Can calculus be used to determine how much total interest I would pay?

[sevensages]

TL;DR Summary

Can calculus be used to determine how much total interest I would pay on a 30-year mortgage with 04.5% interest?

In 2014, I bought a house for $71,000. I paid a down payment of approximately $4,000. Then I paid for the rest of the $67,000 with a 30 year mortgage with a fixed rate of 4.5%.

Here is how my mortgage works: I have to make monthly payments to my mortgage company. The amount of interest that is due to the mortgage company is calculated by multiplying the principle that I owe by .045 (4.5%) . Then I divide that number by 12. That number is the interest due that month. So on the first month that I had the mortgage, i owed the mortgage company $251.25 in interest.

$67,000 X .045= $3,015.00

$3,015÷12=$251.25

My mortgage payments have always been approximately $600.00 per month. My mortgage payments of $600/month comprise four things: interest on loan, principle of loan, property taxes, and home insurance.

My home insurance is approximately $50/month. My property taxes are approximately $100/month. So my first mortgage payment paid $198.75 off the principle of the loan.

For my second mortgage payment, the interest I paid was 4.5% of a principle of $66,801.25 divided by 12.

$66,801.25 X .045= $3,006.05

$3,006.05÷12= $250.50

So my second mortgage payment was still $600.00. But I only paid $250.50 in interest on my second mortgage payment. And on my second mortgage payment, I paid $199.50 off the principle of the mortgage (as opposed to $198.75 off the principle in the first mortgage payment).

I know how I could use arithmetic to calculate the total interest paid on the 30 year mortgage if I only made the minimum payments each month for 30 years. That would be 360 calculations and take me probably six or eight hours of very tedious work.

Is there a way to use calculus to determine how much the total amount of interest paid on the loan would be? If so, how would a person use calculus to calculate how much the total amount of interest paid on the loan over 30 years would be? I mean, would you use Taylor Series or what?

Please show me the formula if you can.

 

[Andrew Mason]

Each month you pay a sum which multiplied by the amortization period in months gives you the total amount you will pay (assuming no interest rate changes). Then just subtract the principal to determine the interest.

AM

 

[Nugatory]

TL;DR Summary: Can calculus be used to determine how much total interest I would pay on a 30-year mortgage with 04.5% interest?

Not easily, it’s the wrong tool for this job. Calculus works with continuous functions and that’s not what you have here.

I know how I could use arithmetic to calculate the total interest paid on the 30 year mortgage if I only made the minimum payments each month for 30 years. That would be 360 calculations and take me probably six or eight hours of very tedious work.

It will just take a few seconds with Excel, LibreOffice Calc, or any similar spreadsheet. Or Googling for “mortgage amortization schedule” will bring up a bunch of online calculators that will do the 360 calculations for you.

 

[sevensages]

Not easily, it’s the wrong tool for this job. Calculus works with continuous functions and that’s not what you have here.

What's the definition of a continuous function?

 

[sevensages]

When I first got my loan, the bill of sale said how much the total interest on the loan would be if I didn't pay off the principle first.

My purpose in creating this thread was far more about my curiosity about math than actually finding out how much the total interest paid would be per se.

My elderly father earned a bachelor's degree in Mathematics about 50 years ago. I asked my father what type of mathematics would be used, and he is not aware of any. But my father's mathematics skills are very rusty. There still might be a form of mathematics suitable for this calculation.

Nugatory says that calculus is not the right tool for this because it is not a continuous function. Is this a non-continuous function? Or is this not a function at all?

What branch of mathematics would be the right tool for this?

P.S. as I recollect from looking at the bill of sale almost 9 years ago, the total interest would be in the range of $70,000-$90,000.

 

[phinds]

What's the definition of a continuous function?

Google is your friend.

 

[Nugatory]

What's the definition of a continuous function?

You will find that in any intro calculus textbook, or you can look at the Wikipedia article.

What branch of mathematics would be the right tool for this?

