Can You Pass Harvard University Entrance Exam?
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- Опубликовано: 21 окт 2024
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@higher_mathematics
#maths #math
![The most beautiful equation in math, explained visually [Euler’s Formula]](/img/1.gif)
If (x = 1): [ 2^1 + 1 = 2 + 1 = 3 ]
If (x = 2): [ 2^2 + 2 = 4 + 2 = 6 ]
If (x = 1.5):
[ 2^{1.5} + 1.5 \approx 2.828 + 1.5 \approx 4.328 ]
If (x = 1.7):
[ 2^{1.7} + 1.7 \approx 3.249 + 1.7 = 4.949 ]
If (x = 1.75):
[ 2^{1.75} + 1.75 \approx 3.363 + 1.75 = 5.113 ]
If (x = 1.72):
[ 2^{1.72} + 1.72 \approx 3.279 + 1.72 = 4.999 ]
From this, we see that (x) is approximately equal to 1.72.
2^{1.72} + 1.72 ≈ 5
X ≈ 1.72
That's what I did in my head because the complicated answer was baffled by how complex he approached it.
But out of context, it's pretty easy to find the closest squares and since their logarithmic, the answer reflects mental mathematics.
I was out by 0.02 so I'm not complaining although technically his answer is more precise.
@@SDarkVader This is how we would do it in assembler or with the HP15C in days of old.
We can solve it graghicly easier
This is why India is ruling
Some functions behave abnormally at some points.
Your method doesn't always apply.
You must solve it analytically or graphically
I swear to god this could've been finished in 3 steps without raising my blood pressure for 10 minutes straight
Same feeling 😂😂😂😂
Me in college days: "2*+x=5"
Me as an International Professional Marine Engineer today: "1+1=2, 2-1=1, 2÷2=1, 2x2=4, 1/2+1/2=1"
Yeah , that's the reality... Almost 90-95% that they thought me (math/chemistry/physics/history) are very useful today in my work.
Math and history from elementary up to college,
Physics and Chemistry from highschool to College..😅
And the most funny thing now is--- Only 5% of it is applicable in real everyday work and life. Like the "-+x÷" 😅😂😂😂
@@diabloprimordial5735 Well said.
What are those steps, just curious.
@@SNagygellertrying x = 1 … too small. trying x = 2 … too big…
trying x= 1.5 … too small… trying x = 1.75 … very close…
probably.
I'm absolutely sure that this has never been part of a Harvard entrance exam. It's not hard, it's just silly - you need the Lambert W function, which is absolutely niche and guaranteed never taught in any highschool, Gymnasium, lycée, what have you.
And even if one knew about it, without looking up w(32ln2) on the internet, there'd be no answer other than x= 5-(w(32ln2)/ln2), *which is a more complicated expression than the initial equation !!*
You're argument about lambert not being taught in high school is true. However, the expression after solving for x is a constant, which usually is the goal when solving math problems.
It's close to any trigonometric function though. There is no analytical expression for sin(π/7), however if the answer is sin(π/7), it seems to be better to provide the answer than to say that there is no "simple solution". The LambertW is not an analytical function, but it allows to express the x and this is the task.
Can we use desmos in Harward entrance exam?
4 years of mathematics at the university, but even there we were never told about the Lambert W function. How could any high school student know about it ????
@@alberodellapace9880 It is not about analytical or not. The sine function is taught in secondary school. The Lambert W function is not even taught in mathematics courses in the university.
First, let's define a function H(a,b) that is the solution to the equation: a^x + x = b
The solution to this problem is obviously H(2,5)
If you need a numerical value, just find an approximation using an iterative method
1.71562073380367
Me in college days: "2*+x=5"
Me as an International Professional Marine Engineer today: "1+1=2, 2-1=1, 2÷2=1, 2x2=4, 1/2+1/2=1"
Yeah , that's the reality... Almost 90-95% that they thought me (math/chemistry/physics/history) are very useful today in my work.
Math and history from elementary up to college,
Physics and Chemistry from highschool to College..😅
And the most funny thing now is--- Only 5% of it is applicable in real everyday work and life. Like the "-+x÷" 😅😂😂😂
Exactly. What a silly exam task. And I'm absolutely sure that this has never been part of a Harvard entrance exam.
@@gewinnste Too much time wasted on one such question. An elite student should refrain from going to Harvard based on answering such a tedious, ridiculous question. The iterative method is the intelligent solution.
Let's define t as rational number that satisfies t + log_2(t) = 5.
Now I can say the answer is x = log_2(t).
If you need a numerical value, just use calculator.
Eqt, 2^x +x = 5
Putting x=1, LHS=3
Putting x=2, LHS=6
Putting x=1.5, LHS=4.32
Putting x=1.75, LHS=5.11
Putting x=1.725, LHS=5.03
Putting x=1.7125, LHS=4.989
So, my answer x=1.7125
Just KIDDING.... thanks!🤣🤣🤣
Efficient and close enough 👍🏽
I can if I can use python
I'd have done it exactly the same way. Successive approximations till you have an answer that's close enough.
