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STEP Tricks and Hacks Thread

One recurring theme is to exploit the symmetry of trigonometric functions during integrating. Using this in conjunction with:

Unparseable latex formula:

\displaystyle[br]\begin{equation*}\int_a^b f(x) \, \mathrm{d}x = \int_a^b f(a+b-x) \, \mathrm{d}x\end{equation*}

is particularly useful.

Here's an example:

\int_0^{\pi/4} \log (1 + \tan \theta) \, \mathrm{d}\theta

. How is this relevant here? Call our integral

I

, then using the substitution

\theta \mapsto \frac{\pi}{4} - \theta

we have:

Unparseable latex formula:

\displaystyle [br]\begin{equation*} I = \int_0^{\pi/4} \log \frac{2}{1 + \tan \theta} \, \mathrm{d}\theta = \frac{\pi}{4}\ln 2 - I \iff I = \frac{\pi}{8}\ln 2\end{equation*}

Extension to this, using the same idea, roughly - attempt

\displaystyle \int_0^{\pi/2} \log \sin \theta \, \mathrm{d}\theta

Scroll to see replies

Reply 1

Duke Glacia

1. Adding zero creatively (i) (Zacken)
2. Trigonometrical symmetry and substitutions for integration (Zacken)
3. Trigonometrical functions as shifts of others (Zacken)
4. Implicit differentiation and logarithmic differentiation (Zacken)
5. Adding zero creatively (ii) (Zacken)
6. Algebra shenanigans with 'it suffices' (IrrationalRoot)
7. Discriminants as a useful tool for inequalities and in mechanics (Duke Glacia)
8. Weierstrass subs (Shamika)
9. Inequalities and injectivity (Zacken)
10. Summing arc-tangents (16Characters...)
11. L'Hopitals rule for limits (Joostan)
12. Factorising special polynomials and techniques to do so (Aymanzayenmannan)
13. Strong induction (Zacken)

Hopefully we can grow this into a useful and extensive collection of tricks and techniques - feel free to share your own!

So to start with Zacken posted a amazing techniques which could have saved me a lot of time and effort.

Original post by Zacken

For one, do you know about the "adding zero creatively trick"? It works wonders.

Here's an example of it:

\displaystyle \int \frac{x}{1+x} \, \mathrm{d}x

, now you could do long division and what not or use a substitution but it's quite elegant to do just:

Unparseable latex formula:

\displaystyle [br]\begin{equation*} \int \frac{x + 1 - 1}{1+x} \, \mathrm{d}x = \int 1 - \frac{1}{x+1} \, \mathrm{d}x = x - \ln |x+1| + \mathcal{C}\end{equation*}

.

If you want, you can try it with this gem:

\displaystyle \int \frac{x^4}{1+x^2} \, \mathrm{d}x

which was on a previous STEP II question.

It's very elegant to do this if you notice that

x^4 \equiv x^4 + x^2 - x^2

so that you can simplify your integrand. :-)

OP written by Zacken

(edited 8 years ago)

Reply 2

Lucasium

Bump. Not sitting STEP but I'll do some papers this summer for kicks

Reply 3

Zacken

Use the fact that some trigonometric functions are shifts of another, common examples include

\sin \theta = \cos (\pi/2 - \theta)

\cot \theta = \tan (\pi/2 - \theta)

and the likes. This comes in useful and saves a lot of time.

Say you were given

\csc \left(\theta + \frac{\pi}{3}\right) = \sec \left(\theta - \frac{\pi}{6}\right)

- this could be an 8/9/10mark A-Level question easily; want a three liner way of doing it?

The equation is equivalent to

\displaystyle \sin \left(\theta + \frac{\pi}{3}\right) = \sin\left(\frac{\pi}{2} - \theta + \frac{\pi}{6}\right) = \sin \left(\frac{2\pi}{3} - \theta\right)

which now becomes elementary to solve.

Extension to this, using a STEP question, show that

\sin A = \cos B \iff A = (4n+1)\frac{\pi}{2} \pm B

- should be a two or three liner using this method.

This trick has rendered one or two STEP I questions completely inane, especially the

\cot \theta = \tan (\pi/2 - \theta)

that's let me do a STEP I question in 4 minutes.

(edited 8 years ago)

Reply 4

Zacken

Using implicit differentiation can make life easier.

Say you wanted to differentiate a nasty function that might require multiple uses of the product rule or such, take logs!

Let's say you have

f(x) = x^p(x+1)^q(x+2)^r

, do you really want to differentiate this normally? I'd hope you said no. What you'd want to do is:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\log f(x) = p \log x + q \log (x+1) + r \log (x+2) \Rightarrow f'(x) = f(x)\left(\frac{p}{x} + \frac{q}{x+1} + \frac{r}{x+2}\right)\end{equation*}

and leaving your answer in this form, i.e: as

f(x)(\cdots)

is quite useful in many a STEP I question that I've come across.

As an aside, here's another place where it's useful:

\frac{\mathrm{d}}{\mathrm{d}x} |f(x)|

, now the standard thing to do would be to consider intervals of

x

that makes

f

positive or negative and differentiate piecewise.

What'd I'd do is

y = |f(x)| \Rightarrow y^2 = f(x)^2 \Rightarrow 2yy' = 2f'(x)f(x) \Rightarrow y' = \frac{f'(x)f(x)}{|f(x)|}

- this also lets you know where the function is not differentiable.

