Usually yes, but in these problems no. We are representing these functions with geometric series, and we know that geometric series diverge at the endpoints. Specifically, as I show at the beginning of this video, we know a/(1-r) converges for |r|<1. Since it is <, not ≤, the endpoints are not included in the interval of convergence. You could still test the endpoints if you wanted to, but the series will diverge every single time in this case. Hope this helps!

You're welcome! Glad the video has been helpful. A couple tips on clearing values/work: If you want to clear basic number values on your screen, CE|E is your go-to button for clearing. If you are inside a menu of some kind, such as the cashflow menu in the CF button or the interest rate conversion menu, the best way to clear those menus is to hit 2ND and then CE|E which has a second function of CLRWORK (or "clear work"). Finally, if you want to clear out the values for the TVM buttons (the white ones) then you want to use 2ND and then FV which has a second function of CLR TVM (or "clear time value money"). Hope this helps!

You're very welcome!

You cannot. ∞^∞ is not an indeterminate form, we can reasonably say that it just approaches ∞. Indeterminate forms compatible with L'Hopital';s rule are ∞/∞, -∞/∞, ∞/-∞, -∞/-∞, and 0/0. Other forms like 0*∞, ∞-∞, 0^0, 1^∞, and ∞^0, are also indeterminate forms, but are not immediately compatible with L'Hopital's rule as you would need to manipulate the limit into one of the compatible forms listed earlier before using the rule. Hope this helps!

It is the standard notation used for present value factors (discount factors) in Financial Mathematics. You will see it used in many other resources for FM, including the exam FM itself. In other words, I did not invent the notation. In general for different math courses, you will find that many notations are created out of convenience. v^t is a lot less to write compared to (1+i)^-t, and also clearly differentiates it from an accumulation factor, so there is less chance for confusion or error if a negative exponent is forgotten. There are a few instances where you will write (1+i)^-t, such as with PV/price functions later on in FM, but otherwise v^t is more common to see.

You're very welcome! And thank you for the kinds words, I appreciate it :)

I explain where the constant goes in the first 3 minutes of the lesson video for this topic. You can find the link to the lesson in the video description. In short, if you were to write out the work, the constant gets cancelled out and does not effect the answer.

Thank you! Yes, Calculus 3 is a big course so I will be at it for a while, but its also my favorite so I'm looking forward to covering all of it. After Calc 3 I will probably either cover linear algebra or prob & stat. My undergrad degree is in pure math, so I've never taken application courses such as thermo or fluid mechanics. I would have to consult my engineering friends for those! So pure math is the plan for now. Also, I am toying with the idea of doing a lower level course such as algebra at the same time as Calc 3 & the next higher level course. We'll see!

Which example are you referring to?

@@JKMath example 2. With 3x^2 + 1/x^3 + 2x^2 + x

Just checked my work and it is still correct. Can you tell me how you got -3? Maybe I can help figure out the discrepancy.

@@JKMath umm.. okay. You said that we get 3 terms after factoring the denominator. Our denominator was x(x+1)^2. Our three terms were A/x + B/(x+1) + C/(x+1)^2 and they’re equal to 3x^2+1/x(x+1)^2. We are trying to make the denominator on the right hand side equal the one on the left.

Right. When you solve for B can you show me how you get -3?

Never used it personally. I looked it up, and it seems it would be a good calculator for the SAT test, but I think I would still prefer the graphing calculators I show in this video for more complex math such as calculus 3. But I am biased, I used the calculators in this video all through college and they never let me down!

Yes. Optimization problems can be found in my Calculus 1 playlist.

ok thanks @@JKMath

Whoops, that is a typo. My apologies on that. Your answer is correct, it should be 12,862.67. I believe i wrote down the answer to one of the problems in this video by accident. I will definitely fix this. Thank you for letting me know!

You're welcome! Good luck on your exam!

The washer and disk methods are essentially the same except that the disk method is the washer method where the inner radius r(x) is 0. In other words, you only have an outer radius for the solid. Now, when comparing disk/washer methods to the shell method, I go over this comparison in my lesson video on the shell method. I recommend you watch that video in my calc 2 playlist (it is lesson 4), specifically starting at the video chapter around the 39 minute mark. Hope this helps!

@@JKMath Ok, thank you so much!!

No, assuming you are trying to calculate the volume of the same solid, but with different methods, the shell method and washer method when used correctly will give you the same volume. They are just two different ways of slicing up the volume to calculate it. Based on the shape of the enclosed region being revolved around an axis, the shell method might be easier or the washer method might be easier to use. The two methods just provide us with options on how to calculate the volume. Hope this helps!

Thank you for the kind feedback! I really appreciate that. Glad I was able to contribute positively to your studies! Best wishes on any exams you might be taking :)

I explain why we need to use 1.06 instead of just 0.06 starting around 11:10 in the video. If you have further questions after rewatching this, please let me know!

@JKMath Thank you

The radius is neither "always" x or y. It will depend on the axis of revolution. If you are revolving around the x-axis, then the radius is y, but if you are revolving around the y-axis, then the radius is x. BUT, if you are revolving around a line that is not the x or y-axis, then you need to adjust the radius accordingly as I explain starting at 25:02. Hope this helps!

@ yes I got what u mean , may God bless you

L'Hopital's rule is used for evaluating limits in an indeterminate form such as 0/0 or ∞/∞. But we are not working with limits in these problems. so L'Hopital's rule does not apply. Hope this helps!

You're welcome!

Awesome, glad I could help! :)

The notation I use in this video is the standard notation for parametric derivatives. To my knowledge, it is the notation most commonly used in calculus textbooks, and other resources online for this topic. It clearly lays out what parametric equation you are taking the derivative of (x or y) and with respect to what variable (in most cases, t).

@JKMath Got it, thanks.

Yep, I highly recommend the 4 step method as many of the formulas out there for geometric annuities are case-specific, and will not work if the annuity is an annuity-due or if the time at which the inflation rate starts is different than with the second period. Good stuff!

Yes, technically you can see a decreasing perpetuity in a problem if it is a decreasing geometric perpetuity, although they are uncommon to see. Regardless, the formula does not change. You use the geometric perpetuity formula I show in this video except where the inflation rate r is negative rather than positive (for example, decreasing by 5% would be r=-0.05). Decreasing arithmetic perpetuities on the other hand are not a thing, because at some point the payment will become 0 as you decrease by a set amount, which is different than a decreasing geometric annuity where the payments would decrease by a percentage, so technically it could still go on forever. But still, the payments will become very small at some point, becoming practically 0. Hope this helps clear up any confusion!

You're very welcome! Glad I could help you :)

You might have to give me an example of the scenario you are referring to. If i understand your question correctly, it all comes down to the comparison test you use. From the direct comparison test (DCT), if you choose to compare to a convergent p series, you need to show that the comparison series is bigger than your original series. If that does not happen, you cannot conclude anything and need to pick a different comparison series. I'd recommend reviewing the DCT, that may answer your question. It is lesson 25 in my calc 2 playlist. Hope this helps!

​@@JKMath ok thanks, i have another question, lets say, i have a problem where i used integral test to find |an| which i got convergent, do i need to do another test to check if its absolutely convergent?

@@Rexcel-y6h No, referencing the chart at 11:22 in the video, if you can show the the series for |an| is convergent, no matter the test, then you can conclude that the series for just an is convergent as well. You can conclude that the series is absolutely convergent.

@@JKMath ok thank you

Can you give me an example of the type of problem you're referring to? It might be easier to explain if I can point to a specific example.

@JKMath Hello there, this example gives me an interval instead "Approximate the function f(x) = cube root of x by a taylor polynomial of degree 2 at c = 8. Estimate the accuracy of the approximation in the interval [7 , 9]

@@heisenBRO-o4d Ah I see. Interesting problem. The way I would approach this is to check both endpoints of the interval, so both 7 and 9, as your value of z and compare the error they would give you. Whichever gives you the maximum error, would be my choice for the accuracy of the approximation on the interval. By choosing the max error, you are saying that the approximation is going to be at most that far off from the actual value on that interval. It could be closer, but it wont be further away than that max. Hope this helps!

@@JKMath Yes, it makes sense

You're welcome!

You can find more examples of problems on the society of actuaries (SOA) website. Look under Exam FM, and then old exam questions/solutions.

Glad to help! Wishing you the best on your final, you've got this 👊

Thank you, I appreciate that. I am just one person, so I can only do one thing at a time. I will make differential equations videos eventually, but my current focus is completing my calculus 3 series. After its complete, I will consider other higher level math courses. Thanks for your patience!

Yep, I mentioned this error in my pinned comment. My apologies on that mistake!

You're welcome!

It's because x in this problem is defined to be one half of the length of the rectangle. That is the way it is given in the problem. For this problem, x is the distance between the y-axis and a point on the semicircle, which will only account for half of the length of the entire rectangle. So we have to multiply x by 2 to get the total length of the rectangle, which we then multiply by the width, which is y to represent the area of the rectangle. Hope this helps!

At some point I do want to create final exam review videos for each course I cover, but for right now my focus is on new lessons and examples videos for Calculus 3. I am just one person, so I can only really focus on thing at a time. Appreciate the suggestion!

@ Sounds good! Another question: do you intend on doing a differential equation series?

@@SainiSalami Eventually, yes! But it won't be for some time. I want to complete my Calc 3 series before starting a new higher level math course.

@@JKMath Alright. Thank you - you’ve helped me a ton with Calc II and III

@@SainiSalami Sure thing! Glad to help :)

Glad it helped! I have an example of partial fractions where the numerator has a higher power than the denominator in the examples video for this topic. You can find the link for it in the description of this video. I believe it is the last example of that video. Hope that helps!

@JKMath wow already a fan of yours, never expected such a prompt reply, will surely watch it, thanks keep uploading more videos to save our grades , you have our support !!!