I'd say e-1 lol
Yup iterative method is simple and curious 😊
I am glad you explained step by step slowly and also giving reminders in between to not lose track of the solution.
Step 1: Lambert W function
Step 2: ?
Step 3: Stephen Hawking
Step 4 Epstein's Island
???
Step X: ADX Florence
@@moldovanmoldovan7593 Step Y: Austin ACL festival
Step Z: Hollygirl dance
Calculator: and where did that bring you?
back to me☠️
That is what you call boomerang math
Na u can do it with hit and trail method 😂😂😂
To reformulate in terms of the W function is really a circular pseudo-solution.
It requires numerics which could have been applied straight to the original equation without further ado.
Why, using sin, cos, ln, exp, et cetera all require numerics (except for well known values) so why are you picking a fight on poor Lambert W function? ❤
The solution is purely symbolic, and exact. The numerical resolution does not provide any further information
@@HoSza1well, imagine the question was cos(-pi/2+x) = 0.2, find x. What this solution was effective doing was reexpress it as sinx = 0.2 and used the calculater to solve that
Absolutely agree
@HoSza1 that is not the same thing at all. Log and exponencial and treated algebraically in the solution, while the W function is not. The W function is a numerical trick in this demonstration.
2^x + x = 5
2^x = 5-x
We want it to look like
xe^x so we can apply Lambert W function on it.
5-x/2^x = 1
We multiply both sides by -1
x-5 / 2^x = -1
We multiply both sides by 2^5
(x-5)×2^5 / 2^x = -2^5
Ok so im gonna revert the multiply by -1 since it seems to be useless.
(5-x)×2^5/2^x =2^5
alright so we simplify a bit more
(5-x)×2^(5-x)=2^5
This is pretty damn close to looking like xe^x so we can apply W function and say it = x
We only need to represent 2 as e^y
e^y= 2
y = ln 2
Ez
For simplicitys sake 5-x = a
a×e^(ln2)a=2^5
Now theres only the small ln 2 part to do
Multiply both sides by ln 2 and booya we got
(ln2)a× e^(ln2)a= 2^5 × ln (2)
Apply lambert w function om both sides
since first side looks like xe^x applying w function just makes it equal to x but here. x= ln2 × a
so: we get
ln2×a = W(2^5 × ln2)
divide both sides by ln 2
5-x=W(2^5 × ln2) / ln 2
-x=(W(2^5 × ln2) / ln 2) -5
x=5-W(2^5×ln2)ln2
That's equal to aprox 1.715620733275586169380916428210115405349201542402693776216135036789959346078769637168046686169244547097096758431988
Why don’t they just ask shortly: “Do you know the Lambert W?”
Maybe because _RUclips_ value longer videos, even when it's a pure waste or time for humanity, multiplied by the number of viewers.
@@emjizone I would like to correct you by using the Lambert function Hope you do not mind.. here we go:
...... waste or time for humanity, elevated to the cube of the number of viewers... ( that would be more like it)
First off, this is **not** a Harvard U entrance exam question for the simple reason there there is no Harvard U entrance exam. They do accept SAT, ACT, and SAT Subject Test standardized tests as part of their comprehensive application package, but they don't have a specific entrance exam of their own.
Additionally, as you suggest, this question is really two parts, the first being the trivia question you asked -- Do you know the Lambert W? -- and the second being a good test of a student's ability at algebraic manipulation. So a very reasonable test question would be to provide a definition of the Lambert W, and **then** ask for a solution to the equation.
Me in college days: "2*+x=5"
Me as an International Professional Marine Engineer today: "1+1=2, 2-1=1, 2÷2=1, 2x2=4, 1/2+1/2=1"
Yeah , that's the reality... Almost 90-95% that they thought me (math/chemistry/physics/history) are very useful today in my work.
Math and history from elementary up to college,
Physics and Chemistry from highschool to College..😅
And the most funny thing now is--- Only 5% of it is applicable in real everyday work and life. Like the "-+x÷" 😅😂😂😂
"Excuse me, sir. Have you heard of our lord and savior Lamber W function?"
Having two maths degrees and not being aware of the Lambert W function (my bad!), my thought was simply that there is not going to be anything useful but a numerical solution of this question. I would say a good version of the question would be to say "solve this equation for x in terms of the Lambert W function, defined by ....". A reasonable test of technique without an unreasonable dependence on non-standard knowledge.
My math education ended @ diffeq. I've never heard of the Lambert-W function. If I did, I didn't recognize it as such.
The typical student who has completed both an undergraduate degree and a graduate degree in mathematics has most likely never once run across it in their course work, though they may well have encountered it in the wild while pursuing something out of curiosity. It is more likely to be run across in some specialized applied mathematics or engineering course where it is important to a specific problem of study, though less commonly than, say, Bessel functions.
Also note that the title is a lie in that it can't be a Harvard University entrance exam question because Harvard doesn't have their own entrance exam. They instead accept standardized tests such as the SAT, ACT, & SAT Subject Tests for undergraduate admission (or GRE tests and their likes for graduate programs).
