Sergio H. Uni Admissions Test .PAT. tutor, IB Spanish tutor, IB Maths...

Sergio H.

Currently unavailable: until 01/08/2015

Degree: Physics with Theoretical Physics (Bachelors) - Imperial College London University

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About me

Hi there, my name’s Sergio. I’m currently finishing my BSc in Physics with Theoretical Physics at Imperial College London and looking forward to a research career in Cosmology. I’ve been always passionate about Maths and its applications to science and truly enjoy sharing that passion when tutoring Maths and Physics.

One year of experience tutoring face-to-face has proven to me how important it is to empathise with the student, spot their specific difficulties and motivate and transmit confidence. Whether you are struggling with bunches of seemingly nonsensical equations or want to refine your knowledge of a particular subject, I will be glad to tailor my tutorials to meet your specific needs.

If you are studying Spanish I can provide you with the always valuable perspective of a native Spanish speaker. Whilst learning the written language is essential, lessons often lack conversation time which you would certainly benefit from as well in my tutorials.

If you have any questions, get in touch with me through a ‘WebMail’ or book a ‘Meet the Tutor Session’ and let me know what you need help with.

Looking forward to meeting you!

Availability: I’m always around in the morning and afternoon, but drop me an email if any other time suits you better.

Subjects offered

SubjectLevelMy prices
Maths A Level £20 /hr
Physics A Level £20 /hr
Spanish A Level £20 /hr
Maths GCSE £18 /hr
Physics GCSE £18 /hr
Spanish GCSE £18 /hr
Maths IB £20 /hr
Physics IB £20 /hr
Spanish IB £20 /hr
.PAT. Uni Admissions Test £25 /hr

Qualifications

QualificationLevelGrade
Mathematics IBachelors Degree79
Mechanics, Vibrations and WavesBachelors Degree77
Electricity and Magnetism and RelativityBachelors Degree82
Quantum Physics and Structure of MatterBachelors Degree88
Professional SkillsBachelors Degree78
Laboratory and Computing IBachelors Degree89
Project IBachelors Degree90
Mathematical AnalysisBachelors Degree83
Physics Year 1 TotalBachelors Degree82.3
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

Currently unavailable: until

01/08/2015

Questions Sergio has answered

How do I approximate an irrational square root without using a calculator?

Although this might look like a smart trick at the beginning, the truth is that as a physicist this kind of approximations turn out to be useful very often. This is quite likely to come up in the PAT as well! It is important to get used to understanding symbolic expressions (no numbers!),...

Although this might look like a smart trick at the beginning, the truth is that as a physicist this kind of approximations turn out to be useful very often. This is quite likely to come up in the PAT as well!

It is important to get used to understanding symbolic expressions (no numbers!), so I will first derive the maths and then apply the resulting expression to a couple of illustrative examples.

Our aim is to obtain an approximate value for the inexact square root x of an arbitrary number n, i.e. to solve x = n1/2. In other words, we want to solve the equation f(x) = x2 = n. Suppose we know the exact root x0 of another number n0 which is very close to n, i.e. f(x0) = x02 = n0 (this is key!). We can do a Taylor series expansion of f(x) about x0 up to first order (first derivative f'(x)), which is simply:

f(x) = f(x0) + f'(x0)(x - x0).

The first derivative f'(x) = 2x, which evaluated at x0 is f'(x0) = 2x0. Substituting f(x) = n and f(x0) = n0 leads to:

n = n0 + 2x0(x - x0).

We are almost done! We know all the values of n, n0 and x0 and just need to solve for x. Rearranging:

x = x0 + (n - n0)/(2x0)

That's it! Remember that this expression is just an approximation for the actual square root of n, but a quite good one the closer n0 is to n.

Let's plug in some numbers. Suppose you were asked to approximate the square root of n = 66. Despite the solution being irrational, there is a close number n0 = 64 with a nice exact square root x0 = 8. Using the expression above we find:

x = 8 + (66 - 64)/(2*8) = 8 + 2/16 = 8 + 1/8 = 8 + 0.125 = 8.125

How good is this? It's just 0.01% off the actual value 8.12404...!

What if we chose n0 = 81 instead, the square root of which is x0 = 9? This would give x = 8.167 - worse, but still a reasonable result considering how far 81 is from 66. Never forget to quote both the positive and negative square roots in your final solution!

Would you be able to derive a similar expression for approximating a cubic root?

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1 year ago

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