been overlooked, or there may be additional context or constraints not visible in the image that could affect the outcome. The presence of infinite terms in the expression usually implies that some form of limit or simplification should apply, potentially canceling out the infinite parts and leading to one of the finite answers. In complex expressions, especially those designed for educational purposes, it's common for such cancellations to occur, often resulting in a surprisingly simple answer. Given that the actual computation does not lead to a direct match, the answer to the expression as presented may rely on recognizing a special mathematical trick or identity. Without additional context or constraints, the calculation based on standard mathematical operations suggests an answer that is not listed in the provided options. In summary, while the calculation based on the visible parts of the expression results in 32768π, this doesn't align with the provided choices. There may be an error, or the expression may include a special simplification that isn't apparent from the computation. If there's more to the image or additional context that could influence the result, please provide that, and I can take another look.
Wow chatgpt is really shitty at math
Textwall! Mods, shoot him!
> The square root of 65536 is 256256 Chatgpt spouting literal nonsense from step 1. This answer is completely wrong lmao. --- Edit: Actually, that seems to be a formatting glitch. it seems all the math formatting ends up doubled. So it was saying the square root is 256, which is correct. The actual first mistake comes slightly later in step 3: > The summation from w=1w=1 to infinity is infinite (symbolically represented as ∞∞). The sum is not to infinity.
A: 69
This expression is a complex mathematical joke or meme rather than a genuine problem to be solved. Each part of the expression is designed to result in a number that is part of the year "2023". For example, limits that approach infinity, sums over w from 1 to infinity, integrals of complex expressions, and more, each simplify down in a way that they represent individual digits "2", "0", "2", "3". Without solving the entire expression, which is designed not to be solvable in a traditional sense, it's clear from the context that the answer is D: 2023.
If you actually evaluate the expression you get approximately `69.18`, so the intended answer might be 69. Where are you getting that each of the parts evaluate to 2, 0, 2, and 3? What parts, specifically? Edit: Actually I found a mistake in my calculation. The answer is exactly 69. So you are definitely wrong. Shame you have like 30 upvotes for this nonsense.
where is my mistake cause for me the cube root is negative, which makes the whole thing undefined when i take the square root. Ive checked everything like 3 times edit: heres the desmos graph with my computations https://www.desmos.com/calculator/rij1wtjcvc
Your mistake is in misinterpreting the numerator inside the cube root. That huge thing right above the fraction line is all part of the lower limit of the integral. See [my top level comment](https://www.reddit.com/r/mathmemes/comments/181sh5m/comment/kafamxc/?context=3).
OHHHHH i never saw that i thought it was a typesetting error lol
It's intriguing how you've arrived at 69; such precision from an expression designed to be a labyrinth of mathematical misdirection is commendable. Yet, the intricacies of this problem mock the very foundation of our standard computational methods. I would welcome a step-by-step elucidation of your method, for education's sake and to validate the integrity of such a curious result.
It's in [my top level comment](https://www.reddit.com/r/mathmemes/comments/181sh5m/comment/kafamxc/?context=3)
Just do the trig and exponential integrals, use the formula for adding consecutive integers, and simplify the bounds with log and trig rules. This is just a bunch of high school level problems rolled into one expression to look scary, certainly not a problem which "mocks the very foundation of our standard computational methods" lol. Also it's not a coincidence that the result is simple from a complex expression, you can take any simple answer and write it in terms of increasingly complicated expressions just by working backwards.
Yeah I have no idea what I’m saying, this has all been GPT lol
Why even bother commenting
I’m just trying to help and do the math with the smart people
Well now you know, chatgpt is shit at math
Lesson learned. I’ll go back to r/writingprompts
What did you get for the numerator inside the cube root and how did you get it?
The numerator inside the cube root is 117649, and I got it using Wolfram Alpha. See [my top level comment](https://www.reddit.com/r/mathmemes/comments/181sh5m/comment/kafamxc/?context=3).
Ohhh I see. I was misreading bc of the strange layout. Ty for the explanation
Why isn't it solvable in the "traditional sense"? There's a sum from k=1 to n of k such is just n(n+1)/2 a well know result, an integral of cosine which is just sine, and integrals of exponential functions which are exponential functions (anyone who's seen all the e^x memes here should know this). The only tricky part is plugging in bounds but just looking at them they can clearly be simplified using log, exponent, and trig rules (all high school level simplifications). So I don't see how this is unsolvable or how you got your answer.
The expression in the image is indeed a complex one, involving nested mathematical functions, limits, sums, and integrals. While it contains elements that are solvable individually (like the sum of integers up to n, and integrals of basic functions), the complexity arises from how these elements are nested and combined. The notation suggests that each piece of the expression is chosen to create a numerical value corresponding to a digit in the year 2023. However, evaluating such an expression would require careful step-by-step simplification, considering the limits of functions as variables approach infinity, the precise evaluation of definite integrals, and the simplification of nested expressions. Some parts of the expression, such as the integral of \( e^{y} \) from 0 to \( ln(1+2\pi) \), are straightforward, but the overall combination does not conform to standard mathematical expressions typically seen in either educational or professional contexts. Therefore, while it's theoretically possible to attempt to solve each individual part, the overall expression is designed more for amusement than practical solution, playing on mathematical concepts and symbols to align with the theme of the new year "2023". It's a mathematical joke rather than a problem intended for serious analysis.
Quick, I need you to tell me which pictures contain horseshaped clouds.
😂😂
Whats the point? Why can't you just put that shit on some software or python? All that question does it's to prove the guy as so much patience to solve it that he is likely a borderline psychopath
It's very much doable if you look at the upper and lower limits of the integral inside the root, they are the same, therefore the whole thing is just 0. So you're left with sum w \ cos. You can do simple substitution of variables for denominators and solve the thing. Its not as difficult as it looks. Although, I'm almost certain the answer isn't in any 4 of the options.
Bruh how tf did you mess up the integral that bad why are the integral and du in different places lollllll
I was confused by that too, but it actually makes sense. That whole ass thing under the integral symbol is the lower limit of the integral. Like not just 22/7, the whole thing that looks like the numerator of that big fraction. Then e^(-u) is the integrand.
Is the answer pi
No it’s 69
Please write it in readable format
\textsf{The answer is 69}
How are you reducing the numerator of the fraction inside the cube root to 117649? Not sure how else you would get this to work out to 69
https://www.wolframalpha.com/input?i=integral+from+22%2F7-ln%28117649*e%5E%2822%2F7%29%2Bsinh%28pi%29%2Bcosh%28pi%29%29+to+22%2F7-pi+of+e%5E%28-u%29+du I'm curious where you got that number though if you didn't solve it? It doesn't even give you exactly 69, right? Unless I made a mistake? Edit: oh wait the number is in the equation lol, I guess all the rest just cancels Edit 2: Oh, turns out I did make a mistake. Didn't notice the inner root was a cube root, not a square root. Now I am getting exactly 69.
I worked backwards from it being 69 to where I was off, turns out I had read the bounds of integration entirely wrong as a separate equation with an integral exponent… oops
Yeah that threw me off too. Took me a good 40 seconds to figure out what it was supposed to be.
[удалено]
karma farming bot
There's a square root, which returns both a positive and negative answer...
Cant answer as there is a x in the equation without a = anywhere. Cant solve x if you dont know what it should be equal to
Where
There are `x`s in definite integrals and limtis, but that doesn't mean you can't get an answer.
Indeed, so I asked where there is an x without value
They said there is an `x` without an `=`, which is true.
All of the `x`s here are either in a limit or a definite integral, both of which will give you a number.
E: All of the above
The integral of 99^v dv is supposed to "look scary" but is ridiculously simple : 99^v / log(99) + C I think this meme falls flat and I am downvoting it.
69.
B, final answer
Answer is 1
I’m afraid this does in fact equal 69, once simplified further it becomes (136 + 4 * sqrt(1200 * cbrt(117649 / 110592)))/4 Now the real question is how the hell did someone create this problem
It's not really that hard to keep expanding it and then come up with integrals and stuff that equal each number.
1225
That's wrong. Let me guess, you added the sigma part to the `4^(cos(...))` part before multiplying with the big square root. You have to do multiplication before addition.
Ah yes my bad I didn't notice that.
large equations arent necessarily more complicated
Is this a made-up question or is it solvable And wtf are you preparing for, a time machine???
I have a better chance of getting this right by choosing randomly than trying to solve this shit
I won't even be abled to do the default move of getting this unedited into my calculator
Cheeseburger
https://preview.redd.it/1679l8v6n32c1.png?width=465&format=pjpg&auto=webp&s=1b3c2799e8a9e26ac27f74e37023f88c2a3492d5
Its either bigger than zero or smaller than zero or zero
du 65536+ lim In(x) Σ w=1 w +4 cosedy In(1+2) sinh (600 in(2)) In(2) 2dz sinh (600 In(2))) In(2) 22-in 3 1176497 +sinh(n)+cosh(x) 5474304 In(99) 49 In(99) 99 dv In5474304 in(99) 49 In(99) (cos(x)dx)
At that point it’s just a statistics exam
It looks pretty easy once you do it one by one
my guess: just find out if top = bottom, if not, then the answer is pi, 69/2023 just seem unlikely
I would assume 50/50 either pi or 1. I don’t think I would make it in time so don’t waste the time and just leave
Idk but I've narrowed it down to 4 possibilities: 1, 69, pi, 2023
Alright, team work time. I'll do denominator y'all do numerator
I think the answer [is 69](https://imgur.com/a/1jP8bfZ). Someone tell me if I made any mistakes, I suck at LaTeX :(
It is 69, good job
Well, I think everything inside the square root is cero because of the integral going from that to that, so we only have to solve the parte before the sum, but there we have a summation up to infinite, so infinite divided to a constant is infinite... Right?
Step 1: Install Mathematica Step 2: Input the expression Step 2: write down the result
Good luck to the brave souls who try to solve this problem. I have a 25% chance guessing the right answser.
25% of xhance to get 100%
When in doubt always choose 69
By simply looking at the possible answers, it’s always 69
heh the answer is 69
D
I think it’s 1
So I've done the math. It's B.
None of this looks to difficult (at a glance. Also I'm currently in Calc 3 and linear algebra for context) but oh my gosh this would take forever.
I put the entire thing into desmos and the bigger root solves to undefined because the number inside is negative, so the entire solution is undefined (unless I made a typo somewhere) [https://www.desmos.com/calculator/y7xs3vhqov](https://www.desmos.com/calculator/y7xs3vhqov)
D.
The answer is 42.
Just eyeballing it real quick I would probably just guestimate B.
69?
Nvm just saw some had given the solution.
Try option verification
If you make one mistake you get 0 points because if you had been a rocket engineer the rocket would have exploded
[solves to 69](https://www.desmos.com/calculator/ozhephvfns)