Ordinary arithmetic, and you already know how to use it
here - you described it in your first post. The calculation is tedious, but that's what computers are for these days.

Back in the pre-electronic days there were books containing nothing but table after table of mortgage amortization schedules. Your lender would look up the table for your loan terms, multiply each number by the loan amount, and they'd have the total amount for principal and interest after each payment.

 

[Andrew Mason]

Nugatory says that calculus is not the right tool for this because it is not a continuous function. Is this a non-continuous function? Or is this not a function at all?

What branch of mathematics would be the right tool for this?

P.S. as I recollect from looking at the bill of sale almost 9 years ago, the total interest would be in the range of $70,000-$90,000.

It is a fairly straight forward bit of algebra with knowing how to sum a geometric series.

Let
n=the amortization term in payment period for the fixed payment mortgage;
z=the number of payments per year (eg if it is a 30 year monthly payment, n=360 and z=12);
x=the payment per payment period
i=the annual interest rate
p=the principal (note the "al" not "le")
D=the outstanding debt, D, after each payment period.

Initially D=p.
After the first payment period and paying x, you owe $D=p+(i/z)p-x=p(1+i/z)-x$ (ie. principal + interest - payment). The interest will then be calculated on the new principal: $p(1+i/z)-x$

After the second payment: $D=[p(1+i/z)-x](1+i/z)-x=p(1+i/z)^2-x(1+(1+i/z))$

After the third payment: $D=[p(1+i/z)^2-x(1+(1+i/z))](1+i/z)-x=p(1+i/z)^3-x(1+(1+i/z)+(1+i/z)^2)$

You can now see that after n periods:

(1) $D=p(1+i/z)^n-x(1+(1+i/z)+(1+i/z)^2+....+(1+i/z)^{n-1})$

The only trick you need to know is how to sum that geometric series:
$(1+(1+i/z)+(1+i/z)^2+....+(1+i/z)^{n-1})$

This series is equal to: $\frac{(1+i/z)^n-1}{(1+i/z)-1)}=\frac{(1+i/z)^n-1}{i/z}$

So we can rewrite D (the debt after n periods) from (1) as:

(2)* $D=p(1+i/z)^n-\frac{x((1+i/z)^n-1)}{i/z}$

To calculate how much interest you have paid at any time, just subtract the total payments from the reduction of the debt from original principal debt (ie. interest=nx-(p-D)) using (2) to calculate D.

If the payment period is different than the interest calculation period (for historical reasons in Canada, interest is calculated semi-annually and paid monthly) then you have to be careful to use the right values. If interest is calculated semi-annually you have to use z=2 and x=6 x monthly payment.

AM

[*Note: In case you want to determine what x should be, we have set x so that after n payment periods, the debt is paid: ie D=0.

So: $p(1+i/z)^n=\frac{x((1+i/z)^n-1)}{i/z}$ which means:

$x=p\frac{(i/z)(1+i/z)^n}{(1+i/z)^n-1}$

This will enable you to determine whether the payments you are making have been correctly determined for the amortization period.]

 

[sevensages]

It is a fairly straight forward bit of algebra with knowing how to sum a geometric series.

Let
n=the amortization term in payment period for the fixed payment mortgage;
z=the number of payments per year (eg if it is a 30 year monthly payment, n=360 and z=12);
x=the payment per payment period
i=the annual interest rate
p=the principal (note the "al" not "le")
D=the outstanding debt, D, after each payment period.

