SuicideFuel Math thread problem (official)

[Diddy29]

Was this what you wrote:
View: https://www.youtube.com/watch?v=vGQMmpTInPU
?

Yeah but j is dependent on i.

 

[Ahnfeltia]

Find S(n), given that S(1) = 0. S(2) = 1.2, S(3) = 1.2 + 1.3 + 2.3. S(4) = 1.2 + 1.3 + 1.4 + 2.3 + 2.4 + 3.4

(1 + 2 + ... + n)^2 = 1^2 + 2^2 + ... + n^2 + 2S(n). Since 1 + 2 + ... + n = n(n+1)/2 and 1^2 + 2^2 + ... + n^2 = n(n+1)(2n+1)/6, one can easily solve for S(n) and find that S(n) = (n-1)n(n+1)(3n+2)/24.

S(n) = Σ (i = 1 to n-1) ( Σ (j= i+1 to n) i*j)

You have merely rewritten his function in terms of summation notation, which hardly counts as solving the problem.

 

[trying to ascend]

(1 + 2 + ... + n)^2 = 1^2 + 2^2 + ... + n^2 + 2S(n). Since 1 + 2 + ... + n = n(n+1)/2 and 1^2 + 2^2 + ... + n^2 = n(n+1)(2n+1)/6, one can easily solve for S(n) and find that S(n) = (n-1)n(n+1)(3n+2)/24.

You have merely rewritten his function in terms of summation notation, which hardly counts as solving the problem.

Correct

 

[cripplecel]

f(x) = 2x^2 what is f’(x)

 

[Ahnfeltia]

Bruh I thought it was a decimal sign. You should've used the * sign for multplication.

@trying to ascend I concur. While using a "." for multiplication is fine when it comes to variables, using it this way between numbers is very confusing because virtually everyone will interpret it as a radix point. Moreover, I believe using a "." for multiplication is Indian notation, but Wikipedia suggests India uses the "." as the decimal separator as well, so I can't imagine using "." to indicate multiplication between numbers is common practice.

 

[Ahnfeltia]

f(x) = 2x^2 what is f’(x)

Hint: power rule.

 

[fukurou]

if the amount of sentences is x, and each param list is marked as a1...an with y elements per list.
how many senstence options are there?

    public PerChanceTest() {
        super();
        sentences.add("here is a  salad vegi1 #, vegi2 # and herb #.");
        sentences.add("how about this salad vegi1 #, vegi2 # and herb #. it goes well with tuna fish");
        UniqueItemSizeLimitedPriorityQueue temp = new UniqueItemSizeLimitedPriorityQueue();
        temp.setLimit(3);
        wordToList.put("vegi1", temp);
        temp = new UniqueItemSizeLimitedPriorityQueue();
        temp.setLimit(3);
        wordToList.put("vegi2", temp);
        temp = new UniqueItemSizeLimitedPriorityQueue();
        temp.setLimit(3);
        wordToList.put("herb", temp);
    }

output example:
output:
here is a salad cucmber, purple onion and cpriander.
here is a salad tomato, purple onion and irit.
here is a salad lettuce, pickle and cpriander.
how about this salad tomato, purple onion and cpriander. it goes well with tuna fish
how about this salad lettuce, pickle and irit. it goes well with tuna fish
how about this salad lettuce, pickle and parcely. it goes well with tuna fish
how about this salad tomato, onion and cpriander. it goes well with tuna fish
how about this salad cucmber, pickle and parcely. it goes well with tuna fish
here is a salad cucmber, purple onion and parcely.
here is a salad cucmber, purple onion and irit.

Process finished with exit code 0

 

[Ahnfeltia]

if the amount of sentences is x, and each param list is marked as a1...an with y elements per list.
how many senstence options are there?

If I understand you correctly, you have x different sentences, each with n variable slots and each slot permitting y options, correct? In that case, the total number of possible sentences is x*(y^n).

 

[trying to ascend]

f(x) = 2x^2 what is f’(x)

4x

@trying to ascend I concur. While using a "." for multiplication is fine when it comes to variables, using it this way between numbers is very confusing because virtually everyone will interpret it as a radix point. Moreover, I believe using a "." for multiplication is Indian notation, but Wikipedia suggests India uses the "." as the decimal separator as well, so I can't imagine using "." to indicate multiplication between numbers is common practice.

