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Jia Hao L.

Currently unavailable: for regular students

Studying: Pure Mathematics and Mathematical Logic (Masters) - Manchester University

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9 reviews| 16 completed tutorials

Contact Jia Hao

About me

About Me:

Currently pursuing a Masters in Pure Mathematics, I have 7 years of tutoring experience in mathematics. In mathematics, understanding is key and I have created many analogies and mnemonics that will definitely help you strengthen your foundation. I love mathematics, and I teach in the hope of helping others see the beauty of the subject too.

 

The Session

To score well in mathematics, one needs to understand the concepts well. The best indicator of how well one understands is how well one explains. As such, you will drive the session. You start by telling me what you think you know, and I will help you (through diagrams, analogies, etc) prefect your reasoning. When you are confident of explaining the concept to me, I know you are capable of applying that knowledge in the exam questions.

 

Math can be dull, but I hope our session will be fun! I hope to infect you with my love for the subject by sharing with you things that are beyond the textbook. That way, mathematics might become more relevant to you and make the subject more interesting for you.

 

Why do I need to know this..?

I love answering these questions! In fact, I've recently set up a blog to help answer these questions. (www.mathsplained.wordpress.com)

 

I love answering questions, and they need not be restricted to mathematics homework questions. Asking questions (the right kind) sharpens your mind, and I encourage you to do that. 

 

What's next?

If you have any questions, contact me through "WebMail" or "Meet-the-Tutor-Session"! Let me know which examination you are taking and which area(s) you are having problems with, and I'll try my very best to help you. 

 

About Me:

Currently pursuing a Masters in Pure Mathematics, I have 7 years of tutoring experience in mathematics. In mathematics, understanding is key and I have created many analogies and mnemonics that will definitely help you strengthen your foundation. I love mathematics, and I teach in the hope of helping others see the beauty of the subject too.

 

The Session

To score well in mathematics, one needs to understand the concepts well. The best indicator of how well one understands is how well one explains. As such, you will drive the session. You start by telling me what you think you know, and I will help you (through diagrams, analogies, etc) prefect your reasoning. When you are confident of explaining the concept to me, I know you are capable of applying that knowledge in the exam questions.

 

Math can be dull, but I hope our session will be fun! I hope to infect you with my love for the subject by sharing with you things that are beyond the textbook. That way, mathematics might become more relevant to you and make the subject more interesting for you.

 

Why do I need to know this..?

I love answering these questions! In fact, I've recently set up a blog to help answer these questions. (www.mathsplained.wordpress.com)

 

I love answering questions, and they need not be restricted to mathematics homework questions. Asking questions (the right kind) sharpens your mind, and I encourage you to do that. 

 

What's next?

If you have any questions, contact me through "WebMail" or "Meet-the-Tutor-Session"! Let me know which examination you are taking and which area(s) you are having problems with, and I'll try my very best to help you. 

 

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Ratings & Reviews

5from 9 customer reviews
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Elizabeth (Student)

May 22 2015

Great tutor. Really takes his time

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Victor (Student)

May 20 2015

It wwas veryy good. Thank you for preparing me for the test

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Victor (Student)

May 15 2015

The tutorial was really good because he answered all of my queries with the topics I was confused with

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Elizabeth (Student)

May 15 2015

Very good tutor. Really deepened my understanding of substitution by parts

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Qualifications

SubjectQualificationGrade
Applied MathematicsDegree (Bachelors)2:1
MathematicsA-level (A2)A
Pure Mathematics and LogicDegree (Masters)-

Subjects offered

SubjectQualificationPrices
Further MathematicsA Level£30 /hr
MathsA Level£30 /hr
MathsGCSE£30 /hr
MathsIB£30 /hr

Questions Jia Hao has answered

Understanding differentiation from first principle.

What is first principle? Why does it look nothing like the differentiation I had been doing for the past few years?

 

Differentiation from first principle is the main idea behind differentiation, a technique we employ to measure instantaneous rate of change. By now, you probably recognize "rate of change" as being synonymous to the term "gradient" or "average speed/distance" in context of several word problems. It is, therefore, not surprising that the definition of first principle is very similar to the gradient formula!

 

Let us recall: the gradient of line connecting two points (x1,y1) and (x2,y2) is given by the equation grad = (y1-y2)/(x1-x2). The definition of first principle says f'(x) = lim_{dx->0} (f(x)-f(x+dx))/dx. (Due to limitation of software, we'll refer to "delta x" by "dx" here for convenience.)