Initially D=p.
After the first payment period and paying x, you owe $D=p+(i/z)p-x=p(1+i/z)-x$ (ie. principal + interest - payment). The interest will then be calculated on the new principal: $p(1+i/z)-x$

After the second payment: $D=[p(1+i/z)-x](1+i/z)-x=p(1+i/z)^2-x(1+(1+i/z))$

After the third payment: $D=[p(1+i/z)^2-x(1+(1+i/z))](1+i/z)-x=p(1+i/z)^3-x(1+(1+i/z)+(1+i/z)^2)$

You can now see that after n periods:

(1) $D=p(1+i/z)^n-x(1+(1+i/z)+(1+i/z)^2+....+(1+i/z)^{n-1})$

The only trick you need to know is how to sum that geometric series:
$(1+(1+i/z)+(1+i/z)^2+....+(1+i/z)^{n-1})$

This series is equal to: $\frac{(1+i/z)^n-1}{(1+i/z)-1)}=\frac{(1+i/z)^n-1}{i/z}$

So we can rewrite D (the debt after n periods) from (1) as:

(2)* $D=p(1+i/z)^n-\frac{x((1+i/z)^n-1)}{i/z}$

To calculate how much interest you have paid at any time, just subtract the total payments from the reduction of the debt from original principal debt (ie. interest=nx-(p-D)) using (2) to calculate D.

If the payment period is different than the interest calculation period (for historical reasons in Canada, interest is calculated semi-annually and paid monthly) then you have to be careful to use the right values. If interest is calculated semi-annually you have to use z=2 and x=6 x monthly payment.

AM

[*Note: In case you want to determine what x should be, we have set x so that after n payment periods, the debt is paid: ie D=0.

So: $p(1+i/z)^n=\frac{x((1+i/z)^n-1)}{i/z}$ which means:

$x=p\frac{(i/z)(1+i/z)^n-1}{(1+i/z)^n-1}$

This will enable you to determine whether the payments you are making have been correctly determined for the amortization period.]

Isn't that calculus? I believe a geometric series is part of calculus.

 

[hutchphd]

The cheap and dirty answer is pretty easy. Its a thirty year mortgage on $67k. The portion that goes to the loan (sans taxes and hoeowners) is about $450 /month so total payment over 360 months is $162k. So you will pay $95k interest (~within 5% because we swagged the taxes and homeowners). The @Andrew Mason calculation is also correct and should yield this result.

Isn't that calculus? I believe a geometric series is part of calculus.

Not in my lexicon.

 

[Andrew Mason]

Isn't that calculus? I believe a geometric series is part of calculus.

Calculus deals with infinitesimals and limits. If a series of n terms converges to a certain value as n increases without bound (ie. a limit) , then you would be using a calculus concept. In general, for any geometric series the sum of the terms in the series is given by:

$$\sum_{i=0}^{n-1} x^i =\frac{x^n-1}{x-1}$$

This was known in Euclid's time (so it is hardly calculus which was invented in the 17th C).

You can see that if x is greater than 1, this sum keeps increasing without bound as n increases. That does not involve concepts of calculus.

However, if x is a fraction (0<x<1) then

$$\sum_{i=0}^{n-1} x^i =\frac{x^n-1}{x-1}\rightarrow\frac{1}{1-x}$$ as $n\rightarrow\infty$

This involves the concept of a limit, which is part of calculus. So, for example, if x=⅓, the sum converges on a value of 3/2 as $n\rightarrow\infty$ so we say:

$$\lim_{n \to \infty}\sum_{i=0}^n (\frac{1}{3})^i =\frac{3}{2}$$

But, as you have probably experienced, as the months go on, the amount of interest you pay does not reach a limit. This has to do with banking rather than calculus.

AM

 

[sevensages]

Calculus deals with infinitesimals and limits. If a series of n terms converges to a certain value as n increases without bound (ie. a limit) , then you would be using a calculus concept. In general, for any geometric series the sum of the terms in the series is given by:

$$\sum_{i=0}^n x^i =\frac{x^n-1}{x-1}$$

This was known in Euclid's time (so it is hardly calculus which was invented in the 17th C).

Why did I learn about geometric series in Calculus 2 then?

 

[hutchphd]

Because you did.

 

[pbuk]

Why did I learn about geometric series in Calculus 2 then?

