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You're welcome!

I made a mistake. Please ignore my comment above.

Hi! There is only 1 withdrawal of k at times 2 and 3. The way you set up your calculation there implies that there is more than one withdrawal at each of those times. Yes, k is the amount of the withdrawal, but when there is a penalty, we need to take out an additional 6% in that withdrawal. To find that, we need to increase the withdrawal amount by multiplying it by 1.06, not .06. Multiplying k by .06 would decrease the payment, not increase it, so that is not exactly a penalty (think: k(1.06) = k + k*.06). So, we need to accumulate each withdrawal to time 10, but you also need to take the penalties into account where necessary. So that's why we multiply by 1.06 for the withdrawals that have penalties. Having an extra term for the penalty like you did with -k(1.03)^8 -k(0.06) -k(1.03)^7 -k(0.06) creates an issue, as you are now adding more withdrawals that do not exist in the problem. You want to keep the penalty factor of 1.06 with the rest of the withdrawal and its accumulation.

Depends on the series, but yes, there are going to be cases where both the Limit comparison test and direct comparison test work. This lesson video is specifically focused on how to use the limit comparison test, so of course I am going to use it for every example, despite that other tests could be used.

@@JKMath thank you for the quick reply. Your videos really helped me in preparation for my test. I wish I found you sooner !

Glad to help! Thank you so much for your generosity!

Awesome! Congrats on passing the exam, its definitely not an easy thing to do! I'm glad to know the videos were able to help in some way, so thanks for sharing this, it means a lot. I greatly appreciate the kind feedback! :)

Glad I could help! :)

Since we are revolving around the y-axis, the height will be measured in terms of x, so you want to look up and down the area, not left to right. From this we can see that the height is between the x-axis and the curve, so the height is just the function that represents the curve, just like I labeled it in the video. Hope this helps!

@@JKMath I see it’s mainly cause since X is a vertical line we have to look up to down rather than side to side (y=)

@@ritvikindupuri2388 You could think of it that way, sure. But in general when working in terms of x, you look from up to down, and when working in terms of y, you look from right to left.

@@JKMath oh i see

So yes, the BV at time 0 is the same as the PV or price of the bond, but that is the only BV that represents the price of the bond. The other BVs represent the amount of the bond left to be paid out at specific moments in time. That is different than the price of callable bonds. The price of callable bonds are always valued at t=0, never anywhere else. When we "redeem at the earliest" date, we are referring to how many coupon periods we want to work with. So if a bond has maturity at the end of 10 years, or t=20 for semiannual periods, but is callable starting with t=15, then 15 is the number of periods for the bond if it is a premium and you want to call it early. All this means is that the redemption value will be received at t=15 instead of t=20, and 5 coupons are never collected. The price is still valued at t=0, not t=15, so it would not be the same as BV15. Does that make sense?

Ahhh got it, thank you so much! I was super confused and now I have better understanding of it!

Thank you so much! I'm glad to know the calc 1 and 2 videos were able to make things easier for you. Nice job getting through calc 3, its definitely not an easy course. Wish I could have had the series completed for you, but there is so much left to cover! I appreciate your support :)

You're very welcome! :)

Wow, that's awesome that you're learning calculus at 15, impressive! Glad I could help. If you ever have questions feel free to let me know :)

Depending on the series, yes, you could use the direct comparison test, which is easier than the integral test in the majority of cases. This video is focused specifically on the integral test and how to perform it/ when can you perform it, so I don’t use any other tests. (Also within my Calc 2 series, this video comes before I introduced the DCT)

Not quite. The inner radius is still 0, so we do need to use the disk method. There are two functions yes, but both functions form the outer radius. The outer radius is one particular function from y=0 to y=2, but then changes from y=2 to y=3, which is why we need to split it up into two integrals.

@@JKMath I see what about using thf shell method

@@ritvikindupuri2388 The shell method would make that example easier, yes. When revolving around the y-axis for the shell method, you want to work in terms of x, so you would not need to switch the functions into terms of y like we do for the disk method. The height of the shells h(x) would be the top function minus the bottom function, and the radius of each shell would be x. This should result in the same answer if you integrate from x=0 to x=1. I don't show the shell method in this video because this is a video on how to use the disk method. I discuss the advantages of the shell method in its lesson video. Hope this helps!

@@JKMath thanks always wanted to learn how to find height using shell. I also was wondering if you do any tutoring for calculus 2,, you are a great teacher

@@JKMath I see also for the radius is it always just x or y, like when does it change

The disk and washer methods are very similar. In fact, they involve the same formula/process. The disk method is just an easier case of the washer method when the inner radius is 0. So in other words, if you have only an outer radius function, you use the disk method, but if you have both an inner radius and outer radius represented by functions that are not just 0, then you use the washer method.

@@JKMath I see thank you for the explanation !!

