Other avenues of attack?

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#1 17 April, 2016 - 14:34
Joined: 2 years 5 months ago

Other avenues of attack?

I've had a few day off from practice (have been doing overtime at the mcjob). I am noticing, not surprisingly that my rate of improvement (second derivative?) is declining.

Has anyone here found tactics for specific, general, or alternative mental calculation that were productive? I have mostly been practicing squares the last month or two and it has been quite productive but I think I need a better plan of attack.

18 April, 2016 - 12:15
r30's picture
Joined: 3 years 1 month ago

First derivative, I guess. Try writing about those monkey wanting banana ideas, that kind of misanthropy is quite fascinating to read.

18 April, 2016 - 19:39
Joined: 2 years 5 months ago

My wife has been feeding me too well. So the banana trick isn't working. I suppose I could have her stand behind me and throw rocks at me when I'm not fast enough. She might enjoy that. I don't think she likes me very much either. Thanks for the suggestion though. Maybe I can hook up a car battery and some jumper cables with an arduino as motivation.... So many patentable ideas. It's probably best I didn't have children.

19 April, 2016 - 09:59
Joined: 4 years 11 months ago

If your rate of improvement is declining, do something else but related.

How would you calculate:

\( e^{2.3}\)

20 April, 2016 - 03:02
Joined: 2 years 5 months ago

OK... I will go there ;) ... I strongly suspecting I am being dragged down the garden path... Looks like something that might be approximated using natural logs... Am I warm?

20 April, 2016 - 02:17
Joined: 4 years 11 months ago

You indeed could.
However, try doing this the conventional way first.
Square \( e \) , estimate \(e^{.3}\), etc.

20 April, 2016 - 04:41
r30's picture
Joined: 3 years 1 month ago

Trickier question: how would you calculate it if you didn't know the value of e ? (nor how e is derived)

20 April, 2016 - 08:02
Joined: 2 years 5 months ago

I think I caught on to the cruel nature of this idea in the shower this morning. I like it :) Much better than a car battery or rocks.

The non repeating irrational is lovely. Had it been pi I would be familiar with the number but e is not my friend yet.

The decimal on the power is new to me but I like the evil of adding a tenth of a cube to a square.

I have a 3 day weekend coming up shortly so this is a perfect place to start.

I am entirely unsure whether I should thank you or not. This is definitely going to hurt ;)

20 April, 2016 - 10:05
Joined: 2 years 5 months ago

The first bit is easy to start with
2.7 1828 1828 45 90 45 but it will get ugly fast.

1/0! + 1/1! + 1/2!... seems like a pretty good series to use if you didn't know the value of e or how to derive it. The additional cruelty of this path has some merit. I may explore this a bit but it could as easily be a stand alone exercise for me.

Alternately you could divide the circumference of a unit circle by 2 but that takes us to pi.

.... Continuing to read on manual methods for calculating fractional powers.. Euler and his buddies were a dedicated bunch. e by itself is, fun but the expansion of fractional exponents and the additional irrationals that pop out... More reading required....

Quite the rabbit hole.

I may need to dust off my copy of Eulers algebra for stupid people that I've been ignoring in the bottom of my bag for the last few months....

It's a shame I'm not terribly bright. These little problems are a lot more fun than working for a living.

20 April, 2016 - 09:33
r30's picture
Joined: 3 years 1 month ago

If you know series, you know a lot more about it? Gimme the fully-tailored process how you would calculate \( e^{2.3} \), and I'll throw you with rock ;)

21 April, 2016 - 00:54
Joined: 4 years 11 months ago

So will I.
Try to get an estimate. Even accurate to zero or one decimal will work.

22 April, 2016 - 17:16
Joined: 2 years 5 months ago

Clearly (lol) an estimate is all I will be able to calculate but going deep on this exercise is the point rather than going wide with many exercises. ..