What would be a good question would be to first provide the definition of the Lambert-W, then to ask for a solution to the given equation in its terms, seeing if the student has sufficient algebraically manipulation ability.
@@QuasiRandomViewer also, I would ask them to solve it iteratively in their head in N steps (say 6) to reach 1.7157 without using anything else, not even paper. As a minimum. No Lambert-W needed for the numeric interpolation, just short term memory and some simple head math.
@@QuasiRandomViewerexactly
How about using an iterative method to solve this equation?
1. Rearrange the equation:
x = log2(5 - x)
2. We treat it as an iterative process:
x_n+1 = log2(5 - x_n)
3. From the original equation 2^x + x = 5, we conclude 0 < x < 5.
4. Write a function:
def find_root(x0, N):
x = x0
xs = []
for i in range(N):
#print(i, x)
xs.append(x)
x = math.log(5-x, 2)
return x, xs
5. Given x0 in (0, 5), for example, x0 = 1,
r, xs = find_root(1, 100)
6. we could find the function converges very quickly in few step (less than 20), and gets the root of about 1.71562.
Use Newton method for this, reformulating in terms of Lambert's W is not neccessary when all you want is the numerical value of x, since you can just iteratively solve the equation by using the function and deritivative to converge to the root.
Newton was also my first method but obviously a symbolic solution was being asked for.
Şu ck a large cucumber
that's Newton-Raphson to be precise. Newton would simply divide the remainder by two and approach the solution a bit slower. But fair enough, same approach.
You won't understand. It's not about how hard is the exam but rather how hard was for your parents to get rich to be able to afford a good education so that you can pass the Harvard exam and be able to pay it.
But there is no Harvard's exam...
This video demonstrate that education now is almost for all.
@@Usernamd-wn8mx "There is no formula for gaining admission to Harvard. Academic accomplishment in high school is important, but the Admissions Committee also considers many other criteria, such as community involvement, leadership and distinction in extracurricular activities, and personal qualities and character."
That is pretty much their "examination" to go Harvard...
@@AutonomousDecentralisation You forgot that "character" means coming from a wealthy family 😉
@@nahuelastor7522 the question is whether your parents have an internet connection, know how to use RUclips, and think it’s in their kids best interests to watch RUclips math problems. But if your kid is 7 feet tall, just give him a basketball.
Approximately (not to pas exam): if we imagine that x=2, we will get number 6 .
6 is approx 20% more than we need, so we have to decrease 15% number 2.
2 minus 15% is= 1.7
That's it
Not good for exam bad enough good for every day life
Control approach
f(x) = 2^x + x -5
f'(x) = ln(2) 2^x + 1
Chain rule (time derivative):
dot_f(x) = f'(x) dot_x
Proportional controller u = dot_x
dot_x = - f(x) / f'(x)
f(x) converges exponentially to zero for any initial condition x0 > 0.
x(t) converges to 1.7156 and can be found by numerical integration.
The equation \(2^x + x = 5\) is a transcendental equation because it involves both an exponential term \(2^x\) and a linear term \(x\). These types of equations are often not solvable using elementary algebraic methods. However, you can solve it using numerical methods or by graphing the equation to find an approximate solution.
### Steps to Solve the Equation
1. **Understand the Equation:**
The equation is:
\[
2^x + x = 5
\]
Here, \(2^x\) is an exponential function, and \(x\) is a linear function.
2. **Isolate the Exponential Term (if possible):**
It's challenging to isolate \(x\) because \(x\) appears in both the exponential and the linear term. So, isolating \(x\) algebraically isn't possible here.
3. **Graphical Approach:**
A good way to solve this equation is by graphing the two sides of the equation separately and finding their intersection point.
- Let \(f(x) = 2^x + x\).
- Let \(g(x) = 5\).
Graph both \(f(x)\) and \(g(x)\) on the same set of axes.
Where the graphs of \(f(x)\) and \(g(x)\) intersect is the solution to the equation.
4. **Finding the Intersection:**
We are looking for the value of \(x\) where \(f(x) = g(x)\). This can be done either by graphing or using a numerical approach such as the Newton-Raphson method or simply by trial and error.
5. **Approximate Solution:**
Through testing values or using a calculator to graph the functions, you find:
\[
\text{At } x \approx 1, \quad 2^1 + 1 = 2 + 1 = 3 \quad (\text{which is less than 5})
\]
\[
\text{At } x \approx 1.5, \quad 2^{1.5} + 1.5 \approx 2.828 + 1.5 = 4.328 \quad (\text{which is closer but still less than 5})
\]
\[
\text{At } x \approx 1.7, \quad 2^{1.7} + 1.7 \approx 3.249 + 1.7 = 4.949 \quad (\text{very close to 5})
\]
\[
\text{At } x \approx 1.8, \quad 2^{1.8} + 1.8 \approx 3.482 + 1.8 = 5.282 \quad (\text{slightly more than 5})
\]
So, the solution to \(2^x + x = 5\) is approximately \(x \approx 1.7\).
