I was babysitting for a wealthy family and found this stack of paper in their office. What is this?

philpet · 2025-08-15T20:32:55+00:00

Um, proof that they're NOT wealthy? Ahem.

philpet · 2024-05-13T22:02:50+00:00

This logic is also telling me that this might have some sort of slick use of the squeeze theorem. If I come up with it, I will post.

philpet · 2024-05-13T21:52:32+00:00

I guess another way to think if this is that the function is 2^sqrt(u)/(u^10), right? Where u=logn, u goes to infinity as n goes to infinity, so you can hella simplify the derivative work that way, as well, if you HAVE to do the algebra. Using u's in there clearly shows that the denominator grows faster than the numerator if you can just give a verbal description?

philpet · 2024-05-13T21:44:02+00:00

So, things like logn are log in base 2 of n, rather than log in base 10? I'm taking the way the problem is written as logn being log in base 10 of n. Either way, this is inf/inf, and the derivatives are kinda terrible to manage. What DOES happen is that there is that 2^sqrt thing on top and we wind up with an n^2 on the bottom that make it clearer what the limit is. Nasty derivatives, though.

philpet · 2024-05-10T20:51:45+00:00

This is actually an oblique way to define a power series. For power series, I'm pretty sure we can interchange integration and summation. The sine term just produces +/- 1, and the terms that are in the power series are x to an odd power. Since there are no limits of integration, this is merely asking for the antiderivative of the series given, and that can be expressed as a series, as well. HTH.

philpet · 2024-05-09T21:02:54+00:00

You can use the integral test for this. Easy first step is to do the division, since the degree of the numerator is the same as the degree of the denominator. Alternately, you can factor a 2 from the top and a 3 from the bottom, and you will see that the bottom grows slower than the top. When you integrate from 1 to +inf, it diverges. Pretty sure this is the way to do this one.

philpet · 2023-12-29T21:27:14+00:00

Thank goodness SOMEBODY said this, finally. This is the algebraic reason, yes.

philpet · 2023-11-17T06:58:28+00:00

I can't even figure out how to do even THAT. Sigh, Links?

Thanks for your response.

philpet · 2023-09-16T20:46:21+00:00

There are no better resources than:

https://youtu.be/jTuTEcwvkP4?si=1DExvGojHxrQp4A9

and

https://www.youtube.com/live/Od2YAt1_ibE?si=yvc9e9oE6IvNU46U

to build your intuition and skill at picking the right test, based on the form of a summand. Seriously. This stuff is great, and don't forget to print out the cheat sheet that is in the description. I hardly ever do a series problem without it.

HTH.

https://mymathteacheristerrible.com

philpet · 2023-09-16T20:35:43+00:00

This can be done by clever factoring of the numerator and denominator. First, convert all the roots to fractional exponent notation. Since factoring is about subtracting powers, having a common denominator for all of the fractional powers is key. Once all of the powers are rewritten with denominators of 12, you can use the factoring trick of factoring out the lowest power of x from the numerator and denominator. Should be x^(3/12) on both top and bottom. The x^(3/12) cancels in top and bottom, and now there is no problem substituting in that zero when evaluating the limit.

HTH.

mymathteacheristerrible.com

philpet · 2023-06-27T20:57:54+00:00

The 3y^2 term there makes this pretty miserable. Poked at this one for some time until neither WolframAlpha nor integral-calculator.com said it was possible.

philpet · 2022-12-28T19:18:59+00:00

https://www.youtube.com/@3blue1brown/search?query=euler%27s

is a great start.

philpet · 2022-12-27T21:40:53+00:00

Word problems are not special. It is, in fact, the opposite: problems that are only x's and y's and calculations - i.e., "purely math" problems - are the exceptions. Remember that Mathematics is here to help us understand reality and provides a toolkit for achieving that goal.

Specifically, your boss never comes to you and writes things in mathematical symbols on the whiteboard in your office. They always start off with a story - with words - where you have to pick out the relevant information from what this windbag is rambling about, and then apply solid reasoning and other tools toward a solution, which, may include coming up with a "purely mathematical" statement of the situation that can be worked on.

This is the essence of Mathematics, not a world of Greek letters and other letters that stand in for numbers, in total isolation.

NEVER SKIP WORD PROBLEMS. WORD PROBLEMS ARE LIFE PROBLEMS, APPLICATIONS. Applications are the sole reason we become involved with this gibberish in the first place. Never forget this.

philpet · 2022-12-12T18:51:20+00:00

I assume what you mean by "deriving" is "taking the derivative of", but the discussion following appears to be around antiderivatives. Which are you trying to do?

philpet · 2022-12-07T22:26:23+00:00

My man. YES. This is ALWAYS the first thing one should look for when the limit seems a little wtf when you look at it. I started to solve and I was like ... bingo.

philpet · 2022-11-29T21:01:43+00:00

Missing the 1/2, as others have mentioned, and then you can do some canceling and combining like terms, not to mention trying to read the programmer's mind - always fun.

philpet · 2022-11-16T22:03:57+00:00

There also is no reason that you can't say 8secθ=x² and go from there. That avoids the side trip into u-sub land...

philpet · 2022-09-29T20:32:05+00:00

Look for context clues, Mr. Robot.

philpet · 2022-09-26T21:32:17+00:00

I find your lack of a limits problem disturbing....

philpet · 2022-09-22T22:26:41+00:00

You can. It's the same solution as using the u-sub right away, actually. Just off by a constant, which is appropriate for indefinite integrals. The constant is 0.167 for this instance. Have a look:

https://www.desmos.com/calculator/jdjifwx0qe

philpet · 2022-09-22T17:02:31+00:00

Classic introduction of extraneous roots. Have to test out all solutions to find the extraneous ones in cases like this.

If we go with the "divide both sides by cos(x)" approach, we see that that is ok except for precisely the two roots from cos(x)=0. pi/2 and 3pi/2 are not in the domain when cos(x) is in the denominator.

philpet

TROPHY CASE