@@ritvikindupuri2388 No problem! 👍

In that example 65 the final equation is to solve X with『50382.16 =3000ä26 + X (1.04)^-26』 which I happily used the BA II Plus function of solving the future value of bond with a twist of changing the annuity immediate to annuity due, then I realized the X is actually the 27th payment at time 26 and were paid separately than the 26 payments of coupon

Interesting question! I have personally never seen a problem involving a bond with coupon payments made at the beginning of each period. I believe this may be because that's not something that happens with bonds in real life, but I could be wrong about that. Anyway, if for some reason a bond does pay coupons in such a way that an annuity-due calculation is necessary, it is my understanding that this would not effect the date of maturity, which is when the redemption value is received. If the bond is a 10 year bond, the redemption value would still be received at the end of year 10, or t=20 if the coupons are semiannual. That's at least how I would go about it. To me it almost seems like that if this were to be an FM problem, to be completely clear, they would have to tell you where the redemption value is paid (at end of final period or at time of final coupon). Since this is such a rare case, I think that clarification would need to be made. In my experience, bonds always pay coupons at the end of each period. Hope this helps!

Eventually I do want to make a series on probability and statistics, which would be the topics seen on Exam P, but it won’t be for some time. I’m currently just a one man operation so I have to focus on one thing at a time. Right now I’m working to complete my Calc 3 series, and then after that I will consider starting a new one. Thanks for your patience! And glad you find the videos helpful :)

@@JKMath Your videos are fantastic and have helped me a lot. Despite the language barrier, your lectures are often understandable 🌸. I want to take the P exam in January 2025 and the FM exam in June 2025. I have finished studying for the P exam, but I urgently need your excellent lecture series to review.

Thank you! Glad I can help :)

Thank you so much!

To get technical, the “area” calculated by integration is actually what we call “signed area”, which can be positive or negative. In particular, we designate area below the x-axis as negative, and area above the x-axis as positive. But to keep things simple, we just say integrals calculate area under a curve when first learning calculus. When you flip the bounds, you are reversing the order in which you add the infinitely small rectangles that integration is taking the sum of to make up the area, and doing so produces a negative. If you watch my video on Riemann sums and switch a and b, you will see where the negative comes from, it’s too much to explain in a single comment. Bottom line is that a negative comes into play, so you could put it inside the integral of a function, therefore integrating the negative version of a function -f(x), so the graph is flipped, so any positive area above the x axis now becomes negative area below the x axis and vice versa. It’s still the same amount of area, just the sign is different.

Which example are you referring to? I looked at the last example and don't see where -∞ would be squared. Can you point me to a specific part of the video? Would like to help you understand!

You're welcome!

There may be no y's in the actual equation, but in either case since we are revolving around the y-axis we want each radius to be represented by an equation where we solve for x. Whether that's something like x=√y or x=1. Regardless of if y is present, that is what we want for R(y) and r(y). Remember, the equation x=1 tells you that no matter what value of y you have, x is equal to 1, which is why we get a vertical line as the graph. It is still valid to use 1 in the place of R(y) in the formula. It just represents the outer radius.

​@@JKMath yes sir understood

Make sure you are clearing the TVM buttons before every new calculation using them. You want to press 2ND and then FV to clear TVM. If you don’t do this before a new calculation you will most likely get a wrong answer. Hope this helps!

@@JKMath Thank you so much, Got it. Now its work perfectly.

You are very welcome! I'm glad to know these videos are able to help you make better sense of calculus! And thank you too, I really appreciate your generosity :)

You're welcome! Glad it helped :)

Correct, for that formula you must know that F=C, which for the majority of problems will be the case. If C is different than F, you could use the other formula you mentioned. I did not really make that clear in the video, my apologies. I would do that differently if I made this video today.

I do not have a dedicated video for odd periods, but if you have any questions about them feel free to ask. The way I understand it, odd periods are just when you want to accumulate interest for a time that is not a full period. So for example, if an interest rate is given as an annual effective rate and you want to generate interest for just a couple months instead of a full year.. that would be an "odd" period. In a situation like that, I would just choose to convert my interest rate to a monthly effective rate and use that to generate interest for the specified number of months. Hope this helps!

nevermind i had to reset the whole calculator. it works now!

Make sure to clear TVM before doing new calculations with the TVM buttons, press 2ND and then FV to clear them out. If that is not cleared before each new calculation you will most likely get the wrong answer. Let me know if this helps!

Yes, that is an alternative way to show the terms are decreasing. You could rewrite a_n as a function and take its derivative, and if the derivative is negative, that means it is a decreasing function and therefore the terms are decreasing. Sometimes depending on a_n that can be tricky to do as the function you would need to differentiate may be complicated, so I usually opt to use the method of comparing terms that I show in this video. Regardless, both ways are valid and should help you get to the right conclusion/answer.

Nominal rates are for compound interest rates only. Simple interest rates are a lot simpler. If you have an annual simple interest rate, then you use (1+i·t) where t is in years, if the simple interest rate is monthly, then you also use (1+i·t) but where t is measured in months. It does not get more complicated than that. Hope this helps!

Interesting! I did not know this. Nice find. It all comes down to personal preferences for what you want to set the number of decimal places to. 8 is just what I personally use, so that's why I recommend it/use it in the video.