6. **Verification:**
If you plug \(x = 1.7\) back into the original equation:
\[
2^{1.7} + 1.7 \approx 4.949
\]
This is close to 5, confirming that \(x \approx 1.7\) is indeed a good approximate solution.
### Conclusion
The solution to the equation \(2^x + x = 5\) is approximately \(x \approx 1.7\). This was found by evaluating the function at various points and checking which one satisfies the equation most closely.
Well, that's not very satisfying, because you can see it's somewhere around root 3 with a few guesses. Of course, it's not less satisfying then the official solution of “it turns out someone got there first and gave a name to that, I hope you're good at trivia”.
Yankee
Ai
Thanks for sharing.
Do not use ChatGPT here
Very nice problem! ❤❤
The Lambert W function can't be expressed by elementary functions. How is it part of a valid solution? Evaluating the W function is as complicated as the original problem!
Use calculator
@@thunderpokemon2456that’s a worthless comment
Решение ни о чем.
@@mayaq8324 🤣 dude i mean to evaluate W you need calculator
@@thunderpokemon2456 sorry for the evil answer :) however I don’t know about any calculator with lambert w function unless you buy a programmable one, otherwise wolframalfa is the way as I know
And the precise value of knowing how to work this out is a career in teaching students how to work it out...
High school does not teach Lambert W function. The only ways to solve this using high school knowledge is graphing. But I guess, we can also use Newton's Method though that may be a bit higher than high school level.
this is an average jee level question
thanks
This kind of questions never comes in Jee Mains or Adv.
Dont comment on the thing u don't know.
This is a typical sat question@@Glancing_Dagger..
I did learn Lambert w function myself this year and it wasn't difficult, plus I'm from algeria
I don't see the logic behind this video. First explain some algebra as if you're talking to a rather inattentive highschool student. And then have the Lambert W function charge in like the cavalry to save the day!
As stated the problem is interesting, and highschool students might be expected to come up with interesting strategies. Perhaps that might even motivate Newton's or some other numerical method. Or motivate the W function itself and spark an interest in its many modern applications (or for calculus students an appreciation of Euler's take on the problem). The point is that the strategies - the _mathematics_ - is interesting. The _answer_ 1.7 obtained by something akin to magic, not so much.
Avg jee aspirant can solve it
There is no direct infinite series or formula to calculate Lambert function it is calculated using approximation which we can also do in the original equation too. There is no need to complicate the solutuon into Lambert function
What do you mean by "no direct infinite series"? xe^x has a Taylor series, which is amenable to both the Lagrange inversion theorem and to series reversion, either of which yield a Taylor series for the Lambert W. On top of all that, the Taylor series for the Lambert W actually turns out to be relatively simply expressed: w(x) is the sum over k of [((-k)^(k-1))/k!]x^k
Me, in my mind, after thinking 5 seconds:"it has to be close to 1.7". I don't even have a degree so close enough for me
@lolilollolilol7773 5/3
@lolilollolilol7773 assume x=1, the left side would be 3, assume x=2, the left side would be 6. So the answer is between these two. Now assume x=1.5 the left side would be almost 4.3, this tells you you should find the answer between (1.5, 2). Continue the same process. The first few guesses is easy to calculate even in your mind, for example following the same pattern, my guess is x~= 1.75.
@@SekiroEnjoyer123 You are not solving it, you do brut forcing.
@@CodeDaemon what? that is a way of solving the problem! Essentially that's THE way to solve an equation in a general form.
@SekiroEnjoyer123 no it's not. Guessing, or brute forcing, is an automatic fail. Obviously you don't have a higher education.
Very nice explanation. I'm not majoring in anything math but I liked this thanks for the explanation
2^x+x=5
2^(x-5)=(5-x)2^-5
2^(5-x)=2^5/(5-x)
(5-x)2^(5-x)=2⁵
ln2(5-x)e^ln2(5-x)=2⁵ln2
ln2(5-x)=W(2⁵ln2)
5-x=W(2⁵ln2)/ln2
x=5-W(32ln2)/ln2
The solution is too smart, very hard, only for hi tech.
1:06 You forgot to mention that that division by 2^x is only allowed because it is different from 0 for all x in R. Only the limit for x=-infinity is 0.
Explain? I'm confused
@@amberthelostsoul Imagine you want to solve equation_1: 5x=x*y.
If x is different from 0, we can devide by x, yielding the equation_2: 5=y.
But if x is 0, we can't derive from equation_1 that y equals 5, i.e. any value for y satisfies equation_1. That's why deviding by 0 is not allowed.
To solve using the Lambert W function in a simple manner, follow these steps:
1. Rewrite in Exponential Form:
2^x = e^{x \ln 2}
e^{x \ln 2} + x = 5
2. Isolate the Exponential Term:
e^{x \ln 2} = 5 - x
3. Introduce a New Variable: Let . Therefore, . Substitute into the equation:
e^u = 5 - \frac{u}{\ln 2}
4. Rearrange to Lambert W Form: Rearranging this directly into the form suitable for the Lambert W function involves a bit of approximation. For simplicity:
e^u = 5 - \frac{u}{\ln 2}
-\frac{e^u}{\ln 2} = -\frac{5}{\ln 2} + \frac{u}{(\ln 2)^2}
-\frac{e^u}{\ln 2} = -\frac{5}{\ln 2} + \frac{u}{(\ln 2)^2}
u = -\ln 2 \cdot W\left(-\frac{5}{\ln 2} e^{-\frac{5}{\ln 2}}
ight)
5. Solve for : Recall :
x = \frac{u}{\ln 2}
x = 5 - \frac{W(32 \ln 2)}{\ln 2}
So the solution expressed using the Lambert W function is:
x = 5 - \frac{W(32 \ln 2)}{\ln 2}
What calculators have the lambda-W function? How did you evaluate W( 32 ln(2) / ln2 ??
Harvard doesn't have an entrance exam......
I’ve been wondering about that on multiple videos
Actually, the important thing is not the Harvard entrance exam. There are a lot of RUclips titles like that.
I think it's okay as a title to attract attention
@@grayliar147no, not ok to lie.
Right, I thought it was just your grade point average. And your little SAT score. Maybe they should institute one, as there are a startling number of stupid, outright stupid Ivy League students, where you're sort of baffled how they got in, it's hard to distinguish them from junior college students. I guess because getting a 4.0 in HS is super easy. Including all math you might take. Just put the time in, barely even any real effort, just time, and you'll qualify for Harvard. I don't know if you'll get in, but you'll be eligible for consideration. Unless they start adding stuff like this....
well they do, but they examine your family's finances and influence.
Well I did appreciate your thorough and detailed explanation, it may have seemed simple to someone else however I found that it was very interesting and it also introduced me to the Lambda W function which opened up a new vista for me
That Relaxed and gentle voice= easy to listen ... impossible to solve ...
He is frenetic, full of himself, showing off his awful Indian English and impossible to follow at the affected velocity he is going. The quintessence of a horrible math teacher.
By seeing this video, i got motivated to stay happy in the college i got 😅
1) Start with the equation: 2^x + x = 5
2) Subtract x from both sides: 2^x = 5 - x
3) Multiply both sides by 2: 2 * 2^x = 2(5 - x)
4) Substitute y = 2^x: y = 2(5 - log_2(y))
5) Rearrange: y/2 = 5 - log_2(y)
6) Exponentiate both sides: 2^(y/2) = 2^(5 - log_2(y))
7) Simplify the right side: 2^(y/2) = 32/y
8) Multiply both sides by y: y * 2^(y/2) = 32
9) Substitute z = y/2: 2z * e^z = 32
10) Divide both sides by 2: z * e^z = 16
11) This is now in the form of the Lambert W function: W(16) = z
12) Recall that y = 2^x, so z = 2^(x-1)
13) Therefore: W(16) = 2^(x-1)
14) Solve for x: x = 1 + log_2(W(16))
This can be evaluated numerically to get the approximate solution x ≈ 1.7095.
A very nice explanation using the W Lambert Function! Greetings from Venezuela 🇻🇪
Very interesting, but you need a calculater for ln function. Within 5 iteration steps you can come to 1.705 within 90 sec. Or you write a small Programm in even Fortran, Basic or what you want. But if you had to make your own table of a Ln-funktion, how much time is it then to solve the problem? You solve the problem by creating another. 😊
That is the difference between a mathematician and a engineer😅
2^x+x=5
Let 2^x=y
y+log base2 y=5
log base 2 2^y.y=5
2^5=2^y.y
Put y=2^x
2^5=2^2^x 2^x
2^5= 2^3x
x=5/3≈1.7
It this correct I suppose
Jared Kushner wrote 2 + 2 = 5 and his dad slipped a check for 1,500,000 $ in his application. Worked just as well.
And corruption is still harmful. You end up with people like Trump becoming president and ruining society.
Spot on
explain pls@@Well...Whatever
@@jpxtv69 a bribe, he meant a bribe
And kushner is still a better negotiatior than Obama, Biden, or Harris.
totally raise my blood pressure, nice work mate.
As 1
The course code for this class is MATH 55a for the fall semester and MATH 55b for the spring semester. Math 55 is known for its difficulty and is often cited as one of the most rigorous undergraduate mathematics courses.
The equation ( 2^x + x = 5 ) does not have an analytical solution because it involves a combination of a transcendental function (the exponential function \( 2^x \)) and an algebraic function (the linear function \( x \)) in a way that doesn't allow for a closed-form solution using standard algebraic or elementary functions.
