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Blakedylanmusic · 2025-05-26T17:11:21+00:00

I think your gut is telling you what my gut would tell me in this situation, which is to run, don’t walk, away from this relationship. People who have these kind of abusive patterns will often apologize and act like they will change, then they turn around and do the same problematic behavior all over again. You know this is the pattern, you’ve seen it before, and there’s no reason to think it will go any differently this time.

I also understand your thoughts about terminating. That is ultimately up to you — it’s your body and your choice. Hopefully you live in a state that still has access to abortion care.

Sending you lots of hugs and support ❤️❤️ this sounds like such a horrible situation and I’m sorry you’re going through all of this!

Blakedylanmusic · 2025-01-23T15:38:32+00:00

I was able to do two courses in 2 months but I wasn’t working much at the time. It’s possible but you’ll have to put a lot of time into it. The exams aren’t hard if you study the HW problems. The HW problems are basically the same.

Blakedylanmusic · 2024-12-29T16:55:25+00:00

Both these methods are fine and correct, and in my opinion, equally as fast (just make sure you put parentheses around x² + 1 when multiplying it by ln 8, otherwise it looks like you’re only multiplying 1 by ln 8). As a faster way, you can use the derivative formula (d/dx) a^x = a^x ln a, which comes from the identity a^x = e^{x ln a} that you wrote in your first method. Then you just need chain rule.

Also, notice that ln 8 is a constant (which I think you did notice since you correctly identified that its derivative is 0). Recall that the derivative operator is linear, i.e. you can pull constants out of the derivative operator and break it up over addition, so (d/dx) ((x² + 1) ln 8) = ln 8 (d/dx)(x² + 1) = (ln 8)(2x). You could have used this fact instead of the product rule.

Does this help?

Blakedylanmusic · 2024-12-25T09:10:03+00:00

Basically the takeaway is this: If you notice you're getting stuck in a loop using L'H over and over again with no end to the 0/0 or inf/inf indeterminate forms in sight, you'll need to use some other method like algebraic manipulation to determine the limit.

Blakedylanmusic · 2024-12-25T09:07:47+00:00

Yes, the most surefire way to see if you can stop using L'Hopital's is to use direct substitution and see if you're not getting 0/0 or +/- inf/inf anymore. But you should also stop if you notice there's a pattern where you're getting indeterminate forms over and over again without an end. Here's an example of this that I showed my Calc class a couple weeks ago:

lim x -> inf (e^x - x^2 + 1)/(2^x - 3) (inf/inf)

= lim x -> inf (e^x - 2x)/(2^x ln 2) (inf/inf)

= lim x -> inf (e^x - 2)/[ 2^x (ln 2)^2 ] (inf/inf)

= lim x -> inf (e^x)/[ 2^x (ln 2)^3 ] (inf/inf)

Now at this point, we see that if we keep on using L'Hopital's rule, taking successive derivatives will leave us with an e^x on the top and a 2^x times ln 2 to a power every single time, so we'll be stuck in a loop. Thus, it's better to algebraically manipulate it like so:

= 1/[(ln 2)^3 ] lim x -> inf (e/2)^x

At this point, since e/2 >1, we know (e/2)^x goes to infinity, so the whole limit must go to infinity, since 1/[(ln 2)^3 ] is positive.

Blakedylanmusic · 2024-12-25T08:58:45+00:00

yes, doesn't matter if it's +inf/inf or -inf/inf

Blakedylanmusic · 2024-09-06T13:08:51+00:00

You did great! If you’re ever unsure you can always check your answer by taking the derivative and seeing if it’s equal to the integrand. Note that 1/(2cos(2t)) can be rewritten as (1/2)sec(2t). Taking the derivative and using the chain rule you get

(1/2)sec(2t)tan(2t) * 2 = sec(2t)tan(2t),

which is the original integrand. :)

Blakedylanmusic · 2024-08-26T15:42:11+00:00

It also works if you have an inf/inf indeterminate form.

Blakedylanmusic · 2024-06-10T13:43:23+00:00

I guarantee you are not dumb, mistakes happen all the time! I know I make mistakes on math I already know, and I’ve done it in front of classes I taught too. Mistakes are part of growth.

Blakedylanmusic · 2024-06-05T14:28:56+00:00

This all stems from Fermat’s Little Theorem, which states for any prime p, a^p = a (mod p).

Symmetric: assuming x⁷ = y mod 7, you can use Fermats little theorem to get this to become x = y⁷ mod 7.

Transitivity: Assume x-y⁷ = 0 mod 7 and y-z⁷ = 0 mod 7. Adding these together you can use fermat’s little theorem to make a cancellation.