Yes, it indeed looked confusing, though I was just copying the original problem

 

[Ahnfeltia]

Yes, it indeed looked confusing, though I was just copying the original problem

fair enough in that case

 

[Diddy29]

(1 + 2 + ... + n)^2 = 1^2 + 2^2 + ... + n^2 + 2S(n). Since 1 + 2 + ... + n = n(n+1)/2 and 1^2 + 2^2 + ... + n^2 = n(n+1)(2n+1)/6, one can easily solve for S(n) and find that S(n) = (n-1)n(n+1)(3n+2)/24.

You have merely rewritten his function in terms of summation notation, which hardly counts as solving the problem.

I thought that was okay.

 

[Ahnfeltia]

I thought that was okay.

Generally not (although there technically are exceptions). Here's a rather extreme example: would you say that sum from j=2 to 4 of j is a valid solution to "find the value of 2 + 3 + 4"?

 

[Diddy29]

Generally not (although there technically are exceptions). Here's a rather extreme example: would you say that sum from j=2 to 4 of j is a valid solution to "find the value of 2 + 3 + 4"?

No but the upper limit for your example is bounded.

 

[Ahnfeltia]

No but the upper limit for your example is bounded.

Does any of the limits being a variable change anything significant? In that case you could argue that summation notation is more formal as opposed to the ellipses, but that's about it, no? At the end of the day you have merely recast the given expression in a more formal form without really doing any kind of simplification.

 

[fukurou]

If I understand you correctly, you have x different sentences, each with n variable slots and each slot permitting y options, correct? In that case, the total number of possible sentences is x*(y^n).

that looks like a lot when it's put into a formula, maybe it could be used to write visual novel type of games, scripts

 

[Ahnfeltia]

that looks like a lot when it's put into a formula, maybe it could be used to write visual novel type of games, scripts

it is a lot, esp. for large n

 

[uranium235]

Consider 9 dots arranged as 3x3 what is the highest amount of closed figures that can be drawn by connecting the dots one by one in a single trial you can imagine this as drawing pattern lock in your phone i have achieved 10 closed figures so far

 

[Ahnfeltia]

Consider 9 dots arranged as 3x3 what is the highest amount of closed figures that can be drawn by connecting the dots one by one in a single trial you can imagine this as drawing pattern lock in your phone i have achieved 10 closed figures so far

I take it you want the number of essentially distinct ones? Mirror images counting as the same figure etc.

 

[uranium235]

I take it you want the number of essentially distinct ones? Mirror images counting as the same figure etc.

Wait i will upload an image to demonstrate

 

[uranium235]

There are 10 closed figures in this image

 

[uranium235]

I take it you want the number of essentially distinct ones? Mirror images counting as the same figure etc.

Notification

 

[Ahnfeltia]

There are 10 closed figures in this image

Aha. I see what you mean now. 11 is possible.

 

[uranium235]

Let P(x) be a polynomial with positive real coefficients. Prove that
Sqrt(P(a)P(b) )≥ Sqrt(P( ab)),for all positive real numbers a and b.

 

[Ahnfeltia]

Let P(x) be a polynomial with positive real coefficients. Prove that
Sqrt(P(a)P(b) )≥ Sqrt(P( ab)),for all positive real numbers a and b.

I'm not convinced this is true. Take P(x) = ¼ + x. In that case your desideratum amounts to ¼ + ab ≤ (¼ + a)(¼ + b) for positive a & b, but this is evidently false for a = b = ¼.

 

[uranium235]

I'm not convinced this is true. Take P(x) = ¼ + x. In that case your desideratum amounts to ¼ + ab ≤ (¼ + a)(¼ + b) for positive a & b, but this is evidently false for a = b = ¼.

Sorry my mistake it was Sqrt(p(a)p(b))=p(Sqrt(ab))

 

[Ahnfeltia]

Sorry my mistake it was Sqrt(p(a)p(b))=p(Sqrt(ab))

The equality sign should have been a nonstrict inequality. Anyhoo, this reminds me of midpoint (or Jensen) convexity, albeit with the geometric mean instead of the arithmetic one.

Let g = sqrt(ab). We want to prove that p(g)p(g) ≤ p(a)p(b). Let p(x) = c_0 + c_1*x + ... + c_n*x^n. Upon expanding the product p(a)p(b) we get a sum of terms of the form c_i*c_k*a^i*b^k. If we can therefore manage to show that c_i^2*g^(2i) ≤ c_i^2*a^i*b^i & c_i*c_k*2g^(i+k) ≤ c_i*c_k*(a^i*b^k + a^k*b^i) for every i ≠ k, we are done (since a & b & the c_i are all strictly positive). The former inequality is obviously an equality and therefore true. Since the c_i are all strictly positive, the latter inequality is tantamount to showing that 2g^(i+k) ≤ a^i*b^k + a^k*b^i.

Showing that 2g^(i+k) ≤ a^i*b^k + a^k*b^i is easy with Lagrange multipliers. For concreteness, I'll illustrate the case where i = 5 & k = 2. In that case, we want to show that 2g^7 ≤ a^5*b^2 + a^2*b^5 = a^2*b^2*(a^3 + b^3) = g^4*(a^3 + b^3) -- i.e., 2g^3 ≤ a^3 + b^3. Now we use Lagrange multipliers. We want to minimize the bivariate function x^3 + y^3 constrained to the hyperbola xy = g^2. Lagrange multipliers readily yields that 3x^2/y = 3y^2/x -- i.e., x^3 = y^3, so x = y = g (assuming x & y to be positive). I'll leave verifying that x = y = g produces a minimum as opposed to, say, a maximum or saddle point to the reader.

 

[uranium235]

The equality sign should have been a nonstrict inequality. Anyhoo, this reminds me of midpoint (or Jensen) convexity, albeit with the geometric mean instead of the arithmetic one.

Let g = sqrt(ab). We want to prove that p(g)p(g) ≤ p(a)p(b). Let p(x) = c_0 + c_1*x + ... + c_n*x^n. Upon expanding the product p(a)p(b) we get a sum of terms of the form c_i*c_k*a^i*b^k. If we can therefore manage to show that c_i^2*g^(2i) ≤ c_i^2*a^i*b^i & c_i*c_k*2g^(i+k) ≤ c_i*c_k*(a^i*b^k + a^k*b^i) for every i ≠ k, we are done (since a & b & the c_i are all strictly positive). The former inequality is obviously an equality and therefore true. Since the c_i are all strictly positive, the latter inequality is tantamount to showing that 2g^(i+k) ≤ a^i*b^k + a^k*b^i.

Showing that 2g^(i+k) ≤ a^i*b^k + a^k*b^i is easy with Lagrange multipliers. For concreteness, I'll illustrate the case where i = 5 & k = 2. In that case, we want to show that 2g^7 ≤ a^5*b^2 + a^2*b^5 = a^2*b^2*(a^3 + b^3) = g^4*(a^3 + b^3) -- i.e., 2g^3 ≤ a^3 + b^3. Now we use Lagrange multipliers. We want to minimize the bivariate function x^3 + y^3 constrained to the hyperbola xy = g^2. Lagrange multipliers readily yields that 3x^2/y = 3y^2/x -- i.e., x^3 = y^3, so x = y = g (assuming x & y to be positive). I'll leave verifying that x = y = g produces a minimum as opposed to, say, a maximum or saddle point to the reader.

Nice
Here is the latex format for ADHDcels who have difficulty in reading

 

[uranium235]

Nice
Here is the latex format for ADHDcels who have difficulty in reading

 

[Ahnfeltia]

Nice
Here is the latex format for ADHDcels who have difficulty in reading

And here I was wondering what kind of madman would manually transcribe my writings into LaTeX. Nice use of ChatGPT.

 

[trying to ascend]

Solve the following equation: 8x(2x² - 1)(8x( to the power of 4) - 8x² + 1) = 1

 

[Caesercel]

The equality sign should have been a nonstrict inequality. Anyhoo, this reminds me of midpoint (or Jensen) convexity, albeit with the geometric mean instead of the arithmetic one.

Let g = sqrt(ab). We want to prove that p(g)p(g) ≤ p(a)p(b). Let p(x) = c_0 + c_1*x + ... + c_n*x^n. Upon expanding the product p(a)p(b) we get a sum of terms of the form c_i*c_k*a^i*b^k. If we can therefore manage to show that c_i^2*g^(2i) ≤ c_i^2*a^i*b^i & c_i*c_k*2g^(i+k) ≤ c_i*c_k*(a^i*b^k + a^k*b^i) for every i ≠ k, we are done (since a & b & the c_i are all strictly positive). The former inequality is obviously an equality and therefore true. Since the c_i are all strictly positive, the latter inequality is tantamount to showing that 2g^(i+k) ≤ a^i*b^k + a^k*b^i.

Showing that 2g^(i+k) ≤ a^i*b^k + a^k*b^i is easy with Lagrange multipliers. For concreteness, I'll illustrate the case where i = 5 & k = 2. In that case, we want to show that 2g^7 ≤ a^5*b^2 + a^2*b^5 = a^2*b^2*(a^3 + b^3) = g^4*(a^3 + b^3) -- i.e., 2g^3 ≤ a^3 + b^3. Now we use Lagrange multipliers. We want to minimize the bivariate function x^3 + y^3 constrained to the hyperbola xy = g^2. Lagrange multipliers readily yields that 3x^2/y = 3y^2/x -- i.e., x^3 = y^3, so x = y = g (assuming x & y to be positive). I'll leave verifying that x = y = g produces a minimum as opposed to, say, a maximum or saddle point to the reader.

 

[Robtical]

Why did the math book go to the hospital?

Spoiler: Answer

it had too many problems

 

[Ahnfeltia]

Why did the math book go to the hospital?

Spoiler: Answer

it had too many problems

Spoiler

and here I was expecting a l'Hôpital's rule joke

 

[Ahnfeltia]

Did I make a mistake somewhere?

 

[Caesercel]

Did I make a mistake somewhere?

 

[Ahnfeltia]

Solve the following equation: 8x(2x² - 1)(8x( to the power of 4) - 8x² + 1) = 1

In addition to x = ½, there there six other ugly casus irreducibilis roots.

 

[trying to ascend]

In addition to x = ½, there there six other ugly casus irreducibilis roots.

Yes, you had to use a cosine substiution to find those

 

[Robtical]

Spoiler

and here I was expecting a l'Hôpital's rule joke

I used to call it la hospital's rule

 

[Ahnfeltia]

Yes, you had to use a cosine substiution to find those

Like this?

 

[Ahnfeltia]

I used to call it la hospital's rule

Pronouncing mathematicians' names correctly is often a doozy.

 

[trying to ascend]

Like this?

Yes

 

[trying to ascend]

10 cars, each one being a different model, with each 2 belonging to the same team, need to be arranged in two columns, such that no cars of the same team belong to the same line.

Find the number of ways such arrangement can be done

 

[trying to ascend]

Find all the solutions to the following equation: 3/(x - 3) + 5/(x - 5) + 17(x - 17) + 19/(x - 19) = x² - 11x - 4

 

[trying to ascend]

If a + b + c = 0.

Find (a^7 + b^7 + c^7))/((abc)(a^4 + b^4 + c^4))

 

[trying to ascend]

Find the integral of (xsin(x))/(1 + cos²(x)) from 0 to pi/2

 

[trying to ascend]

Find the value of Cos³(20) - Cos³(40) - Cos³(80)

 

[Ahnfeltia]

Find the integral of (xsin(x))/(1 + cos²(x)) from 0 to pi/2

Substituting t = tan(x/2) yields 4 times the integral from 0 to 1 of t*arctan(t)/(1+t^4). The only thing I can think of here is to use the Maclaurin expansion of arctan(t), but that results in ½ times the sum of (-1)^n/(2n+1)*(digamma(n/4+7/8) - digamma(n/4+3/8)) which I don't know how to simplify. Where did I go wrong?

 

[Ahnfeltia]

It should be pi²/16.

Numerically the result of the integral should be about 0.8453, whereas pi^2/16 is approximately 0.6.

Also, what's digamma?

Digamma function - Wikipedia

en.wikipedia.org

 

[Ahnfeltia]

I will try to see, but you were supposed to use the king's property (Integral from a to b of f(x) = f(a + b - x)) and end up with a integral that doesn't have the x.

I thought about doing that, but I quickly abandoned the idea because the math didn't seem to work out nicely in my head.

 

[trying to ascend]

Numerically the result of the integral should be about 0.8453, whereas pi^2/16 is approximately 0.6.

Digamma function - Wikipedia

en.wikipedia.org

You are correct, I wrote the wrong integral.

It should be xsin(2x)/(1 + cos²(2x)).

The king's property can't be applied to the integral I wrote previously