Weekly Math Question

@ReChesster said in #40:

Bruh, where do you live?
india

@ReChesster is maybe asking because he does not know the answer

@MIHIR_KATTI said in #42:

@ReChesster is maybe asking because he does not know the answer

I do know the answer, as you might habe noticed, I post a detailed solution and how I would have conquered the problem every week.

@RishikaPlaysChess said in #41:

india

I live in Austria, and the education system here is pretty bad imo, which is why I am learning in homeschooling.

It’s Saturday again already! It’s really incredible to see how fast time passes by. Let’s see what the answer to our last week’s problem looks like!

Since the pro question is a generalization of the “normal” one, I will first show the solution for the pro question and then substitute the appropriate numbers for a and b (which would be a = e and b = 0). Otherwise, this would get way to long for anyone to read. (It’s already ready long anyway, so go grab some snacks before you begin reading.)

Let’s have a look at our equation:

sin(x) + cos(x) = a + bi

As long as the x is inside the sine- and cosine function, we can’t really do a lot with the equation, so we need to find another way to express sin(x) and cos(x). And we can do so by using Euler’s formula:

e^(ix) = cos(x) + i*sin(x)

If we now put -x instead of x, we get:

e^(-ix) = cos(-x) + i*sin(-x)

Since cos(x) is an even function and sin(x) is an odd function, we can rewrite this as

e^(-ix) = cos(x) – i*sin(x)

If we add e^(ix) and e^(-ix), we get

e^(ix) + e^(-ix) = (cos(x) + isin(x)) + (cos(x) – isin(x))

= cos(x) + cos(x) + isin(x) – isin(x)

= 2cos(x)

Based on that, we can see that

cos(x) = (e^(ix) + e^(-ix))/2

We now have an expression for cos(x) that we can actually work with in the original equation. Something similar can be done to find an expression for sin(x), but instead of adding e^(ix) and e^(-ix), we subtract them:

e^(ix) - e^(-ix) = (cos(x) + isin(x)) - (cos(x) – isin(x))

= cos(x) - cos(x) + isin(x) + isin(x)

= 2i*sin(x)

As we can see now,

sin(x) = (e^(ix) - e^(-ix))/(2i)

So we now found expressions for sin(x) and cos(x) that we can use to solve our problem. Substituting our expressions in the equation gives us

((e^(ix) - e^(-ix))/(2i)) + ((e^(ix) + e^(-ix))/2) = a + bi

This can be simplified a bit:

((e^(ix) - e^(-ix))/(2i)) + ((e^(ix) + e^(-ix))/2) = a + bi |*2

(e^(ix) - e^(-ix))/i) + e^(ix) + e^(-ix) = 2(a + bi)

Since 1/i = -i, we see:

(-i)(e^(ix)) + i(e^(-ix)) + e^(ix) + e^(-ix) = 2(a + bi)

If we now factor out e^(ix) and e^(-ix), we get

(e^(ix))(1 - i) + (e^(-ix))(1 + i) = 2(a + bi)

Now we factor out e^(-ix) (be careful with the exponents):

(e^(-ix))((e^(2ix))(1 - i) + (1 + i)) = 2(a + bi)

Dividing both sides by e^(-ix), which is equivalent to multiplying both sides by e^(ix), gives us

(e^(2ix))(1 - i) + (1 + i) = 2(e^(ix))(a + bi)

e^(2ix) can also be written as (e^(ix))^2. Which means we can rewrite this equation after subtracting 2(e^(ix))(a + bi) from both sides:

((e^(ix))^2)(1 - i) – (e^(ix))(2(a + bi)) + (1 + i) = 0

This is a quadratic equation in terms of e^(ix), so we can apply the quadratic formula:

e^(ix) = (2(a + bi) +- sqrt(((2(a + bi))^2) – 4(1 - i)(1+i)))/(2(1 - i))

Since 2^2 = 4, we can factor out a 4 under the square root, and since we can take the square root of every factor separately, we can take the square root of 4 (which is obviously 2) and the rest separately. We then can factor out a 2 in the numerator and cancel it out with the 2 in the denominator, which eventually gives us the expression

e^(ix) = ((a + bi) +- sqrt(((a + bi)^2) – (1 - i)(1+i)))/(1 - i)

Using binomial formulas, we can see that (a + bi)^2 = (a^2) + 2abi – (b^2) and (1 – i)(1 + i) = 2. This means we can now rewrite this as

e^(ix) = ((a + bi) +- sqrt(((a^2) – (b^2) – 2) + 2abi))/(1 - i)

Since we want to express x in the form u + vi, we need to find a different expression for the square root, or, in other words, solve it. How should we be able to do that since we have variables being involved? Well, I’ll show you later. Let’s just for now say that the square root expression equals some complex number s + ti, with s and t being real numbers (this will be important later on, so don’t forget that):

sqrt(((a^2) – (b^2) – 2) + 2abi) = s + ti

Our equation now simplifies to

e^(ix) = ((a + bi) +- (s + ti))/(1 - i)

Since we want to solve for x, we have to take the natural log. But how can we do that with a complex number? Let me show you:
Let’s say we have some complex number of the form k + di (again, k and d are real numbers). As you might know, complex numbers can either be written in cartesian form or in polar form. To solve our problem, we will need to temporarily convert it to polar, which can be done like this:

k + di = r*e^(ip)

r = sqrt((k^2) + (d^2))

p = arctan(d/k)

To be 100% precise, we should add a +2(pi)n at p, but I don’t want to make things more complex as they already are (pun intended), so we’ll just set n = 0, which gives us the primary solution.
In this from, taking the natural log is easy:

ln(k + di) = ln(r*e^(ip))

= ln(r) + ln(e^(ip))

= ln(r) + ip

Since r can’t be negative, we can safely write this down with a and b now:

ln(k + di) = ln(r) + ip

= ln(sqrt((k^2) + (d^2))) + arctan(d/k)*i

However, be aware that arctan(d/k) has the same result as arctan((d/k) + pi) because tan(x) is a periodic function. So it depends on your original number whether or not you have arctan(d/k) or arctan((d/k) + pi). We will, for simplicity, assume that the correct version is arctan(d/k).

Anyway, now that we can take the ln of any complex number, we can continue with our problem:

e^(ix) = ((a + bi) +- (s + ti))/(1 - i)

ix = ln((a + bi) +- (s + ti)) – ln(1 – i)

= ln((a +- s) + (b +- t)i) – (ln(2) + arctan(-1)*i)

= ln(((a +- s)^2) + ((b +- t)^2)) + arctan((b +- t)/(a +- s))i – ln(2) – ((3pi)/4)*i

= ln((((a +- s)^2) + ((b +- t)^2))/2) + (arctan((b +- t)/(a +- s)) - ((3*pi)/4))*i

Finally, dividing everything by i, which is the same as multiplying everything by -i, will give us the result

x = (arctan((b +- t)/(a +- s)) - ((3*pi)/4)) - ln((((a +- s)^2) + ((b +- t)^2))/2)*i

We are almost finished now, we only need to express s and t in terms of a and b now.

Let’s first recall what we defined to be s + ti:

s + ti = sqrt(((a^2) – (b^2) – 2) + 2abi)

If we square both sides, we will get

(s + ti)^2 = ((a^2) – (b^2) – 2) + (2ab)i

(s^2) + 2sti – (t^2) = ((a^2) – (b^2) – 2) + (2ab)i

((s^2) – (t^2)) + 2sti = ((a^2) – (b^2) – 2) + (2ab)i

The only way for this equation to be true is that

Re(((s^2) – (t^2)) + 2sti) = Re(((a^2) – (b^2) – 2) + (2ab)i)

and

Im(((s^2) – (t^2)) + 2sti) = Im(((a^2) – (b^2) – 2) + (2ab)i)

Based on this fact, we can form a system of equations:

I.) (s^2) – (t^2) = (a^2) – (b^2) – 2

II.) 2st = 2ab

All we need to do know to finish the problem is solve this for s and t.

Solving it in this form is a bit complicated, this is why I will first show you how to do it in general (as I did with the natural log) and then apply the general from on our problem.

I do know that we could work with polar coordinates as well, which would probably be easier, but I want to show you how to do it this way because I don’t think many people know how to take square roots using the cartesian from, and since this post can’t get more complicated anymore anyway, let’s just use the hard way :P

Anyway, let’s look at the general form: We want to find the square root of some complex number r + qi (r and q are reals), and we’re first gonna set that equal to some other complex number o + li:

sqrt(r + qi) = o + li

Squaring both sides gives us

r + qi = (o^2) – (l^2) + 2oli

Using the same approach as before, we get

I.) r = (o^2) – (l^2)

II.) q = 2ol

From II.), we can see that l = q/(2o). Inserting that into I.) gives us

r = (o^2) – ((q/(2o))^2)

However, we also see that o = q/(2l) and thus

r = ((q/(2l))^2) – (l^2)

We know have equations for both o and l that we can solve. This is just basic algebra, so I won’t talk to much about it and just show the steps (we’ve done way harder stuff by now):

r = (o^2) – ((q/(2o))^2) | *(o^2)

r*(o^2) = ((o^2)^2) – ((q/2)^2) | - (r*(o^2))

((o^2)^2) - r*(o^2)) - ((q/2)^2) = 0 | quadratic equation in terms of (o^2)

(o^2) = (r +- sqrt((r^2) + (q^2)))/2 | sqrt(...)

o = sqrt((r +- sqrt((r^2) + (q^2)))/2)

r = ((q/(2l))^2) – (l^2) | * (l^2)

r*(l^2) = ((q/2)^2) – ((l^2)^2) | -(((q/2)^2) – ((l^2)^2))

((l^2)^2) + r*(l^2) – ((q/2)^2) = 0 | quadratic equation in terms of (l^2)

(l^2) = (-r +- sqrt((r^2) + (q^2)))/2 | sqrt(...)

l = sqrt((-r +- sqrt((r^2) + (q^2)))/2)

Since n < sqrt((n^2) + m), with m being positive, and o and l can only be real numbers, the final answers are:

o = sqrt((r + sqrt((r^2) + (q^2)))/2)

l = sqrt((-r + sqrt((r^2) + (q^2)))/2)

We can now use these formulas to solve our square root problem from above:

r = (a^2) – (b^2) – 2

q = 2ab

s = sqrt((((a^2) – (b^2) – 2) + sqrt((((a^2) – (b^2) – 2)^2) + ((2ab)^2)))/2)

t = sqrt((-((a^2) – (b^2) – 2) + sqrt((((a^2) – (b^2) – 2)^2) + ((2ab)^2)))/2)

This means the final answers to our problem are:

x = (arctan((b + (sqrt((-((a^2) – (b^2) – 2) + sqrt((((a^2) – (b^2) – 2)^2) + ((2ab)^2)))/2)))/(a + (sqrt((((a^2) – (b^2) – 2) + sqrt((((a^2) – (b^2) – 2)^2) + ((2ab)^2)))/2)))) - ((3*pi)/4)) - ln((((a + (sqrt((((a^2) – (b^2) – 2) + sqrt((((a^2) – (b^2) – 2)^2) + ((2ab)^2)))/2)))^2) + ((b + (sqrt((-((a^2) – (b^2) – 2) + sqrt((((a^2) – (b^2) – 2)^2) + ((2ab)^2)))/2)))^2))/2)*i

x = (arctan((b - (sqrt((-((a^2) – (b^2) – 2) + sqrt((((a^2) – (b^2) – 2)^2) + ((2ab)^2)))/2)))/(a - (sqrt((((a^2) – (b^2) – 2) + sqrt((((a^2) – (b^2) – 2)^2) + ((2ab)^2)))/2)))) - ((3*pi)/4)) - ln((((a - (sqrt((((a^2) – (b^2) – 2) + sqrt((((a^2) – (b^2) – 2)^2) + ((2ab)^2)))/2)))^2) + ((b - (sqrt((-((a^2) – (b^2) – 2) + sqrt((((a^2) – (b^2) – 2)^2) + ((2ab)^2)))/2)))^2))/2)*i

And that’s it for the pro question! Also, remember: This is only the primary solution xD

For the “normal” question, I could just substitute the values for a and b into our solutions, but I’m having way too much fun right now writing this, so here we go again! (This time I will only show the calculation, as I already did the explanation previously. Otherwise, this will get out of hand... I mean: Even more out of hand!)

sin(x) + cos(x) = e

((e^(ix) - e^(-ix))/(2i)) + ((e^(ix) + e^(-ix))/2) = e

(-i)(e^(ix)) + i(e^(-ix)) + e^(ix) + e^(-ix) = 2e

(e^(ix))(1 - i) + (e^(-ix))(1 + i) = 2e

(e^(-ix))((e^(2ix))(1 - i) + (1 + i)) = 2e

(e^(2ix))(1 - i) + (1 + i) = 2e*(e^(ix))

((e^(ix))^2)(1 - i) – (e^(ix))(2e) + (1 + i) = 0

e^(ix) = (2e +- sqrt(((2e)^2) – 4(1 - i)(1+i)))/(2(1 - i))

e^(ix) = (e +- sqrt((e^2) –2)))/(1 - i)

Since e - sqrt((e^2) –2) is positive, we can just normally take the natural log:

ix = ln(e +- sqrt((e^2) –2))) – ln(1 – i)

ix = ln(e +- sqrt((e^2) –2))) – ln(2) - (7*pi)/4)*i

The final solutions are:

x = -((7*pi)/4) – ln((e + sqrt((e^2) –2)))/2)*i

x = -((7*pi)/4) – ln((e - sqrt((e^2) –2)))/2)*i

Now that was fun! Nobody posted the correct solution, but I think @RhishikaPlaysChess deserves a special mention for getting the idea of solving the equation right!

For this week’s question, I will test the calculus skills of you. Since I couldn’t think of any pro question, there will be none this week.

Given the functions

f(x) = e^x

g(x) = ln(x)

h(x) = x

The linear function j(x) intersects h(x) in the first quadrant at an angle of 90º and has a distance of 5 units from the origin. Find the area that is enclosed by the negative x – axis, the negative y – axis, f(x), g(x) and j(x).

If you struggle, try drawing a sketch to visualize it. You will surely find it a lot easier!

I had a lot of fun writing all of this stuff, and I think this post might even be the longest on Lichess xD

Also btw @MIHIR_KATTI I hope this is proof enough that I didn't ask because I don't know the answer :P

Anyway, good luck everyone, and as always, have fun :)

Since no one seems to habe a solution yet, I'll add another week before I reveal the solution.

i can safely go to my death not knowing how to do any of this.

@ReChesster said in #1:

I'm thinking of posting a new math problem every week here if you guys like the idea of it. Anyway, here is the first Math Problem, I'll probably post the solution to it on Saturday:

Given the equation y = (x^n) + x^(n-1)
Solve this expression for n.

---

Pro Question: If the above equation is to easy for you, try this:

Given the equation f(x) = (a^g(x)) * (ln(a) + 1)
Solve this expression for a.

Anyone is welcome to try and solve these questions, but please don't cheat by using high tech software. Good luck everyone!
can you give something for 12 year olds (I am ten but I can solve grade 6 maths)

Bro you have a big brain you are solving your homework in the forum's

@ZIDAN_S said in #49:

?