I was just looking at e^(pi)i=-1 and wondering to myself if there could be a shape/geometry that represents e^x=y in a weirdly similar way.
Integer multiplication doesn't work well with this idea but there might be some kind of shape in which a fractional power of e makes perfectly good sense as a discrete measure where multiplication and addition are redefined in some manner in relation with the shape/curve/object rather than as an estimate using your favorite method ... (or not).

Just a vague thought... I'll be back to my normal arithmetic practice now :)

I may have met a young fellow at work who likes adding things as well so hopefully I will have someone to motivate me for a bit :)

22 April, 2016 - 17:32
Joined: 2 years 5 months ago

e to the pi i for dummies - YouTube


23 April, 2016 - 01:55
Joined: 4 years 11 months ago

Brilliantly explained! Great find! Here is a banana.
The video has a clue for you to mentally calculate \( e^{2.3}\).

Hint: what happens when Homer goes to the 10th Bank of Springfield?

23 April, 2016 - 20:35
Joined: 2 years 5 months ago

I'm quite enjoying just getting to the point of the wtf of the exponential function
\(e^x\) the rate return on 1 unit compounding with a continuous return of 100% over x periods but it also has an interesting relationship with a strange spiral that could be thought of as a circle in an unevenly stretched plane.

$$e^{2.3} = 9.974182454814718\dots$$

in the fashion of \(e^{\pi{i}} + 1 = 0\).

Wolfram says \(e^{23/10}+\frac{3\pi{i}}{5} \approxeq -3.0822 + 9.4860{i}\)

If you can imagine some kind of continuouly stretchy universe laid on top of a warped 2d plane where a circle is a spiral stretched by the rate \(e\) and we track around this spiral \(2.3\pi\) times or periods then there could be some kind of a vector from x,y that you could describe from a flat 2d perspective or a 1d perpective with an imaginary component. I'm not sure if it's more right or wrong to think about warping imaginary numbers or real ones in an extra dimension ;). Likely equally wrong (ignorance is helpful here). This clearly isn't the right way to think about this problem to practice my multiplication skills but it is strangely fascinating.

to play at

is a fair bit of pain all of it's own accord....
obviously I need to create a set of rules for myself for each attempt.

Rounding to progressively smaller decimal points....


\(2.718281828459045^2\) is fairly straight forward but wtf is \(2.718^{.3}\) and how am I going to calculate it in my head or even on a piece of paper? Disappointed that I don't see a nice decomposition to attack using this method yet as the exploration of roots is quite neat. Still beyond my abilities.

So, \(3^{1/2} = \sqrt{3}\)

or more appropriately, \(2.7^{.3}=2.7^{\frac{3}{10}}=(2.7^{3})^{\frac{1}{10}}=\sqrt[10]{2.7^3}\)

Which is clearly no help at all and took me forever to format with mathjax. Another painful option is \(e^x \approxeq (1+\frac{x}{m})^m\) as explained in the Homer video.. It looks much prettier of course but \(e^{.3} \approxeq (1+\frac{.3}{m})^m\) still doesn't get to something that seems terribly accessible. Where m is reasonably small the approximation looks approachable but I need some pointers on how to cheat on this... Just eyeballing this one it looks the series should approach 1 as the fraction approaches 0 and then when I close one eye the series looks like it should approach infinity as the fraction is greater than zero and the power series approaches infinity. With both eyes closed I can see how a big m might get us somewhere in between 1 and infinity but this leads me to believe I have no idea what is going on. I will try plugging in actual numbers tomorrow and seeing why this converges on something other than 1 or infinity.

A Taylor series is a representation of a function as an infinite sum of terms that are calculated from the values of the function's derivatives at a single point. If a Taylor series is centred at zero it is also called a Maclaurin series because Scottish. By using a finite set of this idea it is possible to come up with an approximation of the function.


Appears to be the most mentally approachable so far and provide some interesting practice despite being a long grind.