I'll edit it a nice problem for you lot to try in a bit:

Define

f(x) = x^m(x-1)^n

with both

m, n

being positive integers greater than

1

. Show that the stationary point of this function in the interval

(0,1)

is a maximum if

n

is even and a minimum when

n

is odd.

(edited 8 years ago)

Reply 5

TLDM

Original post by Zacken

Unparseable latex formula:

\displaystyle[br]\begin{equation*}\int_a^b f(x) \, \mathrm{d}x = \int_a^b f(a+b-x) \, \mathrm{d}x\end{equation*}

Is this something we're supposed to know? I've never seen this before.

Reply 6

Zacken

Original post by TLDM

Is this something we're supposed to know? I've never seen this before.

Nothing I've posted is something you should know, but instead it's something that once pointed out is apparent.

In the case you've pointed out, that equality comes right away from using the substitution

x \mapsto a+b-x

. It's quite literally a one-liner.

Reply 7

Zacken

I'd just like to showcase an astounding example of the adding zero creatively trick (even if it's been quoted already by Duke in the OP), it featured heavily in STEP II, 2005, Q8 for example and the examiners report and official solutions didn't even mention it, they decided to mention ridiculous things like hyperbolic trigonometrical substitutions instead of a simple trick.

So what we have is:

Unparseable latex formula:

\displaystyle[br]\begin{equation*} \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x^3y^2}{(1+x^2)^{5/2}}\end{equation*}

The seperable stuff is easy and you all know how to integrate

y^{-2}

, but the RHS is slightly more tricky, how would you integrate

x^3/(1+x^2)^{5/2}

- what I'd do:

Unparseable latex formula:

\displaystyle [br]\begin{equation*} \frac{x^3 +x - x}{(1+x^2)^{5/2}} = \frac{x(1+x^2) - x}{(1+x^2)^{5/2}} = \frac{x}{(1+x^2)^{3/2}} - \frac{x}{(1+x^2)^{5/2}} \end{equation*}

Whereby integrating is very standard now, since

(1+x^2)' = 2x

. Oh, and somebody remind me to do a post about multiplying by one creatively.

Reply 8

IrrationalRoot

Using 'it suffices' a lot is very useful for proving certain algebraic results (whittling down what is required to be proven).

Taking the last part STEP III 2009 Q1 as an example, I got expressions for st/uv and (s+t)/(u+v) in terms of m and n and needed to prove that p+q (each expressed as somewhat ugly expressions involving m,n,s,t,u,v) is equal to 0.

A very easy way of doing this is to start is to:
Start with the result.
Factor out m-n so that it suffices to prove that the other factor is equal to 0.
Combine the fractions so that it suffices to prove that the numerator is equal to 0.
A bit of algebra combined with double-headed implication arrows between steps generates the expressions st/uv and (s+t)/(u+v) and a clear equality results after substituting for these (thus the result follows due to the backwards implication).

This was a simple example but the general technique should be clear.
Starting with a result and progressively reducing it down to smaller problems is a very widely applicable technique.

(edited 8 years ago)

Reply 9

TLDM

Original post by Zacken

x \mapsto a+b-x

. It's quite literally a one-liner.

So is it not in A level then? We've not finished FP2/3 yet (I know we have some more integration to go), so I wasn't sure whether or not it was in there. We've done things like "adding zero creatively" though.

But I think adding zero is the only thing so far that's not new to me. These are all incredibly helpful; thanks for posting these!

Reply 10

Student403

Am I high or in the @Zacken's quote in the OP should it be 1- 1/(1+x) on the second line?

Reply 11

IrrationalRoot

Original post by Student403

Am I high or in the @Zacken's quote in the OP should it be 1- 1/(1+x) on the second line?

No you're right XD.

Reply 12

Student403

Original post by IrrationalRoot

No you're right XD.

Okay phew! Zain remember to edit the logarithm too

Reply 13

IrrationalRoot

Original post by Zacken

Using implicit differentiation can make life easier.

Say you wanted to differentiate a nasty function that might require multiple uses of the product rule or such, take logs!

Let's say you have

f(x) = x^p(x+1)^q(x+2)^r

, do you really want to differentiate this normally? I'd hope you said no. What you'd want to do is:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\log f(x) = p \log x + q \log (x+1) + r \log (x+2) \Rightarrow f'(x) = f(x)\left(\frac{p}{x} + \frac{q}{x+1} + \frac{r}{x+2}\right)\end{equation*}

and leaving your answer in this form, i.e: as

f(x)(\cdots)

is quite useful in many a STEP I question that I've come across.

Never came across the idea of logarithmic differentiation for products. Was this featured in any STEP questions?

Reply 14

Zacken

Original post by TLDM

Original post by Student403

Am I high or in the @Zacken's quote in the OP should it be 1- 1/(1+x) on the second line?

Corrected - thanks for pointing it out; I'm typing all these out in a rush; so people feel free to quote the bits that are wrong and correct me so we can turn this into a nice collection. :biggrin:

Reply 15

Zacken

Original post by IrrationalRoot

Never came across the idea of logarithmic differentiation for products. Was this featured in any STEP questions?

At least two STEP I questions I can think of off the top of my head.

Reply 16

Student403

Original post by Zacken

None of these things are remotely requisite for A-Level, they're not even requisite for STEP - just nice shortcuts that you pickup by yourself informally, none of it is really meant to be taught explicitly. :-)

Corrected - thanks for pointing it out; I'm typing all these out in a rush; so people feel free to quote the bits that are wrong and correct me so we can turn this into a nice collection. :biggrin: