Solving a 'Harvard' University entrance exam

Comments · 1.2K

• BR

  @brendanward2991 · 1 year ago

  It's called the W function because in the end you need to use Wolfram-Alpha to solve the equation.

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• LU

  @luisfilipe2023 · 1 year ago

  I’ll never not be amazed by mathematicians ability to just make stuff up and call it the day

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• CH

  @chuckw4680 · 1 year ago

  So it still can't be solved by hand and needs a computer/calculator and I still don't know what a Lambert function is. I'll call it a day.

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• LI

  @LiteBulb88 · 1 year ago

  I should've tried this technique on tests where I couldn't figure things out. "The answer is B(5), where B is a function I'm defining right now that will solve this problem."

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• MI

  @michaelz6555 · 1 year ago (edited)

  Learning about the “Goal Seek” feature in Excel alone was worth the cost of admission. Thanks!

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• CG

  @cguy96 · 1 year ago

  I think people are missing the fact that the Lambert W function is not just some arbitrary inverse, otherwise Presh could have just said P(2^x+x) = 5 and stopped there. The Lambert W function has been extensively researched, has a lot of properties, and identities, and is quite useful. This is why Presh went to the trouble to reformulate the problem into the product-log form.

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• KR

  @krabkrabkrab · 1 year ago

  In my head, I tried x=5/3 and realized it's a bit low. SO I went for 1.7. Then Newton's method: x_new= x-  (x log(2)-log(5-x))/(log(2)+1/(5-x)) immediately gives 1.7156 (on a calculator that doesn't have a Lambert function).

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• AS

  @asparkdeity8717 · 1 year ago

  And to those complaining, we got a near identical question in our Cambridge maths entrance exam, the very paper I sat had a question with the lambert-W function. Don’t believe me, look up STEP II 2021 Q4. Not something I had ever learnt in school or heard of at the time, but given its introduction I was still able to do the question.<br><br>It’s not about solving the question for an exact answer using a calculator, but it’s about understanding and applying new techniques to gain an analytic closed form solution to an unseen problem. It actually tests your true mathematical ability.

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• JO

  @JonSebastianF · 1 year ago (edited)

  <i><b>U 2 to the Power of U</b></i><b></b><br>...sounds like a power ballad by Prince💜

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• PE

  @PeapotsDoodles-k7v · 1 year ago

  I used an alternate method to solve it that doesn’t require computers (but it does use derivatives.) !! I used linear approximation in which L(x)= f(a)+f’(a)(x-a), where L(x) is an approximation, a is a similar number to the one you!re trying to find, and x is the value you!re trying to find. Basically, which this formula, you can approximately find any value of any equation as long as you know something of a similar value! <br><br>For example, in this equation, I quickly noticed that 2^2+2=6 , which is pretty close to 5. So, I established that a=2. That’s it !! All you do now is solve the equation.<br>f(x)= 2^x +x <br>L(x)= (2^a+a) + ( ln(a)+2^a )(x-a)<br>5 = 2^2+2 + (ln(2)+(2^2))(x-2)<br>5= 6 + (ln(2)+4)(x-2)<br>-1= (ln(2)+4)(x-2)<br>(-1/(ln(2)+4))+2=x<br>x= 1.78!!<br>The answer is a bit off because its still an approximation, but it’s much better than using computers in my opinion.

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• VE

  @verkuilb · 1 year ago

  Let me get this straight—you follow up a video about whether 3x5 is the same as 5x3…with this???<br>🤯

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• CH

  @chrisarmstrong8198 · 1 year ago

  The Lambert W function was never mentioned in my High School or University maths subjects (in the 1970&apos;s !). &nbsp;Thanks for the info.

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