Doerfler's 2007 paper on approximating tangent, hyperbolic tangent and the exponential function to 4 digits is amazingly slick... Fast Approximation of the Tangent, Hyperbolic Tangent, Exponential and Logarithmic Functions found here http://www.myreckonings.com/Dead_Reckoning/Online/Online_Material.htm.

I'm going to have to read this one a few times but I'm pretty sure it would have made Feynman jealous and once learned would seem pretty magical. I may save this for when I get to mental trigonometry or I may take it on as part of this evil little diversion.

23 April, 2016 - 23:25
Joined: 4 years 11 months ago


wtf is \(2.718^{.3}\) and how am I going to calculate it in my head or even on a piece of paper?

*** spoiler ***
Below I'll explain how I calculate this number. If you still want to find this out all by yourself, don't read it just yet!
*** spoiler ***

Let's see if we can get close to \(2.718^{0.3} \).

We could guesstimate \(2.7^{1/3} \) and work from there. At least we should be close, right? And, this calculation is something one can do while running a McJob.

I do the third root of 2.7 with a binary search.

\(1^3 \) = 1 and \(2^3 \) = 8, so the answer lies between 1 and 2. Let's try in the middle.

\(1.5^3 \) is a calculation that we can do mentally: \(15^2 = 225\) and \(225*15 = 3375\), so \(1.5^3 = 3.375\).

A bit too big, but we are getting close to 2.7.

\(1.4^3 \) is a calculation that - again - we can do: \(14^2=196\) and \(196*14 = 2800 - 4*14 = 2744\), so \(1.4^3 = 2.744\).

Now the difference between \(2.7^{0.3333...}\) and \(2.7^{0.3}\).
As it turns out, and I'll tell you later why this is - I think it is in the video too -, \(2.7^{0.03333...} \approx 1.03333\), so we can subtract 3.3% from the answer of 1.4 to get 1.35. 1.35 is a good estimate for \(2.7^{0.3}\) and thus for \(e^{0.3}\).

Another way of getting a number for \(e^{0.3}\) is using calculus. That will probably be my next post.

24 April, 2016 - 13:38
Joined: 2 years 5 months ago

Earlier I had mistakenly thought I would enjoy the .3 as a cube root which is a nice practice toy that combines squares with roots. Of course I was wrong ;).... Mentally interpolating to get the 3rd root to a couple of decimals is within my reach. The .03333... estimate confuses me and I wouldn't have found it anytime soon as this is definitely pulling me out of my comfort zone... I'm going to have to do a fair bit more studying to get to here. Looking forward to the calculus explanation.

... Certainly a motivating diversion. It is interesting how quickly simple(?) arithmetic can get completely out of hand. With a current grade 9 level of math under my belt and a few months of basic arithmetic practice it appears that I have had a nice splash in the deep end but it hasn't magically transformed me into a swimmer. I think a few months of logs, lns, powers and root functions will help a bit as my concepts of these functions is still rooted in addition. I still have Euler's introductory algebra text that I have been dabbling in and a soft copy of Spivak's Calculus text that I have been saving for my graduation from Khan Academy. I'm hoping Hardy's, A course of pure mathematics becomes accessible about then...

There is something very mathy about the \(e^x\) that is not simply numerical calculation. It is such a simple function with such a smooth easily understandable shape that it feels entirely wrong to approximate it using integers on a line. An analogy might be calculating \(\pi\) by saying that it is roughly \(\frac{22}{7}\). While it doesn't take that many digits of \(\pi\) to be adequate for any earthly engineering task it seems dirty even if it is fast and easy.

If you had said calculate \(42^{2.3}\) I probably wouldn't have noticed how stinky it was to root around in the mud adding approximations. I wouldn't have realized that the function had an interesting or generalizable shape. That it had become an interesting object fundamentally different than \(42^{2}\) by some interesting magic that I still don't quite understand as \(42^{2.0}\) and \(42^{2.3}\) do not seem so different at all.