We have no idea, but possibly

• Because you hadn't learned about them before.
• In order to study their behaviour in the limit as time goes to infinity.
• Because not everything you learn in a course fits in the label of the course: for instance "Calculus 2" courses often include polar coordinates.

 

[Andrew Mason]

The cheap and dirty answer is pretty easy. Its a thirty year mortgage on $67k. The portion that goes to the loan (sans taxes and hoeowners) is about $450 /month so total payment over 360 months is $162k. So you will pay $95k interest (~within 5% because we swagged the taxes and homeowners). The @Andrew Mason calculation is also correct and should yield this result.

The payment of $450/month over 360 month amortization period would be for an interest rate of about 7.1% which is quite a bit more than 4.5% indicated. At 4.5% the monthly payment should be $339.48 using:

$x=p\frac{(i/z)(1+i/z)^n}{(1+i/z)^n-1}$

AM

 

[vela]

Why did I learn about geometric series in Calculus 2 then?

Because infinite series are typically covered in Calc 2, and the geometric series is an important type of series.

The first time the geometric series shows up in a math class is in, I think, junior high. You should have definitely seen them before taking calculus. Summing a geometric series with a finite number of terms only requires basic algebra, but summing an infinite number of terms requires concepts from calculus. Since you're (hopefully) not paying the mortgage in perpetuity, all you need is algebra.

 

[sevensages]

The payment of $450/month over 360 month amortization period would be for an interest rate of about 7.1% which is quite a bit more than 4.5% indicated. At 4.5% the monthly payment should be $339.48 using:

$x=p\frac{(i/z)(1+i/z)^n}{(1+i/z)^n-1}$

AM

Your estimate of my monthly payment being $339.48 is not accounting for my property taxes, mortgage insurance, and home insurance.

My home insurance and my property taxes are each over $100/month. In the OP, I gave an erroneous guesstimate that my home insurance is about $50/month. I just had my mortgage insurance removed a week ago. But my mortgage insurance added $35/month to my mortgage

And home insurance and property taxes are both escrowed into my mortgage payments. When I first bought the house, my mortgage payments were about $500/month. Now, due to inflation of both home insurance and property taxes, my mortgage payments are $600/month.

 

[sevensages]

Because infinite series are typically covered in Calc 2, and the geometric series is an important type of series.

The first time the geometric series shows up in a math class is in, I think, junior high. You should have definitely seen them before taking calculus. Summing a geometric series with a finite number of terms only requires basic algebra, but summing an infinite number of terms requires concepts from calculus. Since you're (hopefully) not paying the mortgage in perpetuity, all you need is algebra.

I don't remember learning about geometric series until calculus 2.

Maybe I learned the concept in junior high, but my math textbook and math teacher did not use the word geometric series.

 

[sevensages]

We have no idea, but possibly

• Because you hadn't learned about them before.
• In order to study their behaviour in the limit as time goes to infinity.
• Because not everything you learn in a course fits in the label of the course: for instance "Calculus 2" courses often include polar coordinates.

Yeah, I did learn about polar coordinates in calculus 2. I also remember learning about parabolas in calculus 2.

 

[Andrew Mason]

Your estimate of my monthly payment being $339.48 is not accounting for my property taxes, mortgage insurance, and home insurance.

Of course. It just gives you the monthly payment of principal and interest. If you want to calculate the interest paid you have to subtract all the payments other than the mortgage payment.

AM

 

[hutchphd]

At 4.5% the monthly payment should be $339.48 using:

Yes I get exactly that.
Oh now I see

And home insurance and property taxes are both escrowed into my mortgage payments. When I first bought the house, my mortgage payments were about $500/month. Now, due to inflation of both home insurance and property taxes, my mortgage payments are $600/month.

So with these corrected figures the cheap and dirty calculation comports exactly with the exact geometrical sum. When in doubt go cheap and dirty.
But I remember being taught in seventh grade algebra how to sum the geometric series (my teacher saw I was bored). Thank you Mr Davis, it lit up my brain.