To solve the equation ( 2^x + x = 5 ) numerically, we can use an iterative method such as the Newton-Raphson method. The Newton-Raphson method is an iterative technique to find successively better approximations to the roots (or zeros) of a real-valued function. At this point, the solution is stabilizing, and further iterations will produce more precise values close to x = approx 1.723
Thus, the root of the equation ( 2^x + x = 5 ) is approximately 1.723. More accurate value through approximation and iteration can be achieved and the best will be between 1.7156 and 1.7157. It is breaking the symmetry at 1.7156 , where the result is entering the domain of 4.99 , keeping 1.7156 a limit but using 1.7157 is having an answer of 5.000259726995649. Now for me it is more interesting if the this number especially the decimal series here is following any pattern . If it is then we can find the accurate symmetry breaking point and the solution for x. Best for your endeavours from Cambridge, MA
This is interesting info, even though it's above my math pay grade. I'll have to find the textbook used in Math 55b and start reading. I love a good challenge.
It's more accurate to say that x ~ 1.716 (to 4 sig figs).
@@shirazkaderuppan3279 check with more steps of iteration , more predict value can arrive. 👍🏻👍🏻
@@quantumgravity639 Yes, it's 1.716 (correct to 4 significant figures). Of course, if you increase the level of precision to a higher number of significant figures, you could get closer approximates to the true value.
@@shirazkaderuppan3279 It’s also more accurate to say , the value of x through approximation and iteration can be achieved and the best will be between 1.7156 and 1.7157. It is breaking the symmetry at 1.7156 , where the result is entering the domain of 4.99 , keeping 1.7156 a limit but using 1.7157 is having an answer of 5.000259726995649. Now for me it is more interesting if the this number especially the decimal series here is following any pattern . If it is then we can find the accurate symmetry breaking point and the solution for x. Best from Cambridge , MA, USA
Ans by chatGPT:
The equation \(2^x + x = 5\) is a transcendental equation, and it cannot be solved algebraically. However, it can be solved using numerical methods, such as the Newton-Raphson method or by graphing.
Let's solve it approximately by using a numerical approach.
We start by evaluating the function \(f(x) = 2^x + x - 5\) and find the root where \(f(x) = 0\).
Let's test some values of \(x\) to find an approximate solution:
- For \(x = 1\): \(f(1) = 2^1 + 1 - 5 = 2 + 1 - 5 = -2\)
- For \(x = 2\): \(f(2) = 2^2 + 2 - 5 = 4 + 2 - 5 = 1\)
Since \(f(1)\) is negative and \(f(2)\) is positive, the solution lies between 1 and 2.
To find a more precise solution, we can use interpolation or further narrow down the interval:
- For \(x \approx 1.5\): \(f(1.5) \approx 2^{1.5} + 1.5 - 5 \approx 2.828 + 1.5 - 5 \approx -0.672\)
- For \(x \approx 1.7\): \(f(1.7) \approx 2^{1.7} + 1.7 - 5 \approx 3.249 + 1.7 - 5 \approx -0.051\)
- For \(x \approx 1.75\): \(f(1.75) \approx 2^{1.75} + 1.75 - 5 \approx 3.363 + 1.75 - 5 \approx 0.113\)
Since \(f(1.7)\) is slightly negative and \(f(1.75)\) is slightly positive, the solution is between 1.7 and 1.75.
The approximate value of \(x\) is around **1.74**.
The exact solution would require a numerical solver, but \(x \approx 1.74\) is a good approximation.
in 6 steps, in my head, I get to 1.7157, in about 1/7th the time it takes for this video to explain.. using Newton-Raphson indeed. Using the total video time I think I could reach about 10 decimals precision, using no other means than head.
@@scififan698, I was at 1.716 but got bored and kept it approx
"Answer by ChatGPT" *did not read* thank you
Thanks,
I did this in less steps right in my brain…and then you can actually get more specific by solving it to more decimal points.
1,7157
I would do it loke this :
2^x is a function that rises
x is a function that rises
And there is some kind of law that states if 2 functions are rising then their sum is a rising function as well
So basically we get 2 functions
2^x + x
and
5
If they cross they cross only once
It means there is only 1 solution
Which we have to guess
Nice! But the way you write X confuses me on each line✌️😂
You are using order operations and natural log to converge to the answer subtracted by 5. I completely understand.
there is an easier solution. there is a zutturubut(q) function that gives x for 2^x+x=q. much simpler indeed.
Me: I don't understand what this guy is talking about ... *go back to watching Nicki Minaj twerking*
yeah, but that's German and you can't use that at Harvard.
I love your writing. In South Africa, we also write x like you do to distinguish it from a multiplication x. Beautiful writing.
It's easier to just not use x for multiplication
Don't understand this guy's obsession with the w function. Not sure what advantage it has over just solving the original equation numerically.
Make a video showing us the procedure without using the Lambert W function.
@@mancinieric
log_2(5-log_2(5-log_2(5-log_2(5-log_2(5-log_2(5-log_2(5-log_2(5-log_2(5-log_2(5-log_2(5-log_2(5-log_2(5
~1.716
could not agree more but I do like how he writes the ecs....X )(
And how would you do that?
@@mancinieric
2^x=5-x => x=log_2(5-x)
x=log_2(5-log_2(5-log_2(5...
x=~1.7156
you can also use Banach contraction principle in the form x(n+1)=log_2(5-x(n)) and initial value x(0)=1,5 , convergence is quite good here.
Impossible to focus due to the way he writes "X".
I forgot how to write completely, will sign up for preschool tomorrow.
👏👏- I enjoyed learning this very much. I can appreciate the solution but I have to accept the LW function mechanism just works. A but rusty with logs but will come back with some review. Thanks for explaining - my son who enjoys maths will love this!
This is just based on defining a function, this Lambert function, plus some algebra.
f(x)=2^x+x
g(x)=2^x is a strict increasing function because 2>1
h(x)=x is a strict increasing function
f(x)=g(x)+h(x)=2^x+x is a strict increasing function as sum of increasing functions. So f(x) is an injective function.
This means that equation 2^x+x=5 has only a single solution.
So if we try for x the values 0,1,2,3 we find the solution x=2
Is W actually calculated via a series solution or numerical integration? Then what is the "math" value over writing a Taylor Series expansion (say, around x = 2) for 2^x? You get an estimate 1.735 rather easily with a first-order expansion. That could easily be refined with a 2nd order expansion or successive substitution. Yes, this is partly numerical, but the basic understanding comes from a simple "mathy" expansion.
Didn't know about lambert W function, thanks!
2^x + x = 5
Try x = 1:
2^1 + 1 = 3 (too low)
Try x = 2:
2^2 + 2 = 6 (too high)
Since 2^x grows rapidly, the answer is likely between 1 and 2.
Try x = 1.5:
2^1.5 ≈ 2.83 + 1.5 ≈ 4.33 (getting closer)
Try x = 1.6:
2^1.6 ≈ 2.94 + 1.6 ≈ 4.54 (still a bit low)
Try x = 1.7:
2^1.7 ≈ 3.13 + 1.7 ≈ 4.83 (almost there)
Try x = 1.8:
2^1.8 ≈ 3.25 + 1.8 ≈ 5.05 (slightly high)
So, the value of x is approximately 1.8.
Your calculator seems to have a problem. 2^1.6=3.03..., not 2.94... , values for 2^1.7 and 2^1.8 are also wrong.
Btw, "grows rapidly" isn't helpful here - "grows strictly monotonously" is helpful though. Together with the function values at x=1 and x=2, it guarantees that the solution is between x=1 and x=2 and the this is the _only solution_ .
I told that approximately 1.8.
@@saikatchatterjee2837 So? It's wrong.
No. Why it's wrong?
x=1.66666666666666666666~
Keep 'em coming.
The only Harvard entrance exam is looking at daddy’s portfolio
Or skin color and gender
@@Seagaltalk they/them
@@SeagaltalkYou are not a victim🤡
@@Seagaltalk You are not a victim🤡
@Mariajbh2 now i am even more victimized.. thanks
Fun. Gets your analytical head straight. But it's management and investment skills that will make you wealthy, and arts and literature that will make your life richer and worth living.
Why not use numerical approximation instead of doing it the hardest way imaginable?
I watched this at 2x speed, and it still felt slow. Added this to my sleep playlist.
I would answer, somewhere between 1 and 2.
and a parking lot...
@@Dr.Birkenmeier 🤣🤣🤣🤣
And closer to 2 as a guess. Obviously closer to 2 isn't the answer, but it was obvious it would be.
The answer is somewhere between one and two. Easier to test via reductive methods.
Maybe the log is a matrix of options.
Here is the similar solution. Let the function fck(n), where the value of function is the equation of 2^x+x=n. So, our answer is fck(5)
If you had used ln and differentiation
It would be so easy
I think the easiest way to solve this is to make graphs between y=2^x y=5-x
2^x + x = 5
1 < x < 2
👍
Most high schoolers have a graphics display calculator that can graph equations. Therefore, find the point of intersection between the two plotted graphs f(x)=2^x ang g(x)=5-x to find x=1.716
I eyeballed it and said ‘about 1.75’. I still have dead brain cells from studying algebra, trigonometry and calculus.
Come on,,, maybe the dead brain cells were cause by some smoky elixir....
My answer is x = 3. Is it the right answer? No! But it's MY answer since it still technically proves the theorem, just not from the original. Let me explain. 2^x + x = 5 -> 2^x = 5 - x -> 2^x/2 = 1 and (5-x)/2 -> 1 = (5-x)/2 -> multiply both sides by 2 -> 5-x = 2 -> add x to both sides and subtract 2 from both sides -> 5-2 = x so x = 3. Again, not the correct answer, but i did show my work
I think the whole point of a math brain twister is that it should be solved without additional resources, how are you gonna compute what u call the Lambert function without a computer
I agree But naming the Lambert what ever function increases the mathematical status of the teacher showing to the rest of us that not using it we remain morons
@@Dr.Birkenmeier i did my whole education calling it "x exponential of x". It does have some properties, but none were particularly useful in this case, and surely not so pertinent that it needs a name of its own.
I found the same answer. If you had to use a calculator to figure out the answer, so you dont need to appeal to W function. Just do try and error, just like I did. Unless you only want the answer in terms of ln2, but in that case you want the decimal number though
Using log on both the sides we can solve more easily
That is right back in my log cabin I solved this in no time at all
How do you solve it by log
x = U(5), where U(x) - Cucumber U-function: U(2^x + x) = x.