It leads me to the idea that although \(x^4=x*x*x*x\) that there is something more to the idea of both power functions and irrationals that I do not fully grok.

I need to continue working in the mud with basic calculation and developing rudimentary skill but it has been interesting getting a bit bruised with this function that I can almost but not quite understand. I will keep taking it out and playing with it while I catch up to it in skill. I still need to learn how to use complex numbers. Wolfram graphs the roots of this function against a circle but I don't really understand what I'm looking at yet and it continues to seem to me that the geometric description of this function is the most understandable (and potentially mentally calculable//Assuming one can transcribe the appropriate visual position to an appropriate number)... I think this should/could/might work in a manner similar to the visual logarithms that were presented on github earlier this year. It would still result in an approximation but one that is conceptually meaningful to the estimator rather than simply a trick of calculation. Doerfler's method approximation method appears use a related strategy.

24 April, 2016 - 12:53
Joined: 4 years 11 months ago

Before doing calculus, let's do another way of getting to the same answer.

\((1+1/n)^n \approx e \)

If you don't recall this, see Roberts video above.

This is true for big n's. However; let's apply this for n=10:

\(e \approx (1+1/10)^{10}\)

then also:

\(e^{0.1} \approx ((1+1/10)^{10})^{0.1} = (1 + 1/10) = 1.1\)

In short:
\(e^{0.1} \approx 1.1\)

\((e^{0.1})^3 = e^{0.3} \approx 1.1^3 = 1.331\)
In short:
\(e^{0.3} \approx 1.331\)

\(e^{0.3}\) is actually closer to 1.35, but that is because n=10 is a bit too small here. If you want to know how to get from 1.31 to 1.35, let me know.

In general:
\(e^n \approx 1+n\)
this is true for small n's.

Let's dive into what \(e^x\) really means. Because it helps to bring irrational numbers and irrational functions back to things in real life that we can understand:
Recall the story of the First and Second Bank of Springfield from the video.
We go from one interest payment per year to continuous payment using compound interest.
\(e^{0.1}\) is what 10% interest will give you using continuous, compound interest.

The effect of compound interest is small if the interest is small:

\(e^{0.1} = 1.105...\). This means 10% becomes 10.5%.
\(e^{0.3} = 1.3499...\). This means 30% becomes 35%.
\(e^{0.7} = 2.01...\). This means 70% becomes 101%.
This last one is also called the rule of 70.

If you get a feel for these numbers, then natural logarithms will be easy.
Example. What is the natural logarithm of 3? Well, 3 is about 10% more than \(e\), so it will be around 1.1.


The .03333... estimate confuses me and I wouldn't have found it anytime soon as this is definitely pulling me out of my comfort zone.

I hope you now understand that:
\(e^n \approx 1+n\)
This is true for small n's and 0.03333 qualifies as a small enough n, so therefore:
\(e^{0.03333} \approx 1.03333\)

In English: 3.333% interest, continuously compounded, is still not much more than 3.333%.
That is why an exponent of 0.3 can be calculated using 1/3 or the third root.
It requires a small deduction of 3.3%.

24 April, 2016 - 13:28
Joined: 2 years 5 months ago

... I added some more text to my previous post while you were adding yours :)

... am reading yours now... I find I have to reread and edit my posts multiple times in these discussions to have any chance of communicating as poorly as I do and assume you generally only look in once a day.

24 April, 2016 - 13:56
Joined: 4 years 11 months ago

I see. No problem. Let me read your previous post first and I will respond to it.

24 April, 2016 - 14:21
Joined: 4 years 11 months ago


I still need to learn how to use complex numbers. Wolfram graphs the roots of this function against a circle but I don't really understand what I'm looking at yet and it continues to seem to me that the geometric description of this function is the most understandable (and potentially mentally calculable//Assuming one can transcribe the appropriate visual position to an appropriate number).

I'll write a separate post about complex numbers in a separate thread.
Complex numbers sound difficult, but I think I can explain them in a simple to grasp way.