 

Let's examine my claim that both formulas are very similar. Suppose I have a graph y=f(x). Pick a point on the graph and name it (x1,y1). Now, to find the instantaneous rate of change at point (x1,y1), what can we do? Without the knowledge of the equation of the graph, we cannot perform differentiation using the rules that we'd learned. So, instead of trying to get the correct answer, let's approximate. Let's pick a point as cloes to (x1,y1) as we can and name this point (x2,y2). Because these points are very close together, it is safe to say that the line connecting them is a good approximation of the tangent line to point (x1,y1). In that case, we can use the gradient formula to find an approximate solution to the instantaneous gradient at (x1,y1)! By letting the distance between x1 and x2 be dx, we have: x2=x1+dx. Plug this into the gradient formula and we will get

approx. grad = (y1-y2)/(x1-x2) = (f(x1) - f(x2))/(x1 - (x1+dx)) = (f(x1) - f(x1+dx))/dx.

 

There is still something different - the "lim_{dx -> 0}" notation. What does it mean anyway? Reading the symbols out it says "the limit of the expression as "dx" approaches zero". You see, in order to improve accuracy of our gradient's approximation, we only need to pick (x2,y2) to be even closer to (x1,y1). Theoretically, then, if the distance between x1 and x2 is so close that it is almost zero, then our approximation would be the exact solution. Hence, if we allow "dx" to approach zero, then we can confidently change the left-hand-side of the approximation to the exact value, in other words:

f'(x1) = lim_{dx->0} (f(x1)-f(x1+dx))/dx. 

 

Since (x1,y1) is just a name we gave to the point that we are interested in, we may substitute it with (x,y) (to make the formula applicable in a generic x-y plot) to obtain the definition of first principle as presented in our textbooks. 

What is first principle? Why does it look nothing like the differentiation I had been doing for the past few years?

 

Differentiation from first principle is the main idea behind differentiation, a technique we employ to measure instantaneous rate of change. By now, you probably recognize "rate of change" as being synonymous to the term "gradient" or "average speed/distance" in context of several word problems. It is, therefore, not surprising that the definition of first principle is very similar to the gradient formula!

 

Let us recall: the gradient of line connecting two points (x1,y1) and (x2,y2) is given by the equation grad = (y1-y2)/(x1-x2). The definition of first principle says f'(x) = lim_{dx->0} (f(x)-f(x+dx))/dx. (Due to limitation of software, we'll refer to "delta x" by "dx" here for convenience.)

 

Let's examine my claim that both formulas are very similar. Suppose I have a graph y=f(x). Pick a point on the graph and name it (x1,y1). Now, to find the instantaneous rate of change at point (x1,y1), what can we do? Without the knowledge of the equation of the graph, we cannot perform differentiation using the rules that we'd learned. So, instead of trying to get the correct answer, let's approximate. Let's pick a point as cloes to (x1,y1) as we can and name this point (x2,y2). Because these points are very close together, it is safe to say that the line connecting them is a good approximation of the tangent line to point (x1,y1). In that case, we can use the gradient formula to find an approximate solution to the instantaneous gradient at (x1,y1)! By letting the distance between x1 and x2 be dx, we have: x2=x1+dx. Plug this into the gradient formula and we will get

approx. grad = (y1-y2)/(x1-x2) = (f(x1) - f(x2))/(x1 - (x1+dx)) = (f(x1) - f(x1+dx))/dx.

 

There is still something different - the "lim_{dx -> 0}" notation. What does it mean anyway? Reading the symbols out it says "the limit of the expression as "dx" approaches zero". You see, in order to improve accuracy of our gradient's approximation, we only need to pick (x2,y2) to be even closer to (x1,y1). Theoretically, then, if the distance between x1 and x2 is so close that it is almost zero, then our approximation would be the exact solution. Hence, if we allow "dx" to approach zero, then we can confidently change the left-hand-side of the approximation to the exact value, in other words:

f'(x1) = lim_{dx->0} (f(x1)-f(x1+dx))/dx. 

 

Since (x1,y1) is just a name we gave to the point that we are interested in, we may substitute it with (x,y) (to make the formula applicable in a generic x-y plot) to obtain the definition of first principle as presented in our textbooks. 

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