Imagine spending so much effort to learn maths and physics only to hear a doctor telling you just have an untreatable paralysis with two years of life expectancy 😮😮
in 6 fast steps, I arrive at 1.7157 without breaking a sweat. Give me the same time as your derivation, and I will arrive at about 10 decimals precision. Easy peasy. This without knowing about the Lambert-W function. But.. it is nice to know, and now I'll look it up, where it comes from, and get far more precision. Thanks!
What is w(32ln2) then? Can it not be simplified further?
Calculator
Without W function you could have immediately have mentioned that this is an increasing and continuous function - estimate te value through Newton-Raphson. Don’t see the point of ‘solving’ when you need approximation anyways.
It is easy to see that the solution is between 1 and 2. Both 2^x + x-5, and its derivate are smooth in the range 1 to 2. We can use one or two iterations of Haley's method to get to x~1.716 (en.wikipedia.org/wiki/Halley%27s_method)
While not a closed form solution, the solution is more generic and a practical numerical method for a broad set of problems.
I think realy this exam purpose is not about solving it completely, instead to show the examiner the way you think and the steps you followed to solve this problem, even if you couldn't solve it
Indeed
As it is monotonically increasing function, just use binary search with epsilon 10^-6 to get accuracy and we know that answer lies between 1 and 2 so make this end point of range
The Harvard entrance exam consist of only one question: “Are you a white straight man”? You don’t want to get that one wrong.
Nothing wrong with that Imagine if they asked if you are an eliptical one....
@@Dr.Birkenmeier😂😂😂😂
That will be so right, we dont need to lower our standards.
@@cirjanionutsim So, being "white" is the highest standard?! Even without a high school diploma?!!!
This explanation is essentially no different from defining the solution to 2^x + x = 5 as y and then claiming that y is the answer.
At the very least, there should have been an attempt to express the solution in the form of an infinite series.
I am quite disappointed in myself for expecting something fun or creative.
Just use approximations for x value then lol. Whatever comes closest to 5 is the answer
If you don't use multiply by Ln 2 you still get an answer using lambda W function. The answer is ≈2.4. there is no same reason for employing that method other than if you already know the answer.
8:21 idk what is that mean
This is not clever. Much faster to convert the equation to the form x=log_2(5-x), and to solve it graphically or numerically. Immediately one gets to x=1.716. You dont need a Lambert function to do such simple things.
How many freaking Harvard applicants know what the Lambert W function is?
Everyone with Rich Parents ²
im really basic into math and i was concerned heavily. im so relieved the comments agree
Now I know why i went to Technical College! 😂😂😂
And you should be proud of it, it's not because you didn't make to Harvard that you are less than the people that are there or that you doesn't have the same skills they have or even better ones.
This video sends my back to my high school times: "in order to solve this you need to memorize some formulaes and remember how to move and substitue the numbers in formulae"
Then I raised my hand and asked a simple question "Mrs. Teacher, what are we calculating ?"
The hardest question that day turned out to be mine..
Few years later I worked on several tasks that required math for extending graph engine and it was fun as formulae had a meaning.
2² +1 =5 😅 Harvard University pass 😂😂
Bhai tune 2 alag alag value di hai x ke liye 1,2
@GaminGg483YT
Bro you did not even pass middle school I think.
@@pacivalmuller9333😂😂
😂😂😂
Use quadratic formula
It's so easy to solve
Class 10 ncert question
An answer in terms of W(32Ln(2) ) is not an answer; and if any student of mine had presented it as such I would have given them marks for the working out only (let's say half marks). Introducing a "magic" function is not helpful, even if W(k) has a value.
Suppose I said: 2^x + x + 5
x =J(5), where J(k) has a value.
me: used casio calculator to calculate and its result is 2+2=5 lol
2^x + x = 5
so:
5-x > 0
so:
x < 5
and in this equation we know that if x would be smaller than 0 then answers is smaller than zero , so x is a number between 0 and 5!
if we choose x = 2 answer is 6 and if we choos x= 1 answer is 3 , so x exactly is a number between 1 and 2 and we can calculate x just with som simple and basic ways of approximation in numerical calculations or even try few number between 1 and 2 ! and its exactly what you calculated, approximately 1.7
You can bypass your entrance exam for Harvard with black skin or 🌈 identification
Aha! Always the argumentative political statement coming “straight” from the right 😂
@@JohnSmith-xx9th lots of deluded people today being mislead that “diversity is their strength”, or that there is any inherent value in diversity at all. And let’s not kid ourselves, Harvard math & science is filled with foreign students, subsidized by 🇺🇸 taxes; you know, people who study useful things, and who are too busy for banal, unfruitful pursuits like activism and political arguments with social media addicts. They’re not lowering standards because of the qualified applicants, they’re lowering standards because their target student profile is below the standard.
-class of 05
You won pass because you are stvpid😂
I learn a lot from this. Thank you for your great solution. :)