Doerfler's method approximation method appears use a related strategy.

I love his methods and his book.
I highly recommend his site and book!

I have a very busy week coming up, but I'll try to write the approximation via calculus post soon.

24 April, 2016 - 15:10
Joined: 2 years 5 months ago

No rush... There is more than enough here to chew on for a few weeks/months. The reintroduction to calculus is greatly appreciated but I suspect it will be early fall by the time I have caught up to this point. Shooting for basic calculus and linear algebra done by the new year at this point. It's a shame there aren't 5 or 6 of us as a little coffee group and saturday morning chat's on a video conference might be fun in an odd kind of way. Need to find a few more people to play along.

24 April, 2016 - 18:25
Joined: 2 years 5 months ago

... And yet another video that helped me with convergent and divergent series this evening... http://m.youtube.com/watch?v=jcKRGpMiVTw

25 April, 2016 - 12:37
Joined: 4 years 11 months ago

How to calculate \(e^{0.3}\) using calculus.

Calculus is all about rate of change. How fast does something change?
If I know f(x) and I know how quickly it changes I can calculate \(f(x+\Delta x)\) by looking at the difference \((\Delta x)\), multiplying this by the rate of change (slope) and adding this to the first number or \(f(x)\).

In general, this is what we are doing. We draw a triangle into the graph, an x displacement, the slope, and we calculate the y displacement:
slope of graph

If I know the value of \(e^0\) to be 1 then we can calculate \(f(0.3)\) by seeing how fast the graph climbs or goes down. \(e^x\) only rises if we move x to the right, so there is no going down in this case.

If I know how quickly the graph rises, I can calculate the value of \(e^{0.3}\).

\(e^x\) has the wonderful property of being its own derivative:

\( \frac{d(e^{x})}{dx}=e^{x} \)

We will use this.

In English, the rate of change at \(e^{0.3} = e^{0.3}\). Now we don’t have exact numbers for \(e^{0.3}\) yet, but we do know that it a bit more than 1.3.

At x=0, \(e^0 = 1\) and at x=0.3, \(e^{0.3}\) is more than 1.3.
The rate of change climbs from 1 to more than 1.3. How much more? We will see.
On average the rate of climb is thus more than 1.15.
So a low guesstimate would be 1 + 0.3 * 1.15 = 1.345.
As it turns out, this is already close.

But we can do better.
On average the rate of change is more than 1.15.
This is the derivative at x=0.15, or just in the middle of the range x=0 to x=0.3.
Again, we can do the same sequence for x=0.15:

At x=0, \(e^0 = 1\) and at x=0.15, \(e^{0.15}\) is more than 1.15.
The rate of change climbs from 1 to more than 1.075.
So a better calculation for the rate of change is 1 + 0.15 * 1.075 = 1.16125 (instead of 1.15).

Going back to our previous calculation, a better calculation for \(e^{0.3}\) is then:
1 + 0.3 * 1.16125 = 1.348375.

The calculator gives 1.3499, so we are indeed very close!

25 April, 2016 - 16:54
r30's picture
Joined: 3 years 1 month ago

I'm a bit bored, and I mentioned before that one could calculate \( e^{2.3} \) without knowing the value of \( e \) nor how \( e \) is derived (the formula \( (1+\frac{1}{n})^{n} \) ).

Your approach, Kinma, intuitively covers the \( e^{0.3} \) part very well (I have never thought about using derivatives plus comparative style estimation).
For the \( e^2 \) part, Robert already mentioned Taylor's series, which's special case Maclaurin series only requires the knowledge \( \frac{d(e^{x})}{dx}=e^{x} \) and that \( e^0 = 1 \) , and with that we have eliminated constant \( e \) from the series.