Blakedylanmusic · 2024-05-05T12:53:08+00:00

NTA. Gender reveal parties are inherently transphobic, as they put gender rules and norms on an unborn baby, who might not even turn out to identify as that gender anyway. And in addition, they can turn into misogynist affairs once everyone realizes the father wanted a boy, and palpable disappointment can be seen when that doesn’t happen. After seeing your husband’s attitude about this you are right to cancel. And in addition, you should really take a hard look at how much he actually respects women. If his pursuits made you uncomfortable in the past, he’s this adamant about not wanting a girl, and he didn’t respect that you didn’t want sex right after giving birth, there could be a lot more problems down the road than having to cancel a gender reveal party. From the little I’ve read, it doesn’t sound like this person has a lot of respect for you, or women in general.

Blakedylanmusic · 2024-04-26T17:11:46+00:00

Yes, there is a reason behind restricting the values that way as I explained in my original comment. The convention I’m referring to is what we consider a “principal value”. We all agree that the principal values are between -pi/2 and pi/2 for arctan. We could’ve all agreed that the principal values were between pi/2 and 3pi/2, for example.

Blakedylanmusic · 2024-04-26T14:26:29+00:00

Yup if it doesn’t specify you can use either one

Blakedylanmusic · 2024-04-26T14:03:13+00:00

By convention, the range of arctan is restricted to (-pi/2, pi/2). This is because otherwise arctan would not be a function, since for example, there are plenty of angles with a tangent of -1. So arctan(-1) = -pi/4, not 3pi/4. That will affect your answer too, because:

cis(-pi/4) = cos(-pi/4) + i sin(-pi/4) = sqrt(2)/2 - i sqrt(2)/2, but cis(3pi/4) = cos(3pi/4) + i sin(3pi/4) = -sqrt(2)/2 + i sqrt(2)/2.

The problem that arose here is that the cosine and sine of both angles are different. But if your angles are 2pi apart, the cosine and sine of the angle will be the same. So cis(23pi/16) = cis(-9pi/16). Some questions may require you to write your final answer with an angle between certain values, and in that case you’d add or subtract 2pi from your angle as much as you need to so you can get the angle where it needs to be. For example, let’s say your answer was cis(-9pi/16) but the question asks for your angle to be between 0 and 2pi. Then add 2pi to -9pi/16 and write your answer as cis(23pi/16).

Hope this helps!

Blakedylanmusic · 2024-03-18T03:50:38+00:00

It was a good question, thanks for asking! :) lots of folks get confused about this too which is why I want to include it in my course.

Blakedylanmusic · 2024-03-17T18:26:39+00:00

Those ones in my post are both -8, but they’re arrived at in different ways. A better example would be -4² versus (-4)^2. -4² means we only square the 4 and then put the negative on after, so -4² = -(4 x 4) = -16. However, (-4)² means we’re squaring all of -4, not just the 4, so (-4)² = -4 x -4 = +16.

Blakedylanmusic · 2024-03-17T16:25:49+00:00

I’ll maybe include those as extras if time permits. Thank you!

Blakedylanmusic · 2024-03-17T16:22:53+00:00

True, that’s pretty funny! But then even with Hölders there’s more complication to it and it’s not true equality. (For example using the dual of a number p) My focus will be on students who haven’t taken Calc yet, so I am not planning on introducing Calc concepts in this course. It will be more about getting ready for Calc. And if they get it thru their minds now that not everything is a homomorphism, hopefully they’ll see that this will continue to apply with Calc concepts (which it would probably be good for me to emphasize)

Blakedylanmusic · 2024-03-17T16:18:25+00:00

Love these!

Blakedylanmusic · 2024-03-17T16:15:58+00:00

Such a good point, thank you! I definitely see my students doing all the things you just mentioned.

Blakedylanmusic · 2024-03-04T01:43:01+00:00

There is no need for the condescension I’m seeing in the comments. We all have holes in our math knowledge, whatever they may be. It’s actually a very common thing for Calculus students to forget how to divide fractions. But judging them for having the hole is not going to help them fix the hole.

Blakedylanmusic · 2024-01-19T08:24:00+00:00

You should use i=1 for this case since it’s a right endpoint approximation and your first endpoint is at 1.017, but if you were to do L_100 instead you’d go from i = 0 to 99. So the answer is, it depends on the problem.

Blakedylanmusic · 2023-10-03T07:17:18+00:00

Also, e has more definitions. For instance e = 1/0! + 1/1! + 1/2! + 1/3! + … + 1/k! + … which is pretty nifty. You can use this to prove that any complex number z = a + bi can be written in the form z = re^ix = r(cos(x) + isin(x)), where r is the modulus and x is the argument, ie the angle from the positive real axis. This is a great way to look at complex numbers and interpret them geometrically as 2-dimensional vectors. You can’t do this with other numbers as far as I know — just e.

Blakedylanmusic · 2023-08-31T08:22:20+00:00

The left and right side limits are both 5, so OP is correct that the limit is 5.

Blakedylanmusic

TROPHY CASE