It is interesting that historically the constant that we now know as \( e \) was not known for its two most important properties: retain its exponential form through differentiating and describe complex numbers in exponential form. They knew that the area under rectangular hyperbola y=1/x from 1 to \( e \) is equal to 1, but couldn't comprehend the deeper meaning. Only now that we encounter differential equation under every corner the necessity for such a constant becomes crystal clear.

I tried to derive e from the knowledge of it remaining its own exponential derivative.
We know that: \( {e^{x}}'=e^{x} \) and \( e^{0}=1 \)

\( d(f(x))={f}'(x)dx \)
\( f(x+dx)=f(x)+{f}'(x)dx \)
\( e^{0+dx}=e^{0}+e^{0}dx=1+dx \)
\( e=e^{1}=(e^{0+dx})^{1/dx}=(1+dx)^{1/dx}=\lim_{n \to \infty }(1+\frac{1}{n})^{n} \) , which is its first historical form.

Much more fun would be deriving the value of e from the knowledge \( e^{i\varphi }=cos\varphi +isin\varphi \) , it seems kind of a harder problem. I wonder if one could do it without using none of the knowledge above, only the previous equation.

P.S. Here is the part where I throw Robert with a rock: if the previous statement is true, one could really estimate \( e^{2.3} \) without knowing how \( e \) is derived (both derived and taken derivative of :P ... A trick question, meaning one couln't use his calculator, which uses Taylor's formula for everything, nor the proof I gave above, nor \( \lim_{n \to \infty }(1+\frac{1}{n})^{n} \) )

25 April, 2016 - 16:33
Joined: 2 years 5 months ago

This thread will keep me coming back for several months (or more). There is more here than I understand but enough that I do that it keeps tricking me.

I will need more time with functions, convergent series, irrationals, imaginaries, the unit circle, trig, limits, calculus, algebra, and maybe a little number theory but other than that I have it well in hand :)


It looks so innocent. Well played.

5 May, 2016 - 22:30
Joined: 4 years 11 months ago



have been doing overtime at the mcjob

Can you use the job to calculate?
Like add up all the items and tell the total before the computer does it?

6 May, 2016 - 09:13
Joined: 2 years 5 months ago

Mostly not. Need to switch to a more mentally challenging job or work as a parking lot attendant. Currently working quote hard for very little with no upside.

11 May, 2016 - 20:06
Joined: 2 years 5 months ago


I think this is what I meant by it feels more natural as a circle but I have to admit it hurts when I try to follow the explanation. I simply don't have the prerequisites yet.

20 May, 2016 - 11:33
Joined: 5 years 3 months ago

I have a rather unusual technique for calculating powers of e, which I teach here: http://headinside.blogspot.com/2014/03/calculate-powers-of-e-in-your-hea...

I turn ex into 10y simply by using the fact that log10(e)≈0.4343, so we can calculate y=x×43×101⁄10,000.

How do we do e2.3? Later on, I'm going to have to move the decimal place 4 places to the left anyway (for the division by 10,000), so I might as well calculate e23, and just remember to move the decimal place 5 places.

Let's see 23×4=92 and 23×3=69, so 23×43=989, and 989×101=99889. See the linked article above, if you're wondering how I handle all this mentally.

Now, I remember I need to move the decimal over 5 places, so we now have 0.99889, which means e2.3≈100.99889. Considering the actual answer is e2.3=100.998877, that's not bad for a mental estimate!

22 May, 2016 - 14:16
Joined: 4 years 11 months ago

Hi Greymatters,

That is an excellent way of calculating!
I use the same way and also use it for calculating the ln or natural logarithm.

If I need ln 2 for example, I know from memory that log 2 = 0.30103.
If I divide 0.30103 by 0.4343 I get ln 2.
This might sound like a difficult calculation for some, but if I multiply 43 by 7 I get 280 + 21 = 301.
Using your method we see that 7 X 4343 = 30401.
This is roughly 1% more than 30103.

This means that ln 2 = 0.7 - 1% or 0.7 - 0.07 = 0.693.
This number is actually correct to 3 decimal places.

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